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%     (Warning: These details have been written several years
%     ago, back when I was a grad student, and are probably
%     completely useless, seeing that most of the solutions
%     are already more detailed than necessary.)
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% replace them by
%   \excludecomment{verlong}
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% (3), replace them by
%   \includecomment{verlong}
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% [DG][v76] Added the switch described above. The material
% in version (3) is from my old drafts.

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\begin{verlong}
\title[Hopf algebras in combinatorics (detailed version containing solutions)]{Hopf algebras in combinatorics (detailed version containing solutions)}
\end{verlong}
\begin{vershort}
\title[Hopf algebras in combinatorics (version containing solutions)]{Hopf algebras in combinatorics (version containing solutions)}
\end{vershort}
\begin{nosolutions}
\title[Hopf algebras in combinatorics]{Hopf algebras in combinatorics}
\end{nosolutions}

%\dedicatory{To ...}

\author{Darij Grinberg}
\email{darijgrinberg@gmail.com}
\address{Drexel University,
Korman Center, Room 263,
15 S 33rd Street,
Philadelphia PA, 19104,
USA
// (temporary)
Mathematisches Forschungsinstitut Oberwolfach,
Schwarzwaldstrasse 9--11, 77709 Oberwolfach,
Germany
}
\author{Victor Reiner}
\email{reiner@math.umn.edu}
\address{School of Mathematics\\
University of  Minnesota\\
Minneapolis, MN 55455\\
USA}

% \thanks{NSF ??}

% \subjclass{05A15, 05A10}

\date{July 27, 2020 (with minor corrections \today)}
\keywords{Hopf algebra, combinatorics, symmetric functions, quasisymmetric functions}

\makeindex

\begin{document}


\maketitle

\tableofcontents

\doclicenseThis

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\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Introduction}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Certain Hopf algebras arise in combinatorics because they have bases
%naturally parametrized by combinatorial objects (partitions, compositions,
%permutations, tableaux, graphs, trees, posets, polytopes, etc).
%The rigidity in the structure of
%a Hopf algebra can lead to enlightening proofs, and many
%interesting invariants of combinatorial objects turn out to be evaluations
%of Hopf morphisms.

The concept of a Hopf algebra % is an algebraic notion which
crystallized out of algebraic topology and the study of
algebraic groups in the 1940s and 1950s (see
\cite{AndruskiewitschSantos-history} and \cite{Cartier} for
its history). Being a fairly elementary algebraic notion
itself, it subsequently found applications in other
mathematical disciplines, and is now particularly commonplace in
representation theory\footnote{where it provides explanations
for similarities between group representations and Lie
algebra representations}.

These notes concern themselves
(after a brief introduction into the algebraic foundations of
Hopf algebra theory in Chapter~\ref{Hopf-intro-section})
with the Hopf algebras that appear in combinatorics.
These Hopf algebras tend to have bases naturally parametrized
by combinatorial objects (partitions, compositions,
permutations, tableaux, graphs, trees, posets, polytopes, etc.),
and their Hopf-algebraic operations often encode basic
operations on these objects\footnote{such as concatenating two
compositions, or taking the disjoint union of two graphs
-- but, more often, operations which return a multiset
of results, such as cutting a composition into two pieces
at all possible places, or partitioning a poset into two
subposets in every way that satisfies a certain axiom}.
Combinatorial results can then be seen as particular cases of
general algebraic properties of Hopf algebras
(e.g., the multiplicativity
of the M\"obius function can be recovered from the fact that
the antipode of a Hopf algebra is an algebra
anti-endomorphism), and many interesting invariants of
combinatorial objects turn out to be evaluations
of Hopf morphisms. In some cases (particularly that of
symmetric functions), the rigidity in the
structure of a Hopf algebra can lead to enlightening proofs.

One of the most elementary interesting examples of a
combinatorial Hopf algebra is that of the symmetric functions.
We will devote all of Chapter~\ref{Sym-section} to studying
it, deviating from the usual treatments (such as in
Stanley \cite[Ch. 7]{Stanley}, Sagan \cite{Sagan} and
Macdonald \cite{Macdonald}) by introducing the Hopf-algebraic
structure early on and using it to obtain combinatorial
results. Chapter~\ref{PSH-section} will underpin the
importance of this algebra by proving Zelevinsky's
main theorem of PSH theory, which (roughly) claims that a
Hopf algebra over $\ZZ$ satisfying a certain set of axioms
must be a tensor product of copies of the Hopf algebra
of symmetric functions. These axioms are fairly restrictive,
so this result is far from curtailing the diversity of
combinatorial Hopf algebras; but they are natural enough
that, as we will see in
Chapter~\ref{representations-section}, they are satisfied
for a Hopf algebra of representations of symmetric groups.
As a consequence, this Hopf algebra will be revealed
isomorphic to the symmetric functions -- this is the
famous Frobenius correspondence between symmetric
functions and characters of symmetric groups, usually
obtained through other ways (\cite[\S 7.3]{Fulton},
\cite[\S 4.7]{Sagan}). We will further elaborate on the
representation theories of wreath products and general
linear groups over finite fields; while Zelevinsky's PSH
theory does not fully explain the latter, it illuminates
it significantly.

In the next chapters, we will study further examples of
combinatorial Hopf algebras: the quasisymmetric functions
and the noncommutative symmetric functions in
Chapter~\ref{Qsym-section}, various other algebras (of
graphs, posets, matroids, etc.) in
Chapter~\ref{sect.ABS}, and the Malvenuto-Reutenauer
Hopf algebra of permutations in Chapter~\ref{sect.MR}.

The main prerequisite for reading these notes is a good
understanding of graduate algebra\footnote{William
Schmitt's expositions \cite{Schmitt-exp} are
tailored to a reader interested in combinatorial Hopf
algebras; his notes on modules and algebras cover a
significant part of what we need from abstract algebra,
whereas those on categories cover all category theory
we will use and much more.}, in particular multilinear
algebra (tensor products, symmetric powers and exterior
powers)\footnote{Keith Conrad's expository notes
\cite{Conrad-blurbs} are useful, even if not
comprehensive, sources for the latter.} and basic
categorical language\footnote{We also will use a few
nonstandard notions from linear algebra that are explained
in the Appendix (Chapter~\ref{chp.appendix}).}. In
Chapter~\ref{representations-section}, familiarity with
representation theory of finite groups (over $\CC$)
is assumed, along
with the theory of finite fields and (at some places) the
rational canonical form of a matrix. Only basic knowledge
of combinatorics is required (except for a few spots in
Chapter~\ref{sect.ABS}), and familiarity with
geometry and topology is needed only to understand some
tangential remarks. The concepts of Hopf algebras and
coalgebras and the basics of symmetric function theory will
be introduced as needed. We will work over a commutative
base ring most of the time, but no commutative algebra
(besides, occasionally, properties of modules over a PID)
will be used.

% [DG][v41] Added footnote with Schmitt reference.

These notes began as an accompanying text for
Fall 2012 Math 8680 Topics in Combinatorics,
a graduate class taught by the second author at
the University of Minnesota.
The first author has since added many exercises
(and solutions%
\begin{nosolutions}
\footnote{The version of the notes you are reading does not contain said solutions. The version that does can be downloaded from \url{http://www.cip.ifi.lmu.de/~grinberg/algebra/HopfComb-sols.pdf} or compiled from the sourcecode.}
\end{nosolutions}
), as well as Chapter~\ref{sect.QSym.lyndon}
on Lyndon words and the polynomiality of $\Qsym$.
The notes might still grow, and any
comments, corrections and complaints are welcome!

The course was an attempt to focus on examples that
we find interesting, but which are hard to find
fully explained currently in books or in one paper.
Much of the subject of combinatorial Hopf algebras is
fairly recent (1990s onwards) and still spread over
research papers, although sets of lecture notes do exist,
such as Foissy's \cite{Foissy}.
A reference which we discovered late, having a great deal of overlap with these notes
is Hazewinkel, Gubareni, and Kirichenko \cite{HazewinkelGubareniKirichenko}.
References for the purely algebraic theory of Hopf
algebras are much more frequent (see the beginning of
Chapter~\ref{Hopf-intro-section} for a list).
Another recent text that has a significant amount of
material in common with ours (but focuses on
representation theory and probability applications)
is M\'eliot's \cite{Meliot}.

% [DG][v71] Added \cite{Meliot} reference.

Be warned that our notes are highly idiosyncratic in choice of topics,
and they steal heavily from the sources in the bibliography.

% [DG][v26] I wrote the above introduction (partly using
% your syllabus page http://www-users.math.umn.edu/~reiner/Classes/8680_Fall2012_syllabus.html ).
% Do you know a better source for tensor algebra than
% Keith Conrad's blurbs?

\vskip.2in
\noindent
\textbf{Warnings:} Unless otherwise specified ...
\begin{enumerate}
\item[$\bullet$]
\dfn{$\kk$} here usually denotes a commutative ring\footnote{As explained below,
``ring'' means ``associative ring with $1$''.
The most important cases are when $\kk$ is a field or when $\kk = \ZZ$.}.
\item[$\bullet$]
all maps between $\kk$-modules are $\kk$-linear.
\item[$\bullet$]
every ring or $\kk$-algebra is associative and has a $1$, and every ring morphism or $\kk$-algebra morphism preserves the $1$'s.
\item[$\bullet$]
all $\kk$-algebras $A$ have the property that $\left(\lambda 1_A\right) a = a \left(\lambda 1_A\right) = \lambda a$ for all $\lambda \in \kk$ and $a \in A$.
\item[$\bullet$]
all tensor products are over $\kk$ (unless a subscript specifies a different base ring).\index{$\otimes$}
\item[$\bullet$]
\dfn{$\one$} will denote the multiplicative identity in some ring like $\kk$ or in some
$\kk$-algebra (sometimes also the identity of a group written multiplicatively).
\item[$\bullet$]
for any set $S$, we denote by \dfn{$\id_S$} (or by $\id$) the identity map on $S$.
\item[$\bullet$]
The symbols \dfn{$\subset$} (for ``subset'') and \dfn{$<$}
(for ``subgroup'') don't imply properness
(so $\ZZ \subset \ZZ$ and $\ZZ < \ZZ$).
\item[$\bullet$]
the $n$-th symmetric group (i.e., the group of all permutations of
$\left\{1,2,\ldots,n\right\}$) is denoted \dfn{$\Symm_n$}.
\item[$\bullet$]
A permutation $\sigma \in \Symm_n$ will often be identified with
the $n$-tuple $\left(\sigma\left(1\right), \sigma\left(2\right),
\ldots, \sigma\left(n\right)\right)$, which will occasionally be
written without commas and parentheses (i.e., as follows:
$\sigma\left(1\right)\sigma\left(2\right)\cdots\sigma\left(n\right)$).
This is called the \dfn{one-line notation} for permutations.
\item[$\bullet$]
The product of permutations $a \in \Symm_n$ and $b \in \Symm_n$ is defined by
$(ab)(i) = a(b(i))$ for all $i$.
\item[$\bullet$]
\emph{Words}\index{word} over (or in) an \dfn{alphabet} $I$ simply mean finite tuples of elements
of a set $I$. It is customary to write such a word $\left(a_1,a_2,\ldots,a_k\right)$ as
$a_1a_2\ldots a_k$ when this is not likely to be confused for multiplication.
\item[$\bullet$]
$\NN := \left\{0,1,2,\ldots\right\}$.\index{$\NN$}
\item[$\bullet$]
if $i$ and $j$ are any two objects, then \dfn{$\delta_{i,j}$} denotes the \dfn{Kronecker
delta} of $i$ and $j$; this is the integer $1$ if $i = j$ and $0$ otherwise.
\item[$\bullet$]
a \dfn{family} of objects indexed by a set $I$ means a choice of an object $f_i$ for each element $i \in I$; this family will be denoted either by \dfn{$\left(f_i\right)_{i \in I}$} or by \dfn{$\left\{f_i\right\}_{i \in I}$} (and sometimes the ``$i \in I$'' will be omitted when the context makes it obvious -- so we just write \dfn{$\left\{f_i\right\}$}).
% [DG][v72] Added this. Note that the $\left(...\right)$ notation is mine while the $\left\{...\right\}$ notation is yours; I'd make them uniform but I'm afraid there are just too many places to edit.
\item[$\bullet$]
several objects $s_1, s_2, \ldots, s_k$ are said to be
\dfn{distinct} if every $i \neq j$ satisfy $s_i \neq s_j$.
\item[$\bullet$]
similarly,
several sets $S_1, S_2, \ldots, S_k$ are said to be \dfn{disjoint}
if every $i \neq j$ satisfy $S_i \cap S_j = \varnothing$.
\item[$\bullet$]
the symbol \dfn{$\sqcup$} (and the corresponding quantifier $\bigsqcup$)
denotes a disjoint union of sets or posets.
For example, if $S_1, S_2, \ldots, S_k$ are $k$ sets, then
$\bigsqcup_{i=1}^k S_i$ is their disjoint union.
This \idx{disjoint union} can mean either of the following two things:
\begin{itemize}
\item It can mean the union $\bigcup_{i=1}^k S_i$ in the case when
      the sets $S_1, S_2, \ldots, S_k$ are disjoint.
      This is called an ``\idx{internal disjoint union}'', and is simply
      a way to refer to the union of sets while simultaneously
      claiming that these sets are disjoint.
      Thus, of course, it is only well-defined if the sets are
      disjoint.
\item It can also mean the union
      $\bigcup_{i=1}^k \left\{i\right\} \times S_i$.
      This is called an ``\idx{external disjoint union}'', and is
      well-defined whether or not the sets
      $S_1, S_2, \ldots, S_k$ are disjoint; it is a way to
      assemble the sets $S_1, S_2, \ldots, S_k$ into a larger
      set which contains a copy of each of their elements
      that ``remembers'' which set this element comes from.
\end{itemize}
The two meanings are different, but in the case when
$S_1, S_2, \ldots, S_k$ are disjoint, they are isomorphic.
We hope the reader will not have a hard time telling which
of them we are trying to evoke.
\par
Similarly, the notion of a direct sum of $\kk$-modules
has two meanings (``\idx{internal direct sum}'' and
``\idx{external direct sum}'').
\item[$\bullet$] A sequence $\left(w_1, w_2, \ldots, w_k\right)$
of numbers (or, more generally, of elements of a poset) is said
to be \dfn{strictly increasing} (or, for short, \dfn{increasing})
if it satisfies $w_1 < w_2 < \cdots < w_k$.
A sequence $\left(w_1, w_2, \ldots, w_k\right)$
of numbers (or, more generally, of elements of a poset) is said
to be \dfn{weakly increasing} (or \dfn{nondecreasing})
if it satisfies $w_1 \leq w_2 \leq \cdots \leq w_k$.
Reversing the inequalities, we obtain the definitions of a
\dfn{strictly decreasing} (a.k.a. \dfn{decreasing}) and of a
\dfn{weakly decreasing} (a.k.a. \dfn{nonincreasing})
sequence.
All these definitions extend in an obvious way to infinite sequences.
Note that ``nondecreasing'' is not the same as ``not decreasing'';
for example, any sequence having at most one entry is both
decreasing and nondecreasing, whereas the sequence $\left(1, 3, 1\right)$
is neither.
\end{enumerate}

% [DG] I added the last three bullets. I couldn't believe my eyes when I first
% saw that Sage and GAP (and a fair share of British literature) disagree with
% me on the product of permutations.

% [DG][v17] Changed default type of $\kk$ from field to commutative ring.

% [DG][v19] The three different notions we used to denote by $1$ now are
% called differently:
% -- $1$ or $\one$ is used (interchangeably) for multiplicative identities of
%    rings and monoids.
% -- $\id$ is used for identity maps.
% -- $\triv$ is used for trivial characters of groups and for indicator
%    functions of subsets. This currently resolves to $\underline{1}$,
%    but can be changed easily.

% [DG][v64] Added explanation for $\sqcup$ and $\bigsqcup$.

% [DG][v71] Added bullet point about ``nondecreasing'' vs ``not decreasing''.

Hopefully context will resolve some of the ambiguities.

% [DG][v26] Moved the above conventions and warnings to the introduction,
% as they are global to the text.

% [DG][v37] The outermost layer of sectioning is now called Chapters,
% the next-outermost Sections, etc. (This used to be the case in some
% places but not others; now it should be homogeneous.)

\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{What is a Hopf algebra?}
\label{Hopf-intro-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The standard references for Hopf algebras are Abe \cite{Abe} and Sweedler \cite{Sweedler},
and some other good ones are
\cite{CaenepeelVercruysse, ChariPressley, DascalescuNastasescuRaianu,
HazewinkelGubareniKirichenko, Kassel, Kytola-hopf, Montgomery,
Radford-Hopf, Schweigert-Hopf, Wisbauer}.
See also Foissy \cite{Foissy} and Manchon \cite{Manchon} for introductions to Hopf algebras
tailored to combinatorial applications.
Most texts only study Hopf algebras over fields (with exceptions such as
\cite{ChariPressley, CaenepeelVercruysse, Wisbauer}). We will work over arbitrary commutative
rings\footnote{and we will profit from this generality in Chapters
\ref{PSH-section} and \ref{representations-section}, where we will
be applying the theory of Hopf algebras to $\kk = \ZZ$ in a way that would
not be possible over $\kk = \QQ$}, which requires some more care at certain
points (but we will not go deep enough into the algebraic theory to witness
the situation over commutative rings diverge seriously from that over
fields).

% [DG][v17] Added reference to Caenepeel-Vercruysse (just discovered today)
% and a couple lines on fields vs. rings.

% [DG][v14] Added Foissy and Manchon references. The combinatorics they do
% is mostly different from ours, but the Hopf algebra theory intersects a
% lot (unavoidably).
% Foissy's work is among the most readable I have seen on this subject;
% we should reference it more...

Let's build up the definition of Hopf algebra
structure bit-by-bit, starting with the more familiar definition of algebras.

\subsection{Algebras}

Recall that an \emph{associative $\kk$-algebra}\index{algebra}\index{associative}
is defined to be
a $\kk$-module $A$ equipped with an associative $\kk$-bilinear map
$\operatorname{mult} : A \times A \to A$ (the
\dfn{multiplication map} of $A$) and an element $\one \in A$
(the \emph{(multiplicative) unity}\index{unity of an algebra}
or \emph{identity}\index{identity of an algebra} of $A$)
that is neutral for this map $\operatorname{mult}$ (that is, it
satisfies $\operatorname{mult} \left(a, \one\right) =
\operatorname{mult} \left(\one, a\right) = a$ for all $a \in A$).
If we recall that
\begin{itemize}
\item $\kk$-bilinear maps $A \times A \to A$
      are in 1-to-1 correspondence with $\kk$-linear maps
      $A \otimes A \to A$ (by the universal property of the tensor
      product), and
\item elements of $A$ are in 1-to-1 correspondence with $\kk$-linear
      maps $\kk \to A$,
\end{itemize}
then we can restate this classical definition of associative
$\kk$-algebras
as follows in terms of $\kk$-linear maps\footnote{Explicitly speaking,
we are replacing the $\kk$-bilinear multiplication map
$\operatorname{mult} : A \times A \to A$ by the $\kk$-linear map
$m : A \otimes A \to A, \  a \otimes b \mapsto \operatorname{mult}\left(a, b\right)$,
and we are replacing the element $\one \in A$ by the $\kk$-linear map
$u : \kk \to A, \  1_\kk \mapsto \one$.}:

% [DG][v14] Added preceding sentence as well as the convention that
% all $\kk$-algebras are central (Damien Calaque, who has some papers on
% Hopf algebras, is one of the minority who disagrees with this). Also
% added $\kk$-linearity assumption on $m$ and $u$ in definition below.

% [DG][v77] Added preceding paragraph and slightly clarified the
% definition below (as well as that of a coalgebra).

\begin{definition}
\label{def.algebra}
An \emph{associative $\kk$-algebra}\index{algebra}\index{associative}
is a $\kk$-module $A$
equipped with a $\kk$-linear \dfn{associative operation}\index{$m$}
$A \otimes A \overset{m}{\rightarrow} A$,
and a $\kk$-linear \dfn{unit}\index{$u$}
$\kk \overset{u}{\rightarrow} A$,
for which the following two diagrams are commutative:

\begin{equation}
\label{associativity-diagram}
\xymatrix{
& A  \otimes A \otimes A \ar[dl]_{m \otimes \id} \ar[dr]^{\id \otimes m} &  \\
A \otimes A \ar[dr]^{m} & &  A \otimes A \ar[dl]_{m} \\
& A &
}
\end{equation}


\begin{equation}
\label{unit-diagram}
\xymatrix{
A \otimes \kk \ar[d]_{\id \otimes u} & A \ar[l] \ar[d]_{\id} \ar[r] & \kk \otimes A \ar[d]^{u \otimes \id} \\
A \otimes A \ar[r]^-{m} & A & A \otimes A \ar[l]_-{m}
}
\end{equation}
where the maps $A \rightarrow A \otimes \kk$ and
$A \rightarrow \kk \otimes A$ are the isomorphisms
sending $a \mapsto  a \otimes \one$ and
$a \mapsto \one \otimes a$.

We abbreviate ``associative $\kk$-algebra'' as ``$\kk$-algebra''
(associativity is assumed unless otherwise specified) or as
``algebra'' (when $\kk$ is clear from the context).
We sometimes refer to $m$ as the ``multiplication map'' of $A$
as well.
\end{definition}

As we said, the multiplication map $m : A \otimes A \to A$ sends
each $a \otimes b$ to the product $ab$, and the unit map $u : \kk \to A$
sends the identity $1_\kk$ of $\kk$ to the identity $1_A$ of $A$.

% [DG][v77] Added the preceding paragraph and the last sentence
% of the definition above.

Well-known examples of $\kk$-algebras are
\emph{tensor} and \emph{symmetric algebras},
which we can think of as algebras of \emph{words} and 
\emph{multisets}, respectively.

\begin{example}
\label{exa.tensor-alg}
If $V$ is a $\kk$-module and $n \in \NN$, then the \emph{$n$-fold
tensor power $V^{\otimes n}$}\index{tensor power}
of $V$ is the $\kk$-module
$\underbrace{V \otimes V \otimes \cdots \otimes V}_{n\text{ times}}$.
(For $n = 0$, this is the $\kk$-module $\kk$, spanned by the
``empty tensor'' $1_\kk$.)

% [DG][v19] Introduced tensor powers here, since you are using them
% here.

The \dfn{tensor algebra} $T(V)=\bigoplus_{n \geq 0} V^{\otimes n}$
on a $\kk$-module $V$ is an associative $\kk$-algebra
spanned (as $\kk$-module) by decomposable tensors
$v_1 v_2 \cdots v_k := v_1 \otimes v_2 \otimes \cdots \otimes v_k$ with
$k\in\NN$ and $v_1, v_2, \ldots, v_k \in V$. Its multiplication is
defined $\kk$-linearly by
\[
m \left( v_1 v_2 \cdots v_k \otimes w_1 w_2 \cdots w_\ell \right)
:= v_1 v_2 \cdots v_k w_1 w_2 \cdots w_\ell
\]
\footnote{Some remarks about our notation (which we are using here
and throughout these notes) are in order.

Since we are working with tensor products of $\kk$-modules
like $T\left(V\right)$ -- which themselves are made of tensors -- here,
we must specify what the $\otimes$ sign means in expressions like
$a \otimes b$ where $a$ and $b$ are elements of $T\left(V\right)$.
Our convention is the following:
When $a$ and $b$ are elements of a tensor algebra $T\left(V\right)$,
we always understand $a \otimes b$ to mean the pure tensor
$a \otimes b \in T\left(V\right) \otimes T\left(V\right)$
rather than the product of $a$ and $b$ inside the tensor algebra
$T\left(V\right)$. The latter product will plainly be written $ab$.

The operator
precedence between $\otimes$ and multiplication in $T\left(V\right)$
is such that multiplication in $T\left(V\right)$ binds more
tightly than the $\otimes$ sign; e.g., the term $ab \otimes cd$ means
$\left(ab\right) \otimes \left(cd\right)$.
The same convention applies to any algebra instead of
$T\left(V\right)$.
}
for all $k,\ell\in\NN$ and
$v_1, v_2, \ldots, v_k, w_1, w_2, \ldots, w_\ell$ in $V$.
The unit map $u : \kk \to T\left(V\right)$ sends $1_\kk$ to the
empty tensor $1_{T\left(V\right)} = 1_\kk \in \kk = V^{\otimes 0}$.

If $V$ is a free $\kk$-module, say with $\kk$-basis $\{x_i\}_{i \in I}$,
then $T(V)$ has a $\kk$-basis of decomposable tensors
$x_{i_1} \cdots x_{i_k} := x_{i_1} \otimes \cdots \otimes x_{i_k}$
indexed by \emph{words} $(i_1,\ldots,i_k)$ in the alphabet $I$,
and the multiplication on this basis is given by concatenation of words:
\[
m( x_{i_1} \cdots x_{i_k} \otimes  x_{j_1} \cdots x_{j_\ell}) =
x_{i_1} \cdots x_{i_k} x_{j_1} \cdots x_{j_\ell}.
\]
\end{example}

% [DG][v17] Separated general and free cases in the above definition.

% [DG][v27] I merged three footnotes about notations into a single one,
% which defines how notation like $v_1 v_2 \cdots v_k \otimes
% w_1 w_2 \cdots w_\ell$ is to be understood. (It was a mistake of mine
% to have three footnotes for this to begin with!)

Recall that a \dfn{two-sided ideal} of a $\kk$-algebra $A$
is defined to be a $\kk$-submodule $J$ of $A$ such that
all $j \in J$ and $a \in A$ satisfy $ja \in J$ and $aj \in J$.
Using tensors, we can restate this as follows:
A \dfn{two-sided ideal} of a $\kk$-algebra $A$
means a $\kk$-submodule $J$ of $A$
satisfying $m(J \otimes A) \subset J$
and $m(A \otimes J) \subset J$.
Often, the word ``two-sided'' is omitted and one just speaks of
an ideal.

It is well-known that if $J$ is a two-sided ideal of a
$\kk$-algebra $A$,
then one can form a \emph{quotient algebra} $A/J$.

% [DG][v77] Expanded the above two paragraphs (previously a
% single paragraph).

\begin{example}
\label{exa.SymV}
Let $V$ be a $\kk$-module.
The \dfn{symmetric algebra} $\Sym(V)=\bigoplus_{n \geq 0} \Sym^n(V)$ is the
quotient of $T(V)$ by the two-sided ideal generated by
all elements $x y - y x$ with $x,y$ in $V$.
When $V$ is a free $\kk$-module with basis $\left\{x_i\right\}_{i \in I}$,
this symmetric algebra $S\left(V\right)$ can be identified
with a (commutative) polynomial algebra $\kk[x_i]_{i \in I}$,
having a $\kk$-basis of (commutative) monomials
$x_{i_1} \cdots x_{i_k}$ as
$\left\{ i_1,\ldots,i_k \right\}_{\operatorname{multiset}}$
runs through all finite multisubsets\footnote{By
a \dfn{multisubset} of a set $S$, we mean a multiset each of whose
elements belongs to $S$ (but can appear arbitrarily often).}
of $I$, and with multiplication defined $\kk$-linearly via
multiset union\footnote{The \dfn{multiset union} of two finite
multisets $A$ and $B$ is defined to be the multiset $C$ with the
property that every $x$ satisfies
\[
\left(\text{multiplicity of $x$ in $C$}\right)
= \left(\text{multiplicity of $x$ in $A$}\right)
+ \left(\text{multiplicity of $x$ in $B$}\right) .
\]
Equivalently, the multiset union of
$\left\{a_1, a_2, \ldots, a_k\right\}_{\operatorname{multiset}}$
and
$\left\{b_1, b_2, \ldots, b_\ell\right\}_{\operatorname{multiset}}$
is
$\left\{a_1, a_2, \ldots, a_k, b_1, b_2, \ldots, b_\ell\right\}_{\operatorname{multiset}}$.
The multiset union is also known as the
\emph{disjoint union}\index{disjoint union of multisets}
of multisets.
}.
\end{example}

% [DG][v25] Added clarifications about basis and footnote about
% the meaning of "multisubsets".

% [DG][v77] Added footnote about multiset union. Unfortunately,
% the definition is NOT standard in the literature; to some
% people, the multiset union takes maxima instead of sums of
% multiplicities.

Note that the $\kk$-module $\kk$ itself canonically becomes a
$\kk$-algebra.
Its associative operation $m : \kk \otimes \kk \to \kk$ is the canonical
isomorphism $\kk \otimes \kk \to \kk$, and its unit $u : \kk \to \kk$
is the identity map.

% [DG][v13] I added the above nugget of pedantry, and a similar one
% about the $\kk$-coalgebra $\kk$, to explain your use of labels like
% $\Delta$ on a map $\kk \otimes \kk \to \kk$ in commutative diagrams.

Topology and group theory give more examples.

\begin{example}
The \emph{cohomology algebra} $H^*(X;\kk)=\bigoplus_{i \geq 0}H^i(X;\kk)$
with coefficients in $\kk$ for
a topological space $X$ has an associative \emph{cup product}.  Its unit
$\kk =H^*(\pt;\kk) \overset{u}{\rightarrow} H^*(X;\kk)$ is induced
from the unique (continuous) map $X \rightarrow \pt$, where $\pt$ is a one-point space.
\end{example}

\begin{example}
\label{group-algebra-example}
For a group $G$, the \dfn{group algebra} \dfn{$\kk G$} has $\kk$-basis
$\{t_g\}_{g \in G}$ and multiplication defined $\kk$-linearly by
$t_g t_h = t_{gh}$, and
unit defined by $u(1) = t_e$, where $e$ is the identity element of $G$.
\end{example}



\subsection{Coalgebras}

In Definition~\ref{def.algebra}, we have defined the notion of an
algebra entirely in terms of linear maps; thus, by reversing
all arrows, we can define a dual notion, which is called a
\emph{coalgebra}.
If we are to think of the multiplication $A \otimes A  \rightarrow A$
in an algebra as \emph{putting together} two basis elements of $A$ to get a
sum of basis elements of $A$, then coalgebra structure should be thought of
as \emph{taking basis elements apart}.

% [DG][v77] Added the first sentence in the above paragraph.

\begin{definition}
\label{def.coalgebra}
A \emph{co-associative $\kk$-coalgebra}\index{coalgebra}\index{coassociative}
is a $\kk$-module $C$ equipped with
a \dfn{comultiplication}\index{$\Delta$}, that is, a $\kk$-linear map
$C \overset{\Delta}{\rightarrow} C \otimes C$,
and a $\kk$-linear \dfn{counit}\index{$\epsilon$}
$C \overset{\epsilon}{\rightarrow} \kk$
for which the following diagrams (which are exactly the diagrams
in \eqref{associativity-diagram} and \eqref{unit-diagram} but with
\emph{all arrows reversed}) are commutative:

% [DG][v77] Reworded the preceding paragraph somewhat.

\begin{equation}
\label{coassociativity-diagram}
\xymatrix{
& C  \otimes C \otimes C &  \\
C \otimes C \ar[ur]^{\Delta \otimes \id} & &  C \otimes C \ar[ul]_{\id \otimes \Delta} \\
& C \ar[ur]^\Delta \ar[ul]_\Delta&
}
\end{equation}

\begin{equation}
\label{counit-diagram}
\xymatrix{
C \otimes \kk \ar[r] & C & \kk \otimes C \ar[l] \\
C \otimes C \ar[u]^{\id \otimes \epsilon} & C \ar[l]^-{\Delta} \ar[u]^\id \ar[r]_-{\Delta} & C \otimes C \ar[u]^{\epsilon \otimes \id}
}
\end{equation}
Here the maps $C \otimes \kk \rightarrow C$ and
$\kk \otimes C \rightarrow C$ are the isomorphisms
sending $c \otimes \one \mapsto c$ and
$\one \otimes c \mapsto c$.

We abbreviate ``co-associative $\kk$-coalgebra'' as ``$\kk$-coalgebra''
(co-associativity, i.e., the commutativity of the diagram
\eqref{coassociativity-diagram}, is assumed unless otherwise specified) or as
``coalgebra'' (when $\kk$ is clear from the context).

Sometimes, the word ``coproduct'' is used as a synonym for
``comultiplication''\footnote{although the word ``coproduct''
already has a different meaning in algebra}.
\end{definition}

% [DG][v25] Added the last sentence about "coproduct". (You do
% use it in these notes.)

One often uses the \dfn{Sweedler notation}
\begin{equation}
\Delta(c) = \sum_{(c)} c_1 \otimes c_2 = \sum c_1 \otimes c_2
\label{eq.sweedler-not.def1}
\end{equation}
to abbreviate formulas involving $\Delta$.
This means that an expression of the form
$\sum_{(c)} f\left(c_1, c_2\right)$ (where $f : C \times C \to M$
is some $\kk$-bilinear map from $C \times C$ to some
$\kk$-module $M$)
has to be understood to mean $\sum_{k=1}^m f\left(d_k, e_k\right)$,
where $k \in \NN$ and $d_1, d_2, \ldots, d_k \in C$ and
$e_1, e_2, \ldots, e_k \in C$ are chosen such that
$\Delta\left(c\right) = \sum_{k=1}^m d_k \otimes e_k$.
(There are many ways to choose such $k$, $d_i$ and $e_i$, but
they all produce the same result
$\sum_{k=1}^m f\left(d_k, e_k\right)$.
Indeed, the result they produce is
$F\left(\Delta\left(c\right)\right)$, where
$F : C \otimes C \to M$ is the $\kk$-linear map induced by
the bilinear map $f$.)
For example, commutativity of the
left square in \eqref{counit-diagram} asserts that
$\sum_{(c)} c_1 \epsilon(c_2) = c$ for each $c \in C$.
Likewise, commutativity of the right square in
\eqref{counit-diagram} asserts that
$\sum_{(c)} \epsilon(c_1) c_2 = c$ for each $c \in C$.
The commutativity of \eqref{coassociativity-diagram} can
be written as
$\sum_{(c)} \Delta(c_1) \otimes c_2 = \sum_{(c)} c_1 \otimes \Delta(c_2)$,
or (using nested Sweedler notation to unravel the two remaining $\Delta$'s)
as
\[
\sum_{(c)} \sum_{(c_1)} (c_1)_1 \otimes (c_1)_2 \otimes c_2
=
\sum_{(c)} \sum_{(c_2)} c_1 \otimes (c_2)_1 \otimes (c_2)_2 .
\]

% [DG][v79] Added a bit of explanation for the Sweedler notation.

The $\kk$-module $\kk$ itself canonically becomes a
$\kk$-coalgebra, with its comultiplication $\Delta : \kk \to \kk \otimes \kk$
being the canonical isomorphism $\kk \to \kk \otimes \kk$, and its counit
$\epsilon : \kk \to \kk$ being the identity map.

\begin{example}
Let $\kk$ be a field.
The \emph{homology} $H_*(X;\kk)=\bigoplus_{i \geq 0} H_i(X;\kk)$
for a topological space $X$ is naturally a coalgebra: the
(continuous) \dfn{diagonal embedding}
$X \rightarrow X \times X$ sending $x \mapsto (x,x)$ induces
a coassociative map
\[
H_*(X; \kk) \rightarrow H_*(X \times X;\kk) \cong H_*(X;\kk) \otimes H_*(X;\kk)
\]
in which the last isomorphism comes from the \emph{K\"unneth theorem}
with field coefficients $\kk$.
As before, the unique (continuous) map $X \rightarrow \pt$ induces the
counit $H_*(X;\kk) \overset{\epsilon}{\rightarrow} H_*(\pt;\kk) \cong \kk$.
\end{example}

\begin{exercise}
\label{exe.counit.unique}
Let $C$ be a $\kk$-module, and let
$\Delta : C \rightarrow C \otimes C$ be a $\kk$-linear map.
Prove that there exists
\emph{at most one} $\kk$-linear map
$\epsilon : C \rightarrow \kk$
such that the diagram \eqref{counit-diagram} commutes.
\end{exercise}

% [DG][v3] Added exercise, 19 Apr 2014.

For us, the notion of a coalgebra serves mostly as a stepping
stone towards that of a Hopf algebra, which will be the focus
of these notes.
However, coalgebras have interesting properties of their own
(see, e.g., \cite{Manetti-voyage}).

\subsection{Morphisms, tensor products, and bialgebras}

Just as we rewrote the definition of an algebra in terms of linear
maps (in Definition~\ref{def.algebra}), we can likewise rephrase
the standard definition of a morphism of algebras:

% [DG][v77] Added the preceding paragraph.
% Also slightly rewrote the first sentence of the following
% definition:

\begin{definition} \label{def.morphism-alg-coalg}
A \dfn{morphism of algebras}
is a $\kk$-linear map $A \overset{\varphi}{\rightarrow} B$
between two $\kk$-algebras $A$ and $B$
that makes the following two diagrams commute:

\begin{equation}
\qquad
\xymatrix{
A  \ar[r]^{\varphi} & B  \\
A \otimes A \ar[u]^{m_A} \ar[r]^{\varphi \otimes \varphi} & B \otimes B \ar[u]_{m_B}
}
\qquad
\xymatrix{
A \ar[rr]^{\varphi}& & B  \\
&\kk \ar[ul]^{u_A} \ar[ur]_{u_B}&
}
\end{equation}
Here the subscripts on $m_A,m_B,u_A,u_B$ indicate for which algebra they are
part of the structure (e.g., the map $u_A$ is the map $u$ of the algebra $A$);
we will occasionally use such conventions from now on.

Similarly, a \dfn{morphism of coalgebras}
is a $\kk$-linear map $C \overset{\varphi}{\rightarrow} D$
between two $\kk$-coalgebras $C$ and $D$
that makes the reverse diagrams commute:

\begin{equation}
\label{coalgebra-morphism-diagrams}
\xymatrix{
C \ar[d]_{\Delta_C} \ar[r]^{\varphi} & D \ar[d]^{\Delta_D} \\
C \otimes C \ar[r]^{\varphi \otimes \varphi} & D \otimes D
}
\qquad
\xymatrix{
C \ar[dr]_{\epsilon_C} \ar[rr]^{\varphi}& & D \ar[dl]^{\epsilon_D}  \\
&\kk&
}
\end{equation}

As usual, we shall use the word ``\emph{homomorphism}''
as a synonym for ``morphism'', and we will say
``$\kk$-coalgebra homomorphism'' for ``homomorphism
of coalgebras'' (and similarly for algebras and other
structures).

As usual, the word ``\dfn{isomorphism}'' (of algebras,
of coalgebras, or of other structures that we will define
further below) means ``invertible morphism whose inverse is
a morphism as well''.
Two algebras (or coalgebras, or other structures) are said
to be \dfn{isomorphic} if there exists an isomorphism
between them.
\end{definition}

% [DG][v80] Added the above paragraph about isomorphisms.

\begin{example}
Let $\kk$ be a field.
Continuous maps $X \overset{f}{\rightarrow} Y$ of topological spaces
induce algebra morphisms $H^*(Y;\kk) \rightarrow H^*(X;\kk)$,
and coalgebra morphisms $H_*(X;\kk) \rightarrow H_*(Y;\kk)$.
\end{example}

Coalgebra morphisms behave similarly to algebra morphisms in
many regards:
For example, the inverse of an invertible coalgebra morphism
is again a coalgebra morphism\footnote{The easy proof
of this fact is left to the reader.}. Thus, the invertible
coalgebra morphisms are precisely the coalgebra isomorphisms.

% [DG][v70] Added the above paragraph.

\begin{definition}
\label{def.tensor-prod-algs}
Given two $\kk$-algebras $A, B$, their tensor product $A \otimes B$ also
becomes a $\kk$-algebra defining the multiplication bilinearly via\index{tensor product of algebras}
\[
m((a \otimes b) \otimes (a' \otimes b')) := aa' \otimes bb' ,
\]
or, in other words, $m_{A \otimes B}$ is the composite map
\[
\xymatrixcolsep{4pc}
\xymatrix{
A \otimes B \otimes A \otimes B
\ar[r]^-{\id \otimes T \otimes \id}
&
A \otimes A \otimes B \otimes B \ar[r]^-{m_A \otimes m_B}
&
A \otimes B
}
\]
where $T$ is the \dfn{twist map} $B \otimes A \rightarrow A \otimes B$
that sends $b \otimes a \mapsto a \otimes b$.
(See Exercise~\ref{exe.tensor-prod-algs}(a) below for a proof that
this $\kk$-algebra $A \otimes B$ is well-defined.)

Here we are omitting the topologist's sign\index{topologist's sign convention}
in the twist map which should be present for graded algebras
and coalgebras that come from cohomology and homology:
For homogeneous elements $a$ and $b$ of two graded modules
$A$ and $B$, the topologist's twist map
$T : B \otimes A \to A \otimes B$ sends
\begin{equation}
\label{topologist-twist}
b \otimes a \longmapsto  (-1)^{\deg(b)\deg(a)} a \otimes b
\end{equation}
instead of $b \otimes a \mapsto a \otimes b$.
This means that, if one is using the topologists' convention,
most of our examples which we later call \emph{graded}
should actually be considered to live in only \emph{even} degrees
(which can be achieved, e.g., by artificially doubling all
degrees in their grading).  We will, however, keep to our own
definitions (so that our twist map $T$ will always send
$b \otimes a \mapsto a \otimes b$) unless otherwise
noted.  Only in parts of Exercise~\ref{exe.Sym.wedge} will we use
the topologist's sign.  Readers interested in the wide world of
algebras defined using the topologist's sign convention
(which is also known as the \dfn{Koszul sign rule})
can consult \cite[Appendix A2]{Eisenbud}; see also
\cite{GrosshansRotaStein} for applications to algebraic combinatorics%
\footnote{To be precise, \cite{GrosshansRotaStein} works with the
related concept of \dfn{superalgebras}, which are graded by elements
of $\ZZ / 2 \ZZ$ rather than $\NN$ but use the same sign convention
as the topologists have for algebras.}.

% old wording:
% We will ignore this issue, and hope that it causes no confusion later!

% [DG][v25] I changed the above paragraph to be less scary to
% the reader and less apologetic towards topologists
% -- it used to sound as though we were cheating,
% while all it is supposed to say (I hope) is that we are missing
% out on some of the variety of Hopf algebras. To compensate, I
% pointed out that the bialgebra is super in the homological
% example below and in the exercise on the exterior algebra.
% I hope I caught all places where this matters.
% 
% Note that if we wanted to deal with the super setting for real,
% I think we would also have to change the definitions of
% (co)commutativity as well, unless we require $2$ invertible.
% (And the Jacobi identity for Lie algebras becomes even worse.)

% [DG][v48] Replaced rightarrows/longrightarrows with oversets by
% xymatrix diagrams. This should prevent the labels of the arrows
% from protruding beyond the arrows.

% [DG][v80] Added more sentences (including references) to the
% above semi-apologetic paragraph about the topologist's sign.

The unit element of $A \otimes B$ is $\one_A \otimes \one_B$,
meaning that the
unit map $\kk \overset{u_{A \otimes B}}{\rightarrow} A \otimes B$ is the composite
\[
\xymatrixcolsep{4pc}
\xymatrix{
\kk \ar[r] &
\kk \otimes \kk \ar[r]^-{u_A \otimes u_B} &
A \otimes B
} .
\]

Similarly, given two coalgebras $C, D$, one can make $C \otimes D$
a coalgebra in which the comultiplication and counit maps are the composites of\index{tensor product of coalgebras}
\[
\xymatrixcolsep{4pc}
\xymatrix{
C \otimes D \ar[r]^-{\Delta_C \otimes \Delta_D} &
C \otimes C \otimes D \otimes D
\ar[r]^-{\id \otimes T \otimes \id} &
C \otimes D \otimes C \otimes D
}
\]
and
\[
\xymatrixcolsep{4pc}
\xymatrix{
C \otimes D \ar[r]^-{\epsilon_C \otimes \epsilon_D} &
\kk \otimes \kk \ar[r] & \kk
} .
\]
(See Exercise~\ref{exe.tensor-prod-algs}(b) below for a proof that
this $\kk$-coalgebra $C \otimes D$ is well-defined.)
\end{definition}

\begin{exercise}
\phantomsection\label{exe.tensor-prod-algs}

\begin{enumerate}
\item[(a)] Let $A$ and $B$ be two $\kk$-algebras. Show that the
$\kk$-algebra $A\otimes B$ introduced in
Definition~\ref{def.tensor-prod-algs} is actually
well-defined (i.e., its multiplication
and unit satisfy the axioms of a $\kk$-algebra).

\item[(b)] Let $C$ and $D$ be two $\kk$-coalgebras. Show that the
$\kk$-coalgebra $C\otimes D$ introduced in
Definition~\ref{def.tensor-prod-algs} is actually
well-defined (i.e., its comultiplication and counit satisfy
the axioms of a $\kk$-coalgebra).
\end{enumerate}
\end{exercise}

% [DG][v68] Added the above exercise for the sake of
% completeness.

It is straightforward to show that the concept of tensor
products of algebras and of coalgebras satisfy the properties
one would expect:

\begin{itemize}
\item For any three $\kk$-coalgebras $C$, $D$ and $E$,
the $\kk$-linear map
\[
\left(C \otimes D\right) \otimes E
\to C \otimes \left(D \otimes E\right),
\qquad \left( c \otimes d \right) \otimes e
\mapsto c \otimes \left( d \otimes e \right)
\]
is a coalgebra isomorphism.
This allows us to speak of the $\kk$-coalgebra
$C \otimes D \otimes E$ without worrying about the
parenthesization.
\item For any two $\kk$-coalgebras $C$ and $D$, the
$\kk$-linear map
\[
T : C \otimes D \to D \otimes C,
\qquad c \otimes d \mapsto d \otimes c
\]
is a coalgebra isomorphism.
\item For any $\kk$-coalgebra $C$, the $\kk$-linear maps
\begin{align*}
C &\to \kk \otimes C, \qquad c \mapsto 1 \otimes c
\qquad \text{and} \\
C &\to C \otimes \kk, \qquad c \mapsto c \otimes 1
\end{align*}
are coalgebra isomorphisms.
\item Similar properties hold for algebras instead of
coalgebras.
\end{itemize}

% [DG][v70] Added the above list.

\noindent
One of the first signs that these definitions interact nicely is the following straightforward proposition.

\begin{proposition}
\label{bialgebra-prop}
When $A$ is both a $\kk$-algebra and a
$\kk$-coalgebra, the following are equivalent:
\begin{itemize}
\item The maps $\Delta$ and $\epsilon$ are morphisms for the algebra structure $(A,m,u)$.
\item The maps $m$ and $u$ are morphisms for the coalgebra structure $(A,\Delta,\epsilon)$.
\item These four diagrams commute:
\begin{equation}
\label{bialgebra-diagrams}
\begin{array}{ccc}
& \xymatrix{
 & A \otimes A \ar[dl]_{\Delta \otimes \Delta} \ar[ddr]_m & \\
A \otimes A \otimes A \otimes A \ar[dd]_{\id \otimes T \otimes \id} &  & \\
 & & A \ar[ddl]_\Delta\\
A \otimes A \otimes A \otimes A \ar[dr]_{m \otimes m} &  & \\
 & A \otimes A &
} & \\
\xymatrix{
A \otimes A \ar[r]^{\epsilon \otimes \epsilon} \ar[d]_{m} &\kk \otimes \kk \ar[d]^m \\
A \ar[r]_{\epsilon}  &\kk
} & &
\xymatrix{
\kk \ar[r]^{u} \ar[d]_{\Delta} &A \ar[d]^\Delta \\
\kk \otimes \kk \ar[r]_{u \otimes u}  &A \otimes A
}\\
& \xymatrix{
\kk \ar[dr]_{u} \ar[rr]^{\id} & & \kk \\
 &A \ar[ur]_{\epsilon} &
} &
\end{array}
\end{equation}
\end{itemize}
\end{proposition}

% [DG][v78] Slight rewording of the first two equivalent
% conditions above.

\begin{exercise} \phantomsection
\label{exe.tensorprod.morphism}
\begin{itemize}
\item[(a)] If $A$, $A^\prime$, $B$ and $B^\prime$ are four
$\kk$-algebras, and $f : A \to A^\prime$ and $g : B \to B^\prime$
are two $\kk$-algebra homomorphisms, then show that $f \otimes g :
A \otimes B \to A^\prime \otimes B^\prime$ is a $\kk$-algebra
homomorphism.
\item[(b)] If $C$, $C^\prime$, $D$ and $D^\prime$ are four
$\kk$-coalgebras, and $f : C \to C^\prime$ and $g : D \to D^\prime$
are two $\kk$-coalgebra homomorphisms, then show that
$f \otimes g : C \otimes D \to C^\prime \otimes D^\prime$ is a
$\kk$-coalgebra homomorphism.
\end{itemize}
\end{exercise}

% [DG][v25] Added the above exercise (as a lemma for another
% exercise further down).

\begin{definition}
Call the $\kk$-module $A$ a \emph{$\kk$-bialgebra}\index{bialgebra} if it is
a $\kk$-algebra and $\kk$-coalgebra satisfying the three
equivalent conditions in Proposition~\ref{bialgebra-prop}.
\end{definition}

\begin{example}
For a group $G$, one can make the \idx{group algebra} \idx{$\kk G$} a coalgebra with
counit $\kk G \overset{\epsilon}{\rightarrow} \kk$ mapping
$t_g \mapsto 1$ for all $g$ in $G$, and with comultiplication
$\kk G \overset{\Delta}{\rightarrow} \kk G \otimes \kk G$ given by
$\Delta(t_g) := t_g \otimes t_g$.  Checking the various diagrams in
\eqref{bialgebra-diagrams} commute is easy.  For example, one
can check the pentagonal diagram on each basis element $t_g \otimes t_h$:
\[
\xymatrix{
 & t_g \otimes t_h \ar[dl]_{\Delta \otimes \Delta} \ar[ddr]_m & \\
t_g \otimes t_g \otimes t_h \otimes t_h \ar[dd]_{\id \otimes T \otimes \id} &  & \\
 & & t_{gh} \ar[ddl]_\Delta\\
t_g \otimes t_h \otimes t_g \otimes t_h \ar[dr]_{m \otimes m} &  & \\
 & t_{gh} \otimes t_{gh} &
}
\]
\end{example}

\begin{remark}
\label{comultiplication-as-tensor-action}
In fact, one can think of adding a bialgebra structure to a $\kk$-algebra $A$
as a way of making $A$-modules $M,N$ have an $A$-module structure on their
tensor product $M \otimes N$:  the algebra $A \otimes A$ already acts naturally
on $M \otimes N$, so one can let $a$ in $A$ act via $\Delta(a)$ in $A \otimes A$. 
In the theory of group representations over $\kk$, that is, $\kk G$-modules $M$,
this is how one defines the \dfn{diagonal action} of $G$ on $M \otimes N$,
namely $t_g$ acts as $t_g \otimes t_g$.
\end{remark}

\begin{definition}
An element $x$ in a coalgebra for which $\Delta(x) = x \otimes x$ and
$\epsilon(x)=1$ is called \dfn{group-like}.

% [DG] The "coalgebra" above used to say "bialgebra", but I've made it a bit
% less wasteful. (You don't ever use it in either generality, though.)

An element $x$ in a bialgebra for which $\Delta(x) = 1 \otimes x+x\otimes 1$
is called \dfn{primitive}.
We shall also sometimes abbreviate
``primitive element'' as ``primitive''.
\end{definition}

% [DG][v64] Added the preceding sentence.

\begin{example}
Let $V$ be a $\kk$-module.
The \dfn{tensor algebra} $T(V)=\bigoplus_{n \geq 0} V^{\otimes n}$ is
a coalgebra, with counit $\epsilon$ equal to
the identity on $V^{\otimes 0} = \kk$ and the zero map on
$V^{\otimes n}$ for $n > 0$, and with comultiplication defined
to make the elements $x$ in $V^{\otimes 1}=V$ all primitive:
\[
\Delta(x) := \one \otimes x + x \otimes \one \text{ for }x \in V^{\otimes 1}.
\]
Since the elements of $V$
generate $T(V)$ as a $\kk$-algebra, and since $T(V) \otimes T(V)$
is also an associative $\kk$-algebra, the universal property of $T(V)$ as
the free associative $\kk$-algebra on the generators $V$
allows one to define $T(V) \overset{\Delta}{\rightarrow} T(V) \otimes T(V)$ arbitrarily on $V$, and extend it as
an algebra morphism.

It may not be obvious that this $\Delta$ is coassociative,
but one can prove this as follows. Note that
\[
\left(
(\id \otimes \Delta) \circ \Delta
\right)(x) =
x \otimes \one \otimes \one +
\one \otimes x \otimes \one +
\one \otimes \one \otimes x
=\left( (\Delta \otimes \id) \circ \Delta \right)(x)
\]
for every $x$ in $V$.  Hence the two
maps $(\id \otimes \Delta) \circ \Delta$ and
$(\Delta \otimes \id) \circ \Delta$, considered
as algebra morphisms
$T(V) \rightarrow T(V) \otimes T(V) \otimes T(V)$,
must coincide on every element of $T(V)$ since they coincide on $V$.
We leave it as an exercise to check the map $\epsilon$ defined as
above satisfies the counit axioms \eqref{counit-diagram}.

Here is a sample calculation in $T(V)$ when $x,y,z$
are three elements of $V$:
\begin{align*}
\Delta(xyz) &= \Delta(x) \Delta(y) \Delta(z) \\
&=(1 \otimes x+x\otimes 1)(1 \otimes y+y\otimes 1)(1 \otimes z+z\otimes 1) \\
&=(1 \otimes xy + x \otimes y+ y \otimes x + xy \otimes 1)(1 \otimes z+z\otimes 1) \\
&=1 \otimes xyz
+ x \otimes yz + y \otimes xz + z \otimes xy\\
&\qquad + xy \otimes z + xz \otimes y + yz \otimes x
+xyz \otimes 1 .
\end{align*}
This illustrates the idea that comultiplication ``takes basis elements apart''
(and, in the case of $T(V)$, not just basis elements, but any decomposable tensors).
Here for any $v_1, v_2, \ldots, v_n$ in $V$ one has
\[
\Delta\left( v_1 v_2 \cdots v_n \right) =
\sum
v_{j_1} \cdots v_{j_r}
\otimes v_{k_1} \cdots v_{k_{n-r}}
\]
where the sum is over ordered pairs
$\left( j_1, j_2, \ldots, j_r \right) , \left( k_1, k_2, \ldots, k_{n-r} \right)$
of complementary subwords of the word $\left( 1, 2, \ldots, n \right)$.
\footnote{More formally speaking, the sum is over all permutations
$\left(j_1, j_2, \ldots, j_r, k_1, k_2, \ldots, k_{n-r}\right)$ of
$\left(1,2,\ldots,n\right)$ satisfying $j_1 < j_2 < \cdots < j_r$
and $k_1 < k_2 < \cdots < k_{n-r}$.} Equivalently (and in a more
familiar language),
\begin{equation}
\label{eq.tensor-alg.m}
\Delta\left( v_1 v_2 \cdots v_n \right) =
\sum_{I \subset \left\{1,2,\ldots,n\right\}}
v_I \otimes v_{\left\{1,2,\ldots,n\right\} \setminus I},
\end{equation}
where $v_J$ (for $J$ a subset of $\left\{1,2,\ldots,n\right\}$)
denotes the product of all $v_j$ with $j \in J$ in the order of
increasing $j$.
\end{example}

% [DG][v25] Added footnote clarifying the use of $\otimes$, and
% added \eqref{eq.tensor-alg.m}.

% [DG][v27] Moved footnote into an earlier example.

We can rewrite the axioms of a $\kk$-bialgebra $A$ using Sweedler
notation.
Indeed, asking for $\Delta : A \to A \otimes A$ to be a
$\kk$-algebra morphism is equivalent to requiring that
\begin{equation}
\sum_{(ab)} (ab)_1 \otimes (ab)_2
= \sum_{(a)} \sum_{(b)} a_1 b_1 \otimes a_2 b_2
\qquad \text{ for all $a, b \in A$}
\label{eq.bialgebra.sweedler-Deltaab}
\end{equation}
and $\sum_{(1)} 1_1 \otimes 1_2 = 1_A \otimes 1_A$.
(The other axioms have already been rewritten or don't
need Sweedler notation.)

% [DG][v79] Added the paragraph above.

Recall one can quotient a $\kk$-algebra $A$ by a two-sided ideal $J$ to
obtain a quotient algebra $A/J$.
An analogous construction can be done for coalgebras using the
following concept, which is dual to that of a two-sided ideal:

\begin{definition}
In a coalgebra $C$, a \dfn{two-sided coideal}\index{coideal} is a
$\kk$-submodule $J \subset C$ for which
\begin{align*}
\Delta(J) &\subset J \otimes C + C \otimes J , \\
\epsilon(J) &= 0 .
\end{align*}
The quotient $\kk$-module $C/J$ then inherits a coalgebra
structure\footnote{Indeed, $J \otimes C + C \otimes J$ is contained in
the kernel of the canonical map $C \otimes C \to
\left(C/J\right) \otimes \left(C/J\right)$;
therefore, the condition $\Delta(J) \subset J \otimes C + C \otimes J$
shows that the map $C \overset{\Delta}{\rightarrow} C \otimes C
\twoheadrightarrow \left(C/J\right) \otimes \left(C/J\right)$ factors through
a map $\overline{\Delta} : C/J \to \left(C/J\right) \otimes \left(C/J\right)$.
Likewise, $\epsilon(J) = 0$ shows that the map $\epsilon : C \to \kk$
factors through a map $\overline{\epsilon} : C/J \to \kk$.
Equipping $C/J$ with these maps $\overline{\Delta}$ and $\overline{\epsilon}$,
we obtain a coalgebra (as the commutativity of the required diagrams
follows from the corresponding property of $C$).}.
Similarly, in a bialgebra $A$, a subset $J \subset A$ which is both
a two-sided ideal and two-sided coideal gives rise to a quotient
bialgebra $A/J$.
\end{definition}

% [DG][v77] More details added to the above footnote.

\begin{exercise}
\label{exe.kernel-is-coideal}
Let $A$ and $C$ be two $\kk$-coalgebras, and $f : A \to C$ a
surjective coalgebra homomorphism.
\begin{itemize}
\item[(a)] If $f$ is surjective, then show that $\ker f$
is a two-sided coideal of $A$.
\item[(b)] If $\kk$ is a field, then show that $\ker f$
is a two-sided coideal of $A$.
\end{itemize}
\end{exercise}

% [DG][v13] Added the above exercise.

% [DG][v23] Extended it by part (b).

\begin{example}
\label{symmetric-algebra-as-bialgebra-example}
Let $V$ be a $\kk$-module.
The \emph{symmetric algebra} $\Sym(V)$ was defined as the
quotient of the tensor algebra $T(V)$
by the two-sided ideal $J$ generated by
all \dfn{commutators} $[x,y]=x y - y x$ for $x,y$ in $V$
(see Example~\ref{exa.SymV}).
Note that $x,y$ are primitive elements in $T(V)$,
and the following very reusable calculation
shows that \emph{the commutator of two primitives is primitive}:
\begin{align}
\Delta [x,y] &= \Delta( xy-yx )
= \Delta\left(x\right) \Delta\left(y\right) - \Delta\left(y\right) \Delta\left(x\right) \nonumber\\
& \qquad \left(\text{since $\Delta$ is an algebra homomorphism}\right) \nonumber\\
&=(\one \otimes x+x\otimes \one)(\one \otimes y+y\otimes \one)
-(\one \otimes y+y\otimes \one)(\one \otimes x+x\otimes \one)\nonumber\\
&=\one \otimes xy - \one \otimes yx +
  xy \otimes \one - yx \otimes \one \nonumber\\
&\qquad +
  x \otimes y + y \otimes x
 - x \otimes y - y \otimes x \nonumber\\
&=\one \otimes (xy-yx) + (xy-yx) \otimes \one \nonumber\\
&=\one \otimes [x,y] + [x,y] \otimes \one .
\label{commutator-of-primitives}
\end{align}
In particular, the commutators $[x,y]$ have $\Delta[x,y]$ in
$J \otimes T(V) + T(V) \otimes J$.  They also satisfy
$\epsilon([x,y])=0$.  Since they are generators for
$J$ as a two-sided ideal, it is not hard to see this implies
$\Delta (J) \subset J \otimes T(V) + T(V) \otimes J$,
and $\epsilon(J) = 0$.  Thus $J$ is
also a two-sided coideal, and
$\Sym(V)=T(V)/J$ inherits a bialgebra structure.
\end{example}

In fact we will see in Section~\ref{PSH-implies-polynomial-section}
that symmetric algebras are the universal example of bialgebras which are
\emph{graded, connected, commutative, cocommutative}.  But first we should
define some of these concepts.


\begin{definition} \phantomsection
\begin{enumerate}

\item[(a)]
A \emph{graded $\kk$-module}\index{graded module}\footnote{%
also known as an ``\emph{$\NN$-graded $\kk$-module}''}
is a $\kk$-module $V$ equipped with a $\kk$-module
direct sum decomposition $V=\bigoplus_{n \geq 0} V_n$.
In this case, the addend $V_n$ (for any given $n \in \NN$)
is called the
\emph{$n$-th homogeneous component}\index{homogeneous component}
(or the \emph{$n$-th graded component}\index{graded component})
of the graded $\kk$-module $V$.
Furthermore, elements $x$ in $V_n$
are said to be \dfn{homogeneous} of degree $n$; occasionally, the
notation $\deg(x)=n$ is used to signify this\footnote{This
notation should not be taken too literally, as it would absurdly
imply that $\deg(0)$ ``equals'' every $n \in \NN$ at the same time,
since $0 \in V_n$ for all $n$.}.
The decomposition $\bigoplus_{n \geq 0} V_n$ of $V$ (that is,
the family of submodules $\left( V_n \right)_{n \in \NN}$)
is called the \dfn{grading} of $V$.

% [DG][v14] Where do you use the freeness assumption? I'd much like to
% get rid of it. (In PSH-theory, freeness follows from the axioms
% anyway. In the section about dual spaces, I think we should be
% maximally explicit about its use.)

% [DG][v17] Moved the freeness assumption to the definition of
% "finite type".

\item[(b)]
The tensor product $V \otimes W$ of two graded $\kk$-modules $V$
and $W$ is, by default, endowed
with the graded module structure in which%
\index{tensor product of graded modules}
\[
(V \otimes W)_n:=\bigoplus_{i+j=n} V_i \otimes W_j .
\]

\item[(c)]
A $\kk$-linear map $V \overset{\varphi}{\rightarrow} W$ between
two graded $\kk$-modules is called
\emph{graded}\index{graded linear map} if $\varphi(V_n) \subset W_n$ for all $n$.
Graded $\kk$-linear maps are also called
\emph{homomorphisms of graded $\kk$-modules}\index{homomorphism of graded $\kk$-modules}.
An \dfn{isomorphism of graded $\kk$-modules} means an
invertible graded $\kk$-linear map whose inverse is also graded.%
\footnote{We shall see in Exercise~\ref{exe.graded.iso} that
the ``whose inverse is also graded'' requirement is actually
superfluous (i.e., it is automatically satisfied for an
invertible graded $\kk$-linear map); we are imposing it only in
order to stick to our tradition of defining ``isomorphisms''
as invertible morphisms whose inverses are morphisms as well.}

\item[(d)]
Say that a $\kk$-algebra (or coalgebra, or bialgebra) is
\emph{graded}\index{graded algebra}\index{graded coalgebra}\index{graded bialgebra}
if it is a graded $\kk$-module and
all of the relevant structure maps ($u,\epsilon,m,\Delta$) are graded.

\item[(e)]
Say that a graded $\kk$-module $V$ is
\emph{connected}\index{connected graded module}
if $V_0 \cong \kk$.

\item[(f)]
Let $V$ be a graded $\kk$-module. Then, a
\emph{graded $\kk$-submodule of $V$}\index{graded submodule}
(sometimes also called a
\dfn{homogeneous $\kk$-submodule of $V$}\index{homogeneous submodule})
means a graded $\kk$-module $W$ such that $W \subset V$
as sets, and such that the inclusion map
$W \hookrightarrow V$ is a graded $\kk$-linear map. \\
Note that if $W$ is a graded $\kk$-submodule of $V$,
then the grading of $W$ is uniquely determined by
the underlying set of $W$ and the grading of $V$ --
namely, the $n$-th graded component $W_n$ of $W$
is $W_n = W \cap V_n$ for each $n \in \NN$.
Thus, we can specify a graded $\kk$-submodule of $V$
without explicitly specifying its grading.
From this point of view, a graded $\kk$-submodule of
$V$ can also be defined as a $\kk$-submodule $W$ of $V$
satisfying
$W = \sum_{n \in \NN} \left(W \cap V_n\right)$.
(This sum is automatically a direct sum, and thus
defines a grading on $W$.)

\end{enumerate}
\end{definition}

\begin{example}
Let $\kk$ be a field.
A path-connected space $X$ has its homology and cohomology
\begin{align*}
H_*(X;\kk)&=\bigoplus_{i \geq 0} H_i(X;\kk) ,\\
H^*(X;\kk)&=\bigoplus_{i \geq 0} H^i(X;\kk)
\end{align*}
carrying the structure of connected graded coalgebras and algebras, respectively.  If in addition, $X$ is a topological group,
or even less strongly, a \emph{homotopy-associative $H$-space} (e.g. the \emph{loop space} $\Omega Y$ on some other space $Y$),
the continuous multiplication map
$
X \times X \rightarrow X
$
induces an algebra structure on $H_*(X;\kk)$ and a coalgebra structure on
$H^*(X;\kk)$, so that each become bialgebras in the topologist's sense
(i.e., with the twist as in \eqref{topologist-twist}),
and these bialgebras are dual to each other in a sense soon to be discussed.
This was Hopf's motivation:  the (co-)homology of a compact Lie group carries bialgebra structure that explains
why it takes a certain form;  see Cartier \cite[\S 2]{Cartier}.
\end{example}

% [DG][v25] Isn't the Pontryagin product non-associative in general
% for $H$-spaces? Or do field coefficients somehow spare us this?

\begin{example}
Let $V$ be a graded $\kk$-module.
Then, its tensor algebra $T(V)$ and its
symmetric algebra $\Sym(V)$ are graded Hopf algebras.
The grading is given as follows:
If $v_1, v_2, \ldots, v_k$ are homogeneous elements of $V$
having degrees $i_1, i_2, \ldots, i_k$, respectively,
then the elements $v_1 v_2 \cdots v_k$ of $T(V)$ and $\Sym(V)$
are homogeneous of degree $i_1 + i_2 + \cdots + i_k$.
That is, we have
\[
\deg\left(v_1 v_2 \cdots v_k\right)
= \deg\left(v_1\right) + \deg\left(v_2\right) + \cdots + \deg\left(v_k\right)
\]
for any homogeneous elements $v_1, v_2, \ldots, v_k$ of $V$.

Assuming that $V_0=0$, the graded algebras $T(V)$ and $\Sym(V)$ are connected.
This is a fairly common situation in combinatorics.
For example, we will often turn a (non-graded) $\kk$-module $V$
into a graded $\kk$-module by declaring that all elements of
$V$ are homogeneous of degree $1$, but at other times, it will
make sense to have $V$ live in different (positive) degrees.
\end{example}

% [DG][v25] Inserted parenthesized "more general" statement.

\begin{exercise}
\label{exe.graded.iso}
Let $V$ and $W$ be two graded $\kk$-modules. Prove that
if $f : V \to W$ is an invertible graded $\kk$-linear map,
then its inverse $f^{-1} : W \to V$ is also graded.
\end{exercise}

% [DG][v80] Added the above exercise (I think it is tacitly
% used in a few places).

\begin{exercise}
\label{exe.primitives.graded-coideal}
Let $A = \bigoplus_{n \geq 0} A_n$ be a graded $\kk$-bialgebra. We denote
by $\Liep$ the set of all primitive elements of $A$.

\begin{itemize}

\item[(a)]
Show that
$\Liep$ is a graded $\kk$-submodule of $A$
(that is, we have
$\Liep = \bigoplus_{n \geq 0} \left(\Liep \cap A_n\right)$).

\item[(b)]
Show that $\Liep$ is a two-sided coideal of $A$.

\end{itemize}
\end{exercise}

% [DG][v14] Added the above nearly trivial exercise.

\begin{exercise}
\label{graded-connected-exercise}
Let $A$ be a connected graded $\kk$-bialgebra. Show the following:
\begin{itemize}
\item[(a)] The $\kk$-submodule $\kk = \kk\cdot 1_A$ of $A$ lies in $A_0$.
\item[(b)] The map $u$ is an isomorphism $\kk \overset{u}{\rightarrow} A_0$.
\item[(c)] We have $A_0 = \kk \cdot 1_A$.
\item[(d)] The two-sided ideal
$\ker \epsilon$
is the $\kk$-module of positive degree elements
$I = \bigoplus_{n > 0} A_n$.
\item[(e)] The map $\epsilon$ restricted to $A_0$ is the inverse isomorphism
$A_0 \overset{\epsilon}{\rightarrow} \kk$ to $u$.
\item[(f)] For every $x \in A$, we have
\[
\Delta(x) \in x \otimes 1 + A \otimes I.
\]
\item[(g)] Every $x$ in $I$ satisfies % has comultiplication of the form
\[
\Delta(x) = 1 \otimes x + x \otimes 1 + \Delta_+(x) ,
\qquad \text{where $\Delta_+(x)$ lies in $I \otimes I$.}
\]
\item[(h)] Every $n > 0$ and every $x \in A_n$ satisfy
\[
\Delta \left( x \right)
= 1 \otimes x + x \otimes 1 + \Delta_+ \left( x \right) ,
\qquad
\text{where $\Delta_+ \left( x \right)$ lies in
$\sum_{k=1}^{n-1} A_k \otimes A_{n-k}$.}
\]
\end{itemize}

% [DG][v77] I'm not fond of talking about the "comultiplication of $x$",
% so I've edited part (g) accordingly.

(Use only the gradedness of the unit $u$ and
counit $\epsilon$ maps, along with commutativity of diagrams
\eqref{counit-diagram}, and
\eqref{bialgebra-diagrams} and the connectedness of $A$.)
\end{exercise}

% [DG][v13] I have added part (e) (a weaker version of (f) which
% holds for all $x$, not just for $x \in I$ -- comes in handy in
% proofs --, and also serves as a stepping stone for (f)). I also
% tried hard to formalize the proof of this exercise down to every
% detail, and never found myself using \eqref{unit-diagram}, so
% I have removed the reference to this diagram (mentioning the
% connectedness instead).
% I also clarified what $\kk$ means in part (a) and rewrote some
% of the exercise.

% [DG][v66] Inserted parts (c) and (h), which follow quickly from
% (b) and (c) but are nevertheless worth spelling out (since they
% are used in this form).

Having discussed graded $\kk$-modules, let us also define
the concept of a \emph{graded basis}, which is the analogue of the
notion of a basis in the graded context.
Roughly speaking, a graded basis of a graded $\kk$-module is
a basis that comprises bases of all its homogeneous components.
More formally:

\begin{definition}
\label{def.graded-basis}
Let $V=\bigoplus_{n \geq 0} V_n$ be a graded $\kk$-module.
A \dfn{graded basis} of the graded $\kk$-module $V$
means a basis
$\left\{v_i\right\}_{i \in I}$ of the $\kk$-module $V$ whose
indexing set $I$ is partitioned into subsets
$I_0, I_1, I_2, \ldots$ (which are allowed to be empty) with the
property that, for every $n \in \NN$, the subfamily
$\left\{v_i\right\}_{i \in I_n}$ is a basis of the $\kk$-module
$V_n$.
\end{definition}

\begin{example}
Consider the polynomial ring $\kk\left[x\right]$ in one
variable $x$ over $\kk$.
This is a graded $\kk$-module (graded by the degree of a
polynomial; thus, each $x^n$ is homogeneous of degree $n$).
Then, the family
$\left(x^n\right)_{n \in \NN} = \left(x^0, x^1, x^2, \ldots\right)$
is a graded basis of $\kk\left[x\right]$
(presuming that its indexing set $\NN$ is partitioned into
the one-element subsets
$\left\{0\right\}, \left\{1\right\}, \left\{2\right\}, \ldots$).
The family
$\left(\left(-x\right)^n\right)_{n \in \NN}
= \left(x^0, -x^1, x^2, -x^3, \ldots\right)$
is a graded basis of $\kk\left[x\right]$ as well.
But the family
$\left(\left(1+x\right)^n\right)_{n \in \NN}$
is not, since it contains non-homogeneous elements.
\end{example}

% [DG][v80] I have extracted the definition of a graded
% basis from a footnote (it's used a few times) and added
% the example following it.

We end this section by discussing morphisms between
bialgebras. They are defined as one would expect:

\begin{definition}
A \dfn{morphism of bialgebras} (also known as a
\dfn{$\kk$-bialgebra homomorphism})
is a $\kk$-linear map $A \overset{\varphi}{\rightarrow} B$
between two $\kk$-bialgebras $A$ and $B$
that is simultaneously a $\kk$-algebra homomorphism
and a $\kk$-coalgebra homomorphism.
\end{definition}

For example, any $\kk$-linear map $f : V \to W$
between two $\kk$-modules $V$ and $W$ induces a $\kk$-linear map
$T\left(f\right) : T\left(V\right) \to T\left(W\right)$
between their tensor algebras
(which sends each $v_1 v_2 \cdots v_k \in T\left(V\right)$ to
$f\left(v_1\right) f\left(v_2\right) \cdots f\left(v_k\right) \in T\left(W\right)$)
as well as a $\kk$-linear map
$\Sym\left(f\right) : \Sym\left(V\right) \to \Sym\left(W\right)$
between their symmetric algebras;
both of these maps $T\left(f\right)$ and $\Sym\left(f\right)$
are morphisms of bialgebras.

% [DG][v80] Added above definition and example.

Graded bialgebras come with a special family of endomorphisms,
as the following exercise shows:

\begin{exercise}
\label{exe.graded.Dq}
Fix $q \in \kk$. Let $A = \bigoplus_{n\in \NN} A_n$ be a graded
$\kk$-bialgebra (where the $A_n$ are the homogeneous components of $A$).
Let $D_q : A \rightarrow A$ be the $\kk$-module endomorphism of $A$
defined by setting
\[
D_q \left(  a\right) = q^n a
\qquad \text{for each } n \in \NN \text{ and each } a \in A_n.
\]
(It is easy to see that this is well-defined; equivalently,
$D_q$ could be defined as the direct sum
$\bigoplus_{n\in \NN } \left(  q^n \cdot \id_{A_n}\right)
: \bigoplus_{n\in \NN } A_n \rightarrow \bigoplus_{n\in \NN } A_n$
of the maps $q^n \cdot \id_{A_n} : A_n \rightarrow A_n$.)

Prove that $D_q$ is a $\kk$-bialgebra homomorphism.
\end{exercise}

% [DG][v80] Added the above exercise, as it is tacitly used when
% transferring properties from the antipode of $\Lambda$ to the
% fundamental involution $\omega$ and vice versa.

The tensor product of two bialgebras is canonically a
bialgebra, as the following proposition shows:

\begin{proposition}
\label{prop.bialg.tensor}
Let $A$ and $B$ be two $\kk$-bialgebras.
Then, $A \otimes B$ is both a $\kk$-algebra and a
$\kk$-coalgebra (by Definition~\ref{def.tensor-prod-algs}).
These two structures, combined, turn $A \otimes B$ into
a $\kk$-bialgebra.
\end{proposition}

\begin{exercise} \phantomsection
\label{exe.bialg.tensor}
\begin{enumerate}
\item[(a)] Prove Proposition~\ref{prop.bialg.tensor}.

\item[(b)] Let $G$ and $H$ be two groups.
Show that the $\kk$-bialgebra $\kk G \otimes \kk H$
(defined as in Proposition~\ref{prop.bialg.tensor})
is isomorphic to the $\kk$-bialgebra
$\kk \left[G \times H\right]$.
(The notation $\kk \left[S\right]$ is a synonym for
$\kk S$.)
\end{enumerate}
\end{exercise}

% [DG][v70] Added the above proposition and exercise.

\subsection{Antipodes and Hopf algebras}

There is one more piece of structure needed to make a bialgebra
a Hopf algebra, although it will come for free in the connected graded case.

\begin{definition}
\label{convolution-algebra}
For any coalgebra $C$ and algebra $A$, one can endow the $\kk$-module
$\Hom(C,A)$ (which consists of all $\kk$-linear maps from $C$ to $A$)
% [DG][v77] Minor rewording here.
with an associative algebra structure called
the \dfn{convolution algebra}:  Define the product \dfn{$f \star g$}\index{$\star$}
of two maps $f,g$ in $\Hom(C,A)$ by
% [DG][v62] Replaced "send $f,g$ in $\Hom(C,A)$ to $f \star g$ defined by"
% by "Define the product $f \star g$ of two maps $f,g$ in $\Hom(C,A)$ by".
$(f \star g)(c) = \sum f(c_1) g(c_2)$, using
the Sweedler notation\footnote{See the paragraph around
\eqref{eq.sweedler-not.def1} for the meaning of this
notation.}
$\Delta(c) = \sum c_1 \otimes c_2$.
Equivalently, $f \star g$ is the composite
\[
\xymatrixcolsep{4pc}
\xymatrix{
C \ar[r]^-{\Delta}
& C \otimes C \ar[r]^-{f \otimes g}
& A \otimes A \ar[r]^-{m}
& A
} .
\]
The associativity of this multiplication $\star$ is easy to
check (see Exercise~\ref{exe.convolution.assoc} below).

The map $u \circ \epsilon$ is a two-sided
identity element for $\star$, meaning that every $f \in \Hom(C,A)$ satisfies
\[
\sum f(c_1) \epsilon(c_2) = f(c) = \sum \epsilon(c_1) f(c_2)
\]
for all $c \in C$.
One sees this by adding a top row to \eqref{counit-diagram}:


\begin{equation}
\label{added-to-counit-diagram}
\xymatrix{
A \otimes \kk \ar[r] & A & \kk \otimes A \ar[l] \\
C \otimes \kk \ar[u]^{f \otimes \id} \ar[r] & C \ar[u]^{f} & \kk \otimes C \ar[u]^{\id \otimes f} \ar[l] \\
C \otimes C \ar[u]^{\id \otimes \epsilon} & C \ar[l]^-{\Delta} \ar[u]^\id \ar[r]_-{\Delta} & C \otimes C \ar[u]^{\epsilon \otimes \id}
}
\end{equation}
\end{definition}

In particular, when one has a bialgebra $A$, the convolution product
$\star$ gives an associative algebra structure on $\End(A):=\Hom(A,A)$.

\begin{exercise}
\label{exe.convolution.assoc}
Let $C$ be a $\kk$-coalgebra and $A$ be a $\kk$-algebra. Show that the
binary operation $\star$ on $\Hom\left(C, A\right)$ is associative.
\end{exercise}

% [DG][v32] Added above exercise.

The product $f \star g$ of two elements $f$ and $g$ in a
convolution algebra $\Hom(C, A)$ is often called their
\dfn{convolution}.

The following simple (but useful) property of convolution algebras
says essentially that the $\kk$-algebra
$\left( \Hom\left(C,A\right), \star \right)$ is a covariant
functor in $A$ and a contravariant functor in $C$,
acting on morphisms by pre- and post-composition:

\begin{proposition}
\label{prop.convolution.functor}
Let $C$ and $C'$ be two $\kk$-coalgebras, and
let $A$ and $A'$ be two $\kk$-algebras.
Let $\gamma : C \to C'$
be a $\kk$-coalgebra morphism.
Let $\alpha : A \to A'$
be a $\kk$-algebra morphism.

The map
\[
\Hom\left(C',A\right) \to \Hom\left(C,A'\right),
\qquad
f \mapsto \alpha \circ f \circ \gamma
\]
is a $\kk$-algebra homomorphism from the convolution
algebra $\left(\Hom\left(C', A\right), \star\right)$ to the
convolution algebra
$\left(\Hom\left(C, A'\right), \star\right)$.
\end{proposition}

\begin{proof}[Proof of Proposition~\ref{prop.convolution.functor}.]
Denote this map by $\varphi$.
We must show that $\varphi$ is a $\kk$-algebra homomorphism.

Recall that $\alpha$ is an algebra morphism; thus,
$\alpha \circ m_A = m_{A'} \circ \left(\alpha \otimes \alpha\right)$
and
$\alpha \circ u_A = u_{A'}$.
Also, $\gamma$ is a coalgebra morphism; thus,
$\Delta_{C'} \circ \gamma = \left(\gamma \otimes \gamma\right) \circ \Delta_C$
and
$\epsilon_{C'} \circ \gamma = \epsilon_C$.

Now, the definition of $\varphi$ yields
$
\varphi(u_A \circ \epsilon_{C'})
= \underbrace{\alpha \circ u_A}_{= u_{A'}}
    \circ \underbrace{\epsilon_{C'} \circ \gamma}_{= \epsilon_C}
= u_{A'} \circ \epsilon_{C}
$;
in other words, $\varphi$ sends the unity of the algebra
$\left(\Hom\left(C', A\right), \star\right)$ to the
unity of the algebra
$\left(\Hom\left(C, A'\right), \star\right)$.

Furthermore, every $f \in \Hom \left(C',A\right)$ and
$g \in \Hom\left(C',A\right)$
satisfy
\begin{align}
\varphi(f \star g)
&=\alpha \circ \underbrace{(f \star g)}_{= m_A \circ \left( f \otimes g \right) \circ \Delta_{C'}} \circ \gamma \nonumber\\
&= \underbrace{\alpha \circ m_A}_{= m_{A'} \circ \left(\alpha \otimes \alpha\right)}
   \circ (f \otimes g)
   \circ \underbrace{\Delta_{C'} \circ \gamma}_{= \left(\gamma \otimes \gamma\right) \circ \Delta_C}
   \nonumber\\
&=m_{A'}
   \circ \underbrace{(\alpha \otimes \alpha) \circ (f \otimes g) \circ (\gamma \otimes \gamma)}_{= (\alpha \circ f \circ \gamma) \otimes
          (\alpha \circ g \circ \gamma)} \circ \Delta_C \nonumber\\
&=m_{A'} \circ \left( (\alpha \circ f \circ \gamma) \otimes
          (\alpha \circ g \circ \gamma) \right) \circ \Delta_C \nonumber\\
&= \underbrace{(\alpha \circ f \circ \gamma)}_{= \varphi(f)} \star \underbrace{(\alpha \circ g \circ \gamma)}_{= \varphi(g)}
= \varphi(f) \star \varphi(g) .
\label{pre-and-post-composition-in-convolution}
\end{align}
Thus, $\varphi$ is a $\kk$-algebra homomorphism
(since $\varphi$ is a $\kk$-linear map and
sends the unity of the algebra
$\left(\Hom\left(C', A\right), \star\right)$ to the
unity of the algebra
$\left(\Hom\left(C, A'\right), \star\right)$).
\end{proof}

% [DG][v65] Created the above proposition, which used to be a step
% in the proof of Proposition \ref{antipodes-give-convolution-inverses}.
% Its uses range far beyond that proof, so it makes much sense to me
% to have it separate.
% Extended the proof with more details, too.

\begin{exercise}
\label{exe.convolution.tensor}
Let $C$ and $D$ be two $\kk$-coalgebras, and let $A$ and $B$ be two
$\kk$-algebras. Prove that:
\begin{itemize}
\item[(a)] If $f:C\rightarrow A$,
$f^{\prime}:C\rightarrow A$, $g:D\rightarrow B$ and
$g^{\prime}:D\rightarrow B$ are four $\kk$-linear maps, then
\[
\left(  f\otimes g\right)  \star\left(  f^{\prime}\otimes g^{\prime}\right)
=\left(  f\star f^{\prime}\right)  \otimes\left(  g\star g^{\prime}\right)
\]
in the convolution algebra $\Hom\left(  C\otimes D,A\otimes
B\right)  $.
\item[(b)] Let $R$ be the $\kk$-linear map
$\left(\Hom\left(C,A\right),\star\right) \otimes
 \left(\Hom\left(D,B\right),\star\right) \to
 \left(\Hom\left(C \otimes D, A \otimes B\right),\star\right)$
which sends every tensor
$f \otimes g \in 
 \left(\Hom\left(C,A\right),\star\right) \otimes
 \left(\Hom\left(D,B\right),\star\right)$
to the map $f \otimes g : C \otimes D \to A \otimes B$. (Notice
that the tensor $f \otimes g$ and the map $f \otimes g$ are different
things which happen to be written in the same way.) Then, $R$ is a
$\kk$-algebra homomorphism.
\end{itemize}
\end{exercise}

% [DG][v14] Added the above exercise (simple computation and useful
% lemma for Dynkin idempotent exercise).

\begin{exercise}
\label{exe.convolution.tensor-curry}
Let $C$ and $D$ be two $\kk$-coalgebras.
Let $A$ be a $\kk$-algebra.
Let $\Phi$ be the canonical $\kk$-module isomorphism
$\Hom \left( C \otimes D, A \right)
\to \Hom \left( C, \Hom \left( D, A \right) \right)$
(defined by
$\left( \left( \Phi \left( f \right) \right) \left( c \right)
\right) \left( d \right)
= f \left( c \otimes d \right)$
for all $f \in \Hom \left( C \otimes D, A \right)$,
$c \in C$ and $d \in D$).
Prove that $\Phi$ is a $\kk$-algebra isomorphism
\[
\left( \Hom \left( C \otimes D, A \right) , \star \right)
\to \left( \Hom \left( C, \left( \Hom \left( D, A \right)
, \star\right) \right) , \star \right)  .
\]
\end{exercise}

% [DG][v68] Added the above simple but (to me) surprising
% exercise.

\begin{definition}
A bialgebra $A$ is called a \dfn{Hopf algebra} if there is an
element \dfn{$S$} (called an \dfn{antipode} for $A$) in $\End(A)$
which is a $2$-sided inverse under $\star$ for the identity map $\id_A$.
In other words, this diagram commutes:
\begin{equation}
\label{antipode-diagram}
\xymatrix{
&A \otimes A \ar[rr]^{S \otimes \id_A}& &A \otimes A \ar[dr]^m& \\
A \ar[ur]^\Delta \ar[rr]^{\epsilon} \ar[dr]_\Delta& &\kk \ar[rr]^{u} & & A\\
&A \otimes A \ar[rr]_{\id_A \otimes S}& &A \otimes A \ar[ur]_m& \\
}
\end{equation}
Or equivalently, if we follow the Sweedler notation
in writing $\Delta(a) = \sum a_1 \otimes a_2$,
then
\begin{equation}
\label{antipode-defining-relation}
\sum_{(a)} S(a_1) a_2 = u(\epsilon(a)) = \sum_{(a)} a_1 S(a_2).
\end{equation}
\end{definition}

\begin{example}
For a group algebra $\kk G$, one can define an antipode $\kk$-linearly
via $S(t_g)=t_{g^{-1}}$.  The top pentagon in the above diagram commutes because
\[
(S \star \id)(t_g) = m ((S \otimes \id)(t_g \otimes t_g)) =
S(t_g) t_g = t_{g^{-1}} t_g = t_e = (u \circ \epsilon)(t_g).
\]
\end{example}


Note that when it exists, the antipode $S$ is unique, as with all 2-sided inverses in associative algebras: if $S, S'$ are both 2-sided $\star$-inverses to $\id_A$ then
\[
S'=(u\circ \epsilon) \star S'
=(S \star \id_A) \star S'
=S \star (\id_A \star S')
=S \star (u \circ \epsilon)= S.
\]
Thus, we can speak of ``\emph{the antipode}'' of a Hopf algebra.

Unlike the comultiplication $\Delta$, the antipode $S$ of a Hopf algebra is not always an algebra homomorphism. It is instead an algebra \emph{anti-homomorphism}, a notion we shall now introduce:

\begin{definition} \phantomsection
\label{def.anti-hom}

\begin{itemize}

\item[(a)] For any two $\kk$-modules $U$ and $V$, we let
$T_{U,V} : U \otimes V \rightarrow V \otimes U$\index{$T_{U,V}$}
be the $\kk$-linear map
$U \otimes V \rightarrow V \otimes U$ sending every
$u \otimes v$ to $v \otimes u$.
This map $T_{U,V}$ is called the \dfn{twist map} for $U$ and $V$.

\item[(b)] A
\emph{$\kk$-algebra anti-homomorphism}\index{anti-homomorphism of algebras}
means a $\kk$-linear
map $f : A \rightarrow B$ between two $\kk$-algebras $A$ and $B$
which satisfies
$f \circ m_A
= m_B \circ \left( f \otimes f \right) \circ T_{A, A}$
and $f \circ u_A = u_B$.

\item[(c)] A
\emph{$\kk$-coalgebra anti-homomorphism}\index{anti-homomorphism of coalgebras}
means a $\kk$-linear
map $f : C \rightarrow D$ between two $\kk$-coalgebras $C$ and $D$
which satisfies
$\Delta_D \circ f
= T_{D,D} \circ \left( f \otimes f \right) \circ \Delta_C$
and $\epsilon_D \circ f = \epsilon_C$.

\item[(d)] A
\emph{$\kk$-algebra anti-endomorphism}\index{anti-endomorphism of an algebra}
of a $\kk$-algebra $A$ means a $\kk$-algebra anti-homomorphism from $A$
to $A$.

\item[(e)] A
\emph{$\kk$-coalgebra anti-endomorphism}\index{anti-endomorphism of a coalgebra}
of a $\kk$-coalgebra $C$ means a $\kk$-coalgebra anti-homomorphism from $C$
to $C$.

\end{itemize}
\end{definition}

Parts (b) and (c) of Definition~\ref{def.anti-hom} can be restated in terms of elements:
\begin{itemize}
\item
A $\kk$-linear map $f : A \rightarrow B$ between two $\kk$-algebras $A$ and $B$
is a $\kk$-algebra anti-homomorphism if and only if it satisfies
$f \left(ab\right) = f \left(b\right) f \left(a\right)$ for all $a, b \in A$
as well as $f \left(1\right) = 1$.
\item
A $\kk$-linear map $f : C \to D$ between two $\kk$-coalgebras $C$  and $D$
is a $\kk$-coalgebra anti-homomorphism if and only if it satisfies
$\sum_{\left(f\left(c\right)\right)} \left(f\left(c\right)\right)_1 \otimes \left(f\left(c\right)\right)_2
= \sum_{\left(c\right)} f\left(c_2\right) \otimes f\left(c_1\right)$
and $\epsilon\left(f\left(c\right)\right) = \epsilon\left(c\right)$
for all $c \in C$.
\end{itemize}

\begin{example}
Let $n \in \NN$, and consider the $\kk$-algebra $\kk^{n \times n}$
of $n \times n$-matrices over $\kk$.
The map $\kk^{n \times n} \to \kk^{n \times n}$ that sends
each matrix $A$ to its transpose $A^T$ is a $\kk$-algebra
anti-endomorphism of $\kk^{n \times n}$.
\end{example}

We warn the reader that the composition of two $\kk$-algebra anti-homomorphisms is not generally a $\kk$-algebra anti-homomorphism again, but rather a $\kk$-algebra homomorphism. The same applies to coalgebra anti-homomorphisms.
Other than that, however, anti-homomorphisms share many of the helpful properties of homomorphisms.
In particular, two $\kk$-algebra anti-homomorphisms are identical if they agree on a generating set of their domain.
Thus, the next proposition is useful when one wants to check that a certain map \emph{is} the antipode in a particular Hopf algebra, by checking it on an algebra generating set.

% [DG][v80] Added the above definition (previously partly included
% in footnotes and exercises) and the paragraph after it.

\begin{proposition}
\label{antipodes-are-antiendomorphisms}
The antipode $S$ in a Hopf algebra $A$ is an algebra anti-endomorphism:
$S(\one)=\one$, and $S(ab)=S(b)S(a)$ for all $a,b$ in $A$.
\end{proposition}
\begin{proof}
This is surprisingly nontrivial; the following argument comes from
\cite[proof of Proposition 4.0.1]{Sweedler}.

Since $\Delta$ is an algebra morphism, one has
$\Delta(1)=1\otimes 1$, and therefore
$
\one= u \epsilon (\one) = S(\one) \cdot \one = S(\one).
$

To show $S(ab)=S(b)S(a)$,
consider $A \otimes A$ as a coalgebra and $A$ as an algebra.
Then $\Hom(A \otimes A,A)$ is an associative algebra with a convolution
product $\starstar$ (to be distinguished from the convolution $\star$
on $\End(A)$), having two-sided identity element $u_A \epsilon_{A \otimes A}$.
We define three elements $f$, $g$, $h$
of $\Hom(A \otimes A,A)$ by
\begin{align*}
f(a \otimes b) &= ab, \\
g(a \otimes b) &=S(b) S(a), \\
h(a \otimes b) &=S(ab).
\end{align*}
We will show that these three elements have the property that
\begin{equation}
\label{three-antipode-calculations}
h \starstar f
= u_A \epsilon_{A \otimes A}
= f \starstar g ,
\end{equation}
which would then show the desired equality $h=g$ via associativity:
\[
h=h \starstar (u_A \epsilon_{A \otimes A})
=h \starstar (f \starstar g)
=(h \starstar f) \starstar g
=(u_A \epsilon_{A \otimes A}) \starstar g=g.
\]
So we evaluate the three elements in \eqref{three-antipode-calculations}
on $a \otimes b$. To do so, we use Sweedler notation -- i.e.,
we assume $\Delta(a)=\sum_{(a)} a_1 \otimes a_2$ and
$\Delta(b) = \sum_{(b)} b_1 \otimes b_2$, and hence
$\Delta(ab)= \sum_{(a),(b)} a_1 b_1 \otimes a_2 b_2$
(by \eqref{eq.bialgebra.sweedler-Deltaab});
then,
\begin{align*}
(u_A \epsilon_{A \otimes A})(a \otimes b)
&=u_A( \epsilon_A(a) \epsilon_A(b))= u_A(\epsilon_A (ab)).\\
& \\
(h \starstar f)(a \otimes b)
&=\sum_{(a),(b)} h(a_1 \otimes b_1) f(a_2 \otimes b_2) \\
&=\sum_{(a),(b)} S(a_1 b_1) a_2 b_2 \\
&= (S \star \id_A) (ab)
= u_A (\epsilon_A (ab)).\\
& \\
(f \starstar g)(a \otimes b)
&=\sum_{(a),(b)} f(a_1 \otimes b_1) g(a_2 \otimes b_2) \\
&=\sum_{(a),(b)} a_1 b_1 S(b_2) S(a_2) \\
&=\sum_{(a)} a_1 \cdot (\id_A \star S)(b) \cdot S(a_2) \\
&= u_A(\epsilon_A(b)) \sum_{(a)} a_1 S(a_2)
= u_A (\epsilon_A (b)) u_A (\epsilon_A (a))
= u_A (\epsilon_A (ab)) .
\end{align*}
These results are equal, so that \eqref{three-antipode-calculations}
holds, and we conclude that $h = g$ as explained above.
\end{proof}

% [DG][v77] Slightly uncrowded a couple sentences and added
% one at the end. The proof didn't get longer because
% the blacksquare at the end of the proof already took a
% whole line for itself :)

\begin{remark}
\label{antipode-as-dual-action}
Recall from Remark~\ref{comultiplication-as-tensor-action}
that the comultiplication on a bialgebra $A$ allows one
to define an $A$-module structure on the tensor product $M \otimes N$
of two $A$-modules $M,N$.  Similarly, the anti-endomorphism $S$
in a Hopf algebra allows one to turn \emph{left} $A$-modules into
\emph{right} $A$-modules, or vice-versa.\footnote{Be warned that
these two transformations are not mutually inverse! Turning a left
$A$-module into a right one and then again into a left one using the
antipode might lead to a non-isomorphic $A$-module, unless the
antipode $S$ satisfies $S^2=\id$.}
 E.g., left $A$-modules
$M$ naturally have a right $A$-module structure on the
dual $\kk$-module $M^*:=\Hom(M,\kk)$, defined via $(fa)(m):=f(am)$
for $f$ in $M^*$ and $a$ in $A$.  The antipode $S$
can be used to turn this back
into a left $A$-module $M^*$, via $(af)(m) = f(S(a)m)$.

% [DG][v17] Added footnote. Is the exclamation mark too alarmist?

For groups $G$ and left $\kk G$-modules
(group representations) $M$, this is
how one defines the \dfn{contragredient action} of $G$ on $M^*$,
namely $t_g$ acts as $(t_g f)(m) = f( t_{g^{-1}} m)$.

More generally, if $A$ is a Hopf algebra and $M$ and $N$ are
two left $A$-modules, then $\Hom\left(M, N\right)$ (the $\Hom$
here means $\Hom_\kk$, not $\Hom_A$) canonically
becomes a left $A$-module by setting
\[
\left(af\right)\left(m\right) = \sum_{\left(a\right)} a_1 f\left(S\left(a_2\right) m\right)
\qquad \qquad \text{for all } a \in A, \ f \in \Hom\left(M, N\right) \text{ and } m \in M.
\]
\footnote{In more abstract terms, this $A$-module structure is given
by the composition
\[
\xymatrixcolsep{3pc}
\xymatrix{
A \ar[r]^-{\Delta} & A \otimes A \ar[r]^{\id_A \otimes S} &
A \otimes A^{\operatorname{op}} \ar[r] & \End\left(\Hom\left(M, N\right)\right)
},
\]
where the last arrow is the morphism
\begin{align*}
\nonumber
A \otimes A^{\operatorname{op}} &\longrightarrow \End\left(\Hom\left(M, N\right)\right), \\
a \otimes b &\longmapsto \left(f \mapsto \left(M \to N, \ m \mapsto a f\left(bm\right)\right)\right).
\end{align*}
Here, $A^{\operatorname{op}}$ denotes the \dfn{opposite algebra}
of $A$, which is the $\kk$-algebra differing from $A$ only in the
multiplication being twisted (the product of $a$ and $b$ in
$A^{\operatorname{op}}$ is defined to be the product of $b$ and $a$
in $A$). As $\kk$-modules, $A^{\operatorname{op}} = A$, but we prefer
to use $A^{\operatorname{op}}$ instead of $A$ here to ensure that all
morphisms in the above composition are algebra morphisms.}
When $A$ is the group algebra $\kk G$ of a group $G$, this leads to
\[
\left(t_g f\right)\left(m\right) = t_g f\left(t_{g^{-1}} m\right)
\qquad \qquad \text{for all } g \in G, \ f \in \Hom\left(M, N\right) \text{ and } m \in M.
\]
This is
precisely how one commonly makes $\Hom\left(M, N\right)$ a
representation of $G$ for two representations $M$ and $N$.

Along the same lines, whenever $A$ is a $\kk$-bialgebra,
we are supposed to think of the counit
$A \overset{\epsilon}{\rightarrow} \kk$ as giving a way to
make $\kk$ into a \emph{trivial}\index{trivial module}
$A$-module. This $A$-module $\kk$
behaves as one would expect: the canonical isomorphisms
$\kk \otimes M \to M$, $M \otimes \kk \to M$ and
(if $A$ is a Hopf algebra)
$\Hom\left(M, \kk\right) \to M^*$ are $A$-module isomorphisms
for any $A$-module $M$.
\end{remark}

% [DG][v23] Added definition of $A$-module $\Hom\left(M, N\right)$.

\begin{corollary}
\label{commutative-implies-involutive-antipode-cor}
Let $A$ be a commutative Hopf algebra.
Then, its antipode is an involution: $S^2=\id_A$.
\end{corollary}
\begin{proof}
One checks that $S^2=S \circ S$ is a right $\star$-inverse to $S$, as follows:
\begin{align*}
(S \star S^2)(a)
  &= \sum_{(a)} S(a_1) S^2(a_2) \\
  &= S \left( \sum_{(a)} S(a_2) a_1 \right)
    \qquad \left( \text{by Proposition~\ref{antipodes-are-antiendomorphisms}} \right) \\
  &= S \left( \sum_{(a)} a_1 S(a_2) \right)
    \qquad \left( \text{by commutativity of } A \right) \\
  &= S \left( u (\epsilon(a)) \right)  \\
  & = u (\epsilon(a))
    \qquad \left( \text{since }S(\one)=\one \text{ by Proposition~\ref{antipodes-are-antiendomorphisms}} \right) .
\end{align*}
Since $S$ itself is the $\star$-inverse to $\id_A$, this shows that
$S^2 = \id_A$.
\end{proof}

% [DG][v63] Added a couple more details to the above proof and
% made the annotations follow my usual style
% (\qquad \left( \text{ ... } \right)).

\begin{remark}
\label{rmk.commutative-implies-involutive-antipode-cor.cocomm}
We won't need it, but it is easy to adapt the above proof to show that
$S^2=\id_A$ also holds for \emph{cocommutative} Hopf algebras
(the dual notion to commutativity; see Definition~\ref{def.cocomm} below for the precise definition);
see \cite[Corollary 1.5.12]{Montgomery} or \cite[Proposition 4.0.1 6)]{Sweedler}
or Exercise~\ref{exe.cocomm-antipode-S2} below.
For a general Hopf algebra which is not finite-dimensional over a field
$\kk$, the antipode
$S$ may not even have finite order, even in the connected graded setting.
E.g., Aguiar and Sottile \cite{AguiarSottile}
show that the Malvenuto-Reutenauer Hopf algebra of permutations has antipode of infinite
order.  In general, antipodes need not even be invertible \cite{Takeuchi}.
\end{remark}

\begin{proposition}
\label{prop.hopf.tensor}
Let $A$ and $B$ be two Hopf algebras.
Then, the $\kk$-bialgebra $A \otimes B$ (defined as in
Proposition~\ref{prop.bialg.tensor}) is a Hopf algebra.
The antipode of this Hopf algebra $A \otimes B$ is the
map $S_A \otimes S_B : A \otimes B \to A \otimes B$,
where $S_A$ and $S_B$ are the antipodes of the Hopf
algebras $A$ and $B$.
\end{proposition}

\begin{exercise}
\label{exe.hopf.tensor}
Prove Proposition~\ref{prop.hopf.tensor}.
\end{exercise}

% [DG][v70] Added the above proposition & exercise.

In our frequent setting of connected graded bialgebras,
antipodes come for free.

\begin{proposition}
\label{graded-connected-bialgebras-have-antipodes}
A connected graded bialgebra $A$ has a unique antipode $S$,
which is a graded map $A \overset{S}{\longrightarrow} A$, endowing it with a Hopf structure.
\end{proposition}
\begin{proof}
Let us try to define a ($\kk$-linear) left $\star$-inverse $S$ to $\id_A$
on each homogeneous component $A_n$, via induction on $n$.

In the base case $n=0$, Proposition~\ref{antipodes-are-antiendomorphisms} and its proof show that one
must define $S(\one)=\one$ so $S$ is the identity on $A_0=\kk$.

In the inductive step, recall from
Exercise~\ref{graded-connected-exercise}(h) that
a homogeneous element $a$ of degree $n > 0$
has $\Delta(a) = a \otimes \one  + \sum a'_1 \otimes a'_2$,
with each $\deg(a'_1) < n$.
(Here $\sum a'_1 \otimes a'_2$ stands for a sum of
tensors $a'_{1,k} \otimes a'_{2,k}$, with each $a'_{1,k}$
being homogeneous of degree $\deg(a'_{1,k}) < n$.
This is a slight variation on Sweedler notation.)
Hence in order to have $S \star \id_A = u \epsilon$,
one must define $S(a)$ in such a way that
$
S(a) \cdot \one + \sum S(a'_1) a'_2 = u \epsilon (a) = 0
$
and hence
$
S(a) := - \sum S(a'_1) a'_2,
$
where $S(a'_1)$ have already been uniquely defined
by induction (since $\deg(a'_{1,k}) < n$).
This does indeed define such a left $\star$-inverse $S$ to
$\id_A$, by induction.  It is also a graded map by induction.

The same argument shows how to define a right $\star$-inverse
$S'$ to $\id_A$.  Then $S=S'$ is a
two-sided $\star$-inverse to $\id_A$ by the associativity of $\star$.
\end{proof}

Here is another consequence of the fact that $S(\one)=\one$.

\begin{proposition}
\label{antipode-on-primitives}
In bialgebras, primitive elements $x$ have $\epsilon(x)=0$,
and in Hopf algebras, they have $S(x)=-x$.
\end{proposition}
\begin{proof}
In a bialgebra, $\epsilon(1)=1$.  Hence
$\Delta(x) = \one \otimes x + x \otimes \one$ implies
via \eqref{counit-diagram} that
$1 \cdot \epsilon(x) + \epsilon(1) x = x$, so $\epsilon(x) = 0$.
It also implies via \eqref{antipode-diagram} that
$S(x) \one + S(\one) x=u \epsilon(x)=u(0)=0$, so $S(x)=-x$.
\end{proof}

\noindent
Thus, whenever $A$ is a Hopf algebra generated as an algebra by its
primitive elements, $S$ is its unique $\kk$-algebra
anti-endomorphism that negates all primitive elements.

\begin{example}
\label{exa.tensor-alg.antipode}
The tensor and symmetric algebras
$T(V)$ and $\Sym(V)$ are each generated by $V$,
and each element of $V$ is primitive when regarded as an element
of either of them.  Hence one
has in $T(V)$ that
\begin{equation}
\label{antipode-in-tensor-algebra}
S(x_{i_1} x_{i_2} \cdots x_{i_k})=(-x_{i_k}) \cdots (-x_{i_2})(-x_{i_1}) =
(-1)^k x_{i_k} \cdots x_{i_2}x_{i_1}
\end{equation}
for each word $(i_1,\ldots,i_k)$ in the alphabet $I$
if $V$ is a free $\kk$-module with basis $\left\{x_i\right\}_{i\in I}$.
The same holds in $\Sym(V)$ for each multiset
$\left\{ i_1,\ldots,i_k \right\}_{\operatorname{multiset}}$,
recalling that the monomials are now commutative.  In other words,
for a commutative polynomial $f(x_1, x_2, \ldots, x_n)$ in
$\Sym(V)$, the antipode $S$ sends $f(x_1, x_2, \ldots, x_n)$
to $f(-x_1, -x_2, \ldots, -x_n)$,
negating all the variables.
\end{example}





The antipode for a connected graded Hopf algebra
has an interesting formula due to
Takeuchi \cite{Takeuchi}, reminiscent of P. Hall's formula for the M\"obius function of a
poset\footnote{In fact, for incidence Hopf algebras, Takeuchi's formula generalizes
Hall's formula-- see Corollary~\ref{P.Hall-formula}.}.  For the sake of stating
this, consider (for every $k \in \NN$) the $k$-fold \emph{tensor power}
$
A^{\otimes k}=A \otimes \cdots \otimes A
$
(defined in Example~\ref{exa.tensor-alg})
and define
\dfn{iterated multiplication and comultiplication} maps
\[
\xymatrixcolsep{3pc}
\xymatrix{
A^{\otimes k} \ar[r]^{m^{(k-1)}} & A
}
\qquad \text{and} \qquad
\xymatrix{
A \ar[r]^{\Delta^{(k-1)}} & A^{\otimes k}
}
\]
by induction over $k$, setting
$m^{(-1)}=u$, $\Delta^{(-1)}=\epsilon$, $m^{(0)}=\Delta^{(0)}=\id_A$,
and
\[
\begin{array}{rclr}
m^{(k)}&=m \circ (\id_A \otimes m^{(k-1)})
%    &=m \circ (m^{(k-1)} \otimes \id_A)
    & \qquad \text{ for every } k \geq 1; \\
\Delta^{(k)}&=(\id_A \otimes \Delta^{(k-1)}) \circ \Delta
%    &=(\Delta^{(k-1)} \otimes \id_A) \circ \Delta
    & \qquad \text{ for every } k \geq 1.
\end{array}
\]
Using associativity and coassociativity, one can see that for $k \geq 1$
these maps also satisfy
\[
\begin{array}{rclr}
m^{(k)} % &=m \circ (\id_A \otimes m^{(k-1)})
    &=m \circ (m^{(k-1)} \otimes \id_A)
    & \qquad \text{ for every } k \geq 1; \\
\Delta^{(k)} % &=(\id_A \otimes \Delta^{(k-1)}) \circ \Delta
    &=(\Delta^{(k-1)} \otimes \id_A) \circ \Delta
    & \qquad \text{ for every } k \geq 1
\end{array}
\]
(so we could just as well have used $\id_A \otimes m^{(k-1)}$ instead of
$m^{(k-1)} \otimes \id_A$ in defining them) and further symmetry properties
(see Exercise~\ref{exe.iterated.m} and Exercise~\ref{exe.iterated.Delta}).
They are how
one gives meaning to the right sides of these equations:
\begin{align*}
m^{(k)}(a^{(1)} \otimes \cdots \otimes a^{(k+1)})
&= a^{(1)} \cdots a^{(k+1)} ;\\
\Delta^{(k)}(b)
&= \sum b_1 \otimes \cdots \otimes b_{k+1} \text{ in Sweedler notation.}
\end{align*}

\begin{exercise}
\label{exe.iterated.m}
Let $A$ be a $\kk$-algebra. Let us define, for every $k \in \NN$, a
$\kk$-linear map
$m^{\left(k\right)} : A^{\otimes \left(k+1\right)} \to A$. Namely, we
define these maps by induction over $k$, with the induction base
$m^{\left(0\right)} = \id_A$, and with the induction step
$m^{\left(k\right)} = m \circ \left(\id_A \otimes m^{\left(k-1\right)}\right)$
for every $k \geq 1$. (This generalizes our definition of
$m^{\left(k\right)}$ for Hopf algebras $A$ given above, except for
$m^{\left(-1\right)}$ which we have omitted.)
\begin{itemize}
\item[(a)] Show that
$m^{\left(k\right)} = m \circ \left(m^{\left(i\right)} \otimes m^{\left(k-1-i\right)}\right)$
for every $k \geq 0$ and $0 \leq i \leq k-1$.
\item[(b)] Show that
$m^{\left(k\right)} = m \circ \left(m^{\left(k-1\right)} \otimes \id_A\right)$
for every $k \geq 1$.
\item[(c)] Show that
$m^{\left(k\right)} = m^{\left(k-1\right)} \circ
 \left(\id_{A^{\otimes i}} \otimes m \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right)$
for every $k \geq 0$ and $0 \leq i \leq k-1$.
\item[(d)] Show that
$m^{\left(k\right)}
= m^{\left(k-1\right)} \circ \left(\id_{A^{\otimes \left(k-1\right)}} \otimes m\right)
= m^{\left(k-1\right)} \circ \left(m \otimes \id_{A^{\otimes \left(k-1\right)}}\right)$
for every $k \geq 1$.
\end{itemize}
\end{exercise}

\begin{exercise}
\label{exe.iterated.Delta}
Let $C$ be a $\kk$-coalgebra. Let us define, for every $k \in \NN$, a
$\kk$-linear map
$\Delta^{\left(k\right)} : C \to C^{\otimes \left(k+1\right)}$. Namely, we
define these maps by induction over $k$, with the induction base
$\Delta^{\left(0\right)} = \id_C$, and with the induction step
$\Delta^{\left(k\right)} = \left(\id_C \otimes \Delta^{\left(k-1\right)}\right) \circ \Delta$
for every $k \geq 1$. (This generalizes our definition of
$\Delta^{\left(k\right)}$ for Hopf algebras $A$ given above, except for
$\Delta^{\left(-1\right)}$ which we have omitted.)
\begin{itemize}
\item[(a)] Show that
$\Delta^{\left(k\right)} = \left(\Delta^{\left(i\right)} \otimes \Delta^{\left(k-1-i\right)}\right) \circ \Delta$
for every $k \geq 0$ and $0 \leq i \leq k-1$.
\item[(b)] Show that
$\Delta^{\left(k\right)} = \left(\Delta^{\left(k-1\right)} \otimes \id_C\right) \circ \Delta$
for every $k \geq 1$.
\item[(c)] Show that
$\Delta^{\left(k\right)} =
 \left(\id_{C^{\otimes i}} \otimes \Delta \otimes \id_{C^{\otimes \left(k-1-i\right)}}\right)
 \circ \Delta^{\left(k-1\right)}$
for every $k \geq 0$ and $0 \leq i \leq k-1$.
\item[(d)] Show that
$\Delta^{\left(k\right)}
= \left(\id_{C^{\otimes \left(k-1\right)}} \otimes \Delta\right) \circ \Delta^{\left(k-1\right)}
= \left(\Delta \otimes \id_{C^{\otimes \left(k-1\right)}}\right) \circ \Delta^{\left(k-1\right)}$
for every $k \geq 1$.
\end{itemize}
\end{exercise}

\begin{remark}
Exercise~\ref{exe.iterated.m} holds more generally for nonunital
associative algebras $A$ (that is, $\kk$-modules $A$ equipped with a
$\kk$-linear map $m : A \otimes A \to A$ such that the diagram
\eqref{associativity-diagram} is commutative, but not necessarily admitting
a unit map $u$). Similarly, Exercise~\ref{exe.iterated.Delta} holds for
non-counital coassociative coalgebras $C$. The existence of a unit in $A$,
respectively a counit in $C$, allows slightly extending these two exercises
by additionally introducing maps $m^{\left(-1\right)} = u : \kk \to A$ and
$\Delta^{\left(-1\right)} = \epsilon : C \to \kk$; however, not much is
gained from this extension.\footnote{The identity
$m^{\left(k\right)} = m \circ \left(\id_A \otimes m^{\left(k-1\right)}\right)$
for a $\kk$-algebra $A$ still holds when $k=0$ if it is interpreted in the
right way (viz., if $A$ is identified with $A \otimes \kk$ using the
canonical homomorphism).}
\end{remark}

% [DG][v19] Detailed the introduction of iterated multiplication and
% comultiplication maps and added the two exercises above to prove their
% properties.

\begin{exercise}
\label{exe.iterated.patras}
For every $k \in \NN$ and every $\kk$-bialgebra $H$, consider the map
$\Delta_H^{\left(k\right)} : H \to H^{\otimes \left(k+1\right)}$ (this
is the map $\Delta^{\left(k\right)}$ defined
as in Exercise~\ref{exe.iterated.Delta} for $C = H$), and the map
$m_H^{\left(k\right)} : H^{\otimes \left(k+1\right)} \to H$ (this is the
map $m^{\left(k\right)}$ defined as in Exercise~\ref{exe.iterated.m}
for $A = H$).

Let $H$ be a $\kk$-bialgebra. Let $k \in \NN$. Show that:\footnote{The
following statements are taken from \cite{Patras-descent}; specifically,
part (c) is \cite[Lem. 1.8]{Patras-descent}.}
\begin{itemize}
\item[(a)] The map
$m_H^{\left(k\right)} : H^{\otimes \left(k+1\right)} \to H$ is a
$\kk$-coalgebra homomorphism.
\item[(b)] The map
$\Delta_H^{\left(k\right)} : H \to H^{\otimes \left(k+1\right)}$ is a
$\kk$-algebra homomorphism.
\item[(c)] We have $m_{H^{\otimes \left(k+1\right)}}^{\left(\ell\right)}
\circ \left(\Delta^{\left(k\right)}_H\right)^{\otimes \left(\ell + 1\right)}
= \Delta^{\left(k\right)}_H \circ m^{\left(\ell\right)}_H$ for
every $\ell \in \NN$.
\item[(d)] We have
$\left(m_H^{\left(\ell\right)}\right)^{\otimes \left(k+1\right)}
\circ \Delta_{H^{\otimes \left(\ell + 1\right)}}^{\left(k\right)}
= \Delta^{\left(k\right)}_H \circ m^{\left(\ell\right)}_H$ for
every $\ell \in \NN$.
\end{itemize}
\end{exercise}

% [DG][v25] Added above exercise (which should chip away a little bit
% at the internal product theory, cf. its use in \cite{Patras-descent},
% although it seems to me that it is the easiest part of the proof).

The iterated multiplication and comultiplication maps allow explicitly
computing the convolution of multiple maps; the following formula will
often be used without explicit mention:

\begin{exercise}
\label{exe.convolution.k}
Let $C$ be a $\kk$-coalgebra, and $A$ be a $\kk$-algebra. Let $k\in\NN$.
Let $f_1, f_2, \ldots, f_k$ be $k$ elements of
$\Hom\left(C,A\right)$. Show that
\[
f_1 \star f_2 \star \cdots \star f_k
= m^{\left(k-1\right)}_A
  \circ \left(f_1 \otimes f_2 \otimes \cdots \otimes f_k\right)
  \circ \Delta^{\left(k-1\right)}_C.
\]
\end{exercise}

% [DG][v19] Added the above exercise and referenced it from the proof of
% Proposition \ref{Takeuchi-prop}.



We are now ready to state Takeuchi's formula for the antipode:

\begin{proposition}
\label{Takeuchi-prop}
In a connected graded Hopf algebra $A$, the
antipode has formula
\begin{equation}
\label{Takeuchi-formula}
\begin{aligned}
S&=\sum_{k \geq 0} (-1)^k m^{(k-1)} f^{\otimes k} \Delta^{(k-1)} \\
 &=u\epsilon - f + m \circ f^{\otimes 2} \circ \Delta
    - m^{(2)} \circ f^{\otimes 3} \circ \Delta^{(2)} + \cdots
\end{aligned}
\end{equation}
where $f:=\id_A - u\epsilon$ in $\End(A)$.
\end{proposition}
\begin{proof}
We argue as in \cite[proof of Lemma 14]{Takeuchi} or \cite[\S 5]{AguiarSottile}.
For any $f$ in $\End(A)$, the following explicit formula expresses
its $k$-fold convolution power $f^{\star k}:=f \star \cdots \star f$
in terms of its tensor powers $f^{\otimes k}:=f \otimes \cdots \otimes f$
(according to Exercise~\ref{exe.convolution.k}):
\[
f^{\star k} = m^{(k-1)} \circ f^{\otimes k} \circ \Delta^{(k-1)}.
\]
Therefore any $f$ annihilating $A_0$ will be
\dfn{locally $\star$-nilpotent} on $A$, meaning that for each $n$ one
has that $A_n$ is annihilated by $f^{\star m}$ for every $m > n$:  homogeneity forces that
for $a$ in $A_n$, every summand of $\Delta^{(m-1)}(a)$ must contain among its
$m$ tensor factors at least one factor lying in $A_0$,
so each summand is annihilated by $f^{\otimes m}$, and
$f^{\star m}(a)=0$.

In particular such $f$ have the property that
$u \epsilon + f$ has as two-sided $\star$-inverse
\begin{align*}
(u\epsilon + f)^{\star (-1)}
&= u\epsilon - f + f \star f - f \star f \star f + \cdots \\
& = \sum_{k \geq 0} (-1)^k f^{\star k}
= \sum_{k \geq 0} (-1)^k m^{(k-1)} \circ f^{\otimes k} \circ \Delta^{(k-1)}.
\end{align*}
The proposition follows upon taking
$f:=\id_A - u\epsilon$, which annihilates $A_0$.
\end{proof}

\begin{remark}
\label{non-graded-Takeuchi-remark}
In fact, one can see that Takeuchi's formula applies more generally to
define an antipode $A \overset{S}{\longrightarrow} A$
in any (not necessarily graded)
bialgebra $A$ where the map $\id_A - u\epsilon$ is locally
$\star$-nilpotent.

It is also worth noting that the proof of Proposition~\ref{Takeuchi-prop}
gives an alternate proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
\end{remark}


To finish our discussion of antipodes, we mention some properties
(taken from \cite[Lemma 4.0.3]{Sweedler}) relating antipodes
to convolutional inverses.

\begin{proposition}
\label{antipodes-give-convolution-inverses}
Let $H$ be a Hopf algebra with antipode $S$.
\begin{enumerate}
\item[(a)] For any algebra $A$ and algebra morphism $H \overset{\alpha}{\rightarrow} A$,
one has $\alpha \circ S = \alpha^{\star -1}$, the convolutional inverse to $\alpha$ in $\Hom(H,A)$.
\item[(b)] For any coalgebra $C$ and coalgebra morphism $C \overset{\gamma}{\rightarrow} H$,
one has $S \circ \gamma = \gamma^{\star -1}$, the convolutional inverse to $\gamma$ in $\Hom(C,H)$.
\end{enumerate}
\end{proposition}
\begin{proof}
We prove (a); the proof of (b) is similar.

For assertion (a), note that 
Proposition~\ref{prop.convolution.functor}
(applied to $H$, $H$, $H$, $A$, $\id_H$ and
$\alpha$ instead of $C$, $C'$, $A$, $A'$, $\gamma$
and $\alpha$) shows that the map
\[
\Hom \left(H, H\right) \to \Hom \left(H, A\right),
\qquad
f \mapsto \alpha \circ f
\]
is a $\kk$-algebra homomorphism from the convolution
algebra $\left( \Hom \left(H, H\right), \star \right)$
to the convolution algebra
$\left( \Hom \left(H, A\right), \star \right)$.
Denoting this homomorphism by $\varphi$, we thus have
$\varphi\left((\id_H)^{\star -1}\right)
= \left(\varphi(\id_H)\right)^{\star -1}$
(since $\kk$-algebra homomorphisms preserve inverses).
Now,
\[
\alpha \circ S = \varphi(S)
               = \varphi\left((\id_H)^{\star -1}\right)
               = \left(\varphi(\id_H)\right)^{\star -1}
               = \left(\alpha \circ \id_H\right)^{\star -1}
               = \alpha^{\star -1}.
\]
\end{proof}

A rather useful consequence of Proposition~\ref{antipodes-give-convolution-inverses}
is the fact (\cite[Lemma 4.0.4]{Sweedler})
that a bialgebra morphism between Hopf algebras automatically respects the
antipodes:

\begin{corollary} \label{cor.bialg-mor-is-Hopf}
Let $H_1$ and $H_2$ be Hopf algebras with antipodes $S_1$ and $S_2$, respectively.
Then, any bialgebra morphism $H_1 \overset{\beta}{\rightarrow} H_2$ is a Hopf morphism\footnote{A \dfn{Hopf morphism} (or, more officially, a \dfn{Hopf algebra morphism}, or \dfn{homomorphism of Hopf algebras}) between two Hopf algebras $A$ and $B$ is defined to be a bialgebra morphism $f : A \to B$ that satisfies $f \circ S_A = S_B \circ f$.}, that is, it commutes with the antipodes (i.e., we have $\beta \circ S_1 = S_2 \circ \beta$).
\end{corollary}

\begin{proof}
Proposition~\ref{antipodes-give-convolution-inverses}(a) (applied to $H = H_1$, $S = S_1$, $A = H_2$ and $\alpha = \beta$)
yields $\beta \circ S_1 = \beta^{\star -1}$.
Proposition~\ref{antipodes-give-convolution-inverses}(b) (applied to $H = H_2$, $S = S_2$, $C = H_1$ and $\gamma = \beta$)
yields $S_2 \circ \beta = \beta^{\star -1}$.
Comparing these equalities shows that $\beta \circ S_1 = S_2 \circ \beta$, qed.
\end{proof}

% [DG][v80] Moved the above corollary out of the proposition.

\begin{exercise}
\label{exe.coalg-anti}
Prove that the antipode $S$ of a Hopf algebra $A$ is a
coalgebra anti-endomorphism, i.e., that it satisfies
$\epsilon \circ S = \epsilon$ and
$\Delta \circ S = T \circ \left( S \otimes S \right) \circ \Delta$,
where $T : A\otimes A \rightarrow A \otimes A$ is the
twist map sending every $a\otimes b$ to $b\otimes a$.
\end{exercise}

\begin{exercise}
\label{exe.coalg-anti.gen}
If $C$ is a $\kk$-coalgebra and if $A$ is a $\kk$-algebra, then a
$\kk$-linear map $f : C \rightarrow A$ is
said to be \dfn{$\star$-invertible} if it is invertible as an
element of the $\kk$-algebra
$\left( \Hom \left( C, A \right) , \star \right)$.
In this case, the multiplicative inverse $f^{\star\left( -1\right) }$
of $f$ in $\left( \Hom \left( C, A \right) , \star \right)$
is called the \dfn{$\star$-inverse} of $f$.

Recall the concepts introduced in Definition~\ref{def.anti-hom}.

\begin{itemize}
\item[(a)] If $C$ is a $\kk$-bialgebra, if $A$ is a $\kk$-algebra,
and if $r : C \rightarrow A$ is a $\star$-invertible $\kk$-algebra
homomorphism, then prove that the $\star$-inverse $r^{\star\left(
-1\right)  }$ of $r$ is a $\kk$-algebra anti-homomorphism.

\item[(b)] If $C$ is a $\kk$-bialgebra, if $A$ is a $\kk$-coalgebra,
and if $r : A\rightarrow C$ is a $\star$-invertible $\kk$-coalgebra
homomorphism, then prove that the $\star$-inverse $r^{\star\left(
-1\right)  }$ of $r$ is a $\kk$-coalgebra anti-homomorphism.

\item[(c)] Derive Proposition~\ref{antipodes-are-antiendomorphisms}
from Exercise~\ref{exe.coalg-anti.gen}(a), and derive
Exercise~\ref{exe.coalg-anti} from
Exercise~\ref{exe.coalg-anti.gen}(b).

\item[(d)] Prove
Corollary~\ref{commutative-implies-involutive-antipode-cor} again
using Proposition~\ref{antipodes-give-convolution-inverses}.

\item[(e)] If $C$ is a graded $\kk$-coalgebra, if $A$ is a graded
$\kk$-algebra, and if $r : C \to A$ is a $\star$-invertible
$\kk$-linear map that is graded, then prove that the
$\star$-inverse $r^{\star\left(  -1\right)  }$ of $r$ is also
graded.

\end{itemize}
\end{exercise}

% [DG][v48] Added preceding exercise (nothing particularly interesting,
% but can help in the rare cases when an antipode does not exist).

% [DG][v49] Added part (d).

% [DG][v80] Added part (e), since I realized I do use it tacitly
% in another solution.

\begin{exercise} \phantomsection
\label{exe.antipode.inverse}
\begin{itemize}

\item[(a)]
Let $A$ be a Hopf algebra. If
$P:A\rightarrow A$ is a $\kk$-linear map such that every $a\in A$
satisfies
\[
\sum_{\left( a\right)} P\left( a_2 \right) \cdot a_1
= u\left( \epsilon\left( a\right) \right) ,
\]
then prove that the antipode $S$ of $A$ is invertible and its inverse is $P$.

\item[(b)]
Let $A$ be a Hopf algebra. If
$P:A\rightarrow A$ is a $\kk$-linear map such that every $a\in A$
satisfies
\[
\sum_{\left( a\right)} a_2 \cdot P\left( a_1 \right)
= u\left( \epsilon\left( a\right) \right) ,
\]
then prove that the antipode $S$ of $A$ is invertible and its inverse is $P$.

\item[(c)]
Show that the antipode of a connected graded Hopf algebra is invertible.

\end{itemize}
\end{exercise}

(Compare this exercise to \cite[Lemma 1.5.11]{Montgomery}.)

% [DG][v3] Added two exercises above.

\begin{definition}
Let $C$ be a $\kk$-coalgebra. A \dfn{subcoalgebra} of $C$ means a
$\kk$-coalgebra $D$ such that $D \subset C$ and such that the canonical
inclusion map $D \to C$ is a $\kk$-coalgebra homomorphism\footnote{In
this definition, we follow \cite[p. 55]{Morris} and
\cite[\S 6.7]{Wisbauer}; other authors may use
other definitions.}. When $\kk$
is a field, we can equivalently define a subcoalgebra of $C$ as a
$\kk$-submodule $D$ of $C$ such that $\Delta_C\left(D\right)$ is
a subset of the $\kk$-submodule $D \otimes D$ of $C \otimes C$;
however, this might no longer be equivalent when $\kk$ is not a
field\footnote{This is because the $\kk$-submodule $D \otimes D$ of
$C \otimes C$ is generally not isomorphic to the $\kk$-module
$D \otimes D$. See \cite[p. 56]{Morris} for specific counterexamples
for the non-equivalence of the two notions of a subcoalgebra. Notice
that the equivalence is salvaged if
$D$ is a direct summand of $C$ as a $\kk$-module (see
Exercise~\ref{exe.subcoalgebra.summand} for this).}.

Similarly, a \dfn{subbialgebra} of a bialgebra $C$ is a
$\kk$-bialgebra $D$ such that $D \subset C$ and such that the canonical
inclusion map $D \to C$ is a $\kk$-bialgebra homomorphism.
Also, a \dfn{Hopf subalgebra} of a Hopf algebra $C$ is a
$\kk$-Hopf algebra $D$ such that $D \subset C$ and such that the canonical
inclusion map $D \to C$ is a $\kk$-Hopf algebra homomorphism.\footnote{By
Corollary~\ref{cor.bialg-mor-is-Hopf}, we can also
define it as a subbialgebra of $C$ that happens to be a Hopf algebra.}
\end{definition}

% [DG][v17] Added definition above.
% I am referencing the proceedings volume edited by Morris. Should
% I reference the Nichols-Sweedler paper instead?

\begin{exercise}
\label{exe.subcoalgebra.summand}
Let $C$ be a $\kk$-coalgebra. Let $D$ be a $\kk$-submodule of $C$
such that $D$ is a direct summand of $C$ as a $\kk$-module (i.e.,
there exists a $\kk$-submodule $E$ of $C$ such that $C = D \oplus E$).
(This is automatically satisfied if $\kk$ is a field.)
Assume that $\Delta\left(D\right) \subset C \otimes D$ and
$\Delta\left(D\right) \subset D \otimes C$. (Here, we are abusing
the notation $C \otimes D$ to denote the $\kk$-submodule of $C \otimes C$
spanned by tensors of the form $c \otimes d$ with $c \in C$ and
$d \in D$; similarly, $D \otimes C$ should be understood.)
Show that there is a
canonically defined $\kk$-coalgebra structure on $D$ which makes $D$
a subcoalgebra of $C$.
\end{exercise}

The next exercise is implicit in \cite[\S 5]{AguiarBergeronSottile}:

\begin{exercise}
\label{exe.subcoalgebra.ABS}
Let $\kk$ be a field. Let $C$ be a $\kk$-coalgebra, and let $U$ be any
$\kk$-module. Let $f : C \to U$ be a $\kk$-linear map. Recall the map
$\Delta^{\left(2\right)} : C \to C^{\otimes 3}$ from
Exercise~\ref{exe.iterated.Delta}. Let
$K = \ker\left(\left(\id_C \otimes f \otimes \id_C\right) \circ
\Delta^{\left(2\right)}\right)$.
\begin{itemize}
\item[(a)] Show that $K$ is a $\kk$-subcoalgebra of $C$.
\item[(b)] Show that every $\kk$-subcoalgebra of $C$ which is a subset
of $\ker f$ must be a subset of $K$.
\end{itemize}
\end{exercise}

% [DG][v19] Added the above two exercises. Feel free to make the second
% one into a theorem/proof if you write the Aguiar-Bergeron-Sottile II
% section.

\begin{exercise} \phantomsection
\label{exe.antipode.connected.general}
\begin{itemize}
\item[(a)]
Let $C = \bigoplus_{n\geq 0} C_n$ be a graded $\kk$-coalgebra, and $A$ be
any $\kk$-algebra. Notice that $C_0$ itself is a $\kk$-subcoalgebra of
$C$. Let $h : C \to A$ be a $\kk$-linear map such that the restriction
$h\mid_{C_0}$ is a $\star$-invertible map in $\Hom\left(C_0,A\right)$.
Prove that $h$ is a $\star$-invertible map in $\Hom\left(C,A\right)$. (This is
a weaker version of Takeuchi's \cite[Lemma 14]{Takeuchi}.)

\item[(b)]
Let $A = \bigoplus_{n\geq 0} A_n$ be a graded $\kk$-bialgebra. Notice
that $A_0$ is a subbialgebra of $A$. Assume that $A_0$ is a Hopf algebra.
Show that $A$ is a Hopf algebra.

\item[(c)]
Obtain yet another proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
\end{itemize}
\end{exercise}


\begin{exercise}
\label{exe.I=0}
Let $A = \bigoplus_{n\geq 0} A_n$ be a connected graded
$\kk$-bialgebra. Let $\Liep$ be the $\kk$-submodule of $A$
consisting of the primitive elements of $A$.
\begin{itemize}
\item[(a)] If $I$ is a two-sided coideal of $A$ such that $I \cap \Liep = 0$ and
such that $I = \bigoplus_{n \geq 0}\left( I\cap A_n \right)$, then prove that
$I=0$.
\item[(b)] Let $f:A \to C$ be a graded surjective coalgebra homomorphism from
$A$ to a graded $\kk$-coalgebra $C$. If $f\mid_{\Liep}$ is
injective, then prove that $f$ is injective.
\item[(c)] Assume that $\kk$ is a field. Show that the claim of
Exercise~\ref{exe.I=0}(b) is valid even without requiring $f$ to be surjective.
\end{itemize}
\end{exercise}

\begin{remark}
Exercise~\ref{exe.I=0} (b) and (c) are often used in order to prove that certain
coalgebra homomorphisms are injective.

The word ``bialgebra'' can be replaced by ``coalgebra'' in Exercise~\ref{exe.I=0},
provided that the
notion of a connected graded coalgebra is defined correctly (namely, as a
graded coalgebra such that the restriction of $\epsilon$ to the $0$-th graded
component is an isomorphism), and the notion of the element $1$ of a
connected graded coalgebra is defined accordingly (namely, as the preimage of
$1\in\kk$ under the restriction of $\epsilon$ to the $0$-th graded component).
\end{remark}

% [DG][v13] Added this exercise. The solution comes from an old mail
% of mine and is really a mess; sorry.

% [DG][v23] Added part (c) to this exercise.



\subsection{Commutativity, cocommutativity}

Recall that a $\kk$-algebra $A$ is \dfn{commutative}
if and only if all $a, b \in A$ satisfy $ab=ba$.
Here is a way to restate this classical definition using
tensors instead of pairs of elements:

\begin{definition} \label{def.comm}
A $\kk$-algebra $A$ is said to be \dfn{commutative} if
% $ab=ba$ for all $a, b \in A$, that is,
the following diagram commutes:
\begin{equation}
\label{commutativity-diagram}
\xymatrix{
A \otimes A \ar[rr]^T \ar[dr]_m& &A \otimes A \ar[dl]^m\\
 &A&
}
\end{equation}
where $T$ is the twist map $T_{A,A}$ (see
Definition~\ref{def.anti-hom}(a) for its definition).
% This, of course, is a mere restatement of the classical definition
% of a commutative $\kk$-algebra using tensors instead of pairs of
% elements.
% % [DG][v62] Added the preceding sentence.
\end{definition}

Having thus redefined commutative algebras in terms of
tensors and linear maps, we can dualize this definition
(reversing all arrows)
and obtain the notion of \emph{cocommutative coalgebras}:

\begin{definition} \label{def.cocomm}
A $\kk$-coalgebra $C$ is said to be \dfn{cocommutative} if
the following diagram commutes:
\begin{equation}
\label{cocommutativity-diagram}
\xymatrix{
C \otimes C \ar[rr]^T & &C \otimes C \\
 &C \ar[ul]^\Delta \ar[ur]_\Delta&
}
\end{equation}
where $T$ is the twist map $T_{C,C}$ (see
Definition~\ref{def.anti-hom}(a) for its definition).
\end{definition}

% [DG][v80] Split the above definition into two.

\begin{example}
Group algebras $\kk G$ are always cocommutative.
They are commutative if and only if $G$ is abelian or $\kk = 0$.

Tensor algebras $T(V)$ are always cocommutative,
but not generally commutative\footnote{If $\kk$ is a field,
then $T(V)$ is commutative if and only if $\dim_\kk V \leq 1$.}.

% [DG][v17] Weakened the "if and only if" since $\kk$ is an arbitrary
% commutative ring now.

Symmetric algebras $\Sym(V)$ are always cocommutative and commutative.

Homology and cohomology of $H$-spaces are always cocommutative and commutative
\emph{in the topologist's sense} where one reinterprets that
twist map $A \otimes A \overset{T}{\rightarrow} A \otimes A$ to have
the extra sign as in \eqref{topologist-twist}.
\end{example}

% [DG][v25] Is the kind of (co)commutativity given by
% your definition (reinterpreted using the topologist's twist)
% really what passes for (co)commutativity in the homological
% setting? I remember people asking for more (for commutativity,
% there is an axiom $xx = 0$ for $x$ odd; for cocommutativity,
% don't know what). Is this still true in the homological
% setting?

Note how the cocommutative Hopf algebras $T(V), \Sym(V)$
have much of their structure controlled by their $\kk$-submodules $V$,
which consist of primitive elements only (although, in general, not of
all their primitive elements).  This is not far from the truth in general,
and closely related to Lie algebras.

% [DG][v17] Replaced "their subspace $V$ of primitive elements" by
% "their $\kk$-submodules $V$, which consist of primitive elements only
% (although, in general, not of all their primitive elements)" to get rid
% of ambiguity.

\begin{exercise}
\label{exe.liealg}
Recall that a \dfn{Lie algebra} over $\kk$ is a $\kk$-module
$\Lieg$ with a $\kk$-bilinear map $[\cdot,\cdot]:\Lieg\times\Lieg\to\Lieg$
that satisfies $[x,x]=0$ for $x$ in $\Lieg$, and the \dfn{Jacobi identity}
\begin{align*}
[x,[y,z]]&=[[x,y],z] + [y,[x,z]], \text{ or equivalently }\\
[x,[y,z]]&+[z,[x,y]]+[y,[z,x]]=0
\end{align*}
for all $x, y, z \in \Lieg$.
This $\kk$-bilinear map $[\cdot,\cdot]$ is called the \dfn{Lie
bracket} of $\Lieg$.
\begin{itemize}
\item[(a)]
Check that any associative algebra $A$ gives rise to a
Lie algebra by means of the commutator operation $[a,b]:=ab-ba$.
\item[(b)]
If $A$ is also a bialgebra, show that the $\kk$-submodule
of primitive elements $\Liep \subset A$ is closed under the Lie
bracket, that is, $[\Liep, \Liep] \subset \Liep$, and hence forms
a Lie subalgebra.
\end{itemize}

% [DG][v9] Replaced "$[\cdot,\cdot]$" by
% "$[\cdot,\cdot]:\Lieg\times\Lieg\to\Lieg$".

Conversely, given a Lie algebra $\Liep$,
one constructs the \dfn{universal enveloping algebra}
$\UUU(\Liep):=T(\Liep)/J$\index{$\UUU(\Liep)$}
as the quotient of the tensor algebra
$T(\Liep)$ by the two-sided ideal $J$ generated by all elements
$xy-yx-[x,y]$ for $x,y$ in $\Liep$.
\begin{itemize}
\item[(c)]
Show that $J$ is also a two-sided coideal in $T(\Liep)$ for
its usual coalgebra structure, and hence the quotient
$\UUU(\Liep)$ inherits the structure of a cocommutative bialgebra.
\item[(d)]
Show that the antipode $S$ on $T(\Liep)$ preserves $J$, meaning
that $S(J) \subset J$, and hence $\UUU(\Liep)$ inherits the
structure of a (cocommutative) Hopf algebra.
\end{itemize}
\end{exercise}

There are theorems,
discussed in \cite[\S3.8]{Cartier}, \cite[Chap. 5]{Montgomery},
\cite[\S 3.2]{DuchampMinhTolluChienNghia}
giving various mild hypotheses in addition to  cocommutativity
which imply that the inclusion of the $\kk$-module $\Liep$ of primitives in
a Hopf algebra $A$ extends to a Hopf isomorphism
$\UUU(\Liep) \cong A$.

% [DG][v13] Added {DuchampMinhTolluChienNghia} reference since it is
% one of the few to cover base rings which are not fields. (Another
% is Fresse's operads book, but it is still in the process of being
% written and the proof of Cartier-Milnor-Moore will be rewritten.)

\begin{exercise}
\label{exe.commutative.convolution}
Let $C$ be a cocommutative $\kk$-coalgebra.
Let $A$ be a commutative $\kk$-algebra.
Show that the convolution algebra
$\left( \Hom \left( C, A \right) , \star \right)$ is commutative
(i.e., every $f, g \in \Hom \left( C, A \right)$
satisfy $f \star g = g \star f$).
\end{exercise}

% [DG][v62] Added the simple exercise above.

\begin{exercise} \phantomsection
\label{exe.cocommutative.coalghom}
\begin{itemize}
\item[(a)] Let $C$ be a $\kk$-coalgebra. Show that $C$ is cocommutative if and
only if its comultiplication $\Delta_C : C \to C \otimes C$ is
a $\kk$-coalgebra homomorphism.
\item[(b)] Let $A$ be a $\kk$-algebra. Show that $A$ is commutative if and
only if its multiplication $m_A : A \otimes A \to A$ is
a $\kk$-algebra homomorphism.
\end{itemize}
\end{exercise}

% [DG][v14] Added this exercise.

\begin{remark}
If $C$ is a $\kk$-coalgebra, then $\epsilon_C : C \to \kk$ is always a
$\kk$-coalgebra homomorphism. Similarly, $u_A : \kk \to A$ is a
$\kk$-algebra homomorphism whenever $A$ is a $\kk$-algebra.
\end{remark}

% [DG][v25] Added above remark to accompany the exercise just above it.

\begin{exercise} \phantomsection
\label{exe.comm-cocomm.anti}
\begin{itemize}

\item[(a)] Let $A$ and $B$ be two $\kk$-algebras, at least
one of which is commutative.
Prove that the $\kk$-algebra anti-homomorphisms from $A$ to $B$
are the same as the $\kk$-algebra homomorphisms from $A$ to $B$.

\item[(b)] State and prove the dual of this result.

\end{itemize}
\end{exercise}

% [DG][v80] Added the above near-trivial exercise as its result has
% been tacitly used a few times.

\begin{exercise}
\label{exe.iterated.comm}
Let $A$ be a commutative $\kk$-algebra, and let $k \in \NN$. The
symmetric group $\Symm_k$ acts on the $k$-fold tensor power $A^{\otimes k}$
by permuting the tensor factors:
$\sigma\left(v_1 \otimes v_2 \otimes \cdots \otimes v_k\right)
= v_{\sigma^{-1}\left(1\right)} \otimes v_{\sigma^{-1}\left(2\right)}
  \otimes \cdots \otimes v_{\sigma^{-1}\left(k\right)}$
for all $v_1, v_2, \ldots, v_k \in A$ and $\sigma \in \Symm_k$.
For every $\pi \in \Symm_k$, denote
by $\rho\left(\pi\right)$ the action of $\pi$ on $A^{\otimes k}$ (this is
an endomorphism of $A^{\otimes k}$).
Show that every $\pi \in \Symm_k$ satisfies
$m^{\left(k-1\right)} \circ \left(\rho\left(\pi\right)\right)
= m^{\left(k-1\right)}$. (Recall that
$m^{\left(k-1\right)} : A^{\otimes k} \to A$
is defined as in Exercise~\ref{exe.iterated.m} for $k \geq 1$, and by
$m^{\left(-1\right)} = u : \kk \to A$ for $k = 0$.)
\end{exercise}

\begin{exercise}
\label{exe.iterated.cocomm}
State and solve the analogue of Exercise~\ref{exe.iterated.comm} for
cocommutative $\kk$-coalgebras.
\end{exercise}

% [DG][v19] Added above two exercises.

\begin{exercise} \phantomsection
\label{exe.alghoms.comm}
\begin{itemize}
\item[(a)] If $H$ is a $\kk$-bialgebra and $A$ is a commutative
$\kk$-algebra, and if $f$ and $g$ are two $\kk$-algebra homomorphisms
$H \to A$, then prove that $f \star g$ also is a $\kk$-algebra
homomorphism $H \to A$.
\item[(b)] If $H$ is a $\kk$-bialgebra and $A$ is a commutative
$\kk$-algebra, and if $f_1, f_2, \ldots, f_k$ are several
$\kk$-algebra homomorphisms $H \to A$, then prove that
$f_1 \star f_2 \star \cdots \star f_k$ also is a $\kk$-algebra
homomorphism $H \to A$.
\item[(c)] If $H$ is a Hopf algebra and $A$ is a commutative
$\kk$-algebra, and if $f : H \to A$ is a $\kk$-algebra
homomorphism, then prove that $f \circ S : H \to A$ (where $S$
is the antipode of $H$) is again a $\kk$-algebra homomorphism,
and is a $\star$-inverse to $f$.
\item[(d)] If $A$ is a commutative $\kk$-algebra, then show that
$m^{\left(k\right)}$ is a $\kk$-algebra homomorphism for every
$k \in \NN$. (The map
$m^{\left(k\right)} : A^{\otimes \left(k+1\right)} \to A$
is defined as in Exercise~\ref{exe.iterated.m}.)
\item[(e)] If $C^\prime$ and $C$ are two $\kk$-coalgebras, if
$\gamma : C \to C^\prime$ is a $\kk$-coalgebra homomorphism,
if $A$ and $A^\prime$ are two $\kk$-algebras, if
$\alpha : A \to A^\prime$ is a $\kk$-algebra homomorphism, and
if $f_1, f_2, \ldots, f_k$ are several $\kk$-linear maps
$C^\prime \to A$, then prove that
\[
\alpha \circ \left(f_1 \star f_2 \star \cdots \star f_k\right)
\circ \gamma
= \left(\alpha \circ f_1 \circ \gamma\right)
\star \left(\alpha \circ f_2 \circ \gamma\right)
\star \cdots
\star \left(\alpha \circ f_k \circ \gamma\right) .
\]
\item[(f)] If $H$ is a commutative $\kk$-bialgebra, and $k$
and $\ell$ are two nonnegative integers, then prove that
$\id_H^{\star k} \circ \id_H^{\star \ell}
= \id_H^{\star \left(k\ell\right)}$.
\item[(g)] If $H$ is a commutative $\kk$-Hopf algebra, and $k$
and $\ell$ are two integers, then prove that
$\id_H^{\star k} \circ \id_H^{\star \ell}
= \id_H^{\star \left(k\ell\right)}$. (These powers
$\id_H^{\star k}$, $\id_H^{\star \ell}$ and 
$\id_H^{\star \left(k\ell\right)}$ are well-defined
since $\id_H$ is $\star$-invertible.)
\item[(h)] State and prove the duals of parts (a)--(g) of this
exercise.
\end{itemize}
\end{exercise}

\begin{remark}
The maps $\id_H^{\star k}$ for $k \in \NN$ are known as the
\dfn{Adams operators} of the bialgebra $H$; they are studied,
inter alia, in \cite{AguiarLauve-2014}. Particular cases
(and variants) of
Exercise~\ref{exe.alghoms.comm}(f) appear in
\cite[Corollaire II.9]{Patras-descent} and
\cite[Theorem 1]{GerstenhaberSchack-shuffle}.
Exercise~\ref{exe.alghoms.comm}(f) and its dual are
\cite[Prop. 1.6]{Loday-serie-Hausdorff}.
\end{remark}

% [DG][v25] Added this exercise (part (a) of which I use later).
% 
% The convolution powers of the identity in part (f) are called the
% Adams operators/-ions; you might have some words to say about why
% (I don't know enough geometry).

\begin{exercise}
\label{exe.cocomm-antipode-S2}
Prove that the antipode $S$ of a cocommutative Hopf algebra
$A$ satisfies $S^2 = \id_A$. (This was a statement made in
Remark~\ref{rmk.commutative-implies-involutive-antipode-cor.cocomm}.)
\end{exercise}

% [DG][v80] Added this exercise. (It used to be part (e) in
% Exercise~\ref{exe.coalg-anti.gen} before, but then I realized
% that "cocommutative" was not defined at the point where
% that exercise was posed, so this part was premature.)

\begin{exercise}
\label{exe.dynkin}
Let $A$ be a cocommutative graded Hopf algebra with antipode
$S$. Define a $\kk$-linear map $E : A\rightarrow A$ by having
$E\left( a \right) = \left( \deg a \right) \cdot a$
for every homogeneous element $a$ of $A$.

\begin{itemize}
\item[(a)] Prove that for every $a\in A$, the elements $\left(  S\star
E\right)  \left(  a\right)  $ and $\left(  E\star S\right)  \left(  a\right)
$ (where $\star$ denotes convolution in $\Hom\left(A,A\right)  $) are
primitive.

\item[(b)] Prove that for every primitive $p\in A$, we have
$ \left(  S\star E\right)  \left(  p\right) 
= \left(  E\star S\right)  \left(  p\right)
= E\left(  p\right)  $.

\item[(c)] Prove that for every $a\in A$ and every primitive $p\in A$, we
have $\left(  S\star E\right)  \left(  ap\right)  =\left[  \left(  S\star
E\right)  \left(  a\right)  ,p\right]  +\epsilon\left(  a\right)  E\left(
p\right)  $, where $\left[  u,v\right] $ denotes the commutator $uv-vu$ of
$u$ and $v$.

\item[(d)] If $A$ is connected and $\QQ$ is a subring of $\kk$,
prove that the $\kk$-algebra $A$
is generated by the $\kk$-submodule $\Liep$ consisting of the
primitive elements of $A$.

\item[(e)] Assume that $A$ is the tensor algebra $T\left( V \right)$
of a $\kk$-module $V$, and that the $\kk$-submodule
$V = V^{\otimes 1}$ of $T\left( V \right)$ is the degree-$1$
homogeneous component of $A$.
Show that
$\left( S\star E\right)  \left( x_1 x_2 \ldots x_n \right)
= \left[ \ldots\left[ \left[ x_1, x_2 \right] , x_3 \right] ,
\ldots, x_n \right]  $ for any $n \geq 1$ and any
$x_1, x_2, \ldots , x_n \in V$.
\end{itemize}
\end{exercise}

% [DG][v13] I added part (d) to the above exercise.

% [DG][v37] Added grading condition in (e).

\begin{remark}
\label{rmk.dynkin}
Exercise~\ref{exe.dynkin} gives rise to a certain idempotent
map $A\rightarrow A$ when $\kk$ is a commutative $\QQ$-algebra
and $A$ is a cocommutative connected graded $\kk$-Hopf algebra.
Namely, the $\kk$-linear map
$A \rightarrow A$ sending every homogeneous $a \in A$ to
$\frac{1}{\deg a} \left( S\star E \right) \left( a \right)$
(or $0$ if $\deg a = 0$) is
idempotent and is a projection on the $\kk$-module of primitive
elements of $A$. It is
called the \dfn{Dynkin idempotent}; see
\cite{PatrasReutenauer} for more of its properties.\footnote{We
will see another such idempotent in
Exercise~\ref{exe.eulerian-idp}.}
Part (c) of the exercise is more or less Baker's identity.
\end{remark}

% [DG][v3] Added exercise above.



\subsection{\label{subsect.duals}Duals}

Recall that for $\kk$-modules $V$, taking the dual $\kk$-module
$V^* :=\Hom(V,\kk)$ reverses $\kk$-linear maps.  That is, every
$\kk$-linear map $V \overset{\varphi}{\rightarrow} W$ induces
an \dfn{adjoint map}
$W^* \overset{\varphi^*}{\rightarrow} V^*$ defined uniquely by
\[
( f, \varphi(v) ) = ( \varphi^*(f), v )
\]
in which $(f,v)$ is the bilinear pairing $V^* \times V \rightarrow \kk$
sending $(f,v) \mapsto f(v)$.  If $V$ and $W$ are finite free
$\kk$-modules\footnote{A $\kk$-module is said to be \dfn{finite
free} if it has a finite basis. If $\kk$ is a field, then a
finite free $\kk$-module is the same as a finite-dimensional
$\kk$-vector space.}, more can be said: When $\varphi$ is
expressed in terms of a basis $\{v_i\}_{i \in I}$ for $V$
and a basis $\{w_j\}_{j \in J}$ for $W$ by some matrix,
the map $\varphi^*$ is expressed by the transpose matrix in terms
of the dual bases of these two bases\footnote{If
$\{v_i\}_{i \in I}$ is a basis of a finite free $\kk$-module $V$,
then the \dfn{dual basis} of this basis is defined as the basis
$\{f_i\}_{i \in I}$ of $V^*$ that satisfies
$(f_i,v_j)=\delta_{i,j}$ for all $i$ and $j$. (Recall that
$\delta_{i,j}$ is the Kronecker delta: $\delta_{i,j}= 1$ if $i=j$
and $0$ else.)}.

% [DG][v19] Revamped the last sentence.

The correspondence $\varphi \mapsto \varphi^*$ between
$\kk$-linear maps $V \overset{\varphi}{\rightarrow} W$ and
$\kk$-linear maps $W^* \overset{\varphi^*}{\rightarrow} V^*$ is
one-to-one when $W$ is finite free. However, this is not the case
in many combinatorial situations (in which $W$ is usually free but
not finite free). Fortunately, many of the good properties of
finite free
modules carry over to a certain class of graded modules as long
as the dual $V^*$ is replaced by a smaller module $V^o$ called the
graded dual. Let us first introduce the latter:

When $V=\bigoplus_{n \geq 0} V_n$ is a graded $\kk$-module,
note that the dual $V^* = \prod_{n\geq 0} (V_n)^*$ can contain
functionals $f$ supported on infinitely many $V_n$. However, we
can consider the
$\kk$-submodule $V^o:=\bigoplus_{n \geq 0} (V_n)^* \subset
\prod_{n\geq 0} (V_n)^* = V^*$\index{$V^o$},
sometimes called the \dfn{graded dual}\footnote{Do not mistake
this for the coalgebraic restricted dual $A^\circ$ of
\cite[\S 6.0]{Sweedler}.},
consisting of the functions $f$ that vanish on all but
finitely many $V_n$.
Notice that $V^o$ is graded, whereas $V^*$ (in general) is not.
If $V \overset{\varphi}{\rightarrow} W$ is a graded $\kk$-linear
map, then the adjoint map
$W^* \overset{\varphi^*}{\rightarrow} V^*$
restricts to a graded $\kk$-linear map
$W^o \rightarrow V^o$, which we (abusively) still denote by
$\varphi^*$.

A graded $\kk$-module $V=\bigoplus_{n \geq 0} V_n$ is said to be
\dfn{of finite type} if each $V_n$ is a finite free
$\kk$-module\footnote{This meaning of ``finite type'' can differ
from the standard one.}.
When the graded $\kk$-module $V$ is of finite type,
the graded $\kk$-module $V^o$ is
again of finite type\footnote{More precisely: Let
$V=\bigoplus_{n \geq 0} V_n$ be of finite type, and let
$\left\{v_i\right\}_{i \in I}$ be a \dfn{graded basis} of $V$,
that is, a basis of the $\kk$-module $V$ such that
the indexing set $I$ is partitioned into subsets
$I_0, I_1, I_2, \ldots$ (which are allowed to be empty) with the
property that, for every $n \in \NN$, the subfamily
$\left\{v_i\right\}_{i \in I_n}$ is a basis of the $\kk$-module
$V_n$. Then, we can define a family $\left\{f_i\right\}_{i \in I}$
of elements of $V^o$ by setting
$\left(f_i, v_j\right) = \delta_{i,j}$ for all $i, j \in I$.
This family $\left\{f_i\right\}_{i \in I}$ is a graded basis of
the graded $\kk$-module $V^o$. (Actually, for every $n \in \NN$,
the subfamily $\left\{f_i\right\}_{i \in I_n}$ is a basis of
the $\kk$-submodule $\left(V_n\right)^\ast$ of $V^o$ -- indeed
the dual basis to the basis $\left\{v_i\right\}_{i \in I_n}$ of
$V_n$.) This basis $\left\{f_i\right\}_{i \in I}$ is said to be
the \dfn{dual basis} to the basis $\left\{v_i\right\}_{i \in I}$
of $V$.}
and satisfies $\left(V^o\right)^o \cong V$.
Many other properties of finite free modules are salvaged in this
situation; most importantly:
The correspondence $\varphi \mapsto \varphi^*$ between graded
$\kk$-linear maps $V \rightarrow W$ and graded
$\kk$-linear maps $W^o \rightarrow V^o$ is
one-to-one when $W$ is of finite type\footnote{Only $W$ has
to be of finite type here; $V$ can be any graded $\kk$-module.}.

% [DG][v7] Replaced "which is again a graded vector space" by
% "This $V^o$ is again a graded vector space" to disambiguate grammar.

% [DG][v17] The above part of this section rewritten in a
% ring-friendly and more explicit (and leisurely?) way.
% 
% Also, renamed "restricted dual" by "graded dual" because I have
% seen the Sweedler "zero-dual" being referred to as restricted
% dual, and this is a different beast.
% 
% Shouldn't the statement about matrices require bases for both
% $V$ and $W$?

% [DG][v48] Introduced dual basis of $V^o$ in the finite-type
% setting in a footnote. (The concept is used a lot.)

Reversing the diagrams should then make it clear that, in the
finite free or finite-type situation, duals of algebras are
coalgebras, and vice-versa,
and duals of bialgebras or Hopf algebras are bialgebras or Hopf algebras.
For example, the product in a Hopf algebra $A$ of finite type uniquely defines
the coproduct of $A^o$ via adjointness:
\[
\left(\Delta_{A^o}(f) , a \otimes b\right)_{A \otimes A}
 = (f , a b)_{A}.
\]
Thus if $A$ has a basis $\{a_i\}_{i \in I}$ with \dfn{product structure
constants} $\{c^i_{j,k}\}$, meaning
\[
a_j a_k = \sum_{i \in I} c^i_{j,k} a_i,
\]
then the dual basis $\{f_i\}_{i \in I}$ has the same  $\{c^i_{j,k}\}$ as its
\dfn{coproduct structure constants}:
\[
\Delta_{A^o}(f_i) = \sum_{(j,k) \in I \times I} c^i_{j,k} f_j \otimes f_k.
\]
The assumption that $A$ be of finite type was indispensable here; in general,
the dual of a $\kk$-algebra does not become a $\kk$-coalgebra. However,
the dual of a $\kk$-coalgebra still becomes a $\kk$-algebra, as shown in the
following exercise:

\begin{exercise}
\label{exe.dual-algebra}
For any two $\kk$-modules $U$ and $V$, let $\rho_{U,V} :
U^{\ast} \otimes V^{\ast} \to \left(U \otimes V\right)^{\ast}$ be the
$\kk$-linear map which sends every tensor
$f \otimes g \in U^{\ast} \otimes V^{\ast}$ to the composition
$U \otimes V \overset{f \otimes g} {\longrightarrow}
\kk \otimes \kk \overset{m_\kk} {\longrightarrow} \kk$ of the
map\footnote{Keep in mind that the \emph{tensor}
$f \otimes g \in U^{\ast} \otimes V^{\ast}$ is not the same as the
\emph{map} $U \otimes V \overset{f \otimes g} {\longrightarrow}
\kk \otimes \kk$.} $f \otimes g$ with the canonical isomorphism
$\kk \otimes \kk \overset{m_\kk} {\longrightarrow} \kk$. When $\kk$ is a
field and $U$ is finite-dimensional, this map
$\rho_{U,V}$ is a $\kk$-vector space isomorphism (and usually regarded
as the identity); more generally, it is injective whenever $\kk$ is a
field\footnote{Over arbitrary rings it does not have to be even that!}.
% http://mathoverflow.net/questions/56255/duals-and-tensor-products has
% an example for the footnote.
Also, let $s : \kk \to \kk^*$ be the canonical isomorphism.
Prove that:
\begin{itemize}
\item[(a)] If $C$ is a $\kk$-coalgebra, then $C^*$ becomes a
$\kk$-algebra if we define its associative operation by
$m_{C^*} = \Delta_C^* \circ \rho_{C,C} : C^* \otimes C^* \to C^*$ and
its unit map to be $\epsilon_C^* \circ s : \kk \to C^*$.
\ \ \ \ \footnote{If $C$ is a finite free $\kk$-module, then this
$\kk$-algebra structure is the same as the one defined above by
adjointness. But the advantage of the new definition is that it
works even if $C$ is not a finite free $\kk$-module.}
\item[(b)] The $\kk$-algebra structure defined on $C^*$ in
part (a) is precisely the one defined on $\Hom\left(C,\kk\right) = C^*$
in Definition~\ref{convolution-algebra} applied to $A=\kk$.
\item[(c)] If $C$ is a graded $\kk$-coalgebra, then $C^o$ is
a $\kk$-subalgebra of the $\kk$-algebra $C^*$ defined in part (a).
\item[(d)] If $f : C \to D$ is a homomorphism of $\kk$-coalgebras, then
$f^* : D^* \to C^*$ is a homomorphism of $\kk$-algebras.
\item[(e)] Let $U$ be a graded $\kk$-module (not necessarily of
finite type), and let $V$ be a graded $\kk$-module of finite type.
Then, there is a 1-to-1 correspondence between graded $\kk$-linear maps
$U \to V$ and graded $\kk$-linear maps $V^o \to U^o$ given by
$f \mapsto f^*$.
\item[(f)] Let $C$ be a graded $\kk$-coalgebra (not necessarily of
finite type), and let $D$ be a graded $\kk$-coalgebra of finite type.
Part (e) of this exercise shows that
there is a 1-to-1 correspondence between graded $\kk$-linear maps
$C \to D$ and graded $\kk$-linear maps $D^o \to C^o$ given by
$f \mapsto f^*$. This correspondence has the property that a given
graded $\kk$-linear map $f : C \to D$ is a $\kk$-coalgebra morphism
if and only if $f^* : D^o \to C^o$ is a $\kk$-algebra morphism.
\end{itemize}
\end{exercise}

% [DG][v14] Added the above exercise, and referenced it from Section 6
% in which it *now* is used.

Another example of a Hopf algebra is provided by the so-called shuffle
algebra. Before we introduce it, let us define the \emph{shuffles} of
two words:

\begin{definition}
\label{shuffles}
Given two words $a=\left(  a_{1},a_{2}, \ldots ,a_{n}\right)  $ and
$b=\left(  b_{1},b_{2}, \ldots ,b_{m}\right)  $, the \emph{multiset of shuffles
of $a$ and $b$}\index{shuffle of words} is defined as the multiset
\[
\left\{  \left(  c_{w\left(  1\right)  },c_{w\left(  2\right)
}, \ldots ,c_{w\left(  n+m\right)  }\right)  \ : 
\ w\in\operatorname{Sh}_{n,m}\right\}_{\text{multiset}}  ,
\]
where $\left(  c_{1},c_{2}, \ldots ,c_{n+m}\right)  $ is the concatenation $a\cdot
b=\left(  a_{1},a_{2}, \ldots ,a_{n},b_{1},b_{2}, \ldots ,b_{m}\right)  $, and where
\dfn{$\operatorname{Sh}_{n,m}$} is the subset\footnote{\textbf{Warning:} This
definition of $\operatorname{Sh}_{n,m}$ is highly nonstandard, and many
authors define $\operatorname{Sh}_{n,m}$ to be the set of the inverses
of the permutations belonging to what we call $\operatorname{Sh}_{n,m}$.}
\[
\left\{  w\in\Symm_{n+m}\ : \ w^{-1}\left(  1\right)  <w^{-1}\left(  2\right)
<\cdots<w^{-1}\left(  n\right)  ;\ w^{-1}\left(  n+1\right)  <w^{-1}\left(  n+2\right)
<\cdots<w^{-1}\left(  n+m\right)  \right\}
\]
of the symmetric group $\Symm_{n+m}$. Informally speaking, the shuffles of
the two words $a$ and $b$ are the words obtained by overlaying the words $a$
and $b$, after first moving their letters apart so that no letters get
superimposed when the words are overlayed\footnote{For instance, if
$a = \left(1,3,2,1\right)$ and $b = \left(2,4\right)$, then
the shuffle $\left(1,2,3,2,4,1\right)$ of $a$ and $b$ can be obtained by
moving the letters of $a$ and $b$ apart as follows:
\[
\begin{array}{ccccccc}
a = & 1 &   & 3 & 2 &   & 1 \\
b = &   & 2 &   &   & 4 &   
\end{array}
\]
and then overlaying them to obtain
$\begin{array}{cccccc} 1 & 2 & 3 & 2 & 4 & 1 \end{array}$. Other ways of
moving letters apart lead to further shuffles (not always distinct).}.
In particular, any shuffle of
$a$ and $b$ contains $a$ and $b$ as subsequences. The multiset of shuffles of $a$ and
$b$ has $\binom{m+n}{n}$ elements
(counted with multiplicity) and is denoted by $\shuf{a}{b}$.
For instance, the shuffles of $\left(1,2,1\right)$
and $\left(3,2\right)$ are
\begin{align*}
\left({\underline 1},{\underline 2},{\underline 1},3,2\right),
\left({\underline 1},{\underline 2},3,{\underline 1},2\right),
\left({\underline 1},{\underline 2},3,2,{\underline 1}\right),
\left({\underline 1},3,{\underline 2},{\underline 1},2\right),
\left({\underline 1},3,{\underline 2},2,{\underline 1}\right), \\
\left({\underline 1},3,2,{\underline 2},{\underline 1}\right),
\left(3,{\underline 1},{\underline 2},{\underline 1},2\right),
\left(3,{\underline 1},{\underline 2},2,{\underline 1}\right),
\left(3,{\underline 1},2,{\underline 2},{\underline 1}\right),
\left(3,2,{\underline 1},{\underline 2},{\underline 1}\right),
\end{align*}
listed here as often as they appear in the multiset
$\shuf{ \left(1,2,1\right) }{ \left(3,2\right) }$. Here we have
underlined the letters taken from $a$ -- that is,
the letters at positions $w^{-1}\left(  1\right)$, $w^{-1}\left(  2\right)$,
$\ldots$, $w^{-1}\left(  n\right)$.
\end{definition}

% [DG][v3] I added the definition above.

% [DG][v25] Added footnote with an example of the overlaying procedure.
% And another footnote warning readers against my nonstandard
% notation (my shuffles seem to be most people's unshuffles).
% Hmm... maybe I should change it?

\begin{example}
\label{exa.shuffle-alg}
When $A=T(V)$ is the tensor algebra for a finite free $\kk$-module
$V$, having $\kk$-basis
$\{x_i\}_{i \in I}$, its graded dual $A^o$ is another Hopf
algebra whose basis $\left\{ y_{(i_1,\ldots,i_\ell)} \right\}$ (the
dual basis of the basis $\left\{ x_{i_1} \cdots x_{i_\ell} \right\}$
of $A = T\left(V\right)$) is indexed by
words in the alphabet $I$. This Hopf algebra $A^o$ could be called the
\emph{shuffle algebra} of $V^*$.
(To be more precise, it is isomorphic to the shuffle algebra of $V^*$
introduced in Proposition~\ref{prop.shuffle-alg} further below; we
prefer not to call $A^o$ itself the shuffle algebra of $V^*$, since
$A^o$ has several disadvantages\footnote{Specifically, $A^o$ has
the disadvantages of being defined only when $V^*$ is the
dual of a finite free $\kk$-module $V$, and depending on a choice of
basis,
whereas Proposition~\ref{prop.shuffle-alg} will define shuffle algebras
in full generality and canonically.}.)
Duality shows that the \emph{cut} coproduct in $A^o$ is defined by
\begin{equation}
\label{eq.exa.shuffle-alg.Delta}
\Delta y_{(i_1,\ldots,i_\ell)} =
\sum_{j=0}^{\ell} y_{(i_1,\ldots,i_j)} \otimes y_{(i_{j+1},i_{j+2},\ldots,i_\ell)}.
\end{equation}
For example,
\[
\Delta y_{abcb}  = y_{\varnothing} \otimes y_{abcb}
+y_{a} \otimes y_{bcb}
+y_{ab} \otimes y_{cb}
+y_{abc} \otimes y_{b}
+y_{abcb} \otimes y_{\varnothing} .
\]
Duality also shows that the
\emph{shuffle} product in $A^o$ will be given by
\begin{equation}
\label{eq.exa.shuffle-alg.m}
y_{(i_1,\ldots,i_\ell)} y_{(j_1,\ldots,j_m)} =
\sum_{\kk=(k_1,\ldots,k_{\ell+m})
        \in \shuf{\ii}{\jj} } y_{(k_1,\ldots,k_{\ell+m})}
\end{equation}
where $\shuf{\ii}{\jj}$ (as in Definition~\ref{shuffles})
denotes the multiset of the $\binom{\ell+m}{\ell}$
words obtained as \emph{shuffles} of the two
words $\ii=(i_1,\ldots,i_\ell)$ and $\jj=(j_1,\ldots,j_m)$.
For example,
\begin{align*}
y_{ab} y_{cb} &= y_{abcb} + y_{acbb} + y_{cabb} + y_{cabb}+ y_{acbb} + y_{cbab}\\
&= y_{abcb} + 2 y_{acbb} + 2 y_{cabb} +  y_{cbab} .
\end{align*}
Equivalently, one has
\begin{align}
\label{eq.exa.shuffle-alg.m2}
y_{(i_1,i_2,\ldots,i_\ell)} y_{(i_{\ell+1},i_{\ell+2},\ldots,i_{\ell+m})}
&=
\sum\limits_{\substack{w \in \Symm_{\ell+m}:\\w(1)<\cdots<w(\ell),\\
       w(\ell+1) < \cdots < w(\ell+m)}}
       y_{\left(i_{w^{-1}(1)},i_{w^{-1}(2)},\ldots,i_{w^{-1}(\ell+m)}\right)} \\
&=
\label{eq.exa.shuffle-alg.m3}
\sum\limits_{\substack{\sigma \in \operatorname{Sh}_{\ell,m}}}
       y_{\left(i_{\sigma\left(1\right)},i_{\sigma\left(2\right)},\ldots,i_{\sigma\left(\ell+m\right)}\right)}
\end{align}
(using the notations of Definition~\ref{shuffles} again).
Lastly, the antipode $S$ of $A^o$ is the adjoint of the antipode of $A=T(V)$ described in \eqref{antipode-in-tensor-algebra}:
\[
S y_{(i_1,i_2,\ldots,i_\ell)}
=(-1)^{\ell} y_{(i_\ell,\ldots,i_2,i_1)}.
\]
Since the coalgebra $T\left(V\right)$ is cocommutative, its graded dual
$T\left(V\right)^o$ is commutative.

\end{example}

% [DG][v25] Added parenthetical remark on more general shuffle algebras,
% and yet one more way to write down the multiplication formula.

% [DG][v31] Fixed the use of two different notations
% $\operatorname{Sh}_{\ell,m}$ and
% $\operatorname{Sh}(\ell,m)$ for one and the same thing.
% This is still a heavily nonstandard notation :/




\begin{exercise}
\label{exe.Sym.1dim}
Let $V$ be a $1$-dimensional free $\kk$-module with basis element $x$,
so $\Sym(V) \cong \kk[x]$, with
$\kk$-basis $\{1=x^0,x^1,x^2,\ldots\}$.

\begin{itemize}

\item[(a)]
Check that the powers $x^i$ satisfy
\begin{align*}
x^i \cdot x^j &= x^{i+j} , \\
\Delta(x^n) &= \sum_{i+j=n} \binom{n}{i} x^i \otimes x^j , \\
S(x^n) &= (-1)^n x^n .
\end{align*}

\item[(b)]
Check that the dual basis elements
$\{f^{(0)},f^{(1)},f^{(2)},\ldots\}$ for $\Sym(V)^o$,
defined by $f^{(i)}(x^j)=\delta_{i,j}$, satisfy
\begin{align*}
f^{(i)} f^{(j)} &= \binom{i+j}{i} f^{(i+j)} , \\
\Delta(f^{(n)}) &= \sum_{i+j=n} f^{(i)} \otimes f^{(j)} , \\
S(f^{(n)}) &= (-1)^n f^{(n)} .
\end{align*}

\item[(c)]
Show that if $\QQ$ is a subring of $\kk$, then the $\kk$-linear
map $\Sym(V)^{o} \rightarrow \Sym(V)$ sending
$f^{(n)} \mapsto \frac{x^n}{n!}$ is a graded Hopf isomorphism.

For this reason, the Hopf structure on $\Sym(V)^o$ is
called a \dfn{divided power algebra}.

\item[(d)]
Show that when $\kk$ is a field of characteristic $p >0$,
one has $(f^{(1)})^p = 0$, and hence why there can
be no Hopf isomorphism $\Sym(V)^{o} \rightarrow \Sym(V)$.
\end{itemize}
\end{exercise}


\begin{exercise}
\label{exe.Sym.wedge}
Let $V$ have $\kk$-basis $\{x_1,\ldots,x_n\}$,  and let $V \oplus V$ have
$\kk$-basis $\{x_1,\ldots,x_n, y_1,\ldots,y_n\}$, so that one has
isomorphisms
\[
\Sym(V \oplus V) \cong \kk[\xx,\yy] \cong \kk[\xx] \otimes \kk[\yy] \cong
\Sym(V) \otimes \Sym(V).
\]
Here we are using the abbreviations
$\xx = \left(x_1, x_2, \ldots, x_n\right)$
and $\yy = \left(y_1, y_2, \ldots, y_n\right)$.

\begin{itemize}

\item[(a)]
Show that our usual coproduct on $\Sym(V)$ can be re-expressed
as follows:
\[
\begin{array}{ccc}
\Sym(V) & &  \Sym(V) \otimes \Sym(V) \\
\Vert & & \Vert \\
\kk[\xx] & \overset{\Delta}{\longrightarrow} & \kk[\xx,\yy] , \\
 f(x_1,\ldots,x_n) & \longmapsto & f(x_1+y_1,\ldots,x_n+y_n) .\\
\end{array}
\]
In other words, it is induced from the diagonal map
\begin{equation}
\label{vector-space-diagonal-map}
\begin{array}{rcl}
V & \longrightarrow & V \oplus V ,\\
x_i & \longmapsto & x_i+y_i .\\
\end{array}
\end{equation}

\item[(b)]
One can similarly define a coproduct
on the \dfn{exterior algebra} $\wedge V$, which is the quotient $T(V)/J$
where $J$ is the two-sided ideal generated by the elements
$\{x^2(=x \otimes x)\}_{x \in V}$ in $T^2(V)$.
The ideal $J$ is a graded $\kk$-submodule of $T(V)$
(this is not obvious!), and
the quotient $T(V) / J$ becomes a graded commutative algebra
\[
\wedge V
 = \bigoplus_{d=0}^n \wedge^d V \left( = \bigoplus_{d=0}^\infty \wedge^d V \right),
\]
if one views the elements of $V=\wedge^1 V$ as having \emph{odd} degree, and
uses the topologist's sign convention (as in
\eqref{topologist-twist}).  One again has
$
\wedge(V \oplus V) = \wedge V \otimes \wedge V
$
as graded algebras.  Show that one can again let the diagonal map
\eqref{vector-space-diagonal-map} induce a map
\begin{equation}
\label{eq.exe.Sym.wedge.b.map}
\begin{array}{ccc}
\wedge(V) & \overset{\Delta}{\longrightarrow}
 & \wedge V \otimes \wedge V , \\
  f(x_1,\ldots,x_n) & \longmapsto & f(x_1+y_1,\ldots,x_n+y_n) \\
 \Vert & & \Vert \\
 \sum c_{i_1,\ldots,i_d} \cdot x_{i_1} \wedge \cdots \wedge x_{i_d} &
  & \sum c_{i_1,\ldots,i_d} \cdot (x_{i_1}+y_{i_1} )\wedge \cdots \wedge (x_{i_d}+y_{i_d}) ,\\
\end{array}
\end{equation}
which makes $\wedge V$ into a connected graded Hopf algebra.

\item[(c)]
Show that in the tensor algebra $T(V)$, if
one views the elements of $V=V^{\otimes 1}$ as having odd degree,
and uses the topologist's sign convention
\eqref{topologist-twist} in
the twist map when defining $T(V)$,
then for any $x$ in $V$ one has
$
\Delta(x^2) = 1 \otimes x^2 + x^2 \otimes 1.
$

\item[(d)]
Let us use the convention \eqref{topologist-twist} as in
part (c).
Show that the two-sided ideal $J \subset T(V)$
generated by $\{x^2\}_{x \in V}$ is also a two-sided coideal
and a graded $\kk$-submodule of $T(V)$,
and hence the quotient $\wedge V = T(V)/J$ inherits the structure
of a graded bialgebra.
Check that the coproduct on $\wedge V$
inherited from $T(V)$
is the same as the one defined in part (b).

% [DG][v80] Added the "graded" part to parts (a) and (d),
% because we cannot avoid keeping track of the grading
% when we are using the topologist's sign convention.
% (I mean, we could, if we defined superalgebras, but we
% haven't.)
% Anyway, this makes the exercise more interesting.

\end{itemize}

[\textbf{Hint:} The ideal $J$ in part (b) is a graded
$\kk$-submodule of $T(V)$, but this is not completely
obvious (not all elements of $V$ have to be
homogeneous!).]

% [DG][v25] Added this hint, instead of the old hint after (c)
% that I've commented out.
\end{exercise}

\begin{exercise}
\label{exe.convolution.dual}
Let $C$ be a $\kk$-coalgebra. As we know
from Exercise~\ref{exe.dual-algebra}(a), this makes $C^\ast$
into a $\kk$-algebra.

Let $A$ be a $\kk$-algebra which is finite free as
$\kk$-module. This makes $A^\ast$ into
a $\kk$-coalgebra.

Let $f : C \to A$ and $g : C \to A$ be two $\kk$-linear maps.
Show that $f^\ast \star g^\ast = \left(f \star g\right)^\ast$.
\end{exercise}

% [DG][v19] Added exercise above.

The above arguments might have created the impression that duals
of bialgebras have good properties only under certain restrictive
conditions (e.g., the dual of a bialgebra $H$ does not generally
become a bialgebra unless $H$ is of finite type), and so they
cannot be used in proofs and constructions unless one is willing
to sacrifice some generality (e.g., we had to require
$V$ to be finite free in Example~\ref{exa.shuffle-alg}). While
the first part of this impression is true, the second is not
always; often there is a way to gain back the generality lost
from using duals. As an example of this, let us define the
shuffle algebra of an arbitrary $\kk$-module (not just of a
dual of a finite free $\kk$-module as in
Example~\ref{exa.shuffle-alg}):

\begin{proposition}
\label{prop.shuffle-alg}
Let $V$ be a $\kk$-module. Define a $\kk$-linear map
$\Delta_\shuffle : T\left(V\right) \to T\left(V\right) \otimes T\left(V\right)$
by setting
\begin{align*}
\Delta_\shuffle \left( v_1 v_2 \cdots v_n \right)
= \sum_{k=0}^n \left( v_1 v_2 \cdots v_k \right) \otimes
\left( v_{k+1} v_{k+2} \cdots v_n \right)
\qquad \qquad
\text{ for all } n \in \NN \text{ and }
v_1, v_2, \ldots, v_n \in V .
\end{align*}
\footnote{This is well-defined, because the right hand side is
$n$-multilinear in $v_1, v_2, \ldots, v_n$, and because
any $n$-multilinear map
$V^{\times n} \to M$ into a $\kk$-module $M$ gives rise to a
unique $\kk$-linear map $V^{\otimes n} \to M$.} Define a
$\kk$-bilinear map $\shufmult : T\left(V\right)
\times T\left(V\right) \to T\left(V\right)$, which will be written
in infix notation (that is, we will write $a \shufmult b$
instead of $\shufmult \left(a, b\right)$), by
setting\footnote{Many authors use the symbol
$\shuffle$ instead of $\shufmult$ here, but we prefer to
reserve the former notation for the shuffle product of words.}
\begin{align*}
\left( v_1 v_2 \cdots v_\ell \right) \shufmult
\left( v_{\ell + 1} v_{\ell + 2} \cdots v_{\ell + m} \right)
&= \sum_{\sigma \in \operatorname{Sh}_{\ell, m}}
v_{\sigma\left(1\right)} v_{\sigma\left(2\right)} \cdots
v_{\sigma\left(\ell + m\right)} \\
& \qquad \qquad
\text{ for all } \ell, m \in \NN \text{ and }
v_1, v_2, \ldots, v_{\ell + m} \in V .
\end{align*}
\footnote{Again, this is well-defined by the $\ell + m$-multilinearity of
the right hand side.}
Consider also the comultiplication
$\epsilon$ of the Hopf algebra $T\left(V\right)$.

Then, the $\kk$-module $T\left(V\right)$, endowed with the
multiplication $\shufmult$, the unit
$1_{T\left(V\right)} \in V^{\otimes 0} \subset T\left(V\right)$,
the comultiplication
$\Delta_\shuffle$ and the counit $\epsilon$, becomes a commutative
Hopf algebra.
This Hopf algebra is called the \dfn{shuffle algebra} of $V$, and
denoted by $\operatorname{Sh}\left(V\right)$.
The antipode of the Hopf algebra $\operatorname{Sh}\left(V\right)$
is precisely the antipode $S$ of $T\left(V\right)$.
\end{proposition}

% [DG][v27] Changed the formatting of this definition.

\begin{exercise}
\label{exe.shuffle-alg}
Prove Proposition~\ref{prop.shuffle-alg}.

[\textbf{Hint:} When $V$ is a finite free $\kk$-module,
Proposition~\ref{prop.shuffle-alg} follows from
Example~\ref{exa.shuffle-alg}. The trick is to derive the general
case from this specific one. Every $\kk$-linear map $f : W \to V$
between two $\kk$-modules $W$ and $V$ induces a map
$T\left(f\right) : T\left(W\right) \to T\left(V\right)$ which
preserves $\Delta_\shuffle$, $\shufmult$,
$1_{T\left(W\right)}$, $\epsilon$ and $S$ (in the appropriate
meanings -- e.g., preserving $\Delta_\shuffle$ means
$\Delta_\shuffle \circ T\left(f\right)
= \left(T\left(f\right) \otimes T\left(f\right)\right)
\circ \Delta_\shuffle$). Show that each of the equalities that need to
be proven in order to verify Proposition~\ref{prop.shuffle-alg} can be
``transported'' along such a map $T\left(f\right)$ from a
$T\left(W\right)$ for a suitably chosen finite free $\kk$-module
$W$.]
\end{exercise}

% [DG][v25] Added above proposition & proof. (I might eventually
% elaborate on shuffle algebras as an aside in the Hazewinkel
% basis section; in fact, one of the simplest consequences of
% the Lyndon shuffle lemma is Radford's theorem on shuffle
% algebras, which proves that $\Qsym$ is a polynomial algebra
% over a base ring of characteristic $0$. This is a warmup to the
% harder result by Hazewinkel that it is always a polynomial
% algebra.)
% 
% I have no idea how this became 5 pages :(

% [DG][v26] This is now an exercise, and the proof is in the
% solutions. Does the hint make sense?

It is also possible to prove Proposition~\ref{prop.shuffle-alg}
``by foot'', as long as one is ready to make combinatorial
arguments about cutting shuffles.

\begin{remark} \phantomsection
\label{rmk.shuffle-alg.dual}

\begin{itemize}

\item[(a)] Let $V$ be a finite free $\kk$-module. The
Hopf algebra $T\left(V\right)^o$ (studied in
Example~\ref{exa.shuffle-alg}) is naturally isomorphic
to the shuffle algebra $\operatorname{Sh}\left(V^\ast\right)$
(defined as in Proposition~\ref{prop.shuffle-alg} but for
$V^\ast$ instead of $V$) as Hopf algebras, by the
obvious isomorphism (namely, the direct sum of the
isomorphisms $\left(V^{\otimes n}\right)^{\ast} \to
\left(V^\ast\right)^{\otimes n}$ over all $n \in \NN$).
\ \ \ \ \footnote{This can be verified
by comparing \eqref{eq.exa.shuffle-alg.Delta} with
the definition of $\Delta_\shuffle$, and comparing
\eqref{eq.exa.shuffle-alg.m3} with the definition of
$\shufmult$.}

\item[(b)] The same statement applies to the case when
$V$ is a graded $\kk$-module of finite type
% (i.e., a graded $\kk$-module such that each
% $V_n$ is finite free)
satisfying $V_0 = 0$
rather than a finite free
$\kk$-module, provided that $V^\ast$
and $\left(V^{\otimes n}\right)^{\ast}$ are replaced by
$V^o$ and $\left(V^{\otimes n}\right)^o$.

\end{itemize}

\end{remark}

% [DG][v31] Added remark above.

We shall return to shuffle algebras in
Section~\ref{subsect.shuffle.radford}, where we will show that
under certain conditions ($\QQ$ being a subring of $\kk$, and
$V$ being a free $\kk$-module) the algebra structure on a
shuffle algebra $\operatorname{Sh}(V)$
is a polynomial algebra in an appropriately chosen set
of generators\footnote{This says nothing about the coalgebra
structure on $\operatorname{Sh}(V)$ -- which is much more
complicated in these generators.}.

% [DG][v36] Added above sentence.

\subsection{\label{subsect.leray}Infinite sums and Leray's theorem}

In this section (which can be skipped, as it will not be used except in a few
exercises), we will see how a Hopf algebra structure on a $\kk$-algebra
reveals knowledge about the $\kk$-algebra itself. Specifically, we will
show that if $\kk$ is a commutative $\QQ$-algebra, and if $A$ is
any commutative connected graded $\kk$-Hopf algebra, then $A$ as a
$\kk$-algebra must be (isomorphic to) a symmetric algebra of a
$\kk$-module\footnote{If $\kk$ is a field, then this simply
means that $A$ as a $\kk$-algebra must be a polynomial ring over
$\kk$.}. This is a specimen of a class of facts which are commonly
called \emph{Leray theorems}; for different specimens, see \cite[Theorem
7.5]{MilnorMoore} or \cite[p. 17, ``Hopf's theorem'']{Cartier} or
\cite[\S 2.5, A, B, C]{Cartier} or
\cite[Theorem 3.8.3]{Cartier}.\footnote{Notice that many of these sources
assume $\kk$ to be a field; some of their proofs rely on this
assumption.} In a sense, these facts foreshadow Zelevinsky's theory of
positive self-dual Hopf algebras, which we shall encounter in
Chapter~\ref{PSH-section}; however, the latter theory works in a much less
general setting (and makes much stronger claims).

We shall first explore the possibilities of applying a formal power series $v$
to a linear map $f:C\rightarrow A$ from a coalgebra $C$ to an algebra $A$.
We have already seen an example of this in the proof of
Proposition~\ref{Takeuchi-formula} above (where the power series
$\sum_{k\geq0}\left(  -1\right)  ^{k}T^{k}
\in\kk \left[  \left[  T\right] \right]  $
was applied to the locally $\star$-nilpotent map
$\id_{A}-u_{A}\epsilon_{A}:A\rightarrow A$); we shall now take a more systematic
approach and establish general criteria for when such applications are
possible. First, we will have to make sense of infinite sums of maps from a
coalgebra to an algebra. This is somewhat technical, but the effort will pay off.

\begin{definition}
\label{def.fin-supp}
Let $A$ be an abelian group (written additively).

We say that a family $\left(  a_{q}\right)  _{q\in Q}\in A^{Q}$ of elements of
$A$ is \dfn{finitely supported} if all but finitely many $q\in Q$ satisfy
$a_{q}=0$. Clearly, if $\left(  a_{q}\right)  _{q\in Q}\in A^{Q}$ is a
finitely supported family, then the sum
$\sum_{q\in Q}a_{q}$\index{infinite sums}
is well-defined
(since all but finitely many of its addends are $0$). Sums like this satisfy
the usual rules for sums, even though their indexing set $Q$ may be infinite.
(For example, if $\left(  a_{q}\right)  _{q\in Q}$ and $\left(  b_{q}\right)
_{q\in Q}$ are two finitely supported families in $A^{Q}$, then the family
$\left(  a_{q}+b_{q}\right)  _{q\in Q}$ is also finitely supported, and we
have $\sum_{q\in Q}a_{q}+\sum_{q\in Q}b_{q}=\sum_{q\in Q}\left(  a_{q}%
+b_{q}\right)  $.)
\end{definition}

\begin{definition}
\label{def.pw-fin-supp}
Let $C$ and $A$ be two $\kk$-modules.

We say that a family $\left(  f_{q}\right)  _{q\in Q} \in \left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ of maps
$f_{q} \in \Hom \left(  C,A\right)  $ is
\dfn{pointwise finitely supported} if for each $x\in C$, the family
$\left(  f_{q}\left(  x\right) \right)  _{q\in Q}\in A^{Q}$%
\index{infinite sums}
of elements of $A$ is finitely supported.\footnote{Here are some
examples of pointwise finitely supported families:
\par
\begin{itemize}
\item If $Q$ is a finite set, then any family $\left(  f_{q}\right)  _{q\in
Q}\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$ is
pointwise finitely supported.
\par
\item More generally, any finitely supported family $\left(  f_{q}\right)
_{q\in Q}\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$ is
pointwise finitely supported.
\par
\item If $C$ is a graded $\kk$-module, and if $\left(  f_{n}\right)
_{n\in\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN}$ is a family of maps such that $f_{n}\left(  C_{m}\right)  =0$
whenever $n\neq m$, then the family $\left(  f_{n}\right)  _{n\in\NN}$
is pointwise finitely supported.
\par
\item If $C$ is a graded $\kk$-coalgebra and $A$ is any
$\kk$-algebra, and if $f\in\Hom \left(  C,A\right)  $ satisfies
$f\left(  C_{0}\right)  =0$, then the family $\left(  f^{\star n}\right)
_{n\in\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN}$ is pointwise finitely supported. (This will be proven in
Proposition~\ref{prop.convolution-series}(h).)
\end{itemize}
} If $\left(  f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is a pointwise finitely supported family, then the
sum $\sum_{q\in Q}f_{q}$ is defined to be the map $C\rightarrow A$ sending
each $x\in C$ to $\sum_{q\in Q}f_{q}\left(  x\right)  $.\ \ \ \ \footnote{This
definition of $\sum_{q\in Q}f_{q}$ generalizes the usual definition of
$\sum_{q\in Q}f_{q}$ when $Q$ is a finite set (because if $Q$ is a finite set,
then any family $\left(  f_{q}\right)  _{q\in Q}\in\left(  \Hom %
\left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely supported).}
\end{definition}

Note that the concept of a ``pointwise finitely supported'' family
$\left( f_q \right) _{q\in Q}
\in \left( \Hom \left( C, A \right) \right)^Q$
is precisely the concept of a ``summable'' family in
\cite[Definition 1]{DuchampMinhTolluChienNghia}.

\begin{definition}
\label{def.subsect.leray.standing1}
For the rest of Section~\ref{subsect.leray},
we shall use the following conventions:

\begin{itemize}
\item Let $C$ be a $\kk$-coalgebra. Let $A$ be a $\kk$-algebra.

\item We shall avoid our standard practice of denoting the unit map
$u_{A}:\kk\rightarrow A$ of a $\kk$-algebra $A$ by $u$; instead,
we will use the letter $u$ (without the subscript $A$) for other purposes.
\end{itemize}
\end{definition}

Definition~\ref{def.pw-fin-supp} allows us to work with infinite sums in
$\Hom \left(  C,A\right)  $, provided that we are summing a
pointwise finitely supported family. We shall next state some properties of
such sums:\footnote{See Exercise~\ref{exe.pw-fin-supp.P1-5} below for the
proofs of these properties.}

\begin{proposition}
\label{prop.pw-fin-supp.P1}
Let $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ be a pointwise finitely
supported family. Then, the map $\sum_{q\in Q}f_{q}$ belongs to
$\Hom \left(  C,A\right)  $.
\end{proposition}

\begin{proposition}
\label{prop.pw-fin-supp.P2}
Let $\left(  f_{q}\right)  _{q\in Q}$ and $\left(
g_{q}\right)  _{q\in Q}$ be two pointwise finitely supported families in
$\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$. Then, the
family $\left(  f_{q}+g_{q}\right)  _{q\in Q}\in\left(  \Hom %
\left(  C,A\right)  \right)  ^{Q}$ is also pointwise finitely supported, and
satisfies
\[
\sum_{q\in Q}f_{q}+\sum_{q\in Q}g_{q}=\sum_{q\in Q}\left(  f_{q}+g_{q}\right)
.
\]

\end{proposition}

\begin{proposition}
\label{prop.pw-fin-supp.P3}
Let $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ and $\left(
g_{r}\right)  _{r\in R}\in\left(  \Hom \left(  C,A\right)
\right)  ^{R}$ be two pointwise finitely supported families. Then, the family
$\left(  f_{q}\star g_{r}\right)  _{\left(  q,r\right)  \in Q\times R}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q\times R}$ is
pointwise finitely supported, and satisfies%
\[
\sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star g_{r}\right)
=\left(  \sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}g_{r}\right)  .
\]

\end{proposition}

Roughly speaking, the above three propositions say that sums of the form
$\sum_{q\in Q}f_{q}$ (where $\left(  f_{q}\right)  _{q\in Q}$ is a pointwise
finitely supported family) satisfy the usual rules for finite sums.
Furthermore, the following properties of pointwise finitely supported families hold:

\begin{proposition}
\label{prop.pw-fin-supp.P4}
Let $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ be a pointwise finitely
supported family. Let $\left(  \lambda_{q}\right)  _{q\in Q}\in\kk^{Q}$
be any family of elements of $\kk$. Then, the family $\left(
\lambda_{q}f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is pointwise finitely supported.
\end{proposition}

\begin{proposition}
\label{prop.pw-fin-supp.P5}
Let $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ and $\left(
g_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ be two families such that $\left(  f_{q}\right)  _{q\in Q}$ is
pointwise finitely supported. Then, the family $\left(  f_{q}\star
g_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is also pointwise finitely supported.
\end{proposition}

\begin{exercise}
\label{exe.pw-fin-supp.P1-5}
Prove Propositions~\ref{prop.pw-fin-supp.P1}%
,~\ref{prop.pw-fin-supp.P2},~\ref{prop.pw-fin-supp.P3}%
,~\ref{prop.pw-fin-supp.P4}~and~\ref{prop.pw-fin-supp.P5}.
\end{exercise}

We can now define the notion of a ``pointwise $\star$-nilpotent'' map.
Roughly speaking, this will mean an element of
$\left(  \Hom \left(  C,A\right)  ,\star\right)  $
that can be substituted into any power series because its powers (with
respect to the convolution $\star$) form a pointwise finitely supported
family. Here is the definition:

\begin{definition}
\phantomsection\label{def.leray.pwnilp}

\begin{enumerate}
\item[(a)] A map $f\in\Hom \left(  C,A\right)  $ is said to be
\dfn{pointwise $\star$-nilpotent} if and only if the family
$\left(  f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is pointwise finitely supported.
Equivalently, a map $f\in\Hom \left(  C,A\right)  $ is pointwise
$\star$-nilpotent if and only if for each $x\in C$, the family $\left(
f^{\star n}\left(  x\right)  \right)  _{n\in\NN}$ of elements of $A$ is
finitely supported.

\item[(b)] If $f\in\Hom \left(  C,A\right)  $ is a pointwise
$\star$-nilpotent map, and if $\left(  \lambda_{n}\right)  _{n\in\NN}
\in\kk^{\NN}$ is any family of scalars, then the family
$\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported\footnote{This follows easily from
Proposition~\ref{prop.pw-fin-supp.P4} above. (In fact, the map $f$ is
pointwise $\star$-nilpotent, and thus the family $\left(  f^{\star n}\right)
_{n\in\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN}$ is pointwise finitely supported (by the definition of
``pointwise $\star$-nilpotent''). Hence,
Proposition~\ref{prop.pw-fin-supp.P4} (applied to $Q = \NN$ and $\left(
f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)  _{n\in\NN}$ and
$\left(  \lambda_{q}\right)  _{q\in Q}=\left(  \lambda_{n}\right)
_{n\in\NN}$) shows that the family $\left(  \lambda_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported.)}, and thus the
infinite sum $\sum_{n\geq0}\lambda_{n}f^{\star n}=\sum_{n\in\NN}%
\lambda_{n}f^{\star n}$ is well-defined and belongs to $\Hom
\left(  C,A\right)  $ (by Proposition~\ref{prop.pw-fin-supp.P1}%
).\footnote{Notice that the concept of ``local $\star$-nilpotence''
we used in the proof of
Proposition~\ref{Takeuchi-prop} serves the same function (viz., ensuring that
the sum $\sum_{n\in\NN}\lambda_{n}f^{\star n}$ is well-defined). But
local $\star$-nilpotence is only defined when a grading is present, whereas
pointwise $\star$-nilpotence is defined in the general case. Also, local
$\star$-nilpotence is more restrictive (i.e., a locally $\star$-nilpotent map
is always pointwise $\star$-nilpotent, but the converse does not always
hold).}

\item[(c)] We let \dfn{$\mathfrak{n}\left(  C,A\right)  $}
be the set of all pointwise $\star$-nilpotent maps
$f\in\Hom \left(  C,A\right) $.
Note that this is not necessarily a $\kk$-submodule of
$\Hom \left(  C,A\right)  $.

\item[(d)] Consider the ring $\kk\left[  \left[  T\right]  \right]  $
of formal power series in an indeterminate $T$ over $\kk$. For any
power series $u\in\kk\left[  \left[  T\right]  \right]  $ and any
$f\in\mathfrak{n}\left(  C,A\right)  $, we define a map
$u^{\star}\left(  f\right)  \in\Hom \left(  C,A\right)  $%
\index{$u^{\star}\left(  f\right)$}
by $u^{\star}\left(  f\right)
=\sum_{n\geq0}u_{n}f^{\star n}$, where $u$ is written in the form
$u=\sum_{n\geq0}u_{n}T^{n}$ with $\left(  u_{n}\right)  _{n\geq0}\in
\kk^{\NN}$. (This sum $\sum_{n\geq0}u_{n}f^{\star n}$ is
well-defined in $\Hom \left(  C,A\right)  $, since $f$ is
pointwise $\star$-nilpotent.)
\end{enumerate}
\end{definition}

The following proposition gathers some properties of pointwise $\star
$-nilpotent maps\footnote{See Exercise~\ref{exe.convolution-series} below for
the proofs of these properties.}:

\begin{proposition}
\phantomsection\label{prop.convolution-series}

\begin{enumerate}
\item[(a)] For any $f\in\mathfrak{n}\left(  C,A\right)  $ and $k\in\NN$,
we have%
\begin{equation}
\left(  T^{k}\right)  ^{\star}\left(  f\right)  =f^{\star k}.
\label{eq.exe.convolution-series.a.monom}%
\end{equation}


\item[(b)] For any $f\in\mathfrak{n}\left(  C,A\right)  $ and $u,v\in
\kk\left[  \left[  T\right]  \right]  $, we have%
\begin{align}
\left(  u+v\right)  ^{\star}\left(  f\right)   &  =u^{\star}\left(  f\right)
+v^{\star}\left(  f\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\label{eq.exe.convolution-series.b.1}\\
\left(  uv\right)  ^{\star}\left(  f\right)   &  =u^{\star}\left(  f\right)
\star v^{\star}\left(  f\right)  . \label{eq.exe.convolution-series.b.2}%
\end{align}
Also, for any $f\in\mathfrak{n}\left(  C,A\right)  $ and $u\in\kk%
\left[  \left[  T\right]  \right]  $ and $\lambda\in\kk$, we have%
\begin{equation}
\left(  \lambda u\right)  ^{\star}\left(  f\right)  =\lambda u^{\star}\left(
f\right)  . \label{eq.exe.convolution-series.b.3}%
\end{equation}
Also, for any $f\in\mathfrak{n}\left(  C,A\right)  $, we have%
\begin{align}
0^{\star}\left(  f\right)   &  =0\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{eq.exe.convolution-series.b.4}\\
1^{\star}\left(  f\right)   &  =u_{A}\epsilon_{C}.
\label{eq.exe.convolution-series.b.5}%
\end{align}


\item[(c)] If $f,g\in\mathfrak{n}\left(  C,A\right)  $ satisfy $f\star
g=g\star f$, then $f+g\in\mathfrak{n}\left(  C,A\right)  $.

\item[(d)] For any $\lambda\in\kk$ and $f\in\mathfrak{n}\left(
C,A\right)  $, we have $\lambda f\in\mathfrak{n}\left(  C,A\right)  $.

\item[(e)] If $f\in\mathfrak{n}\left(  C,A\right)  $ and $g\in
\Hom \left(  C,A\right)  $ satisfy $f\star g=g\star f$, then
$f\star g\in\mathfrak{n}\left(  C,A\right)  $.

\item[(f)] If $v\in\kk\left[  \left[  T\right]  \right]  $ is a power
series whose constant term is $0$, then $v^{\star}\left(  f\right)
\in\mathfrak{n}\left(  C,A\right)  $ for each $f\in\mathfrak{n}\left(
C,A\right)  $.

\item[(g)] If $u,v\in\kk\left[  \left[  T\right]  \right]  $ are two
power series such that the constant term of $v$ is $0$, and if $f\in
\mathfrak{n}\left(  C,A\right)  $ is arbitrary, then%
\begin{equation}
\left(  u\left[  v\right]  \right)  ^{\star}\left(  f\right)  =u^{\star
}\left(  v^{\star}\left(  f\right)  \right)  .
\label{eq.exe.convolution-series.g.1}
\end{equation}
Here, $u\left[  v\right]  $ denotes the
\emph{composition}\index{composition of power series}
of $u$ with $v$;
this is the power series obtained by substituting $v$ for $T$ in $u$. (This
power series is well-defined, since $v$ has constant term $0$.) Furthermore,
notice that the right hand side of (\ref{eq.exe.convolution-series.g.1}) is
well-defined, since Proposition~\ref{prop.convolution-series}(f) shows that
$v^{\star}\left(  f\right)  \in\mathfrak{n}\left(  C,A\right)  $.

\item[(h)] If $C$ is a graded $\kk$-coalgebra, and if $f\in
\Hom \left(  C,A\right)  $ satisfies $f\left(  C_{0}\right)
=0$, then $f\in\mathfrak{n}\left(  C,A\right)  $.

\item[(i)] If $B$ is any $\kk$-algebra, and if $s:A\rightarrow B$ is
any $\kk$-algebra homomorphism, then every $u\in\kk\left[
\left[  T\right]  \right]  $ and $f\in\mathfrak{n}\left(  C,A\right)  $
satisfy%
\[
s\circ f\in\mathfrak{n}\left(  C,B\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ u^{\star}\left(  s\circ f\right)  =s\circ\left(  u^{\star
}\left(  f\right)  \right)  .
\]


\item[(j)] If $C$ is a connected graded $\kk$-bialgebra, and if
$F:C\rightarrow A$ is a $\kk$-algebra homomorphism, then $F-u_{A}%
\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $.
\end{enumerate}
\end{proposition}

\begin{example}
Let $C$ be a graded $\kk$-coalgebra. Let $f\in\Hom \left(  C,A\right)  $
be such that $f\left(  C_{0}\right)  =0$. Then, we claim
that the map $u_{A}\epsilon_{C}+f:C\rightarrow A$ is $\star$-invertible. (This
observation has already been made in the proof of Proposition
\ref{Takeuchi-prop}, at least in the particular case when $C=A$.)

Let us see how this claim follows from
Proposition~\ref{prop.convolution-series}. First,
Proposition~\ref{prop.convolution-series}(h) shows that $f\in\mathfrak{n}%
\left(  C,A\right)  $. Now, define a power series $u\in\kk\left[
\left[  T\right]  \right]  $ by $u=1+T$. Then, the power series $u$ has
constant term $1$, and thus has a multiplicative inverse $v=u^{-1}%
\in\kk\left[  \left[  T\right]  \right]  $. Consider this $v$.
(Explicitly, $v=\sum_{n\geq0}\left(  -1\right)  ^{n}T^{n}$, but this does not
matter for us.) Now, (\ref{eq.exe.convolution-series.b.2}) yields $\left(
uv\right)  ^{\star}\left(  f\right)  =u^{\star}\left(  f\right)  \star
v^{\star}\left(  f\right)  $. Since $uv=1$ (because $v=u^{-1}$), we have
$\left(  uv\right)  ^{\star}\left(  f\right)  =1^{\star}\left(  f\right)
=u_{A}\epsilon_{C}$ (by (\ref{eq.exe.convolution-series.b.5})). Thus,
$u^{\star}\left(  f\right)  \star v^{\star}\left(  f\right)  =\left(
uv\right)  ^{\star}\left(  f\right)  =u_{A}\epsilon_{C}$. Hence, the map
$u^{\star}\left(  f\right)  $ has a right $\star$-inverse.

Also, from $u=1+T$, we obtain%
\begin{align*}
u^{\star}\left(  f\right)   &  =\left(  1+T\right)  ^{\star}\left(  f\right)
=\underbrace{1^{\star}\left(  f\right)  }_{=u_{A}\epsilon_{C}}%
+\underbrace{T^{\star}\left(  f\right)  }_{\substack{=f^{\star1}\\\text{(by
(\ref{eq.exe.convolution-series.a.monom}), applied to }k=1\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.convolution-series.b.1}%
)}\right) \\
&  =u_{A}\epsilon_{C}+\underbrace{f^{\star1}}_{=f}=u_{A}\epsilon_{C}+f.
\end{align*}
Thus, the map $u_{A}\epsilon_{C}+f$ has a right $\star$-inverse (since the map
$u^{\star}\left(  f\right)  $ has a right $\star$-inverse). A similar argument
shows that this map $u_{A}\epsilon_{C}+f$ has a left $\star$-inverse.
Consequently, the map $u_{A}\epsilon_{C}+f$ is $\star$-invertible.
\end{example}

\begin{exercise}
\label{exe.convolution-series}
Prove Proposition~\ref{prop.convolution-series}.
\end{exercise}

\begin{definition}
\phantomsection\label{def.subsect.leray.standing2}

\begin{enumerate}
\item[(a)] For the rest of Section~\ref{subsect.leray}, we assume that
$\kk$ is a commutative $\QQ$-algebra. Thus, the two formal power
series $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}\in\kk\left[  \left[
T\right]  \right]  $\index{$\exp$} and
$\log\left(  1+T\right)  =\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}T^{n}\in\kk\left[  \left[  T\right]  \right]  $%
\index{$\log$}
are well-defined.

\item[(b)] Define two power series $\overline{\exp}\in\kk\left[
\left[  T\right]  \right]  $\index{$\overline{\exp}$}
and $\overline{\log}\in\kk\left[  \left[
T\right]  \right]  $\index{$\overline{\log}$}
by $\overline{\exp} = \exp-1$ and
$\overline{\log} = \log\left(  1+T\right)  $.

\item[(c)] If $u$ and $v$ are two power series in $\kk\left[  \left[
T\right]  \right]  $ such that $v$ has constant term $0$, then
\dfn{$u\left[ v \right]$} denotes the
\emph{composition}\index{composition of power series}
of $u$ with $v$; this is the
power series obtained by substituting $v$ for $T$ in $u$.
\end{enumerate}
\end{definition}

The following proposition is just a formal analogue of the well-known fact
that the exponential function and the logarithm are mutually inverse (on their
domains of definition):\footnote{See Exercise~\ref{exe.exp-log} below for the
proof of this proposition, as well as of the lemma and proposition that follow
afterwards.}

\begin{proposition}
\label{prop.exp-log-inv}
Both power series $\overline{\exp}$ and $\overline{\log}$
have constant term $0$ and satisfy $\overline{\exp}\left[
\overline{\log}\right]  =T$ and $\overline{\log}\left[  \overline{\exp
}\right]  =T$.
\end{proposition}

For any map $f\in\mathfrak{n}\left(  C,A\right)  $, the power series $\exp$,
$\overline{\exp}$ and $\overline{\log}$ give rise to three further maps
$\exp^{\star}f$, $\overline{\exp}^{\star}f$ and $\overline{\log}^{\star}f$. We
can also define a map $\log^{\star}g$ whenever $g$ is a map in
$\Hom \left(  C,A\right)  $ satisfying $g-u_{A}\epsilon_{C}%
\in\mathfrak{n}\left(  C,A\right)  $ (but we cannot define $\log^{\star}f$ for
$f\in\mathfrak{n}\left(  C,A\right)  $, since $\log$ is not per se a power
series); in order to do this, we need a simple lemma:

\begin{lemma}
\label{lem.exp-log-log}Let $g\in\Hom \left(  C,A\right)  $ be
such that $g-u_{A}\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $. Then,
$\overline{\log}^{\star}\left(  g-u_{A}\epsilon_{C}\right)  $ is a
well-defined element of $\mathfrak{n}\left(  C,A\right)  $.
\end{lemma}

\begin{definition}
\label{def.exp-log.log}
If $g\in\Hom \left(  C,A\right)  $ is a map satisfying
$g-u_{A}\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $, then we define a map
$\log^{\star}g\in\mathfrak{n}\left(  C,A\right)  $\index{$\log^{\star}g$}
by $\log^{\star} g
=\overline{\log}^{\star}\left(  g-u_{A}\epsilon_{C}\right)  $. (This is
well-defined, according to Lemma~\ref{lem.exp-log-log}.)
\end{definition}

\begin{proposition}
\phantomsection\label{prop.exp-log}

\begin{enumerate}
\item[(a)] Each $f\in\mathfrak{n}\left(  C,A\right)  $ satisfies $\exp^{\star
}f-u_{A}\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $ and%
\[
\log^{\star}\left(  \exp^{\star}f\right)  =f.
\]


\item[(b)] Each $g\in\Hom \left(  C,A\right)  $ satisfying
$g-u_{A}\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $ satisfies%
\[
\exp^{\star}\left(  \log^{\star}g\right)  =g.
\]


\item[(c)] If $f,g\in\mathfrak{n}\left(  C,A\right)  $ satisfy $f\star
g=g\star f$, then $f+g\in\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star
}\left(  f+g\right)  =\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star
}g\right)  $.

\item[(d)] The $\kk$-linear map $0:C\rightarrow A$ satisfies
$0\in\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}0=u_{A}\epsilon_{C}$.

\item[(e)] If $f\in\mathfrak{n}\left(  C,A\right)  $ and $n\in\NN$,
then $nf\in\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}\left(
nf\right)  =\left(  \exp^{\star}f\right)  ^{\star n}$.

\item[(f)] If $f\in\mathfrak{n}\left(  C,A\right)  $, then%
\begin{equation}
\log^{\star}\left(  f+u_{A}\epsilon_{C}\right)  =\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}f^{\star n}. \label{eq.exe.convolution-series.logstar}%
\end{equation}

\end{enumerate}
\end{proposition}

\begin{example}
\label{exa.convolution-series.polynom-d0}Consider again the Hopf algebra
$\kk\left[  x\right]  $ from Exercise \ref{exe.Sym.1dim}. Let
$c_{1}:\kk\left[  x\right]  \rightarrow\kk$ be the $\kk%
$-linear map sending each polynomial $p\in\kk\left[  x\right]  $ to the
coefficient of $x^{1}$ in $p$. (In other words, $c_{1}$ sends each polynomial
$p\in\kk\left[  x\right]  $ to its derivative at $0$.)

Then, $c_{1}\left(  \left(  \kk\left[  x\right]  \right)  _{0}\right)
=0$ (as can easily be seen). Hence, Proposition~\ref{prop.convolution-series}%
(h) shows that $c_{1}\in\mathfrak{n}\left(  \kk\left[  x\right]
,\kk\right)  $. Thus, a map $\exp^{\star}\left(  c_{1}\right)
:\kk\left[  x\right]  \rightarrow\kk$ is well-defined. It is not
hard to see that this map is explicitly given by%
\[
\left(  \exp^{\star}\left(  c_{1}\right)  \right)  \left(  p\right)  =p\left(
1\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }p\in\kk\left[  x\right]
.
\]
(In fact, this follows easily after showing that each $n\in\NN$
satisfies
\[
\left(  c_{1}\right)  ^{\star n}\left(  p\right)  =n!\cdot\left(  \text{the
coefficient of }x^{n}\text{ in }p\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}p\in\kk\left[  x\right]  ,
\]
which in turn is easily seen by induction.)

Note that the equality $\left(  \exp^{\star}\left(  c_{1}\right)  \right)
\left(  p\right)  =p\left(  1\right)  $ shows that the map $\exp^{\star
}\left(  c_{1}\right)  $ is a $\kk$-algebra homomorphism. This is a
particular case of a fact that we will soon see (Proposition
\ref{prop.leray.exp-algh1}).
\end{example}

\begin{exercise}
\label{exe.exp-log}Prove Proposition~\ref{prop.exp-log-inv},
Lemma~\ref{lem.exp-log-log} and Proposition~\ref{prop.exp-log}.
\end{exercise}

Next, we state another sequence of facts (some of which have nothing to do
with Hopf algebras), beginning with a fact about convolutions which is similar
to Proposition~\ref{prop.convolution.functor}:\footnote{See
Exercise~\ref{exe.leray.exp-alg} below for their proofs.}

\begin{proposition}
\label{prop.convolution.nat} Let $C$ and $C^{\prime}$ be two $\kk%
$-coalgebras, and let $A$ and $A^{\prime}$ be two $\kk$-algebras. Let
$\gamma:C\rightarrow C^{\prime}$ be a $\kk$-coalgebra morphism. Let
$\alpha:A\rightarrow A^{\prime}$ be a $\kk$-algebra morphism.

\begin{enumerate}
\item[(a)] If $f\in\Hom \left(  C,A\right)  $, $g\in
\Hom \left(  C,A\right)  $, $f^{\prime}\in\Hom
\left(  C^{\prime},A^{\prime}\right)  $ and $g^{\prime}\in\Hom
\left(  C^{\prime},A^{\prime}\right)  $ satisfy $f^{\prime}\circ\gamma
=\alpha\circ f$ and $g^{\prime}\circ\gamma=\alpha\circ g$, then $\left(
f^{\prime}\star g^{\prime}\right)  \circ\gamma=\alpha\circ\left(  f\star
g\right)  $.

\item[(b)] If $f\in\Hom \left(  C,A\right)  $ and $f^{\prime}%
\in\Hom \left(  C^{\prime},A^{\prime}\right)  $ satisfy
$f^{\prime}\circ\gamma=\alpha\circ f$, then each $n\in \NN$ satisfies
$\left(  f^{\prime}\right)  ^{\star n}\circ\gamma=\alpha\circ f^{\star n}$.
\end{enumerate}
\end{proposition}

\begin{proposition}
\label{prop.leray.f*nxy}Let $C$ be a $\kk$-bialgebra. Let $A$ be a
commutative $\kk$-algebra. Let $f\in\Hom \left(
C,A\right)  $ be such that $f\left(  \left(  \ker\epsilon\right)  ^{2}\right)
=0$ and $f\left(  1\right)  =0$. Then, any $x,y\in C$ and $n\in \NN$
satisfy%
\[
f^{\star n}\left(  xy\right)  =\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  .
\]

\end{proposition}

\begin{proposition}
\label{prop.leray.exp-algh1}Let $C$ be a $\kk$-bialgebra. Let $A$ be a
commutative $\kk$-algebra. Let $f\in\mathfrak{n}\left(  C,A\right)  $
be such that $f\left(  \left(  \ker\epsilon\right)  ^{2}\right)  =0$ and
$f\left(  1\right)  =0$. Then, $\exp^{\star}f:C\rightarrow A$ is a
$\kk$-algebra homomorphism.
\end{proposition}

\begin{lemma}
\label{lem.sol.eulerian-idp.2.poly1} Let $V$ be any torsionfree abelian group
(written additively). Let $N\in{\NN}$. For every $k\in\left\{
0,1,\ldots,N\right\}  $, let $w_{k}$ be an element of $V$. Assume that
\begin{equation}
\sum_{k=0}^{N}w_{k}n^{k}=0\ \ \ \ \ \ \ \ \ \ \text{for all }n\in{\NN}.
\label{eq.lem.sol.eulerian-idp.2.poly1.ass}%
\end{equation}
Then, $w_{k}=0$ for every $k\in\left\{  0,1,\ldots,N\right\}  $.
\end{lemma}

\begin{lemma}
\label{lem.leray.poly1-inf}Let $V$ be a torsionfree abelian group (written
additively). Let $\left(  w_{k}\right)  _{k\in\NN}\in V^{\NN}$
be a finitely supported family of elements of $V$. Assume that%
\[
\sum_{k\in\NN}w_{k}n^{k}=0\ \ \ \ \ \ \ \ \ \ \text{for all }%
n\in \NN.
\]
Then, $w_{k}=0$ for every $k\in \NN$.
\end{lemma}

\begin{proposition}
\label{prop.leray.exp-algh2}Let $C$ be a graded $\kk$-bialgebra. Let
$A$ be a commutative $\kk$-algebra. Let $f\in\Hom \left(
C,A\right)  $ be such that $f\left(  C_{0}\right)  =0$. Assume
that\footnote{Notice that $\exp^{\star}f$ is well-defined, since
Proposition~\ref{prop.convolution-series}(h) yields $f\in\mathfrak{n}\left(
C,A\right)  $.} $\exp^{\star}f:C\rightarrow A$ is a $\kk$-algebra
homomorphism. Then, $f\left(  \left(  \ker\epsilon\right)  ^{2}\right)  =0$.
\end{proposition}

\begin{proposition}
\label{prop.leray.exp-algh3}Let $C$ be a connected graded $\kk%
$-bialgebra. Let $A$ be a commutative $\kk$-algebra. Let $f\in
\mathfrak{n}\left(  C,A\right)  $ be such that $f\left(  \left(  \ker
\epsilon\right)  ^{2}\right)  =0$ and $f\left(  1\right)  =0$. Assume further
that $f\left(  C\right)  $ generates the $\kk$-algebra $A$. Then,
$\exp^{\star}f:C\rightarrow A$ is a surjective $\kk$-algebra homomorphism.
\end{proposition}

\begin{exercise}
\label{exe.leray.exp-alg}Prove Lemmas~\ref{lem.sol.eulerian-idp.2.poly1}%
~and~\ref{lem.leray.poly1-inf} and Propositions~\ref{prop.convolution.nat}%
,~\ref{prop.leray.f*nxy},~\ref{prop.leray.exp-algh1}%
,~\ref{prop.leray.exp-algh2}~and~\ref{prop.leray.exp-algh3}.

[\textbf{Hint:} For Proposition~\ref{prop.leray.exp-algh2}, show first that
$\exp^{\star}\left(  nf\right)  =\left(  \exp^{\star}f\right)  ^{\star n}$ is
a $\kk$-algebra homomorphism for each $n\in \NN$. Turn this into
an equality between polynomials in $n$, and use
Lemma~\ref{lem.leray.poly1-inf}.]
\end{exercise}

With these preparations, we can state our version of \idx{Leray's theorem}:

\begin{theorem}
\label{thm.leray.leray-e}Let $A$ be a commutative connected graded
$\kk$-bialgebra.\footnote{Keep in mind that $\kk$ is assumed to
be a commutative $\QQ$-algebra.}

\begin{enumerate}
\item[(a)] We have $\id_{A}-u_{A}\epsilon_{A}%
\in\mathfrak{n}\left(  A,A\right)  $; thus, the map $\log^{\star}\left(
\id_{A}\right)  \in\mathfrak{n}\left(  A,A\right)  $
is well-defined. We denote this map
$\log^{\star}\left(  \id_{A}\right)  $ by $\mathfrak{e}$.

\item[(b)] We have $\ker\mathfrak{e}=\kk\cdot1_{A}+\left(  \ker
\epsilon\right)  ^{2}$ and $\mathfrak{e}\left(  A\right)  \cong\left(
\ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ (as $\kk$-modules).

\item[(c)] For each $\kk$-module $V$, let $\iota_{V}$ be the canonical
inclusion $V\rightarrow\Sym V$. Let $\mathfrak{q}$ be the map%
\[
A\overset{\mathfrak{e}}{\longrightarrow}\mathfrak{e}\left(  A\right)
\overset{\iota_{\mathfrak{e}\left(  A\right)  }}{\longrightarrow
}\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  .
\]
Then, $\mathfrak{q}\in\mathfrak{n}\left(  A,\Sym \left(
\mathfrak{e}\left(  A\right)  \right)  \right)  $\ \ \ \ \footnote{Do not
mistake the map $\mathfrak{q}$ for $\mathfrak{e}$. While every $a\in A$
satisfies $\mathfrak{q}\left(  a\right)  =\mathfrak{e}\left(  a\right)  $, the
two maps $\mathfrak{q}$ and $\mathfrak{e}$ have different target sets, and
thus we do \textbf{not} have $\left(  \exp^{\star}\mathfrak{q}\right)  \left(
a\right) = \left(  \exp^{\star}\mathfrak{e}\right)  \left(  a\right)  $ for
every $a\in A$.}.

\item[(d)] Let $\mathbf{i}$ be the canonical inclusion $\mathfrak{e}\left(
A\right)  \rightarrow A$. Recall the universal property of the symmetric
algebra: If $V$ is a $\kk$-module, if $W$ is a commutative $\kk%
$-algebra, and if $\varphi:V\rightarrow W$ is any $\kk$-linear map,
then there exists a unique $\kk$-algebra homomorphism $\Phi
:\Sym V\rightarrow W$ satisfying $\varphi=\Phi\circ\iota_{V}$.
Applying this to $V=\mathfrak{e}\left(  A\right)  $, $W=A$ and $\varphi
=\mathbf{i}$, we conclude that there exists a unique $\kk$-algebra
homomorphism $\Phi:\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ satisfying $\mathbf{i}=\Phi\circ\iota_{\mathfrak{e}%
\left(  A\right)  }$. Denote this $\Phi$ by $\mathfrak{s}$. Then, the maps
$\exp^{\star}\mathfrak{q}:A\rightarrow\Sym \left(
\mathfrak{e}\left(  A\right)  \right)  $ and $\mathfrak{s}:\Sym %
\left(  \mathfrak{e}\left(  A\right)  \right)  \rightarrow A$ are mutually
inverse $\kk$-algebra isomorphisms.

\item[(e)] We have $A\cong\Sym \left(  \left(  \ker
\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}\right)  $ as $\kk$-algebras.

\item[(f)] The map $\mathfrak{e}:A\rightarrow A$ is a projection (i.e., it
satisfies $\mathfrak{e}\circ\mathfrak{e}=\mathfrak{e}$).
\end{enumerate}
\end{theorem}

\begin{remark}
\phantomsection\label{rmk.leray.leray-e}

\begin{enumerate}
\item[(a)] The main upshot of Theorem~\ref{thm.leray.leray-e} is that any
commutative connected graded $\kk$-bialgebra $A$ (where $\kk$ is
a commutative $\QQ$-algebra) is isomorphic \textbf{as a $\kk$-algebra}
to the symmetric algebra $\Sym W$ of some
$\kk$-module $W$. (Specifically, Theorem~\ref{thm.leray.leray-e}(e)
claims this for $W=\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)
^{2}$, whereas Theorem~\ref{thm.leray.leray-e}(d) claims this for
$W=\mathfrak{e}\left(  A\right)  $; these two modules $W$ are isomorphic by
Theorem~\ref{thm.leray.leray-e}(b).) This is a useful statement even without
any specific knowledge about $W$, since symmetric algebras are a far tamer
class of algebras than arbitrary commutative algebras. For example, if
$\kk$ is a field, then symmetric algebras are just polynomial algebras
(up to isomorphism). This can be applied, for example, to the case of the
shuffle algebra $\operatorname*{Sh}\left(  V\right)  $ of a $\kk%
$-module $V$. The consequence is that the shuffle algebra $\operatorname*{Sh}%
\left(  V\right)  $ of any $\kk$-module $V$ (where $\kk$ is a
commutative $\QQ$-algebra) is isomorphic \textbf{as a $\kk$-algebra}
to a symmetric algebra $\Sym W$. When $V$ is
a free $\kk$-module, one can actually show that $\operatorname*{Sh}%
\left(  V\right)  $ is isomorphic \textbf{as a $\kk$-algebra}
to the symmetric algebra of a \textbf{free} $\kk$-module $W$ (that is,
to a polynomial ring over $\kk$); however, this $W$ is not easy to
characterize. Such a characterization is given by \emph{Radford's theorem}
(Theorem~\ref{thm.shuffle.radford} below) using the concept of \emph{Lyndon
words}. Notice that if $V$ has rank $\geq2$, then $W$ is not finitely generated.

\item[(b)] The isomorphism in Theorem~\ref{thm.leray.leray-e}(e) is generally
not an isomorphism of Hopf algebras. However, with a little (rather
straightforward) work, it reveals to be an isomorphism of \textbf{graded}
$\kk$-algebras. Actually, all maps mentioned in
Theorem~\ref{thm.leray.leray-e} are graded, provided that we use the
appropriate gradings for $\mathfrak{e}\left(  A\right)  $ and
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $. (To
define the appropriate grading for $\mathfrak{e}\left(  A\right)  $, we must
show that $\mathfrak{e}$ is a graded map, whence $\mathfrak{e}\left(
A\right)  $ is a homogeneous submodule of $A$; this provides $\mathfrak{e}%
\left(  A\right)  $ with the grading we seek. The grading on
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ then
follows from the usual definition of the grading on the symmetric algebra
$\Sym V$ of a graded $\kk$-module $V$: Namely, if $V$ is
a graded $\kk$-module, then the $n$-th graded component of
$\Sym V$ is defined to be the span of all products of the form
$v_{1}v_{2}\cdots v_{k}\in\Sym V$, where $v_{1},v_{2}%
,\ldots,v_{k}\in V$ are homogeneous elements satisfying $\deg\left(
v_{1}\right)  +\deg\left(  v_{2}\right)  +\cdots+\deg\left(  v_{k}\right)  =n$.)

\item[(c)] The map $\mathfrak{e}:A\rightarrow A$ from
Theorem~\ref{thm.leray.leray-e} is called the \dfn{Eulerian idempotent} of
$A$.

\item[(d)] Theorem~\ref{thm.leray.leray-e} is concerned with commutative
bialgebras. Most of its claims have a ``dual version'',
concerning cocommutative bialgebras. Again, the
Eulerian idempotent plays a crucial role; but the result characterizes not the
$\kk$-algebra structure on $A$, but the $\kk$-coalgebra
structure on $A$. This leads to the Cartier-Milnor-Moore theorem; see
\cite[\S 3.8]{Cartier} and \cite[\S 3.2]{DuchampMinhTolluChienNghia}. We shall
say a bit about the Eulerian idempotent for a cocommutative bialgebra in
Exercises~\ref{exe.eulerian-idp}~and~\ref{exe.eulerian-idp.2}.
\end{enumerate}
\end{remark}

\begin{example}
\label{exa.leray.leray-e.Sym}Consider the symmetric algebra
$\Sym V$ of a $\kk$-module $V$. Then,
$\Sym V$ is a commutative connected graded $\kk%
$-bialgebra, and thus Theorem~\ref{thm.leray.leray-e} can be applied to
$A=\Sym V$. What is the projection $\mathfrak{e}:A\rightarrow A$
obtained in this case?

Theorem~\ref{thm.leray.leray-e}(b) shows that its kernel is%
\begin{equation}
\operatorname*{Ker}\mathfrak{e}=\underbrace{\kk\cdot1_{A}%
}_{=\Sym^{0}V}+\underbrace{\left(  \ker\epsilon
\right)  ^{2}}_{=\sum_{n\geq2}\Sym^{n}V}%
=\Sym^{0}V+\sum_{n\geq2}\Sym^{n}V
=\sum_{n\neq1}\Sym^{n}V.
\label{eq.exa.leray.leray-e.Sym.Ker}
\end{equation}
This does not yet characterize $\mathfrak{e}$ completely, because we have yet
to determine the action of $\mathfrak{e}$ on $\Sym^{1}V$.
Fortunately, the elements of $\Sym^{1}V$ are all primitive
(recall that
$\Delta_{\Sym V}\left(  v\right)  =1\otimes v+v\otimes1$
for each $v\in V$), and it can
easily be shown that the map $\mathfrak{e}$ fixes any primitive element of
$A$\ \ \ \ \footnote{See Exercise~\ref{exe.eulerian-idp}(f) further below for
this proof. (While Exercise~\ref{exe.eulerian-idp} requires $A$ to be
cocommutative, this requirement is not used in the solution to
Exercise~\ref{exe.eulerian-idp}(f). That said, this requirement is actually
satisfied for $A=\Sym V$, so we do not even need to avoid it
here.)}. Therefore, the map $\mathfrak{e}$ fixes all elements of
$\Sym^{1}V$. Since we also know that $\mathfrak{e}$
annihilates all elements of $\sum_{n\neq1}\Sym^{n}V$
(by (\ref{eq.exa.leray.leray-e.Sym.Ker})), we thus conclude that
$\mathfrak{e}$ is the canonical projection from the direct sum
$\Sym V=\bigoplus_{n\in\NN}\Sym^{n}V$ onto its addend
$\Sym^{1}V$.
\end{example}

\begin{example}
\label{exa.leray.leray-e.Sh}For this example, let $A$ be the shuffle algebra
$\operatorname*{Sh}\left(  V\right)  $ of a $\kk$-module $V$. (See
Proposition \ref{prop.shuffle-alg} for its definition, and keep in mind that
its product is being denoted by $\shufmult$, whereas the notation
$uv$ is still being used for the product of two elements $u$ and $v$ in the
\textbf{tensor} algebra $T\left(  V\right)  $.)

Theorem~\ref{thm.leray.leray-e} can be applied to $A=\operatorname*{Sh}\left(
V\right)  $. What is the projection $\mathfrak{e}:A\rightarrow A$ obtained in
this case?

Let us compute $\mathfrak{e}\left(  v_{1}v_{2}\right)  $ for two elements
$v_{1},v_{2}\in V$. Indeed, define a map $\widetilde{\id}:A\rightarrow A$
by $\widetilde{\id}=\id_{A}-u_{A}\epsilon_{A}$. Then, $\widetilde{\id}%
\in\mathfrak{n}\left(  A,A\right)  $ and $\log^{\star}\left(
\underbrace{\widetilde{\id}+ u_A \epsilon_A}%
_{=\id_{A}}\right)  =\log^{\star}\left(
\id_{A}\right)  =\mathfrak{e}$. Hence,
(\ref{eq.exe.convolution-series.logstar}) (applied to $C=A$ and
$f=\widetilde{\id}$) shows that
\begin{equation}
\mathfrak{e}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}. \label{eq.exa.leray.leray-e.Sh.e=}%
\end{equation}
Thus, we need to compute $\widetilde{\id}^{\star n}\left(
v_{1}v_{2}\right)  $ for each $n\geq1$.

Notice that the map $\widetilde{\id}$ annihilates $A_{0}$, but
fixes any element of $A_{k}$ for $k>0$. Thus,
\[
\widetilde{\id}\left(  w_{1}w_{2}\cdots w_{k}\right)  =%
\begin{cases}
w_{1}w_{2}\cdots w_{k}, & \text{if }k>0;\\
0, & \text{if }k=0
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for any }w_{1},w_{2},\ldots,w_{k}\in V.
\]


But it is easy to see that the map $\widetilde{\id}^{\star
n}:A\rightarrow A$ annihilates $A_{k}$ whenever $n>k$. In particular, for
every $n>2$, the map $\widetilde{\id}^{\star n}:A\rightarrow A$
annihilates $A_{2}$, and therefore satisfies%
\begin{equation}
\widetilde{\id}^{\star n}\left(  v_{1}v_{2}\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v_{1}v_{2}\in A_{2}\right)  .
\label{eq.exa.leray.leray-e.Sh.2}%
\end{equation}
It remains to find $\widetilde{\id}^{\star n}\left(  v_{1}%
v_{2}\right)  $ for $n\in\left\{  1,2\right\}  $.

We have $\widetilde{\id}^{\star1}=\widetilde{\id}$ and thus
\[
\widetilde{\id}^{\star1}\left(  v_{1}v_{2}\right)
=\widetilde{\id}\left(  v_{1}v_{2}\right)  =v_{1}v_{2}%
\]
and%
\begin{align*}
\widetilde{\id}^{\star2}\left(  v_{1}v_{2}\right)   &
=\underbrace{\widetilde{\id}\left(  1\right)  }_{=0}
\shufmult \underbrace{\widetilde{\id}\left(
v_{1}v_{2}\right)  }_{=v_{1}v_{2}}+\underbrace{\widetilde{\id}
\left(  v_{1}\right)  }_{=v_{1}} \shufmult
\underbrace{\widetilde{\id}\left(  v_{2}\right)  }_{=v_{2}%
}+\underbrace{\widetilde{\id}\left(  v_{1}v_{2}\right)
}_{=v_{1}v_{2}} \shufmult \underbrace{\widetilde{\id}
\left(  1\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{\operatorname*{Sh}%
V}\left(  v_{1}v_{2}\right)  =1\otimes v_{1}v_{2}+v_{1}\otimes v_{2}%
+v_{1}v_{2}\otimes1\right) \\
&  =\underbrace{0 \shufmult \left(  v_{1}v_{2}\right)  }%
_{=0}+\underbrace{v_{1} \shufmult v_{2}}_{=v_{1}v_{2}+v_{2}v_{1}%
}+\underbrace{\left(  v_{1}v_{2}\right)  \shufmult 0}_{=0}\\
&  =v_{1}v_{2}+v_{2}v_{1}.
\end{align*}


Now, applying both sides of (\ref{eq.exa.leray.leray-e.Sh.e=}) to $v_{1}v_{2}%
$, we find
\begin{align*}
&  \mathfrak{e}\left(  v_{1}v_{2}\right) \\
&  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\left(  v_{1}v_{2}\right)
=\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=1}%
\underbrace{\widetilde{\id}^{\star1}\left(  v_{1}v_{2}\right)
}_{=v_{1}v_{2}}+\underbrace{\dfrac{\left(  -1\right)  ^{2-1}}{2}}_{=\dfrac
{-1}{2}}\underbrace{\widetilde{\id}^{\star2}\left(  v_{1}%
v_{2}\right)  }_{=v_{1}v_{2}+v_{2}v_{1}}+\sum_{n\geq3}\dfrac{\left(
-1\right)  ^{n-1}}{n}\underbrace{\widetilde{\id}^{\star
n}\left(  v_{1}v_{2}\right)  }_{\substack{=0\\\text{(by
(\ref{eq.exa.leray.leray-e.Sh.2}))}}}\\
&  =v_{1}v_{2}+\dfrac{-1}{2}\left(  v_{1}v_{2}+v_{2}v_{1}\right)
+\underbrace{\sum_{n\geq3}\dfrac{\left(  -1\right)  ^{n-1}}{n}0}_{=0}%
=\dfrac{1}{2}\left(  v_{1}v_{2}-v_{2}v_{1}\right)  .
\end{align*}
This describes the action of $\mathfrak{e}$ on the graded component $A_{2}$ of
$A=\operatorname*{Sh}\left(  V\right)  $.

Similarly, we can describe $\mathfrak{e}$ acting on any other graded
component:%
\begin{align*}
\mathfrak{e}\left(  1\right)   &  =0;\\
\mathfrak{e}\left(  v_{1}\right)   &  =v_{1}\ \ \ \ \ \ \ \ \ \ \text{for each
}v_{1}\in V;\\
\mathfrak{e}\left(  v_{1}v_{2}\right)   &  =\dfrac{1}{2}\left(  v_{1}%
v_{2}-v_{2}v_{1}\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }v_{1},v_{2}\in
V;\\
\mathfrak{e}\left(  v_{1}v_{2}v_{3}\right)   &  =\dfrac{1}{6}\left(
2v_{1}v_{2}v_{3}-v_{1}v_{3}v_{2}-v_{2}v_{1}v_{3}-v_{2}v_{3}v_{1}-v_{3}%
v_{1}v_{2}+2v_{3}v_{2}v_{1}\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }%
v_{1},v_{2},v_{3}\in V,\\
&  \ldots.
\end{align*}
With some more work, one can show the following formula for the action of
$\mathfrak{e}$ on any nontrivial pure tensor:%
\begin{align*}
\mathfrak{e}\left(  v_{1}v_{2}\cdots v_{n}\right)   &  =\sum_{\sigma
\in\Symm_{n}}\left(  \sum_{k=1+\operatorname*{des}\left(  \sigma
^{-1}\right)  }^{n}\dfrac{\left(  -1\right)  ^{k-1}}{k}\dbinom
{n-1-\operatorname*{des}\left(  \sigma^{-1}\right)  }{k-1-\operatorname*{des}%
\left(  \sigma^{-1}\right)  }\right)  v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }\\
&= \sum_{\sigma \in \Symm_n}
   \dfrac{\left(-1\right)^{\operatorname{des}\left(\sigma^{-1}\right)}}{\operatorname{des}\left(\sigma^{-1}\right)+1}
   \dbinom{n}{\operatorname{des}\left(\sigma^{-1}\right)+1}^{-1}
v_{\sigma\left(  1\right)  }%
v_{\sigma\left(  2\right)  }\cdots v_{\sigma\left(  n\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \text{for any }n\geq1\text{ and }v_{1},v_{2}%
,\ldots,v_{n}\in V,
\end{align*}
where we use the notation $\operatorname*{des}\pi$ for the number of
descents\footnote{A \emph{descent}\index{descent of a permutation}
of a permutation $\pi\in\Symm_n$
means an $i\in\left\{  1,2,\ldots,n-1\right\}  $ satisfying $\pi\left(
i\right)  >\pi\left(  i+1\right)  $.} of any permutation $\pi\in
\Symm_{n}$.
(A statement essentially dual to this appears in \cite[Theorem 9.5]{Schmitt}.)

Theorem~\ref{thm.leray.leray-e}(b) yields $\ker\mathfrak{e}=\kk
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Notice, however, that $\left(
\ker\epsilon\right)  ^{2}$ means the square of the ideal $\ker\epsilon$ with
respect to the shuffle multiplication $\shufmult$; thus, $\left(
\ker\epsilon\right)  ^{2}$ is the $\kk$-linear span of all shuffle
products of the form $a \shufmult b$ with $a\in\ker\epsilon$ and
$b\in\ker\epsilon$.
\end{example}

\begin{exercise}
\label{exe.leray.leray-e}Prove Theorem~\ref{thm.leray.leray-e}.

[\textbf{Hint:} (a) is easy. For (b), define an element
$\widetilde{\id}$ of $\mathfrak{n}\left(  A,A\right)  $ by
$\widetilde{\id}=\id_{A}-u_{A}%
\epsilon_{A}$. Observe that $\mathfrak{e}=\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\widetilde{\id}^{\star n}$, and draw the
conclusions that $\mathfrak{e}\left(  1_{A}\right)  =0$ and that each $x\in A$
satisfies $\widetilde{\id}\left(  x\right)  -\mathfrak{e}%
\left(  x\right)  \in\left(  \ker\epsilon\right)  ^{2}$ (because
$\widetilde{\id}^{\star n}\left(  x\right)  \in\left(
\ker\epsilon\right)  ^{2}$ for every $n\geq2$). Use this to prove
$\ker\mathfrak{e}\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}$. On the other hand, prove $\mathfrak{e}\left(  \left(  \ker
\epsilon\right)  ^{2}\right)  =0$ by applying
Proposition~\ref{prop.leray.exp-algh2}. Combine to obtain $\ker\mathfrak{e}%
=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Finish (b) by
showing that $A/\left(  \kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}\right)  \cong\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)
^{2}$ as $\kk$-modules. Part (c) is easy again. For (d), first apply
Proposition~\ref{prop.convolution-series}(i) to show that $\exp^{\star}\left(
\mathfrak{s}\circ\mathfrak{q}\right)  =\mathfrak{s}\circ\left(  \exp^{\star
}\mathfrak{q}\right)  $. In light of $\mathfrak{s}\circ\mathfrak{q}%
=\mathfrak{e}$ and $\exp^{\star}\mathfrak{e}=\id_{A}$,
this becomes $\id_{A}=\mathfrak{s}\circ\left(
\exp^{\star}\mathfrak{q}\right)  $. To obtain part (d), it remains to show
that $\exp^{\star}\mathfrak{q}$ is a surjective $\kk$-algebra
homomorphism; but this follows from Proposition~\ref{prop.leray.exp-algh3}.
For (e), combine (d) and (b). For (f), use once again the observation that
each $x\in A$ satisfies $\widetilde{\id}\left(  x\right)
-\mathfrak{e}\left(  x\right)  \in\left(  \ker\epsilon\right)  ^{2}$.]
\end{exercise}

% [DG][v67] Added the above section.

\newpage


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Review of symmetric functions $\Lambda$ as Hopf algebra}
\label{Sym-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Here we review the ring of symmetric functions, borrowing heavily from
standard treatments, such as Macdonald \cite[Chap. I]{Macdonald},
Sagan \cite[Chap. 4]{Sagan}, Stanley \cite[Chap. 7]{Stanley},
and Mendes and Remmel \cite{MendesRemmel},
but emphasizing the Hopf structure early on.
Other recent references for this subject are \cite{Wildon2016},
\cite{Sam-symf}, \cite{Egge}, \cite[Chapters 2--3]{Meliot}
and \cite[Chapter 7]{Sagan2020}.

\subsection{\label{subsect.Sym-section.def}Definition of $\Lambda$}

As before, $\kk$ here is a commutative ring (hence could be a field or
the integers $\ZZ$; these are the usual choices).

Given an infinite variable set $\xx=(x_1,x_2,\ldots)$\index{$\xx$},
a monomial
$\xx^\alpha:=x_1^{\alpha_1} x_2^{\alpha_2} \cdots$\index{$\xx^\alpha$}
is indexed by a sequence
$\alpha=(\alpha_1,\alpha_2,\ldots)$ in $\NN^\infty$ having finite
support\footnote{The \dfn{support} of a sequence
$\alpha = \left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)
\in \NN^\infty$ is defined to be the set of all positive integers
$i$ for which $\alpha_i \neq 0$.};
such sequences $\alpha$ are called \dfn{weak compositions}.
The nonzero entries of the sequence
$\alpha = \left(\alpha_1,\alpha_2,\ldots\right)$ are called the
\emph{parts}\index{part of a weak composition}
of the weak composition $\alpha$.

The sum $\alpha_1+\alpha_2+\alpha_3+\cdots$ of all entries of
a weak composition
$\alpha = \left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)$
(or, equivalently, the sum of all parts of $\alpha$)
is called the \emph{size}\index{size of a weak composition}
of $\alpha$ and denoted by $\left|\alpha\right|$.\index{$\mid\alpha\mid$}

Consider the $\kk$-algebra $\kk\left[\left[\xx\right]\right]
:= \kk\left[\left[x_1, x_2, x_3, \ldots\right]\right]$\index{$\kk\left[\left[\xx\right]\right]$}
of all formal power series in the indeterminates $x_1, x_2, x_3, \ldots$
over $\kk$; these series are infinite $\kk$-linear combinations
$\sum_{\alpha} c_\alpha \xx^\alpha$ (with $c_\alpha$ in $\kk$)
of the monomials $\xx^\alpha$ where $\alpha$ ranges over all
weak compositions.
The product of two such formal power series is well-defined
by the usual multiplication rule.

The \emph{degree}\index{degree of a monomial}
of a monomial $\xx^{\alpha}$ is defined to be the number
$\deg(\xx^{\alpha}) := \sum_i \alpha_i \in \NN$\index{$\deg(\xx^{\alpha})$}.
Given a number $d \in \NN$,
we say that a formal power series
$f(\xx) = \sum_{\alpha} c_\alpha \xx^\alpha \in \kk\left[\left[\xx\right]\right]$
(with $c_\alpha$ in $\kk$)
is \dfn{homogeneous of degree $d$}\index{homogeneous power series}
if every weak composition
$\alpha$ satisfying $\deg(\xx^{\alpha}) \neq d$
must satisfy $c_\alpha = 0$.
In other words, a formal power series is homogeneous of degree $d$
if it is an infinite $\kk$-linear combination of monomials
of degree $d$.
Every formal power series $f \in \kk\left[\left[\xx\right]\right]$
can be uniquely represented as an infinite sum
$f_0 + f_1 + f_2 + \cdots$, where each $f_d$ is homogeneous
of degree $d$; in this case, we refer to each $f_d$ as the
\emph{$d$-th homogeneous component of $f$}\index{homogeneous component}.
Note that this does not make $\kk\left[\left[\xx\right]\right]$ into
a graded $\kk$-module, since these sums $f_0 + f_1 + f_2 + \cdots$
can have infinitely many nonzero addends.
Nevertheless, if $f$ and $g$ are homogeneous power series of
degrees $d$ and $e$, then $fg$ is homogeneous of degree $d+e$.

% [DG][v79] Added the above paragraph.

A formal power series
$f(\xx) = \sum_{\alpha} c_\alpha \xx^\alpha \in \kk\left[\left[\xx\right]\right]$
(with $c_\alpha$ in $\kk$)
is said to be \dfn{of bounded degree}\index{bounded degree}
if there exists some
bound $d=d(f) \in \NN$ such that every weak composition
$\alpha = \left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)$
satisfying $\deg(\xx^{\alpha}) > d$
must satisfy $c_\alpha = 0$.
Equivalently, a formal power series $f \in \kk\left[\left[\xx\right]\right]$
is of bounded degree if all but finitely many of its
homogeneous components are zero.
(For example, $x_1^2 + x_2^2 + x_3^2 + \cdots$ and
$1 + x_1 + x_2 + x_3 + \cdots$ are of bounded
degree, while $x_1 + x_1 x_2 + x_1 x_2 x_3 + \cdots$ and
$1 + x_1 + x_1^2 + x_1^3 + \cdots$ are not.)
It is easy to see that the sum and the product of two power
series of bounded degree also have bounded degree.
Thus, the formal power series of bounded degree form
a $\kk$-subalgebra of $\kk\left[\left[\xx\right]\right]$,
which we call \dfn{$R(\xx)$}.
This subalgebra $R(\xx)$ is graded (by degree).

% [DG] I added the definition of "parts", requiring them to be nonzero
% since otherwise "partition with at most $n$ parts" doesn't make much
% sense. On the other hand, you speak of "nonzero parts" sometimes...

% [DG][v79] Parts are now nonzero everywhere. If zeroes are desired,
% I speak of entries instead.

The symmetric group $\Symm_n$ permuting the first $n$ variables
$x_1,\ldots,x_n$ acts as a group of automorphisms on $R(\xx)$, as does
the union $\Symm_{(\infty)}=\bigcup_{n \geq 0} \Symm_n$%
\index{$\Symm_{(\infty)}$}
of the infinite ascending chain
$
\Symm_0 \subset \Symm_1 \subset \Symm_2 \subset \cdots
$
of symmetric groups\footnote{This ascending chain is constructed as
follows: For every $n \in \NN$, there is an injective group
homomorphism $\iota_n : \Symm_n \to \Symm_{n+1}$ which sends every
permutation $\sigma \in \Symm_n$ to the permutation
$\iota_n \left(\sigma\right) = \tau \in \Symm_{n+1}$
defined by
\[
\tau\left(i\right) =
\begin{cases}
\sigma\left(i\right), &\text{if } i \leq n ; \\
i, &\text{if } i = n+1
\end{cases}
\qquad \text{ for all } i \in \left\{1,2,\ldots,n+1\right\} .
\]
These homomorphisms
$\iota_n$ for all $n$ form a chain
$\Symm_0 \overset{\iota_0}{\longrightarrow}
\Symm_1 \overset{\iota_1}{\longrightarrow}
\Symm_2 \overset{\iota_2}{\longrightarrow} \cdots$,
which is often regarded as a chain of inclusions.}.
This group $\Symm_{(\infty)}$ can also be
described as the group of all permutations of the set
$\left\{1,2,3,\ldots\right\}$ which leave all but finitely many
elements invariant.
It is known as the \dfn{finitary symmetric group}
on $\left\{1,2,3,\ldots\right\}$.

The group $\Symm_{(\infty)}$ also acts on the set of
all weak compositions by permuting their entries:
\begin{align*}
\sigma \left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)
&= \left(\alpha_{\sigma^{-1}\left(1\right)},
\alpha_{\sigma^{-1}\left(2\right)},
\alpha_{\sigma^{-1}\left(3\right)}, \ldots\right)
\\
&\qquad
\text{for any weak composition $\left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)$
and any $\sigma \in \Symm_{(\infty)}$}.
\end{align*}

These two actions are connected by the equality
$\sigma\left(\xx^\alpha\right) = \xx^{\sigma\alpha}$
for any
weak composition $\alpha$ and any $\sigma \in \Symm_{(\infty)}$.

% [DG][v14] Again, added $\Symm_0$. (You already used
% $\bigcup_{n \geq 0} \Symm_n$ in the Malvenuto-Reutenauer chapter,
% so this makes things more consistent.)

% [DG][v25] Added last sentence. (I define $\Symm_{(\infty,\infty)}$
% later, in the same way; the "union of a chain of finite symmetric
% groups" definition would be more complicated in this case.)

% [DG][v54] Added footnotes defining "support" and the chain of
% inclusions of the symmetric group.

% [DG][v75] Added lots of details to the above (e.g., examples
% of bounded-degree and not-bounded-degree power series); split
% some sentences.


\begin{definition}
The \emph{ring of symmetric functions in $\xx$ with coefficients in $\kk$}%
\index{symmetric function}\index{ring of symmetric functions}\index{$\Lambda$},
denoted
$
\Lambda = \Lambda_\kk = \Lambda(\xx) = \Lambda_\kk(\xx),
$
is the $\Symm_{(\infty)}$-invariant subalgebra $R(\xx)^{\Symm_{(\infty)}}$
of $R(\xx)$:
\begin{align*}
\Lambda & :=
\left\{ f \in R(\xx) : \sigma \left(f\right) = f
\text{ for all } \sigma \in \Symm_{(\infty)} \right\}
\\
&=
\left\{ f=\sum_\alpha c_\alpha \xx^\alpha \in R(\xx):
c_\alpha=c_\beta\text{ if }\alpha, \beta\text{ lie in the same }
\Symm_{(\infty)}\text{-orbit} \right\}.
\end{align*}
\end{definition}

We refer to the elements of $\Lambda$ as
\emph{symmetric functions}\index{symmetric function}
(over $\kk$); however, despite this terminology, they are not
functions in the usual sense.\footnote{Being power
series, they can be evaluated at appropriate families of
variables. But this does not make them functions (no more than
polynomials are functions).
The terminology ``symmetric function'' is thus not
well-chosen; but it is standard.}

% [DG][v49] Added above paragraph.

Note that $\Lambda$ is a graded $\kk$-algebra, since
$\Lambda = \bigoplus_{n \geq 0} \Lambda_n$ where $\Lambda_n$ are
the symmetric functions $f=\sum_\alpha c_\alpha \xx^\alpha$
which are \emph{homogeneous of degree $n$},
meaning $\deg(\xx^\alpha)=n$ for all $c_\alpha \neq 0$.

\begin{exercise}
\label{exe.Lambda.subs.fin}
Let $f \in R\left(\xx\right)$.
Let $A$ be a commutative $\kk$-algebra, and
$a_1, a_2, \ldots, a_k$ be finitely many elements of $A$. Show
that substituting $a_1, a_2, \ldots, a_k, 0, 0, 0, \ldots$ for
$x_1, x_2, x_3, \ldots$ in $f$ yields an infinite sum
in which all but finitely many addends are zero. Hence, this
sum has a value in $A$, which is commonly denoted by
$f\left(a_1, a_2, \ldots, a_k\right)$.
\end{exercise}

% [DG][v25] Added above exercise.

\begin{definition}
A \dfn{partition} $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_\ell,0,0,\ldots)$
is a weak composition whose entries weakly decrease:
$\lambda_1 \geq \cdots \geq \lambda_\ell > 0$.
The (uniquely defined) $\ell$ is said to be the
\emph{length}\index{length of a partition} of the partition $\lambda$
and denoted by \dfn{$\ell\left(\lambda\right)$}.
Thus, $\ell\left(\lambda\right)$ is the number of
parts\footnote{Recall that a \emph{part} of a partition means a
nonzero entry of the partition.} of $\lambda$.
One sometimes omits trailing zeroes from a partition:
e.g., one can write the partition
$\left(3,1,0,0,0,\ldots\right)$ as $\left(3,1\right)$.
We will often (but not always) write \dfn{$\lambda_i$}
for the $i$-th entry of the partition $\lambda$ (for instance,
if $\lambda = \left(5,3,1,1\right)$, then
$\lambda_2 = 3$ and $\lambda_5 = 0$).
If $\lambda_i$ is nonzero, we will also call it the
\dfn{$i$-th part} of $\lambda$.
The sum
$\lambda_1+\lambda_2+\cdots+\lambda_\ell = \lambda_1+\lambda_2+\cdots$ (where
$\ell = \ell\left(\lambda\right)$) of
all entries of $\lambda$ (or, equivalently, of all
parts of $\lambda$)
is the size $\left|\lambda\right|$ of $\lambda$\index{size of a partition}%
\index{$\mid\lambda\mid$}. % because | signs in index entries seem to be broken.
For a given integer $n$, the partitions of size $n$
are referred to as the \emph{partitions of $n$}\index{partition of $n$}.
The \idx{empty partition} $() = (0,0,0,\ldots)$
is denoted by \dfn{$\varnothing$}.

% [DG] I added the definitions of "length" and "partition of $n$" since you
% use them later.

% [DG][v25] Added more pedantry to the above definition :P

% [DG][v54] Replaced "whose parts weakly
% decrease" by "whose entries weakly decrease" in the definition of
% a partition, since $(3,0,2,0,0,\ldots)$ is not a partition.

Partitions (as defined above) are sometimes called
\emph{integer partitions}\index{integer partition}
in order to distinguish them from set partitions.

Every weak composition $\alpha$ lies in the  $\Symm_{(\infty)}$-orbit
of a unique partition
$\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_\ell,0,0,\ldots)$
with $\lambda_1 \geq \cdots \geq \lambda_\ell > 0$.
For any partition $\lambda$, define the
\dfn{monomial symmetric function}\index{$m_\lambda$}
\begin{equation}
\label{monomial-symmetric-function-definition}
m_\lambda:=\sum_{\alpha \in \Symm_{(\infty)}\lambda}  \xx^\alpha.
\end{equation}
Letting $\lambda$ run through the set \dfn{$\Par$} of all partitions,
this gives the \emph{monomial $\kk$-basis}\index{monomial basis of $\Lambda$}
$\{m_{\lambda}\}$ of $\Lambda$.
Letting $\lambda$ run only through the set \dfn{$\Par_n$} of partitions of $n$
gives the monomial $\kk$-basis for $\Lambda_n$.
\end{definition}

\begin{example}
For $n=3$, one has
\begin{align*}
m_{(3)} &= x_1^3+x_2^3+x_3^3+ \cdots ,\\
m_{(2,1)} &= x_1^2 x_2 + x_1 x_2^2 + x_1^2 x_3 +x_1 x_3^2 + \cdots ,\\
m_{(1,1,1)} &= x_1 x_2 x_3 +x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 + x_1 x_2 x_5 + \cdots .
\end{align*}
\end{example}

The monomial basis $\{m_{\lambda}\}_{\lambda \in \Par}$ of $\Lambda$
is thus a graded basis\footnote{See Definition~\ref{def.graded-basis}
for the meaning of ``graded basis''.}
of the graded $\kk$-module $\Lambda$.
(Here and in the following, when we say that a basis
$\left\{u_\lambda\right\}_{\lambda \in \Par}$ indexed by $\Par$
is a \idx{graded basis of $\Lambda$},
we tacitly understand that $\Par$ is partitioned into
$\Par_0, \Par_1, \Par_2, \ldots$, so that for each $n \in \NN$,
the subfamily
$\left\{u_\lambda\right\}_{\lambda \in \Par_n}$ should be a
basis for $\Lambda_n$.)

% [DG][v75] Added the above (stupid but necessary) paragraph.

\begin{remark}
\label{rmk.Lambda.symmetric}
We have defined the symmetric functions as the elements of
$R\left(\xx\right)$ invariant under the group $\Symm_{(\infty)}$.
However, they also are the elements of $R\left(\xx\right)$
invariant under the group $\Symm_\infty$ of \emph{all}
permutations of the set $\left\{1, 2, 3, \ldots\right\}$
(which acts on $R\left(\xx\right)$ in the same way as its
subgroup $\Symm_{(\infty)}$ does).%
\footnote{\textit{Proof.} We need to show that
$\Lambda = R\left(\xx\right)^{\Symm_\infty}$. Since
\[
\Lambda
= \left\{ f=\sum_\alpha c_\alpha \xx^\alpha \in R\left(\xx\right):
c_\alpha=c_\beta\text{ if }\alpha, \beta\text{ lie in the same }
\Symm_{(\infty)}\text{-orbit} \right\}
\]
and
\[
R\left(\xx\right)^{\Symm_\infty}
= \left\{ f=\sum_\alpha c_\alpha \xx^\alpha \in R\left(\xx\right):
c_\alpha=c_\beta\text{ if }\alpha, \beta\text{ lie in the same }
\Symm_{\infty}\text{-orbit} \right\} ,
\]
this will follow immediately if we can show that two weak
compositions $\alpha$ and $\beta$ lie in the same
$\Symm_{(\infty)}$-orbit if and only if they lie in the same
$\Symm_{\infty}$-orbit. But this is straightforward to check
(in fact, two weak compositions $\alpha$ and $\beta$ lie in the
same orbit under either group if and only if they have the same
multiset of nonzero entries).}
\end{remark}

% [DG][v39] Added above remark.

% [DG][v40] Replaced proof, since the previous argument wasn't very
% easy to follow.

\begin{remark}
\label{finite-variable-set-remark}
It is sometimes convenient to work with finite variable sets $x_1,\ldots,x_n$,
which one justifies as follows.  Note that the algebra homomorphism
\[
R(\xx) \rightarrow R(x_1,\ldots,x_n)=\kk[x_1,\ldots,x_n]
\]
which sends $x_{n+1},x_{n+2},\ldots$ to $0$ restricts to
an algebra homomorphism
\[
\Lambda_\kk(\xx) \rightarrow \Lambda_\kk(x_1,\ldots,x_n)=\kk[x_1,\ldots,x_n]^{\Symm_n}.
\]
Furthermore, this last homomorphism is a $\kk$-module isomorphism when restricted
to $\Lambda_i$ for $0 \leq i \leq n$, since it sends the monomial basis
elements $m_\lambda(\xx)$ to the monomial basis elements
$m_\lambda(x_1,\ldots,x_n)$.
Thus, when one proves identities in $\Lambda_\kk(x_1,\ldots,x_n)$ for all $n$,
they are valid in $\Lambda$, that is, $\Lambda$ is the inverse limit of
the $\Lambda(x_1,\ldots,x_n)$ in the category of graded
$\kk$-algebras.\footnote{\textit{Warning:} The word ``graded'' here
is crucial. Indeed, $\Lambda$ is \textbf{not}
the inverse limit of the $\Lambda(x_1,\ldots,x_n)$ in the category of
$\kk$-algebras. In fact, the latter limit is the $\kk$-algebra of
all symmetric power series $f$ in $\kk\left[\xx\right]$ with the
following property: For each $g \in \NN$, there exists a
$d \in \NN$ such that every monomial in $f$ that involves exactly $g$
distinct indeterminates has degree at most $d$. For example, the power
series
$\left(1+x_1\right)\left(1+x_2\right)\left(1+x_3\right)\cdots$
and
$m_{\left(1\right)} + m_{\left(2,2\right)} + m_{\left(3,3,3\right)}
+ \cdots$ satisfy this property, although they do not lie in $\Lambda$
(unless $\kk$ is a trivial ring).}

This characterization of $\Lambda$ as an inverse limit of
the graded $\kk$-algebras $\Lambda(x_1,\ldots,x_n)$ can be used as
an alternative definition of $\Lambda$. The definitions used by
Macdonald \cite{Macdonald} and Wildon \cite{Wildon2016} are closely
related (see \cite[\S 1.2, p. 19, Remark 1]{Macdonald},
\cite[\S A.11]{Hazewinkel2} and \cite[\S 1.7]{Wildon2016} for
discussions of this definition). It also suggests that much of the
theory of symmetric functions can be rewritten in terms of the
$\Lambda(x_1,\ldots,x_n)$ (at the cost of extra complexity); and this
indeed is possible\footnote{See, for example,
\cite[Chapter SYM]{LaksovLascouxPragaczThorup},
\cite{Prasad-schur} and
\cite[Chapters 10--11]{Loehr-bij} for various results
of this present chapter rewritten in terms of symmetric polynomials
in finitely many variables.}.
\end{remark}

% [DG][v58] Added stuff to the above remark. Also, added the Wildon
% reference.

One can also define a comultiplication on $\Lambda$ as follows.

Consider the countably infinite set of variables
$(\xx,\yy)=(x_1,x_2,\ldots,y_1,y_2,\ldots)$.\index{$(\xx,\yy)$}
Although it properly contains $\xx$, there are nevertheless
bijections between $\xx$ and $(\xx, \yy)$, since these two
variable sets have the same cardinality.

Let $R(\xx, \yy)$ denote the $\kk$-algebra of formal power series
in $(\xx,\yy)$ of bounded degree.
Let $\Symm_{(\infty,\infty)}$
be the group of all permutations of $\left\{ x_1, x_2, \ldots,
y_1, y_2, \ldots\right\}$ leaving
all but finitely many variables invariant.
Then, $\Symm_{(\infty,\infty)}$ acts on $R(\xx, \yy)$ by
permuting variables, in the same way as $\Symm_{(\infty)}$ acts on
$R(\xx)$.
The fixed space $R(\xx, \yy)^{\Symm_{(\infty,\infty)}}$
is a $\kk$-algebra, which we denote by $\Lambda(\xx, \yy)$.
This $\kk$-algebra $\Lambda(\xx, \yy)$
is isomorphic to $\Lambda = \Lambda(\xx)$, since there is
a bijection between the two sets of variables $(\xx, \yy)$ and
$\xx$.
More explicitly: The map
\begin{equation}
\begin{array}{rcl}
\Lambda=\Lambda(\xx) & \overset{\Delta}{\longrightarrow}
   & \Lambda(\xx,\yy) , \\
f(\xx)=f(x_1,x_2,\ldots) & \longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots)
\end{array}
\label{symmetric-functions-in-two-sets-fxxyy}
\end{equation}
is a graded $\kk$-algebra isomorphism.
Here, $f \left( x_1, x_2, \ldots, y_1, y_2, \ldots \right)$ means
the result of choosing some bijection
\newline
$\phi : \left\{ x_1, x_2, x_3, \ldots \right\}
\to \left\{ x_1, x_2, \ldots, y_1, y_2, \ldots \right\}$ and
substituting $\phi\left(x_i\right)$ for every $x_i$ in $f$.
(The choice of $\phi$ is irrelevant since $f$ is
symmetric.\footnote{To be more precise, the choice of $\phi$ is
irrelevant because $f$ is $\Symm_\infty$-invariant, with the
notations of Remark~\ref{rmk.Lambda.symmetric}.})

The group $\Symm_{(\infty)} \times \Symm_{(\infty)}$ is a subgroup of
the group $\Symm_{(\infty,\infty)}$ (via the obvious injection,
which lets each
$\left(\sigma, \tau\right) \in \Symm_{(\infty)} \times \Symm_{(\infty)}$
act by separately
permuting the $x_1, x_2, x_3, \ldots$ using $\sigma$
and permuting the $y_1, y_2, y_3, \ldots$ using $\tau$),
% where $\Symm_{(\infty)} \times \Symm_{(\infty)}$ denotes
% permutations of (finite subsets of) the
% $\xx$  and separate permutations of (finite subsets of)
% the $\yy$, 
and thus also acts on $R(\xx, \yy)$.
Hence, we have an inclusion of $\kk$-algebras
$\Lambda(\xx,\yy)
= R(\xx, \yy)^{\Symm_{(\infty,\infty)}}
\subset R(\xx, \yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)}}
\subset R(\xx, \yy)$.
The $\kk$-module $R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)}}$
has $\kk$-basis $\{ m_\lambda(\xx) m_\mu(\yy) \}_{\lambda, \mu \in \Par}$,
since $m_\lambda(\xx) m_\mu(\yy)$ is just the sum of
all monomials in the
$\Symm_{(\infty)} \times \Symm_{(\infty)}$-orbit of
$\xx^\lambda \yy^\mu$ (and since any
$\Symm_{(\infty)} \times \Symm_{(\infty)}$-orbit of
monomials has exactly one representative of the form
$\xx^\lambda \yy^\mu$ with $\lambda, \mu \in \Par$).
Here, of course, $\yy$ stands for the set of variables
$\left(y_1, y_2, y_3, \ldots\right)$, and we define
$\yy^\mu$ to be $y_1^{\mu_1} y_2^{\mu_2} \cdots$.

On the other hand, the map
% Note that when one decomposes the variables into two sets
% $(\xx,\yy)=(x_1,x_2,\ldots,y_1,y_2,\ldots)$, one has a ring homomorphism
\[
\begin{array}{rcl}
R(\xx) \otimes R(\xx) & \longrightarrow& R(\xx,\yy) , \\
f(\xx) \otimes g(\xx) & \longmapsto& f(\xx)g(\yy)
\end{array}
\]
is a $\kk$-algebra homomorphism.
Restricting it to $R(\xx)^{\Symm_{(\infty)}} \otimes R(\xx)^{\Symm_{(\infty)}}$,
we obtain a $\kk$-algebra homomorphism
\begin{equation}
\label{symmetric-functions-in-two-sets-iso}
\Lambda \otimes \Lambda
=R(\xx)^{\Symm_{(\infty)}} \otimes R(\xx)^{\Symm_{(\infty)}}
\longrightarrow R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)} } ,
\end{equation}
which is an isomorphism because it sends the basis
$\{ m_\lambda \otimes m_\mu \}_{\lambda, \mu \in \Par}$
of the $\kk$-module $\Lambda \otimes \Lambda$
to the basis $\{ m_\lambda(\xx) m_\mu(\yy) \}_{\lambda, \mu \in \Par}$
of the $\kk$-module $R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)}}$.
% As $\Symm_{(\infty)} \times \Symm_{(\infty)}$ is a subgroup of the
% group $\Symm_{(\infty,\infty)}$
% (the group of all permutations of $\left\{ x_1, x_2, \ldots,
% y_1, y_2, \ldots\right\}$ leaving
% all but finitely many variables invariant)
% acting on all of $(\xx,\yy)$,
Thus, we get an inclusion of graded $\kk$-algebras
\[
\Lambda(\xx,\yy) =  R(\xx,\yy)^{\Symm_{(\infty,\infty)}} \hookrightarrow
     R(\xx,\yy)^{\Symm_{(\infty)} \times \Symm_{(\infty)} } \cong \Lambda \otimes \Lambda
\]
where the last isomorphism is the inverse of the one in \eqref{symmetric-functions-in-two-sets-iso}.
This gives a comultiplication
\[
\begin{array}{rcl}
\Lambda=\Lambda(\xx) & \overset{\Delta}{\longrightarrow}
   & \Lambda(\xx,\yy) \hookrightarrow \Lambda \otimes \Lambda ,\\
f(\xx)=f(x_1,x_2,\ldots) & \longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots).
\end{array}
\]
Here, $f \left( x_1, x_2, \ldots, y_1, y_2, \ldots \right)$
is understood as in \eqref{symmetric-functions-in-two-sets-fxxyy}.

% [DG][v25] Added the last two sentences about what
% $f \left( x_1, x_2, \ldots, y_1, y_2, \ldots \right)$ means.
% Or am I blind and this has already been dealt with?
% Also, replaced the "$\Symm_{(\infty)}$ acting on
% all of $(\xx,\yy)$" by $\Symm_{(\infty,\infty)}$ and
% explained what it means (as there is not really an obviously
% intended way to identify these two groups).

% [DG][v40] Added the footnote, as the permutation that takes
% one bijection to another might not have finite support.

% [DG][v75] Expanded the above explanation to give more details
% and to do fewer things in a single sentence.

\begin{example}
One has
\begin{align*}
\Delta m_{(2,1)} &= m_{(2,1)}(x_1,x_2,\ldots,y_1,y_2,\ldots) \\
&= x_1^2 x_2 + x_1 x_2^2 + \cdots \\
 &\quad + x_1^2 y_1 + x_1^2 y_2 + \cdots \\
 &\quad + x_1 y_1^2 + x_1 y_2^2 + \cdots \\
 &\quad + y_1^2 y_2 + y_1 y_2^2 + \cdots\\
&= m_{(2,1)}(\xx) + m_{(2)}(\xx) m_{(1)}(\yy) +
 m_{(1)}(\xx) m_{(2)}(\yy) + m_{(2,1)}(\yy) \\
&= m_{(2,1)} \otimes \one + m_{(2)} \otimes m_{(1)} +
m_{(1)}\otimes m_{(2)} + \one \otimes m_{(2,1)}.
\end{align*}
This example generalizes easily to the following formula:
\begin{equation}
\label{comultiplication-of-monomial-symmetric-function}
\Delta m_\lambda =
  \sum\limits_{\substack{ (\mu, \nu):\\ \mu \sqcup \nu = \lambda} } m_\mu \otimes m_\nu,
\end{equation}
in which \dfn{$\mu \sqcup \nu$}
is the partition obtained by taking the
multiset union  of the parts of $\mu$ and $\nu$, and then
reordering them to make them weakly decreasing.

\end{example}

Checking that $\Delta$ is coassociative amounts to checking that
\[
(\Delta \otimes \id) \circ \Delta f =
f(\xx,\yy,\zz) =
(\id \otimes \Delta) \circ \Delta f
\]
inside $\Lambda(\xx,\yy,\zz)$ as a subring of
$\Lambda \otimes \Lambda \otimes \Lambda$.

\begin{verlong}
Here is this argument in more detail:
\begin{proof}[First proof of the coassociativity of $\Delta$ on $\Lambda$.] 
Just as we identified the ring
$\Lambda\left(  \xx ,\yy \right)  $ with a subring of $\Lambda\otimes\Lambda$, we can
identify the ring $\Lambda\left(  \xx ,\yy ,\zz \right)  $
with a subring of $\Lambda\otimes\Lambda\otimes\Lambda$ (where $\left(
\xx ,\yy ,\zz \right)  $ denotes the family $\left(
x_{1},x_{2},...,y_{1},y_{2},...,z_{1},z_{2},...\right)  $ of indeterminates).
Let $f\in\Lambda$. We write $\Delta f\in\Lambda\otimes\Lambda$ in the form
$\Delta f=\sum\limits_{i\in I}g_{i}\otimes h_{i}$. Then, due to our
identification of the ring $\Lambda\left(  \xx ,\yy \right)  $
with a subring of $\Lambda\otimes\Lambda$, we have $\sum\limits_{i\in I}%
g_{i}\left(  \xx \right)  h_{i}\left(  \yy \right)
=\sum\limits_{i\in I}g_{i}\otimes h_{i}
=\Delta f=f\left(  \xx ,\yy \right)  $.
If we substitute $\left(  \xx ,\yy \right)  $ and
$\zz $ for $\xx $ and $\yy $ in this identity of formal
power series, then we
obtain $\sum\limits_{i\in I}g_{i}\left(  \xx ,\yy \right)
h_{i}\left(  \zz \right)  =f\left(  \left(  \xx ,\yy %
\right) , \zz\right)  =f\left(  \xx ,\yy ,\zz \right)  $. Now,
\begin{align*}
\left(  \Delta\otimes\id \right)  \underbrace{\left(  \Delta
f\right)  }_{=\sum\limits_{i\in I}g_{i}\otimes h_{i}}  &  =\left(
\Delta\otimes\id \right)  \left(  \sum\limits_{i\in I}%
g_{i}\otimes h_{i}\right)  =\sum\limits_{i\in I}\underbrace{\Delta\left(
g_{i}\right)  }_{=g_{i}\left(  \xx ,\yy \right)  }\otimes
h_{i}=\sum\limits_{i\in I}g_{i}\left(  \xx ,\yy \right)  \otimes
h_{i}\\
&  =\sum\limits_{i\in I}g_{i}\left(  \xx ,\yy \right)
h_{i}\left(  \zz \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to our identification of }%
\Lambda\left(  \xx ,\yy ,\zz \right)  \text{ with a subring
of }\Lambda\otimes\Lambda\otimes\Lambda\right) \\
&  =f\left(  \xx ,\yy ,\zz \right)
\end{align*}
and similarly $\left(  \id \otimes\Delta\right)  \left(  \Delta
f\right)  =f\left(  \xx ,\yy ,\zz \right)  $. Thus, $\left(
\Delta\otimes\id \right)  \left(  \Delta f\right)  =f\left(
\xx ,\yy ,\zz \right)  =\left(  \id \otimes\Delta\right)
\left(  \Delta f\right)  $. Since we have proven this
for every $f\in\Lambda$, we thus obtain $\left(  \Delta\otimes
\id \right)  \circ\Delta=\left(  \id \otimes
\Delta\right)  \circ\Delta$. In other words, the coassociativity of $\Delta$
on $\Lambda$ is proved.
\end{proof}

Alternatively, we can also prove the coassociativity of $\Delta$
using \eqref{comultiplication-of-monomial-symmetric-function}:

\begin{proof}[Second proof of the coassociativity of $\Delta$ on $\Lambda$.]
Every $\lambda\in\Par$ satisfies
\begin{align*}
&  \left(  \Delta\otimes\id \right)  \underbrace{\left(  \Delta
m_{\lambda}\right)  }_{\substack{=\sum\limits_{\substack{\left(  \mu,\gamma\right)
\in\Par \times\Par ;\\\mu\sqcup\gamma=\lambda}}
m_{\mu}\otimes m_{\gamma}
\\ \text{(by \eqref{comultiplication-of-monomial-symmetric-function})}
}}\\
&  =\left(  \Delta\otimes\id \right)  \left(
\sum\limits_{\substack{\left(  \mu,\gamma\right)  \in\Par %
\times\Par ;\\\mu\sqcup\gamma=\lambda}}m_{\mu}\otimes m_{\gamma
}\right)  =\underbrace{\sum\limits_{\substack{\left(  \mu,\gamma\right)
\in\Par \times\Par ;\\\mu\sqcup\gamma=\lambda}%
}}_{=\sum\limits_{\gamma\in\Par }\sum\limits_{\substack{\mu
\in\Par ;\\\mu\sqcup\gamma=\lambda}}}\underbrace{\Delta\left(
m_{\mu}\right)  }_{\substack{
=\sum\limits_{\substack{\left(  \alpha,\beta\right)
\in\Par \times\Par ;\\\alpha\sqcup\beta=\mu
}}m_{\alpha}\otimes m_{\beta}
\\ \text{(by \eqref{comultiplication-of-monomial-symmetric-function})}}}
\otimes m_{\gamma}\\
&  =\sum\limits_{\gamma\in\Par }\underbrace{
\sum\limits_{\substack{\mu\in\Par ;\\\mu\sqcup\gamma=\lambda}%
}\sum\limits_{\substack{\left(  \alpha,\beta\right)  \in\Par %
\times\Par ;\\\alpha\sqcup\beta=\mu}}}_{=\sum\limits_{\mu
\in\Par }\sum\limits_{\substack{\left(  \alpha,\beta\right)
\in\Par \times\Par ;\\\alpha\sqcup\beta=\mu
;\ \mu\sqcup\gamma=\lambda}}}m_{\alpha}\otimes m_{\beta}\otimes m_{\gamma}\\
&  =\sum\limits_{\gamma\in\Par }\sum\limits_{\mu\in
\Par }\underbrace{\sum\limits_{\substack{\left(  \alpha
,\beta\right)  \in\Par \times\Par ;\\\alpha
\sqcup\beta=\mu;\ \mu\sqcup\gamma=\lambda}}}_{\substack{=
\sum\limits_{\substack{\left(  \alpha,\beta\right)  \in\Par %
\times\Par ;\\\alpha\sqcup\beta=\mu;\ \alpha\sqcup\beta
\sqcup\gamma=\lambda}}\\\text{(because the assertion }\left(  \alpha
\sqcup\beta=\mu\text{ and }\mu\sqcup\gamma=\lambda\right)  \\\text{is
equivalent to }\left(  \alpha\sqcup\beta=\mu\text{ and}\ \alpha\sqcup
\beta\sqcup\gamma=\lambda\right)  \text{)}}}
m_{\alpha}\otimes m_{\beta}\otimes m_{\gamma}\\
&  =\sum\limits_{\gamma\in\Par }\underbrace{\sum\limits_{\mu
\in\Par }\sum\limits_{\substack{\left(  \alpha,\beta\right)
\in\Par \times\Par ;\\\alpha\sqcup\beta
=\mu;\ \alpha\sqcup\beta\sqcup\gamma=\lambda}}}_{=
\sum\limits_{\substack{\left(  \alpha,\beta\right)  \in\Par %
\times\Par ;\\\alpha\sqcup\beta\sqcup\gamma=\lambda}%
}\sum\limits_{\substack{\mu\in\Par ;\\\alpha\sqcup\beta=\mu}%
}}m_{\alpha}\otimes m_{\beta}\otimes m_{\gamma}=\sum\limits_{\gamma
\in\Par }\sum\limits_{\substack{\left(  \alpha,\beta\right)
\in\Par \times\Par ;\\\alpha\sqcup\beta
\sqcup\gamma=\lambda}}\underbrace{\sum\limits_{\substack{\mu\in
\Par ;\\\alpha\sqcup\beta=\mu}}m_{\alpha}\otimes m_{\beta
}\otimes m_{\gamma}}_{=m_{\alpha}\otimes m_{\beta}\otimes m_{\gamma}}\\
&  =\underbrace{\sum\limits_{\gamma\in\Par }
\sum\limits_{\substack{\left(  \alpha,\beta\right)  \in\Par %
\times\Par ;\\\alpha\sqcup\beta\sqcup\gamma=\lambda}}}%
_{=\sum\limits_{\substack{\left(  \alpha,\beta,\gamma\right)  \in
\Par \times\Par \times\Par %
;\\\alpha\sqcup\beta\sqcup\gamma=\lambda}}}m_{\alpha}\otimes m_{\beta}\otimes
m_{\gamma}=\sum\limits_{\substack{\left(  \alpha,\beta,\gamma\right)
\in\Par \times\Par \times\Par %
;\\\alpha\sqcup\beta\sqcup\gamma=\lambda}}m_{\alpha}\otimes m_{\beta}\otimes
m_{\gamma}
\end{align*}
and (similarly) $\left(  \id \otimes\Delta\right)  \left(
\Delta m_{\lambda}\right)  =\sum\limits_{\substack{\left(  \alpha,\beta
,\gamma\right)  \in\Par \times\Par \times
\Par ;\\\alpha\sqcup\beta\sqcup\gamma=\lambda}}m_{\alpha}\otimes
m_{\beta}\otimes m_{\gamma}$. Thus, every $\lambda\in\Par$
satisfies
\[
\left(  \Delta\otimes\id \right)  \left(  \Delta m_{\lambda
}\right)  =\sum\limits_{\substack{\left(  \alpha,\beta,\gamma\right)
\in\Par \times\Par \times\Par ;\\\alpha\sqcup\beta\sqcup\gamma=\lambda}}
m_{\alpha}\otimes m_{\beta}\otimes m_{\gamma}
=\left(  \id \otimes\Delta\right)  \left(  \Delta m_{\lambda}\right)  .
\]
Since $\left(  m_{\lambda}\right)  _{\lambda\in\Par }$ is a
basis of $\Lambda$, this yields that the maps $\left(  \Delta\otimes
\id \right)  \circ\Delta$ and
$\left(  \id \otimes\Delta\right)  \circ\Delta$ are equal to each
other on a basis of $\Lambda$. Thus,
$\left(  \Delta\otimes\id \right)  \circ
\Delta=\left(  \id \otimes\Delta\right)  \circ\Delta$. In other
words, the coassociativity of $\Delta$ on $\Lambda$ is proven again.
\end{proof}
\end{verlong}

The counit $\Lambda \overset{\epsilon}{\rightarrow} \kk$
is defined in the usual fashion for connected graded coalgebras,
namely $\epsilon$ annihilates $I=\bigoplus_{n > 0} \Lambda_n$, and
$\epsilon$ is the identity on $\Lambda_0=\kk$;  alternatively $\epsilon$
sends a symmetric function $f(\xx)$ to its constant term $f(0,0,\ldots)$.

Note that $\Delta$ is an algebra morphism
$\Lambda \rightarrow \Lambda \otimes \Lambda$
because it is a composition of maps which are all algebra morphisms.
As the unit and counit axioms are easily checked,
$\Lambda$ becomes a connected graded $\kk$-bialgebra of finite type,
and hence also a Hopf algebra by
Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
We will identify its antipode more explicitly in Section~\ref{symm-antipode-section} below.



\subsection{Other Bases}

We introduce the usual other bases of $\Lambda$,
and explain their significance later.

\begin{definition}
\label{def.Lambda.peh}
Define the families of \dfn{power sum symmetric functions} $p_n$,
\dfn{elementary symmetric functions} $e_n$,
and \dfn{complete homogeneous symmetric functions} $h_n$,
for $n=1,2,3,\ldots$ by\index{$p_n$}\index{$e_n$}\index{$h_n$}
\begin{align}
p_n &:= x_1^n +x_2^n + \cdots = m_{(n)} ,
\label{def.pn} \\
e_n &:= \sum_{i_1 < \cdots < i_n}x_{i_1} \cdots x_{i_n} = m_{(1^n)} ,
\label{def.en} \\
h_n &:= \sum_{i_1 \leq \cdots \leq i_n}x_{i_1} \cdots x_{i_n} = \sum_{\lambda \in \Par_n} m_\lambda .
\label{def.hn}
\end{align}
Here, we are using the \dfn{multiplicative notation} for
partitions: whenever $\left(m_1, m_2, m_3, \ldots\right)$ is a
weak composition, $\left(1^{m_1}2^{m_2}3^{m_3}\cdots\right)$
denotes the partition $\lambda$ such that for every $i$, the
multiplicity of the part $i$ in $\lambda$ is $m_i$. The
$i^{m_i}$ satisfying $m_i = 0$ are often omitted from this notation,
and so the $\left(1^n\right)$ in \eqref{def.en} means
$\left(\underbrace{1, 1, \ldots, 1}_{n \text{ ones}}\right)$.
(For another example, $\left(1^2 3^1 4^3\right) =
\left(1^2 2^0 3^1 4^3 5^0 6^0 7^0 \cdots \right)$
means the partition $\left(4,4,4,3,1,1\right)$.)
% where $(1^n)=(1,1,\ldots,1)$, using a multiplicative notation
% $\lambda=(1^{m_1}2^{m_2}\cdots)$ if the multiplicity of the part $i$ in $\lambda$ is $m_i$.
By convention, also define $h_0=e_0=1$, and $h_n=e_n=0$ if $n < 0$.
Extend these multiplicatively to
partitions $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_\ell)$
with $\lambda_1 \geq \cdots \geq \lambda_\ell > 0$
by setting\index{$p_\lambda$}\index{$e_\lambda$}\index{$h_\lambda$}
\begin{align*}
p_\lambda&:=p_{\lambda_1} p_{\lambda_2} \cdots p_{\lambda_\ell} ,\\
e_\lambda&:=e_{\lambda_1} e_{\lambda_2} \cdots e_{\lambda_\ell} ,\\
h_\lambda&:=h_{\lambda_1} h_{\lambda_2} \cdots h_{\lambda_\ell} .\\
\end{align*}
Also define the \dfn{Schur function}\index{$s_\lambda$}
\begin{equation}
\label{Schur-function-definition}
s_\lambda:=\sum_T \xx^{\cont(T)}
\end{equation}
where $T$ runs through all \emph{column-strict tableaux}\index{column-strict tableau}
of shape $\lambda$, that is,
$T$ is an assignment of entries in $\{1,2,3,\ldots\}$
to the cells of the \emph{Ferrers diagram}\footnote{The
\dfn{Ferrers diagram} of a partition $\lambda$ is defined as
the set of all pairs
$\left(i, j\right) \in \left\{ 1, 2, 3, \ldots \right\}^2$
satisfying $j \leq \lambda_i$.
This is a set of cardinality $\left|\lambda\right|$.
Usually, one visually represents a Ferrers diagram by drawing its
elements $\left(i, j\right)$ as points on the plane, although
(unlike the standard convention for drawing points on the plane)
one lets the x-axis go top-to-bottom (i.e., the point
$\left(i+1, j\right)$ is one step below the point
$\left(i, j\right)$), and the y-axis go left-to-right (i.e.,
the point $\left(i, j+1\right)$ is one step to the right of the
point $\left(i, j\right)$). (This is the so-called
\dfn{English notation}, also known as the
\dfn{matrix notation} because it is precisely the way one
labels the entries of a matrix. Other notations appear in
literature, such as the French notation used, e.g., in
Malvenuto's \cite{Malvenuto}, and the Russian notation used,
e.g., in parts of Kerov's \cite{Kerov}.) These points are
drawn either as dots or as square boxes; in the latter case, the
boxes are centered at the points they represent, and they have
sidelength $1$ so that the boxes centered around
$\left(i, j\right)$ and $\left(i, j+1\right)$ touch each other
along a sideline. For example, the Ferrers diagram of the
partition $\left(3, 2, 2\right)$ is represented as
\[
\begin{matrix}
\bullet & \bullet & \bullet \\
\bullet & \bullet & \\
\bullet & \bullet & \\
\end{matrix}
\text{ (using dots) }
\qquad \qquad \text{ or as } \qquad \qquad
\begin{tabular}{ccc}
\hline
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{} & \multicolumn{1}{c|}{}  \\
\hline
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{} &  \\
\cline{1-2}
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{} &  \\
\cline{1-2}
\end{tabular}
\text{ (using boxes).}
\]
The Ferrers diagram of a partition $\lambda$ uniquely determines
$\lambda$. One refers to the elements of the Ferrers diagram of
$\lambda$ as the \emph{cells}\index{cell of a Ferrers diagram}
(or \emph{boxes}\index{box of a Ferrers diagram}) of this
diagram (which is particularly natural when one represents them
by boxes) or, briefly, as the cells of $\lambda$. Notation like
``west'', ``north'', ``left'', ``right'', ``row'' and
``column'' concerning cells of Ferrers diagrams normally refers
to their visual representation.

Ferrers diagrams are also known as \emph{Young diagrams}\index{Young diagram}.

One can characterize the Ferrers diagrams of partitions as
follows: A finite subset $S$ of $\left\{ 1, 2, 3, \ldots \right\}^2$
is the Ferrers diagram of some partition if and only if for
every $\left(i, j\right) \in S$ and every $\left(i', j'\right)
\in \left\{ 1, 2, 3, \ldots \right\}^2$ satisfying $i' \leq i$
and $j' \leq j$, we have $\left(i', j'\right) \in S$. In other
words, a finite subset $S$ of $\left\{ 1, 2, 3, \ldots \right\}^2$
is the Ferrers diagram of some partition if and only if it is
a lower set of the poset $\left\{ 1, 2, 3, \ldots \right\}^2$
with respect to the componentwise order.}
 for $\lambda$, weakly
increasing left-to-right in rows, and strictly increasing top-to-bottom
in columns.  Here \dfn{$\cont(T)$} denotes the weak composition
$\left(|T^{-1}(1)|, |T^{-1}(2)|, |T^{-1}(3)|, \ldots\right)$, so that
$\xx^{\cont(T)}=\prod_i x_i^{|T^{-1}(i)|}$.  For
example,\footnote{To visually represent a column-strict tableau
$T$ of shape $\lambda$, we draw the same picture as when
representing the Ferrers diagram of $\lambda$, but with a little
difference: a cell $\left(i, j\right)$ is no longer represented
by a dot or box, but instead is represented by the entry
of $T$ assigned to this cell. Accordingly, the entry of $T$
assigned to a given cell $c$ is often referred to as
\emph{the entry of $T$ in $c$}\index{entry of a tableau}.}
\[
T=\begin{matrix}
1 & 1 & 1 & 4 & 7\\
2 & 3 & 3 &   & \\
4 & 4 & 6 &   & \\
6 & 7
\end{matrix}
\]
is a column-strict tableau of shape $\lambda=(5,3,3,2)$
with $\xx^{\cont(T)}=x_1^3 x_2^1 x_3^2 x_4^3 x_5^0 x_6^2 x_7^2$.
If $T$ is a column-strict tableau, then the weak composition $\cont(T)$
is called the \emph{content}\index{content of a tableau} of $T$.
\end{definition}

% [DG][v19] Added explicit definition of $\cont(T)$ (not just through its
% monomial).

% [DG][v25] Made the definition of multiplicative notation clearer.
% Unfortunately, switching from {array} to {align} in the definitions
% of p_n, e_n, h_n took the alignment of the second equality signs with
% it; do you have an idea how to get it back? (I want {align} due to
% the possibility of labelling each equation.)

% [DG][v32] Added two footnotes defining Ferrers diagrams and
% notations around tableaux.

Column-strict tableaux are also known as
\emph{semistandard tableaux}\index{semistandard tableau},
and some authors even
omit the adjective and just call them \emph{tableaux}\index{tableau}
(e.g., Fulton in \cite{Fulton}, a book entirely devoted
to them).

\begin{example}
One has
\begin{align*}
m_{(1)}&=p_{(1)}=e_{(1)}=h_{(1)}=s_{(1)}=x_1+x_2+x_3+\cdots , \\
s_{(n)}&=h_n ,\\
s_{(1^n)}&=e_n .
\end{align*}
\end{example}

\begin{example}
\label{transition-matrices-example}
One has for $\lambda=(2,1)$ that
\[
\begin{array}{rl}
p_{(2,1)}&=p_2 p_1 = (x_1^2+x_2^2+\cdots)(x_1+x_2+\cdots) \\
         &          = m_{(2,1)}+m_{(3)} ,\\
& \\
e_{(2,1)}&=e_2 e_1 = (x_1 x_2+x_1 x_3 + \cdots)(x_1+x_2+\cdots)\\
         &          = m_{(2,1)}+3m_{(1,1,1)} ,\\
& \\
h_{(2,1)}&=h_2 h_1 = (x_1^2 +x_2^2+\cdots + x_1x_2+x_1 x_3 + \cdots)(x_1+x_2+\cdots)\\
         &          = m_{(3)}+2m_{(2,1)}+3m_{(1,1,1)} ,\\

\end{array}
\]
and
\begin{align*}
&\begin{matrix}
s_{(2,1)}&=x_1^2x_2 &+x_1^2x_3&+x_1x_2^2&+x_1x_3^2&+x_1x_2x_3&+x_1x_2x_3&+x_1x_2x_4&+\cdots\\
&11            &11       &12            &13       &12        &13        &12 & \\
&2\phantom{1}        &3\phantom{1}   &2\phantom{1}        &3\phantom{1}   &3\phantom{1}    &2\phantom{1}    &4\phantom{1}  & \\
\end{matrix} \\
& \qquad \quad = m_{(2,1)}+2m_{(1,1,1)} .
\end{align*}
In fact, one has these transition matrices for $n=3$ expressing elements in terms of the monomial basis $m_\lambda$:
\[
\bordermatrix{
~          & p_{(3)} & p_{(2,1)} & p_{(1,1,1)} \cr
m_{(3)}    & 1       & 1         & 1  \cr
m_{(2,1)}  & 0       & 1         & 3\cr
m_{(1,1,1)}& 0       & 0         & 6
}
\ ,
\qquad
\bordermatrix{
~          & e_{(3)} & e_{(2,1)} & e_{(1,1,1)} \cr
m_{(3)}    & 0       & 0         & 1  \cr
m_{(2,1)}  & 0       & 1         & 3\cr
m_{(1,1,1)}& 1       & 3         & 6
}
\ ,
\]
\vskip.1in
\[
\bordermatrix{
~          & h_{(3)} & h_{(2,1)} & h_{(1,1,1)} \cr
m_{(3)}    & 1       & 1         & 1  \cr
m_{(2,1)}  & 1       & 2         & 3\cr
m_{(1,1,1)}& 1       & 3         & 6
}
\ ,
\qquad
\bordermatrix{
~          & s_{(3)} & s_{(2,1)} & s_{(1,1,1)} \cr
m_{(3)}    & 1       & 0         & 0  \cr
m_{(2,1)}  & 1       & 1         & 0\cr
m_{(1,1,1)}& 1       & 2         & 1
}
\ .
\]
\end{example}

Our next goal is to show that $e_\lambda, s_\lambda, h_\lambda$
(and, under some conditions, the $p_\lambda$ as well)
all give bases for $\Lambda$.  However at the moment it is not yet
even clear that $s_\lambda$ are symmetric!

% [DG][v48] Removed the $p_\lambda$ from the
% $p_\lambda, e_\lambda, s_\lambda, h_\lambda$ list and put them
% into parentheses.

\begin{proposition}
\label{Schur-functions-are-symmetric-prop}
Schur functions $s_\lambda$ are symmetric, that is, they lie
in $\Lambda$.
\end{proposition}
\begin{proof}
It suffices to show $s_\lambda$ is symmetric under swapping the variables
$x_i, x_{i+1}$, by providing an involution $\iota$ on the set
of all column-strict tableaux $T$ of shape $\lambda$ which switches
the $\cont(T)$ for $(i,i+1) \cont(T)$.
Restrict attention to the entries $i,i+1$
in $T$, which must look something like this:
\[
\begin{matrix}
   &   & & & & &   &   &i  &i  &i  &i&i+1&i+1 \\
   &i  &i&i&i&i&i+1&i+1&i+1&i+1&i+1& &   & \\
i+1&i+1&i+1& & &   &   &   &   &   & &   &
\end{matrix}
\]
One finds several vertically aligned pairs
$\begin{matrix} i\\ i+1 \end{matrix}$.  If one were to remove all such pairs, the remaining entries would be a sequence of rows, each looking like this:
\begin{equation}
\label{i-i+1-row-to-be-swapped}
\underbrace{i,i,\ldots,i}_{r\text{ occurrences}},
\underbrace{i+1,i+1,\ldots,i+1}_{s\text{ occurrences}} \, .
\end{equation}
An involution due to Bender and Knuth\index{Bender-Knuth involution}
tells us to leave fixed all the vertically aligned pairs
$\begin{matrix} i\\ i+1 \end{matrix}$,
but change each sequence of remaining entries
as in \eqref{i-i+1-row-to-be-swapped} to this:
\[
\underbrace{i,i,\ldots,i}_{s\text{ occurrences}},
\underbrace{i+1,i+1,\ldots,i+1}_{r\text{ occurrences}} \, .
\]
For example, the above configuration in $T$ would change
to
\[
\begin{matrix}
   &   & & & &   &   &   &i  &i  &i  &i  &i  &i+1 \\
   &i  &i&i&i&i+1&i+1&i+1&i+1&i+1&i+1&   &   & \\
i  &i+1&i+1& &   &   &   &   &   &   &   &   &
\end{matrix}
\]
It is easily checked that this map is an involution,
and that it has the effect of swapping $(i,i+1)$ in $\cont(T)$.
\end{proof}

\begin{remark}
\label{other-linear-orders-remark}
The symmetry of Schur functions allows one to
reformulate them via column-strict tableaux defined with respect to
\emph{any} total ordering $\LLL$ on the positive integers, rather than the usual
$1< 2 < 3<\cdots$.  For example, one can use
the \emph{reverse order}\footnote{This reverse order is what one uses when one defines a Schur function as a generating function for \emph{reverse semistandard tableaux} or \emph{column-strict plane partitions}; see Stanley \cite[Proposition 7.10.4]{Stanley}.}
$
\cdots <3<2<1,
$
or even more exotic orders, such as
\[
1<3<5<7<\cdots<2<4<6<8<\cdots.
\]
Say that an assignment $T$ of entries in $\{1,2,3,\ldots\}$ to the cells of
the Ferrers diagram of $\lambda$ is an \dfn{$\LLL$-column-strict tableau} if
it is weakly $\LLL$-increasing left-to-right in rows, and strictly
$\LLL$-increasing top-to-bottom in columns.

\begin{proposition}
\label{prop.other-linear-orders.slambda}
For any total order  $\LLL$ on the positive integers,
\begin{equation}
\label{different-total-order-Schur-function}
s_\lambda = \sum_T \xx^{\cont(T)}
\end{equation}
as $T$ runs through all $\LLL$-column-strict tableaux of shape $\lambda$.
\end{proposition}
\begin{proof}
Given a weak composition $\alpha=(\alpha_1,\alpha_2,\ldots)$  with
$\alpha_{n+1}=\alpha_{n+2}=\cdots=0$,
assume that the integers $1,2,\ldots,n$ are totally ordered by $\LLL$ as
$w(1) <_{\LLL} \cdots <_{\LLL} w(n)$ for some $w$ in $\Symm_n$.
Then the coefficient of
$\xx^{\alpha}=x_1^{\alpha_1} \cdots x_{n}^{\alpha_n}$
on the right side of \eqref{different-total-order-Schur-function}
is the same as the coefficient of $\xx^{w^{-1}(\alpha)}$
on the right side of \eqref{Schur-function-definition}
defining $s_\lambda$,
which by symmetry of $s_\lambda$ is the same
as the coefficient of $\xx^{\alpha}$ on the right side of \eqref{Schur-function-definition}.
\end{proof}
\end{remark}

It is now not hard to show that
$p_\lambda, e_\lambda, s_\lambda$ give bases by a triangularity
argument\footnote{See Section~\ref{sect.STmat} for some notions and
notations that will be used in this argument.}.  For this purpose,
let us introduce a useful partial order on partitions.

\begin{definition}
\label{def.dominance}
The \dfn{dominance} or \dfn{majorization} order on $\Par_n$
is the partial order on the set $\Par_n$ whose
greater-or-equal relation \dfn{$\triangleright$} is defined as follows:
For two partitions $\lambda$ and $\mu$ of $n$, we set
$\lambda \triangleright \mu$ (and say that $\lambda$
\emph{dominates}\index{dominate}, or \emph{majorizes}\index{majorize}, $\mu$)
if and only if
\[
\lambda_1+\lambda_2+ \cdots+\lambda_k \geq \mu_1 + \mu_2 + \cdots+\mu_k
\quad \text{ for } k=1,2,\ldots,n.
\]
\end{definition}

(The definition of dominance would not change if we would
replace ``for $k=1,2,\ldots,n$'' by ``for every positive
integer $k$'' or by ``for every $k \in \NN$''.)

% [DG][v29] Added the previous sentence.

% [DG][v56] Restated definition slightly.

\begin{definition}
\label{def.partition.conjugate}
For a partition $\lambda$, its
\emph{conjugate}\index{conjugate of a partition}
or \emph{transpose}\index{transpose of a partition}
partition \dfn{$\lambda^t$} is the one whose
Ferrers diagram is obtained from that of $\lambda$
by exchanging rows for columns
(i.e., by flipping the diagram across the ``main'', i.e.,
top-right-to-bottom-left, diagonal)\footnote{In more
rigorous terms:
The cells of the Ferrers diagram of $\lambda^t$ are the
pairs $\left(j, i\right)$, where $\left(i, j\right)$ ranges
over all cells of $\lambda$.
It is easy to see that this indeed uniquely determines
a partition $\lambda^t$.}.
% [DG][v62] Added the preceding footnote.
Alternatively, one has this formula
for its $i$-th entry:
\begin{equation}
\label{eq.partition.transpose}
(\lambda^t)_i := |\{ j: \lambda_j \geq i \}|.
\end{equation}
\end{definition}

% [DG][v23] Removed the comma before "is the one"; please let me know if
% it was intentional.

For example, $\left( 4,3,1 \right)^t = \left( 3,2,2,1 \right)$,
which can be easily verified by flipping the Ferrers diagram
of $\left( 4,3,1 \right)$ across the ``main diagonal'':
\[
\underbrace{
\begin{matrix}
\bullet & \bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet \\
\bullet & \\
\end{matrix}
}_{\text{Ferrers diagram of $\left( 4,3,1 \right)$}}
\qquad \longmapsto \qquad
\underbrace{
\begin{matrix}
\bullet & \bullet & \bullet \\
\bullet & \bullet & \\
\bullet & \bullet & \\
\bullet
\end{matrix}
}_{\text{Ferrers diagram of $\left( 4,2,2,1 \right)$}}
\]
(or simply counting the boxes in each column of this diagram).

% [DG][v78] Added above example. Also added the "flipping"
% description of conjugation to the definition.

\begin{exercise}
\label{exe.partition.dominance}
Let $\lambda , \mu \in \Par_n$. Show that
$\lambda \triangleright \mu$ if and only if
$\mu^t \triangleright \lambda^t$.
\end{exercise}

% [DG][v23] Made the above paragraph into an actual exercise.

\begin{proposition}
\label{many-symmetric-function-bases-prop}
The families $\{e_\lambda\}$ and $\{s_\lambda\}$, as $\lambda$ runs through
all partitions, are graded bases for the
graded $\kk$-module $\Lambda_\kk$
whenever $\kk$ is a commutative ring.
The same holds for the family $\{p_\lambda\}$ when $\QQ$ is a
subring of $\kk$.
\end{proposition}

% [DG][v80] Replaced "bases" by "graded bases", since we
% prove this stronger fact in fact (and it is more useful).

Our proof of this proposition will involve three separate
arguments, one for each of the three alleged bases
$\{s_\lambda\}$, $\{e_\lambda\}$ and $\{p_\lambda\}$; however,
all these three arguments fit the same mold: Each one shows
that the alleged basis expands invertibly
triangularly\footnote{i.e., triangularly, with all diagonal
coefficients being invertible} in the
basis $\{m_\lambda\}$ (possibly after reindexing), with an
appropriately chosen partial order on the indexing set.
We will simplify our life by restricting ourselves to $\Par_n$
for a given $n \in \NN$, and by stating the common part of the
three arguments in a greater generality (so that we won't have
to repeat it thrice):

\begin{lemma}
\label{lem.many-symmetric-function-bases-prop.last-step}
Let $S$ be a finite poset. We write $\leq$ for the
smaller-or-equal relation of $S$.

Let $M$ be a free $\kk$-module with
a basis $\left( b_\lambda \right)_{\lambda \in S}$.
Let $\left( a_\lambda \right)_{\lambda \in S}$ be a further
family of elements of $M$.

For each $\lambda \in S$, let
$\left( g_{\lambda, \mu} \right)_{\mu \in S}$
be the family of the coefficients in the
expansion of $a_\lambda \in M$ in the basis
$\left( b_\mu \right)_{\mu \in S}$; in other words,
let $\left( g_{\lambda, \mu} \right)_{\mu \in S} \in \kk^S$
be such that
$a_\lambda = \sum\limits_{\mu \in S} g_{\lambda, \mu} b_\mu$.
Assume that:
\begin{itemize}
\item \textit{Assumption A1:} Any $\lambda \in S$ and $\mu \in S$ satisfy $g_{\lambda, \mu} = 0$ unless $\mu \leq \lambda$.
\item \textit{Assumption A2:} For any $\lambda \in S$, the element $g_{\lambda, \lambda}$ of $\kk$ is invertible.
\end{itemize}

Then, the family $\left( a_\lambda \right)_{\lambda \in S}$
is a basis of the $\kk$-module $M$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}.]
Use the notations of Section~\ref{sect.STmat}.
Assumptions A1 and A2 yield that the
$S \times S$-matrix
$\left(g_{\lambda, \mu}\right)_{\left(\lambda, \mu\right) \in S \times S} \in \kk^{S \times S}$
is invertibly triangular. But the definition of the
$g_{\lambda, \mu}$ yields that the
family $\left( a_\lambda \right)_{\lambda \in S}$ expands in the family
$\left( b_\lambda \right)_{\lambda \in S}$ through this matrix
$\left(g_{\lambda, \mu}\right)_{\left(\lambda, \mu\right) \in S \times S}$.
Since the latter matrix is invertibly triangular, this shows
that the family $\left( a_\lambda \right)_{\lambda \in S}$ expands
invertibly triangularly in the family $\left( b_\lambda \right)_{\lambda \in S}$.
Therefore, Corollary~\ref{cor.STmat.expansion-tria-inv}(e) (applied
to
$\left( e_s \right)_{s \in S} = \left( a_\lambda \right)_{\lambda \in S}$
and
$\left( f_s \right)_{s \in S} = \left( b_\lambda \right)_{\lambda \in S}$)
shows that
$\left( a_\lambda \right)_{\lambda \in S}$ is a basis of the $\kk$-module $M$
(since $\left( b_\lambda \right)_{\lambda \in S}$ is a basis of
the $\kk$-module $M$).
\end{proof}

\begin{proof}[Proof of Proposition~\ref{many-symmetric-function-bases-prop}.]
We can restrict our attention to each homogeneous
component $\Lambda_n$
and partitions $\lambda$ of $n$.  Thus, we have to prove that,
for each $n \in \NN$, the families
$\left( e_\lambda \right)_{\lambda \in \Par_n}$ and
$\left( s_\lambda \right)_{\lambda \in \Par_n}$
are bases of the $\kk$-module $\Lambda_n$, and that the same
holds for
$\left( p_\lambda \right)_{\lambda \in \Par_n}$
if $\QQ$ is a subring of $\kk$.

% We check that in each case,
% Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
% applies (with $S = \Par_n$ for some partial order on $\Par_n$,
% with $M = \Lambda_n$ and with $\left( b_s \right)_{s \in S}
% = \left( m_\lambda \right)_{\lambda \in \Par_n}$). This will
% then yield that the proposed basis is indeed a basis.
% the proposed basis expands invertibly triangularly\footnote{See
% Definition~\ref{def.STmat.expansion-tria}(b) for what this means.}
% in the basis $\left\{ m_\lambda \right\}_{\lambda \in \Par_n}$
% with some choice of orderings on $\Par_n$ indexing the rows and columns,
% as illustrated in Example~\ref{transition-matrices-example}.
% This will then yield (by Corollary~\ref{cor.STmat.expansion-tria-inv}(e))
% that the proposed basis is indeed a basis.

Fix $n \in \NN$. We already know that
$\left( m_\lambda \right)_{\lambda \in \Par_n}$ is a
basis of the $\kk$-module $\Lambda_n$.

\begin{enumerate}

\item[1.] We shall first show that
the family $\left( s_\lambda \right)_{\lambda \in \Par_n}$ is a
basis of the $\kk$-module $\Lambda_n$.

For every partition $\lambda$, we have
$s_\lambda = \sum_{\mu \in \Par} K_{\lambda,\mu} m_\mu$,
where the coefficient $K_{\lambda,\mu}$ is the \dfn{Kostka number} counting
the column-strict tableaux $T$ of shape $\lambda$ having $\cont(T)=\mu$;
this follows because both sides are symmetric functions,
and $K_{\lambda,\mu}$ is the coefficient of $\xx^\mu$ on both
sides\footnote{In general, in order to prove that two
symmetric functions $f$ and $g$ are equal, it suffices
to show that, for every $\mu \in \Par$, the coefficients
of $\xx^\mu$ in $f$ and in $g$ are equal. (Indeed, all
other coefficients are determined by these coefficients
because of the symmetry.)}.
Thus, for every $\lambda \in \Par_n$, one has
\begin{equation}
s_\lambda = \sum_{\mu \in \Par_n} K_{\lambda,\mu} m_\mu
\label{pf.many-symmetric-function-bases-prop.s-through-m}
\end{equation}
(since $s_\lambda$ is homogeneous of degree $n$).
\ \ \ \ \footnote{See Exercise~\ref{exe.Lambda.triangular}(c)
below for a detailed proof of
\eqref{pf.many-symmetric-function-bases-prop.s-through-m}.}
But if $\lambda$ and $\mu$ are partitions satisfying
$K_{\lambda, \mu} \neq 0$, then there exists a column-strict
tableau $T$ of shape $\lambda$ having $\cont(T) = \mu$
(since $K_{\lambda, \mu}$ counts such tableaux), and
therefore we must have
$\lambda_1+\lambda_2+ \cdots+\lambda_k \geq \mu_1 + \mu_2 + \cdots+\mu_k$
for each positive integer $k$
(since the entries $1,2,\ldots,k$ in $T$ must all lie within the
first $k$ rows of $\lambda$); in other words,
$\lambda \triangleright \mu$ (if $K_{\lambda,\mu} \neq 0$)
\ \ \ \ \footnote{See Exercise~\ref{exe.Lambda.triangular}(d)
below for a detailed proof of this fact.}.
In other words,
\begin{equation}
\text{any } \lambda \in \Par_n \text{ and }\mu \in \Par_n
\text{ satisfy }K_{\lambda, \mu} = 0
\text{ unless }\lambda \triangleright \mu .
\label{pf.many-symmetric-function-bases-prop.s-through-m.0}
\end{equation}
One can also check that $K_{\lambda, \lambda}=1$ for any
$\lambda \in \Par_n$
\ \ \ \ \footnote{See Exercise~\ref{exe.Lambda.triangular}(e)
below for a proof of this.}. Hence,
\begin{equation}
\text{for any } \lambda \in \Par_n \text{, the element }
K_{\lambda, \lambda} \text{ of } \kk \text{ is invertible.}
\label{pf.many-symmetric-function-bases-prop.s-through-m.1}
\end{equation}

Now, let us regard
the set $\Par_n$ as a poset, whose greater-or-equal relation
is $\triangleright$.
Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
(applied to $S = \Par_n$, $M = \Lambda_n$,
$a_\lambda = s_\lambda$, $b_\lambda = m_\lambda$ and
$g_{\lambda, \mu} = K_{\lambda, \mu}$) shows that
the family $\left( s_\lambda \right)_{\lambda \in \Par_n}$ is
a basis of the $\kk$-module $\Lambda_n$
(because the Assumptions A1 and A2 of
Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
are satisfied\footnote{Indeed, they follow from
\eqref{pf.many-symmetric-function-bases-prop.s-through-m.0}
and
\eqref{pf.many-symmetric-function-bases-prop.s-through-m.1},
respectively.}).

% Hence, the expansion \eqref{pf.many-symmetric-function-bases-prop.s-through-m}
% becomes
% $s_\lambda=\sum_{\mu: \lambda \triangleright \mu} K_{\lambda,\mu} m_\mu$.

\item[2.]
Before we show that $\left(e_\lambda\right)_{\lambda \in \Par_n}$
is a basis, we define a few
notations regarding integer matrices. A \dfn{$\{0,1\}$-matrix}
means a matrix whose entries belong to the set $\{0,1\}$. If
$A \in \NN^{\ell \times m}$ is a matrix, then the \dfn{row sums}
of $A$ means the $\ell$-tuple $\left(r_1, r_2, \ldots, r_\ell\right)$,
where each $r_i$ is the sum of all entries in the $i$-th row of $A$;
similarly, the \dfn{column sums} of $A$ means the $m$-tuple
$\left(c_1, c_2, \ldots, c_m\right)$, where each $c_j$ is the sum
of all entries in the $j$-th column of $A$.
(For instance, the row sums of the $\left\{0,1\right\}$-matrix
$\left( \begin{matrix} 0 & 1 & 1 & 0 & 0 \\
					   1 & 1 & 0 & 1 & 0 \end{matrix} \right)$
is $\left(2, 3\right)$, whereas its column sums is
$\left(1, 2, 1, 1, 0\right)$.)
We identify any $k$-tuple of nonnegative integers
$\left(a_1, a_2, \ldots, a_k\right)$ with the weak composition
$\left(a_1, a_2, \ldots, a_k, 0, 0, 0, \ldots\right)$;
thus, the row sums and the column sums of a matrix in
$\NN^{\ell \times m}$ can be viewed as weak compositions.
(For example, the column sums of the matrix
$\left( \begin{matrix} 0 & 1 & 1 & 0 & 0 \\
					   1 & 1 & 0 & 1 & 0 \end{matrix} \right)$
is the $5$-tuple $\left(1, 2, 1, 1, 0\right)$, and can be
viewed as the weak composition
$\left(1, 2, 1, 1, 0, 0, 0, \ldots \right)$.)

% [DG][v23] Added preceding paragraph, also replacing "row sum" by
% "row sums" below. I feel that this pedantry is helpful because
% "row sum" could also be understood as "entrywise sum of all rows",
% which is precisely what you call "column sums". (But I also feel
% it could have been done better.)

% [DG][v70] Added example and clarifications
% about how a partition can be the row/column sums of a
% finite matrix.

For every $\lambda \in \Par_n$, one has
\begin{equation}
e_\lambda = \sum_{\mu \in \Par_n} a_{\lambda,\mu} m_\mu ,
\label{pf.many-symmetric-function-bases-prop.e-through-m}
\end{equation}
where $a_{\lambda,\mu}$ counts $\{0,1\}$-matrices (of size
$\ell\left(\lambda\right) \times \ell\left(\mu\right)$) having
row sums $\lambda$ and column sums $\mu$: indeed, when one
expands $e_{\lambda_1} e_{\lambda_2} \cdots $,
choosing the monomial $x_{j_1} \ldots x_{j_{\lambda_i}}$ in the
$e_{\lambda_i}$ factor corresponds to putting
$1$'s in the $i$-th row and columns $j_1,\ldots,j_{\lambda_i}$ of the
$\{0,1\}$-matrix
\footnote{See Exercise~\ref{exe.Lambda.triangular}(g)
below for a detailed proof of
\eqref{pf.many-symmetric-function-bases-prop.e-through-m}.}.
Applying \eqref{pf.many-symmetric-function-bases-prop.e-through-m}
to $\lambda^t$ instead of $\lambda$, we see that
\begin{equation}
e_{\lambda^t} = \sum_{\mu \in \Par_n} a_{\lambda^t,\mu} m_\mu
\label{pf.many-symmetric-function-bases-prop.e-through-m.!}
\end{equation}
for every $\lambda \in \Par_n$. 

It is not hard to check\footnote{See
Exercise~\ref{exe.Lambda.triangular}(h)
below for a proof of this.
This is the easy implication in the \dfn{Gale-Ryser Theorem}.
(The hard implication is the converse: It
says that if $\lambda, \mu \in \Par_n$ satisfy
$\lambda^t \triangleright \mu$, then
there exists a $\left\{0,1\right\}$-matrix having
row sums $\lambda$ and column sums $\mu$, so that
$a_{\lambda,\mu}$ is a positive integer.
This is proven, e.g., in \cite{Krause-Gale}, in
\cite[Theorem 2.4]{Dahl-ComMat}
and in \cite[Section 5.2]{Wildon2016}.)}
that $a_{\lambda,\mu}$ vanishes unless
$\lambda^t \triangleright \mu$. Applying this to $\lambda^t$
instead of $\lambda$, we conclude that
\begin{equation}
\text{any } \lambda \in \Par_n \text{ and }\mu \in \Par_n
\text{ satisfy }a_{\lambda^t, \mu} = 0
\text{ unless }\lambda \triangleright \mu .
\label{pf.many-symmetric-function-bases-prop.e-through-m.0}
\end{equation}
Moreover, one can show that
$a_{\lambda^t, \lambda} = 1$ for each $\lambda \in \Par_n$
\ \ \ \ \footnote{See Exercise~\ref{exe.Lambda.triangular}(i)
below for a proof of this.}.
Hence,
\begin{equation}
\text{for any } \lambda \in \Par_n \text{, the element }
a_{\lambda^t, \lambda} \text{ of } \kk \text{ is invertible.}
\label{pf.many-symmetric-function-bases-prop.e-through-m.1}
\end{equation}

Now, let us regard
the set $\Par_n$ as a poset, whose greater-or-equal relation
is $\triangleright$.
Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
(applied to $S = \Par_n$, $M = \Lambda_n$,
$a_\lambda = e_{\lambda^t}$, $b_\lambda = m_\lambda$ and
$g_{\lambda, \mu} = a_{\lambda^t, \mu}$) shows that
the family $\left( e_{\lambda^t} \right)_{\lambda \in \Par_n}$ is
a basis of the $\kk$-module $\Lambda_n$
(because the Assumptions A1 and A2 of
Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
are satisfied\footnote{Indeed, they follow from
\eqref{pf.many-symmetric-function-bases-prop.e-through-m.0}
and
\eqref{pf.many-symmetric-function-bases-prop.e-through-m.1},
respectively.}).
Hence,
$\left( e_{\lambda} \right)_{\lambda \in \Par_n}$ is a
basis of $\Lambda_n$.

\item[3.]
Assume now that $\QQ$ is a subring of $\kk$.
For every $\lambda \in \Par_n$, one has
\begin{equation}
p_\lambda = \sum_{\mu \in \Par_n} b_{\lambda,\mu} m_\mu ,
\label{pf.many-symmetric-function-bases-prop.p-through-m}
\end{equation}
where $b_{\lambda,\mu}$ counts the ways to partition the nonzero parts
$\lambda_1,\ldots,\lambda_\ell$ (where $\ell = \ell\left(\lambda\right)$)
into blocks such that the sums of the blocks
give $\mu$; more formally, $b_{\lambda,\mu}$ is the number of maps
$\varphi: \{1,2,\ldots,\ell\} \rightarrow \{1,2,3,\ldots\}$ having
\[
\mu_j = \sum_{i: \varphi(i)=j} \lambda_i
\qquad \text{ for all }j=1,2,\ldots
\]
\footnote{See Exercise~\ref{exe.Lambda.triangular}(k)
below for a detailed proof of
\eqref{pf.many-symmetric-function-bases-prop.p-through-m}
(and see Exercise~\ref{exe.Lambda.triangular}(j) for a proof
that the numbers $b_{\lambda,\mu}$ are well-defined).}.
Again it is not hard to check that
\begin{equation}
\text{any } \lambda \in \Par_n \text{ and }\mu \in \Par_n
\text{ satisfy }b_{\lambda, \mu} = 0
\text{ unless }\mu \triangleright \lambda .
\label{pf.many-symmetric-function-bases-prop.p-through-m.0}
\end{equation}
\footnote{See Exercise~\ref{exe.Lambda.triangular}(l)
below for a proof of this.}
Furthermore, for any $\lambda \in \Par_n$, the element
$b_{\lambda, \lambda}$ is a positive integer\footnote{This
is proven in Exercise~\ref{exe.Lambda.triangular}(m)
below.}, and thus
invertible in $\kk$ (since $\QQ$ is a subring of $\kk$).
Thus,
\begin{equation}
\text{for any } \lambda \in \Par_n \text{, the element }
b_{\lambda, \lambda} \text{ of } \kk \text{ is invertible}
\label{pf.many-symmetric-function-bases-prop.p-through-m.1}
\end{equation}
(although we don't always have
$b_{\lambda, \lambda} = 1$ this time).

Now, let us regard
the set $\Par_n$ as a poset, whose smaller-or-equal relation
is $\triangleright$.
Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
(applied to $S = \Par_n$, $M = \Lambda_n$,
$a_\lambda = p_\lambda$, $b_\lambda = m_\lambda$ and
$g_{\lambda, \mu} = b_{\lambda, \mu}$) shows that
the family $\left( p_\lambda \right)_{\lambda \in \Par_n}$ is
a basis of the $\kk$-module $\Lambda_n$
(because the Assumptions A1 and A2 of
Lemma~\ref{lem.many-symmetric-function-bases-prop.last-step}
are satisfied\footnote{Indeed, they follow from
\eqref{pf.many-symmetric-function-bases-prop.p-through-m.0}
and
\eqref{pf.many-symmetric-function-bases-prop.p-through-m.1},
respectively.}).

\end{enumerate}
\end{proof}

\begin{remark}
When $\QQ$ is not a subring of $\kk$, the family
$\left\{ p_\lambda \right\}$ is not (in general) a basis of $\Lambda_{\kk}$;
for instance, $e_2 = \frac12 \left(p_{(1,1)}-p_2\right) \in \Lambda_{\QQ}$
is not in the $\ZZ$-span of this family. However, if we define
$b_{\lambda,\mu}$ as in the above proof, then the $\ZZ$-linear span
of all $p_{\lambda}$ equals the $\ZZ$-linear span of all $b_{\lambda,
\lambda} m_{\lambda}$. Indeed, if
$\mu = \left(\mu_1, \mu_2, \ldots, \mu_k\right)$ with $k = \ell(\mu)$,
then $b_{\mu, \mu}$ is the size of the subgroup of $\Symm_k$
consisting of all permutations $\sigma \in \Symm_k$ having each $i$
satisfy $\mu_{\sigma(i)} = \mu_i$\ \ \ \ \footnote{See
Exercise~\ref{exe.Lambda.triangular}(n)
below for a proof of this.}. As a consequence, $b_{\mu, \mu}$
divides $b_{\lambda, \mu}$ for every partition $\mu$ of the same size as
$\lambda$ (because this group acts\footnote{Specifically, an element
$\sigma$ of the group takes
$\varphi: \{1,2,\ldots,\ell\} \rightarrow \{1,2,3,\ldots\}$ to
$\sigma \circ \varphi$.} freely on the set which is
enumerated by $b_{\lambda, \mu}$)\ \ \ \ \footnote{See
Exercise~\ref{exe.Lambda.triangular}(o)
below for a detailed proof of this.}. Hence, the
$\Par_n \times \Par_n$-matrix
$\left( \dfrac{b_{\lambda, \mu}}{b_{\mu, \mu}} \right)_{\left(\lambda, \mu\right) \in \Par_n \times \Par_n}$
has integer entries. Furthermore, this matrix is
unitriangular\footnote{Here, we are using the terminology defined
in Section~\ref{sect.STmat}, and we are regarding $\Par_n$ as a poset
whose smaller-or-equal relation is $\triangleright$.}
(indeed, \eqref{pf.many-symmetric-function-bases-prop.p-through-m.0}
shows that it is triangular, but its diagonal entries are clearly
$1$) and thus invertibly triangular. But
\eqref{pf.many-symmetric-function-bases-prop.p-through-m} shows
that the family $\left( p_\lambda \right)_{\lambda \in \Par_n}$
expands in the family
$\left( b_{\lambda, \lambda} m_\lambda \right)_{\lambda\in\Par_n}$
through this matrix. Hence, the family
$\left( p_\lambda \right)_{\lambda \in \Par_n}$
expands invertibly triangularly in the family
$\left( b_{\lambda, \lambda} m_\lambda \right)_{\lambda\in\Par_n}$.
Thus, Corollary~\ref{cor.STmat.expansion-tria-inv}(b) (applied
to $\ZZ$, $\Lambda_n$, $\Par_n$, $\left( p_\lambda \right)_{\lambda \in \Par_n}$
and $\left( b_{\lambda, \lambda} m_\lambda \right)_{\lambda \in \Par_n}$
instead of $\kk$,
$M$, $S$, $\left(e_s\right)_{s \in S}$ and $\left(f_s\right)_{s \in S}$)
shows that the $\ZZ$-submodule of $\Lambda_n$ spanned by
$\left( p_\lambda \right)_{\lambda \in \Par_n}$ is the
$\ZZ$-submodule of $\Lambda_n$ spanned by
$\left( b_{\lambda, \lambda} m_\lambda \right)_{\lambda \in \Par_n}$.
\end{remark}

% [DG] I added the remark above for no particular reason -- it just looks
% in place here. If you disagree, just remove it.

% [DG][v55] Rewrote the above proof and remark with less informal
% jargon and more structure and precision. (And more length,
% unfortunately.)

% [DG][v56] Rewrote the above proof according to our discussion.
% It has grown longer again, but hopefully it also reads better
% due to less cramping of different things in the same
% sentence.

The purpose of the following exercise is to fill in some details omitted from
the proof of Proposition~\ref{many-symmetric-function-bases-prop}.

\begin{exercise}
\phantomsection\label{exe.Lambda.triangular}
Let $n\in \NN$.

\begin{enumerate}
\item[(a)] Show that every $f \in \Lambda_n$ satisfies
\[
f = \sum_{\mu \in \Par_n}
\left(  \left[  \xx^{\mu}\right]  f\right)  m_{\mu}.
\]
Here, $\left[  \xx^{\mu}\right]  f$ denotes the coefficient of the
monomial $\xx^{\mu}$ in the power series $f$.
\end{enumerate}

Now, we introduce a notation (which generalizes the notation $K_{\lambda,\mu}$
from the proof of Proposition~\ref{many-symmetric-function-bases-prop}): For
any partition $\lambda$ and any weak composition $\mu$, we let $K_{\lambda
,\mu}$ denote the number of all column-strict tableaux $T$ of shape $\lambda$
having $\cont \left(  T\right)  =\mu$.

\begin{enumerate}
\item[(b)] Prove that this number $K_{\lambda,\mu}$ is well-defined (i.e.,
there are only finitely many column-strict tableaux $T$ of shape $\lambda$
having $\cont \left(  T\right)  =\mu$).

\item[(c)] Show that $s_{\lambda}=\sum_{\mu\in\Par_n}
K_{\lambda,\mu}m_{\mu}$ for every $\lambda\in\Par_n$.

\item[(d)] Show that $K_{\lambda,\mu}=0$ for any partitions $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$
that don't satisfy $\lambda\triangleright\mu$.

\item[(e)] Show that $K_{\lambda,\lambda}=1$ for any $\lambda\in
\Par_{n}$.
\end{enumerate}

Next, we recall a further notation: For any two partitions $\lambda$ and $\mu
$, we let $a_{\lambda,\mu}$ denote the number of all $\left\{  0,1\right\}
$-matrices of size $\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)
$ having row sums $\lambda$ and column sums $\mu$. (See the proof of
Proposition~\ref{many-symmetric-function-bases-prop} for the concepts of
$\left\{  0,1\right\}  $-matrices and of row sums and column sums.)

\begin{enumerate}
\item[(f)] Prove that this number $a_{\lambda,\mu}$ is well-defined (i.e.,
there are only finitely many $\left\{  0,1\right\}  $-matrices of size
$\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)  $ having row sums
$\lambda$ and column sums $\mu$).

\item[(g)] Show that $e_{\lambda}=\sum_{\mu\in\Par_n}
a_{\lambda,\mu}m_{\mu}$ for every $\lambda\in\Par_{n}$.

\item[(h)] Show that $a_{\lambda,\mu}=0$ for any partitions $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$
that don't satisfy $\lambda^{t}\triangleright\mu$.

\item[(i)] Show that $a_{\lambda^{t},\lambda}=1$ for any $\lambda
\in\Par_{n}$.

\end{enumerate}

Next, we introduce a further notation (which generalizes the notation
$b_{\lambda,\mu}$ from the proof of
Proposition~\ref{many-symmetric-function-bases-prop}): For any partition
$\lambda$ and any weak composition $\mu$, we let $b_{\lambda,\mu}$ be the
number of all maps $\varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow
\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(  \mu_{j}=
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,\ell\right\}  ;\\\varphi
(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $, where $\ell=\ell\left(
\lambda\right)  $.

\begin{enumerate}
\item[(j)] Prove that this number $b_{\lambda,\mu}$ is well-defined (i.e.,
there are only finitely many maps $\varphi:\left\{  1,2,\ldots,\ell\right\}
\rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,\ell\right\}  ;\\\varphi
(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $).

\item[(k)] Show that $p_{\lambda}=\sum_{\mu\in\Par_n}
b_{\lambda,\mu}m_{\mu}$ for every $\lambda\in\Par_{n}$.

\item[(l)] Show that $b_{\lambda,\mu}=0$ for any partitions $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$
that don't satisfy $\mu\triangleright\lambda$.

\item[(m)] Show that $b_{\lambda,\lambda}$ is a positive integer for any
$\lambda\in\Par_{n}$.

\item[(n)] Show that for any partition $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{k}\right)  \in\Par_{n}$ with $k=\ell\left(
\mu\right)  $, the integer $b_{\mu,\mu}$ is the size of the subgroup of
$\Symm_k$ consisting of all permutations $\sigma\in\Symm_k$
having each $i$ satisfy $\mu_{\sigma(i)}=\mu_{i}$. (In particular, show that
this subgroup is indeed a subgroup.)

\item[(o)] Show that $b_{\mu,\mu}\mid b_{\lambda,\mu}$ for every $\lambda
\in\Par_{n}$ and $\mu\in\Par_n$.
\end{enumerate}
\end{exercise}

% [DG][v71] Added the above exercise, which fills some DYI
% parts of the proof of Proposition~\ref{many-symmetric-function-bases-prop}.

The bases $\left\{p_\lambda\right\}$ and $\left\{e_\lambda\right\}$
of $\Lambda$ are two examples of \emph{multiplicative bases}\index{multiplicative basis}:
these are bases constructed from a sequence $v_1, v_2, v_3, \ldots$
of symmetric functions by taking all possible finite products.
We will soon encounter another example.
First, let us observe that the finite products of a sequence
$v_1, v_2, v_3, \ldots$ of symmetric functions form a basis of
$\Lambda$ if and only if the sequence is an algebraically independent
generating set of $\Lambda$.
This holds more generally for any commutative algebra, as the following
simple exercise shows:

\begin{exercise}
\label{exe.alggen-via-span}
Let $A$ be a commutative $\kk$-algebra.
Let $v_1, v_2, v_3, \ldots$ be some elements of $A$.

For every partition $\lambda$, define an element $v_{\lambda} \in A$ by
$v_{\lambda} = v_{\lambda_1} v_{\lambda_2} \cdots v_{\lambda_{\ell\left( \lambda\right)  }}$.
Prove the following:

\begin{itemize}
\item[(a)] The $\kk$-subalgebra of $A$ generated by
$v_1, v_2, v_3, \ldots$ is the $\kk$-submodule of $A$ spanned by
the family $\left(  v_{\lambda}\right)  _{\lambda\in\Par}$.

\item[(b)] The elements $v_1, v_2, v_3, \ldots$ generate the
$\kk$-algebra $A$ if and only if the family
$\left(  v_{\lambda}\right) _{\lambda\in\Par}$
spans the $\kk$-module $A$.

\item[(c)] The elements $v_1, v_2, v_3, \ldots$ are algebraically
independent over $\kk$ if and only if the family
$\left(  v_{\lambda}\right)  _{\lambda\in\Par}$
is $\kk$-linearly independent.
\end{itemize}
\end{exercise}

% [DG][v75] Added the above exercise (as it is used tacitly
% several times below) and the paragraph above it.

The next exercise states two well-known identities
for the \emph{generating functions} of the sequences
$\left(e_0, e_1, e_2, \ldots\right)$ and
$\left(h_0, h_1, h_2, \ldots\right)$, which will be used
several times further below:

\begin{exercise}
\label{exe.H(t)-and-E(t)}
In the ring of formal power series
$\powser{\left(\powser{\kk}{\xx}\right)}{t}$, prove the two
identities
\begin{equation}
\prod_{i=1}^{\infty} \left(1 - x_i t\right)^{-1}
= 1 + h_1\left(\xx\right) t + h_2\left(\xx\right) t^2 + \cdots
= \sum_{n \geq 0} h_n\left(\xx\right) t^n
\label{h-generating-function-no-HE}
\end{equation}
and
\begin{equation}
\prod_{i=1}^{\infty} \left(1 + x_i t\right)
= 1 + e_1\left(\xx\right) t + e_2\left(\xx\right) t^2 + \cdots
= \sum_{n \geq 0} e_n\left(\xx\right) t^n .
\label{e-generating-function-no-HE}
\end{equation}
\end{exercise}

% [DG][v75] Added the above exercise.



\subsection{Comultiplications}

Thinking about comultiplication $\Lambda \overset{\Delta}{\rightarrow} \Lambda \otimes \Lambda$
on Schur functions forces us to immediately confront the following.

\begin{definition}
\label{skew-Schur-function-definition}
For partitions $\mu$ and $\lambda$ say that $\mu \subseteq \lambda$ if
$\mu_i \leq \lambda_i$ for $i=1,2,\ldots$.
In other words, two partitions $\mu$ and $\lambda$ satisfy
$\mu \subseteq \lambda$ if and only if the Ferrers diagram for
$\mu$ is a subset of the Ferrers diagram of $\lambda$.
In this case, define the \dfn{skew (Ferrers) diagram} $\lambda/\mu$ to
be their set difference.\footnote{In other words, the skew
Ferrers diagram $\lambda / \mu$ is the set of all
$\left(i, j\right) \in \left\{ 1, 2, 3, \ldots \right\}^2$
satisfying $\mu_i < j \leq \lambda_i$.

While the Ferrers diagram for a
single partition $\lambda$ uniquely determines $\lambda$, the
skew Ferrers diagram $\lambda/\mu$ does not uniquely determine
$\mu$ and $\lambda$. (For instance, it is empty whenever $\lambda
= \mu$.) When one wants to keep $\mu$ and $\lambda$ in memory, one
speaks of the \dfn{skew shape} $\lambda / \mu$; this simply means
the pair $\left(\mu, \lambda\right)$. Every notion defined for
skew Ferrers diagrams also makes sense for skew shapes, because to
any skew shape $\lambda / \mu$ we can assign the skew Ferrers diagram
$\lambda / \mu$ (even if not injectively). For instance, the
\emph{cells}\index{cell of a skew shape}
of the skew shape $\lambda / \mu$ are the cells of the
skew Ferrers diagram $\lambda / \mu$.

One can characterize the skew Ferrers diagrams as
follows: A finite subset $S$ of $\left\{ 1, 2, 3, \ldots \right\}^2$
is a skew Ferrers diagram (i.e., there exist two partitions
$\lambda$ and $\mu$ such that $\mu \subseteq \lambda$ and such
that $S$ is the skew Ferrers diagram $\lambda / \mu$)
if and only if for every $\left(i, j\right) \in S$, every
$\left(i', j'\right) \in \left\{ 1, 2, 3, \ldots \right\}^2$ and
every $\left(i'', j''\right) \in S$ satisfying $i'' \leq i' \leq i$
and $j'' \leq j' \leq j$, we have $\left(i', j'\right) \in S$.}

% [DG][v32] Added the previous footnote because you occasionally speak
% of skew shapes (not just in the context of "tableau of skew shape
% \lambda/\mu").

Then define the \dfn{skew Schur function} $s_{\lambda/\mu}(\xx)$\index{$s_{\lambda/\mu}$}
to be the sum $s_{\lambda/\mu}:=\sum_{T} \xx^{\cont(T)}$, where the sum ranges
over all \emph{column-strict tableaux}\index{column-strict tableau}
$T$ of shape $\lambda/\mu$, that is,
assignments of a value in $\{1,2,3,\ldots\}$ to each cell of $\lambda/\mu$,
weakly increasing left-to-right in rows, and strictly increasing top-to-bottom
in columns.
\end{definition}

\begin{example}
\label{skew-tableau-example}
Let $\lambda = \left(5,3,3,2\right)$ and
$\mu = \left(3,1,1,0\right)$. Then, $\mu \subseteq \lambda$.
The Ferrers diagrams for $\lambda$ and
$\mu$ and the skew Ferrers diagram for $\lambda / \mu$ look as follows:
\[
\underbrace{
\begin{matrix}
\bullet & \bullet & \bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet \\
\bullet & \bullet
\end{matrix}
}_{\text{Ferrers diagram of $\lambda$}}
\qquad
\underbrace{
\begin{matrix}
\bullet & \bullet & \bullet \\
\bullet \\
\bullet \\
\vphantom{\bullet}
\end{matrix}
}_{\text{Ferrers diagram of $\mu$}}
\qquad
\underbrace{
\begin{matrix}
\cdot   & \cdot   &   \cdot & \bullet & \bullet \\
\cdot   & \bullet & \bullet \\
\cdot   & \bullet & \bullet \\
\bullet & \bullet
\end{matrix}
}_{\text{skew Ferrers diagram of $\lambda / \mu$}}
\qquad
\]
(where the small dots represent boxes removed from the
diagram).
The filling
\[
T=\begin{matrix}
\cdot & \cdot & \cdot & 2 & 5\\
\cdot & 1     & 1     &   & \\
2     & 2     & 4     &   & \\
4     & 5
\end{matrix}
\]
is a column-strict tableau of shape $\lambda/\mu=(5,3,3,2)/(3,1,0,0)$
and it has $\xx^{\cont(T)}=x_1^2 x_2^3 x_3^0 x_4^2 x_5^2$.

On the other hand, if we took $\lambda = \left(5, 3, 1\right)$
and $\mu = \left(1, 1, 1, 1\right)$, then we wouldn't
have $\mu \subseteq \lambda$, since
$\mu_4 = 1 > 0 = \lambda_4$.
\end{example}

% [DG][v78] Expanded the above example to show not just the
% tableau, but also the diagram.

\begin{remark}
If $\mu$ and $\lambda$ are partitions such that $\mu \subseteq \lambda$,
then $s_{\lambda / \mu} \in \Lambda$. (This is proven similarly as
Proposition~\ref{Schur-functions-are-symmetric-prop}.) Actually,
if $\mu \subseteq \lambda$, then
$s_{\lambda / \mu} \in \Lambda_{\left|\lambda / \mu\right|}$,
where $\left|\lambda / \mu\right|$ denotes the number of cells of
the skew shape $\lambda / \mu$ (so
$\left|\lambda / \mu\right| = \left|\lambda\right| - \left|\mu\right|$).

It is customary to define $s_{\lambda / \mu}$ to be $0$ if we don't
have $\mu \subseteq \lambda$. This can also be seen by a literal
reading of the definition $s_{\lambda/\mu}:=\sum_{T} \xx^{\cont(T)}$,
as long as we understand that there are no column-strict tableaux of
shape $\lambda / \mu$ when $\lambda / \mu$ is not defined.

Clearly, every partition $\lambda$ satisfies $s_\lambda =
s_{\lambda / \varnothing}$.

It is easy to see that two partitions $\lambda$ and $\mu$ satisfy
$\mu \subseteq \lambda$ if and only if they satisfy
$\mu^t \subseteq \lambda^t$.
\end{remark}

% [DG][v25] Added preceding two paragraphs to this remark.

% [DG][v78] Added last sentence to the above remark.

\begin{exercise} \phantomsection
\label{exe.schur.rotate180}

\begin{itemize}

\item[(a)]
State and prove an analogue of
Proposition~\ref{prop.other-linear-orders.slambda} for skew Schur
functions.

\item[(b)]
Let $\lambda$, $\mu$, $\lambda'$ and $\mu'$ be partitions such that
$\mu \subseteq \lambda$ and $\mu' \subseteq \lambda'$. Assume that the
skew Ferrers diagram $\lambda' / \mu'$ can be obtained from the skew
Ferrers diagram $\lambda / \mu$ by a $180^{\circ}$ rotation.\footnote{For
example, this happens when $\lambda = \left(3,2\right)$,
$\mu = \left(1\right)$, $\lambda' = \left(5,4\right)$ and
$\mu' = \left(3,1\right)$.} Prove that
$s_{\lambda / \mu} = s_{\lambda' / \mu'}$.

\end{itemize}
\end{exercise}

% [DG][v14] Inserted the exercise above.

\begin{exercise}
\label{exe.schur.disconnected}
Let $\lambda$ and $\mu$ be two partitions, and let $k \in \NN$
be such that\footnote{As usual, we write $\nu_k$ for the
$k$-th entry of a partition $\nu$.}
$\mu_k \geq \lambda_{k+1}$. Let $F$ be the skew Ferrers
diagram $\lambda / \mu$. Let
$F_{\operatorname{rows}\leq k}$ denote the subset of $F$
consisting of all $\left(i, j\right) \in F$ satisfying
$i \leq k$. Let
$F_{\operatorname{rows} > k}$ denote the subset of $F$
consisting of all $\left(i, j\right) \in F$ satisfying
$i > k$.
Let $\alpha$ and $\beta$ be
two partitions such that $\beta \subseteq \alpha$ and such
that the skew Ferrers diagram
$\alpha / \beta$ can be obtained from
$F_{\operatorname{rows}\leq k}$
by parallel translation.
Let $\gamma$ and $\delta$ be two partitions such that
$\delta \subseteq \gamma$ and such that the skew
Ferrers diagram $\gamma / \delta$ can be obtained from
$F_{\operatorname{rows} > k}$ by parallel
translation.\footnote{Here is an example of the situation:
$\lambda = \left(6, 5, 5, 2, 2\right)$,
$\mu = \left(4, 4, 3, 1\right)$, $k = 3$ (satisfying
$\mu_k = \mu_3 = 3 \geq 2 = \lambda_4 = \lambda_{k+1}$),
$\alpha = \left(3, 2, 2\right)$,
$\beta = \left(1, 1\right)$,
$\gamma = \left(2, 2\right)$, and
$\delta = \left(1\right)$.} Prove that
$s_{\lambda / \mu}
= s_{\alpha / \beta} s_{\gamma / \delta}$.
\end{exercise}

% [DG][v32] Added the exercise above.

\begin{proposition}
\label{symm-comultiplication-formulas-prop}
The comultiplication $\Lambda \overset{\Delta}{\rightarrow} \Lambda \otimes \Lambda$
has the following effect on the symmetric functions discussed so far\footnote{
The abbreviated summation indexing $\sum_{i+j=n} t_{i,j}$ used here is intended to mean
\[
\sum_{ \substack{ (i,j) \in \NN^2; \\ i+j=n} } t_{i,j}.
\]}:
\begin{enumerate}
\item[ (i) ] $\Delta p_n = \one \otimes p_n + p_n \otimes \one$ for every $n \geq 1$,
that is, the power sums $p_n$ are primitive.
\item[ (ii) ] $\Delta e_n = \sum_{i+j=n} e_i \otimes e_j$ for every $n \in \NN$.
\item[ (iii) ] $\Delta h_n = \sum_{i+j=n} h_i \otimes h_j$ for every $n \in \NN$.
\item[ (iv) ] $\Delta s_{\lambda} = \sum_{\mu \subseteq \lambda} s_{\mu} \otimes s_{\lambda/\mu}$
for any partition $\lambda$.
\vspace{0.2pc}
\item[ (v) ] $\Delta s_{\lambda / \nu} = \sum\limits_{\substack{\mu \in \Par: \\ \nu \subseteq \mu \subseteq \lambda}} s_{\mu/\nu} \otimes s_{\lambda/\mu}$ for any partitions $\lambda$ and $\nu$.
\end{enumerate}

% [DG][v25] Added part (v).

\end{proposition}
\begin{proof}
Recall that $\Delta$ sends $f(\xx) \mapsto f(\xx,\yy)$, and one can easily check that
\begin{enumerate}
\item[ (i) ] $p_n(\xx,\yy) = \sum_i x_i^n + \sum_i y_i^n
= p_n(\xx) \cdot 1 + 1 \cdot p_n(\yy)$ for every $n \geq 1$;
\item[ (ii) ] $e_n(\xx,\yy) = \sum_{i+j=n} e_i(\xx)e_j(\yy)$ for every $n \in \NN$;
\item[ (iii) ] $h_n(\xx,\yy) = \sum_{i+j=n} h_i(\xx)h_j(\yy)$ for every $n \in \NN$.
\end{enumerate}
For assertion (iv), note that by \eqref{different-total-order-Schur-function}, one has
\begin{equation}
s_\lambda(\xx,\yy)=\sum_T (\xx,\yy)^{\cont(T)},
\label{pf.symm-comultiplication-formulas-prop.1}
\end{equation}
where the sum is over column-strict tableaux $T$ of shape $\lambda$ having entries
in the linearly ordered alphabet
%\footnote{
%There is a subtle point here:
%the linear ordering \eqref{two-variable-set-ordering} is not equivalent under
%permutations in $\Symm_{(infty)}$ to some other orderings, such as,
%$x_1 < y_1 < x_2 < y_2 < \cdots$.   Given $\lambda$ in $\Par_n$,
%it would be better to first assume one has
%\emph{finite} variable sets $\xx=(x_1,\ldots,x_n),
%\yy=(y_1,\ldots,y_n)$ in proving the identity
%$s_\lambda(\xx,\yy)=\sum_{\mu \subseteq \lambda} s_\mu(\xx) s_{\lambda/\mu}(\yy%)$
%as above;  the result then follows for infinite variable sets by
%the usual considerations.}
\begin{equation}
\label{two-variable-set-ordering}
x_1 < x_2 < \cdots < y_1 < y_2 < \cdots .
\end{equation}
\footnote{Here, $(\xx,\yy)^{\cont(T)}$ means the monomial
$\prod_{a \in \mathfrak{A}} a^{\left|T^{-1}\left(a\right)\right|}$,
where $\mathfrak{A}$ denotes the totally ordered alphabet
$x_1 < x_2 < \cdots < y_1 < y_2 < \cdots $. In other words,
$(\xx,\yy)^{\cont(T)}$ is the product of all entries of the tableau
$T$ (which is a monomial, since the entries of $T$ are not numbers
but variables).

The following rather formal argument should allay any doubts as to
why \eqref{pf.symm-comultiplication-formulas-prop.1} holds:
Let $\LLL$ denote the totally ordered set which is given by the
set $\left\{1,2,3,\ldots\right\}$ of positive integers, equipped
with the total order
$1 <_\LLL 3 <_\LLL 5 <_\LLL 7 <_\LLL \cdots <_\LLL 2 <_\LLL 4 <_\LLL 6 <_\LLL 8 <_\LLL \cdots$.
Then, \eqref{different-total-order-Schur-function} yields
$s_\lambda = \sum_T \xx^{\cont(T)}$
as $T$ runs through all $\LLL$-column-strict tableaux of shape $\lambda$.
Substituting the variables $x_1, y_1, x_2, y_2, x_3, y_3, \ldots$
for $x_1, x_2, x_3, x_4, x_5, x_6, \ldots$ (that is, substituting
$x_i$ for $x_{2i-1}$ and $y_i$ for $x_{2i}$) in this equality, we
obtain \eqref{pf.symm-comultiplication-formulas-prop.1}.}
For example,
\[
T=\begin{matrix}
x_1   & x_1 & x_1 & y_2 & y_5\\
x_2   & y_1 & y_1 &   & \\
y_2   & y_2 & y_4 &   & \\
y_4   & y_5
\end{matrix}
\]
is such a tableau of shape $\lambda=(5,3,3,2)$.  Note that the restriction of
$T$ to the alphabet $\xx$ gives a column-strict tableau $T_\xx$ of some shape
$\mu \subseteq \lambda$, and the restriction of $T$ to the alphabet $\yy$ gives
a column-strict tableau $T_\yy$ of shape $\lambda/\mu$
(e.g. for $T$ in the example above,
the tableau $T_\yy$ appeared in Example~\ref{skew-tableau-example}).
Consequently, one has
\begin{align}
s_{\lambda}(\xx,\yy) =& \sum_T \xx^{\cont(T_\xx)} \cdot \yy^{\cont(T_\yy)}
\nonumber \\
       =& \sum_{\mu \subseteq \lambda} \left( \sum_{T_\xx} \xx^{\cont(T_\xx)} \right)
                                      \left( \sum_{T_\yy} \yy^{\cont(T_\yy)} \right)
       = \sum_{\mu \subseteq \lambda} s_{\mu}(\xx)  s_{\lambda/\mu}(\yy).
\label{eq.Lambda.s.xy}
\end{align}

Assertion (v) is obvious in the case when we don't have
$\nu \subseteq \lambda$ (in fact, in this case, both
$s_{\lambda / \nu}$ and
$\sum\limits_{\substack{\mu \in \Par: \\ \nu \subseteq \mu \subseteq \lambda}}
s_{\mu/\nu} \otimes s_{\lambda/\mu}$
are clearly zero). In the remaining case, the proof of assertion (v)
is similar to that of (iv). (Of course,
the tableaux $T$ and $T_{\xx}$ now have skew shapes $\lambda/\nu$
and $\mu/\nu$, and instead of
\eqref{different-total-order-Schur-function},
we need to use the answer to Exercise~\ref{exe.schur.rotate180}(a).)
\end{proof}

% [DG][v25] Added footnote and proof of (v).

Notice that parts (ii) and (iii) of
Proposition~\ref{symm-comultiplication-formulas-prop} are particular
cases of part (iv), since $h_n = s_{(n)}$ and $e_n = s_{(1^n)}$.

% [DG][v46] Added previous sentence.

\begin{exercise} \phantomsection
\label{exe.Lambda.cocommutative}
\begin{itemize}
\item[(a)] Show that the Hopf algebra $\Lambda$ is cocommutative.
\item[(b)] Show that $\Delta s_{\lambda / \nu} = \sum\limits_{\substack{\mu \in \Par: \\ \nu \subseteq \mu \subseteq \lambda}} s_{\lambda/\mu} \otimes s_{\mu/\nu}$ for any partitions $\lambda$ and $\nu$.
\end{itemize}
\end{exercise}

% [DG][v25] Added the preceding exercise.

\begin{exercise}
\label{exe.schur.finite}
Let $n \in \NN$. Consider the finite variable set $\left(x_1, x_2, \ldots,
x_n\right)$ as a subset of $\xx = \left(x_1, x_2, x_3, \ldots\right)$.
Recall that $f \left( x_1, x_2, \ldots , x_n \right)$ is a
well-defined element of $\kk\left[x_1, x_2, \ldots, x_n\right]$
for every $f \in R\left(\xx\right)$ (and therefore also for
every $f \in \Lambda$, since $\Lambda \subset R\left(\xx\right)$),
according to Exercise~\ref{exe.Lambda.subs.fin}.
\begin{itemize}
\item[(a)] Show that any two partitions $\lambda$ and $\mu$ satisfy
\begin{vershort}
\[
s_{\lambda / \mu} \left(x_1, x_2, \ldots, x_n\right)
= \sum\limits_{\substack{T\text{ is a column-strict}\\ \text{tableau of shape } \lambda / \mu \text{;} \\ \text{all entries of } T \text{ belong} \\ \text{to } \left\{1,2,\ldots,n\right\}}}
\xx^{\cont\left(T\right)}.
\]
\end{vershort}
\begin{verlong}
\[
s_{\lambda / \mu} \left(x_1, x_2, \ldots, x_n\right)
= \sum\limits_{\substack{T\text{ is a column-strict}\\ \text{tableau of shape } \lambda / \mu \text{;} \\ \text{all entries of } T \text{ belong} \\ \text{to } \left\{1,2,\ldots,n\right\}}}
\left(x_1,x_2,\ldots,x_n\right)^{\cont\left(T\right)}.
\]
Here, $\left(x_1,x_2,\ldots,x_n\right)^{\cont\left(T\right)}$
is defined as
$\prod_{i=1}^{n} x_i ^{\left\vert T^{-1}\left( i\right) \right\vert }$
for any column-strict
tableau $T$ having the property that all entries of $T$ belong to $\left\{
1,2,\ldots,n\right\}  $.
\end{verlong}
\item[(b)] If $\lambda$ is a partition having more than $n$ parts%
\footnote{Recall that the \emph{parts} of a partition
are its nonzero entries.}, then
show that $s_\lambda \left(x_1, x_2, \ldots, x_n\right) = 0$.
\end{itemize}
\end{exercise}

% [DG][v25] Added the exercise above, as preparation for the bialternant
% section. More interesting parts could be added (characterizing the
% ideal of symmetric functions that vanish upon specializing to $n$
% variables in several way); do you think that would be good?

% [DG][v26] Replaced $\left(x_1,x_2,\ldots,x_n\right)^{\cont\left(T\right)}$
% by $\xx^{\cont\left(T\right)}$.

\begin{remark}
\label{rmk.many-symmetric-function-bases-prop.finite}
An analogue of Proposition~\ref{many-symmetric-function-bases-prop}
holds for symmetric polynomials in finitely many variables: Let
$N \in \NN$. Then, we have
\begin{itemize}
\item[(a)] The family
$\left\{ m_\lambda \left(x_1, x_2, \ldots, x_N\right) \right\}$,
as $\lambda$ runs through all partitions having length $\leq N$,
is a graded basis of the graded $\kk$-module
$\Lambda \left(x_1, x_2, \ldots, x_N\right)
= \kk \left[x_1, x_2, \ldots, x_N\right]^{\Symm_N}$.
\item[(b)] For any
partition $\lambda$ having length $> N$, we have
$m_\lambda \left(x_1, x_2, \ldots, x_N\right) = 0$.
\item[(c)] The family
$\left\{ e_\lambda \left(x_1, x_2, \ldots, x_N\right) \right\}$,
as $\lambda$ runs through all partitions whose parts are all
$\leq N$, is a graded basis of the graded $\kk$-module
$\Lambda \left(x_1, x_2, \ldots, x_N\right)$.
\item[(d)] The family
$\left\{ s_\lambda \left(x_1, x_2, \ldots, x_N\right) \right\}$,
as $\lambda$ runs through all partitions having length
$\leq N$, is a graded basis of the graded $\kk$-module
$\Lambda \left(x_1, x_2, \ldots, x_N\right)$.
\item[(e)] If $\QQ$ is a subring of $\kk$, then the family
$\left\{ p_\lambda \left(x_1, x_2, \ldots, x_N\right) \right\}$,
as $\lambda$ runs through all partitions having length
$\leq N$, is a graded basis of the graded $\kk$-module
$\Lambda \left(x_1, x_2, \ldots, x_N\right)$.
\item[(f)] If $\QQ$ is a subring of $\kk$, then the family
$\left\{ p_\lambda \left(x_1, x_2, \ldots, x_N\right) \right\}$,
as $\lambda$ runs through all partitions whose parts are all
$\leq N$, is a graded basis of the graded $\kk$-module
$\Lambda \left(x_1, x_2, \ldots, x_N\right)$.
\end{itemize}
Indeed, the claims (a) and (b) are obvious, while the
claims (c), (d) and (e) are proven similarly to our proof of
Proposition~\ref{many-symmetric-function-bases-prop}. We leave
the proof of (f) to the reader; this proof can also be found in
\cite[Theorem 10.86]{Loehr-bij}\footnote{See
\cite[Remark 10.76]{Loehr-bij} for why \cite[Theorem 10.86]{Loehr-bij}
is equivalent to our claim (f).}.

Claim (c) can be rewritten as follows: The elementary symmetric
polynomials $e_i \left(x_1, x_2, \ldots, x_N\right)$, for
$i \in \left\{1, 2, \ldots, N\right\}$, form an algebraically
independent generating set of
$\Lambda \left(x_1, x_2, \ldots, x_N\right)$. This is precisely
the well-known theorem (due to Gauss)\footnote{See, e.g.,
\cite[\textit{Symmetric Polynomials}, Theorem 5 and Remark 17]{Conrad-blurbs}
or \cite[\S 5.3]{vanderWaerden-1}
or \cite[Theorem 1]{BlumSmithCoskey}. In a slightly different
form, it also appears in \cite[Theorem (5.10)]{LaksovLascouxPragaczThorup}.}
that every symmetric polynomial in $N$
variables $x_1, x_2, \ldots, x_N$ can be written uniquely as a
polynomial in the $N$ elementary symmetric polynomials.
\end{remark}

% [DG][v39] Added remark above (it is strange to have a text on
% symmetric functions not mentioning the place every undergrad
% encounters them in). I have put it here since the meaning of
% evaluating at finitely many indeterminates is explained in the
% preceding exercise.

% [DG][v55] Added two more references for Gauss's theorem.





\subsection{The antipode, the involution $\omega$, and algebra generators}
\label{symm-antipode-section}

Since $\Lambda$ is a connected graded $\kk$-bialgebra, it will have an
antipode $\Lambda \overset{S}{\rightarrow} \Lambda$ making it a Hopf algebra
by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
However, we can identify $S$ more explicitly now.

% [DG][v62] "several issues will be resolved by identifying $S$ more explicitly now"
% --> "we can identify $S$ more explicitly now".


\begin{proposition}
\label{symm-antipode-and-h-basis}
Each of the families $\{e_n\}_{n=1,2,\ldots}$ and $\{h_n\}_{n=1,2,\ldots}$
are algebraically independent, and generate $\Lambda_\kk$ as a polynomial
algebra for any commutative ring $\kk$.  The same holds
for $\{p_n\}_{n=1,2,\ldots}$ when $\QQ$ is a subring of $\kk$.

Furthermore, the antipode $S$ acts as follows:
\begin{enumerate}
\item[ (i) ] $S(p_n) = -p_n$ for every positive integer $n$.
\item[ (ii) ] $S(e_n) = (-1)^n h_n$ for every $n \in \NN$.
\item[ (iii) ] $S(h_n) = (-1)^n e_n$ for every $n \in \NN$.
\end{enumerate}
\end{proposition}

\begin{proof}
The assertion that $\{e_n\}_{n \geq 1}$ are algebraically independent and generate
$\Lambda$ is equivalent to Proposition~\ref{many-symmetric-function-bases-prop}
asserting that $\{e_\lambda\}_{\lambda \in \Par}$ is a basis for $\Lambda$.
(Indeed, this equivalence follows from parts (b) and (c) of
Exercise~\ref{exe.alggen-via-span}, applied to $v_n = e_n$
and $v_\lambda = e_\lambda$.)
Thus, the former assertion is true.
If $\QQ$ is a subring of $\kk$, then a similar argument (using $p_n$ and
$p_\lambda$ instead of $e_n$ and $e_\lambda$)
shows that $\{p_n\}_{n \geq 1}$ are algebraically independent and generate
$\Lambda$.

The assertion $S(p_n) = -p_n$ follows from Proposition~\ref{antipode-on-primitives}
since $p_n$ is primitive by Proposition~\ref{symm-comultiplication-formulas-prop}(i).

For the remaining assertions, start with the easy generating function
identities\footnote{See the solution to Exercise~\ref{exe.H(t)-and-E(t)}
for the proofs of the identities.}
\begin{align}
\label{h-generating-function}
H(t)&:=\prod_{i=1}^\infty (1-x_it)^{-1}
     = 1+h_1(\xx)t+h_2(\xx)t^2+\cdots = \sum_{n \geq 0} h_n(\xx) t^n ; \\
\label{e-generating-function}
E(t)&:=\prod_{i=1}^\infty (1+x_it)
     = 1+e_1(\xx)t+e_2(\xx)t^2+\cdots = \sum_{n \geq 0} e_n(\xx) t^n .
\end{align}
These show that
\begin{equation}
\label{E(t)-H(t)-relation}
1=E(-t)H(t)=\left( \sum_{n \geq 0} e_n(\xx) (-t)^n \right)
            \left( \sum_{n \geq 0} h_n(\xx) t^n \right) .
\end{equation}
Hence, equating coefficients of powers of $t$, we see that for $n=0,1,2,\ldots$
we have
\begin{equation}
\label{e-h-relation}
\sum_{i+j=n} (-1)^i e_i h_j = \delta_{0,n}.
\end{equation}
This lets us recursively express the $e_n$ in terms of $h_n$ and vice-versa:
\begin{align}
e_0&=1=h_0 ; \label{e-h-recursions.1}\\
e_n&= e_{n-1} h_1 - e_{n-2} h_2 + e_{n-3} h_3 - \cdots ; \label{e-h-recursions.2}\\
h_n&= h_{n-1} e_1 - h_{n-2} e_2 + h_{n-3} e_3 - \cdots \label{e-h-recursions.3}
\end{align}
for $n =1,2,3,\ldots$
Now, let us use the algebraic independence of the
generators $\{e_n\}$ for $\Lambda$
to define a $\kk$-algebra endomorphism
\[
\begin{array}{rcl}
\Lambda &\overset{\omega}{\rightarrow} &\Lambda ,\\
e_n&\longmapsto&h_n \qquad \text{ (for positive integers $n$)}.
\end{array}
\]
Then,
\begin{align}
\omega\left(e_n\right) = h_n \qquad \text{ for each }
n \geq 0
\label{the-fundamental-involution-eh}
\end{align}
(indeed, this holds for $n > 0$ by definition, and for $n = 0$
because $\omega\left(e_0\right) = \omega\left(1\right) = 1 = h_0$).
Hence, the identical form of the two recursions \eqref{e-h-recursions.2}
and \eqref{e-h-recursions.3}
shows that
\begin{align}
\omega\left(h_n\right) = e_n \qquad \text{ for each }
n \geq 0
\label{the-fundamental-involution-he}
\end{align}
\footnote{%
Here is this argument in more detail:
We must show that $\omega\left(h_n\right) = e_n$ for each $n \geq 0$.
We shall prove this by strong induction on $n$.
Thus, we fix an $n \geq 0$, and assume as induction hypothesis that
$\omega\left(h_m\right) = e_m$ for each $m < n$.
We must then prove that $\omega\left(h_n\right) = e_n$.
If $n = 0$, then this is obvious; thus, assume WLOG that $n > 0$.
Hence,
\begin{align*}
\omega\left(h_n\right)
&= \omega\left( h_{n-1} e_1 - h_{n-2} e_2 + h_{n-3} e_3 - \cdots \right)
\qquad \left(\text{by \eqref{e-h-recursions.3}}\right) \\
&= \omega\left(h_{n-1}\right) \omega\left(e_1\right) - \omega\left(h_{n-2}\right) \omega\left(e_2\right) + \omega\left(h_{n-3}\right) \omega\left(e_3\right) - \cdots
\qquad \left(\text{since $\omega$ is a $\kk$-algebra homomorphism}\right) \\
&= e_{n-1} \omega\left(e_1\right) - e_{n-2} \omega\left(e_2\right) + e_{n-3} \omega\left(e_3\right) - \cdots
\qquad
\left(\text{since $\omega\left(h_m\right) = e_m$ for each $m < n$}\right) \\
&= e_{n-1} h_1 - e_{n-2} h_2 + e_{n-3} h_3 - \cdots
\qquad
\left(\text{since \eqref{the-fundamental-involution-eh} shows that
$\omega\left(e_m\right) = h_m$ for each $m \geq 0$}\right) \\
&= e_n \qquad \left(\text{by \eqref{e-h-recursions.2}}\right) ,
\end{align*}
as desired. This completes the induction step.}.
Combining this with \eqref{the-fundamental-involution-eh}, we
conclude that $\left(\omega\circ\omega\right) \left(e_n\right)
= e_n$ for each $n \geq 0$.
Therefore, the two $\kk$-algebra homomorphisms
$\omega \circ \omega : \Lambda \to \Lambda$ and $\id : \Lambda \to \Lambda$
agree on each element of the generating set $\{e_n\}$ of $\Lambda$.
Hence, they are equal, i.e., we have $\omega \circ \omega = \id$.
Therefore $\omega$ is an involution and therefore a $\kk$-algebra
automorphism of $\Lambda$.
This, in turn, yields that
the $\{h_n\}$ (being the images of the $\{e_n\}$ under this
automorphism) are another algebraically independent generating set
for $\Lambda$.

% [DG][v80] Added details to the above paragraph.

For the assertion about the antipode $S$ applied to $e_n$ or $h_n$,
note that the coproduct formulas for $e_n, h_n$ in
Proposition~\ref{symm-comultiplication-formulas-prop}(ii),(iii)
show that the defining relations for their
antipodes \eqref{antipode-defining-relation}
will in this case be
\begin{align*}
\sum_{i+j=n} S(e_i) e_j &= \delta_{0,n} = \sum_{i+j=n} e_i S(e_j) , \\
\sum_{i+j=n} S(h_i) h_j &= \delta_{0,n} = \sum_{i+j=n} h_i S(h_j)
\end{align*}
because $u \epsilon(e_n)=u \epsilon(h_n)=\delta_{0,n}$.
Comparing these to \eqref{e-h-relation}, one concludes via induction on $n$
that $S(e_n)=(-1)^n h_n$ and $S(h_n)=(-1)^n e_n$.
\end{proof}

The $\kk$-algebra endomorphism $\omega$ of $\Lambda$ defined in the
proof of Proposition~\ref{symm-antipode-and-h-basis} is
sufficiently important that we record its definition and a selection
of fundamental properties:

\begin{definition} \label{def.Lambda.omega}
Let \dfn{$\omega$} be the $\kk$-algebra homomorphism
\begin{equation}
\label{the-fundamental-involution-definition}
\begin{array}{rcl}
\Lambda &\to &\Lambda ,\\
e_n&\longmapsto&h_n \qquad \text{ (for positive integers $n$)}.
\end{array}
\end{equation}
This homomorphism $\omega$ is known as the
\dfn{fundamental involution} of $\Lambda$.
\end{definition}

\begin{proposition} \label{prop.Lambda.omega-and-S}
Consider the fundamental involution $\omega$ and the
antipode $S$ of the Hopf algebra $\Lambda$.

\begin{itemize}
\item[(a)] We have
\[
\omega \left( e_n \right) = h_n
\qquad \text{ for each $n \in \ZZ$.}
\]

\item[(b)] We have
\[
\omega \left( h_n \right) = e_n
\qquad \text{ for each $n \in \ZZ$.}
\]

\item[(c)] We have
\[
\omega \left( p_n \right) = \left(-1\right)^{n-1} p_n
\qquad \text{ for each positive integer $n$.}
\]

\item[(d)] The map $\omega$ is a $\kk$-algebra automorphism
of $\Lambda$ and an involution.

\item[(e)] If $n \in \NN$, then
\begin{equation}
\label{antipode-versus-omega-in-Sym}
S \left(f\right) = \left(-1\right)^n \omega \left(f\right)
\qquad \text{ for all } f \in \Lambda_n .
\end{equation}

\item[(f)] The map $\omega$ is a Hopf algebra automorphism
of $\Lambda$.

\item[(g)] The map $S$ is a Hopf algebra automorphism
of $\Lambda$.

\item[(h)] Every partition $\lambda$ satisfies the three
equalities
\begin{align}
\omega\left(  h_{\lambda}\right)   &  =e_{\lambda};
\label{eq.exe.Lambda.omegahlam.h}
\\
\omega\left(  e_{\lambda}\right)   &  =h_{\lambda};
\label{eq.exe.Lambda.omegahlam.e}
\\
\omega\left(  p_{\lambda}\right)   &  =\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }p_{\lambda}.
\label{eq.exe.Lambda.omegahlam.p}
\end{align}

\item[(i)] The map $\omega$ is an isomorphism of graded
$\kk$-modules.

\item[(j)] The family $\left(h_\lambda\right)_{\lambda \in \Par}$
is a graded basis of the graded $\kk$-module $\Lambda$.

\end{itemize}
\end{proposition}

\begin{exercise}
\label{exe.Lambda.omega-and-S}
Prove Proposition~\ref{prop.Lambda.omega-and-S}.

[\textbf{Hint:} Parts (a), (b) and (d) have been shown in
the proof of Proposition~\ref{symm-antipode-and-h-basis} above.
For part (e), let $D_{-1} : \Lambda \to \Lambda$ be the
$\kk$-algebra morphism sending each homogeneous $f \in \Lambda_n$
to $\left(-1\right)^n f$; then
argue that $\omega \circ D_{-1}$ and $S$ are two $\kk$-algebra
morphisms that agree on all elements
of the generating set $\{e_n\}$.
Derive part (c) from (d) and
Proposition~\ref{symm-antipode-and-h-basis}.
Part (h) then follows by multiplicativity.
For parts (f) and (g), check the coalgebra homomorphism
axioms on the $e_n$.
Parts (i) and (j) are easy consequences.]
\end{exercise}

% [DG][v80] Added the above proposition and exercise. All
% of it was already been said in some way, but now it is
% explicit and referenceable.

Proposition~\ref{prop.Lambda.omega-and-S}(e) shows that the
antipode $S$ on $\Lambda$ is, up to sign,
the same as the fundamental involution $\omega$.
Thus, studying $\omega$ is essentially equivalent to
studying $S$.

\begin{remark}
Up to now we have not yet derived how the involution $\omega$ and the
antipode $S$ act on (skew) Schur functions, which is quite beautiful:
If $\lambda$ and $\mu$ are partitions satisfying $\mu \subseteq \lambda$,
then
\begin{equation}
\label{skew-Schur-antipodes}
\begin{aligned}
\omega(s_{\lambda/\mu}) &= s_{\lambda^t/\mu^t} , \\
S(s_{\lambda/\mu})      &= (-1)^{|\lambda/\mu|} s_{\lambda^t/\mu^t} \\
\end{aligned}
\end{equation}
where recall that $\lambda^t$ is the transpose or conjugate partition
to $\lambda$, and $|\lambda/\mu|$ is the number of squares in
the skew diagram $\lambda/\mu$, that is, $|\lambda/\mu|=n-k$
if $\lambda, \mu$ lie in $\Par_n, \Par_k$ respectively.


We will deduce this later in three ways (once as an exercise using
the Pieri rules in Exercise~\ref{exe.pieri.omega},
once again using
skewing operators in Exercise~\ref{Darij's-skewing-exercise},
and for the third time from the action of
the antipode in $\Qsym$ on $P$-partition enumerators
in Corollary~\ref{antipode-on-skew-Schur-corollary}).
However, one could also deduce it immediately from our knowledge of the action of
$\omega$ and $S$ on $e_n, h_n$, if we were to prove the following
famous \emph{Jacobi-Trudi}\index{Jacobi-Trudi formula}
and \emph{dual Jacobi-Trudi}\index{dual Jacobi-Trudi formula}
formulas\footnote{The second of the following identities is also
known as the \dfn{von N\"agelsbach-Kostka identity}.}:

\begin{theorem}
\label{Jacobi-Trudi-formulae}
Skew Schur functions are the following polynomials in $\{h_n\}, \{e_n\}$:
\begin{align}
s_{\lambda/\mu} &=
\det( h_{\lambda_i - \mu_j - i + j} )_{i,j=1,2,\ldots,\ell} ,
\label{eq.jacobi-trudi.h} \\
s_{\lambda^t/\mu^t} &=
\det( e_{\lambda_i - \mu_j - i + j} )_{i,j=1,2,\ldots,\ell}
\label{eq.jacobi-trudi.e}
\end{align}
for any two partitions $\lambda$ and $\mu$ and any
$\ell \in \NN$ satisfying $\ell\left(\lambda\right) \leq \ell$ and
$\ell\left(\mu\right) \leq \ell$.
\end{theorem}

% [DG][v25] Replaced "$\lambda$ has at most $\ell$ nonzero parts"
% by "$\ell\left(\lambda\right) \leq \ell$ and
% $\ell\left(\mu\right) \leq \ell$" to avoid making a subtly false
% statement if $\mu$ is not contained in $\lambda$.

\noindent
Since we appear not to need these formulas in the sequel, we will
not prove them right away.
\begin{vershort}
However, a proof is sketched in the
solution to Exercise~\ref{exe.jacobi-trudi.macdonald},
\end{vershort}
\begin{verlong}
However, two proofs are sketched in the
solutions to Exercise~\ref{exe.jacobi-trudi} and to
Exercise~\ref{exe.jacobi-trudi.macdonald},
\end{verlong}
and various proofs are well-explained in
\cite[(39) and (41)]{Leeuwen-altSchur}, \cite[\S I.5]{Macdonald},
\cite[Thm. 7.1]{RosasSagan},
\cite[\S 4.5]{Sagan}, \cite[\S 7.16]{Stanley},
\cite[Thms. 3.5 and 3.5$^\ast$]{Wachs}; also, a
simultaneous generalization of both formulas is shown in
\cite[Theorem 11]{GesselViennot}, and three others in
\cite[1.9]{Remmel}, \cite[Thm. 3.1]{HamelGoulden}
and \cite{Jin-outsidenested}.
An elegant treatment of Schur polynomials taking the
Jacobi-Trudi formula \eqref{eq.jacobi-trudi.h}
as the \emph{definition} of $s_\lambda$
is given by Tamvakis \cite{Tamvakis}.
\end{remark}

% [DG][v14] I added the Gessel-Viennot reference -- while I didn't read it,
% it looks very interesting. The generalization there replaces the
% "weakly increasing" and "strictly increasing" conditions of semistandard
% Young tableaux by two relations satisfying fairly liberal constraints.
% Also, I added the van Leeuwen and Rosas-Sagan references; each of them
% proves more than just Jacobi-Trudi.

% [DG][v19] Added Wachs reference. She uses divided-difference operators
% to prove it. This might be a good thing to look into when writing the
% chapter about 0-Hecke algebras!





\subsection{Cauchy product, Hall inner product, self-duality}

The Schur functions, although a bit unmotivated right now, have
special properties with regard to the Hopf structure.  One property is
intimately connected with the following \dfn{Cauchy identity}.

\begin{theorem}
\label{thm.cauchy-identity}
In the power series ring $\kk\left[\left[\xx, \yy\right]\right] :=
\kk \left[\left[x_1, x_2, \ldots, y_1, y_2, \ldots\right]\right]$, one has
the following expansion:
\begin{equation}
\label{Cauchy-identity}
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy).
\end{equation}
\end{theorem}

\begin{remark}
The left hand side of \eqref{Cauchy-identity} is known as the
\dfn{Cauchy product}, or \dfn{Cauchy kernel}.

An equivalent version of the equality \eqref{Cauchy-identity}
is obtained by replacing each $x_i$ by $x_i t$, and writing the
resulting identity in the power series ring $R(\xx, \yy)[[t]]$:
\begin{equation}
\label{Cauchy-identity-with-t}
\prod_{i,j=1}^\infty (1-tx_i y_j)^{-1}
=\sum_{\lambda \in \Par} t^{|\lambda|} s_{\lambda}(\xx) s_\lambda(\yy).
\end{equation}
(Recall that $|\lambda|=\lambda_1 + \lambda_2 + \cdots +\lambda_\ell$
for any partition $\lambda = \left(\lambda_1,\lambda_2,\ldots,\lambda_\ell\right)$.)
\end{remark}

% [DG][v17] Replaced "a suitable completion of $R(\xx) \otimes R(\yy)$"
% by "$\kk[[\xx, \yy]]$" due to the elusiveness of $R(\xx) \otimes R(\yy)$.
% (I don't have a feeling for this tensor product, and I am not sure
% if anyone has; when $\kk$ is not a field, it may not inject into
% $R(\xx, \yy)$ by the canonical map, as shown in
% http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.pjm/1102959646 .
% A completion *might* deal with the issues, but there are too many
% possible candidates.)

\begin{proof}[Proof of Theorem~\ref{thm.cauchy-identity}]
We follow the standard combinatorial proof
(see \cite[\S 4.8]{Sagan},\cite[\S 7.11,7.12]{Stanley}), which rewrites the
left and right sides of \eqref{Cauchy-identity-with-t}, and then compares them with the
\emph{Robinson-Schensted-Knuth} (RSK) bijection.\footnote{The
RSK bijection has been introduced by Knuth \cite{Knuth-RSK},
where what we call ``biletters'' is referred to as
``two-line arrays''. The most important ingredient of this
algorithm -- the RS-insertion operation -- however goes back to
Schensted. The special case of the RSK algorithm where the
biword has to be a permutation (written in two-line notation)
and the two tableaux have to be \emph{standard} (i.e., each
of them has content $\left(1^n\right)$, where $n$ is the size
of their shape)
is the famous \dfn{Robinson-Schensted correspondence}
\cite{Leeuwen-rsk}. More about these algorithms can be found
in \cite[Chapter 3]{Sagan}, \cite[Chapter 5]{MendesRemmel},
\cite[\S 7.11-7.12]{Stanley}, \cite[Sections 10.9--10.22]{Loehr-bij},
\cite[Chapters 1 and A]{Fulton}, \cite[\S 3, \S 6]{BritzFomin}
and various other places.}
On the left side, expanding out each geometric series
\[
(1-tx_i y_j)^{-1}=1+tx_iy_j+(tx_iy_j)^2 +(tx_iy_j)^3+\cdots
\]
and thinking of $(x_i y_j)^m$ as $m$ occurrences of a
\emph{biletter}\footnote{A \dfn{biletter} here simply means a pair
of letters, written as a column vector. A \dfn{letter} means a
positive integer.} $\binom{i}{j}$,
we see that the left hand side can be rewritten as the sum of
$t^\ell \left(x_{i_1} y_{j_1}\right) \left(x_{i_2} y_{j_2}\right)
\cdots \left(x_{i_\ell} y_{j_\ell}\right)$
over all multisets
$\left\{\binom{i_1}{j_1}, \ldots, \binom{i_\ell}{j_\ell}\right\}_{\operatorname{multiset}}$
of biletters.
Order the biletters in such a multiset in the
lexicographic order $\leq_{lex}$, which is the total order on the
set of all biletters defined by
\[
\binom{i_1}{j_1} \leq_{lex} \binom{i_2}{j_2}
\quad \Longleftrightarrow \quad
\left( \text{we have } i_1 \leq i_2 \text{, and if }
i_1 = i_2 \text{, then } j_1 \leq j_2 \right).
\]
% which first checks if $i_1 \leq i_2$ and
% then if $i_1 = i_2$ checks if $j_1 \leq j_2$.
Defining a \dfn{biword} to be an array
$\binom{\ii}{\jj} = \binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$
in which the biletters are ordered
$\binom{i_1}{j_1} \leq_{lex} \cdots \leq_{lex} \binom{i_\ell}{j_\ell}$,
then the left side of \eqref{Cauchy-identity-with-t}
is the sum $\sum t^{\ell} \xx^{\cont(\ii)} \yy^{\cont(\jj)}$ over
all biwords $\binom{\ii}{\jj}$, where $\ell$ stands for the number of biletters in the biword.
On the right side, expanding out the Schur functions as sums of tableaux
gives $\sum_{(P,Q)} t^{\ell} \xx^{\cont(Q)} \yy^{\cont(P)}$ in which the sum is over all ordered
pairs $(P,Q)$ of column-strict tableaux \emph{having the same shape}\footnote{And
this shape should be the Ferrers diagram of a partition (not just a skew
diagram).}, with $\ell$ cells. (We shall refer to such pairs
as \emph{tableau pairs} from now on.)

The \dfn{Robinson-Schensted-Knuth algorithm} gives us a bijection between
the biwords $\binom{\ii}{\jj}$ and the tableau pairs $(P,Q)$, which has
the property that
\begin{align*}
\cont(\ii) &= \cont(Q),\\
\cont(\jj) &= \cont(P)
\end{align*}
(and that the length $\ell$ of the biword $\binom{\ii}{\jj}$
equals the size $\left|\lambda\right|$ of the common shape of
$P$ and $Q$; but this follows automatically from
$\cont(\ii) = \cont(Q)$). Clearly, once such a bijection is
constructed, the equality \eqref{Cauchy-identity-with-t} will
follow.

Before we define this algorithm, we introduce a simpler operation
known as \dfn{RS-insertion} (short for Robinson-Schensted insertion).
RS-insertion takes as input a column-strict tableau $P$ and a letter
$j$, and returns a new column-strict tableau $P'$ along with a corner
cell\footnote{A \dfn{corner cell} of a tableau or of a Ferrers
diagram is defined to be a cell $c$ which belongs to the tableau
(resp. diagram) but whose immediate neighbors to the east and to the
south don't. For example, the cell $\left(3,2\right)$ is a corner
cell of the Ferrers diagram of the partition $\left(3,2,2,1\right)$,
and thus also of any tableau whose shape is this partition.
But the cell $\left(2,2\right)$ is not a corner cell of this
Ferrers diagram, since its immediate neighbor to the south is
still in the diagram.}
$c$ of $P'$, which is constructed as follows:
Start out by setting $P' = P$. The letter $j$ tries to insert itself
into the first row of $P'$ by either bumping out the leftmost letter in the first row
strictly larger than $j$, or else placing itself at the right end of the row if
no such larger letter exists.  If a letter was bumped from the first row, this letter follows
the same rules to insert itself into the second row, and so on\footnote{Here,
rows are allowed to be empty -- so it is possible that a letter is bumped
from the last nonempty row of $P'$ and settles in the next, initially
empty, row.}.  This series of bumps must eventually come to an
end\footnote{since we can
only bump out entries from nonempty rows}.  At the end of the bumping,
the tableau $P'$ created has an extra corner cell not present in $P$.
If we call this corner cell $c$, then $P'$ (in its final form) and $c$
are what the RS-insertion operation returns. One says that $P'$ is the
result of \emph{inserting}\footnote{This terminology is
reminiscent of insertion into binary search trees, a basic operation
in theoretical computer science. This is more than superficial similarity;
there are, in fact, various analogies between Ferrers diagrams (and
their fillings) and unlabelled plane binary trees (resp. their labellings),
and one of them is the analogy between RS-insertion and binary search
tree insertion. See
\cite[\S 4.1]{Hivert}.} $j$ into the tableau $P$. It is straightforward
to see that this resulting filling $P'$ is a column-strict
tableau\footnote{Indeed, the reader can check that $P'$ remains a
column-strict tableau throughout the algorithm that defines
RS-insertion. (The only part of this that isn't obvious is showing that
when a letter $t$ bumped out of some row $k$ is inserted into row $k+1$,
the property that the letters increase strictly down columns is
preserved. Argue that the bumping-out of $t$ from row
$k$ was caused by the insertion of another letter $u < t$, and that the
cell of row $k+1$ into which $t$ is then being inserted is in the same
column as this $u$, or in a column further left than it.)}.

\begin{example}
To give an example
of this operation, let us insert the letter $j = 3$ into the
column-strict tableau
$\begin{matrix} 1 & 1 & 3 & 3 & 4 \\ 2 & 2 & 4 & 6 & \\
3 & 4 & 7 & & \\ 5 & & & & \end{matrix}$ (we are showing all
intermediate states of $P'$; the underlined letter is always
the one that is going to be bumped out at the next step):
\begin{align*}
\begin{matrix} 1 & 1 & 3 & 3 & \underline{4} \\ 2 & 2 & 4 & 6 & \\
3 & 4 & 7 & & \\ 5 & & & & \end{matrix}
\quad & \overset{\longmapsto}{\substack{\text{insert 3;} \\ \text{bump out 4}}} \quad
\begin{matrix} 1 & 1 & 3 & 3 & 3 \\ 2 & 2 & 4 & \underline{6} & \\
3 & 4 & 7 & & \\ 5 & & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 4;} \\ \text{bump out 6}}} \quad
\begin{matrix} 1 & 1 & 3 & 3 & 3 \\ 2 & 2 & 4 & 4 & \\
3 & 4 & \underline{7} & & \\ 5 & & & & \end{matrix} \\
\quad & \overset{\longmapsto}{\substack{\text{insert 6;} \\ \text{bump out 7}}} \quad
\begin{matrix} 1 & 1 & 3 & 3 & 3 \\ 2 & 2 & 4 & 4 & \\
3 & 4 & 6 & & \\ 5 & & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 7;} \\ \text{done}}} \quad
\begin{matrix} 1 & 1 & 3 & 3 & 3 \\ 2 & 2 & 4 & 4 & \\
3 & 4 & 6 & & \\ 5 & 7 & & & \end{matrix} .
\end{align*}
The last tableau in this sequence is the column-strict
tableau that is returned. The corner cell that is returned
is the second cell of the fourth row (the one containing $7$).
\end{example}

RS-insertion will be used as a step in the RSK algorithm; the
construction will rely on a simple fact known as the
\emph{row bumping lemma}. Let us first define the notion of
a \dfn{bumping path} (or \dfn{bumping route}):
If $P$ is a column-strict tableau, and $j$ is a letter, then
some letters are inserted into some cells when RS-insertion
is applied to $P$ and $j$. The sequence of these cells (in
the order in which they see letters inserted into them) is called
the \emph{bumping path} for $P$ and $j$. This bumping path
always ends with the corner cell $c$ which is returned by
RS-insertion. As an example, when
$j=1$ is inserted into the tableau $P$
shown below, the result $P'$ is shown with all entries on
the bumping path underlined:
\[
P=\begin{matrix}
1&1&2&2&3\\
2&2&4&4&\\
3&4&5& &\\
4&6&6& &
\end{matrix}
\qquad
\underset{j=1}{\overset{\text{insert}}{\longmapsto}}
\qquad
P'=\begin{matrix}
1&1&\underline{1}&2&3\\
2&2&\underline{2}&4&\\
3&4&\underline{4}& &\\
4&\underline{5}&6& & \\
\underline{6}& & & &
\end{matrix}
\]
A first simple observation about bumping paths is that
bumping paths \emph{trend weakly left} -- that is, if the
bumping path of $P$ and $j$ is $\left(c_1,c_2,\ldots,c_k\right)$,
then, for each $1\leq i < k$, the cell $c_{i+1}$ lies in the
same column as $c_i$ or in a column further left.\footnote{This
follows easily from the preservation of column-strictness during
RS-insertion.} A subtler property of bumping paths is the following
\dfn{row bumping lemma} (\cite[p. 9]{Fulton}):

\begin{statement}
\textbf{Row bumping lemma:} Let $P$ be a column-strict tableau,
and let $j$ and $j'$ be two letters. Applying RS-insertion to the
tableau $P$ and the letter $j$ yields a new column-strict tableau
$P'$ and a corner cell $c$. Applying RS-insertion to the tableau
$P'$ and the letter $j'$ yields a new column-strict tableau $P''$
and a corner cell $c'$.
\begin{enumerate}
\item[(a)] Assume that $j \leq j'$. Then, the bumping path
for $P'$ and $j'$
stays strictly to the right, within each row, of the bumping path
for $P$ and $j$.
The cell $c'$ (in which the bumping path for $P'$ and $j'$ ends) is
in the same row as the cell $c$ (in which the bumping path for $P$
and $j$ ends) or in a row further up; it is also in a column further
right than $c$.
\item[(b)] Assume instead that $j > j'$. Then, the bumping path
for $P'$ and $j'$
stays weakly to the left, within each row, of the bumping path
for $P$ and $j$.
The cell $c'$ (in which the bumping path for $P'$ and $j'$ ends) is
in a row further down than the cell $c$ (in which the bumping
path for $P$ and $j$ ends); it is also in the same column as $c$ or
in a column further left.
\end{enumerate}
\end{statement}
This lemma can be easily proven by induction over the
row.\footnote{We leave the details to the reader, only giving the
main idea for (a) (the proof of (b) is similar). To prove the
first claim of (a), it is enough to show that for every $i$,
if any letter is inserted into row $i$ during RS-insertion for
$P'$ and $j'$, then some letter is also inserted into row $i$
during RS-insertion for $P$ and $j$, and the former insertion
happens in a cell strictly to the right of the cell where the
latter insertion happens. This follows by induction over $i$.
In the induction step, we need to show that if, for a positive
integer $i$, we try to consecutively insert two letters $k$ and $k'$, in this order,
into the $i$-th row of a column-strict tableau, possibly bumping out
existing letters in the process, and if we have $k \leq k'$, then the
cell into which $k$ is inserted is strictly to the left of the cell
into which $k'$ is inserted, and the letter bumped out by the insertion
of $k$ is $\leq$ to the letter bumped out by the insertion of $k'$ (or
else the insertion of $k'$ bumps out no letter at all -- but it cannot
happen that $k'$ bumps out a letter but $k$ does not). This statement is
completely straightforward to check (by only studying the $i$-th
row). This way, the first claim of (a) is proven, and this entails
that the cell $c'$ (being the last cell of the bumping path for $P'$
and $j'$) is in the same row as the cell $c$ or in a row further up.
It only remains to show that $c'$ is in a column further right than $c$.
This follows by noticing that, if $k$ is the row in which the cell
$c'$ lies, then $c'$ is in a column further right than
the entry of the bumping path for $P$ and $j$ in row $k$ (by the
first claim of (a)), and this latter entry is further right than
or in the same column as the ultimate entry $c$ of this bumping path
(since bumping paths trend weakly left).}

We can now define the actual RSK algorithm.
Let $\binom{\ii}{\jj}$ be a biword.
Starting with the pair $(P_0,Q_0)=(\varnothing,\varnothing)$ and
$m=0$, the algorithm applies the following steps
(see Example~\ref{RSK-example} below):
\begin{itemize}
\item If $i_{m+1}$ does not exist (that is, $m$ is the length of
$\ii$), stop.
\item Apply RS-insertion to the column-strict tableau $P_m$
and the letter $j_{m+1}$ (the bottom letter of
$\binom{i_{m+1}}{j_{m+1}}$). Let $P_{m+1}$ be the resulting
column-strict tableau, and let $c_{m+1}$ be the resulting
corner cell.
\item Create $Q_{m+1}$ from $Q_m$ by adding the top letter
$i_{m+1}$ of $\binom{i_{m+1}}{j_{m+1}}$ to $Q_m$ in the cell
$c_{m+1}$ (which, as we recall, is the extra corner cell of
$P_{m+1}$ not present in $P_m$).
\item Set $m$ to $m+1$.
\end{itemize}
After all of the biletters have been thus processed,
the result of the RSK algorithm
is $(P_\ell,Q_\ell)=:(P,Q)$.



\begin{example}
\label{RSK-example}
The term in the expansion of the left side of \eqref{Cauchy-identity} corresponding to
\[
(x_1 y_2)^1(x_1 y_4)^1(x_2 y_1)^1(x_4 y_1)^1(x_4 y_3)^2 (x_5 y_2)^1
\]
is the biword
$\binom{\ii}{\jj}=\binom{1124445}{2411332}$, whose RSK algorithm goes
as follows:
\[
\begin{array}{rcl|rcl}
P_0&= &\varnothing & Q_0 &=& \varnothing \\
 & & & & &\\
P_1&= &\begin{matrix} 2 \end{matrix}
 & Q_1 &= &\begin{matrix} 1 \end{matrix} \\
 & & & & &\\
P_2&= &\begin{matrix} 2&4 \end{matrix}
 & Q_2 &= &\begin{matrix} 1&1  \end{matrix} \\
 & & & & &\\
P_3&= &\begin{matrix} 1&4\\ 2 \end{matrix}
 & Q_3 &= &\begin{matrix} 1&1\\ 2 \end{matrix} \\
 & & & & &\\
P_4&= &\begin{matrix} 1&1\\ 2&4 \end{matrix}
 & Q_4 &= &\begin{matrix} 1&1\\ 2&4 \end{matrix} \\
 & & & & &\\
P_5&= &\begin{matrix} 1&1&3\\ 2&4& \end{matrix}
 & Q_5 &= &\begin{matrix} 1&1&4\\ 2&4& \end{matrix} \\
 & & & & &\\
P_6&= &\begin{matrix} 1&1&3&3\\ 2&4& & \end{matrix}
 & Q_6 &= &\begin{matrix} 1&1&4&4\\ 2&4& & \end{matrix} \\
 & & & & &\\
P:=P_7&= &\begin{matrix} 1&1&2&3\\ 2&3& & \\ 4& & & \end{matrix}
 & Q:=Q_7 &= &\begin{matrix} 1&1&4&4\\ 2&4& & \\5& & & \end{matrix} \\
\end{array}
\]
\end{example}

The bumping rule obviously maintains the property
that $P_m$ is a column-strict tableau of some Ferrers shape throughout.
It should be clear that $(P_m,Q_m)$ have the same shape at each stage.
Also, the construction of $Q_m$ shows that it is at least weakly
increasing in rows and weakly increasing in columns throughout.
What is perhaps least clear is
that $Q_m$ remains strictly increasing down columns.
That is, when one has a string of equal
letters on top $i_m=i_{m+1}=\cdots=i_{m+r}$,
so that on bottom one bumps in $j_m \leq j_{m+1} \leq \cdots \leq j_{m+r}$,
one needs to know that the new cells form a \dfn{horizontal strip}, that is,
no two of them lie in the same column\footnote{Actually, each of these
new cells (except for the first one) is in a column further right
than the previous one. We will use this stronger fact further below.}.
This follows from (the last claim of) part (a) of the row bumping lemma.
Hence, the result $\left(P, Q\right)$ of the RSK algorithm
is a tableau pair.


To see that the RSK map is a bijection, we show how to
recover $\binom{\ii}{\jj}$ from $(P,Q)$.  This is done by
\dfn{reverse bumping} from $(P_{m+1},Q_{m+1})$ to recover
both the biletter $\binom{i_{m+1}}{j_{m+1}}$ and the tableaux $(P_m,Q_m)$, as follows.
Firstly, $i_{m+1}$ is the maximum entry of $Q_{m+1}$, and $Q_m$ is obtained by removing
the rightmost occurrence of this letter $i_{m+1}$ from $Q_{m+1}$.
\footnote{It necessarily has to be the rightmost occurrence, since
(according to the previous footnote) the cell into which
$i_{m+1}$ was filled at the step from $Q_m$ to $Q_{m+1}$ lies further
right than any existing cell of $Q_m$ containing the letter $i_{m+1}$.}
To produce $P_m$ and $j_{m+1}$, find the position of the
rightmost occurrence of $i_{m+1}$ in $Q_{m+1}$,
and start \emph{reverse bumping} in $P_{m+1}$ from the entry in
this same position, where reverse bumping an
entry means inserting it into one row higher by having it bump
out the rightmost entry which is strictly smaller.\footnote{Let
us give a few more details on this ``reverse bumping'' procedure.
Reverse bumping (also known as \dfn{RS-deletion} or
\dfn{reverse RS-insertion}) is an operation which takes a
column-strict tableau $P'$ and a corner cell $c$ of $P'$,
and constructs a column-strict tableau $P$ and a letter $j$
such that RS-insertion for $P$ and $j$ yields $P'$ and $c$.
It starts by setting $P = P'$, and removing the entry in the
cell $c$ from $P$. This removed entry is then denoted by $k$,
and is inserted into the row of $P$ above $c$, bumping out the
rightmost entry which is smaller than $k$. The letter which is
bumped out -- say, $\ell$ --, in turn, is inserted into the row
above it, bumping out the rightmost entry which is smaller than
$\ell$. This procedure continues in the same way until an entry
is bumped out of the first row (which will eventually happen).
The reverse bumping operation returns the
resulting tableau $P$ and the entry which is bumped out of the
first row.
\par
It is straightforward to check that the reverse bumping
operation is well-defined (i.e., $P$ does stay a column-strict
tableau throughout the procedure) and is the inverse of the
RS-insertion operation. (In fact, these two operations undo each
other step by step.)}
The entry bumped out of the
first row is $j_{m+1}$, and the resulting tableau is $P_{m}$.

% [DG][v46] More details on reverse bumping.

Finally, to see that the RSK map is surjective, one needs to show
that the reverse bumping procedure can be applied to any pair
$(P,Q)$ of column-strict tableaux of the same shape, and will
result in a (lexicographically ordered) biword $\binom{\ii}{\jj}$.
We leave this verification to the reader.\footnote{It is easy to
see that repeatedly applying reverse bumping to $(P,Q)$ will
result in a sequence
$\binom{i_\ell}{j_\ell}, \binom{i_{\ell-1}}{j_{\ell-1}},
\ldots, \binom{i_1}{j_1}$ of biletters such that applying the
RSK algorithm to
$\binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$ gives back
$(P,Q)$. The question is why we have
$\binom{i_1}{j_1} \leq_{lex} \cdots \leq_{lex} \binom{i_\ell}{j_\ell}$.
Since the chain of inequalities $i_1 \leq i_2 \leq \cdots \leq i_\ell$
is clear from the choice of entry to reverse-bump, it only remains
to show that for every string $i_m=i_{m+1}=\cdots=i_{m+r}$
of equal top letters, the corresponding bottom letters
weakly increase (that is,
$j_m \leq j_{m+1} \leq \cdots \leq j_{m+r}$). One way to see
this is the following:
\par
Assume the contrary; i.e., assume that
the bottom letters corresponding
to some string $i_m=i_{m+1}=\cdots=i_{m+r}$ of equal top
letters do not weakly increase.
Thus, $j_{m+p} > j_{m+p+1}$ for some
$p \in \left\{0, 1, \ldots, r-1\right\}$.
Consider this $p$.
\par
Let us consider the cells containing the equal letters
$i_m=i_{m+1}=\cdots=i_{m+r}$ in the tableau $Q_{m+r}$.
Label these cells as $c_m, c_{m+1}, \ldots, c_{m+r}$
from left to right (noticing that no two of them lie
in the same column, since $Q_{m+r}$ is column-strict).
By the definition of reverse bumping, the first entry
to be reverse bumped from $P_{m+r}$ is the entry in
position $c_{m+r}$ (since this is the rightmost
occurrence of the letter $i_{m+r}$ in $Q_{m+r}$); then, the next entry
to be reverse bumped is the one in position $c_{m+r-1}$,
etc., moving further and further left.
Thus, for each $q \in \left\{0, 1, \ldots, r\right\}$,
the tableau $P_{m+q-1}$ is obtained from $P_{m+q}$ by
reverse bumping the entry in position $c_{m+q}$.
Hence, conversely, the tableau $P_{m+q}$ is obtained
from $P_{m+q-1}$ by RS-inserting the entry $j_{m+q}$,
which creates the corner cell $c_{m+q}$.
\par
But recall that $j_{m+p} > j_{m+p+1}$.
Hence, part (b) of the row bumping lemma (applied to
$P_{m+p-1}$, $j_{m+p}$, $j_{m+p+1}$, $P_{m+p}$,
$c_{m+p}$, $P_{m+p+1}$ and $c_{m+p+1}$
instead of $P$, $j$, $j'$, $P'$, $c$, $P''$ and $c'$)
shows that the cell $c_{m+p+1}$ is in the same
column as the cell $c_{m+p}$ or in a column further
left.
But this contradicts the fact that the cell $c_{m+p+1}$ is
in a column further right than the cell $c_{m+p}$ (since
we have labeled our cells as
$c_m, c_{m+1}, \ldots, c_{m+r}$ from left to right,
and no two of them lied in the same column).
This contradiction completes our proof.}
\end{proof}

% [DG][v45] Revamped the above proof, adding significant amounts of
% detail. What triggered the rewriting was my realization that both
% parts of the row bumping lemma are needed (as opposed to only the
% first one); but now that I'm doing it I've also added an example
% for RS-insertion, separated RS-insertion from the whole RSK
% algorithm (which is no longer called "RSK insertion" to remove
% confusion), inserted some references, and moved out the row
% bumping lemma from the niche it occupied in the proof of RSK.
% I hope this is clearer now...

This is by far not the only known proof of Theorem~\ref{thm.cauchy-identity}.
Two further proofs will be sketched in Exercise~\ref{exe.pieri.cauchy}
and Exercise~\ref{exe.cauchy.cauchy}.



Before we move on to extracting identities in $\Lambda$ from Theorem
\ref{thm.cauchy-identity}, let us state (as an exercise) a simple technical
fact that will be useful:

\begin{exercise}
\label{exe.Lambda.infinite-indep}
Let $\left( q_\lambda \right) _{\lambda \in \Par}$
be a basis of the $\kk$-module $\Lambda$.
Assume that for each partition $\lambda$, the element
$q_\lambda \in \Lambda$
is homogeneous of degree $\left\vert \lambda\right\vert $.

\begin{enumerate}
\item[(a)] If two families $\left(  a_{\lambda}\right)  _{\lambda
\in \Par }\in\kk^{ \Par }$ and $\left(
b_{\lambda}\right)  _{\lambda\in \Par }\in\kk%
^{ \Par }$ satisfy%
\begin{equation}
\sum_{\lambda\in \Par }a_{\lambda}q_{\lambda}\left(
\xx\right)  =\sum_{\lambda\in \Par }b_{\lambda}q_{\lambda
}\left(  \xx\right)  \label{eq.exe.Lambda.infinite-indep.a.ass}%
\end{equation}
in $\kk\left[  \left[  \xx\right]  \right]  $, then $\left(
a_{\lambda}\right)  _{\lambda\in \Par }=\left(  b_{\lambda
}\right)  _{\lambda\in \Par }$.\ \ \ \ \footnote{Note that this
does not immediately follow from the linear independence of the basis $\left(
q_{\lambda}\right)  _{\lambda\in \Par }$. Indeed, linear
independence would help if the sums in
(\ref{eq.exe.Lambda.infinite-indep.a.ass}) were finite, but they are not. A
subtler argument (involving the homogeneity of the $q_{\lambda}$) thus has to
be used.}

\item[(b)] Consider a further infinite family $\mathbf{y}=\left(  y_{1}%
,y_{2},y_{3},\ldots\right)  $ of indeterminates (disjoint from $\xx$).
If two families $\left(  a_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)
\in\operatorname{Par}^{2}}\in\kk^{\operatorname{Par}^{2}}$ and $\left(
b_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}%
\in\kk^{\operatorname{Par}^{2}}$ satisfy
\begin{equation}
\sum_{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}a_{\mu,\nu}q_{\mu
}\left( \xx \right)  q_{\nu}\left( \yy \right)
=\sum_{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}b_{\mu,\nu}q_{\mu
}\left( \xx \right)  q_{\nu}\left( \yy \right)
\label{eq.exe.Lambda.infinite-indep.b.ass}
\end{equation}
in $\kk\left[  \left[  \xx,\mathbf{y}\right]  \right]  $, then
$\left(  a_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}%
^{2}}=\left(  b_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in
\operatorname{Par}^{2}}$.

\item[(c)] Consider a further infinite family $\mathbf{z}=\left(  z_{1}%
,z_{2},z_{3},\ldots\right)  $ of indeterminates (disjoint from $\xx$
and $\mathbf{y}$). If two families $\left(  a_{\lambda,\mu,\nu}\right)
_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}\in\kk%
^{\operatorname{Par}^{3}}$ and $\left(  b_{\lambda,\mu,\nu}\right)  _{\left(
\mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}\in\kk%
^{\operatorname{Par}^{3}}$ satisfy
\begin{equation}
\sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}a_{\lambda
,\mu,\nu}q_{\mu}\left(  \xx\right)  q_{\nu}\left(  {\mathbf{y}%
}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  =\sum_{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3}}b_{\lambda,\mu,\nu}q_{\mu
}\left(  \xx\right)  q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda
}\left(  {\mathbf{z}}\right)  \label{eq.exe.Lambda.infinite-indep.c.ass}%
\end{equation}
in $\kk\left[  \left[  \xx,\mathbf{y},\mathbf{z}\right]
\right]  $, then $\left(  a_{\lambda,\mu,\nu}\right)  _{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3}}=\left(  b_{\lambda,\mu,\nu
}\right)  _{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$.
\end{enumerate}
\end{exercise}

\begin{remark}
Clearly, for any $n\in \NN$, we can state an analogue of Exercise
\ref{exe.Lambda.infinite-indep} for $n$ infinite families $\xx%
_{i}=\left(  x_{i,1},x_{i,2},x_{i,3},\ldots\right)  $ of indeterminates (with
$i\in\left\{  1,2,\ldots,n\right\}  $). The three parts of Exercise
\ref{exe.Lambda.infinite-indep} are the particular cases of this analogue for
$n=1$, for $n=2$ and for $n=3$. We have shied away from stating this analogue
in full generality
because these particular cases are the only ones we will need.
\end{remark}





\begin{corollary}
\label{self-duality-corollary}
In the Schur function basis $\{s_\lambda\}$ for $\Lambda$, the
structure constants for multiplication and comultiplication are the same, that is,
if one defines scalars
$c^\lambda_{\mu,\nu}, \hat{c}^\lambda_{\mu,\nu}$ via the unique expansions
\begin{align}
s_\mu s_\nu &= \sum_{\lambda} c^\lambda_{\mu,\nu} s_\lambda ,
\label{symm-structure-constants.1} \\
\Delta(s_\lambda) &= \sum_{\mu, \nu} \hat{c}^\lambda_{\mu,\nu} s_\mu \otimes s_\nu ,
\label{symm-structure-constants.2}
\end{align}
then $c^\lambda_{\mu,\nu}=\hat{c}^\lambda_{\mu,\nu}$.
\end{corollary}
\begin{proof}
Work in the ring $\kk\left[\left[\xx , \yy , \zz\right]\right]$, where
$\yy = \left(y_1, y_2, y_3, \ldots\right)$ and $\zz = \left(z_1, z_2,
z_3, \ldots\right)$ are two new sets of variables.
The identity \eqref{Cauchy-identity}
lets one interpret both $c^\lambda_{\mu,\nu}, \hat{c}^\lambda_{\mu,\nu}$
as the coefficient\footnote{Let us explain why speaking of
coefficients makes sense here:

We want to use the fact that if a power series
$f \in \kk\left[\left[\xx, \yy, \zz\right]\right]$ is written
in the form $f = \sum_{\left(\mu, \nu, \lambda\right) \in \Par^3}
a_{\lambda, \mu, \nu} s_\mu\left(\xx\right) s_\nu\left(\yy\right)
s_\lambda\left(\zz\right)$ for some coefficients
$a_{\lambda, \mu, \nu} \in \kk$, then these coefficients
$a_{\lambda, \mu, \nu}$ are uniquely determined by $f$.
But this fact is precisely the claim of
Exercise~\ref{exe.Lambda.infinite-indep}(c) above
(applied to $q_\lambda = s_\lambda$).}
 of $s_\mu(\xx) s_\nu(\yy) s_\lambda(\zz)$ in
the product
\begin{align*}
\prod_{i,j=1}^\infty (1-x_i z_j)^{-1}
\prod_{i,j=1}^\infty (1-y_i z_j)^{-1}
&\overset{\eqref{Cauchy-identity}}{=} \left( \sum_\mu s_\mu(\xx) s_\mu(\zz) \right)
\left( \sum_\nu s_\nu(\yy) s_\nu(\zz) \right) \\
&=\sum_{\mu,\nu} s_\mu(\xx) s_\nu(\yy) \cdot s_\mu(\zz) s_\nu(\zz) \\
&=
\sum_{\mu,\nu} s_\mu(\xx) s_\nu(\yy) \left( \sum_\lambda c^\lambda_{\mu,\nu} s_\lambda(\zz) \right)
\end{align*}
since, regarding $x_1,x_2,\ldots,y_1,y_2,\ldots$ as lying in a single variable set
$(\xx,\yy)$, separate from the variables $\zz$, the Cauchy identity \eqref{Cauchy-identity}
expands the same product as
\begin{align*}
\prod_{i,j=1}^\infty (1-x_i z_j)^{-1}
\prod_{i,j=1}^\infty (1-y_i z_j)^{-1}
&=\sum_{\lambda} s_\lambda(\xx,\yy) s_\lambda(\zz)\\
&= \sum_\lambda \left( \sum_{\mu,\nu} \hat{c}^\lambda_{\mu,\nu}
     s_\mu(\xx) s_\nu(\yy) \right) s_\lambda(\zz).
\end{align*}
\end{proof}

% [DG][v17] Added footnote.

% [DG][v25] Added first sentence to the proof.

\begin{definition}
The coefficients $c^\lambda_{\mu,\nu}=\hat{c}^\lambda_{\mu,\nu}$ appearing in
the expansions \eqref{symm-structure-constants.1}
and \eqref{symm-structure-constants.2}
are called \dfn{Littlewood-Richardson coefficients}.
\end{definition}

\begin{remark}
\label{third-L-R-interpretation-remark}
We will interpret $c^\lambda_{\mu,\nu}$
combinatorially in Section~\ref{bialternant-section}. By now,
however, we can already prove some properties of these
coefficients:

We have
\begin{equation}
\label{eq.third-L-R-interpretation-remark.commutativity}
c^\lambda_{\mu,\nu} = c^\lambda_{\nu,\mu}
\qquad \qquad \text{ for all } \lambda, \mu, \nu \in \Par
\end{equation}
(by comparing coefficients in
$\sum_{\lambda} c^\lambda_{\mu,\nu} s_\lambda = s_\mu s_\nu
= s_\nu s_\mu = \sum_{\lambda} c^\lambda_{\nu,\mu} s_\lambda$).
Furthermore, let
$\lambda$ and $\mu$ be two partitions (not necessarily
satisfying $\mu \subseteq \lambda$).
Comparing the expansion
\[
s_\lambda(\xx,\yy) = \Delta(s_\lambda)
   = \sum_{\mu,\nu} c^\lambda_{\mu,\nu} s_\mu(\xx) s_\nu(\yy)
   = \sum_{\mu \in \Par} \left( \sum_{\nu \in \Par} c^\lambda_{\mu,\nu} s_\nu(\yy) \right) s_\mu(\xx)
\]
with
\[
s_\lambda(\xx,\yy) = \sum_{\mu \subseteq \lambda} s_\mu(\xx) s_{\lambda/\mu}(\yy)
= \sum_{\mu \in \Par} s_\mu(\xx) s_{\lambda/\mu}(\yy)
\]
\footnote{In the last equality, we removed the condition
$\mu \subseteq \lambda$ on the addends of the sum; this does not
change the value of the sum (because we have $s_{\lambda/\mu} = 0$
whenever we don't have $\mu \subseteq \lambda$).},
one concludes that
\[
\sum_{\mu \in \Par} \left( \sum_{\nu \in \Par} c^\lambda_{\mu,\nu} s_\nu(\yy) \right) s_\mu(\xx)
= \sum_{\mu \in \Par} s_\mu(\xx) s_{\lambda/\mu}(\yy)
= \sum_{\mu \in \Par} s_{\lambda/\mu}(\yy) s_\mu(\xx) .
\]
Treating the indeterminates $\yy$ as constants, and
comparing coefficients before $s_\mu(\xx)$ on both sides
of this equality\footnote{``Comparing coefficients''
means applying Exercise~\ref{exe.Lambda.infinite-indep}(a)
to $q_\lambda = s_\lambda$
in this case (although the base ring $\kk$ is now replaced
by $\kk\left[\left[\yy\right]\right]$, and the index
$\mu$ is used instead of $\lambda$, since $\lambda$ is
already taken).}, we arrive at
another standard interpretation for $c^\lambda_{\mu,\nu}$:
\[
s_{\lambda/\mu} = \sum_{\nu} c^\lambda_{\mu,\nu} s_\nu .
\]
% whenever $\lambda$ and $\mu$ are two partitions (not necessarily
% satisfying $\mu \subseteq \lambda$).
In particular, $c^\lambda_{\mu,\nu}$ vanishes unless
$\mu \subseteq \lambda$. Consequently,
$c^\lambda_{\mu,\nu}$ vanishes unless $\nu \subseteq \lambda$
as well (since $c^\lambda_{\mu,\nu} = c^\lambda_{\nu,\mu}$)
and furthermore vanishes unless the equality
$\left|\mu\right|+\left|\nu\right| = \left|\lambda\right|$
holds\footnote{In fact, this is clear when we don't have
$\mu \subseteq \lambda$. When we do have $\mu \subseteq \lambda$,
this follows from observing that
$s_{\lambda/\mu} \in \Lambda_{\left|\lambda/\mu\right|}$
has zero coefficient before $s_\nu$ whenever
$\left|\mu\right|+\left|\nu\right| \neq \left|\lambda\right|$.}.
Altogether, we conclude that $c^\lambda_{\mu,\nu}$ vanishes unless
$\mu, \nu \subseteq \lambda$ and $|\mu|+|\nu|=|\lambda|$.
\end{remark}

% [DG][v25] Added many details to the above remark.

\begin{exercise}
\label{exe.Lambda.bialg-c}
Show that any four partitions $\kappa$, $\lambda$, $\varphi$ and
$\psi$ satisfy
\begin{vershort}
\[
\sum_{\rho \in \Par} c^\rho_{\kappa, \lambda} c^\rho_{\varphi, \psi}
= \sum_{\left(\alpha , \beta , \gamma , \delta\right) \in \Par^4}
c^\lambda_{\beta, \delta} c^\varphi_{\alpha, \beta}
c^\psi_{\gamma, \delta}.
\]
\end{vershort}
\begin{verlong}
\[
\sum_{\rho \in \Par} c^\rho_{\kappa, \lambda} c^\rho_{\varphi, \psi}
= \sum_{\alpha \in \Par} \sum_{\beta \in \Par} \sum_{\gamma \in \Par}
\sum_{\delta \in \Par} c^\kappa_{\alpha, \gamma}
c^\lambda_{\beta, \delta} c^\varphi_{\alpha, \beta}
c^\psi_{\gamma, \delta}.
\]
\end{verlong}
\end{exercise}

% [DG][v25] Added the above exercise. Is it too scary? (It is the
% bialgebra axiom, written in Schur coordinates, of course.)

\begin{exercise} \phantomsection
\label{exe.cauchy.skew}
\begin{itemize}
\item[(a)] For any partition $\mu$, prove that
\[
\sum_{\lambda \in \Par} s_\lambda\left(\xx\right) s_{\lambda/\mu}\left(\yy\right)
= s_\mu\left(\xx\right) \cdot \prod_{i,j=1}^\infty \left(1-x_iy_j\right)^{-1}
\]
in the power series ring $\kk\left[\left[\xx, \yy\right]\right]
= \kk\left[\left[x_1, x_2, x_3, \ldots, y_1, y_2, y_3, \ldots\right]\right]$.
\item[(b)] Let $\alpha$ and $\beta$ be two partitions. Show that
\[
\sum_{\lambda \in \Par} s_{\lambda/\alpha}\left(\xx\right) s_{\lambda/\beta}\left(\yy\right)
= \left( \sum_{\rho \in \Par} s_{\beta/\rho}\left(\xx\right) s_{\alpha/\rho}\left(\yy\right) \right)
 \cdot \prod_{i,j=1}^\infty \left(1-x_iy_j\right)^{-1}
\]
in the power series ring $\kk\left[\left[\xx, \yy\right]\right]
= \kk\left[\left[x_1, x_2, x_3, \ldots, y_1, y_2, y_3, \ldots\right]\right]$.
\end{itemize}

[\textbf{Hint:} For (b), expand the product
\[
\prod_{i,j=1}^\infty \left(1-x_i y_j\right)^{-1}
\prod_{i,j=1}^\infty \left(1-x_i w_j\right)^{-1}
\prod_{i,j=1}^\infty \left(1-z_i y_j\right)^{-1}
\prod_{i,j=1}^\infty \left(1-z_i w_j\right)^{-1}
\]
in the power series ring
$\kk\left[\left[x_1,x_2,x_3, \ldots ,y_1,y_2,y_3, \ldots ,z_1,z_2,z_3, \ldots ,w_1,w_2,w_3, \ldots \right]\right]$
in two ways: once by applying Theorem~\ref{thm.cauchy-identity} to the
two variable sets $\left(\zz, \xx\right)$ and $\left(\mathbf{w}, \yy\right)$
and then using \eqref{eq.Lambda.s.xy}; once again by applying
\eqref{Cauchy-identity} to the two variable sets $\zz$ and $\mathbf{w}$ and
then applying Exercise~\ref{exe.cauchy.skew}(a) twice.]
\end{exercise}

The statement of Exercise~\ref{exe.cauchy.skew}(b) is known as the
\dfn{skew Cauchy identity}, and appears in
Sagan-Stanley \cite[Cor. 6.12]{SaganStanley},
Stanley \cite[exercise 7.27(c)]{Stanley}
and Macdonald \cite[\S I.5, example 26]{Macdonald}; it seems to be due to
Zelevinsky. It generalizes the statement of Exercise~\ref{exe.cauchy.skew}(a),
which in turn is a generalization of Theorem~\ref{thm.cauchy-identity}.

% [DG][v25] Added the above exercise (following Macdonald mostly).

\begin{definition}
\label{def.hall-inner-product}
Define the \dfn{Hall inner product} on $\Lambda$ to be the $\kk$-bilinear
form $(\cdot,\cdot)$ which makes $\{s_\lambda\}$ an orthonormal basis,
that is, $(s_\lambda, s_\nu)=\delta_{\lambda,\nu}$.
\end{definition}

\begin{exercise} \phantomsection
\label{exe.Lambda.hall.gr}
\begin{itemize}
\item[(a)] If $n$ and $m$ are two distinct nonnegative integers, and if
$f \in \Lambda_n$ and $g \in \Lambda_m$, then show that
$\left(f, g\right) = 0$.
\item[(b)] If $n \in \NN$ and $f \in \Lambda_n$, then prove that
$\left(h_n, f\right) = f\left(1\right)$ (where $f\left(1\right)$ is
defined as in Exercise~\ref{exe.Lambda.subs.fin}).
\item[(c)] Show that $\left(f, g\right) = \left(g, f\right)$ for all
$f \in \Lambda$ and $g \in \Lambda$.
(In other words, the Hall inner product is symmetric.)
\end{itemize}
\end{exercise}

% [DG][v25] Added the preceding (trivial) exercise.

% [DG][v78] Added the trivial part (c) (for ease of referencing).

The Hall inner product induces a $\kk$-module homomorphism
$\Lambda \to \Lambda^o$ (sending every $f \in \Lambda$ to
the $\kk$-linear map $\Lambda \to \kk,\ g \mapsto \left(f, g\right)$).
This homomorphism is invertible (since the Hall inner product
has an orthonormal basis), so that $\Lambda^o \cong \Lambda$
as $\kk$-modules. But in fact, more can be said:

% [DG][v46] Added the above paragraph to clarify how the isomorphism
% below is obtained.

\begin{corollary}
\label{cor.Lambda.selfdual}
The isomorphism $\Lambda^o \cong \Lambda$ induced by the
Hall inner product is an isomorphism of Hopf algebras.
\end{corollary}

\begin{proof}
We have seen that the orthonormal basis  $\{s_\lambda\}$ of Schur functions is
\emph{self-dual}, in the sense that its
multiplication and comultiplication structure constants
are the same.  Thus the isomorphism $\Lambda^o \cong \Lambda$ induced by the
Hall inner product is an isomorphism of bialgebras\footnote{Here are
some details on the proof:
\par
Let $\gamma : \Lambda \to \Lambda^o$ be the $\kk$-module
isomorphism $\Lambda \to \Lambda^o$ induced by the Hall
inner product.
We want to show that $\gamma$ is an isomorphism of bialgebras.
\par
Let $\left\{s_\lambda^*\right\}$ be the basis of $\Lambda^o$
dual to the basis
$\left\{s_\lambda\right\}$ of $\Lambda$.
Thus, for any partition $\lambda$, we have
\begin{equation}
\gamma\left(s_\lambda\right) = s_\lambda^*
\label{pf.cor.Lambda.selfdual.fn1.0}
\end{equation}
(since any partition $\mu$ satisfies
$\left(\gamma\left(s_\lambda\right)\right)\left(s_\mu\right)
= \left(s_\lambda, s_\mu\right)
= \delta_{\lambda, \mu}
= s_\lambda^* \left(s_\mu\right)$,
and thus the two $\kk$-linear maps
$\gamma\left(s_\lambda\right) : \Lambda \to \kk$
and $s_\lambda^* : \Lambda \to \kk$
are equal to each other on the basis
$\left\{s_\mu\right\}$ of $\Lambda$,
which forces them to be identical).
\par
The coproduct structure constants of the basis $\left\{s_\lambda^*\right\}$
of $\Lambda^o$ equal the product structure constants of the basis
$\left\{s_\lambda\right\}$ of $\Lambda$
(according to our discussion of duals in Section~\ref{subsect.duals}).
Since the latter are the Littlewood-Richardson numbers
$c^\lambda_{\mu,\nu}$ (because of
\eqref{symm-structure-constants.1}), we thus conclude that the former
are $c^\lambda_{\mu,\nu}$ as well.
In other words, every $\lambda \in \Par$ satisfies
\begin{equation}
\Delta_{\Lambda^o} s_\lambda^*
= \sum_{\mu, \nu} c^\lambda_{\mu, \nu} s_\mu^* \otimes s_\nu^*
\label{pf.cor.Lambda.selfdual.fn1.1}
\end{equation}
(where the sum is over all pairs $\left(\mu, \nu\right)$ of partitions).
On the other hand, applying the map
$\gamma \otimes \gamma : \Lambda \otimes \Lambda \to \Lambda^o \otimes \Lambda^o$
to the equality
\eqref{symm-structure-constants.2} yields
\begin{align*}
\left(\gamma \otimes \gamma\right) \left( \Delta(s_\lambda) \right)
&= \left(\gamma \otimes \gamma\right) \left(\sum_{\mu, \nu} \hat{c}^\lambda_{\mu,\nu} s_\mu \otimes s_\nu \right)
= \sum_{\mu, \nu} \underbrace{\hat{c}^\lambda_{\mu,\nu}}_{=c^\lambda_{\mu,\nu}} \underbrace{\gamma\left(s_\mu\right)}_{\substack{= s_\mu^* \\ \text{(by \eqref{pf.cor.Lambda.selfdual.fn1.0})}}} \otimes \underbrace{\gamma\left(s_\nu \right)}_{\substack{= s_\nu^* \\ \text{(by \eqref{pf.cor.Lambda.selfdual.fn1.0})}}}
= \sum_{\mu, \nu} c^\lambda_{\mu, \nu} s_\mu^* \otimes s_\nu^*
\\
&= \Delta_{\Lambda^o} \underbrace{s_\lambda^*}_{\substack{= \gamma\left(s_\lambda\right) \\ \text{(by \eqref{pf.cor.Lambda.selfdual.fn1.0})}}}
\qquad \left(\text{by \eqref{pf.cor.Lambda.selfdual.fn1.1}}\right) \\
&= \Delta_{\Lambda^o} \left( \gamma\left(s_\lambda\right) \right)
\end{align*}
for each $\lambda \in \Par$.
In other words, the two $\kk$-linear maps
$\left(\gamma \otimes \gamma\right) \circ \Delta$ and
$\Delta_{\Lambda^o} \circ \gamma$ are equal to each other on
each $s_\lambda$ with $\lambda \in \Par$.
Hence, these two maps must be identical (since the $s_\lambda$ form a
basis of $\Lambda$). Hence,
$\Delta_{\Lambda^o} \circ \gamma = \left(\gamma \otimes \gamma\right) \circ \Delta$.
\par
Our next goal is to show that $\epsilon_{\Lambda^o} \circ \gamma = \epsilon$.
% Indeed, every partition $\lambda \in \Par$ satisfies
% $s_\varnothing^*\left(s_\lambda\right)
% = \left(s_\varnothing, s_\lambda\right)
% = \delta_{\varnothing, \lambda}
% = \epsilon\left(s_\lambda\right)$.
% Thus, $s_\varnothing^* = \epsilon$ as $\kk$-linear maps
% $\Lambda \to \kk$ (since the $s_\lambda$ form a basis of $\Lambda$).
% \par
Indeed, each $\lambda \in \Par$ satisfies
\begin{align*}
\left( \epsilon_{\Lambda^o} \circ \gamma \right) \left(s_\lambda\right)
&= \epsilon_{\Lambda^o} \left(\gamma\left(s_\lambda\right)\right)
= \left(\gamma\left(s_\lambda\right)\right) \left(1\right)
\qquad \left(\text{by the definition of } \epsilon_{\Lambda^o}\right) \\
&= \left(s_\lambda, \underbrace{1}_{= s_\varnothing}\right)
= \left(s_\lambda, s_\varnothing\right)
= \delta_{\lambda, \varnothing}
= \epsilon \left(s_\lambda\right) .
\end{align*}
Hence, $\epsilon_{\Lambda^o} \circ \gamma = \epsilon$.
Combined with
$\Delta_{\Lambda^o} \circ \gamma = \left(\gamma \otimes \gamma\right) \circ \Delta$,
this shows that $\gamma$ is a $\kk$-coalgebra homomorphism.
Similar reasoning can be used to prove that $\gamma$ is a $\kk$-algebra homomorphism.
Altogether, we thus conclude that $\gamma$ is a bialgebra homomorphism.
Since $\gamma$ is a $\kk$-module isomorphism,
this yields that $\gamma$ is an isomorphism of bialgebras.
Qed.
}, and hence also a Hopf
algebra isomorphism by Corollary~\ref{cor.bialg-mor-is-Hopf}.
\end{proof}

% [DG][v64] Added the long footnote above, since I remember tripping up
% there when I first read the notes.

We next identify two other dual pairs of bases, by expanding
the Cauchy product in two other ways.


\begin{proposition}
\label{prop-Two-other-Cauchy-expansions}
One can also expand
\begin{equation}
\label{Two-other-Cauchy-expansions}
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\sum_{\lambda \in \Par} h_{\lambda}(\xx) m_\lambda(\yy)
=\sum_{\lambda \in \Par} z_\lambda^{-1} p_{\lambda}(\xx) p_\lambda(\yy)
\end{equation}
where $z_\lambda:=m_1! \cdot 1^{m_1} \cdot m_2! \cdot 2^{m_2} \cdots$%
\index{$z_\lambda$}
if $\lambda$ is written in multiplicative notation as
$\lambda=(1^{m_1},2^{m_2},\ldots)$ with multiplicity $m_i$ for the part $i$.
(Here, we assume that $\QQ$ is a subring of $\kk$ for the last equality.)
\end{proposition}

% [DG][v17] Added last sentence.

\begin{remark}
\label{rmk.zlambda.centralizer}
It is relevant later (and explains the notation) that $z_\lambda$ is
the size of the $\Symm_n$-centralizer subgroup for a permutation having
cycle type\footnote{If $\sigma$ is a permutation of a finite set $X$,
then the \dfn{cycle type} of $\sigma$ is defined
as the list of the lengths of all cycles of $\sigma$ (that is,
of all orbits of $\sigma$ acting on $X$) written in decreasing order.
This is clearly a partition of $\left|X\right|$. (Some other authors
write it in increasing order instead, or treat it as a multiset.)

For instance, the permutation of the set $\left\{0, 3, 6, 9, 12\right\}$
that sends $0$ to $3$, $3$ to $9$, $6$ to $6$, $9$ to $0$, and
$12$ to $12$ has cycle type $\left(3, 1, 1\right)$, since
the cycles of this permutation have lengths $3$, $1$ and $1$.

It is known that two permutations in $\Symm_n$ have the same cycle
type if and only if they are conjugate. Thus, for a given partition
$\lambda$ with $\left|\lambda\right| = n$, any two permutations in
$\Symm_n$ having cycle type $\lambda$ are conjugate and therefore
their $\Symm_n$-centralizer subgroups have the same size.}
$\lambda$ with $|\lambda|=n$. This is a classical (and
fairly easy) result (see, e.g., \cite[Prop. 1.1.1]{Sagan} or
\cite[Prop. 7.7.3]{Stanley} for a proof).
\end{remark}

% [DG][v45] Added references to Sagan & Stanley.

% [DG][v46] Added footnote defining cycle type, as most authors do not
% encode it as a partition. (Annoyingly, Keith Conrad of all people
% writes it increasingly; but worse are the sources that leave out
% 1-cycles.)

\begin{proof}[Proof of Proposition~\ref{prop-Two-other-Cauchy-expansions}.]
For the first expansion, note that \eqref{h-generating-function-no-HE} shows
\begin{align*}
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
&=\prod_{j=1}^\infty \sum_{n \geq 0} h_n(\xx) y_j^n\\
&=\sum\limits_{\substack{ \text{weak }\\\text{ compositions } \\ (n_1,n_2,\ldots)}}
     ( h_{n_1}(\xx)h_{n_2}(\xx)\cdots  ) (y_1^{n_1} y_2^{n_2}\cdots )\\
&=\sum_{\lambda \in \Par}
     \sum\limits_{\substack{ \text{weak }\\\text{ compositions } \\ (n_1,n_2,\ldots)
              \\ \text{ satisfying } \\ (n_1,n_2,\ldots) \in \Symm_{(\infty)}\lambda }}
     \underbrace{( h_{n_1}(\xx)h_{n_2}(\xx)\cdots  )}_{\substack{=h_\lambda(\xx) \\
                            \text{ (since } (n_1,n_2,\ldots) \in \Symm_{(\infty)}\lambda \text{)}}}
     \underbrace{(y_1^{n_1} y_2^{n_2}\cdots )}_{=\yy^{(n_1, n_2, \ldots)}} \\
&=\sum_{\lambda \in \Par}
     h_\lambda(\xx)
     \underbrace{\sum\limits_{\substack{ \text{weak }\\\text{ compositions } \\ (n_1,n_2,\ldots)
              \\ \text{ satisfying } \\ (n_1,n_2,\ldots) \in \Symm_{(\infty)}\lambda }}
     \yy^{(n_1, n_2, \ldots)}}_{= m_\lambda(\yy)} \\
&=\sum_{\lambda \in \Par} h_{\lambda}(\xx) m_\lambda(\yy).
\end{align*}
For the second expansion (and for later use in
the proof of Theorem~\ref{Sym-to-Hall-isomorphism-thm})
note that
\begin{equation}
\label{H(t)-logarithm}
\log H(t)
= \log \prod_{i=1}^\infty (1-x_i t)^{-1}
=\sum_{i=1}^\infty -\log(1-x_i t)
=\sum_{i=1}^\infty \sum_{m=1}^\infty \frac{(x_it)^m}{m}
=\sum_{m=1}^\infty \frac{1}{m} p_m(\xx) t^m ,
\end{equation}
so that taking $\frac{d}{dt}$ then shows that
\begin{equation}
\label{E(t)-H(t)-P(t)-relation}
P(t):=\sum_{m \geq 0} p_{m+1}t^m = \frac{H'(t)}{H(t)}= H'(t) E(-t).
\end{equation}
A similar calculation shows that
\begin{equation}
\label{eq.cauchy-via-simple-power-sums}
%\begin{aligned}
\log \prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
%&=\sum_{i,j=1}^\infty -\log(1-x_i y_j) \\
%&=\sum_{i,j=1}^\infty \left( \frac{(x_iy_j)^1}{1}+\frac{(x_iy_j)^2}{2}+\frac{(x_iy_j)^3}{3}+\cdots \right) \\
%&=\sum_{m=1}^\infty \frac{1}{m} \sum_{i,j=1}^\infty (x_iy_j)^m
=\sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy)
%\end{aligned}
\end{equation}
and hence
\begin{align*}
&\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\exp\left( \sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy) \right)
= \prod_{m=1}^\infty \exp \left( \frac{1}{m} p_m(\xx) p_m(\yy) \right) \\
&= \prod_{m=1}^\infty \sum_{k=0}^\infty \frac{1}{k!} \left( \frac{1}{m} p_m(\xx) p_m(\yy) \right)^k
= \sum\limits_{\substack{ \text{weak compositions } \\ \left(k_1,k_2,k_3,\ldots\right)}}
  \prod_{m=1}^\infty \left(\frac{1}{k_m!} \left( \frac{1}{m} p_m(\xx) p_m(\yy) \right)^{k_m}\right) \\
&  \qquad \qquad \left(\text{by the product rule}\right) \\
&= \sum\limits_{\substack{ \text{weak compositions } \\ \left(k_1,k_2,k_3,\ldots\right)}}
  \prod_{m=1}^\infty \frac{\left(p_m(\xx) p_m(\yy)\right)^{k_m}}{k_m! m^{k_m}}
= \sum\limits_{\substack{ \text{weak compositions } \\ \left(k_1,k_2,k_3,\ldots\right)}}
  \frac{\prod_{m=1}^\infty \left(p_m(\xx)\right)^{k_m} \prod_{m=1}^\infty \left(p_m(\yy)\right)^{k_m}}{\prod_{m=1}^\infty \left(k_m! m^{k_m}\right)} \\
&= \sum\limits_{\substack{ \text{weak compositions } \\ \left(k_1,k_2,k_3,\ldots\right)}}
  \frac{p_{\left(1^{k_1}2^{k_2}3^{k_3}\cdots\right)}\left(\xx\right) p_{\left(1^{k_1}2^{k_2}3^{k_3}\cdots\right)}\left(\yy\right)}{z_{\left(1^{k_1}2^{k_2}3^{k_3}\cdots\right)}}
=\sum_{\lambda \in \Par} \frac{ p_\lambda(\xx) p_\lambda(\yy) }{z_\lambda}
\end{align*}
due to the fact that every partition can be uniquely written in the
form $\left(1^{k_1}2^{k_2}3^{k_3}\cdots\right)$ with
$\left(k_1, k_2, k_3, \ldots\right)$ a weak composition.
\begin{commentedout}
[\textbf{Note:} The following alternative proof has been commented
out since it uses the multinomial formula for infinite sums,
which might be a bit less elementary than what the above proof
does.]
\begin{align*}
&\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\exp\left( \sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy) \right)
=\sum_{k=0}^\infty \frac{1}{k!}
      \left( \sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy) \right)^k\\
&=\sum_{k=0}^\infty \frac{1}{k!}
           \sum\limits_{\substack{(m_1,m_2,\ldots) \in \NN^\infty:\\ m_1+m_2+\cdots=k}}
             \binom{k}{m_1,m_2,\ldots}
               \left( \frac{p_1(\xx) p_1(\yy)}{1} \right)^{m_1}
               \left( \frac{p_2(\xx) p_2(\yy)}{2} \right)^{m_2} \cdots \\
& \qquad \qquad \left(\text{by the multinomial formula}\right) \\
&=\sum\limits_{\substack{ \text{weak }\\\text{ compositions } \\ (m_1,m_2,\ldots)}}
             \frac{ \left( p_1(\xx) p_1(\yy) \right)^{m_1} }{ 1^{m_1} m_1! }
             \cdot \frac{ \left( p_2(\xx) p_2(\yy)\right)^{m_2} }{ 2^{m_2} m_2! }
             \cdots
=\sum_{\lambda \in \Par} \frac{ p_\lambda(\xx) p_\lambda(\yy) }{z_\lambda}.
\end{align*}
\end{commentedout}
\end{proof}

% [DG][v25] Added some more detail to the proof of the first expansion.

% [DG][v37] Added "by the multinomial formula".
% Actually, there is an alternative way to do the last computation,
% which proceeds by simplifying
% $\exp\left( \sum_{m=1}^\infty \frac{1}{m} p_m(\xx) p_m(\yy) \right)$
% to
% $\prod_{m=1}^\infty \exp\left( \frac{1}{m} p_m(\xx) p_m(\yy) \right)$
% and then expanding the $\exp$ into an infinite sum and expanding the
% product by the product rule. This avoids the multinomial formula.
% Do you also consider it simpler?

% [DG][v38] Proof replaced by the one avoiding the multinomial
% formula (and in much more detail).

\begin{corollary} \phantomsection
\label{dual-bases-h-m-p}
\begin{itemize}
\item[(a)] With respect to the Hall inner product on $\Lambda$, one also has
dual bases $\{h_\lambda\}$ and $\{m_\lambda\}$.
\item[(b)] If $\QQ$ is a subring of $\kk$, then $\{p_\lambda\}$ and
$\left\{z_\lambda^{-1} p_\lambda\right\}$ are also dual bases with
respect to the Hall inner product on $\Lambda$.
\item[(c)] If $\RR$ is a subring of $\kk$, then
$\left\{\dfrac{p_\lambda}{\sqrt{z_\lambda}} \right\}$ is an orthonormal
basis of $\Lambda$ with respect to the Hall inner product.
\end{itemize}
\end{corollary}

% [DG][v7] I added the "in $\Lambda_\RR$" part, since it is not defined
% over $\ZZ$. IMHO, $\RR$ and the square roots are a red herring here,
% but I trust the reader to translate the claim into a rational form
% if needed...
% 
% I also made the fraction a \dfrac, because an inline-sized
% $\frac{p_\lambda}{\sqrt{z_\lambda}}$ didn't look nice. If you disagree,
% just change it back.

% [DG][v25] Split the corollary into (a), (b) and (c), adding part (b) to
% avoid unnecessary use of reals.

\begin{proof}
Since \eqref{Cauchy-identity} and \eqref{Two-other-Cauchy-expansions}
showed
\[
\prod_{i,j=1}^\infty (1-x_i y_j)^{-1}
=\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy)\\
=\sum_{\lambda \in \Par} h_{\lambda}(\xx) m_\lambda(\yy)\\
=\sum_{\lambda \in \Par} p_{\lambda}(\xx)
                       z_\lambda^{-1} p_{\lambda}(\yy)
=\sum_{\lambda \in \Par} \frac{p_{\lambda}(\xx)}{\sqrt{z_\lambda}}
                       \frac{p_{\lambda}(\yy)}{\sqrt{z_\lambda}} ,
\]
it suffices to show that any pair of graded
bases\footnote{See Definition~\ref{def.graded-basis} for the
concept of a ``graded basis'', and recall our convention that
a graded basis of $\Lambda$
is tacitly assumed to have its indexing set $\Par$ partitioned
into $\Par_0, \Par_1, \Par_2, \ldots$.
Thus, a graded basis of $\Lambda$ means a basis
$\left\{w_\lambda\right\}_{\lambda \in \Par}$ of the $\kk$-module
$\Lambda$ (indexed by the partitions $\lambda \in \Par$)
with the property that, for every $n \in \NN$, the subfamily
$\left\{w_\lambda\right\}_{\lambda \in \Par_n}$ is a basis
of the $\kk$-module $\Lambda_n$.}
 $\{u_\lambda\},\{v_\lambda\}$ of $\Lambda$ having
\[
\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy)=
\sum_{\lambda \in \Par} u_{\lambda}(\xx) v_\lambda(\yy)
\]
will be dual with respect to $(\cdot,\cdot)$.
To show this, consider such a pair of graded bases.
Write transition matrices
$A = \left(a_{\nu,\lambda}\right)_{\left(\nu, \lambda\right) \in \Par \times \Par}$
and
$B = \left(b_{\nu,\lambda}\right)_{\left(\nu, \lambda\right) \in \Par \times \Par}$
uniquely expressing
\begin{align}
u_\lambda&=\sum_\nu a_{\nu,\lambda} s_\nu, \label{eq.pf.dual-bases-h-m-p.a} \\
v_\lambda&=\sum_\nu b_{\nu,\lambda} s_\nu. \label{eq.pf.dual-bases-h-m-p.b}
\end{align}
Recall that $\Par = \bigsqcup_{r \in \NN} \Par_r$. Hence, we
can view $A$ as a block matrix, where the blocks are indexed by
pairs of nonnegative integers, and the $\left(r, s\right)$-th block
is $\left(a_{\nu, \lambda}\right)_{\left(\nu, \lambda\right) \in
\Par_r \times \Par_s}$.
For reasons of homogeneity\footnote{More precisely: The power series
$u_\lambda$ is homogeneous of degree $\left|\lambda\right|$, and the
power series $s_\nu$ is homogeneous of degree $\left|\nu\right|$.},
we have $a_{\nu, \lambda} = 0$ for any
$\left(\nu,\lambda\right) \in \Par^2$ satisfying
$\left|\nu\right| \neq \left|\lambda\right|$. Therefore, the
$\left(r, s\right)$-th block of $A$ is zero whenever $r \neq s$.
In other words, the block
matrix $A$ is block-diagonal. Similarly, $B$ can be viewed as a
block-diagonal matrix. The diagonal blocks of $A$ and $B$ are finite
square matrices (since $\Par_r$ is a finite set for each $r \in \NN$);
therefore, products such as $A^t B$, $B^t A$ and $AB^t$ are
well-defined (since all sums involved in their definition have only
finitely many nonzero addends)
and subject to the law of associativity. Moreover, the
matrix $A$ is invertible (being a transition matrix between two bases),
and its inverse is again block-diagonal (because $A$ is
block-diagonal).

The equalities \eqref{eq.pf.dual-bases-h-m-p.a} and
\eqref{eq.pf.dual-bases-h-m-p.b} show that
$(u_\alpha,v_\beta) = \sum_{\nu} a_{\nu,\alpha} b_{\nu,\beta}$
(by the orthonormality of the $s_\lambda$).
Hence, we want to prove that
$\sum_{\nu} a_{\nu,\alpha} b_{\nu,\beta} = \delta_{\alpha, \beta}$.
In other words, we want to prove that
$A^t B = I$, that is, $B^{-1}=A^t$.
  On the other hand, one has
\[
\sum_{\lambda } s_{\lambda}(\xx) s_\lambda(\yy)
=\sum_{\lambda } u_{\lambda}(\xx) v_\lambda(\yy)
=\sum_\lambda \sum_\nu a_{\nu,\lambda} s_\nu(\xx)
  \sum_\rho b_{\rho,\lambda} s_\rho(\yy).
\]
Comparing coefficients\footnote{Comparing coefficients is legitimate
because if a power series $f \in \kk\left[\left[\xx, \yy\right]\right]$
is written
in the form $f = \sum_{\left(\nu, \rho\right) \in \Par^2}
a_{\rho, \nu} s_\nu\left(\xx\right)
s_\rho\left(\yy\right)$ for some coefficients
$a_{\rho, \nu} \in \kk$, then these coefficients
$a_{\rho, \nu}$ are uniquely determined by $f$. This is
just a restatement of Exercise \ref{exe.Lambda.infinite-indep}(b).}
of $s_\nu(\xx) s_\rho(\yy)$ forces
$\sum_\lambda a_{\nu,\lambda} b_{\rho,\lambda} = \delta_{\nu,\rho}$,
or in other words, $AB^t=I$. Since $A$ is invertible, this
yields $B^tA = I$, and hence $A^tB=I$, as desired.\footnote{In our
argument above, we have obtained the invertibility of $A$ from the
fact that $A$ is a transition matrix between two bases. Here is an
alternative way to prove that $A$ is invertible: \par
Recall that $A$ and $B^t$ are block-diagonal matrices. Hence,
the equality $AB^t = I$ rewrites as
$A_{r,r} \left(B^t\right)_{r,r} = I$ for all $r \in \NN$, where we are
using the notation $C_{r,s}$ for the $\left(r, s\right)$-th block of a
block matrix $C$.
But this shows that each diagonal block $A_{r, r}$ of $A$ is
right-invertible. Therefore, each diagonal block $A_{r, r}$ of $A$ is
invertible (because $A_{r, r}$ is a square matrix of finite size,
and such matrices are always invertible when they are
right-invertible). Consequently, the block-diagonal matrix $A$ is
invertible, and its inverse is again a block-diagonal matrix (whose
diagonal blocks are the inverses of the $A_{r,r}$).}
\end{proof}

% [DG][v17] Added footnote.

% [DG][v23] Added clarification that the two bases should be graded;
% also, changed the use of finiteness to a use of the invertibility of
% $B$ (which is presumed known since $\{u_\lambda\}$ and $\{v_\lambda\}$
% are assumed to be bases). The old argument is now in the footnote.

% [DG][v58] Expanded footnote, and added more details. The concept of
% inverses of infinite matrices is somewhat slippery: Infinite
% matrices are not in 1-to-1 correspondence with homomorphisms between
% modules (because a matrix can only correspond to a homomorphism if
% each of its columns has only finitely many nonzero entries); thus,
% we would have to distinguish between invertibility-as-morphism and
% invertibility-as-matrix, and barring the former to make sure that
% associativity holds. I have introduced the block structure early on
% to get rid of these problems.

Corollary~\ref{dual-bases-h-m-p} is a known and fundamental
fact\footnote{For example, Corollary~\ref{dual-bases-h-m-p}(a)
appears in \cite[Corollary 3.3]{Leeuwen-altSchur} (though the
definition of Schur functions in \cite{Leeuwen-altSchur} is different
from ours; we will meet this alternative definition later on),
and parts (b) and (c) of Corollary~\ref{dual-bases-h-m-p} are
equivalent to \cite[\S I.4, (4.7)]{Macdonald} (though Macdonald
defines the Hall inner product using
Corollary~\ref{dual-bases-h-m-p}(a)).}.
However, our definition of the Hall inner product is unusual;
most authors (e.g., Macdonald in \cite[\S I.4, (4.5)]{Macdonald},
Hazewinkel/Gubareni/Kirichenko in
\cite[Def. 4.1.21]{HazewinkelGubareniKirichenko}, and Stanley in
\cite[(7.30)]{Stanley}) \emph{define} the Hall inner product
as the bilinear form satisfying
$\left(h_\lambda, m_\mu\right) = \delta_{\lambda, \mu}$ (or,
alternatively,
$\left(m_\lambda, h_\mu\right) = \delta_{\lambda, \mu}$), and
only later prove that the basis $\left\{s_\lambda\right\}$ is
orthonormal with respect to this scalar product. (Of course,
the fact that this definition is equivalent to our
Definition~\ref{def.hall-inner-product} follows either from
this orthonormality, or from our
Corollary~\ref{dual-bases-h-m-p}(a).)

% [DG][v45] Added preceding paragraph (closing a gap in the van Leeuwen
% paper and connecting the notes with some older literature).

The tactic applied in the proof of Corollary~\ref{dual-bases-h-m-p} can
not only be used to show that certain bases of $\Lambda$ are dual, but
also, with a little help from linear algebra over rings
(Exercise~\ref{exe.linalg.sur-bij}), it can be strengthened to show that
certain families of symmetric functions are bases to begin with, as we
will see in Exercise~\ref{exe.Lambda.basis-vlam} and
Exercise~\ref{exe.Lambda.bases}.

\begin{exercise} \phantomsection
\label{exe.linalg.sur-bij}
\begin{itemize}
\item[(a)] Prove that if an endomorphism of a finitely generated $\kk$-module
is surjective, then this endomorphism is a $\kk$-module isomorphism.
\item[(b)] Let $A$ be a finite free $\kk$-module with finite
basis $\left(  \gamma_{i}\right)  _{i\in I}$. Let $\left(  \beta_{i}\right)
_{i\in I}$ be a family of elements of $A$ which spans the $\kk$-module
$A$. Prove that $\left(  \beta_{i}\right)  _{i\in I}$ is a $\kk$-basis of
$A$.
\end{itemize}
\end{exercise}

\begin{exercise}
\label{exe.Lambda.basis-vlam}
For each partition $\lambda$, let $v_{\lambda}$
be an element of $\Lambda_{\left\vert \lambda\right\vert }$. Assume that the
family $\left(  v_{\lambda}\right)  _{\lambda \in \Par}$ spans
the $\kk$-module $\Lambda$.
Prove that the family $\left(  v_{\lambda
}\right)  _{\lambda \in \Par}$ is a graded basis of the
graded $\kk$-module $\Lambda$.
\end{exercise}

\begin{exercise} \phantomsection
\label{exe.Lambda.bases}
\begin{itemize}
\item[(a)] Assume that for every partition $\lambda$, two homogeneous elements
$u_\lambda$ and $v_\lambda$ of $\Lambda$, both having degree
$\left|\lambda\right|$, are given. Assume further that
\[
\sum_{\lambda \in \Par} s_\lambda\left(\xx\right) s_\lambda\left(\yy\right)
= \sum_{\lambda \in \Par} u_\lambda\left(\xx\right) v_\lambda\left(\yy\right)
\]
in $\kk\left[\left[\xx, \yy\right]\right]
= \kk\left[\left[x_1, x_2, x_3, \ldots, y_1, y_2, y_3, \ldots\right]\right]$.
Show that $\left(u_\lambda\right)_{\lambda \in \Par}$ and
$\left(v_\lambda\right)_{\lambda \in \Par}$ are $\kk$-bases of
$\Lambda$, and actually are dual bases with respect to the
Hall inner product on $\Lambda$.
\item[(b)] Use this to give a new proof of the fact that
$\left(h_\lambda\right)_{\lambda \in \Par}$ is a $\kk$-basis of
$\Lambda$.
\end{itemize}
\end{exercise}

% [DG][v23] Added the above two exercises, to factor out some steps in
% other solutions (e.g., of Exercise \ref{exe.witt}(h)).

\begin{exercise}
\label{exe.E(t)-H(t)-P(t)-relation}
Prove that
$\sum_{m \geq 0} p_{m+1} t^m = \dfrac{H'\left(t\right)} {H\left(t\right)}$.
(This was proven in \eqref{E(t)-H(t)-P(t)-relation} in the case when
$\QQ$ is a subring of $\kk$, but here we make no requirements on $\kk$.)
\end{exercise}

% [DG][v25] Added above exercise (as it was subtly -- sufficiently so
% that I missed it myself -- used in another's solution).

The following exercises give some useful criteria for algebraic independence
of families of symmetric functions:

\begin{exercise}
\label{exe.Lambda.alggen-ind}
Let $v_1, v_2, v_3, \ldots$ be elements of $\Lambda$.
Assume that $v_n \in \Lambda_n$ for each positive integer $n$.
Assume further that $v_1, v_2, v_3, \ldots$ generate the
$\kk$-algebra $\Lambda$.
Then:

\begin{enumerate}
\item[(a)] Prove that $v_1, v_2, v_3, \ldots$ are algebraically independent
over $\kk$.

\item[(b)] For every partition $\lambda$, define an element $v_{\lambda}
\in\Lambda$ by $v_{\lambda}=v_{\lambda_{1}}v_{\lambda_{2}}\cdots
v_{\lambda_{\ell\left(  \lambda\right)  }}$. Prove that the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par}$ is a graded basis of the
graded $\kk$-module $\Lambda$.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exe.Lambda.alggen-crit1}
For each partition $\lambda$, let $a_{\lambda} \in \kk$.
Assume that the element $a_{\left(  n\right)  } \in \kk$
is invertible for each positive integer $n$.
Let $v_1, v_2, v_3, \ldots$ be elements of $\Lambda$ such that
each positive integer $n$ satisfies
$v_n = \sum_{\lambda \in \Par_n} a_{\lambda} h_{\lambda}$.
Prove that the elements $v_1, v_2, v_3, \ldots$ generate the
$\kk$-algebra $\Lambda$ and are algebraically independent over
$\kk$.
\end{exercise}

\begin{exercise}
\label{exe.Lambda.alggen-crit1hall}
Let $v_1, v_2, v_3, \ldots$ be elements of $\Lambda$.
Assume that $v_n \in \Lambda_n$ for each positive integer $n$.
Assume further that $\left( p_n, v_n \right) \in \kk$ is invertible
for each positive integer $n$.
Prove that the elements $v_1, v_2, v_3, \ldots$ generate the
$\kk$-algebra $\Lambda$ and are
algebraically independent over $\kk$.
\end{exercise}

% [DG][v75] Added the above three exercises, as well as
% Exercise \ref{exe.Lambda.basis-vlam} further above.

\begin{exercise}
\label{exe.Lambda.hall-coefficient}
Let $f \in \Lambda$, and let $\beta$ be a weak composition.
Let $\mu \in \Par$ be the partition consisting of the nonzero
entries of $\beta$ (sorted in decreasing order).%
\footnote{For example, if $\beta=\left(  1,0,3,1,2,3,0,0,0,\ldots
\right)  $, then $\mu=\left(  3,3,2,1,1\right)  $.}
Prove that
\[
\left( f, h_\mu \right)
= \left( h_\mu, f \right)
= \left( \text{the coefficient of $\xx^{\beta}$ in $f$} \right)  .
\]

\end{exercise}

% [DG][v78] Added the above simple exercise, in order to cite it
% in a later proof.

\begin{exercise}
\label{exe.Lambda.h-e-through-p}
Assume that $\QQ$ is a subring of $\kk$.
Define a positive integer $z_{\lambda}$ for each $\lambda \in \Par$
as in Proposition~\ref{prop-Two-other-Cauchy-expansions}.
Prove that every $n \in \NN$ satisfies the two equalities
\begin{equation}
h_n = \sum_{\lambda \in \Par_n} z_{\lambda}^{-1} p_{\lambda}
\label{eq.exe.Lambda.h-e-through-p.h}
\end{equation}
and
\begin{equation}
e_n = \sum_{\lambda \in \Par_n}
\left(  -1\right) ^{\left\vert \lambda\right\vert -\ell\left(  \lambda\right)  }
z_{\lambda}^{-1} p_{\lambda}.
\label{eq.exe.Lambda.h-e-through-p.e}
\end{equation}
\end{exercise}

% [DG][v79] Added the above exercise out of completionism.






\subsection{Bialternants, Littlewood-Richardson:  Stembridge's concise proof}
\label{bialternant-section}

There is a more natural way in which Schur functions arise as a $\kk$-basis
for $\Lambda$, coming from consideration of polynomials in a finite variable set,
and the relation between those which are symmetric and those which are \emph{alternating}.

For the remainder of this section,
fix a nonnegative integer $n$, and let $\xx=(x_1,\ldots,x_n)$ be a finite variable set.
This means that $s_{\lambda/\mu}=s_{\lambda/\mu}(\xx)=\sum_T \xx^{\cont(T)}$ is a
generating function for column-strict tableaux $T$ as in
Definition~\ref{skew-Schur-function-definition}, but with the extra condition
that $T$ have entries in $\{1,2,\ldots,n\}$. \ \ \ \ 
\footnote{See Exercise~\ref{exe.schur.finite}(a) for this.}
As a consequence, $s_{\lambda / \mu}$ is a polynomial in
$\kk \left[x_1, x_2, \ldots, x_n\right]$ (not just a
power series), since there are only finitely many column-strict
tableaux $T$ of shape $\lambda / \mu$ having all their entries in
$\left\{ 1, 2, \ldots , n \right\}$.
We will assume without
further mention that all partitions appearing in the section
have at most $n$ parts.

% [DG][v44] Added footnote and sentence about polynomiality.



\begin{definition}
Let $\kk$ be the ring $\ZZ$ or a field of characteristic not equal to $2$.
(We require this to avoid certain annoyances in the discussion of
alternating polynomials in characteristic $2$.)

Say that a polynomial $f(\xx)=f(x_1,\ldots,x_n)$ is
\emph{alternating}\index{alternating polynomial} if
for every permutation $w$ in $\Symm_n$ one has that
\[
(wf)(\xx) = f(x_{w(1)},\ldots,x_{w(n)}) = \sgn(w) f(\xx).
\]
Let $\Lambda^{\sgn} \subset \kk[x_1,\ldots,x_n]$
denote the subset of alternating polynomials\footnote{When $\kk$ has characteristic $2$
(or, more generally, is an arbitrary commutative ring),
it is probably best to define the alternating polynomials
$\Lambda^{\sgn}_\kk$ as the $\kk$-submodule
$\Lambda^{\sgn} \otimes_{\ZZ} \kk$
of $\ZZ[x_1,\ldots,x_n] \otimes_{\ZZ} \kk  \cong \kk[x_1,\ldots,x_n]$.}.
\end{definition}

As with $\Lambda$ and its monomial basis $\{m_\lambda\}$, there is
an obvious $\kk$-basis for $\Lambda^{\sgn}$, coming from the fact that
a polynomial $f=\sum_\alpha c_\alpha \xx^\alpha$
is alternating if and only if $c_{w(\alpha)} = \sgn(w) c_\alpha$ for every $w$ in $\Symm_n$
and every $\alpha \in \NN^n$.  This means that every alternating $f$
is a $\kk$-linear combination of the following elements.

% [DG][v25] Replaced "weak composition $\alpha$" by "$\alpha \in \NN^n$"
% to avoid having to explain that weak compositions are assumed to
% have length $\leq n$ here.

\begin{definition}
\label{def.alternant}
For $\alpha=(\alpha_1,\ldots,\alpha_n)$ in $\NN^n$, define
the \dfn{alternant}
\[
a_{\alpha}:= \sum_{ w \in \Symm_n} \sgn(w) w(\xx^{\alpha})
= \det \left[ \begin{matrix}
x_1^{\alpha_1} & \cdots & x_1^{\alpha_n} \\
x_2^{\alpha_1} & \cdots & x_2^{\alpha_n} \\
\vdots & \ddots & \vdots \\
x_n^{\alpha_1} & \cdots & x_n^{\alpha_n}
\end{matrix}
\right].
\]
\end{definition}

\begin{example}
One has
\[
a_{(1,5,0)} = x_1^1 x_2^5 x_3^0
               -  x_1^5 x_2^1 x_3^0 -  x_1^1 x_2^0 x_3^5 -  x_1^0 x_2^5 x_3^1
               + x_1^0 x_2^1 x_3^5 + x_1^5 x_2^0 x_3^1
            = - a_{(5,1,0)}.
\]
Similarly, $a_{w\left(\alpha\right)} = \sgn(w) a_\alpha$ for every
$w \in \Symm_n$ and every $\alpha \in \NN^n$.

Meanwhile, $a_{(5,2,2)} = 0$ since the transposition $t=\binom{123}{132}$ fixes
$(5,2,2)$ and hence
\[
a_{(5,2,2)} = t(a_{(5,2,2)}) = \sgn(t) a_{(5,2,2)} = -a_{(5,2,2)}.
\]
\footnote{One subtlety should be addressed:
We want to prove that $a_{(5,2,2)} = 0$ in $\kk\left[x_1,\ldots,x_n\right]$
for every commutative ring $\kk$. It is clearly enough to prove that
$a_{(5,2,2)} = 0$ in $\ZZ\left[x_1,\ldots,x_n\right]$. Since $2$ is not a
zero-divisor in $\ZZ\left[x_1,\ldots,x_n\right]$, we can achieve this by showing that
$a_{(5,2,2)} = -a_{(5,2,2)}$. We would not be able to make this argument
directly over an arbitrary commutative ring $\kk$.}
Alternatively, $a_{(5,2,2)}=0$ as it is a determinant of a matrix with two equal columns.
Similarly, $a_\alpha = 0$ for every $n$-tuple $\alpha \in \NN^n$ having two
equal entries.
\end{example}

\noindent
This example illustrates that, for a $\kk$-basis for  $\Lambda^{\sgn}$,
one can restrict attention to alternants $a_\alpha$ in which $\alpha$ is a
\dfn{strict partition}, i.e., in which $\alpha$ satisfies $\alpha_1 > \alpha_2 > \cdots > \alpha_n$.
One can therefore uniquely express $\alpha = \lambda + \rho$,
where $\lambda$ is a (weak) partition
$\lambda_1 \geq \cdots \geq \lambda_n \geq 0$ and where
$
\rho:=(n-1,n-2,\ldots,2,1,0)
$
is sometimes called the \dfn{staircase partition}\footnote{The
name is owed to its Ferrers shape. For instance, if $n = 5$, then
the Ferrers diagram of $\rho$ (represented using dots) has
the form
\[
\begin{matrix}
\bullet & \bullet & \bullet & \bullet \\
\bullet & \bullet & \bullet & \\
\bullet & \bullet & & \\
\bullet & & & 
\end{matrix}
.
\]
}.
For example $\alpha=(5,1,0)=(3,0,0) + (2,1,0)=\lambda+\rho$.

% [DG][v32] Made a footnote out of your brief remark about staircase
% shapes.

\begin{proposition}
\label{prop.alternants.space}
Let $\kk$ be the ring $\ZZ$ or a field of characteristic not equal to $2$.

The alternants $\{ a_{\lambda +\rho} \}$ as $\lambda$ runs through the partitions
with at most $n$ parts form a $\kk$-basis for $\Lambda^{\sgn}$.
In addition, the \emph{bialternants} $\{ \frac{ a_{\lambda +\rho} }{ a_\rho } \}$
as $\lambda$ runs through the same set form a $\kk$-basis for
$\Lambda(x_1,\ldots,x_n)=\kk[x_1,\ldots,x_n]^{\Symm_n}$.
\end{proposition}
\begin{proof}
The first assertion should be clear from our previous discussion:  the
alternants $\{ a_{\lambda +\rho} \}$ span $\Lambda^{\sgn}$ by definition, and
they are $\kk$-linearly independent because they are supported on disjoint
sets of monomials $\xx^\alpha$.

The second assertion follows from the first,
after proving the following \textbf{Claim}: $f(\xx)$ lies in
$\Lambda^{\sgn}$ if and only if $f(\xx)=a_\rho \cdot g(\xx)$ where
$g(\xx)$ lies in $\kk[\xx]^{\Symm_n}$ and where
\[
a_{\rho} = \det( x_i^{n-j} )_{i,j=1,2,\ldots,n} =\prod_{1 \leq i < j \leq n} (x_i - x_j)
\]
is the \dfn{Vandermonde determinant/product}.
In other words,
\[
\Lambda^{\sgn}= a_{\rho}  \cdot \kk[\xx]^{\Symm_n}
\]
is a free $\kk[\xx]^{\Symm_n}$-module of rank $1$, with $a_\rho$ as its
$\kk[\xx]^{\Symm_n}$-basis element.

To see the Claim, first note the inclusion
\[
\Lambda^{\sgn} \supset a_{\rho} \cdot \kk[\xx]^{\Symm_n}
\]
since the product of a symmetric polynomial and an alternating polynomial is
an alternating polynomial.  For the reverse inclusion, note that since an alternating polynomial $f(\xx)$ changes
sign whenever one exchanges two distinct variables $x_i,x_j$, it must vanish upon setting
$x_i = x_j$, and therefore be divisible by $x_i - x_j$, so divisible by
the entire product $\prod_{1 \leq i < j \leq n} (x_i - x_j)=a_\rho$.  But then
the quotient $g(\xx) = \frac{f(\xx)}{a_\rho}$ is symmetric, as
it is a quotient of two alternating polynomials.
\end{proof}

% [DG][v3] Replaced $\supseteq$ by $\supset$ to match notation in the rest
% of the notes. (Also one more time further below.)

Let us now return to the general setting, where $\kk$ is an arbitrary
commutative ring. We are not requiring that the assumptions
of Proposition~\ref{prop.alternants.space} be valid; we can still study
the $a_\alpha$ of Definition~\ref{def.alternant}, but we cannot use
Proposition~\ref{prop.alternants.space} anymore.
We will show that the fraction
$\frac{ a_{\lambda +\rho} }{ a_\rho }$ is nevertheless a well-defined
polynomial in $\Lambda\left(x_1, \ldots, x_n\right)$ whenever $\lambda$
is a partition\footnote{This
can also be deduced by base change from the $\kk = \ZZ$ case of
Proposition~\ref{prop.alternants.space}.}, and in fact equals the
Schur function $s_\lambda (\xx)$. As a consequence, the mysterious
bialternant basis $\{ \frac{ a_{\lambda +\rho} }{ a_\rho } \}$
of $\Lambda\left(x_1, \ldots, x_n\right)$ defined in
Proposition~\ref{prop.alternants.space} still exists in the general
setting, and is plainly the Schur functions $\{ s_\lambda(\xx) \}$.
Stembridge \cite{Stembridge} noted
that one could give a remarkably concise proof of an even stronger
assertion, which simultaneously gives one of the standard combinatorial
interpretations for the Littlewood-Richardson coefficients $c^{\lambda}_{\mu,\nu}$.
For the purposes of stating it,
we introduce for a tableau $T$ the notation $T|_{\cols \geq j}$ (resp. $T|_{\cols \leq j}$)
to indicate the subtableau which is the restriction
of $T$ to the union of its columns $j,j+1,j+2,\ldots$ (resp. columns $1,2,\ldots,j$).

\begin{example}
If $T = \begin{matrix} & & 1 & 2 \\ & 2 & 2 & 3 \\ 3 & 5 \end{matrix}$,
then
\[
T|_{\cols \geq 3} = \begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}
\qquad \text{and} \qquad
T|_{\cols \leq 2} = \begin{matrix} & \\ & 2 \\ 3 & 5 \end{matrix}
\]
(note that $T|_{\cols \leq 2}$ has an empty first row).
\end{example}

% [DG][v77] Added the above example.

\begin{theorem}
\label{Stembridge-theorem}
For partitions $\lambda,\mu,\nu$ with $\mu \subseteq \lambda$, one has\footnote{Again,
we can drop the requirement that $\mu \subseteq \lambda$, provided that we understand
that there are no column-strict tableaux of shape $\lambda / \mu$ unless $\mu
\subseteq \lambda$.}
\[
a_{\nu+\rho} s_{\lambda/\mu} = \sum_T a_{\nu+\cont(T)+\rho}
\]
where $T$ runs through all column-strict tableaux with entries in $\{1,2,\ldots,n\}$
of shape $\lambda/\mu$ with
the property that for each $j=1,2,3,\ldots$, the weak composition
$\nu + \cont(T|_{\cols\geq j})$ is a partition.
\end{theorem}

% [DG][v77] Replaced "one has $\nu + \cont(T|_{\cols\geq j})$ a partition"
% by "the weak composition $\nu + \cont(T|_{\cols\geq j})$ is a partition".

\noindent
Before proving Theorem~\ref{Stembridge-theorem}, let us see some of its consequences.

\begin{corollary}
\label{cor.stembridge.slambda}
For any partition $\lambda$, we have\footnote{Notice that division
by $a_\rho$ is unambiguous in the ring $\kk\left[x_1,\ldots,x_n\right]$,
since $a_\rho$ is not a zero-divisor (in fact,
$a_\rho = \prod_{1 \leq i < j \leq n} (x_i - x_j)$ is the product of the
binomials $x_i - x_j$, none of which is a zero-divisor). \vspace{0.2pc}}
\[
s_\lambda(\xx) = \frac{ a_{\lambda +\rho} }{ a_\rho }.
\]
\end{corollary}
\begin{proof}
Fix a partition $\lambda$.
Take $\nu=\mu=\varnothing$ in Theorem~\ref{Stembridge-theorem}.  Note
that there is only one column-strict tableau $T$
of shape $\lambda$ such that each $\cont(T|_{\cols\geq j})$ is a partition,
namely the tableau having every entry in row $i$ equal to $i$:
\[
\begin{matrix}
1 & 1 & 1 & 1 & 1\\
2 & 2 & 2 &   & \\
3 & 3 & 3 &   & \\
4 & 4
\end{matrix}
\]
\footnote{\textit{Proof.} It is clear that the tableau
having every entry in row $i$ equal to $i$ indeed satisfies the condition
that each $\cont(T|_{\cols\geq j})$ is a partition.
It remains to show that it is the only column-strict tableau
(of shape $\lambda$) satisfying this condition.

Let $T$ be a column-strict tableau of shape $\lambda$ satisfying
the condition that each $\cont(T|_{\cols\geq j})$ is a partition.
We must show that for each $i$, every entry in row $i$ of $T$ is equal
to $i$.
Assume the contrary.
Thus, there exists some $i$ such that row $i$ of $T$ contains an entry
distinct from $i$.
Consider the smallest such $i$.
Hence, rows $1, 2, \ldots, i-1$ of $T$ are filled with entries
$1, 2, \ldots, i-1$, whereas row $i$ has some entry distinct from $i$.
Choose some $j$ such that the $j$-th entry of row $i$ of $T$ is
distinct from $i$.
This entry cannot be smaller than $i$ (since it has $i-1$ entries
above it in its column, and the entries of $T$ increase strictly
down columns); thus, it has to be larger than $i$.
Therefore, all entries in rows $i, i+1, i+2, \ldots$ of $T|_{\cols\geq j}$
are larger than $i$ as well (since they lie southeast of this entry).
Hence, each entry of $T|_{\cols\geq j}$ is either smaller than $i$
(if it is in one of rows $1, 2, \ldots, i-1$)
or larger than $i$ (if it is in row $i$ or further down).
Thus, $i$ is not an entry of $T|_{\cols\geq j}$.
In other words, $\cont_i (T|_{\cols\geq j}) = 0$.
Since $\cont (T|_{\cols\geq j})$ is a partition, we thus conclude that
$\cont_k (T|_{\cols\geq j}) = 0$ for all $k > i$.
In other words, $T|_{\cols\geq j}$ has no entries larger than $i$.
But this contradicts the fact that the $j$-th entry of row $i$ of $T$
is larger than $i$.
This contradiction completes our proof.}.
Furthermore, this $T$ has $\cont(T)=\lambda$, so the theorem says $a_{\rho} s_\lambda= a_{\lambda+\rho}$.
\end{proof}

% [DG][v77] Added footnote to the above proof.

\begin{example}
For $n=2$, so that $\rho=(1,0)$, if we take $\lambda=(4,2)$, then one
has
\begin{align*}
\frac{a_{\lambda+\rho}}{a_\rho}
&=\frac{a_{(4,2)+(1,0)}}{a_{(1,0)}}
=\frac{a_{(5,2)}}{a_{(1,0)}} \\
&=\frac{x_1^5 x_2^2 - x_1^2 x_2^5}{x_1 - x_2}\\
&= x_1^4 x_2^2 + x_1^3 x_2^3 + x_1^2 x_2^4\\
&= \xx^{\cont{
\left(\begin{matrix}
1111\\
22\phantom{1}\phantom{1}
\end{matrix}\right)
}}
+
\xx^{\cont{
\left(\begin{matrix}
1112\\
22\phantom{1}\phantom{1}
\end{matrix}\right)
}}
+
\xx^{\cont{
\left(\begin{matrix}
1122\\
22\phantom{1}\phantom{1}
\end{matrix}\right)
}}\\
&=s_{(4,2)}
=s_\lambda.
\end{align*}
\end{example}

Some authors use the equality in Corollary~\ref{cor.stembridge.slambda}
to \emph{define} the Schur polynomial
$s_\lambda \left(x_1, x_2, \ldots, x_n\right)$ in $n$ variables;
this definition, however, has the drawback of not generalizing
easily to infinitely many variables or to skew Schur
functions\footnote{With some effort, it is possible to use
Corollary~\ref{cor.stembridge.slambda} in order to define
the Schur function $s_\lambda$ in infinitely many variables.
Indeed, one can define this Schur function as the unique
element of $\Lambda$ whose evaluation at $\left(x_1, x_2,
\ldots, x_n\right)$ equals $\frac{ a_{\lambda +\rho} }{ a_\rho }$
for every $n \in \NN$. If one wants to use such a definition,
however, one needs to check that such an element exists. This is
the approach to defining $s_\lambda$ taken in
\cite[Definition 1.4.2]{Leeuwen-altSchur} and in
\cite[\S I.3]{Macdonald}.}.

% [DG][v45] Added preceding paragraph. (Its purpose is to
% advertise the van Leeuwen paper and explain how it ties in
% with these notes.)



Next divide through by $a_\rho$ on both sides of Theorem~\ref{Stembridge-theorem}
(and use Corollary~\ref{cor.stembridge.slambda}) to give the following.
\begin{corollary}
\label{finite-L-R-rule}
For partitions $\lambda, \mu, \nu$ having at most $n$ parts, one has
\begin{equation}
\label{eq.finite-L-R-rule.1}
s_{\nu} s_{\lambda/\mu} = \sum_T s_{\nu+\cont(T)}
\end{equation}
where $T$ runs through the same set as in Theorem~\ref{Stembridge-theorem}.
In particular, taking $\nu=\varnothing$, we obtain
\begin{equation}
\label{eq.finite-L-R-rule.2}
s_{\lambda/\mu} = \sum_T s_{\cont(T)}
\end{equation}
where in the sum $T$ runs through all column-strict tableaux of shape $\lambda/\mu$
for which each $\cont(T|_{\cols\geq j})$ is a partition.

%Rephrasing, this,  whenever, $\lambda, \mu, \nu$ all have at most $n$ parts
%the Littlewood-Richardson coefficient $c^\lambda_{\mu,\nu}$ is the
%number of such tableaux $T$ having $\cont(T)=\nu$.
\end{corollary}

% [DG][v49] Moved the actual LR rule (Corollary~\ref{L-R-rule})
% further down (meaning after the proof of
% Theorem~\ref{Stembridge-theorem}) and split it into two corollaries,
% the first of which sets the number of variables to infinity and the
% second of which passes from equalities between functions to a formula
% for the coefficients. This adds detail to the proof and also creates
% a referenceable skew-LR rule (Corollary~\ref{infinite-L-R-rule}),
% which I will eventually reference when I do Zelevinsky's LR rule.

\begin{proof}[Proof of Theorem~\ref{Stembridge-theorem}]
Start by rewriting the left side of the theorem:
% , and using the fact that
% $w(s_{\lambda/\mu})=s_{\lambda/\mu}$ for any $w$ in $\Symm_n$:
\begin{align*}
a_{\nu+\rho} s_{\lambda/\mu}
&=\sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\rho)}
     s_{\lambda/\mu}
 =\sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\rho)}
     w(s_{\lambda/\mu}) \\
& \qquad \qquad
    \left(\text{since $w(s_{\lambda/\mu})=s_{\lambda/\mu}$ for any $w \in \Symm_n$}\right) \\
&=\sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\rho)}
     \sum\limits_{\substack{\text{column-strict }T\\\text{ of shape }\lambda/\mu}} \xx^{w(\cont(T))}  \\
&=\sum\limits_{\substack{\text{column-strict }T\\\text{ of shape }\lambda/\mu}}
         \qquad \sum_{w \in \Symm_n} \sgn(w) \xx^{w(\nu+\cont(T)+\rho)}  \\
&= \sum\limits_{\substack{\text{column-strict }T\\\text{ of shape }\lambda/\mu}}
      a_{\nu+\cont(T)+\rho} .
\end{align*}
We wish to cancel out all the summands indexed by column-strict tableaux $T$ which
fail any of the conditions that $\nu + \cont(T|_{\cols\geq j})$ be a partition.
Given such a $T$, find the maximal $j$ for which it fails this
condition\footnote{Such a $j$ exists because $\nu + \cont(T|_{\cols\geq j})$ is a
partition for all sufficiently high $j$ (in fact, $\nu$ itself is a partition).},
and
then find the minimal $k$ for which
\[
\nu_k+\cont_k(T|_{\cols\geq j}) <
\nu_{k+1}+\cont_{k+1}(T|_{\cols\geq j}).
\]
Maximality of $j$ forces
\[
\nu_k+\cont_k(T|_{\cols\geq j+1}) \geq
\nu_{k+1}+\cont_{k+1}(T|_{\cols\geq j+1}).
\]
Since column-strictness implies that column $j$ of $T$ can contain at most one
occurrence of $k$ or of $k+1$ (or neither or both),
the previous two inequalities imply that
column $j$ must contain an occurrence of $k+1$ and no occurrence of $k$, so
that
\[
\nu_k+\cont_k(T|_{\cols\geq j}) + 1
= \nu_{k+1}+\cont_{k+1}(T|_{\cols\geq j}).
\]
This implies that the adjacent transposition $t_{k,k+1}$ swapping $k$ and $k+1$
fixes the vector $\nu+\cont(T|_{\cols\geq j})+\rho$.

Now create a new tableau $T^*$ from $T$ by
applying the Bender-Knuth involution (from the proof of
Proposition~\ref{Schur-functions-are-symmetric-prop})
on letters $k,k+1$,
but \emph{only to columns} $1,2,\ldots,j-1$ of $T$,
leaving columns $j,j+1,j+2,\ldots$ unchanged.%
\footnote{See Example~\ref{exa.pf.Stembridge-theorem.invol}
below for an example of this construction.}
One should check that $T^*$ is still column-strict, but this holds
because column $j$ of $T$ has no occurrences of letter $k$.
Note that
\[
t_{k,k+1} \cont(T|_{\cols\leq j-1}) =
              \cont(T^*|_{\cols\leq j-1})
\]
and hence
\[
t_{k,k+1} ( \nu+\cont(T)+\rho ) = \nu+\cont(T^*)+\rho ,
\]
so that $a_{\nu+\cont(T)+\rho} = - a_{\nu+\cont(T^*)+\rho}$.

Because $T$ and $T^*$ have exactly the same columns  $j,j+1,j+2,\ldots$,
the tableau $T^*$ is also a violator of at least one of the conditions that
$\nu + \cont(T^*|_{\cols\geq j})$ be a partition, and
has the same choice of maximal $j$ and minimal $k$ as did $T$.
Hence the map $T \mapsto T^*$ is an involution on the violators
that lets one cancel their summands
$a_{\nu+\cont(T)+\rho}$ and $a_{\nu+\cont(T^*)+\rho}$ in pairs.%
\footnote{One remark is in order: The tableaux $T$ and $T^*$ may
be equal. In this case, the summands
$a_{\nu+\cont(T)+\rho}$ and $a_{\nu+\cont(T^*)+\rho}$ do not cancel,
as they are the same summand.
However, this summand is zero (because
$t_{k,k+1} ( \nu+\cont(T)+\rho )
= \nu+\cont\left(\underbrace{T^*}_{=T}\right)+\rho
= \nu+\cont(T)+\rho$ shows that the $n$-tuple $\nu+\cont(T)+\rho$
has two equal entries, and thus $a_{\nu+\cont(T)+\rho} = 0$),
and thus does not affect the sum.}
\end{proof}

% [DG][v17] I've done various changes to the preceding (bialternant)
% section. The case of $\kk$ being a commutative ring is now the
% new normal, whereas all requirements on $\kk$ to be $\ZZ$ or not
% to have characteristic $2$ are now explicit. The corollary which
% gives the Schur function as a quotient of two alternants is stated
% in the general case, and a footnote now serves to explain why
% the quotient is well-defined. In contrast, the proposition which
% states that the alternants are a basis for $\Lambda^\sgn$ is only
% stated in the restricted setting (as it was before, but now
% explicitly). The argument that $a_{(5,2,2)} = 0$ in the Example
% has gotten a footnote to explain why it works despite $2$ not
% being invertible.

% [DG][v19] Added footnote in the above proof about why a maximal $j$
% exists. I stumbled upon this subtlety when I tried to generalize
% the proof.

% [DG][v77] Added the footnote at the end of the proof, about the
% case $T^* = T$ (thanks Jang Soo Kim for the pointer).

\begin{example}
\label{exa.pf.Stembridge-theorem.invol}
Here is an example of the construction of $T^*$ in the above
proof.
Let $n = 6$ and
$\lambda = \left(5,4,4\right)$ and $\mu = \left(2,2\right)$
and $\nu = \left(1\right)$.
Let $T$ be the column-strict tableau
\[
\begin{matrix}
  &   & 1 & 2 & 2 \\
  &   & 2 & 3 \\
2 & 2 & 3 & 4
\end{matrix}
\quad \text{ of shape } \lambda / \mu .
\]
Then, % $\cont(T) = \left(1,5,2,1,0,0,0,\ldots\right)$ and
\begin{align*}
\cont(T|_{\cols \geq 5}) &= \left(0,1,0,0,0,\ldots\right)
\quad \text{(since $T|_{\cols \geq 5}$ has a single entry, which is $2$)}, \\
\text{so that} \quad
\nu + \cont(T|_{\cols \geq 5}) &= \left(1,1,0,0,0,\ldots\right)
\text{ is a partition.}
\end{align*}
But
\begin{align*}
\cont(T|_{\cols \geq 4}) &= \left(0,2,1,1,0,0,0,\ldots\right) , \\
\text{and thus} \quad
\nu + \cont(T|_{\cols \geq 4}) &= \left(1,2,1,1,0,0,0,\ldots\right)
\text{ is not a partition.}
\end{align*}
Thus, the $j$ in the above proof of Theorem~\ref{Stembridge-theorem}
is $4$.
Furthermore, the $k$ in the proof is $1$, since
$\nu_1 + \cont_1 (T|_{\cols \geq 4}) = 1 + 0
= 1 < 2 = 0 + 2 = \nu_2 + \cont_2 (T|_{\cols \geq 4})$.
Thus, $T^*$ is obtained from $T$ by applying the Bender-Knuth
involution on letters $1,2$ to columns $1,2,3$ only, leaving
columns $4,5$ unchanged. The result is
\[
T^* =
\begin{matrix}
  &   & 1 & 2 & 2 \\
  &   & 2 & 3 \\
1 & 1 & 3 & 4
\end{matrix}
\quad .
\]
\end{example}

% [DG][v77] Added the above example.

So far (in this section) we have worked with a finite set of
variables $x_1,x_2,\ldots,x_n$ (where $n$ is a fixed nonnegative
integer) and with partitions having at most $n$ parts. We now drop
these conventions and restrictions; thus, partitions again mean
arbitrary partitions, and $\xx$ again means the infinite family
$\left(x_1,x_2,x_3,\ldots\right)$ of variables. In this setting,
we have the following analogue of Corollary~\ref{finite-L-R-rule}:

\begin{corollary}
\label{infinite-L-R-rule}
For partitions $\lambda, \mu, \nu$ (of any lengths), one has
\begin{equation}
\label{eq.infinite-L-R-rule.1}
s_{\nu} s_{\lambda/\mu} = \sum_T s_{\nu+\cont(T)}
\end{equation}
where $T$ runs through all column-strict tableaux
of shape $\lambda/\mu$ with
the property that for each $j=1,2,3,\ldots$, the weak
composition $\nu + \cont(T|_{\cols\geq j})$ is a partition.
In particular, taking $\nu=\varnothing$, we obtain
\begin{equation}
\label{eq.infinite-L-R-rule.2}
s_{\lambda/\mu} = \sum_T s_{\cont(T)}
\end{equation}
where in the sum $T$ runs through all column-strict
tableaux of shape $\lambda/\mu$
for which each $\cont(T|_{\cols\geq j})$ is a partition.
\end{corollary}

\begin{proof}[Proof of Corollary~\ref{infinite-L-R-rule}.]
Essentially, Corollary~\ref{infinite-L-R-rule} is obtained from
Corollary~\ref{finite-L-R-rule} by ``letting $n$ (that is,
the number of variables) tend to $\infty$''. This can be formalized
in different ways: One way is to endow the ring
of power series $\kk\left[\left[\xx\right]\right]
= \kk\left[\left[x_1,x_2,x_3,\ldots\right]\right]$ with the
coefficientwise topology\footnote{This topology is defined as
follows:
\par
We endow the ring $\kk$ with the discrete topology. Then, we can regard
the $\kk$-module $\kk \left[ \left[ \xx \right] \right]$ as a direct
product of infinitely many copies of $\kk$ (by identifying every power
series in $\kk \left[ \left[ \xx \right] \right]$ with the
family of its coefficients). Hence, the product topology is a well-defined
topology on $\kk \left[ \left[ \xx \right] \right]$; this
topology is denoted as the \dfn{coefficientwise topology}.
Its name is due to the fact that a sequence
$\left( a_n \right)_{n\in \NN}$ of power series converges
to a power series $a$ with respect to this topology if and
only if for every monomial $\mathfrak{m}$, all sufficiently
high $n\in \NN$ satisfy
\[
\left(  \text{the coefficient of }\mathfrak{m}\text{ in }a_{n}\right)
=\left(  \text{the coefficient of }\mathfrak{m}\text{ in }a\right)  .
\]
}, and to show that the left hand side of
\eqref{eq.finite-L-R-rule.1} tends to the left hand side of
\eqref{eq.infinite-L-R-rule.1} when $n \to \infty$, and the same
holds for the right hand sides. A different approach proceeds by
regarding $\Lambda$ as the inverse limit of the
$\Lambda\left(x_1,x_2,\ldots,x_n\right)$.
\end{proof}

Comparing coefficients of a given Schur function $s_\nu$
in \eqref{eq.infinite-L-R-rule.2}, we obtain
the following version of the Littlewood-Richardson rule.
\begin{corollary}
\label{L-R-rule}
For partitions $\lambda, \mu, \nu$ (of any lengths),
the Littlewood-Richardson coefficient $c^\lambda_{\mu,\nu}$ counts
column-strict tableaux $T$ of shape $\lambda/\mu$ with $\cont(T)=\nu$ having
the property that each $\cont(T|_{\cols\geq j})$ is a partition.
\end{corollary}

% [DG][v19] Last time, I suggested to add
% http://mathoverflow.net/questions/118225/how-to-show-a-certain-determinant-is-non-zero
% as an exercise. Turns out that the intended solution was false...




\subsection{The Pieri and Assaf-McNamara skew Pieri rule}

The classical \dfn{Pieri rule} refers to two special cases of the Littlewood-Richardson rule.
To state them, recall that a skew shape
is called a \emph{horizontal (resp. vertical) strip}\index{horizontal strip}\index{vertical strip}
if no two of its cells lie in the same column (resp. row).
A \emph{horizontal (resp. vertical) $n$-strip}\index{horizontal $n$-strip}\index{vertical $n$-strip}
(for $n \in \NN$)
shall mean a horizontal (resp. vertical) strip of size $n$
(that is, having exactly $n$ cells).

% [DG][v25] Added definition of "strip"s (before, only "$n$-strips"
% were defined, but "strips" were referred to -- but more importantly
% I want to use "strip"s in exercises).

\begin{theorem}
\label{Pieri-rules}
For every partition $\lambda$ and any $n \in \NN$, we have
\begin{align}
s_\lambda h_n =&
\sum\limits_{\substack{\lambda^+: \lambda^+/\lambda \text{ is a }\\\text{horizontal }n\text{-strip}}}  s_{\lambda^+} ;
\label{eq.pieri.h} \\
s_\lambda e_n =&
\sum\limits_{\substack{\lambda^+: \lambda^+/\lambda \text{ is a }\\\text{vertical }n\text{-strip}}}  s_{\lambda^+} .
\label{eq.pieri.e}
\end{align}
\end{theorem}

\begin{example}
\label{exam.Pieri-rules.1}
In the following equality, we are representing each partition
by its Ferrers diagram\footnote{And we are drawing each Ferrers
diagram with its boxes spaced out, in order to facilitate
counting the boxes.}.
\[
\begingroup % keep the change local
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{aligned}
&\begin{matrix}
s &   &   & \\
  &\sq&\sq&\sq\\
  &\sq&\sq&   \\
  &\sq&\sq&
\end{matrix}
\qquad \bullet \qquad
\begin{matrix}
h_2 &   &   \\
%  &\sq&\sq \\
  &   & \\
  &   & \\
  &   & \\
\end{matrix}
\\
 & \\
&= \ \ \ 
%\qquad
\begin{matrix}
s &   &   & \\
  &\sq&\sq&\sq\\
  &\sq&\sq&   \\
  &\sq&\sq& \\
  &\bsq&\bsq&
\end{matrix}
\quad
+
\quad
\begin{matrix}
s &   &   & \\
  &\sq&\sq&\sq\\
  &\sq&\sq&\bsq\\
  &\sq&\sq& \\
  &\bsq&  &
\end{matrix}
\quad 
+
\quad
\begin{matrix}
s &   &   &   &\\
  &\sq&\sq&\sq&\bsq\\
  &\sq&\sq&   &\\
  &\sq&\sq&   &\\
  &\bsq&  &   &
\end{matrix}
\\
& \\
&
\qquad \qquad \qquad
\ 
+
\quad 
\begin{matrix}
s &   &   &   &\\
  &\sq&\sq&\sq&\bsq\\
  &\sq&\sq&\bsq &\\
  &\sq&\sq&   &\\
\end{matrix}
\quad
+
\quad
\begin{matrix}
s &   &   &   &    &\\
  &\sq&\sq&\sq&\bsq&\bsq\\
  &\sq&\sq&   &    &\\
  &\sq&\sq&   &    &\\
\end{matrix}
\end{aligned}
\endgroup
\]
If $\lambda$ is the partition $\left(3,2,2\right)$ on
the left hand side, then all partitions $\lambda^+$ on the
right hand side visibly have the property that
$\lambda^+ / \lambda$ is a horizontal $2$-strip\footnote{We
have colored the boxes of $\lambda^+ / \lambda$ black.},
as \eqref{eq.pieri.h} predicts.
\end{example}

% [DG][v77] Added explanations to the above example.
% Made the space between columns (\arraycolsep) in the
% Ferrers diagrams smaller so it approximately matches
% the space between rows (is there an exact way to do
% this? LaTeX doesn't know \arrayrowsep).
% Also, replaced h_{[][]} by h_2, since I don't see
% much of a point in drawing the diagram under an h.

\begin{vershort}
\begin{proof}[Proof of Theorem~\ref{Pieri-rules}]
For the first Pieri formula involving $h_n$, as $h_n=s_{(n)}$ one has
\[
s_\lambda h_n = \sum_{\lambda^+} c^{\lambda^+}_{\lambda,(n)} s_{\lambda^+}.
\]
Corollary~\ref{L-R-rule} says $c^{\lambda^+}_{\lambda,(n)}$ counts
column-strict tableaux $T$ of shape $\lambda^+/\lambda$
having $\cont(T)=(n)$ (i.e. all entries of $T$ are $1$'s),
with an extra condition.  Since its entries are all equal,
such a $T$ must certainly have shape being a horizontal strip, and
more precisely a horizontal $n$-strip (since it has $n$ cells).
Conversely,
for any horizontal $n$-strip, there is a unique such filling,
and it will trivially satisfy the extra condition
that $\cont(T|_{\cols\geq j})$ is a partition for each $j$.
Hence $c^{\lambda^+}_{\lambda,(n)}$ is $1$ if $\lambda^+/\lambda$ is
a horizontal $n$-strip, and $0$ else.

For the second Pieri formula involving $e_n$, using $e_n=s_{(1^n)}$ one has
\[
s_\lambda e_n = \sum_{\lambda^+} c^{\lambda^+}_{\lambda,(1^n)} s_{\lambda^+}.
\]
Corollary~\ref{L-R-rule} says $c^{\lambda^+}_{\lambda,(1^n)}$
counts column-strict tableaux $T$ of shape $\lambda^+/\lambda$
having $\cont(T)=(1^n)$, so its entries are $1,2,\ldots,n$ each
occurring once, with the extra condition that $1,2,\ldots,n$
appear from right to left.  Together with
the tableau condition, this forces at most one entry in each row, that is
$\lambda^+/\lambda$ is a vertical strip, and then there is a unique
way to fill it  (maintaining column-strictness and the extra condition
that $1,2,\ldots,n$ appear from right to left).  Thus $c^{\lambda^+}_{\lambda,(1^n)}$
is $1$ if $\lambda^+/\lambda$ is a vertical $n$-strip, and $0$ else.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}[Proof of Theorem~\ref{Pieri-rules}]
For the first Pieri formula involving $h_n$, as $h_n=s_{(n)}$ one has
\begin{equation}
\label{pf.Pieri-rules.h}
s_\lambda h_n = \sum_{\lambda^+} c^{\lambda^+}_{\lambda,(n)} s_{\lambda^+}.
\end{equation}
Corollary~\ref{L-R-rule} says $c^{\lambda^+}_{\lambda,(n)}$ counts
column-strict tableaux $T$ of shape $\lambda^+/\lambda$
having $\cont(T)=(n)$ (i.e. all entries of $T$ are $1$'s),
with an extra condition (stating that
$\cont(T|_{\cols\geq j})$ is a partition for each $j$).  Since
its entries are all equal,
such a $T$ must certainly have shape being a horizontal strip, and
more precisely a horizontal $n$-strip (since it has $n$ cells).
Conversely,
for any horizontal $n$-strip, there is a unique such filling,
and it will trivially satisfy the extra condition
that $\cont(T|_{\cols\geq j})$ is a partition for each $j$.
Hence $c^{\lambda^+}_{\lambda,(n)}$ is $1$ if $\lambda^+/\lambda$ is
a horizontal $n$-strip, and $0$ else. Thus,
\eqref{pf.Pieri-rules.h} becomes
\[
s_\lambda h_n =
\sum\limits_{\substack{\lambda^+: \lambda^+/\lambda \text{ is a }\\\text{horizontal }n\text{-strip}}}  s_{\lambda^+} ,
\]
and \eqref{eq.pieri.h} is proven.

For the second Pieri formula involving $e_n$, using $e_n=s_{(1^n)}$ one has
\begin{equation}
\label{pf.Pieri-rules.e}
s_\lambda e_n = \sum_{\lambda^+} c^{\lambda^+}_{\lambda,(1^n)} s_{\lambda^+}.
\end{equation}
Corollary~\ref{L-R-rule} says $c^{\lambda^+}_{\lambda,(1^n)}$
counts column-strict tableaux $T$ of shape $\lambda^+/\lambda$
having $\cont(T)=(1^n)$, so its entries are $1,2,\ldots,n$ each
occurring once, with an extra condition (stating that
$\cont(T|_{\cols\geq j})$ is a partition for each $j$).
This extra condition forces the entries $1,2,\ldots,n$ to
appear in the tableau $T$ from right to left (more
precisely, if $i$ and $i'$ are two entries of $T$
satisfying $i' \leq i$, then the column in which $i'$ appears
is weakly to the right of the column in which $i$
appears\footnote{\textit{Proof:} Assume the
contrary. Then, there are two entries $i$ and $i'$ of $T$
satisfying $i' \leq i$, such that $i'$ appears in a column
strictly to the left of the column in which $i$ appears.
Consider these $i$ and $i'$. Let $j$ be the column in which
$i$ appears, and $j'$ be the column in which $i'$ appears;
thus, $j' < j$. We have $i' \leq i$, so that
$\left(\cont(T|_{\cols\geq j})\right)_{i'} \geq
\left(\cont(T|_{\cols\geq j})\right)_i$ (due to the extra
condition saying that $\cont(T|_{\cols\geq j})$ is a partition).
Hence,
$\left(\cont(T|_{\cols\geq j})\right)_{i'} \geq
\left(\cont(T|_{\cols\geq j})\right)_i \geq 1$ (since $i$ does
appear in $T|_{\cols\geq j}$, being an entry of column $j$ of
$T$), which means that at least one entry of
$T|_{\cols\geq j}$ must be equal to $i'$. In other words, $i'$
appears in one of the columns $j, j+1, j+2, \ldots$ of the
tableau $T$. Since $i'$ also appears in column $j'$ (by the
definition of $j'$), which is \textbf{not} one of the
columns $j, j+1, j+2, \ldots$ (since $j' < j$), this shows
that $i'$ appears
(at least) twice in the tableau $T$. But this is absurd, since
the entries of $T$ are $1,2,\ldots,n$ each occurring once. This
contradiction proves that our assumption was wrong, qed.});
it also forces the tableau $T$ to have at
most one entry in each row\footnote{\textit{Proof:} Assume the
contrary. Thus, some row of $T$ has (at least) two entries
-- say, an entry $i$ in a column $j$, and an entry $i'$ in a
different column $j'$. Assume WLOG that $j' < j$. Thus,
$i' \leq i$ (since the entries of $T$ increase weakly
left-to-right along rows). Now, a contradiction can be obtained
as in the previous footnote, and thus our assumption was wrong,
qed.}. Thus, for such a tableau $T$ to exist,
$\lambda^+/\lambda$ has to be a vertical strip (because $T$
must have at most one entry in each row), more precisely
a vertical $n$-strip. Conversely, given a partition $\lambda^+$
for which $\lambda^+/\lambda$ is a vertical
$n$-strip, there is a unique way to fill it to obtain a
column-strict tableau $T$ satisfying our extra
condition\footnote{Indeed, one must fill $\lambda^+/\lambda$
with the entries $1,2,\ldots,n$ from top to bottom; this is
the only way to maintain column-strictness and the extra condition
that $1,2,\ldots,n$ appear from right to left.}.
Thus $c^{\lambda^+}_{\lambda,(1^n)}$
is $1$ if $\lambda^+/\lambda$ is a vertical $n$-strip, and $0$ else.
Hence, \eqref{pf.Pieri-rules.e} becomes
\[
s_\lambda e_n =
\sum\limits_{\substack{\lambda^+: \lambda^+/\lambda \text{ is a }\\\text{vertical }n\text{-strip}}}  s_{\lambda^+} ,
\]
and \eqref{eq.pieri.e} is proven.
\end{proof}
\end{verlong}

% [DG][v14] Replaced "maintaining column-strictness" by
% "(maintaining column-strictness and the extra condition
% that $1,2,\ldots,n$ appear from right to left)". Or am I
% wrong here?

In 2009, Assaf and McNamara \cite{AssafMcNamara} proved an elegant
generalization.\index{Assaf-McNamara skew Pieri rule}

\begin{theorem}
\label{Assaf-McNamara-skew-Pieri}
For any partitions $\lambda$ and $\mu$ and any $n \in \NN$,
we have\footnote{Note that $\mu \subseteq \lambda$ is not required.
(The left hand sides are $0$ otherwise, but this does not
trivialize the equalities.)}
\begin{align}
s_{\lambda/\mu} h_n &=
\sum\limits_{\substack{\lambda^+,\mu^-:\\
                  \lambda^+/\lambda \text{ a horizontal strip;}\\
                   \mu/\mu^{-}\text{ a vertical strip;}\\
                   |\lambda^+/\lambda|+|\mu/\mu^-|=n}}
           (-1)^{|\mu/\mu^{-}|} s_{\lambda^+/\mu^-} ;
\label{eq.Assaf-McNamara-skew-Pieri.1}
\\
s_{\lambda/\mu} e_n &=
\sum\limits_{\substack{\lambda^+,\mu^-: \\
                 \lambda^+/\lambda \text{ a vertical strip;}\\
                    \mu/\mu^{-}\text{ a horizontal strip;}\\
                    |\lambda^+/\lambda|+|\mu/\mu^-|=n}}
           (-1)^{|\mu/\mu^{-}|} s_{\lambda^+/\mu^-} .
\label{eq.Assaf-McNamara-skew-Pieri.2}
\end{align}
\end{theorem}

\begin{example}
With the same conventions as in Example~\ref{exam.Pieri-rules.1}%
\footnote{but this time coloring both the boxes in $\lambda^+ / \lambda$
and the boxes in $\mu / \mu^-$ black},
we have
\[
\begingroup % keep the change local
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{aligned}
&\begin{matrix}
s &   &   & \\
  &   &\sq&\sq\\
  &   &\sq&   \\
  &\sq&\sq&
\end{matrix}
\qquad \bullet \qquad
\begin{matrix}
h_2 &   &   \\
%  &\sq&\sq \\
  &   & \\
  &   & \\
  &   & \\
\end{matrix}
\\
 & \\
&=\qquad
\begin{matrix}
s &   &   & \\
  &   &\sq&\sq\\
  &   &\sq&   \\
  &\sq&\sq& \\
  &\bsq&\bsq&
\end{matrix}
\qquad+\qquad
\begin{matrix}
s &   &   & \\
  &   &\sq&\sq\\
  &   &\sq&\bsq\\
  &\sq&\sq& \\
  &\bsq&  &
\end{matrix}
\qquad+\qquad
\begin{matrix}
s &   &   &   &\\
  &   &\sq&\sq&\bsq\\
  &   &\sq&   &\\
  &\sq&\sq&   &\\
  &\bsq&  &   &
\end{matrix}
\\
& \\
&\qquad \qquad \qquad
+\qquad
\begin{matrix}
s &   &   &   &\\
  &   &\sq&\sq&\bsq\\
  &   &\sq&\bsq &\\
  &\sq&\sq&   &\\
\end{matrix}
\qquad+\qquad\begin{matrix}
s &   &   &   &    &\\
  &   &\sq&\sq&\bsq&\bsq\\
  &   &\sq&   &    &\\
  &\sq&\sq&   &    &\\
\end{matrix}
\\
 & \\
&\qquad \qquad
-\quad
\begin{matrix}
s &    &   &   \\
  &    &\sq&\sq\\
  &\bsq&\sq&   \\
  &\sq &\sq&   \\
  &\bsq&   &
\end{matrix}
\qquad-\qquad
\begin{matrix}
s &    &   &   \\
  &    &\sq&\sq\\
  &\bsq&\sq&\bsq\\
  &\sq &\sq&   \\
\end{matrix}
\qquad-\qquad
\begin{matrix}
s &    &   &   &\\
  &    &\sq&\sq&\bsq\\
  &\bsq&\sq&   &\\
  &\sq &\sq&   &
\end{matrix}
\\
 & \\
&\qquad\qquad\qquad\qquad
+\qquad
\begin{matrix}
s &    &   &   \\
  &\bsq&\sq&\sq\\
  &\bsq&\sq&   \\
  &\sq &\sq&
\end{matrix}
\end{aligned}
\endgroup
\]
which illustrates the first equality of
Theorem~\ref{Assaf-McNamara-skew-Pieri}.
\end{example}

% [DG][v77] Same changes to this example as above.

\noindent
Theorem~\ref{Assaf-McNamara-skew-Pieri} is proven
in the next section, using an important Hopf algebra tool.

\begin{exercise}
\label{exe.horistrip.postnikov}
Let $\lambda = \left(\lambda_1, \lambda_2, \lambda_3, \ldots\right)$ and
$\mu = \left(\mu_1, \mu_2, \mu_3, \ldots\right)$
be two partitions such that $\mu \subseteq \lambda$.

\begin{enumerate}

\item[(a)] Show
that $\lambda / \mu$ is a horizontal strip if and only if every
$i \in \left\{ 1,2,3,\ldots \right\}$ satisfies
$\mu_i \geq \lambda_{i+1}$.\ \ \ \ \footnote{In
other words, $\lambda / \mu$ is a horizontal strip if and only if
$\left( \lambda_2, \lambda_3, \lambda_4, \ldots \right)
\subseteq \mu$.
This simple observation has been used by Pak and Postnikov
\cite[\S 10]{PakPostnikov-oscil} for a new approach to RSK-type
algorithms.}

\item[(b)] Show
that $\lambda / \mu$ is a vertical strip if and only if every
$i \in \left\{ 1,2,3,\ldots \right\}$ satisfies
$\lambda_i \leq \mu_i + 1$.

\end{enumerate}
\end{exercise}

% [DG][v25] Added this (near-trivial) exercise.

% [DG][v48] Added part (b) (for the sake of "symmetry" and possibly
% of later reference; the proof is a one-liner).

\begin{exercise} \phantomsection
\label{exe.pieri.h2}
\begin{itemize}
\item[(a)] Let $\lambda$ and $\mu$ be two partitions such that $\mu
\subseteq \lambda$. Let $n \in \NN$.
Show that $\left(h_n, s_{\lambda / \mu}\right)$ equals $1$ if $\lambda / \mu$
is a horizontal $n$-strip, and equals $0$ otherwise.
\item[(b)] Use part (a) to give a new proof of
\eqref{eq.pieri.h}.
\end{itemize}
\end{exercise}

% [DG][v25] Added above exercise showing that the first Pieri rule is
% simpler than Littlewood-Richardson (provided one knows Cauchy).
% (I don't see a similar simple argument for the second Pieri rule,
% though.)

\begin{exercise}
\label{exe.pieri.RSK}
Prove Theorem~\ref{Pieri-rules} again using the ideas of
the proof of Theorem~\ref{thm.cauchy-identity}.
\end{exercise}

% [DG][v47] Added previous exercise (it comes almost for free once RSK is
% introduced).

\begin{verlong}
\begin{exercise}
\label{exe.pieri.alt-e}
Obtain a new proof for \eqref{eq.pieri.e} using
Corollary~\ref{cor.stembridge.slambda} and the formula
\eqref{e-generating-function}.\footnote{Such a proof is
given in \cite[\S I.5, Rmk. 1, pp. 73--74]{Macdonald} and in
\cite[Proposition 2.3]{Leeuwen-altSchur}. A similar, but more
complicated, proof can be given for \eqref{eq.pieri.h}; this is
done, for example, in \cite[Proposition 2.5]{Leeuwen-altSchur}.}
\end{exercise}
\end{verlong}

\begin{exercise}
\label{exe.cauchy.cauchy}
Let $A$ be a commutative ring, and $n \in \NN$.
\begin{itemize}
\item[(a)] Let
$a_1, a_2, \ldots, a_n$ be $n$ elements of $A$.
Let $b_1, b_2, \ldots, b_n$ be $n$ further elements
of $A$.
If $a_i - b_j$ is an invertible element of
$A$ for every $i \in \left\{1,2,\ldots,n\right\}$ and
$j \in \left\{1,2,\ldots,n\right\}$, then prove that
\[
\det\left(\left(\frac{1}{a_i - b_j}\right)_{i,j = 1,2,\ldots,n}\right)
= \frac{\prod_{1 \leq j < i \leq n} \left(\left(a_i - a_j\right)
\left(b_j - b_i\right) \right)}{\prod_{\left(i, j\right)
\in \left\{1,2,\ldots,n\right\}^2}
\left(a_i - b_j\right)}.
\]
\item[(b)] Let
$a_1, a_2, \ldots, a_n$ be $n$ elements of $A$.
Let $b_1, b_2, \ldots, b_n$ be $n$ further elements
of $A$.
If $1 - a_i b_j$ is an invertible element
of $A$ for every $i \in \left\{1,2,\ldots,n\right\}$ and
$j \in \left\{1,2,\ldots,n\right\}$, then prove that
\[
\det\left(\left(\frac{1}{1 - a_i b_j}\right)_{i,j = 1,2,\ldots,n}\right)
= \frac{\prod_{1 \leq j < i \leq n} \left(\left(a_i - a_j\right)
\left(b_i - b_j\right) \right)}{\prod_{\left(i, j\right)
\in \left\{1,2,\ldots,n\right\}^2}
\left(1 - a_i b_j\right)}.
\]
\item[(c)] Use the result of part (b) to give a new
proof for Theorem~\ref{thm.cauchy-identity}.\footnote{This
approach to Theorem~\ref{thm.cauchy-identity} is taken in
\cite[\S 4]{CrawleyBoevey} (except that
\cite{CrawleyBoevey} only works with finitely many
variables).}
\end{itemize}
\end{exercise}

The determinant on the left hand side of
Exercise~\ref{exe.cauchy.cauchy}(a) is known as the \dfn{Cauchy
determinant}.

% [DG][v25] Added above exercise.

\begin{exercise}
\label{exe.jacobi-trudi.len2.0}
Prove that $s_{\left(a,b\right)} = h_a h_b - h_{a+1} h_{b-1}$ for any
two integers $a \geq b \geq 0$ (where we set $h_{-1} = 0$ as usual).

(Note that this is precisely the Jacobi-Trudi formula
\eqref{eq.jacobi-trudi.h} in the case when
$\lambda = \left(a,b\right)$ is a partition with at most two
entries and $\mu = \varnothing$.)
\end{exercise}

\begin{exercise}
\label{exe.pieri.cauchy}
If $\lambda$ is a partition and $\mu$ is a weak composition, let
$K_{\lambda, \mu}$ denote the number of column-strict tableaux $T$ of
shape $\lambda$ having $\cont\left(T\right) = \mu$. (This
$K_{\lambda, \mu}$ is called the
\dfn{$\left(\lambda, \mu\right)$-Kostka number}.)
\begin{itemize}
\item[(a)] Use Theorem~\ref{Pieri-rules} to show that every
partition $\mu$ satisfies
$h_\mu = \sum_\lambda K_{\lambda, \mu} s_\lambda$, where the sum
ranges over all partitions $\lambda$.
\item[(b)] Use this to give a new proof for
Theorem~\ref{thm.cauchy-identity}.\footnote{Of course, this gives
a new proof of Theorem~\ref{thm.cauchy-identity} only when coupled
with a proof of Theorem~\ref{Pieri-rules} which does not rely on
Theorem~\ref{thm.cauchy-identity}. The proof of
Theorem~\ref{Pieri-rules} we gave in the text above did not rely
on Theorem~\ref{thm.cauchy-identity}, whereas the proof of
\eqref{eq.pieri.h} given in Exercise~\ref{exe.pieri.h2}(b) did.}
\item[(c)] Give a new proof of the fact (previously shown as
Proposition~\ref{prop.Lambda.omega-and-S}(j)) that
$\left(h_\lambda\right)_{\lambda\in\Par}$ is a graded basis of
the graded $\kk$-module $\Lambda$.
\end{itemize}
\end{exercise}

% [DG][v19] Added above exercise.

% [DG][v23] Added part (c).

\begin{exercise} \phantomsection
\label{exe.pieri.omega}
\begin{itemize}
\item[(a)] Define a $\kk$-linear map $\mathfrak{Z} : \Lambda \to \Lambda$
by having it send $s_\lambda$ to $s_{\lambda^t}$ for every partition
$\lambda$. (This is clearly well-defined, since $\left(s_\lambda\right)_{\lambda \in \Par}$
is a $\kk$-basis of $\Lambda$.) Show that
\[
\mathfrak{Z}\left(f h_n\right) = \mathfrak{Z}\left(f\right) \cdot \mathfrak{Z}\left(h_n\right)
\qquad \qquad \text{ for every } f \in \Lambda \text{ and every } n \in \NN .
\]
\item[(b)] Show that $\mathfrak{Z} = \omega$.
\item[(c)] Show that $c_{\mu, \nu}^{\lambda} = c_{\mu^t, \nu^t}^{\lambda^t}$ for
any three partitions $\lambda$, $\mu$ and $\nu$.
\item[(d)] Use this to prove \eqref{skew-Schur-antipodes}.\footnote{The
first author learned this approach to \eqref{skew-Schur-antipodes} from
Alexander Postnikov.}
\end{itemize}
\end{exercise}

% [DG][v23] Added the preceding exercise.

\begin{exercise} \phantomsection
\label{exe.cauchy.dual}
\begin{itemize}
\item[(a)] Show that
\[
\prod_{i,j=1}^\infty \left(1+x_iy_j\right)
= \sum_{\lambda \in \Par} s_\lambda\left(\xx\right) s_{\lambda^t}\left(\yy\right)
= \sum_{\lambda \in \Par} e_\lambda\left(\xx\right) m_\lambda\left(\yy\right)
\]
in the power series ring $\kk\left[\left[\xx, \yy\right]\right]
= \kk\left[\left[x_1, x_2, x_3, \ldots, y_1, y_2, y_3, \ldots\right]\right]$.
\item[(b)] Assume that $\QQ$ is a subring of $\kk$. Show that
\[
\prod_{i,j=1}^\infty \left(1+x_iy_j\right)
= \sum_{\lambda \in \Par}
\left(-1\right)^{\left|\lambda\right| - \ell\left(\lambda\right)}
z_\lambda^{-1} p_\lambda\left(\xx\right) p_\lambda\left(\yy\right)
\]
in the power series ring $\kk\left[\left[\xx, \yy\right]\right]
= \kk\left[\left[x_1, x_2, x_3, \ldots, y_1, y_2, y_3, \ldots\right]\right]$,
where $z_\lambda$ is defined as in
Proposition~\ref{prop-Two-other-Cauchy-expansions}.
\end{itemize}
\end{exercise}

The first equality of Exercise~\ref{exe.cauchy.dual}(a) appears in
\cite[Thm. 7.14.3]{Stanley}, \cite[Thm. 4.8.6]{Sagan} and several other
references under the name of the \dfn{dual Cauchy identity}, and is commonly
proven using a ``dual'' analogue of the Robinson-Schensted-Knuth algorithm.

% [DG][v25] Added the above exercise (on the dual Cauchy identity).

% [DG][v45] Added a solution to its part (a), using a variation on RSK
% which uses the same RS-insertion step as usual RSK but processes the
% biletters in a different order.
% Curiously it seems to be unpopular in literature (instead, everyone
% uses dual RSK).

% [DG][v46] Added the dual RSK solution too (mainly to nail down the
% properties of dual RSK, which I need to know for a different reason).

\begin{verlong}
The following two exercises (Exercise~\ref{exe.jacobi-trudi} and
Exercise~\ref{exe.jacobi-trudi.macdonald}) give two different proofs of
the Jacobi-Trudi identities.

\begin{exercise}
\label{exe.jacobi-trudi}
For any two partitions $\lambda$ and $\mu$ and any $k \in \NN$, define a
symmetric function $\widetilde{s}_{\lambda / \mu}^{[k]} \in \Lambda$
by
\[
\widetilde{s}_{\lambda / \mu}^{[k]}
= \det \left(\left( h_{\lambda_i - \mu_j - i + j} \right)_{i,j=1,2,\ldots,k}  \right).
\]
(As usual, we are denoting by $\nu_i$ the $i$-th entry of a partition $\nu$
here.)
\begin{itemize}
\item[(a)] For any partitions $\lambda$ and $\mu$ and any $k \in \NN$
satisfying $k \geq \ell\left(\lambda\right)$ and
$k \geq \ell\left(\mu\right)$, show that the symmetric function
$\widetilde{s}_{\lambda / \mu}^{[k]}$ does not depend on $k$. We thus
can define a symmetric function $\widetilde{s}_{\lambda / \mu}$ for
any two partitions $\lambda$ and $\mu$ by setting
$\widetilde{s}_{\lambda / \mu} = \widetilde{s}_{\lambda / \mu}^{[k]}$,
where $k$ is any element of $\NN$ satisfying
$k \geq \ell\left(\lambda\right)$ and $k \geq \ell\left(\mu\right)$.
In the rest of this exercise, we will study this
$\widetilde{s}_{\lambda / \mu}$ (and show that it equals
$s_{\lambda / \mu}$).
\item[(b)] If two partitions $\lambda$ and $\mu$ do not satisfy
$\mu \subseteq \lambda$, then show that
$\widetilde{s}_{\lambda / \mu} = 0$.
\item[(c)] For any two partitions $\lambda$ and $\nu$, show that
\[
\Delta \widetilde{s}_{\lambda / \nu}
= \sum\limits_{\substack{\mu \in \Par: \\ \nu \subseteq \mu \subseteq \lambda}}
  \widetilde{s}_{\lambda / \mu} \otimes \widetilde{s}_{\mu / \nu}.
\]
\item[(d)] Show that
$\widetilde{s}_{\lambda / \mu} \in \Lambda_{\left|\lambda/\mu\right|}$
for any partitions $\lambda$ and $\mu$ satisfying
$\mu \subseteq \lambda$.
\item[(e)] For any two partitions $\lambda$ and $\mu$ and any
$n \in \NN$, show that
\[
\left( h_n, \widetilde{s}_{\lambda / \mu} \right)
= \left( h_n, s_{\lambda / \mu} \right).
\]
\item[(f)] If $x$ is a primitive element of $\Lambda$ such
that every $n \in \NN$ satisfies $\left( h_n, x \right) = 0$,
then show that $x = 0$.
\item[(g)] Prove that
$\widetilde{s}_{\lambda / \mu} = s_{\lambda / \mu}$ for any
two partitions $\lambda$ and $\mu$ satisfying
$\mu \subseteq \lambda$.
\item[(h)] Use this to obtain a proof of
Theorem~\ref{Jacobi-Trudi-formulae}.
\end{itemize}
\end{exercise}

% [DG][v75] Added the preceding exercise, but only to the verlong
% version. (Not the best proof of Jacobi-Trudi, but Hopfish and
% somewhat new. But it's become a mess when I wrote it up a few
% years ago, so I never felt like including it in the notes.
% But this is the verlong version, so no one will find it here
% except for those who really care.
% Macdonald uses similar arguments for a different result; I
% wasn't aware of % that when I started writing this up.)
\end{verlong}

\begin{exercise}
\label{exe.jacobi-trudi.macdonald}
Prove Theorem~\ref{Jacobi-Trudi-formulae}.

[\textbf{Hint:}\footnote{This is the proof given in
Stanley \cite[\S 7.16, Second Proof of Thm. 7.16.1]{Stanley} and
Macdonald \cite[proof of (5.4)]{Macdonald}.}
Switch $\xx$ and $\yy$ in the formula of
Exercise~\ref{exe.cauchy.skew}(a), and specialize the resulting
equality by replacing $\yy$ by a finite set of variables
$\left(y_1, y_2, \ldots, y_\ell\right)$;
then, set $n = \ell$ and $\rho = \left(n-1,n-2,\ldots,0\right)$,
and multiply with the alternant
$a_\rho\left(y_1, y_2, \ldots, y_\ell\right)$, using
Corollary~\ref{cor.stembridge.slambda} to simplify the result;
finally, extract the coefficient of $\yy^{\lambda + \rho}$.]
\end{exercise}

% [DG][v25] Added the preceding exercise (proving Jacobi-Trudi).

\begin{exercise}
\label{exe.hall-e-f}Prove the following:

\begin{enumerate}
\item[(a)] We have $\left(  S\left(  f\right)  ,S\left(  g\right)  \right)
=\left(  f,g\right)  $ for all $f\in\Lambda$ and $g\in\Lambda$.

\item[(b)] We have $\left(  e_{n},f\right)  =\left(  -1\right)  ^{n}%
\cdot\left(  S\left(  f\right)  \right)  \left(  1\right)  $ for any
$n \in \NN$ and $f \in \Lambda_n$. (See Exercise~\ref{exe.Lambda.subs.fin}
for the meaning of $\left(  S\left(  f\right)  \right)  \left(  1\right)  $.)
\end{enumerate}
\end{exercise}

% [DG][v75] Added the above exercise.



\subsection{Skewing and Lam's proof of the skew Pieri rule}

We codify here the operation $s_\mu^\perp$ of \emph{skewing by $s_\mu$},
acting on Schur functions via
\[
s_\mu^\perp(s_\lambda) = s_{\lambda/\mu}
\]
(where, as before, one defines $s_{\lambda/\mu}=0$ if $\mu \not\subseteq \lambda$).
These operations play a crucial role
\begin{itemize}
\item
in Lam's proof of the
skew Pieri rule,
\item in Lam, Lauve, and Sottile's proof \cite{LamLauveSottile}
of a more general  skew Littlewood-Richardson rule
that had been conjectured by Assaf and McNamara, and
\item in Zelevinsky's structure theory of PSH's to be developed in the next chapter.
\end{itemize}
We are going to define them in the general setting of any graded Hopf
algebra.

% [DG][v9] Here and in the following, "PSH-algebra" replaced by
% "PSH".

\begin{definition}
\label{def.skewing}
Given a graded Hopf algebra $A$, and its (graded) dual $A^o$, let
$(\cdot,\cdot)=(\cdot,\cdot)_A : A^o \times A \to \kk$ be
the pairing defined by
$(f,a):=f(a)$ for $f$ in $A^o$ and $a$ in $A$.
Then define for each
$f$ in $A^o$ an operator
$A \overset{f^\perp}{\rightarrow} A$ as follows\footnote{This $f^\perp(a)$ is
called $a \leftharpoonup f$ in Montgomery \cite[Example 1.6.5]{Montgomery}.}:
for $a$ in $A$
with $\Delta(a) = \sum a_1 \otimes a_2$, let
\[
f^\perp(a) = \sum (f,a_1) a_2.
\]
In other words, $f^\perp$ is the composition
\[
\xymatrix{
A \ar[r]^-{\Delta} & A \otimes A \ar[r]^-{f \otimes \id} & \kk \otimes A \ar[r]^-{\cong} & A ,
}
\]
where the rightmost arrow is the canonical isomorphism $\kk \otimes A \to A$.
This operator $f^\perp$ is called \dfn{skewing by $f$}.
\end{definition}

% [DG][v64] Removed the "finite type" condition from the above
% definition (I shall instead require it whenever it is necessary).
% Also, added the alternative definition through the composition
% of arrows.

\noindent
Now, recall that the Hall inner product induces an isomorphism
$\Lambda^o \cong \Lambda$ (by Corollary~\ref{cor.Lambda.selfdual}).
Hence, we can regard any element $f \in \Lambda$ as an element
of $\Lambda^o$; this allows us to define an operator
$f^\perp : \Lambda \to \Lambda$ for each $f \in \Lambda$
(by regarding $f$ as an element of $\Lambda^o$, and applying
Definition~\ref{def.skewing} to $A = \Lambda$).
Explicitly, this operator is given by
\begin{equation}
f^\perp(a) = \sum (f, a_1) a_2 \qquad \text{ whenever } \qquad
\Delta(a) = \sum a_1 \otimes a_2 ,
\label{eq.skewing.Lambda.general}
\end{equation}
where the inner product
$(f, a_1)$ is now understood as a Hall inner product.

Recall that each partition $\lambda$ satisfies
\[
\Delta s_\lambda
= \sum_{\mu \subseteq \lambda} s_\mu \otimes s_{\lambda/\mu}
= \sum_{\nu \subseteq \lambda} s_\nu \otimes s_{\lambda/\nu}
= \sum_{\nu} s_\nu \otimes s_{\lambda/\nu}
\]
(since $s_{\lambda/\nu} = 0$ unless $\nu \subseteq \lambda$).
Hence, for any two partitions $\lambda$ and $\mu$, we have
\begin{align}
s_\mu^\perp \left(s_\lambda\right)
&= \sum_\nu \underbrace{\left(s_\mu, s_\nu\right)}_{= \delta_{\mu, \nu}} s_{\lambda/\nu}
\qquad
\left(\text{by \eqref{eq.skewing.Lambda.general}, applied to }
      f = s_\mu \text{ and } a = s_\lambda \right)
\nonumber\\
&= \sum_\nu \delta_{\mu, \nu} s_{\lambda/\nu}
= s_{\lambda/\mu} .
\label{eq.skewing.Lambda.s}
\end{align}
Thus, skewing acts on the Schur functions exactly as desired.

% [DG][v46] Added footnote on how $\Lambda$ becomes selfdual.

% [DG][v64] Uncrowded the above paragraph, hopefully making it
% a lot clearer.

\begin{proposition}
\label{skewing-properties-prop}
Let $A$ be a graded Hopf algebra.
The $f^\perp$ operators $A \rightarrow A$ have the following properties.
\begin{enumerate}
\item[(i)] For every $f \in A^o$, the map $f^\perp$ is adjoint to left multiplication $A^o \overset{f \cdot}{\rightarrow} A^o$ in the sense that
\[
(g, f^\perp(a) ) = (fg, a).
\]
\item[(ii)] For every $f, g \in A^o$, we have $(fg)^\perp(a) = g^\perp(f^\perp(a))$, that is,
$A$ becomes a right $A^o$-module via the $f^\perp$ action.\footnote{This
makes sense, since $A^o$ is a $\kk$-algebra (by
Exercise~\ref{exe.dual-algebra}(c), applied to $C=A$).}
\item[(iii)] The unity $1_{A^o}$ of the $\kk$-algebra $A^o$
satisfies $\left(1_{A^o}\right)^\perp = \id_A$.
\item[(iv)] Assume that $A$ is of finite type (so $A^o$ becomes a
Hopf algebra, not just an algebra).
If an $f \in A^o$ satisfies $\Delta(f) = \sum f_1 \otimes f_2$, then
\[
f^\perp (a b) = \sum f_1^\perp(a) f_2^\perp(b).
\]
In particular, if $f$ is primitive in $A^o$,
so that $\Delta(f) = f \otimes \one + \one \otimes f$,
then $f^\perp$ is a \emph{derivation}:
\[
f^\perp (a b) = f^\perp(a) \cdot b + a \cdot f^\perp(b).
\]
\end{enumerate}
\end{proposition}

% [DG][v75] Added part (iii) to this proposition (we
% are already using it tacitly).

\begin{proof}
For (i), note that
\[
(g, f^\perp(a) )
=\sum (f,a_1) (g,a_2)
=( f \otimes g , \Delta_A (a) )
=( m_{A^o}(f \otimes g) , a )
= (fg, a).
\]

For (ii), using (i) and considering any $h$ in $A^o$, one has that
\[
(h, (fg)^\perp(a))
= (fgh, a)
= (gh, f^\perp(a))
= (h, g^\perp(f^\perp(a))).
\]

For (iii), we recall that the unity $1_{A^o}$ of $A^o$ is the counit $\epsilon$ of $A$,
and thus every $a \in A$ satisfies
\begin{align*}
\left(1_{A^o}\right)^\perp \left(a\right)
&= \epsilon^\perp \left(a\right)
= \sum_{\left(a\right)} \underbrace{\left(\epsilon, a_1\right)}_{=\epsilon\left(a_1\right)} a_2
\qquad \left(\text{by the definition of $\epsilon^\perp$}\right) \\
&= \sum_{\left(a\right)} \epsilon\left(a_1\right) a_2
= a \qquad \left(\text{by the axioms of a coalgebra}\right) ,
\end{align*}
so that $\left(1_{A^o}\right)^\perp = \id_A$.

For (iv), noting that
\[
\Delta(ab)= \Delta(a)\Delta(b)
= \left( \sum_{(a)} a_1 \otimes a_2 \right)
\left( \sum_{(b)} b_1 \otimes b_2 \right)
= \sum_{(a),(b)} a_1 b_1 \otimes a_2 b_2,
\]
one has that
\begin{align*}
f^\perp(ab)
&= \sum_{(a),(b)} (f,a_1 b_1)_A \,\, a_2 b_2
= \sum_{(a),(b)} (\Delta(f),a_1 \otimes b_1)_{A \otimes A} \,\, a_2 b_2 \\
&= \sum_{(f),(a),(b)} (f_1,a_1)_A (f_2,b_1)_A \,\, a_2 b_2 \\
&= \sum_{(f)} \left( \sum_{(a)} (f_1,a_1)_A a_2 \right)  \left( \sum_{(b)} (f_2,b_1)_A  b_2 \right)
=\sum_{(f)} f_1^\perp(a) f_2^\perp(b).
\end{align*}

%considering any $g$ in $A^o$, one has that
%\begin{align*}
%(g,f^\perp(ab))
%&= (fg, ab)_A \\
%&= (fg, m_A(a \otimes b))_A \\
%&= ( \Delta_{A^o}(fg), a \otimes b )_{A \otimes A}\\
%&= ( \Delta_{A^o}(f) \Delta_{A^o}(g), a \otimes b )_{A \otimes A}\\
%&= \left( \left( \sum f_1 \otimes f_2 \right) \Delta_{A^o}(g), a \otimes b \right)_{A \otimes A}\\
%&= \left(  \Delta_{A^o}(g), \sum f_1^\perp(a) \otimes f_2^\perp(b) \right)_{A \otimes A}\\
%&= \left(  g, \sum f_1^\perp(a) f_2^\perp(b) \right)_A.\\
%\end{align*}
\end{proof}

The Pieri rules (Theorem~\ref{Pieri-rules}) expressed
multiplication by $h_n$ or by $e_n$ in the basis
$\left(s_\lambda\right)_{\lambda \in \Par}$ of $\Lambda$.
We can similarly express skewing by $h_n$ or by $e_n$:

\begin{proposition}
\label{prop.coPieri-rules}
For every partition $\lambda$ and any $n \in \NN$, we have
\begin{align}
h_n^\perp s_\lambda =&
\sum\limits_{\substack{\lambda^-: \lambda/\lambda^- \text{ is a }\\\text{horizontal }n\text{-strip}}}  s_{\lambda^-} ;
\label{eq.copieri.h} \\
e_n^\perp s_\lambda =&
\sum\limits_{\substack{\lambda^-: \lambda/\lambda^- \text{ is a }\\\text{vertical }n\text{-strip}}}  s_{\lambda^-} .
\label{eq.copieri.e}
\end{align}
\end{proposition}

\begin{exercise}
\label{exe.Lambda.coPieri-rules}
Prove Proposition~\ref{prop.coPieri-rules}.

[\textbf{Hint:} Use Theorem~\ref{Pieri-rules}
and $(s_{\mu^-},e_n^\perp s_\mu) = (e_n s_{\mu^-}, s_\mu)$.]
\end{exercise}

% [DG][v79] Added the above proposition and exercise
% (or, rather, moved them hither from the solutions)
% to better bring the point across that skewing is
% like "multiplication backwards", and to make the
% proof of the skew Pieri rules below a little bit
% clearer.

\noindent
The following interaction between multiplication and $h^\perp$ is the key to
deducing the skew Pieri formula from the usual Pieri formulas.
\begin{lemma}
\label{Lam's-lemma}
For any $f,g$ in $\Lambda$ and any $n \in \NN$, one has
\[
f \cdot h_n^\perp(g)
= \sum_{k=0}^n (-1)^k h_{n-k}^\perp(e_k^\perp(f) \cdot g).
\]
\end{lemma}
\begin{proof}
Starting with the right side, first apply
Proposition~\ref{skewing-properties-prop}(iv):
\begin{align*}
&\sum_{k=0}^n (-1)^k \underbrace{h_{n-k}^\perp(e_k^\perp(f) \cdot g)}_{\substack{= \sum_{j=0}^{n-k} h_j^\perp(e_k^\perp(f)) \cdot h_{n-k-j}^\perp(g) \\ \text{(by Proposition~\ref{skewing-properties-prop}(iv), applied }\\ \text{to } h_{n-k} \text{, } e_k^\perp(f) \text{ and } g \text{ instead of } f \text{, } a \text{ and } b \text{)}}}
\\
&=\sum_{k=0}^n (-1)^{k} \sum_{j=0}^{n-k}
                       h_j^\perp(e_k^\perp(f))
                      \cdot h_{n-k-j}^\perp(g)\\
&=\sum_{j=0}^n \sum_{k=0}^{n-j} (-1)^{k}
                       h_j^\perp(e_k^\perp(f))
                      \cdot h_{n-k-j}^\perp(g)\\
&=\sum_{j=0}^n
       \sum_{i=0}^{n-j} (-1)^{n-i-j}
                      h_j^\perp(e_{n-i-j}^\perp(f))
                      \cdot h_i^\perp(g) \qquad (\text{reindexing }i:=n-k-j \text{ in the inner sum}\,\,) \\
&=\sum_{i=0}^n (-1)^{n-i}
       \left( \sum_{j=0}^{n-i}
                      (-1)^j h_j^\perp(e_{n-i-j}^\perp(f)) \right)
                      \cdot h_i^\perp(g) \\
&=\sum_{i=0}^n (-1)^{n-i}
       \left( \sum_{j=0}^{n-i} (-1)^j e_{n-i-j} h_j\right)^\perp \left( f \right)
                      \cdot h_i^\perp(g) \qquad
                       (\text{by Proposition~\ref{skewing-properties-prop}(ii)}\,\,) \\
&=1^\perp(f) \cdot h_n^\perp(g) = f \cdot h_n^\perp(g)
\end{align*}
where the second-to-last equality used \eqref{e-h-relation}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{Assaf-McNamara-skew-Pieri}]
We prove \eqref{eq.Assaf-McNamara-skew-Pieri.1}; the
equality \eqref{eq.Assaf-McNamara-skew-Pieri.2} is analogous, swapping
$h_i \leftrightarrow e_i$ and swapping the words ``vertical'' $\leftrightarrow$ ``horizontal''.
%Symmetry of $(\cdot,\cdot)_\Lambda$, commutativity of $\Lambda$, and
%Proposition~\ref{skewing-properties-prop}(i) imply for $f$ in $\Lambda$,
For any $f \in \Lambda$, we have
\begin{align}
\left( s_{\lambda/\mu}, f \right)
&= \left( s_\mu^\perp(s_\lambda),f \right)
\qquad \left(\text{by \eqref{eq.skewing.Lambda.s}}\right) \nonumber\\
&= \left( f, s_\mu^\perp(s_\lambda) \right)
\qquad \left(\text{by symmetry of } (\cdot,\cdot)_\Lambda \right) \nonumber\\
&= \left( s_\mu f , s_\lambda \right)
\qquad \left(\text{by Proposition~\ref{skewing-properties-prop}(i)} \right) \nonumber\\
&= \left( s_\lambda ,s_\mu f \right)
\qquad \left(\text{by symmetry of } (\cdot,\cdot)_\Lambda \right) .
\label{skewing-versus-multiplication-in-Sym}
\end{align}
Hence for any $g$ in $\Lambda$, one can compute that
\begin{align}
( h_n s_{\lambda/\mu} \,\, ,\,\, g )
&\underset{~\ref{skewing-properties-prop}(i)}{\overset{Prop.}{=}}
( s_{\lambda/\mu} \,\, ,\,\, h_n^\perp g )
\overset{\eqref{skewing-versus-multiplication-in-Sym}}{=}
 ( s_{\lambda}\,\, , \,\, s_\mu \cdot h_n^\perp g ) \nonumber \\
&\underset{\ref{Lam's-lemma}}{\overset{Lemma}{=}}
 \sum_{k=0}^n (-1)^k ( s_\lambda \,\, , \,\,
                 h_{n-k}^\perp (e_k^\perp(s_\mu) \cdot g ) ) \nonumber \\
&\underset{~\ref{skewing-properties-prop}(i)}{\overset{Prop.}{=}}
\sum_{k=0}^n (-1)^k ( h_{n-k} s_\lambda \,\, , e_k^\perp(s_\mu) \cdot g ) .
\label{Lam-proof-calculation}
\end{align}
The first Pieri rule in Theorem~\ref{Pieri-rules} lets one rewrite
$h_{n-k} s_\lambda = \sum_{\lambda^+} s_{\lambda^+}$, with the sum running through
$\lambda^+$ for which $\lambda^+/\lambda$ is a horizontal $(n-k)$-strip.
Meanwhile, \eqref{eq.copieri.e} lets one rewrite
$e_k^\perp s_\mu = \sum_{\mu^-} s_{\mu^-}$, with the sum running through
$\mu^-$ for which $\mu/\mu^-$ is a vertical $k$-strip.
% since $(s_{\mu^-},e_k^\perp s_\mu) = (e_k s_{\mu^-}, s_\mu)$.
Thus the right hand side of \eqref{Lam-proof-calculation} becomes
\[
\sum_{k=0}^n (-1)^k \left( \sum_{\lambda^+}  s_{\lambda^+} \,\, , \,\,
                         \sum_{\mu^-} s_{\mu^-} \cdot g \right) \\
\overset{\eqref{skewing-versus-multiplication-in-Sym}}{=}
\left( \sum_{k=0}^n  (-1)^k \sum_{(\lambda^+,\mu^-)}  s_{\lambda^+/\mu^-} , g \right)
\]
where the sum is over the pairs $(\lambda^+, \mu^-)$
for which $\lambda^+/\lambda$ is a horizontal $(n-k)$-strip and
$\mu/\mu^-$ is a vertical $k$-strip.
This proves \eqref{eq.Assaf-McNamara-skew-Pieri.1}.
\end{proof}


\begin{exercise}
\label{exe.aguiar.appli-omega}
Let $n \in \NN$.

\begin{itemize}

\item[(a)]
For every $k \in \NN$,
let $p\left(n,k\right)$ denote the number of partitions of $n$ of
length $k$. Let $c\left(n\right)$ denote the number of
\emph{self-conjugate}\index{self-conjugate partition}
partitions of $n$ (that is, partitions $\lambda$ of $n$ satisfying
$\lambda^t = \lambda$). Show that
\[
\left(-1\right)^n c\left(n\right) = \sum_{k=0}^n \left(-1\right)^k p\left(n,k\right).
\]
(This application of Hopf algebras was found by Aguiar and Lauve,
\cite[\S 5.1]{AguiarLauve-2014}. See also
\cite[Chapter 1, Exercise 22(b)]{Stanley} for an elementary proof.)

\item[(b)]
For every partition $\lambda$, let
$C\left(\lambda\right)$ denote the number of corner cells of the
Ferrers diagram of $\lambda$ (these are the cells of the Ferrers
diagram whose neighbors to the east and to the south both lie
outside of the Ferrers diagram). For every partition $\lambda$, let
$\mu_1\left(\lambda\right)$ denote the number of parts of $\lambda$
equal to $1$. Show that
\[
\sum_{\lambda \in \Par_n} C\left(\lambda\right)
= \sum_{\lambda \in \Par_n} \mu_1\left(\lambda\right).
\]
(This is also due to Stanley.)

\end{itemize}
\end{exercise}

% [DG][v9] Aguiar/Lauve give some references for part (a) which don't use
% Hopf algebras. Part (b) works in greater generality: For any given
% integer $k$, the sum over all $\lambda \in \Par_n$ of the number
% of length-$k$ hooks in $\lambda$ equals the sum over all
% $\lambda \in \Par_n$ of the number of parts of $\lambda$ equal to
% $k$. This was an exercise Stanley gave in his class last term which
% I couldn't solve; yesterday I rediscovered it (with a 5-line
% solution) in looking for applications of Hopf algebras...
% (How) Should we reference these sources? (The more general version
% of (b) could be introduced if we ever make a section on the
% Murnaghan-Nakayama rule.)

% [VR][v10] I think a good reference for part (a) of the exercise is 
% Stanley, Enum. Comb. Vol 1 Chapter 1 Exercise 9(b), 
% which he solves with some easy generating functions.
%
% For part (b) of the exercise, what about Proposition 4.7 on page 943 of  
% Stanley's "Differential posets", J. Amer. Math. Soc. 1 (1988) ?

\begin{exercise}
\label{Darij's-skewing-exercise}
The goal of this exercise is to prove
\eqref{skew-Schur-antipodes} using the skewing operators that
we have developed.\footnote{Make sure not to use the results
of Exercise~\ref{exe.pieri.omega} or Exercise~\ref{exe.cauchy.dual}
or Exercise~\ref{exe.hall-e-f} here, or anything else that relied
on \eqref{skew-Schur-antipodes}, in order to avoid circular
reasoning.}
Recall the involution $\omega:\Lambda\rightarrow\Lambda$ defined in
\eqref{the-fundamental-involution-definition}.
\begin{itemize}
\item[(a)]
Show that $\omega\left( p_{\lambda} \right)
 = \left( -1\right)^{\left\vert \lambda\right\vert
       -\ell\left( \lambda\right) } p_{\lambda}$
for any $\lambda\in\Par$, where
$\ell\left( \lambda\right)$ denotes the length of the partition $\lambda$.
\item[(b)]
Show that $\omega$ is an isometry.
\item[(c)]
Show that this same map $\omega:\Lambda\rightarrow\Lambda$ is a
Hopf automorphism.
\item[(d)]
Prove that $\omega \left( a^{\perp} b \right) = \left(
\omega \left( a\right) \right) ^{\perp} \left( \omega \left( b \right)
\right)$ for every $a\in\Lambda$ and $b\in\Lambda$.
\item[(e)] For any partition
$\lambda = \left( \lambda_1, \ldots, \lambda_\ell \right)$
with length $\ell \left( \lambda \right) = \ell$, prove that
\[
e_{\ell}^{\perp} s_{\lambda}
= s_{\left( \lambda_1 - 1, \lambda_2 - 1, \ldots, \lambda_{\ell} - 1 \right)} .
\]
\item[(f)] For any partition
$\lambda = \left( \lambda_1, \lambda_2, \ldots \right)$, prove that
\[
h_{\lambda_1}^{\perp} s_{\lambda}
= s_{\left( \lambda_2 , \lambda_3 , \lambda_4 , \ldots \right)}.
\]
\item[(g)]
Prove \eqref{skew-Schur-antipodes}.
\end{itemize}
\end{exercise}

\begin{exercise}
\label{exe.skew-p-by-e}Let $n$ be a positive integer. Prove the following:

\begin{enumerate}
\item[(a)] We have $\left(  e_{n},p_{n}\right)  =\left(  -1\right)  ^{n-1}$.

\item[(b)] We have $\left(  e_{m},p_{n}\right)  =0$ for each $m\in \NN$
satisfying $m\neq n$.

\item[(c)] We have $e_{n}^{\perp}p_{n}=\left(  -1\right)  ^{n-1}$.

\item[(d)] We have $e_{m}^{\perp}p_{n}=0$ for each positive integer $m$
satisfying $m\neq n$.
\end{enumerate}
\end{exercise}

% [DG][v75] Added the above exercise.

\subsection{Assorted exercises on symmetric functions}

% [DG][v25] Made this a separate subsection. It doesn't really live
% up to its name so far, but it might grow.

Over a hundred exercises on symmetric functions are collected in
Stanley's \cite[chapter 7]{Stanley}, and even more (but without any
hints or references) on his website\footnote{\ 
\url{http://math.mit.edu/~rstan/ec/ch7supp.pdf}}.
Further sources for results related to symmetric functions are
Macdonald's work, including his monograph \cite{Macdonald}
and his expository \cite{Macdonald-variations}. In this section,
we gather a few exercises that are not too difficult to handle
with the material given above.

\begin{exercise} \phantomsection
\label{exe.bernstein}
\begin{itemize}

\item[(a)] Let $m \in \ZZ$. Prove that, for every $f \in \Lambda$,
the infinite sum
$\sum_{i \in \NN} \left(-1\right)^i h_{m+i} e_i^\perp f$ is
convergent in the discrete topology (i.e., all but finitely many
addends of this sum are zero). Hence, we can define a map
$\mathbf{B}_m : \Lambda \to \Lambda$ by setting
\[
\mathbf{B}_m\left(f\right)
= \sum_{i \in \NN} \left(-1\right)^i h_{m+i} e_i^\perp f
\qquad \qquad \text{ for all } f \in \Lambda .
\]
Show that this map $\mathbf{B}_m$ is $\kk$-linear.

\item[(b)] Let $\lambda = \left(\lambda_1, \lambda_2, \lambda_3,
\ldots\right)$ be a partition, and let $m \in \ZZ$ be such
that $m \geq \lambda_1$. Show that
\[
\sum_{i \in \NN} \left(-1\right)^i h_{m+i} e_i^\perp s_\lambda
= s_{\left(m, \lambda_1, \lambda_2, \lambda_3, \ldots\right)} .
\]

\item[(c)] Let $n \in \NN$. For every $n$-tuple
$\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right) \in \ZZ^n$,
we define an element
$\overline{s}_{\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)}
\in \Lambda$ by
\[
\overline{s}_{\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)}
= \det\left( \left( h_{\alpha_i - i + j} \right)_{i, j = 1, 2, \ldots, n} \right) .
\]
Show that
\begin{equation}
\label{eq.exe.bernstein.c.1}
s_\lambda
= \overline{s}_{\left(\lambda_1, \lambda_2, \ldots, \lambda_n\right)}
\end{equation}
for every partition
$\lambda = \left(\lambda_1, \lambda_2, \lambda_3, \ldots\right)$
having at most $n$ parts\footnote{Recall that a \emph{part}
of a partition means a nonzero entry of the partition.}.

Furthermore, show that for
every $n$-tuple
$\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right) \in \ZZ^n$,
the symmetric function
$\overline{s}_{\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)}$
either is $0$ or equals $\pm s_\nu$ for some partition $\nu$ having
at most $n$ parts.

Finally, show that for any $n$-tuples
$\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right) \in \ZZ^n$
and
$\left(\beta_1, \beta_2, \ldots, \beta_n\right) \in \NN^n$,
we have
\begin{equation}
\label{eq.exe.bernstein.c.3}
\overline{s}_{\left(\beta_1, \beta_2, \ldots, \beta_n\right)}^\perp
\overline{s}_{\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)}
= \det\left( \left( h_{\alpha_i - \beta_j - i + j} \right)_{i, j = 1, 2, \ldots, n} \right) .
\end{equation}

\item[(d)] For every $n \in \NN$, every $m \in \ZZ$ and every
$n$-tuple
$\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right) \in \ZZ^n$,
prove that
\begin{equation}
\label{eq.exe.bernstein.d}
\sum_{i \in \NN} \left(-1\right)^i h_{m+i} e_i^\perp
\overline{s}_{\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)}
= \overline{s}_{\left(m, \alpha_1, \alpha_2, \ldots,
\alpha_n\right)} ,
\end{equation}
where we are using the notations of
Exercise~\ref{exe.bernstein}(c).

\item[(e)] For every $n \in \NN$ and every $n$-tuple
$\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right) \in \ZZ^n$,
prove that
\[
\overline{s}_{\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)}
= \left(\mathbf{B}_{\alpha_1} \circ \mathbf{B}_{\alpha_2}
\circ \cdots \circ \mathbf{B}_{\alpha_n}\right) \left(1\right) ,
\]
where we are using the notations of
Exercise~\ref{exe.bernstein}(c) and Exercise~\ref{exe.bernstein}(a).

\item[(f)] For every $m \in \ZZ$ and every positive integer $n$,
prove that
$\mathbf{B}_{m} \left( p_n \right) = h_m p_n - h_{m+n}$.
Here, we are using the notations of Exercise~\ref{exe.bernstein}(a).

\end{itemize}
\end{exercise}

\begin{remark}
The map $\mathbf{B}_m$ defined in Exercise~\ref{exe.bernstein}(a) is
the so-called
\emph{$m$-th Bernstein creation operator}\index{Bernstein creation operator};
it appears in
Zelevinsky \cite[\S 4.20(a)]{Zelevinsky} and has been introduced by
J.N. Bernstein, who found the result of Exercise~\ref{exe.bernstein}(b).
It is called a ``Schur row adder'' in \cite{Garsia-adder}.
Exercise~\ref{exe.bernstein}(e) appears in
Berg/Bergeron/Saliola/Serrano/Zabrocki
\cite[Theorem 2.3]{BergBergeronSaliolaSerranoZabrocki-immac1}, where it
is used as a prototype for defining \emph{noncommutative} analogues
of Schur functions, the so-called \emph{immaculate functions}.
The particular case of Exercise~\ref{exe.bernstein}(e) for
$\left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)$ a partition
of length $n$ (a restatement of Exercise~\ref{exe.bernstein}(b))
is proven in \cite[\S I.5, example 29]{Macdonald}.
\end{remark}

% [DG][v25] Added the above exercise, an occasion to cite the
% immaculate-functions paper (and plug a minor expository gap
% created by its authors referring to Zelevinsky for a theorem
% which is more general than what Zelevinsky states).
% 
% When I started writing the solution to this problem, I was
% expecting that the Jacobi-Trudi solution to part (b) would
% immediately generalize to the less restrictive setting of
% (d). It does, but this proved messier than I expected it to be;
% cases need to be distinguished and the claims of (c) need to
% be used.

% [DG][v75] Added part (f) to the above exercise.

\begin{exercise} \phantomsection
\label{exe.witt}
\begin{itemize}

\item[(a)]
Prove that there exists a unique family
$\left( x_n \right) _{n \geq 1}$ of elements of $\Lambda$
such that
\[
H \left( t \right) = \prod_{n=1}^{\infty} \left( 1 - x_n t^n \right) ^{-1}.
\]
Denote this family $\left( x_n \right) _{n \geq 1}$ by
$\left( w_n \right) _{n \geq 1}$.\index{$w_n$}
For instance,
\begin{align*}
w_{1} &=    s_{\left( 1 \right)},\ \ \ \ \ \ \ \ \ \ 
w_{2}  =  - s_{\left( 1,1 \right)},\ \ \ \ \ \ \ \ \ \ 
w_{3}  =  - s_{\left( 2,1 \right)},\\
w_{4} &=  - s_{\left( 1,1,1,1 \right)} - s_{\left( 2,1,1 \right)}
          - s_{\left( 2,2 \right)} - s_{\left( 3,1 \right)},\ \ \ \ \ \ \ \ \ \ 
w_{5}  =  - s_{\left( 2,1,1,1 \right)} - s_{\left( 2,2,1 \right)}
          - s_{\left( 3,1,1 \right)} - s_{\left( 3,2 \right)}
          - s_{\left( 4,1 \right)}.
\end{align*}

\item[(b)]
Show that $w_n$ is homogeneous of degree $n$ for every positive
integer $n$.

\item[(c)]
For every partition $\lambda$, define $w_{\lambda} \in \Lambda$ by
$w_{\lambda} = w_{\lambda_1} w_{\lambda_2} \cdots w_{\lambda_{\ell}}$
(where $\lambda = \left( \lambda_1, \lambda_2, \ldots,
\lambda_{\ell} \right)$ with $\ell = \ell\left( \lambda \right)$).
Notice that $w_{\lambda}$ is homogeneous of degree
$\left\vert \lambda\right\vert$. Prove that $\sum_{\lambda \in \Par_n}
w_{\lambda} = h_n$ for every $n\in\NN$.

\item[(d)]
Show that $\left\{ w_{\lambda} \right\} _{\lambda\in \Par}$
is a $\kk$-basis of $\Lambda$. (This basis is
called the \dfn{Witt basis}\footnote{This is due to its relation
with Witt vectors in the appropriate sense. Most
of the work on this basis has been done by Reutenauer and
Hazewinkel.}; it is studied in
\cite[\S 9-\S 10]{Hazewinkel2}.\footnote{It also implicitly
appears in \cite[\S 5]{BakerRichter}. Indeed, the $q_n$ of
\cite{BakerRichter} are our $w_n$ (for $\kk = R$).})

\item[(e)]
Prove that $p_n = \sum_{d \mid n} d w_d^{n/d}$ for every positive
integer $n$. (Here, the summation sign $\sum_{d \mid n}$ means a sum
over all positive divisors $d$ of $n$.)

\item[(f)]
We are going to show that $-w_n$ is a sum of Schur functions
(possibly with repetitions, but without signs!) for every $n \geq 2$. (For
$n=1$, the opposite is true: $w_{1}$ is a single Schur function.) This proof
goes back to Doran \cite{Doran}\footnote{See also Stanley
\cite[Exercise 7.46]{Stanley}.}.

For any positive integers $n$ and $k$, define $f_{n,k} \in \Lambda$ by
$f_{n,k} = \sum\limits_{\substack{\lambda\in \Par_n,\\\min\lambda\geq k}}w_{\lambda}$,
where $\min\lambda$ denotes the smallest
part\footnote{Recall that a \emph{part} of a partition means
a nonzero entry of the partition.} of $\lambda$. Show that
\[
-f_{n,k} = s_{\left( n-1,1 \right)} + \sum_{i=2}^{k-1} f_{i,i} f_{n-i,i}
\ \ \ \ \ \ \ \ \ \ \text{for every } n\geq k\geq 2.
\]
Conclude that $-f_{n,k}$ is a sum of Schur functions for every $n\in
\NN$ and $k\geq 2$. Conclude that $-w_n$ is a sum of Schur functions
for every $n\geq 2$.

\item[(g)]
For every partition $\lambda$, define $r_{\lambda}\in\Lambda$ by
$r_{\lambda}=\prod_{i\geq 1}h_{v_{i}}\left(  x_{1}^{i},x_{2}^{i},x_{3}
^{i},\ldots\right)  $, where $v_{i}$ is the number of occurrences of $i$ in
$\lambda$. Show that $\sum_{\lambda\in\Par}w_{\lambda}\left(
\xx\right)  r_{\lambda}\left(  \yy\right)  =\prod_{i,j=1}
^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}$.

\item[(h)]
Show that $\left\{ r_\lambda \right\} _{\lambda \in \Par}$ and
$\left\{ w_\lambda \right\} _{\lambda \in \Par}$ are dual bases of
$\Lambda$.

\end{itemize}
\end{exercise}

% [DG][v7] Added exercise above.

\begin{exercise}
\label{exe.Lambda.maps-on-pn}
For this exercise, set $\kk = \ZZ$, and consider $\Lambda =
\Lambda_\ZZ$ as a subring of $\Lambda_\QQ$. Also,
consider $\Lambda \otimes_\ZZ \Lambda$ as a subring of
$\Lambda_\QQ \otimes_\QQ \Lambda_\QQ$. \ \ \ \ \footnote{Here
is how this works: We have $\Lambda_\QQ \cong \QQ \otimes_\ZZ
\Lambda$. But fundamental properties of tensor products yield
\begin{equation}
\label{eq.exe.Lambda.maps-on-pn.inj}
\QQ \otimes_\ZZ \left(\Lambda \otimes_\ZZ \Lambda\right)
\cong
\underbrace{\left(\QQ \otimes_\ZZ \Lambda\right)}_{\cong \Lambda_\QQ}
\otimes_\QQ
\underbrace{\left(\QQ \otimes_\ZZ \Lambda\right)}_{\cong \Lambda_\QQ}
\cong \Lambda_\QQ \otimes_\QQ \Lambda_\QQ
\end{equation}
as $\QQ$-algebras.
But $\Lambda \otimes_\ZZ \Lambda$ is a free $\ZZ$-module
(since $\Lambda$ is a free $\ZZ$-module), and so the canonical
ring homomorphism $\Lambda \otimes_\ZZ \Lambda \to
\QQ \otimes_\ZZ \left(\Lambda \otimes_\ZZ \Lambda\right)$
sending every $u$ to $1_\QQ \otimes_\ZZ u$ is injective.
Composing this ring homomorphism with the $\QQ$-algebra
isomorphism of \eqref{eq.exe.Lambda.maps-on-pn.inj} gives
an injective ring homomorphism
$\Lambda \otimes_\ZZ \Lambda \to
\Lambda_\QQ \otimes_\QQ \Lambda_\QQ$. We use this latter
homomorphism
to identify $\Lambda \otimes_\ZZ \Lambda$ with a subring of
$\Lambda_\QQ \otimes_\QQ \Lambda_\QQ$.}
Recall that the family
$\left(p_n\right)_{n \geq 1}$ generates the $\QQ$-algebra
$\Lambda_\QQ$, but does not generate the $\ZZ$-algebra
$\Lambda$.
\begin{itemize}

\item[(a)] Define a $\QQ$-linear map $Z : \Lambda_\QQ \to
\Lambda_\QQ$ by setting
\[
Z\left(p_\lambda\right) = z_\lambda p_\lambda
\qquad \qquad \text{for every partition } \lambda ,
\]
where $z_\lambda$ is defined as in
Proposition~\ref{prop-Two-other-Cauchy-expansions}.\footnote{This is
well-defined, since $\left(p_\lambda\right)_{\lambda \in \Par}$ is
a $\QQ$-module basis of $\Lambda_\QQ$.} Show that
$Z \left( \Lambda \right) \subset \Lambda$.

\item[(b)] Define a $\QQ$-algebra homomorphism
$\Delta_\times : \Lambda_\QQ \to \Lambda_\QQ \otimes_\QQ \Lambda_\QQ$
by setting
\[
\Delta_\times \left(p_n\right) = p_n \otimes p_n
\qquad \qquad \text{for every positive integer } n.
\]
\footnote{This is well-defined, since the family
$\left(p_n\right)_{n \geq 1}$ generates the $\QQ$-algebra
$\Lambda_\QQ$ and is algebraically independent.} Show that
$\Delta_\times \left( \Lambda \right) \subset
\Lambda \otimes_\ZZ \Lambda$.

\item[(c)] Let $r \in \ZZ$. Define a $\QQ$-algebra homomorphism
$\epsilon_r : \Lambda_\QQ \to \QQ$
by setting
\[
\epsilon_r \left(p_n\right) = r
\qquad \qquad \text{for every positive integer } n.
\]
\footnote{This is well-defined, since the family
$\left(p_n\right)_{n \geq 1}$ generates the $\QQ$-algebra
$\Lambda_\QQ$ and is algebraically independent.} Show that
$\epsilon_r \left( \Lambda \right) \subset \ZZ$.

\item[(d)] Let $r \in \ZZ$. Define a $\QQ$-algebra homomorphism
$\mathbf{i}_r : \Lambda_\QQ \to \Lambda_\QQ$
by setting
\[
\mathbf{i}_r \left(p_n\right) = r p_n
\qquad \qquad \text{for every positive integer } n.
\]
\footnote{This is well-defined, since the family
$\left(p_n\right)_{n \geq 1}$ generates the $\QQ$-algebra
$\Lambda_\QQ$ and is algebraically independent.} Show that
$\mathbf{i}_r \left( \Lambda \right) \subset \Lambda$.

\item[(e)] Define a $\QQ$-linear map $\operatorname{Sq} : \Lambda_\QQ \to
\Lambda_\QQ$ by setting
\[
\operatorname{Sq}\left(p_\lambda\right) = p_\lambda^2
\qquad \qquad \text{for every partition } \lambda .
\]
\footnote{This is
well-defined, since $\left(p_\lambda\right)_{\lambda \in \Par}$ is
a $\QQ$-module basis of $\Lambda_\QQ$.} Show that
$\operatorname{Sq} \left( \Lambda \right) \subset \Lambda$.

\item[(f)] Let $r \in \ZZ$. Define a $\QQ$-algebra homomorphism
$\Delta_r : \Lambda_\QQ \to \Lambda_\QQ \otimes_\QQ \Lambda_\QQ$
by setting
\[
\Delta_r \left(p_n\right) = \sum_{i=1}^{n-1} \binom{n}{i}
p_i \otimes p_{n-i} + r \otimes p_n + p_n \otimes r
\qquad \qquad \text{for every positive integer } n.
\]
\footnote{This is well-defined, since the family
$\left(p_n\right)_{n \geq 1}$ generates the $\QQ$-algebra
$\Lambda_\QQ$ and is algebraically independent.} Show that
$\Delta_r \left( \Lambda \right) \subset
\Lambda \otimes_\ZZ \Lambda$.

\item[(g)] Consider the map $\Delta_\times$ introduced in
Exercise~\ref{exe.Lambda.maps-on-pn}(b) and the map
$\epsilon_1$ introduced in
Exercise~\ref{exe.Lambda.maps-on-pn}(c). Show that the
$\QQ$-algebra $\Lambda_\QQ$, endowed with the
comultiplication $\Delta_\times$ and the counit
$\epsilon_1$, becomes a cocommutative
$\QQ$-bialgebra.\footnote{But unlike $\Lambda_\QQ$ with the
usual coalgebra structure, it is neither graded nor a Hopf
algebra.}

\item[(h)] Define a $\QQ$-bilinear map $* : \Lambda_\QQ
\times \Lambda_\QQ \to \Lambda_\QQ$, which will be written
in infix notation (that is, we will write $a * b$
instead of $*\left(a, b\right)$), by setting
\[
p_\lambda * p_\mu = \delta_{\lambda, \mu} z_\lambda p_\lambda
\qquad \qquad \text{for any partitions } \lambda \text{ and }
\mu
\]
(where $z_\lambda$ is defined as in
Proposition~\ref{prop-Two-other-Cauchy-expansions}).
\footnote{This is
well-defined, since $\left(p_\lambda\right)_{\lambda \in \Par}$ is
a $\QQ$-module basis of $\Lambda_\QQ$.}
Show that $f * g \in \Lambda$ for any $f \in \Lambda$
and $g \in \Lambda$.

\item[(i)] Show that $\epsilon_1\left(f\right) =
f\left(1\right)$ for every $f \in \Lambda_\QQ$ (where
we are using the notation $\epsilon_r$ defined in
Exercise~\ref{exe.Lambda.maps-on-pn}(c)).
\end{itemize}

[\textbf{Hint:}
\begin{itemize}
\item For (b), show that, for every $f \in \Lambda_\QQ$,
the tensor $\Delta_\times \left(f\right)$ is the
preimage of
$f\left( \left(x_i y_j\right)_{\left(i,j\right) \in
\left\{1,2,3,\ldots\right\}^2} \right)
= f\left( x_1 y_1, x_1 y_2, x_1 y_3, \ldots,
x_2 y_1, x_2 y_2, x_2 y_3, \ldots,
\ldots \right) \in \QQ\left[\left[\xx,\yy\right]\right]$
under the canonical injection
$\Lambda_\QQ \otimes_\QQ \Lambda_\QQ \to
\QQ\left[\left[\xx,\yy\right]\right]$ which maps every
$f \otimes g$ to $f\left(\xx\right) g\left(\yy\right)$.
(This requires making sure that the evaluation
$f\left( \left(x_i y_j\right)_{\left(i,j\right) \in
\left\{1,2,3,\ldots\right\}^2} \right)$ is well-defined
to begin with, i.e., converges as a formal power series.)
\par For an alternative solution to (b), compute
$\Delta_\times \left(h_n\right)$ or
$\Delta_\times \left(e_n\right)$.
\item For (c), compute $\epsilon_r \left(e_n\right)$
or $\epsilon_r \left(h_n\right)$.
\item Reduce (d) to (b) and (c) using
Exercise~\ref{exe.tensorprod.morphism}.
\item Reduce (e) to (b).
\item (f) is the hardest part. It is tempting to try and
interpret the definition of $\Delta_r$ as a convoluted way
of saying that $\Delta_r \left(f\right)$ is the preimage
of
$f\left( \left(x_i + y_j\right)_{\left(i,j\right) \in
\left\{1,2,3,\ldots\right\}^2} \right)$ under the
canonical injection
$\Lambda_\QQ \otimes_\QQ \Lambda_\QQ \to
\QQ\left[\left[\xx,\yy\right]\right]$ which maps every
$f \otimes g$ to $f\left(\xx\right) g\left(\yy\right)$.
However, this does not make sense since the evaluation
$f\left( \left(x_i + y_j\right)_{\left(i,j\right) \in
\left\{1,2,3,\ldots\right\}^2} \right)$ is (in general)
not well-defined\footnote{e.g., it involves summing
infinitely many $x_1$'s if $f = e_1$} (and even if it
was, it would fail to explain the $r$). So we need to get down
to finitely many variables.
For every $N \in \NN$, define a
$\QQ$-algebra homomorphism $\mathcal{E}_N :
\Lambda_\QQ \otimes_\QQ \Lambda_\QQ \to
\QQ\left[x_1, x_2, \ldots, x_N, y_1, y_2, \ldots, y_N\right]$
by sending each $f \otimes g$ to
$f\left(x_1, x_2, \ldots, x_N\right)
g\left(y_1, y_2, \ldots, y_N\right)$. Show that
$\Delta_N \left(\Lambda\right) \subset
\mathcal{E}_N^{-1}\left(
\ZZ\left[x_1, x_2, \ldots, x_N, y_1, y_2, \ldots, y_N\right]
\right)$. This shows that, at least, the coefficients
of $\Delta_r \left(f\right)$ in front of the
$m_\lambda \otimes m_\mu$ with $\ell\left(\lambda\right)
\leq r$ and $\ell\left(\mu\right) \leq r$ (in the
$\QQ$-basis
$\left(m_\lambda \otimes m_\mu\right)_{\lambda, \mu \in \Par}$
of $\Lambda_\QQ \otimes_\QQ \Lambda_\QQ$)
are integral for $f \in \Lambda$. Of course, we want
all coefficients. Show that
$\Delta_a = \Delta_b \star \left(\Delta_{\Lambda_\QQ}
\circ \mathbf{i}_{a-b}\right)$ in
$\Hom\left(\Lambda_\QQ, \Lambda_\QQ \otimes_\QQ
\Lambda_\QQ\right)$ for any integers $a$ and $b$. This
allows ``moving'' the $r$. This approach to (f) was partly
suggested to the first author by Richard Stanley.
\item For (h), notice that Definition~\ref{def.self-duality}(b)
(below) allows us to construct a bilinear form
$\left(\cdot, \cdot\right)_{\Lambda_\QQ \otimes_\QQ \Lambda_\QQ}
: \left(\Lambda_\QQ \otimes_\QQ \Lambda_\QQ\right)
\times \left(\Lambda_\QQ \otimes_\QQ \Lambda_\QQ\right)
\to \QQ$ from the Hall inner product
$\left(\cdot, \cdot\right) : \Lambda_\QQ \times \Lambda_\QQ \to
\QQ$. Show that
\begin{equation}
\label{eq.exe.Lambda.maps-on-pn.adjointness}
\left(a * b, c\right)
= \left(a \otimes b, \Delta_\times \left(c\right) \right)_{\Lambda_\QQ \otimes_\QQ \Lambda_\QQ}
\qquad \qquad \text{ for all } a, b, c \in \Lambda_\QQ ,
\end{equation}
and then use (b).
\end{itemize}
]
\end{exercise}

\begin{remark}
The map $\Delta_\times$ defined in
Exercise~\ref{exe.Lambda.maps-on-pn}(b) is known as the
\dfn{internal comultiplication} (or \dfn{Kronecker
comultiplication}) on $\Lambda_\QQ$. Unlike the standard
comultiplication $\Delta_{\Lambda_\QQ}$, it is not a graded
map, but rather sends every homogeneous component
$\left(\Lambda_\QQ\right)_n$ into
$\left(\Lambda_\QQ\right)_n \otimes \left(\Lambda_\QQ\right)_n$.
The bilinear map $*$ from
Exercise~\ref{exe.Lambda.maps-on-pn}(h) is the so-called
\dfn{internal multiplication} (or \dfn{Kronecker
multiplication}), and is similarly not graded but rather
takes
$\left(\Lambda_\QQ\right)_n \times \left(\Lambda_\QQ\right)_m$
to $\left(\Lambda_\QQ\right)_n$ if $n = m$ and to $0$ otherwise.

The analogy between the two internal structures is not perfect:
While we saw in Exercise~\ref{exe.Lambda.maps-on-pn}(g) how
the internal comultiplication yields another bialgebra structure
on $\Lambda_\QQ$, it is not true that the internal
multiplication (combined with the usual coalgebra structure of
$\Lambda_\QQ$) forms a bialgebra structure as well. What is
missing is a multiplicative unity; if we would take the closure
of $\Lambda_\QQ$ with respect to the grading, then
$1 + h_1 + h_2 + h_3 + \cdots$ would be such a unity.

The structure constants of the internal comultiplication
on the Schur basis $\left(s_\lambda\right)_{\lambda \in \Par}$
are equal to the structure constants
of the internal multiplication on the Schur basis\footnote{This
can be obtained, e.g., from
\eqref{eq.exe.Lambda.maps-on-pn.adjointness}.}, and are
commonly referred to as the
\dfn{Kronecker coefficients}. They are known to be
nonnegative integers (this follows from
Exercise~\ref{exe.repSn.inner-tensor.intprod}(c)\footnote{Their
integrality can also be easily deduced from
Exercise~\ref{exe.Lambda.maps-on-pn}(b).}), but no
combinatorial proof is known for their
nonnegativity. Combinatorial
interpretations for these coefficients akin to the
Littlewood-Richardson rule have been found only in special
cases (cf., e.g., \cite{Rosas-kronprod} and
\cite{Blasiak-kronhook} and \cite{Liu-simplkron}).

The map $\Delta_r$ of
Exercise~\ref{exe.Lambda.maps-on-pn}(f) also has some
classical theory behind it, relating to Chern classes of
tensor products (\cite{Manivel-chern},
\cite[\S I.4, example 5]{Macdonald}).
\end{remark}

% [DG][v25] Added exercise above. Part (f) is the statement of
% http://mathoverflow.net/questions/120924/is-the-renormalized-third-comultiplication-on-mathbfsymm-integral?rq=1
% (but the proof is not the one I mentioned in my answer).
% It took me a while to find this solution and I am not sure
% my hint is anywhere near good...

% [DG][v46] Added part (i) (more or less a triviality).

Parts (b), (c), (d), (e) and (f) of Exercise~\ref{exe.Lambda.maps-on-pn} are
instances of a general phenomenon: Many $\ZZ$-algebra homomorphisms
$\Lambda\rightarrow A$ (with $A$ a commutative ring, usually torsionfree) are
easiest to define by first defining a $\QQ$-algebra homomorphism
$\Lambda_\QQ \rightarrow A \otimes \QQ$ and then showing that
this homomorphism restricts to a $\ZZ$-algebra homomorphism
$\Lambda\rightarrow A$. One might ask for general criteria when this is
possible; specifically, for what choices of $\left( b_n \right)  _{n\geq 1}
\in A^{\left\{  1,2,3,\ldots\right\}  }$ does there exist a
$\ZZ$-algebra homomorphism $\Lambda\rightarrow A$ sending the $p_n$ to
$b_n$ ? Such choices are called \dfn{ghost-Witt vectors} in Hazewinkel
\cite{Hazewinkel2}, and we can give various equivalent conditions for a family
$\left( b_n \right) _{n \geq 1}$ to be a ghost-Witt vector:

\begin{exercise}
\label{exe.witt.ghost-equiv}
Let $A$ be a commutative ring.

For every $n\in\left\{  1,2,3,\ldots\right\} $, let $\varphi_{n}:A\rightarrow
A$ be a ring endomorphism of $A$. Assume that the following properties hold:

\begin{itemize}
\item We have $\varphi_{n}\circ\varphi_{m}=\varphi_{nm}$ for any two positive
integers $n$ and $m$.

\item We have $\varphi_{1}= \id$.

\item We have $\varphi_{p}\left(  a\right)  \equiv a^p \operatorname{mod}pA$
for every $a\in A$ and every prime number $p$.
\end{itemize}

\noindent (For example, when $A = \ZZ$, one can set $\varphi_n = \id$
for all $n$; this simplifies the exercise somewhat. More generally, setting
$\varphi_{n} = \id$ works whenever $A$ is a binomial
ring\footnote{A \dfn{binomial ring} is defined to be a torsionfree (as an
additive group) commutative ring $A$ which has one of the following equivalent
properties:
\par
\begin{itemize}
\item For every $n\in \NN$ and $a\in A$, we have $a\left(  a-1\right)
\cdots\left(  a-n+1\right)  \in n!\cdot A$. (That is, binomial coefficients
$\dbinom{a}{n}$ with $a\in A$ and $n\in \NN$ are defined in $A$.)
\par
\item We have $a^{p}\equiv a\operatorname{mod}pA$ for every $a\in A$ and every
prime number $p$.
\end{itemize}
\par
See \cite{Xantcha} and the references therein for studies of these rings. It
is not hard to check that $\ZZ$ and every localization of $\ZZ$
are binomial rings, and so is any commutative $\QQ$-algebra
as well as the ring
\[
\left\{  P\in \QQ \left[  X\right]  \ \mid\ P\left(  n\right)
\in \ZZ \text{ for every }n\in \ZZ \right\}
\]
(but not the ring $ \ZZ \left[  X\right] $ itself).}. However, the results of
this exercise are at their most useful when $A$ is a multivariate polynomial
ring $ \ZZ \left[  x_{1},x_{2},x_{3},\ldots\right]  $ over $\ZZ$
and the homomorphism $\varphi_{n}$ sends every $P\in A$ to
$P\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)  $.)

Let $\mu$ denote the
\dfn{number-theoretic M\"obius function}\index{M\"obius function}\index{$\mu$};
this is
the function $\left\{  1,2,3,\ldots\right\}  \rightarrow \ZZ$ defined by
\[
\mu\left(  m\right)  = \begin{cases}
0,& \text{if }m\text{ is not squarefree;}\\
\left(  -1\right)  ^{\left(  \text{number of prime factors of }m\right)
},& \text{if }m\text{ is squarefree}
\end{cases}
\qquad \qquad \text{for every positive integer } m.
\]


Let \dfn{$\phi$} denote the \dfn{Euler totient function};
this is the function
$\left\{  1,2,3,\ldots\right\}  \rightarrow \NN$ which sends every
positive integer $m$ to the number of elements of $\left\{  1,2,\ldots
,m\right\}  $ coprime to $m$.

Let $\left(  b_{n}\right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots\right\}  }$ be a
family of elements of $A$. Prove that the following seven assertions are equivalent:

\begin{itemize}
\item \textit{Assertion $\mathcal{C}$:} For every positive integer
$n$ and every prime factor $p$ of $n$, we have
\[
\varphi_{p}\left(  b_{n/p}\right)  \equiv b_{n}
\operatorname{mod} p^{v_{p}\left(  n\right)  }A.
\]
Here, $v_{p}\left(  n\right)  $ denotes the exponent of $p$ in the prime
factorization of $n$.

\item \textit{Assertion $\mathcal{D}$:} There exists a family
$\left(  \alpha_{n}\right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots\right\}  }$
of elements of $A$ such that every positive integer $n$ satisfies
\[
b_{n}=\sum_{d\mid n}d\alpha_{d}^{n/d}.
\]
\footnote{Here and in the following, summations of the form $\sum_{d\mid n}$
range over all \textbf{positive} divisors of $n$.}

\item \textit{Assertion $\mathcal{E}$:} There exists a family
$\left(  \beta_{n}\right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots\right\}  }$
of elements of $A$ such that every positive integer $n$ satisfies
\[
b_{n}=\sum_{d\mid n}d\varphi_{n/d}\left(  \beta_{d}\right)  .
\]


\item \textit{Assertion $\mathcal{F}$:} Every positive integer $n$
satisfies
\[
\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)  \in nA.
\]


\item \textit{Assertion $\mathcal{G}$:} Every positive integer $n$
satisfies
\[
\sum_{d\mid n}\phi\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)  \in
nA.
\]


\item \textit{Assertion $\mathcal{H}$:} Every positive integer $n$
satisfies
\[
\sum_{i=1}^{n}\varphi_{n/\gcd\left(  i,n\right)  }\left(  b_{\gcd\left(
i,n\right)  }\right)  \in nA.
\]


\item \textit{Assertion $\mathcal{J}$:} There exists a ring
homomorphism $\Lambda_{ \ZZ }\rightarrow A$ which, for every positive
integer $n$, sends $p_{n}$ to $b_{n}$.
\end{itemize}

[\textbf{Hint:} The following identities hold for every positive integer $n$:
\begin{align}
\sum_{d\mid n}\phi\left(  d\right)   &
=n;\label{eq.exe.witt.ghost-equiv.phi*1}\\
\sum_{d\mid n}\mu\left(  d\right)   &  =\delta_{n,1}
;\label{eq.exe.witt.ghost-equiv.mu*1}\\
\sum_{d\mid n}\mu\left(  d\right)  \dfrac{n}{d}  &  =\phi\left(  n\right)
;\label{eq.exe.witt.ghost-equiv.mu*n}\\
\sum_{d\mid n}d\mu\left(  d\right)  \phi\left(  \dfrac{n}{d}\right)   &
=\mu\left(  n\right)  . \label{eq.exe.witt.ghost-equiv.nmu*phi}
\end{align}
Furthermore, the following simple lemma is useful: If $k$ is a positive
integer, and if $p\in \NN$, $a\in A$ and $b\in A$ are such that $a\equiv
b\operatorname{mod}p^{k}A$, then $a^{p^{\ell}}\equiv b^{p^{\ell}}
\operatorname{mod}p^{k+\ell}A$ for every $\ell\in \NN$.]
\end{exercise}

\begin{remark}
Much of Exercise \ref{exe.witt.ghost-equiv} is folklore, but it is hard to
pinpoint concrete appearances in literature. The equivalence $\mathcal{C}
\Longleftrightarrow\mathcal{D}$ appears in Hesselholt \cite[Lemma
1]{Hesselholt-witt} and \cite[Lemma 1.1]{Hesselholt-deRhamWitt}
(in slightly greater generality), where it is referred to
as Dwork's lemma and used in the construction of the Witt vector functor. This
equivalence is also \cite[Lemma 9.93]{Hazewinkel2}. The equivalence
$\mathcal{D}\Longleftrightarrow\mathcal{F}\Longleftrightarrow\mathcal{G}
\Longleftrightarrow\mathcal{H}$ in the case $A = \ZZ$ is \cite[Corollary
on p. 10]{DressSiebeneicher-burnside}, where it is put into the context of
Burnside rings and necklace counting. The equivalence $\mathcal{C}
\Longleftrightarrow\mathcal{F}$ for finite families $\left(  b_{n}\right)
_{n\in\left\{  1,2,\ldots,m\right\}  }$ in lieu of $\left(  b_{n}\right)
_{n\geq1}$ is \cite[Exercise 5.2 \textbf{a}]{Stanley}. One of the likely
oldest relevant sources is Schur's \cite{Schur-1937}, which proves the
equivalence $\mathcal{C}\Longleftrightarrow\mathcal{D}\Longleftrightarrow
\mathcal{F}$ for finite families $\left(  b_{n}\right)  _{n\in\left\{
1,2,\ldots,m\right\}  }$, as well as a ``finite
version'' of $\mathcal{C}\Longleftrightarrow\mathcal{J}$
(Schur did not have $\Lambda$, but was working with actual power sums of roots
of polynomials).
\end{remark}

\begin{exercise}
\label{exe.witt.ghost-exa}
Let $A$ denote the ring $\ZZ$. For every
$n\in\left\{  1,2,3,\ldots\right\}  $, let $\varphi_{n}$ denote the identity
endomorphism $\id$ of $A$. Prove that the seven equivalent
assertions $\mathcal{C}$, $\mathcal{D}$, $\mathcal{E}$, $\mathcal{F}$,
$\mathcal{G}$, $\mathcal{H}$ and $\mathcal{J}$ of Exercise
\ref{exe.witt.ghost-equiv} are satisfied for each of the following families
$\left(  b_{n}\right)  _{n\geq1}\in \ZZ ^{\left\{  1,2,3,\ldots\right\}  }$:

\begin{itemize}
\item the family $\left(  b_{n}\right)  _{n\geq1}=\left(  q^{n}\right)
_{n\geq1}$, where $q$ is a given integer.

\item the family
$\left(  b_{n}\right)  _{n\geq1}=\left(  q\right)  _{n\geq1}$, where
$q$ is a given integer.

\item the family $\left(  b_{n}\right)  _{n\geq1}
= \left(  \dbinom{qn}{rn}\right)  _{n\geq1}$, where $r\in \QQ $ and
$q\in \ZZ$ are
given. (Here, a binomial coefficient $\dbinom{a}{b}$ has to be interpreted as
$0$ when $b\notin \NN$.)

\item the family $\left(  b_{n}\right)  _{n\geq1}
= \left(  \dbinom{qn-1}{rn-1}\right)  _{n\geq1}$, where $r\in \ZZ$
and $q\in \ZZ$ are given.
\end{itemize}
\end{exercise}

\begin{exercise}
\label{exe.witt.ghost-app}
For every $n\in\left\{  1,2,3,\ldots\right\}  $,
define a map $\mathbf{f}_{n}:\Lambda\rightarrow\Lambda$ by setting
\[
\mathbf{f}_{n}\left(  a\right) =
a\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)
\qquad \qquad \text{for every }a\in\Lambda.
\]
(So what $\mathbf{f}_{n}$ does to a symmetric function is replacing all
variables $x_{1},x_{2},x_{3},\ldots$ by their $n$-th powers.)

\begin{enumerate}
\item[(a)] Show that $\mathbf{f}_{n}:\Lambda\rightarrow\Lambda$ is a
$\kk$-algebra homomorphism for every $n\in\left\{  1,2,3,\ldots
\right\}  $.

\item[(b)] Show that $\mathbf{f}_{n}\circ\mathbf{f}_{m}=\mathbf{f}_{nm}$ for
any two positive integers $n$ and $m$.

\item[(c)] Show that $\mathbf{f}_{1}= \id$.

\item[(d)] Prove that $\mathbf{f}_n : \Lambda \to \Lambda$ is a Hopf
algebra homomorphism for every $n \in \left\{ 1, 2, 3, \ldots \right\}$.

\item[(e)] Prove that $\mathbf{f}_2 \left( h_m \right)
= \sum_{i=0}^{2m} \left( -1\right) ^i h_i h_{2m-i}$ for every $m \in \NN$.

\item[(f)] Assume that $\kk = \ZZ$. Prove that $\mathbf{f}_{p}\left(
a\right)  \equiv a^{p}\operatorname{mod}p\Lambda$ for every $a\in\Lambda$ and
every prime number $p$.

\item[(g)] Use Exercise \ref{exe.witt.ghost-equiv} to obtain new solutions to
parts (b), (c), (d), (e) and (f) of Exercise~\ref{exe.Lambda.maps-on-pn}.
\end{enumerate}
\end{exercise}

The maps $\mathbf{f}_n$ constructed in Exercise~\ref{exe.witt.ghost-app} are
known as the \dfn{Frobenius endomorphisms} of $\Lambda$. They are a
(deceptively) simple particular case of the notion of \emph{plethysm}
(\cite[Chapter 7, Appendix 2]{Stanley} and \cite[Section I.8]{Macdonald}),
and are often used as intermediate
steps in computing more complicated plethysms\footnote{In the notations of
\cite[(A2.160)]{Stanley}, the value $\mathbf{f}_n\left(a\right)$ for an
$a \in \Lambda$ can be written as $a\left[p_n\right]$ or (when $\kk = \ZZ$)
as $p_n\left[a\right]$.}.

% [DG][v34] Added the above three exercises. They are somewhat
% tangentially related to symmetric functions, but I wanted a
% place I could reference, as the result is useful and too hard
% to find explicitly in any single point in the literature.

% [DG][v38] Added part (d) to the exercise, and comments about the
% relation to plethysm.

% [DG][v40] Added part (e) and solution to part (d).

\begin{exercise}
\label{exe.Lambda.verschiebung}
For every $n \in \left\{ 1, 2, 3, \ldots \right\}$, define a
$\kk$-algebra homomorphism
$\mathbf{v}_n : \Lambda \rightarrow \Lambda$ by
\[
\mathbf{v}_n \left( h_m \right)
=  \begin{cases}
h_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\qquad\qquad\text{for every positive integer }m
\]
\footnote{This is well-defined, since the family
$\left( h_m \right) _{m \geq 1}$ generates the $\kk$-algebra
$\Lambda$ and is algebraically independent.}.

\begin{enumerate}
\item[(a)] Show that any positive integers $n$ and $m$ satisfy
\[
\mathbf{v}_n \left( p_m \right)
= \begin{cases}
n p_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad .
\]


\item[(b)] Show that any positive integers $n$ and $m$ satisfy
\[
\mathbf{v}_n \left( e_m \right)
= \begin{cases}
\left( -1 \right) ^{m - m/n} e_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad .
\]


\item[(c)] Prove that $\mathbf{v}_n \circ \mathbf{v}_m = \mathbf{v}_{nm}$ for
any two positive integers $n$ and $m$.

\item[(d)] Prove that $\mathbf{v}_1 = \id$.

\item[(e)] Prove that $\mathbf{v}_n : \Lambda \rightarrow \Lambda$ is a Hopf
algebra homomorphism for every $n \in \left\{ 1, 2, 3, \ldots \right\}$.
\end{enumerate}

Now, consider also the maps $\mathbf{f}_n : \Lambda \rightarrow \Lambda$ defined
in Exercise~\ref{exe.witt.ghost-app}. Fix a positive integer $n$.

\begin{enumerate}
\item[(f)] Prove that the maps $\mathbf{f}_n : \Lambda \rightarrow \Lambda$ and
$\mathbf{v}_n : \Lambda \rightarrow \Lambda$ are adjoint with respect to the
Hall inner product on $\Lambda$.

\item[(g)] Show that $\mathbf{v}_n \circ \mathbf{f}_n
= \id_{\Lambda}^{\star n}$.

\item[(h)] Prove that
$\mathbf{f}_n \circ \mathbf{v}_m = \mathbf{v}_m \circ \mathbf{f}_n$
whenever $m$ is a positive integer coprime to $n$.
\end{enumerate}

Finally, recall the $w_m \in \Lambda$ defined in Exercise~\ref{exe.witt}.

\begin{enumerate}
\item[(i)] Show that any positive integer $m$ satisfies
\[
\mathbf{v}_n \left( w_m \right)
= \begin{cases}
w_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad .
\]

\end{enumerate}
\end{exercise}

The homomorphisms $\mathbf{v}_n : \Lambda \rightarrow \Lambda$ defined in
Exercise~\ref{exe.Lambda.verschiebung} are called the \dfn{Verschiebung
endomorphisms} of $\Lambda$; this name comes from German, where
``Verschiebung'' means ``shift''.
This terminology, as well as that of Frobenius
endomorphisms, originates in the theory of Witt vectors, and the connection
between the Frobenius and Verschiebung endomorphisms of $\Lambda$ and the
identically named operators on Witt vectors is elucidated in \cite[Chapter
13]{Hazewinkel2}\footnote{which is also where most of the statements of
Exercises \ref{exe.witt.ghost-app} and \ref{exe.Lambda.verschiebung} come
from}.

% [DG][v39] Added above exercise.

% [DG][v75] Corrected egregious error in part (b).

\begin{exercise}
\label{exe.shareshian-wachs.path}
Fix $n \in \NN$. For any $n$-tuple $w = \left(w_1, w_2, \ldots, w_n\right)$ of
integers, define the \emph{descent set}\index{descent set of a permutation}
$\Des\left(w\right)$ of $w$ to be
the set $\left\{i \in \left\{1, 2, \ldots, n-1\right\} : w_i > w_{i+1}
\right\}$.

\begin{itemize}
\item[(a)] We say that an $n$-tuple $\left(w_1, w_2, \ldots, w_n\right)$ is
\emph{Smirnov}\index{Smirnov tuple}
if every $i \in \left\{1, 2, \ldots, n-1\right\}$
satisfies $w_i \neq w_{i+1}$.

Fix $k \in \NN$, and let $X_{n, k} \in \kk\left[\left[\xx\right]\right]$
denote the sum of the monomials $x_{w_1} x_{w_2} \cdots x_{w_n}$ over
all Smirnov $n$-tuples $w = \left(w_1, w_2, \ldots, w_n\right) \in
\left\{1, 2, 3, \ldots\right\}^n$ satisfying
$\left|\Des\left(w\right)\right| = k$. Prove that
$X_{n, k} \in \Lambda$.

\item[(b)] For any $n$-tuple $w = \left(w_1, w_2, \ldots, w_n\right)$,
define the \dfn{stagnation set}
$\operatorname{Stag}\left(w\right)$ of $w$ to be the set
\newline % I hate having to do that :/
$\left\{i \in \left\{1, 2, \ldots, n-1\right\} : w_i = w_{i+1}
\right\}$. (Thus, an $n$-tuple is Smirnov if and only if its stagnation
set is empty.)

For any $d \in \NN$ and $s \in \NN$, define a power
series $X_{n, d, s} \in \kk\left[\left[\xx\right]\right]$
as the sum of the monomials $x_{w_1} x_{w_2} \cdots x_{w_n}$ over
all $n$-tuples $w = \left(w_1, w_2, \ldots, w_n\right)
\in \left\{1, 2, 3, \ldots\right\}^n$ satisfying
$\left|\Des\left(w\right)\right| = d$ and
$\left|\operatorname{Stag}\left(w\right)\right| = s$. Prove that
$X_{n, d, s} \in \Lambda$ for any nonnegative integers $d$ and $s$.

\item[(c)] Assume that $n$ is positive.
For any $d \in \NN$ and $s \in \NN$, define three further
power series $U_{n, d, s}$, $V_{n, d, s}$ and $W_{n, d, s}$ in
$\kk\left[\left[\xx\right]\right]$ by the following formulas:
\begin{align}
\label{eq.exe.shareshian-wachs.path.c.U}
U_{n, d, s}
&= \sum\limits_{\substack{w = \left(w_1, w_2, \ldots, w_n\right)
    \in \left\{1, 2, 3, \ldots\right\}^n ; \\
  \left|\Des\left(w\right)\right| = d ; \ 
  \left|\operatorname{Stag}\left(w\right)\right| = s ; \\
  w_1 < w_n}} x_{w_1} x_{w_2} \cdots x_{w_n} ; \\
\label{eq.exe.shareshian-wachs.path.c.V}
V_{n, d, s}
&= \sum\limits_{\substack{w = \left(w_1, w_2, \ldots, w_n\right)
    \in \left\{1, 2, 3, \ldots\right\}^n ; \\
  \left|\Des\left(w\right)\right| = d ; \ 
  \left|\operatorname{Stag}\left(w\right)\right| = s ; \\
  w_1 = w_n}} x_{w_1} x_{w_2} \cdots x_{w_n} ; \\
\label{eq.exe.shareshian-wachs.path.c.W}
W_{n, d, s}
&= \sum\limits_{\substack{w = \left(w_1, w_2, \ldots, w_n\right)
    \in \left\{1, 2, 3, \ldots\right\}^n ; \\
  \left|\Des\left(w\right)\right| = d ; \ 
  \left|\operatorname{Stag}\left(w\right)\right| = s ; \\
  w_1 > w_n}} x_{w_1} x_{w_2} \cdots x_{w_n} .
\end{align}
Prove that these three power series
$U_{n, d, s}$, $V_{n, d, s}$ and $W_{n, d, s}$ belong to
$\Lambda$.
\end{itemize}
\end{exercise}

\begin{remark}
The function $X_{n, k}$ in Exercise~\ref{exe.shareshian-wachs.path}(a)
is a simple example
(\cite[Example 2.5, Theorem C.3]{ShareshianWachs-2014})
of a chromatic quasisymmetric function that happens to be
symmetric. See Shareshian/Wachs \cite{ShareshianWachs-2014} for more
general criteria for such functions to be symmetric, as well as deeper
results. For example, \cite[Theorem 6.3]{ShareshianWachs-2014} gives
an expansion for a wide class of chromatic quasisymmetric functions in
the Schur basis of $\Lambda$, which, in particular, shows that our
$X_{n, k}$ satisfies
\[
X_{n, k} = \sum_{\lambda \in \Par_n} a_{\lambda, k} s_\lambda ,
\]
where $a_{\lambda, k}$ is the number of all assignments $T$ of entries
in $\left\{1, 2, \ldots, n\right\}$ to the cells of the Ferrers
diagram of $\lambda$ such that the following four conditions are
satisfied:
\begin{itemize}
\item Every element of $\left\{1, 2, \ldots, n\right\}$ is used
precisely once in the assignment (i.e., we have $\cont\left(T\right)
= \left(1^n\right)$).
\item Whenever a cell $y$ of the Ferrers diagram lies immediately to the right of
a cell $x$, we have $T\left(y\right) - T\left(x\right) \geq 2$.
\item Whenever a cell $y$ of the Ferrers diagram lies immediately below
a cell $x$, we have $T\left(y\right) - T\left(x\right) \geq -1$.
\item There exist precisely $k$ elements
$i \in \left\{1, 2, \ldots, n-1\right\}$ such that the cell
$T^{-1}\left(i\right)$ lies in a row below $T^{-1}\left(i+1\right)$.
\end{itemize}
Are there any such rules for the $X_{n, d, s}$ of part (b)?

Smirnov $n$-tuples are more usually called Smirnov words, or
(occasionally) \idx{Carlitz words}.

See \cite[Chapter 6]{Ellzey-thesis} for further properties of
the symmetric functions $U_{n, d, 0}$, $V_{n, d, 0}$ and
$W_{n, d, 0}$ from Exercise~\ref{exe.shareshian-wachs.path}(c)
(or, more precisely, of their generating functions
$\sum_d U_{n, d, 0} t^d$ etc.).
\end{remark}

% [DG][v25] Added above exercise. The second solution to part
% (a) (which I would not have written up if I had known of the other
% solution by then) is a fine example of my inability to write
% combinatorial proofs.
% 
% Part (b) as well as the name "stagnation set" is mine (or I have just
% not skimmed enough of the Shareshian-Wachs paper); do you know
% the "right" notion?

% [DG][v77] Added the last paragraph of the above remark.

\begin{exercise} \phantomsection
\label{exe.pe-ep-determinants}

\begin{itemize}

\item[(a)] Let $n \in \NN$. Define a matrix
$A_n = \left(a_{i,j}\right)_{i,j = 1, 2, \ldots, n} \in \Lambda^{n \times n}$
by
\[
a_{i,j} =
  \begin{cases}
   p_{i-j+1}, & \text{if } i \geq j ; \\
   i,         & \text{if } i = j - 1 ; \\
   0,         & \text{if } i < j - 1
  \end{cases}
\qquad \qquad \text{ for all } \left( i, j \right) \in \left\{ 1, 2, \ldots , n \right\}^2 .
\]
This matrix $A_n$ looks as follows:
\[
A_n
=
\left(
\begin{array}[c]{cccccc}
p_1   & 1 & 0 & \cdots & 0 & 0\\
p_2   & p_1   & 2 & \cdots & 0 & 0\\
p_3   & p_2   & p_1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
p_{n-1} & p_{n-2} & p_{n-3} & \cdots & p_1   & n-1\\
p_n & p_{n-1} & p_{n-2} & \cdots & p_2   & p_1 
\end{array}
\right) .
\]
Show that $\det\left(A_n\right) = n! e_n$.

\item[(b)] Let $n$ be a positive integer. Define a matrix
$B_n = \left(b_{i,j}\right)_{i,j = 1, 2, \ldots, n} \in \Lambda^{n \times n}$
by
\[
b_{i,j} =
  \begin{cases}
   i e_i,     & \text{if } j = 1 ; \\
   e_{i-j+1}, & \text{if } j > 1
% The following is redundant because we have $e_k = 0$ for negative $k$, and $e_0 = 1$:
%   e_{i-j+1}, & \text{if } i \geq j > 1 ; \\
%   1,         & \text{if } i = j - 1 ; \\
%   0,         & \text{if } i < j - 1
  \end{cases}
\qquad \qquad \text{ for all } \left( i, j \right) \in \left\{ 1, 2, \ldots , n \right\}^2 .
\]
The matrix $B_n$ looks as follows:
\begin{align*}
B_n
&=
\left(
\begin{array}[c]{cccccc}
e_1  & e_0 & e_{-1} & \cdots & e_{-n+3} & e_{-n+2} \\
2e_2  & e_1  & e_0 & \cdots & e_{-n+4} & e_{-n+3}\\
3e_3  & e_2  & e_1 & \cdots & e_{-n+5} & e_{-n+4}\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
\left(  n-1\right)  e_{n-1}  & e_{n-2} & e_{n-3}  & \cdots & e_1 & e_0\\
n e_n & e_{n-1}  & e_{n-2} & \cdots & e_2  & e_1
\end{array}
\right) \\
&=
\left(
\begin{array}[c]{cccccc}
e_1  & 1 & 0 & \cdots & 0 & 0\\
2e_2  & e_1  & 1 & \cdots & 0 & 0\\
3e_3  & e_2  & e_1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
\left(  n-1\right)  e_{n-1}  & e_{n-2} & e_{n-3}  & \cdots & e_1 & 1\\
n e_n & e_{n-1}  & e_{n-2} & \cdots & e_2  & e_1
\end{array}
\right) .
\end{align*}
Show that $\det\left(B_n\right) = p_n$.

\end{itemize}

\end{exercise}

The formulas of Exercise~\ref{exe.pe-ep-determinants}, for
finitely many variables, appear in
Prasolov's \cite[\S 4.1]{Prasolov-linalg}\footnote{where our
symmetric functions $e_k, h_k, p_k$, evaluated in finitely many
indeterminates, are denoted $\sigma_k, p_k, s_k$, respectively}.
In \cite[\S 4.2]{Prasolov-linalg}, Prasolov gives four more
formulas, which express $e_n$ as a polynomial in the
$h_1,h_2,h_3,\ldots$, or $h_n$ as a polynomial in the
$e_1,e_2,e_3,\ldots$, or $p_n$ as a polynomial in the
$h_1,h_2,h_3,\ldots$, or $n!h_n$ as a polynomial in the
$p_1,p_2,p_3,\ldots$. These are not novel for us, since the first
two of them are particular cases of
Theorem~\ref{Jacobi-Trudi-formulae}, whereas the latter two
can be derived from Exercise~\ref{exe.pe-ep-determinants} by
applying $\omega$. (Note that $\omega$ is only well-defined
on symmetric functions in infinitely many indeterminates, so
we need to apply $\omega$ \textbf{before} evaluating at
finitely many indeterminates; this explains why Prasolov has
to prove the latter two identities separately.)

% [DG][v30] Added this exercise. I tried not to copy the solution of
% Exercise 9.3 in https://github.com/darijgr/lambda too
% much, but there is only so much I can do when the solution is
% essentially the same.

% [DG][v32] Added paragraph with Prasolov reference.

\begin{exercise}
\label{exe.hookschur}
In the following, if $k \in \NN$, we shall use the notation
$1^k$ for $\underbrace{1,1,\ldots,1}_{k\text{ times}}$ (in
contexts such as $\left( n, 1^m \right)$). So, for example,
$\left( 3, 1^4 \right)$ is the partition
$\left( 3, 1, 1, 1, 1 \right)$.

\begin{itemize}

\item[(a)] Show that
$e_n h_m = s_{\left( m+1, 1^{n-1} \right)} + s_{\left( m, 1^n \right)}$
for any two positive integers $n$ and $m$.

\item[(b)] Show that
\[
\sum_{i=0}^{b} \left(-1\right)^i h_{a+i+1} e_{b-i}
= s_{\left( a+1, 1^b \right)}
\]
for any $a \in \NN$ and $b \in \NN$.

\item[(c)] Show that
\[
\sum_{i=0}^{b} \left(-1\right)^i h_{a+i+1} e_{b-i}
= \left(-1\right)^b \delta_{a+b,-1}
\]
for any negative integer $a$ and every $b \in \NN$.
(As usual, we set $h_j = 0$ for $j < 0$ here.)

\item[(d)] Show that
\begin{align*}
\Delta s_{\left( a+1, 1^b \right)}
&= 1 \otimes s_{\left( a+1, 1^b \right)}
 + s_{\left( a+1, 1^b \right)} \otimes 1 \\
& \ \ \ \ \ \ \ \ \ \
 + \sum\limits_{\substack{\left( c, d, e, f \right) \in \NN^4; \\
         c+e=a-1; \\ d+f=b}}
   s_{\left( c+1, 1^d \right)} \otimes s_{\left( e+1, 1^f \right)}
 + \sum\limits_{\substack{\left( c, d, e, f \right) \in \NN^4; \\
         c+e=a; \\ d+f=b-1}}
   s_{\left( c+1, 1^d \right)} \otimes s_{\left( e+1, 1^f \right)}
\end{align*}
for any $a \in \NN$ and $b \in \NN$.

\end{itemize}
\end{exercise}

% [DG][v30] Added above exercise.

Our next few exercises survey some results on
Littlewood-Richardson coefficients.

\begin{exercise}
\label{exe.liri.180}
Let $m \in \NN$ and $k \in \NN$. Let
$\lambda$ and $\mu$ be two partitions such that $\ell\left(  \lambda\right)
\leq k$ and $\ell\left(  \mu\right)  \leq k$. Assume that all parts of
$\lambda$ and all parts of $\mu$ are $\leq m$. (It is easy to see
that this assumption is equivalent to requiring $\lambda_i \leq m$
and $\mu_i \leq m$ for every positive integer $i$.\ \ \ \ \footnote{As usual,
we are denoting by $\nu_i$ the $i$-th entry of a partition $\nu$ here.}).
Let
$\lambda^{\vee}$ and $\mu^{\vee}$ denote the $k$-tuples $\left(  m-\lambda
_{k},m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  $ and $\left(  m-\mu
_{k},m-\mu_{k-1},\ldots,m-\mu_{1}\right)  $, respectively.

\begin{enumerate}
\item[(a)] Show that $\lambda^{\vee}$ and $\mu^{\vee}$ are partitions, and
that $s_{\lambda/\mu} = s_{\mu^{\vee}/\lambda^{\vee}}$.

\item[(b)] Show that
$c_{\mu,\nu}^{\lambda} = c_{\lambda^{\vee},\nu}^{\mu^{\vee}}$
for any partition $\nu$.

\item[(c)] Let $\nu$ be a partition such that $\ell\left(  \nu\right)  \leq
k$, and such that all parts of $\nu$ are $\leq m$. Let $\nu^{\vee}$ denote the
$k$-tuple $\left(  m-\nu_{k},m-\nu_{k-1},\ldots,m-\nu_{1}\right)  $. Show that
$\nu^{\vee}$ is a partition, and satisfies
\[
c_{\mu,\nu}^{\lambda} = c_{\nu,\mu}^{\lambda}
= c_{\lambda^{\vee},\nu}^{\mu^{\vee}} = c_{\nu,\lambda^{\vee}}^{\mu^{\vee}}
= c_{\mu,\lambda^{\vee}}^{\nu^{\vee}} = c_{\lambda^{\vee},\mu}^{\nu^{\vee}} .
\]

\item[(d)] Show that
\[
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)
=\left(  x_{1} x_{2}\cdots x_{k}\right)  ^{m}
 \cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\]
in the Laurent polynomial ring $\kk \left[  x_{1},x_{2},\ldots
,x_{k},x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right]  $.

\item[(e)] Let $r$ be a nonnegative integer. Show that
$\left( r+\lambda_{1}, r+\lambda_{2}, \ldots, r+\lambda_{k} \right)$ is
a partition and satisfies
\[
s_{\left( r+\lambda_{1}, r+\lambda_{2}, \ldots, r+\lambda_{k} \right)}
\left( x_{1},x_{2},\ldots,x_{k} \right)
= \left(  x_{1}x_{2}\cdots x_{k}\right)^{r}
\cdot s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
\]
in the polynomial ring $\kk \left[  x_{1},x_{2},\ldots,x_{k}\right]  $.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exe.liri.180ab}
Let $m \in \NN$, $n \in \NN$ and $k \in \NN$.
Let $\mu$ and $\nu$ be two partitions such that $\ell\left(
\mu\right)  \leq k$ and $\ell\left(  \nu\right)  \leq k$. Assume that all
parts of $\mu$ are $\leq m$ (that is, $\mu_i \leq m$ for every positive
integer $i$)\ \ \ \ \footnote{As usual, we are denoting by $\nu_i$ the
$i$-th entry of a partition $\nu$ here.},
and that all parts of $\nu$ are $\leq n$ (that is, $\nu_i \leq n$
for every positive integer $i$). Let $\mu
^{\vee\left\{  m\right\}  }$ denote the $k$-tuple $\left(  m-\mu_{k}
,m-\mu_{k-1},\ldots,m-\mu_{1}\right)  $, and let $\nu^{\vee\left\{  n\right\}
}$ denote the $k$-tuple $\left(  n-\nu_{k},n-\nu_{k-1},\ldots,n-\nu
_{1}\right)  $.

\begin{enumerate}
\item[(a)] Show that $\mu^{\vee\left\{  m\right\}  }$ and
$\nu^{\vee\left\{ n\right\}  }$ are partitions.
\end{enumerate}

Now, let $\lambda$ be a further partition such that $\ell\left(
\lambda\right)  \leq k$.

\begin{enumerate}
\item[(b)] If not all parts of $\lambda$ are $\leq m+n$, then show that
$c_{\mu,\nu}^{\lambda} = 0$.

\item[(c)] If all parts of $\lambda$ are $\leq m+n$, then show that
$c_{\mu,\nu}^{\lambda} = c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{
n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}$,
where $\lambda ^{\vee\left\{  m+n\right\}  }$ denotes the $k$-tuple
$\left(  m+n-\lambda_{k},m+n-\lambda_{k-1},\ldots,m+n-\lambda_{1}\right)  $.
\end{enumerate}
\end{exercise}

The results of Exercise~\ref{exe.pieri.omega}(c) and
Exercise~\ref{exe.liri.180}(c) are two
\dfn{symmetries of Littlewood-Richardson
coefficients}\footnote{The result of Exercise~\ref{exe.liri.180ab}(c) can also
be regarded as a symmetry of Littlewood-Richardson coefficients; see
\cite[\S 3.3]{Azenhas-LRsymm24}.}; combining them yields further such
symmetries. While these symmetries were relatively easy consequences of our
algebraic definition of the Littlewood-Richardson coefficients, it is a much
more challenging task to derive them bijectively from a combinatorial
definition of these coefficients (such as the one given in
Corollary~\ref{L-R-rule}). Some such derivations appear in \cite{ThomasYong},
in \cite{AzenhasKingTerada}, in
\cite[Example 3.6, Proposition 5.11 and references therein]{BenkartSottileStroomer},
\cite[\S 5.1, \S A.1, \S A.4]{Fulton} and
\cite[(2.12)]{KirillovBerenstein} (though a different combinatorial
interpretation of $c_{\mu,\nu}^{\lambda}$ is used in the latter three).

% [DG][v44] Added the preceding two exercises.

\begin{exercise}
\label{exe.liri.simple}
Recall our usual notations: For every partition $\lambda$
and every positive integer $i$, the $i$-th entry of
$\lambda$ is denoted by $\lambda_i$. The sign $\triangleright$
stands for dominance order. We let $\lambda^t$ denote
the conjugate partition of a partition $\lambda$.

For any two partitions $\mu$ and $\nu$, we define two new
partitions $\mu + \nu$ and $\mu \sqcup \nu$ of
$\left|\mu\right| + \left|\nu\right|$ as follows:
\begin{itemize}
\item The partition $\mu + \nu$ is defined as
$\left(\mu_1 + \nu_1, \mu_2 + \nu_2, \mu_3 + \nu_3, \ldots\right)$.
\item The partition $\mu \sqcup \nu$ is defined as the
result of sorting the list
$\left(\mu_1, \mu_2, \ldots, \mu_{\ell\left(\mu\right)},
\nu_1, \nu_2, \ldots, \nu_{\ell\left(\nu\right)}\right)$
in decreasing order.
\end{itemize}

\begin{itemize}

\item[(a)] Show that any two partitions $\mu$ and $\nu$
satisfy
$\left(\mu + \nu\right)^t = \mu^t \sqcup \nu^t$
and
$\left(\mu \sqcup \nu\right)^t = \mu^t + \nu^t$.

\item[(b)] Show that any two partitions $\mu$ and $\nu$
satisfy $c^{\mu + \nu}_{\mu, \nu} = 1$ and
$c^{\mu \sqcup \nu}_{\mu, \nu} = 1$.

\item[(c)] If $k \in \NN$ and $n \in \NN$ satisfy
$k \leq n$, and if $\mu \in \Par_k$,
$\nu \in \Par_{n-k}$ and $\lambda \in \Par_n$ are such that
$c^\lambda_{\mu, \nu} \neq 0$, then prove that
$\mu + \nu \triangleright \lambda \triangleright \mu \sqcup \nu$.

\item[(d)] If $n \in \NN$ and $m \in \NN$ and $\alpha, \beta
\in \Par_n$ and $\gamma, \delta \in \Par_m$ are such that
$\alpha \triangleright \beta$ and
$\gamma \triangleright \delta$, then show that
$\alpha + \gamma \triangleright \beta + \delta$ and
$\alpha \sqcup \gamma \triangleright \beta \sqcup \delta$.

\item[(e)] Let $m \in \NN$ and $k \in \NN$, and let $\lambda$
be the partition
$\left(m^k\right) = \left(\underbrace{m, m, \ldots, m}_{k
\text{ times}}\right)$.
Show that any two partitions $\mu$ and $\nu$ satisfy
$c^\lambda_{\mu, \nu} \in \left\{0, 1\right\}$.

\item[(f)] Let $a \in \NN$ and $b \in \NN$, and let $\lambda$
be the partition $\left(a+1, 1^b\right)$ (using the notation
of Exercise~\ref{exe.hookschur}).
Show that any two partitions $\mu$ and $\nu$ satisfy
$c^\lambda_{\mu, \nu} \in \left\{0, 1\right\}$.

\item[(g)] If $\lambda$ is any partition, and if $\mu$ and
$\nu$ are two rectangular partitions\footnote{A partition
is called \emph{rectangular}\index{rectangular partition}
if it has the form
$\left(m^k\right) = \left(\underbrace{m, m, \ldots, m}_{k
\text{ times}}\right)$ for some $m \in \NN$ and $k \in \NN$.},
then show that
$c^\lambda_{\mu, \nu} \in \left\{0, 1\right\}$.

\end{itemize}
\end{exercise}

Exercise~\ref{exe.liri.simple}(g) is part of Stembridge's
\cite[Thm. 2.1]{Stembridge-multfree}; we refer to that
article for further results of its kind.

% [DG][v29] Added this exercise.

% [DG][v30] Added part (f).

% [DG][v32] Added part (g).

The Littlewood-Richardson rule comes in many different
forms, whose equivalence is not always immediate. Our version
(Corollary~\ref{L-R-rule}) has the advantage of being the
simplest to prove and one of the simplest to state. Other
versions can be found in \cite[appendix 1 to Ch. 7]{Stanley},
Fulton's \cite[Ch. 5]{Fulton}
and van Leeuwen's \cite{Leeuwen-lrr}. We restrict ourselves
to proving some very basic equivalences that allow us to
restate parts of Corollary~\ref{L-R-rule}:

\begin{exercise}
\label{exe.liri.yamanouchi}
We shall use the following notations:

\begin{itemize}

\item If $T$ is a column-strict tableau and $j$ is a positive
integer, then we use the notation $T|_{\cols \geq j}$ for
the restriction of $T$ to the union of its columns
$j,j+1,j+2,\ldots$.
%, and we use the notation
%$T|_{\operatorname{rows} \geq j}$
%for the restriction of $T$ to the union of its rows
%$j,j+1,j+2,\ldots$.
(This notation has
already been used in Section~\ref{bialternant-section}.)

\item If $T$ is a column-strict tableau and $S$ is a set of
cells of $T$, then we write $T|_S$ for the
restriction of $T$ to the set $S$ of cells.\footnote{This
restriction $T|_S$
is not necessarily a tableau of skew shape; it is just a map
from $S$ to $\left\{1, 2, 3, \ldots\right\}$. The content
$\cont\left(T|_S\right)$ is nevertheless well-defined (in the
usual way: $\left(\cont\left(T|_S\right)\right)_i =
\left| \left(T|_S\right)^{-1} \left(i\right) \right|$).}

\item If $T$ is a column-strict tableau, then an
\emph{NE-set}\index{NE-set of a tableau}
of $T$ means a set $S$ of cells of $T$ such
that whenever $s \in S$, every cell of $T$ which lies
northeast\footnote{A cell $\left(r,c\right)$ is said to lie
\dfn{northeast} of a cell $\left(r',c'\right)$ if and
only if we have $r \leq r'$ and $c \geq c'$.} of $s$ must
also belong to $S$.

\item The \dfn{Semitic reading word}\footnote{The
notation comes from \cite{Leeuwen-lrr} and is a reference
to the Arabic and Hebrew way of writing.} of a
column-strict tableau $T$ is the concatenation\footnote{If
$s_1, s_2, s_3, \ldots$ are several words (finitely or
infinitely many), then the \dfn{concatenation}
$s_1 s_2 s_3 \cdots$ is defined as the word which is
obtained by starting with the empty word, then appending
$s_1$ to its end, then appending $s_2$ to the end of the
result, then appending $s_3$ to the end of the result, etc.}
$r_1 r_2 r_3 \cdots$, where $r_i$ is the word obtained
by reading the $i$-th row of $T$ from right to
left.\footnote{For example, the Semitic reading word of
the tableau
\[
\begin{matrix}
&   & 3 & 4 & 4 & 5 \\
& 1 & 4 & 6 &   &   \\
& 3 & 5 &   &   &
\end{matrix}
\]
is $5443 641 53$.
\par
The Semitic reading word of a tableau $T$ is what
is called the \dfn{reverse reading word} of $T$ in
\cite[\S A.1.3]{Stanley}.}

\item If $w = \left(w_1, w_2, \ldots, w_n\right)$ is a
word, then a \dfn{prefix} of $w$ means a word of the
form $\left(w_1, w_2, \ldots, w_i\right)$ for some
$i \in \left\{0, 1, \ldots, n\right\}$. (In particular,
both $w$ and the empty word are prefixes of $w$.)

A word $w$ over the set of positive integers is said to
be \emph{Yamanouchi}\index{Yamanouchi word}
if for any prefix $v$ of $w$ and
any positive integer $i$, there are at least as many
$i$'s among the letters of $v$ as there are
$\left(i+1\right)$'s among them.\footnote{For instance,
the words $11213223132$ and $1213$ are Yamanouchi, while
the words $132$, $21$ and $1121322332111$ are not. The
Dyck words (defined as in \cite[Example 6.6.6]{Stanley},
and written using $1$'s and $2$'s instead of $x$'s and
$y$'s) are precisely
the Yamanouchi words whose letters are $1$'s and $2$'s
and in which the letter $1$ appears as often as the
letter $2$.

Yamanouchi words are often called lattice permutations.}

\end{itemize}

Prove the following two statements:

\begin{itemize}

\item[(a)] Let $\mu$ be a partition.
Let $b_{i, j}$ be a nonnegative integer for every two
positive integers $i$ and $j$. Assume that $b_{i, j} = 0$
for all but finitely many pairs $\left(i, j\right)$.

The following two assertions are equivalent:

\begin{itemize}

\item
\textit{Assertion $\mathcal{A}$:}
There exist a partition $\lambda$
and a column-strict tableau $T$ of shape $\lambda / \mu$
such that all
$\left(i, j\right) \in
\left\{1, 2, 3, \ldots \right\}^2$ satisfy
\begin{equation}
\label{eq.exe.liri.yamanouchi.A.bij}
b_{i, j} = \left(\text{the number of all entries } i
\text{ in the } j \text{-th row of } T\right) .
\end{equation}

\item
\textit{Assertion $\mathcal{B}$:}
The inequality
\begin{equation}
\label{eq.exe.liri.yamanouchi.B.bij}
\mu_{j+1} + \left(b_{1, j+1} + b_{2, j+1} + \cdots + b_{i+1, j+1}\right)
\leq
\mu_j + \left(b_{1, j} + b_{2, j} + \cdots + b_{i, j}\right)
\end{equation}
holds for all
$\left(i, j\right) \in \NN \times
\left\{1, 2, 3, \ldots \right\}$.

\end{itemize}

\item[(b)] Let $\lambda$ and $\mu$ be two partitions,
and let $T$ be a column-strict tableau of shape
$\lambda / \mu$. Then, the following five assertions are
equivalent:

\begin{itemize}

\item
\textit{Assertion $\mathcal{C}$:} For every positive integer
$j$, the weak composition
$\cont\left(T|_{\cols \geq j}\right)$ is a partition.

\item
\textit{Assertion $\mathcal{D}$:} For every positive integers
$j$ and $i$, the number of entries $i+1$ in the first $j$
rows\footnote{The ``first $j$ rows'' mean the $1$-st row,
the $2$-nd row, etc., the $j$-th row (even if some of these
rows are empty).} of $T$ is $\leq$ to the number of
entries $i$ in the first $j-1$ rows of $T$.

\item
\textit{Assertion $\mathcal{E}$:} For every NE-set $S$
of $T$, the weak composition $\cont\left(T|_S\right)$ is a
partition.

\item
\textit{Assertion $\mathcal{F}$:} The Semitic reading word
of $T$ is Yamanouchi.

\item
\textit{Assertion $\mathcal{G}$:} There exists a column-strict
tableau $S$ whose shape is a partition and which satisfies
the following property: For any positive integers $i$ and $j$,
the number of entries $i$ in the $j$-th row of $T$ equals the
number of entries $j$ in the $i$-th row of $S$.

\end{itemize}
\end{itemize}

\end{exercise}

\begin{remark}
The equivalence of Assertions $\mathcal{C}$ and $\mathcal{F}$
in Exercise~\ref{exe.liri.yamanouchi}(b) is the
``not-too-difficult exercise'' mentioned in
\cite{Stembridge}. It yields the equivalence between our version
of the Littlewood-Richardson rule (Corollary~\ref{L-R-rule}) and
that in \cite[A1.3.3]{Stanley}.
\end{remark}

% [DG][v29] Added this exercise. It (or rather, the equivalences
% between Assertions A and B and between Assertions C and D) is
% needed for another exercise I plan to add; but it also provides
% bridges to some literature. It is still far from proving any
% nontrivial equivalence results between different LR rules
% (partly because we don't define jeu de taquin, hives or
% Gelfand-Tsetlin triangles in these notes and I don't feel like
% opening these cans of worms right now).
% 
% The solution bears silent witness to my inability to write
% combinatorial proofs; it gained in pedantry and length every
% time I discovered a mistake, and that was quite often...

In the next exercises, we shall restate Corollary~\ref{infinite-L-R-rule} in a
different form. While Corollary~\ref{infinite-L-R-rule} provided a
decomposition of the product of a skew Schur function with a Schur function
into a sum of Schur functions, the different form that we will encounter in
Exercise~\ref{exe.liri.skew.1}(b) will give a combinatorial interpretation for
the Hall inner product between two skew Schur functions. Let us first
generalize Exercise~\ref{exe.liri.yamanouchi}(b):

\begin{exercise}
\label{exe.liri.yamanouchi.kappa}
Let us use the notations of Exercise~\ref{exe.liri.yamanouchi}. Let
$\kappa$, $\lambda$ and $\mu$ be three partitions, and let $T$ be a
column-strict tableau of shape $\lambda/\mu$.

\begin{itemize}
\item[(a)] Prove that the following five assertions are equivalent:

\begin{itemize}
\item \textit{Assertion $\mathcal{C}^{\left(  \kappa\right)  }$:}
For every positive integer $j$, the weak composition
$\kappa + \cont \left( T|_{\operatorname{cols}\geq j}\right)$
is a partition.

\item \textit{Assertion $\mathcal{D}^{\left(  \kappa\right)  }$:}
For every positive integers $j$ and $i$, we have
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}


\item \textit{Assertion $\mathcal{E}^{\left(  \kappa\right)  }$:}
For every NE-set $S$ of $T$, the weak composition $\kappa+ \cont
\left(  T|_{S}\right)  $ is a partition.

\item \textit{Assertion $\mathcal{F}^{\left(  \kappa\right)  }$:}
For every prefix $v$ of the Semitic reading word of $T$, and for every
positive integer $i$, we have
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}


\item \textit{Assertion $\mathcal{G}^{\left(  \kappa\right)  }$:}
There exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ which satisfies the following property: For any positive
integers $i$ and $j$, the number of entries $i$ in the $j$-th row of $T$
equals the number of entries $j$ in the $i$-th row of $S$.
\end{itemize}

\item[(b)] Let $\tau$ be a partition such that $\tau=\kappa
+ \cont T$. Consider the five assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$,
$\mathcal{E}^{\left(  \kappa\right)  }$,
$\mathcal{F}^{\left(  \kappa\right)  }$ and
$\mathcal{G}^{\left(  \kappa\right)  }$ introduced in
Exercise~\ref{exe.liri.yamanouchi.kappa}(a). Let us also consider the
following assertion:

\begin{itemize}
\item \textit{Assertion $\mathcal{H}^{\left(  \kappa\right)  }$:}
There exists a column-strict tableau $S$ of shape $\tau/\kappa$ which
satisfies the following property: For any positive integers $i$ and $j$, the
number of entries $i$ in the $j$-th row of $T$ equals the number of entries
$j$ in the $i$-th row of $S$.
\end{itemize}

Prove that the six assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$,
$\mathcal{E}^{\left(  \kappa\right)  }$,
$\mathcal{F}^{\left(  \kappa\right)  }$, $\mathcal{G}^{\left(
\kappa\right)  }$ and $\mathcal{H}^{\left(  \kappa\right)  }$ are equivalent.
\end{itemize}
\end{exercise}

Clearly, Exercise~\ref{exe.liri.yamanouchi}(b) is the particular case of
Exercise~\ref{exe.liri.yamanouchi.kappa} when $\kappa=\varnothing$.

Using Exercise~\ref{exe.liri.yamanouchi.kappa}, we can restate
Corollary~\ref{infinite-L-R-rule} in several ways:

\begin{exercise}
\label{exe.liri.skew.1}
Let $\lambda$, $\mu$ and $\kappa$ be three partitions.

\begin{enumerate}
\item[(a)] Show that
\[
s_{\kappa}s_{\lambda/\mu}=\sum_{T}s_{\kappa+ \cont T},
\]
where the sum ranges over all column-strict tableaux $T$ of shape $\lambda
/\mu$ satisfying the five equivalent assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$,
$\mathcal{E}^{\left(  \kappa\right)  }$,
$\mathcal{F}^{\left(  \kappa\right)  }$ and
$\mathcal{G}^{\left(  \kappa\right)  }$ introduced in
Exercise~\ref{exe.liri.yamanouchi.kappa}(a).

\item[(b)] Let $\tau$ be a partition. Show that $\left(  s_{\lambda/\mu
},s_{\tau/\kappa}\right)  _{\Lambda}$ is the number of all column-strict
tableaux $T$ of shape $\lambda/\mu$ satisfying $\tau=\kappa
+ \cont T$ and also satisfying the six equivalent assertions
$\mathcal{C}^{\left(  \kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)
}$, $\mathcal{E}^{\left(  \kappa\right)  }$, $\mathcal{F}^{\left(
\kappa\right)  }$, $\mathcal{G}^{\left(  \kappa\right)  }$ and
$\mathcal{H}^{\left(  \kappa\right)  }$ introduced in
Exercise~\ref{exe.liri.yamanouchi.kappa}.
\end{enumerate}
\end{exercise}

Exercise~\ref{exe.liri.skew.1}(a) is merely Corollary~\ref{infinite-L-R-rule},
rewritten in light of Exercise~\ref{exe.liri.yamanouchi.kappa}. Various parts
of it appear in the literature. For instance, \cite[(53)]{Leeuwen-altSchur}
easily reveals to be a restatement of the fact that $s_{\kappa}s_{\lambda/\mu
}=\sum_{T}s_{\nu+ \cont T}$, where the sum ranges over all
column-strict tableaux $T$ of shape $\lambda/\mu$ satisfying Assertion
$\mathcal{D}^{\left(  \kappa\right)  }$.

Exercise~\ref{exe.liri.skew.1}(b) is one version of a ``skew
Littlewood-Richardson rule'' that goes back to Zelevinsky
\cite{Zelevinsky-skewLR} (although Zelevinsky's version uses both a different
language and a combinatorial interpretation which is not obviously equivalent
to ours). It appears in various sources; for instance, \cite[Theorem 5.2,
second formula]{Leeuwen-altSchur} says that $\left(  s_{\lambda/\mu}%
,s_{\tau/\kappa}\right)  _{\Lambda}$ is the number of all column-strict
tableaux $T$ of shape $\lambda/\mu$ satisfying $\tau=\kappa
+ \cont T$ and the assertion $\mathcal{H}^{\left(  \kappa\right)
}$, whereas \cite[Theorem 1.2]{Gasharov} says that $\left(  s_{\lambda/\mu
},s_{\tau/\kappa}\right)  _{\Lambda}$ is the number of all all column-strict
tableaux $T$ of shape $\lambda/\mu$ satisfying $\tau=\kappa
+ \cont T$ and the assertion $\mathcal{F}^{\left(  \kappa\right)
}$. (Notice that Gasharov's proof of \cite[Theorem 1.2]{Gasharov} uses the
same involutions as Stembridge's proof of Theorem~\ref{Stembridge-theorem}; it
can thus be regarded as a close precursor to Stembridge's proof. However, it
uses the Jacobi-Trudi identities, while Stembridge's does not.)

% [DG][v54] Added the above two exercises. As exercises they are
% worthless (straightforward reformulations and generalizations of
% things already done), but I think their statements are worth
% making. If I ever come around to writing up the Zelevinsky
% version of the skew LR rule in terms of pictures, these
% exercises will be the springboard.

\begin{exercise}
\label{exe.liri.jordan}

Let $\mathbb{K}$ be a field.\footnote{This field has no relation
to the ring $\kk$, over which our symmetric functions are defined.}
If $N \in \mathbb{K}^{n \times n}$ is
a nilpotent matrix, then the \dfn{Jordan type} of $N$ is
defined to be the list of the sizes of the Jordan blocks in the
Jordan normal form of $N$, sorted in decreasing order\footnote{The
Jordan normal form of $N$ is well-defined even if $\mathbb{K}$ is
not algebraically closed, because $N$ is nilpotent (so the
characteristic polynomial of $N$ is $X^n$).}. This Jordan type
is a partition of $n$, and uniquely determines $N$ up to
similarity (i.e., two nilpotent $n \times n$-matrices
$N$ and $N'$ are similar if and only if the Jordan types of $N$
and $N'$ are equal). If $f$ is a nilpotent endomorphism of a
finite-dimensional $\mathbb{K}$-vector space $V$, then we
define the \dfn{Jordan type} of $f$ as the Jordan type of any
matrix representing $f$ (the choice of the matrix does not matter,
since the Jordan type of a matrix remains unchanged under
conjugation).

\begin{itemize}

\item[(a)] Let $n \in \NN$. Let $N \in \mathbb{K}^{n \times n}$
be a nilpotent matrix. Let $\lambda \in \Par_n$.
Show that the matrix $N$ has Jordan type $\lambda$ if and only
if every $k \in \NN$ satisfies
\[
\dim \left(\ker \left(N^k\right) \right)
= \left(\lambda^t\right)_1 + \left(\lambda^t\right)_2 +
  \ldots + \left(\lambda^t\right)_k .
\]
(Here, we are using the notation $\lambda^t$ for the transpose
of a partition $\lambda$, and the notation $\nu_i$ for the
$i$-th entry of a partition $\nu$.)

\item[(b)] Let $f$ be a nilpotent endomorphism of a
finite-dimensional $\mathbb{K}$-vector space $V$. Let $U$
be an $f$-stable $\mathbb{K}$-vector subspace of $V$ (that is, a
$\mathbb{K}$-vector subspace of $V$ satisfying
$f\left(U\right) \subset U$). Then, restricting $f$ to $U$ gives
a nilpotent endomorphism $f \mid U$ of $U$, and the endomorphism
$f$ also induces a nilpotent endomorphism $\overline f$ of the
quotient space $V/U$. Let $\lambda$, $\mu$ and $\nu$ be the
Jordan types of $f$, $f \mid U$ and $\overline f$, respectively.
Show that $c^\lambda_{\mu, \nu} \neq 0$ (if $\mathbb Z$ is a
subring of $\kk$).

\end{itemize}

[\textbf{Hint:} For (b), Exercise~\ref{exe.pieri.omega}(c)
shows that it is enough to prove that
$c^{\lambda^t}_{\mu^t, \nu^t} \neq 0$. Due to
Corollary~\ref{L-R-rule}, this only requires constructing a
column-strict tableau $T$ of shape $\lambda^t / \mu^t$ with
$\cont T = \nu^t$ which has the property that each
$\cont\left(T | _{\cols \geq j}\right)$ is a partition.
Construct this tableau by defining
$a_{i, j} = \dim \left( \left(f^i\right)^{-1}\left(U\right)
   \cap \ker\left(f^j\right) \right)$
for all $\left( i, j \right) \in \NN^2$,
and requiring that the number of entries $i$ in the $j$-th
row of $T$ be
$a_{i, j} - a_{i, j-1} - a_{i-1, j} + a_{i-1, j-1}$
for all $\left(i, j\right) \in \left\{1,2,3,\ldots\right\}^2$.
Use Exercise~\ref{exe.liri.yamanouchi}(a) to prove that this
indeed defines a column-strict tableau, and
Exercise~\ref{exe.liri.yamanouchi}(b) to verify that it
satisfies the condition on
$\cont\left(T | _{\cols \geq j}\right)$.]

\end{exercise}

\begin{remark}
Exercise~\ref{exe.liri.jordan} is a taste of the connections
between the combinatorics of partitions and the Jordan normal
form. Much more can, and has, been said. Marc van Leeuwen's
\cite{Leeuwen-geometry} is dedicated to some of these
connections; in particular, our
Exercise~\ref{exe.liri.jordan}(a) is
\cite[Proposition 1.1]{Leeuwen-geometry}, and a far stronger
version of Exercise~\ref{exe.liri.jordan}(b) appears in
\cite[Theorem 4.3 (2)]{Leeuwen-geometry}, albeit only for the
case of an infinite $\mathbb{K}$. One can prove a converse to
Exercise~\ref{exe.liri.jordan}(b) as well: If
$c^\lambda_{\mu, \nu} \neq 0$, then there exist $V$, $f$ and
$U$ satisfying the premises of
Exercise~\ref{exe.liri.jordan}(b). When $\mathbb{K}$ is a
finite field, we can ask enumerative questions, such as how
many $U$'s are there for given $V$, $f$, $\lambda$, $\mu$
and $\nu$; we will see a few answers in
Section~\ref{Hall-algebra-section} (specifically,
Proposition~\ref{prop.hall-alg.product}), and a more
detailed treatment is given in \cite[Ch. 2]{Macdonald}.

The relationship between partitions and Jordan normal forms
can be exploited to provide linear-algebraic proofs of purely
combinatorial facts. See \cite[Sections 6 and 9]{BritzFomin}
for some examples. Note that \cite[Lemma 9.10]{BritzFomin} is
the statement that, under the conditions of
Exercise~\ref{exe.liri.jordan}(b), we have
$\nu \subseteq \lambda$. This is a direct consequence of
Exercise~\ref{exe.liri.jordan}(b) (since
$c^\lambda_{\mu, \nu} \neq 0$ can happen only if
$\nu \subseteq \lambda$).
\end{remark}

% [DG][v29] My proof of the converse of
% Exercise~\ref{exe.liri.jordan}(b) is really ugly: For
% finite $\mathbb{K}$, I derive it from the positivity of
% the Hall polynomial's value at the size of $\mathbb{K}$
% (which follows from Theorem 4.2 in Klein's
% "The multiplication of Schur-functions and extensions
% of $p$-modules", J. London Math. Soc., 43 (1968),
% 280--284), whereas for infinite $\mathbb{K}$ it follows
% from Leeuwen's theorem I have cited (I have not read
% either of these papers, sorry). I still hope there
% is a combinatorial solution, some kind of construction
% of the nilpotent matrix from the Littlewood-Richardson
% tableau, but I wasn't able to make it work.
% 
% The remark likely shows off more ignorance than it
% does knowledge. If you can improve it, please do.

\begin{exercise}
\label{exe.Lambda.staircase-skew-1}
Let $a \in \Lambda$. Prove the following:

\begin{enumerate}
\item[(a)] The set $\left\{ g \in \Lambda \ \mid\ g^{\perp} a = \left(
\omega \left( g \right) \right) ^{\perp} a \right\}  $ is a
$\kk$-subalgebra of $\Lambda$.

\item[(b)] Assume that $e_k^{\perp} a = h_k^{\perp} a$ for each positive
integer $k$. Then, $g^{\perp} a = \left( \omega \left( g \right) \right)
^{\perp} a$ for each $g \in \Lambda$.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exe.Lambda.staircase-skew-2}
Let $n \in \NN$. Let $\rho$ be the partition
$\left( n-1, n-2, \ldots, 1 \right)$. Prove that
$s_{\rho / \mu} = s_{\rho / \mu^{t}}$ for every $\mu \in \Par$.
\end{exercise}

\begin{remark}
Exercise~\ref{exe.Lambda.staircase-skew-2} appears in \cite[Corollary
7.32]{ReinerShawVanWilligenburg2007}, and is due to John Stembridge. Using
Remark~\ref{third-L-R-interpretation-remark}, we can rewrite it as yet another
equality between Littlewood-Richardson coefficients: Namely, $c_{\mu,\nu
}^{\rho} = c_{\mu^{t},\nu}^{\rho}$ for any $\mu \in \Par$ and
$\nu \in \Par$.
\end{remark}

% [DG][v59] Added the above two exercises and remark.

\newpage



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Zelevinsky's structure theory of positive self-dual Hopf algebras}
\label{PSH-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Chapter~\ref{Sym-section} showed that,
as a $\ZZ$-basis for the Hopf algebra $\Lambda=\Lambda_\ZZ$,
the Schur functions $\{s_\lambda\}$ have two special properties:
they have the \emph{same} structure constants $ c^{\lambda}_{\mu,\nu}$
for their multiplication as for their comultiplication
(Corollary~\ref{self-duality-corollary}), and these structure constants
are all \emph{nonnegative} integers (Corollary~\ref{L-R-rule}).
Zelevinsky \cite[\S 2,3]{Zelevinsky} isolated these two properties as crucial.


\begin{definition}
Say that a connected graded Hopf algebra $A$ over $\kk=\ZZ$ with a
distinguished $\ZZ$-basis  $\{\sigma_\lambda\}$ consisting of homogeneous
elements\footnote{not necessarily indexed by partitions}
is a \dfn{positive self-dual Hopf algebra} (or \dfn{PSH})
if it satisfies the two further axioms
\begin{itemize}
\item{\textbf{(self-duality)}} \index{self-duality}
The same structure constants
$a^{\lambda}_{\mu,\nu}$ appear for the product
$\sigma_\mu \sigma_\nu = \sum_{\lambda} a^{\lambda}_{\mu,\nu} \sigma_\lambda$ and
the coproduct
$\Delta \sigma_\lambda = \sum_{\mu,\nu} a^{\lambda}_{\mu,\nu} \sigma_\mu \otimes \sigma_\nu$.
\item{\textbf{(positivity)}} \index{positivity}
The $a^\lambda_{\mu,\nu}$ are all nonnegative (integers).
\end{itemize}
Call $\{\sigma_\lambda\}$ the \dfn{PSH-basis} of $A$.
\end{definition}
He then developed a beautiful structure theory for PSH's, explaining
how they can be uniquely expressed as tensor products of copies of PSH's each
isomorphic to $\Lambda$ after rescaling their grading.
The next few sections explain this, following his exposition closely.

\subsection{Self-duality implies polynomiality}
\label{PSH-implies-polynomial-section}

We begin with a property that forces a Hopf algebra to have
algebra structure which is a \emph{polynomial} algebra, specifically the
symmetric algebra $\Sym(\Liep)$, where $\Liep$ is the $\kk$-submodule
of primitive elements.

Recall from Exercise~\ref{graded-connected-exercise}(g)
that for a connected graded Hopf algebra $A = \bigoplus_{n=0}^\infty A_n$,
every $x$ in the two-sided ideal
$
I:=\ker \epsilon = \bigoplus_{n > 0} A_n
$
has the property that its comultiplication takes the form
\[
\Delta(x) = 1 \otimes x + x \otimes 1 + \Delta_+(x)
\]
where $\Delta_+(x)$ lies in $I \otimes I$.
Recall also that the elements $x$ for which $\Delta_+(x)=0$ are
called the \dfn{primitives}.  Denote by $\Liep$ the $\kk$-submodule
of primitive elements inside $A$.

Given a PSH $A$ (over $\kk = \ZZ$) with a PSH-basis $\{\sigma_\lambda\}$,
we consider the bilinear form $(\cdot,\cdot)_A : A \times A \to \ZZ$ on $A$
that makes this basis orthonormal.
Similarly, the elements $\{\sigma_\lambda \otimes\sigma_\mu\}$ give
an orthonormal basis for a form $(\cdot,\cdot)_{A \otimes A}$ on $A \otimes A$.
The bilinear form $(\cdot,\cdot)_A$ on the PSH $A$ gives rise
to a $\ZZ$-linear map $A \to A^o$, which is easily seen to be
injective and a $\ZZ$-algebra homomorphism. We thus identify
$A$ with a subalgebra of $A^o$. When $A$ is of finite type,
this map is a Hopf algebra isomorphism, thus allowing us to
identify $A$ with $A^o$.
This is an instance of the following notion of self-duality.

% [DG][v17] Rewrote the above paragraph. It used to speak of
% "a graded Hopf algebra $A$ of finite type over $\kk = \ZZ$ with a
% PSH-basis", which made it nebulous whether the basis was supposed
% to be graded. More importantly, you were regarding the injection
% of $A$ into $A^o$ as an inclusion even if $A$ is not finite-type.

\begin{definition} \phantomsection
\label{def.self-duality}
\begin{itemize}
\item[(a)] If $\left(\cdot, \cdot\right) : V \times W \to \kk$
is a bilinear form on the
product $V \times W$ of two graded $\kk$-modules
$V = \bigoplus_{n \geq 0} V_n$ and $W = \bigoplus_{n \geq 0} W_n$,
then we say that this form $\left(\cdot, \cdot\right)$ is
\emph{graded}\index{graded bilinear form}
if every two distinct nonnegative integers $n$ and $m$ satisfy
$\left(V_n, W_m\right) = 0$ (that is,
if every two homogeneous elements $v \in V$ and $w \in W$ having
distinct degrees satisfy $\left(v, w\right) = 0$).
\item[(b)] If $\left(\cdot, \cdot\right)_V : V \times V \to \kk$ and
$\left(\cdot, \cdot\right)_W : W \times W \to \kk$ are two
symmetric bilinear forms on
some $\kk$-modules $V$ and $W$, then we can canonically define a
symmetric bilinear form $\left(\cdot, \cdot\right)_{V\otimes W}$
on the $\kk$-module $V \otimes W$ by letting
\[
\left(v \otimes w, v' \otimes w'\right)_{V\otimes W}
= \left(v, v'\right)_V \left(w, w'\right)_W
\qquad \qquad \text{for all } v, v' \in V \text{ and }
w, w' \in W.
\]
This new bilinear form is graded if the original two forms
$\left(\cdot, \cdot\right)_V$ and $\left(\cdot, \cdot\right)_W$
were graded (presuming that $V$ and $W$ are graded).
\item[(c)] Say that a bialgebra $A$ is
\emph{self-dual}\index{self-dual bialgebra}
with respect to a
given symmetric bilinear form $(\cdot, \cdot) : A \times A \to \kk$ if
one has $(a, m(b \otimes c))_A = (\Delta(a) , b \otimes c)_{A \otimes A}$
and $(\one_A,a)=\epsilon(a)$ for $a,b,c$ in $A$.
If $A$ is a graded Hopf algebra of finite type, and this
form $\left(\cdot, \cdot\right)$ is graded,
then this is equivalent to the $\kk$-module map $A \rightarrow A^o$
induced by $(\cdot, \cdot)_A$ giving a Hopf algebra homomorphism.
\end{itemize}
\end{definition}

% [DG][v17] Revamped this definition, which was subtly wrong (gradedness
% condition on the form was missing, and it was claimed to yield a Hopf
% algebra isomorphism rather than homomorphism). Parts (a) and (b) are
% not very exciting, but probably necessary (linear algebra texts are
% too lazy to deal with these things).

Thus, any PSH $A$ is self-dual with respect to the bilinear form
$\left(\cdot, \cdot\right)_A$ that makes its PSH-basis orthonormal.

Notice also that the injective $\ZZ$-algebra homomorphism $A \to A^o$
obtained from the bilinear form $\left(\cdot, \cdot\right)_A$
on a PSH $A$
allows us to regard each $f \in A$ as an element of $A^o$.
Thus, for any PSH $A$ and any
$f \in A$, an operator $f^\perp : A \to A$
is well-defined (indeed, regard $f$ as an element of $A^o$, and
apply Definition~\ref{def.skewing}).

% [DG][v64] Inserted the above paragraph.

\begin{proposition}
\label{primitives-split-off-prop}
Let $A$ be a Hopf algebra over $\kk=\ZZ$ or $\kk=\QQ$ which is graded, connected, and self-dual
with respect to a positive definite graded\footnote{That is,
$(A_i,A_j)=0$ for $i \neq j$.}  bilinear form. Then:
\begin{itemize}
\item[(a)] Within the ideal $I$,
the $\kk$-submodule of primitives $\Liep$ is the orthogonal complement to the $\kk$-submodule $I^2$.
\item[(b)] In particular, $\Liep \cap I^2=0$.
\item[(c)] When $\kk=\QQ$, one has $I = \Liep \oplus I^2$.
\end{itemize}
\end{proposition}
\begin{proof}
(a) Note that $I^2=m(I \otimes I)$.  Hence an element $x$ in $I$ lies in the perpendicular
space to $I^2$ if and only if one has for all $y$ in $I \otimes I$ that
\[
0= ( x , m(y) )_A = (\Delta(x), y)_{A \otimes A} = (\Delta_+(x), y)_{A \otimes A}
\]
where the second equality uses self-duality, while the third equality uses
the fact that $y$ lies in $I \otimes I$ and the form $(\cdot,\cdot)_{A \otimes A}$
makes distinct homogeneous components orthogonal.
Since $y$ was arbitrary, this means $x$ is perpendicular
to $I^2$ if and only if $\Delta_+(x)=0$, that is, $x$ lies in $\Liep$.

(b) This follows from (a), since the form $\left(\cdot, \cdot\right)_A$ is positive definite.

(c) This follows from (a) using some basic linear algebra\footnote{Specifically,
either the
existence of an orthogonal projection on a subspace of a finite-dimensional
inner-product space over $\QQ$, or the fact that $\dim\left(W^\perp\right) = \dim V
- \dim W$ for a subspace $W$ of a finite-dimensional inner-product space $V$
over $\QQ$ can be used.}
when $A$ is of finite type
(which is the only case we will ever encounter in practice). See
Exercise~\ref{exe.primitive-split-off-prop.infinite} for the general proof.
\end{proof}

% [DG][v13] I added the three sentences under (b) and (c) to this proof
% because I think that any use of positivity is a nontrivial step in
% algebra and that (c) is not trivial at all in the infinite-type case.
% Am I overthinking things? (Here is why I believe (c) is not trivial:
% In a vector space with infinite basis $(a_1, a_2, ...)$, take the
% positive-definite bilinear form given by $(a_i, a_j) = \delta_{i,j}$,
% and consider the subspace spanned by $a_i - a_{i+1}$ for all $i$.
% The orthogonal complement of this subspace is $0$, and there is *no*
% direct sum decomposition anymore. Is there a good reason why it is
% clear that such situations don't happen in our case?)

% [DG][v17] Added footnote in proof of (c) to stress that everything must
% be done over $\QQ$.

\begin{remark}
One might wonder why we didn't just say $I = \Liep \oplus I^2$ even when $\kk=\ZZ$
in Proposition~\ref{primitives-split-off-prop}(c).
However, this is false
%\footnote{Thanks to Darij Grinberg for pointing this out to me.},
even for $A=\Lambda_\ZZ$:
the second homogeneous component $(\Liep \oplus I^2)_2$ is the
index $2$ sublattice of $\Lambda_2$ which is $\ZZ$-spanned by
$\{p_2,e_1^2\}$, containing $2e_2$, but not containing $e_2$ itself.
\end{remark}

Already the fact that $\Liep \cap I^2=0$ has a strong implication.

\begin{lemma}
\label{general-commutativity-lemma}
A connected graded Hopf algebra $A$ over any ring $\kk$ having $\Liep \cap I^2=0$
must necessarily be commutative (as an algebra).
\end{lemma}
% [DG][v62] "has algebra structure which is commutative"
% --> "must necessarily be commutative (as an algebra)".
\begin{proof}
The component $A_0 = \kk$ commutes with all of $A$.  This forms the base case for
an induction on $i+j$ in which one shows that any elements $x$ in $A_i$ and $y$ in $A_j$
with $i,j>0$ will have $[x,y]:=xy-yx = 0$.
Since $[x,y]$ lies in $I^2$, it suffices to show that
$[x,y]$ also lies in $\Liep$:
\begin{align*}
\Delta[x,y]
&= [\Delta(x),\Delta(y)]\\
&= [1 \otimes x + x \otimes 1+\Delta_+(x),
    1 \otimes y + y \otimes 1+\Delta_+(y)]\\
&= [1 \otimes x + x \otimes 1,1 \otimes y + y \otimes 1] \\
&\qquad   +[1 \otimes x + x \otimes 1,\Delta_+(y)]
   +[\Delta_+(x), 1 \otimes y + y \otimes 1]
   +[\Delta_+(x),\Delta_+(y)]\\
&=[1 \otimes x + x \otimes 1,1 \otimes y + y \otimes 1] \\
&=1 \otimes [x,y] + [x,y] \otimes 1
\end{align*}
showing that $[x,y]$ lies in $\Liep$.
Here the second-to-last equality used the inductive hypotheses:
 homogeneity implies that $\Delta_+(x)$ is a sum of
homogeneous tensors of the form $z_1 \otimes z_2$
satisfying $\deg (z_1 ),\deg (z_2 ) < i$, so that by
induction they will commute with $\one \otimes y, y \otimes \one$,
thus proving that $[\Delta_+ (x) , \one \otimes y + y \otimes \one] = 0$;
a symmetric argument shows $[\one \otimes x+x \otimes \one,\Delta_+ (y)] = 0$,
and a similar argument shows $[\Delta_+ (x) , \Delta_+ (y)] = 0$.
The last equality is an easy calculation, and was done already in
the process of proving
\eqref{commutator-of-primitives}.
\end{proof}

\begin{remark}
Zelevinsky actually shows \cite[Proof of A.1.3, p. 150]{Zelevinsky} that the
assumption of $\Liep \cap I^2 = 0$
(along with hypotheses of unit,
counit, graded, connected, and $\Delta$ being a morphism for multiplication)
already implies the \emph{associativity} of the multiplication in $A$ !
One shows by induction on $i+j+k$ that any $x,y,z$ in $A_i,A_j,A_k$
with $i,j,k>0$ have vanishing \dfn{associator}
$\assoc(x,y,z) := x(yz) - (xy)z$.  In the inductive step, one first
notes that $\assoc(x,y,z)$ lies in $I^2$, and then checks
that $\assoc(x,y,z)$ also lies in $\Liep$, by a calculation very similar to the one above,
repeatedly using the fact that $\assoc(x,y,z)$ is multilinear in its three arguments.
\end{remark}

% [DG][v17] Replaced "the assumption of self-duality" by
% "the assumption of $\Liep \cap I^2 = 0$", since the former
% would only be true if the form is positive definite.

\begin{exercise}
\label{exe.primitive-split-off-prop.infinite}
Prove Proposition~\ref{primitives-split-off-prop}(c) in the general case.
\end{exercise}

% [DG][v13] Added this exercise.

This leads to a general structure theorem.
\begin{theorem}
\label{general-structure-theorem}
If a connected graded Hopf algebra $A$ over
a field $\kk$ of characteristic zero has
$I=\Liep \oplus I^2$,
then the inclusion $\Liep \hookrightarrow A$ extends to a
Hopf algebra isomorphism from the symmetric algebra
$\Sym_\kk(\Liep) \rightarrow A$.  In particular, $A$ is
both commutative and cocommutative.
\end{theorem}

\noindent
Note that the hypotheses of Theorem~\ref{general-structure-theorem}
% [DG][v62] "these hypotheses" -> "the hypotheses of Theorem~\ref{general-structure-theorem}".
are valid, using Proposition~\ref{primitives-split-off-prop}(c),
whenever $A$ is obtained from a PSH (over $\ZZ$) by tensoring with $\QQ$.

\begin{proof}[Proof of Theorem~\ref{general-structure-theorem}]
Since Lemma~\ref{general-commutativity-lemma} implies that $A$ is commutative,
the universal property of $\Sym_\kk(\Liep)$ as a free commutative algebra
on generators $\Liep$ shows that the inclusion
$\Liep \hookrightarrow A$ at least extends to an algebra morphism
$\Sym_\kk(\Liep) \overset{\varphi}{\rightarrow} A$.
Since the Hopf structure on $\Sym_\kk(\Liep)$
makes the elements of $\Liep$ primitive (see
Example~\ref{symmetric-algebra-as-bialgebra-example}),
this $\varphi$ is actually a coalgebra morphism
(since $\Delta \circ \varphi = (\varphi \otimes \varphi) \circ \Delta$
and $\epsilon \circ \varphi =\epsilon$
need only to be checked on algebra generators),
hence a bialgebra morphism,
hence a Hopf algebra morphism
(by Corollary~\ref{cor.bialg-mor-is-Hopf}).
It remains to show that $\varphi$ is surjective, and injective.

For the surjectivity of $\varphi$, note that the hypothesis
$I=\Liep \oplus I^2$ implies that the composite $\Liep \hookrightarrow I \rightarrow I/I^2$
gives a $\kk$-vector space isomorphism.
What follows is a standard argument
to deduce that $\Liep$ generates $A$ as a commutative graded $\kk$-algebra.
One shows by induction
on $n$ that any homogeneous element $a$ in $A_n$ lies in the
$\kk$-subalgebra generated
by $\Liep$.  The base case $n=0$ is trivial as $a$ lies in $A_0 = \kk \cdot 1_A$.
% [DG][v62] Replaced "$a$ lies in $\kk$" by "$a$ lies in $A_0 = \kk \cdot 1_A$".
In the inductive step where $a$ lies in $I$, write $a \equiv p \bmod{I^2}$
for some $p$ in $\Liep$.  Thus $a=p + \sum_i b_i c_i$,
where $b_i,c_i$ lie in $I$ but have strictly smaller degree,
so that by induction they lie in the
subalgebra generated by $\Liep$, and hence so does $a$.

Note that the surjectivity argument did not use the assumption that
$\kk$ has characteristic zero, but we will now use it in the injectivity
argument for $\varphi$, to establish the following
\begin{equation}
\label{primitives-in-symmetric-algebra}
\text{\textbf{Claim:} Every primitive element of $\Sym(\Liep)$ lies in $\Liep=\Sym^1(\Liep)$.}
\end{equation}
Note that this claim fails in positive characteristic, e.g. if $\kk$ has characteristic $2$
then $x^2$ lies in $\Sym^2(\Liep)$, however
\[
\Delta (x^2) = 1 \otimes x^2 + 2 x \otimes x + x^2 \otimes 1
= 1 \otimes x^2  + x^2 \otimes 1.
\]
To prove the claim \eqref{primitives-in-symmetric-algebra},
assume not, so that by gradedness, there must exist some
primitive element $y \neq 0$ lying in some $\Sym^n(\Liep)$ with $n \geq 2$.
This would mean that $f(y) = 0$, where the map $f$ is defined
as the composition
\[
\Sym^n(\Liep) \overset{\Delta}{\longrightarrow}
 \bigoplus_{i+j=n} \Sym^i(\Liep) \otimes \Sym^j(\Liep)
   \overset{\text{projection}}{\longrightarrow}
   \Sym^1(\Liep) \otimes \Sym^{n-1}(\Liep)
\]
of the coproduct $\Delta$ with the component projection
of $\bigoplus_{i+j=n} \Sym^i(\Liep) \otimes \Sym^j(\Liep)$
onto $\Sym^1(\Liep) \otimes \Sym^{n-1}(\Liep)$.
However, one can check on a basis
that the multiplication backward
$\Sym^1(\Liep) \otimes \Sym^{n-1}(\Liep) \overset{m}{\rightarrow} \Sym^n(\Liep)$
has the property that $m \circ f = n \cdot \id_{\Sym^n(\Liep)}$:
Indeed,
\[
(m \circ f)(x_1 \cdots x_n)
 = m \left(\sum_{j=1}^n x_j \otimes x_1 \cdots \widehat{x_{j}} \cdots x_n \right)
 = n \cdot x_1 \cdots x_n
\]
for $x_1,\ldots,x_n$ in $\Liep$.
Then $n \cdot y = m(f(y)) = m(0) = 0$ leads to the contradiction that $y=0$,
since $\kk$ has characteristic zero.
Thus, \eqref{primitives-in-symmetric-algebra} is proven.

Now one can argue the injectivity of the (graded) map\footnote{The grading on $\Sym(\Liep)$ is induced from the grading on $\Liep$, a homogeneous subspace of $I \subset A$ as it is the kernel of the graded map $I \overset{\Delta_+}{\longrightarrow} A \otimes A$.} $\varphi$ by assuming that one has a nonzero homogeneous
element $u$ in $\ker(\varphi)$ of minimum degree.
In particular, $\deg(u) \geq 1$.  Also since $\Liep \hookrightarrow A$, one has that $u$ is not in $\Sym^1(\Liep) = \Liep$, and hence $u$ is not primitive
by \eqref{primitives-in-symmetric-algebra}.  Consequently $\Delta_+(u) \neq 0$, and one can find
a nonzero component $u^{(i,j)}$ of $\Delta_+(u)$ lying in
$\Sym(\Liep)_i \otimes \Sym(\Liep)_j$ for some $i,j > 0$.
Since this forces $i,j < \deg(u)$,
one has that $\varphi$ maps both $\Sym(\Liep)_i,\Sym(\Liep)_j$ injectively into $A_i,A_j$.
Hence the tensor product map
\[
\Sym(\Liep)_i \otimes \Sym(\Liep)_j
\overset{\varphi \otimes \varphi}{\longrightarrow}
A_i \otimes A_j
\]
is also injective\footnote{One needs to know that for two
injective maps $V_i \overset{\varphi_i}{\rightarrow} W_i$
of $\kk$-vector spaces $V_i,W_i$ with $i=1,2$, the
tensor product $\varphi_1 \otimes \varphi_2$ is also injective.
Factoring it as
$
\varphi_1 \otimes \varphi_2 =
\left( \id \otimes \varphi_2 \right) \circ
\left( \varphi_1 \otimes \id \right),
$
one sees that it suffices to show that for an injective
map $V \overset{\varphi}{\hookrightarrow} W$ of free $\kk$-modules,
and any free $\kk$-module $U$, the map
$V \otimes U \overset{\varphi \otimes \id}{\longrightarrow} W\otimes U$ is
also injective.  Since tensor products commute with direct sums,
and $U$ is (isomorphic to) a direct sum of copies of $\kk$, this
reduces to the easy-to-check case where $U=\kk$.

Note that some kind of freeness or flatness hypothesis on $U$ is needed here since, e.g. the injective $\ZZ$-module maps
$\ZZ \overset{\varphi_1 = (\cdot \times 2)}{\longrightarrow} \ZZ$ and
$\ZZ/2\ZZ \overset{\varphi_2= \id}{\longrightarrow} \ZZ/2\ZZ$
have $\varphi_1 \otimes \varphi_2=0$ on
$\ZZ \otimes_\ZZ \ZZ/2\ZZ \cong \ZZ/2\ZZ \neq 0$.}.
This implies $(\varphi \otimes \varphi)(u^{(i,j)}) \neq 0$, giving the contradiction that
\[
0=\Delta^A_+(0) = \Delta^A_+(\varphi(u)) = (\varphi \otimes \varphi)(\Delta^{\Sym(\Liep)}_+(u))
\]
contains the nonzero $A_i \otimes A_j$-component
$(\varphi\otimes \varphi)(u^{(i,j)})$.

(An alternative proof of the injectivity of $\varphi$ proceeds as
follows: By \eqref{primitives-in-symmetric-algebra}, the subspace
of primitive elements of $\Sym(\Liep)$ is $\Liep$, and clearly
$\varphi\mid_{\Liep}$ is injective. Hence,
Exercise~\ref{exe.I=0}(b) (applied to the homomorphism $\varphi$)
shows that $\varphi$ is injective.)
\end{proof}

% [DG][v13] I added the paragraph above; feel free to remove it if it
% doesn't fit.

Before closing this section, we mention one nonobvious corollary of
the Claim \eqref{primitives-in-symmetric-algebra}, when applied to
the ring of symmetric functions $\Lambda_\QQ$ with $\QQ$-coefficients,
since Proposition~\ref{symm-antipode-and-h-basis} says that
$\Lambda_\QQ=\QQ[p_1,p_2,\ldots]=\Sym(V)$ where $V=\QQ\{p_1,p_2,\ldots\}$.


\begin{corollary}
\label{primitives-in-Lambda}
The subspace $\Liep$ of primitives in $\Lambda_\QQ$ is
one-dimensional in each degree $n=1,2,\ldots$, and
spanned by $\{p_1,p_2,\ldots\}$.
\end{corollary}

We note in passing that this corollary can also be obtained in a
simpler fashion and a greater generality:

\begin{exercise}
\label{exe.Lambda.primitives}
Let $\kk$ be any commutative ring. Show that the primitive
elements of $\Lambda$ are precisely the elements of the
$\kk$-linear span of $p_1, p_2, p_3, \ldots$.
\end{exercise}

% [DG][v28] Added this exercise (one of the few exercises I can
% find for this section).


\subsection{The decomposition theorem}

Our goal here is Zelevinsky's theorem \cite[Theorem 2.2]{Zelevinsky} giving a canonical
decomposition of any PSH as a tensor product into PSH's that each
have only one primitive element in their PSH-basis.  For the sake
of stating it, we introduce some notation.

\begin{definition}
\label{indecomposable-PSH-factors-defn}
Given a PSH $A$ with
% [DG][v62] "Given $A$ a PSH" -> "Given a PSH $A$".
PSH-basis $\Sigma$,
let $\CCC:=\Sigma \cap \Liep$\index{$\CCC$} be the
primitive elements in  $\Sigma$.
For each $\rho$ in $\CCC$, let
$A(\rho) \subset A$ be the $\ZZ$-span of
\[
\Sigma(\rho):=\{\sigma \in \Sigma:
              \text{ there exists }n \geq 0\text{ with }
                (\sigma, \rho^n) \neq 0\}.
\]
\end{definition}

\begin{definition}
The
tensor product of two PSHs $A_1$ and $A_2$ with PSH-bases
$\Sigma_1$ and $\Sigma_2$ is defined as the graded Hopf algebra
$A_1 \otimes A_2$ with PSH-basis
$\left\{\sigma_1 \otimes \sigma_2\right\}
_{\left(\sigma_1, \sigma_2\right) \in \Sigma_1 \times \Sigma_2}$.
It is easy to see that this is again a PSH.
The tensor product of any finite family of PSHs is defined
similarly\footnote{For the empty family, it is the connected graded
Hopf algebra $\ZZ$ with PSH-basis $\left\{1\right\}$.}.
\end{definition}

% [DG][v62] Moved the above definition into an actual
% definition environment (it used to be in a footnote).

\begin{theorem}
\label{Zelevinsky-decomposition-theorem}
Any PSH $A$ has a canonical tensor product decomposition
\[
A = \bigotimes_{ \rho \in \CCC } A(\rho)
\]
with $A(\rho)$ a PSH, and $\rho$ the only
primitive element in its PSH-basis $\Sigma(\rho)$.
\end{theorem}
\noindent
Although in all the applications, $\CCC$ will be finite,
when $\CCC$ is infinite one should interpret the tensor product
in the theorem as the inductive limit of tensor products over finite subsets of $\CCC$,
that is, linear combinations of basic tensors $\bigotimes_\rho {a_\rho}$
in which there are only finitely many factors $a_\rho \neq 1$.

% [DG][v14] Added footnote about tensor products of PSHs.

The first step toward the theorem uses a certain unique
factorization property.

\begin{lemma}
\label{unique-factorization-lemma}
Let $\PPP$ be a set of pairwise orthogonal primitives in a PSH $A$.
Then,
\[
( \rho_1 \cdots \rho_r , \pi_1 \cdots \pi_s ) = 0
\]
for $\rho_i, \pi_j$ in $\PPP$ unless $r=s$ and
one can reindex so that $\rho_i=\pi_i$.
\end{lemma}
\begin{proof}
%Induct on $\min(r,s)$. 
Induct on $r$.
For $r > 0$, one has
\begin{align*}
(\rho_1 \cdots \rho_r , \pi_1 \cdots \pi_s )
&= (\rho_2 \cdots \rho_r , \rho_1^{\perp}(\pi_1 \cdots \pi_s) ) \\
&= (\rho_2 \cdots \rho_r ,
  \sum_{j=1}^s (\pi_1 \cdots \pi_{j-1} \cdot \rho_1^{\perp}(\pi_j) \cdot \pi_{j+1} \cdots \pi_s) )
\end{align*}
from Proposition~\ref{skewing-properties-prop}(iv)
because $\rho_1$ is primitive\footnote{Strictly speaking, this argument needs
further justification since $A$ might not be of finite type (and if it is
not, Proposition~\ref{skewing-properties-prop}(iv) cannot be applied).
It is more adequate to refer to the proof of
Proposition~\ref{skewing-properties-prop}(iv), which indeed goes through
with $\rho_1$ taking the role of $f$.}. 
 On the other hand, since each $\pi_j$ is
primitive, one has
$
\rho_1^\perp(\pi_j) = (\rho_1,\one) \cdot \pi_j + (\rho_1,\pi_j) \cdot \one
=(\rho_1,\pi_j)
$
which vanishes unless $\rho_1 = \pi_j$.
Hence $(\rho_1 \cdots \rho_r , \pi_1 \cdots \pi_s )=0$ unless $\rho_1 \in
\{\pi_1,\ldots,\pi_s\}$, in which case after reindexing so that
$\pi_1=\rho_1$, it equals
\[
n \cdot (\rho_1,\rho_1) \cdot
(\rho_2 \cdots \rho_r , \pi_2 \cdots \pi_s )
\]
if there are exactly $n$ occurrences of $\rho_1$ among
$\pi_1,\ldots,\pi_s$.  Now apply induction.
\end{proof}

% [DG][v14] Replaced "Induct on $\min(r,s)$" by "Induct on $r$" for minimal
% simplification.

% [DG][v17] Added footnote above.

So far the positivity hypothesis for a PSH has played little role.
Now we use it to introduce a certain partial order on the PSH $A$,
and then a semigroup grading.

\begin{definition}
For a subset $S$ of an abelian group,
let $\ZZ S$ (resp. $\NN S$) denote the subgroup of $\ZZ$-linear combinations
(resp. submonoid of $\NN$-linear combinations\footnote{Recall that
$\NN:=\{0,1,2,\ldots\}$.}) of the elements of $S$.

In a PSH $A$ with PSH-basis $\Sigma$, the subset $\NN \Sigma$ forms
a submonoid, and lets one define a partial order on $A$ via $a \leq b$
if $b-a$ lies in $\NN\Sigma$.
\end{definition}

We note a few trivial properties of this partial order:
\begin{itemize}
\item
The positivity hypothesis implies that
$
\NN \Sigma \cdot \NN \Sigma \subset \NN \Sigma.
$
\item
Hence multiplication by an element $c \geq 0$
(meaning $c$ lies in $\NN \Sigma$) preserves the order:
$a \leq b$ implies $ac \leq bc$ since
$(b-a)c$ lies in $\NN \Sigma$.
\item
Thus $0 \leq c \leq d$ and $0 \leq a \leq b$ together imply
$
ac \leq bc \leq bd
$.
\end{itemize}

\noindent
This allows one to introduce a semigroup grading on $A$.

\begin{definition}
Let $\NN^{\CCC}_{\fin}$ denote the additive submonoid
of $\NN^{\CCC}$ consisting of those $\alpha=(\alpha_\rho)_{\rho \in \CCC}$
with finite support.

Note that for any $\alpha$ in $\NN^{\CCC}_{\fin}$, one has that the product
$\prod_{ \rho \in \CCC } \rho^{\alpha_\rho} \geq 0$.  Define
\[
\Sigma(\alpha):=\{ \sigma \in \Sigma: \sigma \leq \prod_{\rho \in \CCC} \rho^{\alpha_\rho} \},
\]
that is, the subset of $\Sigma$ on which $\prod_{ \rho \in \CCC } \rho^{\alpha_\rho}$
has support.  Also define
\[
A_{(\alpha)} := \ZZ \Sigma(\alpha) \subset A.
\]
\end{definition}

\begin{proposition}
\label{semigroup-grading-prop}
The PSH $A$ has an $\NN^{\CCC}_{\fin}$-semigroup-grading:
one has an orthogonal direct sum decomposition
\[
A = \bigoplus_{ \alpha \in \NN^{\CCC}_{\fin}} A_{(\alpha)}
\]
for which
\begin{align}
\label{product-semigroup-graded}
A_{(\alpha)} A_{(\beta)}  &\subset A_{(\alpha+\beta)} , \\
\label{comultiplication-semigroup-graded}
\Delta A_{(\alpha)} &\subset
  \bigoplus_{\alpha=\beta+\gamma}A_{(\beta)} \otimes A_{(\gamma)}.
\end{align}
\end{proposition}
\begin{proof}
We will make free use of the fact that a PSH $A$ is commutative,
since it embeds in $A \otimes_\ZZ \QQ$, which is commutative by
Theorem~\ref{general-structure-theorem}.

Note that the orthogonality $\left( A_{(\alpha)}, A_{(\beta)} \right) = 0$
for $\alpha \neq \beta$ is equivalent to the assertion that
\[
\left(\prod_{ \rho \in \CCC } \rho^{\alpha_\rho},
\prod_{ \rho \in \CCC } \rho^{\beta_\rho} \right)=0 ,
\]
which follows from Lemma~\ref{unique-factorization-lemma}.

Next let us deal with the assertion \eqref{product-semigroup-graded}.
It suffices to check that when $\tau,\omega$ in $\Sigma$
lie in $A_{(\alpha)}, A_{(\beta)}$, respectively, then $\tau\omega$
lies in $A_{(\alpha+\beta)}$.  But note that any $\sigma$ in $\Sigma$
having $\sigma \leq \tau \omega$ will then have
\[
\sigma \leq \tau \omega
\leq
\prod_{ \rho \in \CCC } \rho^{\alpha_\rho} \cdot
\prod_{ \rho \in \CCC } \rho^{\beta_\rho}
=\prod_{ \rho \in \CCC } \rho^{\alpha_\rho+\beta_\rho}
\]
so that $\sigma$ lies in $A_{(\alpha+\beta)}$.  This means
that $\tau\omega$ lies in $A_{(\alpha+\beta)}$.

This lets us check that $\bigoplus_{ \alpha \in \NN^{\CCC}_{\fin}} A_{(\alpha)}$ exhaust $A$.  It suffices
to check that any $\sigma$ in $\Sigma$ lies in some $A_{(\alpha)}$.  Proceed
by induction on $\deg(\sigma)$, with the case $\sigma=\one$ being
trivial;  the element $\one$ always lies in $\Sigma$, and hence lies
in $A_{(\alpha)}$ for $\alpha=0$.  For $\sigma$ lying in $I$,
one either has
$(\sigma,a) \neq 0$ for some $a$ in $I^2$, or
else $\sigma$ lies in $(I^2)^\perp=\Liep$
(by Proposition~\ref{primitives-split-off-prop}(a)), so that
$\sigma$ is in $\CCC$
and we are done.
If $(\sigma,a) \neq 0$ with $a$ in $I^2$, then
$\sigma$ appears in the support of some
$\ZZ$-linear combination of elements $\tau\omega$ where $\tau,\omega$
lie in $\Sigma$ and have strictly smaller degree than $\sigma$ has.  
There exists at least
one such pair $\tau, \omega$ for which $(\sigma, \tau\omega) \neq 0$,
and therefore $\sigma \leq \tau\omega$.  Then by induction $\tau, \omega$
lie in some $A_{(\alpha)}, A_{(\beta)}$, respectively, so $\tau\omega$ lies in $A_{(\alpha+\beta)}$, and hence $\sigma$ lies
in $A_{(\alpha+\beta)}$ also.

% [DG][v14] Added "than $\sigma$ has".

Self-duality shows that
\eqref{product-semigroup-graded}
implies \eqref{comultiplication-semigroup-graded}:
if $a,b,c$ lie in $A_{(\alpha)}, A_{(\beta)}, A_{(\gamma)}$,
respectively, then
$
(\Delta a, b \otimes c)_{A \otimes A} = (a, bc)_A = 0
$
unless $\alpha=\beta+\gamma$.
\end{proof}


\begin{proposition}
\label{disjoint-support-prop}
For $\alpha, \beta$ in $\NN^\CCC_{\fin}$ with disjoint support, one has a bijection
\[
\begin{array}{rcl}
\Sigma(\alpha) \times \Sigma(\beta) & \longrightarrow & \Sigma(\alpha+\beta) ,\\
(\sigma, \tau) & \longmapsto & \sigma \tau.
\end{array}
\]
Thus, the multiplication map
$
A_{(\alpha)} \otimes A_{(\beta)} \rightarrow A_{(\alpha+\beta)}$
is an isomorphism.
\end{proposition}
\begin{proof}
We first check that for $\sigma_1,\sigma_2$ in  $\Sigma(\alpha)$
and $\tau_1,\tau_2$ in $\Sigma(\beta)$, one has
\begin{equation}
\label{disjoint-support-orthonormality}
(\sigma_1 \tau_1, \sigma_2 \tau_2) = \delta_{(\sigma_1,\tau_1),(\sigma_2,\tau_2)}.
\end{equation}
Note that this is equivalent to showing both
\begin{enumerate}
\item[$\bullet$]
that $\sigma \tau$ lie in $\Sigma(\alpha+\beta)$
so that the map is well-defined, since it shows
$(\sigma \tau, \sigma \tau)=1$, and
\item[$\bullet$]
that the map is injective.
\end{enumerate}
One calculates
\begin{align*}
(\sigma_1 \tau_1, \sigma_2 \tau_2)_A
&=(\sigma_1 \tau_1, m(\sigma_2 \otimes \tau_2))_A \\
&=(\Delta(\sigma_1 \tau_1), \sigma_2 \otimes \tau_2)_{A \otimes A}\\
&=(\Delta(\sigma_1) \Delta(\tau_1), \sigma_2 \otimes \tau_2)_{A \otimes A} .
\end{align*}
Note that due to \eqref{comultiplication-semigroup-graded}, $\Delta(\sigma_1) \Delta(\tau_1)$ lies in $\sum A_{(\alpha'+\beta')} \otimes A_{(\alpha'' + \beta'')}$ where
\begin{align*}
\alpha'+\alpha''&=\alpha ,\\
\beta'+\beta''&=\beta.
\end{align*}
Since $\sigma_2 \otimes \tau_2$
lies in $A_{(\alpha)} \otimes A_{(\beta)}$, the only nonvanishing terms in the inner product come
from those with
\begin{align*}
\alpha'+\beta'&=\alpha ,\\
\alpha''+\beta''&=\beta.
\end{align*}
As $\alpha,\beta$ have disjoint support, this can only happen if
\[
\alpha'=\alpha, \,\, \alpha''=0, \,\, \beta'=0,\,\, \beta''=\beta ;
\]
that is, the only nonvanishing term comes from
$(\sigma_1 \otimes \one)(\one \otimes \tau_1)=\sigma_1 \otimes \tau_1$.
Hence
\[
(\sigma_1 \tau_1, \sigma_2 \tau_2)_A
=(\sigma_1 \otimes \tau_1,\sigma_2 \otimes \tau_2)_{A \otimes A}\\
= \delta_{(\sigma_1,\tau_1),(\sigma_2,\tau_2)}.
\]

To see that the map is surjective, express
\begin{align*}
\prod_{\rho \in \CCC} \rho^{\alpha_\rho}&=\sum_i \sigma_i ,\\
\prod_{\rho \in \CCC} \rho^{\beta_\rho}&=\sum_j \tau_j
\end{align*}
with $\sigma_i \in \Sigma(\alpha)$ and $\tau_j \in \Sigma(\beta)$.
Then each product $\sigma_i \tau_j$ is in $\Sigma(\alpha +\beta)$ by
\eqref{disjoint-support-orthonormality}, and
\[
\prod_{\rho \in \CCC} \rho^{\alpha_\rho+\beta_\rho}=\sum_{i,j} \sigma_i \tau_j
\]
shows that $\{\sigma_i \tau_j\}$ exhausts $\Sigma(\alpha+\beta)$.  This
gives surjectivity.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{Zelevinsky-decomposition-theorem}]
Recall from Definition~\ref{indecomposable-PSH-factors-defn} that
for each $\rho$ in $\CCC$, one defines $A(\rho) \subset A$ to
be the $\ZZ$-span of
\[
\Sigma(\rho):=\{\sigma \in \Sigma:
              \text{ there exists }n \geq 0\text{ with }
                (\sigma, \rho^n) \neq 0\}.
\]
In other words, $A(\rho):=\bigoplus_{n \geq 0} A_{(n \cdot e_\rho)}$
where $e_\rho$ in $\NN^{\CCC}_{\fin}$ is the standard basis element indexed by $\rho$.
Proposition~\ref{semigroup-grading-prop} then shows that $A(\rho)$ is a Hopf subalgebra of $A$.
Since every $\alpha$ in $\NN^{\CCC}_{\fin}$ can be expressed as the
(finite) sum $\sum_{\rho} \alpha_{\rho} e_\rho$, and the $e_\rho$ have disjoint support,
iterating Proposition~\ref{disjoint-support-prop} shows that
$A = \bigotimes_{ \rho \in \CCC } A(\rho)$.  Lastly, $\Sigma(\rho)$ is clearly a PSH-basis
for $A(\rho)$, and if $\sigma$ is any primitive
element in $\Sigma(\rho)$ then $(\sigma,\rho^n) \neq 0$ lets one conclude
via Lemma~\ref{unique-factorization-lemma} that $\sigma=\rho$ (and $n=1$).
\end{proof}




\subsection{$\Lambda$ is the unique indecomposable PSH}

The goal here is to prove the rest of Zelevinsky's structure theory for PSH's.
Namely, if $A$ has only one primitive element $\rho$
in its PSH-basis $\Sigma$, then $A$ must be isomorphic as a PSH to
the ring of symmetric functions $\Lambda$, after one rescales the grading of $A$.
Note that every $\sigma$ in $\Sigma$ has
$\sigma \leq \rho^n$ for some $n$, and hence has degree divisible by
the degree of $\rho$.  Thus one can divide all degrees by that of $\rho$
and assume $\rho$ has degree $1$.

The idea is to find within $A$ and $\Sigma$ a set of elements that
play the role of
\[
\{ h_n=s_{(n)} \}_{n=0,1,2,\ldots}, \qquad
\{ e_n=s_{(1^n)} \}_{n=0,1,2,\ldots}
\]
within $A=\Lambda$ and its PSH-basis of Schur
functions $\Sigma=\{s_\lambda\}$.  Zelevinsky's argument does this by
isolating some properties that turn out to characterize these elements:

\begin{enumerate}
\item[(a)] $h_0=e_0=1$, and $h_1=e_1=:\rho$ has $\rho^2$ a sum of two elements of $\Sigma$, namely
\[
\rho^2=h_2+e_2.
\]
\item[(b)] For all $n=0,1,2,\ldots$, there exist unique elements
$h_n, e_n$ in $A_n \cap \Sigma$ that satisfy
\begin{align*}
h_2^\perp e_n &= 0,\\
e_2^\perp h_n &= 0
\end{align*}
with $h_2,e_2$ being the two elements of $\Sigma$ introduced in (a).
\item[(c)] For $k=0,1,2,\ldots,n$ one has
\begin{align*}
h_k^\perp h_n & = h_{n-k} \text{ and }
  \sigma^\perp h_n = 0 \text{ for } \sigma \in \Sigma \setminus \{h_0,h_1,\ldots,h_n\}, \\
e_k^\perp e_n & = e_{n-k} \text{ and }
  \sigma^\perp e_n = 0 \text{ for } \sigma \in \Sigma \setminus \{e_0,e_1,\ldots,e_n\}. \\
\end{align*}
In particular, $e_k^\perp h_n = 0=h_k^\perp e_n$ for $k \geq 2$.
\item[(d)] Their coproducts are
\begin{align*}
\Delta(h_n) &= \sum_{i+j=n} h_i \otimes h_j,\\
\Delta(e_n) &= \sum_{i+j=n} e_i \otimes e_j.\\
\end{align*}
\end{enumerate}

We will prove Zelevinsky's result \cite[Theorem 3.1]{Zelevinsky}
as a combination of the following two theorems.

\begin{theorem}
\label{uniqueness-of-sym-parts-a-b-c-d}
Let $A$ be a PSH with PSH-basis $\Sigma$ containing only one primitive
$\rho$, and assume that the grading has been rescaled so that
$\rho$ has degree $1$.
Then, after renaming $\rho=e_1=h_1$, one can find unique sequences
$
\{ h_n\}_{n=0,1,2,\ldots}, \{ e_n \}_{n=0,1,2,\ldots}
$
of elements of $\Sigma$ having properties (a),(b),(c),(d) listed above.
\end{theorem}

The second theorem uses the following notion.

\begin{definition}
A \dfn{PSH-morphism} $A \overset{\varphi}{\rightarrow} A'$
between two PSH's $A,A'$ having PSH-bases $\Sigma, \Sigma'$
is a graded Hopf algebra morphism for which
$\varphi(\NN \Sigma) \subset \NN \Sigma'$.  If $A=A'$ and
$\Sigma=\Sigma'$ it will be called a \dfn{PSH-endomorphism}.
If $\varphi$ is an isomorphism and restricts to a bijection
$\Sigma \rightarrow \Sigma'$, it will be called a \dfn{PSH-isomorphism}%
\footnote{This definition is easily seen to be equivalent to
saying that a PSH-isomorphism is an invertible PSH-morphism
whose inverse is again a PSH-morphism.};
if it is both a PSH-isomorphism and an endomorphism, it is
a \dfn{PSH-automorphism}.\footnote{The reader should be warned
that not every invertible PSH-endomorphism is necessarily a
PSH-automorphism. For instance, it is an easy exercise to
check that
$\Lambda \otimes \Lambda \to \Lambda \otimes \Lambda,
\ f \otimes g \mapsto \sum_{(f)} f_1 \otimes f_2 g$
is a well-defined invertible PSH-endomorphism of the PSH
$\Lambda \otimes \Lambda$ with PSH-basis
$\left(s_\lambda \otimes s_\mu\right)_{\left(\lambda,
  \mu\right) \in \Par \times \Par}$,
but not a PSH-automorphism.}
\end{definition}

% [DG][v25] Added above footnote (a counterexample I was looking
% for long ago).

\begin{theorem}
\label{uniqueness-of-sym-parts-e-f-g}
The elements
$
\{ h_n\}_{n=0,1,2,\ldots}, \{ e_n \}_{n=0,1,2,\ldots}
$
in Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d} also satisfy the following.

\begin{enumerate}
\item[(e)] The elements $h_n,e_n$ in $A$ satisfy the same relation \eqref{e-h-relation}
\[
\sum_{i+j=n} (-1)^i e_i h_j = \delta_{0,n}
\]
as their counterparts in $\Lambda$, along with the
property that
\[
A=\ZZ[h_1,h_2,\ldots]=\ZZ[e_1,e_2,\ldots].
\]
\item[(f)]  There is exactly one nontrivial automorphism $A \overset{\omega}{\rightarrow} A$ as
a PSH, swapping $h_n \leftrightarrow e_n$.

\item[(g)]  There are exactly two PSH-isomorphisms $A \rightarrow \Lambda$:
\begin{itemize}
\item one sending $h_n$ to the complete homogeneous symmetric functions $h_n(\xx)$, while sending $e_n$ to the elementary symmetric functions $e_n(\xx)$,
\item the second one (obtained by composing the first with $\omega$) sending
$h_n \mapsto e_n(\xx)$ and $e_n \mapsto h_n(\xx)$.
\end{itemize}
\end{enumerate}
\end{theorem}

Before embarking on the proof, we mention one more bit of convenient
terminology:  say that an element $\sigma$ in $\Sigma$ is a \dfn{constituent}
of $a$ in $\NN \Sigma$ when $\sigma \leq a$, that is, $\sigma$ appears with nonzero coefficient $c_\sigma$
in the  unique expansion $a=\sum_{\tau \in \Sigma} c_\tau \tau$.

\begin{proof}[Proof of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}]
One fact that occurs frequently is this:
\begin{equation}
\label{every-sigma-is-a-constituent-of-powers-of-rho}
\text{Every }\sigma\text{ in }\Sigma \cap A_n\text{ is a constituent of }\rho^n.
\end{equation}
This follows from Theorem~\ref{Zelevinsky-decomposition-theorem},
since $\rho$ is the only primitive element of $\Sigma$:  one has
$A=A(\rho)$ and $\Sigma=\Sigma(\rho)$, so that $\sigma$ is a constituent
of some $\rho^m$, and homogeneity considerations force $m=n$.

Notice that $A$ is of finite type (due to
\eqref{every-sigma-is-a-constituent-of-powers-of-rho}). Thus, $A^o$ is
a graded Hopf algebra isomorphic to $A$.

% [DG][v17] Added previous paragraph. I ended up quoting this in three
% further proofs in Chapter 3 as it's subtly used in several places.
% Not the best solution, I guess.

\vskip.1in
\noindent
{\sf Assertion (a).} Note that
\[
(\rho^2,\rho^2)=(\rho^\perp(\rho^2),\rho)=(2\rho,\rho)=2
\]
using the fact that $\rho^\perp$ is a derivation since $\rho$ is primitive
(Proposition~\ref{skewing-properties-prop}(iv)).  On the other hand, expressing
$\rho^2 = \sum_{ \sigma \in \Sigma } c_\sigma \sigma$ with $c_\sigma$ in $\NN$,
one has
$
(\rho^2,\rho^2)=\sum_\sigma c_\sigma^2.
$
Hence exactly two of the $c_\sigma=1$, so $\rho^2$ has exactly two distinct
constituents.   Denote them by $h_2$ and $e_2$.
One concludes that $\Sigma \cap A_2=\{h_2,e_2\}$ from
\eqref{every-sigma-is-a-constituent-of-powers-of-rho}.

Note also that the same argument shows $\Sigma \cap A_1 = \{\rho\}$,
so that $A_1=\ZZ \rho$.  Since $\rho^\perp h_2$ lies in $A_1=\ZZ\rho$
and $(\rho^\perp h_2,\rho)=(h_2,\rho^2)=1$, we have $\rho^\perp h_2=\rho$.
Similarly $\rho^\perp e_2=\rho$.
\vskip.1in
\noindent
{\sf Assertion (b).}
We will show via induction on $n$ the following three assertions for $n \geq 1$:
\begin{equation}
\label{h_n-inductive-properties}
\begin{aligned}
& \bullet \ \text{There exists an
element }h_n\text{ in }\Sigma \cap A_n \text{ with }e_2^\perp h_n=0.\\
& \bullet \ \text{This element }h_n \text{ is unique.} \\
& \bullet \ \text{Furthermore }\rho^\perp h_n = h_{n-1}.\\
\end{aligned}
\end{equation}
In the base cases $n=1,2$, it is not hard to check that
our previously labelled elements, $h_1, h_2$ (namely $h_1:=\rho$,
and $h_2$ as named in part (a)) really \emph{are}
the unique elements satisfying these hypotheses.

In the inductive step, it turns out that
we will find $h_n$ as a constituent of $\rho h_{n-1}$.
Thus we again use the derivation property of $\rho^\perp$ to compute
that $\rho h_{n-1}$ has exactly two constituents:
\begin{align*}
( \rho h_{n-1}, \rho h_{n-1} )
&=( \rho^\perp (\rho h_{n-1}), h_{n-1} ) \\
&=( h_{n-1} + \rho \cdot \rho^\perp h_{n-1}, h_{n-1} )\\
&=( h_{n-1} + \rho h_{n-2}, h_{n-1} )\\
&=1+ ( h_{n-2}, \rho^\perp h_{n-1} )\\
&=1+ ( h_{n-2}, h_{n-2} )=1+1=2
\end{align*}
where the inductive hypothesis $\rho^\perp h_{n-1}=h_{n-2}$ was used twice.
We next show that exactly one of the two constituents of $\rho h_{n-1}$
is annihilated by $e_2^\perp$.  Note that since $e_2$ lies in $A_2$,
and $A_1$ has $\ZZ$-basis element $\rho$, there is a constant $c$ in $\ZZ$
such that
\begin{equation}
\label{coproduct-of-e2}
\Delta(e_2) = e_2 \otimes \one + c \rho \otimes \rho + \one \otimes e_2.
\end{equation}
On the other hand, (a) showed
\[
1=( e_2, \rho^2 )_A = (\Delta (e_2), \rho \otimes \rho )_{A \otimes A}
\]
so one must have $c=1$.
Therefore by Proposition~\ref{skewing-properties-prop}{(iv)} again,
\begin{equation}
\label{e_2-skewing-calculation}
\begin{array}{rcccccl}
e_2^\perp ( \rho h_{n-1} )
&=& e_2^\perp(\rho) h_{n-1} &+& \rho^\perp(\rho) \rho^\perp(h_{n-1}) &+& \rho e_2^{\perp}(h_{n-1}) \\
&=& 0&+&h_{n-2}&+&0\\
&=&h_{n-2} ,
\end{array}
\end{equation}
where the first term vanished due to degree considerations and the last term
vanished by the inductive hypothesis.
Bearing in mind that $\rho h_{n-1}$ lies in $\NN \Sigma$, and
in a PSH with PSH-basis $\Sigma$, any skewing operator
$\sigma^\perp$ for $\sigma$ in $\Sigma$ will
preserve $\NN \Sigma$, one concludes from
\eqref{e_2-skewing-calculation} that
\begin{enumerate}
\item[$\bullet$]
one of the two distinct constituents of the element $\rho h_{n-1}$ must be sent by $e_2^\perp$ to $h_{n-2}$, and
\item[$\bullet$]
the other constituent of $\rho h_{n-1}$ must be annihilated by
$e_2^\perp$; call this second constituent $h_n$.
\end{enumerate}

Lastly, to see that this $h_n$ is unique, it suffices to show that
any element $\sigma$ of $\Sigma \cap A_n$ which is killed by $e_2^\perp$ must be a constituent of
$\rho h_{n-1}$.  This holds for the following reason.
We know $\sigma \leq \rho^n$ by
\eqref{every-sigma-is-a-constituent-of-powers-of-rho},
and hence $0 \neq (\rho^n,\sigma) = (\rho^{n-1},\rho^\perp \sigma)$,
implying that $\rho^\perp \sigma \neq 0$.
On the other hand, since
$0= \rho^\perp e_2^\perp \sigma = e_2^\perp \rho^\perp \sigma$,
one has that $\rho^\perp \sigma$ is annihilated by $e_2^\perp$, and
hence $\rho^\perp \sigma$ must be a (positive) multiple of $h_{n-1}$ by part of
our inductive hypothesis.  Therefore $(\sigma, \rho h_{n-1}) = (\rho^\perp \sigma, h_{n-1})$
is positive, that is, $\sigma$ is a constituent of $\rho h_{n-1}$.

The preceding argument, applied to $\sigma=h_n$, shows that
$\rho^\perp h_n=ch_{n-1}$ for some $c$ in $\{1,2,\ldots\}$.  Since
$(\rho^\perp h_n,h_{n-1})=(h_n,\rho h_{n-1})=1$, this $c$ must be $1$, so
that $\rho^\perp h_n=h_{n-1}$.  This completes the induction step in
the proof of \eqref{h_n-inductive-properties}.

One can then argue, swapping the roles of $e_n, h_n$ in the
above argument, the existence and uniqueness of
a sequence $\{e_n\}_{n=0}^\infty$ in $\Sigma$ satisfying
the properties analogous to
\eqref{h_n-inductive-properties}, with $e_0:=1, e_1:=\rho$.



\vskip.1in
\noindent
{\sf Assertion (c).}
Iterating the property from (b) that $\rho^\perp h_n = h_{n-1}$ shows
that $(\rho^k)^\perp h_n = h_{n-k}$ for $0 \leq k \leq n$.  However one
also has an expansion
\[
\rho^k = c h_k + \sum\limits_{\substack{\sigma \in \Sigma \cap A_k:\\\sigma \neq h_k}} c_ \sigma \sigma
\]
for some integers $c, c_\sigma > 0$, since every $\sigma$ in $\Sigma \cap A_k$
is a constituent of $\rho^k$.  Hence
\[
1 = (h_{n-k}, h_{n-k}) =
((\rho^k)^\perp h_n,(\rho^k)^\perp h_n)
\geq c^2 (h_k^\perp h_n, h_k^\perp h_n)
\]
using Proposition~\ref{skewing-properties-prop}(ii).
Hence if we knew that $h_k^\perp h_n \neq 0$ this would force
\[
h_k^\perp h_n = (\rho^k)^\perp h_n = h_{n-k}
\]
as well as $\sigma^\perp h_n=0$ for all $\sigma \not\in \{h_0,h_1,\ldots,h_n\}$.  But
\[
(\rho^{n-k})^\perp h_k^\perp h_n =
h_k^\perp (\rho^{n-k})^\perp h_n =
h_k^\perp h_k = 1 \neq 0
\]
so $h_k^\perp h_n \neq 0$, as desired.
The argument for $e_k^\perp e_n = e_{n-k}$ is symmetric.

The last assertion in (c) follows if one checks that
$e_n \neq h_n$ for each $n \geq 2$, but this holds since
$e_2^\perp(h_n)=0$ but $e_2^\perp(e_n) = e_{n-2}$.

\vskip.1in
\noindent
{\sf Assertion (d).}
Part (c) implies that
\[
(\Delta h_n, \sigma \otimes \tau)_{A \otimes A}
=(h_n , \sigma \tau)_A
=(\sigma^\perp h_n, \tau)_A = 0
\]
unless $\sigma = h_k$ for some $k=0,1,2,\ldots,n$
and $\tau = h_{n-k}$.  Also one can compute
\[
(\Delta h_n, h_k \otimes h_{n-k})
= (h_n, h_k h_{n-k})
= (h_k^{\perp} h_n, h_{n-k})
\overset{(c)}{=}(h_{n-k}, h_{n-k}) =1.
\]
This is equivalent to the assertion
for $\Delta h_n$ in (d).  The argument for $\Delta e_n$ is symmetric.
\end{proof}


Before proving Theorem~\ref{uniqueness-of-sym-parts-e-f-g}, we note
some consequences of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}.
Define for each partition
$\lambda=(\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_\ell)$
the following two elements of $A$:
\begin{align*}
h_\lambda & = h_{\lambda_1} h_{\lambda_2} \cdots h_{\lambda_\ell} = h_{\lambda_1} h_{\lambda_2} \cdots, \\
e_\lambda & = e_{\lambda_1} e_{\lambda_2} \cdots e_{\lambda_\ell} = e_{\lambda_1} e_{\lambda_2} \cdots.
\end{align*}
Also, define the
\emph{lexicographic order}\index{lexicographic order on partitions}
on $\Par_n$
by saying $\lambda <_{\lex} \mu$ if $\lambda \neq \mu$ and
the smallest index $i$ for which $\lambda_i \neq \mu_i$ has
$\lambda_i < \mu_i$.  Recall also that $\lambda^t$ denotes the
\emph{conjugate} or \emph{transpose} partition to $\lambda$, obtained by
swapping rows and columns in the Ferrers diagram.

The following unitriangularity lemma will play a role in the proof
of Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(e).
\begin{lemma}
\label{triangularity-lemma}
Under the hypotheses of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d},
for $\lambda, \mu$ in $\Par_n$, one has
\begin{equation}
e_\mu^\perp h_\lambda =
\begin{cases}
1, & \text{ if }\mu=\lambda^t ; \\
0, & \text{ if }\mu >_{\lex} \lambda^t.
\end{cases}
\label{eq.triangularity-lemma.perp}
\end{equation}
Consequently
\begin{equation}
\det\left[ (e_{\mu^t}, h_\lambda) \right]_{\lambda,\mu \in \Par_n}= 1.
\label{eq.triangularity-lemma.det=1}
\end{equation}
\end{lemma}

% [DG][v64] Replaced "=\pm 1" by "= 1" in the last equation, because
% this is a honest square matrix (with rows and columns both indexed
% by the same set) and is unitriangular, so how could its det be -1?

\begin{proof}
Notice that $A$ is of finite type (as shown in the
proof of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}). Thus, $A^o$ is
a graded Hopf algebra isomorphic to $A$.

Also, notice that any $m \in \NN$ and any
$a_1, a_2, \ldots, a_\ell \in A$ satisfy
\begin{equation}
e_m^\perp \left(a_1 a_2 \cdots a_\ell\right)
= \sum_{i_1+\cdots+i_\ell = m} e_{i_1}^\perp\left(a_1\right) \cdots e_{i_\ell}^\perp\left(a_\ell\right) .
\label{pf.triangularity-lemma.1}
\end{equation}
Indeed, this follows by induction over $\ell$ using
Proposition~\ref{skewing-properties-prop}(iv)
(and the coproduct formula for $\Delta(e_n)$ in
Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(d)).

In order to prove \eqref{eq.triangularity-lemma.perp},
induct on the length of $\mu$.  If $\lambda$ has length $\ell$, so that $\lambda_1^t=\ell$, then
\begin{align*}
e_\mu^\perp h_\lambda
&= e_{(\mu_2,\mu_3,\ldots)}^\perp
\left(
e_{\mu_1}^\perp (h_{\lambda_1} \cdots h_{\lambda_\ell})
\right)
\qquad \left(\text{since $e_\mu = e_{\mu_1} e_{(\mu_2,\mu_3,\ldots)}$
and thus $e_\mu^\perp = e_{(\mu_2,\mu_3,\ldots)}^\perp \circ e_{\mu_1}^\perp$}\right)
\\
&=e_{(\mu_2,\mu_3,\ldots)}^\perp
\sum_{i_1+\cdots+i_\ell=\mu_1}
e_{i_1}^\perp(h_{\lambda_1}) \cdots
e_{i_\ell}^\perp(h_{\lambda_\ell})
\qquad \left(\text{by \eqref{pf.triangularity-lemma.1}}\right) \\
&=e_{(\mu_2,\mu_3,\ldots)}^\perp
\sum\limits_{\substack{i_1+\cdots+i_\ell=\mu_1; \\ \text{each of } i_1, \ldots, i_\ell \text{ is } \leq 1}}
e_{i_1}^\perp(h_{\lambda_1}) \cdots
e_{i_\ell}^\perp(h_{\lambda_\ell})
\qquad \left(\text{since } e_k^\perp h_n = 0 \text{ for } k \geq 2 \right) \\
&=\begin{cases}
0, & \text{ if }\mu_1 > \ell=\lambda_1^t ;\\
e_{(\mu_2,\mu_3,\ldots)}^\perp  h_{(\lambda_1-1, \ldots,\lambda_\ell-1)}, & \text{ if }\mu_1 = \ell=\lambda_1^t
\end{cases}
\end{align*}
where the last equality used
\[
e_k^\perp(h_n)
=
\begin{cases}
h_{n-1}, & \text{ if }k=1;\\
0, &\text{ if }k \geq 2 .
\end{cases}
\]
Now apply the induction hypothesis, since
$(\lambda_1-1, \ldots,\lambda_\ell-1)^t = (\lambda_2^t,\lambda_3^t,\ldots)$.

To prove \eqref{eq.triangularity-lemma.det=1}, note that
any $\lambda,\mu$ in $\Par_n$ satisfy
$(e_{\mu^t}, h_\lambda)=(e_{\mu^t}^\perp(h_\lambda), 1)=e_{\mu^t}^\perp(h_\lambda)$
(since degree considerations enforce
$e_{\mu^t}^\perp(h_\lambda) \in A_0 = \kk \cdot 1$),
and thus
\[
(e_{\mu^t}, h_\lambda)
= e_{\mu^t}^\perp(h_\lambda)
= \begin{cases}
1, & \text{ if }\mu^t = \lambda^t ; \\
0, & \text{ if }\mu^t >_{\lex} \lambda^t
\end{cases}
\]
(by \eqref{eq.triangularity-lemma.perp}).
This means that the matrix
$\left[ (e_{\mu^t}, h_\lambda) \right]_{\lambda,\mu \in \Par_n}$
is unitriangular with respect to some total order on $\Par_n$
(namely, the lexicographic order on the conjugate partitions),
and hence has determinant $1$.
\end{proof}

% [DG][v64] Made the above proof more detailed.

The following proposition will be the crux of the
proof of Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(f) and (g),
and turns out to be closely related to Kerov's \emph{asymptotic theory of
characters of the symmetric groups} \cite{Kerov}.

\begin{proposition}
\label{normalized-character-prop}
Given a PSH $A$ with PSH-basis $\Sigma$ containing only one primitive $\rho$,
the two maps $A \rightarrow \ZZ$
defined on $A=\bigoplus_{n \geq 0} A_n$ via
\begin{align*}
\delta_h&=\bigoplus_n h_n^\perp,\\
\delta_e&=\bigoplus_n e_n^\perp
\end{align*}
are characterized as the only two
$\ZZ$-linear maps $A \overset{\delta}{\rightarrow} \ZZ$ with the three
properties of being
\begin{itemize}
\item \textbf{positive}: $\delta(\NN \Sigma) \subset \NN$,
\item \textbf{multiplicative}: $\delta(a_1 a_2) = \delta(a_1) \delta(a_2)$ for all $a_1, a_2 \in A$, and
\item \textbf{normalized}: $\delta(\rho)=1$.
\end{itemize}
\end{proposition}
\begin{proof}
Notice that $A$ is of finite type (as shown in the
proof of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}). Thus, $A^o$ is
a graded Hopf algebra isomorphic to $A$.

It should be clear from their definitions that $\delta_h, \delta_e$ are $\ZZ$-linear,
positive and normalized.  To see that $\delta_h$ is multiplicative, by $\ZZ$-linearity,
it suffices to check that for $a_1,a_2$ in $A_{n_1}, A_{n_2}$ with $n_1+n_2=n$, one has
\[
\delta_h(a_1 a_2) = h_n^\perp(a_1 a_2)
= \sum_{i_1 + i_2=n} h_{i_1}^\perp(a_1) h_{i_2}^\perp(a_2)
= h_{n_1}^\perp(a_1) h_{n_2}^\perp(a_2) = \delta_h(a_1) \delta_h(a_2)
\]
in which the second equality used Proposition~\ref{skewing-properties-prop}(iv)
and Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(d).  The argument for $\delta_e$
is symmetric.

Conversely, given $A \overset{\delta}{\rightarrow} \ZZ$ which is $\ZZ$-linear,
positive, multiplicative, and normalized, note that
\[
\delta(h_2)+\delta(e_2) = \delta(h_2+e_2)
= \delta(\rho^2)
= \delta(\rho)^2
= 1^2 = 1
\]
and hence positivity implies that either $\delta(h_2)=0$ or $\delta(e_2)=0$.
Assume the latter holds, and we will show that $\delta=\delta_h$.

Given any $\sigma$ in $\Sigma \cap A_n \setminus \{h_n\}$, note that
$e_2^\perp \sigma \neq 0$ by Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(b), and hence
$
0 \neq (e_2^\perp \sigma, \rho^{n-2}) =( \sigma, e_2 \rho^{n-2}).
$
Thus $\sigma$ is a constituent of $e_2 \rho^{n-2}$, so positivity implies
\[
0 \leq \delta(\sigma) \leq \delta(e_2 \rho^{n-2}) = \delta(e_2) \delta(\rho^{n-2}) = 0.
\]
Thus $\delta(\sigma)=0$ for $\sigma$ in $\Sigma \cap A_n \setminus \{h_n\}$.  Since
$\delta(\rho^n)=\delta(\rho)^n =1^n=1$, this forces $\delta(h_n)=1$, for each $n \geq 0$
(including $n=0$, as
$1=\delta(\rho)=\delta(\rho \cdot 1)=\delta(\rho)\delta(1)=1 \cdot \delta(1)=\delta(1)$).
Thus $\delta=\delta_h$.  The argument when $\delta(h_2)=0$ showing $\delta=\delta_e$ is
symmetric.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{uniqueness-of-sym-parts-e-f-g}]
Many of the assertions of parts (e) and (f) will come from constructing
the unique nontrivial PSH-automorphism $\omega$ of $A$ from the antipode $S$:
for homogeneous $a$ in $A_n$, define $\omega(a):=(-1)^n S(a)$.  We now study some
of the properties of $S$ and $\omega$.

Notice that $A$ is of finite type (as shown in the
proof of Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}). Thus, $A^o$ is
a graded Hopf algebra isomorphic to $A$.

Since $A$ is a PSH, it is commutative by Theorem~\ref{general-structure-theorem}
(applied to $A \otimes_\ZZ \QQ$).  This implies both that
$S$ is an algebra endomorphism by
Proposition~\ref{antipodes-are-antiendomorphisms} (since
Exercise~\ref{exe.comm-cocomm.anti}(a) shows that the
algebra anti-endomorphisms of a commutative algebra are
the same as its algebra endomorphisms), and that
$S^2=\id_A$ by Corollary~\ref{commutative-implies-involutive-antipode-cor}.
Thus, $\omega$ is an algebra endomorphism and satisfies
$\omega^2 = \id_A$.

Since $A$ is self-dual and the defining diagram \eqref{antipode-diagram}
satisfied by the antipode $S$ is sent to itself when one replaces $A$ by $A^o$
and all maps by their adjoints, one concludes that $S=S^*$ (where $S^*$
means the restricted adjoint $S^* : A^o \to A^o$), i.e., $S$ is self-adjoint.
Since $S$ is an algebra endomorphism, and $S=S^*$, in fact $S$ is also a
coalgebra endomorphism, a bialgebra endomorphism, and a Hopf endomorphism
(by Corollary~\ref{cor.bialg-mor-is-Hopf}).  The same properties
are shared by $\omega$.

Since $\id_A = S^2 = S S^*$, one concludes that $S$ is an isometry, and hence so
is $\omega$.

Since $\rho$ is primitive, one has $S(\rho)=-\rho$ and $\omega(\rho)=\rho$.
Therefore $\omega(\rho^n)=\rho^n$ for $n=1,2,\ldots$.  Use this as follows
to check that $\omega$ is a PSH-automorphism, which amounts to checking that
every $\sigma$ in $\Sigma$ has $\omega(\sigma)$ in $\Sigma$:
\[
( \omega(\sigma),\omega(\sigma) )= (\sigma,\sigma)=1
\]
so that $\pm \omega(\sigma)$ lies in $\Sigma$, but also if $\sigma$ lies in $A_n$, then
\[
( \omega(\sigma), \rho^n )= (\sigma , \omega(\rho^n) )= (\sigma , \rho^n )>0.
\]
In summary, $\omega$ is a PSH-automorphism of $A$, an isometry, and an involution.

Let us try to determine the action of $\omega$ on the $\{h_n\}$.
By similar reasoning as in \eqref{coproduct-of-e2}, one has
\[
\Delta(h_2) = h_2 \otimes \one + \rho \otimes \rho + \one \otimes h_2.
\]
Thus $0 = S(h_2) + S(\rho)\rho + h_2$, and combining this with $S(\rho)=-\rho$,
one has $S(h_2) = e_2$.  Thus also $\omega(h_2)=(-1)^2 S(h_2)=e_2$.

We claim that this forces $\omega(h_n)=e_n$, because $h_2^\perp \omega(h_n)= 0$ via
the following calculation:  for any $a$ in $A$ one has
\begin{align*}
(h_2^\perp \omega(h_n), a)
&=  (\omega(h_n), h_2 a) \\
&=  ( h_n, \omega(h_2 a) ) \\
&=  ( h_n, e_2 \omega(a) ) \\
&=  ( e_2^\perp h_n, \omega(a) ) =  ( 0 , \omega(a) )  =0.
\end{align*}

Consequently the involution $\omega$ swaps $h_n$ and $e_n$, while the
antipode $S$ has $S(h_n)=(-1)^ne_n$ and $S(e_n)=(-1)^n h_n$.  Thus
the coproduct formulas in (d) and definition of the  antipode $S$ imply
the relation \eqref{e-h-relation} between $\{h_n\}$ and $\{e_n\}$.

This relation  \eqref{e-h-relation} also lets one recursively express the
$h_n$ as polynomials with integer coefficients in the $\{e_n\}$,
and vice-versa, so that $\{h_n\}$ and $\{e_n\}$  each generate the
same $\ZZ$-subalgebra $A'$ of $A$.  We wish to show that $A'$ exhausts $A$.

We argue that Lemma~\ref{triangularity-lemma} implies that the \emph{Gram matrix}
$\left[ (h_{\mu},h_{\lambda}) \right]_{\mu,\lambda \in \Par_n}$ has determinant $\pm 1$ as
follows.  Since $\{h_n\}$ and $\{e_n\}$  both generate $A'$,
there exists a $\ZZ$-matrix $(a_{\mu,\lambda})$ expressing
$e_{\mu^t} = \sum_{\lambda} a_{\mu,\lambda} h_\lambda$, and one has
\[
 \left[ (e_{\mu^t}, h_{\lambda}) \right]
= \left[ a_{\mu,\lambda} \right] \cdot
\left[ (h_{\mu},h_{\lambda}) \right].
\]
Taking determinants of these three $\ZZ$-matrices, and
using the fact that the determinant on the left is $1$
(by \eqref{eq.triangularity-lemma.det=1}),
both determinants on the right must also be $\pm 1$.

Now we will show that every $\sigma \in \Sigma \cap A_n$ lies in $A_n'$.
Uniquely express $\sigma=\sigma'+\sigma''$ in which $\sigma'$ lies
in the $\RR$-span $\RR A_n'$ and  $\sigma''$ lies in the real perpendicular
space  $(\RR A_n')^\perp$ inside $\RR \otimes_\ZZ A_n$.
One can compute $\RR$-coefficients $(c_\mu)_{\mu \in \Par_n}$
that express $\sigma' = \sum_\mu c_\mu h_\mu$ by solving the system
\[
\left( \sum_\mu c_\mu h_\mu, h_\lambda \right)
 = (\sigma, h_\lambda) \text{ for }\lambda \in \Par_n.
\]
This linear system is governed by the Gram matrix
$\left[ (h_{\mu},h_{\lambda}) \right]_{\mu,\lambda \in \Par_n}$
with determinant $\pm 1$, and its right side has $\ZZ$-entries since $\sigma, h_\lambda$
lie in $A$.  Hence the solution $(c_\mu)_{\mu \in \Par_n}$ will have $\ZZ$-entries,
so $\sigma'$ lies in $A'$.  Furthermore, $\sigma''= \sigma - \sigma'$ will lie in $A$,
and hence by the orthogonality of $\sigma',\sigma''$,
\[
1=(\sigma,\sigma) = (\sigma',\sigma')+(\sigma'',\sigma'').
\]
One concludes that either $\sigma''=0$, or $\sigma'=0$.  The latter cannot
occur since it would mean that $\sigma=\sigma''$ is perpendicular to all of $A'$.
But $\rho^n=h_1^n$ lies in $A'$, and $(\sigma,\rho^n) \neq 0$.  Thus $\sigma''=0$,
meaning $\sigma=\sigma'$ lies in $A'$.  This completes the proof of assertion (e).
Note that in the process, having shown
$\det(h_{\mu},h_{\lambda})_{\lambda,\mu \in \Par_n}=\pm 1$, one also knows
that $\{h_\lambda\}_{\lambda \in \Par_n}$ are $\ZZ$-linearly
independent, so that $\{h_1,h_2,\ldots\}$ are algebraically independent%
\footnote{by Exercise~\ref{exe.alggen-via-span}(c)},
and $A=\ZZ[h_1,h_2,\ldots]$ is the polynomial algebra generated by
$\{h_1,h_2,\ldots\}$.

For assertion (f), we have seen that $\omega$ gives such a
PSH-automorphism $A \rightarrow A$, swapping
$h_n \leftrightarrow e_n$.  Conversely, given a
PSH-automorphism $A \overset{\varphi}{\rightarrow} A$, consider the
positive, multiplicative, normalized $\ZZ$-linear map
$\delta:=\delta_h \circ \varphi: A \rightarrow \ZZ$.
Proposition~\ref{normalized-character-prop} shows that either
\begin{itemize}
\item $\delta=\delta_h$, which then forces $\varphi(h_n)=h_n$ for all $n$,
so $\varphi=\id_A$, or
\item
$\delta=\delta_e$,
which then forces $\varphi(e_n)=h_n$ for all $n$,
so $\varphi=\omega$.
\end{itemize}

For assertion (g), given a PSH $A$ with PSH-basis $\Sigma$ having
exactly one primitive $\rho$, since we have seen $A=\ZZ[h_1,h_2,\ldots]$,
where $h_n$ in $A$ is as defined in
Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d},
one can uniquely define an algebra morphism
$A \overset{\varphi}{\rightarrow} \Lambda$ that sends the element
$h_n$ to the
complete homogeneous symmetric function $h_n(\xx)$.  Assertions (d) and (e)
show that $\varphi$ is a bialgebra isomorphism, and hence it is a Hopf
isomorphism.  To show that it is a PSH-isomorphism, we first note that
it is an isometry because one can iterate
Proposition~\ref{skewing-properties-prop}(iv) together with assertions (c) and
(d) to compute all inner products
\[
( h_\mu, h_\lambda )_A
=( \one , h_\mu^\perp h_\lambda )_A
=(1, h_{\mu_1}^\perp h_{\mu_2}^\perp\cdots \left( h_{\lambda_1} h_{\lambda_2} \cdots \right) )_A
\]
for $\mu, \lambda$ in $\Par_n$.  Hence
\[
( h_\mu, h_\lambda )_A
= ( h_\mu(\xx), h_\lambda(\xx) )_\Lambda
= ( \varphi(h_\mu), \varphi(h_\lambda) )_\Lambda .
\]
Once one knows $\varphi$ is an isometry, then elements
$\omega$ in $\Sigma \cap A_n$
are characterized in terms of the form $(\cdot,\cdot)$
by $(\omega,\omega)=1$ and $(\omega, \rho^n) > 0$.
Hence $\varphi$ sends each $\sigma$ in $\Sigma$ to a
Schur function $s_\lambda$, and is a PSH-isomorphism.
\end{proof}




\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Complex representations
for $\Symm_n$, wreath products, $GL_n(\FF_q)$}
\label{representations-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

After reviewing the basics that we will need from representation and character theory of finite groups, we give Zelevinsky's three main examples of PSH's arising as spaces of virtual characters for three towers of finite groups:
\begin{enumerate}
\item[$\bullet$]
\emph{symmetric} groups,
\item[$\bullet$]
their \emph{wreath products} with any finite group, and
\item[$\bullet$]
the finite \emph{general linear} groups.
\end{enumerate}

Much in this chapter traces its roots to Zelevinsky's book \cite{Zelevinsky}. The results concerning the symmetric groups, however, are significantly older and spread across the literature: see, e.g., \cite[\S 7.18]{Stanley}, \cite[\S 7.3]{Fulton}, \cite[\S I.7]{Macdonald}, \cite[\S 4.7]{Sagan}, \cite{Knutson}, for proofs using different tools.

% [DG][v14] Added previous two sentences.

\subsection{Review of complex character theory}
\label{character-theory-review-section}

We shall now briefly discuss some basics of representation (and character) theory that will be used below. A good source for this material, including the crucial Mackey formula, is Serre \cite[Chaps. 1-7]{Serre}.\footnote{More advanced treatments of representation theory can be found in \cite{Webb} and \cite{Etingof-et-al}.}

% [DG][v58] Added footnote.

\subsubsection{Basic definitions, Maschke, Schur}

For a group $G$, a \emph{representation of $G$}\index{representation of a group} is a homomorphism $G \overset{\varphi}{\rightarrow} GL(V)$ for some vector space $V$ over a field.  We will take the field to be $\CC$ from now on, and we will also assume that $V$ is finite-dimensional over $\CC$.  Thus a representation of $G$ is the same as a finite-dimensional (left) $\CC G$-module $V$. (We use the notations $\CC G$ and $\CC \left[G\right]$ synonymously for the group algebra of $G$ over $\CC$. More generally, if $S$ is a set, then $\CC S = \CC \left[S\right]$ denotes the free $\CC$-module with basis $S$.)

% [DG][v34] Added preceding two sentences.

We also assume that $G$ is finite, so that
Maschke's Theorem\footnote{... which has a beautiful generalization to finite-dimensional Hopf algebras due to Larson and Sweedler; see Montgomery \cite[\S 2.2]{Montgomery}.} says that $\CC G$ is semisimple, meaning that every $\CC G$-module $U \subset V$ has a $\CC G$-module complement $U'$ with $V = U \oplus U'$.  Equivalently, \emph{indecomposable}\index{indecomposable module} $\CC G$-modules are the same thing as \emph{simple}\index{simple module} (=\emph{irreducible}\index{irreducible module}) $\CC G$-modules.

Schur's Lemma implies that for two simple $\CC G$-modules $V_1,V_2$, one has
\[
\Hom_{\CC G}(V_1,V_2) \cong
\begin{cases}
\CC , & \text{ if }V_1 \cong V_2;\\
0, & \text{ if } V_1 \not\cong V_2.
\end{cases}
\]


\subsubsection{Characters and Hom spaces}

A $\CC G$-module $V$ is completely determined up to isomorphism by its
\emph{character}\index{character of a module}
\[
\begin{array}{rcl}
G & \overset{\chi_V}{\longrightarrow} &\CC , \\
g & \longmapsto &\chi_V(g):=\trace(g: V \rightarrow V).
\end{array}
\]
This character $\chi_V$ is a \dfn{class function}, meaning it is
constant on $G$-conjugacy classes.  The space $R_\CC(G)$ of
class functions $G \rightarrow \CC$ has a Hermitian, positive definite
form
\[
(f_1,f_2)_G:=\frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}.
\]
For any two $\CC G$-modules $V_1, V_2$,
\begin{equation}
\label{character-scalprod}
( \chi_{V_1}, \chi_{V_2} )_G=
\dim_\CC \Hom_{\CC G}(V_1,V_2).
\end{equation}
The set of all \emph{irreducible characters}\index{$\Irr(G)$}\index{irreducible character}
\[
\Irr(G)=\{ \chi_V : V \text{ is a simple }\CC G\text{-module}\}
\]
forms an orthonormal basis of $R_\CC(G)$
with respect to this form, and spans a $\ZZ$-sublattice
\[
R(G):=\ZZ \Irr(G) \subset R_\CC(G)
\]
sometimes called the \emph{virtual characters}\index{$R(G)$}\index{virtual character}
of $G$.
For every $\CC G$-module $V$, the character $\chi_V$
belongs to $R(G)$.

% [DG][v25] Added the previous sentence.

Instead of working with the Hermitian form $\left(\cdot, \cdot\right)_G$
on $G$, we could also (and some authors do) define a $\CC$-bilinear
form $\left<\cdot,\cdot\right>_G$ on $R_\CC (G)$ by
\[
\left<f_1,f_2\right>_G := \frac{1}{|G|} \sum_{g \in G} f_1(g) f_2(g^{-1}).
\]
This form is not identical with $\left(\cdot, \cdot\right)_G$ (indeed,
$\left<\cdot,\cdot\right>_G$ is bilinear while $\left(\cdot,\cdot\right)_G$
is Hermitian), but it still satisfies \eqref{character-scalprod}, and
thus is identical with $\left(\cdot, \cdot\right)_G$ on $R(G) \times R(G)$.
Hence, for all we are going to do until
Section~\ref{Hall-algebra-section}, we could just as well use the form
$\left<\cdot,\cdot\right>_G$ instead of $\left(\cdot,\cdot\right)_G$.

% [DG] I added the previous paragraph for ideological reasons; feel
% free to take it out.

% [DG][v14] Added clarification that the bilinear-vs.-Hermitian
% question becomes important in the Hall algebra subsection. I am not
% fully sure what exactly is a better choice for that section.

\subsubsection{Tensor products}

Given two groups $G_1,G_2$ and $\CC G_i$-modules $V_i$ for $i=1,2$,
their tensor product $V_1 \otimes_\CC V_2$ becomes a $\CC[G_1 \times G_2]$-module
via $(g_1,g_2)(v_1 \otimes v_2)=g_1(v_1) \otimes g_2(v_2)$.  This module
is called the
\emph{(outer) tensor product}\index{outer tensor product}\index{tensor product of representations}
of $V_1$ and $V_2$.
When $V_1, V_2$ are
both simple, then so is $V_1 \otimes V_2$, and every
simple $\CC[G_1 \times G_2]$-module arises this way
(with $V_1$ and $V_2$ determined uniquely up to
isomorphism).\footnote{This is proven in
\cite[\S 3.2, Thm. 10]{Serre}. The fact that $\CC$ is
algebraically closed is essential for this!}
Thus one has identifications and isomorphisms
\begin{align*}
\Irr(G_1 \times G_2) &= \Irr(G_1) \times \Irr(G_2) ,\\
R(G_1 \times G_2) &\cong R(G_1) \otimes_{\ZZ} R(G_2) ;
\end{align*}
here, $\chi_{V_1}\otimes\chi_{V_2}\in R(G_1)\otimes_{\ZZ} R(G_2)$ is being
identified with $\chi_{V_1\otimes V_2}\in R(G_1\times G_2)$ for all
$\CC G_1$-modules $V_1$ and all $\CC G_2$-modules $V_2$.
The latter isomorphism is actually a restriction of the
isomorphism
$R_\CC(G_1 \times G_2) \cong R_\CC(G_1) \otimes_\CC R_\CC(G_2)$
under which every pure tensor
$\phi_1 \otimes \phi_2 \in R_\CC(G_1) \otimes_\CC R_\CC(G_2)$
corresponds to the class function
$G_1 \times G_2 \to \CC, \ \left(g_1, g_2\right) \mapsto
\phi_1\left(g_1\right) \otimes \phi_2\left(g_2\right)$.

% [DG][v45] Added the preceding sentence.

Given two $\CC G_1$-modules $V_1$ and $W_1$ and two $\CC G_2$-modules
$V_2$ and $W_2$, we have
\begin{equation}
\label{character-tensorprod}
\left(\chi_{V_1 \otimes V_2}, \chi_{W_1 \otimes W_2}\right)_{G_1 \times G_2}
= \left(\chi_{V_1}, \chi_{W_1}\right)_{G_1} \left(\chi_{V_2}, \chi_{W_2}\right)_{G_2}.
\end{equation}

% [DG] I added the previous statement for later use. (It is a corollary of
% Proposition 4.0a (b) in reiner-errata12.pdf, but of course it is much
% easier. Maybe Proposition 4.0a would make a nice exercise?)

\subsubsection{Induction and restriction}

Given a subgroup $H < G$ and $\CC H$-module $U$, one can use the fact that $\CC G$ is a $(\CC G, \CC H)$-bimodule to form the \dfn{induced $\CC G$-module}\index{$\Ind_H^G U$}
\[
\Ind_H^G U : = \CC G \otimes_{\CC H} U.
\]
The fact that $\CC G$ is free as a (right-)$\CC H$-module\footnote{... which also has a beautiful generalization to finite-dimensional Hopf algebras due to Nichols and Zoeller; see \cite[\S 3.1]{Montgomery}.}
on basis elements $\{t_{g}\}_{gH \in G/H}$
makes this tensor product easy to analyze.  For example one can
compute its character
\begin{equation}
\label{induced-character-formula}
\chi_{\Ind_H^G U}(g) %= \Ind_H^G(\chi_U)(g)
 =\frac{1}{|H|} \sum\limits_{\substack{k \in G:\\ k g k^{-1} \in H}} \chi_U(k g k^{-1}).
\end{equation}
\footnote{See \cite[\S 7.2, Prop. 20(ii)]{Serre} for the
proof of this equality. (Another proof is given in
\cite[Remark 5.9.2 (the Remark after Theorem 4.32 in the arXiv version)]{Etingof-et-al},
but \cite{Etingof-et-al} uses a different definition of
$\Ind_H^G U$; see Remark~\ref{rmk.induction.ind-as-hom}
for why it is equivalent to ours. Yet another proof of
\eqref{induced-character-formula} is given in
Exercise~\ref{exe.Indrho}(k).)}
One can also recognize when a $\CC G$-module
$V$ is isomorphic to $\Ind_H^G U$ for some $\CC H$-module $U$:  this happens
if and only if there is an $H$-stable subspace $U \subset V$ having the
property that $V = \bigoplus_{gH \in G/H} gU$.

The above construction of a $\CC G$-module $\Ind_H^G U$
corresponding to any $\CC H$-module $U$ is part of a
functor $\Ind_H^G$ from the category of $\CC H$-modules to the
category of $\CC G$-modules\footnote{On morphisms, it sends
any $f : U \to U'$ to $\id_{\CC G} \otimes_{\CC H} f :
\CC G \otimes_{\CC H} U \to \CC G \otimes_{\CC H} U'$.};
this functor is called \emph{induction}\index{induction of a $\CC H$-module}.

Besides induction on $\CC H$-modules, one can define induction
on class functions of $H$:

\begin{exercise}
\label{exe.induction.characters}
Let $G$ be a finite group, and $H$ a subgroup of $G$. Let $f
\in R_\CC \left(H\right)$ be a class function. We define the
\emph{induction}\index{induction of a class function}
$\Ind_H^G f$ of $f$ to be the
function $G \to \CC$ given by
\begin{equation}
\left(\Ind_H^G f\right) \left(g\right)
= \frac{1}{\left|H\right|} \sum\limits_{\substack{k \in G:\\ k g k^{-1} \in H}}
  f \left(k g k^{-1}\right)
\qquad \text{for all } g \in G.
\label{eq.exe.induction.characters.eq}
\end{equation}
%(This definition extends the second equality sign in
%\eqref{induced-character-formula} to arbitrary class functions.)

\begin{itemize}
\item[(a)]
Prove that this induction $\Ind_H^G f$ is a class
function on $G$, hence belongs to $R_\CC \left(G\right)$.

\item[(b)]
Let $J$ be a system of right coset\footnote{A \dfn{right coset}
of a subgroup $H$ in a group $G$ is defined to be a subset of $G$
having the form $Hj$ for some $j \in G$. Similarly, a
\dfn{left coset} has the form $jH$ for some $j \in G$.}
representatives for
$H \backslash G$, so that $G = \bigsqcup_{j \in J} H j$. Prove that
\[
\left(\Ind_H^G f\right)\left(g\right) =
\sum\limits_{\substack{j\in J:\\jgj^{-1}\in H}} f\left(jgj^{-1}\right)
\qquad \text{for all } g \in G.
\]

\end{itemize}
\end{exercise}

% [DG][v14] Added the exercise above.

The induction $\Ind_H^G$ defined in
Exercise~\ref{exe.induction.characters} is a $\CC$-linear map
$R_\CC \left(H\right) \to R_\CC \left(G\right)$. Since every
$\CC H$-module $U$ satisfies
\begin{equation}
\chi_{\Ind_H^G U} = \Ind_H^G(\chi_U)
\label{eq.induction.mods-and-class-functs}
\end{equation}
\footnote{This follows by comparing the value of
$\chi_{\Ind_H^G U}(g)$ obtained from
\eqref{induced-character-formula} with the value of
$\left(\Ind_H^G(\chi_U)\right)(g)$ found using
\eqref{eq.exe.induction.characters.eq}.}, this $\CC$-linear map
$\Ind_H^G$ restricts to a $\ZZ$-linear map $R\left(H\right) \to
R\left(G\right)$ (also denoted $\Ind_H^G$) which sends the
character $\chi_U$ of any
$\CC H$-module $U$ to the character $\chi_{\Ind_H^G U}$ of the
induced $\CC G$-module $\Ind_H^G U$.

% [DG][v46] Separated the definitions of induced modules and
% induced characters, and explicitly stated the relation between
% these two notions. I think identifying characters with
% modules can be rather confusing.
% Similar changes have been done below for restriction,
% inflation and $K$-fixed space construction. (Sometimes this
% boiled down to defining them for characters.)

\begin{exercise}
\label{exe.induction.transitivity}
Let $G$, $H$ and $I$ be three finite groups such that
$I < H < G$. Let $U$ be a $\CC I$-module. Prove that
$\Ind^G_H \Ind^H_I U \cong \Ind^G_I U$. (This fact is
often referred to as the \dfn{transitivity of induction}.)
\end{exercise}

% [DG][v27] Added above exercise.

\begin{exercise}
\label{exe.ind.tensor}
Let $G_1$ and $G_2$ be two groups. Let $H_1 < G_1$ and $H_2 < G_2$ be
two subgroups. Let $U_1$ be a $\CC H_1$-module, and $U_2$ be a
$\CC H_2$-module. Show that
\begin{equation}
\label{induction-vs-tensor-product}
\Ind_{H_1\times H_2}^{G_1\times G_2} \left( U_1 \otimes U_2 \right)
\cong \left(\Ind_{H_1}^{G_1} U_1\right) \otimes \left(\Ind_{H_2}^{G_2} U_2\right)
\end{equation}
as $\CC \left[ G_1 \times G_2 \right]$-modules.
\end{exercise}

% [DG][v3] This is used in the proof of the bialgebra identity.

% [DG][v28] Made this into an exercise.

The \emph{restriction}\index{restriction of a $\CC G$-module}\index{$\Res^G_H V$}
operation $V \mapsto \Res^G_H V$ restricts a
$\CC G$-module $V$ to a $\CC H$-module.
\dfn{Frobenius reciprocity} asserts the adjointness between
$\Ind_H^G$ and $\Res^G_H$
\begin{equation}
\label{Frobenius-reciprocity}
\Hom_{\CC G}( \Ind_H^G U, V ) \cong
\Hom_{\CC H}( U, \Res^G_H V ),
\end{equation}
as a special case ($S=A=\CC G, R=\CC H, B=U,C=V$)
of the general \dfn{adjoint associativity}
\begin{equation}
\label{Hom-tensor-adj}
\Hom_S(A \otimes_R B,C) \cong \Hom_R (B,\Hom_S(A,C))
\end{equation}
for $S,R$ two rings,
$A$ an $(S,R)$-bimodule,
$B$ a left $R$-module,
$C$ a left $S$-module.

We can define not just the restriction of a $\CC G$-module,
but also the \dfn{restriction of a class function} $f \in
R_\CC (G)$. When $H$ is a subgroup of $G$, the
restriction \dfn{$\Res^G_H f$} of an $f \in R_{\CC} (G)$ is defined
as the result of restricting the map $f : G \to \CC$ to $H$.
This $\Res^G_H f$ is easily seen to belong to $R_\CC (H)$, and
so $\Res^G_H$ is a $\CC$-linear map $R_\CC (G) \to R_\CC (H)$.
This map restricts to a $\ZZ$-linear map $R(G) \to R(H)$,
since we have $\Res^G_H \chi_V = \chi_{\Res^G_H V}$
for any $\CC G$-module $V$. Taking characters in
\eqref{Frobenius-reciprocity} (and recalling
$\Res^G_H \chi_V = \chi_{\Res^G_H V}$ and
\eqref{eq.induction.mods-and-class-functs}), we obtain
\begin{equation}
\label{Frobenius-reciprocity-chars}
( \Ind_H^G \chi_U, \chi_V )_G
= ( \chi_U, \Res^G_H \chi_V )_H .
\end{equation}
By bilinearity, this yields the equality
\[
\left( \Ind_H^G \alpha , \beta \right)_G
= \left( \alpha , \Res_H^G \beta \right)_H
\]
for any class functions $\alpha \in R_\CC (H)$ and
$\beta \in R_\CC (G)$ (since $R(G)$ spans $R_\CC(G)$
as a $\CC$-vector space).

\begin{exercise}
\label{exe.induction.ind-as-hom}
Let $G$ be a finite group, and let $H < G$. Let $U$ be a
$\CC H$-module. If $A$ and $B$ are two algebras, $P$ is a
$\left(B,A\right)$-bimodule and $Q$ is a left $B$-module,
then $\Hom_B\left(P, Q\right)$ is a left $A$-module
(since $\CC G$ is a $\left(\CC H, \CC G\right)$-bimodule).
As a
consequence, $\Hom_{\CC H}\left(\CC G, U\right)$ is a
$\CC G$-module. Prove that this $\CC G$-module is
isomorphic to $\Ind^G_H U$.
\end{exercise}

\begin{remark}
\label{rmk.induction.ind-as-hom}
Some texts \emph{define} the induction $\Ind^G_H U$ of a
$\CC H$-module $U$
to be $\Hom_{\CC H}\left(\CC G, U\right)$ (rather than to be
$\CC G \otimes_{\CC H} U$, as we did).\footnote{Or they define
it as a set of morphisms of $H$-sets from $G$ to $U$ (this is
how \cite[Def. 5.8.1 (Def. 4.28 in the arXiv version)]{Etingof-et-al}
defines it); this is easily
seen to be equivalent to $\Hom_{\CC H}\left(\CC G, U\right)$.}
As Exercise~\ref{exe.induction.ind-as-hom} shows, this
definition is equivalent to ours as long as $G$ is finite
(but not otherwise).
\end{remark}

% [DG][v14] Added preceding exercise and remark. (This exercise also
% appears in the Frobenius reciprocity section of
% \cite{Etingof-et-al}.)

Exercise~\ref{exe.induction.ind-as-hom} yields the following
``wrong-way'' version of Frobenius reciprocity:

\begin{exercise}
\label{exe.induction.wrong-way-frob}
Let $G$ be a finite group; let $H < G$. Let $U$ be a $\CC G$-module,
and let $V$ be a $\CC H$-module. Prove that
$\Hom_{\CC G}\left(U, \Ind^G_H V\right)
\cong \Hom_{\CC H}\left(\Res^G_H U, V\right)$.
\end{exercise}

% [DG][v14] Added the above exercise.


\subsubsection{Mackey's formula}

Mackey gave an alternate description of a
module which has been induced and then restricted.  To state it, for a subgroup $H < G$ and $g$ in $G$,
let $H^g:=g^{-1}Hg$ and ${}^gH:=gHg^{-1}$.
Given a $\CC H$-module $U$, say defined by a homomorphism
$H \overset{\varphi}{\rightarrow} GL(U)$,
let $U^{g}$ denote the $\CC[gHg^{-1}]$-module
on the same $\CC$-vector space $U$ defined by
the composite homomorphism
\[
\begin{array}{rcccl}
{}^gH &\longrightarrow&H &\overset{\varphi}{\longrightarrow}&GL(U),\\
h &\longmapsto &g^{-1}hg .& &
\end{array}
\]

\begin{theorem}(Mackey's formula)
\label{Mackey-formula}
Consider subgroups $H, K < G$,  and any $\CC H$-module $U$.
If $\{g_1,\ldots,g_t\}$ are double coset representatives for
$K\backslash G /H$, then
\[
\Res^G_K \Ind_H^G U
\cong \bigoplus_{i=1}^t
\Ind^K_{{}^{g_i}H \cap K} \left( \left( \Res^H_{H \cap K^{g_i}} U \right)^{g_i} \right) .
\]
\end{theorem}
\begin{proof}
In this proof, all tensor product symbols $\otimes$
should be interpreted as $\otimes_{\CC H}$.
Recall $\CC G$ has $\CC$-basis $\{t_g\}_{g \in G}$.
For subsets $S \subset G$,
let $\CC[S]$ denote the $\CC$-span of $\{t_g\}_{g \in S}$ in $\CC G$.

Note that each double coset $K g H$ gives rise to a sub-$(K,H)$-bimodule
$\CC[K g H]$ within $\CC G$, and one has
a $\CC K$-module direct sum decomposition
\[
\Ind_H^G U = \CC G \otimes U
= \bigoplus_{i=1}^t \CC[K g_i H] \otimes U.
\]
Hence it suffices to check for any element $g$ in $G$ that
\[
\CC[K g H] \otimes U
\cong
\Ind^K_{{}^{g}H \cap K} \left( \left( \Res^H_{H \cap K^{g}} U \right)^{g} \right).
\]
Note that ${}^{g}H \cap K$ is the subgroup of $K$ consisting of
the elements $k$ in $K$ for which $kg H =gH$.  Hence by picking $\{k_1,\ldots,k_s\}$ to be coset representatives for
$K/({}^gH \cap K)$,
one disjointly decomposes the double coset
\[
KgH = \bigsqcup_{j=1}^s k_j({}^gH \cap K)gH,
\]
giving a $\CC$-vector space direct sum decomposition
\begin{align*}
\CC[K g H] \otimes U
&=\bigoplus_{j=1}^s \CC\left[k_j \left( {}^gH \cap K\right) g H\right]  \otimes U \\
&\cong
\Ind_{{}^gH \cap K}^K
\left(\CC [\left( {}^gH \cap K \right) g H]  \otimes U\right).
\end{align*}
So it remains to check that one has a $\CC [{}^gH \cap K]$-module isomorphism
\[
\CC [\left( {}^gH \cap K \right) g H]  \otimes U
\cong \left( \Res^H_{H \cap K^g} U \right)^g.
\]
Bearing in mind that, for each $k$ in ${}^g H \cap K$ and $h$ in $H$, one has
$g^{-1}kg$ in $H$ and hence
\[
t_{kgh} \otimes u
= t_g \cdot t_{g^{-1}kg \cdot h}  \otimes u
=t_g \otimes g^{-1}kgh \cdot u ,
\]
one sees that this isomorphism can be defined by mapping
\[
t_{kgh} \otimes u \longmapsto  g^{-1}kgh \cdot u.
\]
\end{proof}


\subsubsection{Inflation and fixed points}


There are two (adjoint) constructions on representations
that apply when one has a normal subgroup $K \triangleleft G$.
Given a $\CC[G/K]$-module $U$, say defined by the
homomorphism $G/K \overset{\varphi}{\rightarrow} GL(U)$,
the \emph{inflation}\index{inflation of a $\CC[G/K]$-module}
of $U$ to a
$\CC G$-module \dfn{$\Inf_{G/K}^G U$} has the same underlying space $U$,
and is defined by the composite
homomorphism $G \rightarrow G/K \overset{\varphi}{\rightarrow} GL(U)$.
We will later use the easily-checked
fact that when $H < G$ is any other subgroup, one has
\begin{equation}
\label{restriction-inflation-commutation}
\Res^G_H \Inf^G_{G/K} U =
\Inf^H_{H/H \cap K} \Res^{G/K}_{H/H \cap K} U.
\end{equation}
(We regard $H / H \cap K$ as a subgroup of $G / K$, since the
canonical homomorphism $H / H \cap K \to G / K$ is injective.)

% [DG][v14] Added previous sentence.

Inflation turns out to be adjoint to the
\dfn{$K$-fixed space construction}\index{fixed space of a $\CC G$-module}
sending a $\CC G$-module $V$ to the $\CC[G/K]$-module\index{$V^K$}
\[
V^K:=\{v \in V: k(v)=v \text{ for }k \in K\} .
\]
Note that $V^K$ is indeed a $G$-stable subspace:
for any $v$ in $V^K$ and $g$ in $G$, one
has that $g(v)$ lies in $V^K$ since an element $k$ in $K$ satisfies
$kg(v) = g \cdot g^{-1}kg(v) = g(v)$ as $g^{-1}kg$ lies in $K$.
One has this adjointness
\begin{equation}
\label{inflation-reciprocity}
\Hom_{\CC G}(\Inf_{G/K}^G U, V) = \Hom_{\CC[G/K]}(U, V^K) ,
\end{equation}
because any $\CC G$-module homomorphism $\varphi$ on the left must
have the property that $k \varphi(u) = \varphi(k(u)) = \varphi(u)$
for all $k$ in $K$, so that $\varphi$ actually lies on the right.

We will also need the following formula for the
character $\chi_{V^K}$ in terms of the character $\chi_V$:
\begin{equation}
\label{deflation-character-formula}
\chi_{V^K}(gK) = \frac{1}{|K|} \sum_{k \in K} \chi_V(gk).
\end{equation}
To see this, note that when one has a $\CC$-linear
endomorphism $\varphi$ on a space $V$ that preserves some
$\CC$-subspace $W \subset V$, if $V \overset{\pi}{\rightarrow} W$
is any idempotent projection onto $W$, then the trace of the restriction
$\varphi|_W$ equals the trace of $\varphi \circ \pi$ on $V$.
Applying this to $W=V^K$ and $\varphi = g$, with
$\pi=\frac{1}{|K|} \sum_{k \in K} k$, gives
\eqref{deflation-character-formula}.\footnote{For
another proof of \eqref{deflation-character-formula}, see
Exercise~\ref{exe.Indrho}(l).}

Another way to restate \eqref{deflation-character-formula} is:
\begin{equation}
\label{deflation-character-formula-v2}
\chi_{V^K}(gK) = \frac{1}{|K|} \sum_{h \in gK} \chi_V(h).
\end{equation}

% [DG][v4] Trivial rewriting, only stated to factor out a very
% simple step from the proof of the bialgebra property below.

Inflation and $K$-fixed space construction can also be defined on
class functions. For inflation, this is particularly easy:
Inflation $\Inf_{G/K}^G f$ of an $f \in R_{\CC} (G/K)$ is defined as
the composition $\xymatrix{G \arsurj[r] & G/K \ar[r]^{f} & \CC}$.
This is a class function of $G$ and thus lies in $R_\CC (G)$.
Thus, inflation $\Inf_{G/K}^G$ is a $\CC$-linear map
$R_{\CC} (G/K) \to R_\CC (G)$. It restricts to a $\ZZ$-linear map
$R(G/K) \to R(G)$, since it is clear that every $\CC (G/K)$-module
$U$ satisfies $\Inf_{G/K}^G \chi_U = \chi_{\Inf_{G/K}^G U}$.

We can also use \eqref{deflation-character-formula}
(or \eqref{deflation-character-formula-v2}) as inspiration for
defining a ``$K$-fixed space construction'' on class
functions. Explicitly, for every
class function $f \in R_\CC (G)$, we define a class function
$f^K \in R_\CC (G/K)$ by
\[
f^K (gK) = \frac{1}{|K|} \sum_{k \in K} f(gk) = \frac{1}{|K|} \sum_{h \in gK} f(h).
\]
The map $\left(\cdot\right)^K : R_\CC (G) \to R_\CC (G/K),
\ f \mapsto f^K$ is $\CC$-linear, and restricts to a $\ZZ$-linear
map $R(G) \to R(G/K)$. Again, we have a compatibility with the
$K$-fixed point construction on modules: We have
$\chi_{V^K} = \left(\chi_V\right)^K$ for every $\CC G$-module
$V$.

Taking characters in \eqref{inflation-reciprocity}, we obtain
\begin{equation}
\label{inflation-reciprocity-chars}
( \Inf_{G/K}^G \chi_U, \chi_V )_G
= ( \chi_U, \chi_V^K )_{G/K}
\end{equation}
for any $\CC \left[G/K\right]$-module $U$ and any $\CC G$-module
$V$ (since $\chi_{\Inf_{G/K}^G U} = \Inf_{G/K}^G \chi_U$
and $\chi_{V^K} = \left(\chi_V\right)^K$). By $\ZZ$-linearity,
this implies that
\[
\left( \Inf_{G/K}^G \alpha , \beta \right)_G
= \left( \alpha, \beta^K \right)_{G/K}
\]
for any class functions $\alpha \in R_\CC \left(G/K\right)$
and $\beta \in R_\CC \left(G\right)$.

There is also an analogue of~\eqref{induction-vs-tensor-product}:

\begin{lemma}
Let $G_1$ and $G_2$ be two groups, and $K_1 < G_1$ and $K_2 < G_2$
be two respective subgroups. Let $U_i$ be a $\CC G_i$-module
for each $i \in \left\{1,2\right\}$. Then,
\begin{equation}
\label{invariants-vs-tensor-product}
\left(U_1 \otimes U_2\right)^{K_1 \times K_2} = U_1^{K_1} \otimes U_2^{K_2}
\end{equation}
(as subspaces of $U_1 \otimes U_2$).
\end{lemma}

\begin{proof}
The subgroup $K_1 = K_1 \times 1$ of $G_1 \times G_2$ acts on $U_1 \otimes U_2$,
and its fixed points are
$\left(U_1 \otimes U_2\right)^{K_1} = U_1^{K_1} \otimes U_2$
(because for a $\CC K_1$-module, tensoring with $U_2$ is the same as
taking a direct power, which clearly commutes with taking fixed points).
Similarly, $\left(U_1 \otimes U_2\right)^{K_2} = U_1 \otimes U_2^{K_2}$.
Now,
\[
\left(U_1 \otimes U_2\right)^{K_1 \times K_2}
= \left(U_1 \otimes U_2\right)^{K_1} \cap \left(U_1 \otimes U_2\right)^{K_2}
= \left(U_1^{K_1} \otimes U_2\right) \cap \left(U_1 \otimes U_2^{K_2}\right)
= U_1^{K_1} \otimes U_2^{K_2}
\]
according to the known linear-algebraic fact stating that if $P$ and $Q$ are
subspaces of two vector spaces $U$ and $V$, respectively, then
$\left(P \otimes V\right) \cap \left(U \otimes Q\right) = P \otimes Q$.
\end{proof}

% [DG][v3] This lemma is used (no longer) tacitly in the proof of
% the bialgebra property of A(GL).

\begin{exercise} \phantomsection
\label{exe.HomxHom}
\begin{itemize}
\item[(a)]
Let $G_1$ and $G_2$ be two groups. Let $V_i$ and $W_i$ be finite-dimensional
$\CC G_i$-modules for every $i \in \left\{1,2\right\}$. Prove that
the $\CC$-linear map
\[
\Hom_{\CC G_1} \left(V_1,W_1\right) \otimes \Hom_{\CC G_2} \left(V_2,W_2\right)
\to \Hom_{\CC \left[G_1\times G_2\right]} \left(V_1 \otimes V_2, W_1 \otimes W_2\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms is a vector space isomorphism.
\item[(b)] Use part (a) to give a new proof
of \eqref{character-tensorprod}.
\end{itemize}
\end{exercise}

% [DG][v23] I have moved this exercise here because its solution uses
% \eqref{invariants-vs-tensor-product}. It used to stand right after
% \eqref{character-tensorprod}, but there is a much simpler proof of
% \eqref{character-tensorprod} by direct computation.

As an aside, \eqref{restriction-inflation-commutation} has a
``dual'' analogue:

\begin{exercise}
\label{exe.ind-fixed-points}
Let $G$ be a finite group, and let $K \triangleleft G$ and $H < G$.
Let $U$ be a $\CC H$-module. As usual, regard $H/\left(H\cap K\right)$
as a subgroup of $G/K$. Show that
$\left(\Ind^G_H U\right)^K
\cong \Ind^{G/K}_{H/\left(H\cap K\right)} \left(U^{H\cap K}\right)$
as $\CC \left[G/K\right]$-modules.
\end{exercise}

% [DG][v14] Added above exercise. (I haven't seen this one anywhere.
% The particular case when $K = G$ is the Proposition on p. 121 of
% \cite{Knutson}, but proven there in a rather ugly way that
% depends on the ground field being a field of char $0$. I assume
% the general case can be found somewhere in group cohomology
% literature...)

Inflation also ``commutes'' with induction:

\begin{exercise}
\label{exe.ind-infl}
Let $G$ be a finite group, and let $K < H < G$ be such that
$K \triangleleft G$. Thus, automatically, $K \triangleleft H$,
and we regard the quotient $H/K$ as a subgroup of $G/K$.
Let $V$ be a $\CC \left[H/K\right]$-module.
Show that
$\Inf^G_{G/K} \Ind^{G/K}_{H/K} V
\cong \Ind^G_H \Inf^H_{H/K} V$
as $\CC G$-modules.
\end{exercise}

% [DG][v27] Added above exercise (to use it later in solving another).
% Note: unlike in Exercise~\ref{exe.ind-fixed-points}, the
% $K < H$ condition is important.

\begin{exercise}
\label{exe.fixed-vs-coinvariants}
Let $G$ be a finite group, and let $K \triangleleft G$.
Let $V$ be a $\CC G$-module. Let $I_{V,K}$ denote the $\CC$-vector
subspace of $V$ spanned by all elements of the form $v-kv$ for
$k \in K$ and $v \in V$.
\begin{itemize}
\item[(a)] Show that $I_{V,K}$ is a $\CC G$-submodule of $V$.
\item[(b)] Let $V_K$ denote the quotient $\CC G$-module
$V / I_{V,K}$. (This module is occasionally called the
\dfn{$K$-coinvariant module of $V$}\index{coinvariant module},
a name it sadly shares with
at least two other non-equivalent constructions in algebra.)
Show that $V_K \cong \Inf^G_{G/K} \left(V^K\right)$ as
$\CC G$-modules. (Use $\operatorname{char} \CC = 0$.)
\end{itemize}
\end{exercise}

% [DG][v14] Added the above exercise, not out of need but because it
% is a piece of culture people should see. The reminder to use
% $\operatorname{char} \CC = 0$ is more of an allusion to the fact
% that many other exercises here do not care about the characteristic
% of the ground field.

In the remainder of this subsection, we shall briefly survey
generalized notions of induction and restriction, defined in terms of a
group homomorphism $\rho$ rather than in terms of a group $G$ and a
subgroup $H$. These generalized notions (defined by van Leeuwen in
\cite[\S 2.2]{Leeuwen-hopf}) will not be used in the rest of
these notes, but they shed some new light on the facts about induction,
restriction, inflation and fixed point construction discussed above. (In
particular, they reveal that some of said facts have common generalizations.)

The reader might have noticed that the definitions of inflation and of
restriction (both for characters and for modules) are similar. In fact, they
both are particular cases of the following construction:

\begin{remark}
\label{rmk.Resrho}
Let $G$ and $H$ be two finite groups, and let $\rho : H \rightarrow G$ be
a group homomorphism.

\begin{itemize}
\item If $f \in R_\CC \left( G \right)$, then the \emph{$\rho$-restriction}\index{$\rho$-restriction of a class function}
$\Res_{\rho} f$ of $f$ is defined as the map
$f \circ \rho : H \rightarrow \CC$. This map is easily
seen to belong to $R_\CC \left( H \right)$.

\item If $V$ is a $\CC G$-module, then the \emph{$\rho$-restriction}\index{$\rho$-restriction of a $\CC G$-module}
$\Res_{\rho} V$ of $V$ is the $\CC H$-module with ground
space $V$ and action given by
\[
h \cdot v = \rho \left( h \right) \cdot v
\ \ \ \ \ \ \ \ \ \ \text{for every } h \in H \text{ and } v\in V .
\]

\end{itemize}

This construction generalizes both inflation and restriction:
If $H$ is a subgroup of $G$, and if $\rho : H \rightarrow G$ is the
inclusion map, then $\Res_{\rho} f = \Res_{H}^{G} f$
(for any $f \in R_\CC \left( G \right)$)
and $\Res_{\rho} V = \Res_{H}^{G} V$
(for any $\CC G$-module $V$). If, instead, we have
$G = H/K$ for a normal subgroup $K$ of $H$, and if
$\rho : H \rightarrow G$ is the projection map, then
$\Res_{\rho} f = \Inf_{H/K}^H f$
(for any $f \in R_\CC \left( H/K \right)$) and
$\Res_{\rho} V = \Inf_{H/K}^H V$
(for any $\CC \left[H/K\right]$-module $V$).
\end{remark}

A subtler observation is that induction and fixed point construction
can be generalized by a common notion. This is the subject of
Exercise~\ref{exe.Indrho} below.

\begin{exercise}
\label{exe.Indrho}
Let $G$ and $H$ be two finite groups, and let $\rho : H \rightarrow G$
be a group homomorphism. We introduce the following notations:

\begin{itemize}
\item If $f \in R_\CC \left( H \right)$, then the
\emph{$\rho$-induction}\index{$\rho$-induction of a class function}
$\Ind_{\rho} f$ of $f$ is a map
$G \rightarrow \CC$ which is defined as follows:
\[
\left( \Ind_{\rho} f \right) \left( g \right)
= \dfrac{1}{\left\vert H \right\vert}
  \sum\limits_{\substack{\left( h, k \right) \in H \times G; \\
                  k \rho\left( h \right) k^{-1} = g}}
  f \left( h \right)
\ \ \ \ \ \ \ \ \ \ \text{for every } g \in G.
\]

\item If $U$ is a $\CC H$-module, then the
\emph{$\rho$-induction}\index{$\rho$-induction of a $\CC H$-module}
$\Ind_{\rho} U$ of $U$ is defined as the $\CC G$-module
$\CC G \otimes_{\CC H} U$, where $\CC G$ is regarded as a
$\left( \CC G , \CC H \right)$-bimodule according to the following
rule: The left $\CC G$-module structure on $\CC G$ is plain
multiplication inside $\CC G$; the right $\CC H$-module structure
on $\CC G$ is induced by the $\CC$-algebra homomorphism
$\CC \left[ \rho \right] : \CC H \rightarrow \CC G$ (thus, it
is explicitly given by
$\gamma \eta
= \gamma \cdot \left( \CC \left[ \rho \right] \right) \eta$
for all $\gamma \in \CC G$ and $\eta \in \CC H$).
\end{itemize}

Prove the following properties of this construction:

\begin{enumerate}
\item[(a)] For every $f \in R_\CC \left( H \right)$, we have
$\Ind_{\rho} f \in R_\CC \left( G \right)$.

\item[(b)] For any finite-dimensional $\CC H$-module $U$, we have
$\chi_{\Ind_{\rho} U} = \Ind_{\rho} \chi_{U}$.

\item[(c)] If $H$ is a subgroup of $G$, and if
$\rho : H \rightarrow G$ is the inclusion map, then
$\Ind_{\rho} f = \Ind_H^G f$ for every
$f\in R_\CC \left( H \right)$.

\item[(d)] If $H$ is a subgroup of $G$, and if
$\rho : H \rightarrow G$ is the inclusion map, then
$\Ind_{\rho} U = \Ind_H^G U$ for every $\CC H$-module $U$.

\item[(e)] If $G = H/K$ for some normal subgroup $K$ of $H$, and if
$\rho : H \rightarrow G$ is the projection map, then
$\Ind_{\rho} f = f^K$ for every $f \in R_\CC \left( H \right)$.

\item[(f)] If $G = H/K$ for some normal subgroup $K$ of $H$, and if
$\rho : H \rightarrow G$ is the projection map, then
$\Ind_{\rho} U \cong U^K$ for every $\CC H$-module $U$.

\item[(g)] Any class functions $\alpha \in R_\CC \left( H \right)$
and $\beta \in R_\CC \left( G \right)$ satisfy
\begin{equation}
\left( \Ind_{\rho} \alpha, \beta \right) _G
= \left( \alpha, \Res_{\rho} \beta \right) _H
\label{eq.exe.Indrho.g.herm}
\end{equation}
and
\begin{equation}
\left\langle \Ind_{\rho} \alpha, \beta \right\rangle _G
= \left\langle \alpha, \Res_{\rho} \beta \right\rangle _H .
\label{eq.exe.Indrho.g.bil}
\end{equation}
(See Remark~\ref{rmk.Resrho} for the definition of
$\Res_{\rho} \beta$.)

\item[(h)] We have
$\Hom_{\CC G} \left( \Ind_{\rho} U, V \right)
\cong \Hom_{\CC H} \left( U, \Res_{\rho} V \right)$ for every
$\CC H$-module $U$ and every $\CC G$-module $V$. (See
Remark~\ref{rmk.Resrho} for the definition of $\Res_{\rho} V$.)

\item[(i)] Similarly to how we made $\CC G$ into a
$\left( \CC G, \CC H \right)$-bimodule, let us make $\CC G$ into a
$\left( \CC H, \CC G \right)$-bimodule (so the right
$\CC G$-module structure is plain multiplication inside $\CC G$,
whereas the left $\CC H$-module structure is induced by the
$\CC$-algebra homomorphism
$\CC \left[ \rho \right] : \CC H \rightarrow \CC G$). If $U$ is any
$\CC H$-module, then the $\CC G$-module
$\Hom_{\CC H} \left( \CC G, U \right)$ (defined as in
Exercise~\ref{exe.induction.ind-as-hom} using the
$\left( \CC H, \CC G \right)$-bimodule structure on
$\CC G$) is isomorphic to $\Ind_{\rho} U$.

\item[(j)] We have
$\Hom_{\CC G} \left( U, \Ind_{\rho} V \right)
\cong \Hom_{\CC H} \left( \Res_{\rho} U, V \right)$
for every $\CC G$-module $U$ and every $\CC H$-module $V$.
(See Remark~\ref{rmk.Resrho} for the definition of $\Res_{\rho} V$.)
\end{enumerate}

Furthermore:

\begin{enumerate}
\item[(k)] Use the above to prove the formula
\eqref{induced-character-formula}.

\item[(l)] Use the above to prove the formula
\eqref{deflation-character-formula}.
\end{enumerate}

[\textbf{Hint:} Part (b) of this exercise is hard. To solve it, it is useful
to have a way of computing the trace of a linear operator without knowing a
basis of the vector space it is acting on. There is a way to do this using a
``finite dual generating system'', which is a
somewhat less restricted notion than that of a basis\footnote{More precisely:
Let $\mathbb{K}$ be a field, and $V$ be a $\mathbb{K}$-vector space. A
\dfn{finite dual generating system} for $V$ means a triple
$\left( I, \left( a_i \right)_{i \in I},
        \left( f_i \right)_{i \in I} \right)$,
where
\par
\begin{itemize}
\item $I$ is a finite set;
\par
\item $\left( a_i \right)_{i \in I}$ is a family of elements of $V$;
\par
\item $\left( f_i \right)_{i \in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means
$\Hom_{\mathbb{K}} \left( V, \mathbb{K} \right)$)
\end{itemize}
\par
such that every $v\in V$ satisfies
$v = \sum_{i \in I} f_i \left( v \right) a_i$.
For example, if $\left( e_j \right) _{j \in J}$ is a
finite basis of the vector space $V$, and if
$\left( e_j^{\ast} \right) _{j \in J}$ is the basis of $V^{\ast}$ dual
to this basis $\left( e_j \right) _{j \in J}$, then
$\left( J, \left( e_j \right) _{j \in J},
\left( e_j^{\ast} \right) _{j \in J} \right)$ is a finite dual
generating system for $V$; however, most finite dual generating
systems are not obtained this way.
\par
The crucial observation is now that if
$\left( I, \left( a_i \right)_{i \in I},
        \left( f_i \right)_{i \in I} \right)$
is a finite dual generating system for a vector space $V$, and if $T$
is an endomorphism of $V$, then
\[
\trace T = \sum_{i \in I} f_i \left( T a_i \right) .
\]
Prove this!}. Try to create a finite dual generating system for
$\Ind_{\rho} U$ from one for $U$ (and from the group
$G$), and then use it to compute $\chi_{\Ind_{\rho} U}$.

The solution of part (i) is a modification of the solution of
Exercise~\ref{exe.induction.ind-as-hom}, but complicated by the fact
that $H$ is no longer (necessarily) a subgroup of $G$. Part (f) can be
solved by similar arguments, or using part (i), or using
Exercise~\ref{exe.fixed-vs-coinvariants}(b).]
\end{exercise}

The result of Exercise~\ref{exe.Indrho}(h) generalizes
\eqref{Frobenius-reciprocity} (because of Exercise~\ref{exe.Indrho}(d)), but
also generalizes \eqref{inflation-reciprocity} (due to
Exercise~\ref{exe.Indrho}(f)).
Similarly, Exercise~\ref{exe.Indrho}(g) generalizes both
\eqref{Frobenius-reciprocity-chars} and \eqref{inflation-reciprocity-chars}.
Similarly, Exercise~\ref{exe.Indrho}(i) generalizes
Exercise~\ref{exe.induction.ind-as-hom}, and
Exercise~\ref{exe.Indrho}(j) generalizes
Exercise~\ref{exe.induction.wrong-way-frob}.

Similarly, Exercise~\ref{exe.ind.tensor} is generalized by the
following exercise:

\begin{exercise}
\label{exe.Indrho.tensor}
Let $G_1$, $G_2$, $H_1$ and $H_2$ be four finite groups.
Let $\rho_1 : H_1 \rightarrow G_1$ and $\rho_2 : H_2 \rightarrow G_2$
be two group homomorphisms. These two homomorphisms clearly induce a
group homomorphism
$\rho_1 \times \rho_2 : H_1 \times H_2 \rightarrow G_1 \times G_2$.
Let $U_1$ be a $\CC H_1$-module, and $U_2$ be a $\CC H_2$-module.
Show that
\[
\Ind_{\rho_1 \times \rho_2} \left( U_1 \otimes U_2 \right)
\cong \left( \Ind_{\rho_1} U_1 \right)
      \otimes \left( \Ind_{\rho_2} U_2 \right)
\]
as $\CC \left[ G_1 \times G_2 \right]$-modules.
\end{exercise}

The $\Ind_{\rho}$ and $\Res_{\rho}$ operators behave ``functorially''
with respect to composition. Here is what this means:

\begin{exercise}
\label{exe.Indrho.transitivity}
Let $G$, $H$ and $I$ be three finite groups.
Let $\rho : H \rightarrow G$ and $\tau : I \rightarrow H$ be two
group homomorphisms.

\begin{enumerate}
\item[(a)] We have
$\Ind_{\rho} \Ind_{\tau} U \cong \Ind_{\rho \circ \tau} U$ for every
$\CC I$-module $U$.

\item[(b)] We have
$\Ind_{\rho} \Ind_{\tau} f = \Ind_{\rho \circ \tau} f$ for every
$f \in R_\CC \left( I \right)$.

\item[(c)] We have
$\Res_{\tau} \Res_{\rho} V = \Res_{\rho \circ \tau} V$ for every
$\CC G$-module $V$.

\item[(d)] We have
$\Res_{\tau} \Res_{\rho} f = \Res_{\rho \circ \tau} f$ for every
$f \in R_\CC \left( G \right)$.
\end{enumerate}
\end{exercise}

Exercise~\ref{exe.Indrho.transitivity}(a), of course, generalizes
Exercise~\ref{exe.induction.transitivity}.

% [DG][v52] Added the above discussion (beginning with the two
% paragraphs above Remark~\ref{rmk.Resrho} and ending before
% this very comment) of $\rho$-restriction and $\rho$-induction.
% We don't really need these operations (though one solution
% would have been simpler if I had had them at my disposal back
% when I was writing it), but they just feel morally relevant
% (particularly because they link induction with fixed point
% construction, and restriction with inflation, thus reducing
% the number of fundamental operations to two). Also I haven't
% seen my solution to Exercise~\ref{exe.Indrho}(b) anywhere
% and I thought it deserved writing up.


\subsubsection{Semidirect products}
Recall that a
\emph{semidirect product}\index{semidirect product of groups}
is a group $G \ltimes K$
having two subgroups $G,K$ with
\begin{enumerate}
\item[$\bullet$]
$K \triangleleft (G \ltimes K)$ is a normal subgroup,
\item[$\bullet$]
$G \ltimes K=GK=KG$, and
\item[$\bullet$]
$G \cap K=\{e\}$.
\end{enumerate}
In this setting one has two interesting adjoint constructions,
applied in Section~\ref{wreath-product-section}.

\begin{proposition}
\label{semidirect-product-adjoint-functors}
Fix a $\CC[G \ltimes K]$-module $V$.
\begin{enumerate}
\item[(i)]
For any $\CC G$-module $U$, one has $\CC[G \ltimes K]$-module
structure
\[
\Phi(U):=U \otimes V,
\]
determined via
\begin{align*}
k(u \otimes v)&= u \otimes k(v),\\
g(u \otimes v)&= g(u) \otimes g(v).\\
\end{align*}
\item[(ii)]
For any $\CC[G \ltimes K]$-module $W$, one has $\CC G$-module
structure
\[
\Psi(W):=
\Hom_{\CC K}(\Res^{G \ltimes K}_K V, \Res^{G \ltimes K}_K W),
\]
determined via
$
g(\varphi)=g \circ \varphi \circ g^{-1}.
$
\item[(iii)]
The maps
\[
\CC G-\mods
\underset{\Psi}{\overset{\Phi}{\rightleftharpoons}}
\CC[G \ltimes K]-\mods
\]
are adjoint in the sense that one has an isomorphism
\[
\begin{array}{ccc}
\Hom_{\CC G}(U,\Psi(W))
  &\longrightarrow &
      \Hom_{\CC[G\ltimes K]}(\Phi(U),W) \\
\Vert &  & \Vert\\
\Hom_{\CC G}(U,\Hom_{\CC K}(\Res^{G \ltimes K}_K V, \Res^{G \ltimes K}_K W)) & &
\Hom_{\CC[G\ltimes K]}(U \otimes V,W) , \\
& & \\
\varphi & \longmapsto & \overline{\varphi}(u \otimes v):=\varphi(u)(v) .
\end{array}
\]
\item[(iv)]
One has a $\CC G$-module isomorphism
\[
(\Psi \circ \Phi)(U) \cong U \otimes \End_{\CC K}(\Res^{G \ltimes K}_K  V).
\]
In particular, if $\Res^{G \ltimes K}_K  V$
is a simple $\CC K$-module, then $(\Psi \circ \Phi)(U) \cong U$.

\end{enumerate}
\end{proposition}
\begin{proof}
These are mostly straightforward exercises in the definitions.
To check assertion (iv), for example, note that $K$ acts only
in the right tensor factor in
$\Res^{G \ltimes K}_K(U \otimes V)$,
and hence as $\CC G$-modules one has
\begin{align*}
(\Psi \circ \Phi)(U)
        &= \Hom_{\CC K}(\Res^{G \ltimes K}_K V, \,\,
             \Res^{G \ltimes K}_K(U \otimes V)) \\
        &= \Hom_{\CC K}(\Res^{G \ltimes K}_K V, \,\,
             U \otimes \Res^{G \ltimes K}_K V) \\
        &= U \otimes \Hom_{\CC K}(\Res^{G \ltimes K}_K V, \,\,
                                  \Res^{G \ltimes K}_K V) \\
        &= U \otimes \End_{\CC K}(\Res^{G \ltimes K}_K V) .
\end{align*}
\end{proof}





\subsection{Three towers of groups}
\label{three-towers-section}

Here we consider three \idx{towers of groups}\index{$G_*$}\index{three towers}
\[
G_*= (G_0 < G_1 < G_2 < G_3 < \cdots)
\]
where either
\begin{enumerate}
\item[$\bullet$] $G_n=\Symm_n$, the \emph{symmetric group}\footnote{The symmetric group $\Symm_0$ is the group of all permutations of the empty set $\left\{1,2,\ldots,0\right\}=\varnothing$. It is a trivial group. (Note that $\Symm_1$ is also a trivial group.)}, or
\item[$\bullet$] $G_n=\Symm_n[\Gamma]$, the \emph{wreath product} of the
symmetric group with some arbitrary finite group $\Gamma$, or
\item[$\bullet$] $G_n=GL_n(\FF_q)$, the finite general linear group\footnote{The group $GL_0(\FF_q)$ is a trivial group, consisting of the empty $0\times 0$ matrix.}.
\end{enumerate}
Here the wreath product $\Symm_n[\Gamma]$ can be thought of informally as the group of \emph{monomial}\index{monomial matrix} $n \times n$ matrices whose nonzero entries lie in $\Gamma$, that is,
$n \times n$ matrices having exactly one nonzero entry in each row and column, and that entry is an element of $\Gamma$.  E.g.
\[
\left[
\begin{matrix}
0 & g_2 & 0 \\
g_1 & 0 & 0 \\
0   & 0 & g_3
\end{matrix}
\right]
\left[
\begin{matrix}
0   & 0   & g_6 \\
0   & g_5 & 0 \\
g_4   & 0 & 0
\end{matrix}
\right]
=
\left[
\begin{matrix}
0        & g_2g_5   & 0 \\
0        & 0        & g_1 g_6 \\
g_3g_4   & 0 & 0
\end{matrix}
\right].
\]
More formally, $\Symm_n[\Gamma]$ is
the semidirect product $\Symm_n \ltimes \Gamma^n$
in which $\Symm_n$ acts on $\Gamma^n$ via
$\sigma(\gamma_1,\ldots,\gamma_n)=
(\gamma_{\sigma^{-1}(1)},\ldots,\gamma_{\sigma^{-1}(n)}).
$

For each of the three towers $G_*$,
there are embeddings $G_i \times G_j \hookrightarrow G_{i+j}$
and we introduce maps \dfn{$\ind^{i+j}_{i,j}$}
taking $\CC[G_i \times G_j]$-modules to
$\CC G_{i+j}$-modules, as well as maps
\dfn{$\res^{i+j}_{i,j}$} carrying modules
in the reverse direction which are adjoint:
\begin{equation}
\label{ind-res-adjointness}
\Hom_{\CC G_{i+j}}( \ind^{i+j}_{i,j} U , V ) = \Hom_{\CC[G_i \times G_j]}( U, \res^{i+j}_{i,j} V ).
\end{equation}

\begin{definition}
\label{ind-res-defn}
For $G_n=\Symm_n$, one embeds $\Symm_i \times \Symm_j$ into $\Symm_{i+j}$
as the permutations that permute $\{1,2,\ldots,i\}$ and $\{i+1,i+2,\ldots,i+j\}$
separately.  Here one defines
\begin{align*}
\ind_{i,j}^{i+j}&:=\Ind_{\Symm_i \times \Symm_j}^{\Symm_{i+j}},\\
\res_{i,j}^{i+j}&:=\Res_{\Symm_i \times \Symm_j}^{\Symm_{i+j}}.
\end{align*}
For $G_n=\Symm_n[\Gamma]$, similarly embed
$\Symm_i[\Gamma] \times \Symm_j[\Gamma]$ into
$\Symm_{i+j}[\Gamma]$ as block monomial matrices whose two diagonal blocks
have sizes $i,j$ respectively, and define
\begin{align*}
\ind_{i,j}^{i+j}&:=\Ind_{\Symm_i[\Gamma] \times \Symm_j[\Gamma]}^{\Symm_{i+j}[\Gamma]},\\
\res_{i,j}^{i+j}&:=\Res_{\Symm_i[\Gamma] \times \Symm_j[\Gamma]}^{\Symm_{i+j}[\Gamma]}.
\end{align*}
For $G_n=GL_n(\FF_q)$, which we will denote just $GL_n$,
similarly embed $GL_i \times GL_j$ into
$GL_{i+j}$ as block diagonal matrices whose two diagonal blocks
have sizes $i,j$ respectively.  However, one also introduces
as an intermediate the
\emph{parabolic subgroup}\index{parabolic subgroup of $GL_n$}
$P_{i,j}$ consisting
of the block upper-triangular matrices of the form
\[
\left[
\begin{matrix}
g_i & \ell\\
0   & g_j
\end{matrix}
\right]
\]
where $g_i, g_j$ lie in $GL_i,GL_j$, respectively, and
$\ell$ in $\FF_q^{i \times j}$ is arbitrary.  One has
a quotient map $P_{i,j} \rightarrow GL_i \times GL_j$ whose kernel $K_{i,j}$
is the set of matrices of the form
\[
\left[
\begin{matrix}
I_i & \ell\\
0   & I_j
\end{matrix}
\right]
\]
with $\ell$ again arbitrary.
Here one defines
\begin{align*}
\ind_{i,j}^{i+j}&:=\Ind_{P_{i,j}}^{GL_{i+j}} \Inf^{P_{i,j}}_{GL_i \times GL_j},\\
\res_{i,j}^{i+j}&:=\left(\Res^{GL_{i+j}}_{P_{i,j}}(-)\right)^{K_{i,j}}.
\end{align*}
\end{definition}

In the case $G_n = GL_n$, the operation $\ind_{i,j}^{i+j}$ is sometimes
called \dfn{parabolic induction} or \dfn{Harish-Chandra induction}.
The operation $\res_{i,j}^{i+j}$ is essentially just the $K_{i,j}$-fixed point construction $V \mapsto V^{K_{i,j}}$.
However writing it as the above two-step composite makes it more obvious,
(via \eqref{Frobenius-reciprocity} and \eqref{inflation-reciprocity})
that $\res_{i,j}^{i+j}$ is again adjoint to $\ind_{i,j}^{i+j}$.

\begin{definition}
For each of the three towers $G_*$, define a graded $\ZZ$-module\index{$A(G_*)$}
\[
A:=A(G_*)=\bigoplus_{n \geq 0} R(G_n)
\]
with a bilinear form $(\cdot,\cdot)_A$ whose restriction to $A_n:=R(G_n)$
is the usual form $(\cdot,\cdot)_{G_n}$, and such that
$\Sigma := \bigsqcup_{n \geq 0} \Irr(G_n)$ gives an orthonormal
$\ZZ$-basis.  Notice that $A_0=\ZZ$ has
its basis element $\one$ equal to the unique irreducible character
of the trivial group $G_0$.
\end{definition}

% [DG] I have removed the $\{\chi\}$ from this definition (it used to
% say $\Sigma=\{\chi\}=\bigsqcup_{n \geq 0} \Irr(G_n)$) as I didn't find
% it very helpful (it would only make the impression that $\Sigma$ is
% a one-element set). Sorry if I missed its purpose!
%
% Also, this is one of the places I've replaced "so that" by "such
% that", as I do believe there is a difference between the two idioms.
% (I left wordings like "reindex so that $i$ becomes $1$" intact;
% I only changed cases like "there exists an [object] so that
% [condition]".)

Bearing in mind that $A_n=R(G_n)$ and
\[
A_i \otimes A_j =
R(G_i) \otimes R(G_j) \cong
R(G_i \times G_j) ,
\]
one then has candidates for product and coproduct
defined by
\[
\begin{array}{rrcl}
m:=\ind_{i,j}^{i+j}:&A_i \otimes A_j & \longrightarrow & A_{i+j} , \\
\Delta:=\bigoplus_{i+j=n} \res_{i,j}^{i+j}: &A_n & \longrightarrow & \bigoplus_{i+j=n} A_i \otimes A_j. \\
\end{array}
\]
The coassociativity of $\Delta$ is an easy consequence of transitivity of the
constructions of restriction and fixed points\footnote{More precisely,
using this transitivity, it is easily reduced to proving that
$K_{i+j, k} \cdot \left(K_{i,j} \times \left\{I_k\right\}\right)
= K_{i, j+k} \cdot \left(\left\{I_i\right\} \times K_{j,k}\right)$
(an equality between subgroups of $GL_{i+j+k}$) for any three
nonnegative integers $i, j, k$. But this equality
can be proven by realizing that both of its sides equal the
set of all block matrices of the form $\left(
\begin{array}{ccc}
I_i & \ell & \ell' \\
0 & I_j & \ell'' \\
0 & 0 & I_k
\end{array} \right)$
with $\ell$, $\ell'$ and $\ell''$ being matrices of sizes $i \times j$,
$i \times k$ and $j \times k$, respectively.}.
We could derive the associativity of $m$ from the transitivity of
induction and inflation, but this would be more
complicated\footnote{See Exercise~\ref{exe.towers.indres}(c) for
such a derivation.}; we will instead prove it differently.

We first show that the maps $m$ and $\Delta$ are adjoint with respect
to the forms $\left(\cdot,\cdot\right)_A$ and $\left(\cdot,\cdot\right)_{A\otimes A}$.
In fact, if $U$, $V$, $W$
are modules over $\CC G_i$, $\CC G_j$, $\CC G_{i+j}$, respectively, then
we can write the $\CC [G_i\times G_j]$-module $\res_{i,j}^{i+j} W$ as a
direct sum $\bigoplus_k X_k \otimes Y_k$ with $X_k$ being $\CC G_i$-modules
and $Y_k$ being $\CC G_j$-modules; we then have
\begin{equation}
\label{ind-res-adjointness-pf1}
\res_{i,j}^{i+j} \chi_W = \sum_k \chi_{X_k} \otimes \chi_{Y_k}
\end{equation}
and
\begin{align*}
\left(m\left(\chi_U \otimes \chi_V\right), \chi_W\right)_A
&= \left(\ind_{i,j}^{i+j} \left(\chi_{U\otimes V}\right), \chi_W\right)_A
= \left(\ind_{i,j}^{i+j} \left(\chi_{U\otimes V}\right), \chi_W\right)_{G_{i+j}} \\
&= \left(\chi_{U\otimes V}, \res_{i,j}^{i+j} \chi_W\right)_{G_i\times G_j}
= \left(\chi_{U\otimes V}, \sum_k \chi_{X_k} \otimes \chi_{Y_k}\right)_{G_i\times G_j} \\
&= \sum_k \left(\chi_{U\otimes V}, \chi_{X_k\otimes Y_k}\right)_{G_i\times G_j}
= \sum_k \left(\chi_U, \chi_{X_k}\right)_{G_i} \left(\chi_V, \chi_{Y_k}\right)_{G_j}
\end{align*}
(the third equality sign follows by taking dimensions in
\eqref{ind-res-adjointness} and recalling \eqref{character-scalprod};
the fourth equality sign follows from \eqref{ind-res-adjointness-pf1}; the
sixth one follows from \eqref{character-tensorprod}) and
\begin{align*}
\left(\chi_U \otimes \chi_V, \Delta\left(\chi_W\right)\right)_{A\otimes A}
&= \left(\chi_U \otimes \chi_V, \res_{i,j}^{i+j} \chi_W\right)_{A\otimes A}
= \left(\chi_U \otimes \chi_V, \sum_k \chi_{X_k} \otimes \chi_{Y_k}\right)_{A\otimes A} \\
&= \sum_k \left(\chi_U, \chi_{X_k}\right)_A \left(\chi_V, \chi_{Y_k}\right)_A
= \sum_k \left(\chi_U, \chi_{X_k}\right)_{G_i} \left(\chi_V, \chi_{Y_k}\right)_{G_j}
\end{align*}
(the first equality sign follows by removing all terms in $\Delta\left(\chi_W\right)$
whose scalar product with $\chi_U \otimes \chi_V$ vanishes for reasons of gradedness;
the second equality sign follows from
\eqref{ind-res-adjointness-pf1}), which in comparison yield
$\left(m\left(\chi_U \otimes \chi_V\right), \chi_W\right)_A
= \left(\chi_U \otimes \chi_V, \Delta\left(\chi_W\right)\right)_{A\otimes A}$,
thus showing that $m$ and $\Delta$ are adjoint maps. Therefore, $m$ is associative
(since $\Delta$ is coassociative).

% [DG] I didn't expect the details of the proof of adjointness to take this
% long. Maybe I am confusing myself here.
%
% The argument for associativity of $m$ using adjointness and coassociativity
% is an alternative to the 3-pages proof of the associativity of $m$ that I
% did in reiner-errata12.pdf. Of course, it doesn't show any functorial
% isomorphisms.

Endowing $A=\bigoplus_{n \geq 0} R(G_n)$ with the
obvious unit and counit maps, it thus becomes a graded, finite-type
$\ZZ$-algebra and $\ZZ$-coalgebra.

The next section addresses the issue of why
they form a bialgebra.  However, assuming this for the moment,
it should be clear that each of these algebras $A$ is a PSH having
$\Sigma=\bigsqcup_{n \geq 0} \Irr(G_n)$ as its PSH-basis.
$\Sigma$ is self-dual because $m, \Delta$ are defined by adjoint maps,
and it is positive because $m, \Delta$ take irreducible representations
to genuine representations not just virtual ones, and
hence have characters which are nonnegative sums of irreducible characters.

\begin{exercise}
\label{exe.Ind-ass}
Let $i$, $j$ and $k$ be three nonnegative integers. Let $U$ be a
$\CC \Symm_i$-module, let $V$ be a $\CC \Symm_j$-module, and let $W$ be a
$\CC \Symm_k$-module. Show that there are canonical
$\CC \left[\Symm_i \times \Symm_j \times \Symm_k\right]$-module
isomorphisms
\begin{align*}
\Ind^{\Symm_{i+j+k}}_{\Symm_{i+j} \times \Symm_k}\left(\Ind^{\Symm_{i+j}}_{\Symm_i \times \Symm_j}\left(U\otimes V\right)\otimes W\right)
&\cong \Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_j \times \Symm_k}\left(U\otimes V\otimes W\right) \\
&\cong \Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_{j+k}}\left(U \otimes \Ind^{\Symm_{j+k}}_{\Symm_j \times \Symm_k}\left(V\otimes W\right)\right).
\end{align*}

(Similar statements hold for the other two towers of groups and their
respective $\ind$ functors, although the one for the $GL_*$ tower is
harder to prove. See Exercise~\ref{exe.towers.indres}(a) for a
more general result.)
\end{exercise}

% [DG][v14] I recycled part of reiner-errata12.pdf in the
% above solution.


\subsection{Bialgebra and double cosets}
\label{double-coset-section}

To show that the algebra and coalgebras $A=A(G_*)$ are bialgebras,
the central issue is checking the pentagonal diagram in \eqref{bialgebra-diagrams}, that is, as maps $A \otimes A \rightarrow A \otimes A$, one has
\begin{equation}
\label{pentagonal-diagram-written-out}
\Delta \circ m = (m \otimes m) \circ (\id \otimes T \otimes \id) \circ (\Delta \otimes \Delta).
\end{equation}

In checking this, it is convenient to have a lighter notation for
various subgroups of the groups $G_n$ corresponding to compositions $\alpha$.

\begin{definition} \phantomsection
\label{def.composition-almost-composition}
\begin{itemize}

\item[(a)]
An \dfn{almost-composition} is a (finite) tuple
$\alpha = (\alpha_1,\alpha_2,\ldots,\alpha_\ell)$
of nonnegative integers. Its
\emph{length}\index{length of an almost-composition}
is defined to be $\ell$ and denoted by \dfn{$\ell(\alpha)$};
its \emph{size}\index{size of an almost-composition} is defined to be
$\alpha_1 + \alpha_2 + \cdots + \alpha_\ell$ and denoted by
$\left|\alpha\right|$;
its \emph{parts}\index{parts of an almost-composition}
are its entries $\alpha_1,\alpha_2,\ldots,\alpha_\ell$.
The almost-compositions of size $n$ are called the
\emph{almost-compositions of $n$}\index{almost-composition of $n$}.

\item[(b)] A \dfn{composition} is a finite tuple of positive
integers. Of course, any composition is an almost-composition,
and so all notions defined for almost-compositions (like
size and length) make sense for compositions.%
\index{length of a composition}\index{size of a composition}\index{parts of a composition}

Note that any partition of $n$ (written without trailing zeroes) is a
composition of $n$. We write $\varnothing$ (and sometimes, sloppily,
$(0)$, when there is no danger of mistaking it for the
almost-composition $(0)$) for the empty composition $()$.
\end{itemize}
\end{definition}

% [DG][v28] Introduced almost-compositions, and replaced the use of
% compositions by almost-compositions below. Indeed, this is needed in
% the proof of the bialgebra axiom (though one could avoid dealing with
% that, but that would be less organic).
% 
% I don't call them weak compositions, because the name is already taken
% for infinite sequences.

% [DG] This doesn't look like a great place to introduce compositions,
% but it's the best one I found, as they're used right below for the
% first time. In Chapter 2 it would probably be distracting.

\begin{definition}
Given an almost-composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$ of $n$,
define a subgroup
\[
G_\alpha \cong G_{\alpha_1} \times \cdots \times G_{\alpha_\ell}
< G_n
\]
via the block-diagonal embedding with diagonal blocks of
sizes $(\alpha_1,\ldots,\alpha_\ell)$.
This $G_\alpha$ is called a \dfn{Young subgroup} $\Symm_\alpha$ when
$G_n =\Symm_n$,
and a \dfn{Levi subgroup} when $G_n=GL_n$.
In the case when $G_n = \Symm_n[\Gamma]$, we also denote $G_\alpha$ by
$\Symm_\alpha[\Gamma]$.
In the case where $G_n=GL_n$, also define the
\emph{parabolic subgroup}\index{parabolic subgroup of $GL_n$}
$
P_\alpha
$
to be the subgroup of $G_n$ consisting of block-upper triangular
matrices whose diagonal blocks have sizes $(\alpha_1,\ldots,\alpha_\ell)$,
and let
$
K_\alpha
$
be the kernel of the obvious surjection $P_\alpha \rightarrow G_\alpha$ which sends
a block upper-triangular matrix to the tuple of its diagonal blocks
whose sizes are $\alpha_1,\alpha_2,\ldots,\alpha_\ell$.  Notice that
$P_{(i,j)} = P_{i,j}$ for any $i$ and $j$ with $i+j = n$; similarly,
$K_{(i,j)} = K_{i,j}$ for any $i$ and $j$ with $i+j = n$. We will also
abbreviate $G_{(i,j)} = G_i \times G_j$ by $G_{i,j}$.
%Also define operators $\ind^n_\alpha,
%\res^n_\alpha$ for the three towers analogous to those defined in
%Definition~\ref{ind-res-defn} when $\alpha=(i,j)$ has only two parts.

When $\left(\alpha_1, \alpha_2, \ldots, \alpha_\ell\right)$ is an
almost-composition, we abbreviate
$G_{\left(\alpha_1, \alpha_2, \ldots, \alpha_\ell\right)}$ by
$G_{\alpha_1, \alpha_2, \ldots, \alpha_\ell}$ (and similarly for the
$P$'s).
\end{definition}

% [DG][v28] Commented out the three lines above because these notations
% are not yet needed (and when they are needed, I'll define them in
% more detail).

\begin{definition}
\label{general-restriction}
Let $K$ and $H$ be two groups, $\tau : K \to H$ a group homomorphism,
and $U$ a $\CC H$-module. Then, $U^\tau$ is defined as the
$\CC K$-module with ground space $U$ and action given by
$ k \cdot u = \tau(k) \cdot u $ for all $k \in K$ and $u \in U$.
\ \ \ \ \footnote{We have already met this $\CC K$-module $U^\tau$
in Remark~\ref{rmk.Resrho}, where it was called $\Res_\tau U$.}
This very simple construction generalizes the definition of $U^g$
for an element $g \in G$, where $G$ is a group containing $H$ as
a subgroup; in fact, in this situation we have $U^g = U^\tau$,
where $K = {}^g H$ and $\tau : K \to H$ is the map $k \mapsto g^{-1}kg$.
\end{definition}


Using homogeneity, checking the bialgebra condition
\eqref{pentagonal-diagram-written-out}
in the homogeneous component $(A \otimes A)_n$
amounts to the following:
for each pair of representations $U_1,U_2$ of
$G_{r_1}, G_{r_2}$ with $r_1+r_2=n$, and
for each $(c_1,c_2)$ with $c_1+c_2=n$, one
must verify that
\begin{equation}
\label{bialgebra-as-Mackey}
\begin{aligned}
&\res_{c_1,c_2}^n\left(  \ind_{r_1,r_2}^n \left( U_1 \otimes U_2 \right) \right) \\
&\quad \cong \bigoplus_A
  \left( \ind_{a_{11},a_{21}}^{c_1} \otimes \ind^{c_2}_{a_{12},a_{22}}\right)
   \left( \left( \res^{r_1}_{a_{11},a_{12}} U_1 \otimes \res^{r_2}_{a_{21},a_{22}} U_2 \right)^{\tau_A^{-1}} \right)
\end{aligned}
\end{equation}
where the direct sum is over all matrices
$
A=\left[
\begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{matrix}
\right]
$
in $\NN^{2 \times 2}$ with row sums $(r_1,r_2)$ and column sums $(c_1,c_2)$,
and where $\tau_A$ is the obvious isomorphism
between the subgroups
\begin{equation}
\label{two-by-two-matrix-Mackey-subgroups}
\begin{aligned}
 G_{a_{11},a_{12},a_{21},a_{22}}
&\quad \left( < G_{r_1, r_2} \right) \qquad \text{and}\\
 G_{a_{11},a_{21},a_{12},a_{22}}
&\quad \left( < G_{c_1,c_2} \right)\\
\end{aligned}
\end{equation}
(we are using the inverse $\tau_A^{-1}$ of this isomorphism $\tau_A$
to identify modules for the first subgroup with modules
for the second subgroup, according to Definition~\ref{general-restriction}).

% [DG][v3] I replaced $\tau_A$ by $\tau_A^{-1}$ in the formula so as to match
% the definition \eqref{general-restriction} I introduced. (The definition
% defines $U^\tau$ as a pullback rather than a pushforward because the
% pushforward is defined only for isomorphisms $\tau$, while the pullback
% makes sense for any morphism.)

As one might guess, \eqref{bialgebra-as-Mackey} comes from the Mackey formula
(Theorem~\ref{Mackey-formula}),
once one identifies the appropriate double coset representatives.
This is just as easy to do in a slightly more general setting.


\begin{definition}
\label{def.compositions.intervaldec}
Given almost-compositions $\alpha,\beta$ of $n$ having
lengths $\ell,m$
and a matrix $A$ in $\NN^{\ell \times m}$ with
row sums $\alpha$ and column sums $\beta$, define a permutation $w_A$
in $\Symm_n$ as follows.  Disjointly decompose
$[n]=\{1,2,\ldots,n\}$ into consecutive intervals of numbers
\[
[n]=I_1 \sqcup \cdots \sqcup I_\ell \qquad \text{such that } |I_i|=\alpha_i
\]
(so the smallest $\alpha_1$ elements of $[n]$ go into $I_1$,
the next-smallest $\alpha_2$ elements of $[n]$ go into $I_2$,
and so on).
Likewise, disjointly decompose $[n]$ into consecutive intervals of numbers
\[
[n]=J_1 \sqcup \cdots \sqcup J_m \qquad \text{such that } |J_j|=\beta_j.
\]
For every $j\in [m]$, disjointly decompose $J_{j}$
into consecutive intervals of numbers $J_{j}=J_{j,1}\sqcup J_{j,2}\sqcup
\cdots\sqcup J_{j,\ell}$ such that every $i\in [\ell]$
satisfies $\left\vert J_{j,i}\right\vert =a_{ij}$.
For every $i\in [\ell]$, disjointly decompose $I_{i}$
into consecutive intervals of numbers $I_{i}=I_{i,1}\sqcup I_{i,2}\sqcup
\cdots\sqcup I_{i,m}$ such that every $j\in [m]$
satisfies $\left\vert I_{i,j}\right\vert =a_{ij}$.
Now, for every $i\in [\ell]$ and $j\in [m]$, let
$\pi_{i,j}$ be the increasing bijection
from $J_{j,i}$ to $I_{i,j}$ (this is well-defined since these two sets both
have cardinality $a_{ij}$). The disjoint union of these bijections $\pi_{i,j}$
over all $i$ and $j$ is a bijection $[n]\to [n]$ (since the disjoint union
of the sets $J_{j,i}$ over all $i$ and $j$ is $[n]$, and so is the disjoint
union of the sets $I_{i,j}$), that is, a permutation of $[n]$; this
permutation is what we call $w_A$.
%[Old version:]
%Then let $w_A$ map a consecutive string of $a_{ij}$ of the letters of
%$J_j$ into a consecutive string of $a_{ij}$ of the letters of $I_i$,
%in an increasing fashion.
\end{definition}

% [DG] Shortened my suggestion and still clumsy :(

% [DG][v80] Detailed the "disjointly decompose" a bit.

\begin{example}
Taking
$n=9$ and $\alpha=(4,5), \beta=(3,4,2)$, one has
\[
\begin{array}{ccc}
I_1=\{1,2,3,4\},&I_2=\{5,6,7,8,9\},& \\
J_1=\{1,2,3\}, &J_2=\{4,5,6,7\},&J_3=\{8,9\}.
\end{array}
\]
Then one possible matrix $A$ having row and column sums $\alpha, \beta$ is
$
A=\left[
\begin{matrix}
2 & 2 & 0\\
1 & 2 & 2
\end{matrix}
\right],
$
and its associated permutation $w_A$ written in two-line notation is
\[
\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{1} & \underline{2} & \underline{\underline{5}}&|
&\underline{3} & \underline{4} & \underline{\underline{6}}
&\underline{\underline{7}} &|
&\underline{\underline{8}}
&\underline{\underline{9}}
\end{matrix}
\right)
\]
with vertical lines dividing the sets $J_j$ on top, and
with elements of $I_i$ underlined $i$ times on the bottom.
\end{example}

\begin{remark}
Given almost-compositions $\alpha$ and $\beta$ of $n$
having lengths $\ell$ and $m$,
and a permutation $w \in \Symm_n$. It is easy to see that there exists a
matrix $A \in \NN^{\ell\times m}$ satisfying $w_A = w$ if and only if
the restriction of $w$ to each $J_j$ and the restriction of $w^{-1}$ to
each $I_i$ are increasing. In this case, the matrix $A$ is determined by
$a_{ij} = \left| w(J_j) \cap I_i \right|$.
\end{remark}

% [DG] I added this remark since it is (tacitly) used in the proof of the
% next proposition.

Among our three towers $G_*$ of groups, the symmetric group tower
($G_n = \Symm_n$) is the simplest one. We will now see that it also embeds
into the two others, in the sense that $\Symm_n$ embeds into $\Symm_n[\Gamma]$
for every $\Gamma$ and into $GL_n(\FF_q)$ for every $q$.

First, for every $n\in\NN$ and any group $\Gamma$, we embed
the group $\Symm_n$ into $\Symm_n[\Gamma]$ by means of the canonical embedding
$\Symm_n\rightarrow\Symm_n\ltimes\Gamma^{n}=\Symm_n[\Gamma] $.
If we regard elements of $\Symm_n[\Gamma] $ as $n\times n$ monomial matrices
with nonzero entries in $\Gamma$, then this boils down to identifying every
$\pi\in\Symm_n$ with the permutation matrix of $\pi$ (in which the $1$'s are
read as the neutral element of $\Gamma$). If $\alpha$ is an
almost-composition of $n$,
then this embedding $\Symm_n \rightarrow \Symm_n\left[  \Gamma\right]$
makes the subgroup $\Symm_{\alpha}$ of $\Symm_n$ become a subgroup of
$\Symm_n\left[ \Gamma\right]$, more precisely a subgroup of
$\Symm_{\alpha} \left[ \Gamma\right] < \Symm_n \left[ \Gamma\right]$.

For every $n\in\NN$ and every $q$, we embed the group $\Symm_n$ into
$GL_n \left(  \FF_{q}\right)  $ by identifying every permutation
$\pi \in \Symm_n$ with its permutation matrix in $GL_n \left(
\FF_{q}\right)  $. If $\alpha$ is an
almost-composition of $n$, then this embedding
makes the subgroup $\Symm_{\alpha}$ of $\Symm_n$ become a subgroup of
$GL_n \left(  \FF_{q}\right) $. If we let $G_n = GL_n \left(
\FF_{q}\right)  $, then $\Symm_{\alpha} < G_{\alpha} < P_{\alpha}$.

The embeddings we have just defined commute with the group embeddings
$G_n < G_{n+1}$ on both sides.

% [DG] Shortened my definitions; still too tedious...

\begin{proposition}
\label{double-coset-prop}
The permutations $\{w_A\}$,
as $A$ runs over all matrices in $\NN^{\ell\times m}$
having row sums $\alpha$ and column sums $\beta$,
give
\begin{itemize}
\item[(a)] a system of double coset representatives
           for $\Symm_\alpha \backslash \Symm_n / \Symm_\beta$;
\item[(b)] a system of double coset representatives
           for $\Symm_\alpha[\Gamma] \backslash \Symm_n[\Gamma] / \Symm_\beta[\Gamma]$;
\item[(c)] a system of double coset representatives
           for $P_\alpha \backslash GL_n / P_\beta$.
\end{itemize}
\end{proposition}

% [DG][v75] Broke this proposition up into (a), (b), (c).

\begin{proof}
(a) We give an algorithm to show
that every double coset $\Symm_\alpha w \Symm_\beta$
contains some $w_A$.  Start by altering
$w$ within its coset $w\Symm_\beta$, that is, by permuting the
\emph{positions} within each set $J_j$,
to obtain a representative $w'$ for $w\Symm_\beta$ in
which each set $w'(J_j)$ appears in increasing order in the second line of the
two-line notation for $w'$.  Then alter $w'$ within its coset
$\Symm_\alpha w'$, that is, by permuting the \emph{values} within each set
$I_i$, to obtain a representative
$w_A$ having the elements of each set $I_i$ appearing
in increasing order in the second line;  because the values within
each set $I_i$ are consecutive, this alteration will not ruin the property
that one had each set $w'(J_j)$ appearing in increasing order.
For example, one might have
\begin{align*}
w&=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{4} & \underline{\underline{8}} & \underline{2}&|
&\underline{\underline{5}} & \underline{3} & \underline{\underline{9}}
&\underline{1} &|
&\underline{\underline{7}}
&\underline{\underline{6}}
\end{matrix}
\right) ,\\
w'&=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{2} & \underline{4} & \underline{\underline{8}}&|
&\underline{1} & \underline{3} & \underline{\underline{5}}
&\underline{\underline{9}} &|
&\underline{\underline{6}}
&\underline{\underline{7}}
\end{matrix}
\right) \in w \Symm_\beta ,\\
w_A&=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{1} & \underline{2} & \underline{\underline{5}}&|
&\underline{3} & \underline{4} & \underline{\underline{6}}
&\underline{\underline{7}} &|
&\underline{\underline{8}}
&\underline{\underline{9}}
\end{matrix}
\right) \in \Symm_\alpha w' \subset \Symm_\alpha w' \Symm_\beta = \Symm_\alpha w \Symm_\beta .\\
\end{align*}

Next note that
$\Symm_\alpha w_A \Symm_\beta=\Symm_\alpha w_B \Symm_\beta$ implies
$A=B$, since the quantities
\[
a_{i,j}(w):=|w(J_j) \cap I_i|
\]
are easily seen to be constant on double cosets
$\Symm_\alpha w \Symm_\beta$.

(b) Double coset representatives for
$\Symm_\alpha \backslash \Symm_n / \Symm_\beta$ should also provide
double coset representatives for
$\Symm_\alpha[\Gamma] \backslash \Symm_n[\Gamma] / \Symm_\beta[\Gamma]$,
since
\[
\Symm_\alpha[\Gamma]=\Symm_\alpha  \Gamma^n=\Gamma^n \Symm_\alpha.
\]
Thus, part (b) follows from part (a).

(c) In our proof of part (a) above, we showed that
$\Symm_\alpha w_A \Symm_\beta=\Symm_\alpha w_B \Symm_\beta$ implies
$A=B$.
A similar argument shows that
$P_\alpha w_A P_\beta=P_\alpha w_B P_\beta$ implies
$A=B$:  for $g$ in $GL_n$,
the rank $r_{ij}(g)$ of the matrix obtained by restricting $g$ to
rows $I_i \sqcup I_{i+1} \sqcup \cdots \sqcup I_\ell$
and columns $J_1 \sqcup J_2 \sqcup \cdots \sqcup J_j$ is
constant on double cosets $P_\alpha g P_\beta$, and
for a permutation matrix $w$ one can recover $a_{i,j}(w)$ from the formula
\[
a_{i,j}(w) =
r_{i,j}(w)-r_{i,j-1}(w)-r_{i+1,j}(w) + r_{i+1,j-1}(w).
\]

Thus it only remains to show that every double coset $P_\alpha g P_\beta$ contains
some $w_A$.  Since $\Symm_\alpha < P_\alpha$, and we have seen already that
every double coset $\Symm_\alpha w \Symm_\beta$ contains some $w_A$,
it suffices to show that every double coset
$P_\alpha g P_\beta$ contains some permutation $w$.  However, we claim that
this is  already true for the smaller double cosets $BgB$ where $B=P_{1^n}$ is
the \emph{Borel subgroup}\index{Borel subgroup of $GL_n$}
of upper triangular invertible matrices, that is,
one has the usual \emph{Bruhat decomposition}\index{Bruhat decomposition of $GL_n$}
\[
GL_n = \bigsqcup_{w \in \Symm_n} BwB.
\]
To prove this decomposition, we show how to
find a permutation $w$ in each double coset $BgB$.
The freedom to alter $g$ within its coset $gB$ allows one to scale columns and
add scalar multiples of earlier columns to later columns.  We claim that
using such column operations, one can
always find a representative $g'$ for coset $gB$ in which
\begin{enumerate}
\item[$\bullet$]
the bottommost nonzero entry of each column is $1$ (call this entry a \emph{pivot}),
\item[$\bullet$]
the entries to right of each pivot within its row are all $0$, and
\item[$\bullet$]
there is one pivot in each row and each column, so that
their positions are the positions of the $1$'s in some
permutation matrix $w$.
% [DG][v49] Replaced the line below by the two lines above.
% they lie in the positions of some permutation matrix $w$.
\end{enumerate}
In fact, we will see below that $BgB=BwB$ in this case.
The algorithm which produces $g'$ from $g$ is simple:  starting with the
leftmost column, find its bottommost nonzero entry, and scale the column
to make this entry a $1$, creating the pivot in this column.  Now use this
pivot to clear out all entries in its row to its right, using
column operations that subtract multiples of this column from later columns.
Having done this, move on to the next column to the right, and repeat, scaling to create a pivot, and using it to eliminate entries to its right.\footnote{To see that this works, we need to check three facts:
\begin{itemize}
\item[\textbf{(a)}] We will find a nonzero entry in every column during our
algorithm.
\item[\textbf{(b)}] Our column operations preserve
the zeroes lying to the right of already existing pivots.
\item[\textbf{(c)}] Every row contains exactly one pivot at the end
of the algorithm.
\end{itemize}
But fact \textbf{(a)} simply says that our matrix can never have an
all-zero column during the algorithm; this is clear (since the rank of
the matrix remains constant during the algorithm and was $n$ at its
beginning). Fact \textbf{(b)} holds because all our operations either
scale columns (which clearly preserves zero entries) or subtract a
multiple of the column $c$ containing the current pivot from a later
column $d$ (which will preserve every zero lying to the right of an
already existing pivot, because any already existing pivot must lie
in a column $b < c$ and therefore both
columns $c$ and $d$ have zeroes in its row). Fact \textbf{(c)} follows
from noticing that there are $n$ pivots altogether at the end of the
algorithm, but no row can contain two of them (since the entries to
the right of a pivot in its row are $0$).}

% [DG][v34] Added footnote. Replaced "has been scaled to $1$" by
% "is $1$" in the properties of $g'$ because we are describing $g'$
% there, not the algorithm that gives us $g'$.

For example, the typical matrix $g$ lying in the double coset $BwB$ where
\[
w=\left(
\begin{matrix}
1 & 2 & 3 &|& 4 & 5 & 6 & 7&|& 8 & 9 \\
\underline{4} & \underline{\underline{8}} & \underline{2}&|
&\underline{\underline{5}} & \underline{3} & \underline{\underline{9}}
&\underline{1} &|
&\underline{\underline{7}}
&\underline{\underline{6}}
\end{matrix}
\right)\\
\]
from before is one that can be altered within its coset $gB$ to look like this:
\[
g'=\left[
\begin{matrix}
* & * & * & * & * & * & 1 & 0 & 0 \\
* & * & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
* & * & 0 & * & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & * & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & * & 0 & 0 & 0 & * & 0 & * & 1 \\
0 & * & 0 & 0 & 0 & * & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{matrix}
\right] \in gB.
\]
Having found this $g'$ in $gB$, a similar algorithm using left multiplication
by $B$ shows that $w$ lies in
$Bg' \subset Bg'B=BgB$.  This time no scalings are required
to create the pivot entries:  starting with the bottom row, one
uses its pivot to eliminate all the entries above it in the same column
(shown by stars $*$ above)
by adding multiples of the bottom row to higher rows. Then do the same using the pivot in the next-to-bottom row, etc.  The result is the permutation matrix for $w$.
\end{proof}

\begin{remark}
The Bruhat decomposition $GL_n = \bigsqcup_{w \in \Symm_n} BwB$ is
related to the so-called \emph{LPU factorization} -- one of a
myriad of matrix factorizations appearing in linear
algebra.\footnote{Specifically, an \dfn{LPU factorization} of a
matrix $A \in GL_n(\FF)$ (for an arbitrary field $\FF$)
means a way to write $A$ as a product
$A = LPU$ with $L \in GL_n(\FF)$ being lower-triangular,
$U \in GL_n(\FF)$ being upper-triangular, and
$P \in \Symm_n \subset GL_n(\FF)$ being a permutation
matrix. Such a factorization always exists (although it is generally
not unique).
This can be derived from the Bruhat decomposition (see
Exercise~\ref{exe.bruhat.variations}(a) for a proof).
See also \cite{Strang-elim} for related discussion.}
It is actually a fairly general phenomenon, and requires neither the
finiteness of $\FF$, nor the invertibility, nor even the squareness
of the matrices (see Exercise~\ref{exe.bruhat.variations}(b) for an
analogue holding in a more general setup).
\end{remark}

\begin{exercise}
\label{exe.bruhat.variations}
Let $\FF$ be any field.

\begin{enumerate}
\item[(a)]
For any $n \in \NN$ and any $A \in GL_n(\FF)$, prove that there exist
a lower-triangular matrix $L \in GL_n(\FF)$, an upper-triangular
matrix $U \in GL_n(\FF)$ and a permutation matrix
$P \in \Symm_n \subset GL_n(\FF)$ (here, we identify permutations with
the corresponding permutation matrices) such that $A = LPU$.

\item[(b)]
Let $n \in \NN$ and $m \in \NN$. Let $F_{n, m}$ denote the set of all
$n \times m$-matrices $B \in \left\{0, 1\right\}^{n \times m}$ such that
each row of $B$ contains at most one $1$ and each column of $B$
contains at most one $1$. We regard $F_{n, m}$ as a subset of
$\FF^{n \times m}$ by means of regarding $\left\{0, 1\right\}$ as a
subset of $\FF$.

For every $k \in \NN$, we let $B_k$ denote the subgroup of
$GL_k(\FF)$ consisting of all upper-triangular matrices.

Prove that
\[
\FF^{n \times m}
= \bigsqcup_{f \in F_{n, m}} B_n f B_m .
\]
\end{enumerate}
\end{exercise}

% [DG][v49] Added the above remark and exercise (a side effect of TAing
% a Linear Algebra class). I have left out any comments on Bruhat
% decompositions for other algebraic groups, as that is beyond my
% paycheck (actually, probably not -- but beyond my competence).

\begin{corollary}
\label{three-towers-give-bialgebras}
For each of the three towers of groups $G_*$, the product and coproduct
structures on $A=A(G_*)$ endow it with a bialgebra structure, and hence
they form PSH's.
\end{corollary}
\begin{proof}
The first two towers $G_n = \Symm_n$ and $G_n=\Symm_n[\Gamma]$ have
product, coproduct defined by induction, restriction along
embeddings $G_i \times G_j < G_{i+j}$.  Hence the desired bialgebra equality
\eqref{bialgebra-as-Mackey} follows from Mackey's Theorem~\ref{Mackey-formula},
taking
$
G=G_n,
H=G_{(r_1,r_2)},
K=G_{(c_1,c_2)},
U=U_1\otimes U_2
$
with double coset representatives\footnote{Proposition~\ref{double-coset-prop} gives
as a system of double coset representatives for $G_{(c_1,c_2)} \backslash G_n / G_{(r_1,r_2)}$
the elements
\begin{align*}
&\left\{w_A \ : \  A\in\NN^{2\times 2}, \ A\text{ has row sums }
(c_1,c_2)\text{ and column sums }(r_1,r_2)\right\} \\
&= \left\{w_{A^t} \ : \  A\in\NN^{2\times 2}, \ A\text{ has row sums }
(r_1,r_2)\text{ and column sums }(c_1,c_2)\right\}
\end{align*}
where $A^t$ denotes the transpose matrix of $A$.}
\[
\{g_1,\ldots,g_t\} = \left\{w_{A^t} \ : \  A\in\NN^{2\times 2}, \ A\text{ has row sums }
(r_1,r_2)\text{ and column sums }(c_1,c_2)\right\}
\]
and checking for a given double coset
\[
KgH=(G_{c_1,c_2})w_{A^t}(G_{r_1,r_2})
\]
indexed by a matrix $A$ in $\NN^{2 \times 2}$ with row sums $(r_1,r_2)$ and
column sums $(c_1,c_2)$, that the two subgroups appearing on the left in
\eqref{two-by-two-matrix-Mackey-subgroups} are exactly
\begin{align*}
H \cap K^{w_{A^t}} &= G_{r_1,r_2} \cap (G_{c_1,c_2})^{w_{A^t}}, \\
{}^{w_{A^t}} H \cap K &= {}^{w_{A^t}}(G_{r_1,r_2}) \cap G_{c_1,c_2},
\end{align*}
respectively.
One should also apply \eqref{induction-vs-tensor-product} and check that
the isomorphism $\tau_A$ between the two subgroups
in \eqref{two-by-two-matrix-Mackey-subgroups} is the conjugation isomorphism
by $w_{A^t}$ (that is, $\tau_A(g) = w_{A^t} g w_{A^t}^{-1}$ for every
$g \in H \cap K^{w_{A^t}}$).  We leave all of these bookkeeping details to the reader to check.
\footnote{It helps to recognize $w_{A^t}$ as the permutation written in
two-line notation as
\[
\left(
\begin{matrix}
1 & 2 & \ldots & a_{11} &|& a_{11}+1 & a_{11}+2 & \ldots & r_1 &|& r_1+1 & r_1+2 & \ldots & a'_{22} &|& a'_{22}+1 & a'_{22}+2 & \ldots & n \\
1 & 2 & \ldots & a_{11} &|& c_1+1 & c_1+2 & \ldots & a'_{22} &|& a_{11}+1 & a_{11}+2 & \ldots & c_1 &|& a'_{22}+1 & a'_{22}+2 & \ldots & n
\end{matrix}
\right),
\]
where $a'_{22} = r_1 + a_{21} = c_1 + a_{12} = n - a_{22}$. In matrix form,
$w_{A^t}$ is the block matrix
$\left[
\begin{array}{cccc}
I_{a_{11}} & 0 & 0 & 0 \\
0 & 0 & I_{a_{21}} & 0 \\
0 & I_{a_{12}} & 0 & 0 \\
0 & 0 & 0 & I_{a_{22}}
\end{array}
\right]$.
}

% [DG] The two footnotes are probably way on the pedantic side. I wrote them in
% order to keep track of my computations, but I'm not sure if the reader will
% have any need for them.

For the tower with $G_n=GL_n$, there is slightly more work to be done
to check the equality \eqref{bialgebra-as-Mackey}.
Via Mackey's Theorem~\ref{Mackey-formula} and
Proposition~\ref{double-coset-prop}(c), the left side is
\begin{align}
\nonumber
&\res_{c_1,c_2}^n\left(  \ind_{r_1,r_2}^n \left( U_1 \otimes U_2 \right) \right) \\
\nonumber
&= \left(
 \Res^{G_n}_{P_{c_1,c_2}} \Ind^{G_n}_{P_{r_1,r_2}} \Inf^{P_{r_1,r_2}}_{G_{r_1,r_2}} \left( U_1 \otimes U_2 \right)
\right)^{K_{c_1,c_2}} \\
&= \bigoplus_A
\left(
 \Ind^{P_{c_1,c_2}}_{{}^{w_{A^t}} P_{r_1,r_2} \cap P_{c_1,c_2}}
   \left(\left( \Res^{P_{r_1,r_2}}_{P_{r_1,r_2} \cap P_{c_1,c_2}^{w_{A^t}}}
      \Inf^{P_{r_1,r_2}}_{G_{r_1,r_2}} \left( U_1 \otimes U_2 \right)
   \right)^{\tau_A^{-1}}\right)
\right)^{K_{c_1,c_2}}
\label{left-side-GL-bialgebra}
\end{align}
where $A$ runs over the usual $2 \times 2$ matrices.
The right side is a direct sum over this same set of matrices $A$:
\begin{align}
\nonumber
&\bigoplus_A
  \left( \ind_{a_{11},a_{21}}^{c_1} \otimes \ind^{c_2}_{a_{12},a_{22}}\right)
   \left( \left( \res^{r_1}_{a_{11},a_{12}} U_1 \otimes \res^{r_2}_{a_{21},a_{22}} U_2 \right)^{\tau_A^{-1}} \right) \\
\nonumber
&= \bigoplus_A
 \left(\Ind^{G_{c_1}}_{P_{a_{11},a_{21}}} \otimes \Ind^{G_{c_2}}_{P_{a_{12},a_{22}}}\right) \circ
  \left(\Inf^{P_{a_{11},a_{21}}}_{G_{a_{11},a_{21}}} \otimes \Inf^{P_{a_{12},a_{22}}}_{G_{a_{12},a_{22}}}\right) \\
\nonumber
&\qquad \qquad
    \left(
     \left(
      \left(\Res^{G_{r_1}}_{P_{a_{11},a_{12}}} U_1\right)^{K_{a_{11},a_{12}}}
      \otimes \left(\Res^{G_{r_2}}_{P_{a_{21},a_{22}}} U_2\right)^{K_{a_{21},a_{22}}}
     \right)^{\tau_A^{-1}} \right)\\
\nonumber
&= \bigoplus_A
 \Ind^{G_{c_1,c_2}}_{P_{a_{11},a_{21}} \times P_{a_{12},a_{22}}}  \\
&\qquad \qquad
  \Inf^{P_{a_{11},a_{21}} \times P_{a_{12},a_{22}}}_{G_{a_{11},a_{21},a_{12},a_{22}}}
    \left(
     \left(\left(
      \Res^{G_{r_1,r_2}}_{P_{a_{11},a_{12}} \times P_{a_{21},a_{22}}}  \left( U_1 \otimes U_2 \right)
     \right)^{K_{a_{11},a_{12}} \times K_{a_{21},a_{22}}}
    \right)^{\tau_A^{-1}}\right)
\label{right-side-GL-bialgebra}
\end{align}
(by \eqref{induction-vs-tensor-product}, \eqref{invariants-vs-tensor-product} and their
obvious analogues for restriction and inflation).
Thus it suffices to check for each $2 \times 2$ matrix $A$
that any $\CC G_{c_1,c_2}$-module of the
form $V_1 \otimes V_2$ has the same inner product with the $A$-summands of
\eqref{left-side-GL-bialgebra} and \eqref{right-side-GL-bialgebra}.
Abbreviate $w:=w_{A^t}$ and $\tau:=\tau_A^{-1}$.  

Notice that ${}^w P_{r_1,r_2}$ is the
group of all matrices having the block form
\begin{equation}
\left[
\begin{matrix}
g_{11} & h    & i & j \\
0     &g_{21} & 0 & k \\
d     & e    & g_{12} & \ell \\
0     & f    & 0 & g_{22}
\end{matrix}
\right]
\end{equation}
in which the diagonal blocks $g_{ij}$ for $i,j=1,2$ are invertible of
size $a_{ij} \times a_{ij}$,
while the blocks $h,i,j,k,\ell,d,e,f$ are all arbitrary matrices\footnote{The
blocks $i$ and $j$ have nothing to do with the indices $i,j$ in $g_{ij}$.}
of the appropriate (rectangular) block sizes.
Hence, ${}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}$ is
the group of all matrices having the block form
\begin{equation}
\label{GL-bialgebra-matrixform-1}
\left[
\begin{matrix}
g_{11} & h    & i & j \\
0     &g_{21} & 0 & k \\
0     & 0    & g_{12} & \ell \\
0     & 0    & 0 & g_{22}
\end{matrix}
\right]
\end{equation}
in which the diagonal blocks $g_{ij}$ for $i,j=1,2$ are invertible of
size $a_{ij} \times a_{ij}$,
while the blocks $h,i,j,k,\ell$ are all arbitrary matrices
of the appropriate (rectangular) block sizes;  then
${}^{w}P_{r_1,r_2} \cap G_{c_1,c_2}$ is the subgroup where the blocks $i,j,k$
all vanish. The canonical projection
${}^{w}P_{r_1,r_2} \cap P_{c_1,c_2} \to {}^{w}P_{r_1,r_2} \cap G_{c_1,c_2}$
(obtained by restricting the projection $P_{c_1,c_2} \to G_{c_1,c_2}$)
has kernel ${}^{w}P_{r_1,r_2} \cap P_{c_1,c_2} \cap K_{c_1,c_2}$.
Consequently,
\begin{equation}
\label{GL-bialgebra-quotientgroup-1}
\left({}^wP_{r_1,r_2} \cap P_{c_1,c_2}\right)
/ \left({}^wP_{r_1,r_2} \cap P_{c_1,c_2} \cap K_{c_1,c_2}\right)
= {}^wP_{r_1,r_2} \cap G_{c_1,c_2}.
\end{equation}

Similarly,
\begin{equation}
\label{GL-bialgebra-quotientgroup-2}
\left(P_{r_1,r_2} \cap P_{c_1,c_2}^{w}\right)
/ \left(P_{r_1,r_2} \cap P_{c_1,c_2}^{w} \cap K_{r_1,r_2}\right)
= G_{r_1,r_2} \cap P_{c_1,c_2}^{w}.
\end{equation}

Computing first the inner product
of $V_1 \otimes V_2$ with the $A$-summand of
\eqref{left-side-GL-bialgebra}, and using adjointness properties, one gets
\begin{align*}
&
\left(
\left( \Res^{P_{r_1,r_2}}_{P_{r_1,r_2} \cap P_{c_1,c_2}^{w}}
      \Inf^{P_{r_1,r_2}}_{G_{r_1,r_2}} \left( U_1 \otimes U_2 \right)
   \right)^{\tau} , \right. \\
& \qquad \qquad \qquad \left.
\Res^{P_{c_1,c_2}}_{{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}}
      \Inf^{P_{c_1,c_2}}_{G_{c_1,c_2}} \left( V_1 \otimes V_2 \right)
\right)_{{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}} \\
&\overset{\eqref{restriction-inflation-commutation}}{=}
\left(
\left(
  \Inf^{P_{r_1,r_2} \cap P_{c_1,c_2}^{w}}_{G_{r_1,r_2} \cap P_{c_1,c_2}^{w}}
   \Res^{G_{r_1,r_2}}_{G_{r_1,r_2} \cap P_{c_1,c_2}^{w}} \left( U_1 \otimes U_2 \right)
   \right)^{\tau} , \right. \\
& \qquad \qquad \qquad \left.
  \Inf^{{}^wP_{r_1,r_2} \cap P_{c_1,c_2}}_{{}^wP_{r_1,r_2} \cap G_{c_1,c_2}}
   \Res^{G_{c_1,c_2}}_{{}^wP_{r_1,r_2} \cap G_{c_1,c_2}} \left( V_1 \otimes V_2 \right)
\right)_{{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}}
\end{align*}
(by \eqref{GL-bialgebra-quotientgroup-2} and \eqref{GL-bialgebra-quotientgroup-1}).
One can compute this inner product
by first recalling that ${}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}$ is
the group of matrices having the block form \eqref{GL-bialgebra-matrixform-1}
%\[
%\left[
%\begin{matrix}
%g_{11} & h    & i & j \\
%0     &g_{21} & 0 & k \\
%0     & 0    & g_{12} & \ell \\
%0     & 0    & 0 & g_{22}
%\end{matrix}
%\right]
%\]
in which the diagonal blocks $g_{ij}$ for $i,j=1,2$ are invertible of
size $a_{ij} \times a_{ij}$,
while the blocks $h,i,j,k,\ell$ are all arbitrary matrices
of the appropriate (rectangular) block sizes;  then ${}^{w}P_{r_1,r_2} \cap G_{c_1,c_2}$ is the subgroup where the blocks $i,j,k$ all vanish.  The inner product above then becomes
\begin{equation}
\label{first-big-inner-product-sum}
\begin{aligned}
&\frac{1}{|{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}|}
\sum\limits_{\substack{(g_{ij})\\(h,i,j,k,\ell)}}
  \chi_{U_1}\left(\begin{matrix} g_{11}& i \\ 0 & g_{12} \end{matrix}\right)
  \chi_{U_2}\left(\begin{matrix} g_{21}& k \\ 0 & g_{22} \end{matrix}\right) \\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad
  \overline{\chi}_{V_1}\left(\begin{matrix} g_{11}& h \\ 0 & g_{21} \end{matrix}\right)
  \overline{\chi}_{V_2}\left(\begin{matrix} g_{12}& \ell \\ 0 & g_{22} \end{matrix}\right).
\end{aligned}
\end{equation}
If one instead computes the inner product of $V_1 \otimes V_2$ with the $A$-summand of
\eqref{right-side-GL-bialgebra}, using adjointness properties and \eqref{deflation-character-formula-v2} one gets
\begin{align*}
&\left(
\left(
 \left(
  \Res^{G_{r_1,r_2}}_{P_{a_{11},a_{12}} \times P_{a_{21},a_{22}}} \left( U_1 \otimes U_2 \right)
 \right)^{K_{a_{11},a_{12}} \times K_{a_{21},a_{22}}}
\right)^{\tau} , \right.\\
&\qquad\qquad
 \left.
 \left(
  \Res^{G_{c_1,c_2}}_{P_{a_{11},a_{21}} \times P_{a_{12},a_{22}}} \left( V_1 \otimes V_2 \right)
 \right)^{K_{a_{11},a_{21}} \times K_{a_{12},a_{22}}}
\right)_{G_{a_{11},a_{21},a_{12},a_{22}}} \\
&=\frac{1}{|G_{a_{11},a_{21},a_{12},a_{22}}|}
   \sum_{(g_{ij})}
     \frac{1}{|K_{a_{11},a_{12}} \times K_{a_{21},a_{22}}|}
       \sum_{(i,k)}
          \chi_{U_1}\left(\begin{matrix} g_{11}& i \\ 0 & g_{12} \end{matrix}\right)
          \chi_{U_2}\left(\begin{matrix} g_{21}& k \\ 0 & g_{22} \end{matrix}\right)\\
&\qquad \qquad \qquad \qquad \qquad
     \frac{1}{|K_{a_{11},a_{21}} \times K_{a_{12},a_{22}}|} \sum_{(h,\ell)}
  \overline{\chi}_{V_1}\left(\begin{matrix} g_{11}& h \\ 0 & g_{21} \end{matrix}\right)
  \overline{\chi}_{V_2}\left(\begin{matrix} g_{12}& \ell \\ 0 & g_{22} \end{matrix}\right).
\end{align*}
But this right hand side can be seen to equal
\eqref{first-big-inner-product-sum}, after one notes that
\[
|{}^{w}P_{r_1,r_2} \cap P_{c_1,c_2}|
=
|G_{a_{11},a_{21},a_{12},a_{22}}|
\cdot
|K_{a_{11},a_{12}} \times K_{a_{21},a_{22}}|
\cdot
|K_{a_{11},a_{21}} \times K_{a_{12},a_{22}}|
\cdot \#\{j \in \FF_q^{a_{11} \times a_{22}} \}
\]
and that the summands in  \eqref{first-big-inner-product-sum} are independent of the
matrix $j$ in the summation.
\end{proof}

We can also define a $\CC$-vector space $A_\CC$ as the direct sum
$\bigoplus_{n \geq 0} R_\CC(G_n)$.
In the same way as we have made $A = \bigoplus_{n \geq 0} R(G_n)$ into
a $\ZZ$-bialgebra, we can turn
$A_\CC = \bigoplus_{n \geq 0} R_\CC(G_n)$ into
a $\CC$-bialgebra\footnote{The definitions of $m$ and $\Delta$ for
this $\CC$-bialgebra look the same as for $A$: For instance, $m$ is
still defined to be $\ind^{i+j}_{i,j}$ on $\left(A_\CC\right)_i
\otimes \left(A_\CC\right)_j$, where
$\ind^{i+j}_{i,j}$ is defined by the same formulas as in
Definition~\ref{ind-res-defn}. However, the operators of induction,
restriction, inflation and $K$-fixed space construction appearing in
these formulas now act on class functions as opposed to modules.
\par
The fact that these maps $m$ and $\Delta$ satisfy the axioms of a
$\CC$-bialgebra is easy to check: they are merely the $\CC$-linear
extensions of the maps $m$ and $\Delta$ of the $\ZZ$-bialgebra
$A$ (this is because, for instance, induction of class functions
and induction of modules are related by the identity
\eqref{eq.induction.mods-and-class-functs}), and thus satisfy the
same axioms as the latter.}.
There is a $\CC$-bilinear form $\left(\cdot,\cdot\right)_{A_\CC}$
on $A_\CC$ which can be defined either as the $\CC$-bilinear
extension of the $\ZZ$-bilinear form
$\left(\cdot,\cdot\right)_A : A \times A \to \ZZ$ to $A_\CC$, or
(equivalently) as the $\CC$-bilinear form on $A_\CC$ which
restricts to
$\left<\cdot,\cdot\right>_{\Symm_n}$ on every homogeneous component
$R_\CC(G_n)$ and makes different homogeneous components mutually
orthogonal.
The obvious embedding of $A$ into the $\CC$-bialgebra $A_\CC$ (obtained
from the embeddings $R(G_n) \to R_\CC(G_n)$ for all $n$) respects the
bialgebra operations\footnote{This is because, for example, induction
of class functions harmonizes with induction of modules (i.e.,
the equality \eqref{eq.induction.mods-and-class-functs} holds).}, and
the $\CC$-bialgebra $A_\CC$ can be identified
with $A \otimes_\ZZ \CC$ (the result of extending scalars to $\CC$ in
$A$), because every finite group $G$ satisfies
$R_\CC (G) \cong R(G) \otimes_\ZZ \CC$.
The embedding of $A$ into $A_\CC$ also respects the bilinear
forms.

% [DG][v46] Added above short paragraph. Not sure if it makes anything
% clearer, but I felt like it should be here to define some things
% used later (hopefully not earlier).

\begin{exercise}
\label{exe.towers.indres}
Let $G_*$ be one of the three towers.

For every almost-composition
$\alpha = \left(\alpha_1,\alpha_2,\ldots,\alpha_\ell\right)$ of
$n \in \NN$, let us define a map \dfn{$\ind^n_\alpha$} which takes
$\CC G_\alpha$-modules to $\CC G_n$-modules as follows:
If $G_* = \Symm_*$ or $G_* = \Symm_*\left[\Gamma\right]$, we
set
\[
\ind^n_\alpha := \Ind^{G_n}_{G_\alpha}.
\]
If $G_* = GL_*$, then we set
\[
\ind^n_\alpha := \Ind^{G_n}_{P_\alpha} \Inf^{P_\alpha}_{G_\alpha}.
\]
(Note that $\ind^n_\alpha = \ind^n_{i,j}$ if $\alpha$ has the
form $\left(i, j\right)$.)

Similarly, for every almost-composition
$\alpha = \left(\alpha_1,\alpha_2,\ldots,\alpha_\ell\right)$ of
$n \in \NN$, let us define a map \dfn{$\res^n_\alpha$} which takes
$\CC G_n$-modules to $\CC G_\alpha$-modules as follows:
If $G_* = \Symm_*$ or $G_* = \Symm_*\left[\Gamma\right]$, we
set
\[
\res^n_\alpha := \Res^{G_n}_{G_\alpha}.
\]
If $G_* = GL_*$, then we set
\[
\res^n_\alpha := \left(\Res^{G_n}_{P_\alpha} \left( - \right)\right)^{K_\alpha}.
\]
(Note that $\res^n_\alpha = \res^n_{i,j}$ if $\alpha$ has the
form $\left(i, j\right)$.)

\begin{itemize}

\item[(a)] If
$\alpha = \left(\alpha_1,\alpha_2,\ldots,\alpha_\ell\right)$
is an almost-composition of an integer $n \in \NN$ satisfying
$\ell \geq 1$, and if $V_i$ is a
$\CC G_{\alpha_i}$-module for every $i \in \left\{1,2,\ldots,\ell\right\}$,
then show that
\begin{align*}
& \ind^n_{\alpha_1+\alpha_2+\cdots+\alpha_{\ell-1}, \alpha_\ell}
\left(
 \ind^{\alpha_1+\alpha_2+\cdots+\alpha_{\ell-1}}_{\left(\alpha_1,\alpha_2,\ldots,\alpha_{\ell-1}\right)}
  \left(V_1 \otimes V_2 \otimes \cdots \otimes V_{\ell-1}\right)
 \otimes V_\ell
\right) \\
&\cong \ind^n_{\alpha} \left(V_1 \otimes V_2 \otimes \cdots \otimes V_\ell\right) \\
&\cong \ind^n_{\alpha_1, \alpha_2+\alpha_3+\cdots+\alpha_\ell}
\left(
 V_1 \otimes
 \ind^{\alpha_2+\alpha_3+\cdots+\alpha_\ell}_{\left(\alpha_2,\alpha_3,\ldots,\alpha_\ell\right)}
  \left(V_2 \otimes V_3 \otimes \cdots \otimes V_\ell\right)
\right) .
\end{align*}

\item[(b)] Solve Exercise~\ref{exe.Ind-ass} again
using Exercise~\ref{exe.towers.indres}(a).

\item[(c)] We proved above that the map $m : A \otimes A \to A$
(where $A = A\left(G_*\right)$) is associative, by using the
adjointness of $m$ and $\Delta$. Give a new proof of this fact,
which makes no use of $\Delta$.

\item[(d)] If
$\alpha = \left(\alpha_1,\alpha_2,\ldots,\alpha_\ell\right)$
is an almost-composition of an $n \in \NN$, and if
$\chi_i \in R\left(G_{\alpha_i}\right)$ for every
$i \in \left\{1,2,\ldots,\ell\right\}$, then show that
\[
\chi_1 \chi_2 \cdots \chi_\ell
= \ind^n_\alpha \left(\chi_1 \otimes \chi_2 \otimes \cdots \otimes \chi_\ell\right)
\]
in $A = A\left(G_*\right)$.

\item[(e)] If $n \in \NN$, $\ell \in \NN$ and
$\chi \in R\left(G_n\right)$, then show that
\[
\Delta^{\left(\ell-1\right)} \chi
= \sum \res^n_\alpha \chi
\]
in $A^{\otimes \ell}$, where $A = A\left(G_*\right)$. Here,
the sum on the right hand side runs over all
almost-compositions $\alpha$ of $n$ having length $\ell$.
\end{itemize}
\end{exercise}

% [DG][v28] Added this exercise.



\subsection{Symmetric groups}
\label{symmeric-group-section}

Finally, some payoff.
Consider the tower of symmetric groups
$G_n=\Symm_n$, and $A=A(G_*)=:A(\Symm)$\index{$A(\Symm)$}.  Denote by
$\triv_{\Symm_n}, \sgn_{\Symm_n}$ the \emph{trivial} and \emph{sign} characters on
$\Symm_n$.  For a partition $\lambda$ of $n$, denote by
$\triv_{\Symm_\lambda}, \sgn_{\Symm_\lambda}$ the trivial and sign characters
restricted to the Young subgroup
$
\Symm_\lambda = \Symm_{\lambda_1} \times \Symm_{\lambda_2 } \times \cdots,
$
and denote by
$\triv_\lambda$ the class function
which is the characteristic function
for the $\Symm_n$-conjugacy class of permutations of cycle type $\lambda$.



\begin{theorem} \phantomsection
\label{symmetric-group-Frobenius-map-theorem}
\begin{itemize}

\item[(a)]
Irreducible complex characters $\{ \chi^\lambda \}$ of
$\Symm_n$ are indexed by partitions $\lambda$ in $\Par_n$,
and one has a PSH-isomorphism, the
\dfn{Frobenius characteristic map}\index{$\ch$}\footnote{It is unrelated
to the Frobenius endomorphisms from
Exercise~\ref{exe.witt.ghost-app}.},
\[
A=A(\Symm) \overset{\ch}{\longrightarrow} \Lambda
\]
that for $n \geq 0$ and $\lambda \in \Par_n$ sends
\[
\begin{array}{rcl}
\colstrut \triv_{\Symm_n}& \longmapsto & h_n , \\
%\qquad \text{(for }n \geq 0\text{)}\\
\colstrut \sgn_{\Symm_n}& \longmapsto & e_n , \\
%\qquad \text{(for }n \geq 0\text{)}\\
\colstrut \chi^\lambda & \longmapsto & s_\lambda , \\
%\qquad \text{(for }n \geq 0\text{ and }\lambda \in \Par_n\text{)}\\
\colstrut \Ind_{\Symm_\lambda}^{\Symm_n} \triv_{\Symm_\lambda} & \longmapsto &h_\lambda , \\
%\qquad \text{(for }n \geq 0\text{ and }\lambda \in \Par_n\text{)}\\
\colstrut \Ind_{\Symm_\lambda}^{\Symm_n} \sgn_{\Symm_\lambda} & \longmapsto &e_\lambda , \\
%\qquad \text{(for }n \geq 0\text{ and }\lambda \in \Par_n\text{)}\\
\colstrut \triv_\lambda & \longmapsto &\frac{p_\lambda}{z_\lambda} 
\end{array}
\]
(where $\ch$ is extended to a $\CC$-linear map
$A_\CC \to \Lambda_\CC$),
and for $n \geq 1$ sends
\[
\begin{array}{rcl}
\triv_{(n)}& \longmapsto &  \frac{p_n}{n} .
\end{array}
\]
Here, $z_\lambda$ is defined as in
Proposition~\ref{prop-Two-other-Cauchy-expansions}.
% [DG][v70] Inserted \colstruts into the above array to
% prevent the subscripts of one line from overlapping with
% the superscripts of the next.

% [DG][v13] I added the bit about $\ch$ being extended; otherwise, the
% image of $\triv_\lambda$ is not defined (neither $\triv_\lambda$ nor
% $\triv_{(n)}$ generally belong to $A$). Similar change in the proof
% below.

\item[(b)]
For each $n \geq 0$,
the involution on class functions $f: \Symm_n \rightarrow \CC$ 
sending $f \longmapsto \sgn_{\Symm_n} * f$ 
%For every $n \geq 0$, let $\omega_n : R(\Symm_n) \to R(\Symm_n)$ be the involution of the virtual character space $R(\Symm_n)$ given by
where 
\[
(\sgn_{\Symm_n} * f)(g):=\sgn(g) f(g)
\]
%\[
%\left(\omega_n f\right)\left(g\right) = \sgn\left(g\right) f\left(g\right) \qquad \text{for }g\in \Symm_n\text{ and }f\in R(\Symm_n).
%\]
%The direct sum $\bigoplus_{n\geq 0} \omega_n : \bigoplus_{n\geq 0} R(\Symm_n) \to \bigoplus_{n\geq 0} R(\Symm_n) $ of these maps $\omega_n$ is an involution of $\bigoplus_{n\geq 0} R(\Symm_n) = A(\Symm)$. This involution $\bigoplus_{n\geq 0} \omega_n$
preserves the $\ZZ$-sublattice $R(\Symm_n)$ of genuine characters.
The direct sum of these involutions induces an involution 
on $A=A(\Symm) = \bigoplus_{n\geq 0} R(\Symm_n)$
that corresponds under $\ch$ to the involution $\omega$ on $\Lambda$.

\end{itemize}
\end{theorem}

% [DG][v9] I replaced $\sgn_{\Symm_n} \cdot f$ by $\sgn_{\Symm_n} * f$ to
% avoid confusion with the usual product.

\begin{proof}
(a) Corollary~\ref{three-towers-give-bialgebras}
implies that the set $\Sigma=\bigsqcup_{n \geq 0} \Irr(\Symm_n)$
gives a PSH-basis for $A$.
Since a character $\chi$ of $\Symm_n$ has
\begin{equation}
\label{character-coproduct-in-symm}
\Delta(\chi) = \bigoplus_{i+j=n} \Res^{\Symm_n}_{\Symm_i \times \Symm_j} \chi,
\end{equation}
such an element $\chi \in \Sigma \cap A_n$ is never primitive for $n \geq 2$.
Hence the unique irreducible character $\rho=\triv_{\Symm_1}$ of $\Symm_1$
is the only element of $\CCC = \Sigma \cap \Liep$.

% [DG] I again removed $\{\chi\}$.

Thus Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(g)
tells us that there are two PSH-isomorphisms
$A \rightarrow \Lambda$, each of which sends $\Sigma$ to the
PSH-basis of Schur functions $\{s_\lambda\}$ for $\Lambda$.
It also tells us that we can pin down one of the two isomorphisms
to call $\ch$, by insisting that it map the two characters
$\triv_{\Symm_2},\sgn_{\Symm_2}$ in $\Irr(\Symm_2)$ to $h_2,e_2$ (and not
$e_2, h_2$).

Bearing in mind the coproduct formula \eqref{character-coproduct-in-symm},
and the fact that $\triv_{\Symm_n}, \sgn_{\Symm_n}$ restrict, respectively,
to trivial and sign characters of $\Symm_i \times \Symm_j$ for
$i+j=n$, one finds that for $n \geq 2$ one has
$\sgn_{\Symm_2}^\perp$ annihilating $\triv_{\Symm_n}$,
and $\triv_{\Symm_2}^\perp$ annihilating $\sgn_{\Symm_n}$.
Therefore Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(b)
(applied to $\Lambda$) implies
$
\triv_{\Symm_n}, \sgn_{\Symm_n}
$
are sent under $\ch$ to $h_n, e_n$.
Then the fact that
$\Ind_{\Symm_\lambda}^{\Symm_n} \triv_{\Symm_\lambda}, \Ind_{\Symm_\lambda}^{\Symm_n} \sgn_{\Symm_\lambda}$
are sent to $h_{\lambda}, e_\lambda$ follows via induction products.

% [DG][v9] Added "(applied to $\Lambda$)".

Recall that the $\CC$-vector space
$A_\CC = \bigoplus_{n\geq 0} R_\CC (\Symm_n)$ is a
$\CC$-bialgebra, and can be identified with $A \otimes_\ZZ \CC$.
The multiplication and the comultiplication of $A_\CC$ are
$\CC$-linear extensions of those of $A$,
and are still given by the same formulas
$m = \ind^{i+j}_{i,j}$ and $\Delta = \bigoplus_{i+j=n} \res^{i+j}_{i,j}$
as those of $A$ (but now, induction and restriction are defined for
class functions, not just for representations).
The $\CC$-bilinear form $\left(\cdot,\cdot\right)_{A_\CC}$ on $A_\CC$
extends both the $\ZZ$-bilinear form
$\left(\cdot,\cdot\right)_A$ on $A$ and the $\CC$-bilinear forms
$\left<\cdot,\cdot\right>_{\Symm_n}$ on all $R_\CC(\Symm_n)$.

% [DG][v45] Added the preceding pedantic paragraph since we talk
% of $\triv_{(n)}$ as being primitive in $A_\CC$ in the next
% paragraph, thus requiring a Hopf structure on $A_\CC$.

% [DG][v46] Rewrote it, as part of now comes earlier.

For the assertion about $\triv_{(n)}$, note that it is primitive in
$A_\CC$ for $n \geq 1$, because as a class function, the indicator function
of $n$-cycles vanishes upon restriction to $\Symm_i \times \Symm_j$
for $i+j=n$ if both $i,j \geq 1$;  these subgroups contain no $n$-cycles.
Hence Corollary~\ref{primitives-in-Lambda} implies that
$\ch(\triv_{(n)})$ is a scalar multiple of $p_n$.  To pin down the scalar,
note $p_n=m_{(n)}$ so
$
(h_n,p_n)_{\Lambda}=(h_n,m_n)_{\Lambda}=1,
$
while $\ch^{-1}(h_n)=\triv_{\Symm_n}$ has
\[
(\triv_{\Symm_n},\triv_{(n)})=\frac{1}{n!} \cdot (n-1)!=\frac{1}{n}.
\]
\footnote{The first equality sign in this computation uses
the fact that the number of all $n$-cycles in $\Symm_n$ is
$(n-1)!$. This is because any $n$-cycle in $\Symm_n$ can
be uniquely written in the form
$\left(i_1,i_2,\ldots,i_{n-1},n\right)$ (in cycle notation)
with $\left(i_1,i_2,\ldots,i_{n-1}\right)$ being a
permutation in $\Symm_{n-1}$ (written in one-line notation).}
Thus $\ch(\triv_{(n)})=\frac{p_n}{n}$.  The fact that
$\ch(\triv_{\lambda})=\frac{p_\lambda}{z_\lambda}$ then follows via induction
product calculations\footnote{For instance, one can use
\eqref{induced-character-formula} to show that
$z_\lambda \triv_\lambda
= \lambda_1 \lambda_2 \cdots \lambda_\ell
\cdot \triv_{\left(\lambda_1\right)} \triv_{\left(\lambda_2\right)} \cdots \triv_{\left(\lambda_\ell\right)}$
if $\lambda = \left(\lambda_1, \lambda_2, \ldots, \lambda_\ell\right)$
with $\ell = \ell\left(\lambda\right)$.
See Exercise~\ref{exe.frob-char.plambda}(d) for the details.}.
Part (b) follows from 
Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise} below.
\end{proof}

% [DG][v14] Added the preceding footnote because IMHO that's a lot of
% proof that is being skipped here, and at least it should be
% explicitly left to the reader.

% [DG][v45] Added yet another footnote.

\begin{remark}
The paper of Liulevicius \cite{Liulevicius} gives
a very elegant alternate approach to the Frobenius map
as a Hopf isomorphism $A(\Symm) \overset{\ch}{\longrightarrow} \Lambda$,
inspired by equivariant $K$-theory and vector bundles over spaces which
are finite sets of points!
\end{remark}

\begin{exercise}
\label{exe.frob-char.plambda}

If $P$ is a subset of a group $G$, we denote by $\triv_{P}$ the map
$G \to \CC$ which sends every element of $P$ to $1$ and all
remaining elements of $G$ to $0$. \ \ \ \ \footnote{This is not in conflict
with the notation $\triv_G$ for the trivial character of $G$, since
$\triv_P = \triv_G$ for $P=G$. Note that $\triv_P$
is a class function when $P$ is a union of conjugacy classes of $G$.} For any
finite group $G$ and any $h\in G$, we introduce the following notations:

\begin{itemize}
\item Let $Z_{G}\left(  h\right)$ denote the centralizer of $h$ in $G$.

\item Let $\operatorname{Conj}_{G}\left(  h\right)  $ denote the conjugacy
class of $h$ in $G$.

\item Define a map $\alpha_{G,h} : G \to \CC$ by $\alpha
_{G,h}=\left\vert Z_{G}\left(  h\right)  \right\vert
\triv_{\operatorname{Conj}_{G}\left(  h\right)  }$. This map
$\alpha_{G,h}$ is a
class function\footnote{In fact, $\triv_{\operatorname{Conj}%
_{G}\left(  h\right)  }$ is a class function (since $\operatorname{Conj}%
_{G}\left(  h\right)  $ is a conjugacy class), and so $\alpha_{G,h}$ (being
the scalar multiple $\left\vert Z_{G}\left(  h\right)  \right\vert
\triv_{\operatorname{Conj}_{G}\left(  h\right)  }$ of
$\triv_{\operatorname{Conj}_{G}\left(  h\right)  }$) must also be a
class function.}.
\end{itemize}

\begin{enumerate}
\item[(a)] Prove that $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[
khk^{-1}=g\right]  $ for every finite group $G$ and any $h\in G$ and $g\in G$.
Here, we are using the Iverson bracket notation (that is, for any statement
$\mathcal{A}$, we define $\left[  \mathcal{A}\right]  $ to be the integer $1$
if $\mathcal{A}$ is true, and $0$ otherwise).

\item[(b)] Prove that if $H$ is a subgroup of a finite group $G$, and if $h\in
H$, then $\Ind_{H}^{G}\alpha_{H,h}=\alpha_{G,h}$.

\item[(c)] Prove that if $G_{1}$ and $G_{2}$ are finite groups, and if
$h_{1}\in G_{1}$ and $h_{2}\in G_{2}$, then the canonical isomorphism
$R_\CC \left(  G_{1}\right)  \otimes R_\CC \left(
G_{2}\right)  \rightarrow R_\CC \left(  G_{1}\times G_{2}\right)  $
sends $\alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}}$ to $\alpha
_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }$.

\item[(d)] Fill in the details of the proof of $\ch
(\underline{1}_{\lambda})=\frac{p_{\lambda}}{z_{\lambda}}$ in the proof of
Theorem~\ref{symmetric-group-Frobenius-map-theorem}.

\item[(e)] Obtain an alternative proof of Remark~\ref{rmk.zlambda.centralizer}.

\item[(f)] If $G$ and $H$ are two finite groups, and if
$\rho : H \rightarrow G$ is a group homomorphism, then prove that
$\Ind_{\rho} \alpha_{H, h} = \alpha_{G, \rho \left( h \right)}$ for
every $h \in H$, where $\Ind_{\rho} \alpha_{H, h}$ is defined as in
Exercise~\ref{exe.Indrho}.

\end{enumerate}
\end{exercise}

% [DG][v47] Added the preceding exercise. (I use parts (b) and (c) in
% the solution of another.)

\begin{exercise}
\label{internal-tensor-product-with-sgn-is-omega-exercise}
If $G$ is a group and $U_1$ and $U_2$ are two $\CC G$-modules, then the
tensor product $U_1 \otimes U_2$ is a $\CC\left[G\times G\right]$-module,
which can be made into a $\CC G$-module by letting $g \in G$ act as
$\left(g,g\right) \in G\times G$. This $\CC G$-module $U_1 \otimes U_2$
is called the \dfn{inner tensor product}\index{tensor product of representations}%
\footnote{Do not confuse this
with the inner product of characters.} of $U_1$ and $U_2$, and is a
restriction of the outer tensor product $U_1 \otimes U_2$ using the
inclusion map $G \to G\times G,\ g\mapsto\left(g,g\right)$.

Let $n\geq 0$, and let $\sgn_{\Symm_n}$ be the $1$-dimensional
$\CC \Symm_n$-module $\CC$ on which every $g \in \Symm_n$ acts as
multiplication by $\sgn(g)$. If $V$ is a $\CC \Symm_n$-module, show
that the involution on $A(\Symm)=\bigoplus_{n \geq 0} R(\Symm_n)$
defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)
sends $\chi_V \mapsto \chi_{\sgn_{\Symm_n} \otimes V}$ 
%$\omega_n \left(\chi_V\right) = \chi_{\sgn_{\Symm_n} \otimes V}$,
where $\sgn_{\Symm_n} \otimes V$ is the inner tensor product of
$\sgn_{\Symm_n}$ and $V$.
Use this to show that this involution 
% $\bigoplus_{n\geq 0} \omega_n$ 
is a nontrivial PSH-automorphism of $A(\Symm)$, and deduce
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b).
%The fact that it corresponds to $\omega$ on $\Lambda$ now follows from
%Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(f), once one notes that the latter
%operation induces a (nontrivial) PSH-automorphism of $A$.
\end{exercise}

\begin{exercise}
\label{exe.frob-char.ptype}
Let $n \in \NN$. For every permutation
$\sigma \in \Symm_n$, we let $\operatorname*{type}\sigma$ denote the
cycle type of $\sigma$. Extend $\operatorname*{ch}:A=A\left(  \Symm
\right)  \rightarrow\Lambda$ to a $\CC$-linear map $A_{\CC}
\rightarrow\Lambda_{\CC}$. We shall call the latter map
$\operatorname*{ch}$, too.

\begin{enumerate}
\item[(a)] Prove that every class function $f\in R_{\CC}\left(
 \Symm_n\right)  $ satisfies
\[
\operatorname*{ch}\left(  f\right)  =\dfrac{1}{n!}\sum_{\sigma\in
 \Symm_n} f\left(  \sigma\right)  p_{\operatorname*{type}\sigma}.
\]


\item[(b)] Let $H$ be a subgroup of $\Symm_n$. Prove that every class
function $f\in R_{\CC}\left(  H\right)  $ satisfies%
\[
\operatorname*{ch}\left(  \operatorname*{Ind}\nolimits_{H}^{ \Symm_{n}%
}f\right)  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}f\left(  h\right)
p_{\operatorname*{type}h}.
\]

\end{enumerate}
\end{exercise}

\begin{exercise} \phantomsection
\label{exe.symmetric-group-reps-rational}
\begin{itemize}
\item[(a)] Show that for every $n \geq 0$, every $g \in \Symm_n$ and every
finite-dimensional $\CC \Symm_n$-module $V$, we have
$\chi_V\left(g\right) \in \ZZ$.
\item[(b)] Show that for every $n \geq 0$ and every
finite-dimensional $\CC \Symm_n$-module $V$, there exists a
$\QQ \Symm_n$-module $W$ such that $V \cong \CC \otimes_{\QQ} W$.
(In the representation theorists' parlance, this says that all
representations of $\Symm_n$ are \emph{defined over $\QQ$}. This
part of the exercise requires some familiarity with representation
theory.)
\end{itemize}
\end{exercise}

\begin{remark}
Parts (a) and (b) of Exercise~\ref{exe.symmetric-group-reps-rational} both
follow from an even stronger result: For every $n\geq 0$ and every
finite-dimensional $\CC \Symm_{n}$-module $V$, there exists a
$\ZZ \Symm_n$-module $W$ which is finitely generated and free
as a $\ZZ$-module and satisfies $V\cong \CC\otimes_\ZZ W$
as $\CC \Symm_n$-modules. This follows from the combinatorial approach to the
representation theory of $\Symm_n$, in which the irreducible
representations of $\CC \Symm_n$ (the \emph{Specht modules}) are constructed
using Young tableaux and tabloids. See the literature on the symmetric group,
e.g., \cite{Sagan}, \cite[\S 7]{Fulton}, \cite{Wildon}
or \cite[Section 2.2]{Krob} for this approach.
\end{remark}

% [DG][v13] Added the above exercise and remark. Haven't done the
% wreath-product analogue so far.

The connection between $\Lambda$ and $A\left( \Symm \right)$ as
established in Theorem~\ref{symmetric-group-Frobenius-map-theorem} benefits
both the study of $\Lambda$ and that of $A\left( \Symm \right)  $. The
following two exercises show some applications to $\Lambda$:

\begin{exercise}
\label{exe.repSn.inner-tensor.intprod}
If $G$ is a group and $U_1$ and $U_2$ are two $\CC G$-modules,
then let $U_1 \boxtimes U_2$ denote the inner tensor product of
$U_1$ and $U_2$ (as defined in
Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}).
Consider also the binary operation $*$ on $\Lambda_\QQ$ defined in
Exercise~\ref{exe.Lambda.maps-on-pn}(h).

\begin{itemize}

\item[(a)] Show that
$\ch \left(\chi_{U_1 \boxtimes U_2}\right)
= \ch\left(\chi_{U_1}\right) * \ch\left(\chi_{U_2}\right)$
for any $n \in \NN$ and any two $\CC \Symm_n$-modules $U_1$ and
$U_2$.

\item[(b)] Use this to obtain a new solution for
Exercise~\ref{exe.Lambda.maps-on-pn}(h).

\item[(c)] Show that $s_\mu * s_\nu \in \sum_{\lambda \in \Par}
\NN s_\lambda$ for any two partitions $\mu$ and $\nu$.

\end{itemize}

[\textbf{Hint:} For any group $G$,
introduce a binary operation $*$ on $R_\CC \left(G\right)$
which satisfies
$\chi_{U_1 \boxtimes U_2} = \chi_{U_1} * \chi_{U_2}$ for any
two $\CC G$-modules $U_1$ and $U_2$.]
\end{exercise}

% [DG][v25] Added the above exercise.

\begin{exercise}
\label{exe.repSn.arithmetic-prod}
Define a $\QQ$-bilinear map
$\boxdot:\Lambda_{\QQ}\times\Lambda_{\QQ}\rightarrow
\Lambda_{\QQ}$, which will be written in infix notation (that is, we
will write $a\boxdot b$ instead of $\boxdot\left(  a,b\right)  $), by setting
\[
p_{\lambda}\boxdot p_{\mu}
= \prod_{i=1}^{\ell\left(  \lambda\right)  }
\prod_{j=1}^{\ell\left(  \mu\right)  }
p_{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  }^{
\gcd\left(  \lambda_{i},\mu_{j}\right)  }
\qquad\qquad \text{for any partitions }\lambda\text{ and }\mu.
\]
\footnote{This is well-defined, since $\left(  p_{\lambda}\right)
_{\lambda\in\Par}$ is a $\QQ$-module basis of $\Lambda_{\QQ}$.}

\begin{enumerate}
\item[(a)] Show that $\Lambda_\QQ$, equipped with the binary
operation $\boxdot$, becomes a commutative $\QQ$-algebra with unity
$p_1$.

\item[(b)] For every $r\in\ZZ$, define the $\QQ$-algebra
homomorphism $\epsilon_{r}:\Lambda_\QQ\rightarrow\QQ$ as in
Exercise~\ref{exe.Lambda.maps-on-pn}(c). Show that $1\boxdot f=\epsilon
_{1}\left(  f\right)  1$ for every $f\in\Lambda_\QQ$ (where $1$
denotes the unity of $\Lambda$).

\item[(c)] Show that $s_{\mu}\boxdot s_{\nu}\in\sum_{\lambda\in
\Par} \NN s_{\lambda}$ for any two partitions $\mu$ and
$\nu$.

\item[(d)] Show that $f\boxdot g\in\Lambda$ for any $f\in\Lambda$ and
$g\in\Lambda$.
\end{enumerate}

[\textbf{Hint:} For every set $X$, let $\Symm_{X}$ denote the group of
all permutations of $X$. For two sets $X$ and $Y$, there is a canonical group
homomorphism $\Symm_{X}\times\Symm_{Y}\rightarrow
\Symm_{X\times Y}$, which is injective if $X$ and $Y$ are
nonempty. For positive
integers $n$ and $m$, this yields an embedding $\Symm_{n}
\times\Symm_{m}\rightarrow\Symm_{\left\{  1,2,\ldots,n\right\}
\times\left\{  1,2,\ldots,m\right\}  }$, which, once $\Symm_{\left\{
1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}  }$ is identified
with $\Symm_{nm}$ (using an arbitrary but fixed bijection $\left\{
1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}  \rightarrow\left\{
1,2,\ldots,nm\right\}  $), can be regarded as an embedding
$\Symm_{n}\times\Symm_{m}\rightarrow\Symm_{nm}$ and thus allows
defining a $\CC \Symm_{nm}$-module
$\Ind_{\Symm_{n}\times\Symm_{m}}^{\Symm_{nm}}\left(
U\otimes V\right)  $ for any $\CC \Symm_{n}$-module $U$ and any
$\CC \Symm_{m}$-module $V$. This gives a binary operation on
$A\left( \Symm \right)$. Show that this operation corresponds to
$\boxdot$ under the PSH-isomorphism $\ch:A\left(
\Symm\right)  \rightarrow\Lambda$.]
\end{exercise}

\begin{remark}
The statements (and the idea of the solution) of Exercise
\ref{exe.repSn.arithmetic-prod} are due to Manuel Maia and Miguel M\'{e}ndez
(see \cite{MaiaMendez} and, more explicitly, \cite{Mendez-MO}), who call the
operation $\boxdot$ the \dfn{arithmetic product}. Li \cite[Thm.
3.5]{Li-primegraphs} denotes it by $\boxtimes$ and relates it to the
enumeration of unlabelled graphs.
\end{remark}

% [DG][v47] Added the above exercise and remark.



\subsection{Wreath products}
\label{wreath-product-section}

Next consider the tower of groups
$G_n=\Symm_n[\Gamma]$ for a finite group $\Gamma$,
and the Hopf algebra $A=A(G_*)=:A(\Symm[\Gamma])$.  Recall
%from the previous section
(from Theorem~\ref{symmetric-group-Frobenius-map-theorem})
that irreducible complex representations $\chi^\lambda$ of $\Symm_n$ are
indexed by partitions $\lambda$ in $\Par_n$.  Index the
irreducible complex representations of $\Gamma$ as
$
\Irr(\Gamma) = \{ \rho_1,\ldots,\rho_d\}.
$

% [DG][v37] Replaced section reference by theorem
% reference above.

\begin{definition}
Define for a partition $\lambda$ in $\Par_n$
and $\rho$ in $\Irr(\Gamma)$
a representation $\chi^{\lambda,\rho}$
of $\Symm_n[\Gamma]$ in which $\sigma$ in $\Symm_n$ and
$\gamma=(\gamma_1,\ldots,\gamma_n)$ in $\Gamma^n$ act
on the space
$\chi^\lambda \otimes \left( \rho^{\otimes n} \right)$
as follows:
\begin{equation}
\label{wreath-irreducible-construction}
\begin{aligned}
\sigma( u \otimes (v_1 \otimes \cdots \otimes v_n) )
&= \sigma(u) \otimes (v_{\sigma^{-1}(1)} \otimes \cdots \otimes v_{\sigma^{-1}(n)}) ; \\
\gamma( u \otimes (v_1 \otimes \cdots \otimes v_n) )
&= u \otimes (\gamma_1 v_1 \otimes \cdots \otimes \gamma_n v_n) . \\
\end{aligned}
\end{equation}
\end{definition}

\begin{theorem}
\label{wreath-product-irreducibles}
The irreducible $\CC \Symm_n[\Gamma]$-modules
are the induced characters
\[
 \chi^{\underline{\lambda}}: =
\Ind_{\Symm_{\operatorname{degs}(\underline{\lambda})}[\Gamma]}^{\Symm_n[\Gamma]}
  \left(
   \chi^{\lambda^{(1)},\rho_1} \otimes \cdots
   \otimes \chi^{\lambda^{(d)},\rho_d}
  \right)
\]
as $\underline{\lambda}$ runs through all functions
\[
\begin{array}{rcl}
\Irr(\Gamma) &\overset{\underline{\lambda}}{\longrightarrow}&   \Par , \\
\rho_i &\longmapsto&  \lambda^{(i)}
\end{array}
\]
with the property that $\sum_{i=1}^d |\lambda^{(i)}|=n$.
Here, $\operatorname{degs}(\underline{\lambda})$
denotes the $d$-tuple
$\left(\left|\lambda^{(1)}\right|,\left|\lambda^{(2)}\right|,
\ldots,\left|\lambda^{(d)}\right|\right) \in \NN^d$, and
$\Symm_{\operatorname{degs}(\underline{\lambda})}$ is
defined as the subgroup
$\Symm_{\left|\lambda^{(1)}\right|} \times
\Symm_{\left|\lambda^{(2)}\right|} \times \cdots \times
\Symm_{\left|\lambda^{(d)}\right|}$ of $\Symm_n$.

% [DG][v14] Introduced $\operatorname{degs}(\underline{\lambda})$ to
% correct the definition of $\chi^{\underline{\lambda}}$ (it used to
% have a $\lambda$ instead, which made no sense).
% Also fixed a typo below ($\Lambda^{\otimes n}$ replaced by
% $\Lambda^{\otimes d}$).

Furthermore, one has a PSH-isomorphism
\[
\begin{array}{rcl}
A(\Symm[\Gamma]) &\longrightarrow &\Lambda^{\otimes d} ,\\
\chi^{\underline{\lambda}} & \longmapsto &
 s_{\lambda^{(1)}} \otimes \cdots \otimes s_{\lambda^{(d)}}.
\end{array}
\]
\end{theorem}
\begin{proof}
We know from Corollary~\ref{three-towers-give-bialgebras} that
$A(\Symm[\Gamma])$ is a PSH, with PSH-basis $\Sigma$
given by the union of all irreducible characters of all
groups $\Symm_n[\Gamma]$.  Therefore
Theorem~\ref{Zelevinsky-decomposition-theorem}
tells us that
$A(\Symm[\Gamma]) \cong \bigotimes_{\rho \in \CCC} A(\Symm[\Gamma])(\rho)$
where $\CCC$ is the set of irreducible characters which are
also primitive.  Just as in the case of $\Symm_n$,
it is clear from the definition of the coproduct that
an irreducible character $\rho$ of $\Symm_n[\Gamma]$ is
primitive if and only if $n=1$, that in this case
$\Symm_n[\Gamma]=\Gamma$,
and $\rho$ lies in $\Irr(\Gamma)=\{\rho_1,\ldots,\rho_d\}$.

The remaining assertions of the theorem will then follow from the definition
of the induction product algebra structure on $A(\Symm[\Gamma])$, once
we have shown that, for every $\rho \in \Irr(\Gamma)$,
 there is a PSH-isomorphism sending
\begin{equation}
\label{desired-wreath-PSH-iso}
\begin{array}{rcl}
A(\Symm) &\longrightarrow &A(\Symm[\Gamma])(\rho) ,\\
\chi^{\lambda} &\longmapsto &\chi^{\lambda,\rho}.
\end{array}
\end{equation}
Such an isomorphism comes from applying
Proposition~\ref{semidirect-product-adjoint-functors}
to the semidirect product $\Symm_n[\Gamma] = \Symm_n \ltimes \Gamma^n$,
so that $K=\Gamma^n, G=\Symm_n$, and fixing $V=\rho^{\otimes n}$ as
$\CC \Symm_n[\Gamma]$-module with structure as defined
in \eqref{wreath-irreducible-construction} (but with $\lambda$ set to
$\left(n\right)$, so that $\chi^\lambda$ is the trivial $1$-dimensional
$\CC \Symm_n$-module).
One obtains for each $n$, maps
\[
R(\Symm_n) \underset{\Psi}{\overset{\Phi}{\rightleftharpoons}}
R(\Symm_n[\Gamma])
\]
where
\[
\begin{array}{rcl}
\chi &\overset{\Phi}{\longmapsto} &\chi \otimes (\rho^{\otimes n}) ,\\
\alpha&\overset{\Psi}{\longmapsto}&
  \Hom_{\CC \Gamma^n}(\rho^{\otimes n},\alpha).
\end{array}
\]
Taking the direct sum of these maps for all $n$ gives maps
$
A(\Symm) \underset{\Psi}{\overset{\Phi}{\rightleftharpoons}}
A(\Symm[\Gamma]).
$

% [DG][v14] Changed $A(\Symm_n)$ to $R(\Symm_n)$ since you seem to
% reserve the notation $A$ for the direct sum of the representation
% rings rather than its specific addends (and, alas, also for a matrix).
% 
% Added "(but with $\lambda$ set to
% $\left(n\right)$, so that $\chi^\lambda$ is the trivial $1$-dimensional
% $\CC \Symm_n$-module)".

These maps are coalgebra morphisms because of
their interaction with restriction to $\Symm_i \times \Symm_j$.
Since Proposition~\ref{semidirect-product-adjoint-functors}(iii)
gives the adjointness property that
\[
(\chi,\Psi(\alpha))_{A(\Symm)} = (\Phi(\chi),\alpha)_{A(\Symm[\Gamma])},
\]
one concludes from the self-duality of $A(\Symm), A(\Symm[\Gamma])$
that $\Phi,\Psi$ are also algebra morphisms.  Since they take genuine
characters to genuine characters, they are PSH-morphisms.
Since $\rho$ being a simple $\CC \Gamma$-module implies that
$V=\rho^{\otimes n}$ is a simple $\CC \Gamma^n$-module,
Proposition~\ref{semidirect-product-adjoint-functors}(iv) shows that
\begin{equation}
\label{Psi-Phi-composite}
(\Psi \circ \Phi)(\chi) = \chi
\end{equation}
for all $\Symm_n$-characters $\chi$.   Hence $\Phi$ is an injective
PSH-morphism.   Using adjointness, \eqref{Psi-Phi-composite} also shows that
$\Phi$ sends $\CC \Symm_n$-simples $\chi$
to $\CC[\Symm_n[\Gamma]]$-simples $\Phi(\chi)$:
\[
(\Phi(\chi),\Phi(\chi))_{A(\Symm[\Gamma])}
 = ((\Psi \circ \Phi)(\chi),\chi)_{A(\Symm)}
 = (\chi,\chi)_{A(\Symm)} = 1.
\]
Since $\Phi(\chi)= \chi \otimes (\rho^{\otimes n})$
has $V=\rho^{\otimes n}$ as a constituent upon restriction to $\Gamma^n$,
Frobenius Reciprocity shows that the irreducible character
$\Phi(\chi)$ is a constituent of
$\Ind_{\Gamma^n}^{\Symm_n[\Gamma]} \rho^{\otimes n}= \rho^n$.  Hence the
entire image of $\Phi$ lies in $A(\Symm[\Gamma])(\rho)$ (due to how
we defined $A(\rho)$ in the proof of
Theorem~\ref{Zelevinsky-decomposition-theorem}), and so $\Phi$
must restrict to an isomorphism as desired in \eqref{desired-wreath-PSH-iso}.
\end{proof}

One of Zelevinsky's sample applications of the theorem is this branching rule.

\begin{corollary}
\label{wreath-product-branching-rule}
Given $\underline{\lambda}=(\lambda^{(1)},\ldots,\lambda^{(d)})$ with $\sum_{i=1}^d |\lambda^{(i)}|=n$,
one has
\[
\Res^{\Symm_n[\Gamma]}_{\Symm_{n-1}[\Gamma] \times \Gamma}  \left( \chi^{\underline{\lambda}} \right)
= \sum_{i=1}^d \sum\limits_{\substack{ \lambda^{(i)}_- \subseteq \lambda^{(i)}: \\ |\lambda^{(i)}/\lambda^{(i)}_-|=1}}
\chi^{(\lambda^{(1)}, \ldots,\lambda^{(i)}_-, \ldots,\lambda^{(d)})} \otimes \rho_i.
\]
(We are identifying functions $\underline{\lambda} : \Irr(\Gamma)
\to \Par$ with the corresponding $d$-tuples
$\left(\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(d)}\right)$
here.)
\end{corollary}

\begin{example}
For $\Gamma$ a two-element group, so $\Irr(\Gamma)=\{\rho_1,\rho_2\}$ and
$d=2$, then
\[
\Res^{\Symm_6[\Gamma]}_{\Symm_{5}[\Gamma] \times \Gamma}
\left( \chi^{((3,1),(1,1))} \right)
=
\chi^{((3),(1,1))} \otimes \rho_1
+\chi^{((2,1),(1,1))} \otimes \rho_1
+\chi^{((3,1),(1))} \otimes \rho_2.
\]
\end{example}

\begin{proof}[Proof of Corollary~\ref{wreath-product-branching-rule}]
By Theorem~\ref{wreath-product-irreducibles}, this is equivalent to
computing in the Hopf algebra $A:=\Lambda^{\otimes d}$ the component of the
coproduct of $s_{\lambda^{(1)}} \otimes \cdots \otimes s_{\lambda^{(d)}}$
that lies in $A_{n-1} \otimes A_1$.  Working within each tensor factor $\Lambda$,
we conclude from Proposition~\ref{symm-comultiplication-formulas-prop}(iv)
% the Pieri formula implies
that the
$\Lambda_{\left|\lambda\right|-1} \otimes \Lambda_1$-component
of $\Delta(s_\lambda)$ is
\[
\sum\limits_{\substack{\lambda_- \subseteq \lambda: \\|\lambda/\lambda_-|=1}} s_{\lambda_-} \otimes \rho.
\]
One must apply this in each of the $d$ tensor factors of $A=\Lambda^{\otimes d}$, then sum on $i$.
\end{proof}






\subsection{General linear groups}
\label{general-linear-section}

We now consider the tower of finite general linear groups
$G_n=GL_n=GL_n(\FF_q)$ and $A=A(G_*)=:A(GL)$.
Corollary~\ref{three-towers-give-bialgebras} tells us that
$A(GL)$ is a PSH, with PSH-basis $\Sigma$
given by the union of all irreducible characters of all
groups $GL_n$.  Therefore Theorem~\ref{Zelevinsky-decomposition-theorem}
tells us that
\begin{equation}
\label{GL-Zelevinsky-decomposition}
A(GL) \cong \bigotimes_{\rho \in \CCC} A(GL)(\rho)
\end{equation}
where $\CCC=\Sigma \cap \Liep$ is the set of primitive irreducible characters.

\begin{definition}
Call an irreducible representation $\rho$ of $GL_n$
\emph{cuspidal}\index{cuspidal representation of $GL_n$}
for $n \geq 1$ if it lies in $\CCC$, that is, its restriction to proper parabolic subgroups $P_{i,j}$ with $i+j=n$
and $i,j>0$ contain no nonzero vectors which are $K_{i,j}$-invariant.
Given an irreducible character $\sigma$ of $GL_n$, say that $d(\sigma)=n$,
and let $\CCC_n:=\{ \rho \in \CCC: d(\rho)=n\}$ for $n \geq 1$
denote the subset of cuspidal characters of $GL_n$.
\end{definition}

Just as was the case for $\Symm_1$ and $\Symm_1[\Gamma]=\Gamma$, \emph{every}
irreducible character $\rho$ of $GL_1(\FF_q)=\FF_q^\times$ is cuspidal.
However, this does not exhaust the cuspidal characters.  In fact, one
can predict the number of cuspidal characters in $\CCC_n$, using
knowledge of the number of conjugacy classes in $GL_n$.
Let $\FFF$ denote the set of all nonconstant monic irreducible polynomials $f(x) \neq x$ in $\FF_q[x]$.
Let $\FFF_n:=\{ f \in \FFF: \deg(f)=n\}$ for $n \geq 1$.

\begin{proposition}
The number $|\CCC_n|$ of cuspidal characters of $GL_n(\FF_q)$
is the number of $|\FFF_n|$ of irreducible monic degree $n$ polynomials $f(x) \neq x$
in $\FF_q[x]$ with nonzero constant term.
\end{proposition}
\begin{proof}
We show  $|\CCC_n|=|\FFF_n|$ for $n \geq 1$ by strong induction on $n$.
For the base case\footnote{Actually, we don't need any base case for our
strong induction. We nevertheless handle the case $n=1$ as a warmup.}
$n=1$, just as with the families $G_n=\Symm_n$ and
$G_n=\Symm_n[\Gamma]$, when $n=1$ any irreducible character
$\chi$ of $G_1=GL_1(\FF_q)$
gives a primitive element of $A=A(GL)$, and hence is cuspidal.
Since $GL_1(\FF_q)=\FF_q^\times$ is abelian, there are $|\FF_q^\times|=q-1$
such cuspidal characters in $\CCC_1$, which agrees with the fact that
there are $q-1$ monic (irreducible)
linear polynomials $f(x) \neq x$ in $\FF_q[x]$, namely
$
\FFF_1:=\{ f(x)=x-c: c \in \FF_q^\times\}.
$


In the inductive step, use the fact that
the number $|\Sigma_n|$ of irreducible complex characters $\chi$
of $GL_n(\FF_q)$ equals its number of conjugacy classes.
These conjugacy classes are
uniquely represented by \emph{rational canonical forms},
which are parametrized
by functions $\underline{\lambda}: \FFF \rightarrow \Par$ with
the property that $\sum_{f \in \FFF} \deg(f) |\underline{\lambda}(f)|=n$.
On the other hand, \eqref{GL-Zelevinsky-decomposition} tells us that
$|\Sigma_n|$ is similarly parametrized by the functions
$\underline{\lambda}: \CCC \rightarrow \Par$ having
the property that $\sum_{\rho \in \CCC} d(\rho) |\underline{\lambda}(\rho)|=n$.
Thus we have parallel disjoint decompositions
\[
\begin{array}{rll}
\CCC &= \bigsqcup_{ n \geq 1} \CCC_n
         & \text{ where }\CCC_n=\left\{\rho \in \CCC:d(\rho)=n\right\},\\
\FFF &= \bigsqcup_{ n \geq 1} \FFF_n
         & \text{ where }\FFF_n=\left\{f \in \FFF:\deg(f)=n\right\},
\end{array}
\]
and hence an equality for all $n \geq 1$
\[
\left|\left\{ \CCC\overset{\underline{\lambda}}{\longrightarrow} \Par:
\quad
\sum_{\rho \in \CCC} d(\rho) |\underline{\lambda}(\rho)| =n\right\}\right| \\
=|\Sigma_n|
=\left|\left\{ \FFF\overset{\underline{\lambda}}{\longrightarrow} \Par:
\quad
\sum_{f \in \FFF} \deg(f) |\underline{\lambda}(f)| =n\right\}\right|.
\]
Since there is only one partition $\lambda$ having $|\lambda|=1$
(namely, $\lambda=(1)$),
this leads to parallel recursions
\begin{align*}
|\CCC_n|
&=|\Sigma_n|-
\left|\left\{ \bigsqcup_{i=1}^{n-1} \CCC_i \overset{\underline{\lambda}}{\longrightarrow} \Par: \quad
\sum_{\rho \in \CCC} d(\rho) |\underline{\lambda}(\rho)| =n\right\}\right|, \\
|\FFF_n|
&=|\Sigma_n|-
\left|\left\{ \bigsqcup_{i=1}^{n-1} \FFF_i \overset{\underline{\lambda}}{\longrightarrow} \Par: \quad
\sum_{f \in \FFF} \deg(f) |\underline{\lambda}(f)| =n\right\}\right|,
\end{align*}
and induction implies that $|\CCC_n|=|\FFF_n|$.
\end{proof}

We shall use the notation $\triv_H$ for the trivial character of a group
$H$ whenever $H$ is a finite group. This generalizes the notations
$\triv_{\Symm_n}$ and $\triv_{\Symm_\lambda}$ introduced above.

% [DG][v19] Added the above paragraph.

\begin{example}
\label{low-degree-cuspidal-labelling-example}
Taking $q=2$, let us list the sets $\FFF_n$ of monic
irreducible polynomials $f(x) \neq x$ in $\FF_2[x]$ of degree $n$
for $n \leq 3$, so that we know how many cuspidal characters
of $GL_n(\FF_q)$ in $\CCC_n$ to expect:
\begin{align*}
\FFF_1&=\{ x+1\};\\
\FFF_2&=\{ x^2+x+1 \};\\
\FFF_3&=\{ x^3+x+1, x^3+x^2+1\}.
\end{align*}
Thus we expect
\begin{enumerate}
\item[$\bullet$]
one cuspidal character of $GL_1(\FF_2)$, namely
$\rho_1(=\triv_{GL_1(\FF_2)})$,
\item[$\bullet$]
one cuspidal character $\rho_2$ of $GL_2(\FF_2)$, and
\item[$\bullet$]
two cuspidal characters $\rho_3,\rho_3'$ of $GL_3(\FF_2)$.
\end{enumerate}
We will say more about $\rho_2, \rho_3,\rho_3'$ in the next section.
\end{example}


\begin{exercise}
\label{exe.polynomials-and-necklaces}
Let $\mu : \left\{1,2,3,\ldots\right\} \to \ZZ$ denote the
\dfn{number-theoretic M\"obius function}\index{M\"obius function}\index{$\mu$},
defined by setting
$\mu(m)=(-1)^d$
if $m=p_1 \cdots p_d$ for $d$ distinct primes $p_1, p_2, \ldots, p_d$,
and $\mu(m)=0$ if $m$ is not squarefree.

\begin{itemize}

\item[(a)]
Show that for $n \geq 2$, we have
\begin{equation}
\label{primitive-necklace-number}
|\CCC_n| (=|\FFF_n|)=
\frac{1}{n} \sum_{d \mid n}
\mu\left(\frac{n}{d} \right) q^{d} .
\end{equation}
(Here, the summation
sign $\sum_{d \mid n}$ means a sum over all positive divisors
$d$ of $n$.)

% [DG][v34] Replaced the "d \text{ dividing } n" by a
% "d \mid n" in the summation subscript.

\item[(b)]
Show that \eqref{primitive-necklace-number} also counts
the \emph{necklaces}\index{necklace} with $n$ beads of $q$ colors
(= the equivalence classes under the $\ZZ/n\ZZ$-action of
cyclic rotation on sequences $(a_1,\ldots,a_n)$ in $\FF_q^n$)
which are \emph{primitive}\index{primitive necklace}
in the sense that no nontrivial rotation
fixes any of the sequences within the equivalence class.
For example, when $q=2$, here are systems of distinct
representatives of these primitive necklaces for $n=2,3,4$:
\begin{align*}
n=2:& \quad \{(0,1)\};\\
n=3:& \quad \{(0,0,1),(0,1,1)\};\\
n=4:& \quad \{(0,0,0,1),(0,0,1,1),(0,1,1,1)\}.
\end{align*}

\end{itemize}
\end{exercise}

The result of Exercise~\ref{exe.polynomials-and-necklaces}(a)
was stated by Gauss for prime $q$, and by Witt
for general $q$; it is discussed in \cite{CheboluMinac},
\cite[Section 7.6.2]{Reutenauer} and (for prime $q$)
\cite[(4.12.3)]{Granville-NTmasterclass}.
Exercise~\ref{exe.polynomials-and-necklaces}(b) is also
well-known.
See \cite[Section 7.6.2]{Reutenauer}
for a bijection explaining why the answers to both parts of
Exercise~\ref{exe.polynomials-and-necklaces} are the same.

\subsection{Steinberg's unipotent characters}

Not surprisingly, the (cuspidal) character $\iota:=\triv_{GL_1}$ of $GL_1(\FF_q)$
plays a distinguished role.  The parabolic subgroup $P_{(1^n)}$ of
$GL_n(\FF_q)$ is the Borel subgroup
$B$ of upper triangular matrices,
and we have
$
\iota^n = \Ind_B^{GL_n} \triv_B = \CC[GL_n/B]
$
(identifying representations with their characters as
usual)\footnote{\textit{Proof.} Exercise~\ref{exe.towers.indres}(d)
(applied to $G_* = GL_*$, $\ell = n$, $\alpha = \left(1^n\right)
= \left(\underbrace{1,1,\ldots,1}_{n \text{ times}}\right)$ and
$\chi_i = \iota$) gives
\[
\iota^n = \ind^n_{\left(1^n\right)} \iota^{\otimes n}
= \underbrace{\Ind^{G_n}_{P_{\left(1^n\right)}}}_{= \Ind^{GL_n}_B}
\underbrace{\Inf^{P_{\left(1^n\right)}}_{G_{\left(1^n\right)}}
   \iota^{\otimes n}}_{= \triv_{P_{\left(1^n\right)}} = \triv_B}
= \Ind^{GL_n}_B \triv_B = \CC \left[ GL_n / B \right] ,
\]
where the last equality follows from the general fact that if
$G$ is a finite group and $H$ is a subgroup of $G$, then
$\Ind^G_H \triv_H \cong \CC \left[G/H\right]$ as
$\CC G$-modules.}.
The subalgebra $A(GL)(\iota)$ of $A(GL)$ is the $\ZZ$-span of
the irreducible characters $\sigma$ that appear as constituents of
$
\iota^n = \Ind_B^{GL_n} \triv_B = \CC[GL_n/B]
$
for some $n$.

% [DG][v25] This $\iota^n = \Ind_B^{GL_n} \triv_B$ equality looks
% anything but trivial to me. I guess I see how to prove it with
% a lot of work, but shouldn't there be more details or an exercise?

% [DG][v28] Done.

\begin{definition}
An irreducible character $\sigma$ of $GL_n$ appearing as a constituent
of $\Ind_B^{GL_n} \triv_B = \CC[GL_n/B]$ is called a
\emph{unipotent character}\index{unipotent character of $GL_n$}.
Equivalently, by Frobenius reciprocity, $\sigma$ is unipotent if it contains a nonzero
$B$-invariant vector.
\end{definition}

In particular, $\triv_{GL_n}$ is a unipotent character of $GL_n$ for each $n$.

\begin{proposition}
\label{Sym-to-Steinberg-isomorphism}
One can choose $\Lambda \cong A(GL)(\iota)$ in
Theorem~\ref{uniqueness-of-sym-parts-e-f-g}(g) so that
$h_n \longmapsto \triv_{GL_n}$.
\end{proposition}
\begin{proof}
Theorem~\ref{uniqueness-of-sym-parts-a-b-c-d}(a)
tells us $\iota^2 = \Ind_B^{GL_2} \triv_B$ must have exactly two irreducible
constituents, one of which is $\triv_{GL_2}$;  call the other one $\St_2$.
Choose the isomorphism so as to send $h_2 \longmapsto \triv_{GL_2}$.
Then $h_n \mapsto \triv_{GL_n}$ follows from the
claim that $\St_2^\perp(\triv_{GL_n})=0$ for $n \geq 2$:
one has
\[
\Delta(\triv_{GL_n})
= \sum_{i+j=n} \left(\Res^{G_n}_{P_{i,j}} \triv_{GL_n}\right)^{K_{i,j}}
=\sum_{i+j=n}\triv_{GL_i} \otimes \triv_{GL_j}
\]
so that $\St_2 ^\perp(\triv_{GL_n})=(\St_2,\triv_{GL_2}) \triv_{GL_{n-2}} = 0$ since
$\St_2 \neq \triv_{GL_2}$.

\end{proof}


This subalgebra $A(GL)(\iota)$, and the unipotent characters $\chi^\lambda_q$ corresponding
under this isomorphism to the Schur functions $s_\lambda$, were introduced
by Steinberg \cite{Steinberg}.  He wrote down $\chi^\lambda_q$ as a virtual sum of induced
characters $\Ind_{P_\alpha}^{GL_n} \triv_{P_\alpha} (=\triv_{G_{\alpha_1}} \cdots \triv_{G_{\alpha_\ell}})$,
modelled on the
Jacobi-Trudi determinantal expression for $s_\lambda=\det(h_{\lambda_i-i+j})$.
Note that  $\Ind_{P_\alpha}^{GL_n} \triv_{P_\alpha}$ is the transitive
permutation representation $\CC[G/P_\alpha]$ for $GL_n$
permuting the \dfn{finite partial flag variety} $G/P_\alpha$, that is, the set
of \emph{$\alpha$-flags of subspaces}\index{flag of subspaces}
\[
\{0\} \subset V_{\alpha_1} \subset V_{\alpha_1+\alpha_2}
 \subset \cdots \subset  V_{\alpha_1+\alpha_2+\cdots+\alpha_{\ell-1}}
\subset \FF_q^n
\]
where $\dim_{\FF_q} V_d = d$ in each case.  This character has dimension equal
to $|G/P_\alpha|$, with formula given by the \dfn{$q$-multinomial coefficient}
(see e.g. Stanley \cite[\S 1.7]{Stanley}):
\[
\qbin{n}{\alpha}{q}=\frac{[n]!_q}{[\alpha_1]!_q \cdots [\alpha_\ell]!_q}
\]
where $[n]!_q:=[n]_q [n-1]_q \cdots [2]_q [1]_q$ and
$[n]_q:=1+q+\cdots+q^{n-1}=\frac{q^n-1}{q-1}$.


Our terminology $\St_2$ is motivated by the $n=2$ special case of
the \dfn{Steinberg character} $\St_n$, which is the
unipotent character corresponding under the isomorphism in
Proposition~\ref{Sym-to-Steinberg-isomorphism} to $e_n = s_{(1^n)}$.
It can be defined by the virtual sum
\[
\St_n:=
\chi^{(1^n)}_q = \sum_{\alpha} (-1)^{n-\ell(\alpha)} \Ind_{P_\alpha}^{GL_n} \triv_{P_\alpha}
\]
in which the sum runs through all compositions $\alpha$ of $n$.
This turns out to be the genuine character for
$GL_n(\FF_q)$ acting on the top homology group of its \dfn{Tits building}:
the simplicial complex whose vertices are
nonzero proper subspaces $V$ of $\FF_q^n$, and whose simplices correspond
to flags of nested subspaces.  One needs to know that this Tits building has
only top homology, so that one can deduce the above character formula from
the Hopf trace formula; see Bj\"orner \cite{Bjorner}.



\subsection{Examples: $GL_2(\FF_2)$ and $GL_3(\FF_2)$}

Let's get our hands dirty.


\begin{example}
For $n=2$, there are two unipotent characters, $\chi^{(2)}_q = \triv_{GL_2}$
and
\begin{equation}
\label{Steinberg-on-lines-description}
\St_2:=\chi^{(1,1)}_q
= \triv_{GL_1}^2 - \triv_{GL_2}
= \Ind_B^{GL_2} \triv_B -   \triv_{GL_2}
\end{equation}
since the Jacobi-Trudi formula \eqref{eq.jacobi-trudi.h} gives
$
s_{(1,1)}= \det \left[
\begin{matrix}
h_1 & h_2 \\
1  & h_1
\end{matrix} \right]
=h_1^2-h_2.
$
The description \eqref{Steinberg-on-lines-description} for
this Steinberg character $\St_2$ shows that it
has dimension
\[
|GL_2/B|-1 = (q+1)-1 = q
\]
and that one can think of it as follows:
consider the permutation action of $GL_2$ on the $q+1$ lines
$\{\ell_0,\ell_1,\ldots,\ell_q\}$
in the projective space $\Proj^1_{\FF_q}=GL_2(\FF_q)/B$, and
take the invariant subspace perpendicular to the sum of basis
elements $e_{\ell_0} + \cdots +e_{\ell_q}$.
\end{example}


\begin{example}
Continuing the previous example, but taking $q=2$,
we find that we have constructed two unipotent characters:
$\triv_{GL_2}=\chi^{(2)}_{q=2}$ of dimension $1$, and
$\St_2=\chi^{(1,1)}_{q=2}$ of dimension $q=2$.
This lets us identify the unique cuspidal character
$\rho_2$ of $GL_2(\FF_2)$, using knowledge of
the character table of $GL_2(\FF_2) \cong \Symm_3$:

\vskip.1in
\begin{tabular}{|c||c|c|c|c|}\hline
 & & & & \\
&
& $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$
&
$\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right]$,
$\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right]$,
$\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]$
&
$\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right]$,
$\left[ \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right]$\\
 & & & & \\\hline\hline
$\triv_{GL_2}=\chi^{(2)}_{q=2}$ &\text{unipotent}& $1$ & $1$ & $1$ \\\hline
$\St_2=\chi^{(1,1)}_{q=2}$ &\text{unipotent}& $2$ & $0$ & $-1$ \\\hline
$\rho_2$ &\text{cuspidal}& $1$ & $-1$ & $1$ \\\hline
\end{tabular}

\vskip.1in
\noindent
In other words, the cuspidal character $\rho_2$ of $GL_2(\FF_2)$
corresponds under the isomorphism $GL_2(\FF_2) \cong \Symm_3$
to the sign character $\sgn_{\Symm_3}$.
\end{example}

\begin{example}
Continuing the previous example to $q=2$ and $n=3$ lets us analyze
the irreducible characters of $GL_3(\FF_2)$.
Recalling our labelling
$\rho_1,\rho_2,\rho_3,\rho_3'$ from Example~\ref{low-degree-cuspidal-labelling-example} of the cuspidal characters of $GL_n(\FF_2)$ for $n=1,2,3$,
Zelevinsky's Theorem~\ref{Zelevinsky-decomposition-theorem}
tells us that the $GL_3(\FF_2)$-irreducible
characters should be labelled by functions
$
\{\rho_1,\rho_2,\rho_3,\rho_3'\}
\overset{\underline{\lambda}}{\longrightarrow} \Par
$
for which
\[
1 \cdot |\underline{\lambda}(\rho_1)| +
2 \cdot |\underline{\lambda}(\rho_2)| +
3 \cdot |\underline{\lambda}(\rho_3)| +
3 \cdot |\underline{\lambda}(\rho'_3)| = 3 .
\]
We will label such an irreducible character
$\chi^{\underline{\lambda}}=\chi^{(\underline{\lambda}(\rho_1),\underline{\lambda}(\rho_2),\underline{\lambda}(\rho_3),\underline{\lambda}(\rho'_3))}$.

Three of these irreducibles will be the unipotent characters,
mapping under the isomorphism
from Proposition~\ref{Sym-to-Steinberg-isomorphism} as follows:
\begin{enumerate}
\item[$\bullet$]
$
s_{(3)} =h_3 \longmapsto
\chi^{((3),\varnothing,\varnothing,\varnothing)} = \triv_{GL_3}
$
of dimension $1$.
\item[$\bullet$]
\[
s_{(2,1)}=\det \left[
\begin{matrix}
h_2 & h_3 \\
1  & h_1
\end{matrix} \right]
=h_2h_1-h_3
\longmapsto \chi^{((2,1),\varnothing,\varnothing,\varnothing)}
            =\Ind_{P_{2,1}}^{GL_3} \triv_{P_{2,1}} -   \triv_{GL_3},
\]
of dimension $\qbin{3}{2,1}{q}-\qbin{3}{3}{q}=[3]_q-1=q^2+q
\overset{q=2}{\rightsquigarrow} 6$.
\item[$\bullet$]
Lastly,
\begin{align*}
s_{(1,1,1)}&=\det \left[
\begin{matrix}
h_1 & h_2 & h_3 \\
1  & h_1 & h_2 \\
0  & 1  & h_1
\end{matrix} \right]
=h_1^3-h_2 h_1 - h_1 h_2 + h_3 \\
&\longmapsto \St_3=\chi^{((1,1,1),\varnothing,\varnothing,\varnothing)}
=\Ind_{B}^{GL_3} \triv_{B} -
\Ind_{P_{2,1}}^{GL_3} \triv_{P_{2,1}} -
\Ind_{P_{1,2}}^{GL_3} \triv_{P_{1,2}} +   \triv_{GL_3}
\end{align*}
of dimension
\begin{align*}
&\qbin{3}{1,1,1}{q}-\qbin{3}{2,1}{q}-
\qbin{3}{1,2}{q}+\qbin{3}{3}{q} \\
&=[3]!_q - [3]_q-[3]_q + 1=q^3
\overset{q=2}{\rightsquigarrow} 8.
\end{align*}
\end{enumerate}

There should also be one non-unipotent, non-cuspidal character, namely
\[
\chi^{((1),(1),\varnothing,\varnothing)}
= \rho_1 \rho_2
= \Ind_{P_{1,2}}^{GL_3} \Inf_{GL_1 \times GL_2}^{P_{1,2}}
      \left( \triv_{GL_1} \otimes \rho_2 \right)
\]
having dimension $\qbin{3}{1,2}{q} \cdot 1 \cdot 1=[3]_q
\overset{q=2}{\rightsquigarrow} 7$.

Finally, we expect cuspidal characters
$\rho_3=\chi^{(\varnothing,\varnothing,(1),\varnothing)},
\rho_3'=\chi^{(\varnothing,\varnothing,\varnothing,(1))}$,
whose dimensions $d_3, d_3'$ can be deduced from the equation
\[
1^2+6^2+8^2+7^2+d_3^2+(d_3')^2=|GL_3(\FF_2)|
=\left[(q^3-q^0)(q^3-q^1)(q^3-q^2)\right]_{q=2}=168.
\]
This forces $d_3^2+(d_3')^2=18$,
whose only solution in positive integers
is $d_3=d_3'=3$.


We can check our predictions of the dimensions
for the various  $GL_3(\FF_2)$-irreducible
characters since $GL_3(\FF_2)$ is the finite simple group of
order $168$ (also isomorphic to $PSL_2(\FF_7)$), with known
character table (see James and Liebeck \cite[p. 318]{JamesLiebeck}):\\
\begin{tabular}{|c||c||c|c|c|c|c|c|}\hline
&\text{centralizer order} & $168$ & $8$ & $4$ & $3$  & $7$ & $7$ \\\hline
 &unipotent?/cuspidal? &   &  &  &   &  &\\\hline\hline
$\triv_{GL_3}=\chi^{((3),\varnothing,\varnothing,\varnothing)}$
  &\text{unipotent}& $1$ & $1$ & $1$ & $1$ & $1$ & $1$ \\\hline
$\phantom{\St_3=}\chi^{((2,1),\varnothing,\varnothing,\varnothing)}$
  &\text{unipotent}& $6$ & $2$ & $0$ & $0$ & $-1$ & $-1$ \\\hline
$\St_3=\chi^{((1,1,1),\varnothing,\varnothing,\varnothing)}$
  &\text{unipotent}& $8$ & $0$ & $0$ & $-1$ & $1$ & $1$ \\\hline
$\phantom{\St_3=}\chi^{((1),(1),\varnothing,\varnothing)}$
  &\text{ }& $7$ & $-1$ & $-1$ & $1$ & $0$ & $0$ \\\hline
$\rho_3=\chi^{(\varnothing,\varnothing,(1),\varnothing)}$
  &\text{cuspidal}& $3$ & $-1$ & $1$ & $0$ & $\alpha$ & $\overline{\alpha}$ \\\hline
$\rho_3'=\chi^{(\varnothing,\varnothing,\varnothing,(1))}$
  &\text{cuspidal}& $3$ & $-1$ & $1$ & $0$ & $\overline{\alpha}$ & $\alpha$ \\\hline
\end{tabular}\\
Here $\alpha:=-1/2+i\sqrt{7}/2$.

\begin{remark}
It is known (see e.g. Bump \cite[Cor. 7.4]{Bump})
that, for $n \geq 2$, the dimension of any cuspidal irreducible character
$\rho$ of $GL_n(\FF_q)$ is
\[
(q^{n-1}-1)(q^{n-2}-1)\cdots(q^2-1)(q-1).
\]
Note that when $q=2$,
\begin{enumerate}
\item[$\bullet$]
for $n=2$ this gives $2^1-1=1$ for the dimension of $\rho_2$, and
\item[$\bullet$]
for $n=3$ it gives $(2^2-1)(2-1)=3$ for the dimensions of $\rho_3,\rho_3'$,
\end{enumerate}
agreeing with our calculations above.
Much more is known about the character table of $GL_n(\FF_q)$; see
Remark~\ref{Green-polynomial-remark} below,
Zelevinsky \cite[Chap. 11]{Zelevinsky}, and
Macdonald \cite[Chap. IV]{Macdonald}.
\end{remark}


\end{example}




\subsection{The Hall algebra}
\label{Hall-algebra-section}

There is another interesting Hopf subalgebra (and quotient Hopf algebra) of $A(GL)$,
related to unipotent conjugacy classes in $GL_n(\FF_q)$.

\begin{definition}
Say that an element $g$ in $GL_n(\FF_q)$ is
\emph{unipotent}\index{unipotent element of $GL_n$} if its eigenvalues are
all equal to $1$. Equivalently, $g \in GL_n(\FF_q)$ is unipotent if and only
if $g - \id_{\FF_q^n}$ is nilpotent. A conjugacy class in $GL_n(\FF_q)$ is
\emph{unipotent}\index{unipotent conjugacy class in $GL_n$}
if its elements are unipotent.

Denote by $\HHH_n$ the $\CC$-subspace of $R_\CC(GL_n)$ consisting
of those class functions which are supported only on unipotent conjugacy classes,
and let $\HHH=\bigoplus_{n \geq 0} \HHH_n$\index{$\HHH$} as a  $\CC$-subspace of
$A_\CC(GL)=\bigoplus_{n \geq 0} R_\CC(GL_n)$.
\end{definition}

% [DG][v29] Added second and third sentence of this definition.




\begin{proposition}
\label{Hall-algebra-is-Hopf-sub-quotient}
The subspace $\HHH$ is
a Hopf subalgebra of $A_\CC(GL)$, which
is graded, connected, and of finite type, and self-dual with
respect to the inner product on class functions inherited
from $A_\CC(GL)$.
It is also a quotient Hopf algebra of $A_\CC(GL)$,
as the $\CC$-linear surjection
$
A_\CC(GL) \twoheadrightarrow \HHH
$
restricting class functions to unipotent classes is a
Hopf algebra homomorphism. This surjection
has kernel $\HHH^\perp$, which is both an ideal
and a two-sided coideal.
\end{proposition}

% [DG][v29] Made it more explicit that the surjection is a Hopf
% morphism (also in the proof).

\begin{proof}
It is immediately clear that $\HHH^\perp$ is a graded
$\CC$-vector subspace of $A_\CC\left(GL\right)$, whose
$n$-th homogeneous component consists of those class
functions on $GL_n$ whose values on all unipotent
classes are $0$.
(This holds no matter whether the perpendicular space
is taken with respect to the Hermitian form
$\left(\cdot,\cdot\right)_G$ or with respect to
the bilinear form $\left<\cdot,\cdot\right>_G$.)
In other words, $\HHH^\perp$ is the kernel of the
surjection $A_\CC(GL) \twoheadrightarrow \HHH$ defined
in the proposition.

Given two class functions $\chi_i, \chi_j$ on $GL_i, GL_j$
and $g$ in $GL_{i+j}$, one has
\begin{equation}
\label{Hall-algebra-product}
\left( \chi_i \cdot \chi_j \right)(g) =
\frac{1}{|P_{i,j}|}
  \sum\limits_{\substack{h \in GL_{i+j}:\\
    h^{-1} g h=\left[ \begin{matrix} g_i & * \\0 & g_j\end{matrix} \right] \in P_{i,j}}}
  \chi_i(g_i) \chi_j(g_j).
\end{equation}
Since $g$ is unipotent if and only if $h^{-1}gh$ is unipotent if and only if both
$g_i,g_j$ are unipotent, the formula \eqref{Hall-algebra-product} shows
both that $\HHH$ is a subalgebra\footnote{Indeed,
if $\chi_i$ and $\chi_j$ are both supported only on unipotent classes,
then the same holds for $\chi_i \cdot \chi_j$.
% [DG][v29] The previous sentence was an "if and only if",
% but the converse is false if one of $\chi_i$ and
% $\chi_j$ is zero, and I am not sure if it is true otherwise.
% (It is also not needed.)
}
and that $\HHH^\perp$ is a two-sided ideal\footnote{In
fact, if one of $\chi_i$ and $\chi_j$ annihilates all
unipotent classes, then so does $\chi_i \cdot \chi_j$.}.
It also shows that
the surjection $A_\CC(GL) \twoheadrightarrow \HHH$
restricting every class function to unipotent classes is an
algebra homomorphism\footnote{because if $g$ is unipotent, then the
only values of $\chi_i$ and $\chi_j$ appearing on the
right hand side of \eqref{Hall-algebra-product} are those
on unipotent elements}.

Similarly, for class functions $\chi$ on $GL_n$ and $(g_i, g_j)$ in $GL_{i,j}=GL_i \times GL_j$, one has
\[
\Delta(\chi)(g_i, g_j) = \frac{1}{q^{ij}} \sum_{k \in \FF_q^{i \times j}}
\chi\left[ \begin{matrix} g_i & k \\0 & g_j\end{matrix} \right]
\]
using \eqref{deflation-character-formula-v2}.
This shows both that $\HHH$ is a sub-coalgebra
of $A=A_\CC(GL)$ (that is, it satisfies
$\Delta \HHH \subset \HHH \otimes \HHH$)
and that $\HHH^\perp$ is a two-sided coideal
(that is, we have
$\Delta(\HHH^\perp) \subset \HHH^\perp \otimes A + A \otimes \HHH^\perp$),
since it shows that if
$\chi$ is supported only on unipotent classes,
then $\Delta(\chi)$ vanishes on
$(g_1,g_2)$ that have either $g_1$ or $g_2$
non-unipotent.
% [DG][v29] Again, removed one direction of an "if and only if".
% The other direction was true (because $\Delta(\chi)$ has
% a $GL_0 \times GL_n$-component which basically is a copy of
% $\chi$), but it was again useless and only confusing.
It also
shows that the surjection $A_\CC(GL) \twoheadrightarrow \HHH$
restricting every class function to unipotent classes is a
coalgebra homomorphism.
The rest follows.
\end{proof}

The subspace $\HHH$ is called the \dfn{Hall algebra}. It has
an obvious orthogonal $\CC$-basis, with interesting structure constants.

\begin{definition}
\label{def.hall-alg.hall-coeffs}
Given a partition $\lambda$ of $n$, let
$J_{\lambda}$ denote the $GL_n$-conjugacy class of
unipotent matrices whose \dfn{Jordan type} (that is, the list of the
sizes of the Jordan blocks, in decreasing order) is given by
$\lambda$.
Furthermore, let $z_\lambda(q)$ denote the size of the centralizer
of any element of this conjugacy class $J_{\lambda}$.

The indicator class functions\footnote{Here we use the following notation:
Whenever $P$ is a subset of a group $G$, we denote by $\triv_P$ the map
$G \to \CC$ which sends every element of $P$ to $1$ and all remaining
elements of $G$ to $0$. This is not in conflict with the notation
$\triv_G$ for the trivial character of $G$, since $\triv_P = \triv_G$
for $P = G$. Note that $\triv_P$ is a class function when $P$ is a union
of conjugacy classes of $G$.}
 $\{ \triv_{J_\lambda} \}_{\lambda \in \Par}$
form a $\CC$-basis for $\HHH$ whose multiplicative structure constants
are called the \dfn{Hall coefficients} $g^{\lambda}_{\mu,\nu}(q)$:
\[
\triv_{J_\mu} \triv_{J_\nu} = \sum_\lambda g^{\lambda}_{\mu,\nu}(q) \,\, \triv_{J_\lambda}.
\]
\end{definition}

% [DG][v19] Added the above footnote.

\noindent
Because the dual basis to $\{\triv_{J_\lambda}\}$ is
$\{ z_\lambda(q) \triv_{J_\lambda} \}$,
self-duality of $\HHH$ shows that the Hall coefficients are
(essentially) also structure constants for the comultiplication:
\[
\Delta \triv_{J_\lambda}
= \sum_{\mu,\nu} g^{\lambda}_{\mu,\nu}(q)
\frac{z_{\mu}(q) z_{\nu}(q)}{z_{\lambda}(q)} \cdot
\triv_{J_\mu} \otimes \triv_{J_\nu}.
\]

% [DG][v29] Fixed the definition of $z_\lambda(q)$ and the
% formulas referencing it.

The Hall coefficient $g^{\lambda}_{\mu,\nu}(q)$ has the following interpretation.

\begin{proposition}
\label{prop.hall-alg.product}
Fix any $g$ in $GL_n(\FF_q)$
acting unipotently on $\FF_q^n$ with Jordan type $\lambda$.
Then $g^{\lambda}_{\mu,\nu}(q)$ counts the $g$-stable $\FF_q$-subspaces $V \subset \FF_q^n$
for which the restriction
$g|V$ acts with Jordan type $\mu$, and the induced map $\bar{g}$
on the quotient space $\FF_q^n/V$ has Jordan type $\nu$.
\end{proposition}
\begin{proof}
Given $\mu,\nu$ partitions of $i,j$ with $i+j=n$, taking $\chi_i, \chi_j$ equal
to $\triv_{J_\mu}, \triv_{J_\nu}$ in
\eqref{Hall-algebra-product}
shows that for any $g$ in $GL_n$, the value of
$\left( \triv_{J_\mu} \cdot \triv_{J_{\nu}} \right)(g)$ is given by
\begin{equation}
\label{Hall-coefficient-crux}
\frac{1}{|P_{i,j}|}
  \left|\left\{ h \in GL_n: h^{-1} g h=\left[ \begin{matrix} g_i & * \\0 & g_j\end{matrix} \right] \text{ with }g_i \in J_{\mu},g_j \in J_{\nu} \right\}\right| .
\end{equation}
Let $S$ denote the set appearing in \eqref{Hall-coefficient-crux},
and let $\FF_q^i$ denote the $i$-dimensional
subspace of $\FF_q^n$ spanned by the first $i$ standard basis vectors.
Note that the condition on an element $h$ in $S$ saying that $h^{-1}gh$ is in block upper-triangular form can be re-expressed by saying that the subspace $V:=h(\FF_q^i)$ is $g$-stable.
One then sees that the map $h \overset{\varphi}{\longmapsto} V=h(\FF_q^i)$ surjects
$S$ onto the set of $i$-dimensional $g$-stable subspaces $V$ of $\FF_q^n$ for which
$g|V$ and $\bar{g}$ are unipotent of types $\mu,\nu$, respectively.  Furthermore, for
any particular such $V$,
its fiber $\varphi^{-1}(V)$ in $S$ is a coset of the stabilizer within
% [DG][v29] Added "a coset of" on the previous line.
$GL_n$ of $V$, which is conjugate to $P_{i,j}$, and hence has cardinality
$|\varphi^{-1}(V)|=|P_{i,j}|$.  This proves the assertion of the proposition.
\end{proof}

The Hall algebra $\HHH$ will turn out to be isomorphic to the ring $\Lambda_\CC$ of symmetric
functions with $\CC$ coefficients, via a composite $\varphi$ of three maps
\[
\Lambda_\CC \longrightarrow
A(GL)(\iota)_\CC \longrightarrow
A(GL)_\CC \longrightarrow
\HHH
\]
in which the first map is the isomorphism from
Proposition~\ref{Sym-to-Steinberg-isomorphism}, the second is inclusion,
and the third is the quotient map from
Proposition~\ref{Hall-algebra-is-Hopf-sub-quotient}.

\begin{theorem}
\label{Sym-to-Hall-isomorphism-thm}
The above composite $\varphi$ is a Hopf algebra isomorphism, sending
\[
\begin{array}{rcl}
h_n &\longmapsto & \sum_{\lambda \in \Par_n} \triv_{J_\lambda}, \\
e_n &\longmapsto & q^{\binom{n}{2}} \triv_{J_{(1^n)}}, \\
p_n &\longmapsto & \sum_{\lambda \in \Par_n} (q;q)_{\ell(\lambda) - 1}\triv_{J_\lambda} \qquad \left(\text{for } n > 0\right),
\end{array}
\]
where we are using the notation
\[
(x;q)_{m}
:= (1-x)(1-qx)(1-q^2x) \cdots (1-q^{m-1}x)
  \qquad \text{ for all } m \in \NN \text{ and } x \text{ in any ring}.
\]
\end{theorem}

\begin{proof}
That $\varphi$ is a graded Hopf morphism follows because it is a composite
of three such morphisms.  We claim that once one shows the formula for the
(nonzero) image $\varphi(p_n)$ given above is correct,
then this will already show $\varphi$ is an isomorphism,
by the following argument.  Note first that $\Lambda_\CC$ and $\HHH$
both have dimension $|\Par_n|$ for their $n$-th homogeneous
components, so it suffices to show that the graded map $\varphi$ is injective.
On the other hand, both $\Lambda_\CC$ and $\HHH$
are (graded, connected, finite type) \emph{self-dual} Hopf algebras
(although with respect to a sesquilinear form),
so Theorem~\ref{general-structure-theorem} says that each is the symmetric algebra
on its space of primitive elements.  Thus it suffices to check that
$\varphi$ is injective when restricted to their subspaces of
primitives.\footnote{An alternative way to see that it suffices to
check this is by recalling Exercise~\ref{exe.I=0}(c).}
For $\Lambda_\CC$, by Corollary~\ref{primitives-in-Lambda} the primitives
are spanned by $\{p_1,p_2,\ldots\}$, with only
one basis element in each degree $n\geq 1$.  Hence
$\varphi$ is injective on the subspace of primitives if and only if
it does not annihilate any $p_n$.

% [DG][v14] There is a subtlety here that even I don't want to deal
% with: The Hall algebra $\HHH$ is self-dual with respect to a
% *Hermitian*, not bilinear, form; this isn't something you have
% defined. I guess the Zelevinsky results still work in this case,
% and I also guess it is easy to identify a $\QQ$-form of $A(GL)_\CC$
% in which the elements $h_n$, $e_n$, $p_n$ of $\HHH$ are contained,
% and thus get around this distinction.
% Anyway I don't care much, because Exercise~\ref{exe.I=0} makes it
% possible to completely avoid using PSH theory here. Would you
% object to me adding a reference to this exercise as an alternative
% proof?

% [DG][v23] Added footnote. I don't really like either argument, to be
% honest. What about the following alternative, which I suggest adding
% as an exercise: check that
% $\varphi(e_\lambda)$ equals $(nonzero scalar) \triv_{J_\lambda^t}$ plus
% dominance-smaller terms? (Whence the map is unitriangular.)

% [DG][v29] Added the exercise which follows just after this proof.

Thus it only remains to show the above formulas for the images of
$h_n,e_n,p_n$ under $\varphi$.
This is clear for $h_n$, since
Proposition~\ref{Sym-to-Steinberg-isomorphism} shows
that it maps under the first two composites to the indicator function
$\triv_{GL_n}$ which then restricts to the sum of indicators
$\sum_{\lambda \in \Par_n} \triv_{J_\lambda}$ in $\HHH$.
For $e_n, p_n$, we resort to generating functions.
Let $\tilde{h}_n,\tilde{e}_n,\tilde{p}_n$ denote the
three putative images in $\HHH$ of $h_n,e_n,p_n$,
appearing on the right side in the theorem, and define
generating functions
\[
\tilde{H}(t):=\sum_{n \geq 0} \tilde{h}_n t^n, \quad
\tilde{E}(t):=\sum_{n \geq 0} \tilde{e}_n t^n, \quad
\tilde{P}(t):=\sum_{n \geq 0} \tilde{p}_{n+1} t^n
\qquad \qquad \text{ in } \HHH[[t]] .
\]
We wish to show that the map
$\varphi[[t]] : \Lambda_\CC [[t]] \to \HHH[[t]]$
(induced by $\varphi$) maps $H(t),E(t),P(t)$ in $\Lambda[[t]]$
to these three generating functions\footnote{See
\eqref{h-generating-function},
\eqref{e-generating-function},
\eqref{E(t)-H(t)-P(t)-relation} for the definitions
of $H(t),E(t),P(t)$.}.  Since we have already shown this is
correct for $H(t)$, by \eqref{E(t)-H(t)-relation},
\eqref{E(t)-H(t)-P(t)-relation}, it suffices to check that in $\HHH[[t]]$ one has
\[
\begin{array}{rccl}
\tilde{H}(t) \tilde{E}(-t) = 1,& \text{ or equivalently, }
  &\sum_{k=0}^n (-1)^k \tilde{e}_k \tilde{h}_{n-k} = \delta_{0,n} ;\\
\tilde{H}'(t) \tilde{E}(-t) = \tilde{P}(t),& \text{  or equivalently, }
  &\sum_{k=0}^n (-1)^k (n-k) \tilde{e}_k \tilde{h}_{n-k} = \tilde{p}_n.
\end{array}
\]
Thus it would be helpful to evaluate the class function
$\tilde{e}_k \tilde{h}_{n-k}$.  Note that a unipotent $g$ in $GL_n$
having $\ell$ Jordan blocks has an $\ell$-dimensional $1$-eigenspace, so that
the number of $k$-dimensional $g$-stable $\FF_q$-subspaces of $\FF_q^n$ on
which $g$ has Jordan type $\left(1^k\right)$ (that is, on which $g$
acts as the identity) is the \dfn{$q$-binomial coefficient}
\[
\qbin{\ell}{k}{q} = \frac{(q;q)_\ell}{(q;q)_k (q;q)_{\ell-k}} ,
\]
counting $k$-dimensional $\FF_q$-subspaces $V$ of
an $\ell$-dimensional $\FF_q$-vector space; see, e.g., \cite[\S 1.7]{Stanley}.
Hence, for a unipotent $g$ in $GL_n$
having $\ell$ Jordan blocks, we have
\[
(\tilde{e}_k \tilde{h}_{n-k})(g)
= q^{\binom{k}{2}} \cdot \left(\triv_{J_{(1^k)}} \cdot \tilde{h}_{n-k}\right) (g)
= q^{\binom{k}{2}} \cdot \sum_{\nu \in \Par_{n-k}} \left(\triv_{J_{(1^k)}} \cdot \triv_{J_\nu}\right) (g)
=  q^{\binom{k}{2}} \qbin{\ell}{k}{q}
\]
(by Proposition \ref{prop.hall-alg.product}).
Thus one needs for $\ell \geq 1$ that
\begin{align}
\sum_{k=0}^\ell (-1)^k
q^{\binom{k}{2}} \qbin{\ell}{k}{q}
&= 0,\label{first-q-binomial-identity}\\
\sum_{k=0}^\ell (-1)^k (n-k)
q^{\binom{k}{2}} \qbin{\ell}{k}{q}
&= (q;q)_{\ell - 1}.\label{second-q-binomial-identity}
\end{align}
Identity \eqref{first-q-binomial-identity}
comes from setting $x=1$ in the \dfn{$q$-binomial theorem} \cite[Exer. 3.119]{Stanley}:
\begin{equation}
\sum_{k=0}^\ell (-1)^k q^{\binom{k}{2}} \qbin{\ell}{k}{q} x^{\ell-k}
=(x-1)(x-q)(x-q^2)\cdots (x-q^{\ell-1}).
\label{q-binomial-theorem}
\end{equation}
Identity \eqref{second-q-binomial-identity}
comes from applying $\frac{d}{dx}$ to \eqref{q-binomial-theorem},
then setting $x=1$, and finally
adding $(n-\ell)$ times \eqref{first-q-binomial-identity}.
\end{proof}

% [DG][v29] Added some more detail.

\begin{exercise}
\label{exe.hall-alg.dominance}
Fix a prime power $q$.
For any $k \in \NN$, and any $k$ partitions
$\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}$, we define
a family
$\left(g_{\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}}^\lambda \left(q\right) \right) _ {\lambda \in \Par}$
of elements of $\CC$ by the equation
\[
\triv_{J_{\lambda^{(1)}}} \triv_{J_{\lambda^{(2)}}} \cdots \triv_{J_{\lambda^{(k)}}}
= \sum_{\lambda \in \Par} g_{\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}}^\lambda \left(q\right) \triv_{J_\lambda}
\]
in $\HHH$.
This notation generalizes the notation $g^\lambda_{\mu, \nu} \left(q\right)$ we introduced in Definition~\ref{def.hall-alg.hall-coeffs}. Note that $g^\lambda_\mu \left(q\right) = \delta_{\lambda, \mu}$ for any two partitions $\lambda$ and $\mu$, and that $g^\lambda \left(q\right) = \delta_{\lambda, \varnothing}$ for any partition $\lambda$ (where $g^\lambda \left(q\right)$ is to be understood as $g_{\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}}^\lambda \left(q\right)$ for $k = 0$).

\begin{itemize}

\item[(a)] Let $\lambda \in \Par$, and let $n = \left| \lambda \right|$. Let $V$ be an $n$-dimensional $\FF_q$-vector space, and let $g$ be a unipotent endomorphism of $V$ having Jordan type $\lambda$. Let $k \in \NN$, and let $\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}$ be $k$ partitions. A \emph{$\left(\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}\right)$-compatible $g$-flag}\index{compatible flag} will mean a sequence $0 = V_0 \subset V_1 \subset V_2 \subset \cdots \subset V_k = V$ of $g$-invariant $\FF_q$-vector subspaces $V_i$ of $V$ such that for every $i \in \left\{1, 2, \ldots, k\right\}$, the endomorphism of $V_i / V_{i-1}$ induced by $g$ \ \ \ \ \footnote{This is well-defined. In fact, both $V_i$ and $V_{i-1}$ are $g$-invariant, so that $g$ restricts to an endomorphism of $V_i$, which further restricts to an endomorphism of $V_{i-1}$, and thus gives rise to an endomorphism of $V_i / V_{i-1}$.} has Jordan type $\lambda^{(i)}$.

Show that
$g_{\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}}^\lambda \left(q\right)$
is the number of $\left(\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}\right)$-compatible $g$-flags.\footnote{This can be seen as a generalization of Proposition~\ref{prop.hall-alg.product}. In fact, if $\mu$ and $\nu$ are two partitions, then a $\left(\mu, \nu\right)$-compatible $g$-flag is a sequence $0 = V_0 \subset V_1 \subset V_2 = V$ of $g$-invariant $\FF_q$-vector subspaces $V_i$ of $V$ such that the endomorphism of $V_1 / V_0 \cong V_1$ induced by $g$ has Jordan type $\mu$, and the endomorphism of $V_2 / V_1 \cong V / V_1$ induced by $g$ has Jordan type $\nu$. Choosing such a sequence amounts to choosing $V_1$ (since there is only one choice for each of $V_0$ and $V_2$), and the conditions on this $V_1$ are precisely the conditions on $V$ in Proposition~\ref{prop.hall-alg.product}.}

\item[(b)] Let $\lambda \in \Par$. Let $k \in \NN$, and let $\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}$ be $k$ partitions. Show that
$g_{\lambda^{(1)}, \lambda^{(2)}, \ldots, \lambda^{(k)}}^\lambda \left(q\right) = 0$
unless
$\left| \lambda^{(1)} \right| + \left| \lambda^{(2)} \right| + \cdots + \left| \lambda^{(k)} \right| = \left| \lambda \right|$
and
$\lambda^{(1)} + \lambda^{(2)} + \cdots + \lambda^{(k)} \triangleright \lambda$.
(Here and in the following, we are using the notations of Exercise~\ref{exe.liri.simple}).

\item[(c)] Let $\lambda \in \Par$, and let us write the transpose partition $\lambda^t$ as $\lambda^t = \left(\left(\lambda^t\right)_1, \left(\lambda^t\right)_2, \ldots, \left(\lambda^t\right)_\ell\right)$. Show that
$g_{\left(1^{\left(\lambda^t\right)_1}\right), \left(1^{\left(\lambda^t\right)_2}\right), \ldots, \left(1^{\left(\lambda^t\right)_\ell}\right)}^\lambda \left(q\right) \neq 0$.

\item[(d)] Let $n \in \NN$ and $\lambda \in \Par_n$. Show that
\[
\varphi\left(e_\lambda\right)
= \sum_{\mu \in \Par_n ; \  \lambda^t \triangleright \mu}
  \alpha_{\lambda, \mu} \triv_{J_\mu}
\]
for some coefficients $\alpha_{\lambda, \mu} \in \CC$ satisfying $\alpha_{\lambda, \lambda^t} \neq 0$.

\item[(e)] Give another proof of the fact that the map $\varphi$ is injective.

\end{itemize}

[\textbf{Hint:} For (b), use Exercise~\ref{exe.liri.jordan}(b).]

\end{exercise}

% [DG][v31] Added the hint (which I had intended to add in v29 but forgot). This
% should clear up the issue from your last email?

We next indicate, without proof, how $\HHH$ relates to the
classical Hall algebra.

\begin{definition}
Let $p$ be a prime.
The usual \dfn{Hall algebra}, or what Schiffmann \cite[\S 2.3]{Schiffmann}
calls \dfn{Steinitz's classical Hall algebra}
(see also Macdonald \cite[Chap. II]{Macdonald}),
has $\ZZ$-basis elements
$\{ u_\lambda \}_{\lambda \in \Par}$, with the multiplicative structure constants
$g^{\lambda}_{\mu,\nu}(p)$ in
\[
u_\mu u_\nu = \sum_\lambda
g^{\lambda}_{\mu,\nu}(p)\,\, u_\lambda
\]
defined as follows: fix a finite abelian $p$-group $L$ of
\emph{type $\lambda$}\index{type of an abelian $p$-group},
meaning that
\[
L \cong \bigoplus_{i=1}^{\ell(\lambda)} \ZZ/p^{\lambda_i}\ZZ,
\]
and let $g^{\lambda}_{\mu,\nu}(p)$ be the number of subgroups $M$ of $L$ of
type $\mu$, for which the quotient $N:=L/M$ is of type $\nu$.  In other words,
$g^{\lambda}_{\mu,\nu}(p)$ counts, for a fixed abelian
$p$-group $L$ of type $\lambda$,  the number of short exact sequences
$
0 \rightarrow M
\rightarrow L
\rightarrow N
\rightarrow 0
$
in which $M,N$ have types $\mu,\nu$, respectively (modulo isomorphism
of short exact sequences restricting to the identity on $L$).
\end{definition}

% [DG][v23] Added "Let $p$ be a prime" to the above definition -- or was
% this in the context?

We claim that when one takes the finite field $\FF_q$ of order
$q=p$ a \emph{prime}, the $\ZZ$-linear map
\begin{equation}
\label{classical-Hall-to-Zelevinsky-Hall}
u_\lambda \longmapsto \triv_{J_\lambda}
\end{equation}
gives an isomorphism from this classical Hall algebra
to the $\ZZ$-algebra $\HHH_\ZZ \subset \HHH$.
The key point is \dfn{Hall's Theorem}, a non-obvious
statement for which Macdonald includes two proofs in
\cite[Chap. II]{Macdonald}, one of them due to
Zelevinsky\footnote{See also \cite[Thm. 2.6, Prop. 2.7]{Schiffmann}
for quick proofs of part of it, similar to Zelevinsky's.
Another proof, based on a recent category-theoretical
paradigm, can be found in \cite[Theorem 3.53]{Dyckerhoff}.}.
To state it, we first recall some notions about discrete valuation rings.

% [DG][v52] Added {Dyckerhoff} reference. This looks fascinating and
% readable -- I'll probably start reading it soon.

\begin{definition}
A \dfn{discrete valuation ring} (short \dfn{DVR}) $\ooo$ is
a principal ideal domain having only one maximal ideal $\mmm \neq 0$,
with quotient $k=\ooo/\mmm$ called its \dfn{residue field}.

% [DG][v29] Added nonzeroness condition on $\mmm$.

The structure theorem for finitely generated modules
over a PID implies that an $\ooo$-module $L$
with finite composition series of composition length $n$
must have $L \cong \bigoplus_{i=1}^{\ell(\lambda)} \ooo/\mmm^{\lambda_i}$
for some partition $\lambda$ of $n$; say $L$ has
\emph{type}\index{type of a module over a DVR} $\lambda$
in this situation.
\end{definition}

Here are the two crucial examples for us.

\begin{example}
\label{DVR-example-F[[t]]}
For any field $\FF$, the power series ring $\ooo=\FF[[t]]$ is a DVR
with maximal ideal $\mmm=(t)$ and
residue field $k=\ooo/\mmm=\FF[[t]]/(t)\cong\FF$.
An $\ooo$-module $L$ of type $\lambda$ is an $\FF$-vector space together
with an $\FF$-linear transformation $T \in \End L$ that acts
on $L$ nilpotently
(so that $g:=T+1$ acts unipotently, where $1 = \id_L$) with
Jordan blocks of sizes given by $\lambda$:  each summand
$\ooo/\mmm^{\lambda_i}=\FF[[t]]/(t^{\lambda_i})$ of $L$ has
an $\FF$-basis $\{1,t,t^2,\ldots,t^{\lambda_i-1}\}$ on
which the map $T$ that multiplies by $t$ acts as a nilpotent Jordan block of
size $\lambda_i$.
Note also that, in this setting, $\ooo$-submodules are the same as
$T$-stable (or $g$-stable) $\FF$-subspaces.
\end{example}

\begin{example}
\label{DVR-example-Z_p}
The ring of $p$-adic integers $\ooo=\ZZ_p$ is a DVR with
maximal ideal $\mmm=(p)$ and
residue field $k=\ooo/\mmm=\ZZ_p/p\ZZ_p \cong \ZZ/p\ZZ$.
An $\ooo$-module $L$ of type $\lambda$ is an abelian $p$-group of
type $\lambda$:  for each summand,
$\ooo/\mmm^{\lambda_i}=\ZZ_p/p^{\lambda_i}\ZZ_p \cong \ZZ/p^{\lambda_i}\ZZ$.
Note also that, in this setting, $\ooo$-submodules are the same as
subgroups.
\end{example}

\noindent
One last notation: $n(\lambda):=\sum_{i \geq 1} (i-1)\lambda_i$,
for $\lambda$ in $\Par$.  Hall's Theorem is as follows.

\begin{theorem}
\label{Hall's-theorem}
Assume $\ooo$ is a DVR with maximal ideal $\mmm$, and that its residue field
$k=\ooo/\mmm$ is finite of cardinality $q$.
Fix an $\ooo$-module $L$ of type $\lambda$.
Then the number of $\ooo$-submodules $M$ of
type $\mu$ for which the quotient $N=L/M$ is of type $\nu$ can be
written as the specialization
\[
[g^{\lambda}_{\mu,\nu}(t)]_{t=q}
\]
of a polynomial $g^{\lambda}_{\mu,\nu}(t)$ in $\ZZ[t]$, called the
\dfn{Hall polynomial}.

Furthermore, the Hall polynomial $g^{\lambda}_{\mu,\nu}(t)$
has degree at most $n(\lambda)-(n(\mu)+n(\nu))$, and its
coefficient of $t^{n(\lambda)-(n(\mu)+n(\nu))}$ is the Littlewood-Richardson
coefficient $c^{\lambda}_{\mu,\nu}$.
\end{theorem}

\noindent
Comparing what Hall's Theorem says in
Examples~\ref{DVR-example-F[[t]]}
and \ref{DVR-example-Z_p}, shows that
the map \eqref{classical-Hall-to-Zelevinsky-Hall}
gives the desired isomorphism from the classical Hall algebra
to $\HHH_\ZZ$.

We close this section with some remarks on the vast literature
on Hall algebras that we will \emph{not} discuss here.

\begin{remark}
Macdonald's version of Hall's Theorem \cite[(4.3)]{Macdonald} is stronger
than Theorem~\ref{Hall's-theorem}, and useful for certain applications:
he shows that $g^{\lambda}_{\mu,\nu}(t)$ is the zero polynomial whenever
the Littlewood-Richardson coefficient $c^{\lambda}_{\mu,\nu}$ is zero.
\end{remark}

\begin{remark}
In general, not all coefficients of the Hall polynomials
$g^\lambda_{\mu, \nu}(t)$ are nonnegative (see Butler/Hales
\cite{ButlerHales} for a study of when they are); it often
happens that $g^\lambda_{\mu, \nu}(1) = 0$ despite
$g^\lambda_{\mu, \nu}(t)$ not being the zero
polynomial\footnote{Actually, Butler/Hales show in
\cite[proof of Prop. 2.4]{ButlerHales} that the values
$g^\lambda_{\mu, \nu}(1)$ are the structure constants of
the ring $\Lambda$ with respect to its basis
$\left(m_\lambda\right)_{\lambda \in \Par}$: we have
\[
m_\mu m_\nu = \sum_{\lambda \in \Par} g^\lambda_{\mu, \nu} (1) m_\lambda
\]
for all partitions $\mu$ and $\nu$.}. However,
in \cite[Thm. 4.2]{Klein-hall}, Klein showed that the polynomial
values $g^\lambda_{\mu, \nu}\left(p\right)$ for $p$ prime
are always positive when $c^\lambda_{\mu, \nu} \neq 0$. (This
easily yields the same result for $p$ a prime power.)
\end{remark}

% [DG][v29] Added above remark.

% [DG][v32] Added footnote. I suspect that the unexpected
% proneness of $g^\lambda_{\mu, \nu}(1)$ to vanish is the reason
% why the converse of Exercise~\ref{exe.liri.jordan}(b) cannot
% be proven by a construction of the kind "consider the free
% vector space with basis the set of cells of $\lambda$, and map
% every cell to some other cell or $0$" (viz., the converse of
% Exercise~\ref{exe.liri.jordan}(b) does not hold over the
% "field" $\FF_1$). I have not checked this in detail.

\begin{remark}
\label{Green-polynomial-remark}
Zelevinsky in \cite[Chaps 10, 11]{Zelevinsky} uses
the isomorphism $\Lambda_\CC \rightarrow \HHH$ to derive J. Green's formula
for the value of any irreducible character $\chi$ of $GL_n$ on any unipotent
class $J_\lambda$.  The answer involves values of irreducible characters
of $\Symm_n$ along with \emph{Green's polynomials}
$Q^\lambda_\mu(q)$
(see Macdonald \cite[\S III.7]{Macdonald};  they are
denoted $Q(\lambda,\mu)$ by Zelevinsky),
which express the images under the isomorphism
of Theorem~\ref{Sym-to-Hall-isomorphism-thm}
of the symmetric function basis $\{p_\mu\}$ in terms of the
basis $\{\triv_{J_\lambda}\}$.
\end{remark}

\begin{remark}
The Hall polynomials $g^{\lambda}_{\mu,\nu}(t)$ also essentially give the
multiplicative structure constants for $\Lambda(\xx)[t]$ with
respect to its basis of \emph{Hall-Littlewood symmetric functions}
$P_\lambda=P_{\lambda}(\xx;t)$:
\[
P_{\mu} P_{\nu}
= \sum_{\lambda} t^{n(\lambda)-(n(\mu)+n(\nu))} g^{\lambda}_{\mu,\nu}(t^{-1}) P_{\lambda}.
\]
See Macdonald \cite[\S III.3]{Macdonald}.
\end{remark}

\begin{remark}
Schiffmann \cite{Schiffmann} discusses self-dual Hopf algebras which
vastly generalize the classical Hall algebra called \emph{Ringel-Hall algebras},
associated to abelian categories which are hereditary.  Examples
come from categories of nilpotent representations of quivers;
the quiver having exactly one node and one arc recovers the
classical Hall algebra $\HHH_\ZZ$ discussed above.
\end{remark}

\begin{remark}
The general linear groups $GL_n(\FF_q)$ are one of four families
of so-called \emph{classical groups}. Progress has been made on
extending Zelevinsky's PSH theory to the other families:

(a) Work of Thiem and Vinroot \cite{ThiemVinroot} shows that
the tower $\{G_*\}$ of
\emph{finite unitary groups} $U_n(\FF_{q^2})$
give rise to another positive self-dual Hopf algebra
$A=\bigoplus_{n \geq 0}R( U_n(\FF_{q^2}) )$,
in which the role of Harish-Chandra induction
is played by \emph{Deligne-Lusztig induction}.  In this theory,
character and degree formulas for $U_n(\FF_{q^2})$ are related
to those of $GL_n(\FF_q)$ by substituting $q \mapsto -q$, along
with appropriate scalings by $\pm 1$, a phenomenon sometimes called
\emph{Ennola duality}.  See also \cite[\S 4]{RStantonThiem}.

(b) van Leeuwen \cite{Leeuwen-hopf}
has studied $\bigoplus_{n\geq0}R\left(  Sp_{2n}\left(
\FF_{q}\right)  \right)  $, $\bigoplus_{n\geq0}R\left(
O_{2n}\left(  \FF_{q}\right)  \right)  $ and $\bigoplus_{n\geq
0}R\left(  U_{n}\left(  \FF_{q^{2}}\right)  \right)  $ not as Hopf
algebras, but rather as so-called \emph{twisted PSH-modules} over the
PSH $A(GL)$ (a ``deformed'' version of the older notion of
Hopf modules). He classified these PSH-modules axiomatically
similarly to Zelevinsky's above classification of PSH's.

(c) In a recent honors thesis \cite{ShelleyAbrahamson-honors},
Shelley-Abrahamson defined yet another variation of the concept of
Hopf modules, named \emph{$2$-compatible Hopf modules}, and
identified $\bigoplus_{n\geq0}R\left(  Sp_{2n}\left(
\FF_{q}\right)  \right)  $ and $\bigoplus_{n\geq0}R\left(
O_{2n+1}\left(  \FF_{q}\right)  \right)  $ as such modules over
$A(GL)$.
\end{remark}

% [DG][v3] Here's hoping I haven't misrepresented either of (b) and (c)
% -- I have not really read any of them... I am wondering if
% (b) and (c) are related, particularly for the $Sp_{2n}$ family
% they have in common. The definitions of van Leeuwen's and
% Shelley-Abrahamson's Hopf modules don't completely seem to
% match, although they are similar.

\newpage




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Quasisymmetric functions and $P$-partitions}
\label{Qsym-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We discuss here our next important example of a Hopf algebra arising in
combinatorics:  the \emph{quasisymmetric functions} of Gessel \cite{Gessel},
with roots in work of Stanley \cite{Stanley-thesis} on $P$-partitions.
Other treatments of quasisymmetric functions
can be found in \cite[Section 7.19]{Stanley} and \cite[Chapter 8]{Sagan2020}
(with focus on their enumerative applications rather than on their Hopf
structure) and in \cite[Chapter 6]{Meliot} (with a focus on their
representation-theoretical meaning).
Quasisymmetric functions have found applications in combinatorial
enumeration (\cite[Chapter 8]{Sagan2020}, \cite[Section 7.19]{Stanley}),
topology (\cite{BakerRichter}) and algebraic geometry
(\cite{Morava}, \cite{Oesinghaus}).

\subsection{Definitions, and Hopf structure}

The definitions of quasisymmetric functions require a totally ordered variable set.
Usually we will use a variable set denoted $\xx=(x_1,x_2,\ldots)$ with the
usual ordering $x_1 < x_2 < \cdots$.  However, it is good to have some flexibility
in changing the ordering, which is why we make the following definition.

\begin{definition}
Given any totally ordered set $I$, create a totally ordered
variable set $\{x_i\}_{i\in I}$, and then let $R(\{x_i\}_{i\in I})$
denote the power series
of bounded degree in $\{x_i\}_{i \in I}$
having coefficients in $\kk$.

The \dfn{ring of quasisymmetric functions}\index{quasisymmetric function}
\dfn{$\Qsym(\{x_i\}_{i \in I})$}\index{$\Qsym$}
\emph{over the alphabet $\{x_i\}_{i \in I}$}
will be the $\kk$-submodule consisting of the elements $f$ in
$R(\{x_i\}_{i\in I})$ that have the same coefficient on
the monomials $x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell}$
and $x_{j_1}^{\alpha_1} \cdots x_{j_\ell}^{\alpha_\ell}$
whenever both $i_1 < \cdots < i_\ell$ and $j_1 < \cdots < j_\ell$
in the total order on $I$. We write
$\Qsym_\kk (\{x_i\}_{i \in I})$ instead of
$\Qsym (\{x_i\}_{i \in I})$
to stress the choice of base ring $\kk$.
\end{definition}

% [DG][v17] I restructured this section, making the distinction between
% $\Qsym(\{x_i\}_{i \in I})$ and the "universal" $\Qsym$ more explicit.
% (They are identified only for $I = \{1,2,3,\ldots\}$ now.)

It immediately follows from this definition that
$\Qsym(\{x_i\}_{i \in I})$ is a free $\kk$-submodule of
$R(\{x_i\}_{i\in I})$, having as $\kk$-basis elements the
\emph{monomial quasisymmetric functions}\index{monomial quasisymmetric function}\index{$M_\alpha$}
\[
M_\alpha(\{x_i\}_{i\in I}):=
\sum_{i_1 < \cdots < i_\ell\text{ in }I}
x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell}
\]
for all compositions\footnote{Recall that compositions
were defined in Definition~\ref{def.composition-almost-composition},
along with related concepts such as length and size.}
$\alpha$ satisfying $\ell(\alpha) \leq \left|I\right|$.
When $I$ is infinite, this means that the $M_\alpha$ for all
compositions $\alpha$ form a basis of $\Qsym(\{x_i\}_{i \in I})$.

Note that
$\Qsym (\{x_i\}_{i \in I})
 = \bigoplus_{n \geq 0} \Qsym_n (\{x_i\}_{i \in I})$ is a
graded $\kk$-module of finite type, where
$\Qsym_n (\{x_i\}_{i \in I})$ is the $\kk$-submodule
of quasisymmetric functions which are homogeneous of degree $n$.
Letting \dfn{$\Comp$} denote the set of all compositions $\alpha$, and
\dfn{$\Comp_n$} the compositions $\alpha$ of $n$ (that is, compositions whose
parts sum to $n$), the subset
$\{ M_\alpha \}_{\alpha \in \Comp_n;\ \ell(\alpha) \leq \left|I\right|}$
gives a $\kk$-basis for $\Qsym_n (\{x_i\}_{i \in I})$.


\begin{example}
Taking the variable set $\xx=(x_1 < x_2 < \cdots)$ to define $\Qsym(\xx)$,
for $n=0,1,2,3$, one has these basis elements in $\Qsym_n(\xx)$:
\[
\begin{array}{rll}
M_{()} =M_\varnothing&= 1 ,&\\
& & \\
M_{(1)} &= x_1+x_2+x_3+ \cdots &= m_{(1)}=s_{(1)}=e_1=h_1=p_1 ,\\
& & \\
M_{(2)} &= x_1^2+x_2^2+x_3^2+ \cdots &=m_{(2)}=p_2 ,\\
M_{(1,1)} &= x_1x_2+x_1x_3+x_2x_3+ \cdots&=m_{(1,1)}=e_2 ,\\
& & \\
M_{(3)} &= x_1^3+x_2^3+x_3^3+ \cdots &=m_{(3)}=p_3 ,\\
M_{(2,1)} &= x_1^2 x_2 + x_1^2 x_3 + x_2^2 x_3 + \cdots ,&\\
M_{(1,2)} &= x_1 x_2^2 + x_1 x_3^2 + x_2 x_3^2 + \cdots ,&\\
M_{(1,1,1)} &= x_1 x_2 x_3 +x_1 x_2 x_4 + x_1 x_3 x_4 +  \cdots &=m_{(1,1,1)}=e_3 .
\end{array}
\]
\end{example}

It is not obvious that $\Qsym(\xx)$ is a subalgebra of $R(\xx)$,
but we will show this momentarily.  For example,
\begin{align*}
M_{(a)}M_{(b, c)}
&=(x_1^a + x_2^a + x_3^a+\cdots)
(x_1^b x_2^c + x_1^b x_3^c + x_2^b x_3^c + \cdots)\\
&=x_1^{a+b} x_2^c + \cdots +x_1^b x_3^{a+c} + \cdots
  + x_1^a x_2^b x_3^c + \cdots
  +x_1^b x_2^a x_3^c + \cdots
  + x_1^b x_2^c x_3^a + \cdots\\
&=M_{(a+b, c)} + M_{(b, a+c)} + M_{(a, b, c)} + M_{(b, a, c)} + M_{(b, c, a)} .
\end{align*}

\begin{proposition}
\label{QSym-multiplication}
For any infinite totally ordered set $I$, one has that
$\Qsym(\{x_i\}_{i \in I})$  is a $\kk$-subalgebra of
$R(\{x_i\}_{i\in I})$, with multiplication in the
$\{M_\alpha\}$-basis as follows:
Fix three disjoint chain posets
$(i_1 < \cdots <i_\ell)$, $(j_1 < \cdots < j_m)$ and
$(k_1 < k_2 < \cdots)$. Now, if
$\alpha=(\alpha_1,\ldots,\alpha_\ell)$
and $\beta=(\beta_1,\ldots,\beta_m)$
are two compositions,
then
\begin{equation}
\label{Qsym-product-on-monomials}
M_\alpha M_\beta = \sum_f M_{\wt (f)}
\end{equation}
in which the sum is over all $p \in \NN$ and all maps $f$ from the
disjoint union of two chains to a chain
\begin{equation}
\label{map-from-two-chains-to-one}
(i_1 < \cdots <i_\ell) \sqcup (j_1 < \cdots < j_m)
\overset{f}{\longrightarrow}
(k_1 < \cdots <k_p)
\end{equation}
which are both surjective and strictly order-preserving (that is,
if $x$ and $y$ are two elements in the domain satisfying $x < y$,
then $f(x) < f(y)$),
and where the composition $\wt(f):=(\wt_1(f),\ldots,\wt_p(f))$ is defined by
$
\wt_s(f):= \sum_{i_u \in f^{-1}(k_s)} \alpha_u + \sum_{j_v \in f^{-1}(k_s)} \beta_v.
$
\end{proposition}

\begin{example}
\label{exa.QSym-multiplication}
For this example, set $\alpha = \left(2,1\right)$ and
$\beta = \left(3,4,2\right)$. Let us compute $M_\alpha M_\beta$ using
\eqref{Qsym-product-on-monomials}. Indeed, the length of $\alpha$ is
$\ell = 2$, and the length of $\beta$ is $m = 3$, so the sum on the
right hand side of \eqref{Qsym-product-on-monomials} is a sum over all
$p \in \NN$ and all surjective strictly order-preserving maps $f$
from the disjoint union $(i_1 < i_2) \sqcup (j_1 < j_2 < j_3)$ of two
chains to the chain $(k_1 < k_2 < \cdots < k_p)$. Such maps can exist
only when $p \leq 5$ (due to having to be surjective) and only for
$p \geq 3$ (since, being strictly order-preserving, they have to
be injective when restricted to $(j_1 < j_2 < j_3)$). Hence,
enumerating them is a finite problem. The reader can check
that the value obtained for $M_\alpha M_\beta$ is
\begin{align*}
& M_{\left(2,1,3,4,2\right)} + M_{\left(2,3,1,4,2\right)} + M_{\left(2,3,4,1,2\right)} + M_{\left(2,3,4,2,1\right)}
+ M_{\left(3,2,1,4,2\right)} \\
& + M_{\left(3,2,4,1,2\right)} + M_{\left(3,2,4,2,1\right)}
+ M_{\left(3,4,2,1,2\right)} + M_{\left(3,4,2,2,1\right)}
+ M_{\left(3,4,2,2,1\right)} \\
& + M_{\left(2, 3, 4, 3\right)} + M_{\left(2, 3, 5, 2\right)} + M_{\left(2, 4, 4, 2\right)} + M_{\left(3, 2, 4, 3\right)} + M_{\left(3, 2, 5, 2\right)} + M_{\left(3, 4, 2, 3\right)} \\
& + M_{\left(3, 4, 4, 1\right)} + M_{\left(3, 6, 1, 2\right)} + M_{\left(3, 6, 2, 1\right)} + M_{\left(5, 1, 4, 2\right)} + M_{\left(5, 4, 1, 2\right)} + M_{\left(5, 4, 2, 1\right)} \\
& + M_{\left(5, 4, 3\right)} + M_{\left(5, 5, 2\right)} + M_{\left(3, 6, 3\right)}.
\end{align*}
Here, we have listed the addends corresponding to $p=5$ on the first
two rows, the addends corresponding to $p=4$ on the next two rows,
and those corresponding to $p=3$ on the fifth row. The reader might
notice that the first two rows (i.e., the addends with $p=5$) are
basically a list of shuffles of $\alpha$ and $\beta$: In general,
the maps \eqref{map-from-two-chains-to-one} for $p = \ell + m$ are in
bijection with the elements of $\operatorname{Sh}_{\ell, m}$
\ \ \ \ \footnote{The bijection takes a map $f$ to the inverse of
the permutation $\sigma \in \Symm_p$ which sends every
$x \in \left\{1, 2, \ldots, \ell\right\}$ to the index $y$ satisfying
$f\left(i_x\right) = k_y$, and sends every
$x \in \left\{\ell + 1, \ell + 2, \ldots, \ell + m\right\}$ to the
index $y$ satisfying $f\left(j_{x-\ell}\right) = k_y$.}, and the
corresponding compositions $\wt (f)$ are the shuffles of $\alpha$
and $\beta$. Therefore the name ``overlapping shuffle product''.
\end{example}

% [DG][v25] Added the above example, as the overlapping shuffle rule
% isn't the easiest thing to understand. I fear my example doesn't
% help it, though...

% [DG][v26] Shortened the above example.

\begin{proof}[Proof of Proposition~\ref{QSym-multiplication}]
It clearly suffices to prove the formula
\eqref{Qsym-product-on-monomials}.
Let $\alpha = \left(\alpha_1, \ldots, \alpha_\ell\right)$
and $\beta = \left(\beta_1, \ldots, \beta_m\right)$ be two
compositions.
Fix three disjoint chain posets
$(i_1 < \cdots <i_\ell)$, $(j_1 < \cdots < j_m)$ and
$(k_1 < k_2 < \cdots)$.

Thus, multiplying
$M_\alpha = \sum_{u_1 < \cdots < u_\ell} x_{u_1}^{\alpha_1} \cdots x_{u_\ell}^{\alpha_\ell}$
with
$M_\beta = \sum_{v_1 < \cdots < v_m} x_{v_1}^{\beta_1} \cdots x_{v_m}^{\beta_m}$,
we obtain
\begin{align}
M_\alpha M_\beta
&= \sum_{u_1 < \cdots < u_\ell} \sum_{v_1 < \cdots < v_m}
\left( x_{u_1}^{\alpha_1} \cdots x_{u_\ell}^{\alpha_\ell} \right)
\left( x_{v_1}^{\beta_1} \cdots x_{v_m}^{\beta_m} \right)
\nonumber \\
&= \sum_{\gamma = \left(\gamma_1, \ldots, \gamma_p\right) \in \Comp}
  \sum_{w_1 < \cdots < w_p \text{ in } I}
  N^\gamma_{w_1, \ldots, w_p}
  x_{w_1}^{\gamma_1} \cdots x_{w_p}^{\gamma_p} ,
\label{pf.QSym-multiplication.1}
\end{align}
where $N^\gamma_{w_1, \ldots, w_p}$ is the number of all pairs
\begin{align}
\left(\left( u_1 < \cdots < u_\ell \right),
       \left( v_1 < \cdots < v_m \right) \right)
 \in I^\ell \times I^m
\label{pf.QSym-multiplication.2a}
\end{align}
of two strictly increasing tuples satisfying
\begin{align}
\left( x_{u_1}^{\alpha_1} \cdots x_{u_\ell}^{\alpha_\ell} \right)
\left( x_{v_1}^{\beta_1} \cdots x_{v_m}^{\beta_m} \right)
= x_{w_1}^{\gamma_1} \cdots x_{w_p}^{\gamma_p} .
\label{pf.QSym-multiplication.2b}
\end{align}
\footnote{In the second equality in \eqref{pf.QSym-multiplication.1},
  we have used the fact that
  each monomial can be uniquely written in the form
  $x_{w_1}^{\gamma_1} \cdots x_{w_p}^{\gamma_p}$ for some
  composition
  $\gamma = \left(\gamma_1, \ldots, \gamma_p\right) \in \Comp$
  and some strictly increasing tuple
  $\left( w_1 < \cdots < w_p \right) \in I^p$.}
Thus, we need to show that $N^\gamma_{w_1, \ldots, w_p}$
(for a given
$\gamma = \left(\gamma_1, \ldots, \gamma_p\right) \in \Comp$
and a given
$\left( w_1 < \cdots < w_p \right) \in I^p$)
is also
the number of all surjective strictly order-preserving maps
\begin{align}
(i_1 < \cdots <i_\ell) \sqcup (j_1 < \cdots < j_m)
\overset{f}{\longrightarrow}
(k_1 < \cdots <k_p)
\text{ satisfying } \wt(f) = \gamma
\label{pf.QSym-multiplication.3}
\end{align}
(because then, \eqref{pf.QSym-multiplication.1} will simplify
to \eqref{Qsym-product-on-monomials}).

In order to show this, it suffices to construct a bijection from
the set of all pairs \eqref{pf.QSym-multiplication.2a}
satisfying \eqref{pf.QSym-multiplication.2b}
to the set of all surjective strictly order-preserving maps
\eqref{pf.QSym-multiplication.3}.
This bijection is easy to construct:
Given a pair \eqref{pf.QSym-multiplication.2a}
satisfying \eqref{pf.QSym-multiplication.2b},
the bijection sends it to the map \eqref{pf.QSym-multiplication.3}
determined by:
\begin{align*}
i_g &\overset{f}{\mapsto} k_h, \text{ where } h \text{ is chosen such that }
                        u_g = w_h; \\
j_g &\overset{f}{\mapsto} k_h, \text{ where } h \text{ is chosen such that }
                        v_g = w_h.
\end{align*}
Proving that this bijection is well-defined and bijective
is straightforward\footnote{
The inverse of this bijection sends each
map \eqref{pf.QSym-multiplication.3} to the
pair \eqref{pf.QSym-multiplication.2a} determined by
\begin{align*}
u_g = w_h, \text{ where } h \text{ is chosen such that } f \left(i_g\right) = k_h; \\
v_g = w_h, \text{ where } h \text{ is chosen such that } f \left(j_g\right) = k_h.
\end{align*}
}.
\end{proof}

% [DG][v62] Made the proof above a lot more detailed.

The multiplication rule \eqref{Qsym-product-on-monomials} shows that the
$\kk$-algebra $\Qsym(\{x_i\}_{i \in I})$ does not depend much on
$I$, as long as $I$ is infinite. More precisely, all such $\kk$-algebras
are mutually isomorphic. We can use this to define a $\kk$-algebra
of quasisymmetric functions without any reference to $I$:

\begin{definition}
\label{def.Qsym}
Let $\Qsym$ be the $\kk$-algebra defined as having $\kk$-basis
$\{M_\alpha\}_{\alpha \in \Comp}$ and with multiplication defined
$\kk$-linearly by \eqref{Qsym-product-on-monomials}. This is called
the \dfn{$\kk$-algebra of quasisymmetric functions}. We write
$\Qsym_\kk$ instead of $\Qsym$ to stress the choice of base ring $\kk$.

The $\kk$-algebra $\Qsym$ is graded, and its $n$-th graded component
$\Qsym_n$ has $\kk$-basis $\{ M_\alpha \}_{\alpha \in \Comp_n}$.

For every infinite totally ordered set $I$, the $\kk$-algebra
$\Qsym$ is isomorphic to the $\kk$-algebra
$\Qsym(\{x_i\}_{i \in I})$.  The isomorphism sends
$M_\alpha \longmapsto M_\alpha(\{x_i\}_{i\in I})$.

In particular, we obtain the isomorphism
$\Qsym \cong \Qsym\left(\xx\right)$ for
$\xx$ being the infinite chain $\left(x_1 < x_2 < x_3 < \cdots\right)$.
We will identify $\Qsym$ with $\Qsym\left(\xx\right)$ along this
isomorphism. This allows us to regard quasisymmetric functions either
as power series in a specific set of variables (``alphabet''), or as
formal linear combinations of $M_\alpha$'s, whatever is more
convenient.

For any infinite alphabet
$\left\{x_i\right\}_{i \in I}$ and any $f \in \Qsym$, we denote
by $f\left(\left\{x_i\right\}_{i \in I}\right)$ the image of $f$
under the algebra isomorphism
$\Qsym \to \Qsym\left(\left\{x_i\right\}_{i \in I}\right)$ defined in
Definition~\ref{def.Qsym}.
\end{definition}

The comultiplication of $\Qsym$ will extend the one that we defined
for $\Lambda$, but we need to take care about the order of the
variables this time.
We consider the linear order from
\eqref{two-variable-set-ordering} on two sets of variables
$
(\xx,\yy)=(x_1 < x_2 < \cdots < y_1 < y_2 < \cdots),
$
and we embed the $\kk$-algebra $\Qsym(\xx) \otimes \Qsym(\yy)$ into
the $\kk$-algebra $R(\xx, \yy)$ by identifying every $f \otimes g \in
\Qsym(\xx) \otimes \Qsym(\yy)$ with $fg \in R(\xx, \yy)$
(this embedding is indeed injective\footnote{This is because it
sends the basis elements $M_\beta(\xx) \otimes M_\gamma(\yy)$ of
the former $\kk$-algebra to the linearly independent power series
$M_\beta(\xx) M_\gamma(\yy)$.}). It can then be seen that
\[
\Qsym(\xx,\yy) \subset \Qsym(\xx) \otimes \Qsym(\yy)
\]
(where the right hand side is viewed as $\kk$-subalgebra of
$R(\xx, \yy)$ via said embedding)\footnote{This is not completely
obvious, but can be easily checked by verifying that
$M_\alpha(\xx, \yy)
= \sum\limits_{\substack{(\beta,\gamma):\\ \beta \cdot \gamma=\alpha}} M_\beta (\xx) \otimes M_\gamma (\yy)$
for every composition $\alpha$ (see the proof of
Proposition~\ref{Qsym-coproduct-on-monomials} for why this holds).},
so that one can define $\Qsym \overset{\Delta}{\longrightarrow} \Qsym\otimes \Qsym$
%\begin{equation}
%\label{Qsym-coproduct-defintion}
%\begin{array}{rcl}
%\Qsym=\Qsym(\xx) & \overset{\Delta}{\longrightarrow}
%   & \Qsym(\xx,\yy) \hookrightarrow \Qsym\otimes \Qsym\\
%f(\xx)=f(x_1,x_2,\ldots) & \longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots).
%\end{array}
%\end{equation}
as the  composite of the maps in the bottom row here:
\begin{equation}
\label{Qsym-coproduct-definition}
\begin{array}{rccccl}
      &                 & R(\xx,\yy)     & = & R(\xx, \yy) \\
      &                 & \cup           &       & \cup\\

\Qsym & \cong & \Qsym(\xx,\yy) & \hookrightarrow &
\Qsym(\xx) \otimes \Qsym(\yy) & \cong \Qsym \otimes \Qsym , \\
f&\longmapsto & f(\xx,\yy)=f(x_1,x_2,\ldots,y_1,y_2,\ldots) .& &
\end{array}
\end{equation}
(Recall that $f(\xx,\yy)$ is formally defined as the image of $f$ under the
algebra isomorphism $\Qsym \to \Qsym(\xx,\yy)$ defined in
Definition~\ref{def.Qsym}.)

% [DG][v17] There used to be a $R(\xx,\yy)$ in \eqref{Qsym-coproduct-defintion},
% but it was claiming that $R(\xx,\yy) \cong R(\xx) \otimes R(\yy)$, which
% is false ($\sum_i x_i \otimes y_i$ is in the former but not in the latter).
% When $\kk$ is not a field, there is not even a guarantee that the latter
% embeds into the former
% ( http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.pjm/1102959646 ).
% 
% I still believe it would be better to define the coproduct on the basis.
% How do you check that
% $\Qsym(\xx,\yy) \subset \Qsym(\xx) \otimes \Qsym(\yy)$
% if not by computing it on $M_\alpha$ ?

% [DG][v19] Edited the definition of the coproduct again, this time
% explaining every step. I might have overswung the pendulum to the
% pedantic side this time...

\begin{example}
For example,
\begin{align*}
\Delta M_{(a,b,c)} &= M_{(a,b,c)}(x_1,x_2,\ldots,y_1,y_2,\ldots) \\
&= x_1^a x_2^b x_3^c+ x_1^a x_2^b x_4^c + \cdots \\
 &\quad + x_1^a x_2^b \cdot y_1^c+ x_1^a x_2^b \cdot y_2^c + \cdots \\
 &\quad + x_1^a \cdot y_1^b y_2^c + x_1^a \cdot y_1^b y_3^c + \cdots \\
 &\quad + y_1^a y_2^b y_3^c + y_1^a y_2^b y_4^c+ \cdots\\
&= M_{(a,b,c)}(\xx) + M_{(a,b)}(\xx) M_{(c)}(\yy) +  M_{(a)}(\xx) M_{(b,c)}(\yy)+ M_{(a,b,c)}(\yy) \\
&= M_{(a,b,c)} \otimes \one + M_{(a,b)} \otimes M_{(c)}
 + M_{(a)} \otimes M_{(b,c)} + \one \otimes M_{(a,b,c)} .
\end{align*}
\end{example}

\noindent
Defining the \emph{concatenation}\index{concatenation of compositions}
$\beta \cdot \gamma$ of
two compositions $\beta=(\beta_1,\ldots,\beta_r),
\gamma=(\gamma_1,\ldots,\gamma_s)$ to be the composition
$(\beta_1,\ldots,\beta_r,\gamma_1,\ldots,\gamma_s)$, one has the following
description of the coproduct in the $\{M_\alpha\}$ basis.

\begin{proposition}
\label{Qsym-coproduct-on-monomials}
For a composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$,
one has
\[
\Delta M_\alpha = \sum_{k=0}^\ell M_{(\alpha_1,\ldots,\alpha_k)} \otimes M_{(\alpha_{k+1},\ldots,\alpha_\ell)} = \sum\limits_{\substack{(\beta,\gamma):\\ \beta \cdot \gamma=\alpha}} M_\beta \otimes M_\gamma .
\]
\end{proposition}
\begin{proof}
We work with the infinite totally ordered set $I = \left\{1<2<3<\cdots\right\}$.
The definition of $\Delta$ yields
\begin{equation}
\Delta M_\alpha = M_\alpha(\xx,\yy) = \sum_{p_1 < p_2 < \cdots < p_\ell \text{ in } (\xx, \yy)} p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_\ell^{\alpha_\ell} ,
\label{pf.Qsym-coproduct-on-monomials.1}
\end{equation}
where the sum runs over strictly increasing $\ell$-tuples $\left(p_1 < p_2 < \cdots < p_\ell\right)$ of variables in the variable set $(\xx, \yy)$.
But every such $\ell$-tuple $\left(p_1 < p_2 < \cdots < p_\ell\right)$ can be expressed uniquely in the form $\left(x_{i_1}, \ldots, x_{i_k}, y_{j_1}, \ldots, y_{j_{\ell-k}}\right)$ for some $k \in \left\{ 0,1,\ldots,\ell \right\}$ and some subscripts $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_{\ell-k}$ in $I$. The corresponding monomial $p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_\ell^{\alpha_\ell}$ then rewrites as $x_{i_1}^{\alpha_1} \cdots x_{i_k}^{\alpha_k} \cdot y_{j_1}^{\alpha_{k+1}} \cdots y_{j_{\ell-k}}^{\alpha_\ell}$. Thus, the sum on the right hand side of \eqref{pf.Qsym-coproduct-on-monomials.1} rewrites as
\begin{align*}
& \sum_{k=0}^{\ell} \ \ \sum_{i_1 < \cdots < i_k} \ \ \sum_{j_1 < \cdots < j_{\ell-k}} x_{i_1}^{\alpha_1} \cdots x_{i_k}^{\alpha_k} \cdot
y_{j_1}^{\alpha_{k+1}} \cdots y_{j_{\ell-k}}^{\alpha_\ell} \\
&= \sum_{k=0}^{\ell} \underbrace{\left(\sum_{i_1 < \cdots < i_k} x_{i_1}^{\alpha_1} \cdots x_{i_k}^{\alpha_k} \right)}_{= M_{(\alpha_1,\ldots,\alpha_k)}(\xx)} \cdot
\underbrace{\left(\sum_{j_1 < \cdots < j_{\ell-k}} y_{j_1}^{\alpha_{k+1}} \cdots y_{j_{\ell-k}}^{\alpha_\ell}\right)}_{= M_{(\alpha_{k+1},\ldots,\alpha_\ell)}(\yy)}
\\
&= \sum_{k=0}^{\ell} M_{(\alpha_1,\ldots,\alpha_k)}(\xx) M_{(\alpha_{k+1},\ldots,\alpha_\ell)}(\yy) .
\end{align*}
Thus, \eqref{pf.Qsym-coproduct-on-monomials.1} becomes
\begin{align*}
\Delta M_\alpha = \sum_{p_1 < p_2 < \cdots < p_\ell \text{ in } (\xx, \yy)} p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_\ell^{\alpha_\ell}
&= \sum_{k=0}^{\ell} M_{(\alpha_1,\ldots,\alpha_k)}(\xx) M_{(\alpha_{k+1},\ldots,\alpha_\ell)}(\yy) \\
&= \sum_{k=0}^{\ell} M_{(\alpha_1,\ldots,\alpha_k)} \otimes M_{(\alpha_{k+1},\ldots,\alpha_\ell)}
= \sum\limits_{\substack{(\beta,\gamma):\\ \beta \cdot \gamma=\alpha}} M_\beta \otimes M_\gamma .
\end{align*}
% Old proof:
% This comes from expressing a monomial in $\Delta M_\alpha = M_\alpha(\xx,\yy)$
% uniquely in the form
% $x_{i_1}^{\alpha_1} \cdots x_{i_k}^{\alpha_k} \cdot
% y_{j_1}^{\alpha_{k+1}} \cdots y_{j_{\ell-k}}^{\alpha_\ell}$
% for some $k \in \left\{ 0,1,\ldots,\ell \right\}$ and some subscripts
% $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_{\ell-k}$.
\end{proof}

% [DG][v72] More details in the above proof (and 1 less typo).

\begin{proposition}
The quasisymmetric functions $\Qsym$ form a connected graded Hopf algebra of finite
type, which is commutative, and contains
the symmetric functions $\Lambda$ as a Hopf subalgebra.
\end{proposition}
\begin{proof}
To prove coassociativity of $\Delta$, we need to be slightly careful.
It seems reasonable to argue by
$
(\Delta \otimes \id) \circ \Delta f =
f(\xx,\yy,\zz)=
(\id \otimes \Delta) \circ \Delta f
$
as in the case of $\Lambda$, but this would now require further
justification, as terms like $f(\xx,\yy)$ and $f(\xx,\yy,\zz)$ are
no longer directly defined as evaluations of $f$ on some sequences
(but rather are defined as images of $f$ under certain
homomorphisms). However, it is very easy to see that $\Delta$ is
coassociative by checking $(\Delta \otimes \id) \circ \Delta
= (\id \otimes \Delta) \circ \Delta $ on the $\left\{M_\alpha\right\}$
basis: Proposition~\ref{Qsym-coproduct-on-monomials} yields
\begin{align*}
\left((\Delta \otimes \id) \circ \Delta\right) M_{\alpha}
&= \sum_{k=0}^\ell \Delta(M_{(\alpha_1,\ldots,\alpha_k)}) \otimes M_{(\alpha_{k+1},\ldots,\alpha_\ell)} \\
&= \sum_{k=0}^\ell \left(\sum_{i=0}^k M_{(\alpha_1,\ldots,\alpha_i)} \otimes M_{(\alpha_{i+1},\ldots,\alpha_k)}\right) \otimes M_{(\alpha_{k+1},\ldots,\alpha_\ell)} \\
&= \sum_{k=0}^\ell \sum_{i=0}^k M_{(\alpha_1,\ldots,\alpha_i)} \otimes M_{(\alpha_{i+1},\ldots,\alpha_k)} \otimes M_{(\alpha_{k+1},\ldots,\alpha_\ell)}
\end{align*}
and the same expression for
$\left((\id \otimes \Delta) \circ \Delta\right) M_{\alpha}$.

The coproduct $\Delta$ of $\Qsym$ is an algebra morphism because
it is defined as a composite of algebra morphisms
in the bottom row of \eqref{Qsym-coproduct-definition}.
To prove that the restriction of
$\Delta$ to the subring $\Lambda$ of $\Qsym$ is the comultiplication
of $\Lambda$, it thus is enough to check that it sends the elementary
symmetric function $e_n$ to $\sum_{i=0}^n e_i \otimes e_{n-i}$ for
every $n\in\NN$. This again follows from
Proposition~\ref{Qsym-coproduct-on-monomials}, since
$e_n = M_{(1,1,\ldots,1)}$
(with $n$ times $1$).

% [DG][v3] The above might be shortenable.

The counit is as usual for a connected graded coalgebra,
and just as in the case of $\Lambda$,
sends a quasisymmetric function
$f(\xx)$ to its constant term $f(0,0,\ldots)$.
This is an evaluation, and hence an algebra morphism.
Hence $\Qsym$ forms a bialgebra, and as it is
graded and connected, also
a Hopf algebra by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
It is clearly of finite type and contains
$\Lambda$ as a Hopf subalgebra.
\end{proof}

We will identify the antipode in $\Qsym$ shortly, but we first
deal with another slightly subtle issue.
In addition to the counit evaluation $\epsilon(f) = f(0,0,\ldots)$,
starting in Section~\ref{ABS-part-I-section}, we will
want to specialize elements in $\Qsym(\xx)$ by making other
variable substitutions, in which all but a finite list of variables
are set to zero.  We justify this here.

\begin{proposition}
\label{prop.Qsym.eval}
Fix a totally ordered set $I$, a commutative $\kk$-algebra $A$,
a finite list of variables $x_{i_1}, \ldots,x_{i_m}$, say
with $i_1 < \cdots < i_m$ in $I$,
and an ordered list of elements $(a_1,\ldots,a_m) \in A^m$.

Then there is a well-defined evaluation homomorphism
\[
\begin{array}{rcl}
\Qsym(\{x_i\}_{i \in I}) &\longrightarrow &A ,\\
f &\longmapsto&
\left[ f
\right]_{\substack{x_{i_1}=a_1,\ldots,x_{i_m}=a_m\\
x_j=0 \text{ for }j \not\in \{i_1,\ldots,i_m\}}}.
\end{array}
\]
Furthermore, this homomorphism
depends only upon the list $(a_1,\ldots,a_m)$, as it coincides
with the following:
\[
\begin{array}{rccl}
\Qsym(\{x_i\}_{i \in I}) &\cong
\Qsym(x_1,x_2,\ldots) &\longrightarrow &A ,\\
&f(x_1,x_2,\ldots)& \longmapsto& f(a_1,\ldots,a_m,0,0\ldots).
\end{array}
\]
(This latter statement is stated for the case when $I$ is infinite; otherwise,
read ``$x_1,x_2,\ldots,x_{\left|I\right|}$'' for
``$x_1,x_2,\ldots$'', and interpret
$(a_1,\ldots,a_m,0,0\ldots)$ as an $\left|I\right|$-tuple.)
\end{proposition}
\begin{proof}
One already can make sense of evaluating
$x_{i_1}=a_1,\ldots,x_{i_m}=a_m$ and $x_j=0$ for $j \not \in \{i_1,\ldots,i_m\}$
in the ambient ring $R(\{x_i\}_{i \in I})$ containing $\Qsym(\{x_i\}_{i \in I})$,
since a power series $f$ of bounded degree will have finitely
many monomials that only involve the variables
$x_{i_1},\ldots,x_{i_m}$.  The last assertion follows from quasisymmetry
of $f$, and is perhaps checked most easily when $f=M_\alpha(\{x_i\}_{i \in I})$
for some $\alpha$.
\end{proof}


The antipode in $\Qsym$ has a reasonably simple expression
in the $\{M_\alpha\}$ basis, but requiring a definition.

\begin{definition}
\label{def.QSym.D}
For $\alpha,\beta$ in $\Comp_n$, say that $\alpha$
\emph{refines}\index{refinement of compositions}\index{refining a composition}
$\beta$
or $\beta$
\emph{coarsens}\index{coarsening of compositions}\index{coarsening a composition}
$\alpha$ if, informally, one can obtain
$\beta$ from $\alpha$ by combining some of its adjacent parts. 
Alternatively, this can be defined as follows:
One has a bijection $\Comp_n \rightarrow 2^{[n-1]}$ where
$[n-1]:=\{1,2,\ldots,n-1\}$ which sends
$\alpha=(\alpha_1,\ldots,\alpha_\ell)$ having
length $\ell(\alpha)=\ell$ to its subset of partial sums\index{$D(\alpha)$}
\[
D(\alpha):= \left\{ \alpha_1,\alpha_1+\alpha_2,\ldots,\alpha_1+\cdots+\alpha_{\ell-1} \right\},
\]
and this sends the refinement ordering to the inclusion ordering
on the Boolean algebra $2^{[n-1]}$ (to be more precise: a composition
$\alpha \in \Comp_n$ refines a composition $\beta \in \Comp_n$
if and only if $D(\alpha) \supset D(\beta)$).

There is also a bijection sending every composition
$\alpha$ to its \dfn{ribbon}\index{ribbon diagram}\index{$\Rib(\alpha)$}
diagram $\Rib\left(\alpha\right)$:
the skew diagram $\lambda/\mu$
having rows of sizes $\alpha_1,\ldots,\alpha_\ell$ read from bottom to top
with exactly one column of overlap between adjacent rows.
These bijections and the refinement partial order are illustrated here for $n=4$:
\[
\begin{array}{c|c|c}
\tiny
\xymatrix{
 & \{1,2,3\} & \\
\{1,2\} \ar@{-}[ur]& \{1,3\} \ar@{-}[u] & \{2,3\} \ar@{-}[ul] \\
\{1\} \ar@{-}[ur]\ar@{-}[u]& \{2\}\ar@{-}[ur]\ar@{-}[ul] & \{3\} \ar@{-}[ul]\ar@{-}[u]\\
&\varnothing \ar@{-}[ur]\ar@{-}[u]\ar@{-}[ul]&
}
\quad \phantom{a} & \quad
\tiny \xymatrix{
 & (1,1,1,1)& \\
(1,1,2) \ar@{-}[ur]& (1,2,1) \ar@{-}[u] & (2,1,1) \ar@{-}[ul] \\
(1,3) \ar@{-}[ur]\ar@{-}[u]& (2,2) \ar@{-}[ur]\ar@{-}[ul] & (3,1) \ar@{-}[ul]\ar@{-}[u]\\
&(4) \ar@{-}[ur]\ar@{-}[u]\ar@{-}[ul]&
}
\quad \phantom{a} & \quad
\tiny \xymatrix{
 &\txt{$\sq$ \\ $\sq$ \\ $\sq$ \\  $\sq$} & \\
  \txt{$\sq \sq$\\$\sq\phantom{\sq}$ \\$\sq\phantom{\sq}$} \ar@{-}[ur]
 & \txt{$\phantom{\sq}\sq$ \\  $\sq \sq$ \\$\sq\phantom{\sq}$ } \ar@{-}[u]
 & \txt{$\phantom{\sq}\sq$ \\ $\phantom{\sq}\sq$ \\  $\sq \sq$} \ar@{-}[ul] \\
 \txt{$\sq \sq \sq$\\$\sq\phantom{\sq}\phantom{\sq}$} \ar@{-}[ur]\ar@{-}[u]
 & \txt{$\phantom{\sq}\sq\sq$\\$\sq \sq \phantom{\sq}$} \ar@{-}[ur]\ar@{-}[ul]
 & \txt{$\phantom{\sq}\phantom{\sq}\sq$\\$\sq \sq \sq$} \ar@{-}[ul]\ar@{-}[u]\\
&\txt{$\sq\sq\sq\sq$} \ar@{-}[ur]\ar@{-}[u]\ar@{-}[ul]&
}
\end{array}
\]
(where we have drawn each ribbon diagram with its boxes spaced
out).

Given $\alpha=(\alpha_1,\ldots,\alpha_\ell)$,
its \emph{reverse}\index{reverse composition}\index{$\rev(\alpha)$}
composition is
$\rev(\alpha)=(\alpha_{\ell},\alpha_{\ell-1},\ldots,\alpha_2,\alpha_1)$.
Note that $\alpha \mapsto \rev(\alpha)$ is a poset automorphism of
$\Comp_n$ for the refinement ordering.
\end{definition}

% [DG][v3] I have replaced "from bottom-to-top" by "from bottom to top".
% Done the same to "from left-to-right".

% [DG][v78] Introduced the notation "\Rib" for the ribbon
% corresponding to a composition (so we no longer identify
% it with the composition itself, thus avoiding a possible
% confusion when the composition is a partition).

\begin{theorem}
\label{Qsym-antipode-on-monomials}
For any composition $\alpha$ in $\Comp$,
\[
S(M_\alpha)= (-1)^{\ell(\alpha)} \sum\limits_{\substack{\gamma \in \Comp:\\ \gamma \text{ coarsens }\rev(\alpha)}} M_{\gamma} .
\]
\end{theorem}

For example,
\[
S(M_{(a,b,c)})=-\left( M_{(c,b,a)} + M_{(b+c,a)} + M_{(c,a+b)} + M_{(a+b+c)} \right) .
\]

\begin{proof}
We give Ehrenborg's proof\footnote{A different proof was
given by Malvenuto and Reutenauer \cite[Cor. 2.3]{MalvenutoReutenauer}, and is
sketched in Remark~\ref{second-antipode-proof} below.} \cite[Prop. 3.4]{Ehrenborg}
via induction on $\ell=\ell(\alpha)$.
One has easy base cases when
$\ell(\alpha)=0$, where
$
S(M_\varnothing) = S(1)=1=(-1)^0 M_{\rev(\varnothing)},
$
and when
$\ell(\alpha)=1$, where $M_{(n)}$ is primitive by
Proposition~\ref{Qsym-coproduct-on-monomials}, so Proposition~\ref{antipode-on-primitives} shows
$
S(M_{(n)})=-M_{(n)}=(-1)^1 M_{\rev((n))}.
$

For the inductive step, apply the inductive definition of $S$
from the proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}:
\begin{align*}
S(M_{(\alpha_1,\ldots,\alpha_\ell)})
&=-\sum_{i=0}^{\ell-1} S(M_{(\alpha_1,\ldots,\alpha_i)})
M_{(\alpha_{i+1},\ldots,\alpha_\ell)}  \\
&=\sum_{i=0}^{\ell-1}
\sum\limits_{\substack{ \beta\text{ coarsening } \\ (\alpha_i,\alpha_{i-1},\ldots,\alpha_1)}}
(-1)^{i+1} M_{\beta} M_{(\alpha_{i+1},\ldots,\alpha_\ell)} .
\end{align*}
The idea will be to cancel terms of opposite sign
that appear in the expansions of the products
$M_\beta  M_{(\alpha_{i+1},\ldots,\alpha_\ell)}$.  Note that
each composition $\beta$ appearing above has
first part $\beta_1$ of the form $\alpha_i+\alpha_{i-1}+\cdots+\alpha_h$ for some
$h \leq i$ (unless $\beta = \varnothing$), and hence each term $M_\gamma$ in the
expansion of the product
$
M_\beta  M_{(\alpha_{i+1},\ldots,\alpha_\ell)}
$
has $\gamma_1$ (that is, the first entry of $\gamma$) a sum that
can take one of these three forms:
\begin{enumerate}
\item[$\bullet$]
$\alpha_i+\alpha_{i-1}+\cdots+\alpha_h$,
\item[$\bullet$]
$\alpha_{i+1}+(\alpha_i+\alpha_{i-1}+\cdots+\alpha_h)$,
\item[$\bullet$]
$\alpha_{i+1}$.
\end{enumerate}
Say that the \emph{type} of $\gamma$
is $i$ in the first case, and $i+1$ in the second two cases\footnote{We
imagine that we label the terms obtained by expanding
$M_\beta M_{(\alpha_{i+1},\ldots,\alpha_\ell)}$ by distinct labels,
so that each term knows how exactly it was created (i.e., which $i$, which $\beta$
and which map $f$ as in \eqref{map-from-two-chains-to-one} gave
rise to it). Strictly speaking, it is these triples $\left(i,\beta,f\right)$
that we should be assigning types to, not terms.};
in other words, the type is the largest subscript
$k$ on a part $\alpha_k$ which was combined in the sum $\gamma_1$.
It is not hard to see that a given $\gamma$ for which the type $k$
is strictly smaller than $\ell$ arises from exactly two
pairs $(\beta,\gamma),(\beta',\gamma)$, having opposite signs $(-1)^k$
and $(-1)^{k+1}$ in the above sum\footnote{Strictly speaking, this
means that we have an involution on the set of our $\left(i,\beta,f\right)$
triples having type smaller than $\ell$, and this involution switches the
sign of $(-1)^i M_{\wt (f)}$.}.
For example, if $\alpha=(\alpha_1,\ldots,\alpha_8)$,
then the composition
$\gamma=(\alpha_6+\alpha_5+\alpha_4,\alpha_3,\alpha_7,
          \alpha_8+\alpha_2+\alpha_1)$
of type $6$
can arise from either of
\begin{align*}
\beta&=(\alpha_6+\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=6\text{ and sign }(-1)^7 ,\\
\beta'&=(\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=5 \text{ and sign }(-1)^6.
\end{align*}
Similarly,
$\gamma=(\alpha_6,\alpha_5+\alpha_4,\alpha_3,\alpha_7,
          \alpha_8+\alpha_2+\alpha_1)$ can arise from either of
\begin{align*}
\beta&=(\alpha_6,\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=6\text{ and sign }(-1)^7,\\
\beta'&=(\alpha_5+\alpha_4,\alpha_3,\alpha_2+\alpha_1)
\text{ with }i=5\text{ and sign }(-1)^6.
\end{align*}
Thus one can cancel almost all the terms, excepting those
with $\gamma$ of type $\ell$ among the terms $M_\gamma$
in the expansion of the last ($i=\ell-1$) summand $M_\beta M_{(\alpha_\ell)}$.
A bit of thought shows that these are the
$\gamma$ coarsening $\rev(\alpha)$, and all have sign $(-1)^\ell$.
\end{proof}

% [DG][v3] I added the last two footnotes in the proof above.









\subsection{The fundamental basis and $P$-partitions}

There is a second important basis for $\Qsym$ which arose originally
in Stanley's $P$-partition theory \cite{Stanley-thesis}.\footnote{See
\cite{Gessel-Ppart} for a history of $P$-partitions; our notations,
however, strongly differ from those in \cite{Gessel-Ppart}.}

\begin{definition}
A \dfn{labelled poset} will here mean
a partially ordered set $P$ whose underlying set
is some finite subset of the integers.  A \dfn{$P$-partition} is a function
$P \overset{f}{\rightarrow} \{1,2,\ldots\}$ with the following
two properties:
\begin{enumerate}
\item[$\bullet$]
If $i \in P$ and $j \in P$ satisfy
$i <_P j$ and $i <_\ZZ j$,
then $f(i) \leq f(j)$.
\item[$\bullet$]
If $i \in P$ and $j \in P$ satisfy
$i <_P j$ and $i >_\ZZ j$, then
$f(i) < f(j)$.
\end{enumerate}
Denote by \dfn{$\AAA(P)$} the set of all $P$-partitions $f$, and let
$
F_P(\xx):=\sum_{f \in \AAA(P)} \xx_f
$
where $\xx_f:=\prod_{i \in P} x_{f(i)}$. This $F_P(\xx)$
is an element of $\kk\left[\left[\xx\right]\right]
:= \kk\left[\left[x_1,x_2,\ldots\right]\right]$.
\end{definition}

% [DG][v72] Made the above definition a bit less jargonish by
% replacing "implies" by proper "if/then"-clauses. Similar
% changes made everywhere the word "implies"/"imply" appeared.

\begin{example}
\label{labelled-poset-example}
Depicted is a labelled poset $P$, along with the relations among the
four values $f=(f(1),f(2),f(3),f(4))$ that define its $P$-partitions $f$:
\[
\xymatrix{
 & &2\\
4& &1\ar@{-}[u]\\
 &3\ar@{-}[ur]\ar@{-}[ul]&
} \qquad
\xymatrix{
    & &f(2)\\
f(4)& &f(1)\ar@{-}[u]^{\leq}\\
    &f(3)\ar@{-}[ul]^{\leq} \ar@{-}[ur]^{<}&
}
\]
\end{example}

\begin{remark}
Stanley's treatment of $P$-partitions in \cite[\S 3.15 and \S 7.19]{Stanley} uses
a language different from ours. First, Stanley works not with labelled posets
$P$, but with pairs $\left(P, \omega\right)$ of a poset $P$ and a bijective
labelling $\omega : P \to [n]$. Thus, the relation $<_\ZZ$ is not given on $P$
a priori, but has to be pulled back from $[n]$ using $\omega$ (and it depends
on $\omega$, whence Stanley speaks of ``$\left(P, \omega\right)$-partitions'').
Furthermore, what we call ``$P$-partition'' is called a ``reverse $P$-partition''
in \cite{Stanley}.
Finally, Stanley uses the notations $F_P$ and $F_{P, \omega}$ for something
different from what we denote by $F_P$, whereas what we call $F_P$ is dubbed
$K_{P, \omega}$ in \cite[\S 7.19]{Stanley}.
\end{remark}

% [DG][v25] Added preceding remark on the babylon in this subject.

The so-called \emph{fundamental quasisymmetric functions}
are an important special case of the $F_P(\xx)$.
We shall first define them directly and then see how they are obtained as
$P$-partition enumerators $F_P(\xx)$ for some special labelled posets $P$.

\begin{definition}
\label{fundamental-qsym-def}
Let $n\in\NN$ and $\alpha\in\Comp_n$. We define the
\dfn{fundamental quasisymmetric function}\index{$L_\alpha$}
$L_\alpha = L_\alpha(\xx) \in \Qsym$
by
\begin{equation}
\label{fundamental-qsym-fn-defn}
L_\alpha := \sum\limits_{\substack{\beta \in \Comp_n:\\
                               \beta \text{ refines }\alpha}} M_\beta.
\end{equation}
\end{definition}

\begin{example}
\label{the-two-extreme-fundamentals-example}
The extreme cases for $\alpha$ in $\Comp_n$ give quasisymmetric functions
$L_\alpha$ which are symmetric:
\begin{align*}
L_{(1^n)}&=M_{(1^n)}=e_n,\\
L_{(n)}&=\sum_{\alpha \in \Comp_n} M_{\alpha}= h_n .\\
\end{align*}
\end{example}

Before studying the $L_\alpha$ in earnest, we recall a basic fact
about finite sets, which is sometimes known as the ``principle of
inclusion and exclusion'' (although it is more general than the
formula for the size of a union of sets that commonly
goes by this name):

\begin{lemma}
\label{lem.moebius-bool.subsets}
Let $G$ be a finite set.
Let $V$ be a $\kk$-module.
For each subset $A$ of $G$, we let $f_A$ and $g_A$ be two
elements of $V$.

\begin{enumerate}
\item[(a)] If
\[
\text{every } A \subset G \text{ satisfies } g_A = \sum_{B \subset A} f_B ,
\]
then
\[
\text{every } A \subset G \text{ satisfies } f_A = \sum_{B \subset A} \left(-1\right)^{\left|A \setminus B\right|} g_B .
\]

\item[(b)] If
\[
\text{every } A \subset G \text{ satisfies } g_A = \sum_{B \subset G; \  B \supset A} f_B ,
\]
then
\[
\text{every } A \subset G \text{ satisfies } f_A = \sum_{B \subset G; \  B \supset A} \left(-1\right)^{\left|B \setminus A\right|} g_B .
\]
\end{enumerate}
\end{lemma}

\begin{proof}
This can be proven by elementary arguments (easy exercise).
Alternatively, Lemma~\ref{lem.moebius-bool.subsets} can be
viewed as a particular case of the M\"obius inversion principle
(see, e.g., \cite[Propositions 3.7.1 and 3.7.2]{Stanley})
applied to the Boolean lattice $2^G$
(whose M\"obius function is very simple: see
\cite[Example 3.8.3]{Stanley}).
(This is spelled out in \cite[Example 4.52]{Loehr-bij}, for example.)
\end{proof}

Lemma~\ref{lem.moebius-bool.subsets} can be translated into the language
of compositions:

\begin{lemma}
\label{lem.moebius-bool.comps}
Let $n \in \NN$.
Let $V$ be a $\kk$-module.
For each $\alpha \in \Comp_n$, we let $f_\alpha$ and $g_\alpha$ be two
elements of $V$.

\begin{enumerate}
\item[(a)] If
\[
\text{every } \alpha \in \Comp_n \text{ satisfies } g_\alpha = \sum_{\beta \text{ coarsens } \alpha} f_\beta ,
\]
then
\[
\text{every } \alpha \in \Comp_n \text{ satisfies } f_\alpha = \sum_{\beta \text{ coarsens } \alpha} \left(-1\right)^{\ell\left(\alpha\right) - \ell\left(\beta\right)} g_\beta .
\]

\item[(b)] If
\[
\text{every } \alpha \in \Comp_n \text{ satisfies } g_\alpha = \sum_{\beta \text{ refines } \alpha} f_\beta ,
\]
then
\[
\text{every } \alpha \in \Comp_n \text{ satisfies } f_\alpha = \sum_{\beta \text{ refines } \alpha} \left(-1\right)^{\ell\left(\beta\right) - \ell\left(\alpha\right)} g_\beta .
\]
\end{enumerate}
\end{lemma}

\begin{proof}%[Proof of Lemma~\ref{lem.moebius-bool.comps}.]
Set $\left[n-1\right] = \left\{1,2,\ldots,n-1\right\}$.
Recall (from Definition~\ref{def.QSym.D}) that there is a bijection
$D : \Comp_n \rightarrow 2^{\left[n-1\right]}$ that sends each
$\alpha \in \Comp_n$ to $D \left(\alpha\right) \subset \left[n-1\right]$.
This bijection $D$ has the properties that:
\begin{itemize}
\item a composition $\beta$ refines a composition $\alpha$ if and only if
      $D\left(\beta\right) \supset D\left(\alpha\right)$;
\item a composition $\beta$ coarsens a composition $\alpha$ if and only if
      $D\left(\beta\right) \subset D\left(\alpha\right)$;
\item any composition $\alpha \in \Comp_n$ satisfies
      $\left|D\left(\alpha\right)\right| = \ell\left(\alpha\right)-1$
      (unless $n=0$), and thus
\item any compositions $\alpha$ and $\beta$ in $\Comp_n$ satisfy
      $\left|D\left(\alpha\right)\right| - \left|D\left(\beta\right)\right|
      = \ell\left(\alpha\right) - \ell\left(\beta\right)$.
\end{itemize}
This creates a dictionary between compositions in $\Comp_n$ and
subsets of $\left[n-1\right]$.
Now, apply Lemma~\ref{lem.moebius-bool.subsets} to
$G = \left[n-1\right]$, $f_A = f_{D^{-1}\left(A\right)}$ and
$g_A = g_{D^{-1}\left(A\right)}$, and translate using the
dictionary.
\end{proof}

% [DG][v70] Added the above two lemmas, and replaced the references
% to "inclusion-exclusion" by references to these lemmas.
% The reason is that "inclusion-exclusion" means a rather different
% thing for beginners in combinatorics.

Now, we can see the following about the fundamental quasisymmetric
functions:

\begin{proposition}
\label{prop.QSym.L-basis}
The family $\{L_\alpha\}_{\alpha \in \Comp}$
is a $\kk$-basis for $\Qsym$, and
each $n \in \NN$ and $\alpha \in \Comp_n$ satisfy
\begin{equation}
M_\alpha = \sum\limits_{\substack{\beta \in \Comp_n:\\
           \beta \text{ refines }\alpha}} (-1)^{\ell(\beta)-\ell(\alpha)} L_\beta.
\label{eq.QSym.M-through-L}
\end{equation}
\end{proposition}

\begin{proof}%[Proof of Proposition~\ref{prop.QSym.L-basis}.]
Fix $n \in \NN$.
Recall the equality \eqref{fundamental-qsym-fn-defn}.
Thus, Lemma~\ref{lem.moebius-bool.comps}(b) (applied to
$V = \Qsym$, $f_\alpha = M_\alpha$ and $g_\alpha = L_\alpha$)
yields \eqref{eq.QSym.M-through-L}.

Recall that the family $\left(M_\alpha\right)_{\alpha \in \Comp_n}$
is a basis of the $\kk$-module $\Qsym_n$.
The equality \eqref{fundamental-qsym-fn-defn} shows that the
family $\left(L_\alpha\right)_{\alpha \in \Comp_n}$ expands
invertibly
triangularly\footnote{See Section~\ref{sect.STmat} for a definition
of this concept.} with respect to the family
$\left(M_\alpha\right)_{\alpha \in \Comp_n}$
(where $\Comp_n$ is equipped with the refinement
order).\footnote{In fact, it expands unitriangularly with respect
to the latter family.}
Thus, Corollary~\ref{cor.STmat.expansion-tria-inv}(e) (applied to
$\Qsym_n$, $\Comp_n$, $\left(M_\alpha\right)_{\alpha \in \Comp_n}$
and $\left(L_\alpha\right)_{\alpha \in \Comp_n}$ instead of
$M$, $S$, $\left(e_s\right)_{s\in S}$ and
$\left(f_s\right)_{s\in S}$) shows that the
family $\left(L_\alpha\right)_{\alpha \in \Comp_n}$ is a basis
of the $\kk$-module $\Qsym_n$.
Combining this fact for all $n \in \NN$, we conclude that
the family $\left(L_\alpha\right)_{\alpha \in \Comp}$ is a basis
of the $\kk$-module $\Qsym$.
This completes the proof of Proposition~\ref{prop.QSym.L-basis}.
\end{proof}

% [DG][v70] Promoted the above to a proposition, with an
% explicit proof.

\begin{proposition}
\label{Qsym-technical-lemma-Lxy}
Let $n\in\NN$. Let $\alpha$ be a composition of $n$. Let $I$ be an
infinite totally ordered set. Then,
\[
L_\alpha \left(  \left\{  x_{i}\right\}_{i\in I}\right)
=\sum\limits_{\substack{i_1 \leq i_2 \leq \cdots \leq i_n \text{ in } I; \\ 
i_{j} < i_{j+1} \text{ if } j\in D\left(  \alpha\right) }}
x_{i_1} x_{i_2} \cdots x_{i_n},
\]
where $L_\alpha\left(  \left\{  x_i \right\}_{i\in I}\right)$ is
defined as the image of $L_\alpha$ under the isomorphism
$\Qsym \to \Qsym\left( \left\{  x_i \right\}_{i\in I}\right)$ obtained
in Definition~\ref{def.Qsym}.
In particular, for the standard (totally ordered) variable set
$\xx = \left(x_1 < x_2 < \cdots\right)$, we obtain
\begin{equation}
\label{fundamental-qsym-fn-defn-alternative}
L_\alpha = L_\alpha\left(\xx\right)
=\sum\limits_{\substack{(1 \leq ) i_1 \leq i_2 \leq \cdots \leq i_n; \\
i_{j} < i_{j+1} \text{ if } j\in D\left(  \alpha\right) }}
x_{i_1} x_{i_2} \cdots x_{i_n}.
\end{equation}
\end{proposition}

\begin{proof}
Every composition $\beta =
\left(\beta_1,\ldots,\beta_\ell\right)$ of $n$ satisfies
%% begin pedantry
%\footnote{For the second equality of \eqref{Qsym-technical-lemma-Lxy-proof}, we use the bijection between:
%\begin{enumerate}
%\item[$\bullet$] the sequences $k_1 < \cdots < k_\ell$ in $I$,
%\item[$\bullet$] the sequences $i_{1}\leq i_{2}\leq\cdots \leq i_{n}$ in $I$
%which satisfy $i_j < i_{j+1}$ precisely for those $j$
%satisfying $j\in D\left(  \beta\right)$.
%\end{enumerate}
%This bijection sends $k_1 < \cdots < k_\ell$ to
%$\underbrace{k_1 \leq \cdots \leq k_1}_{\beta_1 \text{ times}} \leq
%\underbrace{k_2 \leq \cdots \leq k_2}_{\beta_2 \text{ times}} \leq \cdots \leq
%\underbrace{k_\ell \leq \cdots \leq k_\ell}_{\beta_\ell \text{ times}}$,
%and thus replaces the monomial $x_{k_1}^{\beta_1} \cdots x_{k_\ell}^{\beta_\ell}$
%by $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$.}
%% end pedantry
\begin{equation}
\label{Qsym-technical-lemma-Lxy-proof}
M_\beta\left(\{x_i\}_{i\in I}\right)
= \sum_{k_1 < \cdots < k_\ell\text{ in }I}
x_{k_1}^{\beta_1} \cdots x_{k_\ell}^{\beta_\ell}
=\sum\limits_{\substack{i_{1}\leq
i_{2}\leq\cdots \leq i_{n}\text{ in } I;\\
i_{j}<i_{j+1}\text{ if and only if }j\in D\left(  \beta\right)
}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}.
\end{equation}
Applying the ring homomorphism
$\Qsym \to \Qsym\left( \left\{  x_{i}\right\}_{i\in I}\right)$
to \eqref{fundamental-qsym-fn-defn}, we obtain
\begin{align*}
L_{\alpha}\left(  \left\{  x_{i}\right\}_{i\in I}\right)
&= \sum\limits_{\substack{\beta \in \Comp_n:\\
                               \beta \text{ refines }\alpha}} M_\beta \left(  \left\{  x_{i}\right\}_{i\in I}\right) 
\overset{\eqref{Qsym-technical-lemma-Lxy-proof}}{=} \sum\limits_{\substack{\beta \in \Comp_n:\\
                               \beta \text{ refines }\alpha}} 
\quad \sum\limits_{\substack{i_{1}\leq
i_{2}\leq\cdots \leq i_{n}\text{ in } I;\\
i_{j}<i_{j+1}\text{ if and only if }j\in D\left(  \beta\right)
}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}} \\
&= \sum\limits_{\substack{\beta \in \Comp_n:\\    D(\alpha) \subset D(\beta)}} \quad
\sum\limits_{\substack{i_{1}\leq
i_{2}\leq\cdots \leq i_{n}\text{ in } I;\\
i_{j}<i_{j+1}\text{ if and only if }j\in D\left(  \beta\right)
}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}} \\
&= \sum\limits_{\substack{Z \subset [n-1]:\\
                               D(\alpha) \subset Z}} 
\quad \sum\limits_{\substack{i_{1}\leq
i_{2}\leq\cdots \leq i_{n}\text{ in } I;\\
i_{j}<i_{j+1}\text{ if and only if }j\in Z
}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}
= \sum\limits_{\substack{i_1 \leq i_2 \leq \cdots \leq i_n \text{ in } I;\\
i_{j}<i_{j+1} \text{ if } j\in D\left(  \alpha\right)  }}
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}.
\end{align*}
\end{proof}

\begin{proposition}
\label{fundamental-P-partition-prop}
Assume that the labelled poset $P$ is a total or linear order
$w=(w_1 < \cdots < w_n)$
(that is, $P = \left\{w_1,w_2,\ldots,w_n\right\}$ as sets,
and the order $<_P$ is given by $w_1 <_P w_2 <_P \cdots <_P w_n$).
Let $\Des(w)$ be the
\emph{descent set}\index{descent set of a permutation} of $w$,
defined by
\[
\Des(w) := \{ i : w_i >_{\ZZ} w_{i+1} \} \subset \{1,2,\ldots,n-1\} .
\]
Let $\alpha \in \Comp_n$ be the unique composition in $\Comp_n$ having
partial sums $D(\alpha)=\Des(w)$.
Then, the generating function
$F_w(\xx)$ equals the fundamental quasisymmetric function
$L_\alpha$.
In particular, $F_w(\xx)$ depends only upon the descent set $\Des(w)$.
\end{proposition}
% [DG][v72] Restated the above proposition so it's no longer one
% long convoluted sentence; also, added the "that is" parenthetical
% to clarify what the $<$ signs in "$(w_1 < \cdots < w_n)$" mean.
\noindent
E.g., total order $w=35142$ has $\Des(w)=\{2,4\}$ and
composition $\alpha=(2,2,1)$, so
\begin{align*}
F_{35142}(\xx)
&=\sum_{f(3) \leq f(5) < f(1) \leq f(4) < f(2)}
         x_{f(3)}x_{f(5)}x_{f(1)}x_{f(4)}x_{f(2)}\\
&=\sum_{i_1 \leq i_2 < i_3 \leq i_4 < i_5}
         x_{i_1}x_{i_2}x_{i_3}x_{i_4}x_{i_5}\\
&= L_{(2,2,1)} = M_{(2,2,1)}+M_{(2,1,1,1)}
                               +M_{(1,1,2,1)}+M_{(1,1,1,1,1)}.
\end{align*}

\begin{proof}[Proof of Proposition~\ref{fundamental-P-partition-prop}.]
Write $F_w(\xx)$ as a sum of monomials $x_{f(w_1)} \cdots x_{f(w_n)}$
over all $w$-partitions $f$. These $w$-partitions are exactly the
maps $f : w \to \left\{1,2,3,\ldots\right\}$ satisfying
$f(w_1) \leq \cdots \leq f(w_n)$ and having
strict inequalities $f(w_i) < f(w_{i+1})$ whenever $i$ is in $\Des(w)$
(because if two elements $w_a$ and $w_b$ of $w$ satisfy $w_a <_w w_b$
and $w_a >_{\ZZ} w_b$, then they must satisfy $a < b$ and
$i \in \Des(w)$ for some $i \in \left\{a, a+1, \ldots, b-1\right\}$; thus, the
conditions ``$f(w_1) \leq \cdots \leq f(w_n)$'' and
``$f(w_i) < f(w_{i+1})$ whenever $i$ is in $\Des(w)$''
ensure that $f\left(w_a\right) < f\left(w_b\right)$ in this case).
Therefore,
they are in bijection with the weakly increasing sequences
$\left(i_1 \leq i_2 \leq \cdots \leq i_n\right)$ of positive integers
having strict inequalities $i_j < i_{j+1}$ whenever $i \in \Des(w)$
(namely, the bijection sends any $w$-partition $f$ to the sequence
$\left(f\left(w_1\right) \leq f\left(w_2\right) \leq \cdots \leq f\left(w_n\right)\right)$).
Hence,
\[
F_w(\xx)
= \sum_{f \in \mathcal{A}(w)} \xx_f
= \sum\limits_{\substack{(1 \leq ) i_1 \leq i_2 \leq \cdots \leq i_n; \\
i_{j} < i_{j+1} \text{ if } j\in \Des(w) }}
x_{i_1} x_{i_2} \cdots x_{i_n}
= \sum\limits_{\substack{(1 \leq ) i_{1} \leq i_{2} \leq \cdots \leq i_n; \\
i_{j} < i_{j+1} \text{ if } j\in D\left(  \alpha\right) }}
x_{i_1} x_{i_2} \cdots x_{i_n}
\]
(since $\Des(w) = D(\alpha)$).
Comparing this with
\eqref{fundamental-qsym-fn-defn-alternative}, we conclude that
$F_w(\xx) = L_\alpha$.
%% Old argument:
%The underlying set $\{f(w_i)\}_{i=1}^n$ will equal $\{j_1 < \cdots < j_\ell\}$
%with indices in increasing order having
%a multiplicity sequence $\beta=(\beta_1,\ldots,\beta_\ell)$
%that gives a composition $\beta$ refining $\alpha$.
\end{proof}

% [DG][v72] Added more details to the above proof.

The next proposition (\cite[Cor. 7.19.5]{Stanley},
\cite[Cor. 3.3.24]{LuotoMykytiukvanWilligenburg}) is an algebraic shadow of
Stanley's main lemma \cite[Thm. 7.19.4]{Stanley} in $P$-partition theory.
  It expands any $F_P(\xx)$ in the $\{L_\alpha\}$ basis,
as a sum over the set \dfn{$\LLL(P)$} of all \emph{linear extensions} $w$
of $P$\ \ \ \ \footnote{Let us explain what we mean by linear extensions
and how we represent them.
\par
If $\mathbf{P}$ is a finite poset, then a \dfn{linear extension} of
$\mathbf{P}$ denotes a total order $w$ on the set $\mathbf{P}$ having the
property that every two elements $i$ and $j$ of $\mathbf{P}$ satisfying
$i <_{\mathbf{P}} j$ satisfy $i <_w j$.
(In other words, it is a linear order on
the ground set $\mathbf{P}$ which extends $\mathbf{P}$ as a poset;
therefore the name.) We identify such a total order $w$ with the % (unique)
list $\left(\mathbf{p}_1, \mathbf{p}_2, \ldots, \mathbf{p}_n\right)$
containing all elements of $\mathbf{P}$ in $w$-increasing order (that
is, $\mathbf{p}_1 <_w \mathbf{p}_2 <_w \cdots <_w \mathbf{p}_n$).
\par
(Stanley, in \cite[\S 3.5]{Stanley}, defines linear extensions in a
slightly different way: For him, a linear extension of a finite
poset $\mathbf{P}$ is an order-preserving bijection from $\mathbf{P}$
to the subposet $\left\{ 1, 2, \ldots, \left|\mathbf{P}\right| \right\}$
of $\ZZ$. But this is equivalent to our definition, since a bijection
like this can be used to transport the order relation of
$\left\{ 1, 2, \ldots, \left|\mathbf{P}\right| \right\}$
back to $\mathbf{P}$, thus resulting in a total order on $\mathbf{P}$
which is a linear extension of $\mathbf{P}$ in our sense.)
}. %, that is, the set of all extensions of $P$ to a linear order.
E.g., the poset $P$ from Example~\ref{labelled-poset-example} has
$\LLL(P) = \{3124,3142,3412\}$.

% [DG][v41] Added footnote about the definition of linear extension.
% Is it any good? (I don't want to leave this lingo undefined, but I am
% not good at defining things :/ .)

% [DG][v42] Shortened as discussed in email.


\begin{theorem}
\label{Stanley's-P-partition-theorem}
For any labelled poset $P$,
\[
F_P(\xx) = \sum_{w \in \LLL(P)} F_w(\xx).
\]
\end{theorem}
\begin{proof}
We give Gessel's proof \cite[Thm. 1]{Gessel}, via induction on the
number of pairs $i,j$ which are incomparable in $P$.  When this quantity
is $0$, then $P$ is itself a linear order $w$, so that $\LLL(P)=\{w\}$
and there is nothing to prove.

In the inductive step,
let $i,j$ be incomparable elements.  Consider the two
posets $P_{i<j}$ and $P_{j<i}$ which are obtained
from $P$ by adding in an order relation between $i$ and $j$, and then taking
the transitive closure;  it is not hard to see that these transitive closures
cannot contain a cycle, so that these really do define two posets.
The result then follows by induction applied to  $P_{i<j}, P_{j<i}$, once
one notices that
$
\LLL(P) = \LLL(P_{i<j}) \sqcup  \LLL(P_{j<i})
$
since every linear extension $w$ of $P$ either has $i$ before $j$ or vice-versa, and
$
\AAA(P) = \AAA(P_{i<j}) \sqcup \AAA(P_{j<i})
$
since, assuming that $i <_\ZZ j$ without loss of generality, every
$f$ in $\AAA(P)$ either satisfies $f(i) \leq f(j)$ or $f(i) > f(j)$.
\end{proof}

\begin{example}
To illustrate the induction in the above proof, consider the poset
$P$ from Example~\ref{labelled-poset-example},
having $\LLL(P)=\{3124,3142,3412\}$.  Then choosing as incomparable pair
$(i,j)=(1,4)$, one has
\[
\xymatrix{
        &4& &2\\
P_{i<j}=& &1\ar@{-}[ur]\ar@{-}[ul]& \\
        & &3\ar@{-}[u]&}
\qquad
\xymatrix{
f(4)& &f(2)\\
 &f(1)\ar@{-}[ur]^{\leq} \ar@{-}[ul]^{\leq}& \\
 &f(3)\ar@{-}[u]^{<}&}
\qquad
\xymatrix{ \\  \text{, thus } \LLL(P_{i<j})=\{3124,3142\}\\ }
\]
and % \vskip.1in
\[
\xymatrix{
 &2\\
P_{j<i}=&1\ar@{-}[u]\\
 &4\ar@{-}[u]\\
 &3\ar@{-}[u]}
\qquad \qquad \qquad \qquad \qquad %\qquad
\xymatrix{
  f(2)\\
  f(1)\ar@{-}[u]^{\leq} \\
  f(4)\ar@{-}[u]^{<} \\
  f(3)\ar@{-}[u]^{\leq}}
\qquad
\xymatrix{ \\  \text{, thus } \LLL(P_{j<i})=\{3412\} .\\ }
\]
\end{example}

\begin{exercise}
\label{exe.P-partition.stanley.altproof}
Give an alternative proof for
Theorem~\ref{Stanley's-P-partition-theorem}.

[\textbf{Hint:} For every
$f : P \rightarrow \left\{ 1, 2, 3, \ldots \right\}$, we
can define a binary relation $\prec_{f}$ on the set $P$ by
letting $i \prec_{f} j$ hold if and only if
\[
\left( f \left( i \right) < f \left( j \right)
\text{ or }
\left( f \left( i \right) = f\left( j \right)
  \text{ and } i <_{\ZZ} j \right) \right)  .
\]
Show that this binary relation $\prec_{f}$ is (the smaller
relation of) a total order. When $f$ is a $P$-partition, then
endowing the set $P$ with this total order yields a linear
extension of $P$. Use this to show that the set
$\mathcal{A}\left( P \right)$ is the union of its
disjoint subsets $\mathcal{A}\left( w \right)$ with
$w \in \mathcal{L}\left( P \right)$.]
\end{exercise}

% [DG][v50] Added above exercise. It is an alternative to the
% "induction over the number of pairs of incomparable elements"
% argument that you and Gessel use.

Various other properties of the quasisymmetric functions
$F_P\left(\xx\right)$ are studied, e.g., in
\cite{McNamaraWard}.

We next wish to describe the
structure maps for the Hopf algebra $\Qsym$ in the basis $\{L_\alpha\}$
of fundamental quasisymmetric functions.  For this purpose, two more
definitions are useful.

\begin{definition}
\label{def.compositions.near-conc}
Given two nonempty compositions
$\alpha=(\alpha_1,\ldots,\alpha_\ell)$ and
$\beta=(\beta_1,\ldots,\beta_m)$, their
\emph{near-concatenation}\index{near-concatenation of compositions}
is
\[
\alpha \odot \beta:=
(\alpha_1,\ldots,\alpha_{\ell-1},\alpha_\ell+\beta_1,\beta_2,\ldots,\beta_m) .
\]
For example, the figure below depicts for $\alpha=(1,3,3)$
(black squares) and $\beta=(4,2)$ (white squares) the concatenation and
near-concatenation as ribbons:\footnote{The ribbons are drawn with their
boxes spaced out in order to facilitate counting.}
\begin{align*}
\Rib\left(\alpha \cdot \beta\right)&=\ %
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &   &   &   &   &   &\sq&\sq\\
   &   &   &   &\sq&\sq&\sq&\sq&   \\
   &   &\bsq&\bsq&\bsq&   &   &   &   \\
\bsq&\bsq&\bsq&   &   &   &   &   &   \\
\bsq&   &   &   &   &   &   &   &
\end{matrix}
\endgroup
\\
\Rib\left(\alpha \odot \beta\right)&=\ %
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &   &   &   &   &   &   &\sq&\sq\\
   &   &\bsq&\bsq&\bsq&\sq&\sq&\sq&\sq&\\
\bsq&\bsq&\bsq&   &   &   &   &   &   &\\
\bsq&   &   &   &   &   &   &   &   &
\end{matrix}
\endgroup
\end{align*}

Lastly, given $\alpha$ in $\Comp_n$, let $\omega(\alpha)$ be the unique composition in $\Comp_n$ whose partial sums $D(\omega(\alpha))$
form the complementary set within $[n-1]$ to the partial sums $D(\rev(\alpha))$;  alternatively, one can check this means that the ribbon for $\omega(\alpha)$ is obtained from that of $\alpha$
by conjugation or transposing, that is, if $\Rib\left(\alpha\right) =\lambda/\mu$ then
$\Rib\left(\omega(\alpha)\right) =\lambda^t/\mu^t$.
E.g. if $\alpha=(4,2,2)$ so that $n=8$,
then $\rev(\alpha)=(2,2,4)$ has $D(\rev(\alpha))=\{2,4\} \subset [7]$,
complementary to the set $\{1,3,5,6,7\}$ which are the partial sums
for $\omega(\alpha)=(1,2,2,1,1,1)$, and the ribbon diagrams of $\alpha$
and $\omega(\alpha)$ are
\[
\begin{matrix}
\Rib\left(\alpha\right) =\ &
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &   &   &\sq&\sq\\
   &   &   &\sq&\sq& \\
\sq&\sq&\sq&\sq&   & \\
\end{matrix}
\endgroup
\qquad \text{ and } \qquad
\Rib\left(\omega(\alpha)\right) =\ &
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &\sq\\
   &   &\sq\\
   &   &\sq\\
   &\sq&\sq\\
\sq&\sq&   \\
\sq&   &   \\
\end{matrix}
\endgroup
\end{matrix}
\]

\end{definition}

\begin{proposition}
\label{fundamental-basis-structure-maps}
The structure maps for the Hopf algebra $\Qsym$ in the basis $\{L_\alpha\}$
of fundamental quasisymmetric functions are as follows:
\begin{align}
\Delta L_\alpha &= \sum\limits_{\substack{(\beta,\gamma):\\
\beta \cdot \gamma=\alpha \text{ or }
\beta \odot \gamma=\alpha}} L_\beta \otimes L_\gamma ,
\label{Qsym-coproduct-on-fundamentals}\\
L_\alpha L_\beta &= \sum_{w \in \shuf{w_\alpha}{w_\beta}} L_{\gamma(w)} ,
\label{Qsym-product-on-fundamentals}\\
S(L_\alpha)&= (-1)^{|\alpha|} L_{\omega(\alpha)} .
\label{Qsym-antipode-on-fundamentals}
\end{align}
Here we are making use of the following notations in
\eqref{Qsym-product-on-fundamentals} (recall also
Definition~\ref{shuffles}):
\begin{enumerate}
\item[$\bullet$] A \dfn{labelled linear order}
will mean a labelled poset $P$ whose order
$<_{P}$ is a total order. We will identify any labelled linear order $P$ with
the word (over the alphabet $\ZZ$) obtained by
writing down the elements of $P$ in increasing order (with respect to the
total order $<_{P}$). This way, every word (over the alphabet $\ZZ$)
which has no two equal letters becomes identified with a labelled linear order.
\item[$\bullet$] $w_{\alpha}$ is any labelled linear order with underlying set
$\left\{  1,2,\ldots,\left\vert \alpha\right\vert \right\}  $ such that
$\Des\left(  w_{\alpha}\right)  =D\left(  \alpha\right)  $.
\item[$\bullet$] $w_{\beta}$ is any labelled linear order with underlying set $\left\{
\left\vert \alpha\right\vert +1,\left\vert \alpha\right\vert +2,\ldots,\left\vert
\alpha\right\vert +\left\vert \beta\right\vert \right\}  $ such that
$\Des\left(  w_{\beta}\right)  =D\left(  \beta\right)  $.
\item[$\bullet$] \dfn{$\gamma(w)$} is the unique composition of
$\left|\alpha\right| + \left|\beta\right|$ with $D(\gamma(w))=\Des(w)$.
\end{enumerate}
(The right hand side of \eqref{Qsym-product-on-fundamentals} is
to be read as a sum over all $w$, for a fixed choice of $w_\alpha$
and $w_\beta$.)
\end{proposition}

% [DG][v63] Added the preceding parenthetical sentence.

\noindent
At first glance the formula \eqref{Qsym-coproduct-on-fundamentals} for $\Delta L_\alpha$
might seem more complicated than the formula of
Proposition~\ref{Qsym-coproduct-on-monomials}
for $\Delta M_\alpha$.  However, it is equally simple
when viewed in terms of ribbon
diagrams:  it cuts the ribbon diagram $\Rib\left(\alpha\right)$
into two smaller ribbons $\Rib\left(\beta\right)$ and
$\Rib\left(\gamma\right)$, in all $|\alpha|+1$ possible ways,
via \emph{horizontal} cuts ($\beta \cdot \gamma = \alpha$)
or \emph{vertical} cuts ($\beta \odot \gamma = \alpha$).
For example,
\[
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{array}{rcccccl}
& \Delta L_{(3,2)} \\
&=1 \otimes L_{(3,2)}
&+L_{(1)} \otimes L_{(2,2)}
&+L_{(2)} \otimes L_{(1,2)}
&+L_{(3)} \otimes L_{(2)}
&+L_{(3,1)} \otimes L_{(1)}
&+L_{(3,2)} \otimes 1 .\\
    &\begin{matrix}  & & \sq & \sq \\ \underline{\sq} & \sq & \sq & \end{matrix}
    &\begin{matrix}  & & \sq & \sq \\ \sq & |\sq & \sq & \end{matrix}
    &\begin{matrix}  & & \sq & \sq \\ \sq & \sq & |\sq & \end{matrix}
    &\begin{matrix}  & & \underline{\sq} & \sq \\ \sq & \sq & \sq & \end{matrix}
    &\begin{matrix}  & & \sq & |\sq \\ \sq & \sq & \sq & \end{matrix}
    &\begin{matrix}  & & \sq & \overline{\sq} \\ \sq & \sq & \sq & \end{matrix}
\end{array}
\endgroup
\]

% [DG] I have edited the first and the last of these pictures so as
% to show that $1 \otimes L_{(3,2)}$ and $L_{(3,2)} \otimes 1$ correspond
% to horizontal (rather than vertical) cuts.

\begin{example}
To multiply $L_{(1,1)} L_{(2)}$, one could pick $w_\alpha = 21$ and $w_\beta=34$,
and then
\[
\begingroup
\setlength\arraycolsep{0.2pc} % Smaller spacing.
\begin{array}{rcccccccccccl}
L_{(1,1)} L_{(2)}
=\sum\limits_{w \in \shuf{21}{34}} L_{\gamma(w)}
&=&L_{\gamma(2134)}
&+&L_{\gamma(2314)}
&+&L_{\gamma(3214)}
&+&L_{\gamma(2341)}
&+&L_{\gamma(3241)}
&+&L_{\gamma(3421)}\\
&=&L_{(1,3)}
&+&L_{(2,2)}
&+&L_{(1,1,2)}
&+&L_{(3,1)}
&+&L_{(1,2,1)}
&+&L_{(2,1,1)}.
\end{array}
\endgroup
\]
\end{example}

Before we prove Proposition~\ref{fundamental-basis-structure-maps},
we state a simple lemma:

\begin{lemma}
\label{lem.QSym.ppar.djun}
Let $Q$ and $R$ be two labelled posets whose underlying
sets are disjoint.
Let $Q \sqcup R$ be the disjoint union of these posets
$Q$ and $R$; this is again a labelled poset.
Then,
\[
F_Q \left( \xx \right) F_R \left( \xx \right)
= F_{Q \sqcup R} \left( \xx \right) .
\]
\end{lemma}

\begin{proof} %[Proof of Lemma~\ref{lem.QSym.ppar.djun}.]
We identify the underlying set of $Q \sqcup R$ with $Q \cup R$
(since the sets $Q$ and $R$ are already disjoint).
If $f : Q \sqcup R \to \left\{1, 2, 3, \ldots\right\}$ is a
$Q \sqcup R$-partition, then its restrictions $f\mid_Q$ and
$f\mid_R$ are a $Q$-partition and an $R$-partition,
respectively.
Conversely, any pair of a $Q$-partition and an $R$-partition
can be combined to form a $Q \sqcup R$-partition.
Thus, there is a bijective correspondence between the addends
in the expanded sum
$F_Q \left( \xx \right) F_R \left( \xx \right)$ and the
addends in $F_{Q \sqcup R} \left( \xx \right)$.
\end{proof}

% [DG][v70] Added the above lemma (mostly in order to be
% able to cite it separately from the proof in which it
% appears).

\begin{proof}[Proof of Proposition~\ref{fundamental-basis-structure-maps}.]
To prove formula \eqref{Qsym-coproduct-on-fundamentals}
for $\alpha$ in $\Comp_n$, note that
\begin{equation}
\Delta L_\alpha = L_\alpha(\xx,\yy)
= \sum_{k=0}^n
\sum\limits_{\substack{
1 \leq i_1 \leq \cdots \leq i_k,\\
1 \leq i_{k+1} \leq \cdots \leq i_n:\\
i_r < i_{r+1} \text{ for }r \in D(\alpha) \setminus \left\{k\right\}
}}
 x_{i_1} \cdots x_{i_k} \cdot y_{i_{k+1}} \cdots y_{i_n}
\label{pf.fundamental-basis-structure-maps.DeltaLalpha}
\end{equation}
by Proposition~\ref{Qsym-technical-lemma-Lxy}
(where we identify $\Qsym \otimes \Qsym$ with a
$\kk$-subalgebra of $R\left(\xx, \yy\right)$ by means of
the embedding
$\Qsym \otimes \Qsym \overset{\cong}{\rightarrow}
\Qsym\left(\xx\right) \otimes \Qsym\left(\yy\right)
\hookrightarrow R\left(\xx, \yy\right)$ as in the definition
of the comultiplication on $\Qsym$).
One then realizes that the inner sums corresponding
to values of $k$ that lie (resp. do not lie) in $D(\alpha) \cup \left\{0,n\right\}$
correspond to the terms $L_\beta(\xx) L_\gamma(\yy)$
for pairs $(\beta,\gamma)$ in which
$\beta \cdot \gamma =\alpha$ (resp. $\beta \odot \gamma =\alpha$).

% [DG][v54] Added the "where we identify..." parenthetical.

%Expanding
%in the monomial basis lets one write both sides as
%sums $\sum_{(\beta,\gamma)} M_\beta \otimes M_\gamma$
%over certain pairs $(\beta,\gamma)$.  On the left side,
%one ranges over all such pairs with $\beta \cdot \gamma$ refining $\alpha$.
%On the right side, one ranges over such pairs in which
%$(\beta,\gamma)$ have coarsenings $(\hat{\beta},\hat{\gamma})$
%for which either $\hat{\beta} \cdot \hat{\gamma} =\alpha$
%or $\hat{\beta} \odot \hat{\gamma} =\alpha$.
%
%One can see that these are the same pairs $(\beta,\gamma)$
%by thinking of them as equivalent to a choice
%of extra dividing lines between
%some of the squares adjacent in the rows of the ribbon for $\alpha$,
%and one distinguished double-dividing line designating the end of
%$\beta$ and the beginning of $\gamma$.  If this double-dividing line
%occurs at the end of row, then $\hat{\beta} \cdot \hat{\gamma} =\alpha$;
%if it occurs in the middle of a row,
%then $\hat{\beta} \cdot \hat{\gamma} =\alpha$.
%For example, if
%$\alpha=(4,1,2,3)$, here are two
%choices of dividing lines:
%\[
%\begin{aligned}
%(\beta,\gamma)&=((1,3,1),(1,1,2,1))\\
%(\hat\beta,\hat\gamma)&=((4,1),(2,3))
%\end{aligned}
%\begin{matrix}
%   &   &    &                  &\phantom{|}\sq &\sq&|\sq \\
%   &   &    &\sq\phantom{\Vert}&|\sq&   &\\
%   &   &    &\sq\Vert&   &   &\\
%\sq&|\sq&\sq&\sq\phantom{\Vert}&   &   &\\
%\end{matrix}
%\]
%\[
%\begin{aligned}
%(\beta,\gamma)&=((1,3,1,1),(1,2,1))\\
%(\hat\beta,\hat\gamma)&=((4,1,1),(1,3))
%\end{aligned}
%\begin{matrix}
%   &   &    &       &\phantom{\Vert}\sq &\sq&|\sq \\
%   &   &    &\sq    &\Vert\sq&   &\\
%   &   &    &\sq&   &   &\\
%\sq&|\sq&\sq&\sq&   &   &\\
%\end{matrix}
%\]
%
For formula \eqref{Qsym-product-on-fundamentals}, let
$P$ be the labelled poset which is the disjoint union of linear orders $w_\alpha, w_\beta$.
Then
\[
L_\alpha L_\beta
=F_{w_\alpha}(\xx) F_{w_\beta}(\xx)
=F_P(\xx) = \sum_{w \in \LLL(P)} F_w(\xx)
=\sum_{w \in \shuf{w_\alpha}{w_\beta}} L_{\gamma(w)}
\]
where the first equality used Proposition~\ref{fundamental-P-partition-prop},
the second equality comes from Lemma~\ref{lem.QSym.ppar.djun},
%the definition of a $P$-partition,
the third equality from Theorem~\ref{Stanley's-P-partition-theorem},
and the fourth from the equality $\LLL(P)=\shuf{w_\alpha}{w_\beta}$.

To prove formula \eqref{Qsym-antipode-on-fundamentals}, compute
using Theorem~\ref{Qsym-antipode-on-monomials} that
\[
%\begin{aligned}
S(L_\alpha)
=\sum_{\beta \text{ refining } \alpha} S(M_\beta)
=\sum\limits_{\substack{(\beta,\gamma):\\
         \beta \text{ refines } \alpha,\\
         \gamma \text{ coarsens }\rev(\beta)}} (-1)^{\ell(\beta)} M_\gamma
=\sum_{\gamma} M_\gamma
     \sum_{\beta}  (-1)^{\ell(\beta)}
%&=(-1)^{|\alpha|}
%    \sum_{\gamma: D(\alpha) \cap D(\rev(\gamma))= \varnothing} M_\gamma \\
%&=(-1)^{|\alpha|} \sum_{\gamma: D(\gamma) \supset D(\omega(\alpha))} M_\gamma% \\
%&=(-1)^{|\alpha|} L_{\omega(\alpha)} \\
%\end{aligned}
\]
in which the last inner sum is over $\beta$ for which
\[
D(\beta) \supset D(\alpha) \cup D(\rev(\gamma)).
\]
The alternating signs make such inner sums vanish
unless they have only the single
term where $D(\beta)=[n-1]$ (that is, $\beta=(1^n)$).  This happens
exactly when
$
D(\rev(\gamma)) \cup D(\alpha)=[n-1]
$
or equivalently, when $D(\rev(\gamma))$ contains the complement
of $D(\alpha)$, that is, when $D(\gamma)$ contains the complement of
$D(\rev(\alpha))$, that is, when
$\gamma$ refines $\omega(\alpha)$.  Thus
\[
S(L_\alpha)=
\sum\limits_{\substack{\gamma \in \Comp_n:\\\gamma \text{ refines }\omega(\alpha)}}
M_\gamma  \cdot (-1)^n
=(-1)^{|\alpha|} L_{\omega(\alpha)}.
\]
\end{proof}

The antipode formula \eqref{Qsym-antipode-on-fundamentals}
for $L_\alpha$ leads to a general interpretation for the antipode
of $\Qsym$ acting on $P$-partition enumerators $F_P(\xx)$.

\begin{definition}
Given a labelled poset $P$ on $\{1,2,\ldots,n\}$, let the
\emph{opposite}\index{opposite labelled poset}
or \emph{dual}\index{dual labelled poset}
labelled poset \dfn{$P^{\opp}$}
be the labelled poset on $\{1,2,\ldots,n\}$ that has
$i <_{P^{\opp}} j$ if and only if $j <_P i$.
\end{definition}
For example,
\[
P=
\xymatrix{
 & &2\\
4& &1\ar@{-}[u]\\
 &3\ar@{-}[ur]\ar@{-}[ul]&
} \qquad \qquad
P^{\opp}=
\xymatrix{
    &3\ar@{-}[dl]\ar@{-}[dr]&\\
  4& & 1\ar@{-}[d]\\
    & &2
}
\]

The following observation is straightforward.

\begin{proposition}
\label{rotating-ribbons-is-reversing-permutations}
When $P$ is a linear order
corresponding to some permutation $w=(w_1,\ldots,w_n)$ in $\Symm_n$,
then $w^{\opp}=w w_0$ where $w_0 \in \Symm_n$ is the permutation that
swaps $i \leftrightarrow n+1-i$ (this is the so-called \dfn{longest
permutation}, thus named due to it having the highest ``Coxeter
length'' among all permutations in $\Symm_n$).  Furthermore, in this situation
one has $F_w(\xx)=L_\alpha$, that is, $\Des(w) = D(\alpha)$ if and only if
$\Des(w^{\opp})= D(\omega(\alpha))$, that is $F_{w^{\opp}}(\xx) = L_{\omega(\alpha)}$.
Thus,
\[
S(F_w(\xx)) =(-1)^n F_{w^{\opp}}(\xx).
\]
\end{proposition}

For example, given the compositions considered earlier:
\[
\alpha =(4,2,2)=
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &   &   &\sq&\sq\\
   &   &   &\sq&\sq& \\
\sq&\sq&\sq&\sq&   & \\
\end{matrix}
\endgroup
\qquad \text{ and } \qquad
\omega(\alpha) =(1,2,2,1,1,1)=
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &\sq\\
   &   &\sq\\
   &   &\sq\\
   &\sq&\sq\\
\sq&\sq&   \\
\sq&   &   \\
\end{matrix}
\endgroup
\]
if one picks $w=1235\cdot47\cdot68$
(with descent positions marked by dots) having
$\Des(w)=\{4,6\}=D(\alpha)$, then
$w^{\opp}=ww_0=8\cdot67\cdot45\cdot3\cdot2\cdot1$
has $\Des(w^{\opp})=\{1,3,5,6,7\}=D(\omega(\alpha))$.


\begin{corollary}
\label{antipode-in-Qsym-as-reciprocity-corollary}
For any labelled poset $P$ on $\{1,2,\ldots,n\}$, one has
\[
S \left( F_P(\xx) \right)= (-1)^n F_{P^{\opp}}(\xx).
\]
\end{corollary}
\begin{proof}
Since $S$ is linear, one can apply Theorem~\ref{Stanley's-P-partition-theorem}
and Proposition~\ref{rotating-ribbons-is-reversing-permutations},
obtaining
\[
S \left( F_P(\xx) \right)=
\sum_{w \in \LLL(P)} S( F_w(\xx) )
=\sum_{w \in \LLL(P)} (-1)^{n} F_{w^{\opp}}(\xx)=
(-1)^n F_{P^{\opp}}(\xx),
\]
as $\LLL(P^{\opp}) = \{w^{\opp}: w \in \LLL(P)\}$.
\end{proof}

\begin{remark}
Malvenuto and Reutenauer, in \cite[Theorem 3.1]{MalvenutoReutenauer-plethysm},
prove an even more general antipode formula, which encompasses our
Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary},
Proposition~\ref{rotating-ribbons-is-reversing-permutations},
Theorem~\ref{Qsym-antipode-on-monomials}
and \eqref{Qsym-antipode-on-fundamentals}.
See \cite[Theorem 4.2]{Grinberg-dp} for a restatement and
a self-contained proof of this theorem (and
\cite[Theorem 4.7]{Grinberg-dp} for an even further
generalization).
\end{remark}

We remark on a special case of Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary}
to which we alluded earlier, related to skew Schur functions.

\begin{corollary}
\label{antipode-on-skew-Schur-corollary}
In $\Lambda$, the action of $\omega$ and
the antipode $S$ on skew Schur functions $s_{\lambda/\mu}$
are as follows:
\begin{align}
\omega(s_{\lambda/\mu})&=s_{\lambda^t/\mu^t} ,\label{omega-on-skew-Schur}\\
S(s_{\lambda/\mu})&=(-1)^{|\lambda/\mu|} s_{\lambda^t/\mu^t} \label{antipode-on-skew-Schur}.
\end{align}
\end{corollary}
\begin{proof}
Given a skew shape $\lambda/\mu$, one can always create a labelled
poset $P$ which is its \dfn{skew Ferrers poset}, together with one of many
\emph{column-strict labellings}\index{column-strict labelling},
in such a way that
$F_P(\xx)=s_{\lambda/\mu}(\xx)$.  An example is shown here
for $\lambda/\mu=(4,4,2)/(1,1,0)$:
\[
\lambda/\mu=
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &\sq&\sq&\sq\\
   &\sq&\sq&\sq\\
\sq&\sq&  &
\end{matrix}
\endgroup
\qquad \qquad
P=\xymatrix@C=8pt@R=8pt{
                      &5&                         &       &       & \\
8\ar@{-}[ur]&       &4\ar@{-}[ul]&       &2& \\
                      &7\ar@{-}[ul]\ar@{-}[ur]&       &3\ar@{-}[ul]\ar@{-}[ur]&       &1\ar@{-}[ul]\\
       &       &6\ar@{-}[ul]\ar@{-}[ur]&       &       &   \\
}
\qquad
\xymatrix@C=8pt@R=8pt{
                      &f(5)&                         &       &       & \\
f(8)\ar@{-}[ur]^{<}&       &f(4)\ar@{-}[ul]^{\leq}&       &f(2)& \\
                      &f(7)\ar@{-}[ul]^{\leq}\ar@{-}[ur]^{<}&       &f(3)\ar@{-}[ul]^{\leq}\ar@{-}[ur]^{<}&       &f(1)\ar@{-}[ul]^{\leq}\\
       &       &f(6)\ar@{-}[ul]^{\leq}\ar@{-}[ur]^{<}&       &       &   \\
}
\]
The general definition is as follows: Let $P$ be the set of all boxes of the skew diagram
$\lambda / \mu$. Label these boxes by the numbers $1,2,\ldots,n$ (where $n=\left\vert
\lambda / \mu\right\vert $) row by row from bottom to top (reading every
row from left to right), and then define an order relation $<_{P}$ on $P$ by
requiring that every box be smaller (in $P$) than its right neighbor and smaller (in
$P$) than its lower neighbor.
It is not hard to see that in this situation, $F_{P^{\opp}}(\xx)=\sum_T \xx^{\cont(T)}$ as $T$ ranges over all
\emph{reverse semistandard tableaux}\index{reverse semistandard tableau}
or \emph{column-strict plane partitions}\index{column-strict plane partition}
of $\lambda^t/\mu^t$:
\[
\lambda^t/\mu^t=
\begingroup
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &\sq\\
\sq&\sq&\sq&\\
\sq&\sq&  &\\
\sq&\sq&  &
\end{matrix}
\endgroup
\qquad \qquad
P^{\opp}=\xymatrix@C=8pt@R=8pt{
 & &6& & &\\
 &7\ar@{-}[ur]& &3\ar@{-}[ul]& &1\\
8\ar@{-}[ur]& &4\ar@{-}[ur]\ar@{-}[ul]& &2\ar@{-}[ur]\ar@{-}[ul]& \\
 &5\ar@{-}[ur]\ar@{-}[ul]& & & &
}
\qquad
\xymatrix@C=8pt@R=8pt{
 & &f(6)& & &\\
 &f(7)\ar@{-}[ur]^{<}& &f(3)\ar@{-}[ul]^{\leq}& &f(1)\\
f(8)\ar@{-}[ur]^{<}& &f(4)\ar@{-}[ur]^{<}\ar@{-}[ul]^{\leq}& &f(2)\ar@{-}[ur]^{<}\ar@{-}[ul]^{\leq}& \\
 &f(5)\ar@{-}[ur]^{<}\ar@{-}[ul]^{\leq}& & & &
}
\]
But this means that $F_{P^{\opp}}(\xx)=s_{\lambda^t/\mu^t}(\xx)$, since
the fact that skew Schur functions lie in $\Lambda$ implies that they
can be defined either as generating functions for
column-strict tableaux or reverse semistandard tableaux;
see Remark~\ref{other-linear-orders-remark} above,
or \cite[Prop. 7.10.4]{Stanley}.

% [DG] I have added a semi-formal definition of the poset $P$. It is
% somewhat clumsy.

% [DG][v3] Replaced "reverse column-strict tableaux" by "reverse
% semistandard tableaux" to match the notations in a footnote in Chapter 2.
% Maybe I should have done it the other way round...

Thus we have
\begin{align*}
F_P(\xx)&=s_{\lambda/\mu}(\xx) ,\\
F_{P^{\opp}}(\xx)&=s_{\lambda^t/\mu^t}(\xx).
\end{align*}
Corollary~\ref{cor.bialg-mor-is-Hopf}
tells us that the antipode for $\Qsym$ must specialize
to the antipode for $\Lambda$ (see also Remark~\ref{antipode-comparison-remark} below), so \eqref{antipode-on-skew-Schur}
is a special case of
Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary}.
Then \eqref{omega-on-skew-Schur}
follows from the relation \eqref{antipode-versus-omega-in-Sym} that
$S(f)=(-1)^n \omega(f)$ for $f$ in $\Lambda_n$.
\end{proof}



\begin{remark}
Before leaving $P$-partitions temporarily, we mention two
open questions about them.


The first is a conjecture of Stanley from his thesis \cite{Stanley-thesis}.
As mentioned in the proof of Corollary~\ref{antipode-on-skew-Schur-corollary},
each skew Schur function $s_{\lambda/\mu}(\xx)$ is a
special instance of $P$-partition enumerator $F_P(\xx)$.

\begin{conjecture}
A labelled poset $P$ has $F_P(\xx)$ symmetric, and not just quasisymmetric,
\emph{if and only if} $P$ is a column-strict labelling of some
skew Ferrers poset $\lambda/\mu$.
\end{conjecture}
\noindent
A somewhat weaker result in this direction
was proven by Malvenuto in her thesis \cite[Thm. 6.4]{Malvenuto}, showing
that if a labelled poset $P$ has the stronger property that its set of linear extensions
$\LLL(P)$ is a union of \emph{plactic} or \emph{Knuth equivalence classes}, then
$P$ must be a column-strict labelling of a skew Ferrers poset.

The next question is due to P. McNamara, and is suggested by the
obvious factorizations of $P$-partition enumerators
$F_{P_1 \sqcup P_2}(\xx) = F_{P_1}(\xx) F_{P_2}(\xx)$ (Lemma~\ref{lem.QSym.ppar.djun}).

\begin{question}
\label{McNamara's-question}
If $\kk$ is a field,
does a \emph{connected} labelled poset $P$ always have
$F_P(\xx)$ \emph{irreducible} within the ring $\Qsym$?
\end{question}

The phrasing of this question requires further comment.
It is assumed here that $\xx=(x_1,x_2,\ldots)$ is infinite;
for example when $P$ is a $2$-element chain labelled ``against
the grain'' (i.e., the bigger element of the chain has the
smaller label),
then $F_P(\xx)=e_2(\xx)$ is irreducible, but its specialization
to two variables $\xx=(x_1,x_2)$ is $e_2(x_1,x_2)=x_1 x_2$,
which is reducible.  If one wishes to work in finitely many variables
$\xx=(x_1,\ldots,x_m)$ one can perhaps assume that $m$ is at least
$|P|+1$.

% [DG] I specified the labelling in the example.

When working in $\Qsym=\Qsym(\xx)$ in
infinitely many variables, it is perhaps not so clear where factorizations occur.
For example, if $f$ lies in $\Qsym$ and factors
$f=g \cdot h$ with $g, h$ in $R(\xx)$, does this imply that $g, h$ also lie in $\Qsym$?
The answer is ``Yes'' (for $\kk = \ZZ$), but this is not obvious, and
was proven by P. Pylyavskyy in \cite[Chap. 11]{Pylyavskyy-thesis}.


One also might wonder whether $\Qsym_\ZZ$ is a unique factorization domain, but
this follows from the result of M. Hazewinkel (\cite{Hazewinkel}
and \cite[Thm. 6.7.5]{HazewinkelGubareniKirichenko}, and
Theorem~\ref{thm.QSym.lyndon} further below) who
proved a conjecture of Ditters that $\Qsym_\ZZ$
is a polynomial algebra;  earlier
Malvenuto and Reutenauer \cite[Cor. 2.2]{MalvenutoReutenauer} had shown
that $\Qsym_\QQ$ is a polynomial algebra.  In fact,
one can find polynomial generators $\{P_{\alpha}\}$ for $\Qsym_\QQ$ as a
subset of the dual basis to the $\QQ$-basis $\{\xi_\alpha\}$ for $\Nsym_\QQ$
which comes from taking
products $\xi_{\alpha}:= \xi_{\alpha_1} \cdots \xi_{\alpha_\ell}$
of the elements $\{\xi_n\}$ defined in Remark~\ref{second-antipode-proof} below.  Specifically,
one takes those $P_\alpha$ for which the composition $\alpha$ is a \emph{Lyndon composition};
see the First proof of Proposition~\ref{prop.QSym.lyndon} for a mild variation
on this construction.

Hazewinkel's proof \cite[Thm. 6.7.5]{HazewinkelGubareniKirichenko}
of the polynomiality of $\Qsym_\ZZ$ also shows that $\Qsym$ is a
polynomial ring over $\Lambda$ (see
Corollary~\ref{cor.QSym.lyndon.free}); in particular, this yields that
$\Qsym$ is a free $\Lambda$-module.\footnote{The latter statement
has an analogue in finitely many indeterminates, proven by Lauve
and Mason in \cite[Corollary 13]{LauveMason-stable}: The
quasisymmetric functions $\Qsym\left(\left\{x_i\right\}_{i\in I}\right)$
are free as a $\Lambda\left(\left\{x_i\right\}_{i\in I}\right)$-module
for any totally ordered set $I$, infinite or not. In the case of
finite $I$, this cannot be derived by Hazewinkel's arguments, as
the ring $\Qsym\left(\left\{x_i\right\}_{i\in I}\right)$ is not
in general a polynomial ring (e.g., when $\kk = \QQ$ and
$I = \left\{1, 2\right\}$, this ring is not even a UFD, as
witnessed by $\left(x_1^2 x_2\right) \cdot \left(x_1 x_2^2\right)
= \left(x_1 x_2\right)^3$).}

% [DG][v34] Added above paragraph on Lauve-Mason work.

% [DG][v36] Updated with references.

An affirmative answer to Question~\ref{McNamara's-question}
is known at least in the special case where $P$ is a connected column-strict labelling
of a skew Ferrers diagram, that is, when $F_P(\xx)=s_{\lambda/\mu}(\xx)$
for some connected skew diagram $\lambda/\mu$;  see \cite{BarekatRWilligenburg}.
\end{remark}



\subsection{\label{subsect.Lalpha-std}Standardization of $n$-tuples and the
fundamental basis}

Another equivalent description of the fundamental quasisymmetric functions
$L_{\alpha}$ (Lemma~\ref{lem.Lalpha-std.std-L} below) relies on the concept of
words and of their standardizations. We shall study words in detail in Chapter
\ref{sect.QSym.lyndon}; at this point, we merely introduce the few notions
that we will need:

\begin{definition}
\label{def.words.1}
We fix a totally ordered set $\mathfrak{A}$, which we call
the \dfn{alphabet}.

We recall that a \emph{word over $\mathfrak{A}$}\index{word}
is just a (finite) tuple of
elements of $\mathfrak{A}$. A word $\left(  w_{1},w_{2},\ldots,w_{n}\right)  $
can be written as $w_{1}w_{2}\cdots w_{n}$ when this incurs no ambiguity.

If $w\in\mathfrak{A}^{n}$ is a word and $i\in\left\{  1,2,\ldots,n\right\}  $,
then the \dfn{$i$-th letter} of $w$ means the $i$-th entry of the $n$-tuple
$w$. This $i$-th letter will be denoted by \dfn{$w_i$}.
\end{definition}

Our next definition relies on a simple fact about permutations and
words:\footnote{See Exercise \ref{exe.Lalpha-std.proofs} below for a proof of
Proposition \ref{prop.words.std-exists}.}

\begin{proposition}
\label{prop.words.std-exists}Let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\mathfrak{A}^{n}$ be any word. Then, there exists a unique permutation
$\sigma\in\Symm_n$ such that for every two elements $a$ and $b$ of
$\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, we have $\left(
\sigma\left(  a\right)  <\sigma\left(  b\right)  \text{ if and only if }%
w_{a}\leq w_{b}\right)  $.
\end{proposition}

\begin{definition}
\label{def.words.std} Let $w\in\mathfrak{A}^{n}$ be any word. The unique
permutation $\sigma\in\Symm_n$ defined in Proposition
\ref{prop.words.std-exists} is called the \dfn{standardization} of $w$, and
is denoted by \dfn{$\operatorname*{std}w$}.
\end{definition}

\begin{example}
If $\mathfrak{A}$ is the alphabet $\left\{  1<2<3<\cdots\right\}  $, then
$\operatorname*{std}\left(  41211424\right)  $ is the permutation which is
written (in one-line notation) as $61423758$.
\end{example}

A simple method to compute the standardization of a word $w\in\mathfrak{A}%
^{n}$ is the following: Replace all occurrences of the smallest letter
appearing in $w$ by the numbers $1,2,\ldots,m_{1}$ (where $m_{1}$ is the
number of these occurrences); then replace all occurrences of the
second-smallest letter appearing in $w$ by the numbers $m_{1}+1,m_{1}%
+2,\ldots,m_{1}+m_{2}$ (where $m_{2}$ is the number of these occurrences), and
so on, until all letters are replaced by numbers.\footnote{Here, a number is
not considered to be a letter; thus, a number that replaces a letter will
always be left in peace afterwards.} The result is the standardization of $w$,
in one-line notation.

Another method to compute the standardization $\operatorname*{std}w$ of a word
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n}$ is based on
sorting. Namely, consider the total order on the set $\mathfrak{A}%
\times \ZZ$ given by%
\[
\left(  a,i\right)  \leq\left(  b,j\right)  \text{ if and only if }\left(
\text{either }a<b\text{ or }\left(  a=b\text{ and }i\leq j\right)  \right)  .
\]
(In other words, two pairs in $\mathfrak{A}\times \ZZ$ are compared by
first comparing their first entries, and then, in the case of a tie, using the
second entries as tiebreakers.) Now, in order to compute $\operatorname*{std}%
w$, we sort the $n$-tuple $\left(  \left(  w_{1},1\right)  ,\left(
w_{2},2\right)  ,\ldots,\left(  w_{n},n\right)  \right)  \in\left(
\mathfrak{A}\times \ZZ\right)  ^{n}$ into increasing order (with respect
to the total order just described), thus obtaining a new $n$-tuple of the form
$\left(  \left(  w_{\tau\left(  1\right)  },\tau\left(  1\right)  \right)
,\left(  w_{\tau\left(  2\right)  },\tau\left(  2\right)  \right)
,\ldots,\left(  w_{\tau\left(  n\right)  },\tau\left(  n\right)  \right)
\right)  $ for some $\tau\in\Symm_n$; the standardization
$\operatorname*{std}w$ is then $\tau^{-1}$.

\begin{definition}
\label{def.QSym.gamma(sigma)} Let $n\in \NN$. Let $\sigma\in
\Symm_n$. Define a subset $\Des \sigma$ of $\left\{
1,2,\ldots,n-1\right\}  $ by
\[
\Des \sigma = \left\{  i\in\left\{  1,2,\ldots,n-1\right\}
\ \mid\ \sigma\left(  i\right)  >\sigma\left(  i+1\right)  \right\}  .
\]
(This is a particular case of the definition of $\Des w$ in
Exercise \ref{exe.shareshian-wachs.path}, if we identify $\sigma$ with the
$n$-tuple $\left(  \sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  n\right)  \right)  $. It is also a particular case of
the definition of $\Des w$ in Proposition
\ref{fundamental-P-partition-prop}, if we identify $\sigma$ with the total
order $\left(  \sigma\left(  1\right)  <\sigma\left(  2\right)  <\cdots
<\sigma\left(  n\right)  \right)  $ on the set $\left\{  1,2,\ldots,n\right\}
$.)

There is a unique composition $\alpha$ of $n$ satisfying $D\left(
\alpha\right)  = \Des \sigma$ (where $D\left(  \alpha\right)  $
is defined as in Definition \ref{def.QSym.D}). This composition will be
denoted by $\gamma\left(  \sigma\right)  $.
\end{definition}

The following lemma (equivalent to \cite[Lemma 9.39]{Reutenauer}) yields
another description of the fundamental quasisymmetric functions:

\begin{lemma}
\label{lem.Lalpha-std.std-L}Let $\mathfrak{A}$ denote the totally ordered set
$\left\{  1<2<3<\cdots\right\}  $ of positive integers. For each word
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n}$, we define a
monomial $\xx_{w}$ in $\kk\left[  \left[  \xx\right]
\right]  $ by $\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$.

Let $n\in \NN$ and $\sigma\in\Symm_n$. Then,%
\[
L_{\gamma\left(  \sigma\right)  }=\sum\limits_{\substack{w\in\mathfrak{A}%
^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}.
\]

\end{lemma}

\begin{exercise}
\label{exe.Lalpha-std.proofs}Prove Proposition \ref{prop.words.std-exists} and
Lemma \ref{lem.Lalpha-std.std-L}.
\end{exercise}





\subsection{The Hopf algebra $\Nsym$ dual to $\Qsym$}

We introduce here the (graded) dual Hopf algebra to $\Qsym$. This is
well-defined, as $\Qsym$ is connected graded of finite type.

\begin{definition}
Let $\Nsym:=\Qsym^o$\index{$\Nsym$}, with dual pairing
$\Nsym \otimes \Qsym \overset{(\cdot,\cdot)}{\longrightarrow} \kk$.
Let $\{H_\alpha\}$\index{$H_\alpha$}
be the $\kk$-basis of $\Nsym$ dual to the $\kk$-basis
$\{M_\alpha\}$ of $\Qsym$, so that
\[
(H_\alpha,M_\beta)=\delta_{\alpha,\beta}.
\]
When the base ring $\kk$ is not clear from the context, we write
$\Nsym_\kk$ in lieu of $\Nsym$.

The Hopf algebra $\Nsym$ is known as the \dfn{Hopf algebra of
noncommutative symmetric functions}\index{noncommutative symmetric
function}. Its study goes back to
\cite{GelfandKrobLascouxLeclercRetakhThibon}.
\end{definition}

% [DG][v7] Added the previous two sentences.

% [DG][v3] Here and in the following, all $\ZZ$s replaced by $\kk$s
% (except when necessary in the proof).

\begin{theorem}
\label{thm.NSym.free}
Letting $H_n:=H_{(n)}$ for $n=0,1,2,\ldots$, with $H_0=1$,
one has that
\begin{equation}
\Nsym \cong \kk\langle H_1,H_2,\ldots \rangle,
\label{eq.thm.NSym.free}
\end{equation}
the free associative (but not commutative) algebra on
generators $\{H_1,H_2,\ldots\}$ with coproduct determined
by\footnote{The abbreviated summation indexing
$\sum_{i+j=n} t_{i,j}$ used here is intended to mean
\[
\sum_{ \substack{ (i,j) \in \NN^2; \\ i+j=n } } t_{i,j}.
\] }
\begin{equation}
\label{Nsym-coproduct-on-H}
\Delta H_n = \sum_{i+j=n} H_i \otimes H_j.
\end{equation}
\end{theorem}
\begin{proof}
Since Proposition~\ref{Qsym-coproduct-on-monomials} asserts that
$\Delta M_\alpha
      = \sum_{(\beta,\gamma): \beta \cdot \gamma=\alpha}
       M_\beta \otimes M_\gamma$, and since $\{H_\alpha\}$ are
dual to $\{M_\alpha\}$, one concludes that for any compositions $\beta, \gamma$, one has
\[
H_\beta H_\gamma= H_{\beta \cdot \gamma} .
\]
Iterating this gives
\begin{equation}
\label{eq.Nsym.Halpha}
H_{\alpha}=H_{(\alpha_1,\ldots,\alpha_\ell)}
    =H_{\alpha_1} \cdots H_{\alpha_\ell}.
\end{equation}
Since the $H_\alpha$ are a $\kk$-basis for $\Nsym$, this shows
$\Nsym \cong \kk\langle H_1,H_2,\ldots \rangle$.

Note that $H_n=H_{(n)}$ is dual to $M_{(n)}$, so to
understand $\Delta H_n$, one should understand how  $M_{(n)}$ can appear
as a term in the product $M_{\alpha} M_{\beta}$.
By \eqref{Qsym-product-on-monomials} this occurs only
if $\alpha=(i), \beta=(j)$ where $i+j=n$, where
\[
M_{(i)} M_{(j)}= M_{(i+j)}+M_{(i,j)}+M_{(j,i)}
\]
(where the $M_{(i,j)}$ and $M_{(j,i)}$ addends have to be
disregarded if one of $i$ and $j$ is $0$).
By duality, this implies the formula \eqref{Nsym-coproduct-on-H}.
\end{proof}


\begin{corollary}
\label{cor.NSym.pi}
The algebra homomorphism defined by
\[
\begin{array}{rcl}
\Nsym &\overset{\pi}{\longrightarrow} &\Lambda ,\\
H_n &\longmapsto &h_n
\end{array}
\]
is a Hopf algebra surjection, and adjoint to the
inclusion $\Lambda \overset{i}{\hookrightarrow} \Qsym$
(with respect to the dual pairing
$\Nsym \otimes \Qsym \overset{(\cdot,\cdot)}{\longrightarrow} \kk$).
\end{corollary}

% [DG][v43] Added
% "(with respect to the dual pairing
% $\Nsym \otimes \Qsym \overset{(\cdot,\cdot)}{\longrightarrow} \kk$)."

\begin{proof}
As an algebra morphism, $\pi$ may be identified with the surjection
$T(V) \rightarrow \Sym(V)$ from the tensor algebra on a graded
free $\kk$-module $V$ with basis $\{H_1, H_2,\ldots\}$ to the
symmetric algebra on $V$, since
\begin{align*}
\Nsym &\cong \kk\langle H_1,H_2,\ldots\rangle ,\\
\Lambda &\cong \kk[h_1,h_2,\ldots] .
\end{align*}
As \eqref{Nsym-coproduct-on-H} and Proposition~\ref{symm-comultiplication-formulas-prop}(iii)
assert that
\begin{align*}
\Delta H_n &= \sum_{i+j=n} H_i \otimes H_j ,\\
\Delta h_n &= \sum_{i+j=n} h_i \otimes h_j ,\\
\end{align*}
this map $\pi$ is also a bialgebra morphism, and hence a Hopf morphism by
Corollary~\ref{cor.bialg-mor-is-Hopf}.

To check $\pi$ is adjoint to $i$, let $\lambda(\alpha)$
denote the partition which is the weakly decreasing rearrangement of the composition $\alpha$,
and note that the bases $\{H_\alpha\}$
of $\Nsym$ and $\{m_\lambda\}$ of $\Lambda$ satisfy
\[
(\pi(H_\alpha),m_\lambda)
=(h_{\lambda(\alpha)},m_\lambda)
=\left\{ \begin{matrix}
1 & \text{ if }\lambda(\alpha)=\lambda \\
0 & \text{ otherwise }
\end{matrix} \right\}
=\left( H_{\alpha},\sum_{\beta: \lambda(\beta)=\lambda} M_\beta \right)
=(H_\alpha, i(m_\lambda)).
\]
\end{proof}

\begin{remark}
\label{second-antipode-proof}
For those who prefer generating functions to sign-reversing involutions,
we sketch here  Malvenuto and Reutenauer's elegant proof
\cite[Cor. 2.3]{MalvenutoReutenauer} of the antipode
formula (Theorem~\ref{Qsym-antipode-on-monomials}).  One needs to
know that when $\QQ$ is a subring of $\kk$, and $A$ is a $\kk$-algebra
(possibly noncommutative),
in the ring of power series $A[[t]]$ where $t$ commutes with
all of $A$, one still has familiar facts, such as
\[
a(t) = \log b(t) \quad \text{ if and only if }\quad b(t) = \exp a(t)
\]
and whenever $a(t), b(t)$ commute in $A[[t]]$, one has
\begin{align}
\exp \left( a(t)+b(t) \right) &= \exp a(t) \exp b(t) , \label{exp-of-sum} \\
\log \left( a(t) b(t) \right) &= \log a(t) + \log b(t) . \label{log-of-product}
\end{align}
Start by assuming WLOG that $\kk=\ZZ$ (as $\Nsym_\kk
= \Nsym_\ZZ \otimes_\ZZ \kk$ in the general case).
Now, define in $\Nsym_\QQ=\Nsym \otimes_\ZZ \QQ$ the elements
$\{\xi_1,\xi_2,\ldots\}$\index{$\xi_n$} via generating functions
in $\Nsym_\QQ[[t]]$:
\begin{equation}
\begin{aligned}
\label{relation-definining-pi's}
\widetilde{H}(t)&:=\sum_{n \geq 0} H_n t^n,\\
\xi(t)&:=\sum_{n \geq 1} \xi_n t^n=\log \widetilde{H}(t) .
\end{aligned}
\end{equation}
One first checks that this makes each $\xi_n$ primitive, via
a computation in the ring $(\Nsym_\QQ \otimes \Nsym_\QQ)[[t]]$ (into which
we ``embed'' the ring $(\Nsym_{\QQ}[[t]]) \otimes_{\QQ[[t]]} (\Nsym_{\QQ}[[t]])$
via the canonical ring homomorphism from the latter into the former
\footnote{This ring homomorphism might fail to be injective, whence the ``embed''
stands in quotation marks. This does not need to worry us, since we will
not draw any conclusions in
$(\Nsym_{\QQ}[[t]]) \otimes_{\QQ[[t]]} (\Nsym_{\QQ}[[t]])$
from our computation.
\par
We are also somewhat cavalier with the notation $\Delta$: we use it both
for the comultiplication $\Delta : \Nsym_\QQ \to \Nsym_\QQ \otimes \Nsym_\QQ$
of the Hopf algebra $\Nsym_\QQ$ and for the continuous $\kk$-algebra
homomorphism $\Nsym_\QQ\left[\left[t\right]\right] \to
\left(\Nsym_\QQ \otimes \Nsym_\QQ\right) \left[\left[t\right]\right]$ it
induces.}):
\begin{align*}
\Delta \xi(t) &= \Delta \left( \log \sum_{n \geq 0} H_n t^n\right)
= \log \sum_{n \geq 0} \Delta(H_n) t^n
= \log \sum_{n \geq 0} \left( \sum_{i+j=n} H_i \otimes H_j \right) t^n\\
&= \log \left( \left(\sum_{i \geq 0} H_i t^i\right)  \otimes
                \left(\sum_{j \geq 0} H_j t^j\right) \right)
= \log \left( \left(\sum_{i \geq 0} H_i t^i \otimes \one  \right)
                \left( \one \otimes \sum_{j \geq 0} H_j t^j\right)  \right)\\
&\overset{\eqref{log-of-product}}{=}\log \widetilde{H}(t) \otimes \one + \one \otimes \log \widetilde{H}(t)
= \xi(t) \otimes \one + \one \otimes \xi(t).
\end{align*}
Comparing coefficients in this equality yields
$\Delta(\xi_n) = \xi_n \otimes \one + \one \otimes \xi_n$.
Thus $S(\xi_n) = -\xi_n$, by Proposition~\ref{antipode-on-primitives}.
This allows one to determine $S(H_n)$ and $S(H_\alpha)$, after
one first inverts the relation \eqref{relation-definining-pi's}
to get that $\widetilde{H}(t) = \exp \xi(t)$, and hence
\begin{align*}
S(\widetilde{H}(t)) &= S(\exp \xi(t))
=\exp S(\xi(t))
=\exp\left( -\xi(t) \right)
\overset{\eqref{exp-of-sum}}{=}(\exp \xi(t)) ^{-1} \\
&=
\widetilde{H}(t)^{-1}=\left( 1 + H_1 t + H_2 t^2 + \cdots \right)^{-1}.
\end{align*}
Upon expanding the right side, and comparing coefficients of $t^n$, this gives
\[
S(H_n)= \sum_{\beta \in \Comp_n} (-1)^{\ell(\beta)} H_\beta
\]
and hence
\[
S(H_\alpha) = S(H_{\alpha_\ell}) \cdots S(H_{\alpha_2}) S(H_{\alpha_1})
=\sum\limits_{\substack{\gamma: \\ \gamma \text{ refines }\rev(\alpha)}}
(-1)^{\ell(\gamma)} H_\gamma
=\sum\limits_{\substack{\gamma: \\ \rev(\gamma) \text{ refines }\alpha}}
(-1)^{\ell(\gamma)} H_\gamma
\]
(because if $\mu$ and $\nu$ are two compositions, then $\mu$ refines $\nu$
if and only if $\rev(\mu)$ refines $\rev(\nu)$).
As $S_{\Nsym}, S_{\Qsym}$ are adjoint, and $\{H_\alpha\},
\{M_\alpha\}$ are dual bases, this is equivalent to
saying that
\[
S(M_\alpha)= (-1)^{\ell(\alpha)} \sum\limits_{\substack{\gamma:\\ \rev(\alpha) \text{ refines } \gamma}} M_{\gamma}
\qquad \text{for all $\alpha \in \Comp$.}
\]
But this is precisely the claim of
Theorem~\ref{Qsym-antipode-on-monomials}.
Thus,
Theorem~\ref{Qsym-antipode-on-monomials} is proven once again.

Let us say a bit more about the elements $\xi_n$ defined in
\eqref{relation-definining-pi's} above. The elements $n\xi_n$
are noncommutative
analogues of the power sum symmetric functions $p_n$ (and, indeed, are
lifts of the latter to $\Nsym$, as Exercise~\ref{exe.xi-lifts-p} below
shows). They are called the \dfn{noncommutative power sums of the second
kind} in \cite{GelfandKrobLascouxLeclercRetakhThibon}\footnote{See
Exercise~\ref{exe.Psi} for the ones of the first kind.},
and their products form a basis of $\Nsym$. They are furthermore
useful in studying the so-called \emph{Eulerian idempotent} of a
cocommutative Hopf algebra, as shown in Exercise~\ref{exe.eulerian-idp}
below.
\end{remark}

% [DG][v43] Renamed $H(t)$ as $\widetilde{H}(t)$ to avoid conflict with
% the $H(t)$ over the symmetric functions. Also expanded the footnote.

\begin{exercise}
\label{exe.xi-lifts-p}
Assume that $\QQ$ is a subring of $\kk$.
Define a sequence
of elements $\xi_1,\xi_2,\xi_3,\ldots$ of $\Nsym = \Nsym_\kk$ by
\eqref{relation-definining-pi's}.
\begin{itemize}
\item[(a)] For every $n \geq 1$, show that $\xi_n$ is a primitive
homogeneous element of $\Nsym$ of degree $n$.
\item[(b)] For every $n \geq 1$, show that $\pi\left(n\xi_n\right)$
is the $n$-th power sum symmetric function $p_n \in \Lambda$.
\item[(c)] For every $n \geq 1$, show that
\begin{equation}
\label{eq.exe.xi-lifts-p.c.1}
\xi_n = \sum_{\alpha \in \Comp_n}
 \left(-1\right)^{\ell\left(\alpha\right)-1}
 \frac{1}{\ell\left(\alpha\right)} H_\alpha .
\end{equation}
\item[(d)] For every
composition $\alpha$, define an element $\xi_\alpha$ of $\Nsym$
by $\xi_\alpha = \xi_{\alpha_1} \xi_{\alpha_2} \cdots
\xi_{\alpha_\ell}$, where $\alpha$ is written in the form
$\alpha = \left(\alpha_1, \alpha_2, \ldots, \alpha_\ell\right)$
with $\ell = \ell\left(\alpha\right)$.
Show that
\begin{equation}
\label{eq.exe.xi-lifts-p.d.1}
H_n = \sum_{\alpha \in \Comp_n} \frac{1}{\ell\left(\alpha\right)!} \xi_\alpha
\end{equation}
for every $n \in \NN$.

Use this to prove that
$\left(\xi_\alpha\right)_{\alpha \in \Comp_n}$ is a $\kk$-basis
of $\Nsym_n$ for every $n \in \NN$.
\end{itemize}
\end{exercise}

% [DG][v19] Added parts (c) and (d) to the above exercise.

\begin{exercise}
\label{exe.eulerian-idp}
Assume that $\QQ$ is a subring of $\kk$. 
Let $A$ be a cocommutative connected graded $\kk$-bialgebra.
Let $A = \bigoplus_{n\geq 0} A_n$ be the decomposition of $A$
into homogeneous components.
If $f$ is any $\kk$-linear map $A \to A$ annihilating $A_0$, then $f$
is locally $\star$-nilpotent\footnote{See the proof of
Proposition~\ref{Takeuchi-prop} for what this means.}, and so the
sum $\log^\star \left(f + u\epsilon\right)
:= \sum_{n \geq 1} \left(-1\right)^{n-1} \frac{1}{n} f^{\star n}$
is a well-defined endomorphism of $A$\ \ \ \ \footnote{%
This definition of $\log^{\star}\left(  f+u\epsilon\right)  $ is actually a
particular case of Definition~\ref{def.exp-log.log}. This can be seen as follows:

We have $f\left(  A_{0}\right)  =0$. Thus,
Proposition~\ref{prop.convolution-series}(h) (applied to $C=A$) yields
$f\in\mathfrak{n}\left(  A,A\right)  $ (where $\mathfrak{n}\left(  A,A\right)
$ is defined as in Section~\ref{subsect.leray}), so that $\left(
f+u\epsilon\right)  -u\epsilon=f\in\mathfrak{n}\left(  A,A\right)  $.
Therefore, Definition~\ref{def.exp-log.log} defines a map $\log^{\star}\left(
f+u\epsilon\right)  \in\mathfrak{n}\left(  A,A\right)  $. This map is
identical to the map $\log^{\star}\left(  f+u\epsilon\right)  :=
\sum_{n\geq1}\left(  -1\right)  ^{n-1}\frac{1}{n}f^{\star n}$ we have just
defined, because Proposition~\ref{prop.exp-log}(f) (applied to $C=A$) shows
that the map $\log^{\star}\left(  f+u\epsilon\right)  $ defined using
Definition~\ref{def.exp-log.log} satisfies%
\[
\log^{\star}\left(  f+u\epsilon\right)  =\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}f^{\star n}=\sum_{n\geq1}\left(  -1\right)  ^{n-1}%
\frac{1}{n}f^{\star n}.
\]
}. Let $\mathfrak{e}$ denote the
endomorphism $\log^\star \left(\id_A\right)$ of $A$ (obtained by setting
$f = \id_A - u\epsilon : A \to A$). Show that $\mathfrak{e}$
is a projection from $A$ to the $\kk$-submodule
$\Liep$ of all primitive elements of $A$ (and thus, in particular, is
idempotent).

\textbf{Hint:}
For every $n \geq 0$, let $\pi_n : A \to A$ be the projection
onto the $n$-th homogeneous component $A_n$.
Since $\Nsym$ is the free $\kk$-algebra with generators
$H_1, H_2, H_3, \ldots$, we can define a $\kk$-algebra homomorphism
$\mathfrak{W} : \Nsym \to \left(\End A, \star\right)$ by sending
$H_n$ to $\pi_n$. Show that:
\begin{itemize}
\item[(a)] The map $\mathfrak{e} : A \to A$ is graded.
For every $n \geq 0$, we will denote the map
$\pi_n \circ \mathfrak{e} = \mathfrak{e} \circ \pi_n : A \to A$ by
$\mathfrak{e}_n$.
\item[(b)] We have $\mathfrak{W}\left(\xi_n\right) = \mathfrak{e}_n$
for all $n \geq 1$, where $\xi_n$ is defined as in
Exercise~\ref{exe.xi-lifts-p}.
\item[(c)] If $w$ is an element of $\Nsym$, and if we write
$\Delta\left(w\right) = \sum_{(w)} w_1 \otimes w_2$ using the Sweedler
notation, then $\Delta \circ \left(\mathfrak{W}\left(w\right)\right)
= \left(\sum_{(w)} \mathfrak{W}\left(w_1\right) \otimes
\mathfrak{W}\left(w_2\right) \right) \circ \Delta$.
\item[(d)] We have $\mathfrak{e}_n\left(A\right) \subset \Liep$ for every
$n \geq 0$.
\item[(e)] We have $\mathfrak{e}\left(A\right) \subset \Liep$.
\item[(f)] The map $\mathfrak{e}$ fixes any element of $\Liep$.
\end{itemize}
\end{exercise}

\begin{remark}
The endomorphism $\mathfrak{e}$ of Exercise~\ref{exe.eulerian-idp}
is known as the \dfn{Eulerian idempotent} of $A$, and can be contrasted
with the Dynkin idempotent of Remark~\ref{rmk.dynkin}. It has been
studied in \cite{Patras}, \cite{PatrasReutenauer-hli},
\cite{Burgunder} and \cite{DuchampMinhTolluChienNghia}, and relates to
the Hochschild cohomology of commutative algebras \cite[\S 4.5.2]{Loday}.
\end{remark}

% [DG][v14] Added the two exercises above and this remark. I also
% renamed the $\pi_n$ as $\xi_n$ to avoid confusion with the $\pi$
% that projects $\Nsym$ on $\Lambda$ and with the projections $\pi_m$
% in the Aguiar-Bergeron-Sottile universal property (which also is
% related to this all...).
% 
% Exercise \ref{exe.eulerian-idp} originates in my feeling that the
% argument proving the primitivity of the $\xi_n$ in Remark
% \ref{second-antipode-proof} looked similar to the proof of the
% properties of the Eulerian idempotent in
% \cite{DuchampMinhTolluChienNghia}. I am likely not the first to
% have this feeling (can it be that one of the proofs inspired the
% other?), but it is not clear who should be cited for it.
% 
% The map $\mathfrak{W}$ constructed in the hint to the exercise
% becomes more important if one starts studying the internal product
% on $\Nsym$; it transports knowledge both ways (from $\Nsym$ to
% the endomorphism algebra of a cocommutative connected graded
% bialgebra and back).
% 
% Also, I should really stop using fraktur...

\begin{exercise}
\label{exe.eulerian-idp.2}
Assume that $\QQ$ is a subring of $\kk$. Let $A$, $A_n$ and
$\mathfrak{e}$ be as in Exercise~\ref{exe.eulerian-idp}.

\begin{itemize}
\item[(a)] Show that
$\mathfrak{e}^{\star n} \circ \mathfrak{e}^{\star m}
= n! \delta_{n,m} \mathfrak{e}^{\star n}$
for all $n \in \NN$ and $m \in \NN$.

\item[(b)] Show that
$\mathfrak{e}^{\star n} \circ \id_A^{\star m}
= \id_{A}^{\star m} \circ \mathfrak{e}^{\star n}
= m^n \mathfrak{e}^{\star n}$
for all $n \in \NN$ and $m \in \NN$.
\end{itemize}
\end{exercise}

% [DG][v28] Added exercise above.


We next explore the basis for $\Nsym$ dual to the $\{L_\alpha\}$ in $\Qsym$.
\begin{definition}
Define the
\emph{noncommutative ribbon functions}\index{noncommutative ribbon function}\index{$R_\alpha$}
$\{R_\alpha\}_{\alpha \in \Comp}$ to be the
$\kk$-basis of $\Nsym$ dual to
the fundamental basis $\{L_\alpha\}_{\alpha \in \Comp}$
of $\Qsym$,
so that
\[
(R_\alpha,L_\beta)=\delta_{\alpha,\beta}
\qquad \text{ for all $\alpha, \beta \in \Comp$.}
\]
\end{definition}

\begin{theorem} \phantomsection
\label{Nsym-structure-on-ribbons-theorem}
\begin{itemize}
\item[(a)] One has that
\begin{align}
H_\alpha&=\sum_{\beta \text{ coarsens }\alpha} R_\beta ;
\label{H-as-sum-of-ribbons}\\
R_\alpha&=\sum_{\beta \text{ coarsens }\alpha} (-1)^{\ell(\beta)-\ell(\alpha)} H_\beta .
\label{ribbon-as-sum-of-Hs}
\end{align}
\item[(b)] The surjection $\Nsym \overset{\pi}{\longrightarrow} \Lambda$
sends $R_\alpha \longmapsto s_{\Rib\left(\alpha\right)}$,
the skew Schur function associated
to the ribbon $\Rib\left(\alpha\right)$.
\item[(c)] Furthermore,
\begin{align}
R_\alpha R_\beta &= R_{\alpha \cdot \beta} + R_{\alpha \odot \beta} \qquad \text{ if }\alpha\text{ and }\beta\text{ are nonempty} ;
\label{Nsym-product-in-ribbons}\\
S(R_\alpha) &= (-1)^{|\alpha|} R_{\omega(\alpha)} .
\label{Nsym-antipode-in-ribbons}
\end{align}
Finally, $R_\varnothing$ is the multiplicative identity of $\Nsym$.
\end{itemize}
\end{theorem}

\begin{proof}
(a) For \eqref{H-as-sum-of-ribbons}, note that
\[
H_\alpha
= \sum_{\beta} (H_\alpha,L_\beta) R_\beta
= \sum_{\beta}
 \left( H_\alpha, \sum\limits_{\substack{\gamma:\\\gamma \text{ refines }\beta}} M_\gamma \right)
    R_\beta
=  \sum\limits_{\substack{\beta:\\ \beta \text{ coarsens }\alpha}}     R_\beta.
\]
The equality \eqref{ribbon-as-sum-of-Hs} follows from
\eqref{H-as-sum-of-ribbons}
by Lemma~\ref{lem.moebius-bool.comps}(a).

(b) Write $\alpha$ as $\left(\alpha_1,\ldots,\alpha_\ell\right)$.
To show that $\pi(R_\alpha)=s_{\Rib\left(\alpha\right)}$,
we instead examine $\pi(H_\alpha)$:
\[
\pi(H_\alpha)
=\pi(H_{\alpha_1} \cdots H_{\alpha_\ell})
=h_{\alpha_1} \cdots h_{\alpha_\ell}
= s_{(\alpha_1)} \cdots s_{(\alpha_\ell)}
= s_{(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)}
\]
where $(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)$ is some
skew shape which is a horizontal strip having rows of lengths $\alpha_1,\ldots,\alpha_\ell$
from bottom to top.  We claim
\[
s_{(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)}
=\sum\limits_{\substack{\beta:\\\beta\text{ coarsens }\alpha}}
s_{\Rib\left(\beta\right)} ,
\]
because column-strict tableaux $T$ of shape
$(\alpha_1) \oplus \cdots \oplus (\alpha_\ell)$ biject to column-strict
tableaux $T'$ of some ribbon $\Rib\left(\beta\right)$
with $\beta$ coarsening $\alpha$, as follows:
Let $a_i,b_i$ denote the leftmost, rightmost entries of the $i$-th row from the
bottom in $T$, of length $\alpha_i$, and
\begin{enumerate}
\item[$\bullet$] if $b_i \leq a_{i+1}$,
merge parts $\alpha_i,\alpha_{i+1}$ in $\beta$,
and concatenate the rows of length $\alpha_i,\alpha_{i+1}$ in $T'$, or
\item[$\bullet$] if $b_i > a_{i+1}$,
do not merge parts $\alpha_i,\alpha_{i+1}$ in $\beta$,
and let these two rows overlap in one column in $T'$.
\end{enumerate}
E.g., if $\alpha=(3,3,2,3,2)$, then
\begin{align*}
\text{the tableau } &
T =
\begin{matrix}
 & & & & & & & & & & &3&4 \\
 & & & & & & & &4&4&5& &\\
 & & & & & &4&4& & & & &\\
 & & &2&2&3& & & & & & &\\
1&1&3& & & & & & & & & &
\end{matrix}
\text{ of shape }
(\alpha_1) \oplus \cdots \oplus (\alpha_\ell) \\
\\
\text{maps to the tableau } &
T' =
\begin{matrix}
 & & & & & & & & &3&4 \\
 & &2&2&3&4&4&4&4&5& &\\
1&1&3& & & & & & & & & &
\end{matrix}
\text{ of shape }
\Rib\left(\beta\right)
\text{ for } \beta=(3,8,2) .
\end{align*}
% [DG][v80] Reorganized the above example.
The reverse bijection breaks the rows of $T'$ into the rows of $T$ of lengths dictated by
the parts of $\alpha$.  Having shown
$\pi(H_\alpha) =\sum_{\beta:\\\beta\text{ coarsens }\alpha} s_{\Rib\left(\beta\right)}$,
we can now apply Lemma~\ref{lem.moebius-bool.comps}(a) to obtain
\[
s_{\Rib\left(\alpha\right)}
= \sum_{\beta:\\\beta\text{ coarsens }\alpha} \left(-1\right)^{\ell\left(\alpha\right)-\ell\left(\beta\right)} \pi\left(H_\beta\right)
= \pi\left(R_\alpha\right)
\qquad \left(\text{by \eqref{ribbon-as-sum-of-Hs}}\right) ;
\]
thus, $\pi(R_\alpha) = s_{\Rib\left(\alpha\right)}$ is proven.

(c) Finally, \eqref{Nsym-product-in-ribbons} and \eqref{Nsym-antipode-in-ribbons}
follow from \eqref{Qsym-coproduct-on-fundamentals} and
\eqref{Qsym-antipode-on-fundamentals} by duality.
\end{proof}

\begin{remark}
\label{antipode-comparison-remark}
Since the maps
\[
\xymatrix{
\Nsym \arsurj[dr]_\pi & & \Qsym \\
      &\Lambda \arinjrev[ur]_i&
}
\]
are Hopf morphisms, they must respect the antipodes $S_{\Lambda}, S_{\Qsym}, S_{\Nsym}$,
but it is interesting to compare them explicitly using the fundamental basis for
$\Qsym$ and the ribbon basis for $\Nsym$.

On one hand \eqref{Qsym-antipode-on-fundamentals}
shows that $S_{\Qsym}(L_\alpha)= (-1)^{|\alpha|} L_{\omega(\alpha)}$
extends the map $S_{\Lambda}$ since $L_{(1^n)} = e_n$ and $L_{(n)}=h_n$, as observed in
Example~\ref{the-two-extreme-fundamentals-example}, and $\omega((n))=(1^n)$.

On the other hand, \eqref{Nsym-antipode-in-ribbons} shows that
$S_{\Nsym}(R_\alpha)= (-1)^{|\alpha|} R_{\omega(\alpha)}$  lifts the map $S_{\Lambda}$ to $S_{\Nsym}$:
Theorem~\ref{Nsym-structure-on-ribbons-theorem}(b) showed that
$R_\alpha$ lifts the skew Schur function $s_{\Rib\left(\alpha\right)}$,
while \eqref{skew-Schur-antipodes} asserted that
$S(s_{\lambda/\mu}) = (-1)^{|\lambda/\mu|} s_{\lambda^t/\mu^t}$,
and a ribbon $\Rib\left(\alpha\right)=\lambda/\mu$ has
$\Rib\left(\omega(\alpha)\right)=\lambda^t/\mu^t$.
\end{remark}

\begin{exercise} \phantomsection
\label{exe.Psi}
\begin{itemize}
\item[(a)] Show that any integers $n$ and $i$ with $0\leq
i<n$ satisfy
\[
R_{\left(  1^{i},n-i\right)  }=\sum_{j=0}^{i}\left(  -1\right)  ^{i-j}
R_{\left(  1^{j}\right)  }H_{n-j}.
\]
(Here, as usual, $1^i$ stands for the number $1$ repeated $i$ times.)

\item[(b)] Show that any integers $n$ and $i$ with $0\leq i<n$ satisfy
\[
\left(  -1\right)  ^{i}R_{\left(  1^{i},n-i\right)  }=\sum_{j=0}^{i}S\left(
H_{j}\right)  H_{n-j}.
\]

\item[(c)] For every positive integer $n$, define an element $\Psi_{n}$ of
$\Nsym$ by
\[
\Psi_{n}=\sum_{i=0}^{n-1}\left(  -1\right)  ^{i}R_{\left(  1^{i},n-i\right)
}.
\]
Show that $\Psi_{n}=\left(  S\star E\right)  \left(  H_{n}\right)  $, where
the map $E : \Nsym \to \Nsym$ is defined as in
Exercise~\ref{exe.dynkin} (for $A=\Nsym$).
Conclude that $\Psi_{n}$ is primitive.

\item[(d)] Prove that
\[
\sum_{k=0}^{n-1}H_{k}\Psi_{n-k} = nH_{n}
\]
for every $n\in\NN$.

\item[(e)] Define two power series $\psi\left(  t\right)  $ and
$\widetilde{H}\left(  t\right)  $ in $\Nsym\left[  \left[
t\right]  \right]  $ by
\begin{align*}
\psi\left(  t\right)   &  =\sum_{n\geq 1}\Psi_{n}t^{n-1};\\
\widetilde{H}\left(  t\right)   &  =\sum_{n\geq0}H_{n}t^{n}.
\end{align*}
Show that\footnote{The derivative $\frac{d}{dt}Q\left(t\right)$ of
a power series $Q\left(t\right) \in R\left[\left[t\right]\right]$
over a noncommutative ring $R$ is defined just as in the case of
$R$ commutative: by setting
$\frac{d}{dt}Q\left(t\right) = \sum_{i \geq 1} iq_i t^{i-1}$,
where $Q\left(t\right)$ is written in the form
$Q\left(t\right) = \sum_{i \geq 0} q_i t^i$.}
$\dfrac{d}{dt}\widetilde{H}\left(  t\right)  =\widetilde{H}\left(
t\right)  \cdot\psi\left(  t\right)  $.

(The functions $\Psi_n$ are called \dfn{noncommutative power sums of the
first kind}; they are studied in
\cite{GelfandKrobLascouxLeclercRetakhThibon}. The power sums of the second
kind are the $n\xi_n$ in Remark~\ref{second-antipode-proof}.)

\item[(f)] Show that $\pi\left(  \Psi_n \right)  $ equals the power sum
symmetric function $p_n$ for every positive integer $n$.

\item[(g)] Show that every positive integer $n$ satisfies
\[
p_n = \sum\limits_{i=0}^{n-1} \left(-1\right)^i s_{\left(n-i, 1^i\right)}
  \qquad \text{ in }\Lambda .
\]

\item[(h)] For every nonempty
composition $\alpha$, define a positive integer
$\operatorname{lp}\left(\alpha\right)$ by
$\operatorname{lp}\left(\alpha\right) = \alpha_\ell$,
where $\alpha$ is written in the form
$\alpha = \left(\alpha_1, \alpha_2, \ldots, \alpha_\ell\right)$
with $\ell = \ell\left(\alpha\right)$.
(Thus, $\operatorname{lp}\left(\alpha\right)$ is the last
part of $\alpha$.) Show that every positive integer $n$
satisfies
\begin{equation}
\label{eq.exe.Psi.h.0}
\Psi_n = \sum_{\alpha \in \Comp_n} \left(-1\right)^{\ell\left(\alpha\right)-1} \operatorname{lp}\left(\alpha\right) H_\alpha .
\end{equation}

\item[(i)] Assume that $\QQ$ is a subring of $\kk$. For every
composition $\alpha$, define an element $\Psi_\alpha$ of $\Nsym$
by $\Psi_\alpha = \Psi_{\alpha_1} \Psi_{\alpha_2} \cdots
\Psi_{\alpha_\ell}$, where $\alpha$ is written in the form
$\alpha = \left(\alpha_1, \alpha_2, \ldots, \alpha_\ell\right)$
with $\ell = \ell\left(\alpha\right)$.
For every composition
$\alpha$, define $\pi_u\left(\alpha\right)$ to be the positive
integer
$\alpha_1 \left(\alpha_1 + \alpha_2\right) \cdots \left(\alpha_1 + \alpha_2 + \cdots + \alpha_\ell\right)$,
where $\alpha$ is written in the form
$\alpha = \left(\alpha_1, \alpha_2, \ldots, \alpha_\ell\right)$
with $\ell = \ell\left(\alpha\right)$.
Show that
\begin{equation}
\label{eq.exe.Psi.h.1}
H_n = \sum_{\alpha \in \Comp_n} \frac{1}{\pi_u\left(\alpha\right)} \Psi_\alpha 
\end{equation}
for every $n \in \NN$.

Use this to prove that
$\left(\Psi_\alpha\right)_{\alpha \in \Comp_n}$ is a $\kk$-basis
of $\Nsym_n$ for every $n \in \NN$.

\item[(j)] Assume that $\QQ$ is a subring of $\kk$.
Let $V$ be the
free $\kk$-module with basis
$\left(\mathfrak{b}_n\right)_{n \in \left\{1, 2, 3, \ldots\right\}}$.
Define a $\kk$-module homomorphism
$f : V \to \Nsym$ by requiring that
$f\left(\mathfrak{b}_n\right) = \Psi_n$ for every
$n \in \left\{1, 2, 3, \ldots\right\}$.
Let $F$ be the $\kk$-algebra homomorphism
$T\left(V\right) \to \Nsym$ induced by this $f$ (using the
universal property of the tensor algebra $T\left(V\right)$).
Show that $F$ is a Hopf algebra isomorphism (where the
Hopf algebra structure on $T\left(V\right)$ is as in
Example~\ref{exa.tensor-alg.antipode}).

\item[(k)] Assume that $\QQ$ is a subring of $\kk$. Let $V$ be as
in Exercise~\ref{exe.Psi}(j). Show that $\Qsym$ is isomorphic to
the shuffle algebra $\operatorname{Sh}\left(V\right)$ (defined as
in Proposition~\ref{prop.shuffle-alg}) as Hopf algebras.

\item[(l)] Solve parts (a) and (b) of Exercise~\ref{exe.hookschur}
again using the ribbon basis functions $R_\alpha$.
\end{itemize}
\end{exercise}

% [DG][v7] Added exercise above.

% [DG][v14] Added part (g), which is an application of $\Nsym$
% to $\Lambda$. Of course, it is not very exciting if one knows
% Murnaghan-Nakayama...

% [DG][v19] Added parts (h) and (i). Copypasted parts of the solution
% from the previous exercise because I was tired.

% [DG][v30] Added part (j).

% [DG][v31] Added parts (j) and (k), moving the old (j) to place (l).

One might wonder whether the Frobenius endomorphisms of $\Lambda$ (defined in
Exercise~\ref{exe.witt.ghost-app}) and the Verschiebung endomorphisms of
$\Lambda$ (defined in Exercise~\ref{exe.Lambda.verschiebung}) generalize to
analogous operators on either $\Qsym$ or $\Nsym$. The next two
exercises (whose claims mostly come from \cite[\S 13]{Hazewinkel2}) answer
this question: The Frobenius endomorphisms extend to $\Qsym$, and
the Verschiebung ones lift to $\Nsym$.

\begin{exercise}
\label{exe.QSym.frobenius}
For every $n \in \left\{ 1, 2, 3, \ldots \right\}$,
define a map $\mathbf{F}_n : \Qsym \to \Qsym$ by setting
\[
\mathbf{F}_n \left( a \right)
= a \left( x_1^n, x_2^n, x_3^n, \ldots \right)
\qquad\qquad\text{for every }a \in \Qsym .
\]
(So what $\mathbf{F}_n$ does to a quasi-symmetric function is replacing all
variables $x_1, x_2, x_3, \ldots$ by their $n$-th powers.)

\begin{enumerate}
\item[(a)] Show that $\mathbf{F}_n : \Qsym \to \Qsym$ is a $\kk$-algebra
homomorphism for every $n \in \left\{ 1, 2, 3, \ldots \right\}$.

\item[(b)] Show that $\mathbf{F}_n \circ \mathbf{F}_m = \mathbf{F}_{nm}$
for any two positive integers $n$ and $m$.

\item[(c)] Show that $\mathbf{F}_1 = \id$.

\item[(d)] Prove that
$\mathbf{F}_n
\left( M_{\left( \beta_1, \beta_2, \ldots, \beta_s \right)} \right)
= M_{\left( n\beta_1, n\beta_2, \ldots, n\beta_s \right)}$
for every $n \in \left\{ 1, 2, 3, \ldots \right\}$
and $\left( \beta_1, \beta_2, \ldots, \beta_s \right) \in \Comp$.

\item[(e)] Prove that $\mathbf{F}_n : \Qsym \to \Qsym$ is a Hopf
algebra homomorphism for every
$n \in \left\{ 1, 2, 3, \ldots \right\}$.

\item[(f)] Consider the maps $\mathbf{f}_n : \Lambda \to \Lambda$
defined in Exercise~\ref{exe.witt.ghost-app}. Show that
$\mathbf{F}_n \mid_{\Lambda} = \mathbf{f}_n$ for every
$n \in \left\{ 1, 2, 3, \ldots \right\}$.

\item[(g)] Assume that $\kk = \ZZ$. Prove that $\mathbf{f}_p
\left( a \right) \equiv a^p \operatorname{mod} p\Qsym$ for
every $a \in \Qsym$ and every prime number $p$.

\item[(h)] Give a new solution to
Exercise~\ref{exe.witt.ghost-app}(d).
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exe.NSym.verschiebung}
For every $n \in \left\{ 1, 2, 3, \ldots \right\}$,
define a $\kk$-algebra homomorphism
$\mathbf{V}_n : \Nsym \rightarrow \Nsym$ by
\[
\mathbf{V}_n \left( H_m \right)
= \begin{cases}
H_{m/n},& \text{if } n\mid m;\\
0,& \text{if } n\nmid m
\end{cases}
\qquad\qquad\text{for every positive integer }m
\]
\footnote{This is well-defined, since $\Nsym$ is (isomorphic
to) the free associative algebra with generators $H_1, H_2, H_3, \ldots$
(according to \eqref{eq.thm.NSym.free}).}.

\begin{enumerate}
\item[(a)] Show that any positive integers $n$ and $m$ satisfy
\[
\mathbf{V}_n \left( \Psi_m \right)
= \begin{cases}
n\Psi_{m/n},& \text{if } n\mid m;\\
0,& \text{if } n\nmid m
\end{cases} \quad ,
\]
where the elements $\Psi_m$ and $\Psi_{m/n}$ of $\Nsym$ are
as defined in Exercise~\ref{exe.Psi}(c).

\item[(b)] Show that if $\QQ$ is a subring of $\kk$, then any
positive integers $n$ and $m$ satisfy
\[
\mathbf{V}_n \left( \xi_m \right)
= \begin{cases}
\xi_{m/n},& \text{if } n\mid m;\\
0,& \text{if } n\nmid m
\end{cases} \quad ,
\]
where the elements $\xi_m$ and $\xi_{m/n}$ of $\Nsym$ are as
defined in Exercise~\ref{exe.xi-lifts-p}.

\item[(c)] Prove that
$\mathbf{V}_n \circ \mathbf{V}_m = \mathbf{V}_{nm}$ for
any two positive integers $n$ and $m$.

\item[(d)] Prove that $\mathbf{V}_1 = \id$.

\item[(e)] Prove that $\mathbf{V}_n : \Nsym \rightarrow
 \Nsym$ is a Hopf algebra homomorphism for every
$n \in \left\{ 1, 2, 3, \ldots \right\}$.
\end{enumerate}

Now, consider also the maps $\mathbf{F}_n : \Qsym
\rightarrow \Qsym$ defined in Exercise~\ref{exe.witt.ghost-app}.
Fix a positive integer $n$.

\begin{enumerate}
\item[(f)] Prove that the maps $\mathbf{F}_n : \Qsym
\rightarrow \Qsym$ and $\mathbf{V}_n : \Nsym
\rightarrow \Nsym$ are adjoint with respect to the dual pairing
$\Nsym \otimes \Qsym
\overset{\left(  \cdot , \cdot\right)  }{\longrightarrow} \kk$.

\item[(g)] Consider the maps
$\mathbf{v}_n : \Lambda \rightarrow \Lambda$
defined in Exercise~\ref{exe.Lambda.verschiebung}. Show that
the surjection $\pi : \Nsym \rightarrow \Lambda$ satisfies
$\mathbf{v}_n \circ \pi = \pi \circ \mathbf{V}_n$ for every
$n \in \left\{ 1, 2, 3, \ldots \right\}$.

\item[(h)] Give a new solution to
Exercise~\ref{exe.Lambda.verschiebung}(f).
\end{enumerate}
\end{exercise}

% [DG][v43] Added the above two exercises.









\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Polynomial generators for $\Qsym$ and Lyndon words}
\label{sect.QSym.lyndon}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


In this chapter, we shall construct an algebraically independent
generating set for $\Qsym$ as a $\kk$-algebra, thus showing that
$\Qsym$ is a polynomial ring over $\kk$. This has been
done by Malvenuto \cite[Cor. 4.19]{Malvenuto} when $\kk$
is a field of characteristic $0$, and by Hazewinkel
\cite{Hazewinkel} in the general case. We will begin by
introducing the notion of \emph{Lyndon words}
(Section~\ref{subsect.lyndon}), on which both of these
constructions rely; we will then
(Section~\ref{subsect.lyndon.shuffles}) elucidate the connection
of Lyndon words with shuffles, and afterwards
(Section~\ref{subsect.shuffle.radford}) apply it to prove
\emph{Radford's theorem} stating that the shuffle algebra of
a free $\kk$-module over a commutative $\QQ$-algebra is a
polynomial algebra (Theorem~\ref{thm.shuffle.radford}). The
shuffle algebra is not yet $\Qsym$, but Radford's theorem on the
shuffle algebra serves
as a natural stepping stone for the study of the more complicated
algebra $\Qsym$. We will prove -- in two ways -- that $\Qsym$
is a polynomial algebra when $\QQ$ is a subring of $\kk$ in
Section~\ref{subsect.shuffle.QSym.1}, and then we will finally
prove the general case in Section~\ref{subsect.shuffle.QSym.2}.
In Section~\ref{subsect.lyndon.gr}, we will explore a different
aspect of the combinatorics of words: the notion of necklaces
(which are in bijection with Lyndon words, as
Exercise~\ref{exe.words.necklaces} will show) and the
\emph{Gessel-Reutenauer bijection}, which help us define and
understand the \emph{Gessel-Reutenauer symmetric functions}.
This will rely on Section~\ref{subsect.lyndon}, but not on any
of the other sections of Chapter~\ref{sect.QSym.lyndon}.

Strictly speaking, this whole Chapter~\ref{sect.QSym.lyndon}
is a digression, as it involves almost no coalgebraic or
Hopf-algebraic structures, and its results will not be used in
further chapters (which means it can be skipped if so desired).
However, it sheds additional light on both quasisymmetric and
symmetric functions,
and serves as an excuse to study Lyndon words, which are a
combinatorial object of independent interest (and are involved
in the study of free algebras and Hopf algebras, apart from
$\Qsym$ -- see \cite{Radford-shuffle} and
\cite{Reutenauer}\footnote{They also are involved in indexing
basis elements of combinatorial Hopf algebras other than
$\Qsym$. See Bergeron/Zabrocki \cite{BergeronZabrocki}.}).

We will take a scenic route to the proof of Hazewinkel's
theorem. A reader only interested in the proof proper can
restrict themselves to reading only the following:

\begin{itemize}

\item from Section~\ref{subsect.lyndon}, everything up to
Corollary~\ref{cor.words.uv=vu.3}, then from
Definition~\ref{def.words.lyndon} up to 
Proposition~\ref{prop.words.lyndon.preorder}, then from
Definition~\ref{def.words.CFL} up to
Lemma~\ref{lem.words.CFLlemma}, and finally
Theorem~\ref{thm.words.lyndon.std}.
(Proposition~\ref{prop.words.lyndon.cutoff} and
Theorem~\ref{thm.words.lyndon.equiv} are also relevant
if one wants to use a different definition of Lyndon words,
as they prove the equivalence of most such definitions.)

\item from Section~\ref{subsect.lyndon.shuffles},
everything except for
Exercise~\ref{exe.shuffle.lyndon.cflcrit}.

\item from Section~\ref{subsect.shuffle.radford},
Definition~\ref{def.shuffle.polyalg},
Lemma~\ref{lem.lyndonbases}, and
Lemma~\ref{lem.shuffle.preserves-lex}.

\item from Section~\ref{subsect.shuffle.QSym.1},
Definition~\ref{def.lyndon-composition},
Theorem~\ref{thm.QSym.lyndon}, then from
Proposition~\ref{prop.QSym.lyndon.MaMb} up to
Definition~\ref{def.wll}, and
Lemma~\ref{lem.QSym.lyndon.triang.wll-lemma}.

\item all of Section~\ref{subsect.shuffle.QSym.2}.

\end{itemize}

\noindent Likewise, Section~\ref{subsect.lyndon.gr} can be
read immediately after Section~\ref{subsect.lyndon}.

\subsection{\label{subsect.lyndon}Lyndon words}

Lyndon words have been independently defined by Shirshov \cite{Shirshov},
Lyndon \cite{Lyndon}, Radford \cite[\S 2]{Radford-shuffle} and de
Bruijn/Klarner \cite{deBruijnKlarner} (though using different and sometimes
incompatible notations). They have since been surfacing in various places in
noncommutative algebra (particularly the study of free Lie algebras);
expositions of their theory can be found in \cite[\S 5]{Lothaire},
\cite[\S 5.1]{Reutenauer} and \cite[\S 1]{Laue} (in German).
We will follow our own approach to the properties
of Lyndon words that we need.

\begin{definition}
\label{def.words}We fix a totally ordered set $\mathfrak{A}$, which we call
the \dfn{alphabet}. Throughout Section~\ref{subsect.lyndon} and
Section~\ref{subsect.lyndon.shuffles}, we will understand
``word'' to mean a word over $\mathfrak{A}$.

We recall that a \dfn{word} is just a (finite) tuple of elements of
$\mathfrak{A}$. In other words, a word is an element of the set $\bigsqcup
_{n\geq0}\mathfrak{A}^{n}$. We denote this set by
\dfn{$\mathfrak{A}^{\ast}$}.

The \dfn{empty word} is the unique tuple with $0$ elements. It is denoted
by $\varnothing$. If $w\in\mathfrak{A}^{n}$ is a word and $i\in\left\{
1,2,\ldots,n\right\}  $, then the \dfn{$i$-th letter} of $w$ means the
$i$-th entry of the $n$-tuple $w$. This $i$-th letter will be denoted by
$w_{i}$.

The \emph{length}\index{length of a word}
$\ell\left(  w\right)  $ of a word $w\in\bigsqcup_{n\geq
0}\mathfrak{A}^{n}$ is defined to be the $n\in \NN$ satisfying
$w\in\mathfrak{A}^{n}$. Thus, $w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(
w\right)  }\right)  $ for every word $w$.

Given two words $u$ and $v$, we say that $u$ is \dfn{longer} than $v$
(or, equivalently, $v$ is \dfn{shorter} than $u$) if and only if
$\ell\left(u\right) > \ell\left(v\right)$.

The \dfn{concatenation} of two words $u$ and $v$ is defined to be the word
$\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right)  $. This concatenation is denoted by
$uv$ or $u\cdot v$. The set $\mathfrak{A}^{\ast}$ of all words is a monoid
with respect to concatenation, with neutral element $\varnothing$. It is
precisely the free monoid on generators $\mathfrak{A}$. If $u$ is a word and
$i\in \NN$, we will understand $u^{i}$ to mean the $i$-th power of $u$
in this monoid (that is, the word $\underbrace{uu \cdots u}_{i\text{ times}}$).

The elements of $\mathfrak{A}$ are called \emph{letters}\index{letter},
and will be
identified with elements of $\mathfrak{A}^{1}\subset\bigsqcup_{n\geq
0}\mathfrak{A}^{n}=\mathfrak{A}^{\ast}$. This identification equates every
letter $u\in\mathfrak{A}$ with the one-letter word $\left(  u\right)
\in\mathfrak{A}^{1}$. Thus, every word $\left(  u_{1},u_{2},\ldots
,u_{n}\right)  \in\mathfrak{A}^{\ast}$ equals the concatenation $u_{1}u_{2}
\cdots u_{n}$ of letters, hence allowing us to use $u_{1}u_{2} \cdots u_{n}$
as a brief notation for the word $\left(  u_{1},u_{2},\ldots,u_{n}\right)  $.

If $w$ is a word, then:

\begin{itemize}
\item a \dfn{prefix} of $w$ means a word of the form $\left(  w_{1}%
,w_{2},\ldots,w_{i}\right)  $ for some $i\in\left\{  0,1,\ldots,\ell\left(
w\right)  \right\}  $;

\item a \dfn{suffix} of $w$ means a word of the form $\left(
w_{i+1},w_{i+2},\ldots,w_{\ell\left(  w\right)  }\right)  $ for some
$i\in\left\{  0,1,\ldots,\ell\left(  w\right)  \right\}  $;

\item a \dfn{proper suffix} of $w$ means a word of the form $\left(
w_{i+1},w_{i+2},\ldots,w_{\ell\left(  w\right)  }\right)  $ for some
$i\in\left\{  1,2,\ldots,\ell\left(  w\right)  \right\}  $.
\end{itemize}

\noindent In other words,

\begin{itemize}
\item a \dfn{prefix} of $w\in\mathfrak{A}^{\ast}$ is a word $u\in
\mathfrak{A}^{\ast}$ such that there exists a $v\in\mathfrak{A}^{\ast}$
satisfying $w=uv$;

\item a \dfn{suffix} of $w\in\mathfrak{A}^{\ast}$ is a word $v\in
\mathfrak{A}^{\ast}$ such that there exists a $u\in\mathfrak{A}^{\ast}$
satisfying $w=uv$;

\item a \dfn{proper suffix} of $w\in\mathfrak{A}^{\ast}$ is a word
$v\in\mathfrak{A}^{\ast}$ such that there exists a nonempty $u\in
\mathfrak{A}^{\ast}$ satisfying $w=uv$.
\end{itemize}

Clearly, any proper suffix of $w\in\mathfrak{A}^{\ast}$ is a suffix
of $w$. Moreover, if $w \in \mathfrak{A}^{\ast}$ is any word, then
a proper suffix of $w$ is the same thing as a suffix of $w$ distinct
from $w$.

We define a relation $\leq$ on the set $\mathfrak{A}^{\ast}$ as follows: For
two words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$, we set
$u\leq v$ to hold if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
This order relation (taken as the smaller-or-equal relation) makes
$\mathfrak{A}^{\ast}$ into a poset (by Proposition \ref{prop.words.lex}(a)
below), and we will always be regarding $\mathfrak{A}^{\ast}$ as endowed with
this poset structure (thus, notations such as $<$, $\leq$, $>$ and $\geq$ will
be referring to this poset structure). This poset is actually totally ordered
(see Proposition \ref{prop.words.lex}(a)).
\end{definition}

Here are some examples of words compared by the relation $\leq$:%
\begin{align*}
113  &  \leq114,\ \ \ \ \ \ \ \ \ \ 113\leq132,\ \ \ \ \ \ \ \ \ \ 19\leq
195,\ \ \ \ \ \ \ \ \ \ 41\leq412,\\
41  &  \leq421,\ \ \ \ \ \ \ \ \ \ 539\leq54,\ \ \ \ \ \ \ \ \ \ \varnothing
\leq21,\ \ \ \ \ \ \ \ \ \ \varnothing\leq\varnothing
\end{align*}
(where $\mathfrak{A}$ is the alphabet $\left\{  1<2<3<\cdots\right\}  $).

Notice that if $u$ and $v$ are two words of the same length (i.e., we have
$u,v\in\mathfrak{A}^{n}$ for one and the same $n$), then $u\leq v$ holds if
and only if $u$ is lexicographically smaller-or-equal to $v$. In other words,
the relation $\leq$ is an extension of the lexicographic order on every
$\mathfrak{A}^{n}$ to $\mathfrak{A}^{\ast}$. This is the reason why this
relation $\leq$ is usually called the
\emph{lexicographic order}\index{lexicographic order on words}
on $\mathfrak{A}^{\ast}$.
In particular, we will be using this name.\footnote{The
relation $\leq$ is also known as the \dfn{dictionary order}, due to the
fact that it is the order in which words appear in a dictionary.} However,
unlike the lexicographic order on $\mathfrak{A}^{n}$, it does not always
respect concatenation from the right: It can happen that $u,v,w\in
\mathfrak{A}^{\ast}$ satisfy $u\leq v$ but not $uw\leq vw$. (For example,
$u=1$, $v=13$ and $w=4$, again with $\mathfrak{A}=\left\{  1<2<3<\cdots
\right\}  $.) We will see in Proposition \ref{prop.words.lex} that this is
rather an exception than the rule and the relation $\leq$ still behaves mostly
predictably with respect to concatenation.

Some basic properties of the order relation $\leq$ just defined are collected
in the following proposition:

\begin{proposition}
\phantomsection
\label{prop.words.lex}

\begin{itemize}
\item[(a)] The order relation $\leq$ is (the smaller-or-equal relation of) a
total order on the set $\mathfrak{A}^{\ast}$.

\item[(b)] If $a,c,d\in\mathfrak{A}^{\ast}$ satisfy $c\leq d$, then $ac\leq
ad$.

\item[(c)] If $a,c,d\in\mathfrak{A}^{\ast}$ satisfy $ac\leq ad$, then $c\leq
d$.

\item[(d)] If $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $a\leq c$, then either
we have $ab\leq cd$ or the word $a$ is a prefix of $c$.

\item[(e)] If $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $ab\leq cd$, then either
we have $a\leq c$ or the word $c$ is a prefix of $a$.

\item[(f)] If $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $ab\leq cd$ and
$\ell\left(  a\right)  \leq\ell\left(  c\right)  $, then $a\leq c$.

\item[(g)] If $a,b,c\in\mathfrak{A}^{\ast}$ satisfy $a\leq b\leq ac$, then $a$
is a prefix of $b$.

\item[(h)] If $a\in\mathfrak{A}^{\ast}$ is a prefix of $b\in\mathfrak{A}%
^{\ast}$, then $a\leq b$.

\item[(i)] If $a$ and $b$ are two prefixes of $c\in\mathfrak{A}^{\ast}$, then
either $a$ is a prefix of $b$, or $b$ is a prefix of $a$.

\item[(j)] If $a,b,c\in\mathfrak{A}^{\ast}$ are such that $a\leq b$ and
$\ell\left(  a\right)  \geq\ell\left(  b\right)  $, then $ac\leq bc$.

\item[(k)] If $a\in\mathfrak{A}^{\ast}$ and $b\in\mathfrak{A}^{\ast}$ are such
that $b$ is nonempty, then $a<ab$.
\end{itemize}
\end{proposition}

\begin{exercise}
\label{exe.words.lex}Prove Proposition \ref{prop.words.lex}.

[\textbf{Hint:} No part of Proposition \ref{prop.words.lex} requires more than
straightforward case analysis. However, the proof of (a) can be simplified by
identifying the order relation $\leq$ on $\mathfrak{A}^{\ast}$ as a
restriction of the lexicographic order on the set $\mathfrak{B}^{\infty}$,
where $\mathfrak{B}$ is a suitable extension of the alphabet $\mathfrak{A}$.
What is this extension, and how to embed $\mathfrak{A}^{\ast}$ into
$\mathfrak{B}^{\infty}$ ?]
\end{exercise}

Proposition \ref{prop.words.lex} provides a set of tools for working with the
lexicographic order without having to refer to its definition; we shall use it
extensively. Proposition \ref{prop.words.lex}(h) (and its equivalent form
stating that $a\leq ac$ for every $a\in\mathfrak{A}^{\ast}$ and $c\in
\mathfrak{A}^{\ast}$) and Proposition \ref{prop.words.lex}(k) will often be
used without explicit mention.

Before we define Lyndon words, let us show two more facts about words which
will be used later. First, when do words commute?

\begin{proposition}
\label{prop.words.uv=vu.2}Let $u,v\in\mathfrak{A}^{\ast}$ satisfy $uv=vu$.
Then, there exist a $t\in\mathfrak{A}^{\ast}$ and two nonnegative integers $n
$ and $m$ such that $u=t^{n}$ and $v=t^{m}$.
\end{proposition}

\begin{proof}
We prove this by strong induction on $\ell\left(  u\right)  +\ell\left(
v\right)  $. We assume WLOG that $\ell\left(  u\right)  $ and $\ell\left(
v\right)  $ are positive (because otherwise, one of $u$ and $v$ is the empty
word, and everything is trivial). It is easy to see that either $u$ is a
prefix of $v$, or $v$ is a prefix of $u$\ \ \ \ \footnote{\textit{Proof.} The
word $u$ is a prefix of $uv$. But the word $v$ is also a prefix of $uv$ (since
$uv=vu$). Hence, Proposition \ref{prop.words.lex}(i) (applied to $a=u$, $b=v$
and $c=uv$) yields that either $u$ is a prefix of $v$, or $v$ is a prefix of
$u$, qed.}. We assume WLOG that $u$ is a prefix of $v$ (since our situation is
symmetric). Thus, we can write $v$ in the form $v=uw$ for some $w\in
\mathfrak{A}^{\ast}$. Consider this $w$. Clearly, $\ell\left(  u\right)
+\ell\left(  w\right)  =\ell\left(  \underbrace{uw}_{=v}\right)  =\ell\left(
v\right)  <\ell\left(  u\right)  +\ell\left(  v\right)  $ (since $\ell\left(
v\right)  $ is positive). Since $v=uw$, the equality $uv=vu$ becomes
$uuw=uwu$. Cancelling $u$ from this equality, we obtain $uw=wu$. Now, we can
apply Proposition \ref{prop.words.uv=vu.2} to $w$ instead of $v$ (by the
induction assumption, since $\ell\left(  u\right)  +\ell\left(  w\right)
<\ell\left(  u\right)  +\ell\left(  v\right)  $), and obtain that there exist
a $t\in\mathfrak{A}^{\ast}$ and two nonnegative integers $n$ and $m$ such that
$u=t^{n}$ and $w=t^{m}$. Consider this $t$ and these $n$ and $m$. Of course,
$u=t^{n}$ and $v=\underbrace{u}_{=t^{n}}\underbrace{w}_{=t^{m}}=t^{n}%
t^{m}=t^{n+m}$. So the induction step is complete, and Proposition
\ref{prop.words.uv=vu.2} is proven.
\end{proof}

\begin{proposition}
\label{prop.words.uv=vu.3}Let $u,v,w\in\mathfrak{A}^{\ast}$ be nonempty words
satisfying $uv\geq vu$, $vw\geq wv$ and $wu\geq uw$. Then, there exist a
$t\in\mathfrak{A}^{\ast}$ and three nonnegative integers $n$, $m$ and $p$ such
that $u=t^{n}$, $v=t^{m}$ and $w=t^{p}$.
\end{proposition}

\begin{proof}
We prove this by strong induction on $\ell\left(  u\right)  +\ell\left(
v\right)  +\ell\left(  w\right)  $. Clearly, $\ell\left(  u\right)  $,
$\ell\left(  v\right)  $ and $\ell\left(  w\right)  $ are positive (since $u$,
$v$ and $w$ are nonempty). We assume WLOG that $\ell\left(  u\right)
=\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  ,\ell\left(
w\right)  \right\}  $ (because there is a cyclic symmetry in our situation).
Thus, $\ell\left(  u\right)  \leq\ell\left(  v\right)  $ and $\ell\left(
u\right)  \leq\ell\left(  w\right)  $. But $vu\leq uv$. Hence, Proposition
\ref{prop.words.lex}(e) (applied to $a=v$, $b=u$, $c=u$ and $d=v$) yields that
either we have $v\leq u$ or the word $u$ is a prefix of $v$. But Proposition
\ref{prop.words.lex}(f) (applied to $a=u$, $b=w$, $c=w$ and $d=u$) yields
$u\leq w$ (since $uw\leq wu$ and $\ell\left(  u\right)  \leq\ell\left(
w\right)  $). Furthermore, $wv\leq vw$. Hence, Proposition
\ref{prop.words.lex}(e) (applied to $a=w$, $b=v$, $c=v$ and $d=w$) yields that
either we have $w\leq v$ or the word $v$ is a prefix of $w$.

From what we have found so far, it is easy to see that $u$ is a prefix of
$v$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $u$ is not a
prefix of $v$. Hence, we must have $v\leq u$ (since either we have $v\leq u$
or the word $u$ is a prefix of $v$), and in fact $v<u$ (because $v=u$ would
contradict to $u$ not being a prefix of $v$). Thus, $v<u\leq w$. But recall
that either we have $w\leq v$ or the word $v$ is a prefix of $w$. Thus, $v$
must be a prefix of $w$ (because $v<w$ rules out $w\leq v$). In other words,
there exists a $q\in\mathfrak{A}^{\ast}$ such that $w=vq$. Consider this $q$.
We have $v<u\leq w=vq$. Thus, Proposition \ref{prop.words.lex}(g) (applied to
$a=v$, $b=u$ and $c=q$) yields that $v$ is a prefix of $u$. In light of
$\ell\left(  u\right)  \leq\ell\left(  v\right)  $, this is only possible if
$v=u$, but this contradicts $v<u$. This contradiction completes this proof.}.
In other words, there exists a $v^{\prime}\in\mathfrak{A}^{\ast}$ such that
$v=uv^{\prime}$. Consider this $v^{\prime}$.

If the word $v^{\prime}$ is empty, then the statement of Proposition
\ref{prop.words.uv=vu.3} can be easily deduced from Proposition
\ref{prop.words.uv=vu.2}\footnote{\textit{Proof.} Assume that the word
$v^{\prime}$ is empty. Then, $v=uv^{\prime}$ becomes $v=u$. Therefore, $vw\geq
wv$ becomes $uw\geq wu$. Combined with $wu\geq uw$, this yields $uw=wu $.
Hence, Proposition \ref{prop.words.uv=vu.2} (applied to $w$ instead of $v$)
yields that there exist a $t\in\mathfrak{A}^{\ast}$ and two nonnegative
integers $n$ and $m$ such that $u=t^{n}$ and $w=t^{m}$. Clearly, $v=u=t^{n}$
as well, and so the statement of Proposition \ref{prop.words.uv=vu.3} is
true.}. Thus, we assume WLOG that this is not the case. Hence, $v^{\prime}$ is nonempty.

Using $v=uv^{\prime}$, we can rewrite $uv\geq vu$ as $uuv^{\prime}\geq
uv^{\prime}u$. That is, $uv^{\prime}u\leq uuv^{\prime}$, so that $v^{\prime
}u\leq uv^{\prime}$ (by Proposition \ref{prop.words.lex}(c), applied to $a=u$,
$c=v^{\prime}u$ and $d=uv^{\prime}$). That is, $uv^{\prime}\geq v^{\prime}u$.
But $\ell\left(  uw\right)  =\ell\left(  u\right)  +\ell\left(  w\right)
=\ell\left(  w\right)  +\ell\left(  u\right)  =\ell\left(  wu\right)  \geq
\ell\left(  wu\right)  $. Hence, Proposition \ref{prop.words.lex}(i) (applied
to $a=uw$, $b=wu$ and $c=v^{\prime}$) yields $uwv^{\prime}\leq wuv^{\prime}$
(since $uw\leq wu$). Now, $\underbrace{uv^{\prime}}_{=v}w=vw\geq
w\underbrace{v}_{=uv^{\prime}}=wuv^{\prime}\geq uwv^{\prime}$ (since
$uwv^{\prime}\leq wuv^{\prime}$), so that $uwv^{\prime}\leq uv^{\prime}w$.
Hence, $wv^{\prime}\leq v^{\prime}w$ (by Proposition \ref{prop.words.lex}(c),
applied to $a=u$, $c=wv^{\prime}$ and $d=v^{\prime}w$), so that $v^{\prime
}w\geq wv^{\prime}$. Now, we can apply Proposition \ref{prop.words.uv=vu.3} to
$v^{\prime}$ instead of $v$ (by the induction hypothesis, because
$\underbrace{\ell\left(  u\right)  +\ell\left(  v^{\prime}\right)
}_{\substack{=\ell\left(  uv^{\prime}\right)  =\ell\left(  v\right)
\\\text{(since }uv^{\prime}=v\text{)}}}+\ell\left(  w\right)  =\ell\left(
v\right)  +\ell\left(  w\right)  <\ell\left(  u\right)  +\ell\left(  v\right)
+\ell\left(  w\right)  $). As a result, we see that there exist a
$t\in\mathfrak{A}^{\ast}$ and three nonnegative integers $n$, $m$ and $p$ such
that $u=t^{n}$, $v^{\prime}=t^{m}$ and $w=t^{p}$. Clearly, this $t$ and these
$n,m,p$ satisfy $v=\underbrace{u}_{=t^{n}}\underbrace{v^{\prime}}_{=t^{m}%
}=t^{n}t^{m}=t^{n+m}$, and so the statement of Proposition
\ref{prop.words.uv=vu.3} is satisfied. The induction step is thus complete.
\end{proof}

\begin{corollary}
\label{cor.words.uv=vu.3}Let $u,v,w\in\mathfrak{A}^{\ast}$ be words satisfying
$uv\geq vu$ and $vw\geq wv$. Assume that $v$ is nonempty. Then, $uw\geq wu$.
\end{corollary}

\begin{proof}
Assume the contrary. Thus, $uw<wu$, so that $wu\geq uw$.

If $u$ or $w$ is empty, then everything is obvious. We thus WLOG assume that
$u$ and $w$ are nonempty. Thus, Proposition \ref{prop.words.uv=vu.3} shows
that there exist a $t\in\mathfrak{A}^{\ast}$ and three nonnegative integers
$n$, $m$ and $p$ such that $u=t^{n}$, $v=t^{m}$ and $w=t^{p}$. But this yields
$wu=t^{p}t^{n}=t^{p+n}=t^{n+p}=\underbrace{t^{n}}_{=u}\underbrace{t^{p}}%
_{=w}=uw$, contradicting $uw<wu$. This contradiction finishes the proof.
\end{proof}

\begin{exercise}
\label{exe.words.uw-wu}Find an alternative proof of Corollary
\ref{cor.words.uv=vu.3} which does not use Proposition
\ref{prop.words.uv=vu.3}.
\end{exercise}

The above results have a curious consequence, which we are not going to use:

\begin{corollary}
\label{cor.words.uv=vu.partord}We can define a preorder on the set
$\mathfrak{A}^{\ast}\setminus\left\{  \varnothing\right\}  $ of all nonempty
words by defining a nonempty word $u$ to be greater-or-equal to a nonempty
word $v$ (with respect to this preorder) if and only if $uv\geq vu$. Two
nonempty words $u,v$ are equivalent with respect to the equivalence relation
induced by this preorder if and only if there exist a $t\in\mathfrak{A}^{\ast
}$ and two nonnegative integers $n$ and $m$ such that $u=t^{n}$ and $v=t^{m}$.
\end{corollary}

\begin{proof}
The alleged preorder is transitive (by Corollary \ref{cor.words.uv=vu.3}) and
reflexive (obviously), and hence is really a preorder. The claim in the second
sentence follows from Proposition \ref{prop.words.uv=vu.2}.
\end{proof}

As another consequence of Proposition \ref{prop.words.uv=vu.3}, we obtain a
classical property of words \cite[Proposition 1.3.1]{Lothaire}:

\begin{exercise}
\label{exe.words.u=pn=qm}Let $u$ and $v$ be words and $n$ and $m$ be positive
integers such that $u^{n}=v^{m}$. Prove that there exists a word $t$ and
positive integers $i$ and $j$ such that $u=t^{i}$ and $v=t^{j}$.
\end{exercise}

Here is another application of Corollary \ref{cor.words.uv=vu.3}:

\begin{exercise}
\label{exe.words.unvm.vmun}Let $n$ and $m$ be positive integers. Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words. Prove
that $uv\geq vu$ holds if and only if $u^{n}v^{m}\geq v^{m}u^{n}$ holds.
\end{exercise}

\begin{exercise}
\label{exe.words.un.vm}Let $n$ and $m$ be positive integers. Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words
satisfying $n\ell\left(  u\right)  =m\ell\left(  v\right)  $. Prove that
$uv\geq vu$ holds if and only if $u^{n}\geq v^{m}$ holds.
\end{exercise}

We can also generalize Propositions \ref{prop.words.uv=vu.2} and
\ref{prop.words.uv=vu.3}:

\begin{exercise}
\label{exe.words.uv=vu.k}Let $u_{1},u_{2},\ldots,u_{k}$ be nonempty words such
that every $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies $u_{i}u_{i+1}\geq
u_{i+1}u_{i}$, where $u_{k+1}$ means $u_{1}$. Show that there exist a word $t
$ and nonnegative integers $n_{1},n_{2},\ldots,n_{k}$ such that $u_{1}%
=t^{n_{1}} $, $u_{2}=t^{n_{2}}$, $\ldots$, $u_{k}=t^{n_{k}}$.
\end{exercise}

Now, we define the notion of a Lyndon word. There are several definitions in
literature, some of which will be proven equivalent in Theorem
\ref{thm.words.lyndon.equiv}.

\begin{definition}
\label{def.words.lyndon}A word $w\in\mathfrak{A}^{\ast}$ is said to be
\emph{Lyndon}\index{Lyndon word} if it is nonempty and satisfies the
following property: Every
nonempty proper suffix $v$ of $w$ satisfies $v>w$.
\end{definition}

For example, the word $113$ is Lyndon (because its nonempty proper suffixes
are $13$ and $3$, and these are both $>113$), and the word $242427$ is Lyndon
(its nonempty proper suffixes are $42427$, $2427$, $427$, $27$ and $7$, and
again these are each $>242427$). The words $2424$ and $35346$ are not Lyndon
(the word $2424$ has a nonempty proper suffix $24\leq2424$, and the word
$35346$ has a nonempty proper suffix $346\leq35346$). Every word of length $1$
is Lyndon (since it has no nonempty proper suffixes). A word $w=\left(
w_{1},w_{2}\right)  $ with two letters is Lyndon if and only if $w_{1}<w_{2}$.
A word $w=\left(  w_{1},w_{2},w_{3}\right)  $ of length $3$ is Lyndon if and
only if $w_{1}<w_{3}$ and $w_{1}\leq w_{2}$. A four-letter word $w=\left(
w_{1},w_{2},w_{3},w_{4}\right)  $ is Lyndon if and only if $w_{1}<w_{4}$,
$w_{1}\leq w_{3}$, $w_{1}\leq w_{2}$ and $\left(  \text{if }w_{1}=w_{3}\text{
then }w_{2}<w_{4}\right)  $. (These rules only get more complicated as the
words grow longer.)

We will show several properties of Lyndon words now. We begin with
trivialities which will make some arguments a bit shorter:

\begin{proposition}
\label{prop.words.lyndon.trivia}Let $w$ be a Lyndon word. Let $u$ and $v$ be
words such that $w=uv$.

\begin{itemize}
\item[(a)] If $v$ is nonempty, then $v\geq w$.

\item[(b)] If $v$ is nonempty, then $v>u$.

\item[(c)] If $u$ and $v$ are nonempty, then $vu>uv$.

\item[(d)] We have $vu\geq uv$.
\end{itemize}
\end{proposition}

\begin{proof}
(a) Assume that $v$ is nonempty. Clearly, $v$ is a suffix of $w$ (since
$w=uv$). If $v$ is a proper suffix of $w$, then the definition of a Lyndon
word yields that $v>w$ (since $w$ is a Lyndon word); otherwise, $v$ must be
$w$ itself. In either case, we have $v\geq w$. Hence, Proposition
\ref{prop.words.lyndon.trivia}(a) is proven.

(b) Assume that $v$ is nonempty. From Proposition
\ref{prop.words.lyndon.trivia}(a), we obtain $v\geq w=uv>u$ (since $v$ is
nonempty). This proves Proposition \ref{prop.words.lyndon.trivia}(b).

(c) Assume that $u$ and $v$ are nonempty. Since $u$ is nonempty, we have
$vu>v\geq w$ (by Proposition \ref{prop.words.lyndon.trivia}(a)). Since $w=uv$,
this becomes $vu>uv$. This proves Proposition \ref{prop.words.lyndon.trivia}(c).

(d) We need to prove that $vu\geq uv$. If either $u$ or $v$ is empty, $vu$ and
$uv$ are obviously equal, and thus $vu\geq uv$ is true in this case. Hence, we
can WLOG assume that $u$ and $v$ are nonempty. Assume this. Then, $vu\geq uv$
follows from Proposition \ref{prop.words.lyndon.trivia}(c). This proves
Proposition \ref{prop.words.lyndon.trivia}(d).
\end{proof}

\begin{corollary}
\label{cor.words.lyndon.trivia}Let $w$ be a Lyndon word. Let $v$ be a nonempty
suffix of $w$. Then, $v\geq w$.
\end{corollary}

\begin{proof}
Since $v$ is a nonempty suffix of $w$, there exists $u\in\mathfrak{A}^{\ast}$
such that $w=uv$. Thus, $v\geq w$ follows from Proposition
\ref{prop.words.lyndon.trivia}(a).
\end{proof}

Our next proposition is \cite[Lemma 6.5.4]{HazewinkelGubareniKirichenko}; its
part (a) is also \cite[(5.1.2)]{Reutenauer}:

\begin{proposition}
\label{prop.words.lyndon.concat}Let $u$ and $v$ be two Lyndon words such that
$u<v$. Then:

\begin{itemize}
\item[(a)] The word $uv$ is Lyndon.

\item[(b)] We have $uv<v$.
\end{itemize}
\end{proposition}

\begin{proof}
(b) The word $u$ is Lyndon and thus nonempty. Hence, $uv\neq v$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $uv=v$. Thus,
$uv=v=\varnothing v$. Cancelling $v$ from this equation, we obtain
$u=\varnothing$. That is, $u$ is empty. This contradicts the fact that $u$ is
nonempty. This contradiction proves that our assumption was wrong, qed.}. If
$uv\leq v\varnothing$, then Proposition \ref{prop.words.lyndon.concat}(b)
easily follows\footnote{\textit{Proof.} Assume that $uv\leq v\varnothing$.
Thus, $uv\leq v\varnothing=v$. Since $uv\neq v$, this becomes $uv<v$, so that
Proposition \ref{prop.words.lyndon.concat}(b) is proven.}. Hence, for the rest
of this proof, we can WLOG assume that we don't have $uv\leq v\varnothing$.
Assume this.

We have $u<v$. Hence, Proposition \ref{prop.words.lex}(d) (applied to $a=u$,
$b=v$, $c=v$ and $d=\varnothing$) yields that either we have $uv\leq
v\varnothing$ or the word $u$ is a prefix of $v$. Since we don't have $uv\leq
v\varnothing$, we thus see that the word $u$ is a prefix of $v$. In other
words, there exists a $t\in\mathfrak{A}^{\ast}$ satisfying $v=ut$. Consider
this $t$. Then, $t$ is nonempty (else we would have $v=u\underbrace{t}%
_{=\varnothing}=u$ in contradiction to $u<v$).

Now, $v=ut$. Hence, $t$ is a proper suffix of $v$ (proper because $u$ is
nonempty). Thus, $t$ is a nonempty proper suffix of $v$. Since every nonempty
proper suffix of $v$ is $>v$ (because $v$ is Lyndon), this shows that $t>v$.
Hence, $v\leq t$. Thus, Proposition \ref{prop.words.lex}(b) (applied to $a=u$,
$c=v$ and $d=t$) yields $uv\leq ut=v$. Combined with $uv\neq v$, this yields
$uv<v$. Hence, Proposition \ref{prop.words.lyndon.concat}(b) is proven.

(a) The word $v$ is nonempty (since it is Lyndon). Hence, $uv$ is nonempty. It
thus remains to check that every nonempty proper suffix $p$ of $uv$ satisfies
$p>uv$.

So let $p$ be a nonempty proper suffix of $uv$. We must show that $p>uv$.
Since $p$ is a nonempty proper suffix of $uv$, we must be in one of the
following two cases (depending on whether this suffix begins before the suffix
$v$ of $uv$ begins or afterwards):

\textit{Case 1:} The word $p$ is a nonempty suffix of $v$. (Note that $p=v$ is allowed.)

\textit{Case 2:} The word $p$ has the form $qv$ where $q$ is a nonempty proper
suffix of $u$.

Let us first handle Case 1. In this case, $p$ is a nonempty suffix of $v$.
Since $v$ is Lyndon, this yields that $p\geq v$ (by Corollary
\ref{cor.words.lyndon.trivia}, applied to $v$ and $p$ instead of $w$ and $v$).
But Proposition \ref{prop.words.lyndon.concat}(b) yields $uv<v$, thus $v>uv$.
Hence, $p\geq v>uv$. We thus have proven $p>uv$ in Case 1.

Let us now consider Case 2. In this case, $p$ has the form $qv$ where $q$ is a
nonempty proper suffix of $u$. Consider this $q$. Clearly, $q>u$ (since $u$ is
Lyndon and since $q$ is a nonempty proper suffix of $u$), so that $u\leq q$.
Thus, Proposition \ref{prop.words.lex}(d) (applied to $a=u$, $b=v$, $c=q$ and
$d=v$) yields that either we have $uv\leq qv$ or the word $u$ is a prefix of
$q$. Since $u$ being a prefix of $q$ is impossible (in fact, $q$ is a proper
suffix of $u$, thus shorter than $u$), we thus must have $uv\leq qv$. Since
$uv\neq qv$ (because otherwise we would have $uv=qv$, thus $u=q$ (because we
can cancel $v$ from the equality $uv=qv$), contradicting $q>u$), this can be
strengthened to $uv<qv=p$. Thus, $p>uv$ is proven in Case 2 as well.

Now that $p>uv$ is shown to hold in both cases, we conclude that $p>uv$ always holds.

Now, let us forget that we fixed $p$. We have thus shown that every nonempty
proper suffix $p$ of $uv$ satisfies $p>uv$. Since $uv$ is nonempty, this
yields that $uv$ is Lyndon (by the definition of a Lyndon word). Thus, the
proof of Proposition \ref{prop.words.lyndon.concat}(a) is complete.
\end{proof}

\begin{commentedout}


\begin{proof}
[Alternative Proof of Proposition~\ref{prop.words.lyndon.concat}%
.][\textbf{Note:} The following proof of
Proposition~\ref{prop.words.lyndon.concat} is the first proof that came into
my mind. It is similar to the proof above, but organized in a less clean way
(and not as easily generalizable to poset alphabets).]

(a) It is clear that $uv$ is nonempty. It thus remains to check that every
nonempty proper suffix $p$ of $uv$ satisfies $p>uv$.

So let $p$ be a nonempty proper suffix of $uv$. We must show that $p>uv$.
Since $p$ is a nonempty proper suffix of $uv$, we must be in one of the
following two cases (depending on whether this suffix begins before the suffix
$v$ of $uv$ begins or afterwards):

\textit{Case 1:} The word $p$ is a nonempty suffix of $v$. (Note that $p=v$ is allowed.)

\textit{Case 2:} The word $p$ has the form $qv$ where $q$ is a nonempty proper
suffix of $u$.

Let us first handle Case 1. In this case, $p$ is a nonempty suffix of $v$.
Since $v$ is Lyndon, this yields that $p\geq v$ (by Corollary
\ref{cor.words.lyndon.trivia}, applied to $v$ and $p$ instead of $w$ and $v$).
We now assume (for the sake of contradiction) that $p\leq uv$. Then, $u<v\leq
p\leq uv$. Proposition \ref{prop.words.lex}(g) (applied to $a=u$, $b=p$ and
$c=v$) thus yields $u$ is a prefix of $p$. Thus, there exists a $t\in
\mathfrak{A}^{\ast}$ satisfying $p=ut$. Consider this $t$. Then, $t$ is
nonempty (else we would have $p=u\underbrace{t}_{=\varnothing}=u$ in
contradiction to $u<p$).

Now, $v=r\underbrace{p}_{=ut}=rut=\left(  ru\right)  t$. Hence, $t$ is a
nonempty proper suffix of $v$ (proper because $u$ is nonempty). Since every
nonempty proper suffix of $v$ is $>v$ (because $v$ is Lyndon), this shows that
$t>v$. Hence, $v\leq t$. Thus, Proposition \ref{prop.words.lex}(b) (applied to
$a=u$, $c=v$ and $d=t$) yields $uv\leq ut=p$. Combined with $p\leq uv$, this
yields $uv=p=ut$. Cancelling $u$ in this equality, we obtain $v=t$,
contradicting $t>v$. This contradiction dispels our assumption that $p\leq
uv$. Hence, $p>uv$. We thus have proven $p>uv$ in Case 1.

Let us now consider Case 2. In this case, $p$ has the form $qv$ where $q$ is a
nonempty proper suffix of $u$. Consider this $q$. Clearly, $q>u$ (since $u$ is
Lyndon), so that $u\leq q$. Thus, Proposition \ref{prop.words.lex}(d) (applied
to $a=u$, $b=v$, $c=q$ and $d=v$) yields that either we have $uv\leq qv$ or
the word $u$ is a prefix of $q$. Since $u$ being a prefix of $q $ is
impossible (in fact, $q$ is a proper suffix of $u$, thus shorter than $u $),
we thus must have $uv\leq qv$. Since $uv\neq qv$ (because otherwise we would
have $uv=qv$, thus $u=q$ (because we can cancel $v$ from the equality
$uv=qv$), contradicting $q>u$), this can be strengthened to $uv<qv=p $. Thus,
$p>uv$ is proven in Case 2 as well.

Now that $p>uv$ is shown to hold in both cases, the proof of Proposition
\ref{prop.words.lyndon.concat}(a) is complete.

(b) We know from Proposition \ref{prop.words.lyndon.concat}(a) that $uv$ is
Lyndon. Hence, Proposition \ref{prop.words.lyndon.trivia}(a) (applied to
$w=uv$) yields $v\geq uv$. Since $v\neq uv$, this yields $v>uv$. This proves
Proposition \ref{prop.words.lyndon.concat}(b).
\end{proof}
\end{commentedout}

Proposition \ref{prop.words.lyndon.concat}(b), combined with Corollary
\ref{cor.words.uv=vu.3}, leads to a technical result which we will find good
use for later:

\begin{corollary}
\label{cor.words.lyndon.uv=vu}Let $u$ and $v$ be two Lyndon words such that
$u<v$. Let $z$ be a word such that $zv\geq vz$ and $uz\geq zu$. Then, $z$ is
the empty word.
\end{corollary}

\begin{proof}
Assume the contrary. Then, $z$ is nonempty. Thus, Corollary
\ref{cor.words.uv=vu.3} (applied to $z$ and $v$ instead of $v$ and $w$) yields
$uv\geq vu$. But Proposition \ref{prop.words.lyndon.concat}(b) yields
$uv<v\leq vu$, contradicting $uv\geq vu$. This contradiction completes our proof.
\end{proof}

We notice that the preorder of Corollary \ref{cor.words.uv=vu.partord} becomes
particularly simple on Lyndon words:

\begin{proposition}
\label{prop.words.lyndon.preorder}Let $u$ and $v$ be two Lyndon words. Then,
$u\geq v$ if and only if $uv\geq vu$.
\end{proposition}

\begin{proof}
We distinguish between three cases:

\textit{Case 1:} We have $u<v$.

\textit{Case 2:} We have $u=v$.

\textit{Case 3:} We have $u>v$.

Let us consider Case 1. In this case, we have $u<v$. Thus,%
\begin{align*}
uv  &  <v\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.words.lyndon.concat}(b)}\right) \\
&  \leq vu.
\end{align*}
Hence, we have neither $u\geq v$ nor $uv\geq vu$ (because we have $u<v$ and
$uv<vu$). Thus, Proposition \ref{prop.words.lyndon.preorder} is proven in Case 1.

In Case 2, we have $u=v$. Therefore, in Case 2, both inequalities $u\geq v$
and $uv\geq vu$ hold (and actually are equalities). Thus, Proposition
\ref{prop.words.lyndon.preorder} is proven in Case 2 as well.

Let us finally consider Case 3. In this case, we have $u>v$. In other words,
$v<u$. Thus,%
\begin{align*}
vu  &  <u\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.words.lyndon.concat}(b), applied to }v\text{ and }u\text{ instead of
}u\text{ and }v\right) \\
&  \leq uv.
\end{align*}
Hence, we have both $u\geq v$ and $uv\geq vu$ (because we have $v<u$ and
$vu<uv$). Thus, Proposition \ref{prop.words.lyndon.preorder} is proven in Case 3.

Proposition \ref{prop.words.lyndon.preorder} is now proven in all three
possible cases.
\end{proof}

\begin{proposition}
\label{prop.words.lyndon.cutoff}Let $w$ be a nonempty word. Let $v$ be the
(lexicographically) smallest nonempty suffix of $w$. Then:

\begin{itemize}
\item[(a)] The word $v$ is a Lyndon word.

\item[(b)] Assume that $w$ is not a Lyndon word. Then there exists a nonempty
$u\in\mathfrak{A}^{\ast}$ such that $w=uv$, $u\geq v$ and $uv\geq vu$.
\end{itemize}
\end{proposition}

\begin{proof}
(a) Every nonempty proper suffix of $v$ is $\geq v$ (since every nonempty
proper suffix of $v$ is a nonempty suffix of $w$, but $v$ is the smallest such
suffix) and therefore $>v$ (since a proper suffix of $v$ cannot be $=v$).
Combined with the fact that $v$ is nonempty, this yields that $v$ is Lyndon.
Proposition \ref{prop.words.lyndon.cutoff}(a) is proven.

(b) Assume that $w$ is not a Lyndon word. Then, $w\neq v$ (since $v$ is Lyndon
(by Proposition \ref{prop.words.lyndon.cutoff}(a)) while $w$ is not). Now, $v$
is a suffix of $w$. Thus, there exists an $u\in\mathfrak{A}^{\ast}$ such that
$w=uv$. Consider this $u$. Clearly, $u$ is nonempty (since $uv=w\neq v$).
Assume (for the sake of contradiction) that $u<v$. Let $v^{\prime}$ be the
(lexicographically) smallest nonempty suffix of $u$. Then, $v^{\prime}$ is a
Lyndon word (by Proposition \ref{prop.words.lyndon.cutoff}(a), applied to $u$
and $v^{\prime}$ instead of $w$ and $v$) and satisfies $v^{\prime}\leq u$
(since $u$ is a nonempty suffix of $u$, whereas $v^{\prime}$ is the smallest
such suffix). Thus, $v^{\prime}$ and $v$ are Lyndon words such that
$v^{\prime}\leq u<v$. Proposition \ref{prop.words.lyndon.concat}(a) (applied
to $v^{\prime}$ instead of $u$) now yields that the word $v^{\prime}v$ is
Lyndon. Hence, every nonempty proper suffix of $v^{\prime}v$ is $>v^{\prime}%
v$. Since $v$ is a nonempty proper suffix of $v^{\prime}v$, this yields that
$v>v^{\prime}v$.

But $v^{\prime}$ is a nonempty suffix of $u$, so that $v^{\prime}v$ is a
nonempty suffix of $uv=w$. Since $v$ is the smallest such suffix, this yields
that $v^{\prime}v\geq v$. This contradicts $v>v^{\prime}v$. Our assumption
(that $u<v$) therefore falls. We conclude that $u\geq v$.

It remains to prove that $uv\geq vu$. Assume the contrary. Then, $uv<vu$.
Thus, there exists at least one suffix $t$ of $u$ such that $tv<vt$ (namely,
$t=u$). Let $p$ be the \textbf{minimum-length} such suffix. Then, $pv<vp$.
Thus, $p$ is nonempty.

Since $p$ is a suffix of $u$, it is clear that $pv$ is a suffix of $uv=w$. So
we know that $pv$ is a nonempty suffix of $w$. Since $v$ is the smallest such
suffix, this yields that $v\leq pv<vp$. Thus, Proposition \ref{prop.words.lex}%
(g) (applied to $a=v$, $b=pv$ and $c=p$) yields that $v$ is a prefix of $pv$.
In other words, there exists a $q\in\mathfrak{A}^{\ast}$ such that $pv=vq$.
Consider this $q$. This $q$ is nonempty (because otherwise we would have
$pv=v\underbrace{q}_{=\varnothing}=v$, contradicting the fact that $p$ is
nonempty). From $vq=pv<vp$, we obtain $q\leq p$ (by Proposition
\ref{prop.words.lex}(c), applied to $a=v$, $c=q$ and $d=p$).

We know that $q$ is a suffix of $pv$ (since $vq=pv$), whereas $pv$ is a suffix
of $w$. Thus, $q$ is a suffix of $w$. So $q$ is a nonempty suffix of $w$.
Since $v$ is the smallest such suffix, this yields that $v\leq q$. We now have
$v\leq q\leq p\leq pv<vp$. Hence, $v$ is a prefix of $p$ (by Proposition
\ref{prop.words.lex}(g), applied to $a=v$, $b=p$ and $c=p$). In other words,
there exists an $r\in\mathfrak{A}^{\ast}$ such that $p=vr$. Consider this $r$.
Clearly, $r$ is a suffix of $p$, while $p$ is a suffix of $u$; therefore, $r$
is a suffix of $u$. Also, $pv<vp$ rewrites as $vrv<vvr$ (because $p=vr$).
Thus, Proposition \ref{prop.words.lex}(c) (applied to $a=v$, $c=rv$ and
$d=vr$) yields $rv\leq vr$. Since $rv\neq vr$ (because otherwise, we would
have $rv=vr$, thus $v\underbrace{rv}_{=vr}=vvr$, contradicting $vrv<vvr$),
this becomes $rv<vr$.

Now, $r$ is a suffix of $u$ such that $rv<vr$. Since $p$ is the minimum-length
such suffix, this yields $\ell\left(  r\right)  \geq\ell\left(  p\right)  $.
But this contradicts the fact that $\ell\left(  \underbrace{p}_{=vr}\right)
=\ell\left(  vr\right)  =\underbrace{\ell\left(  v\right)  }_{>0}+\ell\left(
r\right)  >\ell\left(  r\right)  $. This contradiction proves our assumption
wrong; thus, we have shown that $uv\geq vu$. Proposition
\ref{prop.words.lyndon.cutoff}(b) is proven.
\end{proof}

\begin{theorem}
\label{thm.words.lyndon.equiv}Let $w$ be a nonempty word. The following four
assertions are equivalent:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}$:} The word $w$ is Lyndon.

\item \textit{Assertion $\mathcal{B}$:} Any nonempty words $u$ and
$v$ satisfying $w=uv$ satisfy $v>w$.

\item \textit{Assertion $\mathcal{C}$:} Any nonempty words $u$ and
$v$ satisfying $w=uv$ satisfy $v>u$.

\item \textit{Assertion $\mathcal{D}$:} Any nonempty words $u$ and
$v$ satisfying $w=uv$ satisfy $vu>uv$.
\end{itemize}
\end{theorem}

\begin{proof}
\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{B}%
$:} If Assertion $\mathcal{A}$ holds, then Assertion $\mathcal{B}$
clearly holds (in fact, whenever $u$ and $v$ are nonempty words satisfying
$w=uv$, then $v$ is a nonempty proper suffix of $w$, and therefore $>w$ by the
definition of a Lyndon word).

\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{C}%
$:} This implication follows from Proposition
\ref{prop.words.lyndon.trivia}(b).

\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{D}%
$:} This implication follows from Proposition
\ref{prop.words.lyndon.trivia}(c).

\textit{Proof of the implication $\mathcal{B}\Longrightarrow\mathcal{A}%
$:} Assume that Assertion $\mathcal{B}$ holds. If $v$ is a nonempty
proper suffix of $w$, then there exists an $u\in\mathfrak{A}^{\ast}$
satisfying $w=uv$. This $u$ is nonempty because $v$ is a proper suffix, and
thus Assertion $\mathcal{B}$ yields $v>w$. Hence, every nonempty proper suffix
$v$ of $w$ satisfies $v>w$. By the definition of a Lyndon word, this yields
that $w$ is Lyndon, so that Assertion $\mathcal{A}$ holds.

\textit{Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{A}%
$:} Assume that Assertion $\mathcal{C}$ holds. If $w$ was not Lyndon,
then Proposition \ref{prop.words.lyndon.cutoff}(b) would yield nonempty words
$u$ and $v$ such that $w=uv$ and $u\geq v$; this would contradict Assertion
$\mathcal{C}$. Thus, $w$ is Lyndon, and Assertion $\mathcal{A}$ holds.

\textit{Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{A}%
$:} Assume that Assertion $\mathcal{D}$ holds. If $w$ was not Lyndon,
then Proposition \ref{prop.words.lyndon.cutoff}(b) would yield nonempty words
$u$ and $v$ such that $w=uv$ and $uv\geq vu$; this would contradict Assertion
$\mathcal{D}$. Thus, $w$ is Lyndon, and Assertion $\mathcal{A}$ holds.

Now we have proven enough implications to conclude the equivalence of all four assertions.
\end{proof}

Theorem \ref{thm.words.lyndon.equiv} connects our definition of Lyndon words
with some of the definitions appearing in literature. For example, Lothaire
\cite[\S 5.1]{Lothaire}, Shirshov \cite{Shirshov} and de Bruijn/Klarner
\cite[\S 4]{deBruijnKlarner} define Lyndon words using Assertion $\mathcal{D}$
(note, however, that Shirshov takes $<$ instead of $>$ and calls Lyndon words
``regular words''; also, de Bruijn/Klarner
call Lyndon words ``normal words'').
Chen-Fox-Lyndon \cite[\S 1]{ChenFoxLyndon}, Reutenauer \cite{Reutenauer} and
Radford \cite{Radford-shuffle} use our definition (but Chen-Fox-Lyndon call
the Lyndon words ``standard sequences'', and
Radford calls them ``primes'' and uses $<$
instead of $>$).

Theorem \ref{thm.words.lyndon.equiv} appears (with different notations) in
Zhou-Lu \cite[Proposition 1.4]{ZhouLu}. The equivalence $\mathcal{D}%
\Longleftrightarrow\mathcal{A}$ of our Theorem \ref{thm.words.lyndon.equiv} is
equivalent to \cite[Proposition 5.12]{Lothaire} and to \cite[$\mathfrak{A}%
^{\prime\prime}=\mathfrak{A}^{\prime\prime\prime}$]{ChenFoxLyndon}.

The following exercise provides a different (laborious) approach to Theorem
\ref{thm.words.lyndon.equiv}:

\begin{exercise}
\phantomsection\label{exe.words.lyndon.equiv2}

\begin{enumerate}
\item[(a)] Prove that if $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}%
^{\ast}$ are two words satisfying $uv<vu$, then there exists a nonempty suffix
$s$ of $u$ satisfying $sv<v$.

\item[(b)] Give a new proof of Theorem \ref{thm.words.lyndon.equiv} (avoiding
the use of Proposition \ref{prop.words.lyndon.cutoff}).
\end{enumerate}

[\textbf{Hint:} For (a), perform strong induction on $\ell\left(  u\right)
+\ell\left(  v\right)  $, assume the contrary, and distinguish between the
case when $u\leq v$ and the case when $v$ is a prefix of $u$. For (b), use
part (a) in proving the implication $\mathcal{D}\Longrightarrow\mathcal{B}$,
and factor $v$ as $v=u^{m}v^{\prime}$ with $m$ maximal in the proof of the
implication $\mathcal{C}\Longrightarrow\mathcal{B}$.]
\end{exercise}

The following two exercises are taken from \cite{Hazewinkel-lh}%
\footnote{Exercise \ref{exe.words.lyndon.power-criterion} is more or less
\cite[Lemma 4.3]{Hazewinkel-lh} with a converse added; Exercise
\ref{exe.words.lyndon.concatn} is \cite[Lemma 4.2]{Hazewinkel-lh}.}.

\begin{exercise}
\label{exe.words.lyndon.power-criterion}Let $w$ be a nonempty word. Prove that
$w$ is Lyndon if and only if every nonempty word $t$ and every positive
integer $n$ satisfy $\left(  \text{if }w\leq t^{n}\text{, then }w\leq
t\right)  $.
\end{exercise}

\begin{exercise}
\label{exe.words.lyndon.concatn}Let $w_{1}$, $w_{2}$, $\ldots$, $w_{n}$ be $n$
Lyndon words, where $n$ is a positive integer. Assume that $w_{1}\leq
w_{2}\leq\cdots\leq w_{n}$ and $w_{1}<w_{n}$. Show that $w_{1}w_{2}\cdots
w_{n}$ is a Lyndon word.
\end{exercise}

The following exercise is a generalization (albeit not in an obvious way) of
Exercise \ref{exe.words.lyndon.concatn}:

\begin{exercise}
\label{exe.words.lyndon.concatn-lyndon}Let $w_{1}$, $w_{2}$, $\ldots$, $w_{n}$
be $n$ Lyndon words, where $n$ is a positive integer. Assume that
$w_{i}w_{i+1}\cdots w_{n}\geq w_{1}w_{2}\cdots w_{n}$ for every $i\in\left\{
1,2,\ldots,n\right\}  $. Show that $w_{1}w_{2}\cdots w_{n}$ is a Lyndon word.
\end{exercise}

We are now ready to meet one of the most important features of Lyndon
words: a bijection between all words and multisets of Lyndon
words\footnote{And it is not even the only such bijection: we will see
another in Subsection~\ref{subsub.lyndon.gr.bij}.}; it is clear that
such a bijection is vital for constructing polynomial generating sets
of commutative algebras with bases indexed by words, such as $\Qsym$ or
shuffle algebras. This bijection is given by the
\emph{Chen-Fox-Lyndon factorization}:

\begin{definition}
\label{def.words.CFL}Let $w$ be a word. A \dfn{Chen-Fox-Lyndon
factorization} (in short, \dfn{CFL factorization}) of $w$ means a tuple
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ of Lyndon words satisfying
$w=a_{1}a_{2} \cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$.
\end{definition}

\begin{example}
The tuple $\left(  23,2,14,13323,13,12,12,1\right)  $ is a CFL factorization
of the word $23214133231312121$ over the alphabet $\left\{  1,2,3,\ldots
\right\}  $ (ordered by $1<2<3<\cdots$), since $23$, $2$, $14$, $13323$, $13$,
$12$, $12$ and $1$ are Lyndon words satisfying $23214133231312121=23\cdot
2\cdot14\cdot13323\cdot13\cdot12\cdot12\cdot1$ and $23\geq2\geq14\geq
13323\geq13\geq12\geq12\geq1$.
\end{example}

The bijection is given by the following \dfn{Chen-Fox-Lyndon theorem}
(\cite[Theorem 6.5.5]{HazewinkelGubareniKirichenko}, \cite[Thm. 5.1.5]%
{Lothaire}, \cite[part of Thm. 2.1.4]{Radford-shuffle}):

\begin{theorem}
\label{thm.words.CFL}Let $w$ be a word. Then, there exists a unique CFL
factorization of $w$.
\end{theorem}

Before we prove this, we need to state and prove a lemma
(which is \cite[Proposition 5.1.6]{Lothaire}):

\begin{lemma}
\label{lem.words.CFLlemma}Let $\left( a_1, a_2, \ldots, a_k \right)  $ be a
CFL factorization of a nonempty word $w$. Let $p$ be a nonempty suffix of $w$.
Then, $p\geq a_k$.
\end{lemma}

\begin{proof}%[Proof of Lemma \ref{lem.words.CFLlemma}]
We will prove Lemma \ref{lem.words.CFLlemma} by induction over the
(obviously) positive integer $k$.

\textit{Induction base:} Assume that $k=1$. Thus, $\left(  a_{1},a_{2}%
,\ldots,a_{k}\right)  =\left(  a_{1}\right)  $ is a tuple of Lyndon words
satisfying $w=a_{1}a_{2}\cdots a_{k}$. We have $w=a_{1}a_{2}\cdots a_{k}%
=a_{1}$ (since $k=1$), so that $w$ is a Lyndon word (since $a_{1}$ is a Lyndon
word). Thus, Corollary \ref{cor.words.lyndon.trivia} (applied to $v=p$) yields
$p\geq w=a_{1}=a_{k}$ (since $1=k$). Thus, Lemma \ref{lem.words.CFLlemma} is
proven in the case $k=1$. The induction base is complete.

\textit{Induction step:} Let $K$ be a positive integer. Assume (as the
induction hypothesis) that Lemma \ref{lem.words.CFLlemma} is proven for $k=K$.
We now need to show that Lemma \ref{lem.words.CFLlemma} holds for $k=K+1$.

So let $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ be a CFL factorization of
a nonempty word $w$. Let $p$ be a nonempty suffix of $w$. We need to prove
that $p\geq a_{K+1}$.

By the definition of a CFL factorization, $\left(  a_{1},a_{2},\ldots
,a_{K+1}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{K+1}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{K+1}$. Let $w^{\prime}%
=a_{2}a_{3}\cdots a_{K+1}$; then, $w=a_{1}a_{2}\cdots a_{K+1}=a_{1}%
\underbrace{\left(  a_{2}a_{3}\cdots a_{K+1}\right)  }_{=w^{\prime}}%
=a_{1}w^{\prime}$. Hence, every nonempty suffix of $w$ is either a nonempty
suffix of $w^{\prime}$, or has the form $qw^{\prime}$ for a nonempty suffix
$q$ of $a_{1}$. Since $p$ is a nonempty suffix of $w$, we thus must be in one
of the following two cases:

\textit{Case 1:} The word $p$ is a nonempty suffix of $w^{\prime}$.

\textit{Case 2:} The word $p$ has the form $qw^{\prime}$ for a nonempty suffix
$q$ of $a_{1}$.

Let us first consider Case 1. In this case, $p$ is a nonempty suffix of
$w^{\prime}$. The $K$-tuple $\left(  a_{2},a_{3},\ldots,a_{K+1}\right)  $ of
Lyndon words satisfies $w^{\prime}=a_{2}a_{3} \cdots a_{K+1}$ and $a_{2}\geq
a_{3}\geq\cdots\geq a_{K+1}$; therefore, $\left(  a_{2},a_{3},\ldots
,a_{K+1}\right)  $ is a CFL factorization of $w^{\prime}$. We can thus apply
Lemma \ref{lem.words.CFLlemma} to $K$, $w^{\prime}$ and $\left(  a_{2}%
,a_{3},\ldots,a_{K+1}\right)  $ instead of $k$, $w$ and $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ (because we assumed that Lemma
\ref{lem.words.CFLlemma} is proven for $k=K$). As a result, we obtain that
$p\geq a_{K+1}$. Thus, $p\geq a_{K+1}$ is proven in Case 1.

Let us now consider Case 2. In this case, $p$ has the form $qw^{\prime}$ for a
nonempty suffix $q$ of $a_{1}$. Consider this $q$. Since $a_{1}$ is a Lyndon
word, we have $q\geq a_{1}$ (by Corollary \ref{cor.words.lyndon.trivia},
applied to $a_{1}$ and $q$ instead of $w$ and $v$). Thus, $q\geq a_{1}\geq
a_{2}\geq\cdots\geq a_{K+1}$, so that $p=qw^{\prime}\geq q\geq a_{K+1}$. Thus,
$p\geq a_{K+1}$ is proven in Case 2.

We have now proven $p\geq a_{K+1}$ in all cases. This proves that Lemma
\ref{lem.words.CFLlemma} holds for $k=K+1$. The induction step is thus
finished, and with it the proof of Lemma \ref{lem.words.CFLlemma}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.words.CFL}]Let us first prove that there exists a
CFL factorization of $w$.

Indeed, there clearly exists a tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
$ of Lyndon words satisfying $w=a_{1}a_{2} \cdots a_{k}$\ \ \ \ \footnote{For
instance, the tuple $\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
}\right)  $ of one-letter words is a valid example (recall that one-letter
words are always Lyndon).}. Fix such a tuple with \textbf{minimum} $k$. We
claim that $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$.

Indeed, if some $i\in\left\{  1,2,\ldots,k-1\right\}  $ would satisfy
$a_{i}<a_{i+1}$, then the word $a_{i}a_{i+1}$ would be Lyndon (by Proposition
\ref{prop.words.lyndon.concat}(a), applied to $u=a_{i}$ and $v=a_{i+1}$),
whence $\left(  a_{1},a_{2},\ldots,a_{i-1},a_{i}a_{i+1},a_{i+2},a_{i+3}%
,\ldots,a_{k}\right)  $ would also be a tuple of Lyndon words satisfying
$w=a_{1}a_{2} \cdots a_{i-1}\left(  a_{i}a_{i+1}\right)  a_{i+2}a_{i+3} \cdots
a_{k}$ but having length $k-1<k$, contradicting the fact that $k$ is the
minimum length of such a tuple. Hence, no $i\in\left\{  1,2,\ldots
,k-1\right\}  $ can satisfy $a_{i}<a_{i+1}$. In other words, every
$i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $a_{i}\geq a_{i+1}$. In
other words, $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$. Thus, $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ is a CFL factorization of $w$, so we have shown
that such a CFL factorization exists.

It remains to show that there exists at most one CFL factorization of $w$. We
shall prove this by induction over $\ell\left(  w\right)  $. Thus, we fix a
word $w$ and assume that%
\begin{equation}
\text{for every word }v\text{ with }\ell\left(  v\right)  <\ell\left(
w\right)  \text{, there exists at most one CFL factorization of }v.
\label{pf.thm.words.CFL.uni.1}%
\end{equation}
We now have to prove that there exists at most one CFL factorization of $w$.

Indeed, let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(
b_{1},b_{2},\ldots,b_{m}\right)  $ be two CFL factorizations of $w$. We need
to prove that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(  b_{1}%
,b_{2},\ldots,b_{m}\right)  $. If $w$ is empty, then this is obvious, so we
WLOG assume that it is not; thus, $k>0$ and $m>0$.

Since $\left(  b_{1},b_{2},\ldots,b_{m}\right)  $ is a CFL factorization of
$w$, we have $w=b_{1}b_{2} \cdots b_{m}$, and thus $b_{m}$ is a nonempty
suffix of $w$. Thus, Lemma \ref{lem.words.CFLlemma} (applied to $p=b_{m}$)
yields $b_{m}\geq a_{k}$. The same argument (but with the roles of $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(  b_{1},b_{2},\ldots
,b_{m}\right)  $ switched) shows that $a_{k}\geq b_{m}$. Combined with
$b_{m}\geq a_{k}$, this yields $a_{k}=b_{m}$. Now let $v=a_{1}a_{2} \cdots
a_{k-1}$. Then, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)  $ is a CFL
factorization of $v$ (since $a_{1}\geq a_{2}\geq\cdots\geq a_{k-1}$).

Since $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a CFL factorization of
$w$, we have $w=a_{1}a_{2} \cdots a_{k}=\underbrace{a_{1}a_{2} \cdots a_{k-1}%
}_{=v}\underbrace{a_{k}}_{=b_{m}}=vb_{m}$, so that%
\[
vb_{m}=w=b_{1}b_{2} \cdots b_{m}=b_{1}b_{2} \cdots b_{m-1}b_{m}.
\]
Cancelling $b_{m}$ yields $v=b_{1}b_{2} \cdots b_{m-1}$. Thus, $\left(
b_{1},b_{2},\ldots,b_{m-1}\right)  $ is a CFL factorization of $v$ (since
$b_{1}\geq b_{2}\geq\cdots\geq b_{m-1}$). Since $\ell\left(  v\right)
<\ell\left(  w\right)  $ (because $v=a_{1}a_{2} \cdots a_{k-1}$ is shorter
than $w=a_{1}a_{2} \cdots a_{k}$), we can apply (\ref{pf.thm.words.CFL.uni.1})
to obtain that there exists at most one CFL factorization of $v$. But we
already know two such CFL factorizations: $\left(  a_{1},a_{2},\ldots
,a_{k-1}\right)  $ and $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $. Thus,
$\left(  a_{1},a_{2},\ldots,a_{k-1}\right)  =\left(  b_{1},b_{2}%
,\ldots,b_{m-1}\right)  $, which, combined with $a_{k}=b_{m}$, leads to
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(  b_{1},b_{2},\ldots
,b_{m}\right)  $. This is exactly what we needed to prove. So we have shown
(by induction) that there exists at most one CFL factorization of $w$. This
completes the proof of Theorem \ref{thm.words.CFL}.
\end{proof}

The CFL factorization allows us to count all Lyndon words of a given length if
$\mathfrak{A}$ is finite:

\begin{exercise}
\label{exe.words.lyndon.count}Assume that the alphabet $\mathfrak{A}$ is
finite. Let $q=\left\vert \mathfrak{A}\right\vert $.
Let $\mu$ be the number-theoretic M\"{o}bius
function (defined as in Exercise~\ref{exe.witt.ghost-equiv}).
Show that the number of
Lyndon words of length $n$ equals $\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  q^{n/d}$ for every positive integer $n$ (where ``%
$\sum\limits_{d\mid n}$'' means a sum over all positive
divisors of $n$).\ \ \ \ %
\footnote{In particular, $\dfrac{1}{n}\sum\limits_{d\mid n}%
\mu\left(  d\right)  q^{n/d}$ is an integer.}
\end{exercise}

Exercise \ref{exe.words.lyndon.count} is a well-known result and appears,
e.g., in \cite[Theorem 1.5]{ChenFoxLyndon} or in
\cite[Section 5.1]{Lothaire}.

We will now study another kind of factorization: not of an arbitrary word into
Lyndon words, but of a Lyndon word into two smaller Lyndon words. This
factorization is called \dfn{standard factorization} (\cite[\S 5.1]%
{Lothaire}) or \dfn{canonical factorization} (\cite[Lemma 6.5.33]%
{HazewinkelGubareniKirichenko}); we only introduce it from the viewpoint we
are interested in, namely its providing a way to do induction over Lyndon
words\footnote{e.g., allowing to solve Exercise
\ref{exe.words.lyndon.concatn-lyndon} in a simpler way}. Here is what we need
to know:

\begin{theorem}
\label{thm.words.lyndon.std}Let $w$ be a Lyndon word of length $>1$. Let $v$
be the (lexicographically) smallest nonempty \textbf{proper} suffix of $w$.
Since $v$ is a proper suffix of $w$, there exists a nonempty $u\in
\mathfrak{A}^{\ast}$ such that $w=uv$. Consider this $u$. Then:

\begin{itemize}
\item[(a)] The words $u$ and $v$ are Lyndon.

\item[(b)] We have $u<w<v$.
\end{itemize}
\end{theorem}

\begin{proof}
Every nonempty proper suffix of $v$ is $\geq v$ (since every nonempty proper
suffix of $v$ is a nonempty proper suffix of $w$, but $v$ is the smallest such
suffix) and therefore $>v$ (since a proper suffix of $v$ cannot be $=v$).
Combined with the fact that $v$ is nonempty, this yields that $v$ is Lyndon.

Since $w$ is Lyndon, we know that every nonempty proper suffix of $w$ is $>w$.
Applied to the nonempty proper suffix $v$ of $w$, this yields that $v>w$.
Hence, $w<v$. Since $v$ is nonempty, we have $u<uv=w<v$. This proves Theorem
\ref{thm.words.lyndon.std}(b).

Let $p$ be a nonempty proper suffix of $u$. Then, $pv$ is a nonempty proper
suffix of $uv=w$. Thus, $pv>w$ (since every nonempty proper suffix of $w$ is
$>w$). Thus, $pv>w=uv$, so that $uv<pv$. Thus, Proposition
\ref{prop.words.lex}(e) (applied to $a=u$, $b=v$, $c=p$ and $d=v$) yields that
either we have $u\leq p$ or the word $p$ is a prefix of $u$.

Let us assume (for the sake of contradiction) that $p\leq u$. Then, $p<u$
(because $p$ is a proper suffix of $u$, and therefore $p\neq u$). Hence, we
cannot have $u\leq p$. Thus, the word $p$ is a prefix of $u$ (since either we
have $u\leq p$ or the word $p$ is a prefix of $u$). In other words, there
exists a $q\in\mathfrak{A}^{\ast}$ such that $u=pq$. Consider this $q$. We
have $w=\underbrace{u}_{=pq}v=pqv=p\left(  qv\right)  $, and thus $qv$ is a
proper suffix of $w$ (proper because $p$ is nonempty). Moreover, $qv$ is
nonempty (since $v$ is nonempty). Hence, $qv$ is a nonempty proper suffix of
$w$. Since $v$ is the smallest such suffix, this entails that $v\leq qv$.
Proposition \ref{prop.words.lex}(b) (applied to $a=p$, $c=v$ and $d=qv$) thus
yields $pv\leq pqv$. Hence, $pv\leq pqv=w$, which contradicts $pv>w$. This
contradiction shows that our assumption (that $p\leq u$) was false. We thus
have $p>u$.

We now have shown that $p>u$ whenever $p$ is a nonempty proper suffix of $u$.
Combined with the fact that $u$ is nonempty, this shows that $u$ is a Lyndon
word. This completes the proof of Theorem \ref{thm.words.lyndon.std}(a).
\end{proof}

Another approach to the standard factorization is given in the following exercise:

\begin{exercise}
\label{exe.words.lyndon.std2}Let $w$ be a Lyndon word of length $>1$. Let $v$
be the longest proper suffix of $w$ such that $v$ is Lyndon\footnote{This is
well-defined, because there exists at least one proper suffix $v$ of $w$ such
that $v$ is Lyndon. (Indeed, the last letter of $w$ forms such a suffix,
because it is a proper suffix of $w$ (since $w$ has length $>1$) and is Lyndon
(since it is a one-letter word, and since every one-letter word is Lyndon).)}.
Since $v$ is a proper suffix of $w$, there exists a nonempty $u\in
\mathfrak{A}^{\ast}$ such that $w=uv$. Consider this $u$. Prove that:

\begin{enumerate}
\item[(a)] The words $u$ and $v$ are Lyndon.

\item[(b)] We have $u<w<v$.

\item[(c)] The words $u$ and $v$ are precisely the words $u$ and $v$
constructed in Theorem \ref{thm.words.lyndon.std}.
\end{enumerate}
\end{exercise}

Notice that a well-known recursive characterization of Lyndon words
\cite[$\mathfrak{A}^{\prime}=\mathfrak{A}^{\prime\prime}$]{ChenFoxLyndon} can
be easily derived from Theorem \ref{thm.words.lyndon.std} and Proposition
\ref{prop.words.lyndon.concat}(a). We will not dwell on it.

The following exercise surveys some variations on the characterizations of
Lyndon words\footnote{Compare this with
\cite[\S 7.2.11, Theorem Q]{Knuth-TAoCP4a}.}:

\begin{exercise}
\label{exe.words.powerlyndon}Let $w$ be a nonempty word. Consider the
following nine assertions:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime}$:} The word $w$ is a
power of a Lyndon word.

\item \textit{Assertion $\mathcal{B}^{\prime}$:} If $u$ and $v$ are
nonempty words satisfying $w=uv$, then either we have $v\geq w$ or the word
$v$ is a prefix of $w$.

\item \textit{Assertion $\mathcal{C}^{\prime}$:} If $u$ and $v$ are
nonempty words satisfying $w=uv$, then either we have $v\geq u$ or the word
$v$ is a prefix of $u$.

\item \textit{Assertion $\mathcal{D}^{\prime}$:} If $u$ and $v$ are
nonempty words satisfying $w=uv$, then we have $vu\geq uv$.

\item \textit{Assertion $\mathcal{E}^{\prime}$:} If $u$ and $v$ are
nonempty words satisfying $w=uv$, then either we have $v\geq u$ or the word
$v$ is a prefix of $w$.

\item \textit{Assertion $\mathcal{F}^{\prime}$:} The word $w$ is a
prefix of a Lyndon word in $\mathfrak{A}^{\ast}$.

\item \textit{Assertion $\mathcal{F}^{\prime\prime}$:} Let $m$ be an
object not in the alphabet $\mathfrak{A}$. Let us equip the set $\mathfrak{A}%
\cup\left\{  m\right\}  $ with a total order which extends the total order on
the alphabet $\mathfrak{A}$ and which satisfies $\left(  a<m\text{ for every
}a\in\mathfrak{A}\right)  $. Then, the word $wm\in\left(  \mathfrak{A}%
\cup\left\{  m\right\}  \right)  ^{\ast}$ (the concatenation of the word $w$
with the one-letter word $m$) is a Lyndon word.

\item \textit{Assertion $\mathcal{G}^{\prime}$:} There exists a
Lyndon word $t\in\mathfrak{A}^{\ast}$, a positive integer $\ell$ and a prefix
$p$ of $t$ (possibly empty) such that $w=t^{\ell}p$.

\item \textit{Assertion $\mathcal{H}^{\prime}$:} There exists a
Lyndon word $t\in\mathfrak{A}^{\ast}$, a nonnegative integer $\ell$ and a
prefix $p$ of $t$ (possibly empty) such that $w=t^{\ell}p$.

\item[(a)] Prove the equivalence $\mathcal{A}^{\prime}\Longleftrightarrow
\mathcal{D}^{\prime}$.

\item[(b)] Prove the equivalence $\mathcal{B}^{\prime}\Longleftrightarrow
\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime
}\Longleftrightarrow\mathcal{H}^{\prime}$.

\item[(c)] Prove the implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$.

\item[(d)] Prove the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$. (The implication $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$ is false, as witnessed by the word $11211$.)

\item[(e)] Prove that if there exists a letter $\mu\in\mathfrak{A}$ such that
$\left(  \mu>a\text{ for every letter }a\text{ of }w\right)  $, then the
equivalence $\mathcal{F}^{\prime}\Longleftrightarrow\mathcal{F}^{\prime\prime
}$ holds.

\item[(f)] Prove that if there exists a letter $\mu\in\mathfrak{A}$ such that
$\left(  \mu>a\text{ for some letter }a\text{ of }w\right)  $, then the
equivalence $\mathcal{F}^{\prime}\Longleftrightarrow\mathcal{F}^{\prime\prime
}$ holds.
\end{itemize}
\end{exercise}

The next exercise (based on work of Hazewinkel \cite{Hazewinkel-poset})
extends some of the above properties of Lyndon words (and words in general) to
a more general setting, in which the alphabet $\mathfrak{A}$ is no longer
required to be totally ordered, but only needs to be a poset:

\begin{exercise}
\label{exe.words.poset}In this exercise, we shall loosen the requirement that
the alphabet $\mathfrak{A}$ be a totally ordered set: Instead, we will only
require $\mathfrak{A}$ to be a poset. The resulting more general setting will
be called the
\emph{partial-order setting}\index{partial-order setting for words},
to distinguish it from the
\emph{total-order setting}\index{total-order setting for words}
in which $\mathfrak{A}$ is required to be a
totally ordered set. All results in Chapter \ref{sect.QSym.lyndon} so far
address the total-order setting. In this exercise, we will generalize some of
them to the partial-order setting.

All notions that we have defined in the total-order setting (the notion of a
word, the relation $\leq$, the notion of a Lyndon word, etc.) are defined in
precisely the same way in the partial-order setting. However, the poset
$\mathfrak{A}^{\ast}$ is no longer totally ordered in the partial-order setting.

\begin{enumerate}
\item[(a)] Prove that Proposition \ref{prop.words.lex} holds in the
partial-order setting, as long as one replaces ``a total
order'' by ``a partial order'' in part (a) of this Proposition.

\item[(b)] Prove (in the partial-order setting) that if $a,b,c,d\in
\mathfrak{A}^{\ast}$ are four words such that the words $ab$ and $cd$ are
comparable (with respect to the partial order $\leq$), then the words $a$ and
$c$ are comparable.

\item[(c)] Prove that Proposition \ref{prop.words.uv=vu.2}, Proposition
\ref{prop.words.uv=vu.3}, Corollary \ref{cor.words.uv=vu.3}, Corollary
\ref{cor.words.uv=vu.partord}, Exercise \ref{exe.words.u=pn=qm}, Exercise
\ref{exe.words.unvm.vmun}, Exercise \ref{exe.words.un.vm}, Exercise
\ref{exe.words.uv=vu.k}, Proposition \ref{prop.words.lyndon.trivia}, Corollary
\ref{cor.words.lyndon.trivia}, Proposition \ref{prop.words.lyndon.concat},
Corollary \ref{cor.words.lyndon.uv=vu}, Proposition
\ref{prop.words.lyndon.preorder}, Theorem \ref{thm.words.lyndon.equiv},
Exercise \ref{exe.words.lyndon.equiv2}(a), Exercise
\ref{exe.words.lyndon.concatn}, Exercise \ref{exe.words.lyndon.concatn-lyndon}%
, Exercise \ref{exe.words.lyndon.std2}(a) and Exercise
\ref{exe.words.lyndon.std2}(b) still hold in the partial-order setting.

\item[(d)] Find a counterexample to Exercise
\ref{exe.words.lyndon.power-criterion} in the partial-order setting.

\item[(e)] Salvage Exercise \ref{exe.words.lyndon.power-criterion} in the
partial-order setting (i.e., find a statement which is easily equivalent to
this exercise in the total-order setting, yet true in the partial-order setting).

\item[(f)] In the partial-order setting, a \dfn{Hazewinkel-CFL
factorization} of a word $w$ will mean a tuple $\left(  a_{1},a_{2}%
,\ldots,a_{k}\right)  $ of Lyndon words such that $w=a_{1}a_{2}\cdots a_{k}$
and such that no $i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies
$a_{i}<a_{i+1}$. Prove that every word $w$ has a unique Hazewinkel-CFL
factorization (in the partial-order setting).\footnote{This result, as well as
the validity of Proposition \ref{prop.words.lyndon.concat} in the
partial-order setting, are due to Hazewinkel \cite{Hazewinkel-poset}.}

\item[(g)] Prove that Exercise~\ref{exe.words.powerlyndon} still holds in the
partial-order setting.
\end{enumerate}
\end{exercise}

% [DG][v45] Added part (g) to this exercise.

The reader is invited to try extending other results to the partial-order
setting (it seems that no research has been done on this except for
Hazewinkel's \cite{Hazewinkel-poset}). We shall now, however, return to
the total-order setting (which has the most known applications).

Another extension of the notion of Lyndon words has been introduced
in 2018 by Dolce, Restivo and Reutenauer
\cite{DolceRestivoReutenauer-oglw}; it is based on a generalized version
of the lexicographic order, in which different letters are compared
differently depending on their positions in the word
(i.e., there is one total order for comparing
first letters, another for comparing second letters, etc.).

Lyndon words are related to various other objects in mathematics, such as
free Lie algebras (Subsection \ref{subsect.lyndon.freelie} below),
shuffles and shuffle algebras (Sections
\ref{subsect.lyndon.shuffles} and \ref{subsect.shuffle.radford} below),
$\Qsym$ (Sections \ref{subsect.shuffle.QSym.1} and
\ref{subsect.shuffle.QSym.2}), Markov chains on combinatorial Hopf algebras
(\cite{DiaconisPangRam}), de Bruijn sequences (\cite{FredricksenMaiorana},
\cite{Moreno}, \cite{Moreno-corr}, \cite[\S 7.2.11, Algorithm F]{Knuth-TAoCP4a}),
symmetric functions (specifically, the transition matrices
between the bases $\left(h_\lambda\right)_{\lambda \in \Par}$,
$\left(e_\lambda\right)_{\lambda \in \Par}$ and
$\left(m_\lambda\right)_{\lambda \in \Par}$; see
\cite{KulikauskasRemmel} for this),
and the Burrows-Wheeler algorithm for data compression
(see Remark~\ref{rmk.lyndon.gr.bw} below for a quick
idea, and \cite{CrochemoreDesarmenienPerrin},
\cite{GesselRestivoReutenauer}, \cite{Kufleitner} for more).
They are also connected to \emph{necklaces} (in the
combinatorial sense) -- a combinatorial object that also happens to be related
to a lot of algebra (\cite[Chapter 5]{Rota}, \cite{Dayton-ghosts}). Let us
survey the basics of this latter classical connection in an exercise:

% [DG][v52] Added reference to \cite{KulikauskasRemmel}, which seems
% very appropriate seeing how much space we have dedicated to
% symmetric functions. (No, I have not read that paper.)

\begin{exercise}
\label{exe.words.necklaces}
Let $\mathfrak{A}$ be any set (not necessarily
totally ordered).
Let \dfn{$C$} denote the infinite cyclic group, written multiplicatively.
Fix a generator \dfn{$c$} of $C$.\ \ \ \ \footnote{So $C$ is a
group isomorphic to $\left( \ZZ ,+\right)  $, and the isomorphism
$\left( \ZZ ,+\right)  \rightarrow C$ sends every $n\in \ZZ$ to
$c^n$. (Recall that we write the binary operation of $C$ as $\cdot$ instead
of $+$.)} Fix a positive integer $n$. The group $C$ acts on $\mathfrak{A}^{n}$
from the left according to the rule
\[
c\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{2},a_{3}%
,\ldots,a_{n},a_{1}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in \mathfrak{A}^{n}.
\]
\footnote{In other words, $c$ rotates any $n$-tuple of elements of
$\mathfrak{A}$ cyclically to the left. Thus, $c^{n}\in C$ acts trivially on
$\mathfrak{A}^{n}$, and so this action of $C$ on $\mathfrak{A}^{n}$ factors
through $C/\left\langle c^{n}\right\rangle $ (a cyclic group of order $n$).}
The orbits of this $C$-action will be called
\emph{$n$-necklaces}\index{$n$-necklace}\footnote{Classically, one
visualizes them as necklaces of $n$ beads of
$\left\vert \mathfrak{A} \right\vert$ colors. (The colors are the elements of
$\mathfrak{A}$.) For example, the necklace containing an $n$-tuple
$\left( w_1, w_2, \ldots, w_n \right)$ is visualized as follows:
\[
\xymatrix@R=2.0pc@C=.8pc{
& w_1 \ar@/^1ex/[rr] & & w_2 \ar@/^1ex/[dr] \\
w_n \ar@/^1ex/[ur] & & \;\; & & w_3 \ar@/^1ex/[dl] \\
& w_{n-1} \ar@/^1ex/[ul] & & \iddots \ar@/^1ex/[ll]
}
\]
with $w_1, w_2, \ldots, w_n$ being the colors of the respective beads.
The intuition behind this is that a necklace is an object
that doesn't really change when we rotate it in its plane. However, to make this
intuition match the definition, we need to think of a necklace as being stuck
in its (fixed) plane, so that we cannot lift it up and turn it around,
dropping it back to its plane in a reflected state.}; they form a set
partition of the set $\mathfrak{A}^{n}$.

The $n$-necklace containing a given $n$-tuple $w\in\mathfrak{A}^{n}$ will be
denoted by $\left[  w\right]  $.

\begin{enumerate}
\item[(a)] Prove that
every $n$-necklace $N$ is a finite nonempty set and satisfies
$\left\vert N\right\vert \mid n$. (Recall that $N$ is an orbit, thus a set; as
usual, $\left\vert N\right\vert $ denotes the cardinality of this set.)
\end{enumerate}

The \emph{period}\index{period of an $n$-necklace}
of an $n$-necklace $N$ is defined as the positive integer
$\left\vert N\right\vert $. (This $\left\vert N\right\vert $ is indeed a
positive integer, since $N$ is a finite nonempty set.)\footnote{For example,
the $6$-necklace $\left[ 232232 \right]$ -- or, visually,
\[
\xymatrix@R=2.0pc@C=.8pc{
& 2 \ar@/^1ex/[rr] & & 3 \ar@/^1ex/[dr] \\
2 \ar@/^1ex/[ur] & & \;\; & & 2 \ar@/^1ex/[dl] \\
& 3 \ar@/^1ex/[ul] & & 2 \ar@/^1ex/[ll]
}
\]
-- has period $3$, as it is a set of size $3$
(with elements $232232$, $322322$ and $223223$).
The word ``period'' hints at the geometric meaning: If an $n$-necklace $N$
is represented by coloring the vertices of a regular $n$-gon, then
its period is the smallest positive integer $d$ such that the colors
are preserved when the $n$-gon is rotated by $2\pi d/n$.}

An $n$-necklace is said to be
\emph{aperiodic}\index{aperiodic $n$-necklace} if its period is $n$.

\begin{enumerate}
\item[(b)] Given any $n$-tuple $w = \left(w_1, w_2, \ldots, w_n\right)
\in \mathfrak{A}^n$, prove that the $n$-necklace $\left[w\right]$ is
aperiodic if and only if every $k \in \left\{ 1, 2, \ldots, n-1 \right\}$
satisfies $\left(w_{k+1}, w_{k+2}, \ldots, w_n, w_1, w_2, \ldots, w_k\right)
\neq w$.
\end{enumerate}

From now on, we assume that the set $\mathfrak{A}$ is totally ordered. We use
$\mathfrak{A}$ as our alphabet to define the notions of words, the
lexicographic order, and Lyndon words. All notations that we introduced for
words will thus be used for elements of $\mathfrak{A}^{n}$.

\begin{enumerate}
\item[(c)] Prove that every aperiodic $n$-necklace contains exactly one Lyndon word.

\item[(d)] If $N$ is an $n$-necklace which is not aperiodic, then prove that
$N$ contains no Lyndon word.

\item[(e)] Show that the aperiodic $n$-necklaces are in bijection with Lyndon
words of length $n$.
\end{enumerate}

From now on, we assume that the set $\mathfrak{A}$ is finite. Define the
number-theoretic M\"{o}bius function $\mu$ and the Euler totient function
$\phi$ as in Exercise \ref{exe.witt.ghost-equiv}.

\begin{enumerate}
\item[(f)] Prove that the number of all aperiodic $n$-necklaces is%
\[
\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}.
\]


\item[(g)] Prove that the number of all $n$-necklaces is%
\[
\dfrac{1}{n}\sum_{d\mid n}\phi\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}.
\]


\item[(h)] Solve Exercise~\ref{exe.words.lyndon.count} again.

\item[(i)] Forget that we fixed $\mathfrak{A}$. Show that every $q\in
 \ZZ$ satisfies $n\mid\sum_{d\mid n}\mu\left(  d\right)  q^{n/d}$ and
$n\mid\sum_{d\mid n}\phi\left(  d\right)  q^{n/d}$.
\end{enumerate}

[\textbf{Hint:} For (c), use Theorem~\ref{thm.words.lyndon.equiv}. For (i),
either use parts (f) and (g) and a trick to extend to $q$ negative; or recall
Exercise~\ref{exe.witt.ghost-exa}.]
\end{exercise}

% [DG][v37] Added this exercise (and made some minor corrections to my
% section).

We will pick up the topic of necklaces again in
Section~\ref{subsect.lyndon.gr}, where we will connect it back
to symmetric functions.

\subsubsection{\label{subsect.lyndon.freelie}Free Lie algebras}

In this brief subsection, we shall review the connection between Lyndon words
and free Lie algebras (following \cite[Kap. 4]{Laue}, but avoiding the
generality of Hall sets in favor of just using Lyndon words). None of this
material shall be used in the rest of these notes. We will only prove some
basic results; for more thorough and comprehensive treatments of free Lie
algebras, see \cite{Reutenauer}, \cite[Chapter 2]{Bourbaki-Lie2-3-fr} and
\cite[Kap. 4]{Laue}.

We begin with some properties of Lyndon words.

\begin{exercise}
\label{exe.words.lyndon.stf.props1}Let $w\in\mathfrak{A}^{\ast}$ be a nonempty
word. Let $v$ be the longest Lyndon suffix of $w$\ \ \ \ \footnote{Of course,
a Lyndon suffix of $w$ just means a suffix $p$ of $w$ such that $p$ is
Lyndon.}. Let $t$ be a Lyndon word. Then, $t$ is the longest Lyndon suffix of
$wt$ if and only if we do not have $v<t$.
\end{exercise}

(We have written ``we do not have $v<t$'' instead of ``$v\geq t$'' in Exercise
\ref{exe.words.lyndon.stf.props1} for reasons of generalizability: This way,
Exercise \ref{exe.words.lyndon.stf.props1} generalizes to the partial-order
setting introduced in Exercise \ref{exe.words.poset}, whereas the version with
``$v\geq t$'' does not.)

\begin{exercise}
\label{exe.words.lyndon.stf.props2}Let $w\in\mathfrak{A}^{\ast}$ be a word of
length $>1$. Let $v$ be the longest Lyndon proper suffix of $w$%
\ \ \ \ \footnote{Of course, a Lyndon proper suffix of $w$ just means a proper
suffix $p$ of $w$ such that $p$ is Lyndon.}. Let $t$ be a Lyndon word. Then,
$t$ is the longest Lyndon proper suffix of $wt$ if and only if we do not have
$v<t$.
\end{exercise}

(Exercise \ref{exe.words.lyndon.stf.props2}, while being a trivial consequence
of Exercise \ref{exe.words.lyndon.stf.props1}, is rather useful in the study
of free Lie algebras. It generalizes both \cite[Lemma (1.6)]{ChenFoxLyndon}
(which is obtained by taking $w=c$, $v=b$ and $t=d$) and \cite[Proposition
5.1.4]{Lothaire} (which is obtained by taking $v=m$ and $t=n$).)

\begin{definition}
For the rest of Subsection \ref{subsect.lyndon.freelie}, we let
\dfn{$\mathfrak{L}$} be the
set of all Lyndon words (over the alphabet $\mathfrak{A}$).
\end{definition}

\begin{definition}
\label{def.words.lyndon.stf}Let $w$ be a Lyndon word of length $>1$. Let $v$
be the longest proper suffix of $w$ such that $v$ is Lyndon. (This is
well-defined, as we know from Exercise \ref{exe.words.lyndon.std2}.) Since $v$
is a proper suffix of $w$, there exists a nonempty $u\in\mathfrak{A}^{\ast}$
such that $w=uv$. Consider this $u$. (Clearly, this $u$ is unique.) Theorem
\ref{thm.words.lyndon.std}(a) shows that the words $u$ and $v$ are Lyndon. In
other words, $u\in\mathfrak{L}$ and $v\in\mathfrak{L}$. Hence, $\left(
u,v\right)  \in\mathfrak{L}\times\mathfrak{L}$. The pair $\left(  u,v\right)
\in\mathfrak{L}\times\mathfrak{L}$ is called the \dfn{standard
factorization} of $w$, and is denoted by \dfn{$\operatorname*{stf}w$}.
\end{definition}

For the sake of easier reference, we gather a few basic properties of the
standard factorization:

\begin{exercise}
\label{exe.words.lyndon.stf.basic}Let $w$ be a Lyndon word of length $>1$. Let
$\left(  g,h\right)  =\operatorname*{stf}w$. Prove the following:

\begin{enumerate}
\item[(a)] The word $h$ is the longest Lyndon proper suffix of $w$.

\item[(b)] We have $w=gh$.

\item[(c)] We have $g<gh<h$.

\item[(d)] The word $g$ is Lyndon.

\item[(e)] We have $g\in\mathfrak{L}$, $h\in\mathfrak{L}$, $\ell\left(
g\right)  <\ell\left(  w\right)  $ and $\ell\left(  h\right)  <\ell\left(
w\right)  $.

\item[(f)] Let $t$ be a Lyndon word. Then, $t$ is the longest Lyndon proper
suffix of $wt$ if and only if we do not have $h<t$.
\end{enumerate}
\end{exercise}

\begin{exercise}
\label{exe.freelie.1}Let $\mathfrak{g}$ be a Lie algebra. For every Lyndon
word $w$, let $b_{w}$ be an element of $\mathfrak{g}$. Assume that for every
Lyndon word $w$ of length $>1$, we have%
\begin{equation}
b_{w}=\left[  b_{u},b_{v}\right]  ,\ \ \ \ \ \ \ \ \ \ \text{where }\left(
u,v\right)  =\operatorname*{stf}w. \label{eq.exe.freelie.1.bw}%
\end{equation}


Let $B$ be the $\kk$-submodule of $\mathfrak{g}$ spanned by the family
$\left(  b_{w}\right)  _{w\in\mathfrak{L}}$.

\begin{enumerate}
\item[(a)] Prove that $B$ is a Lie subalgebra of $\mathfrak{g}$.

\item[(b)] Let $\mathfrak{h}$ be a $\kk$-Lie algebra. Let
$f:B\rightarrow\mathfrak{h}$ be a $\kk$-module homomorphism. Assume
that whenever $w$ is a Lyndon word of length $>1$, we have%
\begin{equation}
f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(  b_{u}\right)
,f\left(  b_{v}\right)  \right]  ,\ \ \ \ \ \ \ \ \ \ \text{where }\left(
u,v\right)  =\operatorname*{stf}w. \label{eq.exe.freelie.1.bcond}%
\end{equation}
Prove that $f$ is a Lie algebra homomorphism.
\end{enumerate}

[\textbf{Hint:} Given two words $w$ and $w^{\prime}$, write $w\sim w^{\prime}$
if and only if $w^{\prime}$ is a permutation of $w$. Part (a) follows from the
fact that for any $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$
satisfying $p<q$, we have $\left[  b_{p},b_{q}\right]  \in B_{pq,q}$, where
$B_{h,s}$ denotes the $\kk$-linear span of $\left\{  b_{w}\ \mid
\ w\in\mathfrak{L}\text{, }w\sim h\text{ and }w<s\right\}  $ for any two words
$h$ and $s$. Prove this fact by a double induction, first inducting over
$\ell\left(  pq\right)  $, and then (for fixed $\ell\left(  pq\right)  $)
inducting over the rank of $q$ in lexicographic order (i.e., assume that the
fact is already proven for every $q^{\prime}<q$ instead of $q$). In the
induction step, assume that $\left(  p,q\right)  \neq\operatorname*{stf}%
\left(  pq\right)  $ (since otherwise the claim is rather obvious) and
conclude that $p$ has length $>1$; thus, set $\left(  u,v\right)
=\operatorname*{stf}p$, so that $\left[  \underbrace{b_{p}}_{=\left[
b_{u},b_{v}\right]  },b_{q}\right]  =\left[  \left[  b_{u},b_{v}\right]
,b_{q}\right]  =\left[  \left[  b_{u},b_{q}\right]  ,b_{v}\right]  -\left[
\left[  b_{v},b_{q}\right]  ,b_{u}\right]  $, and use Exercise
\ref{exe.words.lyndon.stf.props2} to obtain $v<q$.

The proof of (b) proceeds by a similar induction, piggybacking on the $\left[
b_{p},b_{q}\right]  \in B_{pq,q}$ claim.]
\end{exercise}

\begin{exercise}
\label{exe.freelie.2}Let $V$ be the free $\kk$-module with basis
$\left(  x_{a}\right)  _{a\in\mathfrak{A}}$. For every word $w\in
\mathfrak{A}^{\ast}$, let $x_{w}$ be the tensor $x_{w_{1}}\otimes x_{w_{2}%
}\otimes\cdots\otimes x_{w_{\ell\left(  w\right)  }}$. As we know from Example
\ref{exa.tensor-alg}, the tensor algebra $T\left(  V\right)  $ is a free
$\kk$-module with basis $\left(  x_{w}\right)  _{w\in\mathfrak{A}%
^{\ast}}$. We regard $V$ as a $\kk$-submodule of $T\left(  V\right)  $.

The tensor algebra $T\left(  V\right)  $ becomes a Lie algebra via the
commutator (i.e., its Lie bracket is defined by $\left[  \alpha,\beta\right]
=\alpha\beta-\beta\alpha$ for all $\alpha\in T\left(  V\right)  $ and
$\beta\in T\left(  V\right)  $).

We define a sequence $\left(  \mathfrak{g}_{1},\mathfrak{g}_{2},\mathfrak{g}%
_{3},\ldots\right)  $ of $\kk$-submodules of $T\left(  V\right)  $ as
follows: Recursively, we set $\mathfrak{g}_{1}=V$, and for every $i\in\left\{
2,3,4,\ldots\right\}  $, we set $\mathfrak{g}_{i}=\left[  V,\mathfrak{g}%
_{i-1}\right]  $. Let $\mathfrak{g}$ be the $\kk$-submodule
$\mathfrak{g}_{1}+\mathfrak{g}_{2}+\mathfrak{g}_{3}+\cdots$ of $T\left(
V\right)  $.

Prove the following:

\begin{enumerate}
\item[(a)] The $\kk$-submodule $\mathfrak{g}$ is a Lie subalgebra of
$T\left(  V\right)  $.

\item[(b)] If $\mathfrak{k}$ is any Lie subalgebra of $T\left(  V\right)  $
satisfying $V\subset\mathfrak{k}$, then $\mathfrak{g}\subset\mathfrak{k}$.
\end{enumerate}

Now, for every $w\in\mathfrak{L}$, we define an element $b_{w}$ of $T\left(
V\right)  $ as follows: We define $b_{w}$ by recursion on the length of $w$.
If the length of $w$ is $1$\ \ \ \ \footnote{The length of any $w\in
\mathfrak{L}$ must be at least $1$. (Indeed, if $w\in\mathfrak{L}$, then the
word $w$ is Lyndon and thus nonempty, and hence its length must be at least
$1$.)}, then we have $w=\left(  a\right)  $ for some letter $a\in\mathfrak{A}%
$, and we set $b_{w}=x_{a}$ for this letter $a$. If the length of $w$ is $>1$,
then we set $b_{w}=\left[  b_{u},b_{v}\right]  $, where $\left(  u,v\right)
=\operatorname*{stf}w$\ \ \ \ \footnote{This is well-defined, because $b_{u}$
and $b_{v}$ have already been defined. [\textit{Proof.} Let $\left(
u,v\right)  =\operatorname*{stf}w$. Then, Exercise
\ref{exe.words.lyndon.stf.basic}(e) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that $u\in\mathfrak{L}$, $v\in\mathfrak{L}$, $\ell\left(
u\right)  <\ell\left(  w\right)  $ and $\ell\left(  v\right)  <\ell\left(
w\right)  $. Recall that we are defining $b_{w}$ by recursion on the length of
$w$. Hence, $b_{p}$ is already defined for every $p\in\mathfrak{L}$ satisfying
$\ell\left(  p\right)  <\ell\left(  w\right)  $. Applying this to $p=u$, we
see that $b_{u}$ is already defined (since $u\in\mathfrak{L}$ and $\ell\left(
u\right)  <\ell\left(  w\right)  $). The same argument (but applied to $v$
instead of $u$) shows that $b_{v}$ is already defined. Hence, $b_{u}$ and
$b_{v}$ have already been defined. Thus, $b_{w}$ is well-defined by
$b_{w}=\left[  b_{u},b_{v}\right]  $, qed.]}.

Prove the following:

\begin{enumerate}
\item[(c)] For every $w\in\mathfrak{L}$, we have%
\[
b_{w}\in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}.
\]


\item[(d)] The family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is a basis
of the $\kk$-module $\mathfrak{g}$.

\item[(e)] Let $\mathfrak{h}$ be any $\kk$-Lie algebra. Let
$\xi:\mathfrak{A}\rightarrow\mathfrak{h}$ be any map. Then, there exists a
unique Lie algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such
that every $a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(
a\right)  $.
\end{enumerate}
\end{exercise}

\begin{remark}
\label{rmk.freelie.2.rmk}Let $V$ and $\mathfrak{g}$ be as in Exercise
\ref{exe.freelie.2}. In the language of universal algebra, the statement of
Exercise \ref{exe.freelie.2}(e) says that $\mathfrak{g}$ (or, to be more
precise, the pair $\left(  \mathfrak{g},f\right)  $, where $f:\mathfrak{A}%
\rightarrow\mathfrak{g}$ is the map sending each $a\in\mathfrak{A}$ to
$x_{a}\in\mathfrak{g}$) satisfies the universal property of the free Lie
algebra on the set $\mathfrak{A}$. Thus, this exercise allows us to call
$\mathfrak{g}$ the \dfn{free Lie algebra} on $\mathfrak{A}$. Most authors
define the free Lie algebra differently, but all reasonable definitions of a
free Lie algebra\footnote{Here, we call a definition ``reasonable'' if the
``free Lie algebra'' it defines satisfies the universal property.} lead
to isomorphic Lie algebras (because the universal property determines the free
Lie algebra uniquely up to canonical isomorphism).

Notice that the Lie algebra $\mathfrak{g}$ does not depend on the total order
on the alphabet $\mathfrak{A}$, but the basis $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ constructed in Exercise \ref{exe.freelie.2}(d) does.
There is no known basis of $\mathfrak{g}$ defined without ordering
$\mathfrak{A}$.

It is worth noticing that our construction of $\mathfrak{g}$ proves not only
that the free Lie algebra on $\mathfrak{A}$ exists, but also that this free
Lie algebra can be realized as a Lie subalgebra of the (associative) algebra
$T\left(  V\right)  $. Therefore, if we want to prove that a certain identity
holds in every Lie algebra, we only need to check that this identity holds in
every associative algebra (if all Lie brackets are replaced by commutators);
the universal property of the free Lie algebra (i.e., Exercise
\ref{exe.freelie.2}(e)) will then ensure that this identity also holds in
every Lie algebra $\mathfrak{h}$.

There is much more to say about free Lie algebras than what we have said
here; in particular, there are connections to symmetric functions,
necklaces, representations of symmetric groups and $\Nsym$. See
\cite[\S 5.3]{Lothaire}, \cite{Reutenauer},
\cite[Chapter 2]{Bourbaki-Lie2-3-fr}, \cite[\S 4]{Laue} and
\cite{BlessenohlLaue} for further developments\footnote{The claim made
in \cite[page 2]{BlessenohlLaue} that ``$\left\{x_1, \ldots, x_n\right\}$
generates freely a Lie subalgebra of $A_R$'' is essentially our
Exercise~\ref{exe.freelie.2}(e).}.
\end{remark}

% [DG][v54] Added this subsection on free Lie algebras.

\subsection{\label{subsect.lyndon.shuffles}Shuffles and Lyndon words}

We will now connect the theory of Lyndon words with the notion of shuffle
products. We have already introduced the latter notion in Definition
\ref{shuffles}, but we will now study it more closely and introduce some more
convenient notations (e.g., we will need a notation for single shuffles, not
just the whole multiset).\footnote{Parts (a) and (c) of the below Definition
\ref{def.shuffle.sigma} define notions which have already been introduced in
Definition \ref{shuffles}. Of course, the definitions of these notions are
equivalent; however, the variables are differently labelled in the two
definitions (for example, the variables $u$, $v$, $w$ and $\sigma$ of
Definition \ref{def.shuffle.sigma}(c) correspond to the variables $a$, $b$,
$c$ and $w$ of Definition \ref{shuffles}). The labels in Definition
\ref{def.shuffle.sigma} have been chosen to match with the rest of Section
\ref{subsect.lyndon.shuffles}.}

\begin{definition}
\phantomsection
\label{def.shuffle.sigma}

\begin{itemize}
\item[(a)] Let $n\in \NN$ and $m\in \NN$. Then,
\dfn{$\operatorname{Sh}_{n,m}$} denotes the subset
\[
\left\{  \sigma\in\Symm_{n+m}\ :\ \sigma^{-1}\left(  1\right)
<\sigma^{-1}\left(  2\right)  <\cdots<\sigma^{-1}\left(  n\right)
;\ \sigma^{-1}\left(  n+1\right)  <\sigma^{-1}\left(  n+2\right)
<\cdots<\sigma^{-1}\left(  n+m\right)  \right\}
\]
of the symmetric group $\Symm_{n+m}$.

\item[(b)] Let $u=\left(  u_{1},u_{2},\ldots,u_{n}\right)  $ and $v=\left(
v_{1},v_{2},\ldots,v_{m}\right)  $ be two words. If $\sigma\in
\operatorname{Sh}_{n,m}$, then, $u\underset{\sigma}{\shuffle}v$
will denote
the word $\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)
},\ldots,w_{\sigma\left(  n+m\right)  }\right)  $, where $\left(  w_{1}%
,w_{2},\ldots,w_{n+m}\right)  $ is the concatenation $u\cdot v=\left(
u_{1},u_{2},\ldots,u_{n},v_{1},v_{2},\ldots,v_{m}\right)  $. We notice that
the multiset of all letters of $u\underset{\sigma}{\shuffle}v$ is the disjoint
union of the multiset of all letters of $u$ with the multiset of all letters
of $v$. As a consequence, $\ell\left(  u\underset{\sigma}{\shuffle}v\right)
=\ell\left(  u\right)  +\ell\left(  v\right)  $.

\item[(c)] Let $u=\left(  u_{1},u_{2},\ldots,u_{n}\right)  $ and $v=\left(
v_{1},v_{2},\ldots,v_{m}\right)  $ be two words. The \dfn{multiset of
shuffles of $u$ and $v$}\index{shuffle of words}
is defined as the multiset $\left\{  \left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  n+m\right)  }\right)  \ :\ \sigma\in
\operatorname{Sh}_{n,m}\right\}  _{\text{multiset}}$, where $\left(  w_{1}%
,w_{2},\ldots,w_{n+m}\right)  $ is the concatenation $u\cdot v=\left(
u_{1},u_{2},\ldots,u_{n},v_{1},v_{2},\ldots,v_{m}\right)  $. In other words,
the multiset of shuffles of $u$ and $v$ is the multiset
\[
\left\{  u\underset{\sigma}{\shuffle}v\ :\ \sigma\in\operatorname{Sh}%
_{n,m}\right\}  _{\text{multiset}}.
\]
It is denoted by \dfn{$u\shuffle v$}.
\end{itemize}
\end{definition}

The next fact provides the main connection between Lyndon words and shuffles:

\begin{theorem}
\label{thm.shuffle.lyndon.used}Let $u$ and $v$ be two words.

Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
$u$. Let $\left(  b_{1},b_{2},\ldots,b_{q}\right)  $ be the CFL factorization
of $v$.

\begin{itemize}
\item[(a)] Let $\left(  c_{1},c_{2},\ldots,c_{p+q}\right)  $ be the result of
sorting the list $\left(  a_{1},a_{2},\ldots,a_{p},b_{1},b_{2},\ldots
,b_{q}\right)  $ in decreasing order\footnote{with respect to the total order
on $\mathfrak{A}^{\ast}$ whose greater-or-equal relation is $\geq$}. Then, the
lexicographically highest element of the multiset $u\shuffle v$ is $c_{1}c_{2}
\cdots c_{p+q}$ (and $\left(  c_{1},c_{2},\ldots,c_{p+q}\right)  $ is the CFL
factorization of this element).

\item[(b)] Let $\mathfrak{L}$ denote the set of all Lyndon words. If $w$ is a
Lyndon word and $z$ is any word, let $\operatorname{mult}_{w}z$
denote the number of terms in the CFL factorization of $z$ which are equal to
$w$. The multiplicity with which the lexicographically highest element of the
multiset $u\shuffle v$ appears in the multiset $u\shuffle v$ is
$\prod_{w\in\mathfrak{L}}
  \dbinom{\operatorname{mult}_{w} u + \operatorname{mult}_{w}v}
         {\operatorname{mult}_{w}u}$.
(This product is well-defined because almost all of its factors are $1$.)

\item[(c)] If $a_{i}\geq b_{j}$ for every $i\in\left\{  1,2,\ldots,p\right\}
$ and $j\in\left\{  1,2,\ldots,q\right\}  $, then the lexicographically
highest element of the multiset $u\shuffle v$ is $uv$.

\item[(d)] If $a_{i} > b_{j}$ for every $i\in\left\{  1,2,\ldots,p\right\}  $
and $j\in\left\{  1,2,\ldots,q\right\}  $, then the multiplicity with which
the word $uv$ appears in the multiset $u\shuffle v$ is $1$.

\item[(e)] Assume that $u$ is a Lyndon word. Also, assume that $u\geq b_{j}$
for every $j\in\left\{  1,2,\ldots,q\right\}  $. Then, the lexicographically
highest element of the multiset $u\shuffle v$ is $uv$, and the multiplicity
with which this word $uv$ appears in the multiset $u\shuffle v$ is
$\operatorname{mult}_{u}v+1$.
\end{itemize}
\end{theorem}

\begin{comment}
Here is an old (equivalent, but messy) version of the statement of
Theorem~\ref{thm.shuffle.lyndon.used}(a):
(a) Let $C$ be the disjoint union of the multisets $\left\{  a_{1}%
,a_{2},\ldots,a_{p}\right\}  _{\text{multiset}}$ and $\left\{  b_{1}%
,b_{2},\ldots,b_{q}\right\}  _{\text{multiset}}$ (where ``disjoint union''
means that the multiplicities of each
element are added). Let $c_{1},c_{2},\ldots,c_{p+q}$ be the elements of $C$
listed in decreasing order (as often as they appear in $C$). Then, the
lexicographically highest element of the multiset $u\shuffle v$ is $c_{1}%
c_{2} \cdots c_{p+q}$ (and $\left(  c_{1},c_{2},\ldots,c_{p+q}\right)  $ is the CFL
factorization of this element).
\end{comment}


\begin{example}
For this example, let $u$ and $v$ be the words $u=23232$ and $v=323221$ over
the alphabet $\mathfrak{A}=\left\{  1,2,3, \ldots\right\}  $ with total order
given by $1<2<3< \cdots$. The CFL factorizations of $u$ and $v$ are $\left(
23,23,2\right)  $ and $\left(  3,23,2,2,1\right)  $, respectively. Thus, using
the notations of Theorem \ref{thm.shuffle.lyndon.used}, we have $p=3$,
$\left(  a_{1},a_{2},\ldots,a_{p}\right)  =\left(  23,23,2\right)  $, $q=5$
and $\left(  b_{1},b_{2},\ldots,b_{q}\right)  =\left(  3,23,2,2,1\right)  $.
Thus, Theorem \ref{thm.shuffle.lyndon.used}(a) predicts that the
lexicographically highest element of the multiset $u\shuffle v$ is $c_{1}%
c_{2}c_{3}c_{4}c_{5}c_{6}c_{7}c_{8}$, where $c_{1},c_{2},c_{3},c_{4}%
,c_{5},c_{6},c_{7},c_{8}$ are the words $23,23,2,3,23,2,2,1$ listed in
decreasing order (in other words, $\left(  c_{1},c_{2},c_{3},c_{4},c_{5}%
,c_{6},c_{7},c_{8}\right)  =\left(  3,23,23,23,2,2,2,1\right)  $). In other
words, Theorem \ref{thm.shuffle.lyndon.used}(a) predicts that the
lexicographically highest element of the multiset $u\shuffle v$ is
$32323232221$. We could verify this by brute force, but this would be
laborious since the multiset $u\shuffle v$ has $\dbinom{5+6}{5}=462$ elements
(with multiplicities). Theorem \ref{thm.shuffle.lyndon.used}(b) predicts that
this lexicographically highest element $32323232221$ appears in the multiset
$u\shuffle v$ with a multiplicity of $\prod_{w\in\mathfrak{L}}\dbinom
{\operatorname{mult}_{w}u+\operatorname{mult}_{w}%
v}{\operatorname{mult}_{w}u}$. This product $\prod_{w\in
\mathfrak{L}}\dbinom{\operatorname{mult}_{w}u+
\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}$ is infinite, but all but
finitely many of its factors are $1$ and therefore can be omitted; the only
factors which are not $1$ are those corresponding to Lyndon words $w$ which
appear both in the CFL factorization of $u$ and in the CFL factorization of
$v$ (since for any other factor, at least one of the numbers
$\operatorname{mult}_{w}u$ or $\operatorname{mult}_{w}v$
equals $0$, and therefore the binomial coefficient $\dbinom
{\operatorname{mult}_{w}u+\operatorname{mult}_{w}%
v}{\operatorname{mult}_{w}u}$ equals $1$). Thus, in order to compute
the product $\prod_{w\in\mathfrak{L}}\dbinom{
\operatorname{mult}_{w}u+\operatorname{mult}_{w}v}{
\operatorname{mult}_{w}u}$, we only need to multiply these factors. In our example,
these are the factors for $w=23$ and for $w=2$ (these are the only Lyndon
words which appear both in the CFL factorization $\left(  23,23,2\right)  $ of
$u$ and in the CFL factorization $\left(  3,23,2,2,1\right)  $ of $v$). So we
have%
\[
\prod_{w\in\mathfrak{L}}\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}%
u}=\underbrace{\dbinom{\operatorname{mult}_{23}%
u+\operatorname{mult}_{23}v}{\operatorname{mult}_{23}u}%
}_{=\dbinom{2+1}{2}=3}\underbrace{\dbinom{\operatorname{mult}_{2}%
u+\operatorname{mult}_{2}v}{\operatorname{mult}_{2}u}%
}_{=\dbinom{1+2}{1}=3}=3\cdot3=9.
\]
The word $32323232221$ must thus appear in the multiset $u\shuffle v$ with a
multiplicity of $9$. This, too, could be checked by brute force.
\end{example}

Theorem \ref{thm.shuffle.lyndon.used} (and Theorem
\ref{thm.shuffle.lyndon.full} further below, which describes more precisely
how the lexicographically highest element of $u\shuffle v$ emerges by
shuffling $u$ and $v$) is fairly close to \cite[Theorem 2.2.2]%
{Radford-shuffle} (and will be used for the same purposes), the main
difference being that we are talking about the shuffle product of two (not
necessarily Lyndon) words, while Radford (and most other authors) study the
shuffle product of many Lyndon words.

In order to prove Theorem \ref{thm.shuffle.lyndon.used}, we will need to make
some stronger statements, for which we first have to introduce some more notation:

\begin{definition}
\begin{itemize}


\item[(a)] If $p$ and $q$ are two integers, then $\left[  p:q\right]  ^{+}$
denotes the interval $\left\{  p+1,p+2,\ldots,q\right\}  $ of $\ZZ$.
Note that $\left\vert \left[  p:q\right]  ^{+}\right\vert =q-p$ if $q\geq p$.

\item[(b)] If $I$ and $J$ are two nonempty intervals of $\ZZ$, then we
say that $I<J$ if and only if every $i\in I$ and $j\in J$ satisfy $i<j$. This
defines a partial order on the set of nonempty intervals of $\ZZ$.
(Roughly speaking, $I<J$ if the interval $I$ ends before $J$ begins.)

\item[(c)] If $w$ is a word with $n$ letters (for some $n\in \NN$), and
$I$ is an interval of $\ZZ$ such that $I\subset\left[  0:n\right]
^{+}$, then $w\left[  I\right]  $ will denote the word $\left(  w_{p+1}%
,w_{p+2},\ldots,w_{q}\right)  $, where $I$ is written in the form $I=\left[
p:q\right]  ^{+}$ with $q \geq p$. Obviously, $\ell\left(  w\left[  I\right]
\right)  =\left\vert I\right\vert =q-p$. A word of the form $w\left[
I\right]  $ for an interval $I\subset\left[  0:n\right]  ^{+}$ (equivalently,
a word which is a prefix of a suffix of $w$) is called a
\emph{factor}\index{factor of a word} of $w$.

\item[(d)] Let $\alpha$ be a composition. Then, we define a tuple
\dfn{$\operatorname*{intsys}\alpha$}
of intervals of $\ZZ$ as follows: Write
$\alpha$ in the form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $ (so that $\ell= \ell\left(  \alpha\right)  $). Then, set
$\operatorname*{intsys}\alpha=\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $,
where
\[
I_{i}=\left[  \sum_{k=1}^{i-1}\alpha_{k}:\sum_{k=1}^{i}\alpha_{k}\right]
^{+}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,\ell\right\}
.
\]
This $\ell$-tuple $\operatorname*{intsys}\alpha$ is a tuple of nonempty
intervals of $\ZZ$. This tuple $\operatorname{intsys}\alpha$ is called
the \dfn{interval system corresponding to $\alpha$}. (This is precisely the
$\ell$-tuple $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $ constructed in
Definition \ref{def.compositions.intervaldec}.) The length of the tuple
$\operatorname*{intsys}\alpha$ is $\ell\left(  \alpha\right)  $.
\end{itemize}
\end{definition}

\begin{example}
\begin{itemize}


\item[(a)] We have $\left[  2:4\right]  ^{+}=\left\{  3,4\right\}  $ and
$\left[  3:3\right]  ^{+}=\varnothing$.

\item[(b)] We have $\left[  2:4\right]  ^{+}<\left[  4:5\right]  ^{+}<\left[
6:8\right]  ^{+}$, but we have neither $\left[  2:4\right]  ^{+}<\left[
3:5\right]  ^{+}$ nor $\left[  3:5\right]  ^{+}<\left[  2:4\right]  ^{+}$.

\item[(c)] If $w$ is the word $915352$, then $w\left[  \left[  0:3\right]
^{+}\right]  =\left(  w_{1},w_{2},w_{3}\right)  =915$ and $w\left[  \left[
2:4\right]  ^{+}\right]  =\left(  w_{3},w_{4}\right)  =53$.

\item[(d)] If $\alpha$ is the composition $\left(  4,1,4,2,3\right)  $, then
the interval system corresponding to $\alpha$ is
\begin{align*}
\operatorname*{intsys}\alpha &  =\left(  \left[  0:4\right]  ^{+},\left[
4:5\right]  ^{+},\left[  5:9\right]  ^{+},\left[  9:11\right]  ^{+},\left[
11:14\right]  ^{+}\right) \\
&  =\left(  \left\{  1,2,3,4\right\}  ,\left\{  5\right\}  ,\left\{
6,7,8,9\right\}  ,\left\{  10,11\right\}  ,\left\{  12,13,14\right\}  \right)
.
\end{align*}

\end{itemize}
\end{example}

The following properties of the notions introduced in the preceding definition
are easy to check:

\begin{remark}
\phantomsection
\label{rmk.intsys.trivia}

\begin{itemize}
\item[(a)] If $I$ and $J$ are two nonempty intervals of $\ZZ$
satisfying $I < J$, then $I$ and $J$ are disjoint.

\item[(b)] If $I$ and $J$ are two disjoint nonempty intervals of
$\ZZ$, then either $I < J$ or $J < I$.

\item[(c)] Let $\alpha$ be a composition. Write $\alpha$ in the form $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ (so that $\ell=
\ell\left(  \alpha\right)  $). The interval system $\operatorname{intsys}%
\alpha$ can be described as the unique $\ell$-tuple $\left(  I_{1}%
,I_{2},\ldots,I_{\ell}\right)  $ of nonempty intervals of $\ZZ$
satisfying the following three properties:

\begin{itemize}
\item The intervals $I_{1}$, $I_{2}$, $\ldots$, $I_{\ell}$ form a set
partition of the set $\left[  0:n\right]  ^{+}$, where $n=\left\vert
\alpha\right\vert $.

\item We have $I_{1}<I_{2}< \cdots<I_{\ell}$.

\item We have $\left\vert I_{i}\right\vert =\alpha_{i}$ for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $.
\end{itemize}
\end{itemize}
\end{remark}

\begin{exercise}
\label{exe.rmk.intsys.trivia}Prove Remark \ref{rmk.intsys.trivia}.
\end{exercise}

The following two lemmas are collections of more or less trivial consequences
of what it means to be an element of $\operatorname{Sh}_{n,m}$ and
what it means to be a shuffle:

\begin{lemma}
\label{lem.shuffle.lyndon.easystep}Let $n\in \NN$ and $m\in \NN$.
Let $\sigma\in\operatorname{Sh}_{n,m}$.

\begin{itemize}
\item[(a)] If $I$ is an interval of $\ZZ$ such that $I\subset\left[
0:n+m\right]  ^{+}$, then $\sigma\left(  I\right)  \cap\left[  0:n\right]
^{+}$ and $\sigma\left(  I\right)  \cap\left[  n:n+m\right]  ^{+}$ are intervals.

\item[(b)] Let $K$ and $L$ be nonempty intervals of $\ZZ$ such that
$K\subset\left[  0:n\right]  ^{+}$ and $L\subset\left[  0:n\right]  ^{+}$ and
$K<L$ and such that $K\cup L$ is an interval. Assume that $\sigma^{-1}\left(
K\right)  $ and $\sigma^{-1}\left(  L\right)  $ are intervals, but
$\sigma^{-1}\left(  K\right)  \cup\sigma^{-1}\left(  L\right)  $ is not an
interval. Then, there exists a nonempty interval $P\subset\left[
n:n+m\right]  ^{+}$ such that $\sigma^{-1}\left(  P\right)  $, $\sigma
^{-1}\left(  K\right)  \cup\sigma^{-1}\left(  P\right)  $ and $\sigma
^{-1}\left(  P\right)  \cup\sigma^{-1}\left(  L\right)  $ are intervals and
such that $\sigma^{-1}\left(  K\right)  <\sigma^{-1}\left(  P\right)
<\sigma^{-1}\left(  L\right)  $.

\item[(c)] Lemma \ref{lem.shuffle.lyndon.easystep}(b) remains valid if
``$K\subset\left[  0:n\right]  ^{+}$ and
$L\subset\left[  0:n\right]  ^{+}$''
and ``$P\subset\left[ n:n+m\right]  ^{+}$''
are replaced by ``$K\subset\left[  n:n+m\right]  ^{+}$ and
$L\subset\left[  n:n+m\right]  ^{+}$'' and
``$P\subset\left[  0:n\right]  ^{+}$'', respectively.
\end{itemize}
\end{lemma}

\begin{exercise}
\label{exe.lem.shuffle.lyndon.easystep}Prove Lemma
\ref{lem.shuffle.lyndon.easystep}.
\end{exercise}

\begin{lemma}
\label{lem.shuffle.lyndon.step}Let $u$ and $v$ be two words. Let
$n=\ell\left(  u\right)  $ and $m=\ell\left(  v\right)  $. Let $\sigma
\in\operatorname{Sh}_{n,m}$.

\begin{itemize}
\item[(a)] If $I$ is an interval of $\ZZ$ satisfying either
$I\subset\left[  0:n\right]  ^{+}$ or $I\subset\left[  n:n+m\right]  ^{+}$,
and if $\sigma^{-1}\left(  I\right)  $ is an interval, then%
\begin{equation}
\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  \sigma^{-1}\left(
I\right)  \right]  =\left(  uv\right)  \left[  I\right]  .
\label{pf.lem.shuffle.lyndon.step.gamma}%
\end{equation}


\item[(b)] Assume that $u\underset{\sigma}{\shuffle}v$ is the
lexicographically highest element of the multiset $u\shuffle v$. Let
$I\subset\left[  0:n\right]  ^{+}$ and $J\subset\left[  n:n+m\right]  ^{+}$ be
two nonempty intervals. Assume that $\sigma^{-1}\left(  I\right)  $ and
$\sigma^{-1}\left(  J\right)  $ are also intervals, that $\sigma^{-1}\left(
I\right)  <\sigma^{-1}\left(  J\right)  $, and that $\sigma^{-1}\left(
I\right)  \cup\sigma^{-1}\left(  J\right)  $ is an interval as well. Then,
$\left(  uv\right)  \left[  I\right]  \cdot\left(  uv\right)  \left[
J\right]  \geq\left(  uv\right)  \left[  J\right]  \cdot\left(  uv\right)
\left[  I\right]  $.

\item[(c)] Lemma \ref{lem.shuffle.lyndon.step}(b) remains valid if
``$I\subset\left[  0:n\right]  ^{+}$ and $J\subset\left[
n:n+m\right]  ^{+}$'' is replaced by
``$I\subset\left[  n:n+m\right]  ^{+}$ and
$J\subset\left[  0:n\right]  ^{+}$''.
\end{itemize}
\end{lemma}

\begin{exercise}
\label{exe.lem.shuffle.lyndon.step}Prove Lemma \ref{lem.shuffle.lyndon.step}.

[\textbf{Hint:} For (b), show that there exists a $\tau\in
\operatorname{Sh}_{n,m}$ such that $u\underset{\tau}{\shuffle}v$ differs from
$u\underset{\sigma}{\shuffle}v$ only in the order of the subwords $\left(
uv\right)  \left[  I\right]  $ and $\left(  uv\right)  \left[  J\right]  $.]
\end{exercise}

We are still a few steps away from stating our results in a way that allows
comfortably proving Theorem \ref{thm.shuffle.lyndon.used}. For the latter aim,
we introduce the notion of \emph{$\alpha$-clumping permutations}, and
characterize them in two ways:

\begin{definition}
\label{def.intervalperm} Let $n\in{ \NN }$. Let $\alpha$ be a composition
of $n$. Let $\ell=\ell\left(  \alpha\right)  $.

\begin{itemize}
\item[(a)] For every set $S$ of positive integers, let $\overrightarrow{S}$
denote the list of all elements of $S$ in increasing order (with each element
appearing exactly once). Notice that this list $\overrightarrow{S}$ is a word
over the set of positive integers.

\item[(b)] For every $\tau\in\Symm_{\ell}$, we define a permutation
$\operatorname{iper}\left(  \alpha,\tau\right)  \in\Symm_n$ as follows:

The interval system corresponding to $\alpha$ is an $\ell$-tuple of intervals
(since $\ell\left(  \alpha\right)  =\ell$); denote this $\ell$-tuple by
$\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $. Now, define
$\operatorname{iper}\left(  \alpha,\tau\right)  $ to be the permutation in
$\Symm_n$ which (in one-line notation) is the word
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  \ell\right)  }}$ (a
concatenation of $\ell$ words). This is well-defined\footnote{In fact, from
the properties of interval systems, we know that the intervals $I_{1}$,
$I_{2}$, $\ldots$, $I_{\ell}$ form a set partition of the set $\left[
0:n\right]  ^{+}$. Hence, the intervals $I_{\tau\left(  1\right)  }$,
$I_{\tau\left(  2\right)  }$, $\ldots$, $I_{\tau\left(  \ell\right)  }$ form a
set partition of the set $\left[  0:n\right]  ^{+}$. As a consequence, the
word $\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau
\left(  2\right)  }}\cdots\overrightarrow{I_{\tau\left(  \ell\right)  }}$ is a
permutation of the word $12\ldots n$, and so there exists a permutation in
$\Symm_n$ which (in one-line notation) is this word, qed.}; hence,
$\operatorname{iper}\left(  \alpha,\tau\right)  \in\Symm_n$ is defined.

\item[(c)] The interval system corresponding to $\alpha$ is an $\ell$-tuple of
intervals (since $\ell\left(  \alpha\right)  =\ell$); denote this $\ell$-tuple
by $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $.

A permutation $\sigma\in\Symm_n$ is said to be
\emph{$\alpha$-clumping}\index{$\alpha$-clumping permutation}\index{clumping}
if every $i\in\left\{  1,2,\ldots,\ell\right\}  $ has the two
properties that:

\begin{itemize}
\item the set $\sigma^{-1}\left(  I_{i}\right)  $ is an interval;

\item the restriction of the map $\sigma^{-1}$ to the interval $I_{i}$ is increasing.
\end{itemize}
\end{itemize}
\end{definition}

\begin{example}
For this example, let $n=7$ and $\alpha=\left(  2,1,3,1\right)  $. Then,
$\ell=\ell\left(  \alpha\right)  =4$ and $\left(  I_{1},I_{2},I_{3}%
,I_{4}\right)  =\left(  \left\{  1,2\right\}  ,\left\{  3\right\}  ,\left\{
4,5,6\right\}  ,\left\{  7\right\}  \right)  $ (where we are using the
notations of Definition \ref{def.intervalperm}). Hence, $\overrightarrow{I_{1}%
}=12$, $\overrightarrow{I_{2}}=3$, $\overrightarrow{I_{3}}=456$ and
$\overrightarrow{I_{4}}=7$.

\begin{itemize}
\item[(a)] If $\tau\in\Symm_{\ell}=\Symm_4$ is the permutation
$\left(  2,3,1,4\right)  $, then $\operatorname*{iper}\left(  \alpha
,\tau\right)  $ is the permutation in $\Symm_7$ which (in one-line
notation) is the word $\overrightarrow{I_{\tau\left(  1\right)  }%
}\overrightarrow{I_{\tau\left(  2\right)  }}\overrightarrow{I_{\tau\left(
3\right)  }}\overrightarrow{I_{\tau\left(  4\right)  }}=\overrightarrow{I_{2}%
}\overrightarrow{I_{3}}\overrightarrow{I_{1}}\overrightarrow{I_{4}}=3456127$.

If $\tau\in\Symm_{\ell}=\Symm_4$ is the permutation $\left(
3,1,4,2\right)  $, then $\operatorname*{iper}\left(  \alpha,\tau\right)  $ is
the permutation in $\Symm_7$ which (in one-line notation) is the word
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\overrightarrow{I_{\tau\left(  3\right)  }}\overrightarrow{I_{\tau
\left(  4\right)  }}=\overrightarrow{I_{3}}\overrightarrow{I_{1}%
}\overrightarrow{I_{4}}\overrightarrow{I_{2}}=4561273$.

\item[(b)] The permutation $\sigma=\left(  3,7,4,5,6,1,2\right)
\in\Symm_7$ (given here in one-line notation) is $\alpha$-clumping, because:

\begin{itemize}
\item every $i\in\left\{  1,2,\ldots,\ell\right\}  =\left\{  1,2,3,4\right\}
$ has the property that $\sigma^{-1}\left(  I_{i}\right)  $ is an interval
(namely, $\sigma^{-1}\left(  I_{1}\right)  =\sigma^{-1}\left(  \left\{
1,2\right\}  \right)  =\left\{  6,7\right\}  $, $\sigma^{-1}\left(
I_{2}\right)  =\sigma^{-1}\left(  \left\{  3\right\}  \right)  =\left\{
1\right\}  $, $\sigma^{-1}\left(  I_{3}\right)  =\sigma^{-1}\left(  \left\{
4,5,6\right\}  \right)  =\left\{  3,4,5\right\}  $ and $\sigma^{-1}\left(
I_{4}\right)  =\sigma^{-1}\left(  \left\{  7\right\}  \right)  =\left\{
2\right\}  $), and

\item the restrictions of the map $\sigma^{-1}$ to the intervals $I_{i}$ are
increasing (this means that $\sigma^{-1}\left(  1\right)  <\sigma^{-1}\left(
2\right)  $ and $\sigma^{-1}\left(  4\right)  <\sigma^{-1}\left(  5\right)
<\sigma^{-1}\left(  6\right)  $, since the one-element intervals $I_{2}$ and
$I_{4}$ do not contribute anything to this condition).
\end{itemize}
\end{itemize}
\end{example}

Here is a more or less trivial observation:

\begin{proposition}
\label{prop.iper.intinverse}Let $n\in{ \NN }$. Let $\alpha$ be a
composition of $n$. Let $\ell=\ell\left(  \alpha\right)  $. Write $\alpha$ in
the form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. The
interval system corresponding to $\alpha$ is an $\ell$-tuple of intervals
(since $\ell\left(  \alpha\right)  =\ell$); denote this $\ell$-tuple by
$\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $. Let $\tau\in\Symm_{\ell}$.
Set $\sigma=\operatorname*{iper}\left(  \alpha,\tau\right)  $.

\begin{itemize}
\item[(a)] We have $\sigma^{-1}\left(  I_{\tau\left(  j\right)  }\right)
=\left[  \sum_{k=1}^{j-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{j}%
\alpha_{\tau\left(  k\right)  }\right]  ^{+}$ for every $j\in\left\{
1,2,\ldots,\ell\right\}  $.

\item[(b)] For every $j\in\left\{  1,2,\ldots,\ell\right\}  $, the restriction
of the map $\sigma^{-1}$ to the interval $I_{\tau\left(  j\right)  }$ is increasing.

\item[(c)] The permutation $\operatorname*{iper}\left(  \alpha,\tau\right)  $
is $\alpha$-clumping.

\item[(d)] Let $i\in\left\{  1,2,\ldots,\ell-1\right\}  $. Then, the sets
$\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  $, $\sigma^{-1}\left(
I_{\tau\left(  i+1\right)  }\right)  $ and $\sigma^{-1}\left(  I_{\tau\left(
i\right)  }\right)  \cup\sigma^{-1}\left(  I_{\tau\left(  i+1\right)
}\right)  $ are nonempty intervals. Also, $\sigma^{-1}\left(  I_{\tau\left(
i\right)  }\right)  <\sigma^{-1}\left(  I_{\tau\left(  i+1\right)  }\right)  $.
\end{itemize}
\end{proposition}

\begin{exercise}
\label{exe.prop.iper.intinverse}Prove Proposition \ref{prop.iper.intinverse}.
\end{exercise}

\begin{proposition}
\label{prop.iper.biject} Let $n\in{ \NN }$. Let $\alpha$ be a composition
of $n$. Let $\ell=\ell\left(  \alpha\right)  $.

\begin{itemize}
\item[(a)] Define a map
\begin{align*}
\operatorname{iper}_{\alpha}:\Symm_{\ell}  &  \longrightarrow\left\{
\omega\in\Symm_n\mid\omega\text{ is }\alpha\text{-clumping}\right\}
,\\
\tau &  \longmapsto\operatorname{iper}\left(  \alpha,\tau\right)
\end{align*}
\footnote{This map is well-defined because for every $\tau\in\Symm_{\ell}$,
the permutation $\operatorname{iper}\left(  \alpha,\tau\right)  $ is
$\alpha$-clumping (according to Proposition \ref{prop.iper.intinverse}(c)).}.
This map $\operatorname{iper}_{\alpha}$ is bijective.

\item[(b)] Let $\sigma\in\Symm_n$ be an $\alpha$-clumping
permutation. Then, there exists a unique $\tau\in\Symm_{\ell}$
satisfying $\sigma=\operatorname{iper}\left(  \alpha,\tau\right)  $.
\end{itemize}
\end{proposition}

\begin{exercise}
\label{exe.prop.iper.biject}Prove Proposition \ref{prop.iper.biject}.
\end{exercise}

Next, we recall that the concatenation $\alpha\cdot\beta$ of two compositions
$\alpha$ and $\beta$ is defined in the same way as the concatenation of two
words; if we regard compositions as words over the alphabet $\left\{
1,2,3,\ldots\right\}  $, then the concatenation $\alpha\cdot\beta$ of two
compositions $\alpha$ and $\beta$ \textbf{is} the concatenation $\alpha\beta$
of the words $\alpha$ and $\beta$. Thus, we are going to write $\alpha\beta$
for the concatenation $\alpha\cdot\beta$ of two compositions $\alpha$ and
$\beta$ from now on.

\begin{proposition}
\label{prop.iper.shuffle}Let $n\in \NN$ and $m\in \NN$. Let
$\alpha$ be a composition of $n$, and $\beta$ be a composition of $m$. Let
$p=\ell\left(  \alpha\right)  $ and $q=\ell\left(  \beta\right)  $. Let
$\tau\in\Symm_{p+q}$. Notice that $\operatorname*{iper}\left(
\alpha\beta,\tau\right)  \in\Symm_{n+m}$ (since $\alpha\beta$ is a
composition of $n+m$ having length $\ell\left(  \alpha\beta\right)
=\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  =p+q$). Then, $\tau
\in\operatorname{Sh}_{p,q}$ if and only if $\operatorname*{iper}%
\left(  \alpha\beta,\tau\right)  \in\operatorname{Sh}_{n,m}$.
\end{proposition}

\begin{exercise}
\label{exe.prop.iper.shuffle}Prove Proposition \ref{prop.iper.shuffle}.
\end{exercise}

Here is one more simple fact:

\begin{lemma}
\label{lem.shuffle.lyndon.step2}Let $u$ and $v$ be two words. Let
$n=\ell\left(  u\right)  $ and $m=\ell\left(  v\right)  $. Let $\alpha$ be a
composition of $n$, and let $\beta$ be a composition of $m$. Let
$p=\ell\left(  \alpha\right)  $ and $q=\ell\left(  \beta\right)  $. The
concatenation $\alpha\beta$ is a composition of $n+m$ having length
$\ell\left(  \alpha\beta\right)  =\ell\left(  \alpha\right)  +\ell\left(
\beta\right)  =p+q$. Thus, the interval system corresponding to $\alpha\beta$
is a $\left(  p+q\right)  $-tuple of intervals which covers $\left[
0:n+m\right]  ^{+}$. Denote this $\left(  p+q\right)  $-tuple by $\left(
I_{1},I_{2},\ldots,I_{p+q}\right)  $.

Let $\tau\in\operatorname{Sh}_{p,q}$. Set $\sigma
=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $. Then,%
\[
u\underset{\sigma}{\shuffle}v=\left(  uv\right)  \left[  I_{\tau\left(
1\right)  }\right]  \cdot\left(  uv\right)  \left[  I_{\tau\left(  2\right)
}\right]  \cdot\cdots\cdot\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]  .
\]

\end{lemma}

\begin{exercise}
\label{exe.lem.shuffle.lyndon.step2}Prove Lemma \ref{lem.shuffle.lyndon.step2}.
\end{exercise}

Having these notations and trivialities in place, we can say a bit more about
the lexicographically highest element of a shuffle product than what was said
in Theorem \ref{thm.shuffle.lyndon.used}:

\begin{theorem}
\label{thm.shuffle.lyndon.full}Let $u$ and $v$ be two words. Let
$n=\ell\left(  u\right)  $ and $m=\ell\left(  v\right)  $.

Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
$u$. Let $\left(  b_{1},b_{2},\ldots,b_{q}\right)  $ be the CFL factorization
of $v$.

Let $\alpha$ be the $p$-tuple $\left(  \ell\left(  a_{1}\right)  ,\ell\left(
a_{2}\right)  ,\ldots,\ell\left(  a_{p}\right)  \right)  $. Then, $\alpha$ is
a composition\footnote{since Lyndon words are nonempty, and thus $\ell\left(
a_{i}\right)  >0$ for every $i$} of length $p$ and size $\sum_{k=1}^{p}%
\ell\left(  a_{k}\right)  =\ell\left(  \underbrace{a_{1}a_{2} \cdots a_{p}%
}_{=u}\right)  =\ell\left(  u\right)  =n$.

Let $\beta$ be the $q$-tuple $\left(  \ell\left(  b_{1}\right)  ,\ell\left(
b_{2}\right)  ,\ldots,\ell\left(  b_{q}\right)  \right)  $. Then, $\beta$ is a
composition of length $q$ and size $\sum_{k=1}^{q}\ell\left(  b_{k}\right)
=m$.\ \ \ \ \footnote{The proof of this is the same as the proof of the fact
that $\alpha$ is a composition of length $p$ and size $\sum_{k=1}^{p}%
\ell\left(  \alpha_{k}\right)  =n$.}

Now, $\alpha$ is a composition of length $p$ and size $n$, and $\beta$ is a
composition of length $q$ and size $m$. Thus, the concatenation $\alpha\beta$
of these two tuples is a composition of length $p+q$ and size $n+m$. The
interval system corresponding to this composition $\alpha\beta$ is a $\left(
p+q\right)  $-tuple (since said composition has length $p+q$); denote this
$\left(  p+q\right)  $-tuple by $\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  $.

\begin{itemize}
\item[(a)] If $\tau\in\operatorname{Sh}_{p,q}$ satisfies $\left(
uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)
\left[  I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)
\left[  I_{\tau\left(  p+q\right)  }\right]  $, and if we set $\sigma
=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $, then $\sigma
\in\operatorname{Sh}_{n,m}$, and the word $u\underset{\sigma
}{\shuffle}v$ is the lexicographically highest element of the multiset
$u\shuffle v$.

\item[(b)] Let $\sigma\in\operatorname{Sh}_{n,m}$ be a permutation
such that $u\underset{\sigma}{\shuffle}v$ is the lexicographically highest
element of the multiset $u\shuffle v$. Then, there exists a unique permutation
$\tau\in\operatorname{Sh}_{p,q}$ satisfying $\left(  uv\right)
\left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)  \left[
I_{\tau\left(  p+q\right)  }\right]  $ and $\sigma=\operatorname*{iper}\left(
\alpha\beta,\tau\right)  $.
\end{itemize}
\end{theorem}

\begin{proof}
Before we step to the actual proof, we need to make some preparation. First of
all, $\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  $ is the interval system
corresponding to the composition $\alpha\beta$. In other words,%
\begin{equation}
\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  =\operatorname*{intsys}\left(
\alpha\beta\right)  . \label{pf.thm.shuffle.lyndon.full.0}%
\end{equation}
But since $\alpha=\left(  \ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)
,\ldots,\ell\left(  a_{p}\right)  \right)  $ and $\beta=\left(  \ell\left(
b_{1}\right)  ,\ell\left(  b_{2}\right)  ,\ldots,\ell\left(  b_{q}\right)
\right)  $, we have%
\[
\alpha\beta=\left(  \ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)
,\ldots,\ell\left(  a_{p}\right)  ,\ell\left(  b_{1}\right)  ,\ell\left(
b_{2}\right)  ,\ldots,\ell\left(  b_{q}\right)  \right)  .
\]
Thus, (\ref{pf.thm.shuffle.lyndon.full.0}) rewrites as%
\[
\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  =\operatorname*{intsys}\left(
\ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(
a_{p}\right)  ,\ell\left(  b_{1}\right)  ,\ell\left(  b_{2}\right)
,\ldots,\ell\left(  b_{q}\right)  \right)  .
\]
By the definition of $\operatorname*{intsys}\left(  \ell\left(  a_{1}\right)
,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{p}\right)  ,\ell\left(
b_{1}\right)  ,\ell\left(  b_{2}\right)  ,\ldots,\ell\left(  b_{q}\right)
\right)  $, we thus have%
\[
I_{i}=\left[  \sum_{k=1}^{i-1}\ell\left(  a_{k}\right)  :\sum_{k=1}^{i}%
\ell\left(  a_{k}\right)  \right]  ^{+}\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in\left\{  1,2,\ldots,p\right\}  ,
\]
and besides
\[
I_{p+j}=\left[  n+\sum_{k=1}^{j-1}\ell\left(  b_{k}\right)  :n+\sum_{k=1}%
^{j}\ell\left(  b_{k}\right)  \right]  ^{+}\ \ \ \ \ \ \ \ \ \ \text{for every
}j\in\left\{  1,2,\ldots,q\right\}
\]
(since $\sum_{k=1}^{p}\ell\left(  a_{k}\right)  =n$). Moreover,
Remark~\ref{rmk.intsys.trivia}(c) shows that $\left(  I_{1},I_{2}%
,\ldots,I_{p+q}\right)  $ is a $\left(  p+q\right)  $-tuple of nonempty
intervals of $\ZZ$ and satisfies the following three properties:

\begin{itemize}
\item The intervals $I_{1}$, $I_{2}$, $\ldots$, $I_{p+q}$ form a set partition
of the set $\left[  0:n+m\right]  ^{+}$.

\item We have $I_{1}<I_{2}< \cdots<I_{p+q}$.

\item We have $\left\vert I_{i}\right\vert =\ell\left(  a_{i}\right)  $ for
every $i\in\left\{  1,2,\ldots,p\right\}  $ and $\left\vert I_{p+j}\right\vert
=\ell\left(  b_{j}\right)  $ for every $j\in\left\{  1,2,\ldots,q\right\}  $.
\end{itemize}

Of course, every $i\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
I_{i}\subset\left[  0:n\right]  ^{+}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  uv\right)  \left[  I_{i}\right]  =u\left[
I_{i}\right]  =a_{i}. \label{pf.thm.shuffle.lyndon.full.prep1}%
\end{equation}
Meanwhile, every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $ satisfies%
\begin{equation}
I_{i}\subset\left[  n:n+m\right]  ^{+}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  uv\right)  \left[  I_{i}\right]  =v\left[
I_{i}-n\right]  =b_{i-p} \label{pf.thm.shuffle.lyndon.full.prep1'}%
\end{equation}
(where $I_{i}-n$ denotes the interval $\left\{  k-n\ \mid\ k\in I_{i}\right\}
$). We thus see that%
\begin{equation}
\left(  uv\right)  \left[  I_{i}\right]  \text{ is a Lyndon word}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,p+q\right\}
\label{pf.thm.shuffle.lyndon.full.prep1lynd}%
\end{equation}
\footnote{Indeed, when $i\leq p$, this follows from
(\ref{pf.thm.shuffle.lyndon.full.prep1}) and the fact that $a_{i}$ is Lyndon;
whereas in the other case, this follows from
(\ref{pf.thm.shuffle.lyndon.full.prep1'}) and the fact that $b_{i-p}$ is
Lyndon.}.

By the definition of a CFL factorization, we have $a_{1}\geq a_{2}\geq
\cdots\geq a_{p}$ and $b_{1}\geq b_{2}\geq\cdots\geq b_{q}$.

We have $\sigma\in\operatorname{Sh}_{n,m}$, so that $\sigma
^{-1}\left(  1\right)  <\sigma^{-1}\left(  2\right)  <\cdots<\sigma
^{-1}\left(  n\right)  $ and $\sigma^{-1}\left(  n+1\right)  <\sigma
^{-1}\left(  n+2\right)  <\cdots<\sigma^{-1}\left(  n+m\right)  $. In other
words, the restriction of the map $\sigma^{-1}$ to the interval $\left[
0:n\right]  ^{+}$ is strictly increasing, and so is the restriction of the map
$\sigma^{-1}$ to the interval $\left[  n:n+m\right]  ^{+}$.

(b) We will first show that%
\begin{equation}
\text{if }J\subset\left[  0:n\right]  ^{+}\text{ is an interval such that the
word }\left(  uv\right)  \left[  J\right]  \text{ is Lyndon, then }\sigma
^{-1}\left(  J\right)  \text{ is an interval.}
\label{pf.thm.shuffle.lyndon.full.1}%
\end{equation}


\textit{Proof of (\ref{pf.thm.shuffle.lyndon.full.1}):} We will prove
(\ref{pf.thm.shuffle.lyndon.full.1}) by strong induction over $\left\vert
J\right\vert $.

So, fix some $N\in \NN$. Assume (as the induction hypothesis) that
(\ref{pf.thm.shuffle.lyndon.full.1}) has been proven whenever $\left\vert
J\right\vert <N$. We now need to prove (\ref{pf.thm.shuffle.lyndon.full.1})
when $\left\vert J\right\vert =N$.

Let $J\subset\left[  0:n\right]  ^{+}$ be an interval such that the word
$\left(  uv\right)  \left[  J\right]  $ is Lyndon and such that $\left\vert
J\right\vert =N$. We have to prove that $\sigma^{-1}\left(  J\right)  $ is an
interval. This is obvious if $\left\vert J\right\vert =1$ (because in this
case, $\sigma^{-1}\left(  J\right)  $ is a one-element set, thus trivially an
interval). Hence, we WLOG assume that we don't have $\left\vert J\right\vert
=1$. We also don't have $\left\vert J\right\vert =0$, because $\left(
uv\right)  \left[  J\right]  $ has to be Lyndon (and the empty word is not).
So we have $\left\vert J\right\vert >1$. Now, $\ell\left(  \left(  uv\right)
\left[  J\right]  \right)  =\left\vert J\right\vert >1$, and thus $\left(
uv\right)  \left[  J\right]  $ is a Lyndon word of length $>1$. Let
$v^{\prime}$ be the (lexicographically) smallest nonempty \textbf{proper}
suffix of $\left(  uv\right)  \left[  J\right]  $. Since $v^{\prime}$ is a
proper suffix of $w$, there exists a nonempty $u^{\prime}\in\mathfrak{A}%
^{\ast}$ such that $\left(  uv\right)  \left[  J\right]  =u^{\prime}v^{\prime
}$. Consider this $u^{\prime}$.

Now, Theorem \ref{thm.words.lyndon.std}(a) (applied to $\left(  uv\right)
\left[  J\right]  $, $u^{\prime}$ and $v^{\prime}$ instead of $w$, $u$ and
$v$) yields that the words $u^{\prime}$ and $v^{\prime}$ are Lyndon. Also,
Theorem \ref{thm.words.lyndon.std}(b) (applied to $\left(  uv\right)  \left[
J\right]  $, $u^{\prime}$ and $v^{\prime}$ instead of $w$, $u$ and $v$) yields
that $u^{\prime} < \left(uv\right) \left[ J \right] < v^{\prime}$.

But from the fact that $\left(  uv\right)  \left[  J\right]  =u^{\prime
}v^{\prime}$ with $u^{\prime}$ and $v^{\prime}$ both being nonempty, it
becomes immediately clear that we can write $J$ as a union of two disjoint
nonempty intervals $K$ and $L$ such that $K<L$, $u^{\prime}=\left(  uv\right)
\left[  K\right]  $ and $v^{\prime}=\left(  uv\right)  \left[  L\right]  $.
Consider these $K$ and $L$. The intervals $K$ and $L$ are nonempty and have
their sizes add up to $\left\vert J\right\vert$
(since they are disjoint and their union is $J$),
and hence both must have size smaller than $\left\vert J\right\vert =N$. So
$K\subset\left[  0:n\right]  ^{+}$ is an interval of size $\left\vert
K\right\vert <N$ having the property that $\left(  uv\right)  \left[
K\right]  $ is Lyndon (since $\left(  uv\right)  \left[  K\right]  =u^{\prime
}$ is Lyndon). Thus, we can apply (\ref{pf.thm.shuffle.lyndon.full.1}) to $K$
instead of $J$ (because of the induction hypothesis). As a result, we conclude
that $\sigma^{-1}\left(  K\right)  $ is an interval. Similarly, we can apply
(\ref{pf.thm.shuffle.lyndon.full.1}) to $L$ instead of $J$ (we know that
$\left(  uv\right)  \left[  L\right]  $ is Lyndon since $\left(  uv\right)
\left[  L\right]  =v^{\prime}$), and learn that $\sigma^{-1}\left(  L\right)
$ is an interval. The intervals $\sigma^{-1}\left(  K\right)  $ and
$\sigma^{-1}\left(  L\right)  $ are both nonempty (since $K$ and $L$ are
nonempty), and their union is $\sigma^{-1}\left(  J\right)  $ (because the
union of $K$ and $L$ is $J$). The nonempty intervals $K$ and $L$ both are
subsets of $\left[  0:n\right]  ^{+}$ (since their union is $J\subset\left[
0:n\right]  ^{+}$), and their union $K\cup L$ is an interval (since their
union $K\cup L$ is $J$, and we know that $J$ is an interval).

Now, assume (for the sake of contradiction) that $\sigma^{-1}\left(  J\right)
$ is not an interval. Since $J$ is the union of $K$ and $L$, we have $J=K\cup
L$ and thus $\sigma^{-1}\left(  J\right)  =\sigma^{-1}\left(  K\cup L\right)
=\sigma^{-1}\left(  K\right)  \cup\sigma^{-1}\left(  L\right)  $ (since
$\sigma$ is a bijection). Therefore, $\sigma^{-1}\left(  K\right)  \cup
\sigma^{-1}\left(  L\right)  $ is not an interval (since $\sigma^{-1}\left(
J\right)  $ is not an interval). Thus, Lemma \ref{lem.shuffle.lyndon.easystep}%
(b) yields that there exists a nonempty interval $P\subset\left[
n:n+m\right]  ^{+}$ such that $\sigma^{-1}\left(  P\right)  $, $\sigma
^{-1}\left(  K\right)  \cup\sigma^{-1}\left(  P\right)  $ and $\sigma
^{-1}\left(  P\right)  \cup\sigma^{-1}\left(  L\right)  $ are intervals and
such that $\sigma^{-1}\left(  K\right)  <\sigma^{-1}\left(  P\right)
<\sigma^{-1}\left(  L\right)  $. Consider this $P$. Since $P$ is nonempty, we
have $\left\vert P\right\vert \neq0$.

Lemma \ref{lem.shuffle.lyndon.step}(b) (applied to $K$ and $P$ instead of $I$
and $J$) yields
\begin{equation}
\left(  uv\right)  \left[  K\right]  \cdot\left(  uv\right)  \left[  P\right]
\geq\left(  uv\right)  \left[  P\right]  \cdot\left(  uv\right)  \left[
K\right]  . \label{pf.thm.shuffle.lyndon.full.5a}%
\end{equation}
Since $\left(  uv\right)  \left[  K\right]  =u^{\prime}$, this rewrites as%
\begin{equation}
u^{\prime}\cdot\left(  uv\right)  \left[  P\right]  \geq\left(  uv\right)
\left[  P\right]  \cdot u^{\prime}. \label{pf.thm.shuffle.lyndon.full.5a1}%
\end{equation}


But Lemma \ref{lem.shuffle.lyndon.step}(c) (applied to $P$ and $L$ instead of
$I$ and $J$) yields
\begin{equation}
\left(  uv\right)  \left[  P\right]  \cdot\left(  uv\right)  \left[  L\right]
\geq\left(  uv\right)  \left[  L\right]  \cdot\left(  uv\right)  \left[
P\right]  . \label{pf.thm.shuffle.lyndon.full.5b}%
\end{equation}
Since $\left(  uv\right)  \left[  L\right]  =v^{\prime}$, this rewrites as%
\begin{equation}
\left(  uv\right)  \left[  P\right]  \cdot v^{\prime}\geq v^{\prime}%
\cdot\left(  uv\right)  \left[  P\right]  .
\label{pf.thm.shuffle.lyndon.full.5b1}%
\end{equation}


Recall also that $u^{\prime}<v^{\prime}$, and that both words $u^{\prime}$ and
$v^{\prime}$ are Lyndon. Now, Corollary \ref{cor.words.lyndon.uv=vu} (applied
to $u^{\prime}$, $v^{\prime}$ and $\left(  uv\right)  \left[  P\right]  $
instead of $u$, $v$ and $z$) yields that $\left(  uv\right)  \left[  P\right]
$ is the empty word (because of (\ref{pf.thm.shuffle.lyndon.full.5a1}) and
(\ref{pf.thm.shuffle.lyndon.full.5b1})), so that $\ell\left(  \left(
uv\right)  \left[  P\right]  \right)  =0$. This contradicts $\ell\left(
\left(  uv\right)  \left[  P\right]  \right)  =\left\vert P\right\vert \neq0$.
This contradiction shows that our assumption (that $\sigma^{-1}\left(
J\right)  $ is not an interval) was wrong. Hence, $\sigma^{-1}\left(
J\right)  $ is an interval. This completes the induction step, and thus
(\ref{pf.thm.shuffle.lyndon.full.1}) is proven.

Similarly to (\ref{pf.thm.shuffle.lyndon.full.1}), we can show that%
\begin{equation}
\text{if }J\subset\left[  n:n+m\right]  ^{+}\text{ is an interval such that
the word }\left(  uv\right)  \left[  J\right]  \text{ is Lyndon, then }%
\sigma^{-1}\left(  J\right)  \text{ is an interval.}
\label{pf.thm.shuffle.lyndon.full.1'}%
\end{equation}


Now, let $i\in\left\{  1,2,\ldots,p+q\right\}  $ be arbitrary. We are going to
prove that%
\begin{equation}
\sigma^{-1}\left(  I_{i}\right)  \text{ is an interval.}
\label{pf.thm.shuffle.lyndon.full.21}%
\end{equation}


\textit{Proof of (\ref{pf.thm.shuffle.lyndon.full.21}):} We must be in one of
the following two cases:

\textit{Case 1:} We have $i\in\left\{  1,2,\ldots,p\right\}  $.

\textit{Case 2:} We have $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $.

Let us first consider Case 1. In this case, we have $i\in\left\{
1,2,\ldots,p\right\}  $. Thus, $I_{i}\subset\left[  0:n\right]  ^{+}$ (by
(\ref{pf.thm.shuffle.lyndon.full.prep1})). Also,
(\ref{pf.thm.shuffle.lyndon.full.prep1}) yields that $\left(  uv\right)
\left[  I_{i}\right]  =a_{i}$ is a Lyndon word. Hence,
(\ref{pf.thm.shuffle.lyndon.full.1}) (applied to $J=I_{i}$) yields that
$\sigma^{-1}\left(  I_{i}\right)  $ is an interval. Thus,
(\ref{pf.thm.shuffle.lyndon.full.21}) is proven in Case 1.

Similarly, we can prove (\ref{pf.thm.shuffle.lyndon.full.21}) in Case 2, using
(\ref{pf.thm.shuffle.lyndon.full.prep1'}) and
(\ref{pf.thm.shuffle.lyndon.full.1'}) instead of
(\ref{pf.thm.shuffle.lyndon.full.prep1}) and
(\ref{pf.thm.shuffle.lyndon.full.1}), respectively. Hence,
(\ref{pf.thm.shuffle.lyndon.full.21}) is proven.

So we know that $\sigma^{-1}\left(  I_{i}\right)  $ is an interval. But we
also know that either $I_{i}\subset\left[  0:n\right]  ^{+}$ or $I_{i}%
\subset\left[  n:n+m\right]  ^{+}$ (depending on whether $i\leq p$ or $i>p$).
As a consequence, the restriction of the map $\sigma^{-1}$ to the interval
$I_{i}$ is increasing (because the restriction of the map $\sigma^{-1}$ to the
interval $\left[  0:n\right]  ^{+}$ is strictly increasing, and so is the
restriction of the map $\sigma^{-1}$ to the interval $\left[  n:n+m\right]
^{+}$).

Now, let us forget that we fixed $i$. We thus have shown that every
$i\in\left\{  1,2,\ldots,p+q\right\}  $ has the two properties that:

\begin{itemize}
\item the set $\sigma^{-1}\left(  I_{i}\right)  $ is an interval;

\item the restriction of the map $\sigma^{-1}$ to the interval $I_{i}$ is increasing.
\end{itemize}

\noindent In other words, the permutation $\sigma$ is
$\left(  \alpha\beta\right)$-clumping
(since $\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  $ is the
interval system corresponding to the composition $\alpha\beta$). Hence,
Proposition \ref{prop.iper.biject}(b) (applied to $n+m$, $\alpha\beta$ and
$p+q$ instead of $n$, $\alpha$ and $\ell$) shows that there exists a unique
$\tau\in\Symm_{p+q}$ satisfying $\sigma=\operatorname*{iper}\left(
\alpha\beta,\tau\right)  $. Thus, the uniqueness part of Theorem
\ref{thm.shuffle.lyndon.full}(b) (i.e., the claim that the $\tau$ in Theorem
\ref{thm.shuffle.lyndon.full}(b) is unique if it exists) is proven.

It now remains to prove the existence part of Theorem
\ref{thm.shuffle.lyndon.full}(b), i.e., to prove that there exists at least
one permutation $\tau\in\operatorname{Sh}_{p,q}$ satisfying $\left(
uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)
\left[  I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)
\left[  I_{\tau\left(  p+q\right)  }\right]  $ and $\sigma
=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $. We already know that
there exists a unique $\tau\in\Symm_{p+q}$ satisfying $\sigma
=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $. Consider this $\tau$.
We will now prove that $\left(  uv\right)  \left[  I_{\tau\left(  1\right)
}\right]  \geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]
\geq\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]
$ and $\tau\in\operatorname{Sh}_{p,q}$. Once this is done, the
existence part of Theorem \ref{thm.shuffle.lyndon.full}(b) will be proven, and
thus the proof of Theorem \ref{thm.shuffle.lyndon.full}(b) will be complete.

Proposition \ref{prop.iper.shuffle} yields that $\tau\in
\operatorname{Sh}_{p,q}$ if and only if $\operatorname*{iper}\left(  \alpha\beta
,\tau\right)  \in\operatorname{Sh}_{n,m}$. Since we know that
$\operatorname*{iper}\left(  \alpha\beta,\tau\right)  =\sigma\in
\operatorname{Sh}_{n,m}$, we thus conclude that $\tau\in
\operatorname{Sh}_{p,q}$. The only thing that remains to be proven
now is that%
\begin{equation}
\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \geq\left(
uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  .
\label{pf.thm.shuffle.lyndon.full.25}%
\end{equation}


\textit{Proof of (\ref{pf.thm.shuffle.lyndon.full.25}):} We have $\tau
\in\operatorname{Sh}_{p,q}$. In other words, $\tau^{-1}\left(
1\right)  <\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(  p\right)  $ and
$\tau^{-1}\left(  p+1\right)  <\tau^{-1}\left(  p+2\right)  <\cdots<\tau
^{-1}\left(  p+q\right)  $. In other words, the restriction of the map
$\tau^{-1}$ to the interval $\left[  0:p\right]  ^{+}$ is strictly increasing,
and so is the restriction of the map $\tau^{-1}$ to the interval $\left[
p:p+q\right]  ^{+}$.

Let $i\in\left\{  1,2,\ldots,p+q-1\right\}  $. We will show that
\begin{equation}
\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  \geq\left(
uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]  .
\label{pf.thm.shuffle.lyndon.full.26}%
\end{equation}


Clearly, both $\tau\left(  i\right)  $ and $\tau\left(  i+1\right)  $ belong
to $\left\{  1,2,\ldots,p+q\right\}  =\left\{  1,2,\ldots,p\right\}
\cup\left\{  p+1,p+2,\ldots,p+q\right\}  $. Thus, we must be in one of the
following four cases:

\textit{Case 1:} We have $\tau\left(  i\right)  \in\left\{  1,2,\ldots
,p\right\}  $ and $\tau\left(  i+1\right)  \in\left\{  1,2,\ldots,p\right\}  $.

\textit{Case 2:} We have $\tau\left(  i\right)  \in\left\{  1,2,\ldots
,p\right\}  $ and $\tau\left(  i+1\right)  \in\left\{  p+1,p+2,\ldots
,p+q\right\}  $.

\textit{Case 3:} We have $\tau\left(  i\right)  \in\left\{  p+1,p+2,\ldots
,p+q\right\}  $ and $\tau\left(  i+1\right)  \in\left\{  1,2,\ldots,p\right\}
$.

\textit{Case 4:} We have $\tau\left(  i\right)  \in\left\{  p+1,p+2,\ldots
,p+q\right\}  $ and $\tau\left(  i+1\right)  \in\left\{  p+1,p+2,\ldots
,p+q\right\}  $.

Let us consider Case 1 first. In this case, we have $\tau\left(  i\right)
\in\left\{  1,2,\ldots,p\right\}  $ and $\tau\left(  i+1\right)  \in\left\{
1,2,\ldots,p\right\}  $. From the fact that the restriction of the map
$\tau^{-1}$ to the interval $\left[  0:p\right]  ^{+}$ is strictly increasing,
we can easily deduce $\tau\left(  i\right)  <\tau\left(  i+1\right)
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $\tau\left(
i\right)  \geq\tau\left(  i+1\right)  $. Since both $\tau\left(  i\right)  $
and $\tau\left(  i+1\right)  $ belong to $\left\{  1,2,\ldots,p\right\}
=\left[  0:p\right]  ^{+}$, this yields $\tau^{-1}\left(  \tau\left(
i\right)  \right)  \geq\tau^{-1}\left(  \tau\left(  i+1\right)  \right)  $
(since the restriction of the map $\tau^{-1}$ to the interval $\left[
0:p\right]  ^{+}$ is strictly increasing), which contradicts $\tau^{-1}\left(
\tau\left(  i\right)  \right)  =i<i+1=\tau^{-1}\left(  \tau\left(  i+1\right)
\right)  $. This contradiction proves the assumption wrong, qed.}. Therefore,
$a_{\tau\left(  i\right)  }\geq a_{\tau\left(  i+1\right)  }$ (since
$a_{1}\geq a_{2}\geq\cdots\geq a_{p}$).

But $\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]
=a_{\tau\left(  i\right)  }$ (by (\ref{pf.thm.shuffle.lyndon.full.prep1}),
applied to $\tau\left(  i\right)  $ instead of $i$) and $\left(  uv\right)
\left[  I_{\tau\left(  i+1\right)  }\right]  =a_{\tau\left(  i+1\right)  }$
(similarly). In view of these equalities, the inequality $a_{\tau\left(
i\right)  }\geq a_{\tau\left(  i+1\right)  }$ rewrites as $\left(  uv\right)
\left[  I_{\tau\left(  i\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  i+1\right)  }\right]  $. Thus,
(\ref{pf.thm.shuffle.lyndon.full.26}) is proven in Case 1.

Similarly, we can show (\ref{pf.thm.shuffle.lyndon.full.26}) in Case 4
(observing that $\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]
=b_{\tau\left(  i\right)  -p}$ and $\left(  uv\right)  \left[  I_{\tau\left(
i+1\right)  }\right]  =b_{\tau\left(  i+1\right)  -p}$ in this case).

Let us now consider Case 2. In this case, we have $\tau\left(  i\right)
\in\left\{  1,2,\ldots,p\right\}  $ and $\tau\left(  i+1\right)  \in\left\{
p+1,p+2,\ldots,p+q\right\}  $. From $\tau\left(  i\right)  \in\left\{
1,2,\ldots,p\right\}  $, we conclude that $I_{\tau\left(  i\right)  }%
\subset\left[  0:n\right]  ^{+}$. From $\tau\left(  i+1\right)  \in\left\{
p+1,p+2,\ldots,p+q\right\}  $, we conclude that $I_{\tau\left(  i+1\right)
}\subset\left[  n:n+m\right]  ^{+}$. The intervals $I_{\tau\left(  i\right)
}$ and $I_{\tau\left(  i+1\right)  }$ are clearly nonempty.

Proposition \ref{prop.iper.intinverse}(d) (applied to $n+m$, $\alpha\beta$,
$p+q$ and $\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  $ instead of $n$,
$\alpha$, $\ell$ and $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $) yields
that the sets $\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  $,
$\sigma^{-1}\left(  I_{\tau\left(  i+1\right)  }\right)  $ and $\sigma
^{-1}\left(  I_{\tau\left(  i\right)  }\right)  \cup\sigma^{-1}\left(
I_{\tau\left(  i+1\right)  }\right)  $ are nonempty intervals, and that we
have $\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  <\sigma
^{-1}\left(  I_{\tau\left(  i+1\right)  }\right)  $. Hence, Lemma
\ref{lem.shuffle.lyndon.step}(b) (applied to $I=I_{\tau\left(  i\right)  }$
and $J=I_{\tau\left(  i+1\right)  }$) yields%
\[
\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  \cdot\left(
uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]  \geq\left(  uv\right)
\left[  I_{\tau\left(  i+1\right)  }\right]  \cdot\left(  uv\right)  \left[
I_{\tau\left(  i\right)  }\right]  .
\]
But $\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  $ and
$\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]  $ are Lyndon
words (as a consequence of (\ref{pf.thm.shuffle.lyndon.full.prep1lynd})).
Thus, Proposition \ref{prop.words.lyndon.preorder} (applied to $\left(
uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  $ and $\left(
uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]  $ instead of $u$ and
$v$) shows that $\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]  $ if and
only if $\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]
\cdot\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]
\cdot\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  $. Since we
know that $\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]
\cdot\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]
\cdot\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  $ holds, we
thus conclude that $\left(  uv\right)  \left[  I_{\tau\left(  i\right)
}\right]  \geq\left(  uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]
$. Thus, (\ref{pf.thm.shuffle.lyndon.full.26}) is proven in Case 2.

The proof of (\ref{pf.thm.shuffle.lyndon.full.26}) in Case 3 is analogous to
that in Case 2 (the main difference being that Lemma
\ref{lem.shuffle.lyndon.step}(c) is used in lieu of Lemma
\ref{lem.shuffle.lyndon.step}(b)).

Thus, (\ref{pf.thm.shuffle.lyndon.full.26}) is proven in all possible cases.
So we always have (\ref{pf.thm.shuffle.lyndon.full.26}). In other words,
$\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]  \geq\left(
uv\right)  \left[  I_{\tau\left(  i+1\right)  }\right]  $.

Now, forget that we fixed $i$. We hence have shown that $\left(  uv\right)
\left[  I_{\tau\left(  i\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  i+1\right)  }\right]  $ for all $i\in\left\{  1,2,\ldots
,p+q-1\right\}  $. This proves (\ref{pf.thm.shuffle.lyndon.full.25}), and thus
completes our proof of Theorem \ref{thm.shuffle.lyndon.full}(b).

(a) Let $\tau\in\operatorname{Sh}_{p,q}$ be such that
\begin{equation}
\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \geq\left(
uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  .
\label{pf.thm.shuffle.lyndon.full.a.1}%
\end{equation}
Set $\sigma=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $. Then,
Proposition \ref{prop.iper.shuffle} yields that $\tau\in
\operatorname{Sh}_{p,q}$ if and only if $\operatorname*{iper}\left(  \alpha\beta
,\tau\right)  \in\operatorname{Sh}_{n,m}$. Since we know that
$\tau\in\operatorname{Sh}_{p,q}$, we can deduce from this that
$\operatorname*{iper}\left(  \alpha\beta,\tau\right)  \in
\operatorname{Sh}_{n,m}$, so that $\sigma=\operatorname*{iper}\left(  \alpha\beta
,\tau\right)  \in\operatorname{Sh}_{n,m}$.

It remains to prove that the word $u\underset{\sigma}{\shuffle}v$ is the
lexicographically highest element of the multiset $u\shuffle v$.

It is clear that the multiset $u\shuffle v$ has \textbf{some}
lexicographically highest element. This element has the form
$u\underset{\widetilde{\sigma}}{\shuffle}v$ for some $\widetilde{\sigma}%
\in\operatorname{Sh}_{n,m}$ (because any element of this multiset
has such a form). Consider this $\widetilde{\sigma}$. Theorem
\ref{thm.shuffle.lyndon.full}(b) (applied to $\widetilde{\sigma}$ instead of
$\sigma$) yields that there exists a unique permutation $\widetilde{\tau}%
\in\operatorname{Sh}_{p,q}$ satisfying $\left(  uv\right)  \left[
I_{\widetilde{\tau}\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\widetilde{\tau}\left(  2\right)  }\right]  \geq\cdots\geq\left(
uv\right)  \left[  I_{\widetilde{\tau}\left(  p+q\right)  }\right]  $ and
$\widetilde{\sigma}=\operatorname*{iper}\left(  \alpha\beta,\widetilde{\tau
}\right)  $. (What we call $\widetilde{\tau}$ here is what has been called
$\tau$ in Theorem \ref{thm.shuffle.lyndon.full}(b).)

Now, the chain of inequalities $\left(  uv\right)  \left[  I_{\widetilde{\tau
}\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[  I_{\widetilde{\tau
}\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)  \left[
I_{\widetilde{\tau}\left(  p+q\right)  }\right]  $ shows that the list
$\left(  \left(  uv\right)  \left[  I_{\widetilde{\tau}\left(  1\right)
}\right]  ,\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(  2\right)
}\right]  ,\ldots,\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(
p+q\right)  }\right]  \right)  $ is the result of sorting the list $\left(
\left(  uv\right)  \left[  I_{1}\right]  ,\left(  uv\right)  \left[
I_{2}\right]  ,\ldots,\left(  uv\right)  \left[  I_{p+q}\right]  \right)  $ in
decreasing order. But the chain of inequalities
(\ref{pf.thm.shuffle.lyndon.full.a.1}) shows that the list \newline$\left(
\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  ,\left(
uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  ,\ldots,\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  \right)  $ is the
result of sorting the same list $\left(  \left(  uv\right)  \left[
I_{1}\right]  ,\left(  uv\right)  \left[  I_{2}\right]  ,\ldots,\left(
uv\right)  \left[  I_{p+q}\right]  \right)  $ in decreasing order. So each of
the two lists $\left(  \left(  uv\right)  \left[  I_{\widetilde{\tau}\left(
1\right)  }\right]  ,\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(
2\right)  }\right]  ,\ldots,\left(  uv\right)  \left[  I_{\widetilde{\tau
}\left(  p+q\right)  }\right]  \right)  $ and \newline$\left(  \left(
uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  ,\left(  uv\right)
\left[  I_{\tau\left(  2\right)  }\right]  ,\ldots,\left(  uv\right)  \left[
I_{\tau\left(  p+q\right)  }\right]  \right)  $ is the result of sorting one
and the same list \newline$\left(  \left(  uv\right)  \left[  I_{1}\right]
,\left(  uv\right)  \left[  I_{2}\right]  ,\ldots,\left(  uv\right)  \left[
I_{p+q}\right]  \right)  $ in decreasing order. Since the result of sorting a
given list in decreasing order is unique, this yields%
\[
\left(  \left(  uv\right)  \left[  I_{\widetilde{\tau}\left(  1\right)
}\right]  ,\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(  2\right)
}\right]  ,\ldots,\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(
p+q\right)  }\right]  \right)  =\left(  \left(  uv\right)  \left[
I_{\tau\left(  1\right)  }\right]  ,\left(  uv\right)  \left[  I_{\tau\left(
2\right)  }\right]  ,\ldots,\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]  \right)  .
\]
Hence,%
\begin{equation}
\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(  1\right)  }\right]
\cdot\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(  2\right)  }\right]
\cdot\cdots\cdot\left(  uv\right)  \left[  I_{\widetilde{\tau}\left(
p+q\right)  }\right]  =\left(  uv\right)  \left[  I_{\tau\left(  1\right)
}\right]  \cdot\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]
\cdot\cdots\cdot\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)
}\right]  . \label{pf.thm.shuffle.lyndon.full.a.5}%
\end{equation}
But Lemma \ref{lem.shuffle.lyndon.step2} yields
\begin{equation}
u\underset{\sigma}{\shuffle}v=\left(  uv\right)  \left[  I_{\tau\left(
1\right)  }\right]  \cdot\left(  uv\right)  \left[  I_{\tau\left(  2\right)
}\right]  \cdot\cdots\cdot\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]  . \label{pf.thm.shuffle.lyndon.full.a.6}%
\end{equation}
Meanwhile, Lemma \ref{lem.shuffle.lyndon.step2} (applied to $\widetilde{\tau}$
and $\widetilde{\sigma}$ instead of $\tau$ and $\sigma$) yields
\begin{align*}
u\underset{\widetilde{\sigma}}{\shuffle}v  &  =\left(  uv\right)  \left[
I_{\widetilde{\tau}\left(  1\right)  }\right]  \cdot\left(  uv\right)  \left[
I_{\widetilde{\tau}\left(  2\right)  }\right]  \cdot\cdots\cdot\left(
uv\right)  \left[  I_{\widetilde{\tau}\left(  p+q\right)  }\right] \\
&  =\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \cdot\left(
uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \cdot\cdots\cdot\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.shuffle.lyndon.full.a.5}%
)}\right) \\
&  =u\underset{\sigma}{\shuffle}v\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.shuffle.lyndon.full.a.6})}\right)  .
\end{align*}
Thus, $u\underset{\sigma}{\shuffle}v$ is the lexicographically highest element
of the multiset $u\shuffle v$ (since we know that
$u\underset{\widetilde{\sigma}}{\shuffle}v$ is the lexicographically highest
element of the multiset $u\shuffle v$). This proves Theorem
\ref{thm.shuffle.lyndon.full}(a).
\end{proof}

Now, in order to prove Theorem \ref{thm.shuffle.lyndon.used}, we record a very
simple fact about counting shuffles:

\begin{proposition}
\label{prop.shuffle.decrmax}Let $p\in \NN$ and $q\in \NN$. Let
$\mathfrak{W}$ be a totally ordered set, and let $h:\left\{  1,2,\ldots
,p+q\right\}  \rightarrow\mathfrak{W}$ be a map. Assume that $h\left(
1\right)  \geq h\left(  2\right)  \geq\cdots\geq h\left(  p\right)  $ and
$h\left(  p+1\right)  \geq h\left(  p+2\right)  \geq\cdots\geq h\left(
p+q\right)  $.

For every $w\in\mathfrak{W}$, let $\mathfrak{a}\left(  w\right)  $ denote the
number of all $i\in\left\{  1,2,\ldots,p\right\}  $ satisfying $h\left(
i\right)  =w$, and let $\mathfrak{b}\left(  w\right)  $ denote the number of
all $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $ satisfying $h\left(  i\right)
=w$.

Then, the number of $\tau\in\operatorname{Sh}_{p,q}$ satisfying
$h\left(  \tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)
\right)  \geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $ is
$\prod_{w\in\mathfrak{W}}\dbinom{\mathfrak{a}\left(  w\right)  +\mathfrak{b}%
\left(  w\right)  }{\mathfrak{a}\left(  w\right)  }$. (Of course, all but
finitely many factors of this product are $1$.)
\end{proposition}

\begin{exercise}
\label{exe.prop.shuffle.decrmax}Prove Proposition \ref{prop.shuffle.decrmax}.
\end{exercise}

\begin{proof}
[Proof of Theorem \ref{thm.shuffle.lyndon.used}.]Let $n=\ell\left(  u\right)
$ and $m=\ell\left(  v\right)  $. Define $\alpha$, $\beta$ and $\left(
I_{1},I_{2},\ldots,I_{p+q}\right)  $ as in Theorem
\ref{thm.shuffle.lyndon.full}.

Since $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is the CFL factorization of
$u$, we have $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$ and $a_{1}a_{2} \cdots
a_{p}=u$. Similarly, $b_{1}\geq b_{2}\geq\cdots\geq b_{q}$ and $b_{1}b_{2}
\cdots b_{q}=v$.

From (\ref{pf.thm.shuffle.lyndon.full.prep1}), we see that $\left(  uv\right)
\left[  I_{i}\right]  =a_{i}$ for every $i\in\left\{  1,2,\ldots,p\right\}  $.
From (\ref{pf.thm.shuffle.lyndon.full.prep1'}), we see that $\left(
uv\right)  \left[  I_{i}\right]  =b_{i-p}$ for every $i\in\left\{
p+1,p+2,\ldots,p+q\right\}  $. Combining these two equalities, we obtain%
\begin{equation}
\left(  uv\right)  \left[  I_{i}\right]  = \begin{cases}
a_{i},& \text{if }i\leq p;\\
b_{i-p},& \text{if }i>p
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,p+q\right\}  .
\label{pf.thm.shuffle.lyndon.used.uvIi}
\end{equation}
In other words,
\begin{equation}
\left(  \left(  uv\right)  \left[  I_{1}\right]  ,\left(  uv\right)  \left[
I_{2}\right]  ,\ldots,\left(  uv\right)  \left[  I_{p+q}\right]  \right)
=\left(  a_{1},a_{2},\ldots,a_{p},b_{1},b_{2},\ldots,b_{q}\right)  .
\label{pf.thm.shuffle.lyndon.used.a.1}%
\end{equation}


(a) Let $z$ be the lexicographically highest element of the multiset
$u\shuffle v$. We must prove that $z=c_{1}c_{2} \cdots c_{p+q}$.

Since $z\in u\shuffle v$, we can write $z$ in the form $u\underset{\sigma
}{\shuffle}v$ for some $\sigma\in\operatorname{Sh}_{n,m}$ (since we
can write any element of $u\shuffle v$ in this form). Consider this $\sigma$.
Then, $u\underset{\sigma}{\shuffle}v=z$ is the lexicographically highest
element of the multiset $u\shuffle v$. Hence, Theorem
\ref{thm.shuffle.lyndon.full}(b) yields that there exists a unique permutation
$\tau\in\operatorname{Sh}_{p,q}$ satisfying $\left(  uv\right)
\left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)  \left[
I_{\tau\left(  p+q\right)  }\right]  $ and $\sigma=\operatorname*{iper}\left(
\alpha\beta,\tau\right)  $. Consider this $\tau$.

Now, $\tau\in\operatorname{Sh}_{p,q}\subset\Symm_{p+q}$ is a
permutation, and thus the list $\left(  \left(  uv\right)  \left[
I_{\tau\left(  1\right)  }\right]  ,\left(  uv\right)  \left[  I_{\tau\left(
2\right)  }\right]  ,\ldots,\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]  \right)  $ is a rearrangement of the list $\left(
\left(  uv\right)  \left[  I_{1}\right]  ,\left(  uv\right)  \left[
I_{2}\right]  ,\ldots,\left(  uv\right)  \left[  I_{p+q}\right]  \right)  $.
Due to (\ref{pf.thm.shuffle.lyndon.used.a.1}), this rewrites as follows: The
list $\left(  \left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]
,\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  ,\ldots,\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  \right)  $ is a
rearrangement of the list $\left(  a_{1},a_{2},\ldots,a_{p},b_{1},b_{2}%
,\ldots,b_{q}\right)  $. Hence, $\left(  \left(  uv\right)  \left[
I_{\tau\left(  1\right)  }\right]  ,\left(  uv\right)  \left[  I_{\tau\left(
2\right)  }\right]  ,\ldots,\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]  \right)  $ is the result of sorting the list $\left(
a_{1},a_{2},\ldots,a_{p},b_{1},b_{2},\ldots,b_{q}\right)  $ in decreasing
order (since $\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \geq
\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  $).
But since the result of sorting the list $\left(  a_{1},a_{2},\ldots
,a_{p},b_{1},b_{2},\ldots,b_{q}\right)  $ in decreasing order is $\left(
c_{1},c_{2},\ldots,c_{p+q}\right)  $, this becomes%
\[
\left(  \left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  ,\left(
uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  ,\ldots,\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  \right)  =\left(
c_{1},c_{2},\ldots,c_{p+q}\right)  .
\]
Hence,%
\[
\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \cdot\left(
uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \cdot\cdots\cdot\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  =c_{1}\cdot c_{2}%
\cdot\cdots\cdot c_{p+q}.
\]
But Lemma \ref{lem.shuffle.lyndon.step2} yields
\[
u\underset{\sigma}{\shuffle}v=\left(  uv\right)  \left[  I_{\tau\left(
1\right)  }\right]  \cdot\left(  uv\right)  \left[  I_{\tau\left(  2\right)
}\right]  \cdot\cdots\cdot\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]  .
\]
Altogether, we have
\begin{align*}
z = u\underset{\sigma}{\shuffle}v=\left(  uv\right)  \left[  I_{\tau\left(
1\right)  }\right]  \cdot\left(  uv\right)  \left[  I_{\tau\left(  2\right)
}\right]  \cdot\cdots\cdot\left(  uv\right)  \left[  I_{\tau\left(
p+q\right)  }\right]
= c_{1}\cdot c_{2}\cdot\cdots\cdot c_{p+q}=c_{1}c_{2} \cdots c_{p+q}.
\end{align*}
This proves Theorem \ref{thm.shuffle.lyndon.used}(a).

(b) Recall that $u\shuffle v=\left\{  u\underset{\sigma}{\shuffle}
v\ :\ \sigma\in\operatorname{Sh}_{n,m}\right\}  _{\text{multiset}}$. Hence,
\begin{align*}
&  \left(  \text{the multiplicity with which the lexicographically highest
element of the multiset}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  u\shuffle v\text{ appears in the multiset
}u\shuffle v\right) \\
&  =\left(  \text{the number of all }\sigma\in
\operatorname{Sh}_{n,m}\text{ such that }u\underset{\sigma}{\shuffle}v\text{ is
the}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{lexicographically highest element of the
multiset }u\shuffle v\right)  .
\end{align*}


However, for a given $\sigma\in\operatorname{Sh}_{n,m}$, we know that
$u\underset{\sigma}{\shuffle}v$ is the lexicographically highest element of
the multiset $u\shuffle v$ if and only if $\sigma$ can be written in the form
$\sigma=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $ for some
$\tau\in\operatorname{Sh}_{p,q}$ satisfying $\left(  uv\right)
\left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)  \left[
I_{\tau\left(  p+q\right)  }\right]  $.\ \ \ \ \footnote{In fact, the
``if'' part of this assertion follows from
Theorem \ref{thm.shuffle.lyndon.full}(a), whereas its ``only
if'' part follows from Theorem \ref{thm.shuffle.lyndon.full}(b).}
Hence,
\begin{align*}
&  \left(  \text{the number of all }\sigma\in\operatorname{Sh}_{n,m}%
\text{ such that }u\underset{\sigma}{\shuffle}v\text{ is the}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{lexicographically highest element of the
multiset }u\shuffle v\right) \\
&  =\left(  \text{the number of all }\sigma\in
\operatorname{Sh}_{n,m}\text{ which can be written in the form }\sigma
=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{for some }\tau\in
\operatorname{Sh}_{p,q}\text{ satisfying }\left(  uv\right)  \left[  I_{\tau\left(
1\right)  }\right]  \geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)
}\right]  \geq\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)
}\right]  \right) \\
&  =\left(  \text{the number of all }\tau\in\operatorname{Sh}_{p,q}%
\text{ satisfying }\left(  uv\right)  \left[  I_{\tau\left(  1\right)
}\right]  \geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]
\geq\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]
\right)
\end{align*}
(because if a $\sigma\in\operatorname{Sh}_{n,m}$ can be written in
the form $\sigma=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $ for
some $\tau\in\operatorname{Sh}_{p,q}$ satisfying $\left(  uv\right)
\left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)  \left[
I_{\tau\left(  p+q\right)  }\right]  $, then $\sigma$ can be written
\textbf{uniquely} in this form\footnote{\textit{Proof.} Let $\sigma
\in\operatorname{Sh}_{n,m}$ be such that $\sigma$ can be written in
the form $\sigma=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $ for
some $\tau\in\operatorname{Sh}_{p,q}$ satisfying $\left(  uv\right)
\left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)  \left[
I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)  \left[
I_{\tau\left(  p+q\right)  }\right]  $. Then, the word $u\underset{\sigma
}{\shuffle}v$ is the lexicographically highest element of the multiset
$u\shuffle v$ (according to Theorem \ref{thm.shuffle.lyndon.full}(a)). Hence,
there exists a unique permutation $\tau\in\operatorname{Sh}_{p,q}$
satisfying $\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \geq
\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  $
and $\sigma=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $ (according
to Theorem \ref{thm.shuffle.lyndon.full}(b)). In other words, $\sigma$ can be
written \textbf{uniquely} in the form $\sigma=\operatorname*{iper}\left(
\alpha\beta,\tau\right)  $ for some $\tau\in\operatorname{Sh}_{p,q}$
satisfying $\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \geq
\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  $,
qed.}). Thus,%
\begin{align}
&  \left(  \text{the multiplicity with which the lexicographically highest
element of the multiset}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  u\shuffle v\text{ appears in the multiset
}u\shuffle v\right) \nonumber\\
&  =\left(  \text{the number of all }\sigma\in
\operatorname{Sh}_{n,m}\text{ such that }u\underset{\sigma}{\shuffle}v\text{ is
the}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{lexicographically highest element of the
multiset }u\shuffle v\right) \nonumber\\
&  =\left(  \text{the number of all }\tau\in\operatorname{Sh}_{p,q}%
\text{ satisfying }\left(  uv\right)  \left[  I_{\tau\left(  1\right)
}\right]  \geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]
\geq\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]
\right)  . \label{pf.thm.shuffle.lyndon.used.b.3}%
\end{align}


Now, define a map $h:\left\{  1,2,\ldots,p+q\right\}  \rightarrow\mathfrak{L}$
by%
\[
h\left(  i\right)  = \begin{cases}
a_{i},& \text{if }i\leq p;\\
b_{i-p},& \text{if }i>p
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,p+q\right\}  .
\]
Then, $h\left(  1\right)  \geq h\left(  2\right)  \geq\cdots\geq h\left(
p\right)  $ (because this is just a rewriting of $a_{1}\geq a_{2}\geq
\cdots\geq a_{p}$) and $h\left(  p+1\right)  \geq h\left(  p+2\right)
\geq\cdots\geq h\left(  p+q\right)  $ (since this is just a rewriting of
$b_{1}\geq b_{2}\geq\cdots\geq b_{q}$). For every $w\in\mathfrak{L}$, the
number of all $i\in\left\{  1,2,\ldots,p\right\}  $ satisfying $h\left(
i\right)  =w$ is
\begin{align*}
&  \left\vert \left\{  i\in\left\{  1,2,\ldots,p\right\}  \ \mid
\ \underbrace{h\left(  i\right)  }_{=a_{i}}=w\right\}  \right\vert \\
&  =\left\vert \left\{  i\in\left\{  1,2,\ldots,p\right\}  \ \mid
\ a_{i}=w\right\}  \right\vert \\
&  =\left(  \text{the number of terms in the list }\left(  a_{1},a_{2}%
,\ldots,a_{p}\right)  \text{ which are equal to }w\right) \\
&  =\left(  \text{the number of terms in the CFL factorization of }u\text{
which are equal to }w\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the list }\left(  a_{1}%
,a_{2},\ldots,a_{p}\right)  \text{ is the CFL factorization of }u\right) \\
&  =\operatorname{mult}_{w}u
\end{align*}
(because $\operatorname{mult}_{w}u$ is defined as the number of
terms in the CFL factorization of $u$ which are equal to $w$). Similarly, for
every $w\in\mathfrak{L}$, the number of all $i\in\left\{  p+1,p+2,\ldots
,p+q\right\}  $ satisfying $h\left(  i\right)  =w$ equals
$\operatorname{mult}_{w}v$. Thus, we can apply Proposition
\ref{prop.shuffle.decrmax} to $\mathfrak{W}=\mathfrak{L}$, $\mathfrak{a}%
\left(  w\right)  =\operatorname{mult}_{w}u$ and $\mathfrak{b}%
\left(  w\right)  =\operatorname{mult}_{w}v$. As a result, we see
that the number of $\tau\in\operatorname{Sh}_{p,q}$ satisfying
$h\left(  \tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)
\right)  \geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $ is
$\prod_{w\in\mathfrak{L}}\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}$. In
other words,%
\begin{align}
&  \left(  \text{the number of all }\tau\in\operatorname{Sh}_{p,q}%
\text{ satisfying }h\left(  \tau\left(  1\right)  \right)  \geq h\left(
\tau\left(  2\right)  \right)  \geq\cdots\geq h\left(  \tau\left(  p+q\right)
\right)  \right) \nonumber\\
&  =\prod_{w\in\mathfrak{L}}\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}.
\label{pf.thm.shuffle.lyndon.used.b.8}%
\end{align}


However, for every $i\in\left\{  1,2,\ldots,p+q\right\}  $, we have%
\[
h\left(  i\right)  =  \begin{cases}
a_{i},& \text{if }i\leq p;\\
b_{i-p},& \text{if }i>p
\end{cases}
\quad =\left(  uv\right)  \left[  I_{i}\right]  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.thm.shuffle.lyndon.used.uvIi})}\right)  .
\]
Hence, for any $\tau\in\operatorname{Sh}_{p,q}$, the condition
$h\left(  \tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)
\right)  \geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $ is
equivalent to $\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]
\geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]  \geq
\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  $.
Thus,
\begin{align*}
&  \left(  \text{the number of all }\tau\in\operatorname{Sh}_{p,q}%
\text{ satisfying }\underbrace{h\left(  \tau\left(  1\right)  \right)  \geq
h\left(  \tau\left(  2\right)  \right)  \geq\cdots\geq h\left(  \tau\left(
p+q\right)  \right)  }_{\substack{\text{this is equivalent to}\\\left(
uv\right)  \left[  I_{\tau\left(  1\right)  }\right]  \geq\left(  uv\right)
\left[  I_{\tau\left(  2\right)  }\right]  \geq\cdots\geq\left(  uv\right)
\left[  I_{\tau\left(  p+q\right)  }\right]  }}\right) \\
&  =\left(  \text{the number of all }\tau\in\operatorname{Sh}_{p,q}%
\text{ satisfying }\left(  uv\right)  \left[  I_{\tau\left(  1\right)
}\right]  \geq\left(  uv\right)  \left[  I_{\tau\left(  2\right)  }\right]
\geq\cdots\geq\left(  uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]
\right) \\
&  =\left(  \text{the multiplicity with which the lexicographically highest
element of the multiset}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  u\shuffle v\text{ appears in the multiset
}u\shuffle v\right)
\end{align*}
(by (\ref{pf.thm.shuffle.lyndon.used.b.3})). Compared with
(\ref{pf.thm.shuffle.lyndon.used.b.8}), this yields%
\begin{align*}
&  \left(  \text{the multiplicity with which the lexicographically highest
element of the multiset}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  u\shuffle v\text{ appears in the multiset
}u\shuffle v\right) \\
&  =\prod_{w\in\mathfrak{L}}\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}.
\end{align*}
This proves Theorem \ref{thm.shuffle.lyndon.used}(b).

(c) We shall use the notations of Theorem \ref{thm.shuffle.lyndon.used}(a) and
Theorem \ref{thm.shuffle.lyndon.used}(b).

Assume that $a_{i}\geq b_{j}$ for every $i\in\left\{  1,2,\ldots,p\right\}  $
and $j\in\left\{  1,2,\ldots,q\right\}  $. This, combined with $a_{1}\geq
a_{2}\geq\cdots\geq a_{p}$ and $b_{1}\geq b_{2}\geq\cdots\geq b_{q}$, yields
that $a_{1}\geq a_{2}\geq\cdots\geq a_{p}\geq b_{1}\geq b_{2}\geq\cdots\geq
b_{q}$. Thus, the list $\left(  a_{1},a_{2},\ldots,a_{p},b_{1},b_{2}%
,\ldots,b_{q}\right)  $ is weakly decreasing. Thus, the result of sorting the
list $\left(  a_{1},a_{2},\ldots,a_{p},b_{1},b_{2},\ldots,b_{q}\right)  $ in
decreasing order is the list $\left(  a_{1},a_{2},\ldots,a_{p},b_{1}%
,b_{2},\ldots,b_{q}\right)  $ itself. But since this result is $\left(
c_{1},c_{2},\ldots,c_{p+q}\right)  $, this shows that $\left(  c_{1}%
,c_{2},\ldots,c_{p+q}\right)  =\left(  a_{1},a_{2},\ldots,a_{p},b_{1}%
,b_{2},\ldots,b_{q}\right)  $. Hence, $c_{1}c_{2} \cdots c_{p+q}%
=\underbrace{a_{1}a_{2} \cdots a_{p}}_{=u}\underbrace{b_{1}b_{2} \cdots b_{q}%
}_{=v}=uv$. Now, Theorem \ref{thm.shuffle.lyndon.used}(a) yields that the
lexicographically highest element of the multiset $u\shuffle v$ is $c_{1}c_{2}
\cdots c_{p+q}=uv$. This proves Theorem \ref{thm.shuffle.lyndon.used}(c).

(d) We shall use the notations of Theorem \ref{thm.shuffle.lyndon.used}(a) and
Theorem \ref{thm.shuffle.lyndon.used}(b).

Assume that $a_{i}>b_{j}$ for every $i\in\left\{  1,2,\ldots,p\right\}  $ and
$j\in\left\{  1,2,\ldots,q\right\}  $. Thus, $a_{i}\geq b_{j}$ for every
$i\in\left\{  1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}
$. Hence, Theorem \ref{thm.shuffle.lyndon.used}(c) yields that the
lexicographically highest element of the multiset $u\shuffle v$ is $uv$.
Therefore, Theorem \ref{thm.shuffle.lyndon.used}(b) shows that the
multiplicity with which this word $uv$ appears in the multiset $u\shuffle v$
is $\prod_{w\in\mathfrak{L}}\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}$.

Now, every $w\in\mathfrak{L}$ satisfies $\dbinom{
\operatorname{mult}_{w}u+\operatorname{mult}_{w}v}{
\operatorname{mult}_{w}u}=1$\ \ \ $\ $\footnote{\textit{Proof.} Assume the contrary.
Then, there exists at least one $w\in\mathfrak{L}$ such that $\dbinom
{\operatorname{mult}_{w}u+\operatorname{mult}_{w}%
v}{\operatorname{mult}_{w}u}\neq1$. Consider this $w$. Both
$\operatorname{mult}_{w}u$ and $\operatorname{mult}_{w}v$
must be positive (since $\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}%
\neq1$). Since $\operatorname{mult}_{w}u$ is positive, there must be
at least one term in the CFL factorization of $u$ which is equal to $w$. In
other words, there is at least one $i\in\left\{  1,2,\ldots,p\right\}  $
satisfying $a_{i}=w$ (since $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is the
CFL factorization of $u$). Similarly, there is at least one $j\in\left\{
1,2,\ldots,q\right\}  $ satisfying $b_{j}=w$. These $i$ and $j$ satisfy
$a_{i}=w=b_{j}$, which contradicts $a_{i}>b_{j}$. This contradiction shows
that our assumption was false, qed.}. Thus, as we know, the multiplicity with
which this word $uv$ appears in the multiset $u\shuffle v$ is $\prod
_{w\in\mathfrak{L}}\underbrace{\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}%
}_{=1}=\prod_{w\in\mathfrak{L}}1=1$. This proves Theorem
\ref{thm.shuffle.lyndon.used}(d).

(e) We shall use the notations of Theorem \ref{thm.shuffle.lyndon.used}(a) and
Theorem \ref{thm.shuffle.lyndon.used}(b).

Since $u$ is a Lyndon word, the $1$-tuple $\left(  u\right)  $ is the CFL
factorization of $u$. Hence, we can apply Theorem
\ref{thm.shuffle.lyndon.used}(c) to $1$ and $\left(  u\right)  $ instead of
$p$ and $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $. As a result, we conclude
that the lexicographically highest element of the multiset $u\shuffle v$ is
$uv$. It remains to prove that the multiplicity with which this word $uv$
appears in the multiset $u\shuffle v$ is $\operatorname{mult}_{u}%
v+1$.

For every $w\in\mathfrak{L}$ satisfying $w\neq u$, we have
\begin{equation}
\operatorname{mult}_{w}u=0 \label{pf.thm.shuffle.lyndon.used.e.0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.shuffle.lyndon.used.e.0}):} Let
$w\in\mathfrak{L}$ be such that $w\neq u$. Then, the number of terms in the
list $\left(  u\right)  $ which are equal to $w$ is $0$. Since $\left(
u\right)  $ is the CFL factorization of $u$, this rewrites as follows: The
number of terms in the CFL factorization of $u$ which are equal to $w$ is $0$.
In other words, $\operatorname{mult}_{w}u=0$. This proves
(\ref{pf.thm.shuffle.lyndon.used.e.0}).}. Also, $
\operatorname{mult}_{u}u=1$ (for a similar reason). But $uv$ is the lexicographically
highest element of the multiset $u\shuffle v$. Hence, the multiplicity with
which the word $uv$ appears in the multiset $u\shuffle v$ is the multiplicity
with which the lexicographically highest element of the multiset $u\shuffle v$
appears in the multiset $u\shuffle v$. According to Theorem
\ref{thm.shuffle.lyndon.used}(b), the latter multiplicity is%
\begin{align*}
&  \prod_{w\in\mathfrak{L}}\dbinom{\operatorname{mult}_{w}%
u+\operatorname{mult}_{w}v}{\operatorname{mult}_{w}u}\\
&  =\underbrace{\dbinom{\operatorname{mult}_{u}%
u+\operatorname{mult}_{u}v}{\operatorname{mult}_{u}u}%
}_{\substack{=\dbinom{1+\operatorname{mult}_{u}v}{1}\\\text{(since
}\operatorname{mult}_{u}u=1\text{)}}}\cdot\prod\limits_{\substack{w\in
\mathfrak{L;}\\w\neq u}}\underbrace{\dbinom{
\operatorname{mult}_{w}u+\operatorname{mult}_{w}v}{
\operatorname{mult}_{w}u}}_{\substack{=\dbinom{0+\operatorname{mult}_{w}v}%
{0}\\\text{(since }\operatorname{mult}_{w}u=0\text{ (by
(\ref{pf.thm.shuffle.lyndon.used.e.0})))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }u\in\mathfrak{L}\right) \\
&  =\underbrace{\dbinom{1+\operatorname{mult}_{u}v}{1}%
}_{=1+\operatorname{mult}_{u}v=\operatorname{mult}_{u}%
v+1}\cdot\prod\limits_{\substack{w\in\mathfrak{L;}\\w\neq u}}\underbrace{\dbinom
{0+\operatorname{mult}_{w}v}{0}}_{=1}=\left(  
\operatorname{mult}_{u}v+1\right)  \cdot\underbrace{\prod\limits_{\substack{w\in\mathfrak{L;}%
\\w\neq u}}1}_{=1}=\operatorname{mult}_{u}v+1.
\end{align*}
This proves Theorem \ref{thm.shuffle.lyndon.used}(e).
\end{proof}

As an application of our preceding results, we can prove a further necessary
and sufficient criterion for a word to be Lyndon; this criterion is due to
Chen/Fox/Lyndon \cite[$\mathfrak{A}^{\prime\prime}=\mathfrak{A}^{\prime
\prime\prime\prime}$]{ChenFoxLyndon}:

\begin{exercise}
\label{exe.shuffle.lyndon.cflcrit}Let $w\in\mathfrak{A}^{\ast}$ be a nonempty
word. Prove that $w$ is Lyndon if and only if for any two nonempty words
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ satisfying $w=uv$,
there exists at least one $s\in u\shuffle v$ satisfying $s>w$.
\end{exercise}

\subsection{\label{subsect.shuffle.radford}Radford's theorem on the shuffle
algebra}

We recall that our goal in Chapter \ref{sect.QSym.lyndon} is to exhibit an
algebraically independent generating set of the $\kk$-algebra
$\Qsym$. Having the notion of Lyndon words -- which will, to
some extent, but not literally, parametrize this generating set -- in place,
we could start the construction of this generating set immediately. However,
it might come off as rather unmotivated this way, and so we begin with some
warmups. First, we shall prove Radford's theorem on the shuffle algebra.

\begin{definition}
\label{def.shuffle.polyalg}
A \dfn{polynomial algebra} will mean a $\kk$-algebra which is
isomorphic to the polynomial ring $\kk \left[  x_{i}\ \mid\ i\in
I\right]  $ as a $\kk$-algebra (for some indexing set $I$). Note that
$I$ need not be finite.

Equivalently, a polynomial algebra can be defined as a $\kk$-algebra
which has an algebraically independent (over $\kk$) generating set. Yet
equivalently, a polynomial algebra can be defined as a $\kk$-algebra
which is isomorphic to the symmetric algebra of a free $\kk$-module.
\end{definition}

Keep in mind that when we say that a certain bialgebra $A$ is a polynomial
algebra, we are making no statement about the coalgebra structure on $A$.
The isomorphism from $A$ to the symmetric algebra of a free $\kk$-module
need not be a coalgebra isomorphism, and the algebraically independent
generating set of $A$ need not consist of primitives. Thus, showing that
a bialgebra $A$ is a polynomial algebra does not trivialize the study of
its bialgebraic structure.

\begin{remark}
\label{rmk.shuffle.shufflealg}
Let $V$ be a $\kk$-module, and let
$\mathfrak{A}$ be a totally ordered set. Let $b_{a}$ be an element of $V$ for
every $a\in\mathfrak{A}$. Consider the shuffle algebra $\operatorname*{Sh}%
\left(  V\right)  $ (defined in Definition \ref{prop.shuffle-alg}).

For every word $w\in\mathfrak{A}^{\ast}$ over the alphabet $\mathfrak{A}$, let
us define an element $b_{w}$ of $\operatorname*{Sh}\left(  V\right)  $ by
$b_{w}=b_{w_{1}}b_{w_{2}} \cdots b_{w_{\ell}}$, where $\ell$ is the length of
$w$. (The multiplication used here is that of $T\left(  V\right)  $, not that
of $\operatorname*{Sh}\left(  V\right)  $; the latter is denoted by
$\shufmult$.)

Let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words over
the alphabet $\mathfrak{A}$. Let $n=\ell\left(  u\right)  $ and $m=\ell\left(
v\right)  $. Then,
\[
b_{u} \shufmult b_{v}=\sum_{\sigma\in
\operatorname{Sh}_{n,m}}b_{u\underset{\sigma}{\shuffle}v}.
\]

\end{remark}

\begin{exercise}
\label{exe.rmk.shuffle.shufflealg}Prove Remark \ref{rmk.shuffle.shufflealg}.

[\textbf{Hint:} This follows from the definition of $\shufmult$.]
\end{exercise}

We can now state Radford's theorem \cite[Theorem 3.1.1(e)]{Radford-shuffle}:

\begin{theorem}
\label{thm.shuffle.radford}Assume that $\QQ$ is a subring of
$\kk$. Let $V$ be a free $\kk$-module with a basis $\left(
b_{a}\right)  _{a\in\mathfrak{A}}$, where $\mathfrak{A}$ is a totally ordered
set. Then, the shuffle algebra $\operatorname*{Sh}\left(  V\right)  $ (defined
in Definition \ref{prop.shuffle-alg}) is a polynomial $\kk$-algebra. An
algebraically independent generating set of $\operatorname*{Sh}\left(
V\right)  $ can be constructed as follows:

For every word $w\in\mathfrak{A}^{\ast}$ over the alphabet $\mathfrak{A}$, let
us define an element $b_{w}$ of $\operatorname*{Sh}\left(  V\right)  $ by
$b_{w}=b_{w_{1}}b_{w_{2}} \cdots b_{w_{\ell}}$, where $\ell$ is the length of
$w$. (The multiplication used here is that of $T\left(  V\right)  $, not that
of $\operatorname*{Sh}\left(  V\right)  $; the latter is denoted by
$\shufmult$.) Let $\mathfrak{L}$ denote the set of all Lyndon words
over the alphabet $\mathfrak{A}$. Then, $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$ is an algebraically independent generating set of the
$\kk$-algebra $\operatorname*{Sh}\left(  V\right)  $.
\end{theorem}

\begin{example}
For this example, let $\mathfrak{A}$ be the alphabet $\left\{  1,2,3,\ldots
\right\}  $ with total order given by $1<2<3< \cdots$, and assume that
$\QQ$ is a subring of $\kk$. Let $V$ be the free $\kk$-module with basis
$\left(  b_{a}\right)  _{a\in\mathfrak{A}}$. We use the
notations of Theorem \ref{thm.shuffle.radford}. Then, Theorem
\ref{thm.shuffle.radford} yields that $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$ is an algebraically independent generating set of the
$\kk$-algebra $\operatorname*{Sh}\left(  V\right)  $. Here are some
examples of elements of $\operatorname*{Sh}\left(  V\right)  $ written as
polynomials in this generating set:
\begin{align*}
b_{12}  &  =b_{12}\ \ \ \ \ \ \ \ \ \ \left(  \text{the word }12\text{ itself
is Lyndon}\right)  ;\\
b_{21}  &  =b_{1} \shufmult b_{2}-b_{12};\\
b_{11}  &  =\dfrac{1}{2}b_{1} \shufmult b_{1};\\
b_{123}  &  =b_{123}\ \ \ \ \ \ \ \ \ \ \left(  \text{the word }123\text{
itself is Lyndon}\right)  ;\\
b_{132}  &  =b_{132}\ \ \ \ \ \ \ \ \ \ \left(  \text{the word }132\text{
itself is Lyndon}\right)  ;\\
b_{213}  &  =b_{2} \shufmult b_{13}-b_{123}-b_{132};\\
b_{231}  &  =b_{23} \shufmult b_{1}-b_{2} \shufmult
b_{13}+b_{132};\\
b_{312}  &  =b_{3} \shufmult b_{12}-b_{123}-b_{132};\\
b_{321}  &  =b_{1} \shufmult b_{2} \shufmult b_{3}%
-b_{23} \shufmult b_{1}-b_{3} \shufmult b_{12}+b_{123};\\
b_{112}  &  =b_{112}\ \ \ \ \ \ \ \ \ \ \left(  \text{the word }112\text{
itself is Lyndon}\right)  ;\\
b_{121}  &  =b_{12} \shufmult b_{1}-2b_{112};\\
b_{1212}  &  =\dfrac{1}{2}b_{12} \shufmult b_{12}-2b_{1122};\\
b_{4321}  &  =b_{1} \shufmult b_{2} \shufmult
b_{3} \shufmult b_{4}-b_{1} \shufmult b_{2}
\shufmult b_{34}-b_{1} \shufmult b_{23} \shufmult
b_{4}-b_{12} \shufmult b_{3} \shufmult b_{4}\\
&  \ \ \ \ \ \ \ \ \ \ +b_{1} \shufmult b_{234}+b_{12}
\shufmult b_{34}+b_{123} \shufmult b_{4}-b_{1234}.
\end{align*}
\footnote{A pattern emerges in the formulas for $b_{21}$, $b_{321}$ and
$b_{4321}$: for every $n\in \NN$, we have
\[
b_{\left(  n,n-1,\ldots,1\right)  }=\sum_{\alpha\in
\Comp_{n}}\left(  -1\right)  ^{n-\ell\left(  \alpha\right)  }%
b_{\mathbf{d}_{1}\left(  \alpha\right)  } \shufmult b_{\mathbf{d}%
_{2}\left(  \alpha\right)  } \shufmult \cdots \shufmult%
b_{\mathbf{d}_{\ell\left(  \alpha\right)  }\left(  \alpha\right)  },
\]
where $\left(  \mathbf{d}_{1}\left(  \alpha\right)  \right)  \cdot\left(
\mathbf{d}_{2}\left(  \alpha\right)  \right)  \cdot\cdots\cdot\left(
\mathbf{d}_{\ell\left(  \alpha\right)  }\left(  \alpha\right)  \right)  $ is
the factorization of the word $\left(  1,2,\ldots,n\right)  $ into factors of
length $\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{\ell}$ (where
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $). This
can be proved by an application of Lemma~\ref{lem.moebius-bool.comps}(a)
(as it is easy to see that for any composition $\alpha$ of $n$, we have
\begin{align*}
b_{\mathbf{d}_{1}\left(  \alpha\right)  } \shufmult b_{\mathbf{d}%
_{2}\left(  \alpha\right)  } \shufmult \cdots \shufmult%
b_{\mathbf{d}_{\ell\left(  \alpha\right)  }\left(  \alpha\right)  }
&= \left(\text{the sum of } b_\pi \text{ for all words }
     \pi \in \Symm_n \text{ satisfying }
     \operatorname{Des}\left(  \pi^{-1}\right)  \subset D\left(  \alpha\right)
   \right) \\
&= \sum\limits_{\substack{\beta \in \Comp_n; \\  \beta \text{ coarsens } \alpha}}
   \ \ %
   \sum\limits_{\substack{\pi \in \Symm_n; \\  \gamma\left(  \pi^{-1}\right) = \beta}}
   b_\pi ,
\end{align*}
where $\gamma\left(\pi^{-1}\right)$ denotes the composition $\tau$
of $n$ satisfying $D \left( \tau \right) = \operatorname{Des}
\left( \pi^{-1} \right)$).}
\end{example}

Note that Theorem \ref{thm.shuffle.radford} cannot survive without the
condition that $\QQ$ be a subring of $\kk$. For instance, for
any $v\in V$, we have $v \shufmult v=2vv$ in $\operatorname*{Sh}%
\left(  V\right)  $, which vanishes if $2=0$ in $\kk$; this stands in
contrast to the fact that polynomial $\kk$-algebras are integral
domains when $\kk$ itself is one. We will see that
$\Qsym$ is less sensitive towards the base ring in this regard
(although proving that $\Qsym$ is a polynomial algebra is much
easier when $\QQ$ is a subring of $\kk$).

\begin{remark}
Theorem~\ref{thm.shuffle.radford} can be contrasted with the
following fact:
If $\QQ$ is a subring of $\kk$,
then the shuffle algebra $\operatorname{Sh}\left(V\right)$
of \textbf{any} $\kk$-module $V$ (not necessarily free!)
is isomorphic (as a $\kk$-algebra) to the symmetric algebra
$\Sym \left( \left(\ker \epsilon\right)
             / \left(\ker \epsilon\right)^2 \right)$
(by Theorem~\ref{thm.leray.leray-e}(e), applied to
$A = \operatorname{Sh}\left(V\right)$).
This fact is closely related to
Theorem~\ref{thm.shuffle.radford}, but neither follows from
it (since Theorem~\ref{thm.shuffle.radford} only considers the
case of free $\kk$-modules $V$) nor yields it (since this
fact does not provide explicit generators for the
$\kk$-module
$\left(\ker \epsilon\right) / \left(\ker \epsilon\right)^2$
and thus for the $\kk$-algebra $\operatorname{Sh}\left(V\right)$).
\end{remark}

In our proof of Theorem \ref{thm.shuffle.radford} (but not only there), we
will use part (a) of the following lemma\footnote{And in a later proof, we
will also use its part (c) (which is tailored for application to
$\Qsym$).}, which makes proving that certain families indexed
by Lyndon words generate certain $\kk$-algebras more comfortable:

\begin{lemma}
\label{lem.lyndonbases}Let $A$ be a commutative $\kk$-algebra. Let
$\mathfrak{A}$ be a totally ordered set. Let $\mathfrak{L}$ be the set of all
Lyndon words over the alphabet $\mathfrak{A}$. Let $b_{w}$ be an element of
$A$ for every $w\in\mathfrak{L}$. For every word $u\in\mathfrak{A}^{\ast}$,
define an element $\mathbf{b}_{u}$ of $A$ by $\mathbf{b}_{u}=b_{a_{1}}%
b_{a_{2}} \cdots b_{a_{p}}$, where $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $
is the CFL factorization of $u$.

\begin{itemize}
\item[(a)] The family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is an
algebraically independent generating set of the $\kk$-algebra $A$ if
and only if the family $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}%
^{\ast}}$ is a basis of the $\kk$-module $A$.

\item[(b)] The family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ generates
the $\kk$-algebra $A$ if and only if the family $\left(  \mathbf{b}%
_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ spans the $\kk$-module $A$.

\item[(c)] Assume that the $\kk$-algebra $A$ is graded. Let
$\operatorname*{wt}:\mathfrak{A}\rightarrow\left\{  1,2,3,\ldots\right\}  $ be
any map such that for every $N\in\left\{  1,2,3,\ldots\right\}  $, the set
$\operatorname{wt}^{-1}\left(  N\right)  $ is finite.

For every word $w\in\mathfrak{A}^{\ast}$, define an element
$\operatorname*{Wt}\left(  w\right)  \in \NN$ by $\operatorname*{Wt}%
\left(  w\right)  =\operatorname*{wt}\left(  w_{1}\right)  +\operatorname*{wt}%
\left(  w_{2}\right)  + \cdots+\operatorname*{wt}\left(  w_{k}\right)  $,
where $k$ is the length of $w$.

Assume that for every $w\in\mathfrak{L}$, the element $b_{w}$ of $A$ is
homogeneous of degree $\operatorname*{Wt}\left(  w\right)  $.

Assume further that the $\kk$-module $A$ has a basis $\left(
g_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ having the property that for every
$u\in\mathfrak{A}^{\ast}$, the element $g_{u}$ of $A$ is homogeneous of degree
$\operatorname*{Wt}\left(  u\right)  $.

Assume also that the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$
generates the $\kk$-algebra $A$.

Then, this family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is an
algebraically independent generating set of the $\kk$-algebra $A$.
\end{itemize}
\end{lemma}

\begin{exercise}
\label{exe.lem.lyndonbases}Prove Lemma \ref{lem.lyndonbases}.

[\textbf{Hint:} For (a) and (b), notice that the $\mathbf{b}_{u}$ are the
``monomials'' in the $b_{w}$. For (c), use
Exercise \ref{exe.linalg.sur-bij}(b) in every homogeneous component of $A$.]
\end{exercise}

The main workhorse of our proof of Theorem \ref{thm.shuffle.radford} will be
the following consequence of Theorem \ref{thm.shuffle.lyndon.used}(c):

\begin{proposition}
\label{prop.shuffle.radford.triang}Let $V$ be a free $\kk$-module with
a basis $\left(  b_{a}\right)  _{a\in\mathfrak{A}}$, where $\mathfrak{A}$ is a
totally ordered set.

For every word $w\in\mathfrak{A}^{\ast}$ over the alphabet $\mathfrak{A}$, let
us define an element $b_{w}$ of $\operatorname*{Sh}\left(  V\right)  $ by
$b_{w}=b_{w_{1}}b_{w_{2}} \cdots b_{w_{\ell}}$, where $\ell$ is the length of
$w$. (The multiplication used here is that of $T\left(  V\right)  $, not that
of $\operatorname*{Sh}\left(  V\right)  $; the latter is denoted by
$\shufmult$.)

For every word $u\in\mathfrak{A}^{\ast}$, define an element $\mathbf{b}_{u}$
by $\mathbf{b}_{u}=b_{a_{1}} \shufmult b_{a_{2}} \shufmult
\cdots \shufmult b_{a_{p}}$, where $\left(  a_{1},a_{2},\ldots
,a_{p}\right)  $ is the CFL factorization of $u$.

If $\ell\in \NN$ and if $x\in\mathfrak{A}^{\ell}$ is a word, then there
is a family $\left(  \eta_{x,y}\right)  _{y\in\mathfrak{A}^{\ell}}%
\in \NN ^{\mathfrak{A}^{\ell}}$ of elements of $\NN$ satisfying
\[
\mathbf{b}_{x}=\sum\limits_{\substack{y\in\mathfrak{A}^{\ell};\\y\leq x}}\eta
_{x,y}b_{y}
\]
and $\eta_{x,x}\neq0$ (in $\NN$).
\end{proposition}

Before we prove this, let us show a very simple lemma:

\begin{lemma}
\label{lem.shuffle.preserves-lex}Let $\mathfrak{A}$ be a totally ordered set.
Let $n\in \NN$ and $m\in \NN$. Let $\sigma\in
\operatorname{Sh}_{n,m}$.

(a) If $u$, $v$ and $v^{\prime}$ are three words satisfying $\ell\left(
u\right)  =n$, $\ell\left(  v\right)  =m$, $\ell\left(  v^{\prime}\right)  =m$
and $v^{\prime}<v$, then $u\underset{\sigma}{\shuffle}v^{\prime}%
<u\underset{\sigma}{\shuffle}v$.

(b) If $u$, $u^{\prime}$ and $v$ are three words satisfying $\ell\left(
u\right)  =n$, $\ell\left(  u^{\prime}\right)  =n$, $\ell\left(  v\right)  =m$
and $u^{\prime}<u$, then $u^{\prime}\underset{\sigma}{\shuffle}%
v<u\underset{\sigma}{\shuffle}v$.

(c) If $u$, $v$ and $v^{\prime}$ are three words satisfying $\ell\left(
u\right)  =n$, $\ell\left(  v\right)  =m$, $\ell\left(  v^{\prime}\right)  =m$
and $v^{\prime}\leq v$, then $u\underset{\sigma}{\shuffle}v^{\prime}\leq
u\underset{\sigma}{\shuffle}v$.
\end{lemma}

\begin{exercise}
\label{exe.shuffle.preserves-lex.pf}Prove Lemma
\ref{lem.shuffle.preserves-lex}.
\end{exercise}

\begin{exercise}
\label{exe.prop.shuffle.radford.triang} Prove Proposition
\ref{prop.shuffle.radford.triang}.

[\textbf{Hint:} Proceed by induction over $\ell$. In the induction step, apply
Theorem \ref{thm.shuffle.lyndon.used}(c)\footnote{Or Theorem
\ref{thm.shuffle.lyndon.used}(e), if you prefer.} to $u = a_{1}$ and $v =
a_{2} a_{3} \cdots a_{p}$, where $\left(  a_{1}, a_{2}, \ldots, a_{p}\right)
$ is the CFL factorization of $x$. Use Lemma \ref{lem.shuffle.preserves-lex}
to get rid of smaller terms.]
\end{exercise}

\begin{exercise}
\label{exe.thm.shuffle.radford} Prove Theorem \ref{thm.shuffle.radford}.

[\textbf{Hint:} According to Lemma \ref{lem.lyndonbases}(a), it suffices to
show that the family $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast
}}$ defined in Proposition \ref{prop.shuffle.radford.triang} is a basis of the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $. When
$\mathfrak{A}$ is finite, the latter can be proven by triangularity using
Proposition \ref{prop.shuffle.radford.triang}. Reduce the general case to that
of finite $\mathfrak{A}$.]
\end{exercise}

\subsection{\label{subsect.shuffle.QSym.1}Polynomial freeness of
$\Qsym$: statement and easy parts}

\begin{definition}
\label{def.lyndon-composition}
For the rest of Section \ref{subsect.shuffle.QSym.1} and
for Section \ref{subsect.shuffle.QSym.2}, we introduce the following
notations: We let $\mathfrak{A}$ be the totally ordered set $\left\{
1,2,3,\ldots\right\}  $ with its natural order (that is, $1<2<3<\cdots$.)
Thus, the words over $\mathfrak{A}$ are precisely the compositions. That is,
$\mathfrak{A}^{\ast}= \Comp $. We let $\mathfrak{L}$ denote the
set of all Lyndon words over $\mathfrak{A}$. These Lyndon words are also
called \emph{Lyndon compositions}\index{Lyndon composition}.
\end{definition}

A natural question is how many Lyndon compositions of a given size exist.
While we will not use the answer, we nevertheless record it:

\begin{exercise}
\label{exe.words.lyndon.compositions.count}Show that the number of Lyndon
compositions of size $n$ equals
\[
\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  \left(  2^{n/d}%
-1\right)  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)
2^{n/d}-\delta_{n,1}
\]
for every positive integer $n$ (where ``$\sum\limits_{d\mid n}$''
means a sum over all positive divisors of $n$, and where
$\mu$ is the number-theoretic M\"{o}bius function).

[\textbf{Hint:} One solution is similar to the solution of Exercise
\ref{exe.words.lyndon.count} using CFL factorization. Another proceeds by
defining a bijection between Lyndon compositions and Lyndon words over a
two-letter alphabet $\left\{  \mathbf{0},\mathbf{1}\right\}  $ (with
$\mathbf{0}<\mathbf{1}$) which are $\neq\mathbf{1}$.\ \ \ \ \footnote{This
bijection is obtained by restricting the bijection
\begin{align*}
\Comp  &  \rightarrow
\left\{ w \in \left\{  \mathbf{0},\mathbf{1}\right\} ^{\ast}
\mid w \text{ does not start with } \mathbf{1} \right\}
,\\
\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)   &  \mapsto
\mathbf{01}^{\alpha_{1}-1}\mathbf{01}^{\alpha_{2}-1}\cdots\mathbf{01}%
^{\alpha_{\ell}-1}%
\end{align*}
(where $\mathbf{01}^{k}$ is to be read as $\mathbf{0}\left(  \mathbf{1}%
^{k}\right)  $, not as $\left(  \mathbf{01}\right)  ^{k}$) to the set of
Lyndon compositions. The idea behind this bijection is well-known in the
Grothendieck-Teichm\"{u}ller community: see, e.g., \cite[\S 3.1]{Henderson}
(and see \cite[Note 5.16]{GelfandKrobLascouxLeclercRetakhThibon} for a
different appearance of this idea).}]
\end{exercise}

Let us now state Hazewinkel's result (\cite[Theorem 8.1]{Hazewinkel},
\cite[\S 6.7]{HazewinkelGubareniKirichenko}) which is the main goal of Chapter
\ref{sect.QSym.lyndon}:

\begin{theorem}
\label{thm.QSym.lyndon}The $\kk$-algebra $\Qsym$ is a
polynomial algebra. It is isomorphic, as a graded $\kk$-algebra, to the
$\kk$-algebra $\kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]
$. Here, the grading on $\kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}%
\right]  $ is defined by setting $\deg\left(  x_{w}\right)  =\sum_{i=1}%
^{\ell\left(  w\right)  }w_{i}$ for every $w\in\mathfrak{L}$.
\end{theorem}

We shall prove Theorem \ref{thm.QSym.lyndon} in the next section
(Section \ref{subsect.shuffle.QSym.2}). But the particular case of Theorem
\ref{thm.QSym.lyndon} when $\QQ$ is a subring of $\kk$ can be
proven more easily; we state it as a proposition:

\begin{proposition}
\label{prop.QSym.lyndon}Assume that $\QQ$ is a subring of $\kk$.
Then, Theorem \ref{thm.QSym.lyndon} holds.
\end{proposition}

We will give two proofs of Proposition \ref{prop.QSym.lyndon} in this
Section \ref{subsect.shuffle.QSym.1}; a third proof of Proposition
\ref{prop.QSym.lyndon} will immediately result from the proof of Theorem
\ref{thm.QSym.lyndon} in Section \ref{subsect.shuffle.QSym.2}. (There
\textbf{is} virtue in giving three different proofs, as they all construct
different isomorphisms $\kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}%
\right]  \rightarrow\Qsym$.)

Our first proof -- originating in Malvenuto's \cite[Corollaire 4.20]%
{Malvenuto} -- can be given right away; it relies on Exercise \ref{exe.Psi}:

\begin{proof}
[First proof of Proposition \ref{prop.QSym.lyndon}.]Let $V$ be the free
$\kk$-module with basis $\left(  \mathfrak{b}_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }$. Endow the $\kk$-module $V$
with a grading by assigning to each basis vector $\mathfrak{b}_{n}$ the degree
$n$. Exercise \ref{exe.Psi}(k) shows that $\Qsym$ is isomorphic
to the shuffle algebra $\operatorname{Sh}\left(  V\right)  $ (defined as in
Proposition~\ref{prop.shuffle-alg}) as Hopf algebras. By being a bit more
careful, we can obtain the slightly stronger result that $\Qsym$
is isomorphic to the shuffle algebra $\operatorname{Sh}\left(  V\right)  $ as
\textbf{graded} Hopf algebras\footnote{\textit{Proof.} In the solution of
Exercise \ref{exe.Psi}(k), we have shown that $\Qsym\cong
T\left(  V\right)  ^{o}$ as graded Hopf algebras. But Remark
\ref{rmk.shuffle-alg.dual}(b) shows that the Hopf algebra $T\left(  V\right)
^{o}$ is naturally isomorphic to the shuffle algebra $\operatorname*{Sh}%
\left(  V^{o}\right)  $ as Hopf algebras; it is easy to see that the natural
isomorphism $T\left(  V\right)  ^{o}\rightarrow\operatorname*{Sh}\left(
V^{o}\right)  $ is graded (because it is the direct sum of the isomorphisms
$\left(  V^{\otimes n}\right)  ^{o}\rightarrow\left(  V^{o}\right)  ^{\otimes
n}$ over all $n\in \NN$, and each of these isomorphisms is graded).
Hence, $T\left(  V\right)  ^{o}\cong\operatorname*{Sh}\left(  V^{o}\right)  $
as graded Hopf algebras. But $V^{o}\cong V$ as graded $\kk$-modules
(since $V$ is of finite type), and thus $\operatorname*{Sh}\left(
V^{o}\right)  \cong\operatorname*{Sh}\left(  V\right)  $ as graded Hopf
algebras. Altogether, we obtain $\Qsym\cong T\left(  V\right)
^{o}\cong\operatorname*{Sh}\left(  V^{o}\right)  \cong\operatorname*{Sh}%
\left(  V\right)  $ as graded Hopf algebras, qed.}. In particular,
$\Qsym\cong\operatorname*{Sh}\left(  V\right)  $ as graded
$\kk$-algebras.

Theorem \ref{thm.shuffle.radford} (applied to $b_{a}=\mathfrak{b}_{a}$) yields
that the shuffle algebra $\operatorname*{Sh}\left(  V\right)  $ is a
polynomial $\kk$-algebra, and that an algebraically independent
generating set of $\operatorname*{Sh}\left(  V\right)  $ can be constructed as follows:

For every word $w\in\mathfrak{A}^{\ast}$ over the alphabet $\mathfrak{A}$, let
us define an element $\mathfrak{b}_{w}$ of $\operatorname*{Sh}\left(
V\right)  $ by $\mathfrak{b}_{w}=\mathfrak{b}_{w_{1}}\mathfrak{b}_{w_{2}}
\cdots\mathfrak{b}_{w_{\ell}}$, where $\ell$ is the length of $w$. (The
multiplication used here is that of $T\left(  V\right)  $, not that of
$\operatorname*{Sh}\left(  V\right)  $; the latter is denoted by
$\shufmult$.) Then, $\left(  \mathfrak{b}_{w}\right)
_{w\in\mathfrak{L}}$ is an algebraically independent generating set of the
$\kk$-algebra $\operatorname*{Sh}\left(  V\right)  $.

For every $w\in\mathfrak{A}^{\ast}$, we have $\mathfrak{b}_{w}=\mathfrak{b}%
_{w_{1}}\mathfrak{b}_{w_{2}} \cdots\mathfrak{b}_{w_{\ell\left(  w\right)  }}$
(by the definition of $\mathfrak{b}_{w}$). For every $w\in\mathfrak{A}^{\ast}%
$, the element $\mathfrak{b}_{w}=\mathfrak{b}_{w_{1}}\mathfrak{b}_{w_{2}}
\cdots\mathfrak{b}_{w_{\ell\left(  w\right)  }}$ of $\operatorname*{Sh}\left(
V\right)  $ is homogeneous of degree $\sum_{i=1}^{\ell\left(  w\right)
}\underbrace{\deg\left(  \mathfrak{b}_{w_{i}}\right)  }_{=w_{i}}=
\sum_{i=1}^{\ell\left(  w\right)  }w_{i}$.

Now, define a grading on the $\kk$-algebra $\kk \left[
x_{w}\ \mid\ w\in\mathfrak{L}\right]  $ by setting $\deg\left(  x_{w}\right)
=\sum_{i=1}^{\ell\left(  w\right)  }w_{i}$ for every $w\in\mathfrak{L}$. By
the universal property of the polynomial algebra $\kk \left[
x_{w}\ \mid\ w\in\mathfrak{L}\right]  $, we can define a $\kk$-algebra
homomorphism $\Phi: \kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]
\rightarrow\operatorname*{Sh}\left(  V\right)  $ by setting%
\[
\Phi\left(  x_{w}\right)  =\mathfrak{b}_{w}\ \ \ \ \ \ \ \ \ \ \text{for every
}w\in\mathfrak{L}.
\]
This homomorphism $\Phi$ is a $\kk$-algebra isomorphism (since $\left(
\mathfrak{b}_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent
generating set of the $\kk$-algebra $\operatorname*{Sh}\left(
V\right)  $) and is graded (because for every $w\in\mathfrak{L}$, the element
$\mathfrak{b}_{w}$ of $\operatorname*{Sh}\left(  V\right)  $ is homogeneous of
degree $\sum_{i=1}^{\ell\left(  w\right)  }w_{i}=\deg\left(  x_{w}\right)  $).
Thus, $\Phi$ is an isomorphism of graded $\kk$-algebras. Hence,
$\operatorname*{Sh}\left(  V\right)  \cong \kk \left[  x_{w}\ \mid
\ w\in\mathfrak{L}\right]  $ as graded $\kk$-algebras. Altogether,
$\Qsym\cong\operatorname*{Sh}\left(  V\right)  \cong%
 \kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]  $ as graded
$\kk$-algebras. Thus, $\Qsym$ is a polynomial algebra.
This proves Theorem \ref{thm.QSym.lyndon} under the assumption that
$\QQ$ be a subring of $\kk$. In other words, this proves
Proposition \ref{prop.QSym.lyndon}.
\end{proof}

Our second proof of Proposition \ref{prop.QSym.lyndon} comes from
Hazewinkel/Gubareni/Kirichenko \cite{HazewinkelGubareniKirichenko} (where
Proposition \ref{prop.QSym.lyndon} appears as \cite[Theorem 6.5.13]%
{HazewinkelGubareniKirichenko}). This proof will construct an explicit
algebraically independent family generating the $\kk$-algebra
$\Qsym$.\ \ \ \ \footnote{We could, of course, obtain such a
family from our above proof as well (this is done by Malvenuto in
\cite[Corollaire 4.20]{Malvenuto}), but it won't be a very simple one.} The
generating set will be very unsophisticated: it will be $\left(  M_{\alpha
}\right)  _{\alpha\in\mathfrak{L}}$, where $\mathfrak{A}$ and $\mathfrak{L}$
are as in Theorem \ref{thm.QSym.lyndon}. Here, we are using the fact that
words over the alphabet $\left\{  1,2,3, \ldots\right\}$ are the same thing
as compositions, so, in particular, a monomial quasisymmetric function
$M_{\alpha}$ is defined for every such word $\alpha$.

It takes a bit of work to show that this family indeed fits the bill. We begin
with a corollary of Proposition \ref{QSym-multiplication} that is essentially
obtained by throwing away all non-bijective maps $f$:

\begin{proposition}
\label{prop.QSym.lyndon.MaMb}Let $\alpha\in\mathfrak{A}^{\ast}$ and $\beta
\in\mathfrak{A}^{\ast}$. Then,%
\begin{align*}
&  M_{\alpha}M_{\beta}\\
&  =\sum_{\gamma\in\alpha\shuffle\beta}M_{\gamma}+\left(  \text{a sum of terms
of the form }M_{\delta}\text{ with }\delta\in\mathfrak{A}^{\ast}\text{
satisfying }\ell\left(  \delta\right)  <\ell\left(  \alpha\right)
+\ell\left(  \beta\right)  \right)  .
\end{align*}
\footnote{The sum $\sum_{\gamma\in\alpha\shuffle\beta}M_{\gamma}$ ranges over
the \textbf{multiset} $\alpha\shuffle\beta$; if an element appears several
times in $\alpha\shuffle\beta$, then it has accordingly many addends
corresponding to it.}
\end{proposition}

\begin{exercise}
\label{exe.prop.QSym.lyndon.MaMb}Prove Proposition \ref{prop.QSym.lyndon.MaMb}.

[\textbf{Hint:} Recall what was said about the $p=\ell+m$ case in Example
\ref{exa.QSym-multiplication}.]
\end{exercise}

\begin{corollary}
\label{cor.QSym.lyndon.MaMb.leq}Let $\alpha\in\mathfrak{A}^{\ast}$ and
$\beta\in\mathfrak{A}^{\ast}$. Then, $M_{\alpha}M_{\beta}$ is a sum of terms
of the form $M_{\delta}$ with $\delta\in\mathfrak{A}^{\ast}$ satisfying
$\ell\left(  \delta\right)  \leq\ell\left(  \alpha\right)  +\ell\left(
\beta\right)  $.
\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.MaMb.leq}Prove Corollary
\ref{cor.QSym.lyndon.MaMb.leq}.
\end{exercise}

We now define a partial order on the compositions of a given nonnegative integer:

\begin{definition}
\label{def.wll}Let $n\in \NN$. We define a binary relation
$\underset{\operatorname*{wll}}{\leq}$ on the set $
\Comp_{n}$ as follows: For two compositions $\alpha$ and $\beta$ in
$\Comp_{n}$, we set $\alpha
\underset{\operatorname*{wll}}{\leq}\beta$ if and only if
\[
\text{either }\ell\left(  \alpha\right)  <\ell\left(  \beta\right)  \text{ or
}\left(  \ell\left(  \alpha\right)  =\ell\left(  \beta\right)  \text{ and
}\alpha\leq\beta\text{ in lexicographic order}\right)  \text{.}%
\]
This binary relation $\underset{\operatorname*{wll}}{\leq}$ is the
smaller-or-equal relation of a total order on $
\Comp_{n}$; we refer to said total order as the \dfn{wll-order} on
$\Comp_{n}$, and we denote by
$\underset{\operatorname*{wll}}{<}$ the smaller relation of this total order.
\end{definition}

Notice that if $\alpha$ and $\beta$ are two compositions satisfying
$\ell\left(  \alpha\right)  =\ell\left(  \beta\right)  $, then $\alpha
\leq\beta$ in lexicographic order if and only if $\alpha\leq\beta$ with
respect to the relation $\leq$ defined in Definition \ref{def.words}.

A remark about the name ``wll-order'' is in
order. We have taken this notation from \cite[Definition 6.7.14]{Hazewinkel},
where it is used for an extension of this order to the whole set
$\Comp$. We will never use this extension, as we will only ever
compare two compositions of the same integer.\footnote{In \cite[Definition
6.7.14]{Hazewinkel}, the name ``wll-order'' is introduced as an
abbreviation for ``\textbf{w}eight first,
then \textbf{l}ength, then \textbf{l}exicographic'' (in the
sense that two compositions are first compared by their weights, then, if the
weights are equal, by their lengths, and finally, if the lengths are also
equal, by the lexicographic order). For us, the alternative explanation
``\textbf{w}ord \textbf{l}ength, then \textbf{l}exicographic'' serves
just as well.}

We now state a fact which is similar (and plays a similar role) to Proposition
\ref{prop.shuffle.radford.triang}:

\begin{proposition}
\label{prop.QSym.lyndon.triang}For every composition $u\in \Comp %
=\mathfrak{A}^{\ast}$, define an element $\mathbf{M}_{u}\in
\Qsym$ by $\mathbf{M}_{u}=M_{a_{1}}M_{a_{2}} \cdots M_{a_{p}}$,
where $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is the CFL factorization of
the word $u$.

If $n\in \NN$ and if $x\in\Comp_{n}$, then there
is a family $\left(  \eta_{x,y}\right)  _{y\in
\Comp_{n}}\in \NN ^{\Comp_{n}}$ of elements
of $\NN$ satisfying
\[
\mathbf{M}_{x}=\sum\limits_{\substack{y\in\Comp_{n};\\
y\underset{\operatorname*{wll}}{\leq}x}}\eta_{x,y}M_{y}
\]
and $\eta_{x,x}\neq0$ (in $\NN$).
\end{proposition}

Before we prove it, let us show the following lemma:

\begin{lemma}
\label{lem.QSym.lyndon.triang.wll-lemma}Let $n\in \NN$ and
$m\in \NN$. Let $u\in\Comp_{n}$ and
$v\in\Comp_{m}$. Let $z$ be the lexicographically
highest element of the multiset $u\shuffle v$.

(a) We have $z\in\Comp_{n+m}$.

(b) There exists a positive integer $h$ such that
\[
M_{u}M_{v}=hM_{z}+\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\]


(c) Let $v^{\prime}\in\Comp_{m}$ be such that
$v^{\prime}\underset{\operatorname*{wll}}{<}v$. Then,%
\[
M_{u}M_{v^{\prime}}=\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\]

\end{lemma}

\begin{exercise}
\label{exe.lem.QSym.lyndon.triang.wll-lemma}Prove Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma}.

[\textbf{Hint:} For (b), set $h$ to be the multiplicity
with which the word $z$ appears in the multiset $u\shuffle v$,
then use Proposition \ref{prop.QSym.lyndon.MaMb} and
notice that $M_{u}M_{v}$ is homogeneous of degree $n+m$.
For (c), use (b) for
$v^{\prime}$ instead of $v$ and notice that Lemma
\ref{lem.shuffle.preserves-lex}(a) shows that the lexicographically highest
element of the multiset $u\shuffle v^{\prime}$ is
$\underset{\operatorname*{wll}}{<}z$.]
\end{exercise}

\begin{exercise}
\label{exe.prop.QSym.lyndon.triang}Prove Proposition
\ref{prop.QSym.lyndon.triang}.

[\textbf{Hint:} Proceed by strong induction over $n$. In the induction step,
let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
$x$, and set $u=a_{1}$ and $v=a_{2}a_{3}\cdots a_{p}$; then apply Proposition
\ref{prop.QSym.lyndon.triang} to $v$ instead of $x$, and multiply the
resulting equality $\mathbf{M}_{v}=\sum\limits_{\substack{y\in
\Comp_{\left\vert v\right\vert };\\y\underset{\operatorname*{wll}}{\leq}%
v}}\eta_{v,y}M_{y}$ with $M_{u}$ to obtain an expression for $M_{u}%
\mathbf{M}_{v}=\mathbf{M}_{x}$. Use Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma} to show that this expression has the
form $\sum\limits_{\substack{y\in\Comp_{n}%
;\\y\underset{\operatorname*{wll}}{\leq}x}}\eta_{x,y}M_{y}$ with $\eta
_{x,x}\neq0$; here it helps to remember that the lexicographically highest
element of the multiset $u\shuffle v$ is $uv=x$ (by Theorem
\ref{thm.shuffle.lyndon.used}(c)).]
\end{exercise}

We are almost ready to give our second proof of Proposition
\ref{prop.QSym.lyndon}; our last step is the following proposition:

\begin{proposition}
\label{prop.QSym.lyndon.M-genset}Assume that $\QQ$ is a subring of
$\kk$. Then, $\left(  M_{w}\right)  _{w\in\mathfrak{L}}$ is an
algebraically independent generating set of the $\kk$-algebra
$\Qsym$.
\end{proposition}

\begin{exercise}
\label{exe.prop.QSym.lyndon.M-genset}Prove Proposition
\ref{prop.QSym.lyndon.M-genset}.

[\textbf{Hint:} Define $\mathbf{M}_{u}$ for every $u\in \Comp $
as in Proposition \ref{prop.QSym.lyndon.triang}. Conclude from Proposition
\ref{prop.QSym.lyndon.triang} that, for every $n\in \NN$, the family
$\left(  \mathbf{M}_{u}\right)  _{u\in\Comp_{n}}$
expands invertibly triangularly\footnote{See
Definition~\ref{def.STmat.expansion-tria}(b) for the
meaning of this.} (with respect to the total order
$\underset{\operatorname*{wll}}{\leq}$ on $\Comp_{n}$)
with respect to the basis $\left(  M_{u}\right)  _{u\in
\Comp_{n}}$ of $\Qsym_{n}$. Conclude that this family $\left(
\mathbf{M}_{u}\right)  _{u\in\Comp_{n}}$ is a basis of
$\Qsym_{n}$ itself, and so the whole family $\left(
\mathbf{M}_{u}\right)  _{u\in \Comp }$ is a basis of
$\Qsym$. Conclude using Lemma \ref{lem.lyndonbases}(a).]
\end{exercise}

\begin{proof}
[Second proof of Proposition \ref{prop.QSym.lyndon}.]Proposition
\ref{prop.QSym.lyndon.M-genset} yields that $\left(  M_{w}\right)
_{w\in\mathfrak{L}}$ is an algebraically independent generating set of the
$\kk$-algebra $\Qsym$.

Define a grading on the $\kk$-algebra $\kk \left[  x_{w}%
\ \mid\ w\in\mathfrak{L}\right]  $ by setting $\deg\left(  x_{w}\right)
=\sum_{i=1}^{\ell\left(  w\right)  }w_{i}$ for every $w\in\mathfrak{L}$. By
the universal property of the polynomial algebra $\kk \left[
x_{w}\ \mid\ w\in\mathfrak{L}\right]  $, we can define a $\kk$-algebra
homomorphism $\Phi: \kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]
\rightarrow\Qsym$ by setting%
\[
\Phi\left(  x_{w}\right)  =M_{w}\ \ \ \ \ \ \ \ \ \ \text{for every }%
w\in\mathfrak{L}.
\]
This homomorphism $\Phi$ is a $\kk$-algebra isomorphism (since $\left(
M_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent generating
set of the $\kk$-algebra $\Qsym$) and is graded (because
for every $w\in\mathfrak{L}$, the element $M_{w}$ of $\Qsym$ is
homogeneous of degree $\left\vert w\right\vert =\sum_{i=1}^{\ell\left(
w\right)  }w_{i}=\deg\left(  x_{w}\right)  $). Thus, $\Phi$ is an isomorphism
of graded $\kk$-algebras. Hence, $\Qsym\cong%
 \kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]  $ as graded
$\kk$-algebras. In particular, this shows that $\Qsym$
is a polynomial algebra. This proves Theorem \ref{thm.QSym.lyndon} under the
assumption that $\QQ$ be a subring of $\kk$. Proposition
\ref{prop.QSym.lyndon} is thus proven again.
\end{proof}

\subsection{\label{subsect.shuffle.QSym.2}Polynomial freeness of
$\Qsym$: the general case}

We now will prepare for proving Theorem \ref{thm.QSym.lyndon} without any
assumptions on $\kk$. In our proof, we follow \cite{Hazewinkel} and
\cite[\S 6.7]{HazewinkelGubareniKirichenko}, but without using the language of
plethysm and Frobenius maps. We start with the following definition:

\begin{definition}
\label{def.QSym.Malphas}Let $\alpha$ be a composition. Write $\alpha$ in the
form $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ with
$\ell=\ell\left(  \alpha\right)  $.

(a) Let $\operatorname*{SIS}\left(  \ell\right)  $ denote the set of all
strictly increasing $\ell$-tuples $\left(  i_{1},i_{2},\ldots,i_{\ell}\right)
$ of positive integers.\footnote{``Strictly
increasing'' means that $i_{1}<i_{2}<\cdots<i_{\ell}$ here.
Of course, the elements of $\operatorname*{SIS}\left(  \ell\right)  $ are in
1-to-1 correspondence with $\ell$-element subsets of $\left\{  1,2,3,\ldots
\right\}  $.} For every $\ell$-tuple $\mathbf{i}=\left(  i_{1},i_{2}%
,\ldots,i_{\ell}\right)  \in\operatorname*{SIS}\left(  \ell\right)  $, we
denote the monomial $x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots
x_{i_{\ell}}^{\alpha_{\ell}}$ by $\xx_{\mathbf{i}}^{\alpha}$. This
$\xx_{\mathbf{i}}^{\alpha}$ is a monomial of degree $\alpha_{1}%
+\alpha_{2}+\cdots+\alpha_{\ell}=\left\vert \alpha\right\vert $. Then,
\begin{equation}
M_{\alpha}=\sum_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\xx_{\mathbf{i}}^{\alpha}. \label{eq.def.QSym.Malphas.a.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{eq.def.QSym.Malphas.a.1}):} By the definition
of $M_{\alpha}$, we have
\begin{align*}
M_{\alpha}  &  =\underbrace{\sum_{i_{1}<i_{2}<\cdots<i_{\ell}\text{ in
}\left\{  1,2,3,\ldots\right\}  }}_{=\sum_{\left(  i_{1},i_{2},\ldots,i_{\ell
}\right)  \in\operatorname*{SIS}\left(  \ell\right)  }}x_{i_{1}}^{\alpha_{1}%
}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}=\sum_{\left(
i_{1},i_{2},\ldots,i_{\ell}\right)  \in\operatorname*{SIS}\left(  \ell\right)
}x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha
_{\ell}}=\sum_{\mathbf{i}=\left(  i_{1},i_{2},\ldots,i_{\ell}\right)
\in\operatorname*{SIS}\left(  \ell\right)  }\underbrace{x_{i_{1}}^{\alpha_{1}%
}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}}%
_{\substack{=\xx_{\mathbf{i}}^{\alpha}\\\text{(by the definition of
}\xx_{\mathbf{i}}^{\alpha}\text{)}}}\\
&  =\sum_{\mathbf{i}=\left(  i_{1},i_{2},\ldots,i_{\ell}\right)
\in\operatorname*{SIS}\left(  \ell\right)  }\xx_{\mathbf{i}}^{\alpha
}=\sum_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }
\xx_{\mathbf{i}}^{\alpha},
\end{align*}
qed.}

(b) Consider the ring $\kk \left[  \left[  \xx\right]  \right]  $
endowed with the coefficientwise topology\footnote{This topology is defined as
follows:
\par
We endow the ring $\kk$ with the discrete topology. Then, we can regard
the $\kk$-module $\kk \left[  \left[  \xx\right]  \right]  $ as a direct
product of infinitely many copies of $\kk$ (by identifying every power
series in $\kk \left[  \left[  \xx\right]  \right]  $ with the
family of its coefficients). Hence, the product topology is a well-defined
topology on $\kk \left[  \left[  \xx\right]  \right]  $; this
topology is denoted as the \dfn{coefficientwise topology}. A sequence
$\left(  a_{n}\right)  _{n\in \NN }$ of power series converges to a power
series $a$ with respect to this topology if and only if for every monomial
$\mathfrak{m}$, all sufficiently high $n\in \NN$ satisfy
\[
\left(  \text{the coefficient of }\mathfrak{m}\text{ in }a_{n}\right)
=\left(  \text{the coefficient of }\mathfrak{m}\text{ in }a\right)  .
\]
Note that this is \textbf{not} the topology obtained by taking the completion
of $\kk \left[  x_{1},x_{2},x_{3},\ldots\right]  $ with respect to the
standard grading (in which all $x_{i}$ have degree $1$). (The latter
completion is actually a smaller ring than $\kk\left[\left[\xx\right]\right]$.)}.
The family $\left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  }$ of elements of $\kk \left[  \left[
\xx\right]  \right]  $ is power-summable\footnote{Let us define what
``power-summable'' means for us:
\par
A family $\left(  n_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}%
\in \NN ^{\mathbf{I}}$ (where $\mathbf{I}$ is some set) is said to be
\dfn{finitely supported} if all but finitely many $\mathbf{i}\in\mathbf{I}$
satisfy $n_{\mathbf{i}}=0$.
\par
If $\left(  n_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in
 \NN ^{\mathbf{I}}$ is a finitely supported family, then
$\sum_{\mathbf{i}\in\mathbf{I}}n_{\mathbf{i}}$ is a well-defined element of
$\NN$. If $N\in \NN$, then a family $\left(  n_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\in \NN ^{\mathbf{I}}$ will be called
\emph{$\left(  \leq N\right)  $-supported} if it is finitely supported and
satisfies $\sum_{\mathbf{i}\in\mathbf{I}}n_{\mathbf{i}}\leq N$.
\par
We say that a family $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in
\mathbf{I}}\in R^{\mathbf{I}}$ of elements of a topological commutative
$\kk$-algebra $R$ is \emph{power-summable}\index{power-summable family}
if it satisfies the following property: For every $N\in \NN$, the sum
\[
\sum\limits_{\substack{\left(  n_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}%
\in \NN ^{\mathbf{I}};\\\left(  n_{\mathbf{i}}\right)  _{\mathbf{i}%
\in\mathbf{I}}\text{ is }\left(  \leq N\right)  \text{-supported }}%
}\alpha_{\left(  n_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}}%
\prod_{\mathbf{i}\in\mathbf{I}}s_{\mathbf{i}}^{n_{\mathbf{i}}}%
\]
converges in the topology on $R$ for every choice of scalars $\alpha_{\left(
n_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}}\in \kk$ corresponding
to all $\left(  \leq N\right)  $-supported $\left(  n_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\in \NN ^{\mathbf{I}}$. In our specific case,
we consider $\kk \left[  \left[  \xx\right]  \right]  $ as a
topological commutative $\kk$-algebra,
where the topology is the coefficientwise
topology. The fact that the family $\left(  \xx_{\mathbf{i}}^{\alpha
}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }$ is
power-summable then can be proven as follows:
\par
\begin{itemize}
\item If $\alpha\neq\varnothing$, then this fact follows from the
(easily-verified) observation that every given monomial in the variables
$x_{1},x_{2},x_{3},\ldots$ can be written as a product of monomials of the
form $\xx_{\mathbf{i}}^{\alpha}$ (with $\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  $) in only finitely many ways.
\par
\item If $\alpha=\varnothing$, then this fact follows by noticing that
$\left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  }$ is a finite family (indeed,
$\operatorname*{SIS}\left(  \ell\right)  =\operatorname*{SIS}\left(  0\right)
=\left\{  \left(  \right)  \right\}  $), and every finite family is
power-summable.
\end{itemize}
}. Hence, for every $f\in\Lambda$, there is a well-defined power series
$f\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  }\right)  \in \kk \left[
\left[  \xx\right]  \right]  $ obtained by ``evaluating'' $f$
at $\left(  \xx_{\mathbf{i}}^{\alpha
}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }%
$\ \ \ \ \footnote{Here is how this power series $f\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  }\right)  $ is formally defined:
\par
Let $R$ be any topological commutative $\kk$-algebra, and let $\left(
s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$ be any
power-summable family of elements of $R$. Assume that the indexing set
$\mathbf{I}$ is countably infinite, and fix a bijection $\mathfrak{j}:\left\{
1,2,3,\ldots\right\}  \rightarrow\mathbf{I}$. Let $g\in R\left(
\xx\right)  $ be arbitrary. Then, we can substitute $s_{\mathfrak{j}%
\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$, $s_{\mathfrak{j}%
\left(  3\right)  }$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$,
$\ldots$ in $g$, thus obtaining an infinite sum which converges in $R$ (in
fact, its convergence follows from the fact that the family $\left(
s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$ is
power-summable). The value of this sum will be denoted by $g\left(  \left(
s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $. In general, this
value depends on the choice of the bijection $\mathfrak{j}$, so the notation
$g\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $
is unambiguous only if this bijection $\mathfrak{j}$ is chosen once and for
all. However, when $g\in\Lambda$, one can easily see that the choice of
$\mathfrak{j}$ has no effect on $g\left(  \left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\right)  $.
\par
We can still define $g\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}%
\in\mathbf{I}}\right)  $ when the set $\mathbf{I}$ is finite instead of being
countably infinite. In this case, we only need to modify our above definition
as follows: Instead of fixing a bijection $\mathfrak{j}:\left\{
1,2,3,\ldots\right\}  \rightarrow\mathbf{I}$, we now fix a bijection
$\mathfrak{j}:\left\{  1,2,\ldots,\left\vert \mathbf{I}\right\vert \right\}
\rightarrow\mathbf{I}$, and instead of substituting $s_{\mathfrak{j}\left(
1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$, $s_{\mathfrak{j}\left(
3\right)  }$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$
in $g$, we now substitute $s_{\mathfrak{j}\left(  1\right)  }$,
$s_{\mathfrak{j}\left(  2\right)  }$, $\ldots$, $s_{\mathfrak{j}\left(
\left\vert \mathbf{I}\right\vert \right)  }$, $0$, $0$, $0$, $\ldots$ for the
variables $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$ in $g$. Again, the same
observations hold as before: $g\left(  \left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\right)  $ is independent on $\mathfrak{j}$ if
$g\in\Lambda$.
\par
Hence, $g\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right)  $ is well-defined for every $g\in R\left(  \xx\right)  $,
every countable (i.e., finite or countably infinite) set $\mathbf{I}$, every
topological commutative $\kk$-algebra $R$ and every power-summable
family $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in
R^{\mathbf{I}}$ of elements of $R$, as long as a bijection $\mathfrak{j}$ is
chosen. In particular, we can apply this to $g=f$, $\mathbf{I}%
=\operatorname*{SIS}\left(  \ell\right)  $, $R= \kk \left[  \left[
\xx\right]  \right]  $ and $\left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}=\left(  \xx_{\mathbf{i}}^{\alpha}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }$, choosing
$\mathfrak{j}$ to be the bijection which sends every positive integer $k$ to
the $k$-th smallest element of $\operatorname*{SIS}\left(  \ell\right)  $ in
the lexicographic order. (Of course, since $f\in\Lambda$, the choice of
$\mathfrak{j}$ is irrelevant.)}. In particular, for every $s\in \ZZ$, we
can evaluate the symmetric function $e_s \in \Lambda$\ \ \ \ \footnote{Recall
that $e_0 = 1$, and that $e_s = 0$ for $s<0$.} at $\left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}
\left(  \ell\right)  }$. The resulting power series $e_{s}\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  }\right)  \in \kk \left[  \left[  \xx\right]
\right]  $ will be denoted $M_{\alpha}^{\left\langle s\right\rangle }$. Thus,%
\[
M_{\alpha}^{\left\langle s\right\rangle }
= e_{s}\left(  \left(  \xx_{\mathbf{i}}^{\alpha}
\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\right)  .
\]

\end{definition}

The power series $M_{\alpha}^{\left\langle s\right\rangle }$ are the power
series $e_{s}\left(  \alpha\right)  $ in \cite{HazewinkelGubareniKirichenko}.
We will shortly (in Corollary \ref{cor.QSym.Malphas.QSym}(a)) see that
$M_{\alpha}^{\left\langle s\right\rangle }\in\Qsym$ (although
this is also easy to prove by inspection). Here are some examples of
$M_{\alpha}^{\left\langle s\right\rangle }$:

\begin{example}
If $\alpha$ is a composition and $\ell$ denotes its length
$\ell\left(\alpha\right)$, then
\[
M_{\alpha}^{\left\langle 0\right\rangle }=\underbrace{e_{0}}_{=1}\left(
\left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  }\right)  =1\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  }\right)  =1
\]
and%
\[
M_{\alpha}^{\left\langle 1\right\rangle }=e_{1}\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\right)  =\sum_{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\xx_{\mathbf{i}}^{\alpha}=M_{\alpha}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.def.QSym.Malphas.a.1})}\right)
\]
and\footnote{This is not completely
obvious, but easy to check (see Exercise~\ref{exe.QSym.Malphas.ps}(b)).}
\[
M_{\alpha}^{\left\langle 2\right\rangle }=e_{2}\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}
\left(  \ell\right)  }\right)
=\sum\limits_{\substack{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  ,\ \mathbf{j}\in\operatorname*{SIS}\left(  \ell\right)
;\\\mathbf{i}<\mathbf{j}}}\xx_{\mathbf{i}}^{\alpha}
\xx_{\mathbf{j}}^{\alpha}
\]
(where the notation ``$\mathbf{i}<\mathbf{j}$'' should be interpreted
with respect to an arbitrary but fixed total order on
the set $\operatorname*{SIS}\left(  \ell\right)  $ -- for example, the
lexicographic order). Applying the last of these three equalities to
$\alpha=\left(  2,1\right)  $, we obtain
\begin{align*}
M_{\left(  2,1\right)  }^{\left\langle 2\right\rangle }  &  =
\sum_{\substack{\mathbf{i}\in\operatorname*{SIS}\left(  2\right)  ,
\ \mathbf{j} \in\operatorname*{SIS}\left(  2\right)  ,\\
\mathbf{i}<\mathbf{j}}}
\xx_{\mathbf{i}}^{\left(  2,1\right)  }\xx_{\mathbf{j}%
}^{\left(  2,1\right)  }=\sum\limits_{\substack{\left(  i_{1},i_{2}\right)
\in\operatorname*{SIS}\left(  2\right)  ,\ \left(  j_{1},j_{2}\right)
\in\operatorname*{SIS}\left(  2\right)  ;\\\left(  i_{1},i_{2}\right)
<\left(  j_{1},j_{2}\right)  }}\underbrace{\xx_{\left(  i_{1}%
,i_{2}\right)  }^{\left(  2,1\right)  }}_{=x_{i_{1}}^{2}x_{i_{2}}^{1}%
}\underbrace{\xx_{\left(  j_{1},j_{2}\right)  }^{\left(  2,1\right)  }%
}_{=x_{j_{1}}^{2}x_{j_{2}}^{1}}\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2}\right)  \in\operatorname*{SIS}\left(
2\right)  ,\ \left(  j_{1},j_{2}\right)  \in\operatorname*{SIS}\left(
2\right)  ;\\\left(  i_{1},i_{2}\right)  <\left(  j_{1},j_{2}\right)
}}x_{i_{1}}^{2}x_{i_{2}}^{1}x_{j_{1}}^{2}x_{j_{2}}^{1}\\
&  =\underbrace{\sum\limits_{\substack{i_{1}<i_{2};\ j_{1}<j_{2};\\i_{1}<j_{1}%
}}x_{i_{1}}^{2}x_{i_{2}}^{1}x_{j_{1}}^{2}x_{j_{2}}^{1}}_{=M_{\left(
2,1,2,1\right)  }+M_{\left(  2,3,1\right)  }+2M_{\left(  2,2,1,1\right)
}+M_{\left(  2,2,2\right)  }}+\underbrace{\sum\limits_{\substack{i_{1}<i_{2}%
;\ j_{1}<j_{2};\\i_{1}=j_{1};\ i_{2}<j_{2}}}x_{i_{1}}^{2}x_{i_{2}}^{1}%
x_{j_{1}}^{2}x_{j_{2}}^{1}}_{=M_{\left(  4,1,1\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have WLOG assumed that the order
on }\operatorname*{SIS}\left(  2\right)  \text{ is lexicographic}\right) \\
&  =M_{\left(  2,1,2,1\right)  }+M_{\left(  2,3,1\right)  }+2M_{\left(
2,2,1,1\right)  }+M_{\left(  2,2,2\right)  }+M_{\left(  4,1,1\right)  }.
\end{align*}


Of course, every negative integer $s$ satisfies $M_{\alpha}^{\left\langle
s\right\rangle }=\underbrace{e_{s}}_{=0}\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}
\left(  \ell\right)  }\right)  =0$.
\end{example}

There is a determinantal formula for the $s!M_{\alpha}^{\left\langle
s\right\rangle }$ (and thus also for $M_{\alpha}^{\left\langle s\right\rangle
}$ when $s!$ is invertible in $\kk$), but in order to state it, we need
to introduce one more notation:

\begin{definition}
Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be a
composition, and let $k$ be a positive integer. Then, $\alpha\left\{
k\right\}  $ will denote the composition $\left(  k\alpha_{1},k\alpha
_{2},\ldots,k\alpha_{\ell}\right)  $. Clearly, $\ell\left(  \alpha\left\{
k\right\}  \right)  =\ell\left(  \alpha\right)  $ and $\left\vert
\alpha\left\{  k\right\}  \right\vert =k\left\vert \alpha\right\vert $.
\end{definition}

\begin{exercise}
\label{exe.QSym.Malphas.ps}Let $\alpha$ be a composition. Write the
composition $\alpha$ in the form $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  $ with $\ell=\ell\left(  \alpha\right)  $.

\begin{enumerate}
\item[(a)] Show that the $s$-th power-sum symmetric function $p_{s}\in\Lambda$
satisfies%
\[
p_{s}\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  }\right)  =M_{\alpha\left\{
s\right\}  }%
\]
for every positive integer $s$.

\item[(b)] Let us fix a total order on the set $\operatorname*{SIS}\left(
\ell\right)  $ (for example, the lexicographic order). Show that the $s$-th
elementary symmetric function $e_{s}\in\Lambda$ satisfies%
\[
M_{\alpha}^{\left\langle s\right\rangle }=e_{s}\left(  \left(  \xx%
_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\right)  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}%
_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}%
\]
for every $s\in \NN$.

\item[(c)] Let $s\in \NN$, and let $n$ be a positive integer. Let
$e_{s}^{\left\langle n\right\rangle }$ be the symmetric function
$\sum_{i_{1}<i_{2}<\cdots<i_{s}}
x_{i_{1}}^{n}x_{i_{2}}^{n}\cdots x_{i_{s}}^{n} \in\Lambda$.
Then, show that
\[
M_{\alpha\left\{  n\right\}  }^{\left\langle s\right\rangle }=e_{s}%
^{\left\langle n\right\rangle }\left(  \left(  \xx_{\mathbf{i}}%
^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\right)  .
\]


\item[(d)] Let $s\in \NN$, and let $n$ be a positive integer. Prove that
there exists a polynomial $P\in \kk \left[  z_{1},z_{2},z_{3}%
,\ldots\right]  $ such that $M_{\alpha\left\{  n\right\}  }^{\left\langle
s\right\rangle }=P\left(  M_{\alpha}^{\left\langle 1\right\rangle },M_{\alpha
}^{\left\langle 2\right\rangle },M_{\alpha}^{\left\langle 3\right\rangle
},\ldots\right)  $.
\end{enumerate}

[\textbf{Hint:} For (a), (b) and (c), apply the definition of $f\left(
\left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  }\right)  $ with $f$ a symmetric
function\footnote{There are two subtleties that need to be addressed:
\par
\begin{itemize}
\item the fact that the definition of $f\left(  \left(  \xx%
_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\right)  $ distinguishes between two cases depending on whether
or not $\operatorname*{SIS}\left(  \ell\right)  $ is finite;
\par
\item the fact that the total order on the set $\left\{  1,2,3,\ldots\right\}
$ (which appears in the summation subscript in the equality $e_{s}%
=\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{s}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{s};\\i_{1}<i_{2}<\cdots<i_{s}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{s}}$) has nothing to do with the total order on the set
$\operatorname*{SIS}\left(  \ell\right)  $ (which appears in the summation
subscript in $\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2}%
,\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}%
_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}$). For instance, the
former total order is well-founded, whereas the latter may and may not be. So
there is (generally) no bijection between $\left\{  1,2,3,\ldots\right\}  $
and $\operatorname*{SIS}\left(  \ell\right)  $ preserving these orders (even
if $\operatorname*{SIS}\left(  \ell\right)  $ is infinite). Fortunately, this
does not matter much, because the total order is only being used to ensure
that every product of $s$ distinct elements appears exactly once in
the sum.
\end{itemize}
}. For (d), recall that $\Lambda$ is generated by $e_{1},e_{2},e_{3},\ldots$.]
\end{exercise}

\begin{exercise}
\label{exe.QSym.Malphas.M1s}Let $s\in \NN$. Show that the composition
$\left(  1\right)  $ satisfies $M_{\left(  1\right)  }^{\left\langle
s\right\rangle }=e_{s}$.
\end{exercise}

\begin{proposition}
\label{prop.QSym.Malphas.dets}Let $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  $ be a composition.

(a) Let $n\in \NN$. Define a matrix $A_{n}^{\left\langle \alpha
\right\rangle }=\left(  a_{i,j}^{\left\langle \alpha\right\rangle }\right)
_{i,j=1,2,\ldots,n}$ by
\[
a_{i,j}^{\left\langle \alpha\right\rangle }=
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\qquad\qquad\text{ for all }\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}.
\]
This matrix $A_{n}^{\left\langle \alpha\right\rangle }$ looks as follows:
\[
A_{n}^{\left\langle \alpha\right\rangle }=\left(
\begin{array}[c]{cccccc}
M_{\alpha\left\{  1\right\}  } & 1 & 0 & \cdots & 0 & 0\\
M_{\alpha\left\{  2\right\}  } & M_{\alpha\left\{  1\right\}  } & 2 & \cdots &
0 & 0\\
M_{\alpha\left\{  3\right\}  } & M_{\alpha\left\{  2\right\}  } &
M_{\alpha\left\{  1\right\}  } & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
M_{\alpha\left\{  n-1\right\}  } & M_{\alpha\left\{  n-2\right\}  } &
M_{\alpha\left\{  n-3\right\}  } & \cdots & M_{\alpha\left\{  1\right\}  } &
n-1\\
M_{\alpha\left\{  n\right\}  } & M_{\alpha\left\{  n-1\right\}  } &
M_{\alpha\left\{  n-2\right\}  } & \cdots & M_{\alpha\left\{  2\right\}  } &
M_{\alpha\left\{  1\right\}  }%
\end{array}
\right)  .
\]
Then, $\det\left(  A_{n}^{\left\langle \alpha\right\rangle }\right)
=n!M_{\alpha}^{\left\langle n\right\rangle }$.

(b) Let $n$ be a positive integer. Define a matrix $B_{n}^{\left\langle
\alpha\right\rangle }=\left(  b_{i,j}^{\left\langle \alpha\right\rangle
}\right)  _{i,j=1,2,\ldots,n}$ by
\[
b_{i,j}^{\left\langle \alpha\right\rangle }=
\begin{cases}
iM_{\alpha}^{\left\langle i\right\rangle }, & \text{if }j=1;\\
M_{\alpha}^{\left\langle i-j+1\right\rangle }, & \text{if }j>1
\end{cases}
\qquad\qquad\text{ for all }\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}.
\]
The matrix $B_{n}^{\left\langle \alpha\right\rangle }$ looks as follows:
\begin{align*}
B_{n}^{\left\langle \alpha\right\rangle }  &  =\left(
\begin{array}[c]{cccccc}
M_{\alpha}^{\left\langle 1\right\rangle } & M_{\alpha}^{\left\langle
0\right\rangle } & M_{\alpha}^{\left\langle -1\right\rangle } & \cdots &
M_{\alpha}^{\left\langle -n+3\right\rangle } & M_{\alpha}^{\left\langle
-n+2\right\rangle }\\
2M_{\alpha}^{\left\langle 2\right\rangle } & M_{\alpha}^{\left\langle
1\right\rangle } & M_{\alpha}^{\left\langle 0\right\rangle } & \cdots &
M_{\alpha}^{\left\langle -n+4\right\rangle } & M_{\alpha}^{\left\langle
-n+3\right\rangle }\\
3M_{\alpha}^{\left\langle 3\right\rangle } & M_{\alpha}^{\left\langle
2\right\rangle } & M_{\alpha}^{\left\langle 1\right\rangle } & \cdots &
M_{\alpha}^{\left\langle -n+5\right\rangle } & M_{\alpha}^{\left\langle
-n+4\right\rangle }\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
\left(  n-1\right)  M_{\alpha}^{\left\langle n-1\right\rangle } & M_{\alpha
}^{\left\langle n-2\right\rangle } & M_{\alpha}^{\left\langle n-3\right\rangle
} & \cdots & M_{\alpha}^{\left\langle 1\right\rangle } & M_{\alpha
}^{\left\langle 0\right\rangle }\\
nM_{\alpha}^{\left\langle n\right\rangle } & M_{\alpha}^{\left\langle
n-1\right\rangle } & M_{\alpha}^{\left\langle n-2\right\rangle } & \cdots &
M_{\alpha}^{\left\langle 2\right\rangle } & M_{\alpha}^{\left\langle
1\right\rangle }%
\end{array}
\right) \\
&  =\left(
\begin{array}[c]{cccccc}
M_{\alpha}^{\left\langle 1\right\rangle } & 1 & 0 & \cdots & 0 & 0\\
2M_{\alpha}^{\left\langle 2\right\rangle } & M_{\alpha}^{\left\langle
1\right\rangle } & 1 & \cdots & 0 & 0\\
3M_{\alpha}^{\left\langle 3\right\rangle } & M_{\alpha}^{\left\langle
2\right\rangle } & M_{\alpha}^{\left\langle 1\right\rangle } & \cdots & 0 &
0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
\left(  n-1\right)  M_{\alpha}^{\left\langle n-1\right\rangle } & M_{\alpha
}^{\left\langle n-2\right\rangle } & M_{\alpha}^{\left\langle n-3\right\rangle
} & \cdots & M_{\alpha}^{\left\langle 1\right\rangle } & 1\\
nM_{\alpha}^{\left\langle n\right\rangle } & M_{\alpha}^{\left\langle
n-1\right\rangle } & M_{\alpha}^{\left\langle n-2\right\rangle } & \cdots &
M_{\alpha}^{\left\langle 2\right\rangle } & M_{\alpha}^{\left\langle
1\right\rangle }
\end{array}
\right)  .
\end{align*}
Then, $\det\left(  B_{n}^{\left\langle \alpha\right\rangle }\right)
=M_{\alpha\left\{  n\right\}  }$.
\end{proposition}

\begin{exercise}
\label{exe.prop.QSym.Malphas.dets}Prove Proposition
\ref{prop.QSym.Malphas.dets}.

[\textbf{Hint:} Substitute $\left(  \xx_{\mathbf{i}}^{\alpha}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }$ for the variable set
in Exercise \ref{exe.pe-ep-determinants}, and recall Exercise
\ref{exe.QSym.Malphas.ps}(a).]
\end{exercise}

\begin{corollary}
\label{cor.QSym.Malphas.QSym}Let $\alpha$ be a composition. Let $s\in
 \ZZ$.

\begin{enumerate}
\item[(a)] We have $M_{\alpha}^{\left\langle s\right\rangle }\in
\Qsym$.

\item[(b)] We have $M_{\alpha}^{\left\langle s\right\rangle }\in
\Qsym_{s\left\vert \alpha\right\vert }$.
\end{enumerate}
\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.Malphas.QSym}Prove Corollary \ref{cor.QSym.Malphas.QSym}.
\end{exercise}

We make one further definition:

\begin{definition}
\label{def.QSym.lyndon.red}Let $\alpha$ be a nonempty composition. Then, we
denote by $\gcd\alpha$ the greatest common divisor of the parts of $\alpha$.
(For instance, $\gcd\left(  8,6,4\right)  =2$.) We also define
$\operatorname*{red}\alpha$ to be the composition $\left(  \dfrac{\alpha_{1}%
}{\gcd\alpha},\dfrac{\alpha_{2}}{\gcd\alpha},\ldots,\dfrac{\alpha_{\ell}}%
{\gcd\alpha}\right)  $, where $\alpha$ is written in the form $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $.

We say that a nonempty composition $\alpha$ is
\emph{reduced}\index{reduced composition} if $\gcd \alpha = 1$.

We define $\mathfrak{RL}$ to be the set of all reduced Lyndon compositions. In
other words, $\mathfrak{RL}=\left\{  w\in\mathfrak{L}\ \mid\ w\text{ is
reduced}\right\}  $ (since $\mathfrak{L}$ is the set of all Lyndon compositions).
\end{definition}

Hazewinkel, in
\cite[proof of Thm. 6.7.5]{HazewinkelGubareniKirichenko},
denotes $\mathfrak{RL}$ by $eLYN$, calling reduced Lyndon compositions
``elementary Lyndon words''.

\begin{remark}
\label{rmk.QSym.lyndon.red}Let $\alpha$ be a nonempty composition.

(a) We have $\alpha=\left(  \operatorname*{red}\alpha\right)  \left\{
\gcd\alpha\right\}  $.

(b) The composition $\alpha$ is Lyndon if and only if the composition
$\operatorname*{red}\alpha$ is Lyndon.

(c) The composition $\operatorname*{red}\alpha$ is reduced.

(d) If $\alpha$ is reduced, then $\operatorname*{red}\alpha=\alpha$.

(e) If $s\in\left\{  1,2,3,\ldots\right\}  $, then the composition
$\alpha\left\{  s\right\}  $ is nonempty and satisfies $\operatorname*{red}%
\left(  \alpha\left\{  s\right\}  \right)  =\operatorname*{red}\alpha$ and
$\gcd\left(  \alpha\left\{  s\right\}  \right)  =s\gcd\alpha$.

(f) We have $\left(  \gcd\alpha\right)  \left\vert \operatorname*{red}%
\alpha\right\vert =\left\vert \alpha\right\vert $.
\end{remark}

\begin{exercise}
\label{exe.rmk.QSym.lyndon.red}Prove Remark \ref{rmk.QSym.lyndon.red}.
\end{exercise}

Our goal in this section is now to prove the following result of Hazewinkel:

\begin{theorem}
\label{thm.QSym.lyndon.Z-genset}The family $\left(  M_{w}^{\left\langle
s\right\rangle }\right)  _{\left(  w,s\right)  \in\mathfrak{RL}\times\left\{
1,2,3,\ldots\right\}  }$ is an algebraically independent generating set of the
$\kk$-algebra $\Qsym$.
\end{theorem}

This will (almost) immediately yield Theorem \ref{thm.QSym.lyndon}.

Our first step towards proving Theorem \ref{thm.QSym.lyndon.Z-genset} is the
following observation:

\begin{lemma}
\label{lem.QSym.lyndon.Z-genset.reindex}The family $\left(  M_{w}%
^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)  \in
\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a reindexing of the
family $\left(  M_{\operatorname*{red}\alpha}^{\left\langle \gcd
\alpha\right\rangle }\right)  _{\alpha\in\mathfrak{L}}$.
\end{lemma}

\begin{exercise}
\label{exe.lem.QSym.lyndon.Z-genset.reindex}Prove Lemma
\ref{lem.QSym.lyndon.Z-genset.reindex}.
\end{exercise}

Next, we show a lemma:

\begin{lemma}
\label{lem.QSym.lyndon.Malphas1}Let $\alpha$ be a nonempty composition. Let
$s\in \NN$. Then,%
\begin{equation}
s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\in
\sum\limits_{\substack{\beta\in\Comp_{s\left\vert \alpha
\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell\left(
\alpha\right)  }} \kk M_{\beta}.
\label{eq.lem.QSym.lyndon.Malphas1}
\end{equation}
(That is, $s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}$ is a
$\kk$-linear combination of terms of the form $M_{\beta}$ with $\beta$
ranging over the compositions of $s\left\vert \alpha\right\vert $ satisfying
$\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell\left(  \alpha\right)
$.)
\end{lemma}

\begin{exercise}
\label{exe.lem.QSym.lyndon.Malphas1}Prove Lemma~\ref{lem.QSym.lyndon.Malphas1}.

[\textbf{Hint:} There are two approaches: One is to apply Proposition
\ref{prop.QSym.Malphas.dets}(a) and expand the determinant; the other is to
argue which monomials can appear in $s!M_{\alpha}^{\left\langle s\right\rangle
}-M_{\alpha}^{s}$.]
\end{exercise}

We now return to studying products of monomial quasisymmetric functions:

\begin{lemma}
\label{lem.QSym.lyndon.triangl.wll-lemma.precise}Let $n\in \NN$ and
$m\in \NN$. Let $u\in\Comp_{n}$ and
$v\in\Comp_{m}$. Let $z$ be the lexicographically
highest element of the multiset $u\shuffle v$. Let $h$ be the multiplicity
with which the word $z$ appears in the multiset $u\shuffle v$.
Then,\footnote{The following equality makes sense because we have
$z\in\Comp_{n+m}$ (by
Lemma~\ref{lem.QSym.lyndon.triang.wll-lemma}(a)).}
\[
M_{u}M_{v}=hM_{z}+\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.QSym.lyndon.triangl.wll-lemma.precise}.]Lemma
\ref{lem.QSym.lyndon.triangl.wll-lemma.precise} was shown during the proof of
Lemma \ref{lem.QSym.lyndon.triang.wll-lemma}(b).
\end{proof}

\begin{corollary}
\label{cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}Let $n\in \NN$ and
$m\in \NN$. Let $u\in\Comp_{n}$ and
$v\in\Comp_{m}$. Regard $u$ and $v$ as words in
$\mathfrak{A}^{\ast}$. Assume that $u$ is a Lyndon word. Let $\left(
b_{1},b_{2},\ldots,b_{q}\right)  $ be the CFL factorization of the word $v$.

Assume that $u\geq b_{j}$ for every $j\in\left\{  1,2,\ldots,q\right\}  $. Let%
\[
h=1+\left\vert \left\{  j\in\left\{  1,2,\ldots,q\right\}  \ \mid
\ b_{j}=u\right\}  \right\vert .
\]
Then,%
\[
M_{u}M_{v}=hM_{uv}+\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}uv\right)  .
\]

\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}Prove Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}.

[\textbf{Hint:} Apply Lemma \ref{lem.QSym.lyndon.triangl.wll-lemma.precise},
and notice that $uv$ is the lexicographically highest element of the multiset
$u\shuffle v$ (by Theorem \ref{thm.shuffle.lyndon.used}(e)), and that $h$ is
the multiplicity with which this word $uv$ appears in the multiset
$u\shuffle
v$ (this is a rewriting of Theorem \ref{thm.shuffle.lyndon.used}(e)).]
\end{exercise}

\begin{corollary}
\label{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}Let $k\in \NN$ and
$s\in \NN$. Let $x\in\Comp_{k}$ be such that $x$
is a Lyndon word. Then:

\begin{enumerate}
\item[(a)] The lexicographically highest element of the multiset $x\shuffle
x^{s}$ is $x^{s+1}$.

\item[(b)] We have%
\[
M_{x}M_{x^{s}}=\left(  s+1\right)  M_{x^{s+1}}+\left(  \text{a sum of terms of
the form }M_{w}\text{ with }w\in\Comp_{\left(
s+1\right)  k}\text{ satisfying }w\underset{\operatorname*{wll}}{<}%
x^{s+1}\right)  .
\]


\item[(c)] Let $t\in\Comp_{sk}$ be such that
$t\underset{\operatorname*{wll}}{<}x^{s}$. Then,%
\[
M_{x}M_{t}=\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{\left(  s+1\right)  k}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}x^{s+1}\right)  .
\]

\end{enumerate}
\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}Prove Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}.

[\textbf{Hint:} Notice that $\left(  \underbrace{x,x,\ldots,x}_{s\text{
times}}\right)  $ is the CFL factorization of the word $x^{s}$. Now, part (a)
of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs} follows from
Theorem \ref{thm.shuffle.lyndon.used}(c), part (b) follows from Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}, and part (c) from Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma}(c) (using part (a)).]
\end{exercise}

\begin{corollary}
\label{cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}Let $n\in \NN$
and $m\in \NN$. Let $u\in\Comp_{n}$ and
$v\in\Comp_{m}$. Regard $u$ and $v$ as words in
$\mathfrak{A}^{\ast}$. Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the
CFL factorization of $u$. Let $\left(  b_{1},b_{2},\ldots,b_{q}\right)  $ be
the CFL factorization of the word $v$. Assume that $a_{i}>b_{j}$ for every
$i\in\left\{  1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}
$. Then,%
\[
M_{u}M_{v}=M_{uv}+\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}uv\right)  .
\]

\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}Prove Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}.

[\textbf{Hint:} Combine Lemma \ref{lem.QSym.lyndon.triangl.wll-lemma.precise}
with the parts (c) and (d) of Theorem \ref{thm.shuffle.lyndon.used}.]
\end{exercise}

\begin{corollary}
\label{cor.QSym.lyndon.triangl.wll-lemma.BC}Let $n\in \NN$. Let
$u\in\Comp_{n}$ be a nonempty composition. Regard $u$
as a word in $\mathfrak{A}^{\ast}$. Let $\left(  a_{1},a_{2},\ldots
,a_{p}\right)  $ be the CFL factorization of $u$. Let $k\in\left\{
1,2,\ldots,p-1\right\}  $ be such that $a_{k}>a_{k+1}$. Let $x$ be the word
$a_{1}a_{2}\cdots a_{k}$, and let $y$ be the word $a_{k+1}a_{k+2}\cdots a_{p}%
$. Then,%
\[
M_{u}=M_{x}M_{y}-\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}u\right)  .
\]

\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.triangl.wll-lemma.BC}Prove Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.BC}.

[\textbf{Hint:} Apply Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict} to $x$, $y$,
$\left\vert x\right\vert $, $\left\vert y\right\vert $, $k$, $p-k$, $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(  a_{k+1},a_{k+2},\ldots
,a_{p}\right)  $ instead of $u$, $v$, $n$, $m$, $p$, $q$, $\left(  a_{1}%
,a_{2},\ldots,a_{p}\right)  $ and $\left(  b_{1},b_{2},\ldots,b_{q}\right)  $;
then, notice that $xy=u$ and $\left\vert x\right\vert +\left\vert y\right\vert
=n$.]
\end{exercise}

\begin{corollary}
\label{cor.QSym.lyndon.triangl.wll-lemma.As}Let $k\in \NN$. Let
$x\in\Comp_{k}$ be a composition. Assume that $x$ is a
Lyndon word. Let $s\in \NN$. Then,%
\[
M_{x}^{s}-s!M_{x^{s}}\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}.
\]
(Recall that $x^{s}$ is defined to be the word $\underbrace{xx\cdots
x}_{s\text{ times}}$.)
\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.triangl.wll-lemma.As}Prove Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As}.

[\textbf{Hint:} Rewrite the claim of Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As} in the form $M_{x}^{s}\in
s!M_{x^{s}}+\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$. This can be
proven by induction over $s$, where in the induction step we need the
following two observations:

\begin{enumerate}
\item We have $M_{x}M_{x^{s}}\in\left(  s+1\right)  M_{x^{s+1}}+
\sum\limits_{\substack{w\in\Comp_{\left(  s+1\right)
k};\\w\underset{\operatorname*{wll}}{<}x^{s+1}}} \kk M_{w}$.

\item For every $t\in\Comp_{sk}$ satisfying
$t\underset{\operatorname*{wll}}{<}x^{s}$, we have $M_{x}M_{t}\in
\sum\limits_{\substack{w\in\Comp_{\left(  s+1\right)
k};\\w\underset{\operatorname*{wll}}{<}x^{s+1}}} \kk M_{w}$.
\end{enumerate}

These two observations follow from parts (b) and (c) of Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}.]
\end{exercise}

\begin{corollary}
\label{cor.QSym.lyndon.triangl.wll-lemma.As2}Let $k\in \NN$. Let
$x\in\Comp_{k}$ be a composition. Assume that $x$ is a
Lyndon word. Let $s\in \NN$. Then,%
\[
M_{x}^{\left\langle s\right\rangle }-M_{x^{s}}\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}} \kk M_{w}.
\]
(Recall that $x^{s}$ is defined to be the word $\underbrace{xx\cdots
x}_{s\text{ times}}$.)
\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.triangl.wll-lemma.As2}Prove Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As2}.

[\textbf{Hint:} Lemma \ref{lem.QSym.lyndon.Malphas1} (applied to $\alpha=x$)
yields%
\[
s!M_{x}^{\left\langle s\right\rangle }-M_{x}^{s}\in\sum\limits_{\substack{\beta
\in\Comp_{sk};\\\ell\left(  \beta\right)  \leq\left(
s-1\right)  \ell\left(  x\right)  }}M_{\beta}=\sum\limits_{\substack{w\in
\Comp_{sk};\\\ell\left(  w\right)  \leq\left(
s-1\right)  \ell\left(  x\right)  }} \kk M_{w}\subset\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}} \kk M_{w}%
\]
\footnote{since every $w\in\Comp_{sk}$ with the
property that $\ell\left(  w\right)  \leq\left(  s-1\right)  \ell\left(
x\right)  $ must satisfy $w\underset{\operatorname*{wll}}{<}x^{s}$}. Adding
this to the claim of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As},
obtain $s!M_{x}^{\left\langle s\right\rangle }-s!M_{x^{s}}\in
\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$, that is,
$s!\left(  M_{x}^{\left\langle s\right\rangle }-M_{x^{s}}\right)  \in
\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$. It remains to
get rid of the $s!$ on the left hand side. Assume WLOG that $\kk %
= \ZZ$, and argue that every $f\in\Qsym$ satisfying
$s!\cdot f\in\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$ must itself lie
in $\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$.]
\end{exercise}

We are now ready to prove Theorem \ref{thm.QSym.lyndon.Z-genset}:

\begin{exercise}
\label{exe.thm.QSym.lyndon.Z-genset}Prove Theorem
\ref{thm.QSym.lyndon.Z-genset}.

[\textbf{Hint:} Lemma \ref{lem.QSym.lyndon.Z-genset.reindex} yields that the
family $\left(  M_{w}^{\left\langle s\right\rangle }\right)  _{\left(
w,s\right)  \in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a
reindexing of the family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}$. Hence, it is enough to prove
that the family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}$ is an algebraically independent
generating set of the $\kk$-algebra $\Qsym$. The latter
claim, in turn, will follow from Lemma \ref{lem.lyndonbases}%
(c)\footnote{applied to $A=\Qsym$, $b_{w}%
=M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }$,
$\operatorname*{wt}\left(  N\right)  =N$ and $g_{u}=M_{u}$} once it is proven
that the family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}$ generates the
$\kk$-algebra $\Qsym$. So it remains to show that the family
$\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }\right)
_{w\in\mathfrak{L}}$ generates the $\kk$-algebra $\Qsym$.

Let $U$ denote the $\kk$-subalgebra of $\Qsym$ generated
by $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$. It then suffices to prove that
$U=\Qsym$. To this purpose, it is enough to prove that
\begin{equation}
M_{\beta}\in U\ \ \ \ \ \ \ \ \ \ \text{for every composition }\beta.
\label{exe.thm.QSym.lyndon.Z-genset.hint1}%
\end{equation}


For every reduced Lyndon composition $\alpha$ and every $j\in\left\{
1,2,3,\ldots\right\}  $, the quasisymmetric function $M_{\alpha}^{\left\langle
j\right\rangle }$ is an element of the family $\left(  M_{\operatorname*{red}%
w}^{\left\langle \gcd w\right\rangle }\right)  _{w\in\mathfrak{L}}$ and thus
belongs to $U$. Combine this with Exercise \ref{exe.QSym.Malphas.ps}(d) to see
that%
\begin{equation}
M_{\beta}^{\left\langle s\right\rangle }\in U\ \ \ \ \ \ \ \ \ \ \text{for
every Lyndon composition }\beta\text{ and every }s\in\left\{  1,2,3,\ldots
\right\}  \label{exe.thm.QSym.lyndon.Z-genset.hint2}%
\end{equation}
(because every Lyndon composition $\beta$ can be written as $\alpha\left\{
n\right\}  $ for a reduced Lyndon composition $\alpha$ and an $n\in\left\{
1,2,3,\ldots\right\}  $). Now, prove (\ref{exe.thm.QSym.lyndon.Z-genset.hint1}%
) by strong induction: first, induct on $\left\vert \beta\right\vert $, and
then, for fixed $\left\vert \beta\right\vert $, induct on $\beta$ in the
wll-order. The induction step looks as follows: Fix some composition $\alpha$,
and assume (as induction hypothesis) that:

\begin{itemize}
\item (\ref{exe.thm.QSym.lyndon.Z-genset.hint1}) holds for every composition
$\beta$ satisfying $\left\vert \beta\right\vert <\left\vert \alpha\right\vert
$;

\item (\ref{exe.thm.QSym.lyndon.Z-genset.hint1}) holds for every composition
$\beta$ satisfying $\left\vert \beta\right\vert =\left\vert \alpha\right\vert
$ and $\beta\underset{\operatorname*{wll}}{<}\alpha$.
\end{itemize}

\noindent It remains to prove that (\ref{exe.thm.QSym.lyndon.Z-genset.hint1})
holds for $\beta=\alpha$.
In other words, it remains to prove that $M_{\alpha}\in U$.

Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
the word $\alpha$. Assume WLOG that $p\neq0$ (else, all is trivial). We are in
one of the following two cases:

\textit{Case 1:} All of the words $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are equal.

\textit{Case 2:} Not all of the words $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are equal.

In Case 2, there exists a $k\in\left\{  1,2,\ldots,p-1\right\}  $ satisfying
$a_{k}>a_{k+1}$ (since $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$), and thus
Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.BC} (applied to $u=\alpha$,
$n=\left\vert \alpha\right\vert $, $x=a_{1}a_{2}\cdots a_{k}$ and
$y=a_{k+1}a_{k+2}\cdots a_{p}$) shows that
\begin{align*}
M_{\alpha}  &  =\underbrace{M_{a_{1}a_{2}\cdots a_{k}}}_{\substack{\in
U\\\text{(by the induction}\\\text{hypothesis)}}}
\underbrace{M_{a_{k+1}a_{k+2}\cdots a_{p}}}_{\substack{\in U\\
\text{(by the induction}\\\text{hypothesis)}}}\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \text{a sum of terms of the form
}\underbrace{M_{w}}_{\substack{\in U\\\text{(by the induction}%
\\\text{hypothesis)}}}\text{ with }w\in
\Comp_{\left\vert \alpha\right\vert }\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}\alpha\right) \\
&  \in UU-\left(  \text{a sum of terms in }U\right)  \subset U.
\end{align*}


Hence, it only remains to deal with Case 1. In this case, set $x=a_{1}%
=a_{2}=\cdots=a_{p}$. Thus, $\alpha=a_{1}a_{2}\cdots a_{p}=x^{p}$, whence
$\left\vert \alpha\right\vert =p\left\vert x\right\vert $. But Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As2} (applied to $s=p$ and
$k=\left\vert x\right\vert $) yields%
\begin{align*}
M_{x}^{\left\langle p\right\rangle }-M_{x^{p}}  &  \in\sum\limits_{\substack{w\in
\Comp_{p\left\vert x\right\vert }%
;\\w\underset{\operatorname*{wll}}{<}x^{p}}} \kk M_{w}=
\sum\limits_{\substack{w\in\Comp_{\left\vert \alpha\right\vert
};\\w\underset{\operatorname*{wll}}{<}\alpha}} \kk \underbrace{M_{w}%
}_{\substack{\in U\\\text{(by the induction}\\\text{hypothesis)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p\left\vert x\right\vert
=\left\vert \alpha\right\vert \text{ and }x^{p}=\alpha\right) \\
&  \subset\sum\limits_{\substack{w\in\Comp_{N}%
;\\w\underset{\operatorname*{wll}}{<}\alpha}} \kk U\subset U,
\end{align*}
so that $M_{x^{p}}\in\underbrace{M_{x}^{\left\langle p\right\rangle }%
}_{\substack{\in U\\\text{(by (\ref{exe.thm.QSym.lyndon.Z-genset.hint2}))}%
}}-U\subset U-U\subset U$. This rewrites as $M_{\alpha}\in U$ (since
$\alpha=x^{p}$). So $M_{\alpha}\in U$ is proven in both Cases 1 and 2, and
thus the induction proof of (\ref{exe.thm.QSym.lyndon.Z-genset.hint1}) is finished.]
\end{exercise}

\begin{exercise}
\label{exe.thm.QSym.lyndon}Prove Theorem \ref{thm.QSym.lyndon}.
\end{exercise}

Of course, this proof of Theorem \ref{thm.QSym.lyndon} yields a new (third)
proof for Proposition \ref{prop.QSym.lyndon}.

We notice the following corollary of our approach to Theorem
\ref{thm.QSym.lyndon}:

\begin{corollary}
\label{cor.QSym.lyndon.free}The $\Lambda$-algebra $\Qsym$ is a
polynomial algebra (over $\Lambda$).
\end{corollary}

\begin{exercise}
\label{exe.cor.QSym.lyndon.free}Prove Corollary \ref{cor.QSym.lyndon.free}.

[\textbf{Hint:} The algebraically independent generating set $\left(
M_{w}^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ of
$\Qsym$ contains the elements $M_{\left(  1\right)
}^{\left\langle s\right\rangle }=e_{s}\in\Lambda$ for all $s\in\left\{
1,2,3,\ldots\right\}  $.]
\end{exercise}

\subsection{\label{subsect.lyndon.gr}The Gessel-Reutenauer bijection and
symmetric functions}

In this section, we shall discuss the Gessel-Reutenauer bijection between
words and multisets of aperiodic necklaces, and use it to study another family
of symmetric functions.

The Gessel-Reutenauer bijection was studied in \cite{GesselReutenauer}, where
it was applied to various enumeration problems (e.g., counting permutations in
$\Symm_{n}$ with given descent set and given cycle type); it is also
closely related to the Burrows-Wheeler bijection used in data compression
(\cite{CrochemoreDesarmenienPerrin}), and to the structure of free Lie algebras
(\cite{GesselRestivoReutenauer}, \cite{Reutenauer}).
We shall first introduce the Gessel-Reutenauer
bijection and study it combinatorially in Subsection
\ref{subsub.lyndon.gr.bij}; then, in the following Subsection
\ref{subsub.lyndon.grfun}, we shall apply it to symmetric functions.

\subsubsection{\label{subsub.lyndon.gr.bij}Necklaces and the Gessel-Reutenauer
bijection}

We begin with definitions, some of which have already been made in Exercise
\ref{exe.words.necklaces}:

\begin{definition}
\label{def.lyndon.gr.cyclic}Throughout Section \ref{subsect.lyndon.gr}, we
shall freely use Definition \ref{def.words} and Definition
\ref{def.words.lyndon}. We fix a totally ordered alphabet $\mathfrak{A}$.
(This alphabet can be arbitrary, although most examples will use
$\mathfrak{A}=\left\{  1<2<3<\cdots\right\}  $.)

Let \dfn{$C$} denote the infinite cyclic group, written multiplicatively.
Fix a generator \dfn{$c$} of $C$.\ \ \ \ \footnote{So $C$ is a group
isomorphic to $\left(  \ZZ ,+\right)  $, and the isomorphism
$\left(  \ZZ ,+\right) \rightarrow C$ sends every $n\in \ZZ$ to $c^{n}$.
(Recall that we write the binary operation of $C$ as $\cdot$ instead of
$+$.)}

For any positive integer $n$, the group $C$ acts on $\mathfrak{A}^{n}$ from
the left according to the rule
\[
c\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{2},a_{3}%
,\ldots,a_{n},a_{1}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(
a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}.
\]
\footnote{In other words, $c$ rotates any $n$-tuple of elements of
$\mathfrak{A}$ cyclically to the left. Thus, $c^{n}\in C$ acts trivially on
$\mathfrak{A}^{n}$, and so this action of $C$ on $\mathfrak{A}^{n}$ factors
through $C/\left\langle c^{n}\right\rangle $ (a cyclic group of order $n$).}
The orbits of this $C$-action will be called
\emph{$n$-necklaces}\index{$n$-necklace}%
\footnote{See Exercise \ref{exe.words.necklaces} for the motivation behind
this word.
\par
Notice that there are no $0$-necklaces, because we required $n$ to be positive
in the definition of a necklace. This is intentional.}; they form a set
partition of the set $\mathfrak{A}^{n}$.

The $n$-necklace containing a given $n$-tuple $w\in\mathfrak{A}^{n}$ will be
denoted by $\left[  w\right]  $.

A \dfn{necklace} shall mean an $n$-necklace for some positive integer $n$.
Thus, for each nonempty word $w$, there is a well-defined necklace $\left[
w\right]  $ (namely, $\left[  w\right]  $ is an $n$-necklace, where
$n=\ell\left(  w\right)  $).

The \emph{period}\index{period of a necklace}
of a necklace $N$ is defined as the positive integer
$\left\vert N\right\vert $. (This $\left\vert N\right\vert $ is indeed a
positive integer, since $N$ is a finite nonempty set\footnote{by
Exercise \ref{exe.words.necklaces}(a), because $N$ is an $n$-necklace
for some positive integer $n$}.)

An $n$-necklace is said to be \emph{aperiodic}\index{aperiodic $n$-necklace}
if its period is $n$.
\end{definition}

\begin{example}
Let $\mathfrak{A}$ be the alphabet $\left\{  1<2<3<\cdots\right\}  $. The
orbit of the word $223$ under the $C$-action is the $3$-necklace $\left\{
223,232,322\right\}  $; it is an aperiodic $3$-necklace. The orbit of the word
$223223$ under the $C$-action is the $6$-necklace $\left\{
223223,232232,322322\right\}  $; it is not aperiodic (since it has period
$3$). The orbit of any nonempty word $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in\mathfrak{A}^{n}$ is the $n$-necklace
\[
\left\{  \left(  w_{i},w_{i+1},\ldots,w_{n},w_{1},w_{2},\ldots,w_{i-1}\right)
\ \mid\ i\in\left\{  1,2,\ldots,n\right\}  \right\}  .
\]
We can draw this $n$-necklace on the plane as follows:
\[
\xymatrix@R=2.0pc@C=.8pc{
& w_1 \ar@/^1ex/[rr] & & w_2 \ar@/^1ex/[dr] \\
w_n \ar@/^1ex/[ur] & & \;\; & & w_3 \ar@/^1ex/[dl] \\
& w_{n-1} \ar@/^1ex/[ul] & & \iddots \ar@/^1ex/[ll]
}
\]
\end{example}

It is easy to see that the notion of an ``aperiodic
necklace'' we just defined is equivalent to the notion of a
``primitive necklace'' used in Exercise
\ref{exe.polynomials-and-necklaces}(b).

Exercise \ref{exe.words.necklaces}(a) shows that any $n$-necklace for any
positive integer $n$ is a finite nonempty set.
In other words, any necklace is a finite nonempty set.

Let us next introduce some notations regarding words and permutations. We
recall that a cycle of a permutation $\tau\in\Symm_{n}$ is an orbit
under the action of $\tau$ on $\left\{  1,2,\ldots,n\right\}  $. (This orbit
can be a $1$-element set, when $\tau$ has fixed points.) We begin with a basic definition:

\begin{definition}
\label{def.words.perm-ord}Let $\tau\in\Symm_{n}$ be a permutation. Let
$h\in\left\{  1,2,\ldots,n\right\}  $.

\begin{enumerate}
\item[(a)] We let $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $
denote the smallest positive integer $i$ such that $\tau^{i}\left(  h\right)
=h$. (Basic properties of permutations show that this $i$ exists.)

\item[(b)] Let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}%
^{n}$ be a word. Then, $w_{\tau,h}$ shall denote the word $w_{\tau^{1}\left(
h\right)  }w_{\tau^{2}\left(  h\right)  }\cdots w_{\tau^{k}\left(  h\right)
}$, where $k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $.
\end{enumerate}
\end{definition}

\begin{example}
Let $\tau$ be the permutation $3142765\in\Symm_{7}$ (in one-line
notation). Then, $\operatorname*{ord}\nolimits_{\tau}\left(  1\right)  =4$
(since $\tau^{4}\left(  1\right)  =1$, but $\tau^{i}\left(  1\right)  \neq1$
for every positive integer $i<4$). Likewise, $\operatorname*{ord}%
\nolimits_{\tau}\left(  2\right)  =4$ and $\operatorname*{ord}\nolimits_{\tau
}\left(  3\right)  =4$ and $\operatorname*{ord}\nolimits_{\tau}\left(
4\right)  =4$ and $\operatorname*{ord}\nolimits_{\tau}\left(  5\right)  =2$
and $\operatorname*{ord}\nolimits_{\tau}\left(  6\right)  =1$ and
$\operatorname*{ord}\nolimits_{\tau}\left(  7\right)  =2$.

Now, let $w$ be the word $4112524\in\mathfrak{A}^{7}$. Then,%
\begin{align*}
w_{\tau,3}  &  =w_{\tau^{1}\left(  3\right)  }w_{\tau^{2}\left(  3\right)
}w_{\tau^{3}\left(  3\right)  }w_{\tau^{4}\left(  3\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{ord}\nolimits_{\tau
}\left(  3\right)  =4\right) \\
&  =w_{4}w_{2}w_{1}w_{3}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau^{1}\left(  3\right)  =4\text{
and }\tau^{2}\left(  3\right)  =\tau\left(  4\right)  =2\text{ and }\tau
^{3}\left(  3\right)  =\tau\left(  2\right)  =1\text{ and }\tau^{4}\left(
3\right)  =\tau\left(  1\right)  =3\right) \\
&  =2141.
\end{align*}
Likewise, we can check that $w_{\tau,1}=w_{3}w_{4}w_{2}w_{1}=1214$ and
$w_{\tau,5}=w_{7}w_{5}=45$ and $w_{\tau,6}=w_{6}=2$.
\end{example}

We begin the study of the words $w_{\tau,h}$ by stating some of their simplest
properties:\footnote{See Exercise \ref{exe.lyndon.gr.basics} below for the
proof of Proposition \ref{prop.words.read1}, as well as for the proofs of all
other propositions stated before Exercise \ref{exe.lyndon.gr.basics}.}

\begin{proposition}
\label{prop.words.read1}Let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\mathfrak{A}^{n}$ be a word. Let $\tau\in\Symm_{n}$. Let
$h\in\left\{  1,2,\ldots,n\right\}  $. Then:

\begin{enumerate}
\item[(a)] The word $w_{\tau,h}$ is nonempty and has length
$\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $.

\item[(b)] The first letter of the word $w_{\tau,h}$ is $w_{\tau\left(
h\right)  }$.

\item[(c)] The last letter of the word $w_{\tau,h}$ is $w_{h}$.

\item[(d)] We have $w_{\tau,\tau\left(  h\right)  }=c\cdot w_{\tau,h}$.

\item[(e)] We have $w_{\tau,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$
for each $i\in \ZZ $.
\end{enumerate}
\end{proposition}

Recall that if $n\in \NN$ and if $w\in\mathfrak{A}^{n}$ is a word, then
a permutation $\operatorname*{std}w\in\Symm_{n}$ was defined in
Definition \ref{def.words.std}. The words $w_{\tau,h}$ have particularly nice
properties when $\tau=\left(  \operatorname*{std}w\right)  ^{-1}$:

\begin{lemma}
\label{lem.sol.lyndon.gr.basics.1}Let $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in\mathfrak{A}^{n}$ be a word. Let $\tau$ be the permutation
$\left(  \operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$. Let $\alpha$
and $\beta$ be two elements of $\left\{  1,2,\ldots,n\right\}  $ such that
$\alpha<\beta$. Then:

\begin{enumerate}
\item[(a)] If $\tau^{-1}\left(  \alpha\right)  <\tau^{-1}\left(  \beta\right)
$, then $w_{\alpha}\leq w_{\beta}$.

\item[(b)] If $\tau^{-1}\left(  \alpha\right)  \geq\tau^{-1}\left(
\beta\right)  $, then $w_{\alpha}>w_{\beta}$.

\item[(c)] We have $w_{\tau\left(  \alpha\right)  }\leq w_{\tau\left(
\beta\right)  }$.

\item[(d)] If $\tau\left(  \alpha\right)  \geq\tau\left(  \beta\right)  $,
then $w_{\tau\left(  \alpha\right)  }<w_{\tau\left(  \beta\right)  }$.

\item[(e)] If $w_{\tau\left(  \alpha\right)  }=w_{\tau\left(  \beta\right)  }%
$, then $\tau\left(  \alpha\right)  <\tau\left(  \beta\right)  $.

\item[(f)] If $w_{\tau,\alpha}=w_{\tau,\beta}$, then $\tau\left(
\alpha\right)  <\tau\left(  \beta\right)  $ and $w_{\tau,\tau\left(
\alpha\right)  }=w_{\tau,\tau\left(  \beta\right)  }$.

\item[(g)] If $w_{\tau,\alpha}=w_{\tau,\beta}$, then $\tau^{i}\left(
\alpha\right)  <\tau^{i}\left(  \beta\right)  $ for each $i\in \NN $.

\item[(h)] Let $j\in \NN $ be such that every $i\in\left\{
0,1,\ldots,j-1\right\}  $ satisfies $w_{\tau^{i+1}\left(  \alpha\right)
}=w_{\tau^{i+1}\left(  \beta\right)  }$. Then, $w_{\tau^{j+1}\left(
\alpha\right)  }\leq w_{\tau^{j+1}\left(  \beta\right)  }$.
\end{enumerate}
\end{lemma}

\begin{proposition}
\label{prop.words.read-std2}Let $w\in\mathfrak{A}^{n}$ be a word. Let $\tau$
be the permutation $\left(  \operatorname*{std}w\right)  ^{-1}\in
\Symm_{n}$. Let $z$ be a cycle of $\tau$. Then:

\begin{enumerate}
\item[(a)] For each $h\in z$, we have $\left[  w_{\tau,h}\right]  =\left\{
w_{\tau,i}\ \mid\ i\in z\right\}  $.

\item[(b)] If $\alpha$ and $\beta$ are two distinct elements of $z$, then
$w_{\tau,\alpha}\neq w_{\tau,\beta}$.

\item[(c)] We have $\left|  \left\{  w_{\tau,i}\ \mid\ i\in z\right\}
\right|  = \left|  z\right|  $.

\item[(d)] The set $\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $ is an
aperiodic necklace.
\end{enumerate}
\end{proposition}

\begin{exercise}
\label{exe.lyndon.gr.basics}Prove Proposition \ref{prop.words.read1}, Lemma
\ref{lem.sol.lyndon.gr.basics.1} and Proposition \ref{prop.words.read-std2}.
\end{exercise}

\begin{definition}
Let $w \in\mathfrak{A}^{n}$ be a
word. Let $\tau$ be the permutation $\left(  \operatorname*{std}w\right)
^{-1}\in\Symm_{n}$. Let $z$ be a cycle of $\tau$. Then, we define an
aperiodic necklace $\left[  w\right]  _{z}$ by $\left[  w\right]
_{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $. (This is indeed an
aperiodic necklace, according to Proposition \ref{prop.words.read-std2}(d).)
\end{definition}

\begin{example}
Let $\mathfrak{A}$ be the alphabet $\left\{ 1<2<3<\cdots \right\}$, and let
$w$ be the word $2511321\in\mathfrak{A}^{7}$. Let $\tau$ be the
permutation $\left(  \operatorname*{std}w\right)  ^{-1}\in\Symm_{7}$;
this is the permutation $3471652$ (in one-line notation). One cycle of $\tau$
is $z=\left\{  1,3,7,2,4\right\}  $. The corresponding aperiodic necklace
$\left[  w\right]  _{z}$ is%
\begin{align*}
\left[  w\right]  _{z}  &  =\left\{  w_{\tau,i}\ \mid\ i\in z\right\}
=\left\{  w_{\tau,1},w_{\tau,3},w_{\tau,7},w_{\tau,2},w_{\tau,4}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }z=\left\{  1,3,7,2,4\right\}  \right)
\\
&  =\left\{  11512,15121,51211,12115,21151\right\}  =\left[  11512\right]  .
\end{align*}

\end{example}

\begin{definition}
We let \dfn{$\mathfrak{N}$} be the set of all necklaces.
We let \dfn{$\mathfrak{N}^{\mathfrak{a}}$} be the set of all aperiodic
necklaces.
We let \dfn{$\mathfrak{MN}^{\mathfrak{a}}$} be the set of all finite
multisets of aperiodic necklaces.
\end{definition}

\begin{definition}
\label{def.lyndon.gr.GR}We define a map $\operatorname*{GR}:\mathfrak{A}%
^{\ast}\rightarrow\mathfrak{MN}^{\mathfrak{a}}$\index{$\operatorname{GR}$}
as follows:

Let $w\in\mathfrak{A}^{\ast}$. Let $n=\ell\left(  w\right)  $ (so that
$w\in\mathfrak{A}^{n}$). Let $\tau$ be the permutation $\left(
\operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$. Then, we define the
multiset $\operatorname*{GR}w\in\mathfrak{MN}^{\mathfrak{a}}$ by setting%
\[
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}.
\]
(This multiset $\operatorname*{GR}w$ is indeed a finite multiset of aperiodic
necklaces\footnote{Indeed, this multiset $\operatorname*{GR}w$ is finite
(since $\tau$ has only finitely many cycles), and its elements $\left[
w\right]  _{z}$ are aperiodic necklaces (as we have seen in the definition of
$\left[  w\right]  _{z}$).}, and thus belongs to $\mathfrak{MN}^{\mathfrak{a}%
}$.)
\end{definition}

\begin{example}
\label{exa.lyndon.gr.GR}Let $\mathfrak{A}$ be the alphabet $\left\{
1<2<3<\cdots\right\}  $, and let $w=33232112\in\mathfrak{A}^{8}$.

To compute $\operatorname*{GR}w$, we first notice that $\operatorname*{std}%
w=67384125$ (in one-line notation). Hence, the
permutation $\tau$ from Definition \ref{def.lyndon.gr.GR} satisfies
$\tau=\left(  \operatorname*{std}w\right)  ^{-1}=
67358124$. The cycles of $\tau$ are $\left\{  1,6\right\}  $,
$\left\{  2,7\right\}  $, $\left\{  3\right\}  $ and $\left\{  4,5,8\right\}
$. Thus,%
\begin{align*}
\operatorname*{GR}w  &  =\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a
cycle of }\tau\right\}  _{\operatorname*{multiset}}=\left\{  \left[  w\right]
_{\left\{  1,6\right\}  },\left[  w\right]  _{\left\{  2,7\right\}  },\left[
w\right]  _{\left\{  3\right\}  },\left[  w\right]  _{\left\{  4,5,8\right\}
}\right\}  _{\operatorname*{multiset}}\\
&  =\left\{  \left[  31\right]  ,\left[  31\right]  ,\left[  2\right]
,\left[  322\right]  \right\}  _{\operatorname*{multiset}}=\left\{  \left[
13\right]  ,\left[  13\right]  ,\left[  2\right]  ,\left[  223\right]
\right\}  _{\operatorname*{multiset}}%
\end{align*}
(since $\left[  31\right]  =\left[  13\right]  $ and $\left[  322\right]
=\left[  223\right]  $ as necklaces).
Drawn on the plane, the necklaces in $\operatorname*{GR}w$ look as follows:
\[
\begin{array}{cccc}
\xymatrix@R=1.4pc@C=.8pc{
\\
1 \ar@/^3ex/[rr] & & 3 \ar@/^3ex/[ll]
}
&
\qquad
\xymatrix@R=1.4pc@C=.8pc{
\\
1 \ar@/^3ex/[rr] & & 3 \ar@/^3ex/[ll]
}
&\quad\qquad
\xymatrix@R=1.4pc@C=.8pc{
\\ 2
}
&\quad\qquad
\xymatrix@R=1.4pc@C=1.0pc{
2 \ar@/^2ex/[drr] \\
& & 2 \ar@/^2ex/[dll] \\
3 \ar@/^4ex/[uu]
}
\end{array}
\]
\end{example}

The map $\operatorname*{GR}$ is called the \dfn{Gessel-Reutenauer
bijection}. In order to show that it indeed is a bijection, we shall construct
its inverse. First, we introduce some further objects.

\begin{definition}
A nonempty word $w$ is said to be \emph{aperiodic}\index{aperiodic word}
if there exist no
$m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$.

Let \dfn{$\mathfrak{A}^{\mathfrak{a}}$} be the set of all aperiodic words in
$\mathfrak{A}^{\ast}$.
\end{definition}

For example, the word $132231$ is aperiodic, but the word $132132$ is not
(since $132132 = u^m$ for $u = 132$ and $m = 2$).

Aperiodic words are directly connected to aperiodic necklaces, as the
following facts show:\footnote{See Exercise \ref{exe.lyndon.gr.aper-rel} for
the proofs of all unproved statements made until Exercise
\ref{exe.lyndon.gr.aper-rel}.}

\begin{proposition}
\label{prop.words.aper-neck}Let $w\in\mathfrak{A}^{\ast}$ be a nonempty word.
Then, the word $w$ is aperiodic if and only if the necklace $\left[  w\right]
$ is aperiodic.
\end{proposition}

\begin{corollary}
\label{cor.words.aper-neck-cw}Let $w\in\mathfrak{A}^{\ast}$ be an aperiodic
word. Then, the word $c\cdot w$ is aperiodic.\footnote{See Definition
\ref{def.lyndon.gr.cyclic} for the definition of $c$ and its
action on words.}
\end{corollary}

\begin{corollary}
\label{cor.words.aper-neck-ms}Each aperiodic necklace is a set of aperiodic words.
\end{corollary}

Let us now introduce a new total order on the set $\mathfrak{A}^{\mathfrak{a}%
}$ of all aperiodic words:

\begin{definition}
Let $u$ and $v$ be two aperiodic words. Then, we write $u\leq_{\omega}v$ if
and only if $uv\leq vu$. Thus, we have defined a binary relation
\dfn{$\leq_{\omega}$}
on the set $\mathfrak{A}^{\mathfrak{a}}$ of all aperiodic words.
\end{definition}

\begin{proposition}
\label{prop.words.aper-rel.tord}The relation $\leq_{\omega}$ on the set
$\mathfrak{A}^{\mathfrak{a}}$ is the smaller-or-equal relation of a total order.
\end{proposition}

For the next proposition, we should recall Definition
\ref{def.lyndon.gr.cyclic} (and, in particular, the meaning of $c$ and its
action on words).

\begin{proposition}
\label{prop.words.aper-rel.1}Let $u$ and $v$ be two aperiodic words.

\begin{enumerate}
\item[(a)] We have $u\leq_{\omega}v$ if and only if either $u_{1}<v_{1}$ or
$\left(  u_{1}=v_{1}\text{ and }c\cdot u\leq_{\omega}c\cdot v\right)
$.\ \ \ \footnote{The relation ``$c\cdot u\leq_{\omega}c\cdot
v$'' here makes sense because the words $c\cdot u$ and
$c\cdot v$ are aperiodic (by Corollary \ref{cor.words.aper-neck-cw}).}

\item[(b)] If $u\neq v$, then there exists some $i\in \NN $ satisfying
$\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$.

\item[(c)] We have $u\leq_{\omega}v$ if and only if the smallest
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$ \textbf{either} does not exist \textbf{or}
satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot v\right)
_{1}$.

\item[(d)] Let $n$ and $m$ be positive integers such that $n\ell\left(
u\right)  =m\ell\left(  v\right)  $. We have $u\leq_{\omega}v$ if and only if
$u^{n}\leq v^{m}$.
\end{enumerate}
\end{proposition}

\begin{remark}
We are avoiding the use of infinite words here; if we didn't, we could restate
the relation $\leq_{\omega}$ in a simpler way (which is easily seen to be
equivalent to Proposition \ref{prop.words.aper-rel.1}(c)): Two aperiodic words
$u$ and $v$ satisfy $u\leq_{\omega}v$ if and only if $u^{\infty}\leq
v^{\infty}$. Here, for any nonempty word $w$, we are letting $w^{\infty}$
denote the infinite word
\[
\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  },w_{1},w_{2}%
,\ldots,w_{\ell\left(  w\right)  },w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
},\ldots\right)
\]
(that is, the word $w$ repeated endlessly), and the symbol ``$\leq$'' in
``$u^{\infty}\leq v^{\infty}$'' refers to the lexicographic order on
$\mathfrak{A}^{\infty}$.

Other equivalent descriptions of the relation $\leq_{\omega}$ (or, more
precisely, of the ``strictly less'' relation corresponding to it) can be
found in \cite[Corollary 11]{DolceRestivoReutenauer-svl}.
\end{remark}

\begin{proposition}
\label{prop.words.read-std3}Let $w
\in\mathfrak{A}^{n}$ be a word. Let $\tau$ be the permutation $\left(
\operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$. Then:

\begin{enumerate}
\item[(a)] The words $w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}$ are aperiodic.

\item[(b)] We have $w_{\tau,1}\leq_{\omega}w_{\tau,2}\leq_{\omega}\cdots
\leq_{\omega}w_{\tau,n}$.
\end{enumerate}
\end{proposition}

\begin{exercise}
\label{exe.lyndon.gr.aper-rel}Prove Proposition \ref{prop.words.aper-neck},
Corollary \ref{cor.words.aper-neck-cw}, Corollary \ref{cor.words.aper-neck-ms}%
, Proposition \ref{prop.words.aper-rel.tord}, Proposition
\ref{prop.words.aper-rel.1} and Proposition \ref{prop.words.read-std3}.
\end{exercise}

We need two more notations about multisets:

\begin{definition}
Let $T$ be a totally ordered set, and let $\leq_T$ be the smaller-or-equal
relation of $T$. Let $M$ be a finite multiset of elements of $T$.
Then, there is a unique list $\left( m_1, m_2, \ldots, m_n \right)$
such that
\[
\left\{ m_1, m_2, \ldots, m_n \right\}_{\operatorname*{multiset}}
= M
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ %
m_1 \leq_T m_2 \leq_T \cdots \leq_T m_n .
\]
This list $\left( m_1, m_2, \ldots, m_n \right)$ is obtained by listing
all elements of $M$ (with their multiplicities) in increasing order
(increasing with respect to $\leq_T$). We shall refer to this list
$\left( m_1, m_2, \ldots, m_n \right)$ as the
\dfn{$\leq_T$-increasing list}\index{increasing list of a multiset}
of $M$.

(For example, the $\leq_{\ZZ}$-increasing list of $\left\{
1,2,3,2,1\right\}  _{\operatorname*{multiset}}$ is
$\left( 1,1,2,2,3 \right) $.)
\end{definition}

\begin{definition}
Let $S$ be a finite multiset.

\begin{enumerate}
\item[(a)] The \emph{support}\index{support of a multiset}
\dfn{$\operatorname*{Supp}S$} is defined to be the
set of all elements of $S$.
Thus, if
$S = \left\{ m_1, m_2, \ldots, m_n \right\}_{\operatorname*{multiset}}$,
then
$\operatorname*{Supp} S = \left\{ m_1, m_2, \ldots, m_n \right\}  $.

\item[(b)] For each $s\in S$, let $M_{s}$ be a finite multiset. Then, we
define the \dfn{multiset union} \dfn{$\biguplus_{s\in S}M_{s}$} to be
the finite multiset $M$ with the following property:
For any object $x$, we have
\begin{align*}
\left(  \text{multiplicity of }x\text{ in }M\right)
&  =\sum_{s\in\operatorname*{Supp}S}\left(  \text{multiplicity of }s\text{ in
}S\right)  \cdot\left(  \text{multiplicity of }x\text{ in }M_{s}\right)  .
\end{align*}


For example:

\begin{itemize}
\item If $S=\left\{  1,2,3\right\}  _{\operatorname*{multiset}}$ and
$M_{s}=\left\{  s,s+1\right\}  _{\operatorname*{multiset}}$ for each
$s\in\operatorname*{Supp}S$, then $\biguplus_{s\in S}M_{s}=\left\{
1,2,2,3,3,4\right\}  _{\operatorname*{multiset}}$.

\item If $S=\left\{  1,1,2\right\}  _{\operatorname*{multiset}}$ and
$M_{s}=\left\{  s,s+1\right\}  _{\operatorname*{multiset}}$ for each
$s\in\operatorname*{Supp}S$, then $\biguplus_{s\in S}M_{s}=\left\{
1,1,2,2,2,3\right\}  _{\operatorname*{multiset}}$.
\end{itemize}

We regard each set as a multiset; thus, the multiset union $\biguplus_{s\in
S}M_{s}$ is also defined when the $M_{s}$ are sets.
\end{enumerate}
\end{definition}

Now, we can construct the inverse of the Gessel-Reutenauer bijection:

\begin{definition}
\label{def.lyndon.gr.RG}We define a map $\operatorname*{RG}:\mathfrak{MN}%
^{\mathfrak{a}}\rightarrow\mathfrak{A}^{\ast}$\index{$\operatorname{RG}$}
as follows:

Let $M\in\mathfrak{MN}^{\mathfrak{a}}$ be a finite multiset of aperiodic
necklaces. Let $M^{\prime}=\biguplus_{N\in M}N$. (We are here using the fact
that each necklace $N\in M$ is a finite set, thus a finite multiset.) Notice
that $M^{\prime}$ is a finite multiset of aperiodic words\footnote{Indeed:
\par
\begin{itemize}
\item Each $N\in M$ is an aperiodic necklace (since $M$ is a multiset
of aperiodic necklaces), and thus (by Corollary \ref{cor.words.aper-neck-ms})
a set of aperiodic words.
Therefore, $\biguplus_{N\in M}N$ is a multiset of aperiodic
words.
\par
\item Each $N\in M$ is a necklace, and thus is a finite set (since any
necklace is a finite set). Since the multiset $M$ is also finite, this shows
that $\biguplus_{N\in M}N$ is finite.
\end{itemize}
\par
Thus, $\biguplus_{N\in M}N$ is a finite multiset of aperiodic words. In other
words, $M^{\prime}$ is a finite multiset of aperiodic words (since $M^{\prime
}=\biguplus_{N\in M}N$).}. Let $\left(  m_{1},m_{2},\ldots,m_{n}\right)  $ be
the $\leq_{\omega}$-increasing list of $M^{\prime}$. For each $i\in\left\{
1,2,\ldots,n\right\}  $, let $\ell_{i}$ be the last letter of the nonempty
word $m_{i}$. Then, $\operatorname*{RG}\left(  M\right)  $ is defined to be
the word $\left(  \ell_{1},\ell_{2},\ldots,\ell_{n}\right)  \in\mathfrak{A}%
^{\ast}$.
\end{definition}

\begin{example}
\label{exa.lyndon.gr.RG}Let $\mathfrak{A}$ be the alphabet $\left\{
1<2<3<\cdots\right\}  $, and let $M=\left\{  \left[  13\right]  ,\left[
13\right]  ,\left[  2\right]  ,\left[  223\right]  \right\}
_{\operatorname*{multiset}}$. Clearly, $M\in\mathfrak{M}\mathfrak{N}%
^{\mathfrak{a}}$ (since $M$ is a finite multiset of aperiodic necklaces).
(Actually, $M$ is the multiset of aperiodic necklaces drawn in
Example~\ref{exa.lyndon.gr.GR}.)
In order to
compute the word $\operatorname*{RG}\left(  M\right)  $, let us first compute
the multiset $M^{\prime}$ from Definition \ref{def.lyndon.gr.RG}. Indeed, the
definition of $M^{\prime}$ yields%
\begin{align*}
M^{\prime}  &  =\biguplus_{N\in M}N=\underbrace{\left[  13\right]
}_{=\left\{  13,31\right\}  }\uplus\underbrace{\left[  13\right]  }_{=\left\{
13,31\right\}  }\uplus\underbrace{\left[  2\right]  }_{=\left\{  2\right\}
}\uplus\underbrace{\left[  223\right]  }_{=\left\{  223,232,322\right\}  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\text{where we are using the notation }M_{1}\uplus M_{2}\uplus\cdots\uplus
M_{k}
\text{ for a multiset union }\biguplus\limits_{s\in\left\{  1,2,\ldots
,k\right\}  }M_{s}
\right) \\
&  =\left\{  13,31\right\}  \uplus\left\{  13,31\right\}  \uplus\left\{
2\right\}  \uplus\left\{  223,232,322\right\} \\
&  =\left\{  13,31,13,31,2,223,232,322\right\}  _{\operatorname*{multiset}}.
\end{align*}
Hence, the $\leq_{\omega}$-increasing list of $M^{\prime}$ is $\left(
13,13,2,223,232,31,31,322\right)  $ (since $13\leq_{\omega}13\leq_{\omega
}2\leq_{\omega}223\leq_{\omega}232\leq_{\omega}31\leq_{\omega}31\leq_{\omega
}322$). The last letters of the words in this list are $3,3,2,3,2,1,1,2$ (in
this order). Hence, Definition \ref{def.lyndon.gr.RG} shows that
\[
\operatorname*{RG}\left(  M\right)  =\left(  3,3,2,3,2,1,1,2\right)
=33232112.
\]

\end{example}

\begin{remark}
The $\leq_{\omega}$-increasing list of a multiset $M^{\prime}$ of aperiodic
words is not always the same as its $\leq$-increasing list. For example, the
$\leq_{\omega}$-increasing list of $\left\{  2,21\right\}  $ is $\left(
21,2\right)  $ (since $21\leq_{\omega}2$), whereas its $\leq$-increasing list
is $\left(  2,21\right)  $ (since $2\leq21$).
\end{remark}

A comparison of Examples \ref{exa.lyndon.gr.GR} and \ref{exa.lyndon.gr.RG}
suggests that the maps $\operatorname*{GR}$ and $\operatorname*{RG}$ undo
one another. This is indeed true, as the following theorem (due to Gessel
and Reutenauer \cite[Lemma 3.4 and Example 3.5]{GesselReutenauer}; also
proved in \cite[Theorem 7.20]{Reutenauer}, \cite[Theorem 3.1 and
Proposition 3.1]{DiaconisMcGrathPitman} and
\cite[\S 2]{GesselRestivoReutenauer}) shows:

\begin{theorem}
\label{thm.lyndon.gr.bij}The maps $\operatorname*{GR}:\mathfrak{A}^{\ast
}\rightarrow\mathfrak{MN}^{\mathfrak{a}}$ and $\operatorname*{RG}%
:\mathfrak{MN}^{\mathfrak{a}}\rightarrow\mathfrak{A}^{\ast}$ are mutually
inverse bijections.
\end{theorem}

\begin{exercise}
\label{exe.lyndon.gr.bij}Prove Theorem \ref{thm.lyndon.gr.bij}.

[\textbf{Hint:} First, use Proposition \ref{prop.words.read-std3} to show that
$\operatorname*{RG}\circ\operatorname*{GR} = \id$. Then recall
the fact that any injective map between two finite sets of the same sizes is a
bijection. This does not directly apply here, since the sets $\mathfrak{A}%
^{\ast}$ and $\mathfrak{MN}^{\mathfrak{a}}$ are usually not finite. However,
$\operatorname*{GR}$ can be restricted to a map between two appropriate finite
subsets, obtained by focussing on a finite sub-alphabet of $\mathfrak{A}$
and fixing the length of the words; these subsets can be shown to have equal
size using the Chen-Fox-Lyndon factorization (see the following paragraph for
the connection).\footnote{This argument roughly follows
\cite{GesselRestivoReutenauer}.}]
\end{exercise}

Theorem \ref{thm.lyndon.gr.bij} shows that the sets $\mathfrak{A}^{\ast}$ and
$\mathfrak{MN}^{\mathfrak{a}}$ are in bijection. This bijection is in some
sense similar to the Chen-Fox-Lyndon factorization\footnote{The
Chen-Fox-Lyndon factorization (Theorem \ref{thm.words.CFL}) provides a
bijection between words in $\mathfrak{A}^{\ast}$ and multisets of Lyndon words
(because the factors in the CFL factorization of a word $w\in\mathfrak{A}%
^{\ast}$ can be stored in a multiset), whereas the Gessel-Reutenauer bijection
$\operatorname*{GR}:\mathfrak{A}^{\ast}\rightarrow\mathfrak{MN}^{\mathfrak{a}%
}$ is a bijection between words in $\mathfrak{A}^{\ast}$ and multisets of
aperiodic necklaces. Since the Lyndon words are in bijection with the
aperiodic necklaces (by Exercise \ref{exe.words.necklaces}(e)), we can thus
view the two bijections as having the same targets (and the same domains).
That said, they are not the same bijection.}, and preserves various quantities
(for example, the number of times a given letter $a$ appears in a word
$w\in\mathfrak{A}^{\ast}$ equals the number of times this letter $a$ appears
in the words in the corresponding multiset $\operatorname*{GR}w\in
\mathfrak{MN}^{\mathfrak{a}}$, provided that we pick one representative
of each necklace in $\operatorname*{GR}w$),
and predictably affects other quantities (for
example, the cycles of the standardization $\operatorname*{std}w$ of a word
$w\in\mathfrak{A}^{\ast}$ have the same lengths as the aperiodic necklaces in
the corresponding multiset $\operatorname*{GR}w\in\mathfrak{MN}^{\mathfrak{a}%
}$); these properties have ample applications to enumerative questions
(discussed in \cite{GesselReutenauer}).

\begin{remark}
\label{rmk.lyndon.gr.bw}
The Gessel-Reutenauer bijection relates to the \dfn{Burrows-Wheeler
transformation} (e.g., \cite[\S 2]{CrochemoreDesarmenienPerrin}). Indeed, the
latter sends an aperiodic word $w\in\mathfrak{A}^{\mathfrak{a}}$ to the word
$\operatorname*{RG}\left(  \left\{  \left[  w\right]  \right\}
_{\operatorname*{multiset}}\right)  $ obtained by applying $\operatorname*{RG}%
$ to the multiset consisting of the single aperiodic necklace $\left[
w\right]  $. This transformation is occasionally applied in (lossless) data
compression, as the word $\operatorname*{RG}\left(  \left\{  \left[  w\right]
\right\}  _{\operatorname*{multiset}}\right)  $ tends to have many strings of
consecutive equal letters when $w$ has substrings occurring multiple times
(for example, if $\mathfrak{A}=\left\{  a<b<c<d<\cdots\right\}  $ and
$w=bananaban$, then $\operatorname*{RG}\left(  \left\{  \left[  w\right]
\right\}  _{\operatorname*{multiset}}\right)  =nnbbnaaaa$), and strings of
consecutive equal letters can easily be compressed. (In order to guarantee
that $w$ can be recovered from the result, one can add a new letter $\zeta$ --
called a ``sentinel symbol'' -- to the
alphabet $\mathfrak{A}$, and apply the Burrows-Wheeler transformation to the
word $w\zeta$ instead of $w$. This also ensures that $w\zeta$ is an aperiodic
word, so the Burrows-Wheeler transformation can be applied to $w\zeta$ even
if it cannot be applied to $w$.)

Kufleitner, in \cite[\S 4]{Kufleitner}, suggests a bijective variant of the
Burrows-Wheeler transformation. In our notations, it sends a word
$w\in\mathfrak{A}^{\ast}$ to the word $\operatorname*{RG}\left(  \left\{
\left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]
\right\}  _{\operatorname*{multiset}}\right)  $, where $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of $w$.
\end{remark}

For variants and generalizations of the Gessel-Reutenauer bijection,
see \cite{Kufleitner}, \cite{Steinhardt}, \cite{ShareshianWachs-euler},
\cite{DressSiebeneicher-diophant} and \cite{Reiner-signedperm}.

\subsubsection{\label{subsub.lyndon.grfun}The Gessel-Reutenauer symmetric
functions}

In this subsection, we shall study a certain family of symmetric functions.
First, we recall that every word $w\in\mathfrak{A}^{\ast}$ has a unique CFL
factorization (see Theorem \ref{thm.words.CFL}). Based on this fact, we can
make the following definition:

\begin{definition}
For the rest of Subsection \ref{subsub.lyndon.grfun}, we let $\mathfrak{A}$ be
the alphabet $\left\{  1<2<3<\cdots\right\}  $.

Let $w\in\mathfrak{A}^{\ast}$ be a word. The
\emph{CFL type}\index{CFL type of a word} of $w$ is
defined to be the partition whose parts are the positive integers $\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $
(listed in decreasing order), where $\left(  a_{1},a_{2},\ldots,a_{k}\right)
$ is the CFL factorization of $w$. This CFL type is denoted by
$\operatorname*{CFLtype}w$.
\end{definition}

\begin{example}
Let $w$ be the word $213212412112$. Then, the tuple $\left(
2,132,124,12,112\right)  $ is the CFL factorization of $w$. Hence, the CFL
type of $w$ is the partition whose parts are the positive integers
$\ell\left(  2\right)  ,\ell\left(  132\right)  ,\ell\left(  124\right)
,\ell\left(  12\right)  ,\ell\left(  112\right)  $ (listed in decreasing
order). In other words, the CFL type of $w$ is the partition $\left(
3,3,3,2,1\right)  $ (since the positive integers $\ell\left(  2\right)
,\ell\left(  132\right)  ,\ell\left(  124\right)  ,\ell\left(  12\right)
,\ell\left(  112\right)  $ are $1,3,3,2,3$).
\end{example}

\begin{definition}
\label{def.lyndon.grfun.GRfun}For each word $w=\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in\mathfrak{A}^{\ast}$, we define a monomial
$\xx_{w}$ in $\kk \left[  \left[  \xx\right]  \right]  $
by setting $\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$. (For example,
$\xx_{\left(  1,3,2,1\right)  }=x_{1}x_{3}x_{2}x_{1}=x_{1}^{2}%
x_{2}x_{3}$.)

For any partition $\lambda$, we define a power series
$\mathbf{GR}_{\lambda}\in \kk\left[  \left[  \xx\right]  \right]  $%
\index{$\mathbf{GR}_{\lambda}$} by
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast
};\\\operatorname*{CFLtype}w=\lambda}}\xx_{w}.
\]

\end{definition}

\begin{example}
Let us compute $\mathbf{GR}_{\left(  2,1\right)  }$. Indeed, the words
$w\in\mathfrak{A}^{\ast}$ satisfying $\operatorname*{CFLtype}w=\left(
2,1\right)  $ are the words whose CFL factorization consists of two words, one
of which has length $1$ and the other has length $2$. In other words, these
words $w\in\mathfrak{A}^{\ast}$ must have the form $w=a_{1}a_{2}$ for two
Lyndon words $a_{1}$ and $a_{2}$ satisfying $a_{1}\geq a_{2}$ and $\left(
\ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  \right)  \in\left\{
\left(  1,2\right)  ,\left(  2,1\right)  \right\}  $. A straightforward
analysis of possibilities reveals that these are precisely the $3$-letter
words $w=\left(  w_{1},w_{2},w_{3}\right)  $ satisfying either $\left(
w_{1}<w_{2}\text{ and }w_{1}\geq w_{3}\right)  $ or $\left(  w_{1}>w_{2}\text{
and }w_{2}<w_{3}\right)  $. Hence,%
\begin{align*}
\mathbf{GR}_{\left(  2,1\right)  }  &  =\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\left(  2,1\right)
}}\xx_{w}=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\w_{1}%
<w_{2}\text{ and }w_{1}\geq w_{3}}}\xx_{w}+\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\w_{1}>w_{2}\text{ and }w_{2}<w_{3}}}\xx_{w}\\
&  =\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\w_{1}<w_{2}\text{ and
}w_{1}\geq w_{3}}}\xx_{w}+\sum\limits_{\substack{w\in\mathfrak{A}%
^{\ast};\\w_{1}>w_{2}\text{ and }w_{2}<w_{3}\text{ and }w_{1}\leq w_{3}%
}}\xx_{w}+\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\w_{1}%
>w_{2}\text{ and }w_{2}<w_{3}\text{ and }w_{1}>w_{3}}}\xx_{w}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have split the second sum according to the relation between }w_{1}\text{ and }w_{3}
\right) \\
&  =\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\w_{3}\leq w_{1}<w_{2}%
}}\xx_{w}+\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\w_{2}%
<w_{1}\leq w_{3}}}\xx_{w}+\sum\limits_{\substack{w\in\mathfrak{A}%
^{\ast};\\w_{2}<w_{3}<w_{1}}}\xx_{w}%
\end{align*}
(here, we rewrote the conditions under the summation signs). The three sums on
the right hand side are clearly quasisymmetric functions. Using
(\ref{fundamental-qsym-fn-defn-alternative}), we can rewrite them as
$L_{\left(  2,1\right)  }$, $L_{\left(  1,2\right)  }$ and $L_{\left(
1,1,1\right)  }$, respectively. Thus, we obtain%
\begin{align*}
\mathbf{GR}_{\left(  2,1\right)  }  &  =L_{\left(  2,1\right)  }+L_{\left(
1,2\right)  }+L_{\left(  1,1,1\right)  }=3M_{\left(  1,1,1\right)
}+M_{\left(  1,2\right)  }+M_{\left(  2,1\right)  }\\
&  =3m_{\left(  1,1,1\right)  }+m_{\left(  2,1\right)  }.
\end{align*}
Thus, $\mathbf{GR}_{\left(  2,1\right)  }$ is actually a symmetric function!
We shall soon (in Proposition \ref{prop.lyndon.grfun.symm}) see that this is
not a coincidence.
\end{example}

We shall now state various properties of the power series $\mathbf{GR}%
_{\lambda}$; their proofs are all part of Exercise \ref{exe.lyndon.grfun.all}.

\begin{proposition}
\label{prop.lyndon.grfun.n}Let $n$ be a positive integer. Then:

\begin{enumerate}
\item[(a)] The partition $\left(  n\right)  $ satisfies%
\[
\mathbf{GR}_{\left(  n\right)  }=\sum\limits_{\substack{w\in\mathfrak{A}%
^{n};\\w\text{ is Lyndon}}}\xx_{w}.
\]


\item[(b)] Assume that $\kk$ is a $\QQ$-algebra. Then,%
\[
\mathbf{GR}_{\left(  n\right)  }=\dfrac{1}{n}\sum_{d\mid n}\mu\left(
d\right)  p_{d}^{n/d}.
\]
Here, $\mu$ denotes the number-theoretical M\"{o}bius function (defined as in
Exercise \ref{exe.witt.ghost-equiv}), and the summation sign
``$\sum_{d\mid n}$'' is understood to range over all
\textbf{positive} divisors $d$ of $n$.
\end{enumerate}
\end{proposition}

\begin{proposition}
\label{prop.lyndon.grfun.symm}Let $\lambda$ be a partition. Then, the power
series $\mathbf{GR}_{\lambda}$ belongs to $\Lambda$.
\end{proposition}

Thus, $\left(  \mathbf{GR}_{\lambda}\right)  _{\lambda\in \Par}$
is a family of symmetric functions.\footnote{Several sources, including
\cite{GesselReutenauer}, \cite[Exercise 7.89]{Stanley} and
\cite{ElizaldeTroyka},
write $L_\lambda$ for what we call $\mathbf{GR}_{\lambda}$.
(So would we if $L_\alpha$ didn't already have another meaning here.)}
Unlike many other such families we have
studied, it is not a basis of $\Lambda$; it is not linearly independent (e.g.,
it satisfies $\mathbf{GR}_{\left(  2,1,1\right)  }=\mathbf{GR}_{\left(
4\right)  }$). Nevertheless, it satisfies a Cauchy-kernel-like
identity\footnote{Recall that $\mathfrak{L}$ denotes the set of Lyndon words
in $\mathfrak{A}^{\ast}$.}:

\begin{proposition}
\label{prop.lyndon.grfun.cauchy}Consider two countable sets of indeterminates
$\xx=\left(  x_{1},x_{2},x_{3},\ldots\right)  $ and $\mathbf{y}=\left(
y_{1},y_{2},y_{3},\ldots\right)  $.

\begin{enumerate}
\item[(a)] In the power series ring $\kk \left[  \left[  \xx%
,\mathbf{y}\right]  \right]  = \kk \left[  \left[  x_{1},x_{2}%
,x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right]  \right]  $, we have%
\[
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\lambda
\in \Par}p_{\lambda}\left(  \xx\right)  \mathbf{GR}%
_{\lambda}\left(  \mathbf{y}\right)  .
\]


\item[(b)] For each word $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\mathfrak{A}^{\ast}$, we define a monomial $\mathbf{y}_{w}$ in $\kk
\left[  \left[  \mathbf{y}\right]  \right]  $ by setting
$\mathbf{y}_{w}=y_{w_{1}}y_{w_{2}}\cdots y_{w_{n}}$. Then,%
\begin{align*}
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)   &  =\sum
_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}w}\left(
\mathbf{y}\right)  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac
{1}{1-\xx_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(
w\right)  }}\\
&  =\sum_{\lambda\in \Par}p_{\lambda}\left(  \xx\right)
\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  .
\end{align*}

\end{enumerate}
\end{proposition}

The proof of this proposition rests upon the following simple
equality\footnote{Recall that $\mathfrak{L}$ denotes the set of Lyndon words
in $\mathfrak{A}^{\ast}$. Also, recall that $\mathfrak{A}=\left\{
1<2<3<\cdots\right\}  $. Thus, $p_{1}=\sum_{i\geq1}x_{i}=\sum_{a\in
\mathfrak{A}}x_{a}$.}:

\begin{proposition}
\label{prop.lyndon.grfun.CFLps}In the power series ring $\left(
\kk \left[  \left[  \xx\right]  \right]  \right)  \left[  \left[
t\right]  \right]  $, we have%
\[
\dfrac{1}{1-p_{1}t}=\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}%
t^{\ell\left(  w\right)  }}.
\]

\end{proposition}

We can furthermore represent the symmetric functions $\mathbf{GR}_{\lambda}$
in terms of the fundamental basis $\left(  L_{\alpha}\right)  _{\alpha
\in\operatorname*{Comp}}$ of $\operatorname*{QSym}$; here, the
Gessel-Reutenauer bijection from Theorem \ref{thm.lyndon.gr.bij} reveals its
usefulness. We will use Definition~\ref{def.QSym.gamma(sigma)}.

\begin{proposition}
\label{prop.lyndon.grfun.F}Let $\lambda$ be a partition. Let $n=\left\vert
\lambda\right\vert $. Then,%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{\sigma\in\Symm_n;
\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)
}.
\]

\end{proposition}

The proof of this relies on Lemma~\ref{lem.Lalpha-std.std-L} (see
Exercise~\ref{exe.lyndon.grfun.all} below for the details).

\begin{definition}
\label{def.lyndon.grfun.type}
Let $\Symm = \bigsqcup_{n\in \NN }\Symm_{n}$ (an external
disjoint union). For each $\sigma \in \Symm$, we let
$\operatorname*{type}\sigma$ denote the cycle type of $\sigma$.
\end{definition}

\begin{proposition}
\label{prop.lyndon.grfun.cauchy2}Consider two countable sets of indeterminates
$\xx=\left(  x_{1},x_{2},x_{3},\ldots\right)  $ and $\mathbf{y}=\left(
y_{1},y_{2},y_{3},\ldots\right)  $.

In the power
series ring $\kk \left[  \left[  \xx,\mathbf{y}\right]  \right]
$, we have%
\[
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\lambda
\in \Par}p_{\lambda}\left(  \xx\right)  \mathbf{GR}%
_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\sigma\in\Symm}
L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)
p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)  .
\]

\end{proposition}

Let us finally give two alternative descriptions of the $\mathbf{GR}_{\lambda
}$ that do not rely on the notion of CFL factorization. First, we state a fact
that is essentially trivial:

\begin{proposition}
\label{prop.lyndon.grfun.xN}Let $N$ be a necklace. Let $w$ and $w^{\prime}$ be
two elements of $N$. Then:

\begin{enumerate}
\item[(a)] There exist words $u$ and $v$ such that $w=uv$ and $w^{\prime}=vu$.

\item[(b)] We have $\xx_{w}=\xx_{w^{\prime}}$.
\end{enumerate}
\end{proposition}

\begin{definition}
Let $N\in\mathfrak{N}$ be a necklace. Then, we define a monomial
$\xx_{N}$ in $\kk \left[  \left[  \xx\right]  \right]  $
by setting $\xx_{N}=\xx_{w}$, where $w$ is any element of $N$.
(This is well-defined, because Proposition \ref{prop.lyndon.grfun.xN}(b) shows
that $\xx_{w}$ does not depend on the choice of $w$.)
\end{definition}

\begin{definition}
Let $M$ be a finite multiset of necklaces. Then, we define a monomial
$\xx_{M}$ in $\kk \left[  \left[  \xx\right]  \right]  $
by setting $\xx_{M}=\xx_{N_{1}}\xx_{N_{2}}\cdots
\xx_{N_{k}}$, where $M$ is written in the form $M=\left\{  N_{1}%
,N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$.
\end{definition}

\begin{definition}
Let $M$ be a finite multiset of necklaces. Then, we can obtain a partition by
listing the sizes of the necklaces in $M$ in decreasing order. This partition
will be called the \emph{type}\index{type of a finite multiset of necklaces}
of $M$, and will be denoted by \dfn{$\operatorname*{type}M$}.
\end{definition}

\begin{example}
If $M=\left\{  \left[  13\right]  ,\left[  13\right]  ,\left[  2\right]
,\left[  223\right]  \right\}  _{\operatorname*{multiset}}$, then the type of
$M$ is $\left(  3,2,2,1\right)  $ (because the sizes of the necklaces in $M$
are $2,2,1,3$).
\end{example}

\begin{proposition}
\label{prop.lyndon.grfun.alt1}Let $\lambda$ be a partition. Then,%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}%
};\\\operatorname*{type}M=\lambda}}\xx_{M}.
\]

\end{proposition}

This was our first alternative description of $\mathbf{GR}_{\lambda}$. Note
that it is used as a definition of $\mathbf{GR}_{\lambda}$ in \cite[(2.1)]%
{GesselReutenauer} (where $\mathbf{GR}_{\lambda}$ is denoted by $L_{\lambda}%
$). Using the Gessel-Reutenauer bijection, we can restate it as follows:

\begin{proposition}
\label{prop.lyndon.grfun.alt2}Let $\lambda$ be a partition. Then,%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast
};\\\operatorname*{type}\left(  \operatorname*{GR}w\right)  =\lambda
}}\xx_{w}.
\]

\end{proposition}

Let us finally give a second alternative description of $\mathbf{GR}_{\lambda
}$:

\begin{proposition}
\label{prop.lyndon.grfun.alt3}Let $\lambda$ be a partition. Then,%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast
};\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda
}}\xx_{w}.
\]

\end{proposition}

\begin{exercise}
\label{exe.lyndon.grfun.all}Prove all statements made in Subsection
\ref{subsub.lyndon.grfun}.

[\textbf{Hint:} Here is one way to proceed:

\begin{itemize}
\item First prove Proposition \ref{prop.lyndon.grfun.CFLps}, by using the CFL
factorization to argue that both sides equal $\sum_{w\in\mathfrak{A}^{\ast}%
}\xx_{w}t^{\ell\left(  w\right)  }$.

\item Use a similar argument to derive Proposition
\ref{prop.lyndon.grfun.cauchy} (starting with part (b)).

\item Proposition \ref{prop.lyndon.grfun.xN} is almost trivial.

\item Derive Proposition \ref{prop.lyndon.grfun.alt1} from the definition of
$\mathbf{GR}_{\lambda}$ using the uniqueness of the CFL factorization.

\item Derive Proposition \ref{prop.lyndon.grfun.alt2} from Proposition
\ref{prop.lyndon.grfun.alt1} using the bijectivity of $\operatorname*{GR}$.

\item Derive Proposition \ref{prop.lyndon.grfun.alt3} from Proposition
\ref{prop.lyndon.grfun.alt2}.

\item Obtain Proposition \ref{prop.lyndon.grfun.F} by combining Proposition
\ref{prop.lyndon.grfun.alt3} with Lemma \ref{lem.Lalpha-std.std-L}.

\item Derive Proposition \ref{prop.lyndon.grfun.cauchy2} from Propositions
\ref{prop.lyndon.grfun.F} and \ref{prop.lyndon.grfun.cauchy}.

\item Derive Proposition \ref{prop.lyndon.grfun.symm} either from Proposition
\ref{prop.lyndon.grfun.alt1} or from Proposition
\ref{prop.lyndon.grfun.cauchy}. (In the latter case, make sure to work with
$\kk = \QQ$ first, and then extend to all other $\kk$, as
the proof will rely on the $\kk$-linear independence of $\left(
p_{\lambda}\right)  _{\lambda\in \Par}$, which doesn't hold for
all $\kk$.)

\item Prove Proposition \ref{prop.lyndon.grfun.n}(a) directly using the
definition of $\mathbf{GR}_{\left(  n\right)  }$.

\item Show that each positive integer $n$ satisfies $p_{1}^{n}=\sum_{d\mid
n}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  $ by taking logarithms in Proposition
\ref{prop.lyndon.grfun.CFLps}. Use this and
(\ref{eq.exe.witt.ghost-equiv.mu*1}) to prove Proposition
\ref{prop.lyndon.grfun.n}(b) recursively.
\end{itemize}

\noindent Other approaches are, of course, possible.]
\end{exercise}

\begin{remark}
\label{rmk.lyndon.grfun.char}
Let $n$ be a positive integer. The symmetric function $\mathbf{GR}_{\left(
n\right)  }$ has a few more properties:

\begin{enumerate}
\item[(a)] It is an $\NN$-linear combination of Schur functions. To
state the precise rule, we need a few more notations: A \dfn{standard
tableau} can be defined as a column-strict tableau $T$ with
$\cont \left(  T\right)  =\left(  1^{m}\right)  $, where $m$ is
the number of boxes of $T$. (That is, each of the numbers $1,2,\ldots,m$
appears exactly once in $T$, and no other numbers appear.) If $T$ is a
standard tableau with $m$ boxes, then a
\emph{descent}\index{descent of a standard tableau} of $T$ means an
$i\in\left\{  1,2,\ldots,m-1\right\}  $ such that the entry $i+1$ appears in
$T$ in a row further down than $i$ does. The
\emph{major index $\operatorname*{maj}T$}\index{major index of a standard tableau}
of a standard tableau $T$ is defined to be the sum
of its descents.\footnote{For example, the tableau
\[
\begin{array}[c]{cccc}
1 & 3 & 4 & 8\\
2 & 5 & 6 & 9\\
7 &  &  &
\end{array}
\]
is standard and has descents $1,4,6,8$ and major index $1+4+6+8=19$.} Now,
\[
\mathbf{GR}_{\left(  n\right)  }
=\sum_{\lambda\in \Par_{n}} a_{\lambda,1}s_{\lambda},
\]
where $a_{\lambda,1}$ is the number of standard tableaux $T$ of shape
$\lambda$ satisfying $\operatorname*{maj}T\equiv1\operatorname{mod}n$. (See
\cite[Exercise 7.89 (c)]{Stanley}.)

\item[(b)] Assume that $\kk = \CC$.
Recall the map $\ch : A\left(  \Symm \right) \rightarrow \Lambda$
from Theorem \ref{symmetric-group-Frobenius-map-theorem}.
Embed the cyclic group $C_n = \ZZ / n \ZZ$ as a subgroup in the
symmetric group $\Symm_n$ by identifying some generator $g$ of
$C_n$ with some $n$-cycle in $\Symm_n$. Let $\omega$ be a primitive
$n$-th root of unity in $\CC$ (for instance, $\exp\left(2\pi i/n\right)$).
Let $\gamma : C_n \rightarrow \CC$ be the character of $C_n$ that sends
each $g^i \in C_n$ to $\omega^i$. Then,
\[
\mathbf{GR}_{\left( n \right)}
= \ch \left( \Ind_{C_n}^{\Symm_n} \gamma \right)  .
\]
(See \cite[Exercise 7.89 (b)]{Stanley}.)

\item[(c)] The character $\Ind_{C_n}^{\Symm_n} \gamma$ of $\Symm_n$
is actually the character of a representation.
To construct it, set $\kk = \CC$, and
recall the notations from Exercise~\ref{exe.freelie.2}
(while keeping $\mathfrak{A} = \left\{1,2,3,\ldots\right\}$).
Let $\mathfrak{m}_n$ be the $\CC$-vector subspace of
$T\left(V\right)$ spanned by the products
$x_{\sigma\left(1\right)} x_{\sigma\left(2\right)}
 \cdots x_{\sigma\left(n\right)}$
with $\sigma \in \Symm_n$.
The symmetric group $\Symm_n$ acts on $T\left(V\right)$ by
algebra homomorphisms, with $\sigma \in \Symm_n$
sending each $x_i$ to $x_{\sigma\left(i\right)}$ when $i \leq n$
and to $x_i$ otherwise.
Both $\mathfrak{g}_n$ and $\mathfrak{m}_n$ are
$\CC \Symm_n$-submodules of $T\left(V\right)$.
Thus, so is the intersection $\mathfrak{g}_n \cap \mathfrak{m}_n$.
It is not hard to see that this intersection is spanned by all
``nested commutators''
$\left[x_{\sigma\left(1\right)},
       \left[x_{\sigma\left(2\right)},
             \left[x_{\sigma\left(3\right)},
                   \ldots
             \right]
       \right]
 \right]$ (in $T\left(V\right)$) with $\sigma \in \Symm_n$.
The character of this $\CC \Symm_n$-module
$\mathfrak{g}_n \cap \mathfrak{m}_n$ is precisely the
$\Ind_{C_n}^{\Symm_n} \gamma$ from Remark~\ref{rmk.lyndon.grfun.char}(b),
so applying the Frobenius characteristic map $\ch$ to it yields
the symmetric function $\mathbf{GR}_{\left(n\right)}$.
(See \cite[Theorem 9.41(i)]{Reutenauer}. There are similar
ways to obtain $\mathbf{GR}_\lambda$ for all $\lambda \in \Par$.)
\end{enumerate}
\end{remark}

\begin{exercise}
\label{exe.lyndon.grfun.ch-cyclic}
Prove the claim of Remark \ref{rmk.lyndon.grfun.char}(b).

[\textbf{Hint:} It helps to recall (or prove) that for any positive integer
$m$, the sum of all primitive $m$-th roots of unity in $\mathbb{C}$ is
$\mu\left(  m\right)  $.]
\end{exercise}

% [DG][v79] Added the above exercise.

The symmetric functions $\mathbf{GR}_{\lambda}$ for more general partitions
$\lambda$ can be expressed in terms of the symmetric functions
$\mathbf{GR}_{\left( n \right)}$ (which, as we recall from
Proposition \ref{prop.lyndon.grfun.n}(b), have a simple expression
in terms of the $p_m$) using the concept of
\emph{plethysm}; see \cite[Theorem 3.6]{GesselReutenauer}.

In \cite{GesselReutenauer}, Gessel and Reutenauer apply the symmetric
functions $\mathbf{GR}_{\lambda}$ to questions of permutation enumeration
via the following result%
\footnote{Proposition~\ref{prop.lyndon.grfun.count-perms}(a) is
\cite[Corollary 2.2]{GesselReutenauer};
Proposition~\ref{prop.lyndon.grfun.count-perms}(b) is
\cite[Theorem 2.1]{GesselReutenauer}.}:

\begin{proposition}
\label{prop.lyndon.grfun.count-perms}
Let $n \in \NN$. Let $\lambda \in \Par_n$ and
$\beta = \left( \beta_1, \beta_2, \ldots, \beta_k \right) \in \Comp_n$.
We shall use the notations introduced in Definition~\ref{def.QSym.D}.
Definition~\ref{def.QSym.gamma(sigma)} and
Definition~\ref{def.lyndon.grfun.type}.

\begin{enumerate}
\item[(a)] Let $\mu \in \Par_n$ be the partition
obtained by sorting the entries of $\beta$ into decreasing order. Then,%
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\Symm_n \text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\beta\text{
refines }\gamma\left(  \sigma\right)  \right) \\
&  =\left(  \text{the number of permutations }\sigma\in\Symm_n \text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }
\Des \sigma \subset D\left(  \beta\right)  \right) \\
&  =\left(  \text{the coefficient of }x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots
x_{k}^{\beta_{k}}\text{ in }\mathbf{GR}_{\lambda}\right)  =\left(  \text{the
coefficient of }\xx^{\mu}\text{ in }\mathbf{GR}_{\lambda}\right) \\
&  =\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{this is the Hall inner product of }\mathbf{GR}_{\lambda}\in\Lambda\text{
and }h_{\mu}\in\Lambda\right)  .
\end{align*}

\item[(b)] Recall the ribbon diagram $\Rib\left(\beta\right)$
corresponding to the composition $\beta$
(defined as in Definition~\ref{def.QSym.D}). Then,
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\Symm_n \text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right) \\
&  =\left(  \text{the number of permutations }\sigma\in\Symm_n \text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }
\Des \sigma =D\left(  \beta\right)  \right) \\
&  =\left(  \mathbf{GR}_{\lambda},s_{\Rib\left(\beta\right)}\right)
\ \ \ \ \ \ \ \ \ \ \left(
\text{this is the Hall inner product of }\mathbf{GR}_{\lambda}\in\Lambda\text{
and }s_{\Rib\left(\beta\right)}\in\Lambda\right)  .
\end{align*}
\end{enumerate}
\end{proposition}

\begin{exercise}
\label{exe.lyndon.grfun.count-perms}
Prove Proposition~\ref{prop.lyndon.grfun.count-perms}.

[\textbf{Hint:} Use Proposition \ref{prop.lyndon.grfun.F}, Theorem
\ref{Nsym-structure-on-ribbons-theorem}, the equality \eqref{eq.Nsym.Halpha}
and the adjointness between $\pi$ and $i$ in Corollary \ref{cor.NSym.pi}.]
\end{exercise}

By strategic application of Proposition~\ref{prop.lyndon.grfun.count-perms},
Gessel and Reutenauer arrive at several enumerative consequences, such as
the following:

\begin{itemize}
\item (\cite[Theorem 8.3]{GesselReutenauer}) If $A$ is a proper subset of
$\left\{  1,2,\ldots,n-1\right\}  $, then
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\Symm_{n}\text{
satisfying }\left\vert \operatorname*{Fix}\sigma\right\vert =0
\text{ and } \Des \sigma =A\right) \\
&  =\left(  \text{the number of permutations }\sigma\in\Symm_{n}\text{
satisfying }\left\vert \operatorname*{Fix}\sigma\right\vert =1
\text{ and } \Des \sigma =A\right)  ,
\end{align*}
where $\operatorname*{Fix}\sigma$ denotes the set of all fixed points of a
permutation $\sigma$.
This can also be proved bijectively; such a bijective proof can be
obtained by combining \cite[Theorems 5.1 and 6.1]{DesarmenienWachs}.

\item (\cite[Theorem 9.4]{GesselReutenauer}) If $i\in\left\{  1,2,\ldots
,n-1\right\}  $, then
\[
\left(  \text{the number of }n\text{-cycles }\sigma\in\Symm_n \text{
satisfying } \Des \sigma = \left\{  i\right\}  \right)
=\sum_{d\mid\gcd\left(  n,i\right)  }\mu\left(  d\right)  \dbinom{n/d}{i/d}.
\]
Note that this also equals the number of necklaces $\left[  \left(
w_{1},w_{2},\ldots,w_{n}\right)  \right]  $ (or, equivalently, Lyndon words
$\left(  w_{1},w_{2},\ldots,w_{n}\right)  $) with $w_{1},w_{2},\ldots,w_{n}%
\in\left\{  0,1\right\}  $ and $w_{1}+w_{2}+\cdots+w_{n}=i$. This suggests
that there should be a bijection between
$\left\{ n\text{-cycles }\sigma\in\Symm_n \text{
satisfying } \Des \sigma = \left\{  i\right\} \right\}$ and the set
of such necklaces; and indeed, such a bijection can be found in
\cite[Theorem 1]{CrochemoreDesarmenienPerrin}.
\end{itemize}

\noindent See \cite{GesselReutenauer} and \cite{ElizaldeTroyka} for more
such applications.

% [DG][v78] Added the above section.


% \subsection{Application:  Multiple zeta values and
% Hoffman's stuffle conjecture}
%
% Perhaps to be filled in later....

% [DG][v36] Commented out.





\newpage





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aguiar-Bergeron-Sottile character theory Part I: $\Qsym$ as a terminal object}
\label{sect.ABS}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

It turns out that the
universal mapping property of $\Nsym$ as a free associative algebra leads
via duality to a universal property for its dual $\Qsym$, elegantly explaining several
combinatorial invariants that take the form of quasisymmetric or symmetric functions:

\begin{enumerate}
\item[$\bullet$]  Ehrenborg's quasisymmetric function of a \emph{ranked poset} \cite{Ehrenborg},
\item[$\bullet$]  Stanley's \emph{chromatic} symmetric function of a \emph{graph} \cite{Stanley-chromatic-qsym-fn},
\item[$\bullet$]  the quasisymmetric function of a \emph{matroid} considered in \cite{BilleraJiaR}.
\end{enumerate}


\subsection{Characters and the universal property}
\label{ABS-part-I-section}


\begin{definition}
Given a Hopf algebra $A$ over $\kk$, a
\emph{character}\index{character of a Hopf algebra} is an algebra morphism
$A \overset{\zeta}{\longrightarrow} \kk$, that is,
\begin{enumerate}
\item[$\bullet$]
$\zeta(\one_A)=\one_\kk$,
\item[$\bullet$]
$\zeta$ is $\kk$-linear, and
\item[$\bullet$]
$\zeta(a b)=\zeta(a)\zeta(b)$ for $a,b$ in $A$.
\end{enumerate}
\end{definition}

\begin{example}
\label{Qsym-zeta-character-example}
A particularly important character for $A=\Qsym$ is defined as follows:\footnote{We
are using the notation of Proposition~\ref{prop.Qsym.eval} here, and we are still
identifying $\Qsym$ with $\Qsym\left(\xx\right)$, where $\xx$
denotes the infinite chain $\left(x_1 < x_2 < \cdots\right)$.}
\[
\begin{array}{rcl}
\Qsym & \overset{\zeta_Q}{\longrightarrow} & \kk , \\
f(\xx)& \longmapsto & f(1,0,0,\ldots) = \left[ f(\xx) \right]_{x_1=1,x_2=x_3=\cdots=0}.
\end{array}
\]
Hence,
\[
\zeta_Q(M_\alpha)=
\zeta_Q(L_\alpha)=
\begin{cases}
1, & \text{ if } \alpha=(n) \text{ for some }n;\\
0, & \text{ otherwise.}
\end{cases}
\]
In other words, the restriction $\zeta_Q|_{\Qsym_n}$ coincides with the
functional $H_n$ in $\Nsym_n=\Hom_\kk(\Qsym_n,\kk)$:  one has for $f$ in $\Qsym_n$ that
\begin{equation}
\label{Qsym-zeta-character-example.zetaQf}
\zeta_Q(f)=(H_n,f).
\end{equation}

% [DG][v17] Added footnote mostly as a reminder.

It is worth remarking that there is nothing special about setting $x_1=1$
and $x_2=x_3=\cdots=0$:  for quasisymmetric $f$, we could have defined the same character
$\zeta_Q$ by picking any variable, say $x_n$, and sending
\[
f(\xx) \longmapsto \left[ f(\xx) \right]_{\substack{x_n=1,\text{ and}\\x_m=0 \text{ for }m\neq n}}.
\]
\end{example}

This character  $\Qsym \overset{\zeta_Q}{\longrightarrow} \kk$
has a certain universal property, known as the
\dfn{Aguiar-Bergeron-Sottile universality theorem}
(part of \cite[Theorem 4.1]{AguiarBergeronSottile}):

\begin{theorem}
\label{Qsym-as-terminal-object-theorem}
Let $A$ be a connected graded Hopf algebra, and let
$A \overset{\zeta}{\longrightarrow} \kk$ be a character.
Then, there is a unique graded Hopf morphism
$A \overset{\Psi}{\longrightarrow} \Qsym$
making the following diagram commute:
\begin{equation}
\label{ABS-terminal-morphism-diagram}
\xymatrix{
A\ar[rr]^{\Psi} \ar[dr]_{\zeta} & & \Qsym \ar[dl]^{\zeta_Q}\\
  &\kk&
}
\end{equation}
Furthermore, $\Psi$ is given by the following formula on
homogeneous elements:
\begin{equation}
\label{ABS-terminal-morphism-formula}
\Psi(a)=\sum_{\alpha \in \Comp_n} \zeta_\alpha(a) M_\alpha
\qquad \text{ for all $n \in \NN$ and $a \in A_n$,}
\end{equation}
where for $\alpha=(\alpha_1,\ldots,\alpha_\ell)$, the map $\zeta_\alpha$ is the composite
\[
A_n \overset{\Delta^{(\ell-1)}}{\longrightarrow} A^{\otimes \ell}
    \overset{\pi_\alpha}{\longrightarrow} A_{\alpha_1} \otimes \cdots \otimes A_{\alpha_\ell}
     \overset{\zeta^{\otimes \ell}}{\longrightarrow} \kk
\]
in which
$
A^{\otimes \ell}
\overset{\pi_\alpha}{\longrightarrow}
A_{\alpha_1} \otimes \cdots \otimes A_{\alpha_\ell}
$
is the canonical projection.
\end{theorem}

% [DG] Replaced "is a projection" by "is the canonical projection" so that it
% doesn't sound like a hidden existence quantor.

% [DG][v14] Removed "finite type" condition and added a footnote about this
% in the proof. It turns out that not just the proposition but also its
% proof has always worked in this generality, provided one handles
% dualization correctly (Exercise \ref{exe.dual-algebra}(e)).
%
% I also removed "finite type" from some corollaries below.

\begin{proof}
One argues that $\Psi$ is unique, and has formula \eqref{ABS-terminal-morphism-formula},
using only that $\zeta$ is $\kk$-linear and sends $\one$ to $\one$ and that
$\Psi$ is a graded \emph{$\kk$-coalgebra} map making \eqref{ABS-terminal-morphism-diagram} commute.
Equivalently, consider the adjoint \emph{$\kk$-algebra} map\footnote{Here we
are using the fact that there is a 1-to-1 correspondence between
graded $\kk$-linear maps $A \to \Qsym$ and graded $\kk$-linear maps
$\Qsym^o \to A^o$ given by $f \mapsto f^*$, and this correspondence
has the property that a given graded map $f : A \to \Qsym$ is a $\kk$-coalgebra morphism if and only if
$f^*$ is a $\kk$-algebra morphism. This is a particular case of
Exercise~\ref{exe.dual-algebra}(f).}
\[
\Nsym = \Qsym^o \overset{\Psi^*}{\longrightarrow} A^o.
\]
Commutativity of \eqref{ABS-terminal-morphism-diagram} implies that for $a$ in $A_n$,
\[
(\Psi^*(H_n),a) = (H_n,\Psi(a)) \overset{\eqref{Qsym-zeta-character-example.zetaQf}}{=} \zeta_Q( \Psi(a) ) = \zeta(a),
\]
whereas gradedness of $\Psi^*$ yields that $(\Psi^*(H_m),a) = 0$ whenever $a \in A_n$ and
$m \neq n$.
In other words, $\Psi^*(H_n)$ is the element of $A^o$ defined as the following functional on $A$:
\begin{equation}
\label{image-of-Hn-as-functional}
\Psi^*(H_n)(a) =
\begin{cases}
\zeta(a), & \text{ if } a \in A_n;\\
0, & \text{ if }a \in A_m\text{ for some }m \neq n.
\end{cases}
\end{equation}
By the universal property for $\Nsym \cong \kk\langle H_1,H_2,\ldots \rangle$
as free associative $\kk$-algebra, we see that
any choice of a $\kk$-linear map $A \overset{\zeta}{\rightarrow} \kk$
uniquely produces a
$\kk$-algebra morphism $\Psi^* : \Qsym^o \to A^o$
which satisfies \eqref{image-of-Hn-as-functional} for all $n \geq 1$.
It is easy to see that this $\Psi^*$ then automatically satisfies
\eqref{image-of-Hn-as-functional} for $n = 0$ as well if $\zeta$ sends
$\one$ to $\one$ (it is here that
we use $\zeta(\one)=\one$ and the connectedness of $A$). Hence, any
given $\kk$-linear map $A \overset{\zeta}{\rightarrow} \kk$
sending $\one$ to $\one$ uniquely produces a
$\kk$-algebra morphism $\Psi^* : \Qsym^o \to A^o$
which satisfies \eqref{image-of-Hn-as-functional} for all $n \geq 0$.
Formula
\eqref{ABS-terminal-morphism-formula} follows as
\[
\Psi(a) = \sum_{\alpha \in \Comp} (H_\alpha, \Psi(a)) \,\, M_\alpha
\]
and for a composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$, one has
\begin{align*}
(H_\alpha, \Psi(a))
=(\Psi^*(H_\alpha),a)
&=\left( \Psi^*(H_{\alpha_1}) \cdots \Psi^*(H_{\alpha_\ell}), a \right)\\
%&=( \Psi^*(H_{\alpha_1}) \otimes \cdots \otimes \Psi^*(H_{\alpha_\ell}),( \pi_\alpha \circ \Delta^{(\ell-1)})(a))
&= \left( \Psi^*(H_{\alpha_1}) \otimes \cdots \otimes \Psi^*(H_{\alpha_\ell}), \Delta^{(\ell-1)}(a) \right) \\
&\overset{\eqref{image-of-Hn-as-functional}}{=} \left(\zeta^{\otimes \ell} \circ \pi_\alpha \right) \left(\Delta^{(\ell-1)}(a)\right)
=\zeta_{\alpha}(a),
\end{align*}
where the definition of $\zeta_\alpha$ was used in the last equality.

% [DG][v14] I sprayed some more detail into this proof.

We wish to show that if, in addition, $A$ is a Hopf algebra and $A \overset{\zeta}{\longrightarrow} \kk$ is a character (i.e., an algebra morphism), then  $A \overset{\Psi}{\longrightarrow} \Qsym$ will be an algebra morphism, that is, the two maps $A \otimes A \longrightarrow \Qsym$
given by $\Psi \circ m$ and $m \circ (\Psi \otimes \Psi)$ coincide.  To see this, consider
these two diagrams having the two maps in question as the composites of their top rows:
\begin{equation}
\xymatrix{
A \otimes A \ar[r]^-{m}\ar[dr]_{\zeta \otimes \zeta} & A\ar[r]^\Psi \ar[d]_{\zeta} &\Qsym \ar[dl]^{\zeta_Q}\\
  &\kk&
}
\qquad
\xymatrix@C=55pt{
A \otimes A \ar[r]^{\Psi \otimes \Psi} \ar[dr]_{\zeta \otimes \zeta}
  & \Qsym^{\otimes 2} \ar[r]^-{m} \ar[d]_{\zeta_Q \otimes \zeta_Q}
    &\Qsym \ar[dl]^{\zeta_Q}\\
  &\kk&
}
\end{equation}
The fact that $\zeta, \zeta_Q$ are algebra morphisms makes the above diagrams commute, so that
applying the uniqueness in the first part of the proof to the
character $A \otimes A \overset{\zeta \otimes \zeta}{\longrightarrow} \kk$ proves
the desired equality $\Psi \circ m=m \circ (\Psi \otimes \Psi)$.
\end{proof}

\begin{remark}
\label{Sym-as-terminal-object-remark}
When one assumes in addition that $A$ is cocommutative,
it follows that the image of $\Psi$ will lie
in the subalgebra $\Lambda \subset \Qsym$, e.g. from the explicit
formula \eqref{ABS-terminal-morphism-formula} and the fact
that one will have $\zeta_\alpha = \zeta_\beta$ whenever $\beta$
is a rearrangement of $\alpha$.  In other words, the character
$\Lambda \overset{\zeta_\Lambda}{\longrightarrow} \kk$ defined by
restricting $\zeta_Q$ to $\Lambda$, or by
\[
\zeta_\Lambda(m_\lambda)=
\begin{cases}
1, &\text{ if }\lambda=(n) \text{ for some }n;\\
0, &\text{ otherwise,}
\end{cases}
\]
has a universal property as terminal object
with respect to characters on cocommutative %co- or 
Hopf algebras.
\end{remark}

% [DG][v78] Removed "co- or" from the previous remark.
% Restated the theorem to uncrowd the first sentence,
% while adding the fiollowing sentence in order to
% fix the "induced" terminology:

The graded Hopf morphism $\Psi$ in
Theorem~\ref{Qsym-as-terminal-object-theorem}
will be called the \dfn{map $A \to \Qsym$ induced
by the character $\zeta$}.

We close this section by discussing a well-known polynomiality and reciprocity
phenomenon; see, e.g., Humpert and Martin \cite[Prop. 2.2]{HumpertMartin},
Stanley \cite[\S 4]{Stanley-chromatic-qsym-fn}.

\begin{definition}
The \dfn{binomial Hopf algebra} (over the commutative ring $\kk$)
is the polynomial algebra $\kk\left[  m\right]  $ in a single
variable $m$, with a Hopf algebra structure transported from the symmetric
algebra $\Sym\left(  \kk^1\right)  $ (which is a Hopf
algebra by virtue of Example~\ref{symmetric-algebra-as-bialgebra-example},
applied to $V=\kk^1$) along the
isomorphism $\Sym\left(  \kk^1\right)  \rightarrow
\kk\left[  m\right]  $ which sends the standard basis element of
$\kk^1$ to $m$. Thus the element $m$ is primitive; that is,
$\Delta m = \one \otimes m + m \otimes \one$ and $S(m) = -m$.
As $S$ is an algebra anti-endomorphism by
Proposition~\ref{antipodes-are-antiendomorphisms} and $\kk[m]$ is commutative,
one has $S(g)(m) = g(-m)$ for all polynomials $g(m)$ in $\kk[m]$.
\end{definition}

% [DG][v3] Not the best way to state it, I fear.


\begin{definition}
\label{principal-specialization-at-1-defn}
For an element $f(\xx)$ in $\Qsym$ and a nonnegative integer $m$,
let $\ps^1(f)(m)$ denote the element of $\kk$
obtained by \dfn{principal specialization at $q=1$}
\begin{align*}
\ps^1(f)(m)&=\left[ f(\xx)
\right]_{\substack{x_1=x_2=\cdots =x_m=1,\\x_{m+1}=x_{m+2}= \cdots =0}} \\
&=f(\underbrace{1,1,\ldots,1}_{m\text{ ones}},0,0,\ldots).
\end{align*}
\end{definition}



\begin{proposition}
\label{zeta-polynomiality-proposition}
Assume that $\QQ$ is a subring of $\kk$.
The map $\ps^1$ has the following properties.
\begin{enumerate}
\item[(i)]
Let $f \in \Qsym$.
There is a unique
polynomial in $\kk[m]$ which agrees for each nonnegative integer $m$
with $\ps^1(f)(m)$, and which, by abuse of notation, we will also denote
$\ps^1(f)(m)$.  If $f$ lies in $\Qsym_n$, then $\ps^1(f)(m)$ is a polynomial
of degree at most $n$, taking these values on $M_\alpha, L_\alpha$
for $\alpha=(\alpha_1,\ldots,\alpha_\ell)$ in $\Comp_n$:
\begin{align*}
\ps^1(M_\alpha)(m) &=\binom{m}{\ell}, \\
\ps^1(L_\alpha)(m) &=\binom{m-\ell+n}{n}.
\end{align*}
\item[(ii)] The map $\Qsym \overset{\ps^1}{\longrightarrow} \kk[m]$
is a Hopf morphism into the binomial Hopf algebra.
\item[(iii)] For all $m$ in $\ZZ$ and $f$ in $\Qsym$ one has
\[
\zeta_Q^{\star m}(f) = \ps^1(f)(m).
\]
In particular, one also has
\[
\zeta_Q^{\star (-m)}(f) = \ps^1(S(f))(m) =\ps^1(f)(-m).
\]
\item[(iv)]
For a graded Hopf algebra $A$ with a character
$A \overset{\zeta}{\longrightarrow} \kk$, and any element $a$ in $A_n$,
the polynomial $\ps^1(\Psi(a))(m)$ in $\kk[m]$ has degree at most $n$,
and when specialized to $m$ in $\ZZ$ satisfies
\[
\zeta^{\star m}(a) = \ps^1(\Psi(a))(m).
\]
\end{enumerate}
\end{proposition}

% [DG] I required characteristic $0$ to make sense of the polynomial
% $\binom{m}{\ell}$.

\begin{proof}
To prove assertion (i), note that one has
\begin{align*}
\ps^1(M_\alpha)(m)
=M_\alpha(1,1,\dots,1,0,0,\ldots)
&=\sum_{1 \leq i_1 < \cdots < i_\ell \leq m}
 \left[ x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell} \right]_{x_j=1}
= \binom{m}{\ell},\\
\ps^1(L_\alpha)(m)
=L_\alpha(1,1,\dots,1,0,0,\ldots)
&=\sum\limits_{\substack{1 \leq i_1 \leq \cdots \leq i_n \leq m : \\ i_k < i_{k+1} \text{ if }k \in D(\alpha)}}  \left[ x_{i_1} \cdots x_{i_n} \right]_{x_j=1}\\
&=|\{1 \leq j_1 \leq j_2 \leq \cdots \leq j_n \leq m-\ell+1\}|
=\binom{m-\ell+n}{n}.
\end{align*}
As $\{M_\alpha\}_{\alpha \in \Comp_n}$ form a basis for $\Qsym_n$, and
$\binom{m}{\ell}$ is a polynomial function in $m$ of degree $\ell (\leq n)$,
one concludes that for $f$ in $\Qsym_n$ one has that $\ps^1(f)(m)$ is a polynomial function in $m$ of
degree at most $n$. The polynomial giving rise to this function is unique,
since infinitely many of its values are fixed.

To prove assertion (ii), note that $\ps^1$ is an algebra morphism because it
is an evaluation homomorphism.
To check that it is a coalgebra morphism, it suffices to check
$\Delta \circ \ps^1 = (\ps^1 \otimes \ps^1) \circ \Delta$ on each $M_\alpha$ for
$\alpha=(\alpha_1,\ldots,\alpha_\ell)$ in $\Comp_n$. Using the  Vandermonde
summation $\binom{A+B}{\ell} = \sum_{k} \binom{A}{k} \binom{B}{\ell-k}$, one has
\[
(\Delta \circ \ps^1)(M_\alpha)
= \Delta \binom{m}{\ell}
= \binom{ m \otimes \one + \one \otimes m }{\ell}
=\sum_{k=0}^\ell \binom{ m \otimes \one }{k}  \binom{ \one \otimes m }{\ell-k}
=\sum_{k=0}^\ell \binom{ m }{k} \otimes \binom{  m }{\ell-k}
\]
while at the same time
\[
\left((\ps^1 \otimes \ps^1) \circ \Delta\right)(M_\alpha)
= \sum_{k=0}^{\ell} \ps^1(M_{(\alpha_1,\ldots,\alpha_k)})
                     \otimes \ps^1(M_{(\alpha_{k+1},\ldots,\alpha_\ell)})
=\sum_{k=0}^{\ell} \binom{m}{k} \otimes \binom{m}{\ell-k}.
\]
Thus $\ps^1$ is a bialgebra morphism, and hence also a Hopf morphism, by
Corollary~\ref{cor.bialg-mor-is-Hopf}.

For assertion (iii), first assume $m$ lies in $\{0,1,2,\ldots\}$.
Since $\zeta_Q(f)=f(1,0,0,\ldots)$, one has
\begin{align*}
\zeta_Q^{\star m}(f)
&= \zeta_Q ^{\otimes m}\circ \Delta^{(m-1)} f(\xx)
=  \zeta_Q ^{\otimes m} \left( f(\xx^{(1)},\xx^{(2)},\ldots,\xx^{(m)}) \right) \\
&=  \left[ f(\xx^{(1)},\xx^{(2)},\ldots,\xx^{(m)})
         \right]_{\substack{x^{(1)}_1=x^{(2)}_1=\cdots=x^{(m)}_1=1,
                    \\x^{(j)}_2=x^{(j)}_3=\cdots=0 \text{ for all }j}}\\
&=f(1,0,0,\ldots,1,0,0,\ldots,\cdots,1,0,0,\ldots)
=f(\underbrace{1,1,\ldots,1}_{m\text{ ones}},0,0,\ldots) = \ps^1(f)(m).
\end{align*}
\footnote{See Exercise~\ref{exe.zeta-polynomiality.altpf} for an alternative
way to prove this, requiring less thought to verify its soundness.}
But then Proposition~\ref{antipodes-give-convolution-inverses}(a) also implies
\begin{align*}
\zeta_Q^{\star (-m)}(f)
&=\left( \zeta_Q^{\star (-1)} \right)^{\star m}(f)
= \left( \zeta_Q \circ S \right)^{\star m}(f)
= \zeta_Q^{\star m}(S(f))\\
&= \ps^1(S(f))(m)
= S(\ps^1(f))(m)
=\ps^1(f)(-m).
\end{align*}
For assertion (iv), note that
\[
\zeta^{\star m}(a)
= (\zeta_Q \circ \Psi)^{\star m}(a)
= (\zeta_Q^{\star m})(\Psi(a))
= \ps^1(\Psi(a))(m),
\]
where the three equalities come from
\eqref{ABS-terminal-morphism-diagram},
Proposition~\ref{antipodes-give-convolution-inverses}(a),
and assertion (iii) above, respectively.
\end{proof}

% [DG][v17] I added the alternative proof in the footnote to avoid
% formalizing the argument you gave (which is rather tricky imho).
% Maybe better as an exercise?

% [DG][v19] Now it's an exercise.


\begin{remark}
Aguiar, Bergeron and Sottile give
a very cute (third) proof
of the $\Qsym$ antipode formula
Theorem~\ref{Qsym-antipode-on-monomials},
via Theorem~\ref{Qsym-as-terminal-object-theorem},
in \cite[Example 4.8]{AguiarBergeronSottile}.
They apply Theorem~\ref{Qsym-as-terminal-object-theorem}
to the \emph{coopposite coalgebra} $\Qsym^{cop}$ and its
character $\zeta_Q^{\star (-1)}$.  One can show that the map
$\Qsym^{cop} \overset{\Psi}{\rightarrow} \Qsym$ induced by $\zeta_Q^{\star (-1)}$
is $\Psi=S$, the antipode of $\Qsym$, because
$S : \Qsym \to \Qsym$ is
a coalgebra anti-endomorphism (by Exercise~\ref{exe.coalg-anti})
satisfying
$\zeta_Q^{\star (-1)} = \zeta_Q \circ S$.  They then use
the formula \eqref{ABS-terminal-morphism-formula}
for $\Psi=S$ (together with the polynomiality
Proposition~\ref{zeta-polynomiality-proposition})
to derive Theorem~\ref{Qsym-antipode-on-monomials}.
\end{remark}

% [DG][v14] Replaced
% $\Qsym \overset{\Psi}{\rightarrow} \Qsym^{cop}$
% by
% $\Qsym^{cop} \overset{\Psi}{\rightarrow} \Qsym$.
% Replaced ^{\star ^{-1}} by ^{\star (-1)}.

\begin{exercise}
\label{exe.zeta-polynomiality.altpf}
Show that
$\zeta_Q^{\star m}(f) = \ps^1(f)(m)$ for all $f \in \Qsym$ and
$m \in \left\{0,1,2,\ldots\right\}$. (This was already
proven in Proposition~\ref{zeta-polynomiality-proposition}(iii);
give an alternative proof using
Proposition~\ref{Qsym-coproduct-on-monomials}.)
\end{exercise}





\subsection{Example: Ehrenborg's quasisymmetric function of a ranked poset}

Here we consider incidence algebras, coalgebras and Hopf algebras generally,
and then particularize to the case of graded posets, to recover Ehrenborg's interesting quasisymmetric function invariant
via Theorem~\ref{Qsym-as-terminal-object-theorem}.





\subsubsection{Incidence algebras, coalgebras, Hopf algebras}


\begin{definition}
Given a family $\PPP$ of finite partially ordered sets $P$,
let $\kk[\PPP]$ denote the free $\kk$-module
whose basis consists of symbols $[P]$ corresponding
to isomorphism classes of posets $P$ in $\PPP$.

We will assume throughout that each $P$ in $\PPP$ is
\emph{bounded}\index{bounded poset},
that is, it has a unique minimal element $\hat{0}:=\hat{0}_P$ and a unique
maximal element $\hat{1}:=\hat{1}_P$.  In particular, $P \neq \varnothing$,
although it is allowed that $|P|=1$, so that $\hat{0}=\hat{1}$;
denote this isomorphism class of posets with one element by $[o]$.

If $\PPP$ is closed under taking intervals
\[
[x,y]:=[x,y]_P:=\{ z \in P: x \leq_P z \leq_P y \} ,
\]
then one can easily see that the following coproduct and counit
endow $\kk[\PPP]$ with the structure of a coalgebra,
called the
\emph{(reduced) incidence coalgebra}\index{reduced incidence coalgebra}:
\begin{align*}
\Delta [P]&:= \sum_{x \in P} [\hat{0},x] \otimes [x,\hat{1}],\\
\epsilon [P]&:=
\begin{cases}
1, & \text{ if }|P|=1 ; \\
0, & \text{ otherwise.}
\end{cases}
\end{align*}
The dual algebra $\kk[\PPP]^\ast$ is generally called the
\emph{reduced incidence algebra (modulo isomorphism)}
for the family $\PPP$ (see, e.g., \cite{Schmitt-antipodes}).
It contains the important element
$\kk[\PPP] \overset{\zeta}{\longrightarrow} \kk$,
called the
\emph{$\zeta$-function}\index{$\zeta$-function in an incidence algebra}
that takes the value $\zeta[P]=1$ for all $P$.

% [DG][v17] Wait: haven't you defined the graded dual only for graded
% Hopf algebras? This one is only filtered. Should I add a definition
% of filtered Hopf algebras? Or do you want the full dual? Or
% Sweedler's zero-dual?

% [DG][v19] Correction made.

If $\PPP$ (is not empty and) satisfies the further property of being
\emph{hereditary}\index{hereditary class of posets}
in the sense that for every $P_1,P_2$ in $\PPP$, the
\dfn{Cartesian product poset} $P_1 \times P_2$ with componentwise
partial order is also in $\PPP$, then one can check that
the following product and unit endow $\kk[\PPP]$ with
the structure of a (commutative) algebra:
\begin{align*}
[P_1] \cdot [P_2]&:=m([P_1] \otimes [P_2]) := [P_1 \times P_2],\\
\one_{\kk[\PPP]}&:=[o].
\end{align*}
\end{definition}

\begin{proposition}
\label{incidence-coalgebras-are-Hopf}
For any hereditary family $\PPP$ of finite posets,
$\kk[\PPP]$ is a bialgebra, and even a Hopf algebra
with antipode $S$ given as in \eqref{Takeuchi-formula} (Takeuchi's formula):
\[
S[P]=\sum_{k \geq 0} (-1)^k
     \sum_{\hat{0}=x_0 < \cdots <x_k=\hat{1}}
        [x_0,x_1] \cdots [x_{k-1},x_k].
\]
\end{proposition}
\begin{proof}
Checking the commutativity of the pentagonal diagram in
\eqref{bialgebra-diagrams}
amounts to the fact that, for any
$(x_1,x_2)  <_{P_1 \times P_2} (y_1,y_2)$,
one has a poset isomorphism
\[
\left[ (x_1,x_2) \,\, , \,\, (y_1,y_2) \right]_{P_1 \times P_2}
\cong [x_1,y_1]_{P_1} \times [x_2,y_2]_{P_2}.
\]
Commutativity of the remaining diagrams in
\eqref{bialgebra-diagrams} is straightforward, and so $\kk[\PPP]$
is a bialgebra.  But then Remark~\ref{non-graded-Takeuchi-remark}
implies that it is a Hopf algebra, with antipode $S$ as in \eqref{Takeuchi-formula},
because the map $f:=\id_{\kk[\PPP]} - u\epsilon$ (sending the class $[o]$ to $0$,
and fixing all other $[P]$) is locally $\star$-nilpotent:
\[
f^{\star k}[P]
=\sum_{\hat{0}=x_0 < \cdots <x_{k}=\hat{1}}
[x_0,x_1] \cdots [x_{k-1},x_{k}]
\]
will vanish due to an empty sum whenever $k$ exceeds the maximum length of
a chain in the finite poset $P$.
\end{proof}


It is perhaps worth remarking how this generalizes the M\"obius function
formula of P. Hall.  Note that the zeta function
$\kk[\PPP] \overset{\zeta}{\longrightarrow} \kk$ is a \emph{character},
that is, an algebra morphism.
Proposition~\ref{antipodes-give-convolution-inverses}(a)
then tells us that $\zeta$ should have a convolutional
inverse $\kk[\PPP] \overset{\mu=\zeta^{\star -1}}{\longrightarrow} \kk$,
traditionally called the
\emph{M\"obius function}\index{M\"obius function of a poset}, with
the formula
$\mu = \zeta^{\star -1} = \zeta \circ S$.  Rewriting this via
the antipode formula for $S$ given
in Proposition~\ref{incidence-coalgebras-are-Hopf} yields
P. Hall's formula.

\begin{corollary}
\label{P.Hall-formula}
For a finite bounded poset $P$, one has
\[
\mu[P] =
\sum_{k \geq 0} (-1)^k
|\{ \text{chains }\hat{0}=x_0 < \cdots <x_k=\hat{1}\text{ in }P\}|.
\]
\end{corollary}

We can also notice that $S$ is an algebra anti-endomorphism
(by Proposition~\ref{antipodes-are-antiendomorphisms}), thus an
algebra endomorphism (since $\kk[\PPP]$ is commutative, so
Exercise~\ref{exe.comm-cocomm.anti}(a) shows that the
algebra anti-endomorphisms of $\kk[\PPP]$ are
the same as the algebra endomorphisms of $\kk[\PPP]$). Hence,
$\mu = \zeta \circ S$ is a composition of two algebra
homomorphisms, thus an algebra homomorphism itself. We
therefore obtain the following classical fact:

\begin{corollary}
For two finite bounded posets $P$ and $Q$, we have
$\mu[P \times Q] = \mu[P] \cdot \mu[Q]$.
\end{corollary}


\subsubsection{The incidence Hopf algebras for ranked posets and Ehrenborg's function}

\begin{definition}
Take $\PPP$ to be the class of bounded
\emph{ranked}\index{ranked poset} finite
posets $P$, that is, those for which all maximal chains from
$\hat{0}$ to $\hat{1}$ have the same length $r(P)$.  This is a hereditary
class, as it implies that any interval is $[x,y]_P$ is also ranked,
and the product of two bounded ranked posets is also bounded and ranked.
It also uniquely defines a
\emph{rank function}\index{rank function of a poset}
$P \overset{r}{\longrightarrow} \NN$
in which $r(\hat{0})=0$ and $r(x)$ is the length of any maximal chain
from $\hat{0}$ to $x$.
\end{definition}

\begin{example}
\label{polytope-example}
Consider a pyramid with apex vertex $a$ over a square base with vertices $b,c,d,e$:
\[
\xymatrix{
 & &a\ar@{-}[dll]\ar@{-}[ddl]\ar@{-}[drr]\ar@{-}[ddrrr]&&& & \\
b\ar@{--}[rrrr]\ar@{-}[dr]& & &&e\ar@{-}[dr]& \\
 &c\ar@{-}[rrrr]& & &&d
}
\]
Ordering its faces by inclusion gives a bounded ranked poset $P$,
where the rank of an element is one more than the dimension
of the face it represents:
\[
\xymatrix{
 & & & & & & & & \text{rank:}\\
   &   &   &abcd\ar@{-}[drrrr]\ar@{-}[d]\ar@{-}[dl]\ar@{-}[dll]\ar@{-}[dlll]&  &  &  & & 4\\
abc\ar@{-}[d]\ar@{-}[dr]&acd\ar@{-}[d]\ar@{-}[dr]&ade\ar@{-}[d]\ar@{-}[dr]&abe\ar@{-}[d]\ar@{-}[dlll] &  &  &  &bcde\ar@{-}[d]\ar@{-}[dl]\ar@{-}[dll]\ar@{-}[dlll]& 3\\
ab &ac &ad &ae  &be\ar@{-}[ul]&bc\ar@{-}[ulllll]&cd\ar@{-}[ulllll]&de\ar@{-}[ulllll]& 2\\
a\ar@{-}[u]\ar@{-}[ur]\ar@{-}[urr]\ar@{-}[urrr]  &   &   &    &b\ar@{-}[u]\ar@{-}[ur]\ar@{-}[ullll] &c\ar@{-}[u]\ar@{-}[ur]\ar@{-}[ullll] &d\ar@{-}[u]\ar@{-}[ur]\ar@{-}[ullll] &e\ar@{-}[u]\ar@{-}[ulll]\ar@{-}[ullll] & 1\\
   &   &   &    &\varnothing\ar@{-}[ullll]\ar@{-}[u]\ar@{-}[ur]\ar@{-}[urr]\ar@{-}[urrr]& & & & 0\\
}
\]
\end{example}

\begin{definition}
\dfn{Ehrenborg's quasisymmetric function} $\Psi[P]$ for a bounded ranked poset $P$
is the image of $[P]$ under the map
$\kk[\PPP] \overset{\Psi}{\longrightarrow} \Qsym$
induced by the zeta function
$\kk[\PPP] \overset{\zeta}{\longrightarrow} \kk$ as
a character, via Theorem~\ref{Qsym-as-terminal-object-theorem}.
\end{definition}

The quasisymmetric function $\Psi[P]$ captures several interesting
combinatorial invariants of $P$; see Stanley \cite[Chap. 3]{Stanley}
for more background on these notions.

\begin{definition}
Let $P$ be a bounded ranked poset $P$ of rank $r(P):=r(\hat{1})$.
Define its \emph{rank-generating function}\index{rank-generating function of a poset}\index{$RGF(P,q)$}
\[
RGF(P,q):=\sum_{p \in P} q^{r(p)} \in \ZZ\left[q\right],
\]
its \emph{characteristic polynomial}\index{characteristic polynomial of a poset}\index{$\chi(P,q)$}
\[
\chi(P,q):=\sum_{p \in P} \mu(\hat{0},p)q^{r(p)} \in \ZZ\left[q\right]
\]
(where $\mu(u,v)$ is shorthand for $\mu([u,v])$),
and its \emph{zeta polynomial}\index{zeta polynomial of a poset}\index{$Z(P,m)$}
\begin{align}
\label{zeta-polynomial-def}
Z(P,m)&=|\{\text{multichains }
\hat{0} \leq_P  p_1 \leq_P \cdots \leq_P p_{m-1}  \leq_P \hat{1} \} |\\
 &= \sum_{s=0}^{r(P)-1} \binom{m}{s+1}
          |\{\text{chains }\hat{0} < p_1 < \cdots < p_s < \hat{1}\}| \in \QQ\left[m\right]
\label{zeta-polynomial-as-chain-sum}
\end{align}
\footnote{Actually, \eqref{zeta-polynomial-as-chain-sum} is false
if $\left|P\right| = 1$ (but only then). We use
\eqref{zeta-polynomial-def} to define $Z(P,m)$ in this case.}.
Also, for each subset $S \subset \{1,2,\ldots,r(P)-1\}$,
define
the \emph{flag number}\index{flag number of a poset}\index{$f_S$}
$f_S$ of $P$ by
\[
f_S=|\{\text{chains } \hat{0} <_P p_1 <_P \cdots <_P p_s  <_P \hat{1}
\text{ with }\{r(p_1),\ldots,r(p_s)\}=S\}|.
\]
These flag numbers are the components of the
\emph{flag $f$-vector}\index{flag $f$-vector of a poset}
$(f_S)_{S \subset [r-1]}$ of $P$.
Further define the \emph{flag $h$-vector}\index{flag $h$-vector of a poset}
$(h_T)_{T \subset [r-1]}$ of $P$,
whose entries $h_T$ are given by
$f_S=\sum_{ T \subset S} h_T$, or, equivalently%
\footnote{The equivalence follows from inclusion-exclusion
(more specifically, from
the converse of Lemma~\ref{lem.moebius-bool.subsets}(a)).},
by
$h_S=\sum_{ T \subset S} (-1)^{|S \setminus T|} f_T$.
\end{definition}

% [DG][v13] Replaced "r" by "r(P)" in
% "and for a subset  $S \subset \{1,2,\ldots,r-1\}$,".

% [DG][v19] Added footnote about border case when $P$ has
% one element only.

% [DG][v80] Replaced run-on sentence by several shorter ones.


\begin{example}
For the poset $P$ in Example~\ref{polytope-example}, one
has $RGF(P,q)=1+5q+8q^2+5q^3+q^4$.  Since $P$ is the poset of
faces of a polytope, the M\"obius function values for its
intervals are easily predicted:  $\mu(x,y)=(-1)^{r[x,y]}$,
that is, $P$ is an \dfn{Eulerian ranked poset}; see Stanley
\cite[\S 3.16]{Stanley}.  Hence
its characteristic polynomial is trivially related to the
rank generating function, sending $q \mapsto -q$, that is,
\[
\chi(P,q)= RGF(P,-q)=1-5q+8q^2-5q^3+q^4.
\]
Its flag $f$-vector and $h$-vector entries are given in the following
table.
\vskip.1in
\begin{tabular}{|c|c|cl|}\hline
$S$           & $f_S$ &       &$h_S$  \\ \hline\hline
$\varnothing$ & $1$   &       &$1$ \\ \hline
$\{1\}$       & $5$   & $5-1=$&$4$ \\ \hline
$\{2\}$       & $8$   & $8-1=$&$7$ \\ \hline
$\{3\}$       & $5$   & $5-1=$&$4$ \\ \hline
$\{1,2\}$     & $16$  & $16-(5+8)+1=$&$4$ \\ \hline
$\{1,3\}$     & $16$  & $16-(5+5)+1=$&$7$ \\ \hline
$\{2,3\}$     & $16$  & $16-(5+8)+1=$&$4$ \\ \hline
$\{1,2,3\}$   & $32$  & $32-(16+16+16)+(5+8+5)-1=$&$1$ \\ \hline
\end{tabular}
\vskip.1in
\noindent
and using \eqref{zeta-polynomial-as-chain-sum}, its zeta polynomial is
\[
Z(P,m)=1 \binom{m}{1} + (5+8+5) \binom{m}{2} + (16+16+16) \binom{m}{3} +
32 \binom{m}{4}=\frac{m^2(2m-1)(2m+1)}{3}.
\]
\end{example}

\begin{theorem}
\label{Ehrenborg's-theorem}
Assume that $\QQ$ is a subring of $\kk$.
Ehrenborg's quasisymmetric function $\Psi[P]$ for a bounded ranked poset $P$
encodes
\begin{enumerate}
\item[(i)]
the flag $f$-vector entries $f_S$ and flag $h$-vector entries $h_S$
as its $M_\alpha$ and $L_\alpha$ expansion coefficients\footnote{In fact,
Ehrenborg \emph{defined} $\Psi[P]$  in \cite[Defn. 4.1]{Ehrenborg}
via this $M_\alpha$ expansion, and then showed that it gave a Hopf morphism.} :
\[
\Psi[P] = \sum_\alpha f_{D(\alpha)}(P) \,\, M_\alpha
= \sum_\alpha h_{D(\alpha)}(P) \,\, L_\alpha ,
\]
\item[(ii)]
the zeta polynomial as the specialization from
Definition~\ref{principal-specialization-at-1-defn}
\[
Z(P,m)=\ps^1(\Psi[P])(m)
=\left[ \Psi[P] \right]_{\substack{x_1=x_2=\cdots =x_m=1,\\ x_{m+1}=x_{m+2}= \cdots =0}} ,
\]
\item[(iii)]
the rank-generating function as the specialization
\[
RGF(P,q)
=\left[ \Psi[P] \right]_{\substack{x_1=q,x_2=1,\\x_3=x_4= \cdots =0}} ,
\]
\item[(iv)]
the characteristic polynomial as the convolution
\[
\chi(P,q)= ((\psi_q \circ S) \star \zeta_Q) \circ \Psi[P] ,
\]
where $\Qsym \overset{\psi_q}{\longrightarrow} \kk[q]$
maps $f(\xx) \longmapsto f(q,0,0,\ldots)$.
\end{enumerate}
\end{theorem}
\begin{proof}
In assertion (i), the
expansion $\Psi[P]=\sum_\alpha f_{D(\alpha)}(P) \, M_\alpha$ is
\eqref{ABS-terminal-morphism-formula}, since
$\zeta_\alpha[P]=f_{D(\alpha)}(P)$.  The $L_\alpha$ expansion
follows from this, as
$L_\alpha=\sum_{\beta: D(\beta) \supset D(\alpha)} M_\beta$
and $f_S(P)=\sum_{T \subset S} h_T$.


Assertion (ii) is immediate from
Proposition~\ref{zeta-polynomiality-proposition}(iv),
since $Z(P,m)=\zeta^{\star m}[P]$.

Assertion (iii) can be deduced from assertion (i),
but it is perhaps more fun and in the spirit of things to proceed
as follows. Note that
$\psi_q(M_\alpha)=q^{n}$ for $\alpha=(n)$, and $\psi_q(M_\alpha)$ vanishes for
all other $\alpha \neq (n)$ in $\Comp_n$.  Hence
for a bounded ranked poset $P$ one has
\begin{equation}
\label{q-rank-functional-equation}
(\psi_q \circ \Psi)[P]= q^{r(P)}.
\end{equation}
But if we treat $\zeta_Q : \Qsym \to \kk$ as a map
$\Qsym \to \kk\left[q\right]$, then
\eqref{pre-and-post-composition-in-convolution}
(applied to $\kk\left[\PPP\right]$, $\Qsym$, $\kk\left[q\right]$,
$\kk\left[q\right]$, $\Psi$, $\id_{\kk\left[q\right]}$,
$\psi_q$ and $\zeta_Q$
instead of $C$, $C'$, $A$, $A'$, $\gamma$, $\alpha$,
$f$ and $g$) shows that
\begin{equation}
\left( \psi_q \star \zeta_Q \right) \circ \Psi
= \left( \psi_q \circ \Psi \right) \star \left( \zeta_Q \circ \Psi \right) ,
\label{pf.Ehrenborg's-theorem.compose-star}
\end{equation}
since $\Psi : \kk\left[\PPP\right] \to \Qsym$ is a $\kk$-coalgebra
homomorphism.
Consequently, one can compute
\begin{align*}
RGF(P,q) & = \sum_{p \in P} q^{r(p)} \cdot 1 = \sum_{p \in P} q^{r([\hat{0},p])} \cdot \zeta[p,\hat{1}]
\overset{\substack{\eqref{q-rank-functional-equation}, \\ \eqref{ABS-terminal-morphism-diagram}}}{=}
\sum_{p \in P} (\psi_q \circ \Psi)[\hat{0},p] \cdot
               (\zeta_Q \circ \Psi)[p,\hat{1}] \\
&= \left( \left( \psi_q \circ \Psi \right) \star \left( \zeta_Q \circ \Psi \right) \right) [P]
\overset{\eqref{pf.Ehrenborg's-theorem.compose-star}}{=}(\psi_q \star \zeta_Q)(\Psi[P])
=(\psi_q \otimes \zeta_Q)\left( \Delta \Psi[P]\right) \\
&=\left[ \Psi[P](\xx,\yy) \right]_{\substack{x_1=q,x_2=x_3=\cdots=0\\y_1=1,y_2=y_3=\cdots=0}}
=\left[ \Psi[P](\xx) \right]_{\substack{x_1=q,x_2=1,\\x_3=x_4=\cdots=0}} .
\end{align*}

%Assertion (ii) follows from observing the evaluation for $\alpha$ in $\Comp_n$
%\[
%\left[ M_\alpha\right]_{\substack{x_1=q,x_2=1,\\x_3=x_4= \cdots =0}}
%=\begin{cases}
%1+q^n & \text{ if }\alpha=(n),\\
%q^k & \text{ if } \alpha=(k,n-k), 0 < k < n\\
%0 & \text{ if }\ell(\alpha) \geq 3.
%\end{cases}
%\]
%This implies that
%\[
%\left[ \Psi[P] \right]_{\substack{x_1=q,x_2=1,\\x_3=x_4= \cdots =0}}
%= 1+q^{r(P)} + \sum_{r=1}^{r(P)-1} f_{(k,n-k)}(P) q^k
%=RGF(P,q).
%\]

Similarly, for assertion (iv) first note  that
\begin{equation}
\left( \left( \psi_q \circ S \right) \star \zeta_Q \right) \circ \Psi
= \left( \psi_q \circ S \circ \Psi \right) \star \left( \zeta_Q \circ \Psi \right) ,
\label{pf.Ehrenborg's-theorem.compose-star2}
\end{equation}
(this is proven similarly to
\eqref{pf.Ehrenborg's-theorem.compose-star}, but now
using the map $\psi_q \circ S$ instead of $\psi_q$).
Now,
Proposition~\ref{incidence-coalgebras-are-Hopf} and
Corollary~\ref{P.Hall-formula} let one calculate that
\begin{align*}
(\psi_q \circ \Psi \circ S)[P]
&= \sum_k (-1)^k \sum_{\hat{0}=x_0 < \cdots < x_k=\hat{1}}
            (\psi_q \circ \Psi)([x_0,x_1]) \cdots (\psi_q \circ \Psi)([x_{k-1},x_k]) \\
&
\overset{\eqref{q-rank-functional-equation}}{=} \sum_k (-1)^k \sum_{\hat{0}=x_0 < \cdots < x_k=\hat{1}} q^{r(P)}
= \mu(\hat{0},\hat{1}) q^{r(P)}.
\end{align*}
This is used in the penultimate equality here:
\begin{align*}
((\psi_q \circ S) \star \zeta_Q) \circ \Psi[P]
%&= m \circ ((\psi_q \circ S) \otimes \zeta_Q) \circ \Delta \circ \Psi[P] \\
%&= m \circ ((\psi_q \circ S) \otimes \zeta_Q) \circ (\Psi \otimes \Psi) \circ\Delta [P] \\
&\overset{\eqref{pf.Ehrenborg's-theorem.compose-star2}}{=}
  ((\psi_q \circ S \circ \Psi) \star (\zeta_Q \circ \Psi)) [P]
= ((\psi_q \circ \Psi \circ S) \star \zeta) [P] \\
&= \sum_{p \in P}  (\psi_q \circ \Psi \circ S)[\hat{0},p] \cdot \zeta[p,\hat{1}]
=\sum_{p \in P} \mu[\hat{0},p] q^{r(p)} = \chi(P,q).
\end{align*}
\end{proof}









\subsection{Example: Stanley's chromatic symmetric function of a graph}

We introduce the \emph{chromatic Hopf algebra of graphs} and an associated
character $\zeta$ so that the map $\Psi$ from
Theorem~\ref{Qsym-as-terminal-object-theorem}
sends a graph $G$ to Stanley's \emph{chromatic symmetric function}
of $G$.  Then principal specialization $\ps^1$ sends this
to the \emph{chromatic polynomial} of the graph.

\subsubsection{The chromatic Hopf algebra of graphs}


\begin{definition}
The \dfn{chromatic Hopf algebra} (see Schmitt \cite[\S 3.2]{Schmitt-structures})
\dfn{$\GGG$} is a free $\kk$-module whose
$\kk$-basis elements $[G]$ are indexed by isomorphism classes of
(finite) simple graphs $G=(V,E)$.
Define for $G_1=(V_1,E_1), G_2=(V_2,E_2)$ the multiplication
\[
[G_1] \cdot [G_2]:=[G_1 \sqcup G_2]
\]
where $[G_1 \sqcup G_2]$ denote the isomorphism class of the
disjoint union, on vertex set $V=V_1 \sqcup V_2$ which
is a disjoint union of copies of their vertex sets $V_1,V_2$, with
edge set $E=E_1 \sqcup E_2$.  For example,
\[
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right] \cdot
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]& &\bullet\ar@{-}[u]
}
\right]
\]
  Thus the class $[\varnothing]$
of the empty graph $\varnothing$ having $V=\varnothing, E=\varnothing$
is a unit element.

Given a graph $G = \left(V, E\right)$ and a subset $V' \subset V$,
the \dfn{subgraph induced on vertex set $V'$} is defined as
the graph $G|_{V'}:=(V',E')$ with
edge set $E'=\{e \in E: e= \left\{v_1,v_2\right\} \subset V'\}$.
This lets one define a comultiplication
$\Delta : \GGG \to \GGG \otimes \GGG$ by setting
\[
\Delta[G]:=\sum_{(V_1,V_2): V_1 \sqcup V_2=V} [G|_{V_1}] \otimes [G|_{V_2}].
\]
Define a counit $\epsilon : \GGG \to \kk$ by
\[
\epsilon [G]:=
\begin{cases}
1, & \text{ if }G=\varnothing \text{;} \\
0, & \text{ otherwise.}
\end{cases}
\]
\end{definition}

\begin{proposition}
\label{prop.GGG.hopf}
The above maps endow $\GGG$ with the structure of a connected graded
finite type Hopf algebra over $\kk$, which is both commutative and cocommutative.
\end{proposition}

\begin{example}
Here are some examples of these structure maps:
\begin{align*}
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right] \cdot
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
&=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]& &\bullet\ar@{-}[u]
}
\right] ;\\
%%%%%%%%%%%%%%
\Delta \left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
&= \one
\otimes
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
+
2
\left[
\xymatrix@C=4pt@R=4pt{\bullet}
\right]
\otimes
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
+
2
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\\
\bullet\ar@{-}[u]
}
\right]
\otimes
\left[
\xymatrix@C=4pt@R=4pt{\bullet}
\right]
+
\left[
\xymatrix@C=4pt@R=4pt{\bullet & \bullet}
\right]
\otimes
\left[
\xymatrix@C=4pt@R=4pt{\bullet}
\right] \\
& \qquad
+
\left[
\xymatrix@C=4pt@R=4pt{\bullet}
\right]
\otimes
\left[
\xymatrix@C=4pt@R=4pt{\bullet & \bullet}
\right]
+
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
\otimes \one
\end{align*}
\end{example}

\begin{proof}[Proof of Proposition~\ref{prop.GGG.hopf}.]
The associativity of the multiplication and comultiplication
should be clear as
\begin{align*}
m^{(2)}([G_1] \otimes [G_2] \otimes [G_3])
&=[G_1 \sqcup G_2 \sqcup G_3] ,\\
\Delta^{(2)}[G]
&=\sum\limits_{\substack{(V_1,V_2,V_3):\\V=V_1 \sqcup V_2 \sqcup V_3}}
       [G|_{V_1}] \otimes [G|_{V_2}] \otimes [G|_{V_3}].
\end{align*}
Checking the unit and counit conditions are straightforward.
Commutativity of the pentagonal bialgebra diagram
in \eqref{bialgebra-diagrams} comes down to check that,
given graphs $G_1,G_2$ on disjoint vertex
sets $V_1,V_2$ , when one applies to $[G_1] \otimes [G_2]$
either the composite $\Delta \circ m$ or the composite
$(m \otimes m) \circ (\id \otimes T \otimes \id) \circ (\Delta \otimes \Delta)$,
the result is the same:
\[
\sum\limits_{\substack{(V_{11},V_{12},V_{21},V_{22}):\\
        V_1=V_{11} \sqcup V_{12}\\
        V_2=V_{21} \sqcup V_{22}}}
[G_1|_{V_{11}} \sqcup G_2|_{V_{21}}] \otimes
[G_1|_{V_{12}} \sqcup G_2|_{V_{22}}].
\]
Letting $\GGG_n$ be the $\kk$-span of $[G]$ having $n$ vertices makes $\GGG$
a bialgebra which is graded and connected, and hence also a
Hopf algebra by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
Cocommutativity should be clear, and commutativity follows from the
graph isomorphism $G_1 \sqcup G_2 \cong G_2 \sqcup G_1$.
Finally, $\GGG$ is of finite type since there are only finitely
many isomorphism classes of simple graphs on $n$ vertices for
every given $n$.
\end{proof}

\begin{remark}
\label{Humpert-Martin-antipode-remark}
Humpert and Martin \cite[Theorem 3.1]{HumpertMartin}
gave the following expansion for the antipode in the
chromatic Hopf algebra, containing fewer terms than Takeuchi's general
formula \eqref{Takeuchi-formula}:  given a graph $G=(V,E)$, one has
\begin{equation}
\label{eq.Humpert-Martin-antipode-remark.S(G)}
S[G]=\sum_{F} (-1)^{|V|-\rank(F)} \acyc(G/F) [G_{V,F}].
\end{equation}
Here $F$ runs over all subsets of edges that form
\emph{flats}\index{flat in a graphic matroid}
in the graphic matroid for $G$, meaning that if $e=\{v,v'\}$ is an
edge in $E$ for which one has a path of edges in $F$ connecting $v$ to $v'$,
then $e$ also lies in $F$.  Here $G/F$ denotes the quotient graph in which all of the
edges of $F$ have been \emph{contracted}, while $\acyc(G/F)$ denotes its number
of \emph{acyclic orientations}, and $G_{V,F}:=(V,F)$ as a simple
graph.\footnote{The notation $\rank(F)$ denotes the
\emph{rank}\index{rank in a graphic matroid}
of $F$ in the graphic matroid of $G$. We can define it without
reference to matroid theory as the maximum cardinality of a
subset $F'$ of $F$ such that the graph $G_{V,F'}$ is acyclic.
Equivalently, $\rank(F)$ is $|V|-c(F)$, where $c(F)$ denotes the
number of connected components of the graph $G_{V,F}$. Thus,
the equality \eqref{eq.Humpert-Martin-antipode-remark.S(G)}
can be rewritten as
$S[G]=\sum_{F} (-1)^{c(F)} \acyc(G/F) [G_{V,F}]$.
In this form, this equality is also proven in
\cite[Thm. 7.1]{BenedettiSagan}.}
\end{remark}

% [DG][v44] Added footnote about the meaning of $\rank(F)$, and also
% referenced \cite{BenedettiSagan} for another proof.

\begin{remark}
In \cite{BenedettiHallamMachacek}, Benedetti, Hallam and Machacek
define a Hopf algebra of simplicial complexes, which contains
$\GGG$ as a Hopf subalgebra (and also has $\GGG$ as a quotient Hopf
algebra). They compute a formula for its antipode similar to
(and generalizing) \eqref{eq.Humpert-Martin-antipode-remark.S(G)}.
\end{remark}

% [DG][v50] Added above remark.

\begin{remark}
The chromatic Hopf algebra $\GGG$ is used in \cite{Lando} and
\cite[\S 14.4]{ChmutovDuzhinMostovoy} to study
\emph{Vassiliev invariants of knots}. In fact, a certain quotient
of $\GGG$ (named $\mathcal{F}$ in \cite{Lando} and $\mathcal{L}$ in
\cite[\S 14.4]{ChmutovDuzhinMostovoy}) is shown to naturally
host invariants of \emph{chord diagrams} and therefore Vassiliev
invariants of knots.
\end{remark}

% [DG][v41] Above remark added.

\begin{remark}
\label{rmk.GGG.free}
The $\kk$-algebra $\GGG$ is isomorphic to a polynomial algebra (in
infinitely many indeterminates) over $\kk$. Indeed, every finite
graph can be uniquely written as a disjoint union of finitely
many connected finite graphs (up to order). Therefore, the basis
elements $\left[G\right]$ of $\GGG$ corresponding to connected
finite graphs $G$ are algebraically independent in $\GGG$ and
generate the whole $\kk$-algebra $\GGG$ (indeed, the disjoint
unions of connected finite graphs are precisely the monomials in
these elements). Thus, $\GGG$ is isomorphic to a polynomial
$\kk$-algebra with countably many generators (one for each
isomorphism class of connected finite graphs). As a consequence,
for example, we see that $\GGG$ is an integral domain if $\kk$ is
an integral domain.
\end{remark}

% [DG][v51] Added above remark.

\subsubsection{A ``ribbon basis'' for $\GGG$ and self-duality}

In this subsection, we shall explore a second basis of $\GGG$
and a bilinear form on $\GGG$. This material will not be used
in the rest of these notes (except in
Exercise~\ref{exe.GGG.chromatic-sharp}), but it is of some
interest and provides an example of how a commutative
cocommutative Hopf algebra can be studied.

First, let us define a second basis of $\GGG$, which is obtained
by M\"obius inversion (in an appropriate sense) from the standard
basis
$\left( \left[ G \right] \right) _{\left[ G \right]
\text{ is an isomorphism class of finite graphs}}$:

\begin{definition}
\label{def.GGG.moebius-basis}
For every finite graph $G = \left(V, E\right)$, set
\[
\left[G\right]^\sharp = \sum\limits_{\substack{H = \left(V, E^\prime\right);\\
                                        E^\prime \supset E^c}}
                        \left(-1\right)^{\left|E^\prime \setminus E^c\right|}
                        \left[H\right]
                      \in \GGG,
\]
where $E^c$ denotes the complement of the subset $E$ in the set of
all two-element subsets of $V$. Clearly, $\left[G\right]^\sharp$
depends only on the isomorphism class $\left[G\right]$ of $G$, not on
$G$ itself.
\end{definition}

\begin{proposition} \phantomsection
\label{prop.GGG.moebius-basis}
\begin{itemize}
\item[(a)] Every finite graph $G = \left(V, E\right)$
satisfies
\[
\left[G\right] = \sum\limits_{\substack{H = \left(V, E^\prime\right);\\
                                        E^\prime \cap E = \varnothing}}
                        \left[H\right]^\sharp .
\]

\item[(b)] The elements $\left[G\right]^\sharp$, where
$\left[G\right]$ ranges over all isomorphism classes of finite graphs,
form a basis of the $\kk$-module $\GGG$.

\item[(c)] For any graph $H = \left(V, E\right)$, we have
\begin{equation}
\label{eq.prop.GGG.moebius-basis.c}
\Delta\left[H\right]^\sharp
=\sum\limits_{\substack{\left(V_1,V_2\right);\\ V=V_1 \sqcup V_2 ;\\ H=H|_{V_1} \sqcup H|_{V_2}}}
\left[H|_{V_1}\right]^\sharp \otimes \left[H|_{V_2}\right]^\sharp .
\end{equation}

\item[(d)] For any two graphs $H_1 = \left(V_1, E_1\right)$ and
$H_2 = \left(V_2, E_2\right)$, we have
\begin{equation}
\label{eq.prop.GGG.moebius-basis.d}
\left[H_1\right]^\sharp \left[H_2\right]^\sharp
 = \sum\limits_{\substack{H=\left(V_1 \sqcup V_2,E\right) ;\\H|_{V_1}=H_1;\\ H|_{V_2}=H_2}} \left[H\right]^\sharp.
\end{equation}
\end{itemize}
\end{proposition}

For example,
\begin{align*}
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]^\sharp
&=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
-
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet\ar@{-}[ur]&
}
\right]
- \left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet\ar@{-}[ul]&
}
\right]
+ \left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet&
}
\right]
\\
&=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
-
2 \left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet\ar@{-}[ur]&
}
\right]
+ \left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet&
}
\right] .
\end{align*}

Proving Proposition~\ref{prop.GGG.moebius-basis} is part of
Exercise~\ref{exe.GGG.moebius-basis} further below.

The equalities that express the elements $\left[G\right]^\sharp$
in terms of the elements $\left[H\right]$ (as in
Definition~\ref{def.GGG.moebius-basis}), and vice versa
(Proposition~\ref{prop.GGG.moebius-basis}(a)), are reminiscent
of the relations \eqref{ribbon-as-sum-of-Hs} and
\eqref{H-as-sum-of-ribbons} between the bases
$\left(R_\alpha\right)$ and $\left(H_\alpha\right)$ of $\Nsym$.
In this sense, we can call the basis of $\GGG$ formed by the
$\left[G\right]^\sharp$ a ``ribbon basis'' of $\GGG$.

We now define a $\kk$-bilinear form on $\GGG$:

\begin{definition}
\label{def.GGG.bilform}
For any two graphs $G$ and $H$, let
$\operatorname*{Iso} \left( G, H \right)$
denote the set of all isomorphisms from $G$ to
$H$\ \ \ \ \footnote{We recall that if $G = \left( V, E \right)$ and
$H = \left( W, F \right)$ are two graphs, then an
\emph{isomorphism}\index{isomorphism of graphs}
from $G$ to $H$ means a bijection $\varphi : V \rightarrow W$ such
that $\varphi_{\ast} \left( E \right) = F$. Here, $\varphi_{\ast}$
denotes the map from the powerset of $V$ to the powerset of $W$ which
sends every $T \subset V$ to $\varphi \left( T \right) \subset W$.}.
Let us now define a $\kk$-bilinear form $\left( \cdot, \cdot \right)
: \GGG \times \GGG \rightarrow \kk$ on $\GGG$ by
setting
\[
\left( \left[ G \right] ^{\sharp}, \left[ H \right] \right)
= \left\vert \operatorname*{Iso} \left( G, H \right) \right\vert .
\]
\footnote{This is well-defined, because:
\par
\begin{itemize}
\item the number $\left\vert \operatorname*{Iso} \left( G, H \right)
\right\vert$ depends only on the isomorphism classes $\left[ G \right]$
and $\left[ H \right]$ of $G$ and $H$, but not on $G$ and $H$
themselves;
\par
\item the elements $\left[ G \right] ^{\sharp}$, where
$\left[ G \right]$ ranges over all isomorphism classes of finite
graphs, form a basis of the $\kk$-module $\GGG$ (because of
Proposition~\ref{prop.GGG.moebius-basis}(b));
\par
\item the elements $\left[ G \right]$, where
$\left[ G \right]$ ranges over all isomorphism classes of finite
graphs, form a basis of the $\kk$-module $\GGG$.
\end{itemize}
}
\end{definition}

\begin{proposition}
\label{prop.GGG.bilform.sym}
The form $\left( \cdot, \cdot \right)
: \GGG \times \GGG \rightarrow \kk$ is symmetric.
\end{proposition}

Again, we refer to Exercise~\ref{exe.GGG.moebius-basis} for a proof
of Proposition~\ref{prop.GGG.bilform.sym}.

The basis of $\GGG$ constructed in
Proposition~\ref{prop.GGG.moebius-basis}(b) and the bilinear form
$\left(\cdot, \cdot\right)$ defined in
Definition~\ref{def.GGG.bilform} can be used to construct a Hopf
algebra homomorphism from $\GGG$ to its graded dual $\GGG^o$:

\begin{definition}
\label{def.GGG.selfdual}
For any finite graph $G$, let $\operatorname*{aut} \left( G \right)$
denote the number
$\left\vert \operatorname*{Iso} \left( G, G \right) \right\vert$.
Notice that this is a positive integer, since the set
$\operatorname*{Iso} \left( G, G \right)$ is nonempty (it contains
$\id_G$).

Now, recall that the Hopf algebra $\GGG$ is a connected graded Hopf
algebra of finite type. The $n$-th homogeneous component is spanned by
the $\left[ G \right]$ where $G$ ranges over the graphs with $n$
vertices. Since $\GGG$ is of finite type, its graded dual $\GGG^{o}$
is defined.
Let $\left( \left[ G \right] ^{\ast} \right) _{\left[ G \right]
\text{ is an isomorphism class of finite graphs}}$
be the basis of $\GGG^{o}$ dual to the basis
$\left( \left[ G \right] \right) _{\left[ G \right]
\text{ is an isomorphism class of finite graphs}}$ of $\GGG$. Define
a $\kk$-linear map $\psi : \GGG \rightarrow \GGG^{o}$ by
\[
\psi \left( \left[ G \right] ^{\sharp} \right)
= \operatorname*{aut} \left( G \right) \cdot \left[ G \right] ^{\ast}
\ \ \ \ \ \ \ \ \ \ \text{for every finite graph } G \text{.}
\]
\footnote{This is well-defined, since
$\left( \left[ G \right] ^{\sharp} \right) _{\left[ G \right]
\text{ is an isomorphism class of finite graphs}}$
is a basis of the $\kk$-module $\GGG$ (because of
Proposition~\ref{prop.GGG.moebius-basis}(b)).}
\end{definition}

\begin{proposition}
\label{prop.GGG.selfdual}
Consider the map $\psi : \GGG \to \GGG^{o}$ defined in
Definition~\ref{def.GGG.selfdual}.

\begin{itemize}
\item[(a)] This map $\psi$ satisfies
$\left( \psi \left( a \right) \right) \left( b \right)
= \left( a, b \right)$ for all $a \in \GGG$ and $b \in \GGG$.

\item[(b)] The map $\psi : \GGG \rightarrow \GGG^{o}$ is a Hopf
algebra homomorphism.

\item[(c)] If $\QQ$ is a subring of $\kk$, then the map $\psi$
is a Hopf algebra isomorphism $\GGG \rightarrow \GGG^{o}$.
\end{itemize}
\end{proposition}

\begin{exercise}
\label{exe.GGG.moebius-basis}
Prove Proposition~\ref{prop.GGG.moebius-basis},
Proposition~\ref{prop.GGG.bilform.sym} and
Proposition~\ref{prop.GGG.selfdual}.
\end{exercise}

\begin{remark}
Proposition~\ref{prop.GGG.selfdual}(c) shows that the Hopf algebra
$\GGG$ is self-dual when $\QQ$ is a subring of $\kk$. On
the other hand, if $\kk$ is a field of positive characteristic, then
$\GGG$ is never self-dual. Here is a quick way to see this: The
elements $\left[ G \right] ^{\ast}$ of $\GGG^{o}$ defined in
Definition~\ref{def.GGG.selfdual} have the property that
\[
\left( \left[ \circ \right] ^{\ast} \right) ^{n}
= n! \cdot \sum\limits_{\substack{\left[ G \right] \text{ is an isomorphism}
\\\text{class of finite graphs on}\\ n \text{ vertices}}}
\left[ G \right] ^{\ast}
\]
for every $n \in \NN$, where $\circ$ denotes the graph with one
vertex.\footnote{To see this, observe that the tensor
$\left[ \circ \right] ^{\otimes n}$ appears in the iterated coproduct
$\Delta^{\left( n-1 \right)} \left( \left[ G \right] \right)$
exactly $n!$ times whenever $G$ is a graph on $n$ vertices.} Thus,
if $p$ is a prime and $\kk$ is a field of characteristic $p$, then
$\left( \left[ \circ \right] ^{\ast} \right) ^{p} = 0$. Hence, the
$\kk$-algebra $\GGG^{o}$ has nilpotents in this situation. However,
the $\kk$-algebra $\GGG$ does not (indeed, Remark~\ref{rmk.GGG.free}
shows that it is an integral domain whenever $\kk$ is an integral
domain). Thus, when $\kk$ is a field of characteristic $p$, then
$\GGG$ and $\GGG^{o}$ are not isomorphic as $\kk$-algebras (let alone
as Hopf algebras).
\end{remark}

\begin{commentedout}
\textbf{Update:} The following claims (specifically, the equations
\eqref{dual-chromatic-Hopf-structure-maps-1} and
\eqref{dual-chromatic-Hopf-structure-maps-2} and
Proposition \ref{prop.GGG.selfdual.OLD-AND-FALSE}) are FALSE.

It turns out that the chromatic Hopf algebra $\GGG$ is self-dual.
In the dual Hopf algebra $\GGG^o$,
let $\{[G]^*\}$ denote the dual basis elements, so that $([H]^*,[G])=\delta_{[H],[G]}$.
To describe the structure maps in $\GGG^o$ explicitly, for any
graph $H = \left(V, E\right)$ one has
\begin{equation}
\label{dual-chromatic-Hopf-structure-maps-1}
\Delta[H]^*
=\sum\limits_{\substack{(V_1,V_2):\\ V=V_1 \sqcup V_2\\H=H|_{V_1} \sqcup H|_{V_2}}}
[H|_{V_1}]^* \otimes [H|_{V_2}]^*,
\end{equation}
and for any two graphs $H_1 = \left(V_1, E_1\right)$ and
$H_2 = \left(V_2, E_2\right)$ one has
\begin{equation}
\label{dual-chromatic-Hopf-structure-maps-2}
[H_1]^* [H_2]^*
 = \sum\limits_{\substack{H=(V_1 \sqcup V_2,E)\\H|_{V_1}=H_1\\H|_{V_2}=H_2}} [H]^*.
\end{equation}

% [DG][v14] I have split the two aligned equations, because $V_1$
% means a different thing in the first one than it means in the
% second.

\begin{proposition}
\label{prop.GGG.selfdual.OLD-AND-FALSE}
One has a Hopf isomorphism
$\GGG \overset{\varphi}{\longrightarrow} \GGG^o$
defined by
\[
[G] \longmapsto \sum\limits_{\substack{H=(V,E'):\\ E' \cap E = \varnothing}} [H]^*
\qquad \text{ for all graphs } G = \left(V,E\right).
\]
\end{proposition}
For example, this isomorphism maps
\[
\varphi \left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet\ar@{-}[ul]\ar@{-}[ur]&
}
\right]
=
\left[
\xymatrix@C=4pt@R=4pt{
\bullet\ar@{-}[rr]& &\bullet\\
       &\bullet&
}
\right]^*
+
\left[
\xymatrix@C=4pt@R=4pt{
\bullet& &\bullet\\
       &\bullet&
}
\right]^*.
\]

\begin{proof}[Proof of Proposition~\ref{prop.GGG.selfdual.OLD-AND-FALSE}.]
First note that $\varphi$ is a $\kk$-module
isomorphism via triangularity:  one has $H=(V,E')$
with $E' \cap E =\varnothing$ if and only
if $H$ is an edge subgraph of the
\emph{complementary graph} $\overline{G}$
to $G$ on the same vertex set $V$.

One can then check that for graphs $G_1,G_2$ on vertex sets $V_1, V_2$,
the fact that $\varphi([G_1][G_2])=
\varphi[G_1]\varphi[G_2]$
amounts, using
\eqref{dual-chromatic-Hopf-structure-maps-2},
to both being a sum of $[H]^*$
over graphs $H$ on $V_1 \sqcup V_2$ that share no
edges with $G_1 \sqcup G_2$.

Similarly, one can check that the fact
that $\Delta \varphi[G]=(\varphi \otimes \varphi)(\Delta[G])$
amounts, using
\eqref{dual-chromatic-Hopf-structure-maps-1},
to both being a sum of $[H_1]^* \otimes [H_2]^*$
over triples $(H,H_1,H_2)$ where
$H$ is a graph on the same vertex
set $V$ as $G$ but sharing no edges with $G$, and
with $H=H_1 \sqcup H_2$.
\end{proof}
\end{commentedout}



% [DG][v50] Replaced false claim of selfduality by a proposition giving
% a new basis.

% [DG][v51] Extended the proposition and added a remark. Together,
% they now show that $\GGG$ is selfdual when $\kk$ is a $\QQ$-algebra,
% and not selfdual when $\kk$ is a field of characteristic $p$.

% [DG][v53] Restructured the statements and proofs around the
% ribbon basis and the bilinear form on $\GGG$.


\subsubsection{Stanley's chromatic symmetric function of a graph}

\begin{definition}
\label{def.GGG.chromatic}
\dfn{Stanley's chromatic symmetric function}\index{chromatic symmetric function}
$\Psi[G]$ for a simple graph $G=(V,E)$
is the image of $[G]$ under the map
$\GGG \overset{\Psi}{\longrightarrow} \Qsym$
induced via Theorem~\ref{Qsym-as-terminal-object-theorem}
from the \dfn{edge-free character}
$\GGG \overset{\zeta}{\longrightarrow} \kk$ defined by
\begin{equation}
\label{edge-free-character-defn}
\zeta[G]=\begin{cases}
1, & \text{ if }G\text{ has no edges, that is, }G\text{ is an independent/stable set of vertices};\\
0, & \text{ otherwise.}
\end{cases}
\end{equation}
Note that, because $\GGG$ is cocommutative, $\Psi[G]$ is symmetric and not just quasisymmetric;
see Remark~\ref{Sym-as-terminal-object-remark}.
\end{definition}

Recall that for a graph $G=(V,E)$,
a (vertex-)coloring $f: V \rightarrow \{1,2,\ldots\}$
is called \emph{proper}\index{proper coloring}
if no edge $e=\{v,v'\}$ in $E$ has $f(v)=f(v')$.

\begin{proposition}
\label{prop.GGG.chromatic}
For a graph $G=(V,E)$, the symmetric function $\Psi[G]$ has the expansion
\footnote{In fact, Stanley \emph{defined} $\Psi[G]$  in
\cite[Defn. 2.1]{Stanley-chromatic-qsym-fn}
via this expansion.}
\[
\Psi[G]=\sum\limits_{\substack{\text{proper colorings}\\f:V \rightarrow \{1,2,\ldots\}}}
         \xx_f
\]
where $\xx_f:=\prod_{v \in V} x_{f(v)}$.
In particular, its specialization from Proposition~\ref{principal-specialization-at-1-defn}
gives the chromatic polynomial of $G$:
\[
\ps^1 \Psi[G](m) = \chi_G(m)
= \left|\{\text{proper colorings }f:V \rightarrow \{1,2,\ldots,m\}\}\right| .
\]
\end{proposition}
\begin{proof}
The iterated coproduct
$\GGG \overset{\Delta^{(\ell-1)}}{\longrightarrow} \GGG^{\otimes \ell}$
sends
\[
[G] \longmapsto \sum\limits_{\substack{(V_1,\ldots,V_{\ell}):\\ V=V_1 \sqcup \cdots \sqcup V_\ell}}
[G|_{V_1}] \otimes \cdots \otimes [G|_{V_\ell}]
\]
and the map $\zeta^{\otimes \ell}$ sends each addend on the right to
$1$ or $0$, depending upon whether each $V_i \subset V$ is a stable set or not,
that is, whether the assignment of color $i$ to the vertices in $V_i$ gives
a proper coloring of $G$.  Thus formula \eqref{ABS-terminal-morphism-formula}
shows that the coefficient $\zeta_\alpha$ of $x_1^{\alpha_1} \cdots x_\ell^{\alpha_\ell}$ in $\Psi[G]$ counts the
proper colorings $f$ in which $|f^{-1}(i)|=\alpha_i$ for each $i$.
\end{proof}

% [DG][v25] Replaced "the element on the right" by "each
% addend on the right".

\begin{example}
For the complete graph $K_n$ on $n$ vertices,
one has
\begin{align*}
\Psi[K_n]&=n! e_n , \qquad \text{thus} \\
\ps^1 (\Psi[K_n])(m)
&=n! e_n(\underbrace{1,1,\ldots,1}_{m\text{ ones}})
=n! \binom{m}{n}\\
& =m(m-1) \cdots (m-(n-1))=\chi_{K_n}(m).
\end{align*}
In particular, the single vertex graph $K_1$ has
$\Psi[K_1]=e_1$, and since the Hopf morphism $\Psi$ is in particular
an algebra morphism, a graph $K_1^{\sqcup n}$ having $n$ isolated
vertices and no edges will have $\Psi[K_1^{\sqcup n}]=e_1^n$.

As a slightly more interesting example, the graph $P_3$ which is
a path having three vertices and two edges will have
\[
\Psi[P_3]=m_{(2,1)}+6m_{(1,1,1)}=e_2 e_1 + 3 e_3 .
\]
\end{example}

One might wonder, based on the previous examples, when $\Psi[G]$ is
\dfn{$e$-positive}, that is, when does its unique expansion in the $\{e_\lambda\}$
basis for $\Lambda$ have nonnegative coefficients?  This is an even stronger
assertion than \emph{$s$-positivity}\index{$s$-positive},
that is, having nonnegative coefficients for the expansion in terms
of Schur functions $\{s_\lambda\}$, since each $e_\lambda$ is $s$-positive.
This weaker property fails, starting with the
\dfn{claw graph $K_{3,1}$}, which has
\[
\Psi[K_{3,1}]=s_{(3,1)}-s_{(2,2)}+5s_{(2,1,1)}+8s_{(1,1,1,1)}.
\]
On the other hand, a result of Gasharov \cite[Theorem 2]{Gasharov} shows that one at
least has $s$-positivity for $\Psi[\inc(P)]$ where
$\inc(P)$ is the \emph{incomparability graph} of a poset
which is $(\mathbf{3}+\mathbf{1})$-free;  we refer the
reader to Stanley \cite[\S 5]{Stanley-chromatic-qsym-fn} for a discussion
of the following conjecture, which remains open\footnote{A recent
refinement for incomparability graphs of posets which are both $(\mathbf{3}+\mathbf{1})$- and
$(\mathbf{2}+\mathbf{2})$-free, also known as \emph{unit interval orders}
is discussed by Shareshian and Wachs \cite{ShareshianWachs}.}:

\begin{conjecture}
For any $(\mathbf{3}+\mathbf{1})$-free poset $P$, the incomparability graph
$\inc(P)$ has $\Psi[\inc(P)]$ an $e$-positive symmetric function.
\end{conjecture}

Here is another question about $\Psi[G]$:  how well does it distinguish nonisomorphic
graphs? Stanley gave this example of two graphs $G_1, G_2$ having $\Psi[G_1]=\Psi[G_2]$:
\[
\begin{array}{c|c}
G_1=
\xymatrix@C=6pt@R=6pt{
\bullet\ar@{-}[rr] &  &\bullet\\
        &\bullet\ar@{-}[ul]\ar@{-}[ur]\ar@{-}[dr]\ar@{-}[dl]& \\
\bullet\ar@{-}[rr] &  &\bullet
}
\qquad \phantom{a} & \qquad
G_2=
\xymatrix{
\bullet\ar@{-}[r] \ar@{-}[dr]\ar@{-}[d]&\bullet\ar@{-}[d]\ar@{-}[dr]  &\\
\bullet\ar@{-}[r] &\bullet &\bullet
}
\end{array}
\]
\noindent
At least $\Psi[G]$ appears to do better at distinguishing \emph{trees}, much
better than its specialization, the chromatic polynomial $\chi_G(m)$, which takes
the same value $m(m-1)^{n-1}$ on all trees with $n$ vertices.
\begin{question}
Does the chromatic symmetric function (for $\kk = \ZZ$) distinguish trees?
\end{question}
\noindent
It has been checked that the answer is affirmative for trees on $23$ vertices or less.
There are also interesting partial results on this question
by Martin, Morin and Wagner \cite{MartinMorinWagner}.

\vskip.1in
We close this section with a few other properties of $\Psi[G]$ proven by Stanley
which follow easily from the theory we have developed.
For example, his work makes
no explicit mention of the chromatic Hopf algebra $\GGG$, and the fact that $\Psi$ is
a Hopf morphism (although he certainly notes the trivial algebra morphism property
$\Psi[G_1 \sqcup G_2]=\Psi[G_1] \Psi[G_2]$).  One property he proves is implicitly related to
$\Psi$ as a coalgebra morphism:  he considers (in the case when $\QQ$ is a subring
of $\kk$) the effect on $\Psi$ of the operator
$
\frac{\partial}{\partial p_1}:
\Lambda_\QQ \longrightarrow \Lambda_\QQ
$
which acts by first expressing a symmetric function
$f \in \Lambda_\QQ$ as a polynomial in the power sums $\{ p_n \}$, and then applies
the partial derivative operator $\frac{\partial}{\partial p_1}$ of the
polynomial ring $\QQ \left[p_1, p_2, p_3, \ldots \right]$.
It is not hard to see that  $\frac{\partial}{\partial p_1}$
is the same as the skewing operator $s_{(1)}^\perp=p_1^\perp$:  both act as  derivations
on $\Lambda_\QQ=\QQ[p_1,p_2,\ldots]$
(since $p_1 \in \Lambda_\QQ$ is primitive),
and agree in their effect on each $p_n$, in that
both send $p_1 \mapsto  1$, and both annihilate $p_2,p_3,\ldots$.


\begin{proposition}(Stanley \cite[Cor. 2.12(a)]{Stanley-chromatic-qsym-fn})
For any graph $G=(V,E)$, one has
\[
\frac{\partial}{\partial p_1} \Psi[G] = \sum_{v \in V} \Psi[G|_{V\setminus v}].
\]
\end{proposition}
\begin{proof}
Since $\Psi$ is a coalgebra homomorphism, we have
\[
\Delta \Psi[G] =
 (\Psi \otimes \Psi) \Delta[G]
=\sum\limits_{\substack{(V_1,V_2):\\V=V_1 \sqcup V_2}} \Psi[G|_{V_1}] \otimes \Psi[G|_{V_2}].
\]
Using this expansion (and the equality
$\frac{\partial}{\partial p_1} = s_{\left(1\right)}^\perp$), we now compute
\[
\frac{\partial}{\partial p_1} \Psi[G] = s_{(1)}^\perp \Psi[G] =
\sum\limits_{\substack{(V_1,V_2):\\V=V_1 \sqcup V_2}} (s_{(1)},\Psi[G|_{V_1}])  \cdot \Psi[G|_{V_2}]
=\sum_{v \in V} \Psi[G|_{V\setminus v}]
\]
(since degree considerations force $(s_{(1)}, \Psi[G|_{V_1}]) = 0$ unless $|V_1|=1$,
in which case $\Psi[G|_{V_1}]=s_{(1)}$).
\end{proof}


\begin{definition}
Given a graph $G=(V,E)$, an acyclic orientation $\Omega$ of the edges $E$
(that is, an orientation of each edge such that the resulting directed graph
has no cycles),
and a vertex-coloring $f: V \rightarrow \{1,2,\ldots\}$,
say that the pair $(\Omega,f)$ are \dfn{weakly compatible}
if whenever $\Omega$ orients an edge $\{v,v'\}$ in $E$ as $v \rightarrow v'$,
one has $f(v) \leq f(v')$.   Note that a
\emph{proper} vertex-coloring $f$ of a graph $G=(V,E)$
is weakly compatible with a unique acyclic orientation $\Omega$.
\end{definition}

% [DG] I added a parenthesized definition of acyclic orientations, and
% replaced "compatible" by "weakly compatible" here and in the proof (since
% you don't define just "compatible").

\begin{proposition}(Stanley \cite[Prop. 4.1, Thm. 4.2]{Stanley-chromatic-qsym-fn})
\label{Stanley-chromatic-reciprocity}
The involution $\omega$ of $\Lambda$ sends $\Psi[G]$ to
$
\omega \left( \Psi[G] \right) = \sum_{(\Omega,f)} \xx_f
$
in which the sum runs over weakly compatible
pairs $(\Omega,f)$ of an acyclic orientation $\Omega$ and
vertex-coloring $f$.

Furthermore, the chromatic polynomial $\chi_G(m)$ has the property that
$(-1)^{|V|} \chi_G(-m)$ counts all such weakly compatible pairs $(\Omega,f)$
in which $f: V \rightarrow \{1,2,\ldots,m\}$ is a vertex-$m$-coloring.
\end{proposition}
\begin{proof}
As observed above, a proper coloring $f$ is weakly compatible with a unique
acyclic orientation $\Omega$ of $G$.  Denote by $P_\Omega$ the
poset on $V$ which is the transitive closure of $\Omega$, endowed
with a \dfn{strict labelling} by integers, that is,
every $i \in P_\Omega$ and $j \in P_\Omega$ satisfying $i<_{P_\Omega} j$ must satisfy $i >_\ZZ j$.
Then proper colorings $f$ that induce $\Omega$ are the same as $P_{\Omega}$-partitions,
so that
\begin{equation}
\label{chromatic-symm-fn-as-sum-of-posets}
\Psi[G] = \sum_{\Omega} F_{P_{\Omega}}(\xx).
\end{equation}
Applying the antipode $S$ and
using Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary} gives
\[
\omega \left( \Psi[G] \right)
=(-1)^{|V|} S \left( \Psi[G] \right)
= \sum_{\Omega}  F_{P_{\Omega}^{\opp}}(\xx)
= \sum_{(\Omega,f)} \xx_{f}
\]
where in the last line one sums over weakly compatible pairs as in the proposition.  The last equality comes from the fact that since each $P_\Omega$ has been given a strict labelling, $P_\Omega^{\opp}$ acquires a
\emph{weak (or natural) labelling}\index{weak labelling}\index{natural labelling},
that is,
every $i \in P_\Omega$ and $j \in P_\omega$ satisfying $i <_{P_\Omega^{\opp}} j$ must satisfy $i <_\ZZ j$.

The last assertion follows from Proposition~\ref{zeta-polynomiality-proposition}(iii).
\end{proof}

\begin{remark}
The interpretation
of $\chi_G(-m)$ in Proposition~\ref{Stanley-chromatic-reciprocity}
is a much older result of Stanley \cite{Stanley-negative-colors}.
The special case interpreting $\chi_G(-1)$ as $(-1)^{|V|}$ times
the number of acyclic orientations of $G$
has sometimes been called Stanley's \dfn{(-1)-color theorem}.
It also follows (via Proposition~\ref{zeta-polynomiality-proposition})
from Humpert and Martin's antipode formula for $\GGG$ discussed in
Remark~\ref{Humpert-Martin-antipode-remark}: taking $\zeta$ to be
the character of $\GGG$ given in \eqref{edge-free-character-defn},
\[
\chi_G(-1)=\zeta^{\star (-1)}[G] = \zeta(S[G])
=\sum_{F} (-1)^{|V|-\rank(F)} \acyc(G/F) \zeta[G_{V,F}]
=(-1)^{|V|}\acyc(G)
\]
where the last equality uses the vanishing of $\zeta$
on graphs that have edges, so only the $F=\varnothing$ term survives.
\end{remark}

\begin{exercise}
\label{exe.GGG.chromatic-sharp}
If $V$ and $X$ are two sets, and if $f : V \rightarrow X$ is any map,
then $\operatorname*{eqs} f$ will denote the set
\[
\left\{ \left\{ u, u^{\prime} \right\} \ \mid\ u \in V,
\ u^{\prime} \in V, \ u \neq u^{\prime} \text{ and }
f \left( u \right) = f \left( u^{\prime} \right) \right\} .
\]
This is a subset of the set of all two-element subsets of $V$.

If $G = \left( V, E \right)$ is a finite graph, then show that the
map $\Psi$ introduced in Definition~\ref{def.GGG.chromatic} satisfies
\[
\Psi \left( \left[ G \right] ^{\sharp} \right)
= \sum\limits_{\substack{f : V \rightarrow \left\{ 1, 2, 3, \ldots \right\}
; \\ \operatorname*{eqs} f = E}} \xx_{f},
\]
where $\xx_{f} := \prod_{v \in V} x_{f \left( v \right)}$.
Here, $\left[ G\right] ^{\sharp}$ is defined as in
Definition~\ref{def.GGG.moebius-basis}.
\end{exercise}

% [DG][v51] Added the above exercise.



\subsection{Example: The quasisymmetric function of a matroid}

We introduce the \emph{matroid-minor Hopf algebra} of Schmitt \cite{Schmitt},
and studied extensively by
Crapo and Schmitt \cite{CrapoSchmitt1, CrapoSchmitt2, CrapoSchmitt3}.   A very simple character
$\zeta$ on this Hopf algebra will then give rise, via the map $\Psi$ from
Theorem~\ref{Qsym-as-terminal-object-theorem}, to the quasisymmetric function
invariant of matroids from the work of Billera, Jia and the second
author \cite{BilleraJiaR}.

\subsubsection{The matroid-minor Hopf algebra}

We begin by reviewing some notions from matroid theory; see Oxley \cite{Oxley} for background,
undefined terms and unproven facts.

\begin{definition}
A \dfn{matroid} $M$ of rank $r$ on a (finite) ground set $E$
is specified by a nonempty collection
\dfn{$\BBB(M)$} of $r$-element subsets of $E$ with the following
\dfn{exchange property}:
\begin{statement}
For any $B, B'$ in $\BBB(M)$ and $b$ in $B$,
there exists $b'$ in $B'$ with $(B \setminus \{b\}) \cup \{b'\}$ in
$\BBB(M)$.
\end{statement}
The elements of $\BBB\left(M\right)$ are called the
\emph{bases}\index{basis of a matroid} of the matroid $M$.
\end{definition}

\begin{example}
\label{vector-matroid-example}
\begin{comment}
\textbf{Old version:}
A matroid $M$ is \emph{represented} by a collection of vectors $E=\{e_1,\ldots,e_n\}$
in a vector space if $\BBB(M)$ is the collection of subsets $B=\{e_{i_1},\ldots,e_{i_r}\}$
having the property that $B$ forms a basis for the span of all of the vectors in $E$.

For example, if $E=\{a,b,c,d\}$ are the four vectors $a=(1,0),b=(1,1),c=(0,1)=d$ in $\RR^2$
depicted here
\[
\xymatrix{
c,d & b \\
  \ar[r] \arsurj[u] \ar[ur]     & a
}
\]
then $\BBB(M)=\{\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\}\}$.
\end{comment}
A matroid $M$ with ground set $E$ is
\emph{represented}\index{represented matroid} by
a family of vectors $S = \left(v_e\right)_{e \in E}$
in a vector space if $\BBB\left(M\right)$ is the collection
of subsets $B \subset E$ having the property that
the subfamily $\left(v_e\right)_{e \in B}$ is a basis for
the span of all of the vectors in $S$.

For example, if $M$ is the matroid with
$\BBB(M)=\{\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\}\}$ on
the ground set $E=\{a,b,c,d\}$, then $M$ is represented by
the family $S = \left(v_a, v_b, v_c, v_d\right)$ of
the four vectors $v_a=(1,0), v_b=(1,1), v_c=(0,1)=v_d$ in
$\RR^2$ depicted here
\[
\xymatrix{
v_c,v_d & v_b \\
  \ar[r] \arsurj[u] \ar[ur]     & v_a
} .
\]

% [DG][v29] Rewrote the above example to distinguish between the elements
% of the matroids and the vectors representing them (the correspondence
% is not 1-to-1, so I would rather not identify them).

Conversely, whenever $E$ is a finite set and
$S = \left(v_e\right)_{e \in E}$ is a family of vectors
in a vector space, then the set
\[
\left\{ B \subset E :
\text{the subfamily } \left(v_e\right)_{e \in B}
\text{ is a basis for the span of all of the vectors in } S \right\}
\]
is a matroid on the ground set $E$.

A matroid is said to be \emph{linear}\index{linear matroid}
if there exists a family of vectors in a vector space
representing it. Not all matroids are linear, but many
important ones are.

% [DG][v29] Added preceding two paragraphs.
\end{example}

\begin{example}
\label{graphic-matroid-example}
A special case of matroids $M$ represented by vectors are
\emph{graphic matroids}\index{graphic matroid}, coming
from a graph $G=(V,E)$, with parallel edges and self-loops allowed.  One represents
these by vectors in $\RR^V$ with standard basis $\{ \epsilon_v\}_{v \in V}$ by
associating the vector $\epsilon_v -\epsilon_{v'}$ to any edge connecting a vertex
$v$ with a vertex $v'$.
% [DG][v29] Replaced "to the edge $e=\{v,v'\}$" by "to any edge connecting a vertex
% $v$ with a vertex $v'$" to fit the model of a graph allowing multiple edges.
One can check (or see \cite[\S 1.2]{Oxley}) that the bases $B$ in $\BBB(M)$ correspond to the edge
sets of \emph{spanning forests}\index{spanning forest}
for $G$, that is, edge sets which are acyclic and contain
one spanning tree for each connected component of $G$.
%For example, the graph $G=(V,E)$ shown below has the same matroid $\BBB(M)$
%as the one represented by the vectors in Example~\ref{vector-matroid-example}:
For example, the matroid $\BBB(M)$ corresponding to the graph $G=(V,E)$ shown below:
\[
\xymatrix{
&\bullet \ar@{-}[dl]_{a}\ar@{-}[dr]^{b}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{c}  \ar@(dr,dl)@{-}[rr]_{d}& &\bullet
}
\]
%whose spanning trees are the edge sets $\BBB(M)=\{ \{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\}\}$.
is exactly the matroid represented by the vectors in
Example~\ref{vector-matroid-example}; indeed, the spanning forests of
this graph $G$ are the edge sets
$\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\}$.
% [DG][v29] Rewrote preceding sentence.
(In this example, spanning forests are the same as spanning trees,
since $G$ is connected.)
\end{example}


To define the matroid-minor Hopf algebra one needs the basic matroid
operations of \emph{deletion} and \emph{contraction}.
These  model the operations of deleting or contracting an edge in a graph.
For configurations of vectors they model the deletion of a vector, or the passage to
images in the quotient space modulo the span of a vector.

\begin{definition}
Given a matroid $M$ of rank $r$ and an element $e$ of its ground set $E$,  say that
$e$ is \emph{loop}\index{loop of a matroid}
(resp. \emph{coloop}\index{coloop of a matroid})
of $M$ if $e$ lies in no basis (resp. every basis) $B$ in $\BBB(M)$.
If $e$ is not a coloop, the
\emph{deletion}\index{deletion in a matroid}\index{$M \setminus e$}
$M \setminus e$ is a matroid of rank $r$
on ground set $E \setminus \{e\}$ having bases
\begin{equation}
\label{deletion-bases}
\BBB(M \setminus e):=\{ B \in \BBB(M): e \not\in B\}.
\end{equation}
If $e$ is not a loop, the
\emph{contraction}\index{contraction in a matroid}\index{$M / e$}
$M / e$ is a matroid of rank $r-1$
on ground set $E \setminus \{e\}$ having bases
\begin{equation}
\label{contraction-bases}
\BBB(M/ e):=\{ B \setminus \{e\}: e \in B \in \BBB(M)\}.
\end{equation}
When $e$ is a loop of $M$, then $M/e$ has rank $r$ instead of $r-1$ and one defines its
bases as in \eqref{deletion-bases} rather than \eqref{contraction-bases}; similarly, if $e$ is a coloop
of $M$ then $M \setminus e$ has rank $r-1$ instead of $r$ and one defines its
bases as in \eqref{contraction-bases} rather than \eqref{deletion-bases}.
\end{definition}

\begin{example}
Starting with the graph $G$ and its graphic matroid $M$ from Example ~\ref{graphic-matroid-example}, the deletion $M\setminus a$ and contraction $M/c$
correspond to the graphs $G \setminus a$ and $G/c$ shown here:
\[
\begin{array}{c|c}
G \setminus a=
\xymatrix{
&\bullet \ar@{-}[dr]^{b}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{c}  \ar@(dr,dl)@{-}[rr]_{d}& &\bullet
}
\qquad \phantom{a}
&
\qquad
G/c=
\xymatrix{
\bullet \ar@(dl,ul)@{-}[d]^{a}\ar@(dr,ur)@{-}[d]^{b}\\
\bullet\ar@(dr,dl) @{-}_{d}
}
\end{array}
\]
One has
\begin{enumerate}
\item[$\bullet$]
$\BBB(M \setminus a)=\{\{b,c\},\{b,d\}\}$,
so that $b$ has become a coloop in $M \setminus a$, and
\item[$\bullet$]
$\BBB(M/c)=\{\{a\},\{b\}\}$,
so that $d$ has become a loop in $M/c$.
\end{enumerate}
\end{example}

\begin{definition}
Deletions and contractions commute with each other. Thus,
given a matroid $M$ with ground set $E$, and a subset
$A \subset E$, two well-defined matroids can be constructed:
\begin{enumerate}
\item[$\bullet$]
the \emph{restriction}\index{restriction of a matroid}\index{$M\mid_A$}
$M|_A$, which is a matroid on ground set $A$,
obtained from $M$ by deleting all
$e \in E \setminus A$ in any order, and
\item[$\bullet$]
the \emph{quotient/contraction}\index{quotient of a matroid}\index{contraction of a matroid}\index{$M/A$}
$M/A$, which is a matroid
on ground set $E \setminus A$,
obtained from $M$ by contracting all $e \in A$ in any order.
\end{enumerate}

% [DG][v59] Reworded the above definition to make it clearer
% (or so I like to think).

We will also need the
\emph{direct sum}\index{direct sum of matroids} \dfn{$M_1 \oplus M_2$}
of two matroids $M_1$ and $M_2$. This is the matroid
whose ground set $E=E_1 \sqcup E_2$ is the disjoint union of a copy of the
ground sets $E_1,E_2$ for $M_1,M_2$, and whose bases
are
\[
\BBB(M_1\oplus M_2):=\{B_1 \sqcup B_2: B_i \in \BBB(M_i) \text{ for }i=1,2\}.
\]

Lastly, say that two matroids $M_1, M_2$ are
\dfn{isomorphic}\index{isomorphic matroids} if there is a bijection
of their ground sets $E_1 \overset{\varphi}{\longrightarrow} E_2$ having the property
that $\varphi \BBB(M_1) = \BBB(M_2)$.
\end{definition}

Now one can define the matroid-minor Hopf algebra, originally introduced by Schmitt \cite[\S 15]{Schmitt}, and studied further by Crapo and Schmitt \cite{CrapoSchmitt1, CrapoSchmitt2, CrapoSchmitt3}.
\begin{definition}
Let \dfn{$\MMM$} have $\kk$-basis  elements $[M]$ indexed by isomorphism classes of
matroids. Define the multiplication via
\[
[M_1] \cdot [M_2]:=[M_1 \oplus M_2] ,
\]
so that the class $[\varnothing]$ of the \dfn{empty matroid} $\varnothing$ having
empty ground set gives a unit.  Define the comultiplication for $M$ a matroid on ground set $E$ via
\[
\Delta[M]:=\sum_{A \subset E} [M|_A] \otimes [M/A],
\]
and a counit
\[
\epsilon [M]:=
\begin{cases}
1, & \text{ if }M=\varnothing \text{;} \\
0, & \text{ otherwise.}
\end{cases}
\]
\end{definition}


\begin{proposition}
The above maps endow $\MMM$ with the structure of a connected graded
finite type Hopf algebra over $\kk$, which is commutative.
\end{proposition}
\begin{proof}
Checking the unit and counit conditions are straightforward.
Associativity and commutativity of the multiplication follow because the direct sum operation
$\oplus$ for matroids is associative and commutative up to isomorphism.  Coassociativity
follows because for a matroid $M$ on ground set $E$,
one has the following equality between the two candidates
for $\Delta^{(2)}[M]$:
\begin{align*}
&\sum_{\varnothing \subset A_1 \subset A_2 \subset E}
[M|_{A_1}] \otimes [(M|_{A_2})/A_1] \otimes [M/A_2] \\
&=
\sum_{\varnothing \subset A_1 \subset A_2 \subset E}[M|_{A_1}] \otimes [(M/A_1)|_{A_2 \setminus A_1}] \otimes [M/A_2]
\end{align*}
due to the matroid isomorphism $(M|_{A_2})/A_1 \cong (M/A_1)|_{A_2 \setminus A_1}$.
Commutativity of the bialgebra diagram in \eqref{bialgebra-diagrams} amounts
to the fact that for a pair of matroids $M_1,M_2$ and subsets $A_1,A_2$ of their
(disjoint) ground sets $E_1,E_2$, one has isomorphisms
\begin{align*}
M_1|_{A_1} \oplus M_2|_{A_2}
  & \cong \left( M_1 \oplus M_2\right)|_{A_1 \sqcup A_2}, \\
M_1/A_1 \oplus M_2/A_2
  & \cong \left( M_1 \oplus M_2\right)/(A_1 \sqcup A_2). \\
\end{align*}
Letting $\MMM_n$ be the $\kk$-span of $[M]$ for matroids whose ground set $E$
has cardinality $|E|=n$, one can then easily check that $\MMM$
becomes a bialgebra which is
graded, connected, and of finite type, hence also a
Hopf algebra by Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
\end{proof}

See \cite{DuchampHoangNghiaKrajewskiTanasa} for an application
of $\MMM$ (and the operator $\exp^{\star}$ from
Section~\ref{subsect.leray}) to proving the \emph{Tutte recipe
theorem}, a ``universal'' property of the Tutte polynomial
of a matroid.











\subsubsection{A quasisymmetric function for matroids}

\begin{definition}
Define  a character $\MMM \overset{\zeta}{\longrightarrow} \kk$ by
\[
\zeta[M]=\begin{cases}
1, & \text{ if }M\text{ has only one basis};\\
0, & \text{ otherwise.}
\end{cases}
\]
It is easily checked that this is a character, that is, an algebra morphism
$\MMM \overset{\zeta}{\longrightarrow} \kk$.
Note that if $M$ has only one basis, say $\BBB(M)=\{B\}$, then
$B:=\coloops(M)$ is the set of coloops of $M$, and
$E\setminus B=\loops(M)$ is the set of loops of $M$.  Equivalently,
$M = \bigoplus_{e \in E} M|_{\{e\}}$ is the direct sum of matroids each
having one element, each a coloop or loop.

Define $\Psi[M]$ for a matroid $M$ to be
the image of $[M]$ under the map
$\MMM\overset{\Psi}{\longrightarrow} \Qsym$
induced via Theorem~\ref{Qsym-as-terminal-object-theorem}
from the above character $\zeta$.
\end{definition}

It turns out that $\Psi[M]$ is intimately related with greedy algorithms and
finding minimum cost bases.
A fundamental property of matroids (and one that characterizes them, in fact; see \cite[\S 1.8]{Oxley})
is that no matter how one assigns costs $f: E \rightarrow \RR$ to the elements of $E$,
the following \emph{greedy algorithm} (generalizing \emph{Kruskal's algorithm} for finding
minimum cost spanning trees) always succeeds
in finding one basis $B$ in $\BBB(M)$ achieving the minimum \emph{total cost} $f(B):=\sum_{b \in B} f(b)$:
\begin{algorithm}
\label{greedy-algorithm}
Start with the empty subset $I_0=\varnothing$ of $E$.
For $j=1,2,\ldots,r$, having already defined the set $I_{j-1}$, let
$e$ be the element of $E \setminus I_{j-1}$ having the lowest cost $f(e)$ among all those
for which $I_{j-1} \cup \{e\}$ is \emph{independent}, that is, still a subset of at least one
basis $B$ in $\BBB(M)$.  Then define $I_j:=I_{j-1} \cup \{e\}$.
Repeat this until $j=r$, and $B=I_r$ will be among the bases that achieve the minimum cost.
\end{algorithm}

\begin{definition}
Say that a cost function $f: E \rightarrow \{1,2,\ldots\}$ is
\emph{$M$-generic}\index{$M$-generic cost function}
if there is a \emph{unique} basis $B$ in $\BBB(M)$
achieving the minimum cost $f(B)$.
\end{definition}

\begin{example}
For the graphic matroid $M$ of Example~\ref{graphic-matroid-example},
this cost function $f_1: E \rightarrow \{1,2,\ldots\}$
\[
\xymatrix{
&\bullet \ar@{-}[dl]_{f_1(a)=1}\ar@{-}[dr]^{f_1(b)=3}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{f_1(c)={3}}  \ar@(dr,dl)@{-}[rr]_{f_1(d)=2}& &\bullet
}
\]
is $M$-generic, as it minimizes uniquely on the basis $\{a,d\}$,
whereas this cost function $f_2: E \rightarrow \{1,2,\ldots\}$
\[
\xymatrix{
&\bullet \ar@{-}[dl]_{f_2(a)=1}\ar@{-}[dr]^{f_2(b)=3}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{f_2(c)=2}  \ar@(dr,dl)@{-}[rr]_{f_2(d)=2}& &\bullet
}
\]
is \emph{not} $M$-generic, as it achieves its minimum value on the two
bases $\{a,c\}, \{a,d\}$.
\end{example}

\begin{proposition}
\label{matroid-qsym-expansion}
For a matroid $M$ on ground set $E$,
one has this expansion\footnote{In fact,
this expansion was the original definition of $\Psi[M]$  in
\cite[Defn. 1.1]{BilleraJiaR}.}
\[
\Psi[M]=\sum\limits_{\substack{M \text{-generic}\\f:E \rightarrow \{1,2,\ldots\}}}
         \xx_f
\]
where $\xx_f:=\prod_{e \in E} x_{f(e)}$.
In particular, for $m \geq 0$, its specialization $ps^1$
from Definition~\ref{principal-specialization-at-1-defn}
has this interpretation:
\[
\ps^1 \Psi[M](m)
=|\{M\text{-generic }f:E \rightarrow \{1,2,\ldots,m\}\}|.
\]
\end{proposition}
\begin{proof}
The iterated coproduct
$\MMM \overset{\Delta^{(\ell-1)}}{\longrightarrow} \MMM^{\otimes \ell}$
sends
\[
[M] \longmapsto
\sum
[M|_{A_1}] \otimes [(M|_{A_2})/A_1] \otimes \cdots \otimes [(M|_{A_{\ell}})/A_{\ell-1}]
\]
where the sum is over flags of nested subsets
\begin{equation}
\label{flag-of-nested-subsets}
\varnothing = A_0 \subset A_1 \subset \cdots
\subset A_{\ell-1} \subset A_{\ell}=E.
\end{equation}
The map $\zeta^{\otimes \ell}$ sends each summand to
$1$ or $0$, depending upon whether each $(M|_{A_j})/A_{j-1}$ has
a unique basis or not.  Thus formula \eqref{ABS-terminal-morphism-formula}
shows that the coefficient $\zeta_\alpha$ of
$x_{i_1}^{\alpha_1} \cdots x_{i_\ell}^{\alpha_\ell}$ in $\Psi[M]$
counts the flags of subsets in \eqref{flag-of-nested-subsets} for
which $|A_j \setminus A_{j-1}|=\alpha_j$ and $(M|_{A_j})/A_{j-1}$
has a unique basis, for each $j$.

Given a flag as in  \eqref{flag-of-nested-subsets},
associate the cost function $f: E \rightarrow \{1,2,\ldots\}$
whose value on each element of $A_j \setminus A_{j-1}$ is $i_j$;  conversely, given
any cost function $f$, say whose distinct values are $i_1 < \cdots < i_\ell$,
one associates the flag having $A_j \setminus A_{j-1}=f^{-1}(i_j)$ for each $j$.

Now, apply the greedy algorithm (Algorithm~\ref{greedy-algorithm})
to find a minimum-cost basis of $M$ for such a cost function $f$.
At each step of the greedy
algorithm, one new element is added to the independent set; these
elements weakly increase in cost as the algorithm
progresses\footnote{\textit{Proof.} Let $e$ be the element added
at step $i$, and let $e'$ be the element added at step $i+1$.
We want to show that $f\left(e\right) \leq f\left(e'\right)$.
But the element $e'$ could already have been added at step $i$.
Since it wasn't, we thus conclude that the element $e$ that was
added instead must have been cheaper or equally expensive.
In other words,
$f\left(e\right) \leq f\left(e'\right)$, qed.}. Thus,
the algorithm first adds some elements of cost $i_1$, then
adds some elements of cost $i_2$, then adds some elements of
cost $i_3$, and so on. We can therefore subdivide the execution
of the algorithm into phases $1, 2, \ldots, \ell$, where each
phase consists of some finite number of steps, such that all
elements added in phase $k$ have cost $i_k$. (A phase may be
empty.) For each $k \in \left\{1, 2, \ldots, \ell\right\}$, we
let $\beta_k$ be the number of steps in phase $k$; in other words,
$\beta_k$ is the number of elements of elements of cost $i_k$
added during the algorithm.

We will prove below, using induction on $s=0,1,2,\ldots,\ell$
the following \textbf{claim}:
After having completed phases $1, 2, \ldots, s$ in the
greedy algorithm (Algorithm~\ref{greedy-algorithm}), there is
\emph{a unique choice}
for the independent set produced thus far,
namely
\begin{equation}
\label{t-step-unique-greedy-basis}
I_{\beta_1+\beta_2+\cdots+\beta_s}=\bigsqcup_{j=1}^{s} \coloops( (M|_{A_j})/A_{j-1} ),
\end{equation}
if and only if each of the matroids $(M|_{A_j})/A_{j-1}$ for $j=1,2,\ldots,s$
\emph{has a unique basis}.

The case $s=\ell$ in this claim would show what we want, namely that
$f$ is $M$-generic, minimizing uniquely on the basis
shown in \eqref{t-step-unique-greedy-basis} with $s=\ell$,
if and only if each $(M|_{A_j})/A_{j-1}$ has a unique basis.

The assertion of the claim is trivially true for $s=0$.
In the inductive step, one may assume that
\begin{enumerate}
\item[$\bullet$]
the independent set $I_{\beta_1+\beta_2+\cdots+\beta_{s-1}}$ takes the form in
\eqref{t-step-unique-greedy-basis}, replacing $s$ by $s-1$,
\item[$\bullet$]
it is the unique $f$-minimizing basis
for $M|_{A_{s-1}}$, and
\item[$\bullet$]
$(M|_{A_j})/A_{j-1}$ has a unique basis for $j=1,2,\ldots,s-1$.
\end{enumerate}
Since $A_{s-1}$ exactly consists of all of the elements $e$ of $E$ whose
costs $f(e)$ lie in the range
$\{i_1,i_2,\ldots,i_{s-1}\}$, in phase $s$ the algorithm
will work in the quotient matroid $M/A_{s-1}$ and
attempt to augment $I_{\beta_1+\beta_2+\cdots+\beta_{s-1}}$
using the next-cheapest elements, namely the
elements of $A_{s} \setminus A_{s-1}$, which all have cost $f$ equal to $i_s$.
Thus the algorithm will have no choices about how to do this augmentation
if and only if $(M|_{A_s})/A_{s-1}$ has a unique basis, namely its set of
coloops, in which case the algorithm will choose to add all of these coloops,
giving $I_{\beta_1+\beta_2+\cdots+\beta_s}$ as described in \eqref{t-step-unique-greedy-basis}.
This completes the induction.

% [DG][v19] Replaced
% "will work in the quotient matroid $M/A_{j-1}$ and" by
% "will work in the quotient matroid $M/A_{s-1}$ and".

% [DG][v59] Corrected the proof, introducing the concept of phases
% and the $\beta_k$. The old version used $\alpha_k$ instead of
% $\beta_k$; but this would mean that the independent set would
% eventually grow to size $\left|E\right|$, which is clearly false.

The last assertion follows from Proposition~\ref{zeta-polynomiality-proposition}.
\end{proof}

\begin{example}
If $M$ has one basis then every function $f:E \rightarrow \{1,2,\ldots\}$ is $M$-generic, and
\[
\Psi[M]=\sum_{f: E \rightarrow \{1,2,\ldots\}} \xx_f = (x_1+x_2+ \cdots)^{|E|} =
M_{(1)}^{|E|}.
\]
\end{example}
\begin{example}
Let $U_{r,n}$ denote the \dfn{uniform matroid} of rank $r$ on $n$ elements $E$, having
$\BBB(U_{r,n})$ equal to all of the $r$-element subsets of $E$.

As $U_{1,2}$ has $E=\{1,2\}$ and $\BBB=\{\{1\},\{2\}\}$,
genericity means $f(1) \neq f(2)$, so
\[
\Psi[U_{1,2}]=\sum\limits_{\substack{(f(1),f(2)):\\f(1) \neq f(2)}} x_{f(1)} x_{f(2)}
=x_1 x_2 + x_2 x_1 + x_1 x_3+x_3 x_1 + \cdots = 2M_{(1,1)}.
\]

Similarly $U_{1,3}$ has $E=\{1,2,3\}$ with $\BBB=\{\{1\},\{2\},\{3\}\}$,
and genericity means either that $f(1),f(2),f(3)$
are all distinct, or that
two of them are the same and the third is smaller.
This shows
\begin{align*}
\Psi[U_{1,3}]&=3\sum_{i<j} x_i x_j^2+6\sum_{i<j<k} x_i x_j x_k\\
&=3M_{(1,2)}+6M_{(1,1,1)} ;\\
\ps^1 \Psi[U_{1,3}](m) &= 3 \binom{m}{2} + 6 \binom{m}{3}=\frac{m(m-1)(2m-1)}{2} .
\end{align*}
One can similarly analyze $U_{2,3}$ and check that
\begin{align*}
\Psi[U_{2,3}]&=3M_{(2,1)}+6M_{(1,1,1)} ;\\
\ps^1 \Psi[U_{2,3}](m) &= 3 \binom{m}{2} + 6 \binom{m}{3}=\frac{m(m-1)(2m-1)}{2} .
\end{align*}
\end{example}

These last examples illustrate the behavior of $\Psi$ under
the duality operation on matroids.

\begin{definition}
Given a matroid $M$ of rank $r$ on ground set $E$, its
\emph{dual}\index{dual matroid} or \dfn{orthogonal matroid}
$M^\perp$ is a matroid of
rank $|E|-r$ on the same ground set $E$, having
\[
\BBB(M^\perp):=\{ E \setminus B \}_{B \in \BBB(M)}.
\]
\end{definition}

See \cite[Theorem 2.1.1]{Oxley} or \cite[Section 4]{Cameron-matroids}
for a proof of the fact that this is well-defined (i.e., that
the collection $\{ E \setminus B \}_{B \in \BBB(M)}$
really satisfies the exchange property).
Here are a few examples of dual matroids.

\begin{example}
The dual of a uniform matroid is another uniform matroid:
\[
U_{r,n}^\perp = U_{n-r,n}.
\]
\end{example}

\begin{example}
If $M$ is matroid of rank $r$ represented by family of vectors $\{e_1,\ldots,e_n\}$ in a vector space over some field $\kk$,
one can find a family of vectors $\{e_1^\perp,\ldots,e_n^\perp\}$ that represent $M^\perp$ in the following way.  Pick a basis for the span of the vectors $\{e_i\}_{i=1}^n$, and create a matrix $A$ in $\kk^{r \times n}$ whose columns express the $e_i$ in terms of this basis.  Then pick any matrix $A^\perp$ whose row space is the null space of $A$, and one finds that the columns $\{e_i^\perp\}_{i=1}^n$ of $A^\perp$ represent $M^\perp$.  See Oxley \cite[\S 2.2]{Oxley}.
\end{example}

\begin{example}
\label{graphical-duality}
Let $G=(V,E)$ be a graph embedded in the plane with edge set $E$, giving rise to a graphic matroid $M$ on ground set $E$.  Let $G^\perp$ be a planar dual of $G$, so that, in particular, for each edge $e$ in $E$, the graph
$G^\perp$ has one edge $e^\perp$, crossing $e$ transversely.
Then the graphic matroid of $G^\perp$ is $M^\perp$.  See Oxley \cite[\S 2.3]{Oxley}.
\end{example}

\begin{proposition}
If $\Psi[M]=\sum_\alpha c_\alpha M_\alpha$ then
$\Psi[M^\perp] = \sum_\alpha c_\alpha M_{\rev(\alpha)}$.

Consequently, $\ps^1 \Psi[M](m) = \ps^1 \Psi[M^\perp](m)$.
\end{proposition}
\begin{proof}
First, let us prove that if
$\Psi[M]=\sum_\alpha c_\alpha M_\alpha$ then
$\Psi[M^\perp] = \sum_\alpha c_\alpha M_{\rev(\alpha)}$.
In other words, let us show that for any given
composition $\alpha$, the coefficient of $M_\alpha$
in $\Psi[M]$ (when $\Psi[M]$ is expanded in the basis
$\left(M_\beta\right)_{\beta \in \Comp}$ of $\Qsym$)
equals the coefficient of $M_{\rev(\alpha)}$ in $\Psi[M^\perp]$.
This amounts to showing that
for any composition $\alpha=(\alpha_1,\ldots,\alpha_\ell)$,
the cardinality of the set of $M$-generic $f$ having $\xx_f=\xx^{\alpha}$
is the same as the cardinality of the set of
$M^\perp$-generic $f^\perp$ having $\xx_{f^\perp}=\xx^{\rev(\alpha)}$.
We claim that the map $f \longmapsto f^\perp$ in which
$f^\perp(e)=\ell+1-f(e)$ gives a bijection between these sets.
To see this, note that any
basis $B$ of $M$ satisfies
\begin{align}
f(B) + f(E \setminus B) &= \sum_{e\in E}f(e) , \label{first-dual-base-relation}\\
f(E \setminus B) + f^\perp(E \setminus B) & = (\ell+1)(|E|-r),
\label{second-dual-base-relation}
\end{align}
where $r$ denotes the rank of $M$.
Thus $B$ is $f$-minimizing
if and only if $E\setminus B$ is $f$-maximizing
(by \eqref{first-dual-base-relation})
if and only if $E \setminus B$ is $f^\perp$-minimizing
(by \eqref{second-dual-base-relation}).
Consequently $f$ is $M$-generic if and only if $f^\perp$ is $M^\perp$-generic.

The last assertion follows, for example, from the calculation in
Proposition~\ref{zeta-polynomiality-proposition}(i) that
$\ps^1 (M_\alpha)(m) = \binom{m}{\ell(\alpha)}$ together with the fact that
$\ell(\rev(\alpha))= \ell(\alpha)$.
\end{proof}

Just as \eqref{chromatic-symm-fn-as-sum-of-posets} showed that
Stanley's chromatic symmetric function of a graph has an expansion as a sum
of $P$-partition enumerators for certain strictly labelled posets\footnote{%
A labelled poset $P$ is said to be
\emph{strictly labelled}\index{strictly labelled poset}
if every two
elements $i$ and $j$ of $P$ satisfying $i <_P j$ satisfy $i >_{\ZZ} j$.}
$P$, the same
holds for $\Psi[M]$.

\begin{definition}
Given a matroid $M$ on ground set $E$, and a basis $B$ in $\BBB(M)$,
define the \dfn{base-cobase poset} \dfn{$P_B$} to have $b < b'$ whenever
$b$ lies in $B$ and $b'$ lies in $E \setminus B$ and
$\left(B  \setminus \{b\}\right) \cup \{b'\}$ is in $\BBB(M)$.
\end{definition}

\begin{proposition}
\label{matroid-qsym-fn-as-sum-of-posets-prop}
For any matroid $M$, one has
$\Psi[M]=\sum_{B \in \BBB(M)} F_{(P_B,\strict)}(\xx)$
where $F_{(P_,\strict)}(\xx)$ for a poset $P$ means the $P$-partition
enumerator for any strict labelling of $P$, i.e.
a labelling such that
the $P$-partitions satisfy $f(i) < f(j)$ whenever $i<_P j$.

In particular, $\Psi[M]$ expands nonnegatively in the
$\{L_\alpha\}$ basis.
\end{proposition}
\begin{proof}
A basic result about matroids, due to Edmonds \cite{Edmonds}, describes
the \emph{edges} in the \emph{matroid base polytope} which is
the convex hull of all vectors $\{ \sum_{b \in B} \epsilon_b\}_{B \in \BBB(M)}$
inside $\RR^E$ with standard basis $\{\epsilon_e\}_{e \in E}$.  He shows
that all such edges connect two bases $B, B'$ that differ by a single
\emph{basis exchange}, that is, $B'=
\left(B  \setminus \{b\}\right) \cup \{b'\}$ for some $b$ in $B$
and $b'$ in $E \setminus B$.

Polyhedral theory then says that
a cost function $f$ on $E$ will minimize uniquely at $B$ if and only if
one has a strict increase
$f(B)<f(B')$ along each such edge $B \rightarrow B'$ emanating from $B$,
that is, if and only if $f(b) < f(b')$ whenever $b<_{P_B} b'$ in the
base-cobase poset $P_B$, that is, $f$ lies in $\AAA(P_B,\strict)$.
\end{proof}

\begin{example}
\label{graphic-matroid-example-continued}
The graphic matroid from Example~\ref{graphic-matroid-example} has
this matroid base polytope, with the bases $B$ in $\BBB(M)$ labelling the
vertices:
\[
\xymatrix{
 & &cd\ar@{-}[dll]\ar@{-}[ddl]\ar@{-}[drr]\ar@{-}[ddrrr]&&& & \\
ac\ar@{--}[rrrr]\ar@{-}[dr]& & &&ad\ar@{-}[dr]& \\
 &bc\ar@{-}[rrrr]& & &&bd
}
\]
The base-cobase posets $P_B$ for its five vertices $B$ are as follows:
\[
\xymatrix@C=6pt@R=6pt{
a\ar@{-}[d]\ar@{-}[dr] & b\ar@{-}[d] \ar@{-}[dl] \\
c & d
}
\]
\[
\xymatrix@C=6pt@R=6pt{
b\ar@{-}[d]& d\ar@{-}[d] \ar@{-}[dl] \\
a & c
}\qquad
\xymatrix@C=6pt@R=6pt{
a\ar@{-}[d]& d\ar@{-}[d] \ar@{-}[dl] \\
b & c
}\qquad
\xymatrix@C=6pt@R=6pt{
a\ar@{-}[d]& c\ar@{-}[d] \ar@{-}[dl] \\
b & d
}\qquad
\xymatrix@C=6pt@R=6pt{
b\ar@{-}[d]& c\ar@{-}[d] \ar@{-}[dl] \\
a & d
}
\]
One can label the first of these five strictly as
\[
\xymatrix@C=6pt@R=6pt{
1\ar@{-}[d]\ar@{-}[dr] & 2\ar@{-}[d] \ar@{-}[dl] \\
3 & 4
}
\]
and compute its strict $P$-partition enumerator from the linear
extensions $\{3412,3421,4312,4321\}$ as
\[
L_{(2,2)}+L_{(2,1,1)}+L_{(1,1,2)}+L_{(1,1,1,1)} ,
\]
while any of the last four can be labelled strictly as
\[
\xymatrix@C=6pt@R=6pt{
1\ar@{-}[d]& 2\ar@{-}[d] \ar@{-}[dl] \\
3 & 4
}
\]
and they each have an extra linear extension $3142$ giving their
strict $P$-partition enumerators as
\[
L_{(2,2)}+L_{(2,1,1)}+L_{(1,1,2)}+L_{(1,1,1,1)}+L_{(1,2,1)}.
\]
Hence one has
\[
\Psi[M]=5L_{(2,2)}+5L_{(1,1,2)}+4L_{(1,2,1)}+5L_{(2,1,1)}+5L_{(1,1,1,1)}.
\]
As $M$ is a graphic matroid for a self-dual planar graph, one has a matroid isomorphism $M \cong M^\perp$ (see Example~\ref{graphical-duality}), reflected in the fact that $\Psi[M]$ is invariant under the symmetry
swapping $M_\alpha \leftrightarrow M_{\rev(\alpha)}$ (and simultaneously
swapping  $L_\alpha \leftrightarrow L_{\rev(\alpha)}$).
\end{example}


This $P$-partition expansion for $\Psi[M]$ also allows us to identify
its image under the antipode of $\Qsym$.

\begin{proposition}
\label{matroid-qsym-reciprocity}
For a matroid $M$ on ground set $E$, one has
\[
S(\Psi[M]) = (-1)^{|E|} \sum_{f: E \rightarrow \{1,2,\ldots\}}
                  |\{f\text{-maximizing bases }B\}| \cdot \xx_f
\]
and
\[
\ps^1\Psi[M](-m) = (-1)^{|E|} \sum_{f: E \rightarrow \{1,2,\ldots,m\}}
                  |\{f\text{-maximizing bases }B\}|.
\]
In particular, the expected number of $f$-maximizing bases among
all cost functions $f:E\rightarrow\left\{  1,2, \ldots ,m\right\}  $
is $(-m)^{-|E|} \ps^1\Psi[M](-m)$.
\end{proposition}

% [DG][v3] I replaced "all cost functions that take at most $m$ values"
% by "all cost functions $f:E\rightarrow\left\{  1,2, \ldots ,m\right\}  $".
% The new wording is clearer, but less expressive (the old one
% presumably implied that the expected number does not depend on the
% choice of $m$-element set to which the cost functions are required
% to go).

\begin{proof}
Corollary~\ref{antipode-in-Qsym-as-reciprocity-corollary} implies
\[
S(\Psi[M])
=\sum_{B \in \BBB(M)} S(F_{(P_B,\strict)}(\xx)) \\
=(-1)^{|E|} \sum_{B \in \BBB(M)} F_{(P_B^{\opp},\weak)}(\xx) ,
\]
where $F_{(P,\weak)}(\xx)$ is the enumerator
for $P$-partitions in which $P$ has been \emph{naturally} labelled,
so that they satisfy $f(i) \leq f(j)$ whenever $i <_P j$.
When $P=P_B^{\opp}$, this is exactly the condition for $f$ to achieve
its maximum value at $f(B)$ (possibly not uniquely), that is,
for $f$ to lie in the \emph{closed} normal cone to the vertex indexed
by $B$ in the matroid base polytope;  compare this with the
discussion in the proof of
Proposition~\ref{matroid-qsym-fn-as-sum-of-posets-prop}.  Thus one has
\[
S(\Psi[M])=(-1)^{|E|} \sum\limits_{\substack{(B,f):\\B \in \BBB(M)\\f\text{ maximizing at }B}}
\xx_f ,
\]
which agrees with the statement of the proposition, after reversing
the order of the summation.

The rest follows from Proposition~\ref{zeta-polynomiality-proposition}.
\end{proof}

\begin{example}
We saw in Example~\ref{graphic-matroid-example-continued}
that the matroid $M$ from Example~\ref{graphic-matroid-example} has
\[
\Psi[M]=5L_{(2,2)}+5L_{(1,1,2)}+4L_{(1,2,1)}+5L_{(2,1,1)}+5L_{(1,1,1,1)},
\]
and therefore will have
\[
%\begin{aligned}
\ps^1 \Psi[M](m)= 5 \binom{m-2+4}{4} + (5+4+5) \binom{m-3+4}{4} + 5 \binom{m-4+4}{4}
            = \frac{m(m-1)(2m^2-2m+1)}{2}
%\end{aligned}
\]
using $\ps^1(L_\alpha)(m) =\binom{m-\ell+|\alpha|}{|\alpha|}$ from
Proposition~\ref{zeta-polynomiality-proposition} (i).
Let us first do a reality-check on a few of its values with
$m \geq 0$ using Proposition~\ref{matroid-qsym-expansion},
and for negative $m$ using Proposition~\ref{matroid-qsym-reciprocity}:
\vskip.1in
\begin{tabular}{|c||c|c|c|c|}\hline
$m$ & $-1$ & $0$ & $1$ & $2$ \\\hline
$\ps^1 \Psi[M](m)$ & $5$ & $0$ & $0$ & $5$\\ \hline
\end{tabular}
\vskip.1in
When $m=0$, interpreting the set of cost functions $f:E \rightarrow \{1,2,\ldots,m\}$
as being empty explains why the value shown is $0$.
When $m=1$, there is only one function $f:E \rightarrow \{1\}$, and it is
not $M$-generic; any of the $5$ bases in $\BBB(M)$ will minimize $f(B)$, explaining
both why the value for $m=1$ is $0$, but also explaining the
value of $5$ for $m=-1$.
The value of $5$ for $m=2$ counts these
$M$-generic cost functions $f:E \rightarrow \{1,2\}$:
\[
\xymatrix{
&\bullet \ar@{-}[dl]_{1}\ar@{-}[dr]^{1}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{2}  \ar@(dr,dl)@{-}[rr]_{2}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{1}\ar@{-}[dr]^{2}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{1}  \ar@(dr,dl)@{-}[rr]_{2}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{1}\ar@{-}[dr]^{2}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{2}  \ar@(dr,dl)@{-}[rr]_{1}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{2}\ar@{-}[dr]^{1}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{1}  \ar@(dr,dl)@{-}[rr]_{2}& &\bullet
}\quad
\xymatrix{
&\bullet \ar@{-}[dl]_{2}\ar@{-}[dr]^{1}& \\
\bullet \ar@(ur,ul)@{-}[rr]_{2}  \ar@(dr,dl)@{-}[rr]_{1}& &\bullet
}
\]
\noindent
Lastly, Proposition~\ref{matroid-qsym-reciprocity} predicts
the expected number of $f$-minimizing bases for
$f: E \rightarrow \{1,2,\ldots,m\}$ as
\[
(-m)^{-|E|}\ps^1 \Psi[M](-m)= (-m)^{-4}\frac{m(m+1)(2m^2+2m+1)}{2}
=\frac{(m+1)(2m^2+2m+1)}{2m^3},
\]
whose limit as $m \rightarrow \infty$ is $1$, consistent with the notion
that ``most'' cost functions should be generic with respect to the bases of
$M$, and maximize/minimize on a unique basis.
\end{example}


\begin{remark}
It is not coincidental that there is a
similarity of results for Stanley's chromatic symmetric
function of a graph $\Psi[G]$ and for the matroid quasisymmetric
function $\Psi[M]$, such as the $P$-partition expansions
\eqref{chromatic-symm-fn-as-sum-of-posets} versus Proposition~\ref{matroid-qsym-fn-as-sum-of-posets-prop}, and the reciprocity results
Proposition~\ref{Stanley-chromatic-reciprocity} versus
Proposition~\ref{matroid-qsym-reciprocity}.
It was noted in \cite[\S 9]{BilleraJiaR} that one can associate a similar
quasisymmetric function invariant to any \emph{generalized permutohedra}
in the sense of Postnikov \cite{Postnikov}.  Furthermore, recent work of
Ardila and Aguiar \cite{AguiarArdila} has shown that there is a Hopf algebra
of such generalized permutohedra, arising from a \emph{Hopf monoid} in the sense of
Aguiar and Mahajan \cite{AguiarMahajan}.  This Hopf algebra
generalizes the chromatic Hopf algebra of
graphs\footnote{Aguiar and Ardila actually work with a larger
Hopf algebra of graphs.
Namely, their concept of graphs allows parallel edges,
and it also allows ``half-edges'', which have only one endpoint.
If $G = \left(V, E\right)$ is such a graph (where $E$ is the
set of its edges and its half-edges), and if $V'$ is a subset
of $V$, then they define $G/_{V'}$ to be
the graph on vertex set $V'$ obtained from $G$ by
\begin{itemize}
\item removing all vertices that are not in $V'$,
\item removing all edges that have no endpoint in $V'$, and
all half-edges that have no endpoint in $V'$, and
\item replacing all edges that have
only one endpoint in $V'$ by half-edges.
\end{itemize}
(This is to be contrasted with the induced subgraph $G\mid_{V'}$,
which is constructed in the same way but with the edges that
have only one endpoint in $V'$ getting removed as well.)
The comultiplication they define on the Hopf algebra of such
graphs sends the isomorphism class $\left[G\right]$ of a
graph $G = \left(V, E\right)$ to
$\sum\limits_{\left(V_1, V_2\right) : V_1 \sqcup V_2 = V}
\left[G\mid_{V_1}\right] \otimes \left[G/_{V_2}\right]$.
This is no longer a cocommutative Hopf algebra;
our Hopf algebra $\GGG$ is a quotient of it.
In \cite[Corollary 13.10]{AguiarArdila}, Ardila and Aguiar
compute the antipode of the Hopf monoid of such graphs;
this immediately leads to a formula for the antipode of the
corresponding Hopf algebra, because what they call the
Fock functor $\overline{\mathcal{K}}$ preserves antipodes
\cite[Theorem 2.18]{AguiarArdila}.}
and the matroid-minor Hopf algebra, and its quasisymmetric function
invariant derives as usual from Theorem~\ref{Qsym-as-terminal-object-theorem}.
Their work \cite{AguiarArdila} also provides a generalization of the chromatic Hopf algebra
antipode formula of Humpert and Martin \cite{HumpertMartin} discussed
in Remark~\ref{Humpert-Martin-antipode-remark} above.
\end{remark}









\newpage




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Malvenuto-Reutenauer Hopf algebra of permutations}
\label{sect.MR}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Like so many Hopf algebras we have seen, the \dfn{Malvenuto-Reutenauer
Hopf algebra} $\FQsym$ can be
thought of fruitfully in more than one way.  One is that it gives
a natural noncommutative lift of the quasisymmetric
$P$-partition enumerators and the
fundamental basis $\{L_\alpha\}$ of $\Qsym$, rendering their
product and coproduct formulas even more natural.


\subsection{Definition and Hopf structure}

\begin{definition}
We shall regard permutations as words (over the alphabet
$\left\{1,2,3,\ldots\right\}$) by identifying
every permutation $\pi \in \Symm_n$
with the word $\left(\pi(1), \pi(2), \ldots, \pi(n)\right)$.

Define $\FQsym=\bigoplus_{n \geq 0} \FQsym_n$\index{$\FQsym$}
to be a graded $\kk$-module
in which $\FQsym_n$ has $\kk$-basis $\{F_w\}_{w \in \Symm_n}$ indexed by
the permutations $w=(w_1,\ldots,w_n)$ in $\Symm_n$.
\end{definition}

We first attempt to lift the product and coproduct formulas
\eqref{Qsym-product-on-fundamentals},
\eqref{Qsym-coproduct-on-fundamentals}
in the $\{L_\alpha\}$ basis of $\Qsym$.
We attempt to define a product for $u \in \Symm_k$ and
$v \in \Symm_{\ell}$ as follows\footnote{Recall that we regard
permutations as words.}:
\begin{equation}
\label{FQsym-product-on-F's}
F_u F_v:= \sum_{w \in \shuf{u}{v[k]}} F_w ,
\end{equation}
where for any word $v=(v_1,\ldots,v_\ell)$
we set $v[k]:=(k+v_1,\ldots,k+v_\ell)$. Note that the multiset
$\shuf{u}{v[k]}$ is an actual set in this situation (i.e., has
each element appear only once) and is a subset of $\Symm_{k+\ell}$.

% [DG][v14] Added the previous sentence. Also added "as follows:"
% before the equation.

% [DG][v37] Added "where we regard permutations as words..." part.

The coproduct will be defined using the notation of standardization
of $\std(w)$ a word $w$ in some linearly ordered alphabet (see
Definition~\ref{def.words.std}).

\begin{example}
\label{standardization-example}
Considering words in the Roman alphabet $a<b<c< \cdots$, we have
\[
\begin{array}{rccccccccccl}
\std(b&a&c&c &b&a&a&b&a&c &b)\\
=   (5&1&9&10&6&2&3&7&4&11&8).
\end{array}
\]
\end{example}

Using this, define for $w=(w_1,\ldots,w_n)$ in $\Symm_n$
the element $\Delta F_w \in \FQsym \otimes \FQsym$ by
\begin{equation}
\label{FQsym-coproduct-on-F's}
\Delta F_w := \sum_{k=0}^n
F_{\std(w_1,w_2,\ldots,w_k)} \otimes F_{\std(w_{k+1},w_{k+2},\ldots,w_n)}.
\end{equation}

It is possible to check directly that the maps defined in
\eqref{FQsym-product-on-F's} and
\eqref{FQsym-coproduct-on-F's}
endow $\FQsym$ with the structure of a connected graded finite type Hopf
algebra; see Hazewinkel, Gubareni, Kirichenko \cite[Thm. 7.1.8]{HazewinkelGubareniKirichenko}.
However in justifying this here, we
will follow the approach of Duchamp, Hivert and Thibon
\cite[\S 3]{NCSF6}, which exhibits $\FQsym$ as a subalgebra of a larger
ring of (noncommutative) power series of bounded degree
in a totally ordered alphabet.

\begin{definition}
Given a totally ordered set $I$, create a totally ordered variable
set $\{X_i\}_{i \in I}$, and the ring
\emph{$R\langle \{X_i\}_{i\in I} \rangle$}\index{$R<\{X_i\}_{i \in I}>$}
of \dfn{noncommutative power series of bounded degree}
in this alphabet\footnote{Let us recall the definition of
$R\langle \{X_i\}_{i\in I} \rangle$. \par
Let $N$ denote the \idx{free monoid}
on the alphabet $\{X_i\}_{i\in I}$; it
consists of words $X_{i_1} X_{i_2} \cdots X_{i_k}$. We define a
topological $\kk$-module $\ncps{\kk}{\{X_i\}_{i\in I}}$ to be the
Cartesian product $\kk^N$ (equipped with the product topology),
but we identify its element $\left(\delta_{w, u}\right)_{u \in N}$
with the word $w$ for every $w \in N$. Thus, every element
$\left(\lambda_w\right)_{w \in N} \in \kk^N =
\ncps{\kk}{\{X_i\}_{i\in I}}$ can be rewritten as the convergent sum
$\sum_{w \in N} \lambda_w w$. We call
$\lambda_w$ the \emph{coefficient of $w$} in this element
(or the \emph{coefficient of this element before $w$}). The
elements of $\ncps{\kk}{\{X_i\}_{i\in I}}$ will be referred to as
\dfn{noncommutative power series}.
We define
a multiplication on $\ncps{\kk}{\{X_i\}_{i\in I}}$ by the formula
\[
 \left(\sum_{w \in N} \lambda_w w\right)
 \left(\sum_{w \in N} \mu_w w\right)
 = \sum_{w \in N}
   \left(\sum_{\left(u, v\right) \in N^2; \  w = uv} \lambda_u \mu_v\right)
   w .
\]
(This is well-defined thanks to
the fact that, for each $w \in N$, there are only finitely many
$\left(u, v\right) \in N^2$ satisfying $w = uv$.)
Thus, $\ncps{\kk}{\{X_i\}_{i\in I}}$ becomes a $\kk$-algebra with unity
$1$ (the empty word). (It is similar to the monoid algebra $\kk N$
of $N$ over $\kk$, with the only difference that infinite sums are
allowed.)
\par
Now, we define $R\langle \{X_i\}_{i\in I} \rangle$ to be the
$\kk$-subalgebra of $\ncps{\kk}{\{X_i\}_{i\in I}}$ consisting of all
noncommutative power series
$\sum_{w \in N} \lambda_w w \in \ncps{\kk}{\{X_i\}_{i\in I}}$
\dfn{of bounded degree} (i.e., such that all words $w \in N$ of
sufficiently high length satisfy $\lambda_w = 0$).}.  Many times,
we will use a variable
set $\XX:=(X_1 < X_2 < \cdots)$, and call the ring $R \langle \XX \rangle$.
\end{definition}

% [DG][v50] Added footnote defining noncommutative power series.
% Noone likes to do this...

We first identify the algebra structure for $\FQsym$ as the subalgebra
of finite type within $R\langle \{X_i\}_{i\in I} \rangle$ spanned
by the elements
\begin{equation}
F_w = F_w( \{X_i\}_{i\in I}):= \sum\limits_{\substack{\ii=(i_1,\ldots,i_n):\\ \std(\ii) = w^{-1}}} \XX_\ii ,
\label{eq.FQSym.Fw-as-series}
\end{equation}
where $\XX_{\ii}:=X_{i_1} \cdots X_{i_n}$,
as $w$ ranges over $\bigcup_{n \geq 0} \Symm_n$ .

\begin{example}
\label{F's-example}
For the alphabet $\XX=(X_1 < X_2 < \cdots)$, in $R\langle \XX \rangle$ one has
\begin{align*}
F_{1}&=\sum_{1 \leq i} X_i=X_1 + X_2 + \cdots , \\
F_{12}&=\sum_{1 \leq i \leq j} X_i X_j=X_1^2 + X_2^2 + \cdots + X_1 X_2 + X_1 X_3 + X_2 X_3 + X_1 X_4+\cdots , \\
F_{21}&=\sum_{1 \leq i < j} X_j X_i =X_2 X_1 + X_3 X_1 + X_3 X_2 + X_4 X_1 + \cdots , \\
F_{312}&=\sum_{\ii: \std(\ii)=231} \XX_\ii
        =\sum_{1 \leq i < j \leq k} X_j X_k X_i \\
        &=X_2^2 X_1 + X_3^2 X_1 + X_3^2 X_2 + \cdots
         + X_2 X_3 X_1 + X_2 X_4 X_1 + \cdots .
\end{align*}
\end{example}

\begin{proposition}
\label{prop.FQsym.iso}
For any totally ordered infinite set $I$,
the elements $\{F_w\}$ as $w$ ranges over $\bigcup_{n \geq 0} \Symm_n$
form a $\kk$-basis for a subalgebra $\FQsym( \{X_i\}_{i\in I})$
of $R \langle \XX \rangle$, which is connected graded and of finite type,
having multiplication defined $\kk$-linearly by
\eqref{FQsym-product-on-F's}.

Consequently all such algebras are isomorphic
to a single algebra $\FQsym$, having basis $\{F_w\}$ and
multiplication given by the rule \eqref{FQsym-product-on-F's},
with the isomorphism mapping $F_w \longmapsto F_w (\{X_i\}_{i\in I})$.
\end{proposition}

For example,
\begin{align*}
F_1 F_{21}
&= (X_1 + X_2 + X_3 + \cdots)
    (X_2 X_1 + X_3 X_1 + X_3 X_2 + X_4 X_1 + \cdots) \\
&= X_1 \cdot X_3 X_2 + X_1 \cdot X_4 X_2 + \cdots
 +X_1 \cdot X_2 X_1 + X_2 \cdot X_3 X_2 + X_2 \cdot X_4 X_2 + \cdots \\
&\qquad   +X_2 \cdot X_3 X_1 + X_2 \cdot X_4 X_1 + \cdots
   +X_2 \cdot X_2 X_1 + X_3 \cdot X_3 X_1 + X_3 \cdot X_3 X_2 + \cdots \\
&\qquad   +X_3 \cdot X_2 X_1 + X_4 \cdot X_2 X_1 + \cdots \\
& = \sum_{\ii:\std(\ii)=132} \XX_\ii
+\sum_{\ii:\std(\ii)=231} \XX_\ii
+\sum_{\ii:\std(\ii)=321} \XX_\ii
= F_{132}+F_{312}+F_{321} = \sum_{w \in \shuf{1}{32}} F_w .
\end{align*}

\begin{proof}[Proof of Proposition~\ref{prop.FQsym.iso}]
The elements $\{F_w(\{X_i\}_{i\in I})\}$ are linearly independent as
they are supported on disjoint monomials, and so
form a $\kk$-basis for their span.  The fact that they multiply
via rule \eqref{FQsym-product-on-F's} is the equivalence of
conditions (i) and (iii) in the following Lemma~\ref{standardization-lemma},
from which all the remaining assertions follow.
\end{proof}


\begin{lemma}
\label{standardization-lemma}
For a triple of permutations
\begin{align*}
u&=(u_1,\ldots,u_k) \text{ in }\Symm_k,\\
v&=(v_1,\ldots,v_{n-k}) \text{ in }\Symm_{n-k},\\
w&=(w_1,\ldots,w_n) \text{ in }\Symm_n,
\end{align*}
the following conditions are equivalent:
\begin{enumerate}
\item[(i)]
$w^{-1}$ lies in the set $\shuf{u^{-1}}{v^{-1}[k]}$.
\item[(ii)]
$u=\std(w_1,\ldots,w_k)$ and
$v=\std(w_{k+1},\ldots,w_n)$,
\item[(iii)]
for some word $\ii=(i_1,\ldots,i_n)$
with $\std(\ii)=w$ one has
$u=\std(i_1,\ldots,i_k)$ and
$v=\std(i_{k+1},\ldots,i_n)$.
\end{enumerate}
\end{lemma}
\begin{proof} %[Proof of Lemma~\ref{standardization-lemma}.]
The implication (ii) $\Rightarrow$ (iii) is clear since $\std(w)=w$.
The reverse implication (iii) $\Rightarrow$ (ii) is best illustrated
by example, e.g. considering Example~\ref{standardization-example}
as concatenated, with $n=11$ and $k=6$ and $n-k=5$:
\[
\begin{tabu}{rccccccrrcccccl}
%\begin{array}{rccccccccccccccl}
w=\std&(b&a&c&c&b&a& | &            &a&b&a&c &b)\\
 =   &(5&1&9&10&6&2& | &            &3&7&4&11&8)\\
       & & &  & & &          &    & & & &  & \\
\hline
u=\std&(5&1&9&10 &6&2)& \multicolumn{1}{c||}{\qquad} & v = \std &(3&7&4&11&8)\\
  =&  (3&1&5&6 &4&2)& \multicolumn{1}{c||}{\qquad} & =    &(1&3&2&5 &4)\\
 =\std&(b&a&c&c &b&a)& \multicolumn{1}{c||}{\qquad} & = \std& (a&b&a&c &b)
%\end{array}
\end{tabu}
\]

% [DG][v14] I changed this to a tabu environment (using the "tabu"
% package) and modified the layout. Is it any better? (My main issue
% was that the equalities for $u$ and $v$ seemed to form one common
% equality chain.)

The equivalence of (i) and (ii) is a fairly standard consequence
of unique parabolic factorization $W=W^J W_J$
where $W=\Symm_n$ and $W_J=\Symm_k \times \Symm_{n-k}$,
so that $W^J$ are the  minimum-length coset representatives for cosets
$xW_J$ (that is, the permutations $x \in \Symm_n$ satisfying
$x_1 < \cdots < x_k$ and $x_{k+1} < \cdots < x_n$).
One can uniquely express any $w$ in $W$ as
$w=x y$ with $x$ in $W^J$ and $y$ in $W_J$, which here means
that $y=u \cdot v[k]=v[k] \cdot u$ for some $u$ in $\Symm_k$
and $v$ in $\Symm_{n-k}$.  Therefore $w=x u v[k]$,
if and only if $w^{-1}=u^{-1} v^{-1}[k] x^{-1}$, which means that
$w^{-1}$ is the shuffle of the sequences $u^{-1}$ in positions
$\{x_1,\ldots,x_k\}$ and $v^{-1}[k]$ in positions $\{x_{k+1},\ldots,x_n\}$.
\end{proof}

\begin{example}
To illustrate the equivalence of (i) and (ii) and the parabolic
factorization in the preceding proof, let $n=9$ and $k=5$ with
\begin{align*}
w&=\left(
\begin{matrix}
1 & 2& 3 & 4  & 5 & | & 6 & 7 & 8 & 9 \\
4 & 9 & 6 & 1 & 5 & | & 8 & 2 & 3 & 7
\end{matrix}
\right)\\
&=
\left(
\begin{matrix}
1 & 2& 3 & 4  & 5 & | & 6 & 7 & 8 & 9 \\
1 & 4 & 5 & 6 & 9 & | & 2 & 3 & 7 & 8
\end{matrix}
\right)
\left(\begin{matrix}
1 & 2& 3 & 4  & 5 \\
2 & 5 & 4 & 1 & 3
\end{matrix}
\right)
\left(\begin{matrix}
6 & 7& 8 & 9 \\
9 & 6 & 7 & 8
\end{matrix}
\right)\\
&=x \cdot u \cdot v[k] ;\\
\text{then} &  \\
w^{-1}&=\left(
\begin{matrix}
1 & 2& 3 & 4  & 5  & 6 & 7 & 8 & 9 \\
4 & 9 & 6 & 1 & 5  & 8 & 2 & 3 & 7
\end{matrix}
\right)\\
&=
\left(\begin{matrix}
1 & 2& 3 & 4  & 5 \\
4 & 1 & 5 & 3 & 2
\end{matrix}
\right)
\left(\begin{matrix}
6 & 7& 8 & 9 \\
7 & 8 & 9 & 6
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 2& 3 & 4  & 5 & 6 & 7 & 8 & 9 \\
\underline{1} & \underline{\underline{6}} & \underline{\underline{7}} & \underline{2} & \underline{3}  & \underline{4} & \underline{\underline{8}} & \underline{\underline{9}} & \underline{5}
\end{matrix}
\right) \\
&= u^{-1} \cdot v^{-1}[k] \cdot x^{-1} .
\end{align*}
\end{example}




Proposition~\ref{prop.FQsym.iso} yields that
$\FQsym$ is isomorphic to the $\kk$-subalgebra
$\FQsym\left(\XX\right)$ of the $\kk$-algebra $R\left<\XX\right>$
when $\XX$ is the variable set
$\left(X_1 < X_2 < \cdots\right)$. We identify
$\FQsym$ with $\FQsym\left(\XX\right)$ along this isomorphism.
% [DG][v17] Captain Obvious was here and inserted the above paragraph.
For any infinite alphabet
$\left\{X_i\right\}_{i \in I}$ and any $f \in \FQsym$, we denote
by $f\left(\left\{X_i\right\}_{i \in I}\right)$ the image of $f$
under the algebra isomorphism
$\FQsym \to \FQsym\left(\left\{X_i\right\}_{i \in I}\right)$ defined in
Proposition~\ref{prop.FQsym.iso}.

% [DG][v19] This is copypasted from something I just added to the
% QSym section. I need it below to define the coproduct.

%[VR][v20] Starting here, Vic began rewriting the alphabet-doubling argument...

One can now use this to define a coalgebra structure on $\FQsym$.
Roughly speaking, one wants to first 
evaluate an element
$f$ in $\FQsym \cong \FQsym\left(\XX\right) \cong \FQsym\left(\XX,\YY\right)$
as $f(\XX,\YY)$, using the linearly ordered 
variable set $(\XX,\YY) :=(X_1 < X_2 < \cdots < Y_1 < Y_2 < \cdots )$.
Then one should take the image of $f(\XX,\YY)$ after
imposing the partial commutativity relations 
\begin{equation}
\label{partial-commutativity-relations}
X_i Y_j = Y_j X_i \text{ for every pair }(X_i,Y_j) \in \XX \times \YY,
\end{equation}
and hope that this image lies
in a subalgebra isomorphic to 
\[
\FQsym\left(\XX\right) 
\otimes 
\FQsym\left(\YY\right) 
\cong \FQsym \otimes \FQsym.
\]
We argue this somewhat carefully.   Start by 
considering the canonical monoid epimorphism 
\begin{equation}
\label{semigroup-epimorphism}
F\langle \XX,\YY \rangle \overset{\rho}{\twoheadrightarrow} M,
\end{equation}
where $F\langle \XX,\YY \rangle$ denotes the \emph{free monoid} on 
the alphabet $(\XX,\YY)$ and $M$ denotes its 
quotient monoid imposing the partial commutativity relations 
\eqref{partial-commutativity-relations}.
Let $\kk^M$ denote the $\kk$-module of all functions
$f: M \rightarrow \kk$, with pointwise addition and scalar multiplication;
similarly define $\kk^{F\langle \XX,\YY \rangle}$.
As both monoids $F\langle \XX,\YY \rangle$ and $M$
enjoy the property that an element 
$m$ has only finitely many factorizations as $m=m_1 m_2$,
one can define a convolution algebra structure on 
both $\kk^{F\langle \XX,\YY \rangle}$ and $\kk^M$ via
\[
(f_1 \star f_2)(m) 
= \sum\limits_{\substack{(m_1,m_2) \in N \times N:\\ m=m_1m_2}}
f_1(m_1)f_2(m_2),
\]
where $N$ is respectively $F\langle \XX,\YY \rangle$ or $M$.
As fibers of the map $\rho$ 
in \eqref{semigroup-epimorphism} are finite, it induces a map
of convolution algebras,
which we also call $\rho$:
\begin{equation}
\label{semigroup-algebra-epimorphism}
\kk^{F\langle \XX,\YY\rangle} \overset{\rho}{\twoheadrightarrow} \kk^M.
\end{equation}
Now recall that $R\langle \XX\rangle$ denotes the algebra of
noncommutative formal power
series in the variable set $\XX$, of bounded degree, with coefficients in $\kk$.  One similarly has
the ring $R\langle \XX,\YY \rangle$, which can be identified with
the subalgebra of $\kk^{F\langle \XX,\YY\rangle}$ consisting of the
functions $f: F\langle \XX,\YY \rangle \rightarrow \kk$ having a 
bound on the length of the words
in their support (the value of $f$ on a word in $(\XX,\YY)$ gives
its power series coefficient corresponding to said word). 
We let $R\langle M\rangle$ denote
the analogous subalgebra of $\kk^M$; this can be thought of as the
algebra of bounded degree ``partially commutative power series''
in the variable sets $\XX$ and $\YY$.
Note that $\rho$ restricts to a map 
\begin{equation}
\label{bounded-power-series-morphism}
R\langle \XX,\YY \rangle \overset{\rho}{\rightarrow} R\langle M\rangle.
\end{equation}
Finally, we claim (and see Proposition~\ref{prop.FQsym.coproduct.wd} below 
for a proof) that this further restricts to a map
\begin{equation}
\label{FQsym-coproduct-defining-morphism}
\FQsym\left(\XX,\YY\right) \overset{\rho}{\rightarrow} 
\FQsym\left(\XX\right) \otimes \FQsym\left(\YY\right)
\end{equation}
in which the target is identified with its image under the
(injective\footnote{as images of the basis 
$F_u(\XX) \otimes F_v(\YY)$ of
$\FQsym(\XX) \otimes \FQsym(\YY)$ are supported on disjoint monomials in
$R\langle M\rangle$, so linearly independent.}) multiplication map 
\[
\begin{array}{rcl}
\FQsym\left(\XX\right) \otimes \FQsym\left(\YY\right) 
& \hookrightarrow & R\langle M\rangle , \\
f(\XX) \otimes g(\YY) &\mapsto& f(\XX)g(\YY).
\end{array}
\]
Using the identification of $\FQsym$ with all three of
$\FQsym\left(\XX \right), \FQsym\left(\YY\right), \FQsym\left(\XX,\YY\right)$,
the map $\rho$ in \eqref{FQsym-coproduct-defining-morphism} will then define
a coproduct structure on $\FQsym$.  Abusing notation, for $f$ in $\FQsym$, we will simply write $\Delta(f) = f(\XX,\YY)$ instead of $\rho(f(\XX,\YY))$.


\begin{comment}
% Below is Darij's old version of the definition of the coproduct.

Before we can do so, we need to go afield and define a few notions.
We have to define a notion of formal power series in ``partially
commutative'' variables (specifically, we will need power series in
a variable set $(\XX,\YY) :=(X_1 < X_2 < \cdots < Y_1 < Y_2 < \cdots )$
with commutativity relations $X_i Y_j = Y_j X_i$ but no relations
among the $X_i$ and no relations among the $Y_j$). To do so, we
introduce a generalization of the notion of formal power series.

A monoid $M$ (written multiplicatively) is said to be
\emph{coinductive}\index{coinductive monoid}
if for every $m \in M$, there exist only finitely many
pairs $\left(p,q\right) \in M \times M$ satisfying $m = pq$.
For instance, the monoid of all words over a given alphabet is
coinductive (since every word can be cut only at finitely many spots).
Similarly, the monoid of all monomials (= words modulo permutation
of letters) over a given alphabet is coinductive, and so is every
finite monoid.

If $M$ is a coinductive monoid, then we can define a $\kk$-algebra
$\kk\left[\left[M\right]\right]$ as the $\kk$-module $\kk^M$ with
multiplication given by
\begin{equation}
\label{FQSym.coinductive.multiplication}
\left(a_m\right)_{m \in M} \cdot \left(b_m\right)_{m \in M}
= \left( \sum_{\left(p, q\right) \in M \times M \ : \ m = pq}
         a_p b_q \right)_{m \in M}
\end{equation}
(the sums here are finite since $M$ is coinductive). As usual, there is
a pointwise topology on this $\kk$-algebra
$\kk\left[\left[M\right]\right]$, and if $n \in M$, then we denote
the element $\left(\delta_{m,n}\right)_{m \in M}$ of
$\kk\left[\left[M\right]\right]$ by $n$; then, any element
$\left(a_m\right)_{m \in M} \in \kk\left[\left[M\right]\right]$
equals $\sum_{m \in M} a_m m$. If we write the elements of
$\kk\left[\left[M\right]\right]$ in this latter form, the
equality \eqref{FQSym.coinductive.multiplication} transforms into
\[
\left(\sum_{m \in M} a_m m \right)
 \cdot \left(\sum_{m \in M} b_m m\right)
= \sum_{m \in M} \left( \sum_{\left(p, q\right) \in M \times M \ : \ m = pq}
         a_p b_q \right) m,
\]
which is precisely the rule according to which one multiplies
elements of a monoid ring, polynomials or power series.
This is not just a similarity: If $\Xi$ is a set
of variables, then the ring $\kk\left<\left<\Xi\right>\right>$ of
noncommutative power series in the variables $\Xi$ is (canonically
isomorphic to) the ring $\kk\left[\left[M_\Xi\right]\right]$,
where $M_\Xi$ is the free monoid with generators $\Xi$.

We notice one further property of this construction. Namely, if
$M$ and $N$ are two cofinite monoids, and if $\phi : M \to N$ is
a monoid homomorphism such that every fiber of $\phi$ is finite,
then we can define a continuous\footnote{with respect to the
pointwise topology} $\kk$-algebra homomorphism
\begin{align*}
\nonumber
\kk\left[\left[\phi\right]\right] : \kk\left[\left[M\right]\right] &\to \kk\left[\left[N\right]\right], \\
                                    \sum_{m \in M} a_m m &\mapsto \sum_{m \in M} a_m \phi\left(m\right) = \sum_{n \in N} \left(\sum_{m \in \phi^{-1}\left(n\right)} a_m\right) n.
\end{align*}

Let us now apply this to $\FQsym$. Given the ordered variable set
\[
(\XX,\YY) :=(X_1 < X_2 < \cdots < Y_1 < Y_2 < \cdots ),
\]
we can form two monoids: the free monoid $M_{\XX,\YY}$ with
generators $(\XX, \YY)$, and the monoid $M_{\XX \sim \YY}$
defined as the quotient of $M_{\XX,\YY}$ by all relations of the
form $X_i Y_j = Y_j X_i$ with $i \geq 1$ and $j \geq 1$. There is
clearly a canonical projection map $\rho : M_{\XX,\YY} \to
M_{\XX \sim \YY}$ (sending $X_i$ to $X_i$, and $Y_j$ to $Y_j$).
Both of the monoids $M_{\XX,\YY}$ and
$M_{\XX \sim \YY}$ are coinductive (in fact, $M_{\XX \sim \YY}$
is isomorphic to the direct product $M_{\XX} \times M_{\YY}$ of
two free monoids), and every fiber of $\rho$ is finite. Hence,
a continuous $\kk$-algebra homomorphism
$\powser{\kk}{\rho} : \powser{\kk}{M_{\XX,\YY}} \to \powser{\kk}{M_{\XX \sim \YY}}$
is well-defined. This homomorphism is surjective (since $\rho$
is surjective).

Now, consider the ring $\kk\left<\left<\XX, \YY\right>\right>$ of
noncommutative formal power series in the variables
$\left(\XX, \YY\right)$. Composing the surjective $\kk$-algebra
homomorphism
$\powser{\kk}{\rho} : \powser{\kk}{M_{\XX,\YY}} \to \powser{\kk}{M_{\XX \sim \YY}}$
with the canonical isomorphism
$\ncps{\kk}{\XX, \YY} \to \powser{\kk}{M_{\XX,\YY}}$,
we obtain a surjective $\kk$-algebra homomorphism
$\ncps{\kk}{\XX, \YY} \to \powser{\kk}{M_{\XX \sim \YY}}$,
which we denote by $\overline{\rho}$.
The $\kk$-algebra $\powser{\kk}{M_{\XX \sim \YY}}$ can be understood
as a ring of formal power series in the partially commutative
variables $\left(\XX, \YY\right)$, in which each $X_i$ commutes
with each $Y_j$ but no relations are imposed on the $X_i$ alone or
on the $Y_j$ alone. The map $\overline{\rho}$, roughly speaking,
``forgets'' the noncommutativity between the $X_i$ and $Y_j$.

Consider the $\kk$-linear map
$\FQsym \otimes \FQsym \to \powser{\kk}{M_{\XX \sim \YY}}$ sending
every $a \otimes b \in \FQsym \otimes \FQsym$ to the element
$\overline{\rho}\left(a\left(\XX\right) b\left(\YY\right)\right)$ of
$ \powser{\kk}{M_{\XX \sim \YY}}$ \ \ \ \ \footnote{This latter
element is well-defined since
$\underbrace{a\left(\XX\right)}_{\substack{\in \FQsym\left(\XX\right) \\
\subset R\left<\XX\right> \\ \subset R\left<\XX, \YY\right>}}
\underbrace{b\left(\YY\right)}_{\substack{\in \FQsym\left(\YY\right) \\
\subset R\left<\YY\right> \\ \subset R\left<\XX, \YY\right>}}
\in R\left<\XX, \YY\right> \subset \ncps{\kk}{\XX, \YY}$.}.
This map is a $\kk$-algebra homomorphism\footnote{since any power
series in the variables $\XX$ alone commutes with any power series
in the variables $\YY$ alone in $\powser{\kk}{M_{\XX \sim \YY}}$}
and injective\footnote{because the images of the basis elements
$F_u \otimes F_v$ of
$\FQsym \otimes \FQsym$ are supported on disjoint monomials in
$\powser{\kk}{M_{\XX \sim \YY}}$, and thus linearly independent}, and
thus allows us to identify $\FQsym \otimes \FQsym$ with a
$\kk$-subalgebra of $\powser{\kk}{M_{\XX \sim \YY}}$.

On the other hand, consider the $\kk$-algebra homomorphism
$\FQsym \to \powser{\kk}{M_{\XX \sim \YY}}$ sending every $f \in \FQsym$
to $\overline{\rho}\left(f\left(\XX, \YY\right)\right)
\in \powser{\kk}{M_{\XX \sim \YY}}$\ \ \ \ \footnote{This is well-defined
because $f\left(\XX, \YY\right) \in \FQsym\left(\XX, \YY\right) \subset
R\left<\XX, \YY\right> \subset \ncps{\kk}{\XX, \YY}$ (recall that
$f\left(\XX, \YY\right)$ is formally defined as the image of $f$
under the algebra isomorphism
$\FQsym \to \FQsym\left(\left\{X_i\right\}_{i \in I}\right)$ defined in
Proposition~\ref{prop.FQsym.iso}).}. It can be seen that the image of
this homomorphism is contained in $\FQsym \otimes \FQsym$ (see
Proposition~\ref{prop.FQsym.coproduct.wd} for a proof), and so the
homomorphism restricts to a $\kk$-algebra homomorphism
\begin{align*}
\nonumber
\FQsym & \overset{\Delta}{\longrightarrow} \FQsym \otimes \FQsym , \\
f & \longmapsto \overline{\rho}\left(f\left(\XX, \YY\right)\right)
\in \FQsym \otimes \FQsym \subset \powser{\kk}{M_{\XX \sim \YY}}.
\end{align*}
This homomorphism will be the coproduct of our $\kk$-bialgebra $\FQsym$.

We shall abuse notation and omit the map $\overline{\rho}$ from
our notation. More precisely, any expression containing
variables from the variable set $\left(\XX, \YY\right)$ will have to be
understood to mean its projection (from $\ncps{\kk}{\XX,\YY}$) onto
$\powser{\kk}{M_{\XX \sim \YY}}$ via the surjective homomorphism
$\overline{\rho}$ (rather than its actual value in
$\ncps{\kk}{\XX,\YY}$, which is uninteresting for us). For instance,
$f\left(\XX, \YY\right)$ (where $f \in \FQsym$) will have to be
understood as $\overline{\rho}\left(f\left(\XX, \YY\right)\right)$, and
$X_3 Y_2$ will mean the element $\overline{\rho}\left(X_3 Y_2\right)
= X_3 Y_2$ of $\powser{\kk}{M_{\XX \sim \YY}}$ rather than the element
$X_3 Y_2$ of $\ncps{\kk}{\XX,\YY}$ (as a consequence, we have
$X_3 Y_2 = Y_2 X_3$).

\end{comment}

% [DG][v17] This has technical issues... First of all, the isomorphism
% $R \langle \XX,\YY \rangle/[\XX,\YY]
% \cong R \langle \XX \rangle \otimes R \langle \YY \rangle$
% is false. What would the (residue class of)
% $\sum_i X_i \otimes Y_i$ on the left hand side correspond to on the
% right hand side?
% Second, you want to factor out the closure of the ideal
% $[X_i,Y_j]$ rather than the ideal itself, or else how do you prove
% that $\sum_i X_i$ commutes with $\sum_j Y_j$ ?

% [DG][v19] Fixed all the problems of this definition, but this is
% now seriously TL;DR. I feel particularly stupid introducing these
% "completed" monoid algebras only to ever use two simple cases
% -- but I know of no good source (noone even seems to bother
% defining noncommutative formal power series). The word
% "coinductive" is my own, and might be not very appropriate (in
% TCS it means something like "an object which can be infinite,
% but all information you will ever want to get from it can be
% obtained by biting off small chunks of the object").
% 
% Is my last paragraph about abuse of notation understandable?
% 
% The changes go on after this comment as well.

\begin{example}
Recall from Example~\ref{F's-example} that one has
\[
F_{312}=\sum_{\ii: \std(\ii)=231} \XX_\ii
        =\sum_{1 \leq i < j \leq k} X_j X_k X_i ,
\]
and therefore its coproduct is
\begin{align*}
\Delta F_{312}
%&=\overline{\rho}\left(F_{312}(X_1,X_2,\ldots,Y_1,Y_2,\ldots)\right) \\
&=F_{312}(X_1,X_2,\ldots,Y_1,Y_2,\ldots) \qquad \qquad \left(\text{by our abuse of notation}\right)\\
&=\sum_{i < j \leq k} X_j X_k X_i
+\sum\limits_{\substack{i < j,\\ k}} X_j Y_k X_i
+\sum\limits_{\substack{i,\\ j \leq k}} Y_j Y_k X_i
+\sum_{i < j \leq k} Y_j Y_k Y_i\\
&= \sum_{i < j \leq k} X_j X_k X_i \cdot 1
+\sum\limits_{\substack{i < j,\\ k}} X_j X_i \cdot Y_k
+\sum\limits_{\substack{i,\\ j \leq k}} X_i \cdot Y_j Y_k
+\sum_{i < j \leq k} 1 \cdot Y_j Y_k Y_i \\
&=F_{312}(\XX)\cdot 1
+ F_{21}(\XX) \cdot F_1(\YY)
+ F_1(\XX) \cdot F_{12}(\YY)
+ 1 \cdot F_{312}(\YY) \\
&=F_{312} \otimes \one
+ F_{21} \otimes F_1
+ F_1 \otimes F_{12}
+ \one \otimes F_{312}.
\end{align*}
\end{example}


\begin{proposition}
\label{prop.FQsym.coproduct.wd}

\begin{comment}
The image of the $\kk$-algebra homomorphism
$\FQsym \to \powser{\kk}{M_{\XX \sim \YY}}$ sending every $f \in \FQsym$
to $\overline{\rho}\left(f\left(\XX, \YY\right)\right)
\in \powser{\kk}{M_{\XX \sim \YY}}$
lies in $\FQsym \otimes \FQsym$, giving rise to a coproduct
\begin{align*}
\nonumber
\FQsym & \overset{\Delta}{\longrightarrow} \FQsym \otimes \FQsym , \\
f & \longmapsto \overline{\rho}\left(f\left(\XX, \YY\right)\right)
\in \FQsym \otimes \FQsym \subset \powser{\kk}{M_{\XX \sim \YY}}.
\end{align*}
\end{comment}

The map $\rho$ in \eqref{bounded-power-series-morphism}
does restrict as claimed to a map 
as in \eqref{FQsym-coproduct-defining-morphism},
and hence defines a coproduct on $\FQsym$,
acting on the $\{F_w\}$ basis by
the rule \eqref{FQsym-coproduct-on-F's}.  
This endows $\FQsym$ with
the structure of a connected graded finite type Hopf algebra.
\end{proposition}
\begin{proof}

\begin{comment}
The $\kk$-algebra homomorphism
$\FQsym \to \powser{\kk}{M_{\XX \sim \YY}}$ sending every $f \in \FQsym$
to $\overline{\rho}\left(f\left(\XX, \YY\right)\right)
\in \powser{\kk}{M_{\XX \sim \YY}}$ sends every $F_w$ to
\end{comment}

Let $I$ be the totally ordered set
$\left\{1 < 2 < 3 < \cdots\right\}$.
Let $J$ be the totally ordered set \newline
$\left\{1 < 2 < 3 < \cdots < \widetilde 1 < \widetilde 2 < \widetilde 3 < \cdots \right\}$.
We set $X_{\widetilde i} = Y_i$ for every positive integer $i$. Then,
the alphabet $(\XX, \YY)$ can be written as $\left\{X_i\right\}_{i \in J}$.

If $\ii$ is a word over the alphabet $I = \left\{1 < 2 < 3 < \cdots\right\}$,
then we denote by $\widetilde{\ii}$ the word over $J$
obtained from $\ii$ by replacing every letter $i$ by $\widetilde i$.

For the first assertion of Proposition~\ref{prop.FQsym.coproduct.wd},
it suffices to check that $F_w$ indeed has the image under $\Delta$
claimed in \eqref{FQsym-coproduct-on-F's}. Let $n \in \NN$ and
$w \in \Symm_n$. Then,
\begin{align}
\Delta F_w
% \overline{\rho}\left(F_w(\XX,\YY)\right)
 &= F_w\left(\XX, \YY\right)  \qquad \qquad \left(\text{by our abuse of notation}\right) \nonumber \\
 &= \sum_{\ii \in J^n : \std(\ii)=w^{-1} } (\XX,\YY)_{\ii} = \sum_{\mathbf{t} \in J^n : \std(\mathbf{t})=w^{-1} } (\XX,\YY)_{\mathbf{t}} \nonumber \\
 &= \sum_{k=0}^n \sum_{(\ii,\jj) \in I^k \times I^{n-k}} \sum\limits_{\substack{\mathbf{t} \in J^n : \\ \std(\mathbf{t})=w^{-1} ; \\ \mathbf{t} \in \shuf{\ii}{\widetilde{\jj}}}} (\XX,\YY)_{\mathbf{t}}
\label{pf.prop.FQsym.coproduct.wd.1}
\end{align}
(since for every $\mathbf{t} \in J^n$, there exists exactly one choice of $k \in \left\{0,1,\ldots,n\right\}$ and $\left(\ii,\jj\right) \in I^k \times I^{n-k}$ satisfying $\mathbf{t} \in \shuf{\ii}{\widetilde{\jj}}$; namely, $\ii$ is the restriction of $\mathbf{t}$ to the subalphabet $I$ of $J$, whereas $\jj$ is the restriction of $\mathbf{t}$ to $J \setminus I$, and $k$ is the length of $\ii$).

We now fix $k$ and $\left(\ii, \jj\right)$, and try to simplify the inner sum
$\sum\limits_{\substack{\mathbf{t} \in J^n : \\ \std(\mathbf{t})=w^{-1} ; \\ \mathbf{t} \in \shuf{\ii}{\widetilde{\jj}}}} (\XX,\YY)_{\mathbf{t}}$ on the right hand side of \eqref{pf.prop.FQsym.coproduct.wd.1}. First we notice that this sum is nonempty if and only if there exists some $\mathbf{t} \in \shuf{\ii}{\widetilde{\jj}}$ satisfying $\std(\mathbf{t})=w^{-1}$. This existence is easily seen to be equivalent to $w^{-1} \in \shuf{\std(\ii)}{\std(\jj)[k]}$ (since the standardization of any shuffle in $\shuf{\ii}{\widetilde{\jj}}$ is the corresponding shuffle in $\shuf{\std(\ii)}{\std(\jj)[k]}$). This, in turn, is equivalent to $\std(\ii)=(\std(w_1,\ldots,w_k))^{-1}$ and $\std(\jj)=(\std(w_{k+1},\ldots,w_n))^{-1}$ (according to the equivalence (i) $\Longleftrightarrow $ (ii) in Lemma~\ref{standardization-lemma}). Hence, the inner sum on the right hand side of \eqref{pf.prop.FQsym.coproduct.wd.1} is nonempty if and only if $\std(\ii)=(\std(w_1,\ldots,w_k))^{-1}$ and $\std(\jj)=(\std(w_{k+1},\ldots,w_n))^{-1}$. When it is nonempty, it has only one addend\footnote{In fact, the elements $\std\left(\mathbf{t}\right)$ for $\mathbf{t} \in \shuf{\ii}{\widetilde{\jj}}$ are distinct, and thus only one of them can equal $w^{-1}$.}, and this addend is $\left(\XX,\YY\right)_{\mathbf{t}} = \XX_\ii \YY_\jj$ (since $\mathbf{t} \in \shuf{\ii}{\widetilde{\jj}}$). Summarizing, we see that the inner sum on the right hand side of \eqref{pf.prop.FQsym.coproduct.wd.1} equals $\XX_\ii \YY_\jj$ when $\std(\ii)=(\std(w_1,\ldots,w_k))^{-1}$ and $\std(\jj)=(\std(w_{k+1},\ldots,w_n))^{-1}$, and is empty otherwise. Thus, \eqref{pf.prop.FQsym.coproduct.wd.1} simplifies to 
\begin{align*}
\Delta F_w
 &= \sum_{k=0}^n \sum\limits_{\substack{(\ii,\jj) \in I^k \times I^{n-k}:\\
          \std(\ii)=(\std(w_1,\ldots,w_k))^{-1}\\
            \std(\jj)=(\std(w_{k+1},\ldots,w_n))^{-1}}} \XX_\ii \YY_\jj \\
 &= \sum_{k=0}^n F_{\std(w_1,\ldots,w_k)}(\XX) F_{\std(w_{k+1},\ldots,w_n)}(\YY) \\
 &= \sum_{k=0}^n F_{\std(w_1,\ldots,w_k)} \otimes F_{\std(w_{k+1},\ldots,w_n)} \in \FQsym \otimes \FQsym .
\end{align*}
This proves \eqref{FQsym-coproduct-on-F's}, and thus the first assertion
of Proposition~\ref{prop.FQsym.coproduct.wd}.

% [DG][v28] Expanded the above argument (I need the details in front
% of my eyes since I am checking an analogous proof for the
% $m$-free quasisymmetric functions).

\begin{comment}
This proves the first statement of the proposition, in particularly
showing that $\Delta$ is well-defined. 
Of course, we now have
\[
\Delta F_w = \overline{\rho}\left(F_w(\XX,\YY)\right) = \sum_{k=0}^n F_{\std(w_1,\ldots,w_k)} \otimes F_{\std(w_{k+1},\ldots,w_n)},
\]
so that the action of $\Delta$ is indeed given by \eqref{FQsym-coproduct-on-F's}.
\end{comment}

From this, it is easy to derive that $\Delta$ satisfies coassociativity
(i.e., the diagram \eqref{coassociativity-diagram} holds for
$C = \FQsym$). (Alternatively, one can obtain this from the
associativity of multiplication using Corollary~\ref{FQsym-self-dual}.)
We have already verified the rule \eqref{FQsym-coproduct-on-F's}.
The connected graded structure on $\FQsym$ gives a counit and an antipode
for free.
\end{proof}

%[VR][v20] This ends the area where Vic made changes to the alphabet-doubling arguments. 

\begin{exercise}
\label{exe.FQSym.free}
We say that a permutation $w\in \Symm_{n}$ is
\emph{connected}\index{connected permutation}
if $n$ is a positive integer and if there exists no
$i\in\left\{  1,2,\ldots,n-1\right\}  $ satisfying $f\left(  \left\{
1,2,\ldots,i\right\}  \right)  =\left\{  1,2,\ldots,i\right\}  $. Let $\mathfrak{CS}%
$ denote the set of all connected permutations of all $n\in \NN$. Show
that $ \FQsym $ is a free (noncommutative) $\kk$-algebra
with generators $\left(  F_{w}\right)  _{w\in\mathfrak{CS}}$. (This statement
means that $\left(  F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}\right)  _{k\in
 \NN ;\ \left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}}$
is a basis of the $\kk$-module $\FQsym$.)

[\textbf{Hint:} This is a result of Poirier and Reutenauer \cite[Theorem
2.1]{PoirierReutenauer}; it is much easier than the similar Theorem
\ref{thm.QSym.lyndon}.]
\end{exercise}

% [DG][v37] Added above exercise.

\begin{corollary}
\label{FQsym-self-dual}
The Hopf algebra $\FQsym$ is self-dual:
Let $\{G_w\}$ be the dual $\kk$-basis
to the $\kk$-basis $\{F_w\}$ for $\FQsym$. Then,
the $\kk$-linear map sending $G_w \longmapsto F_{w^{-1}}$
is a Hopf algebra isomorphism $\FQsym^{o} \longrightarrow \FQsym$.
\end{corollary}
\begin{proof}
For any $0 \leq k \leq n$, any $u \in \Symm_k$ and any
$v \in \Symm_{n-k}$, one has
\[
F_{u^{-1}} F_{v^{-1}}
= \sum_{w^{-1} \in \shuf{u^{-1}}{v^{-1}[k]}} F_{w^{-1}}
= \sum\limits_{\substack{w \in \Symm_n:\\
           \std(w_1,\ldots,w_k)=u \\
           \std(w_{k+1},\ldots,w_n)=v }} F_{w^{-1}}
\]
via the equivalence of (i) and (ii) in Lemma~\ref{standardization-lemma}.
On the other hand, in $\FQsym^o$, the dual $\kk$-basis $\{G_w\}$
to the $\kk$-basis $\{F_w\}$ for $\FQsym$ should have product
formula
\[
G_u G_v = \sum\limits_{\substack{w \in \Symm_n:\\
           \std(w_1,\ldots,w_k)=u \\
           \std(w_{k+1},\ldots,w_n)=v }} G_w
\]
coming from the coproduct formula \eqref{FQsym-coproduct-on-F's}
for $\FQsym$ in the $\{F_w\}$-basis. Comparing these equalities,
we see that the $\kk$-linear map $\tau$ sending $G_w \longmapsto F_{w^{-1}}$
is an isomorphism $\FQsym^{o} \longrightarrow \FQsym$ of
$\kk$-algebras. Hence, the adjoint $\tau^\ast : \FQsym^{o}
\to \left(\FQsym^{o}\right)^{o}$ of this map is an isomorphism
of $\kk$-coalgebras. But identifying $\left(\FQsym^{o}\right)^{o}$
with $\FQsym$ in the natural way (since $\FQsym$ is of finite type),
we easily see that $\tau^\ast = \tau$, whence $\tau$ itself is
an isomorphism of both $\kk$-algebras and $\kk$-coalgebras, hence
of $\kk$-bialgebras, hence of Hopf algebras.
\end{proof}

% [DG][v25] Revamped the above corollary. It used to speak of a map
% $\FQsym \longrightarrow \FQsym^{o}$ sending $F_w$ to $F_{w^{-1}}$,
% which was more than confusing as there was no identification of
% $\FQsym$ with its dual given a priori. Added some more details to
% the proof.

We can now be a bit more precise about the relations between
the various algebras
\[
\Lambda, \Qsym, \Nsym, \FQsym, R \langle \XX \rangle, R(\xx).
\]
Not only does $\FQsym$ allow one to \emph{lift} the Hopf structure
of $\Qsym$, it dually allows one to \emph{extend} the Hopf structure
of $\Nsym$.  To set up this duality,
note that Corollary~\ref{FQsym-self-dual}
motivates the choice of an inner product on $\FQsym$ in which
\[
(F_u, F_v):=\delta_{u^{-1},v}.
\]
We wish to identify the images of the ribbon basis
$\{R_\alpha\}$ of $\Nsym$ when included in $\FQsym$.


\begin{definition}
For any composition $\alpha$, define an element
\dfn{$\RRR_\alpha$} of $\FQsym$ by
% the \dfn{free quasi-ribbon function}
\[
\RRR_\alpha
:=\sum_{\substack{w \in \Symm_{\left|\alpha\right|}: \\ \Des(w)=D(\alpha) }} F_{w^{-1}}
=\sum\limits_{\substack{(w,\ii): \\ w \in \Symm_{\left|\alpha\right|}; \\ \Des(w)=D(\alpha);\\ \std(\ii)=w}} \XX_\ii
=\sum_{\ii: \Des(\ii)=D(\alpha)} \XX_\ii ,
\]
% where the $w$ in the sums are supposed to belong to $\Symm_{\left|\alpha\right|}$, and
where the \emph{descent set}\index{descent set of a sequence}
of a sequence $\ii=(i_1,\ldots,i_n)$
is defined by
\[
\Des(\ii):=\{j \in \{1,2,\ldots,n-1\}: i_j > i_{j+1} \}=
\Des(\std(\ii)).
\]
Alternatively,
\begin{equation}
\RRR_\alpha= \sum_T \XX_T
\label{eq.FQSym.ribbon.tableau-sum}
\end{equation}
in which the sum is over column-strict tableaux of
the ribbon skew shape $\Rib\left(\alpha\right)$,
and $\XX_T=\XX_{\ii}$
in which $\ii$ is the sequence of entries of $T$ read in
order from the southwest toward the northeast.

\end{definition}

% [DG][v14] I commented out "the \dfn{free quasi-ribbon function}" because
% it seems to me that this notation is used for the $F_w$ in NCSF6 and in
% other sources, but never for the $\RRR_\alpha$.


\begin{example}
Taking $\alpha=(1,3,2)$, with ribbon shape and column-strict fillings $T$ as
shown:
\[
\Rib\left(\alpha\right) =
\begingroup % keep the change local
\setlength\arraycolsep{0.2pc} % Make the spacing equal in rows and in columns.
\begin{matrix}
   &   &\sq&\sq\\
\sq&\sq&\sq& \\
\sq&   &   &
\end{matrix}
\endgroup
\qquad\qquad \text{and} \qquad\qquad
T=\quad
\begin{matrix}
   &   &    &   &i_5&\leq&i_6\\
   &   &    &   &\wedge&    &   \\
i_2&\leq&i_3&\leq&i_4& \\
\wedge&   &   &     &  & \\
i_1&   &   &     &  &
\end{matrix}
\]
one has that
\[
\RRR_{(1,3,2)}=
\sum\limits_{\substack{ \ii=(i_1,i_2,i_3,i_4,i_5,i_6):\\
                   \Des(\ii)=D(\alpha)=\{1,4\}}} \XX_\ii
=\sum_{i_1 > i_2 \leq i_3 \leq i_4 > i_5 \leq i_6}
   X_{i_1}  X_{i_2} X_{i_3}  X_{i_4}  X_{i_5} X_{i_6}
=\sum_T \XX_T .
\]
\end{example}




\begin{corollary}
\label{cor.FQsym.pi-iota}
For every $n \in \NN$ and $w \in \Symm_n$, we let \dfn{$\gamma(w)$}
denote the unique composition $\alpha$ of $n$ satisfying
$D\left(\alpha\right) = \Des\left(w\right)$.
% [DG][v50] Added the preceding sentence; the $\gamma(w)$ notation
% is used a few lines below.
% Also, slightly reorganized the next sentence.
\begin{enumerate}
\item[(a)]
The $\kk$-linear map\index{$\pi : \FQsym \to \Qsym$}
\[
\begin{array}{rcl}
\FQsym &\overset{\pi}{\twoheadrightarrow}& \Qsym , \\
F_w & \longmapsto &L_{\gamma(w)} \\
\end{array}
\]
is a surjective Hopf algebra homomorphism.
\item[(b)]
The $\kk$-linear map\index{$\iota : \Nsym \to \FQsym$}
\[
\begin{array}{rcl}
\Nsym &\overset{\iota}{\hookrightarrow}& \FQsym , \\
R_\alpha & \longmapsto & \RRR_\alpha
\end{array}
\]
is an injective Hopf algebra homomorphism.
\item[(c)]
The linear maps $\pi$ and $\iota$ are adjoint maps with respect to the
above choice of inner product on $\FQsym$
and the usual dual pairing between $\Nsym$ and $\Qsym$.
\end{enumerate}
Now, consider the abelianization map
$\operatorname{ab} : R\langle \XX \rangle \twoheadrightarrow R(\xx)$
defined as the continuous $\kk$-algebra homomorphism
sending the noncommutative variable $X_i$ to the commutative $x_i$.
\begin{enumerate}
\item[(d)]
The map $\pi$ is a restriction of $\operatorname{ab}$.
\item[(e)]
The map $\iota$ lets one factor
the surjection $\Nsym \twoheadrightarrow \Lambda$ as follows:
\[
\begin{array}{rcccl}
\Nsym & \rightarrow & \FQsym \hookrightarrow R\langle \XX \rangle & \overset{\operatorname{ab}}{\rightarrow} & R(\xx) , \\
 R_\alpha &\longmapsto & \RRR_\alpha & \longmapsto & s_{\Rib\left(\alpha\right)}(\xx) .
\end{array}
\]
\end{enumerate}
\end{corollary}

\begin{proof}
Given $n \in \NN$, each composition $\alpha$ of $n$ can be written in
the form $\gamma\left(w\right)$ for some $w \in \Symm_n$.
\ \ \ \ \footnote{Indeed, write our composition $\alpha$
as $\left(\alpha_1, \alpha_2, \ldots, \alpha_k\right)$.
Then, we can pick $w$ to be the permutation whose first
$\alpha_1$ entries are the largest $\alpha_1$ elements of
$\left\{1,2,\ldots,n\right\}$ in increasing order;
whose next $\alpha_2$ entries are the next-largest $\alpha_2$
elements of $\left\{1,2,\ldots,n\right\}$ in increasing order;
and so on.
This permutation $w$ will satisfy
$\Des\left(w\right) = \left\{\alpha_1, \alpha_1+\alpha_2, \ldots, \alpha_1+\alpha_2+\cdots+\alpha_{k-1}\right\}
= D\left(\alpha\right)$ and thus
$\gamma\left(w\right) = \alpha$.}
Hence, each fundamental quasisymmetric function $L_\alpha$ lies
in the image of $\pi$. Thus, $\pi$ is surjective.

Also, for each $n \in \NN$ and $\alpha \in \Comp_n$,
the element $\RRR_\alpha$ is a nonempty sum of noncommutative
monomials (nonempty because $\alpha$ can be written in
the form $\gamma\left(w\right)$ for some $w \in \Symm_n$).
Moreover, the elements $\RRR_\alpha$ for varying $n$ and
$\alpha$ are supported on disjoint monomials.
Thus, these elements are linearly independent.
Hence, the map $\iota$ is injective.

(d) Let $\mathfrak{A}$ denote the totally ordered set
$\left\{  1<2<3<\cdots\right\}  $ of positive integers.
For each word
$w = \left( w_1, w_2, \ldots, w_n \right) \in
\mathfrak{A}^{n}$, we define a monomial
$\xx_w$ in $\kk \left[\left[\xx\right]\right]$
by $\xx_w = x_{w_1} x_{w_2} \cdots x_{w_n}$.

Let $n \in \NN$ and $\sigma \in \Symm_n$. Then,
\[
L_{\gamma\left( \sigma \right)  }
= \sum\limits_{\substack{w\in\mathfrak{A}^{n};
\\ \std w=\sigma^{-1}}} \xx_w
\]
(by Lemma~\ref{lem.Lalpha-std.std-L}).
But \eqref{eq.FQSym.Fw-as-series} (applied to $w = \sigma$) yields
\begin{align*}
F_{\sigma}
&= \sum\limits_{\substack{\ii=(i_1,\ldots,i_n):\\ \std(\ii) = \sigma^{-1}}} \XX_\ii
= \sum\limits_{\substack{w\in\mathfrak{A}^{n};\\ \std w=\sigma^{-1}}}\XX_{w}
\end{align*}
and thus
\begin{align*}
\operatorname{ab}\left(F_{\sigma}\right)
&= \operatorname{ab}\left(\sum\limits_{\substack{w\in\mathfrak{A}^{n};
\\ \std w=\sigma^{-1}}}\XX_{w}\right)
= \sum\limits_{\substack{w\in\mathfrak{A}^{n};\\ \std w=\sigma^{-1}}} \underbrace{\operatorname{ab}\left(\XX_w\right)}_{=\xx_w}
= \sum\limits_{\substack{w\in\mathfrak{A}^{n}; \\ \std w=\sigma^{-1}}} \xx_w
= L_{\gamma\left( \sigma \right)  }
= \pi\left(F_\sigma\right) .
\end{align*}
We have shown this for all $n \in \NN$ and $\sigma \in \Symm_n$.
Thus, $\pi$ is a restriction of $\operatorname{ab}$.
This proves Corollary~\ref{cor.FQsym.pi-iota}(d).

(a) Let $n \in \NN$ and $w = \left(w_1, w_2, \ldots, w_n\right) \in \Symm_n$.
Let $\alpha$ be the composition $\gamma\left(w\right)$ of $n$.
Thus, the definition of $\pi$ yields
$\pi\left(F_w\right) = L_\alpha$.
But applying the map $\pi \otimes \pi$ to the equality
\eqref{FQsym-coproduct-on-F's}, we obtain
\begin{align}
\left(\pi \otimes \pi\right) \left(\Delta F_w\right)
&= \left(\pi \otimes \pi\right)
    \left( \sum_{k=0}^n F_{\std(w_1,w_2,\ldots,w_k)}
                \otimes F_{\std(w_{k+1},w_{k+2},\ldots,w_n)}
   \right) \nonumber\\
&= \sum_{k=0}^n \pi\left(F_{\std(w_1,w_2,\ldots,w_k)}\right) \otimes \pi\left(F_{\std(w_{k+1},w_{k+2},\ldots,w_n)}\right) \nonumber\\
&= \sum_{k=0}^n L_{\gamma\left(\std(w_1,w_2,\ldots,w_k)\right)} \otimes L_{\gamma\left(\std(w_{k+1},w_{k+2},\ldots,w_n)\right)}
\label{pf.cor.FQsym.pi-iota.a.1}
\end{align}
(by the definition of $\pi$).
Now, for each $k \in \left\{0,1,\ldots, n\right\}$, the two
compositions $\gamma\left(\std(w_1,w_2,\ldots,w_k)\right)$
$\gamma\left(\std(w_{k+1},w_{k+2},\ldots,w_n)\right)$ form
a pair $\left(\beta, \gamma\right)$ of compositions
satisfying\footnote{See Definition~\ref{def.compositions.near-conc}
for the notation we are using.}
either $\beta \cdot \gamma=\alpha$ or $\beta \odot \gamma=\alpha$,
and in fact they form the only such pair satisfying
$\left|\beta\right| = k$ and $\left|\gamma\right| = n-k$.
Thus, the right hand side of \eqref{pf.cor.FQsym.pi-iota.a.1}
can be rewritten as
\[
\sum\limits_{\substack{(\beta,\gamma):\\
\beta \cdot \gamma=\alpha \text{ or }
\beta \odot \gamma=\alpha}} L_\beta \otimes L_\gamma .\]
But this sum is $\Delta L_\alpha$, as we know from
\eqref{Qsym-coproduct-on-fundamentals}.
Hence, \eqref{pf.cor.FQsym.pi-iota.a.1} becomes
\[
\left(\pi \otimes \pi\right) \left(\Delta F_w\right)
= \Delta L_\alpha = \Delta \left(\pi\left(F_w\right)\right)
\qquad\qquad \left(\text{since $L_\alpha = \pi\left(F_w\right)$}\right).
\]

We have proven this for each $n \in \NN$ and $w \in \Symm_n$.
Thus, we have proven that
$\left(\pi \otimes \pi\right) \circ \Delta_{\FQsym}
= \Delta_{\Qsym} \circ \pi$.
Combined with $\epsilon_{\FQsym} = \epsilon_{\Qsym} \circ \pi$
(which is easy to check), this shows that $\pi$ is a
coalgebra homomorphism.

We can similarly see that $\pi$ is an algebra homomorphism
by checking that it respects the product
(compare \eqref{Qsym-product-on-fundamentals}
and \eqref{FQsym-product-on-F's}).
However, this also follows trivially from
Corollary~\ref{cor.FQsym.pi-iota}(d).

Thus, $\pi$ is a bialgebra morphism, and therefore a Hopf
algebra morphism (by
Corollary~\ref{cor.bialg-mor-is-Hopf}).
This proves Corollary~\ref{cor.FQsym.pi-iota}(a).

(c) For any composition $\alpha$ and any $w \in \Symm$,
we have
\begin{align*}
(\iota(R_\alpha), F_w)
&=(\RRR_\alpha,F_w)
=\sum_{u: \Des(u)=D(\alpha)} (F_{u^{-1}}, F_w)
=
\begin{cases}
1, &\text{ if }\Des(w)=D(\alpha) ; \\
0, &\text{ otherwise }
\end{cases}
\quad
=
\begin{cases}
1, &\text{ if }\gamma(w)=\alpha ; \\
0, &\text{ otherwise }
\end{cases}
\\
&=(R_\alpha, L_{\gamma(w)})
=(R_\alpha,\pi(F_w)).
\end{align*}
Thus, the maps $\pi$ and $\iota$ are adjoint.
This proves Corollary~\ref{cor.FQsym.pi-iota}(c).

(b) Again, there are several ways to prove this.
Here is one:

First, note that $\iota\left(1\right) = 1$
(because $R_{\varnothing} = 1$ and $\RRR_{\varnothing} = 1$).
Next, let
$\alpha$ and $\beta$ be two nonempty compositions.
Let $m = \left|\alpha\right|$ and $n = \left|\beta\right|$.
Then,
$R_\alpha R_\beta = R_{\alpha \cdot \beta} + R_{\alpha \odot \beta}$
(by \eqref{Nsym-product-in-ribbons}) and thus
\begin{align}
\iota\left(R_\alpha R_\beta\right)
 &= \iota\left(R_{\alpha \cdot \beta} + R_{\alpha \odot \beta}\right)
 = \underbrace{\iota\left(R_{\alpha \cdot \beta}\right)}_{= \RRR_{\alpha \cdot \beta} = \sum_{\ii: \Des(\ii)=D(\alpha \cdot \beta)} \XX_\ii}
  + \underbrace{\iota\left(R_{\alpha \odot \beta}\right)}_{= \RRR_{\alpha \odot \beta} = \sum_{\ii: \Des(\ii)=D(\alpha \odot \beta)} \XX_\ii} \nonumber\\
 &= \sum_{\ii: \Des(\ii)=D(\alpha \cdot \beta)} \XX_\ii + \sum_{\ii: \Des(\ii)=D(\alpha \odot \beta)} \XX_\ii
 = \sum_{\ii: \Des(\ii)=D(\alpha \cdot \beta) \text{ or } \Des(\ii)=D(\alpha \odot \beta)} \XX_\ii \nonumber\\
 &= \sum\limits_{\substack{\ii = \left(i_1, i_2, \ldots, i_{m+n}\right): \\ \Des\left(i_1, i_2, \ldots, i_m\right) = D\left(\alpha\right) \text{ and} \\ \Des\left(i_{m+1}, i_{m+2}, \ldots, i_{m+n}\right) = D\left(\beta\right)}} \XX_\ii
\label{pf.cor.FQsym.pi-iota.b.1}
\end{align}
(since the words $\ii$ of length $m+n$ satisfying
$\Des(\ii)=D(\alpha \cdot \beta)$ or $\Des(\ii)=D(\alpha \odot \beta)$
are precisely the words $\ii = \left(i_1, i_2, \ldots, i_{m+n}\right)$
satisfying
$\Des\left(i_1, i_2, \ldots, i_m\right) = D\left(\alpha\right)$ and
$\Des\left(i_{m+1}, i_{m+2}, \ldots, i_{m+n}\right) = D\left(\beta\right)$).
But choosing a word $\ii = \left(i_1, i_2, \ldots, i_{m+n}\right)$
satisfying $\Des\left(i_1, i_2, \ldots, i_m\right) = D\left(\alpha\right)$
and
$\Des\left(i_{m+1}, i_{m+2}, \ldots, i_{m+n}\right) = D\left(\beta\right)$
is tantamount to choosing a pair $\left(\mathbf{u}, \mathbf{v}\right)$
of a word $\mathbf{u} = \left(i_1, i_2, \ldots, i_m\right)$ satisfying
$\Des\mathbf{u} = D\left(\alpha\right)$ and a word
$\mathbf{v} = \left(i_{m+1}, i_{m+2}, \ldots, i_{m+n}\right)$ satisfying
$\Des\mathbf{v} = D\left(\beta\right)$.
Thus, \eqref{pf.cor.FQsym.pi-iota.b.1} becomes
\begin{align*}
\iota\left(R_\alpha R_\beta\right)
 &= \sum\limits_{\substack{\ii = \left(i_1, i_2, \ldots, i_{m+n}\right): \\ \Des\left(i_1, i_2, \ldots, i_m\right) = D\left(\alpha\right) \text{ and} \\ \Des\left(i_{m+1}, i_{m+2}, \ldots, i_{m+n}\right) = D\left(\beta\right)}} \XX_\ii
 =      \sum_{\mathbf{u} : \Des\mathbf{u} = D\left(\alpha\right)}
    \ \ \sum_{\mathbf{v} : \Des\mathbf{v} = D\left(\beta\right)}
        \XX_{\mathbf{u}} \XX_{\mathbf{v}} \\
 &= \underbrace{\left(\sum_{\mathbf{u} : \Des\mathbf{u} = D\left(\alpha\right)} \XX_{\mathbf{u}}\right)}_{= \RRR_\alpha = \iota\left(R_\alpha\right)}
    \underbrace{\left(\sum_{\mathbf{v} : \Des\mathbf{v} = D\left(\beta\right)} \XX_{\mathbf{v}}\right)}_{= \RRR_\beta = \iota\left(R_\beta\right)}
 = \iota\left(R_\alpha\right) \iota\left(R_\beta\right) .
\end{align*}
Thus, we have proven the equality
$\iota\left(R_\alpha R_\beta\right)
= \iota\left(R_\alpha\right) \iota\left(R_\beta\right)$
whenever $\alpha$ and $\beta$ are two nonempty compositions.
It also holds if we drop the ``nonempty'' requirement
(since $R_{\varnothing} = 1$ and $\iota\left(1\right) = 1$).
Thus, the $\kk$-linear map $\iota$ respects the multiplication.
Since $\iota\left(1\right) = 1$, this shows that $\iota$
is a $\kk$-algebra homomorphism.

For each $n \in \NN$, we let $\id_n$ be the identity
permutation in $\Symm_n$.
Next, we observe that each $n \in \NN$ satisfies
$H_n = R_{\left(n\right)}$ (this follows, e.g., from
\eqref{H-as-sum-of-ribbons}, because the composition
$\left(n\right)$ is coarsened only by itself).
Hence, each $n \in \NN$ satisfies
\begin{align}
\iota\left(H_n\right)
&= \iota\left(R_{\left(n\right)}\right)
= \RRR_{\left(n\right)}
= \sum_{\substack{w \in \Symm_n:\\ \Des(w)=D\left(\left(n\right)\right) }} F_{w^{-1}} \nonumber\\
&= F_{\id_n^{-1}}
\qquad \left(\text{since the only $w \in \Symm_n$ satisfying $\Des(w)=D\left(\left(n\right)\right)$ is $\id_n$}\right) \nonumber\\
&= F_{\id_n} .
\label{pf.cor.FQsym.pi-iota.b.iotaH}
\end{align}

In order to show that $\iota$ is a $\kk$-coalgebra
homomorphism, it suffices to check the equalities
$\left(\iota \otimes \iota\right) \circ \Delta_{\Nsym}
= \Delta_{\FQsym} \circ \iota$
and
$\epsilon_{\Nsym} = \epsilon_{\FQsym} \circ \iota$.
We shall only prove the first one, since the second is easy.
Since $\iota$, $\Delta_{\Nsym}$ and $\Delta_{\FQsym}$
are $\kk$-algebra homomorphisms, it suffices to check it
on the generators $H_1, H_2, H_3, \ldots$ of $\Nsym$.
But on these generators, it follows from comparing
\begin{align*}
\left(\left(\iota \otimes \iota\right) \circ \Delta_{\Nsym}\right) \left(H_n\right)
&= \left(\iota \otimes \iota\right) \left( \Delta_{\Nsym} H_n \right)
= \left(\iota \otimes \iota\right) \left( \sum_{i+j=n} H_i \otimes H_j \right)
\qquad \left(\text{by \eqref{Nsym-coproduct-on-H}}\right) \\
&= \sum_{i+j=n} \underbrace{\iota\left(H_i\right)}_{\substack{= F_{\id_i} \\ \text{(by \eqref{pf.cor.FQsym.pi-iota.b.iotaH})}}}
  \otimes \underbrace{\iota\left(H_j\right)}_{\substack{= F_{\id_j} \\ \text{(by \eqref{pf.cor.FQsym.pi-iota.b.iotaH})}}}
= \sum_{i+j=n} F_{\id_i} \otimes F_{\id_j}
= \sum_{k=0}^n F_{\id_k} \otimes F_{\id_{n-k}}
\end{align*}
with
\begin{align*}
\left(\Delta_{\FQsym} \circ \iota\right) \left(H_n\right)
&= \Delta_{\FQsym} \left(\iota\left(H_n\right)\right)
= \Delta_{\FQsym} \left( F_{\id_n} \right)
\qquad \left(\text{by \eqref{pf.cor.FQsym.pi-iota.b.iotaH}}\right) \\
&= \sum_{k=0}^n F_{\id_k} \otimes F_{\id_{n-k}}
\qquad \left(\text{by \eqref{FQsym-coproduct-on-F's}}\right) .
\end{align*}

Thus, we know that $\iota$ is a $\kk$-algebra homomorphism
and a $\kk$-coalgebra homomorphism.
Hence, $\iota$ is a bialgebra morphism, and therefore a Hopf
algebra morphism
(by Corollary~\ref{cor.bialg-mor-is-Hopf}).
This proves Corollary~\ref{cor.FQsym.pi-iota}(b).

An alternative proof of Corollary~\ref{cor.FQsym.pi-iota}(b)
can be obtained by adjointness from
Corollary~\ref{cor.FQsym.pi-iota}(a).
Both the inner product on $\FQsym$ and the dual pairing
$\left(\cdot, \cdot\right) : \Nsym \otimes \Qsym \to \kk$
respect the Hopf structures (i.e., the maps
$\Delta_{\Nsym}$ and $m_{\Qsym}$ are mutually adjoint with
respect to these forms, and so are the maps
$m_{\Nsym}$ and $\Delta_{\Qsym}$, and the maps
$\Delta_{\FQsym}$ and $m_{\FQsym}$, and so on).
Corollary~\ref{cor.FQsym.pi-iota}(c) shows that the map
$\iota$ is adjoint to the map $\pi$ with respect to
these two bilinear forms.
Hence, we have a commutative diagram
\[
\xymatrix{
\Nsym \arinjrev[r]^{\iota} \ar[d]^{\cong} & \FQsym \ar[d]^{\cong} \\
\Qsym^o \ar[r]_{\pi^{\ast}} & \FQsym^o
}
\]
of Hopf algebras (where the two vertical arrows are the isomorphisms
induced by the two bilinear forms).
Thus, Corollary~\ref{cor.FQsym.pi-iota}(b) follows from
Corollary~\ref{cor.FQsym.pi-iota}(a) by duality.

(e) For each composition $\alpha$,
the abelianization map $\operatorname{ab}$
sends the noncommutative tableau monomial $\XX_T$ to the commutative tableau
monomial $\xx_T$
whenever $T$ is a tableau of ribbon shape $\Rib\left(\alpha\right)$.
Thus, $\operatorname{ab}$ sends $\RRR_\alpha$ to
$s_{\Rib\left(\alpha\right)}(\xx)$ (because of the formula
\eqref{eq.FQSym.ribbon.tableau-sum}).
Hence, the composition
$\Nsym \rightarrow \FQsym \hookrightarrow R\langle \XX \rangle \overset{\operatorname{ab}}{\rightarrow} R(\xx)$
does indeed send $R_\alpha$ to $s_{\Rib\left(\alpha\right)}(\xx)$.
But so does the projection $\pi : \Nsym \to \Lambda$,
according to Theorem~\ref{Nsym-structure-on-ribbons-theorem}(b).
Hence, the composition factors the projection.
This proves Corollary~\ref{cor.FQsym.pi-iota}(e).
\end{proof}

We summarize some of this picture as follows:
\[
\xymatrix@M+2pt{
\FQsym \ar@{.}[rr]^{\text{dual}} &         & \FQsym \arsurj[d]^{\pi}\\
\Nsym \ar@{.}[rr]^{\text{dual}} \arsurj[d]^{\pi} \arinjrev[u]_{\iota} &         & \Qsym \\
\Lambda \ar@{.}[rr]^{\text{dual}} & & \Lambda \arinjrev[u]
}
\]
Furthermore, if we denote by $\iota$ the canonical inclusion
$\Lambda \to \Qsym$ as well, then the diagram
\[
\xymatrix{
& \FQsym \arsurj[dr]^{\pi}\\
\Nsym \arsurj[dr]^{\pi} \arinjrev[ur]_{\iota} &         & \Qsym \\
       &\Lambda\arinjrev[ur]_{\iota}& 
}
\]
is commutative (according to
Corollary~\ref{cor.FQsym.pi-iota}(e)).

\begin{remark}
Different notations for $\FQsym$ appear in the literature. In the book
\cite{BlessenohlLaue} (which presents an unusual approach to the
character theory of the symmetric group using $\FQsym$), the Hopf
algebra $\FQsym$ is called $\mathcal{P}$, and its basis that we call
$\left\{G_w\right\}_{w\in\Symm_n}$ is denoted
$\left\{w\right\}_{w\in\Symm_n}$. In
\cite[Chapter 7]{HazewinkelGubareniKirichenko}, the Hopf algebra
$\FQsym$ and its basis $\left\{F_w\right\}_{w\in\Symm_n}$ are denoted
$MPR$ and $\left\{w\right\}_{w\in\Symm_n}$, respectively.
\end{remark}

% [DG][v58] Added the above remark.




% [DG][v25] What do you think about moving all the following (skeletal)
% sections inside "Further topics" before arXiving (lowering them by 1
% level, so they become subsections)?

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Further topics}
\label{further-topics-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The following is a list of topics that were, at one point, planned to
be touched in class, but did not make the cut. They might get
elaborated upon in a future version of these notes.

% [DG][v58] Added the above paragraph, replacing the drafty
% "Some of these we may touch on in class, others are appropriate for student talks.".
% Also, added references for the topics below.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{$0$-Hecke algebras}
\label{0-Hecke-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{itemize}

\item \textbf{Review of representation theory of finite-dimensional algebras.}

Review the notions of indecomposables, simples, projectives, along
with the theorems of Krull-Remak-Schmidt, of Jordan-H\"older,
and the two kinds of Grothendieck groups dual to each other.

\item \textbf{$0$-Hecke algebra representation theory.}

Describe  the simples and projectives, following
Denton, Hivert, Schilling, Thiery \cite{DentonHivertSchillingThiery}
on $\JJJ$-trivial monoids.

\item \textbf{Nsym and Qsym as Grothendieck groups.}

Give Krob and Thibon's interpretation (see \cite[\S 5]{Thibon} for a
brief summary) of
\begin{enumerate}
\item[$\bullet$]
$\Qsym$ and the Grothendieck group of
composition series, and
\item[$\bullet$]
$\Nsym$ and the Grothendieck group of projectives.
\end{enumerate}

\begin{remark}
Mention P. McNamara's interpretation, in the case of \emph{supersolvable lattices}, of the Ehrenborg quasisymmetric function as the composition series enumerator for an $H_n(0)$-action on the maximal chains
\end{remark}

\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Aguiar-Bergeron-Sottile character theory Part II: Odd and even
characters, subalgebras}
\label{ABS-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Face enumeration, Eulerian posets, and cd-indices}
\label{cd-index-section}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Borrowing from Billera's ICM notes \cite{Billera}.

\begin{itemize}

\item f-vectors, h-vectors

\item flag f-vectors, flag h-vectors

\item ab-indices and cd-indices

\end{itemize}

% [DG][v37] I've put the unwritten parts here, into the "Further
% topics" list.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Other topics}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{itemize}
\item Loday-Ronco Hopf algebra of planar binary trees \cite{LodayRonco-trees}
\item Poirier-Reutenauer Hopf algebra of tableaux
\item Reading Hopf algebra of Baxter permutations
\item Hopf monoids, e.g. of Hopf algebra of generalized permutohedra, of matroids, of graphs, Stanley chromatic symmetric functions and Tutte polynomials
\item Lam-Pylyavskyy Hopf algebra of set-valued tableaux
\item Connes-Kreimer Hopf algebra and renormalization
\item Noncommutative symmetric functions and $\Omega \Sigma \CC P^\infty$
\item Maschke's theorem and ``integrals'' for Hopf algebras
\item Nichols-Zoeller structure theorem and group-like elements
\item Cartier-Milnor-Moore structure theorem and primitive elements
\item Quasi-triangular Hopf algebras and quantum groups
\item The Steenrod algebra, its dual, and tree Hopf algebras
\item Ringel-Hall algebras of quivers
\item Ellis-Khovanov odd symmetric function Hopf algebras \cite{EllisKhovanov}
(see also Lauda-Russell \cite{LaudaRussell})
\end{itemize}

Student talks given in class were:
\begin{enumerate}
\item Al Garver, on Maschke's theorem for finite-dimensional Hopf algebras
\item Jonathan Hahn, on the paper by Humpert and Martin.
\item Emily Gunawan, on the paper by Lam, Lauve and Sottile.
\item Jonas Karlsson, on the paper by Connes and Kreimer
\item Thomas McConville, on Butcher's group and generalized Runge-Kutta methods.
\item Cihan Bahran, on universal enveloping algebras and the Poincar\'e-Birkhoff-Witt theorem.
\item Theodosios Douvropolos, on the Cartier-Milnor-Moore theorem.
\item Alex Csar, on the Loday-Ronco Hopf algebra of binary trees
\item Kevin Dilks, on Reading's Hopf algebra of (twisted) Baxter permutations
\item Becky Patrias, on the paper by Lam and Pylyavskyy
\item Meng Wu, on multiple zeta values and Hoffman's homomorphism from $\Qsym$
\end{enumerate}




\section{Some open problems and conjectures}
\begin{itemize}
\item Is there a proof of the Assaf-McNamara skew Pieri rule that
gives a resolution of Specht or Schur/Weyl modules whose character
corresponds to $s_{\lambda/\mu} h_n$, whose terms model their alternating sum?

\item Explicit antipodes in the Lam-Pylyavskyy Hopf algebras? (Answered by Patrias in \cite{Patrias-antipode}.)

\item P. McNamara's question \cite[Question 7.1]{McNamaraWard}:  are $P$-partition enumerators irreducible for connected posets $P$?

\item Stanley's question:  are the only $P$-partition enumerators
which are symmetric (not just quasisymmetric) those for which $P$ is
a skew shape with a column-strict labelling?

\item Does Stanley's chromatic symmetric function distinguish trees?

\item Hoffman's stuffle conjecture

\item Billera-Brenti's nonnegativity conjecture for
the total $cd$-index of  Bruhat intervals
(\cite[Conjecture 6.1]{BilleraBrenti})

\end{itemize}

\newpage

\section{\label{chp.appendix}Appendix: Some basics}

In this appendix, we briefly discuss some basic notions from linear algebra
and elementary combinatorics that are used in these notes.

\subsection{\label{sect.STmat}Linear expansions and triangularity}

In this Section, we shall recall some fundamental results from linear algebra
(most importantly, the notions of a change-of-basis matrix and of a
unitriangular matrix), but in greater generality than how it is usually done
in textbooks. We shall use these results later when studying bases of
combinatorial Hopf algebras; but per se, this section has nothing to do with
Hopf algebras.

\subsubsection{Matrices}

Let us first define the notion of a matrix whose rows and columns are indexed
by arbitrary objects (as opposed to numbers):\footnote{As before, $\kk$
denotes a commutative ring.}

\begin{definition}
\label{def.STmat.STmat}
Let $S$ and $T$ be two sets. An \dfn{$S\times T$-matrix over $\kk$}
shall mean a family
$\left(  a_{s,t} \right)  _{\left(  s,t\right)  \in S\times T}
\in \kk^{S\times T}$ of
elements of $\kk$ indexed by elements of $S\times T$. Thus, the set of
all $S\times T$-matrices over $\kk$ is $\kk^{S\times T}$.

We shall abbreviate ``$S\times T$-matrix over $\kk$'' by
``$S\times T$-matrix'' when the value of $\kk$ is clear from the context.
\end{definition}

This definition of $S\times T$-matrices generalizes the usual notion of
matrices (i.e., the notion of $n\times m$-matrices): Namely, if $n\in
 \NN$ and $m\in \NN$, then the $\left\{  1,2,\ldots,n\right\}
\times\left\{  1,2,\ldots,m\right\}  $-matrices are precisely the $n\times
m$-matrices (in the usual meaning of this word). We shall often use the word
``\dfn{matrix}'' for both the usual notion
of matrices and for the more general notion of $S\times T$-matrices.

Various concepts defined for $n\times m$-matrices (such as addition and
multiplication of matrices, or the notion of a row) can be generalized to
$S\times T$-matrices in a straightforward way. The following four definitions
are examples of such generalizations:

\begin{definition}
\label{def.STmat.+}Let $S$ and $T$ be two sets.

\begin{enumerate}
\item[(a)] The sum of two $S\times T$-matrices is defined by $\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}+\left(  b_{s,t}\right)
_{\left(  s,t\right)  \in S\times T}=\left(  a_{s,t}+b_{s,t}\right)  _{\left(
s,t\right)  \in S\times T}$.

\item[(b)] If $u\in\kk$ and if $\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times T}\in\kk^{S\times T}$, then we define $u\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$ to be the $S\times
T$-matrix $\left(  ua_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$.

\item[(c)] Let $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times
T}$ be an $S\times T$-matrix. For every $s\in S$, we define the
\emph{$s$-th row of $A$}\index{$s$-th row of an $S \times T$-matrix}
to be the $\left\{  1\right\}  \times T$-matrix $\left(
a_{s,t}\right)  _{\left(  i,t\right)  \in\left\{  1\right\}  \times T}$.
(Notice that $\left\{  1\right\}  \times T$-matrices are a generalization of
row vectors.) Similarly, for every $t\in T$, we define the
\emph{$t$-th column of $A$}\index{$t$-th column of an $S \times T$-matrix}
to be the $S\times\left\{  1\right\}  $-matrix $\left(
a_{s,t}\right)  _{\left(  s,i\right)  \in S\times\left\{  1\right\}  }$.
\end{enumerate}
\end{definition}

\begin{definition}
\label{def.STmat.1}Let $S$ be a set.

\begin{enumerate}
\item[(a)] The $S\times S$ \dfn{identity matrix} is defined to be the
$S\times S$-matrix $\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$. This $S\times S$-matrix is denoted by $I_{S}$. (Notice that the
$n\times n$ identity matrix $I_{n}$ is $I_{\left\{  1,2,\ldots,n\right\}  }$
for each $n\in \NN$.)

\item[(b)] An $S\times S$-matrix $\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$ is said to be
\emph{diagonal}\index{diagonal $S\times S$-matrix}
if every $\left(  s,t\right)  \in S\times T$ satisfying
$s\neq t$ satisfies $a_{s,t}=0$.

\item[(c)] Let $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$ be an $S\times S$-matrix. The
\emph{diagonal}\index{diagonal of an $S\times S$-matrix}
of $A$ means the family
$\left(  a_{s,s}\right)  _{s\in S}$. The
\emph{diagonal entries}\index{diagonal entries of an $S\times S$-matrix}
of $A$ are
the entries of this diagonal $\left(  a_{s,s}\right)  _{s\in S}$.
\end{enumerate}
\end{definition}

\begin{definition}
\label{def.STmat.*}Let $S$, $T$ and $U$ be three sets. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$ be an $S\times
T$-matrix, and let $B=\left(  b_{t,u}\right)  _{\left(  t,u\right)  \in
T\times U}$ be a $T\times U$-matrix. Assume that the sum $\sum_{t\in T}%
a_{s,t}b_{t,u}$ is well-defined for every $\left(  s,u\right)  \in S\times U$.
(For example, this is guaranteed to hold if the set $T$ is finite. For
infinite $T$, it may and may not hold.) Then, the $S\times U$-matrix $AB$ is
defined by%
\[
AB=\left(  \sum_{t\in T}a_{s,t}b_{t,u}\right)  _{\left(  s,u\right)  \in
S\times U}.
\]

\end{definition}

\begin{definition}
\label{def.STmat.inv}Let $S$ and $T$ be two finite sets. We say that an
$S\times T$-matrix $A$ is
\emph{invertible}\index{invertible $S \times T$-matrix}
if and only if there exists a
$T\times S$-matrix $B$ satisfying $AB=I_{S}$ and $BA=I_{T}$. In this case,
this matrix $B$ is unique; it is denoted by $A^{-1}$ and is called the
\emph{inverse}\index{inverse of an $S \times T$-matrix} of $A$.
\end{definition}

The definitions that we have just given are straightforward generalizations of
the analogous definitions for $n\times m$-matrices; thus, unsurprisingly, many
properties of $n\times m$-matrices still hold for $S\times T$-matrices. For example:

\begin{proposition}
\phantomsection\label{prop.STmat.ass}

\begin{enumerate}
\item[(a)] Let $S$ and $T$ be two sets. Let $A$ be an $S\times T$-matrix.
Then, $I_{S}A=A$ and $AI_{T}=A$.

\item[(b)] Let $S$, $T$ and $U$ be three sets such that $T$ is finite. Let $A$
and $B$ be two $S\times T$-matrices. Let $C$ be a $T\times U$-matrix. Then,
$\left(  A+B\right)  C=AC+BC$.

\item[(c)] Let $S$, $T$, $U$ and $V$ be four sets such that $T$ and $U$ are
finite. Let $A$ be an $S\times T$-matrix. Let $B$ be a $T\times U$-matrix. Let
$C$ be a $U\times V$-matrix. Then, $\left(  AB\right)  C=A\left(  BC\right)  $.
\end{enumerate}
\end{proposition}

The proof of Proposition \ref{prop.STmat.ass} (and of similar properties that
will be left unstated) is analogous to the proofs of the corresponding
properties of $n\times m$-matrices.\footnote{A little \textbf{warning}: In
Proposition \ref{prop.STmat.ass}(c), the condition that $T$ and $U$ be finite
can be loosened (we leave this to the interested reader), but cannot be
completely disposed of. It can happen that both $\left(  AB\right)  C$ and
$A\left(  BC\right)  $ are defined, but $\left(  AB\right)  C=A\left(
BC\right)  $ does not hold (if we remove this condition). For example, this
happens if $S=\ZZ$, $T=\ZZ$, $U=\ZZ$, $V=\ZZ$,
$A=\left(
\begin{cases}
1, & \text{if }i\geq j;\\
0, & \text{if }i<j
\end{cases}
\right)  _{\left(  i,j\right)  \in \ZZ \times \ZZ}$, $B=\left(
\delta_{i,j}-\delta_{i,j+1}\right)  _{\left(  i,j\right)  \in \ZZ%
\times \ZZ}$ and $C=\left(
\begin{cases}
0, & \text{if }i\geq j;\\
1, & \text{if }i<j
\end{cases}
\right)  _{\left(  i,j\right)  \in \ZZ \times \ZZ}$. (Indeed, in
this example, it is easy to check that $AB=I_{\ZZ}$ and
$BC=-I_{\ZZ}$ and thus $\underbrace{\left(  AB\right)  }%
_{=I_{\ZZ}}C=I_{\ZZ}C=C\neq-A=A\underbrace{\left(
-I_{\ZZ}\right)  }_{=BC}=A\left(  BC\right)  $.)
\par
This seeming paradox is due to the subtleties of rearranging infinite sums
(similarly to how a conditionally convergent series of real numbers can change
its value when its entries are rearranged).} As a consequence of these
properties, it is easy to see that if $S$ is any finite set, then
$\kk^{S\times S}$ is a $\kk$-algebra.

In general, $S\times T$-matrices (unlike $n\times m$-matrices) do not have a
predefined order on their rows and their columns. Thus, the classical notion
of a triangular $n\times n$-matrix cannot be generalized to a notion of a
``triangular $S\times S$-matrix'' when $S$ is
just a set with no additional structure. However, when $S$ is a poset, such a
generalization can be made:

\begin{definition}
\label{def.STmat.tria}Let $S$ be a poset. Let $A=\left(  a_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$ be an $S\times S$-matrix.

\begin{enumerate}
\item[(a)] The matrix $A$ is said to be
\emph{triangular}\index{triangular $S\times S$-matrix}
if and only if
every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s$
must satisfy $a_{s,t}=0$. (Here, $\leq$ denotes the smaller-or-equal relation
of the poset $S$.)

\item[(b)] The matrix $A$ is said to be
\emph{unitriangular}\index{unitriangular $S\times S$-matrix}
if and only if
$A$ is triangular and has the further property that, for every $s\in S$, we
have $a_{s,s}=1$.

\item[(c)] The matrix $A$ is said to be
\emph{invertibly triangular}\index{invertibly triangular $S\times S$-matrix}
if and
only if $A$ is triangular and has the further property that, for every $s\in
S$, the element $a_{s,s}$ of $\kk$ is invertible.
\end{enumerate}

Of course, all three notions of ``triangular'', ``unitriangular'' and
``invertibly triangular'' depend on the partial order on $S$.

Clearly, every invertibly triangular $S\times S$-matrix is triangular. Also,
every unitriangular $S\times S$-matrix is invertibly triangular (because the
element $1$ of $\kk$ is invertible).

We can restate the definition of ``invertibly
triangular'' as follows: The matrix $A$ is said to be
\emph{invertibly triangular} if and only if it is triangular and its
diagonal entries are invertible. Similarly, we can restate the definition of
``unitriangular'' as follows: The matrix $A$
is said to be \emph{unitriangular} if and only if it is triangular and all
its diagonal entries equal $1$.
\end{definition}

Definition \ref{def.STmat.tria}(a) generalizes both the notion of
upper-triangular matrices and the notion of lower-triangular matrices. To wit:

\begin{example}
Let $n\in \NN$. Let $N_{1}$ be the poset whose ground set is $\left\{
1,2,\ldots,n\right\}  $ and whose smaller-or-equal relation $\leq_{1}$ is
given by%
\[
s\leq_{1}t\ \Longleftrightarrow\ s\leq t\text{ (as integers).}%
\]
(This is the usual order relation on this set.) Let $N_{2}$ be the poset whose
ground set is $\left\{  1,2,\ldots,n\right\}  $ and whose order relation
$\leq_{2}$ is given by%
\[
s\leq_{2}t\ \Longleftrightarrow\ s\geq t\text{ (as integers).}%
\]
Let $A\in\kk^{n\times n}$.

\begin{enumerate}
\item[(a)] The matrix $A$ is upper-triangular if and only if $A$ is triangular
when regarded as an $N_{1}\times N_{1}$-matrix.

\item[(b)] The matrix $A$ is lower-triangular if and only if $A$ is triangular
when regarded as an $N_{2}\times N_{2}$-matrix.
\end{enumerate}
\end{example}

More interesting examples of triangular matrices are obtained when the order
on $S$ is not a total order:

\begin{example}
Let $S$ be the poset whose ground set is $\left\{  1,2,3\right\}  $ and whose
smaller relation $<_{S}$ is given by $1<_{S}2$ and $3<_{S}2$. Then, the
triangular $S\times S$-matrices are precisely the $3\times3$-matrices of the
form $\left(
\begin{array}
[c]{ccc}%
a_{1,1} & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3}\\
0 & 0 & a_{3,3}%
\end{array}
\right)  $ with $a_{1,1},a_{2,1},a_{2,2},a_{2,3},a_{3,3}\in\kk$.
\end{example}

We shall now state some basic properties of triangular matrices:

\begin{proposition}
\label{prop.STmat.triangular}Let $S$ be a finite poset.

\begin{enumerate}
\item[(a)] The triangular $S\times S$-matrices form a subalgebra of the
$\kk$-algebra $\kk^{S\times S}$.

\item[(b)] The invertibly triangular $S\times S$-matrices form a group with
respect to multiplication.

\item[(c)] The unitriangular $S\times S$-matrices form a group with respect to multiplication.

\item[(d)] Any invertibly triangular $S\times S$-matrix is invertible, and its
inverse is again invertibly triangular.

\item[(e)] Any unitriangular $S\times S$-matrix is invertible, and its inverse
is again unitriangular.
\end{enumerate}
\end{proposition}

\begin{exercise}
\label{exe.STmat.mat}Prove Proposition~\ref{prop.STmat.triangular}.
\end{exercise}

\subsubsection{Expansion of a family in another}

We will often study situations where two families $\left(  e_{s}\right)
_{s\in S}$ and $\left(  f_{t}\right)  _{t\in T}$ of vectors in a $\kk%
$-module $M$ are given, and the vectors $e_{s}$ can be written as linear
combinations of the vectors $f_{t}$. In such situations, we can form an
$S\times T$-matrix out of the coefficients of these linear combinations; this
is one of the ways how matrices arise in the theory of modules. Let us define
the notations we are going to use in such situations:

\begin{definition}
\label{def.STmat.expansion}Let $M$ be a $\kk$-module. Let $\left(
e_{s}\right)  _{s\in S}$ and $\left(  f_{t}\right)  _{t\in T}$ be two families
of elements of $M$. (The sets $S$ and $T$ may and may not be finite.)

Let $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$ be an
$S\times T$-matrix. Assume that, for every $s\in S$, all but finitely many
$t\in T$ satisfy $a_{s,t}=0$. (This assumption is automatically satisfied if
$T$ is finite.)

We say that the family $\left(  e_{s}\right)  _{s\in S}$ \dfn{expands in
the family $\left(  f_{t}\right)  _{t\in T}$ through the matrix $A$}
if
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in T}a_{s,t}f_{t}.
\label{eq.def.STmat.expansion.es=}%
\end{equation}
In this case, we furthermore say that the matrix $A$ is a
\dfn{change-of-basis matrix} (or \dfn{transition matrix}) from the
family $\left(  e_{s}\right)  _{s\in S}$ to the family $\left(  f_{t}\right)
_{t\in T}$.
\end{definition}

\begin{remark}
The notation in Definition \ref{def.STmat.expansion} is not really standard;
even we ourselves will occasionally deviate in its use. In the formulation
``the family $\left(  e_{s}\right)  _{s\in S}$ expands in the
family $\left(  f_{t}\right)  _{t\in T}$ through the matrix $A$'', the word
``in'' can be replaced by ``with respect to'', and the word
``through'' can be replaced by ``using''.

The notion of a ``change-of-basis matrix'' is
slightly misleading, because neither of the families $\left(  e_{s}\right)
_{s\in S}$ and $\left(  f_{t}\right)  _{t\in T}$ has to be a basis. Our use of
the words ``transition matrix'' should not be
confused with the different meaning that these words have in the theory of
Markov chains. The indefinite article in ``a change-of-basis
matrix'' is due to the fact that, for given families $\left(
e_{s}\right)  _{s\in S}$ and $\left(  f_{t}\right)  _{t\in T}$, there might be
more than one change-of-basis matrix from $\left(  e_{s}\right)  _{s\in S}$ to
$\left(  f_{t}\right)  _{t\in T}$. (There also might be no such matrix.) When
$\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{t}\right)  _{t\in T}$ are
bases of the $\kk$-module $M$, there exists precisely one
change-of-basis matrix from $\left(  e_{s}\right)  _{s\in S}$ to $\left(
f_{t}\right)  _{t\in T}$.
\end{remark}

So a change-of-basis matrix $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)
\in S\times T}$ from one family $\left(  e_{s}\right)  _{s\in S}$ to another
family $\left(  f_{t}\right)  _{t\in T}$ allows us to write the elements of
the former family as linear combinations of the elements of the latter (using
(\ref{eq.def.STmat.expansion.es=})). When such a matrix $A$ is invertible (and
the sets $S$ and $T$ are finite\footnote{We are requiring the finiteness of
$S$ and $T$ mainly for the sake of simplicity. We could allow $S$ and $T$ to
be infinite, but then we would have to make some finiteness requirements on
$A$ and $A^{-1}$.}), it also (indirectly) allows us to do the opposite: i.e.,
to write the elements of the latter family as linear combinations of the
elements of the former. This is because if $A$ is an invertible
change-of-basis matrix from $\left(  e_{s}\right)  _{s\in S}$ to $\left(
f_{t}\right)  _{t\in T}$, then $A^{-1}$ is a change-of-basis matrix from
$\left(  f_{t}\right)  _{t\in T}$ to $\left(  e_{s}\right)  _{s\in S}$. This
is part (a) of the following theorem:

\begin{theorem}
\label{thm.STmat.expansion-inv}Let $M$ be a $\kk$-module. Let $S$ and
$T$ be two finite sets. Let $\left(  e_{s}\right)  _{s\in S}$ and $\left(
f_{t}\right)  _{t\in T}$ be two families of elements of $M$.

Let $A$ be an invertible $S\times T$-matrix. Thus, $A^{-1}$ is a $T\times S$-matrix.

Assume that the family $\left(  e_{s}\right)  _{s\in S}$ expands in the family
$\left(  f_{t}\right)  _{t\in T}$ through the matrix $A$. Then:

\begin{enumerate}
\item[(a)] The family $\left(  f_{t}\right)  _{t\in T}$ expands in the family
$\left(  e_{s}\right)  _{s\in S}$ through the matrix $A^{-1}$.

\item[(b)] The $\kk$-submodule of $M$ spanned by the family $\left(
e_{s}\right)  _{s\in S}$ is the $\kk$-submodule of $M$ spanned by the
family $\left(  f_{t}\right)  _{t\in T}$.

\item[(c)] The family $\left(  e_{s}\right)  _{s\in S}$ spans the $\kk%
$-module $M$ if and only if the family $\left(  f_{t}\right)  _{t\in T}$ spans
the $\kk$-module $M$.

\item[(d)] The family $\left(  e_{s}\right)  _{s\in S}$ is $\kk%
$-linearly independent if and only if the family $\left(  f_{t}\right)  _{t\in
T}$ is $\kk$-linearly independent.

\item[(e)] The family $\left(  e_{s}\right)  _{s\in S}$ is a basis of the
$\kk$-module $M$ if and only if the family $\left(  f_{t}\right)
_{t\in T}$ is a basis of the $\kk$-module $M$.
\end{enumerate}
\end{theorem}

\begin{exercise}
\label{exe.STmat.expansion-inv}Prove Theorem~\ref{thm.STmat.expansion-inv}.
\end{exercise}

\begin{definition}
\label{def.STmat.expansion-tria}Let $M$ be a $\kk$-module. Let $S$ be a
finite poset. Let $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)
_{s\in S}$ be two families of elements of $M$.

\begin{enumerate}
\item[(a)] We say that the family $\left(  e_{s}\right)  _{s\in S}$
\dfn{expands triangularly in the family $\left(  f_{s}\right)  _{s\in S}$}
if and only if there exists a triangular $S\times S$-matrix $A$ such that the
family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$.

\item[(b)] We say that the family $\left(  e_{s}\right)  _{s\in S}$
\dfn{expands invertibly triangularly in the family $\left(  f_{s}\right)
_{s\in S}$} if and only if there exists an invertibly triangular $S\times
S$-matrix $A$ such that the family $\left(  e_{s}\right)  _{s\in S}$ expands
in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $A$.

\item[(c)] We say that the family $\left(  e_{s}\right)  _{s\in S}$
\dfn{expands unitriangularly in the family $\left(  f_{s}\right)  _{s\in
S}$} if and only if there exists a unitriangular $S\times S$-matrix $A$
such that the family $\left(  e_{s}\right)  _{s\in S}$ expands in the
family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $A$.
\end{enumerate}

Clearly, if the family $\left(  e_{s}\right)  _{s\in S}$ expands
unitriangularly in the family $\left(  f_{s}\right)  _{s\in S}$, then it also
expands invertibly triangularly in the family $\left(  f_{s}\right)  _{s\in
S}$ (because any unitriangular matrix is an invertibly triangular matrix).
\end{definition}

We notice that in Definition \ref{def.STmat.expansion-tria}, the two families
$\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)  _{s\in S}$ must
be indexed by one and the same set $S$.

The concepts of ``expanding triangularly'', ``expanding invertibly
triangularly'' and ``expanding unitriangularly'' can also be characterized
without referring to matrices, as follows:

\begin{remark}
\label{rmk.STmat.expansion-tria-shortcuts}
Let $M$ be a $\kk$-module.
Let $S$ be a finite poset. Let $\left( e_s \right)_{s \in S}$ and
$\left( f_s \right)_{s \in S}$ be two families of elements of $M$.
Let $<$ denote the smaller relation of the poset $S$, and let
$\leq$ denote the smaller-or-equal relation of the poset $S$.
Then:

\begin{enumerate}

\item[(a)] The family $\left( e_s \right)_{s \in S}$ expands triangularly
in the family $\left( f_s \right)_{s \in S}$ if and only if every $s\in S$
satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\]

\item[(b)] The family $\left( e_s \right)_{s \in S}$ expands invertibly
triangularly in the family $\left( f_s \right)_{s \in S}$ if and only if
every $s\in S$ satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$.

\item[(c)] The family $\left( e_s \right)_{s \in S}$ expands
unitriangularly in the family $\left( f_s \right)_{s \in S}$ if and only if
every $s\in S$ satisfies
\[
e_{s}=f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\]

\end{enumerate}
\end{remark}

All three parts of Remark~\ref{rmk.STmat.expansion-tria-shortcuts} follow
easily from the definitions.

\begin{example}
Let $n\in \NN$. For this example, let $S$ be the poset $\left\{
1,2,\ldots,n\right\}  $ (with its usual order). Let $M$ be a $\kk%
$-module, and let $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)
_{s\in S}$ be two families of elements of $M$. We shall identify these
families $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)  _{s\in
S}$ with the $n$-tuples $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ and
$\left(  f_{1},f_{2},\ldots,f_{n}\right)  $. Then, the family $\left(
e_{s}\right)  _{s\in S}=\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ expands
triangularly in the family $\left(  f_{s}\right)  _{s\in S}=\left(
f_{1},f_{2},\ldots,f_{n}\right)  $ if and only if, for every $s\in\left\{
1,2,\ldots,n\right\}  $, the vector $e_{s}$ is a $\kk$-linear
combination of $f_{1},f_{2},\ldots,f_{s}$. Moreover, the family $\left(
e_{s}\right)  _{s\in S}=\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ expands
unitriangularly in the family $\left(  f_{s}\right)  _{s\in S}=\left(
f_{1},f_{2},\ldots,f_{n}\right)  $ if and only if, for every $s\in\left\{
1,2,\ldots,n\right\}  $, the vector $e_{s}$ is a sum of $f_{s}$ with a
$\kk$-linear combination of $f_{1},f_{2},\ldots,f_{s-1}$.
\end{example}

\begin{corollary}
\label{cor.STmat.expansion-tria-inv}Let $M$ be a $\kk$-module. Let $S$
be a finite poset. Let $\left(  e_{s}\right)  _{s\in S}$ and $\left(
f_{s}\right)  _{s\in S}$ be two families of elements of $M$. Assume that the
family $\left(  e_{s}\right)  _{s\in S}$ expands invertibly triangularly in
the family $\left(  f_{s}\right)  _{s\in S}$. Then:

\begin{enumerate}
\item[(a)] The family $\left(  f_{s}\right)  _{s\in S}$ expands invertibly
triangularly in the family $\left(  e_{s}\right)  _{s\in S}$.

\item[(b)] The $\kk$-submodule of $M$ spanned by the family $\left(
e_{s}\right)  _{s\in S}$ is the $\kk$-submodule of $M$ spanned by the
family $\left(  f_{s}\right)  _{s\in S}$.

\item[(c)] The family $\left(  e_{s}\right)  _{s\in S}$ spans the $\kk%
$-module $M$ if and only if the family $\left(  f_{s}\right)  _{s\in S}$ spans
the $\kk$-module $M$.

\item[(d)] The family $\left(  e_{s}\right)  _{s\in S}$ is $\kk%
$-linearly independent if and only if the family $\left(  f_{s}\right)  _{s\in
S}$ is $\kk$-linearly independent.

\item[(e)] The family $\left(  e_{s}\right)  _{s\in S}$ is a basis of the
$\kk$-module $M$ if and only if the family $\left(  f_{s}\right)
_{s\in S}$ is a basis of the $\kk$-module $M$.
\end{enumerate}
\end{corollary}

\begin{exercise}
\label{exe.STmat.expansion-tria-inv}Prove
Remark~\ref{rmk.STmat.expansion-tria-shortcuts} and
Corollary~\ref{cor.STmat.expansion-tria-inv}.
\end{exercise}

An analogue of Corollary \ref{cor.STmat.expansion-tria-inv} can be stated for
unitriangular expansions, but we leave this to the reader.

% [DG][v55] Added the above appendix, and referenced it in the text
% accordingly.

\newpage

\section{Further hints to the exercises (work in progress)}

The following pages contain hints to (some of\footnote{Currently only
the ones from Chapter~\ref{Hopf-intro-section}.}) the exercises in the text
(beyond the hints occasionally included in the exercises themselves). Some of
the hints rise to the level of outlined solutions.

\newcommand\hintsboilerplate{Detailed solutions (sometimes different
from the solutions hinted at) can be found in Chapter \ref{chp.solutions}.}

\begin{nosolutions}
\renewcommand\hintsboilerplate{Note that there is also a version of
this text that contains detailed
solutions for all the exercises; this version can be downloaded from
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/HopfComb-sols.pdf} (or
compiled from the sourcecode of the text).}
\end{nosolutions}

\hintsboilerplate

\textbf{Warning:} The hints below are new and have never been proofread. Typos
(or worse) are likely. In case of doubt, consult the detailed solutions.

\subsection{Hints for Chapter \ref{Hopf-intro-section}}

\textit{Hint to Exercise \ref{exe.counit.unique}.} The claim of the exercise
is dual to the classical fact that if $A$ is a $\kk$-module and
$m:A\otimes A\rightarrow A$ is a $\kk$-linear map, then there exists
\textit{at most one} $\kk$-linear map $u:\mathbf{k}\rightarrow A$ such
that the diagram \eqref{unit-diagram} commutes\footnote{This fact is just the
linearization of the known fact that any binary operation has at most one
neutral element.}. Take any proof of this latter fact, rewrite it in an
``element-free'' fashion\footnote{This means
rewriting it completely in terms of linear maps rather than elements. For
example, instead of talking about $m\left(  m\left(  a\otimes b\right)
\otimes c\right)  $ for three elements $a,b,c\in A$, you should talk about the
map $m\circ\left(  m\otimes \id_{A}\right)  :A\otimes
A\otimes A\rightarrow A$ (which is, of course, the map that sends each
$a\otimes b\otimes c$ to $m\left(  m\left(  a\otimes b\right)  \otimes
c\right)  $). Instead of computing with elements, you should compute with maps
(and commutative diagrams).}, and ``reverse all
arrows''. This will yield a solution to Exercise
\ref{exe.counit.unique}.

For an alternative solution, use Sweedler notation (as in
\eqref{eq.sweedler-not.def1}) as follows: The commutativity of the diagram
\eqref{counit-diagram} says that%
\[
c=\sum_{\left(  c\right)  }\epsilon\left(  c_{1}\right)  c_{2}=\sum_{\left(
c\right)  }\epsilon\left(  c_{2}\right)  c_{1}\qquad\text{for each }c\in C.
\]
Thus, if $\epsilon_{1}$ and $\epsilon_{2}$ are two $\kk$-linear maps
$\epsilon:C\rightarrow\mathbf{k}$ such that the diagram \eqref{counit-diagram}
commutes, then each $c\in C$ satisfies%
\[
c=\sum_{\left(  c\right)  }\epsilon_{1}\left(  c_{1}\right)  c_{2}%
=\sum_{\left(  c\right)  }\epsilon_{1}\left(  c_{2}\right)  c_{1}%
\]
and%
\[
c=\sum_{\left(  c\right)  }\epsilon_{2}\left(  c_{1}\right)  c_{2}%
=\sum_{\left(  c\right)  }\epsilon_{2}\left(  c_{2}\right)  c_{1}.
\]
Apply $\epsilon_{2}$ to both sides of the equality $c=\sum_{\left(  c\right)
}\epsilon_{1}\left(  c_{2}\right)  c_{1}$, and apply $\epsilon_{1}$ to both
sides of the equality $c=\sum_{\left(  c\right)  }\epsilon_{2}\left(
c_{1}\right)  c_{2}$. Compare the results, and conclude that $\epsilon
_{1}=\epsilon_{2}$.

\bigskip

\textit{Hint to Exercise \ref{exe.tensor-prod-algs}.} Part (a) is well-known,
and part (b) is dual to part (a). So the trick is (again) to rewrite the
classical proof of part (a) in an ``element-free'' way, and then
``reversing all arrows''. Alternatively, part (b) can be solved using
Sweedler notation.

\bigskip

\textit{Hint to Exercise \ref{exe.tensorprod.morphism}.} Same method as for
Exercise \ref{exe.tensor-prod-algs} above.

\bigskip

\textit{Hint to Exercise \ref{exe.kernel-is-coideal}.} (a) Use the following
fact from linear algebra: If $U$, $V$, $U^{\prime}$ and $V^{\prime}$ are four
$\kk$-modules, and $\phi:U\rightarrow U^{\prime}$ and $\psi
:V\rightarrow V^{\prime}$ are two surjective $\kk$-linear maps, then
the kernel of $\phi\otimes\psi:U\otimes V\rightarrow U^{\prime}\otimes
V^{\prime}$ is
\[
\ker\left(  \phi\otimes\psi\right)  =\left(  \ker\phi\right)  \otimes
V+U\otimes\left(  \ker\psi\right)  .
\]


(b) The fact just mentioned also holds if we no longer require $\phi$ and
$\psi$ to be surjective, but instead require $\kk$ to be a field.

\bigskip

\textit{Hint to Exercise \ref{exe.graded.iso}.} Let $f : V \to W$
be an invertible graded $\kk$-linear map. Let $n \in \NN$
and $w \in W_n$. Show that the $n$-th homogeneous component of
$f^{-1}\left(w\right)$ is also a preimage of $w$ under $f$, and
thus must equal $f^{-1}\left(w\right)$. Therefore,
$f^{-1}\left(w\right) \in W_n$.

\bigskip

\textit{Hint to Exercise \ref{exe.primitives.graded-coideal}.} (a) Define the
$\kk$-linear map $\widetilde{\Delta}:A\rightarrow A\otimes A$ by
$\widetilde{\Delta}\left(  x\right)  =\Delta\left(  x\right)  -\left(
x\otimes1+1\otimes x\right)  $. Argue that $\widetilde{\Delta}$ is graded, so
its kernel $\ker\widetilde{\Delta}$ is a graded $\kk$-submodule of $A$.
But this kernel is precisely $\mathfrak{p}$.

(b) The hard part is to show that $\epsilon\left(  \mathfrak{p}\right)  =0$.
To do so, consider any $x\in\mathfrak{p}$, and apply the map $\epsilon
\otimes \id$ to both sides of the equality $\Delta\left(
x\right)  =x\otimes1+1\otimes x$. The result simplifies to $x=\epsilon\left(
x\right)  \cdot1_{A}+x$. Thus, $\epsilon\left(  x\right)  \cdot1_{A}=0$. Now
apply $\epsilon$ to this, thus obtaining $\epsilon\left(  x\right)  =0$.

\bigskip

\textit{Hint to Exercise \ref{graded-connected-exercise}.} (a) This follows
from $1_{A}\in A_{0}$, which is part of what it means for $A$ to be a graded
$\kk$-algebra.

(b) Let $\epsilon^{\prime}:A_{0}\rightarrow\mathbf{k}$ be the restriction of
the map $\epsilon$ to $A_{0}$. We know that $\epsilon^{\prime}$ is surjective
(since $\epsilon^{\prime}\left(  1_{A}\right)  =1_{\mathbf{k}}$), and that
both $A_{0}$ and $\kk$ are free $\kk$-modules of rank $1$ (since
connectedness of $A$ means $A_{0}\cong\mathbf{k}$ as $\kk$-modules). It
is an an easy exercise in linear algebra to conclude from these facts that
$\epsilon^{\prime}$ is an isomorphism. Since $\epsilon^{\prime}\circ
u= \id_{\kk}$, we thus conclude that
$u:\mathbf{k}\rightarrow A_{0}$ is an isomorphism as well (from $\kk$
to $A_{0}$).

(c) This follows from part (b).

(e) This follows from how we solved part (b).

(d) Since the bialgebra $A$ is graded, the map $\epsilon$ must be graded.
Thus, for each positive integer $n$, we have $\epsilon\left(  A_{n}\right)
\subset\mathbf{k}_{n}=0$. This quickly yields $\epsilon\left(  I\right)  =0$
(where $I=\bigoplus_{n>0}A_{n}$), hence $I\subset\ker\epsilon$. On the other
hand, $\ker\epsilon\subset I$ can be shown as follows: Let $a\in\ker\epsilon$;
write $a$ in the form $a=a^{\prime}+a^{\prime\prime}$ for some $a^{\prime}\in
A_{0}$ and some $a^{\prime\prime}\in I$, and then argue that $0=\epsilon
\left(  a\right)  =\epsilon\left(  a^{\prime}+a^{\prime\prime}\right)
=\epsilon\left(  a^{\prime}\right)  +\underbrace{\epsilon\left(
a^{\prime\prime}\right)  }_{\substack{=0\\\text{(since }a^{\prime\prime}\in
I\subset\ker\epsilon\text{)}}}=\epsilon\left(  a^{\prime}\right)  $, so that
$a^{\prime}=0$ by part (e) and therefore $a\in I$.

(f) This is most intuitive with Sweedler notation: Let $x\in A$. Then,
$\Delta\left(  x\right)  =\sum_{\left(  x\right)  }x_{1}\otimes x_{2}$.
Applying $\id \otimes\epsilon$ and recalling the commutativity
of \eqref{counit-diagram}, we thus get $x=\sum_{\left(  x\right)  }%
\epsilon\left(  x_{2}\right)  x_{1}$. Thus,%
\begin{align*}
\underbrace{\Delta\left(  x\right)  }_{=\sum_{\left(  x\right)  }x_{1}\otimes
x_{2}}-\underbrace{x}_{=\sum_{\left(  x\right)  }\epsilon\left(  x_{2}\right)
x_{1}}\otimes1  &  =\sum_{\left(  x\right)  }x_{1}\otimes x_{2}-\sum_{\left(
x\right)  }\epsilon\left(  x_{2}\right)  x_{1}\otimes1\\
&  =\sum_{\left(  x\right)  }\underbrace{x_{1}}_{\in A}\otimes
\underbrace{\left(  x_{2}-\epsilon\left(  x_{2}\right)  \cdot1\right)
}_{\substack{\in\ker\epsilon=I\\\text{(by part (d))}}}\in A\otimes I.
\end{align*}


(g) Let $x\in I$. Proceeding similarly to part (f), show that%
\[
\Delta\left(  x\right)  -1\otimes x-x\otimes1+\epsilon\left(  x\right)
1\otimes1=\sum_{\left(  x\right)  }\underbrace{\left(  x_{1}-\epsilon\left(
x_{1}\right)  \cdot1\right)  }_{\substack{\in\ker\epsilon=I\\\text{(by part
(d))}}}\otimes\underbrace{\left(  x_{2}-\epsilon\left(  x_{2}\right)
\cdot1\right)  }_{\substack{\in\ker\epsilon=I\\\text{(by part (d))}}}\in
I\otimes I.
\]
Since $x\in I=\ker\epsilon$, the $\epsilon\left(  x\right)  1\otimes1$ term on
the left hand side vanishes.

(h) This follows from part (g), since a simple homogeneity argument shows that
$\left(  I\otimes I\right)  _{n}=\sum_{k=1}^{n-1}A_{k}\otimes A_{n-k}$.

\bigskip

\textit{Hint to Exercise \ref{exe.graded.Dq}.} We need to check the four
equalities $D_{q} \circ m=m\circ\left(  D_{q} \otimes D_{q} \right)  $ and
$D_{q} \circ u=u$ and $\left(  D_{q} \otimes D_{q} \right)  \circ\Delta
=\Delta\circ D_{q} $ and $\epsilon\circ D_{q} =\epsilon$. This can easily be
done by hand (just check everything on homogeneous elements); a more erudite
proof proceeds as follows: Generalize the map $D_{q}$ to a map $D_{q,V}%
:V\rightarrow V$ defined (in the same way as $D_{q}$) for every graded
$\kk$-module $V$, and show that these maps $D_{q,V}$ are functorial
(i.e., if $f:V\rightarrow W$ is a graded $\kk$-linear map between two
graded $\kk$-modules $V$ and $W$, then $D_{q,W}\circ f=f\circ D_{q,V}$)
and ``respect tensor products'' (i.e., we
have $D_{q,V\otimes W}=D_{q,V}\otimes D_{q,W}$ for any two graded $\mathbf{k}%
$-modules $V$ and $W$). The four equalities are then easily obtained from
these two facts, without having to introduce elements.

\bigskip

\textit{Hint to Exercise \ref{exe.bialg.tensor}.} (a) Our definition of the
$\kk$-coalgebra $A\otimes B$ yields%
\[
\Delta_{A\otimes B}=\left(  \id_{A}\otimes T_{A,B}%
\otimes \id_{B}\right)  \circ\left(  \Delta_{A}\otimes\Delta
_{B}\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \epsilon
_{A\otimes B}=\theta\circ\left(  \epsilon_{A}\otimes\epsilon_{B}\right)  ,
\]
where $\theta$ is the canonical $\kk$-module isomorphism $\mathbf{k}%
\otimes\mathbf{k}\rightarrow\mathbf{k}$. All maps on the right hand sides are
$\kk$-algebra homomorphisms (see Exercise \ref{exe.tensorprod.morphism}%
(a)); thus, so are $\Delta_{A\otimes B}$ and $\epsilon_{A\otimes B}$.

(b) Straightforward.

\bigskip

\textit{Hint to Exercise \ref{exe.convolution.assoc}.} Simple computation
(either element-free or with Sweedler notation).

\bigskip

\textit{Hint to Exercise \ref{exe.convolution.tensor}.} Simple computation
(either element-free or with Sweedler notation).

\bigskip

\textit{Hint to Exercise \ref{exe.convolution.tensor-curry}.} Straightforward
computation, best done using Sweedler notation.

\bigskip

\textit{Hint to Exercise \ref{exe.hopf.tensor}.} Use Exercise
\ref{exe.convolution.assoc}.

\bigskip

\textit{Hint to Exercise \ref{exe.iterated.m}.} The following is more context
than hint (see the last paragraph for an actual hint).

It is easiest to prove this by calculating with elements. To wit, in order to
prove that two $\kk$-linear maps from $A^{\otimes\left(  k+1\right)  }$
are identical, it suffices to show that they agree on all pure tensors
$a_{1}\otimes a_{2}\otimes\cdots\otimes a_{k+1}\in A^{\otimes\left(
k+1\right)  }$. But the recursive definition of $m^{\left(  k\right)  }$ shows
that%
\begin{equation}
m^{\left(  k\right)  }\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{k+1}\right)  =a_{1}\left(  a_{2}\left(  a_{3}\left(  \cdots\left(
a_{k}a_{k+1}\right)  \cdots\right)  \right)  \right)
\label{hint.iterated.m.1}%
\end{equation}
for all $a_{1},a_{2},\ldots,a_{k+1}\in A$. Now, the ``general
associativity'' law (a fundamental result in abstract
algebra, commonly used without mention) says that, because the multiplication
of $A$ is associative, the parentheses in the product $a_{1}\left(
a_{2}\left(  a_{3}\left(  \cdots\left(  a_{k}a_{k+1}\right)  \cdots\right)
\right)  \right)  $ can be omitted without making it ambiguous -- i.e., any
two ways of parenthesizing the product $a_{1}a_{2}\cdots a_{k+1}$ evaluate to
the same result. (For example, for $k=4$, this says that
\[
a_{1}\left(  a_{2}\left(  a_{3}a_{4}\right)  \right)  =a_{1}\left(  \left(
a_{2}a_{3}\right)  a_{4}\right)  =\left(  a_{1}a_{2}\right)  \left(
a_{3}a_{4}\right)  =\left(  a_{1}\left(  a_{2}a_{3}\right)  \right)
a_{4}=\left(  \left(  a_{1}a_{2}\right)  a_{3}\right)  a_{4}%
\]
for all $a_{1},a_{2},a_{3},a_{4}\in A$.) Thus, we can rewrite
\eqref{hint.iterated.m.1} as
\[
m^{\left(  k\right)  }\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{k+1}\right)  =a_{1}a_{2}\cdots a_{k+1}.
\]
Using this formula, all four parts of the exercise become trivial: For
example, part (a) simply says that
\[
a_{1}a_{2}\cdots a_{k+1}=\left(  a_{1}a_{2}\cdots a_{i+1}\right)  \left(
a_{i+2}a_{i+3}\cdots a_{k+1}\right)
\]
for all $a_{1},a_{2},\ldots,a_{k+1}\in A$, because we have
\[
\left(  m\circ\left(  m^{\left(  i\right)  }\otimes m^{\left(  k-1-i\right)
}\right)  \right)  \left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{k+1}\right)  =\left(  a_{1}a_{2}\cdots a_{i+1}\right)  \left(
a_{i+2}a_{i+3}\cdots a_{k+1}\right)  .
\]
Likewise, part (c) simply says that%
\[
a_{1}a_{2}\cdots a_{k+1}=a_{1}a_{2}\cdots a_{i}\left(  a_{i+1}a_{i+2}\right)
a_{i+3}a_{i+4}\cdots a_{k+1}%
\]
for all $a_{1},a_{2},\ldots,a_{k+1}\in A$. Parts (b) and (d) are particular
cases of parts (a) and (c), respectively.

Of course, in order for this to be a complete solution, you have to prove the
``general associativity'' law used above. It
turns out that doing so is not much easier than solving the exercise from
scratch (in fact, part (a) of the exercise is an equivalent form of the
``general associativity'' law). So we can
just as well start from scratch and solve part (a) directly by induction on
$k$, then derive part (b) as its particular case, then solve part (c) by
induction on $k$ using the result of part (b), then derive part (d) as a
particular case of (c).

\bigskip

\textit{Hint to Exercise~\ref{exe.iterated.Delta}.} If you have solved
Exercise \ref{exe.iterated.m} in an ``element-free'' way, then you can
reverse all arrows in said
solution and thus obtain a solution to Exercise~\ref{exe.iterated.Delta}.

\bigskip

\textit{Hint to Exercise~\ref{exe.iterated.patras}.} (a) Induction on $k$,
using Exercise \ref{exe.tensorprod.morphism}(b).

(b) This is dual to (a).

(d) For every $\kk$-coalgebra $C$, consider the map $\Delta
_{C}^{\left(  k\right)  }:C\rightarrow C^{\otimes\left(  k+1\right)  }$ (this
is the map $\Delta^{\left(  k\right)  }$ defined in
Exercise~\ref{exe.iterated.Delta}). This map $\Delta_{C}^{\left(  k\right)  }$
is clearly functorial in $C$. By this we mean that if $C$ and $D$ are any two
$\kk$-coalgebras, and $f:C\rightarrow D$ is any $\kk$-coalgebra
homomorphism, then the diagram
\[
\xymatrixcolsep{4pc}
\xymatrix{
C \ar[r]^{f} \ar[d]^{\Delta^{\left(k\right)}_C} & D \ar[d]^{\Delta
^{\left(k\right)}_D} \\
C^{\otimes\left(k+1\right)} \ar[r]_{f^{\otimes\left(k+1\right)}}
& D^{\otimes\left(k+1\right)}
}
\]
commutes. Now, apply this to $C=H^{\otimes\left(  \ell+1\right)  }$, $D=H$ and
$f=m_{H}^{\left(  \ell\right)  }$ (using part (a)).

(c) This is dual to (d).

\bigskip

\textit{Hint to Exercise~\ref{exe.convolution.k}.} Induction on $k$.

\bigskip

\textit{Hint to Exercise~\ref{exe.coalg-anti}.} This is dual to
Proposition~\ref{antipodes-are-antiendomorphisms}, so the usual strategy
(viz., rewriting element-free and reversing all arrows) applies.

\bigskip

\textit{Hint to Exercise~\ref{exe.coalg-anti.gen}.} (a) A straightforward
generalization of the proof of
Proposition~\ref{antipodes-are-antiendomorphisms} (which corresponds to the
particular case when $C=A$ and $r = \id$) does the trick.

(b) This is dual to (a).

(c) Easy.

(d) Apply Exercise~\ref{exe.coalg-anti.gen}(a) to $C=A$ and
$r = \id_{A}$; then, apply
Proposition~\ref{antipodes-give-convolution-inverses}(a) to $H=A$ and
$\alpha=S$.

(e) Let $s:C\rightarrow A$ be the $\kk$-linear map that sends every
homogeneous element $c\in C_{n}$ (for every $n\in \NN$) to the $n$-th
homogeneous component of $r^{\star\left(  -1\right)  }\left(  c\right)  $.
Then, $s$ is graded, and (this takes some work) is also a $\star$-inverse to
$r$. But $r$ has only one $\star$-inverse.

\bigskip

\textit{Hint to Exercise~\ref{exe.antipode.inverse}.} (a) Rewrite the
assumption as $m\circ\left(  P\otimes \id \right)  \circ
T\circ\Delta=u\circ\epsilon$, where $T$ is the twist map $T_{A,A}$.
Proposition~\ref{antipodes-are-antiendomorphisms} leads to $m\circ\left(
S\otimes S\right)  =S\circ m\circ T$ and $u=S\circ u$.
Exercise~\ref{exe.coalg-anti} leads to $\left(  S\otimes S\right)  \circ
\Delta=T\circ\Delta\circ S$ and $\epsilon\circ S=\epsilon$. Use these to show
that $\left(  P\circ S\right)  \star S=u\circ\epsilon$, so that $P\circ
S= \id$. Also, show that $S\star\left(  S\circ P\right)
=u\circ\epsilon$, so that $S\circ P= \id$.

(b) Similar to (a).

(c) Let $A$ be a connected graded Hopf algebra. Just as a left $\star$-inverse
$S$ to $\id_{A}$ has been constructed in the proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}, we could
construct a $\kk$-linear map $P:A\rightarrow A$ such that every $a\in
A$ satisfies $\sum_{\left(  a\right)  }P\left(  a_{2}\right)  \cdot
a_{1}=u\left(  \epsilon\left(  a\right)  \right)  $. Now apply part (a).

\bigskip

\textit{Hint to Exercise~\ref{exe.subcoalgebra.summand}.} Since $D$ is a
direct summand of $C$, we can identify the tensor products $D\otimes C$,
$C\otimes D$ and $D\otimes D$ with their canonical images inside $C\otimes C$.
Now, we can show that $\Delta\left(  D\right)  \subset D\otimes D$ as follows:
Let $p:C\rightarrow D$ be the canonical projection from $C$ onto its direct
summand $D$; then, $\Delta\left(  D\right)  \subset D\otimes C$ shows that
$\left(  p\otimes \id \right)  \circ\Delta=\Delta$, and
$\Delta\left(  D\right)  \subset C\otimes D$ shows that $\left(
\id \otimes p\right)  \circ\Delta=\Delta$. Hence,
\[
\underbrace{\left(  p\otimes p\right)  }_{=\left(  p\otimes\id%
\right)  \circ\left(  \id \otimes p\right)  }\circ\Delta=\left(
p\otimes \id \right)  \circ\underbrace{\left(  \id \otimes p
\right)  \circ\Delta}_{=\Delta}=\left(  p \otimes \id
\right)  \circ\Delta=\Delta.
\]
This yields $\Delta\left(  D\right)  \subset D\otimes D$. Hence, we get a map
$\Delta_{D}:D\rightarrow D\otimes D$ by restricting $\Delta$. Obviously, the
map $\epsilon:C\rightarrow\mathbf{k}$ restricts to a map $\epsilon
_{D}:D\rightarrow\mathbf{k}$ as well. It remains to check the commutativity of
the diagrams \eqref{coassociativity-diagram} and
\eqref{counit-diagram} for $D$ instead of $C$; but this is inherited
from $C$.

\bigskip

\textit{Hint to Exercise~\ref{exe.subcoalgebra.ABS}.} (a) Let $\widetilde{f}%
=\left(  \id_{C}\otimes f\otimes \id_{C}\right)
\circ\Delta^{\left(  2\right)  }:C\rightarrow C\otimes U\otimes C$; then,
$K=\ker\widetilde{f}$. Show (by manipulation of maps, using
Exercise~\ref{exe.iterated.Delta}(b)) that $\left(  \id_{C}
\otimes \id_{U}\otimes\Delta\right)  \circ\widetilde{f}%
=\left(  \widetilde{f}\otimes \id_{C}\right)  \circ\Delta$. Now,%
\[
K=\ker\widetilde{f}\subset\ker\left(  \underbrace{\left(  \id_{C}
\otimes \id_{U}\otimes\Delta\right)  \circ\widetilde{f}%
}_{=\left(  \widetilde{f}\otimes \id_{C}\right)  \circ\Delta
}\right)  =\ker\left(  \left(  \widetilde{f}\otimes \id_{C}
\right)  \circ\Delta\right)  =\Delta^{-1}\left(  \ker\left(
\widetilde{f}\otimes \id_{C}\right)  \right)
\]
and therefore%
\begin{align*}
\Delta\left(  K\right)   &  \subset\ker\left(  \widetilde{f}\otimes
\id_{C}\right)  =\underbrace{\left(  \ker\widetilde{f}\right)
}_{=K}\otimes C\ \ \ \ \ \ \ \ \ \ \left(  \text{since tensoring over a field
is left-exact}\right) \\
&  =K\otimes C.
\end{align*}
Similarly, $\Delta\left(  K\right)  \subset C\otimes K$. Now, apply
Exercise~\ref{exe.subcoalgebra.summand} to $D=K$.

(b) Let $E$ be a $\kk$-subcoalgebra of $C$ which is a subset of $\ker
f$. Then, $\Delta^{\left(  2\right)  }\left(  E\right)  \subset E\otimes
E\otimes E$ (since $E$ is a subcoalgebra) and $f\left(  E\right)  =0$ (since
$E\subset\ker f$). Now,
\begin{align*}
\left(  \left(  \id_{C}\otimes f\otimes \id_{C}\right)  \circ
\Delta^{\left(  2\right)  }\right)  \left(  E\right)   &
=\left(  \id_{C}\otimes f\otimes \id_{C}\right)
\left(  \underbrace{\Delta^{\left(  2\right)  }\left(  E\right)  }_{\subset
E\otimes E\otimes E}\right) \\
&  \subset\left(  \id_{C}\otimes f\otimes \id_{C}\right)
\left(  E\otimes E\otimes E\right) \\
&  = \id_{C}\left(  E\right)  \otimes\underbrace{f\left(
E\right)  }_{=0}\otimes \id_{C}\left(  E\right)  =0.
\end{align*}
Hence, $E\subset\ker\left(  \left(  \id_{C}\otimes
f\otimes \id_{C}\right)  \circ\Delta^{\left(  2\right)  }\right)
=K$.

[\textit{Remark:} Exercise~\ref{exe.subcoalgebra.ABS}(a) would not hold if we
allowed $\kk$ to be an arbitrary commutative ring rather than a field.]

\bigskip

\textit{Hint to Exercise~\ref{exe.antipode.connected.general}.} (a) Here is
Takeuchi's argument: We know that the map $h\mid_{C_{0}}\in\Hom %
\left(  C_{0},A\right)  $ is $\star$-invertible; let $\widetilde{g}$ be its
$\star$-inverse. Extend $\widetilde{g}$ to a $\kk$-linear map
$g:C\rightarrow A$ by defining it as $0$ on every $C_{n}$ for $n>0$. It is
then easy to see that $\left(  h\star g\right)  \mid_{C_{0}}=\left(  g\star
h\right)  \mid_{C_{0}}=\left(  u\epsilon\right)  \mid_{C_{0}}$. This allows us
to assume WLOG that $h\mid_{C_{0}}=\left(  u\epsilon\right)  \mid_{C_{0}}$
(because once we know that $h\star g$ and $g\star h$ are $\star$-invertible,
it follows that so is $h$). Assuming this, we conclude that $h-u\epsilon$
annihilates $C_{0}$. Define $f$ as $h-u\epsilon$. Now, we can proceed as in
the proof of Proposition~\ref{Takeuchi-prop} to show that $\sum_{k\geq
0}\left(  -1\right)  ^{k}f^{\star k}$ is a well-defined linear map
$C\rightarrow A$ and a two-sided $\star$-inverse for $h$. Thus, $h$ is $\star
$-invertible, and part (a) of the exercise is proven. (An alternative proof
proceeds by mimicking the proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}, again by first
assuming WLOG that $h\mid_{C_{0}}=\left(  u\epsilon\right)  \mid_{C_{0}}$.)

(b) Apply part (a) to $C=A$ and the map $\id_{A}:A\rightarrow A$.

(c) Applying part (b), we see that $A$ is a Hopf algebra (since $A_{0}%
=\mathbf{k}$ is a Hopf algebra) in the setting of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}. This yields the
existence of the antipode. Its uniqueness is trivial, and its gradedness
follows from Exercise \ref{exe.coalg-anti.gen}(e).

\bigskip

\textit{Hint to Exercise~\ref{exe.I=0}.} (a) Let $I$ be a two-sided coideal of
$A$ such that $I\cap\mathfrak{p}=0$ and such that $I=\bigoplus_{n\geq0}\left(
I\cap A_{n}\right)  $. Let $I_{n}=I\cap A_{n}$ for every $n\in \NN$.
Then, $I=\bigoplus_{n\geq0}I_{n}$. Since $I$ is a two-sided coideal, we have
$\epsilon\left(  I\right)  =0$.

We want to prove that $I=0$. It clearly suffices to show that every
$n\in \NN$ satisfies $I_{n}=0$ (since $I=\bigoplus_{n\geq0}I_{n}$). We
shall show this by strong induction: We fix an $N\in \NN$, and we assume
(as induction hypothesis) that $I_{n}=0$ for all $n<N$. We must prove that
$I_{N}=0$.

Fix $i\in I_{N}$; we aim to show that $i=0$. We have $i\in I_{N}\subset A_{N}$
and thus $\Delta\left(  i\right)  \in\left(  A\otimes A\right)  _{N}$ (since
$\Delta$ is a graded map). On the other hand, from $i\in I_{N}\subset I$, we
obtain
\begin{align*}
\Delta\left(  i\right)   &  \in\Delta\left(  I\right)  \subset\underbrace{I}%
_{=\bigoplus_{n\geq0}I_{n}}\otimes\underbrace{A}_{=\bigoplus_{m\geq0}A_{m}%
}+\underbrace{A}_{=\bigoplus_{m\geq0}A_{m}}\otimes\underbrace{I}%
_{=\bigoplus_{n\geq0}I_{n}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }I\text{ is
a two-sided coideal}\right) \\
&  =\sum\limits_{\left(  m,n\right)  \in \NN^{2}}I_{n}\otimes
A_{m}+\sum\limits_{\left(  m,n\right)  \in \NN^{2}}A_{m}\otimes I_{n}.
\end{align*}
Combining this with $\Delta\left(  i\right)  \in\left(  A\otimes A\right)
_{N}$, we obtain%
\begin{align*}
\Delta\left(  i\right)   &  \in\sum\limits_{\substack{\left(  m,n\right)
\in \NN^{2};\\m+n=N}}I_{n}\otimes A_{m}+\sum\limits_{\substack{\left(
m,n\right)  \in \NN^{2};\\m+n=N}}A_{m}\otimes I_{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }I_{n}\otimes A_{m}\text{ and }%
A_{m}\otimes I_{n}\text{ are subsets of }\left(  A\otimes A\right)
_{n+m}\right) \\
&  =\sum_{n=0}^{N}I_{n}\otimes A_{N-n}+\sum_{n=0}^{N}A_{N-n}\otimes I_{n}\\
&  =I_{N}\otimes\underbrace{A_{0}}_{=\mathbf{k}\cdot1_{A}}+\sum_{n=0}%
^{N-1}\underbrace{I_{n}}_{\substack{=0\\\text{(by the induction}%
\\\text{hypothesis)}}}\otimes A_{N-n}+\underbrace{A_{0}}_{=\mathbf{k}%
\cdot1_{A}}\otimes I_{N}+\sum_{n=0}^{N-1}A_{N-n}\otimes\underbrace{I_{n}%
}_{\substack{=0\\\text{(by the induction}\\\text{hypothesis)}}}\\
&  =I_{N}\otimes\left(  \mathbf{k}\cdot1_{A}\right)  +\left(  \mathbf{k}%
\cdot1_{A}\right)  \otimes I_{N}.
\end{align*}
In other words,
\begin{equation}
\Delta\left(  i\right)  =j\otimes1_{A}+1_{A}\otimes k \label{hint.I=0.deltai}%
\end{equation}
for some $j,k\in I_{N}$. By applying $\epsilon\otimes \id$ to
both sides of this equality, and recalling the commutativity of
\eqref{counit-diagram}, we obtain $i=\epsilon\left(  j\right)  1_{A}+k$. But
$\epsilon\left(  j\right)  =0$ (since $j\in I_{N}\subset I$, so $\epsilon
\left(  j\right)  \in\epsilon\left(  I\right)  =0$), so this simplifies to
$i=k$. Similarly, $i=j$. Hence, \eqref{hint.I=0.deltai} rewrites as
$\Delta\left(  i\right)  =i\otimes1_{A}+1_{A}\otimes i$, which shows that
$i\in\mathfrak{p}$, hence $i\in I\cap\mathfrak{p}=0$ and thus $i=0$. This was
for proved for each $i\in I_{N}$, so we obtain $I_{N}=0$. This completes the
induction step, and so part (a) is solved.

(b) Exercise \ref{exe.kernel-is-coideal}(a) shows that $\ker f$ is a two-sided
coideal of $C$. If $f\mid_{\mathfrak{p}}$ is injective, then $\left(  \ker
f\right)  \cap\mathfrak{p}=0$. Now, apply part (a) of the current exercise to
$I=\ker f$.

(c) Proceed as in part (b), but use Exercise~\ref{exe.kernel-is-coideal}(b)
instead of Exercise~\ref{exe.kernel-is-coideal}(a).

\bigskip

\textit{Hint to Exercise~\ref{exe.liealg}.} (a) Straightforward (if slightly
laborious) computations.

(b) Direct verification (the hard part of which has been done in
\eqref{commutator-of-primitives} already).

(c) For every subset $S$ of a $\kk$-module $U$, we let $\left\langle
S\right\rangle $ denote the $\kk$-submodule of $U$ spanned by $S$. Our
definition of $J$ thus becomes%
\begin{equation}
J=T\left(  \mathfrak{p}\right)  \cdot C\cdot T\left(  \mathfrak{p}\right)  ,
\label{hint.liealg.1}%
\end{equation}
where $C=\left\langle xy-yx-\left[  x,y\right]  \ \mid\ x,y\in\mathfrak{p}%
\right\rangle $. A simple computation shows that each element of $C$ is
primitive. Hence,
\[
\Delta\left(  C\right)  \subset C\otimes T\left(  \mathfrak{p}\right)
+T\left(  \mathfrak{p}\right)  \otimes C.
\]
Applying $\Delta$ to both sides of \eqref{hint.liealg.1}, and recalling that
$\Delta$ is a $\kk$-algebra homomorphism, we find%
\begin{align*}
\Delta\left(  J\right)   &  =\underbrace{\Delta\left(  T\left(  \mathfrak{p}%
\right)  \right)  }_{\subset T\left(  \mathfrak{p}\right)  \otimes T\left(
\mathfrak{p}\right)  }\cdot\underbrace{\Delta\left(  C\right)  }_{\subset
C\otimes T\left(  \mathfrak{p}\right)  +T\left(  \mathfrak{p}\right)  \otimes
C}\cdot\underbrace{\Delta\left(  T\left(  \mathfrak{p}\right)  \right)
}_{\subset T\left(  \mathfrak{p}\right)  \otimes T\left(  \mathfrak{p}\right)
}\\
&  \subset\left(  T\left(  \mathfrak{p}\right)  \otimes T\left(
\mathfrak{p}\right)  \right)  \cdot\left(  C\otimes T\left(  \mathfrak{p}%
\right)  +T\left(  \mathfrak{p}\right)  \otimes C\right)  \cdot\left(
T\left(  \mathfrak{p}\right)  \otimes T\left(  \mathfrak{p}\right)  \right) \\
&  =J\otimes T\left(  \mathfrak{p}\right)  +T\left(  \mathfrak{p}\right)
\otimes J.
\end{align*}
A similar (but simpler) argument shows $\epsilon\left(  J\right)  =0$. Thus,
$J$ is a two-sided coideal of $T\left(  \mathfrak{p}\right)  $. This yields
that $T\left(  \mathfrak{p}\right)  /J$ is a $\kk$-bialgebra.

(d) We need to show that $S\left(  J\right)  \subset J$. This can be done in a
similar way as we proved $\Delta\left(  J\right)  \subset J\otimes T\left(
\mathfrak{p}\right)  +T\left(  \mathfrak{p}\right)  \otimes J$ in part (c),
once you know (from Proposition~\ref{antipodes-are-antiendomorphisms}) that
the antipode $S$ of $T\left(  \mathfrak{p}\right)  $ is a $\kk$-algebra anti-homomorphism.

\bigskip

\textit{Hint to Exercise~\ref{exe.commutative.convolution}.} Straightforward
and easy verification.

\bigskip

\textit{Hint to Exercise~\ref{exe.cocommutative.coalghom}.} Straightforward
and easy verification. Parts (a) and (b) are dual, of course.

\bigskip

\textit{Hint to Exercise~\ref{exe.comm-cocomm.anti}.} (a) Straightforward and
easy verification.

(b) The dual says the following: Let $A$ and $B$ be two $\mathbf{k}%
$-coalgebras, at least one of which is cocommutative. Prove that the
$\kk$-coalgebra anti-homomorphisms from $A$ to $B$ are the same as the
$\kk$-coalgebra homomorphisms from $A$ to $B$.

\bigskip

\textit{Hint to Exercise~\ref{exe.iterated.comm}.} For every $1\leq i<j\leq
k$, let $t_{i,j}$ be the transposition in $\mathfrak{S}_{k}$ which transposes
$i$ with $j$. It is well-known that the symmetric group $\mathfrak{S}_{k}$ is
generated by the transpositions $t_{i,i+1}$ with $i$ ranging over $\left\{
1,2,\ldots,k-1\right\}  $. However, we have $\left(  \rho\left(  \pi\right)
\right)  \circ\left(  \rho\left(  \psi\right)  \right)  =\rho\left(  \pi
\psi\right)  $ for any two elements $\pi$ and $\psi$ of $\mathfrak{S}_{k}$.
Thus, it suffices to check that
\[
m^{\left(  k-1\right)  }\circ\left(  \rho\left(  t_{i,i+1}\right)  \right)
=m^{\left(  k-1\right)  }\qquad\qquad\text{for all }i\in\left\{
1,2,\ldots,k-1\right\}  .
\]
But this is not hard to check using $m^{\left(  k-1\right)  }=m^{\left(
k-2\right)  }\circ\left(  \id_{A^{\otimes\left(  i-1\right)  }%
}\otimes m\otimes \id_{A^{\otimes\left(  k-1-i\right)  }}\right)
$ (a consequence of Exercise~\ref{exe.iterated.m}(c)) and $m\circ T=m$.

\bigskip

\textit{Hint to Exercise~\ref{exe.iterated.cocomm}.} Here is the dual
statement: Let $C$ be a cocommutative $\kk$-coalgebra, and let $k
\in \NN$. The symmetric group $\mathfrak{S}_{k}$ acts on the $k$-fold
tensor power $C^{\otimes k}$ by permuting the tensor factors: $\sigma\left(
v_{1} \otimes v_{2} \otimes\cdots\otimes v_{k}\right)  = v_{\sigma^{-1}\left(
1\right)  } \otimes v_{\sigma^{-1}\left(  2\right)  } \otimes\cdots\otimes
v_{\sigma^{-1}\left(  k\right)  }$ for all $v_{1}, v_{2}, \ldots, v_{k} \in C$
and $\sigma\in\mathfrak{S}_{k}$. For every $\pi\in\mathfrak{S}_{k}$, denote by
$\rho\left(  \pi\right)  $ the action of $\pi$ on $C^{\otimes k}$ (this is an
endomorphism of $C^{\otimes k}$). Show that every $\pi\in\mathfrak{S}_{k}$
satisfies $\left(  \rho\left(  \pi\right)  \right)  \circ\Delta^{\left(
k-1\right)  } = \Delta^{\left(  k-1\right)  }$. (Recall that $\Delta^{\left(
k-1\right)  } : C \to C^{\otimes k}$ is defined as in
Exercise~\ref{exe.iterated.Delta} for $k \geq1$, and by $\Delta^{\left(
-1\right)  } = \epsilon: C \to\mathbf{k}$ for $k = 0$.)

\bigskip

\textit{Hint to Exercise~\ref{exe.alghoms.comm}.} (a) Use
Exercise~\ref{exe.cocommutative.coalghom}(b) and
Exercise~\ref{exe.tensorprod.morphism}(a) to represent $f\star g$ as a
composition of three $\kk$-algebra homomorphisms.

(b) Induction on $k$, using part (a).

(c) Use Proposition~\ref{antipodes-are-antiendomorphisms}, Proposition
\ref{antipodes-give-convolution-inverses}(a) and the easy fact that a
composition of a $\kk$-algebra homomorphism with a $\kk$-algebra
anti-homomorphism (in either order) always is a $\kk$-algebra anti-homomorphism.

(d) Use Exercise~\ref{exe.cocommutative.coalghom}(b). Then, proceed by
induction on $k$ as in the solution of Exercise~\ref{exe.iterated.patras}(a).

(e) Use Proposition~\ref{prop.convolution.functor}.

(f) Let $H$ be a commutative $\kk$-bialgebra. Let $k$ and $\ell$ be two
nonnegative integers. Then, Exercise~\ref{exe.alghoms.comm}(b) (applied to
$A=H$ and $f_{i}= \id_{H}$) yields that $\id_{H}^{\star k}$ is a
$\kk$-algebra homomorphism $H\rightarrow H$. Now,
apply Exercise~\ref{exe.alghoms.comm}(e) to $H$, $H$, $H$, $H$, $\ell$,
$\id_{H}$, $\id_{H}^{\star k}$ and
$\id_{H}$ instead of $C$, $C^{\prime}$, $A$, $A^{\prime}$, $k$,
$f_{i}$, $\alpha$ and $\gamma$.

(g) This is an exercise in bootstrapping. First, let $k\in \NN$. Then,
part (b) of this exercise shows that $\id_{H}^{\star k}$ is a
$\kk$-algebra homomorphism. Use this together with part (c) to conclude
that $\id_{H}^{\star k}\circ S$ is again a $\kk$-algebra
homomorphism and a $\star$-inverse to $\id_{H}^{\star k}$; thus,
$\id_{H}^{\star k}\circ S
=\left(  \id_{H}^{\star k}\right)  ^{\star\left(  -1\right)  }
=\id_{H}^{\star\left(  -k\right)  }$, and this map
$\id_{H}^{\star\left(  -k\right)  }$ is a $\kk$-algebra homomorphism.

Now forget that we fixed $k$. We thus have shown that
$\id_{H}^{\star k}$ and $\id_{H}^{\star\left(  -k\right)  }$ are
$\kk$-algebra homomorphisms for each $k\in \NN$. In other words,
\begin{equation}
\id_{H}^{\star k}\text{ is a }\mathbf{k}\text{-algebra
homomorphism}\qquad\qquad\text{ for every }k\in \ZZ.
\label{hint.alghoms.comm.g.alghom}%
\end{equation}


Furthermore, we have proved the equality $\id_{H}^{\star k}\circ
S= \id_{H}^{\star\left(  -k\right)  }$ for each $k\in \NN$.
Repeating the proof of this, but now taking $k\in \ZZ$ instead of
$k\in \NN$, we conclude that it also holds for each $k\in \ZZ$
(since we already have proved \eqref{hint.alghoms.comm.g.alghom}). In other
words,%
\begin{equation}
\id_{H}^{\star\left(  -k\right)  }= \id_{H}^{\star
k}\circ S\qquad\qquad\text{ for every }k\in \ZZ.
\label{hint.alghoms.comm.g.starinv}%
\end{equation}


Now, fix two integers $k$ and $\ell$. From \eqref{hint.alghoms.comm.g.alghom},
we know that $\id_{H}^{\star k}$ is a $\kk$-algebra
homomorphism. Hence, if $\ell$ is nonnegative, then we can prove
$\id_{H}^{\star k}\circ \id_{H}^{\star\ell
}= \id_{H}^{\star\left(  k\ell\right)  }$ just as we did in the
solution to Exercise~\ref{exe.alghoms.comm}(f). But the case when $\ell$ is
negative can be reduced to the previous case by applying
\eqref{hint.alghoms.comm.g.starinv} (once to $-\ell$ instead of $k$, and once
again to $-k\ell$ instead of $k$). Thus, in each case, we obtain
$\id_{H}^{\star k}\circ \id_{H}^{\star\ell}
= \id_{H}^{\star\left(  k\ell\right)  }$.

(h) The dual of Exercise~\ref{exe.alghoms.comm}(a) is the following exercise:

\begin{statement}
If $H$ is a $\kk$-bialgebra and $C$ is a cocommutative $\mathbf{k}%
$-coalgebra, and if $f$ and $g$ are two $\kk$-coalgebra homomorphisms
$C \to H$, then prove that $f \star g$ also is a $\kk$-coalgebra
homomorphism $C \to H$.
\end{statement}

The dual of Exercise~\ref{exe.alghoms.comm}(b) is the following exercise:

\begin{statement}
If $H$ is a $\kk$-bialgebra and $C$ is a cocommutative $\mathbf{k}%
$-coalgebra, and if $f_{1}, f_{2}, \ldots, f_{k}$ are several $\mathbf{k}%
$-coalgebra homomorphisms $C \to H$, then prove that $f_{1} \star f_{2}
\star\cdots\star f_{k}$ also is a $\kk$-coalgebra homomorphism $C \to
H$.
\end{statement}

The dual of Exercise~\ref{exe.alghoms.comm}(c) is the following exercise:

\begin{statement}
If $H$ is a Hopf algebra and $C$ is a cocommutative $\kk$-coalgebra,
and if $f : C \to H$ is a $\kk$-coalgebra homomorphism, then prove that
$S \circ f : C \to H$ (where $S$ is the antipode of $H$) is again a
$\kk$-coalgebra homomorphism, and is a $\star$-inverse to $f$.
\end{statement}

The dual of Exercise~\ref{exe.alghoms.comm}(d) is the following exercise:

\begin{statement}
If $C$ is a cocommutative $\kk$-coalgebra, then show that
$\Delta^{\left(  k\right)  }$ is a $\kk$-coalgebra homomorphism for
every $k \in \NN$. (The map $\Delta^{\left(  k\right)  } : C \to
C^{\otimes\left(  k+1\right)  }$ is defined as in
Exercise~\ref{exe.iterated.Delta}.)
\end{statement}

The dual of Exercise~\ref{exe.alghoms.comm}(e) is
Exercise~\ref{exe.alghoms.comm}(e) itself (up to renaming objects and maps).

The dual of Exercise~\ref{exe.alghoms.comm}(f) is the following exercise:

\begin{statement}
If $H$ is a cocommutative $\kk$-bialgebra, and $k$ and $\ell$ are two
nonnegative integers, then prove that $\id_{H}^{\star\ell}
\circ \id_{H}^{\star k} = \id_{H}^{\star\left(  \ell k\right)  }$.
\end{statement}

The dual of Exercise~\ref{exe.alghoms.comm}(g) is the following exercise:

\begin{statement}
If $H$ is a cocommutative $\kk$-Hopf algebra, and $k$ and $\ell$ are
two integers, then prove that $\id_{H}^{\star\ell}
\circ \id_{H}^{\star k} = \id_{H}^{\star\left(  \ell k\right)  }$.
\end{statement}

\bigskip

\textit{Hint to Exercise~\ref{exe.cocomm-antipode-S2}.} This is dual to
Corollary \ref{commutative-implies-involutive-antipode-cor} (but can also
easily be shown using Exercise~\ref{exe.coalg-anti.gen}(b),
Exercise~\ref{exe.comm-cocomm.anti}(b) and
Proposition~\ref{antipodes-give-convolution-inverses}(b)).

\bigskip

\textit{Hint to Exercise~\ref{exe.dynkin}.} (a) This can be proved
computationally (using Sweedler notation), but there is a nicer argument as well:

A \emph{coderivation} of a $\kk$-coalgebra $\left(  C,\Delta
,\epsilon\right)  $ is defined as a $\kk$-linear map $F:C\rightarrow C$
such that $\Delta\circ F=\left(  F\otimes \id+ \id\otimes F\right)
\circ\Delta$. (The reader can check that this axiom is the
result of writing the axiom for a derivation in element-free terms and
reversing all arrows. Nothing less should be expected.) It is easy to see that
$E$ is a coderivation. Hence, it will be enough to check that $\left(  S\star
f\right)  \left(  a\right)  $ and $\left(  f\star S\right)  \left(  a\right)
$ are primitive whenever $f:A\rightarrow A$ is a coderivation and $a\in A$. So
fix a coderivation $f:A\rightarrow A$. Notice that the antipode $S$ of $A$ is
a coalgebra anti-endomorphism (by Exercise~\ref{exe.coalg-anti}), thus a
coalgebra endomorphism (by Exercise~\ref{exe.comm-cocomm.anti}(b)). Thus,
$\Delta\circ S=\left(  S\otimes S\right)  \circ\Delta$. Moreover,
$\Delta:A\rightarrow A\otimes A$ is a coalgebra homomorphism (by
Exercise~\ref{exe.cocommutative.coalghom}(a)) and an algebra homomorphism
(since $A$ is a bialgebra). Applying
\eqref{pre-and-post-composition-in-convolution} to $A\otimes A$, $A$, $A$,
$\Delta$, $\id_{A}$, $S$ and $f$ instead of $A^{\prime}$, $C$,
$C^{\prime}$, $\alpha$, $\gamma$, $f$ and $g$, we obtain
\begin{align*}
\Delta\circ\left(  S\star f\right)   &  =\underbrace{\left(  \Delta\circ
S\right)  }_{=\left(  S\otimes S\right)  \circ\Delta}\star\underbrace{\left(
\Delta\circ f\right)  }_{\substack{=\left(  f\otimes \id
+ \id \otimes f\right)  \circ\Delta\\\text{(since }f\text{ is a
coderivation)}}}\\
&  =\left(  \left(  S\otimes S\right)  \circ\Delta\right)  \star\left(
\left(  f\otimes \id + \id \otimes f\right)
\circ\Delta\right)  =\left(  \left(  S\otimes S\right)  \star\left(
f\otimes \id + \id \otimes f\right)  \right)
\circ\Delta\\
&  =\underbrace{\left(  \left(  S\otimes S\right)  \star\left(  f\otimes
\id \right)  \right)  }_{\substack{=\left(  S\star f\right)
\otimes\left(  S\star \id \right)  \\\text{(by
Exercise~\ref{exe.convolution.tensor}(a))}}}\circ\Delta+\underbrace{\left(
\left(  S\otimes S\right)  \star\left(  \id \otimes f\right)
\right)  }_{\substack{=\left(  S\star \id \right)  \otimes\left(
S\star f\right)  \\\text{(by Exercise~\ref{exe.convolution.tensor}(a))}}%
}\circ\Delta\\
&  =\left(  \left(  S\star f\right)  \otimes\underbrace{\left(  S\star
\id \right)  }_{=u\epsilon}\right)  \circ\Delta+\left(
\underbrace{\left(  S\star \id \right)  }_{=u\epsilon}%
\otimes\left(  S\star f\right)  \right)  \circ\Delta\\
&  =\left(  \left(  S\star f\right)  \otimes u\epsilon\right)  \circ
\Delta+\left(  u\epsilon\otimes\left(  S\star f\right)  \right)  \circ\Delta.
\end{align*}
Hence, every $a\in A$ satisfies
\begin{align*}
\left(  \Delta\circ\left(  S\star f\right)  \right)  \left(  a\right)   &
=\left(  \left(  \left(  S\star f\right)  \otimes u\epsilon\right)
\circ\Delta+\left(  u\epsilon\otimes\left(  S\star f\right)  \right)
\circ\Delta\right)  \left(  a\right) \\
&  =\left(  S\star f\right)  \left(  a\right)  \otimes1+1\otimes\left(  S\star
f\right)  \left(  a\right)
\end{align*}
(after some brief computations using \eqref{counit-diagram}). In other words,
for every $a\in A$, the element $\left(  S\star f\right)  \left(  a\right)  $
is primitive. Similarly the same can be shown for $\left(  f\star S\right)
\left(  a\right)  $, and so we are done.

(b) is a very simple computation. (Alternatively, the $\left(  S\star
E\right)  \left(  p\right)  =E\left(  p\right)  $ part follows from applying
part (c) to $a=1$, and similarly one can show $\left(  E\star S\right)
\left(  p\right)  =E\left(  p\right)  $.)

(c) This is another computation, using
Proposition~\ref{antipode-on-primitives} and the (easy) observation that $E$
is a derivation of the algebra $A$.

(d) Assume that the graded algebra $A=\bigoplus_{n\geq0}A_{n}$ is connected
and that ${\mathbb{Q}}$ is a subring of $\kk$. Let $B$ be the
$\kk$-subalgebra of $A$ generated by $\mathfrak{p}$. In order to prove
part (d), we need to show that $A\subset B$. Clearly, it suffices to show that
$A_{n}\subset B$ for every $n\in \NN$. We prove this by strong
induction on $n$; thus, we fix some $n\in \NN$, and assume as induction
hypothesis that $A_{m}\subset B$ for every $m<n$. Our goal is then to show
that $A_{n}\subset B$. This being trivial for $n=0$ (since $A$ is connected),
we WLOG assume that $n>0$. Let $a\in A_{n}$. Part (a) of this exercise yields
$\left(  S\star E\right)  \left(  a\right)  \in\mathfrak{p}\subset B$. On the
other hand, Exercise \ref{graded-connected-exercise}(h) (applied to $x=a$)
yields%
\[
\Delta\left(  a\right)  \in1\otimes a+a\otimes1+\sum_{k=1}^{n-1}A_{k}\otimes
A_{n-k}.
\]
Hence, from the definition of convolution, we obtain%
\begin{align*}
\left(  S\star E\right)  \left(  a\right)   &  \in\underbrace{S\left(
1\right)  }_{=1}E\left(  a\right)  +S\left(  a\right)  \underbrace{E\left(
1\right)  }_{=0}+\underbrace{\left(  m\circ\left(  S\otimes E\right)  \right)
\left(  \sum_{k=1}^{n-1}A_{k}\otimes A_{n-k}\right)  }_{=\sum_{k=1}%
^{n-1}S\left(  A_{k}\right)  E\left(  A_{n-k}\right)  }\\
&  =E\left(  a\right)  +\sum_{k=1}^{n-1}\underbrace{S\left(  A_{k}\right)
}_{\substack{\subset A_{k}\\\text{(since }S\text{ is graded)}}%
}\underbrace{E\left(  A_{n-k}\right)  }_{\substack{\subset A_{n-k}\subset
B\\\text{(by the induction}\\\text{hypothesis)}}}\subset E\left(  a\right)
+\sum_{k=1}^{n-1}\underbrace{A_{k}}_{\substack{\subset B\\\text{(by the
induction}\\\text{hypothesis)}}}B\subset E\left(  a\right)  +B
\end{align*}
(since $B$ is a subalgebra). Hence, $E\left(  a\right)  \in\left(  S\star
E\right)  \left(  a\right)  +B=B$ (since $\left(  S\star E\right)  \left(
a\right)  \in B$). Since $E\left(  a\right)  =na$, this becomes $na\in B$,
thus $a\in B$ (since $\mathbb{Q}$ is a subring of $\kk$). Since we have
shown this for each $a\in A_{n}$, we thus obtain $A_{n}\subset B$, and our
induction is complete.

This solution of part (d) is not the most generalizable one -- for instance,
(d) also holds if $A$ is connected filtered instead of connected graded, and
then a different argument is necessary. This is a part of the
Cartier-Milnor-Moore theorem, and appears e.g. in \cite[\S 3.2]%
{DuchampMinhTolluChienNghia}.

(e) If $a\in T\left(  V\right)  $ is homogeneous of positive degree and $p\in
V$, then part (c) quickly yields $\left(  S\star E\right)  \left(  ap\right)
=\left[  \left(  S\star E\right)  \left(  a\right)  ,p\right]  $. This allows
proving (e) by induction over $n$, with the induction base $n=1$ being a
consequence of part (b).

\bigskip

\textit{Hint to Exercise~\ref{exe.dual-algebra}.} (a) This can be done by
diagram chasing. For example, if $\mathfrak{m}$ denotes the map $\Delta
_{C}^{\ast}\circ\rho_{C,C}:C^{\ast}\otimes C^{\ast}\rightarrow C^{\ast}$, then
the diagram
\[
\xymatrix{
& & C^* \otimes C^* \otimes C^* \ar@/_5pc/[ddll]^{\mathfrak{m}\otimes\id}
\ar[dl]^{\rho_{C,C}\otimes\id} \ar[dr]^{\id\otimes\rho_{C,C}} \ar@
/^5pc/[ddrr]^{\id\otimes\mathfrak{m}} & & \\
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id}
\ar[dr]^{\rho_{C\otimes C,C}} & & C^* \otimes\left(C\otimes C\right
)^* \ar[dr]_{\id\otimes\Delta_C^*} \ar[dl]^{\rho_{C,C\otimes C}} & \\
C^* \otimes C^* \ar[dr]^{\rho_{C,C}} \ar@/_5pc/[ddrr]^{\mathfrak{m}}
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes\id
\right)^*} \ar[dr]^{\left(\id\otimes\Delta_C\right)^*} & & C^* \otimes
C^* \ar[dl]_{\rho_{C,C}} \ar@/^5pc/[ddll]^{\mathfrak{m}} \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & & \left(C\otimes
C\right)^* \ar[dl]^{\Delta_C^*} & \\
& & C^* & &
}
\]
is commutative (since each of its little triangles and squares is); thus,
$\mathfrak{m}\circ\left(  \mathfrak{m}\otimes \id \right)
=\mathfrak{m}\circ\left(  \id \otimes\mathfrak{m}\right)  $ for
$\mathfrak{m}$. This proves that the diagram \eqref{associativity-diagram}
commutes for our algebra $C^{\ast}$. The commutativity of \eqref{unit-diagram}
is obtained similarly.

Alternatively, we could also solve part (a) trivially by first solving part
(b) and then recalling Exercise \ref{exe.convolution.assoc}.

(b) Straightforward verification on pure tensors.

(c) Let $C=\bigoplus_{n\geq0}C_{n}$ be a graded $\kk$-coalgebra. For
every $n\in \NN$, we identify $\left(  C_{n}\right)  ^{\ast}$ with a
$\kk$-submodule of $C^{\ast}$, namely with the $\kk$-submodule
$\left\{  f\in C^{\ast}\ \mid\ f\left(  C_{p}\right)  =0\text{ for all }%
p\in \NN \text{ satisfying }p\neq n\right\}  $. By the definition of
$C^{o}$, we have $C^{o}=\bigoplus_{n\geq0}\left(  C_{n}\right)  ^{\ast}$.
Hence, it remains to show that $\left(  C_{a}\right)  ^{\ast}\left(
C_{b}\right)  ^{\ast}\subset\left(  C_{a+b}\right)  ^{\ast}$ for all
$a,b\in \NN$, and that $1_{C^{\ast}}\in\left(  C_{0}\right)  ^{\ast}$.
But this is straightforward using the gradedness of $\Delta$ and $\epsilon$.

(d) Diagram chasing or simple element-wise verification.

(e) Simple linear algebra (no Hopf algebras involved here).

(f) The ``only if'' direction is proved in
the same way as part (d) (or as a corollary of part (d), since $D^{\circ}$ and
$C^{\circ}$ are subalgebras of $D^{\ast}$ and $C^{\ast}$). It remains to prove
the ``if'' direction.

Assume that $f^{\ast}:D^{o}\rightarrow C^{o}$ is a $\kk$-algebra
morphism. We want to show that $f:C\rightarrow D$ is a $\kk$-coalgebra
morphism. In other words, we want to show that the two diagrams
\begin{equation}
\xymatrix{ C \ar[d]_{\Delta_C} \ar[r]^{f}
& D \ar[d]^{\Delta_D} \\ C \otimes C \ar[r]^{f \otimes f} & D \otimes
D }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ C \ar[dr]_{\epsilon_C} \ar[rr]^{f}%
& & D \ar[dl]^{\epsilon_D} \\ &\kk& }
\label{hint.dual-algebra.e.1}
\end{equation}
commute. Let us start with the left one of these diagrams. The graded
$\kk$-module $D$ is of finite type, and therefore the map $\rho
_{D,D}:D^{o}\otimes D^{o}\rightarrow\left(  D\otimes D\right)  ^{o}$ (a
restriction of the map $\rho_{D,D}:D^{\ast}\otimes D^{\ast}\rightarrow\left(
D\otimes D\right)  ^{\ast}$) is an isomorphism. Its inverse $\rho_{D,D}%
^{-1}:\left(  D\otimes D\right)  ^{o}\rightarrow D^{o}\otimes D^{o}$ is
therefore well-defined\footnote{Beware: we don't have an inverse of the
non-restricted map $\rho_{D,D}:D^{\ast}\otimes D^{\ast}\rightarrow\left(
D\otimes D\right)  ^{\ast}$.}. We can thus form the (asymmetric!) diagram
\begin{equation}
{\xymatrixcolsep{5pc}
\xymatrix{ D^o \ar[rrr]^{f^*}
& & & C^o \\ & D^o \otimes D^o \ar[ul]^{m_{D^*}} \ar[r]^{f^* \otimes f^*}
& C^o \otimes C^o \ar[ur]^{m_{C^*}} \ar[dr]^{\rho_{C,C}} & \\ \left(D\otimes
D\right)^o \ar[uu]^{\Delta_D^*} \ar[ur]^{\rho_{D,D}^{-1}} \ar[rrr]_{\left
(f\otimes f\right)^*} & & & \left(C\otimes C\right)^o \ar[uu]^{\Delta_C^*} }%
}.
\label{hint.dual-algebra.e.2}
\end{equation}
(The arrows labelled $m_{C^{\ast}}$ and $m_{D^{\ast}}$ could just as well have
been labelled $m_{C^{o}}$ and $m_{D^{o}}$, since the multiplication maps
$m_{C^{o}}$ and $m_{D^{o}}$ are restrictions of $m_{C^{\ast}}$ and
$m_{D^{\ast}}$.) Argue that the diagram \eqref{hint.dual-algebra.e.2}
commutes. Thus, $f^{\ast}\circ\Delta_{D}^{\ast}=\Delta_{C}^{\ast}\circ\left(
f\otimes f\right)  ^{\ast}$ as maps from $\left(  D\otimes D\right)  ^{o}$ to
$C^{o}$. In other words, $\left(  \Delta_{D}\circ f\right)  ^{\ast}=\left(
\left(  f\otimes f\right)  \circ\Delta_{C}\right)  ^{\ast}$ as maps from
$\left(  D\otimes D\right)  ^{o}$ to $C^{o}$. But a general linear-algebraic
fact states that if $U$ and $V$ are two graded $\kk$-modules such that
$V$ is of finite type, and if $\alpha$ and $\beta$ are two graded $\mathbf{k}%
$-linear maps $U\rightarrow V$ such that $\alpha^{\ast}=\beta^{\ast}$ as maps
from $V^{o}$ to $U^{o}$, then $\alpha=\beta$\ \ \ \ \footnote{This follows
immediately from Exercise~\ref{exe.dual-algebra} (e).}. Hence, $\left(
\Delta_{D}\circ f\right)  ^{\ast}=\left(  \left(  f\otimes f\right)
\circ\Delta_{C}\right)  ^{\ast}$ leads to $\Delta_{D}\circ f=\left(  f\otimes
f\right)  \circ\Delta_{C}$. In other words, the first diagram in
\eqref{hint.dual-algebra.e.1} commutes. The second is similar but easier.
Thus, $f$ is a $\kk$-coalgebra morphism, and the ``if'' direction is proved.

\bigskip

\textit{Hint to Exercise~\ref{exe.Sym.1dim}.} Straightforward computations.
For part (d), first show (independently of whether $\kk$ is a field and
its characteristic) that $\left(  f^{\left(  1\right)  }\right)
^{m}=m!f^{\left(  m\right)  }$ for every $m\in \NN$.

\bigskip

\textit{Hint to Exercise~\ref{exe.Sym.wedge}.} It is best to solve parts (c)
and (d) before approaching (b).

(a) Both maps $\Delta_{\operatorname*{Sym}V}$ and%
\[%
\begin{array}
[c]{ccc}%
\mathbf{k}[{\mathbf{x}}] & \overset{\Delta}{\longrightarrow} & \mathbf{k}%
[{\mathbf{x}},{\mathbf{y}}],\\
f(x_{1},\ldots,x_{n}) & \longmapsto & f(x_{1}+y_{1},\ldots,x_{n}+y_{n})
\end{array}
\]
are $\kk$-algebra homomorphisms. Thus, in order to check that they are
equal, it suffices to verify that they agree on $V$ (since $V$ generates
$\operatorname*{Sym}V$).

(c) This is a straightforward computation unless you get confused with the
topologist's sign convention. The latter convention affects the twist map
$T=T_{T\left(  V\right)  ,T\left(  V\right)  }:T\left(  V\right)  \otimes
T\left(  V\right)  \rightarrow T\left(  V\right)  \otimes T\left(  V\right)  $
(in particular, we now have $T\left(  x\otimes x\right)  =-x\otimes x$ instead
of $T\left(  x\otimes x\right)  =x\otimes x$), and thus also affects the
multiplication in the $\kk$-algebra $T\left(  V\right)  \otimes
T\left(  V\right)  $, because this multiplication is given by%
\[
m_{T\left(  V\right)  \otimes T\left(  V\right)  }=\left(  m_{T\left(
V\right)  }\otimes m_{T\left(  V\right)  }\right)  \circ\left(
\id \otimes T \otimes \id \right)  .
\]
Make sure you understand why this leads to $\left(  1\otimes x\right)
\cdot\left(  x\otimes1\right)  =-x\otimes x$ (whereas $\left(  x\otimes
1\right)  \cdot\left(  1\otimes x\right)  =x\otimes x$).

(d) The trickiest part is showing that $J$ is a graded $\kk$-submodule
of $T\left(  V\right)  $. It suffices to check that $J$ is generated (as a
two-sided ideal) by homogeneous elements\footnote{Make sure you understand
why.}; however, this is not completely trivial, as the designated generators
$x^{2}$ for $x\in V$ need not be homogeneous. However, it helps to observe
that $J$ is also the two-sided ideal generated by the set%
\[
\left\{  x\otimes x\right\}  _{x\in V\text{ is homogeneous}}\cup\left\{
x\otimes y+y\otimes x\right\}  _{x,y\in V\text{ are homogeneous}}%
\]
(why?), which set does consist of homogeneous elements. Thus, $J$ is a graded
$\kk$-submodule of $T\left(  V\right)  $. From part (c), it is easy to
observe that $J$ is a two-sided coideal of $T\left(  V\right)  $ as well.
Hence, $T\left(  V\right)  /J$ inherits a graded $\kk$-bialgebra
structure from $T\left(  V\right)  $. The rest is easy.

(b) is now a consequence of what has been done in (d).

\bigskip

\textit{Hint to Exercise~\ref{exe.convolution.dual}.} Easy and straightforward.

\bigskip

\textit{Hint to Exercise~\ref{exe.shuffle-alg}.} The hint after the exercise
shows the way; here are a few more pointers. The solution proceeds in two steps:

\begin{itemize}
\item \textit{Step 1:} Show that Proposition~\ref{prop.shuffle-alg} holds when
$V$ is a finite free $\kk$-module.

\item \textit{Step 2:} Use this to conclude that
Proposition~\ref{prop.shuffle-alg} always holds.
\end{itemize}

The trick to Step 1 is to reduce the proof to Example \ref{exa.shuffle-alg}.
In a bit more detail: If $V$ is a finite free $\kk$-module with basis
$\left(  v_{1},v_{2},\ldots,v_{n}\right)  $, then we know from Example
\ref{exa.shuffle-alg} that the graded dual $A^{o}$ of its tensor algebra
$A:=T\left(  V\right)  $ is a Hopf algebra whose basis $\left\{  y_{\left(
i_{1},i_{2},\ldots,i_{\ell}\right)  }\right\}  $ is indexed by words in the
alphabet $I:=\left\{  1,2,\ldots,n\right\}  $. This allows us to define a
$\kk$-linear map $\phi:A^{o}\rightarrow T\left(  V\right)  $ by
setting
\[
\phi\left(  y_{\left(  i_{1},i_{2},\ldots,i_{\ell}\right)  }\right)
=v_{i_{1}}v_{i_{2}}\cdots v_{i_{\ell}}\ \ \ \ \ \ \ \ \ \ \text{for every
}\ell\in \NN \text{ and }\left(  i_{1},i_{2},\ldots,i_{\ell}\right)
\in I^{\ell}.
\]
This $\kk$-linear map $\phi$ then is an isomorphism from the Hopf
algebra $A^{o}$ to the putative Hopf algebra $\left(  \operatorname*{Sh}%
\left(  V\right)  ,\shufmult,1_{T\left(  V\right)  },\Delta_{\shuffle}%
,\epsilon,S\right)  $, in the sense that it is invertible (since it sends a
basis to a basis) and satisfies the five equalities%
\begin{align*}
\phi\circ m_{A^{o}}  &  =m_{\shuffle}\circ\left(  \phi\otimes\phi\right)  ,\\
\phi\circ u_{A^{o}}  &  =u,\\
\left(  \phi\otimes\phi\right)  \circ\Delta_{A^{o}}  &  =\Delta_{\shuffle}%
\circ\phi,\\
\epsilon_{A^{o}}  &  =\epsilon\circ\phi,\\
\phi\circ S_{A^{o}}  &  =S\circ\phi
\end{align*}
(check all these -- for instance, the first of these equalities follows by
comparing \eqref{eq.exa.shuffle-alg.m3} with the definition of $\shufmult$).
Thus, the latter putative Hopf algebra is an actual Hopf algebra (since the
former is). This proves Proposition~\ref{prop.shuffle-alg} for our finite free
$V$, and thus completes Step 1.

Step 2 demonstrates the power of functoriality. We want to prove
Proposition~\ref{prop.shuffle-alg} in the general case, knowing that it holds
when $V$ is finite free. So let $V$ be an arbitrary $\kk$-module. For
the sake of brevity, we shall write $\mathbf{V}$ for $T\left(  V\right)  $.
Let $m_{\shuffle}$ denote the $\kk$-linear map $\mathbf{V}%
\otimes\mathbf{V}\rightarrow\mathbf{V}$ which sends every $a\otimes b$ to
$a\shufmult b$. One of the things that need to be shown is the commutativity
of the diagram
\begin{equation}
\xymatrix{
& \mathbf{V} \otimes\mathbf{V} \ar[dl]_{\Delta_\shuffle\otimes\Delta_\shuffle}
\ar[ddr]_{m_\shuffle} & \\
\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V}
\ar[dd]_{\id\otimes T \otimes\id} & & \\
& & \mathbf{V} \ar[ddl]_{\Delta_\shuffle}\\
\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V}
\ar[dr]_{m_\shuffle\otimes m_\shuffle} & & \\
& \mathbf{V} \otimes\mathbf{V} &
}
,
\label{hint.shuffle-alg.diag-to-prove}
\end{equation}
where $T$ is the twist map $T_{\mathbf{V},\mathbf{V}}$. By linearity, it is
clearly enough to verify this only on the pure tensors; that is, it is enough
to check that every $a\in\mathbf{V}$ and $b\in\mathbf{V}$ satisfy
\begin{equation}
\left(  \left(  m_{\shuffle}\otimes m_{\shuffle}\right)  \circ\left(
\id \otimes T \otimes \id \right)  \circ\left(
\Delta_{\shuffle}\otimes\Delta_{\shuffle}\right)  \right)  \left(  a\otimes
b\right)  =\left(  \Delta_{\shuffle}\circ m_{\shuffle}\right)  \left(
a\otimes b\right)  . \label{hint.shuffle-alg.diag-toprove}%
\end{equation}
So let $a,b\in\mathbf{V}$ be arbitrary. WLOG assume that $a=v_{1}v_{2}\cdots
v_{p}$ and $b=v_{p+1}v_{p+2}\cdots v_{p+q}$ for some $p,q\in \NN$ and
$v_{1},v_{2},\ldots,v_{p+q}\in V$. Define $W$ to be the free $\mathbf{k}%
$-module with basis $\left(  x_{1},x_{2},\ldots,x_{p+q}\right)  $, and let
$\mathbf{W}$ be its tensor algebra $T\left(  W\right)  $. Then, $W$ is a
finite free $\kk$-module, and so we know from Step 1 that
Proposition~\ref{prop.shuffle-alg} holds for $W$ instead of $V$. But we can
define a $\kk$-linear map $f:W\rightarrow V$ that sends $x_{1}%
,x_{2},\ldots,x_{p+q}$ to $v_{1},v_{2},\ldots,v_{p+q}$, respectively. This map
$f:W\rightarrow V$ clearly induces a $\kk$-algebra homomorphism
$\mathbf{f}:=T\left(  f\right)  :\mathbf{W}\rightarrow\mathbf{V}$ that
respects all relevant shuffle-algebraic structure (i.e., it satisfies
$\mathbf{f}\circ m_{\shuffle}=m_{\shuffle}\circ\left(  \mathbf{f}%
\otimes\mathbf{f}\right)  $ and $\left(  \mathbf{f}\otimes\mathbf{f}\right)
\circ\Delta_{\shuffle}=\Delta_{\shuffle}\circ\mathbf{f}$ and so on), simply
because this structure has been defined canonically in terms of each of $V$
and $W$. Thus, in the diagram%
\[
\xymatrixcolsep{3.5pc}
\xymatrix{
& &  \mathbf{W} \otimes\mathbf{W} \ar[ddll]_{\Delta_\shuffle\otimes
\Delta_\shuffle}
\ar[dddrr]^{m_\shuffle} \ar[d]_{\mathbf{f}\otimes\mathbf{f}} & \\
& & \mathbf{V} \otimes\mathbf{V} \ar[dl]_{\Delta_\shuffle\otimes
\Delta_\shuffle} \ar[ddr]_{m_\shuffle} & \\
\mathbf{W} \otimes\mathbf{W} \otimes\mathbf{W} \otimes\mathbf{W} \ar
[dd]_{\id\otimes T \otimes\id}
\ar[r]_-{\mathbf{f}\otimes\mathbf{f}\otimes\mathbf{f}\otimes\mathbf{f}} &
\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V}
\ar[dd]_{\id\otimes T \otimes\id}
& & \\
& & &  \mathbf{V} \ar[ddl]_{\Delta_\shuffle} &  \mathbf{W} \ar[dddll]^{\Delta
_\shuffle}
\ar[l]^{\mathbf{f}} \\
\mathbf{W} \otimes\mathbf{W} \otimes\mathbf{W} \otimes\mathbf{W}
\ar[ddrr]_{m_\shuffle\otimes m_\shuffle}
\ar[r]^-{\mathbf{f}\otimes\mathbf{f}\otimes\mathbf{f}\otimes\mathbf{f}} &
\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V} \otimes\mathbf{V}
\ar[dr]_{m_\shuffle\otimes m_\shuffle}
& & \\
& &  \mathbf{V} \otimes\mathbf{V} & \\
& &  \mathbf{W} \otimes\mathbf{W} \ar[u]^{\mathbf{f}\otimes\mathbf{f}} &
}
,
\]
all the little quadrilaterals commute. The outer pentagon also commutes, since
Proposition~\ref{prop.shuffle-alg} holds for $W$ instead of $V$. If
$\mathbf{f}$ was surjective, then we would be able to conclude that the inner
pentagon also commutes, so we would immediately get the commutativity of
\eqref{hint.shuffle-alg.diag-to-prove}. But even if $\mathbf{f}$ is not
surjective, we are almost there: The inner pentagon commutes on the image of
the map $\mathbf{f}\otimes\mathbf{f}:\mathbf{W}\otimes\mathbf{W}%
\rightarrow\mathbf{V}\otimes\mathbf{V}$ (because when we start at
$\mathbf{W}\otimes\mathbf{W}$, we can walk around the outer pentagon instead,
which is known to commute), but this image contains $a\otimes b$ (since
$a=v_{1}v_{2}\cdots v_{p}=\mathbf{f}\left(  x_{1}x_{2}\cdots x_{p}\right)  $
and similarly $b=\mathbf{f}\left(  x_{p+1}x_{p+2}\cdots x_{p+q}\right)  $), so
we conclude that \eqref{hint.shuffle-alg.diag-toprove} holds, as we wanted to show.

This is only one of the diagrams we need to prove in order to prove
Proposition~\ref{prop.shuffle-alg}, but the other diagrams are done in the
exact same way.

\bigskip

\textit{Hint to Exercise~\ref{exe.pw-fin-supp.P1-5}.} Straightforward
reasoning using facts like ``a union of finitely many finite
sets is finite'' and ``a tensor is a sum of
finitely many pure tensors''.

\bigskip

\textit{Hint to Exercise~\ref{exe.convolution-series}.} Parts (a), (b), (d)
and (e) of Proposition~\ref{prop.convolution-series} are easy. (In proving
\eqref{eq.exe.convolution-series.b.2} and later, it helps to first establish
an extension of \eqref{eq.exe.convolution-series.b.1} to infinite
sums\footnote{Namely: Let $\left(  r_{q}\right)  _{q\in Q}\in\left(
\mathbf{k}\left[  \left[  T\right]  \right]  \right)  ^{Q}$ be a family of
power series such that the (possibly infinite) sum $\sum_{q\in Q}r_{q}$
converges in $\mathbf{k}\left[  \left[  T\right]  \right]  $. Let
$f\in\mathfrak{n}\left(  C,A\right)  $. Then, the family $\left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported and satisfies $\left(  \sum_{q\in Q}r_{q}\right)  ^{\star}\left(
f\right)  =\sum_{q\in Q}\left(  r_{q}\right)  ^{\star}\left(  f\right)  $.}.)
For part (c), recall that the binomial formula $\left(  a+b\right)  ^{n}%
=\sum_{k=0}^{n}\dbinom{n}{k}a^{k}b^{n-k}$ holds for any two commuting elements
$a$ and $b$ of any ring (such as $f$ and $g$ in the convolution algebra
$\Hom \left(  C,A\right)  $). Part (f) follows from (e) using
\eqref{eq.exe.convolution-series.b.2}. Part (g) is best proved in two steps:
First, use induction to prove part (g) in the case when $u=T^{k}$ for some
$k\in \NN$ (this relies on \eqref{eq.exe.convolution-series.b.2}); then,
notice that both sides of \eqref{eq.exe.convolution-series.g.1} depend
$\kk$-linearly on $u$, whence the general case follows (up to some
mudfighting with infinite sums). Part (h) is an instance of the
``local $\star$-nilpotence'' already observed
in the proof of Proposition \ref{Takeuchi-formula}. Part (j) follows from (h).
Part (i) follows from Proposition \ref{prop.convolution.functor} (applied to
$C^{\prime}=C$, $A^{\prime}=B$, $\gamma= \id_{C}$ and
$\alpha=s$) in a similar way as part (g) followed from \eqref{eq.exe.convolution-series.b.2}.

\bigskip

\textit{Hint to Exercise~\ref{exe.exp-log}.}
Proposition~\ref{prop.exp-log-inv} is a classical result, often proved by a
lazy reference to the mythical complex analysis class the reader has surely
seen it in. Here is a do-it-yourself purely algebraic proof:

\begin{itemize}
\item \textit{Step 1:} If $u,v\in\mathbf{k}\left[  \left[  T\right]  \right]
$ are two power series having the same constant term and satisfying $\dfrac
{d}{dT}u=\dfrac{d}{dT}v$, then $u=v$. This simple lemma (whose analogue for
differentiable functions is a fundamental fact of real analysis) is easily
proved by comparing coefficients in $\dfrac{d}{dT}u=\dfrac{d}{dT}v$ and
recalling that $\kk$ is a $\mathbb{Q}$-algebra (so $1,2,3,\ldots$ are
invertible in $\kk$).

\item \textit{Step 2:} If $u,v\in\mathbf{k}\left[  \left[  T\right]  \right]
$ are two power series having constant term $1$ and satisfying $\left(
\dfrac{d}{dT}u\right)  \cdot v=\left(  \dfrac{d}{dT}v\right)  \cdot u$, then
$u=v$. This can be proved by applying Step 1 to $uv^{-1}$ and $1$ instead of
$u$ and $v$.

\item \textit{Step 3:} The power series $\overline{\log}\left[  \overline
{\exp}\right]  $ and $\overline{\exp}\left[  \overline{\log}\right]  $ are
well-defined and have constant term $0$. (Easy.)

\item \textit{Step 4:} If $w\in\mathbf{k}\left[  \left[  T\right]  \right]  $
is a power series having constant term $0$, then%
\begin{align*}
\dfrac{d}{dT}\left(  \overline{\exp}\left[  w\right]  \right)   &  =\left(
\dfrac{d}{dT}w\right)  \cdot\exp\left[  w\right]
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\dfrac{d}{dT}\left(  \overline{\log}\left[  w\right]  \right)   &  =\left(
\dfrac{d}{dT}w\right)  \cdot\dfrac{1}{1+w}.
\end{align*}
These formulas can be derived from the chain rule, or more directly from
$\overline{\exp}\left[  w\right]  =\sum_{n\geq1}\dfrac{1}{n!}w^{n}$ and
$\overline{\log}\left[  w\right]  =\sum_{n\geq1}\dfrac{\left(  -1\right)
^{n-1}}{n}w^{n}$.

\item \textit{Step 5:} Show $\overline{\exp}\left[  \overline{\log}\right]
=T$ by applying Step 2 to $u=\exp\left[  \overline{\log}\right]  $ and $v=1+T$.

\item \textit{Step 6:} Show $\overline{\log}\left[  \overline{\exp}\right]
=T$ by applying Step 1 to $u=\overline{\log}\left[  \overline{\exp}\right]  $
and $v=T$.
\end{itemize}

Lemma~\ref{lem.exp-log-log} easily follows from
Proposition~\ref{prop.convolution-series}(f).

Remains to prove Proposition~\ref{prop.exp-log}. It is easy to see that
$\log^{\star}\left(  \exp^{\star}f\right)  =\overline{\log}^{\star}\left(
\overline{\exp}^{\star}f\right)  $ for each $f\in\mathfrak{n}\left(
C,A\right)  $; thus, Proposition~\ref{prop.exp-log}(a) follows from
(\ref{eq.exe.convolution-series.g.1}) using Proposition~\ref{prop.exp-log-inv}
and Proposition~\ref{prop.convolution-series}(f) (since $T^{\star}\left(
f\right)  =f$). A similar argument yields Proposition~\ref{prop.exp-log}(b)
(this time, we need to observe that $\exp^{\star}\left(  \log^{\star}g\right)
=\overline{\exp}^{\star}\left(  \overline{\log}^{\star}\left(  g-u_{A}%
\epsilon_{C}\right)  \right)  +u_{A}\epsilon_{C}$ first). To prove
Proposition~\ref{prop.exp-log}(c), first use
Proposition~\ref{prop.convolution-series}(c) to show that $\exp^{\star}\left(
f+g\right)  $ is well-defined; then, apply the well-known fact that
$\exp\left(  x+y\right)  =\exp x\cdot\exp y$ for any two commuting elements
$x$ and $y$ of a ring (provided the exponentials are well-defined; some
yak-shaving is required here to convince oneself that the infinite sums behave
well)\footnote{If you have not seen this well-known fact, prove it by a quick
computation using the binomial formula.}. Part (d) is trivial. Part (e) is an
induction on $n$. Part (f) is a rehash of the definition of $\log^{\star
}\left(  f+u_{A}\epsilon_{C}\right)  =\overline{\log}^{\star}f$.

\bigskip

\textit{Hint to Exercise~\ref{exe.leray.exp-alg}.}
Proposition~\ref{prop.convolution.nat}(a) is easily proved by unpacking the
definition of convolution (just like
Proposition~\ref{prop.convolution.functor}). Part (b) follows from (a) by induction.

The trick to Proposition~\ref{prop.leray.f*nxy} is to realize that if
$f\in\Hom \left(  C,A\right)  $ is as in
Proposition~\ref{prop.leray.f*nxy}, then every $x,y\in C$ satisfy%
\begin{equation}
f\left(  xy\right)  =\epsilon\left(  y\right)  f\left(  x\right)
+\epsilon\left(  x\right)  f\left(  y\right)  ,
\label{hint.leray.exp-alg.f*nxy.1}%
\end{equation}
because $xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x=\epsilon\left(  x\right)  \epsilon\left(  y\right)  \cdot
1+\underbrace{\left(  x-\epsilon\left(  x\right)  \right)  }_{\in\ker\epsilon
}\underbrace{\left(  y-\epsilon\left(  y\right)  \right)  }_{\in\ker\epsilon}$
is annihilated by $f$. Once this equality is known, it is not hard to prove
Proposition~\ref{prop.leray.f*nxy} ``by hand'' by induction on $n$
(using Sweedler notation). Alternatively, for a cleaner
proof, the equality \eqref{hint.leray.exp-alg.f*nxy.1} can be restated in an
element-free way as
\[
f\circ m_{C}=m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f\right)  ,
\]
where $\mathfrak{i}=u_{A}\circ\epsilon_{C}$ is the unity of the $\mathbf{k}%
$-algebra $\left(  \Hom \left(  C,A\right)  ,\star\right)  $;
then, an application of Proposition~\ref{prop.convolution.nat}(b) shows that
every $n\in \NN$ satisfies%
\begin{align*}
f^{\star n}\circ m_{C}  &  =m_{A}\circ\underbrace{\left(  f\otimes
\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star n}}_{\substack{=\sum
_{i=0}^{n}\dbinom{n}{i}\left(  f\otimes\mathfrak{i}\right)  ^{\star i}%
\star\left(  \mathfrak{i}\otimes f\right)  ^{\star\left(  n-i\right)
}\\\text{(by the binomial formula,}\\\text{since }f\otimes\mathfrak{i}\text{
and }\mathfrak{i}\otimes f\text{ commute in}\\\text{the convolution algebra
}\Hom \left(  C\otimes C,A\otimes A\right)  \text{)}}%
}=m_{A}\circ\left(  \sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(
f\otimes\mathfrak{i}\right)  ^{\star i}\star\left(  \mathfrak{i}\otimes
f\right)  ^{\star\left(  n-i\right)  }}_{\substack{=f^{\star i}\otimes
f^{\star\left(  n-i\right)  }\\\text{(by repeated application of
Exercise~\ref{exe.convolution.tensor}(a))}}}\right) \\
&  =m_{A}\circ\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes
f^{\star\left(  n-i\right)  }\right)  ,
\end{align*}
which is precisely Proposition~\ref{prop.leray.f*nxy} (restated in an
element-free way).

Proposition~\ref{prop.leray.exp-algh1} is an easy consequence of
Proposition~\ref{prop.leray.f*nxy}, since $\left(  \exp^{\star}f\right)
\left(  xy\right)  =\sum_{n\in \NN}\dfrac{1}{n!}f^{\star n}\left(
xy\right)  $. (Again, fighting infinite sums is probably the most laborious
part of the proof.)

Lemma \ref{lem.sol.eulerian-idp.2.poly1} can be reduced to the fact that the
matrix $\left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\in{\mathbb{Q}%
}^{\left(  N+1\right)  \times\left(  N+1\right)  }$ is invertible (since its
determinant is the Vandermonde determinant $\prod_{1\leq i<j\leq
N+1}\underbrace{\left(  i-j\right)  }_{\neq0}\neq0$) and thus has trivial
kernel (not just over $\mathbb{Q}$, but on any torsionfree abelian group).

Lemma~\ref{lem.leray.poly1-inf} follows from Lemma
\ref{lem.sol.eulerian-idp.2.poly1}, because a finitely supported family
indexed by nonnegative integers must become all zeroes from some point on.

The proof of Proposition~\ref{prop.leray.exp-algh2} is rather surprising: It
suffices to show that $f\left(  xy\right)  =0$ for all $x,y\in\ker\epsilon$.
So let us fix $x,y\in\ker\epsilon$. Proposition~\ref{prop.convolution-series}%
(h) yields $f\in\mathfrak{n}\left(  C,A\right)  $. Let $t\in \NN$ be
arbitrary. Then, Proposition~\ref{prop.exp-log}(e) (applied to $n=t$) shows
that $tf\in\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}\left(
tf\right)  =\left(  \exp^{\star}f\right)  ^{\star t}$. But
Exercise~\ref{exe.alghoms.comm}(b) shows that $\left(  \exp^{\star}f\right)
^{\star t}$ is a $\kk$-algebra homomorphism $C\rightarrow A$. Hence,
$\left(  \exp^{\star}f\right)  ^{\star t}\left(  xy\right)  =\left(
\exp^{\star}f\right)  ^{\star t}\left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  ^{\star t}\left(  y\right)  $. Rewriting $\left(  \exp^{\star
}f\right)  ^{\star t}$ as $\exp^{\star}\left(  tf\right)  =\sum_{n\in
 \NN}\dfrac{1}{n!}f^{\star n}t^{n}$ on both sides, and multiplying out
the right hand side, we can rewrite this as%
\[
\sum_{k\in \NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)  t^{k}%
=\sum_{k\in \NN}\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)
}{\left(  k-i\right)  !}\right)  t^{k}.
\]
In other words,%
\[
\sum_{k\in \NN} w_k t^{k}=0,
\ \ \ \ \ \ \ \ \ \ \text{where we set }
w_k = \dfrac{1}{k!}f^{\star k}\left(  xy\right)
-\sum_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac
{f^{\star\left(  k-i\right)  }\left(  y\right)  }{\left(  k-i\right)
!} .
\]
But we have proved this for all $t\in \NN$. Thus,
Lemma~\ref{lem.leray.poly1-inf} shows that%
\[
w_k = 0\ \ \ \ \ \ \ \ \ \ \text{for every }%
k\in \NN.
\]
Applying this to $k=1$ and simplifying, we obtain $f\left(  xy\right)
-\epsilon\left(  x\right)  f\left(  y\right)  -f\left(  x\right)
\epsilon\left(  y\right)  =0$. Since $x,y\in\ker\epsilon$, this simplifies
even further to $f\left(  xy\right)  =0$, which proves
Proposition~\ref{prop.leray.exp-algh2}.

Finally, we need to prove Proposition~\ref{prop.leray.exp-algh3}. Set
$F=\exp^{\star}f$ and $\widetilde{F}=F-u_{A}\epsilon_{C}$, so that
$\widetilde{F}\in\mathfrak{n}\left(  C,A\right)  $. Then,
Proposition~\ref{prop.leray.exp-algh1} shows that $F:C\rightarrow A$ is a
$\kk$-algebra homomorphism, so it remains to show that $F$ is
surjective. But it is easy to see using Proposition~\ref{prop.exp-log}(a) that
$f=\overline{\log}^{\star}\widetilde{F}$.

Define $\widetilde{\id} \in \mathfrak{n}\left(  C,C\right)  $ by
$\widetilde{\id} = \id_{C}-u_{C}\epsilon_{C}$.
Then, it is not hard to see that $F\circ
\widetilde{\id}=\widetilde{F}$. Hence, $f=\overline{\log}%
^{\star}\underbrace{\widetilde{F}}_{=F\circ\widetilde{\id}}
=\overline{\log}^{\star}\left(  F\circ\widetilde{\id}\right)
=F\circ\left(  \overline{\log}^{\star}\left(  \widetilde{\id}
\right)  \right)  $ (by Proposition~\ref{prop.convolution-series}(i), since
$F$ is a $\kk$-algebra homomorphism). Therefore, $f\left(  C\right)
\subset F\left(  C\right)  $. Since $F$ is a $\kk$-algebra
homomorphism, this entails that $F\left(  C\right)  $ is a $\mathbf{k}%
$-subalgebra of $A$ that contains $f\left(  C\right)  $ as a subset. But this
causes $F\left(  C\right)  $ to be the whole $A$ (since $f\left(  C\right)  $
generates $A$). Thus, $F$ is surjective, so
Proposition~\ref{prop.leray.exp-algh3} is proven.

\bigskip

\textit{Hint to Exercise~\ref{exe.leray.leray-e}.} We must prove
Theorem~\ref{thm.leray.leray-e}. Part (a) is easy. For the remainder of the
proof, we set $\widetilde{\id}= \id_{A}%
-u_{A}\epsilon_{A}\in\operatorname*{End}A$, and equip ourselves with some
simple lemmas:

\begin{itemize}
\item The kernel $\ker\epsilon$ is an ideal of $A$.

\item We have $\widetilde{\id}\in\mathfrak{n}\left(
A,A\right)  $ and $\ker\widetilde{\id} = \kk \cdot1_{A}$ and
$\widetilde{\id}\left(  A\right)  =\ker\epsilon$.

\item We have $A/\left(  \mathbf{k}\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}\right)  \cong\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)
^{2}$ as $\kk$-modules.
\end{itemize}

Now, to the proof of Theorem~\ref{thm.leray.leray-e}(b). Using $\mathfrak{e}%
=\log^{\star}\left(  \id_{A}\right)  =\overline{\log
}^{\star}\widetilde{\id}$ and $\widetilde{\id}
\left(  1_{A}\right)  =0$, it is easy to see that $\mathfrak{e}\left(
1_{A}\right)  =0$. Hence, $\mathfrak{e}\left(  A_{0}\right)  =0$ since $A$ is
connected. Thus, Proposition~\ref{prop.leray.exp-algh2} shows that
$\mathfrak{e}\left(  \left(  \ker\epsilon\right)  ^{2}\right)  =0$ (since
$\exp^{\star}\mathfrak{e}= \id_{A}$ is a $\mathbf{k}%
$-algebra homomorphism). Combined with $\mathfrak{e}\left(  1_{A}\right)  =0$,
this yields $\mathbf{k}\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset
\ker\mathfrak{e}$. But this inclusion is actually an equality, as we can show
by the following computation: We have $\mathfrak{e}=\overline{\log}^{\star
}\widetilde{\id}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}%
}{n}\widetilde{\id}^{\star n}$, and therefore each $x\in A$
satisfies%
\begin{align*}
\mathfrak{e}\left(  x\right)   &  =\sum_{n\geq1}\dfrac{\left(  -1\right)
^{n-1}}{n}\widetilde{\id}^{\star n}\left(  x\right)
=\underbrace{\widetilde{\id}\left(  x\right)  }%
_{\substack{=x-\epsilon\left(  x\right)  1_{A}\\\text{(by the definition of
}\widetilde{\id}\text{)}}}+\sum_{n\geq2}\dfrac{\left(
-1\right)  ^{n-1}}{n}\underbrace{\widetilde{\id}^{\star
n}\left(  x\right)  }_{\substack{\in\left(  \widetilde{\id}
\left(  A\right)  \right)  ^{n}\\\text{(by induction on }n\text{,}%
\\\text{using the definition}\\\text{of convolution)}}}\\
&  \in x-\epsilon\left(  x\right)  1_{A}+\sum_{n\geq2}\dfrac{\left(
-1\right)  ^{n-1}}{n}\left(  \underbrace{\widetilde{\id}\left(
A\right)  }_{=\ker\epsilon}\right)  ^{n}=x-\underbrace{\epsilon\left(
x\right)  }_{\in\mathbf{k}}1_{A}+\underbrace{\sum_{n\geq2}\dfrac{\left(
-1\right)  ^{n-1}}{n}\left(  \ker\epsilon\right)  ^{n}}_{\subset\left(
\ker\epsilon\right)  ^{2}}\subset x-\mathbf{k}\cdot1_{A}+\left(  \ker
\epsilon\right)  ^{2},
\end{align*}
so that
\begin{equation}
x-\mathfrak{e}\left(  x\right)  \in\mathbf{k}\cdot1_{A}+\left(  \ker
\epsilon\right)  ^{2}. \label{hint.leray.leray-e.b.2}%
\end{equation}
If $x\in\ker\mathfrak{e}$, then this simplifies to $x\in\mathbf{k}\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Thus, $\ker\mathfrak{e}%
\subset\mathbf{k}\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Combining this
with $\mathbf{k}\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset
\ker\mathfrak{e}$, we obtain $\ker\mathfrak{e}=\mathbf{k}\cdot1_{A}+\left(
\ker\epsilon\right)  ^{2}$. But the homomorphism theorem yields%
\[
\mathfrak{e}\left(  A\right)  \cong A/\underbrace{\ker\mathfrak{e}%
}_{=\mathbf{k}\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}}=A/\left(
\mathbf{k}\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)  \cong\left(
\ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{as seen above}\right)
\]
as $\kk$-modules. This completes the proof of
Theorem~\ref{thm.leray.leray-e}(b).

Theorem~\ref{thm.leray.leray-e}(c) just requires showing that $\mathfrak{q}%
\left(  A_{0}\right)  =0$, which is a consequence of $\mathfrak{e}\left(
A_{0}\right)  =0$.

Next, we shall prove Theorem~\ref{thm.leray.leray-e}(d). We have
$\mathfrak{q}\in\mathfrak{n}\left(  A,\operatorname{Sym}\left(  \mathfrak{e}%
\left(  A\right)  \right)  \right)  $. Furthermore, $\mathfrak{q}\left(
A\right)  $ generates the $\kk$-algebra $\operatorname{Sym}\left(
\mathfrak{e}\left(  A\right)  \right)  $ (since $\mathfrak{q}\left(  A\right)
=\operatorname{Sym}^{1}\left(  \mathfrak{e}\left(  A\right)  \right)  $). From
Theorem~\ref{thm.leray.leray-e}(b), we get $\ker\mathfrak{e}=\mathbf{k}%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$, from which we easily obtain
$\mathfrak{q}\left(  1_{A}\right)  =0$ and $\mathfrak{q}\left(  \left(
\ker\epsilon\right)  ^{2}\right)  =0$. Thus,
Proposition~\ref{prop.leray.exp-algh3} (applied to $A$, $\operatorname{Sym}%
\left(  \mathfrak{e}\left(  A\right)  \right)  $ and $\mathfrak{q}$ instead of
$C$, $A$ and $f$) shows that $\exp^{\star}\mathfrak{q}:A\rightarrow
\operatorname{Sym}\left(  \mathfrak{e}\left(  A\right)  \right)  $ is a
surjective $\kk$-algebra homomorphism. But $\mathfrak{s}$ is a
$\kk$-algebra homomorphism $\operatorname{Sym}\left(  \mathfrak{e}%
\left(  A\right)  \right)  \rightarrow A$ and satisfies $\mathbf{i}%
=\mathfrak{s}\circ\iota_{\mathfrak{e}\left(  A\right)  }$ (by its definition).
Thus, Proposition~\ref{prop.convolution-series}(i) (applied to $A$,
$\operatorname{Sym}\left(  \mathfrak{e}\left(  A\right)  \right)  $, $A$,
$\mathfrak{s}$, $\exp$ and $\mathfrak{q}$ instead of $C$, $A$, $B$, $s$, $u$
and $f$) shows that $\mathfrak{s}\circ\mathfrak{q}\in\mathfrak{n}\left(
A,A\right)  $ and $\exp^{\star}\left(  \mathfrak{s}\circ\mathfrak{q}\right)
=\mathfrak{s}\circ\left(  \exp^{\star}\mathfrak{q}\right)  $. However, it is
easy to see that $\mathfrak{s}\circ\mathfrak{q}=\mathfrak{e}$ (since
$\mathbf{i}=\mathfrak{s}\circ\iota_{\mathfrak{e}\left(  A\right)  }$); this
lets us rewrite the equality $\exp^{\star}\left(  \mathfrak{s}\circ
\mathfrak{q}\right)  =\mathfrak{s}\circ\left(  \exp^{\star}\mathfrak{q}%
\right)  $ as $\exp^{\star}\mathfrak{e}=\mathfrak{s}\circ\left(  \exp^{\star
}\mathfrak{q}\right)  $. Comparing this with $\exp^{\star}\mathfrak{e}%
= \id_{A}$, we obtain $\mathfrak{s}\circ\left(
\exp^{\star}\mathfrak{q}\right)  = \id_{A}$. Since $\exp^{\star
}\mathfrak{q}$ is surjective, this entails that the maps $\exp^{\star
}\mathfrak{q}$ and $\mathfrak{s}$ are mutually inverse. This proves
Theorem~\ref{thm.leray.leray-e}(d).

Theorem~\ref{thm.leray.leray-e}(d) shows that $A\cong\operatorname{Sym}\left(
\mathfrak{e}\left(  A\right)  \right)  $ as $\kk$-algebras, but
Theorem~\ref{thm.leray.leray-e}(b) shows that $\mathfrak{e}\left(  A\right)
\cong\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ as
$\kk$-modules. Combining these, we obtain
Theorem~\ref{thm.leray.leray-e}(e).

Finally, to prove Theorem~\ref{thm.leray.leray-e}(f), we notice that each
$x\in A$ satisfies%
\begin{align*}
x-\mathfrak{e}\left(  x\right)   &  \in\mathbf{k}\cdot1_{A}+\left(
\ker\epsilon\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{hint.leray.leray-e.b.2}}\right) \\
&  =\ker\mathfrak{e}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem~\ref{thm.leray.leray-e}(b)}\right)
\end{align*}
and thus $0=\mathfrak{e}\left(  x-\mathfrak{e}\left(  x\right)  \right)
=\mathfrak{e}\left(  x\right)  -\left(  \mathfrak{e}\circ\mathfrak{e}\right)
\left(  x\right)  $.

% [DG][v80] Added the above chapter (mostly pulling from the
% solutions, but omitting all but the most important milestones).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Acknowledgements}
The authors thank the following for helpful comments and/or teaching
them about Hopf algebras:
Marcelo Aguiar, Federico Ardila, Lou Billera, Richard Ehrenborg,
Mark Haiman, Florent Hivert,
Christophe Hohlweg, Jia Huang, Jang Soo Kim, Aaron Lauve,
Dominique Manchon, John Palmieri, Alexander Postnikov,
Margie Readdy, Nathan Reading, Christophe Reutenauer,
Hans-J\"urgen Schneider,
Richard Stanley, Josh Swanson, Muge Taskin, Jean-Yves Thibon.

Parts of this text have been written during stays at the
Mathematisches Forschungsinstitut Oberwolfach (2019 and 2020)%
\footnote{This research was supported through the programme
``Oberwolfach Leibniz Fellows'' by the
Mathematisches Forschungsinstitut Oberwolfach in 2019 and 2020.}
and at the Institut Mittag--Leffler Djursholm (Spring 2020,
supported by the Swedish Research Council under grant
no. 2016-06596);
DG thanks both for their hospitality.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% [DG][v29] Added Stanley to the list.

\begin{thebibliography}{70}

\bibitem{Abe}
Eiichi Abe.
Hopf algebras.
\textit{Cambridge Tracts in Mathematics} \textbf{74}.
Cambridge University Press, Cambridge-New York, 1980.

\bibitem{Aguiar-Andre-plus-26}
Marcelo Aguiar, et al. (28 authors).
Supercharacters, symmetric functions in noncommuting variables, and related Hopf algebras.
\textit{Adv. Math.} \textbf{229} (2012), 2310--2337.
\url{https://doi.org/10.1016/j.aim.2011.12.024} .
Also available as \arxiv{1009.4134v2}.

\bibitem{AguiarArdila}
Marcelo Aguiar and Federico Ardila.
Hopf monoids and generalized permutahedra.
\arxiv{1709.07504v1}.

\bibitem{AguiarBergeronSottile}
Marcelo Aguiar, Nantel Bergeron, and Frank Sottile.
Combinatorial Hopf algebras and generalized Dehn-Sommerville relations.
\textit{Compos. Math.} \textbf{142} (2006), pp. 1--30.
A newer version of this paper appears at \url{http://pi.math.cornell.edu/~maguiar/CHalgebra.pdf}.

% [DG] The version on the website is indeed newer, despite an old date stamp.
% I know it because I sent some corrections to Aguiar and they have been
% incorporated in the file...

\bibitem{AguiarLauve-2014}
Marcelo Aguiar, Aaron Lauve.
The characteristic polynomial of the Adams operators on graded connected
Hopf algebras.
\textit{Algebra \& Number Theory} \textbf{9-3} (2015), 547--583.
Also available at \url{http://pi.math.cornell.edu/~maguiar/adams.pdf}
and as \arxiv{1403.7584v2}.

\bibitem{AguiarMahajan}
Marcelo Aguiar and Swapneel Mahajan.
Monoidal functors, species and Hopf algebras.
\textit{CRM Monograph Series} \textbf{29}.
American Mathematical Society, Providence, RI, 2010.
Available at \url{http://pi.math.cornell.edu/~maguiar/a.pdf}

\bibitem{AguiarSottile}
Marcelo Aguiar and Frank Sottile.
Structure of the Malvenuto-Reutenauer Hopf algebra of permutations.
\textit{Adv. Math.} \textbf{191} (2005), 225--275.
\url{https://doi.org/10.1016/j.aim.2004.03.007}. \newline
A preprint is available at \url{http://pi.math.cornell.edu/~maguiar/MR.pdf}

\bibitem{AndruskiewitschSantos-history}
Nicolas Andruskiewitsch, Walter Ferrer Santos.
The beginnings of the theory of Hopf algebras.
\textit{Acta Appl Math} \textbf{108} (2009), 3--17.
See also a corrected postprint published on arXiv
as \arxiv{0901.2460v3}.

\bibitem{AssafMcNamara}
Sami H. Assaf and Peter R.W. McNamara.
A Pieri rule for skew shapes.
\textit{J. Combin. Theory, Ser. A} \textbf{118} (2011), 277--290.
\url{https://doi.org/10.1016/j.jcta.2010.03.010}

\bibitem{Azenhas-LRsymm24}
Olga Azenhas.
Littlewood-Richardson fillings and their symmetries.
\textit{Matrices and group representations} (Coimbra, 1998),
81--92, Textos Mat. Ser. B, 19, Univ. Coimbra, Coimbra,
1999.
\url{http://www.mat.uc.pt/~oazenhas/graciano+.pdf}.

\bibitem{AzenhasKingTerada}
Olga Azenhas, Ronald C. King, Itaru Terada.
The involutive nature of the Littlewood-Richardson commutativity bijection.
\arxiv{1603.05037v1}.

\bibitem{BakerRichter}
Andrew Baker, and Birgit Richter.
Quasisymmetric functions from a topological point of view.
\textit{Math. Scand.} \textbf{103} (2008), 208--242.
\url{http://dx.doi.org/10.7146/math.scand.a-15078}

\bibitem{BarekatRWilligenburg}
Farzin Barekat, Victor Reiner, Stephanie van Willigenburg.
Corrigendum to ``Coincidences among skew Schur functions''
[\textit{Adv. Math.} \textbf{216} (2007),  118--152].
\textit{Adv. Math.} \textbf{220} (2009), 1655--1656.
See also a corrected version of this paper on \arxiv{math/0602634v4}.

\bibitem{BenedettiHallamMachacek}
Carolina Benedetti, Joshua Hallam, John Machacek.
Combinatorial Hopf Algebras of Simplicial Complexes.
\arxiv{1505.04458v2}.
(Published in:
\textit{SIAM J. Discrete Math.} \textbf{30} (3), 1737--1757.)

\bibitem{BenedettiSagan}
Carolina Benedetti, Bruce Sagan.
Antipodes and involutions.
\arxiv{1410.5023v4}.
(Published in:
\textit{Journal of Combinatorial Theory, Series A} \textbf{148} (2017), 275--315.)

\bibitem{BenkartSottileStroomer}
Georgia Benkart, Frank Sottile, Jeffrey Stroomer.
Tableau Switching: Algorithms and Applications.
\textit{Journal of Combinatorial Theory, Series A}
\textbf{76}, 1, October 1996, 11--43.
\url{https://doi.org/10.1006/jcta.1996.0086} \newline
Preprint available at
\url{http://www.math.tamu.edu/~sottile/research/pdf/switching.pdf}

\bibitem{BergBergeronSaliolaSerranoZabrocki-immac1}
Chris Berg, Nantel Bergeron, Franco Saliola, Luis Serrano, Mike Zabrocki.
A lift of the Schur and Hall-Littlewood bases to non-commutative symmetric functions.
\textit{Canad. J. Math.} \textbf{66} (2014), 525--565.
\url{http://dx.doi.org/10.4153/CJM-2013-013-0}. \\
A preprint is \arxiv{1208.5191v3}.

\bibitem{BergeronZabrocki}
Nantel Bergeron, Mike Zabrocki.
The Hopf algebras of symmetric functions and quasi-symmetric
functions in non-commutative variables are free and co-free.
\textit{Journal of Algebra and Its Applications} \textbf{08},
Issue 04, August 2009, 581--600.
A preprint also appears at
\arxiv{math/0509265v3}.

\bibitem{Billera}
Louis J. Billera.
Flag enumeration in polytopes,
Eulerian partially ordered sets and Coxeter groups.
\textit{Proceedings of the International Congress of Mathematicians} \textbf{IV},
2389--2415, Hindustan Book Agency, New Delhi, 2010.
\url{http://pi.math.cornell.edu/~billera/papers/eulericm.pdf}

\bibitem{BilleraBrenti}
Louis J. Billera, Francesco Brenti.
Quasisymmetric functions and Kazhdan-Lusztig polynomials.
\arxiv{0710.3965v2}.
Published in:
Israel Journal of Mathematics,
August 2011, 184, pp. 317--348.
\url{https://doi.org/10.1007/s11856-011-0070-0}

\bibitem{BilleraJiaR}
Louis J. Billera, Ning Jia, and Victor Reiner.
A quasisymmetric function for matroids.
\textit{European J. Combin.} \textbf{30} (2009), pp. 1727--1757.
\url{https://doi.org/10.1016/j.ejc.2008.12.007} .
A preprint also appears at
\arxiv{math/0606646v3}.

\bibitem{Bjorner}
Anders Bj\"orner.
Some combinatorial and algebraic properties
of Coxeter complexes and Tits buildings.
\textit{Adv. in Math.} \textbf{52} (1984), 173--212.
\url{https://doi.org/10.1016/0001-8708(84)90021-5}

\bibitem{Blasiak-kronhook}
Jonah Blasiak.
Kronecker coefficients for one hook shape.
\textit{Seminaire Lotharingien de Combinatoire} \textbf{77} (2017),
B77c.
\url{https://www.emis.de/journals/SLC/wpapers/s77blasiak.html}

\bibitem{BlessenohlLaue}
D. Blessenohl, H. Laue.
Algebraic combinatorics related to the free Lie algebra.
\textit{Seminaire Lotharingien de Combinatoire} \textbf{29}
(1992), B29e.
\url{https://www.emis.de/journals/SLC/opapers/s29laue.html}

\bibitem{BlessenohlSchocker}
Dieter Blessenohl, Manfred Schocker.
Noncommutative character theory of the symmetric group.
Imperial College Press 2005.
\url{https://www.worldscientific.com/worldscibooks/10.1142/p369}

\bibitem{BlumSmithCoskey}
Ben Blum-Smith, Samuel Coskey.
The Fundamental Theorem on Symmetric Polynomials: History's First Whiff of Galois Theory.
\arxiv{1301.7116v4}.
An updated version was published in:
The College Mathematics Journal Vol. 48, No. 1 (January 2017), pp. 18--29.
\url{https://doi.org/10.4169/college.math.j.48.1.18}

\bibitem{Bourbaki-Lie2-3-fr}
N. Bourbaki.
\'El\'ements de Math\'ematique:
Groupes et alg\`ebres de Lie, Chapitres 2 et 3.
Springer, Heidelberg 2006.

\bibitem{BritzFomin}
Thomas Britz, Sergey Fomin.
Finite posets and Ferrers shapes.
\textit{Adv. in Math.} \textbf{158}, Issue 1, 1 March 2001, 86--127.
Better version to be found on arXiv as \arxiv{math/9912126v1}.

\bibitem{deBruijnKlarner}
N.G. de Bruijn, D.A. Klarner.
Multisets of aperiodic cycles.
\textit{SIAM J. Alg. Disc. Math.} \textbf{3} (1982), no. 3, 359--368.
\url{https://pure.tue.nl/ws/files/1674487/597568.pdf}

\bibitem{Bump}
Daniel Bump.
Notes on representations of $GL(r)$ over a finite field.
Available at \url{http://math.stanford.edu/~bump/}.

\bibitem{Burgunder}
Emily Burgunder.
Eulerian idempotent and Kashiwara-Vergne conjecture.
\textit{Annales de l'institut Fourier} \textbf{58} (2008), Issue 4, 1153--1184.
\url{https://eudml.org/doc/10345}.

\bibitem{ButlerHales}
Lynne M. Butler, Alfred W. Hales.
Nonnegative Hall polynomials.
\textit{Journal of Algebraic Combinatorics} \textbf{2} (1993), Issue 2, 125--135.
\url{https://www.emis.de/journals/JACO/Volume2_2/l42886q158156k2u.html}

\bibitem{CaenepeelVercruysse}
Stefaan Caenepeel, J. Vercruysse.
Hopf algebras.
\textit{Lecture notes, Vrije Universiteit Brussel} \textbf{2013}.
\url{http://homepages.ulb.ac.be/~scaenepe/Hopfalgebra.pdf}

\bibitem{Cameron-matroids}
Peter J. Cameron.
Notes on matroids and codes.
\textit{Lecture notes}, 2000.
\url{http://www.maths.qmul.ac.uk/~pjc/comb/matroid.pdf}

\bibitem{Cartier}
Pierre F. Cartier.
A primer of Hopf algebras.
Frontiers in number theory, physics, and geometry. II, 537--615,
Springer, Berlin, 2007. \newline
A preprint is available at
\url{http://preprints.ihes.fr/2006/M/M-06-40.pdf}

\bibitem{ChariPressley}
Vyjayanthi Chari, and Andrew N. Pressley.
A guide to quantum groups.
Cambridge University Press, Cambridge, 1994.

\bibitem{CheboluMinac}
Sunil K. Chebolu, Jan Minac.
Counting irreducible polynomials over finite fields using the inclusion-exclusion principle.
\textit{Math. Mag.} \textbf{84} (2011), 369--371.
A preprint is \arxiv{1001.0409v6}.

\bibitem{ChenFoxLyndon}
K.T. Chen, R.H. Fox, R.C. Lyndon.
Free Differential Calculus, IV: The Quotient Groups of the Lower
Central Series.
\textit{Annals of Mathematics} \textbf{68} (1), 81--95.
\url{https://doi.org/10.2307/1970044}

\bibitem{ChmutovDuzhinMostovoy}
Sergei Chmutov, Sergei V. Duzhin, Jacob Mostovoy.
Introduction to Vassiliev Knot Invariants.
CUP 2012.
\newline
Various preprint versions can be found at
\url{https://people.math.osu.edu/chmutov.1/preprints/} ,
at \url{http://www.pdmi.ras.ru/~duzhin/papers/cdbook/} and
at \arxiv{1103.5628v3}.

\bibitem{Conrad-blurbs}
Keith Conrad.
Expository papers (``Blurbs''), specifically
\textit{Tensor Products I}, \textit{Tensor Products II},
\textit{Exterior Powers}.
\url{http://www.math.uconn.edu/~kconrad/blurbs}

\bibitem{CrapoSchmitt1}
Henry Crapo and William Schmitt.
Primitive elements in the matroid-minor Hopf algebra.
\textit{J. Algebraic Combin.} \textbf{28} (2008), 43--64.
\url{https://doi.org/10.1007/s10801-007-0066-3}. \\
A preprint is \arxiv{math/0511033v1}.

\bibitem{CrapoSchmitt2}
\bysame.
A unique factorization theorem for matroids.
\textit{J. Combin. Theory Ser. A} \textbf{112} (2005), 222--249.
\url{https://doi.org/10.1016/j.jcta.2005.02.004}

\bibitem{CrapoSchmitt3}
\bysame.
A free subalgebra of the algebra of matroids.
\textit{European J. Combin.} \textbf{26} (2005), 1066--1085.
\url{https://doi.org/10.1016/j.ejc.2004.05.006}

\bibitem{CrawleyBoevey}
William Crawley-Boevey.
\textit{Lectures on representation theory and invariant theory},
Bielefeld 1989/90.
Available from \url{https://www.math.uni-bielefeld.de/~wcrawley/}.

\bibitem{CrochemoreDesarmenienPerrin}
Maxime Crochemore, Jacques D\'{e}sarm\'{e}nien, Dominique Perrin.
A note on the Burrows--Wheeler transformation.
\textit{Theoretical Computer Science} \textbf{332} (2005), pp. 567--572.
\url{https://doi.org/10.1016/j.tcs.2004.11.014} \newline
A preprint is \arxiv{cs/0502073}.

\bibitem{Dahl-ComMat}
Geir Dahl.
Network flows and combinatorial matrix theory.
Lecture notes, 4 September 2013.
\url{http://www.uio.no/studier/emner/matnat/math/MAT-INF4110/h13/lecturenotes/combmatrix.pdf}

\bibitem{DascalescuNastasescuRaianu}
Sorin Dascalescu, Constantin Nastasescu, Serban Raianu.
Hopf algebras. An introduction.
\textit{Monographs and Textbooks in Pure and Applied Mathematics} \textbf{235}.
Marcel Dekker, Inc., New York, 2001.

\bibitem{Dayton-ghosts}
Barry Dayton.
Witt vectors, the Grothendieck Burnside ring, and
Necklaces.
\url{http://orion.neiu.edu/~bhdayton/necksum.htm}

\bibitem{DentonHivertSchillingThiery}
Tom Denton, Florent Hivert, Anne Schilling, and Nicolas M. Thi\'ery.
On the representation theory of finite J-trivial monoids.
\textit{S\'em. Lothar. Combin.} \textbf{64} (2010/11), Art. B64d, 44 pp.
\url{https://www.emis.de/journals/SLC/wpapers/s64dehiscth.html}

\bibitem{DesarmenienWachs}
Jacques D\'{e}sarm\'{e}nien, Michelle L. Wachs.
Descent classes of permutations with a given number of fixed points.
\textit{Journal of Combinatorial Theory, Series A} \textbf{64},
Issue 2, pp. 311--328.
\url{https://doi.org/10.1016/0097-3165(93)90100-M}

\bibitem{DiaconisMcGrathPitman}
Persi Diaconis, Michael Mc Grath, Jim Pitman.
Riffle shuffles, cycles, and descents.
\textit{Combinatorica} \textbf{15}(1), 1995, pp. 11--29.
\url{https://doi.org/10.1007/bf01294457}

\bibitem{DiaconisPangRam}
Persi Diaconis, C.Y. Amy Pang and Arun Ram.
Hopf algebras and Markov chains: Two examples and a theory.
\textit{J. Algebraic Combin.} \textbf{39}, Issue 3, May 2014, 527--585.
% \textit{arXiv:1206.3620}.
A newer version is available at
\url{https://amypang.github.io/papers/hpmc.pdf}

\bibitem{DolceRestivoReutenauer-oglw}
Francesco Dolce, Antonio Restivo, Christophe Reutenauer.
On generalized Lyndon words.
Theoretical Computer Science \textbf{777} (2019), 232--242.
Also available at \arxiv{1812.04515v1}.

\bibitem{DolceRestivoReutenauer-svl}
Francesco Dolce, Antonio Restivo, Christophe Reutenauer.
Some variations on Lyndon words.
\arxiv{1904.00954v1}.

\bibitem{Doran}
William F. Doran IV.
A Proof of Reutenauer's $-q_{\left( n \right)}$ Conjecture.
\textit{J. Combin. Theory, Ser. A} \textbf{74} (1996), 342--344.
\url{https://doi.org/10.1006/jcta.1996.0056}

\bibitem{DressSiebeneicher-diophant}
Andreas W. M. Dress, and Christian Siebeneicher.
On the number of solutions of certain linear diophantine equations.
\textit{Hokkaido Math. J.} \textbf{19} (1990), pp. 385--401.
\url{http://www.math.sci.hokudai.ac.jp/hmj/page/19-3/pdf/HMJ_19_3_1990_385-401.pdf}

\bibitem{DressSiebeneicher-burnside}
Andreas W. M. Dress, and Christian Siebeneicher.
The Burnside Ring of the Infinite Cyclic Group and Its Relations to the Necklace Algebra, $\lambda$-Rings, and the Universal Ring of Witt Vectors.
\textit{Advances in Mathematics} \textbf{78} (1989), 1--41.
\url{https://doi.org/10.1016/0001-8708(89)90027-3}

\bibitem{NCSF6}
G\'erard Duchamp, Florent Hivert, and Jean-Yves Thibon.
Noncommutative symmetric functions VI.
Free quasi-symmetric functions and related algebras.
\textit{Internat. J. Algebra Comput.} \textbf{12} (2002), 671--717.
A preprint is available at
\url{http://monge.univ-mlv.fr/~hivert/PAPER/NCSF6.ps}.

\bibitem{DuchampHoangNghiaKrajewskiTanasa}
G\'erard H. E. Duchamp, Nguyen Hoang-Nghia, Thomas Krajewski, Adrian Tanasa.
Recipe theorem for the Tutte polynomial for matroids,
renormalization group-like approach.
\textit{Advances in Applied Mathematics} \textbf{51}(3), 345–-358.
\url{https://doi.org/10.1016/j.aam.2013.04.006}

\bibitem{DuchampMinhTolluChienNghia}
G\'erard Henry Edmond Duchamp, Vincel Hoang Ngoc Minh, Christophe Tollu, B\`ui Chi\^en, Nguyen Hoang Nghia.
Combinatorics of $\varphi$-deformed stuffle Hopf algebras.
\arxiv{1302.5391v7}.

\bibitem{Dyckerhoff}
Tobias Dyckerhoff.
Hall Algebras - Bonn, Wintersemester 14/15.
Lecture notes, version February 5, 2015.
\url{https://web.archive.org/web/20150601115158/http://www.math.uni-bonn.de/people/dyckerho/notes.pdf}

\bibitem{Edmonds}
Jack Edmonds.
Submodular functions, matroids, and certain polyhedra.
In: Combinatorial Structures and their Applications
(Proc. Calgary Internat. Conf., Calgary, Alta., 1969),
Gordon and Breach, New York, 1970, pp. 66--87; reprinted in
Combinatorial optimization: Eureka, you shrink!,
pp. 11--26, \textit{Lecture Notes in Comput. Sci.} \textbf{2570}, Springer, Berlin, 2003.
\url{https://doi.org/10.1007/3-540-36478-1_2}

\bibitem{Egge}
Eric S. Egge.
\textit{An Introduction to Symmetric Functions and Their Combinatorics}.
Student Mathematical Library \textbf{91},
American Mathematical Society,
2019.
\url{https://bookstore.ams.org/stml-91}

\bibitem{Ehrenborg}
Richard Ehrenborg.
On posets and Hopf algebras.
\textit{Adv. Math.} \textbf{119} (1996), 1--25.
\url{https://doi.org/10.1006/aima.1996.0026}

\bibitem{Eisenbud}
David Eisenbud.
Commutative Algebra
with a View Toward Algebraic Geometry.
\textit{Graduate Texts in Mathematics} \textbf{150},
Springer 1995.
\url{https://doi.org/10.1007/978-1-4612-5350-1}

\bibitem{ElizaldeTroyka}
Sergi Elizalde, Justin M. Troyka.
Exact and asymptotic enumeration of cyclic permutations according to descent set.
\textit{J. Combin. Theory, Ser. A} \textbf{165} (2019), 360--391.
Also available at \arxiv{1710.05103v3}.

\bibitem{EllisKhovanov}
Alexander P. Ellis, and Mikhail Khovanov.
The Hopf algebra of odd symmetric functions.
\textit{Adv. Math.} \textbf{231} (2012), 965--999.
A newer version is available as \arxiv{1107.5610v2}.

% [DG] The arXiv version is not just newer but also bigger, in particular defining
% an "odd plactic monoid" which doesn't appear in the published version. I don't
% know the story behind that.

\bibitem{Ellzey-thesis}
Brittney Ellzey.
On Chromatic Quasisymmetric Functions of
Directed Graphs.
PhD thesis, University of Miami,
2018.
\url{https://scholarlyrepository.miami.edu/oa_dissertations/2091}

\bibitem{Etingof-et-al}
Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina.
Introduction to representation theory.
\textit{Student Mathematical Library} \textbf{59},
Amer. Math. Soc., Providence, RI, 2011.
\url{http://www-math.mit.edu/~etingof/repb.pdf} .
(Parts of this book appear in \arxiv{0901.0827v5}.)
A newer version is available at \url{http://www-math.mit.edu/~etingof/reprbook.pdf} .

\bibitem{Foissy}
Loic Foissy.
Alg\`ebres de Hopf combinatoires.
\url{http://loic.foissy.free.fr/pageperso/Hopf.pdf}

\bibitem{Foissy-freecofree}
Loic Foissy.
Free and cofree Hopf algebras.
\textit{Journal of Pure and Applied Algebra} \textbf{216},
Issue 2, February 2012, 480--494.
\url{https://doi.org/10.1016/j.jpaa.2011.07.010} .
A preprint is \arxiv{1010.5402v3}.

\bibitem{FredricksenMaiorana}
Harold Fredricksen, James Maiorana.
Necklaces of beads in $k$ colors and $k$-ary de Bruijn
sequences.
\textit{Discrete Mathematics} \textbf{23} (1978), 207--210.
\url{https://doi.org/10.1016/0012-365X(78)90002-X}

\bibitem{Fulton}
William Fulton.
Young Tableaux.
\textit{London Mathematical Society Student Texts} \textbf{35},
Cambridge University Press, Cambridge-New York, 1997.
\url{https://doi.org/10.1017/CBO9780511626241}

\bibitem{Garsia-adder}
Adriano M. Garsia.
Permutation q-enumeration with the Schur row adder.
\textit{PU. M. A. (Pure Mathematics and Applications)} \textbf{21} (2010), No. 2, 233--248.
\url{http://puma.dimai.unifi.it/21_2/7_Garsia.pdf} (also mirrored at \url{http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.432.8196&rep=rep1&type=pdf} ).

\bibitem{Gasharov}
Vesselin Gasharov.
Incomparability graphs of (3+1)-free posets are s-positive.
Proceedings of the 6th Conference on Formal Power Series and Algebraic Combinatorics
(New Brunswick, NJ, 1994).
\textit{Discrete Math.} \textbf{157} (1996), 193--197.
\url{https://doi.org/10.1016/S0012-365X(96)83014-7}

\bibitem{Gasharov-LR}
Vesselin Gasharov.
A Short Proof of the Littlewood-Richardson Rule.
\textit{European Journal of Combinatorics},
Volume 19, Issue 4, May 1998, Pages 451--453.
\url{https://doi.org/10.1006/eujc.1998.0212}

\bibitem{GelfandKrobLascouxLeclercRetakhThibon}
Israel M. Gelfand, Daniel Krob, Alain Lascoux, Bernard Leclerc, Vladimir S. Retakh, Jean-Yves Thibon.
Noncommutative symmetric functions.
\textit{Adv. Math.} \textbf{112} (1995), 218--348.
\url{https://doi.org/10.1006/aima.1995.1032}
\newline
A preprint is available as \arxiv{hep-th/9407124v1}.

\bibitem{GerstenhaberSchack-shuffle}
M. Gerstenhaber, S.D. Schack.
The shuffle bialgebra and the cohomology of commutative algebras.
\textit{Journal of Pure and Applied Algebra} \textbf{70} (1991), 263--272.
\url{https://doi.org/10.1016/0022-4049(91)90073-B}

\bibitem{Gessel}
Ira M. Gessel.
Multipartite P-partitions and inner products of skew Schur functions.
Combinatorics and algebra (Boulder, Colo., 1983), 289--317,
\textit{Contemp. Math.} \textbf{34},
Amer. Math. Soc., Providence, RI, 1984.
\url{http://people.brandeis.edu/~gessel/homepage/papers/multipartite.pdf}

\bibitem{Gessel-Ppart}
Ira M. Gessel.
A Historical Survey of P-Partitions.
2015,
\arxiv{1506.03508v1}.
Published in:
Patricia Hersh, Thomas Lam, Pavlo Pylyavskyy and Victor Reiner (eds.),
\textit{The Mathematical Legacy of Richard P. Stanley},
Amer. Math. Soc., Providence, RI, 2016, pp. 169--188.

\bibitem{GesselRestivoReutenauer}
Ira M. Gessel, Antonio Restivo, Christophe Reutenauer.
A Bijection between Words and Multisets of Necklaces.
\textit{European Journal of Combinatorics} \textbf{33} (2012), pp. 1537--1546.
\url{https://doi.org/10.1016/j.ejc.2012.03.016}

\bibitem{GesselReutenauer}
Ira M. Gessel, Christophe Reutenauer.
Counting Permutations with Given Cycle Structure and Descent Set.
\textit{Journal of Combinatorial Theory, Series A} \textbf{64} (1993), 189--215.
\url{https://doi.org/10.1016/0097-3165(93)90095-P}

\bibitem{GesselViennot}
Ira M. Gessel, X.G. Viennot.
Determinants, Paths, and Plane Partitions.
preprint, 1989,
\url{http://people.brandeis.edu/~gessel/homepage/papers/pp.pdf}
% Previously: \url{http://contscience.xavierviennot.org/xavier/articles_files/determinant_89.pdf}

\bibitem{Granville-NTmasterclass}
Andrew Granville.
Number Theory Revealed:
A Masterclass.
\textit{Number Theory Revealed: The Series} \textbf{\#1B},
American Mathematical Society 2019.

\begin{verlong}
\bibitem{Grinberg-19-9}
Darij Grinberg.
\textit{Solution to: Problems from the Book -- Problem 19.9}. \\
Available online at
\url{http://www.cip.ifi.lmu.de/~grinberg/19-9long.pdf}.
% This reference serves a detailed version of a solution.
\end{verlong}

\bibitem{Grinberg-dp}
Darij Grinberg.
Double posets and the antipode of $\Qsym$.
\arxiv{1509.08355v3}.

% [DG][v55] Added this reference (for a QSym antipode formula).

\bibitem{Grinberg-orzech}
Darij Grinberg.
A constructive proof of Orzech's theorem.
Preprint, 20 November 2016. \newline
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/orzech.pdf}

\bibitem{GrosshansRotaStein}
Frank D. Grosshans, Gian-Carlo Rota, Joel A. Stein.
Invariant Theory and Superalgebras.
\textit{CBMS Regional Conference Series in Mathematics} \textbf{69},
American Mathematical Society, 1987.
\url{https://bookstore.ams.org/cbms-69}

\bibitem{HamelGoulden}
A.M. Hamel, I.P. Goulden.
Planar Decompositions of Tableaux and Schur Function Determinants.
\textit{Europ. J. Combinatorics} \textbf{16} (1995), 461--477.
\url{https://doi.org/10.1016/0195-6698(95)90002-0}

\bibitem{Hazewinkel}
Michiel Hazewinkel.
The algebra of quasi-symmetric functions is free over the integers.
\textit{Adv. Math.} \textbf{164} (2001), 283--300.
\url{https://doi.org/10.1006/aima.2001.2017}

\bibitem{Hazewinkel2}
Michiel Hazewinkel.
Witt vectors. Part 1.
In: M. Hazewinkel (ed.), \textit{Handbook of Algebra} \textbf{6}, Elsevier 2009.
Also available at \arxiv{0804.3888v1}.

\bibitem{Hazewinkel-lh}
Michiel Hazewinkel.
The Leibniz-Hopf Algebra and Lyndon Words.
\textit{Preprint AM CWI} \textbf{9612} (1996).
\url{http://oai.cwi.nl/oai/asset/4828/04828D.pdf}

\bibitem{Hazewinkel-poset}
Michiel Hazewinkel.
Chen-Fox-Lyndon Factorization for Words over Partially Ordered Sets.
\textit{Journal of Mathematical Sciences} \textbf{131} (12-2005), Issue 6, 6027--6031.
\url{https://doi.org/10.1007/s10958-005-0458-7}

\bibitem{HazewinkelGubareniKirichenko}
Michiel Hazewinkel, Nadiya Gubareni, and Vladimir V. Kirichenko.
Algebras, rings and modules.
Lie algebras and Hopf algebras.
\textit{Mathematical Surveys and Monographs} \textbf{168}.
American Mathematical Society, Providence, RI, 2010.

\bibitem{Henderson}
Robert Henderson.
The Algebra Of Multiple Zeta Values.
\url{http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.227.5432}

\bibitem{Hesselholt-witt}
Lars Hesselholt.
Lecture notes on Witt vectors.
\url{http://www.math.nagoya-u.ac.jp/~larsh/papers/s03/wittsurvey.ps}

\bibitem{Hesselholt-deRhamWitt}
Lars Hesselholt.
The big de Rham--Witt complex.
\textit{Acta Math.} \textbf{214} (2015), 135--207.
\url{https://doi.org/10.1007/s11511-015-0124-y}

\bibitem{Hivert}
Florent Hivert.
An introduction to combinatorial Hopf algebras: examples and realizations.
\textit{Nato Advanced Study Institute School on Physics and Computer Science, 2005, october, 17--29, Cargese, France}.
\url{http://www-igm.univ-mlv.fr/~hivert/PAPER/Cargese.pdf}

\bibitem{HivertNovelliThibon-commutative}
Florent Hivert, Jean-Christophe Novelli and Jean-Yves Thibon.
Commutative combinatorial Hopf algebras.
\textit{J. Algebraic Combin.} \textbf{28} (2008), no. 1, 65--95.
\url{https://doi.org/10.1007/s10801-007-0077-0}
\newline
Also available as \arxiv{math/0605262v1}.

\bibitem{HivertNovelliThibon-binary-trees}
\bysame.
The algebra of binary search trees.
\textit{Theoret. Comput. Sci.} \textbf{339} (2005), no. 1, 129--165.
\url{https://doi.org/10.1016/j.tcs.2005.01.012}
\newline
A preprint appears as \arxiv{math/0401089v2}.

\bibitem{HivertNovelliThibon-bilinear}
\bysame.
Trees, functional equations, and combinatorial Hopf algebras.
\textit{European J. Combin.}  \textbf{29}  (2008),  no. 7, 1682--1695.
%\url{http://www.sciencedirect.com/science/article/pii/S0195669807001783}
\url{https://doi.org/10.1016/j.ejc.2007.09.005}
\newline
A preprint appears as \arxiv{math/0701539v1}.

\bibitem{Hoffman-trees}
Michael E. Hoffman.
Combinatorics of rooted trees and Hopf algebras.
\textit{Trans. AMS} \textbf{355} (2003), 3795--3811.
\url{https://doi.org/10.1090/S0002-9947-03-03317-8}

\bibitem{Hoffman-multizeta}
\bysame.
A character on the quasi-symmetric functions coming from multiple zeta values.
\textit{The Electronic Journal of Combinatorics} \textbf{15} (2008), R97.
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v15i1r97}

\bibitem{HumpertMartin}
Brandon Humpert, and Jeremy L. Martin.
The incidence Hopf algebra of graphs.
\textit{SIAM Journal on Discrete Mathematics} \textbf{26}, no. 2 (2012), 555--570.
Also available as \arxiv{1012.4786v3}.

\bibitem{JamesLiebeck}
Gordon James and Martin Liebeck.
Representations and characters of groups.
2nd edition, Cambridge University Press, Cambridge-New York, 2001.

\bibitem{Jin-outsidenested}
Emma Yu Jin.
Outside nested decompositions of skew
diagrams and Schur function determinants.
\textit{European Journal of Combinatorics} \textbf{67} (2018), 239--267.
\url{https://doi.org/10.1016/j.ejc.2017.08.007} .
A preprint is available at
\url{http://www.emmayujin.at/Pubs/Jin18.pdf}.

\bibitem{Joni-Rota}
S.A. Joni and Gian-Carlo Rota.
Coalgebras and bialgebras in combinatorics.
\textit{Studies in Applied Mathematics} \textbf{61} (1979), 93--139.
\url{https://doi.org/10.1002/sapm197961293}

\bibitem{Kassel}
Christian Kassel.
Quantum groups.
\textit{Graduate Texts in Mathematics} \textbf{155}.
Springer, Berlin, 1995.

\bibitem{Kerov}
Sergei V. Kerov.
Asymptotic representation theory of the symmetric group and its applications in analysis.
\textit{Translations of Mathematical Monographs} \textbf{219}.
American Mathematical Society, Providence, RI, 2003.

\bibitem{KirillovBerenstein}
Anatol N. Kirillov, Arkadiy D. Berenstein.
Groups generated by involutions, Gelfand-Tsetlin patterns and the
combinatorics of Young tableaux.
\textit{Algebra i Analiz} \textbf{7} (1995), issue 1, 92--152.
A preprint is available at
\url{http://pages.uoregon.edu/arkadiy/bk1.pdf}

\bibitem{Klein-hall}
T. Klein.
The multiplication of Schur-functions and extensions of $p$-modules.
\textit{J. London Math. Soc.} \textbf{43} (1968), 280--284.
\url{https://doi.org/10.1112/jlms/s1-43.1.280}

\bibitem{Knuth-RSK}
Donald E. Knuth.
Permutations, matrices, and generalized Young tableaux.
\textit{Pacific J. Math.} \textbf{34}, Number 3 (1970), 709--727.
\url{https://projecteuclid.org/euclid.pjm/1102971948}

\bibitem{Knuth-TAoCP4a}
Donald E. Knuth.
The Art of Computer Programming, Volume 4A:
Combinatorial Algorithms, Part 1.
Pearson 2011.
See \url{https://www-cs-faculty.stanford.edu/~knuth/taocp.html} for errata.

\bibitem{Knutson}
Donald Knutson.
$\lambda$-Rings and the Representation Theory of the
Symmetric Group.
\textit{Lecture Notes in Mathematics} \textbf{308},
Springer, Berlin-Heidelberg-New York 1973.
\url{https://doi.org/10.1007/BFb0069217}

\bibitem{Krause-Gale}
Manfred Krause.
A Simple Proof of the Gale-Ryser Theorem.
\textit{The American Mathematical Monthly} \textbf{103} (1996), 335--337.
\url{https://doi.org/10.2307/2975191}

\bibitem{Krob}
Daniel Krob.
El\'ements de combinatoire.
Magist\`ere 1-\`ere ann\'ee, Ecole Normale Sup\'erieure,
version 1.0, Novembre 1995.
\url{http://krob.cesames.net/IMG/ps/combi.ps}

\bibitem{Kufleitner}
Manfred Kufleitner.
On Bijective Variants of the Burrows-Wheeler Transform.
Presented at the Prague Stringology Conference 2009 (PSC 2009).
\arxiv{0908.0239v1}.

\bibitem{KulikauskasRemmel}
Andrius Kulikauskas, Jeffrey Remmel.
Lyndon words and transition matrices between
elementary, homogeneous and monomial symmetric
functions.
\textit{Electronic Journal of Combinatorics} \textbf{13} (2006),
Research Paper $\sharp$R18.
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v13i1r18}

\bibitem{Kytola-hopf}
Kalle Kyt\"ol\"a.
Introduction to Hopf algebras and representations.
Lecture notes, Spring 2011.
\url{https://math.aalto.fi/~kkytola/files_KK/lectures_files_KK/Hopf-lecture_notes.pdf}

\bibitem{LaksovLascouxPragaczThorup}
Dan Laksov, Alain Lascoux, Piotr Pragacz, and Anders Thorup.
The LLPT Notes.
Edited by A. Thorup, 1995--2018.
\url{http://web.math.ku.dk/noter/filer/sympol.pdf}

\bibitem{LamLauveSottile}
Thomas Lam, Aaron Lauve, and Frank Sottile.
Skew Littlewood-Richardson rules from Hopf Algebras.
\textit{DMTCS Proceedings, 22nd International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2010)}
\textbf{2010}, 355--366.
A preprint can also be found at \arxiv{0908.3714v3}.

\bibitem{LamPylyavskyy}
Thomas Lam, and Pavlo Pylyavskyy.
Combinatorial Hopf algebras and K-homology of Grassmanians.
\textit{International Mathematics Research Notices}, \textbf{2007} (2007), rnm 125, 48 pages.
A preprint is \arxiv{0705.2189v1}.

% [DG] The typo appears in both arXiv *and* published versions, so I figured it's
% better to keep it...

\bibitem{Lando}
Sergei K. Lando.
On a Hopf Algebra in Graph Theory.
\textit{Journal of Combinatorial Theory, Series B} \textbf{80} (2000), 104--121.
\url{https://doi.org/10.1006/jctb.2000.1973}

\bibitem{LaudaRussell}
Aaron D. Lauda, Heather M. Russell.
Oddification of the cohomology of type A Springer varieties.
\textit{International Math Research Notices} \textbf{2014}, No. 17, 4822--4854.
A preprint is \arxiv{1203.0797v1}.

\bibitem{Laue}
Hartmut Laue.
Freie algebraische Strukturen.
Lecture notes,
Mathematisches Seminar der Universit\"at Kiel 2013,
version 16 Sep 2013.
\url{http://www.uni-kiel.de/math/algebra/laue/vorlesungen/frei/freiealgstr.pdf}

\bibitem{LauveMason-stable}
Aaron Lauve and Sarah K. Mason.
QSym over Sym has a stable basis.
FPSAC 2010, San Francisco, USA.
\textit{DMTCS proc. AN} \textbf{2010}, 367--378.
Also available as \arxiv{1003.2124v1}.

\bibitem{Leeuwen-altSchur}
Marc van Leeuwen.
Schur functions and alternating sums.
\textit{Electronic Journal of Combinatorics} \textbf{11(2)} A5 (2006).
Also available at \url{http://www-math.univ-poitiers.fr/~maavl/pdf/alt-Schur.pdf}.

\bibitem{Leeuwen-geometry}
Marc van Leeuwen.
Flag varieties, and interpretations of Young tableau algorithms.
\textit{Journal of Algebra} \textbf{224} (2000).
Also available at
\url{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/geometry.pdf}

\bibitem{Leeuwen-hopf}
Marc van Leeuwen.
An application of Hopf-Algebra techniques
to representations of finite Classical Groups.
\textit{Journal of Algebra} \textbf{140}, Issue 1, 15 June 1991, pp. 210--246.
Also available at
\url{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/Hopf.pdf}

\bibitem{Leeuwen-lrr}
Marc van Leeuwen.
The Littlewood-Richardson rule, and related combinatorics.
\textit{Math. Soc. of Japan Memoirs} \textbf{11}, Interaction of Combinatorics and Representation Theory.
Also available at
\url{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/lrr.pdf}

\bibitem{Leeuwen-rsk}
Marc van Leeuwen.
The Robinson-Schensted and Sch\"utzenberger algorithms, an elementary approach.
\textit{Electronic Journal of Combinatorics}, Foata Festschrift, \textbf{3} (no. 2), R15 (1996).
Also available at
\url{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/foata-fest.pdf}

\bibitem{Li-primegraphs}
Ji Li.
Prime Graphs and Exponential Composition of Species.
\textit{Journal of Combinatorial Theory, Series A} \textbf{115},
Issue 8, November 2008, 1374--1401. See
\arxiv{0705.0038v4} for a preprint.

\bibitem{Liu-simplkron}
Ricky Ini Liu.
A simplified Kronecker rule for one hook shape.
\textit{Proc. Amer. Math. Soc.} \textbf{145} (2017), pp. 3657--3664.
\url{https://doi.org/10.1090/proc/13692}
See \arxiv{1412.2180v1} for a preprint.

\bibitem{Liulevicius}
Arunas Liulevicius.
Arrows, symmetries and representation rings.
\textit{Journal of Pure and Applied Algebra} \textbf{19} (1980), 259--273.
\url{https://doi.org/10.1016/0022-4049(80)90103-6}

% [DG] I am not abbreviating journal names I don't know how to abbreviate. (This also
% pertains to some references above and below.)

\bibitem{Loday}
Jean-Louis Loday.
Cyclic Homology.
\textit{Grundlehren der mathematischen Wissenschaften} \textbf{301}, 2nd edition,
Springer, Berlin-Heidelberg 1998.

\bibitem{Loday-serie-Hausdorff}
Jean-Louis Loday.
S\'erie de Hausdorff, idempotents Eul\'eriens et alg\`ebres de Hopf.
\textit{Expo. Math.} \textbf{12} (1994), 165--178.
\url{http://www-irma.u-strasbg.fr/~loday/PAPERS/94Loday%28Eulerien%29.pdf}

\bibitem{LodayRonco-comb-hopf}
Jean-Louis Loday and Mar\'ia O.~Ronco.
Combinatorial Hopf algebras.
Quanta of maths, \textit{Clay Math. Proc.} \textbf{11}, 347--383, Amer. Math. Soc., Providence, RI, 2010.
\url{http://www-irma.u-strasbg.fr/~loday/PAPERS/2011LodayRonco(CHA).pdf}

\bibitem{LodayRonco-trees}
\bysame.
Hopf algebra of the planar binary trees.
\textit{Adv. Math.} \textbf{139} (1998), no. 2, 293--309.
\url{https://doi.org/10.1006/aima.1998.1759}

\bibitem{Loehr-bij}
Nicholas A. Loehr.
Bijective Combinatorics.
CRC Press, 2011.
See \url{http://www.math.vt.edu/people/nloehr/bijbook.html} for errata.

\bibitem{Lothaire}
M. Lothaire.
\textit{Combinatorics on words}.
Corrected printing, Cambridge University Press, 1997.

\bibitem{LuotoMykytiukvanWilligenburg}
Kurt Luoto, Stefan Mykytiuk, Stephanie van Willigenburg.
An introduction
to quasisymmetric Schur functions -- Hopf algebras, quasisymmetric functions,
and Young composition tableaux.
Springer, May 23, 2013.
\url{http://www.math.ubc.ca/~steph/papers/QuasiSchurBook.pdf}

\bibitem{Lyndon}
R.C. Lyndon.
On Burnside's Problem.
\textit{Transactions of the AMS} \textbf{77}, 202--215.
\url{https://doi.org/10.1090/S0002-9947-1954-0064049-X}

\bibitem{Macdonald}
Ian Grant Macdonald.
Symmetric functions and Hall polynomials.
2nd edition, Oxford University Press, Oxford-New York, 1995.

\bibitem{Macdonald-variations}
I.G. Macdonald.
Schur functions : theme and variations.
\textit{Publ. I.R.M.A. Strasbourg}, \textbf{1992}, 498/S-27,
Actes 28 e Seminaire Lotharingien, 5--39.
\url{https://www.emis.de/journals/SLC/opapers/s28macdonald.html}

\bibitem{MaiaMendez}
Manuel Maia, Miguel M\'endez.
On the arithmetic product of combinatorial species.
\textit{Discrete Mathematics} \textbf{308}, Issue 23, 6
December 2008, 5407--5427.
%\url{http://www.sciencedirect.com/science/article/pii/S0012365X07007960} .
\url{https://doi.org/10.1016/j.disc.2007.09.062} .
\newline
See \arxiv{math/0503436v2} for a preprint.

\bibitem{Malvenuto}
Claudia Malvenuto.
Produits et coproduits des fonctions quasi-sym\'etriques et de l'alg\`ebre des descents.
PhD dissertation, Univ. du Qu\'eb\'ec \`a Montreal, 1993.
\url{http://lacim.uqam.ca/wp-content/uploads/Publications/16.pdf}

\bibitem{MalvenutoReutenauer}
Clauda [sic] Malvenuto and Christophe Reutenauer.
Duality between quasi-symmetric functions and the Solomon descent algebra.
\textit{J. Algebra} \textbf{177} (1995), 967--982.
\url{https://doi.org/10.1006/jabr.1995.1336}

\bibitem{MalvenutoReutenauer-plethysm}
Claudia Malvenuto and Christophe Reutenauer.
Plethysm and conjugation of quasi-symmetric functions.
\textit{Discrete Mathematics} \textbf{193}, Issues 1--3, 28 November 1998,
 225--233.
\url{https://doi.org/10.1016/S0012-365X(98)00142-3}

\bibitem{MalvenutoReutenauer-doubleposets}
Claudia Malvenuto and Christophe Reutenauer.
A self paired Hopf algebra on double posets and
a Littlewood-Richardson rule.
\textit{Journal of Combinatorial Theory, Series A} \textbf{118} (2011), 1322--1333.
\url{https://doi.org/10.1016/j.jcta.2010.10.010}

% [DG][v3] This reference is so far uncited. I have added it merely to
% have it if I am going to write a section on the double-poset Hopf
% algebra (which, so far, seems to be one of the most useful Hopf
% algebras above FQSym to me).

\bibitem{Manchon}
Dominique Manchon.
 Hopf algebras, from basics to applications to renormalization.
\textit{Comptes Rendus des Rencontres Mathematiques de Glanon} 2001
(published in 2003). \arxiv{math/0408405v2}.

\bibitem{Manetti-voyage}
Marco Manetti.
A voyage round coalgebras.
27 June 2016.
\url{https://www1.mat.uniroma1.it/people/manetti/dispense/voyage.pdf}

\bibitem{Manivel-chern}
Laurent Manivel.
Chern classes of tensor products.
\arxiv{1012.0014v1}.

\bibitem{McNamaraWard}
Peter R.W. McNamara, Ryan E. Ward.
Equality of $P$-partition generating functions.
\arxiv{1210.2412v2}.

\bibitem{Meliot}
Pierre-Lo\"ic M\'eliot.
Representation Theory of Symmetric Groups.
\textit{Discrete Mathematics and its Applications},
CRC Press 2017.

\bibitem{MendesRemmel}
Anthony Mendes, Jeffrey Remmel.
Counting with Symmetric Functions.
\textit{Developments in Mathematics} \textbf{43},
Springer 2015.

\bibitem{Mendez-MO}
Miguel Mendez.
\textit{MathOverflow answer \#139482}.
\url{http://mathoverflow.net/a/139482/}.

\bibitem{MilnorMoore}
John W. Milnor and John C. Moore.
On the structure of Hopf algebras.
\textit{The Annals of Mathematics, Second Series} \textbf{81}, No. 2 (Mar., 1965), 211--264.
\url{https://doi.org/10.2307/1970615}

% [DG] There is also a different preprint version of the Milnor-Moore paper in
% Milnor's collected works. Quoting that book quoting Jim Stasheff:
% "Each has its virtues, and I hope both will be preserved for posterity."
% That doesn't mean both need to be cited here; just saying.

\bibitem{Montgomery}
Susan Montgomery.
Hopf algebras and their actions on rings.
\textit{Regional Conference Series in Mathematics} \textbf{82}, Amer. Math.
Soc., Providence, RI, 2010.
\url{https://bookstore.ams.org/cbms-82}

\bibitem{Morava}
Jack Morava.
Homotopy-theoretically enriched categories of noncommutative motives.
\textit{Research in the Mathematical Sciences} \textbf{2} (2015), no. 8.
\url{https://doi.org/10.1186/s40687-015-0028-7}

\bibitem{Moreno}
Eduardo Moreno.
On the theorem of Fredricksen and Maiorana
about de Bruijn sequences.
\textit{Advances in Applied Mathematics} \textbf{33}, Issue 2,
August 2004, 413--415.
\url{https://doi.org/10.1016/j.aam.2003.10.002}

\bibitem{Moreno-corr}
Eduardo Moreno, Dominique Perrin.
Corrigendum to ``On the theorem of Fredricksen and Maiorana
about de Bruijn sequences''
[Adv. in Appl. Math. 33 (2) (2004) 413--415].
\textit{Advances in Applied Mathematics}
\textbf{62}, January 2015, Pages 184--187.
\url{http://www.sciencedirect.com/science/article/pii/S0196885814000918}

\bibitem{MartinMorinWagner}
Jeremy L. Martin, Matthew Morin, Jennifer D. Wagner.
On distinguishing trees by their chromatic symmetric functions.
\textit{Journal of Combinatorial Theory, Series A} \textbf{115}, Issue 2,
February 2008, 237--253.
\url{https://doi.org/10.1016/j.jcta.2007.05.008}

\bibitem{Morris}
Robert Morris.
Umbral Calculus and Hopf Algebras.
\textit{Contemporary Mathematics} \textbf{6},
AMS, Providence 1982.
\url{https://bookstore.ams.org/conm-6}

\bibitem{Oesinghaus}
Jakob Oesinghaus.
Quasisymmetric functions and the Chow ring of the stack of expanded pairs.
\textit{Res. Math. Sci.} \textbf{6} (2019), no. 5.
\url{https://doi.org/10.1007/s40687-018-0168-7} .
A preprint is \arxiv{1806.10700v1}.

\bibitem{Oxley}
James Oxley.
Matroid theory.
Oxford University Press, Oxford-New York, 1992.

\bibitem{PakPostnikov-oscil}
Igor Pak, Alexander Postnikov.
Oscillating Tableaux, $S_p\times S_q$-modules, and
Robinson-Schensted-Knuth Correspondence.
Updated (yet unfinished) version of FPSAC 1996 abstract,
January 15, 1994.
\url{http://math.mit.edu/~apost/papers/osc.pdf}

\bibitem{Patras}
Fr\'ed\'eric Patras.
La d\'ecomposition en poids des alg\`ebres de Hopf.
\textit{Annales de l'institut Fourier} \textbf{43}, no 4 (1993), 1067--1087.
\url{https://eudml.org/doc/75026}

\bibitem{Patras-descent}
F. Patras.
L'alg\`ebre des descentes d'une big\`ebre gradu\'ee.
\textit{Journal of Algebra} \textbf{170} (1994), 547--566.
\url{https://doi.org/10.1006/jabr.1994.1352}

\bibitem{PatrasReutenauer}
Fr\'ed\'eric Patras, Christophe Reutenauer.
On Dynkin and Klyachko idempotents in graded bialgebras.
\textit{Advanced in Applied Mathematics} \textbf{28}, Issues
3--4, April 2002, 560--579.
\url{https://doi.org/10.1006/aama.2001.0795}

\bibitem{PatrasReutenauer-hli}
Fr\'ed\'eric Patras, Christophe Reutenauer.
Higher Lie idempotents.
\textit{J. Algebra} \textbf{222} (1999), no. 1, 51--64.
\url{https://doi.org/10.1006/jabr.1999.7887}

\bibitem{Patrias-antipode}
Rebecca Patrias.
Antipode formulas for combinatorial Hopf algebras.
\arxiv{1501.00710v2}.
Published in:
\textit{The Electronic Journal of Combinatorics} \textbf{23}, Issue 4 (2016), P4.30.
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v23i4p30/}

\bibitem{Prasolov-linalg}
Victor Prasolov.
Problems and theorems in linear algebra.
\textit{Translations of mathematical monographs} \textbf{134}, 1st edition 1994, AMS.
\url{http://www2.math.su.se/~mleites/books/prasolov-1994-problems.pdf}

\bibitem{PoirierReutenauer}
St\'ephane Poirier, Christophe Reutenauer.
Alg\`ebres de Hopf de tableaux.
\textit{Ann. Sci. Math. Qu\'ebec} \textbf{19} (1995), no. 1, 79--90.
\url{http://www.lacim.uqam.ca/~christo/Publi%C3%A9s/1995/Alg%C3%A8bres%20de%20Hopf%20de%20tableaux.pdf}

\bibitem{Postnikov}
Alexander Postnikov.
Permutohedra, associahedra, and beyond.
\textit{Int. Math. Res. Notices} \textbf{2009}, No. 6, pp. 1026--1106.
A preprint appears at
\url{https://math.mit.edu/~apost/papers/permutohedron.pdf}
and as \arxiv{math/0507163v1}.

\bibitem{Prasad-schur}
Amritanshu Prasad.
An Introduction to Schur Polynomials.
\textit{Graduate J. Math.} \textbf{4} (2019), 62--84.
\url{https://www.gradmath.org/wp-content/uploads/2020/01/Prasad-GJM2019.pdf} .
A preprint appears at \arxiv{1802.06073v2}.

\bibitem{Pylyavskyy-thesis}
Pavlo Pylyavskyy.
Comparing products of Schur functions and
quasisymmetric functions.
PhD dissertation, MIT, 2007.
\url{https://dspace.mit.edu/handle/1721.1/38957}

\bibitem{Radford-Hopf}
David E. Radford.
Hopf algebras.
\textit{Series on Knots and Everything} \textbf{49}.
World Scientific, 2012.
\url{https://doi.org/10.1142/8055}

\bibitem{Radford-shuffle}
David E. Radford.
A Natural Ring Basis for the Shuffle Algebra and an Application to Group Schemes.
\textit{Journal of Algebra} \textbf{58} (1979), 432--454.
\url{https://doi.org/10.1016/0021-8693(79)90171-6}

\bibitem{Reading}
Nathan Reading.
Lattice congruences, fans and Hopf algebras.
\textit{Journal of Combinatorial Theory, Series A} \textbf{110}, Issue 2,
May 2005, pp. 237--273.
\url{https://doi.org/10.1016/j.jcta.2004.11.001} .
A preprint is \arxiv{math/0402063v1}.

\bibitem{Reiner-signedperm}
Victor Reiner.
Signed permutation statistics and cycle type.
\textit{European J. Combin}. \textbf{14} (1993), no. 6, 569--579.
\url{https://doi.org/10.1006/eujc.1993.1059}

\bibitem{ReinerShawVanWilligenburg2007}
Victor Reiner, Kristin M. Shaw, and Stephanie van Willigenburg.
Coincidences among skew Schur functions.
\arxiv{math/0602634v4}.
(Update of a paper published in
\textit{Advances in Mathematics}, \textbf{216(1)}:118--152, 2007.)

\bibitem{Remmel}
Jeffrey B. Remmel.
The combinatorics of $\left(k,\ell\right)$-hook Schur functions.
In: C. Greene (ed.), \textit{Combinatorics and algebra},
Proceedings of the AMS-IMS-SIAM joint summer research
conference in the mathematical sciences on combinatorics
and algebra, Colorado, Boulder, 1983,
\textit{Contemporary Mathematics} \textbf{34}, 1984,
253--287.

\bibitem{Reutenauer}
Christophe Reutenauer.
Free Lie Algebras.
\textit{London Mathematical Society Monographs, New Series} \textbf{7}.
Clarendon Press, Oxford 1993.

\bibitem{Rosas-kronprod}
Mercedes H. Rosas.
The Kronecker Product of Schur Functions Indexed
by Two-Row Shapes or Hook Shapes.
\textit{Journal of Algebraic Combinatorics} \textbf{14} (2001), 153--173.
\url{https://doi.org/10.1023/A:1011942029902} \\
A preprint is \arxiv{math/0001084v1}.

\bibitem{RosasSagan}
Mercedes H. Rosas, Bruce E. Sagan.
Symmetric functions in noncommuting variables.
\textit{Transactions of the American Mathematical Society} \textbf{358},
183--214.
\url{https://doi.org/10.1090/S0002-9947-04-03623-2}

\bibitem{Rota}
Joseph P.S. Kung, Gian-Carlo Rota.
Gian-Carlo Rota on Combinatorics: Introductory Papers and
Commentaries.
Birkh\"auser 1995.

\bibitem{Sagan}
Bruce E. Sagan.
The symmetric group:  representations, combinatorial algorithms,
and symmetric functions.
2nd edition, Springer, New York-Berlin-Heidelberg 2001.
See \url{https://users.math.msu.edu/users/bsagan/Books/Sym/errata.pdf}
for errata.

\bibitem{Sagan2020}
Bruce E. Sagan.
Combinatorics: The Art of Counting.
Draft of a textbook, 2020.
\url{https://users.math.msu.edu/users/bsagan/Books/Aoc/aocAMS.pdf}

\bibitem{SaganStanley}
Bruce E. Sagan, Richard P. Stanley.
Robinson-Schensted Algorithms for Skew Tableaux.
\textit{Journal of Combinatorial Theory, Series A} \textbf{55} (1990), 161--193.
%\url{http://www.sciencedirect.com/science/article/pii/0097316590900666}
\url{https://doi.org/10.1016/0097-3165(90)90066-6}

\bibitem{Sam-symf}
Steven V. Sam.
Notes for Math 740 (Symmetric Functions),
27 April 2017.
\url{https://www.math.wisc.edu/~svs/740/notes.pdf}

\bibitem{Schiffmann}
Olivier Schiffmann.
Lectures on Hall algebras.
\arxiv{math/0611617v2}.

% [DG] I wouldn't be surprised if this has been published, but I couldn't find
% where.

\bibitem{Schmitt}
William R. Schmitt.
Incidence Hopf algebras.
\textit{Journal of Pure and Applied Algebra} \textbf{96} (1994), 299--330.
\url{https://doi.org/10.1016/0022-4049(94)90105-8} .
A preprint appears at
\url{http://home.gwu.edu/~wschmitt/papers/iha.pdf}

\bibitem{Schmitt-antipodes}
William R. Schmitt.
Antipodes and Incidence Coalgebras.
\textit{Journal of Combinatorial Theory, Series A} \textbf{46} (1987), 264--290.
\url{https://doi.org/10.1016/0097-3165(87)90006-9}

\bibitem{Schmitt-exp}
William R. Schmitt.
Expository notes, specifically
``A concrete introduction to category theory'' and
``Notes on modules and algebras''.
\url{http://home.gwu.edu/~wschmitt/}

\bibitem{Schmitt-structures}
\bysame.
Hopf algebras of combinatorial structures.
\textit{Canadian Journal of Mathematics} \textbf{45} (1993), 412--428.
\url{https://doi.org/10.4153/CJM-1993-021-5} .
A preprint appears at
\url{http://home.gwu.edu/~wschmitt/papers/hacs.pdf}

\bibitem{Schur-1937}
I. Schur.
Arithmetische Eigenschaften der Potenzsummen einer algebraischen Gleichung.
\textit{Compositio Mathematica} \textbf{4} (1937), 432--444.
\url{http://www.numdam.org/item?id=CM_1937__4__432_0}

\bibitem{Schweigert-Hopf}
Christoph Schweigert.
Hopf algebras, quantum groups and topological
field theory.
Lecture notes, Winter term 2014/15, Hamburg.
Version of 16 May 2015.
\url{http://www.math.uni-hamburg.de/home/schweigert/ws12/hskript.pdf}

\bibitem{Serre}
Jean-Pierre Serre.
Linear representations of finite groups.
Springer, Berlin-Heidelberg-New York, 1977.
\url{https://doi.org/10.1007/978-1-4684-9458-7}

\bibitem{ShareshianWachs}
John Shareshian and Michelle L. Wachs.
Chromatic quasisymmetric functions and Hessenberg varieties.
In: A. Bj\"orner, F. Cohen, C. De Concini, C. Procesi,
M. Salvetti (Eds.), Configuration Spaces,
\textit{Publications of the Scuola Normale Superiore}
\textbf{14}, Springer, Berlin-Heidelberg-New York 2013.
A preprint is \arxiv{1106.4287v3}.

\bibitem{ShareshianWachs-2014}
John Shareshian and Michelle L. Wachs.
Chromatic quasisymmetric functions.
\textit{Advances in Mathematics} \textbf{295} (2016), pp. 497--551.
A preprint is \arxiv{1405.4629v2}.

\bibitem{ShareshianWachs-euler}
John Shareshian and Michelle L. Wachs.
Eulerian quasisymmetric functions.
\textit{Advances in Mathematics} \textbf{225} (2010), pp. 2921--2966.
\url{https://doi.org/10.1016/j.aim.2010.05.009} .
A preprint is \arxiv{0812.0764v2}

\bibitem{ShelleyAbrahamson-honors}
Seth Shelley-Abrahamson.
Hopf Modules and Representations of Finite Groups of Lie Type.
Honors thesis, Stanford, May 2013.
\url{http://mathematics.stanford.edu/wp-content/uploads/2013/08/Shelley-Abrahamson-Honors-Thesis-2013.pdf}

\bibitem{Shirshov}
Anatolii I. Shirshov.
On Free Lie Rings.
\textit{Mat. Sbornik N.S.} \textbf{45} (87), (1958), no. 2, 113--122.
Original at:
\url{http://mi.mathnet.ru/msb4963}.
Translation in:
L.A. Bokut, V. Latyshev, I. Shestakov, E. Zelmanov (eds.),
\textit{Selected works of A.I. Shirshov}, Birkh\"auser 2009.

\bibitem{Stanley-thesis}
Richard P. Stanley.
Ordered structures and partitions.
\textit{Memoirs of the Amer. Math. Soc.} \textbf{119},
American Mathematical Society, Providence, R.I., 1972.
\url{http://www-math.mit.edu/~rstan/pubs/pubfiles/9.pdf}

\bibitem{Stanley-negative-colors}
\bysame .
Acyclic orientations of graphs.
\textit{Discrete Math.} \textbf{5} (1973), 171--178.
Reprinted in:
\textit{Discrete Math.} \textbf{306} (2006), 905--909.
%\url{http://www.sciencedirect.com/science/article/pii/S0012365X06001385}
\url{https://doi.org/10.1016/j.disc.2006.03.010}

\bibitem{Stanley-chromatic-qsym-fn}
\bysame.
A symmetric function generalization of the chromatic polynomial of a graph.
\textit{Adv. Math.} \textbf{111} (1995), 166--194.
\url{https://doi.org/10.1006/aima.1995.1020}

\bibitem{Stanley}
\bysame.
Enumerative Combinatorics, Volumes 1 and 2.
\textit{Cambridge Studies in Advanced Mathematics}, \textbf{49} and \textbf{62}.
Cambridge University Press, Cambridge, 2nd edition 2011 (volume 1)
and 1st edition 1999 (volume 2).

% [DG][v9] Updated the date for volume 1, after I've already updated the
% citations in the text a while ago.

\bibitem{RStantonThiem}
Shishuo Fu, Victor Reiner, Dennis Stanton, Nathaniel Thiem.
The negative $q$-binomial.
\textit{The Electronic Journal of Combinatorics} \textbf{19}, Issue 1 (2012), P36.
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v19i1p36}

% [DG] Is this the paper (and version) you meant?

\bibitem{Steinberg}
R. Steinberg.
A geometric approach to the representations of the full linear group over a
Galois field.
\textit{Trans. Amer. Math. Soc.} \textbf{71}, (1951), 274--282.
\url{https://doi.org/10.1090/S0002-9947-1951-0043784-0}
%This is his original paper where he first defined the unipotent characters of
%$GL_n(\FF_q)$ ??

\bibitem{Steinhardt}
Jacob Steinhardt.
Permutations with Ascending and Descending Blocks.
\textit{The Electronic Journal of Combinatorics} \textbf{17} (2010), \#R14.
\url{https://www.combinatorics.org/ojs/index.php/eljc/article/view/v17i1r14}

\bibitem{Stembridge}
John R. Stembridge.
A concise proof of the Littlewood-Richardson rule.
\textit{The Electronic Journal of Combinatorics} \textbf{9}, 2002, N5.
\url{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v9i1n5}

\bibitem{Stembridge-multfree}
John Stembridge.
Multiplicity-Free Products of Schur Functions.
\textit{Annals of Combinatorics} \textbf{5} (2001), 113--121.
\url{http://www.math.lsa.umich.edu/~jrs/papers/mfree.ps.gz}

\bibitem{Strang-elim}
Gilbert Strang.
The algebra of Elimination.
\url{http://www-math.mit.edu/~gs/papers/Paper7_ver8.pdf}.

\bibitem{Sweedler}
Moss E. Sweedler.
Hopf algebras.
W.A. Benjamin, New York, 1969.

\bibitem{Takeuchi}
Mitsuhiro Takeuchi.
Free Hopf algebras generated by coalgebras.
\textit{J. Math. Soc. Japan} \textbf{23} (1971), 561--582.
\url{http://projecteuclid.org/euclid.jmsj/1259849779}

\bibitem{Tamvakis}
Harry Tamvakis.
The theory of Schur polynomials revisited.
\textit{Enseign. Math.} \textbf{58} (2012), 147--163.
A preprint appears at
\url{http://www2.math.umd.edu/~harryt/papers/schurrev.pdf}

\bibitem{Thibon}
Jean-Yves Thibon.
An Introduction to Noncommutative Symmetric Functions.
Cargese lecture, October 2005.
J.-P. Gazeau, J. Nesetril, B. Rovan (eds.): From Numbers and Languages to (Quantum) Cryptography,
\textit{NATO Security through Science Series: Information and Communication Security} \textbf{7},
IOS Press, 2007.
Available at \url{http://igm.univ-mlv.fr/~jyt/ARTICLES/cargese_thibon.ps}.

\bibitem{ThiemVinroot}
Nathaniel Thiem and C. Ryan Vinroot.
On the characteristic map of finite unitary groups.
\textit{Advances in Mathematics} \textbf{210}, Issue 2, 1 April 2007, pp. 707--732.
\url{https://doi.org/10.1016/j.aim.2006.07.018} .
A preprint is
\url{http://www.math.wm.edu/~vinroot/charunitary.pdf}

% [DG] There are different versions of this floating around the internet,
% one of them ( http://www.math.wm.edu/~vinroot/charunitary.pdf )
% possibly being a postprint.

\bibitem{ThomasYong}
Hugh Thomas, Alexander Yong.
An $S_3$-symmetric Littlewood-Richardson rule.
\textit{Math. Res. Lett.} \textbf{15} (2008), no. 5, 1027--1037.
\arxiv{0704.0817v1}.

\bibitem{Vermeeren-nets}
Stijn Vermeeren.
Sequences and nets in topology.
Version of 11 September 2013.
\url{http://stijnvermeeren.be/download/mathematics/nets.pdf}

% [DG][v64] Added this reference, since I mention nets in a few footnotes in solutions (though I don't really use nets).

\bibitem{Wachs}
Michelle L. Wachs.
Flagged Schur Functions, Schubert Polynomials, and Symmetrizing Operators.
\textit{Journal of Combinatorial Theory, Series A} \textbf{40} (1985), 276--289.
\url{https://doi.org/10.1016/0097-3165(85)90091-3}

\bibitem{vanderWaerden-1}
Bartel Leendert van der Waerden.
Algebra, Volume I.
Translation of the 7th (German) edition.
Springer 2003.

\bibitem{Webb}
Peter Webb.
A Course in Finite Group Representation Theory.
23 February 2016.
\url{http://www-users.math.umn.edu/~webb/RepBook/}

\bibitem{Wildon}
Mark Wildon.
Representation theory of the symmetric group.
5 April 2018.
\url{http://www.ma.rhul.ac.uk/~uvah099/teaching.html}

\bibitem{Wildon2016}
Mark Wildon.
An involutive introduction to symmetric functions.
8 May 2020.
\url{http://www.ma.rhul.ac.uk/~uvah099/teaching.html}

\bibitem{Wisbauer}
Robert Wisbauer.
Coalgebras and Bialgebras.
\textit{The Egyptian Mathematical Society, The Mathematical Sciences Research Centre (MSRC) Technical Reports} \textbf{No. 1}, 2004.
\url{http://www.math.uni-duesseldorf.de/~wisbauer/}

\bibitem{Xantcha}
Qimh Richey Xantcha.
Binomial Rings: Axiomatisation, Transfer, and Classification.
\arxiv{1104.1931v4}.

\bibitem{Zelevinsky}
Andrey V. Zelevinsky.
Representations of finite classical groups: a Hopf algebra approach.
\textit{Lecture Notes in Mathematics} \textbf{869}. Springer-Verlag, Berlin-New York, 1981.

\bibitem{Zelevinsky-skewLR}
Andrey V. Zelevinsky.
A Generalization of the Littlewood-Richardson Rule and the
Robinson-Schensted-Knuth Correspondence.
\textit{Journal of Algebra} \textbf{69} (1981), 82--94.
\url{https://doi.org/10.1016/0021-8693(81)90128-9}

\bibitem{ZhouLu}
G.-S. Zhou, D.-M. Lu.
Lyndon words for Artin-Schelter regular algebras.
\arxiv{1403.0385v1}.

\end{thebibliography}

\horiline

\IfFileExists{./HopfComb-v80.ind}{\input{HopfComb-v80.ind}}{\newpage An index goes here!}


\begin{nosolutions}
\end{document}
\end{nosolutions}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Solutions to exercises
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

 %\begin{solutions}

\section{Solutions to the exercises} \label{chp.solutions}

This chapter contains solutions to the exercises scattered
throughout the text. These solutions vary in level of detail (some
of them are detailed, some only outline the most important steps,
and many lie inbetween these two extremes), and
sometimes in notation (as they have been written over a long
timespan). They also have seen far less quality control than the
main text, so typos and worse are to be expected.
Comments and alternative solutions are welcome!

%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.counit.unique}}

\textit{First solution to Exercise~\ref{exe.counit.unique}.}
This is analogous to the well-known fact that
any nonunital nonassociative\footnote{The word ``nonassociative''
does not prohibit associativity; it simply
means that associativity is not required. Similarly, ``nonunital''
does not force the nonexistence of a multiplicative identity.}
$\kk$-algebra has at most one multiplicative identity. If you can prove the latter fact
by pure abstract nonsense (i.e., without referring to elements), then the same
proof serves as a solution to Exercise~\ref{exe.counit.unique} once all arrows
are reversed (and all $m$'s and $u$'s are replaced by $\Delta$'s and
$\epsilon$'s). Let us see how this works.

How does one classically prove that every nonunital nonassociative
$\kk$-algebra has at most one multiplicative identity?
Let $A$ be a nonunital
nonassociative $\kk$-algebra, and let $1$ and $1^{\prime}$ be two
elements of $A$ which could both serve as multiplicative identities.
That is, every $a\in A$
satisfies both $1a=a1=a$ and $1^{\prime}a=a1^{\prime}=a$.
Now, applying $a1=a$
to $a=1^{\prime}$ yields $1^{\prime}\cdot 1=1^{\prime}$. But applying
$1^{\prime}a=a$ to $a=1$ yields $1^{\prime}\cdot 1=1$.
Comparing
$1^{\prime}\cdot 1=1$ with $1^{\prime}\cdot 1=1^{\prime}$
yields $1=1^{\prime}$, and
thus the multiplicative identity is unique.

This argument made use of elements, which we need to get rid of in order to be
able to reverse the arrows. The idea is to replace every element $\alpha$ of
$A$ by the linear map $\kk\rightarrow A$ which sends $1\in\kk$
to $\alpha$. In terms of commutative diagrams, a nonunital nonassociative
$\kk$-algebra is a $\kk$-module $A$ endowed with a $\kk$-linear map
$m:A\otimes A\rightarrow A$ which is not a priori required to
satisfy any properties. A multiplicative identity of $A$ then corresponds to a $\kk$-linear map
$u:\kk\rightarrow A$ making the diagram \eqref{unit-diagram} commute. Let $u$ and
$u^{\prime}$ be two such $\kk$-linear maps
$\kk\rightarrow A$. Instead of applying $a1=a$ to $a=1^{\prime}$,
we now need to take the commutative diagram
\[
\xymatrix{
A \otimes\kk\ar[d]_{\id\otimes u} & A \ar[l] \ar[d]_{\id} \\
A \otimes A \ar[r]^m & A
}
\]
(which commutes because $u$ makes \eqref{unit-diagram} commute, and corresponds to the axiom $a1=a$)
and pre-compose it with the morphism $u^{\prime}$ (which corresponds to the
multiplicative identity $1^{\prime}$), thus obtaining
\begin{equation}
\xymatrix{
& & \kk\ar[dl]^{u'} \\
A \otimes\kk\ar[d]_{\id\otimes u} & A \ar[l] \ar[d]_{\id} & \\
A \otimes A \ar[r]^m & A &
}.
\label{sol.counit.unique.1}
\end{equation}
Similarly, the element-free version of applying $1^{\prime}a=a$ to $a=1$
yields the commutative diagram
\begin{equation}
\xymatrix{
\kk\ar[dr]^{u} & & \\
& A \ar[d]_{\id} \ar[r] & \kk\otimes A \ar[d]^{u' \otimes\id} \\
& A & A \otimes A \ar[l]_m
}.
\label{sol.counit.unique.2}
\end{equation}
The path $\kk\overset{u^{\prime}}{\longrightarrow}A\longrightarrow
A\otimes\kk\overset{\id \otimes u}{\longrightarrow}A\otimes A$ through the
diagram \eqref{sol.counit.unique.1} and the path
$\kk\overset{u}{\longrightarrow}A\longrightarrow\kk\otimes
A\overset{u^{\prime}\otimes \id}{\longrightarrow}A\otimes A$ through the diagram
\eqref{sol.counit.unique.2} give the same map, since they both correspond to
the element $1\otimes 1^{\prime}$ of $A\otimes A$. How can this be seen without
referring to elements? Being a tautology at the level of elements, this must
follow from formal properties of tensor products (without using axioms like
the commutativity of \eqref{unit-diagram}). And so it does: In fact, the
three small quadrilaterals in the diagram
\[
\xymatrix{
& \kk\ar[dl]_{u'} \ar[dr]^{u} \ar[d] & \\
A \ar[d] & \kk\otimes\kk\ar[dl]^{u'\otimes \id} \ar[dr]_{\id \otimes u}
& A \ar[d] \\
A \otimes\kk\ar[dr]_{\id\otimes u} & & \kk\otimes A \ar[dl]^{u'\otimes\id} \\
& A \otimes A &
}
\]
commute, and therefore so does the whole diagram. We can now piece this
diagram together with the diagrams \eqref{sol.counit.unique.1} and
\eqref{sol.counit.unique.2} (this is the diagrammatic equivalent of comparing
$1\cdot1^{\prime}=1$ with $1\cdot1^{\prime}=1^{\prime}$), and obtain
\[
\xymatrix{
& \kk\ar[dl]_{u'} \ar[dr]_{u} \ar[d] & \\
A \ar@/_6pc/[dddr]^{\id} \ar[d] & \kk\otimes\kk\ar[dl]^{u'\otimes\id}
\ar[dr]_{\id\otimes u} & A \ar[d] \ar@/^6pc/[dddl]^{\id} \\
A \otimes\kk\ar[dr]_{\id\otimes u} & & \kk\otimes A \ar[dl]^{u'\otimes\id} \\
& A \otimes A \ar[d]^{m} &\\
& A &
}.
\]
Following the outside quadrilateral of this commutative diagram yields $\id\circ
u=\id\circ u^{\prime}$, so that $u=u^{\prime}$, which shows that the two
maps $u$ and $u^{\prime}$ are equal. The purely diagrammatic argument
that we made can now be easily transformed into a solution of
Exercise~\ref{exe.counit.unique} by merely reversing arrows and
replacing $m$ and $u$ by $\Delta$ and $\epsilon$.

\textit{Second solution to Exercise~\ref{exe.counit.unique}.}
The exercise can also be solved using Sweedler notation, which
makes some good practice in using the latter.
Here is how the solution goes:

We must prove that there exists \emph{at most one} $\kk$-linear map
$\epsilon : C\rightarrow \kk$ such that the diagram \eqref{counit-diagram}
commutes. In other words, we must show that if $\epsilon_{1}$ and
$\epsilon_{2}$ are two such maps, then $\epsilon_{1}=\epsilon_{2}$.

So let $\epsilon_{1}$ and $\epsilon_{2}$ be two such maps. We must then show
that $\epsilon_{1}=\epsilon_{2}$.

We know that $\epsilon_{1}$ is a $\kk$-linear map $\epsilon
: C \rightarrow \kk$ such that the diagram \eqref{counit-diagram}
commutes. Thus, the diagram%
\begin{equation}
\xymatrix{
C \otimes\kk\ar[r] & C & \kk\otimes C \ar[l] \\
C \otimes C \ar[u]^{\id\otimes\epsilon_1} & C \ar[l]^-{\Delta} \ar[u]^\id
\ar[r]_-{\Delta} & C \otimes C \ar[u]^{\epsilon_1 \otimes\id}
}
\label{sol.counit.unique.2nd.1}
\end{equation}
commutes. Let us denote the top-left horizontal arrow of this diagram by
$\kappa$; thus, $\kappa : C \otimes \kk \rightarrow C$ is the canonical
isomorphism (sending each $c\otimes\lambda \in C \otimes \kk$ to $\lambda
c \in \kk$). The commutativity of the left square in the commutative
diagram \eqref{sol.counit.unique.2nd.1} thus says that
\begin{equation}
\operatorname*{id}=\kappa\circ\left(  \operatorname*{id}\otimes\epsilon
_{1}\right)  \circ\Delta.\label{sol.counit.unique.2nd.1b}%
\end{equation}


We shall use Sweedler notation as in \eqref{eq.sweedler-not.def1} to
abbreviate formulas involving $\Delta$. For each $c\in C$, we have%
\begin{align}
c  & =\underbrace{\operatorname*{id}}_{\substack{=\kappa\circ\left(
\operatorname*{id}\otimes\epsilon_{1}\right)  \circ\Delta\\\text{(by
\eqref{sol.counit.unique.2nd.1b})}}}\left(  c\right)  =\left(  \kappa
\circ\left(  \operatorname*{id}\otimes\epsilon_{1}\right)  \circ\Delta\right)
\left(  c\right)  =\kappa\left(  \left(  \operatorname*{id}\otimes\epsilon
_{1}\right)  \left(  \underbrace{\Delta\left(  c\right)  }_{=\sum_{\left(
c\right)  }c_{1}\otimes c_{2}}\right)  \right)  \nonumber\\
& =\kappa\left(  \underbrace{\left(  \operatorname*{id}\otimes\epsilon
_{1}\right)  \left(  \sum_{\left(  c\right)  }c_{1}\otimes c_{2}\right)
}_{=\sum_{\left(  c\right)  }\operatorname*{id}\left(  c_{1}\right)
\otimes\epsilon_{1}\left(  c_{2}\right)  }\right)  =\kappa\left(
\sum_{\left(  c\right)  }\underbrace{\operatorname*{id}\left(  c_{1}\right)
}_{=c_{1}}\otimes\epsilon_{1}\left(  c_{2}\right)  \right)  \nonumber\\
& =\kappa\left(  \sum_{\left(  c\right)  }c_{1}\otimes\epsilon_{1}\left(
c_{2}\right)  \right)  \nonumber\\
& =\sum_{\left(  c\right)  }\epsilon_{1}\left(  c_{2}\right)  c_{1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\kappa\right)
.\label{sol.counit.unique.2nd.4}%
\end{align}


We have obtained this equality from the commutativity of the left square in
the commutative diagram \eqref{sol.counit.unique.2nd.1}. Similarly, from the
commutativity of the right square in the same diagram, we can obtain the
equality%
\[
c=\sum_{\left(  c\right)  }\epsilon_{1}\left(  c_{1}\right)  c_{2}.
\]
The same argument (with the roles of $\epsilon_{1}$ and $\epsilon_{2}$
interchanged) yields%
\[
c=\sum_{\left(  c\right)  }\epsilon_{2}\left(  c_{1}\right)  c_{2}.
\]
Applying the map $\epsilon_{1}$ to both sides of this equality, we find%
\[
\epsilon_{1}\left(  c\right)  =\epsilon_{1}\left(  \sum_{\left(  c\right)
}\epsilon_{2}\left(  c_{1}\right)  c_{2}\right)  =\sum_{\left(  c\right)
}\epsilon_{2}\left(  c_{1}\right)  \epsilon_{1}\left(  c_{2}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
} \kk\text{-linear}\right)
\]
On the other hand, applying the map $\epsilon_{2}$ to both sides of the
equality \eqref{sol.counit.unique.2nd.4}, we find%
\begin{align*}
\epsilon_{2}\left(  c\right)    & =\epsilon_{2}\left(  \sum_{\left(  c\right)
}\epsilon_{1}\left(  c_{2}\right)  c_{1}\right)  =\sum_{\left(  c\right)
}\underbrace{\epsilon_{1}\left(  c_{2}\right)  \epsilon_{2}\left(
c_{1}\right)  }_{\substack{=\epsilon_{2}\left(  c_{1}\right)  \epsilon
_{1}\left(  c_{2}\right)  \\\text{(since the multiplication}\\\text{in
} \kk \text{ is commutative)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
the map }\epsilon_{2}\text{ is } \kk\text{-linear}\right)  \\
& =\sum_{\left(  c\right)  }\epsilon_{2}\left(  c_{1}\right)  \epsilon
_{1}\left(  c_{2}\right)  .
\end{align*}
Comparing these two equalities, we find $\epsilon_{1}\left(  c\right)
=\epsilon_{2}\left(  c\right)  $.

Forget that we fixed $c$. We thus have shown that $\epsilon_{1}\left(
c\right)  =\epsilon_{2}\left(  c\right)  $ for each $c\in C$. In other words,
$\epsilon_{1}=\epsilon_{2}$. This is precisely what we set out to prove. Thus,
Exercise \ref{exe.counit.unique} is solved again.

%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.tensor-prod-algs}}

\textit{Solution to Exercise \ref{exe.tensor-prod-algs}.} (a) Exercise
\ref{exe.tensor-prod-algs}(a) is a classical result about algebras, and proven
in various textbooks. Nevertheless, we shall give two solutions to it: one
solution by elementwise computation, and another by formally manipulating
homomorphisms (this is essentially what is called ``diagram
chasing'', except that we are not going to draw any diagrams).
Both solutions have their advantages, and it is useful to see them both.

\textit{First solution to Exercise \ref{exe.tensor-prod-algs}(a).} Let $\xi$
be the canonical $\kk$-module isomorphism $\kk\rightarrow
\kk\otimes\kk$. We have defined the multiplication map
$m_{A\otimes B}$ of the $\kk$-algebra $A\otimes B$ by%
\[
m_{A\otimes B}=\left(  m_{A}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T\otimes\id_{B}\right)  ,
\]
and the unit map $u_{A\otimes B}$ of the $\kk$-algebra $A\otimes B$ by%
\[
u_{A\otimes B}=\left(  u_{A}\otimes u_{B}\right)  \circ\xi.
\]
We now must prove that the $\kk$-module $A\otimes B$, equipped with
these two maps $m_{A\otimes B}$ and $u_{A\otimes B}$, is indeed a $\kk%
$-algebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes B \otimes A \otimes B \otimes A \otimes B \ar[dl]_{m_{A\otimes B} \otimes \id_{A\otimes B}} \ar[dr]^{\id_{A\otimes B} \otimes m_{A\otimes B}} & \\ A \otimes B \otimes A \otimes B \ar[dr]^{m_{A \otimes B}} & & A \otimes B \otimes A \otimes B \ar[dl]_{m_{A \otimes B}} \\ & A \otimes B & }
\label{sol.exe.tensor-prod-algs.a.1.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes B \otimes \kk \ar[d]_{\id_{A\otimes B} \otimes u_{A \otimes B}} & A \otimes B \ar[l] \ar[d]_{\id_{A \otimes B}} \ar[r] & \kk \otimes A \otimes B \ar[d]^{u_{A \otimes B} \otimes \id_{A \otimes B}} \\ A \otimes B\otimes A \otimes B\ar[r]^-{m_{A \otimes B}} & A \otimes B & A \otimes B \otimes A \otimes B\ar[l]_-{m_{A \otimes B}} }
\label{sol.exe.tensor-prod-algs.a.1.diag2}%
\end{equation}
\footnote{where the maps $A\otimes B\rightarrow A\otimes B\otimes\kk$
and $A\otimes B\rightarrow\kk\otimes A\otimes B$ are the isomorphisms
sending each $a\in A\otimes B$ to $a\otimes1$ and to $1\otimes a$,
respectively} commute (since Definition \ref{def.algebra} yields that
$A\otimes B$ is a $\kk$-algebra if and only if these two diagrams commute).

We shall only prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.1.diag1}%
) commutes. The commutativity of the diagram
(\ref{sol.exe.tensor-prod-algs.a.1.diag2}) is proven similarly (but with less
work), and so is left to the reader.

So we need to prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.1.diag1}%
) commutes. In other words, we need to prove that
\[
m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\id_{A\otimes B} \right)
=
m_{A\otimes B}\circ\left(  \id_{A\otimes B}\circ m_{A\otimes B}\right)  .
\]


We first notice that%
\begin{equation}
m_{A\otimes B}\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime}\right)
=pp^{\prime}\otimes qq^{\prime} \label{sol.exe.tensor-prod-algs.a.1.m1}%
\end{equation}
for any $p\in A$, $q\in B$, $p^{\prime}\in A$ and $q^{\prime}\in
B$\ \ \ \ \footnote{\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.1.m1}):}
Let $p\in A$, $q\in B$, $p^{\prime}\in A$ and $q^{\prime}\in B$. Then,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T\otimes\id_{B}\right)  }
\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime
}\right) \\
&  =\left(  \left(  m_{A}\otimes m_{B}\right)  \circ\left(  \id_{A}
\otimes T\otimes\id_{B}\right)  \right)
\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(  \underbrace{\left(
\id_{A}\otimes T\otimes\id_{B}\right)
\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime
}\right)  }_{=\id_{A}\left(  p\right)  \otimes
T\left(  q\otimes p^{\prime}\right)  \otimes\id_{B}%
\left(  q^{\prime}\right)  }\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(
\underbrace{\id_{A}\left(  p\right)  }_{=p}\otimes
\underbrace{T\left(  q\otimes
p^{\prime}\right)  }_{\substack{=p^{\prime}\otimes q\\\text{(by the definition
of }T\text{)}}}\otimes\underbrace{\id_{B}\left(
q^{\prime}\right)  }_{=q^{\prime}}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(  p\otimes p^{\prime}\otimes
q\otimes q^{\prime}\right)  =\underbrace{m_{A}\left(  p\otimes p^{\prime
}\right)  }_{\substack{=pp^{\prime}\\\text{(since }m_{A}\text{ is
the}\\\text{multiplication map of }A\text{)}}}\otimes\underbrace{m_{B}\left(
q\otimes q^{\prime}\right)  }_{\substack{=qq^{\prime}\\\text{(since }%
m_{B}\text{ is the}\\\text{multiplication map of }B\text{)}}}\\
&  =pp^{\prime}\otimes qq^{\prime}.
\end{align*}
Qed.}.

Let $x\in A\otimes B\otimes A\otimes B\otimes A\otimes B$. Thus, $x$ (like any
tensor in $A\otimes B\otimes A\otimes B\otimes A\otimes B$) must be a
$\kk$-linear combination of pure tensors.

We want to show the equality%
\begin{equation}
\left(  m_{A\otimes B}\circ
\left(  m_{A\otimes B}\otimes\id_{A\otimes B}\right)  \right)
\left(  x\right)
= \left(  m_{A\otimes B}\circ\left(  \id_{A\otimes B}\circ
m_{A\otimes B}\right)  \right)  \left(  x\right)  .
\label{sol.exe.tensor-prod-algs.a.1.eq-x1}%
\end{equation}
This equality is $\kk$-linear in $x$ (since all maps that appear in it
are $\kk$-linear). Hence, we can WLOG assume that $x$ is a pure tensor
(since $x$ is a $\kk$-linear combination of pure tensors). Assume this.
Thus, $x=a\otimes b\otimes a^{\prime}\otimes b^{\prime}\otimes a^{\prime
\prime}\otimes b^{\prime\prime}$ for some $a\in A$, $b\in B$, $a^{\prime}\in
A$, $b^{\prime}\in B$, $a^{\prime\prime}\in A$ and $b^{\prime\prime}\in B$.
Consider these $a$, $b$, $a^{\prime}$, $b^{\prime}$, $a^{\prime\prime}$ and
$b^{\prime\prime}$.

Now,%
\begin{align*}
\left(  m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes
\id_{A\otimes B}\right)  \right)  \left(  x\right)   &  =m_{A\otimes
B}\left(  \left(  m_{A\otimes B}\otimes\id_{A\otimes
B}\right)  \left(  \underbrace{x}_{=a\otimes b\otimes a^{\prime}\otimes
b^{\prime}\otimes a^{\prime\prime}\otimes b^{\prime\prime}}\right)  \right)
\\
&  =m_{A\otimes B}\left(  \underbrace{\left(  m_{A\otimes B}\otimes
\id_{A\otimes B}\right)  \left(  a\otimes b\otimes
a^{\prime}\otimes b^{\prime}\otimes a^{\prime\prime}\otimes b^{\prime\prime
}\right)  }_{=m_{A\otimes B}\left(  a\otimes b\otimes a^{\prime}\otimes
b^{\prime}\right)  \otimes\id_{A\otimes B}\left(
a^{\prime\prime}\otimes b^{\prime\prime}\right)  }\right)  \\
&  =m_{A\otimes B}\left(  \underbrace{m_{A\otimes B}\left(  a\otimes b\otimes
a^{\prime}\otimes b^{\prime}\right)  }_{\substack{=aa^{\prime}\otimes
bb^{\prime}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.1.m1}))}}%
}\otimes\underbrace{\id_{A\otimes B}\left(
a^{\prime\prime}\otimes b^{\prime\prime}\right)  }_{=a^{\prime\prime}\otimes
b^{\prime\prime}}\right)  \\
&  =m_{A\otimes B}\left(  aa^{\prime}\otimes bb^{\prime}\otimes a^{\prime
\prime}\otimes b^{\prime\prime}\right)  =\underbrace{\left(  aa^{\prime
}\right)  a^{\prime\prime}}_{=aa^{\prime}a^{\prime\prime}}\otimes
\underbrace{\left(  bb^{\prime}\right)  b^{\prime\prime}}_{=bb^{\prime
}b^{\prime\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.tensor-prod-algs.a.1.m1})}\right)  \\
&  =aa^{\prime}a^{\prime\prime}\otimes bb^{\prime}b^{\prime\prime}.
\end{align*}
A similar computation shows that%
\[
\left(  m_{A\otimes B}\circ\left(  \id_{A\otimes
B}\otimes m_{A\otimes B}\right)  \right)  \left(  x\right)  =aa^{\prime
}a^{\prime\prime}\otimes bb^{\prime}b^{\prime\prime}.
\]
Comparing these two equalities, we obtain $\left(  m_{A\otimes B}\circ\left(
m_{A\otimes B}\otimes\id_{A\otimes B}\right)  \right)
\left(  x\right)  =\left(  m_{A\otimes B}\circ\left(
\id_{A\otimes B}\otimes m_{A\otimes B}\right)  \right)  \left(
x\right)  $. Thus, the equality (\ref{sol.exe.tensor-prod-algs.a.1.eq-x1}) is proven.

Now, forget that we fixed $x$. We thus have proven the equality
(\ref{sol.exe.tensor-prod-algs.a.1.eq-x1}) for every $x\in A\otimes B\otimes
A\otimes B\otimes A\otimes B$. In other words, we have $m_{A\otimes B}%
\circ\left(  m_{A\otimes B}\otimes\id_{A\otimes
B}\right)  =m_{A\otimes B}\circ\left(  \id_{A\otimes
B}\circ m_{A\otimes B}\right)  $. In other words, the diagram
(\ref{sol.exe.tensor-prod-algs.a.1.diag1}) commutes.

It remains to prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.1.diag2}%
) commutes. We leave this to the reader, as the proof is similar to (but
simpler than) the proof of the commutativity of
(\ref{sol.exe.tensor-prod-algs.a.1.diag1}) given above. Hence, both diagrams
(\ref{sol.exe.tensor-prod-algs.a.1.diag1}) and
(\ref{sol.exe.tensor-prod-algs.a.1.diag2}) commute. In other words, the
$\kk$-module $A\otimes B$, equipped with the two maps $m_{A\otimes B}$
and $u_{A\otimes B}$, is a $\kk$-algebra. In other words, the
$\kk$-algebra $A\otimes B$ introduced in Definition
\ref{def.tensor-prod-algs} is actually well-defined. This solves Exercise
\ref{exe.tensor-prod-algs}(a).

\textit{Second solution to Exercise \ref{exe.tensor-prod-algs}(a).} For any
two $\kk$-modules $U$ and $V$, let $T_{U,V}:U\otimes V\rightarrow
V\otimes U$ be the twist map (i.e., the $\kk$-linear map $U\otimes
V\rightarrow V\otimes U$ sending every $u\otimes v$ to $v\otimes u$). A simple
linear-algebraic fact says that if $U$, $V$, $U^{\prime}$ and $V^{\prime}$ are
four $\kk$-modules and $x:U\rightarrow U^{\prime}$ and $y:V\rightarrow
V^{\prime}$ are two $\kk$-linear maps, then
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}=T_{U^{\prime},V^{\prime}}\circ\left(
x\otimes y\right)  .\label{sol.exe.tensor-prod-algs.a.2.twistcomm}%
\end{equation}


For every $\kk$-module $U$, let $\operatorname{kan}_{1,U}:U\rightarrow
U\otimes\kk$ and $\operatorname{kan}_{2,U}:U\rightarrow\kk\otimes U$
be the canonical $\kk$-module isomorphisms. Every
$\kk$-modules $U$ and $V$ satisfy the identities%
\begin{align}
\operatorname{id}_{U}\otimes\operatorname{kan}_{1,V}  &  =\operatorname{kan}%
_{1,U\otimes V},\label{sol.exe.tensor-prod-algs.a.2.kans.1}\\
\operatorname{kan}_{2,V}\otimes\operatorname{id}_{U}  &  =\operatorname{kan}%
_{2,V\otimes U},\label{sol.exe.tensor-prod-algs.a.2.kans.2}\\
\operatorname{id}_{U}\otimes\operatorname{kan}_{1,V}^{-1}  &
=\operatorname{kan}_{1,U\otimes V}^{-1}%
,\label{sol.exe.tensor-prod-algs.a.2.kans.3}\\
\operatorname{kan}_{2,V}^{-1}\otimes\operatorname{id}_{U}  &
=\operatorname{kan}_{2,V\otimes U}^{-1}.
\label{sol.exe.tensor-prod-algs.a.2.kans.4}%
\end{align}
(These identities are well-known and straightforward to check.)

Recall that the $\kk$-module $A$, equipped with the maps $m_{A}$ and
$u_{A}$, is a $\kk$-algebra. In other words, the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes A \otimes A \ar[dl]_{m_A \otimes \id_A} \ar[dr]^{\id_{A} \otimes m_A} & \\ A \otimes A \ar[dr]^{m_{A}} & & A \otimes A \ar[dl]_{m_{A}} \\ & A & }
\label{sol.exe.tensor-prod-algs.a.2.Adiag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes \kk \ar[d]_{\id_{A} \otimes u_{A}} & A \ar[l]_{\operatorname{kan}_{1,A}} \ar[d]_{\id_{A}} \ar[r]^{\operatorname{kan}_{2,A}} & \kk \otimes A \ar[d]^{u_{A} \otimes \id_{A}} \\ A \otimes A \ar[r]^-{m_{A}} & A & A \otimes A \ar[l]_-{m_{A }} }
\label{sol.exe.tensor-prod-algs.a.2.Adiag2}%
\end{equation}
commute (since Definition \ref{def.algebra} yields that $A$ is a $\kk%
$-algebra if and only if these two diagrams commute).

The diagram (\ref{sol.exe.tensor-prod-algs.a.2.Adiag1}) commutes. In other
words, we have%
\begin{equation}
m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Adiag1.eq}%
\end{equation}
The same argument (applied to $B$ instead of $A$) shows that%
\begin{equation}
m_{B}\circ\left(  m_{B}\otimes\id_{B}\right)
=m_{B}\circ\left(  \id_{B}\otimes m_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Bdiag1.eq}%
\end{equation}


The diagram (\ref{sol.exe.tensor-prod-algs.a.2.Adiag2}) commutes. In other
words, we have%
\begin{align}
\id_{A}  &  =m_{A}\circ\left(  \id_{A}\otimes u_{A}\right)
\circ\operatorname*{kan}\nolimits_{1,A}%
\ \ \ \ \ \ \ \ \ \ \text{and} \label{sol.exe.tensor-prod-algs.a.2.Adiag2.eq1}%
\\
\id_{A}  &  =m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  \circ\operatorname*{kan}%
\nolimits_{2,A}. \label{sol.exe.tensor-prod-algs.a.2.Adiag2.eq2}%
\end{align}
The same argument (applied to $B$ instead of $A$) shows that%
\begin{align}
\id_{B}  &  =m_{B}\circ\left(  \id_{B}\otimes u_{B}\right)
\circ\operatorname*{kan}\nolimits_{1,B}%
\ \ \ \ \ \ \ \ \ \ \text{and} \label{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq1}%
\\
\id_{B}  &  =m_{B}\circ\left(  u_{B}\otimes
\id_{B}\right)  \circ\operatorname*{kan}%
\nolimits_{2,B}. \label{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq2}%
\end{align}


Let $\xi$ be the canonical $\kk$-module isomorphism $\kk%
\rightarrow\kk\otimes\kk$. According to Definition
\ref{def.tensor-prod-algs}, we define the map $m_{A\otimes B}:A\otimes
B\otimes A\otimes B\rightarrow A\otimes B$ by%
\[
m_{A\otimes B}=\left(  m_{A}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A}\otimes\id_{B}\right)  ,
\]
and we define the map $u_{A\otimes B}:\kk\rightarrow A\otimes B$ by%
\[
u_{A\otimes B}=\left(  u_{A}\otimes u_{B}\right)  \circ\xi.
\]
We now must prove that the $\kk$-module $A\otimes B$, equipped with
these two maps $m_{A\otimes B}$ and $u_{A\otimes B}$, is indeed a $\kk%
$-algebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes B \otimes A \otimes B \otimes A \otimes B \ar[dl]_{m_{A\otimes B} \otimes \id_{A\otimes B}} \ar[dr]^{\id_{A\otimes B} \otimes m_{A\otimes B}} & \\ A \otimes B \otimes A \otimes B \ar[dr]^{m_{A \otimes B}} & & A \otimes B \otimes A \otimes B \ar[dl]_{m_{A \otimes B}} \\ & A \otimes B & }
\label{sol.exe.tensor-prod-algs.a.2.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes B \otimes \kk \ar[d]_{\id_{A\otimes B} \otimes u_{A \otimes B}} & A \otimes B \ar[l]_{\operatorname{kan}_{1,A\otimes B}} \ar[d]_{\id_{A \otimes B}} \ar[r]^{\operatorname{kan}_{2,A,\otimes B}} & \kk \otimes A \otimes B \ar[d]^{u_{A \otimes B} \otimes \id_{A \otimes B}} \\ A \otimes B\otimes A \otimes B\ar[r]^-{m_{A \otimes B}} & A \otimes B & A \otimes B \otimes A \otimes B\ar[l]_-{m_{A \otimes B}} }
\label{sol.exe.tensor-prod-algs.a.2.diag2}%
\end{equation}
commute (since Definition \ref{def.algebra} yields that $A\otimes B$ is a
$\kk$-algebra if and only if these two diagrams commute).

Let us first prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag1})
commutes. In other words, let us prove that%
\[
m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes
\id_{A\otimes B}\right)  =m_{A\otimes B}\circ
\left(  \id_{A\otimes B}\otimes m_{A\otimes B}\right)  .
\]


Define a $\kk$-linear map $Q:B\otimes A\otimes B\otimes A\rightarrow
A\otimes A\otimes B\otimes B$ by%
\begin{equation}
Q=\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  T_{B,A}\otimes
T_{B,A}\right)  . \label{sol.exe.tensor-prod-algs.a.2.Q=}%
\end{equation}


It is easy to see that%
\begin{equation}
Q=\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Q1}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q1}):} Let $b\in B$,
$a\in A$, $b^{\prime}\in B$ and $a^{\prime}\in A$ be arbitrary. Applying both
sides of the equality (\ref{sol.exe.tensor-prod-algs.a.2.Q=}) to $b\otimes
a\otimes b^{\prime}\otimes a^{\prime}$, we obtain%
\begin{align}
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)   &  =\left(
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  T_{B,A}\otimes
T_{B,A}\right)  \right)  \left(  b\otimes a\otimes b^{\prime}\otimes
a^{\prime}\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \left(  \underbrace{\left(
T_{B,A}\otimes T_{B,A}\right)  \left(  b\otimes a\otimes b^{\prime}\otimes
a^{\prime}\right)  }_{=T_{B,A}\left(  b\otimes a\right)  \otimes T_{B,A}
\left(  b^{\prime}\otimes a^{\prime}\right)  }\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \left(  \underbrace{T_{B,A}\left(
b\otimes a\right)  }_{\substack{=a\otimes b\\\text{(by the definition of
}T_{B,A}\text{)}}}\otimes\underbrace{T_{B,A}\left(
b^{\prime}\otimes a^{\prime}\right)  }_{\substack{=a^{\prime}\otimes
b^{\prime}\\\text{(by the definition of }T_{B,A}\text{)}%
}}\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \left(  a\otimes b\otimes a^{\prime
}\otimes b^{\prime}\right)  =\underbrace{\id_{A}%
\left(  a\right)  }_{=a}\otimes\underbrace{T_{B,A}\left(  b\otimes a^{\prime
}\right)  }_{\substack{=a^{\prime}\otimes b\\\text{(by the definition of
}T_{B,A}\text{)}}}\otimes\underbrace{\id_{B}\left(
b^{\prime}\right)  }_{=b^{\prime}}\nonumber\\
&  =a\otimes a^{\prime}\otimes b\otimes b^{\prime}.
\label{sol.exe.tensor-prod-algs.a.2.Q1.pf.1}%
\end{align}


Comparing this with%
\begin{align*}
&  \left(  \left(  \id_{A}\otimes T_{B\otimes
B,A}\right)  \circ\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \right)  \left(  b\otimes
a\otimes b^{\prime}\otimes a^{\prime}\right)  \\
&  =\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\left(  \underbrace{\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)  }_{=T_{B,A}\left(  b\otimes a\right)
\otimes\id_{B}\left(  b^{\prime}\right)
\otimes\id_{A}\left(  a^{\prime}\right)  }\right)  \\
&  =\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\left(  \underbrace{T_{B,A}\left(  b\otimes a\right)  }_{\substack{=a\otimes
b\\\text{(by the definition of }T_{B,A}\text{)}}}\otimes
\underbrace{\id_{B}\left(  b^{\prime}\right)
}_{=b^{\prime}}\otimes\underbrace{\id_{A}\left(
a^{\prime}\right)  }_{=a^{\prime}}\right)  \\
&  =\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\left(  a\otimes b\otimes b^{\prime}\otimes a^{\prime}\right)
=\underbrace{\id_{A}\left(  a\right)  }_{=a}%
\otimes\underbrace{T_{B\otimes B,A}\left(  b\otimes b^{\prime}\otimes
a^{\prime}\right)  }_{\substack{=a^{\prime}\otimes b\otimes b^{\prime
}\\\text{(by the definition of }T_{B\otimes B,A}\text{)}}}\\
&  =a\otimes a^{\prime}\otimes b\otimes b^{\prime},
\end{align*}
we obtain%
\begin{equation}
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)  =\left(
\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  \right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)
.\label{sol.exe.tensor-prod-algs.a.2.Q1.pf.2}%
\end{equation}


Now, forget that we fixed $b,a,b^{\prime},a^{\prime}$. We thus have shown that
every $b\in B$, $a\in A$, $b^{\prime}\in B$ and $a^{\prime}\in A$ satisfy
(\ref{sol.exe.tensor-prod-algs.a.2.Q1.pf.2}). In other words, the two maps $Q$
and $\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  $ are equal to each other on each pure
tensor. Since these two maps are $\kk$-linear, we thus conclude that
these two maps are identical (because if two $\kk$-linear maps from a
tensor product are equal to each other on each pure tensor, then these two
maps are identical). In other words,
\[
Q=\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  .
\]
This proves (\ref{sol.exe.tensor-prod-algs.a.2.Q1}).]

Now,%
\begin{align}
&  \underbrace{\id_{A}}_{=\id_{A}\circ\id_{A}}
\otimes\underbrace{Q}%
_{\substack{=\left(  \id_{A}\otimes T_{B\otimes
B,A}\right)  \circ\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Q1}))}}}\otimes
\underbrace{\id_{B}}_{=\id_{B}\circ\id_{B}}\nonumber\\
&  =\left(  \id_{A}\circ\id_{A}\right)  \otimes
\left(  \left(  \id_{A}%
\otimes T_{B\otimes B,A}\right)  \circ
\left(  T_{B,A}\otimes\id_{B}\otimes\id_{A}\right)  \right)
\otimes\left(  \id_{B}\circ\id_{B}\right)  \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
\id_{A}\otimes T_{B\otimes B,A}\right)  \otimes
\id_{B}\right)  }_{=\id_{A}%
\otimes\id_{A}\otimes T_{B\otimes B,A}\otimes
\id_{B}}\circ\underbrace{\left(  \id_{A}\otimes
\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \otimes\id_{B}\right)
}_{=\id_{A}\otimes
T_{B,A}\otimes\id_{B}\otimes\id_{A}\otimes\id_{B}}\nonumber\\
&  =\left(  \id_{A}\otimes\id_{A}\otimes
T_{B\otimes B,A}\otimes\id_{B}%
\right)  \circ\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\otimes\id_{A}%
\otimes\id_{B}\right)
.\label{sol.exe.tensor-prod-algs.a.2.Q1'}%
\end{align}


Now, we shall show that%
\begin{equation}
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A\otimes B}%
\otimes\id_{A\otimes B}\right)  =\left(  m_{A}%
\otimes\id_{A}\otimes m_{B}\otimes\id_{B}\right)
\circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)
\label{sol.exe.tensor-prod-algs.a.2.1}%
\end{equation}
(as maps from $A\otimes B\otimes A\otimes B\otimes A\otimes B$ to $A\otimes
A\otimes B\otimes B$).

[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.1}):} We have%
\begin{align}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\otimes
\underbrace{\id_{A\otimes B}}_{=\id_{A\otimes B}%
\circ\id_{A\otimes B}}\nonumber\\
&  =\left(  \left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes\id_{B}\right)
\right)  \otimes\left(  \id_{A\otimes B}%
\circ\id_{A\otimes B}\right)  \nonumber\\
&  =\underbrace{\left(  \left(  m_{A}\otimes m_{B}\right)  \otimes
\id_{A\otimes B}\right)  }_{=m_{A}\otimes m_{B}%
\otimes\id_{A\otimes B}}\circ\underbrace{\left(
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \otimes
\id_{A\otimes B}\right)  }_{=\id_{A}\otimes
T_{B,A}\otimes\id_{B}\otimes
\id_{A\otimes B}}\nonumber\\
&  =\left(  m_{A}\otimes m_{B}\otimes\id_{A\otimes
B}\right)  \circ\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\otimes\id_{A\otimes B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.1.pf.1}%
\end{align}


The equality (\ref{sol.exe.tensor-prod-algs.a.2.twistcomm}) (applied to
$B\otimes B$, $A$, $B$, $A$, $m_{B}$ and $\id_{A}$
instead of $U$, $V$, $U^{\prime}$, $V^{\prime}$, $x$ and $y$) yields
\[
\left(  \id_{A}\otimes m_{B}\right)  \circ T_{B\otimes
B,A}=T_{B,A}\circ\left(  m_{B}\otimes\id_{A}\right)  .
\]
Hence,%
\begin{align*}
&  \left(  \id_{A}\circ m_{A}\right)  \otimes
\underbrace{\left(  \left(  \id_{A}\otimes
m_{B}\right)  \circ T_{B\otimes B,A}\right)  }_{=T_{B,A}\circ\left(
m_{B}\otimes\id_{A}\right)  }\otimes\left(
\id_{B}\circ\id_{B}\right)  \\
&  =\left(  \id_{A}\circ m_{A}\right)  \otimes\left(
T_{B,A}\circ\left(  m_{B}\otimes\id_{A}\right)
\right)  \otimes\left(  \id_{B}\circ\id_{B}\right)  \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(  m_{A}%
\otimes\left(  m_{B}\otimes\id_{A}\right)
\otimes\id_{B}\right)  }_{\substack{=m_{A}\otimes
m_{B}\otimes\id_{A}\otimes\id_{B}
=m_{A}\otimes m_{B}\otimes\id_{A\otimes
B}\\\text{(since }\id_{A}\otimes
\id_{B}=\id_{A}\otimes B\text{)}}}\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A}\otimes m_{B}%
\otimes\id_{A\otimes B}\right)  .
\end{align*}
Thus,%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A}\otimes m_{B}%
\otimes\id_{A\otimes B}\right)  \nonumber\\
&  =\underbrace{\left(  \id_{A}\circ m_{A}\right)
}_{=m_{A}=m_{A}\circ\id_{A\otimes A}}\otimes\left(
\left(  \id_{A}\otimes m_{B}\right)  \circ T_{B\otimes
B,A}\right)  \otimes\left(  \id_{B}\circ
\id_{B}\right)  \nonumber\\
&  =\left(  m_{A}\circ\id_{A\otimes A}\right)
\otimes\left(  \left(  \id_{A}\otimes m_{B}\right)
\circ T_{B\otimes B,A}\right)  \otimes
\left(  \id_{B}\circ\id_{B}\right)  \nonumber\\
&  =\underbrace{\left(  m_{A}\otimes\left(  \id_{A}%
\otimes m_{B}\right)  \otimes\id_{B}\right)  }%
_{=m_{A}\otimes\id_{A}\otimes m_{B}\otimes
\id_{B}}\circ
\left(  \underbrace{\id_{A\otimes A}}_{=\id_{A}\otimes
\id_{A}}\otimes T_{B\otimes B,A}\otimes
\id_{B}\right)  \nonumber\\
&  =\left(  m_{A}\otimes\id_{A}\otimes m_{B}%
\otimes\id_{B}\right)
\circ\left(  \id_{A}\otimes\id_{A}\otimes T_{B\otimes
B,A}\otimes\id_{B}\right)
.\label{sol.exe.tensor-prod-algs.a.2.1.pf.2}%
\end{align}


Now,%
\begin{align*}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(  m_{A\otimes
B}\otimes\id_{A\otimes B}\right)  }%
_{\substack{=\left(  m_{A}\otimes m_{B}\otimes
\id_{A\otimes B}\right)  \circ\left(  \id_{A}%
\otimes T_{B,A}\otimes\id_{B}\otimes
\id_{A\otimes B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.1.pf.1}))}}}\\
&  =\underbrace{\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  \circ\left(  m_{A}\otimes
m_{B}\otimes\id_{A\otimes B}\right)  }%
_{\substack{=\left(  m_{A}\otimes\id_{A}\otimes
m_{B}\otimes\id_{B}\right)  \circ\left(
\id_{A}\otimes\id_{A}\otimes
T_{B\otimes B,A}\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.1.pf.2}))}}}
\circ\left(  \id_{A}\otimes T_{B,A}\otimes\id_{B}%
\otimes\underbrace{\id_{A\otimes B}}%
_{=\id_{A}\otimes\id_{B}%
}\right) \\
&  =\left(  m_{A}\otimes\id_{A}\otimes m_{B}%
\otimes\id_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes\id_{A}\otimes
T_{B\otimes B,A}\otimes\id_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A}\otimes\id_{B}
\otimes\id_{A}\otimes\id_{B}\right)
}_{\substack{=\id_{A}\otimes
Q\otimes\id_{B}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Q1'}))}}}\\
&  =\left(  m_{A}\otimes\id_{A}\otimes m_{B}%
\otimes\id_{B}\right)
\circ\left(  \id_{A}\otimes Q\otimes\id_{B}\right)  .
\end{align*}
This proves (\ref{sol.exe.tensor-prod-algs.a.2.1}).]

Now,%
\begin{align}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ\left(  m_{A\otimes B}%
\otimes\id_{A\otimes B}\right)  \nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\id_{B}\right)  \circ
\left(  m_{A\otimes B}\otimes\id_{A\otimes B}\right)
}_{\substack{=\left(  m_{A}\otimes
\id_{A}\otimes m_{B}\otimes\id_{B}\right)  \circ
\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.1}))}}}\nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(  m_{A}%
\otimes\id_{A}\otimes m_{B}\otimes\id_{B}\right)
}_{=\left(  m_{A}\otimes\id_{A}\right)  \otimes
\left(  m_{B}\otimes\id_{B}\right)  }\circ
\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \nonumber\\
&  =\underbrace{\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(
m_{A}\otimes\id_{A}\right)  \otimes\left(
m_{B}\otimes\id_{B}\right)  \right)  }_{=\left(
m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \right)
\otimes\left(  m_{B}\circ\left(  m_{B}\otimes\id_{B}%
\right)  \right)  }\circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \nonumber\\
&  =\left(  \underbrace{\left(  m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \right)  }_{\substack{=m_{A}%
\circ\left(  \id_{A}\otimes m_{A}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Adiag1.eq}))}}}\otimes\underbrace{\left(
m_{B}\circ\left(  m_{B}\otimes\id_{B}\right)  \right)
}_{\substack{=m_{B}\circ\left(  \id_{B}\otimes
m_{B}\right)  \\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Bdiag1.eq}))}%
}}\right)  \circ\left(  \id_{A}\otimes Q\otimes
\id_{B}\right)  \nonumber\\
&  =\underbrace{\left(  \left(  m_{A}\circ\left(  \id_{A}
\otimes m_{A}\right)  \right)  \otimes\left(  m_{B}\circ\left(
\id_{B}\otimes m_{B}\right)  \right)  \right)
}_{=\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  \id_{A}
\otimes m_{A}\right)  \otimes\left(  \id_{B}\otimes m_{B}\right)
\right)  }\circ\left(  \id_{A}\otimes Q\otimes\id_{B}\right)
\nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(  \left(
\id_{A}\otimes m_{A}\right)  \otimes\left(
\id_{B}\otimes m_{B}\right)  \right)  }%
_{=\id_{A}\otimes m_{A}\otimes\id_{B}\otimes m_{B}}\circ
\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ
\left(  \id_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)
.\label{sol.exe.tensor-prod-algs.a.2.half}%
\end{align}


On the other hand, it is easy to see that%
\begin{equation}
Q=\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\circ\left(  \id_{B}\otimes\id_{A}\otimes T_{B,A}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Q2}
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q2}):} This proof is
similar to the proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q1}), so we omit it.]

Now,%
\begin{align}
&  \underbrace{\id_{A}}_{=\id_{A}\circ\id_{A}}\otimes
\underbrace{Q}_{\substack{=\left(  T_{B,A\otimes A}\otimes\id_{B}%
\right)  \circ\left(  \id_{B}\otimes\id_{A}\otimes T_{B,A}\right)
\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Q2}))}}}\otimes
\underbrace{\id_{B}}_{=\id_{B}\circ\id_{B}}\nonumber\\
&  =\left(  \id_{A}\circ\id_{A}\right)  \otimes
\left(  \left(  T_{B,A\otimes A}\otimes
\id_{B}\right)  \circ
\left(  \id_{B}\otimes\id_{A}\otimes T_{B,A}\right)
\right)  \otimes\left(  \id_{B}\circ\id_{B}\right)  \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
T_{B,A\otimes A}\otimes\id_{B}\right)  \otimes
\id_{B}\right)  }_{=\id_{A}%
\otimes T_{B,A\otimes A}\otimes\id_{B}\otimes
\id_{B}}\circ\underbrace{\left(  \id_{A}\otimes
\left(  \id_{B}\otimes
\id_{A}\otimes T_{B,A}\right)  \otimes
\id_{B}\right)  }_{=\id_{A}%
\otimes\id_{B}\otimes\id_{A}%
\otimes T_{B,A}\otimes\id_{B}}\nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A\otimes A}%
\otimes\id_{B}\otimes\id_{B}%
\right)  \circ
\left(  \id_{A}\otimes\id_{B}\otimes\id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)
.\label{sol.exe.tensor-prod-algs.a.2.Q2'}%
\end{align}


Now, we shall show that%
\begin{equation}
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ
\left(  \id_{A\otimes B}\otimes m_{A,B}\right)
=\left(  \id_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)
\label{sol.exe.tensor-prod-algs.a.2.2}%
\end{equation}
(as maps from $A\otimes B\otimes A\otimes B\otimes A\otimes B$ to $A\otimes
A\otimes B\otimes B$).

[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.2}):} We have%
\begin{align}
&  \underbrace{\id_{A\otimes B}}_{=\id_{A\otimes B}
\circ\id_{A\otimes B}}%
\otimes\underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\nonumber\\
&  =\left(  \id_{A\otimes B}\circ\id_{A\otimes B}\right)
\otimes\left(  \left(  m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \right)  \nonumber\\
&  =\underbrace{\left(  \id_{A\otimes B}\otimes\left(
m_{A}\otimes m_{B}\right)  \right)  }_{=\id_{A\otimes
B}\otimes m_{A}\otimes m_{B}}\circ
\underbrace{\left(  \id_{A\otimes B}\otimes\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \right)  }%
_{=\id_{A\otimes B}\otimes\id_{A}\otimes T_{B,A}\otimes\id_{B}}
\nonumber\\
&  =\left(  \id_{A\otimes B}\otimes m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A\otimes B}%
\otimes\id_{A}\otimes T_{B,A}\otimes\id_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.2.pf.1}
\end{align}


The equality (\ref{sol.exe.tensor-prod-algs.a.2.twistcomm}) (applied to $B$,
$A\otimes A$, $B$, $A$, $\id_{B}$ and $m_{A}$ instead
of $U$, $V$, $U^{\prime}$, $V^{\prime}$, $x$ and $y$) yields
\[
\left(  m_{A}\otimes\id_{B}\right)  \circ
T_{B,A\otimes A}=T_{B,A}\circ\left(  \id_{B}\otimes
m_{A}\right)  .
\]
Hence,%
\begin{align*}
&  \left(  \id_{A}\circ\id_{A}\right)  \otimes
\underbrace{\left(  \left(  m_{A}\otimes
\id_{B}\right)  \circ T_{B,A\otimes A}\right)
}_{=T_{B,A}\circ\left(  \id_{B}\otimes m_{A}\right)
}\otimes\left(  \id_{B}\circ m_{B}\right)  \\
&  =\left(  \id_{A}\circ\id_{A}\right)  \otimes
\left(  T_{B,A}\circ\left(  \id_{B}\otimes m_{A}\right)
\right)  \otimes\left(  \id_{B}\circ m_{B}\right)  \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes\left(  \id_{B}\otimes m_{A}\right)
\otimes m_{B}\right)  }%
_{\substack{=\id_{A}\otimes\id_{B}\otimes m_{A}\otimes m_{B}
=\id_{A\otimes B}\otimes m_{A}\otimes m_{B}\\
\text{(since }\id_{A}\otimes\id_{B}=\id_{A\otimes B}\text{)}}}\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ
\left(  \id_{A\otimes B}\otimes m_{A}\otimes m_{B}\right)  .
\end{align*}
Thus,%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ
\left(  \id_{A\otimes B}\otimes m_{A}\otimes m_{B}\right)  \nonumber\\
&  =\left(  \id_{A}\circ\id_{A}\right)  \otimes
\left(  \left(  m_{A}\otimes\id_{B}\right)  \circ
T_{B,A\otimes A}\right)  \otimes
\underbrace{\left(  \id_{B}\circ m_{B}\right)
}_{=m_{B}=m_{B}\circ\id_{B\otimes B}}\nonumber\\
&  =\left(  \id_{A}\circ\id_{A}\right)  \otimes
\left(  \left(  m_{A}\otimes\id_{B}\right)  \circ
T_{B,A\otimes A}\right)  \otimes\left(  m_{B}%
\circ\id_{B\otimes B}\right)  \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
m_{A}\otimes\id_{B}\right)  \otimes m_{B}\right)
}_{=\id_{A}\otimes m_{A}\otimes\id_{B}\otimes m_{B}}
\circ\left(  \id_{A}\otimes
T_{B,A\otimes A}\otimes\underbrace{\id_{B\otimes B}%
}_{=\id_{B}\otimes\id_{B}%
}\right)  \nonumber\\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{B}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A\otimes A}\otimes
\id_{B}\otimes\id_{B}\right)
.\label{sol.exe.tensor-prod-algs.a.2.2.pf.2}%
\end{align}


Now,%
\begin{align*}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(
\id_{A\otimes B}\otimes m_{A\otimes B}\right)
}_{\substack{=\left(  \id_{A\otimes B}\otimes
m_{A}\otimes m_{B}\right)  \circ\left(  \id_{A\otimes
B}\otimes\id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.2.pf.1}))}}}\\
&  =\underbrace{\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  \circ
\left(  \id_{A\otimes B}\otimes m_{A}\otimes m_{B}\right)
}_{\substack{=\left(
\id_{A}\otimes m_{A}\otimes\id_{B}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A\otimes A}\otimes\id_{B}
\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.2.pf.2}))}}}\circ\left(
\underbrace{\id_{A\otimes B}}_{=\id_{A}\otimes\id_{B}}
\otimes\id_{A}\otimes T_{B,A}\otimes\id_{B}\right) \\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{B}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A\otimes A}\otimes
\id_{B}\otimes\id_{B}\right)
\circ\left(  \id_{A}\otimes\id_{B}\otimes\id_{A}
\otimes T_{B,A} \otimes\id_{B}\right)  }%
_{\substack{=\id_{A}\otimes Q\otimes\id_{B}\\
\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Q2'}))}}}\\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{B}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes Q\otimes\id_{B}\right)  .
\end{align*}
This proves (\ref{sol.exe.tensor-prod-algs.a.2.2}).]

Now,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ
\left(  \id_{A\otimes B}\otimes m_{A\otimes B}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{A\otimes
B}\otimes m_{A\otimes B}\right)  }_{\substack{=\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.2}))}}}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  .
\end{align*}
Comparing this with (\ref{sol.exe.tensor-prod-algs.a.2.half}), we obtain%
\[
m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  =m_{A\otimes B}\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A\otimes B}\right)  .
\]
In other words, the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag1}) commutes.

Let us next prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

We first observe that%
\begin{equation}
\left(  \id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}=\operatorname*{kan}\nolimits_{1,A}%
\otimes\operatorname*{kan}\nolimits_{1,B}%
.\label{sol.exe.tensor-prod-algs.a.2.k1}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.k1}):} This can be proven
along the same lines as the proof of proof of
(\ref{sol.exe.tensor-prod-algs.a.2.Q1}): We fix $a\in A$ and $b\in B$, and
apply both sides of (\ref{sol.exe.tensor-prod-algs.a.2.k1}) to $a\otimes b$,
then check that the results are the same.]

The equality (\ref{sol.exe.tensor-prod-algs.a.2.kans.1}) (applied to $U=A$ and
$V=B$) yields%
\begin{equation}
\operatorname{id}_{A}\otimes\operatorname{kan}_{1,B}=\operatorname{kan}%
_{1,A\otimes B}. \label{sol.exe.tensor-prod-algs.a.2.u0}%
\end{equation}


From%
\begin{align*}
&  \left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\underbrace{\left(  \left(  u_{A}\otimes
\id_{B}\right)  \circ T_{B,\kk}\right)
}_{\substack{=T_{B,A}\circ\left(  \id_{B}\otimes
u_{A}\right)  \\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.twistcomm})
(applied}\\\text{to }B\text{, }\kk\text{, }B\text{, }A\text{,
}\id_{B}\text{ and }u_{A}\\\text{instead of }U\text{,
}V\text{, }U^{\prime}\text{, }V^{\prime}\text{, }x\text{ and }y\text{))}%
}}\otimes\underbrace{\left(  u_{B}\circ\id_{\kk%
}\right)  }_{=u_{B}=\id_{B}\circ u_{B}}\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  T_{B,A}\circ\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{A}\right)  \right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\circ u_{B}\right)  =\left(  \id_{A}%
\otimes T_{B,A}\otimes\id_{B}\right)  \circ
\underbrace{\left(  \id_{A}\otimes\left(
\id_{B}\otimes u_{A}\right)  \otimes u_{B}\right)
}_{=\id_{A}\otimes\id_{B}%
\otimes u_{A}\otimes u_{B}}\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(
\underbrace{\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}}_{=\id_{A\otimes B}}\otimes u_{A}\otimes
u_{B}\right)  =\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes u_{A}\otimes u_{B}\right)  ,
\end{align*}
we obtain%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes u_{A}\otimes u_{B}\right)  \nonumber\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  \left(  u_{A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ T_{B,\kk}\right)  \otimes\left(  u_{B}%
\circ\id_{\kk}\right)  \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
u_{A}\otimes\id_{B}\right)  \otimes u_{B}\right)
}_{=\id_{A}\otimes u_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes u_{B}}\circ\left(  \id_{A}\otimes
T_{B,\kk}\otimes\id_{\kk}\right)
\nonumber\\
&  =\left(  \id_{A}\otimes u_{A}\otimes
\id_{B}\otimes u_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)
.\label{sol.exe.tensor-prod-algs.a.2.u1}%
\end{align}


Next, we observe that%
\begin{align*}
&  \underbrace{\id_{A\otimes B}}_{=\operatorname*{id}%
\nolimits_{A\otimes B}\circ\id_{A\otimes B}}%
\otimes\underbrace{u_{A\otimes B}}_{=\left(  u_{A}\otimes u_{B}\right)
\circ\xi}\\
&  =\left(  \id_{A\otimes B}\circ\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \otimes\left(  \left(  u_{A}\otimes
u_{B}\right)  \circ\xi\right)  =\underbrace{\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes\left(  u_{A}\otimes u_{B}\right)  \right)
}_{=\id_{A\otimes B}\otimes u_{A}\otimes u_{B}}%
\circ\left(  \id_{A\otimes B}\otimes\xi\right)  \\
&  =\left(  \id_{A\otimes B}\otimes u_{A}\otimes
u_{B}\right)  \circ\left(  \id_{A\otimes B}\otimes
\xi\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ\underbrace{\left(
\id_{A\otimes B}\otimes u_{A\otimes B}\right)
}_{=\left(  \id_{A\otimes B}\otimes u_{A}\otimes
u_{B}\right)  \circ\left(  \id_{A\otimes B}\otimes
\xi\right)  }\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{A\otimes
B}\otimes u_{A}\otimes u_{B}\right)  }_{\substack{=\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\otimes\id_{B}\otimes
u_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,\kk}\otimes\id_{\kk}\right)
\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.u1}))}}}\circ\left(
\id_{A\otimes B}\otimes\xi\right)  \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\otimes\id_{B}\otimes
u_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,\kk}\otimes\id_{\kk}\right)
\circ\left(  \id_{A\otimes B}\otimes\xi\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A\otimes B}\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes u_{A\otimes B}\right)  }_{=\left(  m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes u_{A}%
\otimes\id_{B}\otimes u_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  }\circ
\operatorname*{kan}\nolimits_{1,A\otimes B}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes u_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  }_{=\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  }\circ\underbrace{\left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}}_{\substack{=\operatorname*{kan}%
\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.k1}))}}}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  \right)  \circ\left(  \operatorname*{kan}%
\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\id_{A\otimes B} &  =\underbrace{\operatorname*{id}%
\nolimits_{A}}_{\substack{=m_{A}\circ\left(  \id_{A}%
\otimes u_{A}\right)  \circ\operatorname*{kan}\nolimits_{1,A}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Adiag2.eq1}))}}}\otimes
\underbrace{\id_{B}}_{\substack{=m_{B}\circ\left(
\id_{B}\otimes u_{B}\right)  \circ\operatorname*{kan}%
\nolimits_{1,B}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq1}))}%
}}\\
&  =\left(  m_{A}\circ\left(  \id_{A}\otimes
u_{A}\right)  \circ\operatorname*{kan}\nolimits_{1,A}\right)  \otimes\left(
m_{B}\circ\left(  \id_{B}\otimes u_{B}\right)
\circ\operatorname*{kan}\nolimits_{1,B}\right)  \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  \right)  \circ\left(  \operatorname*{kan}%
\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}\right)  ,
\end{align*}
we obtain%
\[
m_{A\otimes B}\circ\left(  \id_{A\otimes B}\otimes
u_{A\otimes B}\right)  \circ\operatorname*{kan}\nolimits_{1,A\otimes
B}=\id_{A\otimes B}.
\]
In other words, the left rectangle of the diagram
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

A similar argument shows that the right rectangle of the diagram
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes\footnote{We leave
the details of this argument to the reader.
Let us just mention that it uses the following equality
(analogous to the equality \eqref{sol.exe.tensor-prod-algs.a.2.k1}
used in the proof of the commutativity of the left diagram):
\begin{align}
& \left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \circ\left(  \xi
\otimes\id_{A\otimes B}\right)  \circ
\operatorname*{kan}\nolimits_{2,A\otimes B}
=\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}%
\nolimits_{2,B}. \label{sol.exe.tensor-prod-algs.a.2.k2}
\end{align}
}. Thus, the whole diagram
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

We have now shown that the two diagrams
(\ref{sol.exe.tensor-prod-algs.a.2.diag1}) and
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commute. Thus, the $\kk%
$-module $A\otimes B$, equipped with the two maps $m_{A\otimes B}$ and
$u_{A\otimes B}$, is a $\kk$-algebra (since Definition
\ref{def.algebra} yields that $A\otimes B$ is a $\kk$-algebra if and
only if the two diagrams (\ref{sol.exe.tensor-prod-algs.a.2.diag1}) and
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commute). In other words, the
$\kk$-algebra $A\otimes B$ introduced in Definition
\ref{def.tensor-prod-algs} is actually well-defined. This solves Exercise
\ref{exe.tensor-prod-algs}(a) again.

(b) Exercise \ref{exe.tensor-prod-algs}(b) is the ``dual'' statement to
Exercise \ref{exe.tensor-prod-algs}(a).
We shall sketch two solutions to it: one solution by elementwise computation
(similar to the first solution to Exercise \ref{exe.tensor-prod-algs}(a), but
somewhat more complicated due to the many sums involved), and another by
formally manipulating homomorphisms. The second solution will be very brief,
because we will not elaborate on it; we will merely explain how it can be
obtained by ``reversing arrows'' from the
second solution to Exercise \ref{exe.tensor-prod-algs}(a).

\textit{First solution to Exercise \ref{exe.tensor-prod-algs}(b).} Let
$\theta$ be the canonical $\kk$-module isomorphism $\kk%
\otimes\kk\rightarrow\kk$. We have defined the comultiplication
map $\Delta_{C\otimes D}$ of the $\kk$-coalgebra $C\otimes D$ by%
\begin{equation}
\Delta_{C\otimes D}=\left(  \id_{C}\otimes
T\otimes\id_{D}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{D}\right)  , \label{sol.exe.tensor-prod-algs.b.1.DelCD=}%
\end{equation}
and the counit map $\epsilon_{C\otimes D}$ of the $\kk$-coalgebra
$C\otimes D$ by%
\[
\epsilon_{C\otimes D}=\theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  .
\]
We now must prove that the $\kk$-module $C\otimes D$, equipped with
these two maps $\Delta_{C\otimes D}$ and $\epsilon_{C\otimes D}$, is indeed a
$\kk$-coalgebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & C \otimes D \otimes C \otimes D \otimes C \otimes D \arback[dl]_{\Delta_{C\otimes D} \otimes \id_{C\otimes D}} \arback[dr]^{\id_{C\otimes D} \otimes \Delta_{C\otimes D}} & \\ C \otimes D \otimes C \otimes D \arback[dr]^{\Delta_{C \otimes D}} & & C \otimes D \otimes C \otimes D \arback[dl]_{\Delta_{C \otimes D}} \\ & C \otimes D & }
\label{sol.exe.tensor-prod-algs.b.1.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ C \otimes D \otimes \kk \arback[d]_{\id_{C\otimes D} \otimes \epsilon_{C \otimes D}} & C \otimes D \arback[l] \arback[d]_{\id_{C \otimes D}} \arback[r] & \kk \otimes C \otimes D \arback[d]^{\epsilon_{C \otimes D} \otimes \id_{C \otimes D}} \\ C \otimes D\otimes C \otimes D\arback[r]^-{\Delta_{C \otimes D}} & C \otimes D & C \otimes D \otimes C \otimes D\arback[l]_-{\Delta_{C \otimes D}} }
\label{sol.exe.tensor-prod-algs.b.1.diag2}%
\end{equation}
\footnote{where the maps $C\otimes D\otimes\kk\rightarrow C\otimes D$
and $\kk\otimes C\otimes D\rightarrow C\otimes D$ are the isomorphisms
sending each $a\otimes\lambda\in C\otimes D\otimes\kk$ with $a\in
C\otimes D$ and $\lambda\in\kk$ (resp., each $\lambda\otimes
a\in\kk\otimes C\otimes D$ with $\lambda\in\kk$ and $a\in
C\otimes D$) to $\lambda a$.} commute (since Definition \ref{def.coalgebra}
yields that $C\otimes D$ is a $\kk$-coalgebra if and only if these two
diagrams commute).

We shall only prove that the diagram (\ref{sol.exe.tensor-prod-algs.b.1.diag1}%
) commutes. The commutativity of the diagram
(\ref{sol.exe.tensor-prod-algs.b.1.diag2}) is proven similarly (but with less
work), and so is left to the reader.

So we need to prove that the diagram (\ref{sol.exe.tensor-prod-algs.b.1.diag1}%
) commutes. In other words, we need to prove that%
\[
\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \circ\Delta_{C\otimes D}=\left(  \operatorname*{id}%
\nolimits_{C\otimes D}\otimes\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes
D}.
\]


We shall use the Sweedler notation, by writing $\sum_{\left(  x\right)  }%
x_{1}\otimes x_{2}$ for $\Delta\left(  x\right)  $ whenever $x$ is an element
of a coalgebra. This is a neat opportunity to practice the use of the
Sweedler notation. But if you are uncomfortable with the Sweedler notation,
you can easily exorcise it from the following argument as follows:

\begin{itemize}
\item Whenever an element $e\in C$ is defined, fix a decomposition $\Delta
_{C}\left(  e\right)  =\sum_{a=1}^{b}r_{a}\otimes s_{a}$ of $\Delta_{C}\left(
e\right)  $ into a sum of pure tensors.

\item Whenever an element $f\in D$ is defined, fix a decomposition $\Delta
_{D}\left(  f\right)  =\sum_{a^{\prime}=1}^{b^{\prime}}r_{a^{\prime}}^{\prime
}\otimes s_{a^{\prime}}^{\prime}$ of $\Delta_{D}\left(  f\right)  $ into a sum
of pure tensors.

\item Whenever an element $c\in C$ is defined, fix a decomposition $\Delta
_{C}\left(  c\right)  =\sum_{i=1}^{n}p_{i}\otimes q_{i}$ of $\Delta_{C}\left(
c\right)  $ into a sum of pure tensors, and furthermore:

\begin{itemize}
\item For each $i\in\left\{  1,2,\ldots,n\right\}  $, fix a decomposition
$\Delta_{C}\left(  p_{i}\right)  =\sum_{j=1}^{k_{i}}p_{i,j}^{\prime}\otimes
p_{i,j}^{\prime\prime}$ of $\Delta_{C}\left(  p_{i}\right)  $ into a sum of
pure tensors.

\item For each $i\in\left\{  1,2,\ldots,n\right\}  $, fix a decomposition
$\Delta_{C}\left(  q_{i}\right)  =\sum_{h=1}^{\ell_{i}}q_{i,h}^{\prime}\otimes
q_{i,h}^{\prime\prime}$ of $\Delta_{C}\left(  q_{i}\right)  $ into a sum of
pure tensors.
\end{itemize}

\item Whenever an element $d\in D$ is defined, fix a decomposition $\Delta
_{D}\left(  d\right)  =\sum_{i^{\prime}=1}^{n^{\prime}}x_{i^{\prime}}\otimes
y_{i^{\prime}}$ of $\Delta_{D}\left(  d\right)  $ into a sum of pure tensors,
and furthermore:

\begin{itemize}
\item For each $i^{\prime}\in\left\{  1,2,\ldots,n^{\prime}\right\}  $, fix a
decomposition $\Delta_{D}\left(  x_{i^{\prime}}\right)  =\sum_{j^{\prime}%
=1}^{k_{i^{\prime}}^{\prime}}x_{i^{\prime},j^{\prime}}^{\prime}\otimes
x_{i^{\prime},j^{\prime}}^{\prime\prime}$ of $\Delta_{D}\left(  x_{i^{\prime}%
}\right)  $ into a sum of pure tensors.

\item For each $i^{\prime}\in\left\{  1,2,\ldots,n^{\prime}\right\}  $, fix a
decomposition $\Delta_{D}\left(  y_{i^{\prime}}\right)  =\sum_{h^{\prime}%
=1}^{\ell_{i^{\prime}}^{\prime}}y_{i^{\prime},h^{\prime}}^{\prime}\otimes
y_{i^{\prime},h^{\prime}}^{\prime\prime}$ of $\Delta_{D}\left(  y_{i^{\prime}%
}\right)  $ into a sum of pure tensors.
\end{itemize}
\end{itemize}

Once these decompositions are chosen, it remains to replace each appearance of
one of the symbols%
\begin{align*}
&  \sum_{\left(  e\right)  },\ \ \ e_{1},\ \ \ e_{2},\ \ \ \sum_{\left(
f\right)  },\ \ \ f_{1},\ \ \ f_{2},\\
&  \sum_{\left(  c\right)  },\ \ \ c_{1},\ \ \ c_{2},\ \ \ \sum_{\left(
c_{1}\right)  },\ \ \ \left(  c_{1}\right)  _{1},\ \ \ \left(  c_{1}\right)
_{2},\ \ \ \sum_{\left(  c_{2}\right)  },\ \ \ \left(  c_{2}\right)
_{1},\ \ \ \left(  c_{2}\right)  _{2},\\
&  \sum_{\left(  d\right)  },\ \ \ d_{1},\ \ \ d_{2},\ \ \ \sum_{\left(
d_{1}\right)  },\ \ \ \left(  d_{1}\right)  _{1},\ \ \ \left(  d_{1}\right)
_{2},\ \ \ \sum_{\left(  d_{2}\right)  },\ \ \ \left(  d_{2}\right)
_{1},\ \ \ \left(  d_{2}\right)  _{2}%
\end{align*}
by the symbol%
\begin{align*}
&  \sum_{a=1}^{b},\ \ \ r_{a},\ \ \ s_{a},\ \ \ \sum_{a^{\prime}=1}%
^{b^{\prime}},\ \ \ r_{a^{\prime}}^{\prime},\ \ \ s_{a^{\prime}}^{\prime},\\
&  \sum_{i=1}^{n},\ \ \ p_{i},\ \ \ q_{i},\ \ \ \sum_{j=1}^{k_{i}%
},\ \ \ p_{i,j}^{\prime},\ \ \ p_{i,j}^{\prime\prime},\ \ \ \sum_{h=1}%
^{\ell_{i}},\ \ \ q_{i,h}^{\prime},\ \ \ q_{i,h}^{\prime\prime},\\
&  \sum_{i^{\prime}=1}^{n^{\prime}},\ \ \ x_{i^{\prime}},\ \ \ y_{i^{\prime}%
},\ \ \ \sum_{j^{\prime}=1}^{k_{i^{\prime}}^\prime},\ \ \ x_{i^{\prime},j^{\prime}%
}^{\prime},\ \ \ x_{i^{\prime},j^{\prime}}^{\prime\prime},\ \ \ %
\sum_{h^{\prime}=1}^{\ell_{i^{\prime}}^\prime},\ \ \ %
y_{i^{\prime},h^{\prime}}^{\prime},\ \ \ %
y_{i^{\prime},h^{\prime}}^{\prime\prime},
\end{align*}
respectively\footnote{Some care must be taken here: For example, if the symbol
``$c_{1}$'' appears inside ``$\left(  c_{1}\right)  _{1}$'', then it
should not be replaced by ``$p_{i}$'', but instead the whole
``$\left(  c_{1}\right)  _{1}$'' should be
replaced by ``$p_{i,j}^{\prime}$''.}. For
example, these replacements transform the expression%
\[
\sum_{\left(  c\right)  }\sum_{\left(  c_{2}\right)  }\sum_{\left(  d\right)
}\sum_{\left(  d_{2}\right)  }c_{1}\otimes d_{1}\otimes\left(  c_{2}\right)
_{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(  c_{2}\right)  _{2}%
\otimes\left(  d_{2}\right)  _{2}%
\]
into%
\[
\sum_{i=1}^{n}\sum_{h=1}^{\ell_{i}}\sum_{i^{\prime}=1}^{n^{\prime}}%
\sum_{h^{\prime}=1}^{\ell_{i^{\prime}}^\prime}p_{i}\otimes x_{i^{\prime}}\otimes
q_{i,h}^{\prime}\otimes y_{i^{\prime},h^{\prime}}^{\prime}\otimes
q_{i,h}^{\prime\prime}\otimes y_{i^{\prime},h^{\prime}}^{\prime\prime}.
\]
Once these replacements are all done, the argument we give below becomes a
perfectly valid argument that does not use the Sweedler notation.

So let us come to the actual argument.

We first notice that%
\begin{equation}
\Delta_{C\otimes D}\left(  e\otimes f\right)  =\sum_{\left(  e\right)  }%
\sum_{\left(  f\right)  }e_{1}\otimes f_{1}\otimes e_{2}\otimes f_{2}
\label{sol.exe.tensor-prod-algs.b.1.m1}%
\end{equation}
for every $e\in C$ and $f\in D$.

[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.b.1.m1}):} Let $e\in C$ and
$f\in D$. Applying both sides of the equality
(\ref{sol.exe.tensor-prod-algs.b.1.DelCD=}) to $e\otimes f$, we obtain
\begin{align*}
\Delta_{C\otimes D}\left(  e\otimes f\right)   &  =\left(  \left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \right)  \left(  e\otimes
f\right)  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}%
_{D}\right)  \underbrace{\left(  \left(  \Delta_{C}\otimes\Delta_{D}\right)
\left(  e\otimes f\right)  \right)  }_{=\Delta_{C}\left(  e\right)
\otimes\Delta_{D}\left(  f\right)  }\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\Delta_{C}\left(  e\right)  }_{=\sum_{\left(  e\right)
}e_{1}\otimes e_{2}}\otimes\underbrace{\Delta_{D}\left(  f\right)  }%
_{=\sum_{\left(  f\right)  }f_{1}\otimes f_{2}}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\left(  \sum_{\left(  e\right)  }e_{1}\otimes
e_{2}\right)  \otimes\left(  \sum_{\left(  f\right)  }f_{1}\otimes
f_{2}\right)  }_{=\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }%
e_{1}\otimes e_{2}\otimes f_{1}\otimes f_{2}}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \sum_{\left(  e\right)  }\sum_{\left(  f\right)  }e_{1}\otimes
e_{2}\otimes f_{1}\otimes f_{2}\right) \\
&  =\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }\underbrace{\left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)  \left(
e_{1}\otimes e_{2}\otimes f_{1}\otimes f_{2}\right)  }_{=\operatorname*{id}%
\nolimits_{C}\left(  e_{1}\right)  \otimes T\left(  e_{2}\otimes f_{1}\right)
\otimes\id_{D}\left(  f_{2}\right)  }\\
&  =\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }%
\underbrace{\id_{C}\left(  e_{1}\right)  }_{=e_{1}%
}\otimes\underbrace{T\left(  e_{2}\otimes f_{1}\right)  }_{\substack{=f_{1}%
\otimes e_{2}\\\text{(by the definition of }T\text{)}}}\otimes
\underbrace{\id_{D}\left(  f_{2}\right)  }_{=f_{2}}\\
&  =\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }e_{1}\otimes
f_{1}\otimes e_{2}\otimes f_{2}.
\end{align*}
This proves (\ref{sol.exe.tensor-prod-algs.b.1.m1}).]

Furthermore, every $c\in C$ satisfies%
\begin{equation}
\sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(  c_{1}\right)
_{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}=\sum_{\left(  c\right)
}\sum_{\left(  c_{2}\right)  }c_{1}\otimes\left(  c_{2}\right)  _{1}%
\otimes\left(  c_{2}\right)  _{2}. \label{sol.exe.tensor-prod-algs.b.1.DelC}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.b.1.DelC}):} Let $c\in C$.
Recall that $C$ is a $\kk$-coalgebra. Thus, the diagram
(\ref{coassociativity-diagram}) commutes (by the definition of a $\kk%
$-coalgebra). In other words, we have $\left(  \Delta_{C}\otimes
\id_{C}\right)  \circ\Delta_{C}=\left(
\id_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}$.
Applying both sides of this equality to $c$, we obtain%
\[
\left(  \left(  \Delta_{C}\otimes\id_{C}\right)
\circ\Delta_{C}\right)  \left(  c\right)  =\left(  \left(  \operatorname*{id}%
\nolimits_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\right)  \left(
c\right)  .
\]
In light of
\begin{align*}
\left(  \left(  \Delta_{C}\otimes\id_{C}\right)
\circ\Delta_{C}\right)  \left(  c\right)   &  =\left(  \Delta_{C}%
\otimes\id_{C}\right)  \left(  \underbrace{\Delta
_{C}\left(  c\right)  }_{=\sum_{\left(  c\right)  }c_{1}\otimes c_{2}}\right)
=\left(  \Delta_{C}\otimes\id_{C}\right)  \left(
\sum_{\left(  c\right)  }c_{1}\otimes c_{2}\right) \\
&  =\sum_{\left(  c\right)  }\underbrace{\Delta_{C}\left(  c_{1}\right)
}_{=\sum_{\left(  c_{1}\right)  }\left(  c_{1}\right)  _{1}\otimes\left(
c_{1}\right)  _{2}}\otimes\underbrace{\id_{C}\left(
c_{2}\right)  }_{=c_{2}}=\sum_{\left(  c\right)  }\left(  \sum_{\left(
c_{1}\right)  }\left(  c_{1}\right)  _{1}\otimes\left(  c_{1}\right)
_{2}\right)  \otimes c_{2}\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(
c_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}%
\end{align*}
and%
\begin{align*}
\left(  \left(  \id_{C}\otimes\Delta_{C}\right)
\circ\Delta_{C}\right)  \left(  c\right)   &  =\left(  \operatorname*{id}%
\nolimits_{C}\otimes\Delta_{C}\right)  \left(  \underbrace{\Delta_{C}\left(
c\right)  }_{=\sum_{\left(  c\right)  }c_{1}\otimes c_{2}}\right)  =\left(
\id_{C}\otimes\Delta_{C}\right)  \left(  \sum_{\left(
c\right)  }c_{1}\otimes c_{2}\right) \\
&  =\sum_{\left(  c\right)  }\underbrace{\id_{C}%
\left(  c_{1}\right)  }_{=c_{1}}\otimes\underbrace{\Delta_{C}\left(
c_{2}\right)  }_{=\sum_{\left(  c_{2}\right)  }\left(  c_{2}\right)
_{1}\otimes\left(  c_{2}\right)  _{2}}=\sum_{\left(  c\right)  }c_{1}%
\otimes\left(  \sum_{\left(  c_{2}\right)  }\left(  c_{2}\right)  _{1}%
\otimes\left(  c_{2}\right)  _{2}\right) \\
&  =\sum_{\left(  c\right)  }\sum_{\left(  c_{2}\right)  }c_{1}\otimes\left(
c_{2}\right)  _{1}\otimes\left(  c_{2}\right)  _{2},
\end{align*}
this rewrites as%
\[
\sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(  c_{1}\right)
_{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}=\sum_{\left(  c\right)
}\sum_{\left(  c_{2}\right)  }c_{1}\otimes\left(  c_{2}\right)  _{1}%
\otimes\left(  c_{2}\right)  _{2}.
\]
This proves (\ref{sol.exe.tensor-prod-algs.b.1.DelC}).]

Also, every $d\in D$ satisfies%
\begin{equation}
\sum_{\left(  d\right)  }\sum_{\left(  d_{1}\right)  }\left(  d_{1}\right)
_{1}\otimes\left(  d_{1}\right)  _{2}\otimes d_{2}=\sum_{\left(  d\right)
}\sum_{\left(  d_{2}\right)  }d_{1}\otimes\left(  d_{2}\right)  _{1}%
\otimes\left(  d_{2}\right)  _{2}. \label{sol.exe.tensor-prod-algs.b.1.DelD}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.b.1.DelD}):} This proof is
analogous to the proof of (\ref{sol.exe.tensor-prod-algs.b.1.DelC}).]

Let $z\in C\otimes D$. Thus, $z$ (like any tensor in $C\otimes D$) must be a
$\kk$-linear combination of pure tensors.

We want to show the equality%
\begin{equation}
\left(  \left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  =\left(  \left(  \id_{C\otimes D}%
\otimes\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  . \label{sol.exe.tensor-prod-algs.b.1.eq-x1}%
\end{equation}
This equality is $\kk$-linear in $z$ (since all maps that appear in it
are $\kk$-linear). Hence, we can WLOG assume that $z$ is a pure tensor
(since $z$ is a $\kk$-linear combination of pure tensors). Assume this.
Thus, $z=c\otimes d$ for some $c\in C$ and $d\in D$. Consider these $c$ and
$d$.

Taking the tensor product of the equalities
(\ref{sol.exe.tensor-prod-algs.b.1.DelC}) and
(\ref{sol.exe.tensor-prod-algs.b.1.DelD}), we obtain%
\begin{align*}
&  \left(  \sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(
c_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}\right)
\otimes\left(  \sum_{\left(  d\right)  }\sum_{\left(  d_{1}\right)  }\left(
d_{1}\right)  _{1}\otimes\left(  d_{1}\right)  _{2}\otimes d_{2}\right) \\
&  =\left(  \sum_{\left(  c\right)  }\sum_{\left(  c_{2}\right)  }c_{1}%
\otimes\left(  c_{2}\right)  _{1}\otimes\left(  c_{2}\right)  _{2}\right)
\otimes\left(  \sum_{\left(  d\right)  }\sum_{\left(  d_{2}\right)  }%
d_{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(  d_{2}\right)
_{2}\right)  .
\end{align*}
In other words,%
\begin{align}
&  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}\otimes\left(  d_{1}\right)
_{1}\otimes\left(  d_{1}\right)  _{2}\otimes d_{2}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{2}\right)  }\sum_{\left(  d_{2}\right)  }c_{1}\otimes\left(  c_{2}\right)
_{1}\otimes\left(  c_{2}\right)  _{2}\otimes d_{1}\otimes\left(  d_{2}\right)
_{1}\otimes\left(  d_{2}\right)  _{2}. \label{sol.exe.tensor-prod-algs.b.1.3}%
\end{align}
Applying the $\kk$-linear map%
\begin{align*}
C\otimes C\otimes C\otimes D\otimes D\otimes D  &  \rightarrow C\otimes
D\otimes C\otimes D\otimes C\otimes D,\\
\gamma_{1}\otimes\gamma_{2}\otimes\gamma_{3}\otimes\delta_{1}\otimes\delta
_{2}\otimes\delta_{3}  &  \mapsto\gamma_{1}\otimes\delta_{1}\otimes\gamma
_{2}\otimes\delta_{2}\otimes\gamma_{3}\otimes\delta_{3}%
\end{align*}
to both sides of this equality, we obtain%
\begin{align}
&  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\otimes c_{2}\otimes d_{2}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{2}\right)  }\sum_{\left(  d_{2}\right)  }c_{1}\otimes d_{1}\otimes\left(
c_{2}\right)  _{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(
c_{2}\right)  _{2}\otimes\left(  d_{2}\right)  _{2}.
\label{sol.exe.tensor-prod-algs.b.1.4}%
\end{align}


Now,%
\begin{align}
&  \left(  \left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right) \nonumber\\
&  =\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \left(  \Delta_{C\otimes D}\left(  \underbrace{z}_{=c\otimes
d}\right)  \right) \nonumber\\
& =\left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \left(  \underbrace{\Delta_{C\otimes D}\left(
c\otimes d\right)  }_{\substack{=\sum_{\left(  c\right)  }\sum_{\left(
d\right)  }c_{1}\otimes d_{1}\otimes c_{2}\otimes d_{2}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.b.1.m1}) (applied to }e=c\text{ and
}f=d\text{))}}}\right) \nonumber\\
&  =\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \left(  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }%
c_{1}\otimes d_{1}\otimes c_{2}\otimes d_{2}\right)  \nonumber \\
&=\sum_{\left(  c\right)
}\sum_{\left(  d\right)  }\underbrace{\left(  \Delta_{C\otimes D}%
\otimes\id_{C\otimes D}\right)  \left(  c_{1}\otimes
d_{1}\otimes c_{2}\otimes d_{2}\right)  }_{=\Delta_{C\otimes D}\left(
c_{1}\otimes d_{1}\right)  \otimes\id_{C\otimes
D}\left(  c_{2}\otimes d_{2}\right)  }\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\underbrace{\Delta
_{C\otimes D}\left(  c_{1}\otimes d_{1}\right)  }_{\substack{=\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.b.1.m1}) (applied to }e=c_{1}\text{ and
}f=d_{1}\text{))}}}\otimes\underbrace{\id_{C\otimes
D}\left(  c_{2}\otimes d_{2}\right)  }_{=c_{2}\otimes d_{2}}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\left(  \sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\right)  \otimes c_{2}\otimes d_{2}%
\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\otimes c_{2}\otimes d_{2};
\label{sol.exe.tensor-prod-algs.b.1.5}%
\end{align}
an analogous computation shows that%
\begin{align}
&  \left(  \left(  \id_{C\otimes D}\otimes
\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(  z\right)
\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{2}\right)  }\sum_{\left(  d_{2}\right)  }c_{1}\otimes d_{1}\otimes\left(
c_{2}\right)  _{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(
c_{2}\right)  _{2}\otimes\left(  d_{2}\right)  _{2}.
\label{sol.exe.tensor-prod-algs.b.1.6}%
\end{align}


In light of (\ref{sol.exe.tensor-prod-algs.b.1.5}) and
(\ref{sol.exe.tensor-prod-algs.b.1.6}), the equality
(\ref{sol.exe.tensor-prod-algs.b.1.4}) rewrites as
\[
\left(  \left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  =\left(  \left(  \id_{C\otimes D}%
\otimes\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  .
\]
Thus, the equality (\ref{sol.exe.tensor-prod-algs.b.1.eq-x1}) is proven.

Now, forget that we fixed $z$. We thus have proven the equality
(\ref{sol.exe.tensor-prod-algs.b.1.eq-x1}) for every $z\in C\otimes D$. In
other words, we have $\left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}=\left(
\id_{C\otimes D}\otimes\Delta_{C\otimes D}\right)
\circ\Delta_{C\otimes D}$. In other words, the diagram
(\ref{sol.exe.tensor-prod-algs.b.1.diag1}) commutes.

It remains to prove that the diagram (\ref{sol.exe.tensor-prod-algs.b.1.diag2}%
) commutes. We leave this to the reader, as the proof is similar to (but
simpler than) the proof of the commutativity of
(\ref{sol.exe.tensor-prod-algs.b.1.diag1}) given above. Hence, both diagrams
(\ref{sol.exe.tensor-prod-algs.b.1.diag1}) and
(\ref{sol.exe.tensor-prod-algs.b.1.diag2}) commute. In other words, the
$\kk$-module $C\otimes D$, equipped with the two maps $\Delta_{C\otimes
D}$ and $\epsilon_{C\otimes D}$, is a $\kk$-coalgebra. In other words,
the $\kk$-coalgebra $C\otimes D$ introduced in Definition
\ref{def.tensor-prod-algs} is actually well-defined. This solves Exercise
\ref{exe.tensor-prod-algs}(b).

\textit{Second solution to Exercise \ref{exe.tensor-prod-algs}(b).} Let
$\theta$ be the canonical $\kk$-module isomorphism $\kk%
\otimes\kk\rightarrow\kk$.

A solution to Exercise \ref{exe.tensor-prod-algs}(b) can now be obtained in a
straightforward fashion from the Second solution to Exercise
\ref{exe.tensor-prod-algs}(a), by making the following modifications:

\begin{enumerate}
\item Replace every appearance of any of the terms%
\[
A,\ B,\ m_{A},\ m_{B},\ m_{A\otimes B},\ u_{A},\ u_{B},\ u_{A\otimes B}%
,\ \xi,\ \operatorname*{kan}\nolimits_{1,U},\ \operatorname*{kan}%
\nolimits_{2,U},\ T_{U,V}%
\]
(for any $\kk$-modules $U$ and $V$) by%
\[
C,\ D,\ \Delta_{C},\ \Delta_{D},\ \Delta_{C\otimes D},\ u_{C},\ u_{D}%
,\ u_{C\otimes D},\ \theta,\ \operatorname*{kan}\nolimits_{1,U}^{-1}%
,\ \operatorname*{kan}\nolimits_{2,U}^{-1},\ T_{V,U},
\]
respectively.

(For example, the equation (\ref{sol.exe.tensor-prod-algs.a.2.1}) becomes%
\begin{equation}
\left(  \id_{C}\otimes T_{C,D}\otimes
\id_{D}\right)  \circ\left(  \Delta_{C\otimes
D}\otimes\id_{C\otimes D}\right)  =\left(  \Delta
_{C}\otimes\id_{C}\otimes\Delta_{D}\otimes
\id_{D}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{C}\otimes Q\otimes\id_{D}\right)  .
\label{sol.exe.tensor-prod-algs.b.2.1.wrong}%
\end{equation}
This new equation makes no sense, because (for example) the maps
$\id_{C}\otimes T_{D,C}\otimes\operatorname*{id}%
\nolimits_{D}$ and $\Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}$ cannot be composed; but this is fine, since we shall
make a further modification which will turn this equation into a meaningful one.

For another example, the map $Q:B\otimes A\otimes B\otimes A\rightarrow
A\otimes A\otimes B\otimes B$ becomes a map $Q:D\otimes C\otimes D\otimes
C\rightarrow C\otimes C\otimes D\otimes D$.)

\item Reverse the direction of all arrows\footnote{This includes both the
arrows in the description of maps and the arrows in commutative diagrams.}. In
other words, any map which used to go from a set $X$ to a set $Y$ shall now go
from $Y$ to $X$.

(For example, the map $Q:D\otimes C\otimes D\otimes C\rightarrow C\otimes
C\otimes D\otimes D$ becomes a map $Q:C\otimes C\otimes D\otimes D\rightarrow
D\otimes C\otimes D\otimes C$.)

\item In any composition of $\kk$-linear maps, reverse the order of the
maps being composed. In other words, replace any composition $f_{1}\circ
f_{2}\circ\cdots\circ f_{k}$ by $f_{k}\circ f_{k-1}\circ\cdots\circ f_{1}$.

(For example, the meaningless equality
(\ref{sol.exe.tensor-prod-algs.b.2.1.wrong}) thus becomes%
\[
\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \circ\left(  \id_{C}\otimes T_{C,D}%
\otimes\id_{D}\right)  =\left(  \operatorname*{id}%
\nolimits_{C}\otimes Q\otimes\id_{D}\right)
\circ\left(  \Delta_{C}\otimes\id_{C}\otimes\Delta
_{D}\otimes\id_{D}\right)  .
\]
This equality is meaningful and correct.)

\item Any part of our solution that involved elements of $A$ and $B$ (as
opposed to mere computations with maps) must be redone from scratch. For
example, the proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q1}) involved
elements of $A$ and $B$ (because we picked $b\in B$, $a\in A$, $b^{\prime}\in
B$ and $a^{\prime}\in A$ in that proof), and thus must be redone, whereas the
proof of (\ref{sol.exe.tensor-prod-algs.a.2.2}) did not involve elements of
$A$ and $B$ and therefore needs not be modified any further.

Fortunately, very few parts of our solution involved elements of $A$ and $B$.
To wit, these parts are the proofs of the equalities
(\ref{sol.exe.tensor-prod-algs.a.2.Q1}),
(\ref{sol.exe.tensor-prod-algs.a.2.Q2}),
(\ref{sol.exe.tensor-prod-algs.a.2.k1}) and
(\ref{sol.exe.tensor-prod-algs.a.2.k2}).
After the above modifications, these
equalities have become%
\begin{align*}
&  Q=\left(  T_{C,D}\otimes\id_{D}\otimes
\id_{C}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{C}\otimes T_{C,D\otimes D}\right)  ,\\
&  Q=\left(  \id_{D}\otimes\operatorname*{id}%
\nolimits_{C}\otimes T_{C,D}\right)  \circ\left(  T_{C\otimes C,D}%
\otimes\id_{D}\right)  ,\\
&  \operatorname*{kan}\nolimits_{1,C\otimes D}^{-1}\circ\left(
\id_{C\otimes D}\otimes\theta\right)  \circ\left(
\id_{C}\otimes T_{\kk,D}\otimes
\id_{\kk}\right)  =\operatorname*{kan}%
\nolimits_{1,C}^{-1}\otimes\operatorname*{kan}\nolimits_{1,D}^{-1}%
,\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \operatorname*{kan}\nolimits_{2,C\otimes D}^{-1}\circ\left(  \theta
\otimes\id_{C\otimes D}\right)  \circ\left(
\id_{\kk}\otimes T_{C,\kk}%
\otimes\id_{D}\right)  =\operatorname*{kan}%
\nolimits_{2,C}^{-1}\otimes\operatorname*{kan}\nolimits_{2,D}^{-1},
\end{align*}
respectively. So these four equalities must be proven. Fortunately, these
proofs are completely straightforward\footnote{These equalities are properties
of tensor products of $\kk$-modules; they make no use of the coalgebra
structures on $C$ and $D$.}; the reader can easily come up with them.
\end{enumerate}

These four modifications are sufficient to transform our Second solution to
Exercise \ref{exe.tensor-prod-algs}(a) into a solution to Exercise
\ref{exe.tensor-prod-algs}(b). Thus, Exercise \ref{exe.tensor-prod-algs}(b) is
solved again.

[\textit{Remark:} The second and the third modifications made above are
usually subsumed under the concept of ``reversing all
arrows''.]
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.tensor-prod-algs}}

\textit{Solution to Exercise \ref{exe.tensor-prod-algs}.} (a) Exercise
\ref{exe.tensor-prod-algs}(a) is a classical result about algebras, and proven
in various textbooks. Nevertheless, we shall give two solutions to it: one
solution by elementwise computation, and another by formally manipulating
homomorphisms (this is essentially what is called ``diagram
chasing'', except that we are not going to draw any diagrams).
Both solutions have their advantages, and it is useful to see them both.

\textit{First solution to Exercise \ref{exe.tensor-prod-algs}(a).} Let $\xi$
be the canonical $\kk$-module isomorphism $\kk\rightarrow
\kk\otimes\kk$. We have defined the multiplication map
$m_{A\otimes B}$ of the $\kk$-algebra $A\otimes B$ by%
\[
m_{A\otimes B}=\left(  m_{A}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T\otimes\operatorname*{id}%
\nolimits_{B}\right)  ,
\]
and the unit map $u_{A\otimes B}$ of the $\kk$-algebra $A\otimes B$ by%
\[
u_{A\otimes B}=\left(  u_{A}\otimes u_{B}\right)  \circ\xi.
\]
We now must prove that the $\kk$-module $A\otimes B$, equipped with
these two maps $m_{A\otimes B}$ and $u_{A\otimes B}$, is indeed a $\kk%
$-algebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes B \otimes A \otimes B \otimes A \otimes B \ar[dl]_{m_{A\otimes B} \otimes \id_{A\otimes B}} \ar[dr]^{\id_{A\otimes B} \otimes m_{A\otimes B}} & \\ A \otimes B \otimes A \otimes B \ar[dr]^{m_{A \otimes B}} & & A \otimes B \otimes A \otimes B \ar[dl]_{m_{A \otimes B}} \\ & A \otimes B & }
\label{sol.exe.tensor-prod-algs.a.1.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes B \otimes \kk \ar[d]_{\id_{A\otimes B} \otimes u_{A \otimes B}} & A \otimes B \ar[l] \ar[d]_{\id_{A \otimes B}} \ar[r] & \kk \otimes A \otimes B \ar[d]^{u_{A \otimes B} \otimes \id_{A \otimes B}} \\ A \otimes B\otimes A \otimes B\ar[r]^-{m_{A \otimes B}} & A \otimes B & A \otimes B \otimes A \otimes B\ar[l]_-{m_{A \otimes B}} }
\label{sol.exe.tensor-prod-algs.a.1.diag2}%
\end{equation}
\footnote{where the maps $A\otimes B\rightarrow A\otimes B\otimes\kk$
and $A\otimes B\rightarrow\kk\otimes A\otimes B$ are the isomorphisms
sending each $a\in A\otimes B$ to $a\otimes1$ and to $1\otimes a$,
respectively} commute (since Definition \ref{def.algebra} yields that
$A\otimes B$ is a $\kk$-algebra if and only if these two diagrams commute).

We shall only prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.1.diag1}%
) commutes. The commutativity of the diagram
(\ref{sol.exe.tensor-prod-algs.a.1.diag2}) is proven similarly (but with less
work), and so is left to the reader.

So we need to prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.1.diag1}%
) commutes. In other words, we need to prove that%
\[
m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  =m_{A\otimes B}\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\circ m_{A\otimes B}\right)  .
\]


We first notice that%
\begin{equation}
m_{A\otimes B}\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime}\right)
=pp^{\prime}\otimes qq^{\prime} \label{sol.exe.tensor-prod-algs.a.1.m1}%
\end{equation}
for any $p\in A$, $q\in B$, $p^{\prime}\in A$ and $q^{\prime}\in
B$\ \ \ \ \footnote{\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.1.m1}):}
Let $p\in A$, $q\in B$, $p^{\prime}\in A$ and $q^{\prime}\in B$. Then,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T\otimes\operatorname*{id}%
\nolimits_{B}\right)  }\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime
}\right) \\
&  =\left(  \left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes T\otimes\id_{B}\right)  \right)
\left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(  \underbrace{\left(
\id_{A}\otimes T\otimes\operatorname*{id}%
\nolimits_{B}\right)  \left(  p\otimes q\otimes p^{\prime}\otimes q^{\prime
}\right)  }_{=\id_{A}\left(  p\right)  \otimes
T\left(  q\otimes p^{\prime}\right)  \otimes\id_{B}%
\left(  q^{\prime}\right)  }\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(  \underbrace{\operatorname*{id}%
\nolimits_{A}\left(  p\right)  }_{=p}\otimes\underbrace{T\left(  q\otimes
p^{\prime}\right)  }_{\substack{=p^{\prime}\otimes q\\\text{(by the definition
of }T\text{)}}}\otimes\underbrace{\id_{B}\left(
q^{\prime}\right)  }_{=q^{\prime}}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(  p\otimes p^{\prime}\otimes
q\otimes q^{\prime}\right)  =\underbrace{m_{A}\left(  p\otimes p^{\prime
}\right)  }_{\substack{=pp^{\prime}\\\text{(since }m_{A}\text{ is
the}\\\text{multiplication map of }A\text{)}}}\otimes\underbrace{m_{B}\left(
q\otimes q^{\prime}\right)  }_{\substack{=qq^{\prime}\\\text{(since }%
m_{B}\text{ is the}\\\text{multiplication map of }B\text{)}}}\\
&  =pp^{\prime}\otimes qq^{\prime}.
\end{align*}
Qed.}.

Let $x\in A\otimes B\otimes A\otimes B\otimes A\otimes B$. Thus, $x$ (like any
tensor in $A\otimes B\otimes A\otimes B\otimes A\otimes B$) must be a
$\kk$-linear combination of pure tensors.

We want to show the equality%
\begin{equation}
\left(  m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \right)  \left(  x\right)  =\left(  m_{A\otimes
B}\circ\left(  \id_{A\otimes B}\circ m_{A\otimes
B}\right)  \right)  \left(  x\right)  .
\label{sol.exe.tensor-prod-algs.a.1.eq-x1}%
\end{equation}
This equality is $\kk$-linear in $x$ (since all maps that appear in it
are $\kk$-linear). Hence, we can WLOG assume that $x$ is a pure tensor
(since $x$ is a $\kk$-linear combination of pure tensors). Assume this.
Thus, $x=a\otimes b\otimes a^{\prime}\otimes b^{\prime}\otimes a^{\prime
\prime}\otimes b^{\prime\prime}$ for some $a\in A$, $b\in B$, $a^{\prime}\in
A$, $b^{\prime}\in B$, $a^{\prime\prime}\in A$ and $b^{\prime\prime}\in B$.
Consider these $a$, $b$, $a^{\prime}$, $b^{\prime}$, $a^{\prime\prime}$ and
$b^{\prime\prime}$.

Now,%
\begin{align*}
\left(  m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \right)  \left(  x\right)   &  =m_{A\otimes
B}\left(  \left(  m_{A\otimes B}\otimes\id_{A\otimes
B}\right)  \left(  \underbrace{x}_{=a\otimes b\otimes a^{\prime}\otimes
b^{\prime}\otimes a^{\prime\prime}\otimes b^{\prime\prime}}\right)  \right) \\
&  =m_{A\otimes B}\left(  \underbrace{\left(  m_{A\otimes B}\otimes
\id_{A\otimes B}\right)  \left(  a\otimes b\otimes
a^{\prime}\otimes b^{\prime}\otimes a^{\prime\prime}\otimes b^{\prime\prime
}\right)  }_{=m_{A\otimes B}\left(  a\otimes b\otimes a^{\prime}\otimes
b^{\prime}\right)  \otimes\id_{A\otimes B}\left(
a^{\prime\prime}\otimes b^{\prime\prime}\right)  }\right) \\
&  =m_{A\otimes B}\left(  \underbrace{m_{A\otimes B}\left(  a\otimes b\otimes
a^{\prime}\otimes b^{\prime}\right)  }_{\substack{=aa^{\prime}\otimes
bb^{\prime}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.1.m1}))}}%
}\otimes\underbrace{\id_{A\otimes B}\left(
a^{\prime\prime}\otimes b^{\prime\prime}\right)  }_{=a^{\prime\prime}\otimes
b^{\prime\prime}}\right) \\
&  =m_{A\otimes B}\left(  aa^{\prime}\otimes bb^{\prime}\otimes a^{\prime
\prime}\otimes b^{\prime\prime}\right)  =\underbrace{\left(  aa^{\prime
}\right)  a^{\prime\prime}}_{=aa^{\prime}a^{\prime\prime}}\otimes
\underbrace{\left(  bb^{\prime}\right)  b^{\prime\prime}}_{=bb^{\prime
}b^{\prime\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.tensor-prod-algs.a.1.m1})}\right) \\
&  =aa^{\prime}a^{\prime\prime}\otimes bb^{\prime}b^{\prime\prime}.
\end{align*}
Comparing this with%
\begin{align*}
\left(  m_{A\otimes B}\circ\left(  \id_{A\otimes
B}\otimes m_{A\otimes B}\right)  \right)  \left(  x\right)   &  =m_{A\otimes
B}\left(  \left(  \id_{A\otimes B}\otimes m_{A\otimes
B}\right)  \left(  \underbrace{x}_{=a\otimes b\otimes a^{\prime}\otimes
b^{\prime}\otimes a^{\prime\prime}\otimes b^{\prime\prime}}\right)  \right) \\
&  =m_{A\otimes B}\left(  \underbrace{\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A\otimes B}\right)  \left(  a\otimes b\otimes
a^{\prime}\otimes b^{\prime}\otimes a^{\prime\prime}\otimes b^{\prime\prime
}\right)  }_{=\id_{A\otimes B}\left(  a\otimes
b\right)  \otimes m_{A\otimes B}\left(  a^{\prime}\otimes b^{\prime}\otimes
a^{\prime\prime}\otimes b^{\prime\prime}\right)  }\right) \\
&  =m_{A\otimes B}\left(  \underbrace{\id_{A\otimes
B}\left(  a\otimes b\right)  }_{=a\otimes b}\otimes\underbrace{m_{A\otimes
B}\left(  a^{\prime}\otimes b^{\prime}\otimes a^{\prime\prime}\otimes
b^{\prime\prime}\right)  }_{\substack{=a^{\prime}a^{\prime\prime}\otimes
b^{\prime}b^{\prime\prime}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.1.m1}%
))}}}\right) \\
&  =m_{A\otimes B}\left(  a\otimes b\otimes a^{\prime}a^{\prime\prime}\otimes
b^{\prime}b^{\prime\prime}\right)  =\underbrace{a\left(  a^{\prime}%
a^{\prime\prime}\right)  }_{=aa^{\prime}a^{\prime\prime}}\otimes
\underbrace{b\left(  b^{\prime}b^{\prime\prime}\right)  }_{=bb^{\prime
}b^{\prime\prime}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.tensor-prod-algs.a.1.m1})}\right) \\
&  =aa^{\prime}a^{\prime\prime}\otimes bb^{\prime}b^{\prime\prime},
\end{align*}
we obtain $\left(  m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes
\id_{A\otimes B}\right)  \right)  \left(  x\right)
=\left(  m_{A\otimes B}\circ\left(  \id_{A\otimes
B}\otimes m_{A\otimes B}\right)  \right)  \left(  x\right)  $. Thus, the
equality (\ref{sol.exe.tensor-prod-algs.a.1.eq-x1}) is proven.

Now, forget that we fixed $x$. We thus have proven the equality
(\ref{sol.exe.tensor-prod-algs.a.1.eq-x1}) for every $x\in A\otimes B\otimes
A\otimes B\otimes A\otimes B$. In other words, we have $m_{A\otimes B}%
\circ\left(  m_{A\otimes B}\otimes\id_{A\otimes
B}\right)  =m_{A\otimes B}\circ\left(  \id_{A\otimes
B}\circ m_{A\otimes B}\right)  $. In other words, the diagram
(\ref{sol.exe.tensor-prod-algs.a.1.diag1}) commutes.

It remains to prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.1.diag2}%
) commutes. We leave this to the reader, as the proof is similar to (but
simpler than) the proof of the commutativity of
(\ref{sol.exe.tensor-prod-algs.a.1.diag1}) given above. Hence, both diagrams
(\ref{sol.exe.tensor-prod-algs.a.1.diag1}) and
(\ref{sol.exe.tensor-prod-algs.a.1.diag2}) commute. In other words, the
$\kk$-module $A\otimes B$, equipped with the two maps $m_{A\otimes B}$
and $u_{A\otimes B}$, is a $\kk$-algebra. In other words, the
$\kk$-algebra $A\otimes B$ introduced in Definition
\ref{def.tensor-prod-algs} is actually well-defined. This solves Exercise
\ref{exe.tensor-prod-algs}(a).

\textit{Second solution to Exercise \ref{exe.tensor-prod-algs}(a).} In this
solution, we will often use the following classical formula from linear algebra:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$ and $W^{\prime}$ are six
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$ and $\beta^{\prime
}:V^{\prime}\rightarrow W^{\prime}$ are four $\kk$-linear maps, then
$\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

\noindent
We will refer to this formula as the \dfn{composition-tensor relation}.
Using two applications of this formula, we can obtain the following formula,
which we will call the \dfn{composition-triple-tensor relation}:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$U^{\prime\prime}$, $V^{\prime\prime}$, $W^{\prime\prime}$ are nine
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$, $\beta^{\prime}:V^{\prime
}\rightarrow W^{\prime}$, $\alpha^{\prime\prime}:U^{\prime\prime}\rightarrow
V^{\prime\prime}$, $\beta^{\prime\prime}:V^{\prime\prime}\rightarrow
W^{\prime\prime}$ are six $\kk$-linear maps, then $\left(  \beta
\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ\alpha^{\prime}\right)
\otimes\left(  \beta^{\prime\prime}\circ\alpha^{\prime\prime}\right)  =\left(
\beta\otimes\beta^{\prime}\otimes\beta^{\prime\prime}\right)  \circ\left(
\alpha\otimes\alpha^{\prime}\otimes\alpha^{\prime\prime}\right)  $.
\end{itemize}

For any two $\kk$-modules $U$ and $V$, let $T_{U,V}:U\otimes
V\rightarrow V\otimes U$ be the twist map (i.e., the $\kk$-linear map
$U\otimes V\rightarrow V\otimes U$ sending every $u\otimes v$ to $v\otimes
u$). A simple linear-algebraic fact says that if $U$, $V$, $U^{\prime}$ and
$V^{\prime}$ are four $\kk$-modules and $x:U\rightarrow U^{\prime}$ and
$y:V\rightarrow V^{\prime}$ are two $\kk$-linear maps, then
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}=T_{U^{\prime},V^{\prime}}\circ\left(
x\otimes y\right)  . \label{sol.exe.tensor-prod-algs.a.2.twistcomm}%
\end{equation}


For every $\kk$-module $U$, let $\operatorname{kan}_{1,U}:U\rightarrow
U\otimes\kk$ and $\operatorname{kan}_{2,U}:U\rightarrow\kk\otimes U$
be the canonical $\kk$-module isomorphisms. Every
$\kk$-modules $U$ and $V$ satisfy the identities%
\begin{align}
\operatorname{id}_{U}\otimes\operatorname{kan}_{1,V}  &  =\operatorname{kan}%
_{1,U\otimes V},\label{sol.exe.tensor-prod-algs.a.2.kans.1}\\
\operatorname{kan}_{2,V}\otimes\operatorname{id}_{U}  &  =\operatorname{kan}%
_{2,V\otimes U},\label{sol.exe.tensor-prod-algs.a.2.kans.2}\\
\operatorname{id}_{U}\otimes\operatorname{kan}_{1,V}^{-1}  &
=\operatorname{kan}_{1,U\otimes V}^{-1}%
,\label{sol.exe.tensor-prod-algs.a.2.kans.3}\\
\operatorname{kan}_{2,V}^{-1}\otimes\operatorname{id}_{U}  &
=\operatorname{kan}_{2,V\otimes U}^{-1}.
\label{sol.exe.tensor-prod-algs.a.2.kans.4}%
\end{align}
(These identities are well-known and straightforward to check.)

Recall that the $\kk$-module $A$, equipped with the maps $m_{A}$ and
$u_{A}$, is a $\kk$-algebra. In other words, the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes A \otimes A \ar[dl]_{m_A \otimes \id_A} \ar[dr]^{\id_{A} \otimes m_A} & \\ A \otimes A \ar[dr]^{m_{A}} & & A \otimes A \ar[dl]_{m_{A}} \\ & A & }
\label{sol.exe.tensor-prod-algs.a.2.Adiag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes \kk \ar[d]_{\id_{A} \otimes u_{A}} & A \ar[l]_{\operatorname{kan}_{1,A}} \ar[d]_{\id_{A}} \ar[r]^{\operatorname{kan}_{2,A}} & \kk \otimes A \ar[d]^{u_{A} \otimes \id_{A}} \\ A \otimes A \ar[r]^-{m_{A}} & A & A \otimes A \ar[l]_-{m_{A }} }
\label{sol.exe.tensor-prod-algs.a.2.Adiag2}%
\end{equation}
commute (since Definition \ref{def.algebra} yields that $A$ is a $\kk%
$-algebra if and only if these two diagrams commute).

The diagram (\ref{sol.exe.tensor-prod-algs.a.2.Adiag1}) commutes. In other
words, we have%
\begin{equation}
m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Adiag1.eq}%
\end{equation}
The same argument (applied to $B$ instead of $A$) shows that%
\begin{equation}
m_{B}\circ\left(  m_{B}\otimes\id_{B}\right)
=m_{B}\circ\left(  \id_{B}\otimes m_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Bdiag1.eq}%
\end{equation}


The diagram (\ref{sol.exe.tensor-prod-algs.a.2.Adiag2}) commutes. In other
words, we have%
\begin{align}
\id_{A}  &  =m_{A}\circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \circ\operatorname*{kan}\nolimits_{1,A}%
\ \ \ \ \ \ \ \ \ \ \text{and} \label{sol.exe.tensor-prod-algs.a.2.Adiag2.eq1}%
\\
\id_{A}  &  =m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  \circ\operatorname*{kan}%
\nolimits_{2,A}. \label{sol.exe.tensor-prod-algs.a.2.Adiag2.eq2}%
\end{align}
The same argument (applied to $B$ instead of $A$) shows that%
\begin{align}
\id_{B}  &  =m_{B}\circ\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  \circ\operatorname*{kan}\nolimits_{1,B}%
\ \ \ \ \ \ \ \ \ \ \text{and} \label{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq1}%
\\
\id_{B}  &  =m_{B}\circ\left(  u_{B}\otimes
\id_{B}\right)  \circ\operatorname*{kan}%
\nolimits_{2,B}. \label{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq2}%
\end{align}


Let $\xi$ be the canonical $\kk$-module isomorphism $\kk%
\rightarrow\kk\otimes\kk$. According to Definition
\ref{def.tensor-prod-algs}, we define the map $m_{A\otimes B}:A\otimes
B\otimes A\otimes B\rightarrow A\otimes B$ by%
\[
m_{A\otimes B}=\left(  m_{A}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  ,
\]
and we define the map $u_{A\otimes B}:\kk\rightarrow A\otimes B$ by%
\[
u_{A\otimes B}=\left(  u_{A}\otimes u_{B}\right)  \circ\xi.
\]
We now must prove that the $\kk$-module $A\otimes B$, equipped with
these two maps $m_{A\otimes B}$ and $u_{A\otimes B}$, is indeed a $\kk%
$-algebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes B \otimes A \otimes B \otimes A \otimes B \ar[dl]_{m_{A\otimes B} \otimes \id_{A\otimes B}} \ar[dr]^{\id_{A\otimes B} \otimes m_{A\otimes B}} & \\ A \otimes B \otimes A \otimes B \ar[dr]^{m_{A \otimes B}} & & A \otimes B \otimes A \otimes B \ar[dl]_{m_{A \otimes B}} \\ & A \otimes B & }
\label{sol.exe.tensor-prod-algs.a.2.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes B \otimes \kk \ar[d]_{\id_{A\otimes B} \otimes u_{A \otimes B}} & A \otimes B \ar[l]_{\operatorname{kan}_{1,A\otimes B}} \ar[d]_{\id_{A \otimes B}} \ar[r]^{\operatorname{kan}_{2,A,\otimes B}} & \kk \otimes A \otimes B \ar[d]^{u_{A \otimes B} \otimes \id_{A \otimes B}} \\ A \otimes B\otimes A \otimes B\ar[r]^-{m_{A \otimes B}} & A \otimes B & A \otimes B \otimes A \otimes B\ar[l]_-{m_{A \otimes B}} }
\label{sol.exe.tensor-prod-algs.a.2.diag2}%
\end{equation}
commute (since Definition \ref{def.algebra} yields that $A\otimes B$ is a
$\kk$-algebra if and only if these two diagrams commute).

Let us first prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag1})
commutes. In other words, let us prove that%
\[
m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  =m_{A\otimes B}\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A\otimes B}\right)  .
\]


Define a $\kk$-linear map $Q:B\otimes A\otimes B\otimes A\rightarrow
A\otimes A\otimes B\otimes B$ by%
\begin{equation}
Q=\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  T_{B,A}\otimes
T_{B,A}\right)  . \label{sol.exe.tensor-prod-algs.a.2.Q=}%
\end{equation}


It is easy to see that%
\begin{equation}
Q=\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Q1}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q1}):} Let $b\in B$,
$a\in A$, $b^{\prime}\in B$ and $a^{\prime}\in A$ be arbitrary. Applying both
sides of the equality (\ref{sol.exe.tensor-prod-algs.a.2.Q=}) to $b\otimes
a\otimes b^{\prime}\otimes a^{\prime}$, we obtain%
\begin{align}
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)   &  =\left(
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  T_{B,A}\otimes
T_{B,A}\right)  \right)  \left(  b\otimes a\otimes b^{\prime}\otimes
a^{\prime}\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \left(  \underbrace{\left(
T_{B,A}\otimes T_{B,A}\right)  \left(  b\otimes a\otimes b^{\prime}\otimes
a^{\prime}\right)  }_{=T_{B,A}\left(  b\otimes a\right)  \otimes T_{B,A}
\left(  b^{\prime}\otimes a^{\prime}\right)  }\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \left(  \underbrace{T_{B,A}\left(
b\otimes a\right)  }_{\substack{=a\otimes b\\\text{(by the definition of
}T_{B,A}\text{)}}}\otimes\underbrace{T_{B,A}\left(
b^{\prime}\otimes a^{\prime}\right)  }_{\substack{=a^{\prime}\otimes
b^{\prime}\\\text{(by the definition of }T_{B,A}\text{)}%
}}\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \left(  a\otimes b\otimes a^{\prime
}\otimes b^{\prime}\right)  =\underbrace{\id_{A}%
\left(  a\right)  }_{=a}\otimes\underbrace{T_{B,A}\left(  b\otimes a^{\prime
}\right)  }_{\substack{=a^{\prime}\otimes b\\\text{(by the definition of
}T_{B,A}\text{)}}}\otimes\underbrace{\id_{B}\left(
b^{\prime}\right)  }_{=b^{\prime}}\nonumber\\
&  =a\otimes a^{\prime}\otimes b\otimes b^{\prime}.
\label{sol.exe.tensor-prod-algs.a.2.Q1.pf.1}%
\end{align}


Comparing this with%
\begin{align*}
&  \left(  \left(  \id_{A}\otimes T_{B\otimes
B,A}\right)  \circ\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \right)  \left(  b\otimes
a\otimes b^{\prime}\otimes a^{\prime}\right) \\
&  =\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\left(  \underbrace{\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)  }_{=T_{B,A}\left(  b\otimes a\right)
\otimes\id_{B}\left(  b^{\prime}\right)
\otimes\id_{A}\left(  a^{\prime}\right)  }\right) \\
&  =\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\left(  \underbrace{T_{B,A}\left(  b\otimes a\right)  }_{\substack{=a\otimes
b\\\text{(by the definition of }T_{B,A}\text{)}}}\otimes
\underbrace{\id_{B}\left(  b^{\prime}\right)
}_{=b^{\prime}}\otimes\underbrace{\id_{A}\left(
a^{\prime}\right)  }_{=a^{\prime}}\right) \\
&  =\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\left(  a\otimes b\otimes b^{\prime}\otimes a^{\prime}\right)
=\underbrace{\id_{A}\left(  a\right)  }_{=a}%
\otimes\underbrace{T_{B\otimes B,A}\left(  b\otimes b^{\prime}\otimes
a^{\prime}\right)  }_{\substack{=a^{\prime}\otimes b\otimes b^{\prime
}\\\text{(by the definition of }T_{B\otimes B,A}\text{)}}}\\
&  =a\otimes a^{\prime}\otimes b\otimes b^{\prime},
\end{align*}
we obtain%
\[
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)  =\left(
\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  \right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)  .
\]


Now, forget that we fixed $b,a,b^{\prime},a^{\prime}$. We thus have shown that
every $b\in B$, $a\in A$, $b^{\prime}\in B$ and $a^{\prime}\in A$ satisfy%
\[
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)  =\left(
\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  \right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)  .
\]
In other words, the two maps $Q$ and $\left(  \operatorname*{id}%
\nolimits_{A}\otimes T_{B\otimes B,A}\right)  \circ\left(  T_{B,A}%
\otimes\id_{B}\otimes\id_{A}%
\right)  $ are equal to each other on each pure tensor. Since these two maps
are $\kk$-linear, we thus conclude that these two maps are identical
(because if two $\kk$-linear maps from a tensor product are equal to
each other on each pure tensor, then these two maps are identical). In other
words,
\[
Q=\left(  \id_{A}\otimes T_{B\otimes B,A}\right)
\circ\left(  T_{B,A}\otimes\id_{B}\otimes
\id_{A}\right)  .
\]
This proves (\ref{sol.exe.tensor-prod-algs.a.2.Q1}).]

Now,%
\begin{align}
&  \underbrace{\id_{A}}_{=\operatorname*{id}%
\nolimits_{A}\circ\id_{A}}\otimes\underbrace{Q}%
_{\substack{=\left(  \id_{A}\otimes T_{B\otimes
B,A}\right)  \circ\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Q1}))}}}\otimes
\underbrace{\id_{B}}_{=\operatorname*{id}%
\nolimits_{B}\circ\id_{B}}\nonumber\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  \left(  \id_{A}%
\otimes T_{B\otimes B,A}\right)  \circ\left(  T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes\id_{A}\right)  \right)
\otimes\left(  \id_{B}\circ\operatorname*{id}%
\nolimits_{B}\right) \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
\id_{A}\otimes T_{B\otimes B,A}\right)  \otimes
\id_{B}\right)  }_{=\id_{A}%
\otimes\id_{A}\otimes T_{B\otimes B,A}\otimes
\id_{B}}\circ\underbrace{\left(  \operatorname*{id}%
\nolimits_{A}\otimes\left(  T_{B,A}\otimes\id_{B}%
\otimes\id_{A}\right)  \otimes\operatorname*{id}%
\nolimits_{B}\right)  }_{=\id_{A}\otimes
T_{B,A}\otimes\id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes\id_{B}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \nonumber\\
&  =\left(  \id_{A}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B\otimes B,A}\otimes\id_{B}%
\right)  \circ\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\otimes\id_{A}%
\otimes\id_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Q1'}%
\end{align}


Now, we shall show that%
\begin{equation}
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A\otimes B}%
\otimes\id_{A\otimes B}\right)  =\left(  m_{A}%
\otimes\id_{A}\otimes m_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)
\label{sol.exe.tensor-prod-algs.a.2.1}%
\end{equation}
(as maps from $A\otimes B\otimes A\otimes B\otimes A\otimes B$ to $A\otimes
A\otimes B\otimes B$).

[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.1}):} We have%
\begin{align}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\otimes\underbrace{\operatorname*{id}%
\nolimits_{A\otimes B}}_{=\id_{A\otimes B}%
\circ\id_{A\otimes B}}\nonumber\\
&  =\left(  \left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\otimes\id_{B}\right)
\right)  \otimes\left(  \id_{A\otimes B}%
\circ\id_{A\otimes B}\right) \nonumber\\
&  =\underbrace{\left(  \left(  m_{A}\otimes m_{B}\right)  \otimes
\id_{A\otimes B}\right)  }_{=m_{A}\otimes m_{B}%
\otimes\id_{A\otimes B}}\circ\underbrace{\left(
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  }_{=\id_{A}\otimes
T_{B,A}\otimes\id_{B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \nonumber\\
&  =\left(  m_{A}\otimes m_{B}\otimes\id_{A\otimes
B}\right)  \circ\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  . \label{sol.exe.tensor-prod-algs.a.2.1.pf.1}%
\end{align}


The equality (\ref{sol.exe.tensor-prod-algs.a.2.twistcomm}) (applied to
$B\otimes B$, $A$, $B$, $A$, $m_{B}$ and $\id_{A}$
instead of $U$, $V$, $U^{\prime}$, $V^{\prime}$, $x$ and $y$) yields
\[
\left(  \id_{A}\otimes m_{B}\right)  \circ T_{B\otimes
B,A}=T_{B,A}\circ\left(  m_{B}\otimes\id_{A}\right)
.
\]
Hence,%
\begin{align*}
&  \left(  \id_{A}\circ m_{A}\right)  \otimes
\underbrace{\left(  \left(  \id_{A}\otimes
m_{B}\right)  \circ T_{B\otimes B,A}\right)  }_{=T_{B,A}\circ\left(
m_{B}\otimes\id_{A}\right)  }\otimes\left(
\id_{B}\circ\id_{B}\right) \\
&  =\left(  \id_{A}\circ m_{A}\right)  \otimes\left(
T_{B,A}\circ\left(  m_{B}\otimes\id_{A}\right)
\right)  \otimes\left(  \id_{B}\circ\operatorname*{id}%
\nolimits_{B}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(  m_{A}%
\otimes\left(  m_{B}\otimes\id_{A}\right)
\otimes\id_{B}\right)  }_{=m_{A}\otimes m_{B}%
\otimes\id_{A}\otimes\id_{B}%
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A}\otimes m_{B}%
\otimes\underbrace{\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}}_{=\id_{A\otimes B}}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A}\otimes m_{B}%
\otimes\id_{A\otimes B}\right)  .
\end{align*}
Thus,%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  m_{A}\otimes m_{B}%
\otimes\id_{A\otimes B}\right) \nonumber\\
&  =\underbrace{\left(  \id_{A}\circ m_{A}\right)
}_{=m_{A}=m_{A}\circ\id_{A\otimes A}}\otimes\left(
\left(  \id_{A}\otimes m_{B}\right)  \circ T_{B\otimes
B,A}\right)  \otimes\left(  \id_{B}\circ
\id_{B}\right) \nonumber\\
&  =\left(  m_{A}\circ\id_{A\otimes A}\right)
\otimes\left(  \left(  \id_{A}\otimes m_{B}\right)
\circ T_{B\otimes B,A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\circ\id_{B}\right) \nonumber\\
&  =\underbrace{\left(  m_{A}\otimes\left(  \id_{A}%
\otimes m_{B}\right)  \otimes\id_{B}\right)  }%
_{=m_{A}\otimes\id_{A}\otimes m_{B}\otimes
\id_{B}}\circ\left(  \underbrace{\operatorname*{id}%
\nolimits_{A\otimes A}}_{=\id_{A}\otimes
\id_{A}}\otimes T_{B\otimes B,A}\otimes
\id_{B}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \nonumber\\
&  =\left(  m_{A}\otimes\id_{A}\otimes m_{B}%
\otimes\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes\id_{A}\otimes T_{B\otimes
B,A}\otimes\id_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.1.pf.2}%
\end{align}


Now,%
\begin{align*}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(  m_{A\otimes
B}\otimes\id_{A\otimes B}\right)  }%
_{\substack{=\left(  m_{A}\otimes m_{B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \circ\left(  \id_{A}%
\otimes T_{B,A}\otimes\id_{B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.1.pf.1}))}}}\\
&  =\underbrace{\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  \circ\left(  m_{A}\otimes
m_{B}\otimes\id_{A\otimes B}\right)  }%
_{\substack{=\left(  m_{A}\otimes\id_{A}\otimes
m_{B}\otimes\id_{B}\right)  \circ\left(
\id_{A}\otimes\id_{A}\otimes
T_{B\otimes B,A}\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.1.pf.2}))}}}\circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\otimes\id_{B}%
\otimes\underbrace{\id_{A\otimes B}}%
_{=\id_{A}\otimes\id_{B}%
}\right) \\
&  =\left(  m_{A}\otimes\id_{A}\otimes m_{B}%
\otimes\id_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes\id_{A}\otimes
T_{B\otimes B,A}\otimes\id_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  }_{\substack{=\id_{A}\otimes
Q\otimes\id_{B}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Q1'}))}}}\\
&  =\left(  m_{A}\otimes\id_{A}\otimes m_{B}%
\otimes\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes Q\otimes\id_{B}\right)  .
\end{align*}
This proves (\ref{sol.exe.tensor-prod-algs.a.2.1}).]

Now,%
\begin{align}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ\left(  m_{A\otimes B}%
\otimes\id_{A\otimes B}\right) \nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  }_{\substack{=\left(  m_{A}\otimes
\id_{A}\otimes m_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.1}))}}}\nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(  m_{A}%
\otimes\id_{A}\otimes m_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  }_{=\left(  m_{A}\otimes\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  m_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  }\circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right) \nonumber\\
&  =\underbrace{\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(
m_{A}\otimes\id_{A}\right)  \otimes\left(
m_{B}\otimes\id_{B}\right)  \right)  }%
_{\substack{=\left(  m_{A}\circ\left(  m_{A}\otimes\operatorname*{id}%
\nolimits_{A}\right)  \right)  \otimes\left(  m_{B}\circ\left(  m_{B}%
\otimes\id_{B}\right)  \right)  \\\text{(since the
composition-tensor relation yields}\\\left(  m_{A}\circ\left(  m_{A}%
\otimes\id_{A}\right)  \right)  \otimes\left(
m_{B}\circ\left(  m_{B}\otimes\id_{B}\right)  \right)
\\=\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  m_{A}%
\otimes\id_{A}\right)  \otimes\left(  m_{B}%
\otimes\id_{B}\right)  \right)  \text{)}}}\circ\left(
\id_{A}\otimes Q\otimes\operatorname*{id}%
\nolimits_{B}\right) \nonumber\\
&  =\left(  \underbrace{\left(  m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \right)  }_{\substack{=m_{A}%
\circ\left(  \id_{A}\otimes m_{A}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Adiag1.eq}))}}}\otimes\underbrace{\left(
m_{B}\circ\left(  m_{B}\otimes\id_{B}\right)  \right)
}_{\substack{=m_{B}\circ\left(  \id_{B}\otimes
m_{B}\right)  \\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Bdiag1.eq}))}%
}}\right)  \circ\left(  \id_{A}\otimes Q\otimes
\id_{B}\right) \nonumber\\
&  =\underbrace{\left(  \left(  m_{A}\circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\right)  \right)  \otimes\left(  m_{B}\circ\left(
\id_{B}\otimes m_{B}\right)  \right)  \right)
}_{\substack{=\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(
\id_{A}\otimes m_{A}\right)  \otimes\left(
\id_{B}\otimes m_{B}\right)  \right)  \\\text{(by the
composition-tensor relation)}}}\circ\left(  \id_{A}%
\otimes Q\otimes\id_{B}\right) \nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(  \left(
\id_{A}\otimes m_{A}\right)  \otimes\left(
\id_{B}\otimes m_{B}\right)  \right)  }%
_{=\id_{A}\otimes m_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes m_{B}}\circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right) \nonumber\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.half}%
\end{align}


On the other hand, it is easy to see that%
\begin{equation}
Q=\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\circ\left(  \id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\right)  . \label{sol.exe.tensor-prod-algs.a.2.Q2}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q2}):} Let $b\in B$,
$a\in A$, $b^{\prime}\in B$ and $a^{\prime}\in A$ be arbitrary. We have%
\[
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)  =a\otimes
a^{\prime}\otimes b\otimes b^{\prime}%
\]
(as we already have shown during the proof of
(\ref{sol.exe.tensor-prod-algs.a.2.Q1})). Comparing this with%
\begin{align*}
&  \left(  \left(  T_{B,A\otimes A}\otimes\id_{B}%
\right)  \circ\left(  \id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\right)  \right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right) \\
&  =\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\left(  \underbrace{\left(  \id_{B}\otimes
\id_{A}\otimes T_{B,A}\right)  \left(  b\otimes
a\otimes b^{\prime}\otimes a^{\prime}\right)  }_{=\operatorname*{id}%
\nolimits_{B}\left(  b\right)  \otimes\id_{A}\left(
a\right)  \otimes T_{B,A}\left(  b^{\prime}\otimes a^{\prime}\right)  }\right)
\\
&  =\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\left(  \underbrace{\id_{B}\left(  b\right)  }%
_{=b}\otimes\underbrace{\id_{A}\left(  a\right)
}_{=a}\otimes\underbrace{T_{B,A}\left(  b^{\prime}\otimes a^{\prime}\right)
}_{\substack{=a^{\prime}\otimes b^{\prime}\\\text{(by the definition of
}T_{B,A}\text{)}}}\right) \\
&  =\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\left(  b\otimes a\otimes a^{\prime}\otimes b^{\prime}\right)
=\underbrace{T_{B,A\otimes A}\left(  b\otimes a\otimes a^{\prime}\right)
}_{\substack{=a\otimes a^{\prime}\otimes b\\\text{(by the definition of
}T_{B,A\otimes A}\text{)}}}\otimes\underbrace{\operatorname*{id}%
\nolimits_{B}\left(  b^{\prime}\right)  }_{=b^{\prime}}\\
&  =a\otimes a^{\prime}\otimes b\otimes b^{\prime},
\end{align*}
we obtain%
\[
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)  =\left(
\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\circ\left(  \id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\right)  \right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)  .
\]


Now, forget that we fixed $b,a,b^{\prime},a^{\prime}$. We thus have shown that
every $b\in B$, $a\in A$, $b^{\prime}\in B$ and $a^{\prime}\in A$ satisfy%
\[
Q\left(  b\otimes a\otimes b^{\prime}\otimes a^{\prime}\right)  =\left(
\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\circ\left(  \id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\right)  \right)  \left(  b\otimes a\otimes
b^{\prime}\otimes a^{\prime}\right)  .
\]
In other words, the two maps $Q$ and $\left(  T_{B,A\otimes A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{B}\otimes\id_{A}\otimes T_{B,A}\right)  $
are equal to each other on each pure tensor. Since these two maps are
$\kk$-linear, we thus conclude that these two maps are identical
(because if two $\kk$-linear maps from a tensor product are equal to
each other on each pure tensor, then these two maps are identical). In other
words,
\[
Q=\left(  T_{B,A\otimes A}\otimes\id_{B}\right)
\circ\left(  \id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\right)  .
\]
This proves (\ref{sol.exe.tensor-prod-algs.a.2.Q2}).]

Now,%
\begin{align}
&  \underbrace{\id_{A}}_{=\operatorname*{id}%
\nolimits_{A}\circ\id_{A}}\otimes\underbrace{Q}%
_{\substack{=\left(  T_{B,A\otimes A}\otimes\id_{B}%
\right)  \circ\left(  \id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Q2}))}}}\otimes
\underbrace{\id_{B}}_{=\operatorname*{id}%
\nolimits_{B}\circ\id_{B}}\nonumber\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  \left(  T_{B,A\otimes A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{B}\otimes\id_{A}\otimes T_{B,A}\right)
\right)  \otimes\left(  \id_{B}\circ\operatorname*{id}%
\nolimits_{B}\right) \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
T_{B,A\otimes A}\otimes\id_{B}\right)  \otimes
\id_{B}\right)  }_{=\id_{A}%
\otimes T_{B,A\otimes A}\otimes\id_{B}\otimes
\id_{B}}\circ\underbrace{\left(  \operatorname*{id}%
\nolimits_{A}\otimes\left(  \id_{B}\otimes
\id_{A}\otimes T_{B,A}\right)  \otimes
\id_{B}\right)  }_{=\id_{A}%
\otimes\id_{B}\otimes\id_{A}%
\otimes T_{B,A}\otimes\id_{B}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \nonumber\\
&  =\left(  \id_{A}\otimes T_{B,A\otimes A}%
\otimes\id_{B}\otimes\id_{B}%
\right)  \circ\left(  \id_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes\id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.Q2'}%
\end{align}


Now, we shall show that%
\begin{equation}
\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A,B}\right)  =\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)
\label{sol.exe.tensor-prod-algs.a.2.2}%
\end{equation}
(as maps from $A\otimes B\otimes A\otimes B\otimes A\otimes B$ to $A\otimes
A\otimes B\otimes B$).

[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.2}):} We have%
\begin{align}
&  \underbrace{\id_{A\otimes B}}_{=\operatorname*{id}%
\nolimits_{A\otimes B}\circ\id_{A\otimes B}}%
\otimes\underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\nonumber\\
&  =\left(  \id_{A\otimes B}\circ\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \otimes\left(  \left(  m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \right) \nonumber\\
&  =\underbrace{\left(  \id_{A\otimes B}\otimes\left(
m_{A}\otimes m_{B}\right)  \right)  }_{=\id_{A\otimes
B}\otimes m_{A}\otimes m_{B}}\circ\underbrace{\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \right)  }%
_{=\id_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\otimes\id_{B}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \nonumber\\
&  =\left(  \id_{A\otimes B}\otimes m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A\otimes B}%
\otimes\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  . \label{sol.exe.tensor-prod-algs.a.2.2.pf.1}%
\end{align}


The equality (\ref{sol.exe.tensor-prod-algs.a.2.twistcomm}) (applied to $B$,
$A\otimes A$, $B$, $A$, $\id_{B}$ and $m_{A}$ instead
of $U$, $V$, $U^{\prime}$, $V^{\prime}$, $x$ and $y$) yields
\[
\left(  m_{A}\otimes\id_{B}\right)  \circ
T_{B,A\otimes A}=T_{B,A}\circ\left(  \id_{B}\otimes
m_{A}\right)  .
\]
Hence,%
\begin{align*}
&  \left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\underbrace{\left(  \left(  m_{A}\otimes
\id_{B}\right)  \circ T_{B,A\otimes A}\right)
}_{=T_{B,A}\circ\left(  \id_{B}\otimes m_{A}\right)
}\otimes\left(  \id_{B}\circ m_{B}\right) \\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  T_{B,A}\circ\left(  \operatorname*{id}%
\nolimits_{B}\otimes m_{A}\right)  \right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\circ m_{B}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes m_{A}\right)  \otimes m_{B}\right)  }%
_{=\id_{A}\otimes\id_{B}%
\otimes m_{A}\otimes m_{B}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(
\underbrace{\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}}_{=\id_{A\otimes B}}\otimes m_{A}\otimes
m_{B}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A}\otimes m_{B}\right)  .
\end{align*}
Thus,%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A}\otimes m_{B}\right) \nonumber\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  \left(  m_{A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ T_{B,A\otimes A}\right)  \otimes
\underbrace{\left(  \id_{B}\circ m_{B}\right)
}_{=m_{B}=m_{B}\circ\id_{B\otimes B}}\nonumber\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  \left(  m_{A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ T_{B,A\otimes A}\right)  \otimes\left(  m_{B}%
\circ\id_{B\otimes B}\right) \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
m_{A}\otimes\id_{B}\right)  \otimes m_{B}\right)
}_{=\id_{A}\otimes m_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes m_{B}}\circ\left(  \id_{A}\otimes
T_{B,A\otimes A}\otimes\underbrace{\id_{B\otimes B}%
}_{=\id_{B}\otimes\id_{B}%
}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \nonumber\\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{B}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A\otimes A}\otimes
\id_{B}\otimes\id_{B}\right)
. \label{sol.exe.tensor-prod-algs.a.2.2.pf.2}%
\end{align}


Now,%
\begin{align*}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\underbrace{\left(
\id_{A\otimes B}\otimes m_{A\otimes B}\right)
}_{\substack{=\left(  \id_{A\otimes B}\otimes
m_{A}\otimes m_{B}\right)  \circ\left(  \id_{A\otimes
B}\otimes\id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.2.pf.1}))}}}\\
&  =\underbrace{\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A}\otimes m_{B}\right)  }_{\substack{=\left(
\id_{A}\otimes m_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes T_{B,A\otimes A}\otimes\id_{B}%
\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.2.pf.2}))}}}\circ\left(
\underbrace{\id_{A\otimes B}}_{=\operatorname*{id}%
\nolimits_{A}\otimes\id_{B}}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\otimes\id_{B}\right) \\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{B}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A\otimes A}\otimes
\id_{B}\otimes\id_{B}\right)
\circ\left(  \id_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes\id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  }%
_{\substack{=\id_{A}\otimes Q\otimes\operatorname*{id}%
\nolimits_{B}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Q2'}))}}}\\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{B}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes Q\otimes\operatorname*{id}%
\nolimits_{B}\right)  .
\end{align*}
This proves (\ref{sol.exe.tensor-prod-algs.a.2.2}).]

Now,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A\otimes B}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{A\otimes
B}\otimes m_{A\otimes B}\right)  }_{\substack{=\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.2}))}}}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes m_{A}\otimes\id_{B}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
Q\otimes\id_{B}\right)  .
\end{align*}
Comparing this with (\ref{sol.exe.tensor-prod-algs.a.2.half}), we obtain%
\[
m_{A\otimes B}\circ\left(  m_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  =m_{A\otimes B}\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes m_{A\otimes B}\right)  .
\]
In other words, the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag1}) commutes.

Let us next prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

We first observe that%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}\nonumber\\
&  =\operatorname*{kan}\nolimits_{1,A}\otimes\operatorname*{kan}%
\nolimits_{1,B}. \label{sol.exe.tensor-prod-algs.a.2.k1}%
\end{align}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.k1}):} Let $a\in A$ and
$b\in B$. Then,%
\begin{align*}
&  \left(  \left(  \id_{A}\otimes T_{B,\kk%
}\otimes\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}\right)  \left(  a\otimes b\right)
\\
&  =\left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \left(  \left(
\id_{A\otimes B}\otimes\xi\right)  \left(
\underbrace{\operatorname*{kan}\nolimits_{1,A\otimes B}\left(  a\otimes
b\right)  }_{\substack{=a\otimes b\otimes1\\\text{(by the definition of
}\operatorname*{kan}\nolimits_{1,A\otimes B}\text{)}}}\right)  \right) \\
&  =\left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \left(
\underbrace{\left(  \id_{A\otimes B}\otimes\xi\right)
\left(  a\otimes b\otimes1\right)  }_{=\id_{A\otimes
B}\left(  a\otimes b\right)  \otimes\xi\left(  1\right)  }\right) \\
&  =\left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \left(
\underbrace{\id_{A\otimes B}\left(  a\otimes b\right)
}_{=a\otimes b}\otimes\underbrace{\xi\left(  1\right)  }_{\substack{=1\otimes
1\\\text{(by the definition of } \xi\text{)}}}\right) \\
&  =\left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \left(  a\otimes
b\otimes1\otimes1\right)  =\underbrace{\id_{A}\left(
a\right)  }_{=a}\otimes\underbrace{T_{B,\kk}\left(  b\otimes1\right)
}_{\substack{=1\otimes b\\\text{(by the definition of }T_{B,\kk%
}\text{)}}}\otimes\underbrace{\id_{\kk}\left(
1\right)  }_{=1}\\
&  =a\otimes1\otimes b\otimes1.
\end{align*}
Comparing this with%
\[
\left(  \operatorname*{kan}\nolimits_{1,A}\otimes\operatorname*{kan}%
\nolimits_{1,B}\right)  \left(  a\otimes b\right)
=\underbrace{\operatorname*{kan}\nolimits_{1,A}\left(  a\right)
}_{\substack{=a\otimes1\\\text{(by the definition of }\operatorname*{kan}%
\nolimits_{1,A}\text{)}}}\otimes\underbrace{\operatorname*{kan}\nolimits_{1,B}%
\left(  b\right)  }_{\substack{=b\otimes1\\\text{(by the definition of
}\operatorname*{kan}\nolimits_{1,B}\text{)}}}=a\otimes1\otimes b\otimes1,
\]
we obtain%
\[
\left(  \operatorname*{kan}\nolimits_{1,A}\otimes\operatorname*{kan}%
\nolimits_{1,B}\right)  \left(  a\otimes b\right)  =\left(  \left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}\right)  \left(  a\otimes b\right)
.
\]


Now, forget that we fixed $a$ and $b$. We thus have proven that%
\[
\left(  \operatorname*{kan}\nolimits_{1,A}\otimes\operatorname*{kan}%
\nolimits_{1,B}\right)  \left(  a\otimes b\right)  =\left(  \left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}\right)  \left(  a\otimes b\right)
\]
for every $a\in A$ and $b\in B$. In other words, the two maps
$\operatorname*{kan}\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}$
and $\left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}$ are equal to each other on each
pure tensor. Since these two maps are $\kk$-linear, we thus conclude
that these two maps are identical (because if two $\kk$-linear maps
from a tensor product are equal to each other on each pure tensor, then these
two maps are identical). In other words,
\[
\operatorname*{kan}\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}%
=\left(  \id_{A}\otimes T_{B,\kk}%
\otimes\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}.
\]
This proves (\ref{sol.exe.tensor-prod-algs.a.2.k1}).]

The equality (\ref{sol.exe.tensor-prod-algs.a.2.kans.1}) (applied to $U=A$ and
$V=B$) yields%
\begin{equation}
\operatorname{id}_{A}\otimes\operatorname{kan}_{1,B}=\operatorname{kan}%
_{1,A\otimes B}. \label{sol.exe.tensor-prod-algs.a.2.u0}%
\end{equation}


From%
\begin{align*}
&  \left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\underbrace{\left(  \left(  u_{A}\otimes
\id_{B}\right)  \circ T_{B,\kk}\right)
}_{\substack{=T_{B,A}\circ\left(  \id_{B}\otimes
u_{A}\right)  \\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.twistcomm})
(applied}\\\text{to }B\text{, }\kk\text{, }B\text{, }A\text{,
}\id_{B}\text{ and }u_{A}\\\text{instead of }U\text{,
}V\text{, }U^{\prime}\text{, }V^{\prime}\text{, }x\text{ and }y\text{))}%
}}\otimes\underbrace{\left(  u_{B}\circ\id_{\kk%
}\right)  }_{=u_{B}=\id_{B}\circ u_{B}}\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  T_{B,A}\circ\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{A}\right)  \right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\circ u_{B}\right)  =\left(  \id_{A}%
\otimes T_{B,A}\otimes\id_{B}\right)  \circ
\underbrace{\left(  \id_{A}\otimes\left(
\id_{B}\otimes u_{A}\right)  \otimes u_{B}\right)
}_{=\id_{A}\otimes\id_{B}%
\otimes u_{A}\otimes u_{B}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(
\underbrace{\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}}_{=\id_{A\otimes B}}\otimes u_{A}\otimes
u_{B}\right)  =\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes u_{A}\otimes u_{B}\right)  ,
\end{align*}
we obtain%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes u_{A}\otimes u_{B}\right) \nonumber\\
&  =\left(  \id_{A}\circ\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  \left(  u_{A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ T_{B,\kk}\right)  \otimes\left(  u_{B}%
\circ\id_{\kk}\right) \nonumber\\
&  =\underbrace{\left(  \id_{A}\otimes\left(
u_{A}\otimes\id_{B}\right)  \otimes u_{B}\right)
}_{=\id_{A}\otimes u_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes u_{B}}\circ\left(  \id_{A}\otimes
T_{B,\kk}\otimes\id_{\kk}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \nonumber\\
&  =\left(  \id_{A}\otimes u_{A}\otimes
\id_{B}\otimes u_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  .
\label{sol.exe.tensor-prod-algs.a.2.u1}%
\end{align}


Next, we observe that%
\begin{align*}
&  \underbrace{\id_{A\otimes B}}_{=\operatorname*{id}%
\nolimits_{A\otimes B}\circ\id_{A\otimes B}}%
\otimes\underbrace{u_{A\otimes B}}_{=\left(  u_{A}\otimes u_{B}\right)
\circ\xi}\\
&  =\left(  \id_{A\otimes B}\circ\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \otimes\left(  \left(  u_{A}\otimes
u_{B}\right)  \circ\xi\right)  =\underbrace{\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes\left(  u_{A}\otimes u_{B}\right)  \right)
}_{=\id_{A\otimes B}\otimes u_{A}\otimes u_{B}}%
\circ\left(  \id_{A\otimes B}\otimes\xi\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  \id_{A\otimes B}\otimes u_{A}\otimes
u_{B}\right)  \circ\left(  \id_{A\otimes B}\otimes
\xi\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ\underbrace{\left(
\id_{A\otimes B}\otimes u_{A\otimes B}\right)
}_{=\left(  \id_{A\otimes B}\otimes u_{A}\otimes
u_{B}\right)  \circ\left(  \id_{A\otimes B}\otimes
\xi\right)  }\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{A\otimes
B}\otimes u_{A}\otimes u_{B}\right)  }_{\substack{=\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\otimes\id_{B}\otimes
u_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,\kk}\otimes\id_{\kk}\right)
\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.u1}))}}}\circ\left(
\id_{A\otimes B}\otimes\xi\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\otimes\id_{B}\otimes
u_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,\kk}\otimes\id_{\kk}\right)
\circ\left(  \id_{A\otimes B}\otimes\xi\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A\otimes B}\circ\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes u_{A\otimes B}\right)  }_{=\left(  m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes u_{A}%
\otimes\id_{B}\otimes u_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  }\circ
\operatorname*{kan}\nolimits_{1,A\otimes B}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes u_{A}\otimes\operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  }_{=\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  }\circ\underbrace{\left(
\id_{A}\otimes T_{B,\kk}\otimes
\id_{\kk}\right)  \circ\left(
\id_{A\otimes B}\otimes\xi\right)  \circ
\operatorname*{kan}\nolimits_{1,A\otimes B}}_{\substack{=\operatorname*{kan}%
\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.k1}))}}}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  \right)  \circ\left(  \operatorname*{kan}%
\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\id_{A\otimes B}  &  =\underbrace{\operatorname*{id}%
\nolimits_{A}}_{\substack{=m_{A}\circ\left(  \id_{A}%
\otimes u_{A}\right)  \circ\operatorname*{kan}\nolimits_{1,A}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Adiag2.eq1}))}}}\otimes
\underbrace{\id_{B}}_{\substack{=m_{B}\circ\left(
\id_{B}\otimes u_{B}\right)  \circ\operatorname*{kan}%
\nolimits_{1,B}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq1}))}%
}}\\
&  =\underbrace{\left(  m_{A}\circ\left(  \id_{A}%
\otimes u_{A}\right)  \circ\operatorname*{kan}\nolimits_{1,A}\right)
}_{=\left(  m_{A}\circ\left(  \id_{A}\otimes
u_{A}\right)  \right)  \circ\operatorname*{kan}\nolimits_{1,A}}\otimes
\underbrace{\left(  m_{B}\circ\left(  \id_{B}\otimes
u_{B}\right)  \circ\operatorname*{kan}\nolimits_{1,B}\right)  }_{=\left(
m_{B}\circ\left(  \id_{B}\otimes u_{B}\right)
\right)  \circ\operatorname*{kan}\nolimits_{1,B}}\\
&  =\left(  \left(  m_{A}\circ\left(  \id_{A}\otimes
u_{A}\right)  \right)  \circ\operatorname*{kan}\nolimits_{1,A}\right)
\otimes\left(  \left(  m_{B}\circ\left(  \id_{B}%
\otimes u_{B}\right)  \right)  \circ\operatorname*{kan}\nolimits_{1,B}\right)
\\
&  =\underbrace{\left(  \left(  m_{A}\circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \right)  \otimes\left(  m_{B}\circ\left(
\id_{B}\otimes u_{B}\right)  \right)  \right)
}_{\substack{=\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(
\id_{A}\otimes u_{A}\right)  \otimes\left(
\id_{B}\otimes u_{B}\right)  \right)  \\\text{(by the
composition-tensor relation)}}}\circ\left(  \operatorname*{kan}\nolimits_{1,A}%
\otimes\operatorname*{kan}\nolimits_{1,B}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  \operatorname*{id}%
\nolimits_{A}\otimes u_{A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\otimes u_{B}\right)  \right)  \circ\left(  \operatorname*{kan}%
\nolimits_{1,A}\otimes\operatorname*{kan}\nolimits_{1,B}\right)  ,
\end{align*}
we obtain%
\[
m_{A\otimes B}\circ\left(  \id_{A\otimes B}\otimes
u_{A\otimes B}\right)  \circ\operatorname*{kan}\nolimits_{1,A\otimes
B}=\id_{A\otimes B}.
\]
In other words, the left rectangle of the diagram
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

Next, we shall prove the same for the right rectangle. The proof is similar to
the one for the left rectangle; but we nevertheless show it for the sake of completeness.

We first observe that%
\begin{align}
&  \left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \circ\left(  \xi
\otimes\id_{A\otimes B}\right)  \circ
\operatorname*{kan}\nolimits_{2,A\otimes B}\nonumber\\
&  =\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}%
\nolimits_{2,B}. \label{sol.exe.tensor-prod-algs.a.2.k2}%
\end{align}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.a.2.k1}):} Let $a\in A$ and
$b\in B$. Then,%
\begin{align*}
&  \left(  \left(  \id_{\kk}\otimes
T_{\kk,A}\otimes\id_{B}\right)  \circ\left(
\xi\otimes\id_{A\otimes B}\right)  \circ
\operatorname*{kan}\nolimits_{2,A\otimes B}\right)  \left(  a\otimes b\right)
\\
&  =\left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \left(  \left(  \xi
\otimes\id_{A\otimes B}\right)  \left(
\underbrace{\operatorname*{kan}\nolimits_{2,A\otimes B}\left(  a\otimes
b\right)  }_{\substack{=1\otimes a\otimes b\\\text{(by the definition of
}\operatorname*{kan}\nolimits_{2,A\otimes B}\text{)}}}\right)  \right) \\
&  =\left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \left(  \underbrace{\left(
\xi\otimes\id_{A\otimes B}\right)  \left(  1\otimes
a\otimes b\right)  }_{= \xi\left(  1\right)  \otimes\operatorname*{id}%
\nolimits_{A\otimes B}\left(  a\otimes b\right)  }\right) \\
&  =\left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \left(  \underbrace{
\xi\left(  1\right)  }_{\substack{=1\otimes1\\\text{(by the definition of }
\xi\text{)}}}\otimes\underbrace{\id_{A\otimes
B}\left(  a\otimes b\right)  }_{=a\otimes b}\right) \\
&  =\left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \left(  1\otimes1\otimes
a\otimes b\right)  =\underbrace{\id_{\kk%
}\left(  1\right)  }_{=1}\otimes\underbrace{T_{\kk,A}\left(  1\otimes
a\right)  }_{\substack{=a\otimes1\\\text{(by the definition of }%
T_{\kk,A}\text{)}}}\otimes\underbrace{\id_{B}%
\left(  b\right)  }_{=b}\\
&  =1\otimes a\otimes1\otimes b.
\end{align*}
Comparing this with%
\[
\left(  \operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}%
\nolimits_{2,B}\right)  \left(  a\otimes b\right)
=\underbrace{\operatorname*{kan}\nolimits_{2,A}\left(  a\right)
}_{\substack{=1\otimes a\\\text{(by the definition of }\operatorname*{kan}%
\nolimits_{2,A}\text{)}}}\otimes\underbrace{\operatorname*{kan}\nolimits_{2,B}%
\left(  b\right)  }_{\substack{=1\otimes b\\\text{(by the definition of
}\operatorname*{kan}\nolimits_{2,B}\text{)}}}=1\otimes a\otimes1\otimes b,
\]
we obtain%
\[
\left(  \operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}%
\nolimits_{2,B}\right)  \left(  a\otimes b\right)  =\left(  \left(
\id_{\kk}\otimes T_{\kk,A}%
\otimes\id_{B}\right)  \circ\left(  \xi\otimes
\id_{A\otimes B}\right)  \circ\operatorname*{kan}%
\nolimits_{2,A\otimes B}\right)  \left(  a\otimes b\right)  .
\]


Now, forget that we fixed $a$ and $b$. We thus have proven that%
\[
\left(  \operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}%
\nolimits_{2,B}\right)  \left(  a\otimes b\right)  =\left(  \left(
\id_{\kk}\otimes T_{\kk,A}%
\otimes\id_{B}\right)  \circ\left(  \xi\otimes
\id_{A\otimes B}\right)  \circ\operatorname*{kan}%
\nolimits_{2,A\otimes B}\right)  \left(  a\otimes b\right)
\]
for every $a\in A$ and $b\in B$. In other words, the two maps
$\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}\nolimits_{2,B}$
and $\left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \circ\left(  \xi
\otimes\id_{A\otimes B}\right)  \circ
\operatorname*{kan}\nolimits_{2,A\otimes B}$ are equal to each other on each
pure tensor. Since these two maps are $\kk$-linear, we thus conclude
that these two maps are identical (because if two $\kk$-linear maps
from a tensor product are equal to each other on each pure tensor, then these
two maps are identical). In other words,
\[
\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}\nolimits_{2,B}%
=\left(  \id_{\kk}\otimes T_{\kk%
,A}\otimes\id_{B}\right)  \circ\left(  \xi
\otimes\id_{A\otimes B}\right)  \circ
\operatorname*{kan}\nolimits_{2,A\otimes B}.
\]
This proves (\ref{sol.exe.tensor-prod-algs.a.2.k2}).]

The equality (\ref{sol.exe.tensor-prod-algs.a.2.kans.2}) (applied to $U=B$ and
$V=A$) yields%
\begin{equation}
\operatorname*{kan}\nolimits_{2,A}\otimes\id_{B}%
=\operatorname{kan}_{2,A\otimes B}. \label{sol.exe.tensor-prod-algs.a.2.u2}%
\end{equation}


From%
\begin{align*}
&  \underbrace{\left(  u_{A}\circ\id_{\kk%
}\right)  }_{=u_{A}=\id_{A}\circ u_{A}}\otimes
\underbrace{\left(  \left(  \id_{A}\otimes
u_{B}\right)  \circ T_{\kk,A}\right)  }_{\substack{=T_{B,A}\circ\left(
u_{B}\otimes\id_{A}\right)  \\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.twistcomm}) (applied}\\\text{to }%
\kk\text{, }A\text{, }B\text{, }A\text{, }u_{B}\text{ and
}\id_{A}\\\text{instead of }U\text{, }V\text{,
}U^{\prime}\text{, }V^{\prime}\text{, }x\text{ and }y\text{))}}}\otimes\left(
\id_{B}\circ\id_{B}\right) \\
&  =\left(  \id_{A}\circ u_{A}\right)  \otimes\left(
T_{B,A}\circ\left(  u_{B}\otimes\id_{A}\right)
\right)  \otimes\left(  \id_{B}\circ\operatorname*{id}%
\nolimits_{B}\right)  =\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \circ\underbrace{\left(
u_{A}\otimes\left(  u_{B}\otimes\id_{A}\right)
\otimes\id_{B}\right)  }_{=u_{A}\otimes u_{B}%
\otimes\id_{A}\otimes\id_{B}%
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \\
&  =\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  u_{A}\otimes u_{B}%
\otimes\underbrace{\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}}_{=\id_{A\otimes B}}\right)  =\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  u_{A}\otimes u_{B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  ,
\end{align*}
we obtain%
\begin{align}
&  \left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  \circ\left(  u_{A}\otimes u_{B}%
\otimes\id_{A\otimes B}\right) \nonumber\\
&  =\left(  u_{A}\circ\id_{\kk}\right)
\otimes\left(  \left(  \id_{A}\otimes u_{B}\right)
\circ T_{\kk,A}\right)  \otimes\left(  \operatorname*{id}%
\nolimits_{B}\circ\id_{B}\right) \nonumber\\
&  =\underbrace{\left(  u_{A}\otimes\left(  \id_{A}%
\otimes u_{B}\right)  \otimes\id_{B}\right)  }%
_{=u_{A}\otimes\id_{A}\otimes u_{B}\otimes
\id_{B}}\circ\left(  \operatorname*{id}%
\nolimits_{\kk}\otimes T_{\kk,A}\otimes\operatorname*{id}%
\nolimits_{B}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-triple-tensor
relation}\right) \nonumber\\
&  =\left(  u_{A}\otimes\id_{A}\otimes u_{B}%
\otimes\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{\kk}\otimes T_{\kk,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  . \label{sol.exe.tensor-prod-algs.a.2.u3}%
\end{align}


Next, we observe that%
\begin{align*}
&  \underbrace{u_{A\otimes B}}_{=\left(  u_{A}\otimes u_{B}\right)  \circ\xi
}\otimes\underbrace{\id_{A\otimes B}}%
_{=\id_{A\otimes B}\circ\operatorname*{id}%
\nolimits_{A\otimes B}}\\
&  =\left(  \left(  u_{A}\otimes u_{B}\right)  \circ\xi\right)  \otimes\left(
\id_{A\otimes B}\circ\operatorname*{id}%
\nolimits_{A\otimes B}\right)  =\underbrace{\left(  \left(  u_{A}\otimes
u_{B}\right)  \otimes\id_{A\otimes B}\right)
}_{=u_{A}\otimes u_{B}\otimes\id_{A\otimes B}}%
\circ\left(  \xi\otimes\id_{A\otimes B}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  u_{A}\otimes u_{B}\otimes\id_{A\otimes
B}\right)  \circ\left(  \xi\otimes\id_{A\otimes
B}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id_{A}\otimes T_{B,A}\otimes
\id_{B}\right)  }\circ\underbrace{\left(  u_{A\otimes
B}\otimes\id_{A\otimes B}\right)  }_{=\left(
u_{A}\otimes u_{B}\otimes\id_{A\otimes B}\right)
\circ\left(  \xi\otimes\id_{A\otimes B}\right)  }\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  u_{A}\otimes u_{B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  }_{\substack{=\left(  u_{A}\otimes
\id_{A}\otimes u_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{\kk%
}\otimes T_{\kk,A}\otimes\id_{B}\right)
\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.u3}))}}}\circ\left(  \xi
\otimes\id_{A\otimes B}\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  u_{A}\otimes
\id_{A}\otimes u_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \id_{\kk%
}\otimes T_{\kk,A}\otimes\id_{B}\right)
\circ\left(  \xi\otimes\id_{A\otimes B}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A\otimes B}\circ\left(  u_{A\otimes B}\otimes
\id_{A\otimes B}\right)  }_{=\left(  m_{A}\otimes
m_{B}\right)  \circ\left(  u_{A}\otimes\id_{A}\otimes
u_{B}\otimes\id_{B}\right)  \circ\left(
\id_{\kk}\otimes T_{\kk,A}%
\otimes\id_{B}\right)  \circ\left(  \xi\otimes
\id_{A\otimes B}\right)  }\circ\operatorname*{kan}%
\nolimits_{2,A\otimes B}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\underbrace{\left(  u_{A}%
\otimes\id_{A}\otimes u_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  }_{=\left(  u_{A}\otimes\operatorname*{id}%
\nolimits_{A}\right)  \otimes\left(  u_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  }\circ\underbrace{\left(  \operatorname*{id}%
\nolimits_{\kk}\otimes T_{\kk,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\left(  \xi\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \circ\operatorname*{kan}\nolimits_{2,A\otimes
B}}_{\substack{=\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}%
\nolimits_{2,B}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.k2}))}}}\\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  u_{A}%
\otimes\id_{A}\right)  \otimes\left(  u_{B}%
\otimes\id_{B}\right)  \right)  \circ\left(
\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}\nolimits_{2,B}%
\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\id_{A\otimes B}  &  =\underbrace{\operatorname*{id}%
\nolimits_{A}}_{\substack{=m_{A}\circ\left(  u_{A}\otimes\operatorname*{id}%
\nolimits_{A}\right)  \circ\operatorname*{kan}\nolimits_{2,A}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.a.2.Adiag2.eq2}))}}}\otimes
\underbrace{\id_{B}}_{\substack{=m_{B}\circ\left(
u_{B}\otimes\id_{B}\right)  \circ\operatorname*{kan}%
\nolimits_{2,B}\\\text{(by (\ref{sol.exe.tensor-prod-algs.a.2.Bdiag2.eq2}))}%
}}\\
&  =\underbrace{\left(  m_{A}\circ\left(  u_{A}\otimes\operatorname*{id}%
\nolimits_{A}\right)  \circ\operatorname*{kan}\nolimits_{2,A}\right)
}_{=\left(  m_{A}\circ\left(  u_{A}\otimes\id_{A}%
\right)  \right)  \circ\operatorname*{kan}\nolimits_{2,A}}\otimes
\underbrace{\left(  m_{B}\circ\left(  u_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \circ\operatorname*{kan}\nolimits_{2,B}\right)
}_{=\left(  m_{B}\circ\left(  u_{B}\otimes\id_{B}%
\right)  \right)  \circ\operatorname*{kan}\nolimits_{2,B}}\\
&  =\left(  \left(  m_{A}\circ\left(  u_{A}\otimes\operatorname*{id}%
\nolimits_{A}\right)  \right)  \circ\operatorname*{kan}\nolimits_{2,A}\right)
\otimes\left(  \left(  m_{B}\circ\left(  u_{B}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \right)  \circ\operatorname*{kan}\nolimits_{2,B}\right)
\\
&  =\underbrace{\left(  \left(  m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  \right)  \otimes\left(  m_{B}%
\circ\left(  u_{B}\otimes\id_{B}\right)  \right)
\right)  }_{\substack{=\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(
u_{A}\otimes\id_{A}\right)  \otimes\left(
u_{B}\otimes\id_{B}\right)  \right)  \\\text{(by the
composition-tensor relation)}}}\circ\left(  \operatorname*{kan}\nolimits_{2,A}%
\otimes\operatorname*{kan}\nolimits_{2,B}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \circ\left(  \left(  u_{A}%
\otimes\id_{A}\right)  \otimes\left(  u_{B}%
\otimes\id_{B}\right)  \right)  \circ\left(
\operatorname*{kan}\nolimits_{2,A}\otimes\operatorname*{kan}\nolimits_{2,B}%
\right)  ,
\end{align*}
we obtain%
\[
m_{A\otimes B}\circ\left(  u_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A\otimes B}\right)  \circ\operatorname*{kan}\nolimits_{2,A\otimes
B}=\id_{A\otimes B}.
\]
In other words, the right rectangle of the diagram
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

We now know that both the left rectangle and the right rectangle of the
diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commute. Hence, the diagram
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commutes.

We have now shown that the two diagrams
(\ref{sol.exe.tensor-prod-algs.a.2.diag1}) and
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commute. Thus, the $\kk%
$-module $A\otimes B$, equipped with the two maps $m_{A\otimes B}$ and
$u_{A\otimes B}$, is a $\kk$-algebra (since Definition
\ref{def.algebra} yields that $A\otimes B$ is a $\kk$-algebra if and
only if the two diagrams (\ref{sol.exe.tensor-prod-algs.a.2.diag1}) and
(\ref{sol.exe.tensor-prod-algs.a.2.diag2}) commute). In other words, the
$\kk$-algebra $A\otimes B$ introduced in Definition
\ref{def.tensor-prod-algs} is actually well-defined. This solves Exercise
\ref{exe.tensor-prod-algs}(a) again.

\begin{commentedout}
[The following third solution is unfinished.]
\usepackage{lscape}
\newcommand*\numcircledmod[1]{\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {#1}}}}
\textit{Third solution to Exercise \ref{exe.tensor-prod-algs}(a).} The
following solution to Exercise \ref{exe.tensor-prod-algs}(a) is essentially
the same as the Second solution given above, merely translated into the
language of commutative diagrams (which shortens it significantly).

Let $\xi$ be the canonical $\kk$-module isomorphism $\kk%
\rightarrow\kk\otimes\kk$. According to Definition
\ref{def.tensor-prod-algs}, we define the map $m_{A\otimes B}:A\otimes
B\otimes A\otimes B\rightarrow A\otimes B$ by%
\[
m_{A\otimes B}=\left(  m_{A}\otimes m_{B}\right)  \circ\left(
\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  ,
\]
and we define the map $u_{A\otimes B}:\kk\rightarrow A\otimes B$ by%
\[
u_{A\otimes B}=\left(  u_{A}\otimes u_{B}\right)  \circ\xi.
\]
We now must prove that the $\kk$-module $A\otimes B$, equipped with
these two maps $m_{A\otimes B}$ and $u_{A\otimes B}$, is indeed a $\kk%
$-algebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & A \otimes B \otimes A \otimes B \otimes A \otimes B \ar[dl]_{m_{A\otimes B} \otimes \id_{A\otimes B}} \ar[dr]^{\id_{A\otimes B} \otimes m_{A\otimes B}} & \\ A \otimes B \otimes A \otimes B \ar[dr]^{m_{A \otimes B}} & & A \otimes B \otimes A \otimes B \ar[dl]_{m_{A \otimes B}} \\ & A \otimes B & }
\label{sol.exe.tensor-prod-algs.a.3.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ A \otimes B \otimes \kk \ar[d]_{\id_{A\otimes B} \otimes u_{A \otimes B}} & A \otimes B \ar[l]_{\operatorname{kan}_{1,A\otimes B}} \ar[d]_{\id_{A \otimes B}} \ar[r]^{\operatorname{kan}_{2,A,\otimes B}} & \kk \otimes A \otimes B \ar[d]^{u_{A \otimes B} \otimes \id_{A \otimes B}} \\ A \otimes B\otimes A \otimes B\ar[r]^-{m_{A \otimes B}} & A \otimes B & A \otimes B \otimes A \otimes B\ar[l]_-{m_{A \otimes B}} }
\label{sol.exe.tensor-prod-algs.a.3.diag2}%
\end{equation}
commute (since Definition \ref{def.algebra} yields that $A\otimes B$ is a
$\kk$-algebra if and only if these two diagrams commute).

Let us first prove that the diagram (\ref{sol.exe.tensor-prod-algs.a.2.diag1}) commutes.

Consider the diagram \begin{landscape}
\[
\xymatrixrowsep{4.5pc}
\xymatrixcolsep{1pc}
\xymatrix{
& & \numcircledmod{a} & & A \otimes B \otimes A \otimes B \otimes A \otimes B \ar@/_10pc/[llllddd]|-{\id_{A \otimes B} \otimes m_{A \otimes B}} \ar[lld]|-{\id_A \otimes \id_B \otimes \id_A \otimes T_{B,A} \otimes \id_B} \ar[rrd]|-{\id_A \otimes T_{B,A} \otimes \id_B \otimes \id_A \otimes \id_B} \ar@/^10pc/[rrrrddd]|-{m_{A \otimes B} \otimes \id_{A \otimes B}} & & \numcircledmod{b} \\
& & A \otimes B \otimes A \otimes A \otimes B \otimes B \ar[lldd]|-(0.7){\id_{A \otimes B} \otimes m_A \otimes m_B} \ar[d]|-(0.7){\qquad \id_A \otimes \id_B \otimes \id_A \otimes \id_A \otimes m_B} \ar[rrd]|-(0.4){\qquad \qquad \id_A \otimes T_{B, A \otimes A} \otimes \id_B \otimes \id_B} & & \numcircledmod{c} & & A \otimes A \otimes B \otimes B \otimes A \otimes B \ar[lld]|-(0.4){\id_A \otimes \id_A \otimes T_{B \otimes B, A} \otimes \id_B \qquad \qquad} \ar[d]|-(0.7){m_A \otimes \id_B \otimes \id_B \otimes \id_A \otimes \id_B \qquad} \ar[rrdd]|-(0.7){m_A \otimes m_B \otimes \id_{A \otimes B} \qquad} \\
& & A \otimes B \otimes A \otimes A \otimes B \ar[d]|-(0.7){\id_A \otimes T_{B, A \otimes A} \otimes \id_B \qquad} & \numcircledmod{e} & A \otimes A \otimes A \otimes B \otimes B \otimes B \ar[lld]|-(0.4){\id_A \otimes \id_A \otimes \id_A \otimes \id_B \otimes m_B \qquad \qquad} \ar[llddd]|-(0.5){\id_A \otimes m_A \otimes \id_B \otimes m_B} \ar[rrddd]|-(0.5){m_A \otimes \id_A \otimes m_B \otimes \id_B} \ar[rrd]|-(0.4){\qquad \qquad m_A \otimes \id_A \otimes \id_B \otimes \id_B \otimes \id_B} & \numcircledmod{f} & A \otimes B \otimes B \otimes A \otimes B \ar[d]|-(0.7){\id_A \otimes T_{B \otimes B, A} \otimes \id_B} \\
A \otimes B \otimes A \otimes B \ar@/_10pc/[rrrrddd]|-{m_{A \otimes B}} \ar[rrdd]|-(0.6){\id_A \otimes T_{B,A} \otimes \id_B} & \numcircledmod{d} & A \otimes A \otimes A \otimes B \otimes B \ar[dd]|-(0.4){\id_A \otimes m_A \otimes \id_B \otimes \id_B} & \numcircledmod{h} & & \numcircledmod{i} & A \otimes A \otimes B \otimes B \otimes B \ar[dd]|-(0.4){\id_A \otimes \id_A \otimes m_B \otimes \id_B} & \numcircledmod{g} & A \otimes B \otimes A \otimes B \ar[lldd]|-(0.6){\id_A \otimes T_{B,A} \otimes \id_B} \ar@/^10pc/[llllddd]|-{m_{A \otimes B}} \\
& & & & \numcircledmod{j} \\
& \numcircledmod{k} & A \otimes A \otimes B \otimes B \ar[rrd]|-{m_A \otimes m_B} & & & & A \otimes A \otimes B \otimes B \ar[lld]|-{m_A \otimes m_B} & \numcircledmod{l} \\
& & & & A \otimes B
}
\]
\end{landscape}
in which the letters in circles ($\numcircledmod{a}, \numcircledmod{b},
\ldots, \numcircledmod{l}$) are just labels for subdiagrams (such as triangles
and rectangles).

I claim that this diagram is commutative. In order to prove this, we need to
check that each of the $12$ subdiagrams labeled
$\numcircledmod{a},\numcircledmod{b},\ldots,\numcircledmod{l}$ is commutative.
This can be checked as follows:

\begin{itemize}
\item The subdiagram labeled $\numcircledmod{a}$ (a triangle) is commutative
because
\begin{align*}
\id_{A\otimes B}\circ m_{A\otimes B}  &
=\id_{A\otimes B}\circ\left(  \left(  m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \right)  \\
& =\underbrace{\left(  \id_{A\otimes B}\otimes\left(
m_{A}\otimes m_{B}\right)  \right)  }_{=\id_{A\otimes
B}\otimes m_{A}\otimes m_{B}}\circ\underbrace{\left(  \operatorname*{id}%
\nolimits_{A\otimes B}\otimes\left(  \id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  \right)  }%
_{\substack{=\id_{A\otimes B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\otimes\id_{B}%
\\=\id_{A}\otimes\id_{B}%
\otimes\id_{A}\otimes T_{B,A}\otimes\operatorname*{id}%
\nolimits_{B}}}\\
& =\left(  \id_{A\otimes B}\otimes m_{A}\otimes
m_{B}\right)  \circ\left(  \id_{A}\otimes
\id_{B}\otimes\id_{A}\otimes
T_{B,A}\otimes\id_{B}\right)  .
\end{align*}


\item The subdiagram labeled $\numcircledmod{b}$ is commutative for a similar reason.

\item The subdiagram labeled $\numcircledmod{c}$ is commutative by a trivial
computation (indeed, both maps $\left(  \id_{A}\otimes
T_{B,A\otimes A}\otimes\id_{B}\otimes
\id_{B}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{A}\otimes\id_{B}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B,A}\otimes\id_{B}\right)  $
and $\left(  \id_{A}\otimes\operatorname*{id}%
\nolimits_{A}\otimes T_{B\otimes B,A}\otimes\id_{B}%
\right)  \circ\left(  \id_{A}\otimes T_{B,A}%
\otimes\id_{B}\otimes\id_{A}%
\otimes\id_{B}\right)  $ send a pure tensor $a\otimes
b\otimes a^{\prime}\otimes b^{\prime}\otimes a^{\prime\prime}\otimes
b^{\prime\prime}\in A\otimes B\otimes A\otimes B\otimes
A\otimes B$ to $a\otimes a^{\prime}\otimes
a^{\prime\prime}\otimes b\otimes b^{\prime}\otimes b^{\prime\prime}$, and
therefore are identical).

\item The subdiagram labeled $\numcircledmod{d}$ is commutative [...]
\end{itemize}

[To be continued!]
\end{commentedout}

(b) Exercise \ref{exe.tensor-prod-algs}(b) is the ``dual'' statement to
Exercise \ref{exe.tensor-prod-algs}(a).
We shall sketch two solutions to it: one solution by elementwise computation
(similar to the first solution to Exercise \ref{exe.tensor-prod-algs}(a), but
somewhat more complicated due to the many sums involved), and another by
formally manipulating homomorphisms. The second solution will be very brief,
because we will not elaborate on it; we will merely explain how it can be
obtained by ``reversing arrows'' from the
second solution to Exercise \ref{exe.tensor-prod-algs}(a).

\textit{First solution to Exercise \ref{exe.tensor-prod-algs}(b).} Let
$\theta$ be the canonical $\kk$-module isomorphism $\kk%
\otimes\kk\rightarrow\kk$. We have defined the comultiplication
map $\Delta_{C\otimes D}$ of the $\kk$-coalgebra $C\otimes D$ by%
\begin{equation}
\Delta_{C\otimes D}=\left(  \id_{C}\otimes
T\otimes\id_{D}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{D}\right)  , \label{sol.exe.tensor-prod-algs.b.1.DelCD=}%
\end{equation}
and the counit map $\epsilon_{C\otimes D}$ of the $\kk$-coalgebra
$C\otimes D$ by%
\[
\epsilon_{C\otimes D}=\theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  .
\]
We now must prove that the $\kk$-module $C\otimes D$, equipped with
these two maps $\Delta_{C\otimes D}$ and $\epsilon_{C\otimes D}$, is indeed a
$\kk$-coalgebra. In other words, we must show that the two diagrams%
\begin{equation}
\xymatrix{ & C \otimes D \otimes C \otimes D \otimes C \otimes D \arback[dl]_{\Delta_{C\otimes D} \otimes \id_{C\otimes D}} \arback[dr]^{\id_{C\otimes D} \otimes \Delta_{C\otimes D}} & \\ C \otimes D \otimes C \otimes D \arback[dr]^{\Delta_{C \otimes D}} & & C \otimes D \otimes C \otimes D \arback[dl]_{\Delta_{C \otimes D}} \\ & C \otimes D & }
\label{sol.exe.tensor-prod-algs.b.1.diag1}%
\end{equation}
and%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ C \otimes D \otimes \kk \arback[d]_{\id_{C\otimes D} \otimes \epsilon_{C \otimes D}} & C \otimes D \arback[l] \arback[d]_{\id_{C \otimes D}} \arback[r] & \kk \otimes C \otimes D \arback[d]^{\epsilon_{C \otimes D} \otimes \id_{C \otimes D}} \\ C \otimes D\otimes C \otimes D\arback[r]^-{\Delta_{C \otimes D}} & C \otimes D & C \otimes D \otimes C \otimes D\arback[l]_-{\Delta_{C \otimes D}} }
\label{sol.exe.tensor-prod-algs.b.1.diag2}%
\end{equation}
\footnote{where the maps $C\otimes D\otimes\kk\rightarrow C\otimes D$
and $\kk\otimes C\otimes D\rightarrow C\otimes D$ are the isomorphisms
sending each $a\otimes\lambda\in C\otimes D\otimes\kk$ with $a\in
C\otimes D$ and $\lambda\in\kk$ (resp., each $\lambda\otimes
a\in\kk\otimes C\otimes D$ with $\lambda\in\kk$ and $a\in
C\otimes D$) to $\lambda a$.} commute (since Definition \ref{def.coalgebra}
yields that $C\otimes D$ is a $\kk$-coalgebra if and only if these two
diagrams commute).

We shall only prove that the diagram (\ref{sol.exe.tensor-prod-algs.b.1.diag1}%
) commutes. The commutativity of the diagram
(\ref{sol.exe.tensor-prod-algs.b.1.diag2}) is proven similarly (but with less
work), and so is left to the reader.

So we need to prove that the diagram (\ref{sol.exe.tensor-prod-algs.b.1.diag1}%
) commutes. In other words, we need to prove that%
\[
\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \circ\Delta_{C\otimes D}=\left(  \operatorname*{id}%
\nolimits_{C\otimes D}\otimes\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes
D}.
\]


We shall use the Sweedler notation, by writing $\sum_{\left(  x\right)  }%
x_{1}\otimes x_{2}$ for $\Delta\left(  x\right)  $ whenever $x$ is an element
of a coalgebra. This is a neat opportunity to practice the use of the
Sweedler notation. But if you are uncomfortable with the Sweedler notation,
you can easily exorcise it from the following argument as follows:

\begin{itemize}
\item Whenever an element $e\in C$ is defined, fix a decomposition $\Delta
_{C}\left(  e\right)  =\sum_{a=1}^{b}r_{a}\otimes s_{a}$ of $\Delta_{C}\left(
e\right)  $ into a sum of pure tensors.

\item Whenever an element $f\in D$ is defined, fix a decomposition $\Delta
_{D}\left(  f\right)  =\sum_{a^{\prime}=1}^{b^{\prime}}r_{a^{\prime}}^{\prime
}\otimes s_{a^{\prime}}^{\prime}$ of $\Delta_{D}\left(  f\right)  $ into a sum
of pure tensors.

\item Whenever an element $c\in C$ is defined, fix a decomposition $\Delta
_{C}\left(  c\right)  =\sum_{i=1}^{n}p_{i}\otimes q_{i}$ of $\Delta_{C}\left(
c\right)  $ into a sum of pure tensors, and furthermore:

\begin{itemize}
\item For each $i\in\left\{  1,2,\ldots,n\right\}  $, fix a decomposition
$\Delta_{C}\left(  p_{i}\right)  =\sum_{j=1}^{k_{i}}p_{i,j}^{\prime}\otimes
p_{i,j}^{\prime\prime}$ of $\Delta_{C}\left(  p_{i}\right)  $ into a sum of
pure tensors.

\item For each $i\in\left\{  1,2,\ldots,n\right\}  $, fix a decomposition
$\Delta_{C}\left(  q_{i}\right)  =\sum_{h=1}^{\ell_{i}}q_{i,h}^{\prime}\otimes
q_{i,h}^{\prime\prime}$ of $\Delta_{C}\left(  q_{i}\right)  $ into a sum of
pure tensors.
\end{itemize}

\item Whenever an element $d\in D$ is defined, fix a decomposition $\Delta
_{D}\left(  d\right)  =\sum_{i^{\prime}=1}^{n^{\prime}}x_{i^{\prime}}\otimes
y_{i^{\prime}}$ of $\Delta_{D}\left(  d\right)  $ into a sum of pure tensors,
and furthermore:

\begin{itemize}
\item For each $i^{\prime}\in\left\{  1,2,\ldots,n^{\prime}\right\}  $, fix a
decomposition $\Delta_{D}\left(  x_{i^{\prime}}\right)  =\sum_{j^{\prime}%
=1}^{k_{i^{\prime}}^{\prime}}x_{i^{\prime},j^{\prime}}^{\prime}\otimes
x_{i^{\prime},j^{\prime}}^{\prime\prime}$ of $\Delta_{D}\left(  x_{i^{\prime}%
}\right)  $ into a sum of pure tensors.

\item For each $i^{\prime}\in\left\{  1,2,\ldots,n^{\prime}\right\}  $, fix a
decomposition $\Delta_{D}\left(  y_{i^{\prime}}\right)  =\sum_{h^{\prime}%
=1}^{\ell_{i^{\prime}}^{\prime}}y_{i^{\prime},h^{\prime}}^{\prime}\otimes
y_{i^{\prime},h^{\prime}}^{\prime\prime}$ of $\Delta_{D}\left(  y_{i^{\prime}%
}\right)  $ into a sum of pure tensors.
\end{itemize}
\end{itemize}

Once these decompositions are chosen, it remains to replace each appearance of
one of the symbols%
\begin{align*}
&  \sum_{\left(  e\right)  },\ \ \ e_{1},\ \ \ e_{2},\ \ \ \sum_{\left(
f\right)  },\ \ \ f_{1},\ \ \ f_{2},\\
&  \sum_{\left(  c\right)  },\ \ \ c_{1},\ \ \ c_{2},\ \ \ \sum_{\left(
c_{1}\right)  },\ \ \ \left(  c_{1}\right)  _{1},\ \ \ \left(  c_{1}\right)
_{2},\ \ \ \sum_{\left(  c_{2}\right)  },\ \ \ \left(  c_{2}\right)
_{1},\ \ \ \left(  c_{2}\right)  _{2},\\
&  \sum_{\left(  d\right)  },\ \ \ d_{1},\ \ \ d_{2},\ \ \ \sum_{\left(
d_{1}\right)  },\ \ \ \left(  d_{1}\right)  _{1},\ \ \ \left(  d_{1}\right)
_{2},\ \ \ \sum_{\left(  d_{2}\right)  },\ \ \ \left(  d_{2}\right)
_{1},\ \ \ \left(  d_{2}\right)  _{2}%
\end{align*}
by the symbol%
\begin{align*}
&  \sum_{a=1}^{b},\ \ \ r_{a},\ \ \ s_{a},\ \ \ \sum_{a^{\prime}=1}%
^{b^{\prime}},\ \ \ r_{a^{\prime}}^{\prime},\ \ \ s_{a^{\prime}}^{\prime},\\
&  \sum_{i=1}^{n},\ \ \ p_{i},\ \ \ q_{i},\ \ \ \sum_{j=1}^{k_{i}%
},\ \ \ p_{i,j}^{\prime},\ \ \ p_{i,j}^{\prime\prime},\ \ \ \sum_{h=1}%
^{\ell_{i}},\ \ \ q_{i,h}^{\prime},\ \ \ q_{i,h}^{\prime\prime},\\
&  \sum_{i^{\prime}=1}^{n^{\prime}},\ \ \ x_{i^{\prime}},\ \ \ y_{i^{\prime}%
},\ \ \ \sum_{j^{\prime}=1}^{k_{i^{\prime}}^\prime},\ \ \ x_{i^{\prime},j^{\prime}%
}^{\prime},\ \ \ x_{i^{\prime},j^{\prime}}^{\prime\prime},\ \ \ %
\sum_{h^{\prime}=1}^{\ell_{i^{\prime}}^\prime},\ \ \ %
y_{i^{\prime},h^{\prime}}^{\prime},\ \ \ %
y_{i^{\prime},h^{\prime}}^{\prime\prime},
\end{align*}
respectively\footnote{Some care must be taken here: For example, if the symbol
``$c_{1}$'' appears inside ``$\left(  c_{1}\right)  _{1}$'', then it
should not be replaced by ``$p_{i}$'', but instead the whole
``$\left(  c_{1}\right)  _{1}$'' should be
replaced by ``$p_{i,j}^{\prime}$''.}. For
example, these replacements transform the expression%
\[
\sum_{\left(  c\right)  }\sum_{\left(  c_{2}\right)  }\sum_{\left(  d\right)
}\sum_{\left(  d_{2}\right)  }c_{1}\otimes d_{1}\otimes\left(  c_{2}\right)
_{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(  c_{2}\right)  _{2}%
\otimes\left(  d_{2}\right)  _{2}%
\]
into%
\[
\sum_{i=1}^{n}\sum_{h=1}^{\ell_{i}}\sum_{i^{\prime}=1}^{n^{\prime}}%
\sum_{h^{\prime}=1}^{\ell_{i^{\prime}}^\prime}p_{i}\otimes x_{i^{\prime}}\otimes
q_{i,h}^{\prime}\otimes y_{i^{\prime},h^{\prime}}^{\prime}\otimes
q_{i,h}^{\prime\prime}\otimes y_{i^{\prime},h^{\prime}}^{\prime\prime}.
\]
Once these replacements are all done, the argument we give below becomes a
perfectly valid argument that does not use the Sweedler notation.

So let us come to the actual argument.

We first notice that%
\begin{equation}
\Delta_{C\otimes D}\left(  e\otimes f\right)  =\sum_{\left(  e\right)  }%
\sum_{\left(  f\right)  }e_{1}\otimes f_{1}\otimes e_{2}\otimes f_{2}
\label{sol.exe.tensor-prod-algs.b.1.m1}%
\end{equation}
for every $e\in C$ and $f\in D$.

[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.b.1.m1}):} Let $e\in C$ and
$f\in D$. Applying both sides of the equality
(\ref{sol.exe.tensor-prod-algs.b.1.DelCD=}) to $e\otimes f$, we obtain
\begin{align*}
\Delta_{C\otimes D}\left(  e\otimes f\right)   &  =\left(  \left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \right)  \left(  e\otimes
f\right)  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}%
_{D}\right)  \underbrace{\left(  \left(  \Delta_{C}\otimes\Delta_{D}\right)
\left(  e\otimes f\right)  \right)  }_{=\Delta_{C}\left(  e\right)
\otimes\Delta_{D}\left(  f\right)  }\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\Delta_{C}\left(  e\right)  }_{=\sum_{\left(  e\right)
}e_{1}\otimes e_{2}}\otimes\underbrace{\Delta_{D}\left(  f\right)  }%
_{=\sum_{\left(  f\right)  }f_{1}\otimes f_{2}}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\left(  \sum_{\left(  e\right)  }e_{1}\otimes
e_{2}\right)  \otimes\left(  \sum_{\left(  f\right)  }f_{1}\otimes
f_{2}\right)  }_{=\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }%
e_{1}\otimes e_{2}\otimes f_{1}\otimes f_{2}}\right) \\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \sum_{\left(  e\right)  }\sum_{\left(  f\right)  }e_{1}\otimes
e_{2}\otimes f_{1}\otimes f_{2}\right) \\
&  =\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }\underbrace{\left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)  \left(
e_{1}\otimes e_{2}\otimes f_{1}\otimes f_{2}\right)  }_{=\operatorname*{id}%
\nolimits_{C}\left(  e_{1}\right)  \otimes T\left(  e_{2}\otimes f_{1}\right)
\otimes\id_{D}\left(  f_{2}\right)  }\\
&  =\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }%
\underbrace{\id_{C}\left(  e_{1}\right)  }_{=e_{1}%
}\otimes\underbrace{T\left(  e_{2}\otimes f_{1}\right)  }_{\substack{=f_{1}%
\otimes e_{2}\\\text{(by the definition of }T\text{)}}}\otimes
\underbrace{\id_{D}\left(  f_{2}\right)  }_{=f_{2}}\\
&  =\sum_{\left(  e\right)  }\sum_{\left(  f\right)  }e_{1}\otimes
f_{1}\otimes e_{2}\otimes f_{2}.
\end{align*}
This proves (\ref{sol.exe.tensor-prod-algs.b.1.m1}).]

Furthermore, every $c\in C$ satisfies%
\begin{equation}
\sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(  c_{1}\right)
_{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}=\sum_{\left(  c\right)
}\sum_{\left(  c_{2}\right)  }c_{1}\otimes\left(  c_{2}\right)  _{1}%
\otimes\left(  c_{2}\right)  _{2}. \label{sol.exe.tensor-prod-algs.b.1.DelC}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.b.1.DelC}):} Let $c\in C$.
Recall that $C$ is a $\kk$-coalgebra. Thus, the diagram
(\ref{coassociativity-diagram}) commutes (by the definition of a $\kk%
$-coalgebra). In other words, we have $\left(  \Delta_{C}\otimes
\id_{C}\right)  \circ\Delta_{C}=\left(
\id_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}$.
Applying both sides of this equality to $c$, we obtain%
\[
\left(  \left(  \Delta_{C}\otimes\id_{C}\right)
\circ\Delta_{C}\right)  \left(  c\right)  =\left(  \left(  \operatorname*{id}%
\nolimits_{C}\otimes\Delta_{C}\right)  \circ\Delta_{C}\right)  \left(
c\right)  .
\]
In light of
\begin{align*}
\left(  \left(  \Delta_{C}\otimes\id_{C}\right)
\circ\Delta_{C}\right)  \left(  c\right)   &  =\left(  \Delta_{C}%
\otimes\id_{C}\right)  \left(  \underbrace{\Delta
_{C}\left(  c\right)  }_{=\sum_{\left(  c\right)  }c_{1}\otimes c_{2}}\right)
=\left(  \Delta_{C}\otimes\id_{C}\right)  \left(
\sum_{\left(  c\right)  }c_{1}\otimes c_{2}\right) \\
&  =\sum_{\left(  c\right)  }\underbrace{\Delta_{C}\left(  c_{1}\right)
}_{=\sum_{\left(  c_{1}\right)  }\left(  c_{1}\right)  _{1}\otimes\left(
c_{1}\right)  _{2}}\otimes\underbrace{\id_{C}\left(
c_{2}\right)  }_{=c_{2}}=\sum_{\left(  c\right)  }\left(  \sum_{\left(
c_{1}\right)  }\left(  c_{1}\right)  _{1}\otimes\left(  c_{1}\right)
_{2}\right)  \otimes c_{2}\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(
c_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}%
\end{align*}
and%
\begin{align*}
\left(  \left(  \id_{C}\otimes\Delta_{C}\right)
\circ\Delta_{C}\right)  \left(  c\right)   &  =\left(  \operatorname*{id}%
\nolimits_{C}\otimes\Delta_{C}\right)  \left(  \underbrace{\Delta_{C}\left(
c\right)  }_{=\sum_{\left(  c\right)  }c_{1}\otimes c_{2}}\right)  =\left(
\id_{C}\otimes\Delta_{C}\right)  \left(  \sum_{\left(
c\right)  }c_{1}\otimes c_{2}\right) \\
&  =\sum_{\left(  c\right)  }\underbrace{\id_{C}%
\left(  c_{1}\right)  }_{=c_{1}}\otimes\underbrace{\Delta_{C}\left(
c_{2}\right)  }_{=\sum_{\left(  c_{2}\right)  }\left(  c_{2}\right)
_{1}\otimes\left(  c_{2}\right)  _{2}}=\sum_{\left(  c\right)  }c_{1}%
\otimes\left(  \sum_{\left(  c_{2}\right)  }\left(  c_{2}\right)  _{1}%
\otimes\left(  c_{2}\right)  _{2}\right) \\
&  =\sum_{\left(  c\right)  }\sum_{\left(  c_{2}\right)  }c_{1}\otimes\left(
c_{2}\right)  _{1}\otimes\left(  c_{2}\right)  _{2},
\end{align*}
this rewrites as%
\[
\sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(  c_{1}\right)
_{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}=\sum_{\left(  c\right)
}\sum_{\left(  c_{2}\right)  }c_{1}\otimes\left(  c_{2}\right)  _{1}%
\otimes\left(  c_{2}\right)  _{2}.
\]
This proves (\ref{sol.exe.tensor-prod-algs.b.1.DelC}).]

Also, every $d\in D$ satisfies%
\begin{equation}
\sum_{\left(  d\right)  }\sum_{\left(  d_{1}\right)  }\left(  d_{1}\right)
_{1}\otimes\left(  d_{1}\right)  _{2}\otimes d_{2}=\sum_{\left(  d\right)
}\sum_{\left(  d_{2}\right)  }d_{1}\otimes\left(  d_{2}\right)  _{1}%
\otimes\left(  d_{2}\right)  _{2}. \label{sol.exe.tensor-prod-algs.b.1.DelD}%
\end{equation}


[\textit{Proof of (\ref{sol.exe.tensor-prod-algs.b.1.DelD}):} This proof is
analogous to the proof of (\ref{sol.exe.tensor-prod-algs.b.1.DelC}).]

Let $z\in C\otimes D$. Thus, $z$ (like any tensor in $C\otimes D$) must be a
$\kk$-linear combination of pure tensors.

We want to show the equality%
\begin{equation}
\left(  \left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  =\left(  \left(  \id_{C\otimes D}%
\otimes\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  . \label{sol.exe.tensor-prod-algs.b.1.eq-x1}%
\end{equation}
This equality is $\kk$-linear in $z$ (since all maps that appear in it
are $\kk$-linear). Hence, we can WLOG assume that $z$ is a pure tensor
(since $z$ is a $\kk$-linear combination of pure tensors). Assume this.
Thus, $z=c\otimes d$ for some $c\in C$ and $d\in D$. Consider these $c$ and
$d$.

Taking the tensor product of the equalities
(\ref{sol.exe.tensor-prod-algs.b.1.DelC}) and
(\ref{sol.exe.tensor-prod-algs.b.1.DelD}), we obtain%
\begin{align*}
&  \left(  \sum_{\left(  c\right)  }\sum_{\left(  c_{1}\right)  }\left(
c_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}\right)
\otimes\left(  \sum_{\left(  d\right)  }\sum_{\left(  d_{1}\right)  }\left(
d_{1}\right)  _{1}\otimes\left(  d_{1}\right)  _{2}\otimes d_{2}\right) \\
&  =\left(  \sum_{\left(  c\right)  }\sum_{\left(  c_{2}\right)  }c_{1}%
\otimes\left(  c_{2}\right)  _{1}\otimes\left(  c_{2}\right)  _{2}\right)
\otimes\left(  \sum_{\left(  d\right)  }\sum_{\left(  d_{2}\right)  }%
d_{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(  d_{2}\right)
_{2}\right)  .
\end{align*}
In other words,%
\begin{align}
&  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  c_{1}\right)  _{2}\otimes c_{2}\otimes\left(  d_{1}\right)
_{1}\otimes\left(  d_{1}\right)  _{2}\otimes d_{2}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{2}\right)  }\sum_{\left(  d_{2}\right)  }c_{1}\otimes\left(  c_{2}\right)
_{1}\otimes\left(  c_{2}\right)  _{2}\otimes d_{1}\otimes\left(  d_{2}\right)
_{1}\otimes\left(  d_{2}\right)  _{2}. \label{sol.exe.tensor-prod-algs.b.1.3}%
\end{align}
Applying the $\kk$-linear map%
\begin{align*}
C\otimes C\otimes C\otimes D\otimes D\otimes D  &  \rightarrow C\otimes
D\otimes C\otimes D\otimes C\otimes D,\\
\gamma_{1}\otimes\gamma_{2}\otimes\gamma_{3}\otimes\delta_{1}\otimes\delta
_{2}\otimes\delta_{3}  &  \mapsto\gamma_{1}\otimes\delta_{1}\otimes\gamma
_{2}\otimes\delta_{2}\otimes\gamma_{3}\otimes\delta_{3}%
\end{align*}
to both sides of this equality, we obtain%
\begin{align}
&  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\otimes c_{2}\otimes d_{2}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{2}\right)  }\sum_{\left(  d_{2}\right)  }c_{1}\otimes d_{1}\otimes\left(
c_{2}\right)  _{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(
c_{2}\right)  _{2}\otimes\left(  d_{2}\right)  _{2}.
\label{sol.exe.tensor-prod-algs.b.1.4}%
\end{align}


Now,%
\begin{align}
&  \left(  \left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right) \nonumber\\
&  =\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \left(  \Delta_{C\otimes D}\left(  \underbrace{z}_{=c\otimes
d}\right)  \right) \nonumber\\
& =\left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \left(  \underbrace{\Delta_{C\otimes D}\left(
c\otimes d\right)  }_{\substack{=\sum_{\left(  c\right)  }\sum_{\left(
d\right)  }c_{1}\otimes d_{1}\otimes c_{2}\otimes d_{2}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.b.1.m1}) (applied to }e=c\text{ and
}f=d\text{))}}}\right) \nonumber\\
&  =\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \left(  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }%
c_{1}\otimes d_{1}\otimes c_{2}\otimes d_{2}\right)  \nonumber \\
&=\sum_{\left(  c\right)
}\sum_{\left(  d\right)  }\underbrace{\left(  \Delta_{C\otimes D}%
\otimes\id_{C\otimes D}\right)  \left(  c_{1}\otimes
d_{1}\otimes c_{2}\otimes d_{2}\right)  }_{=\Delta_{C\otimes D}\left(
c_{1}\otimes d_{1}\right)  \otimes\id_{C\otimes
D}\left(  c_{2}\otimes d_{2}\right)  }\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\underbrace{\Delta
_{C\otimes D}\left(  c_{1}\otimes d_{1}\right)  }_{\substack{=\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\\\text{(by
(\ref{sol.exe.tensor-prod-algs.b.1.m1}) (applied to }e=c_{1}\text{ and
}f=d_{1}\text{))}}}\otimes\underbrace{\id_{C\otimes
D}\left(  c_{2}\otimes d_{2}\right)  }_{=c_{2}\otimes d_{2}}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\left(  \sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\right)  \otimes c_{2}\otimes d_{2}%
\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{1}\right)  }\sum_{\left(  d_{1}\right)  }\left(  c_{1}\right)  _{1}%
\otimes\left(  d_{1}\right)  _{1}\otimes\left(  c_{1}\right)  _{2}%
\otimes\left(  d_{1}\right)  _{2}\otimes c_{2}\otimes d_{2};
\label{sol.exe.tensor-prod-algs.b.1.5}%
\end{align}
an analogous computation shows that%
\begin{align}
&  \left(  \left(  \id_{C\otimes D}\otimes
\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(  z\right)
\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\sum_{\left(
c_{2}\right)  }\sum_{\left(  d_{2}\right)  }c_{1}\otimes d_{1}\otimes\left(
c_{2}\right)  _{1}\otimes\left(  d_{2}\right)  _{1}\otimes\left(
c_{2}\right)  _{2}\otimes\left(  d_{2}\right)  _{2}.
\label{sol.exe.tensor-prod-algs.b.1.6}%
\end{align}


In light of (\ref{sol.exe.tensor-prod-algs.b.1.5}) and
(\ref{sol.exe.tensor-prod-algs.b.1.6}), the equality
(\ref{sol.exe.tensor-prod-algs.b.1.4}) rewrites as
\[
\left(  \left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  =\left(  \left(  \id_{C\otimes D}%
\otimes\Delta_{C\otimes D}\right)  \circ\Delta_{C\otimes D}\right)  \left(
z\right)  .
\]
Thus, the equality (\ref{sol.exe.tensor-prod-algs.b.1.eq-x1}) is proven.

Now, forget that we fixed $z$. We thus have proven the equality
(\ref{sol.exe.tensor-prod-algs.b.1.eq-x1}) for every $z\in C\otimes D$. In
other words, we have $\left(  \Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}\right)  \circ\Delta_{C\otimes D}=\left(
\id_{C\otimes D}\otimes\Delta_{C\otimes D}\right)
\circ\Delta_{C\otimes D}$. In other words, the diagram
(\ref{sol.exe.tensor-prod-algs.b.1.diag1}) commutes.

It remains to prove that the diagram (\ref{sol.exe.tensor-prod-algs.b.1.diag2}%
) commutes. We leave this to the reader, as the proof is similar to (but
simpler than) the proof of the commutativity of
(\ref{sol.exe.tensor-prod-algs.b.1.diag1}) given above. Hence, both diagrams
(\ref{sol.exe.tensor-prod-algs.b.1.diag1}) and
(\ref{sol.exe.tensor-prod-algs.b.1.diag2}) commute. In other words, the
$\kk$-module $C\otimes D$, equipped with the two maps $\Delta_{C\otimes
D}$ and $\epsilon_{C\otimes D}$, is a $\kk$-coalgebra. In other words,
the $\kk$-coalgebra $C\otimes D$ introduced in Definition
\ref{def.tensor-prod-algs} is actually well-defined. This solves Exercise
\ref{exe.tensor-prod-algs}(b).

\textit{Second solution to Exercise \ref{exe.tensor-prod-algs}(b).} Let
$\theta$ be the canonical $\kk$-module isomorphism $\kk%
\otimes\kk\rightarrow\kk$.

A solution to Exercise \ref{exe.tensor-prod-algs}(b) can now be obtained in a
straightforward fashion from the Second solution to Exercise
\ref{exe.tensor-prod-algs}(a), by making the following modifications:

\begin{enumerate}
\item Replace every appearance of any of the terms%
\[
A,\ B,\ m_{A},\ m_{B},\ m_{A\otimes B},\ u_{A},\ u_{B},\ u_{A\otimes B}%
,\ \xi,\ \operatorname*{kan}\nolimits_{1,U},\ \operatorname*{kan}%
\nolimits_{2,U},\ T_{U,V}%
\]
(for any $\kk$-modules $U$ and $V$) by%
\[
C,\ D,\ \Delta_{C},\ \Delta_{D},\ \Delta_{C\otimes D},\ u_{C},\ u_{D}%
,\ u_{C\otimes D},\ \theta,\ \operatorname*{kan}\nolimits_{1,U}^{-1}%
,\ \operatorname*{kan}\nolimits_{2,U}^{-1},\ T_{V,U},
\]
respectively.

(For example, the equation (\ref{sol.exe.tensor-prod-algs.a.2.1}) becomes%
\begin{equation}
\left(  \id_{C}\otimes T_{C,D}\otimes
\id_{D}\right)  \circ\left(  \Delta_{C\otimes
D}\otimes\id_{C\otimes D}\right)  =\left(  \Delta
_{C}\otimes\id_{C}\otimes\Delta_{D}\otimes
\id_{D}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{C}\otimes Q\otimes\id_{D}\right)  .
\label{sol.exe.tensor-prod-algs.b.2.1.wrong}%
\end{equation}
This new equation makes no sense, because (for example) the maps
$\id_{C}\otimes T_{D,C}\otimes\operatorname*{id}%
\nolimits_{D}$ and $\Delta_{C\otimes D}\otimes\operatorname*{id}%
\nolimits_{C\otimes D}$ cannot be composed; but this is fine, since we shall
make a further modification which will turn this equation into a meaningful one.

For another example, the map $Q:B\otimes A\otimes B\otimes A\rightarrow
A\otimes A\otimes B\otimes B$ becomes a map $Q:D\otimes C\otimes D\otimes
C\rightarrow C\otimes C\otimes D\otimes D$.)

\item Reverse the direction of all arrows\footnote{This includes both the
arrows in the description of maps and the arrows in commutative diagrams.}. In
other words, any map which used to go from a set $X$ to a set $Y$ shall now go
from $Y$ to $X$.

(For example, the map $Q:D\otimes C\otimes D\otimes C\rightarrow C\otimes
C\otimes D\otimes D$ becomes a map $Q:C\otimes C\otimes D\otimes D\rightarrow
D\otimes C\otimes D\otimes C$.)

\item In any composition of $\kk$-linear maps, reverse the order of the
maps being composed. In other words, replace any composition $f_{1}\circ
f_{2}\circ\cdots\circ f_{k}$ by $f_{k}\circ f_{k-1}\circ\cdots\circ f_{1}$.

(For example, the meaningless equality
(\ref{sol.exe.tensor-prod-algs.b.2.1.wrong}) thus becomes%
\[
\left(  \Delta_{C\otimes D}\otimes\id_{C\otimes
D}\right)  \circ\left(  \id_{C}\otimes T_{C,D}%
\otimes\id_{D}\right)  =\left(  \operatorname*{id}%
\nolimits_{C}\otimes Q\otimes\id_{D}\right)
\circ\left(  \Delta_{C}\otimes\id_{C}\otimes\Delta
_{D}\otimes\id_{D}\right)  .
\]
This equality is meaningful and correct.)

\item Any part of our solution that involved elements of $A$ and $B$ (as
opposed to mere computations with maps) must be redone from scratch. For
example, the proof of (\ref{sol.exe.tensor-prod-algs.a.2.Q1}) involved
elements of $A$ and $B$ (because we picked $b\in B$, $a\in A$, $b^{\prime}\in
B$ and $a^{\prime}\in A$ in that proof), and thus must be redone, whereas the
proof of (\ref{sol.exe.tensor-prod-algs.a.2.2}) did not involve elements of
$A$ and $B$ and therefore needs not be modified any further.

Fortunately, very few parts of our solution involved elements of $A$ and $B$.
To wit, these parts are the proofs of the equalities
(\ref{sol.exe.tensor-prod-algs.a.2.Q1}),
(\ref{sol.exe.tensor-prod-algs.a.2.Q2}),
(\ref{sol.exe.tensor-prod-algs.a.2.k1}) and
(\ref{sol.exe.tensor-prod-algs.a.2.k2}). After the above modifications, these
equalities have become%
\begin{align*}
&  Q=\left(  T_{C,D}\otimes\id_{D}\otimes
\id_{C}\right)  \circ\left(  \operatorname*{id}%
\nolimits_{C}\otimes T_{C,D\otimes D}\right)  ,\\
&  Q=\left(  \id_{D}\otimes\operatorname*{id}%
\nolimits_{C}\otimes T_{C,D}\right)  \circ\left(  T_{C\otimes C,D}%
\otimes\id_{D}\right)  ,\\
&  \operatorname*{kan}\nolimits_{1,C\otimes D}^{-1}\circ\left(
\id_{C\otimes D}\otimes\theta\right)  \circ\left(
\id_{C}\otimes T_{\kk,D}\otimes
\id_{\kk}\right)  =\operatorname*{kan}%
\nolimits_{1,C}^{-1}\otimes\operatorname*{kan}\nolimits_{1,D}^{-1}%
,\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \operatorname*{kan}\nolimits_{2,C\otimes D}^{-1}\circ\left(  \theta
\otimes\id_{C\otimes D}\right)  \circ\left(
\id_{\kk}\otimes T_{C,\kk}%
\otimes\id_{D}\right)  =\operatorname*{kan}%
\nolimits_{2,C}^{-1}\otimes\operatorname*{kan}\nolimits_{2,D}^{-1},
\end{align*}
respectively. So these four equalities must be proven. Fortunately, these
proofs are completely straightforward\footnote{These equalities are properties
of tensor products of $\kk$-modules; they make no use of the coalgebra
structures on $C$ and $D$.}; the reader can easily come up with them.
\end{enumerate}

These four modifications are sufficient to transform our Second solution to
Exercise \ref{exe.tensor-prod-algs}(a) into a solution to Exercise
\ref{exe.tensor-prod-algs}(b). Thus, Exercise \ref{exe.tensor-prod-algs}(b) is
solved again.

[\textit{Remark:} The second and the third modifications made above are
usually subsumed under the concept of ``reversing all
arrows''.]
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.tensorprod.morphism}}

\textit{Solution to Exercise~\ref{exe.tensorprod.morphism}.}

(b) Let $C$, $C^{\prime}$, $D$ and $D^{\prime}$ be four $\kk$-coalgebras.
Let $f:C\rightarrow C^{\prime}$ and $g:D\rightarrow D^{\prime}$
be two $\kk$-coalgebra homomorphisms. We need to prove that $f\otimes
g:C\otimes D\rightarrow C^{\prime}\otimes D^{\prime}$ is a
$\kk$-coalgebra homomorphism.

Recall that (by the definition of a ``$\kk$-coalgebra
homomorphism'') the map $f\otimes g:C\otimes D\rightarrow
C^{\prime}\otimes D^{\prime}$ is a $\kk$-coalgebra homomorphism if and
only if the two diagrams
\begin{equation}
\xymatrixcolsep{5pc}
\xymatrix{ C \otimes D \ar[d]_{\Delta_{C \otimes D}} \ar[r]^{f \otimes g} & C^\prime \otimes D^\prime\ar[d]^{\Delta_{C^\prime\otimes D^\prime}} \\ \left(C \otimes D\right) \otimes\left(C \otimes D\right) \ar[r]^-{\left (f \otimes g\right) \otimes\left(f \otimes g\right)} & \left(C^\prime\otimes D^\prime\right) \otimes\left(C^\prime\otimes D^\prime\right) }
\label{sol.tensorprod.morphism.a.goal1}%
\end{equation}
and
\begin{equation}
\xymatrix{ C \otimes D \ar[dr]_{\epsilon_{C \otimes D}} \ar[rr]^{f \otimes g} & & C^\prime\otimes D^\prime\ar[dl]^{\epsilon_{C^\prime\otimes D^\prime}} \\ & \kk& }
\label{sol.tensorprod.morphism.a.goal2}%
\end{equation}
commute. We shall now prove that these diagrams indeed commute.

We know that the map $f:C\rightarrow C^{\prime}$ is a $\kk$-coalgebra
homomorphism. By the definition of a $\kk$-coalgebra homomorphism, this
means that the two diagrams%
\begin{equation}
\xymatrixcolsep{5pc}\xymatrix{ C \ar[d]_{\Delta_C} \ar[r]^{f} & C^\prime\ar[d]^{\Delta_{C^\prime}} \\ C \otimes C \ar[r]^-{f \otimes f} & C^\prime\otimes C^\prime}\label{sol.tensorprod.morphism.a.f1}%
\end{equation}
and
\begin{equation}
\xymatrix{ C \ar[dr]_{\epsilon_C} \ar[rr]^{f} & & C^\prime\ar[dl]^{\epsilon_{C^\prime}} \\ & \kk& }\label{sol.tensorprod.morphism.a.f2}%
\end{equation}
commute.

For any two $\kk$-modules $U$ and $V$, let $T_{U,V}:U\otimes
V\rightarrow V\otimes U$ be the twist map (i.e., the $\kk$-linear map
$U\otimes V\rightarrow V\otimes U$ sending every $u\otimes v$ to $v\otimes
u$). A simple linear-algebraic fact says that if $U$, $V$, $U^{\prime}$ and
$V^{\prime}$ are four $\kk$-modules and $x:U\rightarrow U^{\prime}$ and
$y:V\rightarrow V^{\prime}$ are two $\kk$-linear maps, then
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}=T_{U^{\prime},V^{\prime}}\circ\left(
x\otimes y\right)  . \label{sol.tensorprod.morphism.a.twist}%
\end{equation}


The definition of the $\kk$-coalgebra $C\otimes D$ yields
\begin{equation}
\Delta_{C\otimes D}=\left(   \id_{C}\otimes T_{C,D}%
\otimes \id_{D}\right)  \circ\left(  \Delta_{C}\otimes\Delta
_{D}\right)  . \label{sol.tensorprod.morphism.a.Delta1}%
\end{equation}
Similarly,
\begin{equation}
\Delta_{C^{\prime}\otimes D^{\prime}} = \left(   \id_{C^{\prime}%
}\otimes T_{C^{\prime},D^{\prime}}\otimes \id_{D^{\prime}%
}\right)  \circ\left(  \Delta_{C^{\prime}}\otimes\Delta_{D^{\prime}}\right)  .
\label{sol.tensorprod.morphism.a.Delta2}%
\end{equation}
But $\left(  f\otimes f\right)  \circ\Delta_{C}=\Delta_{C^{\prime}}\circ f$
(since the diagram \eqref{sol.tensorprod.morphism.a.f1} commutes), and
similarly $\left(  g\otimes g\right)  \circ\Delta_{D}=\Delta_{D^{\prime}}\circ
g$.

Now,
\begin{align*}
&  \underbrace{\left(  \left(  f\otimes g\right)  \otimes\left(  f\otimes
g\right)  \right)  }_{=f\otimes\left(  g\otimes f\right)  \otimes g}%
\circ\underbrace{\Delta_{C\otimes D}}_{=\left(   \id_{C}\otimes
T_{C,D}\otimes \id_{D}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{D}\right)  }\\
&  =\underbrace{\left(  f\otimes\left(  g\otimes f\right)  \otimes g\right)
\circ\left(   \id_{C}\otimes T_{C,D}\otimes\id_{D} \right)
 }_{=\left(  f\circ \id_{C}\right)  \otimes\left(
\left(  g\otimes f\right)  \circ T_{C,D}\right)  \otimes\left(  g\circ
 \id_{D}\right)  }\circ\left(  \Delta_{C}\otimes\Delta
_{D}\right)  \\
&  =\left(  \underbrace{\left(  f\circ \id_{C}\right)
}_{=f= \id_{C^{\prime}}\circ f}\otimes\underbrace{\left(  \left(
g\otimes f\right)  \circ T_{C,D}\right)  }_{\substack{=T_{C^{\prime}%
,D^{\prime}}\circ\left(  f\otimes g\right)  \\\text{(by
\eqref{sol.tensorprod.morphism.a.twist})}}}\otimes\underbrace{\left(
g\circ \id_{D}\right)  }_{=g= \id_{D^{\prime}}\circ
g}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \\
&  =\underbrace{\left(  \left(   \id_{C^{\prime}}\circ f\right)
\otimes\left(  T_{C^{\prime},D^{\prime}}\circ\left(  f\otimes g\right)
\right)  \otimes\left(   \id_{D^{\prime}}\circ g\right)  \right)
}_{=\left(   \id_{C^{\prime}}\otimes T_{C^{\prime},D^{\prime}%
}\otimes \id_{D^{\prime}}\right)  \circ\left(  f\otimes\left(
f\otimes g\right)  \otimes g\right)  }\circ\left(  \Delta_{C}\otimes\Delta
_{D}\right)  \\
&  =\left(   \id_{C^{\prime}}\otimes T_{C^{\prime},D^{\prime}%
}\otimes \id_{D^{\prime}}\right)  \circ\underbrace{\left(
f\otimes\left(  f\otimes g\right)  \otimes g\right)  }_{=\left(  f\otimes
f\right)  \otimes\left(  g\otimes g\right)  }\circ\left(  \Delta_{C}%
\otimes\Delta_{D}\right)  \\
&  =\left(   \id_{C^{\prime}}\otimes T_{C^{\prime},D^{\prime}%
}\otimes \id_{D^{\prime}}\right)  \circ\underbrace{\left(
\left(  f\otimes f\right)  \otimes\left(  g\otimes g\right)  \right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  }_{\substack{=\left(  \left(
f\otimes f\right)  \circ\Delta_{C}\right)  \otimes\left(  \left(  g\otimes
g\right)  \circ\Delta_{D}\right)  }}\\
&  =\left(   \id_{C^{\prime}}\otimes T_{C^{\prime},D^{\prime}%
}\otimes \id_{D^{\prime}}\right)  \circ\left(
\underbrace{\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)
}_{=\Delta_{C^{\prime}}\circ f}\otimes\underbrace{\left(  \left(  g\otimes
g\right)  \circ\Delta_{D}\right)  }_{=\Delta_{D^{\prime}}\circ g}\right)  \\
&  =\left(   \id_{C^{\prime}}\otimes T_{C^{\prime},D^{\prime}%
}\otimes \id_{D^{\prime}}\right)  \circ\underbrace{\left(
\left(  \Delta_{C^{\prime}}\circ f\right)  \otimes\left(  \Delta_{D^{\prime}%
}\circ g\right)  \right)  }_{=\left(  \Delta_{C^{\prime}}\otimes
\Delta_{D^{\prime}}\right)  \circ\left(  f\otimes g\right)  }\\
&  =\underbrace{\left(   \id_{C^{\prime}}\otimes T_{C^{\prime
},D^{\prime}}\otimes \id_{D^{\prime}}\right)  \circ\left(
\Delta_{C^{\prime}}\otimes\Delta_{D^{\prime}}\right)  }_{\substack{=\Delta
_{C^{\prime}\otimes D^{\prime}}\\\text{(by
\eqref{sol.tensorprod.morphism.a.Delta2})}}}\circ\left(  f\otimes g\right)
=\Delta_{C^{\prime}\otimes D^{\prime}}\circ\left(  f\otimes g\right)  .
\end{align*}
In other words, the diagram \eqref{sol.tensorprod.morphism.a.goal1} commutes.

We have $\epsilon_{C^{\prime}}\circ f=\epsilon_{C}$ (since the diagram
\eqref{sol.tensorprod.morphism.a.f2} commutes) and $\epsilon_{D^{\prime}}\circ
g=\epsilon_{D}$ (similarly).

Now, let $\theta$ be the canonical $\kk$-module isomorphism $\kk 
\otimes \kk  \rightarrow \kk$. Then, the definition of the
$\kk$-coalgebra $C\otimes D$ yields
\begin{equation}
\epsilon_{C\otimes D}=\theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  . \label{sol.tensorprod.morphism.a.epsilon1}%
\end{equation}
Similarly,
\begin{equation}
\epsilon_{C^{\prime}\otimes D^{\prime}}=\theta\circ\left(  \epsilon
_{C^{\prime}}\otimes\epsilon_{D^{\prime}}\right)  .
\label{sol.tensorprod.morphism.a.epsilon2}%
\end{equation}


Now,
\begin{align*}
\underbrace{\epsilon_{C^{\prime}\otimes D^{\prime}}}_{=\theta\circ\left(
\epsilon_{C^{\prime}}\otimes\epsilon_{D^{\prime}}\right)  }\circ\left(
f\otimes g\right)   &  = \theta\circ\underbrace{\left(  \epsilon_{C^{\prime}%
}\otimes\epsilon_{D^{\prime}}\right)  \circ\left(  f\otimes g\right)
}_{=\left(  \epsilon_{C^{\prime}}\circ f\right)  \otimes\left(
\epsilon_{D^{\prime}}\circ g\right)}\\
&  =\theta\circ\left(  \underbrace{\left(  \epsilon_{C^{\prime}}\circ
f\right)  }_{=\epsilon_{C}}\otimes\underbrace{\left(  \epsilon_{D^{\prime}%
}\circ g\right)  }_{=\epsilon_{D}}\right)  =\theta\circ\left(  \epsilon_{C}
\otimes\epsilon_{D}\right)  =\epsilon_{C\otimes D}%
\end{align*}
(by \eqref{sol.tensorprod.morphism.a.epsilon1}). In other words, the diagram
\eqref{sol.tensorprod.morphism.a.goal2} commutes.

We now know that the two diagrams \eqref{sol.tensorprod.morphism.a.goal1} and
\eqref{sol.tensorprod.morphism.a.goal2} commute. Thus, the map $f\otimes
g:C\otimes D\rightarrow C^{\prime}\otimes D^{\prime}$ is a $\kk %
$-coalgebra homomorphism (because we know that the map $f \otimes g : C\otimes
D \rightarrow C^{\prime} \otimes D^{\prime}$ is a $\kk$-coalgebra
homomorphism if and only if the two diagrams
\eqref{sol.tensorprod.morphism.a.goal1} and
\eqref{sol.tensorprod.morphism.a.goal2} commute). This solves Exercise
\ref{exe.tensorprod.morphism}(b).

(a) In order to obtain a solution to Exercise~\ref{exe.tensorprod.morphism}%
(a), it is enough to reverse all arrows in the above solution to
Exercise~\ref{exe.tensorprod.morphism}(b) (and, of course, replace $C$, $D$,
$\Delta_{C}$ etc. by $A$, $B$, $m_{A}$ etc.).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.tensorprod.morphism}}

\textit{Solution to Exercise~\ref{exe.tensorprod.morphism}.} In this
solution, we will often use the following classical formula from linear algebra:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$ and $W^{\prime}$ are six
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$ and $\beta^{\prime
}:V^{\prime}\rightarrow W^{\prime}$ are four $\kk$-linear maps, then
$\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

\noindent
We will refer to this formula as the \dfn{composition-tensor relation}.
Using two applications of this formula, we can obtain the following formula,
which we will call the \dfn{composition-triple-tensor relation}:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$U^{\prime\prime}$, $V^{\prime\prime}$, $W^{\prime\prime}$ are nine
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$, $\beta^{\prime}:V^{\prime
}\rightarrow W^{\prime}$, $\alpha^{\prime\prime}:U^{\prime\prime}\rightarrow
V^{\prime\prime}$, $\beta^{\prime\prime}:V^{\prime\prime}\rightarrow
W^{\prime\prime}$ are six $\kk$-linear maps, then $\left(  \beta
\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ\alpha^{\prime}\right)
\otimes\left(  \beta^{\prime\prime}\circ\alpha^{\prime\prime}\right)  =\left(
\beta\otimes\beta^{\prime}\otimes\beta^{\prime\prime}\right)  \circ\left(
\alpha\otimes\alpha^{\prime}\otimes\alpha^{\prime\prime}\right)  $.
\end{itemize}

(b) Let $C$, $C^{\prime}$, $D$ and $D^{\prime}$ be four $\kk$-coalgebras.
Let $f:C\rightarrow C^{\prime}$ and $g:D\rightarrow D^{\prime}$
be two $\kk$-coalgebra homomorphisms. We need to prove that $f\otimes
g:C\otimes D\rightarrow C^{\prime}\otimes D^{\prime}$ is a
$\kk$-coalgebra homomorphism.

Recall that (by the definition of a ``$\kk$-coalgebra
homomorphism'') the map $f\otimes g:C\otimes D\rightarrow
C^{\prime}\otimes D^{\prime}$ is a $\kk$-coalgebra homomorphism if and
only if the two diagrams
\begin{equation}
\xymatrixcolsep{5pc}
\xymatrix{
C \otimes D \ar[d]_{\Delta_{C \otimes D}} \ar[r]^{f \otimes g} & C^\prime
\otimes D^\prime\ar[d]^{\Delta_{C^\prime\otimes D^\prime}} \\
\left(C \otimes D\right) \otimes\left(C \otimes D\right) \ar[r]^-{\left
(f \otimes g\right) \otimes\left(f \otimes g\right)} & \left(C^\prime\otimes
D^\prime\right) \otimes\left(C^\prime\otimes D^\prime\right)
}
\label{sol.tensorprod.morphism.a.goal1}
\end{equation}
and
\begin{equation}
\xymatrix{
C \otimes D \ar[dr]_{\epsilon_{C \otimes D}} \ar[rr]^{f \otimes g}
& & C^\prime\otimes D^\prime\ar[dl]^{\epsilon_{C^\prime\otimes D^\prime}} \\
& \kk&
}
\label{sol.tensorprod.morphism.a.goal2}
\end{equation}
commute. We shall now prove that these diagrams indeed commute.

According to the definition of a ``$\kk$-coalgebra
homomorphism'', the map $f:C\rightarrow C^{\prime}$ is a
$\kk$-coalgebra homomorphism if and only if the two diagrams
\begin{equation}
\xymatrixcolsep{5pc}
\xymatrix{
C \ar[d]_{\Delta_C} \ar[r]^{f} & C^\prime\ar[d]^{\Delta_{C^\prime}} \\
C \otimes C \ar[r]^-{f \otimes f} & C^\prime\otimes C^\prime}
\label{sol.tensorprod.morphism.a.f1}
\end{equation}
and
\begin{equation}
\xymatrix{
C \ar[dr]_{\epsilon_C} \ar[rr]^{f}
& & C^\prime\ar[dl]^{\epsilon_{C^\prime}} \\
& \kk&
}
\label{sol.tensorprod.morphism.a.f2}
\end{equation}
commute. Since we know that the map  $f:C\rightarrow C^{\prime}$ is a
$\kk$-coalgebra homomorphism, we thus conclude that the two diagrams
\eqref{sol.tensorprod.morphism.a.f1} and \eqref{sol.tensorprod.morphism.a.f2}
indeed commute.

For any two $\kk$-modules $U$ and $V$, let $T_{U,V}:U\otimes
V\rightarrow V\otimes U$ be the twist map (i.e., the $\kk$-linear map
$U\otimes V\rightarrow V\otimes U$ sending every $u\otimes v$ to $v\otimes
u$). A simple linear-algebraic fact says that if $U$, $V$, $U^{\prime}$ and
$V^{\prime}$ are four $\kk$-modules and $x:U\rightarrow U^{\prime}$ and
$y:V\rightarrow V^{\prime}$ are two $\kk$-linear maps, then
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}=T_{U^{\prime},V^{\prime}}\circ\left(
x\otimes y\right)  .
\label{sol.tensorprod.morphism.a.twist}
\end{equation}


The definition of the $\kk$-coalgebra $C\otimes D$ yields
\begin{equation}
\Delta_{C\otimes D}=\left(  \id_{C}\otimes
T_{C,D}\otimes\id_{D}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{D}\right)  .
\label{sol.tensorprod.morphism.a.Delta1}
\end{equation}
Similarly,
\begin{equation}
\Delta_{C^{\prime}\otimes D^{\prime}}
= \left(  \id_{C^{\prime}}\otimes T_{C^{\prime},D^{\prime}}\otimes
\id_{D^{\prime}}\right)  \circ\left(  \Delta
_{C^{\prime}}\otimes\Delta_{D^{\prime}}\right)  .
\label{sol.tensorprod.morphism.a.Delta2}
\end{equation}
But $\left(  f\otimes f\right)  \circ\Delta_{C}=\Delta_{C^{\prime}}\circ f$
(since the diagram \eqref{sol.tensorprod.morphism.a.f1} commutes), and
similarly $\left(  g\otimes g\right)  \circ\Delta_{D}=\Delta_{D^{\prime}}\circ
g$.

Now,
\begin{align*}
& \underbrace{\left(  \left(  f\otimes g\right)  \otimes\left(  f\otimes
g\right)  \right)  }_{=f\otimes\left(  g\otimes f\right)  \otimes g}
\circ\underbrace{\Delta_{C\otimes D}}_{=\left(  \id_{C}\otimes
T_{C,D}\otimes\id_{D}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  }\\
& =\underbrace{\left(  f\otimes\left(  g\otimes f\right)  \otimes g\right)
\circ\left(  \id_{C}\otimes T_{C,D}\otimes
\id_{D}\right)  }_{\substack{=\left(  f\circ
\id_{C}\right)  \otimes\left(  \left(  g\otimes
f\right) \circ T_{C,D}\right) \otimes\left( g \circ \id_D \right) 
 \\\text{(since the composition-triple-tensor relation
yields}\\\left(  f\circ\id_{C}\right)  \otimes\left(
\left(  g\otimes f\right)  \circ T_{C,D}\right)  \otimes\left(  g\circ
\id_{D}\right)  \\=\left(  f\otimes\left(  g\otimes
f\right)  \otimes g\right) 
 \circ\left(  \id_{C} \otimes T_{C,D}\otimes\id_{D}\right) 
 \text{)}}}\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \\
& =\left(  \underbrace{\left(  f\circ\id_{C}\right)
}_{=f=\id_{C^{\prime}}\circ f}\otimes
\underbrace{\left(  \left(  g\otimes f\right)  \circ T_{C,D}\right)
}_{\substack{=T_{C^{\prime},D^{\prime}}\circ\left(  f\otimes g\right)
\\\text{(by \eqref{sol.tensorprod.morphism.a.twist}, applied}\\\text{to
}U=C\text{, }V=D\text{, }U^{\prime}=C^{\prime}\text{, }V^{\prime}=D^{\prime
}\text{, }x=f\text{ and }y=g\text{)}}}\otimes\underbrace{\left(
g\circ\id_{D}\right)  }_{= g = \id_{D^{\prime}}\circ g}\right) 
 \circ\left(  \Delta_{C}\otimes \Delta_{D}\right)  \\
& =\underbrace{\left(  \left(  \id_{C^{\prime}}\circ
f\right)  \otimes\left(  T_{C^{\prime},D^{\prime}}\circ\left(  f\otimes
g\right)  \right)  \otimes\left(  \id_{D^{\prime}} \circ g\right) 
 \right)  }_{=\left(  \id_{C^{\prime}}
\otimes T_{C^{\prime},D^{\prime}} \otimes \id_{D^{\prime}}\right) 
 \circ\left(  f\otimes\left(  f\otimes g\right)
\otimes g\right)  }\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \\
& =\left(  \id_{C^{\prime}}\otimes T_{C^{\prime
},D^{\prime}}\otimes\id_{D^{\prime}}\right)
\circ\underbrace{\left(  f\otimes\left(  f\otimes g\right)  \otimes g\right)
}_{=\left(  f\otimes f\right)  \otimes\left(  g\otimes g\right)  }\circ\left(
\Delta_{C}\otimes\Delta_{D}\right)  \\
& =\left(  \id_{C^{\prime}}\otimes T_{C^{\prime
},D^{\prime}}\otimes\id_{D^{\prime}}\right)
\circ\underbrace{\left(  \left(  f\otimes f\right)  \otimes\left(  g\otimes
g\right)  \right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)
}_{\substack{=\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)
\otimes\left(  \left(  g\otimes g\right)  \circ\Delta_{D}\right)
\\\text{(because the composition-tensor relation yields}\\\left(  \left(
f\otimes f\right)  \circ\Delta_{C}\right)  \otimes\left(  \left(  g\otimes
g\right)  \circ\Delta_{D}\right)  \\=\left(  \left(  f\otimes f\right)
\otimes\left(  g\otimes g\right)  \right)  \circ\left(  \Delta_{C}
\otimes\Delta_{D}\right)  \text{)}}}\\
& =\left(  \id_{C^{\prime}}\otimes T_{C^{\prime
},D^{\prime}}\otimes\id_{D^{\prime}}\right)
\circ\left(  \underbrace{\left(  \left(  f\otimes f\right)  \circ\Delta
_{C}\right)  }_{=\Delta_{C^{\prime}}\circ f}\otimes\underbrace{\left(  \left(
g\otimes g\right)  \circ\Delta_{D}\right)  }_{=\Delta_{D^{\prime}}\circ
g}\right)  \\
& =\left(  \id_{C^{\prime}}\otimes T_{C^{\prime
},D^{\prime}}\otimes\id_{D^{\prime}}\right)
\circ\underbrace{\left(  \left(  \Delta_{C^{\prime}}\circ f\right)
\otimes\left(  \Delta_{D^{\prime}}\circ g\right)  \right)  }_{=\left(
\Delta_{C^{\prime}}\otimes\Delta_{D^{\prime}}\right)  \circ\left(  f\otimes
g\right)  }\\
& =\underbrace{\left(  \id_{C^{\prime}}\otimes
T_{C^{\prime},D^{\prime}}\otimes\id_{D^{\prime}}
\right)  \circ\left(  \Delta_{C^{\prime}}\otimes\Delta_{D^{\prime}}\right)
}_{\substack{=\Delta_{C^{\prime}\otimes D^{\prime}}\\\text{(by
\eqref{sol.tensorprod.morphism.a.Delta2})}}}\circ\left(  f\otimes g\right)
=\Delta_{C^{\prime}\otimes D^{\prime}}\circ\left( f\otimes g \right) .
\end{align*}
In other words, the diagram \eqref{sol.tensorprod.morphism.a.goal1} commutes.

We have $\epsilon_{C^{\prime}}\circ f=\epsilon_{C}$ (since the diagram
\eqref{sol.tensorprod.morphism.a.f2} commutes) and $\epsilon_{D^{\prime}}\circ
g=\epsilon_{D}$ (similarly).

Now, let $\theta$ be the canonical $\kk$-module isomorphism
$\kk \otimes \kk \rightarrow \kk$. Then, the definition of
the $\kk$-coalgebra $C\otimes D$ yields
\begin{equation}
\epsilon_{C\otimes D}=\theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  .
\label{sol.tensorprod.morphism.a.epsilon1}
\end{equation}
Similarly,
\begin{equation}
\epsilon_{C^{\prime}\otimes D^{\prime}}=\theta\circ\left(  \epsilon
_{C^{\prime}}\otimes\epsilon_{D^{\prime}}\right) .
\label{sol.tensorprod.morphism.a.epsilon2}
\end{equation}


Now,
\begin{align*}
\underbrace{\epsilon_{C^{\prime}\otimes D^{\prime}}}_{=\theta\circ\left(
\epsilon_{C^{\prime}}\otimes\epsilon_{D^{\prime}}\right)  }\circ\left(
f\otimes g\right)
& = \theta\circ\underbrace{\left( \epsilon_{C^{\prime}}\otimes
\epsilon_{D^{\prime}}\right)  \circ\left(  f\otimes g\right)
}_{\substack{=\left(  \epsilon_{C^{\prime}}\circ f\right)  \otimes\left(
\epsilon_{D^{\prime}}\circ g\right)  \\\text{(since the composition-tensor
relation yields}\\\left(  \epsilon_{C^{\prime}}\circ f\right)  \otimes\left(
\epsilon_{D^{\prime}}\circ g\right)  =\left(  \epsilon_{C^{\prime}}
\otimes\epsilon_{D^{\prime}}\right)  \circ\left(  f\otimes g\right)
 \text{)}}}\\
& =\theta\circ\left(  \underbrace{\left(  \epsilon_{C^{\prime}}\circ f\right)
}_{=\epsilon_{C}}\otimes\underbrace{\left(  \epsilon_{D^{\prime}}\circ
g\right)  }_{=\epsilon_{D}}\right)  =\theta\circ\left(  \epsilon_{C}
\otimes\epsilon_{D}\right)  =\epsilon_{C\otimes D}
\end{align*}
(by \eqref{sol.tensorprod.morphism.a.epsilon1}). In other words, the diagram
\eqref{sol.tensorprod.morphism.a.goal2} commutes.

We now know that the two diagrams \eqref{sol.tensorprod.morphism.a.goal1} and
\eqref{sol.tensorprod.morphism.a.goal2} commute. Thus, the map $f\otimes
g:C\otimes D\rightarrow C^{\prime}\otimes D^{\prime}$ is a
$\kk$-coalgebra homomorphism (because we know that the map
$f \otimes g : C\otimes D \rightarrow C^{\prime} \otimes D^{\prime}$
is a $\kk$-coalgebra
homomorphism if and only if the two diagrams
\eqref{sol.tensorprod.morphism.a.goal1} and
\eqref{sol.tensorprod.morphism.a.goal2} commute). This solves Exercise
\ref{exe.tensorprod.morphism}(b).

(a) In order to obtain a solution to
Exercise~\ref{exe.tensorprod.morphism}(a), it is enough to reverse
all arrows in the above solution to
Exercise~\ref{exe.tensorprod.morphism}(b) (and, of course,
replace $C$, $D$, $\Delta_{C}$ etc. by $A$, $B$, $m_{A}$ etc.).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.kernel-is-coideal}}

\textit{Solution to Exercise~\ref{exe.kernel-is-coideal}.}
(a) Assume that $f$ is surjective.
It is a known fact from linear algebra\footnote{proven, e.g., in Keith
Conrad's \cite[``Tensor Products II'', Thm. 2.19]{Conrad-blurbs}}
that if $U$, $V$, $U^{\prime}$ and
$V^{\prime}$ are four $\kk$-modules, and $\phi:U\rightarrow U^{\prime}$
and $\psi:V\rightarrow V^{\prime}$ are two surjective $\kk$-linear
maps, then the kernel of $\phi\otimes\psi:U\otimes V\rightarrow U^{\prime
}\otimes V^{\prime}$ is
\[
\ker\left(  \phi\otimes\psi\right)  =\left(  \ker\phi\right)  \otimes
V+U\otimes\left(  \ker\psi\right)  .
\]
Applying this to $U=A$, $U^{\prime}=C$, $V=A$, $V^{\prime}=C$,
$\phi=f$ and $\psi=f$, we obtain $\ker\left(  f\otimes f\right)  =\left(  \ker
f\right)  \otimes A+A\otimes\left(  \ker f\right)  $.

But $f$ is a coalgebra homomorphism, so that $\epsilon=\epsilon\circ f$.
Hence, every $x\in\ker f$ satisfies $\underbrace{\epsilon}_{=\epsilon\circ
f}\left(  x\right)  =\left(  \epsilon\circ f\right)  \left(  x\right)
=\epsilon\left(  \underbrace{f\left(  x\right)  }_{\substack{=0\\\text{(since
}x\in\ker f\text{)}}}\right)  =\epsilon\left(  0\right)  =0$. In other words,
$\epsilon\left(  \ker f\right)  =0$. Also,
\[
\ker f\subset\ker\underbrace{\left(  \Delta\circ f\right)  }
_{\substack{=\left(  f\otimes f\right)  \circ\Delta\\\text{(since }f\text{ is
a coalgebra}\\\text{homomorphism)}}}=\ker\left(  \left(  f\otimes f\right)
\circ\Delta\right)  =\Delta^{-1}\left(  \ker\left(  f\otimes f\right)
\right),
\]
so that
\[
\Delta\left(  \ker f\right)  \subset\ker\left(  f\otimes f\right)  =\left(
\ker f\right)  \otimes A+A\otimes\left(  \ker f\right)  .
\]
Combined with $\epsilon\left(  \ker f\right)  =0$, this shows that $\ker f$ is
a two-sided coideal of $A$. Thus, Exercise~\ref{exe.kernel-is-coideal}(a) is
solved.

(b) Assume that $\kk$ is a field. Then, it is a known fact from linear
algebra\footnote{proven, e.g., in Keith
Conrad's \cite[``Tensor Products II'', Thm. 5.5]{Conrad-blurbs}}
that if $U$, $V$, $U^{\prime}$ and
$V^{\prime}$ are four $\kk$-modules, and $\phi:U\rightarrow U^{\prime}$
and $\psi:V\rightarrow V^{\prime}$ are two $\kk$-linear
maps, then the kernel of $\phi\otimes\psi:U\otimes V\rightarrow U^{\prime
}\otimes V^{\prime}$ is
\[
\ker\left(  \phi\otimes\psi\right)  =\left(  \ker\phi\right)  \otimes
V+U\otimes\left(  \ker\psi\right)  .
\]
Starting from this point, we can continue arguing as in the solution of part
(a). Thus, Exercise~\ref{exe.kernel-is-coideal}(b) is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.graded.iso}}

\textit{Solution to Exercise~\ref{exe.graded.iso}.}

It is not hard to solve Exercise~\ref{exe.graded.iso}
directly; we shall take a longer but somewhat more
elegant approach. We begin with a definition:

\begin{definition}
\label{def.graded.mod.pi}
Let $V=\bigoplus_{n\in\NN }V_{n}$ be a
graded $\kk$-module (where the $V_{n}$ are the homogeneous components
of $V$). Let $n\in\NN $. Then, we shall let
$\pi_{n,V}:V\rightarrow V_{n}$ denote the canonical
projection from $V$ to its $n$-th graded component $V_{n}$.
\end{definition}

Let us show a few properties of these projections:

\begin{lemma}
\label{lem.graded.pi.commute.hg}Let $V$ and $W$ be two graded
$\kk$-modules. Let $f:V\rightarrow W$ be a graded $\kk$-linear map.
Let $m\in\NN$. Let $v$ be a homogeneous element of $V$. Then,
$\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.graded.pi.commute.hg}.]Write $V$ as
$V=\bigoplus_{n\in{\NN }}V_{n}$ (where the $V_{n}$ are the homogeneous
components of $V$). Write $W$ as $W=\bigoplus_{n\in{\NN }}W_{n}$ (where
the $W_{n}$ are the homogeneous components of $W$).

The map $\pi_{m,V}$ is the canonical projection from $V$ to its $m$-th graded
component $V_{m}$. Hence, $\pi_{m,V}$ fixes every element of $V_{m}$. In other
words, we have%
\begin{equation}
\pi_{m,V}\left(  p\right)  =p\ \ \ \ \ \ \ \ \ \ \text{for each }p\in V_{m}.
\label{pf.lem.graded.pi.commute.hg.1}%
\end{equation}
The same argument (applied to $W$ and $W_m$ instead of $V$ and $V_m$) yields
\begin{equation}
\pi_{m,W}\left(  p\right)  =p\ \ \ \ \ \ \ \ \ \ \text{for each }p\in W_{m}.
\label{pf.lem.graded.pi.commute.hg.2}%
\end{equation}


The map $\pi_{m,V}$ is the canonical projection from $V$ to its $m$-th graded
component $V_{m}$. Hence, $\pi_{m,V}$ annihilates all graded components of $V$
other than $V_{m}$. In other words,
\begin{equation}
\pi_{m,V}\left(  V_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\NN \text{ satisfying }n\neq m. \label{pf.lem.graded.pi.commute.hg.3}%
\end{equation}
The same argument (applied to $W$ and $W_m$ instead of $V$ and $V_m$) yields
\begin{equation}
\pi_{m,W}\left(  W_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\NN \text{ satisfying }n\neq m. \label{pf.lem.graded.pi.commute.hg.4}%
\end{equation}

Recall that $v$ is a homogeneous element of $V$. Thus, there exists some
$n\in \NN$ such that $v\in V_{n}$. Consider this $n$. From $v\in
V_{n}$, we obtain $f\left(  v\right)  \in f\left(  V_{n}\right)  \subset
W_{n}$ (since the map $f$ is graded). We must show that
$\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$.

We are in one of the following two cases:

\textit{Case 1:} We have $n=m$.

\textit{Case 2:} We have $n\neq m$.

Let us first consider Case 1 first. In this case, we have $n=m$. Now, $v\in
V_{n}=V_{m}$ (since $n=m$). Thus, \eqref{pf.lem.graded.pi.commute.hg.1}
(applied to $p = v$) yields
$\pi_{m,V}\left(  v\right)  =v$. Hence, $v=\pi_{m,V}\left(  v\right)  $. Also,
$f\left(  v\right)  \in W_{n}=W_{m}$ (since $n=m$). Hence,
\eqref{pf.lem.graded.pi.commute.hg.2} (applied to $p = f\left(  v\right)  $)
yields
\[
\pi_{m,W}\left(  f\left(  v\right)  \right)
=f\left(  \underbrace{v}_{=\pi_{m,V}\left(  v\right)
}\right)  =f\left(  \pi_{m,V}\left(  v\right)  \right) .
\]
Thus, $\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$
is proved in Case 1.

Let us next consider Case 2. In this case, we have $n\neq m$. But $v\in V_{n}%
$, so that $\pi_{m,V}\left(  v\right)  \in\pi_{m,V}\left(  V_{n}\right)  =0$
(by \eqref{pf.lem.graded.pi.commute.hg.3}). Hence, $\pi_{m,V}\left(  v\right)
=0$. Therefore, $f\left(  \underbrace{\pi_{m,V}\left(  v\right)  }%
_{=0}\right)  =f\left(  0\right)  =0$ (since the map $f$ is $\kk %
$-linear). On the other hand, $\pi_{m,W}\left(  \underbrace{f\left(  v\right)
}_{\in W_{n}}\right)  \in\pi_{m,W}\left(  W_{n}\right)  =0$ (by
\eqref{pf.lem.graded.pi.commute.hg.4}), so that
\begin{align*}
\pi_{m,W}\left(
f\left(  v\right)  \right)  =0=f\left(  \pi_{m,V}\left(  v\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\left(  \pi_{m,V}\left(  v\right)
\right)  =0\right) .
\end{align*}
Thus, $\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$
is proved in Case 2.

We have now proved $\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$ in each of these two
Cases 1 and 2. Since these two Cases cover all possibilities, we thus conclude
that $\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$ always holds.

This proves Lemma~\ref{lem.graded.pi.commute.hg}.
\end{proof}

\begin{proposition}
\label{prop.graded.pi.commute}Let $V$ and $W$ be two graded
$\kk$-modules. Let $f:V\rightarrow W$ be a graded $\kk$-linear map.
Let $m\in\NN$. Let $v \in V$ be arbitrary. Then,
$\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.pi.commute}.]
Every element of $V$ is a sum of homogeneous elements
(since $V$ is graded).
Thus, $v$ is a sum of homogeneous elements
(since $v$ is an element of $V$).
In other words, we can write $v$ in the form
$v = \sum_{i=1}^k v_i$, where $k \in \NN$, and where
$v_1, v_2, \ldots, v_k$ are homogeneous elements of $V$.
Consider this $k$ and these $v_1, v_2, \ldots, v_k$.

For each $i \in \left\{1,2,\ldots,k\right\}$, the element
$v_i$ of $V$ is homogeneous (since
$v_1, v_2, \ldots, v_k$ are homogeneous elements of $V$)
and therefore satisfies
\begin{equation}
\pi_{m,W}\left( f\left(v_i\right) \right)
= f\left( \pi_{m,V} \left(v_i\right) \right)
\label{pf.prop.graded.pi.commute.vi}
\end{equation}
(by Lemma~\ref{lem.graded.pi.commute.hg}, applied to
$v_i$ instead of $v$).
Now,
\begin{align*}
\pi_{m,W}\left(  f\left(  \underbrace{v}_{=\sum_{i=1}^{k}v_{i}}\right)
\right)    & =\pi_{m,W}\left(  \underbrace{f\left(  \sum_{i=1}^{k}%
v_{i}\right)  }_{\substack{=\sum_{i=1}^{k}f\left(  v_{i}\right)
\\\text{(since the map }f\text{ is }\mathbf{k}\text{-linear)}}}\right)
=\pi_{m,W}\left(  \sum_{i=1}^{k}f\left(  v_{i}\right)  \right)  \\
& =\sum_{i=1}^{k}\underbrace{\pi_{m,W}\left(  f\left(  v_{i}\right)  \right)
}_{\substack{=f\left(  \pi_{m,V}\left(  v_{i}\right)  \right)  \\\text{(by
\eqref{pf.prop.graded.pi.commute.vi})}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\pi_{m,W}\text{ is }\mathbf{k}\text{-linear}\right)  \\
& =\sum_{i=1}^{k}f\left(  \pi_{m,V}\left(  v_{i}\right)  \right)  .
\end{align*}
Comparing this with%
\begin{align*}
f\left(  \pi_{m,V}\left(  \underbrace{v}_{=\sum_{i=1}^{k}v_{i}}\right)
\right)    & =f\left(  \underbrace{\pi_{m,V}\left(  \sum_{i=1}^{k}%
v_{i}\right)  }_{\substack{=\sum_{i=1}^{k}\pi_{m,V}\left(  v_{i}\right)
\\\text{(since the map }\pi_{m,V}\text{ is }\mathbf{k}\text{-linear)}%
}}\right)  =f\left(  \sum_{i=1}^{k}\pi_{m,V}\left(  v_{i}\right)  \right)  \\
& =\sum_{i=1}^{k}f\left(  \pi_{m,V}\left(  v_{i}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\mathbf{k}%
\text{-linear}\right)  ,
\end{align*}
we obtain
$\pi_{m,W}\left( f\left(v\right) \right)
= f\left( \pi_{m,V} \left(v\right) \right)$.
This proves Proposition \ref{prop.graded.pi.commute}.
\end{proof}

Let us now come to the solution of Exercise \ref{exe.graded.iso}.
Let $f : V \to W$ be an invertible graded $\kk$-linear map.
We must prove that its inverse $f^{-1} : W \to V$ is also graded.
In other words, we must prove that $f^{-1}\left(W_n\right)
\subset V_n$ for each $n \in \NN$.

So let $n \in \NN$.
Let $v \in f^{-1}\left(W_n\right)$.
Thus, $v \in V$ and $f\left(v\right) \in W_n$.

The map $\pi_{n,W}$ is the canonical projection from $W$ to its $n$-th graded
component $W_n$. Hence, $\pi_{n,W}$ fixes every element of $W_n$. In other
words, we have $\pi_{n,W}\left(p\right) = p$ for each $p \in W_n$.
Applying this to $p = f\left(v\right)$, we obtain
$\pi_{n,W}\left( f\left(v\right) \right)
= f\left(v\right)$ (since $f\left(v\right) \in W_n$).
But Proposition \ref{prop.graded.pi.commute} (applied to $m = n$)
yields
$\pi_{n,W}\left( f\left(v\right) \right)
= f\left( \pi_{n,V} \left(v\right) \right)$.
Comparing these two equalities, we obtain
$f\left( v \right) = f\left( \pi_{n,V} \left(v\right) \right)$.

But the map $f$ is invertible, thus injective. In other words,
if $a, b \in V$ are two elements satisfying $f\left(a\right)
= f\left(b\right)$, then $a = b$. Applying this to
$a = v$ and $b = \pi_{n,V} \left(v\right)$,
we find $v = \pi_{n,V} \left(v\right)$
(since $f\left( v \right) = f\left( \pi_{n,V} \left(v\right) \right)$).

But the map $\pi_{n,V}$ is the canonical projection from $V$
to its $n$-th graded component $V_n$. Thus,
$\pi_{n,V} \left( V \right) \subset V_n$.
Hence, $\pi_{n,V} \left(v\right) \in V_n$
(since $v \in V$). Thus,
$v = \pi_{n,V} \left(v\right) \in V_n$.

Forget that we fixed $v$. We thus have shown that
$v \in V_n$ for each $v \in f^{-1}\left(W_n\right)$.
In other words, $f^{-1}\left(W_n\right) \subset V_n$.

Forget that we fixed $n$. We thus have shown that
$f^{-1}\left(W_n\right) \subset V_n$ for each $n \in \NN$.
In other words, the map $f^{-1}$ is graded.
This solves Exercise~\ref{exe.graded.iso}.

%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.primitives.graded-coideal}}

\textit{Solution to Exercise~\ref{exe.primitives.graded-coideal}.}

(a) Define a map $\widetilde{\Delta} : A \rightarrow A \otimes A$ by
\[
\widetilde{\Delta}\left(  x\right)  =\Delta\left(  x\right)  -\left(
x \otimes 1 + 1 \otimes x\right)
 \ \ \ \ \ \ \ \ \ \ \text{for all }x \in A.
\]


\begin{vershort}
It is easily seen that $\widetilde{\Delta}$ is a homomorphism of graded
$\kk$-modules.
\end{vershort}
\begin{verlong}
This map $\widetilde{\Delta}$ is $\kk$-linear (because the tensor
$\Delta\left(  x\right)  -\left(  x \otimes 1 + 1 \otimes x\right)  $ depends
$\kk$-linearly on $x$) and graded\footnote{\textit{Proof.} Let
$n\in\NN$. Let $x\in A_n$. Then, $\Delta\left(  \underbrace{x}_{\in
A_n}\right)  \in\Delta\left(  A_n\right)  \subset\left(  A\otimes
A\right)  _n$ (since the map $\Delta$ is graded). Also, $\underbrace{x}_{\in
A_n}\otimes\underbrace{1}_{\in A_0}\in A_n\otimes A_0\subset\left(
A\otimes A\right)  _n$ and $\underbrace{1}_{\in A_0}\otimes\underbrace{x}
_{\in A_n}\in A_0\otimes A_n\subset\left(  A\otimes A\right)  _n$.
Hence,
\[
\underbrace{x\otimes1}_{\in\left(  A\otimes A\right)  _n}
+\underbrace{1\otimes x}_{\in\left(  A\otimes A\right)  _n}\in\left(
A\otimes A\right)  _n+\left(  A\otimes A\right)  _n\subset\left(  A\otimes
A\right)  _n.
\]
Now,
\[
\widetilde{\Delta}\left(  x\right)  =\underbrace{\Delta\left(  x\right)
}_{\in\left(  A\otimes A\right)  _n}-\underbrace{\left(  x\otimes1+1\otimes
x\right)  }_{\in\left(  A\otimes A\right)  _n}\in\left(  A\otimes A\right)
_n-\left(  A\otimes A\right)  _n \subset \left(  A\otimes A\right)  _n .
\]
\par
Now, forget that we fixed $x$. We thus have shown that every $x\in A_n$
satisfies $\widetilde{\Delta}\left(  x\right)  \in\left(  A\otimes A\right)
_n$. In other words, $\widetilde{\Delta}\left(  A_n\right)  \subset\left(
A\otimes A\right)  _n$.
\par
Now, forget that we fixed $n$. We thus have proven that $\widetilde{\Delta
}\left(  A_n\right)  \subset\left(  A\otimes A\right)  _n$ for every
$n\in\NN$. In other words, the map $\widetilde{\Delta}$ is graded,
qed.}.
In other words, $\widetilde{\Delta}$ is a homomorphism of graded
$\kk$-modules.
\end{verlong}
Hence, the kernel $\ker\widetilde{\Delta}$ of
$\widetilde{\Delta}$ is a graded $\kk$-submodule of $A$ (since the
kernel of a homomorphism of graded $\kk$-modules always is a graded
$\kk$-submodule of the domain).

We have defined $\Liep$ as the set of all primitive elements of $A$.
\begin{verlong}
In other words, $\Liep$ is the set of all $x\in A$ such that
$\Delta\left(  x\right)  =x\otimes1+1\otimes x$ (because the primitive
elements of $A$ are precisely the elements $x\in A$ such that
$\Delta\left( x\right)  =x\otimes1+1\otimes x$).
\end{verlong}
In other words,
\begin{align*}
\Liep   & =\left\{  x\in A\ \mid\ \Delta\left(  x\right)  =x\otimes
1+1\otimes x\right\}  \\
& =\left\{  x\in A\ \mid\ \underbrace{\Delta\left(  x\right)  -\left(
x\otimes1+1\otimes x\right)  }_{\substack{=\widetilde{\Delta}\left(  x\right)
\\\text{(since }\widetilde{\Delta}\left(  x\right)  =\Delta\left(  x\right)
-\left(  x\otimes1+1\otimes x\right)  \text{)}}}=0\right\}  \\
& =\left\{  x\in A\ \mid\ \widetilde{\Delta}\left(  x\right)  =0\right\}
=\ker\widetilde{\Delta}.
\end{align*}
Thus, $\Liep$ is a graded $\kk$-submodule of $A$ (since we know
that $\ker\widetilde{\Delta}$ is a graded $\kk$-submodule of $A$). This
solves Exercise~\ref{exe.primitives.graded-coideal} (a).

(b) We notice first that
\begin{equation}
\label{sol.primitives.graded-coideal.b}
\epsilon\left(x\right) = 0 \qquad \text{ for every } x \in \Liep .
\end{equation}

[\textit{Proof of \eqref{sol.primitives.graded-coideal.b}.}
We know that $A$ is a $\kk$-bialgebra. Thus, $\epsilon$ is a
$\kk$-algebra homomorphism (by the definition of a $\kk$-bialgebra).
Hence, $\epsilon\left(1_A\right) = 1_{\kk}$
and $\epsilon\left(0\right) = 0$.

Let $x\in\Liep$. Thus, $x$ is primitive, so that $\Delta\left(
x\right)  =x\otimes1+1\otimes x$ (by the definition of a
primitive element). Applying the map $\epsilon\otimes
\id$ to both sides of this equality, we obtain
\[
\left(  \epsilon\otimes\id\right)  \left(  \Delta\left(
x\right)  \right)  =\left(  \epsilon\otimes\id \right)  \left(
x\otimes1+1\otimes x\right)  =\epsilon\left(  x\right)  \cdot
\id \left(  1\right)  +\epsilon\left(  1\right)  \cdot
\id\left(x\right)
\]
(where we are identifying $\kk \otimes A$ with $A$ along the canonical
isomorphism). Compared with $\left(  \epsilon\otimes\id \right)
\left(  \Delta\left(  x\right)  \right)  =x$ (this is a consequence of the
axioms of a coalgebra), this yields
\[
x=\epsilon\left(  x\right)  \cdot\underbrace{\id \left(
1\right)  }_{=1=1_A}+\underbrace{\epsilon\left(  1\right)  }_{
= \epsilon\left(1_A\right) =1_{\kk}}\cdot
\underbrace{\id\left(x\right)}_{=x}
=\epsilon\left(  x\right) \cdot 1_A +x.
\]
Subtracting $x$ from this equality, we obtain
$0=\epsilon\left(  x\right) \cdot 1_A$.
Applying the map $\epsilon$ to this equality, we find
$\epsilon\left(0\right)
= \epsilon\left(\epsilon\left(  x\right) \cdot 1_A\right)
= \epsilon\left(  x\right) \cdot
\underbrace{\epsilon\left(1_A\right)}_{=1_{\kk}}
= \epsilon\left(  x\right)$.
Hence, $\epsilon\left(  x\right) = \epsilon\left(0\right) = 0$.
This proves \eqref{sol.primitives.graded-coideal.b}.]

(\textit{Note:} We will reprove \eqref{sol.primitives.graded-coideal.b}
below in Proposition~\ref{antipode-on-primitives}.)

Now, for every $x\in\Liep$, we have
\begin{align*}
\Delta\left(  x\right)    & =\underbrace{x}_{\in\Liep }\otimes
\underbrace{1}_{\in A}+\underbrace{1}_{\in A}\otimes\underbrace{x}
_{\in\Liep }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }x\in\Liep 
\text{, so that }x\text{ is primitive}\right)  \\
& \in\Liep \otimes A+A\otimes\Liep .
\end{align*}
In other words, $\Delta\left(  \Liep \right)  \subset\Liep \otimes
A+A\otimes\Liep$. Combined with $\epsilon\left(  \Liep \right)
=0$ (this follows from \eqref{sol.primitives.graded-coideal.b}), this
yields that $\Liep$ is a two-sided coideal of $A$. This solves
Exercise~\ref{exe.primitives.graded-coideal} (b).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{graded-connected-exercise}}

\textit{Solution to Exercise~\ref{graded-connected-exercise}.}
(a) The unit map $u:\kk\rightarrow A$ is
graded (since $A$ is a graded algebra). Hence,
$u\left( \kk_0\right)  \subset A_0$, where $\kk_0$ denotes the $0$-th graded
component of $\kk$. But the grading on $\kk$ is such that
$\kk_{0}=\kk$. Thus, $u\left(  \kk_0\right)  =u\left(
\kk\right)  $, so that $u\left(  \kk\right)  =u\left(
\kk_{0}\right)  \subset A_{0}$. But
\[
u\left(  \kk\right) = \left\{  \underbrace{u\left(  \lambda\right)
}_{\substack{=\lambda\cdot 1_A \\ \text{(by the definition of }u\text{)}}}
\ :\ \lambda\in\kk\right\}  = \left\{  \lambda\cdot 1_A
\ :\ \lambda\in\kk\right\}  = \kk\cdot 1_A.
\]
Hence, $\kk\cdot 1_A = u\left(  \kk\right)  \subset A_{0}$, so
that part (a) is solved.

(b) Since $A$ is connected, we have $A_{0}\cong\kk$. In other words,
there exists a $\kk$-module isomorphism $\phi:A_{0}\rightarrow
\kk$. Consider this $\phi$. Since $\phi$ is a $\kk$-module
isomorphism $A_{0}\rightarrow\kk$, the inverse $\phi^{-1}$ of $\phi$ is
a well-defined $\kk$-module isomorphism $\kk\rightarrow A_{0}$.

We saw in the proof of part (a) that $u\left(  \kk\right)  \subset
A_{0} $. Hence, $u$ restricts to a map $\kk\rightarrow A_{0}$. Denote
this map $\kk\rightarrow A_{0}$ by $u^{\prime}$. Then, $u^{\prime}$ is
a restriction of $u$ (more precisely, a corestriction of $u$, because we are
restricting the target rather than the domain).

Let also $\epsilon^{\prime}$ denote the restriction of $\epsilon$ to $A_{0}$.
Since $\epsilon^{\prime}$ and $u^{\prime}$ are restrictions of $\epsilon$ and
$u$, we have $\epsilon^{\prime}\circ u^{\prime}=\epsilon\circ u = \id_{\kk}$
(by the axioms of a coalgebra). Now, $\left(  \epsilon^{\prime}\circ
\phi^{-1}\right)  \circ\left(  \phi\circ u^{\prime}\right)  =\epsilon^{\prime
}\circ u^{\prime} = \id_{\kk}$. Hence, the $\kk$-linear map
$\epsilon^{\prime}\circ\phi^{-1}:\kk\rightarrow\kk$ has a right
inverse. Thus, the $\kk$-linear map $\epsilon^{\prime}\circ\phi
^{-1}:\kk\rightarrow\kk$ is surjective. Since every surjective
$\kk$-linear map $\kk\rightarrow\kk$ is an
isomorphism\footnote{\textit{Proof.} Let $\alpha$ be a surjective
$\kk$-linear map $\kk\rightarrow\kk$. We need to show that $\alpha$
is an isomorphism.
\par
Let $\lambda\in\ker\alpha$. Then, $\lambda\in\kk$ satisfies
$\alpha\left(  \lambda\right)  =0$. But $\alpha$ is surjective, so that
$\kk=\alpha\left(  \kk\right)  $. Hence,
$1\in\kk =\alpha\left(  \kk\right)  $. Thus, there exists a $\mu\in\kk$
such that $1=\alpha\left(  \mu\right)  $. Consider this $\mu$. Then, the
$\kk$-linearity of $\alpha$ yields $\alpha\left(  \lambda\mu\right)
=\lambda\underbrace{\alpha\left(  \mu\right)  }_{=1}=\lambda$. But the
$\kk$-linearity of $\alpha$ also shows that $\alpha\left(  \mu
\lambda\right)  =\mu\underbrace{\alpha\left(  \lambda\right)  }_{=0}=0$. Thus,
$0=\alpha\left(  \mu\lambda\right)  =\alpha\left(  \lambda\mu\right)
=\lambda$, so that $\lambda=0$. We thus have shown that every $\lambda\in
\ker\alpha$ satisfies $\lambda=0$. Hence, $\ker\alpha=0$, so that the map
$\alpha$ is injective. Since $\alpha$ is injective and surjective, we see that
$\alpha$ is bijective, thus an isomorphism, qed.}, this shows that the
$\kk$-linear map $\epsilon^{\prime}\circ\phi^{-1} : \kk\rightarrow\kk$ is
an isomorphism. Since $\phi$ also is an isomorphism,
the map $\epsilon^{\prime}\circ\phi^{-1}\circ\phi$ is a composition of two
isomorphisms, and thus an isomorphism. In other words, $\epsilon^{\prime}$ is
an isomorphism (since $\epsilon^{\prime}\circ\phi^{-1}\circ\phi=\epsilon
^{\prime}$). Hence, the inverse map of $\epsilon^{\prime}$ is well-defined.
This inverse map must be $u^{\prime}$ (since $\epsilon^{\prime}\circ
u^{\prime}=\id_{\kk}$), and so we conclude that $u^{\prime}$ is an
isomorphism. In other words, the restriction of $u$ to a map
$\kk\rightarrow A_{0}$ is an isomorphism. This solves part (b).

(c) Part (b) shows that the map $\kk \overset{u}{\rightarrow} A_0$
is an isomorphism.
Hence, this map $\kk \overset{u}{\rightarrow} A_0$ is bijective,
and thus also surjective.
In other words, we have $A_0 = u \left( \kk \right)$.
Hence,
\begin{align*}
A_0
&= u \left( \underbrace{\kk}_{= \kk \cdot 1} \right)
= u \left( \kk \cdot 1 \right)
= \kk \cdot \underbrace{u \left( 1 \right)}_{\substack{= 1 \cdot 1_A \\ \text{(by the definition of } u \text{)}}}
\qquad \left(\text{since the map } u \text{ is } \kk\text{-linear}\right)
\\
&= \underbrace{\kk \cdot 1}_{=\kk} \cdot 1_A = \kk \cdot 1_A .
\end{align*}
This proves part (c).

(e) In the proof of part (b), we showed that $u^{\prime}$ is the inverse map
of $\epsilon^{\prime}$. Hence, $\epsilon^{\prime}$ is the inverse map of
$u^{\prime}$. In other words, the restriction of $\epsilon$ to $A_{0}$ is the
inverse map of the restriction of $u$ to a map $\kk\rightarrow A_{0}$.
This solves part (e).

(d) The counit map $\epsilon$ is graded (since $A$ is a graded coalgebra).
Hence, every $n\geq 0$ satisfies $\epsilon\left( A_n\right) \subset
\kk_n$ (where $\kk_n$ denotes the $n$-th graded component of
$\kk$). For every positive $n$, this shows that $A_n \subset
\ker\epsilon$\ \ \ \ \footnote{\textit{Proof.} Let $n$ be a positive integer.
Then, $\epsilon\left(  A_{n}\right)  \subset\kk_{n}=0$ (because of how
the grading on $\kk$ is constructed), so that $\epsilon\left(
A_{n}\right)  =0$ and thus $A_{n}\subset\ker\epsilon$, qed.}. Hence,
$\bigoplus_{n>0} A_n
= \sum_{n>0} \underbrace{A_n}_{\subset\ker\epsilon}
\subset \sum_{n>0} \ker\epsilon
\subset \ker\epsilon$ (since $\ker\epsilon$ is a $\kk$-submodule of $A$).

Now, let $a\in\ker\epsilon$ be arbitrary. Then, $a\in A$ satisfies
$\epsilon\left(  a\right)  =0$. We have $a\in A=\bigoplus_{n\geq0}A_{n}
=A_{0}\oplus\bigoplus_{n>0}A_{n}$. Hence, we can write $a$ in the form
$a=a^{\prime}+a^{\prime\prime}$ for $a^{\prime}\in A_{0}$ and $a^{\prime
\prime}\in\bigoplus_{n>0}A_{n}$. Consider these $a^{\prime}$ and
$a^{\prime\prime}$. We have $a^{\prime\prime} \in \bigoplus_{n>0}A_n
\subset \ker\epsilon$, so that
$\epsilon\left(  a^{\prime\prime}\right)  =0$. Since
$a=a^{\prime}+a^{\prime\prime}$, we have $\epsilon\left(  a\right)
=\epsilon\left(  a^{\prime}+a^{\prime\prime}\right)  =\epsilon\left(
a^{\prime}\right)  +\underbrace{\epsilon\left(  a^{\prime\prime}\right)
}_{=0}=\epsilon\left(  a^{\prime}\right)  $, thus $\epsilon\left(  a^{\prime
}\right)  =\epsilon\left(  a\right)  =0$. Since $\epsilon$ restricted to
$A_{0}$ is injective (in fact, part (e) of this problem shows that $\epsilon$
restricted to $A_{0}$ is an isomorphism), this yields that $a^{\prime}=0$
(because $a^{\prime}\in A_{0}$). Hence, $a=\underbrace{a^{\prime}}_{=0}
+a^{\prime\prime}=a^{\prime\prime}\in\bigoplus_{n>0}A_{n}$.

Now forget that we fixed $a$. We thus have seen that every $a\in\ker\epsilon$
satisfies $a\in\bigoplus_{n>0}A_{n}$. In other words, $\ker\epsilon
\subset\bigoplus_{n>0}A_{n}$. Combined with $\bigoplus_{n>0}A_{n}\subset
\ker\epsilon$, this yields $\ker\epsilon=\bigoplus_{n>0}A_{n}$. This solves
part (d).

(f) Let $x\in A$. We have $A = \bigoplus_{n\geq0} A_n
= \underbrace{A_{0}}_{=\kk\cdot 1_{A}}
\oplus\underbrace{\bigoplus_{n>0} A_n}_{=I}
=\kk\cdot 1_{A}\oplus I = \kk\cdot 1_{A}+I$ and
\[
\Delta\left(  x\right)  \in A\otimes\underbrace{A}_{=\kk\cdot1_{A}+I}
=A\otimes\left(  \kk\cdot1_{A}+I\right)  =A\otimes\left(
\kk\cdot1_{A}\right)  +A\otimes I.
\]
Hence, there exist $y\in A\otimes\left(  \kk\cdot 1_{A}\right)  $ and
$z\in A\otimes I$ such that $\Delta\left(  x\right)  =y+z$. Consider these $y$
and $z$. We will show that $y=x\otimes1_{A}$.

Since $y \in A\otimes\left(  \kk\cdot 1_{A}\right) 
 = \kk \cdot \left(A \otimes 1_A\right) = A \otimes 1_A$, we can write $y$ in
the form $y=y^{\prime}\otimes 1_{A}$ for some $y^{\prime}\in A$. Consider this
$y^{\prime}$.

By the commutativity of \eqref{counit-diagram}, we have $\left(  \id_{A}
\otimes\epsilon\right)  \left(  \Delta\left(  x\right)  \right)  =x$ (where we
identify $A\otimes\kk$ with $A$). Hence,
\begin{align*}
x  & =\left(  \id_{A}\otimes\epsilon\right)  \left(  \underbrace{\Delta\left(
x\right)  }_{=y+z}\right)  =\left(  \id_{A}\otimes\epsilon\right)  \left(
y+z\right)  
 =\left(  \id_{A}\otimes\epsilon\right)  \left(  \underbrace{y}_{=y^{\prime
}\otimes1_{A}}\right)  +\left(  \id_{A}\otimes\epsilon\right)  \left(
\underbrace{z}_{\in A\otimes I}\right)  \\
& \in\underbrace{\left(  \id_{A}\otimes\epsilon\right)  \left(  y^{\prime
}\otimes1_{A}\right)  }_{=y^{\prime}\epsilon\left(  1_{A}\right)
}+\underbrace{\left(  \id_{A}\otimes\epsilon\right)  \left(  A\otimes I\right)
}_{=\id_{A}\left(  A\right)  \epsilon\left(  I\right)  }
 =y^{\prime}\underbrace{\epsilon\left(  1_{A}\right)  }_{=1}+\id_{A}\left(
A\right)  \underbrace{\epsilon\left(  I\right)  }_{\substack{=0\\\text{(since
}I=\ker\epsilon\text{ by part (d))}}} \\
&= y^{\prime}1+\id_{A}\left(  A\right)0
=y^{\prime}+0 = y^{\prime},
\end{align*}
which shows that $x = y^{\prime}$. Now, $y = \underbrace{y^{\prime}}_{=x}
\otimes 1_{A} = x\otimes 1_{A}$ and
\[
x=\underbrace{y}_{=x\otimes 1_{A}}+\underbrace{z}_{\in A\otimes I}\in
x\otimes\underbrace{1_{A}}_{=1}+A\otimes I=x\otimes1+A\otimes I.
\]
This solves part (f).

(g) Let $x\in I$. Thus, $x\in I=\ker\epsilon$ (by part (d)), so that
$\epsilon\left(  x\right)  =0$.

Let us introduce $y$ and $z$ as in the solution to part (f). As we saw in that
solution, we have $y=x\otimes1_{A}$.

We have
\[
z\in\underbrace{A}_{= \kk\cdot1_{A}+I}\otimes I=\left(
\kk\cdot1_{A}+I\right)  \otimes I=\left(  \kk\cdot1_{A}\right)
\otimes I+I\otimes I.
\]
Hence, there exist $u\in\left(  \kk\cdot1_{A}\right)  \otimes I$ and
$v\in I\otimes I$ such that $z=u+v$. Consider these $u$ and $v$. We will show
that $u=1_{A}\otimes x$.

We have $v\in I\otimes I$, so that $\left(  \epsilon\otimes\id_{A}\right)
\left(  v\right)  \in\left(  \epsilon\otimes\id_{A}\right)  \left(  I\otimes
I\right)  =\underbrace{\epsilon\left(  I\right)  }_{\substack{=0\\\text{(since
}I=\ker\epsilon\text{)}}}\id_{A}\left(  I\right)  =0$. Thus, $\left(
\epsilon\otimes\id_{A}\right)  \left(  v\right)  =0$.

Since $u\in\left(  \kk\cdot1_{A}\right)  \otimes I 
= \kk \cdot \left(1_A \otimes I\right) = 1_A \otimes I$, we can write $u$
in the form $u=1_{A}\otimes u^{\prime}$ for some $u^{\prime}\in I$. Consider
this $u^{\prime}$.

By the commutativity of \eqref{counit-diagram}, we have $\left(
\epsilon\otimes\id_{A}\right)  \left(  \Delta\left(  x\right)  \right)  =x$
(where we identify $\kk\otimes A$ with $A$). Hence,
\begin{align*}
x  & =\left(  \epsilon\otimes\id_{A}\right)  \left(  \underbrace{\Delta\left(
x\right)  }_{=y+z}\right)  =\left(  \epsilon\otimes\id_{A}\right)  \left(
y+z\right) 
 =\left(  \epsilon\otimes\id_{A}\right)  \left(  \underbrace{y}_{=x\otimes
1_{A}}\right)  +\left(  \epsilon\otimes\id_{A}\right) 
 \left(  \underbrace{z}_{=u+v}\right)  \\
& =\underbrace{\left(  \epsilon\otimes\id_{A}\right)  \left(  x\otimes
1_{A}\right)  }_{\substack{=\epsilon\left(  x\right)  \id_{A}\left(
1_{A}\right)  =0\\\text{(since }\epsilon\left(  x\right)  =0\text{)}}}+\left(
\epsilon\otimes\id_{A}\right)  \left(  u+v\right) 
 =\left(  \epsilon\otimes\id_{A}\right)  \left(  u+v\right)  =\left(
\epsilon\otimes\id_{A}\right)  \left(  u\right)  +\underbrace{\left(
\epsilon\otimes\id_{A}\right)  \left(  v\right)  }_{=0} \\
& =\left(  \epsilon
\otimes\id_{A}\right)  \left(  \underbrace{u}_{=1_{A}\otimes u^{\prime}}\right)
 =\left(  \epsilon\otimes\id_{A}\right)  \left(  1_{A}\otimes u^{\prime
}\right)  =\underbrace{\epsilon\left(  1_{A}\right)  }_{=1}u^{\prime
}=u^{\prime}.
\end{align*}
Now, $u=1_{A}\otimes\underbrace{u^{\prime}}_{=x}=1_{A}\otimes x$. Now,
\begin{align*}
\Delta\left(  x\right)    & =\underbrace{y}_{=x\otimes1_{A}}+
\underbrace{z}_{=u+v}=x\otimes1_{A}+\underbrace{u}_{=1_{A}\otimes x}+
\underbrace{v}_{\in
I\otimes I}\\
& \in x\otimes\underbrace{1_{A}}_{=1}+\underbrace{1_{A}}_{=1}\otimes
x+I\otimes I=x\otimes1+1\otimes x+I\otimes I
 =1\otimes x+x\otimes1+I\otimes I.
\end{align*}
In other words, $\Delta\left(  x\right)  =1\otimes x+x\otimes1+\Delta
_{+}\left(  x\right)  $ for some $\Delta_{+}\left(  x\right)  \in I\otimes I$.
This solves part (g).

(h) The definition of $I$ yields%
\begin{equation}
I=\bigoplus_{n>0}A_{n}=\bigoplus_{\ell>0}A_{\ell}=\sum_{\ell>0}A_{\ell}.
\label{sol.graded-connected-exercise.h.AlinI}%
\end{equation}


Now, let $n>0$ and $x\in A_{n}$. We must show that $\Delta\left(  x\right)
=1\otimes x+x\otimes1+\Delta_{+}\left(  x\right)  $, where $\Delta_{+}\left(
x\right)  $ lies in $\sum_{k=1}^{n-1}A_{k}\otimes A_{n-k}$.

Part (g) yields $\Delta\left(  x\right)  =1\otimes x+x\otimes1+\Delta
_{+}\left(  x\right)  $ for some $\Delta_{+}\left(  x\right)  \in I\otimes I$.
Consider this $\Delta_{+}\left(  x\right)  $. It clearly suffices to show that
this $\Delta_{+}\left(  x\right)  $ satisfies $\Delta_{+}\left(  x\right)
\in\sum_{k=1}^{n-1}A_{k}\otimes A_{n-k}$.

Regard $I\otimes I$ as a $\kk$-submodule of $A\otimes A$. From
$I=\sum_{\ell>0}A_{\ell}$, we obtain%
\begin{equation}
I\otimes I=\left(  \sum_{\ell>0}A_{\ell}\right)  \otimes\left(  \sum_{\ell
>0}A_{\ell}\right)  =\sum_{i>0,\ j>0}A_{i}\otimes A_{j}%
.\label{sol.graded-connected-exercise.h.2}%
\end{equation}


Let $\pi_{n}:A\otimes A\rightarrow A\otimes A$ be the projection of the graded
$\kk$-module $A\otimes A$ onto its $n$-th graded component $\left(
A\otimes A\right)  _{n}$. Then:

\begin{itemize}
\item The map $\pi_{n}$ annihilates the $k$-th graded component $\left(
A\otimes A\right)  _{k}$ for every $k\neq n$. In other words,
\begin{equation}
\pi_{n}\left(  \left(  A\otimes A\right)  _{k}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in \NN \text{ satisfying }k\neq
n\text{.}\label{sol.graded-connected-exercise.h.ann1}%
\end{equation}
Hence, every $i\in \NN $ and $j\in \NN $ satisfying $i+j\neq n$
satisfy%
\begin{equation}
\pi_{n}\left(  A_{i}\otimes A_{j}\right)
=0\label{sol.graded-connected-exercise.h.ann2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.graded-connected-exercise.h.ann2}):} Let
$i\in \NN $ and $j\in \NN $ be such that $i+j\neq n$. The definition
of the grading on $A\otimes A$ yields $A_{i}\otimes A_{j}\subset\left(
A\otimes A\right)  _{i+j}$. Applying the map $\pi_{n}$ to both sides of this
relation, we find%
\[
\pi_{n}\left(  A_{i}\otimes A_{j}\right)  \subset\pi_{n}\left(  \left(
A\otimes A\right)  _{i+j}\right)  =0
\]
(by (\ref{sol.graded-connected-exercise.h.ann1}), applied to $k=i+j$). Hence,
$\pi_{n}\left(  A_{i}\otimes A_{j}\right)  =0$. This proves
(\ref{sol.graded-connected-exercise.h.ann2}).}.

\item The map $\pi_{n}$ acts as the identity on the $n$-th graded component
$\left(  A\otimes A\right)  _{n}$. In other words,%
\begin{equation}
\pi_{n}\left(  z\right)  =z\ \ \ \ \ \ \ \ \ \ \text{for each }z\in\left(
A\otimes A\right)  _{n}.\label{sol.graded-connected-exercise.h.pres3}%
\end{equation}
Therefore, every $i\in \NN$ and $j\in \NN$ satisfying $i+j=n$
satisfy%
\begin{equation}
\pi_{n}\left(  A_{i}\otimes A_{j}\right)  =A_{i}\otimes A_{j}%
\label{sol.graded-connected-exercise.h.pres4}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.graded-connected-exercise.h.pres4}):} Let
$i\in \NN$ and $j\in \NN$ be such that $i+j=n$. The definition of
the grading on $A\otimes A$ yields $A_{i}\otimes A_{j}\subset\left(  A\otimes
A\right)  _{i+j}=\left(  A\otimes A\right)  _{n}$ (since $i+j=n$). Hence, each
$z\in A_{i}\otimes A_{j}$ satisfies $z\in\left(  A\otimes A\right)  _{n}$ and
therefore $\pi_{n}\left(  z\right)  =z$ (by
(\ref{sol.graded-connected-exercise.h.pres3})). In other words, the map
$\pi_{n}$ acts as the identity on the set $A_{i}\otimes A_{j}$. Therefore,
$\pi_{n}\left(  A_{i}\otimes A_{j}\right)  =A_{i}\otimes A_{j}$. This proves
(\ref{sol.graded-connected-exercise.h.pres4}).}.
\end{itemize}

Now, $x\in A_{n}$, so that $\Delta\left(  x\right)  \in\Delta\left(
A_{n}\right)  \subset\left(  A\otimes A\right)  _{n}$ (since the map $\Delta$
is graded). Also, $\underbrace{1}_{\in A_{0}}\otimes\underbrace{x}_{\in A_{n}%
}\in A_{0}\otimes A_{n}\subset\left(  A\otimes A\right)  _{0+n}$ (by the
definition of the grading on $A\otimes A$); this rewrites as $1\otimes
x\in\left(  A\otimes A\right)  _{n}$. Similarly, $x\otimes1\in\left(  A\otimes
A\right)  _{n}$. From $\Delta\left(  x\right)  =1\otimes x+x\otimes
1+\Delta_{+}\left(  x\right)  $, we obtain%
\[
\Delta_{+}\left(  x\right)  =\underbrace{\Delta\left(  x\right)  }_{\in\left(
A\otimes A\right)  _{n}}-\underbrace{1\otimes x}_{\in\left(  A\otimes
A\right)  _{n}}-\underbrace{x\otimes1}_{\in\left(  A\otimes A\right)  _{n}}%
\in\left(  A\otimes A\right)  _{n}-\left(  A\otimes A\right)  _{n}-\left(
A\otimes A\right)  _{n}\subset\left(  A\otimes A\right)  _{n}.
\]
Hence, (\ref{sol.graded-connected-exercise.h.pres3}) (applied to $z=\Delta
_{+}\left(  x\right)  $) yields $\pi_{n}\left(  \Delta_{+}\left(  x\right)
\right)  =\Delta_{+}\left(  x\right)  $. Thus,%
\begin{align*}
\Delta_{+}\left(  x\right)   &  =\pi_{n}\left(  \underbrace{\Delta_{+}\left(
x\right)  }_{\in I\otimes I}\right)  \in\pi_{n}\left(  I\otimes I\right)
=\pi_{n}\left(  \sum_{i>0,\ j>0}A_{i}\otimes A_{j}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.graded-connected-exercise.h.2}%
)}\right) \\
&  =\sum_{i>0,\ j>0}\pi_{n}\left(  A_{i}\otimes A_{j}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi_{n}\text{ is }%
\kk\text{-linear}\right) \\
&  =\sum\limits_{\substack{i>0,\ j>0;\\i+j=n}}\underbrace{\pi_{n}\left(  A_{i}\otimes
A_{j}\right)  }_{\substack{=A_{i}\otimes A_{j}\\\text{(by
(\ref{sol.graded-connected-exercise.h.pres4}))}}}+
\sum\limits_{\substack{i>0,\ j>0;\\i+j\neq n}}\underbrace{\pi_{n}\left(  A_{i}\otimes
A_{j}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.graded-connected-exercise.h.ann2}))}}}\\
&  =\sum\limits_{\substack{i>0,\ j>0;\\i+j=n}}A_{i}\otimes A_{j}+\underbrace{
\sum\limits_{\substack{i>0,\ j>0;\\i+j\neq n}}0}_{=0}
=\sum\limits_{\substack{i>0,\ j>0;\\i+j=n}%
}A_{i}\otimes A_{j}=\sum_{k=1}^{n-1}A_{k}\otimes A_{n-k}.
\end{align*}
We thus have shown that $\Delta\left(  x\right)  =1\otimes x+x\otimes
1+\Delta_{+}\left(  x\right)  $, where $\Delta_{+}\left(  x\right)  $ lies in
$\sum_{k=1}^{n-1}A_{k}\otimes A_{n-k}$. This solves
Exercise~\ref{graded-connected-exercise}(h).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.graded.Dq}}

\textit{Solution to Exercise \ref{exe.graded.Dq}.}

It is easy to solve Exercise \ref{exe.graded.Dq} by direct verification of all
the axioms on homogeneous elements. However, we shall take a different (more
``functorial'') route instead. The first step
on that route is to generalize the definition of $D_q $ given in the exercise:

\begin{definition}
\label{def.sol.graded.Dq.DqV}Fix $q\in \kk $. Let $V=\bigoplus
_{n\in \NN }V_n$ be a graded $\kk$-module (where the $V_n$ are
the homogeneous components of $V$). Let $D_{q,V}:V\rightarrow V$ be the
$\kk$-module endomorphism of $V$ defined by setting%
\[
D_{q,V}\left(  a\right)  =q^n a\qquad\text{for each }n\in \NN \text{ and
each }a\in V_{n}.
\]
(It is easy to see that this is well-defined; equivalently, $D_{q,V}$ could be
defined as the direct sum $\bigoplus_{n\in \NN }\left(  q^n %
\cdot\id_{V_{n}}\right)  :\bigoplus_{n\in \NN %
}V_{n}\rightarrow\bigoplus_{n\in \NN }V_n$ of the maps $q^n %
\cdot\id_{V_{n}}:V_{n}\rightarrow V_n$.)
\end{definition}

Let us now state a few basic properties of these endomorphisms $D_{q,V}$:

\begin{proposition}
\label{prop.sol.graded.DqV.f-comm}Fix $q\in \kk $. Let $V$ and $W$ be two
graded $\kk$-modules. Let $f:V\rightarrow W$ be a graded $ \kk %
$-linear map. Then, $D_{q,W}\circ f=f\circ D_{q,V}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.graded.DqV.f-comm}.]Write $V$ as
$V=\bigoplus_{n\in \NN }V_n$ (where the $V_n$ are the homogeneous
components of $V$). Write $W$ as
$W=\bigoplus_{n\in \NN }W_{n}$ (where the $W_{n}$ are the homogeneous
components of $W$).

Every element of $V$ is a $\kk$-linear combination
of homogeneous elements of $V$ (since $V$ is graded). Thus, if two
$\kk$-linear maps from $V$ agree on each homogeneous element of $V$,
then these two maps must be equal.

Let $v$ be a homogeneous element of $V$. Thus, there exists some
$n\in \NN $ such that $v\in V_n$. Consider this $n$. From $v\in V_n$,
we obtain $D_{q,V}\left(  v\right)  =q^n v$ (by the definition of $D_{q,V}$).

But from $v\in V_n$, we also obtain $f\left(  v\right)  \in f\left(
V_{n}\right)  \subset W_{n}$ (since the map $f$ is graded).

But the definition of $D_{q,W}$ shows that $D_{q,W}\left(  a\right)  =q^n a$
for each $a\in W_{n}$. Applying this to $a=f\left(  v\right)  $, we find
$D_{q,W}\left(  f\left(  v\right)  \right)  =q^n f\left(  v\right)  $ (since
$f\left(  v\right)  \in W_{n}$). Now, comparing%
\[
\left(  D_{q,W}\circ f\right)  \left(  v\right)  =D_{q,W}\left(  f\left(
v\right)  \right)  =q^n f\left(  v\right)
\]
with%
\[
\left(  f\circ D_{q,V}\right)  \left(  v\right)  =f\left(  \underbrace{D_{q,V}%
\left(  v\right)  }_{=q^n v}\right)  =f\left(  q^n v\right)  =q^n f\left(
v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is
} \kk \text{-linear}\right)  ,
\]
we obtain $\left(  D_{q,W}\circ f\right)  \left(  v\right)  =\left(  f\circ
D_{q,V}\right)  \left(  v\right)  $.

Let us forget that we fixed $v$. We thus have proved that $\left(
D_{q,W}\circ f\right)  \left(  v\right)  =\left(  f\circ D_{q,V}\right)
\left(  v\right)  $ for each homogeneous element $v$ of $V$. In other words,
the two maps $D_{q,W}\circ f$ and $f\circ D_{q,V}$ (from $V$ to $W$) agree on
each homogeneous element of $V$. Since these two maps are $\kk$-linear,
we can thus conclude that they must be equal (because if two $ \kk %
$-linear maps from $V$ agree on each homogeneous element of $V$, then these
two maps must be equal). In other words, we have $D_{q,W}\circ f=f\circ
D_{q,V}$. This proves Proposition \ref{prop.sol.graded.DqV.f-comm}.
\end{proof}

\begin{proposition}
\label{prop.sol.graded.DqV.tensor}Fix $q\in \kk $. Let $V$ and $W$ be two
graded $\kk$-modules. Then, $D_{q,V\otimes W}=D_{q,V}\otimes D_{q,W}$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.sol.graded.DqV.tensor}.]Let us first show the following:

\begin{statement}
\textit{Claim 1:} Let $v \in V$ and $w \in W$.
Then, $D_{q,V\otimes W}\left(  v\otimes w\right)
=\left(  D_{q,V}\otimes D_{q,W}\right)  \left(  v\otimes w\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} We must prove the equality
$D_{q,V\otimes W}\left(  v\otimes w\right)
=\left(  D_{q,V}\otimes D_{q,W}\right)  \left(  v\otimes w\right)  $.
This equality is clearly $\kk$-linear in $v$ (that is, both of its sides
depend $\kk$-linearly on $v$).
Thus, we can WLOG assume that $v$ is a homogeneous element of $V$
(since $v$ is always a $\kk$-linear combination of homogeneous elements
of $V$ (because $V$ is graded)).
Assume this.

For the same reason, we can WLOG assume that $w$ is a homogeneous element
of $W$. Assume this.

The element $v\in V$ is homogeneous. Thus, there
exists some $i\in \NN $ such that $v\in V_{i}$. Consider this $i$. From
$v\in V_{i}$, we obtain $D_{q,V}\left(  v\right)  =q^{i}v$ (by the definition
of $D_{q,V}$).

The element $w\in W$ is homogeneous. Thus, there exists some $j\in \NN $
such that $w\in W_{j}$. Consider this $j$. From $w\in W_{j}$, we obtain
$D_{q,W}\left(  w\right)  =q^{j}w$ (by the definition of $D_{q,W}$).

The definition of the grading on $V\otimes W$ yields $V_{i}\otimes
W_{j}\subset\left(  V\otimes W\right)  _{i+j}$. Hence,%
\[
\underbrace{v}_{\in V_{i}}\otimes\underbrace{w}_{\in W_{j}}\in V_{i}\otimes
W_{j}\subset\left(  V\otimes W\right)  _{i+j}.
\]


But the definition of $D_{q,V\otimes W}$ shows that $D_{q,V\otimes W}\left(
a\right)  =q^n a$ for each $n\in \NN $ and each $a\in\left(  V\otimes
W\right)  _{n}$. Applying this to $n=i+j$ and $a=v\otimes w$, we obtain
\begin{align*}
D_{q,V\otimes W}\left(  v\otimes w\right)   &  =\underbrace{q^{i+j}}%
_{=q^{i}q^{j}}\left(  v\otimes w\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }v\otimes w\in\left(  V\otimes W\right)  _{i+j}\right) \\
&  =q^{i}q^{j}\left(  v\otimes w\right)  =q^{i}v\otimes q^{j}w=\left(
D_{q,V}\otimes D_{q,W}\right)  \left(  v\otimes w\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  D_{q,V}\otimes
D_{q,W}\right)  \left(  v\otimes w\right)  =\underbrace{D_{q,V}\left(
v\right)  }_{=q^{i}v}\otimes\underbrace{D_{q,W}\left(  w\right)  }_{=q^{j}%
w}=q^{i}v\otimes q^{j}w\right)  .
\end{align*}
This proves Claim 1.]

Now, recall that the $\kk$-module $V\otimes W$ is spanned by the pure
tensors. Thus, if two $\kk$-linear maps from $V\otimes W$ agree on each
pure tensor, then these two maps must be equal.

But Claim 1 shows precisely that the two maps
$D_{q,V\otimes W}$ and $D_{q,V}\otimes D_{q,W}$ (from $V\otimes W$ to
$V\otimes W$) agree on each pure tensor. Since these two maps are $ \kk %
$-linear, we can thus conclude that these two maps must be equal (because if
two $\kk$-linear maps from $V\otimes W$ agree on each pure tensor, then
these two maps must be equal). In other words, $D_{q,V\otimes W}%
=D_{q,V}\otimes D_{q,W}$. This proves Proposition
\ref{prop.sol.graded.DqV.tensor}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.sol.graded.DqV.tensor}.]Let us first show the following:

\begin{statement}
\textit{Claim 1:} Let $v$ be a homogeneous element of $V$, and let $w$ be a
homogeneous element of $W$. Then, $D_{q,V\otimes W}\left(  v\otimes w\right)
=\left(  D_{q,V}\otimes D_{q,W}\right)  \left(  v\otimes w\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} The element $v\in V$ is homogeneous. Thus, there
exists some $i\in \NN $ such that $v\in V_{i}$. Consider this $i$. From
$v\in V_{i}$, we obtain $D_{q,V}\left(  v\right)  =q^{i}v$ (by the definition
of $D_{q,V}$).

The element $w\in W$ is homogeneous. Thus, there exists some $j\in \NN $
such that $w\in W_{j}$. Consider this $j$. From $w\in W_{j}$, we obtain
$D_{q,W}\left(  w\right)  =q^{j}w$ (by the definition of $D_{q,W}$).

The definition of the grading on $V\otimes W$ yields $V_{i}\otimes
W_{j}\subset\left(  V\otimes W\right)  _{i+j}$. Hence,%
\[
\underbrace{v}_{\in V_{i}}\otimes\underbrace{w}_{\in W_{j}}\in V_{i}\otimes
W_{j}\subset\left(  V\otimes W\right)  _{i+j}.
\]


But the definition of $D_{q,V\otimes W}$ shows that $D_{q,V\otimes W}\left(
a\right)  =q^n a$ for each $n\in \NN $ and each $a\in\left(  V\otimes
W\right)  _{n}$. Applying this to $n=i+j$ and $a=v\otimes w$, we obtain
\begin{align*}
D_{q,V\otimes W}\left(  v\otimes w\right)   &  =\underbrace{q^{i+j}}%
_{=q^{i}q^{j}}\left(  v\otimes w\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }v\otimes w\in\left(  V\otimes W\right)  _{i+j}\right) \\
&  =q^{i}q^{j}\left(  v\otimes w\right)  =q^{i}v\otimes q^{j}w=\left(
D_{q,V}\otimes D_{q,W}\right)  \left(  v\otimes w\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  D_{q,V}\otimes
D_{q,W}\right)  \left(  v\otimes w\right)  =\underbrace{D_{q,V}\left(
v\right)  }_{=q^{i}v}\otimes\underbrace{D_{q,W}\left(  w\right)  }_{=q^{j}%
w}=q^{i}v\otimes q^{j}w\right)  .
\end{align*}
This proves Claim 1.]

Now, recall that the $\kk$-module $V\otimes W$ is spanned by the pure
tensors. Thus, if two $\kk$-linear maps from $V\otimes W$ agree on each
pure tensor, then these two maps must be equal.

Let $p\in V\otimes W$ be a pure tensor. We shall show that $D_{q,V\otimes
W}\left(  p\right)  =\left(  D_{q,V}\otimes D_{q,W}\right)  \left(  p\right)
$.

Indeed, we can write $p$ in the form $p=v\otimes w$ for some $v\in V$ and
$w\in W$ (since $p$ is a pure tensor). Consider these $v$ and $w$.

Every element of $V$ is a $\kk$-linear combination of homogeneous
elements of $V$ (since $V$ is graded). Thus, $v$ is a $\kk$-linear
combination of homogeneous elements of $V$ (since $v$ is an element of $V$).
In other words, we can write $v$ in the form $v=\sum_{i=1}^{k}v_{i}$ for some
$k\in \NN $ and some homogeneous elements $v_{1},v_{2},\ldots,v_{k}$ of
$V$. Consider this $k$ and these $v_{1},v_{2},\ldots,v_{k}$.

Every element of $W$ is a $\kk$-linear combination of homogeneous
elements of $W$ (since $W$ is graded). Thus, $w$ is a $\kk$-linear
combination of homogeneous elements of $W$ (since $w$ is an element of $W$).
In other words, we can write $w$ in the form $w=\sum_{j=1}^{\ell}w_{j}$ for
some $\ell\in \NN $ and some homogeneous elements $w_{1},w_{2}%
,\ldots,w_{\ell}$ of $W$. Consider this $\ell$ and these $w_{1},w_{2}%
,\ldots,w_{\ell}$.

If $i\in\left\{  1,2,\ldots,k\right\}  $ and $j\in\left\{  1,2,\ldots
,\ell\right\}  $, then Claim 1 (applied to $v_{i}$ and $w_{j}$ instead of $v$
and $w$) shows that%
\begin{equation}
D_{q,V\otimes W}\left(  v_{i}\otimes w_{j}\right)  =\left(  D_{q,V}\otimes
D_{q,W}\right)  \left(  v_{i}\otimes w_{j}\right)
\label{pf.prop.sol.graded.DqV.tensor.3}%
\end{equation}
(since $v_{i}$ is a homogeneous element of $V$, whereas $w_{j}$ is a
homogeneous element of $W$).

Now,%
\begin{equation}
p=\underbrace{v}_{=\sum_{i=1}^{k}v_{i}}\otimes\underbrace{w}_{=\sum
_{j=1}^{\ell}w_{j}}=\left(  \sum_{i=1}^{k}v_{i}\right)  \otimes\left(
\sum_{j=1}^{\ell}w_{j}\right)  =\sum_{i=1}^{k}\sum_{j=1}^{\ell}v_{i}\otimes
w_{j}. \label{pf.prop.sol.graded.DqV.tensor.4}%
\end{equation}
Applying the map $D_{q,V\otimes W}$ to both sides of this equality, we obtain%
\begin{align}
D_{q,V\otimes W}\left(  p\right)   &  =D_{q,V\otimes W}\left(  \sum_{i=1}%
^{k}\sum_{j=1}^{\ell}v_{i}\otimes w_{j}\right) \nonumber\\
&  =\sum_{i=1}^{k}\sum_{j=1}^{\ell}\underbrace{D_{q,V\otimes W}\left(
v_{i}\otimes w_{j}\right)  }_{\substack{=\left(  D_{q,V}\otimes D_{q,W}%
\right)  \left(  v_{i}\otimes w_{j}\right)  \\\text{(by
\eqref{pf.prop.sol.graded.DqV.tensor.3})}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }D_{q,V\otimes W}\text{ is } \kk \text{-linear}%
\right) \nonumber\\
&  =\sum_{i=1}^{k}\sum_{j=1}^{\ell}\left(  D_{q,V}\otimes D_{q,W}\right)
\left(  v_{i}\otimes w_{j}\right)  . \label{pf.prop.sol.graded.DqV.tensor.5}%
\end{align}
On the other hand, applying the map $D_{q,V}\otimes D_{q,W}$ to both sides of
the equality \eqref{pf.prop.sol.graded.DqV.tensor.4}, we obtain%
\[
\left(  D_{q,V}\otimes D_{q,W}\right)  \left(  p\right)  =\left(
D_{q,V}\otimes D_{q,W}\right)  \left(  \sum_{i=1}^{k}\sum_{j=1}^{\ell}%
v_{i}\otimes w_{j}\right)  =\sum_{i=1}^{k}\sum_{j=1}^{\ell}\left(
D_{q,V}\otimes D_{q,W}\right)  \left(  v_{i}\otimes w_{j}\right)
\]
(since the map $D_{q,V}\otimes D_{q,W}$ is $\kk$-linear). Comparing
this with \eqref{pf.prop.sol.graded.DqV.tensor.5}, we find%
\[
D_{q,V\otimes W}\left(  p\right)  =\left(  D_{q,V}\otimes D_{q,W}\right)
\left(  p\right)  .
\]


Now, forget that we fixed $p$. We thus have proved that $D_{q,V\otimes
W}\left(  p\right)  =\left(  D_{q,V}\otimes D_{q,W}\right)  \left(  p\right)
$ for each pure tensor $p\in V\otimes W$. In other words, the two maps
$D_{q,V\otimes W}$ and $D_{q,V}\otimes D_{q,W}$ (from $V\otimes W$ to
$V\otimes W$) agree on each pure tensor. Since these two maps are $ \kk %
$-linear, we can thus conclude that these two maps must be equal (because if
two $\kk$-linear maps from $V\otimes W$ agree on each pure tensor, then
these two maps must be equal). In other words, $D_{q,V\otimes W}%
=D_{q,V}\otimes D_{q,W}$. This proves Proposition
\ref{prop.sol.graded.DqV.tensor}.
\end{proof}
\end{verlong}

We are now ready to solve Exercise \ref{exe.graded.Dq}:

In Definition \ref{def.sol.graded.Dq.DqV}, we have defined a map $D_{q,V}$ for
any graded $\kk$-module $V$. Applying this to $V=A$, we obtain a map
$D_{q,A}$. This map $D_{q,A}$ equals our map $D_q $ (because these two maps
are defined in the exact same way). In other words, we have $D_{q,A}=D_q $.

Let $m,u,\Delta,\epsilon$ be the structure maps of the bialgebra $A$, labelled
as usual (so $m:A\otimes A\rightarrow A$ is the multiplication map,
$u: \kk \rightarrow A$ is the unit, $\Delta:A\rightarrow A\otimes A$ is
the comultiplication, and $\epsilon:A\rightarrow \kk $ is the counit).
These structure maps $m,u,\Delta,\epsilon$ are graded (since $A$ is a graded bialgebra).

The map $D_q $ is a $\kk$-algebra homomorphism if and only if it is
$\kk$-linear and makes the diagrams%
\[
\xymatrix@C=5pc{
A \ar[r]^{D_q }  &  A \\
A \otimes A \ar[u]^{m}
\ar[r]^{D_q  \otimes D_q }  &  A \otimes A \ar[u]_{m}
}
\qquad
\xymatrix{ A \ar[rr]^{D_q}& & A \\ &\kk \ar[ul]^{u} \ar[ur]_{u}& }
\]
commute (by the definition of a $\kk$-algebra homomorphism). In other
words, the map $D_q $ is a $\kk$-algebra homomorphism if and only if
it is $\kk$-linear and satisfies $D_q \circ m=m\circ\left(
D_q \otimes D_q \right)  $ and $D_q \circ u=u$.

But Proposition \ref{prop.sol.graded.DqV.tensor} (applied to $V=A$ and $W=A$)
yields $D_{q,A\otimes A}=D_{q,A}\otimes D_{q,A}$.

The map $m:A\otimes A\rightarrow A$ is graded and $\kk$-linear; hence,
Proposition \ref{prop.sol.graded.DqV.f-comm} (applied to $V=A\otimes A$, $W=A$
and $f=m$) yields $D_{q,A}\circ m=m\circ D_{q,A\otimes A}$. In view of
$D_{q,A\otimes A}=D_{q,A}\otimes D_{q,A}$, this rewrites as $D_{q,A}\circ
m=m\circ\left(  D_{q,A}\otimes D_{q,A}\right)  $. In other words, $D_q \circ
m=m\circ\left(  D_q \otimes D_q \right)  $ (since $D_{q,A}=D_q $).

Also, the graded $\kk$-module $\kk$ satisfies $D_{q, \kk %
}=\id_{ \kk }$\ \ \ \ \footnote{\textit{Proof.}
Let $\lambda\in \kk $. Then, $\lambda\in \kk = \kk _{0}$ (by
the definition of the grading on $\kk$). But the definition of
$D_{q, \kk }$ yields%
\[
D_{q, \kk }\left(  a\right)  =q^n a\qquad\text{for each }n\in
 \NN \text{ and each }a\in \kk _{n}.
\]
Applying this to $n=0$ and $a=\lambda$, we obtain $D_{q, \kk }\left(
\lambda\right)  =q^{0}\lambda$ (since $\lambda\in \kk _{0}$). Hence,
$D_{q, \kk }\left(  \lambda\right)  =\underbrace{q^{0}}_{=1}%
\lambda=\lambda=\id_{ \kk }\left(  \lambda
\right)  $.
\par
Forget that we fixed $\lambda$. We thus have shown that $D_{q, \kk %
}\left(  \lambda\right)  =\id_{ \kk }\left(
\lambda\right)  $ for each $\lambda\in \kk $. In other words,
$D_{q, \kk }=\id_{ \kk }$.}.

The map $u: \kk \rightarrow A$ is graded and $\kk$-linear; hence,
Proposition \ref{prop.sol.graded.DqV.f-comm} (applied to $V= \kk $, $W=A$
and $f=u$) yields $D_{q,A}\circ u=u\circ\underbrace{D_{q, \kk }%
}_{=\id_{ \kk }}=u$. In other words, $D_q \circ
u=u$ (since $D_{q,A}=D_q $).

Recall that the map $D_q $ is a $\kk$-algebra homomorphism if and only
if it is $\kk$-linear and satisfies $D_q \circ m=m\circ\left(
D_q \otimes D_q \right)  $ and $D_q \circ u=u$. Thus, the map $D_q $ is a
$\kk$-algebra homomorphism (since it is $\kk$-linear and
satisfies $D_q \circ m=m\circ\left(  D_q \otimes D_q \right)  $ and
$D_q \circ u=u$).

The map $D_q $ is a $\kk$-coalgebra homomorphism if and only if it is
$\kk$-linear and makes the diagrams%
\[
\xymatrix@C=5pc{
A \ar[d]_{\Delta} \ar[r]^{D_q} & A \ar[d]^{\Delta} \\
A \otimes A \ar[r]^{D_q \otimes D_q} & A \otimes A
}\qquad
\xymatrix{
A \ar[dr]_{\epsilon} \ar[rr]^{D_q} & & A \ar[dl]^{\epsilon}  \\
&\kk&
}
\]
commute (by the definition of a $\kk$-coalgebra homomorphism). In other
words, the map $D_q $ is a $\kk$-coalgebra homomorphism if and only if
it is $\kk$-linear and satisfies $\left(  D_q \otimes D_q \right)
\circ\Delta=\Delta\circ D_q $ and $\epsilon\circ D_q =\epsilon$.

The map $\Delta:A\rightarrow A\otimes A$ is graded and $\kk$-linear;
hence, Proposition \ref{prop.sol.graded.DqV.f-comm} (applied to $V=A$,
$W=A\otimes A$ and $f=\Delta$) yields $D_{q,A\otimes A}\circ\Delta=\Delta\circ
D_{q,A}$. In view of $D_{q,A\otimes A}=D_{q,A}\otimes D_{q,A}$, this rewrites
as $\left(  D_{q,A}\otimes D_{q,A}\right)  \circ\Delta=\Delta\circ D_{q,A}$.
In other words, $\left(  D_q \otimes D_q \right)  \circ\Delta=\Delta\circ
D_q $ (since $D_{q,A}=D_q $).

The map $\epsilon:A\rightarrow \kk $ is graded and $\kk$-linear;
hence, Proposition \ref{prop.sol.graded.DqV.f-comm} (applied to $V=A$,
$W= \kk $ and $f=\epsilon$) yields $D_{q, \kk }\circ\epsilon
=\epsilon\circ\underbrace{D_{q,A}}_{=D_q }=\epsilon\circ D_q $. Hence,
$\epsilon\circ D_q =\underbrace{D_{q, \kk }}_{=\id_{\kk}}
\circ\epsilon=\epsilon$.

Recall that the map $D_q $ is a $\kk$-coalgebra homomorphism if and
only if it is $\kk$-linear and satisfies $\left(  D_q \otimes
D_q \right)  \circ\Delta=\Delta\circ D_q $ and $\epsilon\circ D_q %
=\epsilon$. Hence, the map $D_q $ is a $\kk$-coalgebra homomorphism
(since it is $\kk$-linear and satisfies $\left(  D_q \otimes
D_q \right)  \circ\Delta=\Delta\circ D_q $ and $\epsilon\circ D_q %
=\epsilon$).

Now, we know that the map $D_q $ is a $\kk$-algebra homomorphism and a
$\kk$-coalgebra homomorphism. In other words, $D_q $ is a $ \kk %
$-bialgebra homomorphism (by the definition of a $\kk$-bialgebra
homomorphism). This solves Exercise \ref{exe.graded.Dq}.

%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.bialg.tensor}}

\textit{Solution to Exercise \ref{exe.bialg.tensor}.} (a)

\begin{proof}
[Proof of Proposition~\ref{prop.bialg.tensor}.]Let $\theta$ be the canonical
$\kk$-module isomorphism $\kk\otimes\kk\rightarrow
\kk$ (sending each $\lambda\otimes\mu$ to $\lambda\mu$). Thus, $\theta$
is a $\kk$-algebra isomorphism. Our definition of the $\kk%
$-coalgebra $A\otimes B$ yields%
\begin{align}
\Delta_{A\otimes B}  &  =\left(  \id_{A}\otimes
T\otimes\id_{B}\right)  \circ\left(  \Delta_{A}%
\otimes\Delta_{B}\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\label{pf.prop.bialg.tensor.1}\\
\epsilon_{A\otimes B}  &  =\theta\circ\left(  \epsilon_{A}\otimes\epsilon
_{B}\right)  . \label{pf.prop.bialg.tensor.2}%
\end{align}


But recall that $A$ is a $\kk$-bialgebra. Thus, the maps $\Delta
_{A}:A\rightarrow A\otimes A$ and $\epsilon_{A}:A\rightarrow\kk$ are
$\kk$-algebra homomorphisms (by the definition of a bialgebra). The
same argument (applied to $B$ instead of $A$) shows that the maps $\Delta
_{B}:B\rightarrow B\otimes B$ and $\epsilon_{B}:B\rightarrow\kk$ are
$\kk$-algebra homomorphisms.

Now, Exercise~\ref{exe.tensorprod.morphism}(a) (applied to $A^{\prime
}=A\otimes A$, $B^{\prime}=B\otimes B$, $f=\Delta_{A}$ and $g=\Delta_{B}$)
shows that $\Delta_{A}\otimes\Delta_{B}:A\otimes B\rightarrow A\otimes
A\otimes B\otimes B$ is a $\kk$-algebra homomorphism.

Also, it is easy to show that if $\mathfrak{A}$ and $\mathfrak{B}$ are any two
$\kk$-algebras, then the map $T:\mathfrak{A}\otimes\mathfrak{B}%
\rightarrow\mathfrak{B}\otimes\mathfrak{A}$ is a $\kk$-algebra
homomorphism. Applying this to $\mathfrak{A}=A$ and $\mathfrak{B}=B$, we
conclude that the map $T:A\otimes B\rightarrow B\otimes A$ is a $\kk%
$-algebra homomorphism. Also, the maps $\id_{A}%
:A\rightarrow A$ and $\id_{B}:B\rightarrow B$ are
$\kk$-algebra homomorphisms. Now,
Exercise~\ref{exe.tensorprod.morphism}(a) (applied to $A$, $A$, $A\otimes B$,
$B\otimes A$, $\id_{A}$ and $T$ instead of $A$,
$A^{\prime}$, $B$, $B^{\prime}$, $f$ and $g$) shows that $\operatorname*{id}%
\nolimits_{A}\otimes T:A\otimes A\otimes B\rightarrow A\otimes B\otimes A$ is
a $\kk$-algebra homomorphism. Hence,
Exercise~\ref{exe.tensorprod.morphism}(a) (applied to $A\otimes A\otimes B$,
$A\otimes B\otimes A$, $B$, $B$, $\id_{A}\otimes T$
and $\id_{B}$ instead of $A$, $A^{\prime}$, $B$,
$B^{\prime}$, $f$ and $g$) shows that $\id_{A}\otimes
T\otimes\id_{B}:A\otimes A\otimes B\otimes
B\rightarrow A\otimes B\otimes A\otimes B$ is a $\kk$-algebra homomorphism.

We now know that the two maps $\Delta_{A}\otimes\Delta_{B}:A\otimes
B\rightarrow A\otimes A\otimes B\otimes B$ and $\operatorname*{id}%
\nolimits_{A}\otimes T\otimes\id_{B}:A\otimes A\otimes
B\otimes B\rightarrow A\otimes B\otimes A\otimes B$ are $\kk$-algebra
homomorphisms. Hence, their composition $\left(  \operatorname*{id}%
\nolimits_{A}\otimes T\otimes\id_{B}\right)
\circ\left(  \Delta_{A}\otimes\Delta_{B}\right)  $ must also be a $\kk%
$-algebra homomorphism. In light of (\ref{pf.prop.bialg.tensor.1}), this
rewrites as follows: The map $\Delta_{A\otimes B}$ is a $\kk$-algebra homomorphism.

Furthermore, Exercise~\ref{exe.tensorprod.morphism}(a) (applied to $A^{\prime
}=\kk$, $B^{\prime}=\kk$, $f=\epsilon_{A}$ and $g=\epsilon_{B}$)
shows that $\epsilon_{A}\otimes\epsilon_{B}:A\otimes B\rightarrow
\kk\otimes\kk$ is a $\kk$-algebra homomorphism. We now
know that the two maps $\epsilon_{A}\otimes\epsilon_{B}:A\otimes
B\rightarrow\kk\otimes\kk$ and $\theta:\kk\otimes
\kk\rightarrow\kk$ are $\kk$-algebra homomorphisms.
Hence, their composition $\theta\circ\left(  \epsilon_{A}\otimes\epsilon
_{B}\right)  $ must also be a $\kk$-algebra homomorphism. In light of
(\ref{pf.prop.bialg.tensor.2}), this rewrites as follows: The map
$\epsilon_{A\otimes B}$ is a $\kk$-algebra homomorphism.

Now, we know that the maps $\Delta_{A\otimes B}$ and $\epsilon_{A\otimes B} $
are $\kk$-algebra homomorphisms. In other words, $A\otimes B$ is a
$\kk$-bialgebra (by the definition of a bialgebra). This proves
Proposition~\ref{prop.bialg.tensor}.
\end{proof}

Hence, Exercise \ref{exe.bialg.tensor}(a) is solved.

(b) We know that $\left(  t_{g}\right)  _{g\in G}$ is a basis of the
$\kk$-module $\kk G$, whereas $\left(  t_{h}\right)  _{h\in H}$
is a basis of the $\kk$-module $\kk H$. Thus, $\left(
t_{g}\otimes t_{h}\right)  _{\left(  g,h\right)  \in G\times H}$ is a basis of
the $\kk$-module $\kk G\otimes\kk H$. Hence, we can define
a $\kk$-linear map $\Phi:\kk G\otimes\kk H\rightarrow
\kk\left[  G\times H\right]  $ by setting
\begin{equation}
\left(  \Phi\left(  t_{g}\otimes t_{h}\right)  =t_{\left(  g,h\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }\left(  g,h\right)  \in G\times H\right)
. \label{sol.bialg.tensor.b.Phi=}%
\end{equation}
Consider this map $\Phi$.

The family $\left(  t_{\left(  g,h\right)  }\right)  _{\left(  g,h\right)  \in
G\times H}$ is a basis of the $\kk$-module $\kk\left[  G\times
H\right]  $. Now, the map $\Phi$ is $\kk$-linear, and sends the basis
$\left(  t_{g}\otimes t_{h}\right)  _{\left(  g,h\right)  \in G\times H}$ of
$\kk G\otimes\kk H$ to the basis $\left(  t_{\left(  g,h\right)
}\right)  _{\left(  g,h\right)  \in G\times H}$ of $\kk\left[  G\times
H\right]  $ (by (\ref{sol.bialg.tensor.b.Phi=})). Therefore, this map $\Phi$
is an isomorphism of $\kk$-modules. Thus, in particular, the map $\Phi$
is invertible.

It is easy to see that
\begin{equation}
\Phi\left(  ab\right)  =\Phi\left(  a\right)  \Phi\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\kk G\otimes\kk H\text{
and }b\in\kk G\otimes\kk H \label{sol.bialg.tensor.b.mult}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bialg.tensor.b.mult}):} Let
$a\in\kk G\otimes\kk H$ and $b\in\kk G\otimes\kk H$ be
arbitrary. We must prove the equality $\Phi\left(  ab\right)  =\Phi\left(
a\right)  \Phi\left(  b\right)  $.
\par
Since this equality is $\kk$-linear in $a$, we can WLOG assume that $a
$ belongs to the basis $\left(  t_{g}\otimes t_{h}\right)  _{\left(
g,h\right)  \in G\times H}$ of the $\kk$-module $\kk%
G\otimes\kk H$. Assume this. Thus, $a=t_{g_{1}}\otimes t_{h_{1}}$ for
some $\left(  g_{1},h_{1}\right)  \in G\times H$. Consider this $\left(
g_{1},h_{1}\right)  $.
\par
Since the equality $\Phi\left(  ab\right)  =\Phi\left(  a\right)  \Phi\left(
b\right)  $ is $\kk$-linear in $b$, we can WLOG assume that $b$ belongs
to the basis $\left(  t_{g}\otimes t_{h}\right)  _{\left(  g,h\right)  \in
G\times H}$ of the $\kk$-module $\kk G\otimes\kk H$. Assume
this. Thus, $b=t_{g_{2}}\otimes t_{h_{2}}$ for some $\left(  g_{2}%
,h_{2}\right)  \in G\times H$. Consider this $\left(  g_{2},h_{2}\right)  $.
\par
Multiplying the equalities $a=t_{g_{1}}\otimes t_{h_{1}}$ and $b=t_{g_{2}%
}\otimes t_{h_{2}}$, we obtain%
\[
ab=\left(  t_{g_{1}}\otimes t_{h_{1}}\right)  \left(  t_{g_{2}}\otimes
t_{h_{2}}\right)  =\underbrace{t_{g_{1}}t_{g_{2}}}_{=t_{g_{1}g_{2}}}%
\otimes\underbrace{t_{h_{1}}t_{h_{2}}}_{=t_{h_{1}h_{2}}}=t_{g_{1}g_{2}}\otimes
t_{h_{1}h_{2}}.
\]
Applying the map $\Phi$ to both sides of this equality, we find%
\[
\Phi\left(  ab\right)  =\Phi\left(  t_{g_{1}g_{2}}\otimes t_{h_{1}h_{2}%
}\right)  =t_{\left(  g_{1}g_{2},h_{1}h_{2}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\right)  .
\]
Comparing this with%
\begin{align*}
\Phi\left(  \underbrace{a}_{=t_{g_{1}}\otimes t_{h_{1}}}\right)  \Phi\left(
\underbrace{b}_{=t_{g_{2}}\otimes t_{h_{2}}}\right)   &  =\underbrace{\Phi
\left(  t_{g_{1}}\otimes t_{h_{1}}\right)  }_{\substack{=t_{\left(
g_{1},h_{1}\right)  } \\\text{(by the definition of }\Phi\text{)}%
}}\underbrace{\Phi\left(  t_{g_{2}}\otimes t_{h_{2}}\right)  }%
_{\substack{=t_{\left(  g_{2},h_{2}\right)  } \\\text{(by the definition of
}\Phi\text{)} }}\\
&  =t_{\left(  g_{1},h_{1}\right)  }t_{\left(  g_{2},h_{2}\right)
}=t_{\left(  g_{1},h_{1}\right)  \left(  g_{2},h_{2}\right)  }=t_{\left(
g_{1}g_{2},h_{1}h_{2}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left(  g_{1},h_{1}\right)  \left(  g_{2},h_{2}\right)  =\left(  g_{1}%
g_{2},h_{1}h_{2}\right)  \right)  ,
\end{align*}
we obtain $\Phi\left(  ab\right)  =\Phi\left(  a\right)  \Phi\left(  b\right)
$. This proves (\ref{sol.bialg.tensor.b.mult}).}. It is also easy to see that
$\Phi\left(  1_{\kk G\otimes\kk H}\right)  =1_{\kk\left[
G\times H\right]  }$ (since $1_{\kk G\otimes\kk H}=1_{\kk%
G}\otimes1_{\kk H}=t_{1_{G}}\otimes t_{1_{H}}$ and $1_{\kk\left[
G\times H\right]  }=t_{1_{G\times H}}=t_{\left(  1_{G},1_{H}\right)  }$).
Combining this with (\ref{sol.bialg.tensor.b.mult}), we conclude that $\Phi$
is a $\kk$-algebra homomorphism (since the map $\Phi$ is $\kk%
$-linear). Hence, $\Phi$ is a $\kk$-algebra isomorphism (since $\Phi$
is invertible).

Furthermore,%
\begin{equation}
\Delta_{\kk\left[  G\times H\right]  }\circ\Phi=\left(  \Phi\otimes
\Phi\right)  \circ\Delta_{\kk G\otimes\kk H}%
\label{sol.bialg.tensor.b.comult}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bialg.tensor.b.comult}):} Let $\left(
g,h\right)  \in G\times H$. Then,%
\begin{equation}
\left(  \Delta_{\kk\left[  G\times H\right]  }\circ\Phi\right)  \left(
t_{g}\otimes t_{h}\right)  =\Delta_{\kk\left[  G\times H\right]
}\left(  \underbrace{\Phi\left(  t_{g}\otimes t_{h}\right)  }%
_{\substack{=t_{\left(  g,h\right)  }\\\text{(by the definition of }%
\Phi\text{)}}}\right)  =\Delta_{\kk\left[  G\times H\right]  }\left(
t_{\left(  g,h\right)  }\right)  =t_{\left(  g,h\right)  }\otimes t_{\left(
g,h\right)  }\label{sol.bialg.tensor.b.comult.pf.1}%
\end{equation}
(by the definition of the coalgebra structure on $\kk\left[  G\times
H\right]  $).
\par
On the other hand, the definition of the coalgebra structure on $\kk G$
shows that $\Delta_{\kk G}\left(  t_{g}\right)  =t_{g}\otimes t_{g}$.
Similarly, $\Delta_{\kk H}\left(  t_{h}\right)  =t_{h}\otimes t_{h}$.
But the definition of the coalgebra $\kk G\otimes\kk H$ shows that
$\Delta_{\kk G\otimes\kk H}=\left(  \operatorname*{id}%
\nolimits_{\kk G}\otimes T\otimes\id_{\kk%
H}\right)  \circ\left(  \Delta_{\kk G}\otimes\Delta_{\kk%
H}\right)  $. Applying both sides of this equality to $t_{g}\otimes t_{h}$, we
find%
\begin{align*}
\Delta_{\kk G\otimes\kk H}\left(  t_{g}\otimes t_{h}\right)   &
=\left(  \left(  \id_{\kk G}\otimes
T\otimes\id_{\kk H}\right)  \circ\left(
\Delta_{\kk G}\otimes\Delta_{\kk H}\right)  \right)  \left(
t_{g}\otimes t_{h}\right)  =\left(  \id_{\kk%
G}\otimes T\otimes\id_{\kk H}\right)  \left(
\underbrace{\left(  \Delta_{\kk G}\otimes\Delta_{\kk H}\right)
\left(  t_{g}\otimes t_{h}\right)  }_{=\Delta_{\kk G}\left(
t_{g}\right)  \otimes\Delta_{\kk H}\left(  t_{h}\right)  }\right)  \\
&  =\left(  \id_{\kk G}\otimes T\otimes
\id_{\kk H}\right)  \left(  \underbrace{\Delta
_{\kk G}\left(  t_{g}\right)  }_{=t_{g}\otimes t_{g}}\otimes
\underbrace{\Delta_{\kk H}\left(  t_{h}\right)  }_{=t_{h}\otimes t_{h}%
}\right)  =\left(  \id_{\kk G}\otimes
T\otimes\id_{\kk H}\right)  \left(  t_{g}\otimes
t_{g}\otimes t_{h}\otimes t_{h}\right)  \\
&  =\underbrace{\id_{\kk G}\left(  t_{g}\right)
}_{=t_{g}}\otimes\underbrace{T\left(  t_{g}\otimes t_{h}\right)
}_{\substack{=t_{h}\otimes t_{g}\\\text{(by the definition of }T\text{)}%
}}\otimes\underbrace{\id_{\kk H}\left(
t_{h}\right)  }_{=t_{h}}=t_{g}\otimes t_{h}\otimes t_{g}\otimes t_{h}.
\end{align*}
Applying the map $\Phi\otimes\Phi$ to both sides of this equality, we find%
\begin{align*}
\left(  \Phi\otimes\Phi\right)  \left(  \Delta_{\kk G\otimes\kk%
H}\left(  t_{g}\otimes t_{h}\right)  \right)   &  =\left(  \Phi\otimes
\Phi\right)  \left(  t_{g}\otimes t_{h}\otimes t_{g}\otimes t_{h}\right)
=\underbrace{\Phi\left(  t_{g}\otimes t_{h}\right)  }_{\substack{=t_{\left(
g,h\right)  }\\\text{(by the definition of }\Phi\text{)}}}\otimes
\underbrace{\Phi\left(  t_{g}\otimes t_{h}\right)  }_{\substack{=t_{\left(
g,h\right)  }\\\text{(by the definition of }\Phi\text{)}}}\\
&  =t_{\left(  g,h\right)  }\otimes t_{\left(  g,h\right)  }.
\end{align*}
Comparing this with (\ref{sol.bialg.tensor.b.comult.pf.1}), we obtain%
\[
\left(  \Delta_{\kk\left[  G\times H\right]  }\circ\Phi\right)  \left(
t_{g}\otimes t_{h}\right)  =\left(  \Phi\otimes\Phi\right)  \left(
\Delta_{\kk G\otimes\kk H}\left(  t_{g}\otimes t_{h}\right)
\right)  =\left(  \left(  \Phi\otimes\Phi\right)  \circ\Delta_{\kk%
G\otimes\kk H}\right)  \left(  t_{g}\otimes t_{h}\right)  .
\]
\par
Now, forget that we fixed $\left(  g,h\right)  $. We thus have shown that
$\left(  \Delta_{\kk\left[  G\times H\right]  }\circ\Phi\right)
\left(  t_{g}\otimes t_{h}\right)  =\left(  \left(  \Phi\otimes\Phi\right)
\circ\Delta_{\kk G\otimes\kk H}\right)  \left(  t_{g}\otimes
t_{h}\right)  $ for every $\left(  g,h\right)  \in G\times H$. In other words,
the two maps $\Delta_{\kk\left[  G\times H\right]  }\circ\Phi$ and
$\left(  \Phi\otimes\Phi\right)  \circ\Delta_{\kk G\otimes\kk H}$
are equal to each other on the basis $\left(  t_{g}\otimes t_{h}\right)
_{\left(  g,h\right)  \in G\times H}$ of the $\kk$-module
$\kk G\otimes\kk H$. Therefore, these two maps must be identical
(because they are $\kk$-linear). In other words, $\Delta_{\kk%
\left[  G\times H\right]  }\circ\Phi=\left(  \Phi\otimes\Phi\right)
\circ\Delta_{\kk G\otimes\kk H}$. This proves
(\ref{sol.bialg.tensor.b.comult}).}. A similar argument shows that
\[
\epsilon_{\kk\left[  G\times H\right]  }\circ\Phi=\epsilon
_{\kk G\otimes\kk H}.
\]
Combining this with (\ref{sol.bialg.tensor.b.comult}), we conclude that $\Phi$
is a $\kk$-coalgebra homomorphism (since the map $\Phi$ is $\kk%
$-linear). Hence, $\Phi$ is a $\kk$-coalgebra isomorphism (since $\Phi$
is invertible).

Thus, the map $\Phi$ is both a $\kk$-algebra isomorphism and a
$\kk$-coalgebra isomorphism. Hence, $\Phi$ is a $\kk$-bialgebra
isomorphism. Therefore, the $\kk$-bialgebra $\kk G\otimes
\kk H$ is isomorphic to the $\kk$-bialgebra $\kk\left[
G\times H\right]  $ (through the map $\Phi$). This solves Exercise
\ref{exe.bialg.tensor}(b).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.assoc}}

\textit{Solution to Exercise~\ref{exe.convolution.assoc}.} We have to
prove that the binary operation $\star$ on $\Hom\left(
C,A\right)  $ is associative. In other words, we have to prove that any three
elements $f$, $g$ and $h$ of $\Hom\left(  C,A\right)  $ satisfy
$f\star\left(  g\star h\right)  =\left(  f\star g\right)  \star h$.

So let $f$, $g$ and $h$ be three elements of $\Hom\left(
C,A\right)  $. As usual, denote by $m:A\otimes A\rightarrow A$ the
multiplication of $A$, and by $\Delta:C\rightarrow C\otimes C$ the
comultiplication of $C$.

By the definition of $g\star h$, we have $g\star h=m\circ\left(  g\otimes
h\right)  \circ\Delta$, so that
\begin{align*}
\underbrace{f}_{=\operatorname{id}_{A}\circ f\circ
\id_{C}}\otimes\underbrace{\left(  g\star h\right)  }_{=m\circ\left(
g\otimes h\right)  \circ\Delta} &  =\left(  \operatorname{id}_{A}%
\circ f\circ\operatorname{id}_{C}\right)  \otimes\left(
m\circ\left(  g\otimes h\right)  \circ\Delta\right)  \\
&  =\left(  \operatorname{id}_{A}\otimes m\right)  \circ\left(
f\otimes\left(  g\otimes h\right)  \right)  \circ\left(  
\id_{C}\otimes\Delta\right)  \\
&  =\left(  \operatorname{id}_{A}\otimes m\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\Delta\right)  .
\end{align*}
But the definition of $f\star\left(  g\star h\right)  $ yields%
\begin{equation}
f\star\left(  g\star h\right)  =m\circ\underbrace{\left(  f\otimes\left(
g\star h\right)  \right)  }_{=\left(  \operatorname{id}_{A}\otimes
m\right)  \circ\left(  f\otimes g\otimes h\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta\right)  }\circ\Delta
=m\circ\left(  \operatorname{id}_{A}\otimes m\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \operatorname{id}_{C}%
\otimes\Delta\right)  \circ\Delta.\label{sol.convolution.assoc.4}%
\end{equation}


On the other hand, $f\star g=m\circ\left(  f\otimes g\right)  \circ\Delta$ (by
the definition of $f\star g$), so that%
\begin{align*}
\underbrace{\left(  f\star g\right)  }_{=m\circ\left(  f\otimes g\right)
\circ\Delta}\circ\underbrace{h}_{=\operatorname{id}_{A}\circ
h\circ\operatorname{id}_{C}} &  =\left(  m\circ\left(  f\otimes
g\right)  \circ\Delta\right)  \otimes\left(  \operatorname{id}_{C}%
\circ h\circ\operatorname{id}_{C}\right)  \\
&  =\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
\left(  f\otimes g\right)  \otimes h\right)  \circ\left(  \Delta
\otimes\operatorname{id}_{C}\right)  \\
&  =\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \Delta\otimes
\id_{C}\right)  .
\end{align*}
Now, the definition of $\left(  f\star g\right)  \star h$ yields%
\begin{equation}
\left(  f\star g\right)  \star h=m\circ\underbrace{\left(  \left(  f\star
g\right)  \otimes h\right)  }_{=\left(  m\otimes
\id_{A}\right)  \circ\left(  f\otimes g\otimes h\right)  \circ\left(
\Delta\otimes\operatorname{id}_{C}\right)  }\circ\Delta
=m\circ\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta.\label{sol.convolution.assoc.14}%
\end{equation}


Now, recall that $A$ is a $\kk$-algebra, and hence the diagram
(\ref{associativity-diagram}) commutes (by the definition of a $\kk %
$-algebra). Thus, $m\circ\left(  \operatorname{id}_{A}\otimes
m\right)  =m\circ\left(  m\otimes\operatorname{id}_{A}\right)  $.
Also, $C$ is a $\kk$-coalgebra, and thus the diagram
(\ref{coassociativity-diagram}) commutes (by the definition of a $\kk %
$-coalgebra). Hence, $\left(  \operatorname{id}_{C}\otimes
\Delta\right)  \circ\Delta=\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta$. Now, (\ref{sol.convolution.assoc.4})
becomes%
\begin{align*}
f\star\left(  g\star h\right)   &  =\underbrace{m\circ\left(
\operatorname{id}_{A}\otimes m\right)  }_{=m\circ\left(
m\otimes\operatorname{id}_{A}\right)  }\circ\left(  f\otimes
g\otimes h\right)  \circ\underbrace{\left(  \operatorname{id}_{C}%
\otimes\Delta\right)  \circ\Delta}_{=\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta}\\
&  =m\circ\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta=\left(  f\star g\right)  \star h
\end{align*}
(by (\ref{sol.convolution.assoc.14})). Thus, $f\star\left(  g\star h\right)
=\left(  f\star g\right)  \star h$ is proven, and the solution of Exercise
\ref{exe.convolution.assoc} is complete.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.assoc}}

\textit{Solution to Exercise~\ref{exe.convolution.assoc}.} In this
solution, we will often use the following classical formula from linear algebra:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$ and $W^{\prime}$ are six
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$ and $\beta^{\prime
}:V^{\prime}\rightarrow W^{\prime}$ are four $\kk$-linear maps, then
$\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

\noindent
We will refer to this formula as the \dfn{composition-tensor relation}.

We have to prove that the binary operation $\star$ on $\Hom%
\left(  C, A\right)  $ is associative. In other words, we have to prove that
any three elements $f$, $g$ and $h$ of $\Hom\left(  C, A\right)
$ satisfy $f \star\left(  g \star h\right)  = \left(  f \star g\right)  \star
h$.

So let $f$, $g$ and $h$ be three elements of $\Hom\left(
C,A\right)  $. As usual, denote by $m:A\otimes A\rightarrow A$ the
multiplication of $A$, and by $\Delta:C\rightarrow C\otimes C$ the
comultiplication of $C$.

By the definition of $g\star h$, we have $g\star h=m\circ\left(  g\otimes
h\right)  \circ\Delta$, so that
\begin{align}
f\otimes\underbrace{\left(  g\star h\right)  }_{=m\circ\left(  g\otimes
h\right)  \circ\Delta} &  =\underbrace{f}_{=\operatorname{id}_{A}%
\circ f}\otimes\underbrace{\left(  m\circ\left(  g\otimes h\right)
\circ\Delta\right)  }_{=\left(  m\circ\left(  g\otimes h\right)  \right)
\circ\Delta}=\left(  \operatorname{id}_{A}\circ f\right)
\otimes\left(  \left(  m\circ\left(  g\otimes h\right)  \right)  \circ
\Delta\right)  \nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes\left(  m\circ\left(
g\otimes h\right)  \right)  \right)  \circ\left(  f\otimes\Delta\right)
\label{sol.convolution.assoc.1}%
\end{align}
(by the composition-tensor relation (applied to $U=C$, $V=A$, $W=A$,
$U^{\prime}=C$, $V^{\prime}=C\otimes C$, $W^{\prime}=A$, $\alpha=f$,
$\beta=\operatorname{id}_{A}$, $\alpha^{\prime}=\Delta$ and
$\beta^{\prime}=m\circ\left(  g\otimes h\right)  $)). But we have%
\[
\underbrace{\operatorname{id}_{A}}_{=
\id_{A}\circ\operatorname{id}_{A}}\otimes\left(  m\circ\left(
g\otimes h\right)  \right)  =\left(  \operatorname{id}_{A}%
\circ\operatorname{id}_{A}\right)  \otimes\left(  m\circ\left(
g\otimes h\right)  \right)  =\left(  \operatorname{id}_{A}\otimes
m\right)  \circ\left(  \operatorname{id}_{A}\otimes\left(  g\otimes
h\right)  \right)
\]
(by the composition-tensor relation (applied to $U=A$, $V=A$, $W=A$,
$U^{\prime}=C\otimes C$, $V^{\prime}=A\otimes A$, $W^{\prime}=A$,
$\alpha=\operatorname{id}_{A}$, $\beta=
\id_{A}$, $\alpha^{\prime}=g\otimes h$ and $\beta^{\prime}=m$)) and%
\[
\underbrace{f}_{=f\circ\operatorname{id}_{C}}\otimes
\underbrace{\Delta}_{=\operatorname{id}_{C\otimes C}\circ\Delta
}=\left(  f\circ\operatorname{id}_{C}\right)  \otimes\left(
\operatorname{id}_{C\otimes C}\circ\Delta\right)  =\left(
f\otimes\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta\right)
\]
(by the composition-tensor relation (applied to $U=C$, $V=C$, $W=A$,
$U^{\prime}=C$, $V^{\prime}=C\otimes C$, $W^{\prime}=C\otimes C$,
$\alpha=\operatorname{id}_{C}$, $\beta=f$, $\alpha^{\prime}=\Delta$
and $\beta^{\prime}=\operatorname{id}_{C\otimes C}$)). Hence,
(\ref{sol.convolution.assoc.1}) becomes%
\begin{align}
f\otimes\left(  g\star h\right)   &  =\underbrace{\left(  
\id_{A}\otimes\left(  m\circ\left(  g\otimes h\right)  \right)  \right)
}_{=\left(  \operatorname{id}_{A}\otimes m\right)  \circ\left(
\operatorname{id}_{A}\otimes\left(  g\otimes h\right)  \right)
}\circ\underbrace{\left(  f\otimes\Delta\right)  }_{=\left(  f\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta\right)  }\nonumber\\
&  =\left(  \operatorname{id}_{A}\otimes m\right)  \circ\left(
\operatorname{id}_{A}\otimes\left(  g\otimes h\right)  \right)
\circ\left(  f\otimes\operatorname{id}_{C\otimes C}\right)
\circ\left(  \operatorname{id}_{C}\otimes\Delta\right)
.\label{sol.convolution.assoc.2}%
\end{align}


But the composition-tensor relation (applied to $U=C$, $V=A$, $W=A$,
$U^{\prime}=C\otimes C$, $V^{\prime}=C\otimes C$, $W^{\prime}=A\otimes A$,
$\alpha=f$, $\beta=\operatorname{id}_{A}$, $\alpha^{\prime
}=\operatorname{id}_{C\otimes C}$ and $\beta^{\prime}=g\otimes h$)
yields%
\[
\left(  \operatorname{id}_{A}\circ f\right)  \otimes\left(  \left(
g\otimes h\right)  \circ\operatorname{id}_{C\otimes C}\right)
=\left(  \operatorname{id}_{A}\otimes\left(  g\otimes h\right)
\right)  \circ\left(  f\otimes\operatorname{id}_{C\otimes C}\right)
,
\]
thus%
\begin{equation}
\left(  \operatorname{id}_{A}\otimes\left(  g\otimes h\right)
\right)  \circ\left(  f\otimes\operatorname{id}_{C\otimes C}\right)
=\underbrace{\left(  \operatorname{id}_{A}\circ f\right)  }%
_{=f}\otimes\underbrace{\left(  \left(  g\otimes h\right)  \circ
\operatorname{id}_{C\otimes C}\right)  }_{=g\otimes h}=f\otimes
g\otimes h. \label{sol.convolution.assoc.3}%
\end{equation}
Now, the definition of $f\star\left(  g\star h\right)  $ yields%
\begin{align}
f\star\left(  g\star h\right)   &  =m\circ\underbrace{\left(  f\otimes\left(
g\star h\right)  \right)  }_{\substack{=\left(  
\id_{A}\otimes m\right)  \circ\left(  \operatorname{id}_{A}%
\otimes\left(  g\otimes h\right)  \right)  \circ\left(  f\otimes
\operatorname{id}_{C\otimes C}\right)  \circ\left(
\operatorname{id}_{C}\otimes\Delta\right)  \\\text{(by
(\ref{sol.convolution.assoc.2}))}}}\circ\Delta\nonumber\\
&  =m\circ\left(  \operatorname{id}_{A}\otimes m\right)
\circ\underbrace{\left(  \operatorname{id}_{A}\otimes\left(
g\otimes h\right)  \right)  \circ\left(  f\otimes
\id_{C\otimes C}\right)  }_{\substack{=f\otimes g\otimes h\\\text{(by
(\ref{sol.convolution.assoc.3}))}}}\circ\left(  
\id_{C}\otimes\Delta\right)  \circ\Delta\nonumber\\
&  =m\circ\left(  \operatorname{id}_{A}\otimes m\right)
\circ\left(  f\otimes g\otimes h\right)  \circ\left(  
\id_{C}\otimes\Delta\right)  \circ\Delta.
\label{sol.convolution.assoc.4}%
\end{align}


Our next goal is to obtain a similar formula for $\left(  f\star g\right)
\star h$. In fact, we start out by noticing that $f\star g=m\circ\left(
f\otimes g\right)  \circ\Delta$ (by the definition of $f\star g$). Hence,%
\begin{align}
\underbrace{\left(  f\star g\right)  }_{=m\circ\left(  f\otimes g\right)
\circ\Delta}\circ h &  =\underbrace{\left(  m\circ\left(  f\otimes g\right)
\circ\Delta\right)  }_{=\left(  m\circ\left(  f\otimes g\right)  \right)
\circ\Delta}\otimes\underbrace{h}_{=\operatorname{id}_{A}\circ
h}=\left(  \left(  m\circ\left(  f\otimes g\right)  \right)  \circ
\Delta\right)  \otimes\left(  \operatorname{id}_{A}\circ h\right)
\nonumber\\
&  =\left(  \left(  m\circ\left(  f\otimes g\right)  \right)  \otimes
\operatorname{id}_{A}\right)  \circ\left(  \Delta\otimes h\right)
\label{sol.convolution.assoc.11}%
\end{align}
(by the composition-tensor relation (applied to $U^{\prime}=C$, $V^{\prime}%
=A$, $W^{\prime}=A$, $U=C$, $V=C\otimes C$, $W=A$, $\alpha^{\prime}=h$,
$\beta^{\prime}=\operatorname{id}_{A}$, $\alpha=\Delta$ and
$\beta=m\circ\left(  f\otimes g\right)  $)). But we have%
\[
\left(  m\circ\left(  f\otimes g\right)  \right)  \otimes
\underbrace{\operatorname{id}_{A}}_{=
\id_{A}\circ\operatorname{id}_{A}}=\left(  m\circ\left(
f\otimes g\right)  \right)  \otimes\left(  \operatorname{id}_{A}%
\circ\operatorname{id}_{A}\right)  =\left(  m\otimes
\operatorname{id}_{A}\right)  \circ\left(  \left(  f\otimes
g\right)  \otimes\operatorname{id}_{A}\right)
\]
(by the composition-tensor relation (applied to $U^{\prime}=A$, $V^{\prime}%
=A$, $W^{\prime}=A$, $U=C\otimes C$, $V=A\otimes A$, $W=A$, $\alpha^{\prime
}=\operatorname{id}_{A}$, $\beta^{\prime}=
\id_{A}$, $\alpha=f\otimes g$ and $\beta=m$)) and%
\[
\underbrace{\Delta}_{=\operatorname{id}_{C\otimes C}\circ\Delta
}\otimes\underbrace{h}_{=h\circ\operatorname{id}_{C}}=\left(
\operatorname{id}_{C\otimes C}\circ\Delta\right)  \otimes\left(
h\circ\operatorname{id}_{C}\right)  =\left(  
\id_{C\otimes C}\otimes h\right)  \circ\left(  \Delta\otimes
\operatorname{id}_{C}\right)
\]
(by the composition-tensor relation (applied to $U^{\prime}=C$, $V^{\prime}%
=C$, $W^{\prime}=A$, $U=C$, $V=C\otimes C$, $W=C\otimes C$, $\alpha^{\prime
}=\operatorname{id}_{C}$, $\beta^{\prime}=h$, $\alpha=\Delta$ and
$\beta=\operatorname{id}_{C\otimes C}$)). Hence,
(\ref{sol.convolution.assoc.11}) becomes%
\begin{align}
\left(  f\star g\right)  \otimes h &  =\underbrace{\left(  \left(
m\circ\left(  f\otimes g\right)  \right)  \otimes
\id_{A}\right)  }_{=\left(  m\otimes\operatorname{id}_{A}%
\right)  \circ\left(  \left(  f\otimes g\right)  \otimes
\id_{A}\right)  }\circ\underbrace{\left(  \Delta\otimes h\right)
}_{=\left(  \operatorname{id}_{C\otimes C}\otimes h\right)
\circ\left(  \Delta\otimes\operatorname{id}_{C}\right)  }\nonumber\\
&  =\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
\left(  f\otimes g\right)  \otimes\operatorname{id}_{A}\right)
\circ\left(  \operatorname{id}_{C\otimes C}\otimes h\right)
\circ\left(  \Delta\otimes\operatorname{id}_{C}\right)
.\label{sol.convolution.assoc.12}%
\end{align}


But the composition-tensor relation (applied to $U^{\prime}=C$, $V^{\prime}%
=A$, $W^{\prime}=A$, $U=C\otimes C$, $V=C\otimes C$, $W=A\otimes A$,
$\alpha^{\prime}=h$, $\beta^{\prime}=\operatorname{id}_{A}$,
$\alpha=\operatorname{id}_{C\otimes C}$ and $\beta=f\otimes g$)
yields%
\[
\left(  \left(  f\otimes g\right)  \circ\operatorname{id}_{C\otimes
C}\right)  \otimes\left(  \operatorname{id}_{A}\circ h\right)
=\left(  \left(  f\otimes g\right)  \otimes\operatorname{id}_{A}%
\right)  \circ\left(  \operatorname{id}_{C\otimes C}\otimes
h\right)  ,
\]
thus%
\begin{equation}
\left(  \left(  f\otimes g\right)  \otimes\operatorname{id}_{A}%
\right)  \circ\left(  \operatorname{id}_{C\otimes C}\otimes
h\right)  =\underbrace{\left(  \left(  f\otimes g\right)  \circ
\operatorname{id}_{C\otimes C}\right)  }_{=f\otimes g}%
\otimes\underbrace{\left(  \operatorname{id}_{A}\circ h\right)
}_{=h}=f\otimes g\otimes h.\label{sol.convolution.assoc.13}%
\end{equation}
Now, the definition of $\left(  f\star g\right)  \star h$ yields%
\begin{align}
\left(  f\star g\right)  \star h &  =m\circ\underbrace{\left(  \left(  f\star
g\right)  \otimes h\right)  }_{\substack{=\left(  m\otimes
\id_{A}\right)  \circ\left(  \left(  f\otimes g\right)  \otimes
\operatorname{id}_{A}\right)  \circ\left(  
\id_{C\otimes C}\otimes h\right)  \circ\left(  \Delta\otimes
\operatorname{id}_{C}\right)  \\\text{(by
(\ref{sol.convolution.assoc.12}))}}}\circ\Delta\nonumber\\
&  =m\circ\left(  m\otimes\operatorname{id}_{A}\right)
\circ\underbrace{\left(  \left(  f\otimes g\right)  \otimes
\id_{A}\right)  \circ\left(  \operatorname{id}_{C\otimes
C}\otimes h\right)  }_{\substack{=f\otimes g\otimes h\\\text{(by
(\ref{sol.convolution.assoc.13}))}}}\circ\left(  \Delta\otimes
\operatorname{id}_{C}\right)  \circ\Delta\nonumber\\
&  =m\circ\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta.\label{sol.convolution.assoc.14}%
\end{align}


Now, recall that $A$ is a $\kk$-algebra, and hence the diagram
(\ref{associativity-diagram}) commutes (by the definition of a $\kk %
$-algebra). Thus, $m\circ\left(  \operatorname{id}_{A}\otimes
m\right)  =m\circ\left(  m\otimes\operatorname{id}_{A}\right)  $.
Also, $C$ is a $\kk$-coalgebra, and thus the diagram
(\ref{coassociativity-diagram}) commutes (by the definition of a $\kk %
$-coalgebra). Hence, $\left(  \operatorname{id}_{C}\otimes
\Delta\right)  \circ\Delta=\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta$. Now, (\ref{sol.convolution.assoc.4})
becomes%
\begin{align*}
f\star\left(  g\star h\right)    & =\underbrace{m\circ\left(
\operatorname{id}_{A}\otimes m\right)  }_{=m\circ\left(
m\otimes\operatorname{id}_{A}\right)  }\circ\left(  f\otimes
g\otimes h\right)  \circ\underbrace{\left(  \operatorname{id}_{C}%
\otimes\Delta\right)  \circ\Delta}_{=\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta}\\
& =m\circ\left(  m\otimes\operatorname{id}_{A}\right)  \circ\left(
f\otimes g\otimes h\right)  \circ\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta=\left(  f\star g\right)  \star h
\end{align*}
(by (\ref{sol.convolution.assoc.14})). Thus, $f\star\left(  g\star h\right)
=\left(  f\star g\right)  \star h$ is proven, and the solution of
Exercise~\ref{exe.convolution.assoc} is complete.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.tensor}}

\textit{Solution to Exercise~\ref{exe.convolution.tensor}.} (a) There are
two ways to solve an exercise like this: either by explicitly evaluating the
two sides on elements of their domain (in this case, it is of course enough
to only evaluate them on pure tensors) or by diagram chasing. In
the case of this particular exercise, both solutions are very easy, so let
us show them both.

\textit{First solution:} Here is the solution by explicit computations:

We need to prove that $\left( f\otimes g\right) \star \left( f^{\prime
}\otimes g^{\prime }\right) =\left( f\star f^{\prime }\right) \otimes \left(
g\star g^{\prime }\right) $. For this, it is clearly enough to show that
$\left( \left( f\otimes g\right) \star \left( f^{\prime }\otimes g^{\prime
}\right) \right) \left( c\otimes d\right) =\left( \left( f\star f^{\prime
}\right) \otimes \left( g\star g^{\prime }\right) \right) \left( c\otimes
d\right) $ for every $c\in C$ and $d\in D$. But this can be done directly:
Since $\Delta \left( c\otimes d\right) =\sum_{\left( c\right) ,\left(
d\right) }\left( c_{1}\otimes d_{1}\right) \otimes \left( c_{2}\otimes
d_{2}\right) $ (we are using the Sweedler notation here), we have
\begin{align*}
\left( \left( f\otimes g\right) \star \left( f^{\prime }\otimes g^{\prime}
\right) \right) \left( c\otimes d\right)
&=\sum_{\left( c\right) ,\left( d\right) }\underbrace{\left( f\otimes
g\right) \left( c_{1}\otimes d_{1}\right) }_{=f\left( c_{1}\right) \otimes
g\left( d_{1}\right) }\cdot \underbrace{\left( f^{\prime }\otimes g^{\prime
}\right) \left( c_{2}\otimes d_{2}\right) }_{=f^{\prime }\left( c_{2}\right)
\otimes g^{\prime }\left( d_{2}\right) } \\
&=\sum_{\left( c\right) ,\left( d\right) }\left( f\left( c_{1}\right)
\otimes g\left( d_{1}\right) \right) \left( f^{\prime }\left( c_{2}\right)
\otimes g^{\prime }\left( d_{2}\right) \right)  \\
&=\sum_{\left( c\right) ,\left( d\right) }f\left( c_{1}\right) f^{\prime
}\left( c_{2}\right) \otimes g\left( d_{1}\right) g^{\prime }\left(
d_{2}\right) .
\end{align*}
Compared to
\begin{align*}
\left( \left( f\star f^{\prime }\right) \otimes \left( g\star g^{\prime}
\right) \right) \left( c\otimes d\right)
&=\underbrace{\left( f\star f^{\prime }\right) \left( c\right) }
_{=\sum_{\left( c\right) }f\left( c_{1}\right) f^{\prime }\left(
c_{2}\right) }\otimes \underbrace{\left( g\star g^{\prime }\right) \left(
d\right) }_{=\sum_{\left( d\right) }g\left( d_{1}\right) g^{\prime }\left(
d_{2}\right) } \\
&=\left( \sum_{\left( c\right) }f\left( c_{1}\right) f^{\prime }\left(
c_{2}\right) \right) \otimes \left( \sum_{\left( d\right) }g\left(
d_{1}\right) g^{\prime }\left( d_{2}\right) \right)  \\
&=\sum_{\left( c\right) ,\left( d\right) }f\left( c_{1}\right) f^{\prime
}\left( c_{2}\right) \otimes g\left( d_{1}\right) g^{\prime }\left(
d_{2}\right) ,
\end{align*}
this yields $\left( \left( f\otimes g\right) \star \left( f^{\prime }\otimes
g^{\prime }\right) \right) \left( c\otimes d\right) =\left( \left( f\star
f^{\prime }\right) \otimes \left( g\star g^{\prime }\right) \right) \left(
c\otimes d\right) $, which completes our solution of
Exercise~\ref{exe.convolution.tensor}(a).

\textit{Second solution:} Now comes the solution by diagram chasing. Actually, we will not see any
diagrams here because in this particular case it would be a waste of space
to draw them; everything can be done by a short computation. Denote by $T$
the twist map $U\otimes V\rightarrow V\otimes U$ for any two $\kk$-modules $U$
and $V$. (We leave $U$ and $V$ out of the notation since these will always
be clear from the context.) By the definition of convolution, we have
\begin{align*}
\left( f\otimes g\right) \star \left( f^{\prime }\otimes g^{\prime }\right) 
&=\underbrace{m_{A\otimes B}}_{=\left( m_{A}\otimes m_{B}\right) \circ
\left( \id\otimes T\otimes \id\right) }\circ \left( \left( f\otimes g\right)
\otimes \left( f^{\prime }\otimes g^{\prime }\right) \right) \circ 
\underbrace{\Delta _{C\otimes D}}_{=\left( \id\otimes T\otimes \id\right) \circ
\left( \Delta _{C}\otimes \Delta _{D}\right) } \\
&=\left( m_{A}\otimes m_{B}\right) \circ \underbrace{\left( \id\otimes
T\otimes \id\right) \circ \left( \left( f\otimes g\right) \otimes \left(
f^{\prime }\otimes g^{\prime }\right) \right) }_{=\left( f\otimes f^{\prime
}\otimes g\otimes g^{\prime }\right) \circ \left( \id\otimes T\otimes \id\right)
}\circ \left( \id\otimes T\otimes \id\right) \circ \left( \Delta _{C}\otimes
\Delta _{D}\right)  \\
&=\left( m_{A}\otimes m_{B}\right) \circ \left( f\otimes f^{\prime }\otimes
g\otimes g^{\prime }\right) \circ \underbrace{\left( \id\otimes T\otimes
\id\right) \circ \left( \id\otimes T\otimes \id\right) }_{\substack{ =\id \\ 
\text{(since }T^{2}=\id\text{)}}}\circ \left( \Delta _{C}\otimes
\Delta _{D}\right) 
\\
&=\left( m_{A}\otimes m_{B}\right) \circ \left( f\otimes f^{\prime }\otimes
g\otimes g^{\prime }\right) \circ \left( \Delta _{C}\otimes \Delta
_{D}\right)  \\
&=\underbrace{\left( m_{A}\circ \left( f\otimes f^{\prime }\right) \circ
\Delta _{C}\right) }_{\substack{ =f\star f^{\prime } \\ \text{(by the
definition of convolution)}}}\otimes \underbrace{\left( m_{B}\circ \left(
g\otimes g^{\prime }\right) \circ \Delta _{D}\right) }_{\substack{ =g\star
g^{\prime } \\ \text{(by the definition of convolution)}}} \\
&=\left( f\star f^{\prime }\right) \otimes \left( g\star g^{\prime }\right)
.
\end{align*}
This solves Exercise~\ref{exe.convolution.tensor}(a) again.

(b) Recall that the products in the $\kk$-algebras
$\left(\Hom\left(C,A\right),\star\right)$,
$\left(\Hom\left(D,B\right),\star\right)$ and
$\left(\Hom\left(C \otimes D, A \otimes B\right),\star\right)$ are the
convolution products $\star$. For the sake of consistency, we will also
denote by $\star$ the product in the $\kk$-algebra
$\left(\Hom\left(C,A\right),\star\right) \otimes
 \left(\Hom\left(D,B\right),\star\right)$.

We need to prove that $R$ is a $\kk$-algebra homomorphism. For this, it
is clearly enough to show that $R$ preserves products and sends the unity
$\left(u_A \epsilon_C\right) \otimes \left(u_B \epsilon_D\right)$ of the
$\kk$-algebra
$\left(\Hom\left(C,A\right),\star\right) \otimes
 \left(\Hom\left(D,B\right),\star\right)$
to the unity $u_{A\otimes B} \epsilon_{C\otimes D}$ of the $\kk$-algebra
$\left(\Hom\left(C \otimes D, A \otimes B\right),\star\right)$.

Let us check this. First, let us verify that $R$ preserves products.
So we need to show that $R\left(F \star F^\prime\right) = R\left(F\right)
\star R\left(F^\prime\right)$ for all $F$ and $F^\prime$ in
$\left(\Hom\left(C,A\right),\star\right) \otimes
 \left(\Hom\left(D,B\right),\star\right)$. In proving this, we can WLOG
assume that $F$ and $F^\prime$ are pure tensors (since the claim
$R\left(F \star F^\prime\right) = R\left(F\right)
\star R\left(F^\prime\right)$
is linear in $F$ and $F^\prime$). Assume this, and write $F$ and $F^\prime$
as $F = f \otimes g$ and $F^\prime = f^\prime \otimes g^\prime$,
respectively. Then,
\begin{align*}
R\left(  F\star F^{\prime}\right)   &  =R\left(  \underbrace{\left(  f\otimes
g\right)  \star\left(  f^{\prime}\otimes g^{\prime}\right)  }
_{\substack{=\left(  f\star f^{\prime}\right)  \otimes\left(  g\star
g^{\prime}\right)  \\\text{(by the definition of the product in}\\\left(
\Hom \left(  C,A\right)  ,\star\right)  \otimes\left(
\Hom \left(  D,B\right)  ,\star\right)  \text{)}}}\right)
=R\left(  \left(  f\star f^{\prime}\right)  \otimes\left(  g\star g^{\prime
}\right)  \right)  =\left(  f\star f^{\prime}\right)  \otimes\left(  g\star
g^{\prime}\right)  \\
&  \qquad\qquad\left(  \text{by the definition of }R\text{; note that the
tensor sign changed its meaning here}\right)  \\
&  =\underbrace{\left(  f\otimes g\right)  }_{\substack{=R\left(  f\otimes
g\right)  \\\text{(by the definition of }R\text{)}}}\star\underbrace{\left(
f^{\prime}\otimes g^{\prime}\right)  }_{\substack{=R\left(  f^{\prime}\otimes
g^{\prime}\right)  \\\text{(by the definition of }R\text{)}}}\qquad
\qquad\left(  \text{by Exercise~\ref{exe.convolution.tensor}(a)}\right)  \\
&  =R\left(  \underbrace{f\otimes g}_{=F}\right)  \star R\left(
\underbrace{f^{\prime}\otimes g^{\prime}}_{=F^{\prime}}\right)  =R\left(
F\right)  \star R\left(  F^{\prime}\right)  ,
\end{align*}
where the meaning of the tensor sign (each time standing either for a tensor
or for a tensor product of maps) should be clear from the context. Thus,
we are done checking that $R$ preserves products.

It thus remains to verify that $R$ sends
$\left(u_A \epsilon_C\right) \otimes \left(u_B \epsilon_D\right)$
to $u_{A\otimes B} \epsilon_{C\otimes D}$. This is straightforward:
If $s$ denotes the canonical isomorphism $\kk \to \kk \otimes \kk$, then
\begin{align*}
R\left(\left(u_A \epsilon_C\right) \otimes \left(u_B \epsilon_D\right)\right)
&= \left(u_A \epsilon_C\right) \otimes \left(u_B \epsilon_D\right)
= \underbrace{\left(u_A \otimes u_B\right)}_{= u_{A\otimes B} \circ s^{-1}}
  \circ \underbrace{\left(\epsilon_C \otimes \epsilon_D\right)}
          _{= s \circ \epsilon_{C \otimes D}}
= u_{A\otimes B} \circ s^{-1} \circ s \circ \epsilon_{C \otimes D}
= u_{A\otimes B} \epsilon_{C\otimes D}.
\end{align*}

The solution of Exercise~\ref{exe.convolution.tensor}(b) is thus complete.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.tensor-curry}}

\textit{Solution to Exercise~\ref{exe.convolution.tensor-curry}.} Let $f$ and
$g$ be two elements of $\Hom \left(  C\otimes D,A\right)  $. We
are going to show that $\Phi\left(  f\right)  \star\Phi\left(  g\right)
=\Phi\left(  f\star g\right)  $.

Let $c\in C$. Let $d\in D$.

We shall use the Sweedler notation, namely writing
$\Delta_{C}\left(  c\right) =\sum_{\left(  c\right)  }c_{1}\otimes c_{2}$
and
$\Delta_{D}\left(  d\right) =\sum_{\left(  d\right)  }d_{1}\otimes d_{2}$.
(This is a neat opportunity to practice the use of the Sweedler notation
in a particularly simple setting. If
you are uncomfortable with the Sweedler notation, you are invited to fix a
decomposition $\Delta_{C}\left(  c\right)  =\sum_{i=1}^{n}p_{i}\otimes q_{i}$
of $\Delta_{C}\left(  c\right)  $ into a sum of pure tensors, as well as a
similar decomposition $\Delta_{D}\left(  d\right)  =\sum_{j=1}^{k}%
x_{j}\otimes y_{j}$ for
$\Delta_{D}\left(  d\right)$, and to replace each appearance of one of the
symbols%
\[
\sum_{\left(  c\right)  },\ \ \ c_{1},\ \ \ c_{2},\ \ \ \sum_{\left(
d\right)  },\ \ \ d_{1},\ \ \ d_{2}%
\]
by the symbol%
\[
\sum_{i=1}^{n},\ \ \ p_{i},\ \ \ q_{i},\ \ \ \sum_{j=1}^{k},\ \ \ x_{j}%
,\ \ \ \ y_{j},
\]
respectively.
This will translate our argument into a perfectly valid argument that does not
use the Sweedler notation.)

The definition of the $\kk$-coalgebra $C\otimes D$ yields
\[
\Delta_{C\otimes D}=\left(  \operatorname{id}_{C}\otimes T\otimes
\operatorname{id}_{D}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)
.
\]
Applying both sides of this equality to $c\otimes d$, we obtain%
\begin{align}
\Delta_{C\otimes D}\left(  c\otimes d\right)   &  =\left(  \left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \right)  \left(  c\otimes
d\right)  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}%
_{D}\right)  \underbrace{\left(  \left(  \Delta_{C}\otimes\Delta_{D}\right)
\left(  c\otimes d\right)  \right)  }_{=\Delta_{C}\left(  c\right)
\otimes\Delta_{D}\left(  d\right)  }\nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\Delta_{C}\left(  c\right)  }_{=\sum_{\left(  c\right)
}c_{1}\otimes c_{2}}\otimes\underbrace{\Delta_{D}\left(  d\right)  }%
_{=\sum_{\left(  d\right)  }d_{1}\otimes d_{2}}\right)  \nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\left(  \sum_{\left(  c\right)  }c_{1}\otimes
c_{2}\right)  \otimes\left(  \sum_{\left(  d\right)  }d_{1}\otimes
d_{2}\right)  }_{=\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }%
c_{1}\otimes c_{2}\otimes d_{1}\otimes d_{2}}\right)  \nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \sum_{\left(  c\right)  }\sum_{\left(  d\right)  }c_{1}\otimes
c_{2}\otimes d_{1}\otimes d_{2}\right)  \nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\underbrace{\left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)  \left(
c_{1}\otimes c_{2}\otimes d_{1}\otimes d_{2}\right)  }_{=\operatorname*{id}%
\nolimits_{C}\left(  c_{1}\right)  \otimes T\left(  c_{2}\otimes d_{1}\right)
\otimes\id_{D}\left(  d_{2}\right)  }\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }%
\underbrace{\id_{C}\left(  c_{1}\right)  }_{=c_{1}%
}\otimes\underbrace{T\left(  c_{2}\otimes d_{1}\right)  }_{\substack{=d_{1}%
\otimes c_{2}\\\text{(by the definition of }T\text{)}}}\otimes
\underbrace{\id_{D}\left(  d_{2}\right)  }_{=d_{2}%
}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }c_{1}\otimes
d_{1}\otimes c_{2}\otimes d_{2}.\label{sol.convolution.tensor-curry.Delcd}%
\end{align}
Now, the definition of convolution yields%
\[
\left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)  \left(
c\right)  =\sum_{\left(  c\right)  }\left(  \Phi\left(  f\right)  \right)
\left(  c_{1}\right)  \star\left(  \Phi\left(  g\right)  \right)  \left(
c_{2}\right)
\]
(since the multiplication in the $\kk$-algebra $\left(
\Hom \left(  D,A\right)  ,\star\right)  $ is $\star$). Applying
both sides of this equality to $d$, we obtain%
\begin{align}
&  \left(  \left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)
\left(  c\right)  \right)  \left(  d\right)  \nonumber\\
&  =\left(  \sum_{\left(  c\right)  }\left(  \Phi\left(  f\right)  \right)
\left(  c_{1}\right)  \star\left(  \Phi\left(  g\right)  \right)  \left(
c_{2}\right)  \right)  \left(  d\right)  \nonumber\\
&  =\sum_{\left(  c\right)  }\underbrace{\left(  \left(  \Phi\left(  f\right)
\right)  \left(  c_{1}\right)  \star\left(  \Phi\left(  g\right)  \right)
\left(  c_{2}\right)  \right)  \left(  d\right)  }_{\substack{=\sum_{\left(
d\right)  }\left(  \left(  \Phi\left(  f\right)  \right)  \left(
c_{1}\right)  \right)  \left(  d_{1}\right)  \cdot\left(  \left(  \Phi\left(
g\right)  \right)  \left(  c_{2}\right)  \right)  \left(  d_{2}\right)
\\\text{(by the definition of convolution)}}}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }\underbrace{\left(
\left(  \Phi\left(  f\right)  \right)  \left(  c_{1}\right)  \right)  \left(
d_{1}\right)  }_{\substack{=f\left(  c_{1}\otimes d_{1}\right)  \\\text{(by
the definition of }\Phi\text{)}}}\cdot\underbrace{\left(  \left(  \Phi\left(
g\right)  \right)  \left(  c_{2}\right)  \right)  \left(  d_{2}\right)
}_{\substack{=g\left(  c_{2}\otimes d_{2}\right)  \\\text{(by the definition
of }\Phi\text{)}}}\nonumber\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }f\left(  c_{1}\otimes
d_{1}\right)  g\left(  c_{2}\otimes d_{2}\right)
.\label{sol.convolution.tensor-curry.left}%
\end{align}


On the other hand, the definition of $\Phi$ yields%
\begin{align*}
&  \left(  \left(  \Phi\left(  f\star g\right)  \right)  \left(  c\right)
\right)  \left(  d\right)  \\
&  =\underbrace{\left(  f\star g\right)  }_{\substack{=m_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C\otimes D}\\\text{(by the definition of
convolution)}}}\left(  c\otimes d\right)  =\left(  m_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C\otimes D}\right)  \left(  c\otimes d\right)  \\
&  =m_{A}\left(  \left(  f\otimes g\right)  \left(  \underbrace{\Delta
_{C\otimes D}\left(  c\otimes d\right)  }_{\substack{=\sum_{\left(  c\right)
}\sum_{\left(  d\right)  }c_{1}\otimes d_{1}\otimes c_{2}\otimes
d_{2}\\\text{(by (\ref{sol.convolution.tensor-curry.Delcd}))}}}\right)
\right)  =m_{A}\left(  \left(  f\otimes g\right)  \left(  \sum_{\left(
c\right)  }\sum_{\left(  d\right)  }c_{1}\otimes d_{1}\otimes c_{2}\otimes
d_{2}\right)  \right)  \\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }m_{A}\left(
\underbrace{\left(  f\otimes g\right)  \left(  c_{1}\otimes d_{1}\otimes
c_{2}\otimes d_{2}\right)  }_{=f\left(  c_{1}\otimes d_{1}\right)  \otimes
g\left(  c_{2}\otimes d_{2}\right)  }\right)  =\sum_{\left(  c\right)  }%
\sum_{\left(  d\right)  }\underbrace{m_{A}\left(  f\left(  c_{1}\otimes
d_{1}\right)  \otimes g\left(  c_{2}\otimes d_{2}\right)  \right)
}_{\substack{=f\left(  c_{1}\otimes d_{1}\right)  g\left(  c_{2}\otimes
d_{2}\right)  \\\text{(by the definition of }m_{A}\text{)}}}\\
&  =\sum_{\left(  c\right)  }\sum_{\left(  d\right)  }f\left(  c_{1}\otimes
d_{1}\right)  g\left(  c_{2}\otimes d_{2}\right)  .
\end{align*}
Comparing this with (\ref{sol.convolution.tensor-curry.left}), we obtain%
\[
\left(  \left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)
\left(  c\right)  \right)  \left(  d\right)  =\left(  \left(  \Phi\left(
f\star g\right)  \right)  \left(  c\right)  \right)  \left(  d\right)  .
\]


Now, forget that we fixed $d$. We thus have shown that $\left(  \left(
\Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)  \left(  c\right)
\right)  \left(  d\right)  =\left(  \left(  \Phi\left(  f\star g\right)
\right)  \left(  c\right)  \right)  \left(  d\right)  $ for each $d\in D$. In
other words, we have $\left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)
\right)  \left(  c\right)  =\left(  \Phi\left(  f\star g\right)  \right)
\left(  c\right)  $.

Now, forget that we fixed $c$. We thus have shown that $\left(  \Phi\left(
f\right)  \star\Phi\left(  g\right)  \right)  \left(  c\right)  =\left(
\Phi\left(  f\star g\right)  \right)  \left(  c\right)  $ for each $c\in C$.
In other words, we have $\Phi\left(  f\right)  \star\Phi\left(  g\right)
=\Phi\left(  f\star g\right)  $.

Now, forget that we fixed $f$ and $g$. We thus have proven that every two
elements $f$ and $g$ of $\Hom \left(  C\otimes D,A\right)  $
satisfy%
\begin{equation}
\Phi\left(  f\right)  \star\Phi\left(  g\right)  =\Phi\left(  f\star g\right)
. \label{sol.convolution.tensor-curry.*}%
\end{equation}


Now, let $\theta$ be the canonical $\kk$-module isomorphism
$\kk\otimes\kk\rightarrow\kk$. Then, the definition of
the $\kk$-coalgebra $C\otimes D$ yields
\[
\epsilon_{C\otimes D}=\theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  .
\]


The unity of the $\kk$-algebra $\left(  \Hom \left(
D,A\right)  ,\star\right)  $ is $u_{A}\circ\epsilon_{D}$. In other words,%
\begin{equation}
1_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }=u_{A}%
\circ\epsilon_{D}.\label{sol.convolution.tensor-curry.u1}%
\end{equation}
Similarly,%
\begin{equation}
1_{\left(  \Hom \left(  C,\left(  \Hom \left(
D,A\right)  ,\star\right)  \right)  ,\star\right)  }=u_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\circ\epsilon
_{C}\label{sol.convolution.tensor-curry.u2}%
\end{equation}
and%
\begin{equation}
1_{\left(  \Hom \left(  C\otimes D,A\right)  ,\star\right)
}=u_{A}\circ\epsilon_{C\otimes D}.\label{sol.convolution.tensor-curry.u3}%
\end{equation}


Let $c\in C$. Let $d\in D$. The definition of $\Phi$ yields%
\begin{align*}
&  \left(  \left(  \Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)
\right)  \left(  c\right)  \right)  \left(  d\right) \\
&  =\left(  u_{A}\circ\epsilon_{C\otimes D}\right)  \left(  c\otimes d\right)
=u_{A}\left(  \underbrace{\epsilon_{C\otimes D}}_{=\theta\circ\left(
\epsilon_{C}\otimes\epsilon_{D}\right)  }\left(  c\otimes d\right)  \right)
=u_{A}\left(  \left(  \theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  \right)  \left(  c\otimes d\right)  \right) \\
&  =u_{A}\left(  \theta\left(  \underbrace{\left(  \epsilon_{C}\otimes
\epsilon_{D}\right)  \left(  c\otimes d\right)  }_{=\epsilon_{C}\left(
c\right)  \otimes\epsilon_{D}\left(  d\right)  }\right)  \right)
=u_{A}\left(  \underbrace{\theta\left(  \epsilon_{C}\left(  c\right)
\otimes\epsilon_{D}\left(  d\right)  \right)  }_{\substack{=\epsilon
_{C}\left(  c\right)  \epsilon_{D}\left(  d\right)  \\\text{(by the definition
of }\theta\text{)}}}\right)  =u_{A}\left(  \epsilon_{C}\left(  c\right)
\epsilon_{D}\left(  d\right)  \right) \\
&  =\epsilon_{C}\left(  c\right)  \epsilon_{D}\left(  d\right)  \cdot
1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }u_{A}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \underbrace{\left(  u_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\circ\epsilon_{C}\right)  \left(  c\right)  }_{=u_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\left(  \epsilon
_{C}\left(  c\right)  \right)  }\right)  \left(  d\right)   &  =\left(
\underbrace{u_{\left(  \Hom \left(  D,A\right)  ,\star\right)
}\left(  \epsilon_{C}\left(  c\right)  \right)  }_{\substack{=\epsilon
_{C}\left(  c\right)  \cdot1_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\\\text{(by the definition of }u_{\left(  \Hom
\left(  D,A\right)  ,\star\right)  }\text{)}}}\right)  \left(  d\right)
=\left(  \epsilon_{C}\left(  c\right)  \cdot1_{\left(  \Hom
\left(  D,A\right)  ,\star\right)  }\right)  \left(  d\right) \\
&  =\epsilon_{C}\left(  c\right)  \cdot\underbrace{1_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }}_{=u_{A}\circ
\epsilon_{D}}\left(  d\right)  =\epsilon_{C}\left(  c\right)  \cdot
\underbrace{\left(  u_{A}\circ\epsilon_{D}\right)  \left(  d\right)
}_{\substack{=u_{A}\left(  \epsilon_{D}\left(  d\right)  \right)
=\epsilon_{D}\left(  d\right)  \cdot1_{A}\\\text{(by the definition of }%
u_{A}\text{)}}}=\epsilon_{C}\left(  c\right)  \epsilon_{D}\left(  d\right)
\cdot1_{A},
\end{align*}
we obtain $\left(  \left(  \Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)
\right)  \left(  c\right)  \right)  \left(  d\right)  =\left(  \left(
u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}\right)  \left(  c\right)  \right)  \left(  d\right)  $.

Now, forget that we fixed $d$. We thus have shown that $\left(  \left(
\Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)  \right)  \left(  c\right)
\right)  \left(  d\right)  =\left(  \left(  u_{\left(  \Hom
\left(  D,A\right)  ,\star\right)  }\circ\epsilon_{C}\right)  \left(
c\right)  \right)  \left(  d\right)  $ for each $d\in D$. In other words, we
have $\left(  \Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)  \right)
\left(  c\right)  =\left(  u_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\circ\epsilon_{C}\right)  \left(  c\right)  $.

Now, forget that we fixed $c$. We thus have shown that $\left(  \Phi\left(
u_{A}\circ\epsilon_{C\otimes D}\right)  \right)  \left(  c\right)  =\left(
u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}\right)  \left(  c\right)  $ for each $c\in C$. In other
words, we have $\Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)
=u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}$.

Now,%
\[
\Phi\left(  \underbrace{1_{\left(  \Hom \left(  C\otimes
D,A\right)  ,\star\right)  }}_{\substack{=u_{A}\circ\epsilon_{C\otimes
D}\\\text{(by (\ref{sol.convolution.tensor-curry.u3}))}}}\right)  =\Phi\left(
u_{A}\circ\epsilon_{C\otimes D}\right)  =u_{\left(  \Hom \left(
D,A\right)  ,\star\right)  }\circ\epsilon_{C}=1_{\left(  \Hom
\left(  C,\left(  \Hom \left(  D,A\right)  ,\star\right)
\right)  ,\star\right)  }%
\]
(by (\ref{sol.convolution.tensor-curry.u2})).

We now know that the map $\Phi$ is $\kk$-linear and satisfies
(\ref{sol.convolution.tensor-curry.*}) and $\Phi\left(  1_{\left(
\Hom \left(  C\otimes D,A\right)  ,\star\right)  }\right)
=1_{\left(  \Hom \left(  C,\left(  \Hom \left(
D,A\right)  ,\star\right)  \right)  ,\star\right)  }$. Thus, the map $\Phi$ is
a $\kk$-algebra homomorphism%
\[
\left(  \Hom \left(  C\otimes D,A\right)  ,\star\right)
\rightarrow\left(  \Hom \left(  C,\left(  \Hom
\left(  D,A\right)  ,\star\right)  \right)  ,\star\right)  .
\]
Since the map $\Phi$ is furthermore invertible (because $\Phi$ is a
$\kk$-module isomorphism), we thus conclude that $\Phi$ is a
$\kk$-algebra isomorphism%
\[
\left(  \Hom \left(  C\otimes D,A\right)  ,\star\right)
\rightarrow\left(  \Hom \left(  C,\left(  \Hom
\left(  D,A\right)  ,\star\right)  \right)  ,\star\right)  .
\]
This solves Exercise \ref{exe.convolution.tensor-curry}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.tensor-curry}}

\textit{Solution to Exercise \ref{exe.convolution.tensor-curry}.} Let $f$ and
$g$ be two elements of $\Hom \left(  C\otimes D,A\right)  $. We
are going to show that $\Phi\left(  f\right)  \star\Phi\left(  g\right)
=\Phi\left(  f\star g\right)  $.

Let $c\in C$. Let $d\in D$.

The element $\Delta_{C}\left(  c\right)  \in C\otimes C$ can be written as a
sum of pure tensors (since every element of $C\otimes C$ can be written as a
sum of pure tensors). In other words, $\Delta_{C}\left(  c\right)  $ can be
written as $\Delta_{C}\left(  c\right)  =\sum_{i=1}^{n}p_{i}\otimes q_{i}$ for
some $n\in \NN$, some elements $p_{1},p_{2},\ldots,p_{n}$ of $C$ and
some elements $q_{1},q_{2},\ldots,q_{n}$ of $C$. Consider this $n$, these
$p_{1},p_{2},\ldots,p_{n}$ and these $q_{1},q_{2},\ldots,q_{n}$.

The element $\Delta_{D}\left(  d\right)  \in D\otimes D$ can be written as a
sum of pure tensors (since every element of $D\otimes D$ can be written as a
sum of pure tensors). In other words, $\Delta_{D}\left(  d\right)  $ can be
written as $\Delta_{D}\left(  d\right)  =\sum_{j=1}^{k}x_{j}\otimes y_{j}$ for
some $k\in \NN$, some elements $x_{1},x_{2},\ldots,x_{k}$ of $D$ and
some elements $y_{1},y_{2},\ldots,y_{k}$ of $D$. Consider this $k$, these
$x_{1},x_{2},\ldots,x_{k}$ and these $y_{1},y_{2},\ldots,y_{k}$.

For each $i\in\left\{  1,2,\ldots,n\right\}  $, we define an element $P_{i}%
\in\Hom \left(  D,A\right)  $ by%
\begin{equation}
P_{i}=\left(  \Phi\left(  f\right)  \right)  \left(  p_{i}\right)  .
\label{sol.convolution.tensor-curry.Pi=}%
\end{equation}
Then, each $i\in\left\{  1,2,\ldots,n\right\}  $ and $z\in D$ satisfy%
\begin{equation}
\underbrace{P_{i}}_{=\left(  \Phi\left(  f\right)  \right)  \left(
p_{i}\right)  }\left(  z\right)  =\left(  \left(  \Phi\left(  f\right)
\right)  \left(  p_{i}\right)  \right)  \left(  z\right)  =f\left(
p_{i}\otimes z\right)  \label{sol.convolution.tensor-curry.Piz=}%
\end{equation}
(by the definition of the map $\Phi$).

For each $i\in\left\{  1,2,\ldots,n\right\}  $, we define an element $Q_{i}%
\in\Hom \left(  D,A\right)  $ by%
\begin{equation}
Q_{i}=\left(  \Phi\left(  g\right)  \right)  \left(  q_{i}\right)  .
\label{sol.convolution.tensor-curry.Qi=}%
\end{equation}
Then, each $i\in\left\{  1,2,\ldots,n\right\}  $ and $z\in D$ satisfy%
\begin{equation}
\underbrace{Q_{i}}_{=\left(  \Phi\left(  g\right)  \right)  \left(
q_{i}\right)  }\left(  z\right)  =\left(  \left(  \Phi\left(  g\right)
\right)  \left(  q_{i}\right)  \right)  \left(  z\right)  =g\left(
q_{i}\otimes z\right)  \label{sol.convolution.tensor-curry.Qiz=}%
\end{equation}
(by the definition of the map $\Phi$).

The definition of the $\kk$-coalgebra $C\otimes D$ yields
\[
\Delta_{C\otimes D}=\left(  \operatorname{id}_{C}\otimes T\otimes
\operatorname{id}_{D}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{D}\right)
.
\]
Applying both sides of this equality to $c\otimes d$, we obtain%
\begin{align}
\Delta_{C\otimes D}\left(  c\otimes d\right)   &  =\left(  \left(
\operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{D}\right)  \right)  \left(  c\otimes
d\right)  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}%
_{D}\right)  \underbrace{\left(  \left(  \Delta_{C}\otimes\Delta_{D}\right)
\left(  c\otimes d\right)  \right)  }_{=\Delta_{C}\left(  c\right)
\otimes\Delta_{D}\left(  d\right)  }\nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\Delta_{C}\left(  c\right)  }_{=\sum_{i=1}^{n}p_{i}\otimes
q_{i}}\otimes\underbrace{\Delta_{D}\left(  d\right)  }_{=\sum_{j=1}^{k}%
x_{j}\otimes y_{j}}\right) \nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \underbrace{\left(  \sum_{i=1}^{n}p_{i}\otimes q_{i}\right)
\otimes\left(  \sum_{j=1}^{k}x_{j}\otimes y_{j}\right)  }_{\substack{=
\sum_{i=1}^{n}\sum_{j=1}^{k}\left(  p_{i}\otimes q_{i}\right)  \otimes\left(
x_{j}\otimes y_{j}\right)  \\\text{(since the tensor product is }%
\kk\text{-bilinear)}}}\right) \nonumber\\
&  =\left(  \operatorname{id}_{C}\otimes T\otimes\operatorname{id}_{D}\right)
\left(  \sum_{i=1}^{n}\sum_{j=1}^{k}\left(  p_{i}\otimes q_{i}\right)
\otimes\left(  x_{j}\otimes y_{j}\right)  \right) \nonumber\\
&  =\sum_{i=1}^{n}\underbrace{\left(  \operatorname{id}_{C}\otimes
T\otimes\operatorname{id}_{D}\right)  \left(  \sum_{j=1}^{k}\left(
p_{i}\otimes q_{i}\right)  \otimes\left(  x_{j}\otimes y_{j}\right)  \right)
}_{\substack{=\sum_{j=1}^{k}\left(  \operatorname{id}_{C}\otimes
T\otimes\operatorname{id}_{D}\right)  \left(  \left(  p_{i}\otimes
q_{i}\right)  \otimes\left(  x_{j}\otimes y_{j}\right)  \right)
\\\text{(since the map }\operatorname{id}_{C}\otimes T\otimes\operatorname{id}%
_{D}\text{ is }\kk\text{-linear)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\operatorname{id}%
_{C}\otimes T\otimes\operatorname{id}_{D}\text{ is }\kk\text{-linear}%
\right) \nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}\left(  \operatorname{id}_{C}\otimes
T\otimes\operatorname{id}_{D}\right)  \left(  \underbrace{\left(  p_{i}\otimes
q_{i}\right)  \otimes\left(  x_{j}\otimes y_{j}\right)  }_{=p_{i}\otimes
q_{i}\otimes x_{j}\otimes y_{j}}\right) \nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}\underbrace{\left(  \operatorname{id}%
_{C}\otimes T\otimes\operatorname{id}_{D}\right)  \left(  p_{i}\otimes
q_{i}\otimes x_{j}\otimes y_{j}\right)  }_{=\id_{C}%
\left(  p_{i}\right)  \otimes T\left(  q_{i}\otimes x_{j}\right)
\otimes\id_{D}\left(  y_{j}\right)  }\nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}\underbrace{\id_{C}%
\left(  p_{i}\right)  }_{=p_{i}}\otimes\underbrace{T\left(  q_{i}\otimes
x_{j}\right)  }_{\substack{=x_{j}\otimes q_{i}\\\text{(by the definition of
}T\text{)}}}\otimes\underbrace{\id_{D}\left(
y_{j}\right)  }_{=y_{j}}\nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}p_{i}\otimes x_{j}\otimes q_{i}\otimes y_{j}.
\label{sol.convolution.tensor-curry.Delcd}%
\end{align}
Now, the definition of convolution yields%
\[
\Phi\left(  f\right)  \star\Phi\left(  g\right)  =m_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\circ\left(
\Phi\left(  f\right)  \otimes\Phi\left(  g\right)  \right)  \circ\Delta_{C}.
\]
Applying both sides of this equality to $c$, we obtain%
\begin{align*}
\left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)  \left(
c\right)   &  =\left(  m_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\circ\left(  \Phi\left(  f\right)  \otimes\Phi\left(
g\right)  \right)  \circ\Delta_{C}\right)  \left(  c\right) \\
&  =m_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }\left(
\left(  \Phi\left(  f\right)  \otimes\Phi\left(  g\right)  \right)  \left(
\underbrace{\Delta_{C}\left(  c\right)  }_{=\sum_{i=1}^{n}p_{i}\otimes q_{i}%
}\right)  \right) \\
&  =m_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }\left(
\underbrace{\left(  \Phi\left(  f\right)  \otimes\Phi\left(  g\right)
\right)  \left(  \sum_{i=1}^{n}p_{i}\otimes q_{i}\right)  }_{\substack{=
\sum_{i=1}^{n}\left(  \Phi\left(  f\right)  \otimes
\Phi\left(  g\right)  \right)
\left(  p_{i}\otimes q_{i}\right)  \\\text{(since the map }\Phi\left(
f\right)  \otimes\Phi\left(  g\right)  \text{ is }\kk\text{-linear)}%
}}\right) \\
&  =m_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }\left(
\sum_{i=1}^{n}\underbrace{\left(  \Phi\left(  f\right)  \otimes\Phi\left(
g\right)  \right)  \left(  p_{i}\otimes q_{i}\right)  }_{=\left(  \Phi\left(
f\right)  \right)  \left(  p_{i}\right)  \otimes\left(  \Phi\left(  g\right)
\right)  \left(  q_{i}\right)  }\right) \\
&  =m_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }\left(
\sum_{i=1}^{n}\underbrace{\left(  \Phi\left(  f\right)  \right)  \left(
p_{i}\right)  }_{\substack{=P_{i}\\\text{(by
(\ref{sol.convolution.tensor-curry.Pi=}))}}}\otimes\underbrace{\left(
\Phi\left(  g\right)  \right)  \left(  q_{i}\right)  }_{\substack{=Q_{i}%
\\\text{(by (\ref{sol.convolution.tensor-curry.Qi=}))}}}\right)  =m_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\left(  \sum_{i=1}%
^{n}P_{i}\otimes Q_{i}\right) \\
&  =\sum_{i=1}^{n}\underbrace{m_{\left(  \Hom \left(
D,A\right)  ,\star\right)  }\left(  P_{i}\otimes Q_{i}\right)  }%
_{\substack{=P_{i}\star Q_{i}\\\text{(by the definition of }m_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\\\text{(since the
multiplication in the }\kk\text{-algebra }\left(  \Hom
\left(  D,A\right)  ,\star\right)  \text{ is }\star\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }m_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\text{ is }%
\kk\text{-linear}\right) \\
&  =\sum_{i=1}^{n}\underbrace{P_{i}\star Q_{i}}_{\substack{=m_{A}\circ\left(
P_{i}\otimes Q_{i}\right)  \circ\Delta_{D}\\\text{(by the definition of
convolution)}}}=\sum_{i=1}^{n}m_{A}\circ\left(  P_{i}\otimes Q_{i}\right)
\circ\Delta_{D}.
\end{align*}
Applying both sides of this equality to $d$, we obtain%
\begin{align}
&  \left(  \left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)
\left(  c\right)  \right)  \left(  d\right) \nonumber\\
&  =\left(  \sum_{i=1}^{n}m_{A}\circ\left(  P_{i}\otimes Q_{i}\right)
\circ\Delta_{D}\right)  \left(  d\right)  =\sum_{i=1}^{n}\underbrace{\left(
m_{A}\circ\left(  P_{i}\otimes Q_{i}\right)  \circ\Delta_{D}\right)  \left(
d\right)  }_{=m_{A}\left(  \left(  P_{i}\otimes Q_{i}\right)  \left(
\Delta_{D}\left(  d\right)  \right)  \right)  }\nonumber\\
&  =\sum_{i=1}^{n}m_{A}\left(  \left(  P_{i}\otimes Q_{i}\right)  \left(
\underbrace{\Delta_{D}\left(  d\right)  }_{=\sum_{j=1}^{k}x_{j}\otimes y_{j}%
}\right)  \right) \nonumber\\
&  =\sum_{i=1}^{n}m_{A}\left(  \underbrace{\left(  P_{i}\otimes Q_{i}\right)
\left(  \sum_{j=1}^{k}x_{j}\otimes y_{j}\right)  }_{\substack{=\sum_{j=1}%
^{k}\left(  P_{i}\otimes Q_{i}\right)  \left(  x_{j}\otimes y_{j}\right)
\\\text{(since the map }P_{i}\otimes Q_{i}\text{ is }\kk\text{-linear)}%
}}\right) \nonumber\\
&  =\sum_{i=1}^{n}m_{A}\left(  \sum_{j=1}^{k}\underbrace{\left(  P_{i}\otimes
Q_{i}\right)  \left(  x_{j}\otimes y_{j}\right)  }_{=P_{i}\left(
x_{j}\right)  \otimes Q_{i}\left(  y_{j}\right)  }\right) \nonumber\\
&  =\sum_{i=1}^{n}\underbrace{m_{A}\left(  \sum_{j=1}^{k}P_{i}\left(
x_{j}\right)  \otimes Q_{i}\left(  y_{j}\right)  \right)  }_{\substack{=
\sum_{j=1}^{k}m_{A}\left(  P_{i}\left(  x_{j}\right)  \otimes Q_{i}\left(
y_{j}\right)  \right)  \\\text{(since the map }m_{A}\text{ is }\kk%
\text{-linear)}}}\nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}\underbrace{m_{A}\left(  P_{i}\left(
x_{j}\right)  \otimes Q_{i}\left(  y_{j}\right)  \right)  }_{\substack{=P_{i}%
\left(  x_{j}\right)  Q_{i}\left(  y_{j}\right)  \\\text{(by the definition of
}m_{A}\text{)}}}\nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}\underbrace{P_{i}\left(  x_{j}\right)
}_{\substack{=f\left(  p_{i}\otimes x_{j}\right)  \\\text{(by
(\ref{sol.convolution.tensor-curry.Piz=}) (applied to }z=x_{j}\text{))}%
}}\underbrace{Q_{i}\left(  y_{j}\right)  }_{\substack{=g\left(  q_{i}\otimes
y_{j}\right)  \\\text{(by (\ref{sol.convolution.tensor-curry.Qiz=}) (applied
to }z=y_{j}\text{))}}}\nonumber\\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}f\left(  p_{i}\otimes x_{j}\right)  g\left(
q_{i}\otimes y_{j}\right)  . \label{sol.convolution.tensor-curry.left}%
\end{align}


On the other hand, the definition of $\Phi$ yields%
\begin{align*}
&  \left(  \left(  \Phi\left(  f\star g\right)  \right)  \left(  c\right)
\right)  \left(  d\right) \\
&  =\underbrace{\left(  f\star g\right)  }_{\substack{=m_{A}\circ\left(
f\otimes g\right)  \circ\Delta_{C\otimes D}\\\text{(by the definition of
convolution)}}}\left(  c\otimes d\right)  =\left(  m_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C\otimes D}\right)  \left(  c\otimes d\right) \\
&  =m_{A}\left(  \left(  f\otimes g\right)  \left(  \underbrace{\Delta
_{C\otimes D}\left(  c\otimes d\right)  }_{\substack{=\sum_{i=1}^{n}
\sum_{j=1}^{k}p_{i}\otimes x_{j}\otimes q_{i}\otimes y_{j}\\\text{(by
(\ref{sol.convolution.tensor-curry.Delcd}))}}}\right)  \right)  =m_{A}\left(
\underbrace{\left(  f\otimes g\right)  \left(  \sum_{i=1}^{n}\sum_{j=1}%
^{k}p_{i}\otimes x_{j}\otimes q_{i}\otimes y_{j}\right)  }_{\substack{=
\sum_{i=1}^{n}\left(  f\otimes g\right)  \left(  \sum_{j=1}^{k}p_{i}\otimes
x_{j}\otimes q_{i}\otimes y_{j}\right)  \\\text{(since the map }f\otimes
g\text{ is }\kk\text{-linear)}}}\right) \\
&  =m_{A}\left(  \sum_{i=1}^{n}\underbrace{\left(  f\otimes g\right)  \left(
\sum_{j=1}^{k}p_{i}\otimes x_{j}\otimes q_{i}\otimes y_{j}\right)
}_{\substack{=\sum_{j=1}^{k}\left(  f\otimes g\right)  \left(  p_{i}\otimes
x_{j}\otimes q_{i}\otimes y_{j}\right)  \\\text{(since the map }f\otimes
g\text{ is }\kk\text{-linear)}}}\right)  =m_{A}\left(  \sum_{i=1}%
^{n}\sum_{j=1}^{k}\underbrace{\left(  f\otimes g\right)  \left(  p_{i}\otimes
x_{j}\otimes q_{i}\otimes y_{j}\right)  }_{=f\left(  p_{i}\otimes
x_{j}\right)  \otimes g\left(  q_{i}\otimes y_{j}\right)  }\right) \\
&  =m_{A}\left(  \sum_{i=1}^{n}\sum_{j=1}^{k}f\left(  p_{i}\otimes
x_{j}\right)  \otimes g\left(  q_{i}\otimes y_{j}\right)  \right)  =
\sum_{i=1}^{n}\underbrace{m_{A}\left(  \sum_{j=1}^{k}f\left(  p_{i}\otimes
x_{j}\right)  \otimes g\left(  q_{i}\otimes y_{j}\right)  \right)
}_{\substack{=\sum_{j=1}^{k}m_{A}\left(  f\left(  p_{i}\otimes x_{j}\right)
\otimes g\left(  q_{i}\otimes y_{j}\right)  \right)  \\\text{(since the map
}m_{A}\text{ is }\kk\text{-linear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }m_{A}\text{ is }%
\kk\text{-linear}\right) \\
&  =\sum_{i=1}^{n}\sum_{j=1}^{k}\underbrace{m_{A}\left(  f\left(  p_{i}\otimes
x_{j}\right)  \otimes g\left(  q_{i}\otimes y_{j}\right)  \right)
}_{\substack{=f\left(  p_{i}\otimes x_{j}\right)  g\left(  q_{i}\otimes
y_{j}\right)  \\\text{(by the definition of }m_{A}\text{)}}}=\sum_{i=1}%
^{n}\sum_{j=1}^{k}f\left(  p_{i}\otimes x_{j}\right)  g\left(  q_{i}\otimes
y_{j}\right)  .
\end{align*}
Comparing this with (\ref{sol.convolution.tensor-curry.left}), we obtain%
\[
\left(  \left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)
\left(  c\right)  \right)  \left(  d\right)  =\left(  \left(  \Phi\left(
f\star g\right)  \right)  \left(  c\right)  \right)  \left(  d\right)  .
\]


Now, forget that we fixed $d$. We thus have shown that $\left(  \left(
\Phi\left(  f\right)  \star\Phi\left(  g\right)  \right)  \left(  c\right)
\right)  \left(  d\right)  =\left(  \left(  \Phi\left(  f\star g\right)
\right)  \left(  c\right)  \right)  \left(  d\right)  $ for each $d\in D$. In
other words, we have $\left(  \Phi\left(  f\right)  \star\Phi\left(  g\right)
\right)  \left(  c\right)  =\left(  \Phi\left(  f\star g\right)  \right)
\left(  c\right)  $.

Now, forget that we fixed $c$. We thus have shown that $\left(  \Phi\left(
f\right)  \star\Phi\left(  g\right)  \right)  \left(  c\right)  =\left(
\Phi\left(  f\star g\right)  \right)  \left(  c\right)  $ for each $c\in C$.
In other words, we have $\Phi\left(  f\right)  \star\Phi\left(  g\right)
=\Phi\left(  f\star g\right)  $.

Now, forget that we fixed $f$ and $g$. We thus have proven that every two
elements $f$ and $g$ of $\Hom \left(  C\otimes D,A\right)  $
satisfy%
\begin{equation}
\Phi\left(  f\right)  \star\Phi\left(  g\right)  =\Phi\left(  f\star g\right)
. \label{sol.convolution.tensor-curry.*}%
\end{equation}


Now, let $\theta$ be the canonical $\kk$-module isomorphism
$\kk\otimes\kk\rightarrow\kk$. Then, the definition of
the $\kk$-coalgebra $C\otimes D$ yields
\[
\epsilon_{C\otimes D}=\theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  .
\]


We have%
\begin{align}
1_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }  &
=\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  D,A\right)  ,\star\right)  \right) \nonumber\\
&  =u_{A}\circ\epsilon_{D}. \label{sol.convolution.tensor-curry.u1}%
\end{align}
Also,%
\begin{align}
1_{\left(  \Hom \left(  C,\left(  \Hom \left(
D,A\right)  ,\star\right)  \right)  ,\star\right)  }  &  =\left(  \text{the
unity of the }\kk\text{-algebra }\left(  \Hom \left(
C,\left(  \Hom \left(  D,A\right)  ,\star\right)  \right)
,\star\right)  \right) \nonumber\\
&  =u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}. \label{sol.convolution.tensor-curry.u2}%
\end{align}
Also,%
\begin{align}
1_{\left(  \Hom \left(  C\otimes D,A\right)  ,\star\right)  }
&  =\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C\otimes D,A\right)  ,\star\right)  \right)
\nonumber\\
&  =u_{A}\circ\epsilon_{C\otimes D}. \label{sol.convolution.tensor-curry.u3}%
\end{align}


Let $c\in C$. Let $d\in D$. The definition of $\Phi$ yields%
\begin{align*}
&  \left(  \left(  \Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)
\right)  \left(  c\right)  \right)  \left(  d\right) \\
&  =\left(  u_{A}\circ\epsilon_{C\otimes D}\right)  \left(  c\otimes d\right)
=u_{A}\left(  \underbrace{\epsilon_{C\otimes D}}_{=\theta\circ\left(
\epsilon_{C}\otimes\epsilon_{D}\right)  }\left(  c\otimes d\right)  \right)
=u_{A}\left(  \left(  \theta\circ\left(  \epsilon_{C}\otimes\epsilon
_{D}\right)  \right)  \left(  c\otimes d\right)  \right) \\
&  =u_{A}\left(  \theta\left(  \underbrace{\left(  \epsilon_{C}\otimes
\epsilon_{D}\right)  \left(  c\otimes d\right)  }_{=\epsilon_{C}\left(
c\right)  \otimes\epsilon_{D}\left(  d\right)  }\right)  \right)
=u_{A}\left(  \underbrace{\theta\left(  \epsilon_{C}\left(  c\right)
\otimes\epsilon_{D}\left(  d\right)  \right)  }_{\substack{=\epsilon
_{C}\left(  c\right)  \epsilon_{D}\left(  d\right)  \\\text{(by the definition
of }\theta\text{)}}}\right)  =u_{A}\left(  \epsilon_{C}\left(  c\right)
\epsilon_{D}\left(  d\right)  \right) \\
&  =\epsilon_{C}\left(  c\right)  \epsilon_{D}\left(  d\right)  \cdot
1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }u_{A}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \underbrace{\left(  u_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\circ\epsilon_{C}\right)  \left(  c\right)  }_{=u_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }\left(  \epsilon
_{C}\left(  c\right)  \right)  }\right)  \left(  d\right)   &  =\left(
\underbrace{u_{\left(  \Hom \left(  D,A\right)  ,\star\right)
}\left(  \epsilon_{C}\left(  c\right)  \right)  }_{\substack{=\epsilon
_{C}\left(  c\right)  \cdot1_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\\\text{(by the definition of }u_{\left(  \Hom
\left(  D,A\right)  ,\star\right)  }\text{)}}}\right)  \left(  d\right)
=\left(  \epsilon_{C}\left(  c\right)  \cdot1_{\left(  \Hom
\left(  D,A\right)  ,\star\right)  }\right)  \left(  d\right) \\
&  =\epsilon_{C}\left(  c\right)  \cdot\underbrace{1_{\left(
\Hom \left(  D,A\right)  ,\star\right)  }}_{=u_{A}\circ
\epsilon_{D}}\left(  d\right)  =\epsilon_{C}\left(  c\right)  \cdot
\underbrace{\left(  u_{A}\circ\epsilon_{D}\right)  \left(  d\right)
}_{\substack{=u_{A}\left(  \epsilon_{D}\left(  d\right)  \right)
=\epsilon_{D}\left(  d\right)  \cdot1_{A}\\\text{(by the definition of }%
u_{A}\text{)}}}=\epsilon_{C}\left(  c\right)  \epsilon_{D}\left(  d\right)
\cdot1_{A},
\end{align*}
we obtain $\left(  \left(  \Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)
\right)  \left(  c\right)  \right)  \left(  d\right)  =\left(  \left(
u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}\right)  \left(  c\right)  \right)  \left(  d\right)  $.

Now, forget that we fixed $d$. We thus have shown that $\left(  \left(
\Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)  \right)  \left(  c\right)
\right)  \left(  d\right)  =\left(  \left(  u_{\left(  \Hom
\left(  D,A\right)  ,\star\right)  }\circ\epsilon_{C}\right)  \left(
c\right)  \right)  \left(  d\right)  $ for each $d\in D$. In other words, we
have $\left(  \Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)  \right)
\left(  c\right)  =\left(  u_{\left(  \Hom \left(  D,A\right)
,\star\right)  }\circ\epsilon_{C}\right)  \left(  c\right)  $.

Now, forget that we fixed $c$. We thus have shown that $\left(  \Phi\left(
u_{A}\circ\epsilon_{C\otimes D}\right)  \right)  \left(  c\right)  =\left(
u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}\right)  \left(  c\right)  $ for each $c\in C$. In other
words, we have $\Phi\left(  u_{A}\circ\epsilon_{C\otimes D}\right)
=u_{\left(  \Hom \left(  D,A\right)  ,\star\right)  }%
\circ\epsilon_{C}$.

Now,%
\[
\Phi\left(  \underbrace{1_{\left(  \Hom \left(  C\otimes
D,A\right)  ,\star\right)  }}_{\substack{=u_{A}\circ\epsilon_{C\otimes
D}\\\text{(by (\ref{sol.convolution.tensor-curry.u3}))}}}\right)  =\Phi\left(
u_{A}\circ\epsilon_{C\otimes D}\right)  =u_{\left(  \Hom \left(
D,A\right)  ,\star\right)  }\circ\epsilon_{C}=1_{\left(  \Hom
\left(  C,\left(  \Hom \left(  D,A\right)  ,\star\right)
\right)  ,\star\right)  }%
\]
(by (\ref{sol.convolution.tensor-curry.u2})).

We now know that the map $\Phi$ is $\kk$-linear and satisfies
(\ref{sol.convolution.tensor-curry.*}) and $\Phi\left(  1_{\left(
\Hom \left(  C\otimes D,A\right)  ,\star\right)  }\right)
=1_{\left(  \Hom \left(  C,\left(  \Hom \left(
D,A\right)  ,\star\right)  \right)  ,\star\right)  }$. Thus, the map $\Phi$ is
a $\kk$-algebra homomorphism%
\[
\left(  \Hom \left(  C\otimes D,A\right)  ,\star\right)
\rightarrow\left(  \Hom \left(  C,\left(  \Hom
\left(  D,A\right)  ,\star\right)  \right)  ,\star\right)  .
\]
Since the map $\Phi$ is furthermore invertible (because $\Phi$ is a
$\kk$-module isomorphism), we thus conclude that $\Phi$ is a
$\kk$-algebra isomorphism%
\[
\left(  \Hom \left(  C\otimes D,A\right)  ,\star\right)
\rightarrow\left(  \Hom \left(  C,\left(  \Hom
\left(  D,A\right)  ,\star\right)  \right)  ,\star\right)  .
\]
This solves Exercise \ref{exe.convolution.tensor-curry}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.hopf.tensor}}

\textit{Solution to Exercise~\ref{exe.hopf.tensor}.}

\begin{proof}
[Proof of Proposition~\ref{prop.hopf.tensor}.]Let $S_{A}$ and $S_{B}$ be the
antipodes of the Hopf algebras $A$ and $B$. Recall that the antipode $S_{A}$
of the Hopf algebra $A$ is the $2$-sided inverse under $\star$ for the
identity map $\id_{A}\in\Hom \left(
A,A\right)  $. In other words, $S_{A}$ is the multiplicative inverse of
$\id_{A}$ in the convolution algebra $\left(
\Hom \left(  A,A\right)  ,\star\right)  $. Therefore,%
\[
S_{A}\star\id_{A}=1_{\left(  \Hom
\left(  A,A\right)  ,\star\right)  }\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \id_{A}\star S_{A}=1_{\left(
\Hom \left(  A,A\right)  ,\star\right)  }.
\]
The same argument (applied to $B$ instead of $A$) shows that%
\[
S_{B}\star\id_{B}=1_{\left(  \Hom
\left(  B,B\right)  ,\star\right)  }\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \id_{B}\star S_{B}=1_{\left(
\Hom \left(  B,B\right)  ,\star\right)  }.
\]


Now, it is easy to see that
\[
1_{\left(  \Hom \left(  A,A\right)  ,\star\right)  }%
\otimes1_{\left(  \Hom \left(  B,B\right)  ,\star\right)
}=1_{\left(  \Hom \left(  A\otimes B,A\otimes B\right)
,\star\right)  }%
\]
(as maps from $A\otimes B$ to $A\otimes B$)\ \ \ \ \footnote{\textit{Proof.}
We know that the unity of the convolution algebra $\left(  \Hom
\left(  A,A\right)  ,\star\right)  $ is $u_{A}\epsilon_{A}$. In other words,
$1_{\left(  \Hom \left(  A,A\right)  ,\star\right)  }%
=u_{A}\epsilon_{A}$. Similarly, $1_{\left(  \Hom \left(
B,B\right)  ,\star\right)  }=u_{B}\epsilon_{B}$ and $1_{\left(
\Hom \left(  A\otimes B,A\otimes B\right)  ,\star\right)
}=u_{A\otimes B}\epsilon_{A\otimes B}$.
\par
Now, let $s$ denote the canonical isomorphism $\kk\rightarrow
\kk\otimes\kk$. Then, the definition of the $\kk$-algebra
$A\otimes B$ yields $u_{A\otimes B}=\left(  u_{A}\otimes u_{B}\right)  \circ
s$. On the other hand, the definition of the $\kk$-coalgebra $A\otimes
B$ yields $\epsilon_{A\otimes B}=s^{-1}\circ\left(  \epsilon_{A}%
\otimes\epsilon_{B}\right)  $ (since $s^{-1}$ is the canonical isomorphism
$\kk\otimes\kk\rightarrow\kk$). Thus, $\epsilon
_{A}\otimes\epsilon_{B}=s\circ\epsilon_{A\otimes B}$. Now,%
\begin{align*}
\underbrace{1_{\left(  \Hom \left(  A,A\right)  ,\star\right)
}}_{=u_{A}\epsilon_{A}}\otimes\underbrace{1_{\left(  \Hom
\left(  B,B\right)  ,\star\right)  }}_{=u_{B}\epsilon_{B}}  & =\left(
u_{A}\epsilon_{A}\right)  \otimes\left(  u_{B}\epsilon_{B}\right)  =\left(
u_{A}\otimes u_{B}\right)  \circ\underbrace{\left(  \epsilon_{A}%
\otimes\epsilon_{B}\right)  }_{=s\circ\epsilon_{A\otimes B}}%
=\underbrace{\left(  u_{A}\otimes u_{B}\right)  \circ s}_{=u_{A\otimes B}%
}\circ\epsilon_{A\otimes B}=u_{A\otimes B}\epsilon_{A\otimes B}\\
& =1_{\left(  \Hom \left(  A\otimes B,A\otimes B\right)
,\star\right)  }.
\end{align*}
Qed.}.

Exercise \ref{exe.convolution.tensor}(a) (applied to $C=A$, $D=B$, $f=S_{A}$,
$f^{\prime}=\id_{A}$, $g=S_{B}$ and $g^{\prime
}=\id_{B}$) shows that%
\begin{align}
\left(  S_{A}\otimes S_{B}\right)  \star\left(  \operatorname*{id}%
\nolimits_{A}\otimes\id_{B}\right)    &
=\underbrace{\left(  S_{A}\star\id_{A}\right)
}_{=1_{\left(  \Hom \left(  A,A\right)  ,\star\right)  }}%
\otimes\underbrace{\left(  S_{B}\star\id_{B}\right)
}_{=1_{\left(  \Hom \left(  B,B\right)  ,\star\right)  }%
}\nonumber\\
& =1_{\left(  \Hom \left(  A,A\right)  ,\star\right)  }%
\otimes1_{\left(  \Hom \left(  B,B\right)  ,\star\right)
}=1_{\left(  \Hom \left(  A\otimes B,A\otimes B\right)
,\star\right)  }\label{pf.prop.hopf.tensor.1}%
\end{align}
in the convolution algebra $\Hom\left(  A\otimes B,A\otimes
B\right)  $.

Exercise \ref{exe.convolution.tensor}(a) (applied to $C=A$, $D=B$,
$f=\id_{A}$, $f^{\prime}=S_{A}$, $g=\operatorname*{id}%
\nolimits_{B}$ and $g^{\prime}=S_{B}$) shows that%
\begin{align*}
\left(  \id_{A}\otimes\operatorname*{id}%
\nolimits_{B}\right)  \star\left(  S_{A}\otimes S_{B}\right)    &
=\underbrace{\left(  \id_{A}\star S_{A}\right)
}_{=1_{\left(  \Hom \left(  A,A\right)  ,\star\right)  }}%
\otimes\underbrace{\left(  \id_{B}\star S_{B}\right)
}_{=1_{\left(  \Hom \left(  B,B\right)  ,\star\right)  }}\\
& =1_{\left(  \Hom \left(  A,A\right)  ,\star\right)  }%
\otimes1_{\left(  \Hom \left(  B,B\right)  ,\star\right)
}=1_{\left(  \Hom \left(  A\otimes B,A\otimes B\right)
,\star\right)  }%
\end{align*}
in the convolution algebra $\Hom\left(  A\otimes B,A\otimes
B\right)  $. Combining this with (\ref{pf.prop.hopf.tensor.1}), we conclude
that the two elements $S_{A}\otimes S_{B}$ and $\operatorname*{id}%
\nolimits_{A}\otimes\id_{B}$ of the convolution
algebra $\left(  \Hom \left(  A\otimes B,A\otimes B\right)
,\star\right)  $ are mutually inverse. In other words, $S_{A}\otimes S_{B}$ is
a $2$-sided inverse for $\id_{A}\otimes
\id_{B}$ under $\star$.  In other words, $S_{A}\otimes
S_{B}$ is a $2$-sided inverse for $\id_{A\otimes B}$
under $\star$ (since $\id_{A}\otimes\operatorname*{id}%
\nolimits_{B}=\id_{A\otimes B}$). Hence, the element
$\id_{A\otimes B}\in\Hom \left(  A\otimes
B,A\otimes B\right)  $ has a $2$-sided inverse under $\star$ (namely,
$S_{A}\otimes S_{B}$).

We recall that a bialgebra $D$ is a Hopf algebra if and only if the element
$\id_{D}\in\Hom \left(  D,D\right)  $
has a $2$-sided inverse under $\star$. Applying this to $D=A\otimes B$, we
conclude that the bialgebra $A\otimes B$ is a Hopf algebra if and only if the
element $\id_{A\otimes B}\in\Hom \left(
A\otimes B,A\otimes B\right)  $ has a $2$-sided inverse under $\star$.
Therefore, the bialgebra $A\otimes B$ is a Hopf algebra (since the element
$\id_{A\otimes B}\in\Hom \left(  A\otimes
B,A\otimes B\right)  $ has a $2$-sided inverse under $\star$).

The antipode of any Hopf algebra $D$ is the $2$-sided inverse for
$\id_{D}$ under $\star$. Applying this to $D=A\otimes
B$, we conclude that the antipode of $A\otimes B$ is the $2$-sided inverse for
$\id_{A\otimes B}$ under $\star$. In other words, the
antipode of $A\otimes B$ is the map $S_{A}\otimes S_{B}:A\otimes B\rightarrow
A\otimes B$ (since the $2$-sided inverse for $\operatorname*{id}%
\nolimits_{A\otimes B}$ under $\star$ is the map $S_{A}\otimes S_{B}:A\otimes
B\rightarrow A\otimes B$ (since $S_{A}\otimes S_{B}$ is a $2$-sided inverse
for $\id_{A\otimes B}$ under $\star$)). This completes
the proof of Proposition~\ref{prop.hopf.tensor}.
\end{proof}

Thus, Exercise~\ref{exe.hopf.tensor} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.iterated.m}}

\textit{Solution to Exercise~\ref{exe.iterated.m}.}
Let us start with an observation which is irrelevant to our solution of the exercise.
Namely, let us notice that
\[
m^{\left(k\right)} \left(a_1 \otimes a_2 \otimes ... \otimes a_{k+1}\right)
= a_1 \left(a_2 \left(a_3 \left( ... \left(a_k a_{k+1}\right) ... \right) \right) \right)
\]
for any $k \geq 0$ and any $k+1$ elements $a_1, a_2, ..., a_{k+1}$ of $A$.
The statements of Exercise~\ref{exe.iterated.m} are nothing but different aspects of
what is known as ``general associativity''\footnote{that is, the rule stating that the
product of several elements of a $\kk$-algebra does not depend on the bracketing}
(although they all fall short of defining an ``arbitrary bracketing'' of a
$\left(k+1\right)$-fold product), written in an element-free fashion (that is, written without
any reference to elements of $A$, but only in terms of maps). For instance,
part (a) of the exercise says that any $k+1$ elements $a_1, a_2, ..., a_{k+1}$
of $A$ satisfy
\[
a_1 \left(a_2 \left(a_3 \left( ... \left(a_k a_{k+1}\right) ... \right) \right) \right)
= \left( a_1 \left(a_2 \left(a_3 \left(... \left(a_i a_{i+1}\right) ... \right) \right) \right) \right)
\cdot \left(a_{i+2} \left(a_{i+3} \left(a_{i+4} \left( ... \left(a_k a_{k+1}\right) ... \right) \right) \right) \right).
\]
However, there is virtue in solving Exercise~\ref{exe.iterated.m} in an
element-free way (i.e., without referring to elements, but only referring to
maps), because such a solution will automatically yield a solution of
Exercise~\ref{exe.iterated.Delta} by reversing all arrows. So let us
show an element-free solution of Exercise~\ref{exe.iterated.m}.

(a) We will solve Exercise~\ref{exe.iterated.m}(a) by induction over $k$.

The induction base ($k=0$) is vacuously true, since there exists no
$0 \leq i \leq k-1$ for $k=0$. So let us proceed to the induction step.
Let $K$ be a positive integer. We want to prove that the claim of
Exercise~\ref{exe.iterated.m}(a) holds for $k=K$, assuming (as the
induction hypothesis) that the claim of
Exercise~\ref{exe.iterated.m}(a) holds for $k=K-1$.

We first notice that
\begin{equation}
\label{sol.iterated.m.a.indhyp1}
m^{\left(K-1\right)} = m \circ \left(m^{\left(i\right)} \otimes m^{\left(\left(K-1\right)-1-i\right)}\right)
\end{equation}
for all $0 \leq i \leq \left(K-1\right)-1$. (This is merely a restatement
of the induction hypothesis.)

Now fix $0 \leq i \leq K-1$. We need to show that
\begin{equation}
\label{sol.iterated.m.a.goal}
m^{\left(K\right)} = m \circ \left(m^{\left(i\right)} \otimes m^{\left(K-1-i\right)}\right).
\end{equation}

If $i = 0$, then \eqref{sol.iterated.m.a.goal} is obviously true\footnote{because
if $i = 0$, then
$m \circ \left(\underbrace{m^{\left(i\right)}}_{=m^{\left(0\right)} = \id_A} \otimes \underbrace{m^{\left(K-1-i\right)}}_{=m^{\left(K-1-0\right)} = m^{\left(K-1\right)}}\right)
= m \circ \left(\id_A \otimes m^{\left(K-1\right)}\right) = m^{\left(K\right)}
$
(by the inductive definition of $m^{\left(K\right)}$)}. Hence, we can WLOG assume
that we don't have $i = 0$. Assume this. Then, $i > 0$. Hence,
the recursive definition of $m^{\left(i\right)}$ yields
$m^{\left(i\right)} = m \circ \left(\id_A \otimes m^{\left(i-1\right)}\right)$.
Thus,
\begin{align}
\nonumber
\underbrace{m^{\left(i\right)}}_{= m \circ \left(\id_A \otimes m^{\left(i-1\right)}\right)} \otimes \underbrace{m^{\left(K-1-i\right)}}_{= \id_A \circ m^{\left(K-1-i\right)} }
&= \left(m \circ \left(\id_A \otimes m^{\left(i-1\right)}\right) \right)  \otimes \left(\id_A \circ m^{\left(K-1-i\right)}\right)\\
\nonumber
&= \left(m \otimes \id_A\right) \circ \left(\left(\id_A \otimes m^{\left(i-1\right)}\right) \otimes m^{\left(K-1-i\right)}\right) \\
\label{sol.iterated.m.a.indhyp2}
&= \left(m \otimes \id_A\right) \circ \left(\id_A \otimes m^{\left(i-1\right)} \otimes m^{\left(K-1-i\right)}\right)
\end{align}

On the other hand,
\begin{align}
\nonumber
\id_A \otimes \underbrace{m^{\left(K-1\right)}}_{\substack{
   = m \circ \left(m^{\left(i-1\right)} \otimes m^{\left(\left(K-1\right)-1-\left(i-1\right)\right)}\right) \\
   \text{ (by \eqref{sol.iterated.m.a.indhyp1}, applied to } i - 1 \text{ instead of } i\text{)}}}
&= \id_A \otimes \left(m \circ \left(m^{\left(i-1\right)} \otimes m^{\left(\left(K-1\right)-1-\left(i-1\right)\right)}\right)\right) \\
\nonumber
&= \id_A \otimes \left(m \circ \left(m^{\left(i-1\right)} \otimes m^{\left(K-1-i\right)}\right)\right) \\
\nonumber
&= \left(\id_A \otimes m\right) \circ \left(\id_A \otimes \left(m^{\left(i-1\right)} \otimes m^{\left(K-1-i\right)}\right)\right) \\
\label{sol.iterated.m.a.indhyp1a}
&= \left(\id_A \otimes m\right) \circ \left(\id_A \otimes m^{\left(i-1\right)} \otimes m^{\left(K-1-i\right)}\right)
\end{align}

Now, the upper left triangle in the diagram
\begin{equation}
\label{sol.iterated.m.a.diagr}
\xymatrixcolsep{5pc}
\xymatrix{
A^{\otimes \left(K+1\right)} \ar@/^3pc/[rrd]^{\id_A \otimes m^{\left(K-1\right)}} \ar[rd]^{\ \ \qquad \id_A \otimes m^{\left(i-1\right)} \otimes m^{\left(K-1-i\right)}}
\ar@/_2pc/[rdd]_{m^{\left(i\right)} \otimes m^{\left(K-1-i\right)}} & & \\
& A \otimes A \otimes A \ar[r]_{\id_A \otimes m} \ar[d]_{m \otimes \id_A} & A \otimes A \ar[d]^{m} \\
& A \otimes A \ar[r]_{m} & A
}
\end{equation}
is commutative (by \eqref{sol.iterated.m.a.indhyp1a}), and so is the lower left
triangle (according to \eqref{sol.iterated.m.a.indhyp2}). Since the square in
the diagram \eqref{sol.iterated.m.a.diagr} is also commutative (by the
commutativity of \eqref{associativity-diagram}), we thus
conclude that the whole diagram \eqref{sol.iterated.m.a.diagr} is commutative.
Hence, following the outermost arrows in this diagram, we obtain
$m \circ \left(\id_A \otimes m^{\left(K-1\right)}\right)
= m \circ \left(m^{\left(i\right)} \otimes m^{\left(K-1-i\right)}\right)$.
Now, the recursive definition of $m^{\left(K\right)}$ yields
$m^{\left(K\right)}
= m \circ \left(\id_A \otimes m^{\left(K-1\right)}\right)
= m \circ \left(m^{\left(i\right)} \otimes m^{\left(K-1-i\right)}\right)$.
Hence, \eqref{sol.iterated.m.a.goal} is proven.

We thus have shown that Exercise~\ref{exe.iterated.m}(a) holds for
$k=K$. This completes the induction step, and thus
Exercise~\ref{exe.iterated.m}(a) is solved by induction.

(b) Let $k \geq 1$. Then, Exercise~\ref{exe.iterated.m}(a) (applied to
$i = k-1$) yields
$m^{\left(k\right)} = m \circ \left(m^{\left(k-1\right)} \otimes \underbrace{m^{\left(k-1-\left(k-1\right)\right)}}_{=m^{\left(0\right)} = \id_A}\right)
= m \circ \left(m^{\left(k-1\right)} \otimes \id_A\right)$.
Thus, Exercise~\ref{exe.iterated.m}(b) is solved.

(c) We will solve Exercise~\ref{exe.iterated.m}(c) by induction over $k$.

The induction base ($k=0$) is vacuously true, since there exists no $0\leq
i\leq k-1$ for $k=0$. So let us proceed to the induction step. Let $K$ be a
positive integer. We want to prove that the claim of
Exercise~\ref{exe.iterated.m}(c) holds for $k=K$, assuming (as the induction
hypothesis) that the claim of Exercise~\ref{exe.iterated.m}(c) holds for
$k=K-1$.

So let $0\leq i\leq K-1$ be arbitrary. Thus, $K-1\geq0$, so that $K\geq1$. We
are going to prove that
\begin{equation}
m^{\left(  K\right)  }
=m^{\left(  K-1\right)  }\circ
\left(  \id_{A^{\otimes i}}\otimes m\otimes\id_{A^{\otimes\left(  K-1-i\right)  }}\right)
.\label{sol.iterated.m.c.goal}
\end{equation}


We must be in one of the following three cases:

\textit{Case 1:} We have $i\neq0$.

\textit{Case 2:} We have $i\neq K-1$.

\textit{Case 3:} We have neither $i\neq0$ nor $i\neq K-1$.

Let us consider Case 1 first. In this case, we have $i\neq0$. Hence, $i\geq1$
(because $0\leq i$), so that $i-1\geq0$. Thus, we can apply
Exercise~\ref{exe.iterated.m}(c) to $K-1$ and $i-1$ instead of $k$ and $i$
(because we have assumed that the claim of Exercise~\ref{exe.iterated.m}(c)
holds for $k=K-1$). As a result, we obtain
\begin{align*}
m^{\left(  K-1\right)  }  & =\underbrace{m^{\left(  \left(  K-1\right)
-1\right)  }}_{=m^{\left(  K-2\right)  }}\circ\left( 
 \id_{A^{\otimes\left(  i-1\right)  }}\otimes m\otimes
\underbrace{\id_{A^{\otimes\left(  \left(  K-1\right)
-1-\left(  i-1\right)  \right)  }}}_{=\id_{A^{\otimes
\left(  K-1-i\right)  }}}\right)  \\
& =m^{\left(  K-2\right)  }\circ\left(  
\id_{A^{\otimes\left(  i-1\right)  }}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  .
\end{align*}


But $K\geq1$. Hence, the recursive definition of $m^{\left(  K\right)  }$
yields
\begin{align}
m^{\left(  K\right)  }  & =m\circ\left(  \id _{A}
\otimes\underbrace{m^{\left(  K-1\right)  }}_{=m^{\left(  K-2\right)  }
\circ\left(  \id _{A^{\otimes\left(  i-1\right)  }}
\otimes m\otimes\id _{A^{\otimes\left(  K-1-i\right)
}}\right)  }\right)  \nonumber\\
& =m\circ\underbrace{\left(  \id _{A}\otimes\left(
m^{\left(  K-2\right)  }\circ\left(  \id _{A^{\otimes
\left(  i-1\right)  }}\otimes m\otimes\id _{A^{\otimes
\left(  K-1-i\right)  }}\right)  \right)  \right)  }_{=\left(
\id _{A}\otimes m^{\left(  K-2\right)  }\right)
\circ\left(  \id _{A}\otimes\left( 
 \id_{A^{\otimes\left(  i-1\right)  }}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  \right)  }\nonumber\\
& =m\circ\left(  \id _{A}\otimes m^{\left(  K-2\right)
}\right)  \circ\underbrace{\left(  \id _{A}
\otimes\left(  \id _{A^{\otimes\left(  i-1\right)  }}
\otimes m\otimes\id _{A^{\otimes\left(  K-1-i\right)
}}\right)  \right)  }_{=\id _{A}\otimes
\id _{A^{\otimes\left(  i-1\right)  }}\otimes
m\otimes\id _{A^{\otimes\left(  K-1-i\right)  }}}
\nonumber\\
& =m\circ\left(  \id _{A}\otimes m^{\left(  K-2\right)
}\right)  \circ\left(  \underbrace{\id _{A}\otimes
\id _{A^{\otimes\left(  i-1\right)  }}}
_{=\id _{A^{\otimes i}}}\otimes m\otimes
\id _{A^{\otimes\left(  K-1-i\right)  }}\right)
\nonumber\\
& =m\circ\left(  \id _{A}\otimes m^{\left(  K-2\right)
}\right)  \circ\left(  \id _{A^{\otimes i}}\otimes
m\otimes\id _{A^{\otimes\left(  K-1-i\right)  }}\right)  .
\label{sol.iterated.m.c.case1.1}
\end{align}
But $i\geq1$ and $i\leq K-1$ together yield $1\leq i\leq K-1$, so that
$K-1\geq1$. Thus, the recursive definition of $m^{\left(  K-1\right)  }$
yields
\begin{equation}
m^{\left(  K-1\right)  }=m\circ\left(  \id _{A}
\otimes\underbrace{m^{\left(  \left(  K-1\right)  -1\right)  }}_{=m^{\left(
K-2\right)  }}\right)  =m\circ\left(  \id _{A}\otimes
m^{\left(  K-2\right)  }\right)  .\label{sol.iterated.m.c.case1.2}
\end{equation}
Hence, \eqref{sol.iterated.m.c.case1.1} becomes
\begin{align*}
m^{\left(  K\right)  }  & =\underbrace{m\circ\left(  \id_{A}\otimes 
m^{\left(  K-2\right)  }\right)  }_{\substack{=m^{\left(
K-1\right)  }\\\text{(by \eqref{sol.iterated.m.c.case1.2})}}}\circ\left(
\id _{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  \\
& =m^{\left(  K-1\right)  }\circ\left( 
 \id_{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  .
\end{align*}
Thus, \eqref{sol.iterated.m.c.goal} is proven in Case 1.

Let us next consider Case 2. In this case, we have $i\neq K-1$. Hence, $i<K-1$
(since $0\leq k-1$), so that $i\leq\left(  K-1\right)  -1$ (since $i$ and
$K-1$ are integers). Thus, we can apply Exercise~\ref{exe.iterated.m}(c) to
$K-1$ instead of $k$ (because we have assumed that the claim of
Exercise~\ref{exe.iterated.m}(c) holds for $k=K-1$). As a result, we obtain
\begin{align*}
m^{\left(  K-1\right)  }  & =\underbrace{m^{\left(  \left(  K-1\right)
-1\right)  }}_{=m^{\left(  K-2\right)  }}\circ\left( 
 \id_{A^{\otimes i}}\otimes m\otimes\underbrace{
\id_{A^{\otimes\left(  \left(  K-1\right)  -1-i\right)  }}}
_{=\id _{A^{\otimes\left(  K-1-i-1\right)  }}}\right)
\\
& =m^{\left(  K-2\right)  }\circ\left(  \id_{A^{\otimes i}}
\otimes m\otimes\id_{A^{\otimes\left(  K-1-i-1\right)  }}\right)  .
\end{align*}


But $K\geq1$. Hence, Exercise~\ref{exe.iterated.m}(b) (applied to $k=K$)
yields
\begin{align}
m^{\left(  K\right)  }  & =m\circ\left( 
 \underbrace{m^{\left(  K-1\right)  }}_{=m^{\left(  K-2\right)  }\circ\left( 
 \id_{A^{\otimes i}}\otimes m\otimes\id_{A^{\otimes\left(  K-1-i-1\right)  }}
\right)  }\otimes
\id _{A}\right)  \nonumber\\
& =m\circ\underbrace{\left(  \left(  m^{\left(  K-2\right)  }\circ\left(
\id _{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i-1\right)  }}\right)  \right)
\otimes\id _{A}\right)  }_{=\left(  m^{\left(
K-2\right)  }\otimes\id _{A}\right)  \circ\left(
\left(  \id _{A^{\otimes i}}\otimes m\otimes
\id _{A^{\otimes\left(  K-1-i-1\right)  }}\right)
\otimes\id _{A}\right)  }\nonumber\\
& =m\circ\left(  m^{\left(  K-2\right)  }\otimes\id_{A}\right) 
 \circ\underbrace{\left(  \left( 
 \id_{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i-1\right)  }}\right)  \otimes
\id _{A}\right)  }_{=\id_{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i-1\right)  }}\otimes
\id_{A}}\nonumber\\
& =m\circ\left(  m^{\left(  K-2\right)  }\otimes
\id_{A}\right)  \circ\left(  \id _{A^{\otimes i}}
\otimes m\otimes\underbrace{\id _{A^{\otimes\left(
K-1-i-1\right)  }}\otimes\id _{A}}_{=
\id_{A^{\otimes\left(  K-1-i\right)  }}}\right)  \nonumber\\
& =m\circ\left(  m^{\left(  K-2\right)  }\otimes
\id_{A}\right)  \circ\left(  \id _{A^{\otimes i}}
\otimes m\otimes\id _{A^{\otimes\left(  K-1-i\right)
}}\right)  .\label{sol.iterated.m.c.case2.1}
\end{align}
But $0\leq i\leq\left(  K-1\right)  -1$ yields $1\leq K-1$, so that $K-1\geq
1$. Thus, Exercise~\ref{exe.iterated.m}(b) (applied to $k=K-1$) yields
\begin{equation}
m^{\left(  K-1\right)  }=m\circ\left(  \underbrace{m^{\left(  \left(
K-1\right)  -1\right)  }}_{=m^{\left(  K-2\right)  }}\otimes
\id_{A}\right)  =m\circ\left(  m^{\left(  K-2\right)  }\otimes
\id _{A}\right)  .\label{sol.iterated.m.c.case2.2}
\end{equation}
Hence, \eqref{sol.iterated.m.c.case2.1} becomes
\begin{align*}
m^{\left(  K\right)  }  & =\underbrace{m\circ\left(  m^{\left(  K-2\right)
}\otimes\id _{A}\right)  }_{\substack{=m^{\left(
K-1\right)  }\\\text{(by \eqref{sol.iterated.m.c.case2.2})}}}\circ\left(
\id _{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  \\
& =m^{\left(  K-1\right)  }\circ\left( 
 \id_{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  .
\end{align*}
Thus, \eqref{sol.iterated.m.c.goal} is proven in Case 2.

Let us finally consider Case 3. In this case, we have neither $i\neq0$ nor
$i\neq K-1$. Hence, we have both $i=0$ and $i=K-1$. Thus, $K-1=i=0$, so that
$K=1$. Thus,
\begin{align*}
m^{\left(  K\right)  }  & =m^{\left(  1\right)  }=m\circ\left(
\id _{A}\otimes\underbrace{m^{\left(  1-1\right)  }}
_{=m^{\left(  0\right)  }=\id _{A}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the recursive definition of }m^{\left(
1\right)  }\right)  \\
& =m\circ\underbrace{\left(  \id _{A}\otimes
\id _{A}\right)  }_{=\id_{A\otimes A}}=m.
\end{align*}
Compared with
\begin{align*}
& m^{\left(  K-1\right)  }\circ\left(  \id _{A^{\otimes
i}}\otimes m\otimes\id _{A^{\otimes\left(
K-1-i\right)  }}\right)  \\
& =\underbrace{m^{\left(  0\right)  }}_{=\id _{A}}
\circ\underbrace{\left(  \id _{A^{\otimes0}}\otimes
m\otimes\id _{A^{\otimes0}}\right)  }_{=m}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }K-1=0\text{ and }i=0
\text{ and }\underbrace{K-1}_{=0}-\underbrace{i}_{=0}
=0-0=0\right)  \\
& =\id _{A}\circ m=m,
\end{align*}
this yields $m^{\left(  K\right)  }=m^{\left(  K-1\right)  }\circ\left(
\id _{A^{\otimes i}}\otimes m\otimes
\id_{A^{\otimes\left(  K-1-i\right)  }}\right)  $. Thus,
\eqref{sol.iterated.m.c.goal} is proven in Case 3.

We have now proven \eqref{sol.iterated.m.c.goal} in each of the three Cases 1,
2 and 3. Since these three Cases cover all possibilities, this shows that
\eqref{sol.iterated.m.c.goal} always holds (for all $0\leq i\leq K-1$). In
other words, the claim of Exercise~\ref{exe.iterated.m}(c) holds for $k=K$.
This completes the induction step. The induction proof of the claim of
Exercise~\ref{exe.iterated.m}(c) is therefore complete.

(d) Let $k\geq 1$. Applying Exercise~\ref{exe.iterated.m}(c) to $i=0$, we
obtain
\[
m^{\left(  k\right)  }=m^{\left(  k-1\right)  }\circ\underbrace{\left(
\id _{A^{\otimes0}}\otimes m\otimes
\id_{A^{\otimes\left(  k-1-0\right)  }}\right)  }_{=m\otimes
\id _{A^{\otimes\left(  k-1-0\right)  }}=m\otimes
\id _{A^{\otimes\left(  k-1\right)  }}}=m^{\left(
k-1\right)  }\circ\left(  m\otimes\id _{A^{\otimes
\left(  k-1\right)  }}\right)  .
\]
Applying Exercise~\ref{exe.iterated.m}(c) to $i=k-1$, we obtain
\begin{align*}
\nonumber
m^{\left(  k\right)  }&=m^{\left(  k-1\right)  }\circ\left( 
 \id_{A^{\otimes\left(  k-1\right)  }}\otimes m\otimes
\underbrace{\id _{A^{\otimes\left(  k-1-\left(
k-1\right)  \right)  }}}_{=\id _{A^{\otimes0}}}\right)
=m^{\left(  k-1\right)  }\circ\left(  \id _{A^{\otimes
\left(  k-1\right)  }}\otimes\underbrace{m\otimes
\id_{A^{\otimes0}}}_{=m}\right)  \\
\nonumber
&=m^{\left(  k-1\right)  }\circ\left(
\id _{A^{\otimes\left(  k-1\right)  }}\otimes m\right)
.
\end{align*}
This solves Exercise~\ref{exe.iterated.m}(d).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.iterated.Delta}}

\textit{Solution to Exercise~\ref{exe.iterated.Delta}.} A solution
for Exercise~\ref{exe.iterated.Delta} can be obtained by reversing all
arrows (and renaming $A$, $m$ and $m^{\left(k\right)}$ by $C$, $\Delta$
and $\Delta^{\left(k\right)}$) in the element-free solution of
Exercise~\ref{exe.iterated.m} that we gave above.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.iterated.patras}}

\textit{Solution to Exercise~\ref{exe.iterated.patras}.} (a) We will
solve Exercise~\ref{exe.iterated.patras}(a) by induction over $k$:

The induction base ($k=0$) requires us to prove that the map $m_{H}^{\left(
0\right)  }:H^{\otimes\left(  0+1\right)  }\rightarrow H$ is a
$\kk$-coalgebra homomorphism. But this is obvious, because $m_{H}^{\left(
0\right)  }=\id_{H}$ (by the definition of
$m_{H}^{\left(  0\right)  }$).

Now, let us proceed to the induction step. Let $K$ be a positive integer. We
want to prove that the claim of Exercise~\ref{exe.iterated.patras}(a) holds
for $k=K$, assuming (as the induction hypothesis) that the claim of Exercise
\ref{exe.iterated.patras}(a) holds for $k=K-1$.

By the axioms of a bialgebra, we know that $m_{H}$ is a $\kk$-coalgebra
homomorphism (since $H$ is a $\kk$-bialgebra).

We have $m_{H}^{\left(  K\right)  }=m_{H}\circ\left(  \id _{H}
\otimes m_{H}^{\left(  K-1\right)  }\right)  $ (by the recursive
definition of $m_{H}^{\left(  K\right)  }$). We know that $m_{H}^{\left(
K-1\right)  }:H^{\otimes\left(  \left(  K-1\right)  +1\right)  }\rightarrow H$
is a $\kk$-coalgebra homomorphism (since the claim of Exercise
\ref{exe.iterated.patras}(a) holds for $k=K-1$). Of course,
$\id_{H}:H\rightarrow H$ is also a $\kk$-coalgebra homomorphism.
Exercise~\ref{exe.tensorprod.morphism}(b) (applied
to $C=H$, $C^{\prime}=H$, $D=H^{\otimes\left(  \left(  K-1\right)  +1\right)
}$, $D^{\prime}=H$, $f=\id_{H}$ and $g=m_{H}^{\left(
K-1\right)  }$) thus yields that the map $\id _{H}\otimes
m_{H}^{\left(  K-1\right)  }:H\otimes H^{\otimes\left(  \left(  K-1\right)
+1\right)  }\rightarrow H\otimes H$ is a $\kk$-coalgebra homomorphism.
Since $H\otimes H^{\otimes\left(  \left(  K-1\right)  +1\right)  }
=H^{\otimes\left(  \left(  \left(  K-1\right)  +1\right)  +1\right)
}=H^{\otimes\left(  K+1\right)  }$, this rewrites as follows: The map
$\id _{H}\otimes m_{H}^{\left(  K-1\right)  }:H^{\otimes\left(
K+1\right)  }\rightarrow H\otimes H$ is a $\kk$-coalgebra homomorphism.
Thus, $m_{H}\circ\left(  \id _{H}\otimes m_{H}^{\left(
K-1\right)  }\right)  $ is the composition of two $\kk$-coalgebra
homomorphisms (namely, of $m_{H}$ and $\id _{H}\otimes
m_{H}^{\left(  K-1\right)  }$), hence a $\kk$-coalgebra homomorphism
itself (since the composition of any two $\kk$-coalgebra homomorphisms
is a $\kk$-coalgebra homomorphism). In other words, $m_{H}^{\left(
K\right)  }$ is a $\kk$-coalgebra homomorphism (since $m_{H}^{\left(
K\right)  }=m_{H}\circ\left(  \id _{H}\otimes m_{H}^{\left(
K-1\right)  }\right)  $). In other words, the claim of Exercise
\ref{exe.iterated.patras}(a) holds for $k=K$. This completes the induction
step. Thus, Exercise~\ref{exe.iterated.patras}(a) is solved by induction.

(b) The solution of Exercise~\ref{exe.iterated.patras}(b) can be obtained from
the solution of Exercise~\ref{exe.iterated.patras}(a) by
``reversing all arrows''. (The details of this are left to the reader.)

(d) Let $\ell\in\NN$. Exercise~\ref{exe.iterated.patras}(a) (applied to
$\ell$ instead of $k$) yields that the map $m_{H}^{\left(  \ell\right)
}:H^{\otimes\left(  \ell+1\right)  }\rightarrow H$ is a $\kk$-coalgebra homomorphism.

For every $\kk$-coalgebra $C$, consider the map $\Delta_{C}^{\left(
k\right)  }:C\rightarrow C^{\otimes\left(  k+1\right)  }$ (this is the map
$\Delta^{\left(  k\right)  }$ defined in Exercise~\ref{exe.iterated.Delta}).
This map $\Delta_{C}^{\left(  k\right)  }$ is clearly functorial in $C$. By
this we mean that if $C$ and $D$ are any two $\kk$-coalgebras, and
$f:C\rightarrow D$ is any $\kk$-coalgebra homomorphism, then the
diagram
\[
\xymatrixcolsep{4pc}
\xymatrix{
C \ar[r]^{f} \ar[d]^{\Delta^{\left(k\right)}_C} & D \ar[d]^{\Delta
^{\left(k\right)}_D} \\
C^{\otimes\left(k+1\right)} \ar[r]_{f^{\otimes\left(k+1\right)}}
& D^{\otimes\left(k+1\right)}
}
\]
commutes.\footnote{This can be proven by induction over $k$ in a completely
straightforward manner.} We can apply this to $C=H^{\otimes\left(
\ell+1\right)  }$, $D=H$ and $f=m_{H}^{\left(  \ell\right)  }$ (since
$m_{H}^{\left(  \ell\right)  }$ is a $\kk$-coalgebra homomorphism). As
a result, we conclude that the diagram
\[
\xymatrixcolsep{5pc}
\xymatrix{
H^{\otimes\left(\ell+1\right)} \ar[r]^{m_H^{\left(\ell\right)}} \ar
[d]^{\Delta^{\left(k\right)}_{H^{\otimes\left(\ell+1\right)}}} & H \ar
[d]^{\Delta^{\left(k\right)}_H} \\
\left(H^{\otimes\left(\ell+1\right)}\right)^{\otimes\left(k+1\right)}
\ar[r]_-{\left(m_H^{\left(\ell\right)}\right)^{\otimes\left(k+1\right)}}
& H^{\otimes\left(k+1\right)}
}
\]
commutes. In other words,
$\left( m_{H}^{\left( \ell \right) } \right) ^{\otimes \left( k+1 \right)}
\circ \Delta_{H^{\otimes \left( \ell+1 \right)}}^{\left(  k\right)  }
=\Delta_{H}^{\left(  k\right)  }\circ m_{H}^{\left(
\ell\right)  }$. This solves Exercise~\ref{exe.iterated.patras}(d).

(c) The solution of Exercise~\ref{exe.iterated.patras}(c) can be obtained from
the solution of Exercise~\ref{exe.iterated.patras}(d) by
``reversing all arrows''.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.k}}

\textit{Solution to Exercise~\ref{exe.convolution.k}.} If $k=0$, then
the statement of Exercise~\ref{exe.convolution.k} is clearly
true\footnote{\textit{Proof.} Let $k=0$. Then,
\begin{align*}
f_{1}\star f_{2}\star\cdots\star f_{k}  & =f_{1}\star f_{2}\star\cdots\star
f_{0}=\left(  \text{empty product in }\left(  \Hom\left(
C,A\right)  ,\star\right)  \right)  \\
& =1_{\left(  \Hom\left(  C,A\right)  ,\star\right)  }
=u_{A}\epsilon_{C}
\end{align*}
and
\begin{align*}
m_{A}^{\left(  k-1\right)  }\circ\left(  f_{1}\otimes f_{2}\otimes
\cdots\otimes f_{k}\right)  \circ \Delta_C^{\left(  k-1\right)  }
& =\underbrace{m_{A}^{\left(  0-1\right)  }}_{=m_{A}^{\left(  -1\right)
}=u_{A}}\circ\underbrace{\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{0}\right)  }_{=\left(  \text{empty tensor product}\right)
=\id_{\kk }}\circ\underbrace{\Delta_C^{\left(
k-1\right)  }}_{=\Delta_{C}^{\left(  -1\right)  }=\epsilon_{C}}\\
& =u_{A}\circ\id_{\kk }\circ\epsilon_{C}
=u_{A}\epsilon_{C}.
\end{align*}
Hence, $f_{1}\star f_{2}\star\cdots\star f_{k}=u_{A}\epsilon_{C}
=m_{A}^{\left(  k-1\right)  }\circ\left(  f_{1}\otimes f_{2}\otimes
\cdots\otimes f_{k}\right)  \circ\Delta_{C}^{\left(  k-1\right)  }$, so that
Exercise~\ref{exe.convolution.k} is solved in the case when $k=0$.}. Hence,
for the rest of this solution, we can WLOG assume that we don't have $k=0$.
Assume this.

We shall show that
\begin{equation}
f_{1}\star f_{2}\star\cdots\star f_{i}=m_{A}^{\left(  i-1\right)  }
\circ\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes f_{i}\right)  \circ
\Delta_{C}^{\left(  i-1\right)  }\label{sol.convolution.k.driver}
\end{equation}
for every $i\in\left\{  1,2,...,k\right\}  $.

\textit{Proof of \eqref{sol.convolution.k.driver}:} We will prove
\eqref{sol.convolution.k.driver} by induction over $i$:

\textit{Induction base:} If $i=1$, then \eqref{sol.convolution.k.driver}
holds.\footnote{\textit{Proof.} Assume that $i=1$. Then, since $i=1$, we have
\[
f_{1}\star f_{2}\star\cdots\star f_{i}
= f_{1}\star f_{2}\star\cdots\star f_{1} = f_{1}.
\]
Also, since $i=1$, we have
\begin{align*}
m_{A}^{\left(  i-1\right)  }\circ\left(  f_{1}\otimes f_{2}\otimes
\cdots\otimes f_{i}\right)  \circ\Delta_{C}^{\left(  i-1\right)  }
& =\underbrace{m_{A}^{\left(  1-1\right)  }}_{=m_{A}^{\left(  0\right)
}=\id_{A}}\circ\underbrace{\left(  f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{1}\right)  }_{=f_{1}}\circ\underbrace{\Delta
_{C}^{\left(  1-1\right)  }}_{=\Delta_{C}^{\left(  0\right)  }
=\id_{C}}=\id_{A}\circ
f_{1}\circ\id_{C}=f_{1}.
\end{align*}
Thus, $f_{1}\star f_{2}\star\cdots\star f_{i}=f_{1}=m_{A}^{\left(  i-1\right)
}\circ\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes f_{i}\right)  \circ
\Delta_{C}^{\left(  i-1\right)  }$. Hence, \eqref{sol.convolution.k.driver}
holds, qed.} This completes the induction base.

\textit{Induction step:} Let $I\in\left\{  1,2,...,k\right\}  $ be such that
$I<k$. Assume that \eqref{sol.convolution.k.driver} holds for $i=I$. We must
prove that \eqref{sol.convolution.k.driver} holds for $i=I+1$.

Denote the $\kk$-linear map $f_{1}\otimes f_{2}\otimes\cdots\otimes
f_{I}:C^{\otimes I}\rightarrow A^{\otimes I}$ by $g$. Then, $g=f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{I}$. But we assumed that
\eqref{sol.convolution.k.driver} holds for $i=I$. In other words,
\[
f_{1}\star f_{2}\star\cdots\star f_{I}=m_{A}^{\left(  I-1\right)  }
\circ\underbrace{\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes f_{I}\right)
}_{=g}\circ\Delta_{C}^{\left(  I-1\right)  }=m_{A}^{\left(  I-1\right)  }\circ
g\circ\Delta_{C}^{\left(  I-1\right)  }.
\]
Now,
\begin{align*}
f_{1}\star f_{2}\star\cdots\star f_{I+1}  & =\left(  f_{1}\star f_{2}
\star\cdots\star f_{I}\right)  \star f_{I+1}\\
& =m_{A}\circ\left(  \underbrace{\left(  f_{1}\star f_{2}\star\cdots\star
f_{I}\right)  }_{=m_{A}^{\left(  I-1\right)  }\circ g\circ\Delta_{C}^{\left(
I-1\right)  }}\otimes\underbrace{f_{I+1}}_{=\id_{A}
\circ f_{I+1}\circ\id_{C}}\right)  \circ\Delta_{C}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of convolution}\right)
\\
& =m_{A}\circ\underbrace{\left(  \left(  m_{A}^{\left(  I-1\right)  }\circ
g\circ\Delta_{C}^{\left(  I-1\right)  }\right)  \otimes\left(
\id_{A}\circ f_{I+1}\circ
\id_{C}\right)  \right)  }_{=\left(  m_{A}^{\left(  I-1\right)  }
\otimes\id_{A}\right)  \circ\left(  g\otimes
f_{I+1}\right)  \circ\left(  \Delta_{C}^{\left(  I-1\right)  }\otimes
\id_{C}\right)  }\circ\Delta_{C}\\
& =\underbrace{m_{A}\circ\left(  m_{A}^{\left(  I-1\right)  }\otimes
\id_{A}\right)  }_{\substack{=m_{A}^{\left(  I\right)
}\\\text{(since }m_{A}^{\left(  I\right)  }=m_{A}\circ\left(  m_{A}^{\left(
I-1\right)  }\otimes\id_{A}\right)  \\\text{(by
Exercise~\ref{exe.iterated.m}(b), applied to }k=I\text{))}}}\circ\left(
\underbrace{g}_{=f_{1}\otimes f_{2}\otimes\cdots\otimes f_{I}}\otimes
f_{I+1}\right)  \circ\underbrace{\left(  \Delta_{C}^{\left(  I-1\right)
}\otimes\id_{C}\right)  \circ\Delta_{C}}
_{\substack{=\Delta_{C}^{\left(  I\right)  }\\\text{(since }\Delta
_{C}^{\left(  I\right)  }=\left(  \Delta_{C}^{\left(  I-1\right)  }
\otimes\id_{C}\right)  \circ\Delta_{C}\\\text{(by
Exercise~\ref{exe.iterated.Delta}(b), applied to }k=I\text{))}}}\\
& =\underbrace{m_{A}^{\left(  I\right)  }}_{=m_{A}^{\left(  \left(
I+1\right)  -1\right)  }}\circ\underbrace{\left(  \left(  f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{I}\right)  \otimes f_{I+1}\right)  }
_{=f_{1}\otimes f_{2}\otimes\cdots\otimes f_{I+1}}\circ\underbrace{\Delta
_{C}^{\left(  I\right)  }}_{=\Delta_{C}^{\left(  \left(  I+1\right)
-1\right)  }}\\
& =m_{A}^{\left(  \left(  I+1\right)  -1\right)  }\circ\left(  f_{1}\otimes
f_{2}\otimes\cdots\otimes f_{I+1}\right)  \circ\Delta_{C}^{\left(  \left(
I+1\right)  -1\right)  }.
\end{align*}
In other words, \eqref{sol.convolution.k.driver} holds for $i=I+1$. Thus, the
induction step is complete. Therefore, \eqref{sol.convolution.k.driver} is
proven by induction.

Recall that we don't have $k=0$. Hence, $k>0$, so that $k\in\left\{
1,2,...,k\right\}  $. Therefore, \eqref{sol.convolution.k.driver} (applied to
$i=k$) yields
\[
f_{1}\star f_{2}\star\cdots\star f_{k}=m_{A}^{\left(  k-1\right)  }
\circ\left(  f_{1}\otimes f_{2}\otimes\cdots\otimes f_{k}\right)  \circ
\Delta_{C}^{\left(  k-1\right)  }.
\]
This solves Exercise~\ref{exe.convolution.k}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.coalg-anti}}

\textit{Solution to Exercise~\ref{exe.coalg-anti}.}
The solution of this exercise, of course, proceeds by inverting all
the arrows in the proof of
Proposition~\ref{antipodes-are-antiendomorphisms}, but this
is complicated by the fact that said proof first has to be rewritten
in terms of arrows (i.e., freed of all uses of
elements)\footnote{Rewriting a proof in terms of arrows is usually
an exercise in category theory (or, rather, category practice); see
the First solution to Exercise~\ref{exe.counit.unique} (or the
solution to Exercise~\ref{exe.cocommutative.coalghom} further below)
for how this is done (in a simple case).}. We give such a solution
to Exercise~\ref{exe.coalg-anti} further below -- see the solution of
Exercise~\ref{exe.coalg-anti.gen}(c).

Alternatively, solutions of Exercise~\ref{exe.coalg-anti} can be found
in many places: e.g., \cite[Thm. 2.1.4(iv)]{Abe},
\cite[III.(3.5)]{Kassel}, \cite[Prop. I.7.1 2)]{Manchon}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.coalg-anti.gen}}

\textit{Solution to Exercise~\ref{exe.coalg-anti.gen}.} (a) We shall give two
solutions of Exercise~\ref{exe.coalg-anti.gen}(a). The first of these
solutions will be a (completely straightforward) generalization of the proof
of Proposition~\ref{antipodes-are-antiendomorphisms} (which proposition, as we
will see in solving Exercise~\ref{exe.coalg-anti.gen}(c), is a particular case
of this exercise). The second solution will be element-free (i.e., it will
only manipulate linear maps, but never refer to elements of $C$ or $A$), but
essentially is just a result of rewriting the first solution in element-free
terms. One of the advantages of an element-free solution is the ease of
applying it to more general contexts (such as tensor categories), but also the
possibility of straightforwardly ``reversing the arrows'' in such a solution
and thus obtaining a solution to Exercise~\ref{exe.coalg-anti.gen}(b). (We
will elaborate on this when we solve Exercise~\ref{exe.coalg-anti.gen}(b).)

\textit{First solution to Exercise~\ref{exe.coalg-anti.gen}(a).} Let
$\overline{r}=r^{\star\left(  -1\right)  }$. Then, $\overline{r}%
=r^{\star\left(  -1\right)  }$ is the $\star$-inverse of $f$. Thus,
$\overline{r}\star r=1_{\left(  \Hom \left(  C,A\right)
,\star\right)  }=u_{A}\circ\epsilon_{C}$ and similarly $r\star
\overline{r}=u_{A}\circ\epsilon_{C}$.

Since $\Delta_{C}$ is an algebra morphism, one has $\Delta_{C}\left(  1_{C}\right)
=1_{C}\otimes 1_{C}$, and thus
\[
\left(  r\star\overline{r}\right)  \left(
1_{C}\right)  =\underbrace{r\left(  1_{C}\right)  }_{\substack{=1_{A}%
\\\text{(since }r\text{ is a }\kk\text{-algebra}\\\text{homomorphism)}%
}}\overline{r}\left(  1_{C}\right)  =\overline{r}\left(  1_{C}\right)  ,
\]
so
that $\overline{r}\left(  1_{C}\right)  =\underbrace{\left(  r\star
\overline{r}\right)  }_{=u_{A}\circ\epsilon_{C}}\left(  1_{C}\right)  =\left(
u_{A}\circ\epsilon_{C}\right)  \left(  1_{C}\right)  =1_{A}$.

But we need to prove that $r^{\star\left(  -1\right)  }$ is a $\kk$-algebra
anti-homomorphism. In other words, we need to prove that
$\overline{r}$ is a $\kk$-algebra anti-homomorphism (because
$\overline{r}=r^{\star\left(  -1\right)  }$). In other words, we need to prove
that $\overline{r}\left(  1_{C}\right)  =1_{A}$ and that every $a\in C$ and
$b\in C$ satisfy $\overline{r}\left(  ab\right)  =\overline{r}\left(
b\right)  \overline{r}\left(  a\right)  $. Since $\overline{r}\left(
1_{C}\right)  =1_{A}$ is already proven, it thus remains to prove that every
$a\in C$ and $b\in C$ satisfy $\overline{r}\left(  ab\right)  =
\overline{r}\left(  b\right)  \overline{r}\left(  a\right)  $.

For this purpose, we shall \textbf{not} fix $a$ and $b$. Instead, we consider
$C\otimes C$ as a $\kk$-coalgebra, and $A$ as a $\kk$-algebra.
Then, $\Hom \left(  C\otimes C,A\right)  $ is an associative
algebra with a convolution product $\circledast$ (to be distinguished from
the convolution $\star$ on $\Hom \left(  C,A\right)  $), having
two-sided identity element $u_{A}\epsilon_{C\otimes C}$. We define three
$\kk$-linear maps $f$, $g$ and $h$ from $C\otimes C$ to $A$ as follows:%
\begin{align*}
f\left(  a\otimes b\right)   &  =r\left(  a\right)  r\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in C\text{ and }b\in C;\\
g\left(  a\otimes b\right)   &  =\overline{r}\left(  b\right)
\overline{r}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in C\text{ and }b\in C;\\
h\left(  a\otimes b\right)   &  =\overline{r}\left(  ab\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in C\text{ and }b\in C.
\end{align*}
(These definitions make sense, since each of $r\left(  a\right)  r\left(
b\right)  $, $\overline{r}\left(  b\right)  \overline{r}\left(  a\right)  $
and $\overline{r}\left(  ab\right)  $ depends $\kk$-bilinearly on
$\left(  a,b\right)  $.) Thus, $f$, $g$ and $h$ are three elements of
$\Hom \left(  C\otimes C,A\right)  $. We shall now prove that%
\begin{equation}
h\circledast f=u_{A}\epsilon_{C\otimes C}=f\circledast g.
\label{sol.coalg-anti.gen.a.sol1.hffg}%
\end{equation}
Once this equality is proven, we will then obtain%
\begin{align*}
h  &  =h\circledast\left(  u_{A}\epsilon_{C\otimes C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\epsilon_{C\otimes C}\text{ is
the identity element of }\left(  \Hom \left(  C\otimes
C,A\right)  ,\circledast\right)  \right) \\
&  =h\circledast\left(  f\circledast g\right)  =\left(  h\circledast f\right)
\circledast g\ \ \ \ \ \ \ \ \ \ \left(  \text{by the associativity of the
convolution }\circledast\right) \\
&  =\left(  u_{A}\epsilon_{C\otimes C}\right)  \circledast
g=g\ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\epsilon_{C\otimes C}\text{
is the identity element of }\left(  \Hom \left(  C\otimes
C,A\right)  ,\circledast\right)  \right)  .
\end{align*}


In order to prove (\ref{sol.coalg-anti.gen.a.sol1.hffg}), we evaluate the
three maps $h\circledast f$, $u_{A}\epsilon_{C\otimes C}$ and $f\circledast g$
on pure tensors $a\otimes b\in C\otimes C$. We use the Sweedler
notation in the form $\Delta(a)=\sum_{(a)}a_{1}\otimes a_{2}$ and
$\Delta(b)=\sum_{(b)}b_{1}\otimes b_{2}$; thus, $\Delta(ab)=
\sum_{(a),(b)}a_{1}b_{1}\otimes a_{2}b_{2}$ (since $\Delta$ is a
$\kk$-algebra homomorphism). We then obtain
\[
\left(  u_{A}\epsilon_{C\otimes C}\right)  \left(  a\otimes b\right)
=u_{A}\left(  \epsilon_{C}\left(  a\right)  \epsilon_{C}\left(  b\right)
\right)  =u_{A}\left(  \epsilon_{C}\left(  ab\right)  \right)  .
\]
Furthermore,
\begin{align}
\left(  h\circledast f\right)  \left(  a\otimes b\right)
&  =\sum_{(a),(b)}h\left(  a_{1}\otimes b_{1}\right)  f\left(  a_{2}\otimes
b_{2}\right)  =\sum_{(a),(b)}\overline{r}\left(  a_{1}b_{1}\right)
\underbrace{r\left(  a_{2}\right)  r\left(  b_{2}\right)  }%
_{\substack{=r\left(  a_{2}b_{2}\right)  \\\text{(since }r\text{ is a
}\kk\text{-algebra}\\\text{homomorphism)}}}\nonumber\\
&  =\sum_{(a),(b)}\overline{r}\left(  a_{1}b_{1}\right)  r\left(  a_{2}%
b_{2}\right)  =\underbrace{\left(  \overline{r}\star r\right)  }_{=u_{A}%
\circ\epsilon_{C}}\left(  ab\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\sum_{(a),(b)}a_{1}b_{1}\otimes a_{2}b_{2}=\Delta\left(  ab\right)  \right)
\nonumber\\
&  =\left(  u_{A}\circ\epsilon_{C}\right)  \left(  ab\right)  =u_{A}\left(
\epsilon_{C}\left(  ab\right)  \right)  \label{sol.coalg-anti.gen.a.sol1.hf}%
\end{align}
and%
\begin{align}
\left(  f\circledast g\right)  \left(  a\otimes b\right)
&  = \sum_{(a),(b)}f\left(  a_{1}\otimes b_{1}\right)  g\left(  a_{2}\otimes
b_{2}\right)  =\sum_{(a),(b)}r\left(  a_{1}\right)  r\left(  b_{1}\right)
\overline{r}\left(  b_{2}\right)  \overline{r}\left(  a_{2}\right) \nonumber\\
&  =\sum_{(a)}r\left(  a_{1}\right)  \underbrace{\left(  \sum_{(b)}r\left(
b_{1}\right)  \overline{r}\left(  b_{2}\right)  \right)  }_{\substack{=\left(
r\star\overline{r}\right)  \left(  b\right)  =\left(  u_{A}\circ\epsilon
_{C}\right)  \left(  b\right)  \\\text{(since }r\star\overline{r}=u_{A}%
\circ\epsilon_{C}\text{)}}}\overline{r}\left(  a_{2}\right)
= \sum_{(a)}r\left(  a_{1}\right)  \underbrace{\left(  u_{A}\circ\epsilon
_{C}\right)  \left(  b\right)  \overline{r}\left(  a_{2}\right)
}_{\substack{=\epsilon_{C}\left(  b\right)  \overline{r}\left(  a_{2}\right)
}}\nonumber\\
&  =\sum_{(a)}r\left(  a_{1}\right)  \epsilon_{C}\left(  b\right)
\overline{r}\left(  a_{2}\right)  =\underbrace{\left(  \sum_{(a)}r\left(
a_{1}\right)  \overline{r}\left(  a_{2}\right)  \right)  }_{\substack{=\left(
r\star\overline{r}\right)  \left(  a\right)  =\left(  u_{A}\circ\epsilon
_{C}\right)  \left(  a\right)  \\\text{(since }r\star\overline{r}=u_{A}%
\circ\epsilon_{C}\text{)}}}\epsilon_{C}\left(  b\right) \nonumber\\
&  =\left(  u_{A}\circ\epsilon_{C}\right)  \left(  a\right)  \epsilon
_{C}\left(  b\right)  =\left(  u_{A}\circ\epsilon_{C}\right)  \left(
a\right)  \left(  u_{A}\circ\epsilon_{C}\right)  \left(  b\right)  =\left(
u_{A}\circ\epsilon_{C}\right)  \left(  ab\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\circ\epsilon_{C}\text{ is a
}\kk\text{-algebra homomorphism}\right) \nonumber\\
&  =u_{A}\left(  \epsilon_{C}\left(  ab\right)  \right)  .
\label{sol.coalg-anti.gen.a.sol1.fg}%
\end{align}
These three results being all equal, we thus have shown that the maps
$h\circledast f$, $u_{A}\epsilon_{C\otimes C}$ and $f\circledast g$ are equal
on each pure tensor. Correspondingly, these maps must be identical (since they
are $\kk$-linear). In other words,
(\ref{sol.coalg-anti.gen.a.sol1.hffg}) holds. As explained above, this yields
$h=g$. Thus, every $a\in C$ and $b\in C$ satisfy%
\begin{align*}
\overline{r}\left(  ab\right)   &  =\underbrace{h}_{=g}\left(  a\otimes
b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }h\left(  a\otimes b\right)
\text{ is defined to be }\overline{r}\left(  ab\right)  \right) \\
&  =g\left(  a\otimes b\right)
=  \overline{r}\left(  b\right)  \overline{r}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}g\left(  a\otimes b\right)  \right)  .
\end{align*}
As we already know, this completes the solution of
Exercise~\ref{exe.coalg-anti.gen}(a).

\begin{vershort}

\textit{Second solution to Exercise~\ref{exe.coalg-anti.gen}(a).} Let us first
recall some linear-algebraic facts. One such fact states that if $U$, $V$,
$U^{\prime}$ and $V^{\prime}$ are four $\kk$-modules and
$x:U\rightarrow U^{\prime}$ and $y:V\rightarrow V^{\prime}$ are two
$\kk$-linear maps, then
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}=T_{U^{\prime},V^{\prime}}\circ\left(
x\otimes y\right)  . \label{sol.coalg-anti.gen.a.sol2.twist}%
\end{equation}


For every $\kk$-module $U$, let $\operatorname{kan}_{1,U}%
:U\rightarrow U\otimes\kk$ and $\operatorname{kan}_{2,U}%
:U\rightarrow\kk\otimes U$ be the canonical $\kk$-module
isomorphisms. These two isomorphisms are related to each other via the
equalities%
\begin{align}
\operatorname{kan}_{2,U}^{-1}\circ T_{U,\kk}  &
=\operatorname{kan}_{1,U}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kan12.1}\\
\operatorname{kan}_{1,U}^{-1}\circ T_{\kk,U}  &
=\operatorname{kan}_{2,U}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kan12.2}\\
T_{\kk,U}\circ\operatorname{kan}_{2,U}  &
=\operatorname{kan}_{1,U},\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{sol.coalg-anti.gen.a.sol2.kan12.3}\\
T_{U,\kk}\circ\operatorname{kan}_{1,U}  &
=\operatorname{kan}_{2,U} \label{sol.coalg-anti.gen.a.sol2.kan12.4}%
\end{align}
for every $\kk$-module $U$. Moreover, every $\kk$-modules $U$
and $V$ satisfy%
\begin{align}
\id_{U}\otimes\operatorname{kan}_{1,V}  &
=\operatorname{kan}_{1,U\otimes V}%
,\label{sol.coalg-anti.gen.a.sol2.kans.1}\\
\operatorname{kan}_{2,V}\otimes\id_{U}  &
=\operatorname{kan}_{2,V\otimes U}%
,\label{sol.coalg-anti.gen.a.sol2.kans.2}\\
\id_{U}\otimes\operatorname{kan}_{1,V}^{-1}
&  =\operatorname{kan}_{1,U\otimes V}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kans.3}\\
\operatorname{kan}_{2,V}^{-1}\otimes\id_{U}
&  =\operatorname{kan}_{2,V\otimes U}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kans.4}\\
\id_{U}\otimes\operatorname{kan}_{2,V}  &
=\operatorname{kan}_{1,U}\otimes\id_{V}%
,\label{sol.coalg-anti.gen.a.sol2.kans.5}\\
\id_{U}\otimes\operatorname{kan}_{2,V}^{-1}
&  =\operatorname{kan}_{1,U}^{-1}\otimes\id_{V}.
\label{sol.coalg-anti.gen.a.sol2.kans.6}
\end{align}


Furthermore, if $U$ and $V$ are two $\kk$-modules and $\alpha
:U\rightarrow V$ is a $\kk$-linear map, then%
\begin{align}
\operatorname{kan}_{1,V}\circ\alpha &  =\left(  \alpha
\otimes\id_{\kk}\right)  \circ
\operatorname{kan}_{1,U},\label{sol.coalg-anti.gen.a.sol2.kan1}\\
\operatorname{kan}_{2,V}\circ\alpha &  =\left(  \id_{\kk}
\otimes\alpha\right)  \circ\operatorname{kan}_{2,U},
\label{sol.coalg-anti.gen.a.sol2.kan2}\\
\operatorname{kan}_{1,V}^{-1}\circ\left(  \alpha\otimes
\id_{\kk}\right)   &  =\alpha\circ
\operatorname{kan}_{1,U}^{-1},\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{sol.coalg-anti.gen.a.sol2.kan1a}\\
\operatorname{kan}_{2,V}^{-1}\circ\left(  \id_{\kk}
\otimes\alpha\right)   &  =\alpha\circ
\operatorname{kan}_{2,U}^{-1}.
\label{sol.coalg-anti.gen.a.sol2.kan2a}
\end{align}


Now, let us step to the actual solution of Exercise~\ref{exe.coalg-anti.gen}(a).

Let $\overline{r}=r^{\star\left(  -1\right)  }$. Then, $\overline{r}%
=r^{\star\left(  -1\right)  }$ is the $\star$-inverse of $f$. Thus,
$\overline{r}\star r=1_{\left(  \Hom \left(  C,A\right)
,\star\right)  }=u_{A}\circ\epsilon_{C}$ and similarly
$r\star\overline{r}=u_{A}\circ\epsilon_{C}$.

Recall that $C$ is a $\kk$-coalgebra. By the axioms of a $\kk%
$-coalgebra, this shows that
\begin{align*}
\left(  \Delta_{C}\otimes\id_{C}\right)  \circ
\Delta_{C}  &  =\left(  \id_{C}\otimes\Delta
_{C}\right)  \circ\Delta_{C},\\
\left(  \id_{C}\otimes\epsilon_{C}\right)  \circ
\Delta_{C}  &  =\operatorname{kan}_{1,C};\\
\left(  \epsilon_{C}\otimes\id_{C}\right)  \circ
\Delta_{C}  &  =\operatorname{kan}_{2,C}.
\end{align*}
Also, recall that $C$ is a $\kk$-algebra. By the axioms of a
$\kk$-algebra, this shows that%
\begin{align*}
m_{C}\circ\left(  m_{C}\otimes\id_{C}\right)   &
=m_{C}\circ\left(  \id_{C}\otimes m_{C}\right)  ;\\
m_{C}\circ\left(  \id_{C}\otimes u_{C}\right)   &
=\operatorname{kan}_{1,C}^{-1};\\
m_{C}\circ\left(  u_{C}\otimes\id_{C}\right)   &
=\operatorname{kan}_{2,C}^{-1}.
\end{align*}
Also, recall that $C$ is a $\kk$-bialgebra. Due to the axioms of a
$\kk$-bialgebra, this shows that $\Delta_{C}$ and $\epsilon_{C}$ are
$\kk$-algebra homomorphisms, and that $m_{C}$ and $u_{C}$ are
$\kk$-coalgebra homomorphisms.

Furthermore, $A$ is a $\kk$-algebra. By the axioms of a
$\kk$-algebra, this shows that
\begin{align*}
m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)   &
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  ;\\
m_{A}\circ\left(  \id_{A}\otimes u_{A}\right)   &
=\operatorname{kan}_{1,A}^{-1};\\
m_{A}\circ\left(  u_{A}\otimes\id_{A}\right)   &
=\operatorname{kan}_{2,A}^{-1}.
\end{align*}


Since $\epsilon_{C}$ is a $\kk$-algebra homomorphism, we have
$\epsilon_{C}\circ u_{C}=u_{\kk}=
\id_{\kk}$.

Recall that $\operatorname{kan}_{1, \kk}$ is the canonical
$\kk$-module isomorphism $\kk\rightarrow\kk\otimes\kk$.
Hence, $u_{C\otimes C}=\left(  u_{C}\otimes u_{C}\right)
\circ\operatorname{kan}_{1, \kk}$ (by the definition of the
$\kk$-algebra $C\otimes C$). Also, $\operatorname*{kan}%
\nolimits_{1, \kk}=\operatorname{kan}_{2, \kk}$ (since
each of $\operatorname{kan}_{1, \kk}$ and $\operatorname*{kan}%
\nolimits_{2, \kk}$ is the canonical $\kk$-module isomorphism
$\kk\rightarrow\kk\otimes\kk$).

We know that $r:C\rightarrow A$ is a $\kk$-algebra homomorphism. In
other words, $r$ is a $\kk$-linear map satisfying $r\circ m_{C}%
=m_{A}\circ\left(  r\otimes r\right)  $ and $r\circ u_{C}=u_{A}$.

Now, we need to prove that $r^{\star\left(  -1\right)  }$ is a $\kk%
$-algebra anti-homomorphism. In other words, we need to prove that
$\overline{r}$ is a $\kk$-algebra anti-homomorphism (since
$\overline{r}=r^{\star\left(  -1\right)  }$). In other words, we need to prove
that $\overline{r}$ satisfies the two equations $\overline{r}\circ m_{C}%
=m_{A}\circ\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$ and
$\overline{r}\circ u_{C}=u_{A}$ (because these two equations are what makes
$\overline{r}$ a $\kk$-algebra anti-homomorphism, according to the
definition of a ``$\kk$-algebra anti-homomorphism'').

Let us first prove the equality $\overline{r}\circ u_{C}=u_{A}$.

We have $r\star\overline{r}=u_{A}\circ\epsilon_{C}$, so that
$\underbrace{\left(  r\star\overline{r}\right)  }_{=u_{A}\circ\epsilon_{C}%
}\circ u_{C}=u_{A}\circ\underbrace{\epsilon_{C}\circ u_{C}}%
_{=\id_{\kk}}=u_{A}$. Hence,%
\begin{align*}
u_{A}  &  =\underbrace{\left(  r\star\overline{r}\right)  }_{\substack{=m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\\\text{(by the
definition}\\\text{of convolution)}}}\circ u_{C}=m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\underbrace{\Delta_{C}\circ u_{C}%
}_{\substack{=u_{C\otimes C}\\\text{(since }\Delta_{C}\text{ is a }%
\kk\text{-algebra}\\\text{homomorphism)}}}\\
&  =m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ
\underbrace{u_{C\otimes C}}_{\substack{=\left(  u_{C}\otimes u_{C}\right)
\circ\operatorname{kan}_{1, \kk}}}=m_{A}\circ
\underbrace{\left(  r\otimes\overline{r}\right)  \circ\left(  u_{C}\otimes
u_{C}\right)  }_{\substack{=\left(  r\circ u_{C}\right)  \otimes\left(
\overline{r}\circ u_{C}\right)  }}\circ\operatorname*{kan}%
\nolimits_{1, \kk}\\
&  =m_{A}\circ\left(  \underbrace{\left(  r\circ u_{C}\right)  }_{=u_{A}%
=u_{A}\circ\id_{\kk}}\otimes\underbrace{\left(
\overline{r}\circ u_{C}\right)  }_{=\id_{A}%
\circ\left(  \overline{r}\circ u_{C}\right)  }\right)  \circ
\underbrace{\operatorname{kan}_{1, \kk}}_{=\operatorname*{kan}%
\nolimits_{2, \kk}}=m_{A}\circ\underbrace{\left(  \left(  u_{A}%
\circ\id_{\kk}\right)  \otimes\left(
\id_{A}\circ\left(  \overline{r}\circ u_{C}\right)
\right)  \right)  }_{\substack{=\left(  u_{A}\otimes
\id_{A}\right)  \circ\left(  \id_{\kk%
}\otimes\left(  \overline{r}\circ u_{C}\right)  \right)  }}\circ
\operatorname{kan}_{2, \kk}\\
&  =\underbrace{m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  }_{=\operatorname{kan}_{2,A}^{-1}}%
\circ\underbrace{\left(  \id_{\kk}%
\otimes\left(  \overline{r}\circ u_{C}\right)  \right)  \circ
\operatorname{kan}_{2, \kk}}_{\substack{=\operatorname*{kan}%
\nolimits_{2,A}\circ\left(  \overline{r}\circ u_{C}\right)  \\\text{(since
(\ref{sol.coalg-anti.gen.a.sol2.kan2}) (applied to }U=\kk%
\text{,}\\V=A\text{ and }\alpha=\overline{r}\circ u_{C}\text{) yields}%
\\\operatorname{kan}_{2,A}\circ\left(  \overline{r}\circ
u_{C}\right)  =\left(  \id_{\kk}\otimes\left(
\overline{r}\circ u_{C}\right)  \right)  \circ\operatorname*{kan}%
\nolimits_{2, \kk}\text{)}}}=\underbrace{\operatorname*{kan}%
\nolimits_{2,A}^{-1}\circ\operatorname{kan}_{2,A}}%
_{=\id_{A}}\circ\left(  \overline{r}\circ u_{C}\right)
\\
&  =\overline{r}\circ u_{C}.
\end{align*}
Hence, $\overline{r}\circ u_{C}=u_{A}$.

Now, it remains to prove the equality $\overline{r}\circ m_{C}=m_{A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$. This is
harder. Let us first make some preparations.

Recall that $C\otimes C$ is a $\kk$-coalgebra (since $C$ is a
$\kk$-coalgebra), and $A$ is an $\kk$-algebra. Thus,
$\Hom \left(  C\otimes C,A\right)  $ is an associative algebra
with respect to convolution. We shall denote the convolution on
$\Hom \left(  C\otimes C,A\right)  $ by $\circledast$ rather
than by $\star$ (in order to distinguish it from the convolution $\star$ on
$\Hom \left(  C,A\right)  $). The algebra $\left(
\Hom \left(  C\otimes C,A\right)  ,\star\right)  $ has two-sided
identity element $u_{A}\circ\epsilon_{C\otimes C}$.

We define a $\kk$-linear map $f:C\otimes C\rightarrow A$ by
$f=m_{A}\circ\left(  r\otimes r\right)  $.

We define a $\kk$-linear map $g:C\otimes C\rightarrow A$ by
$g=m_{A}\circ T_{A,A}\circ\left(  \overline{r}\otimes\overline{r}\right)  $.

We define a $\kk$-linear map $h:C\otimes C\rightarrow A$ by
$h=\overline{r}\circ m_{C}$.

Clearly, $f$, $g$ and $h$ are three elements of $\Hom \left(
C\otimes C,A\right)  $. Our next goal is to prove that%
\begin{equation}
h\circledast f=u_{A}\circ\epsilon_{C\otimes C}=f\circledast g.
\label{sol.coalg-anti.gen.a.sol2.hffg}%
\end{equation}
Once this equality is proven, we will then obtain%
\begin{align}
h  &  =h\circledast\underbrace{\left(  u_{A}\circ\epsilon_{C\otimes C}\right)
}_{\substack{=f\circledast g\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.hffg}%
))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\circ\epsilon_{C\otimes
C}\text{ is the identity element of }\left(  \Hom \left(
C\otimes C,A\right)  ,\circledast\right)  \right) \nonumber\\
&  =h\circledast\left(  f\circledast g\right)  =\underbrace{\left(
h\circledast f\right)  }_{\substack{=u_{A}\circ\epsilon_{C\otimes
C}\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.hffg}))}}}\circledast
g\ \ \ \ \ \ \ \ \ \ \left(  \text{by the associativity of the convolution
}\circledast\right) \nonumber\\
&  =\left(  u_{A}\circ\epsilon_{C\otimes C}\right)  \circledast
g=g\label{sol.coalg-anti.gen.a.sol2.h=g}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\circ\epsilon_{C\otimes
C}\text{ is the identity element of }\left(  \Hom \left(
C\otimes C,A\right)  ,\circledast\right)  \right)  .\nonumber
\end{align}


So let us concentrate on proving (\ref{sol.coalg-anti.gen.a.sol2.hffg}). Let
us first notice that%
\begin{equation}
u_{A}\circ\epsilon_{C}\circ m_{C}=u_{A}\circ\epsilon_{C\otimes C}.
\label{sol.coalg-anti.gen.a.sol2.hf.pf.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7}):} We know
that $m_{C}$ is a $\kk$-coalgebra homomorphism. Hence, $\epsilon
_{C}\circ m_{C}=\epsilon_{C\otimes C}$. Thus, $u_{A}\circ\underbrace{\epsilon
_{C}\circ m_{C}}_{=\epsilon_{C\otimes C}}=u_{A}\circ\epsilon_{C\otimes C}$.
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7}).}

Next, we make the following observations:

\begin{itemize}
\item First, we notice that%
\begin{equation}
h\circledast f=m_{A}\circ\left(  h\otimes f\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.hf.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}):} By the
definition of the convolution $h\circledast f$, we have
\[
h\circledast f=m_{A}\circ\left(  h\otimes f\right)  \circ\underbrace{\Delta
_{C\otimes C}}_{\substack{=\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \\\text{(by the definition of the }%
\kk\text{-coalgebra }C\otimes C\text{)}}}=m_{A}\circ\left(  h\otimes
f\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\otimes\id_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  .
\]
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}).}

\item Next, we notice that%
\begin{align}
&  m_{A}\circ\left(  h\otimes f\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.hf.pf.2}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}):} We
have
\begin{align*}
&  m_{A}\circ\left(  \underbrace{h}_{=\overline{r}\circ m_{C}}\otimes
\underbrace{f}_{=m_{A}\circ\left(  r\otimes r\right)  }\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}).}

\item Next, we have%
\begin{align}
&  m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\nonumber\\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
r\circ m_{C}\right)  \right)  \circ\left(  \id_{C}%
\otimes T_{C,C}\otimes\id_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  \label{sol.coalg-anti.gen.a.sol2.hf.pf.3}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}):} We have%
\begin{align*}
&  m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes
\underbrace{\left(  r\circ m_{C}\right)  }_{=m_{A}\circ\left(  r\otimes
r\right)  }\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}).}

\item Furthermore, we have%
\begin{align}
&  m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
r\circ m_{C}\right)  \right)  \circ\left(  \id_{C}%
\otimes T_{C,C}\otimes\id_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ m_{C}
\label{sol.coalg-anti.gen.a.sol2.hf.pf.4}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}):} We have%
\begin{align*}
&  m_{A}\circ\underbrace{\left(  \left(  \overline{r}\circ m_{C}\right)
\otimes\left(  r\circ m_{C}\right)  \right)  }_{=\left(  \overline{r}\otimes
r\right)  \circ\left(  m_{C}\otimes m_{C}\right)  }\circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\underbrace{\left(
m_{C}\otimes m_{C}\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  }%
_{\substack{=m_{C\otimes C}\\\text{(since the definition of the }%
\kk\text{-algebra }C\otimes C\\\text{yields }m_{C\otimes C}=\left(
m_{C}\otimes m_{C}\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \text{)}}}\circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ
\underbrace{m_{C\otimes C}\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
}_{\substack{=\Delta_{C}\circ m_{C}\\\text{(since }\Delta_{C}:C\rightarrow
C\otimes C\text{ is a }\kk\text{-algebra homomorphism)}}}=m_{A}%
\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ m_{C}.
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}).}

\item Moreover, we have%
\begin{equation}
m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ
m_{C}=\left(  \overline{r}\star r\right)  \circ m_{C}.
\label{sol.coalg-anti.gen.a.sol2.hf.pf.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}):} The
definition of the convolution $\overline{r}\star r$ yields $\overline{r}\star
r=m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}$. Hence,
$\underbrace{\left(  \overline{r}\star r\right)  }_{=m_{A}\circ\left(
\overline{r}\otimes r\right)  \circ\Delta_{C}}\circ m_{C}=m_{A}\circ\left(
\overline{r}\otimes r\right)  \circ\Delta_{C}\circ m_{C}$. This proves
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}).}

\item Finally, we have%
\begin{equation}
\left(  \overline{r}\star r\right)  \circ m_{C}=u_{A}\circ\epsilon_{C}\circ
m_{C}. \label{sol.coalg-anti.gen.a.sol2.hf.pf.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}):} We have
$\underbrace{\left(  \overline{r}\star r\right)  }_{=u_{A}\circ\epsilon_{C}%
}\circ m_{C}=u_{A}\circ\epsilon_{C}\circ m_{C}$. Thus,
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}) is proven.}
\end{itemize}

Now, we have%
\begin{align*}
h\circledast f  &  =m_{A}\circ\left(  h\otimes f\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}%
)}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}%
)}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
r\circ m_{C}\right)  \right)  \circ\left(  \id_{C}%
\otimes T_{C,C}\otimes\id_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3})}\right) \\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ
m_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4})}\right) \\
&  =\left(  \overline{r}\star r\right)  \circ m_{C}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5})}\right) \\
&  =u_{A}\circ\epsilon_{C}\circ m_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6})}\right) \\
&  =u_{A}\circ\epsilon_{C\otimes C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7})}\right)  .
\end{align*}
The first equality in (\ref{sol.coalg-anti.gen.a.sol2.hffg}) is thus proven.

We shall next show the second equality in
(\ref{sol.coalg-anti.gen.a.sol2.hffg}).

For every $k\in \NN$, let us define the map $m^{\left(  k\right)
}:A^{\otimes\left(  k+1\right)  }\rightarrow A$ as in
Exercise~\ref{exe.iterated.m}. We recall that these maps $m^{\left(  k\right)
}$ are defined by induction over $k$, with the induction base $m^{\left(
0\right)  }=\operatorname{id}_{A}$, and with the induction step%
\begin{equation}
m^{\left(  k\right)  }=m_{A}\circ\left(  \operatorname{id}_{A}\otimes
m^{\left(  k-1\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }k\geq1.
\label{sol.coalg-anti.gen.a.sol2.mk}%
\end{equation}
\footnote{This is precisely the definition of these maps given in
Exercise~\ref{exe.iterated.m}, with the only difference that $m_{A}$ was
denoted by $m$ in that exercise.} Out of these maps, we shall only need
$m^{\left(  0\right)  }$, $m^{\left(  1\right)  }$, $m^{\left(  2\right)  }$
and $m^{\left(  3\right)  }$. These satisfy the following formulae:%
\begin{align}
m^{\left(  0\right)  }  &  =\id_{A}%
;\label{sol.coalg-anti.gen.a.sol2.m0}\\
m^{\left(  1\right)  }  &  =m_{A};\label{sol.coalg-anti.gen.a.sol2.m1}\\
m^{\left(  2\right)  }  &  =m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  ;\label{sol.coalg-anti.gen.a.sol2.m2a}\\
m^{\left(  2\right)  }  &  =m_{A}\circ\left(  
\id_{A}\otimes m_{A}\right)  ;\label{sol.coalg-anti.gen.a.sol2.m2b}\\
m^{\left(  3\right)  }  &  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right)
;\label{sol.coalg-anti.gen.a.sol2.m3}\\
m^{\left(  3\right)  }  &  =m^{\left(  2\right)  }\circ\left(
\id_{A}\otimes m_{A}\otimes
\id_{A}\right)  . \label{sol.coalg-anti.gen.a.sol2.m3o}%
\end{align}
\footnote{Here are proofs for these formulae:
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m0}):} The formula
(\ref{sol.coalg-anti.gen.a.sol2.m0}) follows immediately from the definition
of $m^{\left(  0\right)  }$.
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m1}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=1$, we obtain $m^{\left(  1\right)
}=m_{A}\circ\left(  \id_{A}\otimes
\underbrace{m^{\left(  1-1\right)  }}_{=m^{\left(  0\right)  }%
=\id_{A}}\right)  =m_{A}\circ\underbrace{\left(
\id_{A}\otimes\id_{A}\right)
}_{=\id_{A\otimes A}}=m_{A}$. This proves
(\ref{sol.coalg-anti.gen.a.sol2.m1}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m2b}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=2$, we obtain $m^{\left(  2\right)
}=m_{A}\circ\left(  \id_{A}\otimes
\underbrace{m^{\left(  2-1\right)  }}_{\substack{=m^{\left(  1\right)  }%
=m_{A}\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.m1}))}}}\right)  =m_{A}%
\circ\left(  \id_{A}\otimes m_{A}\right)  $. This
proves (\ref{sol.coalg-anti.gen.a.sol2.m2b}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m2a}):} From
(\ref{sol.coalg-anti.gen.a.sol2.m2b}), we obtain $m^{\left(  2\right)  }%
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)
=m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  $
(since $m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  $).
This proves (\ref{sol.coalg-anti.gen.a.sol2.m2a}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m3}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=3$, we obtain%
\begin{align*}
m^{\left(  3\right)  }  &  =m_{A}\circ\left(  \underbrace{
\id_{A}}_{=\id_{A}\circ
\id_{A}}\otimes\underbrace{m^{\left(  3-1\right)  }}%
_{\substack{=m^{\left(  2\right)  }=m_{A}\circ\left(  
\id_{A}\otimes m_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2b}))}}}\right)  =m_{A}\circ
\underbrace{\left(  \left(  \id_{A}\circ
\id_{A}\right)  \otimes\left(  m_{A}\circ\left(
\id_{A}\otimes m_{A}\right)  \right)  \right)
}_{=\left(  \id_{A}\otimes m_{A}\right)  \circ\left(
\id_{A}\otimes\left(  
\id_{A}\otimes m_{A}\right)  \right)  }\\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  }_{\substack{=m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \\\text{(since }m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  =m_{A}\circ\left(  
\id_{A}\otimes m_{A}\right)  \text{)}}}\circ\underbrace{\left(
\id_{A}\otimes\left(  
\id_{A}\otimes m_{A}\right)  \right)  }_{\substack{=
\id_{A}\otimes\id_{A}\otimes m_{A}%
\\=\id_{A\otimes A}\otimes m_{A}}}=m_{A}%
\circ\underbrace{\left(  m_{A}\otimes\id_{A}\right)
\circ\left(  \id_{A\otimes A}\otimes m_{A}\right)
}_{\substack{=\left(  m_{A}\circ\id_{A\otimes
A}\right)  \otimes\left(  \id_{A}\circ m_{A}\right)
}}\\
&  =m_{A}\circ\left(  \underbrace{\left(  m_{A}\circ
\id_{A\otimes A}\right)  }_{=m_{A}}\otimes\underbrace{\left(
\id_{A}\circ m_{A}\right)  }_{=m_{A}}\right)
=m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.m3}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m3o}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=3$, we obtain%
\begin{align*}
m^{\left(  3\right)  }  &  =m_{A}\circ\left(  \underbrace{
\id_{A}}_{=\id_{A}\circ
\id_{A}}\otimes\underbrace{m^{\left(  3-1\right)  }}%
_{\substack{=m^{\left(  2\right)  }=m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2a}))}}}\right)  =m_{A}\circ
\underbrace{\left(  \left(  \id_{A}\circ
\id_{A}\right)  \otimes\left(  m_{A}\circ\left(
m_{A}\otimes\id_{A}\right)  \right)  \right)
}_{=\left(  \id_{A}\otimes m_{A}\right)  \circ\left(
\id_{A}\otimes\left(  m_{A}\otimes
\id_{A}\right)  \right)  }\\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  }_{\substack{=m^{\left(  2\right)  }\\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2b}))}}}\circ\underbrace{\left(
\id_{A}\otimes\left(  m_{A}\otimes
\id_{A}\right)  \right)  }_{=\id_{A}\otimes
m_{A}\otimes\id_{A}}=m^{\left(  2\right)  }%
\circ\left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.m3o}).}

Now, we make the following observations:

\begin{itemize}
\item We have%
\begin{equation}
f\circledast g=m_{A}\circ\left(  f\otimes g\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.1}):} By the
definition of the convolution $f\circledast g$, we have%
\[
f\circledast g=m_{A}\circ\left(  f\otimes g\right)  \circ\underbrace{\Delta
_{C\otimes C}}_{\substack{=\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \\\text{(by the definition of the }%
\kk\text{-coalgebra }C\otimes C\text{)}}}=m_{A}\circ\left(  f\otimes
g\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\otimes\id_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  .
\]
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.1}).}

\item We have
\begin{align}
&  m_{A}\circ\left(  f\otimes g\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \nonumber\\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
.\label{sol.coalg-anti.gen.a.sol2.fg.pf.2}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.2}):} The
equality (\ref{sol.coalg-anti.gen.a.sol2.twist}) (applied to $U=C$, $V=C$,
$U^{\prime}=A$, $V^{\prime}=A$, $x=\overline{r}$ and $y=\overline{r}$) yields
$\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}=T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  $. Hence, $g=m_{A}%
\circ\underbrace{T_{A,A}\circ\left(  \overline{r}\otimes\overline{r}\right)
}_{=\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}}=m_{A}%
\circ\left(  \left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}\right)  $. Now,%
\begin{align*}
&  m_{A}\circ\left(  \underbrace{f}_{=m_{A}\circ\left(  r\otimes r\right)
}\otimes\underbrace{g}_{=m_{A}\circ\left(  \left(  \overline{r}\otimes
\overline{r}\right)  \circ T_{C,C}\right)  }\right)  \\
&  =m_{A}\circ\underbrace{\left(  \left(  m_{A}\circ\left(  r\otimes r\right)
\right)  \otimes\left(  m_{A}\circ\left(  \left(  \overline{r}\otimes
\overline{r}\right)  \circ T_{C,C}\right)  \right)  \right)  }_{=\left(
m_{A}\otimes m_{A}\right)  \circ\left(  \left(  r\otimes r\right)
\otimes\left(  \left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}\right)  \right)  }=\underbrace{m_{A}\circ\left(  m_{A}\otimes
m_{A}\right)  }_{\substack{=m^{\left(  3\right)  }\\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m3}))}}}\circ\left(  \underbrace{\left(
r\otimes r\right)  }_{=\left(  r\otimes r\right)  \circ
\id_{C\otimes C}}\otimes\left(  \left(  \overline{r}\otimes
\overline{r}\right)  \circ T_{C,C}\right)  \right)  \\
&  =m^{\left(  3\right)  }\circ\underbrace{\left(  \left(  \left(  r\otimes
r\right)  \circ\id_{C\otimes C}\right)  \otimes\left(
\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}\right)  \right)
}_{=\left(  \left(  r\otimes r\right)  \otimes\left(  \overline{r}%
\otimes\overline{r}\right)  \right)  \circ\left(  
\id_{C\otimes C}\otimes T_{C,C}\right)  }=m^{\left(  3\right)  }%
\circ\underbrace{\left(  \left(  r\otimes r\right)  \otimes\left(
\overline{r}\otimes\overline{r}\right)  \right)  }_{=r\otimes r\otimes
\overline{r}\otimes\overline{r}}\circ\left(  \underbrace{
\id_{C\otimes C}}_{=\id_{C}\otimes
\id_{C}}\otimes T_{C,C}\right)  \\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A}\circ\left(  f\otimes g\right)  }_{=m^{\left(  3\right)
}\circ\left(  r\otimes r\otimes\overline{r}\otimes\overline{r}\right)
\circ\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  }\circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.2}).}

\item We have%
\begin{align}
&  m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\nonumber\\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
.\label{sol.coalg-anti.gen.a.sol2.fg.pf.3}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3}):} We
first notice that%
\begin{equation}
\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
=\id_{C}\otimes T_{C,C\otimes C}%
.\label{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}%
\end{equation}
In fact, the two maps $\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  $ and $\id_{C}\otimes
T_{C,C\otimes C}$ are $\kk$-linear, and are equal to each other on pure
tensors (in fact, each of them sends every pure tensor $a\otimes b\otimes
c\otimes d\in C\otimes C\otimes C\otimes C$ to $a\otimes c\otimes d\otimes
b$); therefore, they must be identical, so that
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}) holds.
\par
Now,%
\begin{align*}
&  \underbrace{m^{\left(  3\right)  }}_{\substack{=m^{\left(  2\right)  }%
\circ\left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m3o}))}}}\circ\left(  r\otimes r\otimes
\overline{r}\otimes\overline{r}\right)  \circ\underbrace{\left(
\id_{C}\otimes\id_{C}\otimes
T_{C,C}\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  }%
_{\substack{=\id_{C}\otimes T_{C,C\otimes
C}\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}))}}}\circ\left(
\underbrace{\Delta_{C}}_{=\id_{C\otimes C}\circ
\Delta_{C}}\otimes\underbrace{\Delta_{C}}_{=\Delta_{C}\circ
\id_{C}}\right)  \\
&  =m^{\left(  2\right)  }\circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  r\otimes
r\otimes\overline{r}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ
\underbrace{\left(  \left(  \id_{C\otimes C}%
\circ\Delta_{C}\right)  \otimes\left(  \Delta_{C}\circ
\id_{C}\right)  \right)  }_{=\left(  
\id_{C\otimes C}\otimes\Delta_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\id_{C}\right)  }\\
&  =m^{\left(  2\right)  }\circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  r\otimes
r\otimes\overline{r}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ\left(
\id_{C\otimes C}\otimes\Delta_{C}\right)  \circ\left(
\Delta_{C}\otimes\id_{C}\right)  .
\end{align*}
Since%
\begin{align*}
&  \left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \circ\underbrace{\left(  r\otimes
r\otimes\overline{r}\otimes\overline{r}\right)  }_{=r\otimes\left(
r\otimes\overline{r}\right)  \otimes\overline{r}}\\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \circ\left(  r\otimes\left(
r\otimes\overline{r}\right)  \otimes\overline{r}\right)  =\underbrace{\left(
\id_{A}\circ r\right)  }_{=r}\otimes\left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \right)  \otimes\underbrace{\left(
\id_{A}\circ\overline{r}\right)  }_{=\overline{r}%
}=r\otimes\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)
\otimes\overline{r}%
\end{align*}
and%
\begin{align*}
&  \left(  \id_{C}\otimes T_{C,C\otimes C}\right)
\circ\left(  \underbrace{\id_{C\otimes C}%
}_{=\id_{C}\otimes\id_{C}%
}\otimes\Delta_{C}\right)  \\
&  =\left(  \id_{C}\otimes T_{C,C\otimes C}\right)
\circ\left(  \id_{C}\otimes
\id_{C}\otimes\Delta_{C}\right)  =\left(  
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ\left(  
\id_{C}\otimes\left(  \id_{C}\otimes\Delta
_{C}\right)  \right)  \\
&  =\left(  \id_{C}\circ
\id_{C}\right)  \otimes\underbrace{\left(  T_{C,C\otimes C}\circ\left(
\id_{C}\otimes\Delta_{C}\right)  \right)
}_{\substack{=\left(  \Delta_{C}\otimes\id_{C}\right)
\circ T_{C,C}\\\text{(since }\left(  \Delta_{C}\otimes
\id_{C}\right)  \circ T_{C,C}=T_{C,C\otimes C}\circ\left(
\id_{C}\otimes\Delta_{C}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.twist}), applied to }U=C\text{, }%
V=C\text{,}\\U^{\prime}=C\text{, }V^{\prime}=C\otimes C\text{, }%
x=\id_{C}\text{ and }y=\Delta_{C}\text{))}}}\\
&  =\left(  \id_{C}\circ
\id_{C}\right)  \otimes\left(  \left(  \Delta_{C}\otimes
\id_{C}\right)  \circ T_{C,C}\right)
=\underbrace{\left(  \id_{C}\otimes\left(  \Delta
_{C}\otimes\id_{C}\right)  \right)  }%
_{=\id_{C}\otimes\Delta_{C}\otimes
\id_{C}}\circ\left(  \id_{C}\otimes
T_{C,C}\right)  =\left(  \id_{C}\otimes\Delta
_{C}\otimes\id_{C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  ,
\end{align*}
this becomes%
\begin{align*}
&  m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  \\
&  =m^{\left(  2\right)  }\circ\underbrace{\left(  
\id_{A}\otimes m_{A}\otimes\id_{A}\right)
\circ\left(  r\otimes r\otimes\overline{r}\otimes\overline{r}\right)
}_{=r\otimes\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)
\otimes\overline{r}}\circ\underbrace{\left(  \id_{C}%
\otimes T_{C,C\otimes C}\right)  \circ\left(  
\id_{C\otimes C}\otimes\Delta_{C}\right)  }_{=\left(  
\id_{C}\otimes\Delta_{C}\otimes\id_{C}\right)
\circ\left(  \id_{C}\otimes T_{C,C}\right)  }%
\circ\left(  \Delta_{C}\otimes\id_{C}\right)  \\
&  =m^{\left(  2\right)  }\circ\underbrace{\left(  r\otimes\left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \right)  \otimes\overline{r}\right)
\circ\left(  \id_{C}\otimes\Delta_{C}\otimes
\id_{C}\right)  }_{\substack{=\left(  r\circ
\id_{C}\right)  \otimes\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \right)  \circ\Delta_{C}\right)
\otimes\left(  \overline{r}\circ\id_{C}\right)
}}\circ\left(  \id_{C}\otimes T_{C,C}\right)
\circ\left(  \Delta_{C}\otimes\id_{C}\right)  \\
&  =m^{\left(  2\right)  }\circ\left(  \underbrace{\left(  r\circ
\id_{C}\right)  }_{=r}\otimes\underbrace{\left(
\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)  \circ
\Delta_{C}\right)  }_{=m_{A}\circ\left(  r\otimes\overline{r}\right)
\circ\Delta_{C}}\otimes\underbrace{\left(  \overline{r}\circ
\id_{C}\right)  }_{=\overline{r}}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta
_{C}\otimes\id_{C}\right)  \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
.
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3}).}

\item We have%
\begin{align}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
\nonumber\\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.4}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.4}):} We have
$r\star\overline{r}=m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ
\Delta_{C}$ (according to the definition of convolution). Compared with
$r\star\overline{r}=u_{A}\circ\epsilon_{C}$, this yields $m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}=u_{A}\circ\epsilon_{C}$. Thus,%
\begin{align*}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\underbrace{\left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  }%
_{=u_{A}\circ\epsilon_{C}}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta
_{C}\otimes\id_{C}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.4}).}

\item We have%
\begin{align}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \nonumber\\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  \circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)
.\label{sol.coalg-anti.gen.a.sol2.fg.pf.5}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.5}):} We have%
\[
\underbrace{r}_{=\id_{A}\circ r}\otimes\left(
u_{A}\circ\epsilon_{C}\right)  \otimes\underbrace{\overline{r}}%
_{=\id_{A}\circ\overline{r}}=\left(
\id_{A}\circ r\right)  \otimes\left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\left(  \id_{A}%
\circ\overline{r}\right)  =\left(  \id_{A}\otimes
u_{A}\otimes\id_{A}\right)  \circ\left(
r\otimes\epsilon_{C}\otimes\overline{r}\right)  ,
\]
so that%
\begin{align*}
&  \underbrace{m^{\left(  2\right)  }}_{\substack{=m_{A}\circ\left(
\id_{A}\otimes m_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2b}))}}}\circ\underbrace{\left(
r\otimes\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\overline{r}\right)
}_{=\left(  \id_{A}\otimes u_{A}\otimes
\id_{A}\right)  \circ\left(  r\otimes\epsilon
_{C}\otimes\overline{r}\right)  }\\
&  =m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)
\circ\underbrace{\left(  \id_{A}\otimes u_{A}%
\otimes\id_{A}\right)  }_{=
\id_{A}\otimes\left(  u_{A}\otimes\id_{A}%
\right)  }\circ\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)
=m_{A}\circ\underbrace{\left(  \id_{A}\otimes
m_{A}\right)  \circ\left(  \id_{A}\otimes\left(
u_{A}\otimes\id_{A}\right)  \right)  }%
_{\substack{=\left(  \id_{A}\circ
\id_{A}\right)  \otimes\left(  m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  \right)  }}\circ\left(  r\otimes
\epsilon_{C}\otimes\overline{r}\right)  \\
&  =m_{A}\circ\left(  \underbrace{\left(  \id_{A}%
\circ\id_{A}\right)  }_{=
\id_{A}}\otimes\underbrace{\left(  m_{A}\circ\left(  u_{A}%
\otimes\id_{A}\right)  \right)  }%
_{=\operatorname{kan}_{2,A}^{-1}}\right)  \circ\left(
r\otimes\epsilon_{C}\otimes\overline{r}\right)  =m_{A}\circ\left(
\id_{A}\otimes\operatorname{kan}_{2,A}%
^{-1}\right)  \circ\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\overline{r}\right)  }_{=m_{A}\circ\left(
\id_{A}\otimes\operatorname{kan}_{2,A}%
^{-1}\right)  \circ\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)
}\circ\left(  \id_{C}\otimes T_{C,C}\right)
\circ\left(  \Delta_{C}\otimes\id_{C}\right)  \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{2,A}^{-1}\right)  \circ\left(  r\otimes
\epsilon_{C}\otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\end{align*}
Since%
\begin{align*}
\underbrace{\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)
}_{=r\otimes\left(  \epsilon_{C}\otimes\overline{r}\right)  }\circ\left(
\id_{C}\otimes T_{C,C}\right)   &  =\left(
r\otimes\left(  \epsilon_{C}\otimes\overline{r}\right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  =\underbrace{\left(
r\circ\id_{C}\right)  }_{=r=
\id_{A}\circ r}\otimes\underbrace{\left(  \left(  \epsilon_{C}%
\otimes\overline{r}\right)  \circ T_{C,C}\right)  }%
_{\substack{=T_{A,\kk}\circ\left(  \overline{r}\otimes\epsilon
_{C}\right)  \\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.twist}), applied to
}U=C\text{, }V=C\text{, }U^{\prime}=A\text{, }V^{\prime}=\kk%
\text{,}\\x=\overline{r}\text{ and }y=\epsilon_{C}\text{)}}}\\
&  =\left(  \id_{A}\circ r\right)  \otimes\left(
T_{A,\kk}\circ\left(  \overline{r}\otimes\epsilon_{C}\right)  \right)
=\left(  \id_{A}\otimes T_{A,\kk}\right)
\circ\underbrace{\left(  r\otimes\left(  \overline{r}\otimes\epsilon
_{C}\right)  \right)  }_{=r\otimes\overline{r}\otimes\epsilon_{C}}=\left(
\id_{A}\otimes T_{A,\kk}\right)  \circ\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  ,
\end{align*}
this becomes%
\begin{align*}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{2,A}^{-1}\right)  \circ\underbrace{\left(
r\otimes\epsilon_{C}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  }_{=\left(
\id_{A}\otimes T_{A,\kk}\right)  \circ\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  }\circ\left(  \Delta
_{C}\otimes\id_{C}\right)  \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{2,A}^{-1}\right)  \circ\left(
\id_{A}\otimes T_{A,\kk}\right)  \circ\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\id_{C}\right)  .
\end{align*}
Since%
\[
\left(  \id_{A}\otimes\operatorname*{kan}%
\nolimits_{2,A}^{-1}\right)  \circ\left(  \id_{A}%
\otimes T_{A,\kk}\right)  =\underbrace{\left(  
\id_{A}\circ\id_{A}\right)  }%
_{=\id_{A}}\otimes\underbrace{\left(
\operatorname{kan}_{2,A}^{-1}\circ T_{A,\kk}\right)
}_{\substack{=\operatorname{kan}_{1,A}^{-1}\\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.kan12.1}), applied to }U=A\text{)}%
}}=\id_{A}\otimes\operatorname{kan}_{1,A}%
^{-1},
\]
this becomes%
\begin{align*}
m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)   &  =m_{A}\circ\underbrace{\left(
\id_{A}\otimes\operatorname{kan}_{2,A}%
^{-1}\right)  \circ\left(  \id_{A}\otimes
T_{A,\kk}\right)  }_{=\id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}}\circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ
\left(  \Delta_{C}\otimes \id_{C}\right)  \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  \circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.5}).}

\item We have%
\begin{align}
&  m_{A}\circ\left(  \id_{A}\otimes\operatorname*{kan}%
\nolimits_{1,A}^{-1}\right)  \circ\left(  r\otimes\overline{r}\otimes
\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \nonumber\\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\epsilon_{C}\right)  .\label{sol.coalg-anti.gen.a.sol2.fg.pf.6}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.6}):} The
equality (\ref{sol.coalg-anti.gen.a.sol2.kan1a}) (applied to $U=A\otimes A$,
$V=A$ and $\alpha=m_{A}$) yields $\operatorname{kan}_{1,A}^{-1}%
\circ\left(  m_{A}\otimes\id_{\kk}\right)
=m_{A}\circ\operatorname{kan}_{1,A\otimes A}^{-1}$. But
$\id_{A}\otimes\operatorname{kan}_{1,A}%
^{-1}=\operatorname{kan}_{1,A\otimes A}^{-1}$ (according to
(\ref{sol.coalg-anti.gen.a.sol2.kans.3}), applied to $U=A$ and $V=A$), and
thus $m_{A}\circ\underbrace{\left(  \id_{A}%
\otimes\operatorname{kan}_{1,A}^{-1}\right)  }_{=\operatorname*{kan}%
\nolimits_{1,A\otimes A}^{-1}}=m_{A}\circ\operatorname*{kan}%
\nolimits_{1,A\otimes A}^{-1}=\operatorname{kan}_{1,A}^{-1}%
\circ\left(  m_{A}\otimes\id_{\kk}\right)  $.
Hence,%
\begin{align*}
&  \underbrace{m_{A}\circ\left(  \id_{A}%
\otimes\operatorname{kan}_{1,A}^{-1}\right)  }_{=\operatorname*{kan}%
\nolimits_{1,A}^{-1}\circ\left(  m_{A}\otimes
\id_{\kk}\right)  }\circ\left(  r\otimes\overline{r}%
\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  m_{A}\otimes
\id_{\kk}\right)  \circ\underbrace{\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  }_{=\left(  r\otimes
\overline{r}\right)  \otimes\epsilon_{C}}\circ\left(  \Delta_{C}%
\otimes\id_{C}\right)  =\operatorname*{kan}%
\nolimits_{1,A}^{-1}\circ\underbrace{\left(  m_{A}\otimes
\id_{\kk}\right)  \circ\left(  \left(  r\otimes
\overline{r}\right)  \otimes\epsilon_{C}\right)
\circ\left(  \Delta_{C}\otimes
\id_{C}\right)  }_{\substack{=\left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\left(  \id_{\kk}\circ\epsilon_{C}%
\circ\id_{C}\right)  }}\\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\underbrace{\left(  \id_{\kk}%
\circ\epsilon_{C}\circ\id_{C}\right)  }_{=\epsilon
_{C}}\right)  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\epsilon_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.6}).}

\item We have%
\begin{equation}
\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes\epsilon
_{C}\right)  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)
.\label{sol.coalg-anti.gen.a.sol2.fg.pf.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.7}):} The
definition of convolution yields $r\star\overline{r}=m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}$. Compared with $r\star
\overline{r}=u_{A}\circ\epsilon_{C}$, this yields $m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}=u_{A}\circ\epsilon_{C}$. Hence,
$\operatorname{kan}_{1,A}^{-1}\circ\left(  \underbrace{\left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
}_{=u_{A}\circ\epsilon_{C}}\otimes\epsilon_{C}\right)  =\operatorname*{kan}%
\nolimits_{1,A}^{-1}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)
\otimes\epsilon_{C}\right)  $. This proves
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.7}).}

\item We have%
\begin{equation}
\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  u_{A}\circ
\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  =m_{A}\circ\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon_{C}\right)
\right)  \label{sol.coalg-anti.gen.a.sol2.fg.pf.8}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.8}):} We have%
\[
\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\underbrace{\epsilon_{C}%
}_{=\id_{\kk}\circ\epsilon_{C}}=\left(
u_{A}\circ\epsilon_{C}\right)  \otimes\left(  
\id_{\kk}\circ\epsilon_{C}\right)  =\left(  u_{A}\otimes
\id_{\kk}\right)  \circ\left(  \epsilon
_{C}\otimes\epsilon_{C}\right)  ,
\]
so that%
\[
\operatorname{kan}_{1,A}^{-1}\circ\underbrace{\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  }_{=\left(
u_{A}\otimes\id_{\kk}\right)  \circ\left(
\epsilon_{C}\otimes\epsilon_{C}\right)  }=\operatorname{kan}_{1,A}%
^{-1}\circ\left(  u_{A}\otimes\id_{\kk}\right)
\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)  .
\]
Compared with
\begin{align*}
m_{A}\circ\left(  \left(  \underbrace{u_{A}}_{=
\id_{A}\circ u_{A}}\circ\epsilon_{C}\right)  \otimes\left(
\underbrace{u_{A}}_{=u_{A}\circ\id_{\kk}}%
\circ\epsilon_{C}\right)  \right)   &  =m_{A}\circ\underbrace{\left(  \left(
\id_{A}\circ u_{A}\circ\epsilon_{C}\right)
\otimes\left(  u_{A}\circ\id_{\kk}\circ
\epsilon_{C}\right)  \right)  }_{=\left(  \id_{A}%
\otimes u_{A}\right)  \circ\left(  u_{A}\otimes
\id_{\kk}\right)  \circ\left(  \epsilon_{C}\otimes\epsilon
_{C}\right)  }\\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes
u_{A}\right)  }_{=\operatorname{kan}_{1,A}^{-1}}\circ\left(
u_{A}\otimes\id_{\kk}\right)  \circ\left(
\epsilon_{C}\otimes\epsilon_{C}\right)  =\operatorname{kan}_{1,A}%
^{-1}\circ\left(  u_{A}\otimes\id_{\kk}\right)
\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)  ,
\end{align*}
this yields $\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  =m_{A}\circ\left(
\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \right)  $. Thus, (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.8}) is
solved.}

\item We have%
\begin{equation}
m_{A}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(
u_{A}\circ\epsilon_{C}\right)  \right)  =u_{A}\circ\epsilon_{C}\circ m_{C}.
\label{sol.coalg-anti.gen.a.sol2.fg.pf.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.9}):} We know
that $u_{A}$ is a $\kk$-algebra homomorphism (indeed, this is an easy
fact that holds whenever $A$ is a $\kk$-algebra). Since the two maps
$u_{A}$ and $\epsilon_{C}$ are $\kk$-algebra homomorphisms, their
composition $u_{A}\circ\epsilon_{C}$ is a $\kk$-algebra homomorphism.
Consequently, $\left(  u_{A}\circ\epsilon_{C}\right)  \circ m_{C}=m_{A}%
\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}%
\circ\epsilon_{C}\right)  \right)  $ (in fact, this is one of the axioms a
$\kk$-algebra homomorphism has to satisfy), so that $m_{A}\circ\left(
\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \right)  =\left(  u_{A}\circ\epsilon_{C}\right)  \circ
m_{C}=u_{A}\circ\epsilon_{C}\circ m_{C}$. This proves
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.9}).}
\end{itemize}

Now,%
\begin{align*}
f\circledast g  &  =m_{A}\circ\left(  f\otimes g\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.1}%
)}\right) \\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.2}%
)}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3}%
)}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.4})}\right) \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  \circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.5})}\right) \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\epsilon_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.6})}\right) \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.7}%
)}\right) \\
&  =m_{A}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(
u_{A}\circ\epsilon_{C}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.8})}\right) \\
&  =u_{A}\circ\epsilon_{C}\circ m_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.9})}\right) \\
&  =u_{A}\circ\epsilon_{C\otimes C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7})}\right)  .
\end{align*}
The second equality in (\ref{sol.coalg-anti.gen.a.sol2.hffg}) is thus proven.

Thus, both equalities in (\ref{sol.coalg-anti.gen.a.sol2.hffg}) are proven.
Hence, (\ref{sol.coalg-anti.gen.a.sol2.hffg}) is proven. As we have already
seen, this yields (\ref{sol.coalg-anti.gen.a.sol2.h=g}). In other words,
$h=g$. But $h=\overline{r}\circ m_{C}$, so that
$\overline{r}\circ m_{C}=h=g
=m_{A}\circ T_{A,A}\circ\left(  \overline{r}\otimes
\overline{r}\right)  $.

Finally, (\ref{sol.coalg-anti.gen.a.sol2.twist}) (applied to $U=C$, $V=C$,
$U^{\prime}=A$, $V^{\prime}=A$, $x=\overline{r}$ and $y=\overline{r}$) yields
$\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}=T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  $. Thus, $T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  =\left(
\overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$, so that
$\overline{r}\circ
m_{C}=m_{A}\circ\underbrace{T_{A,A}\circ\left(  \overline{r}\otimes
\overline{r}\right)  }_{=\left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}}=m_{A}\circ\left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}$. As we know, this completes the solution of
Exercise~\ref{exe.coalg-anti.gen}(a).

[\textit{Remark:} The second solution of Exercise~\ref{exe.coalg-anti.gen}(a)
has been obtained more or less straightforwardly from the first solution by
rewriting it in an element-free fashion. First of all, the maps $f$, $g$ and
$h$ introduced in the second solution are precisely the maps $f$, $g$ and $h$
introduced in the first solution, just rewritten in an element-free way. Also,
for example, the equalities (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}) and
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}) have been found by rewriting the six
equality signs in the computation (\ref{sol.coalg-anti.gen.a.sol1.hf}) in an
element-free way: For instance, the second equality sign in
(\ref{sol.coalg-anti.gen.a.sol1.hf}) stands for the equality
\[
\sum_{(a),(b)}h\left(  a_{1}\otimes b_{1}\right)  f\left(  a_{2}\otimes
b_{2}\right)  =\sum_{(a),(b)}\overline{r}\left(  a_{1}b_{1}\right)  r\left(
a_{2}\right)  r\left(  b_{2}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a\in
C\text{ and }b\in C,
\]
i.e., for the following equality of maps:%
\begin{align*}
&  \left(  \text{the }\kk\text{-linear map }C\otimes C\rightarrow
A\text{ sending every }a\otimes b\text{ to }\sum_{(a),(b)}h\left(
a_{1}\otimes b_{1}\right)  f\left(  a_{2}\otimes b_{2}\right)  \right) \\
&  =\left(  \text{the }\kk\text{-linear map }C\otimes C\rightarrow
A\text{ sending every }a\otimes b\text{ to }\sum_{(a),(b)}\overline{r}\left(
a_{1}b_{1}\right)  r\left(  a_{2}\right)  r\left(  b_{2}\right)  \right)  .
\end{align*}
But upon rewriting these two maps without referring to elements, this takes
the form%
\begin{align*}
&  m_{A}\circ\left(  h\otimes f\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\end{align*}
(because the $\kk$-linear map $C\otimes C\rightarrow A$ sending every
$a\otimes b$ to $\sum_{(a),(b)}h\left(  a_{1}\otimes b_{1}\right)  f\left(
a_{2}\otimes b_{2}\right)  $ is $m_{A}\circ\left(  h\otimes f\right)
\circ\left(  \id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes
\Delta_{C}\right)  $, and the $\kk$-linear map $C\otimes C\rightarrow
A$ sending every $a\otimes b$ to $\sum_{(a),(b)}\overline{r}\left(  a_{1}%
b_{1}\right)  r\left(  a_{2}\right)  r\left(  b_{2}\right)  $ is $m_{A}%
\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(  m_{A}%
\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  $); and
this is precisely the equality (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}). The
process of rewriting a map in an element-free way is not completely
deterministic (e.g., we could also have rewritten the $\kk$-linear map
$C\otimes C\rightarrow A$ sending every $a\otimes b$ to $\sum_{(a),(b)}%
\overline{r}\left(  a_{1}b_{1}\right)  r\left(  a_{2}\right)  r\left(
b_{2}\right)  $ as $m^{\left(  3\right)  }\circ\left(  \left(
\overline{r}\circ m_{C}\right)  \otimes r\otimes r\right)  \circ
\left(  \id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  $,
since $m^{\left(  3\right)  }$ is the $\kk$-linear map $A\otimes
A\otimes A \to A$ sending every $a\otimes b\otimes c$ to $abc$),
but all possible
results (for a given map) can be reduced to each other using purely
linear-algebraic formulae\footnote{By ``purely linear-algebraic
formulae'', I mean formulae such as
(\ref{sol.coalg-anti.gen.a.sol2.twist}) and
(\ref{sol.coalg-anti.gen.a.sol2.kan12.1}) and the
identity $\left(\beta \circ \alpha\right) \otimes \left(\beta^\prime
\circ \alpha^\prime\right) = \left(\beta \otimes \beta^\prime\right)
\circ \left(\alpha \otimes \alpha^\prime\right)$ for tensor products
of compositions of maps. This kind of formulae hold in any
tensor category (at least if we follow the abuse of notation that
allows us to treat the tensor product as associative).}
and various forms of the general
associativity law (such as (\ref{sol.coalg-anti.gen.a.sol2.m3}) and
(\ref{sol.coalg-anti.gen.a.sol2.m3o}))\footnote{The reason why we
need to use the general associativity law is that, in the first
solution, we used unparenthesized products of more than one element
of $A$ (for example, the expression
$\sum_{(a),(b)}\overline{r}\left(  a_{1} b_{1}\right)
r\left(  a_{2}\right)  r\left(  b_{2}\right)$
contains the unparenthesized product
$\overline{r}\left(  a_{1} b_{1}\right)
r\left(  a_{2}\right)  r\left(  b_{2}\right)$ of three factors).
If we would parenthesize all such products in such a way that no more
than two factors are ever multiplied at the same time (e.g., we could
replace $\overline{r}\left(  a_{1} b_{1}\right)
r\left(  a_{2}\right)  r\left(  b_{2}\right)$ by
$\overline{r}\left(  a_{1} b_{1}\right)
\left(r\left(  a_{2}\right)  r\left(  b_{2}\right) \right)$), and
if we would explicitly use the (non-general) associativity law
$\left(xy\right) z = x\left(yz\right)$ to switch between these
parenthesizations, then we would not have to use general
associativity any more when rewriting the proof in an element-free
fashion (but, of course, the proof would be longer).}. The equalities
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}) and
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}) are somewhat more complicated to
check than the corresponding equality signs in
(\ref{sol.coalg-anti.gen.a.sol1.hf}), because of the fact that one and the
same map can be written in different forms whose equivalence needs to be
proven. However, this additional complexity is straightforward to surmount:
each equality, when rewritten in element-free terms, can be proven by the same
arguments as the corresponding equality that uses elements, combined with
purely linear-algebraic formulae like (\ref{sol.coalg-anti.gen.a.sol2.twist})
and (\ref{sol.coalg-anti.gen.a.sol2.kan12.1}) and various forms of the general
associativity law (such as (\ref{sol.coalg-anti.gen.a.sol2.m3}) and
(\ref{sol.coalg-anti.gen.a.sol2.m3o})). (There is also a way how to do
element-free computations without such added difficulty, using \emph{string
diagrams}\footnote{See
\url{http://ncatlab.org/nlab/show/string+diagram}
and the references therein.}.)

See the First solution of Exercise~\ref{exe.counit.unique} as well
as the solution of Exercise~\ref{exe.cocommutative.coalghom} for
other examples of how a proof that uses
elements can be rewritten in an element-free fashion.]

\end{vershort}

\begin{verlong}

\textit{Second solution to Exercise~\ref{exe.coalg-anti.gen}(a).} In this
second solution of Exercise~\ref{exe.coalg-anti.gen}(a), we shall often use
the following classical formula from linear algebra:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$ and $W^{\prime}$ are six
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$ and $\beta^{\prime
}:V^{\prime}\rightarrow W^{\prime}$ are four $\kk$-linear maps, then
$\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

\noindent
We will refer to this formula as the \dfn{composition-tensor relation}.
Using two applications of this formula, we can obtain the following formula,
which we will call the \dfn{composition-triple-tensor relation}:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$U^{\prime\prime}$, $V^{\prime\prime}$, $W^{\prime\prime}$ are nine
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$, $\beta^{\prime}:V^{\prime
}\rightarrow W^{\prime}$, $\alpha^{\prime\prime}:U^{\prime\prime}\rightarrow
V^{\prime\prime}$, $\beta^{\prime\prime}:V^{\prime\prime}\rightarrow
W^{\prime\prime}$ are six $\kk$-linear maps, then $\left(  \beta
\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ\alpha^{\prime}\right)
\otimes\left(  \beta^{\prime\prime}\circ\alpha^{\prime\prime}\right)  =\left(
\beta\otimes\beta^{\prime}\otimes\beta^{\prime\prime}\right)  \circ\left(
\alpha\otimes\alpha^{\prime}\otimes\alpha^{\prime\prime}\right)  $.
\end{itemize}

On the other hand, by applying the composition-tensor relation twice, we can
also prove the following formula, which we will call the
\dfn{triple-composition-tensor relation}:

\begin{itemize}
\item If $U$, $V$, $W$, $T$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$ and
$T^{\prime}$ are eight $\kk$-modules and $\alpha:U\rightarrow V$,
$\beta:V\rightarrow W$, $\gamma:W\rightarrow T$, $\alpha^{\prime}:U^{\prime
}\rightarrow V^{\prime}$, $\beta^{\prime}:V^{\prime}\rightarrow W^{\prime}$
and $\gamma^{\prime}:W^{\prime}\rightarrow T^{\prime}$ are six $\kk$-linear
maps, then $\left(  \gamma\circ\beta\circ\alpha\right)
\otimes\left(  \gamma^{\prime}\circ\beta^{\prime}\circ\alpha^{\prime}\right)
=\left(  \gamma\otimes\gamma^{\prime}\right)  \circ\left(  \beta\otimes
\beta^{\prime}\right)  \circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

Another elementary linear-algebraic fact says that if $U$, $V$, $U^{\prime}$
and $V^{\prime}$ are four $\kk$-modules and $x:U\rightarrow U^{\prime}$
and $y:V\rightarrow V^{\prime}$ are two $\kk$-linear maps, then
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}=T_{U^{\prime},V^{\prime}}\circ\left(
x\otimes y\right)  . \label{sol.coalg-anti.gen.a.sol2.twist}%
\end{equation}


For every $\kk$-module $U$, let $\operatorname{kan}_{1,U}%
:U\rightarrow U\otimes\kk$ and $\operatorname{kan}_{2,U}%
:U\rightarrow\kk\otimes U$ be the canonical $\kk$-module
isomorphisms. These two isomorphisms are related to each other via the
equalities%
\begin{align}
\operatorname{kan}_{2,U}^{-1}\circ T_{U,\kk}  &
=\operatorname{kan}_{1,U}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kan12.1}\\
\operatorname{kan}_{1,U}^{-1}\circ T_{\kk,U}  &
=\operatorname{kan}_{2,U}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kan12.2}\\
T_{\kk,U}\circ\operatorname{kan}_{2,U}  &
=\operatorname{kan}_{1,U},\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{sol.coalg-anti.gen.a.sol2.kan12.3}\\
T_{U,\kk}\circ\operatorname{kan}_{1,U}  &
=\operatorname{kan}_{2,U} \label{sol.coalg-anti.gen.a.sol2.kan12.4}%
\end{align}
for every $\kk$-module $U$. Moreover, every $\kk$-modules $U$
and $V$ satisfy%
\begin{align}
\id_{U}\otimes\operatorname{kan}_{1,V}  &
=\operatorname{kan}_{1,U\otimes V}%
,\label{sol.coalg-anti.gen.a.sol2.kans.1}\\
\operatorname{kan}_{2,V}\otimes\id_{U}  &
=\operatorname{kan}_{2,V\otimes U}%
,\label{sol.coalg-anti.gen.a.sol2.kans.2}\\
\id_{U}\otimes\operatorname{kan}_{1,V}^{-1}
&  =\operatorname{kan}_{1,U\otimes V}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kans.3}\\
\operatorname{kan}_{2,V}^{-1}\otimes\id_{U}
&  =\operatorname{kan}_{2,V\otimes U}^{-1}%
,\label{sol.coalg-anti.gen.a.sol2.kans.4}\\
\id_{U}\otimes\operatorname{kan}_{2,V}  &
=\operatorname{kan}_{1,U}\otimes\id_{V}%
,\label{sol.coalg-anti.gen.a.sol2.kans.5}\\
\id_{U}\otimes\operatorname{kan}_{2,V}^{-1}
&  =\operatorname{kan}_{1,U}^{-1}\otimes\id_{V}.
\label{sol.coalg-anti.gen.a.sol2.kans.6}
\end{align}


A very simple fact from linear algebra says that if $U$ and $V$ are two
$\kk$-modules and $\alpha:U\rightarrow V$ is a $\kk$-linear map,
then
\begin{align}
\operatorname{kan}_{1,V}\circ\alpha &  =\left(  \alpha
\otimes\id_{\kk}\right)  \circ
\operatorname{kan}_{1,U},\label{sol.coalg-anti.gen.a.sol2.kan1}\\
\operatorname{kan}_{2,V}\circ\alpha &  =\left(  \id_{\kk}
\otimes\alpha\right)  \circ\operatorname{kan}_{2,U},
\label{sol.coalg-anti.gen.a.sol2.kan2}\\
\operatorname{kan}_{1,V}^{-1}\circ\left(  \alpha\otimes
\id_{\kk}\right)   &  =\alpha\circ
\operatorname{kan}_{1,U}^{-1},\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{sol.coalg-anti.gen.a.sol2.kan1a}\\
\operatorname{kan}_{2,V}^{-1}\circ\left(  \id_{\kk}
\otimes\alpha\right)   &  =\alpha\circ
\operatorname{kan}_{2,U}^{-1}.
\label{sol.coalg-anti.gen.a.sol2.kan2a}
\end{align}


Now, let us step to the actual solution of Exercise~\ref{exe.coalg-anti.gen}(a).

Let $\overline{r}=r^{\star\left(  -1\right)  }$. Then, $\overline{r}%
=r^{\star\left(  -1\right)  }$ is the $\star$-inverse of $f$. Thus,
$\overline{r}\star r=1_{\left(  \Hom \left(  C,A\right)
,\star\right)  }=u_{A}\circ\epsilon_{C}$ and similarly
$r\star\overline{r}=u_{A}\circ\epsilon_{C}$.

Recall that $C$ is a $\kk$-coalgebra. By the axioms of a $\kk%
$-coalgebra, this shows that
\begin{align*}
\left(  \Delta_{C}\otimes\id_{C}\right)  \circ
\Delta_{C}  &  =\left(  \id_{C}\otimes\Delta
_{C}\right)  \circ\Delta_{C},\\
\left(  \id_{C}\otimes\epsilon_{C}\right)  \circ
\Delta_{C}  &  =\operatorname{kan}_{1,C};\\
\left(  \epsilon_{C}\otimes\id_{C}\right)  \circ
\Delta_{C}  &  =\operatorname{kan}_{2,C}.
\end{align*}
Also, recall that $C$ is a $\kk$-algebra. By the axioms of a
$\kk$-algebra, this shows that%
\begin{align*}
m_{C}\circ\left(  m_{C}\otimes\id_{C}\right)   &
=m_{C}\circ\left(  \id_{C}\otimes m_{C}\right)  ;\\
m_{C}\circ\left(  \id_{C}\otimes u_{C}\right)   &
=\operatorname{kan}_{1,C}^{-1};\\
m_{C}\circ\left(  u_{C}\otimes\id_{C}\right)   &
=\operatorname{kan}_{2,C}^{-1}.
\end{align*}
Also, recall that $C$ is a $\kk$-bialgebra. Due to the axioms of a
$\kk$-bialgebra, this shows that $\Delta_{C}$ and $\epsilon_{C}$ are
$\kk$-algebra homomorphisms, and that $m_{C}$ and $u_{C}$ are
$\kk$-coalgebra homomorphisms.

Furthermore, $A$ is a $\kk$-algebra. By the axioms of a
$\kk$-algebra, this shows that
\begin{align*}
m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)   &
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  ;\\
m_{A}\circ\left(  \id_{A}\otimes u_{A}\right)   &
=\operatorname{kan}_{1,A}^{-1};\\
m_{A}\circ\left(  u_{A}\otimes\id_{A}\right)   &
=\operatorname{kan}_{2,A}^{-1}.
\end{align*}


We have $\epsilon_{C}\circ u_{C}=u_{\kk}$ (since $\epsilon_{C}$ is a
$\kk$-algebra homomorphism). Since $u_{\kk}
=\id_{\kk}$, this rewrites as $\epsilon_{C}\circ u_{C}%
=\id_{\kk}$.

Recall that $\operatorname{kan}_{1, \kk}$ is the canonical
$\kk$-module isomorphism $\kk\rightarrow\kk\otimes\kk$.
Hence, $u_{C\otimes C}=\left(  u_{C}\otimes u_{C}\right)
\circ\operatorname{kan}_{1, \kk}$ (by the definition of the
$\kk$-algebra $C\otimes C$). Also, $\operatorname{kan}_{1, \kk}
=\operatorname{kan}_{2, \kk}$ (since
each of $\operatorname{kan}_{1, \kk}$ and
$\operatorname{kan}_{2, \kk}$ is the canonical $\kk$-module
isomorphism $\kk\rightarrow\kk\otimes\kk$).

We know that $r:C\rightarrow A$ is a $\kk$-algebra homomorphism. In
other words, $r$ is a $\kk$-linear map satisfying $r\circ m_{C}%
=m_{A}\circ\left(  r\otimes r\right)  $ and $r\circ u_{C}=u_{A}$.

Now, we need to prove that $r^{\star\left(  -1\right)  }$ is a $\kk%
$-algebra anti-homomorphism. In other words, we need to prove that
$\overline{r}$ is a $\kk$-algebra anti-homomorphism (since
$\overline{r}=r^{\star\left(  -1\right)  }$). In other words, we need to prove
that $\overline{r}$ satisfies the two equations $\overline{r}\circ m_{C}%
=m_{A}\circ\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$ and
$\overline{r}\circ u_{C}=u_{A}$ (because $\overline{r}$ is a $\kk%
$-algebra anti-homomorphism if and only if $\overline{r}$ satisfies the two
equations $\overline{r}\circ m_{C}=m_{A}\circ\left(  \overline{r}%
\otimes\overline{r}\right)  \circ T_{C,C}$ and $\overline{r}\circ u_{C}=u_{A}$
(according to the definition of a $\kk$-algebra anti-homomorphism)).

Let us first prove the equality $\overline{r}\circ u_{C}=u_{A}$.

We have $r\star\overline{r}=u_{A}\circ\epsilon_{C}$, so that
$\underbrace{\left(  r\star\overline{r}\right)  }_{=u_{A}\circ\epsilon_{C}%
}\circ u_{C}=u_{A}\circ\underbrace{\epsilon_{C}\circ u_{C}}%
_{=\id_{\kk}}=u_{A}$. Hence,%
\begin{align*}
u_{A}  &  =\underbrace{\left(  r\star\overline{r}\right)  }_{\substack{=m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\\\text{(by the
definition}\\\text{of convolution)}}}\circ u_{C}=m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\underbrace{\Delta_{C}\circ u_{C}%
}_{\substack{=u_{C\otimes C}\\\text{(since }\Delta_{C}\text{ is a }%
\kk\text{-algebra}\\\text{homomorphism)}}}\\
&  =m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ
\underbrace{u_{C\otimes C}}_{\substack{=\left(  u_{C}\otimes u_{C}\right)
\circ\operatorname{kan}_{1, \kk}}}=m_{A}\circ
\underbrace{\left(  r\otimes\overline{r}\right)  \circ\left(  u_{C}\otimes
u_{C}\right)  }_{\substack{=\left(  r\circ u_{C}\right)  \otimes\left(
\overline{r}\circ u_{C}\right)  \\\text{(since the composition-tensor}%
\\\text{relation yields}\\\left(  r\circ u_{C}\right)  \otimes\left(
\overline{r}\circ u_{C}\right)  =\left(  r\otimes\overline{r}\right)
\circ\left(  u_{C}\otimes u_{C}\right)  \text{)}}}\circ
\operatorname{kan}_{1, \kk}\\
&  =m_{A}\circ\left(  \underbrace{\left(  r\circ u_{C}\right)  }_{=u_{A}%
=u_{A}\circ\id_{\kk}}\otimes\underbrace{\left(
\overline{r}\circ u_{C}\right)  }_{=\id_{A}%
\circ\left(  \overline{r}\circ u_{C}\right)  }\right)  \circ
\underbrace{\operatorname{kan}_{1, \kk}}_{=
\operatorname{kan}_{2, \kk}}
=m_{A}\circ\underbrace{\left(  \left(  u_{A}%
\circ\id_{\kk}\right)  \otimes\left(
\id_{A}\circ\left(  \overline{r}\circ u_{C}\right)
\right)  \right)  }_{\substack{=\left(  u_{A}\otimes
\id_{A}\right)  \circ\left(  \id_{\kk%
}\otimes\left(  \overline{r}\circ u_{C}\right)  \right)  \\\text{(by the
composition-tensor}\\\text{relation)}}}\circ\operatorname*{kan}%
\nolimits_{2, \kk}\\
&  =\underbrace{m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  }_{=\operatorname{kan}_{2,A}^{-1}}%
\circ\underbrace{\left(  \id_{\kk}%
\otimes\left(  \overline{r}\circ u_{C}\right)  \right)  \circ
\operatorname{kan}_{2, \kk}}_{\substack{=\operatorname*{kan}%
\nolimits_{2,A}\circ\left(  \overline{r}\circ u_{C}\right)  \\\text{(since
(\ref{sol.coalg-anti.gen.a.sol2.kan2}) (applied to }U=\kk%
\text{,}\\V=A\text{ and }\alpha=\overline{r}\circ u_{C}\text{) yields}%
\\\operatorname{kan}_{2,A}\circ\left(  \overline{r}\circ
u_{C}\right)  =\left(  \id_{\kk}\otimes\left(
\overline{r}\circ u_{C}\right)  \right)  \circ\operatorname*{kan}%
\nolimits_{2, \kk}\text{)}}}=\underbrace{\operatorname*{kan}%
\nolimits_{2,A}^{-1}\circ\operatorname{kan}_{2,A}}%
_{=\id_{A}}\circ\left(  \overline{r}\circ u_{C}\right)
\\
&  =\overline{r}\circ u_{C}.
\end{align*}
Hence, $\overline{r}\circ u_{C}=u_{A}$.

Now, it remains to prove the equality $\overline{r}\circ m_{C}=m_{A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$. This is
harder. Let us first make some preparations.

Recall that $C\otimes C$ is a $\kk$-coalgebra (since $C$ is a
$\kk$-coalgebra), and $A$ is an $\kk$-algebra. Thus,
$\Hom \left(  C\otimes C,A\right)  $ is an associative algebra
with respect to convolution. We shall denote the convolution on
$\Hom \left(  C\otimes C,A\right)  $ by $\circledast$ rather
than by $\star$ (in order to distinguish it from the convolution $\star$ on
$\Hom \left(  C,A\right)  $). The algebra $\left(
\Hom \left(  C\otimes C,A\right)  ,\star\right)  $ has two-sided
identity element $u_{A}\circ\epsilon_{C\otimes C}$.

We define a $\kk$-linear map $f:C\otimes C\rightarrow A$ by
$f=m_{A}\circ\left(  r\otimes r\right)  $.

We define a $\kk$-linear map $g:C\otimes C\rightarrow A$ by
$g=m_{A}\circ T_{A,A}\circ\left(  \overline{r}\otimes\overline{r}\right)  $.

We define a $\kk$-linear map $h:C\otimes C\rightarrow A$ by
$h=\overline{r}\circ m_{C}$.

Clearly, $f$, $g$ and $h$ are three elements of $\Hom \left(
C\otimes C,A\right)  $. Our next goal is to prove that%
\begin{equation}
h\circledast f=u_{A}\circ\epsilon_{C\otimes C}=f\circledast g.
\label{sol.coalg-anti.gen.a.sol2.hffg}%
\end{equation}
Once this equality is proven, we will then obtain%
\begin{align}
h  &  =h\circledast\underbrace{\left(  u_{A}\circ\epsilon_{C\otimes C}\right)
}_{\substack{=f\circledast g\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.hffg}%
))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\circ\epsilon_{C\otimes
C}\text{ is the identity element of }\left(  \Hom \left(
C\otimes C,A\right)  ,\circledast\right)  \right) \nonumber\\
&  =h\circledast\left(  f\circledast g\right)  =\underbrace{\left(
h\circledast f\right)  }_{\substack{=u_{A}\circ\epsilon_{C\otimes
C}\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.hffg}))}}}\circledast
g\ \ \ \ \ \ \ \ \ \ \left(  \text{by the associativity of the convolution
}\circledast\right) \nonumber\\
&  =\left(  u_{A}\circ\epsilon_{C\otimes C}\right)  \circledast
g=g\label{sol.coalg-anti.gen.a.sol2.h=g}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }u_{A}\circ\epsilon_{C\otimes
C}\text{ is the identity element of }\left(  \Hom \left(
C\otimes C,A\right)  ,\circledast\right)  \right)  .\nonumber
\end{align}


So let us concentrate on proving (\ref{sol.coalg-anti.gen.a.sol2.hffg}). Let
us first notice that%
\begin{equation}
u_{A}\circ\epsilon_{C}\circ m_{C}=u_{A}\circ\epsilon_{C\otimes C}.
\label{sol.coalg-anti.gen.a.sol2.hf.pf.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7}):} We know
that $m_{C}$ is a $\kk$-coalgebra homomorphism. Hence, $\epsilon
_{C}\circ m_{C}=\epsilon_{C\otimes C}$. Thus, $u_{A}\circ\underbrace{\epsilon
_{C}\circ m_{C}}_{=\epsilon_{C\otimes C}}=u_{A}\circ\epsilon_{C\otimes C}$.
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7}).}

Next, we make the following observations:

\begin{itemize}
\item First, we notice that%
\begin{equation}
h\circledast f=m_{A}\circ\left(  h\otimes f\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.hf.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}):} By the
definition of the convolution $h\circledast f$, we have
\[
h\circledast f=m_{A}\circ\left(  h\otimes f\right)  \circ\underbrace{\Delta
_{C\otimes C}}_{\substack{=\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \\\text{(by the definition of the }%
\kk\text{-coalgebra }C\otimes C\text{)}}}=m_{A}\circ\left(  h\otimes
f\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\otimes\id_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  .
\]
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}).}

\item Next, we notice that%
\begin{align}
&  m_{A}\circ\left(  h\otimes f\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.hf.pf.2}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}):} We
have
\begin{align*}
&  m_{A}\circ\left(  \underbrace{h}_{=\overline{r}\circ m_{C}}\otimes
\underbrace{f}_{=m_{A}\circ\left(  r\otimes r\right)  }\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}).}

\item Next, we have%
\begin{align}
&  m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\nonumber\\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
r\circ m_{C}\right)  \right)  \circ\left(  \id_{C}%
\otimes T_{C,C}\otimes\id_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  \label{sol.coalg-anti.gen.a.sol2.hf.pf.3}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}):} We have%
\begin{align*}
&  m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes
\underbrace{\left(  r\circ m_{C}\right)  }_{=m_{A}\circ\left(  r\otimes
r\right)  }\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}).}

\item Furthermore, we have%
\begin{align}
&  m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
r\circ m_{C}\right)  \right)  \circ\left(  \id_{C}%
\otimes T_{C,C}\otimes\id_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ m_{C}
\label{sol.coalg-anti.gen.a.sol2.hf.pf.4}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}):} We have%
\begin{align*}
&  m_{A}\circ\underbrace{\left(  \left(  \overline{r}\circ m_{C}\right)
\otimes\left(  r\circ m_{C}\right)  \right)  }_{\substack{=\left(
\overline{r}\otimes r\right)  \circ\left(  m_{C}\otimes m_{C}\right)
\\\text{(by the composition-tensor relation)}}}\circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\underbrace{\left(
m_{C}\otimes m_{C}\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  }%
_{\substack{=m_{C\otimes C}\\\text{(since the definition of the }%
\kk\text{-algebra }C\otimes C\\\text{yields }m_{C\otimes C}=\left(
m_{C}\otimes m_{C}\right)  \circ\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \text{)}}}\circ\left(
\Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ
\underbrace{m_{C\otimes C}\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
}_{\substack{=\Delta_{C}\circ m_{C}\\\text{(since }\Delta_{C}:C\rightarrow
C\otimes C\text{ is a }\kk\text{-algebra homomorphism)}}}=m_{A}%
\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ m_{C}.
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}).}

\item Moreover, we have%
\begin{equation}
m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ
m_{C}=\left(  \overline{r}\star r\right)  \circ m_{C}.
\label{sol.coalg-anti.gen.a.sol2.hf.pf.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}):} The
definition of the convolution $\overline{r}\star r$ yields $\overline{r}\star
r=m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}$. Hence,
$\underbrace{\left(  \overline{r}\star r\right)  }_{=m_{A}\circ\left(
\overline{r}\otimes r\right)  \circ\Delta_{C}}\circ m_{C}=m_{A}\circ\left(
\overline{r}\otimes r\right)  \circ\Delta_{C}\circ m_{C}$. This proves
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}).}

\item Finally, we have%
\begin{equation}
\left(  \overline{r}\star r\right)  \circ m_{C}=u_{A}\circ\epsilon_{C}\circ
m_{C}. \label{sol.coalg-anti.gen.a.sol2.hf.pf.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}):} We have
$\underbrace{\left(  \overline{r}\star r\right)  }_{=u_{A}\circ\epsilon_{C}%
}\circ m_{C}=u_{A}\circ\epsilon_{C}\circ m_{C}$. Thus,
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}) is proven.}
\end{itemize}

Now, we have%
\begin{align*}
h\circledast f  &  =m_{A}\circ\left(  h\otimes f\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}%
)}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}%
)}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
r\circ m_{C}\right)  \right)  \circ\left(  \id_{C}%
\otimes T_{C,C}\otimes\id_{C}\right)  \circ\left(
\Delta_{C}\otimes\Delta_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3})}\right) \\
&  =m_{A}\circ\left(  \overline{r}\otimes r\right)  \circ\Delta_{C}\circ
m_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4})}\right) \\
&  =\left(  \overline{r}\star r\right)  \circ m_{C}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5})}\right) \\
&  =u_{A}\circ\epsilon_{C}\circ m_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6})}\right) \\
&  =u_{A}\circ\epsilon_{C\otimes C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7})}\right)  .
\end{align*}
The first equality in (\ref{sol.coalg-anti.gen.a.sol2.hffg}) is thus proven.

We shall next show the second equality in
(\ref{sol.coalg-anti.gen.a.sol2.hffg}).

For every $k\in \NN$, let us define the map $m^{\left(  k\right)
}:A^{\otimes\left(  k+1\right)  }\rightarrow A$ as in
Exercise~\ref{exe.iterated.m}. We recall that these maps $m^{\left(  k\right)
}$ are defined by induction over $k$, with the induction base $m^{\left(
0\right)  }=\operatorname{id}_{A}$, and with the induction step%
\begin{equation}
m^{\left(  k\right)  }=m_{A}\circ\left(  \operatorname{id}_{A}\otimes
m^{\left(  k-1\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }k\geq1.
\label{sol.coalg-anti.gen.a.sol2.mk}%
\end{equation}
\footnote{This is precisely the definition of these maps given in
Exercise~\ref{exe.iterated.m}, with the only difference that $m_{A}$ was
denoted by $m$ in that exercise.} Out of these maps, we shall only need
$m^{\left(  0\right)  }$, $m^{\left(  1\right)  }$, $m^{\left(  2\right)  }$
and $m^{\left(  3\right)  }$. These satisfy the following formulae:%
\begin{align}
m^{\left(  0\right)  }  &  =\id_{A}%
;\label{sol.coalg-anti.gen.a.sol2.m0}\\
m^{\left(  1\right)  }  &  =m_{A};\label{sol.coalg-anti.gen.a.sol2.m1}\\
m^{\left(  2\right)  }  &  =m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  ;\label{sol.coalg-anti.gen.a.sol2.m2a}\\
m^{\left(  2\right)  }  &  =m_{A}\circ\left(  
\id_{A}\otimes m_{A}\right)  ;\label{sol.coalg-anti.gen.a.sol2.m2b}\\
m^{\left(  3\right)  }  &  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right)
;\label{sol.coalg-anti.gen.a.sol2.m3}\\
m^{\left(  3\right)  }  &  =m^{\left(  2\right)  }\circ\left(
\id_{A}\otimes m_{A}\otimes
\id_{A}\right)  . \label{sol.coalg-anti.gen.a.sol2.m3o}%
\end{align}
\footnote{Here are proofs for these formulae:
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m0}):} The formula
(\ref{sol.coalg-anti.gen.a.sol2.m0}) follows immediately from the definition
of $m^{\left(  0\right)  }$.
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m1}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=1$, we obtain $m^{\left(  1\right)
}=m_{A}\circ\left(  \id_{A}\otimes
\underbrace{m^{\left(  1-1\right)  }}_{=m^{\left(  0\right)  }%
=\id_{A}}\right)  =m_{A}\circ\underbrace{\left(
\id_{A}\otimes\id_{A}\right)
}_{=\id_{A\otimes A}}=m_{A}$. This proves
(\ref{sol.coalg-anti.gen.a.sol2.m1}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m2b}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=2$, we obtain $m^{\left(  2\right)
}=m_{A}\circ\left(  \id_{A}\otimes
\underbrace{m^{\left(  2-1\right)  }}_{\substack{=m^{\left(  1\right)  }%
=m_{A}\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.m1}))}}}\right)  =m_{A}%
\circ\left(  \id_{A}\otimes m_{A}\right)  $. This
proves (\ref{sol.coalg-anti.gen.a.sol2.m2b}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m2a}):} From
(\ref{sol.coalg-anti.gen.a.sol2.m2b}), we obtain $m^{\left(  2\right)  }%
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)
=m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  $
(since $m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)
=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  $).
This proves (\ref{sol.coalg-anti.gen.a.sol2.m2a}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m3}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=3$, we obtain%
\begin{align*}
m^{\left(  3\right)  }  &  =m_{A}\circ\left(  \underbrace{
\id_{A}}_{=\id_{A}\circ
\id_{A}}\otimes\underbrace{m^{\left(  3-1\right)  }}%
_{\substack{=m^{\left(  2\right)  }=m_{A}\circ\left(  
\id_{A}\otimes m_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2b}))}}}\right)  =m_{A}\circ
\underbrace{\left(  \left(  \id_{A}\circ
\id_{A}\right)  \otimes\left(  m_{A}\circ\left(
\id_{A}\otimes m_{A}\right)  \right)  \right)
}_{\substack{=\left(  \id_{A}\otimes m_{A}\right)
\circ\left(  \id_{A}\otimes\left(  
\id_{A}\otimes m_{A}\right)  \right)  \\\text{(by the composition-tensor
relation)}}}\\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  }_{\substack{=m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \\\text{(since }m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  =m_{A}\circ\left(  
\id_{A}\otimes m_{A}\right)  \text{)}}}\circ\underbrace{\left(
\id_{A}\otimes\left(  
\id_{A}\otimes m_{A}\right)  \right)  }_{=
\id_{A}\otimes\id_{A}\otimes m_{A}}=m_{A}%
\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(
\underbrace{\id_{A}\otimes
\id_{A}}_{=\id_{A\otimes A}}\otimes m_{A}\right)
\\
&  =m_{A}\circ\underbrace{\left(  m_{A}\otimes
\id_{A}\right)  \circ\left(  \id_{A\otimes
A}\otimes m_{A}\right)  }_{\substack{=\left(  m_{A}\circ
\id_{A\otimes A}\right)  \otimes\left(  
\id_{A}\circ m_{A}\right)  \\\text{(since the composition-tensor
relation yields}\\\left(  m_{A}\circ\id_{A\otimes
A}\right)  \otimes\left(  \id_{A}\circ m_{A}\right)
=\left(  m_{A}\otimes\id_{A}\right)  \circ\left(
\id_{A\otimes A}\otimes m_{A}\right)  \text{)}}%
}=m_{A}\circ\left(  \underbrace{\left(  m_{A}\circ
\id_{A\otimes A}\right)  }_{=m_{A}}\otimes\underbrace{\left(
\id_{A}\circ m_{A}\right)  }_{=m_{A}}\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.m3}).
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.m3o}):} Applying
(\ref{sol.coalg-anti.gen.a.sol2.mk}) to $k=3$, we obtain%
\begin{align*}
m^{\left(  3\right)  }  &  =m_{A}\circ\left(  \underbrace{
\id_{A}}_{=\id_{A}\circ
\id_{A}}\otimes\underbrace{m^{\left(  3-1\right)  }}%
_{\substack{=m^{\left(  2\right)  }=m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2a}))}}}\right)  =m_{A}\circ
\underbrace{\left(  \left(  \id_{A}\circ
\id_{A}\right)  \otimes\left(  m_{A}\circ\left(
m_{A}\otimes\id_{A}\right)  \right)  \right)
}_{\substack{=\left(  \id_{A}\otimes m_{A}\right)
\circ\left(  \id_{A}\otimes\left(  m_{A}%
\otimes\id_{A}\right)  \right)  \\\text{(by the
composition-tensor relation)}}}\\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  }_{\substack{=m^{\left(  2\right)  }\\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2b}))}}}\circ\underbrace{\left(
\id_{A}\otimes\left(  m_{A}\otimes
\id_{A}\right)  \right)  }_{=\id_{A}\otimes
m_{A}\otimes\id_{A}}=m^{\left(  2\right)  }%
\circ\left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.m3o}).}

Now, we make the following observations:

\begin{itemize}
\item We have%
\begin{equation}
f\circledast g=m_{A}\circ\left(  f\otimes g\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.1}):} By the
definition of the convolution $f\circledast g$, we have%
\[
f\circledast g=m_{A}\circ\left(  f\otimes g\right)  \circ\underbrace{\Delta
_{C\otimes C}}_{\substack{=\left(  \id_{C}\otimes
T_{C,C}\otimes\id_{C}\right)  \circ\left(  \Delta
_{C}\otimes\Delta_{C}\right)  \\\text{(by the definition of the }%
\kk\text{-coalgebra }C\otimes C\text{)}}}=m_{A}\circ\left(  f\otimes
g\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\otimes\id_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\Delta_{C}\right)  .
\]
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.1}).}

\item We have
\begin{align}
&  m_{A}\circ\left(  f\otimes g\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \nonumber\\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.2}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.2}):} The
equality (\ref{sol.coalg-anti.gen.a.sol2.twist}) (applied to $U=C$, $V=C$,
$U^{\prime}=A$, $V^{\prime}=A$, $x=\overline{r}$ and $y=\overline{r}$) yields
$\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}=T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  $. Hence, $T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  =\left(
\overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$. Thus,
\[
g=m_{A}\circ\underbrace{T_{A,A}\circ\left(  \overline{r}\otimes
\overline{r}\right)  }_{=\left(  \overline{r}\otimes\overline{r}\right)
\circ T_{C,C}}
=m_{A}\circ\left(  \left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}\right)  .
\]
Now,%
\begin{align*}
&  m_{A}\circ\left(  \underbrace{f}_{=m_{A}\circ\left(  r\otimes r\right)
}\otimes\underbrace{g}_{=m_{A}\circ\left(  \left(  \overline{r}\otimes
\overline{r}\right)  \circ T_{C,C}\right)  }\right) \\
&  =m_{A}\circ\underbrace{\left(  \left(  m_{A}\circ\left(  r\otimes r\right)
\right)  \otimes\left(  m_{A}\circ\left(  \left(  \overline{r}\otimes
\overline{r}\right)  \circ T_{C,C}\right)  \right)  \right)  }%
_{\substack{=\left(  m_{A}\otimes m_{A}\right)  \circ\left(  \left(  r\otimes
r\right)  \otimes\left(  \left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}\right)  \right)  \\\text{(by the composition-tensor relation)}}}\\
&  =\underbrace{m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  }%
_{\substack{=m^{\left(  3\right)  }\\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m3}))}}}\circ\left(  \underbrace{\left(
r\otimes r\right)  }_{=\left(  r\otimes r\right)  \circ
\id_{C\otimes C}}\otimes\left(  \left(  \overline{r}\otimes
\overline{r}\right)  \circ T_{C,C}\right)  \right) \\
&  =m^{\left(  3\right)  }\circ\underbrace{\left(  \left(  \left(  r\otimes
r\right)  \circ\id_{C\otimes C}\right)  \otimes\left(
\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}\right)  \right)
}_{\substack{=\left(  \left(  r\otimes r\right)  \otimes\left(
\overline{r}\otimes\overline{r}\right)  \right)  \circ\left(  
\id_{C\otimes C}\otimes T_{C,C}\right)  \\\text{(by the
composition-tensor relation)}}}\\
&  =m^{\left(  3\right)  }\circ\underbrace{\left(  \left(  r\otimes r\right)
\otimes\left(  \overline{r}\otimes\overline{r}\right)  \right)  }_{=r\otimes
r\otimes\overline{r}\otimes\overline{r}}\circ\left(
\underbrace{\id_{C\otimes C}}_{=
\id_{C}\otimes\id_{C}}\otimes T_{C,C}\right) \\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m_{A}\circ\left(  f\otimes g\right)  }_{=m^{\left(  3\right)
}\circ\left(  r\otimes r\otimes\overline{r}\otimes\overline{r}\right)
\circ\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  }\circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.2}).}

\item We have%
\begin{align}
&  m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\nonumber\\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
. \label{sol.coalg-anti.gen.a.sol2.fg.pf.3}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3}):} We
first notice that%
\begin{equation}
\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
=\id_{C}\otimes T_{C,C\otimes C}.
\label{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}%
\end{equation}
\par
\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}):} For every
$\left(  a,b,c,d\right)  \in C\times C\times C\times C$, we have%
\begin{align*}
&  \left(  \left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\right)  \left(  a\otimes b\otimes c\otimes d\right) \\
&  =\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \left(  \underbrace{\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \left(  a\otimes b\otimes c\otimes d\right)
}_{\substack{=\id_{C}\left(  a\right)  \otimes
T_{C,C}\left(  b\otimes c\right)  \otimes\id_{C}%
\left(  d\right)  \\\text{(by the definition of the map }
\id_{C}\otimes T_{C,C}\otimes\id_{C}\text{)}%
}}\right) \\
&  =\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \left(  \underbrace{
\id_{C}\left(  a\right)  }_{=a}\otimes\underbrace{T_{C,C}\left(
b\otimes c\right)  }_{\substack{=c\otimes b\\\text{(by the definition}%
\\\text{of the map }T_{C,C}\text{)}}}\otimes\underbrace{
\id_{C}\left(  d\right)  }_{=d}\right) \\
&  =\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \left(  a\otimes c\otimes b\otimes
d\right)  =\underbrace{\id_{C}\left(  a\right)  }%
_{=a}\otimes\underbrace{\id_{C}\left(  c\right)
}_{=c}\otimes\underbrace{T_{C,C}\left(  b\otimes d\right)  }%
_{\substack{=d\otimes b\\\text{(by the definition}\\\text{of the map }%
T_{C,C}\text{)}}}\\
&  =a\otimes c\otimes d\otimes b=\underbrace{a}_{=
\id_{C}\left(  a\right)  }\otimes\underbrace{\left(  c\otimes d\right)
\otimes b}_{\substack{=T_{C,C\otimes C}\left(  b\otimes\left(  c\otimes
d\right)  \right)  \\\text{(since }T_{C,C\otimes C}\left(  b\otimes\left(
c\otimes d\right)  \right)  \\=\left(  c\otimes d\right)  \otimes b\\\text{(by
the definition of }T_{C,C\otimes C}\text{))}}}=
\id_{C}\left(  a\right)  \otimes T_{C,C\otimes C}\left(  b\otimes\left(
c\otimes d\right)  \right) \\
&  =\left(  \id_{C}\otimes T_{C,C\otimes C}\right)
\left(  \underbrace{a\otimes\left(  b\otimes\left(  c\otimes d\right)
\right)  }_{=a\otimes b\otimes c\otimes d}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  \id_{C}\otimes T_{C,C\otimes
C}\right)  \left(  a\otimes\left(  b\otimes\left(  c\otimes d\right)  \right)
\right)  =\id_{C}\left(  a\right)  \otimes
T_{C,C\otimes C}\left(  b\otimes\left(  c\otimes d\right)  \right) \\
\text{(by the definition of the map }\id_{C}\otimes
T_{C,C\otimes C}\text{)}%
\end{array}
\right) \\
&  =\left(  \id_{C}\otimes T_{C,C\otimes C}\right)
\left(  a\otimes b\otimes c\otimes d\right)  .
\end{align*}
In other words, the two maps $\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  $ and $\id_{C}\otimes
T_{C,C\otimes C}$ are equal to each other on each pure tensor. Since these two
maps are $\kk$-linear, this yields that these two maps are identical
(because any two $\kk$-linear maps from a tensor product which are
equal to each other on each pure tensor must be identical). In other words, we
have $\left(  \id_{C}\otimes
\id_{C}\otimes T_{C,C}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
=\id_{C}\otimes T_{C,C\otimes C}$. This proves
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}).
\par
Now,%
\begin{align*}
&  m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\underbrace{\left(  
\id_{C}\otimes\id_{C}\otimes T_{C,C}\right)
\circ\left(  \id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  }_{\substack{=
\id_{C}\otimes T_{C,C\otimes C}\\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}))}}}\circ\left(  \Delta
_{C}\otimes\Delta_{C}\right) \\
&  =\underbrace{m^{\left(  3\right)  }}_{\substack{=m^{\left(  2\right)
}\circ\left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m3o}))}}}\circ\left(  r\otimes r\otimes
\overline{r}\otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ\left(  \underbrace{\Delta
_{C}}_{=\id_{C\otimes C}\circ\Delta_{C}}%
\otimes\underbrace{\Delta_{C}}_{=\Delta_{C}\circ
\id_{C}}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  r\otimes
r\otimes\overline{r}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ
\underbrace{\left(  \left(  \id_{C\otimes C}%
\circ\Delta_{C}\right)  \otimes\left(  \Delta_{C}\circ
\id_{C}\right)  \right)  }_{\substack{=\left(  
\id_{C\otimes C}\otimes\Delta_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\id_{C}\right)  \\\text{(by the
composition-tensor relation)}}}\\
&  =m^{\left(  2\right)  }\circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  r\otimes
r\otimes\overline{r}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ\left(
\id_{C\otimes C}\otimes\Delta_{C}\right)  \circ\left(
\Delta_{C}\otimes\id_{C}\right)  .
\end{align*}
Since%
\begin{align*}
&  \left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \circ\underbrace{\left(  r\otimes
r\otimes\overline{r}\otimes\overline{r}\right)  }_{=r\otimes\left(
r\otimes\overline{r}\right)  \otimes\overline{r}}\\
&  =\left(  \id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \circ\left(  r\otimes\left(
r\otimes\overline{r}\right)  \otimes\overline{r}\right) \\
&  =\underbrace{\left(  \id_{A}\circ r\right)  }%
_{=r}\otimes\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)
\otimes\underbrace{\left(  \id_{A}\circ\overline{r}
\right)  }_{=\overline{r}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the composition-triple-tensor relation yields}\\
\left(  \id_{A}\circ r\right)  \otimes\left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)  \otimes\left(
\id_{A}\circ\overline{r}\right)  =\left(
\id_{A}\otimes m_{A}\otimes
\id_{A}\right)  \circ\left(  r\otimes\left(  r\otimes
\overline{r}\right)  \otimes\overline{r}\right)
\end{array}
\right) \\
&  =r\otimes\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)
\otimes\overline{r}%
\end{align*}
and%
\begin{align*}
&  \left(  \id_{C}\otimes T_{C,C\otimes C}\right)
\circ\left(  \underbrace{\id_{C\otimes C}%
}_{=\id_{C}\otimes\id_{C}%
}\otimes\Delta_{C}\right) \\
&  =\left(  \id_{C}\otimes T_{C,C\otimes C}\right)
\circ\left(  \id_{C}\otimes
\id_{C}\otimes\Delta_{C}\right)  =\left(  
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ\left(  
\id_{C}\otimes\left(  \id_{C}\otimes\Delta
_{C}\right)  \right) \\
&  =\left(  \id_{C}\circ
\id_{C}\right)  \otimes\underbrace{\left(  T_{C,C\otimes C}\circ\left(
\id_{C}\otimes\Delta_{C}\right)  \right)
}_{\substack{=\left(  \Delta_{C}\otimes\id_{C}\right)
\circ T_{C,C}\\\text{(since }\left(  \Delta_{C}\otimes
\id_{C}\right)  \circ T_{C,C}=T_{C,C\otimes C}\circ\left(
\id_{C}\otimes\Delta_{C}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.twist}), applied to }U=C\text{, }%
V=C\text{,}\\U^{\prime}=C\text{, }V^{\prime}=C\otimes C\text{, }%
x=\id_{C}\text{ and }y=\Delta_{C}\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the composition-tensor relation yields}\\
\left(  \id_{C}\circ\id_{C}%
\right)  \otimes\left(  T_{C,C\otimes C}\circ\left(  
\id_{C}\otimes\Delta_{C}\right)  \right)  =\left(  
\id_{C}\otimes T_{C,C\otimes C}\right)  \circ\left(  
\id_{C}\otimes\left(  \id_{C}\otimes\Delta
_{C}\right)  \right)
\end{array}
\right) \\
&  =\left(  \id_{C}\circ
\id_{C}\right)  \otimes\left(  \left(  \Delta_{C}\otimes
\id_{C}\right)  \circ T_{C,C}\right)
=\underbrace{\left(  \id_{C}\otimes\left(  \Delta
_{C}\otimes\id_{C}\right)  \right)  }%
_{=\id_{C}\otimes\Delta_{C}\otimes
\id_{C}}\circ\left(  \id_{C}\otimes
T_{C,C}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  \id_{C}\otimes\Delta_{C}\otimes
\id_{C}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  ,
\end{align*}
this becomes%
\begin{align*}
&  m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m^{\left(  2\right)  }\circ\underbrace{\left(  
\id_{A}\otimes m_{A}\otimes\id_{A}\right)
\circ\left(  r\otimes r\otimes\overline{r}\otimes\overline{r}\right)
}_{=r\otimes\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)
\otimes\overline{r}}\circ\underbrace{\left(  \id_{C}%
\otimes T_{C,C\otimes C}\right)  \circ\left(  
\id_{C\otimes C}\otimes\Delta_{C}\right)  }_{=\left(  
\id_{C}\otimes\Delta_{C}\otimes\id_{C}\right)
\circ\left(  \id_{C}\otimes T_{C,C}\right)  }%
\circ\left(  \Delta_{C}\otimes\id_{C}\right) \\
&  =m^{\left(  2\right)  }\circ\underbrace{\left(  r\otimes\left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \right)  \otimes\overline{r}\right)
\circ\left(  \id_{C}\otimes\Delta_{C}\otimes
\id_{C}\right)  }_{\substack{=\left(  r\circ
\id_{C}\right)  \otimes\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \right)  \circ\Delta_{C}\right)
\otimes\left(  \overline{r}\circ\id_{C}\right)
\\\text{(since the composition-triple-tensor relation yields}\\\left(
r\circ\id_{C}\right)  \otimes\left(  \left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)  \circ\Delta
_{C}\right)  \otimes\left(  \overline{r}\circ\id_{C}%
\right)  \\=\left(  r\otimes\left(  m_{A}\circ\left(  r\otimes
\overline{r}\right)  \right)  \otimes\overline{r}\right)  \circ
\left(  \id_{C}\otimes\Delta_{C}\otimes
\id_{C}\right)  \text{)}}}\circ\left(  \id_{C}%
\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  \underbrace{\left(  r\circ
\id_{C}\right)  }_{=r}\otimes\underbrace{\left(
\left(  m_{A}\circ\left(  r\otimes\overline{r}\right)  \right)  \circ
\Delta_{C}\right)  }_{=m_{A}\circ\left(  r\otimes\overline{r}\right)
\circ\Delta_{C}}\otimes\underbrace{\left(  \overline{r}\circ
\id_{C}\right)  }_{=\overline{r}}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta
_{C}\otimes\id_{C}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
.
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3}).}

\item We have%
\begin{align}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
\nonumber\\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.4}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.4}):} We have
$r\star\overline{r}=m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ
\Delta_{C}$ (according to the definition of convolution). Compared with
$r\star\overline{r}=u_{A}\circ\epsilon_{C}$, this yields $m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}=u_{A}\circ\epsilon_{C}$. Thus,%
\begin{align*}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\underbrace{\left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  }%
_{=u_{A}\circ\epsilon_{C}}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta
_{C}\otimes\id_{C}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.4}).}

\item We have%
\begin{align}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right) \nonumber\\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  \circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.5}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.5}):} We have%
\[
\underbrace{r}_{=\id_{A}\circ r}\otimes\left(
u_{A}\circ\epsilon_{C}\right)  \otimes\underbrace{\overline{r}}%
_{=\id_{A}\circ\overline{r}}=\left(
\id_{A}\circ r\right)  \otimes\left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\left(  \id_{A}%
\circ\overline{r}\right)  =\left(  \id_{A}\otimes
u_{A}\otimes\id_{A}\right)  \circ\left(
r\otimes\epsilon_{C}\otimes\overline{r}\right)
\]
(by the composition-triple-tensor relation). Thus,%
\begin{align*}
&  \underbrace{m^{\left(  2\right)  }}_{\substack{=m_{A}\circ\left(
\id_{A}\otimes m_{A}\right)  \\\text{(by
(\ref{sol.coalg-anti.gen.a.sol2.m2b}))}}}\circ\underbrace{\left(
r\otimes\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\overline{r}\right)
}_{=\left(  \id_{A}\otimes u_{A}\otimes
\id_{A}\right)  \circ\left(  r\otimes\epsilon
_{C}\otimes\overline{r}\right)  }\\
&  =m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)
\circ\underbrace{\left(  \id_{A}\otimes u_{A}%
\otimes\id_{A}\right)  }_{=
\id_{A}\otimes\left(  u_{A}\otimes\id_{A}%
\right)  }\circ\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right) \\
&  =m_{A}\circ\underbrace{\left(  \id_{A}\otimes
m_{A}\right)  \circ\left(  \id_{A}\otimes\left(
u_{A}\otimes\id_{A}\right)  \right)  }%
_{\substack{=\left(  \id_{A}\circ
\id_{A}\right)  \otimes\left(  m_{A}\circ\left(  u_{A}\otimes
\id_{A}\right)  \right)  \\\text{(since the
composition-tensor relation}\\\text{yields }\left(  
\id_{A}\circ\id_{A}\right)  \otimes\left(
m_{A}\circ\left(  u_{A}\otimes\id_{A}\right)  \right)
\\=\left(  \id_{A}\otimes m_{A}\right)  \circ\left(
\id_{A}\otimes\left(  u_{A}\otimes
\id_{A}\right)  \right)  \text{)}}}\circ\left(  r\otimes\epsilon
_{C}\otimes\overline{r}\right) \\
&  =m_{A}\circ\left(  \underbrace{\left(  \id_{A}%
\circ\id_{A}\right)  }_{=
\id_{A}}\otimes\underbrace{\left(  m_{A}\circ\left(  u_{A}%
\otimes\id_{A}\right)  \right)  }%
_{=\operatorname{kan}_{2,A}^{-1}}\right)  \circ\left(
r\otimes\epsilon_{C}\otimes\overline{r}\right)  =m_{A}\circ\left(
\id_{A}\otimes\operatorname{kan}_{2,A}%
^{-1}\right)  \circ\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\overline{r}\right)  }_{=m_{A}\circ\left(
\id_{A}\otimes\operatorname{kan}_{2,A}%
^{-1}\right)  \circ\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)
}\circ\left(  \id_{C}\otimes T_{C,C}\right)
\circ\left(  \Delta_{C}\otimes\id_{C}\right) \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{2,A}^{-1}\right)  \circ\left(  r\otimes
\epsilon_{C}\otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\end{align*}
Since%
\begin{align*}
&  \underbrace{\left(  r\otimes\epsilon_{C}\otimes\overline{r}\right)
}_{=r\otimes\left(  \epsilon_{C}\otimes\overline{r}\right)  }\circ\left(
\id_{C}\otimes T_{C,C}\right) \\
&  =\left(  r\otimes\left(  \epsilon_{C}\otimes\overline{r}\right)  \right)
\circ\left(  \id_{C}\otimes T_{C,C}\right)
=\underbrace{\left(  r\circ\id_{C}\right)
}_{=r=\id_{A}\circ r}\otimes\underbrace{\left(
\left(  \epsilon_{C}\otimes\overline{r}\right)  \circ T_{C,C}\right)
}_{\substack{=T_{A,\kk}\circ\left(  \overline{r}\otimes\epsilon
_{C}\right)  \\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.twist}), applied to
}U=C\text{, }V=C\text{, }U^{\prime}=A\text{, }V^{\prime}=\kk%
\text{,}\\x=\overline{r}\text{ and }y=\epsilon_{C}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  \id_{A}\circ r\right)  \otimes\left(
T_{A,\kk}\circ\left(  \overline{r}\otimes\epsilon_{C}\right)  \right)
=\left(  \id_{A}\otimes T_{A,\kk}\right)
\circ\underbrace{\left(  r\otimes\left(  \overline{r}\otimes\epsilon
_{C}\right)  \right)  }_{=r\otimes\overline{r}\otimes\epsilon_{C}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the composition-tensor relation}%
\right) \\
&  =\left(  \id_{A}\otimes T_{A,\kk}\right)
\circ\left(  r\otimes\overline{r}\otimes\epsilon_{C}\right)  ,
\end{align*}
this becomes%
\begin{align*}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right) \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{2,A}^{-1}\right)  \circ\underbrace{\left(
r\otimes\epsilon_{C}\otimes\overline{r}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\right)  }_{=\left(
\id_{A}\otimes T_{A,\kk}\right)  \circ\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  }\circ\left(  \Delta
_{C}\otimes\id_{C}\right) \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{2,A}^{-1}\right)  \circ\left(
\id_{A}\otimes T_{A,\kk}\right)  \circ\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}%
\otimes\id_{C}\right)  .
\end{align*}
Since%
\begin{align*}
&  \left(  \id_{A}\otimes\operatorname*{kan}%
\nolimits_{2,A}^{-1}\right)  \circ\left(  \id_{A}%
\otimes T_{A,\kk}\right) \\
&  =\underbrace{\left(  \id_{A}\circ
\id_{A}\right)  }_{=\id_{A}}\otimes
\underbrace{\left(  \operatorname{kan}_{2,A}^{-1}\circ
T_{A,\kk}\right)  }_{\substack{=\operatorname{kan}_{1,A}%
^{-1}\\\text{(by (\ref{sol.coalg-anti.gen.a.sol2.kan12.1}), applied to
}U=A\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the composition-tensor relation yields}\\
\left(  \id_{A}\circ\id_{A}%
\right)  \otimes\left(  \operatorname{kan}_{2,A}^{-1}\circ
T_{A,\kk}\right)  =\left(  \id_{A}%
\otimes\operatorname{kan}_{2,A}^{-1}\right)  \circ\left(
\id_{A}\otimes T_{A,\kk}\right)
\end{array}
\right) \\
&  =\id_{A}\otimes\operatorname{kan}_{1,A}%
^{-1},
\end{align*}
this becomes%
\begin{align*}
&  m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right) \\
&  =m_{A}\circ\underbrace{\left(  \id_{A}%
\otimes\operatorname{kan}_{2,A}^{-1}\right)  \circ\left(
\id_{A}\otimes T_{A,\kk}\right)  }%
_{=\id_{A}\otimes\operatorname{kan}_{1,A}%
^{-1}}\circ\left(  r\otimes\overline{r}\otimes\epsilon_{C}\right)
\circ\left(  \Delta_{C}\otimes\id_{C}\right) \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  \circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.5}).}

\item We have%
\begin{align}
&  m_{A}\circ\left(  \id_{A}\otimes\operatorname*{kan}%
\nolimits_{1,A}^{-1}\right)  \circ\left(  r\otimes\overline{r}\otimes
\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right) \nonumber\\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\epsilon_{C}\right)  . \label{sol.coalg-anti.gen.a.sol2.fg.pf.6}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.6}):} The
equality (\ref{sol.coalg-anti.gen.a.sol2.kan1a}) (applied to $U=A\otimes A$,
$V=A$ and $\alpha=m_{A}$) yields $\operatorname{kan}_{1,A}^{-1}%
\circ\left(  m_{A}\otimes\id_{\kk}\right)
=m_{A}\circ\operatorname{kan}_{1,A\otimes A}^{-1}$. Thus,
$m_{A}\circ\operatorname{kan}_{1,A\otimes A}^{-1}%
=\operatorname{kan}_{1,A}^{-1}\circ\left(  m_{A}\otimes
\id_{\kk}\right)  $. But $
\id_{A}\otimes\operatorname{kan}_{1,A}^{-1}%
=\operatorname{kan}_{1,A\otimes A}^{-1}$ (according to
(\ref{sol.coalg-anti.gen.a.sol2.kans.3}), applied to $U=A$ and $V=A$), and
thus%
\[
m_{A}\circ\underbrace{\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  }_{=\operatorname*{kan}%
\nolimits_{1,A\otimes A}^{-1}}=m_{A}\circ\operatorname*{kan}%
\nolimits_{1,A\otimes A}^{-1}=\operatorname{kan}_{1,A}^{-1}%
\circ\left(  m_{A}\otimes\id_{\kk}\right)  .
\]
Hence,%
\begin{align*}
&  \underbrace{m_{A}\circ\left(  \id_{A}%
\otimes\operatorname{kan}_{1,A}^{-1}\right)  }_{=\operatorname*{kan}%
\nolimits_{1,A}^{-1}\circ\left(  m_{A}\otimes
\id_{\kk}\right)  }\circ\left(  r\otimes\overline{r}%
\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right) \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  m_{A}\otimes
\id_{\kk}\right)  \circ\underbrace{\left(
r\otimes\overline{r}\otimes\epsilon_{C}\right)  }_{=\left(  r\otimes
\overline{r}\right)  \otimes\epsilon_{C}}\circ\left(  \Delta_{C}%
\otimes\id_{C}\right) \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\underbrace{\left(
m_{A}\otimes\id_{\kk}\right)  \circ\left(
\left(  r\otimes\overline{r}\right)  \otimes\epsilon_{C}\right)  \circ\left(
\Delta_{C}\otimes\id_{C}\right)  }_{\substack{=\left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\left(  \id_{\kk}\circ\epsilon_{C}%
\circ\id_{C}\right)  \\\text{(since the
triple-composition-tensor relation}\\\text{yields }\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes\left(
\id_{\kk}\circ\epsilon_{C}\circ
\id_{C}\right)  \\=\left(  m_{A}\otimes
\id_{\kk}\right)  \circ\left(  \left(
r\otimes\overline{r}\right)  \otimes\epsilon_{C}\right)  \circ\left(
\Delta_{C}\otimes\id_{C}\right)  \text{)}}}\\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\underbrace{\left(  \id_{\kk}%
\circ\epsilon_{C}\circ\id_{C}\right)  }_{=\epsilon
_{C}}\right)  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\epsilon_{C}\right)  .
\end{align*}
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.6}).}

\item We have%
\begin{equation}
\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes\epsilon
_{C}\right)  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  .
\label{sol.coalg-anti.gen.a.sol2.fg.pf.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.7}):} The
definition of convolution yields $r\star\overline{r}=m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}$. Compared with $r\star
\overline{r}=u_{A}\circ\epsilon_{C}$, this yields $m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}=u_{A}\circ\epsilon_{C}$. Hence,%
\[
\operatorname{kan}_{1,A}^{-1}\circ\left(  \underbrace{\left(
m_{A}\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
}_{=u_{A}\circ\epsilon_{C}}\otimes\epsilon_{C}\right)  =\operatorname*{kan}%
\nolimits_{1,A}^{-1}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)
\otimes\epsilon_{C}\right)  .
\]
This proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.7}).}

\item We have%
\begin{equation}
\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  u_{A}\circ
\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  =m_{A}\circ\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon_{C}\right)
\right)  \label{sol.coalg-anti.gen.a.sol2.fg.pf.8}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.8}):} We have%
\[
\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\underbrace{\epsilon_{C}%
}_{=\id_{\kk}\circ\epsilon_{C}}=\left(
u_{A}\circ\epsilon_{C}\right)  \otimes\left(  
\id_{\kk}\circ\epsilon_{C}\right)  =\left(  u_{A}\otimes
\id_{\kk}\right)  \circ\left(  \epsilon
_{C}\otimes\epsilon_{C}\right)
\]
(by the composition-tensor relation), so that%
\[
\operatorname{kan}_{1,A}^{-1}\circ\underbrace{\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  }_{=\left(
u_{A}\otimes\id_{\kk}\right)  \circ\left(
\epsilon_{C}\otimes\epsilon_{C}\right)  }=\operatorname{kan}_{1,A}%
^{-1}\circ\left(  u_{A}\otimes\id_{\kk}\right)
\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)  .
\]
Compared with
\begin{align*}
&  m_{A}\circ\left(  \left(  \underbrace{u_{A}}_{=
\id_{A}\circ u_{A}}\circ\epsilon_{C}\right)  \otimes\left(
\underbrace{u_{A}}_{=u_{A}\circ\id_{\kk}}%
\circ\epsilon_{C}\right)  \right) \\
&  =m_{A}\circ\underbrace{\left(  \left(  \id_{A}\circ
u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ
\id_{\kk}\circ\epsilon_{C}\right)  \right)  }_{\substack{=\left(
\id_{A}\otimes u_{A}\right)  \circ\left(  u_{A}%
\otimes\id_{\kk}\right)  \circ\left(
\epsilon_{C}\otimes\epsilon_{C}\right)  \\\text{(by the
triple-composition-tensor relation)}}}\\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes
u_{A}\right)  }_{=\operatorname{kan}_{1,A}^{-1}}\circ\left(
u_{A}\otimes\id_{\kk}\right)  \circ\left(
\epsilon_{C}\otimes\epsilon_{C}\right)  =\operatorname{kan}_{1,A}%
^{-1}\circ\left(  u_{A}\otimes\id_{\kk}\right)
\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)  ,
\end{align*}
this yields $\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(
u_{A}\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)  =m_{A}\circ\left(
\left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \right)  $. Thus, (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.8}) is
solved.}

\item We have%
\begin{equation}
m_{A}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(
u_{A}\circ\epsilon_{C}\right)  \right)  =u_{A}\circ\epsilon_{C}\circ m_{C}.
\label{sol.coalg-anti.gen.a.sol2.fg.pf.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.9}):} We know
that $u_{A}$ is a $\kk$-algebra homomorphism (indeed, this is an easy
fact that holds whenever $A$ is a $\kk$-algebra). Since the two maps
$u_{A}$ and $\epsilon_{C}$ are $\kk$-algebra homomorphisms, their
composition $u_{A}\circ\epsilon_{C}$ is a $\kk$-algebra homomorphism.
In other words, $u_{A}\circ\epsilon_{C}$ satisfies $\left(  u_{A}\circ
\epsilon_{C}\right)  \circ m_{C}=m_{A}\circ\left(  \left(  u_{A}\circ
\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon_{C}\right)  \right)  $
and $\left(  u_{A}\circ\epsilon_{C}\right)  \circ u_{C}=u_{A}$ (because
$u_{A}\circ\epsilon_{C}$ is a $\kk$-algebra homomorphism if and only if
$u_{A}\circ\epsilon_{C}$ satisfies $\left(  u_{A}\circ\epsilon_{C}\right)
\circ m_{C}=m_{A}\circ\left(  u_{A}\circ\epsilon_{C}\right)  $ and $\left(
u_{A}\circ\epsilon_{C}\right)  \circ u_{C}=u_{A}$ (according to the definition
of a $\kk$-algebra homomorphism)). Thus, we have $\left(  u_{A}%
\circ\epsilon_{C}\right)  \circ m_{C}=m_{A}\circ\left(  \left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\left(  u_{A}\circ\epsilon_{C}\right)
\right)  $, so that $m_{A}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)
\otimes\left(  u_{A}\circ\epsilon_{C}\right)  \right)  =\left(  u_{A}%
\circ\epsilon_{C}\right)  \circ m_{C}=u_{A}\circ\epsilon_{C}\circ m_{C}$. This
proves (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.9}).}
\end{itemize}

Now,%
\begin{align*}
f\circledast g  &  =m_{A}\circ\left(  f\otimes g\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.1}%
)}\right) \\
&  =m^{\left(  3\right)  }\circ\left(  r\otimes r\otimes\overline{r}%
\otimes\overline{r}\right)  \circ\left(  \id_{C}%
\otimes\id_{C}\otimes T_{C,C}\right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.2}%
)}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  m_{A}\circ\left(
r\otimes\overline{r}\right)  \circ\Delta_{C}\right)  \otimes
\overline{r}\right)  \circ\left(  \id_{C}\otimes T_{C,C}%
\right)  \circ\left(  \Delta_{C}\otimes\id_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3}%
)}\right) \\
&  =m^{\left(  2\right)  }\circ\left(  r\otimes\left(  u_{A}\circ\epsilon
_{C}\right)  \otimes\overline{r}\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.4})}\right) \\
&  =m_{A}\circ\left(  \id_{A}\otimes
\operatorname{kan}_{1,A}^{-1}\right)  \circ\left(  r\otimes
\overline{r}\otimes\epsilon_{C}\right)  \circ\left(  \Delta_{C}\otimes
\id_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.5})}\right) \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  m_{A}%
\circ\left(  r\otimes\overline{r}\right)  \circ\Delta_{C}\right)
\otimes\epsilon_{C}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.6})}\right) \\
&  =\operatorname{kan}_{1,A}^{-1}\circ\left(  \left(  u_{A}%
\circ\epsilon_{C}\right)  \otimes\epsilon_{C}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.7}%
)}\right) \\
&  =m_{A}\circ\left(  \left(  u_{A}\circ\epsilon_{C}\right)  \otimes\left(
u_{A}\circ\epsilon_{C}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.8})}\right) \\
&  =u_{A}\circ\epsilon_{C}\circ m_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.9})}\right) \\
&  =u_{A}\circ\epsilon_{C\otimes C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.7})}\right)  .
\end{align*}
The second equality in (\ref{sol.coalg-anti.gen.a.sol2.hffg}) is thus proven.

Thus, both equalities in (\ref{sol.coalg-anti.gen.a.sol2.hffg}) are proven.
Hence, (\ref{sol.coalg-anti.gen.a.sol2.hffg}) is proven. As we have already
seen, this yields (\ref{sol.coalg-anti.gen.a.sol2.h=g}). In other words,
$h=g$. But $h=\overline{r}\circ m_{C}$, so that $\overline{r}\circ
m_{C}=h=g=m_{A}\circ T_{A,A}\circ\left(  \overline{r}\otimes
\overline{r}\right)  $.

Finally, (\ref{sol.coalg-anti.gen.a.sol2.twist}) (applied to $U=C$, $V=C$,
$U^{\prime}=A$, $V^{\prime}=A$, $x=\overline{r}$ and $y=\overline{r}$) yields
$\left(  \overline{r}\otimes\overline{r}\right)  \circ T_{C,C}=T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  $. Thus, $T_{A,A}%
\circ\left(  \overline{r}\otimes\overline{r}\right)  =\left(
\overline{r}\otimes\overline{r}\right)  \circ T_{C,C}$, so that
$\overline{r}\circ
m_{C}=m_{A}\circ\underbrace{T_{A,A}\circ\left(  \overline{r}\otimes
\overline{r}\right)  }_{=\left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}}=m_{A}\circ\left(  \overline{r}\otimes\overline{r}\right)  \circ
T_{C,C}$. As we know, this completes the solution of
Exercise~\ref{exe.coalg-anti.gen}(a).

[\textit{Remark:} The second solution of Exercise~\ref{exe.coalg-anti.gen}(a)
has been obtained more or less straightforwardly from the first solution by
rewriting it in an element-free fashion. First of all, the maps $f$, $g$ and
$h$ introduced in the second solution are precisely the maps $f$, $g$ and $h$
introduced in the first solution, just rewritten in an element-free
way. Also, for example, the
equalities (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}) and
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}) have been found by rewriting the six
equality signs in the computation (\ref{sol.coalg-anti.gen.a.sol1.hf}) in an
element-free way: For instance, the second equality sign in
(\ref{sol.coalg-anti.gen.a.sol1.hf}) stands for the equality
\[
\sum_{(a),(b)}h\left(  a_{1}\otimes b_{1}\right)  f\left(  a_{2}\otimes
b_{2}\right)  =\sum_{(a),(b)}\overline{r}\left(  a_{1}b_{1}\right)  r\left(
a_{2}\right)  r\left(  b_{2}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a\in
C\text{ and }b\in C,
\]
i.e., for the following equality of maps:%
\begin{align*}
&  \left(  \text{the }\kk\text{-linear map }C\otimes C\rightarrow
A\text{ sending every }a\otimes b\text{ to }\sum_{(a),(b)}h\left(
a_{1}\otimes b_{1}\right)  f\left(  a_{2}\otimes b_{2}\right)  \right) \\
&  =\left(  \text{the }\kk\text{-linear map }C\otimes C\rightarrow
A\text{ sending every }a\otimes b\text{ to }\sum_{(a),(b)}\overline{r}\left(
a_{1}b_{1}\right)  r\left(  a_{2}\right)  r\left(  b_{2}\right)  \right)  .
\end{align*}
But upon rewriting these two maps without referring to elements, this takes
the form%
\begin{align*}
&  m_{A}\circ\left(  h\otimes f\right)  \circ\left(  
\id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right) \\
&  =m_{A}\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(
m_{A}\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)
\end{align*}
(because the $\kk$-linear map $C\otimes C\rightarrow A$ sending every
$a\otimes b$ to $\sum_{(a),(b)}h\left(  a_{1}\otimes b_{1}\right)  f\left(
a_{2}\otimes b_{2}\right)  $ is $m_{A}\circ\left(  h\otimes f\right)
\circ\left(  \id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes
\Delta_{C}\right)  $, and the $\kk$-linear map $C\otimes C\rightarrow
A$ sending every $a\otimes b$ to $\sum_{(a),(b)}\overline{r}\left(  a_{1}%
b_{1}\right)  r\left(  a_{2}\right)  r\left(  b_{2}\right)  $ is $m_{A}%
\circ\left(  \left(  \overline{r}\circ m_{C}\right)  \otimes\left(  m_{A}%
\circ\left(  r\otimes r\right)  \right)  \right)  \circ\left(
\id_{C}\otimes T_{C,C}\otimes
\id_{C}\right)  \circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  $); and
this is precisely the equality (\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}). The
process of rewriting a map in an element-free way is not completely
deterministic (e.g., we could also have rewritten the $\kk$-linear map
$C\otimes C\rightarrow A$ sending every $a\otimes b$ to $\sum_{(a),(b)}%
\overline{r}\left(  a_{1}b_{1}\right)  r\left(  a_{2}\right)  r\left(
b_{2}\right)  $ as $m^{\left(  3\right)  }\circ\left(  \left(
\overline{r}\circ m_{C}\right)  \otimes r\otimes r\right)  \circ
\left(  \id_{C}\otimes T_{C,C}\otimes\id_{C}\right)
\circ\left(  \Delta_{C}\otimes\Delta_{C}\right)  $,
since $m^{\left(  3\right)  }$ is the $\kk$-linear map $A\otimes
A\otimes A \to A$ sending every $a\otimes b\otimes c$ to $abc$),
but all possible
results (for a given map) can be reduced to each other using purely
linear-algebraic formulae\footnote{By ``purely linear-algebraic
formulae'', I mean formulae such as
(\ref{sol.coalg-anti.gen.a.sol2.twist}) and
(\ref{sol.coalg-anti.gen.a.sol2.kan12.1}) and the
composition-tensor relation. This kind of formulae hold in any
tensor category (at least if we follow the abuse of notation that
allows us to treat the tensor product as associative).}
and various forms of the general
associativity law (such as (\ref{sol.coalg-anti.gen.a.sol2.m3}) and
(\ref{sol.coalg-anti.gen.a.sol2.m3o}))\footnote{The reason why we
need to use the general associativity law is that, in the first
solution, we used unparenthesized products of more than one element
of $A$ (for example, the expression
$\sum_{(a),(b)}\overline{r}\left(  a_{1} b_{1}\right)
r\left(  a_{2}\right)  r\left(  b_{2}\right)$
contains the unparenthesized product
$\overline{r}\left(  a_{1} b_{1}\right)
r\left(  a_{2}\right)  r\left(  b_{2}\right)$ of three factors).
If we would parenthesize all such products in such a way that no more
than two factors are ever multiplied at the same time (e.g., we could
replace $\overline{r}\left(  a_{1} b_{1}\right)
r\left(  a_{2}\right)  r\left(  b_{2}\right)$ by
$\overline{r}\left(  a_{1} b_{1}\right)
\left(r\left(  a_{2}\right)  r\left(  b_{2}\right) \right)$), and
if we would explicitly use the (non-general) associativity law
$\left(xy\right) z = x\left(yz\right)$ to switch between these
parenthesizations, then we would not have to use general
associativity any more when rewriting the proof in an element-free
fashion (but, of course, the proof would be longer).}. The equalities
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.1}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.2}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.3}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.4}),
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.5}) and
(\ref{sol.coalg-anti.gen.a.sol2.hf.pf.6}) are somewhat more complicated to
check than the corresponding equality signs in
(\ref{sol.coalg-anti.gen.a.sol1.hf}), because of the fact that one and the
same map can be written in different forms whose equivalence needs to be
proven. However, this additional complexity is straightforward to surmount:
each equality, when rewritten in element-free terms, can be proven by the same
arguments as the corresponding equality that uses elements, combined with
purely linear-algebraic formulae like (\ref{sol.coalg-anti.gen.a.sol2.twist})
and (\ref{sol.coalg-anti.gen.a.sol2.kan12.1}) and various forms of the general
associativity law (such as (\ref{sol.coalg-anti.gen.a.sol2.m3}) and
(\ref{sol.coalg-anti.gen.a.sol2.m3o})). (There is also a way how to do
element-free computations without such added difficulty, using \emph{string
diagrams}\footnote{See
\url{http://ncatlab.org/nlab/show/string+diagram}
and the references therein.}.)

See the First solution of Exercise~\ref{exe.counit.unique} as well
as the solution of Exercise~\ref{exe.cocommutative.coalghom} for
other examples of how a proof that uses
elements can be rewritten in an element-free fashion.]

\end{verlong}

(b) We have solved Exercise~\ref{exe.coalg-anti.gen}(a) in an element-free
fashion (in the Second solution to Exercise~\ref{exe.coalg-anti.gen}%
(a) given above)\footnote{To be fully honest, this second solution was not entirely
element-free, since we proved the auxiliary equality
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}) using elements. However,
for the purposes of what we are going to do (reversing arrows), this is not
problematic, since this auxiliary equality
(\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1}) can be easily shown to hold
with all arrows reversed (the proof is more or less the same as for the
original (\ref{sol.coalg-anti.gen.a.sol2.fg.pf.3.pf.1})).}. Thus, by
``reversing all arrows'' in this solution of
Exercise~\ref{exe.coalg-anti.gen}(a), we can obtain a solution to the dual of
Exercise~\ref{exe.coalg-anti.gen}(a). Consequently, the dual of
Exercise~\ref{exe.coalg-anti.gen}(a) holds.

But it is easy to see that the notion of a $\kk$-coalgebra
anti-homomorphism is dual to the notion of a $\kk$-algebra
anti-homomorphism (i.e., is obtained from the latter notion by
``reversing all arrows''), and the notion of
convolution is dual to itself. Hence, the dual of
Exercise~\ref{exe.coalg-anti.gen}(a) is the following exercise:

\begin{statement}
\textit{Exercise~A:} If $C$ is a $\kk$-bialgebra, if $A$ is a
$\kk$-coalgebra, and if $r:A\rightarrow C$ is a $\star$-invertible
$\kk$-coalgebra homomorphism, then prove that the $\star$-inverse
$r^{\star\left(  -1\right)  }$ of $r$ is a $\kk$-coalgebra
anti-homomorphism.
\end{statement}

Exercise~\ref{exe.coalg-anti.gen}(b) immediately follows
from this Exercise~A
(applied to $A$ and $C$ instead of $C$ and $A$).
Exercise~\ref{exe.coalg-anti.gen}(b) is thus solved.

(c) \textit{Alternative proof of Proposition
\ref{antipodes-are-antiendomorphisms} using
Exercise~\ref{exe.coalg-anti.gen}(a).}
Let $A$ be a Hopf algebra.
The antipode $S$ of this Hopf algebra $A$ is
defined as the $\star$-inverse of the identity map
$\id_{A}$; thus,
$S=\id_{A}^{\star\left(  -1\right)  }$.
Hence, the $\kk$-algebra homomorphism
$\id_{A}:A\rightarrow A$ is $\star$-invertible. Therefore,
Exercise~\ref{exe.coalg-anti.gen}(a) (applied to $C=A$ and
$r = \id_{A}$) yields that the $\star$-inverse
$\id_{A}^{\star\left(  -1\right)  }$ of $\id_{A}$
is a $\kk$-algebra anti-homomorphism. In other words, $S$ is a
$\kk$-algebra anti-homomorphism (since $S$ is the $\star$-inverse
$\id_{A}^{\star\left(  -1\right)  }$ of
$\id_{A}$). In other words, $S$ is a
$\kk$-algebra anti-endomorphism of $A$. This proves
Proposition~\ref{antipodes-are-antiendomorphisms}.

\begin{vershort}

\textit{Alternative solution of Exercise~\ref{exe.coalg-anti} using
Exercise~\ref{exe.coalg-anti.gen}(b).}
This is analogous to the proof of
Proposition~\ref{antipodes-are-antiendomorphisms} using
Exercise~\ref{exe.coalg-anti.gen}(a) just shown,
but instead of Exercise~\ref{exe.coalg-anti.gen}(a) we now
need to use Exercise~\ref{exe.coalg-anti.gen}(b).

\end{vershort}

\begin{verlong}

\textit{Alternative solution of Exercise~\ref{exe.coalg-anti} using
Exercise~\ref{exe.coalg-anti.gen}(b).}
Let $A$ be a Hopf algebra. The antipode $S$ of
this Hopf algebra $A$ is defined as the $\star$-inverse of the identity map
$\id_{A}$; thus, $S = \id_{A}^{\star\left(  -1\right)  }$.
Hence, the $\kk$-coalgebra homomorphism
$\id_{A}:A\rightarrow A$ is $\star$-invertible.
Therefore, Exercise~\ref{exe.coalg-anti.gen}(b) (applied to $C=A$ and
$r=\id_{A}$) yields that the $\star$-inverse
$\id_{A}^{\star\left(  -1\right)  }$ of
$\id_{A}$ is a $\kk$-coalgebra
anti-homomorphism. In other words, $S$ is a $\kk$-coalgebra
anti-homomorphism (since $S$ is the $\star$-inverse
$\id_{A}^{\star\left(  -1\right)  }$ of $\id_{A}$).
In other words, $S$ is a $\kk$-coalgebra anti-endomorphism of $A$.
This solves Exercise~\ref{exe.coalg-anti}.

\end{verlong}

(d) \textit{Alternative proof of
Corollary~\ref{commutative-implies-involutive-antipode-cor}
using Proposition~\ref{antipodes-give-convolution-inverses}.} Let $A$ be a
commutative Hopf algebra.
Then, a $\kk$-algebra anti-homomorphism
$A \to A$ is the same thing as a $\kk$-algebra homomorphism $A \to A$
(since $A$ is commutative).
(This follows from Exercise~\ref{exe.comm-cocomm.anti}(a), but it is also
nearly trivial to check by hand.)

The $\kk$-linear map $\id_A : A \to A$ is
$\star$-invertible (since $A$ is a Hopf algebra), and its $\star$-inverse
$\id_A^{\star \left(-1\right)}$ is the antipode $S$ of $A$. That is,
$\id_A^{\star \left(-1\right)} = S$. Applying
Exercise~\ref{exe.coalg-anti.gen}(a) to $C = A$ and $r = \id_A$, we now
conclude that $\id_A^{\star \left(-1\right)}$ is a $\kk$-algebra
anti-homomorphism $A \to A$ (since $\id_A$ is a $\kk$-algebra
homomorphism $A \to A$). In other words,
$\id_A^{\star \left(-1\right)}$ is a $\kk$-algebra homomorphism $A \to A$
(since a $\kk$-algebra anti-homomorphism
$A \to A$ is the same thing as a $\kk$-algebra homomorphism $A \to A$).
In other words, $S$ is a $\kk$-algebra homomorphism $A \to A$ (since
$\id_A^{\star \left(-1\right)} = S$). Now,
Proposition~\ref{antipodes-give-convolution-inverses}(a) (applied to
$H = A$ and $\alpha = S$) yields $S \circ S = S^{\star \left(-1\right)}$.
But $S^{\star \left(-1\right)} = \id_A$ (since
$S = \id_A^{\star \left(-1\right)}$). Hence,
$S^2 = S \circ S = S^{\star \left(-1\right)} = \id_A$. This proves
Corollary~\ref{commutative-implies-involutive-antipode-cor}.
Exercise~\ref{exe.coalg-anti.gen}(d) is solved.

(e) Let us first prepare some general properties of maps between graded
$\kk$-modules. We begin with two definitions:

\begin{definition}
\label{def.graded.mod.pi-iota}Let $V=\bigoplus_{n\in\NN }V_{n}$ be a
graded $\kk$-module (where the $V_{n}$ are the homogeneous components
of $V$). Let $n\in\NN $. Then:

\begin{enumerate}
\item[(a)] We shall let $\pi_{n,V}:V\rightarrow V_{n}$ denote the canonical
projection from $V$ to its $n$-th graded component $V_{n}$.
(This has already been defined in Definition~\ref{def.graded.mod.pi};
we are just repeating it here for convenience.)

\item[(b)] We shall let $\iota_{n,V}:V_{n}\rightarrow V$ denote the inclusion
map from $V_{n}$ to $V$.
\end{enumerate}
\end{definition}

\begin{definition}
\label{def.graded.overlinef}
Let $V=\bigoplus_{n\in\NN }V_{n}$ and
$W=\bigoplus_{n\in\NN }W_{n}$ be two graded $\kk$-modules (where
the $V_{n}$ and the $W_{n}$ are the homogeneous components of $V$ and $W$,
respectively). Let $f:V\rightarrow W$ be a $\kk$-linear map. For each
$n\in\NN $, let $\overline{f}_{n}:V_{n}\rightarrow W_{n}$ be the
$\kk$-linear map $\pi_{n,W}\circ f\circ\iota_{n,V}:V_{n}\rightarrow
W_{n}$. We let $\overline{f}$ be the direct sum $\bigoplus_{n\in\NN %
}\overline{f}_{n}:\bigoplus_{n\in\NN }V_{n}\rightarrow\bigoplus
_{n\in\NN }W_{n}$ of these $\kk$-linear maps $\overline{f}%
_{n}:V_{n}\rightarrow W_{n}$ over all $n\in\NN $; thus, $\overline{f}$
is a $\kk$-linear map from $V$ to $W$ (since $V=\bigoplus
_{n\in\NN }V_{n}$ and $W=\bigoplus_{n\in\NN }W_{n}$).
\end{definition}

Let us now state a litany of basic properties of this map $\overline{f}$:

\begin{proposition}
\label{prop.graded.overlinef.redef}Let $V$ and $W$ be two graded $\kk %
$-modules. Write $V$ as $V=\bigoplus_{n\in{\NN }}V_{n}$ (where the
$V_{n}$ are the homogeneous components of $V$). Let $f:V\rightarrow W$ be a
$\kk$-linear map. Then,%
\[
\overline{f}\left(  v\right)  =\pi_{n,W}\left(  f\left(  v\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\NN \text{ and each }v\in
V_{n}.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.redef}.]Write $W$ in the form
$W=\bigoplus_{n\in\NN }W_{n}$ (where the $W_{n}$ are the homogeneous
components of $W$). Consider the $\kk$-linear maps $\overline{f}_{n}$
defined in Definition \ref{def.graded.overlinef}; thus, $\overline{f}_{n}%
=\pi_{n,W}\circ f\circ\iota_{n,V}$ for each $n\in\NN $. The map
$\overline{f}$ was defined to be the direct sum $\bigoplus_{n\in\NN %
}\overline{f}_{n}$. Thus, we have%
\begin{equation}
\overline{f}\left(  v\right)  =\overline{f}_{n}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\NN \text{ and each }v\in V_{n}.
\label{pf.prop.graded.overlinef.itself.1}%
\end{equation}
Now, let $n\in\NN $ and $v\in V_{n}$. Then, $\iota_{n,V}\left(
v\right)  =v$ (since $\iota_{n,V}$ is the inclusion map from $V_{n}$ to $V$).
Now, \eqref{pf.prop.graded.overlinef.itself.1} yields%
\[
\overline{f}\left(  v\right)  =\underbrace{\overline{f}_{n}}_{\substack{=\pi
_{n,W}\circ f\circ\iota_{n,V}\\\text{(by the definition of }\overline{f}%
_{n}\text{)}}}\left(  v\right)  =\left(  \pi_{n,W}\circ f\circ\iota
_{n,V}\right)  \left(  v\right)  =\pi_{n,W}\left(  f\left(  \underbrace{\iota
_{n,V}\left(  v\right)  }_{=v}\right)  \right)  =\pi_{n,W}\left(  f\left(
v\right)  \right)  .
\]
This proves Proposition \ref{prop.graded.overlinef.redef}.
\end{proof}

\begin{proposition}
\label{prop.graded.overlinef.itself}Let $V$ and $W$ be two graded $\kk %
$-modules. Let $f:V\rightarrow W$ be a graded $\kk$-linear map. Then,
$\overline{f}=f$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.itself}.]Write $V$ as
$V=\bigoplus_{n\in{\NN }}V_{n}$ (where the $V_{n}$ are the homogeneous
components of $V$). Write $W$ as $W=\bigoplus_{n\in{\NN }}W_{n}$ (where
the $W_{n}$ are the homogeneous components of $W$).

Let $v$ be a homogeneous element of $V$. Thus, there exists some
$n\in \NN$ such that $v\in V_{n}$. Consider this $n$. We have $f\left(
V_{n}\right)  \subset W_{n}$ (since the map $f$ is graded). Hence, $f\left(
\underbrace{v}_{\in V_{n}}\right)  \in f\left(  V_{n}\right)  \subset W_{n}$.

But $\pi_{n,W}\left(  w\right)  =w$ for each $w\in W_{n}$ (since $\pi_{n,W}$
is the canonical projection from $W$ to its $n$-th graded component $W_{n}$).
Applying this to $w=f\left(  v\right)  $, we obtain $\pi_{n,W}\left(  f\left(
v\right)  \right)  =f\left(  v\right)  $ (since $f\left(  v\right)  \in W_{n}%
$). Now, recall that $v\in V_{n}$; hence, Proposition
\ref{prop.graded.overlinef.redef} yields%
\[
\overline{f}\left(  v\right)  =\pi_{n,W}\left(  f\left(  v\right)  \right)
=f\left(  v\right)  .
\]


Forget that we fixed $v$. We thus have shown that $\overline{f}\left(
v\right)  =f\left(  v\right)  $ for each homogeneous element $v$ of $V$.
In other words, the two $\kk$-linear maps $\overline{f}$ and $f$ (from
$V$ to $W$) agree on each homogeneous element of $V$.

But $V$ is a graded $\kk$-module. Hence, each element of $V$ is a
$\kk$-linear combination of homogeneous elements of $V$. Therefore, if
two $\kk$-linear maps from $V$ agree on each homogeneous element of
$V$, then these two maps must be equal. Hence, the two $\kk$-linear
maps $\overline{f}$ and $f$ (from $V$ to $W$) must be equal (because we have
shown that they agree on each homogeneous element of $V$). In other words,
$\overline{f}=f$. This proves Proposition \ref{prop.graded.overlinef.itself}.
\end{proof}

\begin{proposition}
\label{prop.graded.overlinef.graded}Let $V$ and $W$ be two graded $\kk %
$-modules. Let $f:V\rightarrow W$ be a $\kk$-linear map. Then, the
$\kk$-linear map $\overline{f}:V\rightarrow W$ is graded.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.graded}.]Write $V$ and $W$ in
the form $V=\bigoplus_{n\in\NN }V_{n}$ and $W=\bigoplus_{n\in\NN %
}W_{n}$ (where the $V_{n}$ and the $W_{n}$ are the homogeneous components of
$V$ and $W$, respectively). Consider the $\kk$-linear maps
$\overline{f}_{n}$ defined in Definition \ref{def.graded.overlinef}; thus,
$\overline{f}_{n}=\pi_{n,W}\circ f\circ\iota_{n,V}$ for each $n\in\NN $.
The map $\overline{f}$ was defined to be the direct sum $\bigoplus
_{n\in\NN }\overline{f}_{n}$. Thus, we have%
\begin{equation}
\overline{f}\left(  v\right)  =\overline{f}_{n}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\NN \text{ and each }v\in V_{n}.
\label{pf.prop.graded.overlinef.graded.1}%
\end{equation}


Let $n\in\NN $. We have $\pi_{n,W}\left(  W\right)  \subset W_{n}$
(since $\pi_{n,W}$ is the canonical projection from $W$ to its $n$-th graded
component $W_{n}$). Now,
\begin{align*}
\overline{f}\left(  V_{n}\right)
&= \left\{ \underbrace{\overline{f}\left(v\right)}_{\substack{
            = \overline{f}_n\left(v\right) \\
            \text{(by \eqref{pf.prop.graded.overlinef.graded.1})} }}
           \mid v \in V_n \right\}
= \left\{ \overline{f}_n\left(v\right) \mid v \in V_n \right\}
=\underbrace{\overline{f}_{n}}%
_{\substack{=\pi_{n,W}\circ f\circ\iota_{n,V}\\\text{(by the definition of
}\overline{f}_{n}\text{)}}}\left(  V_{n}\right)  \\
&= \left(  \pi_{n,W}\circ
f\circ\iota_{n,V}\right)  \left(  V_{n}\right)  =\pi_{n,W}\left(
\underbrace{f\left(  \iota_{n,V}\left(  V_{n}\right)  \right)  }_{\subset
W}\right)  \subset\pi_{n,W}\left(  W\right)  \subset W_{n}.
\end{align*}


Now, forget that we fixed $n$. We thus have shown that $\overline{f}\left(
V_{n}\right)  \subset W_{n}$ for every $n\in\NN $. In other words, the
$\kk$-linear map $\overline{f}:V\rightarrow W$ is graded. This proves
Proposition \ref{prop.graded.overlinef.graded}.
\end{proof}

\begin{proposition}
\label{prop.graded.overlinef.circ1}Let $U$, $V$ and $W$ be three graded
$\kk$-modules. Let $f:V\rightarrow W$ and $g:U\rightarrow V$ be two
$\kk$-linear maps such that $f$ is graded. Then, $\overline{f\circ
g}=f\circ\overline{g}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.circ1}.]Write $U$ in the form
$U=\bigoplus_{n\in\NN }U_{n}$ (where the $U_{n}$ are the homogeneous
components of $U$).

Every element of $U$ is a $\kk$-linear combination of homogeneous
elements of $U$ (since $U$ is graded). Thus, if two $\kk$-linear maps
from $U$ agree on each homogeneous element of $U$, then these two maps must be equal.

Let $u$ be a homogeneous element of $U$. Thus, there exists some
$n\in \NN$ such that $u\in U_{n}$. Consider this $n$. Then,
Proposition \ref{prop.graded.overlinef.redef} (applied to $U$, $V$, $U_{m}$,
$g$ and $u$ instead of $V$, $W$, $V_{m}$, $f$ and $v$) yields
\begin{equation}
\overline{g}\left(  u\right)  =\pi_{n,V}\left(  g\left(  u\right)  \right)  .
\label{pf.prop.graded.overlinef.circ1.1}%
\end{equation}
Also, $f\circ g:U\rightarrow W$ is a $\kk$-linear map; thus,
Proposition \ref{prop.graded.overlinef.redef} (applied to $U$, $U_{m}$,
$f\circ g$ and $u$ instead of $V$, $V_{m}$, $f$ and $v$) yields
$\overline{f\circ g}\left(  u\right)
=\pi_{n,W}\left(  \left(  f\circ g\right)  \left(  u\right)  \right)  $.
Hence,
\begin{align*}
\overline{f\circ g}\left(  u\right)   &  =\pi_{n,W}\left(  \left(  f\circ
g\right)  \left(  u\right)  \right)
= \pi_{n,W} \left( f\left( g\left( u\right)\right)\right)
= f\left( \underbrace{\pi_{n,V}\left(  g\left(  u\right)  \right)
}_{\substack{=\overline{g}\left(  u\right)  \\\text{(by
\eqref{pf.prop.graded.overlinef.circ1.1})}}} \right)
\\
& \ \ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.graded.pi.commute},
applied to $v = g\left(u\right)$ and $m = n$}\right) \\
&  = f\left(  \overline{g}\left(  u\right)  \right)
=\left(  f\circ\overline{g}\right)  \left(  u\right)  .
\end{align*}


Now, forget that we fixed $u$. We thus have proved that $\overline{f\circ
g}\left(  u\right)  =\left(  f\circ\overline{g}\right)  \left(  u\right)  $
for each homogeneous element $u$ of $U$. In other words, the two maps
$\overline{f\circ g}$ and $f\circ\overline{g}$ (from $U$ to $W$) agree on each
homogeneous element of $U$. Since these two maps are $\kk$-linear, we
can thus conclude that they must be equal (because if two $\kk$-linear
maps from $U$ agree on each homogeneous element of $U$, then these two maps
must be equal). In other words, we have $\overline{f\circ g}=f\circ
\overline{g}$. This proves Proposition \ref{prop.graded.overlinef.circ1}.
\end{proof}

\begin{proposition}
\label{prop.graded.overlinef.circ2}Let $U$, $V$ and $W$ be three graded
$\kk$-modules. Let $f:V\rightarrow W$ and $g:U\rightarrow V$ be two
$\kk$-linear maps such that $g$ is graded. Then, $\overline{f\circ
g}=\overline{f}\circ g$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.circ2}.]Write $U$ and $V$ in
the forms $U=\bigoplus_{n\in\NN }U_{n}$ and $V=\bigoplus_{n\in
\NN }V_{n}$ (where the $U_{n}$ and $V_{n}$ are the homogeneous
components of $U$ and $V$, respectively).

Every element of $U$ is a $\kk$-linear combination of homogeneous
elements of $U$ (since $U$ is graded). Thus, if two $\kk$-linear maps
from $U$ agree on each homogeneous element of $U$, then these two maps must be equal.

Let $u$ be a homogeneous element of $U$. Thus, there exists some
$n\in \NN$ such that $u\in U_{n}$. Consider this $n$. From $u\in
U_{n}$, we obtain $g\left(  u\right)  \in g\left(  U_{n}\right)  \subset
V_{n}$ (since $g$ is a graded map). Hence, Proposition
\ref{prop.graded.overlinef.redef} (applied to $v=g\left(  u\right)  $) yields%
\begin{equation}
\overline{f}\left(  g\left(  u\right)  \right)  =\pi_{n,W}\left(  f\left(
g\left(  u\right)  \right)  \right)  .
\label{pf.prop.graded.overlinef.circ2.1}%
\end{equation}
Also, $f\circ g:U\rightarrow W$ is a $\kk$-linear map; thus,
Proposition \ref{prop.graded.overlinef.redef} (applied to $U$, $U_{m}$,
$f\circ g$ and $u$ instead of $V$, $V_{m}$, $f$ and $v$) yields
$\overline{f\circ g}\left(  u\right)
=\pi_{n,W}\left(  \left(  f\circ g\right)  \left(
u\right)  \right)  =\pi_{n,W}\left(  f\left(  g\left(  u\right)  \right)
\right)  $. Comparing this with \eqref{pf.prop.graded.overlinef.circ2.1}, we
find $\overline{f\circ g}\left(  u\right)  =\overline{f}\left(  g\left(
u\right)  \right)  =\left(  \overline{f}\circ g\right)  \left(  u\right)  $.

Now, forget that we fixed $u$. We thus have proved that $\overline{f\circ
g}\left(  u\right)  =\left(  \overline{f}\circ g\right)  \left(  u\right)  $
for each homogeneous element $u$ of $U$. In other words, the two maps
$\overline{f\circ g}$ and $\overline{f}\circ g$ (from $U$ to $W$) agree on
each homogeneous element of $U$. Since these two maps are $\kk$-linear,
we can thus conclude that they must be equal (because if two $\kk %
$-linear maps from $U$ agree on each homogeneous element of $U$, then these
two maps must be equal). In other words, we have $\overline{f\circ
g}=\overline{f}\circ g$. This proves Proposition
\ref{prop.graded.overlinef.circ2}.
\end{proof}

\begin{remark}
Proposition \ref{prop.graded.overlinef.circ1} and Proposition
\ref{prop.graded.overlinef.circ2} cannot be generalized to \textquotedblleft%
$\overline{f\circ g}=\overline{f}\circ\overline{g}$ for any two (not
necessarily graded) $\kk$-linear maps $f:V\rightarrow W$ and
$g:U\rightarrow V$\textquotedblright. Counterexamples are easy to find (e.g.,
take $U=\kk \left[  x\right]  $, $V=\kk \left[  x\right]  $,
$W=\kk \left[  x\right]  $, $f\left(  x^{i}\right)  =x^{i+1}$, $g\left(
x^{j}\right)  =%
\begin{cases}
x^{j-1}, & \text{if }j>0;\\
0, & \text{if }j=0
\end{cases}
$).
\end{remark}

\begin{proposition}
\label{prop.graded.overlinef.tens2}Let $U$, $V$, $X$ and $Y$ be four graded
$\kk$-modules. Let $f:U\rightarrow V$ and $g:X\rightarrow Y$ be two
$\kk$-linear maps such that $f$ is graded. Then, $\overline{g\otimes
f}=\overline{g}\otimes f$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.tens2}.]Write $U$, $V$, $X$
and $Y$ in the forms $U=\bigoplus_{n\in\NN }U_{n}$, $V=\bigoplus
_{n\in\NN }V_{n}$, $X=\bigoplus_{n\in\NN }X_{n}$ and
$Y=\bigoplus_{n\in\NN }Y_{n}$ (where the $U_{n}$, $V_{n}$, $X_{n}$ and
$Y_{n}$ are the homogeneous components of $U$, $V$, $X$ and $Y$,
respectively). Write $X\otimes U$ and $Y\otimes V$ in the forms $X\otimes
U=\bigoplus_{n\in\NN }\left(  X\otimes U\right)  _{n}$ and $Y\otimes
V=\bigoplus_{n\in\NN }\left(  Y\otimes V\right)  _{n}$ (where the
$\left(  X\otimes U\right)  _{n}$ and $\left(  Y\otimes V\right)  _{n}$ are
the homogeneous components of $X\otimes U$ and $Y\otimes V$, respectively).
Recall that the grading on $X\otimes U$ is defined in such a way that%
\begin{equation}
\left(  X\otimes U\right)  _{n}=\bigoplus_{\substack{\left(  i,j\right)
\in\NN ^{2};\\i+j=n}}X_{i}\otimes U_{j}%
\label{pf.prop.graded.overlinef.tens2.XUn}%
\end{equation}
for each $n\in\NN $. Likewise, the grading on $Y\otimes V$ is defined in
such a way that%
\begin{equation}
\left(  Y\otimes V\right)  _{n}=\bigoplus_{\substack{\left(  i,j\right)
\in\NN ^{2};\\i+j=n}}Y_{i}\otimes V_{j}%
\label{pf.prop.graded.overlinef.tens2.YVn}%
\end{equation}
for each $n\in\NN $.

If $n\in\NN $, then%
\begin{equation}
\pi_{n,X}\left(  X_{m}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
m\in\NN \text{ satisfying }m\neq
n\label{pf.prop.graded.overlinef.tens2.4}%
\end{equation}
(since the map $\pi_{n,X}$ is the canonical projection from $X$ to its $n$-th
graded component $X_{n}$, and thus annihilates all graded components of $X$
other than $X_{n}$). The same reasoning (applied to $X\otimes U$ and
$\left(X \otimes U\right)_m$ instead of
$X$ and $X_m$) shows that if $n\in\NN $, then%
\begin{equation}
\pi_{n,X\otimes U}\left(  \left(  X\otimes U\right)  _{m}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\NN \text{ satisfying }m\neq
n.\label{pf.prop.graded.overlinef.tens2.5}%
\end{equation}


If $n\in\NN $, then%
\begin{equation}
\pi_{n,Y}\left(  Y_{m}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
m\in\NN \text{ satisfying }m\neq
n\label{pf.prop.graded.overlinef.tens2.4b}%
\end{equation}
(since the map $\pi_{n,Y}$ is the canonical projection from $Y$ to its $n$-th
graded component $Y_{n}$, and thus annihilates all graded components of $Y$
other than $Y_{n}$). The same reasoning (applied to $Y\otimes V$ and
$\left(Y \otimes V\right)_m$ instead of $Y$ and $Y_m$)
shows that if $n\in\NN $, then
\begin{equation}
\pi_{n,Y\otimes V}\left(  \left(  Y\otimes V\right)  _{m}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\NN \text{ satisfying }m\neq
n.\label{pf.prop.graded.overlinef.tens2.5b}%
\end{equation}


We shall now prove the following:

\begin{statement}
\textit{Claim 1:} Let $q\in\NN $. Let $v\in V_{q}$. Let $y\in Y$. Let
$p\in\NN $. Then,
\[
\pi_{p+q,Y\otimes V}\left(  y\otimes v\right)  =\pi_{p,Y}\left(  y\right)
\otimes v.
\]

\end{statement}

[\textit{Proof of Claim 1:} We have $y\in Y=\bigoplus_{n\in\NN }Y_{n}$.
Hence, we can write $y$ in the form $y=\sum_{n\in\NN }y_{n}$ for some
family $\left(  y_{n}\right)  _{n\in\NN }\in\prod_{n\in\NN }Y_{n}$
with the property that all but finitely many $n\in\NN $ satisfy
$y_{n}=0$. Consider this family $\left(  y_{n}\right)  _{n\in\NN }$.
Note that%
\begin{equation}
y_{n}\in Y_{n}\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\NN %
\label{pf.prop.graded.overlinef.tens2.c1.pf.1}%
\end{equation}
(since $\left(  y_{n}\right)  _{n\in\NN }\in\prod_{n\in\NN }Y_{n}$).

It is easy to see that%
\begin{equation}
y_{n}\otimes v\in\left(  Y\otimes V\right)  _{n+q}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\NN %
\label{pf.prop.graded.overlinef.tens2.c1.pf.2}%
\end{equation}
\footnote{\textit{Proof of \eqref{pf.prop.graded.overlinef.tens2.c1.pf.2}:}
Let $n\in\NN $. Applying \eqref{pf.prop.graded.overlinef.tens2.YVn} to
$n+q$ instead of $n$, we find%
\begin{equation}
\left(  Y\otimes V\right)  _{n+q}=\bigoplus_{\substack{\left(  i,j\right)
\in\NN ^{2};\\i+j=n+q}}Y_{i}\otimes V_{j}%
.\label{pf.prop.graded.overlinef.tens2.c1.pf.2.pf.1}%
\end{equation}
But $Y_{n}\otimes V_{q}$ is an addend of the direct sum $\bigoplus
_{\substack{\left(  i,j\right)  \in\NN ^{2};\\i+j=n+q}}Y_{i}\otimes
V_{j}$ (namely, the addend for $\left(  i,j\right)  =\left(  n,q\right)  $),
since $\left(  n,q\right)  \in\NN ^{2}$ and $n+q=n+q$. Hence,
$Y_{n}\otimes V_{q}\subset\bigoplus_{\substack{\left(  i,j\right)
\in\NN ^{2};\\i+j=n+q}}Y_{i}\otimes V_{j}$. In view of
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.2.pf.1}, this rewrites as
$Y_{n}\otimes V_{q}\subset\left(  Y\otimes V\right)  _{n+q}$. But
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.1} yields $y_{n}\in Y_{n}$.
Combining this with $v\in V_{q}$, we find $y_{n}\otimes v\in Y_{n}\otimes
V_{q}\subset\left(  Y\otimes V\right)  _{n+q}$. This proves
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.2}.}.

It is easy to see that%
\begin{equation}
\pi_{p,Y}\left(  y\right)  =y_{p}%
\label{pf.prop.graded.overlinef.tens2.c1.pf.3}%
\end{equation}
\ \footnote{\textit{Proof.} Applying the map $\pi_{p,Y}$ to both sides of the
equality $y=\sum_{n\in\NN }y_{n}$, we obtain
\begin{align}
\pi_{p,Y}\left(  y\right)   &  =\pi_{p,Y}\left(  \sum_{n\in\NN }%
y_{n}\right)  =\sum_{n\in\NN }\pi_{p,Y}\left(  y_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi_{p,Y}\text{ is
}\kk \text{-linear}\right)  \nonumber\\
&  =\pi_{p,Y}\left(  y_{p}\right)  +\sum_{\substack{n\in\NN ;\\n\neq
p}}\pi_{p,Y}\left(  y_{n}\right)
\label{pf.prop.graded.overlinef.tens2.c1.pf.fn1.1}%
\end{align}
(here, we have split off the addend for $n=p$ from the sum).
\par
But \eqref{pf.prop.graded.overlinef.tens2.c1.pf.1} (applied to $n=p$) yields
$y_{p}\in Y_{p}$. Recall that $\pi_{p,Y}$ is the canonical projection from $Y$
to its $p$-th graded component $Y_{p}$. Hence, $\pi_{p,Y}$ fixes every element
of $Y_{p}$. In other words, we have $\pi_{p,Y}\left(  a\right)  =a$ for each
$a\in Y_{p}$. Applying this to $a=y_{p}$, we obtain $\pi_{p,Y}\left(
y_{p}\right)  =y_{p}$.
\par
The map $\pi_{p,Y}$ is the canonical projection from $Y$ to its $p$-th graded
component $Y_{p}$. Hence, $\pi_{p,Y}$ annihilates all graded components of $Y$
other than $Y_{p}$. In other words,
\begin{equation}
\pi_{p,Y}\left(  Y_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\NN \text{ satisfying }n\neq
p.\label{pf.prop.graded.overlinef.tens2.c1.pf.fn1.2}%
\end{equation}
Now, each $n\in\NN $ satisfying $n\neq p$ satisfies%
\[
\pi_{p,Y}\left(  \underbrace{y_{n}}_{\substack{\in Y_{n}\\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.1})}}}\right)  \in\pi_{p,Y}\left(
Y_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn1.2}}\right)
\]
and thus%
\begin{equation}
\pi_{p,Y}\left(  y_{n}\right)
=0.\label{pf.prop.graded.overlinef.tens2.c1.pf.fn1.3}%
\end{equation}
\par
Now, \eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn1.1} becomes%
\[
\pi_{p,Y}\left(  y\right)  =\underbrace{\pi_{p,Y}\left(  y_{p}\right)
}_{=y_{p}}+\sum_{\substack{n\in\NN ;\\n\neq p}}\underbrace{\pi
_{p,Y}\left(  y_{n}\right)  }_{\substack{=0\\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn1.3})}}}=y_{p}+\underbrace{\sum
_{\substack{n\in\NN ;\\n\neq p}}0}_{=0}=y_{p}.
\]
}. We can also see that
\[
\pi_{p+q,Y\otimes V}\left(  y\otimes v\right)  =y_{p}\otimes v
\]
\footnote{\textit{Proof.} We have%
\[
\underbrace{y}_{=\sum_{n\in\NN }y_{n}}\otimes v=\left(  \sum
_{n\in\NN }y_{n}\right)  \otimes v=\sum_{n\in\NN }y_{n}\otimes v.
\]
Applying the map $\pi_{p+q,Y\otimes V}$ to both sides of this equality, we
obtain
\begin{align}
\pi_{p+q,Y\otimes V}\left(  y\otimes v\right)   &  =\pi_{p+q,Y\otimes
V}\left(  \sum_{n\in\NN }y_{n}\otimes v\right)  =\sum_{n\in\NN %
}\pi_{p+q,Y\otimes V}\left(  y_{n}\otimes v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi_{p+q,Y\otimes V}\text{ is
}\kk \text{-linear}\right)  \nonumber\\
&  =\pi_{p+q,Y\otimes V}\left(  y_{p}\otimes v\right)  +\sum_{\substack{n\in
\NN ;\\n\neq p}}\pi_{p+q,Y\otimes V}\left(  y_{n}\otimes v\right)
\label{pf.prop.graded.overlinef.tens2.c1.pf.fn2.1}%
\end{align}
(here, we have split off the addend for $n=p$ from the sum).
\par
But \eqref{pf.prop.graded.overlinef.tens2.c1.pf.2} (applied to $n=p$) yields
$y_{p}\otimes v\in\left(  Y\otimes V\right)  _{p+q}$. Recall that
$\pi_{p+q,Y\otimes V}$ is the canonical projection from $Y\otimes V$ to its
$\left(  p+q\right)  $-th graded component $\left(  Y\otimes V\right)  _{p+q}%
$. Hence, $\pi_{p+q,Y\otimes V}$ fixes every element of $\left(  Y\otimes
V\right)  _{p+q}$. In other words, we have $\pi_{p+q,Y\otimes V}\left(
a\right)  =a$ for each $a\in\left(  Y\otimes V\right)  _{p+q}$. Applying this
to $a=y_{p}\otimes v$, we obtain $\pi_{p+q,Y\otimes V}\left(  y_{p}\otimes
v\right)  =y_{p}\otimes v$.
\par
The map $\pi_{p+q,Y\otimes V}$ is the canonical projection from $Y\otimes V$
to its $\left(  p+q\right)  $-th graded component $\left(  Y\otimes V\right)
_{p+q}$. Hence, $\pi_{p+q,Y\otimes V}$ annihilates all graded components of
$Y\otimes V$ other than $\left(  Y\otimes V\right)  _{p+q}$. In other words,
\begin{equation}
\pi_{p+q,Y\otimes V}\left(  \left(  Y\otimes V\right)  _{m}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\NN \text{ satisfying }m\neq
p+q.\label{pf.prop.graded.overlinef.tens2.c1.pf.fn2.2a}%
\end{equation}
Now, if $n\in\NN $ satisfies $n\neq p$, then $n+q\neq p+q$ (since $n\neq
p$) and therefore%
\begin{equation}
\pi_{p+q,Y\otimes V}\left(  \left(  Y\otimes V\right)  _{n+q}\right)
=0\label{pf.prop.graded.overlinef.tens2.c1.pf.fn2.2b}%
\end{equation}
(by \eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn2.2a}, applied to $m=n+q$).
\par
Now, each $n\in\NN $ satisfying $n\neq p$ satisfies%
\[
\pi_{p+q,Y\otimes V}\left(  \underbrace{y_{n}\otimes v}_{\substack{\in\left(
Y\otimes V\right)  _{n+q}\\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.2})}}}\right)  \in\pi
_{p+q,Y\otimes V}\left(  \left(  Y\otimes V\right)  _{n+q}\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn2.2b}}\right)
\]
and thus%
\begin{equation}
\pi_{p+q,Y\otimes V}\left(  y_{n}\otimes v\right)
=0.\label{pf.prop.graded.overlinef.tens2.c1.pf.fn2.3}%
\end{equation}
\par
Now, \eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn2.1} becomes%
\[
\pi_{p+q,Y\otimes V}\left(  y\otimes v\right)  =\underbrace{\pi_{p+q,Y\otimes
V}\left(  y_{p}\otimes v\right)  }_{=y_{p}\otimes v}+\sum_{\substack{n\in
\NN ;\\n\neq p}}\underbrace{\pi_{p+q,Y\otimes V}\left(  y_{n}\otimes
v\right)  }_{\substack{=0\\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.fn2.3})}}}=y_{p}\otimes
v+\underbrace{\sum_{\substack{n\in\NN ;\\n\neq p}}0}_{=0}=y_{p}\otimes
v.
\]
}. Hence,%
\[
\pi_{p+q,Y\otimes V}\left(  y\otimes v\right)  =\underbrace{y_{p}%
}_{\substack{=\pi_{p,Y}\left(  y\right)  \\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c1.pf.3})}}}\otimes v=\pi_{p,Y}\left(
y\right)  \otimes v.
\]
This proves Claim 1.]

\begin{statement}
\textit{Claim 2:} Let $p\in\NN $ and $q\in\NN $. Let $x\in X_{p}$
and $u\in U_{q}$. Then,
\[
\overline{g\otimes f}\left(  x\otimes u\right)  =\left(  \overline{g}\otimes
f\right)  \left(  x\otimes u\right)  .
\]

\end{statement}

[\textit{Proof of Claim 2:} We have $u\in U_{q}$, thus $f\left(  u\right)  \in
f\left(  U_{q}\right)  \subset V_{q}$ (since $f$ is graded).

We have $x\in X_{p}$. Hence, Proposition \ref{prop.graded.overlinef.redef}
(applied to $X$, $Y$, $X_{m}$, $g$, $p$ and $x$ instead of $V$, $W$, $V_{m}$,
$f$, $n$ and $v$) yields $\overline{g}\left(  x\right)  =\pi_{p,Y}\left(
g\left(  x\right)  \right)  $. Hence, $\pi_{p,Y}\left(  g\left(  x\right)
\right)  =\overline{g}\left(  x\right)  $.

Applying \eqref{pf.prop.graded.overlinef.tens2.XUn} to $n=p+q$, we find%
\begin{equation}
\left(  X\otimes U\right)  _{p+q}=\bigoplus_{\substack{\left(  i,j\right)
\in\NN ^{2};\\i+j=p+q}}X_{i}\otimes U_{j}%
.\label{pf.prop.graded.overlinef.tens2.c2.pf.1}%
\end{equation}
But $X_{p}\otimes U_{q}$ is an addend of the direct sum $\bigoplus
_{\substack{\left(  i,j\right)  \in\NN ^{2};\\i+j=p+q}}X_{i}\otimes
U_{j}$ (namely, the addend for $\left(  i,j\right)  =\left(  p,q\right)  $),
since $\left(  p,q\right)  \in\NN ^{2}$ and $p+q=p+q$. Hence,
$X_{p}\otimes U_{q}\subset\bigoplus_{\substack{\left(  i,j\right)
\in\NN ^{2};\\i+j=p+q}}X_{i}\otimes U_{j}$. In view of
\eqref{pf.prop.graded.overlinef.tens2.c2.pf.1}, this rewrites as $X_{p}\otimes
U_{q}\subset\left(  X\otimes U\right)  _{p+q}$.

From $x\in X_{p}$ and $u\in U_{q}$, we obtain $x\otimes u\in X_{p}\otimes
U_{q}\subset\left(  X\otimes U\right)  _{p+q}$. Hence, Proposition
\ref{prop.graded.overlinef.redef} (applied to $X\otimes U$, $Y\otimes V$,
$\left(  X\otimes U\right)  _{m}$, $g\otimes f$, $p+q$ and $x\otimes u$
instead of $V$, $W$, $V_{m}$, $f$, $n$ and $v$) yields
\begin{align*}
\overline{g\otimes f}\left(  x\otimes u\right)    & =\pi_{p+q,Y\otimes
V}\left(  \underbrace{\left(  g\otimes f\right)  \left(  x\otimes u\right)
}_{=g\left(  x\right)  \otimes f\left(  u\right)  }\right)  =\pi_{p+q,Y\otimes
V}\left(  g\left(  x\right)  \otimes f\left(  u\right)  \right)  \\
& =\underbrace{\pi_{p,Y}\left(  g\left(  x\right)  \right)  }_{=
\overline{g}\left(  x\right)  }\otimes f\left(  u\right)
\ \ \ \ \ \ \ \ \ \ \left(
\text{by Claim 1, applied to }y=g\left(  x\right)  \text{ and }v=f\left(
u\right)  \right)  \\
& =\overline{g}\left(  x\right)  \otimes f\left(  u\right)  =\left(
\overline{g}\otimes f\right)  \left(  x\otimes u\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \overline{g}\otimes f\right)
\left(  x\otimes u\right)  =\overline{g}\left(  x\right)  \otimes f\left(
u\right)  \right)  .
\end{align*}
This proves Claim 2.]

\begin{statement}
\textit{Claim 3:} Let $p\in\NN $ and $q\in\NN $. Let $t\in
X_{p}\otimes U_{q}$. Then,
\[
\overline{g\otimes f}\left(  t\right)  =\left(  \overline{g}\otimes f\right)
\left(  t\right)  .
\]

\end{statement}

[\textit{Proof of Claim 3:} We have $t\in X_{p}\otimes U_{q}$. Hence, $t$ is a
$\kk$-linear combination of pure tensors in $X_{p}\otimes U_{q}$. In
other words, we can write $t$ in the form
\begin{equation}
t=\sum_{i=1}^{k}x_{i}\otimes u_{i}%
\label{pf.prop.graded.overlinef.tens2.c3.pf.1}%
\end{equation}
for some $k\in\NN $, some elements $x_{1},x_{2},\ldots,x_{k}$ of $X_{p}$
and some elements $u_{1},u_{2},\ldots,u_{k}$ of $U_{q}$. Consider this $k$,
these $x_{1},x_{2},\ldots,x_{k}$ and these $u_{1},u_{2},\ldots,u_{k}$. For
each $i\in\left\{  1,2,\ldots,k\right\}  $, we have
\begin{equation}
\overline{g\otimes f}\left(  x_{i}\otimes u_{i}\right)
= \left(  \overline{g}\otimes f\right)  \left(  x_{i}\otimes u_{i}\right)
\label{pf.prop.graded.overlinef.tens2.c3.pf.2}%
\end{equation}
(by Claim 2, applied to $x=x_{i}$ and $u=u_{i}$), since $x_{i}\in X_{p}$ and
$u_{i}\in U_{q}$.

Applying the map $\overline{g\otimes f}$ to both sides of the equality
\eqref{pf.prop.graded.overlinef.tens2.c3.pf.1}, we find%
\begin{align*}
\overline{g\otimes f}\left(  t\right)    & =\overline{g\otimes f}\left(
\sum_{i=1}^{k}x_{i}\otimes u_{i}\right)  =\sum_{i=1}^{k}\underbrace{
\overline{g\otimes f}\left(  x_{i}\otimes u_{i}\right)  }_{\substack{=\left(
\overline{g}\otimes f\right)  \left(  x_{i}\otimes u_{i}\right)  \\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c3.pf.2})}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\overline{g\otimes f}\text{ is }\kk \text{-linear}%
\right)  \\
& =\sum_{i=1}^{k}\left(  \overline{g}\otimes f\right)  \left(  x_{i}\otimes
u_{i}\right)  =\left(  \overline{g}\otimes f\right)  \left(  \underbrace{\sum
_{i=1}^{k}x_{i}\otimes u_{i}}_{\substack{=t\\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c3.pf.1})}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\overline{g}\otimes f\text{
is }\kk \text{-linear}\right)  \\
& =\left(  \overline{g}\otimes f\right)  \left(  t\right)  .
\end{align*}
This proves Claim 3.]

\begin{statement}
\textit{Claim 4:} Let $t\in X\otimes U$. Then,%
\[
\overline{g\otimes f}\left(  t\right)  =\left(  \overline{g}\otimes f\right)
\left(  t\right)  .
\]

\end{statement}

[\textit{Proof of Claim 4:} Let $t\in X\otimes U$. Then,%
\begin{align*}
t  & \in\underbrace{X}_{\substack{=\bigoplus_{n\in\NN }X_{n}=\sum
_{n\in\NN }X_{n}=\sum_{p\in\NN }X_{p}\\\text{(here, we have
renamed the}\\\text{summation index }n\text{ as }p\text{)}}}\otimes
\underbrace{U}_{\substack{=\bigoplus_{n\in\NN }U_{n}=\sum_{n\in
\NN }U_{n}=\sum_{q\in\NN }U_{q}\\\text{(here, we have renamed
the}\\\text{summation index }n\text{ as }q\text{)}}}=\left(  \sum
_{p\in\NN }X_{p}\right)  \otimes\left(  \sum_{q\in\NN }%
U_{q}\right)  \\
& =\sum_{p\in\NN }\sum_{q\in\NN }X_{p}\otimes U_{q}=\sum_{\left(
p,q\right)  \in\NN ^{2}}X_{p}\otimes U_{q}.
\end{align*}
Hence, we can write $t$ in the form%
\begin{equation}
t=\sum_{\left(  p,q\right)  \in\NN ^{2}}t_{\left(  p,q\right)
}\label{pf.prop.graded.overlinef.tens2.c5.pf.2}%
\end{equation}
for some family $\left(  t_{\left(  p,q\right)  }\right)  _{\left(
p,q\right)  \in\NN ^{2}}\in\prod_{\left(  p,q\right)  \in\NN ^{2}%
}X_{p}\otimes U_{q}$ such that all but finitely many $\left(  p,q\right)
\in\NN ^{2}$ satisfy $t_{\left(  p,q\right)  }=0$. Consider this family
$\left(  t_{\left(  p,q\right)  }\right)  _{\left(  p,q\right)  \in
\NN ^{2}}$.

Now, we have $\left(  t_{\left(  p,q\right)  }\right)  _{\left(  p,q\right)
\in\NN ^{2}}\in\prod_{\left(  p,q\right)  \in\NN ^{2}}X_{p}\otimes
U_{q}$. In other words, $t_{\left(  p,q\right)  }\in X_{p}\otimes U_{q}$ for
each $\left(  p,q\right)  \in\NN ^{2}$. Hence, for each $\left(
p,q\right)  \in\NN ^{2}$, we have%
\begin{equation}
\overline{g\otimes f}\left(  t_{\left(  p,q\right)  }\right)  =\left(
\overline{g}\otimes f\right)  \left(  t_{\left(  p,q\right)  }\right)
\label{pf.prop.graded.overlinef.tens2.c5.pf.3}%
\end{equation}
(by Claim 3, applied to $t=t_{\left(  p,q\right)  }$). Now, applying the map
$\overline{g\otimes f}$ to both sides of the equality
\eqref{pf.prop.graded.overlinef.tens2.c5.pf.2}, we obtain
\begin{align*}
\overline{g\otimes f}\left(  t\right)    & =\overline{g\otimes f}\left(
\sum_{\left(  p,q\right)  \in\NN ^{2}}t_{\left(  p,q\right)  }\right)
=\sum_{\left(  p,q\right)  \in\NN ^{2}}\underbrace{\overline{g\otimes
f}\left(  t_{\left(  p,q\right)  }\right)  }_{=\left(  \overline{g}\otimes
f\right)  \left(  t_{\left(  p,q\right)  }\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\overline{g\otimes f}\text{
is }\kk \text{-linear}\right)  \\
& =\sum_{\left(  p,q\right)  \in\NN ^{2}}\left(  \overline{g}\otimes
f\right)  \left(  t_{\left(  p,q\right)  }\right)  =\left(
\overline{g}\otimes f\right)  \left(  \underbrace{\sum_{\left(  p,q\right)
\in\NN ^{2}}t_{\left(  p,q\right)  }}_{\substack{=t\\\text{(by
\eqref{pf.prop.graded.overlinef.tens2.c5.pf.2})}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\overline{g}\otimes f\text{
is }\kk \text{-linear}\right)  \\
& =\left(  \overline{g}\otimes f\right)  \left(  t\right)  .
\end{align*}
This proves Claim 4.]

Now, Claim 4 says that $\overline{g\otimes f}\left(  t\right)  =\left(
\overline{g}\otimes f\right)  \left(  t\right)  $ for every $t\in X\otimes U$.
In other words, $\overline{g\otimes f}=\overline{g}\otimes f$. This proves
Proposition \ref{prop.graded.overlinef.tens2}.
\end{proof}

\begin{proposition}
\label{prop.graded.overlinef.tens1}Let $U$, $V$, $X$ and $Y$ be four graded
$\kk$-modules. Let $f:U\rightarrow V$ and $g:X\rightarrow Y$ be two
$\kk$-linear maps such that $f$ is graded. Then, $\overline{f\otimes
g}=f\otimes\overline{g}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.graded.overlinef.tens1}.]This proof is
analogous to the proof of Proposition \ref{prop.graded.overlinef.tens2}. (The
only difference is that the order of the tensorands has been swapped.)
\end{proof}

\begin{remark}
Again, Proposition \ref{prop.graded.overlinef.tens1} and Proposition
\ref{prop.graded.overlinef.tens2} cannot be generalized to the case when
neither $f$ nor $g$ is graded.
\end{remark}

We can now step to the solution of Exercise~\ref{exe.coalg-anti.gen}(e). Let
$C$ be a graded $\kk$-coalgebra; let $A$ be a graded $\kk %
$-algebra; let $r:C\rightarrow A$ be a $\star$-invertible $\kk$-linear
map that is graded. We must prove that the $\star$-inverse $r^{\star\left(
-1\right)  }$ of $r$ is also graded.

The $\star$-inverse $r^{\star\left(  -1\right)  }$ of $r$ exists (since $r$ is
$\star$-invertible), and is a $\kk$-linear map from $C$ to $A$. Let
$s=\overline{r^{\star\left(  -1\right)  }}$ (defined according to Definition
\ref{def.graded.overlinef}). Thus, $\overline{r^{\star\left(  -1\right)  }}=s$.

The $\kk$-linear map $\overline{r^{\star\left(  -1\right)  }%
}:C\rightarrow A$ is graded (by Proposition \ref{prop.graded.overlinef.graded}%
, applied to $V=C$, $W=A$ and $f=r^{\star\left(  -1\right)  }$). In other
words, the $\kk$-linear map $s:C\rightarrow A$ is graded (since
$s=\overline{r^{\star\left(  -1\right)  }}$).

We know that $r^{\star\left(  -1\right)  }$ is the $\star$-inverse of $r$.
Thus, $r^{\star\left(  -1\right)  }\star r=r\star r^{\star\left(  -1\right)
}=u_{A}\circ\epsilon_{C}$ (since $u_{A}\circ\epsilon_{C}$ is the unity of the
convolution algebra $\operatorname*{Hom}\left(  C,A\right)  $). Thus,%
\begin{equation}
u_{A}\circ\epsilon_{C}=r^{\star\left(  -1\right)  }\star r=m_{A}\circ\left(
r^{\star\left(  -1\right)  }\otimes r\right)  \circ\Delta_{C}
\label{sol.coalg-anti.gen.e.1}%
\end{equation}
(by the definition of convolution).

Proposition \ref{prop.graded.overlinef.tens2} (applied to $U=C$, $V=A$, $X=C$,
$Y=A$, $f=r$ and $g=r^{\star\left(  -1\right)  }$) yields
$\overline{r^{\star\left(  -1\right)  }\otimes r}
=\underbrace{\overline{r^{\star\left(
-1\right)  }}}_{=s}\otimes r=s\otimes r$.

The map $m_{A}:A\otimes A\rightarrow A$ is graded (since $A$ is a graded
$\kk$-algebra). Hence, Proposition \ref{prop.graded.overlinef.circ1}
(applied to $U=C\otimes C$, $V=A\otimes A$, $W=A$, $f=m_{A}$ and
$g=r^{\star\left(  -1\right)  }\otimes r$) yields $\overline{m_{A}\circ\left(
r^{\star\left(  -1\right)  }\otimes r\right)  }=m_{A}\circ
\underbrace{\overline{r^{\star\left(  -1\right)  }\otimes r}}_{=s\otimes
r}=m_{A}\circ\left(  s\otimes r\right)  $.

The map $\Delta_{C}:C\rightarrow C\otimes C$ is graded (since $C$ is a graded
$\kk$-coalgebra). Thus, Proposition \ref{prop.graded.overlinef.circ2}
(applied to $U=C$, $V=C\otimes C$, $W=A$, $f=m_{A}\circ\left(  r^{\star\left(
-1\right)  }\otimes r\right)  $ and $g=\Delta_{C}$) yields
\[
\overline{m_{A}\circ\left(  r^{\star\left(  -1\right)  }\otimes r\right)
\circ\Delta_{C}}=\underbrace{\overline{m_{A}\circ\left(  r^{\star\left(
-1\right)  }\otimes r\right)  }}_{=m_{A}\circ\left(  s\otimes r\right)  }%
\circ\Delta_{C}=m_{A}\circ\left(  s\otimes r\right)  \circ\Delta_{C}=s\star r
\]
(since the definition of convolution yields $s\star r=m_{A}\circ\left(
s\otimes r\right)  \circ\Delta_{C}$). In view of
\eqref{sol.coalg-anti.gen.e.1}, we can rewrite this as%
\begin{equation}
\overline{u_{A}\circ\epsilon_{C}}=s\star r. \label{sol.coalg-anti.gen.e.4}%
\end{equation}


But the map $\epsilon_{C}:C\rightarrow\kk $ is graded (since $C$ is a
graded $\kk$-coalgebra), and the map $u_{A}:\kk \rightarrow A$ is
graded as well (since $A$ is a graded $\kk$-algebra). Hence, the
composition $u_{A}\circ\epsilon_{C}:C\rightarrow A$ of these two maps is also
graded. Thus, Proposition \ref{prop.graded.overlinef.itself} (applied to
$V=C$, $W=A$ and $f=u_{A}\circ\epsilon_{C}$) yields $\overline{u_{A}%
\circ\epsilon_{C}}=u_{A}\circ\epsilon_{C}$. Comparing this with
\eqref{sol.coalg-anti.gen.e.4}, we find
\[
s\star r=u_{A}\circ\epsilon_{C}=\left(  \text{the unity of the convolution
algebra }\operatorname*{Hom}\left(  C,A\right)  \right)  .
\]


\begin{commentedout}
Similarly, we can show that $r\star s=\left(  \text{the unity of the
convolution algebra }\operatorname*{Hom}\left(  C,A\right)  \right)  $. (Of
course, in order to show this, we need to use $r\star r^{\star\left(
-1\right)  }=u_{A}\circ\epsilon_{C}$ instead of $r^{\star\left(  -1\right)
}\star r=u_{A}\circ\epsilon_{C}$, and we need to apply Proposition
\ref{prop.graded.overlinef.tens1} instead of Proposition
\ref{prop.graded.overlinef.tens2}.)
\end{commentedout}

This yields that $s$ is the $\star$-inverse of $r$ (since we know that $r$ is
$\star$-invertible). In other words, $s=r^{\star\left(  -1\right)  }$. Since
$s$ is graded, we thus conclude that $r^{\star\left(  -1\right)  }$ is graded.
This solves Exercise~\ref{exe.coalg-anti.gen}(e).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.antipode.inverse}}

\textit{Solution to Exercise~\ref{exe.antipode.inverse}.}
(a) Consider a map $P$ satisfying the
given assumption. Consider also the antipode $S$ of $A$.
Let $T$ be the twist map
$T_{A,A} : A\otimes A\rightarrow A\otimes A,\ c\otimes d\mapsto d\otimes c$.
Then, $T \circ T = \id$.
Moreover, if $f : A \to A$ and $g : A \to A$ are any two
$\kk$-linear maps, then
\begin{align}
T \circ \left(f \otimes g\right) = \left(g \otimes f\right) \circ T .
\label{sol.antipode.inverse.twist-fg}
\end{align}
(Indeed, this is a basic property of the twist map\footnote{%
actually, a particular case of \eqref{sol.coalg-anti.gen.a.sol2.twist}},
and can easily be checked.)

We know from
Exercise~\ref{exe.coalg-anti} that $S$ is a coalgebra anti-endomorphism;
thus, $\Delta \circ S = T \circ \left( S\otimes S\right) \circ \Delta$ and
$\epsilon\circ S=\epsilon$.
Hence,
\[
T \circ \underbrace{\Delta \circ S}_{= T \circ \left( S\otimes S\right) \circ \Delta }
= \underbrace{T \circ T}_{= \id} \circ \left( S\otimes S\right) \circ \Delta
= \left( S\otimes S\right) \circ \Delta ,
\]
so that $\left( S\otimes S\right) \circ \Delta = T \circ \Delta \circ S$.
We also know from Proposition~\ref{antipodes-are-antiendomorphisms} that $S$
is an algebra anti-endomorphism; this can be rewritten as
$S \circ m = m \circ \left(S \otimes S\right) \circ T$ and $S\circ u=u$.
Hence,
\[
\underbrace{S \circ m}_{= m \circ \left(S \otimes S\right) \circ T} \circ T
= m \circ \left(S \otimes S\right) \circ \underbrace{T \circ T}_{= \id}
= m \circ \left(S \otimes S\right) ,
\]
so that $m \circ \left(S \otimes S\right) = S \circ m \circ T$.

On the other hand,
we know that $m\circ\left(  P\otimes\id\right)  \circ T\circ\Delta
=u\circ\epsilon$ (indeed, this is just an element-free rewriting of the
assumption that every $a\in A$ satisfies $\sum_{\left(  a\right)  }P\left(
a_{2}\right)  \cdot a_{1}=u\left(  \epsilon\left(  a\right)  \right)  $).
Now, the definition of convolution shows that
\begin{align*}
\left(  P\circ S\right)  \star S  & =m\circ\underbrace{\left(  \left(  P\circ
S\right)  \otimes S\right)  }_{=\left(  P\otimes\id\right)  \circ\left(
S\otimes S\right)  }\circ\Delta=m\circ\left(  P\otimes\id\right)  \circ
\underbrace{\left(  S\otimes S\right)  \circ\Delta}_{=T\circ\Delta\circ S}\\
& =\underbrace{m\circ\left(  P\otimes\id\right)  \circ T\circ\Delta}
_{=u\circ\epsilon}\circ S=u\circ\underbrace{\epsilon\circ S}_{=\epsilon
}=u\circ\epsilon,
\end{align*}
so that $P\circ S$ is a left $\star$-inverse to $S$. Since $S$ is
$\star$-invertible with $\star$-inverse $\id$, this yields that
$P\circ S=\id$.
Furthermore, the definition of convolution shows that
\begin{align*}
S\star\left(  S\circ P\right)    & =m\circ\underbrace{\left(  S\otimes\left(
S\circ P\right)  \right)  }_{=\left(  S\otimes S\right)  \circ\left( \id\otimes
P\right)  }\circ\Delta=\underbrace{m\circ\left(  S\otimes S\right)  }_{=S\circ
m\circ T}\circ\left(  \id\otimes P\right)  \circ\Delta\\
& =S\circ m\circ\underbrace{T\circ\left(  \id\otimes P\right)  }
_{\substack{=\left(  P\otimes\id\right)  \circ T\\
\text{(by \eqref{sol.antipode.inverse.twist-fg})}}}\circ\Delta
=S\circ\underbrace{m\circ\left(  P\otimes\id\right)
\circ T\circ\Delta}_{=u\circ\epsilon} 
=\underbrace{S\circ u}_{=u}\circ\epsilon=u\circ\epsilon,
\end{align*}
whence $S\circ P$ is a right $\star$-inverse to $S$. Since $S$ is
$\star$-invertible with $\star$-inverse $\id$, this yields that
$S\circ P=\id$. Combined
with $P\circ S=\id$, this yields that $S$ is invertible and its inverse is $P$.

(b) Similar to the solution for (a), the details being left to the reader.

(c) Let $A$ be a connected graded Hopf algebra. Just as a left
$\star$-inverse $S$ to $\id_{A}$ has been constructed in the proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}, we could
construct a $\kk$-linear map $P:A\rightarrow
A$ such that every $a\in A$ satisfies
\[
\sum_{\left(  a\right)  }P\left(  a_{2}\right)  \cdot a_{1}=u\left(
\epsilon\left(  a\right)  \right)  .
\]
By part (a), this yields that the antipode of $A$ is invertible.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.subcoalgebra.summand}}

\textit{Solution to Exercise~\ref{exe.subcoalgebra.summand}.} We will
be maximally explicit in this solution; in particular, we will not regard
inclusions as identities even in cases where we would usually do that. We
will only use the notations $\Delta$ and $\epsilon$ to denote the maps
$\Delta_{C}$ and $\epsilon_{C}$ (not the maps $\Delta_{D}$ and $\epsilon_{D}$,
which we will introduce later).

We know that $D$ is a direct summand of $C$ as a $\kk$-module. In other
words, there exists a $\kk$-submodule $E$ of $C$ such that $C=D\oplus
E$. Fix such an $E$.

Let $i:D\rightarrow C$ be the canonical inclusion map. Let $p:D\oplus
E\rightarrow D$ be the canonical projection from the direct sum $D\oplus E$
onto its summand $D$. Notice that $p$ is a $\kk$-linear map from
$D\oplus E=C$ to $D$, and satisfies $p\circ i=\id_{D}$. Hence,
$i\circ\underbrace{p\circ i}_{=\id_{D}}=i$.

In the statement of the exercise, we have assumed that $\Delta\left(
D\right)  \subset C\otimes D$. Since we don't want to abuse notation, we have
to rewrite this as
$\Delta\left(  D\right)  \subset\left(  \id_{C}\otimes i\right) 
 \left(  C\otimes D\right)  $ (because the $\kk$-submodule of $C\otimes C$
spanned by tensors of the form $c\otimes d$ with
$c\in C$ and $d\in D$ is precisely $\left(  \id_{C}\otimes
i\right)  \left(  C\otimes D\right)  $). Similarly, $\Delta\left(  D\right)
\subset\left(  i\otimes\id_{C}\right)  \left(  D\otimes
C\right)  $.

Now,
\begin{equation}
\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
x\right)  \right)  =x\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\Delta\left(
D\right)  . \label{sol.subcoalgebra.summand.1}
\end{equation}


\textit{Proof of \eqref{sol.subcoalgebra.summand.1}:} Let $x\in\Delta\left(
D\right)  $. Then, $x\in\Delta\left(  D\right)  \subset\left(
\id_{C}\otimes i\right)  \left(  C\otimes D\right)  $. Hence,
there exists some $y\in C\otimes D$ such that
$x=\left(  \id_{C}\otimes i\right)  \left(  y\right)  $. Consider this
$y$. We have
\begin{align*}
\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
\underbrace{x}_{=\left(  \id_{C}\otimes i\right)  \left(
y\right)  }\right)  \right)   &  =\left(  i\otimes i\right)  \left(  \left(
p\otimes p\right)  \left(  \left(  \id_{C}\otimes i\right)
\left(  y\right)  \right)  \right)
=\underbrace{\left(  \left(  i\otimes i\right)  \circ\left(  p\otimes
p\right)  \circ\left(  \id_{C}\otimes i\right)  \right)
}_{=\left(  i\circ p\circ\id_{C}\right)  \otimes\left(  i\circ
p\circ i\right)  }\left(  y\right) \\
&  =\left(  \underbrace{\left(  i\circ p\circ\id_{C}\right)
}_{=\left(  \left(  i\circ p\right)  \circ\id_{C}\right)
}\otimes\underbrace{\left(  i\circ p\circ i\right)  }
_{=i=\id_{C}\circ i}\right)  \left(  y\right)
=\underbrace{\left(  \left(  \left(  i\circ p\right)  \circ
\id_{C}\right)  \otimes\left(  \id_{C}\circ
i\right)  \right)  }_{=\left(  \left(  i\circ p\right)  \otimes
\id_{C}\right)  \circ\left(  \id_{C}\otimes
i\right)  }\left(  y\right) \\
&  =\left(  \left(  \left(  i\circ p\right)  \otimes\id_{C}\right) 
 \circ\left(  \id_{C}\otimes i\right)  \right)
\left(  y\right)
=\left(  \left(  i\circ p\right)  \otimes\id_{C}\right)
\left(  \underbrace{\left(  \id_{C}\otimes i\right)  \left(
y\right)  }_{=x}\right) \\
&  =\left(  \left(  i\circ p\right)  \otimes\id_{C}\right)
\left(  x\right)  .
\end{align*}
On the other hand, $x\in\Delta\left(  D\right)  \subset\left(  i\otimes
\id_{C}\right)  \left(  D\otimes C\right)  $. Thus, there exists
some $z\in D\otimes C$ such that $x=\left(  i\otimes\id_{C}\right) 
 \left(  z\right)  $. Consider this $z$. We have
\begin{align*}
\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
x\right)  \right)   &  =\left(  \left(  i\circ p\right)  \otimes
\id_{C}\right)  \left(  \underbrace{x}_{=\left(  i\otimes
\id_{C}\right)  \left(  z\right)  }\right)
=\left(  \left(  i\circ p\right)  \otimes\id_{C}\right)
\left(  \left(  i\otimes\id_{C}\right)  \left(  z\right)
\right) \\
&  =\underbrace{\left(  \left(  \left(  i\circ p\right)  \otimes
\id_{C}\right)  \circ\left(  i\otimes\id_{C}\right) 
 \right)  }_{=\left(  \left(  i\circ p\right)  \circ i\right)
\otimes\left(  \id_{C}\circ\id_{C}\right)
}\left(  z\right)
=\left(  \underbrace{\left(  \left(  i\circ p\right)  \circ i\right)
}_{=i \circ p \circ i = i}\otimes
\underbrace{\left(  \id_{C}\circ\id_{C}\right)  }
_{=\id_{C}}\right)  \left(  z\right) \\
&  =\left(  i\otimes\id_{C}\right)  \left(  z\right)  =x.
\end{align*}
This proves \eqref{sol.subcoalgebra.summand.1}.

Next, let us define a $\kk$-linear map $\Delta_{D}:D\rightarrow
D\otimes D$ by
\[
\Delta_{D}=\left(  p\otimes p\right)  \circ\Delta\circ i.
\]
Let us also define a $\kk$-linear map $\epsilon_{D}:D\rightarrow
\kk$ by
\[
\epsilon_{D}=\epsilon\circ i.
\]
We will show that $\left(  D,\Delta_{D},\epsilon_{D}\right)  $ is a
$\kk$-coalgebra.\footnote{Recall that we are not going to abbreviate
$\Delta_{D}$ and $\epsilon_{D}$ by $\Delta$ and $\epsilon$; thus, $\Delta$ and
$\epsilon$ still mean the maps $\Delta_{C}$ and $\epsilon_{C}$.} But first,
let us see that
\begin{equation}
\left(  i\otimes i\right)  \circ\Delta_{D}=\Delta\circ i.
\label{sol.subcoalgebra.summand.2}
\end{equation}


\textit{Proof of \eqref{sol.subcoalgebra.summand.2}:} Let $d\in D$. Then,
$i\left(  d\right)  =d$ (since $i$ is just an inclusion map), so that
$\Delta\left(  \underbrace{i\left(  d\right)  }_{=d\in D}\right)  \in
\Delta\left(  D\right)  $. Hence, $\left(  i\otimes i\right)  \left(  \left(
p\otimes p\right)  \left(  \Delta\left(  i\left(  d\right)  \right)  \right)
\right)  =\Delta\left(  i\left(  d\right)  \right)  $ (by
\eqref{sol.subcoalgebra.summand.1}, applied to $x=\Delta\left(  i\left(
d\right)  \right)  $). Now,
\begin{align*}
\left(  \left(  i\otimes i\right)  \circ\underbrace{\Delta_{D}}_{=\left(
p\otimes p\right)  \circ\Delta\circ i}\right)  \left(  d\right)   &  =\left(
\left(  i\otimes i\right)  \circ\left(  p\otimes p\right)  \circ\Delta\circ
i\right)  \left(  d\right) \\
&  =\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
\Delta\left(  i\left(  d\right)  \right)  \right)  \right)  =\Delta\left(
i\left(  d\right)  \right)  =\left(  \Delta\circ i\right)  \left(  d\right)  .
\end{align*}


Now, forget that we fixed $d$. We thus have shown that every $d\in D$
satisfies $\left(  \left(  i\otimes i\right)  \circ\Delta_{D}\right)  \left(
d\right)  =\left(  \Delta\circ i\right)  \left(  d\right)  $. In other words,
$\left(  i\otimes i\right)  \circ\Delta_{D}=\Delta\circ i$. This proves \eqref{sol.subcoalgebra.summand.2}.

Now, let us check that $\left(  D,\Delta_{D},\epsilon_{D}\right)  $ is a
$\kk$-coalgebra. In order to do so, we must check that the diagrams
\begin{equation}
\xymatrix{
& D \otimes D \otimes D & \\
D \otimes D \ar[ur]^{\Delta_D \otimes\id_D} & & D \otimes D \ar[ul]_{\id _D \otimes\Delta_D} \\
& D \ar[ur]^{\Delta_D} \ar[ul]_{\Delta_D} &
}
\label{sol.subcoalgebra.axiom1}
\end{equation}
and
\begin{equation}
\xymatrixcolsep{4pc}\xymatrix{
D \otimes\kk\ar[r]^{\cong} & D & \kk\otimes D \ar [l]_{\cong} \\
D \otimes D \ar[u]^{\id_D \otimes\epsilon_D} & D \ar[l]^{\Delta_D} \ar [u]^{\id_D} \ar[r]_{\Delta_D} & D \otimes D \ar[u]_{\epsilon_D \otimes\id_D}
}
\label{sol.subcoalgebra.axiom2}
\end{equation}
are commutative (where the canonical isomorphisms $\kk \otimes
D\rightarrow D$ and $D\otimes\kk \rightarrow D$ are used in
\eqref{sol.subcoalgebra.axiom2}). Let us start with the first
diagram.

Since the diagram \eqref{coassociativity-diagram} for $C$ is commutative (as
$C$ is a $\kk$-coalgebra), we have $\left(  \Delta\otimes
\id_{C}\right)  \circ\Delta=\left(  \id_{C}
\otimes\Delta\right)  \circ\Delta$. But
\begin{align*}
\underbrace{\left(  i\otimes i\otimes i\right)  \circ\left(  \Delta
_{D}\otimes\id_{D}\right)  }_{\substack{=\left(  \left(
i\otimes i\right)  \otimes i\right)  \circ\left(  \Delta_{D}\otimes
\id_{D}\right)  \\=\left(  \left(  i\otimes i\right)
\circ\Delta_{D}\right)  \otimes\left(  i\circ\id_{D}\right)
}}\circ\Delta_{D}
&= \left(  \underbrace{\left(  \left(  i\otimes i\right)  \circ\Delta
_{D}\right)  }_{\substack{=\Delta\circ i\\\text{(by
\eqref{sol.subcoalgebra.summand.2})}}}\otimes\underbrace{\left(
i\circ\id_{D}\right)  }_{=i=\id_{C}\circ
i}\right)  \circ\Delta_{D}=\underbrace{\left(  \left(  \Delta\circ i\right)
\otimes\left(  \id_{C}\circ i\right)  \right)  }_{=\left(
\Delta\otimes\id_{C}\right)  \circ\left(  i\otimes i\right)
}\circ\Delta_{D}\\
&  =\left(  \Delta\otimes\id_{C}\right)  \circ
\underbrace{\left(  i\otimes i\right)  \circ\Delta_{D}}_{\substack{=\Delta
\circ i\\\text{(by \eqref{sol.subcoalgebra.summand.2})}}}=\left(
\Delta\otimes\id_{C}\right)  \circ\Delta\circ i.
\end{align*}
Using this and the analogously provable
identity $\left(  i\otimes i\otimes
i\right)  \circ\left(  \id_{D}\otimes\Delta_{D}\right)
\circ\Delta_{D}=\left(  \id_{C}\otimes\Delta\right)  \circ
\Delta\circ i$, we obtain
\[
\left(  i\otimes i\otimes i\right)  \circ\left(  \Delta_{D}\otimes
\id_{D}\right)  \circ\Delta_{D}=\underbrace{\left(
\Delta\otimes\id_{C}\right)  \circ\Delta}_{=\left(
\id_{C}\otimes\Delta\right)  \circ\Delta}\circ i=\left(
\id_{C}\otimes\Delta\right)  \circ\Delta\circ i=\left(  i\otimes
i\otimes i\right)  \circ\left(  \id_{D}\otimes\Delta_{D}\right)
\circ\Delta_{D}.
\]
We can cancel the left $i\otimes i\otimes i$ factor from this
equation\footnote{because $i\otimes i\otimes i$ is left-invertible:
\[
\left(  p\otimes p\otimes p\right)  \circ\left(  i\otimes i\otimes i\right)
=\underbrace{\left(  p\circ i\right)  }_{=\id_{D}}
\otimes\underbrace{\left(  p\circ i\right)  }_{=\id_{D}}
\otimes\underbrace{\left(  p\circ i\right)  }_{=\id_{D}}
=\id_{D}\otimes\id_{D}\otimes\id_{D}
=\id_{D\otimes D\otimes D}.
\]
}, and thus obtain $\left(  \Delta_{D}\otimes\id_{D}\right)
\circ\Delta_{D}=\left(  \id_{D}\otimes\Delta_{D}\right)
\circ\Delta_{D}$. In other words, the diagram \eqref{sol.subcoalgebra.axiom1}
is commutative.

Let us now check that the diagram \eqref{sol.subcoalgebra.axiom2} is
commutative. We will only prove this for its left square, leaving the
(completely analogous) right square to the reader. For every $\kk$-module
$V$, we let $\operatorname{can}_{V}$ denote the canonical $\kk$-module
isomorphism $V \otimes \kk \to V$. The upper left horizontal arrow
on diagram \eqref{sol.subcoalgebra.axiom2} is precisely
$\operatorname{can}_D$. Notice that
$\operatorname{can}_{C}\circ\left(
\id_{C}\otimes\epsilon\right)  \circ\Delta = \id_C$ due to the
commutativity of the diagram \eqref{counit-diagram} (which, of course,
commutes since $C$ is a coalgebra).

Now, any two $\kk$-modules $V$ and $W$ and any
$\kk$-linear map $f:V\rightarrow W$ satisfy
\[
f\circ\operatorname{can}_V = \operatorname{can}_W
\circ\left( f\otimes\id_{\kk} \right)
\]
(this is just trivial linear algebra).
This (applied to $V=D$, $W=C$ and $f=i$) yields
$i\circ\operatorname{can}_{D}=\operatorname{can}_{C}\circ\left(  i\otimes
\id_{\kk }\right)  $. Hence,
\begin{align*}
\underbrace{i\circ\operatorname{can}_{D}}
_{=\operatorname{can}_{C}\circ\left(  i\otimes\id_{\kk}\right)  }\circ
\left(  \id_{D}\otimes\epsilon
_{D}\right)  \circ\Delta_{D}
&  =\operatorname{can}_{C}\circ\underbrace{\left(  i\otimes
\id_{\kk }\right)  \circ\left(
\id_{D}\otimes\epsilon_{D}\right)  }
_{=\left(  i\circ\id_{D}\right)
\otimes\left(  \id_{\kk }\circ\epsilon
_{D}\right)}\circ\Delta_{D}
=\operatorname{can}_{C}\circ\left(  \underbrace{\left(
i\circ\id_{D}\right)  }_{=i=\id_{C}\circ i}
\otimes\underbrace{\left(  \id_{\kk }\circ\epsilon_{D}\right)  }
_{=\epsilon_{D}=\epsilon\circ
i}\right)  \circ\Delta_{D}\\
&  =\operatorname{can}_{C}\circ\underbrace{\left(  \left(
\id_{C}\circ i\right)  \otimes\left(  \epsilon\circ
i\right)  \right)  }_{=\left(  \id_{C}
\otimes\epsilon\right)  \circ\left(  i\otimes i\right)}\circ\Delta_{D}
=\operatorname{can}_{C} \circ\left(  \id_{C}\otimes\epsilon\right) 
 \circ\underbrace{\left(  \left(  i\otimes
i\right)  \circ\Delta_{D}\right)  }_{\substack{=\Delta\circ i\\\text{(by
\eqref{sol.subcoalgebra.summand.2})}}}\\
&  =\underbrace{\operatorname{can}_{C}\circ\left(
\id_{C}\otimes\epsilon\right)  \circ\Delta }_{=\id_{C}}\circ i
= \id_{C} \circ i = i.
\end{align*}
We can cancel the $i$ factors from this equation (because $i$ is
left-invertible: $p \circ i = \id_D$), and thus are left with
$\operatorname{can}_{D}
 \circ\left(  \id_{D}\otimes\epsilon_{D}\right) \circ\Delta_{D}=\id_{D}$.
This means precisely that the left square of
\eqref{sol.subcoalgebra.axiom2} is commutative. As we said, this
completes the verification of the fact that $\left(  D,\Delta
_{D},\epsilon_{D}\right)  $ must be a $\kk$-coalgebra. We will denote
this $\kk$-coalgebra simply by $D$.

Now, the diagrams
\[
\xymatrix{ D \ar[d]_{\Delta_D} \ar[r]^{i} & C \ar[d]^{\Delta} \\ D \otimes D \ar[r]^{i \otimes i} & C \otimes C }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ D \ar[dr]_{\epsilon_D} \ar[rr]^{i} & & C \ar[dl]^{\epsilon} \\ & \kk & }
\]
are commutative\footnote{In fact, the commutativity of the first of these
diagrams follows from \eqref{sol.subcoalgebra.summand.2}, whereas the
commutativity of the second diagram follows from $\epsilon_{D}=\epsilon\circ
i$.}. Hence, $i$ is a $\kk$-coalgebra homomorphism (by the definition
of a $\kk$-coalgebra homomorphism). In other words, the canonical
inclusion map $D\rightarrow C$ is a $\kk$-coalgebra homomorphism (since
$i$ is the canonical inclusion map $D\rightarrow C$).

So we know that $D$ is a $\kk$-coalgebra such that $D\subset C$ and
such that the canonical inclusion map $D\rightarrow C$ is a $\kk$-coalgebra
homomorphism. In other words, $D$ is a $\kk$-subcoalgebra
of $C$ (by the definition of a $\kk$-subcoalgebra). This solves the exercise.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.subcoalgebra.summand}}

\textit{Solution to Exercise~\ref{exe.subcoalgebra.summand}.} We will
be maximally explicit in this solution. In particular, we will not regard
inclusions as identities even in cases where we would usually do that. We
will only use the notations $\Delta$ and $\epsilon$ to denote the maps
$\Delta_{C}$ and $\epsilon_{C}$ (not the maps $\Delta_{D}$ and $\epsilon_{D}$,
which we will introduce later).

We will often use the following classical formula from linear algebra:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$ and $W^{\prime}$ are six
$\kk$-modules and  $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$ and $\beta^{\prime
}:V^{\prime}\rightarrow W^{\prime}$ are four $\kk$-linear maps, then
$\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

\noindent
We will refer to this formula as the \dfn{composition-tensor relation}.
Applying this formula twice, we obtain the following formula, which we will
call the \dfn{triple-composition-tensor relation}:

\begin{itemize}
\item If $U$, $V$, $W$, $T$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$ and
$T^{\prime}$
are eight $\kk$-modules and $\alpha:U\rightarrow V$, $\beta
:V\rightarrow W$, $\gamma:W\rightarrow T$, $\alpha^{\prime}:U^{\prime
}\rightarrow V^{\prime}$, $\beta^{\prime}:V^{\prime}\rightarrow W^{\prime}$
and $\gamma^{\prime}:W^{\prime}\rightarrow T^{\prime}$ are six
$\kk$-linear maps, then $\left(  \gamma\circ\beta\circ\alpha\right)
\otimes\left(  \gamma^{\prime}\circ\beta^{\prime}\circ\alpha^{\prime}\right)
=\left(  \gamma\otimes\gamma^{\prime}\right)  \circ\left(  \beta\otimes
\beta^{\prime}\right)  \circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

On the other hand, by a repeated application of the composition-tensor
relation, we can also prove the following formula, which we will
call the \dfn{composition-triple-tensor relation}:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$, $W^{\prime}$,
$U^{\prime\prime}$, $V^{\prime\prime}$, $W^{\prime\prime}$
are nine $\kk$-modules and $\alpha:U\rightarrow V$, $\beta
:V\rightarrow W$, $\alpha^{\prime}:U^{\prime
}\rightarrow V^{\prime}$, $\beta^{\prime}:V^{\prime}\rightarrow W^{\prime}$,
$\alpha^{\prime\prime} : U^{\prime\prime} \to V^{\prime\prime}$,
$\beta^{\prime\prime} : V^{\prime\prime} \to W^{\prime\prime}$
are six $\kk$-linear maps, then
$\left(\beta \circ \alpha\right)
\otimes \left(\beta^\prime \circ \alpha^\prime\right)
\otimes \left(\beta^{\prime\prime} \circ \alpha^{\prime\prime}\right)
= \left(\beta \otimes \beta^\prime \otimes \beta^{\prime\prime}\right)
\circ \left(\alpha \otimes \alpha^\prime \otimes \alpha^{\prime\prime}\right)$.
\end{itemize}

We know that $D$ is a direct summand of $C$ as a $\kk$-module. In other
words, there exists a $\kk$-submodule $E$ of $C$ such that $C=D\oplus
E$. Fix such an $E$.

Let $i:D\rightarrow C$ be the canonical inclusion map. Let $p:D\oplus
E\rightarrow D$ be the canonical projection from the direct sum $D\oplus E$
onto its summand $D$. Notice that $p$ is a $\kk$-linear map from
$D\oplus E=C$ to $D$. Moreover, $p\circ i=\id_{D}$
\ \ \ \ \footnote{This is because $p$ is the canonical projection from the
direct sum $D\oplus E$ onto its summand $D$, while $i$ is the canonical
injection from the summand $D$ into the direct sum $D\oplus E$ (since $i$ is
the canonical injection $D\rightarrow C=D\oplus E$).}. Hence, $i\circ
\underbrace{p\circ i}_{=\id_{D}}=i\circ
\id_{D}=i$.

In the statement of the exercise, we have assumed that $\Delta\left(
D\right)  \subset C\otimes D$. Since we don't want to abuse notation, we have
to rewrite this as
$\Delta\left(  D\right)  \subset
\left(  \id_{C}\otimes i\right)  \left(  C\otimes D\right)  $ (because what we
denoted $C\otimes D$ in the statement of the exercise really means the
$\kk$-submodule of $C\otimes C$ spanned by tensors of the form
$c\otimes d$ with $c\in C$ and $d\in D$; but this $\kk$-submodule is
precisely $\left(  \id_{C}\otimes i\right)  \left(
C\otimes D\right)  $). Similarly, $\Delta\left(  D\right)  \subset\left(
i\otimes\id_{C}\right)  \left(  D\otimes C\right)  $.

Now,
\begin{equation}
\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
x\right)  \right)  =x\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\Delta\left(
D\right)  .\label{sol.subcoalgebra.summand.1}
\end{equation}


\textit{Proof of \eqref{sol.subcoalgebra.summand.1}:} Let $x\in\Delta\left(
D\right)  $. Then, $x\in\Delta\left(  D\right)  \subset\left(
\id_{C}\otimes i\right)  \left(  C\otimes D\right)  $.
Hence, there exists some $y\in C\otimes D$ such that $x=\left(
\id_{C}\otimes i\right)  \left(  y\right)  $. Consider
this $y$. We have
\begin{align*}
\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
\underbrace{x}_{=\left(  \id_{C}\otimes i\right)
\left(  y\right)  }\right)  \right)   &  =\left(  i\otimes i\right)  \left(
\left(  p\otimes p\right)  \left(  \left(  \id_{C}
\otimes i\right)  \left(  y\right)  \right)  \right)  \\
&  =\underbrace{\left(  \left(  i\otimes i\right)  \circ\left(  p\otimes
p\right)  \circ\left(  \id_{C}\otimes i\right)
\right)  }_{\substack{=\left(  i\circ p\circ\id_{C}
\right)  \otimes\left(  i\circ p\circ i\right)  \\\text{(by the
triple-composition-tensor relation)}}}\left(  y\right)  \\
&  =\left(  \underbrace{\left(  i\circ p\circ\id_{C}
\right)  }_{=\left(  \left(  i\circ p\right)  \circ\id_{C}\right)  }
\otimes\underbrace{\left(  i\circ p\circ i\right)
}_{=i=\id_{C}\circ i}\right)  \left(  y\right)  \\
&  =\underbrace{\left(  \left(  \left(  i\circ p\right)  \circ
\id_{C}\right)  \otimes\left(  \id_{C}\circ i\right)
  \right)  }_{\substack{=\left(  \left(  i\circ
p\right)  \otimes\id_{C}\right)  \circ\left(
\id_{C}\otimes i\right)  \\\text{(by the
composition-tensor relation)}}}\left(  y\right)  \\
&  =\left(  \left(  \left(  i\circ p\right)  \otimes\id_{C}\right) 
 \circ\left(  \id_{C}\otimes
i\right)  \right)  \left(  y\right)  \\
&  =\left(  \left(  i\circ p\right)  \otimes\id_C
\right)  \left(  \underbrace{\left(  \id_{C}\otimes
i\right)  \left(  y\right)  }_{=x}\right)  \\
&  =\left(  \left(  i\circ p\right)  \otimes\id_C
\right)  \left(  x\right)  .
\end{align*}
On the other hand, $x\in\Delta\left(  D\right)  \subset\left(  i\otimes
\id_{C}\right)  \left(  D\otimes C\right)  $. Thus,
there exists some $z\in D\otimes C$ such that $x=\left(  i\otimes
\id_{C}\right)  \left(  z\right)  $. Consider this
$z$. We have
\begin{align*}
\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
x\right)  \right)   &  =\left(  \left(  i\circ p\right)  \otimes
\id_{C}\right)  \left(  \underbrace{x}_{=\left(
i\otimes\id_{C}\right)  \left(  z\right)  }\right)  \\
&  =\left(  \left(  i\circ p\right)  \otimes\id_{C}
\right)  \left(  \left(  i\otimes\id_{C}\right)
\left(  z\right)  \right)  \\
&  =\underbrace{\left(  \left(  \left(  i\circ p\right)  \otimes
\id_{C}\right)  \circ\left(  i\otimes
\id_{C}\right)  \right)  }_{\substack{=\left(  \left(
i\circ p\right)  \circ i\right)  \otimes\left(  \id_{C}\circ\id_{C}\right)
 \\\text{(by the composition-tensor relation)}}}\left(  z\right)  \\
&  =\left(  \underbrace{\left(  \left(  i\circ p\right)  \circ i\right)
}_{=i \circ p \circ i = i}\otimes\underbrace{\left(  \id_{C}
\circ\id_{C}\right)  }_{=\id_{C}}\right) 
 \left(  z\right)  \\
&  =\left(  i\otimes\id_{C}\right)  \left(  z\right)
=x.
\end{align*}
This proves \eqref{sol.subcoalgebra.summand.1}.

We know that $C$ is a $\kk$-coalgebra. Thus, the diagrams
\eqref{coassociativity-diagram} and \eqref{counit-diagram} are commutative.

For every $\kk$-module $V$, denote by $\operatorname{can}_{1,V}$
the canonical $\kk$-module isomorphism $V\otimes
\kk \rightarrow V$. For every $\kk$-module $V$, denote by
$\operatorname{can}_{2,V}$ the canonical $\kk$-module
isomorphism $\kk \otimes V\rightarrow V$. Recall that the diagram
\eqref{counit-diagram} is commutative. In other words,
\[
\operatorname{can}_{1,C}\circ
\left(  \id_{C}\otimes\epsilon\right)  \circ\Delta=\id_{C}
\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \operatorname{can}_{2,C}\circ\left(
\epsilon\otimes\id_{C}\right)  \circ\Delta
=\id_{C}.
\]


Also, the diagram \eqref{coassociativity-diagram} is commutative. In other
words,
\[
\left(  \Delta\otimes\id_{C}\right)  \circ
\Delta=\left(  \id_{C}\otimes\Delta\right)
\circ\Delta.
\]


Next, let us define a $\kk$-linear map $\Delta_{D}:D\rightarrow
D\otimes D$ by
\[
\Delta_{D}=\left(  p\otimes p\right)  \circ\Delta\circ i.
\]
Let us also define a $\kk$-linear map $\epsilon_{D}:D\rightarrow
\kk$ by
\[
\epsilon_{D}=\epsilon\circ i.
\]
We are going to show that $\left(  D,\Delta_{D},\epsilon_{D}\right)  $ is a
$\kk$-coalgebra. (Recall that we are not going to abbreviate
$\Delta_{D}$ and $\epsilon_{D}$ by $\Delta$ and $\epsilon$; thus, $\Delta$ and
$\epsilon$ still mean the maps $\Delta_{C}$ and $\epsilon_{C}$.)

We can easily see that
\begin{equation}
\left(  i\otimes i\right)  \circ\Delta_{D}=\Delta\circ
i.\label{sol.subcoalgebra.summand.2}
\end{equation}


\textit{Proof of \eqref{sol.subcoalgebra.summand.2}:} Let $d\in D$. Then,
$i\left(  d\right)  =d$ (since $i$ is just an inclusion map), so that
$\Delta\left(  \underbrace{i\left(  d\right)  }_{=d\in D}\right)  \in
\Delta\left(  D\right)  $. Hence, $\left(  i\otimes i\right)  \left(  \left(
p\otimes p\right)  \left(  \Delta\left(  i\left(  d\right)  \right)  \right)
\right)  =\Delta\left(  i\left(  d\right)  \right)  $ (by
\eqref{sol.subcoalgebra.summand.1}, applied to $x=\Delta\left(  i\left(
d\right)  \right)  $). Now,
\begin{align*}
\left(  \left(  i\otimes i\right)  \circ\underbrace{\Delta_{D}}_{=\left(
p\otimes p\right)  \circ\Delta\circ i}\right)  \left(  d\right)   &  =\left(
\left(  i\otimes i\right)  \circ\left(  p\otimes p\right)  \circ\Delta\circ
i\right)  \left(  d\right)  \\
&  =\left(  i\otimes i\right)  \left(  \left(  p\otimes p\right)  \left(
\Delta\left(  i\left(  d\right)  \right)  \right)  \right)  =\Delta\left(
i\left(  d\right)  \right)  =\left(  \Delta\circ i\right)  \left(  d\right)  .
\end{align*}


Now, forget that we fixed $d$. We thus have shown that every $d\in D$
satisfies $\left(  \left(  i\otimes i\right)  \circ\Delta_{D}\right)  \left(
d\right)  =\left(  \Delta\circ i\right)  \left(  d\right)  $. In other words,
$\left(  i\otimes i\right)  \circ\Delta_{D}=\Delta\circ i$. This proves
\eqref{sol.subcoalgebra.summand.2}.

The composition-triple-tensor relation yields
$\left(  p\circ i\right) \circ \left(  p\circ i\right) \circ \left(  p\circ i\right)
= \left(  p\otimes p\otimes p\right)  \circ\left(  i\otimes i\otimes i\right)$,
so that
\begin{align}
\left(  p\otimes p\otimes p\right)  \circ\left(  i\otimes i\otimes i\right)
&  =\underbrace{\left(  p\circ i\right)  }_{=\id_{D}}\otimes
\underbrace{\left(  p\circ i\right)  }_{=\id_{D}}
\otimes\underbrace{\left(  p\circ i\right)  }_{=\id_{D}}
\nonumber\\
&  =\id_{D}\otimes\id_{D} \otimes\id_{D} = \id_{D\otimes D\otimes D}.
\label{sol.subcoalgebra.summand.iii}
\end{align}
Now,
\begin{align*}
&  \underbrace{\left(  i\otimes i\otimes i\right)  \circ\left(  \Delta
_{D}\otimes\id_{D}\right)  }_{\substack{=\left(
\left(  i\otimes i\right)  \otimes i\right)  \circ\left(  \Delta_{D}
\otimes\id_{D}\right)  \\=\left(  \left(  i\otimes
i\right)  \circ\Delta_{D}\right)  \otimes\left(  i\circ\id_{D}\right) 
 \\\text{(by the composition-tensor relation)}}} \circ\Delta_{D}\\
&  =\left(  \underbrace{\left(  \left(  i\otimes i\right)  \circ\Delta
_{D}\right)  }_{\substack{=\Delta\circ i\\\text{(by
\eqref{sol.subcoalgebra.summand.2})}}}\otimes\underbrace{\left(
i\circ\id_{D}\right)  }_{=i=\id_{C}\circ i}\right) 
 \circ\Delta_{D}\\
&  =\underbrace{\left(  \left(  \Delta\circ i\right)  \otimes\left(
\id_{C}\circ i\right)  \right)  }_{\substack{=\left(
\Delta\otimes\id_{C}\right)  \circ\left(  i\otimes
i\right)  \\\text{(by the composition-tensor relation)}}}\circ\Delta
_{D}=\left(  \Delta\otimes\id_{C}\right)
\circ\underbrace{\left(  i\otimes i\right)  \circ\Delta_{D}}
_{\substack{=\Delta\circ i\\\text{(by \eqref{sol.subcoalgebra.summand.2})}}}\\
&  =\underbrace{\left(  \Delta\otimes\id_{C}\right)
\circ\Delta}_{=\left(  \id_{C}\otimes\Delta\right) \circ\Delta}
\circ i
=\left(  \id_{C}\otimes \Delta\right)  \circ\Delta\circ i.
\end{align*}
Compared with
\begin{align*}
&  \underbrace{\left(  i\otimes i\otimes i\right)  \circ\left(
\id_{D}\otimes\Delta_{D}\right)  }_{\substack{=\left(
i\otimes\left(  i\otimes i\right)  \right)  \circ\left( 
\id_{D}\otimes\Delta_{D}\right)  \\=\left(  i\circ\id_{D}\right) 
 \otimes\left(  \left(  i\otimes i\right)  \circ
\Delta_{D}\right)  \\\text{(by the composition-tensor relation)}}}\circ
\Delta_{D}\\
&  =\left(  \underbrace{\left(  i\circ\id_{D}\right)
}_{=i=\id_{C}\circ i}\otimes\underbrace{\left(
\left(  i\otimes i\right)  \circ\Delta_{D}\right)  }_{\substack{=\Delta\circ
i\\\text{(by \eqref{sol.subcoalgebra.summand.2})}}}\right)  \circ\Delta_{D}\\
&  =\underbrace{\left(  \left(  \id_{C}\circ i\right)
\otimes\left(  \Delta\circ i\right)  \right)  }_{\substack{=\left(
\id_{C}\otimes\Delta\right)  \circ\left(  i\otimes
i\right)  \\\text{(by the composition-tensor relation)}}}\circ\Delta
_{D}=\left(  \id_{C}\otimes\Delta\right)
\circ\underbrace{\left(  i\otimes i\right)  \circ\Delta_{D}}
_{\substack{=\Delta\circ i\\\text{(by \eqref{sol.subcoalgebra.summand.2})}}}\\
&  =\left(  \id_{C}\otimes\Delta\right)  \circ
\Delta\circ i,
\end{align*}
this yields
\[
\left(  i\otimes i\otimes i\right)  \circ\left(  \Delta_{D}\otimes
\id_{D}\right)  \circ\Delta_{D}=\left(  i\otimes
i\otimes i\right)  \circ\left(  \id_{D}\otimes
\Delta_{D}\right)  \circ\Delta_{D}.
\]
Hence,
\begin{align*}
&  \left(  p\otimes p\otimes p\right)  \circ\underbrace{\left(  i\otimes
i\otimes i\right)  \circ\left(  \Delta_{D}\otimes\id_{D}\right) 
 \circ\Delta_{D}}_{=\left(  i\otimes i\otimes i\right)
\circ\left(  \id_{D}\otimes\Delta_{D}\right)
\circ\Delta_{D}}\\
&  =\underbrace{\left(  p\otimes p\otimes p\right)  \circ\left(  i\otimes
i\otimes i\right)  }_{\substack{=\id_{D\otimes
D\otimes D}\\\text{(by \eqref{sol.subcoalgebra.summand.iii})}}}\circ\left(
\id_{D}\otimes\Delta_{D}\right)  \circ\Delta
_{D}=\left(  \id_{D}\otimes\Delta_{D}\right)
\circ\Delta_{D}.
\end{align*}
Compared with
\[
\underbrace{\left(  p\otimes p\otimes p\right)  \circ\left(  i\otimes i\otimes
i\right)  }_{\substack{=\id_{D\otimes D\otimes
D}\\\text{(by \eqref{sol.subcoalgebra.summand.iii})}}}\circ\left(  \Delta
_{D}\otimes\id_{D}\right)  \circ\Delta_{D}=\left(
\Delta_{D}\otimes\id_{D}\right)  \circ\Delta_{D},
\]
this yields $\left(  \Delta_{D}\otimes\id_{D}\right)
\circ\Delta_{D}=\left(  \id_{D}\otimes\Delta
_{D}\right)  \circ\Delta_{D}$. In other words, the diagram
\begin{equation}
\xymatrix{
& D \otimes D \otimes D & \\
D \otimes D \ar[ur]^{\Delta_D \otimes\id_D} & & D \otimes D \ar[ul]_{\id
_D \otimes\Delta_D} \\
& D \ar[ur]^{\Delta_D} \ar[ul]_{\Delta_D} &
}
\label{sol.subcoalgebra.summand.axiom1}
\end{equation}
is commutative.

Moreover, let us notice that any two $\kk$-modules $V$ and $W$ and any
$\kk$-linear map $f:V\rightarrow W$ satisfy
\[
f\circ\operatorname{can}_{1,V} = \operatorname{can}_{1,W}
\circ\left( f\otimes\id_{\kk} \right)  .
\]
This (applied to $V=D$, $W=C$ and $f=i$) yields
$i\circ\operatorname{can}_{1,D}=\operatorname{can}_{1,C}\circ\left(  i\otimes
\id_{\kk }\right)  $. Hence,
\begin{align*}
&  \underbrace{i\circ\operatorname{can}_{1,D}}
_{=\operatorname{can}_{1,C}\circ\left(  i\otimes\id_{\kk}\right)  }\circ
\left(  \id_{D}\otimes\epsilon
_{D}\right)  \circ\Delta_{D}\\
&  =\operatorname{can}_{1,C}\circ\underbrace{\left(  i\otimes
\id_{\kk }\right)  \circ\left(
\id_{D}\otimes\epsilon_{D}\right)  }
_{\substack{=\left(  i\circ\id_{D}\right)
\otimes\left(  \id_{\kk }\circ\epsilon
_{D}\right)  \\\text{(by the composition-tensor relation)}}}\circ\Delta_{D}\\
&  =\operatorname{can}_{1,C}\circ\left(  \underbrace{\left(
i\circ\id_{D}\right)  }_{=i=\id_{C}\circ i}
\otimes\underbrace{\left(  \id_{\kk }\circ\epsilon_{D}\right)  }
_{=\epsilon_{D}=\epsilon\circ
i}\right)  \circ\Delta_{D}\\
&  =\operatorname{can}_{1,C}\circ\underbrace{\left(  \left(
\id_{C}\circ i\right)  \otimes\left(  \epsilon\circ
i\right)  \right)  }_{\substack{=\left(  \id_{C}
\otimes\epsilon\right)  \circ\left(  i\otimes i\right)  \\\text{(by the
composition-tensor relation)}}}\circ\Delta_{D}\\
&  =\operatorname{can}_{1,C} \circ\left(  \id_{C}\otimes\epsilon\right) 
 \circ\underbrace{\left(  \left(  i\otimes
i\right)  \circ\Delta_{D}\right)  }_{\substack{=\Delta\circ i\\\text{(by
\eqref{sol.subcoalgebra.summand.2})}}}\\
&  =\underbrace{\operatorname{can}_{1,C}\circ\left(
\id_{C}\otimes\epsilon\right)  \circ\Delta }_{=\id_{C}}\circ i
= \id_{C} \circ i = i.
\end{align*}
Hence,
\[
p\circ\underbrace{i\circ\operatorname{can}_{1,D}\circ\left(
\id_{D}\otimes\epsilon_{D}\right)  \circ\Delta_{D}}
_{=i}=p\circ i=\id_{D}.
\]
Compared with
\begin{align*}
&  \underbrace{p\circ i}_{=\id_{D}}\circ
\operatorname{can}_{1,D}
 \circ\left(  \id_{D}\otimes\epsilon_{D}\right)  \circ\Delta_{D}\\
&  =\id_{D}\circ\operatorname{can}_{1,D}
\circ\left(  \id_{D}\otimes\epsilon_{D}\right)
\circ\Delta_{D}=\operatorname{can}_{1,D}\circ\left(
\id_{D}\otimes\epsilon_{D}\right)  \circ\Delta_{D},
\end{align*}
this yields
\begin{equation}
\operatorname{can}_{1,D}
 \circ\left(  \id_{D}\otimes\epsilon_{D}\right) \circ\Delta_{D}=\id_{D}.
\label{sol.subcoalgebra.summand.epsi1}
\end{equation}


On the other hand, let us notice that any two $\kk$-modules $V$ and $W$
and any $\kk$-linear map $f:V\rightarrow W$ satisfy
\[
f\circ\operatorname{can}_{2,V} = \operatorname{can}_{2,W}
\circ\left( \id_{\kk} \otimes f\right)  .
\]
This (applied to $V=D$, $W=C$ and $f=i$) yields
$i\circ\operatorname{can}_{2,D}=\operatorname{can}_{2,C}\circ\left(
\id_{\kk}\otimes i\right)  $. Hence,
\begin{align*}
&  \underbrace{i\circ\operatorname{can}_{2,D}}
_{=\operatorname{can}_{2,C}\circ\left(  \id_{\kk }\otimes
i\right)  }\circ\left(  \epsilon_{D}\otimes\id_{D}
\right)  \circ\Delta_{D}\\
&  =\operatorname{can}_{2,C}\circ\underbrace{\left(
\id_{\kk }\otimes i\right)  \circ\left(
\epsilon_{D}\otimes\id_{D}\right)  }
_{\substack{=\left(  \id_{\kk }\circ\epsilon
_{D}\right)  \otimes\left(  i\circ\id_{D}\right)
\\\text{(by the composition-tensor relation)}}}\circ\Delta_{D}\\
&  =\operatorname{can}_{2,C}\circ\left(  \underbrace{\left(
\id_{\kk }\circ\epsilon_{D}\right)  }
_{=\epsilon_{D}=\epsilon\circ i}\otimes\underbrace{\left(  i\circ
\id_{D}\right)  }_{=i=\id_{C}\circ i}\right)  \circ\Delta_{D}\\
&  =\operatorname{can}_{2,C}\circ\underbrace{\left(  \left(
\epsilon\circ i\right)  \otimes\left(  \id_{C}\circ
i\right)  \right)  }_{\substack{=\left(  \epsilon\otimes\id_{C}\right) 
 \circ\left(  i\otimes i\right)  \\\text{(by the
composition-tensor relation)}}}\circ\Delta_{D}\\
&  =\operatorname{can}_{2,C}\circ\left(  \epsilon\otimes
\id_{C}\right)  \circ\underbrace{\left(  \left(
i\otimes i\right)  \circ\Delta_{D}\right)  }_{\substack{=\Delta\circ
i\\\text{(by \eqref{sol.subcoalgebra.summand.2})}}}\\
&  =\underbrace{\operatorname{can}_{2,C}\circ\left(  \epsilon
\otimes\id_{C}\right)  \circ\Delta}
_{=\id_{C}}\circ i=\id_{C}
\circ i=i.
\end{align*}
Hence,
\[
p\circ\underbrace{i\circ\operatorname{can}_{2,D}\circ\left(
\epsilon_{D}\otimes\id_{D}\right)  \circ\Delta_{D}}_{=i}
=p\circ i=\id_{D}.
\]
Compared with
\begin{align*}
&  \underbrace{p\circ i}_{=\id_{D}}\circ
\operatorname{can}_{2,D}\circ\left(  \epsilon_{D}\otimes
\id_{D}\right)  \circ\Delta_{D}\\
&  =\id_{D}\circ\operatorname{can}_{2,D}
\circ\left(  \epsilon_{D}\otimes\id_{D}\right)
\circ\Delta_{D}=\operatorname{can}_{2,D}\circ\left(  \epsilon
_{D}\otimes\id_{D}\right)  \circ\Delta_{D},
\end{align*}
this yields
\begin{equation}
\operatorname{can}_{2,D}\circ\left(  \epsilon_{D}\otimes
\id_{D}\right)  \circ\Delta_{D}
=\id_{D}.
\label{sol.subcoalgebra.summand.epsi2}
\end{equation}


Now, the diagram
\begin{equation}
\xymatrixcolsep{4pc}
\xymatrix{
D \otimes\kk\ar[r]^{\operatorname{can}_{1,D}} & D & \kk\otimes D \ar
[l]_{\operatorname{can}_{2,D}} \\
D \otimes D \ar[u]^{\id_D \otimes\epsilon_D} & D \ar[l]^{\Delta_D} \ar
[u]^{\id_D} \ar[r]_{\Delta_D} & D \otimes D \ar[u]_{\epsilon_D \otimes\id_D}
}
\label{sol.subcoalgebra.summand.axiom2}
\end{equation}
is commutative (because of \eqref{sol.subcoalgebra.summand.epsi1} and
\eqref{sol.subcoalgebra.summand.epsi2}). Hence, we now know that the diagrams
\eqref{sol.subcoalgebra.summand.axiom1} and \eqref{sol.subcoalgebra.summand.axiom2} are
commutative. Hence, $\left(  D,\Delta_{D},\epsilon_{D}\right) $ must be a
$\kk$-coalgebra (by the definition of a $\kk$-coalgebra). We
will denote this $\kk$-coalgebra simply by $D$.

Now, the diagrams
\[
\xymatrix{ D \ar[d]_{\Delta_D} \ar[r]^{i} & C \ar[d]^{\Delta} \\ D \otimes D \ar[r]^{i \otimes i} & C \otimes C }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ D \ar[dr]_{\epsilon_D} \ar[rr]^{i} & & C \ar[dl]^{\epsilon} \\ & \kk & }
\]
are commutative\footnote{In fact, the commutativity of the first of these
diagrams follows from \eqref{sol.subcoalgebra.summand.2}, whereas the
commutativity of the second diagram follows from $\epsilon_{D}=\epsilon\circ
i$.}. Hence, $i$ is a $\kk$-coalgebra homomorphism (by the definition
of a $\kk$-coalgebra homomorphism). In other words, the canonical
inclusion map $D\rightarrow C$ is a $\kk$-coalgebra homomorphism (since
$i$ is the canonical inclusion map $D\rightarrow C$).

So we know that $D$ is a $\kk$-coalgebra such that $D\subset C$ and
such that the canonical inclusion map $D\rightarrow C$ is a
$\kk$-coalgebra homomorphism. In other words, $D$ is a $\kk$-subcoalgebra
of $C$ (by the definition of a $\kk$-subcoalgebra). This solves the exercise.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.subcoalgebra.ABS}}

\textit{Solution to Exercise~\ref{exe.subcoalgebra.ABS}.} We will
identify the tensor products $K\otimes K$, $C\otimes K$ and $K\otimes C$ with
the corresponding $\kk$-submodules of the tensor product $C\otimes C$.
(We can afford to do this since $\kk$ is a field.)

We will only use the notations $\Delta$ and $\epsilon$ to denote the maps
$\Delta_{C}$ and $\epsilon_{C}$.

Notice that $C$ is a free $\kk$-module. Hence, tensoring with $C$ is an
exact functor, thus a left-exact functor.

By the recursive definition of $\Delta^{\left(  1\right)  }$, we have
$\Delta^{\left(  1\right)  }=\left(  \id_{C}
\otimes\underbrace{\Delta^{\left(  1-1\right)  }}_{=\Delta^{\left(  0\right)
}=\id_{C}}\right)  \circ\Delta=\underbrace{\left(
\id_{C}\otimes\id_{C}\right)
}_{=\id_{C\otimes C}}\circ\Delta=\Delta$.

By the recursive definition of $\Delta^{\left(  2\right)  }$, we have
$\Delta^{\left(  2\right)  }=\left(  \id_{C}
\otimes\underbrace{\Delta^{\left(  1\right)  }}_{=\Delta}\right)  \circ
\Delta=\left(  \id_{C}\otimes\Delta\right)
\circ\Delta=\left(  \Delta\otimes\id_{C}\right)
\circ\Delta$ (by the axioms of a coalgebra).

(a) Applying Exercise~\ref{exe.iterated.Delta}(b) to $k=3$, we obtain
$\Delta^{\left(  3\right)  }=\left(  \Delta^{\left(  2\right)  }
\otimes\id_{C}\right)  \circ\Delta$, so that
\begin{equation}
\left(  \Delta^{\left(  2\right)  }\otimes\id_{C}
\right)  \circ\Delta=\Delta^{\left(  3\right)  }
=\left(  \id_{C}\otimes\Delta^{\left(  2\right)  }\right)  \circ\Delta
\label{sol.subcoalgebra.ABS.1}
\end{equation}
(by the recursive definition of $\Delta^{\left(  3\right)  }$).

Let $\widetilde{f}=\left(  \id_{C}\otimes f\otimes
\id_{C}\right)  \circ\Delta^{\left(  2\right)  }:C\rightarrow
C\otimes U\otimes C$. Then, $\ker\widetilde{f}=\ker\left(  \left(
\id_{C}\otimes f\otimes\id_{C}\right)  \circ
\Delta^{\left(  2\right)  }\right)  =K$. But
\begin{align*}
&  \underbrace{\widetilde{f}}_{=\left(  \id_{C}\otimes
f\otimes\id_{C}\right)  \circ\Delta^{\left(  2\right)  }}
\otimes\id_{C}\\
&  =\left(  \left(  \id_{C}\otimes f\otimes\id_{C}\right) 
 \circ\Delta^{\left(  2\right)  }\right)  \otimes
\id_{C}=\underbrace{\left( 
 \left(  \id_{C}\otimes f\otimes\id_{C}\right)  \otimes\id_{C}\right) }
_{\substack{=\id_{C}\otimes
f\otimes\id_{C}\otimes\id_{C}\\
=\id_{C}\otimes f\otimes
\id_{C\otimes C}}}\circ\left(  \Delta^{\left(
2\right)  }\otimes\id_{C}\right) \\
&  =\left(  \id_{C}\otimes f\otimes
\id_{C\otimes C}\right)  \circ\left(  \Delta^{\left(  2\right)  }
\otimes\id_{C}\right)  ,
\end{align*}
so that
\begin{align*}
&  \underbrace{\left(  \widetilde{f}\otimes\id_{C}
\right)  }_{=\left(  \id_{C}\otimes f\otimes
\id_{C\otimes C}\right)  \circ\left(  \Delta^{\left(
2\right)  }\otimes\id_{C}\right)  }\circ\Delta\\
&  =\left(  \id_{C}\otimes f\otimes
\id_{C\otimes C}\right)  \circ\underbrace{\left(  \Delta^{\left(
2\right)  }\otimes\id_{C}\right)  \circ\Delta
}_{\substack{=\left(  \id_{C}\otimes\Delta^{\left(
2\right)  }\right)  \circ\Delta\\
\text{(by \eqref{sol.subcoalgebra.ABS.1})}}}\\
&  =\underbrace{\left(  \id_{C}\otimes f\otimes
\id_{C\otimes C}\right)  \circ\left(
\id_{C}\otimes\Delta^{\left(  2\right)  }\right)
}_{=\id_{C}\otimes\left(  \left(  f\otimes
\id_{C\otimes C}\right)  \circ\Delta^{\left(
2\right)  }\right)  }\circ\Delta\\
&  =\left(  \id_{C}\otimes\left(  \left(
f\otimes\id_{C\otimes C}\right)  \circ
\underbrace{\Delta^{\left(  2\right)  }}
_{=\left(  \id_{C}\otimes\Delta\right)  \circ\Delta}\right)  \right) 
 \circ\Delta\\
&  =\left(  \id_{C}\otimes\left(  \underbrace{\left(
f\otimes\id_{C\otimes C}\right)  \circ\left(
\id_{C}\otimes\Delta\right)  }_{\substack{=f\otimes
\Delta\\=\left(  \id_{U}\otimes\Delta\right)
\circ\left(  f\otimes\id_{C}\right)  }}\circ
\Delta\right)  \right)  \circ\Delta\\
&  =\underbrace{\left(  \id_{C}\otimes\left(  \left(
\id_{U}\otimes\Delta\right)  \circ\left(
f\otimes\id_{C}\right)  \circ\Delta\right)  \right)
}_{=\left(  \id_{C}\otimes
\left(  \id_{U}\otimes\Delta\right)  \right)  \circ
\left(  \id_{C}\otimes\left(  f\otimes\id_{C}\right)
\right)  \circ\left(  \id_{C}\otimes\Delta\right)
}\circ\Delta\\
&  =\underbrace{\left(  \id_{C}\otimes\left(
\id_{U}\otimes\Delta\right)  \right)  }
_{=\id_{C}\otimes\id_{U}
\otimes\Delta}\circ\underbrace{\left(  \id_{C}
\otimes\left(  f\otimes\id_{C}\right)  \right)
}_{=\id_{C}\otimes f\otimes\id_{C}}\circ
\underbrace{\left(  \id_{C}
\otimes\Delta\right)  \circ\Delta}_{=\Delta^{\left(  2\right)  }}\\
&  =\left(  \id_{C}\otimes\id_{U}\otimes\Delta\right) 
 \circ\underbrace{\left(  \id_{C}\otimes f\otimes\id_{C}\right)
\circ\Delta^{\left(  2\right)  }}_{=\widetilde{f}}\\
&  =\left(  \id_{C}\otimes\id_{U}\otimes\Delta\right) 
 \circ\widetilde{f}.
\end{align*}
Hence,
\begin{align*}
\ker\left(  \left(  \widetilde{f}\otimes\id_{C}
\right)  \circ\Delta\right)   &  
=\ker\left(  \left(  \id_{C}\otimes\id_{U}\otimes\Delta\right)
\circ\widetilde{f}\right) \supset\ker\widetilde{f}=K,
\end{align*}
so that
\[
K\subset\ker\left(  \left(  \widetilde{f}\otimes\id_{C}\right) 
 \circ\Delta\right)  =\Delta^{-1}\left(  \ker\left(
\widetilde{f}\otimes\id_{C}\right)  \right)  ,
\]
so that
\begin{align*}
\Delta\left(  K\right)   &  \subset\ker\left(  \widetilde{f}\otimes
\id_{C}\right)  =\underbrace{\left(  \ker
\widetilde{f}\right)  }_{=K}\otimes C\ \ \ \ \ \ \ \ \ \ \left(  \text{since
tensoring with }C\text{ is a left-exact functor}\right) \\
&  =K\otimes C.
\end{align*}
Similarly, $\Delta\left(  K\right)  \subset C\otimes K$. But $K$ is a direct
summand of $C$ as a $\kk$-module (since $\kk$ is a field).
Hence, Exercise~\ref{exe.subcoalgebra.summand} (applied to $D=K$) yields that
there is a canonically defined $\kk$-coalgebra structure on $K$ which
makes $K$ a subcoalgebra of $C$. In other words, $K$ is a $\kk$-subcoalgebra
of $C$. This solves Exercise~\ref{exe.subcoalgebra.ABS}(a).

(b) Let $E$ be a $\kk$-subcoalgebra of $C$ which is a subset of $\ker
f$. We must show that $E$ is a subset of $K$.

We have $E\subset\ker f$, so that $f\left(  E\right)  =0$. Since $E$ is a
$\kk$-subcoalgebra of $C$, we have $\Delta\left(  E\right)  \subset
E\otimes E$, and
\begin{align*}
\underbrace{\Delta^{\left(  2\right)  }}
_{=\left(  \id_{C}\otimes\Delta\right)  \circ\Delta}\left(  E\right) 
  &  =\left(
\left(  \id_{C}\otimes\Delta\right)  \circ
\Delta\right)  \left(  E\right)  =\left(  \id_{C}
\otimes\Delta\right)  \left(  \underbrace{\Delta\left(  E\right)  }_{\subset
E\otimes E}\right)  \\
&  \subset\left(  \id_{C}\otimes\Delta\right)  \left(
E\otimes E\right)  =\underbrace{\id_{C}\left(
E\right)  }_{=E}\otimes\underbrace{\Delta\left(  E\right)  }_{\subset E\otimes
E}\subset E\otimes E\otimes E
\end{align*}
and thus
\begin{align*}
\left(  \left(  \id_{C}\otimes f\otimes\id_{C}\right) 
 \circ\Delta^{\left(  2\right)  }\right)  \left(  E\right)   &
=\left(  \id_{C}\otimes f\otimes\id_{C}\right)
\left(  \underbrace{\Delta^{\left(  2\right)  }\left(  E\right)  }_{\subset
E\otimes E\otimes E}\right)  \\
&  \subset\left(  \id_{C}\otimes f\otimes\id_{C}\right) 
 \left(  E\otimes E\otimes E\right)  \\
&  =\id_{C}\left(  E\right)  \otimes
\underbrace{f\left(  E\right)  }_{=0}\otimes\id_{C}
\left(  E\right)  =0.
\end{align*}
Hence, $E\subset\ker\left(  \left(  \id_{C}\otimes
f\otimes\id_{C}\right)  \circ\Delta^{\left(  2\right)  }\right)
=K$, which means that $E$ is a subset of $K$. This solves
Exercise~\ref{exe.subcoalgebra.ABS}(b).

\textit{Remark:} Exercise~\ref{exe.subcoalgebra.ABS}(a) would not hold if we
allowed $\kk$ to be an arbitrary commutative ring rather than a field.
(For a stupid counterexample, try $\kk = \ZZ$, $C = \kk$ and
$U=\ZZ / 2$, with $f$ being the canonical projection.) It might be an
interesting question to figure out how much freedom we can allow without
breaking correctness. Here is one case which definitely works: If $\kk$
is a principal ideal domain and $C$ and $U$ are finite free $\kk$-modules,
then Exercise~\ref{exe.subcoalgebra.ABS} is correct. (In fact, our
above solution works in this case, after one notices that $\left(
\id_{C}\otimes f\otimes\id_{C}\right)  \circ
\Delta^{\left(  2\right)  }$ is a homomorphism of finite free $\kk$-modules,
and the kernel of every homomorphism of finite free $\kk$-modules over a
principal ideal domain is a direct summand of its domain.)
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.antipode.connected.general}}

\textit{Solution to Exercise~\ref{exe.antipode.connected.general}.}

\begin{lemma}
\label{lem.convolution.restrict-coalg}
Let $C$ be a $\kk$-coalgebra, and
let $A$ be a $\kk$-algebra. Let $D$ be a subcoalgebra of $C$. Let
$f:C\rightarrow A$ and $g:C\rightarrow A$ be two $\kk$-linear maps.
Then, $\left.  \left(  f\star g\right)  \mid_{D}\right.  =\left(  f\mid
_{D}\right)  \star\left(  g\mid_{D}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.convolution.restrict-coalg}.]Recall the following
classical formula from linear algebra:

\begin{itemize}
\item If $U$, $V$, $W$, $U^{\prime}$, $V^{\prime}$ and $W^{\prime}$ are six
$\kk$-modules and $\alpha:U\rightarrow V$, $\beta:V\rightarrow W$,
$\alpha^{\prime}:U^{\prime}\rightarrow V^{\prime}$ and $\beta^{\prime
}:V^{\prime}\rightarrow W^{\prime}$ are four $\kk$-linear maps, then
$\left(  \beta\circ\alpha\right)  \otimes\left(  \beta^{\prime}\circ
\alpha^{\prime}\right)  =\left(  \beta\otimes\beta^{\prime}\right)
\circ\left(  \alpha\otimes\alpha^{\prime}\right)  $.
\end{itemize}

\noindent We will refer to this formula as the \emph{composition-tensor
relation}.

Let $i:D\rightarrow C$ be the inclusion map from $D$ to $C$. Then, $i$ is a
$\kk$-coalgebra homomorphism (since $D$ is a subcoalgebra of $C$). In
other words, $i$ is a $\kk$-linear map that makes the diagrams
\[
\xymatrix{
D \ar[d]_{\Delta_D} \ar[r]^i & C \ar[d]^{\Delta_C} \\
D \otimes D \ar[r]^{i \otimes i} & C \otimes C
}
\qquad
\xymatrix{
D \ar[dr]_{\epsilon_D} \ar[rr]^{i}& & C \ar[dl]^{\epsilon_C}  \\
&\kk&
}
\]
commute. In other words, $i$ is a $\kk$-linear map satisfying $\left(
i\otimes i\right)  \circ\Delta_{D}=\Delta_{C}\circ i$ and $\epsilon
_{D}=\epsilon_{C}\circ i$.

Recall that $i$ is the inclusion map from $D$ to $C$.
Hence, if $h:C\rightarrow A$ is any map, then
\begin{equation}
\left.  h\mid_{D}\right.  = h \circ i.
\label{pf.lem.convolution.restrict-coalg.restrict}
\end{equation}


Now, the definition of convolution yields%
\begin{align*}
\left(  f\mid_{D}\right)  \star\left(  g\mid_{D}\right)   &  =m_{A}%
\circ\left(  \underbrace{\left(  f\mid_{D}\right)  }_{\substack{=f\circ
i\\\text{(by \eqref{pf.lem.convolution.restrict-coalg.restrict},}%
\\\text{applied to }h=f\text{)}}}\otimes\underbrace{\left(  g\mid_{D}\right)
}_{\substack{=g\circ i\\\text{(by
\eqref{pf.lem.convolution.restrict-coalg.restrict},}\\\text{applied to
}h=g\text{)}}}\right)  \circ\Delta_{D}\\
&  =m_{A}\circ\underbrace{\left(  \left(  f\circ i\right)  \otimes\left(
g\circ i\right)  \right)  }_{\substack{=\left(  f\otimes g\right)
\circ\left(  i\otimes i\right)  \\\text{(by the composition-tensor relation)}%
}}\circ\Delta_{D}\\
&  =m_{A}\circ\left(  f\otimes g\right)  \circ\underbrace{\left(  i\otimes
i\right)  \circ\Delta_{D}}_{=\Delta_{C}\circ i}=m_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}\circ i.
\end{align*}
Comparing this with%
\begin{align*}
\left.  \left(  f\star g\right)  \mid_{D}\right.    & =\underbrace{\left(
f\star g\right)  }_{\substack{=m_{A}\circ\left(  f\otimes g\right)
\circ\Delta_{C}\\\text{(by the definition}\\\text{of convolution)}}}\circ
i\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.lem.convolution.restrict-coalg.restrict}, applied to }h=f\star
g\right)  \\
& =m_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}\circ i,
\end{align*}
we obtain $\left.  \left(  f\star g\right)  \mid_{D}\right.  =\left(
f\mid_{D}\right)  \star\left(  g\mid_{D}\right)  $. This proves Lemma
\ref{lem.convolution.restrict-coalg}.
\end{proof}

Our next lemma is a particular case of
Exercise~\ref{exe.antipode.connected.general}(a), which we will use
as a stepping stone towards solving the exercise in the general case:

\begin{lemma}
\label{lem.sol.antipode.connected.general.a.1}
Let $C = \bigoplus_{n\geq 0} C_n$ be a graded $\kk$-coalgebra, and $A$ be
any $\kk$-algebra. Let $h : C \to A$ be a $\kk$-linear map such that
$\left. h \mid_{C_0} \right.
= \left(u_A \circ \epsilon_C\right) \mid_{C_0} $.
Then, $h$ is a $\star$-invertible map in $\Hom\left(C,A\right)$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem.sol.antipode.connected.general.a.1}.]
Define a $\kk$-linear map $f : C \to A$ as $h - u_A \circ \epsilon_C$.
Thus, $h = u_A \circ \epsilon_C + f = u \epsilon + f$.
Furthermore, the map $f$ annihilates $C_0$, because
\[
\left. \underbrace{f}_{= h - u_A \circ \epsilon_C} \mid_{C_0} \right.
= \left( h - u_A \circ \epsilon_C \right) \mid_{C_0}
= \left(h\mid_{C_0}\right)
   - \left( \left(u_A \circ \epsilon_C\right) \mid_{C_0} \right)
= 0
\ \ \ \ \ \ \ \ \ \ \left(\text{since
$\left. h \mid_{C_0} \right. = \left(u_A \circ \epsilon_C\right) \mid_{C_0} $}\right) .
\]
Thus, we can proceed as in
the proof of Proposition~\ref{Takeuchi-prop} to show that $\sum_{k\geq 0}
\left( -1\right) ^k f^{\star k}$ is a well-defined linear map
$C\rightarrow A$ and a two-sided $\star$-inverse for $u \epsilon + f$.
Thus, $u \epsilon + f$ is $\star$-invertible.
In other words, $h$ is $\star$-invertible (since $h = u \epsilon + f$).
This proves Lemma~\ref{lem.sol.antipode.connected.general.a.1}.

An alternative proof of Lemma~\ref{lem.sol.antipode.connected.general.a.1}
proceeds by mimicking the proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
\end{proof}

We now step to the solution of the exercise.

(a) Here is Takeuchi's argument: We know that the map
$\left. h \mid_{C_0} \right.
\in \Hom \left( C_0, A \right)$ is $\star$-invertible; let
$\widetilde{g}$ be its $\star$-inverse. Thus,
$\widetilde{g} : C_0 \to A$ is a $\kk$-linear map satisfying
\[
\widetilde{g} \star \left( h \mid_{C_0} \right)
= \left( h \mid_{C_0} \right) \star \widetilde{g}
= u_A \circ \underbrace{\epsilon_{C_0}}_{= \epsilon_C \mid_{C_0}}
= u_A \circ \left(\epsilon_C \mid_{C_0}\right)
= \left(u_A \circ \epsilon_C\right) \mid_{C_0} .
\]
Extend $\widetilde{g}$ to a
$\kk$-linear map $g : C \to A$ by defining it to be $0$
on every $C_n$ for $n>0$.
Then, $\left. g \mid_{C_0} \right. = \widetilde{g}$, so that
\begin{align*}
\left. \left( g \star h \right) \mid_{C_0} \right.
&= \underbrace{\left( g \mid_{C_0} \right)}_{=\widetilde{g}}
  \star \left( h \mid_{C_0} \right)
\ \ \ \ \ \ \ \ \ \ \left(\text{by
Lemma \ref{lem.convolution.restrict-coalg},
applied to $C_0$, $g$ and $h$ instead of $D$, $f$ and $g$}\right)
\\
&= \widetilde{g} \star \left( h \mid_{C_0} \right)
= \left(u_A \circ \epsilon_C\right) \mid_{C_0}
\end{align*}
and similarly
\[
\left. \left( h \star g \right) \mid_{C_0} \right.
= \left(u_A \circ \epsilon_C\right) \mid_{C_0} .
\]
Now, Lemma~\ref{lem.sol.antipode.connected.general.a.1}
(applied to $g \star h$ instead of $h$) shows that
$g \star h$ is a $\star$-invertible map in
$\Hom\left(C, A\right)$
(since $\left. \left( g \star h \right) \mid_{C_0} \right.
= \left(u_A \circ \epsilon_C\right) \mid_{C_0}$).
In other words, there exists a
map $p \in \Hom\left(C, A\right)$ such that
$\left(g \star h\right) \star p = p \star \left(g \star h\right)
= u_A \epsilon_C$.
Consider this $p$.
Also, Lemma~\ref{lem.sol.antipode.connected.general.a.1}
(applied to $h \star g$ instead of $h$) shows that
$h \star g$ is a $\star$-invertible map in
$\Hom\left(C, A\right)$
(since $\left. \left( h \star g \right) \mid_{C_0} \right.
= \left(u_A \circ \epsilon_C\right) \mid_{C_0}$).
In other words, there exists a
map $q \in \Hom\left(C, A\right)$ such that
$\left(h \star g\right) \star q = q \star \left(h \star g\right)
= u_A \epsilon_C$.
Consider this $q$.
We conclude that $p \star g$ is a left $\star$-inverse to
$h$ (since
$\left(p \star g\right) \star h = p \star \left(g \star h\right)
= u_A \epsilon_C$), so that the map $h$ has a left
$\star$-inverse.
We also conclude that $g \star q$ is a right $\star$-inverse
to $h$ (since
$h \star \left(g \star q\right) = \left(h \star g\right) \star q
= u_A \epsilon_C$), so that the map $h$ has a right
$\star$-inverse.
Altogether, we now know that the map $h$ has a left
$\star$-inverse and a right $\star$-inverse; therefore, it is
$\star$-invertible (because an element of an algebra that has a
left inverse and a right inverse must always be invertible).
This solves part (a) of the exercise.

(b) Apply part (a) to $C=A$ and the map $\id_A : A \rightarrow A$.

(c) Let $A$ be a connected graded bialgebra.
We must show that $A$ has a unique antipode $S$, which is a
graded map $A \overset{S}{\to} A$, and that this endows $A$
with a Hopf structure.

The map $u : \kk \to A$ is a $\kk$-algebra homomorphism
and a $\kk$-coalgebra homomorphism. Hence, $u$ is a
$\kk$-bialgebra homomorphism.
Hence, $\kk \overset{u}{\rightarrow} A_0$ is a
$\kk$-bialgebra homomorphism.
Since we know from
Exercise~\ref{graded-connected-exercise}(b) that the
map $u$ is an isomorphism $\kk \overset{u}{\rightarrow} A_0$,
we thus conclude that $\kk \overset{u}{\rightarrow} A_0$ is a
$\kk$-bialgebra isomorphism.
Hence, $A_0 \cong \kk$ as $\kk$-bialgebras.
Since $\kk$ is a Hopf algebra (with antipode $\id_{\kk}$),
we thus conclude that $A_0$ is a Hopf algebra.
Hence, part (b) of this exercise shows that $A$ is a Hopf algebra.
Therefore, $A$ has an antipode.
It is clear that this antipode is unique (since an antipode
of $A$ is the same thing as a $\star$-inverse to the map
$\id : A \to A$, but it is clear that the $\star$-inverse
of any given map is unique).
Thus, we know that $A$ has a unique antipode $S$, endowing
it with a Hopf structure.
In order to conclude the (new) proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes},
it thus remains to show that this antipode $S$ is graded.

The antipode $S$ is the $\star$-inverse of the map
$\id : A \to A$. Hence, the map $\id : A \to A$ is
$\star$-invertible. This map $\id$ is also graded
(obviously). Thus, Exercise~\ref{exe.coalg-anti.gen}(e)
(applied to $C = A$ and $r = \id$) shows that the
$\star$-inverse $\id^{\star\left(  -1\right)  }$ of $\id$ is also
graded. In other words, $S$ is graded (since the
$\star$-inverse $\id^{\star\left(  -1\right)  }$ of $\id$ is
$S$). This concludes the (new) proof of
Proposition~\ref{graded-connected-bialgebras-have-antipodes}.
Thus, part (c) of the exercise is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.I=0}}

\textit{Solution to Exercise~\ref{exe.I=0}.}
(a) Let $I$ be a two-sided coideal of $A$ such that $I \cap \Liep = 0$ and
such that $I = \bigoplus_{n \geq 0}\left( I\cap A_n \right)$.
Let $I_n = I \cap A_n$ for every $n\in \NN$. Then,
$I = \bigoplus_{n \geq 0} \underbrace{\left( I\cap A_n\right) }_{= I_n}
= \bigoplus_{n \geq 0} I_n$.

Since $I$ is a two-sided coideal, we have $\epsilon\left( I \right) = 0$.
%Thus, $\epsilon\left( I_0 \right) = 0$. Since $\epsilon \mid_{A_0}$ is an
%isomorphism (this is one of the many properties characterizing a connected
%graded bialgebra), this yields $I_0 = 0$.

We now will prove that
\begin{equation}
\text{every } n \in \NN \text{ satisfies } I_n = 0.
\label{exe.I=0.sol.1}
\end{equation}


\textit{Proof of \eqref{exe.I=0.sol.1}:} Let us prove \eqref{exe.I=0.sol.1} by
strong induction over $n$:

Let $N \in \NN$. Assume that \eqref{exe.I=0.sol.1} holds for every
$n \in \NN$ satisfying $n < N$. We must then prove that
\eqref{exe.I=0.sol.1} holds for $n = N$.

We know that \eqref{exe.I=0.sol.1} holds for every $n \in \NN$ satisfying
$n < N$. In other words,
\begin{equation}
\text{for every } n \in \NN \text{ satisfying } n < N \text{, we have } I_n=0.
\label{exe.I=0.sol.2}
\end{equation}


%If $N = 0$, then \eqref{exe.I=0.sol.1} holds for $n = N$ (because $I_0 = 0$).
%Thus, for the rest of this proof, we can WLOG assume that $N \neq 0$. Assume
%this. Then, $N \geq 1$. Thus, $\epsilon\left( A_N \right) = 0$, so that
%$\epsilon\left( I_N \right) = 0$ (since $I_N \subset A_N$).

We have $I_N = I \cap A_N \subset I$ and thus
$\epsilon \left( \underbrace{I_N}_{\subset I} \right) \subset
\epsilon \left( I \right) = 0$, hence $\epsilon \left( I_N \right) = 0$.

Let $\pi_{N}:A\otimes A\rightarrow\left(  A\otimes A\right)  _{N}$ be the
projection from the graded $\kk$-module $A\otimes A$ to its $N$-th
graded component $\left(  A\otimes A\right)  _{N}$. Then,
\begin{equation}
\pi_{N}\left(  t\right)  =t\ \ \ \ \ \ \ \ \ \ \text{for every }t\in\left(
A\otimes A\right)  _{N}, \label{exe.I=0.sol.3}
\end{equation}
and
\begin{equation}
\pi_{N}\left(  t\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }\ell
\in\NN\setminus\left\{  N\right\}  \text{ and every }t\in\left(
A\otimes A\right)  _{\ell}. \label{exe.I=0.sol.4}
\end{equation}
As a consequence,
\begin{equation}
\text{every }\left(  n,m\right)  \in\NN^{2}\text{ such that }n+m\neq
N\text{ satisfy }\pi_{N}\left(  A_{n}\otimes A_{m}\right)  =0.
\label{exe.I=0.sol.5}
\end{equation}


Let $i\in I_N$ be arbitrary. Then, $i\in A_N$, so that $\Delta\left(
i\right)  \in\Delta\left(  A_{N}\right)  \subset \left(  A\otimes A\right)
_{N}$ (since $\Delta$ is a graded map). Thus, $\pi_{N}\left(  \Delta\left(
i\right)  \right)  =\Delta\left(  i\right)  $ (by \eqref{exe.I=0.sol.3},
applied to $t=\Delta\left(  i\right)  $). On the other hand, since $i\in
I_{N}\subset I$, we have
\begin{align*}
\Delta\left(  i\right)   &  \in\Delta\left(  I\right)  \subset
\underbrace{I}_{=\bigoplus_{n\geq0}I_{n}}\otimes\underbrace{A}_{=\bigoplus
_{m\geq0}A_{m}}+\underbrace{A}_{=\bigoplus_{m\geq0}A_{m}}\otimes
\underbrace{I}_{=\bigoplus_{n\geq0}I_{n}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }I\text{ is a two-sided coideal}\right) \\
&  =\underbrace{\left(  \bigoplus_{n\geq0}I_{n}\right)  \otimes\left(
\bigoplus_{m\geq0}A_{m}\right)  }_{=\sum\limits_{\left(  m,n\right)
\in\NN^{2}}I_{n}\otimes A_{m}}+\underbrace{\left(  \bigoplus_{m\geq
0}A_{m}\right)  \otimes\left(  \bigoplus_{n\geq0}I_{n}\right)  }
_{=\sum\limits_{\left(  m,n\right)  \in\NN^{2}}A_{m}\otimes I_{n}}
  =\sum\limits_{\left(  m,n\right)  \in\NN^{2}}I_{n}\otimes A_{m}
+\sum\limits_{\left(  m,n\right)  \in\NN^{2}}A_{m}\otimes I_{n}.
\end{align*}
Thus,
\begin{align*}
&  \pi_{N}\left(  \Delta\left(  i\right)  \right) \\
&  \in\pi_{N}\left(  \sum\limits_{\left(  m,n\right)  \in\NN^2}
I_{n}\otimes A_{m}+\sum\limits_{\left(  m,n\right)  \in\NN^2}
A_{m}\otimes I_{n}\right) \\
&  =\underbrace{\sum\limits_{\left(  m,n\right)  \in\NN^{2}}\pi
_{N}\left(  I_{n}\otimes A_{m}\right)  }_{=\sum\limits_{\substack{\left(
m,n\right)  \in\NN^{2};\\ n+m=N}}\pi_{N}\left(  I_{n}\otimes
A_{m}\right)  +\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};
\\ n+m\neq N}}\pi_{N}\left(  I_{n}\otimes A_{m}\right)  }+\underbrace{
\sum\limits_{\left(  m,n\right)  \in\NN^{2}}\pi_{N}\left(  A_{m}\otimes
I_{n}\right)  }_{=\sum\limits_{\substack{\left(  m,n\right)  \in
\NN^{2};\\m+n=N}}\pi_{N}\left(  A_{m}\otimes I_{n}\right)  +
\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\m+n\neq N}}\pi
_{N}\left(  A_{m}\otimes I_{n}\right)  }\\
&  =\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\ n+m=N}}
\pi_{N}\left(  I_{n}\otimes A_{m}\right)  +\sum\limits_{\substack{\left(
m,n\right)  \in\NN^{2};\\ n+m\neq N}}\pi_{N}\left(  \underbrace{I_{n}}
_{\subset A_{n}}\otimes A_{m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  m,n\right)
\in\NN^{2};\\m+n=N}}\pi_{N}\left(  A_{m}\otimes I_{n}\right)
+\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\m+n\neq N}}
\pi_{N}\left(  A_{m}\otimes\underbrace{I_{n}}_{\subset A_{n}}\right) \\
&  \subset\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};
\\ n+m=N}}\pi_{N}\left(  I_{n}\otimes A_{m}\right)  +
\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\ n+m\neq
N}}\underbrace{\pi_{N}\left(  A_{n}\otimes A_{m}\right)  }
_{\substack{=0\\\text{(by \eqref{exe.I=0.sol.5})}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  m,n\right)
\in\NN^{2};\\m+n=N}}\pi_{N}\left(  A_{m}\otimes I_{n}\right)
+\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\m+n\neq
N}}\underbrace{\pi_{N}\left(  A_{m}\otimes A_{n}\right)  }
_{\substack{=0\\\text{(by \eqref{exe.I=0.sol.5}, applied}\\\text{to }\left(
m,n\right)  \text{ instead of }\left(  n,m\right)  \text{)}}}\\
&  =\underbrace{\sum\limits_{\substack{\left(  m,n\right)  \in
\NN^{2};\\ n+m=N}}\pi_{N}\left(  I_{n}\otimes A_{m}\right)  }_{=
\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\ n+m=N;\\ n<N}}
\pi_{N}\left(  I_{n}\otimes A_{m}\right)  +\sum\limits_{\substack{\left(
m,n\right)  \in\NN^{2};\\ n+m=N;\\ n\geq N}}\pi_{N}\left(  I_{n}\otimes
A_{m}\right)  }+\underbrace{\sum\limits_{\substack{\left(  m,n\right)
\in\NN^{2};\\m+n=N}}\pi_{N}\left(  A_{m}\otimes I_{n}\right)  }
_{=\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\m+n=N;\\ n<N}}
\pi_{N}\left(  A_{m}\otimes I_{n}\right)  +\sum\limits_{\substack{\left(
m,n\right)  \in\NN^{2};\\m+n=N;\\ n\geq N}}\pi_{N}\left(  A_{m}\otimes
I_{n}\right)  } \\
&  =\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\ n+m=N;
\\ n<N}}\pi_{N}\left(  \underbrace{I_{n}}_{\substack{=0\\
\text{(by \eqref{exe.I=0.sol.2})}}}\otimes A_{m}\right) 
+\underbrace{\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\
n+m=N;\\ n\geq N}}\pi_{N}\left(  I_{n}\otimes A_{m}\right)  }
_{\substack{=I_N\otimes A_0\\\text{(since the only pair }\left( m,n\right)
\in\NN^{2} \\\text{satisfying }n+m=N\text{ and }n\geq N\text{ is }
\left( 0,N\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  m,n\right)
\in\NN^{2};\\m+n=N;\\ n<N}}\pi_{N}\left(  A_{m}\otimes\underbrace{I_{n}}
_{\substack{=0\\\text{(by \eqref{exe.I=0.sol.2})}}}\right)  +\underbrace{
\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};\\m+n=N;\\ n\geq
N}}\pi_{N}\left(  A_{m}\otimes I_{n}\right)  }_{\substack{=A_{0}\otimes
I_{N}\\\text{(since the only pair }\left(  m,n\right)  \in\NN^{2}
\\\text{satisfying }m+n=N\text{ and }n\geq N\text{ is }\left(
0,N\right)  \text{)}}}\\
&  =\sum\limits_{\substack{\left(  m,n\right)  \in\NN^{2};
\\ n+m=N;\\ n<N}}\underbrace{\pi_{N}\left(  0\otimes A_{m}\right)  }_{=0}
+I_{N}\otimes\underbrace{A_{0}}_{\substack{=\kk 1_{A}\\\text{(since
 }A\text{ is connected)}}}+\sum\limits_{\substack{\left(  m,n\right)
\in\NN^{2};\\m+n=N;\\ n<N}}\underbrace{\pi_{N}\left(  A_{m}
\otimes0\right)  }_{=0}+\underbrace{A_{0}}_{\substack{=\kk 1_{A}
\\\text{(since }A\text{ is connected)}}}\otimes I_{N}\\
&  =I_{N}\otimes\kk 1_{A} + \kk 1_{A}\otimes I_{N}.
\end{align*}
Since $\pi_{N}\left(  \Delta\left(  i\right)  \right)  =\Delta\left(
i\right)  $, this rewrites as $\Delta\left(  i\right)  \in I_{N}
\otimes \kk 1_{A} + \kk 1_{A}\otimes I_{N}$. In other words, there
exists some $y\in I_{N}\otimes \kk 1_{A}$ and some
$z\in \kk 1_A \otimes I_{N}$ such that $\Delta\left(  i\right)  =y+z$. Consider these
$y$ and $z$.

Since $y\in I_{N} \otimes \kk 1_{A} = I_N \otimes 1_A$,
there exists a $j\in I_{N}$ such
that $y=j\otimes 1_{A}$. Consider this $j$.

Since $z\in \kk 1_{A}\otimes I_{N} = 1_A \otimes I_N$,
there exists a $k\in I_{N}$ such
that $z=1_{A}\otimes k$. Consider this $k$.

Now,
\[
\Delta\left(  i\right)  =\underbrace{y}_{=j\otimes1_{A}}+\underbrace{z}
_{=1_{A}\otimes k}=j\otimes1_{A}+1_{A}\otimes k.
\]
Applying the map $\epsilon\otimes\id :A\otimes A\rightarrow A$
to both sides of this equation, we obtain
\[
\left(  \epsilon\otimes\id \right)  \left(  \Delta\left(
i\right)  \right)  =\left(  \epsilon\otimes\id \right)  \left(
j\otimes1_{A}+1_{A}\otimes k\right)  =\underbrace{\epsilon\left(  j\right)
}_{\substack{=0\\\text{(since }j\in I_{N}\text{, so that}\\\epsilon\left(
j\right)  \in\epsilon\left(  I_{N}\right)  =0\text{)}}}\id
\left(  1_{A}\right)  +\underbrace{\epsilon\left(  1_{A}\right)  }
_{=1}\underbrace{\id \left(  k\right)  }_{=k}
=\underbrace{0\id \left(  1_{A}\right)  }_{=0}+k=k.
\]
Since $\left(  \epsilon\otimes\id \right)  \left(  \Delta\left(
i\right)  \right)  =i$ (by the axioms of a coalgebra), this rewrites as $i=k$.

Similarly, applying $\id \otimes\epsilon$ to both sides of the
equation $\Delta\left(  i\right)  =j\otimes1_{A}+1_{A}\otimes k$ and
simplifying, we obtain $i=j$.

Now,
\[
\Delta\left(  i\right)  =\underbrace{j}_{=i}\otimes1_{A}+1_{A}\otimes
\underbrace{k}_{=i}=i\otimes1_{A}+1_{A}\otimes i.
\]
Hence, $i$ is primitive. In other words, $i\in\Liep$. Combined with
$i\in I_{N}=I\cap A_{N}\subset I$, this yields $i\in I\cap\Liep =0$,
so that $i=0$.

Now, forget that we fixed $i$. We thus have shown that every $i\in I_{N}$
satisfies $i=0$. In other words, $I_{N}=0$. In other words,
\eqref{exe.I=0.sol.1} holds for $n=N$. This completes the induction proof of
\eqref{exe.I=0.sol.1}.

Now, $I=\bigoplus_{n\geq0}\underbrace{I_{n}}_{\substack{=0\\\text{(by
\eqref{exe.I=0.sol.1})}}}=\bigoplus_{n\geq0}0=0$. This solves part (a) of the exercise.

(b) By Exercise~\ref{exe.kernel-is-coideal}(a), we know that $\ker f$ is a
two-sided coideal of $A$. It further satisfies $\ker f=\bigoplus_{n\geq
0}\left(  \left(  \ker f\right)  \cap A_{n}\right)  $ (since $f$ is graded).
If $f\mid_{\Liep}$ is injective, then $\left(  \ker f\right)
\cap\Liep=0$, and thus part (a) of the current exercise (applied to
$I=\ker f$) yields that $\ker f=0$, so that $f$ is injective. This solves part
(b) of the exercise.

(c) The solution of part (c) proceeds precisely as the solution of part (b),
except that instead of using Exercise~\ref{exe.kernel-is-coideal}(a) we now
must use Exercise~\ref{exe.kernel-is-coideal}(b).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liealg}}

\textit{Solution to Exercise~\ref{exe.liealg}.}

(a) Let $A$ be any associative $\kk$-algebra. Define a $\kk$-bilinear map
$\left[ \cdot, \cdot \right] : A\times A \rightarrow A$ by
setting
\[
\left[ a,b \right] = ab-ba\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in A.
\]


We must prove that this $\kk$-bilinear map $\left[ \cdot, \cdot \right]$
makes $A$ into a Lie algebra. In order to do so, it is clearly enough to
prove that
\begin{equation}
\left[  x,x\right]  =0\ \ \ \ \ \ \ \ \ \ \text{for all }x\in A,
\label{sol.liealg.1}
\end{equation}
and that
\begin{equation}
\left[  x,\left[  y,z\right]  \right]  +\left[  z,\left[  x,y\right]  \right]
+\left[  y,\left[  z,x\right]  \right]  =0
\ \ \ \ \ \ \ \ \ \ \text{for all }x,y,z\in A.
\label{sol.liealg.2}
\end{equation}


\textit{Proof of \eqref{sol.liealg.1}:} Every $x\in A$ satisfies
\begin{align*}
\left[  x,x\right]   &  =xx-xx\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\left[  x,x\right]  \right) \\
&  =0.
\end{align*}
This proves \eqref{sol.liealg.1}.

\textit{Proof of \eqref{sol.liealg.2}:} Every $x,y,z\in A$ satisfy
\begin{align*}
&  \underbrace{\left[  x,\left[  y,z\right]  \right]  }_{\substack{=x\left[
y,z\right]  -\left[  y,z\right]  x\\\text{(by the definition}\\\text{of
}\left[  x,\left[  y,z\right]  \right]  \text{)}}}+\underbrace{\left[
z,\left[  x,y\right]  \right]  }_{\substack{=z\left[  x,y\right]  -\left[
x,y\right]  z\\\text{(by the definition}\\\text{of }\left[  z,\left[
x,y\right]  \right]  \text{)}}}+\underbrace{\left[  y,\left[  z,x\right]
\right]  }_{\substack{=y\left[  z,x\right]  -\left[  z,x\right]  y\\\text{(by
the definition}\\\text{of }\left[  y,\left[  z,x\right]  \right]  \text{)}}}\\
&  =x\underbrace{\left[  y,z\right]  }_{\substack{=yz-zy\\\text{(by the
definition}\\\text{of }\left[  y,z\right]  \text{)}}}-\underbrace{\left[
y,z\right]  }_{\substack{=yz-zy\\\text{(by the definition}\\\text{of }\left[
y,z\right]  \text{)}}}x+z\underbrace{\left[  x,y\right]  }
_{\substack{=xy-yx\\\text{(by the definition}\\\text{of }\left[  x,y\right]
\text{)}}}-\underbrace{\left[  x,y\right]  }_{\substack{=xy-yx\\\text{(by the
definition}\\\text{of }\left[  x,y\right]  \text{)}}}z+y\underbrace{\left[
z,x\right]  }_{\substack{=zx-xz\\\text{(by the definition}\\\text{of }\left[
z,x\right]  \text{)}}}-\underbrace{\left[  z,x\right]  }
_{\substack{=zx-xz\\\text{(by the definition}\\\text{of }\left[  z,x\right]
\text{)}}}y\\
&  =x\left(  yz-zy\right)  -\left(  yz-zy\right)  x+z\left(  xy-yx\right)
-\left(  xy-yx\right)  z+y\left(  zx-xz\right)  -\left(  zx-xz\right)  y\\
&  =xyz-xzy-yzx+zyx+zxy-zyx-xyz+yxz+yzx-yxz-zxy+xzy\\
&  =0.
\end{align*}
This proves \eqref{sol.liealg.2}.

Now, both \eqref{sol.liealg.1} and \eqref{sol.liealg.2} are proven. Hence, the
$\kk$-module $A$ endowed with the $\kk$-bilinear map
$\left[ \cdot, \cdot \right]$ satisfies the axioms of a Lie algebra, and
therefore is a Lie algebra. This solves part (a) of the exercise.

(b) Let $A$ be a bialgebra. Let $\Liep$ be the set of all primitive
elements of $A$. Then, $\Liep$ is a $\kk$-submodule of $A$
(because it is easy to see that $0\in\Liep$, that $\lambda
a\in\Liep$ for every $\lambda\in\kk$ and $a\in\Liep$, and
that $a+b\in\Liep$ for all $a\in\Liep$ and $b\in\Liep$).
A simple computation shows that every $x\in\Liep$ and $y\in
\Liep$ satisfy $\left[  x,y\right]  \in\Liep$
\ \ \ \ \footnote{\textit{Proof.} Let $x\in\Liep$ and $y\in
\Liep$. Then, $\Delta\left[  x,y\right]  =1\otimes\left[  x,y\right]
+\left[  x,y\right]  \otimes1$ (by \eqref{commutator-of-primitives}). In other
words, the element $\left[  x,y\right]  $ of $A$ is primitive. In other words,
$\left[  x,y\right]  \in\Liep$ (since $\Liep$ is the set of all
primitive elements of $A$), qed.}. Hence, $\left[  \Liep ,\Liep 
\right]  \subset\Liep$ (since $\Liep$ is a
$\kk$-submodule of $A$). Therefore, $\Liep$ is a Lie subalgebra of $A$.
This solves part (b) of the exercise.

(c) For every subset $S$ of a $\kk$-module $U$, we let $\left\langle
S\right\rangle$ denote the $\kk$-submodule of $U$ spanned by $S$.

We have defined $J$ as the two-sided ideal of $T\left(  \Liep \right)  $
generated by all elements $xy-yx-\left[  x,y\right]  $ for $x,y$ in
$\Liep$. In other words,
\begin{align}
J &  =\underbrace{T\left(  \Liep \right)  }_{\supset\kk }
\cdot\left\langle xy-yx-\left[  x,y\right]  \ \mid\ x,y\in\Liep 
\right\rangle \cdot\underbrace{T\left(  \Liep \right)  }_{\supset
\kk }\label{sol.liealg.c.1}\\
&  \supset\kk \cdot\left\langle xy-yx-\left[  x,y\right]  \ \mid
\ x,y\in\Liep \right\rangle \cdot\kk =\left\langle xy-yx-\left[
x,y\right]  \ \mid\ x,y\in\Liep \right\rangle \label{sol.liealg.c.1a}\\
&  \supset\left\{  xy-yx-\left[  x,y\right]  \ \mid\ x,y\in\Liep 
\right\}  .\nonumber
\end{align}
Thus,
\begin{equation}
xy-yx-\left[  x,y\right]  \in J\ \ \ \ \ \ \ \ \ \ \text{for all }
x,y\in\Liep .\label{sol.liealg.c.2.pf.1}
\end{equation}
It is also easy to show that
\begin{equation}
xy-yx-\left[  x,y\right]  \text{ is a primitive element of }T\left(
\Liep \right)  \text{ for all }x,y\in\Liep 
.\label{sol.liealg.c.prim}
\end{equation}
\footnote{\textit{Proof.} Let $x,y\in\Liep$. Then, $\Delta\left(
xy-yx\right)  =1\otimes\left(  xy-yx\right)  +\left(  xy-yx\right)  \otimes1$
(this can be proven just as in the proof of \eqref{commutator-of-primitives}).
But $\left[  x,y\right]  $ is an element of $\Liep$, and thus (by the
definition of the comultiplication of $T\left(  \Liep \right)  $)
satisfies $\Delta\left[  x,y\right]  =1\otimes\left[  x,y\right]  +\left[
x,y\right]  \otimes1$ in $T\left(  \Liep \right)  \otimes T\left(
\Liep \right)  $. Now, since $\Delta$ is a $\kk$-linear map, we
have
\begin{align*}
\Delta\left(  xy-yx-\left[  x,y\right]  \right)   &  =\underbrace{\Delta
\left(  xy-yx\right)  }_{=1\otimes\left(  xy-yx\right)  +\left(  xy-yx\right)
\otimes1}-\underbrace{\Delta\left[  x,y\right]  }_{=1\otimes\left[
x,y\right]  +\left[  x,y\right]  \otimes1}\\
&  =\left(  1\otimes\left(  xy-yx\right)  +\left(  xy-yx\right)
\otimes1\right)  -\left(  1\otimes\left[  x,y\right]  +\left[  x,y\right]
\otimes1\right)  \\
&  =\left(  \underbrace{\left(  xy-yx\right)  \otimes1-\left[  x,y\right]
\otimes1}_{=\left(  xy-yx-\left[  x,y\right]  \right)  \otimes1}\right)
+\left(  \underbrace{1\otimes\left(  xy-yx\right)  -1\otimes\left[
x,y\right]  }_{=1\otimes\left(  xy-yx-\left[  x,y\right]  \right)  }\right)
\\
&  =\left(  xy-yx-\left[  x,y\right]  \right)  \otimes1+1\otimes\left(
xy-yx-\left[  x,y\right]  \right)  \\
&  =1\otimes\left(  xy-yx-\left[  x,y\right]  \right)  +\left(  xy-yx-\left[
x,y\right]  \right)  \otimes1.
\end{align*}
In other words, the element $xy-yx-\left[  x,y\right]  $ of $T\left(
\Liep \right)  $ is primitive. This proves \eqref{sol.liealg.c.prim}.}

From this it is easy to obtain
\begin{equation}
\Delta\left(  \left\langle xy-yx-\left[  x,y\right]  \ \mid\ x,y\in
\Liep \right\rangle \right)  \subset J\otimes T\left(  \Liep 
\right)  +T\left(  \Liep \right)  \otimes J.\label{sol.liealg.c.2}
\end{equation}
\footnote{\textit{Proof.} Let $x,y\in\Liep$ be arbitrary. Then,
$xy-yx-\left[  x,y\right]  $ is a primitive element of $T\left(
\Liep \right)  $ (by \eqref{sol.liealg.c.prim}). In other words,
\begin{align*}
\Delta\left(  xy-yx-\left[  x,y\right]  \right)   &  =1\otimes\left(
xy-yx-\left[  x,y\right]  \right)  +\left(  xy-yx-\left[  x,y\right]  \right)
\otimes1\\
&  =\underbrace{\left(  xy-yx-\left[  x,y\right]  \right)  }_{\substack{\in
J\\\text{(by \eqref{sol.liealg.c.2.pf.1})}}}\otimes\underbrace{1}_{\in
T\left(  \Liep \right)  }+\underbrace{1}_{\in T\left(  \Liep 
\right)  }\otimes\underbrace{\left(  xy-yx-\left[  x,y\right]  \right)
}_{\substack{\in J\\\text{(by \eqref{sol.liealg.c.2.pf.1})}}}\\
&  \in J\otimes T\left(  \Liep \right)  +T\left(  \Liep \right)
\otimes J.
\end{align*}
\par
Now forget that we fixed $x,y$. We thus have proven that $\Delta\left(
xy-yx-\left[  x,y\right]  \right)  \in J\otimes T\left(  \Liep \right)
+T\left(  \Liep \right)  \otimes J$ for all $x,y\in\Liep$. Since
$J\otimes T\left(  \Liep \right)  +T\left(  \Liep \right)  \otimes
J$ is a $\kk$-submodule of $T\left(  \Liep \right)  \otimes
T\left(  \Liep \right)  $, this yields that $\Delta\left(  \left\langle
xy-yx-\left[  x,y\right]  \ \mid\ x,y\in\Liep \right\rangle \right)
\subset J\otimes T\left(  \Liep \right)  +T\left(  \Liep \right)
\otimes J$. Thus, \eqref{sol.liealg.c.2} is proven.} But applying $\Delta$ to
both sides of the equality \eqref{sol.liealg.c.1}, we obtain
\begin{align}
\Delta\left(  J\right)   &  =\Delta\left(  T\left(  \Liep \right)
\cdot\left\langle xy-yx-\left[  x,y\right] \ \mid
\ x,y\in\Liep \right\rangle \cdot T\left(
\Liep \right)  \right) \nonumber \\
& \subset\underbrace{\Delta\left(  T\left(
\Liep \right)  \right)  }_{\subset T\left(  \Liep \right)  \otimes
T\left(  \Liep \right)  }\cdot\underbrace{\Delta\left(  \left\langle
xy-yx-\left[  x,y\right] \ \mid
\ x,y\in\Liep  \right\rangle \right)  }_{\substack{\subset J\otimes
T\left(  \Liep \right)  +T\left(  \Liep \right)  \otimes
J\\\text{(by \eqref{sol.liealg.c.2})}}}\cdot\underbrace{\Delta\left(  T\left(
\Liep \right)  \right)  }_{\subset T\left(  \Liep \right)  \otimes
T\left(  \Liep \right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a }\kk 
\text{-algebra homomorphism}\right)  \nonumber\\
&  \subset\left(  T\left(  \Liep \right)  \otimes T\left(
\Liep \right)  \right)  \cdot\left(  J\otimes T\left(  \Liep 
\right)  +T\left(  \Liep \right)  \otimes J\right)  \cdot\left(
T\left(  \Liep \right)  \otimes T\left(  \Liep \right)  \right)
\nonumber\\
&  =\underbrace{\left(  T\left(  \Liep \right)  \otimes T\left(
\Liep \right)  \right)  \cdot\left(  J\otimes T\left(  \Liep 
\right)  \right)  \cdot\left(  T\left(  \Liep \right)  \otimes T\left(
\Liep \right)  \right)  }_{=\left(  T\left(  \Liep \right)  \cdot
J\cdot T\left(  \Liep \right)  \right)  \otimes\left(  T\left(
\Liep \right)  \cdot T\left(  \Liep \right)  \cdot T\left(
\Liep \right)  \right)  }+\underbrace{\left(  T\left(  \Liep 
\right)  \otimes T\left(  \Liep \right)  \right)  \cdot\left(  T\left(
\Liep \right)  \otimes J\right)  \cdot\left(  T\left(  \Liep 
\right)  \otimes T\left(  \Liep \right)  \right)  }_{=\left(  T\left(
\Liep \right)  \cdot T\left(  \Liep \right)  \cdot T\left(
\Liep \right)  \right)  \otimes\left(  T\left(  \Liep \right)
\cdot J\cdot T\left(  \Liep \right)  \right)  }\nonumber\\
&  =\underbrace{\left(  T\left(  \Liep \right)  \cdot J\cdot T\left(
\Liep \right)  \right)  }_{\substack{=J\\\text{(since }J\text{ is a
two-sided ideal of }T\left(  \Liep \right)  \text{)}}}\otimes
\underbrace{\left(  T\left(  \Liep \right)  \cdot T\left(
\Liep \right)  \cdot T\left(  \Liep \right)  \right)  }_{=T\left(
\Liep \right)  } \nonumber \\
& \qquad \qquad +\underbrace{\left(  T\left(  \Liep \right)  \cdot
T\left(  \Liep \right)  \cdot T\left(  \Liep \right)  \right)
}_{=T\left(  \Liep \right)  }\otimes\underbrace{\left(  T\left(
\Liep \right)  \cdot J\cdot T\left(  \Liep \right)  \right)
}_{\substack{=J\\\text{(since }J\text{ is a two-sided ideal of }T\left(
\Liep \right)  \text{)}}}\nonumber\\
&  =J\otimes T\left(  \Liep \right)  +T\left(  \Liep \right)
\otimes J.\label{sol.liealg.c.Delta}
\end{align}


Also,
\begin{equation}
\epsilon\left(  \left\langle xy-yx-\left[  x,y\right]  \ \mid\ x,y\in
\Liep \right\rangle \right)  =0.\label{sol.liealg.c.3}
\end{equation}
\footnote{\textit{Proof.} Let $x,y\in\Liep$. Then, $xy-yx\in
\Liep ^{\otimes2}$ and thus $\epsilon\left(  xy-yx\right)  =0$ (by the
definition of the comultiplication $\epsilon$ on $T\left(  \Liep 
\right)  $). Also, $\left[  x,y\right]  \in\Liep ^{\otimes1}$ and thus
$\epsilon\left(  \left[  x,y\right]  \right)  =0$ (again by the definition of
the comultiplication $\epsilon$ on $T\left(  \Liep \right)  $). Since
$\epsilon$ is $\kk$-linear, we have
\[
\epsilon\left(  xy-yx-\left[  x,y\right]  \right)  =\underbrace{\epsilon
\left(  xy-yx\right)  }_{=0}-\underbrace{\epsilon\left(  \left[  x,y\right]
\right)  }_{=0}=0-0=0.
\]
\par
Now, forget that we fixed $x,y$. We thus have shown that $\epsilon\left(
xy-yx-\left[  x,y\right]  \right)  =0$ for all $x,y\in\Liep$. By linearity,
this yields
$\epsilon\left(  \left\langle xy-yx-\left[  x,y\right]  \ \mid
\ x,y\in\Liep \right\rangle \right)  =0$, so that \eqref{sol.liealg.c.3}
is proven.} Applying the map $\epsilon$ to both sides of the equality
\eqref{sol.liealg.c.1}, we obtain
\begin{align*}
\epsilon\left(  J\right)    & =\epsilon\left(  T\left(  \Liep \right)
\cdot\left\langle xy-yx-\left[  x,y\right] \ \mid
\ x,y\in\Liep \right\rangle \cdot T\left(
\Liep \right)  \right)  \subset\epsilon\left(  T\left(  \Liep 
\right)  \right)  \cdot\underbrace{\epsilon\left(  \left\langle xy-yx-\left[
x,y\right] \ \mid
\ x,y\in\Liep \right\rangle \right)  }_{\substack{=0\\\text{(by
\eqref{sol.liealg.c.3})}}}\cdot\epsilon\left(  T\left(  \Liep \right)
\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(\text{since } \epsilon \text{ is a }\kk\text{-algebra homomorphism}\right) \\
& =0.
\end{align*}
Thus, $\epsilon\left(  J\right)  =0$. Combined with
\eqref{sol.liealg.c.Delta}, this yields that $J$ is a two-sided coideal of
$T\left(  \Liep \right)  $. Thus, the quotient
$T\left(  \Liep \right)  /J$ becomes a
$\kk$-coalgebra. Also, $T\left(  \Liep \right)  /J$ is a
$\kk$-algebra, since $J$ is a two-sided ideal of $T\left(
\Liep \right)  $.

Now we want to show that $T\left(  \Liep \right)  /J$ has a structure of
a cocommutative $\kk$-bialgebra inherited from $T\left(  \Liep 
\right)  $. We already know that $T\left(  \Liep \right) /J$ is a
$\kk$-algebra and a $\kk$-coalgebra, with both structures being
inherited from $T\left(  \Liep \right)  $. The remaining axioms of a
cocommutative $\kk$-bialgebra that need to be checked for $T\left(
\Liep \right) /J$ are the commutativity of the diagrams
\eqref{bialgebra-diagrams} and the commutativity of the diagram
\eqref{cocommutativity-diagram}; these axioms are clearly preserved under
taking quotients. Hence, $T\left(  \Liep \right)  /J$ is a cocommutative
$\kk$-bialgebra, with its structure being inherited from $T\left(
\Liep \right)  $. In other words, $\mathcal{U}\left(  \Liep 
\right)  $ is a cocommutative $\kk$-bialgebra, with its structure being
inherited from $T\left(  \Liep \right)  $ (since $\mathcal{U}\left(
\Liep \right)  =T\left(  \Liep \right)  /J$). This solves part (c)
of the exercise.

(d) Proposition~\ref{antipodes-are-antiendomorphisms} shows that
the antipode $S$ of $T\left(  \Liep \right)  $ is a $\kk$-algebra
anti-homomorphism from $T\left(  \Liep \right)  $ to
$T\left(  \Liep \right)  $.

It is easy to see that
\begin{equation}
S\left(  \left\langle xy-yx-\left[  x,y\right]  \ \mid\ x,y\in\Liep 
\right\rangle \right)  \subset J.\label{sol.liealg.d.1}
\end{equation}
\footnote{\textit{Proof.} Let $x,y\in\Liep$. Then, $xy-yx-\left[
x,y\right]  $ is a primitive element of $T\left(  \Liep \right)  $ (by
\eqref{sol.liealg.c.prim}). Hence,
Proposition~\ref{antipode-on-primitives} (applied to
$xy-yx-\left[x,y\right]$ instead of $x$) yields
\[
S\left(  xy-yx-\left[  x,y\right]  \right)  =-\underbrace{\left(
xy-yx-\left[  x,y\right]  \right)  }_{\substack{\in J\\\text{(by
\eqref{sol.liealg.c.2.pf.1})}}}\in-J=J
\]
(since $J$ is a two-sided ideal).
\par
Now, forget that we fixed $x,y$. We thus have shown that
$S\left( xy-yx-\left[  x,y\right] \right) \in J$ for all
$x,y\in\Liep$. Since $J$ is a $\kk$-module, this yields that
$S\left(  \left\langle xy-yx-\left[  x,y\right]  \ \mid\ x,y\in\Liep 
\right\rangle \right)  \subset J$. Thus, \eqref{sol.liealg.d.1} is proven.}
But recall that if $A$ and $B$ are two $\kk$-algebras and
$\varphi:A\rightarrow B$ is a $\kk$-algebra homomorphism, then any
$\kk$-submodules $V_{1}$, $V_{2}$, $...$, $V_{n}$ of $A$ satisfy
\[
\varphi\left(  V_{1}V_{2}...V_{n}\right)  =\varphi\left(  V_{1}\right)
\cdot\varphi\left(  V_{2}\right)  \cdot...\cdot\varphi\left(  V_{n}\right)  .
\]
Similarly, if $A$ and $B$ are two $\kk$-algebras and $\varphi
:A\rightarrow B$ is a $\kk$-algebra anti-homomorphism, then any
$\kk$-submodules $V_{1}$, $V_{2}$, $...$, $V_{n}$ of $A$ satisfy
\[
\varphi\left(  V_{1}V_{2}...V_{n}\right)  =\varphi\left(  V_{n}\right)
\cdot\varphi\left(  V_{n-1}\right)  \cdot...\cdot\varphi\left(  V_{1}\right)
.
\]
Applying this to $A=T\left(  \Liep \right)  $, $B=T\left(
\Liep \right)  $, $\varphi=S$, $n=3$, $V_{1}=T\left(  \Liep 
\right)  $, $V_{2}=\left\langle xy-yx-\left[  x,y\right] \ \mid
\ x,y\in\Liep \right\rangle $ and
$V_{3}=T\left(  \Liep \right)  $, we obtain
\begin{align*}
S\left(  T\left(  \Liep \right)  \cdot\left\langle xy-yx-\left[
x,y\right] \ \mid
\ x,y\in\Liep \right\rangle \cdot T\left(  \Liep \right)  \right)    &
=\underbrace{S\left(  T\left(  \Liep \right)  \right)  }_{\subset
T\left(  \Liep \right)  }\cdot
\underbrace{S\left(\left\langle xy-yx-\left[  x,y\right]
\ \mid \ x,y\in\Liep
\right\rangle\right)}
_{\subset J} \cdot\underbrace{S\left(  T\left(  \Liep \right)  \right)
}_{\subset T\left(  \Liep \right)  }\\
& \subset T\left(  \Liep \right)  \cdot J
 \cdot T\left(  \Liep \right) \subset J
\end{align*}
(since $J$ is a two-sided ideal).
Due to \eqref{sol.liealg.c.1}, this rewrites as
$S\left(  J\right)  \subset J$. Hence, the $\kk$-linear map $S:T\left(
\Liep \right)  \rightarrow T\left(  \Liep \right)  $ induces a
$\kk$-linear map $\overline{S}:T\left(  \Liep \right)
/J\rightarrow T\left(  \Liep \right)  /J$ on the quotient
$\kk$-modules. This resulting map $\overline{S}:T\left(  \Liep \right)
/J\rightarrow T\left(  \Liep \right)  /J$ is an antipode for the
$\kk$-bialgebra $T\left(  \Liep \right)  /J$ (in fact, the
diagram \eqref{antipode-diagram} with $A$ and $S$ replaced by $T\left(
\Liep \right)  /J$ and $\overline{S}$ commutes, because the diagram
\eqref{antipode-diagram} with $A$ replaced by $T\left(  \Liep \right)  $
commutes). Hence, the $\kk$-bialgebra $T\left(  \Liep \right)
/J$ has an antipode, and thus is a Hopf algebra. In other words,
$\mathcal{U}\left(  \Liep \right)  $ is a Hopf algebra (since
$\mathcal{U}\left(  \Liep \right)  =T\left(  \Liep \right)  /J$).
Since we already know that $\mathcal{U}\left(  \Liep \right)  $ is
cocommutative, this yields that $\mathcal{U}\left(  \Liep \right)  $ is
a cocommutative Hopf algebra. This solves part (d) of the exercise.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.commutative.convolution}}

\textit{Solution to Exercise \ref{exe.commutative.convolution}.} Let
$f, g \in \Hom \left( C, A \right)$. We must show that
$f \star g = g \star f$.

There are several ways to do this. The slickest one is perhaps the following:

The $\kk$-algebra $A$ is commutative. In other words, the diagram
\eqref{commutativity-diagram} commutes. In other words, $m_{A}=m_{A}\circ
T_{A}$, where $T_{A}$ denotes the twist map $A\otimes A\rightarrow A\otimes
A,\ b\otimes a\mapsto a\otimes b$.

The $\kk$-coalgebra $C$ is cocommutative. In other words, the diagram
\eqref{cocommutativity-diagram} commutes. In other words, $\Delta_{C}%
=T_{C}\circ\Delta_{C}$, where $T_{C}$ denotes the twist map $C\otimes
C\rightarrow C\otimes C,\ c\otimes d\mapsto d\otimes c$.

It is straightforward to see that
\begin{equation}
T_{A}\circ\left(  f\otimes g\right)  =\left(  g\otimes f\right)  \circ
T_{C}.\label{sol.commutative.convolution.twist-comm}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.commutative.convolution.twist-comm}:}
Let $z\in C\otimes C$. We shall prove the equality $\left(  T_{A}\circ\left(
f\otimes g\right)  \right)  \left(  z\right)  =\left(  \left(  g\otimes
f\right)  \circ T_{C}\right)  \left(  z\right)  $.
\par
Since this equality is $\kk$-linear in $z$, we can WLOG assume that $z$
is a pure tensor (since each tensor $C\otimes C$ is a $\kk$-linear
combination of pure tensors). Assume this. Hence, $z=c\otimes d$ for some
$c\in C$ and $d\in C$. Consider these $c$ and $d$.
\par
Now,%
\begin{align*}
\left(  T_{A}\circ\left(  f\otimes g\right)  \right)  \left(  \underbrace{z}%
_{=c\otimes d}\right)    & =\left(  T_{A}\circ\left(  f\otimes g\right)
\right)  \left(  c\otimes d\right)  =T_{A}\left(  \underbrace{\left(  f\otimes
g\right)  \left(  c\otimes d\right)  }_{=f\left(  c\right)  \otimes g\left(
d\right)  }\right)  =T_{A}\left(  f\left(  c\right)  \otimes g\left(
d\right)  \right)  \\
& =g\left(  d\right)  \otimes f\left(  c\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the map }T_{A}\right)  .
\end{align*}
Comparing this with%
\[
\left(  \left(  g\otimes f\right)  \circ T_{C}\right)  \left(  \underbrace{z}%
_{=c\otimes d}\right)  =\left(  \left(  g\otimes f\right)  \circ T_{C}\right)
\left(  c\otimes d\right)  =\left(  g\otimes f\right)  \left(
\underbrace{T_{C}\left(  c\otimes d\right)  }_{\substack{=d\otimes
c\\\text{(by the definition of the map }T_{C}\text{)}}}\right)  =\left(
g\otimes f\right)  \left(  d\otimes c\right)  =g\left(  d\right)  \otimes
f\left(  c\right)  ,
\]
we obtain $\left(  T_{A}\circ\left(  f\otimes g\right)  \right)  \left(
z\right)  =\left(  \left(  g\otimes f\right)  \circ T_{C}\right)  \left(
z\right)  $.
\par
Now, forget that we fixed $z$. We thus have shown that $\left(  T_{A}%
\circ\left(  f\otimes g\right)  \right)  \left(  z\right)  =\left(  \left(
g\otimes f\right)  \circ T_{C}\right)  \left(  z\right)  $ for each $z\in
C\otimes C$. In other words, $T_{A}\circ\left(  f\otimes g\right)  =\left(
g\otimes f\right)  \circ T_{C}$. This proves
\eqref{sol.commutative.convolution.twist-comm}.}

Now, the definition of convolution yields $f\star g=m_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$ and $g\star f=m_{A}\circ\left(  g\otimes f\right)
\circ\Delta_{C}$. Hence,%
\begin{align*}
f\star g  & =\underbrace{m_{A}}_{=m_{A}\circ T_{A}}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}=m_{A}\circ\underbrace{T_{A}\circ\left(  f\otimes
g\right)  }_{=\left(  g\otimes f\right)  \circ T_{C}}\circ\Delta_{C}\\
& =m_{A}\circ\left(  g\otimes f\right)  \circ\underbrace{T_{C}\circ\Delta_{C}%
}_{=\Delta_{C}}=m_{A}\circ\left(  g\otimes f\right)  \circ\Delta_{C}=g\star f.
\end{align*}
This solves Exercise~\ref{exe.commutative.convolution}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cocommutative.coalghom}}

\textit{Solution to Exercise~\ref{exe.cocommutative.coalghom}.} Recall
that our abstract definition of a $\kk$-algebra (Definition~\ref{def.algebra})
and our definition of a $\kk$-coalgebra (Definition~\ref{def.coalgebra})
differ from each other only in the directions of the
arrows. More precisely, reversing all arrows in the former definition yields the
latter definition. Similarly, our definition of a cocommutative $\kk$-coalgebra
is obtained by reversing all arrows in our abstract definition of a commutative
$\kk$-algebra, and our definition of a $\kk$-coalgebra homomorphism is
obtained by reversing all arrows in our abstract definition of a
$\kk$-algebra homomorphism. Hence, the statements of parts (a)
and (b) of this exercise can be obtained from each other by reversing all
arrows. Therefore, if we can solve part (b) of this exercise in an element-free
way\footnote{By an ``element-free'' argument,
we mean an argument which only talks about
linear maps, but never talks about elements of modules such as $A$
and $A \otimes A$. For instance, the Second solution of
Exercise~\ref{exe.convolution.tensor}(a) that we gave above
was element-free, whereas the First solution of
Exercise~\ref{exe.convolution.tensor}(a) (which we
also showed above) was not element-free (since it involved elements $c$ and
$d$ of $C$ and $D$). A synonym for ``element-free argument'' is ``argument
by pure diagram chasing'', although it is not required that one actually
draws any diagrams in the argument (commutative diagrams are just shortcuts
for identities between maps).}, then we can clearly apply the same argument
``with all arrows reversed'' (and, of course, with $A$, $m_A$ and $u_A$
replaced by $C$, $\Delta_C$ and $\epsilon_C$) to solve part
(a). Hence, in order to solve this exercise, it is enough to find an
element-free solution to its part (b). Let us do this now.

Let us first show how to solve part (b) using computations with elements
(i.e., not in an element-free way). This is very easy. We need to prove the
following statements:

\begin{statement}
\textit{Statement 1:} If a $\kk$-algebra $A$ is commutative, then its
multiplication $m_{A}:A\otimes A\rightarrow A$ is a $\kk$-algebra homomorphism.
\end{statement}

\begin{statement}
\textit{Statement 2:} If a $\kk$-algebra $A$ has the property that its
multiplication $m_{A}:A\otimes A\rightarrow A$ is a $\kk$-algebra
homomorphism, then $A$ is commutative.
\end{statement}

\textit{Proof of Statement 1 using computations with elements:} Assume that a
$\kk$-algebra $A$ is commutative. We need to prove that $m_{A}:A\otimes
A\rightarrow A$ is a $\kk$-algebra homomorphism. To do so, it is enough
to show that $m_{A}$ preserves products and that $m_{A}$ maps the unity of
$A\otimes A$ to the unity of $A$.

Let us show that $m_{A}$ preserves products. This means proving that
$m_{A}\left(  uv\right)  =m_{A}\left(  u\right)  \cdot m_{A}\left(  v\right)
$ for all $u\in A\otimes A$ and $v\in A\otimes A$. So let $u\in A\otimes A$
and $v\in A\otimes A$. Since the equality in question ($m_{A}\left(
uv\right)  =m_{A}\left(  u\right)  \cdot m_{A}\left(  v\right)  $) is linear
in each of $u$ and $v$, we can WLOG assume that $u$ and $v$ are pure tensors.
Having made this assumption, we can write $u=a\otimes b$ for some $a\in A$ and
$b\in A$, and we can write $v=c\otimes d$ for some $c\in A$ and $d\in A$. Now,
\[
m_{A}\left(  \underbrace{u}_{=a\otimes b}\underbrace{v}_{=c\otimes d}\right)
=m_{A}\left(  \underbrace{\left(  a\otimes b\right)  \left(  c\otimes
d\right)  }_{=ac\otimes bd}\right)  =m_{A}\left(  ac\otimes bd\right)
=\left(  ac\right)  \left(  bd\right)
\]
and
\[
m_{A}\left(  \underbrace{u}_{=a\otimes b}\right)  \cdot m_{A}\left(
\underbrace{v}_{=c\otimes d}\right)  =\underbrace{m_{A}\left(  a\otimes
b\right)  }_{=ab}\cdot\underbrace{m_{A}\left(  c\otimes d\right)  }
_{=cd}=\left(  ab\right)  \left(  cd\right)  .
\]
But\footnote{In the following computation, we are deliberately being
painstakingly slow and writing
down every single step, including every application of associativity. This is
to simplify our job later on (when we will translate this computation to an
element-free argument).}
\begin{equation}
\left(  ac\right)  \left(  bd\right)  =\left(  \left(  ac\right)  b\right)
d=\left(  a\underbrace{\left(  cb\right)  }_{\substack{=bc\\\text{(since
}A\text{ is}\\\text{commutative)}}}\right)  d=\left(  a\left(  bc\right)
\right)  d=\left(  \left(  ab\right)  c\right)  d=\left(  ab\right)  \left(
cd\right)  . \label{sol.cocommutative.coalghom.pf.s1.comp}
\end{equation}
Hence, altogether,
\[
m_{A}\left(  uv\right)  =\left(  ac\right)  \left(  bd\right)  =\left(
ab\right)  \left(  cd\right)  =m_{A}\left(  u\right)  \cdot m_{A}\left(
v\right)  .
\]
Thus, we have proven that $m_{A}$ preserves products. In order to prove that
$m_{A}$ maps the unity of $A\otimes A$ to the unity of $A$, we recall that the
former unity is $1_{A}\otimes 1_{A}$ and the latter unity is $1_{A}$, which
satisfy $m_{A}\left(  1_{A}\otimes 1_{A}\right)  =1_{A}\cdot 1_{A}=1_{A}$. We
have thus shown that $m_{A}$ preserves products and that $m_{A}$ maps the
unity of $A\otimes A$ to the unity of $A$. In other words, $m_{A}$ is a
$\kk$-algebra homomorphism, and Statement 1 is proven.

\textit{Proof of Statement 2 using computations with elements:} Assume that a
$\kk$-algebra $A$ has the property that its multiplication
$m_{A}:A\otimes A\rightarrow A$ is a $\kk$-algebra homomorphism. Let
$a$ and $b$ be elements of $A$. Then,
\[
m_{A}\left(  \underbrace{\left(  1_{A}\otimes b\right)  \left(  a\otimes
1_{A}\right)  }_{=\left(  1_{A}a\right)  \otimes\left(  b1_{A}\right)
=a\otimes b}\right)  =m_{A}\left(  a\otimes b\right)  =ab.
\]
Comparing this with
\begin{align*}
m_{A}\left(  \left(  1_{A}\otimes b\right)  \left(  a\otimes1_{A}\right)
\right) & = \underbrace{m_{A}\left(  1_{A}\otimes b\right)  }_{=1_{A}
b=b}\underbrace{m_{A}\left(  a\otimes1_{A}\right)  }_{=a1_{A}=a}
\qquad \qquad \left(\text{since }m_A \text{ is a }\kk\text{-algebra homomorphism}\right)
\\ 
&= ba,
\end{align*}
this becomes $ab=ba$. We thus have shown that $ab=ba$ for all $a\in A$ and
$b\in A$. In other words, $A$ is commutative, and thus Statement 2 is proven.

We thus have solved part (b) of the exercise using computations with elements.
But we want an element-free solution of part (b). It turns out
that we can obtain such a solution from our above solution by a more or less
straightforward rewriting procedure. Let us show how this works.

Again, we need to prove Statements 1 and 2 made above.

\textit{Element-free proof of Statement 1:} Assume that a
$\kk$-algebra $A$ is commutative. We need to prove that $m_{A}:A\otimes
A\rightarrow A$ is a $\kk$-algebra homomorphism. To do so, it is enough
to show that the diagrams
\[
\xymatrixcolsep{5pc}
\xymatrix{
A \otimes A \ar[r]^{m_A} & A \\
A \otimes A \otimes A \otimes A \ar[u]^{m_{A\otimes A}} \ar[r]^-{m_A \otimes
m_A} & A \otimes A \ar[u]^{m_A}
}
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\xymatrix{
A \otimes A \ar[rr]^{m_A} & & A \\
& \kk\ar[ul]^{u_{A\otimes A}} \ar[ur]^{u_A}
}
\]
are commutative. In other words, it is enough to show that
\begin{equation}
m_{A}\circ m_{A\otimes A}=m_{A}\circ\left(  m_{A}\otimes m_{A}\right)
\label{sol.cocommutative.coalghom.pf.s1.d1}
\end{equation}
and
\begin{equation}
u_{A}=m_{A}\circ u_{A\otimes A}. \label{sol.cocommutative.coalghom.pf.s1.d2}
\end{equation}


Let us prove \eqref{sol.cocommutative.coalghom.pf.s1.d1}
first. If we were allowed to compute with elements, then we could prove
\eqref{sol.cocommutative.coalghom.pf.s1.d1} by evaluating both sides of
\eqref{sol.cocommutative.coalghom.pf.s1.d1} at a pure tensor $a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A$; this would leave us with
the task of showing that $\left(  ac\right)  \left(  bd\right)  =\left(
ab\right)  \left(  cd\right)  $, which we already have done in the computation
which proved \eqref{sol.cocommutative.coalghom.pf.s1.comp}. However, we are
not allowed to do this, because we want this proof to be element-free.
But what we \emph{can} do is computing with maps instead of
elements. We just need to replace the computation which proved
\eqref{sol.cocommutative.coalghom.pf.s1.comp} by a computation which uses maps
instead of elements. If we replace every expression in
\eqref{sol.cocommutative.coalghom.pf.s1.comp} by the $\kk$-linear map
which sends every $a\otimes b\otimes c\otimes d\in A\otimes A\otimes A\otimes
A$ to said expression, then we obtain:
\begin{align}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(
ac\right)  \left(  bd\right)  \right) \nonumber\\
&  =\left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  \left(
ac\right)  b\right)  d\right) \nonumber\\
&  =\left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  a\left(
cb\right)  \right)  d\right) \nonumber\\
&  =\left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  a\left(
bc\right)  \right)  d\right) \nonumber\\
&  =\left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  \left(
ab\right)  c\right)  d\right) \nonumber\\
&  =\left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(
ab\right)  \left(  cd\right)  \right)  .
\label{sol.cocommutative.coalghom.pf.s1.comp2}
\end{align}
We are not yet done, because we still are using elements (in describing the
maps). So we should rewrite the maps appearing in the computation
\eqref{sol.cocommutative.coalghom.pf.s1.comp2} in such a way that no elements
occur in them anymore. Denoting by $T$ the twist map $A\otimes A\rightarrow
A\otimes A$ (sending every $a\otimes b$ to $b\otimes a$)\ \ \ \ \footnote{We
do not count the use of this map $T$ as a use of elements (even though we
just defined it using elements). Twist maps like $T$ are one of the basic
features of tensor products (along with associativity isomorphisms
$\left(U \otimes V\right) \otimes W \to U \otimes \left(V \otimes W\right)$,
which we are suppressing, and with trivial isomorphisms of the form
$\kk \otimes U \to U$), and their use is allowed in element-free arguments.
They don't interfere with ``reversing the arrows'' because arrows like $T$
are very easy to reverse.}, we have
\begin{align*}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(
ac\right)  \left(  bd\right)  \right) \\
&  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right) 
 \circ\left( \id_{A}\otimes T\otimes \id_{A}\right)  ;
\end{align*}
\begin{align*}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  \left(
ac\right)  b\right)  d\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  m_{A}\otimes
\id_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)  ;
\end{align*}
\begin{align*}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  a\left(
cb\right)  \right)  d\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)  ;
\end{align*}
\begin{align*}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  a\left(
bc\right)  \right)  d\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  ;
\end{align*}
\begin{align*}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(  \left(
ab\right)  c\right)  d\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  m_{A}\otimes
\id_{A}\otimes\id_{A}\right)  ;
\end{align*}
\begin{align*}
&  \left(  \text{the }\kk \text{-linear map sending every }a\otimes
b\otimes c\otimes d\in A\otimes A\otimes A\otimes A\text{ to }\left(
ab\right)  \left(  cd\right)  \right) \\
&  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right) .
\end{align*}
Hence, the computation \eqref{sol.cocommutative.coalghom.pf.s1.comp2} rewrites
as
\begin{align}
&  m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  \circ\left(  \id_{A}\otimes
T\otimes\id_{A}\right) \nonumber\\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  m_{A}\otimes
\id_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)
\nonumber\\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)
\nonumber\\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right) \nonumber\\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  m_{A}\otimes
\id_{A}\otimes\id_{A}\right) \nonumber\\
&  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  .
\label{sol.cocommutative.coalghom.pf.s1.comp3}
\end{align}
However, now that we are no longer talking about elements, our computation has
become significantly harder to follow. For example, the first equality in the
computation \eqref{sol.cocommutative.coalghom.pf.s1.comp3} is far less obvious
than the corresponding equality $\left(  ac\right)  \left(  bd\right)
=\left(  \left(  ac\right)  b\right)  d$ in the computation
\eqref{sol.cocommutative.coalghom.pf.s1.comp}. So, in order to justify the
computation \eqref{sol.cocommutative.coalghom.pf.s1.comp3}, we need to recall
where exactly we used associativity or commutativity in
\eqref{sol.cocommutative.coalghom.pf.s1.comp2}, and translate these uses into
element-free language. Let us do this step by step:

\textit{The first equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}:}
The first equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}
is simply an element-free way to state $\left(  ac\right)  \left(
bd\right)  =\left(  \left(  ac\right)  b\right)  d$ for all $a,b,c,d\in A$. On
the level of elements, this follows from applying associativity to the
elements $ac$, $b$ and $d$ of $A$. In other words, this follows from applying
the associativity law $m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)
=m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  $ to the tensor $ac\otimes
b\otimes d=\left(  \left(  m_{A}\otimes\id_{A}\otimes\id_{A}\right)  \circ\left(
\id_{A}\otimes T\otimes\id_{A}\right)  \right)  \left(  a\otimes b\otimes c\otimes
d\right)  $. Therefore, the first equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3} should follow from $m_{A}
\circ\left(  \id_{A}\otimes m_{A}\right)  =m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  $ by composition with the map $\left(  m_{A}\otimes\id_{A}
\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)  $ on the
right. And indeed, this is how it is proven:
\begin{align*}
&  m_{A}\circ\underbrace{\left(  m_{A}\otimes m_{A}\right)  }
_{\substack{=\left(  \id_{A}\otimes m_{A}\right)  \circ\left(  m_{A}
\otimes\id_{A\otimes A}\right)  \\=\left(  \id_{A}\otimes m_{A}\right)
\circ\left(  m_{A}\otimes\id_{A}\otimes\id_{A}\right)  }}
\circ\left(  \id_{A}\otimes T\otimes\id_{A}\right) \\
&  =\underbrace{m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  }_{=m_{A}
\circ\left(  m_{A}\otimes\id_{A}\right)  }\circ\left(  m_{A}\otimes\id_{A}
\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  m_{A}\otimes
\id_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)  .
\end{align*}
Thus, we have found an element-free proof of the first equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}.

\textit{The second equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}:}
The second equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}
is simply an element-free way to state $\left(  \left(  ac\right)
b\right)  d=\left(  a\left(  cb\right)  \right)  d$ for all $a,b,c,d\in A$. On
the level of elements, this follows from applying associativity to the
elements $a$, $c$ and $b$ of $A$, and then multiplying with $d$ on the right.
In other words, this follows from applying the associativity law $m_{A}
\circ\left(  m_{A}\otimes\id_{A}\right)  =m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  $ to the first three tensorands of the tensor $a\otimes c\otimes
b\otimes d=\left(  \id_{A}\otimes T\otimes\id_{A}\right)  \left(  a\otimes
b\otimes c\otimes d\right)  $, and then applying $m_{A}$. In other words, this
follows from applying the equality $\left(  m_{A}\circ\left(  m_{A}
\otimes\id_{A}\right)  \right)  \otimes\id_{A}=\left(  m_{A}\circ\left(
\id_{A}\otimes m_{A}\right)  \right)  \otimes\id_{A}$ to the tensor $\left(
\id_{A}\otimes T\otimes\id_{A}\right)  \left(  a\otimes b\otimes c\otimes
d\right)  $, and then applying $m_{A}$. Therefore, the second equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3} should follow from $m_{A}
\circ\left(  m_{A}\otimes \id_{A}\right)  =m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  $ by tensoring both sides with $\id_{A}$ on the right and then
composing them with the map $\id_{A}\otimes T\otimes\id_{A}$ on the right and with
$m_{A}$ on the left. And indeed, this is how it is proven:
\begin{align*}
&  m_{A}\circ\underbrace{\left(  m_{A}\otimes\id_{A}\right)  \circ\left(
m_{A}\otimes\id_{A}\otimes\id_{A}\right)  }_{=\left(  m_{A}\circ\left(
m_{A}\otimes\id_{A}\right)  \right)  \otimes\id_{A}}\circ\left(  \id_{A}\otimes
T\otimes\id_{A}\right) \\
&  =m_{A}\circ\left(  \underbrace{\left(  m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \right)  }_{=m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)
}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right) \\
&  =m_{A}\circ\underbrace{\left(  \left(  m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  \right)  \otimes\id_{A}\right)  }_{=\left(  m_{A}\otimes
\id_{A}\right)  \circ\left(  \id_{A}\otimes m_{A}\otimes\id_{A}\right)  }
\circ\left(  \id_{A}\otimes T\otimes\id_{A}\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)  .
\end{align*}
Thus, we have obtained an element-free proof of the second equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}.

\textit{The third equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}:}
The third equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}
is simply an element-free way to state $\left(  a\left(  cb\right)
\right)  d=\left(  a\left(  bc\right)  \right)  d$ for all $a,b,c,d\in A$. On
the level of elements, this follows from applying commutativity to the
elements $c$ and $b$ of $A$, then multiplying with $a$ on the left, and then
multiplying with $d$ on the right. In other words, this follows from applying
the commutativity law $m_{A}=m_{A}\circ T$ to the second and third tensorands
of the tensor $a\otimes c\otimes b\otimes d=\left(  \id_{A}\otimes T\otimes
\id_{A}\right)  \left(  a\otimes b\otimes c\otimes d\right)  $, and then
applying $m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  $. In other words, this
follows from applying the equality $\id_{A}\otimes m_{A}\otimes\id_{A}
=\id_{A}\otimes\left(  m_{A}\circ T\right)  \otimes\id_{A}$ to the tensor $\left(
\id_{A}\otimes T\otimes\id_{A}\right)  \left(  a\otimes b\otimes c\otimes
d\right)  $, and then applying $m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  $.
Therefore, the third equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}
should follow from $m_{A}=m_{A}\circ T$ by tensoring both sides with $\id_{A}$
on the left and on the right and then composing them with the map
$\id_{A}\otimes T\otimes\id_{A}$ on the right and with the map $m_{A}\circ\left(
m_{A}\otimes\id_{A}\right)  $ on the left. And indeed, this is how it is proven:
\begin{align*}
&  m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}
\otimes\underbrace{m_{A}}_{=m_{A}\circ T}\otimes\id_{A}\right)  \circ\left(
\id_{A}\otimes T\otimes\id_{A}\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\underbrace{\left(
\id_{A}\otimes\left(  m_{A}\circ T\right)  \otimes\id_{A}\right)  }_{=\left(
\id_{A}\otimes m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes
\id_{A}\right)  }\circ\left(  \id_{A}\otimes T\otimes\id_{A}\right) \\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  \circ\underbrace{\left(  \id_{A}\otimes T\otimes
\id_{A}\right)  \circ\left(  \id_{A}\otimes T\otimes\id_{A}\right)  }
_{\substack{=\id_{A\otimes A\otimes A\otimes A}\\\text{(since }T\circ
T=\id_{A\otimes A}\text{)}}}\\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  \id_{A}\otimes
m_{A}\otimes\id_{A}\right)  .
\end{align*}
Thus, we have obtained an element-free proof of the third equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}.

\textit{The fourth equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}:} The fourth equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}
is simply an element-free way to state $\left(  a\left(  bc\right)
\right)  d=\left(  \left(  ab\right)  c\right)  d$ for all $a,b,c,d\in A$. On
the level of elements, this follows from applying associativity to the
elements $a$, $b$ and $c$ of $A$, and then multiplying with $d$ on the right.
In other words, this follows from applying the associativity law $m_{A}
\circ\left(  \id_{A}\otimes m_{A}\right)  =m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  $ to the first three tensorands of the tensor $a\otimes b\otimes
c\otimes d$, and then applying the map $m_{A}$. In other words, this follows
from applying the equality $\left(  m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  \right)  \otimes\id_{A}=\left(  m_{A}\circ\left(  m_{A}
\otimes\id_{A}\right)  \right)  \otimes\id_{A}$ to the tensor $a\otimes b\otimes
c\otimes d$, and then applying the map $m_{A}$. Therefore, the fourth equality
in \eqref{sol.cocommutative.coalghom.pf.s1.comp3} should follow from
$m_{A}\circ\left(  \id_{A}\otimes m_{A}\right)  =m_{A}\circ\left(  m_{A}
\otimes\id_{A}\right)  $ by tensoring both sides with $\id_{A}$ on the right and
then composing them with the map $m_{A}$ on the left. And indeed, this is how
it is proven:
\begin{align*}
&  m_{A}\circ\underbrace{\left(  m_{A}\otimes\id_{A}\right)  \circ\left(
\id_{A}\otimes m_{A}\otimes\id_{A}\right)  }_{=\left(  m_{A}\circ\left(
\id_{A}\otimes m_{A}\right)  \right)  \otimes\id_{A}}\\
&  =m_{A}\circ\left(  \underbrace{\left(  m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  \right)  }_{=m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)
}\otimes\id_{A}\right) \\
&  =m_{A}\circ\underbrace{\left(  \left(  m_{A}\circ\left(  m_{A}\otimes
\id_{A}\right)  \right)  \otimes\id_{A}\right)  }_{=\left(  m_{A}\otimes
\id_{A}\right)  \circ\left(  m_{A}\otimes\id_{A}\otimes\id_{A}\right)  }\\
&  =m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  \circ\left(  m_{A}\otimes
\id_{A}\otimes\id_{A}\right)  .
\end{align*}
Thus, we have obtained an element-free proof of the fourth equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}.

\textit{The fifth equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}:}
The fifth equality in \eqref{sol.cocommutative.coalghom.pf.s1.comp3}
is simply an element-free way to state $\left(  \left(  ab\right)
c\right)  d=\left(  ab\right)  \left(  cd\right)  $ for all $a,b,c,d\in A$. On
the level of elements, this follows from applying associativity to the
elements $ab$, $c$ and $d$ of $A$. In other words, this follows from applying
the associativity law $m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  =m_{A}
\circ\left(  \id_{A}\otimes m_{A}\right)  $ to the tensor $ab\otimes c\otimes
d=\left(  m_{A}\otimes\id_{A}\otimes\id_{A}\right)  \left(  a\otimes b\otimes
c\otimes d\right)  $. Therefore, the fifth equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3} should follow from $m_{A}
\circ\left(  m_{A}\otimes\id_{A}\right)  =m_{A}\circ\left(  \id_{A}\otimes
m_{A}\right)  $ by composing both sides of this equality with $m_{A}
\otimes\id_{A}\otimes\id_{A}$ on the right. And indeed, this is how it is proven:
\begin{align*}
&  \underbrace{m_{A}\circ\left(  m_{A}\otimes\id_{A}\right)  }_{=m_{A}
\circ\left(  \id_{A}\otimes m_{A}\right)  }\circ\left(  m_{A}\otimes\id_{A}
\otimes\id_{A}\right) \\
&  =m_{A}\circ\underbrace{\left(  \id_{A}\otimes m_{A}\right)  \circ\left(
m_{A}\otimes\id_{A}\otimes\id_{A}\right)  }_{=\left(  \id_{A}\otimes m_{A}\right)
\circ\left(  m_{A}\otimes\id_{A\otimes A}\right)  =m_{A}\otimes m_{A}}\\
&  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  .
\end{align*}
Thus, we have obtained an element-free proof of the fifth equality in
\eqref{sol.cocommutative.coalghom.pf.s1.comp3}.

Now, all five equalities in the computation
\eqref{sol.cocommutative.coalghom.pf.s1.comp3} are proven without reference to
elements. Hence, we have shown
$ m_{A}\circ\left(  m_{A}\otimes m_{A}\right)  \circ\left(  \id_{A}\otimes
T\otimes\id_{A}\right)  =m_{A}\circ\left(  m_{A}\otimes m_{A}\right)$
in an element-free way. In other words,
$m_{A}\circ m_{A\otimes A}=m_{A}\circ\left(  m_{A}\otimes
m_{A}\right)  $ is proven in an element-free way
(because the definition of $m_{A\otimes A}$ yields
$m_{A\otimes A}=\left(  m_{A}\otimes m_{A}\right)  \circ\left(  \id_{A}\otimes
T\otimes\id_{A}\right)  $). In other words,
\eqref{sol.cocommutative.coalghom.pf.s1.d1} is proven.

It remains to prove \eqref{sol.cocommutative.coalghom.pf.s1.d2}. We leave this
very simple proof to the reader (noticing that it does not require the
commutativity of $A$).

Now we know that both \eqref{sol.cocommutative.coalghom.pf.s1.d1} and
\eqref{sol.cocommutative.coalghom.pf.s1.d2} hold. Statement 1 is thus proven
in an element-free way.

We leave it to the reader to prove Statement 2 in an element-free way.
Altogether, Statements 1 and 2 have now been proven. Therefore, part (b) of
the exercise has been solved in an element-free way. Therefore, a solution
of part (a) can be obtained mechanically by ``reversing all arrows''.

Of course, part (a) can alternatively be solved using the Sweedler notation.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.comm-cocomm.anti}}

\textit{Solution to Exercise~\ref{exe.comm-cocomm.anti}.}

We shall use the notations introduced in
Definition~\ref{def.anti-hom}.
Let us first state two simple facts from linear algebra:

\begin{proposition}
\label{prop.linalg.twist-vs-xy}
Let $U$, $V$, $U^{\prime}$ and $V^{\prime}$ be
four $\kk$-modules.
Let $x:U\rightarrow U^{\prime}$ and $y:V\rightarrow
V^{\prime}$ be two $\kk$-linear maps.
Then,
\begin{equation}
\left(  y\otimes x\right)  \circ T_{U,V}
= T_{U^{\prime},V^{\prime}} \circ \left(  x\otimes y\right)  .
\label{eq.prop.linalg.twist-vs-xy.eq}
\end{equation}
\end{proposition}

\begin{proposition}
\label{prop.linalg.twist-inv}
Let $U$ and $V$ be two $\kk$-modules.
Then,
\begin{equation}
T_{V, U} \circ T_{U, V} = \id_{U \otimes V}.
\label{eq.prop.linalg.twist-inv.eq}
\end{equation}
\end{proposition}

We shall further need the following lemma, which follows easily
from the definition of commutativity:

\begin{lemma}
\label{lem.sol.comm-cocomm.anti.1}
Let $A$ be a commutative $\kk$-algebra.
Then, $m_{A} = m_{A}\circ T_{A,A}$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem.sol.comm-cocomm.anti.1}.]
Let $T$ denote the twist map $T_{A,A}$.
The $\kk$-algebra $A$ is commutative if and only if
the diagram \eqref{commutativity-diagram} commutes
(by Definition~\ref{def.comm}).
Hence, the diagram \eqref{commutativity-diagram} commutes
(since $A$ is commutative).
In other words, we have $m = m \circ T$.
In other words, we have $m_{A} = m_{A}\circ T_{A,A}$
(since $m = m_A$ and $T = T_{A,A}$).
This proves Lemma~\ref{lem.sol.comm-cocomm.anti.1}.
\end{proof}

We now come to the solution of the exercise.

(a) This is easy to prove by computing with elements, but let us give an
``element-free'' proof.

Let $f:A\rightarrow B$ be a $\kk$-linear map. We shall prove the
logical equivalence%
\begin{equation}
\left(  f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \circ T_{A,A}%
\right)  \ \Longleftrightarrow\ \left(  f\circ m_{A}=m_{B}\circ\left(
f\otimes f\right)  \right)  .
\label{sol.comm-cocomm.anti.crux}
\end{equation}


[\textit{Proof of \eqref{sol.comm-cocomm.anti.crux}:} We have assumed that at
least one of the $\kk$-algebras $A$ and $B$ is commutative. In other
words, the $\kk$-algebra $A$ is commutative or the $\kk$-algebra
$B$ is commutative. Thus, we are in one of the following two cases:

\textit{Case 1:} The $\kk$-algebra $A$ is commutative.

\textit{Case 2:} The $\kk$-algebra $B$ is commutative.

Let us first consider Case 1. In this case, the $\kk$-algebra $A$ is
commutative. Thus, Lemma \ref{lem.sol.comm-cocomm.anti.1} yields $m_{A}%
=m_{A}\circ T_{A,A}$.

But Proposition \ref{prop.linalg.twist-inv} (applied to $U=A$ and $V=A$)
yields $T_{A,A}\circ T_{A,A}=\operatorname*{id}\nolimits_{A\otimes A}$. Thus,
if we have $f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \circ T_{A,A}$,
then we also have%
\[
f\circ\underbrace{m_{A}}_{=m_{A}\circ T_{A,A}}=\underbrace{f\circ m_{A}%
}_{=m_{B}\circ\left(  f\otimes f\right)  \circ T_{A,A}}\circ T_{A,A}%
=m_{B}\circ\left(  f\otimes f\right)  \circ\underbrace{T_{A,A}\circ T_{A,A}%
}_{=\operatorname*{id}\nolimits_{A\otimes A}}=m_{B}\circ\left(  f\otimes
f\right)  .
\]
Thus, we have proved the implication%
\begin{equation}
\ \left(  f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \circ
T_{A,A}\right)  \ \Longrightarrow\ \left(  f\circ m_{A}=m_{B}\circ\left(
f\otimes f\right)  \right)  .\label{sol.comm-cocomm.anti.crux.c1.pf.1}%
\end{equation}
On the other hand, if we have $f\circ m_{A}=m_{B}\circ\left(  f\otimes
f\right)  $, then we also have%
\[
f\circ\underbrace{m_{A}}_{=m_{A}\circ T_{A,A}}=\underbrace{f\circ m_{A}%
}_{=m_{B}\circ\left(  f\otimes f\right)  }\circ T_{A,A}=m_{B}\circ\left(
f\otimes f\right)  \circ T_{A,A}.
\]
Hence, we have proved the implication%
\[
\ \left(  f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \right)
\ \Longrightarrow\ \left(  f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)
\circ T_{A,A}\right)  .
\]
Combining this implication with \eqref{sol.comm-cocomm.anti.crux.c1.pf.1}, we
obtain the logical equivalence
\[
\left(  f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \circ T_{A,A}%
\right)  \ \Longleftrightarrow\ \left(  f\circ m_{A}=m_{B}\circ\left(
f\otimes f\right)  \right)  .
\]
Thus, \eqref{sol.comm-cocomm.anti.crux} is proved in Case 1.

Let us next consider Case 2. In this case, the $\kk$-algebra $B$ is
commutative. Thus, Lemma \ref{lem.sol.comm-cocomm.anti.1} (applied to $B$
instead of $A$) yields $m_{B}=m_{B}\circ T_{B,B}$, so that $m_{B}\circ
T_{B,B}=m_{B}$. On the other hand, Proposition \ref{prop.linalg.twist-vs-xy}
(applied to $U=A$, $V=A$, $U^{\prime}=B$, $V^{\prime}=B$, $x=f$ and $y=f$)
yields%
\[
\left(  f\otimes f\right)  \circ T_{A,A}=T_{B,B}\circ\left(  f\otimes
f\right)  .
\]
Now, we have the following chain of logical equivalences:%
\[
\ \left(  f\circ m_{A}=m_{B}\circ\underbrace{\left(  f\otimes f\right)  \circ
T_{A,A}}_{=T_{B,B}\circ\left(  f\otimes f\right)  }\right)
\ \Longleftrightarrow\ \left(  f\circ m_{A}=\underbrace{m_{B}\circ T_{B,B}%
}_{=m_{B}}\circ\left(  f\otimes f\right)  \right)  \ \Longleftrightarrow
\ \left(  f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \right)  .
\]
Hence, the equivalence \eqref{sol.comm-cocomm.anti.crux} holds. Thus,
\eqref{sol.comm-cocomm.anti.crux} is proved in Case 2.

We have now proved \eqref{sol.comm-cocomm.anti.crux} in both Cases 1 and 2.
Hence, the proof of \eqref{sol.comm-cocomm.anti.crux} is complete (since Cases
1 and 2 cover all possibilities).]

Now, forget that we fixed $f$. We thus have proved the equivalence
\eqref{sol.comm-cocomm.anti.crux} for any $\kk$-linear map
$f:A\rightarrow B$.

Now, the $\kk$-algebra homomorphisms from $A$ to $B$ are precisely the
$\kk$-linear maps $f:A\rightarrow B$ that make the two diagrams%
\[
\xymatrix{ A \ar[r]^{f} & B \\ A \otimes A \ar[u]^{m_A} \ar[r]^{f \otimes f} & B \otimes B \ar[u]_{m_B} }\qquad
\text{and}\qquad
\xymatrix{ A \ar[rr]^f & & B \\ &\kk \ar[ul]^{u_A} \ar[ur]_{u_B}& }
\]
commute (by Definition \ref{def.morphism-alg-coalg}). In other words, the
$\kk$-algebra homomorphisms from $A$ to $B$ are the $\kk$-linear
maps $f:A\rightarrow B$ that satisfy $f\circ m_{A}=m_{B}\circ\left(  f\otimes
f\right)  $ and $f\circ u_{A}=u_{B}$. Thus,%
\begin{align}
&  \left\{  \text{the } \kk \text{-algebra homomorphisms from }A\text{ to
}B\right\} \nonumber\\
&  =\left\{  \text{the } \kk \text{-linear maps }f:A\rightarrow B\text{
that satisfy }f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \text{ and
}f\circ u_{A}=u_{B}\right\}  . \label{sol.comm-cocomm.anti.4}%
\end{align}


On the other hand, the $\kk$-algebra anti-homomorphisms from $A$ to $B$
are the $\kk$-linear maps $f:A\rightarrow B$ that satisfy $f\circ
m_{A}=m_{B}\circ\left(  f\otimes f\right)  \circ T_{A,A}$ and $f\circ
u_{A}=u_{B}$ (by Definition \ref{def.anti-hom}(b)). Hence,%
\begin{align*}
&  \left\{  \text{the } \kk \text{-algebra anti-homomorphisms from
}A\text{ to }B\right\} \\
&  =\left\{  \text{the } \kk \text{-linear maps }f:A\rightarrow B\text{
that satisfy }\underbrace{f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)
\circ T_{A,A}}_{\substack{\Longleftrightarrow\ \left(  f\circ m_{A}=m_{B}%
\circ\left(  f\otimes f\right)  \right)  \\\text{(by
\eqref{sol.comm-cocomm.anti.crux})}}}\text{ and }f\circ u_{A}=u_{B}\right\} \\
&  =\left\{  \text{the } \kk \text{-linear maps }f:A\rightarrow B\text{
that satisfy }f\circ m_{A}=m_{B}\circ\left(  f\otimes f\right)  \text{ and
}f\circ u_{A}=u_{B}\right\} \\
&  =\left\{  \text{the } \kk \text{-algebra homomorphisms from }A\text{
to }B\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.comm-cocomm.anti.4}}\right)  .
\end{align*}
In other words, the $\kk$-algebra anti-homomorphisms from $A$ to $B$
are the same as the $\kk$-algebra homomorphisms from $A$ to $B$. This
solves Exercise \ref{exe.comm-cocomm.anti}(a).

(b) We have solved Exercise~\ref{exe.comm-cocomm.anti}(a) in an element-free
fashion. Thus, by
``reversing all arrows'' in this solution of
Exercise~\ref{exe.comm-cocomm.anti}(a) (and replacing ``algebra'' and
``commutative'' by ``coalgebra'' and ``cocommutative'', respectively),
we can obtain a solution to the dual of
Exercise~\ref{exe.comm-cocomm.anti}(a). Consequently, the dual of
Exercise~\ref{exe.comm-cocomm.anti}(a) holds.

The notion of a cocommutative $\kk$-coalgebra is dual to the
notion of a commutative $\kk$-algebra
(i.e., is obtained from the latter notion by
``reversing all arrows'').
Furthermore, it is easy to see that the notion of a $\kk$-coalgebra
anti-homomorphism is dual to the notion of a $\kk$-algebra
anti-homomorphism, whereas the notion of a $\kk$-coalgebra
homomorphism is dual to the notion of a $\kk$-algebra
homomorphism. Thus, the dual of
Exercise~\ref{exe.comm-cocomm.anti}(a) is the following exercise:

\begin{statement}
\textit{Exercise~A:} Let $A$ and $B$ be two $\kk$-coalgebras, at least
one of which is cocommutative.
Prove that the $\kk$-coalgebra anti-homomorphisms from $A$ to $B$
are the same as the $\kk$-coalgebra homomorphisms from $A$ to $B$.
\end{statement}

As we have seen above, this dual holds.
Thus, Exercise~\ref{exe.comm-cocomm.anti}(b) is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.iterated.comm}}

\textit{Solution to Exercise~\ref{exe.iterated.comm}.}
For every $1 \leq i < j \leq k$, let $t_{i, j}$ be the transposition
in $\Symm_k$ which transposes $i$ with $j$. It is well-known that the
symmetric group $\Symm_k$ is generated by the transpositions
$t_{i, i+1}$ with $i$ ranging over $\left\{1, 2, \ldots, k-1\right\}$.

But let us notice that
$\left(\rho\left(\pi\right)\right) \circ \left(\rho\left(\psi\right)\right)
= \rho\left(\pi \psi\right)$
for any two elements $\pi$ and $\psi$ of $\Symm_k$. Hence, the set of
all $\pi \in \Symm_k$ satisfying
$m^{\left(k-1\right)} \circ \left(\rho\left(\pi\right)\right)
= m^{\left(k-1\right)}$
is closed under multiplication. Since this set also contains the trivial
permutation $\id = \one_{\Symm_k} \in \Symm_k$ and is closed under taking
inverses (this is easy to check), this shows that this set is a subgroup
of $\Symm_k$. Therefore, if this set contains a set of generators of
$\Symm_k$, then this set must be the whole $\Symm_k$. Hence, if we can
show that this set contains the transposition $t_{i, i+1}$ for every
$i \in \left\{1, 2, \ldots, k-1\right\}$, then it will follow that this
set must be the whole $\Symm_k$ (because the group $\Symm_k$ is
generated by the transpositions $t_{i, i+1}$ with $i$ ranging over
$\left\{1, 2, \ldots, k-1\right\}$), and this will entail that every
$\pi \in \Symm_k$ satisfies
$m^{\left(k-1\right)} \circ \left(\rho\left(\pi\right)\right)
= m^{\left(k-1\right)}$,
which will solve the problem. Hence, in order to complete this solution,
it is enough to check that the set of all $\pi \in \Symm_k$ satisfying
$m^{\left(k-1\right)} \circ \left(\rho\left(\pi\right)\right)
= m^{\left(k-1\right)}$ contains the transposition $t_{i, i+1}$ for
every $i \in \left\{1, 2, \ldots, k-1\right\}$. In other words, it is
enough to check that
\begin{equation}
\label{sol.iterated.comm.goal}
m^{\left(k-1\right)} \circ \left(\rho\left(t_{i, i+1}\right)\right)
= m^{\left(k-1\right)}
\qquad \qquad \text{for all } i \in \left\{ 1,2,\ldots,k-1 \right\}.
\end{equation}

\textit{Proof of \eqref{sol.iterated.comm.goal}:} Let
$i \in \left\{ 1,2,\ldots,k-1 \right\}$. Let $T$ denote the twist
map $A \otimes A \to A \otimes A$ sending every pure tensor
$a \otimes b$ to $b \otimes a$. Any $k$ vectors $v_1, v_2, \ldots, v_k$
in $A$ satisfy
\begin{align*}
\nonumber
& \left(\rho\left(t_{i, i+1}\right)\right)
  \left(v_1 \otimes v_2 \otimes \cdots \otimes v_k\right) \\
&= t_{i, i+1} \left(v_1 \otimes v_2 \otimes \cdots \otimes v_k\right)
= \underbrace{v_1 \otimes v_2 \otimes \cdots \otimes v_{i-1}}_{= \id_{A^{\otimes \left(i-1\right)}}\left(v_1 \otimes v_2 \otimes \cdots \otimes v_{i-1}\right)}
  \otimes \underbrace{v_{i+1} \otimes v_i}_{= T\left(v_i \otimes v_{i+1}\right)}
  \otimes \underbrace{v_{i+2} \otimes v_{i+3} \otimes \cdots \otimes v_k}_{= \id_{A^{\otimes \left(k-1-i\right)}}\left(v_{i+2} \otimes v_{i+3} \otimes \cdots \otimes v_k\right)} \\
&= \id_{A^{\otimes \left(i-1\right)}}\left(v_1 \otimes v_2 \otimes \cdots \otimes v_{i-1}\right)
  \otimes T\left(v_i \otimes v_{i+1}\right)
  \otimes \id_{A^{\otimes \left(k-1-i\right)}}\left(v_{i+2} \otimes v_{i+3} \otimes \cdots \otimes v_k\right) \\
&= \left( \id_{A^{\otimes \left(i-1\right)}} \otimes T \otimes \id_{A^{\otimes \left(k-1-i\right)}} \right)
  \left(\underbrace{v_1 \otimes v_2 \otimes \cdots \otimes v_{i-1} \otimes v_i \otimes v_{i+1} \otimes v_{i+2} \otimes v_{i+3} \otimes \cdots \otimes v_k}_{= v_1 \otimes v_2 \otimes \cdots \otimes v_k}\right) \\
&= \left( \id_{A^{\otimes \left(i-1\right)}} \otimes T \otimes \id_{A^{\otimes \left(k-1-i\right)}} \right)
  \left(v_1 \otimes v_2 \otimes \cdots \otimes v_k\right).
\end{align*}
In other words, the two maps $\rho\left(t_{i, i+1}\right)$ and
$\id_{A^{\otimes \left(i-1\right)}} \otimes T \otimes \id_{A^{\otimes \left(k-1-i\right)}}$
are equal to each other on every pure tensor. Being $\kk$-linear maps,
these two maps must therefore be identical, i.e., we have
$\rho\left(t_{i, i+1}\right) =
\id_{A^{\otimes \left(i-1\right)}} \otimes T \otimes \id_{A^{\otimes \left(k-1-i\right)}}$.

But recall that $A$ is commutative, whence the diagram
\eqref{commutativity-diagram} commutes. Thus, $m \circ T = m$.

We have $i \in \left\{ 1, 2, \ldots, k-1 \right\}$, so that
$1 \leq i \leq k-1$ and therefore $0 \leq i-1 \leq \left(k-1\right)-1$.
Hence, in particular, $k-1 \geq 1 \geq 0$. We can thus apply
Exercise~\ref{exe.iterated.m}(c) to $k-1$ and $i-1$ instead
of $k$ and $i$. As a result, we obtain
\[
m^{\left(k-1\right)} = m^{\left(\left(k-1\right)-1\right)} \circ
 \left(\id_{A^{\otimes \left(i-1\right)}} \otimes m \otimes \id_{A^{\otimes \left(\left(k-1\right)-1-\left(i-1\right)\right)}}\right)
= m^{\left(\left(k-1\right)-1\right)} \circ
 \left(\id_{A^{\otimes \left(i-1\right)}} \otimes m \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right).
\]
Hence,
\begin{align*}
& \underbrace{m^{\left(k-1\right)}}
  _{= m^{\left(\left(k-1\right)-1\right)} \circ
    \left(\id_{A^{\otimes \left(i-1\right)}} \otimes m \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right)}
  \circ
  \underbrace{\left(\rho\left(t_{i, i+1}\right)\right)}
  _{= \id_{A^{\otimes \left(i-1\right)}} \otimes T \otimes \id_{A^{\otimes \left(k-1-i\right)}}} \\
&= m^{\left(\left(k-1\right)-1\right)} \circ
  \underbrace{\left(\id_{A^{\otimes \left(i-1\right)}} \otimes m \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right)
  \circ \left(\id_{A^{\otimes \left(i-1\right)}} \otimes T \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right)}
  _{= \left(\id_{A^{\otimes \left(i-1\right)}} \otimes \left(m \circ T\right) \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right)} \\
&= m^{\left(\left(k-1\right)-1\right)} \circ
  \left(\id_{A^{\otimes \left(i-1\right)}} \otimes \underbrace{\left(m \circ T\right)}_{=m} \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right) \\
&= m^{\left(\left(k-1\right)-1\right)} \circ
  \left(\id_{A^{\otimes \left(i-1\right)}} \otimes m \otimes \id_{A^{\otimes \left(k-1-i\right)}}\right) \\
&= m^{\left(k-1\right)}.
\end{align*}
Thus, \eqref{sol.iterated.comm.goal} is proven. This completes the
solution of Exercise~\ref{exe.iterated.comm}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.iterated.cocomm}}

\textit{Solution to Exercise~\ref{exe.iterated.cocomm}.}
Here is the statement:

\begin{statement}
\textbf{Exercise.}
Let $C$ be a cocommutative $\kk$-coalgebra, and let $k \in \NN$. The
symmetric group $\Symm_k$ acts on the $k$-fold tensor power $C^{\otimes k}$
by permuting the tensor factors:
$\sigma\left(v_1 \otimes v_2 \otimes \cdots \otimes v_k\right)
= v_{\sigma^{-1}\left(1\right)} \otimes v_{\sigma^{-1}\left(2\right)}
  \otimes \cdots \otimes v_{\sigma^{-1}\left(k\right)}$
for all $v_1, v_2, \ldots, v_k \in C$ and $\sigma \in \Symm_k$.
For every $\pi \in \Symm_k$, denote
by $\rho\left(\pi\right)$ the action of $\pi$ on $C^{\otimes k}$ (this is
an endomorphism of $C^{\otimes k}$).
Show that every $\pi \in \Symm_k$ satisfies
$\left(\rho\left(\pi\right)\right) \circ \Delta^{\left(k-1\right)}
= \Delta^{\left(k-1\right)}$. (Recall that
$\Delta^{\left(k-1\right)} : C \to C^{\otimes k}$
is defined as in Exercise~\ref{exe.iterated.Delta} for $k \geq 1$, and by
$\Delta^{\left(-1\right)} = \epsilon : C \to \kk$ for $k = 0$.)
\end{statement}

The solution of this exercise can be obtained from the above solution
of Exercise~\ref{exe.iterated.comm} by reversing all arrows (and replacing
$A$, $m$ and $m^{\left(i\right)}$ by $C$, $\Delta$ and
$\Delta^{\left(i\right)}$).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.alghoms.comm}}

\textit{Solution to Exercise~\ref{exe.alghoms.comm}.}
(a) Let $H$ be a $\kk$-bialgebra and $A$ be a commutative
$\kk$-algebra. Let $f$ and $g$ be two $\kk$-algebra homomorphisms
$H \to A$.

Since $A$ is commutative, the map $m_A : A \otimes A \to A$
is a $\kk$-algebra homomorphism (according to
Exercise~\ref{exe.cocommutative.coalghom}(b)). Also, $f \otimes g$
is a $\kk$-algebra homomorphism (by
Exercise~\ref{exe.tensorprod.morphism}(a), since $f$ and $g$ are
$\kk$-algebra homomorphisms). Finally, the axioms of a
$\kk$-bialgebra show that $\Delta_H : H \to H \otimes H$ is a
$\kk$-algebra homomorphism (since $H$ is a $\kk$-bialgebra). Now,
the definition of convolution yields that
$f \star g = m_A \circ \left(f \otimes g\right) \circ \Delta_H$.
Thus, $f \star g$ is a composition of three $\kk$-algebra
homomorphisms (namely, of $m_A$, $f \otimes g$ and $\Delta_H$), and
hence a $\kk$-algebra homomorphism itself. This solves
Exercise~\ref{exe.alghoms.comm}(a).

(b) Exercise~\ref{exe.alghoms.comm}(b) can be solved by
straightforward induction over $k$ using (in the induction step)
the result of Exercise~\ref{exe.alghoms.comm}(a) and (in the
induction base) the fact that $u_A \circ \epsilon_H$ is a
$\kk$-algebra homomorphism (since $u_A$ and $\epsilon_H$ are
$\kk$-algebra homomorphisms). The details are left to the reader.

(c) Let $H$ be a Hopf algebra, and let $A$ be a commutative
$\kk$-algebra. Let $f : H \to A$ be a $\kk$-algebra
homomorphism.

Proposition~\ref{antipodes-are-antiendomorphisms} shows that the antipode
$S$ of $H$ is an algebra anti-endomorphism. Combined with the
fact that $f$ is a $\kk$-algebra homomorphism, this yields
that the composition $f \circ S$ is a $\kk$-algebra
anti-homomorphism\footnote{Here, we have used the (easily proved)
fact that a composition of a $\kk$-algebra homomorphism with
a $\kk$-algebra anti-homomorphism (in either order) always is
a $\kk$-algebra anti-homomorphism.}. But since
algebra anti-homomorphisms $H \to A$ are the same as algebra
homomorphisms
$H \to A$ (since $A$ is commutative), this yields that $f \circ S$
is a $\kk$-algebra homomorphism. It remains to prove that
$f \circ S$ is $\star$-inverse to $f$. But this follows from
Proposition \ref{antipodes-give-convolution-inverses}(a)
(applied to $\alpha = f$).\ \ \ \ \footnote{Just for the sake
of it, here is a standalone proof of this claim: The
definition of convolution yields
\begin{align*}
\left(f \circ S\right) \star f
&= m_A \circ \left(\left(f \circ S\right) \otimes \underbrace{f}_{= f \circ \id}\right) \circ \Delta_H
= m_A \circ \underbrace{\left(\left(f \circ S\right) \otimes \left(f \circ \id\right)\right)}_{=\left(f \otimes f\right) \circ \left(S \otimes \id\right)} \circ \Delta_H \\
&= \underbrace{m_A \circ \left(f \otimes f\right)}_{\substack{= f \circ m_H \\ \text{(since } f \text{ is a } \kk\text{-algebra homomorphism)}}}
   \circ \left(S \otimes \id\right) \circ \Delta_H
 = f \circ \underbrace{m_H \circ \left(S \otimes \id\right) \circ \Delta_H}_{\substack{= u_H \circ \epsilon_H \\ \text{(by the commutativity of \eqref{antipode-diagram})}}} \\
&= \underbrace{f \circ u_H}_{\substack{= u_A \\ \text{(since } f \text{ is a } \kk\text{-algebra homomorphism)}}} \circ \epsilon_H = u_A \circ \epsilon_H,
\end{align*}
and similarly $f \star \left(f \circ S\right) = u_A \circ \epsilon_H$,
so that $f \circ S$ is $\star$-inverse to $f$.} This solves
Exercise~\ref{exe.alghoms.comm}(c).

(d) Let $A$ be a commutative $\kk$-algebra. Then, the map
$m_A : A \otimes A \to A$ is a $\kk$-algebra homomorphism (according to
Exercise~\ref{exe.cocommutative.coalghom}(b)).

We need to prove that $m_A^{\left(k\right)}$ is a $\kk$-algebra
homomorphism (since $m^{\left(k\right)} = m_A^{\left(k\right)}$).
To do so, it suffices to adapt the solution of
Exercise~\ref{exe.iterated.patras}(a) with only very minor changes
(mainly we have to change $H$ into $A$, replace ``coalgebra'' by
``algebra'' and use Exercise~\ref{exe.tensorprod.morphism}(a)
instead of using Exercise~\ref{exe.tensorprod.morphism}(b)). The
details of this adaptation are left to the reader.

(e) Let $C^\prime$ and $C$ be two $\kk$-coalgebras. Let
$\gamma : C \to C^\prime$ be a $\kk$-coalgebra homomorphism.
Let $A$ and $A^\prime$ be two $\kk$-algebras. Let
$\alpha : A \to A^\prime$ be a $\kk$-algebra homomorphism.
Let $f_1, f_2, \ldots, f_k$ be several $\kk$-linear maps
$C^\prime \to A$.

Proposition~\ref{prop.convolution.functor} shows that the
map
\[
\Hom\left(C',A\right) \to \Hom\left(C,A'\right),
\qquad
f \mapsto \alpha \circ f \circ \gamma
\]
is a $\kk$-algebra homomorphism
$\left(\Hom\left(C^\prime, A\right), \star\right) \to
\left(\Hom\left(C, A^\prime\right), \star\right)$.
Denote this $\kk$-algebra homomorphism by $\varphi$.

Since $\varphi$ is a $\kk$-algebra homomorphism,
we have
\begin{align*}
\varphi\left(f_1 \star f_2 \star \cdots \star f_k\right)
&=    \underbrace{\varphi\left(f_1\right)}_{\substack{
     = \alpha \circ f_1 \circ \gamma \\ \text{(by the
          definition of } \varphi \text{)}}}
\star \underbrace{\varphi\left(f_2\right)}_{\substack{
     = \alpha \circ f_2 \circ \gamma \\ \text{(by the
          definition of } \varphi \text{)}}}
\star \cdots
\star \underbrace{\varphi\left(f_k\right)}_{\substack{
     = \alpha \circ f_k \circ \gamma \\ \text{(by the
          definition of } \varphi \text{)}}} \\
&= \left(\alpha \circ f_1 \circ \gamma\right)
\star \left(\alpha \circ f_2 \circ \gamma\right)
\star \cdots
\star \left(\alpha \circ f_k \circ \gamma\right) .
\end{align*}
Hence,
\[
\left(\alpha \circ f_1 \circ \gamma\right)
\star \left(\alpha \circ f_2 \circ \gamma\right)
\star \cdots
\star \left(\alpha \circ f_k \circ \gamma\right)
= \varphi\left(f_1 \star f_2 \star \cdots \star f_k\right)
= \alpha \circ \left(f_1 \star f_2 \star \cdots \star f_k\right)
  \circ \gamma
\]
(by the definition of $\varphi$). This solves
Exercise~\ref{exe.alghoms.comm}(e).

(f) Let $H$ be a commutative $\kk$-bialgebra. Let $k$ and $\ell$ be
two nonnegative integers. Then, Exercise~\ref{exe.alghoms.comm}(b)
(applied to $A=H$ and $f_i = \id_H$) yields that
$\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
_{k\text{ times}}$
is a $\kk$-algebra homomorphism $H \to H$. Since
$\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
_{k\text{ times}} = \id_H^{\star k}$, this shows that
$\id_H^{\star k}$ is a $\kk$-algebra homomorphism $H \to H$.
We can thus apply Exercise~\ref{exe.alghoms.comm}(e) to
$H$, $H$, $H$, $H$, $\ell$, $\id_H$, $\id_H^{\star k}$ and $\id_H$
instead of
$C$, $C^\prime$, $A$, $A^\prime$, $k$, $f_i$, $\alpha$ and $\gamma$.
As a result, we obtain
\[
\id_H^{\star k} \circ
\left(\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
  _{\ell\text{ times}}\right)
\circ \id_H
= \underbrace{
\left(\id_H^{\star k} \circ \id_H \circ \id_H\right)
\star \left(\id_H^{\star k} \circ \id_H \circ \id_H\right)
\star \cdots
\star \left(\id_H^{\star k} \circ \id_H \circ \id_H\right)
}_{\ell \text{ times}} .
\]
Since $\id_H^{\star k} \circ \id_H \circ \id_H
= \id_H^{\star k}$ and
$\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
  _{\ell\text{ times}}
= \id_H^{\star \ell}$,
this rewrites as
\[
\id_H^{\star k} \circ \id_H^{\star \ell}
= \underbrace{
\id_H^{\star k} \star \id_H^{\star k} \star \cdots
\star \id_H^{\star k}
}_{\ell \text{ times}}
= \left(\id_H^{\star k}\right)^{\star \ell}
= \id_H^{\star \left(k\ell\right)} .
\]
This solves Exercise~\ref{exe.alghoms.comm}(f).

(g) Let $H$ be a commutative $\kk$-Hopf algebra.

First, it is easy to see that
\begin{equation}
\label{sol.alghoms.comm.g.alghom}
\id_H^{\star k} \text{ is a } \kk\text{-algebra
homomorphism } H \to H
\qquad \qquad \text{ for every } k \in \ZZ .
\end{equation}
\footnote{\textit{Proof of \eqref{sol.alghoms.comm.g.alghom}:}
Let $k \in \ZZ$. If $k$ is nonnegative,
then \eqref{sol.alghoms.comm.g.alghom} can be proven just as in the solution to
Exercise~\ref{exe.alghoms.comm}(f). Hence, for the rest of this
proof of \eqref{sol.alghoms.comm.g.alghom}, we assume WLOG that $k$ is not nonnegative. Thus,
$k < 0$, so that $-k$ is nonnegative. Hence,
Exercise~\ref{exe.alghoms.comm}(b)
(applied to $H$, $-k$ and $\id_H$ instead of $A$, $k$
and $f_i$) yields that
$\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
_{-k\text{ times}}$
is a $\kk$-algebra homomorphism $H \to H$. Since
$\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
_{-k\text{ times}} = \id_H^{\star \left(-k\right)}$, this
shows that $\id_H^{\star \left(-k\right)}$ is a
$\kk$-algebra homomorphism $H \to H$. Thus,
Exercise~\ref{exe.alghoms.comm}(c) (applied to $A = H$ and
$f = \id_H^{\star \left(-k\right)}$) yields that
$\id_H^{\star \left(-k\right)} \circ S : H \to H$ is again
a $\kk$-algebra homomorphism, and is a $\star$-inverse
to $\id_H^{\star \left(-k\right)}$.

Since $\id_H^{\star \left(-k\right)} \circ S$ is a
$\star$-inverse to $\id_H^{\star \left(-k\right)}$, we have
$\id_H^{\star \left(-k\right)} \circ S
= \left(\id_H^{\star \left(-k\right)}\right)^{\star\left(-1\right)}
= \id_H^{\star \left(\left(-k\right)\left(-1\right)\right)}
= \id_H^{\star k}$.
Hence, $\id_H^{\star k}$ is a $\kk$-algebra homomorphism
(since $\id_H^{\star \left(-k\right)} \circ S$ is a
$\kk$-algebra homomorphism), and thus
\eqref{sol.alghoms.comm.g.alghom} is proven.}

Furthermore, we have
\begin{equation}
\label{sol.alghoms.comm.g.starinv}
\id_H^{\star \left(-k\right)} = \id_H^{\star k} \circ S
\qquad \qquad \text{ for every } k \in \ZZ .
\end{equation}
\footnote{\textit{Proof of \eqref{sol.alghoms.comm.g.starinv}:}
Let $k \in \ZZ$. We know that $\id_H^{\star k}$ is
a $\kk$-algebra homomorphism $H \to H$ (according to
\eqref{sol.alghoms.comm.g.alghom}). Thus,
Exercise~\ref{exe.alghoms.comm}(c) (applied to $A = H$ and
$f = \id_H^{\star k}$) yields that
$\id_H^{\star k} \circ S : H \to H$ is again
a $\kk$-algebra homomorphism, and is a $\star$-inverse
to $\id_H^{\star k}$.

Since $\id_H^{\star k} \circ S$ is a
$\star$-inverse to $\id_H^{\star k}$, we have
$\id_H^{\star k} \circ S
= \left(\id_H^{\star k}\right)^{\star\left(-1\right)}
= \id_H^{\star \left(k\left(-1\right)\right)}
= \id_H^{\star \left(-k\right)}$.
This proves
\eqref{sol.alghoms.comm.g.starinv}.}

Now, fix two integers $k$ and $\ell$. Due to
\eqref{sol.alghoms.comm.g.alghom}, we know that $\id_H^{\star k}$
is a $\kk$-algebra homomorphism $H \to H$.
Hence, if $\ell$ is nonnegative, then we can prove
$\id_H^{\star k} \circ \id_H^{\star \ell}
= \id_H^{\star \left(k\ell\right)}$ just as we did in the
solution to Exercise~\ref{exe.alghoms.comm}(f) (and thus
finish the solution to Exercise~\ref{exe.alghoms.comm}(g)).
Hence, for the rest of this solution, we WLOG assume that $\ell$
is not nonnegative. Thus, $\ell < 0$, so that $-\ell$ is
nonnegative.
We can thus apply Exercise~\ref{exe.alghoms.comm}(e) to
$H$, $H$, $H$, $H$, $-\ell$, $\id_H$, $\id_H^{\star k}$ and $\id_H$
instead of
$C$, $C^\prime$, $A$, $A^\prime$, $k$, $f_i$, $\alpha$ and $\gamma$.
As a result, we obtain
\[
\id_H^{\star k} \circ
\left(\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
  _{-\ell\text{ times}}\right)
\circ \id_H
= \underbrace{
\left(\id_H^{\star k} \circ \id_H \circ \id_H\right)
\star \left(\id_H^{\star k} \circ \id_H \circ \id_H\right)
\star \cdots
\star \left(\id_H^{\star k} \circ \id_H \circ \id_H\right)
}_{-\ell \text{ times}} .
\]
Since $\id_H^{\star k} \circ \id_H \circ \id_H
= \id_H^{\star k}$ and
$\underbrace{\id_H \star \id_H \star \cdots \star \id_H}
  _{-\ell\text{ times}}
= \id_H^{\star \left(-\ell\right)}$,
this rewrites as
\[
\id_H^{\star k} \circ \id_H^{\star \left(-\ell\right)}
\circ \id_H
= \underbrace{
\id_H^{\star k} \star \id_H^{\star k} \star \cdots
\star \id_H^{\star k}
}_{-\ell \text{ times}}
= \left(\id_H^{\star k}\right)^{\star \left(-\ell\right)}
= \id_H^{\star \left(k\left(-\ell\right)\right)}
= \id_H^{\star \left(-k \ell\right)} .
\]
In view of $\id_H^{\star \left(-\ell\right)}
\circ \id_H = \id_H^{\star \left(-\ell\right)}$, this rewrites as
\[
\id_H^{\star k} \circ \id_H^{\star \left(-\ell\right)}
= \id_H^{\star \left(-k \ell\right)} .
\]
Now,
$\id_H^{\star \ell} = \id_H^{\star \left(- \left(-\ell\right) \right)}
= \id_H^{\star \left(-\ell\right)} \circ S$
(by \eqref{sol.alghoms.comm.g.starinv}, applied to $-\ell$
instead of $k$), and thus
\[
\id_H^{\star k} \circ \underbrace{\id_H^{\star \ell}}_{= \id_H^{\star \left(-\ell\right)} \circ S}
= \underbrace{\id_H^{\star k} \circ \id_H^{\star \left(-\ell\right)}}_{=\id_H^{\star \left(-k \ell\right)}} \circ S
= \id_H^{\star \left(-k \ell\right)} \circ S .
\]
Compared with
$\id_H^{\star \left(k\ell\right)} = \id_H^{\star \left(- \left(-k\ell\right) \right)}
= \id_H^{\star \left(-k\ell\right)} \circ S$
(by \eqref{sol.alghoms.comm.g.starinv}, applied to $-k\ell$
instead of $k$), this yields
$\id_H^{\star k} \circ \id_H^{\star \ell}
= \id_H^{\star \left(k\ell\right)}$.
This solves Exercise~\ref{exe.alghoms.comm}(g).

(h) The dual of Exercise~\ref{exe.alghoms.comm}(a) is the following
exercise:
\begin{statement}
If $H$ is a $\kk$-bialgebra and $C$ is a cocommutative
$\kk$-coalgebra, and if $f$ and $g$ are two $\kk$-coalgebra homomorphisms
$C \to H$, then prove that $f \star g$ also is a $\kk$-coalgebra
homomorphism $C \to H$.
\end{statement}
The solution of this exercise is obtained from our above solution of
Exercise~\ref{exe.alghoms.comm}(a) by ``reversing arrows'' (and replacing
``algebra'' by ``coalgebra'', and applying
Exercise~\ref{exe.cocommutative.coalghom}(a) instead of
Exercise~\ref{exe.cocommutative.coalghom}(b), and using
Exercise~\ref{exe.tensorprod.morphism}(b) instead of
Exercise~\ref{exe.tensorprod.morphism}(a)).

The dual of Exercise~\ref{exe.alghoms.comm}(b) is the following
exercise:
\begin{statement}
If $H$ is a $\kk$-bialgebra and $C$ is a cocommutative
$\kk$-coalgebra, and if $f_1, f_2, \ldots, f_k$ are several
$\kk$-coalgebra homomorphisms $C \to H$, then prove that
$f_1 \star f_2 \star \cdots \star f_k$ also is a $\kk$-coalgebra
homomorphism $C \to H$.
\end{statement}
This can be solved by induction over $k$ in the same way as
Exercise~\ref{exe.alghoms.comm}(b) (but now using the dual of
Exercise~\ref{exe.alghoms.comm}(a) instead of
Exercise~\ref{exe.alghoms.comm}(a) itself).

The dual of Exercise~\ref{exe.alghoms.comm}(c) is the following
exercise:
\begin{statement}
If $H$ is a Hopf algebra and $C$ is a cocommutative
$\kk$-coalgebra, and if $f : C \to H$ is a $\kk$-coalgebra
homomorphism, then prove that $S \circ f : C \to H$ (where $S$
is the antipode of $H$) is again a $\kk$-coalgebra homomorphism,
and is a $\star$-inverse to $f$.
\end{statement}
A solution of this can be obtained by reversing all arrows in
the above solution of Exercise~\ref{exe.alghoms.comm}(c) (and
using Exercise~\ref{exe.coalg-anti} in lieu of
Proposition~\ref{antipodes-are-antiendomorphisms},
and using Proposition \ref{antipodes-give-convolution-inverses}(b)
in lieu of Proposition \ref{antipodes-give-convolution-inverses}(a)).

The dual of Exercise~\ref{exe.alghoms.comm}(d) is the following
exercise:
\begin{statement}
If $C$ is a cocommutative $\kk$-coalgebra, then show that
$\Delta^{\left(k\right)}$ is a $\kk$-coalgebra homomorphism for every
$k \in \NN$. (The map
$\Delta^{\left(k\right)} : C \to C^{\otimes \left(k+1\right)}$
is defined as in Exercise~\ref{exe.iterated.Delta}.)
\end{statement}
This can be solved just as we solved
Exercise~\ref{exe.alghoms.comm}(d), but again with all arrows
reversed (and referring to
Exercise~\ref{exe.cocommutative.coalghom}(a) and
Exercise~\ref{exe.iterated.patras}(b) instead of
Exercise~\ref{exe.cocommutative.coalghom}(b) and
Exercise~\ref{exe.iterated.patras}(a), respectively).

The dual of Exercise~\ref{exe.alghoms.comm}(e) is
Exercise~\ref{exe.alghoms.comm}(e) itself (up to renaming objects
and maps).

The dual of Exercise~\ref{exe.alghoms.comm}(f) is the following
exercise:
\begin{statement}
If $H$ is a cocommutative $\kk$-bialgebra, and $k$ and $\ell$
are two nonnegative integers, then prove that
$\id_H^{\star \ell} \circ \id_H^{\star k}
= \id_H^{\star \left(\ell k\right)}$.
\end{statement}
This can be solved just as we solved
Exercise~\ref{exe.alghoms.comm}(f), but again with all arrows
reversed.

The dual of Exercise~\ref{exe.alghoms.comm}(g) is the following
exercise:
\begin{statement}
If $H$ is a cocommutative $\kk$-Hopf algebra, and $k$ and $\ell$
are two integers, then prove that
$\id_H^{\star \ell} \circ \id_H^{\star k}
= \id_H^{\star \left(\ell k\right)}$.
\end{statement}
This can be solved just as we solved
Exercise~\ref{exe.alghoms.comm}(g), but again with all arrows
reversed.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cocomm-antipode-S2}}

\textit{Solution to Exercise~\ref{exe.cocomm-antipode-S2}.}
We will use the concepts of ``$\star$-invertible'' maps
and their ``$\star$-inverses''
as defined in Exercise~\ref{exe.coalg-anti.gen}.
We will also use the notations introduced in
Definition~\ref{def.anti-hom}.

Let $A$ be a cocommutative Hopf algebra.
Then, the $\kk$-linear map $\id_A : A \to A$ is
$\star$-invertible (since $A$ is a Hopf algebra), and its $\star$-inverse
$\id_A^{\star \left(-1\right)}$ is the antipode $S$ of $A$. That is,
$\id_A^{\star \left(-1\right)} = S$. Applying
Exercise~\ref{exe.coalg-anti.gen}(b) to $C = A$ and $r = \id_A$, we now
conclude that $\id_A^{\star \left(-1\right)}$ is a $\kk$-coalgebra
anti-homomorphism $A \to A$ (since $\id_A$ is a $\kk$-coalgebra
homomorphism $A \to A$). Since a $\kk$-coalgebra anti-homomorphism
$A \to A$ is the same thing as a $\kk$-coalgebra homomorphism $A \to A$
(by Exercise~\ref{exe.comm-cocomm.anti}(b),
because $A$ is cocommutative), this yields that
$\id_A^{\star \left(-1\right)}$ is a $\kk$-coalgebra homomorphism
$A \to A$.
In other words, $S$ is a $\kk$-coalgebra homomorphism $A \to A$ (since
$\id_A^{\star \left(-1\right)} = S$). Now,
Proposition~\ref{antipodes-give-convolution-inverses}(b) (applied to
$H = A$, $C = A$ and $\gamma = S$) yields
$S \circ S = S^{\star \left(-1\right)}$.
But $S^{\star \left(-1\right)} = \id_A$ (since
$S = \id_A^{\star \left(-1\right)}$). Hence,
$S^2 = S \circ S = S^{\star \left(-1\right)} = \id_A$. This solves
Exercise~\ref{exe.cocomm-antipode-S2}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.dynkin}}

\textit{Solution to Exercise~\ref{exe.dynkin}.}
(a) We shall solve Exercise~\ref{exe.dynkin}(a) in two ways.

First, here is a messy computational solution:

Fix $a\in A$, and assume WLOG that $a$ is homogeneous of degree $n\in\NN$.
Let us prove that $\left( S\star E \right) \left( a \right)$
is primitive.

In fact, using the Sweedler notation,
we can write $\Delta\left(  a\right)  =\sum_{\left(  a\right)  }
a_{1}\otimes a_{2}$ with all $a_{1}$ and $a_{2}$ homogeneous. Thus,
\[
\left(  S\star E\right)  \left(  a\right)  =\sum_{\left(  a\right)  }S\left(
a_{1}\right)  \cdot\underbrace{E\left(  a_{2}\right)  }_{=\left(  \deg
a_{2}\right)  \cdot a_{2}}=\sum_{\left(  a\right)  }\left(  \deg a_{2}\right)
S\left(  a_{1}\right)  \cdot a_{2}.
\]
Hence,
\begin{align*}
\Delta\left(  \left(  S\star E\right)  \left(  a\right)  \right)   &
=\sum_{\left(  a\right)  }\left(  \deg a_{2}\right)  \Delta\left(  S\left(
a_{1}\right)  \cdot a_{2}\right)  =\sum_{\left(  a\right)  }\left(  \deg
a_{2}\right)  \underbrace{\sum_{\left(  a_{1}\right)  }\sum_{\left(
a_{2}\right)  }\left(  S\left(  a_{1}\right)  \right)  _{1}\cdot\left(
a_{2}\right)  _{1}\otimes\left(  S\left(  a_{1}\right)  \right)  _{2}
\cdot\left(  a_{2}\right)  _{2}}_{\substack{=\sum_{\left(  a_{1}\right)  }
\sum_{\left(  a_{2}\right)  }\left(  S\left(  \left(  a_{1}\right)
_{2}\right)  \right)  \cdot\left(  a_{2}\right)  _{1}\otimes\left(  S\left(
\left(  a_{1}\right)  _{1} \right) \right)  \cdot\left(  a_{2}\right)
_{2}\\\text{(here we used }\left(  S\left(  a_{1}\right)  \right)  _{1}
\otimes\left(  S\left(  a_{1}\right)  \right)  _{2}=S\left(  \left(
a_{1}\right)  _{2}\right)  \otimes S\left(  \left(  a_{1}\right) _{1}
\right)
\text{,}\\\text{which is a consequence of Exercise~\ref{exe.coalg-anti})}
}}\\
&  =\sum_{\left(  a\right)  }\left(  \deg a_{2}\right)  \sum_{\left(
a_{1}\right)  }\sum_{\left(  a_{2}\right)  }\left(  S\left(  \left(
a_{1}\right)  _{2}\right)  \right)  \cdot\left(  a_{2}\right)  _{1}
\otimes\left(  S\left(  \left(  a_{1}\right) _{1} \right) \right)
\cdot\left(  a_{2}\right)  _{2}\\
&  =\sum_{\left(  a\right)  }\left(  \deg a_{3}+\deg a_{4}\right)  \left(
S\left(  a_{2}\right)  \right)  \cdot a_{3}\otimes\left(  S\left(
a_{1}\right)  \right)  \cdot a_{4}\\
&  =\sum_{\left(  a\right)  }\left(  \deg a_{3}\right)  \left(  S\left(
a_{2}\right)  \right)  \cdot a_{3}\otimes\left(  S\left(  a_{1}\right)
\right)  \cdot a_{4} +\sum_{\left(  a\right)  }\left(  \deg a_{4}\right)
\left(  S\left(  a_{2}\right)  \right)  \cdot a_{3}\otimes\left(  S\left(
a_{1}\right)  \right)  \cdot a_{4}\\
&  =\sum_{\left(  a\right)  }\left(  \deg a_{2}\right)  \left(  S\left(
a_{1}\right)  \right)  \cdot a_{2}\otimes\underbrace{\left(  S\left(
a_{3}\right)  \right)  \cdot a_{4}}_{=\epsilon\left(  a_{3}\right)  } +\sum_{\left(  a\right)  }\left(  \deg a_{2}\right)
\underbrace{\left(  S\left(  a_{3}\right)  \right)  \cdot a_{4}}
_{=\epsilon\left(  a_{3}\right)  }\otimes\left(  S\left(  a_{1}\right)
\right)  \cdot a_{2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the cocommutativity of }A\right)  \\
&  =\underbrace{\sum_{\left(  a\right)  }\left(  \deg a_{2}\right)  \left(
S\left(  a_{1}\right)  \right)  \cdot a_{2}}_{=\left(  S\star E\right)
\left(  a\right)  }\otimes1+1\otimes\underbrace{\sum_{\left(  a\right)
}\left(  \deg a_{2}\right)  \left(  S\left(  a_{1}\right)  \right)  \cdot
a_{2}}_{=\left(  S\star E\right)  \left(  a\right)  }\\
&  =\left(  S\star E\right)  \left(  a\right)  \otimes1+1\otimes\left(  S\star
E\right)  \left(  a\right)  .
\end{align*}
This (slightly unclean but easily formalizable) computation shows that
$\left(  S\star E\right)  \left(  a\right)  $ is primitive, and similarly the
same can be shown for $\left(  E\star S\right)  \left(  a\right)  $. This
proves (a).

There is, however, a nicer proof: A \dfn{coderivation} of a
$\kk$-coalgebra $\left(  C,\Delta,\epsilon\right)  $ is defined as a
$\kk$-linear map $F:C\rightarrow C$ such that $\Delta\circ F=\left(
F\otimes\id + \id\otimes F\right)  \circ\Delta$. (The reader can check that this
axiom is the result of writing the axiom for a derivation in element-free
terms and reversing all arrows. Nothing less should be expected.) It is easy
to see (by checking on each homogeneous component) that $E$ is a coderivation.
Hence, it will be enough to check that $\left(  S\star f\right)  \left(
a\right)  $ and $\left(  f\star S\right)  \left(  a\right)  $ are primitive
whenever $f:A\rightarrow A$ is a coderivation and $a\in A$. So fix a
coderivation $f:A\rightarrow A$. Notice that the antipode $S$ of $A$ is a
coalgebra anti-endomorphism (by Exercise~\ref{exe.coalg-anti}), thus a
coalgebra endomorphism (because coalgebra anti-endomorphisms of a
cocommutative coalgebra are precisely the same as its coalgebra endomorphisms%
\footnote{This is the result of Exercise~\ref{exe.comm-cocomm.anti}(b).}).
Thus, $\Delta \circ S = \left(S \otimes S\right) \circ \Delta$.
Moreover, $\Delta:A\rightarrow A\otimes A$ is a coalgebra
homomorphism\footnote{In fact, Exercise~\ref{exe.cocommutative.coalghom}(a)
(applied to $C=A$) shows that $A$ is cocommutative
if and only if $\Delta:A\rightarrow A\otimes A$ is a coalgebra
homomorphism. But we know that $A$ is
cocommutative, so that $\Delta:A\rightarrow A\otimes A$ is a coalgebra
homomorphism.} and an algebra homomorphism (since $A$ is a bialgebra).
Applying
\eqref{pre-and-post-composition-in-convolution} to
$A\otimes A$, $A$, $A$, $\Delta$, $\id_{A}$, $S$ and $f$ instead of
$A^{\prime}$, $C$, $C^{\prime}$, $\alpha$, $\gamma$, $f$ and $g$, we obtain
\begin{align}
\Delta\circ\left(  S\star f\right)   &  =\underbrace{\left(  \Delta\circ
S\right)  }_{=\left(  S\otimes S\right)  \circ\Delta}\star\underbrace{\left(
\Delta\circ f\right)  }_{\substack{=\left(  f\otimes\id+\id\otimes f\right)
\circ\Delta\\\text{(since }f\text{ is a coderivation)}}}\nonumber\\
&  =\left(  \left(  S\otimes S\right)  \circ\Delta\right)  \star\left(
\left(  f\otimes\id+\id\otimes f\right)  \circ\Delta\right)
=\left(  \left(  S\otimes S\right)  \star\left(  f\otimes\id+\id\otimes
f\right)  \right)  \circ\Delta\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by \eqref{pre-and-post-composition-in-convolution}, applied to }
A\otimes A\text{, }A\otimes A\text{, }A\otimes
A\otimes A\otimes A\text{, }A\otimes A\text{,}\\
\id \text{, }\Delta\text{, }S\otimes S\text{ and }
f\otimes\id+\id\otimes f\text{ instead of }A\text{, }A^{\prime}\text{, }C\text{,
}C^{\prime}\text{, }\alpha\text{, }\gamma\text{, }f\text{ and }g
\end{array}
\right) \nonumber\\
&  =\left(  \left(  S\otimes S\right)  \star\left(  f\otimes\id\right)  \right)
\circ\Delta+\left(  \left(  S\otimes S\right)  \star\left(  \id\otimes f\right)
\right)  \circ\Delta . \label{sol.dynkin.1}
\end{align}
But Exercise~\ref{exe.convolution.tensor}(a) yields
$\left(  S\otimes S\right)  \star\left(  f\otimes\id\right)
=\left(  S\star f\right)  \otimes\underbrace{\left(  S\star\id\right)
}_{=u\epsilon}=\left(  S\star f\right)  \otimes u\epsilon$ and similarly
$\left(  S\otimes S\right)  \star\left(  \id\otimes f\right)  =u\epsilon
\otimes\left(  S\star f\right)  $. Now, \eqref{sol.dynkin.1} becomes
\begin{align*}
\Delta\circ\left(  S\star f\right)   &  =\underbrace{\left(  \left(  S\otimes
S\right)  \star\left(  f\otimes\id\right)  \right)  }_{=\left(  S\star f\right)
\otimes u\epsilon}\circ\Delta+\underbrace{\left(  \left(  S\otimes S\right)
\star\left(  \id\otimes f\right)  \right)  }_{=u\epsilon\otimes\left(  S\star
f\right)  }\circ\Delta\\
&  =\left(  \left(  S\star f\right)  \otimes u\epsilon\right)  \circ
\Delta+\left(  u\epsilon\otimes\left(  S\star f\right)  \right)  \circ\Delta.
\end{align*}
Hence, every $a\in A$ satisfies (using the Sweedler notation)
\begin{align*}
\left(  \Delta\circ\left(  S\star f\right)  \right)  \left(  a\right)   &
=\left(  \left(  \left(  S\star f\right)  \otimes u\epsilon\right)
\circ\Delta+\left(  u\epsilon\otimes\left(  S\star f\right)  \right)
\circ\Delta\right)  \left(  a\right) \\
&  =\left(  \left(  S\star f\right)  \otimes u\epsilon\right)  \left(
\Delta\left(  a\right)  \right)  +\left(  u\epsilon\otimes\left(  S\star
f\right)  \right)  \left(  \Delta\left(  a\right)  \right) \\
&  =\sum_{\left(  a\right)  }\left(  S\star f\right)  \left(  a_{1}\right)
\otimes\left(  u\epsilon\right)  \left(  a_{2}\right)  +\sum_{\left(
a\right)  }\left(  u\epsilon\right)  \left(  a_{1}\right)  \otimes\left(
S\star f\right)  \left(  a_{2}\right) \\
&  =\underbrace{\sum_{\left(  a\right)  }\left(  S\star f\right)  \left(
a_{1}\right)  \epsilon\left(  a_{2}\right)  }_{=\left(  S\star f\right)
\left(  a\right)  }\otimes 1 + 1\otimes\underbrace{\sum_{\left(  a\right)
}\epsilon\left(  a_{1}\right)  \left(  S\star f\right)  \left(  a_{2}\right)
}_{=\left(  S\star f\right)  \left(  a\right)  }\\
&  =\left(  S\star f\right)  \left(  a\right)  \otimes 1 +
 1\otimes\left(  S\star f\right)  \left(  a\right)  .
\end{align*}
In other words, for every $a\in A$, the element $\left(  S\star f\right)
\left(  a\right)  $ is primitive. Similarly the same can be shown for $\left(
f\star S\right)  \left(  a\right)  $, and so we are done.

(b) is a very simple computation. (Alternatively, the $\left(  S\star
E\right)  \left(  p\right)  =E\left(  p\right)  $ part follows from applying
part (c) to $a=1$, and similarly one can show $\left(  E\star
S\right)  \left(  p\right)  =E\left(  p\right)  $.)

(c) This is computational again: It is straightforward to check that
$E$ is a derivation of the algebra $A$. Now,
\begin{align*}
\Delta\left(  ap\right)   &  =\underbrace{\Delta\left(  a\right)  }
_{=\sum_{\left(  a\right)  }a_{1}\otimes a_{2}}\underbrace{\Delta\left(
p\right)  }_{=p\otimes1+1\otimes p}=\left(  \sum_{\left(  a\right)  }
a_{1}\otimes a_{2}\right)  \left(  p\otimes1+1\otimes p\right) \\
&  =\sum_{\left(  a\right)  }a_{1}p\otimes a_{2}+\sum_{\left(  a\right)
}a_{1}\otimes a_{2}p,
\end{align*}
so that
\begin{align*}
\left(  S\star E\right)  \left(  ap\right)   &  =\sum_{\left(  a\right)
}\underbrace{S\left(  a_{1}p\right)  }_{\substack{=S\left(  p\right)  S\left(
a_{1}\right)  \\\text{(since }S\text{ is an algebra}
\\\text{anti-endomorphism)}}}E\left(  a_{2}\right)  +\sum_{\left(  a\right)
}S\left(  a_{1}\right)  \underbrace{E\left(  a_{2}p\right)  }
_{\substack{=E\left(  a_{2}\right)  p+a_{2}E\left(  p\right)  \\\text{(since
}E\text{ is a derivation)}}}\\
&  =\sum_{\left(  a\right)  }\underbrace{S\left(  p\right)  }
_{\substack{=-p\\\text{(by Proposition~\ref{antipode-on-primitives})}}}
S\left(  a_{1}\right)  E\left(
a_{2}\right)  +\sum_{\left(  a\right)  }S\left(  a_{1}\right)  \left(
E\left(  a_{2}\right)  p+a_{2}E\left(  p\right)  \right) \\
&  =-p\underbrace{\sum_{\left(  a\right)  }S\left(  a_{1}\right)  E\left(
a_{2}\right)  }_{=\left(  S\star E\right)  \left(  a\right)  }
+\underbrace{\sum_{\left(  a\right)  }S\left(  a_{1}\right)  E\left(
a_{2}\right)  }_{=\left(  S\star E\right)  \left(  a\right)  }
p+\underbrace{\sum_{\left(  a\right)  }S\left(  a_{1}\right)  a_{2}
}_{=u\left(  \epsilon\left(  a\right)  \right)  }E\left(  p\right) \\
&  =\underbrace{-p\left(  S\star E\right)  \left(  a\right)  +\left(  S\star
E\right)  \left(  a\right)  p}_{=\left[  \left(  S\star E\right)  \left(
a\right)  ,p\right]  }+\underbrace{u\left(  \epsilon\left(  a\right)  \right)
E\left(  p\right)  }_{=\epsilon\left(  a\right)  E\left(  p\right)  }\\
&  =\left[  \left(  S\star E\right)  \left(  a\right)  ,p\right]
+\epsilon\left(  a\right)  E\left(  p\right)  ,
\end{align*}
thus proving part (c).

(d) Assume that $A$ is connected and that $\QQ$ is a subring of $\kk$.
Let $B$ be the $\kk$-subalgebra of $A$ generated by $\Liep$. In order to
prove part (d), we need to show that $A \subset B$. Clearly,
$\Liep \subset B$.

Consider the grading $A = \bigoplus_{n\geq 0} A_n$ of $A$. Now, we need to
prove that $A \subset B$. In order to prove this, it is clearly enough to
show that $A_n \subset B$ for every $n \in \NN$. We will prove this by
strong induction over $n$.
Let $n \in \NN$, and assume that we have shown that
$A_m \subset B$ for every nonnegative integer $m < n$. We need to prove
that $A_n \subset B$.

If $n = 0$, then this is obvious (since $A_0 = \kk \cdot 1_A \subset B$).
Thus, for the rest of this proof, we WLOG assume that $n \neq 0$.
Hence, $n$ is a positive integer.

Let $a \in A_n$. We know that $\left(  S\star E\right)  \left(  a\right)  $
is primitive (by part (a)), so that $\left(  S\star E\right)  \left(  a\right)  \in
\Liep \subset B$. On the other hand, $a\in A_{n}$ shows that
\begin{align*}
\Delta\left(  a\right)    & \in\Delta\left(  A_{n}\right)  \subset\left(
A\otimes A\right)  _{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{
is a graded map}\right)  \\
& =\sum_{i=0}^{n}A_{i}\otimes A_{n-i}=A_{0}\otimes A_{n}+\sum_{i=1}^{n}
A_{i}\otimes A_{n-i}.
\end{align*}
We can thus write $\Delta\left(  a\right)  $ in the form $\Delta\left(
a\right)  =u+v$ for some $u\in A_{0}\otimes A_{n}$ and $v\in\sum_{i=1}
^{n}A_{i}\otimes A_{n-i}$. Consider these $u$ and $v$.

We are first going to prove that $u=1\otimes a$. Indeed, applying
$\epsilon\otimes\id$ to the equality $\Delta\left(  a\right)  =u+v$, we obtain
$\left(  \epsilon\otimes\id\right)  \left(  \Delta\left(  a\right)  \right)
=\left(  \epsilon\otimes\id\right)  \left(  u+v\right)  =\left(  \epsilon
\otimes\id\right)  \left(  u\right)  +\left(  \epsilon\otimes\id\right)  \left(
v\right)  $. But since $\left(  \epsilon\otimes\id\right)  \left(  \Delta\left(
a\right)  \right)  =a$ (by the commutativity of the diagram
\eqref{counit-diagram}) and $\left(  \epsilon\otimes\id\right)  \left(  v\right)  =0$ (because
\begin{align*}
\left(  \epsilon\otimes\id\right)  \left(  v\right)    & \in\left(
\epsilon\otimes\id\right)  \left(  \sum_{i=1}^{n}A_{i}\otimes A_{n-i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v\in\sum_{i=1}^{n}A_{i}\otimes
A_{n-i}\right)  \\
& =\sum_{i=1}^{n}\underbrace{\epsilon\left(  A_{i}\right)  }
_{\substack{=0\\\text{(since }i>0\text{)}}}\otimes A_{n-i}=\sum_{i=1}^{n}
0 \otimes A_{n-i} = 0
\end{align*}
), this rewrites as $a=\left(  \epsilon\otimes\id\right)  \left(  u\right)  +0$.
In other words, $a=\left(  \epsilon\otimes\id\right)  \left(  u\right)  $. But
the element $u$ has the form $u=1_{A}\otimes u^{\prime}$ for some $u^{\prime
}\in A_{n}$ (because $u\in\underbrace{A_{0}}_{=\kk \cdot 1_{A}}\otimes
A_n = \kk \cdot 1_A \otimes A_n = 1_A \otimes A_n$). This
$u^{\prime}\in A_n$ can be
recovered by $u^{\prime}=\left(  \epsilon\otimes\id\right)  \left(  u\right)  $
(since it satisfies $\left(  \epsilon\otimes\id\right)  \left(  \underbrace{u}
_{=1_{A}\otimes u^{\prime}}\right)  =\left(  \epsilon\otimes\id\right)  \left(
1_{A}\otimes u^{\prime}\right)  =\underbrace{\epsilon\left(  1_{A}\right)
}_{=1}u^{\prime}=u^{\prime}$), and thus simply equals $a$ (since $a=\left(
\epsilon\otimes\id\right)  \left(  u\right)  $). Thus, $u=1_{A}\otimes a$.

Now,
\begin{align*}
\underbrace{\left(  S\star E\right)  }_{=m\circ\left(  S\otimes E\right)
\circ\Delta}\left(  a\right)   & =\left(  m\circ\left(  S\otimes E\right)
\circ\Delta\right)  \left(  a\right)  =\left(  m\circ\left(  S\otimes
E\right)  \right)  \left(  \underbrace{\Delta\left(  a\right)  }_{=u+v}\right)
\\
& =\left(  m\circ\left(  S\otimes E\right)  \right)  \left(  u+v\right) 
 =\left(  m\circ\left(  S\otimes E\right)  \right)  \left(  u\right)
+\left(  m\circ\left(  S\otimes E\right)  \right)  \left(  v\right)  .
\end{align*}
Since
\begin{align*}
\left(  m\circ\left(  S\otimes E\right)  \right)  \left(  \underbrace{u}
_{=1_{A}\otimes a}\right)   & =\left(  m\circ\left(  S\otimes E\right)
\right)  \left(  1_{A}\otimes a\right)  =m\left(  \underbrace{\left(  S\otimes
E\right)  \left(  1_{A}\otimes a\right)  }_{=S\left(  1_{A}\right)  \otimes
E\left(  a\right)  }\right) \\
& =m\left(  S\left(  1_{A}\right)  \otimes E\left(  a\right)  \right)
=\underbrace{S\left(  1_{A}\right)  }_{=1_{A}}\underbrace{E\left(  a\right)  }
_{\substack{=na \\\text{(since }a\in A_{n}\text{)}}}=na
\end{align*}
and
\begin{align*}
& \left(  m\circ\left(  S\otimes E\right)  \right)  \left(  \underbrace{v}
_{\in\sum_{i=1}^{n}A_{i}\otimes A_{n-i}}\right) \\
& \in\left(  m\circ\left(  S\otimes E\right)  \right)  \left(  \sum_{i=1}
^{n}A_{i}\otimes A_{n-i}\right)  =\sum_{i=1}^{n}\left(  m\circ\left(  S\otimes
E\right)  \right)  \left(  A_{i}\otimes A_{n-i}\right) \\
& =\sum_{i=1}^{n}m\underbrace{\left(  \left(  S\otimes E\right)  \left(
A_{i}\otimes A_{n-i}\right)  \right)  }_{=S\left(  A_{i}\right)  \otimes
E\left(  A_{n-i}\right)  }=\sum_{i=1}^{n}m\left(  S\left(  A_{i}\right)
\otimes E\left(  A_{n-i}\right)  \right) \\
& =\sum_{i=1}^{n}S\left(  A_{i}\right)  \cdot E\left(  A_{n-i}\right)
=\sum_{i=1}^{n-1}S\left(  A_{i}\right)  \cdot E\left(  A_{n-i}\right)
+S\left(  A_{n}\right)  \cdot\underbrace{E\left(  A_{0}\right)  }
_{\substack{=0 \\\text{(by the definition} \\\text{of }E\text{)}}}\\
& =\sum_{i=1}^{n-1}\underbrace{S\left(  A_{i}\right)  }_{\substack{\subset
A_{i} \\\text{(since the map }S\text{ is graded)}}}\cdot\underbrace{E\left(
A_{n-i}\right)  }_{\substack{\subset A_{n-i} \\\text{(since the map }E\text{
is graded)}}}\\
& \subset\sum_{i=1}^{n-1}\underbrace{A_{i}}_{\substack{\subset B
\\\text{(since }A_{m}\subset B\text{ for every} \\\text{nonnegative integer
}m<n\text{)}}}\cdot\underbrace{A_{n-i}}_{\substack{\subset B \\\text{(since
}A_{m}\subset B\text{ for every} \\\text{nonnegative integer }m<n\text{)}
}}\subset\sum_{i=1}^{n-1}B\cdot B\subset B
\end{align*}
(since $B$ is an algebra), this becomes
\[
\left(  S\star E\right)  \left(  a\right)  =\underbrace{\left(  m\circ\left(
S\otimes E\right)  \right)  \left(  u\right)  }_{=na}+\underbrace{\left(
m\circ\left(  S\otimes E\right)  \right)  \left(  v\right)  }_{\in B}\in na+B,
\]
so that
\[
na\in\underbrace{\left(  S\star E\right)  \left(  a\right)  }_{\in
B}+B\subset B+B\subset B.
\]
Since $n$ is positive and $\QQ$ is a subring of $\kk$, we can divide
this by $n$ and obtain $a\in B$.

We have thus shown that $a\in B$ for every $a\in A_{n}$. Hence, $A_{n}\subset
B$. This completes the induction step. Thus, we know that $A_{n}\subset B$ for
every $n\in\NN$. Consequently, $A\subset B$, and the proof of (d) is
complete.

This solution of part (d) is not the most generalizable one -- for instance,
(d) also holds if $A$ is connected filtered instead of connected graded,
and then a different argument is necessary. This is a part of the
Cartier-Milnor-Moore theorem, and appears e.g. in
\cite[\S 3.2]{DuchampMinhTolluChienNghia}.

(e) If $a\in T\left(  V\right)  $ is homogeneous of positive degree
and $p\in V$, then part (c) yields
\begin{align*}
\left(  S\star E\right)  \left(  ap\right)   &  =\left[  \left(  S\star
E\right)  \left(  a\right)  ,p\right]  +\underbrace{\epsilon\left(  a\right)
}_{=0}E\left(  p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }p\text{ is
primitive in }T\left(  V\right)  \right) \\
&  =\left[  \left(  S\star E\right)  \left(  a\right)  ,p\right]  .
\end{align*}
This allows proving (e) by induction over $n$, with the induction
base $n=1$ being a consequence of part (b).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.dual-algebra}}

\textit{Solution to Exercise~\ref{exe.dual-algebra}.} (a) Let
$\mathfrak{u}$ be the map $\epsilon_{C}^{\ast}\circ s: \kk \rightarrow
C^{\ast}$. Let $\mathfrak{m}$ be the map $\Delta_{C}^{\ast}\circ\rho
_{C,C}:C^{\ast}\otimes C^{\ast}\rightarrow C^{\ast}$. Our goal is to prove
that $C^{\ast}$, endowed with $\mathfrak{m}$ as the associative operation and
$\mathfrak{u}$ as the unity map, is a $\kk$-algebra. In order to
achieve this, we need to prove that the diagrams
\eqref{coassociativity-diagram} and \eqref{counit-diagram} with $A$, $m$ and
$u$ replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ are commutative.

Let us first prove that the diagram \eqref{coassociativity-diagram} with $A$,
$m$ and $u$ replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ is
commutative. In other words, let us prove that the diagram
\begin{equation}
\xymatrix{ & C^* \otimes C^* \otimes C^* \ar[dl]_{\mathfrak{m} \otimes \id} \ar[dr]^{\id \otimes \mathfrak{m}} & \\ C^* \otimes C^* \ar[dr]^{\mathfrak{m}} & & C^* \otimes C^* \ar[dl]_{\mathfrak{m}} \\ & C^* & }
\label{sol.dual-algebra.a.assoc}
\end{equation}
is commutative.

Indeed, consider the diagram
\begin{equation}
\xymatrix{
& & C^* \otimes C^* \otimes C^* \ar@/_5pc/[ddll]^{\mathfrak{m}\otimes\id}
\ar[dl]^{\rho_{C,C}\otimes\id} \ar[dr]^{\id\otimes\rho_{C,C}} \ar@
/^5pc/[ddrr]^{\id\otimes\mathfrak{m}} & & \\
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id}
\ar[dr]^{\rho_{C\otimes C,C}} & & C^* \otimes\left(C\otimes C\right
)^* \ar[dr]_{\id\otimes\Delta_C^*} \ar[dl]^{\rho_{C,C\otimes C}} & \\
C^* \otimes C^* \ar[dr]^{\rho_{C,C}} \ar@/_5pc/[ddrr]^{\mathfrak{m}}
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes
\id\right)^*} \ar[dr]^{\left(\id\otimes\Delta_C\right)^*} & & C^* \otimes
C^* \ar[dl]_{\rho_{C,C}} \ar@/^5pc/[ddll]^{\mathfrak{m}} \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & & \left(C\otimes
C\right)^* \ar[dl]^{\Delta_C^*} & \\
& & C^* & &
}
. \label{sol.dual-algebra.a.assoc.pf}
\end{equation}
We are going to show that this diagram is commutative. In order to do so, we
will show that its little squares and triangles are commutative.

The triangle
\[
\xymatrix{
& & C^* \otimes C^* \otimes C^* \ar@/_5pc/[ddll]^{\mathfrak{m}\otimes\id}
\ar[dl]^{\rho_{C,C}\otimes\id} \\
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id} &
\\
C^* \otimes C^* & &
}
\]
is commutative, since $\mathfrak{m}=\Delta_{C}^{\ast}\circ\rho_{C,C}$. For the
same reason, the triangle
\[
\xymatrix{
C^* \otimes C^* \otimes C^* \ar[dr]^{\id\otimes\rho_{C,C}} \ar@
/^5pc/[ddrr]^{\id\otimes\mathfrak{m}} & & \\
& C^* \otimes\left(C\otimes C\right)^* \ar[dr]_{\id\otimes\Delta_C^*} & \\
& & C^* \otimes C^*
}
\]
is commutative. The commutativity of the triangles
\[
\xymatrix{
C^* \otimes C^* \ar[dr]^{\rho_{C,C}} \ar@/_5pc/[ddrr]^{\mathfrak{m}}
& & \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & \\
& & C^*
}
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\xymatrix{
& & C^* \otimes C^* \ar[dl]_{\rho_{C,C}} \ar@/^5pc/[ddll]^{\mathfrak{m}}\\
& \left(C\otimes C\right)^* \ar[dl]^{\Delta_C^*} & \\
C^* & &
}
\]
also clearly follows from $\mathfrak{m}=\Delta_{C}^{\ast}\circ\rho_{C,C}$. The
commutativity of the square
\[
\xymatrix{
& & C^* \otimes C^* \otimes C^*
\ar[dl]^{\rho_{C,C}\otimes\id} \ar[dr]^{\id\otimes\rho_{C,C}} & & \\
& \left(C\otimes C\right)^* \otimes C^*
\ar[dr]^{\rho_{C\otimes C,C}} & & C^* \otimes\left(C\otimes C\right
)^* \ar[dl]^{\rho_{C,C\otimes C}} & \\
& & \left(C\otimes C\otimes C\right)^* & &
}
\]
is a basic linear-algebraic fact. The commutativity of the square
\[
\xymatrix{
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id}
\ar[dr]^{\rho_{C\otimes C,C}} & \\
C^* \otimes C^* \ar[dr]^{\rho_{C,C}}
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes
\id\right)^*} \\
& \left(C\otimes C\right)^* &
}
\]
is a particular case of the following linear-algebraic fact: If $X$, $Y$ and
$B$ are three $\kk$-modules and $f:Y\rightarrow X$ is a
$\kk$-linear map, then the diagram
\[
\xymatrix{
& X^* \otimes B^* \ar[dl]^{f^*\otimes\id} \ar[dr]^{\rho_{X,B}} & \\
Y^* \otimes B^* \ar[dr]^{\rho_{Y,B}}
& & \left(X \otimes B\right)^* \ar[dl]^{\left(f\otimes\id\right)^*} \\
& \left(Y\otimes B\right)^* &
}
\]
is commutative.\footnote{This is part of the reason why $\rho_{U,V}$ is
functorial in $U$.} Similarly, the commutativity of the square
\[
\xymatrix{
& C^* \otimes\left(C\otimes C\right)^* \ar[dr]_{\id\otimes\Delta_C^*}
\ar[dl]^{\rho_{C,C\otimes C}} & \\
\left(C\otimes C\otimes C\right)^* \ar[dr]^{\left(\id\otimes\Delta
_C\right)^*}
& & C^* \otimes C^* \ar[dl]_{\rho_{C,C}} \\
& \left(C\otimes C\right)^* &
}
\]
can be shown. Finally, the square
\[
\xymatrix{
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes
\id\right)^*} \ar[dr]^{\left(\id\otimes\Delta_C\right)^*} & & \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & & \left(C\otimes
C\right)^* \ar[dl]^{\Delta_C^*} & \\
& & C^* & &
}
\]
is commutative, because it is obtained by dualizing the commutative diagram
\eqref{coassociativity-diagram}.

We thus have shown that every little square and every little triangle of the
diagram \eqref{sol.dual-algebra.a.assoc.pf} is commutative. Hence, the whole
diagram \eqref{sol.dual-algebra.a.assoc.pf} is commutative. In particular, the
square formed by the four long curved arrows in
\eqref{sol.dual-algebra.a.assoc.pf} is commutative. But this square is
precisely the diagram \eqref{sol.dual-algebra.a.assoc}. Hence, we have shown
that the diagram \eqref{sol.dual-algebra.a.assoc} is commutative. In other
words, the diagram \eqref{coassociativity-diagram} with $A$, $m$ and $u$
replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ is commutative.

It now remains to prove that the diagram \eqref{counit-diagram} with $A$, $m$
and $u$ replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ is
commutative. In other words, it remains to prove that the diagram
\[
\xymatrix{
C^* \otimes \kk \ar[d]_{\id \otimes \mathfrak{u}} & C^* \ar[l] \ar[d]_{\id} \ar[r]
& \kk \otimes C^* \ar[d]^{\mathfrak{u} \otimes \id} \\
C^* \otimes C^* \ar[r]^{\mathfrak{m}} & C^* & C^* \otimes C^* \ar[l]_{\mathfrak{m} }
}
\]
is commutative. We will only prove the commutativity of the left square of
this diagram (since the right square is analogous). That is, we will only
prove the commutativity of the square
\begin{equation}
\xymatrix{ C^* \otimes \kk \ar[d]_{\id \otimes \mathfrak{u}} & C^* \ar[l] \ar[d]_{\id} \\ C^* \otimes C^* \ar[r]^{\mathfrak{m}} & C^* }.
\label{sol.dual-algebra.a.assoc.pfu}
\end{equation}


Indeed, the proof is similar to our proof of \eqref{sol.dual-algebra.a.assoc},
but instead of the big diagram \eqref{sol.dual-algebra.a.assoc.pf} we now have
the diagram
\begin{equation}
\xymatrix{
C^* \otimes\kk\ar[d]^{\id\otimes s} \ar@/_5pc/[dd]^{\id\otimes\mathfrak{u}}
& & C^* \ar[ll]^{\cong} \ar[dl] \ar[dd]^{\id^*} \ar@/^5pc/[dd]^{\id} \\
C^* \otimes\kk^* \ar[r]^{\rho_{C,\kk}} \ar[d]^{\id\otimes\epsilon_C^*}
& \left(C\otimes\kk\right)^* \ar[d]^{\left(\id\otimes\epsilon_C\right)^*}
& \\
C^* \otimes C^* \ar[r]^{\rho_{C,C}} \ar@/_2pc/[rr]_{\mathfrak{m}}
& \left(C\otimes C\right)^* \ar[r]^-{\Delta_C^*} & C^*}
\label{sol.dual-algebra.a.assoc.pfu.pf}
\end{equation}
in which the arrow $C^{\ast}\rightarrow\left(  C\otimes \kk \right)
^{\ast}$ is the adjoint map of the canonical isomorphism $C\otimes
\kk \rightarrow C$ (and in which we identify $\kk^*$ with $\kk$).
The commutativity of the little triangles and
squares is again easily proven (the square
\[
\xymatrix{
& C^* \ar[dl] \ar[dd]^{\id^*} \\
\left(C\otimes\kk\right)^* \ar[d]_{\left(\id\otimes\epsilon_C\right)^*} & \\
\left(C\otimes C\right)^* \ar[r]^-{\Delta_C^*} & C^*}
\]
is commutative by virtue of being the dual of the left square in
\eqref{counit-diagram}). The commutativity of the ``$2$-gon''
\[
\xymatrix{
C^* \ar[d]^{\id^*} \ar@/^3pc/[d]^{\id} \\
C^*}
\]
simply says that $\id^{\ast}=\id$, which is obvious. Thus, everything in
\eqref{sol.dual-algebra.a.assoc.pfu.pf} commutes. By following the
``outer quadrilateral'' of
\eqref{sol.dual-algebra.a.assoc.pfu.pf}, we obtain precisely the commutativity
of \eqref{sol.dual-algebra.a.assoc.pfu}. This completes our solution of
Exercise~\ref{exe.dual-algebra} (a).

\textit{Remark:} Our solution was not the simplest one (by far). We could have
saved much work by doing certain abuses of notation (such as identifying $s$
with the identity map). We could also solve part (a) very easily if we had
solved part (b) first. In the above solution, we have avoided all such shortcuts.

(b) Let $C$ be a $\kk$-coalgebra. Let us notice that
\begin{equation}
\rho_{C,C}\left(  f\otimes g\right)  = m_{\kk}\circ\left(  f\otimes
g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }f\in C^{\ast}\text{ and }g\in
C^{\ast}. \label{sol.dual-algebra.b.1}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.dual-algebra.b.1}:} Let $f\in C^{\ast}$
and $g\in C^{\ast}$. Then, the definition of $\rho_{C,C}$ shows that
$\rho_{C,C}\left(  f\otimes g\right)  $ is the composition $C\otimes
C\overset{f\otimes g}{\longrightarrow}\kk\otimes\kk
\overset{m_{\kk}}{\longrightarrow}\kk$ of the map $f\otimes g$
with the canonical isomorphism $\kk\otimes\kk
\overset{m_{\kk}}{\longrightarrow}\kk$. In other words,
$\rho_{C,C}\left(  f\otimes g\right)  =m_{\kk}\circ\left(  f\otimes
g\right)  $. This proves \eqref{sol.dual-algebra.b.1}.}

Now, we want to prove that the $\kk$-algebra structure defined on
$C^{\ast}$ in part (a) is precisely the one defined on
$\Hom \left(  C,\kk \right)  =C^{\ast}$ in
Definition~\ref{convolution-algebra} applied to $A=\kk$. In order to do
this, it is clearly enough to show that the product of any two elements of
$C^{\ast}$ with respect to the former $\kk$-algebra structure equals
their product with respect to the latter $\kk$-algebra structure. In
other words, it is enough to prove that for any $f\in C^{\ast}$ and $g\in
C^{\ast}$, we have
\begin{align*}
&  \left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }C^{\ast}\text{ defined in part (a)}\right) \\
&  =\left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }\Hom \left(  C,k\right)  \text{ defined in
Definition~\ref{convolution-algebra} applied to }A=\kk\right)  .
\end{align*}
But this follows from the following computation:
\begin{align*}
&  \left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }C^{\ast}\text{ defined in part (a)}\right) \\
&  =\underbrace{m_{C^{\ast}}}_{= \Delta_{C}^{\ast}\circ\rho
_{C,C} }\left(  f\otimes g\right)  =\left(  \Delta_{C}^{\ast}\circ
\rho_{C,C}\right)  \left(  f\otimes g\right)  =\Delta_{C}^{\ast}\left(
\underbrace{\rho_{C,C}\left(  f\otimes g\right)  }_{\substack{=m
_{\kk}\circ\left(  f\otimes g\right)  \\\text{(by
\eqref{sol.dual-algebra.b.1})}}}\right) \\
&  =\Delta_{C}^{\ast}\left(  m_{\kk}\circ\left(  f\otimes g\right)
\right)  =m_{\kk}\circ\left(  f\otimes g\right)  \circ\Delta
_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Delta_{C}^{\ast
}\right) \\
&  =f\star g\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\star g=m_{\kk}
\circ\left(  f\otimes g\right)  \circ\Delta_{C}\text{ (by the definition of
}f\star g\text{)}\right) \\
&  =\left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }\Hom \left(  C,k\right)  \text{ defined in
Definition~\ref{convolution-algebra} applied to }A=\kk\right)  .
\end{align*}
Thus, part (b) of the exercise is solved.

(c) Let $C$ be a graded $\kk$-coalgebra. Let $C=\bigoplus_{n\geq0} C_n$
be its decomposition into homogeneous components. Then, for every
$n\in\NN$, we identify $\left(  C_{n}\right)  ^{\ast}$ with a
$\kk$-submodule of $C^{\ast}$, namely with the $\kk$-submodule $\left\{
f\in C^{\ast}\ \mid\ f\left(  C_{p}\right)  =0\text{ for all }p\in
\NN\text{ satisfying }p\neq n\right\}  $. Thus,
\begin{equation}
\left(  C_{n}\right)  ^{\ast}=\left\{  f\in C^{\ast}\ \mid\ f\left(
C_{p}\right)  =0\text{ for all }p\in \NN \text{ satisfying }p\neq
n\right\}  \label{sol.dual-algebra.c.1}
\end{equation}
for every $n\in\NN$.

By the definition of $C^{o}$, we have $C^{o}=\bigoplus_{n\geq0}\left(
C_{n}\right)  ^{\ast}=\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}$.

Now let $a$ and $b$ be two nonnegative integers. We shall show that $\left(
C_{a}\right)  ^{\ast}\left(  C_{b}\right)  ^{\ast}\subset\left(
C_{a+b}\right)  ^{\ast}$.

Indeed, let $x\in\left(  C_{a}\right)  ^{\ast}$ and $y\in\left(  C_{b}\right)
^{\ast}$ be arbitrary. We have
\[
x\in\left(  C_{a}\right)  ^{\ast}=\left\{  f\in C^{\ast}\ \mid\ f\left(
C_{p}\right)  =0\text{ for all }p\in \NN \text{ satisfying }p\neq
a\right\}
\]
(by \eqref{sol.dual-algebra.c.1}, applied to $n=a$). In other words, $x$ is an
element of $C^{\ast}$ such that
\begin{equation}
x\left(  C_{p}\right)  =0\text{ for all }p\in \NN \text{ satisfying
}p\neq a. \label{sol.dual-algebra.c.1x}
\end{equation}
Similarly, $y$ is an element of $C^{\ast}$ such that
\begin{equation}
y\left(  C_{p}\right)  =0\text{ for all }p\in\NN \text{ satisfying
}p\neq b. \label{sol.dual-algebra.c.1y}
\end{equation}
Also, \eqref{sol.dual-algebra.c.1} (applied to $n=a+b$) yields
\begin{equation}
\left(  C_{a+b}\right)  ^{\ast}=\left\{  f\in C^{\ast}\ \mid\ f\left(
C_{p}\right)  =0\text{ for all }p\in\NN \text{ satisfying }p\neq
a+b\right\}  . \label{sol.dual-algebra.c.3}
\end{equation}


Now, part (b) of this exercise yields that the $\kk$-algebra structure
defined on $C^{\ast}$ in part (a) is precisely the one defined on
$\Hom \left(  C,\kk \right)  =C^{\ast}$ in
Definition~\ref{convolution-algebra} applied to $A=\kk$. Thus, the
product of the $\kk$-algebra $C^{\ast}$ is precisely the convolution
product $\star$ on $\Hom\left(  C,\kk \right)  =C^{\ast}$. Hence,
\[
xy=x\star y=m_{\kk}\circ\left(  x\otimes y\right)  \circ\Delta_{C}.
\]


Now, fix $p\in\NN$ such that $p\neq a+b$. Then, $\Delta_{C}\left(
C_{p}\right)  \subset\sum_{q=0}^{p}C_{q}\otimes C_{p-q}$ (since the coalgebra
$C$ is graded). Now, every $q\in\left\{  0,1,...,p\right\}  $ satisfies
\begin{equation}
\left(  x\otimes y\right)  \left(  C_{q}\otimes C_{p-q}\right)  =0.
\label{sol.dual-algebra.c.2}
\end{equation}
(\textit{Proof of \eqref{sol.dual-algebra.c.2}:} Let
$q \in \left\{  0,1,...,p\right\}$. Then we must have
either $q \neq a$ or $p-q \neq b$ (or both), because otherwise we would
have $p = \underbrace{q}_{=a}+\underbrace{\left(p-q\right)}_{=b} = a+b$
which would contradict $p \neq a+b$.
If $q \neq a$, then $x\left( C_{q}\right)  =0$ (by
\eqref{sol.dual-algebra.c.1x}) and therefore
$\left(  x\otimes y\right)  \left(  C_{q}\otimes C_{p-q}\right) 
= \underbrace{x\left(  C_{q}\right)  }_{=0}\otimes y\left(
C_{p-q}\right)  =0$. If $p-q \neq b$, then
\eqref{sol.dual-algebra.c.1y} yields
$y\left(  C_{p-q}\right)  =0$, and thus $\left(  x\otimes y\right)  \left(
C_{q}\otimes C_{p-q}\right)  =x\left(  C_{q}\right)  \otimes
\underbrace{y\left(  C_{p-q}\right)  }_{=0}=0$.
Hence, in either case, we have $\left(  x\otimes
y\right)  \left(  C_{q}\otimes C_{p-q}\right)  =0$, and so
\eqref{sol.dual-algebra.c.2} is proven.)

Now,
\begin{align*}
\left(  \underbrace{xy}_{=m_{\kk}\circ\left(  x\otimes y\right)
\circ\Delta_{C}}\right)  \left(  C_{p}\right)   &  =\left(  m_{\kk}
\circ\left(  x\otimes y\right)  \circ\Delta_{C}\right)  \left(  C_{p}\right)
=m_{\kk}\left(  \left(  x\otimes y\right)  \left(  \underbrace{\Delta
_{C}\left(  C_{p}\right)  }_{\subset\sum_{q=0}^{p}C_{q}\otimes C_{p-q}}
\right)  \right) \\
&  \subset m_{\kk}\left(  \left(  x\otimes y\right)  \left(
\sum_{q=0}^{p}C_{q}\otimes C_{p-q}\right)  \right)  =\sum_{q=0}^{p}
m_{\kk}\left(  \underbrace{\left(  x\otimes y\right)  \left(
C_{q}\otimes C_{p-q}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.dual-algebra.c.2})}}}\right) \\
&  =\sum_{q=0}^{p}m_{\kk}0 = 0.
\end{align*}


In other words, $\left(  xy\right)  \left(  C_{p}\right)  =0$.

Now, forget that we fixed $p$. We thus have shown that $\left(  xy\right)
\left(  C_{p}\right)  =0$ for all $p\in\NN$ satisfying $p\neq a+b$. In
other words,
\[
xy\in\left\{  f\in C^{\ast}\ \mid\ f\left(  C_{p}\right)  =0\text{ for all
}p\in\NN \text{ satisfying }p\neq a+b\right\}  =\left(  C_{a+b}\right)
^{\ast}
\]
(by \eqref{sol.dual-algebra.c.3}).

Now, forget that we fixed $x$ and $y$. We thus have shown that $xy\in\left(
C_{a+b}\right)  ^{\ast}$ for all $x\in\left(  C_{a}\right)  ^{\ast}$ and
$y\in\left(  C_{b}\right)  ^{\ast}$. This yields $\left(  C_{a}\right)
^{\ast}\left(  C_{b}\right)  ^{\ast}\subset\left(  C_{a+b}\right)  ^{\ast}$
(since $\left(  C_{a+b}\right)  ^{\ast}$ is a $\kk$-module).

But this has been proven for all $a \in \NN$ and $b \in \NN$. From this,
it is easy to conclude that $C^o C^o \subset C^o$ (since
$C^{o}=\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}$). Hence, the
$\kk$-submodule $C^o$ of $C^{\ast}$ is closed under multiplication. Since
we also have $1_{C^{\ast}}\in C^{o}$ (in fact, it is very easy to see that
$1_{C^{\ast}}=\epsilon_{C}\in\left(  C_{0}\right)  ^{\ast}\subset\sum_{n\geq
0}\left(  C_{n}\right)  ^{\ast}=C^{o}$), this shows that $C^{o}$ is a
$\kk$-subalgebra of $C^{\ast}$. This solves part (c) of the exercise.

(d) Let $C$ and $D$ be two $\kk$-coalgebras. Let $f:C\rightarrow D$ be
a homomorphism of $\kk$-coalgebras. Then, the two diagrams
\begin{equation}
\xymatrix{ C \ar[d]_{\Delta_C} \ar[r]^{f} & D \ar[d]^{\Delta_D} \\ C \otimes C \ar[r]^{f \otimes f} & D \otimes D }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ C \ar[dr]_{\epsilon_C} \ar[rr]^{f}& & D \ar[dl]^{\epsilon_D} \\ &\kk& }
\label{sol.dual-algebra.d.start}
\end{equation}
are commutative (since $f$ is a homomorphism of $\kk$-coalgebras).

We need to prove that $f^{\ast}:D^{\ast}\rightarrow C^{\ast}$ is a
homomorphism of $\kk$-algebras. In other words, we need to prove that
the two diagrams
\begin{equation}
\xymatrix{ D^* \ar[r]^{f^*} & C^* \\ D^* \otimes D^* \ar[u]^{m_{D^*}} \ar[r]^{f^* \otimes f^*} & C^* \otimes C^* \ar[u]_{m_{C^*}} }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ D^* \ar[rr]^{f^*} & & C^* \\ &\kk \ar[ul]^{u_{D^*}} \ar[ur]_{u_{C^*}}& }
\label{sol.dual-algebra.d.goal}
\end{equation}
are commutative.

By the definition of the $\kk$-algebra $D^{\ast}$, we have $m_{D^{\ast
}}=\Delta_{D}^{\ast}\circ\rho_{D,D}$. Thus, the left triangle (with two
vertical edges and one curved edge) in the diagram
\begin{equation}
\xymatrix{
D^* \ar[r]^{f^*} & C^* \\
\left(D\otimes D\right)^* \ar[u]^{\Delta_D^*} \ar[r]^{\left(f\otimes
f\right)^*} & \left(C\otimes C\right)^* \ar[u]^{\Delta_C^*} \\
D^* \otimes D^* \ar[u]^{\rho_{D,D}} \ar@/^4pc/[uu]^{m_{D^*}} \ar
[r]^{f^* \otimes f^*} & C^* \otimes C^* \ar[u]^{\rho_{C,C}} \ar@
/_4pc/[uu]_{m_{C^*}}
}
\label{sol.dual-algebra.d.2}
\end{equation}
commutes. Similarly, the right triangle in \eqref{sol.dual-algebra.d.2} also
commutes. The upper rectangle commutes because it is obtained from the first
diagram in \eqref{sol.dual-algebra.d.start} by dualizing. The lower triangle
in \eqref{sol.dual-algebra.d.2} commutes by basic linear algebra. Hence, the
whole diagram \eqref{sol.dual-algebra.d.2} commutes. But the outer rim of the
diagram \eqref{sol.dual-algebra.d.2} is exactly the first diagram in
\eqref{sol.dual-algebra.d.goal}. Thus, the first diagram in
\eqref{sol.dual-algebra.d.goal} commutes. The (even simpler) task of proving
the commutativity of the second diagram in \eqref{sol.dual-algebra.d.goal} is
left to the reader. Thus, $f^{\ast}$ is a homomorphism of $\kk$-algebras.
Part (d) of the exercise is solved.

(e) This is an exercise in linear algebra (it has nothing to do with Hopf
algebras). Here is a rough sketch of how it is solved: Let $U=\bigoplus
_{n\geq0}U_{n}$ be the decomposition of $U$ into its homogeneous components,
and let $V=\bigoplus_{n\geq0}V_{n}$ be the decomposition of $V$ into its
homogeneous components. Notice that every $V_{n}$ is finite free (since
$V$ is of finite type). Graded $\kk$-linear maps $U\rightarrow V$ can
be regarded as elements of $\prod_{n\geq0}\Hom\left(U_{n},V_{n}\right)  $
(because a graded $\kk$-linear map $U\rightarrow
V$ restricts to a $\kk$-linear map $U_{n}\rightarrow V_{n}$ for every
$n\in\NN$, and is uniquely determined by the totality of these
restrictions). Similarly, graded $\kk$-linear maps $V^{o}\rightarrow
U^{o}$ can be regarded as elements of
$\prod_{n\geq0}\Hom\left(  \left(  V_{n}\right)  ^{\ast},
\left(  U_{n}\right)  ^{\ast}\right)  $.
For every $n\in\NN$, there is a canonical $\kk$-module
isomorphism $\Hom\left(  U_{n},V_{n}\right)  \rightarrow
\Hom\left(  \left(  V_{n}\right)  ^{\ast},\left(  U_{n}\right)
^{\ast}\right)  $ which sends every $\varphi\in\Hom\left(
U_{n},V_{n}\right)  $ to $\varphi^{\ast}\in\Hom\left(  \left(
V_{n}\right)  ^{\ast},\left(  U_{n}\right)  ^{\ast}\right)  $ (since $V_{n}$
is finite free). Taking the product of these isomorphisms, we obtain a
$\kk$-module isomorphism from $\prod_{n\geq0}\Hom\left(
U_{n},V_{n}\right)  $ to $\prod_{n\geq0}\Hom\left(  \left(
V_{n}\right)  ^{\ast},\left(  U_{n}\right)  ^{\ast}\right)  $, that is, a
$\kk$-module isomorphism from the $\kk$-module of all graded
$\kk$-linear maps $U\rightarrow V$ to the $\kk$-module of all
graded $\kk$-linear maps $V^{o}\rightarrow U^{o}$. It is easy to see
that this isomorphism sends every $f:U\rightarrow V$ to
$f^{\ast}:V^{o}\rightarrow U^{o}$. Hence, there is a 1-to-1
correspondence between
graded $\kk$-linear maps $U\rightarrow V$ and graded $\kk$-linear maps
$V^{o}\rightarrow U^{o}$ given by $f\mapsto f^{\ast}$ (namely,
this isomorphism). This solves Exercise~\ref{exe.dual-algebra} (e).

(f) Let $f:C\rightarrow D$ be a graded $\kk$-linear map. We need to
prove that $f:C\rightarrow D$ is a $\kk$-coalgebra morphism if and only if
$f^{\ast}:D^{o}\rightarrow C^{o}$ is a $\kk$-algebra morphism. In other
words, we need to prove the following two assertions:

\begin{statement}
\textit{Assertion 1:} If $f:C\rightarrow D$ is a $\kk$-coalgebra morphism,
then $f^{\ast}:D^{o}\rightarrow C^{o}$ is a $\kk$-algebra morphism.
\end{statement}

\begin{statement}
\textit{Assertion 2:} If $f^{\ast}:D^{o}\rightarrow C^{o}$ is a
$\kk$-algebra morphism, then $f:C\rightarrow D$ is a $\kk$-coalgebra morphism.
\end{statement}

We start by proving Assertion 1. One way to prove it proceeds by repeating the
solution of Exercise~\ref{exe.dual-algebra} (d), except that $D^{\ast}$,
$\left(  D\otimes D\right)  ^{\ast}$, $C^{\ast}$ and $\left(  C\otimes
C\right)  ^{\ast}$ are replaced by $D^{o}$, $\left(  D\otimes D\right)  ^{o}$,
$C^{o}$ and $\left(  C\otimes C\right)  ^{o}$ (where, of course, the map
$\rho_{D,D}:D^{o}\rightarrow\left(  D\otimes D\right)  ^{o}$ has to be
interpreted as the restriction of the map $\rho_{D,D}:D^{\ast}\rightarrow
\left(  D\otimes D\right)  ^{\ast}$ to the $\kk$-submodule $D^{o}$ of $D^{\ast}$, and
similarly for the other maps). This replacement can be done completely
robotically, and thus is left to the reader. Another way to prove Assertion 1
is by realizing that it follows immediately from
Exercise~\ref{exe.dual-algebra} (d) (since $f^{\ast}:D^{o}\rightarrow C^{o}$
is a restriction of the map $f^{\ast}:D^{\ast}\rightarrow C^{\ast}$).
Either way,
we do not end up using the assumption that $D$ is of finite type. However, we
will need this assumption in our proof of Assertion 2.

Now, let us prove Assertion 2. Assume that $f^{\ast}:D^{o}\rightarrow C^{o}$
is a $\kk$-algebra morphism. We want to show that $f:C\rightarrow D$ is a
$\kk$-coalgebra morphism. In other words, we want to show that the two
diagrams \eqref{sol.dual-algebra.d.start} commute. Let us start with the left
one of these diagrams.

The graded $\kk$-module $D$ is of finite type, and therefore the map
$\rho_{D,D}:D^{o}\otimes D^{o}\rightarrow\left(  D\otimes D\right)  ^{o}$ (a
restriction of the map $\rho_{D,D}:D^{\ast}\otimes D^{\ast}\rightarrow\left(
D\otimes D\right)  ^{\ast}$) is an isomorphism. Its inverse
$\rho_{D,D}^{-1} : \left(  D\otimes D\right) ^o \rightarrow D^o \otimes D^o$
is therefore well-defined\footnote{Beware: we don't have an inverse of the
non-restricted map $\rho_{D,D}:D^{\ast}\otimes D^{\ast}\rightarrow\left(
D\otimes D\right)  ^{\ast}$.}. We can thus form the (asymmetric!) diagram
\begin{equation}
{
\xymatrixcolsep{5pc}
\xymatrix{
D^o \ar[rrr]^{f^*} & & & C^o \\
& D^o \otimes D^o \ar[ul]^{m_{D^*}} \ar[r]^{f^* \otimes f^*} & C^o \otimes
C^o \ar[ur]^{m_{C^*}}  \ar[dr]^{\rho_{C,C}} & \\
\left(D\otimes D\right)^o \ar[uu]^{\Delta_D^*} \ar[ur]^{\rho_{D,D}^{-1}}
\ar[rrr]_{\left(f\otimes f\right)^*} & & & \left(C\otimes C\right
)^o \ar[uu]^{\Delta_C^*}
}
}
.\label{sol.dual-algebra.e.2}
\end{equation}
(The arrows labelled $m_{C^{\ast}}$ and $m_{D^{\ast}}$ could just as well have
been labelled $m_{C^{o}}$ and $m_{D^{o}}$, since the multiplication maps
$m_{C^{o}}$ and $m_{D^{o}}$ are restrictions of $m_{C^{\ast}}$ and
$m_{D^{\ast}}$.) The two triangles in \eqref{sol.dual-algebra.e.2} commute due
to $m_{D^{\ast}}=\Delta_{D}^{\ast}\circ\rho_{D,D}$ and $m_{C^{\ast}}
=\Delta_{C}^{\ast}\circ\rho_{C,C}$. The upper quadrilateral in
\eqref{sol.dual-algebra.e.2} commutes because $f^{\ast}$ is a
$\kk$-algebra homomorphism, and the lower quadrilateral in
\eqref{sol.dual-algebra.e.2} commutes because of the commutativity of the
diagram
\[
\xymatrix{
D^o \otimes D^o \ar[d]^{\rho_{D,D}} \ar[r]^{f^* \otimes f^*} & C^o \otimes
C^o \ar[d]^{\rho_{C,C}} \\
\left(D\otimes D\right)^o \ar[r]_{\left(f\otimes f\right)^*} & \left(C\otimes
C\right)^o
}
\]
(which follows from standard linear algebra). Hence, the whole diagram
\eqref{sol.dual-algebra.e.2} commutes. Removing the two interior nodes of this
diagram, we obtain the commutative diagram
\begin{equation}
\xymatrixcolsep{5pc}
\xymatrix{
D^o \ar[r]^{f^*} & C^o \\
\left(D\otimes D\right)^o \ar[u]^{\Delta_D^*} \ar[r]_{\left(f\otimes
f\right)^*} & \left(C\otimes C\right)^o \ar[u]^{\Delta_C^*}
}
.\label{sol.dual-algebra.e.close}
\end{equation}
This does not immediately yield the commutativity of the first diagram in
\eqref{sol.dual-algebra.d.start} (because we cannot revert taking dual
$\kk$-modules), so we are not yet done. But we are close.

We have
\begin{align*}
\left(  \Delta_{D}\circ f\right)  ^{\ast}  & =f^{\ast}\circ\Delta_{D}^{\ast
}=\Delta_{C}^{\ast}\circ\left(  f\otimes f\right)  ^{\ast}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the diagram
\eqref{sol.dual-algebra.e.close} commutes}\right)  \\
& =\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)  ^{\ast}
\end{align*}
as maps from $\left(  D\otimes D\right)  ^{o}$ to $C^{o}$. But a general
linear-algebraic fact states that if $U$ and $V$ are two graded
$\kk$-modules such that $V$ is of finite type, and if $\alpha$
and $\beta$ are two graded $\kk$-linear maps $U\rightarrow V$ such that
$\alpha^{\ast}=\beta^{\ast}$ as maps from $V^{o}$ to $U^{o}$, then
$\alpha=\beta$\ \ \ \ \footnote{This follows immediately from
Exercise~\ref{exe.dual-algebra} (e).}. Applying this to $U=C$,
$V=D\otimes D$,
$\alpha=\Delta_{D}\circ f$ and $\beta=\left(  f\otimes f\right)  \circ
\Delta_{C}$, we obtain $\Delta_{D}\circ f=\left(  f\otimes f\right)
\circ\Delta_{C}$ (since $D\otimes D$ is of finite type\footnote{This is
because $D$ is of finite type.} and since $\left(  \Delta_{D}\circ f\right)
^{\ast}=\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)  ^{\ast}$).
In other words, the first diagram in \eqref{sol.dual-algebra.d.start} is
commutative. The reader can verify that so is the second diagram in
\eqref{sol.dual-algebra.d.start} (once again, this is the easier part). So we
have shown that both diagrams in \eqref{sol.dual-algebra.d.start} are
commutative, and thus $f$ is a $\kk$-coalgebra morphism. This proves
Assertion 2.

Now that Assertions 1 and 2 are both proven, part (f) of the exercise is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.dual-algebra}}

\textit{Solution to Exercise~\ref{exe.dual-algebra}.} (a) Let
$\mathfrak{u}$ be the map $\epsilon_{C}^{\ast}\circ s: \kk \rightarrow
C^{\ast}$. Let $\mathfrak{m}$ be the map $\Delta_{C}^{\ast}\circ\rho
_{C,C}:C^{\ast}\otimes C^{\ast}\rightarrow C^{\ast}$. Our goal is to prove
that $C^{\ast}$, endowed with $\mathfrak{m}$ as the associative operation and
$\mathfrak{u}$ as the unity map, is a $\kk$-algebra. In order to
achieve this, we need to prove that the diagrams
\eqref{coassociativity-diagram} and \eqref{counit-diagram} with $A$, $m$ and
$u$ replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ are commutative.

Let us first prove that the diagram \eqref{coassociativity-diagram} with $A$,
$m$ and $u$ replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ is
commutative. In other words, let us prove that the diagram
\begin{equation}
\xymatrix{ & C^* \otimes C^* \otimes C^* \ar[dl]_{\mathfrak{m} \otimes \id} \ar[dr]^{\id \otimes \mathfrak{m}} & \\ C^* \otimes C^* \ar[dr]^{\mathfrak{m}} & & C^* \otimes C^* \ar[dl]_{\mathfrak{m}} \\ & C^* & }
\label{sol.dual-algebra.a.assoc}
\end{equation}
is commutative.

Indeed, consider the diagram
\begin{equation}
\xymatrix{
& & C^* \otimes C^* \otimes C^* \ar@/_5pc/[ddll]^{\mathfrak{m}\otimes\id}
\ar[dl]^{\rho_{C,C}\otimes\id} \ar[dr]^{\id\otimes\rho_{C,C}} \ar@
/^5pc/[ddrr]^{\id\otimes\mathfrak{m}} & & \\
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id}
\ar[dr]^{\rho_{C\otimes C,C}} & & C^* \otimes\left(C\otimes C\right
)^* \ar[dr]_{\id\otimes\Delta_C^*} \ar[dl]^{\rho_{C,C\otimes C}} & \\
C^* \otimes C^* \ar[dr]^{\rho_{C,C}} \ar@/_5pc/[ddrr]^{\mathfrak{m}}
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes
\id\right)^*} \ar[dr]^{\left(\id\otimes\Delta_C\right)^*} & & C^* \otimes
C^* \ar[dl]_{\rho_{C,C}} \ar@/^5pc/[ddll]^{\mathfrak{m}} \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & & \left(C\otimes
C\right)^* \ar[dl]^{\Delta_C^*} & \\
& & C^* & &
}
. \label{sol.dual-algebra.a.assoc.pf}
\end{equation}
We are going to show that this diagram is commutative. In order to do so, we
will show that its little squares and triangles are commutative.

The triangle
\[
\xymatrix{
& & C^* \otimes C^* \otimes C^* \ar@/_5pc/[ddll]^{\mathfrak{m}\otimes\id}
\ar[dl]^{\rho_{C,C}\otimes\id} \\
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id} &
\\
C^* \otimes C^* & &
}
\]
is commutative, since $\mathfrak{m}=\Delta_{C}^{\ast}\circ\rho_{C,C}$. For the
same reason, the triangle
\[
\xymatrix{
C^* \otimes C^* \otimes C^* \ar[dr]^{\id\otimes\rho_{C,C}} \ar@
/^5pc/[ddrr]^{\id\otimes\mathfrak{m}} & & \\
& C^* \otimes\left(C\otimes C\right)^* \ar[dr]_{\id\otimes\Delta_C^*} & \\
& & C^* \otimes C^*
}
\]
is commutative. The commutativity of the triangles
\[
\xymatrix{
C^* \otimes C^* \ar[dr]^{\rho_{C,C}} \ar@/_5pc/[ddrr]^{\mathfrak{m}}
& & \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & \\
& & C^*
}
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\xymatrix{
& & C^* \otimes C^* \ar[dl]_{\rho_{C,C}} \ar@/^5pc/[ddll]^{\mathfrak{m}}\\
& \left(C\otimes C\right)^* \ar[dl]^{\Delta_C^*} & \\
C^* & &
}
\]
also clearly follows from $\mathfrak{m}=\Delta_{C}^{\ast}\circ\rho_{C,C}$. The
commutativity of the square
\[
\xymatrix{
& & C^* \otimes C^* \otimes C^*
\ar[dl]^{\rho_{C,C}\otimes\id} \ar[dr]^{\id\otimes\rho_{C,C}} & & \\
& \left(C\otimes C\right)^* \otimes C^*
\ar[dr]^{\rho_{C\otimes C,C}} & & C^* \otimes\left(C\otimes C\right
)^* \ar[dl]^{\rho_{C,C\otimes C}} & \\
& & \left(C\otimes C\otimes C\right)^* & &
}
\]
is a basic linear-algebraic fact. The commutativity of the square
\[
\xymatrix{
& \left(C\otimes C\right)^* \otimes C^* \ar[dl]^{\Delta_C^*\otimes\id}
\ar[dr]^{\rho_{C\otimes C,C}} & \\
C^* \otimes C^* \ar[dr]^{\rho_{C,C}}
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes
\id\right)^*} \\
& \left(C\otimes C\right)^* &
}
\]
is a particular case of the following linear-algebraic fact: If $X$, $Y$ and
$B$ are three $\kk$-modules and $f:Y\rightarrow X$ is a
$\kk$-linear map, then the diagram
\[
\xymatrix{
& X^* \otimes B^* \ar[dl]^{f^*\otimes\id} \ar[dr]^{\rho_{X,B}} & \\
Y^* \otimes B^* \ar[dr]^{\rho_{Y,B}}
& & \left(X \otimes B\right)^* \ar[dl]^{\left(f\otimes\id\right)^*} \\
& \left(Y\otimes B\right)^* &
}
\]
is commutative.\footnote{This is part of the reason why $\rho_{U,V}$ is
functorial in $U$.} Similarly, the commutativity of the square
\[
\xymatrix{
& C^* \otimes\left(C\otimes C\right)^* \ar[dr]_{\id\otimes\Delta_C^*}
\ar[dl]^{\rho_{C,C\otimes C}} & \\
\left(C\otimes C\otimes C\right)^* \ar[dr]^{\left(\id\otimes\Delta
_C\right)^*}
& & C^* \otimes C^* \ar[dl]_{\rho_{C,C}} \\
& \left(C\otimes C\right)^* &
}
\]
can be shown. Finally, the square
\[
\xymatrix{
& & \left(C\otimes C\otimes C\right)^* \ar[dl]^{\left(\Delta_C\otimes
\id\right)^*} \ar[dr]^{\left(\id\otimes\Delta_C\right)^*} & & \\
& \left(C\otimes C\right)^* \ar[dr]^{\Delta_C^*} & & \left(C\otimes
C\right)^* \ar[dl]^{\Delta_C^*} & \\
& & C^* & &
}
\]
is commutative, because it is obtained by dualizing the commutative diagram
\eqref{coassociativity-diagram}.

We thus have shown that every little square and every little triangle of the
diagram \eqref{sol.dual-algebra.a.assoc.pf} is commutative. Hence, the whole
diagram \eqref{sol.dual-algebra.a.assoc.pf} is commutative. In particular, the
square formed by the four long curved arrows in
\eqref{sol.dual-algebra.a.assoc.pf} is commutative. But this square is
precisely the diagram \eqref{sol.dual-algebra.a.assoc}. Hence, we have shown
that the diagram \eqref{sol.dual-algebra.a.assoc} is commutative. In other
words, the diagram \eqref{coassociativity-diagram} with $A$, $m$ and $u$
replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ is commutative.

It now remains to prove that the diagram \eqref{counit-diagram} with $A$, $m$
and $u$ replaced by $C^{\ast}$, $\mathfrak{m}$ and $\mathfrak{u}$ is
commutative. In other words, it remains to prove that the diagram
\[
\xymatrix{
C^* \otimes \kk \ar[d]_{\id \otimes \mathfrak{u}} & C^* \ar[l] \ar[d]_{\id} \ar[r]
& \kk \otimes C^* \ar[d]^{\mathfrak{u} \otimes \id} \\
C^* \otimes C^* \ar[r]^{\mathfrak{m}} & C^* & C^* \otimes C^* \ar[l]_{\mathfrak{m} }
}
\]
is commutative. We will only prove the commutativity of the left square of
this diagram (since the right square is analogous). That is, we will only
prove the commutativity of the square
\begin{equation}
\xymatrix{ C^* \otimes \kk \ar[d]_{\id \otimes \mathfrak{u}} & C^* \ar[l] \ar[d]_{\id} \\ C^* \otimes C^* \ar[r]^{\mathfrak{m}} & C^* }.
\label{sol.dual-algebra.a.assoc.pfu}
\end{equation}


Indeed, the proof is similar to our proof of \eqref{sol.dual-algebra.a.assoc},
but instead of the big diagram \eqref{sol.dual-algebra.a.assoc.pf} we now have
the diagram
\begin{equation}
\xymatrix{
C^* \otimes\kk\ar[d]^{\id\otimes s} \ar@/_5pc/[dd]^{\id\otimes\mathfrak{u}}
& & C^* \ar[ll]^{\cong} \ar[dl] \ar[dd]^{\id^*} \ar@/^5pc/[dd]^{\id} \\
C^* \otimes\kk^* \ar[r]^{\rho_{C,\kk}} \ar[d]^{\id\otimes\epsilon_C^*}
& \left(C\otimes\kk\right)^* \ar[d]^{\left(\id\otimes\epsilon_C\right)^*}
& \\
C^* \otimes C^* \ar[r]^{\rho_{C,C}} \ar@/_2pc/[rr]_{\mathfrak{m}}
& \left(C\otimes C\right)^* \ar[r]^-{\Delta_C^*} & C^*}
\label{sol.dual-algebra.a.assoc.pfu.pf}
\end{equation}
in which the arrow $C^{\ast}\rightarrow\left(  C\otimes \kk \right)
^{\ast}$ is the adjoint map of the canonical isomorphism $C\otimes
\kk \rightarrow C$ (and in which we identify $\kk^*$ with $\kk$).
The commutativity of the little triangles and
squares is again easily proven (the square
\[
\xymatrix{
& C^* \ar[dl] \ar[dd]^{\id^*} \\
\left(C\otimes\kk\right)^* \ar[d]_{\left(\id\otimes\epsilon_C\right)^*} & \\
\left(C\otimes C\right)^* \ar[r]^-{\Delta_C^*} & C^*}
\]
is commutative by virtue of being the dual of the left square in
\eqref{counit-diagram}). The commutativity of the ``$2$-gon''
\[
\xymatrix{
C^* \ar[d]^{\id^*} \ar@/^3pc/[d]^{\id} \\
C^*}
\]
simply says that $\id^{\ast}=\id$, which is obvious. Thus, everything in
\eqref{sol.dual-algebra.a.assoc.pfu.pf} commutes. By following the
``outer quadrilateral'' of
\eqref{sol.dual-algebra.a.assoc.pfu.pf}, we obtain precisely the commutativity
of \eqref{sol.dual-algebra.a.assoc.pfu}. This completes our solution of
Exercise~\ref{exe.dual-algebra} (a).

\textit{Remark:} Our solution was not the simplest one (by far). We could have
saved much work by doing certain abuses of notation (such as identifying $s$
with the identity map). We could also solve part (a) very easily if we had
solved part (b) first (since we already know from
Definition~\ref{convolution-algebra} and
Exercise~\ref{exe.convolution.assoc} that
$\Hom\left(C, \kk\right)$ is a $\kk$-algebra).
In the above solution, we have avoided all such shortcuts.

(b) Let $C$ be a $\kk$-coalgebra. Let us notice that
\begin{equation}
\rho_{C,C}\left(  f\otimes g\right)  = m_{\kk}\circ\left(  f\otimes
g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }f\in C^{\ast}\text{ and }g\in
C^{\ast}. \label{sol.dual-algebra.b.1}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.dual-algebra.b.1}:} Let $f\in C^{\ast}$
and $g\in C^{\ast}$. Then, the definition of $\rho_{C,C}$ shows that
$\rho_{C,C}\left(  f\otimes g\right)  $ is the composition $C\otimes
C\overset{f\otimes g}{\longrightarrow}\kk\otimes\kk
\overset{m_{\kk}}{\longrightarrow}\kk$ of the map $f\otimes g$
with the canonical isomorphism $\kk\otimes\kk
\overset{m_{\kk}}{\longrightarrow}\kk$. In other words,
$\rho_{C,C}\left(  f\otimes g\right)  =m_{\kk}\circ\left(  f\otimes
g\right)  $. This proves \eqref{sol.dual-algebra.b.1}.}

Now, we want to prove that the $\kk$-algebra structure defined on
$C^{\ast}$ in part (a) is precisely the one defined on
$\Hom \left(  C,\kk \right)  =C^{\ast}$ in
Definition~\ref{convolution-algebra} applied to $A=\kk$. In order to do
this, it is clearly enough to show that the product of any two elements of
$C^{\ast}$ with respect to the former $\kk$-algebra structure equals
their product with respect to the latter $\kk$-algebra structure. In
other words, it is enough to prove that for any $f\in C^{\ast}$ and $g\in
C^{\ast}$, we have
\begin{align*}
&  \left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }C^{\ast}\text{ defined in part (a)}\right) \\
&  =\left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }\Hom \left(  C,k\right)  \text{ defined in
Definition \ref{convolution-algebra} applied to }A=\kk\right)  .
\end{align*}
But this follows from the following computation:
\begin{align*}
&  \left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }C^{\ast}\text{ defined in part (a)}\right) \\
&  =\underbrace{m_{C^{\ast}}}_{= \Delta_{C}^{\ast}\circ\rho
_{C,C} }\left(  f\otimes g\right)  =\left(  \Delta_{C}^{\ast}\circ
\rho_{C,C}\right)  \left(  f\otimes g\right)  =\Delta_{C}^{\ast}\left(
\underbrace{\rho_{C,C}\left(  f\otimes g\right)  }_{\substack{=m
_{\kk}\circ\left(  f\otimes g\right)  \\\text{(by
\eqref{sol.dual-algebra.b.1})}}}\right) \\
&  =\Delta_{C}^{\ast}\left(  m_{\kk}\circ\left(  f\otimes g\right)
\right)  =m_{\kk}\circ\left(  f\otimes g\right)  \circ\Delta
_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Delta_{C}^{\ast
}\right) \\
&  =f\star g\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\star g=m_{\kk}
\circ\left(  f\otimes g\right)  \circ\Delta_{C}\text{ (by the definition of
}f\star g\text{)}\right) \\
&  =\left(  \text{the product }fg\text{ with respect to the }
\kk\text{-algebra }\Hom \left(  C,k\right)  \text{ defined in
Definition \ref{convolution-algebra} applied to }A=\kk\right)  .
\end{align*}
Thus, part (b) of the exercise is solved.

(c) Let $C$ be a graded $\kk$-coalgebra. Let $C=\bigoplus_{n\geq0} C_n$
be its decomposition into homogeneous components. Then, for every
$n\in\NN$, we identify $\left(  C_{n}\right)  ^{\ast}$ with a
$\kk$-submodule of $C^{\ast}$, namely with the $\kk$-submodule $\left\{
f\in C^{\ast}\ \mid\ f\left(  C_{p}\right)  =0\text{ for all }p\in
\NN\text{ satisfying }p\neq n\right\}  $. Thus,
\begin{equation}
\left(  C_{n}\right)  ^{\ast}=\left\{  f\in C^{\ast}\ \mid\ f\left(
C_{p}\right)  =0\text{ for all }p\in \NN \text{ satisfying }p\neq
n\right\}  \label{sol.dual-algebra.c.1}
\end{equation}
for every $n\in\NN$.

By the definition of $C^{o}$, we have $C^{o}=\bigoplus_{n\geq0}\left(
C_{n}\right)  ^{\ast}=\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}$.

Now let $a$ and $b$ be two nonnegative integers. We shall show that $\left(
C_{a}\right)  ^{\ast}\left(  C_{b}\right)  ^{\ast}\subset\left(
C_{a+b}\right)  ^{\ast}$.

Indeed, let $x\in\left(  C_{a}\right)  ^{\ast}$ and $y\in\left(  C_{b}\right)
^{\ast}$ be arbitrary. We have
\[
x\in\left(  C_{a}\right)  ^{\ast}=\left\{  f\in C^{\ast}\ \mid\ f\left(
C_{p}\right)  =0\text{ for all }p\in \NN \text{ satisfying }p\neq
a\right\}
\]
(by \eqref{sol.dual-algebra.c.1}, applied to $n=a$). In other words, $x$ is an
element of $C^{\ast}$ such that
\begin{equation}
x\left(  C_{p}\right)  =0\text{ for all }p\in \NN \text{ satisfying
}p\neq a. \label{sol.dual-algebra.c.1x}
\end{equation}
Similarly, $y$ is an element of $C^{\ast}$ such that
\begin{equation}
y\left(  C_{p}\right)  =0\text{ for all }p\in\NN \text{ satisfying
}p\neq b. \label{sol.dual-algebra.c.1y}
\end{equation}
Also, \eqref{sol.dual-algebra.c.1} (applied to $n=a+b$) yields
\begin{equation}
\left(  C_{a+b}\right)  ^{\ast}=\left\{  f\in C^{\ast}\ \mid\ f\left(
C_{p}\right)  =0\text{ for all }p\in\NN \text{ satisfying }p\neq
a+b\right\}  . \label{sol.dual-algebra.c.3}
\end{equation}


Now, part (b) of this exercise yields that the $\kk$-algebra structure
defined on $C^{\ast}$ in part (a) is precisely the one defined on
$\Hom \left(  C,\kk \right)  =C^{\ast}$ in
Definition~\ref{convolution-algebra} applied to $A=\kk$. Thus, the
product of the $\kk$-algebra $C^{\ast}$ is precisely the convolution
product $\star$ on $\Hom\left(  C,\kk \right)  =C^{\ast}$. Hence,
\[
xy=x\star y=m_{\kk}\circ\left(  x\otimes y\right)  \circ\Delta_{C}.
\]


Now, fix $p\in\NN$ such that $p\neq a+b$. Then, $\Delta_{C}\left(
C_{p}\right)  \subset\sum_{q=0}^{p}C_{q}\otimes C_{p-q}$ (since the coalgebra
$C$ is graded). Now, every $q\in\left\{  0,1,...,p\right\}  $ satisfies
\begin{equation}
\left(  x\otimes y\right)  \left(  C_{q}\otimes C_{p-q}\right)  =0.
\label{sol.dual-algebra.c.2}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.dual-algebra.c.2}:} Let $q\in\left\{
0,1,...,p\right\}  $. Then, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $q=a$.
\par
\textit{Case 2:} We have $q\neq a$.
\par
Let us first consider Case 1. In this case, we have $q=a$. Hence,
$p-\underbrace{q}_{=a}=p-a\neq b$ (since $p\neq a+b$). Hence,
\eqref{sol.dual-algebra.c.1y} (applied to $p-q$ instead of $p$) yields
$y\left(  C_{p-q}\right)  =0$. Now, $\left(  x\otimes y\right)  \left(
C_{q}\otimes C_{p-q}\right)  =x\left(  C_{q}\right)  \otimes
\underbrace{y\left(  C_{p-q}\right)  }_{=0}=0$. Thus, $\left(  x\otimes
y\right)  \left(  C_{q}\otimes C_{p-q}\right)  =0$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $q\neq a$. Therefore,
\eqref{sol.dual-algebra.c.1x} (applied to $q$ instead of $p$) yields $x\left(
C_{q}\right)  =0$. Now, $\left(  x\otimes y\right)  \left(  C_{q}\otimes
C_{p-q}\right)  =\underbrace{x\left(  C_{q}\right)  }_{=0}\otimes y\left(
C_{p-q}\right)  =0$. Thus, $\left(  x\otimes y\right)  \left(  C_{q}\otimes
C_{p-q}\right)  =0$ is proven in Case 2.
\par
We have now proven $\left(  x\otimes y\right)  \left(  C_{q}\otimes
C_{p-q}\right)  =0$ in each of the two Cases 1 and 2. Hence, $\left(  x\otimes
y\right)  \left(  C_{q}\otimes C_{p-q}\right)  =0$ always holds (since Cases 1
and 2 cover all possibilities). This proves \eqref{sol.dual-algebra.c.2}.}
Now,
\begin{align*}
\left(  \underbrace{xy}_{=m_{\kk}\circ\left(  x\otimes y\right)
\circ\Delta_{C}}\right)  \left(  C_{p}\right)   &  =\left(  m_{\kk}
\circ\left(  x\otimes y\right)  \circ\Delta_{C}\right)  \left(  C_{p}\right)
=m_{\kk}\left(  \left(  x\otimes y\right)  \left(  \underbrace{\Delta
_{C}\left(  C_{p}\right)  }_{\subset\sum_{q=0}^{p}C_{q}\otimes C_{p-q}}
\right)  \right) \\
&  \subset m_{\kk}\left(  \left(  x\otimes y\right)  \left(
\sum_{q=0}^{p}C_{q}\otimes C_{p-q}\right)  \right)  =\sum_{q=0}^{p}
m_{\kk}\left(  \underbrace{\left(  x\otimes y\right)  \left(
C_{q}\otimes C_{p-q}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.dual-algebra.c.2})}}}\right) \\
&  =\sum_{q=0}^{p}m_{\kk}0 = 0.
\end{align*}


In other words, $\left(  xy\right)  \left(  C_{p}\right)  =0$.

Now, forget that we fixed $p$. We thus have shown that $\left(  xy\right)
\left(  C_{p}\right)  =0$ for all $p\in\NN$ satisfying $p\neq a+b$. In
other words,
\[
xy\in\left\{  f\in C^{\ast}\ \mid\ f\left(  C_{p}\right)  =0\text{ for all
}p\in\NN \text{ satisfying }p\neq a+b\right\}  =\left(  C_{a+b}\right)
^{\ast}
\]
(by \eqref{sol.dual-algebra.c.3}).

Now, forget that we fixed $x$ and $y$. We thus have shown that $xy\in\left(
C_{a+b}\right)  ^{\ast}$ for all $x\in\left(  C_{a}\right)  ^{\ast}$ and
$y\in\left(  C_{b}\right)  ^{\ast}$. This yields $\left(  C_{a}\right)
^{\ast}\left(  C_{b}\right)  ^{\ast}\subset\left(  C_{a+b}\right)  ^{\ast}$
(since $\left(  C_{a+b}\right)  ^{\ast}$ is a $\kk$-module).

Now, forget that we fixed $a$ and $b$. We thus have proven that $\left(
C_{a}\right)  ^{\ast}\left(  C_{b}\right)  ^{\ast}\subset\left(
C_{a+b}\right)  ^{\ast}$ for every $a\in\NN$ and $b\in\NN$.
Thus,
\begin{equation}
\left(  C_{a}\right)  ^{\ast}\left(  C_{b}\right)  ^{\ast}\subset\left(
C_{a+b}\right)  ^{\ast}\subset\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast
} = C^{o}
\label{sol.dual-algebra.c.5}
\end{equation}
for every $a\in\NN$ and $b\in\NN$.


Now, for all $x\in C^{o}$ and $y\in C^{o}$, we have
\begin{align*}
\underbrace{x}_{\in C^{o}=\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}}
\cdot\underbrace{y}_{\in C^{o}=\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}}  &
\in\underbrace{\left(  \sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}\right)
}_{=\sum_{a\geq0}\left(  C_{a}\right)  ^{\ast}}\cdot\underbrace{\left(
\sum_{n\geq0}\left(  C_{n}\right)  ^{\ast}\right)  }_{=\sum_{b\geq0}\left(
C_{b}\right)  ^{\ast}}\\
&  =\left(  \sum_{a\geq0}\left(  C_{a}\right)  ^{\ast}\right)  \cdot\left(
\sum_{b\geq0}\left(  C_{b}\right)  ^{\ast}\right) \\
&  =\sum_{a\geq0}\sum_{b\geq0}\underbrace{\left(  C_{a}\right)  ^{\ast}\left(
C_{b}\right)  ^{\ast}}_{\substack{\subset C^{o}\\\text{(by
\eqref{sol.dual-algebra.c.5})}}}\subset\sum_{a\geq0}\sum_{b\geq0}C^{o}\subset
C^{o}
\end{align*}
(since $C^{o}$ is a $\kk$-module). Hence, the $\kk$-submodule $C^{o}$ of
$C^{\ast}$ is closed under multiplication. Since we also
have $1_{C^{\ast}}\in C^{o}$ (in fact, it is very easy to see that
$1_{C^{\ast}}=\epsilon_{C}\in\left(  C_{0}\right)  ^{\ast}\subset\sum_{n\geq
0}\left(  C_{n}\right)  ^{\ast}=C^{o}$), this shows that $C^{o}$ is a
$\kk$-subalgebra of $C^{\ast}$. This solves part (c) of the exercise.

(d) Let $C$ and $D$ be two $\kk$-coalgebras. Let $f:C\rightarrow D$ be
a homomorphism of $\kk$-coalgebras. Then, the two diagrams
\begin{equation}
\xymatrix{ C \ar[d]_{\Delta_C} \ar[r]^{f} & D \ar[d]^{\Delta_D} \\ C \otimes C \ar[r]^{f \otimes f} & D \otimes D }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ C \ar[dr]_{\epsilon_C} \ar[rr]^{f}& & D \ar[dl]^{\epsilon_D} \\ &\kk& }
\label{sol.dual-algebra.d.start}
\end{equation}
are commutative (since $f$ is a homomorphism of $\kk$-coalgebras).

We need to prove that $f^{\ast}:D^{\ast}\rightarrow C^{\ast}$ is a
homomorphism of $\kk$-algebras. In other words, we need to prove that
the two diagrams
\begin{equation}
\xymatrix{ D^* \ar[r]^{f^*} & C^* \\ D^* \otimes D^* \ar[u]^{m_{D^*}} \ar[r]^{f^* \otimes f^*} & C^* \otimes C^* \ar[u]_{m_{C^*}} }\ \ \ \ \ \ \ \ \ \ \text{and}
\ \ \ \ \ \ \ \ \ \ \xymatrix{ D^* \ar[rr]^{f^*} & & C^* \\ &\kk \ar[ul]^{u_{D^*}} \ar[ur]_{u_{C^*}}& }
\label{sol.dual-algebra.d.goal}
\end{equation}
are commutative.

By the definition of the $\kk$-algebra $D^{\ast}$, we have $m_{D^{\ast
}}=\Delta_{D}^{\ast}\circ\rho_{D,D}$. Thus, the left triangle (with two
vertical edges and one curved edge) in the diagram
\begin{equation}
\xymatrix{
D^* \ar[r]^{f^*} & C^* \\
\left(D\otimes D\right)^* \ar[u]^{\Delta_D^*} \ar[r]^{\left(f\otimes
f\right)^*} & \left(C\otimes C\right)^* \ar[u]^{\Delta_C^*} \\
D^* \otimes D^* \ar[u]^{\rho_{D,D}} \ar@/^4pc/[uu]^{m_{D^*}} \ar
[r]^{f^* \otimes f^*} & C^* \otimes C^* \ar[u]^{\rho_{C,C}} \ar@
/_4pc/[uu]_{m_{C^*}}
}
\label{sol.dual-algebra.d.2}
\end{equation}
commutes. Similarly, the right triangle in \eqref{sol.dual-algebra.d.2} also
commutes. The upper rectangle commutes because it is obtained from the first
diagram in \eqref{sol.dual-algebra.d.start} by dualizing. The lower triangle
in \eqref{sol.dual-algebra.d.2} commutes by basic linear algebra. Hence, the
whole diagram \eqref{sol.dual-algebra.d.2} commutes. But the outer rim of the
diagram \eqref{sol.dual-algebra.d.2} is exactly the first diagram in
\eqref{sol.dual-algebra.d.goal}. Thus, the first diagram in
\eqref{sol.dual-algebra.d.goal} commutes. The (even simpler) task of proving
the commutativity of the second diagram in \eqref{sol.dual-algebra.d.goal} is
left to the reader. Thus, $f^{\ast}$ is a homomorphism of $\kk$-algebras.
Part (d) of the exercise is solved.

(e) This is an exercise in linear algebra (it has nothing to do with Hopf
algebras). Here is a rough sketch of how it is solved: Let $U=\bigoplus
_{n\geq0}U_{n}$ be the decomposition of $U$ into its homogeneous components,
and let $V=\bigoplus_{n\geq0}V_{n}$ be the decomposition of $V$ into its
homogeneous components. Notice that every $V_{n}$ is finite free (since
$V$ is of finite type). Graded $\kk$-linear maps $U\rightarrow V$ can
be regarded as elements of $\prod_{n\geq0}\Hom\left(U_{n},V_{n}\right)  $
(because a graded $\kk$-linear map $U\rightarrow
V$ restricts to a $\kk$-linear map $U_{n}\rightarrow V_{n}$ for every
$n\in\NN$, and is uniquely determined by the totality of these
restrictions). Similarly, graded $\kk$-linear maps $V^{o}\rightarrow
U^{o}$ can be regarded as elements of
$\prod_{n\geq0}\Hom\left(  \left(  V_{n}\right)  ^{\ast},
\left(  U_{n}\right)  ^{\ast}\right)  $.
For every $n\in\NN$, there is a canonical $\kk$-module
isomorphism $\Hom\left(  U_{n},V_{n}\right)  \rightarrow
\Hom\left(  \left(  V_{n}\right)  ^{\ast},\left(  U_{n}\right)
^{\ast}\right)  $ which sends every $\varphi\in\Hom\left(
U_{n},V_{n}\right)  $ to $\varphi^{\ast}\in\Hom\left(  \left(
V_{n}\right)  ^{\ast},\left(  U_{n}\right)  ^{\ast}\right)  $ (since $V_{n}$
is finite free). Taking the product of these isomorphisms, we obtain a
$\kk$-module isomorphism from $\prod_{n\geq0}\Hom\left(
U_{n},V_{n}\right)  $ to $\prod_{n\geq0}\Hom\left(  \left(
V_{n}\right)  ^{\ast},\left(  U_{n}\right)  ^{\ast}\right)  $, that is, a
$\kk$-module isomorphism from the $\kk$-module of all graded
$\kk$-linear maps $U\rightarrow V$ to the $\kk$-module of all
graded $\kk$-linear maps $V^{o}\rightarrow U^{o}$. It is easy to see
that this isomorphism sends every $f:U\rightarrow V$ to
$f^{\ast}:V^{o}\rightarrow U^{o}$. Hence, there is a 1-to-1
correspondence between graded $\kk$-linear maps $U\rightarrow V$ and
graded $\kk$-linear maps $V^{o}\rightarrow U^{o}$ given by
$f\mapsto f^{\ast}$ (namely, this isomorphism). This solves
Exercise~\ref{exe.dual-algebra} (e).

(f) Let $f:C\rightarrow D$ be a graded $\kk$-linear map. We need to
prove that $f:C\rightarrow D$ is a $\kk$-coalgebra morphism if and only if
$f^{\ast}:D^{o}\rightarrow C^{o}$ is a $\kk$-algebra morphism. In other
words, we need to prove the following two assertions:

\begin{statement}
\textit{Assertion 1:} If $f:C\rightarrow D$ is a $\kk$-coalgebra morphism,
then $f^{\ast}:D^{o}\rightarrow C^{o}$ is a $\kk$-algebra morphism.
\end{statement}

\begin{statement}
\textit{Assertion 2:} If $f^{\ast}:D^{o}\rightarrow C^{o}$ is a
$\kk$-algebra morphism, then $f:C\rightarrow D$ is a $\kk$-coalgebra morphism.
\end{statement}

We start by proving Assertion 1. One way to prove it proceeds by repeating the
solution of Exercise~\ref{exe.dual-algebra} (d), except that $D^{\ast}$,
$\left(  D\otimes D\right)  ^{\ast}$, $C^{\ast}$ and $\left(  C\otimes
C\right)  ^{\ast}$ are replaced by $D^{o}$, $\left(  D\otimes D\right)  ^{o}$,
$C^{o}$ and $\left(  C\otimes C\right)  ^{o}$ (where, of course, the map
$\rho_{D,D}:D^{o}\rightarrow\left(  D\otimes D\right)  ^{o}$ has to be
interpreted as the restriction of the map $\rho_{D,D}:D^{\ast}\rightarrow
\left(  D\otimes D\right)  ^{\ast}$ to the $\kk$-submodule $D^{o}$ of $D^{\ast}$, and
similarly for the other maps). This replacement can be done completely
robotically, and thus is left to the reader. Another way to prove Assertion 1
is by realizing that it follows immediately from
Exercise~\ref{exe.dual-algebra} (d) (since
$f^{\ast}:D^{o}\rightarrow C^{o}$ is a
restriction of the map $f^{\ast}:D^{\ast}\rightarrow C^{\ast}$). Either way,
we do not end up using the assumption that $D$ is of finite type. However, we
will need this assumption in our proof of Assertion 2.

Now, let us prove Assertion 2. Assume that $f^{\ast}:D^{o}\rightarrow C^{o}$
is a $\kk$-algebra morphism. We want to show that $f:C\rightarrow D$ is a
$\kk$-coalgebra morphism. In other words, we want to show that the two
diagrams \eqref{sol.dual-algebra.d.start} commute. Let us start with the left
one of these diagrams.

The graded $\kk$-module $D$ is of finite type, and therefore the map
$\rho_{D,D}:D^{o}\otimes D^{o}\rightarrow\left(  D\otimes D\right)  ^{o}$ (a
restriction of the map $\rho_{D,D}:D^{\ast}\otimes D^{\ast}\rightarrow\left(
D\otimes D\right)  ^{\ast}$) is an isomorphism. Its inverse
$\rho_{D,D}^{-1} : \left(  D\otimes D\right) ^o \rightarrow D^o \otimes D^o$
is therefore well-defined\footnote{Beware: we don't have an inverse of the
non-restricted map $\rho_{D,D}:D^{\ast}\otimes D^{\ast}\rightarrow\left(
D\otimes D\right)  ^{\ast}$.}. We can thus form the (asymmetric!) diagram
\begin{equation}
{
\xymatrixcolsep{5pc}
\xymatrix{
D^o \ar[rrr]^{f^*} & & & C^o \\
& D^o \otimes D^o \ar[ul]^{m_{D^*}} \ar[r]^{f^* \otimes f^*} & C^o \otimes
C^o \ar[ur]^{m_{C^*}}  \ar[dr]^{\rho_{C,C}} & \\
\left(D\otimes D\right)^o \ar[uu]^{\Delta_D^*} \ar[ur]^{\rho_{D,D}^{-1}}
\ar[rrr]_{\left(f\otimes f\right)^*} & & & \left(C\otimes C\right
)^o \ar[uu]^{\Delta_C^*}
}
}
.\label{sol.dual-algebra.e.2}
\end{equation}
(The arrows labelled $m_{C^{\ast}}$ and $m_{D^{\ast}}$ could just as well have
been labelled $m_{C^{o}}$ and $m_{D^{o}}$, since the multiplication maps
$m_{C^{o}}$ and $m_{D^{o}}$ are restrictions of $m_{C^{\ast}}$ and
$m_{D^{\ast}}$.) The two triangles in \eqref{sol.dual-algebra.e.2} commute due
to $m_{D^{\ast}}=\Delta_{D}^{\ast}\circ\rho_{D,D}$ and $m_{C^{\ast}}
=\Delta_{C}^{\ast}\circ\rho_{C,C}$. The upper quadrilateral in
\eqref{sol.dual-algebra.e.2} commutes because $f^{\ast}$ is a
$\kk$-algebra homomorphism, and the lower quadrilateral in
\eqref{sol.dual-algebra.e.2} commutes because of the commutativity of the
diagram
\[
\xymatrix{
D^o \otimes D^o \ar[d]^{\rho_{D,D}} \ar[r]^{f^* \otimes f^*} & C^o \otimes
C^o \ar[d]^{\rho_{C,C}} \\
\left(D\otimes D\right)^o \ar[r]_{\left(f\otimes f\right)^*} & \left(C\otimes
C\right)^o
}
\]
(which follows from standard linear algebra). Hence, the whole diagram
\eqref{sol.dual-algebra.e.2} commutes. Removing the two interior nodes of this
diagram, we obtain the commutative diagram
\begin{equation}
\xymatrixcolsep{5pc}
\xymatrix{
D^o \ar[r]^{f^*} & C^o \\
\left(D\otimes D\right)^o \ar[u]^{\Delta_D^*} \ar[r]_{\left(f\otimes
f\right)^*} & \left(C\otimes C\right)^o \ar[u]^{\Delta_C^*}
}
.\label{sol.dual-algebra.e.close}
\end{equation}
This does not immediately yield the commutativity of the first diagram in
\eqref{sol.dual-algebra.d.start} (because we cannot revert taking dual
$\kk$-modules), so we are not yet done. But we are close.

We have
\begin{align*}
\left(  \Delta_{D}\circ f\right)  ^{\ast}  & =f^{\ast}\circ\Delta_{D}^{\ast
}=\Delta_{C}^{\ast}\circ\left(  f\otimes f\right)  ^{\ast}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the diagram
\eqref{sol.dual-algebra.e.close} commutes}\right)  \\
& =\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)  ^{\ast}
\end{align*}
as maps from $\left(  D\otimes D\right)  ^{o}$ to $C^{o}$. But a general
linear-algebraic fact states that if $U$ and $V$ are two graded
$\kk$-modules such that $V$ is of finite type, and if $\alpha$
and $\beta$ are two graded $\kk$-linear maps $U\rightarrow V$ such that
$\alpha^{\ast}=\beta^{\ast}$ as maps from $V^{o}$ to $U^{o}$, then
$\alpha=\beta$\ \ \ \ \footnote{This follows immediately from
Exercise~\ref{exe.dual-algebra} (e).}. Applying this to $U=C$,
$V=D\otimes D$, $\alpha=\Delta_{D}\circ f$ and
$\beta=\left(  f\otimes f\right)  \circ \Delta_{C}$, we obtain
$\Delta_{D}\circ f=\left(  f\otimes f\right)
\circ\Delta_{C}$ (since $D\otimes D$ is of finite type\footnote{This is
because $D$ is of finite type.} and since $\left(  \Delta_{D}\circ f\right)
^{\ast}=\left(  \left(  f\otimes f\right)  \circ\Delta_{C}\right)  ^{\ast}$).
In other words, the first diagram in \eqref{sol.dual-algebra.d.start} is
commutative. The reader can verify that so is the second diagram in
\eqref{sol.dual-algebra.d.start} (once again, this is the easier part). So we
have shown that both diagrams in \eqref{sol.dual-algebra.d.start} are
commutative, and thus $f$ is a $\kk$-coalgebra morphism. This proves
Assertion 2.

Now that Assertions 1 and 2 are both proven, part (f) of the exercise is solved.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Sym.1dim}}

\textit{Solution to Exercise \ref{exe.Sym.1dim}.} We have
$\Sym \left(V\right) \cong \kk\left[  x\right]  $. Thus, $\left(  x^{k}\right)
_{k\in\NN}$ is a basis of the $\kk$-module $\Sym \left(  V\right)  $.
Therefore, $\left(  x^{k}\otimes x^{\ell}\right)
_{\left(  k,\ell\right)  \in\NN^{2}}$ is a basis of the $\kk$-module
$\Sym \left(  V\right)  \otimes\Sym \left(  V\right)  $.

On the other hand, $\left(  f^{\left(  k\right)  }\right)  _{k\in\NN}$
is a basis of the $\kk$-module $\left(  \Sym \left(
V\right)  \right)  ^{o}$ (namely, the dual basis to the graded basis $\left(
x^{k}\right)  _{k\in\NN}$ of $\Sym \left(  V\right)  $).
Hence, $\left(  f^{\left(  k\right)  }\otimes f^{\left(  \ell\right)
}\right)  _{\left(  k,\ell\right)  \in\NN^{2}}$ is a basis of the
$\kk$-module $\left(  \Sym \left(  V\right)  \right)
^{o}\otimes\left(  \Sym \left(  V\right)  \right)  ^{o}$.

We denote by $m$ the multiplication map of $\Sym \left(
V\right)  $. We denote by $u$ the unity map of $\Sym \left(
V\right)  $ (that is, the map $\kk\rightarrow\Sym \left(
V\right)  $ which sends $1_{\kk}$ to $1_{\Sym \left(
V\right)  }$). We denote by $\Delta$ the comultiplication of
$\Sym \left(  V\right)  $. We denote by $\epsilon$ the counit of
$\Sym \left(  V\right)  $. We denote by $S$ the antipode of
$\Sym \left(  V\right)  $.

We denote by $m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$,
$u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$,
$\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$,
$\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$ and
$S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$ the maps
analogous to $m$, $u$, $\Delta$, $\epsilon$ and $S$ but defined for the Hopf
algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ instead
of $\Sym \left(  V\right)  $.

(a) Clearly, $x^{i}\cdot x^{j}=x^{i+j}$ for all $i\in \NN$ and
$j\in \NN$.

We have $\Delta_{T\left(  V\right)  }\left(  x\right)  =1\otimes x+x\otimes1$
in $T\left(  V\right)  \otimes T\left(  V\right)  $ (by the definition of
$\Delta_{T\left(  V\right)  }$). Projecting this equality down onto
$\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  $, we obtain $\Delta_{\Sym \left(  V\right)  }\left(
x\right)  =x\otimes1+1\otimes x$ (since $\Sym \left(  V\right)
$ is a quotient Hopf algebra of $T\left(  V\right)  $). But we defined
$\Delta$ to mean the comultiplication of $\Sym \left(  V\right)
$. We thus have $\Delta=\Delta_{\Sym \left(  V\right)  }$, so
that $\Delta\left(  x\right)  =\Delta_{\Sym \left(  V\right)
}\left(  x\right)  =x\otimes1+1\otimes x$. Thus, $x$ is a primitive element of
$\Sym \left(  V\right)  $. Proposition
\ref{antipode-on-primitives} thus yields $S\left(  x\right)  =-x$.

Proposition \ref{antipodes-are-antiendomorphisms} yields that the antipode $S$
of $\Sym \left(  V\right)  $ is an algebra anti-endomorphism of
$\Sym \left(  V\right)  $. Since an algebra anti-endomorphism of
$\Sym \left(  V\right)  $ means the same as an algebra
endomorphism of $\Sym \left(  V\right)  $ (by
Exercise~\ref{exe.comm-cocomm.anti}(a), because
$\Sym \left(  V\right)  $ is commutative), this yields that $S$
is an algebra endomorphism of $\Sym \left(  V\right)  $.
Consequently, for any $n\in \NN$, we have%
\begin{equation}
S\left(  x^{n}\right)  =\left(  \underbrace{S\left(  x\right)  }_{=-x}\right)
^{n}=\left(  -x\right)  ^{n}=\left(  -1\right)  ^{n}x^{n}.
\label{sol.Sym.1dim.a.claimS}%
\end{equation}


It remains to prove that $\Delta\left(  x^{n}\right)  =\sum_{i+j=n}\dbinom
{n}{i}x^{i}\otimes x^{j}$ for every $n\in \NN$ (where the summation sign
``$\sum_{i+j=n}$'' is shorthand for
``$\sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}$''). Let $n\in \NN$. The $\kk$-algebra
$\Sym \left(  V\right)  \otimes\Sym \left(  V\right)  $
is commutative (since $\Sym \left(  V\right)  $ is commutative),
and thus the binomial formula can be applied in
it. We know that $\Sym \left(  V\right)  $ is a bialgebra, and
thus $\Delta$ is a $\kk$-algebra homomorphism (by the axioms of a
bialgebra). Hence,
\begin{align}
\Delta\left(  x^{n}\right)   &  =\left(  \underbrace{\Delta\left(  x\right)
}_{=x\otimes1+1\otimes x}\right)  ^{n}=\left(  x\otimes1+1\otimes x\right)
^{n}\nonumber\\
&  =\sum_{\ell=0}^{n}\dbinom{n}{\ell}\underbrace{\left(  x\otimes1\right)
^{\ell}}_{=x^{\ell}\otimes1}\underbrace{\left(  1\otimes x\right)  ^{n-\ell}%
}_{=1\otimes x^{n-\ell}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the binomial
formula}\right) \nonumber\\
&  =\sum_{\ell=0}^{n}\dbinom{n}{\ell}\underbrace{\left(  x^{\ell}%
\otimes1\right)  \left(  1\otimes x^{n-\ell}\right)  }_{=\left(  x^{\ell
}1\right)  \otimes\left(  1x^{n-\ell}\right)  =x^{\ell}\otimes x^{n-\ell}%
}=\sum_{\ell=0}^{n}\dbinom{n}{\ell}x^{\ell}\otimes x^{n-\ell}=
\sum_{i+j=n}\dbinom{n}{i}x^{i}\otimes x^{j}
\label{sol.Sym.1dim.a.3}
\end{align}
(here, we have substituted $\left(  i,j\right)  $ for $\left(  \ell
,n-\ell\right)  $ in the sum). This completes the solution of Exercise
\ref{exe.Sym.1dim}(a).

(b) The definition of $f^{\left(  i\right)  }$ shows that%
\begin{equation}
f^{\left(  i\right)  }\left(  x^{j}\right)  =\delta_{i,j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\NN \text{ and }j\in\NN .
\label{sol.Sym.1dim.b.delta}
\end{equation}


But the definition of the $\kk$-algebra
$\left(  \Sym \left(  V\right)  \right)  ^{o}$ shows that the
multiplication map $m_{\left( \Sym \left(  V\right)  \right)  ^{o}}$
of the $\kk$-algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$
is adjoint to the comultiplication map $\Delta$ of the
$\kk$-coalgebra $\Sym \left(  V\right)  $. In other words, we have
\begin{equation}
\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  ,b\right)  _{\Sym \left(  V\right)  }=\left(
a,\Delta\left(  b\right)  \right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  } \label{sol.Sym.1dim.b.mD}%
\end{equation}
for any $a\in\left(  \Sym \left(  V\right)  \right)  ^{o}%
\otimes\left(  \Sym \left(  V\right)  \right)  ^{o}$ and any
$b\in\Sym \left(  V\right)  $. Thus, any $p\in\left(
\Sym \left(  V\right)  \right)  ^{o}$, $q\in\left(
\Sym \left(  V\right)  \right)  ^{o}$ and $b\in
\Sym \left(  V\right)  $ satisfy%
\begin{equation}
\left(  pq,b\right)  _{\Sym \left(  V\right)  }=\left(  p\otimes
q,\Delta\left(  b\right)  \right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }. \label{sol.Sym.1dim.b.mD2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.mD2}):} Let $p\in\left(
\Sym \left(  V\right)  \right)  ^{o}$, $q\in\left(
\Sym \left(  V\right)  \right)  ^{o}$ and $b\in
\Sym \left(  V\right)  $. Then,
$m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}
\left(  p\otimes q\right)  =pq$ (since
$m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$ is the
multiplication map of the algebra $\left(  \Sym \left(
V\right)  \right)  ^{o}$). Thus,%
\[
\left(  \underbrace{pq}_{=m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  p\otimes q\right)  },b\right)  _{\Sym %
\left(  V\right)  }=\left(  m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  p\otimes q\right)  ,b\right)  _{\Sym %
\left(  V\right)  }=\left(  p\otimes q,\Delta\left(  b\right)  \right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }%
\]
(by (\ref{sol.Sym.1dim.b.mD}), applied to $a=p\otimes q$). This proves
(\ref{sol.Sym.1dim.b.mD2}).}

On the other hand, the definition of the $\kk$-coalgebra $\left(
\Sym \left(  V\right)  \right)  ^{o}$ shows that the
comultiplication map $\Delta_{\left(  \Sym \left(  V\right)
\right)  ^{o}}$ of the $\kk$-coalgebra
$\left(  \Sym \left(  V\right)  \right)  ^{o}$ is adjoint to the
multiplication map $m$ of
the $\kk$-algebra $\Sym \left(  V\right)  $. In other
words, we have%
\begin{equation}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,b\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }=\left(  a,m\left(  b\right)
\right)  _{\Sym \left(  V\right)  } \label{sol.Sym.1dim.b.Dm}%
\end{equation}
for any $a\in\left(  \Sym \left(  V\right)  \right)  ^{o}$ and
any $b\in\Sym \left(  V\right)  \otimes\Sym %
\left(  V\right)  $. Thus, any $a\in\left(  \Sym \left(
V\right)  \right)  ^{o}$, $p\in\Sym \left(  V\right)  $ and
$q\in\Sym \left(  V\right)  $ satisfy%
\begin{equation}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,p\otimes q\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }=\left(  a,pq\right)
_{\Sym \left(  V\right)  }. \label{sol.Sym.1dim.b.Dm2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.Dm2}):} Let $a\in\left(
\Sym \left(  V\right)  \right)  ^{o}$,
$p\in\Sym \left(  V\right)  $ and $q\in\Sym \left(  V\right)  $.
Then,
$m\left(  p\otimes q\right)  =pq$ (since $m$ is the multiplication map of the
algebra $\Sym \left(  V\right)  $). But (\ref{sol.Sym.1dim.b.Dm}%
) (applied to $b=p\otimes q$) yields%
\[
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,p\otimes q\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }=\left(  a,\underbrace{m\left(
p\otimes q\right)  }_{=pq}\right)  _{\Sym \left(  V\right)
}=\left(  a,pq\right)  _{\Sym \left(  V\right)  }.
\]
This proves (\ref{sol.Sym.1dim.b.Dm2}).}

The definition of the $\kk$-Hopf algebra $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$ shows that the antipode $S_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ of the $\kk$-Hopf
algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ is
adjoint to the antipode $S$ of the $\kk$-Hopf algebra
$\Sym \left(  V\right)  $. In other words, we have%
\begin{equation}
\left(  S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  ,b\right)  _{\Sym \left(  V\right)  }=\left(
a,S\left(  b\right)  \right)  _{\Sym \left(  V\right)  }
\label{sol.Sym.1dim.b.S}%
\end{equation}
for any $a\in\left(  \Sym \left(  V\right)  \right)  ^{o}$ and
$b\in\Sym \left(  V\right)  $.

Let us also notice a simple fact about sums: If $u$, $v$ and $n$ are three
nonnegative integers, if $\mathfrak{A}$ is an additive abelian group, and if
$\left(  t_{i,j}\right)  _{\left(  i,j\right)  \in\NN^{2}}$ is a family
of elements of $\mathfrak{A}$, then%
\begin{equation}
\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta_{v,j}=t_{u,v}\delta_{u+v,n}
\label{sol.Sym.1dim.b.t}%
\end{equation}
(where, again, ``$\sum_{i+j=n}$'' is shorthand
for ``$\sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}$'').\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.Sym.1dim.b.t}):} Let $u$, $v$ and $n$ be three nonnegative integers.
Let $\mathfrak{A}$ be an additive abelian group. Let $\left(  t_{i,j}\right)
_{\left(  i,j\right)  \in\NN^{2}}$ be a family of elements of
$\mathfrak{A}$.
\par
Let us first assume that $u+v\neq n$. Each addend of the sum
$\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta_{v,j}$ contains the factor $\delta
_{u,i}\delta_{v,j}$, which is $0$ unless we have both $u=i$ and $v=j$. Thus,
each addend of this sum vanishes unless it satisfies both $u=i$ and $v=j$ at
the same time. Since no addend of this sum can satisfy both $u=i$ and $v=j$ at
the same time (because such an addend would then also satisfy $\underbrace{u}%
_{=i}+\underbrace{v}_{=j}=i+j=n$, which would contradict $u+v\neq n$), this
shows that each addend of this sum vanishes. Consequently, the sum itself must
vanish. That is, we have $\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta_{v,j}=0$.
Compared with $t_{u,v}\underbrace{\delta_{u+v,n}}_{\substack{=0\\\text{(since
}u+v\neq n\text{)}}}=0$, this yields $\sum_{i+j=n}t_{i,j}\delta_{u,i}%
\delta_{v,j}=t_{u,v}\delta_{u+v,n}$. Hence, (\ref{sol.Sym.1dim.b.t}) is proven
under the assumption that $u+v\neq n$.
\par
Therefore, for the rest of our proof of (\ref{sol.Sym.1dim.b.t}), we can WLOG
assume that we \textbf{don't} have $u+v\neq n$. Thus, $u+v=n$. Again, each
addend of the sum $\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta_{v,j}$ contains the
factor $\delta_{u,i}\delta_{v,j}$, which is $0$ unless we have both $u=i$ and
$v=j$. Thus, each addend of this sum vanishes unless it satisfies both $u=i$
and $v=j$ at the same time. But there exists exactly one addend of the sum
$\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta_{v,j}$ which satisfies both $u=i$ and
$v=j$: namely, the addend for $\left(  i,j\right)  =\left(  u,v\right)  $ (and
this is indeed an addend because we have $u+v=n$). This addend equals
$t_{u,v}\underbrace{\delta_{u,u}}_{=1}\underbrace{\delta_{v,v}}_{=1}=t_{u,v}$.
Because of this, and because all other addends of our sum vanish, we thus
conclude that the whole sum must equal $t_{u,v}$. That is, we have
$\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta_{v,j}=t_{u,v}$. Compared with
$t_{u,v}\underbrace{\delta_{u+v,n}}_{\substack{=1\\\text{(since }%
u+v=n\text{)}}}=t_{u,v}$, this yields $\sum_{i+j=n}t_{i,j}\delta_{u,i}\delta
_{v,j}=t_{u,v}\delta_{u+v,n}$. Hence, (\ref{sol.Sym.1dim.b.t}) is proven.}

Now, we have%
\begin{equation}
f^{\left(  u\right)  }f^{\left(  v\right)  }=\dbinom{u+v}{u}f^{\left(
u+v\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\NN \text{ and
}v\in\NN . \label{sol.Sym.1dim.b.1}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.b.1}):} Let $u\in \NN$ and
$v\in \NN$.

Let $k\in \NN$. Then,
\begin{align*}
\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }\right)  \left(
x^{k}\right)   &  =\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }%
,x^{k}\right)  _{\Sym \left(  V\right)  }=\left(  f^{\left(
u\right)  }\otimes f^{\left(  v\right)  },\underbrace{\Delta\left(
x^{k}\right)  }_{\substack{=\sum_{i+j=k}\dbinom{k}{i}x^{i}\otimes
x^{j}\\\text{(by (\ref{sol.Sym.1dim.a.3}), applied to }n=k\text{)}}}\right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.mD2}), applied to
}p=f^{\left(  u\right)  }\text{, }q=f^{\left(  v\right)  }\text{ and }%
b=x^{k}\right) \\
&  =\left(  f^{\left(  u\right)  }\otimes f^{\left(  v\right)  },
\sum_{i+j=k}\dbinom{k}{i}x^{i}\otimes x^{j}
\right)  _{\Sym \left(
V\right)  \otimes\Sym \left(  V\right)  }\\
&  =\sum_{i+j=k}\dbinom{k}{i}\underbrace{\left(  f^{\left(  u\right)  }\otimes
f^{\left(  v\right)  },x^{i}\otimes x^{j}\right)  _{\Sym \left(
V\right)  \otimes\Sym \left(  V\right)  }}_{\substack{=\left(
f^{\left(  u\right)  },x^{i}\right)  _{\Sym \left(  V\right)
}\left(  f^{\left(  v\right)  },x^{j}\right)  _{\Sym \left(
V\right)  }\\\text{(by the definition of the}\\\text{bilinear form }\left(
\cdot,\cdot\right)  _{\Sym \left(  V\right)  \otimes
\Sym \left(  V\right)  }\text{)}}}\\
&  =\sum_{i+j=k}\dbinom{k}{i}\underbrace{\left(  f^{\left(  u\right)  }%
,x^{i}\right)  _{\Sym \left(  V\right)  }}%
_{\substack{=f^{\left(  u\right)  }\left(  x^{i}\right)  =\delta
_{u,i}\\\text{(by (\ref{sol.Sym.1dim.b.delta}), applied}\\\text{to }u\text{
and }i\text{ instead of }i\text{ and }j\text{)}}}\underbrace{\left(
f^{\left(  v\right)  },x^{j}\right)  _{\Sym \left(  V\right)  }%
}_{\substack{=f^{\left(  v\right)  }\left(  x^{j}\right)  =\delta
_{v,j}\\\text{(by (\ref{sol.Sym.1dim.b.delta}), applied}\\\text{to }v\text{
instead of }i\text{)}}}\\
&  =\sum_{i+j=k}\dbinom{k}{i}\delta_{u,i}\delta_{v,j} \\
&=\dbinom{k}{u}%
\delta_{u+v,k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.t}),
applied to }n=k\text{, }\mathfrak{A}=\kk\text{ and }t_{i,j}=\dbinom
{k}{i}\right) \\
&  =\dbinom{u+v}{u}\delta_{u+v,k}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the equality }\dbinom{k}{u}\delta_{u+v,k}=\dbinom{u+v}{u}%
\delta_{u+v,k}\text{ holds}\\
\text{in the case when }u+v=k\text{ (obviously) and in the case}\\
\text{when }u+v\neq k\text{ (since both sides of this equality are }0\text{
in}\\
\text{this case (due to }\delta_{u+v,k}=0\text{))}%
\end{array}
\right)  .
\end{align*}
Comparing this with%
\[
\left(  \dbinom{u+v}{u}f^{\left(  u+v\right)  }\right)  \left(  x^{k}\right)
=\dbinom{u+v}{u}\underbrace{f^{\left(  u+v\right)  }\left(  x^{k}\right)
}_{\substack{=\delta_{u+v,k}\\\text{(by (\ref{sol.Sym.1dim.b.delta}),
applied}\\\text{to }u+v\text{ and }k\text{ instead of }i\text{ and }j\text{)}%
}}=\dbinom{u+v}{u}\delta_{u+v,k},
\]
we obtain $\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }\right)
\left(  x^{k}\right)  =\left(  \dbinom{u+v}{u}f^{\left(  u+v\right)  }\right)
\left(  x^{k}\right)  $.

Let us now forget that we fixed $k$. We thus have shown that $\left(
f^{\left(  u\right)  }f^{\left(  v\right)  }\right)  \left(  x^{k}\right)
=\left(  \dbinom{u+v}{u}f^{\left(  u+v\right)  }\right)  \left(  x^{k}\right)
$ for every $k\in \NN$. In other words, the two $\kk$-linear maps
$f^{\left(  u\right)  }f^{\left(  v\right)  }$ and $\dbinom{u+v}{u}f^{\left(
u+v\right)  }$ are equal to each other on the basis $\left(  x^{k}\right)
_{k\in\NN}$ of the $\kk$-module $\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $f^{\left(  u\right)  }f^{\left(
v\right)  }=\dbinom{u+v}{u}f^{\left(  u+v\right)  }$. Thus,
(\ref{sol.Sym.1dim.b.1}) is proven.

Thus, we have shown that%
\begin{equation}
f^{\left(  i\right)  }f^{\left(  j\right)  }=\dbinom{i+j}{i}f^{\left(
i+j\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\NN \text{ and
}j\in\NN . \label{sol.Sym.1dim.b.1.ij}%
\end{equation}


(In fact, (\ref{sol.Sym.1dim.b.1.ij}) follows from (\ref{sol.Sym.1dim.b.1}) by
renaming the variables $u$ and $v$ as $i$ and $j$.)

We shall now show that
\begin{equation}
\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  =\sum_{i+j=n}f^{\left(  i\right)  }\otimes
f^{\left(  j\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\NN
\label{sol.Sym.1dim.b.2}%
\end{equation}
(where, again, ``$\sum_{i+j=n}$'' is shorthand
for ``$\sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}$'').

\textit{Proof of (\ref{sol.Sym.1dim.b.2}):} Let $n\in \NN$. We identify
$\left(  \Sym \left(  V\right)  \right)  ^{o}\otimes\left(
\Sym \left(  V\right)  \right)  ^{o}$ with $\left(
\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  \right)  ^{o}$ (since $\Sym \left(  V\right)  $ is a
graded $\kk$-module of finite type); thus, an element of $\left(
\Sym \left(  V\right)  \right)  ^{o}\otimes\left(
\Sym \left(  V\right)  \right)  ^{o}$ can be regarded as a
$\kk$-linear map $\Sym \left(  V\right)  \otimes
\Sym \left(  V\right)  \rightarrow\kk$. In particular,
$\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  $ thus becomes a $\kk$-linear map
$\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  \rightarrow\kk$.

Fix $\left(  k,\ell\right)  \in\NN^{2}$. Thus, $k\in \NN$ and
$\ell\in \NN$. Then,%
\begin{align*}
&  \left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right) \\
&  =\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)
^{o}}\left(  f^{\left(  n\right)  }\right)  ,x^{k}\otimes x^{\ell}\right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }\\
&  =\left(  f^{\left(  n\right)  },\underbrace{x^{k}x^{\ell}}_{=x^{k+\ell}%
}\right)  _{\Sym \left(  V\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Sym.1dim.b.Dm2}), applied to }a=f^{\left(  n\right)
}\text{, }p=x^{k}\text{ and }q=x^{\ell}\right) \\
&  =\left(  f^{\left(  n\right)  },x^{k+\ell}\right)  _{\Sym %
\left(  V\right)  }=\delta_{n,k+\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.delta}), applied to }i=n\text{ and }j=k+\ell\right)  .
\end{align*}
Compared with%
\begin{align*}
\left(  \sum_{i+j=n}f^{\left(  i\right)  }\otimes f^{\left(  j\right)
}\right)  \left(  x^{k}\otimes x^{\ell}\right)   &  =\sum_{i+j=n}%
\underbrace{\left(  f^{\left(  i\right)  }\otimes f^{\left(  j\right)
}\right)  \left(  x^{k}\otimes x^{\ell}\right)  }_{=f^{\left(  i\right)
}\left(  x^{k}\right) f^{\left(  j\right)  }\left(  x^{\ell}\right)
}=\sum_{i+j=n}\underbrace{f^{\left(  i\right)  }\left(  x^{k}\right)
}_{\substack{=\delta_{i,k}\\\text{(by (\ref{sol.Sym.1dim.b.delta}),
applied}\\\text{to }k\text{ instead of }j\text{)}}}\underbrace{f^{\left(
j\right)  }\left(  x^{\ell}\right)  }_{\substack{=\delta_{j,\ell}\\\text{(by
(\ref{sol.Sym.1dim.b.delta}), applied to }j\\\text{and }\ell\text{ instead of
}i\text{ and }j\text{)}}}\\
&  =\sum_{i+j=n}\underbrace{\delta_{i,k}}_{=\delta_{k,i}}\underbrace{\delta
_{j,\ell}}_{=\delta_{\ell,j}}=\sum_{i+j=n}\delta_{k,i}\delta_{\ell,j}%
=\sum_{i+j=n}1\delta_{k,i}\delta_{\ell,j}=1\delta_{k+\ell,n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.t}), applied to
}k\text{, }\ell\text{, }\kk\text{ and }1\text{ instead of }u\text{,
}v\text{, }\mathfrak{A}\text{ and }t_{i,j}\right) \\
&  =\delta_{k+\ell,n}=\delta_{n,k+\ell},
\end{align*}
this yields%
\begin{equation}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  =\left(  \sum_{i+j=n}f^{\left(  i\right)  }\otimes f^{\left(
j\right)  }\right)  \left(  x^{k}\otimes x^{\ell}\right)  .
\label{sol.Sym.1dim.b.2.pf.5}%
\end{equation}


Let us now forget that we fixed $\left(  k,\ell\right)  $. We thus have shown
that (\ref{sol.Sym.1dim.b.2.pf.5}) holds for every $\left(  k,\ell\right)
\in\NN^{2}$. In other words, the two $\kk$-linear maps
$\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  $ and $\sum_{i+j=n}f^{\left(  i\right)
}\otimes f^{\left(  j\right)  }$ (from $\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  $ to $\kk$) are equal to
each other on the basis $\left(  x^{k}\otimes x^{\ell}\right)  _{\left(
k,\ell\right)  \in\NN^{2}}$ of the $\kk$-module
$\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  f^{\left(
n\right)  }\right)  =\sum_{i+j=n}f^{\left(  i\right)  }\otimes f^{\left(
j\right)  }$. Thus, (\ref{sol.Sym.1dim.b.2}) is proven.

Finally, let us show that
\begin{equation}
S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  =\left(  -1\right)  ^{n}f^{\left(  n\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\NN . \label{sol.Sym.1dim.b.3}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.b.3}):} Let $n\in \NN$. Let
$k\in \NN$. Then,%
\begin{align*}
\left(  S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)   &  =\left(
S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  ,x^{k}\right)  _{\Sym \left(
V\right)  }=\left(  f^{\left(  n\right)  },\underbrace{S\left(  x^{k}\right)
}_{\substack{=\left(  -1\right)  ^{k}x^{k}\\\text{(by
(\ref{sol.Sym.1dim.a.claimS}), applied}\\\text{to }k\text{ instead of
}n\text{)}}}\right)  _{\Sym \left(  V\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.S}), applied to
}a=f^{\left(  n\right)  }\text{ and }b=x^{k}\right) \\
&  =\left(  f^{\left(  n\right)  },\left(  -1\right)  ^{k}x^{k}\right)
_{\Sym \left(  V\right)  }=f^{\left(  n\right)  }\left(  \left(
-1\right)  ^{k}x^{k}\right) \\
&  =\left(  -1\right)  ^{k}\underbrace{f^{\left(  n\right)  }\left(
x^{k}\right)  }_{\substack{=\delta_{n,k}\\\text{(by
(\ref{sol.Sym.1dim.b.delta}), applied to}\\i=n\text{ and }j=k\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f^{\left(  n\right)
}\text{ is }\kk\text{-linear}\right) \\
&  =\left(  -1\right)  ^{k}\delta_{n,k}.
\end{align*}
Since $\left(  -1\right)  ^{k}\delta_{n,k}=\left(  -1\right)  ^{n}\delta
_{n,k}$\ \ \ \ \footnote{This equality is obvious when $n=k$, and elsewise it
follows from $\delta_{n,k}=0$.}, this becomes%
\[
\left(  S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)  =\left(
-1\right)  ^{n}\delta_{n,k}.
\]
Compared with%
\[
\left(  \left(  -1\right)  ^{n}f^{\left(  n\right)  }\right)  \left(
x^{k}\right)  =\left(  -1\right)  ^{n}\underbrace{f^{\left(  n\right)
}\left(  x^{k}\right)  }_{\substack{=\delta_{n,k}\\\text{(by
(\ref{sol.Sym.1dim.b.delta}), applied to}\\i=n\text{ and }j=k\text{)}%
}}=\left(  -1\right)  ^{n}\delta_{n,k},
\]
this yields $\left(  S_{\left(  \Sym \left(  V\right)  \right)
^{o}}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)
=\left(  \left(  -1\right)  ^{n}f^{\left(  n\right)  }\right)  \left(
x^{k}\right)  $.

Let us now forget that we fixed $k$. We thus have shown that $\left(
S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)  =\left(  \left(
-1\right)  ^{n}f^{\left(  n\right)  }\right)  \left(  x^{k}\right)  $ for
every $k\in \NN$. In other words, the two $\kk$-linear maps
$S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  $ and $\left(  -1\right)  ^{n}f^{\left(
n\right)  }$ are equal to each other on the basis $\left(  x^{k}\right)
_{k\in\NN}$ of the $\kk$-module $\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $S_{\left(  \Sym %
\left(  V\right)  \right)  ^{o}}\left(  f^{\left(  n\right)  }\right)
=\left(  -1\right)  ^{n}f^{\left(  n\right)  }$. Thus, (\ref{sol.Sym.1dim.b.3}%
) is proven.

Now, all of the identities (\ref{sol.Sym.1dim.b.1.ij}),
(\ref{sol.Sym.1dim.b.2}) and (\ref{sol.Sym.1dim.b.3}) are proven; thus,
Exercise \ref{exe.Sym.1dim}(b) is solved.

Before we start the solution of Exercise \ref{exe.Sym.1dim}(c), let us make a
few more simple observations.

We have
\begin{equation}
\epsilon\left(  x^{k}\right)  =\delta_{k,0}\ \ \ \ \ \ \ \ \ \ \text{for every
}k\in\NN .
\label{sol.Sym.1dim.b.epsi}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.epsi}):} Let $k\in \NN$.
Then, the counit $\epsilon$ of $\Sym \left(  V\right)  $ is a
$\kk$-algebra homomorphism (by the axioms of a $\kk$-bialgebra
(since $\Sym \left(  V\right)  $ is a $\kk$-bialgebra)).
Hence, $\epsilon\left(  x^{k}\right)  =\left(  \underbrace{\epsilon\left(
x\right)  }_{=0}\right)  ^{k}= 0^k = \begin{cases}
1,& \text{if }k=0;\\
0,& \text{if }k\neq0
\end{cases} \quad =\delta_{k,0}$. This proves (\ref{sol.Sym.1dim.b.epsi}).}

The definition of the $\kk$-algebra $\left(  \Sym \left(
V\right)  \right)  ^{o}$ shows that the unity map $u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ of the $\kk%
$-algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ is
adjoint to the counit map $\epsilon$ of the $\kk$-coalgebra
$\Sym \left(  V\right)  $. In other words, we have%
\begin{equation}
\left(  u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  ,b\right)  _{\Sym \left(  V\right)  }=\left(
a,\epsilon\left(  b\right)  \right)  _{\kk} \label{sol.Sym.1dim.b.ue}%
\end{equation}
for any $a\in\kk$ and any $b\in\Sym \left(  V\right)  $.
Thus,%
\begin{equation}
u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  =af^{\left(  0\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }%
a\in\kk. \label{sol.Sym.1dim.b.ue2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.ue2}):} Let $a\in\kk$.
Then, every $b\in\Sym \left(  V\right)  $ satisfies%
\begin{align}
\left(  u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  \right)  \left(  b\right)   &  =\left(  u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  a\right)
,b\right)  _{\Sym \left(  V\right)  }=\left(  a,\epsilon\left(
b\right)  \right)  _{\kk}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.ue})}\right) \nonumber\\
&  =a\cdot\epsilon\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of the form }\left(  \cdot,\cdot\right)  _{\kk}\right)  .
\label{sol.Sym.1dim.b.ue2.pf.1}%
\end{align}
Now, every $k\in \NN$ satisfies%
\begin{align*}
\left(  u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  \right)  \left(  x^{k}\right)   &  =a\cdot\underbrace{\epsilon
\left(  x^{k}\right)  }_{\substack{=\delta_{k,0}\\\text{(by
(\ref{sol.Sym.1dim.b.epsi}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.ue2.pf.1}), applied to }b=x^{k}\right) \\
&  =a\cdot\underbrace{\delta_{k,0}}_{\substack{=f^{\left(  0\right)  }\left(
x^{k}\right)  \\\text{(since (\ref{sol.Sym.1dim.b.delta}) (applied
to}\\i=0\text{ and }j=k\text{) yields }f^{\left(  0\right)  }\left(
x^{k}\right)  =\delta_{0,k}=\delta_{k,0}\text{)}}}=a\cdot f^{\left(  0\right)
}\left(  x^{k}\right)
= \left(  af^{\left(  0\right)  }\right)  \left( x^{k} \right)  .
\end{align*}
In other words, the two $\kk$-linear maps $u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  a\right)  $ and
$af^{\left(  0\right)  }$ are equal to each other on the basis $\left(
x^{k}\right)  _{k\in\NN}$ of the $\kk$-module
$\Sym \left(  V\right)  $. Therefore, these two $\kk%
$-linear maps must be identical (because if two $\kk$-linear maps from
one and the same domain are equal to each other on a basis of this domain,
then these $\kk$-linear maps must be identical). In other words,
$u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  =af^{\left(  0\right)  }$. Thus, (\ref{sol.Sym.1dim.b.ue2}) is
proven.}

The definition of the $\kk$-coalgebra $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$ shows that the counit map $\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ of the $\kk%
$-coalgebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ is
adjoint to the unity map $u$ of the $\kk$-algebra $\Sym %
\left(  V\right)  $. In other words, we have%
\begin{equation}
\left(  \epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,b\right)  _{\kk}=\left(  a,u\left(  b\right)
\right)  _{\Sym \left(  V\right)  } \label{sol.Sym.1dim.b.eu}%
\end{equation}
for any
$a \in \left(\Sym \left( V \right)\right)^o$ and any $b \in \kk$.
Thus,
\begin{equation}
\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  k\right)  }\right)  =\delta_{k,0}\ \ \ \ \ \ \ \ \ \ \text{for
every }k\in\NN . \label{sol.Sym.1dim.b.eu2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.eu2}):} Let $k\in \NN$.
The map $u$ is the unity map of the $\kk$-algebra $\Sym %
\left(  V\right)  $; thus, $u\left(  1\right)  =1_{\Sym \left(
V\right)  }=x^{0}$.
\par
The definition of the bilinear form $\left(  \cdot,\cdot\right)  _{\kk%
}$ yields $\left(  \epsilon_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  ,1\right)  _{\kk%
}=\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  k\right)  }\right)  \cdot1=\epsilon_{\left(  \Sym %
\left(  V\right)  \right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  $, so
that%
\begin{align*}
\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  k\right)  }\right)   &  =\left(  \epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  f^{\left(
k\right)  }\right)  ,1\right)  _{\kk}=\left(  f^{\left(  k\right)
},\underbrace{u\left(  1\right)  }_{=x^{0}}\right)  _{\Sym %
\left(  V\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.eu}), applied to }a=f^{\left(  k\right)  }\text{ and
}b=1\right) \\
&  =\left(  f^{\left(  k\right)  },x^{0}\right)  _{\Sym \left(
V\right)  }=\delta_{k,0}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.delta}), applied to }i=k\text{ and }j=0\right)  .
\end{align*}
This proves (\ref{sol.Sym.1dim.b.eu2}).}

(c) For the time being, let us \textbf{not} assume that $\QQ$ is a
subring of $\kk$. Instead, we shall introduce a map and prove some of
its properties which hold for every $\kk$.

We define a $\kk$-linear map $\Phi:\Sym \left(  V\right)
\rightarrow\left(  \Sym \left(  V\right)  \right)  ^{o}$ by
\begin{equation}
\left(  \Phi\left(  x^{k}\right)  =k!f^{\left(  k\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\NN \right)
.\label{sol.Sym.1dim.c.Phi.def}%
\end{equation}
(This is well-defined, since $\left(  x^{k}\right)  _{k\in\NN}$ is a
basis of the $\kk$-module $\Sym \left(  V\right)  $.) We
can now see that this $\kk$-linear map $\Phi$ satisfies $\Phi\circ
m=m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\circ\left(
\Phi\otimes\Phi\right)  $\ \ \ \ \footnote{\textit{Proof.} Fix $\left(
k,\ell\right)  \in\NN^{2}$. Recall that $m$ is the multiplication map
of the $\kk$-algebra $\Sym \left(  V\right)  $. Thus,
$m\left(  x^{k}\otimes x^{\ell}\right)  =x^{k}x^{\ell}=x^{k+\ell}$. Also,
$m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$ is the
multiplication map of the $\kk$-algebra $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$. Thus, $m_{\left(  \Sym \left(
V\right)  \right)  ^{o}}\left(  f^{\left(  k\right)  }\otimes f^{\left(
\ell\right)  }\right)  =f^{\left(  k\right)  }f^{\left(  \ell\right)
}=\dbinom{k+\ell}{k}f^{\left(  k+\ell\right)  }$ (by (\ref{sol.Sym.1dim.b.1}),
applied to $u=k$ and $v=\ell$). Now,%
\[
\left(  \Phi\circ m\right)  \left(  x^{k}\otimes x^{\ell}\right)  =\Phi\left(
\underbrace{m\left(  x^{k}\otimes x^{\ell}\right)  }_{=x^{k+\ell}}\right)
=\Phi\left(  x^{k+\ell}\right)  =\left(  k+\ell\right)  !f^{\left(
k+\ell\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.c.Phi.def}), applied to }k+\ell\text{ instead of }k\right)
.
\]
Compared with%
\begin{align*}
\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}%
\circ\left(  \Phi\otimes\Phi\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)   &  =m_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  \underbrace{\left(  \Phi\otimes\Phi\right)  \left(  x^{k}\otimes
x^{\ell}\right)  }_{=\Phi\left(  x^{k}\right)  \otimes\Phi\left(  x^{\ell
}\right)  }\right)  =m_{\left(  \Sym \left(  V\right)  \right)
^{o}}\left(  \underbrace{\Phi\left(  x^{k}\right)  }_{\substack{=k!f^{\left(
k\right)  }\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}))}}}\otimes
\underbrace{\Phi\left(  x^{\ell}\right)  }_{\substack{=\ell!f^{\left(
\ell\right)  }\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}),}\\\text{applied to
}\ell\text{ instead of }k\text{)}}}\right)  \\
&  =m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
\underbrace{\left(  k!f^{\left(  k\right)  }\right)  \otimes\left(
\ell!f^{\left(  \ell\right)  }\right)  }_{=k!\ell!f^{\left(  k\right)
}\otimes f^{\left(\ell\right)}}\right)  =m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  k!\ell!f^{\left(  k\right)  }\otimes f^{\left(\ell\right)}\right)  \\
&  =k!\ell!\underbrace{m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\otimes f^{\left(\ell\right)}\right)
}_{=\dbinom{k+\ell}{k}f^{\left(  k+\ell\right)  }}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }m_{\left(  \Sym \left(  V\right)  \right)
^{o}}\text{ is }\kk\text{-linear}\right)  \\
&  =\underbrace{k!\ell!\dbinom{k+\ell}{k}}_{\substack{=\left(  k+\ell\right)
!\\\text{(since }\dbinom{k+\ell}{k}=\dfrac{\left(  k+\ell\right)  !}{k!\ell
!}\text{)}}}f^{\left(  k+\ell\right)  }=\left(  k+\ell\right)  !f^{\left(
k+\ell\right)  },
\end{align*}
this yields $\left(  \Phi\circ m\right)  \left(  x^{k}\otimes x^{\ell}\right)
=\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}%
\circ\left(  \Phi\otimes\Phi\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  $.
\par
Let us now forget that we fixed $\left(  k,\ell\right)  $. We thus have shown
that $\left(  \Phi\circ m\right)  \left(  x^{k}\otimes x^{\ell}\right)
=\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}%
\circ\left(  \Phi\otimes\Phi\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  $ for every $\left(  k,\ell\right)  \in\NN^{2}$. In other
words, the two $\kk$-linear maps $\Phi\circ m$ and $m_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\left(  \Phi
\otimes\Phi\right)  $ (from $\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  $ to $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$) are equal to each other on the basis
$\left(  x^{k}\otimes x^{\ell}\right)  _{\left(  k,\ell\right)  \in
\NN^{2}}$ of the $\kk$-module $\Sym \left(
V\right)  \otimes\Sym \left(  V\right)  $. Therefore, these two
$\kk$-linear maps must be identical (because if two $\kk$-linear
maps from one and the same domain are equal to each other on a basis of this
domain, then these $\kk$-linear maps must be identical). In other
words, $\Phi\circ m=m_{\left(  \Sym \left(  V\right)  \right)
^{o}}\circ\left(  \Phi\otimes\Phi\right)  $. Qed.} and $\Phi\circ u=u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$%
\ \ \ \ \footnote{\textit{Proof.} The map $u$ is the unity map of the
$\kk$-algebra $\Sym \left(  V\right)  $; thus, $u\left(
1\right)  =1_{\Sym \left(  V\right)  }=x^{0}$. Now, every
$a\in\kk$ satisfies%
\begin{align*}
\left(  \Phi\circ u\right)  \left(  \underbrace{a}_{=a\cdot1}\right)   &
=\left(  \Phi\circ u\right)  \left(  a\cdot1\right)  =a\cdot
\underbrace{\left(  \Phi\circ u\right)  \left(  1\right)  }_{=\Phi\left(
u\left(  1\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\Phi\circ u\text{ is }\kk\text{-linear}\right)  \\
&  =a\cdot\Phi\left(  \underbrace{u\left(  1\right)  }_{=x^{0}}\right)
=a\cdot\underbrace{\Phi\left(  x^{0}\right)  }_{\substack{=0!f^{\left(
0\right)  }\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}), applied to
}0\\\text{instead of }k\text{)}}}=a\cdot\underbrace{0!}_{=1}f^{\left(
0\right)  }=af^{\left(  0\right)  }=u_{\left(  \Sym \left(
V\right)  \right)  ^{o}}\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.ue2})}\right)  .
\end{align*}
Thus, $\Phi\circ u=u_{\left(  \Sym \left(  V\right)  \right)
^{o}}$, qed.}. Hence, $\Phi$ is a $\kk$-algebra homomorphism. Also, the
$\kk$-linear map $\Phi$ satisfies $\left(  \Phi\otimes\Phi\right)
\circ\Delta=\Delta_{\left(  \Sym \left(  V\right)  \right)
^{o}}\circ\Phi$\ \ \ \ \footnote{\textit{Proof.} Let $k\in \NN$. We have%
\begin{align*}
\left(  \left(  \Phi\otimes\Phi\right)  \circ\Delta\right)  \left(
x^{k}\right)   &  =\left(  \Phi\otimes\Phi\right)  \left(  \underbrace{\Delta
\left(  x^{k}\right)  }_{\substack{=\sum_{i+j=k}\dbinom{k}{i}x^{i}\otimes
x^{j}\\\text{(by (\ref{sol.Sym.1dim.a.3}), applied to }n=k\text{)}}}\right)
=\left(  \Phi\otimes\Phi\right)  \left(  \sum_{i+j=k}\dbinom{k}{i}x^{i}\otimes
x^{j}\right)  \\
&  =\sum_{i+j=k}\dbinom{k}{i}\underbrace{\Phi\left(  x^{i}\right)
}_{\substack{=i!f^{\left(  i\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}))}}}\otimes\underbrace{\Phi\left(  x^{j}\right)
}_{\substack{=j!f^{\left(  j\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\Phi\otimes\Phi\right)  \\
&  =\sum_{i+j=k}\dbinom{k}{i}i!f^{\left(  i\right)  }\otimes j!f^{\left(
j\right)  }=\sum_{i+j=k}\dbinom{k}{i}i!\underbrace{j!}_{\substack{=\left(
k-i\right)  !\\\text{(since }j=k-i\\\text{(since }i+j=k\text{))}}}f^{\left(
i\right)  }\otimes f^{\left(  j\right)  }\\
&  =\sum_{i+j=k}\underbrace{\dbinom{k}{i}i!\left(  k-i\right)  !}%
_{\substack{=k!\\\text{(since }\dbinom{k}{i}=\dfrac{k!}{i!\left(  k-i\right)
!}\text{)}}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }%
=k!\sum_{i+j=k}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }.
\end{align*}
Compared with%
\begin{align*}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi\right)  \left(  x^{k}\right)   &  =\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  \underbrace{\Phi
\left(  x^{k}\right)  }_{\substack{=k!f^{\left(  k\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}))}}}\right)  =\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  k!f^{\left(
k\right)  }\right)  =k!\underbrace{\Delta_{\left(  \Sym \left(
V\right)  \right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  }%
_{\substack{=\sum_{i+j=k}f^{\left(  i\right)  }\otimes f^{\left(  j\right)
}\\\text{(by (\ref{sol.Sym.1dim.b.2}), applied to }n=k\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\text{ is }\kk%
\text{-linear}\right)  \\
&  =k!\sum_{i+j=k}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  },
\end{align*}
this yields $\left(  \left(  \Phi\otimes\Phi\right)  \circ\Delta\right)
\left(  x^{k}\right)  =\left(  \Delta_{\left(  \Sym \left(
V\right)  \right)  ^{o}}\circ\Phi\right)  \left(  x^{k}\right)  $.
\par
Let us now forget that we fixed $k$. We thus have shown that $\left(  \left(
\Phi\otimes\Phi\right)  \circ\Delta\right)  \left(  x^{k}\right)  =\left(
\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\circ
\Phi\right)  \left(  x^{k}\right)  $ for every $k\in \NN$. In other
words, the two $\kk$-linear maps $\left(  \Phi\otimes\Phi\right)
\circ\Delta$ and $\Delta_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\circ\Phi$ are equal to each other on the basis $\left(
x^{k}\right)  _{k\in\NN}$ of the $\kk$-module
$\Sym \left(  V\right)  $. Therefore, these two $\kk%
$-linear maps must be identical (because if two $\kk$-linear maps from
one and the same domain are equal to each other on a basis of this domain,
then these $\kk$-linear maps must be identical). In other words,
$\left(  \Phi\otimes\Phi\right)  \circ\Delta=\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\Phi$. Qed.} and
$\epsilon=\epsilon_{\left(  \Sym \left(  V\right)  \right)
^{o}}\circ\Phi$\ \ \ \ \footnote{\textit{Proof.} Let $k\in \NN$. We have%
\begin{align*}
\left(  \epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi\right)  \left(  x^{k}\right)   &  =\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  \underbrace{\Phi
\left(  x^{k}\right)  }_{\substack{=k!f^{\left(  k\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}))}}}\right)  =\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  k!f^{\left(
k\right)  }\right)  =k!\underbrace{\epsilon_{\left(  \Sym %
\left(  V\right)  \right)  ^{o}}\left(  f^{\left(  k\right)  }\right)
}_{\substack{=\delta_{k,0}\\\text{(by (\ref{sol.Sym.1dim.b.eu2}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\text{ is }\kk%
\text{-linear}\right)  \\
&  =k!\delta_{k,0}.
\end{align*}
But it is easy (by treating the cases $k=0$ and $k\neq0$ separately) to see
that $k!\delta_{k,0}=\delta_{k,0}$. Hence,
\[
\left(  \epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi\right)  \left(  x^{k}\right)  =k!\delta_{k,0}=\delta_{k,0}%
=\epsilon\left(  x^{k}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.epsi})}\right)  .
\]
In other words, $\epsilon\left(  x^{k}\right)  =\left(  \epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\Phi\right)  \left(
x^{k}\right)  $.
\par
Let us now forget that we fixed $k$. We thus have shown that $\epsilon\left(
x^{k}\right)  =\left(  \epsilon_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\circ\Phi\right)  \left(  x^{k}\right)  $ for every
$k\in \NN$. In other words, the two $\kk$-linear maps $\epsilon$
and $\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi$ are equal to each other on the basis $\left(  x^{k}\right)
_{k\in\NN}$ of the $\kk$-module $\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $\epsilon=\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\Phi$. Qed.}. Hence,
$\Phi$ is a $\kk$-coalgebra homomorphism. Thus, $\Phi$ is a
$\kk$-bialgebra homomorphism (since $\Phi$ is both a $\kk%
$-algebra homomorphism and a $\kk$-coalgebra homomorphism), and
therefore a Hopf algebra homomorphism (according to
Corollary~\ref{cor.bialg-mor-is-Hopf}, applied to $H_{1}%
=\Sym \left(  V\right)  $, $H_{2}=\left(  \Sym %
\left(  V\right)  \right)  ^{o}$, $S_{1}=S$ and $S_{2}=S_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$). The map $\Phi$ is also
graded\footnote{\textit{Proof.} Let $n\in \NN$. For every graded
$\kk$-module $A$, let $A_{n}$ denote the $n$-th homogeneous component
of $A$. We shall show that $\Phi\left(  \left(  \Sym \left(
V\right)  \right)  _{n}\right)  \subset\left(  \left(  \Sym %
\left(  V\right)  \right)  ^{o}\right)  _{n}$.
\par
Indeed, recall that $\left(  f^{\left(  k\right)  }\right)  _{k\in\NN}$
is a \textbf{graded} basis of the $\kk$-module $\left(
\Sym \left(  V\right)  \right)  ^{o}$. Thus, the one-element
family $\left(  f^{\left(  n\right)  }\right)  $ is a basis of the $n$-th
homogeneous component $\left(  \left(  \Sym \left(  V\right)
\right)  ^{o}\right)  _{n}$. Thus, this family $\left(  f^{\left(  n\right)
}\right)  $ spans the $\kk$-module $\left(  \left(  \Sym %
\left(  V\right)  \right)  ^{o}\right)  _{n}$. In other words, $\left(
\left(  \Sym \left(  V\right)  \right)  ^{o}\right)
_{n}=\kk\cdot f^{\left(  n\right)  }$.
\par
On the other hand, recall that $\left(  x^{k}\right)  _{k\in\NN}$ is a
\textbf{graded} basis of the $\kk$-module $\Sym \left(
V\right)  $. Thus, the one-element family $\left(  x^{n}\right)  $ is a basis
of the $n$-th homogeneous component $\left(  \Sym \left(
V\right)  \right)  _{n}$. Thus, this family $\left(  x^{n}\right)  $ spans the
$\kk$-module $\left(  \Sym \left(  V\right)  \right)
_{n}$. In other words, $\left(  \Sym \left(  V\right)  \right)
_{n}=\kk\cdot x^{n}$. Hence,%
\begin{align*}
\Phi\left(  \underbrace{\left(  \Sym \left(  V\right)  \right)
_{n}}_{=\kk\cdot x^{n}}\right)   &  =\Phi\left(  \kk\cdot
x^{n}\right)  \subset\kk\cdot\underbrace{\Phi\left(  x^{n}\right)
}_{\substack{=n!f^{\left(  n\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}),}\\\text{applied to }k=n\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Phi\text{ is }%
\kk\text{-linear}\right)  \\
&  =\underbrace{\kk\cdot n!}_{\subset\kk}f^{\left(  n\right)
}\subset\kk\cdot f^{\left(  n\right)  }=\left(  \left(
\Sym \left(  V\right)  \right)  ^{o}\right)  _{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \left(  \Sym %
\left(  V\right)  \right)  ^{o}\right)  _{n}=\kk\cdot f^{\left(
n\right)  }\right)  .
\end{align*}
\par
Now, let us forget that we fixed $n$. We thus have shown that $\Phi\left(
\left(  \Sym \left(  V\right)  \right)  _{n}\right)
\subset\left(  \left(  \Sym \left(  V\right)  \right)
^{o}\right)  _{n}$ for every $n\in \NN$. In other words, the map $\Phi$
is graded, qed.}. Thus, $\Phi$ is a homomorphism of graded $\kk$-Hopf
algebras (since $\Phi$ is graded and a Hopf algebra homomorphism).

Now, let us assume that $\QQ$ is a subring of $\kk$. We define a
$\kk$-linear map $\Psi:\left(  \Sym \left(  V\right)
\right)  ^{o}\rightarrow\Sym \left(  V\right)  $ by
\begin{equation}
\left(  \Psi\left(  f^{\left(  k\right)  }\right)  =\dfrac{x^{k}}%
{k!}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\NN \right)  .
\label{sol.Sym.1dim.c.Psi.def}%
\end{equation}
(This is well-defined, since $\left(  f^{\left(  k\right)  }\right)
_{k\in\NN}$ is a basis of the $\kk$-module $\left(
\Sym \left(  V\right)  \right)  ^{o}$.) It is clear that the
maps $\Phi$ and $\Psi$ are mutually inverse\footnote{Indeed, the map
$\Phi\circ\Psi$ sends every $f^{\left(  k\right)  }$ to $f^{\left(  k\right)
}$ and thus is the identity map, whereas the map $\Psi\circ\Phi$ sends every
$x^{k}$ to $x^{k}$ and therefore is the identity map as well.}. Thus, the map
$\Phi$ is invertible, and its inverse is $\Psi$.

The map $\Phi$ is an invertible homomorphism of graded $\kk$-Hopf
algebras and therefore an isomorphism of graded $\kk$-Hopf algebras
(since every invertible homomorphism of graded $\kk$-Hopf algebras is
an isomorphism of graded $\kk$-Hopf algebras). Hence, the inverse of
$\Phi$ is also an isomorphism of graded $\kk$-Hopf algebras. In other
words, $\Psi$ is an isomorphism of graded $\kk$-Hopf algebras (since
the inverse of $\Phi$ is $\Psi$).

But every $n\in \NN$ satisfies $\Psi\left(  f^{\left(  n\right)
}\right)  =\dfrac{x^{n}}{n!}$ (according to (\ref{sol.Sym.1dim.c.Psi.def}),
applied to $k=n$). Hence, $\Psi$ is the $\kk$-linear map $\left(
\Sym \left(  V\right)  \right)  ^{o}\rightarrow
\Sym \left(  V\right)  $ sending $f^{(n)}\mapsto\frac{x^{n}}{n!}$.
Thus, the
$\kk$-linear map $\left(  \Sym \left(  V\right)  \right)
^{o}\rightarrow\Sym \left(  V\right)  $ sending $f^{(n)}%
\mapsto\frac{x^{n}}{n!}$ is an isomorphism of graded $\kk$-Hopf
algebras (since $\Psi$ is an isomorphism of graded $\kk$-Hopf
algebras). This solves Exercise \ref{exe.Sym.1dim}(c).

(d) For the time being, let us \textbf{not} assume that $\kk$ is a
field of characteristic $p > 0$.
Instead, let us first prove a formula which does not
require any restrictions on $\kk$.

Namely, let us show that
\begin{equation}
\left(  f^{\left(  1\right)  }\right)  ^{m}=m!f^{\left(  m\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\NN .
\label{sol.Sym.1dim.d.goal}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.d.goal}):} Let us consider the
$\kk$-linear map $\Phi:\Sym \left(  V\right)
\rightarrow\left(  \Sym \left(  V\right)  \right)  ^{o}$ defined
in our solution to Exercise \ref{exe.Sym.1dim}(c) above. Then, $\Phi$ is a
$\kk$-algebra homomorphism. (In fact, this was proven in our solution
to Exercise \ref{exe.Sym.1dim}(c) above.) Applying
(\ref{sol.Sym.1dim.c.Phi.def}) to $k=1$, we obtain $\Phi\left(  x^{1}\right)
=\underbrace{1!}_{=1}f^{\left(  1\right)  }=f^{\left(  1\right)  }$, so that
$f^{\left(  1\right)  }=\Phi\left(x^1\right)
= \Phi\left(  x\right)  $. Thus, for every
$m\in \NN$, we have%
\begin{align*}
\left(  \underbrace{f^{\left(  1\right)  }}_{=\Phi\left(  x\right)  }\right)
^{m}  &  =\left(  \Phi\left(  x\right)  \right)  ^{m}=\Phi\left(
x^{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\text{ is a
}\kk\text{-algebra homomorphism}\right) \\
&  =m!f^{\left(  m\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.c.Phi.def}), applied to }k=m\right)  .
\end{align*}
This proves (\ref{sol.Sym.1dim.d.goal}).

Now, let us assume that $\kk$ is a field of characteristic $p>0$. Then,
$p\cdot1_{\kk}=0$. But $\underbrace{p!}_{=\left(  p-1\right)  !p}%
\cdot1_{\kk}=\left(  p-1\right)  !\underbrace{p\cdot1_{\kk}%
}_{=0}=0$. Applying (\ref{sol.Sym.1dim.d.goal}) to $m=p$, we obtain%
\[
\left(  f^{\left(  1\right)  }\right)  ^{p}=p!f^{\left(  p\right)
}=\underbrace{p!\cdot1_{\kk}}_{=0}\cdot f^{\left(  p\right)  }=0.
\]


Now, it remains to show that there can be no Hopf isomorphism $\left(
\Sym \left(  V\right)  \right)  ^{o}\rightarrow\Sym \left(  V\right)  $.
In fact, we can prove something stronger: Namely, there
exists no algebra isomorphism $\left(  \Sym \left(  V\right)
\right)  ^{o}\rightarrow\Sym \left(  V\right)  $. This is because
the algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ has a
nonzero nilpotent element (namely, $f^{\left(  1\right)  }$ is nonzero and
satisfies $\left(  f^{\left(  1\right)  }\right)  ^{p}=0$), whereas the
algebra $\Sym \left(  V\right)  $ contains no nonzero nilpotent
elements (in fact, $\Sym \left(  V\right)  \cong\kk%
\left[  x\right]  $ is isomorphic to a polynomial ring over the field
$\kk$, and therefore is an integral domain, which yields that it
contains no nonzero nilpotent elements). This concludes the solution of
Exercise \ref{exe.Sym.1dim}(d).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Sym.1dim}}

\textit{Solution to Exercise \ref{exe.Sym.1dim}.} We have
$\Sym \left(V\right) \cong \kk\left[  x\right]  $. Thus,
$\left(  x^{k}\right)_{k\in\NN}$ is a basis of the $\kk$-module
$\Sym \left(  V\right)  $.
Therefore, $\left(  x^{k}\otimes x^{\ell}\right)
_{\left(  k,\ell\right)  \in\NN^{2}}$ is a basis of the $\kk$-module
$\Sym \left(  V\right)  \otimes\Sym \left(  V\right)  $.

On the other hand, $\left(  f^{\left(  k\right)  }\right)  _{k\in\NN}$
is a basis of the $\kk$-module $\left(  \Sym \left(
V\right)  \right)  ^{o}$ (namely, the dual basis to the graded basis $\left(
x^{k}\right)  _{k\in\NN}$ of $\Sym \left(  V\right)  $).
Hence, $\left(  f^{\left(  k\right)  }\otimes f^{\left(  \ell\right)
}\right)  _{\left(  k,\ell\right)  \in\NN^{2}}$ is a basis of the
$\kk$-module $\left(  \Sym \left(  V\right)  \right)
^{o}\otimes\left(  \Sym \left(  V\right)  \right)  ^{o}$.

We denote by $m$ the multiplication map of $\Sym \left(
V\right)  $. We denote by $u$ the unity map of $\Sym \left(
V\right)  $ (that is, the map $\kk\rightarrow\Sym \left(
V\right)  $ which sends $1_{\kk}$ to $1_{\Sym \left(
V\right)  }$). We denote by $\Delta$ the comultiplication of
$\Sym \left(  V\right)  $. We denote by $\epsilon$ the counit of
$\Sym \left(  V\right)  $. We denote by $S$ the antipode of
$\Sym \left(  V\right)  $.

We denote by $m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$
the multiplication map of $\left(  \Sym \left(  V\right)
\right)  ^{o}$. We denote by $u_{\left(  \Sym \left(  V\right)
\right)  ^{o}}$ the unity map of $\left(  \Sym \left(  V\right)
\right)  ^{o}$ (that is, the map $\kk\rightarrow\left(
\Sym \left(  V\right)  \right)  ^{o}$ which sends $1_{\kk%
}$ to $1_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$). We
denote by $\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}$ the comultiplication of $\left(  \Sym \left(  V\right)
\right)  ^{o}$. We denote by $\epsilon_{\left(  \Sym \left(
V\right)  \right)  ^{o}}$ the counit of $\left(  \Sym \left(
V\right)  \right)  ^{o}$. We denote by $S_{\left(  \Sym \left(
V\right)  \right)  ^{o}}$ the antipode of $\left(  \Sym \left(
V\right)  \right)  ^{o}$.

(a) Clearly, $x^{i}\cdot x^{j}=x^{i+j}$ for all $i\in \NN$ and
$j\in \NN$.

We have $\Delta_{T\left(  V\right)  }\left(  x\right)  =1\otimes x+x\otimes1$
in $T\left(  V\right)  \otimes T\left(  V\right)  $ (by the definition of
$\Delta_{T\left(  V\right)  }$). Projecting this equality down onto
$\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  $, we obtain $\Delta_{\Sym \left(  V\right)  }\left(
x\right)  =x\otimes1+1\otimes x$ (since $\Sym \left(  V\right)
$ is a quotient Hopf algebra of $T\left(  V\right)  $). But we defined
$\Delta$ to mean the comultiplication of $\Sym \left(  V\right)
$. We thus have $\Delta=\Delta_{\Sym \left(  V\right)  }$, so
that $\Delta\left(  x\right)  =\Delta_{\Sym \left(  V\right)
}\left(  x\right)  =x\otimes1+1\otimes x$. Thus, $x$ is a primitive element of
$\Sym \left(  V\right)  $. Proposition
\ref{antipode-on-primitives} thus yields $S\left(  x\right)  =-x$.

Proposition \ref{antipodes-are-antiendomorphisms} yields that the antipode $S$
of $\Sym \left(  V\right)  $ is an algebra anti-endomorphism of
$\Sym \left(  V\right)  $. Since an algebra anti-endomorphism of
$\Sym \left(  V\right)  $ means the same as an algebra
endomorphism of $\Sym \left(  V\right)  $ (by
Exercise~\ref{exe.comm-cocomm.anti}(a), because
$\Sym \left(  V\right)  $ is commutative), this yields that $S$
is an algebra endomorphism of $\Sym \left(  V\right)  $.
Consequently, for any $n\in \NN$, we have%
\begin{equation}
S\left(  x^{n}\right)  =\left(  \underbrace{S\left(  x\right)  }_{=-x}\right)
^{n}=\left(  -x\right)  ^{n}=\left(  -1\right)  ^{n}x^{n}.
\label{sol.Sym.1dim.a.claimS}%
\end{equation}


It remains to prove that $\Delta\left(  x^{n}\right)  =\sum_{i+j=n}\dbinom
{n}{i}x^{i}\otimes x^{j}$ for every $n\in \NN$ (where the summation sign
``$\sum_{i+j=n}$'' is shorthand for
``$\sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}$''). Let $n\in \NN$. The $\kk%
$-algebra $\Sym \left(  V\right)  \otimes\Sym \left(  V\right)  $ is
commutative (since $\Sym \left( V\right)  $ is commutative),
and thus the binomial formula can be applied in
it. We know that $\Sym \left(  V\right)  $ is a bialgebra, and
thus $\Delta$ is a $\kk$-algebra homomorphism (by the axioms of a
bialgebra). Hence,%
\begin{align}
\Delta\left(  x^{n}\right)   &  =\left(  \underbrace{\Delta\left(  x\right)
}_{=x\otimes1+1\otimes x}\right)  ^{n}=\left(  x\otimes1+1\otimes x\right)
^{n}\nonumber\\
&  =\sum_{\ell=0}^{n}\dbinom{n}{\ell}\underbrace{\left(  x\otimes1\right)
^{\ell}}_{=x^{\ell}\otimes1}\underbrace{\left(  1\otimes x\right)  ^{n-\ell}%
}_{=1\otimes x^{n-\ell}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the binomial
formula}\right) \nonumber\\
&  =\sum_{\ell=0}^{n}\dbinom{n}{\ell}\underbrace{\left(  x^{\ell}%
\otimes1\right)  \left(  1\otimes x^{n-\ell}\right)  }_{=\left(  x^{\ell
}1\right)  \otimes\left(  1x^{n-\ell}\right)  =x^{\ell}\otimes x^{n-\ell}%
}=\sum_{\ell=0}^{n}\dbinom{n}{\ell}x^{\ell}\otimes x^{n-\ell}.
\label{sol.Sym.1dim.a.3}%
\end{align}


But%
\begin{align}
&  \underbrace{\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2}%
;\\i+j=n}}}_{=\sum_{i\in\NN}\sum\limits_{\substack{j\in\NN ;\\i+j=n}%
}}\dbinom{n}{i}x^{i}\otimes x^{j}=\sum_{i\in\NN}\sum\limits_{\substack{j\in
\NN ;\\i+j=n}}\dbinom{n}{i}x^{i}\otimes x^{j}\nonumber\\
&  =\sum\limits_{\substack{i\in\NN ;\\i\leq n}}\sum\limits_{\substack{j\in
\NN ;\\i+j=n}}\dbinom{n}{i}x^{i}\otimes x^{j}+\sum\limits_{\substack{i\in
\NN ;\\i>n}}\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\dbinom{n}{i}%
x^{i}\otimes x^{j}\label{sol.Sym.1dim.a.5}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in\NN \text{
satisfies exactly one of the conditions }i\leq n\text{ and }i>n\right)
.\nonumber
\end{align}


But every $i\in \NN$ satisfying $i>n$ satisfies%
\begin{equation}
\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\dbinom{n}{i}x^{i}\otimes x^{j}=0
\label{sol.Sym.1dim.a.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.a.6}):} Let $i\in \NN$ be
such that $i>n$. Then, every $j\in \NN$ satisfies $i+\underbrace{j}%
_{\geq0}\geq i>n$. Hence, every $j\in \NN$ satisfies $i+j\neq n$. In
other words, no $j\in \NN$ satisfies $i+j=n$. Thus, the sum
$\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\dbinom{n}{i}x^{i}\otimes x^{j}$ is
an empty sum. Hence, $\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\dbinom{n}%
{i}x^{i}\otimes x^{j}=\left(  \text{empty sum}\right)  =0$. This proves
(\ref{sol.Sym.1dim.a.6}).}. Also, every $i\in \NN$ satisfying $i\leq n$
satisfies%
\begin{equation}
\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\dbinom{n}{i}x^{i}\otimes
x^{j}=\dbinom{n}{i}x^{i}\otimes x^{n-i} \label{sol.Sym.1dim.a.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.a.7}):} Let $i\in \NN$ be
such that $i\leq n$. Then, $n-i\geq0$ (since $i\leq n$), so that
$n-i\in \NN$. Hence, there exists exactly one $j\in \NN$ such that
$j=n-i$, namely $j=n-i$. Therefore, $\sum\limits_{\substack{j\in\NN %
;\\j=n-i}}\dbinom{n}{i}x^{i}\otimes x^{j}=\dbinom{n}{i}x^{i}\otimes x^{n-i}$.
But
\[
\underbrace{\sum\limits_{\substack{j\in\NN ;\\i+j=n}}}_{\substack{=
\sum\limits_{\substack{j\in\NN ;\\j=n-i}}\\
\text{(since the condition}\\i+j=n\text{
is equivalent to}\\j=n-i\text{)}}}\dbinom{n}{i}x^{i}\otimes x^{j}%
=\sum\limits_{\substack{j\in\NN ;\\j=n-i}}\dbinom{n}{i}x^{i}\otimes
x^{j}=\dbinom{n}{i}x^{i}\otimes x^{n-i}.
\]
This proves (\ref{sol.Sym.1dim.a.7}).}. Now, (\ref{sol.Sym.1dim.a.5}) becomes
\begin{align*}
&  \sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=n}}\dbinom
{n}{i}x^{i}\otimes x^{j}\\
&  =\sum\limits_{\substack{i\in\NN ;\\i\leq n}}
\underbrace{\sum\limits_{\substack{j\in \NN ;\\i+j=n}}
\dbinom{n}{i}x^{i}\otimes x^{j}}_{\substack{=\dbinom{n}%
{i}x^{i}\otimes x^{n-i}\\\text{(by (\ref{sol.Sym.1dim.a.7}))}}}+
\sum\limits_{\substack{i\in\NN ;\\i>n}}
\underbrace{\sum\limits_{\substack{j\in
\NN ;\\i+j=n}}\dbinom{n}{i}x^{i}\otimes x^{j}}_{\substack{=0\\\text{(by
(\ref{sol.Sym.1dim.a.6}))}}}\\
&  =\sum\limits_{\substack{i\in\NN ;\\i\leq n}}\dbinom{n}{i}x^{i}\otimes
x^{n-i}+\underbrace{\sum\limits_{\substack{i\in\NN ;\\i>n}}0}_{=0}%
=\underbrace{\sum\limits_{\substack{i\in\NN ;\\i\leq n}}}_{=\sum_{i=0}^{n}%
}\dbinom{n}{i}x^{i}\otimes x^{n-i}=\sum_{i=0}^{n}\dbinom{n}{i}x^{i}\otimes
x^{n-i}\\
&  =\sum_{\ell=0}^{n}\dbinom{n}{\ell}x^{\ell}\otimes x^{n-\ell}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}i\text{ as }\ell\right) \\
&  =\Delta\left(  x^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.a.3})}\right)  ,
\end{align*}
so that%
\begin{align}
\Delta\left(  x^{n}\right)   &  =\underbrace{\sum\limits_{\substack{\left(
i,j\right)  \in\NN^{2};\\i+j=n}}}_{=\sum_{i+j=n}}\dbinom{n}{i}%
x^{i}\otimes x^{j}\label{sol.Sym.1dim.a.claimDelta.1}\\
&  =\sum_{i+j=n}\dbinom{n}{i}x^{i}\otimes x^{j}.\nonumber
\end{align}


Now, let us forget that we fixed $n$. We thus have proven the identity%
\begin{equation}
\Delta\left(  x^{n}\right)  =\sum_{i+j=n}\dbinom{n}{i}x^{i}\otimes x^{j}
\label{sol.Sym.1dim.a.claimDelta}%
\end{equation}
for every $n\in \NN$. This completes the solution of Exercise
\ref{exe.Sym.1dim}(a).

(b) Let us first state a very elementary algebraic fact: If $u$, $v$ and $k$
are any three elements of $ \NN$, then%

\begin{equation}
\sum\limits_{\substack{j\in\NN ;\\u+j=k}}\delta_{v,j}=\delta_{u+v,k}
\label{sol.Sym.1dim.b.1.pf.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.1.pf.5}):} Let $u$, $v$ and
$k$ be three elements of $ \NN$. We must be in one of the following two
cases:
\par
\textit{Case 1:} We have $u+v=k$.
\par
\textit{Case 2:} We have $u+v\neq k$.
\par
Let us first consider Case 1. In this case, we have $u+v=k$. Thus,
$\delta_{u+v,k}=1$. On the other hand, $k-u=v$ (since $u+v=k$). Now,%
\begin{align*}
\underbrace{\sum\limits_{\substack{j\in\NN ;\\u+j=k}}}_{\substack{=
\sum\limits_{\substack{j\in\NN ;\\j=k-u}}\\
\text{(since the condition}\\u+j=k\text{
is equivalent to}\\j=k-u\text{)}}}\delta_{v,j}  &  =\underbrace{
\sum\limits_{\substack{j\in\NN ;\\j=k-u}}}_{\substack{
=\sum\limits_{\substack{j\in
\NN ;\\j=v}}\\\text{(since }k-u=v\text{)}}}\delta_{v,j}=
\sum\limits_{\substack{j\in\NN ;\\j=v}}\delta_{v,j}=\delta_{v,v}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v\in\NN \right) \\
&  =1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v=v\right) \\
&  =\delta_{u+v,k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\delta
_{u+v,k}\right)  .
\end{align*}
Thus, (\ref{sol.Sym.1dim.b.1.pf.5}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $u+v\neq k$. Thus,
$\delta_{u+v,k}=0$. On the other hand, let $j\in \NN$ be such that
$u+j=k$. Then, $k-u=j$ (since $u+j=k$) but $v\neq k-u$ (since $u+v\neq k$).
Hence, $v\neq k-u=j$, so that $\delta_{v,j}=0$. Now, let us forget that we
fixed $j$. We thus have shown that every $j\in \NN$ satisfying $u+j=k$
satisfies $\delta_{v,j}=0$. Hence, $\sum\limits_{\substack{j\in\NN %
;\\u+j=k}}\underbrace{\delta_{v,j}}_{=0}=\sum\limits_{\substack{j\in\NN %
;\\u+j=k}}0=0=\delta_{u+v,k}$ (since $\delta_{u+v,k}=0$). Thus,
(\ref{sol.Sym.1dim.b.1.pf.5}) is proven in Case 2.
\par
We have now proven (\ref{sol.Sym.1dim.b.1.pf.5}) in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, we therefore conclude
that (\ref{sol.Sym.1dim.b.1.pf.5}) always holds.}.

The definition of $f^{\left(  i\right)  }$ shows that%
\begin{equation}
f^{\left(  i\right)  }\left(  x^{j}\right)  =\delta_{i,j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\NN \text{ and }j\in\NN .
\label{sol.Sym.1dim.b.delta}%
\end{equation}


But the definition of the $\kk$-algebra
$\left(  \Sym \left(  V\right)  \right)  ^{o}$ shows that the
multiplication map $m_{\left( \Sym \left(  V\right)  \right)  ^{o}}$
of the $\kk$-algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$
is adjoint to the comultiplication map $\Delta$ of the
$\kk$-coalgebra $\Sym \left(  V\right)  $. In other words, we have
\begin{equation}
\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  ,b\right)  _{\Sym \left(  V\right)  }=\left(
a,\Delta\left(  b\right)  \right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  } \label{sol.Sym.1dim.b.mD}%
\end{equation}
for any $a\in\left(  \Sym \left(  V\right)  \right)  ^{o}%
\otimes\left(  \Sym \left(  V\right)  \right)  ^{o}$ and any
$b\in\Sym \left(  V\right)  $. Thus, any $p\in\left(
\Sym \left(  V\right)  \right)  ^{o}$, $q\in\left(
\Sym \left(  V\right)  \right)  ^{o}$ and $b\in
\Sym \left(  V\right)  $ satisfy%
\begin{equation}
\left(  pq,b\right)  _{\Sym \left(  V\right)  }=\left(  p\otimes
q,\Delta\left(  b\right)  \right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }. \label{sol.Sym.1dim.b.mD2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.mD2}):} Let $p\in\left(
\Sym \left(  V\right)  \right)  ^{o}$, $q\in\left(
\Sym \left(  V\right)  \right)  ^{o}$ and $b\in
\Sym \left(  V\right)  $. Then, $m_{\left(  \Sym %
\left(  V\right)  \right)  ^{o}}\left(  p\otimes q\right)  =pq$ (since
$m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}$ is the
multiplication map of the algebra $\left(  \Sym \left(
V\right)  \right)  ^{o}$). Thus,%
\[
\left(  \underbrace{pq}_{=m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  p\otimes q\right)  },b\right)  _{\Sym %
\left(  V\right)  }=\left(  m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  p\otimes q\right)  ,b\right)  _{\Sym %
\left(  V\right)  }=\left(  p\otimes q,\Delta\left(  b\right)  \right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }%
\]
(by (\ref{sol.Sym.1dim.b.mD}), applied to $a=p\otimes q$). This proves
(\ref{sol.Sym.1dim.b.mD2}).}

On the other hand, the definition of the $\kk$-coalgebra $\left(
\Sym \left(  V\right)  \right)  ^{o}$ shows that the
comultiplication map $\Delta_{\left(  \Sym \left(  V\right)
\right)  ^{o}}$ of the $\kk$-coalgebra $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$ is adjoint to the multiplication map $m$ of
the $\kk$-algebra $\Sym \left(  V\right)  $. In other
words, we have%
\begin{equation}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,b\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }=\left(  a,m\left(  b\right)
\right)  _{\Sym \left(  V\right)  } \label{sol.Sym.1dim.b.Dm}%
\end{equation}
for any $a\in\left(  \Sym \left(  V\right)  \right)  ^{o}$ and
any $b\in\Sym \left(  V\right)  \otimes\Sym %
\left(  V\right)  $. Thus, any $a\in\left(  \Sym \left(
V\right)  \right)  ^{o}$, $p\in\Sym \left(  V\right)  $ and
$q\in\Sym \left(  V\right)  $ satisfy%
\begin{equation}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,p\otimes q\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }=\left(  a,pq\right)
_{\Sym \left(  V\right)  }. \label{sol.Sym.1dim.b.Dm2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.Dm2}):} Let $a\in\left(
\Sym \left(  V\right)  \right)  ^{o}$, $p\in\Sym %
\left(  V\right)  $ and $q\in\Sym \left(  V\right)  $. Then,
$m\left(  p\otimes q\right)  =pq$ (since $m$ is the multiplication map of the
algebra $\Sym \left(  V\right)  $). But (\ref{sol.Sym.1dim.b.Dm}%
) (applied to $b=p\otimes q$) yields%
\[
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,p\otimes q\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }=\left(  a,\underbrace{m\left(
p\otimes q\right)  }_{=pq}\right)  _{\Sym \left(  V\right)
}=\left(  a,pq\right)  _{\Sym \left(  V\right)  }.
\]
This proves (\ref{sol.Sym.1dim.b.Dm2}).}

The definition of the $\kk$-Hopf algebra $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$ shows that the antipode $S_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ of the $\kk$-Hopf
algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ is
adjoint to the antipode $S$ of the $\kk$-Hopf algebra
$\Sym \left(  V\right)  $. In other words, we have%
\begin{equation}
\left(  S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  ,b\right)  _{\Sym \left(  V\right)  }=\left(
a,S\left(  b\right)  \right)  _{\Sym \left(  V\right)  }
\label{sol.Sym.1dim.b.S}%
\end{equation}
for any $a\in\left(  \Sym \left(  V\right)  \right)  ^{o}$ and
$b\in\Sym \left(  V\right)  $.

Now, we have%
\begin{equation}
f^{\left(  u\right)  }f^{\left(  v\right)  }=\dbinom{u+v}{u}f^{\left(
u+v\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\NN \text{ and
}v\in\NN . \label{sol.Sym.1dim.b.1}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.b.1}):} Let $u\in \NN$ and
$v\in \NN$.

Let $k\in \NN$. Then,%
\begin{align}
\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }\right)  \left(
x^{k}\right)   &  =\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }%
,x^{k}\right)  _{\Sym \left(  V\right)  }=\left(  f^{\left(
u\right)  }\otimes f^{\left(  v\right)  },\underbrace{\Delta\left(
x^{k}\right)  }_{\substack{=\sum_{i+j=k}\dbinom{k}{i}x^{i}\otimes
x^{j}\\\text{(by (\ref{sol.Sym.1dim.a.claimDelta}), applied to }n=k\text{)}%
}}\right)  _{\Sym \left(  V\right)  \otimes\Sym %
\left(  V\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.mD2}), applied to
}p=f^{\left(  u\right)  }\text{, }q=f^{\left(  v\right)  }\text{ and }%
b=x^{k}\right) \nonumber\\
&  =\left(  f^{\left(  u\right)  }\otimes f^{\left(  v\right)  },
\sum_{i+j=k}\dbinom{k}{i}x^{i}\otimes x^{j}
\right)  _{\Sym \left(
V\right)  \otimes\Sym \left(  V\right)  }\nonumber\\
&  =\underbrace{\sum_{i+j=k}}_{=\sum\limits_{\substack{\left(  i,j\right)
\in\NN^{2};\\i+j=k}}}\dbinom{k}{i}\underbrace{\left(  f^{\left(
u\right)  }\otimes f^{\left(  v\right)  },x^{i}\otimes x^{j}\right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }}_{\substack{=\left(  f^{\left(  u\right)  },x^{i}\right)
_{\Sym \left(  V\right)  }\left(  f^{\left(  v\right)  }%
,x^{j}\right)  _{\Sym \left(  V\right)  }\\\text{(by the
definition of the}\\\text{bilinear form }\left(  \cdot,\cdot\right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }\text{)}}}\nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2}%
;\\i+j=k}}}_{=\sum_{i\in\NN}\sum\limits_{\substack{j\in\NN ;\\i+j=k}%
}}\dbinom{k}{i}\underbrace{\left(  f^{\left(  u\right)  },x^{i}\right)
_{\Sym \left(  V\right)  }}_{\substack{=f^{\left(  u\right)
}\left(  x^{i}\right)  =\delta_{u,i}\\\text{(by (\ref{sol.Sym.1dim.b.delta}),
applied}\\\text{to }u\text{ and }i\text{ instead of }i\text{ and }j\text{)}%
}}\underbrace{\left(  f^{\left(  v\right)  },x^{j}\right)
_{\Sym \left(  V\right)  }}_{\substack{=f^{\left(  v\right)
}\left(  x^{j}\right)  =\delta_{v,j}\\\text{(by (\ref{sol.Sym.1dim.b.delta}),
applied}\\\text{to }v\text{ instead of }i\text{)}}}\nonumber\\
&  =\sum_{i\in\NN}\sum\limits_{\substack{j\in\NN ;\\i+j=k}}\dbinom{k}%
{i}\delta_{u,i}\delta_{v,j}=\sum\limits_{\substack{j\in\NN ;\\u+j=k}}\dbinom
{k}{u}\underbrace{\delta_{u,u}}_{\substack{=1\\\text{(since }u=u\text{)}%
}}\delta_{v,j}+\sum\limits_{\substack{i\in\NN ;\\i\neq u}}\sum\limits_{\substack{j\in
\NN ;\\i+j=k}}\dbinom{k}{i}\underbrace{\delta_{u,i}}%
_{\substack{=0\\\text{(since }u\neq i\\\text{(since }i\neq u\text{))}}%
}\delta_{v,j}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the }i=u\text{
addend from the sum (since }u\in\NN \text{)}\right) \nonumber\\
&  =\sum\limits_{\substack{j\in\NN ;\\u+j=k}}\dbinom{k}{u}\delta_{v,j}%
+\underbrace{\sum\limits_{\substack{i\in\NN ;\\i\neq u}}\sum\limits_{\substack{j\in
\NN ;\\i+j=k}}\dbinom{k}{i}0\delta_{v,j}}_{=0}=\sum\limits_{\substack{j\in
\NN ;\\u+j=k}}\dbinom{k}{u}\delta_{v,j}=\dbinom{k}{u}
\sum\limits_{\substack{j\in\NN ;\\u+j=k}}\delta_{v,j}.
\label{sol.Sym.1dim.b.1.pf.3}
\end{align}


But
\begin{equation}
\dbinom{k}{u}\delta_{u+v,k}=\dbinom{u+v}{u}\delta_{u+v,k}
\label{sol.Sym.1dim.b.1.pf.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.1.pf.7}):} We must be in one
of the following two cases:
\par
\textit{Case 1:} We have $u+v=k$.
\par
\textit{Case 2:} We have $u+v\neq k$.
\par
Let us first consider Case 1. In this case, we have $u+v=k$. Thus, $k=u+v$, so
that $\dbinom{k}{u}=\dbinom{u+v}{u}$. Hence, $\underbrace{\dbinom{k}{u}%
}_{=\dbinom{u+v}{u}}\delta_{u+v,k}=\dbinom{u+v}{u}\delta_{u+v,k}$. Thus,
(\ref{sol.Sym.1dim.b.1.pf.7}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $u+v\neq k$. Thus,
$\delta_{u+v,k}=0$. Hence, $\dbinom{k}{u}\underbrace{\delta_{u+v,k}}_{=0}=0$
and $\dbinom{u+v}{u}\underbrace{\delta_{u+v,k}}_{=0}=0$. Thus, $\dbinom{k}%
{u}\delta_{u+v,k}=0=\dbinom{u+v}{u}\delta_{u+v,k}$. Therefore,
(\ref{sol.Sym.1dim.b.1.pf.7}) is proven in Case 2.
\par
We have now proven (\ref{sol.Sym.1dim.b.1.pf.7}) in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, we therefore conclude
that (\ref{sol.Sym.1dim.b.1.pf.7}) always holds.}. Thus,
(\ref{sol.Sym.1dim.b.1.pf.3}) becomes%
\[
\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }\right)  \left(
x^{k}\right)  =\dbinom{k}{u}\underbrace{\sum\limits_{\substack{j\in\NN %
;\\u+j=k}}\delta_{v,j}}_{\substack{=\delta_{u+v,k}\\\text{(by
(\ref{sol.Sym.1dim.b.1.pf.5}))}}}=\dbinom{k}{u}\delta_{u+v,k}=\dbinom{u+v}%
{u}\delta_{u+v,k}%
\]
(by (\ref{sol.Sym.1dim.b.1.pf.7})). Comparing this with%
\[
\left(  \dbinom{u+v}{u}f^{\left(  u+v\right)  }\right)  \left(  x^{k}\right)
=\dbinom{u+v}{u}\underbrace{f^{\left(  u+v\right)  }\left(  x^{k}\right)
}_{\substack{=\delta_{u+v,k}\\\text{(by (\ref{sol.Sym.1dim.b.delta}),
applied}\\\text{to }u+v\text{ and }k\text{ instead of }i\text{ and }j\text{)}%
}}=\dbinom{u+v}{u}\delta_{u+v,k},
\]
we obtain $\left(  f^{\left(  u\right)  }f^{\left(  v\right)  }\right)
\left(  x^{k}\right)  =\left(  \dbinom{u+v}{u}f^{\left(  u+v\right)  }\right)
\left(  x^{k}\right)  $.

Let us now forget that we fixed $k$. We thus have shown that $\left(
f^{\left(  u\right)  }f^{\left(  v\right)  }\right)  \left(  x^{k}\right)
=\left(  \dbinom{u+v}{u}f^{\left(  u+v\right)  }\right)  \left(  x^{k}\right)
$ for every $k\in \NN$. In other words, the two $\kk$-linear maps
$f^{\left(  u\right)  }f^{\left(  v\right)  }$ and $\dbinom{u+v}{u}f^{\left(
u+v\right)  }$ are equal to each other on the basis $\left(  x^{k}\right)
_{k\in\NN}$ of the $\kk$-module $\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $f^{\left(  u\right)  }f^{\left(
v\right)  }=\dbinom{u+v}{u}f^{\left(  u+v\right)  }$. Thus,
(\ref{sol.Sym.1dim.b.1}) is proven.

Thus, we have shown that%
\begin{equation}
f^{\left(  i\right)  }f^{\left(  j\right)  }=\dbinom{i+j}{i}f^{\left(
i+j\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\NN \text{ and
}j\in\NN . \label{sol.Sym.1dim.b.1.ij}%
\end{equation}


(In fact, (\ref{sol.Sym.1dim.b.1.ij}) follows from (\ref{sol.Sym.1dim.b.1}) by
renaming the variables $u$ and $v$ as $i$ and $j$.)

We shall now show that
\begin{equation}
\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  =\sum\limits_{\substack{\left(  i,j\right)
\in\NN^{2};\\i+j=n}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\NN . \label{sol.Sym.1dim.b.2}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.b.2}):} Let $n\in \NN$. We identify
$\left(  \Sym \left(  V\right)  \right)  ^{o}\otimes\left(
\Sym \left(  V\right)  \right)  ^{o}$ with $\left(
\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  \right)  ^{o}$ (since $\Sym \left(  V\right)  $ is a
graded $\kk$-module of finite type); thus, an element of $\left(
\Sym \left(  V\right)  \right)  ^{o}\otimes\left(
\Sym \left(  V\right)  \right)  ^{o}$ can be regarded as a
$\kk$-linear map $\Sym \left(  V\right)  \otimes
\Sym \left(  V\right)  \rightarrow\kk$. In particular,
$\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  $ thus becomes a $\kk$-linear map
$\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  \rightarrow\kk$.

Fix $\left(  k,\ell\right)  \in\NN^{2}$. Thus, $k\in \NN$ and
$\ell\in \NN$. Then,%
\begin{align*}
&  \left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right) \\
&  =\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)
^{o}}\left(  f^{\left(  n\right)  }\right)  ,x^{k}\otimes x^{\ell}\right)
_{\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  }\\
&  =\left(  f^{\left(  n\right)  },\underbrace{x^{k}x^{\ell}}_{=x^{k+\ell}%
}\right)  _{\Sym \left(  V\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Sym.1dim.b.Dm2}), applied to }a=f^{\left(  n\right)
}\text{, }p=x^{k}\text{ and }q=x^{\ell}\right) \\
&  =\left(  f^{\left(  n\right)  },x^{k+\ell}\right)  _{\Sym %
\left(  V\right)  }=\delta_{n,k+\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.delta}), applied to }i=n\text{ and }j=k+\ell\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \left(  \sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2}%
;\\i+j=n}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }\right)  \left(
x^{k}\otimes x^{\ell}\right) \\
&  =\left(  \sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2}%
;\\i+j=n}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  },x^{k}\otimes
x^{\ell}\right)  _{\Sym \left(  V\right)  \otimes
\Sym \left(  V\right)  }\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=n}%
}\underbrace{\left(  f^{\left(  i\right)  }\otimes f^{\left(  j\right)
},x^{k}\otimes x^{\ell}\right)  _{\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  }}_{\substack{
=\left(  f^{\left(i\right)}%
,x^{k}\right)  _{\Sym \left(  V\right)  }\left(  f^{\left(
j\right)  },x^{\ell}\right)  _{\Sym \left(  V\right)
}\\\text{(by the definition of the}\\\text{bilinear form }\left(  \cdot
,\cdot\right)  _{\Sym \left(  V\right)  \otimes
\Sym \left(  V\right)  }\text{)}}}\\
&  =\underbrace{\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2}%
;\\i+j=n}}}_{=\sum_{i\in\NN}\sum\limits_{\substack{j\in\NN ;\\i+j=n}%
}}
\underbrace{\left(  f^{\left(i\right)},x^{k}\right)  _{
\Sym \left(
V\right)  }}_{\substack{=f^{\left(  i\right)  }\left(  x^{k}\right)
=\delta_{i,k}\\\text{(by (\ref{sol.Sym.1dim.b.delta}), applied}\\\text{to
}k\text{ instead of }j\text{)}}}\underbrace{\left(  f^{\left(  j\right)
},x^{\ell}\right)  _{\Sym \left(  V\right)  }}%
_{\substack{=f^{\left(  j\right)  }\left(  x^{\ell}\right)  =\delta_{j,\ell
}\\\text{(by (\ref{sol.Sym.1dim.b.delta}), applied}\\\text{to }j\text{ and
}\ell\text{ instead of }i\text{ and }j\text{)}}}\\
&  =\sum_{i\in\NN}\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\delta
_{i,k}\delta_{j,\ell}=\sum\limits_{\substack{j\in\NN ;\\k+j=n}%
}\underbrace{\delta_{k,k}}_{\substack{=1\\\text{(since }k=k\text{)}}%
}\delta_{j,\ell}+\sum\limits_{\substack{i\in\NN ;\\i\neq k}}
\sum\limits_{\substack{j\in\NN ;\\i+j=n}}\underbrace{\delta_{i,k}}%
_{\substack{=0\\\text{(since }i\neq k\text{)}}}\delta_{j,\ell}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the }i=k\text{
addend from the sum (since }k\in\NN \text{)}\right) \\
&  =\sum\limits_{\substack{j\in\NN ;\\k+j=n}}\delta_{j,\ell}+\underbrace{
\sum\limits_{\substack{i\in\NN ;\\i\neq k}}\sum\limits_{\substack{j\in\NN %
;\\i+j=n}}0\delta_{j,\ell}}_{=0}=\sum\limits_{\substack{j\in\NN ;\\k+j=n}%
}\underbrace{\delta_{j,\ell}}_{=\delta_{\ell,j}}=\sum\limits_{\substack{j\in
\NN ;\\k+j=n}}\delta_{\ell,j}=\delta_{k+\ell,n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.1.pf.5}), applied
to }k\text{, }\ell\text{ and }n\text{ instead of }u\text{, }v\text{ and
}k\right) \\
&  =\delta_{n,k+\ell},
\end{align*}
this yields%
\[
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  =\left(  \sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }\right)
\left(  x^{k}\otimes x^{\ell}\right)  .
\]


Let us now forget that we fixed $\left(  k,\ell\right)  $. We thus have shown
that
\[
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  =\left(  \sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }\right)
\left(  x^{k}\otimes x^{\ell}\right)
\]
for every $\left(  k,\ell\right)  \in\NN^{2}$. In other words, the two
$\kk$-linear maps $\Delta_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  n\right)  }\right)  $ and
$\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=n}}f^{\left(
i\right)  }\otimes f^{\left(  j\right)  }$ (from $\Sym \left(
V\right)  \otimes\Sym \left(  V\right)  $ to $\kk$) are
equal to each other on the basis $\left(  x^{k}\otimes x^{\ell}\right)
_{\left(  k,\ell\right)  \in\NN^{2}}$ of the $\kk$-module
$\Sym \left(  V\right)  \otimes\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  f^{\left(
n\right)  }\right)  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^{2};
\\i+j=n}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }$. Thus,
(\ref{sol.Sym.1dim.b.2}) is proven.

Abbreviating the summation sign ``$\sum\limits_{\substack{\left(
i,j\right)  \in\NN^{2};\\i+j=n}}$'' as
``$\sum_{i+j=n}$'', we can rewrite our result
(\ref{sol.Sym.1dim.b.2}) as follows:%
\begin{equation}
\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  =\sum_{i+j=n}f^{\left(  i\right)  }\otimes
f^{\left(  j\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\NN .
\label{sol.Sym.1dim.b.2.re}%
\end{equation}


Finally, let us show that
\begin{equation}
S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  =\left(  -1\right)  ^{n}f^{\left(  n\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\NN . \label{sol.Sym.1dim.b.3}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.b.3}):} Let $n\in \NN$. Let
$k\in \NN$. Then,%
\begin{align*}
\left(  S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)   &  =\left(
S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  ,x^{k}\right)  _{\Sym \left(
V\right)  }=\left(  f^{\left(  n\right)  },\underbrace{S\left(  x^{k}\right)
}_{\substack{=\left(  -1\right)  ^{k}x^{k}\\\text{(by
(\ref{sol.Sym.1dim.a.claimS}), applied}\\\text{to }k\text{ instead of
}n\text{)}}}\right)  _{\Sym \left(  V\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.S}), applied to
}a=f^{\left(  n\right)  }\text{ and }b=x^{k}\right) \\
&  =\left(  f^{\left(  n\right)  },\left(  -1\right)  ^{k}x^{k}\right)
_{\Sym \left(  V\right)  }=f^{\left(  n\right)  }\left(  \left(
-1\right)  ^{k}x^{k}\right) \\
&  =\left(  -1\right)  ^{k}\underbrace{f^{\left(  n\right)  }\left(
x^{k}\right)  }_{\substack{=\delta_{n,k}\\\text{(by
(\ref{sol.Sym.1dim.b.delta}), applied to}\\i=n\text{ and }j=k\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f^{\left(  n\right)
}\text{ is }\kk\text{-linear}\right) \\
&  =\left(  -1\right)  ^{k}\delta_{n,k}.
\end{align*}
Since $\left(  -1\right)  ^{k}\delta_{n,k}=\left(  -1\right)  ^{n}\delta
_{n,k}$\ \ \ \ \footnote{\textit{Proof.} We must be in one of the following
two cases:
\par
\textit{Case 1:} We have $n=k$.
\par
\textit{Case 2:} We don't have $n=k$.
\par
Let us consider Case 1 first. In this case, we have $n=k$. Hence, $k=n$, so
that $\left(  -1\right)  ^{k}=\left(  -1\right)  ^{n}$ and thus
$\underbrace{\left(  -1\right)  ^{k}}_{=\left(  -1\right)  ^{n}}\delta
_{n,k}=\left(  -1\right)  ^{n}\delta_{n,k}$. Thus, $\left(  -1\right)
^{k}\delta_{n,k}=\left(  -1\right)  ^{n}\delta_{n,k}$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we don't have $n=k$. Hence, we have
$n\neq k$. Thus, $\delta_{n,k}=0$. Hence, $\left(  -1\right)  ^{k}%
\underbrace{\delta_{n,k}}_{=0}=0$ and $\left(  -1\right)  ^{n}%
\underbrace{\delta_{n,k}}_{=0}=0$. Thus, $\left(  -1\right)  ^{k}\delta
_{n,k}=0=\left(  -1\right)  ^{n}\delta_{n,k}$. Therefore, $\left(  -1\right)
^{k}\delta_{n,k}=\left(  -1\right)  ^{n}\delta_{n,k}$ is proven in Case 2.
\par
Now, $\left(  -1\right)  ^{k}\delta_{n,k}=\left(  -1\right)  ^{n}\delta_{n,k}$
is proven in each of the two Cases 1 and 2. Since these Cases cover all
possibilities, this yields that $\left(  -1\right)  ^{k}\delta_{n,k}=\left(
-1\right)  ^{n}\delta_{n,k}$ always holds. Qed.}, this becomes%
\[
\left(  S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)  =\left(
-1\right)  ^{n}\delta_{n,k}.
\]
Compared with%
\[
\left(  \left(  -1\right)  ^{n}f^{\left(  n\right)  }\right)  \left(
x^{k}\right)  =\left(  -1\right)  ^{n}\underbrace{f^{\left(  n\right)
}\left(  x^{k}\right)  }_{\substack{=\delta_{n,k}\\\text{(by
(\ref{sol.Sym.1dim.b.delta}), applied to}\\i=n\text{ and }j=k\text{)}%
}}=\left(  -1\right)  ^{n}\delta_{n,k},
\]
this yields $\left(  S_{\left(  \Sym \left(  V\right)  \right)
^{o}}\left(  f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)
=\left(  \left(  -1\right)  ^{n}f^{\left(  n\right)  }\right)  \left(
x^{k}\right)  $.

Let us now forget that we fixed $k$. We thus have shown that $\left(
S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  \right)  \left(  x^{k}\right)  =\left(  \left(
-1\right)  ^{n}f^{\left(  n\right)  }\right)  \left(  x^{k}\right)  $ for
every $k\in \NN$. In other words, the two $\kk$-linear maps
$S_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  n\right)  }\right)  $ and $\left(  -1\right)  ^{n}f^{\left(
n\right)  }$ are equal to each other on the basis $\left(  x^{k}\right)
_{k\in\NN}$ of the $\kk$-module $\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $S_{\left(  \Sym %
\left(  V\right)  \right)  ^{o}}\left(  f^{\left(  n\right)  }\right)
=\left(  -1\right)  ^{n}f^{\left(  n\right)  }$. Thus, (\ref{sol.Sym.1dim.b.3}%
) is proven.

Now, all of the identities (\ref{sol.Sym.1dim.b.1.ij}),
(\ref{sol.Sym.1dim.b.2.re}) and (\ref{sol.Sym.1dim.b.3}) are proven; thus,
Exercise \ref{exe.Sym.1dim}(b) is solved.

Before we start the solution of Exercise \ref{exe.Sym.1dim}(c), let us make a
few more simple observations.

We have
\begin{equation}
\epsilon\left(  x^{k}\right)  =\delta_{k,0}\ \ \ \ \ \ \ \ \ \ \text{for every
}k\in\NN .
\label{sol.Sym.1dim.b.epsi}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.epsi}):} Let $k\in \NN$.
Then, the counit $\epsilon$ of $\Sym \left(  V\right)  $ is a
$\kk$-algebra homomorphism (by the axioms of a $\kk$-bialgebra
(since $\Sym \left(  V\right)  $ is a $\kk$-bialgebra)).
Hence, $\epsilon\left(  x^{k}\right)  =\left(  \underbrace{\epsilon\left(
x\right)  }_{=0}\right)  ^{k}= 0^k = \begin{cases}
1,& \text{if }k=0;\\
0,& \text{if }k\neq0
\end{cases} \quad =\delta_{k,0}$. This proves (\ref{sol.Sym.1dim.b.epsi}).}

The definition of the $\kk$-algebra $\left(  \Sym \left(
V\right)  \right)  ^{o}$ shows that the unity map $u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ of the $\kk%
$-algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ is
adjoint to the counit map $\epsilon$ of the $\kk$-coalgebra
$\Sym \left(  V\right)  $. In other words, we have%
\begin{equation}
\left(  u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  ,b\right)  _{\Sym \left(  V\right)  }=\left(
a,\epsilon\left(  b\right)  \right)  _{\kk} \label{sol.Sym.1dim.b.ue}%
\end{equation}
for any $a\in\kk$ and any $b\in\Sym \left(  V\right)  $.
Thus,%
\begin{equation}
u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  =af^{\left(  0\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }%
a\in\kk. \label{sol.Sym.1dim.b.ue2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.ue2}):} Let $a\in\kk$.
Then, every $b\in\Sym \left(  V\right)  $ satisfies%
\begin{align}
\left(  u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  \right)  \left(  b\right)   &  =\left(  u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  a\right)
,b\right)  _{\Sym \left(  V\right)  }=\left(  a,\epsilon\left(
b\right)  \right)  _{\kk}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.ue})}\right) \nonumber\\
&  =a\cdot\epsilon\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of the form }\left(  \cdot,\cdot\right)  _{\kk}\right)  .
\label{sol.Sym.1dim.b.ue2.pf.1}%
\end{align}
Now, every $k\in \NN$ satisfies%
\begin{align*}
\left(  u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  \right)  \left(  x^{k}\right)   &  =a\cdot\underbrace{\epsilon
\left(  x^{k}\right)  }_{\substack{=\delta_{k,0}\\\text{(by
(\ref{sol.Sym.1dim.b.epsi}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.ue2.pf.1}), applied to }b=x^{k}\right) \\
&  =a\cdot\underbrace{\delta_{k,0}}_{\substack{=f^{\left(  0\right)  }\left(
x^{k}\right)  \\\text{(since (\ref{sol.Sym.1dim.b.delta}) (applied
to}\\i=0\text{ and }j=k\text{) yields }f^{\left(  0\right)  }\left(
x^{k}\right)  =\delta_{0,k}=\delta_{k,0}\text{)}}}=a\cdot f^{\left(  0\right)
}\left(  x^{k}\right)
= \left(  af^{\left(  0\right)  }\right)  \left( x^{k} \right)  .
\end{align*}
In other words, the two $\kk$-linear maps $u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  a\right)  $ and
$af^{\left(  0\right)  }$ are equal to each other on the basis $\left(
x^{k}\right)  _{k\in\NN}$ of the $\kk$-module
$\Sym \left(  V\right)  $. Therefore, these two $\kk%
$-linear maps must be identical (because if two $\kk$-linear maps from
one and the same domain are equal to each other on a basis of this domain,
then these $\kk$-linear maps must be identical). In other words,
$u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  =af^{\left(  0\right)  }$. Thus, (\ref{sol.Sym.1dim.b.ue2}) is
proven.}

The definition of the $\kk$-coalgebra $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$ shows that the counit map $\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ of the $\kk%
$-coalgebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ is
adjoint to the unity map $u$ of the $\kk$-algebra
$\Sym \left(  V\right)  $. In other words, we have%
\begin{equation}
\left(  \epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  a\right)  ,b\right)  _{\kk}=\left(  a,u\left(  b\right)
\right)  _{\Sym \left(  V\right)  } \label{sol.Sym.1dim.b.eu}%
\end{equation}
for any
$a \in \left(\Sym \left( V \right)\right)^o$ and any $b \in \kk$.
Thus,
\begin{equation}
\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  k\right)  }\right)  =\delta_{k,0}\ \ \ \ \ \ \ \ \ \ \text{for
every }k\in\NN . \label{sol.Sym.1dim.b.eu2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Sym.1dim.b.eu2}):} Let $k\in \NN$.
The map $u$ is the unity map of the $\kk$-algebra
$\Sym \left(  V\right)  $; thus,
$u\left(  1\right)  =1_{\Sym \left( V\right)  }=x^{0}$.
\par
The definition of the bilinear form $\left(  \cdot,\cdot\right)  _{\kk%
}$ yields $\left(  \epsilon_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  ,1\right)  _{\kk%
}=\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  k\right)  }\right)  \cdot1=\epsilon_{\left(  \Sym %
\left(  V\right)  \right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  $, so
that%
\begin{align*}
\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
f^{\left(  k\right)  }\right)   &  =\left(  \epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  f^{\left(
k\right)  }\right)  ,1\right)  _{\kk}=\left(  f^{\left(  k\right)
},\underbrace{u\left(  1\right)  }_{=x^{0}}\right)  _{\Sym %
\left(  V\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.eu}), applied to }a=f^{\left(  k\right)  }\text{ and
}b=1\right) \\
&  =\left(  f^{\left(  k\right)  },x^{0}\right)  _{\Sym \left(
V\right)  }=\delta_{k,0}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.delta}), applied to }i=k\text{ and }j=0\right)  .
\end{align*}
This proves (\ref{sol.Sym.1dim.b.eu2}).}

(c) For the time being, let us \textbf{not} assume that $\QQ$ is a
subring of $\kk$. Instead, we shall introduce a map and prove some of
its properties which hold for every $\kk$.

We define a $\kk$-linear map $\Phi:\Sym \left(  V\right)
\rightarrow\left(  \Sym \left(  V\right)  \right)  ^{o}$ by
\begin{equation}
\left(  \Phi\left(  x^{k}\right)  =k!f^{\left(  k\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\NN \right)
.\label{sol.Sym.1dim.c.Phi.def}%
\end{equation}
(This is well-defined, since $\left(  x^{k}\right)  _{k\in\NN}$ is a
basis of the $\kk$-module $\Sym \left(  V\right)  $.) We
can now see that this $\kk$-linear map $\Phi$ satisfies $\Phi\circ
m=m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\circ\left(
\Phi\otimes\Phi\right)  $\ \ \ \ \footnote{\textit{Proof.} Fix $\left(
k,\ell\right)  \in\NN^{2}$. Thus, $k\in \NN$ and $\ell
\in \NN$. Recall that $m$ is the multiplication map of the $\kk%
$-algebra $\Sym \left(  V\right)  $. Thus, $m\left(
x^{k}\otimes x^{\ell}\right)  =x^{k}x^{\ell}=x^{k+\ell}$. Also, $m_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$ is the multiplication map
of the $\kk$-algebra $\left(  \Sym \left(  V\right)
\right)  ^{o}$. Thus, $m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\otimes f^{\left(  \ell\right)
}\right)  =f^{\left(  k\right)  }f^{\left(  \ell\right)  }=\dbinom{k+\ell}%
{k}f^{\left(  k+\ell\right)  }$ (by (\ref{sol.Sym.1dim.b.1}), applied to $u=k$
and $v=\ell$). Now,%
\[
\left(  \Phi\circ m\right)  \left(  x^{k}\otimes x^{\ell}\right)  =\Phi\left(
\underbrace{m\left(  x^{k}\otimes x^{\ell}\right)  }_{=x^{k+\ell}}\right)
=\Phi\left(  x^{k+\ell}\right)  =\left(  k+\ell\right)  !f^{\left(
k+\ell\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.c.Phi.def}), applied to }k+\ell\text{ instead of }k\right)
.
\]
Compared with%
\begin{align*}
\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}%
\circ\left(  \Phi\otimes\Phi\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)   &  =m_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\left(  \underbrace{\left(  \Phi\otimes\Phi\right)  \left(  x^{k}\otimes
x^{\ell}\right)  }_{\substack{=\Phi\left(  x^{k}\right)  \otimes\Phi\left(
x^{\ell}\right)  \\\text{(by the definition of the map }\Phi\otimes
\Phi\text{)}}}\right)  \\
&  =m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
\underbrace{\Phi\left(  x^{k}\right)  }_{\substack{=k!f^{\left(  k\right)
}\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}))}}}\otimes\underbrace{\Phi\left(
x^{\ell}\right)  }_{\substack{=\ell!f^{\left(  \ell\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}),}\\\text{applied to }\ell\text{ instead of
}k\text{)}}}\right)  \\
&  =m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
\underbrace{\left(  k!f^{\left(  k\right)  }\right)  \otimes\left(
\ell!f^{\left(  \ell\right)  }\right)  }_{=k!\ell!f^{\left(  k\right)
}\otimes f^{\left(\ell\right)}}\right)  =m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  k!\ell!f^{\left(  k\right)  }\otimes f^{\left(\ell\right)}\right)  \\
&  =k!\ell!\underbrace{m_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\otimes f^{\left(\ell\right)}\right)
}_{=\dbinom{k+\ell}{k}f^{\left(  k+\ell\right)  }}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }m_{\left(  \Sym \left(  V\right)  \right)
^{o}}\text{ is }\kk\text{-linear}\right)  \\
&  =\underbrace{k!\ell!\dbinom{k+\ell}{k}}_{\substack{=\left(  k+\ell\right)
!\\\text{(since }\dbinom{k+\ell}{k}=\dfrac{\left(  k+\ell\right)  !}{k!\ell
!}\text{)}}}f^{\left(  k+\ell\right)  }=\left(  k+\ell\right)  !f^{\left(
k+\ell\right)  },
\end{align*}
this yields $\left(  \Phi\circ m\right)  \left(  x^{k}\otimes x^{\ell}\right)
=\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}%
\circ\left(  \Phi\otimes\Phi\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  $.
\par
Let us now forget that we fixed $\left(  k,\ell\right)  $. We thus have shown
that $\left(  \Phi\circ m\right)  \left(  x^{k}\otimes x^{\ell}\right)
=\left(  m_{\left(  \Sym \left(  V\right)  \right)  ^{o}}%
\circ\left(  \Phi\otimes\Phi\right)  \right)  \left(  x^{k}\otimes x^{\ell
}\right)  $ for every $\left(  k,\ell\right)  \in\NN^{2}$. In other
words, the two $\kk$-linear maps $\Phi\circ m$ and $m_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\left(  \Phi
\otimes\Phi\right)  $ (from $\Sym \left(  V\right)
\otimes\Sym \left(  V\right)  $ to $\left(  \Sym %
\left(  V\right)  \right)  ^{o}$) are equal to each other on the basis
$\left(  x^{k}\otimes x^{\ell}\right)  _{\left(  k,\ell\right)  \in
\NN^{2}}$ of the $\kk$-module $\Sym \left(
V\right)  \otimes\Sym \left(  V\right)  $. Therefore, these two
$\kk$-linear maps must be identical (because if two $\kk$-linear
maps from one and the same domain are equal to each other on a basis of this
domain, then these $\kk$-linear maps must be identical). In other
words, $\Phi\circ m=m_{\left(  \Sym \left(  V\right)  \right)
^{o}}\circ\left(  \Phi\otimes\Phi\right)  $. Qed.} and $\Phi\circ u=u_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$%
\ \ \ \ \footnote{\textit{Proof.} The map $u$ is the unity map of the
$\kk$-algebra $\Sym \left(  V\right)  $; thus, $u\left(
1\right)  =1_{\Sym \left(  V\right)  }=x^{0}$. Now, every
$a\in\kk$ satisfies%
\begin{align*}
\left(  \Phi\circ u\right)  \left(  \underbrace{a}_{=a\cdot1}\right)   &
=\left(  \Phi\circ u\right)  \left(  a\cdot1\right)  =a\cdot
\underbrace{\left(  \Phi\circ u\right)  \left(  1\right)  }_{=\Phi\left(
u\left(  1\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\Phi\circ u\text{ is }\kk\text{-linear}\right)  \\
&  =a\cdot\Phi\left(  \underbrace{u\left(  1\right)  }_{=x^{0}}\right)
=a\cdot\underbrace{\Phi\left(  x^{0}\right)  }_{\substack{=0!f^{\left(
0\right)  }\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}), applied to
}0\\\text{instead of }k\text{)}}}=a\cdot\underbrace{0!}_{=1}f^{\left(
0\right)  }=af^{\left(  0\right)  }\\
&  =u_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\left(
a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Sym.1dim.b.ue2}%
)}\right)  .
\end{align*}
Thus, $\Phi\circ u=u_{\left(  \Sym \left(  V\right)  \right)
^{o}}$, qed.}. Hence, $\Phi$ is a $\kk$-algebra homomorphism. Also, the
$\kk$-linear map $\Phi$ satisfies $\left(  \Phi\otimes\Phi\right)
\circ\Delta=\Delta_{\left(  \Sym \left(  V\right)  \right)
^{o}}\circ\Phi$\ \ \ \ \footnote{\textit{Proof.} Let $k\in \NN$. We have%
\begin{align*}
\left(  \left(  \Phi\otimes\Phi\right)  \circ\Delta\right)  \left(
x^{k}\right)   &  =\left(  \Phi\otimes\Phi\right)  \left(  \underbrace{\Delta
\left(  x^{k}\right)  }_{\substack{=\sum\limits_{\substack{\left(  i,j\right)
\in\NN^{2};\\i+j=k}}\dbinom{k}{i}x^{i}\otimes x^{j}\\\text{(by
(\ref{sol.Sym.1dim.a.claimDelta.1}), applied to }n=k\text{)}}}\right)
=\left(  \Phi\otimes\Phi\right)  \left(  \sum\limits_{\substack{\left(  i,j\right)
\in\NN^{2};\\i+j=k}}\dbinom{k}{i}x^{i}\otimes x^{j}\right)  \\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}}\dbinom
{k}{i}\underbrace{\left(  \Phi\otimes\Phi\right)  \left(  x^{i}\otimes
x^{j}\right)  }_{\substack{=\Phi\left(  x^{i}\right)  \otimes\Phi\left(
x^{j}\right)  \\\text{(by the definition}\\\text{of the map }\Phi\otimes
\Phi\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Phi\otimes
\Phi\text{ is }\kk\text{-linear}\right)  \\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}}\dbinom
{k}{i}\underbrace{\Phi\left(  x^{i}\right)  }_{\substack{=i!f^{\left(
i\right)  }\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}),}\\\text{applied to
}i\text{ instead of }k\text{)}}}\otimes\underbrace{\Phi\left(  x^{j}\right)
}_{\substack{=j!f^{\left(  j\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}),}\\\text{applied to }j\text{ instead of
}k\text{)}}}\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}}\dbinom
{k}{i}i!f^{\left(  i\right)  }\otimes j!f^{\left(  j\right)  }=
\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}}\dbinom{k}%
{i}i!\underbrace{j!}_{\substack{=\left(  k-i\right)  !\\\text{(since
}j=k-i\\\text{(since }i+j=k\text{))}}}f^{\left(  i\right)  }\otimes f^{\left(
j\right)  }\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}%
}\underbrace{\dbinom{k}{i}i!\left(  k-i\right)  !}%
_{\substack{=k!\\\text{(since }\dbinom{k}{i}=\dfrac{k!}{i!\left(  k-i\right)
!}\text{)}}}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  }%
=\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}}k!f^{\left(
i\right)  }\otimes f^{\left(  j\right)  }=k!\sum\limits_{\substack{\left(
i,j\right)  \in\NN^{2};\\i+j=k}}f^{\left(  i\right)  }\otimes
f^{\left(  j\right)  }.
\end{align*}
Compared with%
\begin{align*}
\left(  \Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi\right)  \left(  x^{k}\right)   &  =\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  \underbrace{\Phi
\left(  x^{k}\right)  }_{\substack{=k!f^{\left(  k\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}))}}}\right)  =\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  k!f^{\left(
k\right)  }\right)  \\
&  =k!\underbrace{\Delta_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  }_{\substack{
=\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}}f^{\left(
i\right)  }\otimes f^{\left(  j\right)  }\\\text{(by (\ref{sol.Sym.1dim.b.2}),
applied to }n=k\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\text{ is
}\kk\text{-linear}\right)  \\
&  =k!\sum\limits_{\substack{\left(  i,j\right)  \in\NN^{2};\\i+j=k}%
}f^{\left(  i\right)  }\otimes f^{\left(  j\right)  },
\end{align*}
this yields $\left(  \left(  \Phi\otimes\Phi\right)  \circ\Delta\right)
\left(  x^{k}\right)  =\left(  \Delta_{\left(  \Sym \left(
V\right)  \right)  ^{o}}\circ\Phi\right)  \left(  x^{k}\right)  $.
\par
Let us now forget that we fixed $k$. We thus have shown that $\left(  \left(
\Phi\otimes\Phi\right)  \circ\Delta\right)  \left(  x^{k}\right)  =\left(
\Delta_{\left(  \Sym \left(  V\right)  \right)  ^{o}}\circ
\Phi\right)  \left(  x^{k}\right)  $ for every $k\in \NN$. In other
words, the two $\kk$-linear maps $\left(  \Phi\otimes\Phi\right)
\circ\Delta$ and $\Delta_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\circ\Phi$ are equal to each other on the basis $\left(
x^{k}\right)  _{k\in\NN}$ of the $\kk$-module
$\Sym \left(  V\right)  $. Therefore, these two $\kk%
$-linear maps must be identical (because if two $\kk$-linear maps from
one and the same domain are equal to each other on a basis of this domain,
then these $\kk$-linear maps must be identical). In other words,
$\left(  \Phi\otimes\Phi\right)  \circ\Delta=\Delta_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\Phi$. Qed.} and
$\epsilon=\epsilon_{\left(  \Sym \left(  V\right)  \right)
^{o}}\circ\Phi$\ \ \ \ \footnote{\textit{Proof.} Let $k\in \NN$. We have%
\begin{align*}
\left(  \epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi\right)  \left(  x^{k}\right)   &  =\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  \underbrace{\Phi
\left(  x^{k}\right)  }_{\substack{=k!f^{\left(  k\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}))}}}\right)  =\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\left(  k!f^{\left(
k\right)  }\right)  \\
&  =k!\underbrace{\epsilon_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\left(  f^{\left(  k\right)  }\right)  }_{\substack{=\delta
_{k,0}\\\text{(by (\ref{sol.Sym.1dim.b.eu2}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\epsilon_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\text{ is }\kk\text{-linear}\right)  \\
&  =k!\delta_{k,0}.
\end{align*}
\par
We shall now show that $k!\delta_{k,0}=\delta_{k,0}$.
\par
Indeed, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $k=0$.
\par
\textit{Case 2:} We don't have $k=0$.
\par
Let us first consider Case 1. In this case, we have $k=0$. Thus, $k!=0!=1$, so
that $\underbrace{k!}_{=1}\delta_{k,0}=\delta_{k,0}$. Therefore,
$k!\delta_{k,0}=\delta_{k,0}$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we don't have $k=0$. Hence, $k\neq
0$, so that $\delta_{k,0}=0$ and thus $k!\underbrace{\delta_{k,0}}%
_{=0}=0=\delta_{k,0}$. Thus, $k!\delta_{k,0}=\delta_{k,0}$ is proven in Case
2.
\par
We have now proven $k!\delta_{k,0}=\delta_{k,0}$ in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this shows that
$k!\delta_{k,0}=\delta_{k,0}$ always holds.
\par
Now,
\[
\left(  \epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi\right)  \left(  x^{k}\right)  =k!\delta_{k,0}=\delta_{k,0}%
=\epsilon\left(  x^{k}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.b.epsi})}\right)  .
\]
In other words, $\epsilon\left(  x^{k}\right)  =\left(  \epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\Phi\right)  \left(
x^{k}\right)  $.
\par
Let us now forget that we fixed $k$. We thus have shown that $\epsilon\left(
x^{k}\right)  =\left(  \epsilon_{\left(  \Sym \left(  V\right)
\right)  ^{o}}\circ\Phi\right)  \left(  x^{k}\right)  $ for every
$k\in \NN$. In other words, the two $\kk$-linear maps $\epsilon$
and $\epsilon_{\left(  \Sym \left(  V\right)  \right)  ^{o}%
}\circ\Phi$ are equal to each other on the basis $\left(  x^{k}\right)
_{k\in\NN}$ of the $\kk$-module $\Sym \left(
V\right)  $. Therefore, these two $\kk$-linear maps must be identical
(because if two $\kk$-linear maps from one and the same domain are
equal to each other on a basis of this domain, then these $\kk$-linear
maps must be identical). In other words, $\epsilon=\epsilon_{\left(
\Sym \left(  V\right)  \right)  ^{o}}\circ\Phi$. Qed.}. Hence,
$\Phi$ is a $\kk$-coalgebra homomorphism. Thus, $\Phi$ is a
$\kk$-bialgebra homomorphism (since $\Phi$ is both a $\kk%
$-algebra homomorphism and a $\kk$-coalgebra homomorphism), and
therefore a Hopf algebra homomorphism (according to
Corollary~\ref{cor.bialg-mor-is-Hopf}, applied to $H_{1}%
=\Sym \left(  V\right)  $, $H_{2}=\left(  \Sym %
\left(  V\right)  \right)  ^{o}$, $S_{1}=S$ and $S_{2}=S_{\left(
\Sym \left(  V\right)  \right)  ^{o}}$). The map $\Phi$ is also
graded\footnote{\textit{Proof.} Let $n\in \NN$. For every graded
$\kk$-module $A$, let $A_{n}$ denote the $n$-th homogeneous component
of $A$. We shall show that $\Phi\left(  \left(  \Sym \left(
V\right)  \right)  _{n}\right)  \subset\left(  \left(  \Sym %
\left(  V\right)  \right)  ^{o}\right)  _{n}$.
\par
Indeed, recall that $\left(  f^{\left(  k\right)  }\right)  _{k\in\NN}$
is a \textbf{graded} basis of the $\kk$-module $\left(
\Sym \left(  V\right)  \right)  ^{o}$. Thus, for every
$m\in \NN$, the one-element family $\left(  f^{\left(  m\right)
}\right)  $ is a basis of the $m$-th homogeneous component $\left(  \left(
\Sym \left(  V\right)  \right)  ^{o}\right)  _{m}$. Applying
this to $m=n$, we conclude that the one-element family $\left(  f^{\left(
n\right)  }\right)  $ is a basis of the $n$-th homogeneous component $\left(
\left(  \Sym \left(  V\right)  \right)  ^{o}\right)  _{n}$.
Thus, this family $\left(  f^{\left(  n\right)  }\right)  $ spans the
$\kk$-module $\left(  \left(  \Sym \left(  V\right)
\right)  ^{o}\right)  _{n}$. In other words, $\left(  \left(
\Sym \left(  V\right)  \right)  ^{o}\right)  _{n}=\kk%
\cdot f^{\left(  n\right)  }$.
\par
On the other hand, recall that $\left(  x^{k}\right)  _{k\in\NN}$ is a
\textbf{graded} basis of the $\kk$-module $\Sym \left(
V\right)  $. Thus, for every $m\in \NN$, the one-element family $\left(
x^{m}\right)  $ is a basis of the $m$-th homogeneous component $\left(
\Sym \left(  V\right)  \right)  _{m}$. Applying this to $m=n$,
we conclude that the one-element family $\left(  x^{n}\right)  $ is a basis of
the $n$-th homogeneous component $\left(  \Sym \left(  V\right)
\right)  _{n}$. Thus, this family $\left(  x^{n}\right)  $ spans the
$\kk$-module $\left(  \Sym \left(  V\right)  \right)
_{n}$. In other words, $\left(  \Sym \left(  V\right)  \right)
_{n}=\kk\cdot x^{n}$. Hence,%
\begin{align*}
\Phi\left(  \underbrace{\left(  \Sym \left(  V\right)  \right)
_{n}}_{=\kk\cdot x^{n}}\right)   &  =\Phi\left(  \kk\cdot
x^{n}\right)  \subset\kk\cdot\underbrace{\Phi\left(  x^{n}\right)
}_{\substack{=n!f^{\left(  n\right)  }\\\text{(by
(\ref{sol.Sym.1dim.c.Phi.def}),}\\\text{applied to }k=n\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Phi\text{ is }%
\kk\text{-linear}\right)  \\
&  =\underbrace{\kk\cdot n!}_{\subset\kk}f^{\left(  n\right)
}\subset\kk\cdot f^{\left(  n\right)  }=\left(  \left(
\Sym \left(  V\right)  \right)  ^{o}\right)  _{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \left(  \Sym %
\left(  V\right)  \right)  ^{o}\right)  _{n}=\kk\cdot f^{\left(
n\right)  }\right)  .
\end{align*}
\par
Now, let us forget that we fixed $n$. We thus have shown that $\Phi\left(
\left(  \Sym \left(  V\right)  \right)  _{n}\right)
\subset\left(  \left(  \Sym \left(  V\right)  \right)
^{o}\right)  _{n}$ for every $n\in \NN$. In other words, the map $\Phi$
is graded, qed.}. Thus, $\Phi$ is a homomorphism of graded $\kk$-Hopf
algebras (since $\Phi$ is graded and a Hopf algebra homomorphism).

Now, let us assume that $\QQ$ is a subring of $\kk$. We define a
$\kk$-linear map $\Psi:\left(  \Sym \left(  V\right)
\right)  ^{o}\rightarrow\Sym \left(  V\right)  $ by
\begin{equation}
\left(  \Psi\left(  f^{\left(  k\right)  }\right)  =\dfrac{x^{k}}%
{k!}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\NN \right)  .
\label{sol.Sym.1dim.c.Psi.def}%
\end{equation}
(This is well-defined, since $\left(  f^{\left(  k\right)  }\right)
_{k\in\NN}$ is a basis of the $\kk$-module $\left(
\Sym \left(  V\right)  \right)  ^{o}$.) Then, $\Psi\circ
\Phi=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} For every
$k\in \NN$, we have%
\begin{align*}
\left(  \Psi\circ\Phi\right)  \left(  x^{k}\right)   &  =\Psi\left(
\underbrace{\Phi\left(  x^{k}\right)  }_{\substack{=k!f^{\left(  k\right)
}\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}))}}}\right)  =\Psi\left(
k!f^{\left(  k\right)  }\right)  =k!\underbrace{\Psi\left(  f^{\left(
k\right)  }\right)  }_{\substack{=\dfrac{x^{k}}{k!}\\\text{(by
(\ref{sol.Sym.1dim.c.Psi.def}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }\Psi\text{ is }\kk\text{-linear}\right) \\
&  =k!\cdot\dfrac{x^{k}}{k!}=x^{k}=\operatorname*{id}\left(  x^{k}\right)  .
\end{align*}
In other words, the two $\kk$-linear maps $\Psi\circ\Phi$ and
$\operatorname*{id}$ are equal to each other on the basis $\left(
x^{k}\right)  _{k\in\NN}$ of the $\kk$-module
$\Sym \left(  V\right)  $. Therefore, these two $\kk%
$-linear maps must be identical (because if two $\kk$-linear maps from
one and the same domain are equal to each other on a basis of this domain,
then these $\kk$-linear maps must be identical). In other words,
$\Psi\circ\Phi=\operatorname*{id}$. Qed.} and $\Phi\circ\Psi
=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} For every $k\in
 \NN$, we have%
\begin{align*}
\left(  \Phi\circ\Psi\right)  \left(  f^{\left(  k\right)  }\right)   &
=\Phi\left(  \underbrace{\Psi\left(  f^{\left(  k\right)  }\right)
}_{\substack{=\dfrac{x^{k}}{k!}\\\text{(by (\ref{sol.Sym.1dim.c.Psi.def}))}%
}}\right)  =\Phi\left(  \underbrace{\dfrac{x^{k}}{k!}}_{=\dfrac{1}{k!}x^{k}%
}\right)  =\Phi\left(  \dfrac{1}{k!}x^{k}\right) \\
&  =\dfrac{1}{k!}\underbrace{\Phi\left(  x^{k}\right)  }%
_{\substack{=k!f^{\left(  k\right)  }\\\text{(by (\ref{sol.Sym.1dim.c.Phi.def}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Phi\text{ is
}\kk\text{-linear}\right) \\
&  =\dfrac{1}{k!}k!f^{\left(  k\right)  }=f^{\left(  k\right)  }%
=\operatorname*{id}\left(  f^{\left(  k\right)  }\right)  .
\end{align*}
In other words, the two $\kk$-linear maps $\Phi\circ\Psi$ and
$\operatorname*{id}$ are equal to each other on the basis $\left(  f^{\left(
k\right)  }\right)  _{k\in\NN}$ of the $\kk$-module $\left(
\Sym \left(  V\right)  \right)  ^{o}$. Therefore, these two
$\kk$-linear maps must be identical (because if two $\kk$-linear
maps from one and the same domain are equal to each other on a basis of this
domain, then these $\kk$-linear maps must be identical). In other
words, $\Phi\circ\Psi=\operatorname*{id}$. Qed.}. Hence, the maps $\Phi$ and
$\Psi$ are mutually inverse. Thus, the map $\Phi$ is invertible, and its
inverse is $\Psi$.

The map $\Phi$ is an invertible homomorphism of graded $\kk$-Hopf
algebras and therefore an isomorphism of graded $\kk$-Hopf algebras
(since every invertible homomorphism of graded $\kk$-Hopf algebras is
an isomorphism of graded $\kk$-Hopf algebras). Hence, the inverse of
$\Phi$ is also an isomorphism of graded $\kk$-Hopf algebras. In other
words, $\Psi$ is an isomorphism of graded $\kk$-Hopf algebras (since
the inverse of $\Phi$ is $\Psi$).

But every $n\in \NN$ satisfies $\Psi\left(  f^{\left(  n\right)
}\right)  =\dfrac{x^{n}}{n!}$ (according to (\ref{sol.Sym.1dim.c.Psi.def}),
applied to $k=n$). Hence, $\Psi$ is the $\kk$-linear map $\left(
\Sym \left(  V\right)  \right)  ^{o}\rightarrow\Sym \left(  V\right)  $
sending $f^{(n)}\mapsto\frac{x^{n}}{n!}$. Thus, the
$\kk$-linear map $\left(  \Sym \left(  V\right)  \right)
^{o}\rightarrow\Sym \left(  V\right)  $ sending $f^{(n)}%
\mapsto\frac{x^{n}}{n!}$ is an isomorphism of graded $\kk$-Hopf
algebras (since $\Psi$ is an isomorphism of graded $\kk$-Hopf
algebras). This solves Exercise \ref{exe.Sym.1dim}(c).

(d) For the time being, let us \textbf{not} assume that $\kk$ is a
field of characteristic $p > 0$.
Instead, let us first prove a formula which does not
require any restrictions on $\kk$.

Namely, let us show that
\begin{equation}
\left(  f^{\left(  1\right)  }\right)  ^{m}=m!f^{\left(  m\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\NN .
\label{sol.Sym.1dim.d.goal}%
\end{equation}


\textit{Proof of (\ref{sol.Sym.1dim.d.goal}):} Let us consider the
$\kk$-linear map $\Phi:\Sym \left(  V\right)
\rightarrow\left(  \Sym \left(  V\right)  \right)  ^{o}$ defined
in our solution to Exercise \ref{exe.Sym.1dim}(c) above. Then, $\Phi$ is a
$\kk$-algebra homomorphism. (In fact, this was proven in our solution
to Exercise \ref{exe.Sym.1dim}(c) above.) Applying
(\ref{sol.Sym.1dim.c.Phi.def}) to $k=1$, we obtain $\Phi\left(  x^{1}\right)
=\underbrace{1!}_{=1}f^{\left(  1\right)  }=f^{\left(  1\right)  }$, so that
$f^{\left(  1\right)  }=\Phi\left(\underbrace{x^1}_{=x}\right)
= \Phi\left(  x\right)  $. Thus, for every
$m\in \NN$, we have%
\begin{align*}
\left(  \underbrace{f^{\left(  1\right)  }}_{=\Phi\left(  x\right)  }\right)
^{m}  &  =\left(  \Phi\left(  x\right)  \right)  ^{m}=\Phi\left(
x^{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\text{ is a
}\kk\text{-algebra homomorphism}\right) \\
&  =m!f^{\left(  m\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Sym.1dim.c.Phi.def}), applied to }k=m\right)  .
\end{align*}
This proves (\ref{sol.Sym.1dim.d.goal}).

Now, let us assume that $\kk$ is a field of characteristic $p>0$. Then,
$p\cdot1_{\kk}=0$. But $\underbrace{p!}_{=\left(  p-1\right)  !p}%
\cdot1_{\kk}=\left(  p-1\right)  !\underbrace{p\cdot1_{\kk}%
}_{=0}=0$. Applying (\ref{sol.Sym.1dim.d.goal}) to $m=p$, we obtain%
\[
\left(  f^{\left(  1\right)  }\right)  ^{p}=p!f^{\left(  p\right)
}=\underbrace{p!\cdot1_{\kk}}_{=0}\cdot f^{\left(  p\right)  }=0.
\]


Now, it remains to show that there can be no Hopf isomorphism $\left(
\Sym \left(  V\right)  \right)  ^{o}\rightarrow\Sym \left(  V\right)  $.
In fact, we can prove something stronger: Namely, there
exists no algebra isomorphism $\left(  \Sym \left(  V\right)
\right)  ^{o}\rightarrow\Sym \left(  V\right)  $. This is because
the algebra $\left(  \Sym \left(  V\right)  \right)  ^{o}$ has a
nonzero nilpotent element (namely, $f^{\left(  1\right)  }$ is nonzero and
satisfies $\left(  f^{\left(  1\right)  }\right)  ^{p}=0$), whereas the
algebra $\Sym \left(  V\right)  $ contains no nonzero nilpotent
elements (in fact, $\Sym \left(  V\right)  \cong\kk%
\left[  x\right]  $ is isomorphic to a polynomial ring over the field
$\kk$, and therefore is an integral domain, which yields that it
contains no nonzero nilpotent elements). This concludes the solution of
Exercise \ref{exe.Sym.1dim}(d).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Sym.wedge}}

\textit{Solution to Exercise~\ref{exe.Sym.wedge}.} (a) Let
$\Delta^{\prime}: \kk \left[ \xx \right]  \rightarrow
 \kk \left[ \xx , \yy \right]  $ be the map sending every
polynomial $f\left(  x_{1},x_{2},...,x_{n}\right)  \in \kk \left[
 \xx \right]  $ to $f\left(  x_{1}+y_{1},x_{2}+y_{2},...,x_{n}
+y_{n}\right)  \in \kk \left[ \xx , \yy \right]  $. Our goal
is to prove that $\Delta^{\prime}=\Delta_{ \Sym \left(  V\right)
}$, where $\Delta_{ \Sym \left(  V\right)  }$ is the usual
coproduct on $ \Sym \left(  V\right)  $ (part of the coalgebra
structure obtained by regarding $ \Sym \left(  V\right)  $ as a
quotient of $T\left(  V\right)  $ as in Exercise
\ref{symmetric-algebra-as-bialgebra-example}), and where we are identifying
$ \Sym \left(  V\right)  $ with $\kk \left[ \xx \right]  $
and $\Sym \left( V \right)  \otimes  \Sym \left( V \right)$
with $\kk \left[ \xx , \yy \right]$ along the isomorphisms
given at the beginning of the exercise.

Notice first that $\Delta_{ \Sym \left(  V\right)  }$ is a
$\kk$-algebra homomorphism $ \Sym \left(  V\right)
\rightarrow \Sym \left( V \right)  \otimes \Sym \left( V \right)$
(because the axioms of a $\kk$-bialgebra require
that the coproduct of any $\kk$-bialgebra $A$ is a $\kk$-algebra
homomorphism $A\rightarrow A\otimes A$). On the other hand, the map
$\Delta^{\prime}$ is a $\kk$-algebra homomorphism\footnote{In fact, by
its very definition, it is an \emph{evaluation homomorphism}, meaning a map
from a polynomial ring to a commutative $\kk$-algebra $A$ which sends
every polynomial $f$ to the evaluation $f\left(  a_{1},a_{2},...,a_{\ell
}\right)  $ for some given tuple $\left(  a_{1},a_{2},...,a_{\ell}\right)  $
of elements of $A$. Such maps are always $\kk$-algebra homomorphisms.}.
Hence, the equality that we are trying to prove, namely $\Delta^{\prime
}=\Delta_{ \Sym \left(  V\right)  }$, is an equality between two
$\kk$-algebra homomorphisms. It is well-known that in order to prove
such an equality, it is enough to verify it on a generating set of the domain
of these homomorphisms\footnote{In our case, the domain is $\kk \left[
 \xx \right]  $.}; i.e., it is enough to pick out a generating set of its
domain, and check that for every element $s$ of the generating set, the images
of $s$ under the two sides of the equality are equal to each other. In our
case, the $\kk$-algebra homomorphisms $\Delta^{\prime}$ and
$\Delta_{ \Sym \left(  V\right)  }$ have domain $\kk \left[ \xx \right]$,
and as a generating set of this $\kk$-algebra $\kk \left[ \xx \right]$
we can pick the set
$\left\{  x_{1},x_{2},...,x_{n}\right\}  $. We then have to check that for
every element $s$ of this generating set, the images of $s$ under
$\Delta^{\prime}$ and $\Delta_{ \Sym \left(  V\right)  }$ are
equal to each other.

So let $s\in\left\{  x_{1},x_{2},...,x_{n}\right\}  $ be arbitrary. Then,
$s=x_{i}$ for some $i\in\left\{  1,2,...,n\right\}  $. Consider this $i$. We
have $x_{i}\in V$, and thus $x_{i}$ is a primitive element of $T\left(
V\right)  $ (by the definition of the coalgebra structure on $T\left(
V\right)  $). This shows that $\Delta_{T\left(  V\right)  }\left(
x_{i}\right)  =1\otimes x_{i}+x_{i}\otimes1$. Since $\Sym \left( V \right)$
is a quotient bialgebra of $T\left(  V\right)  $, this
shows that $\Delta_{ \Sym \left(  V\right)  }\left(
x_{i}\right)  =1\otimes x_{i}+x_{i}\otimes1$ as well. Under our identification
of $ \Sym \left(  V\right)  \otimes \Sym \left(
V\right)  $ with $\kk \left[ \xx , \yy \right]  $, the
element $1\otimes x_{i}$ of $ \Sym \left(  V\right)
\otimes \Sym \left(  V\right)  $ equals the element $y_{i}$ of
$\kk \left[ \xx , \yy \right]  $, and the element
$x_{i}\otimes1$ of $ \Sym \left(  V\right)  \otimes
 \Sym \left(  V\right)  $ equals the element $x_{i}$ of
$\kk \left[ \xx , \yy \right]  $. Hence, $\Delta
_{ \Sym \left(  V\right)  }\left(  x_{i}\right)  =1\otimes
x_{i}+x_{i}\otimes1$ rewrites as $\Delta_{ \Sym \left(  V\right)
}\left(  x_{i}\right)  =y_{i}+x_{i}$.

But by the definition of $\Delta^{\prime}$, we have $\Delta^{\prime}\left(
x_{i}\right)  =x_{i}\left(  x_{1}+y_{1},x_{2}+y_{2},...,x_{n}+y_{n}\right)
=x_{i}+y_{i}=y_{i}+x_{i}=\Delta_{ \Sym \left(  V\right)  }\left(
x_{i}\right)  $.
In other words,
$\Delta^{\prime}\left( s\right) = \Delta_{ \Sym \left(  V\right)  }\left(
s\right)  $ (since $s = x_i$).
Hence, the images of $s$ under $\Delta^{\prime}$ and
$\Delta_{ \Sym \left(  V\right)  }$ are equal to each other. This
is exactly what we needed to prove, and so the solution to Exercise
\ref{exe.Sym.wedge}(a) is complete.

We will solve the remaining parts of the exercise starting with (c), then
continuing with (d) and finally solving (b). This order has the advantage of
requiring the least amount of work.

(c) We introduce a few notations. As long as we are using the topologist's
sign convention, an element $a$ of a graded $\kk$-module is said to be
\dfn{odd}\index{odd element of a graded module}
if $a$ is a sum of homogeneous elements of odd degree, and an
element $b$ of a graded $\kk$-module is said to be
\dfn{even}\index{even element of a graded module}
if $b$
is a sum of homogeneous elements of even degree. By assumption, every element
of $V$ is odd. Notice that $0$ is both even and odd, and a graded
$\kk$-module can (in general) contain vectors which are neither even
nor odd.

It is easy to see that if $P$ and $Q$ are two graded $\kk$-modules, and
$p\in P$ and $q\in Q$ are two elements, then
\begin{align}
T\left(  p\otimes q\right)   &  =q\otimes p\ \ \ \ \ \ \ \ \ \ \text{if
}p\text{ is even and }q\text{ is even;}
\label{sol.Sym.wedge.c.00}\\
T\left(  p\otimes q\right)   &  =q\otimes p\ \ \ \ \ \ \ \ \ \ \text{if
}p\text{ is even and }q\text{ is odd;}
\label{sol.Sym.wedge.c.01}\\
T\left(  p\otimes q\right)   &  =q\otimes p\ \ \ \ \ \ \ \ \ \ \text{if
}p\text{ is odd and }q\text{ is even;}
\label{sol.Sym.wedge.c.10}\\
T\left(  p\otimes q\right)   &  =-q\otimes p\ \ \ \ \ \ \ \ \ \ \text{if
}p\text{ is odd and }q\text{ is odd.}
\label{sol.Sym.wedge.c.11}
\end{align}
\footnote{\textit{Proof.} We will only prove (\ref{sol.Sym.wedge.c.11}); the
other three identities are analogous.
\par
Let $P$ and $Q$ be two graded $\kk$-modules. Let $p\in P$ and $q\in Q$.
Assume that $p$ is odd and $q$ is odd. Since $p$ is odd, we know that $p$ is a
sum of homogeneous elements of $P$ of odd degree; in other words, we can write
$p$ as $p=\sum_{i \in I}p_{i}$, where $I$ is a finite set and each
$i \in I$ has $p_{i}\in P$ homogeneous of odd degree.
Similarly, we can write $q$ as $q=\sum_{j \in J}q_{j}$, where $J$ is a finite
set and each $j \in J$ has $q_{j}\in Q$ homogeneous of odd degree.
Using these $p_{i}$ and $q_{j}$, we now find
\begin{align*}
T\left(  \underbrace{p}_{=\sum_{i \in I} p_{i}}
\otimes\underbrace{q}_{=\sum_{j \in J}q_{j}}\right)
& = T\left(  \left(  \sum_{i \in I}p_{i}\right) 
 \otimes\left(  \sum_{j \in J}q_{j}\right)  \right)
=T\left(  \sum _{i \in I}\sum_{j \in J} p_{i}\otimes q_{j}\right) \\
&  =\sum_{i \in I}\sum_{j \in J}
\underbrace{T\left(  p_{i}\otimes q_{j}\right)
}_{\substack{
=\left(  -1\right)  ^{\deg\left(  p_{i}\right)  \cdot\deg\left(
q_{j}\right)  }q_{j}\otimes p_{i}
\\
\text{(by the definition of the} \\ \text{topologist's twist map $T$)}}}
=\sum_{i \in I}\sum_{j \in J}
\underbrace{\left( -1\right)
 ^{\deg\left(  p_{i}\right)  \cdot\deg\left(  q_{j}\right)  }}
_{\substack{=-1\\\text{(since }\deg\left( p_i \right) \text{ and}
\\\deg\left( q_j \right) \text{ are odd} \\
\text{(because }p_i\text{ and }q_j \text{ are}\\
\text{homogeneous of odd degree))}}}q_{j}\otimes p_{i}\\
&  =\sum_{i \in I}\sum_{j \in J}\left(  -1\right)  q_{j}\otimes p_{i}
=-\sum_{i \in I}\sum_{j \in J}q_{j}\otimes p_{i}
  =-\underbrace{\left(  \sum_{j \in J}
q_{j}\right)  }_{=q}\otimes\underbrace{\left(  \sum_{i \in I}
p_{i}\right)  }_{=p}=-q\otimes p.
\end{align*}
This proves (\ref{sol.Sym.wedge.c.11}).}

Now, let $x\in V$. Then, $x$ is odd (since every element of $V$ is odd). The
definition of $\Delta$ enforces that $\Delta$ is a $\kk$-bialgebra
homomorphism $T\left(  V\right)  \rightarrow T\left(  V\right)  \otimes
T\left(  V\right)  $, but the meaning of this depends on the
$\kk$-algebra $T\left(  V\right)  \otimes T\left(  V\right)  $,
and therefore on the twist map $T$, because the multiplication map of
the $\kk$-algebra $T\left(  V\right)  \otimes T\left(  V\right)  $ is
\[
m_{T\left(  V\right)  \otimes T\left(  V\right)  }=\left(  m_{T\left(
V\right)  }\otimes m_{T\left(  V\right)  }\right)  \circ\left(
 \id \otimes T\otimes \id \right)  .
\]
The fact that we used the topologist's sign convention
\eqref{topologist-twist} in
this definition means that this twist map $T$ is as in
\eqref{topologist-twist}. Thus, (\ref{sol.Sym.wedge.c.00}),
(\ref{sol.Sym.wedge.c.01}), (\ref{sol.Sym.wedge.c.10}) and
(\ref{sol.Sym.wedge.c.11}) apply. Since $x$ is odd, we can apply
(\ref{sol.Sym.wedge.c.11}) to $P=T\left(  V\right)  $, $Q=T\left(  V\right)
$, $p=x$ and $q=x$, and obtain $T\left(  x\otimes x\right)  =-x\otimes x$.
Thus, in the $\kk$-algebra $T\left(  V\right)  \otimes T\left(
V\right)  $, we have%
\begin{align*}
&  \left(  1\otimes x\right)  \cdot\left(  x\otimes1\right) \\
&  =\underbrace{m_{T\left(  V\right)  \otimes T\left(  V\right)  }}_{=\left(
m_{T\left(  V\right)  }\otimes m_{T\left(  V\right)  }\right)  \circ\left(
 \id \otimes T\otimes \id \right)  }
\underbrace{\left(  \left(  1\otimes x\right)  \otimes\left(  x\otimes
1\right)  \right)  }_{=1\otimes x\otimes x\otimes1}\\
&  =\left(  \left(  m_{T\left(  V\right)  }\otimes m_{T\left(  V\right)
}\right)  \circ\left(   \id \otimes T\otimes \id %
\right)  \right)  \left(  1\otimes x\otimes x\otimes1\right) \\
&  =\left(  m_{T\left(  V\right)  }\otimes m_{T\left(  V\right)  }\right)
\left(  \underbrace{\left(   \id \otimes T\otimes
 \id \right)  \left(  1\otimes x\otimes x\otimes1\right)
}_{\substack{=1\otimes T\left(  x\otimes x\right)  \otimes1\\=-1\otimes
x\otimes x\otimes1\\\text{(since }T\left(  x\otimes x\right)  =-x\otimes
x\text{)}}}\right) \\
&  =\left(  m_{T\left(  V\right)  }\otimes m_{T\left(  V\right)  }\right)
\left(  -1\otimes x\otimes x\otimes1\right)  =-\underbrace{m_{T\left(
V\right)  }\left(  1\otimes x\right)  }_{=1\cdot x=x}\otimes
\underbrace{m_{T\left(  V\right)  }\left(  x\otimes1\right)  }_{=x\cdot1=x}\\
&  =-x\otimes x.
\end{align*}
Similarly, we can compute%
\begin{align*}
\left(  1\otimes x\right)  \cdot\left(  1\otimes x\right)   &  =1\otimes
x^{2},\\
\left(  x\otimes1\right)  \cdot\left(  1\otimes x\right)   &  =x\otimes x,\\
\left(  x\otimes1\right)  \cdot\left(  x\otimes1\right)   &  =x^{2}\otimes1.
\end{align*}
(Since $1\in T\left(  V\right)  $ is even, we have to use
(\ref{sol.Sym.wedge.c.10}), (\ref{sol.Sym.wedge.c.00}) and respectively
(\ref{sol.Sym.wedge.c.01}) instead of (\ref{sol.Sym.wedge.c.11}) here; thus,
we incur no negative signs.)

Since $\Delta$ is a $\kk$-algebra homomorphism, we have%
\begin{align*}
\Delta\left(  x^{2}\right)   &  =\left(  \underbrace{\Delta\left(  x\right)
}_{\substack{=1\otimes x+x\otimes1\\\text{(since }x\text{ is primitive in
}T\left(  V\right)  \text{)}}}\right)  ^{2}=\left(  1\otimes x+x\otimes
1\right)  ^{2}\\
&  =\left(  1\otimes x+x\otimes1\right)  \cdot\left(  1\otimes x+x\otimes
1\right) \\
&  =\underbrace{\left(  1\otimes x\right)  \cdot\left(  1\otimes x\right)
}_{=1\otimes x^{2}}+\underbrace{\left(  1\otimes x\right)  \cdot\left(
x\otimes1\right)  }_{=-x\otimes x}+\underbrace{\left(  x\otimes1\right)
\cdot\left(  1\otimes x\right)  }_{=x\otimes x}+\underbrace{\left(
x\otimes1\right)  \cdot\left(  x\otimes1\right)  }_{=x^{2}\otimes1}\\
&  =1\otimes x^{2}+\left(  -x\otimes x\right)  +x\otimes x+x^{2}
\otimes1=1\otimes x^{2}+x^{2}\otimes1.
\end{align*}
This solves Exercise \ref{exe.Sym.wedge}(c).

(d) Let us first check that the ideal\footnote{By ``ideal'',
we always mean ``two-sided ideal'', unless we explicitly say
``left ideal'' or ``right ideal''.}
$J$ of $T\left(  V\right)  $ is a graded $\kk$-submodule of $T\left(
V\right)  $. This is \textbf{not} obvious! We know that
\begin{equation}
\left(
\begin{array}[c]{c}
\text{if an ideal }I\text{ of a graded } \kk \text{-algebra }A\text{ is
generated by homogeneous}\\
\text{elements, then }I\text{ is a graded } \kk \text{-submodule of }A
\end{array}
\right)  .
\label{sol.Sym.wedge.d.general}
\end{equation}
\footnote{This is a well-known fact in the case when $A$ is commutative (and
the topologist's sign convention is not used), but it is proven exactly the
same way in the general case.} So we should concentrate on showing that the
ideal $J$ is generated by homogeneous elements.

In fact, we know that $J$ is the ideal of $T\left(  V\right)  $ generated by
$\left\{  x^{2}\right\}  _{x\in V}$. Thus,%
\begin{equation}
x^{2}\in J\ \ \ \ \ \ \ \ \ \ \text{for every }x\in V.
\label{sol.Sym.wedge.d.1}
\end{equation}
Also, any two elements $x$ and $y$ of $V$ satisfy%
\begin{align}
xy+yx  &  =\underbrace{\left(  x+y\right)  ^{2}}_{\substack{\in J\\\text{(by
(\ref{sol.Sym.wedge.d.1}), applied}\\\text{to }x+y\text{ instead of }%
x\text{)}}}-\underbrace{x^{2}}_{\substack{\in J\\\text{(by
(\ref{sol.Sym.wedge.d.1}))}}}-\underbrace{y^{2}}_{\substack{\in J\\\text{(by
(\ref{sol.Sym.wedge.d.1}), applied}\\\text{to }y\text{ instead of }x\text{)}%
}}\nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  x+y\right)  ^{2}%
=x^{2}+xy+yx+y^{2}\right) \nonumber\\
&  \in J-J-J\subset J
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }J\text{ is a }
 \kk \text{-module}\right)  .
\label{sol.Sym.wedge.d.2}
\end{align}
Now, define a subset $G$ of $T\left(  V\right)  $ by%
\begin{align}
G  &  =\left\{  x^{2}\ \mid\ x\in V;\ x\text{ is homogeneous}\right\}
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cup\left\{  xy+yx\ \mid\ x\in V;\ y\in V;\ x\text{ and
}y\text{ are homogeneous}\right\}  .
\label{sol.Sym.wedge.d.G}
\end{align}
It is clear that all elements of $G$ are homogeneous elements of $T\left(
V\right)  $. By the definition of $G$, we have%
\begin{equation}
x^{2}\in G\ \ \ \ \ \ \ \ \ \ \text{for every homogeneous }x\in V,
\label{sol.Sym.wedge.d.Gxx}%
\end{equation}
and for the same reason we have%
\begin{equation}
xy+yx\in G\ \ \ \ \ \ \ \ \ \ \text{for any two homogeneous elements }x\text{
and }y\text{ of }V. \label{sol.Sym.wedge.d.Gxy}%
\end{equation}


We shall now show that the ideal generated by $G$ is $J$.

Indeed, let $g\in G$ be arbitrary. We will now show that $g\in J$. We have%
\begin{align*}
g\in G  &  =\left\{  x^{2}\ \mid\ x\in V;\ x\text{ is homogeneous}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \cup\left\{  xy+yx\ \mid\ x\in V;\ y\in V;\ x\text{ and
}y\text{ are homogeneous}\right\}  .
\end{align*}
Hence, either $g$ has the form $g=x^{2}$ for some homogeneous $x\in V$, or $g$
has the form $g=xy+yx$ for two homogeneous elements $x$ and $y$ of $V$. In the
first of these two cases, it is clear that $g=x^{2}\in J$ (by
(\ref{sol.Sym.wedge.d.1})). In the second of these cases, we have $g=xy+yx\in
J$ (by (\ref{sol.Sym.wedge.d.2})). Hence, we have shown that $g\in J$ in
either case. This proves that $g\in J$.

Now, forget that we fixed $g$. We have thus proven that $g\in J$ for every
$g\in G$. Thus, $G\subset J$. In other words, $J$ contains $G$ as a subset.
Since $J$ is an ideal of $T\left(  V\right)  $, we thus have%
\begin{align}
J  &  \supset\left(  \text{the smallest ideal of }T\left(  V\right)  \text{
containing }G\text{ as a subset}\right) \nonumber\\
&  =\left(  \text{the ideal of }T\left(  V\right)  \text{ generated by
}G\right)  . \label{sol.Sym.wedge.d.G.onedir}%
\end{align}


Let us now show the reverse inclusion. Let $x\in V$. Then, $x$ is a sum of
homogeneous elements of $V$ (because every element of a graded
$\kk$-module is a sum of homogeneous elements).
In other words, we can write $x$ in the form $x = \sum_{i=1}^{\ell} x_{i}$,
where $\ell \in \NN$, and where $x_1, x_2, \ldots, x_{\ell}$ are homogeneous
elements of $V$.
Consider this $\ell$ and these $x_1, x_2, \ldots, x_{\ell}$.
Set $I = \left\{ 1, 2, \ldots, \ell \right\}$.
Thus, we have the following equality of summation signs:
$\sum_{i \in I} = \sum_{i \in \left\{ 1, 2, \ldots, \ell \right\}}
= \sum_{i=1}^{\ell}$.
Hence, the equality $x = \sum_{i=1}^{\ell} x_{i}$ rewrites as
$x = \sum_{i \in I} x_{i}$.
Squaring both sides of the equality $x = \sum_{i \in I} x_i$,
we obtain
\begin{vershort}
\begin{align*}
x^{2} &  =\left(  \sum_{i\in I}x_{i}\right)  ^{2}=\sum_{\left(  i,j\right)
\in I^{2}}x_{i}x_{j}=\underbrace{\sum\limits_{\substack{\left(  i,j\right)  \in
I^{2};\\i<j}}x_{i}x_{j}}_{\substack{=\sum\limits_{\substack{\left(  i,j\right)  \in
I^{2};\\i>j}}x_{j}x_{i}\\\text{(here, we substituted }\left(  j,i\right)
\\\text{for }\left(  i,j\right)  \text{ in the sum)}}}+\sum\limits_{\substack{\left(
i,j\right)  \in I^{2};\\i=j}}\underbrace{x_{i}x_{j}}_{\substack{=x_{j}%
^{2}\\\text{(since }i=j\text{)}}}+\sum\limits_{\substack{\left(  i,j\right)  \in
I^{2};\\i>j}}x_{i}x_{j}\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i>j}}x_{j}x_{i}%
+\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i=j}}x_{j}^{2}+
\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i>j}}x_{i}x_{j}=
\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i=j}}x_{j}^{2}+
\underbrace{\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i>j}}x_{i}x_{j}
+\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i>j}}x_{j}x_{i}}_{=
\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i>j}}\left(  x_{i}x_{j}+x_{j}%
x_{i}\right)  }\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i=j}}\underbrace{x_{j}%
^{2}}_{\substack{\in G\\\text{(by (\ref{sol.Sym.wedge.d.Gxx}),}\\\text{applied
to }x_{j}\text{ instead of }x\text{)}}}+\sum\limits_{\substack{\left(  i,j\right)
\in I^{2};\\i>j}}\underbrace{\left(  x_{i}x_{j}+x_{j}x_{i}\right)
}_{\substack{\in G\\\text{(by (\ref{sol.Sym.wedge.d.Gxy}), applied}\\\text{to
}x_{i}\text{ and }x_{j}\text{ instead of }x\text{ and }y\text{)}}}\\
&  \in\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i=j}}G+
\sum\limits_{\substack{\left(  i,j\right)  \in I^{2};\\i>j}} G
\subset\left(  \text{the ideal of }T\left(  V\right)
\text{ generated by }G\right)  .
\end{align*}
\end{vershort}
\begin{verlong}
\begin{align*}
x^{2}  &  =\left(  \sum_{i\in I} x_{i}\right)  ^{2}
=\left(  \sum_{i\in I} x_{i}\right)  \cdot\left(  \sum_{i\in I} x_{i}\right)
=\left(  \sum_{i\in I} x_{i}\right)  \cdot\left(  \sum_{j\in I} x_{j}\right)
\\
&  =\underbrace{\sum_{i\in I }\sum_{j\in I }}_{=\sum_{\left(
i,j\right)  \in I ^{2}}}x_{i}x_{j}=\sum_{\left(  i,j\right)
\in I ^{2}}x_{i}x_{j}=\underbrace{\sum\limits_{\substack{\left(  i,j\right)
\in I ^{2};\\i<j}}x_{i}x_{j}}_{\substack{=\sum\limits_{\substack{\left(
j,i\right)  \in I ^{2};\\j<i}}x_{j}x_{i}\\\text{(here, we renamed
the}\\\text{summation index }\left(  i,j\right)  \text{ as }\left(
i,j\right)  \text{)}}}+\sum\limits_{\substack{\left(  i,j\right)  \in I %
^{2};\\i=j}}\underbrace{x_{i}}_{\substack{=x_{j}\\\text{(since }i=j\text{)}%
}}x_{j}+\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i>j}%
}x_{i}x_{j}\\
&  =\underbrace{\sum\limits_{\substack{\left(  j,i\right)  \in I ^{2};\\j<i}%
}}_{=\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i>j}}}x_{j}%
x_{i}+\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i=j}%
}\underbrace{x_{j}x_{j}}_{=x_{j}^{2}}+\sum\limits_{\substack{\left(  i,j\right)
\in I ^{2};\\i>j}}x_{i}x_{j}\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i>j}}x_{j}%
x_{i}+\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i=j}}x_{j}%
^{2}+\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i>j}}x_{i}%
x_{j}=\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i=j}}x_{j}%
^{2}+\underbrace{\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2}%
;\\i>j}}x_{i}x_{j}+\sum\limits_{\substack{\left(  i,j\right)  \in I %
^{2};\\i>j}}x_{j}x_{i}}_{=\sum\limits_{\substack{\left(  i,j\right)  \in
 I ^{2};\\i>j}}\left(  x_{i}x_{j}+x_{j}x_{i}\right)  }\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i=j}%
}\underbrace{x_{j}^{2}}_{\substack{\in G\\\text{(by (\ref{sol.Sym.wedge.d.Gxx}%
),}\\\text{applied to }x_{j}\text{ instead of }x\text{)}}}+
\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i>j}}
\underbrace{\left(
x_{i}x_{j}+x_{j}x_{i}\right)  }_{\substack{\in G\\\text{(by
(\ref{sol.Sym.wedge.d.Gxy}), applied}\\\text{to }x_{i}\text{ and }x_{j}\text{
instead of }x\text{ and }y\text{)}}}\\
&  \in\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i=j}%
}G+\sum\limits_{\substack{\left(  i,j\right)  \in I ^{2};\\i>j}}G\subset
\left(  \text{the } \kk \text{-submodule of }T\left(  V\right)  \text{
generated by }G\right) \\
&  \subset\left(  \text{the ideal of }T\left(  V\right)  \text{ generated by
}G\right)  .
\end{align*}
\end{verlong}
Now, forget that we fixed $x$. We thus have shown that $x^{2}\in\left(
\text{the ideal of }T\left(  V\right)  \text{ generated by }G\right)  $ for
every $x\in V$. Hence, $\left(  \text{the ideal of }T\left(  V\right)  \text{
generated by }G\right)  $ contains the elements $x^{2}$ for all $x\in V$.
Since $\left(  \text{the ideal of }T\left(  V\right)  \text{ generated by
}G\right)  $ is an ideal, this yields that%
\begin{align*}
&  \left(  \text{the ideal of }T\left(  V\right)  \text{ generated by
}G\right)  \\
&  \supset\left(  \text{the smallest ideal of }T\left(  V\right)  \text{
containing the elements }x^{2}\text{ for all }x\in V\right)  \\
&  =\left(  \text{the ideal generated by }\left\{  x^{2}\right\}  _{x\in
V}\right)  =J.
\end{align*}
Combined with (\ref{sol.Sym.wedge.d.G.onedir}), this yields%
\[
J=\left(  \text{the ideal of }T\left(  V\right)  \text{ generated by
}G\right)  .
\]
The ideal $J$ is thus generated by $G$. Thus, the ideal $J$ of $T\left(
V\right)  $ is generated by homogeneous elements of $T\left(  V\right)  $
(since all elements of $G$ are homogeneous elements of $T\left(  V\right)  $).
Therefore, (\ref{sol.Sym.wedge.d.general}) (applied to $J$ and $T\left(
V\right)  $ instead of $I$ and $A$) yields that $J$ is a graded $\kk %
$-submodule of $T\left(  V\right)  $.
Hence, the quotient $T\left(  V\right)  / J$ is a graded
$\kk$-module.

Now, let us show that $\Delta\left(  J\right)  \subset J\otimes T\left(
V\right)  +T\left(  V\right)  \otimes J$.

If $A$ and $B$ are two graded $\kk$-algebras (in the topologist's
sense) and $P$ and $Q$ are two ideals of $A$ and $B$ which are graded
$\kk$-submodules of $A$ and $B$, then $P\otimes Q$ is an ideal of
$A\otimes B$\ \ \ \ \footnote{\textit{Proof.} Since $P$ is an ideal of $A$, we
see that $P$ is a $\kk$-submodule of $A$ satisfying $m_{A}\left(
P\otimes A\right)  \subset P$ and $m_{A}\left(  A\otimes P\right)  \subset P$.
\par
Since $Q$ is an ideal of $B$, we see that $Q$ is a $\kk$-submodule of
$B$ satisfying $m_{B}\left(  Q\otimes B\right)  \subset Q$ and $m_{B}\left(
B\otimes Q\right)  \subset Q$.
\par
But since $Q$ is a graded $\kk$-submodule of $B$, there is a
topologist's twist map $T:Q\otimes A\rightarrow A\otimes Q$, which is the
restriction of the twist map $T:B\otimes A\rightarrow A\otimes B$. Thus,
$T\left(  Q\otimes A\right)  \subset A\otimes Q$. Now,%
\begin{align*}
&  \underbrace{m_{A\otimes B}}_{=\left(  m_{A}\otimes m_{B}\right)
\circ\left(  \id \otimes T \otimes \id \right)
}\left(  \underbrace{\left(  P\otimes Q\right)  \otimes\left(  A\otimes
B\right)  }_{=P\otimes Q\otimes A\otimes B}\right) \\
&  =\left(  \left(  m_{A}\otimes m_{B}\right)  \circ\left(  \id
\otimes T \otimes \id \right)  \right)  \left(  P\otimes Q\otimes
A\otimes B\right) \\
&  =\left(  m_{A}\otimes m_{B}\right)  \left(  \underbrace{\left(
 \id \otimes T \otimes \id \right)  \left(  P\otimes
Q\otimes A\otimes B\right)  }_{\substack{=P\otimes T\left(  Q\otimes A\right)
\otimes B\\\subset P\otimes A\otimes Q\otimes B\\\text{(since }T\left(
Q\otimes A\right)  \subset A\otimes Q\text{)}}}\right) \\
&  \subset \left(  m_{A}\otimes m_{B}\right)  \left(  P\otimes A\otimes Q\otimes
B\right)  =\underbrace{m_{A}\left(  P\otimes A\right)  }_{\subset P}%
\otimes\underbrace{m_{B}\left(  Q\otimes B\right)  }_{\subset Q}\subset
P\otimes Q.
\end{align*}
Similarly, $m_{A\otimes B}\left(  \left(  A\otimes B\right)  \otimes\left(
P\otimes Q\right)  \right)  \subset P\otimes Q$ (but here, we need to use
$T\left(  B\otimes P\right)  \subset P\otimes B$ instead of $T\left(  Q\otimes
A\right)  \subset A\otimes Q$). These two inclusions prove that $P\otimes Q$
is an ideal of $A\otimes B$, qed.
\par
We could have also proven this by working with elements, but that way we
would have to take care of the fact that the twist $T$ is the topologist's one
and comes with signs. The way we have done it, we were almost entirely
untroubled by this fact.}. Applying this to $A=T\left(  V\right)  $,
$B=T\left(  V\right)  $, $P=J$ and $Q=T\left(  V\right)  $, we conclude that
$J\otimes T\left(  V\right)  $ is an ideal of $T\left(  V\right)  \otimes
T\left(  V\right)  $. Similarly, $T\left(  V\right)  \otimes J$ is an ideal of
$T\left(  V\right)  \otimes T\left(  V\right)  $ as well. The sum $J\otimes
T\left(  V\right)  +T\left(  V\right)  \otimes J$ of these two ideals
therefore is an ideal of $T\left(  V\right)  \otimes T\left(  V\right)  $,
too. As a consequence, $\Delta^{-1}\left(  J\otimes T\left(  V\right)
+T\left(  V\right)  \otimes J\right)  $ is an ideal of $T\left(  V\right)  $
(because $\Delta$ is a $\kk$-algebra homomorphism\footnote{by the
definition of $\Delta$}, and the preimage of an ideal under a
$\kk$-algebra homomorphism is always an ideal).

Now, every $x\in V$ satisfies
\begin{align*}
\Delta\left(  x^{2}\right)   &  =\underbrace{1}_{\in T\left(  V\right)
}\otimes\underbrace{x^{2}}_{\substack{\in J\\\text{(by
(\ref{sol.Sym.wedge.d.1}))}}}+\underbrace{x^{2}}_{\substack{\in J\\\text{(by
(\ref{sol.Sym.wedge.d.1}))}}}\otimes\underbrace{1}_{\in T\left(  V\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.Sym.wedge}(c)}\right)
\\
&  \in T\left(  V\right)  \otimes J+J\otimes T\left(  V\right)  =J\otimes
T\left(  V\right)  +T\left(  V\right)  \otimes J,
\end{align*}
so that $x^{2}\in\Delta^{-1}\left(  J\otimes T\left(  V\right)  +T\left(
V\right)  \otimes J\right)  $. So the ideal $\Delta^{-1}\left(  J\otimes
T\left(  V\right)  +T\left(  V\right)  \otimes J\right)  $ contains the
elements $x^{2}$ for all $x\in V$. Hence,
\begin{align*}
\Delta^{-1}\left(  J\otimes T\left(  V\right)  +T\left(  V\right)  \otimes
J\right)   &  \supset\left(  \text{the smallest ideal which contains the
elements }x^{2}\text{ for all }x\in V\right) \\
&  =\left(  \text{the ideal generated by }\left\{  x^{2}\right\}  _{x\in
V}\right)  =J.
\end{align*}


Thus, $\Delta\left(  J\right)  \subset J\otimes T\left(  V\right)  +T\left(
V\right)  \otimes J$.

\begin{vershort}
It remains to prove that $\epsilon\left(  J\right)  =0$. This is similar to
the above argument but much simpler. Since $\epsilon$ is a $\kk %
$-algebra homomorphism, its kernel $\ker\epsilon$ is an ideal
of $T\left(  V\right)  $. Since $\epsilon$ is a $\kk$-algebra
homomorphism, every $x\in V$ satisfies $\epsilon\left(  x^{2}\right)  =\left(
\underbrace{\epsilon\left(  x\right)  }_{=0}\right)  ^{2}=0$ and thus
$x^{2}\in\ker\epsilon$. Thus, the ideal $\ker\epsilon$ contains
the elements $x^{2}$ for all $x\in V$. Hence,
\end{vershort}

\begin{verlong}
It remains to prove that $\epsilon\left(  J\right)  =0$. This is similar to
the above argument but much simpler. Since $\epsilon$ is a $\kk %
$-algebra homomorphism, its kernel $\ker\epsilon$ is an ideal
of $T\left(  V\right)  $. Every $x\in V$ satisfies
\begin{align*}
\epsilon\left(  x^{2}\right)   &  =\left(  \underbrace{\epsilon\left(
x\right)  }_{=0}\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\epsilon\text{ is a } \kk \text{-algebra homomorphism}\right) \\
&  =0
\end{align*}
and thus $x^{2}\in\ker\epsilon$. Thus, the ideal $\ker\epsilon$
contains the elements $x^{2}$ for all $x\in V$. Hence,
\end{verlong}
\begin{align*}
\ker\epsilon &  \supset\left(  \text{the smallest ideal which
contains the elements }x^{2}\text{ for all }x\in V\right) \\
&  =\left(  \text{the ideal generated by }\left\{  x^{2}\right\}  _{x\in
V}\right)  =J.
\end{align*}


This yields $\epsilon\left(  J\right)  =0$. Combined with $\Delta\left(
J\right)  \subset J\otimes T\left(  V\right)  +T\left(  V\right)  \otimes J$,
this shows that $J$ is a two-sided coideal of $T\left(  V\right)  $. Since $J$
is also a two-sided ideal and a graded $\kk$-submodule,
this shows that the quotient $T\left(  V\right)
/ J$ inherits a graded $\kk$-bialgebra structure from $T\left(
V\right)  $. This quotient $T\left(  V\right)  / J$ is $\wedge V$, and
so we obtain a graded $\kk$-bialgebra structure on $\wedge V$. The
$\kk$-bialgebra $\wedge V$ obtained this way is graded (because it is
the quotient of the graded $\kk$-bialgebra $T\left(  V\right)  $ by the
graded ideal $J$) and connected (since its $0$-th graded component is
$\wedge^{0}V= \kk$\ \ \ \ \footnote{Here we are using the fact that
$V_{0}=0$ (which is a consequence of the fact that $V$ is concentrated in odd
degrees).}), therefore a Hopf algebra (by Proposition
\ref{graded-connected-bialgebras-have-antipodes}\footnote{Here it helps to
notice that our use of the topologist's sign convention does not invalidate
the proof of Proposition \ref{graded-connected-bialgebras-have-antipodes}; in
fact, no changes are necessary to that proof! (This might not be too
surprising, given that said proof made no use of the bialgebra axioms
(\ref{bialgebra-diagrams}).)}). Thus, $\wedge V$ is a connected graded
Hopf algebra, and therefore the comultiplication $\Delta_{\wedge V}$ of
$\wedge V$ is part of a connected graded Hopf algebra structure on
$\wedge V$. In other words, the comultiplication $\Delta_{\wedge V}$ makes
the $\kk$-algebra $\wedge V$ into a connected graded Hopf algebra.

Let us recall that the comultiplication $\Delta_{T\left(  V\right)  }$ of the
$\kk$-bialgebra $T\left(  V\right)  $ satisfies
\[
\Delta_{T\left(  V\right)  }\left(  x\right)  =1\otimes x+x\otimes
1\ \ \ \ \ \ \ \ \ \ \text{for every }x\in V
\]
(by the definition of $\Delta_{T\left(  V\right)  }$). Since the
$\kk$-bialgebra $\wedge V$ was defined as a quotient of
$T\left(  V\right)  $, we can project this equality down on
$\left(  \wedge V\right)  \otimes\left(  \wedge V\right)  $,
and thus conclude that the comultiplication $\Delta_{\wedge V}$
of the $\kk$-bialgebra $\wedge V$ satisfies
\begin{equation}
\Delta_{\wedge V}\left(  x\right)  =1\otimes x+x\otimes
1\ \ \ \ \ \ \ \ \ \ \text{for every }x\in V.
\label{sol.Sym.wedge.d.comul}
\end{equation}
Thus, every $i\in\left\{  1,2,...,n\right\}  $ satisfies
\begin{equation}
\Delta_{\wedge V}\left(  x_{i}\right)  =1\otimes x_{i}+x_{i}\otimes1
\label{sol.Sym.wedge.d.comul2}
\end{equation}
(by (\ref{sol.Sym.wedge.d.comul}), applied to $x=x_{i}$). But our
identification of $\left(  \wedge V\right)  \otimes\left(  \wedge V\right)  $
with $\wedge\left(  V\oplus V\right)  $ equates $1\otimes x_{i}$ with $y_{i}$,
and equates $x_{i}\otimes1$ with $x_{i}$ for every $i\in\left\{
1,2,...,n\right\}  $. Hence, for every $i\in\left\{  1,2,...,n\right\}  $, the
equality (\ref{sol.Sym.wedge.d.comul2}) rewrites as
\begin{equation}
\Delta_{\wedge V}\left(  x_{i}\right)  =\underbrace{1\otimes x_{i}}_{=y_{i}%
}+\underbrace{x_{i}\otimes1}_{=x_{i}}=y_{i}+x_{i}=x_{i}+y_{i}
\label{sol.Sym.wedge.d.comul3}%
\end{equation}
in $\left(  \wedge V\right)  \otimes\left(  \wedge V\right)  =\wedge\left(
V\oplus V\right)  $.

Recall that we made $\wedge V$ into a $\kk$-bialgebra by viewing it as
the quotient $T\left(  V\right)  / J$. Now, it remains to prove that the
coproduct on $\wedge V$ which is part of this $\kk$-bialgebra structure
on $T\left(  V\right)  $ is the same as the one defined in Exercise
\ref{exe.Sym.wedge}(b)\footnote{We should be careful because we have not yet
proven that the latter coproduct satisfies the axioms for a coproduct (we have
not solved Exercise \ref{exe.Sym.wedge}(b) yet).}. In order to do so, we shall
prove the following claim: If $\Delta_{\wedge V}$ denotes the coproduct on
$\wedge V$ obtained by regarding $\wedge V$ as the quotient $\kk %
$-bialgebra $T\left(  V\right)  / J$, then
\begin{equation}
\Delta_{\wedge V}\left(  \sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}%
}\wedge\cdots\wedge x_{i_{d}}\right)  =\sum_{i_{1}<\cdots<i_{d}}%
c_{i_{1},...,i_{d}}\left(  x_{i_{1}}+y_{i_{1}}\right)  \wedge\cdots
\wedge\left(  x_{i_{d}}+y_{i_{d}}\right)  \label{sol.Sym.wedge.d.bigclaim}%
\end{equation}
for every family $\left(  c_{i_{1},...,i_{d}}\right)  _{i_{1}<\cdots<i_{d}}$
of elements of $\kk$ indexed by the strictly increasing sequences
$\left(  i_{1}<\cdots<i_{d}\right)  $ of elements of $\left\{
1,2,...,n\right\}  $.

\textit{Proof of (\ref{sol.Sym.wedge.d.bigclaim}):} Let $\left(
c_{i_{1},...,i_{d}}\right)  _{i_{1}<\cdots<i_{d}}$ be a family of elements of
$\kk$ indexed by the strictly increasing sequences $\left(
i_{1}<\cdots<i_{d}\right)  $ of elements of $\left\{  1,2,...,n\right\}  $.
The multiplication in $\wedge V$ is given by the wedge product, so that we
have $x_{i_{1}}\cdots x_{i_{d}}=x_{i_{1}}\wedge\cdots\wedge x_{i_{d}}$ for
every strictly increasing sequence $\left(  i_{1}<\cdots<i_{d}\right)  $ of
elements of $\left\{  1,2,...,n\right\}  $.

But the comultiplication $\Delta_{\wedge V}$ is a $\kk$-algebra
homomorphism (by the axioms of a $\kk$-bialgebra, since $\wedge V$ is a
$\kk$-bialgebra), and thus we have%
\begin{align*}
\Delta_{\wedge V}\left(  \sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}%
}\cdots x_{i_{d}}\right)   &  =\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}%
}\underbrace{\Delta_{\wedge V}\left(  x_{i_{1}}\right)
}_{\substack{=x_{i_{1}}+y_{i_{1}}\\\text{(by (\ref{sol.Sym.wedge.d.comul3}%
),}\\\text{applied to }i=i_{1}\text{)}}}\cdots\underbrace{
\Delta_{\wedge V}\left(  x_{i_{d}}\right)  }_{\substack{=x_{i_{d}}+y_{i_{d}%
}\\\text{(by (\ref{sol.Sym.wedge.d.comul3}),}\\\text{applied to }%
i=i_{d}\text{)}}}\\
&  =\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}\underbrace{\left(  x_{i_{1}%
}+y_{i_{1}}\right)  \cdots\left(  x_{i_{d}}+y_{i_{d}}\right)  }%
_{\substack{=\left(  x_{i_{1}}+y_{i_{1}}\right)  \wedge\cdots\wedge\left(
x_{i_{d}}+y_{i_{d}}\right)  \\\text{(since multiplication in } \wedge
\left( V \oplus V \right) \\
\text{is given by the wedge product)}}}\\
&  =\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}\left(  x_{i_{1}}+y_{i_{1}%
}\right)  \wedge\cdots\wedge\left(  x_{i_{d}}+y_{i_{d}}\right)  .
\end{align*}
Since $x_{i_{1}}\cdots x_{i_{d}}=x_{i_{1}}\wedge\cdots\wedge x_{i_{d}}$ for
every strictly increasing sequence $\left(  i_{1}<\cdots<i_{d}\right)  $ of
elements of $\left\{  1,2,...,n\right\}  $, this rewrites as follows:%
\[
\Delta_{\wedge V}\left(  \sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}%
}\wedge\cdots\wedge x_{i_{d}}\right)  =\sum_{i_{1}<\cdots<i_{d}}%
c_{i_{1},...,i_{d}}\left(  x_{i_{1}}+y_{i_{1}}\right)  \wedge\cdots
\wedge\left(  x_{i_{d}}+y_{i_{d}}\right)  .
\]
Thus, (\ref{sol.Sym.wedge.d.bigclaim}) is proven.

\begin{vershort}
Now, the equality (\ref{sol.Sym.wedge.d.bigclaim}) shows that the map
$\Delta_{\wedge V}$ satisfies the defining property of the map
(\ref{eq.exe.Sym.wedge.b.map}) (namely, mapping every $\sum_{i_{1}%
<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}}\wedge\cdots\wedge x_{i_{d}}$ to
$\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}\left(  x_{i_{1}}+y_{i_{1}%
}\right)  \wedge\cdots\wedge\left(  x_{i_{d}}+y_{i_{d}}\right)  $). Hence,
there exists a map satisfying this property. Such a map is furthermore unique
(because the defining property determines its value on every element of the
form $\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}}\wedge\cdots\wedge
x_{i_{d}}$, but every element of $\wedge V$ can be written in this form), and
therefore the map (\ref{eq.exe.Sym.wedge.b.map}) is well-defined. This map is
our map $\Delta_{\wedge V}$ (because our map $\Delta_{\wedge V}$ satisfies the
defining property of the map (\ref{eq.exe.Sym.wedge.b.map})), and therefore
makes $\wedge V$ into a connected graded Hopf algebra (since we know that the
comultiplication $\Delta_{\wedge V}$ makes the $\kk$-algebra $\wedge V$
into a connected graded Hopf algebra). This solves
Exercise~\ref{exe.Sym.wedge}(b).
\end{vershort}

\begin{verlong}
Now, the equality (\ref{sol.Sym.wedge.d.bigclaim}) shows that the map
$\Delta_{\wedge V}$ satisfies the defining property of the map
(\ref{eq.exe.Sym.wedge.b.map}) (namely, mapping every $\sum_{i_{1}%
<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}}\wedge\cdots\wedge x_{i_{d}}$ to
$\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}\left(  x_{i_{1}}+y_{i_{1}%
}\right)  \wedge\cdots\wedge\left(  x_{i_{d}}+y_{i_{d}}\right)  $). Hence,
there exists a map satisfying this property. Such a map is furthermore
unique\footnote{In fact, the defining property of the map
(\ref{eq.exe.Sym.wedge.b.map}) determines the value of this map on every
element of the form $\sum_{i_{1}<\cdots<i_{d}}c_{i_{1},...,i_{d}}x_{i_{1}%
}\wedge\cdots\wedge x_{i_{d}}$ with $\left(  c_{i_{1},...,i_{d}}\right)
_{i_{1}<\cdots<i_{d}}$ being a family of elements of $\kk$ indexed by
the strictly increasing sequences $\left(  i_{1}<\cdots<i_{d}\right)  $ of
elements of $\left\{  1,2,...,n\right\}  $. But every element of $\wedge V$ is
of this form, and therefore the value of this map on every element of $\wedge
V$ is determined by the defining property. Thus, the map is uniquely
determined.}. Therefore, there is exactly one map satisfying the defining
property of the map (\ref{eq.exe.Sym.wedge.b.map}). Thus, the map
(\ref{eq.exe.Sym.wedge.b.map}) is well-defined. This map is our map
$\Delta_{\wedge V}$ (because our map $\Delta_{\wedge V}$ satisfies the
defining property of the map (\ref{eq.exe.Sym.wedge.b.map})), and therefore
makes $\wedge V$ into a connected graded Hopf algebra (since we know that
the comultiplication $\Delta_{\wedge V}$ makes
the $\kk$-algebra $\wedge V$ into a connected graded Hopf algebra).
This solves Exercise~\ref{exe.Sym.wedge}(b).
\end{verlong}

But the coproduct on $\wedge V$ inherited from $T\left(  V\right)  $ is
$\Delta_{\wedge V}$, and as we know, this $\Delta_{\wedge V}$ is exactly the
map (\ref{eq.exe.Sym.wedge.b.map}), i.e., the coproduct defined in Exercise
\ref{exe.Sym.wedge}(b). Hence, the coproduct on $\wedge V$ inherited from
$T\left(  V\right)  $ is the coproduct defined in Exercise \ref{exe.Sym.wedge}%
(b). Thus, Exercise \ref{exe.Sym.wedge}(d) is also solved. The solution to
Exercise \ref{exe.Sym.wedge} is thus complete.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution.dual}}

\textit{Solution to Exercise~\ref{exe.convolution.dual}.} Define a
$\kk$-linear map $\rho_{U,V}:U^{\ast}\otimes V^{\ast}\rightarrow\left(
U\otimes V\right)  ^{\ast}$ for any two $\kk$-modules $U$ and $V$ as in
Exercise~\ref{exe.dual-algebra}. Recall that this $\rho_{U,V}$ is a
$\kk$-module isomorphism if both $U$ and $V$ are finite free. Hence,
$\rho_{A,A}$ is a $\kk$-module isomorphism.

Also, basic linear algebra shows that if $U$, $V$, $U^{\prime}$ and
$V^{\prime}$ are four $\kk$-modules and if $\alpha:U\rightarrow
U^{\prime}$ and $\beta:V\rightarrow V^{\prime}$ are two $\kk$-linear
maps, then
\[
\rho_{U,V}\circ\left(  \alpha^{\ast}\otimes\beta^{\ast}\right)  =\left(
\alpha\otimes\beta\right)  ^{\ast}\circ\rho_{U^{\prime},V^{\prime}}.
\]
This (applied to $U=C$, $V=C$, $U^{\prime}=A$, $V^{\prime}=A$, $\alpha=f$ and
$\beta=g$) yields
\[
\rho_{C,C}\circ\left(  f^{\ast}\otimes g^{\ast}\right)  =\left(  f\otimes
g\right)  ^{\ast}\circ\rho_{A,A}.
\]


The definition of convolution yields both
\begin{equation}
f\star g=m_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}
\label{sol.convolution.dual.1}
\end{equation}
and
\begin{align*}
f^{\ast}\star g^{\ast}  & =\underbrace{m_{C^{\ast}}}_{\substack{=\Delta
_{C}^{\ast}\circ\rho_{C,C}\\
\text{(by the definition of }m_{C^{\ast}}\text{)}}}
\circ\left(  f^{\ast}\otimes g^{\ast}\right)  \circ\underbrace{\Delta
_{A^{\ast}}}_{\substack{=\rho_{A,A}^{-1}\circ m_{A}^{\ast}\\\text{(by the
definition of }\Delta_{A^{\ast}}\text{)}}}\\
& =\Delta_{C}^{\ast}\circ\underbrace{\rho_{C,C}\circ\left(  f^{\ast}\otimes
g^{\ast}\right)  }_{=\left(  f\otimes g\right)  ^{\ast}\circ\rho_{A,A}}
\circ\rho_{A,A}^{-1}\circ m_{A}^{\ast}
=\Delta_{C}^{\ast}\circ\left(  f\otimes g\right)  ^{\ast}\circ
\underbrace{\rho_{A,A}\circ\rho_{A,A}^{-1}}
_{=\id_{\left(A\otimes A\right)^{\ast}}}\circ m_{A}^{\ast} \\
& =\Delta_{C}^{\ast}\circ\left(  f\otimes g\right)  ^{\ast
}\circ m_{A}^{\ast}
=\left( \underbrace{m_{A}\circ\left( f\otimes g\right) \circ\Delta_{C}}
_{\substack{=f\star g\\\text{(by \eqref{sol.convolution.dual.1})}}}\right)
^{\ast}=\left(  f\star g\right)  ^{\ast}.
\end{align*}
This solves Exercise~\ref{exe.convolution.dual}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.shuffle-alg}}

\textit{Solution to Exercise~\ref{exe.shuffle-alg}.}
Our goal is to prove Proposition~\ref{prop.shuffle-alg}.

Let $m_{\shuffle}$ denote the $\kk$-linear map $T\left(  V\right)
\otimes T\left(  V\right)  \rightarrow T\left(  V\right)  $ which sends every
$a\otimes b$ to $a \shufmult b$. Let $u$ denote the unit map of
$T\left(  V\right)  $ (that is, the $\kk$-linear map $\kk
\rightarrow T\left(  V\right)  $ sending $1_\kk$ to $1_{T\left(
V\right)  }$). Let $S$ denote the antipode of the Hopf algebra $T\left(
V\right)  $. Then, the result that we have to prove boils down to the
statement that the $\kk$-module $T\left(  V\right)  $, endowed with the
multiplication $m_{\shuffle}$, the unit $u$, the comultiplication
$\Delta_{\shuffle}$ and the counit $\epsilon$, becomes a commutative
Hopf algebra with the
antipode $S$. Since we already know that $m_{\shuffle}$, $u$, $\Delta
_{\shuffle}$, $\epsilon$ and $S$ are $\kk$-linear, this latter
statement will immediately follow once we can show that the following diagrams commute:

\begin{itemize}
\item the diagrams \eqref{associativity-diagram} and \eqref{unit-diagram},
with $A$ and $m$ replaced by $T\left(  V\right)  $ and $m_{\shuffle}$;

\item the diagrams \eqref{coassociativity-diagram} and \eqref{counit-diagram},
with $C$ and $\Delta$ replaced by $T\left(  V\right)  $ and $\Delta
_{\shuffle}$;

\item the diagrams \eqref{bialgebra-diagrams}, with $A$, $m$ and $\Delta$
replaced by $T\left(  V\right)  $, $m_{\shuffle}$ and $\Delta_{\shuffle}$;

\item the diagram \eqref{antipode-diagram}, with $A$, $m$ and $\Delta$
replaced by $T\left(  V\right)  $, $m_{\shuffle}$ and $\Delta_{\shuffle}$;

\item the diagram \eqref{commutativity-diagram}, with $A$ and $m$ replaced
by $T\left(V\right)$ and $m_{\shuffle}$.
\end{itemize}

Out of all these statements, we will only prove the commutativity of the first
diagram in \eqref{bialgebra-diagrams}, since all other diagrams are similar
but only easier.

So we must prove the commutativity of the first diagram in
\eqref{bialgebra-diagrams}, with $A$, $m$ and $\Delta$ replaced by $T\left(
V\right)$, $m_{\shuffle}$ and $\Delta_{\shuffle}$. In other words, we must
show that the diagram
\begin{equation}
\xymatrix{ & T\left(V\right) \otimes T\left(V\right) \ar[dl]_{\Delta_\shuffle \otimes \Delta_\shuffle} \ar[ddr]_{m_\shuffle} & \\ T\left(V\right) \otimes T\left(V\right) \otimes T\left(V\right) \otimes T\left(V\right) \ar[dd]_{\id \otimes T \otimes \id} & & \\ & & T\left(V\right) \ar[ddl]_{\Delta_\shuffle}\\ T\left(V\right) \otimes T\left(V\right) \otimes T\left(V\right) \otimes T\left(V\right) \ar[dr]_{m_\shuffle \otimes m_\shuffle} & & \\ & T\left(V\right) \otimes T\left(V\right) & }
\label{pf.shuffle-alg.diagram-goal}
\end{equation}
commutes, where $T : T\left(  V\right)  \otimes T\left(  V\right)  \rightarrow
T\left(  V\right)  \otimes T\left(  V\right)  $ is the twist map
$T_{T\left(V\right), T\left(V\right)}$ sending every
$a\otimes b$ to $b\otimes a$ (and being $\kk$-linear at that). Let us
prove this now.

We need to prove that the diagram \eqref{pf.shuffle-alg.diagram-goal}
commutes. In other words, we need to prove the identity
\begin{equation}
\left(  m_{\shuffle}\otimes m_{\shuffle}\right)  \circ\left(
\id \otimes T\otimes\id \right)  \circ\left(
\Delta_{\shuffle}\otimes\Delta_{\shuffle}\right)  =\Delta_{\shuffle}\circ
m_{\shuffle}.
\label{pf.shuffle-alg.diagram-goal.eq}
\end{equation}
By linearity, it is clearly enough to verify this identity only on the pure
tensors in $T\left(  V\right)  \otimes T\left(  V\right)  $; that is, it is
enough to check that every $a\in T\left(  V\right)  $ and $b\in T\left(
V\right)  $ satisfy
\begin{equation}
\left(  \left(  m_{\shuffle}\otimes m_{\shuffle}\right)  \circ\left(
\id \otimes T\otimes\id \right)  \circ\left(
\Delta_{\shuffle}\otimes\Delta_{\shuffle}\right)  \right)  \left(  a\otimes
b\right)  =\left(  \Delta_{\shuffle}\circ m_{\shuffle}\right)  \left(
a\otimes b\right)  .
\label{pf.shuffle-alg.diagram-goal.eq2}
\end{equation}
So let $a\in T\left(  V\right)  $ and $b\in T\left(  V\right)  $ be arbitrary.
All we need now is to prove \eqref{pf.shuffle-alg.diagram-goal.eq2}. By
linearity again, we can WLOG assume that $a$ and $b$ have the form
$a=v_{1}v_{2} \cdots v_{p}$ and $b=v_{p+1}v_{p+2} \cdots v_{p+q}$ for some
$p\in\NN$, $q\in\NN$ and $v_{1},v_{2}, \ldots ,v_{p+q}\in V$ (since
$T\left(  V\right)$ is spanned as a $\kk$-module by pure tensors).
Assume this, and define $W$ to be the free $\kk$-module with basis
$\left\{  x_{1},x_{2}, \ldots ,x_{p+q}\right\}  $. Let $A$ be the tensor algebra
$T\left(  W\right)  $ of this $\kk$-module $W$. Then, $W$ is a finite
free $\kk$-module, and so we know from Example \ref{exa.shuffle-alg}
(applied to $W$ instead of $V$) that the graded dual $A^{o}$ of its tensor
algebra $A=T\left(  W\right)  $ is a Hopf algebra whose basis $\left\{
y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right\}  $ is indexed by words
in the alphabet $I:=\left\{  1,2, \ldots ,p+q\right\}  $.

Notice that $\left\{  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\otimes
y_{\left(  j_{1},j_{2}, \ldots ,j_{m}\right)  }\right\}  _{\ell\in\NN
,\ m\in\NN ,\ \left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  \in I^{\ell
},\ \left(  j_{1},j_{2}, \ldots ,j_{m}\right)  \in I^{m}}$ is a $\kk$-module
basis of $A^{o}\otimes A^{o}$ (since $\left\{  y_{\left(  i_{1},i_{2}
, \ldots ,i_{\ell}\right)  }\right\}  $ is a basis of $A^{o}$). Relabelling this
basis, we see that
\[
\left\{  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell
}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell+m}\right)
}\right\}  _{\ell\in\NN ,\ m\in\NN ,\ \left(  i_{1},i_{2}
, \ldots ,i_{\ell+m}\right)  \in I^{\ell+m}}
\text{ is a $\kk$-module basis of $A^{o}\otimes A^{o}$.}
\]

We can define a $\kk$-linear map $\phi:A^{o}\rightarrow T\left(
V\right)  $ by setting
\[
\phi\left(  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right) 
= v_{i_{1}}v_{i_{2}} \cdots v_{i_{\ell}}\ \ \ \ \ \ \ \ \ \ 
\text{for every }\ell \in \NN \text{ and }
\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  \in I^{\ell}
\]
(because $\left\{  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right\}  $ is
a basis of $A^{o}$). Consider this map $\phi$.

Notice that the definition of $\phi$ yields $\phi\left(  y_{\left(
1,2, \ldots ,p\right)  }\right) = v_1 v_2 \cdots v_p = a$, and similarly $\phi\left(
y_{\left(  p+1,p+2, \ldots ,p+q\right)  }\right)  =b$. Thus,
\begin{align*}
\underbrace{a}_{=\phi\left(  y_{\left(  1,2, \ldots ,p\right)  }\right)  }
\otimes\underbrace{b}_{=\phi\left(  y_{\left(  p+1,p+2, \ldots ,p+q\right)
}\right)  }  &  =\phi\left(  y_{\left(  1,2, \ldots ,p\right)  }\right)
\otimes\phi\left(  y_{\left(  p+1,p+2, \ldots ,p+q\right)  }\right) \\
&  =\left(  \phi\otimes\phi\right)  \left(  \underbrace{y_{\left(
1,2, \ldots ,p\right)  }\otimes y_{\left(  p+1,p+2, \ldots ,p+q\right)  }}_{\in
A^{o}\otimes A^{o}}\right)  \in\left(  \phi\otimes\phi\right)  \left(
A^{o}\otimes A^{o}\right)  .
\end{align*}
In other words, $a\otimes b$ lies in the image of the map $\phi\otimes\phi$.

Notice that we already know that $A^{o}$ is a $\kk$-bialgebra, and thus
it satisfies all the axioms of a bialgebra; in particular, the diagrams
\eqref{bialgebra-diagrams}, with $A$, $m$ and $\Delta$ replaced by $A^{o}$,
$m_{A^{o}}$ and $\Delta_{A^{o}}$, commute. In particular, the first of these
diagrams commutes. In other words, the diagram
\begin{equation}
\xymatrix{ & A^o \otimes A^o \ar[dl]_{\Delta_{A^o} \otimes \Delta_{A^o}} \ar[ddr]_{m_{A^o}} & \\ A^o \otimes A^o \otimes A^o \otimes A^o \ar[dd]_{\id \otimes T \otimes \id} & & \\ & & A^o \ar[ddl]_{\Delta_{A^o}} \\ A^o \otimes A^o \otimes A^o \otimes A^o \ar[dr]_{m_{A^o} \otimes m_{A^o}} & & \\ & A^o \otimes A^o & }
\label{pf.shuffle-alg.diagram-known}
\end{equation}
(with $T$ now denoting the twist map $T_{A^o, A^o} :
A^{o}\otimes A^{o}\rightarrow A^{o}\otimes A^{o}$) commutes.

Now, we claim that
\begin{align}
\phi\circ m_{A^{o}}  &  =m_{\shuffle}\circ\left(  \phi\otimes\phi\right)
,
\label{pf.shuffle-alg.equivariance.m}\\
\phi\circ u_{A^{o}}  &  =u,\label{pf.shuffle-alg.equivariance.u}\\
\left(  \phi\otimes\phi\right)  \circ\Delta_{A^{o}}
&  =\Delta_{\shuffle} \circ\phi ,
\label{pf.shuffle-alg.equivariance.Delta}\\
\epsilon_{A^{o}}  &  =\epsilon\circ\phi ,
\label{pf.shuffle-alg.equivariance.epsilon}\\
\phi\circ S_{A^{o}}  &  =S\circ\phi .
\label{pf.shuffle-alg.equivariance.S}
\end{align}
\footnote{If we knew that the $\kk$-module $T\left(  V\right)  $,
endowed with the multiplication $m_{\shuffle}$, the unit $u$, the
comultiplication $\Delta_{\shuffle}$ and the counit $\epsilon$, becomes a Hopf
algebra with the antipode $S$, then these five equalities would be saying that
$\phi:A^{o}\rightarrow T\left(  V\right)  $ is a Hopf algebra homomorphism.}

We are going to only prove the two equalities
\eqref{pf.shuffle-alg.equivariance.m} and
\eqref{pf.shuffle-alg.equivariance.Delta}, leaving the (simpler!) proofs of
the other three equalities \eqref{pf.shuffle-alg.equivariance.u},
\eqref{pf.shuffle-alg.equivariance.epsilon} and
\eqref{pf.shuffle-alg.equivariance.S} to the reader. (Only the equalities
\eqref{pf.shuffle-alg.equivariance.m} and
\eqref{pf.shuffle-alg.equivariance.Delta} will be used in the proof of the
commutativity of the first diagram in \eqref{bialgebra-diagrams}; the other
are used for the other diagrams.)

\textit{Proof of \eqref{pf.shuffle-alg.equivariance.m}:} We need to prove the
equality \eqref{pf.shuffle-alg.equivariance.m}. Since both sides of this
equality \eqref{pf.shuffle-alg.equivariance.m} are $\kk$-linear maps,
it is enough to prove this equality on a $\kk$-basis of $A^{o}\otimes
A^{o}$. Let us pick the basis \newline$\left\{  y_{\left(  i_{1}
,i_{2}, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell
+2}, \ldots ,i_{\ell+m}\right)  }\right\}  _{\ell\in\NN ,\ m\in
\NN ,\ \left(  i_{1},i_{2}, \ldots ,i_{\ell+m}\right)  \in I^{\ell+m}}$; it
thus is enough to prove the equality \eqref{pf.shuffle-alg.equivariance.m}
on this basis, i.e., to prove that
\begin{align}
& \left(  \phi\circ m_{A^{o}}\right)  \left(  y_{\left(  i_{1},i_{2}
, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell
+m}\right)  }\right)  \nonumber\\
&= \left(  m_{\shuffle}\circ\left(  \phi\otimes
\phi\right)  \right)  \left(  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)
}\otimes y_{\left(  i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell+m}\right)  }\right)
\label{pf.shuffle-alg.equivariance.m.pf.1}
\end{align}
for every $\ell \in \NN$, $m \in \NN$ and $\left(  i_{1}
,i_{2}, \ldots ,i_{\ell+m}\right)  \in I^{\ell+m}$. So let us fix $\ell
\in\NN$, $m\in\NN$ and $\left(  i_{1},i_{2}, \ldots ,i_{\ell
+m}\right)  \in I^{\ell+m}$. Comparing
\begin{align*}
& \left(  \phi\circ m_{A^{o}}\right)  \left(  y_{\left(  i_{1},i_{2}
, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell
+m}\right)  }\right)  \\
 &  =\phi\left(  \underbrace{m_{A^{o}}\left(  y_{\left(
i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell
+2}, \ldots ,i_{\ell+m}\right)  }\right)  }_{\substack{=y_{\left(  i_{1}
,i_{2}, \ldots ,i_{\ell}\right)  }y_{\left(  i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell
+m}\right)  }\\=\sum\limits_{\substack{\sigma\in\operatorname{Sh}_{\ell,m}  }}
y_{\left(  i_{\sigma\left(  1\right)  },i_{\sigma\left(
2\right)  },\ldots,i_{\sigma\left(  \ell+m\right)  }\right)  }\\\text{(by
\eqref{eq.exa.shuffle-alg.m3})}}}\right)
 =\phi\left(  \sum\limits_{\substack{\sigma\in\operatorname{Sh}_{\ell,m} }}
y_{\left(  i_{\sigma\left(  1\right)  },i_{\sigma\left(
2\right)  },\ldots,i_{\sigma\left(  \ell+m\right)  }\right)  }\right) \\
&  =\sum\limits_{\substack{\sigma\in\operatorname{Sh}_{\ell,m} }}
\underbrace{\phi\left(  y_{\left(  i_{\sigma\left(  1\right)  }
,i_{\sigma\left(  2\right)  },\ldots,i_{\sigma\left(  \ell+m\right)  }\right)
}\right)  }_{\substack{=v_{i_{\sigma\left(  1\right)  }}v_{i_{\sigma\left(
2\right)  }} \cdots v_{i_{\sigma\left(  \ell+m\right)  }}\\\text{(by the definition
of }\phi\text{)}}}
 =\sum\limits_{\substack{\sigma\in\operatorname{Sh}_{\ell,m} }}
v_{i_{\sigma\left(  1\right)  }}v_{i_{\sigma\left(  2\right)  }}
\cdots v_{i_{\sigma\left(  \ell+m\right)  }}
\end{align*}
with
\begin{align*}
&  \left(  m_{\shuffle}\circ\left(  \phi\otimes\phi\right)  \right)  \left(
y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell
+1},i_{\ell+2}, \ldots ,i_{\ell+m}\right)  }\right) \\
&  =m_{\shuffle}\left(  \underbrace{\left(  \phi\otimes\phi\right)  \left(
y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell
+1},i_{\ell+2}, \ldots ,i_{\ell+m}\right)  }\right)  }_{=\phi\left(  y_{\left(
i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right)  \otimes\phi\left(  y_{\left(
i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell+m}\right)  }\right)  }\right)
 =m_{\shuffle}\left(  \underbrace{\phi\left(  y_{\left(  i_{1}
,i_{2}, \ldots ,i_{\ell}\right)  }\right)  }
_{=v_{i_{1}}v_{i_{2}} \cdots v_{i_{\ell}}}
\otimes\underbrace{\phi\left(  y_{\left(  i_{\ell+1},i_{\ell+2}
, \ldots ,i_{\ell+m}\right)  }\right)  }_{=v_{i_{\ell+1}}v_{i_{\ell+2}}
 \cdots v_{i_{\ell+m}}}\right) \\
&  =m_{\shuffle}\left( \left( v_{i_{1}}v_{i_{2}} \cdots v_{i_{\ell}} \right)
\otimes \left( v_{i_{\ell+1}}v_{i_{\ell+2}} \cdots v_{i_{\ell+m}} \right)
\right)  =\left(  v_{i_{1}}v_{i_{2}} \cdots v_{i_{\ell}}\right) 
 \shufmult \left(  v_{i_{\ell+1}}
v_{i_{\ell+2}} \cdots v_{i_{\ell+m}}\right) \\
&  =\sum\limits_{\substack{\sigma\in\operatorname{Sh}_{\ell,m} }}
v_{i_{\sigma\left(  1\right)  }}v_{i_{\sigma\left(  2\right)  }}
\cdots v_{i_{\sigma\left(  \ell+m\right)  }}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }
\shufmult\right)  ,
\end{align*}
we obtain $\left(  \phi\circ m_{A^{o}}\right)  \left(  y_{\left(  i_{1}
,i_{2}, \ldots ,i_{\ell}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell
+2}, \ldots ,i_{\ell+m}\right)  }\right)  =\left(  m_{\shuffle}\circ\left(
\phi\otimes\phi\right)  \right)  \left(  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell
}\right)  }\otimes y_{\left(  i_{\ell+1},i_{\ell+2}, \ldots ,i_{\ell+m}\right)
}\right)  $. Thus, \eqref{pf.shuffle-alg.equivariance.m.pf.1} is proven, and
this establishes the equality \eqref{pf.shuffle-alg.equivariance.m}.

\textit{Proof of \eqref{pf.shuffle-alg.equivariance.Delta}:} Now we must prove
the equality \eqref{pf.shuffle-alg.equivariance.Delta}. By linearity, it is
enough to verify this equality on a $\kk$-basis of $A^{o}$. We will use
the basis $\left\{  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right\}  $;
then, we need to check the equality
\[
\left(  \left(  \phi\otimes\phi\right)  \circ\Delta_{A^{o}}\right)  \left(
y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right)  =\left(  \Delta
_{\shuffle}\circ\phi\right)  \left(  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell
}\right)  }\right)
\]
holds for every $\ell\in\NN$ and $\left(  i_{1},i_{2}, \ldots ,i_{\ell
}\right)  \in I^{\ell}$. This equality follows by comparing
\begin{align*}
\left(  \left(  \phi\otimes\phi\right)  \circ\Delta_{A^{o}}\right)  \left(
y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right)   &  =\left(  \phi
\otimes\phi\right)  \left(  \underbrace{\Delta_{A^{o}}\left(  y_{\left(
i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right)  }_{\substack{=\sum_{j=0}^{\ell
}y_{\left(  i_{1},\ldots,i_{j}\right)  }\otimes y_{\left(  i_{j+1}
,i_{j+2},\ldots,i_{\ell}\right)  }
\\\text{(by \eqref{eq.exa.shuffle-alg.Delta})}}}\right) \\
&  =\left(  \phi\otimes\phi\right)  \left(  \sum_{j=0}^{\ell}y_{\left(
i_{1},\ldots,i_{j}\right)  }\otimes y_{\left(  i_{j+1},i_{j+2},\ldots,i_{\ell
}\right)  }\right) \\
&  =\sum_{j=0}^{\ell}\underbrace{\phi\left(  y_{\left(  i_{1},\ldots
,i_{j}\right)  }\right)  }_{=v_{i_{1}}v_{i_{2}} \cdots v_{i_{j}}}\otimes
\underbrace{\phi\left(  y_{\left(  i_{j+1},i_{j+2},\ldots,i_{\ell}\right)
}\right)  }_{=v_{i_{j+1}}v_{i_{j+2}} \cdots v_{i_{\ell}}}
 =\sum_{j=0}^{\ell}\left(  v_{i_{1}}v_{i_{2}} \cdots v_{i_{j}}\right)
\otimes\left(  v_{i_{j+1}}v_{i_{j+2}} \cdots v_{i_{\ell}}\right)
\end{align*}
with
\begin{align*}
\left(  \Delta_{\shuffle}\circ\phi\right)  \left(  y_{\left(  i_{1}
,i_{2}, \ldots ,i_{\ell}\right)  }\right)   &  =\Delta_{\shuffle}\left(
\underbrace{\phi\left(  y_{\left(  i_{1},i_{2}, \ldots ,i_{\ell}\right)  }\right)
}_{=v_{i_{1}}v_{i_{2}} \cdots v_{i_{\ell}}}\right)  =\Delta_{\shuffle}\left(
v_{i_{1}}v_{i_{2}} \cdots v_{i_{\ell}}\right) \\
&  =\sum_{k=0}^{\ell}\left(  v_{i_{1}}v_{i_{2}} \cdots v_{i_{k}}\right)
\otimes\left(  v_{i_{k+1}}v_{i_{k+2}} \cdots v_{i_{\ell}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Delta_{\shuffle}
\right) \\
&  =\sum_{j=0}^{\ell}\left(  v_{i_{1}}v_{i_{2}} \cdots v_{i_{j}}\right)
\otimes\left(  v_{i_{j+1}}v_{i_{j+2}} \cdots v_{i_{\ell}}\right)  .
\end{align*}
Thus, \eqref{pf.shuffle-alg.equivariance.Delta} is proven.

Now, we can derive \eqref{pf.shuffle-alg.diagram-goal.eq2} in a very
straightforward way from the commutativity of
\eqref{pf.shuffle-alg.diagram-known} using the equalities
\eqref{pf.shuffle-alg.equivariance.Delta} and
\eqref{pf.shuffle-alg.equivariance.m}: Consider the diagram
\[
\xymatrixcolsep{4.2pc}
\xymatrix{
& & A^o \otimes A^o \ar[ddll]_{\Delta_{A^o} \otimes\Delta_{A^o}}
\ar[dddrr]^{m_{A^o}} \ar[d]_{\phi\otimes\phi} & \\
& & \mathcal{T} \otimes\mathcal{T} \ar[dl]_{\Delta_\shuffle\otimes
\Delta_\shuffle} \ar[ddr]_{m_\shuffle} & \\
A^o \otimes A^o \otimes A^o \otimes A^o \ar[dd]_{\id\otimes T \otimes\id}
\ar[r]_-{\phi\otimes\phi\otimes\phi\otimes\phi} &
\mathcal{T} \otimes\mathcal{T} \otimes\mathcal{T} \otimes\mathcal{T}
\ar[dd]_{\id\otimes T \otimes\id}
& & \\
& & & \mathcal{T} \ar[ddl]_{\Delta_\shuffle} & A^o \ar[dddll]^{\Delta_{A^o}}
\ar[l]^{\phi} \\
A^o \otimes A^o \otimes A^o \otimes A^o \ar[ddrr]_{m_{A^o} \otimes m_{A^o}}
\ar[r]^-{\phi\otimes\phi\otimes\phi\otimes\phi} &
\mathcal{T} \otimes\mathcal{T} \otimes\mathcal{T} \otimes\mathcal{T}
\ar[dr]_{m_\shuffle\otimes m_\shuffle}
& & \\
& & \mathcal{T} \otimes\mathcal{T} & \\
& & A^o \otimes A^o \ar[u]^{\phi\otimes\phi} &
}
,
\]
where $\mathcal{T}$ is shorthand for $T\left(  V\right)  $. The large pentagon
in this diagram is commutative (because it is the diagram
\eqref{pf.shuffle-alg.diagram-known}, which is known to commute), and so are
all five quadrilaterals\footnote{In fact:
\par
\begin{itemize}
\item the northeastern quadrilateral commutes because of
\eqref{pf.shuffle-alg.equivariance.m};
\par
\item the southeastern quadrilateral commutes because of
\eqref{pf.shuffle-alg.equivariance.Delta};
\par
\item the northwestern quadrilateral commutes because
\begin{align*}
\left(  \phi\otimes\phi\otimes\phi\otimes\phi\right)  \circ\left(
\Delta_{A^{o}}\otimes\Delta_{A^{o}}\right)   &  =\underbrace{\left(  \left(
\phi\otimes\phi\right)  \circ\Delta_{A^{o}}\right)  }_{\substack{=\Delta
_{\shuffle}\circ\phi\\
\text{(by \eqref{pf.shuffle-alg.equivariance.Delta})}}}
\otimes\underbrace{\left(  \left(  \phi\otimes\phi\right)  \circ
\Delta_{A^{o}}\right)  }_{\substack{=\Delta_{\shuffle}\circ\phi\\\text{(by
\eqref{pf.shuffle-alg.equivariance.Delta})}}}\\
&  =\left(  \Delta_{\shuffle}\circ\phi\right)  \otimes\left(  \Delta
_{\shuffle}\circ\phi\right)  =\left(  \Delta_{\shuffle}\otimes\Delta
_{\shuffle}\right)  \circ\left(  \phi\otimes\phi\right)  ;
\end{align*}
\par
\item the southwestern quadrilateral commutes for a similar reason (but using
\eqref{pf.shuffle-alg.equivariance.m} instead of
\eqref{pf.shuffle-alg.equivariance.Delta});
\par
\item the western quadrilateral commutes as a consequence of simple linear
algebra.
\end{itemize}
}. This does not automatically yield the commutativity of the small pentagon,
but it yields that all paths from the $A^{o}\otimes A^{o}$ at the top of the
diagram to the $\mathcal{T}\otimes\mathcal{T}$ one row above the very bottom
give the same map; in particular, we have
\[
\left(  m_{\shuffle}\otimes m_{\shuffle}\right)  \circ\left(
\id \otimes T\otimes\id \right)  \circ\left(
\Delta_{\shuffle}\otimes\Delta_{\shuffle}\right)  \circ\left(  \phi\otimes
\phi\right)  =\left(  \Delta_{\shuffle}\circ m_{\shuffle}\right)  \circ\left(
\phi\otimes\phi\right)  .
\]
Thus, the two maps $\left(  m_{\shuffle}\otimes m_{\shuffle}\right)
\circ\left(  \id \otimes T\otimes\id \right)
\circ\left(  \Delta_{\shuffle}\otimes\Delta_{\shuffle}\right)  $ and
$\Delta_{\shuffle}\circ m_{\shuffle}$ are equal to each other on the image of
the map $\phi\otimes\phi$. Since $a\otimes b$ lies in the image of the map
$\phi\otimes\phi$, this yields that these two maps are equal to each other on
$a\otimes b$. In other words, \eqref{pf.shuffle-alg.diagram-goal.eq2} holds.
As we have said above, this completes our proof of
Proposition~\ref{prop.shuffle-alg}.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pw-fin-supp.P1-5}}

\textit{Solution to Exercise~\ref{exe.pw-fin-supp.P1-5}.} We shall use the
following simple fact:

\begin{statement}
\textit{Fact A.0:} Let $\alpha$ and $\beta$ be two maps in
$\Hom \left(  C,A\right)  $. Let $x\in C$. Let some
$k\in \NN$, some elements $y_{1},y_{2},\ldots,y_{k}\in C$ and some
elements $z_{1},z_{2},\ldots,z_{k}\in C$ be chosen such that $\Delta\left(
x\right)  =\sum_{p=1}^{k}y_{p}\otimes z_{p}$. Then,
\[
\left(  \alpha\star\beta\right)  \left(  x\right)  =\sum_{p=1}^{k}%
\alpha\left(  y_{p}\right)  \beta\left(  z_{p}\right)  .
\]

\end{statement}

[\textit{Proof of Fact A.0:} Let $m$ denote the multiplication map $A\otimes
A\rightarrow A$ of the $\kk$-algebra $A$. The definition of convolution
yields $\alpha\star\beta=m\circ\left(  \alpha\otimes\beta\right)  \circ\Delta
$. Hence,%
\begin{align*}
\left(  \alpha\star\beta\right)  \left(  x\right)   &  =\left(  m\circ\left(
\alpha\otimes\beta\right)  \circ\Delta\right)  \left(  x\right)  =m\left(
\left(  \alpha\otimes\beta\right)  \left(  \underbrace{\Delta\left(  x\right)
}_{=\sum_{p=1}^{k}y_{p}\otimes z_{p}}\right)  \right) \\
&  =m\left(  \underbrace{\left(  \alpha\otimes\beta\right)  \left(
\sum_{p=1}^{k}y_{p}\otimes z_{p}\right)  }_{=\sum_{p=1}^{k}\alpha\left(
y_{p}\right)  \otimes\beta\left(  z_{p}\right)  }\right)  =m\left(
\sum_{p=1}^{k}\alpha\left(  y_{p}\right)  \otimes\beta\left(  z_{p}\right)
\right) \\
&  =\sum_{p=1}^{k}\alpha\left(  y_{p}\right)  \beta\left(  z_{p}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }m\right)  .
\end{align*}
This proves Fact A.0.]

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P1}.]Let $g$ be the map
$\sum_{q\in Q}f_{q}$.\ \ \ \ \footnote{This map is well-defined, since the
family $\left(  f_{q}\right)  _{q\in Q}$ is pointwise finitely supported.}
Thus, $g$ is a map $C\rightarrow A$. Moreover, $g=\sum_{q\in Q}f_{q}$. Thus,
each $x\in C$ satisfies%
\begin{equation}
g\left(  x\right)  =\left(  \sum_{q\in Q}f_{q}\right)  \left(  x\right)
=\sum_{q\in Q}f_{q}\left(  x\right)
\label{sol.exe.convolution-series.a.PA.1.pf.1}%
\end{equation}
(by the definition of $\sum_{q\in Q}f_{q}$).

Now, let $c\in C$, $d\in C$, $\lambda\in\kk$ and $\mu\in\kk$ be
arbitrary. Then, (\ref{sol.exe.convolution-series.a.PA.1.pf.1}) (applied to
$x=c$) yields%
\[
g\left(  c\right)  =\sum_{q\in Q}f_{q}\left(  c\right)  .
\]
Moreover, the family $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported. In other words, for each $x\in C$, the family $\left(  f_{q}\left(
x\right)  \right)  _{q\in Q}\in A^{Q}$ of elements of $A$ is finitely
supported (by the definition of ``pointwise finitely
supported''). Applying this to $x=c$, we conclude that the
family $\left(  f_{q}\left(  c\right)  \right)  _{q\in Q}\in A^{Q}$ of
elements of $A$ is finitely supported. In other words, all but finitely many
$q\in Q$ satisfy $f_{q}\left(  c\right)  =0$. Hence, all but finitely many
$q\in Q$ satisfy $\lambda f_{q}\left(  c\right)  =0$ (since every $q\in Q$
that satisfies $f_{q}\left(  c\right)  =0$ must also satisfy $\lambda
\underbrace{f_{q}\left(  c\right)  }_{=0}=0$). In other words, the family
$\left(  \lambda f_{q}\left(  c\right)  \right)  _{q\in Q}\in A^{Q}$ is
finitely supported. The same argument (applied to $d$ and $\mu$ instead of $c$
and $\lambda$) shows that the family $\left(  \mu f_{q}\left(  d\right)
\right)  _{q\in Q}\in A^{Q}$ of elements of $A$ is finitely supported.

Recall that sums of finitely supported families satisfy the same rules as
finite sums. Since the families $\left(  \lambda f_{q}\left(  c\right)
\right)  _{q\in Q}$ and $\left(  \mu f_{q}\left(  d\right)  \right)  _{q\in
Q}$ are finitely supported, we thus have%
\begin{equation}
\sum_{q\in Q}\lambda f_{q}\left(  c\right)  +\sum_{q\in Q}\mu f_{q}\left(
d\right)  =\sum_{q\in Q}\left(  \lambda f_{q}\left(  c\right)  +\mu
f_{q}\left(  d\right)  \right)  \label{sol.exe.convolution-series.a.PA.1.pf.3}%
\end{equation}
(and in particular, the family $\left(  \lambda f_{q}\left(  c\right)  +\mu
f_{q}\left(  d\right)  \right)  _{q\in Q}$ is finitely supported). For the
same reason, we have
\begin{equation}
\lambda\sum_{q\in Q}f_{q}\left(  c\right)  =\sum_{q\in Q}\lambda f_{q}\left(
c\right)  \label{sol.exe.convolution-series.a.PA.1.pf.4a}%
\end{equation}
(since the family $\left(  f_{q}\left(  c\right)  \right)  _{q\in Q}$ is
finitely supported) and%
\begin{equation}
\mu\sum_{q\in Q}f_{q}\left(  d\right)  =\sum_{q\in Q}\mu f_{q}\left(
d\right)  \label{sol.exe.convolution-series.a.PA.1.pf.4b}%
\end{equation}
(for similar reasons).

Now, (\ref{sol.exe.convolution-series.a.PA.1.pf.1}) (applied to $x=\lambda
c+\mu d$) yields%
\begin{align*}
g\left(  \lambda c+\mu d\right)   &  =\sum_{q\in Q}\underbrace{f_{q}\left(
\lambda c+\mu d\right)  }_{\substack{=\lambda f_{q}\left(  c\right)  +\mu
f_{q}\left(  d\right)  \\\text{(since }f_{q}\in\Hom \left(
C,A\right)  \text{)}}}=\sum_{q\in Q}\left(  \lambda f_{q}\left(  c\right)
+\mu f_{q}\left(  d\right)  \right) \\
&  =\underbrace{\sum_{q\in Q}\lambda f_{q}\left(  c\right)  }%
_{\substack{=\lambda\sum_{q\in Q}f_{q}\left(  c\right)  \\\text{(by
(\ref{sol.exe.convolution-series.a.PA.1.pf.4a}))}}}+\underbrace{\sum_{q\in
Q}\mu f_{q}\left(  d\right)  }_{\substack{=\mu\sum_{q\in Q}f_{q}\left(
d\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.1.pf.4b}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.1.pf.3})}\right) \\
&  =\lambda\sum_{q\in Q}f_{q}\left(  c\right)  +\mu\sum_{q\in Q}f_{q}\left(
d\right)  .
\end{align*}
Comparing this with%
\[
\lambda\underbrace{g\left(  c\right)  }_{=\sum_{q\in Q}f_{q}\left(  c\right)
}+\mu\underbrace{g\left(  d\right)  }_{\substack{=\sum_{q\in Q}f_{q}\left(
d\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.1.pf.1}) (applied
to }x=d\text{))}}}=\lambda\sum_{q\in Q}f_{q}\left(  c\right)  +\mu\sum_{q\in
Q}f_{q}\left(  d\right)  ,
\]
we obtain $g\left(  \lambda c+\mu d\right)  =\lambda g\left(  c\right)  +\mu
g\left(  d\right)  $.

Now, forget that we fixed $c$, $d$, $\lambda$ and $\mu$. We thus have proven
that $g\left(  \lambda c+\mu d\right)  =\lambda g\left(  c\right)  +\mu
g\left(  d\right)  $ for every $c\in C$, $d\in C$, $\lambda\in\kk$ and
$\mu\in\kk$. In other words, the map $g$ is $\kk$-linear. In
other words, $g\in\Hom \left(  C,A\right)  $. Thus, $\sum_{q\in
Q}f_{q}=g\in\Hom \left(  C,A\right)  $. This proves
Proposition~\ref{prop.pw-fin-supp.P1}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P2}.]Fix $x\in C$. The family
$\left(  f_{q}\left(  x\right)  \right)  _{q\in Q}\in A^{Q}$ of elements of
$A$ is finitely supported (since the family $\left(  f_{q}\right)  _{q\in
Q}\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$ is
pointwise finitely supported). In other words, all but finitely many $q\in Q$
satisfy $f_{q}\left(  x\right)  =0$. In other words, there exists a finite
subset $Q_{1}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{1}\text{ satisfies }f_{q}\left(  x\right)  =0.
\label{sol.exe.convolution-series.a.PA.2.pf.3}
\end{equation}
Similarly, there exists a finite subset $Q_{2}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{2}\text{ satisfies }g_{q}\left(  x\right)  =0.
\label{sol.exe.convolution-series.a.PA.2.pf.4}
\end{equation}
Consider these two finite subsets $Q_{1}$ and $Q_{2}$.

The set $Q_{1}\cup Q_{2}$ is finite (since it is the union of the two finite
sets $Q_{1}$ and $Q_{2}$). Thus, all but finitely many $q\in Q$ satisfy $q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $.

We have $Q\setminus\underbrace{\left(  Q_{1}\cup Q_{2}\right)  }_{\supset
Q_{1}}\subset Q\setminus Q_{1}$ and $Q\setminus\underbrace{\left(  Q_{1}\cup
Q_{2}\right)  }_{\supset Q_{2}}\subset Q\setminus Q_{2}$.

Also, $Q_{1}\cup Q_{2}$ is a subset of $Q$ (since both $Q_{1}$ and $Q_{2}$ are
subsets of $Q$). In other words, $Q_{1}\cup Q_{2}\subset Q$.

Every $q\in Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $ satisfies%
\begin{equation}
\left(  f_{q}+g_{q}\right)  \left(  x\right)  =\underbrace{f_{q}\left(
x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.3})}\\\text{(since }q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  \subset Q\setminus Q_{1}\text{))}%
}}+\underbrace{g_{q}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.4})}\\\text{(since }q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  \subset Q\setminus Q_{2}\text{))}%
}}=0. \label{sol.exe.convolution-series.a.PA.2.pf.7}%
\end{equation}
Hence, all but finitely many $q\in Q$ satisfy $\left(  f_{q}+g_{q}\right)
\left(  x\right)  =0$ (since all but finitely many $q\in Q$ satisfy $q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $). In other words,
\begin{equation}
\text{the family }\left(  \left(  f_{q}+g_{q}\right)  \left(  x\right)
\right)  _{q\in Q}\in A^{Q}\text{ is finitely supported.}
\label{sol.exe.convolution-series.a.PA.2.pf.fs}%
\end{equation}


The set $Q$ is the union of its two disjoint subsets $Q_{1}\cup Q_{2}$ and
$Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $ (since $Q_{1}\cup Q_{2}\subset
Q$). Thus, the sum $\sum_{q\in Q}f_{q}\left(  x\right)  $ can be split as
follows:%
\begin{align}
\sum_{q\in Q}f_{q}\left(  x\right)   &  =\sum_{q\in Q_{1}\cup Q_{2}}%
f_{q}\left(  x\right)  +\sum_{q\in Q\setminus\left(  Q_{1}\cup Q_{2}\right)
}\underbrace{f_{q}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.3})}\\\text{(since }q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  \subset Q\setminus Q_{1}\text{))}%
}}\nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\underbrace{\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }0}_{=0}=\sum_{q\in Q_{1}\cup Q_{2}%
}f_{q}\left(  x\right)  . \label{sol.exe.convolution-series.a.PA.2.pf.8}%
\end{align}
A similar argument (using (\ref{sol.exe.convolution-series.a.PA.2.pf.4})
instead of (\ref{sol.exe.convolution-series.a.PA.2.pf.3})) yields%
\[
\sum_{q\in Q}g_{q}\left(  x\right)  =\sum_{q\in Q_{1}\cup Q_{2}}g_{q}\left(
x\right)  .
\]
Adding this equality to (\ref{sol.exe.convolution-series.a.PA.2.pf.8}), we
obtain%
\begin{equation}
\sum_{q\in Q}f_{q}\left(  x\right)  +\sum_{q\in Q}g_{q}\left(  x\right)
=\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\sum_{q\in Q_{1}\cup
Q_{2}}g_{q}\left(  x\right)  . \label{sol.exe.convolution-series.a.PA.2.pf.10}%
\end{equation}


But recall again that the set $Q$ is the union of its two disjoint subsets
$Q_{1}\cup Q_{2}$ and $Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $. Hence,
\begin{align}
\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  \left(  x\right)   &  =\sum_{q\in
Q_{1}\cup Q_{2}}\underbrace{\left(  f_{q}+g_{q}\right)  \left(  x\right)
}_{=f_{q}\left(  x\right)  +g_{q}\left(  x\right)  }+\sum_{q\in Q\setminus
\left(  Q_{1}\cup Q_{2}\right)  }\underbrace{\left(  f_{q}+g_{q}\right)
\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.7}))}}}\nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}\left(  f_{q}\left(  x\right)  +g_{q}\left(
x\right)  \right)  +\underbrace{\sum_{q\in Q\setminus\left(  Q_{1}\cup
Q_{2}\right)  }0}_{=0}=\sum_{q\in Q_{1}\cup Q_{2}}\left(  f_{q}\left(
x\right)  +g_{q}\left(  x\right)  \right) \nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\sum_{q\in Q_{1}\cup
Q_{2}}g_{q}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have manipulated a finite sum}\\
\text{(since }Q_{1}\cup Q_{2}\text{ is a finite set)}%
\end{array}
\right) \nonumber\\
&  =\sum_{q\in Q}f_{q}\left(  x\right)  +\sum_{q\in Q}g_{q}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.2.pf.10})}\right)  .
\label{sol.exe.convolution-series.a.PA.2.pf.13}%
\end{align}


Now, let us forget that we fixed $x$. We thus have proven that every $x\in C$
satisfies (\ref{sol.exe.convolution-series.a.PA.2.pf.fs}) and
(\ref{sol.exe.convolution-series.a.PA.2.pf.13}).

In particular, every $x\in C$ satisfies
(\ref{sol.exe.convolution-series.a.PA.2.pf.fs}). In other words, for each
$x\in C$, the family $\left(  \left(  f_{q}+g_{q}\right)  \left(  x\right)
\right)  _{q\in Q}\in A^{Q}$ of elements of $A$ is finitely supported. In
other words, the family $\left(  f_{q}+g_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported (by the definition of ``pointwise finitely
supported'').

Hence, the sum $\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  $ is well-defined.
Also, the sum $\sum_{q\in Q}f_{q}$ is well-defined (since the family $\left(
f_{q}\right)  _{q\in Q}$ is pointwise finitely supported). Similarly, the sum
$\sum_{q\in Q}g_{q}$ is well-defined.

Moreover, each $x\in C$ satisfies%
\begin{align*}
\left(  \sum_{q\in Q}\left(  f_{q}+g_{q}\right)  \right)  \left(  x\right)
&  =\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  \left(  x\right)
=\underbrace{\sum_{q\in Q}f_{q}\left(  x\right)  }_{=\left(  \sum_{q\in
Q}f_{q}\right)  \left(  x\right)  }+\underbrace{\sum_{q\in Q}g_{q}\left(
x\right)  }_{=\left(  \sum_{q\in Q}g_{q}\right)  \left(  x\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.2.pf.13})}\right) \\
&  =\left(  \sum_{q\in Q}f_{q}\right)  \left(  x\right)  +\left(  \sum_{q\in
Q}g_{q}\right)  \left(  x\right)  =\left(  \sum_{q\in Q}f_{q}+\sum_{q\in
Q}g_{q}\right)  \left(  x\right)  .
\end{align*}
In other words, $\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  =\sum_{q\in Q}%
f_{q}+\sum_{q\in Q}g_{q}$. In other words, $\sum_{q\in Q}f_{q}+\sum_{q\in
Q}g_{q}=\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  $. This completes the proof
of Proposition~\ref{prop.pw-fin-supp.P2}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P3}.]Let $x\in C$. Write the
element $\Delta\left(  x\right)  \in C\otimes C$ in the form $\Delta\left(
x\right)  =\sum_{p=1}^{k}y_{p}\otimes z_{p}$ for some $k\in \NN$, some
elements $y_{1},y_{2},\ldots,y_{k}\in C$ and some elements $z_{1},z_{2}%
,\ldots,z_{k}\in C$. (This is possible, because $\Delta\left(  x\right)  $ can
be written as a sum of pure tensors\footnote{This is because any tensor in
$C\otimes C$ can be written as a sum of pure tensors.}.)

For each $p\in\left\{  1,2,\ldots,k\right\}  $, there exists a finite subset
$Q_{p}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{p}\text{ satisfies }f_{q}\left(  y_{p}\right)
=0 \label{sol.exe.convolution-series.a.PA.3.pf.1}%
\end{equation}
\footnote{\textit{Proof:} Let $p\in\left\{  1,2,\ldots,k\right\}  $. The
family $\left(  f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is pointwise finitely supported. Hence, the family
$\left(  f_{q}\left(  y_{p}\right)  \right)  _{q\in Q}\in A^{Q}$ is finitely
supported. In other words, all but finitely many $q\in Q$ satisfy
$f_{q}\left(  y_{p}\right)  =0$. In other words, there exists a finite subset
$Q_{p}$ of $Q$ such that every $q\in Q\setminus Q_{p}$ satisfies $f_{q}\left(
y_{p}\right)  =0$. Qed.}. Consider this $Q_{p}$.

Define a subset $Q^{\prime}$ of $Q$ by $Q^{\prime}=Q_{1}\cup Q_{2}\cup
\cdots\cup Q_{k}$. Thus, $Q^{\prime}$ is the union of the $k$ finite sets
$Q_{1},Q_{2},\ldots,Q_{k}$. Therefore, $Q^{\prime}$ itself is a finite set
(since a union of $k$ finite sets is always finite).

Every $q\in Q\setminus Q^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $
satisfy $f_{q}\left(  y_{p}\right)  =0$\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.exe.convolution-series.a.PA.3.pf.3}):} Let $q\in Q\setminus
Q^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $.
\par
We have $Q_{p}\subset Q_{1}\cup Q_{2}\cup\cdots\cup Q_{k}=Q^{\prime}$ (since
$Q^{\prime}=Q_{1}\cup Q_{2}\cup\cdots\cup Q_{k}$), so that $Q\setminus
\underbrace{Q_{p}}_{\subset Q^{\prime}}\supset Q\setminus Q^{\prime}$. Hence,
$Q\setminus Q^{\prime}\subset Q\setminus Q_{p}$. Now,
(\ref{sol.exe.convolution-series.a.PA.3.pf.1}) yields $f_{q}\left(
y_{p}\right)  =0$ (since $q\in Q\setminus Q^{\prime}\subset Q\setminus Q_{p}%
$). This proves (\ref{sol.exe.convolution-series.a.PA.3.pf.3}).}.

We have thus constructed a finite subset $Q^{\prime}$ of $Q$ with the property
that every $q\in Q\setminus Q^{\prime}$ and $p\in\left\{  1,2,\ldots
,k\right\}  $ satisfy%
\begin{equation}
f_{q}\left(  y_{p}\right)  =0. \label{sol.exe.convolution-series.a.PA.3.pf.3}%
\end{equation}
Similarly, we can construct a finite subset $R^{\prime}$ of $R$ with the
property that every $r\in R\setminus R^{\prime}$ and $p\in\left\{
1,2,\ldots,k\right\}  $ satisfy%
\begin{equation}
g_{r}\left(  z_{p}\right)  =0. \label{sol.exe.convolution-series.a.PA.3.pf.3b}%
\end{equation}
Consider this $R^{\prime}$.

The set $Q^{\prime}\times R^{\prime}$ is a Cartesian product of two finite
sets (since $Q^{\prime}$ and $R^{\prime}$ are finite sets), and thus is itself
finite. Hence, all but finitely many $\left(  q,r\right)  \in Q\times R$
satisfy $\left(  q,r\right)  \in\left(  Q\times R\right)  \setminus\left(
Q^{\prime}\times R^{\prime}\right)  $.

We shall now show that each $\left(  q,r\right)  \in\left(  Q\times R\right)
\setminus\left(  Q^{\prime}\times R^{\prime}\right)  $ satisfies $\left(
f_{q}\star g_{r}\right)  \left(  x\right)  =0$.

Indeed, fix $\left(  q,r\right)  \in\left(  Q\times R\right)  \setminus\left(
Q^{\prime}\times R^{\prime}\right)  $. Thus, $\left(  q,r\right)  \in Q\times
R$ and $\left(  q,r\right)  \notin Q^{\prime}\times R^{\prime}$.

We are in one of the following two cases:

\textit{Case 1:} We have $q\in Q^{\prime}$.

\textit{Case 2:} We have $q\notin Q^{\prime}$.

Let us first consider Case 1. In this case, we have $q\in Q^{\prime}$. If we
had $r\in R^{\prime}$, we thus would have $\left(  q,r\right)  \in Q^{\prime
}\times R^{\prime}$ (since $q\in Q^{\prime}$ and $r\in R^{\prime}$), which
would contradict $\left(  q,r\right)  \notin Q^{\prime}\times R^{\prime}$.
Hence, we cannot have $r\in R^{\prime}$. In other words, we have $r\notin
R^{\prime}$. Combining $r\in R$ with $r\notin R^{\prime}$, we obtain $r\in
R\setminus R^{\prime}$. Now, Fact A.0 (applied to $\alpha=f_{q}$ and
$\beta=g_{r}$) yields%
\[
\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =\sum_{p=1}^{k}f_{q}\left(
y_{p}\right)  \underbrace{g_{r}\left(  z_{p}\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.3b}))}%
}}=\sum_{p=1}^{k}f_{q}\left(  y_{p}\right)  0=0.
\]
Thus, $\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ is proven in
Case 1.

Let us now consider Case 2. In this case, we have $q\notin Q^{\prime}$.
Combining $q\in Q$ with $q\notin Q^{\prime}$, we obtain $q\in Q\setminus
Q^{\prime}$. Now, Fact A.0 (applied to $\alpha=f_{q}$ and $\beta=g_{r}$)
yields%
\[
\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =\sum_{p=1}^{k}%
\underbrace{f_{q}\left(  y_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.3}))}}}g_{r}\left(  z_{p}\right)
=\sum_{p=1}^{k}0g_{r}\left(  z_{p}\right)  =0.
\]
Thus, $\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ is proven in
Case 2.

We have thus proven $\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ in
both Cases 1 and 2. Hence, $\left(  f_{q}\star g_{r}\right)  \left(  x\right)
=0$ always holds.

Now, forget that we fixed $\left(  q,r\right)  $. We thus have proven that
\begin{equation}
\text{each }\left(  q,r\right)  \in\left(  Q\times R\right)  \setminus\left(
Q^{\prime}\times R^{\prime}\right)  \text{ satisfies }\left(  f_{q}\star
g_{r}\right)  \left(  x\right)  =0.
\label{sol.exe.convolution-series.a.PA.3.pf.res1}%
\end{equation}
Thus, all but finitely many $\left(  q,r\right)  \in Q\times R$ satisfy
$\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ (since all but
finitely many $\left(  q,r\right)  \in Q\times R$ satisfy $\left(  q,r\right)
\in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times R^{\prime
}\right)  $). In other words,%
\begin{equation}
\text{the family }\left(  \left(  f_{q}\star g_{r}\right)  \left(  x\right)
\right)  _{\left(  q,r\right)  \in Q\times R}\in A^{Q\times R}\text{ is
finitely supported.} \label{sol.exe.convolution-series.a.PA.3.pf.res3}%
\end{equation}


Define a map $F:C\rightarrow A$ by $F=\sum_{q\in Q}f_{q}$. (This is
well-defined, since the family $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely supported.)

Define a map $G:C\rightarrow A$ by $G=\sum_{r\in R}g_{r}$. (This is
well-defined, since the family $\left(  g_{r}\right)  _{r\in R}\in\left(
\Hom \left(  C,A\right)  \right)  ^{R}$ is pointwise finitely supported.)

We have $F=\sum_{q\in Q}f_{q}\in\Hom \left(  C,A\right)  $ (by
Proposition~\ref{prop.pw-fin-supp.P1}) and $G\in\Hom \left(
C,A\right)  $ (for similar reasons). Hence, the map $F\star G\in
\Hom \left(  C,A\right)  $ is well-defined.

Define a further map $F^{\prime}\in\Hom \left(  C,A\right)  $ by
$F^{\prime}=\sum_{q\in Q^{\prime}}f_{q}$. Notice that this is a finite sum,
since $Q^{\prime}$ is a finite set.

Define a further map $G^{\prime}\in\Hom \left(  C,A\right)  $ by
$G^{\prime}=\sum_{r\in R^{\prime}}g_{r}$. Notice that this is a finite sum,
since $R^{\prime}$ is a finite set.

From $F^{\prime}=\sum_{q\in Q^{\prime}}f_{q}$ and $G^{\prime}=\sum_{r\in
R^{\prime}}g_{r}$, we obtain\footnote{The following manipulations of sums are
legitimate, since all the sums involved are finite (because $Q^{\prime}$ and
$R^{\prime}$ are finite sets).}%
\begin{align}
F^{\prime}\star G^{\prime}  &  =\left(  \sum_{q\in Q^{\prime}}f_{q}\right)
\star\left(  \sum_{r\in R^{\prime}}g_{r}\right) \nonumber\\
&  =\underbrace{\sum_{q\in Q^{\prime}}\sum_{r\in R^{\prime}}}_{=\sum_{\left(
q,r\right)  \in Q^{\prime}\times R^{\prime}}}\left(  f_{q}\star g_{r}\right)
=\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(  f_{q}\star
g_{r}\right)  . \label{sol.exe.convolution-series.a.PA.3.pf.F'G'1}%
\end{align}


But every $p\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
F^{\prime}\left(  y_{p}\right)  =F\left(  y_{p}\right)
\label{sol.exe.convolution-series.a.PA.3.pf.F'1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1}):}
Let $p\in\left\{  1,2,\ldots,k\right\}  $. Recall that $Q^{\prime}\subset Q$;
thus, the set $Q$ is the union of its two disjoint subsets $Q^{\prime}$ and
$Q\setminus Q^{\prime}$.
\par
From $F=\sum_{q\in Q}f_{q}$, we obtain
\begin{align*}
F\left(  y_{p}\right)   &  =\left(  \sum_{q\in Q}f_{q}\right)  \left(
y_{p}\right)  =\sum_{q\in Q}f_{q}\left(  y_{p}\right)  =\sum_{q\in Q^{\prime}%
}f_{q}\left(  y_{p}\right)  +\sum_{q\in Q\setminus Q^{\prime}}%
\underbrace{f_{q}\left(  y_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.3}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the set }Q\text{ is the union of
its disjoint subsets }Q^{\prime}\text{ and }Q\setminus Q^{\prime}\right) \\
&  =\sum_{q\in Q^{\prime}}f_{q}\left(  y_{p}\right)  +\underbrace{\sum_{q\in
Q\setminus Q^{\prime}}0}_{=0}=\sum_{q\in Q^{\prime}}f_{q}\left(  y_{p}\right)
.
\end{align*}
Comparing this with $\underbrace{F^{\prime}}_{=\sum_{q\in Q^{\prime}}f_{q}%
}\left(  y_{p}\right)  =\left(  \sum_{q\in Q^{\prime}}f_{q}\right)  \left(
y_{p}\right)  =\sum_{q\in Q^{\prime}}f_{q}\left(  y_{p}\right)  $, we obtain
$F^{\prime}\left(  y_{p}\right)  =F\left(  y_{p}\right)  $. This proves
(\ref{sol.exe.convolution-series.a.PA.3.pf.F'1}).} and%
\begin{equation}
G^{\prime}\left(  z_{p}\right)  =G\left(  z_{p}\right)
\label{sol.exe.convolution-series.a.PA.3.pf.F'1b}%
\end{equation}
(for similar reasons). Hence,%
\begin{equation}
\left(  F^{\prime}\star G^{\prime}\right)  \left(  x\right)  =\left(  F\star
G\right)  \left(  x\right)  \label{sol.exe.convolution-series.a.PA.3.pf.F'G'}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.F'G'}):}
Fact A.0 (applied to $\alpha=F$ and $\beta=G$) yields%
\begin{equation}
\left(  F\star G\right)  \left(  x\right)  =\sum_{p=1}^{k}F\left(
y_{p}\right)  G\left(  z_{p}\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.F'G'.pf.2}%
\end{equation}
\par
Fact A.0 (applied to $\alpha=F^{\prime}$ and $\beta=G^{\prime}$) yields%
\begin{equation}
\left(  F^{\prime}\star G^{\prime}\right)  \left(  x\right)  =\sum_{p=1}%
^{k}\underbrace{F^{\prime}\left(  y_{p}\right)  }_{\substack{=F\left(
y_{p}\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1}))}%
}}\underbrace{G^{\prime}\left(  z_{p}\right)  }_{\substack{=G\left(
z_{p}\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1b}))}%
}}=\sum_{p=1}^{k}F\left(  y_{p}\right)  G\left(  z_{p}\right)  .\nonumber
\end{equation}
Comparing this with (\ref{sol.exe.convolution-series.a.PA.3.pf.F'G'.pf.2}), we
obtain $\left(  F^{\prime}\star G^{\prime}\right)  \left(  x\right)  =\left(
F\star G\right)  \left(  x\right)  $.}. Thus,%
\begin{align}
\left(  F\star G\right)  \left(  x\right)   &  =\underbrace{\left(  F^{\prime
}\star G^{\prime}\right)  }_{\substack{=\sum_{\left(  q,r\right)  \in
Q^{\prime}\times R^{\prime}}\left(  f_{q}\star g_{r}\right)  \\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.F'G'1}))}}}\left(  x\right)
=\left(  \sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \right)  \left(  x\right) \nonumber\\
&  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \left(  x\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.FGx=}%
\end{align}


On the other hand, $\underbrace{Q^{\prime}}_{\subset Q}\times
\underbrace{R^{\prime}}_{\subset R}\subset Q\times R$. Hence, the set $Q\times
R$ is the union of its two disjoint subsets $Q^{\prime}\times R^{\prime}$ and
$\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times R^{\prime
}\right)  $.

But the sum $\sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star
g_{r}\right)  \left(  x\right)  $ is well-defined (since the family $\left(
\left(  f_{q}\star g_{r}\right)  \left(  x\right)  \right)  _{\left(
q,r\right)  \in Q\times R}\in A^{Q\times R}$ is finitely supported). Since the
set $Q\times R$ is the union of its two disjoint subsets $Q^{\prime}\times
R^{\prime}$ and $\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times
R^{\prime}\right)  $, we can split this sum as follows:
\begin{align}
&  \sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star g_{r}\right)
\left(  x\right) \nonumber\\
&  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \left(  x\right)  +\sum_{\left(  q,r\right)
\in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times R^{\prime
}\right)  }\underbrace{\left(  f_{q}\star g_{r}\right)  \left(  x\right)
}_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.res1}%
))}}}\nonumber\\
&  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \left(  x\right)  +\underbrace{\sum_{\left(
q,r\right)  \in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times
R^{\prime}\right)  }0}_{=0}=\sum_{\left(  q,r\right)  \in Q^{\prime}\times
R^{\prime}}\left(  f_{q}\star g_{r}\right)  \left(  x\right) \nonumber\\
&  =\left(  \underbrace{F}_{=\sum_{q\in Q}f_{q}}\star\underbrace{G}%
_{=\sum_{r\in R}g_{r}}\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.exe.convolution-series.a.PA.3.pf.FGx=})}\right) \nonumber\\
&  =\left(  \left(  \sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}%
g_{r}\right)  \right)  \left(  x\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.res6}%
\end{align}


Now, forget that we fixed $x$. We thus have shown that each $x\in C$ satisfies
(\ref{sol.exe.convolution-series.a.PA.3.pf.res3}) and
(\ref{sol.exe.convolution-series.a.PA.3.pf.res6}).

In particular, each $x\in C$ satisfies
(\ref{sol.exe.convolution-series.a.PA.3.pf.res3}). In other words, for each
$x\in C$, the family $\left(  \left(  f_{q}\star g_{r}\right)  \left(
x\right)  \right)  _{\left(  q,r\right)  \in Q\times R}\in A^{Q\times R}$ is
finitely supported. In other words, the family $\left(  f_{q}\star
g_{r}\right)  _{\left(  q,r\right)  \in Q\times R}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q\times R}$ is pointwise
finitely supported (by the definition of ``pointwise finitely
supported''). Hence, the sum $\sum_{\left(  q,r\right)  \in
Q\times R}\left(  f_{q}\star g_{r}\right)  $ is well-defined. For each $x\in
C$, we have%
\[
\left(  \sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star
g_{r}\right)  \right)  \left(  x\right)  =\sum_{\left(  q,r\right)  \in
Q\times R}\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =\left(  \left(
\sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}g_{r}\right)  \right)
\left(  x\right)
\]
(by (\ref{sol.exe.convolution-series.a.PA.3.pf.res6})). In other words, we
have%
\[
\sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star g_{r}\right)
=\left(  \sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}g_{r}\right)  .
\]
This completes the proof of Proposition~\ref{prop.pw-fin-supp.P3}.
\end{proof}

We shall delay the proof of Proposition~\ref{prop.pw-fin-supp.P4} until after
Proposition~\ref{prop.pw-fin-supp.P5} is proven; the reason is that
Proposition~\ref{prop.pw-fin-supp.P5} yields a quick shortcut to
Proposition~\ref{prop.pw-fin-supp.P4}.

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P5}.]Let $x\in C$. Write the
element $\Delta\left(  x\right)  \in C\otimes C$ in the form $\Delta\left(
x\right)  =\sum_{p=1}^{k}y_{p}\otimes z_{p}$ for some $k\in \NN$, some
elements $y_{1},y_{2},\ldots,y_{k}\in C$ and some elements $z_{1},z_{2}%
,\ldots,z_{k}\in C$. (This is possible, because $\Delta\left(  x\right)  $ can
be written as a sum of pure tensors\footnote{This is because any tensor in
$C\otimes C$ can be written as a sum of pure tensors.}.)

For each $p\in\left\{  1,2,\ldots,k\right\}  $, there exists a finite subset
$Q_{p}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{p}\text{ satisfies }f_{q}\left(  y_{p}\right)
=0 \label{sol.exe.convolution-series.a.PA.5.pf.1}%
\end{equation}
\footnote{This can be shown in the same way as we did it during the proof of
Proposition~\ref{prop.pw-fin-supp.P3}.}. Consider this $Q_{p}$.

Define a subset $Q^{\prime}$ of $Q$ by $Q^{\prime}=Q_{1}\cup Q_{2}\cup
\cdots\cup Q_{k}$. Thus, $Q^{\prime}$ is the union of the $k$ finite sets
$Q_{1},Q_{2},\ldots,Q_{k}$. Hence, $Q^{\prime}$ itself a finite set. Hence,
all but finitely many $q\in Q$ satisfy $q\in Q\setminus Q^{\prime}$.

Every $q\in Q\setminus Q^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $
satisfy%
\begin{equation}
f_{q}\left(  y_{p}\right)  =0 \label{sol.exe.convolution-series.a.PA.5.pf.3}%
\end{equation}
\footnote{This can be shown in the same way as we did it during the proof of
Proposition~\ref{prop.pw-fin-supp.P3}.}.

Now, fix $q\in Q\setminus Q^{\prime}$. Fact A.0 (applied to $\alpha=f_{q}$ and
$\beta=g_{q}$) yields%
\[
\left(  f_{q}\star g_{q}\right)  \left(  x\right)  =\sum_{p=1}^{k}%
\underbrace{f_{q}\left(  y_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.5.pf.3}))}}}g_{q}\left(  z_{p}\right)
=\sum_{p=1}^{k}0g_{q}\left(  z_{p}\right)  =0.
\]


Now, forget that we fixed $q$. We thus have proven that each $q\in Q\setminus
Q^{\prime}$ satisfies $\left(  f_{q}\star g_{q}\right)  \left(  x\right)  =0$.
Hence, all but finitely many $q\in Q$ satisfy $\left(  f_{q}\star
g_{q}\right)  \left(  x\right)  =0$ (since all but finitely many $q\in Q$
satisfy $q\in Q\setminus Q^{\prime}$). In other words, the family $\left(
\left(  f_{q}\star g_{q}\right)  \left(  x\right)  \right)  _{q\in Q}\in
A^{Q}$ is finitely supported.

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
family $\left(  \left(  f_{q}\star g_{q}\right)  \left(  x\right)  \right)
_{q\in Q}\in A^{Q}$ is finitely supported. In other words, the family $\left(
f_{q}\star g_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is pointwise finitely supported. This proves
Proposition~\ref{prop.pw-fin-supp.P5}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P4}.]Let $\mathfrak{i}$ be the
unity of the $\kk$-algebra $\left(  \Hom \left(
C,A\right)  ,\star\right)  $. (This $\mathfrak{i}$ is the map $u_{A}%
\circ\epsilon_{C}:C\rightarrow A$; but this does not matter to us.)

Applying Proposition~\ref{prop.pw-fin-supp.P5} to the family $\left(
g_{q}\right)  _{q\in Q}=\left(  \lambda_{q}\mathfrak{i}\right)  _{q\in Q}$, we
conclude that the family $\left(  f_{q}\star\left(  \lambda_{q}\mathfrak{i}%
\right)  \right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is pointwise finitely supported. Since each $q\in Q$ satisfies%
\[
f_{q}\star\left(  \lambda_{q}\mathfrak{i}\right)  =\lambda_{q}\cdot
\underbrace{\left(  f_{q}\star\mathfrak{i}\right)  }_{\substack{=f_{q}%
\\\text{(since }\mathfrak{i}\text{ is the unity of}\\\text{the }%
\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \text{)}}}=\lambda_{q}f_{q},
\]
this rewrites as follows: The family $\left(  \lambda_{q}f_{q}\right)  _{q\in
Q}\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$ is
pointwise finitely supported. This proves
Proposition~\ref{prop.pw-fin-supp.P4}.
\end{proof}

We have now proven all five Propositions~\ref{prop.pw-fin-supp.P1}%
,~\ref{prop.pw-fin-supp.P2},~\ref{prop.pw-fin-supp.P3}%
,~\ref{prop.pw-fin-supp.P4}~and~\ref{prop.pw-fin-supp.P5}. Thus,
Exercise~\ref{exe.pw-fin-supp.P1-5} is solved.

[\textit{Remark:} We can re-interpret the concept of
``pointwise finitely supported'' families $\left(
f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ and their sums $\sum_{q\in Q}f_{q}$ in topological terms.
To that end, we shall use the concept of a ``net''
(see, e.g., \url{https://en.wikipedia.org/wiki/Net_(mathematics)} or
\cite[\S 4]{Vermeeren-nets} for an introduction).

Recall that a \dfn{preordered set} means a set $Z$ equipped with a
\idx{preorder relation}
(i.e., a binary relation on $Z$ that is both reflexive and
transitive\footnote{but (unlike a partial order) not necessarily
antisymmetric}). This preorder relation is commonly denoted by $\leq$. A
nonempty preordered set $Z$ is said to be a \dfn{directed set} if its
preorder relation $\leq$ has the property that every two elements $x\in Z$ and
$y\in Z$ have an upper bound (i.e., some $z\in Z$ satisfying $x\leq z$ and
$y\leq z$). If $Z$ is a preordered set, and if $\mathcal{A}\left(  z\right)  $
is a logical statement for each $z\in Z$, then we say that
``$\mathcal{A}\left(  z\right)  $ holds
\emph{for all sufficiently high }$z\in Z$''
if and only if there exists a $w\in Z$ such that every
$z\in Z$ satisfying $w\leq z$ satisfies $\mathcal{A}\left(  z\right)  $.

Two important examples of directed sets are the following:

\begin{itemize}
\item The set $\NN$, equipped with the usual less-or-equal relation
$\leq$, is a directed set. This directed set will simply be called
$\NN$.

\item If $Q$ is any set, then the set $\mathcal{P}_{\operatorname*{fin}%
}\left(  Q\right)  $ of all finite subsets of $Q$ is naturally a directed set:
Its preorder relation $\leq$ is defined to be the subset relation $\subset$.
Every two elements $x\in\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  $
and $y\in\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  $ clearly have an
upper bound (for example, $x\cup y$). This directed set will simply be called
$\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  $.
\end{itemize}

A \dfn{net} in a set $X$ is defined to be a family $\left(  x_{z}\right)
_{z\in Z}\in X^{Z}$, where $Z$ is some directed set.

If $X$ is a topological space, if $x\in X$, and if $\left(  x_{z}\right)
_{z\in Z}\in X^{Z}$ is a net in $X$, then the net $\left(  x_{z}\right)
_{z\in Z}$ is said to \emph{converge}\index{convergence of a net}
to $x$ if and only if for each
neighborhood $U$ of $x$, we have%
\[
\left(  x_{z}\in U\text{ for all sufficiently high }z\in Z\right)  .
\]
Thus, in any topological space, we have defined the notion of a convergent
net. This notion generalizes the notion of a convergent sequence (indeed,
convergent sequences are precisely the same as convergent nets whose indexing
set $Z$ is the directed set $\NN$). But it is, in a sense, a more
natural notion than the latter: Unlike the latter, it characterizes the
topological space. That is, we can define a topological space on a set $Y$ by
specifying which nets in $Y$ converge to which elements of $Y$ (provided that
this specification satisfies certain axioms); but we cannot (in general)
define a topological space on a set $Y$ by specifying which sequences in $Y$
converge to which elements of $Y$.

If a net $\left(  x_{z}\right)  _{z\in Z}\in X^{Z}$ in a topological space $X$
converges to an element $x\in X$, then $x$ is called a
\emph{limit}\index{limit of a net} of
$\left(  x_{z}\right)  _{z\in Z}$. If $X$ is Hausdorff, then any convergent
net $\left(  x_{z}\right)  _{z\in Z}\in X^{Z}$ has only one limit, and so we
can call this limit ``\emph{the} limit'' of
$\left(  x_{z}\right)  _{z\in Z}$.

If $X$ is a topological space with the discrete topology, then convergence of
nets can be described very simply: A net $\left(  x_{z}\right)  _{z\in Z}\in
X^{Z}$ in a discrete topological space $X$ converges to an element $x\in X$ if
and only if we have $\left(  x_{z}=x\text{ for all sufficiently high }z\in
Z\right)  $. This behavior is also called \dfn{stabilization}: i.e., we say
that a net $\left(  x_{z}\right)  _{z\in Z}\in X^{Z}$
\emph{stabilizes}\index{stabilizing net} to
an element $x\in X$ if and only if we have $\left(  x_{z}=x\text{ for all
sufficiently high }z\in Z\right)  $.

Let us equip the set $A$ with the discrete topology. Thus, $A$ becomes a
topological $\kk$-algebra (because equipping any $\kk$-algebra
with the discrete topology results in a topological $\kk$-algebra).

Now, we equip the set $\Hom \left(  C,A\right)  $ with a
topology, which can be defined in any of the following two ways:

\begin{itemize}
\item It is the unique topology on the set $\Hom \left(
C,A\right)  $ that has the following property: A net $\left(  f_{z}\right)
_{z\in Z}$ of maps $f_{z}\in\Hom \left(  C,A\right)  $ converges
to a map $f\in\Hom \left(  C,A\right)  $ in this topology if and
only if for each $c\in C$, the net $\left(  f_{z}\left(  c\right)  \right)
_{z\in Z}$ in $A$ stabilizes to $f\left(  c\right)  $.

\item Alternatively, we can define the topology on $\Hom \left(
C,A\right)  $ in the usual way (i.e., via open sets): The set $A^{C}$ of all
maps from $C$ to $A$ is equipped with a product topology (since it is the
product $\prod_{c\in C}A$). The set $\Hom \left(  C,A\right)  $
thus also gets a topology, being a subset of $A^{C}$.
\end{itemize}

\noindent
These two definitions give rise to the same topology. This topology is called
the \dfn{topology of pointwise convergence}. We consider
$\Hom \left(  C,A\right)  $ to be equipped with this topology
from now on. This topology allows us to work with limits of nets in
$\Hom \left(  C,A\right)  $ (as long as these nets converge),
since the topological space $\Hom \left(  C,A\right)  $ is Hausdorff.

\begin{proposition}
\label{prop.sol.pw-fin-supp.P1-5.top}The $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ is a topological
$\kk$-algebra. That is, the maps%
\begin{align*}
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)   &  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto f+g,\\
\Hom \left(  C,A\right)   &  \rightarrow\Hom
\left(  C,A\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto-f,\\
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)   &  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto f\star g,\\
\kk\times\Hom \left(  C,A\right)   &  \rightarrow
\Hom \left(  C,A\right)  ,\ \ \ \ \ \ \ \ \ \ \left(
\lambda,f\right)  \mapsto\lambda f
\end{align*}
are continuous (where the topology on $\kk$ is the discrete topology).
\end{proposition}

We omit the proof of Proposition~\ref{prop.sol.pw-fin-supp.P1-5.top}, since we
shall not use it; it is not hard to prove with some standard techniques from
point-set topology. But let us see how it allows us to re-interpret pointwise
finitely supported families:

\begin{itemize}
\item Any family $\left(  a_{q}\right)  _{q\in Q}\in A^{Q}$ of elements of $A$
gives rise to a net $\left(  \sum_{q\in K}a_{q}\right)  _{K\in\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }\in A^{\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }$ in $A$ (which consists of all sums
of finite subfamilies of $\left(  a_{q}\right)  _{q\in Q}$). It is easy to see
that the family $\left(  a_{q}\right)  _{q\in Q}$ is finitely supported if and
only if the net $\left(  \sum_{q\in K}a_{q}\right)  _{K\in\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }\in A^{\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }$ converges in the discrete space
$A$. In this case, the limit of the net is precisely $\sum_{q\in Q}a_{q}$.

\item Any family $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ of elements of
$\Hom \left(  C,A\right)  $ gives rise to a net $\left(
\sum_{q\in K}f_{q}\right)  _{K\in\mathcal{P}_{\operatorname*{fin}}\left(
Q\right)  }\in\left(  \Hom \left(  C,A\right)  \right)
^{\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  }$ in
$\Hom \left(  C,A\right)  $ (which consists of all sums of
finite subfamilies of $\left(  f_{q}\right)  _{q\in Q}$). It is easy to see
that the family $\left(  f_{q}\right)  _{q\in Q}$ is pointwise finitely
supported if and only if the net $\left(  \sum_{q\in K}f_{q}\right)
_{K\in\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  }\in\left(
\Hom \left(  C,A\right)  \right)  ^{\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }$ converges in $\Hom
\left(  C,A\right)  $. In this case, the limit of the net is precisely
$\sum_{q\in Q}f_{q}$.
\end{itemize}

It is clear that this line of reasoning allows us to generalize the notion of
``pointwise finitely supported families'' to
families in any topological $\kk$-module, and to define the sum of any
such family.]

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pw-fin-supp.P1-5}}

\textit{Solution to Exercise~\ref{exe.pw-fin-supp.P1-5}.} We shall use the
following simple fact:

\begin{statement}
\textit{Fact A.0:} Let $\alpha$ and $\beta$ be two maps in
$\Hom \left(  C,A\right)  $. Let $x\in C$. Let some
$k\in \NN$, some elements $y_{1},y_{2},\ldots,y_{k}\in C$ and some
elements $z_{1},z_{2},\ldots,z_{k}\in C$ be chosen such that $\Delta\left(
x\right)  =\sum_{p=1}^{k}y_{p}\otimes z_{p}$. Then,
\[
\left(  \alpha\star\beta\right)  \left(  x\right)  =\sum_{p=1}^{k}%
\alpha\left(  y_{p}\right)  \beta\left(  z_{p}\right)  .
\]

\end{statement}

[\textit{Proof of Fact A.0:} Let $m$ denote the multiplication map $A\otimes
A\rightarrow A$ of the $\kk$-algebra $A$. The definition of convolution
yields $\alpha\star\beta=m\circ\left(  \alpha\otimes\beta\right)  \circ\Delta
$. Hence,%
\begin{align*}
\left(  \alpha\star\beta\right)  \left(  x\right)   &  =\left(  m\circ\left(
\alpha\otimes\beta\right)  \circ\Delta\right)  \left(  x\right)  =m\left(
\left(  \alpha\otimes\beta\right)  \left(  \underbrace{\Delta\left(  x\right)
}_{=\sum_{p=1}^{k}y_{p}\otimes z_{p}}\right)  \right) \\
&  =m\left(  \underbrace{\left(  \alpha\otimes\beta\right)  \left(
\sum_{p=1}^{k}y_{p}\otimes z_{p}\right)  }_{\substack{=\sum_{p=1}^{k}\left(
\alpha\otimes\beta\right)  \left(  y_{p}\otimes z_{p}\right)  \\\text{(since
the map }\alpha\otimes\beta\text{ is }\kk\text{-linear)}}}\right)
=m\left(  \sum_{p=1}^{k}\underbrace{\left(  \alpha\otimes\beta\right)  \left(
y_{p}\otimes z_{p}\right)  }_{\substack{=\alpha\left(  y_{p}\right)
\otimes\beta\left(  z_{p}\right)  \\\text{(by the definition of }\alpha
\otimes\beta\text{)}}}\right)  =m\left(  \sum_{p=1}^{k}\alpha\left(
y_{p}\right)  \otimes\beta\left(  z_{p}\right)  \right) \\
&  =\sum_{p=1}^{k}\alpha\left(  y_{p}\right)  \beta\left(  z_{p}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }m\right)  .
\end{align*}
This proves Fact A.0.]

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P1}.]Let $g$ be the map
$\sum_{q\in Q}f_{q}$.\ \ \ \ \footnote{This map is well-defined, since the
family $\left(  f_{q}\right)  _{q\in Q}$ is pointwise finitely supported.}
Thus, $g$ is a map $C\rightarrow A$. Moreover, $g=\sum_{q\in Q}f_{q}$. Thus,
each $x\in C$ satisfies%
\begin{equation}
\underbrace{g}_{=\sum_{q\in Q}f_{q}}\left(  x\right)  =\left(  \sum_{q\in
Q}f_{q}\right)  \left(  x\right)  =\sum_{q\in Q}f_{q}\left(  x\right)
\label{sol.exe.convolution-series.a.PA.1.pf.1}%
\end{equation}
(by the definition of $\sum_{q\in Q}f_{q}$).

Now, let $c\in C$, $d\in C$, $\lambda\in\kk$ and $\mu\in\kk$ be
arbitrary. Then, (\ref{sol.exe.convolution-series.a.PA.1.pf.1}) (applied to
$x=c$) yields%
\[
g\left(  c\right)  =\sum_{q\in Q}f_{q}\left(  c\right)  .
\]
Moreover, the family $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported. In other words, for each $x\in C$, the family $\left(  f_{q}\left(
x\right)  \right)  _{q\in Q}\in A^{Q}$ of elements of $A$ is finitely
supported (by the definition of ``pointwise finitely
supported''). Applying this to $x=c$, we conclude that the
family $\left(  f_{q}\left(  c\right)  \right)  _{q\in Q}\in A^{Q}$ of
elements of $A$ is finitely supported. In other words, all but finitely many
$q\in Q$ satisfy $f_{q}\left(  c\right)  =0$. Hence, all but finitely many
$q\in Q$ satisfy $\lambda f_{q}\left(  c\right)  =0$ (since every $q\in Q$
that satisfies $f_{q}\left(  c\right)  =0$ must also satisfy $\lambda
\underbrace{f_{q}\left(  c\right)  }_{=0}=0$). In other words, the family
$\left(  \lambda f_{q}\left(  c\right)  \right)  _{q\in Q}\in A^{Q}$ is
finitely supported. The same argument (applied to $d$ and $\mu$ instead of $c$
and $\lambda$) shows that the family $\left(  \mu f_{q}\left(  d\right)
\right)  _{q\in Q}\in A^{Q}$ of elements of $A$ is finitely supported.

Recall that sums of finitely supported families satisfy the same rules as
finite sums. Since the families $\left(  \lambda f_{q}\left(  c\right)
\right)  _{q\in Q}$ and $\left(  \mu f_{q}\left(  d\right)  \right)  _{q\in
Q}$ are finitely supported, we thus have%
\begin{equation}
\sum_{q\in Q}\lambda f_{q}\left(  c\right)  +\sum_{q\in Q}\mu f_{q}\left(
d\right)  =\sum_{q\in Q}\left(  \lambda f_{q}\left(  c\right)  +\mu
f_{q}\left(  d\right)  \right)  \label{sol.exe.convolution-series.a.PA.1.pf.3}%
\end{equation}
(and in particular, the family $\left(  \lambda f_{q}\left(  c\right)  +\mu
f_{q}\left(  d\right)  \right)  _{q\in Q}$ is finitely supported). For the
same reason, we have
\begin{equation}
\lambda\sum_{q\in Q}f_{q}\left(  c\right)  =\sum_{q\in Q}\lambda f_{q}\left(
c\right)  \label{sol.exe.convolution-series.a.PA.1.pf.4a}%
\end{equation}
(since the family $\left(  f_{q}\left(  c\right)  \right)  _{q\in Q}$ is
finitely supported) and%
\begin{equation}
\mu\sum_{q\in Q}f_{q}\left(  d\right)  =\sum_{q\in Q}\mu f_{q}\left(
d\right)  \label{sol.exe.convolution-series.a.PA.1.pf.4b}%
\end{equation}
(for similar reasons).

Now,%
\begin{align*}
g\left(  \lambda c+\mu d\right)   &  =\sum_{q\in Q}\underbrace{f_{q}\left(
\lambda c+\mu d\right)  }_{\substack{=\lambda f_{q}\left(  c\right)  +\mu
f_{q}\left(  d\right)  \\\text{(since the map }f_{q}\text{ is }\kk%
\text{-linear}\\\text{(since }f_{q}\in\Hom \left(  C,A\right)
\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.1.pf.1}) (applied to }x=\lambda c+\mu
d\text{)}\right) \\
&  =\sum_{q\in Q}\left(  \lambda f_{q}\left(  c\right)  +\mu f_{q}\left(
d\right)  \right)  =\underbrace{\sum_{q\in Q}\lambda f_{q}\left(  c\right)
}_{\substack{=\lambda\sum_{q\in Q}f_{q}\left(  c\right)  \\\text{(by
(\ref{sol.exe.convolution-series.a.PA.1.pf.4a}))}}}+\underbrace{\sum_{q\in
Q}\mu f_{q}\left(  d\right)  }_{\substack{=\mu\sum_{q\in Q}f_{q}\left(
d\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.1.pf.4b}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.1.pf.3})}\right) \\
&  =\lambda\sum_{q\in Q}f_{q}\left(  c\right)  +\mu\sum_{q\in Q}f_{q}\left(
d\right)  .
\end{align*}
Comparing this with%
\[
\lambda\underbrace{g\left(  c\right)  }_{=\sum_{q\in Q}f_{q}\left(  c\right)
}+\mu\underbrace{g\left(  d\right)  }_{\substack{=\sum_{q\in Q}f_{q}\left(
d\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.1.pf.1}) (applied
to }x=d\text{))}}}=\lambda\sum_{q\in Q}f_{q}\left(  c\right)  +\mu\sum_{q\in
Q}f_{q}\left(  d\right)  ,
\]
we obtain $g\left(  \lambda c+\mu d\right)  =\lambda g\left(  c\right)  +\mu
g\left(  d\right)  $.

Now, forget that we fixed $c$, $d$, $\lambda$ and $\mu$. We thus have proven
that $g\left(  \lambda c+\mu d\right)  =\lambda g\left(  c\right)  +\mu
g\left(  d\right)  $ for every $c\in C$, $d\in C$, $\lambda\in\kk$ and
$\mu\in\kk$. In other words, the map $g$ is $\kk$-linear. In
other words, the map $\sum_{q\in Q}f_{q}$ is $\kk$-linear (since
$g=\sum_{q\in Q}f_{q}$). In other words, the map $\sum_{q\in Q}f_{q}$ belongs
to $\Hom \left(  C,A\right)  $ (since $\sum_{q\in Q}f_{q}$ is a
map $C\rightarrow A$). This proves Proposition~\ref{prop.pw-fin-supp.P1}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P2}.]The family $\left(
f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is pointwise finitely supported. In other words, for each $x\in
C$,%
\begin{equation}
\text{the family }\left(  f_{q}\left(  x\right)  \right)  _{q\in Q}\in
A^{Q}\text{ of elements of }A\text{ is finitely supported}
\label{sol.exe.convolution-series.a.PA.2.pf.1}%
\end{equation}
(by the definition of ``pointwise finitely
supported''). The same argument (applied to $\left(
g_{q}\right)  _{q\in Q}$ instead of $\left(  f_{q}\right)  _{q\in Q}$) shows
that for each $x\in C$,%
\begin{equation}
\text{the family }\left(  g_{q}\left(  x\right)  \right)  _{q\in Q}\in
A^{Q}\text{ of elements of }A\text{ is finitely supported.}
\label{sol.exe.convolution-series.a.PA.2.pf.2}%
\end{equation}


Now, fix $x\in C$. From (\ref{sol.exe.convolution-series.a.PA.2.pf.1}), we
know that the family $\left(  f_{q}\left(  x\right)  \right)  _{q\in Q}\in
A^{Q}$ of elements of $A$ is finitely supported. In other words, all but
finitely many $q\in Q$ satisfy $f_{q}\left(  x\right)  =0$. In other words,
there exists a finite subset $Q_{1}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{1}\text{ satisfies }f_{q}\left(  x\right)  =0.
\label{sol.exe.convolution-series.a.PA.2.pf.3}%
\end{equation}
Consider this $Q_{1}$.

From (\ref{sol.exe.convolution-series.a.PA.2.pf.2}), we know that the family
$\left(  g_{q}\left(  x\right)  \right)  _{q\in Q}\in A^{Q}$ of elements of
$A$ is finitely supported. In other words, all but finitely many $q\in Q$
satisfy $g_{q}\left(  x\right)  =0$. In other words, there exists a finite
subset $Q_{2}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{2}\text{ satisfies }g_{q}\left(  x\right)  =0.
\label{sol.exe.convolution-series.a.PA.2.pf.4}%
\end{equation}
Consider this $Q_{2}$.

The two sets $Q_{1}$ and $Q_{2}$ are finite. Hence, their union $Q_{1}\cup
Q_{2}$ is also finite. Thus, all but finitely many $q\in Q$ satisfy $q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $.

Also, $Q_{1}\cup Q_{2}$ is a subset of $Q$ (since both $Q_{1}$ and $Q_{2}$ are
subsets of $Q$). In other words, $Q_{1}\cup Q_{2}\subset Q$.

Every $q\in Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $ satisfies
\begin{equation}
f_{q}\left(  x\right)  =0 \label{sol.exe.convolution-series.a.PA.2.pf.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.2.pf.5}):}
Let $q\in Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $. Thus, $q\in
Q\setminus\underbrace{\left(  Q_{1}\cup Q_{2}\right)  }_{\supset Q_{1}}\subset
Q\setminus Q_{1}$. Hence, $f_{q}\left(  x\right)  =0$ (by
(\ref{sol.exe.convolution-series.a.PA.2.pf.3})). This proves
(\ref{sol.exe.convolution-series.a.PA.2.pf.5}).} and
\begin{equation}
g_{q}\left(  x\right)  =0 \label{sol.exe.convolution-series.a.PA.2.pf.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.2.pf.6}):}
Let $q\in Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $. Thus, $q\in
Q\setminus\underbrace{\left(  Q_{1}\cup Q_{2}\right)  }_{\supset Q_{2}}\subset
Q\setminus Q_{2}$. Hence, $g_{q}\left(  x\right)  =0$ (by
(\ref{sol.exe.convolution-series.a.PA.2.pf.4})). This proves
(\ref{sol.exe.convolution-series.a.PA.2.pf.6}).}. Thus, every $q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $ satisfies%
\begin{equation}
\left(  f_{q}+g_{q}\right)  \left(  x\right)  =\underbrace{f_{q}\left(
x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.5}))}}}+\underbrace{g_{q}\left(
x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.6}))}}}=0+0=0.
\label{sol.exe.convolution-series.a.PA.2.pf.7}%
\end{equation}
Hence, all but finitely many $q\in Q$ satisfy $\left(  f_{q}+g_{q}\right)
\left(  x\right)  =0$ (since all but finitely many $q\in Q$ satisfy $q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  $). In other words,
\begin{equation}
\text{the family }\left(  \left(  f_{q}+g_{q}\right)  \left(  x\right)
\right)  _{q\in Q}\in A^{Q}\text{ is finitely supported.}
\label{sol.exe.convolution-series.a.PA.2.pf.fs}%
\end{equation}


We have%
\begin{align}
\sum_{q\in Q}f_{q}\left(  x\right)   &  =\underbrace{\sum\limits_{\substack{q\in
Q;\\q\in Q_{1}\cup Q_{2}}}}_{\substack{=\sum_{q\in Q_{1}\cup Q_{2}%
}\\\text{(since }Q_{1}\cup Q_{2}\subset Q\text{)}}}f_{q}\left(  x\right)
+\underbrace{\sum\limits_{\substack{q\in Q;\\q\notin Q_{1}\cup Q_{2}}}}_{=\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }}f_{q}\left(  x\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }q\in Q\text{ satisfies either }q\in Q_{1}\cup Q_{2}\text{ or
}q\notin Q_{1}\cup Q_{2}\\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }\underbrace{f_{q}\left(  x\right)
}_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.a.PA.2.pf.5}))}%
}}\nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\underbrace{\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }0}_{=0}=\sum_{q\in Q_{1}\cup Q_{2}%
}f_{q}\left(  x\right)  . \label{sol.exe.convolution-series.a.PA.2.pf.8}%
\end{align}
Also,%
\begin{align*}
\sum_{q\in Q}g_{q}\left(  x\right)   &  =\underbrace{\sum\limits_{\substack{q\in
Q;\\q\in Q_{1}\cup Q_{2}}}}_{\substack{=\sum_{q\in Q_{1}\cup Q_{2}%
}\\\text{(since }Q_{1}\cup Q_{2}\subset Q\text{)}}}g_{q}\left(  x\right)
+\underbrace{\sum\limits_{\substack{q\in Q;\\q\notin Q_{1}\cup Q_{2}}}}_{=\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }}g_{q}\left(  x\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }q\in Q\text{ satisfies either }q\in Q_{1}\cup Q_{2}\text{ or
}q\notin Q_{1}\cup Q_{2}\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum_{q\in Q_{1}\cup Q_{2}}g_{q}\left(  x\right)  +\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }\underbrace{g_{q}\left(  x\right)
}_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.a.PA.2.pf.6}))}%
}}\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}g_{q}\left(  x\right)  +\underbrace{\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }0}_{=0}=\sum_{q\in Q_{1}\cup Q_{2}%
}g_{q}\left(  x\right)  .
\end{align*}
Adding this equality to (\ref{sol.exe.convolution-series.a.PA.2.pf.8}), we
obtain%
\begin{equation}
\sum_{q\in Q}f_{q}\left(  x\right)  +\sum_{q\in Q}g_{q}\left(  x\right)
=\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\sum_{q\in Q_{1}\cup
Q_{2}}g_{q}\left(  x\right)  . \label{sol.exe.convolution-series.a.PA.2.pf.10}%
\end{equation}


But%
\begin{align}
\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  \left(  x\right)   &
=\underbrace{\sum\limits_{\substack{q\in Q;\\q\in Q_{1}\cup Q_{2}}}}_{\substack{
=\sum_{q\in Q_{1}\cup Q_{2}}\\\text{(since }Q_{1}\cup Q_{2}\subset Q\text{)}%
}}\left(  f_{q}+g_{q}\right)  \left(  x\right)  +\underbrace{
\sum\limits_{\substack{q\in Q;\\q\notin Q_{1}\cup Q_{2}}}}_{
=\sum_{q\in Q\setminus\left(
Q_{1}\cup Q_{2}\right)  }}\left(  f_{q}+g_{q}\right)  \left(  x\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }q\in Q\text{ satisfies either }q\in Q_{1}\cup Q_{2}\text{ or
}q\notin Q_{1}\cup Q_{2}\\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}\underbrace{\left(  f_{q}+g_{q}\right)  \left(
x\right)  }_{=f_{q}\left(  x\right)  +g_{q}\left(  x\right)  }+\sum_{q\in
Q\setminus\left(  Q_{1}\cup Q_{2}\right)  }\underbrace{\left(  f_{q}%
+g_{q}\right)  \left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.2.pf.7}))}}}\nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}\left(  f_{q}\left(  x\right)  +g_{q}\left(
x\right)  \right)  +\underbrace{\sum_{q\in Q\setminus\left(  Q_{1}\cup
Q_{2}\right)  }0}_{=0}=\sum_{q\in Q_{1}\cup Q_{2}}\left(  f_{q}\left(
x\right)  +g_{q}\left(  x\right)  \right) \nonumber\\
&  =\sum_{q\in Q_{1}\cup Q_{2}}f_{q}\left(  x\right)  +\sum_{q\in Q_{1}\cup
Q_{2}}g_{q}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have manipulated a finite sum}\\
\text{(since }Q_{1}\cup Q_{2}\text{ is a finite set)}%
\end{array}
\right) \nonumber\\
&  =\sum_{q\in Q}f_{q}\left(  x\right)  +\sum_{q\in Q}g_{q}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.2.pf.10})}\right)  .
\label{sol.exe.convolution-series.a.PA.2.pf.13}%
\end{align}


Now, let us forget that we fixed $x$. We thus have proven that every $x\in C$
satisfies (\ref{sol.exe.convolution-series.a.PA.2.pf.fs}) and
(\ref{sol.exe.convolution-series.a.PA.2.pf.13}).

In particular, every $x\in C$ satisfies
(\ref{sol.exe.convolution-series.a.PA.2.pf.fs}). In other words, for each
$x\in C$, the family $\left(  \left(  f_{q}+g_{q}\right)  \left(  x\right)
\right)  _{q\in Q}\in A^{Q}$ of elements of $A$ is finitely supported. In
other words, the family $\left(  f_{q}+g_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported (by the definition of ``pointwise finitely
supported'').

Hence, the sum $\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  $ is well-defined.
Also, the sum $\sum_{q\in Q}f_{q}$ is well-defined (since the family $\left(
f_{q}\right)  _{q\in Q}$ is pointwise finitely supported). Similarly, the sum
$\sum_{q\in Q}g_{q}$ is well-defined.

Moreover, each $x\in C$ satisfies%
\begin{align*}
\left(  \sum_{q\in Q}\left(  f_{q}+g_{q}\right)  \right)  \left(  x\right)
&  =\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  \left(  x\right)
=\underbrace{\sum_{q\in Q}f_{q}\left(  x\right)  }_{=\left(  \sum_{q\in
Q}f_{q}\right)  \left(  x\right)  }+\underbrace{\sum_{q\in Q}g_{q}\left(
x\right)  }_{=\left(  \sum_{q\in Q}g_{q}\right)  \left(  x\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.convolution-series.a.PA.2.pf.13})}\right) \\
&  =\left(  \sum_{q\in Q}f_{q}\right)  \left(  x\right)  +\left(  \sum_{q\in
Q}g_{q}\right)  \left(  x\right)  =\left(  \sum_{q\in Q}f_{q}+\sum_{q\in
Q}g_{q}\right)  \left(  x\right)  .
\end{align*}
In other words, $\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  =\sum_{q\in Q}%
f_{q}+\sum_{q\in Q}g_{q}$. In other words, $\sum_{q\in Q}f_{q}+\sum_{q\in
Q}g_{q}=\sum_{q\in Q}\left(  f_{q}+g_{q}\right)  $. This completes the proof
of Proposition~\ref{prop.pw-fin-supp.P2}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P3}.]Let $x\in C$. Write the
element $\Delta\left(  x\right)  \in C\otimes C$ in the form $\Delta\left(
x\right)  =\sum_{p=1}^{k}y_{p}\otimes z_{p}$ for some $k\in \NN$, some
elements $y_{1},y_{2},\ldots,y_{k}\in C$ and some elements $z_{1},z_{2}%
,\ldots,z_{k}\in C$. (This is possible, because $\Delta\left(  x\right)  $ can
be written as a sum of pure tensors\footnote{This is because any tensor in
$C\otimes C$ can be written as a sum of pure tensors.}.)

For each $p\in\left\{  1,2,\ldots,k\right\}  $, there exists a finite subset
$Q_{p}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{p}\text{ satisfies }f_{q}\left(  y_{p}\right)
=0 \label{sol.exe.convolution-series.a.PA.3.pf.1}%
\end{equation}
\footnote{\textit{Proof:} Let $p\in\left\{  1,2,\ldots,k\right\}  $. The
family $\left(  f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is pointwise finitely supported. In other words,
for each $c\in C$, the family $\left(  f_{q}\left(  c\right)  \right)  _{q\in
Q}\in A^{Q}$ is finitely supported (by the definition of
``pointwise finitely supported''). Applying this to $c=y_{p}$,
we conclude that the family $\left(  f_{q}\left(  y_{p}\right)  \right)
_{q\in Q}\in A^{Q}$ is finitely supported. In other words, all but finitely
many $q\in Q$ satisfy $f_{q}\left(  y_{p}\right)  =0$. In other words, there
exists a finite subset $Q_{p}$ of $Q$ such that every $q\in Q\setminus Q_{p}$
satisfies $f_{q}\left(  y_{p}\right)  =0$. Qed.}. Consider this $Q_{p}$.

Define a subset $Q^{\prime}$ of $Q$ by $Q^{\prime}=Q_{1}\cup Q_{2}\cup
\cdots\cup Q_{k}$. Thus, $Q^{\prime}$ is the union of the $k$ finite sets
$Q_{1},Q_{2},\ldots,Q_{k}$. Hence, $Q^{\prime}$ is a union of finitely many
finite sets. Thus, $Q^{\prime}$ is itself a finite set (since a union of
finitely many finite sets must itself be a finite set).

Every $q\in Q\setminus Q^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $
satisfy%
\begin{equation}
f_{q}\left(  y_{p}\right)  =0 \label{sol.exe.convolution-series.a.PA.3.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.3}):}
Let $q\in Q\setminus Q^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $.
\par
We have $p\in\left\{  1,2,\ldots,k\right\}  $. Thus, $Q_{p}$ is one of the
terms in the union $Q_{1}\cup Q_{2}\cup\cdots\cup Q_{k}$. Hence, $Q_{p}\subset
Q_{1}\cup Q_{2}\cup\cdots\cup Q_{k}=Q^{\prime}$ (since $Q^{\prime}=Q_{1}\cup
Q_{2}\cup\cdots\cup Q_{k}$), so that $Q\setminus\underbrace{Q_{p}}_{\subset
Q^{\prime}}\supset Q\setminus Q^{\prime}$. Hence, $Q\setminus Q^{\prime
}\subset Q\setminus Q_{p}$. Hence, $q\in Q\setminus Q^{\prime}\subset
Q\setminus Q_{p}$. Thus, (\ref{sol.exe.convolution-series.a.PA.3.pf.1}) yields
$f_{q}\left(  y_{p}\right)  =0$. This proves
(\ref{sol.exe.convolution-series.a.PA.3.pf.3}).}.

For each $p\in\left\{  1,2,\ldots,k\right\}  $, there exists a finite subset
$R_{p}$ of $R$ such that
\begin{equation}
\text{every }r\in R\setminus R_{p}\text{ satisfies }g_{r}\left(  z_{p}\right)
=0 \label{sol.exe.convolution-series.a.PA.3.pf.1b}%
\end{equation}
\footnote{\textit{Proof:} Let $p\in\left\{  1,2,\ldots,k\right\}  $. The
family $\left(  g_{r}\right)  _{r\in R}\in\left(  \Hom \left(
C,A\right)  \right)  ^{R}$ is pointwise finitely supported. In other words,
for each $c\in C$, the family $\left(  g_{r}\left(  c\right)  \right)  _{r\in
R}\in A^{R}$ is finitely supported (by the definition of
``pointwise finitely supported''). Applying this to $c=z_{p}$,
we conclude that the family $\left(  g_{r}\left(  z_{p}\right)  \right)
_{r\in R}\in A^{R}$ is finitely supported. In other words, all but finitely
many $r\in R$ satisfy $g_{r}\left(  z_{p}\right)  =0$. In other words, there
exists a finite subset $R_{p}$ of $R$ such that every $r\in R\setminus R_{p}$
satisfies $g_{r}\left(  z_{p}\right)  =0$. Qed.}. Consider this $R_{p}$.

Define a subset $R^{\prime}$ of $R$ by $R^{\prime}=R_{1}\cup R_{2}\cup
\cdots\cup R_{k}$. Thus, $R^{\prime}$ is the union of the $k$ finite sets
$R_{1},R_{2},\ldots,R_{k}$. Hence, $R^{\prime}$ is a union of finitely many
finite sets. Thus, $R^{\prime}$ is itself a finite set (since a union of
finitely many finite sets must itself be a finite set).

Every $r\in R\setminus R^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $
satisfy%
\begin{equation}
g_{r}\left(  z_{p}\right)  =0 \label{sol.exe.convolution-series.a.PA.3.pf.3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.3b}):}
Let $r\in R\setminus R^{\prime}$ and $p\in\left\{  1,2,\ldots,k\right\}  $.
\par
We have $p\in\left\{  1,2,\ldots,k\right\}  $. Thus, $R_{p}$ is one of the
terms in the union $R_{1}\cup R_{2}\cup\cdots\cup R_{k}$. Hence, $R_{p}\subset
R_{1}\cup R_{2}\cup\cdots\cup R_{k}=R^{\prime}$ (since $R^{\prime}=R_{1}\cup
R_{2}\cup\cdots\cup R_{k}$), so that $R\setminus\underbrace{R_{p}}_{\subset
R^{\prime}}\supset R\setminus R^{\prime}$. Hence, $R\setminus R^{\prime
}\subset R\setminus R_{p}$. Hence, $r\in R\setminus R^{\prime}\subset
R\setminus R_{p}$. Thus, (\ref{sol.exe.convolution-series.a.PA.3.pf.1b})
yields $g_{r}\left(  z_{p}\right)  =0$. This proves
(\ref{sol.exe.convolution-series.a.PA.3.pf.3b}).}.

The set $Q^{\prime}\times R^{\prime}$ is a Cartesian product of two finite
sets (since $Q^{\prime}$ and $R^{\prime}$ are finite sets), and thus is itself
finite (since a Cartesian product of two finite sets is always finite). Hence,
all but finitely many $\left(  q,r\right)  \in Q\times R$ satisfy $\left(
q,r\right)  \in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times
R^{\prime}\right)  $.

We shall now show that each $\left(  q,r\right)  \in\left(  Q\times R\right)
\setminus\left(  Q^{\prime}\times R^{\prime}\right)  $ satisfies $\left(
f_{q}\star g_{r}\right)  \left(  x\right)  =0$.

Indeed, fix $\left(  q,r\right)  \in\left(  Q\times R\right)  \setminus\left(
Q^{\prime}\times R^{\prime}\right)  $. We want to prove that $\left(
f_{q}\star g_{r}\right)  \left(  x\right)  =0$.

From $\left(  q,r\right)  \in\left(  Q\times R\right)  \setminus\left(
Q^{\prime}\times R^{\prime}\right)  $, we obtain $\left(  q,r\right)  \in
Q\times R$ and $\left(  q,r\right)  \notin Q^{\prime}\times R^{\prime}$. From
$\left(  q,r\right)  \in Q\times R$, we obtain $q\in Q$ and $r\in R$.

We are in one of the following two cases:

\textit{Case 1:} We have $q\in Q^{\prime}$.

\textit{Case 2:} We have $q\notin Q^{\prime}$.

Let us first consider Case 1. In this case, we have $q\in Q^{\prime}$. If we
had $r\in R^{\prime}$, we thus would have $\left(  q,r\right)  \in Q^{\prime
}\times R^{\prime}$ (since $q\in Q^{\prime}$ and $r\in R^{\prime}$), which
would contradict $\left(  q,r\right)  \notin Q^{\prime}\times R^{\prime}$.
Hence, we cannot have $r\in R^{\prime}$. In other words, we have $r\notin
R^{\prime}$. Combining $r\in R$ with $r\notin R^{\prime}$, we obtain $r\in
R\setminus R^{\prime}$. Now, Fact A.0 (applied to $\alpha=f_{q}$ and
$\beta=g_{r}$) yields%
\[
\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =\sum_{p=1}^{k}f_{q}\left(
y_{p}\right)  \underbrace{g_{r}\left(  z_{p}\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.3b}))}%
}}=\sum_{p=1}^{k}f_{q}\left(  y_{p}\right)  0=0.
\]
Thus, $\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ is proven in
Case 1.

Let us now consider Case 2. In this case, we have $q\notin Q^{\prime}$.
Combining $q\in Q$ with $q\notin Q^{\prime}$, we obtain $q\in Q\setminus
Q^{\prime}$. Now, Fact A.0 (applied to $\alpha=f_{q}$ and $\beta=g_{r}$)
yields%
\[
\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =\sum_{p=1}^{k}%
\underbrace{f_{q}\left(  y_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.3}))}}}g_{r}\left(  z_{p}\right)
=\sum_{p=1}^{k}0g_{r}\left(  z_{p}\right)  =0.
\]
Thus, $\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ is proven in
Case 2.

We have thus proven $\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ in
both Cases 1 and 2. Hence, $\left(  f_{q}\star g_{r}\right)  \left(  x\right)
=0$ always holds.

Now, forget that we fixed $\left(  q,r\right)  $. We thus have proven that
\begin{equation}
\text{each }\left(  q,r\right)  \in\left(  Q\times R\right)  \setminus\left(
Q^{\prime}\times R^{\prime}\right)  \text{ satisfies }\left(  f_{q}\star
g_{r}\right)  \left(  x\right)  =0.
\label{sol.exe.convolution-series.a.PA.3.pf.res1}%
\end{equation}
Thus, all but finitely many $\left(  q,r\right)  \in Q\times R$ satisfy
$\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =0$ (since all but
finitely many $\left(  q,r\right)  \in Q\times R$ satisfy $\left(  q,r\right)
\in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times R^{\prime
}\right)  $). In other words,%
\begin{equation}
\text{the family }\left(  \left(  f_{q}\star g_{r}\right)  \left(  x\right)
\right)  _{\left(  q,r\right)  \in Q\times R}\in A^{Q\times R}\text{ is
finitely supported.} \label{sol.exe.convolution-series.a.PA.3.pf.res3}%
\end{equation}


Define a map $F:C\rightarrow A$ by $F=\sum_{q\in Q}f_{q}$. (This is
well-defined, since the family $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely supported.)

Define a map $G:C\rightarrow A$ by $G=\sum_{r\in R}g_{r}$. (This is
well-defined, since the family $\left(  g_{r}\right)  _{r\in R}\in\left(
\Hom \left(  C,A\right)  \right)  ^{R}$ is pointwise finitely supported.)

We have $F=\sum_{q\in Q}f_{q}$ (by the definition of $F$). But the map
$\sum_{q\in Q}f_{q}$ belongs to $\Hom \left(  C,A\right)  $ (by
Proposition~\ref{prop.pw-fin-supp.P1}). In other words, the map $F$ belongs to
$\Hom \left(  C,A\right)  $ (since $F=\sum_{q\in Q}f_{q}$).
Similarly, the map $G$ belongs to $\Hom \left(  C,A\right)  $.
Hence, the map $F\star G\in\Hom \left(  C,A\right)  $ is
well-defined (since $F$ and $G$ belong to $\Hom \left(
C,A\right)  $).

Define a further map $F^{\prime}\in\Hom \left(  C,A\right)  $ by
$F^{\prime}=\sum_{q\in Q^{\prime}}f_{q}$. Notice that this is a finite sum,
since $Q^{\prime}$ is a finite set.

Define a further map $G^{\prime}\in\Hom \left(  C,A\right)  $ by
$G^{\prime}=\sum_{r\in R^{\prime}}g_{r}$. Notice that this is a finite sum,
since $R^{\prime}$ is a finite set.

We have%
\begin{align}
\underbrace{F^{\prime}}_{=\sum_{q\in Q^{\prime}}f_{q}}\star
\underbrace{G^{\prime}}_{=\sum_{r\in R^{\prime}}g_{r}}  &  =\left(  \sum_{q\in
Q^{\prime}}f_{q}\right)  \star\left(  \sum_{r\in R^{\prime}}g_{r}\right)
=\sum_{q\in Q^{\prime}}\underbrace{\left(  f_{q}\star\left(  \sum_{r\in
R^{\prime}}g_{r}\right)  \right)  }_{\substack{=\sum_{r\in R^{\prime}}\left(
f_{q}\star g_{r}\right)  \\\text{(by distributivity,}\\\text{since }\sum_{r\in
R^{\prime}}g_{r}\text{ is a finite sum)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by distributivity, since }\sum_{q\in
Q^{\prime}}f_{q}\star\left(  \sum_{r\in R^{\prime}}g_{r}\right)  \text{ is a
finite sum}\right) \nonumber\\
&  =\underbrace{\sum_{q\in Q^{\prime}}\sum_{r\in R^{\prime}}}_{\substack{=
\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\\\text{(since
}Q^{\prime}\text{ and }R^{\prime}\text{ are finite sets)}}}\left(  f_{q}\star
g_{r}\right)  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}%
}\left(  f_{q}\star g_{r}\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.F'G'1}%
\end{align}


But every $p\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
F^{\prime}\left(  y_{p}\right)  =F\left(  y_{p}\right)
\label{sol.exe.convolution-series.a.PA.3.pf.F'1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1}):}
Let $p\in\left\{  1,2,\ldots,k\right\}  $. Then,%
\begin{align*}
\underbrace{F}_{=\sum_{q\in Q}f_{q}}\left(  y_{p}\right)   &  =\left(
\sum_{q\in Q}f_{q}\right)  \left(  y_{p}\right)  =\sum_{q\in Q}f_{q}\left(
y_{p}\right)  =\underbrace{\sum\limits_{\substack{q\in Q;\\q\in Q^{\prime}}%
}}_{\substack{=\sum_{q\in Q^{\prime}}\\\text{(since }Q^{\prime}\subset
Q\text{)}}}f_{q}\left(  y_{p}\right)  +\underbrace{\sum\limits_{\substack{q\in
Q;\\q\notin Q^{\prime}}}}_{=\sum_{q\in Q\setminus Q^{\prime}}}f_{q}\left(
y_{p}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }q\in Q\text{ satisfies either }q\in Q^{\prime}\text{ or
}q\notin Q^{\prime}\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum_{q\in Q^{\prime}}f_{q}\left(  y_{p}\right)  +\sum_{q\in Q\setminus
Q^{\prime}}\underbrace{f_{q}\left(  y_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.3}))}}}=\sum_{q\in Q^{\prime}}%
f_{q}\left(  y_{p}\right)  +\underbrace{\sum_{q\in Q\setminus Q^{\prime}}%
0}_{=0}\\
&  =\sum_{q\in Q^{\prime}}f_{q}\left(  y_{p}\right)  .
\end{align*}
Comparing this with $\underbrace{F^{\prime}}_{=\sum_{q\in Q^{\prime}}f_{q}%
}\left(  y_{p}\right)  =\left(  \sum_{q\in Q^{\prime}}f_{q}\right)  \left(
y_{p}\right)  =\sum_{q\in Q^{\prime}}f_{q}\left(  y_{p}\right)  $, we obtain
$F^{\prime}\left(  y_{p}\right)  =F\left(  y_{p}\right)  $. This proves
(\ref{sol.exe.convolution-series.a.PA.3.pf.F'1}).} and%
\begin{equation}
G^{\prime}\left(  z_{p}\right)  =G\left(  z_{p}\right)
\label{sol.exe.convolution-series.a.PA.3.pf.F'1b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1b}):}
Let $p\in\left\{  1,2,\ldots,k\right\}  $. Then,%
\begin{align*}
\underbrace{G}_{=\sum_{r\in R}g_{r}}\left(  z_{p}\right)   &  =\left(
\sum_{r\in R}g_{r}\right)  \left(  z_{p}\right)  =\sum_{r\in R}g_{r}\left(
z_{p}\right)  =\underbrace{\sum\limits_{\substack{r\in R;\\r\in R^{\prime}}%
}}_{\substack{=\sum_{r\in R^{\prime}}\\\text{(since }R^{\prime}\subset
R\text{)}}}g_{r}\left(  z_{p}\right)  +\underbrace{\sum\limits_{\substack{r\in
R;\\r\notin R^{\prime}}}}_{=\sum_{r\in R\setminus R^{\prime}}}g_{r}\left(
z_{p}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }r\in R\text{ satisfies either }r\in R^{\prime}\text{ or
}r\notin R^{\prime}\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum_{r\in R^{\prime}}g_{r}\left(  z_{p}\right)  +\sum_{r\in R\setminus
R^{\prime}}\underbrace{g_{r}\left(  z_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.3b}))}}}=\sum_{r\in R^{\prime}%
}g_{r}\left(  z_{p}\right)  +\underbrace{\sum_{r\in R\setminus R^{\prime}}%
0}_{=0}\\
&  =\sum_{r\in R^{\prime}}g_{r}\left(  z_{p}\right)  .
\end{align*}
Comparing this with $\underbrace{G^{\prime}}_{=\sum_{r\in R^{\prime}}g_{r}%
}\left(  z_{p}\right)  =\left(  \sum_{r\in R^{\prime}}g_{r}\right)  \left(
z_{p}\right)  =\sum_{r\in R^{\prime}}g_{r}\left(  z_{p}\right)  $, we obtain
$G^{\prime}\left(  z_{p}\right)  =G\left(  z_{p}\right)  $. This proves
(\ref{sol.exe.convolution-series.a.PA.3.pf.F'1b}).}. Hence,%
\begin{equation}
\left(  F^{\prime}\star G^{\prime}\right)  \left(  x\right)  =\left(  F\star
G\right)  \left(  x\right)  \label{sol.exe.convolution-series.a.PA.3.pf.F'G'}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.3.pf.F'G'}):}
Fact A.0 (applied to $\alpha=F$ and $\beta=G$) yields%
\begin{equation}
\left(  F\star G\right)  \left(  x\right)  =\sum_{p=1}^{k}F\left(
y_{p}\right)  G\left(  z_{p}\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.F'G'.pf.2}%
\end{equation}
\par
Fact A.0 (applied to $\alpha=F^{\prime}$ and $\beta=G^{\prime}$) yields%
\begin{equation}
\left(  F^{\prime}\star G^{\prime}\right)  \left(  x\right)  =\sum_{p=1}%
^{k}\underbrace{F^{\prime}\left(  y_{p}\right)  }_{\substack{=F\left(
y_{p}\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1}))}%
}}\underbrace{G^{\prime}\left(  z_{p}\right)  }_{\substack{=G\left(
z_{p}\right)  \\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.F'1b}))}%
}}=\sum_{p=1}^{k}F\left(  y_{p}\right)  G\left(  z_{p}\right)  .\nonumber
\end{equation}
Comparing this with (\ref{sol.exe.convolution-series.a.PA.3.pf.F'G'.pf.2}), we
obtain $\left(  F^{\prime}\star G^{\prime}\right)  \left(  x\right)  =\left(
F\star G\right)  \left(  x\right)  $.}. Thus,%
\begin{align}
\left(  F\star G\right)  \left(  x\right)   &  =\underbrace{\left(  F^{\prime
}\star G^{\prime}\right)  }_{\substack{=\sum_{\left(  q,r\right)  \in
Q^{\prime}\times R^{\prime}}\left(  f_{q}\star g_{r}\right)  \\\text{(by
(\ref{sol.exe.convolution-series.a.PA.3.pf.F'G'1}))}}}\left(  x\right)
=\left(  \sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \right)  \left(  x\right) \nonumber\\
&  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \left(  x\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.FGx=}%
\end{align}


On the other hand, $\underbrace{Q^{\prime}}_{\subset Q}\times
\underbrace{R^{\prime}}_{\subset R}\subset Q\times R$.

But the sum $\sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star
g_{r}\right)  \left(  x\right)  $ is well-defined (since the family $\left(
\left(  f_{q}\star g_{r}\right)  \left(  x\right)  \right)  _{\left(
q,r\right)  \in Q\times R}\in A^{Q\times R}$ is finitely supported). It
satisfies
\begin{align}
&  \sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star g_{r}\right)
\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in Q\times R;\\\left(
q,r\right)  \in Q^{\prime}\times R^{\prime}}}}_{\substack{=\sum_{\left(
q,r\right)  \in Q^{\prime}\times R^{\prime}}\\\text{(since }Q^{\prime}\times
R^{\prime}\subset Q\times R\text{)}}}\left(  f_{q}\star g_{r}\right)  \left(
x\right)  +\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in Q\times
R;\\\left(  q,r\right)  \notin Q^{\prime}\times R^{\prime}}}}_{=\sum_{\left(
q,r\right)  \in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times
R^{\prime}\right)  }}\left(  f_{q}\star g_{r}\right)  \left(  x\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }\left(  q,r\right)  \in Q\times R\text{ satisfies either
}\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}\text{ or }\left(
q,r\right)  \notin Q^{\prime}\times R^{\prime}\\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \left(  x\right)  +\sum_{\left(  q,r\right)
\in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times R^{\prime
}\right)  }\underbrace{\left(  f_{q}\star g_{r}\right)  \left(  x\right)
}_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.a.PA.3.pf.res1}%
))}}}\nonumber\\
&  =\sum_{\left(  q,r\right)  \in Q^{\prime}\times R^{\prime}}\left(
f_{q}\star g_{r}\right)  \left(  x\right)  +\underbrace{\sum_{\left(
q,r\right)  \in\left(  Q\times R\right)  \setminus\left(  Q^{\prime}\times
R^{\prime}\right)  }0}_{=0}=\sum_{\left(  q,r\right)  \in Q^{\prime}\times
R^{\prime}}\left(  f_{q}\star g_{r}\right)  \left(  x\right) \nonumber\\
&  =\left(  \underbrace{F}_{=\sum_{q\in Q}f_{q}}\star\underbrace{G}%
_{=\sum_{r\in R}g_{r}}\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.exe.convolution-series.a.PA.3.pf.FGx=})}\right) \nonumber\\
&  =\left(  \left(  \sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}%
g_{r}\right)  \right)  \left(  x\right)  .
\label{sol.exe.convolution-series.a.PA.3.pf.res6}%
\end{align}


Now, forget that we fixed $x$. We thus have shown that each $x\in C$ satisfies
(\ref{sol.exe.convolution-series.a.PA.3.pf.res3}) and
(\ref{sol.exe.convolution-series.a.PA.3.pf.res6}).

In particular, each $x\in C$ satisfies
(\ref{sol.exe.convolution-series.a.PA.3.pf.res3}). In other words, for each
$x\in C$, the family $\left(  \left(  f_{q}\star g_{r}\right)  \left(
x\right)  \right)  _{\left(  q,r\right)  \in Q\times R}\in A^{Q\times R}$ is
finitely supported. In other words, the family $\left(  f_{q}\star
g_{r}\right)  _{\left(  q,r\right)  \in Q\times R}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q\times R}$ is pointwise
finitely supported (by the definition of ``pointwise finitely
supported''). Hence, the sum $\sum_{\left(  q,r\right)  \in
Q\times R}\left(  f_{q}\star g_{r}\right)  $ is well-defined. For each $x\in
C$, we have%
\[
\left(  \sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star
g_{r}\right)  \right)  \left(  x\right)  =\sum_{\left(  q,r\right)  \in
Q\times R}\left(  f_{q}\star g_{r}\right)  \left(  x\right)  =\left(  \left(
\sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}g_{r}\right)  \right)
\left(  x\right)
\]
(by (\ref{sol.exe.convolution-series.a.PA.3.pf.res6})). In other words, we
have%
\[
\sum_{\left(  q,r\right)  \in Q\times R}\left(  f_{q}\star g_{r}\right)
=\left(  \sum_{q\in Q}f_{q}\right)  \star\left(  \sum_{r\in R}g_{r}\right)  .
\]
This completes the proof of Proposition~\ref{prop.pw-fin-supp.P3}.
\end{proof}

We shall delay the proof of Proposition~\ref{prop.pw-fin-supp.P4} until after
Proposition~\ref{prop.pw-fin-supp.P5} is proven; the reason is that
Proposition~\ref{prop.pw-fin-supp.P5} yields a quick shortcut to
Proposition~\ref{prop.pw-fin-supp.P4}.

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P5}.]Let $x\in C$. Write the
element $\Delta\left(  x\right)  \in C\otimes C$ in the form $\Delta\left(
x\right)  =\sum_{p=1}^{k}y_{p}\otimes z_{p}$ for some $k\in \NN$, some
elements $y_{1},y_{2},\ldots,y_{k}\in C$ and some elements $z_{1},z_{2}%
,\ldots,z_{k}\in C$. (This is possible, because $\Delta\left(  x\right)  $ can
be written as a sum of pure tensors\footnote{This is because any tensor in
$C\otimes C$ can be written as a sum of pure tensors.}.)

For each $p\in\left\{  1,2,\ldots,k\right\}  $, there exists a finite subset
$Q_{p}$ of $Q$ such that
\begin{equation}
\text{every }q\in Q\setminus Q_{p}\text{ satisfies }f_{q}\left(  y_{p}\right)
=0 \label{sol.exe.convolution-series.a.PA.5.pf.1}%
\end{equation}
\footnote{\textit{Proof:} Let $p\in\left\{  1,2,\ldots,k\right\}  $. The
family $\left(  f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is pointwise finitely supported. In other words,
for each $c\in C$, the family $\left(  f_{q}\left(  c\right)  \right)  _{q\in
Q}\in A^{Q}$ is finitely supported (by the definition of
``pointwise finitely supported''). Applying this to $c=y_{p}$,
we conclude that the family $\left(  f_{q}\left(  y_{p}\right)  \right)
_{q\in Q}\in A^{Q}$ is finitely supported. In other words, all but finitely
many $q\in Q$ satisfy $f_{q}\left(  y_{p}\right)  =0$. In other words, there
exists a finite subset $Q_{p}$ of $Q$ such that every $q\in Q\setminus Q_{p}$
satisfies $f_{q}\left(  y_{p}\right)  =0$. Qed.}. Consider this $Q_{p}$.

Define a subset $Q^{\prime}$ of $Q$ by $Q^{\prime}=Q_{1}\cup Q_{2}\cup
\cdots\cup Q_{k}$. Thus, $Q^{\prime}$ is the union of the $k$ finite sets
$Q_{1},Q_{2},\ldots,Q_{k}$. Hence, $Q^{\prime}$ is a union of finitely many
finite sets. Thus, $Q^{\prime}$ is itself a finite set (since a union of
finitely many finite sets must itself be a finite set). Hence, all but
finitely many $q\in Q$ satisfy $q\in Q\setminus Q^{\prime}$.

Let $q\in Q\setminus Q^{\prime}$. Thus, $q\in Q$ and $q\notin Q^{\prime}$.
Each $p\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
f_{q}\left(  y_{p}\right)  =0 \label{sol.exe.convolution-series.a.PA.5.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.a.PA.5.pf.3}):}
Let $p\in\left\{  1,2,\ldots,k\right\}  $.
\par
We have $p\in\left\{  1,2,\ldots,k\right\}  $. Thus, $Q_{p}$ is one of the
terms in the union $Q_{1}\cup Q_{2}\cup\cdots\cup Q_{k}$. Hence, $Q_{p}\subset
Q_{1}\cup Q_{2}\cup\cdots\cup Q_{k}=Q^{\prime}$ (since $Q^{\prime}=Q_{1}\cup
Q_{2}\cup\cdots\cup Q_{k}$), so that $Q\setminus\underbrace{Q_{p}}_{\subset
Q^{\prime}}\supset Q\setminus Q^{\prime}$. Hence, $Q\setminus Q^{\prime
}\subset Q\setminus Q_{p}$. Hence, $q\in Q\setminus Q^{\prime}\subset
Q\setminus Q_{p}$. Thus, (\ref{sol.exe.convolution-series.a.PA.5.pf.1}) yields
$f_{q}\left(  y_{p}\right)  =0$. This proves
(\ref{sol.exe.convolution-series.a.PA.5.pf.3}).}.

Fact A.0 (applied to $\alpha=f_{q}$ and $\beta=g_{q}$) yields%
\[
\left(  f_{q}\star g_{q}\right)  \left(  x\right)  =\sum_{p=1}^{k}%
\underbrace{f_{q}\left(  y_{p}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.convolution-series.a.PA.5.pf.3}))}}}g_{q}\left(  z_{p}\right)
=\sum_{p=1}^{k}0g_{q}\left(  z_{p}\right)  =0.
\]


Now, forget that we fixed $q$. We thus have proven that each $q\in Q\setminus
Q^{\prime}$ satisfies $\left(  f_{q}\star g_{q}\right)  \left(  x\right)  =0$.
Hence, all but finitely many $q\in Q$ satisfy $\left(  f_{q}\star
g_{q}\right)  \left(  x\right)  =0$ (since all but finitely many $q\in Q$
satisfy $q\in Q\setminus Q^{\prime}$). In other words, the family $\left(
\left(  f_{q}\star g_{q}\right)  \left(  x\right)  \right)  _{q\in Q}\in
A^{Q}$ is finitely supported.

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
family $\left(  \left(  f_{q}\star g_{q}\right)  \left(  x\right)  \right)
_{q\in Q}\in A^{Q}$ is finitely supported. In other words, the family $\left(
f_{q}\star g_{q}\right)  _{q\in Q}\in\left(  \Hom \left(
C,A\right)  \right)  ^{Q}$ is pointwise finitely supported. This proves
Proposition~\ref{prop.pw-fin-supp.P5}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.pw-fin-supp.P4}.]Let $\mathfrak{i}$ be the
unity of the $\kk$-algebra $\left(  \Hom \left(
C,A\right)  ,\star\right)  $. (This $\mathfrak{i}$ is the map $u_{A}%
\circ\epsilon_{C}:C\rightarrow A$; but this does not matter to us.)

Applying Proposition~\ref{prop.pw-fin-supp.P5} to the family $\left(
g_{q}\right)  _{q\in Q}=\left(  \lambda_{q}\mathfrak{i}\right)  _{q\in Q}$, we
conclude that the family $\left(  f_{q}\star\left(  \lambda_{q}\mathfrak{i}%
\right)  \right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is pointwise finitely supported. But each $q\in Q$ satisfies%
\[
f_{q}\star\left(  \lambda_{q}\mathfrak{i}\right)  =\lambda_{q}\cdot
\underbrace{\left(  f_{q}\star\mathfrak{i}\right)  }_{\substack{=f_{q}%
\\\text{(since }\mathfrak{i}\text{ is the unity of}\\\text{the }%
\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \text{)}}}=\lambda_{q}f_{q}.
\]
In other words, $\left(  f_{q}\star\left(  \lambda_{q}\mathfrak{i}\right)
\right)  _{q\in Q}=\left(  \lambda_{q}f_{q}\right)  _{q\in Q}$.

We know that the family $\left(  f_{q}\star\left(  \lambda_{q}\mathfrak{i}%
\right)  \right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is pointwise finitely supported. Since $\left(  f_{q}%
\star\left(  \lambda_{q}\mathfrak{i}\right)  \right)  _{q\in Q}=\left(
\lambda_{q}f_{q}\right)  _{q\in Q}$, this rewrites as follows: The family
$\left(  \lambda_{q}f_{q}\right)  _{q\in Q}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely supported. This
proves Proposition~\ref{prop.pw-fin-supp.P4}.
\end{proof}

We have now proven all five Propositions~\ref{prop.pw-fin-supp.P1}%
,~\ref{prop.pw-fin-supp.P2},~\ref{prop.pw-fin-supp.P3}%
,~\ref{prop.pw-fin-supp.P4}~and~\ref{prop.pw-fin-supp.P5}. Thus,
Exercise~\ref{exe.pw-fin-supp.P1-5} is solved.

[\textit{Remark:} We can re-interpret the concept of
``pointwise finitely supported'' families $\left(
f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ and their sums $\sum_{q\in Q}f_{q}$ in topological terms.
To that end, we shall use the concept of a ``net''
(see, e.g., \url{https://en.wikipedia.org/wiki/Net_(mathematics)} or
\cite[\S 4]{Vermeeren-nets} for an introduction).

Recall that a \dfn{preordered set} means a set $Z$ equipped with a
\idx{preorder relation}
(i.e., a binary relation on $Z$ that is both reflexive and
transitive\footnote{but (unlike a partial order) not necessarily
antisymmetric}). This preorder relation is commonly denoted by $\leq$. A
nonempty preordered set $Z$ is said to be a \dfn{directed set} if its
preorder relation $\leq$ has the property that every two elements $x\in Z$ and
$y\in Z$ have an upper bound (i.e., some $z\in Z$ satisfying $x\leq z$ and
$y\leq z$). If $Z$ is a preordered set, and if $\mathcal{A}\left(  z\right)  $
is a logical statement for each $z\in Z$, then we say that
``$\mathcal{A}\left(  z\right)  $ holds
\emph{for all sufficiently high }$z\in Z$''
if and only if there exists a $w\in Z$ such that every
$z\in Z$ satisfying $w\leq z$ satisfies $\mathcal{A}\left(  z\right)  $.

Two important examples of directed sets are the following:

\begin{itemize}
\item The set $\NN$, equipped with the usual less-or-equal relation
$\leq$, is a directed set. This directed set will simply be called
$\NN$.

\item If $Q$ is any set, then the set $\mathcal{P}_{\operatorname*{fin}%
}\left(  Q\right)  $ of all finite subsets of $Q$ is naturally a directed set:
Its preorder relation $\leq$ is defined to be the subset relation $\subset$.
Every two elements $x\in\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  $
and $y\in\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  $ clearly have an
upper bound (for example, $x\cup y$). This directed set will simply be called
$\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  $.
\end{itemize}

A \dfn{net} in a set $X$ is defined to be a family $\left(  x_{z}\right)
_{z\in Z}\in X^{Z}$, where $Z$ is some directed set.

If $X$ is a topological space, if $x\in X$, and if $\left(  x_{z}\right)
_{z\in Z}\in X^{Z}$ is a net in $X$, then the net $\left(  x_{z}\right)
_{z\in Z}$ is said to \emph{converge}\index{convergence of a net}
to $x$ if and only if for each
neighborhood $U$ of $x$, we have%
\[
\left(  x_{z}\in U\text{ for all sufficiently high }z\in Z\right)  .
\]
Thus, in any topological space, we have defined the notion of a convergent
net. This notion generalizes the notion of a convergent sequence (indeed,
convergent sequences are precisely the same as convergent nets whose indexing
set $Z$ is the directed set $\NN$). But it is, in a sense, a more
natural notion than the latter: Unlike the latter, it characterizes the
topological space. That is, we can define a topological space on a set $Y$ by
specifying which nets in $Y$ converge to which elements of $Y$ (provided that
this specification satisfies certain axioms); but we cannot (in general)
define a topological space on a set $Y$ by specifying which sequences in $Y$
converge to which elements of $Y$.

If a net $\left(  x_{z}\right)  _{z\in Z}\in X^{Z}$ in a topological space $X$
converges to an element $x\in X$, then $x$ is called a
\emph{limit}\index{limit of a net} of
$\left(  x_{z}\right)  _{z\in Z}$. If $X$ is Hausdorff, then any convergent
net $\left(  x_{z}\right)  _{z\in Z}\in X^{Z}$ has only one limit, and so we
can call this limit ``\emph{the} limit'' of
$\left(  x_{z}\right)  _{z\in Z}$.

If $X$ is a topological space with the discrete topology, then convergence of
nets can be described very simply: A net $\left(  x_{z}\right)  _{z\in Z}\in
X^{Z}$ in a discrete topological space $X$ converges to an element $x\in X$ if
and only if we have $\left(  x_{z}=x\text{ for all sufficiently high }z\in
Z\right)  $. This behavior is also called \dfn{stabilization}: i.e., we say
that a net $\left(  x_{z}\right)  _{z\in Z}\in X^{Z}$
\emph{stabilizes}\index{stabilizing net} to
an element $x\in X$ if and only if we have $\left(  x_{z}=x\text{ for all
sufficiently high }z\in Z\right)  $.

Let us equip the set $A$ with the discrete topology. Thus, $A$ becomes a
topological $\kk$-algebra (because equipping any $\kk$-algebra
with the discrete topology results in a topological $\kk$-algebra).

Now, we equip the set $\Hom \left(  C,A\right)  $ with a
topology, which can be defined in any of the following two ways:

\begin{itemize}
\item It is the unique topology on the set $\Hom \left(
C,A\right)  $ that has the following property: A net $\left(  f_{z}\right)
_{z\in Z}$ of maps $f_{z}\in\Hom \left(  C,A\right)  $ converges
to a map $f\in\Hom \left(  C,A\right)  $ in this topology if and
only if for each $c\in C$, the net $\left(  f_{z}\left(  c\right)  \right)
_{z\in Z}$ in $A$ stabilizes to $f\left(  c\right)  $.

\item Alternatively, we can define the topology on $\Hom \left(
C,A\right)  $ in the usual way (i.e., via open sets): The set $A^{C}$ of all
maps from $C$ to $A$ is equipped with a product topology (since it is the
product $\prod_{c\in C}A$). The set $\Hom \left(  C,A\right)  $
thus also gets a topology, being a subset of $A^{C}$.
\end{itemize}

\noindent
These two definitions give rise to the same topology. This topology is called
the \dfn{topology of pointwise convergence}. We consider
$\Hom \left(  C,A\right)  $ to be equipped with this topology
from now on. This topology allows us to work with limits of nets in
$\Hom \left(  C,A\right)  $ (as long as these nets converge),
since the topological space $\Hom \left(  C,A\right)  $ is Hausdorff.

\begin{proposition}
\label{prop.sol.pw-fin-supp.P1-5.top}The $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ is a topological
$\kk$-algebra. That is, the maps%
\begin{align}
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)   &  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto
f+g,\label{eq.exe.convolution-series.a.1}\\
\Hom \left(  C,A\right)   &  \rightarrow\Hom
\left(  C,A\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto
-f,\label{eq.exe.convolution-series.a.2}\\
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)   &  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto f\star
g,\label{eq.exe.convolution-series.a.3}\\
\kk\times\Hom \left(  C,A\right)   &  \rightarrow
\Hom \left(  C,A\right)  ,\ \ \ \ \ \ \ \ \ \ \left(
\lambda,f\right)  \mapsto\lambda f \label{eq.exe.convolution-series.a.4}%
\end{align}
are continuous (where the topology on $\kk$ is the discrete topology).
\end{proposition}

For the sake of completeness, we are going to give a (fairly straightforward)
proof of this proposition further below. But let us first see how it allows us
to re-interpret pointwise finitely supported families:

\begin{itemize}
\item Any family $\left(  a_{q}\right)  _{q\in Q}\in A^{Q}$ of elements of $A$
gives rise to a net $\left(  \sum_{q\in K}a_{q}\right)  _{K\in\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }\in A^{\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }$ in $A$ (which consists of all sums
of finite subfamilies of $\left(  a_{q}\right)  _{q\in Q}$). It is easy to see
that the family $\left(  a_{q}\right)  _{q\in Q}$ is finitely supported if and
only if the net $\left(  \sum_{q\in K}a_{q}\right)  _{K\in\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }\in A^{\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }$ converges in the discrete space
$A$. In this case, the limit of the net is precisely $\sum_{q\in Q}a_{q}$.

\item Any family $\left(  f_{q}\right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ of elements of
$\Hom \left(  C,A\right)  $ gives rise to a net $\left(
\sum_{q\in K}f_{q}\right)  _{K\in\mathcal{P}_{\operatorname*{fin}}\left(
Q\right)  }\in\left(  \Hom \left(  C,A\right)  \right)
^{\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  }$ in
$\Hom \left(  C,A\right)  $ (which consists of all sums of
finite subfamilies of $\left(  f_{q}\right)  _{q\in Q}$). It is easy to see
that the family $\left(  f_{q}\right)  _{q\in Q}$ is pointwise finitely
supported if and only if the net $\left(  \sum_{q\in K}f_{q}\right)
_{K\in\mathcal{P}_{\operatorname*{fin}}\left(  Q\right)  }\in\left(
\Hom \left(  C,A\right)  \right)  ^{\mathcal{P}%
_{\operatorname*{fin}}\left(  Q\right)  }$ converges in $\Hom
\left(  C,A\right)  $. In this case, the limit of the net is precisely
$\sum_{q\in Q}f_{q}$.
\end{itemize}

It is clear that this line of reasoning allows us to generalize the notion of
``pointwise finitely supported families'' to
families in any topological $\kk$-module, and to define the sum of any
such family.

Let us now prove Proposition~\ref{prop.sol.pw-fin-supp.P1-5.top}.

\begin{proof}
[Proof of Proposition~\ref{prop.sol.pw-fin-supp.P1-5.top}.]There are various
levels of abstraction at which this proof can be done. We shall approach it
using universal properties of product topologies. To facilitate this, let us
do some preliminary work first.

For each $x\in C$, we let $\operatorname*{ev}\nolimits_{x}$ be the map
$\Hom \left(  C,A\right)  \rightarrow A$ sending each
$f\in\Hom \left(  C,A\right)  $ to $f\left(  x\right)  \in A$.
This map $\operatorname*{ev}\nolimits_{x}$ is $\kk$-linear and
continuous (by the definition of the topology on $\Hom \left(
C,A\right)  $).

Equip the $\kk$-algebra $A$ with the discrete topology. Then, $A$
becomes a topological $\kk$-algebra (since any $\kk$-algebra
becomes a topological $\kk$-algebra when equipped with the discrete topology).

Recall the definition of the topology on $\Hom \left(
C,A\right)  $ via the product topology: The set $A^{C}$ of all maps from $C$
to $A$ is equipped with a product topology (since it is the product
$\prod_{c\in C}A$). The set $\Hom \left(  C,A\right)  $ thus
also gets a topology, being a subset of $A^{C}$. This is the topology of
pointwise convergence.

The universal property of the product topology on $A^{C}$ says the following:

\begin{statement}
\textit{Fact A.1:} Let $X$ be any set, and $\varphi:X\rightarrow
\Hom \left(  C,A\right)  $ be any map. Then, $\varphi$ is
continuous if and only if for each $x\in C$, the map $\operatorname*{ev}%
\nolimits_{x}\circ\varphi:X\rightarrow A$ is continuous.
\end{statement}

We will use mainly the ``if'' direction of
Fact A.1. Thus, let us state this direction as a separate fact:

\begin{statement}
\textit{Fact A.2:} Let $X$ be any set, and $\varphi:X\rightarrow
\Hom \left(  C,A\right)  $ be any map. Assume that for each
$x\in C$, the map $\operatorname*{ev}\nolimits_{x}\circ\varphi:X\rightarrow A$
is continuous. Then, the map $\varphi$ is continuous.
\end{statement}

[\textit{Proof of Fact A.2:} Fact A.2 is just the ``if'' part of Fact A.1.]

Let $\pi_{1}$ be the map%
\[
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto f.
\]
Let $\pi_{2}$ be the map
\[
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto g.
\]
Both maps $\pi_{1}$ and $\pi_{2}$ are continuous (by the definition of the
product topology on $\Hom \left(  C,A\right)  \times
\Hom \left(  C,A\right)  $).

Now, we are going to prove that the maps (\ref{eq.exe.convolution-series.a.1}%
), (\ref{eq.exe.convolution-series.a.2}), (\ref{eq.exe.convolution-series.a.3}%
) and (\ref{eq.exe.convolution-series.a.4}) are continuous:

[\textit{Proof of the continuity of the map
(\ref{eq.exe.convolution-series.a.1}):} Denote the map
(\ref{eq.exe.convolution-series.a.1}) by $\mathbf{a}$. Thus, $\mathbf{a}$ is
the map
\[
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto f+g.
\]
We must prove that $\mathbf{a}$ is continuous.

Let $\operatorname*{add}\nolimits_{A}$ be the map $A\times A\rightarrow
A,\ \left(  a,b\right)  \mapsto a+b$. This map $\operatorname*{add}%
\nolimits_{A}$ is continuous (since $A$ is a topological $\kk$-algebra).

Fix any $x\in C$. Then, the map $\operatorname*{ev}\nolimits_{x}%
:\Hom \left(  C,A\right)  \rightarrow A$ is continuous (as we
know). Thus, the map $\operatorname*{ev}\nolimits_{x}\times\operatorname*{ev}%
\nolimits_{x}:\Hom \left(  C,A\right)  \times\Hom
\left(  C,A\right)  \rightarrow A\times A$ is continuous as well. Now,
$\operatorname*{ev}\nolimits_{x}\circ\mathbf{a}=\operatorname*{add}%
\nolimits_{A}\circ\left(  \operatorname*{ev}\nolimits_{x}\times
\operatorname*{ev}\nolimits_{x}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$\left(  f,g\right)  \in\Hom \left(  C,A\right)  \times
\Hom \left(  C,A\right)  $. Then, the definition of $\mathbf{a}$
yields $\mathbf{a}\left(  f,g\right)  =f+g$. Now,%
\begin{align*}
\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{a}\right)  \left(
f,g\right)   &  =\operatorname*{ev}\nolimits_{x}\left(  \underbrace{\mathbf{a}%
\left(  f,g\right)  }_{=f+g}\right)  =\operatorname*{ev}\nolimits_{x}\left(
f+g\right)  =\left(  f+g\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the map }\operatorname*{ev}\nolimits_{x}\right) \\
&  =f\left(  x\right)  +g\left(  x\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \operatorname*{add}\nolimits_{A}\circ\left(  \operatorname*{ev}%
\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)  \right)  \left(
f,g\right)   &  =\operatorname*{add}\nolimits_{A}\left(  \underbrace{\left(
\operatorname*{ev}\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)
\left(  f,g\right)  }_{=\left(  \operatorname*{ev}\nolimits_{x}\left(
f\right)  ,\operatorname*{ev}\nolimits_{x}\left(  g\right)  \right)  }\right)
=\operatorname*{add}\nolimits_{A}\left(  \operatorname*{ev}\nolimits_{x}%
\left(  f\right)  ,\operatorname*{ev}\nolimits_{x}\left(  g\right)  \right) \\
&  =\underbrace{\operatorname*{ev}\nolimits_{x}\left(  f\right)
}_{\substack{=f\left(  x\right)  \\\text{(by the definition}\\\text{of the map
}\operatorname*{ev}\nolimits_{x}\text{)}}}+\underbrace{\operatorname*{ev}%
\nolimits_{x}\left(  g\right)  }_{\substack{=g\left(  x\right)  \\\text{(by
the definition}\\\text{of the map }\operatorname*{ev}\nolimits_{x}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map
}\operatorname*{add}\nolimits_{A}\right) \\
&  =f\left(  x\right)  +g\left(  x\right)  ,
\end{align*}
we obtain $\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{a}\right)
\left(  f,g\right)  =\left(  \operatorname*{add}\nolimits_{A}\circ\left(
\operatorname*{ev}\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)
\right)  \left(  f,g\right)  $.
\par
Now, forget that we fixed $\left(  f,g\right)  $. We thus have shown that
$\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{a}\right)  \left(
f,g\right)  =\left(  \operatorname*{add}\nolimits_{A}\circ\left(
\operatorname*{ev}\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)
\right)  \left(  f,g\right)  $ for each $\left(  f,g\right)  \in
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  $. In other words, $\operatorname*{ev}\nolimits_{x}\circ
\mathbf{a}=\operatorname*{add}\nolimits_{A}\circ\left(  \operatorname*{ev}%
\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)  $.}. But the map
$\operatorname*{add}\nolimits_{A}\circ\left(  \operatorname*{ev}%
\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)  $ is continuous
(since it is the composition of the two continuous maps $\operatorname*{add}%
\nolimits_{A}$ and $\operatorname*{ev}\nolimits_{x}\times\operatorname*{ev}%
\nolimits_{x}$). Hence, the map $\operatorname*{ev}\nolimits_{x}%
\circ\mathbf{a}$ is continuous (since $\operatorname*{ev}\nolimits_{x}%
\circ\mathbf{a}=\operatorname*{add}\nolimits_{A}\circ\left(
\operatorname*{ev}\nolimits_{x}\times\operatorname*{ev}\nolimits_{x}\right)  $).

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
map $\operatorname*{ev}\nolimits_{x}\circ\mathbf{a}:\Hom \left(
C,A\right)  \times\Hom \left(  C,A\right)  \rightarrow A$ is
continuous. Hence, Fact A.2 (applied to $X=\Hom \left(
C,A\right)  \times\Hom \left(  C,A\right)  $ and $\varphi
=\mathbf{a}$) shows that the map $\mathbf{a}$ is continuous. In other words,
the map (\ref{eq.exe.convolution-series.a.1}) is continuous (since the map
$\mathbf{a}$ is precisely the map (\ref{eq.exe.convolution-series.a.1})).]

[\textit{Proof of the continuity of the map
(\ref{eq.exe.convolution-series.a.2}):} Denote the map
(\ref{eq.exe.convolution-series.a.2}) by $\mathbf{n}$. Thus, $\mathbf{n}$ is
the map
\[
\Hom \left(  C,A\right)  \rightarrow\Hom \left(
C,A\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto-f.
\]
We must prove that $\mathbf{n}$ is continuous.

Let $\operatorname*{neg}\nolimits_{A}$ be the map $A\rightarrow A,\ a\mapsto
-a$. This map $\operatorname*{neg}\nolimits_{A}$ is continuous (since $A$ is a
topological $\kk$-algebra).

Fix any $x\in C$. Then, the map $\operatorname*{ev}\nolimits_{x}%
:\Hom \left(  C,A\right)  \rightarrow A$ is continuous (as we
know). Now, $\operatorname*{ev}\nolimits_{x}\circ\mathbf{n}%
=\operatorname*{neg}\nolimits_{A}\circ\operatorname*{ev}\nolimits_{x}%
$\ \ \ \ \footnote{\textit{Proof.} Let $f\in\Hom \left(
C,A\right)  $. Then, the definition of $\mathbf{n}$ yields $\mathbf{n}\left(
f\right)  =-f$. Now,%
\begin{align*}
\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{n}\right)  \left(
f\right)   &  =\operatorname*{ev}\nolimits_{x}\left(  \underbrace{\mathbf{n}%
\left(  f\right)  }_{=-f}\right)  =\operatorname*{ev}\nolimits_{x}\left(
-f\right)  =\left(  -f\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the map }\operatorname*{ev}\nolimits_{x}\right) \\
&  =-f\left(  x\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \operatorname*{neg}\nolimits_{A}\circ\operatorname*{ev}\nolimits_{x}%
\right)  \left(  f\right)   &  =\operatorname*{neg}\nolimits_{A}\left(
\operatorname*{ev}\nolimits_{x}\left(  f\right)  \right)
=-\underbrace{\operatorname*{ev}\nolimits_{x}\left(  f\right)  }%
_{\substack{=f\left(  x\right)  \\\text{(by the definition}\\\text{of the map
}\operatorname*{ev}\nolimits_{x}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of the map }\operatorname*{neg}\nolimits_{A}\right) \\
&  =-f\left(  x\right)  ,
\end{align*}
we obtain $\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{n}\right)
\left(  f\right)  =\left(  \operatorname*{neg}\nolimits_{A}\circ
\operatorname*{ev}\nolimits_{x}\right)  \left(  f\right)  $.
\par
Now, forget that we fixed $f$. We thus have shown that $\left(
\operatorname*{ev}\nolimits_{x}\circ\mathbf{n}\right)  \left(  f\right)
=\left(  \operatorname*{neg}\nolimits_{A}\circ\operatorname*{ev}%
\nolimits_{x}\right)  \left(  f\right)  $ for each $f\in\Hom
\left(  C,A\right)  $. In other words, $\operatorname*{ev}\nolimits_{x}%
\circ\mathbf{n}=\operatorname*{neg}\nolimits_{A}\circ\operatorname*{ev}%
\nolimits_{x}$.}. But the map $\operatorname*{neg}\nolimits_{A}\circ
\operatorname*{ev}\nolimits_{x}$ is continuous (since it is the composition of
the two continuous maps $\operatorname*{neg}\nolimits_{A}$ and
$\operatorname*{ev}\nolimits_{x}$). Hence, the map $\operatorname*{ev}%
\nolimits_{x}\circ\mathbf{n}$ is continuous (since $\operatorname*{ev}%
\nolimits_{x}\circ\mathbf{n}=\operatorname*{neg}\nolimits_{A}\circ
\operatorname*{ev}\nolimits_{x}$).

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
map $\operatorname*{ev}\nolimits_{x}\circ\mathbf{n}:\Hom \left(
C,A\right)  \rightarrow A$ is continuous. Hence, Fact A.2 (applied to
$X=\Hom \left(  C,A\right)  $ and $\varphi=\mathbf{n}$) shows
that the map $\mathbf{n}$ is continuous. In other words, the map
(\ref{eq.exe.convolution-series.a.2}) is continuous (since the map
$\mathbf{n}$ is precisely the map (\ref{eq.exe.convolution-series.a.2})).]

[\textit{Proof of the continuity of the map
(\ref{eq.exe.convolution-series.a.3}):} Denote the map
(\ref{eq.exe.convolution-series.a.3}) by $\mathbf{m}$. Thus, $\mathbf{m}$ is
the map
\[
\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  \rightarrow\Hom \left(  C,A\right)
,\ \ \ \ \ \ \ \ \ \ \left(  f,g\right)  \mapsto f\star g.
\]
We must prove that $\mathbf{m}$ is continuous.

Set $X=\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  $. Thus, $\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}$ is a map
$X\rightarrow A$ (since $\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}$ is a
map $\Hom \left(  C,A\right)  \times\Hom \left(
C,A\right)  \rightarrow A$).

Let $\operatorname*{mul}\nolimits_{A}$ be the map $A\times A\rightarrow
A,\ \left(  a,b\right)  \mapsto ab$.\ \ \ \ \footnote{Do not mistake this map
$\operatorname*{mul}\nolimits_{A}$ for the map $m_{A}:A\otimes A\rightarrow
A,\ a\otimes b\mapsto ab$. The former acts on pairs, whereas the latter acts
on tensors.} This map $\operatorname*{mul}\nolimits_{A}$ is continuous (since
$A$ is a topological $\kk$-algebra).

Fix any $x\in C$. Write the element $\Delta\left(  x\right)  \in C\otimes C$
in the form $\Delta\left(  x\right)  =\sum_{i=1}^{n}y_{i}\otimes z_{i}$ for
some $n\in \NN$, some elements $y_{1},y_{2},\ldots,y_{n}\in C$ and some
elements $z_{1},z_{2},\ldots,z_{n}\in C$. (This is possible, because
$\Delta\left(  x\right)  $ can be written as a sum of pure
tensors\footnote{This is because any tensor in $C\otimes C$ can be written as
a sum of pure tensors.}.)

For each $i\in\left\{  1,2,\ldots,n\right\}  $, define a map $\rho
_{i}:X\rightarrow A$ by $\rho_{i}=\operatorname*{mul}\nolimits_{A}\circ\left(
\operatorname*{ev}\nolimits_{y_{i}}\times\operatorname*{ev}\nolimits_{z_{i}%
}\right)  $. For each $i\in\left\{  1,2,\ldots,n\right\}  $, the map $\rho
_{i}$ is continuous\footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots
,n\right\}  $. Recall that the map $\operatorname*{ev}\nolimits_{x}%
:\Hom \left(  C,A\right)  \rightarrow A$ is continuous.
Similarly, the maps $\operatorname*{ev}\nolimits_{y_{i}}:\Hom
\left(  C,A\right)  \rightarrow A$ and $\operatorname*{ev}\nolimits_{z_{i}%
}:\Hom \left(  C,A\right)  \rightarrow A$ are continuous. Thus,
the Cartesian product $\operatorname*{ev}\nolimits_{y_{i}}\times
\operatorname*{ev}\nolimits_{z_{i}}:\Hom \left(  C,A\right)
\times\Hom \left(  C,A\right)  \rightarrow A\times A$ of these
two maps is continuous as well. In other words, the map $\operatorname*{ev}%
\nolimits_{y_{i}}\times\operatorname*{ev}\nolimits_{z_{i}}:X\rightarrow
A\times A$ is continuous (since $X=\Hom \left(  C,A\right)
\times\Hom \left(  C,A\right)  $).
\par
Now, the map $\operatorname*{mul}\nolimits_{A}\circ\left(  \operatorname*{ev}%
\nolimits_{y_{i}}\times\operatorname*{ev}\nolimits_{z_{i}}\right)  $ is
continuous (since it is the composition of the two continuous maps
$\operatorname*{mul}\nolimits_{A}$ and $\operatorname*{ev}\nolimits_{y_{i}%
}\times\operatorname*{ev}\nolimits_{z_{i}}$). In other words, the map
$\rho_{i}$ is continuous (since $\rho_{i}=\operatorname*{mul}\nolimits_{A}%
\circ\left(  \operatorname*{ev}\nolimits_{y_{i}}\times\operatorname*{ev}%
\nolimits_{z_{i}}\right)  $). Qed.}. Hence, the map%
\[
\prod_{i=1}^{n}\rho_{i}:\prod_{i=1}^{n}X\rightarrow\prod_{i=1}^{n}A
\]
is also continuous (since it is a Cartesian product of $n$ continuous maps).

Let $\operatorname*{diag}\nolimits_{n}:X\rightarrow\prod_{i=1}^{n}X$ be the
map sending each $z\in X$ to $\underbrace{\left(  z,z,\ldots,z\right)
}_{n\text{ times }z}\in\prod_{i=1}^{n}X$. This map $\operatorname*{diag}%
\nolimits_{n}$ is clearly continuous.

Let $\operatorname{sum}_{A,n}:\prod_{i=1}^{n}A\rightarrow A$ be the
map sending each
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in \prod_{i=1}^{n}A$ to
$a_{1}+a_{2}+\cdots+a_{n}\in A$. This map $\operatorname{sum}_{A,n}$
is continuous (since $A$ is a topological $\kk$-algebra).

Now,%
\[
\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}=\operatorname*{sum}%
\nolimits_{A,n}\circ\left(  \prod_{i=1}^{n}\rho_{i}\right)  \circ
\operatorname*{diag}\nolimits_{n}%
\]
\footnote{\textit{Proof.} Let $\left(  f,g\right)  \in\Hom
\left(  C,A\right)  \times\Hom \left(  C,A\right)  $. Thus,
$\left(  f,g\right)  \in\Hom \left(  C,A\right)  \times
\Hom \left(  C,A\right)  =X$. Thus, for each $i\in\left\{
1,2,\ldots,n\right\}  $, we have%
\begin{align}
\underbrace{\rho_{i}}_{=\operatorname*{mul}\nolimits_{A}\circ\left(
\operatorname*{ev}\nolimits_{y_{i}}\times\operatorname*{ev}\nolimits_{z_{i}%
}\right)  }\left(  f,g\right)   &  =\left(  \operatorname*{mul}\nolimits_{A}%
\circ\left(  \operatorname*{ev}\nolimits_{y_{i}}\times\operatorname*{ev}%
\nolimits_{z_{i}}\right)  \right)  \left(  f,g\right)  =\operatorname*{mul}%
\nolimits_{A}\left(  \underbrace{\left(  \operatorname*{ev}\nolimits_{y_{i}%
}\times\operatorname*{ev}\nolimits_{z_{i}}\right)  \left(  f,g\right)
}_{=\left(  \operatorname*{ev}\nolimits_{y_{i}}\left(  f\right)
,\operatorname*{ev}\nolimits_{z_{i}}\left(  g\right)  \right)  }\right)
\nonumber\\
&  =\operatorname*{mul}\nolimits_{A}\left(  \operatorname*{ev}\nolimits_{y_{i}%
}\left(  f\right)  ,\operatorname*{ev}\nolimits_{z_{i}}\left(  g\right)
\right)  =\underbrace{\operatorname*{ev}\nolimits_{y_{i}}\left(  f\right)
}_{\substack{=f\left(  y_{i}\right)  \\\text{(by the definition}\\\text{of the
map }\operatorname*{ev}\nolimits_{y_{i}}\text{)}}%
}\underbrace{\operatorname*{ev}\nolimits_{z_{i}}\left(  g\right)
}_{\substack{=g\left(  z_{i}\right)  \\\text{(by the definition}\\\text{of the
map }\operatorname*{ev}\nolimits_{z_{i}}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the map }\operatorname*{mul}\nolimits_{A}\right)
\nonumber\\
&  =f\left(  y_{i}\right)  g\left(  z_{i}\right)  .
\label{sol.convolution-series.a.3.3.fn1.2}%
\end{align}
Now,%
\begin{align*}
&  \left(  \left(  \prod_{i=1}^{n}\rho_{i}\right)  \circ\operatorname*{diag}%
\nolimits_{n}\right)  \left(  f,g\right) \\
&  =\left(  \prod_{i=1}^{n}\rho_{i}\right)  \left(
\underbrace{\operatorname*{diag}\nolimits_{n}\left(  f,g\right)
}_{\substack{=\underbrace{\left(  \left(  f,g\right)  ,\left(  f,g\right)
,\ldots,\left(  f,g\right)  \right)  }_{n\text{ times }\left(  f,g\right)
}\\\text{(by the definition of the map }\operatorname*{diag}\nolimits_{n}%
\text{)}}}\right)  =\left(  \prod_{i=1}^{n}\rho_{i}\right)  \left(
\underbrace{\left(  \left(  f,g\right)  ,\left(  f,g\right)  ,\ldots,\left(
f,g\right)  \right)  }_{n\text{ times }\left(  f,g\right)  }\right) \\
&  =\left(  \rho_{1}\left(  f,g\right)  ,\rho_{2}\left(  f,g\right)
,\ldots,\rho_{n}\left(  f,g\right)  \right)  .
\end{align*}
\par
Now, the definition of $\mathbf{m}$ yields $\mathbf{m}\left(  f,g\right)
=f\star g=m\circ\left(  f\otimes g\right)  \circ\Delta$ (by the definition of
convolution). Now,%
\begin{align*}
\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{m}\right)  \left(
f,g\right)   &  =\operatorname*{ev}\nolimits_{x}\left(  \mathbf{m}\left(
f,g\right)  \right)  =\underbrace{\left(  \mathbf{m}\left(  f,g\right)
\right)  }_{=m\circ\left(  f\otimes g\right)  \circ\Delta}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\operatorname*{ev}\nolimits_{x}\right) \\
&  =\left(  m\circ\left(  f\otimes g\right)  \circ\Delta\right)  \left(
x\right)  =m\left(  \left(  f\otimes g\right)  \left(  \underbrace{\Delta
\left(  x\right)  }_{=\sum_{i=1}^{n}y_{i}\otimes z_{i}}\right)  \right) \\
&  =m\left(  \underbrace{\left(  f\otimes g\right)  \left(  \sum_{i=1}%
^{n}y_{i}\otimes z_{i}\right)  }_{=\sum_{i=1}^{n}f\left(  y_{i}\right)
\otimes g\left(  z_{i}\right)  }\right)  =m\left(  \sum_{i=1}^{n}f\left(
y_{i}\right)  \otimes g\left(  z_{i}\right)  \right) \\
&  =\sum_{i=1}^{n}f\left(  y_{i}\right)  g\left(  z_{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }m\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \operatorname*{sum}\nolimits_{A,n}\circ\left(  \prod_{i=1}^{n}\rho
_{i}\right)  \circ\operatorname*{diag}\nolimits_{n}\right)  \left(
f,g\right)   &  =\operatorname*{sum}\nolimits_{A,n}\left(  \underbrace{\left(
\left(  \prod_{i=1}^{n}\rho_{i}\right)  \circ\operatorname*{diag}%
\nolimits_{n}\right)  \left(  f,g\right)  }_{=\left(  \rho_{1}\left(
f,g\right)  ,\rho_{2}\left(  f,g\right)  ,\ldots,\rho_{n}\left(  f,g\right)
\right)  }\right) \\
&  =\operatorname*{sum}\nolimits_{A,n}\left(  \rho_{1}\left(  f,g\right)
,\rho_{2}\left(  f,g\right)  ,\ldots,\rho_{n}\left(  f,g\right)  \right) \\
&  =\rho_{1}\left(  f,g\right)  +\rho_{2}\left(  f,g\right)  +\cdots+\rho
_{n}\left(  f,g\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
the map }\operatorname*{sum}\nolimits_{A,n}\right) \\
&  =\sum_{i=1}^{n}\underbrace{\rho_{i}\left(  f,g\right)  }%
_{\substack{=f\left(  y_{i}\right)  g\left(  z_{i}\right)  \\\text{(by
(\ref{sol.convolution-series.a.3.3.fn1.2}))}}}=\sum_{i=1}^{n}f\left(
y_{i}\right)  g\left(  z_{i}\right)  ,
\end{align*}
we obtain $\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{m}\right)
\left(  f,g\right)  =\left(  \operatorname*{sum}\nolimits_{A,n}\circ\left(
\prod_{i=1}^{n}\rho_{i}\right)  \circ\operatorname*{diag}\nolimits_{n}\right)
\left(  f,g\right)  $.
\par
Now, forget that we fixed $\left(  f,g\right)  $. We thus have shown that
$\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{m}\right)  \left(
f,g\right)  =\left(  \operatorname*{sum}\nolimits_{A,n}\circ\left(
\prod_{i=1}^{n}\rho_{i}\right)  \circ\operatorname*{diag}\nolimits_{n}\right)
\left(  f,g\right)  $ for each $\left(  f,g\right)  \in\Hom
\left(  C,A\right)  \times\Hom \left(  C,A\right)  $. In other
words, $\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}=\operatorname*{sum}%
\nolimits_{A,n}\circ\left(  \prod_{i=1}^{n}\rho_{i}\right)  \circ
\operatorname*{diag}\nolimits_{n}$.}. But the map $\operatorname*{sum}%
\nolimits_{A,n}\circ\left(  \prod_{i=1}^{n}\rho_{i}\right)  \circ
\operatorname*{diag}\nolimits_{n}$ is continuous (since it is the composition
of the three continuous maps $\operatorname*{sum}\nolimits_{A,n}$,
$\prod_{i=1}^{n}\rho_{i}$ and $\operatorname*{diag}\nolimits_{n}$). Hence, the
map $\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}$ is continuous (since
$\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}=\operatorname*{sum}%
\nolimits_{A,n}\circ\left(  \prod_{i=1}^{n}\rho_{i}\right)  \circ
\operatorname*{diag}\nolimits_{n}$).

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
map $\operatorname*{ev}\nolimits_{x}\circ\mathbf{m}:X\rightarrow A$ is
continuous. Hence, Fact A.2 (applied to $\varphi=\mathbf{m}$) shows that the
map $\mathbf{m}$ is continuous. In other words, the map
(\ref{eq.exe.convolution-series.a.3}) is continuous (since the map
$\mathbf{m}$ is precisely the map (\ref{eq.exe.convolution-series.a.3})).]

[\textit{Proof of the continuity of the map
(\ref{eq.exe.convolution-series.a.4}):} Denote the map
(\ref{eq.exe.convolution-series.a.4}) by $\mathbf{s}$. Thus, $\mathbf{s}$ is
the map
\[
\kk\times\Hom \left(  C,A\right)  \rightarrow
\Hom \left(  C,A\right)  ,\ \ \ \ \ \ \ \ \ \ \left(
\lambda,f\right)  \mapsto\lambda f.
\]
We must prove that $\mathbf{s}$ is continuous.

Let $\operatorname*{scal}\nolimits_{A}$ be the map $\kk\times
A\rightarrow A,\ \left(  \lambda,a\right)  \mapsto\lambda a$. This map
$\operatorname*{scal}\nolimits_{A}$ is continuous (since $A$ is a topological
$\kk$-algebra).

Fix any $x\in C$. Then, the map $\operatorname*{ev}\nolimits_{x}%
:\Hom \left(  C,A\right)  \rightarrow A$ is continuous (as we
know). Thus, the map $\id_{\kk}\times
\operatorname*{ev}\nolimits_{x}:\kk\times\Hom \left(
C,A\right)  \rightarrow\kk\times A$ is continuous as well. Now,
$\operatorname*{ev}\nolimits_{x}\circ\mathbf{s}=\operatorname*{scal}%
\nolimits_{A}\circ\left(  \id_{\kk}%
\times\operatorname*{ev}\nolimits_{x}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $\left(  \lambda,f\right)  \in
\kk\times\Hom \left(  C,A\right)  $. Then, the definition
of $\mathbf{s}$ yields $\mathbf{s}\left(  \lambda,f\right)  =\lambda f$. Now,%
\begin{align*}
\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{s}\right)  \left(
\lambda,f\right)   &  =\operatorname*{ev}\nolimits_{x}\left(
\underbrace{\mathbf{s}\left(  \lambda,f\right)  }_{=\lambda f}\right)
=\operatorname*{ev}\nolimits_{x}\left(  \lambda f\right)  =\left(  \lambda
f\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of the map }\operatorname*{ev}\nolimits_{x}\right) \\
&  =\lambda f\left(  x\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \operatorname*{scal}\nolimits_{A}\circ\left(  \operatorname*{id}%
\nolimits_{\kk}\times\operatorname*{ev}\nolimits_{x}\right)  \right)
\left(  \lambda,f\right)   &  =\operatorname*{scal}\nolimits_{A}\left(
\underbrace{\left(  \id_{\kk}\times
\operatorname*{ev}\nolimits_{x}\right)  \left(  \lambda,f\right)  }_{=\left(
\id_{\kk}\left(  \lambda\right)
,\operatorname*{ev}\nolimits_{x}\left(  f\right)  \right)  }\right)
=\operatorname*{scal}\nolimits_{A}\left(  \operatorname*{id}%
\nolimits_{\kk}\left(  \lambda\right)  ,\operatorname*{ev}%
\nolimits_{x}\left(  f\right)  \right) \\
&  =\underbrace{\id_{\kk}\left(  \lambda
\right)  }_{=\lambda}\underbrace{\operatorname*{ev}\nolimits_{x}\left(
f\right)  }_{\substack{=f\left(  x\right)  \\\text{(by the definition}%
\\\text{of the map }\operatorname*{ev}\nolimits_{x}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\operatorname*{scal}\nolimits_{A}\right) \\
&  =\lambda f\left(  x\right)  ,
\end{align*}
we obtain $\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{s}\right)
\left(  \lambda,f\right)  =\left(  \operatorname*{scal}\nolimits_{A}%
\circ\left(  \id_{\kk}\times\operatorname*{ev}%
\nolimits_{x}\right)  \right)  \left(  \lambda,f\right)  $.
\par
Now, forget that we fixed $\left(  \lambda,f\right)  $. We thus have shown
that $\left(  \operatorname*{ev}\nolimits_{x}\circ\mathbf{s}\right)  \left(
\lambda,f\right)  =\left(  \operatorname*{scal}\nolimits_{A}\circ\left(
\id_{\kk}\times\operatorname*{ev}%
\nolimits_{x}\right)  \right)  \left(  \lambda,f\right)  $ for each $\left(
\lambda,f\right)  \in\kk\times\Hom \left(  C,A\right)  $.
In other words, $\operatorname*{ev}\nolimits_{x}\circ\mathbf{s}%
=\operatorname*{scal}\nolimits_{A}\circ\left(  \operatorname*{id}%
\nolimits_{\kk}\times\operatorname*{ev}\nolimits_{x}\right)  $.}. But
the map $\operatorname*{scal}\nolimits_{A}\circ\left(  \operatorname*{id}%
\nolimits_{\kk}\times\operatorname*{ev}\nolimits_{x}\right)  $ is
continuous (since it is the composition of the two continuous maps
$\operatorname*{scal}\nolimits_{A}$ and $\operatorname*{id}%
\nolimits_{\kk}\times\operatorname*{ev}\nolimits_{x}$). Hence, the map
$\operatorname*{ev}\nolimits_{x}\circ\mathbf{s}$ is continuous (since
$\operatorname*{ev}\nolimits_{x}\circ\mathbf{s}=\operatorname*{scal}%
\nolimits_{A}\circ\left(  \id_{\kk}%
\times\operatorname*{ev}\nolimits_{x}\right)  $).

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
map $\operatorname*{ev}\nolimits_{x}\circ\mathbf{s}:\kk\times
\Hom \left(  C,A\right)  \rightarrow A$ is continuous. Hence,
Fact A.2 (applied to $X=\kk\times\Hom \left(  C,A\right)
$ and $\varphi=\mathbf{s}$) shows that the map $\mathbf{s}$ is continuous. In
other words, the map (\ref{eq.exe.convolution-series.a.4}) is continuous
(since the map $\mathbf{s}$ is precisely the map
(\ref{eq.exe.convolution-series.a.4})).]

We now have shown that the maps (\ref{eq.exe.convolution-series.a.1}),
(\ref{eq.exe.convolution-series.a.2}), (\ref{eq.exe.convolution-series.a.3})
and (\ref{eq.exe.convolution-series.a.4}) are continuous. Hence, the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ is a topological $\kk$-algebra. This proves
Proposition~\ref{prop.sol.pw-fin-supp.P1-5.top}.
\end{proof}

]

%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution-series}}

\textit{Solution to Exercise~\ref{exe.convolution-series}.} Before we start
proving Proposition~\ref{prop.convolution-series}, let us prove some facts
which will be useful on several occasions:

\begin{statement}
\textit{Fact B.1:} If $f\in\Hom \left(  C,A\right)  $ is a
pointwise $\star$-nilpotent map, and if $\left(  \lambda_{n}\right)
_{n\in\NN}\in\kk^{\NN}$ is any family of scalars, then
the family $\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported.
\end{statement}

[\textit{Proof of Fact B.1:} Fact B.1 has been stated in
Definition~\ref{def.leray.pwnilp}(b); it was already proven in a footnote.]

\begin{statement}
\textit{Fact B.2:} Let $f\in\mathfrak{n}\left(  C,A\right)  $. Let $\left(
\lambda_{n}\right)  _{n\in\NN}\in\kk^{\NN}$ be any family
of scalars. Then, the family $\left(  \lambda_{n}f^{\star n}\right)
_{n\in\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN}$ is pointwise finitely supported, and its sum $\sum_{n\geq
0}\lambda_{n}f^{\star n}$ belongs to $\Hom \left(  C,A\right)  $.
\end{statement}

[\textit{Proof of Fact B.2:} We have $f\in\mathfrak{n}\left(  C,A\right)  $.
In other words, $f$ is a pointwise $\star$-nilpotent map in
$\Hom \left(  C,A\right)  $ (since $\mathfrak{n}\left(
C,A\right)  $ is the set of all pointwise $\star$-nilpotent maps in
$\Hom \left(  C,A\right)  $). Thus, Fact B.1 shows that the
family $\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported. Hence, the sum $\sum_{n\in\NN}\lambda_{n}f^{\star
n}$ is well-defined. In other words, the sum $\sum_{n\geq0}\lambda_{n}f^{\star
n}$ is well-defined (since $\sum_{n\geq0}=\sum_{n\in\NN}$).

Now, Proposition~\ref{prop.pw-fin-supp.P1} (applied to $\NN$ and
$\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}$ instead of $Q$ and
$\left(  f_{q}\right)  _{q\in Q}$) shows that the map $\sum_{n\in\NN}
\lambda_{n}f^{\star n}$ belongs to $\Hom \left(  C,A\right)  $.
In other words, the map $\sum_{n\geq0}\lambda_{n}f^{\star n}$ belongs to
$\Hom \left(  C,A\right)  $ (since $\sum_{n\geq0}=\sum_{n\in\NN}$).
Thus, Fact B.2 is proven.]

\begin{statement}
\textit{Fact B.3:} Let $\left(  f_{q}\right)  _{q\in Q}$ be a pointwise
finitely supported family in $\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$. Let $\lambda\in\kk$. Then, the family $\left(  \lambda
f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is also pointwise finitely supported, and satisfies%
\[
\lambda\sum_{q\in Q}f_{q}=\sum_{q\in Q}\lambda f_{q}.
\]

\end{statement}

[\textit{Proof of Fact B.3:} Fact B.3 is similar to
Proposition~\ref{prop.pw-fin-supp.P2}, and its proof is analogous to the proof
of the latter (but simpler).]

Before we state the next fact, let us recall how the convergence of a
(possibly infinite) sum of power series in $\kk\left[  \left[
T\right]  \right]  $ is defined:

\begin{statement}
\textit{Definition:} Let us introduce a notation: If $u\in\kk\left[
\left[  T\right]  \right]  $ is any power series, and if $n\in \NN$,
then $\left[  T^{n}\right]  u$ will mean the coefficient of $T^{n}$ in $u$.
Thus, $u=\sum_{n\geq0}\left(  \left[  T^{n}\right]  u\right)  \cdot T^{n}$ for
each $u\in\kk\left[  \left[  T\right]  \right]  $.

Let $\left(  r_{q}\right)  _{q\in Q}\in\left(  \kk\left[  \left[
T\right]  \right]  \right)  ^{Q}$ be a family of power series in
$\kk\left[  \left[  T\right]  \right]  $. We say that the sum
$\sum_{q\in Q}r_{q}$
\emph{converges}\index{convergence of an infinite sum in $\kk[[T]]$}
in $\kk\left[  \left[  T\right]  \right]  $ if and only if
for each $n\in \NN$,
\[
\text{all but finitely many }q\in Q\text{ satisfy }\left[  T^{n}\right]
r_{q}=0.
\]
\footnote{Of course, what precisely ``all but finitely many
$q\in Q$'' means here -- i.e., which $q$ are excluded --
depends on $n$.} In this case, the sum $\sum_{q\in Q}r_{q}$ is defined to be
the power series in $\kk\left[  \left[  T\right]  \right]  $ whose
coefficients are given by the rule%
\[
\left(  \left[  T^{n}\right]  \left(  \sum_{q\in Q}r_{q}\right)  =\sum_{q\in
Q}\left[  T^{n}\right]  r_{q}\ \ \ \ \ \ \ \ \ \ \text{for all }n\in
\NN\right)  .
\]

\end{statement}

Now, we can state a useful fact that relates this notion of convergence to
manipulations of maps in $\Hom \left(  C,A\right)  $:

\begin{statement}
\textit{Fact B.7:} Let $\left(  r_{q}\right)  _{q\in Q}\in\left(
\kk\left[  \left[  T\right]  \right]  \right)  ^{Q}$ be a family of
power series such that the (possibly infinite) sum $\sum_{q\in Q}r_{q}$
converges in $\kk\left[  \left[  T\right]  \right]  $. Let
$f\in\mathfrak{n}\left(  C,A\right)  $. Then, the family $\left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported and satisfies%
\[
\left(  \sum_{q\in Q}r_{q}\right)  ^{\star}\left(  f\right)  =\sum_{q\in
Q}\left(  r_{q}\right)  ^{\star}\left(  f\right)  .
\]

\end{statement}

[\textit{Proof of Fact B.7:} Define a power series $s\in\kk\left[
\left[  T\right]  \right]  $ by $s=\sum_{q\in Q}r_{q}$. (This is well-defined,
since the sum $\sum_{q\in Q}r_{q}$ converges.)

Let us introduce a notation: If $u\in\kk\left[  \left[  T\right]
\right]  $ is any power series, and if $n\in \NN$, then $\left[
T^{n}\right]  u$ will mean the coefficient of $T^{n}$ in $u$. Thus,
$u=\sum_{n\geq0}\left(  \left[  T^{n}\right]  u\right)  \cdot T^{n}$ for each
$u\in\kk\left[  \left[  T\right]  \right]  $. Applying this to $u=s$,
we find $s=\sum_{n\geq0}\left(  \left[  T^{n}\right]  s\right)  \cdot T^{n}$.

We know that the sum $\sum_{q\in Q}r_{q}$ converges in $\kk\left[
\left[  T\right]  \right]  $. In other words, for each $n\in \NN$,
\begin{equation}
\text{all but finitely many }q\in Q\text{ satisfy }\left[  T^{n}\right]
r_{q}=0 \label{sol.convolution-series.b.7.pf.allbut}%
\end{equation}
(by the definition of convergence for an infinite sum in $\kk\left[
\left[  T\right]  \right]  $). (Of course, what precisely
``all but finitely many $q\in Q$'' means here -- i.e., which
$q$ are excluded -- depends on $n$.)

For every $q\in Q$, we have $r_{q}=\sum_{n\geq0}\left(  \left[  T^{n}\right]
r_{q}\right)  \cdot T^{n}$ (since $u=\sum_{n\geq0}\left(  \left[
T^{n}\right]  u\right)  \cdot T^{n}$ for each $u\in\kk\left[  \left[
T\right]  \right]  $) and therefore%
\begin{equation}
\left(  r_{q}\right)  ^{\star}\left(  f\right)  =\sum_{n\geq0}\left(  \left[
T^{n}\right]  r_{q}\right)  f^{\star n}
\label{sol.convolution-series.b.7.pf.rq*f}%
\end{equation}
(by the definition of $\left(  r_{q}\right)  ^{\star}\left(  f\right)  $).

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported (since $f$ is pointwise $\star$-nilpotent). In other words,
for each $x\in C$,%
\begin{equation}
\text{the family }\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}\text{ is finitely supported.}
\label{sol.convolution-series.b.7.pf.fs}%
\end{equation}


Now, let $x\in C$. Then, there exists an $N\in \NN$ such that%
\begin{equation}
\text{every }n\geq N\text{ satisfies }f^{\star n}\left(  x\right)  =0
\label{sol.convolution-series.b.7.pf.0}%
\end{equation}
\footnote{\textit{Proof.} The family $\left(  f^{\star n}\left(  x\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported (by
(\ref{sol.convolution-series.b.7.pf.fs})). In other words, all but finitely
many $n\in \NN$ satisfy $f^{\star n}\left(  x\right)  =0$. In other
words, there exists a finite subset $Z$ of $\NN$ such that
\begin{equation}
\text{each }n \in \NN\setminus Z\text{ satisfies }f^{\star n}\left(
x\right)  =0. \label{sol.convolution-series.b.7.pf.0.pf.1}%
\end{equation}
Consider this $Z$.
\par
The set $Z$ is a finite subset of $\NN$, and thus has an upper bound
(since any finite subset of $\NN$ has an upper bound). In other words,
there exists some $w\in \NN$ such that each $z\in Z$ satisfies $z\leq
w$. Consider this $w$.
\par
Now, let $n\in \NN$ be such that $n\geq w+1$. Assume (for the sake of
contradiction) that $n\in Z$. Recall that each $z\in Z$ satisfies $z\leq w$.
Applying this to $z=n$, we obtain $n\leq w$ (since $n\in Z$), so that $n\leq
w<w+1$. This contradicts $n\geq w+1$. This contradiction shows that our
assumption (that $n\in Z$) was wrong. Hence, we have $n\notin Z$. Combining
$n\in \NN$ with $n\notin Z$, we obtain $n\in \NN \setminus Z$.
Hence, (\ref{sol.convolution-series.b.7.pf.0.pf.1}) yields $f^{\star n}\left(
x\right)  =0$.
\par
Now, forget that we fixed $n$. We thus have shown that every $n\geq w+1$
satisfies $f^{\star n}\left(  x\right)  =0$. Hence, there exists an
$N\in \NN$ such that every $n\geq N$ satisfies $f^{\star n}\left(
x\right)  =0$ (namely, $N=w+1$).}. Consider this $N$.

Every $q\in Q$ satisfies%
\begin{align}
\underbrace{\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)
}_{\substack{=\sum_{n\geq0}\left(  \left[  T^{n}\right]  r_{q}\right)
f^{\star n}\\\text{(by (\ref{sol.convolution-series.b.7.pf.rq*f}))}}}\left(
x\right)   &  =\left(  \sum_{n\geq0}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\right)  \left(  x\right)  =\sum_{n\geq0}\left(
\left[  T^{n}\right]  r_{q}\right)  f^{\star n}\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{n\geq0;\\n<N}}}_{=\sum_{n=0}^{N-1}}\left(
\left[  T^{n}\right]  r_{q}\right)  f^{\star n}\left(  x\right)
+\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(  \left[  T^{n}\right]  r_{q}\right)
\underbrace{f^{\star n}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.0}))}}}\nonumber\\
&  =\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]  r_{q}\right)  f^{\star
n}\left(  x\right)  +\underbrace{\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(
\left[  T^{n}\right]  r_{q}\right)  0}_{=0}\nonumber\\
&  =\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]  r_{q}\right)  f^{\star
n}\left(  x\right)  . \label{sol.convolution-series.b.7.pf.rq*fx}%
\end{align}


For each $n\in \NN$, there exists a finite subset $Q_{n}$ of $Q$ such
that
\begin{equation}
\text{all }q\in Q\setminus Q_{n}\text{ satisfy }\left[  T^{n}\right]  r_{q}=0
\label{sol.convolution-series.b.7.pf.1}%
\end{equation}
(by (\ref{sol.convolution-series.b.7.pf.allbut})). Consider this $Q_{n}$.

Let $Q^{\prime}$ be the subset $Q_{0}\cup Q_{1}\cup\cdots\cup Q_{N-1}$ of $Q$.
Then, $Q^{\prime}$ is the union of the $N$ finite sets $Q_{0},Q_{1}%
,\ldots,Q_{N-1}$. Hence, $Q^{\prime}$ itself is a finite set. Thus, all but
finitely many $q\in Q$ satisfy $q\in Q\setminus Q^{\prime}$. Notice that the
set $Q$ is the union of its two disjoint subsets $Q^{\prime}$ and $Q\setminus
Q^{\prime}$ (since $Q^{\prime}$ is a subset of $Q$).

Moreover, if $n\in\left\{  0,1,\ldots,N-1\right\}  $, then
\begin{equation}
\text{every }q\in Q\setminus Q^{\prime}\text{ satisfies }\left[  T^{n}\right]
r_{q}=0 \label{sol.convolution-series.b.7.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.b.7.pf.2}):} Let
$n\in\left\{  0,1,\ldots,N-1\right\}  $. Let $q\in Q\setminus Q^{\prime}$.
\par
From $n\in\left\{  0,1,\ldots,N-1\right\}  $, we obtain $Q_{n}\subset
Q_{0}\cup Q_{1}\cup\cdots\cup Q_{N-1}=Q^{\prime}$ (since $Q^{\prime}=Q_{0}\cup
Q_{1}\cup\cdots\cup Q_{N-1}$). Hence, $Q\setminus\underbrace{Q_{n}}_{\subset
Q^{\prime}}\supset Q\setminus Q^{\prime}$, so that $Q\setminus Q^{\prime
}\subset Q\setminus Q_{n}$. Hence, $q\in Q\setminus Q^{\prime}\subset
Q\setminus Q_{n}$. Therefore, (\ref{sol.convolution-series.b.7.pf.1}) yields
$\left[  T^{n}\right]  r_{q}=0$. This proves
(\ref{sol.convolution-series.b.7.pf.2}).}. Hence, every $n\in\left\{
0,1,\ldots,N-1\right\}  $ satisfies%
\begin{align}
\left[  T^{n}\right]  \underbrace{s}_{=\sum_{q\in Q}r_{q}}  &  =\left[
T^{n}\right]  \left(  \sum_{q\in Q}r_{q}\right)  =\sum_{q\in Q}\left[
T^{n}\right]  r_{q}=\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}%
+\sum_{q\in Q\setminus Q^{\prime}}\underbrace{\left[  T^{n}\right]  r_{q}%
}_{\substack{=0\\\text{(by (\ref{sol.convolution-series.b.7.pf.2}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the set }Q\text{ is the union of
its two disjoint subsets }Q^{\prime}\text{ and }Q\setminus Q^{\prime}\right)
\nonumber\\
&  =\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}+\underbrace{\sum_{q\in
Q\setminus Q^{\prime}}0}_{=0}=\sum_{q\in Q^{\prime}}\left[  T^{n}\right]
r_{q}. \label{sol.convolution-series.b.7.pf.2cor}%
\end{align}


Now, each $q\in Q\setminus Q^{\prime}$ satisfies%
\begin{align}
\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(
x\right)   &  =\sum_{n=0}^{N-1}\underbrace{\left(  \left[  T^{n}\right]
r_{q}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.2}))}}}f^{\star n}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.convolution-series.b.7.pf.rq*fx})}\right) \nonumber\\
&  =\sum_{n=0}^{N-1}0f^{\star n}\left(  x\right)  =0.
\label{sol.convolution-series.b.7.pf.rq*fx=0}%
\end{align}
Hence, all but finitely many $q\in Q$ satisfy $\left(  \left(  r_{q}\right)
^{\star}\left(  f\right)  \right)  \left(  x\right)  =0$ (since all but
finitely many $q\in Q$ satisfy $q\in Q\setminus Q^{\prime}$). In other words,
\begin{equation}
\text{the family }\left(  \left(  \left(  r_{q}\right)  ^{\star}\left(
f\right)  \right)  \left(  x\right)  \right)  _{q\in Q}\in A^{Q}\text{ is
finitely supported.} \label{sol.convolution-series.b.7.pf.cl1}%
\end{equation}


Recall again that the set $Q$ is the union of its two disjoint subsets
$Q^{\prime}$ and $Q\setminus Q^{\prime}$. Hence,%
\begin{align}
\sum_{q\in Q}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)
\left(  x\right)   &  =\sum_{q\in Q^{\prime}}\underbrace{\left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(  x\right)
}_{\substack{=\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]  r_{q}\right)
f^{\star n}\left(  x\right)  \\\text{(by
(\ref{sol.convolution-series.b.7.pf.rq*fx}))}}}+\sum_{q\in Q\setminus
Q^{\prime}}\underbrace{\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)
\right)  \left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.rq*fx=0}))}}}\nonumber\\
&  =\underbrace{\sum_{q\in Q^{\prime}}\sum_{n=0}^{N-1}}_{\substack{=
\sum_{n=0}^{N-1}\sum_{q\in Q^{\prime}}\\\text{(here, we are interchanging}%
\\\text{two \textbf{finite} sums)}}}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\left(  x\right)  +\underbrace{\sum_{q\in Q\setminus
Q^{\prime}}0}_{=0}\nonumber\\
&  =\sum_{n=0}^{N-1}\sum_{q\in Q^{\prime}}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\left(  x\right)  .
\label{sol.convolution-series.b.7.pf.q1}%
\end{align}


On the other hand, recall that $s=\sum_{n\geq0}\left(  \left[  T^{n}\right]
s\right)  \cdot T^{n}$. Therefore,%
\[
s^{\star}\left(  f\right)  =\sum_{n\geq0}\left(  \left[  T^{n}\right]
s\right)  f^{\star n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}s^{\star}\left(  f\right)  \right)  .
\]
Applying both sides of this equality to $x$, we obtain%
\begin{align}
\left(  s^{\star}\left(  f\right)  \right)  \left(  x\right)   &  =\left(
\sum_{n\geq0}\left(  \left[  T^{n}\right]  s\right)  f^{\star n}\right)
\left(  x\right)  =\sum_{n\geq0}\left(  \left[  T^{n}\right]  s\right)
f^{\star n}\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{n\geq0;\\n<N}}}_{=\sum_{n=0}^{N-1}}\left(
\left[  T^{n}\right]  s\right)  f^{\star n}\left(  x\right)  +
\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(  \left[  T^{n}\right]  s\right)
\underbrace{f^{\star n}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.0}))}}}\nonumber\\
&  =\sum_{n=0}^{N-1}\underbrace{\left(  \left[  T^{n}\right]  s\right)
}_{\substack{=\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}\\\text{(by
(\ref{sol.convolution-series.b.7.pf.2cor}))}}}f^{\star n}\left(  x\right)
+\underbrace{\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(  \left[  T^{n}\right]
s\right)  0}_{=0}\nonumber\\
&  =\sum_{n=0}^{N-1}\left(  \sum_{q\in Q^{\prime}}\left(  \left[
T^{n}\right]  r_{q}\right)  \right)  f^{\star n}\left(  x\right)  =
\sum_{n=0}^{N-1}\sum_{q\in Q^{\prime}}
\left(  \left[  T^{n}\right]  r_{q}\right)
f^{\star n}\left(  x\right) \nonumber\\
&  =\sum_{q\in Q}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)
\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.convolution-series.b.7.pf.q1})}\right)  .
\label{sol.convolution-series.b.7.pf.cl2}%
\end{align}


Now, forget that we fixed $x$. We thus have shown that each $x\in C$ satisfies
(\ref{sol.convolution-series.b.7.pf.cl1}) and
(\ref{sol.convolution-series.b.7.pf.cl2}).

In particular, for each $x\in C$, the family $\left(  \left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(  x\right)  \right)
_{q\in Q}\in A^{Q}$ is finitely supported (since each $x\in C$ satisfies
(\ref{sol.convolution-series.b.7.pf.cl1})). In other words, the family
$\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  _{q\in Q}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise
finitely supported. Hence, the sum $\sum_{q\in Q}\left(  r_{q}\right)
^{\star}\left(  f\right)  $ is well-defined.

Furthermore, (\ref{sol.convolution-series.b.7.pf.cl2}) shows that each $x\in
C$ satisfies%
\[
\left(  s^{\star}\left(  f\right)  \right)  \left(  x\right)  =\sum_{q\in
Q}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(
x\right)  =\left(  \sum_{q\in Q}\left(  r_{q}\right)  ^{\star}\left(
f\right)  \right)  \left(  x\right)  .
\]
In other words, we have $s^{\star}\left(  f\right)  =\sum_{q\in Q}\left(
r_{q}\right)  ^{\star}\left(  f\right)  $. Since $s=\sum_{q\in Q}r_{q}$, this
rewrites as $\left(  \sum_{q\in Q}r_{q}\right)  ^{\star}\left(  f\right)
=\sum_{q\in Q}\left(  r_{q}\right)  ^{\star}\left(  f\right)  $. This
completes the proof of Fact B.7.]

\begin{proof}
[Proof of Proposition~\ref{prop.convolution-series}.](a) Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $k\in \NN$. The power series
$T^{k}\in\kk\left[  \left[  T\right]  \right]  $ can be written in the
form $T^{k}=\sum_{n\geq0}\delta_{n,k}T^{n}$ (since all addends in the sum
$\sum_{n\geq0}\delta_{n,k}T^{n}$ are zero except for the addend for $n=k$).
Hence, the definition of $\left(  T^{k}\right)  ^{\star}\left(  f\right)  $
yields%
\[
\left(  T^{k}\right)  ^{\star}\left(  f\right)  =\sum_{n\geq0}\delta
_{n,k}f^{\star n}=f^{\star k}%
\]
(since all addends in the sum $\sum_{n\geq0}\delta_{n,k}f^{\star n}$ are zero
except for the addend for $n=k$). This proves
Proposition~\ref{prop.convolution-series}(a).

(b) We are going to prove the formulas (\ref{eq.exe.convolution-series.b.1}),
(\ref{eq.exe.convolution-series.b.3}), (\ref{eq.exe.convolution-series.b.2}),
(\ref{eq.exe.convolution-series.b.4}), and
(\ref{eq.exe.convolution-series.b.5}) in this order.

[\textit{Proof of (\ref{eq.exe.convolution-series.b.1}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $u,v\in\kk\left[  \left[
T\right]  \right]  $. We must prove the equality
(\ref{eq.exe.convolution-series.b.1}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $).

Write the power series $v$ in the form $v=\sum_{n\geq0}v_{n}T^{n}$ with
$\left(  v_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $v^{\star
}\left(  f\right)  =\sum_{n\geq0}v_{n}f^{\star n}$ (by the definition of
$v^{\star}\left(  f\right)  $).

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
u_{n}\right)  _{n\in\NN}$) shows that the family $\left(  u_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}u_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $.

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
v_{n}\right)  _{n\in\NN}$) shows that the family $\left(  v_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}v_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $.

Proposition~\ref{prop.pw-fin-supp.P2} (applied to $Q=\NN$, $\left(
f_{q}\right)  _{q\in Q}=\left(  u_{n}f^{\star n}\right)  _{n\in\NN}$
and $\left(  g_{q}\right)  _{q\in Q}=\left(  v_{n}f^{\star n}\right)
_{n\in\NN}$) now shows that the family $\left(  u_{n}f^{\star n}%
+v_{n}f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is also pointwise finitely
supported, and satisfies%
\[
\sum_{n\in\NN}u_{n}f^{\star n}+\sum_{n\in\NN}v_{n}f^{\star
n}=\sum_{n\in\NN}\left(  u_{n}f^{\star n}+v_{n}f^{\star n}\right)  .
\]
Since $\sum_{n\in\NN}=\sum_{n\geq0}$, this rewrites as%
\begin{equation}
\sum_{n\geq0}u_{n}f^{\star n}+\sum_{n\geq0}v_{n}f^{\star n}=\sum_{n\geq
0}\underbrace{\left(  u_{n}f^{\star n}+v_{n}f^{\star n}\right)  }_{=\left(
u_{n}+v_{n}\right)  f^{\star n}}=\sum_{n\geq0}\left(  u_{n}+v_{n}\right)
f^{\star n}. \label{sol.convolution-series.b.1.pf.2}%
\end{equation}


Adding the equalities $u=\sum_{n\geq0}u_{n}T^{n}$ and $v=\sum_{n\geq0}%
v_{n}T^{n}$, we obtain%
\[
u+v=\sum_{n\geq0}u_{n}T^{n}+\sum_{n\geq0}v_{n}T^{n}=\sum_{n\geq0}\left(
u_{n}+v_{n}\right)  T^{n}.
\]
Hence, the definition of $\left(  u+v\right)  ^{\star}\left(  f\right)  $
yields
\begin{align*}
\left(  u+v\right)  ^{\star}\left(  f\right)   &  =\sum_{n\geq0}\left(
u_{n}+v_{n}\right)  f^{\star n}=\underbrace{\sum_{n\geq0}u_{n}f^{\star n}%
}_{=u^{\star}\left(  f\right)  }+\underbrace{\sum_{n\geq0}v_{n}f^{\star n}%
}_{=v^{\star}\left(  f\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.convolution-series.b.1.pf.2})}\right) \\
&  =u^{\star}\left(  f\right)  +v^{\star}\left(  f\right)  .
\end{align*}
This proves (\ref{eq.exe.convolution-series.b.1}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.3}):} This is similar to
the proof of (\ref{eq.exe.convolution-series.b.1}), but this time we need to
apply Fact B.3 (instead of applying Proposition~\ref{prop.pw-fin-supp.P2}).
The straightforward details are left to the reader.]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.2}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $u,v\in\kk\left[  \left[
T\right]  \right]  $. We must prove the equality
(\ref{eq.exe.convolution-series.b.2}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $).

Write the power series $v$ in the form $v=\sum_{n\geq0}v_{n}T^{n}$ with
$\left(  v_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $v^{\star
}\left(  f\right)  =\sum_{n\geq0}v_{n}f^{\star n}$ (by the definition of
$v^{\star}\left(  f\right)  $).

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
u_{n}\right)  _{n\in\NN}$) shows that the family $\left(  u_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}u_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $. Renaming the index $n$ as $q$ in this statement, we obtain the
following: The family $\left(  u_{q}f^{\star q}\right)  _{q\in\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported, and its sum $\sum_{q\geq0}u_{q}f^{\star q}$
belongs to $\Hom \left(  C,A\right)  $.

Similarly, the family $\left(  v_{r}f^{\star r}\right)  _{r\in\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported, and its sum $\sum_{r\geq0}v_{r}f^{\star r}$
belongs to $\Hom \left(  C,A\right)  $.

Proposition~\ref{prop.pw-fin-supp.P3} (applied to $Q=\NN$,
$R=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  u_{q}f^{\star
q}\right)  _{q\in\NN}$ and $\left(  g_{r}\right)  _{r\in R}=\left(
v_{r}f^{\star r}\right)  _{r\in\NN}$) thus shows that the family
$\left(  \left(  u_{q}f^{\star q}\right)  \star\left(  v_{r}f^{\star
r}\right)  \right)  _{\left(  q,r\right)  \in \NN\times\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN\times\NN}$
is pointwise finitely supported, and satisfies%
\[
\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(  \left(
u_{q}f^{\star q}\right)  \star\left(  v_{r}f^{\star r}\right)  \right)
=\left(  \sum_{q\in\NN}u_{q}f^{\star q}\right)  \star\left(
\sum_{r\in\NN}v_{r}f^{\star r}\right)  .
\]
Hence,
\begin{align}
\left(  \sum_{q\in\NN}u_{q}f^{\star q}\right)  \star\left(
\sum_{r\in\NN}v_{r}f^{\star r}\right)   &  =\sum_{\left(  q,r\right)
\in \NN\times\NN}\underbrace{\left(  \left(  u_{q}f^{\star
q}\right)  \star\left(  v_{r}f^{\star r}\right)  \right)  }_{\substack{=u_{q}%
v_{r}f^{\star q}\star f^{\star r}\\=u_{q}v_{r}f^{\star\left(  q+r\right)  }%
}}\nonumber\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}v_{r}%
f^{\star\left(  q+r\right)  }. \label{sol.convolution-series.b.2.pf.5}%
\end{align}


Multiplying the equalities
\[
u=\sum_{n\geq0}u_{n}T^{n}=\sum_{n\in\NN}u_{n}T^{n}=\sum_{q\in
\NN}u_{q}T^{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed
the summation index }n\text{ as }q\right)
\]
and%
\[
v=\sum_{n\geq0}v_{n}T^{n}=\sum_{n\in\NN}v_{n}T^{n}=\sum_{r\in
\NN}v_{r}T^{r}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed
the summation index }n\text{ as }r\right)  ,
\]
we obtain
\begin{align}
uv  &  =\left(  \sum_{q\in\NN}u_{q}T^{q}\right)  \left(
\sum_{r\in\NN}v_{r}T^{r}\right)  =\underbrace{\sum_{q\in\NN}
\sum_{r\in\NN}}_{=\sum_{\left(  q,r\right)  \in \NN\times\NN}}
\underbrace{u_{q}T^{q}v_{r}T^{r}}_{=u_{q}v_{r}T^{q+r}}
\nonumber\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}v_{r}T^{q+r}.
\label{sol.convolution-series.b.2.pf.uv=}%
\end{align}
Thus, in particular, the sum $\sum_{\left(  q,r\right)  \in \NN%
\times\NN}u_{q}v_{r}T^{q+r}$ converges in $\kk\left[  \left[
T\right]  \right]  $. Hence, Fact B.7 (applied to $\NN\times\NN$
and $\left(  u_{q}v_{r}T^{q+r}\right)  _{\left(  q,r\right)  \in
\NN\times\NN}$ instead of $Q$ and $\left(  r_{q}\right)  _{q\in
Q}$) shows that the family $\left(  \left(  u_{q}v_{r}T^{q+r}\right)  ^{\star
}\left(  f\right)  \right)  _{\left(  q,r\right)  \in \NN\times
\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN\times\NN}$ is pointwise finitely supported and satisfies%
\begin{equation}
\left(  \sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}%
v_{r}T^{q+r}\right)  ^{\star}\left(  f\right)  =\sum_{\left(  q,r\right)
\in \NN\times\NN}\left(  u_{q}v_{r}T^{q+r}\right)  ^{\star
}\left(  f\right)  . \label{sol.convolution-series.b.2.pf.2}%
\end{equation}
In light of (\ref{sol.convolution-series.b.2.pf.uv=}), the equality
(\ref{sol.convolution-series.b.2.pf.2}) rewrites as%
\begin{align*}
\left(  uv\right)  ^{\star}\left(  f\right)   &  =\sum_{\left(  q,r\right)
\in \NN\times\NN}\underbrace{\left(  u_{q}v_{r}T^{q+r}\right)
^{\star}\left(  f\right)  }_{\substack{=u_{q}v_{r}\left(  T^{q+r}\right)
^{\star}\left(  f\right)  \\\text{(by (\ref{eq.exe.convolution-series.b.3})
(applied}\\\text{to }u_{q}v_{r}\text{ and }T^{q+r}\text{ instead of }%
\lambda\text{ and }u\text{))}}}=\sum_{\left(  q,r\right)  \in \NN
\times\NN}u_{q}v_{r}\underbrace{\left(  T^{q+r}\right)  ^{\star}\left(
f\right)  }_{\substack{=f^{\star\left(  q+r\right)  }\\\text{(by
(\ref{eq.exe.convolution-series.a.monom}) (applied to }k=q+r\text{))}}}\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}v_{r}%
f^{\star\left(  q+r\right)  }=\left(  \sum_{q\in\NN}u_{q}f^{\star
q}\right)  \star\left(  \sum_{r\in\NN}v_{r}f^{\star r}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.b.2.pf.5}%
)}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\underbrace{u^{\star}\left(  f\right)  }_{=\sum_{n\geq0}u_{n}f^{\star n}}%
\star\underbrace{v^{\star}\left(  f\right)  }_{=\sum_{n\geq0}v_{n}f^{\star
n}}  &  =\left(  \underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}%
u_{n}f^{\star n}\right)  \star\left(  \underbrace{\sum_{n\geq0}}_{=
\sum_{n\in\NN}}v_{n}f^{\star n}\right)  =\underbrace{\left(
\sum_{n\in\NN}u_{n}f^{\star n}\right)  }_{\substack{=\sum_{q\in\NN}
u_{q}f^{\star q}}}\star\underbrace{\left(  \sum_{n\in\NN}v_{n}f^{\star
n}\right)  }_{\substack{=\sum_{r\in\NN}v_{r}f^{\star r}}}\\
&  =\left(  \sum_{q\in\NN}u_{q}f^{\star q}\right)  \star\left(
\sum_{r\in\NN}v_{r}f^{\star r}\right)  ,
\end{align*}
we obtain $\left(  uv\right)  ^{\star}\left(  f\right)  =u^{\star}\left(
f\right)  \star v^{\star}\left(  f\right)  $. This proves
(\ref{eq.exe.convolution-series.b.2}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.4}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $. Applying
(\ref{eq.exe.convolution-series.b.3}) to $u=0$ and $\lambda=0$, we find
$\left(  0\cdot0\right)  ^{\star}\left(  f\right)  =0\cdot0^{\star}\left(
f\right)  =0$. In other words, $0^{\star}\left(  f\right)  =0$. This proves
(\ref{eq.exe.convolution-series.b.4}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.5}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $. Applying
(\ref{eq.exe.convolution-series.a.monom}) to $k=0$, we find
\[
\left(  T^{0}\right)  ^{\star}\left(  f\right)  =f^{\star0}=\left(  \text{the
unity of the }\kk\text{-algebra }\left(  \Hom \left(
C,A\right)  ,\star\right)  \right)  =u_{A}\epsilon_{C}%
\]
(since the unity of the $\kk$-algebra $\left(  \Hom \left(
C,A\right)  ,\star\right)  $ is $u_{A}\epsilon_{C}$). In view of $T^{0}=1$,
this rewrites as $1^{\star}\left(  f\right)  =u_{A}\epsilon_{C}$. This proves
(\ref{eq.exe.convolution-series.b.5}).]

We have now proven all the equalities (\ref{eq.exe.convolution-series.b.1}),
(\ref{eq.exe.convolution-series.b.2}), (\ref{eq.exe.convolution-series.b.3}),
(\ref{eq.exe.convolution-series.b.4}) and (\ref{eq.exe.convolution-series.b.5}%
). Thus, Proposition~\ref{prop.convolution-series}(b) is proven.

(c) Let $f,g\in\mathfrak{n}\left(  C,A\right)  $ be such that $f\star g=g\star
f$. We must prove that $f+g\in\mathfrak{n}\left(  C,A\right)  $.

From $f\star g=g\star f$, we conclude that the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute.

Let $\mathfrak{G}$ be the $\kk$-subalgebra of $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ generated by the two
elements $f$ and $g$. Thus, the $\kk$-algebra $\mathfrak{G}$ is
generated by commuting elements (because the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute), and therefore is commutative (since any
$\kk$-algebra generated by commuting elements must be commutative).
Hence, the binomial formula holds in this $\kk$-algebra $\mathfrak{G}$.
Thus, we have%
\begin{equation}
\left(  f+g\right)  ^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\star
g^{\star\left(  n-i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in \NN \label{sol.convolution-series.c.binom}%
\end{equation}
(since the multiplication in the $\kk$-algebra $\mathfrak{G}$ is
$\star$).

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise finitely
supported. Renaming the index $n$ as $q$ in this statement, we thus conclude
that the family $\left(  f^{\star q}\right)  _{q\in\NN}$ is pointwise
finitely supported. Similarly, the family $\left(  g^{\star r}\right)
_{r\in\NN}$ is pointwise finitely supported.

Thus, Proposition~\ref{prop.pw-fin-supp.P3} (applied to $Q=\NN$,
$R=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star q}\right)
_{q\in\NN}$ and $\left(  g_{r}\right)  _{r\in R}=\left(  g^{\star
r}\right)  _{r\in\NN}$) shows that the family $\left(  f^{\star q}\star
g^{\star r}\right)  _{\left(  q,r\right)  \in \NN\times\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN\times\NN}$
is pointwise finitely supported, and that it satisfies%
\[
\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(  f^{\star
q}\star g^{\star r}\right)  =\left(  \sum_{q\in\NN}f^{\star q}\right)
\star\left(  \sum_{r\in\NN}g^{\star r}\right)  .
\]


In particular, the family $\left(  f^{\star q}\star g^{\star r}\right)
_{\left(  q,r\right)  \in \NN\times\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN\times\NN}$
is pointwise finitely supported. In other words, for each $x\in C$,%
\begin{equation}
\text{the family }\left(  \left(  f^{\star q}\star g^{\star r}\right)  \left(
x\right)  \right)  _{\left(  q,r\right)  \in \NN\times\NN}\in
A^{\NN\times\NN}\text{ is finitely supported.}
\label{sol.convolution-series.c.fs}%
\end{equation}


Let $x\in C$. Then, all but finitely many $\left(  q,r\right)  \in
\NN\times\NN$ satisfy $\left(  f^{\star q}\star g^{\star
r}\right)  \left(  x\right)  =0$ (because of
(\ref{sol.convolution-series.c.fs})). In other words, there exists a finite
subset $K$ of $\NN\times\NN$ such that%
\begin{equation}
\text{each }\left(  q,r\right)  \in\left(  \NN\times\NN\right)
\setminus K\text{ satisfies }\left(  f^{\star q}\star g^{\star r}\right)
\left(  x\right)  =0. \label{sol.convolution-series.c.3}%
\end{equation}
Consider this $K$.

Let $Q=\left\{  u+v\ \mid\ \left(  u,v\right)  \in K\right\}  $. Thus, $Q$ is
a finite set (since $K$ is a finite set). Thus, all but finitely many
$n\in \NN$ satisfy $n\in \NN\setminus Q$.

But every $n\in \NN\setminus Q$ satisfies $\left(  f+g\right)  ^{\star
n}\left(  x\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Let $n\in
\NN\setminus Q$. We must show that $\left(  f+g\right)  ^{\star
n}\left(  x\right)  =0$.
\par
We have $n\in \NN\setminus Q$. In other words, $n\in \NN$ and
$n\notin Q$.
\par
Let $i\in\left\{  0,1,\ldots,n\right\}  $ be arbitrary. We shall show that
$\left(  f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =0$ first.
\par
From $i\in\left\{  0,1,\ldots,n\right\}  $, we obtain $i\in \NN$ and
$n-i\in \NN$. Thus, $\left(  i,n-i\right)  \in \NN\times\NN$.
\par
If we had $\left(  i,n-i\right)  \in K$, then we would have%
\[
i+\left(  n-i\right)  \in\left\{  u+v\ \mid\ \left(  u,v\right)  \in
K\right\}  =Q\ \ \ \ \ \ \ \ \ \ \left(  \text{since }Q=\left\{
u+v\ \mid\ \left(  u,v\right)  \in K\right\}  \right)  ,
\]
which would contradict $i+\left(  n-i\right)  =n\notin Q$. Thus, we cannot
have $\left(  i,n-i\right)  \in K$. In other words, we have $\left(
i,n-i\right)  \notin K$.
\par
Combining $\left(  i,n-i\right)  \in \NN\times\NN$ with $\left(
i,n-i\right)  \notin K$, we obtain $\left(  i,n-i\right)  \in\left(
\NN\times\NN\right)  \setminus K$. Hence,
(\ref{sol.convolution-series.c.3}) (applied to $\left(  q,r\right)  =\left(
i,n-i\right)  $) yields $\left(  f^{\star i}\star g^{\star\left(  n-i\right)
}\right)  \left(  x\right)  =0$.
\par
Now, forget that we fixed $i$. We thus have shown that%
\begin{equation}
\left(  f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  0,1,\ldots
,n\right\}  . \label{sol.convolution-series.c.3.pf.2}%
\end{equation}
\par
Now, applying both sides of the equality (\ref{sol.convolution-series.c.binom}%
) to $x$, we obtain%
\[
\left(  f+g\right)  ^{\star n}\left(  x\right)  =\left(  \sum_{i=0}^{n}%
\dbinom{n}{i}f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\star
g^{\star\left(  n-i\right)  }\right)  \left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.convolution-series.c.3.pf.2}))}}%
}=\sum_{i=0}^{n}\dbinom{n}{i}0=0.
\]
Qed.}. Hence, all but finitely many $n\in \NN$ satisfy $\left(
f+g\right)  ^{\star n}\left(  x\right)  =0$ (since all but finitely many
$n\in \NN$ satisfy $n\in \NN\setminus Q$). In other words, the
family $\left(  \left(  f+g\right)  ^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported.

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
family $\left(  \left(  f+g\right)  ^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported. In other words,
the family $\left(  \left(  f+g\right)  ^{\star n}\right)  _{n\in\NN}
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported. In other words, the map $f+g$ is pointwise
$\star$-nilpotent (by the definition of ``pointwise $\star
$-nilpotent''). In other words, $f+g\in\mathfrak{n}\left(
C,A\right)  $ (since $\mathfrak{n}\left(  C,A\right)  $ is the set of all
pointwise $\star$-nilpotent maps in $\Hom \left(  C,A\right)
$). This proves Proposition~\ref{prop.convolution-series}(c).

(d) Let $\lambda\in\kk$ and $f\in\mathfrak{n}\left(  C,A\right)  $. We
must prove that $\lambda f\in\mathfrak{n}\left(  C,A\right)  $.

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $.
Thus, the family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported. Hence, Proposition~\ref{prop.pw-fin-supp.P4} (applied to
$Q=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)
_{n\in\NN}$ and $\left(  \lambda_{q}\right)  _{q\in Q}=\left(
\lambda^{n}\right)  _{n\in\NN}$) shows that the family $\left(
\lambda^{n}f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is pointwise finitely supported.
In other words, the family $\left(  \left(  \lambda f\right)  ^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported (since $\left(
\lambda f\right)  ^{\star n}=\lambda^{n}f^{\star n}$ for each $n\in \NN$).
In other words, the map $\lambda f$ is pointwise $\star$-nilpotent. In
other words, $\lambda f\in\mathfrak{n}\left(  C,A\right)  $. This proves
Proposition~\ref{prop.convolution-series}(d).

(e) Let $f\in\mathfrak{n}\left(  C,A\right)  $ and $g\in\Hom
\left(  C,A\right)  $ be such that $f\star g=g\star f$. We must prove that
$f\star g\in\mathfrak{n}\left(  C,A\right)  $.

From $f\star g=g\star f$, we conclude that the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute.

Let $\mathfrak{G}$ be the $\kk$-subalgebra of $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ generated by the two
elements $f$ and $g$. Thus, the $\kk$-algebra $\mathfrak{G}$ is
generated by commuting elements (because the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute), and therefore is commutative (since any
$\kk$-algebra generated by commuting elements must be commutative).
Hence, the usual laws for exponentiation hold in this $\kk$-algebra
$\mathfrak{G}$. In particular, we have%
\begin{equation}
\left(  f\star g\right)  ^{\star n}=f^{\star n}\star g^{\star n}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in \NN
\label{sol.convolution-series.e.powerlaw}%
\end{equation}
(since the multiplication in the $\kk$-algebra $\mathfrak{G}$ is the
convolution $\star$).

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $.
Thus, the family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported.

Thus, Proposition~\ref{prop.pw-fin-supp.P5} (applied to $Q=\NN$,
$\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)  _{n\in\NN}$
and $\left(  g_{q}\right)  _{q\in Q}=\left(  g^{\star n}\right)
_{n\in\NN}$) shows that the family $\left(  f^{\star n}\star g^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported. In other words, the
family $\left(  \left(  f\star g\right)  ^{\star n}\right)  _{n\in\NN}
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported (because of
(\ref{sol.convolution-series.e.powerlaw})). In other words, the map $f\star g$
is pointwise $\star$-nilpotent. In other words, $f\star g\in\mathfrak{n}%
\left(  C,A\right)  $. This proves Proposition~\ref{prop.convolution-series}(e).

(f) Let $v\in\kk\left[  \left[  T\right]  \right]  $ be a power series
whose constant term is $0$. Let $f\in\mathfrak{n}\left(  C,A\right)  $. We
must show that $v^{\star}\left(  f\right)  \in\mathfrak{n}\left(  C,A\right)
$.

We know that the constant term of $v$ is $0$. Thus, the power series $v$ is
divisible by $T$ in the ring $\kk\left[  \left[  T\right]  \right]  $.
In other words, there exists a power series $u\in\kk\left[  \left[
T\right]  \right]  $ such that $v=Tu$. Consider this $u$. Define
$g\in\Hom \left(  C,A\right)  $ by $g=u^{\star}\left(  f\right)
$.

Applying (\ref{eq.exe.convolution-series.a.monom}) to $k=1$, we find $\left(
T^{1}\right)  ^{\star}\left(  f\right)  =f^{\star1}=f$. Since $T^{1}=T$, this
rewrites as $T^{\star}\left(  f\right)  =f$.

From $v=Tu$, we obtain%
\begin{align}
v^{\star}\left(  f\right)   &  =\left(  Tu\right)  ^{\star}\left(  f\right)
=\underbrace{T^{\star}\left(  f\right)  }_{=f}\star\underbrace{u^{\star
}\left(  f\right)  }_{=g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.exe.convolution-series.b.2}) (applied to }T\text{ and }u\text{
instead of }u\text{ and }v\text{)}\right) \nonumber\\
&  =f\star g. \label{sol.convolution-series.f.1}%
\end{align}


On the other hand, from $v=Tu=u\cdot T$, we obtain%
\begin{align*}
v^{\star}\left(  f\right)   &  =\left(  u\cdot T\right)  ^{\star}\left(
f\right)  =\underbrace{u^{\star}\left(  f\right)  }_{=g}\star
\underbrace{T^{\star}\left(  f\right)  }_{=f}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.exe.convolution-series.b.2}) (applied to }T\text{ instead of
}v\text{)}\right) \\
&  =g\star f.
\end{align*}
Comparing this with (\ref{sol.convolution-series.f.1}), we obtain $f\star
g=g\star f$. Hence, Proposition~\ref{prop.convolution-series}(e) yields
$f\star g\in\mathfrak{n}\left(  C,A\right)  $. In light of
(\ref{sol.convolution-series.f.1}), this rewrites as $v^{\star}\left(
f\right)  \in\mathfrak{n}\left(  C,A\right)  $. This proves
Proposition~\ref{prop.convolution-series}(f).

(g) Let us first prove the following fact:

\begin{statement}
\textit{Fact G.1:} Let $v\in\kk\left[  \left[  T\right]  \right]  $ be
any power series. Let $f\in\mathfrak{n}\left(  C,A\right)  $. Then,%
\begin{equation}
\left(  v^{n}\right)  ^{\star}\left(  f\right)  =\left(  v^{\star}\left(
f\right)  \right)  ^{\star n}\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in \NN. \label{sol.convolution-series.g.1}%
\end{equation}

\end{statement}

[\textit{Proof of Fact G.1:} We shall prove (\ref{sol.convolution-series.g.1})
by induction over $n$:

\textit{Induction base:} We have $\left(  \underbrace{v^{0}}_{=1}\right)
^{\star}\left(  f\right)  =1^{\star}\left(  f\right)  =u_{A}\epsilon_{C}$ (by
(\ref{eq.exe.convolution-series.b.5})). Comparing this with
\[
\left(  v^{\star}\left(  f\right)  \right)  ^{\star0}=\left(  \text{the unity
of the }\kk\text{-algebra }\left(  \Hom \left(
C,A\right)  ,\star\right)  \right)  =u_{A}\epsilon_{C},
\]
we obtain $\left(  v^{0}\right)  ^{\star}\left(  f\right)  =\left(  v^{\star
}\left(  f\right)  \right)  ^{\star0}$. In other words,
(\ref{sol.convolution-series.g.1}) holds for $n=0$. This completes the
induction base.

\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{sol.convolution-series.g.1}) holds for $n=N$. We must prove that
(\ref{sol.convolution-series.g.1}) holds for $n=N+1$.

We have assumed that (\ref{sol.convolution-series.g.1}) holds for $n=N$. In
other words, we have $\left(  v^{N}\right)  ^{\star}\left(  f\right)  =\left(
v^{\star}\left(  f\right)  \right)  ^{\star N}$.

Now,%
\begin{align*}
\left(  \underbrace{v^{N+1}}_{=v^{N}v}\right)  ^{\star}\left(  f\right)   &
=\left(  v^{N}v\right)  ^{\star}\left(  f\right)  =\underbrace{\left(
v^{N}\right)  ^{\star}\left(  f\right)  }_{=\left(  v^{\star}\left(  f\right)
\right)  ^{\star N}}\star v^{\star}\left(  f\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.convolution-series.b.2})
(applied to }u=v^{N}\text{)}\right) \\
&  =\left(  v^{\star}\left(  f\right)  \right)  ^{\star N}\star v^{\star
}\left(  f\right)  =\left(  v^{\star}\left(  f\right)  \right)  ^{\star\left(
N+1\right)  }.
\end{align*}
In other words, (\ref{sol.convolution-series.g.1}) holds for $n=N+1$. This
completes the induction step. Thus, the induction proof of
(\ref{sol.convolution-series.g.1}) is complete. In other words, Fact G.1 is proven.]

Now, let $u,v\in\kk\left[  \left[  T\right]  \right]  $ be two power
series such that the constant term of $v$ is $0$. Let $f\in\mathfrak{n}\left(
C,A\right)  $ be arbitrary. We must prove (\ref{eq.exe.convolution-series.g.1}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus,
\begin{equation}
u^{\star}\left(  v^{\star}\left(  f\right)  \right)  =\sum_{n\geq0}%
u_{n}\left(  v^{\star}\left(  f\right)  \right)  ^{\star n}
\label{sol.convolution-series.g.3}%
\end{equation}
(by the definition of $u^{\star}\left(  v^{\star}\left(  f\right)  \right)  $).

But the definition of the composition $u\left[  v\right]  $ yields $u\left[
v\right]  =\sum_{n\in\NN}u_{n}v^{n}$ (since $u=\sum_{n\geq0}u_{n}%
T^{n}=\sum_{n\in\NN}u_{n}T^{n}$). In particular, the sum
$\sum_{n\in\NN}u_{n}v^{n}$ converges in $\kk\left[  \left[  T\right]
\right]  $. Hence, Fact B.7 (applied to $Q=\NN$ and $\left(
r_{q}\right)  _{q\in Q}=\left(  u_{n}v^{n}\right)  _{n\in\NN}$) shows
that the family $\left(  \left(  u_{n}v^{n}\right)  ^{\star}\left(  f\right)
\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported and satisfies%
\begin{equation}
\left(  \sum_{n\in\NN}u_{n}v^{n}\right)  ^{\star}\left(  f\right)
=\sum_{n\in\NN}\left(  u_{n}v^{n}\right)  ^{\star}\left(  f\right)  .
\label{sol.convolution-series.g.4}%
\end{equation}


Now, recall that $u\left[  v\right]  =\sum_{n\in\NN}u_{n}v^{n}$. Hence,%
\begin{align*}
\left(  u\left[  v\right]  \right)  ^{\star}\left(  f\right)   &  =\left(
\sum_{n\in\NN}u_{n}v^{n}\right)  ^{\star}\left(  f\right)  =
\sum_{n\in\NN}\underbrace{\left(  u_{n}v^{n}\right)  ^{\star}\left(
f\right)  }_{\substack{=u_{n}\left(  v^{n}\right)  ^{\star}\left(  f\right)
\\\text{(by (\ref{eq.exe.convolution-series.b.3}) (applied}\\\text{to }%
u_{n}\text{ and }v^{n}\text{ instead of }\lambda\text{ and }u\text{))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.g.4}%
)}\right) \\
&  =\sum_{n\in\NN}u_{n}\underbrace{\left(  v^{n}\right)  ^{\star
}\left(  f\right)  }_{\substack{=\left(  v^{\star}\left(  f\right)  \right)
^{\star n}\\\text{(by Fact G.1)}}}=\underbrace{\sum_{n\in\NN}}%
_{=\sum_{n\geq0}}u_{n}\left(  v^{\star}\left(  f\right)  \right)  ^{\star
n}=\sum_{n\geq0}u_{n}\left(  v^{\star}\left(  f\right)  \right)  ^{\star
n}=u^{\star}\left(  v^{\star}\left(  f\right)  \right)
\end{align*}
(by (\ref{sol.convolution-series.g.3})). Thus,
(\ref{eq.exe.convolution-series.g.1}) is proven. This proves
Proposition~\ref{prop.convolution-series}(g).

(h) Let us first prove a simple fact:

\begin{statement}
\textit{Fact H.1:} Let $C$ be a graded $\kk$-coalgebra. Let
$f\in\Hom \left(  C,A\right)  $ be such that $f\left(
C_{0}\right)  =0$. Then, for each $i\in \NN$, we have%
\begin{equation}
f^{\star i}\left(  C_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in \NN \text{ satisfying }i>n. \label{sol.convolution-series.h.1}%
\end{equation}

\end{statement}

[\textit{Proof of Fact H.1:} We shall prove (\ref{sol.convolution-series.h.1})
by induction over $i$:

\textit{Induction base:} There exists no $n\in \NN$ satisfying $0>n$
(since each $n\in \NN$ satisfies $n\geq0$). Hence,
(\ref{sol.convolution-series.h.1}) is vacuously true for $i=0$. This completes
the induction base.

\textit{Induction step:} Let $p\in \NN$. Assume that
(\ref{sol.convolution-series.h.1}) holds for $i=p$. We must prove that
(\ref{sol.convolution-series.h.1}) holds for $i=p+1$.

We have assumed that (\ref{sol.convolution-series.h.1}) holds for $i=p$. In
other words, we have%
\begin{equation}
f^{\star p}\left(  C_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in \NN \text{ satisfying }p>n. \label{sol.convolution-series.h.1.pf.1}%
\end{equation}


Now, let $n\in \NN$ be such that $p+1>n$.

It is easy to see that
\begin{equation}
\left(  f\otimes f^{\star p}\right)  \left(  C_{i}\otimes C_{j}\right)  =0
\label{sol.convolution-series.h.1.pf.2}%
\end{equation}
for every $\left(  i,j\right)  \in \NN^{2}$ satisfying $i+j=n$%
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.convolution-series.h.1.pf.2}):}
Let $\left(  i,j\right)  \in \NN^{2}$ be such that $i+j=n$. We must
prove (\ref{sol.convolution-series.h.1.pf.2}).
\par
From $\left(  i,j\right)  \in \NN^{2}$, we obtain $i\in \NN$ and
$j\in \NN$. If $i=0$, then
\[
\left(  f\otimes f^{\star p}\right)  \left(  C_{i}\otimes C_{j}\right)
=f\left(  \underbrace{C_{i}}_{\substack{=C_{0}\\\text{(since }i=0\text{)}%
}}\right)  \otimes f^{\star p}\left(  C_{j}\right)  =\underbrace{f\left(
C_{0}\right)  }_{=0}\otimes f^{\star p}\left(  C_{j}\right)  =0\otimes
f^{\star p}\left(  C_{j}\right)  =0.
\]
Hence, if $i=0$, then (\ref{sol.convolution-series.h.1.pf.2}) is proven. Thus,
for the rest of the proof of (\ref{sol.convolution-series.h.1.pf.2}), we can
WLOG assume that we don't have $i=0$. Assume this.
\par
We have $i\neq0$ (since we don't have $i=0$), and thus $i\geq1$ (since
$i\in \NN$). But $i+j=n$, so that $j=n-\underbrace{i}_{\geq1}\leq n-1$.
But $p+1>n$; thus, $p>n-1\geq j$ (since $j\leq n-1$). Hence,
(\ref{sol.convolution-series.h.1.pf.1}) (applied to $j$ instead of $n$) shows
that $f^{\star p}\left(  C_{j}\right)  =0$. Now,%
\[
\left(  f\otimes f^{\star p}\right)  \left(  C_{i}\otimes C_{j}\right)
=f\left(  C_{i}\right)  \otimes\underbrace{f^{\star p}\left(  C_{j}\right)
}_{=0}=f\left(  C_{i}\right)  \otimes0=0.
\]
This proves (\ref{sol.convolution-series.h.1.pf.2}).}.

Recall that the $\kk$-coalgebra $C$ is graded. Thus, its
comultiplication $\Delta$ is graded. In other words, $\Delta\left(
C_{k}\right)  \subset\left(  C\otimes C\right)  _{k}$ for each $k\in
\NN$. Applying this to $k=n$, we obtain
\begin{align*}
\Delta\left(  C_{n}\right)   &  \subset\left(  C\otimes C\right)
_{n}=\bigoplus_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}%
}C_{i}\otimes C_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the
grading on }C\otimes C\right) \\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}%
}C_{i}\otimes C_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are
sums}\right)  .
\end{align*}
Applying the map $f\otimes f^{\star p}$ to both sides of this relation, we
obtain%
\begin{align}
\left(  f\otimes f^{\star p}\right)  \left(  \Delta\left(  C_{n}\right)
\right)   &  \subset\left(  f\otimes f^{\star p}\right)  \left(
\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}}C_{i}\otimes
C_{j}\right)  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2%
;\\i+j=n}}\underbrace{\left(  f\otimes f^{\star p}\right)  \left(
C_{i}\otimes C_{j}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.h.1.pf.2}))}}}\nonumber\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}}0=0.
\label{sol.convolution-series.h.1.pf.4}%
\end{align}


But the definition of convolution yields $f\star f^{\star p}=m\circ\left(
f\otimes f^{\star p}\right)  \circ\Delta$ (where $m$ denotes the
multiplication map $A\otimes A\rightarrow A$). We have%
\[
f^{\star\left(  p+1\right)  }=f\star f^{\star p}=m\circ\left(  f\otimes
f^{\star p}\right)  \circ\Delta
\]
and thus%
\begin{align*}
\underbrace{f^{\star\left(  p+1\right)  }}_{=m\circ\left(  f\otimes f^{\star
p}\right)  \circ\Delta}\left(  C_{n}\right)   &  =\left(  m\circ\left(
f\otimes f^{\star p}\right)  \circ\Delta\right)  \left(  C_{n}\right)
=m\left(  \underbrace{\left(  f\otimes f^{\star p}\right)  \left(
\Delta\left(  C_{n}\right)  \right)  }_{\substack{\subset0\\\text{(by
(\ref{sol.convolution-series.h.1.pf.4}))}}}\right) \\
&  \subset m\left(  0\right)  =0.
\end{align*}
In other words, $f^{\star\left(  p+1\right)  }\left(  C_{n}\right)  =0$.

Now, forget that we fixed $n$. We thus have proven that
\[
f^{\star\left(  p+1\right)  }\left(  C_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \NN \text{ satisfying
}p+1>n.
\]
In other words, (\ref{sol.convolution-series.h.1}) holds for $i=p+1$. This
completes the induction step. Thus, (\ref{sol.convolution-series.h.1}) is
proven by induction. Hence, Fact H.1 is proven.]

Now, let us actually prove Proposition~\ref{prop.convolution-series}(h).

Assume that $C$ is a graded $\kk$-coalgebra. Assume that $f\in
\Hom \left(  C,A\right)  $ satisfies $f\left(  C_{0}\right)
=0$. We must show that $f\in\mathfrak{n}\left(  C,A\right)  $.

Let $x\in C$. Thus, $x$ is a sum of finitely many homogeneous elements of $C$
(since $C$ is graded). In other words, there exist some $k\in \NN$ and
some homogeneous elements $x_{1},x_{2},\ldots,x_{k}\in C$ satisfying
$x=\sum_{g=1}^{k}x_{g}$. Consider this $k$ and these $x_{1},x_{2},\ldots
,x_{k}$.

For each $g\in\left\{  1,2,\ldots,k\right\}  $, there exists some $n_{g}%
\in \NN$ satisfying $x_{g}\in C_{n_{g}}$ (since $x_{g}$ is a homogeneous
element of $C$). Consider this $n_{g}$. The set $\left\{  n_{1},n_{2}%
,\ldots,n_{k}\right\}  $ is a finite subset of $\NN$, and thus has an
upper bound (since any finite subset of $\NN$ has an upper bound). In
other words, there exists some $N\in \NN$ such that
\begin{equation}
\text{each }n\in\left\{  n_{1},n_{2},\ldots,n_{k}\right\}  \text{ satisfies
}n\leq N. \label{sol.exe.convolution-series.h.N}%
\end{equation}
Consider this $N$.

Let $Q=\left\{  0,1,\ldots,N\right\}  $. Then, $Q$ is a finite subset of
$\NN$. Thus, all but finitely many $n\in \NN$ satisfy
$n\in \NN \setminus Q$.

Now, let $n\in \NN \setminus Q$ be arbitrary. Thus,
$n\in \NN \setminus Q=\left\{  N+1,N+2,N+3,\ldots\right\}  $
(since $Q=\left\{  0,1,\ldots,N\right\}  $). Hence, $n>N$.

Thus, each $g\in\left\{  1,2,\ldots,k\right\}  $ satisfies
\begin{equation}
f^{\star n}\left(  x_{g}\right)  =0
\label{sol.exe.convolution-series.h.fnxg=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.h.fnxg=0}):} Let
$g\in\left\{  1,2,\ldots,k\right\}  $. Then, $n_{g}\in\left\{  n_{1}%
,n_{2},\ldots,n_{k}\right\}  $. Hence, (\ref{sol.exe.convolution-series.h.N})
(applied to $n=n_{g}$) yields $n_{g}\leq N$. Hence, $N\geq n_{g}$, so that
$n>N\geq n_{g}$. Hence, (\ref{sol.convolution-series.h.1}) (applied to $n$ and
$n_{g}$ instead of $i$ and $n$) yields $f^{\star n}\left(  C_{n_{g}}\right)
=0$.
\par
But $x_{g}\in C_{n_{g}}$ (by the definition of $n_{g}$) and thus $f^{\star
n}\left(  \underbrace{x_{g}}_{\in C_{n_{g}}}\right)  \in f^{\star n}\left(
C_{n_{g}}\right)  =0$. In other words, $f^{\star n}\left(  x_{g}\right)  =0$.
This proves (\ref{sol.exe.convolution-series.h.fnxg=0}).}. Applying the map
$f^{\star n}$ to the equality $x=\sum_{g=1}^{k}x_{g}$, we obtain%
\[
f^{\star n}\left(  x\right)  =f^{\star n}\left(  \sum_{g=1}^{k}x_{g}\right)
=\sum_{g=1}^{k}\underbrace{f^{\star n}\left(  x_{g}\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.h.fnxg=0}))}}%
}=\sum_{g=1}^{k}0=0.
\]


Now, forget that we fixed $n$. We thus have shown that each $n\in
\NN \setminus Q$ satisfies $f^{\star n}\left(  x\right)  =0$. Hence, all
but finitely many $n\in \NN$ satisfy $f^{\star n}\left(  x\right)  =0$
(since all but finitely many $n\in \NN$ satisfy $n\in \NN \setminus
Q$). In other words, the family $\left(  f^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported.

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
family $\left(  f^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
A^{\NN}$ is finitely supported. In other words, the family $\left(
f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom \left(
C,A\right)  \right)  ^{\NN}$ is pointwise finitely supported. In other
words, the map $f$ is pointwise $\star$-nilpotent. In other words,
$f\in\mathfrak{n}\left(  C,A\right)  $. This proves
Proposition~\ref{prop.convolution-series}(h).

(i) Before we prove this, let us state a general fact:

\begin{statement}
\textit{Fact H.7:} Let $V$ and $W$ be two $\kk$-modules. Let
$\varphi:V\rightarrow W$ be a $\kk$-linear map. Let $\left(
v_{q}\right)  _{q\in Q}\in V^{Q}$ be a finitely supported family of elements
of $V$. Then, the family $\left(  \varphi\left(  v_{q}\right)  \right)  _{q\in
Q}\in W^{Q}$ is also finitely supported, and satisfies $\sum_{q\in Q}%
\varphi\left(  v_{q}\right)  =\varphi\left(  \sum_{q\in Q}v_{q}\right)  $.
\end{statement}

[\textit{Proof of Fact H.7:} Fact H.7 is a basic property of linear maps
(essentially, it says that any $\kk$-linear map preserves sums in which
all but finitely many addends are zero), and we omit its simple proof.]

Now, let us start proving Proposition~\ref{prop.convolution-series}(i).

Let $B$ be any $\kk$-algebra. Let $s:A\rightarrow B$ be any
$\kk$-algebra homomorphism.

Proposition~\ref{prop.convolution.functor} (applied to $A^{\prime}=B$,
$C^{\prime}=C$, $\alpha=s$ and $\gamma=\id_{C}$) shows
that the map%
\[
\Hom \left(  C,A\right)  \rightarrow\Hom \left(
C,B\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto s\circ f\circ\operatorname*{id}%
\nolimits_{C}%
\]
is a $\kk$-algebra homomorphism from the $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ to the $\kk%
$-algebra $\left(  \Hom \left(  C,B\right)  ,\star\right)  $.
Since each $f\in\Hom \left(  C,A\right)  $ satisfies $s\circ
f\circ\id_{C}=s\circ f$, this rewrites as follows: The
map%
\[
\Hom \left(  C,A\right)  \rightarrow\Hom \left(
C,B\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto s\circ f
\]
is a $\kk$-algebra homomorphism from the $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ to the $\kk%
$-algebra $\left(  \Hom \left(  C,B\right)  ,\star\right)  $.
Let us denote this $\kk$-algebra homomorphism by $\Phi$.

Now, let $u\in\kk\left[  \left[  T\right]  \right]  $ and
$f\in\mathfrak{n}\left(  C,A\right)  $. We must show that%
\begin{equation}
s\circ f\in\mathfrak{n}\left(  C,B\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ u^{\star}\left(  s\circ f\right)  =s\circ\left(  u^{\star
}\left(  f\right)  \right)  . \label{pf.prop.convolution-series.i.goal}%
\end{equation}


Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $).

The definition of $\Phi$ yields $\Phi\left(  f\right)  =s\circ f$. Each
$n\in \NN$ satisfies%
\begin{align}
s\circ f^{\star n}  &  =\Phi\left(  f^{\star n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\left(  f^{\star n}\right)
=s\circ f^{\star n}\text{ (by the definition of }\Phi\text{)}\right)
\nonumber\\
&  =\left(  \underbrace{\Phi\left(  f\right)  }_{=s\circ f}\right)  ^{\star
n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\text{ is a }\kk%
\text{-algebra homomorphism}\right) \nonumber\\
&  =\left(  s\circ f\right)  ^{\star n}.
\label{pf.prop.convolution-series.i.pow-pres}%
\end{align}
Thus, each $n\in \NN$ satisfies%
\begin{equation}
s\left(  f^{\star n}\left(  x\right)  \right)  =\underbrace{\left(  s\circ
f^{\star n}\right)  }_{\substack{=\left(  s\circ f\right)  ^{\star
n}\\\text{(by (\ref{pf.prop.convolution-series.i.pow-pres}))}}}\left(
x\right)  =\left(  s\circ f\right)  ^{\star n}\left(  x\right)  .
\label{pf.prop.convolution-series.i.pow-pres2}%
\end{equation}


We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $.
Thus, the family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported. Hence, Proposition~\ref{prop.pw-fin-supp.P4} (applied to
$Q=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)
_{n\in\NN}$ and $\left(  \lambda_{q}\right)  _{q\in Q}=\left(
u_{n}\right)  _{n\in\NN}$) shows that the family $\left(  u_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported. In other words, for
each $x\in C$, the family $\left(  \left(  u_{n}f^{\star n}\right)  \left(
x\right)  \right)  _{n\in\NN}\in A^{\NN}$ is finitely supported.
In other words, for each $x\in C$,%
\begin{equation}
\text{the family }\left(  u_{n}f^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}\text{ is finitely supported}
\label{pf.prop.convolution-series.i.fs1}%
\end{equation}
(since each $n\in \NN$ satisfies $\left(  u_{n}f^{\star n}\right)
\left(  x\right)  =u_{n}f^{\star n}\left(  x\right)  $).

Also, recall that the family $\left(  f^{\star n}\right)  _{n\in\NN}$
is pointwise finitely supported. In other words, for each $x\in C$,
\begin{equation}
\text{the family }\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}\text{ is finitely supported.}
\label{pf.prop.convolution-series.i.fs2}%
\end{equation}

Let $x \in C$. Then,
(\ref{pf.prop.convolution-series.i.fs2}) shows that the family
$\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}$ is finitely supported.
Hence, Fact H.7 (applied to $V=A$, $W=B$, $\varphi=s$, $Q=\NN$ and
$\left(  v_{q}\right)  _{q\in Q}=\left(  f^{\star n}\left(  x\right)  \right)
_{n\in\NN}$) shows that the family $\left(  s\left(  f^{\star n}\left(
x\right)  \right)  \right)  _{n\in\NN}\in B^{\NN}$ is also
finitely supported, and that it satisfies $\sum_{n\in\NN}s\left(
f^{\star n}\left(  x\right)  \right)  =s\left(  \sum_{n\in\NN}f^{\star
n}\left(  x\right)  \right)  $. Thus, in particular, the family $\left(
s\left(  f^{\star n}\left(  x\right)  \right)  \right)  _{n\in\NN}\in
B^{\NN}$ is finitely supported. Since each $n\in \NN$ satisfies
(\ref{pf.prop.convolution-series.i.pow-pres2}), this rewrites as follows:%
\begin{equation}
\text{The family }\left(  \left(  s\circ f\right)  ^{\star n}\left(  x\right)
\right)  _{n\in\NN}\in B^{\NN}\text{ is finitely supported.}
\label{pf.prop.convolution-series.i.7}%
\end{equation}

From (\ref{pf.prop.convolution-series.i.fs1}), we know that the
family $\left(  u_{n}f^{\star n}\left(  x\right)  \right)  _{n\in\NN}
\in A^{\NN}$ is finitely supported. Hence, Fact H.7 (applied to $V=A$,
$W=B$, $\varphi=s$, $Q=\NN$ and $\left(  v_{q}\right)  _{q\in
Q}=\left(  u_{n}f^{\star n}\left(  x\right)  \right)  _{n\in\NN}$)
shows that the family $\left(  s\left(  u_{n}f^{\star n}\left(  x\right)
\right)  \right)  _{n\in\NN}\in B^{\NN}$ is also finitely
supported, and that it satisfies
\begin{equation}
\sum_{n\in\NN}s\left(  u_{n}f^{\star n}\left(  x\right)  \right)
=s\left(  \sum_{n\in\NN}u_{n}f^{\star n}\left(  x\right)  \right)  .
\label{pf.prop.convolution-series.i.5}%
\end{equation}
From (\ref{pf.prop.convolution-series.i.5}), we obtain%
\begin{align}
s\left(  \sum_{n\in\NN}u_{n}f^{\star n}\left(  x\right)  \right)   &
=\sum_{n\in\NN}\underbrace{s\left(  u_{n}f^{\star n}\left(  x\right)
\right)  }_{\substack{=u_{n}s\left(  f^{\star n}\left(  x\right)  \right)
\\\text{(since the map }s\text{ is }\kk\text{-linear)}}}=
\sum_{n\in\NN}u_{n}\underbrace{s\left(  f^{\star n}\left(  x\right)
\right)  }_{\substack{=\left(  s\circ f\right)  ^{\star n}\left(  x\right)
\\\text{(by (\ref{pf.prop.convolution-series.i.pow-pres2}))}}}\nonumber\\
&  =\sum_{n\in\NN}u_{n}\left(  s\circ f\right)  ^{\star n}\left(
x\right)  . \label{pf.prop.convolution-series.i.6}%
\end{align}


Now, forget that we fixed $x$. We thus have proven that each $x\in C$
satisfies (\ref{pf.prop.convolution-series.i.6}) and
(\ref{pf.prop.convolution-series.i.7}).

In particular, each $x\in C$ satisfies (\ref{pf.prop.convolution-series.i.7}).
In other words, for each $x\in C$, the family $\left(  \left(  s\circ
f\right)  ^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
B^{\NN}$ is finitely supported. In other words, the family $\left(
\left(  s\circ f\right)  ^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,B\right)  \right)  ^{\NN}$ is pointwise
finitely supported. In other words, the map $s\circ f:C\rightarrow B$ is
pointwise $\star$-nilpotent. In other words, $s\circ f\in\mathfrak{n}\left(
C,B\right)  $. Hence, $u^{\star}\left(  s\circ f\right)  $ is well-defined.

Recall that $u=\sum_{n\geq0}u_{n}T^{n}$. Thus,%
\[
u^{\star}\left(  s\circ f\right)  =\underbrace{\sum_{n\geq0}}_{=
\sum_{n\in\NN}}u_{n}\left(  s\circ f\right)  ^{\star n}=\sum_{n\in
\NN}u_{n}\left(  s\circ f\right)  ^{\star n}.
\]
Hence, each $x\in C$ satisfies%
\begin{align*}
\left(  u^{\star}\left(  s\circ f\right)  \right)  \left(  x\right)   &
=\left(  \sum_{n\in\NN}u_{n}\left(  s\circ f\right)  ^{\star n}\right)
\left(  x\right)  =\sum_{n\in\NN}u_{n}\left(  s\circ f\right)  ^{\star
n}\left(  x\right)  =s\left(  \sum_{n\in\NN}u_{n}f^{\star n}\left(
x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.convolution-series.i.6})}\right) \\
&  =\left(  s\circ\left(  u^{\star}\left(  f\right)  \right)  \right)  \left(
x\right)
\end{align*}
(since
\begin{align*}
\left(  s\circ\left(  u^{\star}\left(  f\right)  \right)  \right)  \left(
x\right)   &  =s\left(  \underbrace{\left(  u^{\star}\left(  f\right)
\right)  }_{=\sum_{n\geq0}u_{n}f^{\star n}}\left(  x\right)  \right)
=s\left(  \underbrace{\left(  \sum_{n\geq0}u_{n}f^{\star n}\right)  \left(
x\right)  }_{=\sum_{n\geq0}u_{n}f^{\star n}\left(  x\right)  }\right) \\
&  =s\left(  \underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}u_{n}f^{\star
n}\left(  x\right)  \right)  =s\left(  \sum_{n\in\NN}u_{n}f^{\star
n}\left(  x\right)  \right)
\end{align*}
). In other words, we have $u^{\star}\left(  s\circ f\right)  =s\circ\left(
u^{\star}\left(  f\right)  \right)  $.

We have now proven that $s\circ f\in\mathfrak{n}\left(  C,B\right)  $ and
$u^{\star}\left(  s\circ f\right)  =s\circ\left(  u^{\star}\left(  f\right)
\right)  $. This proves (\ref{pf.prop.convolution-series.i.goal}). Hence,
Proposition~\ref{prop.convolution-series}(i) is proven.

(j) Let $C$ be a connected graded $\kk$-bialgebra. Let $F:C\rightarrow
A$ be a $\kk$-algebra homomorphism. We must prove that $F-u_{A}%
\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $.

We have $F\left(  1_{C}\right)  =1_{A}$ (since $F$ is a $\kk$-algebra
homomorphism). But the axioms of a $\kk$-bialgebra yield $\epsilon
_{C}\left(  1_{C}\right)  =1$.

The definition of the map $u_{A}$ yields $u_{A}\left(  1\right)  =1\cdot
1_{A}=1_{A}$. Now,%
\[
\left(  F-u_{A}\epsilon_{C}\right)  \left(  1_{C}\right)
=\underbrace{F\left(  1_{C}\right)  }_{=1_{A}}-\underbrace{\left(
u_{A}\epsilon_{C}\right)  \left(  1_{C}\right)  }_{=u_{A}\left(  \epsilon
_{C}\left(  1_{C}\right)  \right)  }=1_{A}-u_{A}\left(  \underbrace{\epsilon
_{C}\left(  1_{C}\right)  }_{=1}\right)  =1_{A}-\underbrace{u_{A}\left(
1\right)  }_{=1_{A}}=1_{A}-1_{A}=0.
\]


But Exercise~\ref{graded-connected-exercise}(c) (applied to $C$ instead of
$A$) shows that $C_{0} =\kk\cdot1_{C}$. Applying the map $F-u_{A}%
\epsilon_{C}$ to both sides of this relation, we obtain%
\begin{align*}
\left(  F-u_{A}\epsilon_{C}\right)  \left(  C_{0}\right)   &  =\left(
F-u_{A}\epsilon_{C}\right)  \left(  \kk\cdot1_{C}\right)
=\kk\cdot\underbrace{\left(  F-u_{A}\epsilon_{C}\right)  \left(
1_{C}\right)  }_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
F-u_{A}\epsilon_{C}\text{ is }\kk\text{-linear}\right) \\
&  =\kk\cdot0=0.
\end{align*}
Hence, Proposition~\ref{prop.convolution-series}(h) (applied to $f=F-u_{A}%
\epsilon_{C}$) shows that $F-u_{A}\epsilon_{C}\in\mathfrak{n}\left(
C,A\right)  $. Thus, Proposition~\ref{prop.convolution-series}(j) is proven.
\end{proof}

Thus, Proposition~\ref{prop.convolution-series} is proven, so that
Exercise~\ref{exe.convolution-series} is solved.

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.convolution-series}}

\textit{Solution to Exercise~\ref{exe.convolution-series}.} Before we start
proving Proposition~\ref{prop.convolution-series}, let us prove some facts
which will be useful on several occasions:

\begin{statement}
\textit{Fact B.1:} If $f\in\Hom \left(  C,A\right)  $ is a
pointwise $\star$-nilpotent map, and if $\left(  \lambda_{n}\right)
_{n\in\NN}\in\kk^{\NN}$ is any family of scalars, then
the family $\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported.
\end{statement}

[\textit{Proof of Fact B.1:} Fact B.1 has been stated in
Definition~\ref{def.leray.pwnilp}(b); it was already proven in a footnote.]

\begin{statement}
\textit{Fact B.2:} Let $f\in\mathfrak{n}\left(  C,A\right)  $. Let $\left(
\lambda_{n}\right)  _{n\in\NN}\in\kk^{\NN}$ be any family
of scalars. Then, the family $\left(  \lambda_{n}f^{\star n}\right)
_{n\in\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN}$ is pointwise finitely supported, and its sum $\sum_{n\geq
0}\lambda_{n}f^{\star n}$ belongs to $\Hom \left(  C,A\right)  $.
\end{statement}

[\textit{Proof of Fact B.2:} We have $f\in\mathfrak{n}\left(  C,A\right)  $.
In other words, $f$ is a pointwise $\star$-nilpotent map in
$\Hom \left(  C,A\right)  $ (since $\mathfrak{n}\left(
C,A\right)  $ is the set of all pointwise $\star$-nilpotent maps in
$\Hom \left(  C,A\right)  $). Thus, Fact B.1 shows that the
family $\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported. Hence, the sum $\sum_{n\in\NN}\lambda_{n}f^{\star
n}$ is well-defined. In other words, the sum $\sum_{n\geq0}\lambda_{n}f^{\star
n}$ is well-defined (since $\sum_{n\geq0}=\sum_{n\in\NN}$).

Now, Proposition~\ref{prop.pw-fin-supp.P1} (applied to $\NN$ and
$\left(  \lambda_{n}f^{\star n}\right)  _{n\in\NN}$ instead of $Q$ and
$\left(  f_{q}\right)  _{q\in Q}$) shows that the map $\sum_{n\in\NN}
\lambda_{n}f^{\star n}$ belongs to $\Hom \left(  C,A\right)  $.
In other words, the map $\sum_{n\geq0}\lambda_{n}f^{\star n}$ belongs to
$\Hom \left(  C,A\right)  $ (since $\sum_{n\geq0}=\sum_{n\in\NN}$).
Thus, Fact B.2 is proven.]

\begin{statement}
\textit{Fact B.3:} Let $\left(  f_{q}\right)  _{q\in Q}$ be a pointwise
finitely supported family in $\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$. Let $\lambda\in\kk$. Then, the family $\left(  \lambda
f_{q}\right)  _{q\in Q}\in\left(  \Hom \left(  C,A\right)
\right)  ^{Q}$ is also pointwise finitely supported, and satisfies%
\[
\lambda\sum_{q\in Q}f_{q}=\sum_{q\in Q}\lambda f_{q}.
\]

\end{statement}

[\textit{Proof of Fact B.3:} Fact B.3 is similar to
Proposition~\ref{prop.pw-fin-supp.P2}, and its proof is analogous to the proof
of the latter (but simpler).]

Before we state the next fact, let us recall how the convergence of a
(possibly infinite) sum of power series in $\kk\left[  \left[
T\right]  \right]  $ is defined:

\begin{statement}
\textit{Definition:} Let us introduce a notation: If $u\in\kk\left[
\left[  T\right]  \right]  $ is any power series, and if $n\in \NN$,
then $\left[  T^{n}\right]  u$ will mean the coefficient of $T^{n}$ in $u$.
Thus, $u=\sum_{n\geq0}\left(  \left[  T^{n}\right]  u\right)  \cdot T^{n}$ for
each $u\in\kk\left[  \left[  T\right]  \right]  $.

Let $\left(  r_{q}\right)  _{q\in Q}\in\left(  \kk\left[  \left[
T\right]  \right]  \right)  ^{Q}$ be a family of power series in
$\kk\left[  \left[  T\right]  \right]  $. We say that the sum
$\sum_{q\in Q}r_{q}$
\emph{converges}\index{convergence of an infinite sum in $\kk[[T]]$}
in $\kk\left[  \left[  T\right]  \right]  $ if and only if
for each $n\in \NN$,
\[
\text{all but finitely many }q\in Q\text{ satisfy }\left[  T^{n}\right]
r_{q}=0.
\]
\footnote{Of course, what precisely ``all but finitely many
$q\in Q$'' means here -- i.e., which $q$ are excluded --
depends on $n$.} In this case, the sum $\sum_{q\in Q}r_{q}$ is defined to be
the power series in $\kk\left[  \left[  T\right]  \right]  $ whose
coefficients are given by the rule%
\[
\left(  \left[  T^{n}\right]  \left(  \sum_{q\in Q}r_{q}\right)  =\sum_{q\in
Q}\left[  T^{n}\right]  r_{q}\ \ \ \ \ \ \ \ \ \ \text{for all }n\in
\NN\right)  .
\]

\end{statement}

Now, we can state a useful fact that relates this notion of convergence to
manipulations of maps in $\Hom \left(  C,A\right)  $:

\begin{statement}
\textit{Fact B.7:} Let $\left(  r_{q}\right)  _{q\in Q}\in\left(
\kk\left[  \left[  T\right]  \right]  \right)  ^{Q}$ be a family of
power series such that the (possibly infinite) sum $\sum_{q\in Q}r_{q}$
converges in $\kk\left[  \left[  T\right]  \right]  $. Let
$f\in\mathfrak{n}\left(  C,A\right)  $. Then, the family $\left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  _{q\in Q}\in\left(
\Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise finitely
supported and satisfies%
\[
\left(  \sum_{q\in Q}r_{q}\right)  ^{\star}\left(  f\right)  =\sum_{q\in
Q}\left(  r_{q}\right)  ^{\star}\left(  f\right)  .
\]

\end{statement}

[\textit{Proof of Fact B.7:} Define a power series $s\in\kk\left[
\left[  T\right]  \right]  $ by $s=\sum_{q\in Q}r_{q}$. (This is well-defined,
since the sum $\sum_{q\in Q}r_{q}$ converges.)

Let us introduce a notation: If $u\in\kk\left[  \left[  T\right]
\right]  $ is any power series, and if $n\in \NN$, then $\left[
T^{n}\right]  u$ will mean the coefficient of $T^{n}$ in $u$. Thus,
$u=\sum_{n\geq0}\left(  \left[  T^{n}\right]  u\right)  \cdot T^{n}$ for each
$u\in\kk\left[  \left[  T\right]  \right]  $. Applying this to $u=s$,
we find $s=\sum_{n\geq0}\left(  \left[  T^{n}\right]  s\right)  \cdot T^{n}$.

We know that the sum $\sum_{q\in Q}r_{q}$ converges in $\kk\left[
\left[  T\right]  \right]  $. In other words, for each $n\in \NN$,
\begin{equation}
\text{all but finitely many }q\in Q\text{ satisfy }\left[  T^{n}\right]
r_{q}=0 \label{sol.convolution-series.b.7.pf.allbut}%
\end{equation}
(by the definition of convergence for an infinite sum in $\kk\left[
\left[  T\right]  \right]  $). (Of course, what precisely
``all but finitely many $q\in Q$'' means here -- i.e., which
$q$ are excluded -- depends on $n$.)

For every $q\in Q$, we have $r_{q}=\sum_{n\geq0}\left(  \left[  T^{n}\right]
r_{q}\right)  \cdot T^{n}$ (since $u=\sum_{n\geq0}\left(  \left[
T^{n}\right]  u\right)  \cdot T^{n}$ for each $u\in\kk\left[  \left[
T\right]  \right]  $) and therefore%
\begin{equation}
\left(  r_{q}\right)  ^{\star}\left(  f\right)  =\sum_{n\geq0}\left(  \left[
T^{n}\right]  r_{q}\right)  f^{\star n}
\label{sol.convolution-series.b.7.pf.rq*f}%
\end{equation}
(by the definition of $\left(  r_{q}\right)  ^{\star}\left(  f\right)  $).

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported (since $f$ is pointwise $\star$-nilpotent). In other words,
for each $x\in C$,%
\begin{equation}
\text{the family }\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}\text{ is finitely supported.}
\label{sol.convolution-series.b.7.pf.fs}%
\end{equation}


Now, let $x\in C$. Then, there exists an $N\in \NN$ such that%
\begin{equation}
\text{every }n\geq N\text{ satisfies }f^{\star n}\left(  x\right)  =0
\label{sol.convolution-series.b.7.pf.0}%
\end{equation}
\footnote{\textit{Proof.} The family $\left(  f^{\star n}\left(  x\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported (by
(\ref{sol.convolution-series.b.7.pf.fs})). In other words, all but finitely
many $n\in \NN$ satisfy $f^{\star n}\left(  x\right)  =0$. In other
words, there exists a finite subset $Z$ of $\NN$ such that
\begin{equation}
\text{each }n\in \NN \setminus Z\text{ satisfies }f^{\star n}\left(
x\right)  =0. \label{sol.convolution-series.b.7.pf.0.pf.1}%
\end{equation}
Consider this $Z$.
\par
The set $Z$ is a finite subset of $\NN$, and thus has an upper bound
(since any finite subset of $\NN$ has an upper bound). In other words,
there exists some $w\in \NN$ such that each $z\in Z$ satisfies $z\leq
w$. Consider this $w$.
\par
Now, let $n\in \NN$ be such that $n\geq w+1$. Assume (for the sake of
contradiction) that $n\in Z$. Recall that each $z\in Z$ satisfies $z\leq w$.
Applying this to $z=n$, we obtain $n\leq w$ (since $n\in Z$), so that $n\leq
w<w+1$. This contradicts $n\geq w+1$. This contradiction shows that our
assumption (that $n\in Z$) was wrong. Hence, we cannot have $n\in Z$. In other
words, we have $n\notin Z$. Combining $n\in \NN$ with $n\notin Z$, we
obtain $n\in \NN \setminus Z$. Hence,
(\ref{sol.convolution-series.b.7.pf.0.pf.1}) yields $f^{\star n}\left(
x\right)  =0$.
\par
Now, forget that we fixed $n$. We thus have shown that every $n\geq w+1$
satisfies $f^{\star n}\left(  x\right)  =0$. Hence, there exists an
$N\in \NN$ such that every $n\geq N$ satisfies $f^{\star n}\left(
x\right)  =0$ (namely, $N=w+1$).}. Consider this $N$.

Every $q\in Q$ satisfies%
\begin{align}
\underbrace{\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)
}_{\substack{=\sum_{n\geq0}\left(  \left[  T^{n}\right]  r_{q}\right)
f^{\star n}\\\text{(by (\ref{sol.convolution-series.b.7.pf.rq*f}))}}}\left(
x\right)   &  =\left(  \sum_{n\geq0}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\right)  \left(  x\right)  =\sum_{n\geq0}\left(
\left[  T^{n}\right]  r_{q}\right)  f^{\star n}\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{n\geq0;\\n<N}}}_{=\sum_{n=0}^{N-1}}\left(
\left[  T^{n}\right]  r_{q}\right)  f^{\star n}\left(  x\right)
+\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(  \left[  T^{n}\right]  r_{q}\right)
\underbrace{f^{\star n}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.0}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }n\in \NN \text{
satisfies either }n<N\text{ or }n\geq N\text{ (but not both)}\right)
\nonumber\\
&  =\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]  r_{q}\right)  f^{\star
n}\left(  x\right)  +\underbrace{\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(
\left[  T^{n}\right]  r_{q}\right)  0}_{=0}\nonumber\\
&  =\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]  r_{q}\right)  f^{\star
n}\left(  x\right)  . \label{sol.convolution-series.b.7.pf.rq*fx}%
\end{align}


For each $n\in \NN$, there exists a finite subset $Q_{n}$ of $Q$ such
that
\begin{equation}
\text{all }q\in Q\setminus Q_{n}\text{ satisfy }\left[  T^{n}\right]  r_{q}=0
\label{sol.convolution-series.b.7.pf.1}%
\end{equation}
\footnote{\textit{Proof:} Let $n\in \NN$. Then, all but finitely many
$q\in Q$ satisfy $\left[  T^{n}\right]  r_{q}=0$ (by
(\ref{sol.convolution-series.b.7.pf.allbut})). In other words, there exists a
finite subset $Q_{n}$ of $Q$ such that all $q\in Q\setminus Q_{n}$ satisfy
$\left[  T^{n}\right]  r_{q}=0$. Qed.}. Consider this $Q_{n}$.

Let $Q^{\prime}$ be the subset $Q_{0}\cup Q_{1}\cup\cdots\cup Q_{N-1}$ of $Q$.
Then, $Q^{\prime}$ is the union of the finite sets $Q_{0},Q_{1},\ldots
,Q_{N-1}$. Hence, $Q^{\prime}$ is a union of finitely many finite sets.
Therefore, $Q^{\prime}$ itself is a finite set (since a union of finitely many
finite sets is always a finite set). Thus, all but finitely many $q\in Q$
satisfy $q\in Q\setminus Q^{\prime}$.

Moreover, if $n\in\left\{  0,1,\ldots,N-1\right\}  $, then
\begin{equation}
\text{every }q\in Q\setminus Q^{\prime}\text{ satisfies }\left[  T^{n}\right]
r_{q}=0 \label{sol.convolution-series.b.7.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.b.7.pf.2}):} Let
$n\in\left\{  0,1,\ldots,N-1\right\}  $. Let $q\in Q\setminus Q^{\prime}$.
\par
From $n\in\left\{  0,1,\ldots,N-1\right\}  $, we conclude that $Q_{n}$ is a
term in the union $Q_{0}\cup Q_{1}\cup\cdots\cup Q_{N-1}$. Thus, $Q_{n}\subset
Q_{0}\cup Q_{1}\cup\cdots\cup Q_{N-1}=Q^{\prime}$ (since $Q^{\prime}=Q_{0}\cup
Q_{1}\cup\cdots\cup Q_{N-1}$ (by the definition of $Q^{\prime}$)). Hence,
$Q\setminus\underbrace{Q_{n}}_{\subset Q^{\prime}}\supset Q\setminus
Q^{\prime}$, so that $Q\setminus Q^{\prime}\subset Q\setminus Q_{n}$. Hence,
$q\in Q\setminus Q^{\prime}\subset Q\setminus Q_{n}$. Therefore,
(\ref{sol.convolution-series.b.7.pf.1}) yields $\left[  T^{n}\right]  r_{q}%
=0$. This proves (\ref{sol.convolution-series.b.7.pf.2}).}. Hence, every
$n\in\left\{  0,1,\ldots,N-1\right\}  $ satisfies%
\begin{equation}
\left[  T^{n}\right]  s=\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}
\label{sol.convolution-series.b.7.pf.2cor}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.b.7.pf.2cor}):} Let
$n\in\left\{  0,1,\ldots,N-1\right\}  $. Then,%
\begin{align*}
\left[  T^{n}\right]  \underbrace{s}_{=\sum_{q\in Q}r_{q}}  &  =\left[
T^{n}\right]  \left(  \sum_{q\in Q}r_{q}\right)  =\sum_{q\in Q}\left[
T^{n}\right]  r_{q}=\underbrace{\sum\limits_{\substack{q\in Q;\\q\in Q^{\prime}}%
}}_{\substack{=\sum_{q\in Q^{\prime}}\\\text{(since }Q^{\prime}\subset
Q\text{)}}}\left[  T^{n}\right]  r_{q}+\underbrace{\sum\limits_{\substack{q\in
Q;\\q\notin Q^{\prime}}}}_{=\sum_{q\in Q\setminus Q^{\prime}}}\left[
T^{n}\right]  r_{q}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }q\in Q\text{ satisfies either }q\in Q^{\prime}\text{ or
}q\notin Q^{\prime}\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}+\sum_{q\in Q\setminus
Q^{\prime}}\underbrace{\left[  T^{n}\right]  r_{q}}_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.2}))}}}=\sum_{q\in Q^{\prime}}\left[
T^{n}\right]  r_{q}+\underbrace{\sum_{q\in Q\setminus Q^{\prime}}0}_{=0}%
=\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}.
\end{align*}
This proves (\ref{sol.convolution-series.b.7.pf.2cor}).}.

Now,%
\begin{equation}
\text{each }q\in Q\setminus Q^{\prime}\text{ satisfies }\left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(  x\right)  =0
\label{sol.convolution-series.b.7.pf.rq*fx=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.b.7.pf.rq*fx=0}):} Let
$q\in Q\setminus Q^{\prime}$. Thus, $q\in Q\setminus Q^{\prime}\subset Q$.
Hence, (\ref{sol.convolution-series.b.7.pf.rq*fx}) yields%
\[
\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(
x\right)  =\sum_{n=0}^{N-1}\underbrace{\left(  \left[  T^{n}\right]
r_{q}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.2}))}}}f^{\star n}\left(  x\right)
=\sum_{n=0}^{N-1}0f^{\star n}\left(  x\right)  =0.
\]
This proves (\ref{sol.convolution-series.b.7.pf.rq*fx=0}).}. Hence, all but
finitely many $q\in Q$ satisfy $\left(  \left(  r_{q}\right)  ^{\star}\left(
f\right)  \right)  \left(  x\right)  =0$ (since all but finitely many $q\in Q$
satisfy $q\in Q\setminus Q^{\prime}$). In other words,
\begin{equation}
\text{the family }\left(  \left(  \left(  r_{q}\right)  ^{\star}\left(
f\right)  \right)  \left(  x\right)  \right)  _{q\in Q}\in A^{Q}\text{ is
finitely supported.} \label{sol.convolution-series.b.7.pf.cl1}%
\end{equation}
Furthermore,%
\begin{align}
\sum_{q\in Q}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)
\left(  x\right)   &  =\underbrace{\sum\limits_{\substack{q\in Q;\\q\in Q^{\prime}}%
}}_{\substack{=\sum_{q\in Q^{\prime}}\\\text{(since }Q^{\prime}\subset
Q\text{)}}}\underbrace{\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)
\right)  \left(  x\right)  }_{\substack{=\sum_{n=0}^{N-1}\left(  \left[
T^{n}\right]  r_{q}\right)  f^{\star n}\left(  x\right)  \\\text{(by
(\ref{sol.convolution-series.b.7.pf.rq*fx}))}}}+\underbrace{
\sum\limits_{\substack{q\in Q;\\q\notin Q^{\prime}}}}_{=
\sum_{q\in Q\setminus Q^{\prime}%
}}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(
x\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }q\in Q\text{ satisfies either }q\in Q^{\prime}\text{ or
}q\notin Q^{\prime}\\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum_{q\in Q^{\prime}}\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\left(  x\right)  +\sum_{q\in Q\setminus Q^{\prime}%
}\underbrace{\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)
\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.rq*fx=0}))}}}\nonumber\\
&  =\sum_{q\in Q^{\prime}}\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\left(  x\right)  +\underbrace{\sum_{q\in Q\setminus
Q^{\prime}}0}_{=0}=\underbrace{\sum_{q\in Q^{\prime}}\sum_{n=0}^{N-1}%
}_{\substack{=\sum_{n=0}^{N-1}\sum_{q\in Q^{\prime}}\\\text{(here, we are
interchanging}\\\text{two \textbf{finite} sums)}}}\left(  \left[
T^{n}\right]  r_{q}\right)  f^{\star n}\left(  x\right) \nonumber\\
&  =\sum_{n=0}^{N-1}\sum_{q\in Q^{\prime}}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\left(  x\right)  .
\label{sol.convolution-series.b.7.pf.q1}%
\end{align}


On the other hand, recall that $s=\sum_{n\geq0}\left(  \left[  T^{n}\right]
s\right)  \cdot T^{n}$. Therefore,%
\[
s^{\star}\left(  f\right)  =\sum_{n\geq0}\left(  \left[  T^{n}\right]
s\right)  f^{\star n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}s^{\star}\left(  f\right)  \right)  .
\]
Applying both sides of this equality to $x$, we obtain%
\begin{align}
\left(  s^{\star}\left(  f\right)  \right)  \left(  x\right)   &  =\left(
\sum_{n\geq0}\left(  \left[  T^{n}\right]  s\right)  f^{\star n}\right)
\left(  x\right)  =\sum_{n\geq0}\left(  \left[  T^{n}\right]  s\right)
f^{\star n}\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{n\geq0;\\n<N}}}_{=\sum_{n=0}^{N-1}}\left(
\left[  T^{n}\right]  s\right)  f^{\star n}\left(  x\right)  +
\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(  \left[  T^{n}\right]  s\right)
\underbrace{f^{\star n}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.b.7.pf.0}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }n\in \NN \text{
satisfies either }n<N\text{ or }n\geq N\text{ (but not both)}\right)
\nonumber\\
&  =\sum_{n=0}^{N-1}\left(  \left[  T^{n}\right]  s\right)  f^{\star n}\left(
x\right)  +\underbrace{\sum\limits_{\substack{n\geq0;\\n\geq N}}\left(  \left[
T^{n}\right]  s\right)  0}_{=0}\nonumber\\
&  =\sum_{n=0}^{N-1}\underbrace{\left(  \left[  T^{n}\right]  s\right)
}_{\substack{=\sum_{q\in Q^{\prime}}\left[  T^{n}\right]  r_{q}\\\text{(by
(\ref{sol.convolution-series.b.7.pf.2cor}))}}}f^{\star n}\left(  x\right)
=\sum_{n=0}^{N-1}\left(  \sum_{q\in Q^{\prime}}\left(  \left[  T^{n}\right]
r_{q}\right)  \right)  f^{\star n}\left(  x\right) \nonumber\\
&  =\sum_{n=0}^{N-1}\sum_{q\in Q^{\prime}}\left(  \left[  T^{n}\right]
r_{q}\right)  f^{\star n}\left(  x\right) \nonumber\\
&  =\sum_{q\in Q}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)
\right)  \left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.convolution-series.b.7.pf.q1})}\right)  .
\label{sol.convolution-series.b.7.pf.cl2}%
\end{align}


Now, forget that we fixed $x$. We thus have shown that each $x\in C$ satisfies
(\ref{sol.convolution-series.b.7.pf.cl1}) and
(\ref{sol.convolution-series.b.7.pf.cl2}).

In particular, for each $x\in C$, the family $\left(  \left(  \left(
r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(  x\right)  \right)
_{q\in Q}\in A^{Q}$ is finitely supported (since each $x\in C$ satisfies
(\ref{sol.convolution-series.b.7.pf.cl1})). In other words, the family
$\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  _{q\in Q}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{Q}$ is pointwise
finitely supported (by the definition of ``pointwise finitely
supported''). Hence, the sum $\sum_{q\in Q}\left(
r_{q}\right)  ^{\star}\left(  f\right)  $ is well-defined.

Furthermore, each $x\in C$ satisfies%
\begin{align*}
\left(  s^{\star}\left(  f\right)  \right)  \left(  x\right)   &  =\sum_{q\in
Q}\left(  \left(  r_{q}\right)  ^{\star}\left(  f\right)  \right)  \left(
x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.convolution-series.b.7.pf.cl2})}\right) \\
&  =\left(  \sum_{q\in Q}\left(  r_{q}\right)  ^{\star}\left(  f\right)
\right)  \left(  x\right)  .
\end{align*}
In other words, we have $s^{\star}\left(  f\right)  =\sum_{q\in Q}\left(
r_{q}\right)  ^{\star}\left(  f\right)  $. Since $s=\sum_{q\in Q}r_{q}$, this
rewrites as $\left(  \sum_{q\in Q}r_{q}\right)  ^{\star}\left(  f\right)
=\sum_{q\in Q}\left(  r_{q}\right)  ^{\star}\left(  f\right)  $. This
completes the proof of Fact B.7.]

\begin{proof}
[Proof of Proposition~\ref{prop.convolution-series}.](a) Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $k\in \NN$. The power series
$T^{k}\in\kk\left[  \left[  T\right]  \right]  $ can be written in the
form $T^{k}=\sum_{n\geq0}\delta_{n,k}T^{n}$\ \ \ \ \footnote{since
\begin{align*}
\sum_{n\geq0}\delta_{n,k}T^{n}  &  =\underbrace{\delta_{k,k}}%
_{\substack{=1\\\text{(since }k=k\text{)}}}T^{k}+\sum\limits_{\substack{n\geq
0;\\n\neq k}}\underbrace{\delta_{n,k}}_{\substack{=0\\\text{(since }n\neq
k\text{)}}}T^{n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend}\\
\text{for }n=k\text{ from the sum}%
\end{array}
\right) \\
&  =T^{k}+\underbrace{\sum\limits_{\substack{n\geq0;\\n\neq k}}0T^{n}}_{=0}=T^{k}%
\end{align*}
}. Hence, the definition of $\left(  T^{k}\right)  ^{\star}\left(  f\right)  $
yields%
\begin{align*}
\left(  T^{k}\right)  ^{\star}\left(  f\right)   &  =\sum_{n\geq0}\delta
_{n,k}f^{\star n}\\
&  =\underbrace{\delta_{k,k}}_{\substack{=1\\\text{(since }k=k\text{)}%
}}f^{\star k}+\sum\limits_{\substack{n\geq0;\\n\neq k}}\underbrace{\delta_{n,k}%
}_{\substack{=0\\\text{(since }n\neq k\text{)}}}f^{\star n}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend}\\
\text{for }n=k\text{ from the sum}%
\end{array}
\right) \\
&  =f^{\star k}+\underbrace{\sum\limits_{\substack{n\geq0;\\n\neq k}}0f^{\star n}%
}_{=0}=f^{\star k}.
\end{align*}
This proves Proposition~\ref{prop.convolution-series}(a).

(b) We are going to prove the formulas (\ref{eq.exe.convolution-series.b.1}),
(\ref{eq.exe.convolution-series.b.3}), (\ref{eq.exe.convolution-series.b.2}),
(\ref{eq.exe.convolution-series.b.4}), and
(\ref{eq.exe.convolution-series.b.5}) in this order.

[\textit{Proof of (\ref{eq.exe.convolution-series.b.1}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $u,v\in\kk\left[  \left[
T\right]  \right]  $. We must prove the equality
(\ref{eq.exe.convolution-series.b.1}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $).

Write the power series $v$ in the form $v=\sum_{n\geq0}v_{n}T^{n}$ with
$\left(  v_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $v^{\star
}\left(  f\right)  =\sum_{n\geq0}v_{n}f^{\star n}$ (by the definition of
$v^{\star}\left(  f\right)  $).

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
u_{n}\right)  _{n\in\NN}$) shows that the family $\left(  u_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}u_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $.

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
v_{n}\right)  _{n\in\NN}$) shows that the family $\left(  v_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}v_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $.

Proposition~\ref{prop.pw-fin-supp.P2} (applied to $Q=\NN$, $\left(
f_{q}\right)  _{q\in Q}=\left(  u_{n}f^{\star n}\right)  _{n\in\NN}$
and $\left(  g_{q}\right)  _{q\in Q}=\left(  v_{n}f^{\star n}\right)
_{n\in\NN}$) now shows that the family $\left(  u_{n}f^{\star n}%
+v_{n}f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is also pointwise finitely
supported, and satisfies%
\[
\sum_{n\in\NN}u_{n}f^{\star n}+\sum_{n\in\NN}v_{n}f^{\star
n}=\sum_{n\in\NN}\left(  u_{n}f^{\star n}+v_{n}f^{\star n}\right)  .
\]
Thus,%
\[
\sum_{n\in\NN}u_{n}f^{\star n}+\sum_{n\in\NN}v_{n}f^{\star
n}=\sum_{n\in\NN}\underbrace{\left(  u_{n}f^{\star n}+v_{n}f^{\star
n}\right)  }_{=\left(  u_{n}+v_{n}\right)  f^{\star n}}
=\sum_{n\in\NN}\left(  u_{n}+v_{n}\right)  f^{\star n}.
\]
This rewrites as%
\begin{equation}
\sum_{n\geq0}u_{n}f^{\star n}+\sum_{n\geq0}v_{n}f^{\star n}=\sum_{n\geq
0}\left(  u_{n}+v_{n}\right)  f^{\star n}
\label{sol.convolution-series.b.1.pf.2}%
\end{equation}
(since $\sum_{n\geq0}=\sum_{n\in\NN}$).

Adding the equalities $u=\sum_{n\geq0}u_{n}T^{n}$ and $v=\sum_{n\geq0}%
v_{n}T^{n}$, we obtain%
\[
u+v=\sum_{n\geq0}u_{n}T^{n}+\sum_{n\geq0}v_{n}T^{n}=\sum_{n\geq0}\left(
u_{n}+v_{n}\right)  T^{n}.
\]
Hence, the definition of $\left(  u+v\right)  ^{\star}\left(  f\right)  $
yields
\begin{align*}
\left(  u+v\right)  ^{\star}\left(  f\right)   &  =\sum_{n\geq0}\left(
u_{n}+v_{n}\right)  f^{\star n}=\underbrace{\sum_{n\geq0}u_{n}f^{\star n}%
}_{=u^{\star}\left(  f\right)  }+\underbrace{\sum_{n\geq0}v_{n}f^{\star n}%
}_{=v^{\star}\left(  f\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.convolution-series.b.1.pf.2})}\right) \\
&  =u^{\star}\left(  f\right)  +v^{\star}\left(  f\right)  .
\end{align*}
This proves (\ref{eq.exe.convolution-series.b.1}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.3}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $u\in\kk\left[  \left[  T\right]
\right]  $ and $\lambda\in\kk$. We must prove the equality
(\ref{eq.exe.convolution-series.b.3}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $). Multiplying this equality by $\lambda$, we
find%
\begin{equation}
\lambda u^{\star}\left(  f\right)  =\lambda\underbrace{\sum_{n\geq0}}%
_{=\sum_{n\in\NN}}u_{n}f^{\star n}=\lambda\sum_{n\in\NN}%
u_{n}f^{\star n}. \label{sol.convolution-series.b.3.pf.0}%
\end{equation}


But Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}
=\left(  u_{n}\right)  _{n\in\NN}$) shows that the family $\left(
u_{n}f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is pointwise finitely supported,
and that its sum $\sum_{n\geq0}u_{n}f^{\star n}$ belongs to
$\Hom \left(  C,A\right)  $. Hence, Fact B.3 (applied to
$Q=\NN$ and $\left(  f_{q}\right)  _{q\in Q}=\left(  u_{n}f^{\star
n}\right)  _{n\in\NN}$) shows that the family $\left(  \lambda
u_{n}f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is also pointwise finitely
supported, and satisfies%
\[
\lambda\sum_{n\in\NN}u_{n}f^{\star n}=\sum_{n\in\NN}\lambda
u_{n}f^{\star n}.
\]
Thus, (\ref{sol.convolution-series.b.3.pf.0}) becomes%
\begin{equation}
\lambda u^{\star}\left(  f\right)  =\lambda\sum_{n\in\NN}u_{n}f^{\star
n}=\underbrace{\sum_{n\in\NN}}_{=\sum_{n\geq0}}\lambda u_{n}f^{\star
n}=\sum_{n\geq0}\lambda u_{n}f^{\star n}.
\label{sol.convolution-series.b.3.pf.1}%
\end{equation}


Multiplying the equality $u=\sum_{n\geq0}u_{n}T^{n}$ by $\lambda$, we obtain
$\lambda u=\lambda\sum_{n\geq0}u_{n}T^{n}=\sum_{n\geq0}\lambda u_{n}T^{n}$.
Hence, $\left(  \lambda u\right)  ^{\star}\left(  f\right)  =\sum_{n\geq
0}\lambda u_{n}f^{\star n}$ (by the definition of $\left(  \lambda u\right)
^{\star}\left(  f\right)  $). Comparing this with
(\ref{sol.convolution-series.b.3.pf.1}), we obtain $\left(  \lambda u\right)
^{\star}\left(  f\right)  =\lambda u^{\star}\left(  f\right)  $. This proves
(\ref{eq.exe.convolution-series.b.3}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.2}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $ and $u,v\in\kk\left[  \left[
T\right]  \right]  $. We must prove the equality
(\ref{eq.exe.convolution-series.b.2}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $).

Write the power series $v$ in the form $v=\sum_{n\geq0}v_{n}T^{n}$ with
$\left(  v_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $v^{\star
}\left(  f\right)  =\sum_{n\geq0}v_{n}f^{\star n}$ (by the definition of
$v^{\star}\left(  f\right)  $).

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
u_{n}\right)  _{n\in\NN}$) shows that the family $\left(  u_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}u_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $. Renaming the index $n$ as $q$ in this statement, we obtain the
following: The family $\left(  u_{q}f^{\star q}\right)  _{q\in\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported, and its sum $\sum_{q\geq0}u_{q}f^{\star q}$
belongs to $\Hom \left(  C,A\right)  $.

Fact B.2 (applied to $\left(  \lambda_{n}\right)  _{n\in\NN}=\left(
v_{n}\right)  _{n\in\NN}$) shows that the family $\left(  v_{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported, and that its sum
$\sum_{n\geq0}v_{n}f^{\star n}$ belongs to $\Hom \left(
C,A\right)  $. Renaming the index $n$ as $r$ in this statement, we obtain the
following: The family $\left(  v_{r}f^{\star r}\right)  _{r\in\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported, and its sum $\sum_{r\geq0}v_{r}f^{\star r}$
belongs to $\Hom \left(  C,A\right)  $.

Proposition~\ref{prop.pw-fin-supp.P3} (applied to $Q=\NN$,
$R=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  u_{q}f^{\star
q}\right)  _{q\in\NN}$ and $\left(  g_{r}\right)  _{r\in R}=\left(
v_{r}f^{\star r}\right)  _{r\in\NN}$) thus shows that the family
$\left(  \left(  u_{q}f^{\star q}\right)  \star\left(  v_{r}f^{\star
r}\right)  \right)  _{\left(  q,r\right)  \in \NN\times\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN\times\NN}$
is pointwise finitely supported, and satisfies%
\[
\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(  \left(
u_{q}f^{\star q}\right)  \star\left(  v_{r}f^{\star r}\right)  \right)
=\left(  \sum_{q\in\NN}u_{q}f^{\star q}\right)  \star\left(
\sum_{r\in\NN}v_{r}f^{\star r}\right)  .
\]
Hence,%
\begin{align}
&  \left(  \sum_{q\in\NN}u_{q}f^{\star q}\right)  \star\left(
\sum_{r\in\NN}v_{r}f^{\star r}\right) \nonumber\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}%
\underbrace{\left(  \left(  u_{q}f^{\star q}\right)  \star\left(
v_{r}f^{\star r}\right)  \right)  }_{=u_{q}v_{r}f^{\star q}\star f^{\star r}%
}=\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}v_{r}%
\underbrace{f^{\star q}\star f^{\star r}}_{=f^{\star\left(  q+r\right)  }%
}\nonumber\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}v_{r}%
f^{\star\left(  q+r\right)  }. \label{sol.convolution-series.b.2.pf.5}%
\end{align}


Multiplying the equalities
\[
u=\underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}u_{n}T^{n}=
\sum_{n\in\NN}u_{n}T^{n}=\sum_{q\in\NN}u_{q}T^{q}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}n\text{ as }q\right)
\]
and%
\[
v=\underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}v_{n}T^{n}=
\sum_{n\in\NN}v_{n}T^{n}=\sum_{r\in\NN}v_{r}T^{r}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}n\text{ as }r\right)  ,
\]
we obtain%
\begin{align}
uv  &  =\left(  \sum_{q\in\NN}u_{q}T^{q}\right)  \left(
\sum_{r\in\NN}v_{r}T^{r}\right)  =\sum_{q\in\NN}u_{q}T^{q}
\sum_{r\in\NN}v_{r}T^{r}=\underbrace{\sum_{q\in\NN}
\sum_{r\in\NN}}_{=\sum_{\left(  q,r\right)  \in \NN\times\NN}%
}\underbrace{u_{q}T^{q}v_{r}T^{r}}_{=u_{q}v_{r}T^{q}T^{r}}\nonumber\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}%
v_{r}\underbrace{T^{q}T^{r}}_{=T^{q+r}}=\sum_{\left(  q,r\right)
\in \NN\times\NN}u_{q}v_{r}T^{q+r}.
\label{sol.convolution-series.b.2.pf.uv=}%
\end{align}
Thus, in particular, the sum $\sum_{\left(  q,r\right)  \in
\NN\times\NN}u_{q}v_{r}T^{q+r}$ converges in $\kk\left[  \left[
T\right]  \right]  $. Hence, Fact B.7 (applied to $\NN\times\NN$
and $\left(  u_{q}v_{r}T^{q+r}\right)  _{\left(  q,r\right)  \in
\NN\times\NN}$ instead of $Q$ and $\left(  r_{q}\right)  _{q\in
Q}$) shows that the family $\left(  \left(  u_{q}v_{r}T^{q+r}\right)  ^{\star
}\left(  f\right)  \right)  _{\left(  q,r\right)  \in \NN\times
\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN\times\NN}$ is pointwise finitely supported and satisfies%
\begin{equation}
\left(  \sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}%
v_{r}T^{q+r}\right)  ^{\star}\left(  f\right)  =\sum_{\left(  q,r\right)
\in \NN\times\NN}\left(  u_{q}v_{r}T^{q+r}\right)  ^{\star
}\left(  f\right)  . \label{sol.convolution-series.b.2.pf.2}%
\end{equation}
In light of (\ref{sol.convolution-series.b.2.pf.uv=}), the equality
(\ref{sol.convolution-series.b.2.pf.2}) rewrites as%
\begin{align*}
\left(  uv\right)  ^{\star}\left(  f\right)   &  =\sum_{\left(  q,r\right)
\in \NN\times\NN}\underbrace{\left(  u_{q}v_{r}T^{q+r}\right)
^{\star}\left(  f\right)  }_{\substack{=u_{q}v_{r}\left(  T^{q+r}\right)
^{\star}\left(  f\right)  \\\text{(by (\ref{eq.exe.convolution-series.b.3})
(applied}\\\text{to }u_{q}v_{r}\text{ and }T^{q+r}\text{ instead of }%
\lambda\text{ and }u\text{))}}}=\sum_{\left(  q,r\right)  \in \NN%
\times\NN}u_{q}v_{r}\underbrace{\left(  T^{q+r}\right)  ^{\star}\left(
f\right)  }_{\substack{=f^{\star\left(  q+r\right)  }\\\text{(by
(\ref{eq.exe.convolution-series.a.monom}) (applied to }k=q+r\text{))}}}\\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}u_{q}v_{r}%
f^{\star\left(  q+r\right)  }=\left(  \sum_{q\in\NN}u_{q}f^{\star
q}\right)  \star\left(  \sum_{r\in\NN}v_{r}f^{\star r}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.b.2.pf.5}%
)}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\underbrace{u^{\star}\left(  f\right)  }_{=\sum_{n\geq0}u_{n}f^{\star n}}%
\star\underbrace{v^{\star}\left(  f\right)  }_{=\sum_{n\geq0}v_{n}f^{\star
n}}  &  =\left(  \underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}%
u_{n}f^{\star n}\right)  \star\left(  \underbrace{\sum_{n\geq0}}_{=
\sum_{n\in\NN}}v_{n}f^{\star n}\right) \\
&  =\underbrace{\left(  \sum_{n\in\NN}u_{n}f^{\star n}\right)
}_{\substack{=\sum_{q\in\NN}u_{q}f^{\star q}\\\text{(here, we have
renamed}\\\text{the summation index }n\text{ as }q\text{)}}}\star
\underbrace{\left(  \sum_{n\in\NN}v_{n}f^{\star n}\right)
}_{\substack{=\sum_{r\in\NN}v_{r}f^{\star r}\\\text{(here, we have
renamed}\\\text{the summation index }n\text{ as }r\text{)}}}\\
&  =\left(  \sum_{q\in\NN}u_{q}f^{\star q}\right)  \star\left(
\sum_{r\in\NN}v_{r}f^{\star r}\right)  ,
\end{align*}
we obtain $\left(  uv\right)  ^{\star}\left(  f\right)  =u^{\star}\left(
f\right)  \star v^{\star}\left(  f\right)  $. This proves
(\ref{eq.exe.convolution-series.b.2}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.4}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $. Applying
(\ref{eq.exe.convolution-series.b.3}) to $u=0$ and $\lambda=0$, we find
$\left(  0\cdot0\right)  ^{\star}\left(  f\right)  =0\cdot0^{\star}\left(
f\right)  =0$. In other words, $0^{\star}\left(  f\right)  =0$. This proves
(\ref{eq.exe.convolution-series.b.4}).]

[\textit{Proof of (\ref{eq.exe.convolution-series.b.5}):} Let $f\in
\mathfrak{n}\left(  C,A\right)  $. Applying
(\ref{eq.exe.convolution-series.a.monom}) to $k=0$, we find
\[
\left(  T^{0}\right)  ^{\star}\left(  f\right)  =f^{\star0}=\left(  \text{the
unity of the }\kk\text{-algebra }\left(  \Hom \left(
C,A\right)  ,\star\right)  \right)  =u_{A}\epsilon_{C}%
\]
(since the unity of the $\kk$-algebra $\left(  \Hom \left(
C,A\right)  ,\star\right)  $ is $u_{A}\epsilon_{C}$). In view of $T^{0}=1$,
this rewrites as $1^{\star}\left(  f\right)  =u_{A}\epsilon_{C}$. This proves
(\ref{eq.exe.convolution-series.b.5}).]

We have now proven all the equalities (\ref{eq.exe.convolution-series.b.1}),
(\ref{eq.exe.convolution-series.b.2}), (\ref{eq.exe.convolution-series.b.3}),
(\ref{eq.exe.convolution-series.b.4}) and (\ref{eq.exe.convolution-series.b.5}%
). Thus, Proposition~\ref{prop.convolution-series}(b) is proven.

(c) Let $f,g\in\mathfrak{n}\left(  C,A\right)  $ be such that $f\star g=g\star
f$. We must prove that $f+g\in\mathfrak{n}\left(  C,A\right)  $.

From $f\star g=g\star f$, we conclude that the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute. Thus, we can easily see that%
\begin{equation}
\left(  f+g\right)  ^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\star
g^{\star\left(  n-i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in \NN \label{sol.convolution-series.c.binom}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.c.binom}):} Let
$n\in \NN$. Let $\mathfrak{G}$ be the $\kk$-subalgebra of
$\left(  \Hom \left(  C,A\right)  ,\star\right)  $ generated by
the two elements $f$ and $g$. Thus, the $\kk$-algebra $\mathfrak{G}$ is
generated by commuting elements (because the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute), and therefore is commutative (since any
$\kk$-algebra generated by commuting elements must be commutative).
Hence, the binomial formula holds in this $\kk$-algebra $\mathfrak{G}$.
In other words, any $\alpha\in\mathfrak{G}$ and $\beta\in\mathfrak{G}$ and
$m\in \NN$ satisfy%
\[
\left(  \alpha+\beta\right)  ^{\star m}=\sum_{i=0}^{m}\dbinom{m}{i}%
\alpha^{\star i}\star\beta^{\star\left(  m-i\right)  }%
\]
(since the multiplication in the $\kk$-algebra $\mathfrak{G}$ is the
convolution $\star$). Applying this to $\alpha=f$, $\beta=g$ and $m=n$, we
obtain $\left(  f+g\right)  ^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\star g^{\star\left(  n-i\right)  }$. This proves
(\ref{sol.convolution-series.c.binom}).}.

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise finitely
supported (since $f$ is pointwise $\star$-nilpotent). The same argument
(applied to $g$ instead of $f$) shows that the family $\left(  g^{\star
n}\right)  _{n\in\NN}$ is pointwise finitely supported.

The family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported. Renaming the index $n$ as $q$ in this statement, we thus
conclude that the family $\left(  f^{\star q}\right)  _{q\in\NN}$ is
pointwise finitely supported.

The family $\left(  g^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported. Renaming the index $n$ as $r$ in this statement, we thus
conclude that the family $\left(  g^{\star r}\right)  _{r\in\NN}$ is
pointwise finitely supported.

Thus, Proposition~\ref{prop.pw-fin-supp.P3} (applied to $Q=\NN$,
$R=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star q}\right)
_{q\in\NN}$ and $\left(  g_{r}\right)  _{r\in R}=\left(  g^{\star
r}\right)  _{r\in\NN}$) shows that the family $\left(  f^{\star q}\star
g^{\star r}\right)  _{\left(  q,r\right)  \in \NN\times\NN}%
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN\times\NN}$
is pointwise finitely supported, and that it satisfies%
\[
\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(  f^{\star
q}\star g^{\star r}\right)  =\left(  \sum_{q\in\NN}f^{\star q}\right)
\star\left(  \sum_{r\in\NN}g^{\star r}\right)  .
\]


In particular, the family $\left(  f^{\star q}\star g^{\star r}\right)
_{\left(  q,r\right)  \in \NN\times\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN\times\NN}$
is pointwise finitely supported. In other words, for each $x\in C$,%
\begin{equation}
\text{the family }\left(  \left(  f^{\star q}\star g^{\star r}\right)  \left(
x\right)  \right)  _{\left(  q,r\right)  \in \NN\times\NN}\in
A^{\NN\times\NN}\text{ is finitely supported.}
\label{sol.convolution-series.c.fs}%
\end{equation}


Let $x\in C$. Then, the family $\left(  \left(  f^{\star q}\star g^{\star
r}\right)  \left(  x\right)  \right)  _{\left(  q,r\right)  \in
\NN\times\NN}\in A^{\NN\times\NN}$ is finitely supported (by
(\ref{sol.convolution-series.c.fs})). In other words, all but finitely many
$\left(  q,r\right)  \in \NN\times\NN$ satisfy $\left(  f^{\star
q}\star g^{\star r}\right)  \left(  x\right)  =0$. In other words, there
exists a finite subset $K$ of $\NN\times\NN$ such that%
\begin{equation}
\text{each }\left(  q,r\right)  \in\left(  \NN\times\NN\right)
\setminus K\text{ satisfies }\left(  f^{\star q}\star g^{\star r}\right)
\left(  x\right)  =0. \label{sol.convolution-series.c.3}%
\end{equation}
Consider this $K$.

Let $Q=\left\{  u+v\ \mid\ \left(  u,v\right)  \in K\right\}  $. Thus, $Q$ is
a finite set (since $K$ is a finite set). Thus, all but finitely many
$n\in \NN$ satisfy $n\in \NN \setminus Q$.

But every $n\in \NN \setminus Q$ satisfies $\left(  f+g\right)  ^{\star
n}\left(  x\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Let $n\in
\NN \setminus Q$. We must show that $\left(  f+g\right)  ^{\star
n}\left(  x\right)  =0$.
\par
We have $n\in \NN \setminus Q$. In other words, $n\in \NN$ and
$n\notin Q$.
\par
Let $i\in\left\{  0,1,\ldots,n\right\}  $ be arbitrary. We shall show that
$\left(  f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =0$ first.
\par
Indeed, $i\in\left\{  0,1,\ldots,n\right\}  $ shows that $i\leq n$, so that
$n-i\in \NN$. Also, $i\in\left\{  0,1,\ldots,n\right\}  \subset
\NN$. Combining $i\in \NN$ with $n-i\in \NN$, we obtain
$\left(  i,n-i\right)  \in \NN\times\NN$.
\par
If we had $\left(  i,n-i\right)  \in K$, then we would have%
\begin{align*}
i+\left(  n-i\right)   &  \in\left\{  u+v\ \mid\ \left(  u,v\right)  \in
K\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i+\left(  n-i\right)
=u+v\text{ for some }\left(  u,v\right)  \in K\text{ (namely, for }\left(
u,v\right)  =\left(  i,n-i\right)  \text{)}\right) \\
&  =Q\ \ \ \ \ \ \ \ \ \ \left(  \text{since }Q=\left\{  u+v\ \mid\ \left(
u,v\right)  \in K\right\}  \right)  ,
\end{align*}
which would contradict $i+\left(  n-i\right)  =n\notin Q$. Thus, we cannot
have $\left(  i,n-i\right)  \in K$. In other words, we have $\left(
i,n-i\right)  \notin K$.
\par
Combining $\left(  i,n-i\right)  \in \NN\times\NN$ with $\left(
i,n-i\right)  \notin K$, we obtain $\left(  i,n-i\right)  \in\left(
\NN\times\NN\right)  \setminus K$. Hence,
(\ref{sol.convolution-series.c.3}) (applied to $\left(  q,r\right)  =\left(
i,n-i\right)  $) yields $\left(  f^{\star i}\star g^{\star\left(  n-i\right)
}\right)  \left(  x\right)  =0$.
\par
Now, forget that we fixed $i$. We thus have shown that%
\begin{equation}
\left(  f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  0,1,\ldots
,n\right\}  . \label{sol.convolution-series.c.3.pf.2}%
\end{equation}
\par
Now, (\ref{sol.convolution-series.c.binom}) yields $\left(  f+g\right)
^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\star g^{\star\left(
n-i\right)  }$. Applying both sides of this equality to $x$, we obtain%
\[
\left(  f+g\right)  ^{\star n}\left(  x\right)  =\left(  \sum_{i=0}^{n}%
\dbinom{n}{i}f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\star
g^{\star\left(  n-i\right)  }\right)  \left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.convolution-series.c.3.pf.2}))}}%
}=\sum_{i=0}^{n}\dbinom{n}{i}0=0.
\]
Qed.}. Hence, all but finitely many $n\in \NN$ satisfy $\left(
f+g\right)  ^{\star n}\left(  x\right)  =0$ (since all but finitely many
$n\in \NN$ satisfy $n\in \NN \setminus Q$). In other words, the
family $\left(  \left(  f+g\right)  ^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported.

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
family $\left(  \left(  f+g\right)  ^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported. In other words,
the family $\left(  \left(  f+g\right)  ^{\star n}\right)  _{n\in\NN}
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported. In other words, the map $f+g$ is pointwise
$\star$-nilpotent (by the definition of ``pointwise $\star
$-nilpotent''). In other words, $f+g\in\mathfrak{n}\left(
C,A\right)  $ (since $\mathfrak{n}\left(  C,A\right)  $ is the set of all
pointwise $\star$-nilpotent maps in $\Hom \left(  C,A\right)
$). This proves Proposition~\ref{prop.convolution-series}(c).

(d) Let $\lambda\in\kk$ and $f\in\mathfrak{n}\left(  C,A\right)  $. We
must prove that $\lambda f\in\mathfrak{n}\left(  C,A\right)  $.

For each $n\in \NN$, we have $\left(  \lambda f\right)  ^{\star
n}=\lambda^{n}f^{\star n}$. In other words, $\left(  \left(  \lambda f\right)
^{\star n}\right)  _{n\in\NN}=\left(  \lambda^{n}f^{\star n}\right)
_{n\in\NN}$.

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise finitely
supported (since $f$ is pointwise $\star$-nilpotent). Thus,
Proposition~\ref{prop.pw-fin-supp.P4} (applied to $Q=\NN$, $\left(
f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)  _{n\in\NN}$ and
$\left(  \lambda_{q}\right)  _{q\in Q}=\left(  \lambda^{n}\right)
_{n\in\NN}$) shows that the family $\left(  \lambda^{n}f^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported. In other words, the
family $\left(  \left(  \lambda f\right)  ^{\star n}\right)  _{n\in\NN}
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported (since $\left(  \left(  \lambda f\right)  ^{\star
n}\right)  _{n\in\NN}=\left(  \lambda^{n}f^{\star n}\right)
_{n\in\NN}$). In other words, the map $\lambda f$ is pointwise $\star
$-nilpotent (by the definition of ``pointwise $\star
$-nilpotent''). In other words, $\lambda f\in\mathfrak{n}%
\left(  C,A\right)  $ (since $\mathfrak{n}\left(  C,A\right)  $ is the set of
all pointwise $\star$-nilpotent maps in $\Hom \left(
C,A\right)  $). This proves Proposition~\ref{prop.convolution-series}(d).

(e) Let $f\in\mathfrak{n}\left(  C,A\right)  $ and $g\in\Hom
\left(  C,A\right)  $ be such that $f\star g=g\star f$. We must prove that
$f\star g\in\mathfrak{n}\left(  C,A\right)  $.

From $f\star g=g\star f$, we conclude that the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute. Thus, we can easily see that%
\begin{equation}
\left(  f\star g\right)  ^{\star n}=f^{\star n}\star g^{\star n}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in \NN
\label{sol.convolution-series.e.powerlaw}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.e.powerlaw}):} Let
$n\in \NN$. Let $\mathfrak{G}$ be the $\kk$-subalgebra of
$\left(  \Hom \left(  C,A\right)  ,\star\right)  $ generated by
the two elements $f$ and $g$. Thus, the $\kk$-algebra $\mathfrak{G}$ is
generated by commuting elements (because the elements $f$ and $g$ of the
$\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $ commute), and therefore is commutative (since any
$\kk$-algebra generated by commuting elements must be commutative).
Hence, the usual laws for exponentiation hold in this $\kk$-algebra
$\mathfrak{G}$. In particular, any $\alpha\in\mathfrak{G}$ and $\beta
\in\mathfrak{G}$ and $m\in \NN$ satisfy%
\[
\left(  \alpha\star\beta\right)  ^{\star m}=\alpha^{\star m}\star\beta^{\star
m}%
\]
(since the multiplication in the $\kk$-algebra $\mathfrak{G}$ is the
convolution $\star$). Applying this to $\alpha=f$, $\beta=g$ and $m=n$, we
obtain $\left(  f\star g\right)  ^{\star n}=f^{\star n}\star g^{\star n}$.
This proves (\ref{sol.convolution-series.e.powerlaw}).}. In other words,
$\left(  \left(  f\star g\right)  ^{\star n}\right)  _{n\in\NN}=\left(
f^{\star n}\star g^{\star n}\right)  _{n\in\NN}$.

We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise finitely
supported (since $f$ is pointwise $\star$-nilpotent).

Thus, Proposition~\ref{prop.pw-fin-supp.P5} (applied to $Q=\NN$,
$\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)  _{n\in\NN}$
and $\left(  g_{q}\right)  _{q\in Q}=\left(  g^{\star n}\right)
_{n\in\NN}$) shows that the family $\left(  f^{\star n}\star g^{\star
n}\right)  _{n\in\NN}\in\left(  \Hom \left(  C,A\right)
\right)  ^{\NN}$ is pointwise finitely supported. In other words, the
family $\left(  \left(  f\star g\right)  ^{\star n}\right)  _{n\in\NN}
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported (since $\left(  \left(  f\star g\right)  ^{\star
n}\right)  _{n\in\NN}=\left(  f^{\star n}\star g^{\star n}\right)
_{n\in\NN}$). In other words, the map $f\star g$ is pointwise $\star
$-nilpotent (by the definition of ``pointwise $\star
$-nilpotent''). In other words, $f\star g\in\mathfrak{n}%
\left(  C,A\right)  $ (since $\mathfrak{n}\left(  C,A\right)  $ is the set of
all pointwise $\star$-nilpotent maps in $\Hom \left(
C,A\right)  $). This proves Proposition~\ref{prop.convolution-series}(e).

(f) Let $v\in\kk\left[  \left[  T\right]  \right]  $ be a power series
whose constant term is $0$. Let $f\in\mathfrak{n}\left(  C,A\right)  $. We
must show that $v^{\star}\left(  f\right)  \in\mathfrak{n}\left(  C,A\right)
$.

We know that the constant term of $v$ is $0$. Thus, there exists a power
series $u\in\kk\left[  \left[  T\right]  \right]  $ such that
$v=Tu$\ \ \ \ \footnote{\textit{Proof.} Write the power series $v$ in the form
$v=\sum_{n\geq0}v_{n}T^{n}$ with $\left(  v_{n}\right)  _{n\geq0}\in
\kk^{\NN}$. Thus, the constant term of $v$ is $v_{0}$. In other
words, $0$ is $v_{0}$ (since the constant term of $v$ is $0$). In other words,
$0=v_{0}$. Now,
\begin{align*}
v  &  =\sum_{n\geq0}v_{n}T^{n}=\underbrace{v_{0}}_{=0}T^{0}+\sum_{n\geq1}%
v_{n}T^{n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }n=0\\
\text{from the sum}%
\end{array}
\right) \\
&  =\underbrace{0T^{0}}_{=0}+\sum_{n\geq1}v_{n}T^{n}=\sum_{n\geq1}v_{n}%
T^{n}=\sum_{n\geq0}v_{n+1}\underbrace{T^{n+1}}_{=T^{n}\cdot T}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }n+1\text{ for }n\\
\text{in the sum}%
\end{array}
\right) \\
&  =\sum_{n\geq0}v_{n+1}T^{n}\cdot T=\left(  \sum_{n\geq0}v_{n+1}T^{n}\right)
\cdot T=T\cdot\left(  \sum_{n\geq0}v_{n+1}T^{n}\right)  .
\end{align*}
Thus, there exists a power series $u\in\kk\left[  \left[  T\right]
\right]  $ such that $v=Tu$ (namely, $u=\sum_{n\geq0}v_{n+1}T^{n}$). Qed.}.
Consider this $u$.

Applying (\ref{eq.exe.convolution-series.a.monom}) to $k=1$, we find $\left(
T^{1}\right)  ^{\star}\left(  f\right)  =f^{\star1}=f$. Since $T^{1}=T$, this
rewrites as $T^{\star}\left(  f\right)  =f$.

Define $g\in\Hom \left(  C,A\right)  $ by $g=u^{\star}\left(
f\right)  $.

From $v=Tu$, we obtain%
\begin{align}
v^{\star}\left(  f\right)   &  =\left(  Tu\right)  ^{\star}\left(  f\right)
=\underbrace{T^{\star}\left(  f\right)  }_{=f}\star\underbrace{u^{\star
}\left(  f\right)  }_{=g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.exe.convolution-series.b.2}) (applied to }T\text{ and }u\text{
instead of }u\text{ and }v\text{)}\right) \nonumber\\
&  =f\star g. \label{sol.convolution-series.f.1}%
\end{align}


On the other hand, from $v=Tu=u\cdot T$, we obtain%
\begin{align*}
v^{\star}\left(  f\right)   &  =\left(  u\cdot T\right)  ^{\star}\left(
f\right)  =\underbrace{u^{\star}\left(  f\right)  }_{=g}\star
\underbrace{T^{\star}\left(  f\right)  }_{=f}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.exe.convolution-series.b.2}) (applied to }T\text{ instead of
}v\text{)}\right) \\
&  =g\star f.
\end{align*}
Comparing this with (\ref{sol.convolution-series.f.1}), we obtain $f\star
g=g\star f$. Hence, Proposition~\ref{prop.convolution-series}(e) yields
$f\star g\in\mathfrak{n}\left(  C,A\right)  $. In light of
(\ref{sol.convolution-series.f.1}), this rewrites as $v^{\star}\left(
f\right)  \in\mathfrak{n}\left(  C,A\right)  $. This proves
Proposition~\ref{prop.convolution-series}(f).

(g) Let us first prove the following fact:

\begin{statement}
\textit{Fact G.1:} Let $v\in\kk\left[  \left[  T\right]  \right]  $ be
any power series. Let $f\in\mathfrak{n}\left(  C,A\right)  $. Then,%
\begin{equation}
\left(  v^{n}\right)  ^{\star}\left(  f\right)  =\left(  v^{\star}\left(
f\right)  \right)  ^{\star n}\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in \NN. \label{sol.convolution-series.g.1}%
\end{equation}

\end{statement}

[\textit{Proof of Fact G.1:} We shall prove (\ref{sol.convolution-series.g.1})
by induction over $n$:

\textit{Induction base:} We have $\left(  \underbrace{v^{0}}_{=1}\right)
^{\star}\left(  f\right)  =1^{\star}\left(  f\right)  =u_{A}\epsilon_{C}$ (by
(\ref{eq.exe.convolution-series.b.5})). Comparing this with
\[
\left(  v^{\star}\left(  f\right)  \right)  ^{\star0}=\left(  \text{the unity
of the }\kk\text{-algebra }\left(  \Hom \left(
C,A\right)  ,\star\right)  \right)  =u_{A}\epsilon_{C},
\]
we obtain $\left(  v^{0}\right)  ^{\star}\left(  f\right)  =\left(  v^{\star
}\left(  f\right)  \right)  ^{\star0}$. In other words,
(\ref{sol.convolution-series.g.1}) holds for $n=0$. This completes the
induction base.

\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{sol.convolution-series.g.1}) holds for $n=N$. We must prove that
(\ref{sol.convolution-series.g.1}) holds for $n=N+1$.

We have assumed that (\ref{sol.convolution-series.g.1}) holds for $n=N$. In
other words, we have $\left(  v^{N}\right)  ^{\star}\left(  f\right)  =\left(
v^{\star}\left(  f\right)  \right)  ^{\star N}$.

Now,%
\begin{align*}
\left(  \underbrace{v^{N+1}}_{=v^{N}v}\right)  ^{\star}\left(  f\right)   &
=\left(  v^{N}v\right)  ^{\star}\left(  f\right)  =\underbrace{\left(
v^{N}\right)  ^{\star}\left(  f\right)  }_{=\left(  v^{\star}\left(  f\right)
\right)  ^{\star N}}\star v^{\star}\left(  f\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.convolution-series.b.2})
(applied to }u=v^{N}\text{)}\right) \\
&  =\left(  v^{\star}\left(  f\right)  \right)  ^{\star N}\star v^{\star
}\left(  f\right)  =\left(  v^{\star}\left(  f\right)  \right)  ^{\star\left(
N+1\right)  }.
\end{align*}
In other words, (\ref{sol.convolution-series.g.1}) holds for $n=N+1$. This
completes the induction step. Thus, the induction proof of
(\ref{sol.convolution-series.g.1}) is complete. In other words, Fact G.1 is proven.]

Now, let $u,v\in\kk\left[  \left[  T\right]  \right]  $ be two power
series such that the constant term of $v$ is $0$. Let $f\in\mathfrak{n}\left(
C,A\right)  $ be arbitrary. We must prove (\ref{eq.exe.convolution-series.g.1}).

Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus,
\begin{equation}
u^{\star}\left(  v^{\star}\left(  f\right)  \right)  =\sum_{n\geq0}%
u_{n}\left(  v^{\star}\left(  f\right)  \right)  ^{\star n}
\label{sol.convolution-series.g.3}%
\end{equation}
(by the definition of $u^{\star}\left(  v^{\star}\left(  f\right)  \right)  $).

But let us again recall that $u=\sum_{n\geq0}u_{n}T^{n}$. Hence, $u\left[
v\right]  =\sum_{n\geq0}u_{n}v^{n}$ (by the definition of the composition
$u\left[  v\right]  $). In particular, the sum $\sum_{n\geq0}u_{n}v^{n}$
converges in $\kk\left[  \left[  T\right]  \right]  $. In other words,
the sum $\sum_{n\in\NN}u_{n}v^{n}$ converges in $\kk\left[
\left[  T\right]  \right]  $ (since $\sum_{n\in\NN}=\sum_{n\geq0}$).
Hence, Fact B.7 (applied to $Q=\NN$ and $\left(  r_{q}\right)  _{q\in
Q}=\left(  u_{n}v^{n}\right)  _{n\in\NN}$) shows that the family
$\left(  \left(  u_{n}v^{n}\right)  ^{\star}\left(  f\right)  \right)
_{n\in\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN}$ is pointwise finitely supported and satisfies%
\begin{equation}
\left(  \sum_{n\in\NN}u_{n}v^{n}\right)  ^{\star}\left(  f\right)
=\sum_{n\in\NN}\left(  u_{n}v^{n}\right)  ^{\star}\left(  f\right)  .
\label{sol.convolution-series.g.4}%
\end{equation}


Now, recall that $u\left[  v\right]  =\underbrace{\sum_{n\geq0}}_{=
\sum_{n\in\NN}}u_{n}v^{n}=\sum_{n\in\NN}u_{n}v^{n}$. Hence,%
\begin{align*}
\left(  u\left[  v\right]  \right)  ^{\star}\left(  f\right)   &  =\left(
\sum_{n\in\NN}u_{n}v^{n}\right)  ^{\star}\left(  f\right)  =
\sum_{n\in\NN}\underbrace{\left(  u_{n}v^{n}\right)  ^{\star}\left(
f\right)  }_{\substack{=u_{n}\left(  v^{n}\right)  ^{\star}\left(  f\right)
\\\text{(by (\ref{eq.exe.convolution-series.b.3}) (applied}\\\text{to }%
u_{n}\text{ and }v^{n}\text{ instead of }\lambda\text{ and }u\text{))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.g.4}%
)}\right) \\
&  =\sum_{n\in\NN}u_{n}\underbrace{\left(  v^{n}\right)  ^{\star
}\left(  f\right)  }_{\substack{=\left(  v^{\star}\left(  f\right)  \right)
^{\star n}\\\text{(by Fact G.1)}}}=\underbrace{\sum_{n\in\NN}}%
_{=\sum_{n\geq0}}u_{n}\left(  v^{\star}\left(  f\right)  \right)  ^{\star
n}=\sum_{n\geq0}u_{n}\left(  v^{\star}\left(  f\right)  \right)  ^{\star
n}=u^{\star}\left(  v^{\star}\left(  f\right)  \right)
\end{align*}
(by (\ref{sol.convolution-series.g.3})). Thus,
(\ref{eq.exe.convolution-series.g.1}) is proven. This proves
Proposition~\ref{prop.convolution-series}(g).

(h) Let us first prove a simple fact:

\begin{statement}
\textit{Fact H.1:} Let $C$ be a graded $\kk$-coalgebra. Let
$f\in\Hom \left(  C,A\right)  $ be such that $f\left(
C_{0}\right)  =0$. Then, for each $i\in \NN$, we have%
\begin{equation}
f^{\star i}\left(  C_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in \NN \text{ satisfying }i>n. \label{sol.convolution-series.h.1}%
\end{equation}

\end{statement}

[\textit{Proof of Fact H.1:} We shall prove (\ref{sol.convolution-series.h.1})
by induction over $i$:

\textit{Induction base:} There exists no $n\in \NN$ satisfying $0>n$
(since each $n\in \NN$ satisfies $n\geq0$). Hence, we have $f^{\star
0}\left(  C_{n}\right)  =0$ for every $n\in \NN$ satisfying $0>n$
(indeed, this is vacuously true). In other words,
(\ref{sol.convolution-series.h.1}) holds for $i=0$. This completes the
induction base.

\textit{Induction step:} Let $p\in \NN$. Assume that
(\ref{sol.convolution-series.h.1}) holds for $i=p$. We must prove that
(\ref{sol.convolution-series.h.1}) holds for $i=p+1$.

We have assumed that (\ref{sol.convolution-series.h.1}) holds for $i=p$. In
other words, we have%
\begin{equation}
f^{\star p}\left(  C_{n}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in \NN \text{ satisfying }p>n. \label{sol.convolution-series.h.1.pf.1}%
\end{equation}


Now, let $n\in \NN$ be such that $p+1>n$.

It is easy to see that
\begin{equation}
\left(  f\otimes f^{\star p}\right)  \left(  C_{i}\otimes C_{j}\right)  =0
\label{sol.convolution-series.h.1.pf.2}%
\end{equation}
for every $\left(  i,j\right)  \in \NN^2$ satisfying $i+j=n$%
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.convolution-series.h.1.pf.2}):}
Let $\left(  i,j\right)  \in \NN^2$ be such that $i+j=n$. We must
prove (\ref{sol.convolution-series.h.1.pf.2}).
\par
From $\left(  i,j\right)  \in \NN^2$, we obtain $i\in \NN$ and
$j\in \NN$. If $i=0$, then
\[
\left(  f\otimes f^{\star p}\right)  \left(  C_{i}\otimes C_{j}\right)
=f\left(  \underbrace{C_{i}}_{\substack{=C_{0}\\\text{(since }i=0\text{)}%
}}\right)  \otimes f^{\star p}\left(  C_{j}\right)  =\underbrace{f\left(
C_{0}\right)  }_{=0}\otimes f^{\star p}\left(  C_{j}\right)  =0\otimes
f^{\star p}\left(  C_{j}\right)  =0.
\]
Hence, if $i=0$, then (\ref{sol.convolution-series.h.1.pf.2}) is proven. Thus,
for the rest of the proof of (\ref{sol.convolution-series.h.1.pf.2}), we can
WLOG assume that we don't have $i=0$. Assume this.
\par
We have $i\neq0$ (since we don't have $i=0$), and thus $i>0$ (since
$i\in \NN$). In other words, $i\geq1$ (since $i$ is an integer). But
$i+j=n$, so that $j=n-\underbrace{i}_{\geq1}\leq n-1$. But $p+1>n$; thus,
$p>n-1\geq j$ (since $j\leq n-1$). Hence,
(\ref{sol.convolution-series.h.1.pf.1}) (applied to $j$ instead of $n$) shows
that $f^{\star p}\left(  C_{j}\right)  =0$. Now,%
\[
\left(  f\otimes f^{\star p}\right)  \left(  C_{i}\otimes C_{j}\right)
=f\left(  C_{i}\right)  \otimes\underbrace{f^{\star p}\left(  C_{j}\right)
}_{=0}=f\left(  C_{i}\right)  \otimes0=0.
\]
This proves (\ref{sol.convolution-series.h.1.pf.2}).}.

Recall that the $\kk$-coalgebra $C$ is graded. Thus, its
comultiplication $\Delta$ is graded. In other words, $\Delta\left(
C_{k}\right)  \subset\left(  C\otimes C\right)  _{k}$ for each $k\in
\NN$. Applying this to $k=n$, we obtain
\begin{align*}
\Delta\left(  C_{n}\right)   &  \subset\left(  C\otimes C\right)
_{n}=\bigoplus_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}%
}C_{i}\otimes C_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the
grading on }C\otimes C\right) \\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}%
}C_{i}\otimes C_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are
sums}\right)  .
\end{align*}
Applying the map $f\otimes f^{\star p}$ to both sides of this relation, we
obtain%
\begin{align}
\left(  f\otimes f^{\star p}\right)  \left(  \Delta\left(  C_{n}\right)
\right)   &  \subset\left(  f\otimes f^{\star p}\right)  \left(
\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}}C_{i}\otimes
C_{j}\right)  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2%
;\\i+j=n}}\underbrace{\left(  f\otimes f^{\star p}\right)  \left(
C_{i}\otimes C_{j}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.h.1.pf.2}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\otimes f^{\star p}\text{
is }\kk\text{-linear}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  i,j\right)  \in \NN^2;\\i+j=n}}0=0.
\label{sol.convolution-series.h.1.pf.4}%
\end{align}


But the definition of convolution yields $f\star f^{\star p}=m\circ\left(
f\otimes f^{\star p}\right)  \circ\Delta$ (where $m$ denotes the
multiplication map $A\otimes A\rightarrow A$). We have%
\[
f^{\star\left(  p+1\right)  }=f\star f^{\star p}=m\circ\left(  f\otimes
f^{\star p}\right)  \circ\Delta
\]
and thus%
\begin{align*}
\underbrace{f^{\star\left(  p+1\right)  }}_{=m\circ\left(  f\otimes f^{\star
p}\right)  \circ\Delta}\left(  C_{n}\right)   &  =\left(  m\circ\left(
f\otimes f^{\star p}\right)  \circ\Delta\right)  \left(  C_{n}\right)
=m\left(  \underbrace{\left(  f\otimes f^{\star p}\right)  \left(
\Delta\left(  C_{n}\right)  \right)  }_{\substack{\subset0\\\text{(by
(\ref{sol.convolution-series.h.1.pf.4}))}}}\right) \\
&  \subset m\left(  0\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }m\text{ is }\kk\text{-linear}\right)  .
\end{align*}
In other words, $f^{\star\left(  p+1\right)  }\left(  C_{n}\right)  =0$.

Now, forget that we fixed $n$. We thus have proven that
\[
f^{\star\left(  p+1\right)  }\left(  C_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \NN \text{ satisfying
}p+1>n.
\]
In other words, (\ref{sol.convolution-series.h.1}) holds for $i=p+1$. This
completes the induction step. Thus, (\ref{sol.convolution-series.h.1}) is
proven by induction. Hence, Fact H.1 is proven.]

Now, let us actually prove Proposition~\ref{prop.convolution-series}(h).

Assume that $C$ is a graded $\kk$-coalgebra. Assume that $f\in
\Hom \left(  C,A\right)  $ satisfies $f\left(  C_{0}\right)
=0$. We must show that $f\in\mathfrak{n}\left(  C,A\right)  $.

Let $x\in C$.

The $\kk$-module $C$ is graded. Thus, any element of $C$ is a sum of
finitely many homogeneous elements of $C$. Hence, $x$ is a sum of finitely
many homogeneous elements of $C$ (since $x$ is an element of $C$). In other
words, there exist some $k\in \NN$ and some homogeneous elements
$x_{1},x_{2},\ldots,x_{k}\in C$ satisfying $x=\sum_{g=1}^{k}x_{g}$. Consider
this $k$ and these $x_{1},x_{2},\ldots,x_{k}$.

For each $g\in\left\{  1,2,\ldots,k\right\}  $, there exists some $n_{g}%
\in \NN$ satisfying $x_{g}\in C_{n_{g}}$ (since $x_{g}$ is a homogeneous
element of $C$). Consider this $n_{g}$. The set $\left\{  n_{1},n_{2}%
,\ldots,n_{k}\right\}  $ is a finite subset of $\NN$, and thus has an
upper bound (since any finite subset of $\NN$ has an upper bound). In
other words, there exists some $N\in \NN$ such that
\begin{equation}
\text{each }n\in\left\{  n_{1},n_{2},\ldots,n_{k}\right\}  \text{ satisfies
}n\leq N. \label{sol.exe.convolution-series.h.N}%
\end{equation}
Consider this $N$.

Let $Q=\left\{  0,1,\ldots,N\right\}  $. Then, $Q$ is a finite subset of
$\NN$. Thus, all but finitely many $n\in \NN$ satisfy
$n\in \NN \setminus Q$.

Now, let $n\in \NN \setminus Q$ be arbitrary. Thus, $n\in \NN
\setminus\underbrace{Q}_{=\left\{  0,1,\ldots,N\right\}  } = \NN
\setminus\left\{  0,1,\ldots,N\right\}  =\left\{  N+1,N+2,N+3,\ldots\right\}
$. Hence, $n>N$.

Thus, each $g\in\left\{  1,2,\ldots,k\right\}  $ satisfies
\begin{equation}
f^{\star n}\left(  x_{g}\right)  =0
\label{sol.exe.convolution-series.h.fnxg=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.convolution-series.h.fnxg=0}):} Let
$g\in\left\{  1,2,\ldots,k\right\}  $. Then, $n_{g}\in\left\{  n_{1}%
,n_{2},\ldots,n_{k}\right\}  $. Hence, (\ref{sol.exe.convolution-series.h.N})
(applied to $n=n_{g}$) yields $n_{g}\leq N$. Hence, $N\geq n_{g}$, so that
$n>N\geq n_{g}$. Hence, (\ref{sol.convolution-series.h.1}) (applied to $n$ and
$n_{g}$ instead of $i$ and $n$) yields $f^{\star n}\left(  C_{n_{g}}\right)
=0$.
\par
But $x_{g}\in C_{n_{g}}$ (by the definition of $n_{g}$) and thus $f^{\star
n}\left(  \underbrace{x_{g}}_{\in C_{n_{g}}}\right)  \in f^{\star n}\left(
C_{n_{g}}\right)  =0$. In other words, $f^{\star n}\left(  x_{g}\right)  =0$.
This proves (\ref{sol.exe.convolution-series.h.fnxg=0}).}. Applying the map
$f^{\star n}$ to the equality $x=\sum_{g=1}^{k}x_{g}$, we obtain%
\begin{align*}
f^{\star n}\left(  x\right)   &  =f^{\star n}\left(  \sum_{g=1}^{k}%
x_{g}\right)  =\sum_{g=1}^{k}\underbrace{f^{\star n}\left(  x_{g}\right)
}_{\substack{=0\\\text{(by (\ref{sol.exe.convolution-series.h.fnxg=0}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f^{\star n}\text{ is
}\kk\text{-linear}\right) \\
&  =\sum_{g=1}^{k}0=0.
\end{align*}


Now, forget that we fixed $n$. We thus have shown that each $n\in
\NN \setminus Q$ satisfies $f^{\star n}\left(  x\right)  =0$. Hence, all
but finitely many $n\in \NN$ satisfy $f^{\star n}\left(  x\right)  =0$
(since all but finitely many $n\in \NN$ satisfy $n\in \NN \setminus
Q$). In other words, the family $\left(  f^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported.

Now, forget that we fixed $x$. We thus have shown that for each $x\in C$, the
family $\left(  f^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
A^{\NN}$ is finitely supported. In other words, the family $\left(
f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom \left(
C,A\right)  \right)  ^{\NN}$ is pointwise finitely supported. In other
words, the map $f$ is pointwise $\star$-nilpotent (by the definition of
``pointwise $\star$-nilpotent''). In other
words, $f\in\mathfrak{n}\left(  C,A\right)  $ (since $\mathfrak{n}\left(
C,A\right)  $ is the set of all pointwise $\star$-nilpotent maps in
$\Hom \left(  C,A\right)  $). This proves
Proposition~\ref{prop.convolution-series}(h).

(i) Before we prove this, let us state a general fact:

\begin{statement}
\textit{Fact H.7:} Let $V$ and $W$ be two $\kk$-modules. Let
$\varphi:V\rightarrow W$ be a $\kk$-linear map. Let $\left(
v_{q}\right)  _{q\in Q}\in V^{Q}$ be a finitely supported family of elements
of $V$. Then, the family $\left(  \varphi\left(  v_{q}\right)  \right)  _{q\in
Q}\in W^{Q}$ is also finitely supported, and satisfies $\sum_{q\in Q}%
\varphi\left(  v_{q}\right)  =\varphi\left(  \sum_{q\in Q}v_{q}\right)  $.
\end{statement}

[\textit{Proof of Fact H.7:} Fact H.7 is a basic property of linear maps
(essentially, it says that any $\kk$-linear map preserves sums in which
all but finitely many addends are zero), and its proof is straightforward. For
the sake of completeness, let us nevertheless show this proof:

The map $\varphi$ is $\kk$-linear. Hence, $\varphi\left(  0\right)  =0$.

We know that the family $\left(  v_{q}\right)  _{q\in Q}$ is finitely
supported. In other words, all but finitely many $q\in Q$ satisfy $v_{q}=0$.
In other words, there exists a finite subset $Q^{\prime}$ of $Q$ such that%
\begin{equation}
\text{every }q\in Q\setminus Q^{\prime}\text{ satisfies }v_{q}=0.
\label{sol.convolution-series.i.1}%
\end{equation}
Consider this $Q^{\prime}$.

Now, every $q\in Q\setminus Q^{\prime}$ satisfies $\varphi\left(
\underbrace{v_{q}}_{\substack{=0\\\text{(by (\ref{sol.convolution-series.i.1}%
))}}}\right)  =\varphi\left(  0\right)  =0$. Hence, all but finitely many
$q\in Q$ satisfy $\varphi\left(  v_{q}\right)  =0$ (since all but finitely
many $q\in Q$ satisfy $q\in Q\setminus Q^{\prime}$ (since the set $Q^{\prime}$
is finite)). In other words, the family $\left(  \varphi\left(  v_{q}\right)
\right)  _{q\in Q}$ is finitely supported.

It remains to prove that $\sum_{q\in Q}\varphi\left(  v_{q}\right)
=\varphi\left(  \sum_{q\in Q}v_{q}\right)  $.

The map $\varphi$ is $\kk$-linear, and thus preserves finite sums.
Hence, $\varphi\left(  \sum_{q\in Q^{\prime}}v_{q}\right)  =\sum_{q\in
Q^{\prime}}\varphi\left(  v_{q}\right)  $ (since $\sum_{q\in Q^{\prime}}v_{q}$
is a finite sum (since $Q^{\prime}$ is a finite set)).

But we have%
\begin{align*}
\sum_{q\in Q}v_{q}  &  =\underbrace{\sum\limits_{\substack{q\in Q;\\q\in Q^{\prime}%
}}}_{\substack{=\sum_{q\in Q^{\prime}}\\\text{(since }Q^{\prime}\subset
Q\text{)}}}v_{q}+\underbrace{\sum\limits_{\substack{q\in Q;\\q\notin Q^{\prime}}%
}}_{=\sum_{q\in Q\setminus Q^{\prime}}}v_{q}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }q\in Q\text{ satisfies either
}q\in Q^{\prime}\text{ or }q\notin Q^{\prime}\text{ (but not both)}\right) \\
&  =\sum_{q\in Q^{\prime}}v_{q}+\sum_{q\in Q\setminus Q^{\prime}%
}\underbrace{v_{q}}_{\substack{=0\\\text{(by (\ref{sol.convolution-series.i.1}%
))}}}=\sum_{q\in Q^{\prime}}v_{q}+\underbrace{\sum_{q\in Q\setminus Q^{\prime
}}0}_{=0}=\sum_{q\in Q^{\prime}}v_{q}.
\end{align*}
Applying the map $\varphi$ to both sides of this equality, we obtain%
\[
\varphi\left(  \sum_{q\in Q}v_{q}\right)  =\varphi\left(  \sum_{q\in
Q^{\prime}}v_{q}\right)  =\sum_{q\in Q^{\prime}}\varphi\left(  v_{q}\right)
.
\]
Comparing this with%
\begin{align*}
\sum_{q\in Q}\varphi\left(  v_{q}\right)   &  =\underbrace{
\sum\limits_{\substack{q\in Q;\\q\in Q^{\prime}}}
}_{\substack{=\sum_{q\in Q^{\prime}%
}\\\text{(since }Q^{\prime}\subset Q\text{)}}}\varphi\left(  v_{q}\right)
+\underbrace{\sum\limits_{\substack{q\in Q;\\q\notin Q^{\prime}}}}_{=\sum_{q\in
Q\setminus Q^{\prime}}}\varphi\left(  v_{q}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }q\in Q\text{ satisfies either
}q\in Q^{\prime}\text{ or }q\notin Q^{\prime}\text{ (but not both)}\right) \\
&  =\sum_{q\in Q^{\prime}}\varphi\left(  v_{q}\right)  +\sum_{q\in Q\setminus
Q^{\prime}}\varphi\left(  \underbrace{v_{q}}_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.i.1}))}}}\right)  =\sum_{q\in Q^{\prime}}%
\varphi\left(  v_{q}\right)  +\sum_{q\in Q\setminus Q^{\prime}}%
\underbrace{\varphi\left(  0\right)  }_{=0}\\
&  =\sum_{q\in Q^{\prime}}\varphi\left(  v_{q}\right)  +\underbrace{\sum_{q\in
Q\setminus Q^{\prime}}0}_{=0}=\sum_{q\in Q^{\prime}}\varphi\left(
v_{q}\right)  ,
\end{align*}
we obtain $\sum_{q\in Q}\varphi\left(  v_{q}\right)  =\varphi\left(
\sum_{q\in Q}v_{q}\right)  $. This completes the proof of Fact H.7.]

Now, let us start proving Proposition~\ref{prop.convolution-series}(i).

Let $B$ be any $\kk$-algebra. Let $s:A\rightarrow B$ be any
$\kk$-algebra homomorphism.

Proposition~\ref{prop.convolution.functor} (applied to $A^{\prime}=B$,
$C^{\prime}=C$, $\alpha=s$ and $\gamma=\id_{C}$) shows
that the map%
\[
\Hom \left(  C,A\right)  \rightarrow\Hom \left(
C,B\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto s\circ f\circ\operatorname*{id}%
\nolimits_{C}%
\]
is a $\kk$-algebra homomorphism from the $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ to the $\kk%
$-algebra $\left(  \Hom \left(  C,B\right)  ,\star\right)  $.
Since each $f\in\Hom \left(  C,A\right)  $ satisfies $s\circ
f\circ\id_{C}=s\circ f$, this rewrites as follows: The
map%
\[
\Hom \left(  C,A\right)  \rightarrow\Hom \left(
C,B\right)  ,\ \ \ \ \ \ \ \ \ \ f\mapsto s\circ f
\]
is a $\kk$-algebra homomorphism from the $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $ to the $\kk%
$-algebra $\left(  \Hom \left(  C,B\right)  ,\star\right)  $.
Let us denote this $\kk$-algebra homomorphism by $\Phi$.

Now, let $u\in\kk\left[  \left[  T\right]  \right]  $ and
$f\in\mathfrak{n}\left(  C,A\right)  $. We must show that%
\begin{equation}
s\circ f\in\mathfrak{n}\left(  C,B\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ u^{\star}\left(  s\circ f\right)  =s\circ\left(  u^{\star
}\left(  f\right)  \right)  . \label{pf.prop.convolution-series.i.goal}%
\end{equation}


Write the power series $u$ in the form $u=\sum_{n\geq0}u_{n}T^{n}$ with
$\left(  u_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $u^{\star
}\left(  f\right)  =\sum_{n\geq0}u_{n}f^{\star n}$ (by the definition of
$u^{\star}\left(  f\right)  $).

Each $n\in \NN$ satisfies%
\begin{align}
s\circ f^{\star n}  &  =\Phi\left(  f^{\star n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\left(  f^{\star n}\right)
=s\circ f^{\star n}\text{ (by the definition of }\Phi\text{)}\right)
\nonumber\\
&  =\left(  \underbrace{\Phi\left(  f\right)  }_{\substack{=s\circ
f\\\text{(by the definition of }\Phi\text{)}}}\right)  ^{\star n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\text{ is a }\kk%
\text{-algebra homomorphism}\right) \nonumber\\
&  =\left(  s\circ f\right)  ^{\star n}.
\label{pf.prop.convolution-series.i.pow-pres}%
\end{align}


We have $f\in\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a
pointwise $\star$-nilpotent map in $\Hom \left(  C,A\right)  $
(since $\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise finitely
supported (since $f$ is pointwise $\star$-nilpotent). Hence,
Proposition~\ref{prop.pw-fin-supp.P4} (applied to $Q=\NN$, $\left(
f_{q}\right)  _{q\in Q}=\left(  f^{\star n}\right)  _{n\in\NN}$ and
$\left(  \lambda_{q}\right)  _{q\in Q}=\left(  u_{n}\right)  _{n\in\NN}$)
shows that the family $\left(  u_{n}f^{\star n}\right)  _{n\in\NN}
\in\left(  \Hom \left(  C,A\right)  \right)  ^{\NN}$ is
pointwise finitely supported. In other words, for each $x\in C$,%
\[
\text{the family }\left(  \left(  u_{n}f^{\star n}\right)  \left(  x\right)
\right)  _{n\in\NN}\in A^{\NN}\text{ is finitely supported}%
\]
(by the definition of ``pointwise finitely
supported''). In other words, for each $x\in C$,%
\begin{equation}
\text{the family }\left(  u_{n}f^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}\text{ is finitely supported}
\label{pf.prop.convolution-series.i.fs1}%
\end{equation}
(since each $n\in \NN$ satisfies $\left(  u_{n}f^{\star n}\right)
\left(  x\right)  =u_{n}f^{\star n}\left(  x\right)  $).

Also, recall that the family $\left(  f^{\star n}\right)  _{n\in\NN}$
is pointwise finitely supported. In other words, for each $x\in C$,
\begin{equation}
\text{the family }\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}\text{ is finitely supported}
\label{pf.prop.convolution-series.i.fs2}%
\end{equation}
(by the definition of ``pointwise finitely supported'').

Let $x\in C$. From (\ref{pf.prop.convolution-series.i.fs1}), we know that the
family $\left(  u_{n}f^{\star n}\left(  x\right)  \right)  _{n\in\NN}
\in A^{\NN}$ is finitely supported. Hence, Fact H.7 (applied to $V=A$,
$W=B$, $\varphi=s$, $Q=\NN$ and $\left(  v_{q}\right)  _{q\in
Q}=\left(  u_{n}f^{\star n}\left(  x\right)  \right)  _{n\in\NN}$)
shows that the family $\left(  s\left(  u_{n}f^{\star n}\left(  x\right)
\right)  \right)  _{n\in\NN}\in B^{\NN}$ is also finitely
supported, and that it satisfies
\begin{equation}
\sum_{n\in\NN}s\left(  u_{n}f^{\star n}\left(  x\right)  \right)
=s\left(  \sum_{n\in\NN}u_{n}f^{\star n}\left(  x\right)  \right)  .
\label{pf.prop.convolution-series.i.5}%
\end{equation}
From (\ref{pf.prop.convolution-series.i.5}), we obtain%
\begin{align}
s\left(  \sum_{n\in\NN}u_{n}f^{\star n}\left(  x\right)  \right)   &
=\sum_{n\in\NN}\underbrace{s\left(  u_{n}f^{\star n}\left(  x\right)
\right)  }_{\substack{=u_{n}s\left(  f^{\star n}\left(  x\right)  \right)
\\\text{(since the map }s\text{ is }\kk\text{-linear)}}}=
\sum_{n\in\NN}u_{n}\underbrace{s\left(  f^{\star n}\left(  x\right)
\right)  }_{=\left(  s\circ f^{\star n}\right)  \left(  x\right)  }=
\sum_{n\in\NN}u_{n}\underbrace{\left(  s\circ f^{\star n}\right)
}_{\substack{=\left(  s\circ f\right)  ^{\star n}\\\text{(by
(\ref{pf.prop.convolution-series.i.pow-pres}))}}}\left(  x\right) \nonumber\\
&  =\sum_{n\in\NN}u_{n}\left(  s\circ f\right)  ^{\star n}\left(
x\right)  . \label{pf.prop.convolution-series.i.6}%
\end{align}


Also, from (\ref{pf.prop.convolution-series.i.fs2}), we know that the family
$\left(  f^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
A^{\NN}$ is finitely supported. Hence, Fact H.7 (applied to $V=A$,
$W=B$, $\varphi=s$, $Q=\NN$ and $\left(  v_{q}\right)  _{q\in
Q}=\left(  f^{\star n}\left(  x\right)  \right)  _{n\in\NN}$) shows
that the family $\left(  s\left(  f^{\star n}\left(  x\right)  \right)
\right)  _{n\in\NN}\in B^{\NN}$ is also finitely supported, and
that it satisfies $\sum_{n\in\NN}s\left(  f^{\star n}\left(  x\right)
\right)  =s\left(  \sum_{n\in\NN}f^{\star n}\left(  x\right)  \right)
$. Thus, in particular,%
\[
\text{the family }\left(  s\left(  f^{\star n}\left(  x\right)  \right)
\right)  _{n\in\NN}\in B^{\NN}\text{ is finitely supported.}%
\]
Since each $n\in \NN$ satisfies $s\left(  f^{\star n}\left(  x\right)
\right)  =\underbrace{\left(  s\circ f^{\star n}\right)  }_{\substack{=\left(
s\circ f\right)  ^{\star n}\\\text{(by
(\ref{pf.prop.convolution-series.i.pow-pres}))}}}\left(  x\right)  =\left(
s\circ f\right)  ^{\star n}\left(  x\right)  $, this rewrites as follows:%
\begin{equation}
\text{The family }\left(  \left(  s\circ f\right)  ^{\star n}\left(  x\right)
\right)  _{n\in\NN}\in B^{\NN}\text{ is finitely supported.}
\label{pf.prop.convolution-series.i.7}%
\end{equation}


Now, forget that we fixed $x$. We thus have proven that each $x\in C$
satisfies (\ref{pf.prop.convolution-series.i.6}) and
(\ref{pf.prop.convolution-series.i.7}).

In particular, each $x\in C$ satisfies (\ref{pf.prop.convolution-series.i.7}).
In other words, for each $x\in C$, the family $\left(  \left(  s\circ
f\right)  ^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
B^{\NN}$ is finitely supported. In other words, the family $\left(
\left(  s\circ f\right)  ^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,B\right)  \right)  ^{\NN}$ is pointwise
finitely supported (by the definition of ``pointwise finitely
supported''). In other words, the map $s\circ f:C\rightarrow
B$ is pointwise $\star$-nilpotent (by the definition of
``pointwise $\star$-nilpotent''). In other words, $s\circ
f\in\mathfrak{n}\left(  C,B\right)  $ (since $\mathfrak{n}\left(  C,B\right)
$ is the set of all pointwise $\star$-nilpotent maps in $\Hom
\left(  C,B\right)  $). Hence, $u^{\star}\left(  s\circ f\right)  $ is well-defined.

Recall that $u=\sum_{n\geq0}u_{n}T^{n}$. Thus,%
\[
u^{\star}\left(  s\circ f\right)  =\underbrace{\sum_{n\geq0}}_{=
\sum_{n\in\NN}}u_{n}\left(  s\circ f\right)  ^{\star n}=\sum_{n\in
\NN}u_{n}\left(  s\circ f\right)  ^{\star n}.
\]
Hence, each $x\in C$ satisfies%
\begin{align*}
\left(  \underbrace{u^{\star}\left(  s\circ f\right)  }_{=\sum_{n\in
\NN}u_{n}\left(  s\circ f\right)  ^{\star n}}\right)  \left(  x\right)
&  =\left(  \sum_{n\in\NN}u_{n}\left(  s\circ f\right)  ^{\star
n}\right)  \left(  x\right)  =\sum_{n\in\NN}u_{n}\left(  s\circ
f\right)  ^{\star n}\left(  x\right) \\
&  =s\left(  \sum_{n\in\NN}u_{n}f^{\star n}\left(  x\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.convolution-series.i.6}%
)}\right) \\
&  =\left(  s\circ\left(  u^{\star}\left(  f\right)  \right)  \right)  \left(
x\right)
\end{align*}
(since
\begin{align*}
\left(  s\circ\left(  u^{\star}\left(  f\right)  \right)  \right)  \left(
x\right)   &  =s\left(  \underbrace{\left(  u^{\star}\left(  f\right)
\right)  }_{=\sum_{n\geq0}u_{n}f^{\star n}}\left(  x\right)  \right)
=s\left(  \underbrace{\left(  \sum_{n\geq0}u_{n}f^{\star n}\right)  \left(
x\right)  }_{=\sum_{n\geq0}u_{n}f^{\star n}\left(  x\right)  }\right) \\
&  =s\left(  \underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}u_{n}f^{\star
n}\left(  x\right)  \right)  =s\left(  \sum_{n\in\NN}u_{n}f^{\star
n}\left(  x\right)  \right)
\end{align*}
). In other words, we have $u^{\star}\left(  s\circ f\right)  =s\circ\left(
u^{\star}\left(  f\right)  \right)  $.

We have now proven that $s\circ f\in\mathfrak{n}\left(  C,B\right)  $ and
$u^{\star}\left(  s\circ f\right)  =s\circ\left(  u^{\star}\left(  f\right)
\right)  $. This proves (\ref{pf.prop.convolution-series.i.goal}). Hence,
Proposition~\ref{prop.convolution-series}(i) is proven.

(j) Let $C$ be a connected graded $\kk$-bialgebra. Let $F:C\rightarrow
A$ be a $\kk$-algebra homomorphism. We must prove that $F-u_{A}%
\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $.

We have $F\left(  1_{C}\right)  =1_{A}$ (since $F$ is a $\kk$-algebra homomorphism).

The axioms of a $\kk$-bialgebra show that the counit $\epsilon_{C}$ of
the $\kk$-coalgebra $C$ is a $\kk$-algebra homomorphism (since
$C$ is a $\kk$-bialgebra). Thus, $\epsilon_{C}\left(  1_{C}\right)  =1$.

The definition of the map $u_{A}$ yields $u_{A}\left(  1\right)  =1\cdot
1_{A}=1_{A}$. Now,%
\begin{align*}
\left(  F-u_{A}\epsilon_{C}\right)  \left(  1_{C}\right)   &
=\underbrace{F\left(  1_{C}\right)  }_{=1_{A}}-\underbrace{\left(
u_{A}\epsilon_{C}\right)  \left(  1_{C}\right)  }_{=u_{A}\left(  \epsilon
_{C}\left(  1_{C}\right)  \right)  }=1_{A}-u_{A}\left(  \underbrace{\epsilon
_{C}\left(  1_{C}\right)  }_{=1}\right) \\
&  =1_{A}-\underbrace{u_{A}\left(  1\right)  }_{=1_{A}}=1_{A}-1_{A}=0.
\end{align*}


But Exercise~\ref{graded-connected-exercise}(c) (applied to $C$ instead of
$A$) shows that $C_{0} =\kk\cdot1_{C}$. Applying the map $F-u_{A}%
\epsilon_{C}$ to both sides of this relation, we obtain%
\begin{align*}
\left(  F-u_{A}\epsilon_{C}\right)  \left(  C_{0}\right)   &  =\left(
F-u_{A}\epsilon_{C}\right)  \left(  \kk\cdot1_{C}\right)
=\kk\cdot\underbrace{\left(  F-u_{A}\epsilon_{C}\right)  \left(
1_{C}\right)  }_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
F-u_{A}\epsilon_{C}\text{ is }\kk\text{-linear}\right) \\
&  =\kk\cdot0=0.
\end{align*}
Hence, Proposition~\ref{prop.convolution-series}(h) (applied to $f=F-u_{A}%
\epsilon_{C}$) shows that $F-u_{A}\epsilon_{C}\in\mathfrak{n}\left(
C,A\right)  $. Thus, Proposition~\ref{prop.convolution-series}(j) is proven.
\end{proof}

Thus, Proposition~\ref{prop.convolution-series} is proven, so that
Exercise~\ref{exe.convolution-series} is solved.

%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.exp-log}}

\textit{Solution to Exercise~\ref{exe.exp-log}.}

\begin{proof}
[Proof of Proposition~\ref{prop.exp-log-inv}.]%
Proposition~\ref{prop.exp-log-inv} is a well-known fact that is often used in
enumerative combinatorics (for computing generating functions). We shall give
a purely algebraic proof (somewhat similar to the one given in \cite[Example
7.67]{Loehr-bij}). Other proofs (some combinatorial, some analytic) can be
found in the literature.

The proof will rely on several simple facts about power series. Keep in mind
that all of the following facts assume that $\kk$ is a commutative
$\QQ$-algebra.

First, we notice that%
\begin{align}
\overline{\exp}  &  =\underbrace{\exp}_{=\sum_{n\geq0}\dfrac{1}{n!}T^{n}%
}-1=\sum_{n\geq0}\dfrac{1}{n!}T^{n}-1=\underbrace{\dfrac{1}{0!}}_{=\dfrac
{1}{1}=1}\underbrace{T^{0}}_{=1}+\sum_{n\geq1}\dfrac{1}{n!}T^{n}-1\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \nonumber\\
&  =1+\sum_{n\geq1}\dfrac{1}{n!}T^{n}-1=\sum_{n\geq1}\dfrac{1}{n!}T^{n}.
\label{sol.convolution-series.i.expbar=}%
\end{align}
Hence, the power series $\overline{\exp}$ has constant term $0$. Hence, the
power series $\overline{\log}\left[  \overline{\exp}\right]  $ is well-defined.

Also,%
\begin{equation}
\overline{\log}=\log\left(  1+T\right)  =\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}T^{n}. \label{sol.convolution-series.i.logbar=}%
\end{equation}
Hence, the power series $\overline{\log}$ has constant term $0$. Hence, the
power series $\overline{\exp}\left[  \overline{\log}\right]  $ is well-defined.

For each $n\geq1$, we have%
\begin{equation}
\left(  \text{the constant term of }\overline{\log}^{n}\right)  =0
\label{sol.convolution-series.i.cstlog0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.i.cstlog0}):} Let
$n\geq1$. The power series $\overline{\log}$ is divisible by $T$ (since it has
constant term $0$). Hence, the power series $\overline{\log}^{n}$ is divisible
by $T^{n}$. Thus, the power series $\overline{\log}^{n}$ is also divisible by
$T$ (since $T^{n}$ is divisible by $T$ (since $n\geq1$)), and therefore has
constant term $0$. In other words, we have $\left(  \text{the constant term of
}\overline{\log}^{n}\right)  =0$. This proves
(\ref{sol.convolution-series.i.cstlog0}).}.

Substituting $\overline{\log}$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.expbar=}), we obtain%
\[
\overline{\exp}\left[  \overline{\log}\right]  =\sum_{n\geq1}\dfrac{1}%
{n!}\overline{\log}^{n}.
\]
Hence,%
\begin{align*}
\left(  \text{the constant term of }\overline{\exp}\left[  \overline{\log
}\right]  \right)   &  =\left(  \text{the constant term of }\sum_{n\geq
1}\dfrac{1}{n!}\overline{\log}^{n}\right) \\
&  =\sum_{n\geq1}\dfrac{1}{n!}\underbrace{\left(  \text{the constant term of
}\overline{\log}^{n}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.i.cstlog0}))}}}=\sum_{n\geq1}\dfrac{1}{n!}0=0.
\end{align*}
In other words, the power series $\overline{\exp}\left[  \overline{\log
}\right]  $ has constant term $0$. A similar argument (with the roles of
$\overline{\exp}$ and $\overline{\log}$ switched) shows that the power series
$\overline{\log}\left[  \overline{\exp}\right]  $ has constant term $0$.

\begin{statement}
\textit{Fact I.1:} Let $u\in\kk\left[  \left[  T\right]  \right]  $ and
$v\in\kk\left[  \left[  T\right]  \right]  $ be two power series having
the same constant term. Assume that $\dfrac{d}{dT}u=\dfrac{d}{dT}v$. Then,
$u=v$.
\end{statement}

[\textit{Proof of Fact I.1:} Write the power series $u$ in the form
$u=\sum_{n\geq0}u_{n}T^{n}$ with $\left(  u_{n}\right)  _{n\geq0}\in
\kk^{\NN}$. Thus, $\dfrac{d}{dT}u=\sum_{n\geq1}nu_{n}T^{n-1}$
(by the definition of the derivative).

Write the power series $v$ in the form $v=\sum_{n\geq0}v_{n}T^{n}$ with
$\left(  v_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $\dfrac
{d}{dT}v=\sum_{n\geq1}nv_{n}T^{n-1}$ (by the definition of the derivative).

Now,%
\[
\sum_{n\geq1}nu_{n}T^{n-1}=\dfrac{d}{dT}u=\dfrac{d}{dT}v=\sum_{n\geq1}%
nv_{n}T^{n-1}.
\]
Comparing coefficients in front of $T^{n-1}$ on both sides of this equality,
we obtain%
\begin{equation}
nu_{n}=nv_{n}\ \ \ \ \ \ \ \ \ \ \text{for each integer }n\geq1.
\label{sol.convolution-series.i.1.pf.1}%
\end{equation}


On the other hand, the power series $u$ has constant term $u_{0}$ (since
$u=\sum_{n\geq0}u_{n}T^{n}$), and the power series $v$ has constant term
$v_{0}$ (similarly). Thus, the constant terms of $u$ and $v$ are $u_{0}$ and
$v_{0}$, respectively. Therefore, $u_{0}=v_{0}$ (since the power series $u$
and $v$ have the same constant term).

Now, each $n\in \NN$ satisfies $u_{n}=v_{n}$%
\ \ \ \ \footnote{\textit{Proof.} Let $n\in \NN$. We must prove that
$u_{n}=v_{n}$.
\par
If $n=0$, then this follows immediately from $u_{0}=v_{0}$. Hence, we WLOG
assume that we don't have $n=0$. Thus, $n\geq1$ (since $n\in \NN$).
Therefore, (\ref{sol.convolution-series.i.1.pf.1}) yields $nu_{n}=nv_{n}$. We
can multiply both sides of this equality by $\dfrac{1}{n}$ (since $\kk$
is a $\QQ$-algebra), and thus obtain $u_{n}=v_{n}$, qed.}. Hence,
$\sum_{n\geq0}\underbrace{u_{n}}_{=v_{n}}T^{n}=\sum_{n\geq0}v_{n}T^{n}$. Thus,
$u=\sum_{n\geq0}u_{n}T^{n}=\sum_{n\geq0}v_{n}T^{n}=v$. This proves Fact I.1.]

\begin{statement}
\textit{Fact I.2:} Let $w\in\kk\left[  \left[  T\right]  \right]  $ be
a power series having constant term $0$. Then,
\begin{equation}
\dfrac{d}{dT}\left(  \overline{\exp}\left[  w\right]  \right)  =\left(
\dfrac{d}{dT}w\right)  \cdot\exp\left[  w\right]
\label{sol.convolution-series.i.2.exp}%
\end{equation}
and%
\begin{equation}
\dfrac{d}{dT}\left(  \overline{\log}\left[  w\right]  \right)  =\left(
\dfrac{d}{dT}w\right)  \cdot\dfrac{1}{1+w}.
\label{sol.convolution-series.i.2.log}%
\end{equation}

\end{statement}

[\textit{Proof of Fact I.2:} It is easy to see that each positive integer $n$
satisfies%
\begin{equation}
\dfrac{d}{dT}\left(  w^{n}\right)  =n\left(  \dfrac{d}{dT}w\right)  w^{n-1}.
\label{sol.convolution-series.i.2.pf.power}%
\end{equation}
(Indeed, (\ref{sol.convolution-series.i.2.pf.power}) can be proven by a
straightforward induction on $n$, using the Leibniz identity $\dfrac{d}%
{dT}\left(  uv\right)  =\left(  \dfrac{d}{dT}u\right)  v+u\dfrac{d}{dT}v$.)

Substituting $w$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.expbar=}), we obtain%
\[
\overline{\exp}\left[  w\right]  =\sum_{n\geq1}\dfrac{1}{n!}w^{n}.
\]
Applying the operator $\dfrac{d}{dT}$ to this equality, we find%
\begin{align*}
\dfrac{d}{dT}\overline{\exp}\left[  w\right]   &  =\dfrac{d}{dT}\sum_{n\geq
1}\dfrac{1}{n!}w^{n}=\sum_{n\geq1}\dfrac{1}{n!}\cdot\underbrace{\dfrac{d}%
{dT}\left(  w^{n}\right)  }_{\substack{=n\left(  \dfrac{d}{dT}w\right)
w^{n-1}\\\text{(by (\ref{sol.convolution-series.i.2.pf.power}))}}}=
\sum_{n\geq1}
\underbrace{\dfrac{1}{n!}\cdot n}_{=\dfrac{1}{\left(  n-1\right)  !}%
}\left(  \dfrac{d}{dT}w\right)  w^{n-1}\\
&  =\sum_{n\geq1}\dfrac{1}{\left(  n-1\right)  !}\left(  \dfrac{d}%
{dT}w\right)  w^{n-1}=\sum_{n\geq0}\dfrac{1}{n!}\left(  \dfrac{d}{dT}w\right)
w^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n\text{ for
}n-1\text{ in the sum}\right)  .
\end{align*}
Comparing this with%
\[
\left(  \dfrac{d}{dT}w\right)  \cdot\underbrace{\exp\left[  w\right]
}_{\substack{=\sum_{n\geq0}\dfrac{1}{n!}w^{n}\\\text{(since }\exp=
\sum_{n\geq0}\dfrac{1}{n!}T^{n}\text{)}}}
=\left(  \dfrac{d}{dT}w\right)  \cdot
\sum_{n\geq0}\dfrac{1}{n!}w^{n}=\sum_{n\geq0}\dfrac{1}{n!}\left(  \dfrac
{d}{dT}w\right)  w^{n},
\]
we obtain $\dfrac{d}{dT}\left(  \overline{\exp}\left[  w\right]  \right)
=\left(  \dfrac{d}{dT}w\right)  \cdot\exp\left[  w\right]  $. This proves
(\ref{sol.convolution-series.i.2.exp}).

Substituting $w$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.logbar=}), we obtain%
\[
\overline{\log}\left[  w\right]  =\sum_{n\geq1}\dfrac{\left(  -1\right)
^{n-1}}{n}w^{n}.
\]
Applying the operator $\dfrac{d}{dT}$ to this equality, we find%
\begin{align*}
\dfrac{d}{dT}\overline{\log}\left[  w\right]   &  =\dfrac{d}{dT}\sum_{n\geq
1}\dfrac{\left(  -1\right)  ^{n-1}}{n}w^{n}=\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\cdot\underbrace{\dfrac{d}{dT}\left(  w^{n}\right)
}_{\substack{=n\left(  \dfrac{d}{dT}w\right)  w^{n-1}\\\text{(by
(\ref{sol.convolution-series.i.2.pf.power}))}}}=\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\cdot n\left(  \dfrac{d}{dT}w\right)  w^{n-1}\\
&  =\sum_{n\geq1}\left(  -1\right)  ^{n-1}\left(  \dfrac{d}{dT}w\right)
w^{n-1}=\sum_{n\geq0}\left(  -1\right)  ^{n}\left(  \dfrac{d}{dT}w\right)
w^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n\text{ for
}n-1\text{ in the sum}\right)  .
\end{align*}
Comparing this with%
\[
\left(  \dfrac{d}{dT}w\right)  \cdot\underbrace{\dfrac{1}{1+w}}_{=
\sum_{n\geq0}\left(  -1\right)  ^{n}w^{n}}=\left(  \dfrac{d}{dT}w\right)
\cdot\sum_{n\geq0}\left(  -1\right)  ^{n}w^{n}=\sum_{n\geq0}\left(  -1\right)
^{n}\left(  \dfrac{d}{dT}w\right)  w^{n},
\]
we obtain $\dfrac{d}{dT}\left(  \overline{\log}\left[  w\right]  \right)
=\left(  \dfrac{d}{dT}w\right)  \cdot\dfrac{1}{1+w}$. This proves
(\ref{sol.convolution-series.i.2.log}). Thus, Fact I.2 is proven.]

\begin{statement}
\textit{Fact I.3:} Let $u\in\kk\left[  \left[  T\right]  \right]  $ and
$v\in\kk\left[  \left[  T\right]  \right]  $ be two power series having
constant term $1$. Assume that $\left(  \dfrac{d}{dT}u\right)  \cdot v=\left(
\dfrac{d}{dT}v\right)  \cdot u$. Then, $u=v$.
\end{statement}

[\textit{Proof of Fact I.3:} The power series $v$ has constant term $1$, and
thus has a multiplicative inverse $v^{-1}$. The Leibniz rule (applied to $v$
and $v^{-1}$) yields%
\[
\dfrac{d}{dT}\left(  v\cdot v^{-1}\right)  =\left(  \dfrac{d}{dT}v\right)
v^{-1}+v\dfrac{d}{dT}\left(  v^{-1}\right)  .
\]
Comparing this with $\dfrac{d}{dT}\underbrace{\left(  v\cdot v^{-1}\right)
}_{=1}=\dfrac{d}{dT}1=0$, we obtain $\left(  \dfrac{d}{dT}v\right)
v^{-1}+v\dfrac{d}{dT}\left(  v^{-1}\right)  =0$. Solving this equality for
$\dfrac{d}{dT}\left(  v^{-1}\right)  $, we find%
\[
\dfrac{d}{dT}\left(  v^{-1}\right)  =-\dfrac{1}{v}\left(  \dfrac{d}%
{dT}v\right)  v^{-1}=-v^{-2}\left(  \dfrac{d}{dT}v\right)  .
\]


Now, the Leibniz rule (applied to $u$ and $v^{-1}$) yields%
\begin{align*}
\dfrac{d}{dT}\left(  uv^{-1}\right)   &  =\left(  \dfrac{d}{dT}u\right)
v^{-1}+u\underbrace{\dfrac{d}{dT}\left(  v^{-1}\right)  }_{=-v^{-2}\left(
\dfrac{d}{dT}v\right)  }=\left(  \dfrac{d}{dT}u\right)  v^{-1}+u\left(
-v^{-2}\left(  \dfrac{d}{dT}v\right)  \right) \\
&  =v^{-2}\underbrace{\left(  \left(  \dfrac{d}{dT}u\right)  \cdot v-\left(
\dfrac{d}{dT}v\right)  \cdot u\right)  }_{\substack{=0\\\text{(since }\left(
\dfrac{d}{dT}u\right)  \cdot v=\left(  \dfrac{d}{dT}v\right)  \cdot u\text{)}%
}}=v^{-2}0=0=\dfrac{d}{dT}1.
\end{align*}
Moreover, the power series $uv^{-1}$ and $1$ have the same constant
term\footnote{\textit{Proof.} The power series $v$ has constant term $1$.
Hence, its inverse $v^{-1}$ has constant term $1^{-1}=1$. Now, both power
series $u$ and $v^{-1}$ have constant term $1$. Hence, their product $uv^{-1}$
has constant term $1\cdot1=1$. Since the power series $1$ also has constant
term $1$, this shows that the power series $uv^{-1}$ and $1$ have the same
constant term (namely, $1$).}. Hence, Fact I.1 (applied to $uv^{-1}$ and $1$
instead of $u$ and $v$) shows that $uv^{-1}=1$. Thus, $u=v$. This proves Fact I.3.]

The equality (\ref{sol.convolution-series.i.2.log}) (applied to $w=T$) yields
$\dfrac{d}{dT}\left(  \overline{\log}\left[  T\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}T\right)  }_{=1}\cdot\dfrac{1}{1+T}%
=\dfrac{1}{1+T}$. In other words, $\dfrac{d}{dT}\overline{\log}=\dfrac{1}%
{1+T}$ (since $\overline{\log}=\overline{\log}\left[  T\right]  $).

Now, (\ref{sol.convolution-series.i.2.exp}) (applied to $w=\overline{\log}$)
shows that%
\[
\dfrac{d}{dT}\left(  \overline{\exp}\left[  \overline{\log}\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}\overline{\log}\right)  }_{=\dfrac{1}{1+T}%
}\cdot\exp\left[  \overline{\log}\right]  =\dfrac{1}{1+T}\cdot\exp\left[
\overline{\log}\right]  .
\]


But $\overline{\exp}=\exp-1$ and thus $\exp=\overline{\exp}+1$. Substituting
$\overline{\log}$ for $T$ in this equality, we find $\exp\left[
\overline{\log}\right]  =\overline{\exp}\left[  \overline{\log}\right]  +1$.
Hence,%
\[
\dfrac{d}{dT}\left(  \exp\left[  \overline{\log}\right]  \right)  =\dfrac
{d}{dT}\left(  \overline{\exp}\left[  \overline{\log}\right]  +1\right)
=\dfrac{d}{dT}\overline{\exp}\left[  \overline{\log}\right]
+\underbrace{\dfrac{d}{dT}1}_{=0}=\dfrac{d}{dT}\overline{\exp}\left[
\overline{\log}\right]  =\dfrac{1}{1+T}\cdot\exp\left[  \overline{\log
}\right]  .
\]
Multiplying this equality by $1+T$, we find%
\[
\left(  \dfrac{d}{dT}\left(  \exp\left[  \overline{\log}\right]  \right)
\right)  \cdot\left(  1+T\right)  =\exp\left[  \overline{\log}\right]  .
\]
Comparing this with $\underbrace{\left(  \dfrac{d}{dT}\left(  1+T\right)
\right)  }_{=1}\cdot\exp\left[  \overline{\log}\right]  =\exp\left[
\overline{\log}\right]  $, we find
\[
\left(  \dfrac{d}{dT}\left(  \exp\left[  \overline{\log}\right]  \right)
\right)  \cdot\left(  1+T\right)  =\left(  \dfrac{d}{dT}\left(  1+T\right)
\right)  \cdot\exp\left[  \overline{\log}\right]  .
\]
Since both power series $\exp\left[  \overline{\log}\right]  $ and $1+T$ have
constant term $1$\ \ \ \ \footnote{\textit{Proof.} It is clear that the power
series $1+T$ has constant term $1$. Thus, it remain to prove that the power
series $\exp\left[  \overline{\log}\right]  $ has constant term $1$.
\par
Recall that the power series $\overline{\exp}\left[  \overline{\log}\right]  $
has constant term $0$. Hence, the power series $\overline{\exp}\left[
\overline{\log}\right]  +1$ has constant term $0+1=1$. In other words, the
power series $\exp\left[  \overline{\log}\right]  $ has constant term $1$
(since $\exp\left[  \overline{\log}\right]  =\overline{\exp}\left[
\overline{\log}\right]  +1$). Qed.}, we can thus apply Fact I.3 to
$u=\exp\left[  \overline{\log}\right]  $ and $v=1+T$. We thus conclude that
$\exp\left[  \overline{\log}\right]  =1+T$. Comparing this with $\exp\left[
\overline{\log}\right]  =\overline{\exp}\left[  \overline{\log}\right]  +1$,
we obtain $\overline{\exp}\left[  \overline{\log}\right]  +1=1+T$. Subtracting
$1$ from this equality, we find $\overline{\exp}\left[  \overline{\log
}\right]  =T$.

The equality (\ref{sol.convolution-series.i.2.exp}) (applied to $w=T$) yields
$\dfrac{d}{dT}\left(  \overline{\exp}\left[  T\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}T\right)  }_{=1}\cdot\underbrace{\exp\left[
T\right]  }_{=\exp}=\exp$. In other words, $\dfrac{d}{dT}\overline{\exp}=\exp$
(since $\overline{\exp}=\overline{\exp}\left[  T\right]  $).

On the other hand, (\ref{sol.convolution-series.i.2.log}) (applied to
$w=\overline{\exp}$) shows that%
\[
\dfrac{d}{dT}\left(  \overline{\log}\left[  \overline{\exp}\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}\overline{\exp}\right)  }_{=\exp
=\overline{\exp}+1=1+\overline{\exp}}\cdot\dfrac{1}{1+\overline{\exp}}=\left(
1+\overline{\exp}\right)  \cdot\dfrac{1}{1+\overline{\exp}}=1=\dfrac{d}{dT}T.
\]
Since the two power series $\overline{\log}\left[  \overline{\exp}\right]  $
and $T$ have the same constant term\footnote{This is because the power series
$\overline{\log}\left[  \overline{\exp}\right]  $ has constant term $0$, and
the power series $T$ also has constant term $0$.}, we can thus apply Fact I.1
to $u=\overline{\log}\left[  \overline{\exp}\right]  $ and $v=T$. We thus
conclude that $\overline{\log}\left[  \overline{\exp}\right]  =T$. The proof
of Proposition~\ref{prop.exp-log-inv} is thus complete.
\end{proof}

\begin{proof}
[Proof of Lemma~\ref{lem.exp-log-log}.]The power series $\overline{\log}$ has
constant term $0$ (by Proposition~\ref{prop.exp-log-inv}). Hence,
Proposition~\ref{prop.convolution-series}(f) (applied to $v=\overline{\log}$
and $f=g-u_{A}\epsilon_{C}$) yields that $\overline{\log}^{\star}\left(
g-u_{A}\epsilon_{C}\right)  \in\mathfrak{n}\left(  C,A\right)  $. This proves
Lemma~\ref{lem.exp-log-log}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.exp-log}.](a) Let $f\in\mathfrak{n}\left(
C,A\right)  $. We must prove that $\exp^{\star}f-u_{A}\epsilon_{C}%
\in\mathfrak{n}\left(  C,A\right)  $ and $\log^{\star}\left(  \exp^{\star
}f\right)  =f$.

The power series $\overline{\exp}$ has constant term $0$ (by
Proposition~\ref{prop.exp-log-inv}). Hence,
Proposition~\ref{prop.convolution-series}(f) (applied to $v=\overline{\exp}$)
yields that $\overline{\exp}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $. But
$\overline{\exp}=\exp-1$, so that $\exp=\overline{\exp}+1$. Hence,%
\begin{align*}
\exp^{\star}\left(  f\right)   &  =\left(  \overline{\exp}+1\right)  ^{\star
}\left(  f\right)  =\underbrace{\overline{\exp}^{\star}\left(  f\right)
}_{=\overline{\exp}^{\star}f}+\underbrace{1^{\star}\left(  f\right)
}_{\substack{=u_{A}\epsilon_{C}\\\text{(by
(\ref{eq.exe.convolution-series.b.5}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.exe.convolution-series.b.1})}\right) \\
&  =\overline{\exp}^{\star}f+u_{A}\epsilon_{C}.
\end{align*}
Solving this equation for $\overline{\exp}^{\star}f$, we obtain
$\overline{\exp}^{\star}f=\exp^{\star}\left(  f\right)  -u_{A}\epsilon_{C}
=\exp^{\star}f-u_{A}\epsilon_{C}$.
Hence, $\exp^{\star}f-u_{A}\epsilon_{C}=\overline{\exp
}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $. It thus remains to prove that
$\log^{\star}\left(  \exp^{\star}f\right)  =f$.

The map $\exp^{\star}f$ satisfies $\exp^{\star}f-u_{A}\epsilon_{C}%
\in\mathfrak{n}\left(  C,A\right)  $. Hence, the map $\log^{\star}\left(
\exp^{\star}f\right)  \in\Hom \left(  C,A\right)  $ is
well-defined, and satisfies
\begin{align}
\log^{\star}\left(  \exp^{\star}f\right)   &  =\overline{\log}^{\star}\left(
\underbrace{\exp^{\star}f-u_{A}\epsilon_{C}}_{=\overline{\exp}^{\star}%
f}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\log
^{\star}\left(  \exp^{\star}f\right)  \right) \nonumber\\
&  =\overline{\log}^{\star}\left(  \overline{\exp}^{\star}f\right)  .
\label{sol.convolution-series.k.2}%
\end{align}


But the power series $\overline{\exp}$ has constant term $0$. Thus,
(\ref{eq.exe.convolution-series.g.1}) (applied to $u=\overline{\log}$ and
$v=\overline{\exp}$) yields $\left(  \overline{\log}\left[  \overline{\exp
}\right]  \right)  ^{\star}\left(  f\right)  =\overline{\log}^{\star}\left(
\overline{\exp}^{\star}\left(  f\right)  \right)  =\overline{\log}^{\star
}\left(  \overline{\exp}^{\star}f\right)  $. Since $\overline{\log}\left[
\overline{\exp}\right]  =T$ (by Proposition~\ref{prop.exp-log-inv}), this
rewrites as $T^{\star}\left(  f\right)  =\overline{\log}^{\star}\left(
\overline{\exp}^{\star}f\right)  $.

Applying (\ref{eq.exe.convolution-series.a.monom}) to $k=1$, we obtain
$\left(  T^{1}\right)  ^{\star}\left(  f\right)  =f^{\star1}=f$. Since
$T^{1}=T$, this rewrites as $T^{\star}\left(  f\right)  =f$. Compared with
$T^{\star}\left(  f\right)  =\overline{\log}^{\star}\left(  \overline{\exp
}^{\star}f\right)  $, this yields $\overline{\log}^{\star}\left(
\overline{\exp}^{\star}f\right)  =f$. Now, (\ref{sol.convolution-series.k.2})
becomes $\log^{\star}\left(  \exp^{\star}f\right)  =\overline{\log}^{\star
}\left(  \overline{\exp}^{\star}f\right)  =f$. This proves
Proposition~\ref{prop.exp-log}(a).

(b) Let $g\in\Hom \left(  C,A\right)  $ be such that
$g-u_{A}\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $. We must prove that
$\exp^{\star}\left(  \log^{\star}g\right)  =g$.

Set $f=g-u_{A}\epsilon_{C}$. Thus, $f=g-u_{A}\epsilon_{C}\in\mathfrak{n}%
\left(  C,A\right)  $.

The power series $\overline{\log}$ has constant term $0$ (by
Proposition~\ref{prop.exp-log-inv}). Hence,
Proposition~\ref{prop.convolution-series}(f) (applied to $v=\overline{\log}$)
yields that $\overline{\log}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $.

The definition of $\log^{\star}g$ yields $\log^{\star}g=\overline{\log}%
^{\star}\underbrace{\left(  g-u_{A}\epsilon_{C}\right)  }_{=f}=\overline{\log
}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $.

But the power series $\overline{\log}$ has constant term $0$. Thus,
(\ref{eq.exe.convolution-series.g.1}) (applied to $u=\overline{\exp}$ and
$v=\overline{\log}$) yields $\left(  \overline{\exp}\left[  \overline{\log
}\right]  \right)  ^{\star}\left(  f\right)  =\overline{\exp}^{\star}\left(
\overline{\log}^{\star}\left(  f\right)  \right)  =\overline{\exp}^{\star
}\left(  \overline{\log}^{\star}f\right)  $. Since $\overline{\exp}\left[
\overline{\log}\right]  =T$ (by Proposition~\ref{prop.exp-log-inv}), this
rewrites as $T^{\star}\left(  f\right)  =\overline{\exp}^{\star}\left(
\overline{\log}^{\star}f\right)  $.

Applying (\ref{eq.exe.convolution-series.a.monom}) to $k=1$, we obtain
$\left(  T^{1}\right)  ^{\star}\left(  f\right)  =f^{\star1}=f$. Since
$T^{1}=T$, this rewrites as $T^{\star}\left(  f\right)  =f$. Compared with
$T^{\star}\left(  f\right)  =\overline{\exp}^{\star}\left(  \overline{\log
}^{\star}f\right)  $, this yields $\overline{\exp}^{\star}\left(
\overline{\log}^{\star}f\right)  =f$.

But $\overline{\exp}=\exp-1$, so that $\exp=\overline{\exp}+1$. Hence,%
\begin{align*}
\exp^{\star}\left(  \log^{\star}g\right)   &  =\left(  \overline{\exp
}+1\right)  ^{\star}\left(  \underbrace{\log^{\star}g}_{=\overline{\log
}^{\star}f}\right)  =\left(  \overline{\exp}+1\right)  ^{\star}\left(
\overline{\log}^{\star}f\right)  =\underbrace{\overline{\exp}^{\star}\left(
\overline{\log}^{\star}f\right)  }_{=f=g-u_{A}\epsilon_{C}}%
+\underbrace{1^{\star}\left(  \overline{\log}^{\star}f\right)  }%
_{\substack{=u_{A}\epsilon_{C}\\\text{(by (\ref{eq.exe.convolution-series.b.5}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by an application of
(\ref{eq.exe.convolution-series.b.1})}\right) \\
&  =\left(  g-u_{A}\epsilon_{C}\right)  +u_{A}\epsilon_{C}=g.
\end{align*}
This proves Proposition~\ref{prop.exp-log}(b).

(c) Let $f,g\in\mathfrak{n}\left(  C,A\right)  $ be such that $f\star g=g\star
f$. Then, Proposition~\ref{prop.convolution-series}(c) shows that
$f+g\in\mathfrak{n}\left(  C,A\right)  $. Hence, $\exp^{\star}\left(
f+g\right)  $ is well-defined.

Let us recall the following facts, which have been proven in the proof of
Proposition~\ref{prop.convolution-series}(c) (during the solution to
Exercise~\ref{exe.convolution-series}):

\begin{itemize}
\item We have%
\begin{equation}
\left(  f+g\right)  ^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\star
g^{\star\left(  n-i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in \NN . \label{sol.convolution-series.m.binom}%
\end{equation}


\item The family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported.

\item For each $x\in C$,
\begin{equation}
\text{the family }\left(  \left(  f^{\star q}\star g^{\star r}\right)  \left(
x\right)  \right)  _{\left(  q,r\right)  \in \NN\times\NN}\in
A^{\NN\times\NN}\text{ is finitely supported.}
\label{sol.convolution-series.m.fs}%
\end{equation}

\end{itemize}

Furthermore, the family $\left(  \dfrac{f^{\star q}}{q!}\right)
_{q\in\NN}$ is pointwise finitely supported\footnote{\textit{Proof.} We
know that the family $\left(  f^{\star n}\right)  _{n\in\NN}$ is
pointwise finitely supported. Hence, Proposition~\ref{prop.pw-fin-supp.P4}
(applied to $Q=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star
n}\right)  _{n\in\NN}$ and $\left(  \lambda_{q}\right)  _{q\in
Q}=\left(  \dfrac{1}{n!}\right)  _{n\in\NN}$) shows that the family
$\left(  \dfrac{1}{n!}f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported. Since $\dfrac{1}{n!}f^{\star n}=\dfrac{f^{\star n}}{n!}$
for each $n\in \NN$, this result rewrites as follows: The family
$\left(  \dfrac{f^{\star n}}{n!}\right)  _{n\in\NN}$ is pointwise
finitely supported. Renaming the index $n$ as $q$ in this statement, we obtain
the following: The family $\left(  \dfrac{f^{\star q}}{q!}\right)
_{q\in\NN}$ is pointwise finitely supported. Qed.}. Similarly, the
family $\left(  \dfrac{g^{\star r}}{r!}\right)  _{r\in\NN}$ is
pointwise finitely supported. Hence, Proposition~\ref{prop.pw-fin-supp.P3}
(applied to $Q=\NN$, $R=\NN$, $\left(  f_{q}\right)  _{q\in
Q}=\left(  \dfrac{f^{\star q}}{q!}\right)  _{q\in\NN}$ and $\left(
g_{r}\right)  _{r\in R}=\left(  \dfrac{g^{\star r}}{r!}\right)  _{r\in
\NN}$) shows that the family $\left(  \dfrac{f^{\star q}}{q!}%
\star\dfrac{g^{\star r}}{r!}\right)  _{\left(  q,r\right)  \in
\NN\times\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN\times\NN}$ is pointwise finitely supported, and that it
satisfies%
\begin{equation}
\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(  \dfrac
{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  =\left(
\sum_{q\in\NN}\dfrac{f^{\star q}}{q!}\right)  \star\left(
\sum_{r\in\NN}\dfrac{g^{\star r}}{r!}\right)  .
\label{sol.convolution-series.m.prodfam}
\end{equation}


Let $x\in C$. Then, the family $\left(  \left(  f^{\star q}\star g^{\star
r}\right)  \left(  x\right)  \right)  _{\left(  q,r\right)  \in
\NN\times\NN}\in A^{\NN\times\NN}$ is finitely supported (by
(\ref{sol.convolution-series.m.fs})). In other words, all but finitely many
$\left(  q,r\right)  \in \NN\times\NN$ satisfy $\left(  f^{\star
q}\star g^{\star r}\right)  \left(  x\right)  =0$. In other words, there
exists a finite subset $K$ of $\NN\times\NN$ such that%
\begin{equation}
\text{each }\left(  q,r\right)  \in\left(  \NN\times\NN\right)
\setminus K\text{ satisfies }\left(  f^{\star q}\star g^{\star r}\right)
\left(  x\right)  =0. \label{sol.convolution-series.m.3}%
\end{equation}
Consider this $K$. Then,%
\begin{equation}
\text{each }\left(  q,r\right)  \in \NN\times\NN \text{ satisfying
}\left(  q,r\right)  \notin K\text{ satisfies }\left(  \dfrac{f^{\star q}}%
{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  =0
\label{sol.convolution-series.m.3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.m.3b}):} Let $\left(
q,r\right)  \in \NN\times\NN$ be such that $\left(  q,r\right)
\notin K$. Then, $\left(  q,r\right)  \in\left(  \NN\times
\NN\right)  \setminus K$ (since $\left(  q,r\right)  \in\NN\times\NN$
but $\left(  q,r\right)  \notin K$). Thus,
\[
\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  =\dfrac{1}{q!}\cdot\dfrac{1}{r!}\cdot\underbrace{\left(  f^{\star
q}\star g^{\star r}\right)  \left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.m.3}))}}}=\dfrac{1}{q!}\cdot\dfrac{1}{r!}%
\cdot0=0.
\]
This proves (\ref{sol.convolution-series.m.3b}).}.

Let $Q=\left\{  u+v\ \mid\ \left(  u,v\right)  \in K\right\}  $. Thus, $Q$ is
a finite set (since $K$ is a finite set). Also, $Q\subset \NN$ (since
each $\left(  u,v\right)  \in K$ satisfies $\left(  u,v\right)  \in
K\subset \NN\times\NN$ and thus $u+v\in \NN$). Thus, the
set $\NN$ is the union of its two disjoint subsets $Q$ and
$\NN \setminus Q$.

But
\begin{equation}
\text{every }n\in \NN \setminus Q\text{ satisfies }\left(  f+g\right)
^{\star n}\left(  x\right)  =0 \label{sol.convolution-series.m.5}%
\end{equation}
\footnote{This has already been proven during our proof of
Proposition~\ref{prop.convolution-series}(c).}.

Let $n\in \NN$. Let us first observe that the map
\begin{align*}
\left\{  0,1,\ldots,n\right\}   &  \rightarrow\left\{  \left(  q,r\right)
\in \NN\times\NN\ \mid\ q+r=n\right\}  ,\\
i  &  \mapsto\left(  i,n-i\right)
\end{align*}
is a bijection. Hence, we can substitute $\left(  i,n-i\right)  $ for $\left(
q,r\right)  $ in the sum $\sum\limits_{\substack{\left(  q,r\right)  \in
\NN\times\NN;\\q+r=n}}\dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}$. We thus obtain%
\begin{equation}
\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times\NN%
;\\q+r=n}}\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}=
\sum_{i\in\left\{  0,1,\ldots,n\right\}  }\dfrac{f^{\star i}}{i!}\star
\dfrac{g^{\star\left(  n-i\right)  }}{\left(  n-i\right)  !}.
\label{sol.convolution-series.m.subs-sum}%
\end{equation}


But (\ref{sol.convolution-series.m.binom}) yields%
\begin{align*}
\left(  f+g\right)  ^{\star n}  &  =\underbrace{\sum_{i=0}^{n}}_{=
\sum_{i\in\left\{  0,1,\ldots,n\right\}  }}
\underbrace{\dbinom{n}{i}}_{=\dfrac
{n!}{i!\left(  n-i\right)  !}}f^{\star i}\star g^{\star\left(  n-i\right)  }\\
&  =\sum_{i\in\left\{  0,1,\ldots,n\right\}  }\dfrac{n!}{i!\left(  n-i\right)
!}f^{\star i}\star g^{\star\left(  n-i\right)  }=n!\cdot\sum_{i\in\left\{
0,1,\ldots,n\right\}  }\dfrac{f^{\star i}}{i!}\star\dfrac{g^{\star\left(
n-i\right)  }}{\left(  n-i\right)  !}.
\end{align*}
Multiplying this equality by $\dfrac{1}{n!}$, we obtain
\begin{equation}
\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}=\sum_{i\in\left\{  0,1,\ldots
,n\right\}  }\dfrac{f^{\star i}}{i!}\star\dfrac{g^{\star\left(  n-i\right)  }%
}{\left(  n-i\right)  !}=\sum\limits_{\substack{\left(  q,r\right)  \in
\NN\times\NN;\\q+r=n}}\dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!} \label{sol.convolution-series.m.8}%
\end{equation}
(by (\ref{sol.convolution-series.m.subs-sum})). Applying both sides of this
equality to $x$, we find%
\begin{align}
\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)   &
=\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times\NN%
;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)
\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in\NN%
\times\NN;\\q+r=n;\\\left(  q,r\right)  \in K}}}_{\substack{=
\sum\limits_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\\
\text{(since }K\subset
\NN\times\NN\text{)}}}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  +\sum\limits_{\substack{\left(
q,r\right)  \in \NN\times\NN;\\q+r=n;\\\left(  q,r\right)  \notin
K}}\underbrace{\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}%
{r!}\right)  \left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.m.3b}))}}}\nonumber\\
&  =\sum\limits_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\left(  \dfrac{f^{\star
q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
+\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times
\NN;\\q+r=n;\\\left(  q,r\right)  \notin K}}0}_{=0}=
\sum\limits_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}
\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)
\left(  x\right)  .
\label{sol.convolution-series.m.9}
\end{align}


Now, forget that we fixed $n$. We thus have proven
(\ref{sol.convolution-series.m.9}) for each $n\in \NN$.

From $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$, we obtain
\begin{align*}
\exp^{\star}\left(  f+g\right)   &  =\underbrace{\sum_{n\geq0}}_{=
\sum_{n\in\NN}}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\exp^{\star}\left(
f+g\right)  \right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}.
\end{align*}
Applying both sides of this equality to $x$, we find%
\begin{align}
\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)   &
=\left(  \sum_{n\in\NN}\dfrac{1}{n!}\left(  f+g\right)  ^{\star
n}\right)  \left(  x\right)  =\sum_{n\in\NN}\dfrac{1}{n!}\left(
f+g\right)  ^{\star n}\left(  x\right) \nonumber\\
&  =\sum_{n\in Q}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)
+\sum_{n\in\NN \setminus Q}\dfrac{1}{n!}\underbrace{\left(  f+g\right)
^{\star n}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.m.5}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the set }\NN \text{ is the
union of its two disjoint subsets }Q\text{ and }\NN \setminus Q\right)
\nonumber\\
&  =\sum_{n\in Q}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)
+\underbrace{\sum_{n\in \NN \setminus Q}\dfrac{1}{n!}0}_{=0}=\sum_{n\in
Q}\underbrace{\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)
}_{\substack{=\sum\limits_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\\\text{(by (\ref{sol.convolution-series.m.9}))}}}\nonumber\\
&  =\underbrace{\sum_{n\in Q}\sum\limits_{\substack{\left(  q,r\right)  \in
K;\\q+r=n}}}_{\substack{=\sum_{\left(  q,r\right)  \in K}\sum\limits_{\substack{n\in
Q;\\q+r=n}}\\\text{(since both sums }\sum_{n\in Q}\text{ and }
\sum\limits_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\\
\text{are finite)}}}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\nonumber\\
&  =\sum_{\left(  q,r\right)  \in K}\sum\limits_{\substack{n\in Q;\\q+r=n}}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
. \label{sol.convolution-series.m.11}%
\end{align}
But each $\left(  q,r\right)  \in K$ satisfies
\begin{equation}
\sum\limits_{\substack{n\in Q;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  =\left(  \dfrac{f^{\star q}%
}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\label{sol.convolution-series.m.14}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.m.14}):} Let $\left(
q,r\right)  \in K$. Thus, $q+r\in\left\{  u+v\ \mid\ \left(  u,v\right)  \in
K\right\}  =Q$ (since $Q=\left\{  u+v\ \mid\ \left(  u,v\right)  \in
K\right\}  $).
\par
Hence, there exists some $n\in Q$ satisfying $q+r=n$ (namely, $n=q+r$).
Furthermore, this $n$ is unique (because the condition $q+r=n$ clearly
determines $n$ uniquely). Thus, there exists a \textbf{unique} $n\in Q$
satisfying $q+r=n$. Therefore, the sum $\sum\limits_{\substack{n\in Q;\\q+r=n}%
}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  $ has exactly one addend. Consequently, this sum simplifies as
follows:%
\[
\sum\limits_{\substack{n\in Q;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  =\left(  \dfrac{f^{\star q}%
}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  .
\]
This proves (\ref{sol.convolution-series.m.14}).}.

Hence, (\ref{sol.convolution-series.m.11}) becomes%
\begin{equation}
\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)
=\sum_{\left(  q,r\right)  \in K}\underbrace{\sum\limits_{\substack{n\in
Q;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)
\left(  x\right)  }_{\substack{=\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  \\\text{(by
(\ref{sol.convolution-series.m.14}))}}}=\sum_{\left(  q,r\right)  \in
K}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  . \label{sol.convolution-series.m.21}%
\end{equation}


From $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$, we obtain
\begin{align*}
\exp^{\star}f  &  =\underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}%
\dfrac{1}{n!}f^{\star n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\exp^{\star}f\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}.
\end{align*}
The same argument (applied to $g$ instead of $f$) shows that $\exp^{\star
}g=\sum_{n\in\NN}\dfrac{1}{n!}g^{\star n}$. Now,%
\begin{align*}
\underbrace{\left(  \exp^{\star}f\right)  }_{\substack{=\sum_{n\in\NN}
\dfrac{1}{n!}f^{\star n}\\=\sum_{q\in\NN}\dfrac{1}{q!}f^{\star q}%
}}\star\underbrace{\left(  \exp^{\star}g\right)  }_{\substack{=
\sum_{n\in\NN}\dfrac{1}{n!}g^{\star n}\\=\sum_{r\in\NN}\dfrac{1}%
{r!}g^{\star r}}}  &  =\left(  \sum_{q\in\NN}\underbrace{\dfrac{1}%
{q!}f^{\star q}}_{=\dfrac{f^{\star q}}{q!}}\right)  \star\left(  \sum
_{r\in\NN}\underbrace{\dfrac{1}{r!}g^{\star r}}_{=\dfrac{g^{\star r}%
}{r!}}\right)  =\left(  \sum_{q\in\NN}\dfrac{f^{\star q}}{q!}\right)
\star\left(  \sum_{r\in\NN}\dfrac{g^{\star r}}{r!}\right) \\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.m.prodfam}%
)}\right)  .
\end{align*}
Applying both sides of this equality to $x$, we obtain%
\begin{align*}
\left(  \left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}g\right)
\right)  \left(  x\right)   &  =\left(  \sum_{\left(  q,r\right)
\in \NN\times\NN}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \right)  \left(  x\right)  =\sum_{\left(
q,r\right)  \in \NN\times\NN}\left(  \dfrac{f^{\star q}}{q!}%
\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right) \\
&  =\underbrace{\sum\limits_{\substack{\left(  q,r\right) \in \NN\times\NN;
\\\left(  q,r\right)  \in K}}}_{\substack{=\sum_{\left(
q,r\right)  \in K}\\\text{(since }K\subset \NN\times\NN\text{)}%
}}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  +\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times
\NN;\\\left(  q,r\right)  \notin K}}\underbrace{\left(  \dfrac{f^{\star
q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.convolution-series.m.3b}))}}}\\
&  =\sum_{\left(  q,r\right)  \in K}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  +\underbrace{\sum\limits
_{\substack{\left(  q,r\right)  \in \NN\times\NN;\\\left(
q,r\right)  \notin K}}0}_{=0}=\sum_{\left(  q,r\right)  \in K}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\\
&  =\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.m.21}%
)}\right)  .
\end{align*}


Now, forget that we fixed $x$. We thus have shown that $\left(  \left(
\exp^{\star}f\right)  \star\left(  \exp^{\star}g\right)  \right)  \left(
x\right)  =\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)
$ for each $x\in C$. In other words, $\left(  \exp^{\star}f\right)
\star\left(  \exp^{\star}g\right)  =\exp^{\star}\left(  f+g\right)  $. In
other words, $\exp^{\star}\left(  f+g\right)  =\left(  \exp^{\star}f\right)
\star\left(  \exp^{\star}g\right)  $. This completes the proof of
Proposition~\ref{prop.exp-log}(c).

(d) Consider the $\kk$-linear map $0:C\rightarrow A$. It satisfies
$0\in\mathfrak{n}\left(  C,A\right)  $\ \ \ \ \footnote{This can
easily be checked by the reader (using the fact that $0^{\star n} = 0$
for every positive integer $n$).}. It remains to show that
$\exp^{\star}0=u_{A}\epsilon_{C}$.

We have $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$. Thus, the definition of
$\exp^{\star}0$ yields%
\begin{align*}
\exp^{\star}0  &  =\sum_{n\geq0}\dfrac{1}{n!}0^{\star n}=\underbrace{\dfrac
{1}{0!}}_{=\dfrac{1}{1}=1}0^{\star0}+\sum_{n\geq1}\dfrac{1}{n!}%
\underbrace{0^{\star n}}_{\substack{=0\\\text{(since }n\geq1\text{)}%
}}=0^{\star0}+\underbrace{\sum_{n\geq1}\dfrac{1}{n!}0}_{=0}\\
&  =0^{\star0}=\left(  \text{the unity of the }\kk\text{-algebra
}\left(  \Hom \left(  C,A\right)  ,\star\right)  \right)
=u_{A}\epsilon_{C}.
\end{align*}
Thus, Proposition~\ref{prop.exp-log}(d) is proven.

(e) Let $f\in\mathfrak{n}\left(  C,A\right)  $. We must prove that each
$n\in \NN$ satisfies%
\begin{equation}
nf\in\mathfrak{n}\left(  C,A\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(  nf\right)  =\left(  \exp^{\star
}f\right)  ^{\star n}. \label{sol.convolution-series.n.goal}%
\end{equation}


[\textit{Proof of (\ref{sol.convolution-series.n.goal}):} We shall prove
(\ref{sol.convolution-series.n.goal}) by induction over $n$:

\textit{Induction base:} We have $0f\in\mathfrak{n}\left(  C,A\right)
$\ \ \ \ \footnote{\textit{Proof.} Proposition~\ref{prop.convolution-series}%
(d) (applied to $\lambda=0\cdot1_{\kk}$) yields $0\cdot1_{\kk%
}f\in\mathfrak{n}\left(  C,A\right)  $. Thus, $0f=0=0\cdot1_{\kk}%
f\in\mathfrak{n}\left(  C,A\right)  $.}. Furthermore,%
\begin{align*}
\exp^{\star}\left(  \underbrace{0f}_{=0}\right)   &  =\exp^{\star}%
0=u_{A}\epsilon_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{prop.exp-log}(d)}\right) \\
&  =\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right)  =\left(
\exp^{\star}f\right)  ^{\star0}.
\end{align*}
Thus, we have shown that $0f\in\mathfrak{n}\left(  C,A\right)  $ and
$\exp^{\star}\left(  0f\right)  =\left(  \exp^{\star}f\right)  ^{\star0}$. In
other words, (\ref{sol.convolution-series.n.goal}) holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Let $k\in \NN$. Assume that
(\ref{sol.convolution-series.n.goal}) holds for $n=k$. We must prove that
(\ref{sol.convolution-series.n.goal}) holds for $n=k+1$.

We have assumed that (\ref{sol.convolution-series.n.goal}) holds for $n=k$. In
other words,%
\[
kf\in\mathfrak{n}\left(  C,A\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(  kf\right)  =\left(  \exp^{\star
}f\right)  ^{\star k}.
\]


Now, $f\star\left(  kf\right)  =kf\star f=\left(  kf\right)  \star f$. Hence,
Proposition~\ref{prop.exp-log}(c) (applied to $g=kf$) yields that
$f+kf\in\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}\left(
f+kf\right)  =\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}\left(
kf\right)  \right)  $. Now, $\left(  k+1\right)  f=f+kf\in\mathfrak{n}\left(
C,A\right)  $ and%
\[
\exp^{\star}\left(  \underbrace{\left(  k+1\right)  f}_{=f+kf}\right)
=\exp^{\star}\left(  f+kf\right)  =\left(  \exp^{\star}f\right)
\star\underbrace{\left(  \exp^{\star}\left(  kf\right)  \right)  }_{=\left(
\exp^{\star}f\right)  ^{\star k}}=\left(  \exp^{\star}f\right)  \star\left(
\exp^{\star}f\right)  ^{\star k}=\left(  \exp^{\star}f\right)  ^{\star\left(
k+1\right)  }.
\]


Thus, we have shown that%
\[
\left(  k+1\right)  f\in\mathfrak{n}\left(  C,A\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(  \left(
k+1\right)  f\right)  =\left(  \exp^{\star}f\right)  ^{\star\left(
k+1\right)  }.
\]
In other words, (\ref{sol.convolution-series.n.goal}) holds for $n=k+1$. This
completes the induction step. Thus, the induction proof of
(\ref{sol.convolution-series.n.goal}) is finished.]

Now, (\ref{sol.convolution-series.n.goal}) is proven. In other words,
Proposition~\ref{prop.exp-log}(e) is proven.

(f) Let $f\in\mathfrak{n}\left(  C,A\right)  $. Define a map $g\in
\Hom \left(  C,A\right)  $ by $g=f+u_{A}\epsilon_{C}$. Thus,
$g-u_{A}\epsilon_{C}=f\in\mathfrak{n}\left(  C,A\right)  $. Hence,
$\log^{\star}g$ is well-defined.

For each $n\in \NN$, define an element $\lambda_{n}\in\kk$ by
$\lambda_{n}=
\begin{cases}
\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}, & \text{if }n\geq1;\\
0, & \text{if }n=0
\end{cases} \quad
$. Then, $\lambda_{0}=0$, whereas
\begin{equation}
\text{every integer }n\geq1\text{ satisfies }\lambda_{n}=\dfrac{\left(
-1\right)  ^{n-1}}{n}1_{\kk}. \label{pf.prop.exp-log.f.1}%
\end{equation}
Hence,
\begin{align*}
\sum_{n\geq0}\lambda_{n}T^{n}  &  =\underbrace{\lambda_{0}}_{=0}T^{0}%
+\sum_{n\geq1}\underbrace{\lambda_{n}}_{\substack{=\dfrac{\left(  -1\right)
^{n-1}}{n}1_{\kk}\\\text{(by (\ref{pf.prop.exp-log.f.1}))}}%
}T^{n}=\underbrace{0T^{0}}_{=0}+\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}%
}{n}\underbrace{1_{\kk}T^{n}}_{=T^{n}}\\
&  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}T^{n}=\log\left(
1+T\right)  =\overline{\log}.
\end{align*}
In other words, $\overline{\log}=\sum_{n\geq0}\lambda_{n}T^{n}$. Hence, the
definition of $\overline{\log}^{\star}f$ yields%
\[
\overline{\log}^{\star}f=\sum_{n\geq0}\lambda_{n}f^{\star n}%
=\underbrace{\lambda_{0}}_{=0}f^{\star0}+\sum_{n\geq1}\underbrace{\lambda_{n}%
}_{\substack{=\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}\\\text{(by
(\ref{pf.prop.exp-log.f.1}))}}}f^{\star n}=\underbrace{0f^{\star0}}_{=0}%
+\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}\underbrace{1_{\kk%
}f^{\star n}}_{=f^{\star n}}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}%
{n}f^{\star n}.
\]
Now, the definition of $\log^{\star}g$ yields
\[
\log^{\star}g=\overline{\log}^{\star}f=\sum_{n\geq1}\dfrac{\left(  -1\right)
^{n-1}}{n}f^{\star n}.
\]
Since $g=f+u_{A}\epsilon_{C}$, this rewrites as $\log^{\star}\left(
f+u_{A}\epsilon_{C}\right)  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}%
{n}f^{\star n}$. This proves Proposition~\ref{prop.exp-log}(f).
\end{proof}

Thus, Proposition~\ref{prop.exp-log-inv}, Lemma~\ref{lem.exp-log-log} and
Proposition~\ref{prop.exp-log} are proven. Hence, Exercise~\ref{exe.exp-log}
is solved.

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.exp-log}}

\textit{Solution to Exercise~\ref{exe.exp-log}.}

\begin{proof}
[Proof of Proposition~\ref{prop.exp-log-inv}.]%
Proposition~\ref{prop.exp-log-inv} is a well-known fact that is often used in
enumerative combinatorics (for computing generating functions). We shall give
a purely algebraic proof (somewhat similar to the one given in \cite[Example
7.67]{Loehr-bij}). Other proofs (some combinatorial, some analytic) can be
found in the literature.

The proof will rely on several simple facts about power series. Keep in mind
that all of the following facts assume that $\kk$ is a commutative
$\QQ$-algebra.

First, we notice that%
\begin{align}
\overline{\exp}  &  =\underbrace{\exp}_{=\sum_{n\geq0}\dfrac{1}{n!}T^{n}%
}-1=\sum_{n\geq0}\dfrac{1}{n!}T^{n}-1=\underbrace{\dfrac{1}{0!}}_{=\dfrac
{1}{1}=1}\underbrace{T^{0}}_{=1}+\sum_{n\geq1}\dfrac{1}{n!}T^{n}-1\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \nonumber\\
&  =1+\sum_{n\geq1}\dfrac{1}{n!}T^{n}-1=\sum_{n\geq1}\dfrac{1}{n!}T^{n}.
\label{sol.convolution-series.i.expbar=}%
\end{align}
Hence, the power series $\overline{\exp}$ has constant term $0$. Hence, the
power series $\overline{\log}\left[  \overline{\exp}\right]  $ is well-defined.

Also,%
\begin{equation}
\overline{\log}=\log\left(  1+T\right)  =\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}T^{n}. \label{sol.convolution-series.i.logbar=}%
\end{equation}
Hence, the power series $\overline{\log}$ has constant term $0$. Hence, the
power series $\overline{\exp}\left[  \overline{\log}\right]  $ is well-defined.

For each $n\geq1$, we have%
\begin{equation}
\left(  \text{the constant term of }\overline{\log}^{n}\right)  =0
\label{sol.convolution-series.i.cstlog0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.i.cstlog0}):} Let
$n\geq1$. The power series $\overline{\log}$ is divisible by $T$ (since it has
constant term $0$). Hence, the power series $\overline{\log}^{n}$ is divisible
by $T^{n}$. Thus, the power series $\overline{\log}^{n}$ is also divisible by
$T$ (since $T^{n}$ is divisible by $T$ (since $n\geq1$)), and therefore has
constant term $0$. In other words, we have $\left(  \text{the constant term of
}\overline{\log}^{n}\right)  =0$. This proves
(\ref{sol.convolution-series.i.cstlog0}).}. A similar argument (with the roles
of $\overline{\exp}$ and $\overline{\log}$ switched) shows that for each
$n\geq1$, we have%
\begin{equation}
\left(  \text{the constant term of }\overline{\exp}^{n}\right)  =0.
\label{sol.convolution-series.i.cstexp0}%
\end{equation}


Substituting $\overline{\log}$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.expbar=}), we obtain%
\[
\overline{\exp}\left[  \overline{\log}\right]  =\sum_{n\geq1}\dfrac{1}%
{n!}\overline{\log}^{n}.
\]
Hence,%
\begin{align*}
\left(  \text{the constant term of }\overline{\exp}\left[  \overline{\log
}\right]  \right)   &  =\left(  \text{the constant term of }\sum_{n\geq
1}\dfrac{1}{n!}\overline{\log}^{n}\right) \\
&  =\sum_{n\geq1}\dfrac{1}{n!}\underbrace{\left(  \text{the constant term of
}\overline{\log}^{n}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.i.cstlog0}))}}}=\sum_{n\geq1}\dfrac{1}{n!}0=0.
\end{align*}
In other words, the power series $\overline{\exp}\left[  \overline{\log
}\right]  $ has constant term $0$.

Substituting $\overline{\exp}$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.logbar=}), we obtain%
\[
\overline{\log}\left[  \overline{\exp}\right]  =\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\overline{\exp}^{n}.
\]
Hence,%
\begin{align*}
\left(  \text{the constant term of }\overline{\log}\left[  \overline{\exp
}\right]  \right)   &  =\left(  \text{the constant term of }\sum_{n\geq
1}\dfrac{\left(  -1\right)  ^{n-1}}{n}\overline{\exp}^{n}\right) \\
&  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}\underbrace{\left(
\text{the constant term of }\overline{\exp}^{n}\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.convolution-series.i.cstexp0}))}}%
}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}0=0.
\end{align*}
In other words, the power series $\overline{\log}\left[  \overline{\exp
}\right]  $ has constant term $0$.

\begin{statement}
\textit{Fact I.1:} Let $u\in\kk\left[  \left[  T\right]  \right]  $ and
$v\in\kk\left[  \left[  T\right]  \right]  $ be two power series having
the same constant term. Assume that $\dfrac{d}{dT}u=\dfrac{d}{dT}v$. Then,
$u=v$.
\end{statement}

[\textit{Proof of Fact I.1:} Write the power series $u$ in the form
$u=\sum_{n\geq0}u_{n}T^{n}$ with $\left(  u_{n}\right)  _{n\geq0}\in
\kk^{\NN}$. Thus, $\dfrac{d}{dT}u=\sum_{n\geq1}nu_{n}T^{n-1}$
(by the definition of the derivative).

Write the power series $v$ in the form $v=\sum_{n\geq0}v_{n}T^{n}$ with
$\left(  v_{n}\right)  _{n\geq0}\in\kk^{\NN}$. Thus, $\dfrac
{d}{dT}v=\sum_{n\geq1}nv_{n}T^{n-1}$ (by the definition of the derivative).

Now,%
\[
\sum_{n\geq1}nu_{n}T^{n-1}=\dfrac{d}{dT}u=\dfrac{d}{dT}v=\sum_{n\geq1}%
nv_{n}T^{n-1}.
\]
Comparing coefficients in front of $T^{n-1}$ on both sides of this equality,
we obtain%
\begin{equation}
nu_{n}=nv_{n}\ \ \ \ \ \ \ \ \ \ \text{for each integer }n\geq1.
\label{sol.convolution-series.i.1.pf.1}%
\end{equation}


On the other hand, the power series $u$ has constant term $u_{0}$ (since
$u=\sum_{n\geq0}u_{n}T^{n}$), and the power series $v$ has constant term
$v_{0}$ (similarly). Thus, the constant terms of $u$ and $v$ are $u_{0}$ and
$v_{0}$, respectively. Therefore, $u_{0}=v_{0}$ (since the power series $u$
and $v$ have the same constant term).

Now, each $n\in \NN$ satisfies $u_{n}=v_{n}$%
\ \ \ \ \footnote{\textit{Proof.} Let $n\in \NN$. We must prove that
$u_{n}=v_{n}$.
\par
If $n=0$, then this follows immediately from $u_{0}=v_{0}$. Hence, we WLOG
assume that we don't have $n=0$. Thus, $n\geq1$ (since $n\in \NN$).
Therefore, (\ref{sol.convolution-series.i.1.pf.1}) yields $nu_{n}=nv_{n}$. We
can multiply both sides of this equality by $\dfrac{1}{n}$ (since $\kk$
is a $\QQ$-algebra), and thus obtain $u_{n}=v_{n}$, qed.}. Hence,
$\sum_{n\geq0}\underbrace{u_{n}}_{=v_{n}}T^{n}=\sum_{n\geq0}v_{n}T^{n}$. Thus,
$u=\sum_{n\geq0}u_{n}T^{n}=\sum_{n\geq0}v_{n}T^{n}=v$. This proves Fact I.1.]

\begin{statement}
\textit{Fact I.2:} Let $w\in\kk\left[  \left[  T\right]  \right]  $ be
a power series having constant term $0$. Then,
\begin{equation}
\dfrac{d}{dT}\left(  \overline{\exp}\left[  w\right]  \right)  =\left(
\dfrac{d}{dT}w\right)  \cdot\exp\left[  w\right]
\label{sol.convolution-series.i.2.exp}%
\end{equation}
and%
\begin{equation}
\dfrac{d}{dT}\left(  \overline{\log}\left[  w\right]  \right)  =\left(
\dfrac{d}{dT}w\right)  \cdot\dfrac{1}{1+w}.
\label{sol.convolution-series.i.2.log}%
\end{equation}

\end{statement}

[\textit{Proof of Fact I.2:} It is easy to see that each positive integer $n$
satisfies%
\begin{equation}
\dfrac{d}{dT}\left(  w^{n}\right)  =n\left(  \dfrac{d}{dT}w\right)  w^{n-1}.
\label{sol.convolution-series.i.2.pf.power}%
\end{equation}
(Indeed, (\ref{sol.convolution-series.i.2.pf.power}) can be proven by a
straightforward induction on $n$, using the Leibniz identity $\dfrac{d}%
{dT}\left(  uv\right)  =\left(  \dfrac{d}{dT}u\right)  v+u\dfrac{d}{dT}v$.)

Substituting $w$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.expbar=}), we obtain%
\[
\overline{\exp}\left[  w\right]  =\sum_{n\geq1}\dfrac{1}{n!}w^{n}.
\]
Applying the operator $\dfrac{d}{dT}$ to this equality, we find%
\begin{align*}
\dfrac{d}{dT}\overline{\exp}\left[  w\right]   &  =\dfrac{d}{dT}\sum_{n\geq
1}\dfrac{1}{n!}w^{n}=\sum_{n\geq1}\dfrac{1}{n!}\cdot\underbrace{\dfrac{d}%
{dT}\left(  w^{n}\right)  }_{\substack{=n\left(  \dfrac{d}{dT}w\right)
w^{n-1}\\\text{(by (\ref{sol.convolution-series.i.2.pf.power}))}}}=\sum
_{n\geq1}\underbrace{\dfrac{1}{n!}\cdot n}_{=\dfrac{1}{\left(  n-1\right)  !}%
}\left(  \dfrac{d}{dT}w\right)  w^{n-1}\\
&  =\sum_{n\geq1}\dfrac{1}{\left(  n-1\right)  !}\left(  \dfrac{d}%
{dT}w\right)  w^{n-1}=\sum_{n\geq0}\dfrac{1}{n!}\left(  \dfrac{d}{dT}w\right)
w^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n\text{ for
}n-1\text{ in the sum}\right)  .
\end{align*}
Comparing this with%
\[
\left(  \dfrac{d}{dT}w\right)  \cdot\underbrace{\exp\left[  w\right]
}_{\substack{=\sum_{n\geq0}\dfrac{1}{n!}w^{n}\\\text{(since }\exp=\sum
_{n\geq0}\dfrac{1}{n!}T^{n}\text{)}}}=\left(  \dfrac{d}{dT}w\right)  \cdot
\sum_{n\geq0}\dfrac{1}{n!}w^{n}=\sum_{n\geq0}\dfrac{1}{n!}\left(  \dfrac
{d}{dT}w\right)  w^{n},
\]
we obtain $\dfrac{d}{dT}\left(  \overline{\exp}\left[  w\right]  \right)
=\left(  \dfrac{d}{dT}w\right)  \cdot\exp\left[  w\right]  $. This proves
(\ref{sol.convolution-series.i.2.exp}).

Substituting $w$ for $T$ on both sides of the equality
(\ref{sol.convolution-series.i.logbar=}), we obtain%
\[
\overline{\log}\left[  w\right]  =\sum_{n\geq1}\dfrac{\left(  -1\right)
^{n-1}}{n}w^{n}.
\]
Applying the operator $\dfrac{d}{dT}$ to this equality, we find%
\begin{align*}
\dfrac{d}{dT}\overline{\log}\left[  w\right]   &  =\dfrac{d}{dT}\sum_{n\geq
1}\dfrac{\left(  -1\right)  ^{n-1}}{n}w^{n}=\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\cdot\underbrace{\dfrac{d}{dT}\left(  w^{n}\right)
}_{\substack{=n\left(  \dfrac{d}{dT}w\right)  w^{n-1}\\\text{(by
(\ref{sol.convolution-series.i.2.pf.power}))}}}=\sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\cdot n\left(  \dfrac{d}{dT}w\right)  w^{n-1}\\
&  =\sum_{n\geq1}\left(  -1\right)  ^{n-1}\left(  \dfrac{d}{dT}w\right)
w^{n-1}=\sum_{n\geq0}\left(  -1\right)  ^{n}\left(  \dfrac{d}{dT}w\right)
w^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n\text{ for
}n-1\text{ in the sum}\right)  .
\end{align*}
Comparing this with%
\[
\left(  \dfrac{d}{dT}w\right)  \cdot\underbrace{\dfrac{1}{1+w}}_{=\sum
_{n\geq0}\left(  -1\right)  ^{n}w^{n}}=\left(  \dfrac{d}{dT}w\right)
\cdot\sum_{n\geq0}\left(  -1\right)  ^{n}w^{n}=\sum_{n\geq0}\left(  -1\right)
^{n}\left(  \dfrac{d}{dT}w\right)  w^{n},
\]
we obtain $\dfrac{d}{dT}\left(  \overline{\log}\left[  w\right]  \right)
=\left(  \dfrac{d}{dT}w\right)  \cdot\dfrac{1}{1+w}$. This proves
(\ref{sol.convolution-series.i.2.log}). Thus, Fact I.2 is proven.]

\begin{statement}
\textit{Fact I.3:} Let $u\in\kk\left[  \left[  T\right]  \right]  $ and
$v\in\kk\left[  \left[  T\right]  \right]  $ be two power series having
constant term $1$. Assume that $\left(  \dfrac{d}{dT}u\right)  \cdot v=\left(
\dfrac{d}{dT}v\right)  \cdot u$. Then, $u=v$.
\end{statement}

[\textit{Proof of Fact I.3:} The power series $v$ has constant term $1$, and
thus has a multiplicative inverse $v^{-1}$. The Leibniz rule (applied to $v$
and $v^{-1}$) yields%
\[
\dfrac{d}{dT}\left(  v\cdot v^{-1}\right)  =\left(  \dfrac{d}{dT}v\right)
v^{-1}+v\dfrac{d}{dT}\left(  v^{-1}\right)  .
\]
Comparing this with $\dfrac{d}{dT}\underbrace{\left(  v\cdot v^{-1}\right)
}_{=1}=\dfrac{d}{dT}1=0$, we obtain $\left(  \dfrac{d}{dT}v\right)
v^{-1}+v\dfrac{d}{dT}\left(  v^{-1}\right)  =0$. Solving this equality for
$\dfrac{d}{dT}\left(  v^{-1}\right)  $, we find%
\[
\dfrac{d}{dT}\left(  v^{-1}\right)  =-\dfrac{1}{v}\left(  \dfrac{d}%
{dT}v\right)  v^{-1}=-v^{-2}\left(  \dfrac{d}{dT}v\right)  .
\]


Now, the Leibniz rule (applied to $u$ and $v^{-1}$) yields%
\begin{align*}
\dfrac{d}{dT}\left(  uv^{-1}\right)   &  =\left(  \dfrac{d}{dT}u\right)
v^{-1}+u\underbrace{\dfrac{d}{dT}\left(  v^{-1}\right)  }_{=-v^{-2}\left(
\dfrac{d}{dT}v\right)  }=\left(  \dfrac{d}{dT}u\right)  v^{-1}+u\left(
-v^{-2}\left(  \dfrac{d}{dT}v\right)  \right) \\
&  =v^{-2}\underbrace{\left(  \left(  \dfrac{d}{dT}u\right)  \cdot v-\left(
\dfrac{d}{dT}v\right)  \cdot u\right)  }_{\substack{=0\\\text{(since }\left(
\dfrac{d}{dT}u\right)  \cdot v=\left(  \dfrac{d}{dT}v\right)  \cdot u\text{)}%
}}=v^{-2}0=0=\dfrac{d}{dT}1.
\end{align*}
Moreover, the power series $uv^{-1}$ and $1$ have the same constant
term\footnote{\textit{Proof.} The power series $v$ has constant term $1$.
Hence, its inverse $v^{-1}$ has constant term $1^{-1}=1$. Now, both power
series $u$ and $v^{-1}$ have constant term $1$. Hence, their product $uv^{-1}$
has constant term $1\cdot1=1$. Since the power series $1$ also has constant
term $1$, this shows that the power series $uv^{-1}$ and $1$ have the same
constant term (namely, $1$).}. Hence, Fact I.1 (applied to $uv^{-1}$ and $1$
instead of $u$ and $v$) shows that $uv^{-1}=1$. Thus, $u=v$. This proves Fact I.3.]

The equality (\ref{sol.convolution-series.i.2.log}) (applied to $w=T$) yields
$\dfrac{d}{dT}\left(  \overline{\log}\left[  T\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}T\right)  }_{=1}\cdot\dfrac{1}{1+T}%
=\dfrac{1}{1+T}$. In other words, $\dfrac{d}{dT}\overline{\log}=\dfrac{1}%
{1+T}$ (since $\overline{\log}=\overline{\log}\left[  T\right]  $).

Now, (\ref{sol.convolution-series.i.2.exp}) (applied to $w=\overline{\log}$)
shows that%
\[
\dfrac{d}{dT}\left(  \overline{\exp}\left[  \overline{\log}\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}\overline{\log}\right)  }_{=\dfrac{1}{1+T}%
}\cdot\exp\left[  \overline{\log}\right]  =\dfrac{1}{1+T}\cdot\exp\left[
\overline{\log}\right]  .
\]


But $\overline{\exp}=\exp-1$ and thus $\exp=\overline{\exp}+1$. Substituting
$\overline{\log}$ for $T$ in this equality, we find $\exp\left[
\overline{\log}\right]  =\overline{\exp}\left[  \overline{\log}\right]  +1$.
Hence,%
\[
\dfrac{d}{dT}\left(  \exp\left[  \overline{\log}\right]  \right)  =\dfrac
{d}{dT}\left(  \overline{\exp}\left[  \overline{\log}\right]  +1\right)
=\dfrac{d}{dT}\overline{\exp}\left[  \overline{\log}\right]
+\underbrace{\dfrac{d}{dT}1}_{=0}=\dfrac{d}{dT}\overline{\exp}\left[
\overline{\log}\right]  =\dfrac{1}{1+T}\cdot\exp\left[  \overline{\log
}\right]  .
\]
Multiplying this equality by $1+T$, we find%
\[
\left(  \dfrac{d}{dT}\left(  \exp\left[  \overline{\log}\right]  \right)
\right)  \cdot\left(  1+T\right)  =\exp\left[  \overline{\log}\right]  .
\]
Comparing this with $\underbrace{\left(  \dfrac{d}{dT}\left(  1+T\right)
\right)  }_{=1}\cdot\exp\left[  \overline{\log}\right]  =\exp\left[
\overline{\log}\right]  $, we find
\[
\left(  \dfrac{d}{dT}\left(  \exp\left[  \overline{\log}\right]  \right)
\right)  \cdot\left(  1+T\right)  =\left(  \dfrac{d}{dT}\left(  1+T\right)
\right)  \cdot\exp\left[  \overline{\log}\right]  .
\]
Since both power series $\exp\left[  \overline{\log}\right]  $ and $1+T$ have
constant term $1$\ \ \ \ \footnote{\textit{Proof.} It is clear that the power
series $1+T$ has constant term $1$. Thus, it remain to prove that the power
series $\exp\left[  \overline{\log}\right]  $ has constant term $1$.
\par
Recall that the power series $\overline{\exp}\left[  \overline{\log}\right]  $
has constant term $0$. Hence, the power series $\overline{\exp}\left[
\overline{\log}\right]  +1$ has constant term $0+1=1$. In other words, the
power series $\exp\left[  \overline{\log}\right]  $ has constant term $1$
(since $\exp\left[  \overline{\log}\right]  =\overline{\exp}\left[
\overline{\log}\right]  +1$). Qed.}, we can thus apply Fact I.3 to
$u=\exp\left[  \overline{\log}\right]  $ and $v=1+T$. We thus conclude that
$\exp\left[  \overline{\log}\right]  =1+T$. Comparing this with $\exp\left[
\overline{\log}\right]  =\overline{\exp}\left[  \overline{\log}\right]  +1$,
we obtain $\overline{\exp}\left[  \overline{\log}\right]  +1=1+T$. Subtracting
$1$ from this equality, we find $\overline{\exp}\left[  \overline{\log
}\right]  =T$.

The equality (\ref{sol.convolution-series.i.2.exp}) (applied to $w=T$) yields
$\dfrac{d}{dT}\left(  \overline{\exp}\left[  T\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}T\right)  }_{=1}\cdot\underbrace{\exp\left[
T\right]  }_{=\exp}=\exp$. In other words, $\dfrac{d}{dT}\overline{\exp}=\exp$
(since $\overline{\exp}=\overline{\exp}\left[  T\right]  $).

On the other hand, (\ref{sol.convolution-series.i.2.log}) (applied to
$w=\overline{\exp}$) shows that%
\[
\dfrac{d}{dT}\left(  \overline{\log}\left[  \overline{\exp}\right]  \right)
=\underbrace{\left(  \dfrac{d}{dT}\overline{\exp}\right)  }_{=\exp
=\overline{\exp}+1=1+\overline{\exp}}\cdot\dfrac{1}{1+\overline{\exp}}=\left(
1+\overline{\exp}\right)  \cdot\dfrac{1}{1+\overline{\exp}}=1=\dfrac{d}{dT}T.
\]
Since the two power series $\overline{\log}\left[  \overline{\exp}\right]  $
and $T$ have the same constant term\footnote{This is because the power series
$\overline{\log}\left[  \overline{\exp}\right]  $ has constant term $0$, and
the power series $T$ also has constant term $0$.}, we can thus apply Fact I.1
to $u=\overline{\log}\left[  \overline{\exp}\right]  $ and $v=T$. We thus
conclude that $\overline{\log}\left[  \overline{\exp}\right]  =T$. The proof
of Proposition~\ref{prop.exp-log-inv} is thus complete.
\end{proof}

\begin{proof}
[Proof of Lemma~\ref{lem.exp-log-log}.]The power series $\overline{\log}$ has
constant term $0$ (by Proposition~\ref{prop.exp-log-inv}). Hence,
Proposition~\ref{prop.convolution-series}(f) (applied to $v=\overline{\log}$
and $f=g-u_{A}\epsilon_{C}$) yields that $\overline{\log}^{\star}\left(
g-u_{A}\epsilon_{C}\right)  \in\mathfrak{n}\left(  C,A\right)  $. Thus,
$\overline{\log}^{\star}\left(  g-u_{A}\epsilon_{C}\right)  $ is a
well-defined element of $\mathfrak{n}\left(  C,A\right)  $. This proves
Lemma~\ref{lem.exp-log-log}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.exp-log}.](a) Let $f\in\mathfrak{n}\left(
C,A\right)  $. We must prove that $\exp^{\star}f-u_{A}\epsilon_{C}%
\in\mathfrak{n}\left(  C,A\right)  $ and $\log^{\star}\left(  \exp^{\star
}f\right)  =f$.

The power series $\overline{\exp}$ has constant term $0$ (by
Proposition~\ref{prop.exp-log-inv}). Hence,
Proposition~\ref{prop.convolution-series}(f) (applied to $v=\overline{\exp}$)
yields that $\overline{\exp}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $. But
$\overline{\exp}=\exp-1$, so that $\exp=\overline{\exp}+1$. Hence,%
\begin{align*}
\exp^{\star}\left(  f\right)   &  =\left(  \overline{\exp}+1\right)  ^{\star
}\left(  f\right)  =\underbrace{\overline{\exp}^{\star}\left(  f\right)
}_{=\overline{\exp}^{\star}f}+\underbrace{1^{\star}\left(  f\right)
}_{\substack{=u_{A}\epsilon_{C}\\\text{(by
(\ref{eq.exe.convolution-series.b.5}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.convolution-series.b.1})
(applied to }\overline{\exp}\text{ and }1\text{ instead of }u\text{ and
}v\text{)}\right) \\
&  =\overline{\exp}^{\star}f+u_{A}\epsilon_{C}.
\end{align*}
Solving this equation for $\overline{\exp}^{\star}f$, we obtain
$\overline{\exp}^{\star}f=\exp^{\star}\left(  f\right)  -u_{A}\epsilon_{C}
=\exp^{\star}f-u_{A}\epsilon_{C}$.
Hence, $\exp^{\star}f-u_{A}\epsilon_{C}=\overline{\exp
}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $. It thus remains to prove that
$\log^{\star}\left(  \exp^{\star}f\right)  =f$.

The map $\exp^{\star}f$ satisfies $\exp^{\star}f-u_{A}\epsilon_{C}%
\in\mathfrak{n}\left(  C,A\right)  $. Hence, the map $\log^{\star}\left(
\exp^{\star}f\right)  \in\Hom \left(  C,A\right)  $ is
well-defined, and satisfies
\begin{align}
\log^{\star}\left(  \exp^{\star}f\right)   &  =\overline{\log}^{\star}\left(
\underbrace{\exp^{\star}f-u_{A}\epsilon_{C}}_{=\overline{\exp}^{\star}%
f}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\log
^{\star}\left(  \exp^{\star}f\right)  \right) \nonumber\\
&  =\overline{\log}^{\star}\left(  \overline{\exp}^{\star}f\right)  .
\label{sol.convolution-series.k.2}%
\end{align}


But the power series $\overline{\exp}$ has constant term $0$. Thus,
(\ref{eq.exe.convolution-series.g.1}) (applied to $u=\overline{\log}$ and
$v=\overline{\exp}$) yields $\left(  \overline{\log}\left[  \overline{\exp
}\right]  \right)  ^{\star}\left(  f\right)  =\overline{\log}^{\star}\left(
\overline{\exp}^{\star}\left(  f\right)  \right)  =\overline{\log}^{\star
}\left(  \overline{\exp}^{\star}f\right)  $. Since $\overline{\log}\left[
\overline{\exp}\right]  =T$ (by Proposition~\ref{prop.exp-log-inv}), this
rewrites as $T^{\star}\left(  f\right)  =\overline{\log}^{\star}\left(
\overline{\exp}^{\star}f\right)  $.

Applying (\ref{eq.exe.convolution-series.a.monom}) to $k=1$, we obtain
$\left(  T^{1}\right)  ^{\star}\left(  f\right)  =f^{\star1}=f$. Since
$T^{1}=T$, this rewrites as $T^{\star}\left(  f\right)  =f$. Compared with
$T^{\star}\left(  f\right)  =\overline{\log}^{\star}\left(  \overline{\exp
}^{\star}f\right)  $, this yields $\overline{\log}^{\star}\left(
\overline{\exp}^{\star}f\right)  =f$. Now, (\ref{sol.convolution-series.k.2})
becomes $\log^{\star}\left(  \exp^{\star}f\right)  =\overline{\log}^{\star
}\left(  \overline{\exp}^{\star}f\right)  =f$. This proves
Proposition~\ref{prop.exp-log}(a).

(b) Let $g\in\Hom \left(  C,A\right)  $ be such that
$g-u_{A}\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $. We must prove that
$\exp^{\star}\left(  \log^{\star}g\right)  =g$.

Set $f=g-u_{A}\epsilon_{C}$. Thus, $f=g-u_{A}\epsilon_{C}\in\mathfrak{n}%
\left(  C,A\right)  $.

The power series $\overline{\log}$ has constant term $0$ (by
Proposition~\ref{prop.exp-log-inv}). Hence,
Proposition~\ref{prop.convolution-series}(f) (applied to $v=\overline{\log}$)
yields that $\overline{\log}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $.

The definition of $\log^{\star}g$ yields $\log^{\star}g=\overline{\log}%
^{\star}\underbrace{\left(  g-u_{A}\epsilon_{C}\right)  }_{=f}=\overline{\log
}^{\star}f\in\mathfrak{n}\left(  C,A\right)  $.

But the power series $\overline{\log}$ has constant term $0$. Thus,
(\ref{eq.exe.convolution-series.g.1}) (applied to $u=\overline{\exp}$ and
$v=\overline{\log}$) yields $\left(  \overline{\exp}\left[  \overline{\log
}\right]  \right)  ^{\star}\left(  f\right)  =\overline{\exp}^{\star}\left(
\overline{\log}^{\star}\left(  f\right)  \right)  =\overline{\exp}^{\star
}\left(  \overline{\log}^{\star}f\right)  $. Since $\overline{\exp}\left[
\overline{\log}\right]  =T$ (by Proposition~\ref{prop.exp-log-inv}), this
rewrites as $T^{\star}\left(  f\right)  =\overline{\exp}^{\star}\left(
\overline{\log}^{\star}f\right)  $.

Applying (\ref{eq.exe.convolution-series.a.monom}) to $k=1$, we obtain
$\left(  T^{1}\right)  ^{\star}\left(  f\right)  =f^{\star1}=f$. Since
$T^{1}=T$, this rewrites as $T^{\star}\left(  f\right)  =f$. Compared with
$T^{\star}\left(  f\right)  =\overline{\exp}^{\star}\left(  \overline{\log
}^{\star}f\right)  $, this yields $\overline{\exp}^{\star}\left(
\overline{\log}^{\star}f\right)  =f$.

But $\overline{\exp}=\exp-1$, so that $\exp=\overline{\exp}+1$. Hence,%
\begin{align*}
\exp^{\star}\left(  \log^{\star}g\right)   &  =\left(  \overline{\exp
}+1\right)  ^{\star}\left(  \underbrace{\log^{\star}g}_{=\overline{\log
}^{\star}f}\right)  =\left(  \overline{\exp}+1\right)  ^{\star}\left(
\overline{\log}^{\star}f\right)  =\overline{\exp}^{\star}\left(
\overline{\log}^{\star}f\right)  +\underbrace{1^{\star}\left(  \overline{\log
}^{\star}f\right)  }_{\substack{=u_{A}\epsilon_{C}\\\text{(by
(\ref{eq.exe.convolution-series.b.5}) (applied}\\\text{to }\overline{\log
}^{\star}f\text{ instead of }f\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.convolution-series.b.1})
(applied to }\overline{\exp}\text{, }1\text{ and }\overline{\log}^{\star
}f\text{ instead of }u\text{, }v\text{ and }f\text{)}\right) \\
&  =\underbrace{\overline{\exp}^{\star}\left(  \overline{\log}^{\star
}f\right)  }_{=f=g-u_{A}\epsilon_{C}}+u_{A}\epsilon_{C}=\left(  g-u_{A}%
\epsilon_{C}\right)  +u_{A}\epsilon_{C}=g.
\end{align*}
This proves Proposition~\ref{prop.exp-log}(b).

(c) Let $f,g\in\mathfrak{n}\left(  C,A\right)  $ be such that $f\star g=g\star
f$. Then, Proposition~\ref{prop.convolution-series}(c) shows that
$f+g\in\mathfrak{n}\left(  C,A\right)  $. Hence, $\exp^{\star}\left(
f+g\right)  $ is well-defined.

Let us recall the following facts, which have been proven in the proof of
Proposition~\ref{prop.convolution-series}(c) (during the solution to
Exercise~\ref{exe.convolution-series}):

\begin{itemize}
\item We have%
\begin{equation}
\left(  f+g\right)  ^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\star
g^{\star\left(  n-i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in \NN . \label{sol.convolution-series.m.binom}%
\end{equation}


\item The family $\left(  f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported.

\item The family $\left(  g^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported.

\item For each $x\in C$,
\begin{equation}
\text{the family }\left(  \left(  f^{\star q}\star g^{\star r}\right)  \left(
x\right)  \right)  _{\left(  q,r\right)  \in \NN\times\NN}\in
A^{\NN\times\NN}\text{ is finitely supported.}
\label{sol.convolution-series.m.fs}%
\end{equation}

\end{itemize}

Furthermore, the family $\left(  \dfrac{f^{\star q}}{q!}\right)
_{q\in\NN}$ is pointwise finitely supported\footnote{\textit{Proof.} We
know that the family $\left(  f^{\star n}\right)  _{n\in\NN}$ is
pointwise finitely supported. Hence, Proposition~\ref{prop.pw-fin-supp.P4}
(applied to $Q=\NN$, $\left(  f_{q}\right)  _{q\in Q}=\left(  f^{\star
n}\right)  _{n\in\NN}$ and $\left(  \lambda_{q}\right)  _{q\in
Q}=\left(  \dfrac{1}{n!}\right)  _{n\in\NN}$) shows that the family
$\left(  \dfrac{1}{n!}f^{\star n}\right)  _{n\in\NN}$ is pointwise
finitely supported. Since $\dfrac{1}{n!}f^{\star n}=\dfrac{f^{\star n}}{n!}$
for each $n\in \NN$, this result rewrites as follows: The family
$\left(  \dfrac{f^{\star n}}{n!}\right)  _{n\in\NN}$ is pointwise
finitely supported. Renaming the index $n$ as $q$ in this statement, we obtain
the following: The family $\left(  \dfrac{f^{\star q}}{q!}\right)
_{q\in\NN}$ is pointwise finitely supported. Qed.}. The same argument
(applied to $g$ instead of $f$) shows that the family $\left(  \dfrac{g^{\star
q}}{q!}\right)  _{q\in\NN}$ is pointwise finitely supported.
Renaming the index $q$ as $r$ in this statement, we obtain the following: The
family $\left(  \dfrac{g^{\star r}}{r!}\right)  _{r\in\NN}$ is
pointwise finitely supported. Hence, Proposition~\ref{prop.pw-fin-supp.P3}
(applied to $Q=\NN$, $R=\NN$, $\left(  f_{q}\right)  _{q\in
Q}=\left(  \dfrac{f^{\star q}}{q!}\right)  _{q\in\NN}$ and $\left(
g_{r}\right)  _{r\in R}=\left(  \dfrac{g^{\star r}}{r!}\right)  _{r\in
\NN}$) shows that the family $\left(  \dfrac{f^{\star q}}{q!}%
\star\dfrac{g^{\star r}}{r!}\right)  _{\left(  q,r\right)  \in
\NN\times\NN}\in\left(  \Hom \left(  C,A\right)  \right)
^{\NN\times\NN}$ is pointwise finitely supported, and that it
satisfies%
\begin{equation}
\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(  \dfrac
{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  =\left(  \sum
_{q\in\NN}\dfrac{f^{\star q}}{q!}\right)  \star\left(  \sum
_{r\in\NN}\dfrac{g^{\star r}}{r!}\right)  .
\label{sol.convolution-series.m.prodfam}%
\end{equation}


Let $x\in C$. Then, the family $\left(  \left(  f^{\star q}\star g^{\star
r}\right)  \left(  x\right)  \right)  _{\left(  q,r\right)  \in\NN%
\times\NN}\in A^{\NN\times\NN}$ is finitely supported (by
(\ref{sol.convolution-series.m.fs})). In other words, all but finitely many
$\left(  q,r\right)  \in \NN\times\NN$ satisfy $\left(  f^{\star
q}\star g^{\star r}\right)  \left(  x\right)  =0$. In other words, there
exists a finite subset $K$ of $\NN\times\NN$ such that%
\begin{equation}
\text{each }\left(  q,r\right)  \in\left(  \NN\times\NN\right)
\setminus K\text{ satisfies }\left(  f^{\star q}\star g^{\star r}\right)
\left(  x\right)  =0. \label{sol.convolution-series.m.3}%
\end{equation}
Consider this $K$. Then,%
\begin{equation}
\text{each }\left(  q,r\right)  \in \NN\times\NN \text{ satisfying
}\left(  q,r\right)  \notin K\text{ satisfies }\left(  \dfrac{f^{\star q}}%
{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  =0
\label{sol.convolution-series.m.3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.m.3b}):} Let $\left(
q,r\right)  \in \NN\times\NN$ be such that $\left(  q,r\right)
\notin K$. Then, $\left(  q,r\right)  \in\left(  \NN\times
\NN\right)  \setminus K$ (since $\left(  q,r\right)  \in \NN\times\NN$
but $\left(  q,r\right)  \notin K$). Thus,
\begin{align*}
\underbrace{\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}%
{r!}\right)  }_{=\dfrac{1}{q!}\cdot\dfrac{1}{r!}\cdot f^{\star q}\star
g^{\star r}}\left(  x\right)   &  =\left(  \dfrac{1}{q!}\cdot\dfrac{1}%
{r!}\cdot f^{\star q}\star g^{\star r}\right)  \left(  x\right)  =\dfrac
{1}{q!}\cdot\dfrac{1}{r!}\cdot\underbrace{\left(  f^{\star q}\star g^{\star
r}\right)  \left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.m.3}))}}}\\
&  =\dfrac{1}{q!}\cdot\dfrac{1}{r!}\cdot0=0.
\end{align*}
This proves (\ref{sol.convolution-series.m.3b}).}.

Let $Q=\left\{  u+v\ \mid\ \left(  u,v\right)  \in K\right\}  $. Thus, $Q$ is
a finite set (since $K$ is a finite set). Also, $Q\subset \NN%
$\ \ \ \ \footnote{\textit{Proof.} Let $w\in Q$. Thus, $w\in Q=\left\{
u+v\ \mid\ \left(  u,v\right)  \in K\right\}  $. In other words, $w=u+v$ for
some $\left(  u,v\right)  \in K$. Consider this $\left(  u,v\right)  $.
\par
We have $\left(  u,v\right)  \in K\subset \NN\times\NN$. In other
words, $u\in \NN$ and $v\in \NN$. Hence, $u+v\in \NN$. Thus,
$w=u+v\in \NN$.
\par
Now, forget that we fixed $w$. We thus have shown that $w\in \NN$ for
each $w\in Q$. In other words, $Q\subset \NN$.}.

But
\begin{equation}
\text{every }n\in \NN \setminus Q\text{ satisfies }\left(  f+g\right)
^{\star n}\left(  x\right)  =0 \label{sol.convolution-series.m.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.m.5}):} Let
$n\in \NN \setminus Q$. We must show that $\left(  f+g\right)  ^{\star
n}\left(  x\right)  =0$.
\par
We have $n\in \NN \setminus Q$. In other words, $n\in \NN$ and
$n\notin Q$.
\par
Let $i\in\left\{  0,1,\ldots,n\right\}  $ be arbitrary. We shall show that
$\left(  f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =0$ first.
\par
Indeed, $i\in\left\{  0,1,\ldots,n\right\}  $ shows that $i\leq n$, so that
$n-i\in \NN$. Also, $i\in\left\{  0,1,\ldots,n\right\}  \subset
\NN$. Combining $i\in \NN$ with $n-i\in \NN$, we obtain
$\left(  i,n-i\right)  \in \NN\times\NN$.
\par
If we had $\left(  i,n-i\right)  \in K$, then we would have%
\begin{align*}
i+\left(  n-i\right)   &  \in\left\{  u+v\ \mid\ \left(  u,v\right)  \in
K\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i+\left(  n-i\right)
=u+v\text{ for some }\left(  u,v\right)  \in K\text{ (namely, for }\left(
u,v\right)  =\left(  i,n-i\right)  \text{)}\right) \\
&  =Q\ \ \ \ \ \ \ \ \ \ \left(  \text{since }Q=\left\{  u+v\ \mid\ \left(
u,v\right)  \in K\right\}  \right)  ,
\end{align*}
which would contradict $i+\left(  n-i\right)  =n\notin Q$. Thus, we cannot
have $\left(  i,n-i\right)  \in K$. In other words, we have $\left(
i,n-i\right)  \notin K$.
\par
Combining $\left(  i,n-i\right)  \in \NN\times\NN$ with $\left(
i,n-i\right)  \notin K$, we obtain $\left(  i,n-i\right)  \in\left(
\NN\times\NN\right)  \setminus K$. Hence,
(\ref{sol.convolution-series.m.3}) (applied to $\left(  q,r\right)  =\left(
i,n-i\right)  $) yields $\left(  f^{\star i}\star g^{\star\left(  n-i\right)
}\right)  \left(  x\right)  =0$.
\par
Now, forget that we fixed $i$. We thus have shown that%
\begin{equation}
\left(  f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  0,1,\ldots
,n\right\}  . \label{sol.convolution-series.m.5.pf.1}%
\end{equation}
\par
Now, (\ref{sol.convolution-series.m.binom}) yields $\left(  f+g\right)
^{\star n}=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\star g^{\star\left(
n-i\right)  }$. Applying both sides of this equality to $x$, we obtain%
\[
\left(  f+g\right)  ^{\star n}\left(  x\right)  =\left(  \sum_{i=0}^{n}%
\dbinom{n}{i}f^{\star i}\star g^{\star\left(  n-i\right)  }\right)  \left(
x\right)  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\star
g^{\star\left(  n-i\right)  }\right)  \left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.convolution-series.m.5.pf.1}))}}%
}=\sum_{i=0}^{n}\dbinom{n}{i}0=0.
\]
Qed.}.

Let $n\in \NN$. Then, (\ref{sol.convolution-series.m.binom}) yields%
\begin{align*}
\left(  f+g\right)  ^{\star n}  &  =\underbrace{\sum_{i=0}^{n}}_{=\sum
_{i\in\left\{  0,1,\ldots,n\right\}  }}\underbrace{\dbinom{n}{i}%
}_{\substack{=\dfrac{n!}{i!\left(  n-i\right)  !}\\=n!\cdot\dfrac{1}{i!}%
\cdot\dfrac{1}{\left(  n-i\right)  !}}}f^{\star i}\star g^{\star\left(
n-i\right)  }\\
&  =\sum_{i\in\left\{  0,1,\ldots,n\right\}  }n!\cdot\dfrac{1}{i!}\cdot
\dfrac{1}{\left(  n-i\right)  !}f^{\star i}\star g^{\star\left(  n-i\right)
}=n!\cdot\sum_{i\in\left\{  0,1,\ldots,n\right\}  }\underbrace{\dfrac{1}%
{i!}\cdot\dfrac{1}{\left(  n-i\right)  !}f^{\star i}\star g^{\star\left(
n-i\right)  }}_{=\dfrac{f^{\star i}}{i!}\star\dfrac{g^{\star\left(
n-i\right)  }}{\left(  n-i\right)  !}}\\
&  =n!\cdot\sum_{i\in\left\{  0,1,\ldots,n\right\}  }\dfrac{f^{\star i}}%
{i!}\star\dfrac{g^{\star\left(  n-i\right)  }}{\left(  n-i\right)  !}%
=n!\cdot\sum_{h\in\left\{  0,1,\ldots,n\right\}  }\dfrac{f^{\star h}}{h!}%
\star\dfrac{g^{\star\left(  n-h\right)  }}{\left(  n-h\right)  !}%
\end{align*}
(here, we have renamed the summation index $i$ as $h$). Multiplying this
equality by $\dfrac{1}{n!}$, we obtain
\begin{equation}
\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}=\dfrac{1}{n!}n!\cdot\sum
_{h\in\left\{  0,1,\ldots,n\right\}  }\dfrac{f^{\star h}}{h!}\star
\dfrac{g^{\star\left(  n-h\right)  }}{\left(  n-h\right)  !}=\sum
_{h\in\left\{  0,1,\ldots,n\right\}  }\dfrac{f^{\star h}}{h!}\star
\dfrac{g^{\star\left(  n-h\right)  }}{\left(  n-h\right)  !}.
\label{sol.convolution-series.m.7}%
\end{equation}


Define a subset $Z_{n}$ of $\NN\times\NN$ by $Z_{n}=\left\{
\left(  q,r\right)  \in \NN\times\NN\ \mid\ q+r=n\right\}  $.
Each $h\in\left\{  0,1,\ldots,n\right\}  $ satisfies $\left(  h,n-h\right)
\in Z_{n}$\ \ \ \ \footnote{\textit{Proof.} Let $h\in\left\{  0,1,\ldots
,n\right\}  $. Then, $h\leq n$ (since $h\in\left\{  0,1,\ldots,n\right\}  $)
and thus $n-h\geq0$. Hence, $n-h\in \NN$. Also, $h\in\left\{
0,1,\ldots,n\right\}  \subset \NN$. Combining this with $n-h\in
\NN$, we obtain $\left(  h,n-h\right)  \in \NN\times\NN$.
Hence, $\left(  h,n-h\right)  $ is an element of $\NN\times\NN$
and satisfies $h+\left(  n-h\right)  =n$. In other words, $\left(
h,n-h\right)  $ is an element $\left(  q,r\right)  $ of $\NN\times\NN$
satisfying $q+r=n$. In other words, $\left(  h,n-h\right)
\in\left\{  \left(  q,r\right)  \in \NN\times\NN\ \mid
\ q+r=n\right\}  $. In other words, $\left(  h,n-h\right)  \in Z_{n}$ (since
$Z_{n}=\left\{  \left(  q,r\right)  \in \NN\times\NN\ \mid\ q+r=n
\right\}  $). Qed.}. Hence, the map%
\[
\left\{  0,1,\ldots,n\right\}  \rightarrow Z_{n},\ \ \ \ \ \ \ \ \ \ h\mapsto
\left(  h,n-h\right)
\]
is well-defined. Denote this map by $p$. This map $p$ is
injective\footnote{\textit{Proof.} Let $h_{1}$ and $h_{2}$ be two elements of
$\left\{  0,1,\ldots,n\right\}  $ satisfying $p\left(  h_{1}\right)  =p\left(
h_{2}\right)  $. We shall show that $h_{1}=h_{2}$.
\par
We have $p\left(  h_{1}\right)  =\left(  h_{1},n-h_{1}\right)  $ (by the
definition of $p$) and $p\left(  h_{2}\right)  =\left(  h_{2},n-h_{2}\right)
$ (by the definition of $p$). Hence, $\left(  h_{1},n-h_{1}\right)  =p\left(
h_{1}\right)  =p\left(  h_{2}\right)  =\left(  h_{2},n-h_{2}\right)  $. In
other words, $h_{1}=h_{2}$ and $n-h_{1}=n-h_{2}$. Hence, we have shown that
$h_{1}=h_{2}$.
\par
Now, forget that we fixed $h_{1}$ and $h_{2}$. We thus have proven that if
$h_{1}$ and $h_{2}$ are two elements of $\left\{  0,1,\ldots,n\right\}  $
satisfying $p\left(  h_{1}\right)  =p\left(  h_{2}\right)  $, then
$h_{1}=h_{2}$. In other words, the map $p$ is injective.} and
surjective\footnote{\textit{Proof.} Let $c\in Z_{n}$ be arbitrary. We shall
show that $c\in p\left(  \left\{  0,1,\ldots,n\right\}  \right)  $.
\par
We have $c\in Z_{n}=\left\{  \left(  q,r\right)  \in \NN\times
\NN\ \mid\ q+r=n\right\}  $. In other words, $c=\left(  q,r\right)  $
for some $\left(  q,r\right)  \in \NN\times\NN$ satisfying
$q+r=n$. Consider this $\left(  q,r\right)  $. We have $\left(  q,r\right)
\in \NN\times\NN$; in other words, $q\in \NN$ and
$r\in \NN$. Now, $n=q+\underbrace{r}_{\substack{\geq0\\\text{(since
}r\in \NN\text{)}}}\geq q$, so that $q\leq n$. Hence, $q\in\left\{
0,1,\ldots,n\right\}  $ (since $q\in \NN$). Thus, $p\left(  q\right)  $
is well-defined. The definition of $p$ yields $p\left(  q\right)  =\left(
q,\underbrace{n-q}_{\substack{=r\\\text{(since }n=q+r\text{)}}}\right)
=\left(  q,r\right)  =c$. Hence, $c=p\left(  \underbrace{q}_{\in\left\{
0,1,\ldots,n\right\}  }\right)  \in p\left(  \left\{  0,1,\ldots,n\right\}
\right)  $.
\par
Now, forget that we fixed $c$. We thus have shown that each $c\in Z_{n}$
satisfies $c\in p\left(  \left\{  0,1,\ldots,n\right\}  \right)  $. In other
words, $Z_{n}\subset p\left(  \left\{  0,1,\ldots,n\right\}  \right)  $. In
other words, the map $p$ is surjective.}. Thus, the map $p$ is bijective. In
other words, the map $p$ is a bijection. In other words, the map
\[
\left\{  0,1,\ldots,n\right\}  \rightarrow Z_{n},\ \ \ \ \ \ \ \ \ \ h\mapsto
\left(  h,n-h\right)
\]
is a bijection\footnote{since the map $p$ is the map
\[
\left\{  0,1,\ldots,n\right\}  \rightarrow Z_{n},\ \ \ \ \ \ \ \ \ \ h\mapsto
\left(  h,n-h\right)
\]
}. Hence, we can substitute $\left(  h,n-h\right)  $ for $\left(  q,r\right)
$ in the sum $\sum_{\left(  q,r\right)  \in Z_{n}}\dfrac{f^{\star q}}{q!}%
\star\dfrac{g^{\star r}}{r!}$. We thus obtain%
\[
\sum_{\left(  q,r\right)  \in Z_{n}}\dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}=\sum_{h\in\left\{  0,1,\ldots,n\right\}  }%
\dfrac{f^{\star h}}{h!}\star\dfrac{g^{\star\left(  n-h\right)  }}{\left(
n-h\right)  !}.
\]
Thus,%
\[
\sum_{h\in\left\{  0,1,\ldots,n\right\}  }\dfrac{f^{\star h}}{h!}\star
\dfrac{g^{\star\left(  n-h\right)  }}{\left(  n-h\right)  !}=\underbrace{\sum
_{\left(  q,r\right)  \in Z_{n}}}_{\substack{=\sum\limits_{\substack{\left(
q,r\right)  \in \NN\times\NN;\\q+r=n}}\\\text{(since }%
Z_{n}=\left\{  \left(  q,r\right)  \in \NN\times\NN%
\ \mid\ q+r=n\right\}  \text{)}}}\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star
r}}{r!}=\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times\NN%
;\\q+r=n}}\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}.
\]
Thus, (\ref{sol.convolution-series.m.7}) becomes%
\begin{equation}
\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}=\sum_{h\in\left\{  0,1,\ldots
,n\right\}  }\dfrac{f^{\star h}}{h!}\star\dfrac{g^{\star\left(  n-h\right)  }%
}{\left(  n-h\right)  !}=\sum\limits_{\substack{\left(  q,r\right)  \in
\NN\times\NN;\\q+r=n}}\dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}. \label{sol.convolution-series.m.8}%
\end{equation}
Applying both sides of this equality to $x$, we find%
\begin{align}
\left(  \dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\right)  \left(  x\right)
&  =\left(  \sum\limits_{\substack{\left(  q,r\right)  \in \NN\times
\NN;\\q+r=n}}\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}%
{r!}\right)  \left(  x\right)  =\sum\limits_{\substack{\left(  q,r\right)
\in \NN\times\NN;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}%
\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in\NN%
\times\NN;\\q+r=n;\\\left(  q,r\right)  \in K}}}_{\substack{=\sum\limits
_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\\\text{(since }K\subset
\NN\times\NN\text{)}}}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  +\sum\limits_{\substack{\left(
q,r\right)  \in \NN\times\NN;\\q+r=n;\\\left(  q,r\right)  \notin
K}}\underbrace{\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}%
{r!}\right)  \left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.m.3b}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }\left(  q,r\right)  \in \NN\times\NN \text{
satisfies either }\left(  q,r\right)  \in K\text{ or }\left(  q,r\right)
\notin K\\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\left(  \dfrac{f^{\star
q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
+\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times
\NN;\\q+r=n;\\\left(  q,r\right)  \notin K}}0}_{=0}=\sum\limits
_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\left(  \dfrac{f^{\star q}}%
{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  .
\label{sol.convolution-series.m.9}%
\end{align}


Now, forget that we fixed $n$. We thus have proven
(\ref{sol.convolution-series.m.9}) for each $n\in \NN$.

From $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$, we obtain
\begin{align*}
\exp^{\star}\left(  f+g\right)   &  =\underbrace{\sum_{n\geq0}}_{=\sum
_{n\in\NN}}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\exp^{\star}\left(
f+g\right)  \right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}.
\end{align*}
Applying both sides of this equality to $x$, we find%
\begin{align}
\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)   &
=\left(  \sum_{n\in\NN}\dfrac{1}{n!}\left(  f+g\right)  ^{\star
n}\right)  \left(  x\right)  =\sum_{n\in\NN}\dfrac{1}{n!}\left(
f+g\right)  ^{\star n}\left(  x\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{n\in\NN;\\n\in Q}}}_{\substack{=\sum
_{n\in Q}\\\text{(since }Q\subset\NN\text{)}}}\dfrac{1}{n!}\left(
f+g\right)  ^{\star n}\left(  x\right)  +\underbrace{\sum\limits_{\substack{n\in
\NN;\\n\notin Q}}}_{=\sum_{n\in \NN \setminus Q}}\dfrac{1}%
{n!}\left(  f+g\right)  ^{\star n}\left(  x\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }n\in \NN \text{
satisfies either }n\in Q\text{ or }n\notin Q\text{ (but not both)}\right)
\nonumber\\
&  =\sum_{n\in Q}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)
+\sum_{n\in \NN \setminus Q}\dfrac{1}{n!}\underbrace{\left(  f+g\right)
^{\star n}\left(  x\right)  }_{\substack{=0\\\text{(by
(\ref{sol.convolution-series.m.5}))}}}\nonumber\\
&  =\sum_{n\in Q}\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)
+\underbrace{\sum_{n\in \NN \setminus Q}\dfrac{1}{n!}0}_{=0}=\sum_{n\in
Q}\underbrace{\dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\left(  x\right)
}_{=\left(  \dfrac{1}{n!}\left(  f+g\right)  ^{\star n}\right)  \left(
x\right)  }\nonumber\\
&  =\sum_{n\in Q}\underbrace{\left(  \dfrac{1}{n!}\left(  f+g\right)  ^{\star
n}\right)  \left(  x\right)  }_{\substack{=\sum\limits_{\substack{\left(  q,r\right)
\in K;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}%
{r!}\right)  \left(  x\right)  \\\text{(by (\ref{sol.convolution-series.m.9}%
))}}}=\underbrace{\sum_{n\in Q}\sum\limits_{\substack{\left(  q,r\right)  \in
K;\\q+r=n}}}_{\substack{=\sum_{\left(  q,r\right)  \in K}\sum\limits_{\substack{n\in
Q;\\q+r=n}}\\\text{(since both sums }\sum_{n\in Q}\text{ and }\sum\limits
_{\substack{\left(  q,r\right)  \in K;\\q+r=n}}\\\text{are finite)}}}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\nonumber\\
&  =\sum_{\left(  q,r\right)  \in K}\sum\limits_{\substack{n\in Q;\\q+r=n}}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
. \label{sol.convolution-series.m.11}%
\end{align}
But each $\left(  q,r\right)  \in K$ satisfies
\begin{equation}
\sum\limits_{\substack{n\in Q;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  =\left(  \dfrac{f^{\star q}%
}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\label{sol.convolution-series.m.14}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.convolution-series.m.14}):} Let $\left(
q,r\right)  \in K$. Thus, $q+r$ has the form $u+v$ for some $\left(
u,v\right)  \in K$ (namely, for $\left(  u,v\right)  =\left(  q,r\right)  $).
In other words, $q+r\in\left\{  u+v\ \mid\ \left(  u,v\right)  \in K\right\}
$. This rewrites as $q+r\in Q$ (since $Q=\left\{  u+v\ \mid\ \left(
u,v\right)  \in K\right\}  $).
\par
Hence, there exists some $n\in Q$ satisfying $q+r=n$ (namely, $n=q+r$).
Furthermore, this $n$ is unique (because the condition $q+r=n$ clearly
determines $n$ uniquely). Thus, there exists a \textbf{unique} $n\in Q$
satisfying $q+r=n$. Therefore, the sum $\sum\limits_{\substack{n\in Q;\\q+r=n}%
}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  $ has exactly one addend. Consequently, this sum simplifies as
follows:%
\[
\sum\limits_{\substack{n\in Q;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  =\left(  \dfrac{f^{\star q}%
}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  .
\]
This proves (\ref{sol.convolution-series.m.14}).}.

Hence, (\ref{sol.convolution-series.m.11}) becomes%
\begin{equation}
\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)
=\sum_{\left(  q,r\right)  \in K}\underbrace{\sum\limits_{\substack{n\in
Q;\\q+r=n}}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)
\left(  x\right)  }_{\substack{=\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  \\\text{(by
(\ref{sol.convolution-series.m.14}))}}}=\sum_{\left(  q,r\right)  \in
K}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  . \label{sol.convolution-series.m.21}%
\end{equation}


From $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$, we obtain
\begin{align*}
\exp^{\star}f  &  =\underbrace{\sum_{n\geq0}}_{=\sum_{n\in\NN}}%
\dfrac{1}{n!}f^{\star n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\exp^{\star}f\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}.
\end{align*}
The same argument (applied to $g$ instead of $f$) shows that $\exp^{\star
}g=\sum_{n\in\NN}\dfrac{1}{n!}g^{\star n}$. Now,%
\begin{align*}
&  \underbrace{\left(  \exp^{\star}f\right)  }_{\substack{=\sum_{n\in
\NN}\dfrac{1}{n!}f^{\star n}\\=\sum_{q\in\NN}\dfrac{1}%
{q!}f^{\star q}\\\text{(here, we have renamed the}\\\text{summation index
}n\text{ as }q\text{)}}}\star\underbrace{\left(  \exp^{\star}g\right)
}_{\substack{=\sum_{n\in\NN}\dfrac{1}{n!}g^{\star n}\\=\sum
_{r\in\NN}\dfrac{1}{r!}g^{\star r}\\\text{(here, we have renamed
the}\\\text{summation index }n\text{ as }r\text{)}}}\\
&  =\left(  \sum_{q\in\NN}\underbrace{\dfrac{1}{q!}f^{\star q}%
}_{=\dfrac{f^{\star q}}{q!}}\right)  \star\left(  \sum_{r\in \NN}
\underbrace{\dfrac{1}{r!}g^{\star r}}_{=\dfrac{g^{\star r}}{r!}}\right)
=\left(  \sum_{q\in\NN}\dfrac{f^{\star q}}{q!}\right)  \star\left(
\sum_{r\in\NN}\dfrac{g^{\star r}}{r!}\right) \\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.m.prodfam}%
)}\right)  .
\end{align*}
Applying both sides of this equality to $x$, we obtain%
\begin{align*}
\left(  \left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}g\right)
\right)  \left(  x\right)   &  =\left(  \sum_{\left(  q,r\right)
\in \NN\times\NN}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \right)  \left(  x\right) \\
&  =\sum_{\left(  q,r\right)  \in \NN\times\NN}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\\
&  =\underbrace{\sum\limits_{\substack{\left(  q,r\right)  \in \NN%
\times\NN;\\\left(  q,r\right)  \in K}}}_{\substack{=\sum_{\left(
q,r\right)  \in K}\\\text{(since }K\subset \NN\times\NN\text{)}%
}}\left(  \dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(
x\right)  +\sum\limits_{\substack{\left(  q,r\right)  \in \NN\times
\NN;\\\left(  q,r\right)  \notin K}}\underbrace{\left(  \dfrac{f^{\star
q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.convolution-series.m.3b}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }\left(  q,r\right)  \in \NN\times\NN \text{
satisfies either }\left(  q,r\right)  \in K\text{ or }\left(  q,r\right)
\notin K\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum_{\left(  q,r\right)  \in K}\left(  \dfrac{f^{\star q}}{q!}\star
\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)  +\underbrace{\sum\limits
_{\substack{\left(  q,r\right)  \in \NN\times\NN;\\\left(
q,r\right)  \notin K}}0}_{=0}=\sum_{\left(  q,r\right)  \in K}\left(
\dfrac{f^{\star q}}{q!}\star\dfrac{g^{\star r}}{r!}\right)  \left(  x\right)
\\
&  =\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.convolution-series.m.21}%
)}\right)  .
\end{align*}


Now, forget that we fixed $x$. We thus have shown that $\left(  \left(
\exp^{\star}f\right)  \star\left(  \exp^{\star}g\right)  \right)  \left(
x\right)  =\left(  \exp^{\star}\left(  f+g\right)  \right)  \left(  x\right)
$ for each $x\in C$. In other words, $\left(  \exp^{\star}f\right)
\star\left(  \exp^{\star}g\right)  =\exp^{\star}\left(  f+g\right)  $. In
other words, $\exp^{\star}\left(  f+g\right)  =\left(  \exp^{\star}f\right)
\star\left(  \exp^{\star}g\right)  $. This completes the proof of
Proposition~\ref{prop.exp-log}(c).

(d) Consider the $\kk$-linear map $0:C\rightarrow A$. It satisfies
$0\in\mathfrak{n}\left(  C,A\right)  $\ \ \ \ \footnote{\textit{Proof.}
The $\kk$-linear map $0:C\rightarrow A$ is the zero of
the convolution algebra $\left(  \Hom\left(  C,A\right)
,\star\right)  $. Hence, it satisfies $0^{\star n}=0$ for each positive
integer $n$.
\par
Now, fix $x\in C$. Then, all positive integers $n$ satisfy
$\underbrace{0^{\star n}}_{=0}\left(  x\right)  =0\left(  x\right)  =0$.
Hence, all but finitely many $n\in\NN$ satisfy $0^{\star n}\left(
x\right)  =0$ (since all but finitely many $n\in\NN$ are positive
integers). In other words, the family $\left(  0^{\star n}\left(  x\right)
\right)  _{n\in\NN}\in A^{\NN}$ of elements of $A$ is finitely
supported (by the definition of ``finitely
supported'').
\par
Forget that we fixed $x$. We thus have showed that for each $x\in C$, the
family $\left(  0^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
A^{\NN}$ of elements of $A$ is finitely supported. In other words, the
family $\left(  0^{\star n}\right)  _{n\in\NN}\in\left(
\operatorname*{Hom}\left(  C,A\right)  \right)  ^{\NN}$ (where $0$
denotes the $\kk$-linear map $0:C\rightarrow A$) is pointwise finitely
supported (by the definition of ``pointwise finitely
supported''). In other words, the map $0\in\Hom \left(  C,A\right)  $
is pointwise $\star$-nilpotent (by the definition of
``pointwise $\star$-nilpotent''). In other
words, this map satisfies $0\in\mathfrak{n}\left(  C,A\right)  $ (since
$\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps $f\in\Hom\left(  C,A\right)  $).}.
It remains to show that $\exp^{\star}0=u_{A}\epsilon_{C}$.

We have $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$. Thus, the definition of
$\exp^{\star}0$ yields%
\begin{align*}
\exp^{\star}0  &  =\sum_{n\geq0}\dfrac{1}{n!}0^{\star n}=\underbrace{\dfrac
{1}{0!}}_{=\dfrac{1}{1}=1}0^{\star0}+\sum_{n\geq1}\dfrac{1}{n!}%
\underbrace{0^{\star n}}_{\substack{=0\\\text{(since }n\geq1\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \\
&  =0^{\star0}+\underbrace{\sum_{n\geq1}\dfrac{1}{n!}0}_{=0}=0^{\star
0}=\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right) \\
&  =u_{A}\epsilon_{C}.
\end{align*}
Thus, Proposition~\ref{prop.exp-log}(d) is proven.

(e) Let $f\in\mathfrak{n}\left(  C,A\right)  $. We must prove that each
$n\in \NN$ satisfies%
\begin{equation}
nf\in\mathfrak{n}\left(  C,A\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(  nf\right)  =\left(  \exp^{\star
}f\right)  ^{\star n}. \label{sol.convolution-series.n.goal}%
\end{equation}


[\textit{Proof of (\ref{sol.convolution-series.n.goal}):} We shall prove
(\ref{sol.convolution-series.n.goal}) by induction over $n$:

\textit{Induction base:} We have $0f\in\mathfrak{n}\left(  C,A\right)
$\ \ \ \ \footnote{\textit{Proof.} Proposition~\ref{prop.convolution-series}%
(d) (applied to $\lambda=0\cdot1_{\kk}$) yields $0\cdot1_{\kk%
}f\in\mathfrak{n}\left(  C,A\right)  $. Thus, $0f=0=0\cdot1_{\kk}%
f\in\mathfrak{n}\left(  C,A\right)  $.}. Furthermore,%
\begin{align*}
\exp^{\star}\left(  \underbrace{0f}_{=0}\right)   &  =\exp^{\star}%
0=u_{A}\epsilon_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{prop.exp-log}(d)}\right) \\
&  =\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right)  =\left(
\exp^{\star}f\right)  ^{\star0}%
\end{align*}
(since $\left(  \exp^{\star}f\right)  ^{\star0}=\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \right)  $).

Thus, we have shown that $0f\in\mathfrak{n}\left(  C,A\right)  $ and
$\exp^{\star}\left(  0f\right)  =\left(  \exp^{\star}f\right)  ^{\star0}$. In
other words, (\ref{sol.convolution-series.n.goal}) holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Let $k\in \NN$. Assume that
(\ref{sol.convolution-series.n.goal}) holds for $n=k$. We must prove that
(\ref{sol.convolution-series.n.goal}) holds for $n=k+1$.

We have assumed that (\ref{sol.convolution-series.n.goal}) holds for $n=k$. In
other words,%
\[
kf\in\mathfrak{n}\left(  C,A\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(  kf\right)  =\left(  \exp^{\star
}f\right)  ^{\star k}.
\]


Now, $f\star\left(  kf\right)  =kf\star f=\left(  kf\right)  \star f$. Hence,
Proposition~\ref{prop.exp-log}(c) (applied to $g=kf$) yields that
$f+kf\in\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}\left(
f+kf\right)  =\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}\left(
kf\right)  \right)  $. Now, $\left(  k+1\right)  f=f+kf\in\mathfrak{n}\left(
C,A\right)  $ and%
\begin{align*}
\exp^{\star}\left(  \underbrace{\left(  k+1\right)  f}_{=f+kf}\right)   &
=\exp^{\star}\left(  f+kf\right)  =\left(  \exp^{\star}f\right)
\star\underbrace{\left(  \exp^{\star}\left(  kf\right)  \right)  }_{=\left(
\exp^{\star}f\right)  ^{\star k}}=\left(  \exp^{\star}f\right)  \star\left(
\exp^{\star}f\right)  ^{\star k}\\
&  =\left(  \exp^{\star}f\right)  ^{\star\left(  k+1\right)  }.
\end{align*}


Thus, we have shown that%
\[
\left(  k+1\right)  f\in\mathfrak{n}\left(  C,A\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(  \left(
k+1\right)  f\right)  =\left(  \exp^{\star}f\right)  ^{\star\left(
k+1\right)  }.
\]
In other words, (\ref{sol.convolution-series.n.goal}) holds for $n=k+1$. This
completes the induction step. Thus, the induction proof of
(\ref{sol.convolution-series.n.goal}) is finished.]

Now, (\ref{sol.convolution-series.n.goal}) is proven. In other words,
Proposition~\ref{prop.exp-log}(e) is proven.

(f) Let $f\in\mathfrak{n}\left(  C,A\right)  $. Define a map $g\in
\Hom \left(  C,A\right)  $ by $g=f+u_{A}\epsilon_{C}$. Thus,
$g-u_{A}\epsilon_{C}=f\in\mathfrak{n}\left(  C,A\right)  $. Hence,
$\log^{\star}g$ is well-defined.

For each $n\in \NN$, define an element $\lambda_{n}\in\kk$ by
$\lambda_{n}=%
\begin{cases}
\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}, & \text{if }n\geq1;\\
0, & \text{if }n=0
\end{cases} \quad
$. Then, the definition of $\lambda_{0}$ yields $\lambda_{0}=%
\begin{cases}
\dfrac{\left(  -1\right)  ^{0-1}}{0}1_{\kk}, & \text{if }0\geq1;\\
0, & \text{if }0=0
\end{cases} \quad
=0$ (since $0=0$). Now,
\begin{align*}
\sum_{n\geq0}\lambda_{n}T^{n}  &  =\underbrace{\lambda_{0}}_{=0}T^{0}%
+\sum_{n\geq1}\lambda_{n}T^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have
split off the addend for }n=0\text{ from the sum}\right) \\
&  =\underbrace{0T^{0}}_{=0}+\sum_{n\geq1}\lambda_{n}T^{n}=\sum_{n\geq
1}\underbrace{\lambda_{n}}_{\substack{=%
\begin{cases}
\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}, & \text{if }n\geq1;\\
0, & \text{if }n=0
\end{cases}
\\=\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}\\\text{(since }%
n\geq1\text{)}}}T^{n}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}%
{n}\underbrace{1_{\kk}T^{n}}_{=T^{n}}\\
&  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}T^{n}=\log\left(
1+T\right)  =\overline{\log}.
\end{align*}
In other words, $\overline{\log}=\sum_{n\geq0}\lambda_{n}T^{n}$. Hence, the
definition of $\overline{\log}^{\star}f$ yields%
\begin{align*}
\overline{\log}^{\star}f  &  =\sum_{n\geq0}\lambda_{n}f^{\star n}\\
&  =\underbrace{\lambda_{0}}_{=0}f^{\star0}+\sum_{n\geq1}\lambda_{n}f^{\star
n}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \\
&  =\underbrace{0f^{\star0}}_{=0}+\sum_{n\geq1}\lambda_{n}f^{\star n}%
=\sum_{n\geq1}\underbrace{\lambda_{n}}_{\substack{=%
\begin{cases}
\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}, & \text{if }n\geq1;\\
0, & \text{if }n=0
\end{cases}
\\=\dfrac{\left(  -1\right)  ^{n-1}}{n}1_{\kk}\\\text{(since }%
n\geq1\text{)}}}f^{\star n}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}%
{n}\underbrace{1_{\kk}f^{\star n}}_{=f^{\star n}}\\
&  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}f^{\star n}.
\end{align*}
Now, the definition of $\log^{\star}g$ yields
\[
\log^{\star}g=\overline{\log}^{\star}f=\sum_{n\geq1}\dfrac{\left(  -1\right)
^{n-1}}{n}f^{\star n}.
\]
Since $g=f+u_{A}\epsilon_{C}$, this rewrites as $\log^{\star}\left(
f+u_{A}\epsilon_{C}\right)  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}%
{n}f^{\star n}$. This proves Proposition~\ref{prop.exp-log}(f).
\end{proof}

Thus, Proposition~\ref{prop.exp-log-inv}, Lemma~\ref{lem.exp-log-log} and
Proposition~\ref{prop.exp-log} are proven. Hence, Exercise~\ref{exe.exp-log}
is solved.

%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.leray.exp-alg}}

\textit{Solution to Exercise~\ref{exe.leray.exp-alg}.}

\begin{proof}
[Proof of Proposition~\ref{prop.convolution.nat}.]We know that $\gamma
:C\rightarrow C^{\prime}$ is a $\kk$-coalgebra morphism. In other
words, $\gamma:C\rightarrow C^{\prime}$ is a $\kk$-linear map
satisfying $\Delta_{C^{\prime}}\circ\gamma=\left(  \gamma\otimes\gamma\right)
\circ\Delta_{C}$ and $\epsilon_{C^{\prime}}\circ\gamma=\epsilon_{C}$.

We know that $\alpha:A\rightarrow A^{\prime}$ is a $\kk$-algebra
morphism. In other words, $\alpha:A\rightarrow A^{\prime}$ is a $\kk%
$-linear map satisfying $\alpha\circ m_{A}=m_{A^{\prime}}\circ\left(
\alpha\otimes\alpha\right)  $ and $\alpha\circ u_{A}=u_{A^{\prime}}$.

(a) Let $f\in\Hom \left(  C,A\right)  $, $g\in
\Hom \left(  C,A\right)  $, $f^{\prime}\in\Hom
\left(  C^{\prime},A^{\prime}\right)  $ and $g^{\prime}\in\Hom
\left(  C^{\prime},A^{\prime}\right)  $ be such that $f^{\prime}\circ
\gamma=\alpha\circ f$ and $g^{\prime}\circ\gamma=\alpha\circ g$. We must prove
that $\left(  f^{\prime}\star g^{\prime}\right)  \circ\gamma=\alpha
\circ\left(  f\star g\right)  $.

The definition of convolution yields $f\star g=m_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$ and $f^{\prime}\star g^{\prime}=m_{A^{\prime}}%
\circ\left(  f^{\prime}\otimes g^{\prime}\right)  \circ\Delta_{C^{\prime}}$.
Now,%
\begin{align*}
\underbrace{\left(  f^{\prime}\star g^{\prime}\right)  }_{=m_{A^{\prime}}%
\circ\left(  f^{\prime}\otimes g^{\prime}\right)  \circ\Delta_{C^{\prime}}%
}\circ\gamma &  =m_{A^{\prime}}\circ\left(  f^{\prime}\otimes g^{\prime
}\right)  \circ\underbrace{\Delta_{C^{\prime}}\circ\gamma}_{=\left(
\gamma\otimes\gamma\right)  \circ\Delta_{C}}=m_{A^{\prime}}\circ
\underbrace{\left(  f^{\prime}\otimes g^{\prime}\right)  \circ\left(
\gamma\otimes\gamma\right)  }_{=\left(  f^{\prime}\circ\gamma\right)
\otimes\left(  g^{\prime}\circ\gamma\right)  }\circ\Delta_{C}\\
&  =m_{A^{\prime}}\circ\left(  \underbrace{\left(  f^{\prime}\circ
\gamma\right)  }_{=\alpha\circ f}\otimes\underbrace{\left(  g^{\prime}%
\circ\gamma\right)  }_{=\alpha\circ g}\right)  \circ\Delta_{C}=m_{A^{\prime}%
}\circ\underbrace{\left(  \left(  \alpha\circ f\right)  \otimes\left(
\alpha\circ g\right)  \right)  }_{=\left(  \alpha\otimes\alpha\right)
\circ\left(  f\otimes g\right)  }\circ\Delta_{C}\\
&  =\underbrace{m_{A^{\prime}}\circ\left(  \alpha\otimes\alpha\right)
}_{=\alpha\circ m_{A}}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
=\alpha\circ\underbrace{m_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
}_{=f\star g}=\alpha\circ\left(  f\star g\right)  .
\end{align*}
This proves Proposition~\ref{prop.convolution.nat}(a).

(b) Let $f\in\Hom \left(  C,A\right)  $ and $f^{\prime}%
\in\Hom \left(  C^{\prime},A^{\prime}\right)  $ be such that
$f^{\prime}\circ\gamma=\alpha\circ f$. We must prove that every $n\in
\NN$ satisfies%
\begin{equation}
\left(  f^{\prime}\right)  ^{\star n}\circ\gamma=\alpha\circ f^{\star n}.
\label{pf.prop.convolution.nat.b.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.convolution.nat.b.goal}):} We shall prove
(\ref{pf.prop.convolution.nat.b.goal}) by induction over $n$:

\textit{Induction base:} We have $f^{\star0}=\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \right)  =u_{A}\circ\epsilon_{C}$ and similarly $\left(
f^{\prime}\right)  ^{\star0}=u_{A^{\prime}}\circ\epsilon_{C^{\prime}}$. Hence,%
\[
\underbrace{\left(  f^{\prime}\right)  ^{\star0}}_{=u_{A^{\prime}}%
\circ\epsilon_{C^{\prime}}}\circ\gamma=\underbrace{u_{A^{\prime}}}%
_{=\alpha\circ u_{A}}\circ\underbrace{\epsilon_{C^{\prime}}\circ\gamma
}_{=\epsilon_{C}}=\alpha\circ\underbrace{u_{A}\circ\epsilon_{C}}_{=f^{\star0}%
}=\alpha\circ f^{\star0}.
\]
In other words, (\ref{pf.prop.convolution.nat.b.goal}) holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Fix $k\in \NN$. Assume that
(\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k$. We must now prove that
(\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k+1$.

We have assumed that (\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k$.
In other words, we have $\left(  f^{\prime}\right)  ^{\star k}\circ
\gamma=\alpha\circ f^{\star k}$.

Now,%
\begin{align*}
\underbrace{\left(  f^{\prime}\right)  ^{\star\left(  k+1\right)  }%
}_{=f^{\prime}\star\left(  f^{\prime}\right)  ^{\star k}}\circ\gamma &
=\left(  f^{\prime}\star\left(  f^{\prime}\right)  ^{\star k}\right)
\circ\gamma=\alpha\circ\underbrace{\left(  f\star f^{\star k}\right)
}_{=f^{\star\left(  k+1\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{prop.convolution.nat}%
(a) (applied to }g=f^{\star k}\text{ and }g^{\prime}=\left(  f^{\prime
}\right)  ^{\star k}\text{)}\right) \\
&  =\alpha\circ f^{\star\left(  k+1\right)  }.
\end{align*}
In other words, (\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k+1$. This
completes the induction step. Thus, the induction proof of
(\ref{pf.prop.convolution.nat.b.goal}) is finished.]

Hence, (\ref{pf.prop.convolution.nat.b.goal}) is proven. In other words, we
have proven Proposition~\ref{prop.convolution.nat}(b).
\end{proof}

Before we move on to the proof of Proposition~\ref{prop.leray.f*nxy}, let us
show a simple lemma (which can also easily be obtained as a consequence of
Exercise~\ref{exe.convolution.tensor}(b)):

\begin{lemma}
\label{lem.sol.leray.exp-alg.F(X)G*k}Let $C$ be a $\kk$-coalgebra. Let
$A$ be a $\kk$-algebra. Let $F\in\Hom \left(  C,A\right)
$ and $G\in\Hom \left(  C,A\right)  $. Then, each $k\in \NN$
satisfies%
\begin{equation}
\left(  F\otimes G\right)  ^{\star k}=F^{\star k}\otimes G^{\star k}
\label{pf.prop.leray.f*nxy.FGk}%
\end{equation}
(as maps $C\otimes C\rightarrow A\otimes A$).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.leray.exp-alg.F(X)G*k}.]We must prove that
(\ref{pf.prop.leray.f*nxy.FGk}) holds for each $k\in \NN$. We shall
prove this by induction over $k$:

\textit{Induction base:} Let $s$ denote the canonical $\kk$-algebra
isomorphism $\kk\rightarrow\kk\otimes\kk$. We have%
\begin{align*}
\left(  F\otimes G\right)  ^{\star0}  &  =\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C\otimes
C,A\otimes A\right)  ,\star\right)  \right) \\
&  =\underbrace{u_{A\otimes A}}_{\substack{=\left(  u_{A}\otimes u_{A}\right)
\circ s\\\text{(by the definition of}\\\text{the }\kk\text{-algebra
}A\otimes A\text{)}}}\circ\underbrace{\epsilon_{C\otimes C}}%
_{\substack{=s^{-1}\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)
\\\text{(by the definition of}\\\text{the }\kk\text{-coalgebra
}C\otimes C\text{)}}}=\left(  u_{A}\otimes u_{A}\right)  \circ
\underbrace{s\circ s^{-1}}_{=\id_{\kk%
\otimes\kk}}\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)
=\left(  u_{A}\otimes u_{A}\right)  \circ\left(  \epsilon_{C}\otimes
\epsilon_{C}\right) \\
&  =\underbrace{\left(  u_{A}\circ\epsilon_{C}\right)  }_{\substack{=\left(
\text{the unity of the }\kk\text{-algebra }\left(  \Hom
\left(  C,A\right)  ,\star\right)  \right)  \\=F^{\star0}}}\otimes
\underbrace{\left(  u_{A}\circ\epsilon_{C}\right)  }_{\substack{=\left(
\text{the unity of the }\kk\text{-algebra }\left(  \Hom
\left(  C,A\right)  ,\star\right)  \right)  \\=G^{\star0}}}=F^{\star0}\otimes
G^{\star0}.
\end{align*}
In other words, (\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=0$. This
completes the induction base.

\textit{Induction step:} Let $\ell\in \NN$. Assume that
(\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell$. We must prove that
(\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell+1$.

We have assumed that (\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell$. In
other words, we have
\begin{equation}
\left(  F\otimes G\right)  ^{\star\ell}=F^{\star\ell}\otimes G^{\star\ell}.
\label{pf.prop.leray.f*nxy.FGk.pf.indhyp}%
\end{equation}


Now, Exercise~\ref{exe.convolution.tensor}(a) (applied to $C$, $A$,
$F^{\star\ell}$, $F$, $G^{\star\ell}$ and $G$ instead of $D$, $B$, $f$,
$f^{\prime}$, $g$ and $g^{\prime}$) shows that%
\[
\left(  F^{\star\ell}\otimes G^{\star\ell}\right)  \star\left(  F\otimes
G\right)  =\left(  F^{\star\ell}\star F\right)  \otimes\left(  G^{\star\ell
}\star G\right)
\]
in the convolution algebra $\Hom \left(  C\otimes C,A\otimes
A\right)  $. Thus,%
\[
\left(  F^{\star\ell}\otimes G^{\star\ell}\right)  \star\left(  F\otimes
G\right)  =\underbrace{\left(  F^{\star\ell}\star F\right)  }_{=F^{\star
\left(  \ell+1\right)  }}\otimes\underbrace{\left(  G^{\star\ell}\star
G\right)  }_{=G^{\star\left(  \ell+1\right)  }}=F^{\star\left(  \ell+1\right)
}\otimes G^{\star\left(  \ell+1\right)  }.
\]
Hence,%
\[
F^{\star\left(  \ell+1\right)  }\otimes G^{\star\left(  \ell+1\right)
}=\underbrace{\left(  F^{\star\ell}\otimes G^{\star\ell}\right)
}_{\substack{=\left(  F\otimes G\right)  ^{\star\ell}\\\text{(by
(\ref{pf.prop.leray.f*nxy.FGk.pf.indhyp}))}}}\star\left(  F\otimes G\right)
=\left(  F\otimes G\right)  ^{\star\ell}\star\left(  F\otimes G\right)
=\left(  F\otimes G\right)  ^{\star\left(  \ell+1\right)  }.
\]
In other words, (\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell+1$. This
completes the induction step. Hence, (\ref{pf.prop.leray.f*nxy.FGk}) is proven
by induction. Thus, Lemma \ref{lem.sol.leray.exp-alg.F(X)G*k} is proven.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.f*nxy}.]Let $\mathfrak{i}$ be the unity
of the $\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $. Thus,%
\[
\mathfrak{i}=\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right)  =u_{A}%
\circ\epsilon_{C}.
\]


The axioms of a $\kk$-bialgebra show that $\epsilon\left(  1_C\right)
=1_{\kk}$ (since $C$ is a $\kk$-bialgebra). In other words,
$\epsilon\left(1\right) = 1$.

Let us first show that every $x\in C$ and $y\in C$ satisfy%
\begin{equation}
f\left(  xy\right)  =\epsilon\left(  y\right)  f\left(  x\right)
+\epsilon\left(  x\right)  f\left(  y\right)  .
\label{pf.prop.leray.f*nxy.fxy}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.leray.f*nxy.fxy}):} Let $x\in C$ and $y\in C$.
The map $\epsilon$ is $\kk$-linear; thus,%
\[
\epsilon\left(  x-\epsilon\left(  x\right)  1\right)  =\epsilon\left(
x\right)  -\epsilon\left(  x\right)  \underbrace{\epsilon\left(  1\right)
}_{=1}=\epsilon\left(  x\right)  -\epsilon\left(  x\right)  =0.
\]
In other words, $x-\epsilon\left(  x\right)  1\in\ker\epsilon$. The same
argument (applied to $y$ instead of $x$) shows that $y-\epsilon\left(
y\right)  1\in\ker\epsilon$. But%
\[
\left(  x-\epsilon\left(  x\right)  1\right)  \left(  y-\epsilon\left(
y\right)  1\right)  =xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1,
\]
so that%
\[
xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)  x+\epsilon\left(
x\right)  \epsilon\left(  y\right)  1=\underbrace{\left(  x-\epsilon\left(
x\right)  1\right)  }_{\in\ker\epsilon}\underbrace{\left(  y-\epsilon\left(
y\right)  1\right)  }_{\in\ker\epsilon}\in\left(  \ker\epsilon\right)  \left(
\ker\epsilon\right)  =\left(  \ker\epsilon\right)  ^{2}.
\]
Applying the map $f$ to both sides of this relation, we find%
\[
f\left(  xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1\right)  \in f\left(
\left(  \ker\epsilon\right)  ^{2}\right)  =0,
\]
so that $f\left(  xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1\right)  =0$. Hence,%
\begin{align*}
0  &  =f\left(  xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1\right) \\
&  =f\left(  xy\right)  -\epsilon\left(  x\right)  f\left(  y\right)
-\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(  x\right)
\epsilon\left(  y\right)  \underbrace{f\left(  1\right)  }_{=0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =f\left(  xy\right)  -\epsilon\left(  x\right)  f\left(  y\right)
-\epsilon\left(  y\right)  f\left(  x\right)  .
\end{align*}
Solving this equality for $f\left(  xy\right)  $, we obtain $f\left(
xy\right)  =\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(
x\right)  f\left(  y\right)  $. This proves (\ref{pf.prop.leray.f*nxy.fxy}).]

Now, we shall show that%
\begin{equation}
\left(  f\circ m_{C}\right)  \left(  z\right)  =\left(  m_{A}\circ\left(
f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  \right)  \left(  z\right)
\label{pf.prop.leray.f*nxy.fm}%
\end{equation}
for each $z\in C\otimes C$.

[\textit{Proof of (\ref{pf.prop.leray.f*nxy.fm}):} Let $z\in C\otimes C$. We
are going to prove the equality (\ref{pf.prop.leray.f*nxy.fm}).

Both sides of the equality (\ref{pf.prop.leray.f*nxy.fm}) are $\kk%
$-linear in $z$. Hence, for the proof of (\ref{pf.prop.leray.f*nxy.fm}), we
can WLOG assume that $z$ is a pure tensor (since any $z\in C\otimes C$ is a
$\kk$-linear combination of pure tensors). Assume this.

There exist $x\in C$ and $y\in C$ satisfying $z=x\otimes y$ (since $z$ is a
pure tensor). Consider these $x$ and $y$. We have
\[
\underbrace{\mathfrak{i}}_{=u_{A}\circ\epsilon_{C}}\left(  x\right)  =\left(
u_{A}\circ\underbrace{\epsilon_{C}}_{=\epsilon}\right)  \left(  x\right)
=\left(  u_{A}\circ\epsilon\right)  \left(  x\right)  =u_{A}\left(
\epsilon\left(  x\right)  \right)  =\epsilon\left(  x\right)  1_{A}%
\]
(by the definition of the map $u_{A}$). The same argument (applied to $y$
instead of $x$) shows that $\mathfrak{i}\left(  y\right)  =\epsilon\left(
y\right)  1_{A}$.

From $z=x\otimes y$, we obtain $m_{C}\left(  z\right)  =m_{C}\left(  x\otimes
y\right)  =xy$ (by the definition of $m_{C}$). Now,%
\[
\left(  f\circ m_{C}\right)  \left(  z\right)  =f\left(  \underbrace{m_{C}%
\left(  z\right)  }_{=xy}\right)  =f\left(  xy\right)  =\epsilon\left(
y\right)  f\left(  x\right)  +\epsilon\left(  x\right)  f\left(  y\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.leray.f*nxy.fxy})}\right)
.
\]
Comparing this with%
\begin{align*}
\left(  m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)
\right)  \left(  z\right)   &  =m_{A}\left(  \left(  f\otimes\mathfrak{i}%
+\mathfrak{i}\otimes f\right)  \left(  \underbrace{z}_{=x\otimes y}\right)
\right)  =m_{A}\left(  \underbrace{\left(  f\otimes\mathfrak{i}+\mathfrak{i}%
\otimes f\right)  \left(  x\otimes y\right)  }_{=\left(  f\otimes
\mathfrak{i}\right)  \left(  x\otimes y\right)  +\left(  \mathfrak{i}\otimes
f\right)  \left(  x\otimes y\right)  }\right) \\
&  =m_{A}\left(  \underbrace{\left(  f\otimes\mathfrak{i}\right)  \left(
x\otimes y\right)  }_{=f\left(  x\right)  \otimes\mathfrak{i}\left(  y\right)
}+\underbrace{\left(  \mathfrak{i}\otimes f\right)  \left(  x\otimes y\right)
}_{=\mathfrak{i}\left(  x\right)  \otimes f\left(  y\right)  }\right)
=m_{A}\left(  f\left(  x\right)  \otimes\mathfrak{i}\left(  y\right)
+\mathfrak{i}\left(  x\right)  \otimes f\left(  y\right)  \right) \\
&  =f\left(  x\right)  \underbrace{\mathfrak{i}\left(  y\right)  }%
_{=\epsilon\left(  y\right)  1_{A}}+\underbrace{\mathfrak{i}\left(  x\right)
}_{=\epsilon\left(  x\right)  1_{A}}f\left(  y\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }m_{A}\right) \\
&  =\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(  x\right)
f\left(  y\right)  ,
\end{align*}
we obtain $\left(  f\circ m_{C}\right)  \left(  z\right)  =\left(  m_{A}%
\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  \right)
\left(  z\right)  $. Thus, (\ref{pf.prop.leray.f*nxy.fm}) is proven.]

Now, we have proven that (\ref{pf.prop.leray.f*nxy.fm}) holds for each $z\in
C$. In other words, we have%
\begin{equation}
f\circ m_{C}=m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f\right)  . \label{pf.prop.leray.f*nxy.fm2}%
\end{equation}


The axioms of a $\kk$-bialgebra show that the comultiplication
$\Delta_{C}:C\rightarrow C\otimes C$ of the $\kk$-coalgebra $C$ is a
$\kk$-algebra homomorphism (since $C$ is a $\kk$-bialgebra).
They also show that the multiplication $m_C : C \otimes C \to C$ of the
$\kk$-algebra $C$ is a $\kk$-coalgebra homomorphism (since $C$ is a
$\kk$-bialgebra).

Exercise~\ref{exe.cocommutative.coalghom}(b) shows that the $\kk%
$-algebra $A$ is commutative if and only if its multiplication $m_{A}:A\otimes
A\rightarrow A$ is a $\kk$-algebra homomorphism. Thus, its
multiplication $m_{A}:A\otimes A\rightarrow A$ is a $\kk$-algebra
homomorphism (since $A$ is commutative).

Thus, Proposition~\ref{prop.convolution.nat}(b) (applied to $C\otimes C$, $C$,
$A\otimes A$, $A$, $m_{C}$, $m_{A}$, $f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f$ and $f$ instead of $C$, $C^{\prime}$, $A$, $A^{\prime}$, $\gamma$, $\alpha
$, $f$ and $f^{\prime}$) shows that each $n\in \NN$ satisfies%
\begin{equation}
f^{\star n}\circ m_{C}=m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}%
\otimes f\right)  ^{\star n} \label{pf.prop.leray.f*nxy.fnm}%
\end{equation}
(because of (\ref{pf.prop.leray.f*nxy.fm2})).

The two $\kk$-linear maps $f\otimes\mathfrak{i}$ and $\mathfrak{i}%
\otimes f$ in $\Hom \left(  C\otimes C,A\otimes A\right)  $
satisfy%
\begin{equation}
\left(  f\otimes\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes
f\right)  =\left(  \mathfrak{i}\otimes f\right)  \star\left(  f\otimes
\mathfrak{i}\right)  \label{pf.prop.leray.f*nxy.fiif}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.leray.f*nxy.fiif}):}
Exercise~\ref{exe.convolution.tensor}(a) (applied to $C$, $A$, $f$,
$\mathfrak{i}$, $\mathfrak{i}$ and $f$ instead of $D$, $B$, $f$, $f^{\prime}$,
$g$ and $g^{\prime}$) shows that%
\begin{equation}
\left(  f\otimes\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes
f\right)  =\underbrace{\left(  f\star\mathfrak{i}\right)  }_{=f}%
\otimes\underbrace{\left(  \mathfrak{i}\star f\right)  }_{=f}=f\otimes f.
\label{pf.prop.leray.f*nxy.fiif.pf.1}%
\end{equation}
But Exercise~\ref{exe.convolution.tensor}(a) (applied to $C$, $A$,
$\mathfrak{i}$, $f$, $f$ and $\mathfrak{i}$ instead of $D$, $B$, $f$,
$f^{\prime}$, $g$ and $g^{\prime}$) shows that%
\[
\left(  \mathfrak{i}\otimes f\right)  \star\left(  f\otimes\mathfrak{i}%
\right)  =\underbrace{\left(  \mathfrak{i}\star f\right)  }_{=f}%
\otimes\underbrace{\left(  f\star\mathfrak{i}\right)  }_{=f}=f\otimes f.
\]
Comparing this with (\ref{pf.prop.leray.f*nxy.fiif.pf.1}), we obtain $\left(
f\otimes\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes f\right)
=\left(  \mathfrak{i}\otimes f\right)  \star\left(  f\otimes\mathfrak{i}%
\right)  $. This proves (\ref{pf.prop.leray.f*nxy.fiif}).}. In other words,
the two elements $f\otimes\mathfrak{i}$ and $\mathfrak{i}\otimes f$ of the
$\kk$-algebra $\left(  \Hom \left(  C\otimes C,A\otimes
A\right)  ,\star\right)  $ commute.

Hence, it is easy to see%
\begin{equation}
\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star n}%
=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes f^{\star\left(  n-i\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }n\in \NN.
\label{pf.prop.leray.f*nxy.binom}%
\end{equation}
[\textit{Proof of (\ref{pf.prop.leray.f*nxy.binom}):} Let $n\in \NN$.
Let $\mathfrak{G}$ be the $\kk$-subalgebra of $\left(
\Hom \left(  C\otimes C,A\otimes A\right)  ,\star\right)  $
generated by the two elements $f\otimes\mathfrak{i}$ and $\mathfrak{i}\otimes
f$. Thus, the $\kk$-algebra $\mathfrak{G}$ is generated by commuting
elements (because the elements $f\otimes\mathfrak{i}$ and $\mathfrak{i}\otimes
f$ of the $\kk$-algebra $\left(  \Hom \left(  C\otimes
C,A\otimes A\right)  ,\star\right)  $ commute), and therefore is commutative
(since any $\kk$-algebra generated by commuting elements must be
commutative). Hence, the binomial formula holds in this $\kk$-algebra
$\mathfrak{G}$. In other words, any $\alpha\in\mathfrak{G}$ and $\beta
\in\mathfrak{G}$ and $m\in \NN$ satisfy%
\[
\left(  \alpha+\beta\right)  ^{\star m}=\sum_{i=0}^{m}\dbinom{m}{i}%
\alpha^{\star i}\star\beta^{\star\left(  m-i\right)  }%
\]
(since the multiplication in the $\kk$-algebra $\mathfrak{G}$ is the
convolution $\star$). Applying this to $\alpha=f\otimes\mathfrak{i}$,
$\beta=\mathfrak{i}\otimes f$ and $m=n$, we obtain
\begin{align*}
&  \left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star n}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f\otimes\mathfrak{i}%
\right)  ^{\star i}}_{\substack{=f^{\star i}\otimes\mathfrak{i}^{\star
i}\\\text{(by (\ref{pf.prop.leray.f*nxy.FGk}) (applied}\\\text{to }f\text{,
}\mathfrak{i}\text{ and }i\text{ instead of }F\text{, }G\text{ and }%
k\text{))}}}\star\underbrace{\left(  \mathfrak{i}\otimes f\right)
^{\star\left(  n-i\right)  }}_{\substack{=\mathfrak{i}^{\star\left(
n-i\right)  }\otimes f^{\star\left(  n-i\right)  }\\\text{(by
(\ref{pf.prop.leray.f*nxy.FGk}) (applied}\\\text{to }\mathfrak{i}\text{,
}f\text{ and }n-i\text{ instead of }F\text{, }G\text{ and }k\text{))}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\star i}\otimes
\underbrace{\mathfrak{i}^{\star i}}_{\substack{=\mathfrak{i}\\\text{(since
}\mathfrak{i}\text{ is the unity}\\\text{of the }\kk\text{-algebra
}\left(  \Hom \left(  C,A\right)  ,\star\right)  \text{)}%
}}\right)  \star\left(  \underbrace{\mathfrak{i}^{\star\left(  n-i\right)  }%
}_{\substack{=\mathfrak{i}\\\text{(since }\mathfrak{i}\text{ is the
unity}\\\text{of the }\kk\text{-algebra }\left(  \Hom
\left(  C,A\right)  ,\star\right)  \text{)}}}\otimes f^{\star\left(
n-i\right)  }\right) \\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\otimes
\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes f^{\star\left(
n-i\right)  }\right)  }_{\substack{=\left(  f^{\star i}\star\mathfrak{i}%
\right)  \otimes\left(  \mathfrak{i}\star f^{\star\left(  n-i\right)
}\right)  \\\text{(by Exercise~\ref{exe.convolution.tensor}(a) (applied}%
\\\text{to }C\text{, }A\text{, }f^{\star i}\text{, }\mathfrak{i}\text{,
}\mathfrak{i}\text{, }f^{\star\left(  n-i\right)  }\\\text{instead of
}D\text{, }B\text{, }f\text{, }f^{\prime}\text{, }g\text{ and }g^{\prime
}\text{))}}}=\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}%
\star\mathfrak{i}\right)  }_{\substack{=f^{\star i}\\\text{(since
}\mathfrak{i}\text{ is the unity}\\\text{of the }\kk\text{-algebra
}\left(  \Hom \left(  C,A\right)  ,\star\right)  \text{)}%
}}\otimes\underbrace{\left(  \mathfrak{i}\star f^{\star\left(  n-i\right)
}\right)  }_{\substack{=f^{\star\left(  n-i\right)  }\\\text{(since
}\mathfrak{i}\text{ is the unity}\\\text{of the }\kk\text{-algebra
}\left(  \Hom \left(  C,A\right)  ,\star\right)  \text{)}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes f^{\star\left(  n-i\right)
}.
\end{align*}
This proves (\ref{pf.prop.leray.f*nxy.binom}).]

Now, let $x,y\in C$ and $n\in \NN$ be arbitrary. Then, the definition of
$m_{C}$ yields $m_{C}\left(  x\otimes y\right)  =xy$. Hence,%
\[
\left(  f^{\star n}\circ m_{C}\right)  \left(  x\otimes y\right)  =f^{\star
n}\left(  \underbrace{m_{C}\left(  x\otimes y\right)  }_{=xy}\right)
=f^{\star n}\left(  xy\right)  .
\]
Hence,%
\begin{align*}
f^{\star n}\left(  xy\right)   &  =\underbrace{\left(  f^{\star n}\circ
m_{C}\right)  }_{\substack{=m_{A}\circ\left(  f\otimes\mathfrak{i}%
+\mathfrak{i}\otimes f\right)  ^{\star n}\\\text{(by
(\ref{pf.prop.leray.f*nxy.fnm}))}}}\left(  x\otimes y\right)  =\left(
m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star
n}\right)  \left(  x\otimes y\right) \\
&  =m_{A}\left(  \underbrace{\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f\right)  ^{\star n}}_{\substack{=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\otimes f^{\star\left(  n-i\right)  }\\\text{(by
(\ref{pf.prop.leray.f*nxy.binom}))}}}\left(  x\otimes y\right)  \right)
=m_{A}\left(  \left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes
f^{\star\left(  n-i\right)  }\right)  \left(  x\otimes y\right)  \right) \\
&  =m_{A}\left(  \underbrace{\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\otimes f^{\star\left(  n-i\right)  }\right)  \left(  x\otimes y\right)
}_{=\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\star i}\otimes f^{\star\left(
n-i\right)  }\right)  \left(  x\otimes y\right)  }\right)  =m_{A}\left(
\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\otimes
f^{\star\left(  n-i\right)  }\right)  \left(  x\otimes y\right)  }_{=f^{\star
i}\left(  x\right)  \otimes f^{\star\left(  n-i\right)  }\left(  y\right)
}\right) \\
&  =m_{A}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(  x\right)
\otimes f^{\star\left(  n-i\right)  }\left(  y\right)  \right)  =\sum
_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)
\end{align*}
(by the definition of $m_{A}$). This proves Proposition~\ref{prop.leray.f*nxy}.
\end{proof}

Before we prove Proposition~\ref{prop.leray.exp-algh1}, let us show two
lemmas:\footnote{Recall that we are still using the conventions that are in
place throughout Section~\ref{subsect.leray}.}

\begin{lemma}
\label{lem.pf.prop.leray.exp-algh1.1}Let $C$ be a $\kk$-coalgebra. Let
$A$ be a $\kk$-algebra. Let $f\in\mathfrak{n}\left(  C,A\right)  $.

\begin{enumerate}
\item[(a)] If $z\in C$ is arbitrary, then the family $\left(  \dfrac{1}%
{n!}f^{\star n}\left(  z\right)  \right)  _{n\in\NN}\in A^{\NN}$
is finitely supported and satisfies%
\[
\left(  \exp^{\star}f\right)  \left(  z\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  z\right)  .
\]


\item[(b)] If $x\in C$ and $y\in C$ are arbitrary, then the family $\left(
\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\[
\left(  \exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  \left(  y\right)  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  .
\]

\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}.]We have $f\in
\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a pointwise $\star
$-nilpotent map in $\Hom \left(  C,A\right)  $. Thus, the family
$\left(  f^{\star n}\right)  _{n\in\NN}\in\left(  \Hom
\left(  C,A\right)  \right)  ^{\NN}$ is pointwise finitely supported.
In other words, for each $x\in C$,%
\begin{equation}
\text{the family }\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}\text{ is finitely supported.}
\label{pf.prop.leray.exp-algh1.fs}%
\end{equation}


(a) Let $z\in C$ be arbitrary. Recall that $\exp=\sum_{n\geq0}\dfrac{1}%
{n!}T^{n}$. Hence, $\exp^{\star}f=\sum_{n\geq0}\dfrac{1}{n!}f^{\star n}$ (by
the definition of $\exp^{\star}f$). Hence,%
\[
\left(  \exp^{\star}f\right)  \left(  z\right)  =\left(  \sum_{n\geq0}%
\dfrac{1}{n!}f^{\star n}\right)  \left(  z\right)  =\underbrace{\sum_{n\geq0}%
}_{=\sum_{n\in\NN}}\dfrac{1}{n!}f^{\star n}\left(  z\right)
=\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}\left(  z\right)  .
\]
In particular, the sum $\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}\left(
z\right)  $ is well-defined. Hence, the family $\left(  \dfrac{1}{n!}f^{\star
n}\left(  z\right)  \right)  _{n\in\NN}\in A^{\NN}$ is finitely
supported. Thus, the proof of Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) is complete.

(b) Let $x\in C$ and $y\in C$ be arbitrary.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=x$) shows that the
family $\left(  \dfrac{1}{n!}f^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies
$\left(  \exp^{\star}f\right)  \left(  x\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  x\right)  $. Renaming the index $n$ as $q$ in
this sentence, we obtain the following: The family $\left(  \dfrac{1}%
{q!}f^{\star q}\left(  x\right)  \right)  _{q\in\NN}\in A^{\NN}$
is finitely supported.and satisfies
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  x\right)  =\sum_{q\in\NN}%
\dfrac{1}{q!}f^{\star q}\left(  x\right)  . \label{pf.prop.leray.exp-algh1.1'}%
\end{equation}
Similarly, the family $\left(  \dfrac{1}{r!}f^{\star r}\left(  y\right)
\right)  _{r\in\NN}\in A^{\NN}$ is finitely supported and
satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  y\right)  =\sum_{r\in\NN}%
\dfrac{1}{r!}f^{\star r}\left(  y\right)  . \label{pf.prop.leray.exp-algh1.2'}%
\end{equation}


Recall that sums of the form $\sum_{q\in Q}f_{q}$ (where $\left(
f_{q}\right)  _{q\in Q}$ is a finitely supported family) satisfy the usual
rules for finite sums, even though their indexing set $Q$ may be infinite.
This pertains, in particular, to the sums $\sum_{q\in\NN}\dfrac{1}%
{q!}f^{\star q}\left(  x\right)  $ and $\sum_{r\in\NN}\dfrac{1}%
{r!}f^{\star r}\left(  y\right)  $ (because the families $\left(  \dfrac
{1}{q!}f^{\star q}\left(  x\right)  \right)  _{q\in\NN}$ and $\left(
\dfrac{1}{r!}f^{\star r}\left(  y\right)  \right)  _{r\in\NN}$ are
finitely supported). Thus, the following manipulations make sense:%
\begin{align*}
&  \left(  \sum_{q\in\NN}\dfrac{1}{q!}f^{\star q}\left(  x\right)
\right)  \left(  \sum_{r\in\NN}\dfrac{1}{r!}f^{\star r}\left(
y\right)  \right) \\
&  =\sum_{q\in\NN}\sum_{r\in\NN}\dfrac{1}{q!}f^{\star q}\left(
x\right)  \dfrac{1}{r!}f^{\star r}\left(  y\right)  =\sum_{q\in\NN}
\underbrace{\sum_{r\in\NN}\dfrac{1}{q!\cdot r!}f^{\star q}\left(
x\right)  f^{\star r}\left(  y\right)  }_{\substack{=\sum\limits_{\substack{n\in
\mathbb{N};\\n\geq q}}\dfrac{1}{q!\cdot\left(  n-q\right)  !}f^{\star
q}\left(  x\right)  f^{\star\left(  n-q\right)  }\left(  y\right)
\\\text{(here, we have substituted }n-q\text{ for }r\text{ in the sum)}}}\\
&  =\underbrace{\sum_{q\in\NN}\sum\limits_{\substack{n\in\mathbb{N};\\n\geq
q}}}_{=\sum_{n\in\NN}\sum\limits_{\substack{q\in\mathbb{N};\\n\geq q}}}%
\dfrac{1}{q!\cdot\left(  n-q\right)  !}f^{\star q}\left(  x\right)
f^{\star\left(  n-q\right)  }\left(  y\right)  =\sum_{n\in\mathbb{N}%
}\underbrace{\sum\limits_{\substack{q\in\mathbb{N};\\n\geq q}}}_{=\sum\limits
_{\substack{q\in\mathbb{N};\\q\leq n}}=\sum_{q=0}^{n}}\dfrac{1}{q!\cdot\left(
n-q\right)  !}f^{\star q}\left(  x\right)  f^{\star\left(  n-q\right)
}\left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{q=0}^{n}\dfrac{1}{q!\cdot\left(  n-q\right)
!}f^{\star q}\left(  x\right)  f^{\star\left(  n-q\right)  }\left(  y\right)
=\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}f^{\star i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)
\end{align*}
(here, we have renamed the summation index $q$ as $i$). Comparing this with%
\[
\underbrace{\left(  \sum_{q\in\NN}\dfrac{1}{q!}f^{\star q}\left(
x\right)  \right)  }_{\substack{=\left(  \exp^{\star}f\right)  \left(
x\right)  \\\text{(by (\ref{pf.prop.leray.exp-algh1.1'}))}}%
}\underbrace{\left(  \sum_{r\in\NN}\dfrac{1}{r!}f^{\star r}\left(
y\right)  \right)  }_{\substack{=\left(  \exp^{\star}f\right)  \left(
y\right)  \\\text{(by (\ref{pf.prop.leray.exp-algh1.2'}))}}}=\left(
\exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star}f\right)
\left(  y\right)  ,
\]
we obtain%
\[
\left(  \exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  \left(  y\right)  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  .
\]
In particular, the sum on the right hand side of this equality is
well-defined. Thus, the family \newline$\left(  \sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  \right)  _{n\in\NN}\in A^{\NN}$
is finitely supported. This completes the proof of
Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b).
\end{proof}

\begin{lemma}
\label{lem.pf.prop.leray.exp-algh1.2}Let $C$ be a $\kk$-bialgebra. Let
$A$ be a $\kk$-algebra. Let $f\in\Hom \left(  C,A\right)
$. Every $n\in \NN$ satisfies%
\begin{equation}
f^{\star n}\left(  1\right)  =\left(  f\left(  1\right)  \right)  ^{n}.
\label{pf.prop.leray.exp-algh1.F1}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.pf.prop.leray.exp-algh1.2}.]We must prove that
(\ref{pf.prop.leray.exp-algh1.F1}) holds for each $n\in \NN$. We shall
prove this by induction over $n$:

\textit{Induction base:} The axioms of a $\kk$-bialgebra yield
$\epsilon_{C}\left(  1\right)  =1$ (since $C$ is a $\kk$-bialgebra).
Now,
\[
f^{\star0}=\left(  \text{the unity of the }\kk\text{-algebra
} \left(\Hom \left(  C,A\right) , \star\right) \right)
=u_{A}\circ\epsilon_{C}.
\]
Thus,%
\[
\underbrace{f^{\star0}}_{=u_{A}\circ\epsilon_{C}}\left(  1\right)  =\left(
u_{A}\circ\epsilon_{C}\right)  \left(  1\right)  =u_{A}\left(
\underbrace{\epsilon_{C}\left(  1\right)  }_{=1}\right)  =u_{A}\left(
1\right)  =1\cdot1_{A}=1_{A}=\left(  f\left(  1\right)  \right)  ^{0}.
\]
In other words, (\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Let $k\in \NN$. Assume that
(\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k$. We must show that
(\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k+1$.

The axioms of a $\kk$-bialgebra yield $\Delta_{C}\left(  1\right)
=1\otimes1$ (since $C$ is a $\kk$-bialgebra).

We have assumed that (\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k$. In
other words, we have $f^{\star k}\left(  1\right)  =\left(  f\left(  1\right)
\right)  ^{k}$. Now, $f^{\star\left(  k+1\right)  }=f\star f^{\star k}%
=m_{A}\circ\left(  f\otimes f^{\star k}\right)  \circ\Delta_{C}$ (by the
definition of convolution). Applying both sides of this equality to $1\in C$,
we obtain%
\begin{align*}
f^{\star\left(  k+1\right)  }\left(  1\right)   &  =\left(  m_{A}\circ\left(
f\otimes f^{\star k}\right)  \circ\Delta_{C}\right)  \left(  1\right)
=m_{A}\left(  \left(  f\otimes f^{\star k}\right)  \left(  \underbrace{\Delta
_{C}\left(  1\right)  }_{=1\otimes1}\right)  \right)  =m_{A}\left(
\underbrace{\left(  f\otimes f^{\star k}\right)  \left(  1\otimes1\right)
}_{=f\left(  1\right)  \otimes f^{\star k}\left(  1\right)  }\right) \\
&  =m_{A}\left(  f\left(  1\right)  \otimes f^{\star k}\left(  1\right)
\right)  =f\left(  1\right)  \underbrace{f^{\star k}\left(  1\right)
}_{=\left(  f\left(  1\right)  \right)  ^{k}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }m_{A}\right) \\
&  =f\left(  1\right)  \cdot\left(  f\left(  1\right)  \right)  ^{k}=\left(
f\left(  1\right)  \right)  ^{k+1}.
\end{align*}
In other words, (\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k+1$. This
completes the induction step. Thus, (\ref{pf.prop.leray.exp-algh1.F1}) is
proven by induction. Hence, Lemma~\ref{lem.pf.prop.leray.exp-algh1.2} is proven.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.exp-algh1}.]Let $x\in C$ and $y\in C$.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=xy$) shows that
the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  xy\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  xy\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  xy\right)  .
\label{pf.prop.leray.exp-algh1.q0}%
\end{equation}
Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b) shows that the family $\left(
\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  \left(  y\right)  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  . \label{pf.prop.leray.exp-algh1.q1}%
\end{equation}


Comparing (\ref{pf.prop.leray.exp-algh1.q1}) with%
\begin{align*}
\left(  \exp^{\star}f\right)  \left(  xy\right)   &  =\sum_{n\in\mathbb{N}%
}\dfrac{1}{n!}\underbrace{f^{\star n}\left(  xy\right)  }_{\substack{=\sum
_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  \\\text{(by Proposition~\ref{prop.leray.f*nxy}%
)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.leray.exp-algh1.q0}%
)}\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  =\sum
_{n\in\NN}\sum_{i=0}^{n}\underbrace{\dfrac{1}{n!}\dbinom{n}{i}%
}_{\substack{=\dfrac{1}{i!\cdot\left(  n-i\right)  !}\\\text{(since }%
\dbinom{n}{i}=\dfrac{n!}{i!\cdot\left(  n-i\right)  !}\text{)}}}f^{\star
i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}f^{\star i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)
,
\end{align*}
we obtain $\left(  \exp^{\star}f\right)  \left(  xy\right)  =\left(
\exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star}f\right)
\left(  y\right)  $.

Now, forget that we fixed $x$ and $y$. We thus have shown that
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  xy\right)  =\left(  \exp^{\star
}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star}f\right)  \left(
y\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }x\in C\text{ and }y\in C.
\label{pf.prop.leray.exp-algh1.res1}%
\end{equation}


Now, $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$. Hence, $\exp^{\star}f=\sum
_{n\geq0}\dfrac{1}{n!}f^{\star n}$ (by the definition of $\exp^{\star}f$).
Hence,%
\begin{align*}
\left(  \exp^{\star}f\right)  \left(  1\right)   &  =\left(  \sum_{n\geq
0}\dfrac{1}{n!}f^{\star n}\right)  \left(  1\right)  =\sum_{n\geq0}\dfrac
{1}{n!}\underbrace{f^{\star n}\left(  1\right)  }_{\substack{=\left(  f\left(
1\right)  \right)  ^{n}\\\text{(by Lemma~\ref{lem.pf.prop.leray.exp-algh1.2}%
)}}}=\sum_{n\geq0}\dfrac{1}{n!}\left(  \underbrace{f\left(  1\right)  }%
_{=0}\right)  ^{n}\\
&  =\sum_{n\geq0}\dfrac{1}{n!}0^{n}=\underbrace{\dfrac{1}{0!}}_{=\dfrac{1}%
{1}=1}\underbrace{0^{0}}_{=1}+\sum_{n\geq1}\dfrac{1}{n!}\underbrace{0^{n}%
}_{\substack{=0\\\text{(since }n\geq1\text{)}}}=1+\underbrace{\sum_{n\geq
1}\dfrac{1}{n!}0}_{=0}=1.
\end{align*}


Thus, we know that $\exp^{\star}f$ is a $\kk$-linear map $C\rightarrow
A$ (since $\exp^{\star}f\in\Hom \left(  C,A\right)  $)
satisfying (\ref{pf.prop.leray.exp-algh1.res1}) and $\left(  \exp^{\star
}f\right)  \left(  1\right)  =1$. In other words, $\exp^{\star}f:C\rightarrow
A$ is a $\kk$-algebra homomorphism. This proves
Proposition~\ref{prop.leray.exp-algh1}.
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.sol.eulerian-idp.2.poly1}.]The matrix $\left(
i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\in \QQ^{\left(  N+1\right)
\times\left(  N+1\right)  }$ is invertible (since its determinant is the
Vandermonde determinant $\prod_{1\leq i<j\leq N+1}\underbrace{\left(
i-j\right)  }_{\neq0}\neq0$). Let $\left(  s_{i,j}\right)  _{i,j=1,2,\ldots
,N+1}\in \QQ^{\left(  N+1\right)  \times\left(  N+1\right)  }$ be its
inverse matrix. Then, $\left(  s_{i,j}\right)  _{i,j=1,2,\ldots,N+1}%
\cdot\left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}=I_{N+1}$ (the $\left(
N+1\right)  \times\left(  N+1\right)  $ identity matrix). Comparing the
entries on both sides of this equality, we see: For every $\left(  u,v\right)
\in\left\{  1,2,\ldots,N+1\right\}  ^{2}$, we have%
\[
\sum_{j=1}^{N+1}s_{u,j}j^{N+1-v}=\delta_{u,v}%
\]
(because $\sum_{j=1}^{N+1}s_{u,j}j^{N+1-v}$ is the $\left(  u,v\right)  $-th
entry of the matrix $\left(  s_{i,j}\right)  _{i,j=1,2,\ldots,N+1}\cdot\left(
i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}$, while $\delta_{u,v}$ is the $\left(
u,v\right)  $-th entry of the matrix $I_{N+1}$).

The entries $s_{i,j}$ of the matrix $\left(  s_{i,j}\right)  _{i,j=1,2,\ldots
,N+1}$ are rational numbers, and therefore there exists a positive integer $M$
such that every $\left(  i,j\right)  \in\left\{  1,2,\ldots,N+1\right\}  ^{2}$
satisfies $Ms_{i,j}\in \ZZ$ (because finitely many rational numbers
always have a common denominator). Consider this $M$. For every $\left(
i,j\right)  \in\left\{  1,2,\ldots,N+1\right\}  ^{2}$, define an element
$t_{i,j}\in \ZZ$ by $t_{i,j}=Ms_{i,j}$. Then,
\begin{equation}
\sum_{j=1}^{N+1}\underbrace{t_{u,j}}_{=Ms_{u,j}}j^{N+1-v}=M\underbrace{\sum
_{j=1}^{N+1}s_{u,j}j^{N+1-v}}_{=\delta_{u,v}}=M\delta_{u,v}
\label{pf.lem.sol.eulerian-idp.2.poly1.2}%
\end{equation}
for every $\left(  u,v\right)  \in\left\{  1,2,\ldots,N+1\right\}  ^{2}$.

Now, let $i\in\left\{  0,1,\ldots,N\right\}  $. Then,
\begin{align*}
\sum_{j=1}^{N+1}t_{N+1-i,j}\sum_{k=0}^{N}w_{k}j^{k}  &  =\sum_{k=0}%
^{N}\underbrace{\left(  \sum_{j=1}^{N+1}t_{N+1-i,j}j^{k}\right)
}_{\substack{=M\delta_{N+1-i,N+1-k}\\\text{(by
(\ref{pf.lem.sol.eulerian-idp.2.poly1.2}),}\\\text{applied to }u=N+1-i\text{
and }v=N+1-k\text{)}}}w_{k}=M\sum_{k=0}^{N}\underbrace{\delta_{N+1-i,N+1-k}%
}_{=\delta_{i,k}}w_{k}\\
&  =M\underbrace{\sum_{k=0}^{N}\delta_{i,k}w_{k}}_{=w_{i}}=Mw_{i}.
\end{align*}
Hence,%
\[
Mw_{i}=\sum_{j=1}^{N+1}t_{N+1-i,j}\underbrace{\sum_{k=0}^{N}w_{k}j^{k}%
}_{\substack{=0\\\text{(by (\ref{eq.lem.sol.eulerian-idp.2.poly1.ass}),
applied}\\\text{to }n=j\text{)}}}=\sum_{j=1}^{N+1}t_{N+1-i,j}0=0.
\]
Since $M$ is a positive integer, we can cancel $M$ from this equality (because
$V$ is torsionfree), and thus obtain $w_{i}=0$.

Now forget that we fixed $i$. We thus have proven that $w_{i}=0$ for every
$i\in\left\{  0,1,\ldots,N\right\}  $. In other words, $w_{k}=0$ for every
$k\in\left\{  0,1,\ldots,N\right\}  $.
Lemma~\ref{lem.sol.eulerian-idp.2.poly1} is thus proven.
\end{proof}

\begin{proof}
[Proof of Lemma~\ref{lem.leray.poly1-inf}.]The family $\left(  w_{k}\right)
_{k\in\NN}$ is finitely supported. In other words, all but finitely
many $k\in \NN$ satisfy $w_{k}=0$. In other words, there exists a finite
subset $K$ of $\NN$ such that%
\begin{equation}
\text{each }k\in\mathbb{N}\setminus K\text{ satisfies }w_{k}=0.
\label{pf.lem.leray.poly1-inf.fs}%
\end{equation}
Consider this $K$.

The set $K$ is a finite subset of $\NN$, and thus has an upper bound
(since any finite subset of $\NN$ has an upper bound). In other words,
there exists some $N\in \NN$ such that
\begin{equation}
\text{each }n\in K\text{ satisfies }n\leq N. \label{pf.lem.leray.poly1-inf.ub}%
\end{equation}
Consider this $N$.

Now,
\begin{equation}
\text{each }k\in\mathbb{N}\text{ satisfying }k>N\text{ satisfies }w_{k}=0
\label{pf.lem.leray.poly1-inf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray.poly1-inf.1}):} Let
$k\in \NN$ be such that $k>N$. We must prove that $w_{k}=0$.
\par
If we had $k\in K$, then we would have $k\leq N$ (by
(\ref{pf.lem.leray.poly1-inf.ub}) (applied to $n=k$)), which would contradict
$k>N$. Hence, we cannot have $k\in K$. Thus, we have $k\notin K$. Combining
$k\in \NN$ with $k\notin K$, we obtain $k\in\mathbb{N}\setminus K$.
Hence, (\ref{pf.lem.leray.poly1-inf.fs}) yields $w_{k}=0$. This proves
(\ref{pf.lem.leray.poly1-inf.1}).}.

We have assumed that $\sum_{k\in\NN}w_{k}n^{k}=0$ for all
$n\in \NN$. Thus, for all $n\in \NN$, we have%
\[
0=\sum_{k\in\NN}w_{k}n^{k}=\underbrace{\sum\limits_{\substack{k\in
\mathbb{N};\\k\leq N}}}_{=\sum_{k=0}^{N}}w_{k}n^{k}+\sum\limits_{\substack{k\in
\mathbb{N};\\k>N}}\underbrace{w_{k}}_{\substack{=0\\\text{(by
(\ref{pf.lem.leray.poly1-inf.1}))}}}n^{k}=\sum_{k=0}^{N}w_{k}n^{k}%
+\underbrace{\sum\limits_{\substack{k\in\mathbb{N};\\k>N}}0n^{k}}_{=0}=\sum_{k=0}%
^{N}w_{k}n^{k}.
\]
Hence, for all $n\in \NN$, we have $\sum_{k=0}^{N}w_{k}n^{k}=0$. Thus,
Lemma~\ref{lem.sol.eulerian-idp.2.poly1} shows that%
\begin{equation}
w_{k}=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,\ldots,N\right\}
. \label{pf.lem.leray.poly1-inf.3}%
\end{equation}


Now, it is easy to see that $w_{k}=0$ for every $k\in \NN$%
\ \ \ \ \footnote{\textit{Proof.} Let $k\in \NN$. We must prove that
$w_{k}=0$.
\par
If $k\in\left\{  0,1,\ldots,N\right\}  $, then this follows immediately from
(\ref{pf.lem.leray.poly1-inf.3}). Hence, for the rest of this proof, we WLOG
assume that we don't have $k\in\left\{  0,1,\ldots,N\right\}  $. Thus,
$k\notin\left\{  0,1,\ldots,N\right\}  $. Hence, $k\in\mathbb{N}%
\setminus\left\{  0,1,\ldots,N\right\}  =\left\{  N+1,N+2,N+3,\ldots\right\}
$. Therefore, $k\geq N+1>N$. Thus, (\ref{pf.lem.leray.poly1-inf.1}) shows that
$w_{k}=0$. Qed.}. This proves Lemma~\ref{lem.leray.poly1-inf}.
\end{proof}

Before we come to the proof of Proposition~\ref{prop.leray.exp-algh2}, we
state another lemma, which is an easy consequence of Lemma~\ref{lem.leray.poly1-inf}:

\begin{lemma}
\label{lem.leray.poly1-inf2}Let $V$ be a torsionfree abelian group (written
additively). Let $\left(  a_{k}\right)  _{k\in\NN}\in V^{\NN}$
and $\left(  b_{k}\right)  _{k\in\NN}\in V^{\NN}$ be two
finitely supported families of elements of $V$. Assume that%
\begin{equation}
\sum_{k\in\NN}a_{k}t^{k}=\sum_{k\in\NN}b_{k}t^{k}%
\ \ \ \ \ \ \ \ \ \ \text{for all }t\in\mathbb{N}.
\label{eq.lem.leray.poly1-inf2.cond}%
\end{equation}
Then, $a_{k}=b_{k}$ for every $k\in \NN$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray.poly1-inf2}.]The families $\left(
a_{k}\right)  _{k\in\NN}$ and $\left(  b_{k}\right)  _{k\in\NN}$
are finitely supported. Hence, (by a straightforward and well-known argument)
it follows that the family $\left(  a_{k}-b_{k}\right)  _{k\in\NN}\in
V^{\NN}$ is also finitely supported.

Now, let $t\in \NN$. Then, the family $\left(  a_{k}t^{k}\right)
_{k\in\NN}\in V^{\NN}$ is finitely supported.\footnote{This is
an easy consequence of the fact that the family $\left(  a_{k}\right)
_{k\in\NN}$ is finitely supported.} Similarly, the family $\left(
b_{k}t^{k}\right)  _{k\in\NN}\in V^{\NN}$ is finitely supported.

Recall that sums of the form $\sum_{q\in Q}f_{q}$ (where $\left(
f_{q}\right)  _{q\in Q}$ is a finitely supported family) satisfy the usual
rules for finite sums, even though their indexing set $Q$ may be infinite. In
particular, this pertains to the sums $\sum_{k\in\NN}a_{k}t^{k}$ and
$\sum_{k\in\NN}b_{k}t^{k}$ (since the families $\left(  a_{k}%
t^{k}\right)  _{k\in\NN}$ and $\left(  b_{k}t^{k}\right)
_{k\in\NN}$ are finitely supported). Hence, the following manipulations
are valid:%
\[
\sum_{k\in\NN}a_{k}t^{k}-\sum_{k\in\NN}b_{k}t^{k}=\sum
_{k\in\NN}\underbrace{\left(  a_{k}t^{k}-b_{k}t^{k}\right)  }_{=\left(
a_{k}-b_{k}\right)  t^{k}}=\sum_{k\in\NN}\left(  a_{k}-b_{k}\right)
t^{k}.
\]
Hence,%
\begin{equation}
\sum_{k\in\NN}\left(  a_{k}-b_{k}\right)  t^{k}=\sum_{k\in\mathbb{N}%
}a_{k}t^{k}-\sum_{k\in\NN}b_{k}t^{k}=0
\label{pf.lem.leray.poly1-inf2.3}%
\end{equation}
(by (\ref{eq.lem.leray.poly1-inf2.cond})).

Now, forget that we fixed $t$. We thus have proven that $\sum_{k\in\mathbb{N}%
}\left(  a_{k}-b_{k}\right)  t^{k}=0$ for all $t\in \NN$. Renaming the
index $t$ as $n$ in this statement, we conclude that $\sum_{k\in\mathbb{N}%
}\left(  a_{k}-b_{k}\right)  n^{k}=0$ for all $n\in \NN$. Hence,
Lemma~\ref{lem.leray.poly1-inf} (applied to $\left(  w_{k}\right)
_{k\in\NN}=\left(  a_{k}-b_{k}\right)  _{k\in\NN}$) shows that
$a_{k}-b_{k}=0$ for every $k\in \NN$. In other words, $a_{k}=b_{k}$ for
every $k\in \NN$. This proves Lemma~\ref{lem.leray.poly1-inf2}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.exp-algh2}.]Any $\kk$-module
naturally becomes a $\QQ$-module (since $\kk$ is a $\QQ$-algebra).
Thus, in particular, the $\kk$-module $A$ becomes a
$\QQ$-module. Hence, the $\kk$-module $A$ is a torsionfree
abelian group (written additively).

Now, fix $x\in\ker\epsilon$ and $y\in\ker\epsilon$. We shall show that
$f\left(  xy\right)  =0$.

We have $f\left(  C_{0}\right)  =0$. Thus,
Proposition~\ref{prop.convolution-series}(h) yields $f\in\mathfrak{n}\left(
C,A\right)  $.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=xy$) shows that
the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  xy\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\[
\left(  \exp^{\star}f\right)  \left(  xy\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  xy\right)  .
\]
In particular, the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  xy\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported. Renaming
the index $n$ as $k$ in this statement, we conclude that the family $\left(
\dfrac{1}{k!}f^{\star k}\left(  xy\right)  \right)  _{k\in\NN}\in
A^{\NN}$ is finitely supported.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b) yields that the family $\left(
\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported. In other words,
the family $\left(  \sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)  }%
{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(
n-i\right)  !}\right)  _{n\in\NN}\in A^{\NN}$ is finitely
supported (since $\dfrac{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  =\dfrac{f^{\star
i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(
y\right)  }{\left(  n-i\right)  !}$ for every $n\in \NN$ and
$i\in\left\{  0,1,\ldots,n\right\}  $). Renaming the index $n$ as $k$ in this
statement, we obtain the following: The family $\left(  \sum_{i=0}^{k}%
\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(
k-i\right)  }\left(  y\right)  }{\left(  k-i\right)  !}\right)  _{k\in
\NN}\in A^{\NN}$ is finitely supported.

On the other hand, let $t\in \NN$ be arbitrary. Then,
Proposition~\ref{prop.exp-log}(e) (applied to $n=t$) shows that $tf\in
\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}\left(  tf\right)  =\left(
\exp^{\star}f\right)  ^{\star t}$.

Exercise~\ref{exe.alghoms.comm}(b) (applied to $C$, $t$ and $\exp^{\star}f$
instead of $H$, $k$ and $f_{i}$) shows that the map \newline%
$\underbrace{\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}f\right)
\star\cdots\star\left(  \exp^{\star}f\right)  }_{t\text{ times}}$ is a
$\kk$-algebra homomorphism $C\rightarrow A$. In light of%
\[
\underbrace{\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}f\right)
\star\cdots\star\left(  \exp^{\star}f\right)  }_{t\text{ times}}=\left(
\exp^{\star}f\right)  ^{\star t}=\exp^{\star}\left(  tf\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\exp^{\star}\left(  tf\right)
=\left(  \exp^{\star}f\right)  ^{\star t}\right)  ,
\]
this rewrites as follows: The map $\exp^{\star}\left(  tf\right)  $ is a
$\kk$-algebra homomorphism $C\rightarrow A$.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $tf$ and $xy$ instead
of $f$ and $z$) shows that the family $\left(  \dfrac{1}{n!}\left(  tf\right)
^{\star n}\left(  xy\right)  \right)  _{n\in\NN}\in A^{\NN}$ is
finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  xy\right)  =\sum
_{n\in\NN}\dfrac{1}{n!}\left(  tf\right)  ^{\star n}\left(  xy\right)
. \label{pf.prop.leray.exp-algh2.etfxy}%
\end{equation}


Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b) (applied to $tf$ instead of $f$)
shows that the family \newline$\left(  \sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(
n-i\right)  !}\left(  tf\right)  ^{\star i}\left(  x\right)  \left(
tf\right)  ^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\[
\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  x\right)  \cdot\left(
\exp^{\star}\left(  tf\right)  \right)  \left(  y\right)  =\sum_{n\in
\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}\left(
tf\right)  ^{\star i}\left(  x\right)  \left(  tf\right)  ^{\star\left(
n-i\right)  }\left(  y\right)  .
\]
Comparing
\begin{align*}
&  \left(  \exp^{\star}\left(  tf\right)  \right)  \left(  x\right)
\cdot\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}\underbrace{\left(  tf\right)  ^{\star i}\left(  x\right)  }_{=t^{i}f^{\star
i}\left(  x\right)  }\underbrace{\left(  tf\right)  ^{\star\left(  n-i\right)
}\left(  y\right)  }_{=t^{n-i}f^{\star\left(  n-i\right)  }\left(  y\right)
}=\sum_{n\in\NN}\sum_{i=0}^{n}\underbrace{\dfrac{1}{i!\cdot\left(
n-i\right)  !}t^{i}f^{\star i}\left(  x\right)  t^{n-i}f^{\star\left(
n-i\right)  }\left(  y\right)  }_{=\dfrac{f^{\star i}\left(  x\right)  }%
{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(
n-i\right)  !}t^{i}t^{n-i}}\\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)
}{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(
n-i\right)  !}\underbrace{t^{i}t^{n-i}}_{=t^{i+\left(  n-i\right)  }=t^{n}%
}=\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)  }%
{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(
n-i\right)  !}t^{n}\\
&  =\sum_{n\in\NN}\left(  \sum_{i=0}^{n}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)
}{\left(  n-i\right)  !}\right)  t^{n}=\sum_{k\in\NN}\left(  \sum
_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(
k-i\right)  }\left(  y\right)  }{\left(  k-i\right)  !}\right)  t^{k}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}n\text{ as }k\right)
\end{align*}
with%
\begin{align*}
&  \left(  \exp^{\star}\left(  tf\right)  \right)  \left(  x\right)
\cdot\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  y\right) \\
&  =\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  xy\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\exp^{\star}\left(  tf\right)  \text{
is a }\kk\text{-algebra homomorphism}\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\underbrace{\left(  tf\right)  ^{\star
n}\left(  xy\right)  }_{\substack{=t^{n}f^{\star n}\left(  xy\right)
\\=f^{\star n}\left(  xy\right)  t^{n}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.leray.exp-algh2.etfxy})}\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}\left(  xy\right)  t^{n}%
=\sum_{k\in\NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)
t^{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation
index }n\text{ as }k\right)  ,
\end{align*}
we obtain%
\[
\sum_{k\in\NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)  t^{k}%
=\sum_{k\in\NN}\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)
}{\left(  k-i\right)  !}\right)  t^{k}.
\]


Now, forget that we fixed $t$. We thus have shown that%
\[
\sum_{k\in\NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)  t^{k}%
=\sum_{k\in\NN}\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)
}{\left(  k-i\right)  !}\right)  t^{k}%
\]
for all $t\in \NN$.

Hence, we can apply Lemma~\ref{lem.leray.poly1-inf2} to $V=A$, $\left(
a_{k}\right)  _{k\in\NN}=\left(  \dfrac{1}{k!}f^{\star k}\left(
xy\right)  \right)  _{k\in\NN}$ and \newline$\left(  b_{k}\right)
_{k\in\NN}=\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)
}{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)  }{\left(
k-i\right)  !}\right)  _{k\in\NN}$ (since $A$ is a torsionfree abelian
group (written additively), and since the families $\left(  \dfrac{1}%
{k!}f^{\star k}\left(  xy\right)  \right)  _{k\in\NN}$ and $\left(
\sum_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac
{f^{\star\left(  k-i\right)  }\left(  y\right)  }{\left(  k-i\right)
!}\right)  _{k\in\NN}$ are finitely supported). As a result, we obtain%
\[
\dfrac{1}{k!}f^{\star k}\left(  xy\right)  =\sum_{i=0}^{k}\dfrac{f^{\star
i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(
y\right)  }{\left(  k-i\right)  !}\ \ \ \ \ \ \ \ \ \ \text{for every }%
k\in\mathbb{N}.
\]
Applying this to $k=1$, we find%
\begin{align}
\dfrac{1}{1!}f^{\star1}\left(  xy\right)   &  =\sum_{i=0}^{1}\dfrac{f^{\star
i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(  1-i\right)  }\left(
y\right)  }{\left(  1-i\right)  !}=\dfrac{f^{\star0}\left(  x\right)  }%
{0!}\cdot\dfrac{f^{\star\left(  1-0\right)  }\left(  y\right)  }{\left(
1-0\right)  !}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot\dfrac
{f^{\star\left(  1-1\right)  }\left(  y\right)  }{\left(  1-1\right)
!}\nonumber\\
&  =\dfrac{f^{\star0}\left(  x\right)  }{0!}\cdot\dfrac{f^{\star1}\left(
y\right)  }{1!}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot\dfrac{f^{\star
0}\left(  y\right)  }{0!} \label{pf.prop.leray.exp-algh2.5}%
\end{align}
(since $1-0=1$ and $1-1=0$).

But recall that $x\in\ker\underbrace{\epsilon}_{=\epsilon_{C}}=\ker\left(
\epsilon_{C}\right)  $ and thus $\epsilon_{C}\left(  x\right)  =0$. But%
\[
f^{\star0}=\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right)  =u_{A}%
\circ\epsilon_{C},
\]
and thus $f^{\star0}\left(  x\right)  =\left(  u_{A}\circ\epsilon_{C}\right)
\left(  x\right)  =u_{A}\left(  \underbrace{\epsilon_{C}\left(  x\right)
}_{=0}\right)  =u_{A}\left(  0\right)  =0$. Hence, $\dfrac{f^{\star0}\left(
x\right)  }{0!}=\dfrac{0}{0!}=0$. Similarly, $\dfrac{f^{\star0}\left(
y\right)  }{0!}=0$. Hence, (\ref{pf.prop.leray.exp-algh2.5}) becomes%
\[
\dfrac{1}{1!}f^{\star1}\left(  xy\right)  =\underbrace{\dfrac{f^{\star
0}\left(  x\right)  }{0!}}_{=0}\cdot\dfrac{f^{\star1}\left(  y\right)  }%
{1!}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot\underbrace{\dfrac
{f^{\star0}\left(  y\right)  }{0!}}_{=0}=0\cdot\dfrac{f^{\star1}\left(
y\right)  }{1!}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot0=0.
\]
Comparing this with $\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}=1}%
\underbrace{f^{\star1}}_{=f}\left(  xy\right)  =f\left(  xy\right)  $, we
obtain $f\left(  xy\right)  =0$.

Now, forget that we fixed $x$ and $y$. We thus have shown that%
\[
f\left(  xy\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\ker
\epsilon\text{ and }y\in\ker\epsilon\text{.}%
\]


Since the map $f$ is $\kk$-linear, this entails that $f\left(  \left(
\ker\epsilon\right)  ^{2}\right)  =0$ (because the $\kk$-module
$\left(  \ker\epsilon\right)  ^{2}$ is spanned by elements of the form $xy$
with $x\in\ker\epsilon$ and $y\in\ker\epsilon$). This proves
Proposition~\ref{prop.leray.exp-algh2}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.exp-algh3}.]We know that $f\left(
C\right)  $ generates the $\kk$-algebra $A$. Thus, any $\kk%
$-subalgebra of $A$ that contains $f\left(  C\right)  $ as a subset must be
the whole $A$. In other words,
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }D\text{ is a }\kk\text{-subalgebra of }A\text{ satisfying
}f\left(  C\right)  \subset D\text{,}\\
\text{then }D=A
\end{array}
\right)  . \label{pf.prop.leray.exp-algh3.gener}%
\end{equation}


Define a $\kk$-linear map $F\in\Hom \left(  C,A\right)  $
by $F=\exp^{\star}f$. (This is well-defined, since $f\in\mathfrak{n}\left(
C,A\right)  $.)

Proposition~\ref{prop.leray.exp-algh1} shows that $\exp^{\star}f:C\rightarrow
A$ is a $\kk$-algebra homomorphism. In other words, $F:C\rightarrow A$
is a $\kk$-algebra homomorphism (since $F=\exp^{\star}f$).

Proposition~\ref{prop.exp-log}(a) yields that $\exp^{\star}f-u_{A}\epsilon
_{C}\in\mathfrak{n}\left(  C,A\right)  $ and $\log^{\star}\left(  \exp^{\star
}f\right)  =f$.

Define an element $\widetilde{F}$ of $\mathfrak{n}\left(  C,A\right)  $ by
$\widetilde{F}=F-u_{A}\epsilon_{C}$. (This is well-defined, since
$\underbrace{F}_{=\exp^{\star}f}-u_{A}\epsilon_{C}=\exp^{\star}f-u_{A}%
\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $.) From $\log^{\star}\left(
\exp^{\star}f\right)  =f$, we obtain%
\begin{align}
f  &  =\log^{\star}\underbrace{\left(  \exp^{\star}f\right)  }_{=F}%
=\log^{\star}F=\overline{\log}^{\star}\underbrace{\left(  F-u_{A}\epsilon
_{C}\right)  }_{=\widetilde{F}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\log^{\star}F\right) \nonumber\\
&  =\overline{\log}^{\star}\widetilde{F}.
\label{pf.prop.leray.exp-algh3.f=log}%
\end{align}


On the other hand, Proposition~\ref{prop.convolution-series}(j) (applied to
$C$ and $\id_{C}$ instead of $A$ and $F$) shows that
$\id_{C}-u_{C}\epsilon_{C}\in\mathfrak{n}\left(
C,C\right)  $ (since $\id_{C}:C\rightarrow C$ is a
$\kk$-algebra homomorphism).

Define an element $\widetilde{\id}$ of $\mathfrak{n}\left(
C,C\right)  $ by $\widetilde{\id}=\operatorname*{id}%
\nolimits_{C}-u_{C}\epsilon_{C}$. (This is well-defined, since
$\id_{C}-u_{C}\epsilon_{C}\in\mathfrak{n}\left(
C,C\right)  $.) Then, $F\circ\widetilde{\id}=\widetilde{F}%
$\ \ \ \ \footnote{\textit{Proof.} Recall that $F$ is a $\kk$-algebra
homomorphism. In other words, $F$ is a $\kk$-linear map satisfying
$F\circ m_{C}=m_{A}\circ\left(  F\otimes F\right)  $ and $F\circ u_{C}=u_{A}$.
Now,%
\begin{align*}
F\circ\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{C}-u_{C}\epsilon_{C}}  &  =F\circ\left(  \operatorname*{id}%
\nolimits_{C}-u_{C}\epsilon_{C}\right)  =\underbrace{F\circ\operatorname*{id}%
\nolimits_{C}}_{=F}-\underbrace{F\circ\left(  u_{C}\epsilon_{C}\right)
}_{=F\circ u_{C}\circ\epsilon_{C}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
composition of }\kk\text{-linear maps is }\kk\text{-bilinear}%
\right) \\
&  =F-\underbrace{F\circ u_{C}}_{=u_{A}}\circ\epsilon_{C}=F-u_{A}\circ
\epsilon_{C}=F-u_{A}\epsilon_{C}=\widetilde{F}.
\end{align*}
}. Hence, (\ref{pf.prop.leray.exp-algh3.f=log}) becomes%
\begin{equation}
f=\overline{\log}^{\star}\underbrace{\widetilde{F}}_{=F\circ
\widetilde{\id}}=\overline{\log}^{\star}\left(  F\circ
\widetilde{\id}\right)  .
\label{pf.prop.leray.exp-algh3.f=log2}%
\end{equation}


Proposition~\ref{prop.convolution-series}(i) (applied to $C$, $A$, $F$,
$\overline{\log}$ and $\widetilde{\id}$ instead of $A$, $B$,
$s$, $u$ and $f$) shows that%
\[
F\circ\widetilde{\id}\in\mathfrak{n}\left(  C,A\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \overline{\log}^{\star
}\left(  F\circ\widetilde{\id}\right)  =F\circ\left(
\overline{\log}^{\star}\left(  \widetilde{\id}\right)  \right)
.
\]


Now, (\ref{pf.prop.leray.exp-algh3.f=log2}) becomes%
\[
f=\overline{\log}^{\star}\left(  F\circ\widetilde{\id}\right)
=F\circ\left(  \overline{\log}^{\star}\left(  \widetilde{\operatorname*{id}%
}\right)  \right)  .
\]
Hence,%
\[
f\left(  C\right)  =\left(  F\circ\left(  \overline{\log}^{\star}\left(
\widetilde{\id}\right)  \right)  \right)  \left(  C\right)
=F\left(  \underbrace{\left(  \overline{\log}^{\star}\left(
\widetilde{\id}\right)  \right)  \left(  C\right)  }_{\subset
C}\right)  \subset F\left(  C\right)  .
\]


But $F\left(  C\right)  $ is a $\kk$-subalgebra of $A$ (since $F$ is a
$\kk$-algebra homomorphism). Hence,
(\ref{pf.prop.leray.exp-algh3.gener}) (applied to $D=F\left(  C\right)  $)
shows that $F\left(  C\right)  =A$ (since $f\left(  C\right)  \subset F\left(
C\right)  $). In other words, the map $F$ is surjective.

Hence, $F$ is a surjective $\kk$-algebra homomorphism. In other words,
$\exp^{\star}f$ is a surjective $\kk$-algebra homomorphism (since
$F=\exp^{\star}f$). This proves Proposition~\ref{prop.leray.exp-algh3}.
\end{proof}

We have now proven Lemmas~\ref{lem.sol.eulerian-idp.2.poly1}%
~and~\ref{lem.leray.poly1-inf} and Propositions~\ref{prop.convolution.nat}%
,~\ref{prop.leray.f*nxy},~\ref{prop.leray.exp-algh1}%
,~\ref{prop.leray.exp-algh2}~and~\ref{prop.leray.exp-algh3}. Thus,
Exercise~\ref{exe.leray.exp-alg} is solved.

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.leray.exp-alg}}

\textit{Solution to Exercise~\ref{exe.leray.exp-alg}.}

\begin{proof}
[Proof of Proposition~\ref{prop.convolution.nat}.]We know that $\gamma
:C\rightarrow C^{\prime}$ is a $\kk$-coalgebra morphism. In other
words, $\gamma:C\rightarrow C^{\prime}$ is a $\kk$-linear map
satisfying $\Delta_{C^{\prime}}\circ\gamma=\left(  \gamma\otimes\gamma\right)
\circ\Delta_{C}$ and $\epsilon_{C^{\prime}}\circ\gamma=\epsilon_{C}$.

We know that $\alpha:A\rightarrow A^{\prime}$ is a $\kk$-algebra
morphism. In other words, $\alpha:A\rightarrow A^{\prime}$ is a $\kk%
$-linear map satisfying $\alpha\circ m_{A}=m_{A^{\prime}}\circ\left(
\alpha\otimes\alpha\right)  $ and $\alpha\circ u_{A}=u_{A^{\prime}}$.

(a) Let $f\in\Hom \left(  C,A\right)  $, $g\in
\Hom \left(  C,A\right)  $, $f^{\prime}\in\Hom
\left(  C^{\prime},A^{\prime}\right)  $ and $g^{\prime}\in\Hom
\left(  C^{\prime},A^{\prime}\right)  $ be such that $f^{\prime}\circ
\gamma=\alpha\circ f$ and $g^{\prime}\circ\gamma=\alpha\circ g$. We must prove
that $\left(  f^{\prime}\star g^{\prime}\right)  \circ\gamma=\alpha
\circ\left(  f\star g\right)  $.

The definition of convolution yields $f\star g=m_{A}\circ\left(  f\otimes
g\right)  \circ\Delta_{C}$ and $f^{\prime}\star g^{\prime}=m_{A^{\prime}}%
\circ\left(  f^{\prime}\otimes g^{\prime}\right)  \circ\Delta_{C^{\prime}}$.
Now,%
\begin{align*}
\underbrace{\left(  f^{\prime}\star g^{\prime}\right)  }_{=m_{A^{\prime}}%
\circ\left(  f^{\prime}\otimes g^{\prime}\right)  \circ\Delta_{C^{\prime}}%
}\circ\gamma &  =m_{A^{\prime}}\circ\left(  f^{\prime}\otimes g^{\prime
}\right)  \circ\underbrace{\Delta_{C^{\prime}}\circ\gamma}_{=\left(
\gamma\otimes\gamma\right)  \circ\Delta_{C}}=m_{A^{\prime}}\circ
\underbrace{\left(  f^{\prime}\otimes g^{\prime}\right)  \circ\left(
\gamma\otimes\gamma\right)  }_{=\left(  f^{\prime}\circ\gamma\right)
\otimes\left(  g^{\prime}\circ\gamma\right)  }\circ\Delta_{C}\\
&  =m_{A^{\prime}}\circ\left(  \underbrace{\left(  f^{\prime}\circ
\gamma\right)  }_{=\alpha\circ f}\otimes\underbrace{\left(  g^{\prime}%
\circ\gamma\right)  }_{=\alpha\circ g}\right)  \circ\Delta_{C}=m_{A^{\prime}%
}\circ\underbrace{\left(  \left(  \alpha\circ f\right)  \otimes\left(
\alpha\circ g\right)  \right)  }_{=\left(  \alpha\otimes\alpha\right)
\circ\left(  f\otimes g\right)  }\circ\Delta_{C}\\
&  =\underbrace{m_{A^{\prime}}\circ\left(  \alpha\otimes\alpha\right)
}_{=\alpha\circ m_{A}}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
=\alpha\circ\underbrace{m_{A}\circ\left(  f\otimes g\right)  \circ\Delta_{C}%
}_{=f\star g}=\alpha\circ\left(  f\star g\right)  .
\end{align*}
This proves Proposition~\ref{prop.convolution.nat}(a).

(b) Let $f\in\Hom \left(  C,A\right)  $ and $f^{\prime}%
\in\Hom \left(  C^{\prime},A^{\prime}\right)  $ be such that
$f^{\prime}\circ\gamma=\alpha\circ f$. We must prove that every $n\in
\mathbb{N}$ satisfies%
\begin{equation}
\left(  f^{\prime}\right)  ^{\star n}\circ\gamma=\alpha\circ f^{\star n}.
\label{pf.prop.convolution.nat.b.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.convolution.nat.b.goal}):} We shall prove
(\ref{pf.prop.convolution.nat.b.goal}) by induction over $n$:

\textit{Induction base:} We have $f^{\star0}=\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \right)  =u_{A}\circ\epsilon_{C}$ and similarly $\left(
f^{\prime}\right)  ^{\star0}=u_{A^{\prime}}\circ\epsilon_{C^{\prime}}$. Hence,%
\[
\underbrace{\left(  f^{\prime}\right)  ^{\star0}}_{=u_{A^{\prime}}%
\circ\epsilon_{C^{\prime}}}\circ\gamma=\underbrace{u_{A^{\prime}}}%
_{=\alpha\circ u_{A}}\circ\underbrace{\epsilon_{C^{\prime}}\circ\gamma
}_{=\epsilon_{C}}=\alpha\circ\underbrace{u_{A}\circ\epsilon_{C}}_{=f^{\star0}%
}=\alpha\circ f^{\star0}.
\]
In other words, (\ref{pf.prop.convolution.nat.b.goal}) holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Fix $k\in \NN$. Assume that
(\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k$. We must now prove that
(\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k+1$.

We have assumed that (\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k$.
In other words, we have $\left(  f^{\prime}\right)  ^{\star k}\circ
\gamma=\alpha\circ f^{\star k}$.

Now,%
\begin{align*}
\underbrace{\left(  f^{\prime}\right)  ^{\star\left(  k+1\right)  }%
}_{=f^{\prime}\star\left(  f^{\prime}\right)  ^{\star k}}\circ\gamma &
=\left(  f^{\prime}\star\left(  f^{\prime}\right)  ^{\star k}\right)
\circ\gamma=\alpha\circ\underbrace{\left(  f\star f^{\star k}\right)
}_{=f^{\star\left(  k+1\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition~\ref{prop.convolution.nat}%
(a) (applied to }g=f^{\star k}\text{ and }g^{\prime}=\left(  f^{\prime
}\right)  ^{\star k}\text{)}\right) \\
&  =\alpha\circ f^{\star\left(  k+1\right)  }.
\end{align*}
In other words, (\ref{pf.prop.convolution.nat.b.goal}) holds for $n=k+1$. This
completes the induction step. Thus, the induction proof of
(\ref{pf.prop.convolution.nat.b.goal}) is finished.]

Hence, (\ref{pf.prop.convolution.nat.b.goal}) is proven. In other words, we
have proven Proposition~\ref{prop.convolution.nat}(b).
\end{proof}

Before we move on to the proof of Proposition~\ref{prop.leray.f*nxy}, let us
show a simple lemma (which can also easily be obtained as a consequence of
Exercise~\ref{exe.convolution.tensor}(b)):

\begin{lemma}
\label{lem.sol.leray.exp-alg.F(X)G*k}Let $C$ be a $\kk$-coalgebra. Let
$A$ be a $\kk$-algebra. Let $F\in\Hom \left(  C,A\right)
$ and $G\in\Hom \left(  C,A\right)  $. Then, each $k\in
\mathbb{N}$ satisfies%
\begin{equation}
\left(  F\otimes G\right)  ^{\star k}=F^{\star k}\otimes G^{\star k}
\label{pf.prop.leray.f*nxy.FGk}%
\end{equation}
(as maps $C\otimes C\rightarrow A\otimes A$).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.leray.exp-alg.F(X)G*k}.]We must prove that
(\ref{pf.prop.leray.f*nxy.FGk}) holds for each $k\in \NN$. We shall
prove this by induction over $k$:

\textit{Induction base:} Let $s$ denote the canonical $\kk$-algebra
isomorphism $\kk\rightarrow\kk\otimes\kk$. We have%
\begin{align*}
\left(  F\otimes G\right)  ^{\star0}  &  =\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C\otimes
C,A\otimes A\right)  ,\star\right)  \right) \\
&  =\underbrace{u_{A\otimes A}}_{\substack{=\left(  u_{A}\otimes u_{A}\right)
\circ s\\\text{(by the definition of}\\\text{the }\kk\text{-algebra
}A\otimes A\text{)}}}\circ\underbrace{\epsilon_{C\otimes C}}%
_{\substack{=s^{-1}\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)
\\\text{(by the definition of}\\\text{the }\kk\text{-coalgebra
}C\otimes C\text{)}}}=\left(  u_{A}\otimes u_{A}\right)  \circ
\underbrace{s\circ s^{-1}}_{=\id_{\kk%
\otimes\kk}}\circ\left(  \epsilon_{C}\otimes\epsilon_{C}\right)
=\left(  u_{A}\otimes u_{A}\right)  \circ\left(  \epsilon_{C}\otimes
\epsilon_{C}\right) \\
&  =\underbrace{\left(  u_{A}\circ\epsilon_{C}\right)  }_{\substack{=\left(
\text{the unity of the }\kk\text{-algebra }\left(  \Hom
\left(  C,A\right)  ,\star\right)  \right)  \\=F^{\star0}}}\otimes
\underbrace{\left(  u_{A}\circ\epsilon_{C}\right)  }_{\substack{=\left(
\text{the unity of the }\kk\text{-algebra }\left(  \Hom
\left(  C,A\right)  ,\star\right)  \right)  \\=G^{\star0}}}=F^{\star0}\otimes
G^{\star0}.
\end{align*}
In other words, (\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=0$. This
completes the induction base.

\textit{Induction step:} Let $\ell\in \NN$. Assume that
(\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell$. We must prove that
(\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell+1$.

We have assumed that (\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell$. In
other words, we have
\begin{equation}
\left(  F\otimes G\right)  ^{\star\ell}=F^{\star\ell}\otimes G^{\star\ell}.
\label{pf.prop.leray.f*nxy.FGk.pf.indhyp}%
\end{equation}


Now, Exercise~\ref{exe.convolution.tensor}(a) (applied to $C$, $A$,
$F^{\star\ell}$, $F$, $G^{\star\ell}$ and $G$ instead of $D$, $B$, $f$,
$f^{\prime}$, $g$ and $g^{\prime}$) shows that%
\[
\left(  F^{\star\ell}\otimes G^{\star\ell}\right)  \star\left(  F\otimes
G\right)  =\left(  F^{\star\ell}\star F\right)  \otimes\left(  G^{\star\ell
}\star G\right)
\]
in the convolution algebra $\Hom \left(  C\otimes C,A\otimes
A\right)  $. Thus,%
\[
\left(  F^{\star\ell}\otimes G^{\star\ell}\right)  \star\left(  F\otimes
G\right)  =\underbrace{\left(  F^{\star\ell}\star F\right)  }_{=F^{\star
\left(  \ell+1\right)  }}\otimes\underbrace{\left(  G^{\star\ell}\star
G\right)  }_{=G^{\star\left(  \ell+1\right)  }}=F^{\star\left(  \ell+1\right)
}\otimes G^{\star\left(  \ell+1\right)  }.
\]
Hence,%
\[
F^{\star\left(  \ell+1\right)  }\otimes G^{\star\left(  \ell+1\right)
}=\underbrace{\left(  F^{\star\ell}\otimes G^{\star\ell}\right)
}_{\substack{=\left(  F\otimes G\right)  ^{\star\ell}\\\text{(by
(\ref{pf.prop.leray.f*nxy.FGk.pf.indhyp}))}}}\star\left(  F\otimes G\right)
=\left(  F\otimes G\right)  ^{\star\ell}\star\left(  F\otimes G\right)
=\left(  F\otimes G\right)  ^{\star\left(  \ell+1\right)  }.
\]
In other words, (\ref{pf.prop.leray.f*nxy.FGk}) holds for $k=\ell+1$. This
completes the induction step. Hence, (\ref{pf.prop.leray.f*nxy.FGk}) is proven
by induction. Thus, Lemma \ref{lem.sol.leray.exp-alg.F(X)G*k} is proven.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.f*nxy}.]Let $\mathfrak{i}$ be the unity
of the $\kk$-algebra $\left(  \Hom \left(  C,A\right)
,\star\right)  $. Thus,%
\[
\mathfrak{i}=\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right)  =u_{A}%
\circ\epsilon_{C}.
\]


The axioms of a $\kk$-bialgebra show that the counit $\epsilon$ of the
$\kk$-coalgebra $C$ is a $\kk$-algebra homomorphism (since $C$
is a $\kk$-bialgebra). Thus, $\epsilon\left(  1\right)  =1$.

Let us first show that every $x\in C$ and $y\in C$ satisfy%
\begin{equation}
f\left(  xy\right)  =\epsilon\left(  y\right)  f\left(  x\right)
+\epsilon\left(  x\right)  f\left(  y\right)  .
\label{pf.prop.leray.f*nxy.fxy}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.leray.f*nxy.fxy}):} Let $x\in C$ and $y\in C$.
The map $\epsilon$ is $\kk$-linear; thus,%
\[
\epsilon\left(  x-\epsilon\left(  x\right)  1\right)  =\epsilon\left(
x\right)  -\epsilon\left(  x\right)  \underbrace{\epsilon\left(  1\right)
}_{=1}=\epsilon\left(  x\right)  -\epsilon\left(  x\right)  =0.
\]
In other words, $x-\epsilon\left(  x\right)  1\in\ker\epsilon$. The same
argument (applied to $y$ instead of $x$) shows that $y-\epsilon\left(
y\right)  1\in\ker\epsilon$. But%
\begin{align*}
\left(  x-\epsilon\left(  x\right)  1\right)  \left(  y-\epsilon\left(
y\right)  1\right)   &  =xy-\epsilon\left(  x\right)  1y-x\epsilon\left(
y\right)  1+\epsilon\left(  x\right)  1\cdot\epsilon\left(  y\right)  1\\
&  =xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)  x+\epsilon\left(
x\right)  \epsilon\left(  y\right)  1,
\end{align*}
so that%
\[
xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)  x+\epsilon\left(
x\right)  \epsilon\left(  y\right)  1=\underbrace{\left(  x-\epsilon\left(
x\right)  1\right)  }_{\in\ker\epsilon}\underbrace{\left(  y-\epsilon\left(
y\right)  1\right)  }_{\in\ker\epsilon}\in\left(  \ker\epsilon\right)  \left(
\ker\epsilon\right)  =\left(  \ker\epsilon\right)  ^{2}.
\]
Applying the map $f$ to both sides of this relation, we find%
\[
f\left(  xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1\right)  \in f\left(
\left(  \ker\epsilon\right)  ^{2}\right)  =0,
\]
so that $f\left(  xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1\right)  =0$. Hence,%
\begin{align*}
0  &  =f\left(  xy-\epsilon\left(  x\right)  y-\epsilon\left(  y\right)
x+\epsilon\left(  x\right)  \epsilon\left(  y\right)  1\right) \\
&  =f\left(  xy\right)  -\epsilon\left(  x\right)  f\left(  y\right)
-\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(  x\right)
\epsilon\left(  y\right)  \underbrace{f\left(  1\right)  }_{=0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =f\left(  xy\right)  -\epsilon\left(  x\right)  f\left(  y\right)
-\epsilon\left(  y\right)  f\left(  x\right)  +\underbrace{\epsilon\left(
x\right)  \epsilon\left(  y\right)  0}_{=0}\\
&  =f\left(  xy\right)  -\epsilon\left(  x\right)  f\left(  y\right)
-\epsilon\left(  y\right)  f\left(  x\right)  .
\end{align*}
Solving this equality for $f\left(  xy\right)  $, we obtain $f\left(
xy\right)  =\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(
x\right)  f\left(  y\right)  $. This proves (\ref{pf.prop.leray.f*nxy.fxy}).]

Now, we shall show that%
\begin{equation}
\left(  f\circ m_{C}\right)  \left(  z\right)  =\left(  m_{A}\circ\left(
f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  \right)  \left(  z\right)
\label{pf.prop.leray.f*nxy.fm}%
\end{equation}
for each $z\in C\otimes C$.

[\textit{Proof of (\ref{pf.prop.leray.f*nxy.fm}):} Let $z\in C\otimes C$. We
are going to prove the equality (\ref{pf.prop.leray.f*nxy.fm}).

Both sides of the equality (\ref{pf.prop.leray.f*nxy.fm}) are $\kk%
$-linear in $z$. Hence, for the proof of (\ref{pf.prop.leray.f*nxy.fm}), we
can WLOG assume that $z$ is a pure tensor (since any $z\in C\otimes C$ is a
$\kk$-linear combination of pure tensors). Assume this.

There exist $x\in C$ and $y\in C$ satisfying $z=x\otimes y$ (since $z$ is a
pure tensor). Consider these $x$ and $y$. We have
\[
\underbrace{\mathfrak{i}}_{=u_{A}\circ\epsilon_{C}}\left(  x\right)  =\left(
u_{A}\circ\underbrace{\epsilon_{C}}_{=\epsilon}\right)  \left(  x\right)
=\left(  u_{A}\circ\epsilon\right)  \left(  x\right)  =u_{A}\left(
\epsilon\left(  x\right)  \right)  =\epsilon\left(  x\right)  1_{A}%
\]
(by the definition of the map $u_{A}$). The same argument (applied to $y$
instead of $x$) shows that $\mathfrak{i}\left(  y\right)  =\epsilon\left(
y\right)  1_{A}$.

Now,%
\begin{align*}
\left(  f\circ m_{C}\right)  \left(  z\right)   &  =f\left(  m_{C}\left(
\underbrace{z}_{=x\otimes y}\right)  \right)  =f\left(  \underbrace{m_{C}%
\left(  x\otimes y\right)  }_{\substack{=xy\\\text{(by the definition of
}m_{C}\text{)}}}\right)  =f\left(  xy\right) \\
&  =\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(  x\right)
f\left(  y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.leray.f*nxy.fxy})}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)
\right)  \left(  z\right)   &  =m_{A}\left(  \left(  f\otimes\mathfrak{i}%
+\mathfrak{i}\otimes f\right)  \left(  \underbrace{z}_{=x\otimes y}\right)
\right)  =m_{A}\left(  \underbrace{\left(  f\otimes\mathfrak{i}+\mathfrak{i}%
\otimes f\right)  \left(  x\otimes y\right)  }_{=\left(  f\otimes
\mathfrak{i}\right)  \left(  x\otimes y\right)  +\left(  \mathfrak{i}\otimes
f\right)  \left(  x\otimes y\right)  }\right) \\
&  =m_{A}\left(  \underbrace{\left(  f\otimes\mathfrak{i}\right)  \left(
x\otimes y\right)  }_{=f\left(  x\right)  \otimes\mathfrak{i}\left(  y\right)
}+\underbrace{\left(  \mathfrak{i}\otimes f\right)  \left(  x\otimes y\right)
}_{=\mathfrak{i}\left(  x\right)  \otimes f\left(  y\right)  }\right) \\
&  =m_{A}\left(  f\left(  x\right)  \otimes\mathfrak{i}\left(  y\right)
+\mathfrak{i}\left(  x\right)  \otimes f\left(  y\right)  \right)
=\underbrace{m_{A}\left(  f\left(  x\right)  \otimes\mathfrak{i}\left(
y\right)  \right)  }_{\substack{=f\left(  x\right)  \mathfrak{i}\left(
y\right)  \\\text{(by the definition of }m_{A}\text{)}}}+\underbrace{m_{A}%
\left(  \mathfrak{i}\left(  x\right)  \otimes f\left(  y\right)  \right)
}_{\substack{=\mathfrak{i}\left(  x\right)  f\left(  y\right)  \\\text{(by the
definition of }m_{A}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }m_{A}\text{ is }%
\kk\text{-linear}\right) \\
&  =f\left(  x\right)  \underbrace{\mathfrak{i}\left(  y\right)  }%
_{=\epsilon\left(  y\right)  1_{A}}+\underbrace{\mathfrak{i}\left(  x\right)
}_{=\epsilon\left(  x\right)  1_{A}}f\left(  y\right)  =\underbrace{f\left(
x\right)  \epsilon\left(  y\right)  1_{A}}_{=\epsilon\left(  y\right)
f\left(  x\right)  1_{A}=\epsilon\left(  y\right)  f\left(  x\right)
}+\epsilon\left(  x\right)  \underbrace{1_{A}f\left(  y\right)  }_{=f\left(
y\right)  }\\
&  =\epsilon\left(  y\right)  f\left(  x\right)  +\epsilon\left(  x\right)
f\left(  y\right)  ,
\end{align*}
we obtain $\left(  f\circ m_{C}\right)  \left(  z\right)  =\left(  m_{A}%
\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  \right)
\left(  z\right)  $. Thus, (\ref{pf.prop.leray.f*nxy.fm}) is proven.]

Now, we have proven that (\ref{pf.prop.leray.f*nxy.fm}) holds for each $z\in
C$. In other words, we have%
\begin{equation}
f\circ m_{C}=m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f\right)  . \label{pf.prop.leray.f*nxy.fm2}%
\end{equation}


The axioms of a $\kk$-bialgebra show that the comultiplication
$\Delta_{C}:C\rightarrow C\otimes C$ of the $\kk$-coalgebra $C$ is a
$\kk$-algebra homomorphism (since $C$ is a $\kk$-bialgebra).
They also show that the multiplication $m_C : C \otimes C \to C$ of the
$\kk$-algebra $C$ is a $\kk$-coalgebra homomorphism (since $C$ is a
$\kk$-bialgebra).

Exercise~\ref{exe.cocommutative.coalghom}(b) shows that the $\kk%
$-algebra $A$ is commutative if and only if its multiplication $m_{A}:A\otimes
A\rightarrow A$ is a $\kk$-algebra homomorphism. Thus, its
multiplication $m_{A}:A\otimes A\rightarrow A$ is a $\kk$-algebra
homomorphism (since $A$ is commutative).

Thus, Proposition~\ref{prop.convolution.nat}(b) (applied to $C\otimes C$, $C$,
$A\otimes A$, $A$, $m_{C}$, $m_{A}$, $f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f$ and $f$ instead of $C$, $C^{\prime}$, $A$, $A^{\prime}$, $\gamma$, $\alpha
$, $f$ and $f^{\prime}$) shows that each $n\in \NN$ satisfies%
\begin{equation}
f^{\star n}\circ m_{C}=m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}%
\otimes f\right)  ^{\star n} \label{pf.prop.leray.f*nxy.fnm}%
\end{equation}
(because of (\ref{pf.prop.leray.f*nxy.fm2})).

Recall that $\mathfrak{i}$ is the unity of the $\kk$-algebra $\left(
\Hom \left(  C,A\right)  ,\star\right)  $. Thus, $\mathfrak{i}%
\star f=f$ and $f\star\mathfrak{i}=f$. Also, for the same reason, we have%
\begin{equation}
\mathfrak{i}^{\star k}=\mathfrak{i}\ \ \ \ \ \ \ \ \ \ \text{for each }%
k\in\mathbb{N}. \label{pf.prop.leray.f*nxy.i*k}%
\end{equation}


The two $\kk$-linear maps $f\otimes\mathfrak{i}$ and $\mathfrak{i}%
\otimes f$ in $\Hom \left(  C\otimes C,A\otimes A\right)  $
satisfy%
\begin{equation}
\left(  f\otimes\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes
f\right)  =\left(  \mathfrak{i}\otimes f\right)  \star\left(  f\otimes
\mathfrak{i}\right)  \label{pf.prop.leray.f*nxy.fiif}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.leray.f*nxy.fiif}):}
Exercise~\ref{exe.convolution.tensor}(a) (applied to $C$, $A$, $f$,
$\mathfrak{i}$, $\mathfrak{i}$ and $f$ instead of $D$, $B$, $f$, $f^{\prime}$,
$g$ and $g^{\prime}$) shows that%
\begin{equation}
\left(  f\otimes\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes
f\right)  =\underbrace{\left(  f\star\mathfrak{i}\right)  }_{=f}%
\otimes\underbrace{\left(  \mathfrak{i}\star f\right)  }_{=f}=f\otimes f.
\label{pf.prop.leray.f*nxy.fiif.pf.1}%
\end{equation}
But Exercise~\ref{exe.convolution.tensor}(a) (applied to $C$, $A$,
$\mathfrak{i}$, $f$, $f$ and $\mathfrak{i}$ instead of $D$, $B$, $f$,
$f^{\prime}$, $g$ and $g^{\prime}$) shows that%
\[
\left(  \mathfrak{i}\otimes f\right)  \star\left(  f\otimes\mathfrak{i}%
\right)  =\underbrace{\left(  \mathfrak{i}\star f\right)  }_{=f}%
\otimes\underbrace{\left(  f\star\mathfrak{i}\right)  }_{=f}=f\otimes f.
\]
Comparing this with (\ref{pf.prop.leray.f*nxy.fiif.pf.1}), we obtain $\left(
f\otimes\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes f\right)
=\left(  \mathfrak{i}\otimes f\right)  \star\left(  f\otimes\mathfrak{i}%
\right)  $. This proves (\ref{pf.prop.leray.f*nxy.fiif}).}. In other words,
the two elements $f\otimes\mathfrak{i}$ and $\mathfrak{i}\otimes f$ of the
$\kk$-algebra $\left(  \Hom \left(  C\otimes C,A\otimes
A\right)  ,\star\right)  $ commute.

Hence, it is easy to see%
\begin{equation}
\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star n}%
=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes f^{\star\left(  n-i\right)
}\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}.
\label{pf.prop.leray.f*nxy.binom}%
\end{equation}
[\textit{Proof of (\ref{pf.prop.leray.f*nxy.binom}):} Let $n\in \NN$.
Let $\mathfrak{G}$ be the $\kk$-subalgebra of $\left(
\Hom \left(  C\otimes C,A\otimes A\right)  ,\star\right)  $
generated by the two elements $f\otimes\mathfrak{i}$ and $\mathfrak{i}\otimes
f$. Thus, the $\kk$-algebra $\mathfrak{G}$ is generated by commuting
elements (because the elements $f\otimes\mathfrak{i}$ and $\mathfrak{i}\otimes
f$ of the $\kk$-algebra $\left(  \Hom \left(  C\otimes
C,A\otimes A\right)  ,\star\right)  $ commute), and therefore is commutative
(since any $\kk$-algebra generated by commuting elements must be
commutative). Hence, the binomial formula holds in this $\kk$-algebra
$\mathfrak{G}$. In other words, any $\alpha\in\mathfrak{G}$ and $\beta
\in\mathfrak{G}$ and $m\in \NN$ satisfy%
\[
\left(  \alpha+\beta\right)  ^{\star m}=\sum_{i=0}^{m}\dbinom{m}{i}%
\alpha^{\star i}\star\beta^{\star\left(  m-i\right)  }%
\]
(since the multiplication in the $\kk$-algebra $\mathfrak{G}$ is the
convolution $\star$). Applying this to $\alpha=f\otimes\mathfrak{i}$,
$\beta=\mathfrak{i}\otimes f$ and $m=n$, we obtain
\begin{align*}
\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star n}  &
=\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f\otimes\mathfrak{i}\right)
^{\star i}}_{\substack{=f^{\star i}\otimes\mathfrak{i}^{\star i}\\\text{(by
(\ref{pf.prop.leray.f*nxy.FGk}) (applied}\\\text{to }f\text{, }\mathfrak{i}%
\text{ and }i\text{ instead of }F\text{, }G\text{ and }k\text{))}}%
}\star\underbrace{\left(  \mathfrak{i}\otimes f\right)  ^{\star\left(
n-i\right)  }}_{\substack{=\mathfrak{i}^{\star\left(  n-i\right)  }\otimes
f^{\star\left(  n-i\right)  }\\\text{(by (\ref{pf.prop.leray.f*nxy.FGk})
(applied}\\\text{to }\mathfrak{i}\text{, }f\text{ and }n-i\text{ instead of
}F\text{, }G\text{ and }k\text{))}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\star i}\otimes
\underbrace{\mathfrak{i}^{\star i}}_{\substack{=\mathfrak{i}\\\text{(by
(\ref{pf.prop.leray.f*nxy.i*k}) (applied to }k=i\text{))}}}\right)
\star\left(  \underbrace{\mathfrak{i}^{\star\left(  n-i\right)  }%
}_{\substack{=\mathfrak{i}\\\text{(by (\ref{pf.prop.leray.f*nxy.i*k}) (applied
to }k=n-i\text{))}}}\otimes f^{\star\left(  n-i\right)  }\right) \\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\otimes
\mathfrak{i}\right)  \star\left(  \mathfrak{i}\otimes f^{\star\left(
n-i\right)  }\right)  }_{\substack{=\left(  f^{\star i}\star\mathfrak{i}%
\right)  \otimes\left(  \mathfrak{i}\star f^{\star\left(  n-i\right)
}\right)  \\\text{(by Exercise~\ref{exe.convolution.tensor}(a) (applied}%
\\\text{to }C\text{, }A\text{, }f^{\star i}\text{, }\mathfrak{i}\text{,
}\mathfrak{i}\text{, }f^{\star\left(  n-i\right)  }\\\text{instead of
}D\text{, }B\text{, }f\text{, }f^{\prime}\text{, }g\text{ and }g^{\prime
}\text{))}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star i}\star
\mathfrak{i}\right)  }_{\substack{=f^{\star i}\\\text{(since }\mathfrak{i}%
\text{ is the unity}\\\text{of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \text{)}}}\otimes
\underbrace{\left(  \mathfrak{i}\star f^{\star\left(  n-i\right)  }\right)
}_{\substack{=f^{\star\left(  n-i\right)  }\\\text{(since }\mathfrak{i}\text{
is the unity}\\\text{of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \text{)}}}\\
&  =\sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes f^{\star\left(  n-i\right)
}.
\end{align*}
This proves (\ref{pf.prop.leray.f*nxy.binom}).]

Now, let $x,y\in C$ and $n\in \NN$ be arbitrary. Then, the definition of
the multiplication map $m_{C}$ yields $m_{C}\left(  x\otimes y\right)  =xy$.
Hence,%
\[
\left(  f^{\star n}\circ m_{C}\right)  \left(  x\otimes y\right)  =f^{\star
n}\left(  \underbrace{m_{C}\left(  x\otimes y\right)  }_{=xy}\right)
=f^{\star n}\left(  xy\right)  .
\]
Hence,%
\begin{align*}
f^{\star n}\left(  xy\right)   &  =\underbrace{\left(  f^{\star n}\circ
m_{C}\right)  }_{\substack{=m_{A}\circ\left(  f\otimes\mathfrak{i}%
+\mathfrak{i}\otimes f\right)  ^{\star n}\\\text{(by
(\ref{pf.prop.leray.f*nxy.fnm}))}}}\left(  x\otimes y\right)  =\left(
m_{A}\circ\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes f\right)  ^{\star
n}\right)  \left(  x\otimes y\right) \\
&  =m_{A}\left(  \underbrace{\left(  f\otimes\mathfrak{i}+\mathfrak{i}\otimes
f\right)  ^{\star n}}_{\substack{=\sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\otimes f^{\star\left(  n-i\right)  }\\\text{(by
(\ref{pf.prop.leray.f*nxy.binom}))}}}\left(  x\otimes y\right)  \right)
=m_{A}\left(  \left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\otimes
f^{\star\left(  n-i\right)  }\right)  \left(  x\otimes y\right)  \right) \\
&  =m_{A}\left(  \underbrace{\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\otimes f^{\star\left(  n-i\right)  }\right)  \left(  x\otimes y\right)
}_{=\sum_{i=0}^{n}\dbinom{n}{i}\left(  f^{\star i}\otimes f^{\star\left(
n-i\right)  }\right)  \left(  x\otimes y\right)  }\right) \\
&  =m_{A}\left(  \sum_{i=0}^{n}\dbinom{n}{i}\underbrace{\left(  f^{\star
i}\otimes f^{\star\left(  n-i\right)  }\right)  \left(  x\otimes y\right)
}_{=f^{\star i}\left(  x\right)  \otimes f^{\star\left(  n-i\right)  }\left(
y\right)  }\right) \\
&  =m_{A}\left(  \sum_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(  x\right)
\otimes f^{\star\left(  n-i\right)  }\left(  y\right)  \right)  =\sum
_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)
\end{align*}
(by the definition of $m_{A}$). This proves Proposition~\ref{prop.leray.f*nxy}.
\end{proof}

Before we prove Proposition~\ref{prop.leray.exp-algh1}, let us show two
lemmas:\footnote{Recall that we are still using the conventions that are in
place throughout Section~\ref{subsect.leray}.}

\begin{lemma}
\label{lem.pf.prop.leray.exp-algh1.1}Let $C$ be a $\kk$-coalgebra. Let
$A$ be a $\kk$-algebra. Let $f\in\mathfrak{n}\left(  C,A\right)  $.

\begin{enumerate}
\item[(a)] If $z\in C$ is arbitrary, then the family $\left(  \dfrac{1}%
{n!}f^{\star n}\left(  z\right)  \right)  _{n\in\NN}\in A^{\NN}$
is finitely supported and satisfies%
\[
\left(  \exp^{\star}f\right)  \left(  z\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  z\right)  .
\]


\item[(b)] If $x\in C$ and $y\in C$ are arbitrary, then the family $\left(
\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\[
\left(  \exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  \left(  y\right)  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  .
\]

\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}.]We have $f\in
\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a pointwise $\star
$-nilpotent map in $\Hom \left(  C,A\right)  $ (since
$\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). Thus, the
family $\left(  f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported (since $f$ is pointwise $\star$-nilpotent). In other words,
for each $x\in C$,%
\begin{equation}
\text{the family }\left(  f^{\star n}\left(  x\right)  \right)  _{n\in
\NN}\in A^{\NN}\text{ is finitely supported}
\label{pf.prop.leray.exp-algh1.fs}%
\end{equation}
(by the definition of ``pointwise finitely supported'').

(a) Let $z\in C$ be arbitrary. Then, (\ref{pf.prop.leray.exp-algh1.fs})
(applied to $x=z$) shows that the family $\left(  f^{\star n}\left(  z\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported. Thus,
\begin{equation}
\text{the family }\left(  \dfrac{1}{n!}f^{\star n}\left(  z\right)  \right)
_{n\in\NN}\in A^{\NN}\text{ is finitely supported}
\label{pf.prop.leray.exp-algh1.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.leray.exp-algh1.3}):} We know that
the family $\left(  f^{\star n}\left(  z\right)  \right)  _{n\in\NN}\in
A^{\NN}$ is finitely supported. In other words, all but finitely many
$n\in \NN$ satisfy $f^{\star n}\left(  z\right)  =0$. In other words,
there exists a finite subset $Q$ of $\NN$ such that
\begin{equation}
\text{every }n\in\mathbb{N}\setminus Q\text{ satisfies }f^{\star n}\left(
z\right)  =0. \label{pf.prop.leray.exp-algh1.fn3.1}%
\end{equation}
Consider this $Q$.
\par
Now, every $n\in\mathbb{N}\setminus Q$ satisfies $\dfrac{1}{n!}%
\underbrace{f^{\star n}\left(  z\right)  }_{\substack{=0\\\text{(by
(\ref{pf.prop.leray.exp-algh1.fn3.1}))}}}=0$. Thus, all but finitely many
$n\in \NN$ satisfy $\dfrac{1}{n!}f^{\star n}\left(  z\right)  =0$ (since
all but finitely many $n\in \NN$ satisfy $n\in\mathbb{N}\setminus Q$
(since $Q$ is a finite set)). In other words, the family $\left(  \dfrac
{1}{n!}f^{\star n}\left(  z\right)  \right)  _{n\in\NN}\in
A^{\NN}$ is finitely supported. Qed.}.

Recall that $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$. Hence, $\exp^{\star}%
f=\sum_{n\geq0}\dfrac{1}{n!}f^{\star n}$ (by the definition of $\exp^{\star}%
f$). Hence,%
\[
\left(  \exp^{\star}f\right)  \left(  z\right)  =\left(  \sum_{n\geq0}%
\dfrac{1}{n!}f^{\star n}\right)  \left(  z\right)  =\underbrace{\sum_{n\geq0}%
}_{=\sum_{n\in\NN}}\dfrac{1}{n!}f^{\star n}\left(  z\right)
=\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}\left(  z\right)  .
\]
Thus, the proof of Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) is complete.

(b) Let $x\in C$ and $y\in C$ be arbitrary.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=x$) shows that the
family $\left(  \dfrac{1}{n!}f^{\star n}\left(  x\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  x\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  x\right)  . \label{pf.prop.leray.exp-algh1.1}%
\end{equation}
Thus,%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  x\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  x\right)  =\sum_{q\in\NN}\dfrac{1}%
{q!}f^{\star q}\left(  x\right)  \label{pf.prop.leray.exp-algh1.1'}%
\end{equation}
(here, we have renamed the summation index $n$ as $q$).

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=y$) shows that the
family $\left(  \dfrac{1}{n!}f^{\star n}\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  y\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  y\right)  . \label{pf.prop.leray.exp-algh1.2}%
\end{equation}
Thus,%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  y\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  y\right)  =\sum_{r\in\NN}\dfrac{1}%
{r!}f^{\star r}\left(  y\right)  \label{pf.prop.leray.exp-algh1.2'}%
\end{equation}
(here, we have renamed the summation index $n$ as $r$).

Recall that the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  x\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported. Renaming
the index $n$ as $q$ in this sentence, we obtain the following: The family
$\left(  \dfrac{1}{q!}f^{\star q}\left(  x\right)  \right)  _{q\in\mathbb{N}%
}\in A^{\NN}$ is finitely supported.

Recall that the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  y\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported. Renaming
the index $n$ as $r$ in this sentence, we obtain the following: The family
$\left(  \dfrac{1}{r!}f^{\star r}\left(  y\right)  \right)  _{r\in\mathbb{N}%
}\in A^{\NN}$ is finitely supported.

Recall that sums of the form $\sum_{q\in Q}f_{q}$ (where $\left(
f_{q}\right)  _{q\in Q}$ is a finitely supported family) satisfy the usual
rules for finite sums, even though their indexing set $Q$ may be infinite.
This pertains, in particular, to the sums $\sum_{q\in\NN}\dfrac{1}%
{q!}f^{\star q}\left(  x\right)  $ and $\sum_{r\in\NN}\dfrac{1}%
{r!}f^{\star r}\left(  y\right)  $ (because the families $\left(  \dfrac
{1}{q!}f^{\star q}\left(  x\right)  \right)  _{q\in\NN}$ and $\left(
\dfrac{1}{r!}f^{\star r}\left(  y\right)  \right)  _{r\in\NN}$ are
finitely supported). Thus, the following manipulations make sense:%
\begin{align*}
&  \left(  \sum_{q\in\NN}\dfrac{1}{q!}f^{\star q}\left(  x\right)
\right)  \left(  \sum_{r\in\NN}\dfrac{1}{r!}f^{\star r}\left(
y\right)  \right) \\
&  =\sum_{q\in\NN}\dfrac{1}{q!}f^{\star q}\left(  x\right)  \sum
_{r\in\NN}\dfrac{1}{r!}f^{\star r}\left(  y\right)  =\sum
_{q\in\NN}\sum_{r\in\NN}\underbrace{\dfrac{1}{q!}f^{\star
q}\left(  x\right)  \dfrac{1}{r!}f^{\star r}\left(  y\right)  }_{=\dfrac
{1}{q!\cdot r!}f^{\star q}\left(  x\right)  f^{\star r}\left(  y\right)  }\\
&  =\sum_{q\in\NN}\underbrace{\sum_{r\in\NN}\dfrac{1}{q!\cdot
r!}f^{\star q}\left(  x\right)  f^{\star r}\left(  y\right)  }%
_{\substack{=\sum\limits_{\substack{n\in\mathbb{N};\\n\geq q}}\dfrac{1}%
{q!\cdot\left(  n-q\right)  !}f^{\star q}\left(  x\right)  f^{\star\left(
n-q\right)  }\left(  y\right)  \\\text{(here, we have substituted }n-q\text{
for }r\text{ in the sum)}}}=\underbrace{\sum_{q\in\NN}\sum\limits
_{\substack{n\in\mathbb{N};\\n\geq q}}}_{=\sum\limits_{\substack{\left(  q,n\right)
\in \NN\times\mathbb{N};\\n\geq q}}=\sum_{n\in\NN}\sum\limits
_{\substack{q\in\mathbb{N};\\n\geq q}}}\dfrac{1}{q!\cdot\left(  n-q\right)
!}f^{\star q}\left(  x\right)  f^{\star\left(  n-q\right)  }\left(  y\right)
\\
&  =\sum_{n\in\NN}\underbrace{\sum\limits_{\substack{q\in\mathbb{N};\\n\geq
q}}}_{=\sum\limits_{\substack{q\in\mathbb{N};\\q\leq n}}=\sum_{q=0}^{n}}\dfrac
{1}{q!\cdot\left(  n-q\right)  !}f^{\star q}\left(  x\right)  f^{\star\left(
n-q\right)  }\left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{q=0}^{n}\dfrac{1}{q!\cdot\left(  n-q\right)
!}f^{\star q}\left(  x\right)  f^{\star\left(  n-q\right)  }\left(  y\right)
\\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}f^{\star i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)
\end{align*}
(here, we have renamed the summation index $q$ as $i$). Comparing this with%
\[
\underbrace{\left(  \sum_{q\in\NN}\dfrac{1}{q!}f^{\star q}\left(
x\right)  \right)  }_{\substack{=\left(  \exp^{\star}f\right)  \left(
x\right)  \\\text{(by (\ref{pf.prop.leray.exp-algh1.1'}))}}%
}\underbrace{\left(  \sum_{r\in\NN}\dfrac{1}{r!}f^{\star r}\left(
y\right)  \right)  }_{\substack{=\left(  \exp^{\star}f\right)  \left(
y\right)  \\\text{(by (\ref{pf.prop.leray.exp-algh1.2'}))}}}=\left(
\exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star}f\right)
\left(  y\right)  ,
\]
we obtain%
\[
\left(  \exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  \left(  y\right)  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  .
\]
In particular, the sum on the right hand side of this equality is
well-defined. Thus, the family \newline$\left(  \sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  \right)  _{n\in\NN}\in A^{\NN}$
is finitely supported. This completes the proof of
Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b).
\end{proof}

\begin{lemma}
\label{lem.pf.prop.leray.exp-algh1.2}Let $C$ be a $\kk$-bialgebra. Let
$A$ be a $\kk$-algebra. Let $f\in\Hom \left(  C,A\right)
$. Every $n\in \NN$ satisfies%
\begin{equation}
f^{\star n}\left(  1\right)  =\left(  f\left(  1\right)  \right)  ^{n}.
\label{pf.prop.leray.exp-algh1.F1}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.pf.prop.leray.exp-algh1.2}.]We must prove that
(\ref{pf.prop.leray.exp-algh1.F1}) holds for each $n\in \NN$. We shall
prove this by induction over $n$:

\textit{Induction base:} The axioms of a $\kk$-bialgebra yield
$\epsilon_{C}\left(  1\right)  =1$ (since $C$ is a $\kk$-bialgebra).
Now,
\[
f^{\star0}=\left(  \text{the unity of the }\kk\text{-algebra
} \left(\Hom \left(  C,A\right) , \star\right) \right)
=u_{A}\circ\epsilon_{C}.
\]
Thus,%
\begin{align*}
\underbrace{f^{\star0}}_{=u_{A}\circ\epsilon_{C}}\left(  1\right)   &
=\left(  u_{A}\circ\epsilon_{C}\right)  \left(  1\right)  =u_{A}\left(
\underbrace{\epsilon_{C}\left(  1\right)  }_{=1}\right)  =u_{A}\left(
1\right)  =1\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}u_{A}\right) \\
&  =1_{A}=\left(  f\left(  1\right)  \right)  ^{0}.
\end{align*}
In other words, (\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=0$. This
completes the induction base.

\textit{Induction step:} Let $k\in \NN$. Assume that
(\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k$. We must show that
(\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k+1$.

The axioms of a $\kk$-bialgebra yield $\Delta_{C}\left(  1\right)
=1\otimes1$ (since $C$ is a $\kk$-bialgebra).

We have assumed that (\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k$. In
other words, we have $f^{\star k}\left(  1\right)  =\left(  f\left(  1\right)
\right)  ^{k}$. Now, $f^{\star\left(  k+1\right)  }=f\star f^{\star k}%
=m_{A}\circ\left(  f\otimes f^{\star k}\right)  \circ\Delta_{C}$ (by the
definition of convolution). Applying both sides of this equality to the
element $1$ of $C$, we obtain%
\begin{align*}
f^{\star\left(  k+1\right)  }\left(  1\right)   &  =\left(  m_{A}\circ\left(
f\otimes f^{\star k}\right)  \circ\Delta_{C}\right)  \left(  1\right)
=m_{A}\left(  \left(  f\otimes f^{\star k}\right)  \left(  \underbrace{\Delta
_{C}\left(  1\right)  }_{=1\otimes1}\right)  \right) \\
&  =m_{A}\left(  \underbrace{\left(  f\otimes f^{\star k}\right)  \left(
1\otimes1\right)  }_{=f\left(  1\right)  \otimes f^{\star k}\left(  1\right)
}\right)  =m_{A}\left(  f\left(  1\right)  \otimes f^{\star k}\left(
1\right)  \right) \\
&  =f\left(  1\right)  \underbrace{f^{\star k}\left(  1\right)  }_{=\left(
f\left(  1\right)  \right)  ^{k}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }m_{A}\right) \\
&  =f\left(  1\right)  \cdot\left(  f\left(  1\right)  \right)  ^{k}=\left(
f\left(  1\right)  \right)  ^{k+1}.
\end{align*}
In other words, (\ref{pf.prop.leray.exp-algh1.F1}) holds for $n=k+1$. This
completes the induction step. Thus, (\ref{pf.prop.leray.exp-algh1.F1}) is
proven by induction. Hence, Lemma~\ref{lem.pf.prop.leray.exp-algh1.2} is proven.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.exp-algh1}.]Let $x\in C$ and $y\in C$.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=xy$) shows that
the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  xy\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  xy\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  xy\right)  .
\label{pf.prop.leray.exp-algh1.q0}%
\end{equation}
Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b) shows that the family $\left(
\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(
x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star
}f\right)  \left(  y\right)  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac
{1}{i!\cdot\left(  n-i\right)  !}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  . \label{pf.prop.leray.exp-algh1.q1}%
\end{equation}


Comparing (\ref{pf.prop.leray.exp-algh1.q1}) with%
\begin{align*}
\left(  \exp^{\star}f\right)  \left(  xy\right)   &  =\sum_{n\in\mathbb{N}%
}\dfrac{1}{n!}\underbrace{f^{\star n}\left(  xy\right)  }_{\substack{=\sum
_{i=0}^{n}\dbinom{n}{i}f^{\star i}\left(  x\right)  f^{\star\left(
n-i\right)  }\left(  y\right)  \\\text{(by Proposition~\ref{prop.leray.f*nxy}%
)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.leray.exp-algh1.q0}%
)}\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\sum_{i=0}^{n}\dbinom{n}{i}f^{\star
i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\underbrace{\dfrac{1}{n!}\dbinom{n}{i}%
}_{\substack{=\dfrac{1}{i!\cdot\left(  n-i\right)  !}\\\text{(since }%
\dbinom{n}{i}=\dfrac{n!}{i!\cdot\left(  n-i\right)  !}=n!\cdot\dfrac
{1}{i!\cdot\left(  n-i\right)  !}\text{)}}}f^{\star i}\left(  x\right)
f^{\star\left(  n-i\right)  }\left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}f^{\star i}\left(  x\right)  f^{\star\left(  n-i\right)  }\left(  y\right)
,
\end{align*}
we obtain $\left(  \exp^{\star}f\right)  \left(  xy\right)  =\left(
\exp^{\star}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star}f\right)
\left(  y\right)  $.

Now, forget that we fixed $x$ and $y$. We thus have shown that
\begin{equation}
\left(  \exp^{\star}f\right)  \left(  xy\right)  =\left(  \exp^{\star
}f\right)  \left(  x\right)  \cdot\left(  \exp^{\star}f\right)  \left(
y\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }x\in C\text{ and }y\in C.
\label{pf.prop.leray.exp-algh1.res1}%
\end{equation}


Now, $\exp=\sum_{n\geq0}\dfrac{1}{n!}T^{n}$. Hence, $\exp^{\star}f=\sum
_{n\geq0}\dfrac{1}{n!}f^{\star n}$ (by the definition of $\exp^{\star}f$).
Hence,%
\begin{align*}
\left(  \exp^{\star}f\right)  \left(  1\right)   &  =\left(  \sum_{n\geq
0}\dfrac{1}{n!}f^{\star n}\right)  \left(  1\right)  =\sum_{n\geq0}\dfrac
{1}{n!}\underbrace{f^{\star n}\left(  1\right)  }_{\substack{=\left(  f\left(
1\right)  \right)  ^{n}\\\text{(by Lemma~\ref{lem.pf.prop.leray.exp-algh1.2}%
)}}}=\sum_{n\geq0}\dfrac{1}{n!}\left(  \underbrace{f\left(  1\right)  }%
_{=0}\right)  ^{n}\\
&  =\sum_{n\geq0}\dfrac{1}{n!}0^{n}=\underbrace{\dfrac{1}{0!}}_{=\dfrac{1}%
{1}=1}\underbrace{0^{0}}_{=1}+\sum_{n\geq1}\dfrac{1}{n!}\underbrace{0^{n}%
}_{\substack{=0\\\text{(since }n\geq1\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=0\text{ from the sum}\right) \\
&  =1+\underbrace{\sum_{n\geq1}\dfrac{1}{n!}0}_{=0}=1.
\end{align*}


Thus, we know that $\exp^{\star}f$ is a $\kk$-linear map $C\rightarrow
A$ (since $\exp^{\star}f\in\Hom \left(  C,A\right)  $)
satisfying (\ref{pf.prop.leray.exp-algh1.res1}) and $\left(  \exp^{\star
}f\right)  \left(  1\right)  =1$. In other words, $\exp^{\star}f:C\rightarrow
A$ is a $\kk$-algebra homomorphism. This proves
Proposition~\ref{prop.leray.exp-algh1}.
\end{proof}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.eulerian-idp.2.poly1}.]The Vandermonde
determinant formula says that
\[
\det\left(  \left(  x_{i}^{n-j}\right)  _{i,j=1,2,\ldots,n}\right)
=\prod_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)
\]
in the polynomial ring $\mathfrak{A}\left[  x_{1},x_{2},\ldots,x_{n}\right]  $
for every commutative ring $\mathfrak{A}$ and every $n\in{\NN}$.
Applying this to $\mathfrak{A}= \ZZ$ and $n=N+1$, we obtain
\[
\det\left(  \left(  x_{i}^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\right)
=\prod_{1\leq i<j\leq N+1}\left(  x_{i}-x_{j}\right)
\]
in $\ZZ \left[  x_{1},x_{2},\ldots,x_{N+1}\right]  $. Evaluating both
sides of this polynomial identity at $\left(  x_{1},x_{2},\ldots
,x_{N+1}\right)  =\left(  1,2,\ldots,N+1\right)  $, we obtain%
\[
\det\left(  \left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\right)
=\prod_{1\leq i<j\leq N+1}\left(  i-j\right)  .
\]
But $\prod_{1\leq i<j\leq N+1}\left(  i-j\right)  \neq0$ in $\ZZ$
(since $i-j\neq0$ for every $i<j$). Thus,%
\[
\det\left(  \left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\right)
=\prod_{1\leq i<j\leq N+1}\left(  i-j\right)  \neq0.
\]
The matrix $\left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\in
\QQ^{\left(  N+1\right)  \times\left(  N+1\right)  }$ is thus invertible in
$\QQ^{\left(  N+1\right)  \times\left(  N+1\right)  }$. In other
words, there exists a matrix $\left(  s_{i,j}\right)  _{i,j=1,2,\ldots,N+1}%
\in \QQ^{\left(  N+1\right)  \times\left(  N+1\right)  }$ satisfying
\[
\left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}\cdot\left(  s_{i,j}\right)
_{i,j=1,2,\ldots,N+1}=I_{N+1}=\left(  s_{i,j}\right)  _{i,j=1,2,\ldots
,N+1}\cdot\left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}.
\]
Consider this matrix $\left(  s_{i,j}\right)  _{i,j=1,2,\ldots,N+1}$. All of
its entries $s_{i,j}$ are rational numbers, and therefore there exists a
positive integer $M$ such that every $\left(  i,j\right)  \in\left\{
1,2,\ldots,N+1\right\}  ^{2}$ satisfies $Ms_{i,j}\in \ZZ$ (because
finitely many rational numbers always have a common denominator). Consider
this $M$. For every $\left(  i,j\right)  \in\left\{  1,2,\ldots,N+1\right\}
^{2}$, define an element $t_{i,j}\in \ZZ$ by $t_{i,j}=Ms_{i,j}$ (this
makes sense because $Ms_{i,j}\in \ZZ$).

For every $\left(  u,v\right)  \in\left\{  1,2,\ldots,N+1\right\}  ^{2}$, we
have%
\begin{align*}
\delta_{u,v}  &  =\left(  \text{the }\left(  u,v\right)  \text{-th entry of
the matrix }\underbrace{I_{N+1}}_{=\left(  s_{i,j}\right)  _{i,j=1,2,\ldots
,N+1}\cdot\left(  i^{N+1-j}\right)  _{i,j=1,2,\ldots,N+1}}\right) \\
&  =\left(  \text{the }\left(  u,v\right)  \text{-th entry of the matrix
}\left(  s_{i,j}\right)  _{i,j=1,2,\ldots,N+1}\cdot\left(  i^{N+1-j}\right)
_{i,j=1,2,\ldots,N+1}\right)  =\sum_{j=1}^{N+1}s_{u,j}j^{N+1-v}.
\end{align*}
Multiplying this by $M$, we obtain
\begin{equation}
M\delta_{u,v}=M\sum_{j=1}^{N+1}s_{u,j}j^{N+1-v}=\sum_{j=1}^{N+1}%
\underbrace{Ms_{u,j}}_{\substack{=t_{u,j}\\\text{(since }t_{u,j}\text{
is}\\\text{defined as }Ms_{u,j}\text{)}}}j^{N+1-v}=\sum_{j=1}^{N+1}%
t_{u,j}j^{N+1-v} \label{pf.lem.sol.eulerian-idp.2.poly1.2}%
\end{equation}
for every $\left(  u,v\right)  \in\left\{  1,2,\ldots,N+1\right\}  ^{2}$.

Now, let $i\in\left\{  0,1,\ldots,N\right\}  $. Then,
\begin{align*}
\sum_{k=0}^{N}\delta_{i,k}w_{k}  &  =\sum_{k\in\left\{  0,1,\ldots,N\right\}
}\delta_{i,k}w_{k}=\underbrace{\sum\limits_{\substack{k\in\left\{  0,1,\ldots
,N\right\}  ;\\k=i}}\delta_{i,k}w_{k}}_{\substack{=\delta_{i,i}w_{i}%
\\\text{(since }i\in\left\{  0,1,\ldots,N\right\}  \text{)}}}+\sum\limits
_{\substack{k\in\left\{  0,1,\ldots,N\right\}  ;\\k\neq i}}\underbrace{\delta
_{i,k}}_{\substack{=0\\\text{(since }k\neq i\text{)}}}w_{k}\\
&  =\delta_{i,i}w_{i}+\underbrace{\sum\limits_{\substack{k\in\left\{  0,1,\ldots
,N\right\}  ;\\k\neq i}}0w_{k}}_{=0}=\underbrace{\delta_{i,i}}_{=1}w_{i}%
=w_{i}.
\end{align*}
Therefore, $w_{i}=\sum_{k=0}^{N}\delta_{i,k}w_{k}$. Multiplying this by $M$,
we obtain
\begin{align*}
Mw_{i}  &  =M\sum_{k=0}^{N}\underbrace{\delta_{i,k}}_{=\delta_{N+1-i,N+1-k}%
}w_{k}=\sum_{k=0}^{N}\underbrace{M\delta_{N+1-i,N+1-k}}_{\substack{=\sum
_{j=1}^{N+1}t_{N+1-i,j}j^{N+1-\left(  N+1-k\right)  }\\\text{(by
(\ref{pf.lem.sol.eulerian-idp.2.poly1.2}),}\\\text{applied to }u=N+1-i\text{
and }v=N+1-k\text{)}}}w_{k}\\
&  =\sum_{k=0}^{N}\left(  \sum_{j=1}^{N+1}t_{N+1-i,j}%
\underbrace{j^{N+1-\left(  N+1-k\right)  }}_{=j^{k}}\right)  w_{k}=\sum
_{k=0}^{N}\left(  \sum_{j=1}^{N+1}t_{N+1-i,j}j^{k}\right)  w_{k}=\sum
_{j=1}^{N+1}t_{N+1-i,j}\underbrace{\sum_{k=0}^{N}w_{k}j^{k}}%
_{\substack{=0\\\text{(by (\ref{eq.lem.sol.eulerian-idp.2.poly1.ass}),
applied}\\\text{to }n=j\text{)}}}\\
&  =\sum_{j=0}^{N}t_{N+1-i,j}0=0.
\end{align*}
Since $M$ is a positive integer, we can cancel $M$ from this equality (because
$V$ is torsionfree), and thus obtain $w_{i}=0$.

Now forget that we fixed $i$. We thus have proven that $w_{i}=0$ for every
$i\in\left\{  0,1,\ldots,N\right\}  $. In other words, $w_{k}=0$ for every
$k\in\left\{  0,1,\ldots,N\right\}  $.
Lemma~\ref{lem.sol.eulerian-idp.2.poly1} is thus proven.
\end{proof}

\begin{proof}
[Proof of Lemma~\ref{lem.leray.poly1-inf}.]The family $\left(  w_{k}\right)
_{k\in\NN}$ is finitely supported. In other words, all but finitely
many $k\in \NN$ satisfy $w_{k}=0$. In other words, there exists a finite
subset $K$ of $\NN$ such that%
\begin{equation}
\text{each }k\in\mathbb{N}\setminus K\text{ satisfies }w_{k}=0.
\label{pf.lem.leray.poly1-inf.fs}%
\end{equation}
Consider this $K$.

The set $K$ is a finite subset of $\NN$, and thus has an upper bound
(since any finite subset of $\NN$ has an upper bound). In other words,
there exists some $N\in \NN$ such that
\begin{equation}
\text{each }n\in K\text{ satisfies }n\leq N. \label{pf.lem.leray.poly1-inf.ub}%
\end{equation}
Consider this $N$.

Now,
\begin{equation}
\text{each }k\in\mathbb{N}\text{ satisfying }k>N\text{ satisfies }w_{k}=0
\label{pf.lem.leray.poly1-inf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.leray.poly1-inf.1}):} Let
$k\in \NN$ be such that $k>N$. We must prove that $w_{k}=0$.
\par
If we had $k\in K$, then we would have $k\leq N$ (by
(\ref{pf.lem.leray.poly1-inf.ub}) (applied to $n=k$)), which would contradict
$k>N$. Hence, we cannot have $k\in K$. Thus, we have $k\notin K$. Combining
$k\in \NN$ with $k\notin K$, we obtain $k\in\mathbb{N}\setminus K$.
Hence, (\ref{pf.lem.leray.poly1-inf.fs}) yields $w_{k}=0$. This proves
(\ref{pf.lem.leray.poly1-inf.1}).}.

We have assumed that $\sum_{k\in\NN}w_{k}n^{k}=0$ for all
$n\in \NN$. Thus, for all $n\in \NN$, we have%
\begin{align*}
0  &  =\sum_{k\in\NN}w_{k}n^{k}=\underbrace{\sum\limits_{\substack{k\in
\mathbb{N};\\k\leq N}}}_{=\sum_{k=0}^{N}}w_{k}n^{k}+\sum\limits_{\substack{k\in
\mathbb{N};\\k>N}}\underbrace{w_{k}}_{\substack{=0\\\text{(by
(\ref{pf.lem.leray.poly1-inf.1}))}}}n^{k}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }k\in\mathbb{N}\text{
satisfies either }k\leq N\text{ or }k>N\text{ (but not both)}\right) \\
&  =\sum_{k=0}^{N}w_{k}n^{k}+\underbrace{\sum\limits_{\substack{k\in\mathbb{N}%
;\\k>N}}0n^{k}}_{=0}=\sum_{k=0}^{N}w_{k}n^{k}.
\end{align*}
Hence, for all $n\in \NN$, we have $\sum_{k=0}^{N}w_{k}n^{k}=0$. Thus,
Lemma~\ref{lem.sol.eulerian-idp.2.poly1} shows that%
\begin{equation}
w_{k}=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,\ldots,N\right\}
. \label{pf.lem.leray.poly1-inf.3}%
\end{equation}


Now, it is easy to see that $w_{k}=0$ for every $k\in \NN$%
\ \ \ \ \footnote{\textit{Proof.} Let $k\in \NN$. We must prove that
$w_{k}=0$.
\par
If $k\in\left\{  0,1,\ldots,N\right\}  $, then this follows immediately from
(\ref{pf.lem.leray.poly1-inf.3}). Hence, for the rest of this proof, we can
WLOG assume that we don't have $k\in\left\{  0,1,\ldots,N\right\}  $. Assume
this.
\par
We have $k\notin\left\{  0,1,\ldots,N\right\}  $ (since we don't have
$k\in\left\{  0,1,\ldots,N\right\}  $). Combining $k\in \NN$ with
$k\notin\left\{  0,1,\ldots,N\right\}  $, we obtain $k\in\mathbb{N}%
\setminus\left\{  0,1,\ldots,N\right\}  =\left\{  N+1,N+2,N+3,\ldots\right\}
$. Hence, $k\geq N+1>N$. Thus, (\ref{pf.lem.leray.poly1-inf.1}) shows that
$w_{k}=0$. Qed.}. This proves Lemma~\ref{lem.leray.poly1-inf}.
\end{proof}

Before we come to the proof of Proposition~\ref{prop.leray.exp-algh2}, we
state another lemma, which is an easy consequence of
Lemma~\ref{lem.leray.poly1-inf}:

\begin{lemma}
\label{lem.leray.poly1-inf2}Let $V$ be a torsionfree abelian group (written
additively). Let $\left(  a_{k}\right)  _{k\in\NN}\in V^{\NN}$
and $\left(  b_{k}\right)  _{k\in\NN}\in V^{\NN}$ be two
finitely supported families of elements of $V$. Assume that%
\begin{equation}
\sum_{k\in\NN}a_{k}t^{k}=\sum_{k\in\NN}b_{k}t^{k}%
\ \ \ \ \ \ \ \ \ \ \text{for all }t\in\mathbb{N}.
\label{eq.lem.leray.poly1-inf2.cond}%
\end{equation}
Then, $a_{k}=b_{k}$ for every $k\in \NN$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.leray.poly1-inf2}.]The family $\left(  a_{k}\right)
_{k\in\NN}$ is finitely supported. In other words, all but finitely
many $k\in \NN$ satisfy $a_{k}=0$. In other words, there exists a finite
subset $X$ of $\NN$ such that%
\begin{equation}
\text{every }k\in\mathbb{N}\setminus X\text{ satisfies }a_{k}=0.
\label{pf.lem.leray.poly1-inf2.ak0}%
\end{equation}
Consider this $X$.

The family $\left(  b_{k}\right)  _{k\in\NN}$ is finitely supported. In
other words, all but finitely many $k\in \NN$ satisfy $b_{k}=0$. In
other words, there exists a finite subset $Y$ of $\NN$ such that%
\begin{equation}
\text{every }k\in\mathbb{N}\setminus Y\text{ satisfies }b_{k}=0.
\label{pf.lem.leray.poly1-inf2.bk0}%
\end{equation}
Consider this $Y$.

The set $X\cup Y$ is the union of two finite sets (namely, of the two finite
sets $X$ and $Y$), and thus is finite (since a union of two finite sets is
always finite). Hence, all but finitely many $k\in \NN$ satisfy
$k\in\mathbb{N}\setminus\left(  X\cup Y\right)  $. But%
\[
\text{every }k\in\mathbb{N}\setminus\left(  X\cup Y\right)  \text{ satisfies
}a_{k}-b_{k}=0
\]
\footnote{\textit{Proof.} Let $k\in\mathbb{N}\setminus\left(  X\cup Y\right)
$.
\par
Thus, $k\in\mathbb{N}\setminus\underbrace{\left(  X\cup Y\right)  }_{\supset
X}\subset\mathbb{N}\setminus X$. Hence, (\ref{pf.lem.leray.poly1-inf2.ak0})
shows that $a_{k}=0$.
\par
Also, $k\in\mathbb{N}\setminus\underbrace{\left(  X\cup Y\right)  }_{\supset
Y}\subset\mathbb{N}\setminus Y$. Hence, (\ref{pf.lem.leray.poly1-inf2.bk0})
shows that $b_{k}=0$.
\par
Now, $\underbrace{a_{k}}_{=0}-\underbrace{b_{k}}_{=0}=0-0=0$. Qed.}. Hence,
all but finitely many $k\in \NN$ satisfy $a_{k}-b_{k}=0$ (since all but
finitely many $k\in \NN$ satisfy $k\in\mathbb{N}\setminus\left(  X\cup
Y\right)  $). In other words, the family $\left(  a_{k}-b_{k}\right)
_{k\in\NN}\in V^{\NN}$ is finitely supported.

Now, let $t\in \NN$. Then, the family $\left(  a_{k}t^{k}\right)
_{k\in\NN}\in V^{\NN}$ is finitely
supported\footnote{\textit{Proof.} Every $k\in\mathbb{N}\setminus Q$ satisfies
$\underbrace{a_{k}}_{\substack{=0\\\text{(by
(\ref{pf.lem.leray.poly1-inf2.ak0}))}}}t^{k}=0t^{k}=0$. Hence, all but
finitely many $k\in \NN$ satisfy $a_{k}t^{k}=0$ (since all but finitely
many $k\in \NN$ satisfy $k\in\mathbb{N}\setminus Q$ (since the set $Q$
is finite)). In other words, the family $\left(  a_{k}t^{k}\right)
_{k\in\NN}\in V^{\NN}$ is finitely supported.}. The same
argument (applied to the family $\left(  b_{k}\right)  _{k\in\NN}$
instead of $\left(  a_{k}\right)  _{k\in\NN}$) shows that the family
$\left(  b_{k}t^{k}\right)  _{k\in\NN}\in V^{\NN}$ is finitely supported.

Recall that sums of the form $\sum_{q\in Q}f_{q}$ (where $\left(
f_{q}\right)  _{q\in Q}$ is a finitely supported family) satisfy the usual
rules for finite sums, even though their indexing set $Q$ may be infinite. In
particular, this pertains to the sums $\sum_{k\in\NN}a_{k}t^{k}$ and
$\sum_{k\in\NN}b_{k}t^{k}$ (since the families $\left(  a_{k}%
t^{k}\right)  _{k\in\NN}$ and $\left(  b_{k}t^{k}\right)
_{k\in\NN}$ are finitely supported). Hence, the following manipulations
are valid:%
\[
\sum_{k\in\NN}a_{k}t^{k}-\sum_{k\in\NN}b_{k}t^{k}=\sum
_{k\in\NN}\underbrace{\left(  a_{k}t^{k}-b_{k}t^{k}\right)  }_{=\left(
a_{k}-b_{k}\right)  t^{k}}=\sum_{k\in\NN}\left(  a_{k}-b_{k}\right)
t^{k}.
\]
Hence,%
\begin{equation}
\sum_{k\in\NN}\left(  a_{k}-b_{k}\right)  t^{k}=\sum_{k\in\mathbb{N}%
}a_{k}t^{k}-\sum_{k\in\NN}b_{k}t^{k}=0
\label{pf.lem.leray.poly1-inf2.3}%
\end{equation}
(by (\ref{eq.lem.leray.poly1-inf2.cond})).

Now, forget that we fixed $t$. We thus have proven that $\sum_{k\in\mathbb{N}%
}\left(  a_{k}-b_{k}\right)  t^{k}=0$ for all $t\in \NN$. Renaming the
index $t$ as $n$ in this statement, we conclude that $\sum_{k\in\mathbb{N}%
}\left(  a_{k}-b_{k}\right)  n^{k}=0$ for all $n\in \NN$. Hence,
Lemma~\ref{lem.leray.poly1-inf} (applied to $\left(  w_{k}\right)
_{k\in\NN}=\left(  a_{k}-b_{k}\right)  _{k\in\NN}$) shows that
$a_{k}-b_{k}=0$ for every $k\in \NN$. In other words, $a_{k}=b_{k}$ for
every $k\in \NN$. This proves Lemma~\ref{lem.leray.poly1-inf2}.
\end{proof}

Next, let us state another simple lemma:

\begin{lemma}
\label{lem.sol.leray.exp-alg.g*nz=0}Let $C$ be a $\kk$-coalgebra. Let
$A$ be a $\kk$-algebra. Let $f\in\mathfrak{n}\left(  C,A\right)  $ and
$z\in C$. Then, there exists an $N\in \NN$ such that every integer $n>N$
satisfies $f^{\star n}\left(  z\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.leray.exp-alg.g*nz=0}.]We have $f\in
\mathfrak{n}\left(  C,A\right)  $. In other words, $f$ is a pointwise $\star
$-nilpotent map in $\Hom \left(  C,A\right)  $ (since
$\mathfrak{n}\left(  C,A\right)  $ is the set of all pointwise $\star
$-nilpotent maps in $\Hom \left(  C,A\right)  $). In other
words, the family $\left(  f^{\star n}\right)  _{n\in\NN}\in\left(
\Hom \left(  C,A\right)  \right)  ^{\NN}$ is pointwise
finitely supported (by the definition of ``pointwise $\star
$-nilpotent''). In other words, for each $x\in C$, the family
$\left(  f^{\star n}\left(  x\right)  \right)  _{n\in\NN}\in
A^{\NN}$ is finitely supported (by the definition of
``pointwise finitely supported''). Applying this to $x=z$, we
conclude that the family $\left(  f^{\star n}\left(  z\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported. In other words,
all but finitely many $n\in \NN$ satisfy $f^{\star n}\left(  z\right)
=0$. In other words, there exists a finite subset $R$ of $\NN$ such
that
\begin{equation}
\text{each }n\in\mathbb{N}\setminus R\text{ satisfies }f^{\star n}\left(
z\right)  =0. \label{pf.prop.leray.exp-algh2.P.1.pf.1}%
\end{equation}
Consider this $R$.

The set $R$ is a finite subset of $\NN$, and thus has an upper bound
(since any finite subset of $\NN$ has an upper bound). In other words,
there exists some $w\in \NN$ such that each $r\in R$ satisfies $r\leq
w$. Consider this $w$.

Now, let $n$ be an integer such that $n>w$. Thus, $n>w\geq0$ (since
$w\in \NN$), so that $n\in \NN$.

Assume (for the sake of contradiction) that $n\in R$. Recall that each $r\in
R$ satisfies $r\leq w$. Applying this to $r=n$, we obtain $n\leq w$ (since
$n\in R$). This contradicts $n>w$. This contradiction shows that our
assumption (that $n\in R$) was wrong. Hence, we cannot have $n\in R$. In other
words, we have $n\notin R$. Combining $n\in \NN$ with $n\notin R$, we
obtain $n\in\mathbb{N}\setminus R$. Hence,
(\ref{pf.prop.leray.exp-algh2.P.1.pf.1}) yields $f^{\star n}\left(  z\right)
=0$.

Now, forget that we fixed $n$. We thus have shown that every integer $n>w$
satisfies $f^{\star n}\left(  z\right)  =0$. Hence, there exists an
$N\in \NN$ such that every integer $n>N$ satisfies $f^{\star n}\left(
z\right)  =0$ (namely, $N=z$). This proves Lemma
\ref{lem.sol.leray.exp-alg.g*nz=0}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.exp-algh2}.]Any $\kk$-module
naturally becomes a $\QQ$-module (since $\kk$ is a $\QQ$-algebra).
Thus, in particular, the $\kk$-module $A$ becomes a
$\QQ$-module. Hence, the $\kk$-module $A$ is a torsionfree
abelian group (written additively).

Now, fix $x\in\ker\epsilon$ and $y\in\ker\epsilon$. We shall show that
$f\left(  xy\right)  =0$.

We have $f\left(  C_{0}\right)  =0$. Thus,
Proposition~\ref{prop.convolution-series}(h) yields $f\in\mathfrak{n}\left(
C,A\right)  $.

Thus, Lemma \ref{lem.sol.leray.exp-alg.g*nz=0} (applied to $z=x$) shows that
there exists an $N\in \NN$ such that every integer $n>N$ satisfies
$f^{\star n}\left(  x\right)  =0$. Consider such an $N$, and denote it by
$N_{1}$. Thus, $N_{1}$ is an element of $\NN$ and has the property that%
\begin{equation}
\text{every integer }n>N_{1}\text{ satisfies }f^{\star n}\left(  x\right)  =0.
\label{pf.prop.leray.exp-algh2.N1}%
\end{equation}


Also, Lemma \ref{lem.sol.leray.exp-alg.g*nz=0} (applied to $z=y$) shows that
there exists an $N\in \NN$ such that every integer $n>N$ satisfies
$f^{\star n}\left(  y\right)  =0$. Consider such an $N$, and denote it by
$N_{2}$. Thus, $N_{2}$ is an element of $\NN$ and has the property that%
\begin{equation}
\text{every integer }n>N_{2}\text{ satisfies }f^{\star n}\left(  y\right)  =0.
\label{pf.prop.leray.exp-algh2.N2}%
\end{equation}


Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $z=xy$) shows that
the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  xy\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\[
\left(  \exp^{\star}f\right)  \left(  xy\right)  =\sum_{n\in\NN}%
\dfrac{1}{n!}f^{\star n}\left(  xy\right)  .
\]
In particular, the family $\left(  \dfrac{1}{n!}f^{\star n}\left(  xy\right)
\right)  _{n\in\NN}\in A^{\NN}$ is finitely supported. Renaming
the index $n$ as $k$ in this statement, we conclude that the family $\left(
\dfrac{1}{k!}f^{\star k}\left(  xy\right)  \right)  _{k\in\NN}\in
A^{\NN}$ is finitely supported.

The set $\left\{  N_{1},N_{2}\right\}  $ is a finite subset of $\NN$
(since $N_{1}$ and $N_{2}$ are elements of $\NN$), and thus has an
upper bound (since every finite subset of $\NN$ has an upper bound). In
other words, there exists some $N^{\prime}\in \NN$ such that
\begin{equation}
\text{every }k\in\left\{  N_{1},N_{2}\right\}  \text{ satisfies }k\leq
N^{\prime}. \label{pf.prop.leray.exp-algh2.N'}%
\end{equation}
Consider this $N^{\prime}$.

Every $n\in\mathbb{N}\setminus\left\{  0,1,\ldots,2N^{\prime}\right\}  $ and
every $i\in\left\{  0,1,\ldots,n\right\}  $ satisfy%
\begin{equation}
\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(
n-i\right)  }\left(  y\right)  }{\left(  n-i\right)  !}=0
\label{pf.prop.leray.exp-algh2.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.leray.exp-algh2.3}):} Let
$n\in\mathbb{N}\setminus\left\{  0,1,\ldots,2N^{\prime}\right\}  $ and
$i\in\left\{  0,1,\ldots,n\right\}  $.
\par
We have $i\in\left\{  0,1,\ldots,n\right\}  $, so that $i\leq n$. Moreover,
$n\in\mathbb{N}\setminus\left\{  0,1,\ldots,2N^{\prime}\right\}  =\left\{
2N^{\prime}+1,2N^{\prime}+2,2N^{\prime}+3,\ldots\right\}  $. Thus,
$n\geq2N^{\prime}+1>2N^{\prime}$.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $i>N^{\prime}$.
\par
\textit{Case 2:} We have $i\leq N^{\prime}$.
\par
Let us first consider Case 1. In this case, we have $i>N^{\prime}$. But
(\ref{pf.prop.leray.exp-algh2.N'}) (applied to $k=N_{1}$) yields $N_{1}\leq
N^{\prime}$ (since $N_{1}\in\left\{  N_{1},N_{2}\right\}  $). Hence,
$N^{\prime}\geq N_{1}$, so that $i>N^{\prime}\geq N_{1}$. Thus,
(\ref{pf.prop.leray.exp-algh2.N1}) (applied to $i$ instead of $n$) shows that
$f^{\star i}\left(  x\right)  =0$. Hence, $\dfrac{f^{\star i}\left(  x\right)
}{i!}=\dfrac{0}{i!}=0$, and therefore%
\[
\underbrace{\dfrac{f^{\star i}\left(  x\right)  }{i!}}_{=0}\cdot
\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)
!}=0.
\]
Thus, (\ref{pf.prop.leray.exp-algh2.3}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i\leq N^{\prime}$. But
(\ref{pf.prop.leray.exp-algh2.N'}) (applied to $k=N_{2}$) yields $N_{2}\leq
N^{\prime}$ (since $N_{2}\in\left\{  N_{1},N_{2}\right\}  $). Hence,
$N^{\prime}\geq N_{2}$. Now, $\underbrace{n}_{>2N^{\prime}}-\underbrace{i}%
_{\leq N^{\prime}}>2N^{\prime}-N^{\prime}=N^{\prime}\geq N_{2}$. Thus,
(\ref{pf.prop.leray.exp-algh2.N2}) (applied to $n-i$ instead of $n$) shows
that $f^{\star\left(  n-i\right)  }\left(  y\right)  =0$. Hence,
$\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)
!}=\dfrac{0}{\left(  n-i\right)  !}=0$, and therefore%
\[
\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\underbrace{\dfrac
{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)  !}}%
_{=0}=0.
\]
Thus, (\ref{pf.prop.leray.exp-algh2.3}) is proven in Case 2.
\par
We have now proven (\ref{pf.prop.leray.exp-algh2.3}) in each of the two Cases
1 and 2. Hence, (\ref{pf.prop.leray.exp-algh2.3}) always holds (since Cases 1
and 2 cover all possibilities).}.

Hence, every $n\in\mathbb{N}\setminus\left\{  0,1,\ldots,2N^{\prime}\right\}
$ satisfies%
\begin{equation}
\sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac
{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)  !}=0
\label{pf.prop.leray.exp-algh2.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.leray.exp-algh2.4}):} Let
$n\in\mathbb{N}\setminus\left\{  0,1,\ldots,2N^{\prime}\right\}  $. Then,
\[
\sum_{i=0}^{n}\underbrace{\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot
\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)
!}}_{\substack{=0\\\text{(by (\ref{pf.prop.leray.exp-algh2.3}))}}}=\sum
_{i=0}^{n}0=0.
\]
This proves (\ref{pf.prop.leray.exp-algh2.4}).}. Thus, all but finitely many
$n\in \NN$ satisfy%
\[
\sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac
{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)  !}=0
\]
(since all but finitely many $n\in \NN$ satisfy $n\in\mathbb{N}%
\setminus\left\{  0,1,\ldots,2N^{\prime}\right\}  $). In other words, the
family%
\[
\left(  \sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot
\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(  n-i\right)
!}\right)  _{n\in\NN}\in A^{\NN}%
\]
is finitely supported. Renaming the index $n$ as $k$ in this statement, we
obtain the following: The family%
\[
\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot
\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)  }{\left(  k-i\right)
!}\right)  _{k\in\NN}\in A^{\NN}%
\]
is finitely supported.

On the other hand, let $t\in \NN$ be arbitrary. Then,
Proposition~\ref{prop.exp-log}(e) (applied to $n=t$) shows that $tf\in
\mathfrak{n}\left(  C,A\right)  $ and $\exp^{\star}\left(  tf\right)  =\left(
\exp^{\star}f\right)  ^{\star t}$.

Exercise~\ref{exe.alghoms.comm}(b) (applied to $C$, $t$ and $\exp^{\star}f$
instead of $H$, $k$ and $f_{i}$) shows that the map \newline%
$\underbrace{\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}f\right)
\star\cdots\star\left(  \exp^{\star}f\right)  }_{t\text{ times}}$ is a
$\kk$-algebra homomorphism $C\rightarrow A$. In light of%
\[
\underbrace{\left(  \exp^{\star}f\right)  \star\left(  \exp^{\star}f\right)
\star\cdots\star\left(  \exp^{\star}f\right)  }_{t\text{ times}}=\left(
\exp^{\star}f\right)  ^{\star t}=\exp^{\star}\left(  tf\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\exp^{\star}\left(  tf\right)
=\left(  \exp^{\star}f\right)  ^{\star t}\right)  ,
\]
this rewrites as follows: The map $\exp^{\star}\left(  tf\right)  $ is a
$\kk$-algebra homomorphism $C\rightarrow A$.

Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(a) (applied to $tf$ and $xy$ instead
of $f$ and $z$) shows that the family $\left(  \dfrac{1}{n!}\left(  tf\right)
^{\star n}\left(  xy\right)  \right)  _{n\in\NN}\in A^{\NN}$ is
finitely supported and satisfies%
\begin{equation}
\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  xy\right)  =\sum
_{n\in\NN}\dfrac{1}{n!}\left(  tf\right)  ^{\star n}\left(  xy\right)
. \label{pf.prop.leray.exp-algh2.etfxy}%
\end{equation}


Lemma~\ref{lem.pf.prop.leray.exp-algh1.1}(b) (applied to $tf$ instead of $f$)
shows that the family \newline$\left(  \sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(
n-i\right)  !}\left(  tf\right)  ^{\star i}\left(  x\right)  \left(
tf\right)  ^{\star\left(  n-i\right)  }\left(  y\right)  \right)
_{n\in\NN}\in A^{\NN}$ is finitely supported and satisfies%
\[
\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  x\right)  \cdot\left(
\exp^{\star}\left(  tf\right)  \right)  \left(  y\right)  =\sum_{n\in
\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)  !}\left(
tf\right)  ^{\star i}\left(  x\right)  \left(  tf\right)  ^{\star\left(
n-i\right)  }\left(  y\right)  .
\]
Comparing
\begin{align*}
&  \left(  \exp^{\star}\left(  tf\right)  \right)  \left(  x\right)
\cdot\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}\underbrace{\left(  tf\right)  ^{\star i}}_{=t^{i}f^{\star i}}\left(
x\right)  \underbrace{\left(  tf\right)  ^{\star\left(  n-i\right)  }%
}_{=t^{n-i}f^{\star\left(  n-i\right)  }}\left(  y\right) \\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{1}{i!\cdot\left(  n-i\right)
!}\underbrace{\left(  t^{i}f^{\star i}\right)  \left(  x\right)  }%
_{=t^{i}f^{\star i}\left(  x\right)  }\underbrace{\left(  t^{n-i}%
f^{\star\left(  n-i\right)  }\right)  \left(  y\right)  }_{=t^{n-i}%
f^{\star\left(  n-i\right)  }\left(  y\right)  }\\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\underbrace{\dfrac{1}{i!\cdot\left(
n-i\right)  !}t^{i}f^{\star i}\left(  x\right)  t^{n-i}f^{\star\left(
n-i\right)  }\left(  y\right)  }_{=\dfrac{f^{\star i}\left(  x\right)  }%
{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(
n-i\right)  !}t^{i}t^{n-i}}\\
&  =\sum_{n\in\NN}\sum_{i=0}^{n}\dfrac{f^{\star i}\left(  x\right)
}{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)  }{\left(
n-i\right)  !}\underbrace{t^{i}t^{n-i}}_{=t^{i+\left(  n-i\right)  }=t^{n}%
}=\sum_{n\in\NN}\underbrace{\sum_{i=0}^{n}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)
}{\left(  n-i\right)  !}t^{n}}_{=\left(  \sum_{i=0}^{n}\dfrac{f^{\star
i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(
y\right)  }{\left(  n-i\right)  !}\right)  t^{n}}\\
&  =\sum_{n\in\NN}\left(  \sum_{i=0}^{n}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  n-i\right)  }\left(  y\right)
}{\left(  n-i\right)  !}\right)  t^{n}=\sum_{k\in\NN}\left(  \sum
_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(
k-i\right)  }\left(  y\right)  }{\left(  k-i\right)  !}\right)  t^{k}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}n\text{ as }k\right)
\end{align*}
with%
\begin{align*}
&  \left(  \exp^{\star}\left(  tf\right)  \right)  \left(  x\right)
\cdot\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  y\right) \\
&  =\left(  \exp^{\star}\left(  tf\right)  \right)  \left(  xy\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\exp^{\star}\left(  tf\right)  \text{
is a }\kk\text{-algebra homomorphism}\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\underbrace{\left(  tf\right)  ^{\star
n}}_{=t^{n}f^{\star n}}\left(  xy\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.leray.exp-algh2.etfxy})}\right) \\
&  =\sum_{n\in\NN}\dfrac{1}{n!}\underbrace{\left(  t^{n}f^{\star
n}\right)  \left(  xy\right)  }_{=t^{n}f^{\star n}\left(  xy\right)  }%
=\sum_{n\in\NN}\dfrac{1}{n!}t^{n}f^{\star n}\left(  xy\right)
=\sum_{n\in\NN}\dfrac{1}{n!}f^{\star n}\left(  xy\right)  t^{n}\\
&  =\sum_{k\in\NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)
t^{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation
index }n\text{ as }k\right)  ,
\end{align*}
we obtain%
\[
\sum_{k\in\NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)  t^{k}%
=\sum_{k\in\NN}\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)
}{\left(  k-i\right)  !}\right)  t^{k}.
\]


Now, forget that we fixed $t$. We thus have shown that%
\[
\sum_{k\in\NN}\dfrac{1}{k!}f^{\star k}\left(  xy\right)  t^{k}%
=\sum_{k\in\NN}\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(
x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)
}{\left(  k-i\right)  !}\right)  t^{k}%
\]
for all $t\in \NN$.

Hence, we can apply Lemma~\ref{lem.leray.poly1-inf2} to $V=A$, $\left(
a_{k}\right)  _{k\in\NN}=\left(  \dfrac{1}{k!}f^{\star k}\left(
xy\right)  \right)  _{k\in\NN}$ and \newline$\left(  b_{k}\right)
_{k\in\NN}=\left(  \sum_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)
}{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(  y\right)  }{\left(
k-i\right)  !}\right)  _{k\in\NN}$ (since $A$ is a torsionfree abelian
group (written additively), and since the families $\left(  \dfrac{1}%
{k!}f^{\star k}\left(  xy\right)  \right)  _{k\in\NN}$ and $\left(
\sum_{i=0}^{k}\dfrac{f^{\star i}\left(  x\right)  }{i!}\cdot\dfrac
{f^{\star\left(  k-i\right)  }\left(  y\right)  }{\left(  k-i\right)
!}\right)  _{k\in\NN}$ are finitely supported). As a result, we obtain%
\[
\dfrac{1}{k!}f^{\star k}\left(  xy\right)  =\sum_{i=0}^{k}\dfrac{f^{\star
i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(  k-i\right)  }\left(
y\right)  }{\left(  k-i\right)  !}\ \ \ \ \ \ \ \ \ \ \text{for every }%
k\in\mathbb{N}.
\]
Applying this to $k=1$, we find%
\begin{align}
\dfrac{1}{1!}f^{\star1}\left(  xy\right)   &  =\sum_{i=0}^{1}\dfrac{f^{\star
i}\left(  x\right)  }{i!}\cdot\dfrac{f^{\star\left(  1-i\right)  }\left(
y\right)  }{\left(  1-i\right)  !}=\dfrac{f^{\star0}\left(  x\right)  }%
{0!}\cdot\dfrac{f^{\star\left(  1-0\right)  }\left(  y\right)  }{\left(
1-0\right)  !}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot\dfrac
{f^{\star\left(  1-1\right)  }\left(  y\right)  }{\left(  1-1\right)
!}\nonumber\\
&  =\dfrac{f^{\star0}\left(  x\right)  }{0!}\cdot\dfrac{f^{\star1}\left(
y\right)  }{1!}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot\dfrac{f^{\star
0}\left(  y\right)  }{0!} \label{pf.prop.leray.exp-algh2.5}%
\end{align}
(since $1-0=1$ and $1-1=0$).

But recall that $x\in\ker\underbrace{\epsilon}_{=\epsilon_{C}}=\ker\left(
\epsilon_{C}\right)  $ and thus $\epsilon_{C}\left(  x\right)  =0$. But%
\[
f^{\star0}=\left(  \text{the unity of the }\kk\text{-algebra }\left(
\Hom \left(  C,A\right)  ,\star\right)  \right)  =u_{A}%
\circ\epsilon_{C},
\]
and thus $f^{\star0}\left(  x\right)  =\left(  u_{A}\circ\epsilon_{C}\right)
\left(  x\right)  =u_{A}\left(  \underbrace{\epsilon_{C}\left(  x\right)
}_{=0}\right)  =u_{A}\left(  0\right)  =0$ (since the map $u_{A}$ is
$\kk$-linear). Hence, $\dfrac{f^{\star0}\left(  x\right)  }{0!}%
=\dfrac{0}{0!}=0$. The same argument (applied to $y$ instead of $x$) shows
that $\dfrac{f^{\star0}\left(  y\right)  }{0!}=0$. Hence,
(\ref{pf.prop.leray.exp-algh2.5}) becomes%
\[
\dfrac{1}{1!}f^{\star1}\left(  xy\right)  =\underbrace{\dfrac{f^{\star
0}\left(  x\right)  }{0!}}_{=0}\cdot\dfrac{f^{\star1}\left(  y\right)  }%
{1!}+\dfrac{f^{\star1}\left(  x\right)  }{1!}\cdot\underbrace{\dfrac
{f^{\star0}\left(  y\right)  }{0!}}_{=0}=\underbrace{0\cdot\dfrac{f^{\star
1}\left(  y\right)  }{1!}}_{=0}+\underbrace{\dfrac{f^{\star1}\left(  x\right)
}{1!}\cdot0}_{=0}=0+0=0.
\]
Comparing this with $\underbrace{\dfrac{1}{1!}}_{=\dfrac{1}{1}=1}%
\underbrace{f^{\star1}}_{=f}\left(  xy\right)  =f\left(  xy\right)  $, we
obtain $f\left(  xy\right)  =0$.

Now, forget that we fixed $x$ and $y$. We thus have shown that%
\begin{equation}
f\left(  xy\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\ker
\epsilon\text{ and }y\in\ker\epsilon\text{.} \label{pf.prop.leray.exp-algh2.9}%
\end{equation}


Now, let $z\in\left(  \ker\epsilon\right)  ^{2}$ be arbitrary. Thus,
$z\in\left(  \ker\epsilon\right)  ^{2}=\left(  \ker\epsilon\right)  \left(
\ker\epsilon\right)  $. In other words, $z$ can be written in the form
$z=\sum_{i=1}^{g}x_{i}y_{i}$ for some $g\in \NN$, some elements
$x_{1},x_{2},\ldots,x_{g}\in\ker\epsilon$ and some elements $y_{1}%
,y_{2},\ldots,y_{g}\in\ker\epsilon$. Consider this $g$, these $x_{1}%
,x_{2},\ldots,x_{g}$ and these $y_{1},y_{2},\ldots,y_{g}$. Applying the map
$f$ to both sides of the equation $z=\sum_{i=1}^{g}x_{i}y_{i}$, we obtain%
\begin{align*}
f\left(  z\right)   &  =f\left(  \sum_{i=1}^{g}x_{i}y_{i}\right)  =\sum
_{i=1}^{g}\underbrace{f\left(  x_{i}y_{i}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.prop.leray.exp-algh2.9}) (applied}\\\text{to }x=x_{i}\text{ and
}y=y_{i}\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is
}\kk\text{-linear}\right) \\
&  =\sum_{i=1}^{g}0=0.
\end{align*}
In other words, $z\in\ker f$.

Now, forget that we fixed $z$. We thus have shown that $z\in\ker f$ for each
$z\in\left(  \ker\epsilon\right)  ^{2}$. In other words, $\left(  \ker
\epsilon\right)  ^{2}\subset\ker f$. Hence, $f\left(  \underbrace{\left(
\ker\epsilon\right)  ^{2}}_{\subset\ker f}\right)  \subset f\left(  \ker
f\right)  =0$. In other words, $f\left(  \left(  \ker\epsilon\right)
^{2}\right)  =0$. This proves Proposition~\ref{prop.leray.exp-algh2}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.leray.exp-algh3}.]We know that $f\left(
C\right)  $ generates the $\kk$-algebra $A$. Thus, any $\kk%
$-subalgebra of $A$ that contains $f\left(  C\right)  $ as a subset must be
the whole $A$. In other words,
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }D\text{ is a }\kk\text{-subalgebra of }A\text{ satisfying
}f\left(  C\right)  \subset D\text{,}\\
\text{then }D=A
\end{array}
\right)  . \label{pf.prop.leray.exp-algh3.gener}%
\end{equation}


Define a $\kk$-linear map $F\in\Hom \left(  C,A\right)  $
by $F=\exp^{\star}f$. (This is well-defined, since $f\in\mathfrak{n}\left(
C,A\right)  $.)

Proposition~\ref{prop.leray.exp-algh1} shows that $\exp^{\star}f:C\rightarrow
A$ is a $\kk$-algebra homomorphism. In other words, $F:C\rightarrow A$
is a $\kk$-algebra homomorphism (since $F=\exp^{\star}f$).

Proposition~\ref{prop.exp-log}(a) yields that $\exp^{\star}f-u_{A}\epsilon
_{C}\in\mathfrak{n}\left(  C,A\right)  $ and $\log^{\star}\left(  \exp^{\star
}f\right)  =f$.

Define an element $\widetilde{F}$ of $\mathfrak{n}\left(  C,A\right)  $ by
$\widetilde{F}=F-u_{A}\epsilon_{C}$. (This is well-defined, since
$\underbrace{F}_{=\exp^{\star}f}-u_{A}\epsilon_{C}=\exp^{\star}f-u_{A}%
\epsilon_{C}\in\mathfrak{n}\left(  C,A\right)  $.) From $\log^{\star}\left(
\exp^{\star}f\right)  =f$, we obtain%
\begin{align}
f  &  =\log^{\star}\underbrace{\left(  \exp^{\star}f\right)  }_{=F}%
=\log^{\star}F=\overline{\log}^{\star}\underbrace{\left(  F-u_{A}\epsilon
_{C}\right)  }_{=\widetilde{F}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\log^{\star}F\right) \nonumber\\
&  =\overline{\log}^{\star}\widetilde{F}.
\label{pf.prop.leray.exp-algh3.f=log}%
\end{align}


On the other hand, Proposition~\ref{prop.convolution-series}(j) (applied to
$C$ and $\id_{C}$ instead of $A$ and $F$) shows that
$\id_{C}-u_{C}\epsilon_{C}\in\mathfrak{n}\left(
C,C\right)  $ (since $\id_{C}:C\rightarrow C$ is a
$\kk$-algebra homomorphism).

Define an element $\widetilde{\id}$ of $\mathfrak{n}\left(
C,C\right)  $ by $\widetilde{\id}=\operatorname*{id}%
\nolimits_{C}-u_{C}\epsilon_{C}$. (This is well-defined, since
$\id_{C}-u_{C}\epsilon_{C}\in\mathfrak{n}\left(
C,C\right)  $.) Then, $F\circ\widetilde{\id}=\widetilde{F}%
$\ \ \ \ \footnote{\textit{Proof.} Recall that $F$ is a $\kk$-algebra
homomorphism. In other words, $F$ is a $\kk$-linear map satisfying
$F\circ m_{C}=m_{A}\circ\left(  F\otimes F\right)  $ and $F\circ u_{C}=u_{A}$.
Now,%
\begin{align*}
F\circ\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{C}-u_{C}\epsilon_{C}}  &  =F\circ\left(  \operatorname*{id}%
\nolimits_{C}-u_{C}\epsilon_{C}\right)  =\underbrace{F\circ\operatorname*{id}%
\nolimits_{C}}_{=F}-\underbrace{F\circ\left(  u_{C}\epsilon_{C}\right)
}_{=F\circ u_{C}\circ\epsilon_{C}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }\kk%
\text{-linear maps is }\kk\text{-bilinear}\right) \\
&  =F-\underbrace{F\circ u_{C}}_{=u_{A}}\circ\epsilon_{C}=F-u_{A}\circ
\epsilon_{C}=F-u_{A}\epsilon_{C}=\widetilde{F}.
\end{align*}
}. Hence, (\ref{pf.prop.leray.exp-algh3.f=log}) becomes%
\begin{equation}
f=\overline{\log}^{\star}\underbrace{\widetilde{F}}_{=F\circ
\widetilde{\id}}=\overline{\log}^{\star}\left(  F\circ
\widetilde{\id}\right)  .
\label{pf.prop.leray.exp-algh3.f=log2}%
\end{equation}


Proposition~\ref{prop.convolution-series}(i) (applied to $C$, $A$, $F$,
$\overline{\log}$ and $\widetilde{\id}$ instead of $A$, $B$,
$s$, $u$ and $f$) shows that%
\[
F\circ\widetilde{\id}\in\mathfrak{n}\left(  C,A\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \overline{\log}^{\star
}\left(  F\circ\widetilde{\id}\right)  =F\circ\left(
\overline{\log}^{\star}\left(  \widetilde{\id}\right)  \right)
.
\]


Now, (\ref{pf.prop.leray.exp-algh3.f=log2}) becomes%
\[
f=\overline{\log}^{\star}\left(  F\circ\widetilde{\id}\right)
=F\circ\left(  \overline{\log}^{\star}\left(  \widetilde{\operatorname*{id}%
}\right)  \right)  .
\]
Hence,%
\[
f\left(  C\right)  =\left(  F\circ\left(  \overline{\log}^{\star}\left(
\widetilde{\id}\right)  \right)  \right)  \left(  C\right)
=F\left(  \underbrace{\left(  \overline{\log}^{\star}\left(
\widetilde{\id}\right)  \right)  \left(  C\right)
}_{\substack{\subset C\\\text{(since }\overline{\log}^{\star}\left(
\widetilde{\id}\right)  \text{ is a map }C\rightarrow
C\text{)}}}\right)  \subset F\left(  C\right)  .
\]


But $F\left(  C\right)  $ is a $\kk$-subalgebra of $A$ (since $F$ is a
$\kk$-algebra homomorphism). Hence,
(\ref{pf.prop.leray.exp-algh3.gener}) (applied to $D=F\left(  C\right)  $)
shows that $F\left(  C\right)  =A$ (since $f\left(  C\right)  \subset F\left(
C\right)  $). In other words, the map $F$ is surjective.

Hence, $F$ is a surjective $\kk$-algebra homomorphism. In other words,
$\exp^{\star}f$ is a surjective $\kk$-algebra homomorphism (since
$F=\exp^{\star}f$). This proves Proposition~\ref{prop.leray.exp-algh3}.
\end{proof}

We have now proven Lemmas~\ref{lem.sol.eulerian-idp.2.poly1}%
~and~\ref{lem.leray.poly1-inf} and Propositions~\ref{prop.convolution.nat}%
,~\ref{prop.leray.f*nxy},~\ref{prop.leray.exp-algh1}%
,~\ref{prop.leray.exp-algh2}~and~\ref{prop.leray.exp-algh3}. Thus,
Exercise~\ref{exe.leray.exp-alg} is solved.

%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.leray.leray-e}}

\textit{Solution to Exercise~\ref{exe.leray.leray-e}.} We begin by proving
some simple lemmas:

\begin{lemma}
\label{lem.sol.leray.leray-e.1}Let $C$ be a $\kk$-coalgebra. Let $A$ be
a $\kk$-algebra. Let $f\in\Hom \left(  C,A\right)  $.
Then, every $n\in \NN$ satisfies%
\begin{equation}
f^{\star n}\left(  C\right)  \subset\left(  f\left(  C\right)  \right)  ^{n}.
\label{eq.lem.sol.leray.leray-e.1.claim}%
\end{equation}
(Here, we set $V^{0}=\kk\cdot1_{A}$ for any $\kk$-submodule $V$
of $A$.)
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.leray.leray-e.1}.]We must prove the relation
(\ref{eq.lem.sol.leray.leray-e.1.claim}). We shall prove it by induction over
$n$:

\textit{Induction base:} We have $f^{\star0}=\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \right)  =u_{A}\epsilon_{C}$. Thus, every $x\in C$ satisfies%
\begin{align*}
\underbrace{f^{\star0}}_{=u_{A}\epsilon_{C}}\left(  x\right)   &  =\left(
u_{A}\epsilon_{C}\right)  \left(  x\right)  =u_{A}\left(  \epsilon_{C}\left(
x\right)  \right)  =\underbrace{\epsilon_{C}\left(  x\right)  }_{\in
\kk}\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}u_{A}\right) \\
&  \in\kk\cdot1_{A}=\left(  f\left(  C\right)  \right)  ^{0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  f\left(  C\right)  \right)
^{0}=\kk\cdot1_{A}\right)  .
\end{align*}
In other words, we have $f^{\star0}\left(  C\right)  \subset\left(  f\left(
C\right)  \right)  ^{0}$. In other words,
(\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=0$.

\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N$. We must prove that
(\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N+1$.

If $X$ and $Y$ are two $\kk$-submodules of $A$, then the multiplication
$m_{A}:A\otimes A\rightarrow A$ of the $\kk$-algebra $A$ satisfies%
\begin{equation}
m_{A}\left(  X\otimes Y\right)  =XY. \label{pf.lem.sol.leray.leray-e.1.XY}%
\end{equation}
(This follows easily from the definitions of $m_{A}$ and of $XY$.)

We have assumed that (\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N$.
In other words, we have $f^{\star N}\left(  C\right)  \subset\left(  f\left(
C\right)  \right)  ^{N}$. But%
\[
f^{\star\left(  N+1\right)  }=f\star f^{\star N}=m_{A}\circ\left(  f\otimes
f^{\star N}\right)  \circ\Delta_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of convolution}\right)  .
\]
Hence,%
\begin{align*}
\underbrace{f^{\star\left(  N+1\right)  }}_{=m_{A}\circ\left(  f\otimes
f^{\star N}\right)  \circ\Delta_{C}}\left(  C\right)   &  =\left(  m_{A}%
\circ\left(  f\otimes f^{\star N}\right)  \circ\Delta_{C}\right)  \left(
C\right)  =m_{A}\left(  \left(  f\otimes f^{\star N}\right)  \left(
\underbrace{\Delta_{C}\left(  C\right)  }_{\subset C\otimes C}\right)  \right)
\\
&  \subset m_{A}\left(  \underbrace{\left(  f\otimes f^{\star N}\right)
\left(  C\otimes C\right)  }_{=f\left(  C\right)  \otimes f^{\star N}\left(
C\right)  }\right)  =m_{A}\left(  f\left(  C\right)  \otimes f^{\star
N}\left(  C\right)  \right) \\
&  =f\left(  C\right)  \cdot\underbrace{f^{\star N}\left(  C\right)
}_{\subset\left(  f\left(  C\right)  \right)  ^{N}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.sol.leray.leray-e.1.XY}) (applied to }X=f\left(
C\right)  \text{ and }Y=f^{\star N}\left(  C\right)  \text{)}\right) \\
&  \subset f\left(  C\right)  \cdot\left(  f\left(  C\right)  \right)
^{N}=\left(  f\left(  C\right)  \right)  ^{N+1}.
\end{align*}
In other words, (\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N+1$.
This completes the induction step. Thus, the proof of
(\ref{eq.lem.sol.leray.leray-e.1.claim}) by induction is complete. In other
words, Lemma~\ref{lem.sol.leray.leray-e.1} is proven.
\end{proof}

\begin{lemma}
\label{lem.sol.leray.leray-e.2}Let $A$ be a $\kk$-bialgebra. Let
$\widetilde{\id}$ be the $\kk$-linear map
$\id_{A}-u_{A}\epsilon_{A}:A\rightarrow A$. Then:

\begin{enumerate}
\item[(a)] We have $\ker\widetilde{\id}=\kk\cdot1_{A}$.

\item[(b)] We have $\widetilde{\id}\left(  A\right)
\subset\ker\epsilon$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.leray.leray-e.2}.]The axioms of a $\kk%
$-bialgebra yield $\epsilon_{A}\left(  1_A\right)  =1$ (since $A$ is a
$\kk$-bialgebra). Also, the definition of $u_{A}$ yields $u_{A}\left(
1\right)  =1\cdot1_{A}=1_{A}$.

We have%
\begin{align*}
\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}}\left(  1_{A}\right)   &  =\left(
\id_{A}-u_{A}\epsilon_{A}\right)  \left(
1_{A}\right)  =\underbrace{\id_{A}\left(
1_{A}\right)  }_{=1_{A}}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(
1_{A}\right)  }_{=u_{A}\left(  \epsilon_{A}\left(  1_{A}\right)  \right)
}=1_{A}-u_{A}\left(  \underbrace{\epsilon_{A}\left(  1_{A}\right)  }%
_{=1}\right) \\
&  =1_{A}-\underbrace{u_{A}\left(  1\right)  }_{=1_{A}}=1_{A}-1_{A}=0.
\end{align*}


Now, the map $\widetilde{\id}=\id_{A}%
-u_{A}\epsilon_{A}$ is $\kk$-linear (since all three maps
$\id_{A}$, $u_{A}$ and $\epsilon_{A}$ are $\kk%
$-linear). Therefore, $\widetilde{\id}\left(  \kk%
\cdot1_{A}\right)  =\kk\cdot\underbrace{\widetilde{\operatorname*{id}%
}\left(  1_{A}\right)  }_{=0}=\kk\cdot0=0$. Hence, $\kk%
\cdot1_{A}\subset\ker\widetilde{\id}$.

On the other hand, let $x\in\ker\widetilde{\id}$. Thus,
$\widetilde{\id}\left(  x\right)  =0$. Comparing this to%
\[
\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}}\left(  x\right)  =\left(  \operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}\right)  \left(  x\right)
=\underbrace{\id_{A}\left(  x\right)  }_{=x}%
-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\epsilon_{A}\left(  x\right)  \cdot1_{A},
\]
we obtain $x-\epsilon_{A}\left(  x\right)  \cdot1_{A}=0$. Thus,
$x=\underbrace{\epsilon_{A}\left(  x\right)  }_{\in\kk}\cdot1_{A}%
\in\kk\cdot1_{A}$.

Now, forget that we fixed $x$. We thus have proven that $x\in\kk%
\cdot1_{A}$ for each $x\in\ker\widetilde{\id}$. In other words,
$\ker\widetilde{\id}\subset\kk\cdot1_{A}$. Combining
this with $\kk\cdot1_{A}\subset\ker\widetilde{\id}$, we
obtain $\ker\widetilde{\id}=\kk\cdot1_{A}$. This proves
Lemma~\ref{lem.sol.leray.leray-e.2}(a).

(b) The axioms of a $\kk$-bialgebra yield $\epsilon_{A}\circ
u_{A}=\id_{\kk}$ (since $A$ is a $\kk%
$-bialgebra). Now,%
\begin{align*}
\epsilon_{A}\circ\underbrace{\widetilde{\id}}%
_{=\id_{A}-u_{A}\epsilon_{A}}  &  =\epsilon_{A}%
\circ\left(  \id_{A}-u_{A}\epsilon_{A}\right)
=\underbrace{\epsilon_{A}\circ\id_{A}}_{=\epsilon_{A}%
}-\underbrace{\epsilon_{A}\circ\left(  u_{A}\epsilon_{A}\right)  }%
_{=\epsilon_{A}\circ u_{A}\circ\epsilon_{A}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }\kk%
\text{-linear maps is }\kk\text{-bilinear}\right) \\
&  =\epsilon_{A}-\underbrace{\epsilon_{A}\circ u_{A}}_{=\operatorname*{id}%
\nolimits_{\kk}}\circ\epsilon_{A}=\epsilon_{A}-\epsilon_{A}=0.
\end{align*}
Hence, $\epsilon_{A}\left(  \widetilde{\id}\left(  A\right)
\right)  =\underbrace{\left(  \epsilon_{A}\circ\widetilde{\operatorname*{id}%
}\right)  }_{=0}\left(  A\right)  =0\left(  A\right)  =0$. Therefore,
$\widetilde{\id}\left(  A\right)  \subset\ker\left(
\underbrace{\epsilon_{A}}_{=\epsilon}\right)  =\ker\epsilon$. This proves
Lemma~\ref{lem.sol.leray.leray-e.2}(b).
\end{proof}

\begin{lemma}
\label{lem.sol.leray.leray-e.3}Let $A$ be a $\kk$-bialgebra. Then,
$A/\left(  \kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)
\cong\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ as
$\kk$-modules.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.leray.leray-e.3}.]The axioms of a $\kk%
$-bialgebra show that the counit $\epsilon$ of $A$ is a $\kk$-algebra
homomorphism (since $A$ is a $\kk$-bialgebra). Thus, $\ker\epsilon$ is
an ideal of $A$. Now, $\left(  \ker\epsilon\right)  ^{2}=\underbrace{\left(
\ker\epsilon\right)  }_{\subset A}\left(  \ker\epsilon\right)  \subset
A\left(  \ker\epsilon\right)  \subset\ker\epsilon$ (since $\ker\epsilon$ is an
ideal of $A$). Hence, the quotient $\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}$ makes sense.

Let $\widetilde{\id}$ be the map $\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}:A\rightarrow A$. Then,
Lemma~\ref{lem.sol.leray.leray-e.2}(a) shows that $\ker
\widetilde{\id}=\kk\cdot1_{A}$.

Lemma~\ref{lem.sol.leray.leray-e.2}(b) shows that
$\widetilde{\id}\left(  A\right)  \subset\ker\epsilon$. Thus,
each $a\in A$ satisfies $\widetilde{\id}\left(  \underbrace{a}%
_{\in A}\right)  \in\widetilde{\id}\left(  A\right)
\subset\ker\epsilon$. Hence, we can define a map $\pi:A\rightarrow\ker
\epsilon$ by%
\[
\left(  \pi\left(  a\right)  =\widetilde{\id}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }a\in A\right)  .
\]
Consider this map $\pi$. This map $\pi$ differs from
$\widetilde{\id}$ only in its target (namely, its target is
$\ker\epsilon$, whereas the target of $\widetilde{\id}$ is
$A$); therefore, this map $\pi$ is $\kk$-linear (since the map
$\widetilde{\id}$ is $\kk$-linear).

Each $a\in\ker\epsilon$ satisfies%
\begin{align*}
\pi\left(  a\right)   &  =\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  a\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
a\right)  =\underbrace{\id_{A}\left(  a\right)  }%
_{=a}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  a\right)
}_{=u_{A}\left(  \epsilon_{A}\left(  a\right)  \right)  }=a-u_{A}\left(
\underbrace{\epsilon_{A}}_{=\epsilon}\left(  a\right)  \right) \\
&  =a-u_{A}\left(  \underbrace{\epsilon\left(  a\right)  }%
_{\substack{=0\\\text{(since }a\in\ker\epsilon\text{)}}}\right)
=a-\underbrace{u_{A}\left(  0\right)  }_{\substack{=0\\\text{(since the map
}u_{A}\text{ is }\kk\text{-linear)}}}=a.
\end{align*}
Hence, each $a\in\ker\epsilon$ satisfies $a=\pi\left(  \underbrace{a}_{\in
A}\right)  \in\pi\left(  A\right)  $. In other words, we have $\ker
\epsilon\subset\pi\left(  A\right)  $. Hence, the map $\pi$ is surjective.

Let $\gamma$ be the canonical projection $\ker\epsilon\rightarrow\left(
\ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$. Thus, the map
$\gamma$ is surjective and satisfies $\ker\gamma=\left(  \ker\epsilon\right)
^{2}$.

Both maps $\gamma$ and $\pi$ are $\kk$-linear. Thus, their composition
$\gamma\circ\pi$ is $\kk$-linear. Hence, its kernel $\ker\left(
\gamma\circ\pi\right)  $ is a $\kk$-submodule of $A$.

The map $\gamma\circ\pi:A\rightarrow\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}$ is the composition of two surjective maps (since
the two maps $\gamma$ and $\pi$ are surjective), and thus is itself
surjective. In other words, $\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}=\left(  \gamma\circ\pi\right)  \left(  A\right)  $.

It is known that if $V$ and $W$ are two $\kk$-modules, and if
$\delta:V\rightarrow W$ is a $\kk$-linear map, then $\delta\left(
V\right)  \cong V/\ker\delta$ as $\kk$-modules. Applying this to $V=A$,
$W=\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ and
$\delta=\gamma\circ\pi$, we conclude that $\left(  \gamma\circ\pi\right)
\left(  A\right)  \cong A/\ker\left(  \gamma\circ\pi\right)  $ as $\kk%
$-modules. Thus,%
\begin{equation}
\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}=\left(
\gamma\circ\pi\right)  \left(  A\right)  \cong A/\ker\left(  \gamma\circ
\pi\right)  \label{pf.lem.sol.leray.leray-e.3.1}%
\end{equation}
as $\kk$-modules.

Now, let us show that $\ker\left(  \gamma\circ\pi\right)  =\kk%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$.

We first observe that $\kk\cdot1_{A}\subset\ker\left(  \gamma\circ
\pi\right)  $\ \ \ \ \footnote{\textit{Proof.} We have $1_{A}=\underbrace{1}%
_{\in\kk}\cdot1_{A}\in\kk\cdot1_{A}=\ker
\widetilde{\id}$ (since $\ker\widetilde{\operatorname*{id}%
}=\kk\cdot1_{A}$) and thus $\widetilde{\id}\left(
1_{A}\right)  =0$. The definition of $\pi$ yields $\pi\left(  1_{A}\right)
=\widetilde{\id}\left(  1_{A}\right)  =0$. Now, $\left(
\gamma\circ\pi\right)  \left(  1_{A}\right)  =\gamma\left(  \underbrace{\pi
\left(  1_{A}\right)  }_{=0}\right)  =\gamma\left(  0\right)  =0$ (since the
map $\gamma$ is $\kk$-linear). Since the map $\gamma\circ\pi$ is
$\kk$-linear, we have $\left(  \gamma\circ\pi\right)  \left(
\kk\cdot1_{A}\right)  =\kk\cdot\underbrace{\left(  \gamma
\circ\pi\right)  \left(  1_{A}\right)  }_{=0}=\kk\cdot0=0$. In other
words, $\kk\cdot1_{A}\subset\ker\left(  \gamma\circ\pi\right)  $.} and
$\left(  \ker\epsilon\right)  ^{2}\subset\ker\left(  \gamma\circ\pi\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left(  \ker\epsilon\right)  ^{2}%
$. Thus, $x\in\left(  \ker\epsilon\right)  ^{2}\subset\ker\epsilon$, so that
$\epsilon\left(  x\right)  =0$.
\par
The definition of $\pi$ yields
\begin{align*}
\pi\left(  x\right)   &  =\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  x\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
x\right)  =\underbrace{\id_{A}\left(  x\right)  }%
_{=x}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\underbrace{\epsilon_{A}}_{=\epsilon}\left(  x\right)
\cdot1_{A}\\
&  =x-\underbrace{\epsilon\left(  x\right)  }_{=0}\cdot1_{A}%
=x-\underbrace{0\cdot1_{A}}_{=0}=x\in\left(  \ker\epsilon\right)  ^{2}%
=\ker\gamma
\end{align*}
(since $\ker\gamma=\left(  \ker\epsilon\right)  ^{2}$). Therefore,
$\gamma\left(  \pi\left(  x\right)  \right)  =0$. Now, $\left(  \gamma\circ
\pi\right)  \left(  x\right)  =\gamma\left(  \pi\left(  x\right)  \right)
=0$, so that $x\in\ker\left(  \gamma\circ\pi\right)  $.
\par
Now, forget that we fixed $x$. We thus have shown that $x\in\ker\left(
\gamma\circ\pi\right)  $ for each $x\in\left(  \ker\epsilon\right)  ^{2}$. In
other words, $\left(  \ker\epsilon\right)  ^{2}\subset\ker\left(  \gamma
\circ\pi\right)  $.}. Hence,%
\begin{equation}
\underbrace{\kk\cdot1_{A}}_{\subset\ker\left(  \gamma\circ\pi\right)
}+\underbrace{\left(  \ker\epsilon\right)  ^{2}}_{\subset\ker\left(
\gamma\circ\pi\right)  }\subset\ker\left(  \gamma\circ\pi\right)  +\ker\left(
\gamma\circ\pi\right)  \subset\ker\left(  \gamma\circ\pi\right)
\label{pf.lem.sol.leray.leray-e.3.5}%
\end{equation}
(since $\ker\left(  \gamma\circ\pi\right)  $ is a $\kk$-submodule of
$A$).

On the other hand, let $z\in\ker\left(  \gamma\circ\pi\right)  $ be arbitrary.
We shall show that $z\in\kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}$.

We have $\gamma\left(  \pi\left(  z\right)  \right)  =\left(  \gamma\circ
\pi\right)  \left(  z\right)  =0$ (since $z\in\ker\left(  \gamma\circ
\pi\right)  $), so that $\pi\left(  z\right)  \in\ker\gamma=\left(
\ker\epsilon\right)  ^{2}$. But the definition of $\pi$ yields%
\[
\pi\left(  z\right)  =\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  z\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
z\right)  =\underbrace{\id_{A}\left(  z\right)  }%
_{=z}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  z\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  z\right)  \right)
\\=\epsilon_{A}\left(  z\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=z-\epsilon_{A}\left(  z\right)  \cdot1_{A}.
\]
Hence,%
\[
z-\epsilon_{A}\left(  z\right)  \cdot1_{A}=\pi\left(  z\right)  \in\left(
\ker\epsilon\right)  ^{2}.
\]
Therefore,%
\[
z\in\underbrace{\epsilon_{A}\left(  z\right)  }_{\in\kk}\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset\kk\cdot1_{A}+\left(
\ker\epsilon\right)  ^{2}.
\]


Now, forget that we fixed $z$. We thus have shown that $z\in\kk%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$ for each $z\in\ker\left(
\gamma\circ\pi\right)  $. In other words, $\ker\left(  \gamma\circ\pi\right)
\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Combining this
with (\ref{pf.lem.sol.leray.leray-e.3.5}), we obtain $\ker\left(  \gamma
\circ\pi\right)  =\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$.
Thus, (\ref{pf.lem.sol.leray.leray-e.3.1}) becomes%
\[
\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}\cong
A/\underbrace{\ker\left(  \gamma\circ\pi\right)  }_{=\kk\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}}=A/\left(  \kk\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)
\]
as $\kk$-modules. This proves Lemma~\ref{lem.sol.leray.leray-e.3}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{thm.leray.leray-e}.]%
Proposition~\ref{prop.convolution-series}(j) (applied to $C=A$ and
$F=\id_{A}$) yields that $\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}\in\mathfrak{n}\left(  A,A\right)  $ (since
$\id_{A}:A\rightarrow A$ is a $\kk$-algebra
homomorphism). Thus, the map $\log^{\star}\left(  \operatorname*{id}%
\nolimits_{A}\right)  \in\mathfrak{n}\left(  A,A\right)  $ is well-defined.
This proves Theorem~\ref{thm.leray.leray-e}(a).

Define an element $\widetilde{\id}$ of $\mathfrak{n}\left(
A,A\right)  $ by $\widetilde{\id}=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}$. (This is well-defined, since
$\id_{A}-u_{A}\epsilon_{A}\in\mathfrak{n}\left(
A,A\right)  $.)

(b) The axioms of a $\kk$-bialgebra show that $\epsilon:A\rightarrow
\kk$ is a $\kk$-algebra homomorphism. Thus, $\ker\epsilon$ is an
ideal of $A$. Now, it is easy to see that if $n\in \NN$ and
$m\in \NN$ satisfy $n\geq m>0$, then\footnote{Here, we set
$V^{0}=\kk\cdot1_{A}$ for any $\kk$-submodule $V$ of $A$.}%
\begin{equation}
\widetilde{\id}^{\star n}\left(  A\right)  \subset\left(
\ker\epsilon\right)  ^{m}. \label{pf.thm.leray.leray-e.b.2}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.leray.leray-e.b.2}):} Let $n\in \NN$ and
$m\in \NN$ be such that $n\geq m>0$. Then, $\left(  \ker\epsilon\right)
^{m}$ is an ideal of $A$ (since $\ker\epsilon$ is an ideal of $A$, and since
$m>0$). But Lemma~\ref{lem.sol.leray.leray-e.1} (applied to $C=A$ and
$f=\widetilde{\id}$) yields
\begin{align*}
\widetilde{\id}^{\star n}\left(  A\right)   &  \subset\left(
\underbrace{\widetilde{\id}\left(  A\right)  }%
_{\substack{\subset\ker\epsilon\\\text{(by Lemma~\ref{lem.sol.leray.leray-e.2}%
(b))}}}\right)  ^{n}\subset\left(  \ker\epsilon\right)  ^{n}%
=\underbrace{\left(  \ker\epsilon\right)  ^{n-m}}_{\subset A}\left(
\ker\epsilon\right)  ^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\geq
m\right) \\
&  \subset A\left(  \ker\epsilon\right)  ^{m}\subset\left(  \ker
\epsilon\right)  ^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\ker\epsilon\right)  ^{m}\text{ is an ideal of }A\right)  .
\end{align*}
This proves (\ref{pf.thm.leray.leray-e.b.2}).]

Proposition~\ref{prop.exp-log}(f) (applied to $C=A$ and
$f=\widetilde{\id}$) yields%
\[
\log^{\star}\left(  \widetilde{\id}+u_{A}\epsilon_{A}\right)
=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}.
\]
Since $\widetilde{\id}+u_{A}\epsilon_{A}=\operatorname*{id}%
\nolimits_{A}$ (because $\widetilde{\id}=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}$), this rewrites as%
\[
\log^{\star}\left(  \id_{A}\right)  =\sum_{n\geq
1}\dfrac{\left(  -1\right)  ^{n-1}}{n}\widetilde{\id}^{\star
n}.
\]
Thus,%
\begin{align}
\mathfrak{e}  &  =\log^{\star}\left(  \id_{A}\right)
=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\label{pf.thm.leray.leray-e.b.5a}\\
&  =\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=1}%
\underbrace{\widetilde{\id}^{\star1}}%
_{=\widetilde{\id}}+\sum_{n\geq2}\dfrac{\left(  -1\right)
^{n-1}}{n}\widetilde{\id}^{\star n}%
=\widetilde{\id}+\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}%
}{n}\widetilde{\id}^{\star n}.
\label{pf.thm.leray.leray-e.b.5b}%
\end{align}


Furthermore,%
\begin{equation}
\mathfrak{e}\left(  1_{A}\right)  =0. \label{pf.thm.leray.leray-e.b.6}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.leray.leray-e.b.6}):}
Lemma~\ref{lem.pf.prop.leray.exp-algh1.2} (applied to $C=A$ and
$f=\widetilde{\id}$) shows that every $n\in \NN$
satisfies%
\begin{equation}
\widetilde{\id}^{\star n}\left(  1_{A}\right)  =\left(
\widetilde{\id}\left(  1_{A}\right)  \right)  ^{n}.
\label{pf.thm.leray.leray-e.b.6.pf.1}%
\end{equation}


But Lemma~\ref{lem.sol.leray.leray-e.2}(a) yields $\ker
\widetilde{\id}=\kk\cdot1_{A}$. Hence, $1_{A}%
=\underbrace{1}_{\in\kk}\cdot1_{A}\in\kk\cdot1_{A}%
=\ker\widetilde{\id}$, so that $\widetilde{\operatorname*{id}%
}\left(  1_{A}\right)  =0$. Now, if $n$ is any positive integer, then
\begin{align}
\widetilde{\id}^{\star n}\left(  1_{A}\right)   &  =\left(
\underbrace{\widetilde{\id}\left(  1_{A}\right)  }_{=0}\right)
^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.leray.leray-e.b.6.pf.1}%
)}\right) \nonumber\\
&  =0^{n}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\text{ is a positive
integer}\right)  . \label{pf.thm.leray.leray-e.b.6.pf.2}%
\end{align}


Applying both sides of the equality (\ref{pf.thm.leray.leray-e.b.5a}) to
$1_{A}$, we obtain%
\[
\mathfrak{e}\left(  1_{A}\right)  =\left(  \sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\widetilde{\id}^{\star n}\right)  \left(
1_{A}\right)  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\underbrace{\widetilde{\id}^{\star n}\left(  1_{A}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.6.pf.2}))}}%
}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}0=0.
\]
This proves (\ref{pf.thm.leray.leray-e.b.6}).]

The map $\mathfrak{e}$ is $\kk$-linear (since $\mathfrak{e}=\log
^{\star}\left(  \id_{A}\right)  \in\mathfrak{n}\left(
A,A\right)  \subset\Hom \left(  A,A\right)  $). Thus,%
\[
\mathfrak{e}\left(  \kk\cdot1_{A}\right)  =\kk\cdot
\underbrace{\mathfrak{e}\left(  1_{A}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.thm.leray.leray-e.b.6}))}}}=\kk\cdot0=0.
\]
But Exercise~\ref{graded-connected-exercise}(c) shows that $A_{0}%
=\kk\cdot1_{A}$. Applying the map $\mathfrak{e}$ to both sides of this
equality, we find%
\begin{equation}
\mathfrak{e}\left(  A_{0}\right)  =\mathfrak{e}\left(  \kk\cdot
1_{A}\right)  =0. \label{pf.thm.leray.leray-e.b.7}%
\end{equation}


Furthermore, recall that $\id_{A}-u_{A}\epsilon_{A}%
\in\mathfrak{n}\left(  A,A\right)  $. Hence, Proposition~\ref{prop.exp-log}(b)
(applied to $g=\id_{A}$) yields $\exp^{\star}\left(
\log^{\star}\left(  \id_{A}\right)  \right)
=\id_{A}$. Thus,%
\begin{equation}
\exp^{\star}\underbrace{\mathfrak{e}}_{=\log^{\star}\left(  \operatorname*{id}%
\nolimits_{A}\right)  }=\exp^{\star}\left(  \log^{\star}\left(
\id_{A}\right)  \right)  =\operatorname*{id}%
\nolimits_{A}. \label{pf.thm.leray.leray-e.b.8}%
\end{equation}
Thus, $\exp^{\star}\mathfrak{e}:A\rightarrow A$ is a $\kk$-algebra
homomorphism (since $\id_{A}:A\rightarrow A$ is a
$\kk$-algebra homomorphism). Hence,
Proposition~\ref{prop.leray.exp-algh2} (applied to $C=A$ and $f=\mathfrak{e}$)
shows that%
\begin{equation}
\mathfrak{e}\left(  \left(  \ker\epsilon\right)  ^{2}\right)  =0.
\label{pf.thm.leray.leray-e.b.9}%
\end{equation}


The map $\mathfrak{e}$ is $\kk$-linear. Thus,%
\[
\mathfrak{e}\left(  \kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}\right)  =\kk\cdot\underbrace{\mathfrak{e}\left(  1_{A}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.6}))}}%
}+\underbrace{\mathfrak{e}\left(  \left(  \ker\epsilon\right)  ^{2}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.9}))}}}=\kk%
\cdot0+0=0.
\]
In other words,%
\begin{equation}
\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset\ker\mathfrak{e}.
\label{pf.thm.leray.leray-e.b.side1}%
\end{equation}


On the other hand, let us prove the reverse inclusion. Let us first observe
that each $x\in A$ satisfies%
\begin{equation}
\widetilde{\id}\left(  x\right)  -\mathfrak{e}\left(  x\right)
\in\left(  \ker\epsilon\right)  ^{2}. \label{pf.thm.leray.leray-e.b.idp1}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.leray.leray-e.b.idp1}):} Let $x\in A$. Applying
both sides of the equality (\ref{pf.thm.leray.leray-e.b.5b}) to $x$, we find%
\[
\mathfrak{e}\left(  x\right)  =\left(  \widetilde{\id}%
+\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\right)  \left(  x\right)
=\widetilde{\id}\left(  x\right)  +\sum_{n\geq2}\dfrac{\left(
-1\right)  ^{n-1}}{n}\widetilde{\id}^{\star n}\left(  x\right)
.
\]
Subtracting $\widetilde{\id}\left(  x\right)  $ from both sides
of this equality, we find%
\begin{align*}
\mathfrak{e}\left(  x\right)  -\widetilde{\id}\left(  x\right)
&  =\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\left(  \underbrace{x}_{\in A}\right)
\in\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\underbrace{\widetilde{\id}^{\star n}\left(  A\right)
}_{\substack{\subset\left(  \ker\epsilon\right)  ^{2}\\\text{(by
(\ref{pf.thm.leray.leray-e.b.2}) (applied to }m=2\text{)}\\\text{(since }%
n\geq2>0\text{))}}}\\
&  \subset\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}\left(
\ker\epsilon\right)  ^{2}\subset\left(  \ker\epsilon\right)  ^{2}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \ker\epsilon\right)
^{2}\text{ is a }\kk\text{-submodule of }A\right)  .
\end{align*}
Hence,%
\[
\widetilde{\id}\left(  x\right)  -\mathfrak{e}\left(  x\right)
=-\underbrace{\left(  \mathfrak{e}\left(  x\right)
-\widetilde{\id}\left(  x\right)  \right)  }_{\in\left(
\ker\epsilon\right)  ^{2}}\in-\left(  \ker\epsilon\right)  ^{2}\subset\left(
\ker\epsilon\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\ker\epsilon\right)  ^{2}\text{ is a }\kk\text{-submodule of }A\right)
.
\]
This proves (\ref{pf.thm.leray.leray-e.b.idp1}).]

Now, let $x\in\ker\mathfrak{e}$. Thus, $\mathfrak{e}\left(  x\right)  =0$.
Now, (\ref{pf.thm.leray.leray-e.b.idp1}) yields $\widetilde{\operatorname*{id}%
}\left(  x\right)  -\mathfrak{e}\left(  x\right)  \in\left(  \ker
\epsilon\right)  ^{2}$.

But%
\[
\widetilde{\id}\left(  x\right)  -\underbrace{\mathfrak{e}%
\left(  x\right)  }_{=0}=\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  x\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
x\right)  =\underbrace{\id_{A}\left(  x\right)  }%
_{=x}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\epsilon_{A}\left(  x\right)  \cdot1_{A}.
\]
Thus,%
\[
x-\epsilon_{A}\left(  x\right)  \cdot1_{A}=\widetilde{\operatorname*{id}%
}\left(  x\right)  -\mathfrak{e}\left(  x\right)  \in\left(  \ker
\epsilon\right)  ^{2}
\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.leray.leray-e.b.idp1})}\right) .
\]
Hence,%
\[
x\in\underbrace{\epsilon_{A}\left(  x\right)  }_{\in\kk}\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset\kk\cdot1_{A}+\left(
\ker\epsilon\right)  ^{2}.
\]


Now, forget that we fixed $x$. We thus have shown that every $x\in
\ker\mathfrak{e}$ satisfies $x\in\kk\cdot1_{A}+\left(  \ker
\epsilon\right)  ^{2}$. In other words, we have $\ker\mathfrak{e}%
\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Combining this
with (\ref{pf.thm.leray.leray-e.b.side1}), we obtain%
\begin{equation}
\ker\mathfrak{e}=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}.
\label{pf.thm.leray.leray-e.b.part1}%
\end{equation}
Thus, one part of Theorem~\ref{thm.leray.leray-e}(b) is proven. It remains to
show that $\mathfrak{e}\left(  A\right)  \cong\left(  \ker\epsilon\right)
/\left(  \ker\epsilon\right)  ^{2}$ (as $\kk$-modules).

It is known that if $V$ and $W$ are two $\kk$-modules, and if
$\delta:V\rightarrow W$ is a $\kk$-linear map, then $\delta\left(
V\right)  \cong V/\ker\delta$ as $\kk$-modules. Applying this to $V=A$,
$W=A$ and $\delta=\mathfrak{e}$, we obtain%
\[
\mathfrak{e}\left(  A\right)  \cong A/\underbrace{\ker\mathfrak{e}%
}_{\substack{=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}%
\\\text{(by (\ref{pf.thm.leray.leray-e.b.part1}))}}}=A/\left(  \kk%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)  \cong\left(  \ker
\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma~\ref{lem.sol.leray.leray-e.3}}\right)
\]
as $\kk$-modules. This completes the proof of
Theorem~\ref{thm.leray.leray-e}(b).

(c) The definition of $\mathfrak{q}$ shows that%
\begin{equation}
\mathfrak{q}\left(  x\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for each
}x\in A. \label{pf.thm.leray.leray-e.c.qx=}%
\end{equation}
Thus, each $x\in A_{0}$ satisfies%
\[
\mathfrak{q}\left(  x\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  \underbrace{x}_{\in A_{0}}\right)  \right)  \in
\iota_{\mathfrak{e}\left(  A\right)  }\left(  \underbrace{\mathfrak{e}\left(
A_{0}\right)  }_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.7}))}%
}}\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(  0\right)  =0.
\]
In other words, we have $\mathfrak{q}\left(  A_{0}\right)  =0$. Hence,
Proposition~\ref{prop.convolution-series}(h) (applied to $A$,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ and
$\mathfrak{q}$ instead of $C$, $A$ and $f$) shows that $\mathfrak{q}%
\in\mathfrak{n}\left(  A,\Sym \left(  \mathfrak{e}\left(
A\right)  \right)  \right)  $. This proves Theorem~\ref{thm.leray.leray-e}(c).

(d) Theorem~\ref{thm.leray.leray-e}(c) yields $\mathfrak{q}\in\mathfrak{n}%
\left(  A,\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
\right)  $. Hence, $\exp^{\star}\mathfrak{q}\in\Hom \left(
A,\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  \right)  $
is well-defined.

Recall that the $\kk$-algebra $\Sym V$ is commutative
whenever $V$ is a $\kk$-module. Applying this to $V=\mathfrak{e}\left(
A\right)  $, we conclude that the $\kk$-algebra $\Sym %
\left(  \mathfrak{e}\left(  A\right)  \right)  $ is commutative.

We have $\mathfrak{q}\left(  A\right)  =\Sym^{1}%
\left(  \mathfrak{e}\left(  A\right)  \right)  $%
\ \ \ \ \footnote{\textit{Proof.} Every $\kk$-module $V$ satisfies
$\iota_{V}\left(  V\right)  =\Sym^{1}V$. Applying this
to $V=\mathfrak{e}\left(  A\right)  $, we obtain $\iota_{\mathfrak{e}\left(
A\right)  }\left(  \mathfrak{e}\left(  A\right)  \right)
= \Sym^{1}\left(  \mathfrak{e}\left(  A\right)  \right)  $.
\par
Now,%
\[
\mathfrak{q}\left(  A\right)  =\left\{  \underbrace{\mathfrak{q}\left(
x\right)  }_{\substack{=\iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  x\right)  \right)  \\\text{(by
(\ref{pf.thm.leray.leray-e.c.qx=}))}}}\ \mid\ x\in A\right\}  =\left\{
\iota_{\mathfrak{e}\left(  A\right)  }\left(  \mathfrak{e}\left(  x\right)
\right)  \ \mid\ x\in A\right\}  =\iota_{\mathfrak{e}\left(  A\right)
}\left(  \underbrace{\left\{  \mathfrak{e}\left(  x\right)  \ \mid\ x\in
A\right\}  }_{=\mathfrak{e}\left(  A\right)  }\right)  =\iota_{\mathfrak{e}%
\left(  A\right)  }\left(  \mathfrak{e}\left(  A\right)  \right)
=\Sym^{1}\left(  \mathfrak{e}\left(  A\right)
\right)  .
\]
}. Thus, $\mathfrak{q}\left(  A\right)  $ generates the $\kk$-algebra
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
$\ \ \ \ \footnote{\textit{Proof.} It is known that if $V$ is a $\kk%
$-module, then $\Sym^{1}V$ generates the $\kk%
$-algebra $\Sym V$. Applying this to $V=\mathfrak{e}\left(
A\right)  $, we conclude that $\Sym^{1}\left(
\mathfrak{e}\left(  A\right)  \right)  $ generates the $\kk$-algebra
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $. In other
words, $\mathfrak{q}\left(  A\right)  $ generates the $\kk$-algebra
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ (since
$\mathfrak{q}\left(  A\right)  =\Sym^{1}\left(
\mathfrak{e}\left(  A\right)  \right)  $).}.

Furthermore, Theorem~\ref{thm.leray.leray-e}(b) yields that $\ker
\mathfrak{e}=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Hence,
$1=1_{A}=\underbrace{1}_{\in\kk}\cdot1_{A}\in\kk\cdot
1_{A}\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}%
=\ker\mathfrak{e}$ and $\left(  \ker\epsilon\right)  ^{2}\subset
\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}=\ker\mathfrak{e}$. Now,
(\ref{pf.thm.leray.leray-e.c.qx=}) (applied to $x=1$) yields%
\[
\mathfrak{q}\left(  1\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(
\underbrace{\mathfrak{e}\left(  1\right)  }_{\substack{=0\\\text{(since }%
1\in\ker\mathfrak{e}\text{)}}}\right)  =\iota_{\mathfrak{e}\left(  A\right)
}\left(  0\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\iota_{\mathfrak{e}\left(  A\right)  }\text{ is }\kk\text{-linear}%
\right)  .
\]
Also, each $x\in\left(  \ker\epsilon\right)  ^{2}$ satisfies%
\begin{align*}
\mathfrak{q}\left(  x\right)   &  =\iota_{\mathfrak{e}\left(  A\right)
}\left(  \underbrace{\mathfrak{e}\left(  x\right)  }%
_{\substack{=0\\\text{(since }x\in\left(  \ker\epsilon\right)  ^{2}\subset
\ker\mathfrak{e}\text{)}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.leray.leray-e.c.qx=})}\right) \\
&  =\iota_{\mathfrak{e}\left(  A\right)  }\left(  0\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\iota_{\mathfrak{e}\left(
A\right)  }\text{ is }\kk\text{-linear}\right)  .
\end{align*}
In other words, $\mathfrak{q}\left(  \left(  \ker\epsilon\right)  ^{2}\right)
\subset0$. Hence, $\mathfrak{q}\left(  \left(  \ker\epsilon\right)
^{2}\right)  =0$.

Thus, Proposition~\ref{prop.leray.exp-algh3} (applied to $A$,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ and
$\mathfrak{q}$ instead of $C$, $A$ and $f$) shows that $\exp^{\star
}\mathfrak{q}:A\rightarrow\Sym \left(  \mathfrak{e}\left(
A\right)  \right)  $ is a surjective $\kk$-algebra homomorphism.

But let us recall that $\mathfrak{s}$ is the unique $\kk$-algebra
homomorphism $\Phi:\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ satisfying $\mathbf{i}=\Phi\circ\iota_{\mathfrak{e}%
\left(  A\right)  }$. Hence, $\mathfrak{s}$ is a $\kk$-algebra
homomorphism $\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ and satisfies $\mathbf{i}=\mathfrak{s}\circ
\iota_{\mathfrak{e}\left(  A\right)  }$.

Thus, Proposition~\ref{prop.convolution-series}(i) (applied to $A$,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $, $A$,
$\mathfrak{s}$, $\exp$ and $\mathfrak{q}$ instead of $C$, $A$, $B$, $s$, $u$
and $f$) shows that%
\[
\mathfrak{s}\circ\mathfrak{q}\in\mathfrak{n}\left(  A,A\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(
\mathfrak{s}\circ\mathfrak{q}\right)  =\mathfrak{s}\circ\left(  \exp^{\star
}\mathfrak{q}\right)  .
\]


Furthermore,
\begin{equation}
\mathfrak{s}\circ\mathfrak{q}=\mathfrak{e} \label{pf.thm.leray.leray-e.c.sq=e}%
\end{equation}
(since each $x\in A$ satisfies%
\begin{align*}
\left(  \mathfrak{s}\circ\mathfrak{q}\right)  \left(  x\right)   &
=\mathfrak{s}\left(  \underbrace{\mathfrak{q}\left(  x\right)  }%
_{\substack{=\iota_{\mathfrak{e}\left(  A\right)  }\left(  \mathfrak{e}\left(
x\right)  \right)  \\\text{(by (\ref{pf.thm.leray.leray-e.c.qx=}))}}}\right)
=\mathfrak{s}\left(  \iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  x\right)  \right)  \right) \\
&  =\underbrace{\left(  \mathfrak{s}\circ\iota_{\mathfrak{e}\left(  A\right)
}\right)  }_{\substack{=\mathbf{i}\\\text{(since }\mathbf{i}=\mathfrak{s}%
\circ\iota_{\mathfrak{e}\left(  A\right)  }\text{)}}}\left(  \mathfrak{e}%
\left(  x\right)  \right)  =\mathbf{i}\left(  \mathfrak{e}\left(  x\right)
\right)  =\mathfrak{e}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathbf{i}\text{ is just an inclusion map}\right)
\end{align*}
). Hence, the equality $\exp^{\star}\left(  \mathfrak{s}\circ\mathfrak{q}%
\right)  =\mathfrak{s}\circ\left(  \exp^{\star}\mathfrak{q}\right)  $ rewrites
as $\exp^{\star}\mathfrak{e}=\mathfrak{s}\circ\left(  \exp^{\star}%
\mathfrak{q}\right)  $. Comparing this with (\ref{pf.thm.leray.leray-e.b.8}),
we obtain $\mathfrak{s}\circ\left(  \exp^{\star}\mathfrak{q}\right)
=\id_{A}$. Thus, the map $\exp^{\star}\mathfrak{q}$
has a left inverse (with respect to composition), and hence is injective.

Now, the map $\exp^{\star}\mathfrak{q}$ is both injective and surjective.
Consequently, $\exp^{\star}\mathfrak{q}$ is bijective, i.e., invertible. Since
$\exp^{\star}\mathfrak{q}$ is an invertible $\kk$-algebra homomorphism,
we conclude that $\exp^{\star}\mathfrak{q}$ is a $\kk$-algebra
isomorphism. Its inverse must be $\mathfrak{s}$ (since $\mathfrak{s}%
\circ\left(  \exp^{\star}\mathfrak{q}\right)  =\operatorname*{id}%
\nolimits_{A}$). Hence, the maps $\exp^{\star}\mathfrak{q}:A\rightarrow
\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ and
$\mathfrak{s}:\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ are mutually inverse $\kk$-algebra
isomorphisms. This proves Theorem~\ref{thm.leray.leray-e}(d).

(e) Theorem~\ref{thm.leray.leray-e}(d) shows that the maps $\exp^{\star
}\mathfrak{q}:A\rightarrow\Sym \left(  \mathfrak{e}\left(
A\right)  \right)  $ and $\mathfrak{s}:\Sym \left(
\mathfrak{e}\left(  A\right)  \right)  \rightarrow A$ are mutually inverse
$\kk$-algebra isomorphisms. Hence, $A\cong\Sym \left(
\mathfrak{e}\left(  A\right)  \right)  $ as $\kk$-algebras (via these
isomorphisms). But Theorem~\ref{thm.leray.leray-e}(b) shows that
$\mathfrak{e}\left(  A\right)  \cong\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}$ (as $\kk$-modules). Hence,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
\cong\Sym \left(  \left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}\right)  $ as $\kk$-algebras. Thus,
$A\cong\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
\cong\Sym \left(  \left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}\right)  $ as $\kk$-algebras. This proves
Theorem~\ref{thm.leray.leray-e}(e).

(f) Let $x\in A$. We have%
\[
\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}}\left(  x\right)  =\left(  \operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}\right)  \left(  x\right)
=\underbrace{\id_{A}\left(  x\right)  }_{=x}%
-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\epsilon_{A}\left(  x\right)  \cdot1_{A},
\]
so that $x=\epsilon_{A}\left(  x\right)  \cdot1_{A}%
+\widetilde{\id}\left(  x\right)  $. Subtracting $\mathfrak{e}%
\left(  x\right)  $ from both sides of this equality, we obtain%
\[
x-\mathfrak{e}\left(  x\right)  =\underbrace{\epsilon_{A}\left(  x\right)
}_{\in\kk}\cdot1_{A}+\underbrace{\widetilde{\id}\left(
x\right)  -\mathfrak{e}\left(  x\right)  }_{\substack{\in\left(  \ker
\epsilon\right)  ^{2}\\\text{(by (\ref{pf.thm.leray.leray-e.b.idp1}))}}%
}\in\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}=\ker\mathfrak{e}%
\]
(by Theorem~\ref{thm.leray.leray-e}(b)). In other words, $\mathfrak{e}\left(
x-\mathfrak{e}\left(  x\right)  \right)  =0$. Comparing this with%
\begin{align*}
\mathfrak{e}\left(  x-\mathfrak{e}\left(  x\right)  \right)   &
=\mathfrak{e}\left(  x\right)  -\underbrace{\mathfrak{e}\left(  \mathfrak{e}%
\left(  x\right)  \right)  }_{=\left(  \mathfrak{e}\circ\mathfrak{e}\right)
\left(  x\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\mathfrak{e}\text{ is }\kk\text{-linear}\right)  \\
&  =\mathfrak{e}\left(  x\right)  -\left(  \mathfrak{e}\circ\mathfrak{e}%
\right)  \left(  x\right)  ,
\end{align*}
we obtain $\mathfrak{e}\left(  x\right)  -\left(  \mathfrak{e}\circ
\mathfrak{e}\right)  \left(  x\right)  =0$. In other words, $\mathfrak{e}%
\left(  x\right)  =\left(  \mathfrak{e}\circ\mathfrak{e}\right)  \left(
x\right)  $.

Now, forget that we fixed $x$. We thus have shown that $\mathfrak{e}\left(
x\right)  =\left(  \mathfrak{e}\circ\mathfrak{e}\right)  \left(  x\right)  $
for each $x\in A$. In other words, $\mathfrak{e}=\mathfrak{e}\circ
\mathfrak{e}$. In other words, $\mathfrak{e}\circ\mathfrak{e}=\mathfrak{e}$.
In other words, the map $\mathfrak{e}:A\rightarrow A$ is a projection. This
proves Theorem~\ref{thm.leray.leray-e}(f).
\end{proof}


%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.leray.leray-e}}

\textit{Solution to Exercise~\ref{exe.leray.leray-e}.} We begin by proving
some simple lemmas:

\begin{lemma}
\label{lem.sol.leray.leray-e.1}Let $C$ be a $\kk$-coalgebra. Let $A$ be
a $\kk$-algebra. Let $f\in\Hom \left(  C,A\right)  $.
Then, every $n\in \NN$ satisfies%
\begin{equation}
f^{\star n}\left(  C\right)  \subset\left(  f\left(  C\right)  \right)  ^{n}.
\label{eq.lem.sol.leray.leray-e.1.claim}%
\end{equation}
(Here, we set $V^{0}=\kk\cdot1_{A}$ for any $\kk$-submodule $V$
of $A$.)
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.leray.leray-e.1}.]We must prove the equality
(\ref{eq.lem.sol.leray.leray-e.1.claim}). We shall prove it by induction over
$n$:

\textit{Induction base:} We have $f^{\star0}=\left(  \text{the unity of the
}\kk\text{-algebra }\left(  \Hom \left(  C,A\right)
,\star\right)  \right)  =u_{A}\epsilon_{C}$. Thus, every $x\in C$ satisfies%
\begin{align*}
\underbrace{f^{\star0}}_{=u_{A}\epsilon_{C}}\left(  x\right)   &  =\left(
u_{A}\epsilon_{C}\right)  \left(  x\right)  =u_{A}\left(  \epsilon_{C}\left(
x\right)  \right)  =\underbrace{\epsilon_{C}\left(  x\right)  }_{\in
\kk}\cdot1_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}u_{A}\right) \\
&  \in\kk\cdot1_{A}=\left(  f\left(  C\right)  \right)  ^{0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  f\left(  C\right)  \right)
^{0}=\kk\cdot1_{A}\right)  .
\end{align*}
In other words, we have $f^{\star0}\left(  C\right)  \subset\left(  f\left(
C\right)  \right)  ^{0}$. In other words,
(\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=0$.

\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N$. We must prove that
(\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N+1$.

If $X$ and $Y$ are two $\kk$-submodules of $A$, then the multiplication
$m_{A}:A\otimes A\rightarrow A$ of the $\kk$-algebra $A$ satisfies%
\begin{equation}
m_{A}\left(  X\otimes Y\right)  =XY. \label{pf.lem.sol.leray.leray-e.1.XY}%
\end{equation}
(This follows easily from the definitions of $m_{A}$ and of $XY$.)

We have assumed that (\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N$.
In other words, we have $f^{\star N}\left(  C\right)  \subset\left(  f\left(
C\right)  \right)  ^{N}$. But%
\[
f^{\star\left(  N+1\right)  }=f\star f^{\star N}=m_{A}\circ\left(  f\otimes
f^{\star N}\right)  \circ\Delta_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of convolution}\right)  .
\]
Hence,%
\begin{align*}
&  \underbrace{f^{\star\left(  N+1\right)  }}_{=m_{A}\circ\left(  f\otimes
f^{\star N}\right)  \circ\Delta_{C}}\left(  C\right) \\
&  =\left(  m_{A}\circ\left(  f\otimes f^{\star N}\right)  \circ\Delta
_{C}\right)  \left(  C\right)  =m_{A}\left(  \left(  f\otimes f^{\star
N}\right)  \left(  \underbrace{\Delta_{C}\left(  C\right)  }_{\subset C\otimes
C}\right)  \right) \\
&  \subset m_{A}\left(  \underbrace{\left(  f\otimes f^{\star N}\right)
\left(  C\otimes C\right)  }_{=f\left(  C\right)  \otimes f^{\star N}\left(
C\right)  }\right)  =m_{A}\left(  f\left(  C\right)  \otimes f^{\star
N}\left(  C\right)  \right) \\
&  =f\left(  C\right)  \cdot\underbrace{f^{\star N}\left(  C\right)
}_{\subset\left(  f\left(  C\right)  \right)  ^{N}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.sol.leray.leray-e.1.XY}) (applied to }X=f\left(
C\right)  \text{ and }Y=f^{\star N}\left(  C\right)  \text{)}\right) \\
&  \subset f\left(  C\right)  \cdot\left(  f\left(  C\right)  \right)
^{N}=\left(  f\left(  C\right)  \right)  ^{N+1}.
\end{align*}
In other words, (\ref{eq.lem.sol.leray.leray-e.1.claim}) holds for $n=N+1$.
This completes the induction step. Thus, the proof of
(\ref{eq.lem.sol.leray.leray-e.1.claim}) by induction is complete. In other
words, Lemma~\ref{lem.sol.leray.leray-e.1} is proven.
\end{proof}

\begin{lemma}
\label{lem.sol.leray.leray-e.2}Let $A$ be a $\kk$-bialgebra. Let
$\widetilde{\id}$ be the $\kk$-linear map
$\id_{A}-u_{A}\epsilon_{A}:A\rightarrow A$. Then:

\begin{enumerate}
\item[(a)] We have $\ker\widetilde{\id}=\kk\cdot1_{A}$.

\item[(b)] We have $\widetilde{\id}\left(  A\right)
\subset\ker\epsilon$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.leray.leray-e.2}.]The axioms of a $\kk%
$-bialgebra yield $\epsilon_{A}\left(  1_A\right)  =1$ (since $A$ is a
$\kk$-bialgebra). Also, the definition of $u_{A}$ yields $u_{A}\left(
1\right)  =1\cdot1_{A}=1_{A}$.

We have%
\begin{align*}
\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}}\left(  1_{A}\right)   &  =\left(
\id_{A}-u_{A}\epsilon_{A}\right)  \left(
1_{A}\right)  =\underbrace{\id_{A}\left(
1_{A}\right)  }_{=1_{A}}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(
1_{A}\right)  }_{=u_{A}\left(  \epsilon_{A}\left(  1_{A}\right)  \right)
}=1_{A}-u_{A}\left(  \underbrace{\epsilon_{A}\left(  1_{A}\right)  }%
_{=1}\right) \\
&  =1_{A}-\underbrace{u_{A}\left(  1\right)  }_{=1_{A}}=1_{A}-1_{A}=0.
\end{align*}


Now, the map $\widetilde{\id}=\id_{A}%
-u_{A}\epsilon_{A}$ is $\kk$-linear (since all three maps
$\id_{A}$, $u_{A}$ and $\epsilon_{A}$ are $\kk%
$-linear). Hence,%
\[
\widetilde{\id}\left(  \kk\cdot1_{A}\right)
=\kk\cdot\underbrace{\widetilde{\id}\left(
1_{A}\right)  }_{=0}=\kk\cdot0=0.
\]
Hence, $\kk\cdot1_{A}\subset\ker\widetilde{\id}$.

On the other hand, let $x\in\ker\widetilde{\id}$. Thus,
$\widetilde{\id}\left(  x\right)  =0$. Comparing this to%
\[
\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}}\left(  x\right)  =\left(  \operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}\right)  \left(  x\right)
=\underbrace{\id_{A}\left(  x\right)  }_{=x}%
-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\epsilon_{A}\left(  x\right)  \cdot1_{A},
\]
we obtain $x-\epsilon_{A}\left(  x\right)  \cdot1_{A}=0$. Thus,
$x=\underbrace{\epsilon_{A}\left(  x\right)  }_{\in\kk}\cdot1_{A}%
\in\kk\cdot1_{A}$.

Now, forget that we fixed $x$. We thus have proven that $x\in\kk%
\cdot1_{A}$ for each $x\in\ker\widetilde{\id}$. In other words,
$\ker\widetilde{\id}\subset\kk\cdot1_{A}$. Combining
this with $\kk\cdot1_{A}\subset\ker\widetilde{\id}$, we
obtain $\ker\widetilde{\id}=\kk\cdot1_{A}$. This proves
Lemma~\ref{lem.sol.leray.leray-e.2}(a).

(b) The maps $\id_{A}$, $u_{A}$ and $\epsilon_{A}$ are
all $\kk$-linear. The axioms of a $\kk$-bialgebra yield
$\epsilon_{A}\circ u_{A}=\id_{\kk}$ (since $A$
is a $\kk$-bialgebra). Now,%
\begin{align*}
\epsilon_{A}\circ\underbrace{\widetilde{\id}}%
_{=\id_{A}-u_{A}\epsilon_{A}}  &  =\epsilon_{A}%
\circ\left(  \id_{A}-u_{A}\epsilon_{A}\right)
=\underbrace{\epsilon_{A}\circ\id_{A}}_{=\epsilon_{A}%
}-\underbrace{\epsilon_{A}\circ\left(  u_{A}\epsilon_{A}\right)  }%
_{=\epsilon_{A}\circ u_{A}\circ\epsilon_{A}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }\kk%
\text{-linear maps is }\kk\text{-bilinear}\right) \\
&  =\epsilon_{A}-\underbrace{\epsilon_{A}\circ u_{A}}_{=\operatorname*{id}%
\nolimits_{\kk}}\circ\epsilon_{A}=\epsilon_{A}-\epsilon_{A}=0.
\end{align*}
Hence,%
\[
\epsilon_{A}\left(  \widetilde{\id}\left(  A\right)  \right)
=\underbrace{\left(  \epsilon_{A}\circ\widetilde{\id}\right)
}_{=0}\left(  A\right)  =0\left(  A\right)  =0.
\]
Therefore, $\widetilde{\id}\left(  A\right)  \subset\ker\left(
\underbrace{\epsilon_{A}}_{=\epsilon}\right)  =\ker\epsilon$. This proves
Lemma~\ref{lem.sol.leray.leray-e.2}(b).
\end{proof}

\begin{lemma}
\label{lem.sol.leray.leray-e.3}Let $A$ be a $\kk$-bialgebra. Then,
$A/\left(  \kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)
\cong\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ as
$\kk$-modules.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.leray.leray-e.3}.]The axioms of a $\kk%
$-bialgebra show that the counit $\epsilon$ of $A$ is a $\kk$-algebra
homomorphism (since $A$ is a $\kk$-bialgebra). Thus, $\ker\epsilon$ is
an ideal of $A$. Now, $\left(  \ker\epsilon\right)  ^{2}=\underbrace{\left(
\ker\epsilon\right)  }_{\subset A}\left(  \ker\epsilon\right)  \subset
A\left(  \ker\epsilon\right)  \subset\ker\epsilon$ (since $\ker\epsilon$ is an
ideal of $A$). Hence, the quotient $\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}$ makes sense.

Let $\widetilde{\id}$ be the map $\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}:A\rightarrow A$. Then,
Lemma~\ref{lem.sol.leray.leray-e.2}(a) shows that $\ker
\widetilde{\id}=\kk\cdot1_{A}$.

Lemma~\ref{lem.sol.leray.leray-e.2}(b) shows that
$\widetilde{\id}\left(  A\right)  \subset\ker\epsilon$. Thus,
each $a\in A$ satisfies $\widetilde{\id}\left(  \underbrace{a}%
_{\in A}\right)  \in\widetilde{\id}\left(  A\right)
\subset\ker\epsilon$. Hence, we can define a map $\pi:A\rightarrow\ker
\epsilon$ by%
\[
\left(  \pi\left(  a\right)  =\widetilde{\id}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }a\in A\right)  .
\]
Consider this map $\pi$. This map $\pi$ differs from
$\widetilde{\id}$ only in its target (namely, its target is
$\ker\epsilon$, whereas the target of $\widetilde{\id}$ is
$A$); therefore, this map $\pi$ is $\kk$-linear (since the map
$\widetilde{\id}$ is $\kk$-linear).

Each $a\in\ker\epsilon$ satisfies%
\begin{align*}
\pi\left(  a\right)   &  =\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\pi\right) \\
&  =\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
a\right)  =\underbrace{\id_{A}\left(  a\right)  }%
_{=a}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  a\right)
}_{=u_{A}\left(  \epsilon_{A}\left(  a\right)  \right)  }=a-u_{A}\left(
\underbrace{\epsilon_{A}}_{=\epsilon}\left(  a\right)  \right) \\
&  =a-u_{A}\left(  \underbrace{\epsilon\left(  a\right)  }%
_{\substack{=0\\\text{(since }a\in\ker\epsilon\text{)}}}\right)
=a-\underbrace{u_{A}\left(  0\right)  }_{\substack{=0\\\text{(since the map
}u_{A}\text{ is }\kk\text{-linear)}}}=a.
\end{align*}
Hence, each $a\in\ker\epsilon$ satisfies $a=\pi\left(  \underbrace{a}_{\in
A}\right)  \in\pi\left(  A\right)  $. In other words, we have $\ker
\epsilon\subset\pi\left(  A\right)  $. Hence, the map $\pi$ is surjective.

Let $\gamma$ be the canonical projection $\ker\epsilon\rightarrow\left(
\ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$. Thus, the map
$\gamma$ is surjective and satisfies $\ker\gamma=\left(  \ker\epsilon\right)
^{2}$.

Both maps $\gamma$ and $\pi$ are $\kk$-linear. Thus, their composition
$\gamma\circ\pi$ is $\kk$-linear. Hence, its kernel $\ker\left(
\gamma\circ\pi\right)  $ is a $\kk$-submodule of $A$.

The map $\gamma\circ\pi:A\rightarrow\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}$ is the composition of two surjective maps (since
the two maps $\gamma$ and $\pi$ are surjective), and thus is itself
surjective. In other words, $\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}=\left(  \gamma\circ\pi\right)  \left(  A\right)  $.

It is known that if $V$ and $W$ are two $\kk$-modules, and if
$\delta:V\rightarrow W$ is a $\kk$-linear map, then $\delta\left(
V\right)  \cong V/\ker\delta$ as $\kk$-modules. Applying this to $V=A$,
$W=\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ and
$\delta=\gamma\circ\pi$, we conclude that $\left(  \gamma\circ\pi\right)
\left(  A\right)  \cong A/\ker\left(  \gamma\circ\pi\right)  $ as $\kk%
$-modules. Thus,%
\begin{equation}
\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}=\left(
\gamma\circ\pi\right)  \left(  A\right)  \cong A/\ker\left(  \gamma\circ
\pi\right)  \label{pf.lem.sol.leray.leray-e.3.1}%
\end{equation}
as $\kk$-modules.

Now, let us show that $\ker\left(  \gamma\circ\pi\right)  =\kk%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$.

We first observe that $\kk\cdot1_{A}\subset\ker\left(  \gamma\circ
\pi\right)  $\ \ \ \ \footnote{\textit{Proof.} We have $1_{A}=\underbrace{1}%
_{\in\kk}\cdot1_{A}\in\kk\cdot1_{A}=\ker
\widetilde{\id}$ (since $\ker\widetilde{\operatorname*{id}%
}=\kk\cdot1_{A}$) and thus $\widetilde{\id}\left(
1_{A}\right)  =0$. The definition of $\pi$ yields $\pi\left(  1_{A}\right)
=\widetilde{\id}\left(  1_{A}\right)  =0$. Now, $\left(
\gamma\circ\pi\right)  \left(  1_{A}\right)  =\gamma\left(  \underbrace{\pi
\left(  1_{A}\right)  }_{=0}\right)  =\gamma\left(  0\right)  =0$ (since the
map $\gamma$ is $\kk$-linear). Since the map $\gamma\circ\pi$ is
$\kk$-linear, we have%
\[
\left(  \gamma\circ\pi\right)  \left(  \kk\cdot1_{A}\right)
=\kk\cdot\underbrace{\left(  \gamma\circ\pi\right)  \left(
1_{A}\right)  }_{=0}=\kk\cdot0=0.
\]
In other words, $\kk\cdot1_{A}\subset\ker\left(  \gamma\circ\pi\right)
$.} and $\left(  \ker\epsilon\right)  ^{2}\subset\ker\left(  \gamma\circ
\pi\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left(  \ker
\epsilon\right)  ^{2}$. Thus, $x\in\left(  \ker\epsilon\right)  ^{2}%
\subset\ker\epsilon$, so that $\epsilon\left(  x\right)  =0$.
\par
The definition of $\pi$ yields
\begin{align*}
\pi\left(  x\right)   &  =\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  x\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
x\right)  =\underbrace{\id_{A}\left(  x\right)  }%
_{=x}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\underbrace{\epsilon_{A}}_{=\epsilon}\left(  x\right)
\cdot1_{A}\\
&  =x-\underbrace{\epsilon\left(  x\right)  }_{=0}\cdot1_{A}%
=x-\underbrace{0\cdot1_{A}}_{=0}=x\in\left(  \ker\epsilon\right)  ^{2}%
=\ker\gamma
\end{align*}
(since $\ker\gamma=\left(  \ker\epsilon\right)  ^{2}$). Therefore,
$\gamma\left(  \pi\left(  x\right)  \right)  =0$. Now, $\left(  \gamma\circ
\pi\right)  \left(  x\right)  =\gamma\left(  \pi\left(  x\right)  \right)
=0$, so that $x\in\ker\left(  \gamma\circ\pi\right)  $.
\par
Now, forget that we fixed $x$. We thus have shown that $x\in\ker\left(
\gamma\circ\pi\right)  $ for each $x\in\left(  \ker\epsilon\right)  ^{2}$. In
other words, $\left(  \ker\epsilon\right)  ^{2}\subset\ker\left(  \gamma
\circ\pi\right)  $.}. Hence,%
\begin{equation}
\underbrace{\kk\cdot1_{A}}_{\subset\ker\left(  \gamma\circ\pi\right)
}+\underbrace{\left(  \ker\epsilon\right)  ^{2}}_{\subset\ker\left(
\gamma\circ\pi\right)  }\subset\ker\left(  \gamma\circ\pi\right)  +\ker\left(
\gamma\circ\pi\right)  \subset\ker\left(  \gamma\circ\pi\right)
\label{pf.lem.sol.leray.leray-e.3.5}%
\end{equation}
(since $\ker\left(  \gamma\circ\pi\right)  $ is a $\kk$-submodule of
$A$).

On the other hand, let $z\in\ker\left(  \gamma\circ\pi\right)  $ be arbitrary.
We shall show that $z\in\kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}$.

We have $\gamma\left(  \pi\left(  z\right)  \right)  =\left(  \gamma\circ
\pi\right)  \left(  z\right)  =0$ (since $z\in\ker\left(  \gamma\circ
\pi\right)  $), so that $\pi\left(  z\right)  \in\ker\gamma=\left(
\ker\epsilon\right)  ^{2}$. But the definition of $\pi$ yields%
\[
\pi\left(  z\right)  =\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  z\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
z\right)  =\underbrace{\id_{A}\left(  z\right)  }%
_{=z}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  z\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  z\right)  \right)
\\=\epsilon_{A}\left(  z\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=z-\epsilon_{A}\left(  z\right)  \cdot1_{A}.
\]
Hence,%
\[
z-\epsilon_{A}\left(  z\right)  \cdot1_{A}=\pi\left(  z\right)  \in\left(
\ker\epsilon\right)  ^{2}.
\]
Therefore,%
\[
z\in\underbrace{\epsilon_{A}\left(  z\right)  }_{\in\kk}\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset\kk\cdot1_{A}+\left(
\ker\epsilon\right)  ^{2}.
\]


Now, forget that we fixed $z$. We thus have shown that $z\in\kk%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$ for each $z\in\ker\left(
\gamma\circ\pi\right)  $. In other words, $\ker\left(  \gamma\circ\pi\right)
\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Combining this
with (\ref{pf.lem.sol.leray.leray-e.3.5}), we obtain $\ker\left(  \gamma
\circ\pi\right)  =\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$.
Thus, (\ref{pf.lem.sol.leray.leray-e.3.1}) becomes%
\[
\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}\cong
A/\underbrace{\ker\left(  \gamma\circ\pi\right)  }_{=\kk\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}}=A/\left(  \kk\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)
\]
as $\kk$-modules. In other words, $A/\left(  \kk\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)  \cong\left(  \ker
\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}$ as $\kk$-modules.
This proves Lemma~\ref{lem.sol.leray.leray-e.3}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{thm.leray.leray-e}.]%
Proposition~\ref{prop.convolution-series}(j) (applied to $C=A$ and
$F=\id_{A}$) yields that $\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}\in\mathfrak{n}\left(  A,A\right)  $ (since
$\id_{A}:A\rightarrow A$ is a $\kk$-algebra
homomorphism). Thus, the map $\log^{\star}\left(  \operatorname*{id}%
\nolimits_{A}\right)  \in\mathfrak{n}\left(  A,A\right)  $ is well-defined.
This proves Theorem~\ref{thm.leray.leray-e}(a).

Define an element $\widetilde{\id}$ of $\mathfrak{n}\left(
A,A\right)  $ by $\widetilde{\id}=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}$. (This is well-defined, since
$\id_{A}-u_{A}\epsilon_{A}\in\mathfrak{n}\left(
A,A\right)  $.)

(b) The axioms of a $\kk$-bialgebra show that $\epsilon:A\rightarrow
\kk$ is a $\kk$-algebra homomorphism (since $A$ is a
$\kk$-bialgebra). Thus, $\ker\epsilon$ is an ideal of the
$\kk$-algebra $A$. Now, it is easy to see that if $n\in \NN$ and
$m\in \NN$ satisfy $n\geq m>0$, then\footnote{Here, we set
$V^{0}=\kk\cdot1_{A}$ for any $\kk$-submodule $V$ of $A$.}%
\begin{equation}
\widetilde{\id}^{\star n}\left(  A\right)  \subset\left(
\ker\epsilon\right)  ^{m}. \label{pf.thm.leray.leray-e.b.2}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.leray.leray-e.b.2}):} Let $n\in \NN$ and
$m\in \NN$ be such that $n\geq m>0$. Then, $\left(  \ker\epsilon\right)
^{m}$ is an ideal of the $\kk$-algebra $A$ (since $\ker\epsilon$ is an
ideal of the $\kk$-algebra $A$, and since $m>0$). But
Lemma~\ref{lem.sol.leray.leray-e.1} (applied to $C=A$ and
$f=\widetilde{\id}$) yields
\begin{align*}
\widetilde{\id}^{\star n}\left(  A\right)   &  \subset\left(
\underbrace{\widetilde{\id}\left(  A\right)  }%
_{\substack{\subset\ker\epsilon\\\text{(by Lemma~\ref{lem.sol.leray.leray-e.2}%
(b))}}}\right)  ^{n}\subset\left(  \ker\epsilon\right)  ^{n}%
=\underbrace{\left(  \ker\epsilon\right)  ^{n-m}}_{\subset A}\left(
\ker\epsilon\right)  ^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\geq
m\right) \\
&  \subset A\left(  \ker\epsilon\right)  ^{m}\subset\left(  \ker
\epsilon\right)  ^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\ker\epsilon\right)  ^{m}\text{ is an ideal of the }\kk\text{-algebra
}A\right)  .
\end{align*}
This proves (\ref{pf.thm.leray.leray-e.b.2}).]

Proposition~\ref{prop.exp-log}(f) (applied to $C=A$ and
$f=\widetilde{\id}$) yields%
\[
\log^{\star}\left(  \widetilde{\id}+u_{A}\epsilon_{A}\right)
=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}.
\]
Since $\widetilde{\id}+u_{A}\epsilon_{A}=\operatorname*{id}%
\nolimits_{A}$ (because $\widetilde{\id}=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}$), this rewrites as%
\[
\log^{\star}\left(  \id_{A}\right)  =\sum_{n\geq
1}\dfrac{\left(  -1\right)  ^{n-1}}{n}\widetilde{\id}^{\star
n}.
\]
Thus,%
\begin{align}
\mathfrak{e}  &  =\log^{\star}\left(  \id_{A}\right)
=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\label{pf.thm.leray.leray-e.b.5a}\\
&  =\underbrace{\dfrac{\left(  -1\right)  ^{1-1}}{1}}_{=\dfrac{\left(
-1\right)  ^{0}}{1}=\dfrac{1}{1}=1}\underbrace{\widetilde{\operatorname*{id}%
}^{\star1}}_{=\widetilde{\id}}+\sum_{n\geq2}\dfrac{\left(
-1\right)  ^{n-1}}{n}\widetilde{\id}^{\star n}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}n=1\text{ from the sum}\right) \nonumber\\
&  =\widetilde{\id}+\sum_{n\geq2}\dfrac{\left(  -1\right)
^{n-1}}{n}\widetilde{\id}^{\star n}.
\label{pf.thm.leray.leray-e.b.5b}%
\end{align}


Furthermore,%
\begin{equation}
\mathfrak{e}\left(  1_{A}\right)  =0. \label{pf.thm.leray.leray-e.b.6}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.leray.leray-e.b.6}):}
Lemma~\ref{lem.pf.prop.leray.exp-algh1.2} (applied to $C=A$ and
$f=\widetilde{\id}$) shows that every $n\in \NN$
satisfies%
\[
\widetilde{\id}^{\star n}\left(  1\right)  =\left(
\widetilde{\id}\left(  1\right)  \right)  ^{n}.
\]
Since $1$ stands for $1_{A}$ in this equality, we can rewrite it as follows:
Every $n\in \NN$ satisfies%
\begin{equation}
\widetilde{\id}^{\star n}\left(  1_{A}\right)  =\left(
\widetilde{\id}\left(  1_{A}\right)  \right)  ^{n}.
\label{pf.thm.leray.leray-e.b.6.pf.1}%
\end{equation}


But Lemma~\ref{lem.sol.leray.leray-e.2}(a) yields $\ker
\widetilde{\id}=\kk\cdot1_{A}$. Hence, $1_{A}%
=\underbrace{1}_{\in\kk}\cdot1_{A}\in\kk\cdot1_{A}%
=\ker\widetilde{\id}$, so that $\widetilde{\operatorname*{id}%
}\left(  1_{A}\right)  =0$. Now, if $n$ is any positive integer, then
\begin{align}
\widetilde{\id}^{\star n}\left(  1_{A}\right)   &  =\left(
\underbrace{\widetilde{\id}\left(  1_{A}\right)  }_{=0}\right)
^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.leray.leray-e.b.6.pf.1}%
)}\right) \nonumber\\
&  =0^{n}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\text{ is a positive
integer}\right)  . \label{pf.thm.leray.leray-e.b.6.pf.2}%
\end{align}


Applying both sides of the equality (\ref{pf.thm.leray.leray-e.b.5a}) to
$1_{A}$, we obtain%
\[
\mathfrak{e}\left(  1_{A}\right)  =\left(  \sum_{n\geq1}\dfrac{\left(
-1\right)  ^{n-1}}{n}\widetilde{\id}^{\star n}\right)  \left(
1_{A}\right)  =\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\underbrace{\widetilde{\id}^{\star n}\left(  1_{A}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.6.pf.2}))}}%
}=\sum_{n\geq1}\dfrac{\left(  -1\right)  ^{n-1}}{n}0=0.
\]
This proves (\ref{pf.thm.leray.leray-e.b.6}).]

The map $\mathfrak{e}$ is $\kk$-linear (since $\mathfrak{e}=\log
^{\star}\left(  \id_{A}\right)  \in\mathfrak{n}\left(
A,A\right)  \subset\Hom \left(  A,A\right)  $). Thus,%
\[
\mathfrak{e}\left(  \kk\cdot1_{A}\right)  =\kk\cdot
\underbrace{\mathfrak{e}\left(  1_{A}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.thm.leray.leray-e.b.6}))}}}=\kk\cdot0=0.
\]
But Exercise~\ref{graded-connected-exercise}(c) shows that $A_{0}%
=\kk\cdot1_{A}$. Applying the map $\mathfrak{e}$ to both sides of this
equality, we find%
\begin{equation}
\mathfrak{e}\left(  A_{0}\right)  =\mathfrak{e}\left(  \kk\cdot
1_{A}\right)  =0. \label{pf.thm.leray.leray-e.b.7}%
\end{equation}


Furthermore, recall that $\id_{A}-u_{A}\epsilon_{A}%
\in\mathfrak{n}\left(  A,A\right)  $. Hence, Proposition~\ref{prop.exp-log}(b)
(applied to $g=\id_{A}$) yields $\exp^{\star}\left(
\log^{\star}\left(  \id_{A}\right)  \right)
=\id_{A}$. Thus,%
\begin{equation}
\exp^{\star}\underbrace{\mathfrak{e}}_{=\log^{\star}\left(  \operatorname*{id}%
\nolimits_{A}\right)  }=\exp^{\star}\left(  \log^{\star}\left(
\id_{A}\right)  \right)  =\operatorname*{id}%
\nolimits_{A}. \label{pf.thm.leray.leray-e.b.8}%
\end{equation}
Thus, $\exp^{\star}\mathfrak{e}:A\rightarrow A$ is a $\kk$-algebra
homomorphism (since $\id_{A}:A\rightarrow A$ is a
$\kk$-algebra homomorphism). Hence,
Proposition~\ref{prop.leray.exp-algh2} (applied to $C=A$ and $f=\mathfrak{e}$)
shows that%
\begin{equation}
\mathfrak{e}\left(  \left(  \ker\epsilon\right)  ^{2}\right)  =0.
\label{pf.thm.leray.leray-e.b.9}%
\end{equation}


The map $\mathfrak{e}$ is $\kk$-linear. Thus,%
\[
\mathfrak{e}\left(  \kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}\right)  =\kk\cdot\underbrace{\mathfrak{e}\left(  1_{A}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.6}))}}%
}+\underbrace{\mathfrak{e}\left(  \left(  \ker\epsilon\right)  ^{2}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.leray.leray-e.b.9}))}}}=\kk%
\cdot0+0=0.
\]
In other words,%
\begin{equation}
\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset\ker\mathfrak{e}.
\label{pf.thm.leray.leray-e.b.side1}%
\end{equation}


On the other hand, let us prove the reverse inclusion. Let us first observe
that each $x\in A$ satisfies%
\begin{equation}
\widetilde{\id}\left(  x\right)  -\mathfrak{e}\left(  x\right)
\in\left(  \ker\epsilon\right)  ^{2}. \label{pf.thm.leray.leray-e.b.idp1}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.leray.leray-e.b.idp1}):} Let $x\in A$. Applying
both sides of the equality (\ref{pf.thm.leray.leray-e.b.5b}) to $x$, we find%
\[
\mathfrak{e}\left(  x\right)  =\left(  \widetilde{\id}%
+\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\right)  \left(  x\right)
=\widetilde{\id}\left(  x\right)  +\sum_{n\geq2}\dfrac{\left(
-1\right)  ^{n-1}}{n}\widetilde{\id}^{\star n}\left(  x\right)
.
\]
Subtracting $\widetilde{\id}\left(  x\right)  $ from both sides
of this equality, we find%
\begin{align*}
\mathfrak{e}\left(  x\right)  -\widetilde{\id}\left(  x\right)
&  =\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\widetilde{\id}^{\star n}\left(  \underbrace{x}_{\in A}\right)
\in\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}%
\underbrace{\widetilde{\id}^{\star n}\left(  A\right)
}_{\substack{\subset\left(  \ker\epsilon\right)  ^{2}\\\text{(by
(\ref{pf.thm.leray.leray-e.b.2}) (applied to }m=2\text{)}\\\text{(since }%
n\geq2>0\text{))}}}\\
&  \subset\sum_{n\geq2}\dfrac{\left(  -1\right)  ^{n-1}}{n}\left(
\ker\epsilon\right)  ^{2}\subset\left(  \ker\epsilon\right)  ^{2}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \ker\epsilon\right)
^{2}\text{ is a }\kk\text{-submodule of }A\right)  .
\end{align*}
Hence,%
\[
\widetilde{\id}\left(  x\right)  -\mathfrak{e}\left(  x\right)
=-\underbrace{\left(  \mathfrak{e}\left(  x\right)
-\widetilde{\id}\left(  x\right)  \right)  }_{\in\left(
\ker\epsilon\right)  ^{2}}\in-\left(  \ker\epsilon\right)  ^{2}\subset\left(
\ker\epsilon\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\ker\epsilon\right)  ^{2}\text{ is a }\kk\text{-submodule of }A\right)
.
\]
This proves (\ref{pf.thm.leray.leray-e.b.idp1}).]

Now, let $x\in\ker\mathfrak{e}$. Thus, $\mathfrak{e}\left(  x\right)  =0$.
Now, (\ref{pf.thm.leray.leray-e.b.idp1}) yields $\widetilde{\operatorname*{id}%
}\left(  x\right)  -\mathfrak{e}\left(  x\right)  \in\left(  \ker
\epsilon\right)  ^{2}$.

But%
\[
\widetilde{\id}\left(  x\right)  -\underbrace{\mathfrak{e}%
\left(  x\right)  }_{=0}=\underbrace{\widetilde{\id}%
}_{=\id_{A}-u_{A}\epsilon_{A}}\left(  x\right)
=\left(  \id_{A}-u_{A}\epsilon_{A}\right)  \left(
x\right)  =\underbrace{\id_{A}\left(  x\right)  }%
_{=x}-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\epsilon_{A}\left(  x\right)  \cdot1_{A}.
\]
Thus,%
\[
x-\epsilon_{A}\left(  x\right)  \cdot1_{A}=\widetilde{\operatorname*{id}%
}\left(  x\right)  -\mathfrak{e}\left(  x\right)  \in\left(  \ker
\epsilon\right)  ^{2}
\ \ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.thm.leray.leray-e.b.idp1})}\right) .
\]
Hence,%
\[
x\in\underbrace{\epsilon_{A}\left(  x\right)  }_{\in\kk}\cdot
1_{A}+\left(  \ker\epsilon\right)  ^{2}\subset\kk\cdot1_{A}+\left(
\ker\epsilon\right)  ^{2}.
\]


Now, forget that we fixed $x$. We thus have shown that every $x\in
\ker\mathfrak{e}$ satisfies $x\in\kk\cdot1_{A}+\left(  \ker
\epsilon\right)  ^{2}$. In other words, we have $\ker\mathfrak{e}%
\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Combining this
with (\ref{pf.thm.leray.leray-e.b.side1}), we obtain%
\begin{equation}
\ker\mathfrak{e}=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}.
\label{pf.thm.leray.leray-e.b.part1}%
\end{equation}
Thus, one part of Theorem~\ref{thm.leray.leray-e}(b) is proven. It remains to
show that $\mathfrak{e}\left(  A\right)  \cong\left(  \ker\epsilon\right)
/\left(  \ker\epsilon\right)  ^{2}$ (as $\kk$-modules).

It is known that if $V$ and $W$ are two $\kk$-modules, and if
$\delta:V\rightarrow W$ is a $\kk$-linear map, then $\delta\left(
V\right)  \cong V/\ker\delta$ as $\kk$-modules. Applying this to $V=A$,
$W=A$ and $\delta=\mathfrak{e}$, we obtain%
\[
\mathfrak{e}\left(  A\right)  \cong A/\underbrace{\ker\mathfrak{e}%
}_{\substack{=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}%
\\\text{(by (\ref{pf.thm.leray.leray-e.b.part1}))}}}=A/\left(  \kk%
\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}\right)  \cong\left(  \ker
\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma~\ref{lem.sol.leray.leray-e.3}}\right)
\]
as $\kk$-modules. This completes the proof of
Theorem~\ref{thm.leray.leray-e}(b).

(c) The definition of $\mathfrak{q}$ shows that%
\begin{equation}
\mathfrak{q}\left(  x\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  x\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for each
}x\in A. \label{pf.thm.leray.leray-e.c.qx=}%
\end{equation}


Now, each $x\in A_{0}$ satisfies%
\begin{align*}
\mathfrak{q}\left(  x\right)   &  =\iota_{\mathfrak{e}\left(  A\right)
}\left(  \mathfrak{e}\left(  \underbrace{x}_{\in A_{0}}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.leray.leray-e.c.qx=}%
)}\right) \\
&  \in\iota_{\mathfrak{e}\left(  A\right)  }\left(  \underbrace{\mathfrak{e}%
\left(  A_{0}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.thm.leray.leray-e.b.7}))}}}\right)  =\iota_{\mathfrak{e}\left(
A\right)  }\left(  0\right)  =0
\end{align*}
(since the map $\iota_{\mathfrak{e}\left(  A\right)  }$ is $\kk%
$-linear). In other words, we have $\mathfrak{q}\left(  A_{0}\right)
\subset0$. Thus, $\mathfrak{q}\left(  A_{0}\right)  =0$. Hence,
Proposition~\ref{prop.convolution-series}(h) (applied to $A$,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ and
$\mathfrak{q}$ instead of $C$, $A$ and $f$) shows that $\mathfrak{q}%
\in\mathfrak{n}\left(  A,\Sym \left(  \mathfrak{e}\left(
A\right)  \right)  \right)  $. This proves Theorem~\ref{thm.leray.leray-e}(c).

(d) Theorem~\ref{thm.leray.leray-e}(c) yields $\mathfrak{q}\in\mathfrak{n}%
\left(  A,\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
\right)  $. Hence, $\exp^{\star}\mathfrak{q}\in\Hom \left(
A,\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  \right)  $
is well-defined.

Recall that the $\kk$-algebra $\Sym V$ is commutative
whenever $V$ is a $\kk$-module. Applying this to $V=\mathfrak{e}\left(
A\right)  $, we conclude that the $\kk$-algebra $\Sym %
\left(  \mathfrak{e}\left(  A\right)  \right)  $ is commutative.

We have $\mathfrak{q}\left(  A\right)  =\Sym^{1}%
\left(  \mathfrak{e}\left(  A\right)  \right)  $%
\ \ \ \ \footnote{\textit{Proof.} Every $\kk$-module $V$ satisfies
$\iota_{V}\left(  V\right)  =\Sym^{1}V$. Applying this
to $V=\mathfrak{e}\left(  A\right)  $, we obtain $\iota_{\mathfrak{e}\left(
A\right)  }\left(  \mathfrak{e}\left(  A\right)  \right)
= \Sym^{1}\left(  \mathfrak{e}\left(  A\right)  \right)  $.
\par
Now,%
\begin{align*}
\mathfrak{q}\left(  A\right)   &  =\left\{  \underbrace{\mathfrak{q}\left(
x\right)  }_{\substack{=\iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  x\right)  \right)  \\\text{(by
(\ref{pf.thm.leray.leray-e.c.qx=}))}}}\ \mid\ x\in A\right\}  =\left\{
\iota_{\mathfrak{e}\left(  A\right)  }\left(  \mathfrak{e}\left(  x\right)
\right)  \ \mid\ x\in A\right\} \\
&  =\iota_{\mathfrak{e}\left(  A\right)  }\left(  \underbrace{\left\{
\mathfrak{e}\left(  x\right)  \ \mid\ x\in A\right\}  }_{=\mathfrak{e}\left(
A\right)  }\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  A\right)  \right)  =\Sym^{1}%
\left(  \mathfrak{e}\left(  A\right)  \right)  .
\end{align*}
}. Thus, $\mathfrak{q}\left(  A\right)  $ generates the $\kk$-algebra
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
$\ \ \ \ \footnote{\textit{Proof.} It is known that if $V$ is a $\kk%
$-module, then $\Sym^{1}V$ generates the $\kk%
$-algebra $\Sym V$. Applying this to $V=\mathfrak{e}\left(
A\right)  $, we conclude that $\Sym^{1}\left(
\mathfrak{e}\left(  A\right)  \right)  $ generates the $\kk$-algebra
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $. In other
words, $\mathfrak{q}\left(  A\right)  $ generates the $\kk$-algebra
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ (since
$\mathfrak{q}\left(  A\right)  =\Sym^{1}\left(
\mathfrak{e}\left(  A\right)  \right)  $).}.

Furthermore, Theorem~\ref{thm.leray.leray-e}(b) yields that $\ker
\mathfrak{e}=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$. Hence,
$1=1_{A}=\underbrace{1}_{\in\kk}\cdot1_{A}\in\kk\cdot
1_{A}\subset\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}%
=\ker\mathfrak{e}$ and $\left(  \ker\epsilon\right)  ^{2}\subset
\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}=\ker\mathfrak{e}$. Now,
(\ref{pf.thm.leray.leray-e.c.qx=}) (applied to $x=1$) yields%
\[
\mathfrak{q}\left(  1\right)  =\iota_{\mathfrak{e}\left(  A\right)  }\left(
\underbrace{\mathfrak{e}\left(  1\right)  }_{\substack{=0\\\text{(since }%
1\in\ker\mathfrak{e}\text{)}}}\right)  =\iota_{\mathfrak{e}\left(  A\right)
}\left(  0\right)  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\iota_{\mathfrak{e}\left(  A\right)  }\text{ is }\kk\text{-linear}%
\right)  .
\]
Also, each $x\in\left(  \ker\epsilon\right)  ^{2}$ satisfies%
\begin{align*}
\mathfrak{q}\left(  x\right)   &  =\iota_{\mathfrak{e}\left(  A\right)
}\left(  \underbrace{\mathfrak{e}\left(  x\right)  }%
_{\substack{=0\\\text{(since }x\in\left(  \ker\epsilon\right)  ^{2}\subset
\ker\mathfrak{e}\text{)}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.leray.leray-e.c.qx=})}\right) \\
&  =\iota_{\mathfrak{e}\left(  A\right)  }\left(  0\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\iota_{\mathfrak{e}\left(
A\right)  }\text{ is }\kk\text{-linear}\right)  .
\end{align*}
In other words, $\mathfrak{q}\left(  \left(  \ker\epsilon\right)  ^{2}\right)
\subset0$. Hence, $\mathfrak{q}\left(  \left(  \ker\epsilon\right)
^{2}\right)  =0$.

Thus, Proposition~\ref{prop.leray.exp-algh3} (applied to $A$,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ and
$\mathfrak{q}$ instead of $C$, $A$ and $f$) shows that $\exp^{\star
}\mathfrak{q}:A\rightarrow\Sym \left(  \mathfrak{e}\left(
A\right)  \right)  $ is a surjective $\kk$-algebra homomorphism.

But let us recall that $\mathfrak{s}$ is the unique $\kk$-algebra
homomorphism $\Phi:\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ satisfying $\mathbf{i}=\Phi\circ\iota_{\mathfrak{e}%
\left(  A\right)  }$. Hence, $\mathfrak{s}$ is a $\kk$-algebra
homomorphism $\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ and satisfies $\mathbf{i}=\mathfrak{s}\circ
\iota_{\mathfrak{e}\left(  A\right)  }$.

Thus, Proposition~\ref{prop.convolution-series}(i) (applied to $A$,
$\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $, $A$,
$\mathfrak{s}$, $\exp$ and $\mathfrak{q}$ instead of $C$, $A$, $B$, $s$, $u$
and $f$) shows that%
\[
\mathfrak{s}\circ\mathfrak{q}\in\mathfrak{n}\left(  A,A\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \exp^{\star}\left(
\mathfrak{s}\circ\mathfrak{q}\right)  =\mathfrak{s}\circ\left(  \exp^{\star
}\mathfrak{q}\right)  .
\]


Furthermore,
\begin{equation}
\mathfrak{s}\circ\mathfrak{q}=\mathfrak{e} \label{pf.thm.leray.leray-e.c.sq=e}%
\end{equation}
(since each $x\in A$ satisfies%
\begin{align*}
\left(  \mathfrak{s}\circ\mathfrak{q}\right)  \left(  x\right)   &
=\mathfrak{s}\left(  \underbrace{\mathfrak{q}\left(  x\right)  }%
_{\substack{=\iota_{\mathfrak{e}\left(  A\right)  }\left(  \mathfrak{e}\left(
x\right)  \right)  \\\text{(by (\ref{pf.thm.leray.leray-e.c.qx=}))}}}\right)
=\mathfrak{s}\left(  \iota_{\mathfrak{e}\left(  A\right)  }\left(
\mathfrak{e}\left(  x\right)  \right)  \right) \\
&  =\underbrace{\left(  \mathfrak{s}\circ\iota_{\mathfrak{e}\left(  A\right)
}\right)  }_{\substack{=\mathbf{i}\\\text{(since }\mathbf{i}=\mathfrak{s}%
\circ\iota_{\mathfrak{e}\left(  A\right)  }\text{)}}}\left(  \mathfrak{e}%
\left(  x\right)  \right)  =\mathbf{i}\left(  \mathfrak{e}\left(  x\right)
\right)  =\mathfrak{e}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathbf{i}\text{ is just an inclusion map}\right)
\end{align*}
). Hence, the equality $\exp^{\star}\left(  \mathfrak{s}\circ\mathfrak{q}%
\right)  =\mathfrak{s}\circ\left(  \exp^{\star}\mathfrak{q}\right)  $ rewrites
as $\exp^{\star}\mathfrak{e}=\mathfrak{s}\circ\left(  \exp^{\star}%
\mathfrak{q}\right)  $. Comparing this with (\ref{pf.thm.leray.leray-e.b.8}),
we obtain $\mathfrak{s}\circ\left(  \exp^{\star}\mathfrak{q}\right)
=\id_{A}$. Thus, the map $\exp^{\star}\mathfrak{q}$
has a left inverse (with respect to composition), and hence is injective.

Now, the map $\exp^{\star}\mathfrak{q}$ is both injective and surjective.
Consequently, $\exp^{\star}\mathfrak{q}$ is bijective, i.e., invertible. Since
$\exp^{\star}\mathfrak{q}$ is an invertible $\kk$-algebra homomorphism,
we conclude that $\exp^{\star}\mathfrak{q}$ is a $\kk$-algebra
isomorphism. Its inverse must be $\mathfrak{s}$ (since $\mathfrak{s}%
\circ\left(  \exp^{\star}\mathfrak{q}\right)  =\operatorname*{id}%
\nolimits_{A}$). Hence, the maps $\exp^{\star}\mathfrak{q}:A\rightarrow
\Sym \left(  \mathfrak{e}\left(  A\right)  \right)  $ and
$\mathfrak{s}:\Sym \left(  \mathfrak{e}\left(  A\right)
\right)  \rightarrow A$ are mutually inverse $\kk$-algebra
isomorphisms. This proves Theorem~\ref{thm.leray.leray-e}(d).

(e) Theorem~\ref{thm.leray.leray-e}(d) shows that the maps $\exp^{\star
}\mathfrak{q}:A\rightarrow\Sym \left(  \mathfrak{e}\left(
A\right)  \right)  $ and $\mathfrak{s}:\Sym \left(
\mathfrak{e}\left(  A\right)  \right)  \rightarrow A$ are mutually inverse
$\kk$-algebra isomorphisms. Hence, $A\cong\Sym \left(
\mathfrak{e}\left(  A\right)  \right)  $ as $\kk$-algebras (via these
isomorphisms). But Theorem~\ref{thm.leray.leray-e}(b) shows that
$\mathfrak{e}\left(  A\right)  \cong\left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}$ (as $\kk$-modules). Hence,
$\Sym \left(  \underbrace{\mathfrak{e}\left(  A\right)  }%
_{\cong\left(  \ker\epsilon\right)  /\left(  \ker\epsilon\right)  ^{2}%
}\right)  \cong\Sym \left(  \left(  \ker\epsilon\right)
/\left(  \ker\epsilon\right)  ^{2}\right)  $ as $\kk$-algebras. Thus,%
\[
A\cong\Sym \left(  \mathfrak{e}\left(  A\right)  \right)
\cong\Sym \left(  \left(  \ker\epsilon\right)  /\left(
\ker\epsilon\right)  ^{2}\right)
\]
as $\kk$-algebras. This proves Theorem~\ref{thm.leray.leray-e}(e).

(f) Theorem~\ref{thm.leray.leray-e}(b) yields that $\ker\mathfrak{e}%
=\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}$.

Let $x\in A$. We have%
\[
\underbrace{\widetilde{\id}}_{=\operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}}\left(  x\right)  =\left(  \operatorname*{id}%
\nolimits_{A}-u_{A}\epsilon_{A}\right)  \left(  x\right)
=\underbrace{\id_{A}\left(  x\right)  }_{=x}%
-\underbrace{\left(  u_{A}\epsilon_{A}\right)  \left(  x\right)
}_{\substack{=u_{A}\left(  \epsilon_{A}\left(  x\right)  \right)
\\=\epsilon_{A}\left(  x\right)  \cdot1_{A}\\\text{(by the definition of
}u_{A}\text{)}}}=x-\epsilon_{A}\left(  x\right)  \cdot1_{A},
\]
so that $x=\epsilon_{A}\left(  x\right)  \cdot1_{A}%
+\widetilde{\id}\left(  x\right)  $. Subtracting $\mathfrak{e}%
\left(  x\right)  $ from both sides of this equality, we obtain%
\[
x-\mathfrak{e}\left(  x\right)  =\underbrace{\epsilon_{A}\left(  x\right)
}_{\in\kk}\cdot1_{A}+\underbrace{\widetilde{\id}\left(
x\right)  -\mathfrak{e}\left(  x\right)  }_{\substack{\in\left(  \ker
\epsilon\right)  ^{2}\\\text{(by (\ref{pf.thm.leray.leray-e.b.idp1}))}}%
}\in\kk\cdot1_{A}+\left(  \ker\epsilon\right)  ^{2}=\ker\mathfrak{e}%
\]
(since $\ker\mathfrak{e}=\kk\cdot1_{A}+\left(  \ker\epsilon\right)
^{2}$). In other words, $\mathfrak{e}\left(  x-\mathfrak{e}\left(  x\right)
\right)  =0$. Comparing this with%
\begin{align*}
\mathfrak{e}\left(  x-\mathfrak{e}\left(  x\right)  \right)   &
=\mathfrak{e}\left(  x\right)  -\underbrace{\mathfrak{e}\left(  \mathfrak{e}%
\left(  x\right)  \right)  }_{=\left(  \mathfrak{e}\circ\mathfrak{e}\right)
\left(  x\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\mathfrak{e}\text{ is }\kk\text{-linear}\right)  \\
&  =\mathfrak{e}\left(  x\right)  -\left(  \mathfrak{e}\circ\mathfrak{e}%
\right)  \left(  x\right)  ,
\end{align*}
we obtain $\mathfrak{e}\left(  x\right)  -\left(  \mathfrak{e}\circ
\mathfrak{e}\right)  \left(  x\right)  =0$. In other words, $\mathfrak{e}%
\left(  x\right)  =\left(  \mathfrak{e}\circ\mathfrak{e}\right)  \left(
x\right)  $.

Now, forget that we fixed $x$. We thus have shown that $\mathfrak{e}\left(
x\right)  =\left(  \mathfrak{e}\circ\mathfrak{e}\right)  \left(  x\right)  $
for each $x\in A$. In other words, $\mathfrak{e}=\mathfrak{e}\circ
\mathfrak{e}$. In other words, $\mathfrak{e}\circ\mathfrak{e}=\mathfrak{e}$.
In other words, the map $\mathfrak{e}:A\rightarrow A$ is a projection. This
proves Theorem~\ref{thm.leray.leray-e}(f).
\end{proof}

%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.subs.fin}}

\textit{Solution to Exercise~\ref{exe.Lambda.subs.fin}.} We
have $f \in R\left(\xx\right)$. Thus, $f$ is a formal
power series of bounded degree. In other words,
$f = \sum_\alpha c_\alpha \xx^\alpha$ (with the sum ranging over all
weak compositions $\alpha$) for some elements $c_\alpha$ in $\kk$ such
that there exists a $d \in \NN$ such that every $\alpha$ satisfying
$\deg(\xx^\alpha) > d$ must satisfy $c_\alpha = 0$. Consider these
$c_\alpha$ and this $d$.

We have written $f$ as the sum $\sum_\alpha c_\alpha \xx^\alpha$. Now,
substituting $a_1, a_2, \ldots, a_k, 0, 0, \ldots$ for
$x_1, x_2, x_3, \ldots$ in $f$ maps all but finitely many of
the terms $c_\alpha \xx^\alpha$ appearing in this sum to $0$.
In fact:
\begin{itemize}
\item all terms $c_\alpha \xx^\alpha$ such that the monomial
$\xx^\alpha$ contains at least one of the variables
$x_{k+1}, x_{k+2}, x_{k+3}, \ldots$ (that is, such that for
some integer $i > k$, the $i$-th entry of $\alpha$ is
nonzero) become $0$ under our substitution
(because the substitution takes each of
the variables $x_{k+1}, x_{k+2}, x_{k+3}, \ldots$ to $0$);
\item all terms $c_\alpha \xx^\alpha$ with
$\deg(\xx^\alpha) > d$ become $0$ under our substitution
(because we know that
$c_\alpha = 0$ for every $\alpha$ satisfying
$\deg(\xx^\alpha) > d$);
\item the remaining terms (that is, the terms
$c_\alpha \xx^\alpha$ satisfying neither of the preceding two
conditions) might not get sent to $0$, but there are only
finitely many such terms\footnote{In
fact, these are the terms for which $\xx^\alpha$ is a monomial
which has degree $\leq d$ and contains no other variables
than $x_1, x_2, \ldots, x_k$. Clearly, there are only finitely
many such monomials, and thus only finitely many such terms.}.
\end{itemize}
Hence, our substitution maps all but finitely many of
the terms $c_\alpha \xx^\alpha$ appearing in the sum
$\sum_\alpha c_\alpha \xx^\alpha$ to $0$. Since
$\sum_\alpha c_\alpha \xx^\alpha = f$, this rewrites as
follows: Our substitution maps all but finitely many of
the terms of $f$ to $0$. This solves
Exercise~\ref{exe.Lambda.subs.fin}.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.partition.dominance}}

\textit{Solution to Exercise~\ref{exe.partition.dominance}.} Let us
first make an auxiliary observation. Namely, let us prove the
following lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.transpose1}Let $\nu \in \Par$ and
$k\in \NN$. Then,
\begin{equation}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k}
= \sum_{j=1}^{\infty}\min\left\{  \nu_{j},k\right\}  .
\label{sol.partition.dominance.lemma}
\end{equation}
(Note that the right hand side of \eqref{sol.partition.dominance.lemma} is
well-defined, because every sufficiently high $j$
satisfies $\min\left\{  \nu_{j},k\right\}  =0$
(since every sufficiently high $j$ satisfies $\nu_{j}=0$)).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.transpose1}.]
This is best seen by
double-counting boxes in a Ferrers diagram. The left hand side of
\eqref{sol.partition.dominance.lemma} counts the boxes in the first $k$ rows
of the Ferrers diagram of $\nu^{t}$. Since the Ferrers diagram of $\nu^{t}$ is
obtained from that of $\nu$ by exchanging rows for columns (i.e., reflecting
the diagram across its main diagonal), it is clear that this is the same as
counting the boxes in the first $k$ columns of the Ferrers diagram of $\nu$.
But these latter boxes can also be counted row-by-row:
For every $j\in\left\{ 1,2,3,\ldots\right\}  $,
the number of boxes in the first $k$ columns of the
Ferrers diagram of $\nu$ that lie in row $j$ is
$\min\left\{  \nu_{j},k\right\}  $.
Thus, the total amount of boxes in the first $k$ columns of the Ferrers
diagram of $\nu$ is $\sum_{j=1}^{\infty} \min\left\{  \nu_{j},k\right\}  $,
which is precisely the right hand side of
\eqref{sol.partition.dominance.lemma}.
Hence, the left hand side of \eqref{sol.partition.dominance.lemma} and
the right hand side of \eqref{sol.partition.dominance.lemma} are equal
(since they both count the same boxes). This proves
\eqref{sol.partition.dominance.lemma}, and thus
Lemma \ref{lem.sol.Lambda.triangular.transpose1} follows.
\end{proof}

Let us now come to the solution of the exercise. We need to prove that
\begin{equation}
\left(  \text{if }\lambda\triangleright\mu\text{, then }\mu^{t}\triangleright
\lambda^{t}\right)
\label{sol.partition.dominance.goal1}
\end{equation}
and
\begin{equation}
\left(  \text{if }\mu^{t}\triangleright\lambda^{t}\text{, then }
\lambda\triangleright\mu\right) .
\label{sol.partition.dominance.goal2}
\end{equation}


\textit{Proof of \eqref{sol.partition.dominance.goal2}:} Assume that $\mu
^{t}\triangleright\lambda^{t}$. By the definition of the dominance order, this
means that
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu
^{t}\right)  _{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+...+\left(  \lambda^{t}\right)  _{k}
\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,...,n\right\}  .
\label{sol.partition.dominance.known}
\end{equation}
This inequality holds for $k=0$ as well (because both sides are $0$ when
$k=0$), and thus it holds for all $k\in\left\{  0,1,...,n\right\}  $. We can
further rewrite this inequality by applying
\eqref{sol.partition.dominance.lemma}: The left hand side becomes $\sum
_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}  $, and the right hand side
becomes $\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},k\right\}  $. As a
result, we obtain
\begin{equation}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}  \geq\sum_{j=1}^{\infty}
\min\left\{  \lambda_{j},k\right\}  \ \ \ \ \ \ \ \ \ \ \text{for all }
k\in\left\{  0,1,...,n\right\}  .
\label{sol.partition.dominance.known.rewr}
\end{equation}


Now, let $k\in\left\{  1,2,...,n\right\}  $ be arbitrary. We are going to show
that $\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu_{2}+...+\mu_{k}$.

First of all, $\lambda\in\Par_{n}$, so that
$n=\left\vert \lambda\right\vert \geq\lambda_{k}$. Thus, $\lambda_{k}\in\left\{
0,1,...,n\right\}  $. Therefore, \eqref{sol.partition.dominance.known.rewr}
(applied to $\lambda_{k}$ instead of $k$) yields that
\begin{equation}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda_{k}\right\}  \geq\sum
_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda_{k}\right\}
.
\label{sol.partition.dominance.known.ineq}
\end{equation}
But the right hand side of this inequality can be rewritten as follows:
\begin{align}
\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda_{k}\right\}   &
=\sum_{j=1}^{k}\underbrace{\min\left\{  \lambda_{j},\lambda_{k}\right\}
}_{\substack{=\lambda_{k}\\\text{(since }j\leq k\text{ and thus }\lambda
_{j}\geq\lambda_{k}\\\text{(because }\lambda\text{ is a partition))}}}
+\sum_{j=k+1}^{\infty}\underbrace{\min\left\{  \lambda_{j},\lambda
_{k}\right\}  }_{\substack{=\lambda_{j}\\\text{(since }j>k\text{ and thus
}\lambda_{j}\leq\lambda_{k}\\
\text{(because }\lambda\text{ is a partition))}}}\nonumber\\
&  =\underbrace{\sum_{j=1}^{k}\lambda_{k}}_{=k\lambda_{k}}+\underbrace{\sum
_{j=k+1}^{\infty}\lambda_{j}}_{\substack{=\lambda_{k+1}+\lambda_{k+2}
+\lambda_{k+3}+...\\=\left\vert \lambda\right\vert -\left(  \lambda
_{1}+\lambda_{2}+...+\lambda_{k}\right)  }}=k\lambda_{k}
+\underbrace{\left\vert \lambda\right\vert }_{=n}-\left(  \lambda_{1}
+\lambda_{2}+...+\lambda_{k}\right)  \nonumber\\
&  =k\lambda_{k}+n-\left(  \lambda_{1}+\lambda_{2}+...+\lambda_{k}\right)
.
\label{sol.partition.dominance.p1}
\end{align}
Meanwhile, the left hand side can be bounded from above:
\begin{align}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda_{k}\right\}   &  =\sum
_{j=1}^{k}\underbrace{\min\left\{  \mu_{j},\lambda_{k}\right\}  }_{\leq
\lambda_{k}}+\sum_{j=k+1}^{\infty}\underbrace{\min\left\{  \mu_{j},\lambda
_{k}\right\}  }_{\leq\mu_{j}}\nonumber\\
&  \leq\underbrace{\sum_{j=1}^{k}\lambda_{k}}_{=k\lambda_{k}}+\underbrace{\sum
_{j=k+1}^{\infty}\mu_{j}}_{\substack{=\mu_{k+1}+\mu_{k+2}+\mu_{k+3}
+...\\=\left\vert \mu\right\vert -\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)
}}=k\lambda_{k}+\underbrace{\left\vert \mu\right\vert }_{=n}-\left(  \mu
_{1}+\mu_{2}+...+\mu_{k}\right)  \nonumber\\
&  =k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)
.
\label{sol.partition.dominance.p2}
\end{align}


Now, \eqref{sol.partition.dominance.known.ineq} yields
\begin{align*}
0 &  \leq\underbrace{\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda
_{k}\right\}  }_{\substack{\leq k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}
+...+\mu_{k}\right)  \\
\text{(by \eqref{sol.partition.dominance.p2})}}}
-\underbrace{\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda
_{k}\right\}  }_{\substack{=k\lambda_{k}+n-\left(  \lambda_{1}+\lambda
_{2}+...+\lambda_{k}\right)  \\
\text{(by \eqref{sol.partition.dominance.p1})}}}\\
&  \leq\left(  k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)
\right)  -\left(  k\lambda_{k}+n-\left(  \lambda_{1}+\lambda_{2}
+...+\lambda_{k}\right)  \right)  \\
&  =\left(  \lambda_{1}+\lambda_{2}+...+\lambda_{k}\right)  -\left(  \mu
_{1}+\mu_{2}+...+\mu_{k}\right)  .
\end{align*}
In other words, $\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu
_{2}+...+\mu_{k}$.

Now, forget that we fixed $k$. We thus have shown that
\[
\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu_{2}+...+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,...,n\right\}
\text{.}
\]
In other words, $\lambda\triangleright\mu$. This proves
\eqref{sol.partition.dominance.goal2}.

\textit{Proof of \eqref{sol.partition.dominance.goal1}:} Assume that
$\lambda\triangleright\mu$. We have $\left(  \lambda^{t}\right)  ^{t}
=\lambda\triangleright\mu=\left(  \mu^{t}\right)  ^{t}$. Thus, we can apply
\eqref{sol.partition.dominance.goal2} to $\mu^{t}$ and $\lambda^{t}$ instead
of $\lambda$ and $\mu$. As a result, we obtain $\mu^{t}\triangleright
\lambda^{t}$. This proves \eqref{sol.partition.dominance.goal1}.

Combining \eqref{sol.partition.dominance.goal1} and
\eqref{sol.partition.dominance.goal2}, we obtain the equivalence of the
two assertions $\lambda\triangleright\mu$ and
$\mu^{t}\triangleright \lambda^{t}$. This solves
Exercise~\ref{exe.partition.dominance}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.partition.dominance}}

Before we begin solving Exercise~\ref{exe.partition.dominance}, let us prove
some basic facts. First, let us recall that (in
Definition~\ref{def.partition.conjugate}) we gave two definitions of a
conjugate partition: a first definition using Ferrers diagrams, and then an
alternative definition by the equality \eqref{eq.partition.transpose}. We
did not show that these two definitions are equivalent. We are going
to rectify this omission soon. More precisely, we shall first show that
the second definition of $\lambda^t$ (i.e., the one using
\eqref{eq.partition.transpose}) gives a well-defined partition; then,
we will prove that the first definition defines the same conjugate partition.
Afterwards, we shall show some properties of conjugate partitions, and finally
proceed to the solution of Exercise~\ref{exe.partition.dominance}.

For this whole section, we shall use the following notation: If $\nu$ is an
infinite sequence of integers (e.g., a weak composition), and if $i$ is a
positive integer, then $\nu_{i}$ will denote the $i$-th entry of $\nu$. Thus,
every infinite sequence $\nu$ of integers satisfies $\nu=\left(  \nu_{1}%
,\nu_{2},\nu_{3},\ldots\right)  $.

We begin with a simple lemma:

\begin{lemma}
\label{lem.sol.partition.dominance.lambdat-is-a-partition}Let $\lambda$ be a
weak composition. Define a sequence $\lambda^{t}$ of nonnegative integers by%
\begin{equation}
\left(  \left(  \lambda^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }i\geq1\right)  .
\label{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}%
\end{equation}
Then, $\lambda^{t}$ is well-defined and is a partition.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.partition.dominance.lambdat-is-a-partition}.]The
sequence $\lambda$ is a weak composition. In other words, the support of
$\lambda$ is finite. Let $I$ be this support. Thus, for every $i\in\left\{
1,2,3,\ldots\right\}  $, the set $\left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq i\right\}  $ is finite\footnote{\textit{Proof.} Let
$i\in\left\{  1,2,3,\ldots\right\}  $.
\par
Let $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i\right\}  $. Thus, $x$ is an element $j$ of $\left\{  1,2,3,\ldots\right\}  $
satisfying $\lambda_{j}\geq i$. In other words, $x$ is an element of $\left\{
1,2,3,\ldots\right\}  $ and satisfies $\lambda_{x}\geq i$. From $\lambda
_{x}\geq i>0$, we obtain $\lambda_{x}\neq0$. Hence, $x$ is an element of
$\left\{  1,2,3,\ldots\right\}  $ and satisfies $\lambda_{x}\neq0$. In other
words, $x$ is an element $j$ of $\left\{  1,2,3,\ldots\right\}  $ satisfying
$\lambda_{j}\neq0$.
\par
But $I$ is the support of $\lambda$. In other words, $I$ is the set of all
elements $j$ of $\left\{  1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}%
\neq0$. Thus, $x$ belongs to $I$ (since $x$ is an element $j$ of $\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}\neq0$). In other words, $x\in
I$.
\par
Now, forget that we fixed $x$. We thus have proven that $x\in I$ for each
$x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i\right\}  $. In other words, $\left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq i\right\}  \subset I$. Hence, the set $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  $ is
finite (since $I$ is finite). Qed.}. Therefore, for every $i\in\left\{
1,2,3,\ldots\right\}  $, the cardinality $\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert $ is a
well-defined nonnegative integer. In other words, for every $i\in\left\{
1,2,3,\ldots\right\}  $, the right hand side of
(\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}) is a
well-defined nonnegative integer. Hence, the sequence $\lambda^{t}$ is well-defined.

For every $i\in\left\{  1,2,3,\ldots\right\}  $, we have $\left(  \lambda
^{t}\right)  _{i}\geq\left(  \lambda^{t}\right)  _{i+1}$%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots\right\}  $.
Then, (\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}) (applied
to $i+1$ instead of $i$) yields%
\begin{equation}
\left(  \lambda^{t}\right)  _{i+1}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i+1\right\}  \right\vert .
\label{pf.lem.sol.partition.dominance.lambdat-is-a-partition.fn2.1}%
\end{equation}
Also, (\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}) yields
\begin{equation}
\left(  \lambda^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert .
\label{pf.lem.sol.partition.dominance.lambdat-is-a-partition.fn2.2}%
\end{equation}
\par
Let $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i+1\right\}  $. Thus, $x$ is an element $j$ of $\left\{  1,2,3,\ldots\right\}
$ satisfying $\lambda_{j}\geq i+1$. In other words, $x$ is an element of
$\left\{  1,2,3,\ldots\right\}  $ and satisfies $\lambda_{x}\geq i+1$. Now,
$x$ is an element of $\left\{  1,2,3,\ldots\right\}  $ and satisfies
$\lambda_{x}\geq i$ (since $\lambda_{x}\geq i+1\geq i$). In other words, $x$
is an element $j$ of $\left\{  1,2,3,\ldots\right\}  $ satisfying $\lambda
_{j}\geq i$. In other words, $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq i\right\}  $.
\par
Now, forget that we fixed $x$. We thus have shown that $x\in\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  $ for
each $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i+1\right\}  $. In other words,%
\[
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i+1\right\}  \subset \left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq i\right\}  .
\]
Hence,%
\[
\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i+1\right\}  \right\vert \leq\left\vert \left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert .
\]
Now, (\ref{pf.lem.sol.partition.dominance.lambdat-is-a-partition.fn2.1})
yields%
\begin{align*}
\left(  \lambda^{t}\right)  _{i+1}  &  =\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i+1\right\}  \right\vert
\leq\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid
\ \lambda_{j}\geq i\right\}  \right\vert =\left(  \lambda^{t}\right)  _{i}%
\end{align*}
(by (\ref{pf.lem.sol.partition.dominance.lambdat-is-a-partition.fn2.2})).
In other words, $\left(  \lambda^{t}\right)  _{i}
\geq \left(  \lambda^{t}\right)  _{i+1}$.
Qed.}. In other words, $\left(  \lambda^{t}\right)  _{1}\geq\left(
\lambda^{t}\right)  _{2}\geq\left(  \lambda^{t}\right)  _{3}\geq\cdots$. In
other words, the sequence $\lambda^{t}$ is weakly decreasing.

Furthermore, the set $\left\{  \lambda_{k}\ \mid\ k\in I\right\}  $ is finite
(since the set $I$ is finite) and is a subset of $ \NN$ (since
$\lambda_{k}\in \NN$ for each $k\in I$). Thus, this set is a finite
subset of $ \NN$. Hence, this set has an upper bound in $ \NN$
(since every finite subset of $ \NN$ has an upper bound in $\NN$).
Fix such an upper bound $u$. Then, $u$ is an upper bound of the set
$\left\{  \lambda_{k}\ \mid\ k\in I\right\}  $. In other words,
\begin{equation}
u\geq p\ \ \ \ \ \ \ \ \ \ \text{for each }p\in\left\{  \lambda_{k}%
\ \mid\ k\in I\right\}  .
\label{pf.lem.sol.partition.dominance.lambdat-is-a-partition.up}%
\end{equation}


Also, $u\in \NN$. Thus, $u+1$ is a positive integer.

For every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfying $i\geq u+1$, we
have
\begin{equation}
\left(  \lambda^{t}\right)  _{i}=0
\label{pf.lem.sol.partition.dominance.lambdat-is-a-partition.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.partition.dominance.lambdat-is-a-partition.3}):} Let
$i\in\left\{  1,2,3,\ldots\right\}  $ be such that $i\geq u+1$. We must show
that $\left(  \lambda^{t}\right)  _{i}=0$.
\par
Let $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i\right\}  $. We shall derive a contradiction.
\par
We have $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda
_{j}\geq i\right\}  $. Thus, $x$ is an element $j$ of $\left\{  1,2,3,\ldots
\right\}  $ satisfying $\lambda_{j}\geq i$. In other words, $x$ is an element
of $\left\{  1,2,3,\ldots\right\}  $ and satisfies $\lambda_{x}\geq i$. From
$\lambda_{x}\geq i>0$, we obtain $\lambda_{x}\neq0$. Hence, $x$ is an element
of $\left\{  1,2,3,\ldots\right\}  $ and satisfies $\lambda_{x}\neq0$. In
other words, $x$ is an element $j$ of $\left\{  1,2,3,\ldots\right\}  $
satisfying $\lambda_{j}\neq0$.
\par
But $I$ is the support of $\lambda$. In other words, $I$ is the set of all
elements $j$ of $\left\{  1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}%
\neq0$. Thus, $x$ belongs to $I$ (since $x$ is an element $j$ of $\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}\neq0$). In other words, $x\in
I$. Thus, $\lambda_{x}\in\left\{  \lambda_{k}\ \mid\ k\in I\right\}  $.
Therefore, (\ref{pf.lem.sol.partition.dominance.lambdat-is-a-partition.up})
(applied to $p=\lambda_{x}$) yields $u\geq\lambda_{x}$. Thus, $i\geq
u+1>u\geq\lambda_{x}$. This contradicts $\lambda_{x}\geq i$.
\par
Now, forget that we fixed $x$. We thus have derived a contradiction for each
$x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i\right\}  $. Hence, there exists no $x\in\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq i\right\}  $. In other words, $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}
=\varnothing$.
\par
Now, (\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}) yields
$\left(  \lambda^{t}\right)  _{i}=\left\vert \underbrace{\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  }_{=\varnothing
}\right\vert =\left\vert \varnothing\right\vert =0$. This proves
(\ref{pf.lem.sol.partition.dominance.lambdat-is-a-partition.3}).}. Hence, the
support of the sequence $\lambda^{t}$ is a subset of $\left\{  1,2,\ldots
,u\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $i$ be an element of the
support of the sequence $\lambda^{t}$. We shall show that $i\in\left\{
1,2,\ldots,u\right\}  $.
\par
We know that $i$ belongs to the support of the sequence $\lambda^{t}$. In
other words, $i$ is an element $j$ of $\left\{  1,2,3,\ldots\right\}  $
satisfying $\left(  \lambda^{t}\right)  _{j}\neq0$ (since the support of the
sequence $\lambda^{t}$ is the set of all elements $j$ of $\left\{
1,2,3,\ldots\right\}  $ satisfying $\left(  \lambda^{t}\right)  _{j}\neq0$).
In other words, $i$ is an element of $\left\{  1,2,3,\ldots\right\}  $ and
satisfies $\left(  \lambda^{t}\right)  _{i}\neq0$.
\par
If we had $i\geq u+1$, then we would have $\left(  \lambda^{t}\right)  _{i}=0$
(by (\ref{pf.lem.sol.partition.dominance.lambdat-is-a-partition.3})), which
would contradict $\left(  \lambda^{t}\right)  _{i}\neq0$. Hence, we cannot
have $i\geq u+1$. Thus, we have $i<u+1$. Since $i$ and $u+1$ are integers,
this shows that $i\leq\left(  u+1\right)  -1=u$. In other words, $i\in\left\{
1,2,\ldots,u\right\}  $ (since $i\in\left\{  1,2,3,\ldots\right\}  $).
\par
Now, forget that we fixed $i$. We thus have shown that if $i$ is an element of
the support of the sequence $\lambda^{t}$, then $i\in\left\{  1,2,\ldots
,u\right\}  $. In other words, the support of the sequence $\lambda^{t}$ is a
subset of $\left\{  1,2,\ldots,u\right\}  $. Qed.}, and thus is finite (since
the set $\left\{  1,2,\ldots,u\right\}  $ is finite). Hence, the sequence
$\lambda^{t}$ is a weak composition. Since we also know that the sequence
$\lambda^{t}$ is weakly decreasing, we thus conclude that $\lambda^{t}$ is a
weakly decreasing weak composition. In other words, $\lambda^{t}$ is a
partition. This completes the proof of
Lemma~\ref{lem.sol.partition.dominance.lambdat-is-a-partition}.
\end{proof}

Let us now define the conjugate partition (from scratch, i.e., ignoring
Definition~\ref{def.partition.conjugate}):

\begin{definition}
\label{def.sol.partition.dominance.lambdat}Let $\lambda$ be a partition. Then,
$\lambda$ is a weak composition. Define a sequence $\lambda^{t}$ of
nonnegative integers by%
\[
\left(  \left(  \lambda^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }i\geq1\right)  .
\]
This sequence $\lambda^{t}$ is well-defined (by Lemma
\ref{lem.sol.partition.dominance.lambdat-is-a-partition}) and a partition
(again by Lemma \ref{lem.sol.partition.dominance.lambdat-is-a-partition}). We
shall refer to $\lambda^{t}$ as the \dfn{conjugate partition} of $\lambda$
(or as the \emph{conjugate} of the partition $\lambda$).
\end{definition}

The definition of $\lambda^{t}$ that we just gave is identical with the second
definition of $\lambda^{t}$ given in Definition~\ref{def.partition.conjugate}
(namely, the definition
via the equality \eqref{eq.partition.transpose}). In order to show that it is
also equivalent to the first definition of $\lambda^{t}$ given in
Definition~\ref{def.partition.conjugate},
we need to prove the following proposition:

\begin{proposition}
\label{prop.sol.partition.dominance.Y}Let $\lambda$ be a partition. Consider
the conjugate partition $\lambda^{t}$ defined in Definition
\ref{def.sol.partition.dominance.lambdat}. Let us denote by $Y\left(
\mu\right)  $ the Ferrers diagram of a partition $\mu$. Let $\left(
u,v\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. Then, $\left(
u,v\right)  \in Y\left(  \lambda^{t}\right)  $ holds if and only if $\left(
v,u\right)  \in Y\left(  \lambda\right)  $.
\end{proposition}

Visually speaking, Proposition \ref{prop.sol.partition.dominance.Y} says that
the cells of $Y\left(  \lambda^{t}\right)  $ are precisely the cells of
$Y\left(  \lambda\right)  $ reflected around the main diagonal\footnote{The
\emph{main diagonal}\index{main diagonal of a Ferrers diagram}
means the line passing through all cells of the form
$\left(  i,i\right)  $.}. In other words, the Ferrers diagram of $\lambda^{t}$
is obtained from that of $\lambda$ by exchanging rows for columns. Thus,
Proposition \ref{prop.sol.partition.dominance.Y} shows that the two
definitions of $\lambda^{t}$ given in Definition~\ref{def.partition.conjugate}
are equivalent.

\begin{proof}
[Proof of Proposition \ref{prop.sol.partition.dominance.Y}.]Recall that the
Ferrers diagram of $\lambda$ is defined as the set of all pairs $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq
\lambda_{i}$. In other words, the Ferrers diagram of $\lambda$ is $\left\{
\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq
\lambda_{i}\right\}  $. In other words, $Y\left(  \lambda\right)  $ is
$\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}%
\ \mid\ j\leq\lambda_{i}\right\}  $ (since $Y\left(  \lambda\right)  $ is the
Ferrers diagram of $\lambda$). In other words,%
\begin{equation}
Y\left(  \lambda\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  .
\label{pf.prop.sol.partition.dominance.Y.Ylambda=}%
\end{equation}
The same argument (applied to $\lambda^{t}$ instead of $\lambda$) yields%
\begin{equation}
Y\left(  \lambda^{t}\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\left(  \lambda^{t}\right)
_{i}\right\}  . \label{pf.prop.sol.partition.dominance.Y.Ylambdat=}%
\end{equation}


We have $\left(  u,v\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. In other
words, $u\in\left\{  1,2,3,\ldots\right\}  $ and $v\in\left\{  1,2,3,\ldots
\right\}  $. In other words, $v\in\left\{  1,2,3,\ldots\right\}  $ and
$u\in\left\{  1,2,3,\ldots\right\}  $. In other words, $\left(  v,u\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$.

The sequence $\lambda$ is a partition, and thus is weakly decreasing. In other
words, we have $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$. In other
words, if $u$ and $v$ are two positive integers satisfying $u\leq v$, then%
\begin{equation}
\lambda_{u}\geq\lambda_{v}. \label{pf.prop.sol.partition.dominance.Y.decrease}%
\end{equation}


The definition of $\lambda^{t}$ yields
\begin{equation}
\left(  \lambda^{t}\right)  _{u}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq u\right\}  \right\vert .
\label{pf.prop.sol.partition.dominance.Y.lambdatu=}%
\end{equation}


We shall now prove two claims:

\begin{statement}
\textit{Claim 1:} If $\left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  $,
then $\left(  v,u\right)  \in Y\left(  \lambda\right)  $.
\end{statement}

\begin{statement}
\textit{Claim 2:} If $\left(  v,u\right)  \in Y\left(  \lambda\right)  $, then
$\left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  $.
\end{statement}

\textit{Proof of Claim 1:} Assume that $\left(  u,v\right)  \in Y\left(
\lambda^{t}\right)  $. We must prove that $\left(  v,u\right)  \in Y\left(
\lambda\right)  $.

We have $\left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  =\left\{
\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\left(
\lambda^{t}\right)  _{i}\right\}  $. In other words, $\left(  u,v\right)  $ is
an element $\left(  i,j\right)  $ of $\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfying $j\leq\left(  \lambda^{t}\right)  _{i}$. In other words, $\left(
u,v\right)  $ is an element of $\left\{  1,2,3,\ldots\right\}  ^{2}$ and
satisfies $v\leq\left(  \lambda^{t}\right)  _{u}$. From $\left(  u,v\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$, we obtain $u\in\left\{
1,2,3,\ldots\right\}  $ and $v\in\left\{  1,2,3,\ldots\right\}  $.

But (\ref{pf.prop.sol.partition.dominance.Y.lambdatu=}) yields%
\begin{equation}
\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
u\right\}  \right\vert =\left(  \lambda^{t}\right)  _{u}\geq v
\label{pf.prop.sol.partition.dominance.Y.c1.1}%
\end{equation}
(since $v\leq\left(  \lambda^{t}\right)  _{u}$).

Now, assume (for the sake of contradiction) that $\lambda_{v}<u$. Then,
$\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
u\right\}  \subset \left\{  1,2,\ldots,v-1\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq u\right\}  $. We will show that
$x\in\left\{  1,2,\ldots,v-1\right\}  $.
\par
We have $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda
_{j}\geq u\right\}  $. In other words, $x$ is an element $j$ of $\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}\geq u$. In other words, $x$ is
an element of $\left\{  1,2,3,\ldots\right\}  $ and satisfies $\lambda_{x}\geq
u$. Thus, $\lambda_{x}\geq u>\lambda_{v}$ (since $\lambda_{v}<u$).
\par
If we had $v\leq x$, then we would have $\lambda_{v}\geq\lambda_{x}$ (by
(\ref{pf.prop.sol.partition.dominance.Y.decrease}) (applied to $v$ and $x$
instead of $u$ and $v$)); but this would contradict $\lambda_{x}>\lambda_{v}$.
Hence, we cannot have $v\leq x$. We thus must have $v>x$. Thus, $x<v$. Since
$x$ and $v$ are integers, this shows that $x\leq v-1$. Since $x\in\left\{
1,2,3,\ldots\right\}  $, this entails that $x\in\left\{  1,2,\ldots
,v-1\right\}  $.
\par
Now, forget that we fixed $x$. We thus have shown that $x\in\left\{
1,2,\ldots,v-1\right\}  $ for every $x\in\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq u\right\}  $. In other words, $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq u\right\}
\subset \left\{  1,2,\ldots,v-1\right\}  $, qed.}. Hence,%
\[
\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
u\right\}  \right\vert \leq\left\vert \left\{  1,2,\ldots,v-1\right\}
\right\vert =v-1<v.
\]
This contradicts (\ref{pf.prop.sol.partition.dominance.Y.c1.1}).

Thus, we have found a contradiction. Hence, our assumption (that $\lambda
_{v}<u$) was false. Hence, we have $\lambda_{v}\geq u$. Therefore,
$u\leq\lambda_{v}$.

Now, $\left(  v,u\right)  $ is an element of $\left\{  1,2,3,\ldots\right\}
^{2}$ and satisfies $u\leq\lambda_{v}$. In other words, $\left(  v,u\right)  $
is an element $\left(  i,j\right)  $ of $\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfying $j\leq\lambda_{i}$. In other words,%
\[
\left(  v,u\right)  \in\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  .
\]
In light of (\ref{pf.prop.sol.partition.dominance.Y.Ylambda=}), this rewrites
as $\left(  v,u\right)  \in Y\left(  \lambda\right)  $. This proves Claim 1.

\textit{Proof of Claim 2:} Assume that $\left(  v,u\right)  \in Y\left(
\lambda\right)  $. We must prove that $\left(  u,v\right)  \in Y\left(
\lambda^{t}\right)  $.

We have $\left(  v,u\right)  \in Y\left(  \lambda\right)  =\left\{  \left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda
_{i}\right\}  $. In other words, $\left(  v,u\right)  $ is an element $\left(
i,j\right)  $ of $\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying
$j\leq\lambda_{i}$. In other words, $\left(  v,u\right)  $ is an element of
$\left\{  1,2,3,\ldots\right\}  ^{2}$ and satisfies $u\leq\lambda_{v}$. From
$\left(  v,u\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$, we obtain
$v\in\left\{  1,2,3,\ldots\right\}  $ and $u\in\left\{  1,2,3,\ldots\right\}
$.

We have $\left\{  1,2,\ldots,v\right\}  \subset \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq u\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in\left\{  1,2,\ldots,v\right\}  $.
We shall show that $x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq u\right\}  $.
\par
We have $x\in\left\{  1,2,\ldots,v\right\}  \subset \left\{  1,2,3,\ldots
\right\}  $. Thus, $x$ is an element of $\left\{  1,2,3,\ldots\right\}  $.
\par
We have $x\in\left\{  1,2,\ldots,v\right\}  $ and thus $x\leq v$. Hence,
(\ref{pf.prop.sol.partition.dominance.Y.decrease}) (applied to $x$ instead of
$u$) yields $\lambda_{x}\geq\lambda_{v}\geq u$ (since $u\leq\lambda_{v}$).
Hence, $x$ is an element of $\left\{  1,2,3,\ldots\right\}  $ and satisfies
$\lambda_{x}\geq u$. In other words, $x$ is an element $j$ of $\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}\geq u$. In other words,
$x\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
u\right\}  $.
\par
Now, forget that we fixed $x$. We thus have proven that $x\in\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq u\right\}  $ for
each $x\in\left\{  1,2,\ldots,v\right\}  $. In other words, $\left\{
1,2,\ldots,v\right\}  \subset \left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq u\right\}  $, qed.}. Hence,%
\[
\left\vert \left\{  1,2,\ldots,v\right\}  \right\vert \leq\left\vert \left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq u\right\}
\right\vert =\left(  \lambda^{t}\right)  _{u}%
\]
(by (\ref{pf.prop.sol.partition.dominance.Y.lambdatu=})). Thus, $\left(
\lambda^{t}\right)  _{u}\geq\left\vert \left\{  1,2,\ldots,v\right\}
\right\vert =v$. In other words, $v\leq\left(  \lambda^{t}\right)  _{u}$.

Now, $\left(  u,v\right)  $ is an element of $\left\{  1,2,3,\ldots\right\}
^{2}$ and satisfies $v\leq\left(  \lambda^{t}\right)  _{u}$. In other words,
$\left(  u,v\right)  $ is an element $\left(  i,j\right)  $ of $\left\{
1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\left(  \lambda^{t}\right)  _{i}%
$. In other words,%
\[
\left(  u,v\right)  \in\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}\ \mid\ j\leq\left(  \lambda^{t}\right)  _{i}\right\}  .
\]
In light of (\ref{pf.prop.sol.partition.dominance.Y.Ylambdat=}), this rewrites
as $\left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  $. This proves Claim 2.

Now, both Claim 1 and Claim 2 are proven. Combining these two claims, we
conclude that $\left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  $ holds if
and only if $\left(  v,u\right)  \in Y\left(  \lambda\right)  $. This proves
Proposition \ref{prop.sol.partition.dominance.Y}.
\end{proof}

Next comes a simple lemma:

\begin{lemma}
\label{lem.sol.partition.dominance.Y=Y}Let us denote by $Y\left(  \mu\right)
$ the Ferrers diagram of a partition $\mu$. Let $\lambda$ and $\mu$ be two
partitions such that $Y\left(  \lambda\right)  =Y\left(  \mu\right)  $. Then,
$\lambda=\mu$.
\end{lemma}

Lemma \ref{lem.sol.partition.dominance.Y=Y} just says that if two partitions
have the same Ferrers diagram, then they are equal. In other words, a
partition is uniquely determined by its Ferrers diagram.

\begin{proof}
[Proof of Lemma \ref{lem.sol.partition.dominance.Y=Y}.]Recall that the Ferrers
diagram of $\lambda$ is defined as the set of all pairs $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$. In
other words, the Ferrers diagram of $\lambda$ is $\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $. In
other words, $Y\left(  \lambda\right)  $ is $\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $
(since $Y\left(  \lambda\right)  $ is the Ferrers diagram of $\lambda$). In
other words,%
\begin{equation}
Y\left(  \lambda\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  .
\label{pf.lem.sol.partition.dominance.Y=Y.1}%
\end{equation}


Let $\pi_{1}:\left\{  1,2,3,\ldots\right\}  ^{2}\rightarrow\left\{
1,2,3,\ldots\right\}  $ be the map that sends each $\left(  u,v\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ to $u\in\left\{  1,2,3,\ldots\right\}
$.

Fix $u\in\left\{  1,2,3,\ldots\right\}  $. We shall show that%
\begin{equation}
Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  =\left\{
u\right\}  \times\left\{  1,2,\ldots,\lambda_{u}\right\}  .
\label{pf.lem.sol.partition.dominance.Y=Y.cl1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.partition.dominance.Y=Y.cl1}):} Let $x\in
Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  $. Then,
\[
x\in Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  \subset
Y\left(  \lambda\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  .
\]
In other words, $x$ is an element $\left(  i,j\right)  $ of $\left\{
1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$. Consider this
$\left(  i,j\right)  $. Thus, $x=\left(  i,j\right)  $.

We have $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. Thus,
$i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in\left\{  1,2,3,\ldots\right\}
$.

We have $x\in Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)
\subset \pi_{1}^{-1}\left(  u\right)  $. In other words, $\pi_{1}\left(
x\right)  =u$.

The definition of $\pi_{1}$ yields $\pi_{1}\left(  i,j\right)  =i$. Now,
$\pi_{1}\left(  \underbrace{x}_{=\left(  i,j\right)  }\right)  =\pi_{1}\left(
i,j\right)  =i$. Comparing this with $\pi_{1}\left(  x\right)  =u$, we obtain
$i=u \in \left\{ u\right\}$.
But $j\leq\lambda_{i}=\lambda_{u}$ (since $i=u$). Since $j\in\left\{
1,2,3,\ldots\right\}  $, this entails $j\in\left\{  1,2,\ldots,\lambda
_{u}\right\}  $. Now, $x=\left(  i,j\right)  \in\left\{  u\right\}
\times\left\{  1,2,\ldots,\lambda_{u}\right\}  $ (since $i\in\left\{
u\right\}  $ and $j\in\left\{  1,2,\ldots,\lambda_{u}\right\}  $).

Now, forget that we fixed $x$. We thus have proven that every $x\in Y\left(
\lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  $ satisfies $x\in\left\{
u\right\}  \times\left\{  1,2,\ldots,\lambda_{u}\right\}  $. In other words,%
\begin{equation}
Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  \subset \left\{
u\right\}  \times\left\{  1,2,\ldots,\lambda_{u}\right\}  .
\label{pf.lem.sol.partition.dominance.Y=Y.cl1.pf.part1}%
\end{equation}


On the other hand, let $y\in\left\{  u\right\}  \times\left\{  1,2,\ldots
,\lambda_{u}\right\}  $. Thus, $y=\left(  p,q\right)  $ for some $p\in\left\{
u\right\}  $ and some $q\in\left\{  1,2,\ldots,\lambda_{u}\right\}  $.
Consider these $p$ and $q$.

We have $p\in\left\{  u\right\}  $, so that $p=u$. But $\pi_{1}\left(
\underbrace{y}_{=\left(  p,q\right)  }\right)  =\pi_{1}\left(  p,q\right)  =p$
(by the definition of $\pi_{1}$). Thus, $\pi_{1}\left(  y\right)  =p=u$, so
that $y\in\pi_{1}^{-1}\left(  u\right)  $.

But $q\in\left\{  1,2,\ldots,\lambda_{u}\right\}  $, so that $q\leq\lambda
_{u}=\lambda_{p}$ (since $u=p$ (since $p=u$)). Now, $\left(  p,q\right)  $ is
an element of $\left\{  1,2,3,\ldots\right\}  ^{2}$ (since $p=u\in\left\{
1,2,3,\ldots\right\}  $ and $q\in\left\{  1,2,\ldots,\lambda_{u}\right\}
\subset \left\{  1,2,3,\ldots\right\}  $) and satisfies $q\leq\lambda_{p}$.
In other words, $\left(  p,q\right)  $ is an element $\left(  i,j\right)  $ of
$\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$. In other
words,%
\[
\left(  p,q\right)  \in\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  .
\]
In view of (\ref{pf.lem.sol.partition.dominance.Y=Y.1}), this rewrites as
$\left(  p,q\right)  \in Y\left(  \lambda\right)  $. Thus, $y=\left(
p,q\right)  \in Y\left(  \lambda\right)  $. Combining this with $y\in\pi
_{1}^{-1}\left(  u\right)  $, we obtain $y\in Y\left(  \lambda\right)  \cap
\pi_{1}^{-1}\left(  u\right)  $.

Now, forget that we fixed $y$. We thus have proven that every $y\in\left\{
u\right\}  \times\left\{  1,2,\ldots,\lambda_{u}\right\}  $ satisfies $y\in
Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  $. In other words,%
\[
\left\{  u\right\}  \times\left\{  1,2,\ldots,\lambda_{u}\right\}  \subset
Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)  .
\]
Combining this with (\ref{pf.lem.sol.partition.dominance.Y=Y.cl1.pf.part1}),
we obtain $Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)
=\left\{  u\right\}  \times\left\{  1,2,\ldots,\lambda_{u}\right\}  $. This
proves (\ref{pf.lem.sol.partition.dominance.Y=Y.cl1}).]

Now, (\ref{pf.lem.sol.partition.dominance.Y=Y.cl1}) yields
\[
\left\vert Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(  u\right)
\right\vert =\left\vert \left\{  u\right\}  \times\left\{  1,2,\ldots
,\lambda_{u}\right\}  \right\vert =\underbrace{\left\vert \left\{  u\right\}
\right\vert }_{=1}\cdot\underbrace{\left\vert \left\{  1,2,\ldots,\lambda
_{u}\right\}  \right\vert }_{=\lambda_{u}}=1\cdot\lambda_{u}=\lambda_{u}.
\]
Thus,%
\[
\lambda_{u}=\left\vert Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(
u\right)  \right\vert .
\]


Now, forget that we fixed $u$. We thus have shown that%
\begin{equation}
\lambda_{u}=\left\vert Y\left(  \lambda\right)  \cap\pi_{1}^{-1}\left(
u\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \text{for every }u\in\left\{
1,2,3,\ldots\right\}  . \label{pf.lem.sol.partition.dominance.Y=Y.lambdau=}%
\end{equation}
The same argument (applied to $\mu$ instead of $\lambda$) yields%
\begin{equation}
\mu_{u}=\left\vert Y\left(  \mu\right)  \cap\pi_{1}^{-1}\left(  u\right)
\right\vert \ \ \ \ \ \ \ \ \ \ \text{for every }u\in\left\{  1,2,3,\ldots
\right\}  . \label{pf.lem.sol.partition.dominance.Y=Y.muu=}%
\end{equation}


Now, every $u\in\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align*}
\lambda_{u}  &  =\left\vert \underbrace{Y\left(  \lambda\right)  }_{=Y\left(
\mu\right)  }\cap\pi_{1}^{-1}\left(  u\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.partition.dominance.Y=Y.lambdau=})}\right) \\
&  =\left\vert Y\left(  \mu\right)  \cap\pi_{1}^{-1}\left(  u\right)
\right\vert =\mu_{u}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.partition.dominance.Y=Y.muu=})}\right)  .
\end{align*}
In other words, $\lambda=\mu$. This proves Lemma
\ref{lem.sol.partition.dominance.Y=Y}.
\end{proof}

The next proposition that we prove is the following.

\begin{proposition}
\label{prop.sol.partition.dominance.lambdatt}Let $\lambda$ be a partition.
Consider the conjugate partition $\lambda^{t}$ defined in Definition
\ref{def.sol.partition.dominance.lambdat}. Then, $\left(  \lambda^{t}\right)
^{t}=\lambda$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.partition.dominance.lambdatt}.]We first
state three claims:

\begin{statement}
\textit{Claim 1:} Let $\left(  u,v\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$. Then, we have the following logical equivalence:
\[
\left(  \left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  \right)
\ \Longleftrightarrow\ \left(  \left(  v,u\right)  \in Y\left(  \lambda
\right)  \right)  .
\]
\end{statement}

\begin{statement}
\textit{Claim 2:} Let $\left(  u,v\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$. Then, we have the following logical equivalence:
\[
\left(  \left(  u,v\right)  \in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  \right)  \ \Longleftrightarrow\ \left(  \left(  v,u\right)  \in
Y\left(  \lambda^{t}\right)  \right)  .
\]
\end{statement}

\begin{statement}
\textit{Claim 3:} Let $\left(  u,v\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$. Then, we have the following logical equivalence:
\[
\left(  \left(  u,v\right)  \in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  \right)  \ \Longleftrightarrow\ \left(  \left(  u,v\right)  \in
Y\left(  \lambda\right)  \right)  .
\]
\end{statement}

\textit{Proof of Claim 1:} Proposition \ref{prop.sol.partition.dominance.Y}
shows that $\left(  u,v\right)  \in Y\left(  \lambda^{t}\right)  $ holds if
and only if $\left(  v,u\right)  \in Y\left(  \lambda\right)  $. In other
words, we have the logical equivalence $\left(  \left(  u,v\right)  \in
Y\left(  \lambda^{t}\right)  \right)  \ \Longleftrightarrow\ \left(  \left(
v,u\right)  \in Y\left(  \lambda\right)  \right)  $. This proves Claim 1.

We thus have proven Claim 1. The same argument (applied to $\lambda^{t}$
instead of $\lambda$) yields Claim 2.

\textit{Proof of Claim 3:} We have $\left(  u,v\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$. In other words, $u\in\left\{  1,2,3,\ldots
\right\}  $ and $v\in\left\{  1,2,3,\ldots\right\}  $. In other words,
$v\in\left\{  1,2,3,\ldots\right\}  $ and $u\in\left\{  1,2,3,\ldots\right\}
$. In other words, $\left(  v,u\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$. Hence, Claim 1 (applied to $\left(  v,u\right)  $ instead of $\left(
u,v\right)  $) yields the following logical equivalence:%
\begin{equation}
\left(  \left(  v,u\right)  \in Y\left(  \lambda^{t}\right)  \right)
\ \Longleftrightarrow\ \left(  \left(  u,v\right)  \in Y\left(  \lambda
\right)  \right)  . \label{pf.prop.sol.partition.dominance.lambdatt.1}%
\end{equation}


Now, we have the following chain of logical equivalences:%
\begin{align*}
\left(  \left(  u,v\right)  \in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  \right)  \  &  \Longleftrightarrow\ \left(  \left(  v,u\right)
\in Y\left(  \lambda^{t}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Claim 2}\right) \\
&  \Longleftrightarrow\ \left(  \left(  u,v\right)  \in Y\left(
\lambda\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.partition.dominance.lambdatt.1})}\right)  .
\end{align*}
This proves Claim 3.

Now, recall that both $Y\left(  \lambda\right)  $ and $Y\left(  \left(
\lambda^{t}\right)  ^{t}\right)  $ are subsets of $\left\{  1,2,3,\ldots
\right\}  ^{2}$ (by the definition of a Ferrers diagram).

Now, we have $Y\left(  \lambda\right)  \subset Y\left(  \left(  \lambda
^{t}\right)  ^{t}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $x\in
Y\left(  \lambda\right)  $. Then, $x\in Y\left(  \lambda\right)  \subset
\left\{
1,2,3,\ldots\right\}  ^{2}$. In other words, $x$ has the form $\left(
u,v\right)  $ for some $\left(  u,v\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$. Consider this $\left(  u,v\right)  $. We have $\left(  u,v\right)
=x\in Y\left(  \lambda\right)  $. But Claim 3 shows that we have the following
logical equivalence: $\left(  \left(  u,v\right)  \in Y\left(  \left(
\lambda^{t}\right)  ^{t}\right)  \right)  \ \Longleftrightarrow\ \left(
\left(  u,v\right)  \in Y\left(  \lambda\right)  \right)  $. Therefore, we
have $\left(  u,v\right)  \in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  $ (since $\left(  u,v\right)  \in Y\left(  \lambda\right)  $).
Thus, $x=\left(  u,v\right)  \in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  $.
\par
Now, forget that we fixed $x$. We thus have shown that every $x\in Y\left(
\lambda\right)  $ satisfies $x\in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  $. In other words, $Y\left(  \lambda\right)  \subset Y\left(
\left(  \lambda^{t}\right)  ^{t}\right)  $. Qed.} and $Y\left(  \left(
\lambda^{t}\right)  ^{t}\right)  \subset Y\left(  \lambda\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $x\in Y\left(  \left(  \lambda
^{t}\right)  ^{t}\right)  $. Then, $x\in Y\left(  \left(  \lambda^{t}\right)
^{t}\right)
\subset \left\{  1,2,3,\ldots\right\}  ^{2}$. In other words, $x$ has
the form $\left(  u,v\right)  $ for some $\left(  u,v\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$. Consider this $\left(  u,v\right)  $. We have
$\left(  u,v\right)  =x\in Y\left(  \left(  \lambda^{t}\right)  ^{t}\right)
$. But Claim 3 shows that we have the following logical equivalence: $\left(
\left(  u,v\right)  \in Y\left(  \left(  \lambda^{t}\right)  ^{t}\right)
\right)  \ \Longleftrightarrow\ \left(  \left(  u,v\right)  \in Y\left(
\lambda\right)  \right)  $. Therefore, we have $\left(  u,v\right)  \in
Y\left(  \lambda\right)  $ (since $\left(  u,v\right)  \in Y\left(  \left(
\lambda^{t}\right)  ^{t}\right)  $). Thus, $x=\left(  u,v\right)  \in Y\left(
\lambda\right)  $.
\par
Now, forget that we fixed $x$. We thus have shown that every $x\in Y\left(
\left(  \lambda^{t}\right)  ^{t}\right)  $ satisfies $x\in Y\left(
\lambda\right)  $. In other words, $Y\left(  \left(  \lambda^{t}\right)
^{t}\right)  \subset Y\left(  \lambda\right)  $. Qed.}. Combining these two
inclusions, we obtain $Y\left(  \lambda\right)  =Y\left(  \left(  \lambda
^{t}\right)  ^{t}\right)  $. Thus, Lemma \ref{lem.sol.partition.dominance.Y=Y}
(applied to $\mu=\left(  \lambda^{t}\right)  ^{t}$) yields $\lambda=\left(
\lambda^{t}\right)  ^{t}$. This proves Proposition
\ref{prop.sol.partition.dominance.lambdatt}.
\end{proof}

Next, we show a further proposition that will come particularly handy in our
solution of Exercise \ref{exe.partition.dominance}:

\begin{proposition}
\label{prop.sol.partition.dominance.lemma}Let $\nu\in \Par $ and
$k\in \NN$. Then,%
\begin{equation}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{  \nu_{j},k\right\}  .
\label{sol.partition.dominance.lemma}%
\end{equation}
(Note that the right hand side of (\ref{sol.partition.dominance.lemma}) is
well-defined because every sufficiently high $j$ satisfies $\min\left\{
\nu_{j},k\right\}  =0$ (since every sufficiently high $j$ satisfies $\nu
_{j}=0$)).
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.partition.dominance.lemma}.]In the
following, we use the so-called \dfn{Iverson bracket notation}: For every
assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $ denote the
integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$. The definition of $\nu^{t}$ (in Definition
\ref{def.sol.partition.dominance.lambdat}) shows that
\begin{equation}
\left(  \nu^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \nu_{j}\geq i\right\}  \right\vert
\label{pf.prop.sol.partition.dominance.lemma.nuti=}
\end{equation}
for every positive integer $i$.

We have
\begin{align}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k} &  =\sum_{i=1}^{k}\underbrace{\left(  \nu^{t}\right)
_{i}}_{\substack{=\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \nu_{j}\geq i\right\}  \right\vert \\\text{(by
(\ref{pf.prop.sol.partition.dominance.lemma.nuti=}))}}}\nonumber\\
&  =\sum_{i=1}^{k}\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \nu_{j}\geq i\right\}  \right\vert
.\label{sol.partition.dominance.lemma.pf.0}%
\end{align}
But every $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies
\begin{align}
\sum_{j\in\left\{  1,2,3,\ldots\right\}  }\left[  \nu_{j}\geq i\right]   &
=\sum\limits_{\substack{j\in\left\{  1,2,3,\ldots\right\}  ;\\\nu_{j}\geq
i}}\underbrace{\left[  \nu_{j}\geq i\right]  }_{\substack{=1\\\text{(since
}\nu_{j}\geq i\text{)}}}+\sum\limits_{\substack{j\in\left\{  1,2,3,\ldots\right\}
;\\\text{we don't have }\nu_{j}\geq i}}\underbrace{\left[  \nu_{j}\geq
i\right]  }_{\substack{=0\\\text{(since we don't}\\\text{have }\nu_{j}\geq
i\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{j\in\left\{  1,2,3,\ldots\right\}  ;\\\nu_{j}\geq
i}}1+\underbrace{\sum\limits_{\substack{j\in\left\{  1,2,3,\ldots\right\}
;\\\text{we don't have }\nu_{j}\geq i}}0}_{=0}=\sum\limits_{\substack{j\in\left\{
1,2,3,\ldots\right\}  ;\\\nu_{j}\geq i}}1\nonumber\\
&  =\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \nu_{j}\geq
i\right\}  \right\vert \cdot1=\left\vert \left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \nu_{j}\geq i\right\}  \right\vert
.\label{sol.partition.dominance.lemma.pf.1}%
\end{align}
(This proves, in particular, that the sum $\sum_{j\in\left\{  1,2,3,\ldots
\right\}  }\left[  \nu_{j}\geq i\right]  $ is well-defined for every
$i\in\left\{  1,2,\ldots,k\right\}  $.) Now,
(\ref{sol.partition.dominance.lemma.pf.0}) becomes
\begin{align}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k} &  =\sum_{i=1}^{k}\underbrace{\left\vert \left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \nu_{j}\geq i\right\}  \right\vert
}_{\substack{=\sum_{j\in\left\{  1,2,3,\ldots\right\}  }\left[  \nu_{j}\geq
i\right]  \\\text{(by (\ref{sol.partition.dominance.lemma.pf.1}))}%
}}\nonumber\\
&  =\underbrace{\sum_{i=1}^{k}\sum_{j\in\left\{  1,2,3,\ldots\right\}  }%
}_{=\sum_{j\in\left\{  1,2,3,\ldots\right\}  }\sum_{i=1}^{k}}\left[  \nu
_{j}\geq i\right]  =\sum_{j\in\left\{  1,2,3,\ldots\right\}  }\sum_{i=1}%
^{k}\left[  \nu_{j}\geq i\right]  .\label{sol.partition.dominance.lemma.pf.2}%
\end{align}
But every $j\in\left\{  1,2,3,\ldots\right\}  $ satisfies
\begin{equation}
\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  =\min\left\{  \nu_{j},k\right\}
\label{sol.partition.dominance.lemma.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.partition.dominance.lemma.pf.3}):} Let
$j\in\left\{  1,2,3,\ldots\right\}  $. We need to prove that
(\ref{sol.partition.dominance.lemma.pf.3}) holds. We distinguish between two
cases:
\par
\textit{Case 1:} We have $\nu_{j}\geq k$.
\par
\textit{Case 2:} We have $\nu_{j}<k$.
\par
Let us first consider Case 1. In this case, we have $\nu_{j}\geq k$. Thus,
\[
\sum_{i=1}^{k}\underbrace{\left[  \nu_{j}\geq i\right]  }%
_{\substack{=1\\\text{(since }\nu_{j}\geq k\geq i\\\text{(because }i\leq
k\\\text{(since }i\in\left\{  1,2,\ldots,k\right\}  \text{)))}}}=\sum
_{i=1}^{k}1=k\cdot1=k.
\]
Compared with $\min\left\{  \nu_{j},k\right\}  =k$ (since $\nu_{j}\geq k$),
this yields $\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  =\min\left\{  \nu
_{j},k\right\}  $. Thus, (\ref{sol.partition.dominance.lemma.pf.3}) is proven
in Case 1.
\par
Let us now consider Case 2. In this case, we have $\nu_{j}<k$. Thus, $\nu
_{j}\leq k$. Hence, $\nu_{j}\in\left\{  0,1,\ldots,k\right\}  $, so that we
can split the sum $\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  $ at $i=\nu
_{j}$. We thus obtain
\begin{align*}
\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]   &  =\sum_{i=1}^{\nu_{j}%
}\underbrace{\left[  \nu_{j}\geq i\right]  }_{\substack{=1\\\text{(since }%
\nu_{j}\geq i\\\text{(because }i\leq\nu_{j}\\\text{(since }i\in\left\{
1,2,\ldots,\nu_{j}\right\}  \text{)))}}}+\sum_{i=\nu_{j}+1}^{k}%
\underbrace{\left[  \nu_{j}\geq i\right]  }_{\substack{=0\\\text{(since we
don't have }\nu_{j}\geq i\\\text{(because }i>\nu_{j}\\\text{(since }%
i\in\left\{  \nu_{j}+1,\nu_{j}+2,\ldots,k\right\}  \text{)))}}}\\
&  =\sum_{i=1}^{\nu_{j}}1+\underbrace{\sum_{i=\nu_{j}+1}^{k}0}_{=0}=\sum
_{i=1}^{\nu_{j}}1=\nu_{j}\cdot1=\nu_{j}.
\end{align*}
Compared with $\min\left\{  \nu_{j},k\right\}  =\nu_{j}$ (since $\nu_{j}<k$),
this yields $\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  =\min\left\{  \nu
_{j},k\right\}  $. Thus, (\ref{sol.partition.dominance.lemma.pf.3}) is proven
in Case 2.
\par
We have thus proven (\ref{sol.partition.dominance.lemma.pf.3}) in both Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
(\ref{sol.partition.dominance.lemma.pf.3}) always holds, qed.}. Hence,
(\ref{sol.partition.dominance.lemma.pf.2}) becomes
\begin{align*}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k} &  =\sum_{j\in\left\{  1,2,3,\ldots\right\}
}\underbrace{\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  }_{\substack{=\min
\left\{  \nu_{j},k\right\}  \\\text{(by
(\ref{sol.partition.dominance.lemma.pf.3}))}}}\\
&  =\underbrace{\sum_{j\in\left\{  1,2,3,\ldots\right\}  }}_{=\sum
_{j=1}^{\infty}}\min\left\{  \nu_{j},k\right\}  =\sum_{j=1}^{\infty}%
\min\left\{  \nu_{j},k\right\}  .
\end{align*}
This proves Proposition \ref{prop.sol.partition.dominance.lemma}.
\end{proof}

Let us now come to the solution of Exercise \ref{exe.partition.dominance}.

\begin{proof}
[Solution to Exercise \ref{exe.partition.dominance}.]We need to prove that
\begin{equation}
\left(  \text{if }\lambda\triangleright\mu\text{, then }\mu^{t}\triangleright
\lambda^{t}\right)  \label{sol.partition.dominance.goal1}%
\end{equation}
and
\begin{equation}
\left(  \text{if }\mu^{t}\triangleright\lambda^{t}\text{, then }%
\lambda\triangleright\mu\right)  .\label{sol.partition.dominance.goal2}%
\end{equation}


\textit{Proof of (\ref{sol.partition.dominance.goal2}):} Assume that $\mu
^{t}\triangleright\lambda^{t}$. By the definition of the dominance order, we
have $\mu^{t}\triangleright\lambda^{t}$ if and only if we have
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}%
\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
.\label{sol.partition.dominance.known}%
\end{equation}


Since we have assumed that $\mu^{t}\triangleright\lambda^{t}$, we thus
conclude that (\ref{sol.partition.dominance.known}) holds.

We can slightly strengthen this result: We have
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}%
\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  0,1,\ldots,n\right\}
.\label{sol.partition.dominance.known.better}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.partition.dominance.known.better}):} Let
$k\in\left\{  0,1,\ldots,n\right\}  $. We need to show that
(\ref{sol.partition.dominance.known.better}) holds.
\par
If $k\in\left\{  1,2,\ldots,n\right\}  $, then
(\ref{sol.partition.dominance.known.better}) follows immediately from
(\ref{sol.partition.dominance.known}). Hence, for the rest of the proof of
(\ref{sol.partition.dominance.known.better}), we can WLOG assume that we don't
have $k\in\left\{  1,2,\ldots,n\right\}  $. Assume this.
\par
We don't have $k\in\left\{  1,2,\ldots,n\right\}  $. Hence, we have
$k\notin\left\{  1,2,\ldots,n\right\}  $. Combined with $k\in\left\{
0,1,\ldots,n\right\}  $, this yields $k\in\left\{  0,1,\ldots,n\right\}
\setminus\left\{  1,2,\ldots,n\right\}  =\left\{  0\right\}  $. Thus, $k=0$.
Hence,
\[
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}=\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)
_{2}+\cdots+\left(  \mu^{t}\right)  _{0}=\left(  \text{empty sum}\right)  =0
\]
and
\[
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}=\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)
_{0}=\left(  \text{empty sum}\right)  =0.
\]
Therefore,
\[
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}=0\geq0=\left(  \lambda^{t}\right)  _{1}+\left(
\lambda^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}%
\]
(since $\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\cdots+\left(  \lambda^{t}\right)  _{k}=0$). This proves
(\ref{sol.partition.dominance.known.better}).} We can rewrite this further: We
have
\begin{equation}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}  \geq\sum_{j=1}^{\infty}%
\min\left\{  \lambda_{j},k\right\}  \ \ \ \ \ \ \ \ \ \ \text{for all }%
k\in\left\{  0,1,\ldots,n\right\}  \label{sol.partition.dominance.known.rewr}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.partition.dominance.known.rewr}):} Let
$k\in\left\{  0,1,\ldots,n\right\}  $. We have $\lambda \in \Par_n
\subset \Par$. Applying Proposition
\ref{prop.sol.partition.dominance.lemma} to $\nu=\lambda$, we obtain
\[
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},k\right\}  .
\]
\par
But we also have $\mu \in \Par_n \subset \Par$. Applying Proposition
\ref{prop.sol.partition.dominance.lemma} to $\nu=\mu$, we obtain
\[
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+..\cdots+\left(
\mu^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}  .
\]
Hence,
\begin{align*}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}   &  =\left(  \mu
^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)
_{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\cdots+\left(  \lambda^{t}\right)  _{k}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.partition.dominance.known.better})}\right)  \\
&  =\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},k\right\}  .
\end{align*}
This proves (\ref{sol.partition.dominance.known.rewr}).}.

Now, let $k\in\left\{  1,2,\ldots,n\right\}  $ be arbitrary. We are going to
show that $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$.

Since $\lambda\in\operatorname{Par}_{n}$, we have $\left\vert \lambda
\right\vert =n$, so that $n=\left\vert \lambda\right\vert =\lambda_{1}%
+\lambda_{2}+\lambda_{3}+\cdots\geq\lambda_{k}$ (since $\lambda_{k}$ is an
addend in the sum $\lambda_{1}+\lambda_{2}+\lambda_{3}+\cdots$). Thus,
$\lambda_{k}\in\left\{  0,1,\ldots,n\right\}  $. Therefore,
(\ref{sol.partition.dominance.known.rewr}) (applied to $\lambda_{k}$ instead
of $k$) yields that
\begin{equation}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda_{k}\right\}  \geq\sum
_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda_{k}\right\}
.\label{sol.partition.dominance.known.ineq}%
\end{equation}


Now, we have $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$ (since
$\lambda$ is a partition). Hence, every $j\in\left\{  1,2,\ldots,k\right\}  $
satisfies $\min\left\{  \lambda_{j},\lambda_{k}\right\}  =\lambda_{k}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,k\right\}  $.
Then, $j\leq k$, so that $\lambda_{j}\geq\lambda_{k}$ (since $\lambda_{1}%
\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$). Thus, $\min\left\{  \lambda
_{j},\lambda_{k}\right\}  =\lambda_{k}$, qed.}. Consequently, $\sum_{j=1}%
^{k}\underbrace{\min\left\{  \lambda_{j},\lambda_{k}\right\}  }_{=\lambda_{k}%
}=\sum_{j=1}^{k}\lambda_{k}=k\lambda_{k}$. Also, every $j\in\left\{
k+1,k+2,k+3,\ldots\right\}  $ satisfies $\min\left\{  \lambda_{j},\lambda
_{k}\right\}  =\lambda_{j}$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{
k+1,k+2,k+3,\ldots\right\}  $. Then, $j>k$, so that $k<j$. Hence, $\lambda
_{k}\geq\lambda_{j}$ (since $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}%
\geq\cdots$). Thus, $\min\left\{  \lambda_{j},\lambda_{k}\right\}
=\lambda_{j}$, qed.}. Therefore,
\begin{align*}
\sum_{j=k+1}^{\infty}\underbrace{\min\left\{  \lambda_{j},\lambda_{k}\right\}
}_{=\lambda_{j}} &  =\sum_{j=k+1}^{\infty}\lambda_{j}=\underbrace{\sum
_{j=1}^{\infty}\lambda_{j}}_{\substack{=\lambda_{1}+\lambda_{2}+\lambda
_{3}+\cdots\\=\left\vert \lambda\right\vert =n\\\text{(since }\lambda
\in\operatorname{Par}_{n}\text{)}}}-\underbrace{\sum_{j=1}^{k}\lambda_{j}%
}_{=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sum_{j=1}^{\infty}\lambda_{j}=\sum_{j=1}^{k}\lambda_{j}%
+\sum_{j=k+1}^{\infty}\lambda_{j}\right)  \\
&  =n-\left(  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\right)  .
\end{align*}
Now,
\begin{align}
\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda_{k}\right\}   &
=\underbrace{\sum_{j=1}^{k}\min\left\{  \lambda_{j},\lambda_{k}\right\}
}_{=k\lambda_{k}}+\underbrace{\sum_{j=k+1}^{\infty}\min\left\{  \lambda
_{j},\lambda_{k}\right\}  }_{=n-\left(  \lambda_{1}+\lambda_{2}+\cdots
+\lambda_{k}\right)  }\nonumber\\
&  =k\lambda_{k}+n-\left(  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\right)
.\label{sol.partition.dominance.p1}%
\end{align}


On the other hand, any two integers $a$ and $b$ satisfy $\min\left\{
a,b\right\}  \leq b$. Hence, every $j\in\left\{  1,2,\ldots,k\right\}  $
satisfies $\min\left\{  \mu_{j},\lambda_{k}\right\}  \leq\lambda_{k}$ (by the
inequality $\min\left\{  a,b\right\}  \leq b$, applied to $a=\mu_{j}$ and
$b=\lambda_{k}$). Therefore, $\sum_{j=1}^{k}\underbrace{\min\left\{  \mu
_{j},\lambda_{k}\right\}  }_{\leq\lambda_{k}}\leq\sum_{j=1}^{k}\lambda
_{k}=k\lambda_{k}$. Also, any two integers $a$ and $b$ satisfy $\min\left\{
a,b\right\}  \leq a$. Hence, every $j\in\left\{  k+1,k+2,k+3,\ldots\right\}  $
satisfies $\min\left\{  \mu_{j},\lambda_{k}\right\}  \leq\mu_{j}$ (by the
inequality $\min\left\{  a,b\right\}  \leq a$, applied to $a=\mu_{j}$ and
$b=\lambda_{k}$). Therefore,
\begin{align*}
\sum_{j=k+1}^{\infty}\underbrace{\min\left\{  \mu_{j},\lambda_{k}\right\}
}_{\leq\mu_{j}} &  \leq\sum_{j=k+1}^{\infty}\mu_{j}=\underbrace{\sum
_{j=1}^{\infty}\mu_{j}}_{\substack{=\mu_{1}+\mu_{2}+\mu_{3}+\cdots
\\=\left\vert \mu\right\vert =n\\\text{(since }\mu\in\operatorname{Par}%
_{n}\text{)}}}-\underbrace{\sum_{j=1}^{k}\mu_{j}}_{=\mu_{1}+\mu_{2}+\cdots
+\mu_{k}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{j=1}^{\infty}\mu
_{j}=\sum_{j=1}^{k}\mu_{j}+\sum_{j=k+1}^{\infty}\mu_{j}\right)  \\
&  =n-\left(  \mu_{1}+\mu_{2}+\cdots+\mu_{k}\right)  .
\end{align*}
Now,
\begin{align}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda_{k}\right\}   &
=\underbrace{\sum_{j=1}^{k}\min\left\{  \mu_{j},\lambda_{k}\right\}  }_{\leq
k\lambda_{k}}+\underbrace{\sum_{j=k+1}^{\infty}\min\left\{  \mu_{j}%
,\lambda_{k}\right\}  }_{\leq n-\left(  \mu_{1}+\mu_{2}+\cdots+\mu_{k}\right)
}\nonumber\\
&  \leq k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}+\cdots+\mu_{k}\right)
.\label{sol.partition.dominance.p2}%
\end{align}


Now, (\ref{sol.partition.dominance.known.ineq}) yields $\sum_{j=1}^{\infty
}\min\left\{  \mu_{j},\lambda_{k}\right\}  -\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},\lambda_{k}\right\}  \geq0$. Thus,
\begin{align*}
0 &  \leq\underbrace{\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda
_{k}\right\}  }_{\substack{\leq k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}%
+\cdots+\mu_{k}\right)  \\\text{(by (\ref{sol.partition.dominance.p2}))}%
}}-\underbrace{\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda
_{k}\right\}  }_{\substack{=k\lambda_{k}+n-\left(  \lambda_{1}+\lambda
_{2}+\cdots+\lambda_{k}\right)  \\\text{(by (\ref{sol.partition.dominance.p1}%
))}}}\\
&  \leq\left(  k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}+\cdots+\mu_{k}\right)
\right)  -\left(  k\lambda_{k}+n-\left(  \lambda_{1}+\lambda_{2}%
+\cdots+\lambda_{k}\right)  \right)  \\
&  =\left(  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\right)  -\left(
\mu_{1}+\mu_{2}+\cdots+\mu_{k}\right)  .
\end{align*}
In other words, $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$.

Now, forget that we fixed $k$. We thus have shown that
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\text{.}%
\]


Now, recall that we have $\lambda\triangleright\mu$ if and only if we have
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\]
(according to the definition of the dominance order). Thus, we have
$\lambda\triangleright\mu$ (because we know that we have
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\]
). This proves (\ref{sol.partition.dominance.goal2}).

\textit{Proof of (\ref{sol.partition.dominance.goal1}):} Assume that
$\lambda\triangleright\mu$. We have $\mu^{t}\in\operatorname{Par}_{n}$ and
$\lambda^{t}\in\operatorname{Par}_{n}$. Also, Proposition
\ref{prop.sol.partition.dominance.lambdatt} yields $\left(  \lambda
^{t}\right)  ^{t}=\lambda$. Furthermore, Proposition
\ref{prop.sol.partition.dominance.lambdatt} (applied to $\mu$ instead of
$\lambda$) yields $\left(  \mu^{t}\right)  ^{t}=\mu$. Now, $\left(
\lambda^{t}\right)  ^{t}=\lambda\triangleright\mu=\left(  \mu^{t}\right)
^{t}$. Thus, we can apply (\ref{sol.partition.dominance.goal2}) to $\mu^{t}$
and $\lambda^{t}$ instead of $\lambda$ and $\mu$. As a result, we obtain
$\mu^{t}\triangleright\lambda^{t}$. This proves
(\ref{sol.partition.dominance.goal1}).

Now, both (\ref{sol.partition.dominance.goal1}) and
(\ref{sol.partition.dominance.goal2}) are proven. In other words, we have
$\lambda\triangleright\mu$ if and only if $\mu^{t}\triangleright\lambda^{t}$.
This solves Exercise~\ref{exe.partition.dominance}.
\end{proof}

\begin{commentedout}
What follows is the OLD solution of Exercise~\ref{exe.partition.dominance}.
It is not wrong, but (unlike the new solution) it does not include proofs
of the basic properties of conjugate partitions (such as
$\left(\lambda^t\right)^t = \lambda$).

\textit{Solution to Exercise~\ref{exe.partition.dominance}.} Let us
first make an auxiliary observation. Namely, let us show that every $\nu
\in\Par_{n}$ and every $k\in\NN$ satisfy
\begin{equation}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+...+\left(  \nu
^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{  \nu_{j},k\right\}  .
\label{sol.partition.dominance.lemma}
\end{equation}
(Note that the right hand side of (\ref{sol.partition.dominance.lemma}) is
well-defined because every sufficiently high $j$ satisfies $\min\left\{
\nu_{j},k\right\}  =0$ (since every sufficiently high $j$ satisfies $\nu
_{j}=0$)).

\textit{Proof of (\ref{sol.partition.dominance.lemma}):} In the following, we
use the so-called \dfn{Iverson bracket notation}: For every assertion
$\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $ denote the integer
$\begin{cases}
1, & \text{ if }\mathcal{A}\text{ is true;}\\
0, & \text{ if }\mathcal{A}\text{ is false}
\end{cases} \quad$. We have
\begin{align}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+...+\left(  \nu
^{t}\right)  _{k} 
&  =\sum_{i=1}^{k}\underbrace{\left(  \nu^{t}\right) _{i}}
_{\substack{=\left\vert \left\{  j\ :\ \nu_{j}\geq i\right\}  \right\vert
\\\text{(by (\ref{eq.partition.transpose}), applied to}\\\nu\text{ instead of
}\lambda\text{)}}}=\sum_{i=1}^{k}\underbrace{\left\vert \left\{  j\ :\ \nu
_{j}\geq i\right\}  \right\vert }_{\substack{=\left\vert \left\{  j\ \mid
\ \nu_{j}\geq i\right\}  \right\vert \\=\left\vert \left\{  j\in\left\{
1,2,3,...\right\}  \ \mid\ \nu_{j}\geq i\right\}  \right\vert }}\nonumber\\
&  =\sum_{i=1}^{k}\left\vert \left\{  j\in\left\{  1,2,3,...\right\}
\ \mid\ \nu_{j}\geq i\right\}  \right\vert .
\label{sol.partition.dominance.lemma.pf.0}
\end{align}
But every $i\in\left\{  1,2,...,k\right\}  $ satisfies
\begin{align}
\sum_{j\in\left\{  1,2,3,...\right\}  }\left[  \nu_{j}\geq i\right]   &
=\sum\limits_{\substack{j\in\left\{  1,2,3,...\right\}  ;\\\nu_{j}\geq i}}
\underbrace{\left[  \nu_{j}\geq i\right]  }_{\substack{=1\\\text{(since }
\nu_{j}\geq i\text{)}}}+\sum\limits_{\substack{j\in\left\{  1,2,3,...\right\}
;\\\text{we don't have }\nu_{j}\geq i}}\underbrace{\left[  \nu_{j}\geq
i\right]  }_{\substack{=0\\\text{(since we don't}\\\text{have }\nu_{j}\geq
i\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{j\in\left\{  1,2,3,...\right\}  ;\\\nu_{j}\geq
i}}1+\underbrace{\sum\limits_{\substack{j\in\left\{  1,2,3,...\right\}  ;\\\text{we
don't have }\nu_{j}\geq i}}0}_{=0}=\sum\limits_{\substack{j\in\left\{
1,2,3,...\right\}  ;\\\nu_{j}\geq i}}1\nonumber\\
&  =\left\vert \left\{  j\in\left\{  1,2,3,...\right\}  \ \mid\ \nu_{j}\geq
i\right\}  \right\vert \cdot1=\left\vert \left\{  j\in\left\{
1,2,3,...\right\}  \ \mid\ \nu_{j}\geq i\right\}  \right\vert .
\label{sol.partition.dominance.lemma.pf.1}
\end{align}
(This proves, in particular, that the sum $\sum_{j\in\left\{
1,2,3,...\right\}  }\left[  \nu_{j}\geq i\right]  $ is well-defined for every
$i\in\left\{  1,2,...,k\right\}  $). Now,
(\ref{sol.partition.dominance.lemma.pf.0}) becomes
\begin{align}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+...+\left(  \nu
^{t}\right)  _{k}  &  =\sum_{i=1}^{k}\underbrace{\left\vert \left\{
j\in\left\{  1,2,3,...\right\}  \ \mid\ \nu_{j}\geq i\right\}  \right\vert
}_{\substack{=\sum_{j\in\left\{  1,2,3,...\right\}  }\left[  \nu_{j}\geq
i\right]  \\\text{(by (\ref{sol.partition.dominance.lemma.pf.1}))}}}
\nonumber\\
&  =\underbrace{\sum_{i=1}^{k}\sum_{j\in\left\{  1,2,3,...\right\}  }}
_{=\sum_{j\in\left\{  1,2,3,...\right\}  }\sum_{i=1}^{k}}\left[  \nu_{j}\geq
i\right]  =\sum_{j\in\left\{  1,2,3,...\right\}  }\sum_{i=1}^{k}\left[
\nu_{j}\geq i\right]  . \label{sol.partition.dominance.lemma.pf.2}
\end{align}
But every $j\in\left\{  1,2,3,...\right\}  $ satisfies
\begin{equation}
\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  =\min\left\{  \nu_{j},k\right\}
\label{sol.partition.dominance.lemma.pf.3}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.partition.dominance.lemma.pf.3}):} Let
$j\in\left\{  1,2,3,...\right\}  $. We need to prove that
(\ref{sol.partition.dominance.lemma.pf.3}) holds. We distinguish between two
cases:
\par
\textit{Case 1:} We have $\nu_{j}\geq k$.
\par
\textit{Case 2:} We have $\nu_{j}<k$.
\par
Let us first consider Case 1. In this case, we have $\nu_{j}\geq k$. Thus,
\[
\sum_{i=1}^{k}\underbrace{\left[  \nu_{j}\geq i\right]  }
_{\substack{=1\\\text{(since }\nu_{j}\geq k\geq i\\\text{(because }i\leq
k\\\text{(since }i\in\left\{  1,2,...,k\right\}  \text{)))}}}
=\sum_{i=1}^{k}1=k\cdot1=k.
\]
Compared with $\min\left\{  \nu_{j},k\right\}  =k$ (since $\nu_{j}\geq k$),
this yields $\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  =\min\left\{  \nu
_{j},k\right\}  $. Thus, (\ref{sol.partition.dominance.lemma.pf.3}) is proven
in Case 1.
\par
Let us now consider Case 2. In this case, we have $\nu_{j}<k$. Thus, $\nu
_{j}\in\left\{  0,1,...,k\right\}  $, so that we can split the sum $\sum
_{i=1}^{k}\left[  \nu_{j}\geq i\right]  $ at $i=\nu_{j}$. We thus obtain
\begin{align*}
\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]   &  =\sum_{i=1}^{\nu_{j}}
\underbrace{\left[  \nu_{j}\geq i\right]  }_{\substack{=1\\\text{(since }
\nu_{j}\geq i\\\text{(because }i\leq\nu_{j}\\\text{(since }i\in\left\{
1,2,...,\nu_{j}\right\}  \text{)))}}}+\sum_{i=\nu_{j}+1}^{k}
\underbrace{\left[  \nu_{j}\geq i\right]  }_{\substack{=0\\\text{(since we
don't have }\nu_{j}\geq i\\\text{(because }i>\nu_{j}\\\text{(since }
i\in\left\{  \nu_{j}+1,\nu_{j}+2,...,k\right\}  \text{)))}}}\\
&  =\sum_{i=1}^{\nu_{j}}1+\underbrace{\sum_{i=\nu_{j}+1}^{k}0}_{=0}=\sum
_{i=1}^{\nu_{j}}1=\nu_{j}\cdot1=\nu_{j}.
\end{align*}
Compared with $\min\left\{  \nu_{j},k\right\}  =\nu_{j}$ (since $\nu_{j}<k$),
this yields $\sum_{i=1}^{k}\left[  \nu_{j}\geq i\right]  =\min\left\{  \nu
_{j},k\right\}  $. Thus, (\ref{sol.partition.dominance.lemma.pf.3}) is proven
in Case 2.
\par
We have thus proven (\ref{sol.partition.dominance.lemma.pf.3}) in both Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
(\ref{sol.partition.dominance.lemma.pf.3}) always holds, qed.}. Hence,
(\ref{sol.partition.dominance.lemma.pf.2}) becomes
\begin{align*}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+...+\left(  \nu
^{t}\right)  _{k}  &  =\sum_{j\in\left\{  1,2,3,...\right\}  }\underbrace{\sum
_{i=1}^{k}\left[  \nu_{j}\geq i\right]  }_{\substack{=\min\left\{  \nu
_{j},k\right\}  \\\text{(by (\ref{sol.partition.dominance.lemma.pf.3}))}}}\\
&  =\underbrace{\sum_{j\in\left\{  1,2,3,...\right\}  }}_{=\sum_{j=1}^{\infty
}}\min\left\{  \nu_{j},k\right\}  =\sum_{j=1}^{\infty}\min\left\{  \nu
_{j},k\right\}  .
\end{align*}
This proves (\ref{sol.partition.dominance.lemma}).

Let us now come to the solution of the exercise. We need to prove that
\begin{equation}
\left(  \text{if }\lambda\triangleright\mu\text{, then }\mu^{t}\triangleright
\lambda^{t}\right)
\label{sol.partition.dominance.goal1}
\end{equation}
and
\begin{equation}
\left(  \text{if }\mu^{t}\triangleright\lambda^{t}\text{, then }
\lambda\triangleright\mu\right).
\label{sol.partition.dominance.goal2}
\end{equation}


\textit{Proof of (\ref{sol.partition.dominance.goal2}):} Assume that $\mu
^{t}\triangleright\lambda^{t}$. By the definition of the dominance order, we
have $\mu^{t}\triangleright\lambda^{t}$ if and only if we have
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu
^{t}\right)  _{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+...+\left(  \lambda^{t}\right)  _{k}
\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,...,n\right\}  .
\label{sol.partition.dominance.known}
\end{equation}
Since we have assumed that $\mu^{t}\triangleright\lambda^{t}$, we thus
conclude that (\ref{sol.partition.dominance.known}) holds.

We can slightly strengthen this result: We have
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu
^{t}\right)  _{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+...+\left(  \lambda^{t}\right)  _{k}
\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  0,1,...,n\right\}
.\label{sol.partition.dominance.known.better}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.partition.dominance.known.better}):} Let
$k\in\left\{  0,1,...,n\right\}  $. We need to show that
(\ref{sol.partition.dominance.known.better}) holds.
\par
If $k\in\left\{  1,2,...,n\right\}  $, then
(\ref{sol.partition.dominance.known.better}) follows immediately from
(\ref{sol.partition.dominance.known}). Hence, for the rest of the proof of
(\ref{sol.partition.dominance.known.better}), we can WLOG assume that we don't
have $k\in\left\{  1,2,...,n\right\}  $. Assume this.
\par
We don't have $k\in\left\{  1,2,...,n\right\}  $. Hence, we have
$k\notin\left\{  1,2,...,n\right\}  $. Combined with $k\in\left\{
0,1,...,n\right\}  $, this yields $k\in\left\{  0,1,...,n\right\}
\setminus\left\{  1,2,...,n\right\}  =\left\{  0\right\}  $. Thus, $k=0$.
Hence,
\[
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu
^{t}\right)  _{k}=\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)
_{2}+...+\left(  \mu^{t}\right)  _{0}=\left(  \text{empty sum}\right)  =0
\]
and
\[
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+...+\left(
\lambda^{t}\right)  _{k}=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+...+\left(  \lambda^{t}\right)  _{0}=\left(  \text{empty
sum}\right)  =0.
\]
Therefore,
\[
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu
^{t}\right)  _{k}=0\geq0=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+...+\left(  \lambda^{t}\right)  _{k}
\]
(since $\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+...+\left(  \lambda^{t}\right)  _{k}=0$). This proves
(\ref{sol.partition.dominance.known.better}).} We can rewrite this further: We
have
\begin{equation}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}  \geq\sum_{j=1}^{\infty}
\min\left\{  \lambda_{j},k\right\}  \ \ \ \ \ \ \ \ \ \ \text{for all }
k\in\left\{  0,1,...,n\right\}  .\label{sol.partition.dominance.known.rewr}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.partition.dominance.known.rewr}):} Let
$k\in\left\{  0,1,...,n\right\}  $. Applying
(\ref{sol.partition.dominance.lemma}) to $\nu=\lambda$, we obtain
\[
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+...+\left(
\lambda^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{  \lambda
_{j},k\right\}  .
\]
\par
Applying (\ref{sol.partition.dominance.lemma}) to $\nu=\mu$, we obtain
\[
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu
^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}  .
\]
Hence,
\begin{align*}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},k\right\}    & =\left(  \mu
^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+...+\left(  \mu^{t}\right)
_{k}\geq\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+...+\left(  \lambda^{t}\right)  _{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.partition.dominance.known.better})}\right)  \\
& =\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},k\right\}  .
\end{align*}
This proves (\ref{sol.partition.dominance.known.rewr}).}

Now, let $k\in\left\{  1,2,...,n\right\}  $ be arbitrary. We are going to show
that $\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu_{2}+...+\mu_{k}$.

Since $\lambda\in\Par_{n}$, we have $\left\vert
\lambda\right\vert =n$, so that $n=\left\vert \lambda\right\vert =\lambda_{1}+\lambda_{2}
+\lambda_{3}+...\geq\lambda_{k}$. Thus, $\lambda_{k}\in\left\{  0,1,...,n\right\}  $.
Therefore, (\ref{sol.partition.dominance.known.rewr}) (applied to $\lambda_{k}$
instead of $k$) yields that
\begin{equation}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda_{k}\right\}  \geq\sum
_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda_{k}\right\}
.\label{sol.partition.dominance.known.ineq}
\end{equation}


Now, we have $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq...$ (since
$\lambda$ is a partition). Hence, every $j\in\left\{  1,2,...,k\right\}  $
satisfies
$\min\left\{  \lambda_{j},\lambda_{k}\right\}  =\lambda_{k}$\ \ \ \ 
\footnote{\textit{Proof.} Let $j\in\left\{  1,2,...,k\right\}  $.
Then, $j\leq k$, so that $\lambda_{j}\geq\lambda_{k}$ (since $\lambda_{1}
\geq\lambda_{2}\geq\lambda_{3}\geq...$). Thus, $\min\left\{  \lambda
_{j},\lambda_{k}\right\}  =\lambda_{k}$, qed.}. Consequently,
$\sum_{j=1}^{k}\underbrace{\min\left\{  \lambda_{j},\lambda_{k}\right\} }
_{=\lambda_{k}}
=\sum_{j=1}^{k}\lambda_{k}=k\lambda_{k}$. Also, every $j\in\left\{
k+1,k+2,k+3,...\right\}  $ satisfies $\min\left\{  \lambda_{j},\lambda
_{k}\right\}  =\lambda_{j}$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{
k+1,k+2,k+3,...\right\}  $. Then, $j>k$, so that $k<j$. Hence, $\lambda
_{k}\geq\lambda_{j}$ (since $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq
...$). Thus, $\min\left\{  \lambda_{j},\lambda_{k}\right\}  =\lambda_{j}$,
qed.}. Therefore,
\begin{align*}
\sum_{j=k+1}^{\infty}\underbrace{\min\left\{  \lambda_{j},\lambda_{k}\right\}
}_{=\lambda_{j}}  & =\sum_{j=k+1}^{\infty}\lambda_{j}=\underbrace{\sum
_{j=1}^{\infty}\lambda_{j}}_{\substack{=\lambda_{1}+\lambda_{2}+\lambda
_{3}+...\\=\left\vert \lambda\right\vert =n\\\text{(since }\lambda
\in\Par_{n}\text{)}}}-\underbrace{\sum_{j=1}^{k}\lambda_{j}}
_{=\lambda_{1}+\lambda_{2}+...+\lambda_{k}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{j=1}^{\infty}\lambda_{j}
=\sum_{j=1}^{k}\lambda_{j}+\sum_{j=k+1}^{\infty}\lambda_{j}\right)  \\
& =n-\left(  \lambda_{1}+\lambda_{2}+...+\lambda_{k}\right)  .
\end{align*}
Now,
\begin{align}
\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda_{k}\right\}    &
=\underbrace{\sum_{j=1}^{k}\min\left\{  \lambda_{j},\lambda_{k}\right\}
}_{=k\lambda_{k}}+\underbrace{\sum_{j=k+1}^{\infty}\min\left\{  \lambda
_{j},\lambda_{k}\right\}  }_{=n-\left(  \lambda_{1}+\lambda_{2}+...+\lambda
_{k}\right)  }\nonumber\\
& =k\lambda_{k}+n-\left(  \lambda_{1}+\lambda_{2}+...+\lambda_{k}\right)
.\label{sol.partition.dominance.p1}
\end{align}


On the other hand, any two integers $a$ and $b$ satisfy $\min\left\{
a,b\right\}  \leq b$. Hence, every $j\in\left\{  1,2,...,k\right\}  $
satisfies $\min\left\{  \mu_{j},\lambda_{k}\right\}  \leq\lambda_{k}$ (by the
inequality $\min\left\{  a,b\right\}  \leq b$, applied to $a=\mu_{j}$ and
$b=\lambda_{k}$). Therefore, $\sum_{j=1}^{k}\underbrace{\min\left\{  \mu
_{j},\lambda_{k}\right\}  }_{\leq\lambda_{k}}\leq\sum_{j=1}^{k}\lambda
_{k}=k\lambda_{k}$. Also, any two integers $a$ and $b$ satisfy $\min\left\{
a,b\right\}  \leq a$. Hence, every $j\in\left\{  k+1,k+2,k+3,...\right\}  $
satisfies $\min\left\{  \mu_{j},\lambda_{k}\right\}  \leq\mu_{j}$ (by the
inequality $\min\left\{  a,b\right\}  \leq a$, applied to $a=\mu_{j}$ and
$b=\lambda_{k}$). Therefore,
\begin{align*}
\sum_{j=k+1}^{\infty}\underbrace{\min\left\{  \mu_{j},\lambda_{k}\right\}
}_{\leq\mu_{j}}  & \leq\sum_{j=k+1}^{\infty}\mu_{j}=\underbrace{\sum
_{j=1}^{\infty}\mu_{j}}_{\substack{=\mu_{1}+\mu_{2}+\mu_{3}+...\\=\left\vert
\mu\right\vert =n\\\text{(since }\mu\in\Par_{n}\text{)}}}
-\underbrace{\sum_{j=1}^{k}\mu_{j}}_{=\mu_{1}+\mu_{2}+...+\mu_{k}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{j=1}^{\infty}\mu_{j}
=\sum_{j=1}^{k}\mu_{j}+\sum_{j=k+1}^{\infty}\mu_{j}\right)  \\
& =n-\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)  .
\end{align*}
Now,
\begin{align}
\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda_{k}\right\}    &
=\underbrace{\sum_{j=1}^{k}\min\left\{  \mu_{j},\lambda_{k}\right\}  }_{\leq
k\lambda_{k}}+\underbrace{\sum_{j=k+1}^{\infty}\min\left\{  \mu_{j},
\lambda_{k}\right\}  }_{\leq n-\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)
}\nonumber\\
& \leq k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)
.\label{sol.partition.dominance.p2}
\end{align}


Now, (\ref{sol.partition.dominance.known.ineq}) yields $\sum_{j=1}^{\infty
}\min\left\{  \mu_{j},\lambda_{k}\right\}  -\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},\lambda_{k}\right\}  \geq0$. Thus,
\begin{align*}
0  & \leq\underbrace{\sum_{j=1}^{\infty}\min\left\{  \mu_{j},\lambda
_{k}\right\}  }_{\substack{\leq k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}
+...+\mu_{k}\right)  \\\text{(by (\ref{sol.partition.dominance.p2}))}}}
-\underbrace{\sum_{j=1}^{\infty}\min\left\{  \lambda_{j},\lambda
_{k}\right\}  }_{\substack{=k\lambda_{k}+n-\left(  \lambda_{1}+\lambda
_{2}+...+\lambda_{k}\right)  \\
\text{(by (\ref{sol.partition.dominance.p1}))}}}\\
& \leq\left(  k\lambda_{k}+n-\left(  \mu_{1}+\mu_{2}+...+\mu_{k}\right)
\right)  -\left(  k\lambda_{k}+n-\left(  \lambda_{1}+\lambda_{2}
+...+\lambda_{k}\right)  \right)  \\
& =\left(  \lambda_{1}+\lambda_{2}+...+\lambda_{k}\right)  -\left(  \mu
_{1}+\mu_{2}+...+\mu_{k}\right)  .
\end{align*}
In other words, $\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu
_{2}+...+\mu_{k}$.

Now, forget that we fixed $k$. We thus have shown that
\[
\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu_{2}+...+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,...,n\right\}
\text{.}
\]


Now, recall that we have $\lambda\triangleright\mu$ if and only if we have
\[
\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu_{2}+...+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,...,n\right\}
\]
(according to the definition of the dominance order). Thus, we have
$\lambda\triangleright\mu$ (because we know that we have
\[
\lambda_{1}+\lambda_{2}+...+\lambda_{k}\geq\mu_{1}+\mu_{2}+...+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,...,n\right\}
\]
). This proves (\ref{sol.partition.dominance.goal2}).

\textit{Proof of (\ref{sol.partition.dominance.goal1}):} Assume that
$\lambda\triangleright\mu$. We have $\mu^{t}\in\Par_{n}$ and
$\lambda^{t}\in\Par_{n}$. Also,
$\left(  \lambda^{t}\right)  ^{t}=\lambda$ and $\left(  \mu^{t}\right)
^{t}=\mu$; now, $\left(  \lambda^{t}\right)  ^{t}=\lambda\triangleright
\mu=\left(  \mu^{t}\right)  ^{t}$. Thus, we can apply
(\ref{sol.partition.dominance.goal2}) to $\mu^{t}$ and $\lambda^{t}$ instead
of $\lambda$ and $\mu$. As a result, we obtain $\mu^{t}\triangleright
\lambda^{t}$. This proves (\ref{sol.partition.dominance.goal1}).

Now, both (\ref{sol.partition.dominance.goal1}) and
(\ref{sol.partition.dominance.goal2}) are proven. In other words, we have
$\lambda\triangleright\mu$ if and only if $\mu^{t}\triangleright\lambda^{t}$.
This solves Exercise~\ref{exe.partition.dominance}.
\end{commentedout}
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.triangular}}

\textit{Solution to Exercise \ref{exe.Lambda.triangular}.} Let us first
introduce some terminology.

\begin{definition}
If $\alpha \in \NN^\infty$ is any sequence, and if $i$ is any positive integer,
then $\alpha_i$ shall denote the $i$-th entry of $\alpha$.
(This generalizes the notation $\lambda_i$ for the $i$-th entry of a
partition $\lambda$.)
Thus, any sequence $\alpha \in \NN^\infty$ satisfies
$\alpha = \left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)$.
\end{definition}

\begin{definition}
Let \dfn{$\operatorname*{WC}$} denote the set of all weak compositions. For
every $f\in\kk\left[  \left[  \xx\right]  \right]  $ and $\mu
\in\operatorname*{WC}$, we let $\left[  \xx^{\mu}\right]  f$ denote the
coefficient of the monomial $\xx^{\mu}$ in the power series $f$. (This
generalizes the notation $\left[  \xx^{\mu}\right]  f$ introduced in
Exercise \ref{exe.Lambda.triangular}(a).)
\end{definition}

We observe the following obvious facts:

\begin{itemize}
\item For any $\alpha\in\operatorname*{WC}$ and $\mu\in\operatorname*{WC}$, we
have%
\begin{equation}
\left[  \xx^{\mu}\right]  \left(  \xx^{\alpha}\right)
=\delta_{\mu,\alpha}. \label{sol.Lambda.triangular.coeff-mon}%
\end{equation}


\item If a power series $f\in\kk\left[  \left[  \xx\right]
\right]  $ is written in the form $f=\sum_{\alpha\in\operatorname*{WC}%
}c_{\alpha} \xx^{\alpha}$ for some family $\left(  c_{\alpha}\right)
_{\alpha\in\operatorname*{WC}}\in\kk^{\operatorname*{WC}}$ of scalars,
then
\begin{equation}
\text{every }\alpha\in\operatorname*{WC}\text{ satisfies }\left[
\mathbf{x}^{\alpha}\right]  f=c_{\alpha}.
\label{sol.Lambda.triangular.coeff-f}%
\end{equation}

\end{itemize}

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(a).
\end{commentedout}

(a) Let $f\in\Lambda_{n}$. Thus,%
\[
f\in\Lambda_{n}\subset\Lambda=\left\{  \sum_{\alpha\in\operatorname*{WC}%
}c_{\alpha}{\mathbf{x}}^{\alpha}\in R\left(  \xx\right)  \ \mid
\ c_{\alpha}=c_{\beta}\text{ if }\alpha,\beta\text{ lie in the same
}\Symm_{\left(  \infty\right)  }\text{-orbit}\right\}
\]
(by the definition of $\Lambda$). In other words, $f$ can be written in the
form $f=\sum_{\alpha\in\operatorname*{WC}}c_{\alpha}{\mathbf{x}}^{\alpha}$ for
some family $\left(  c_{\alpha}\right)  _{\alpha\in\operatorname*{WC}}%
\in\kk^{\operatorname*{WC}}$ of scalars having the property that
\begin{equation}
\left(  c_{\alpha}=c_{\beta}\text{ if }\alpha,\beta\text{ lie in the same
}\Symm_{\left(  \infty\right)  }\text{-orbit}\right)  .
\label{sol.Lambda.triangular.a.c=c}%
\end{equation}
Consider this family $\left(  c_{\alpha}\right)  _{\alpha\in\operatorname*{WC}%
}$. From (\ref{sol.Lambda.triangular.coeff-f}), we conclude that every
$\alpha\in\operatorname*{WC}$ satisfies%
\begin{equation}
\left[  \mathbf{x}^{\alpha}\right]  f=c_{\alpha}.
\label{sol.Lambda.triangular.a.coeff}%
\end{equation}


Now, fix $\alpha\in\operatorname*{WC}$. Thus, $\alpha$ is a weak composition
(since $\operatorname*{WC}$ is the set of all weak compositions). Hence, there
exists a \textbf{unique} partition $\lambda$ that is a permutation of $\alpha
$\ \ \ \ \footnote{Namely, this partition $\lambda$ is the result of sorting
the entries of $\alpha$ into decreasing order (or, more precisely: moving the
positive entries of $\alpha$ to the left of all zero entries, and then sorting
the former into decreasing order). For example, if $\alpha=\left(
0,2,5,0,3,1,0,0,1,0,0,\ldots\right)  $, then this partition $\lambda$ is
$\left(  5,3,2,1,1,0,0,\ldots\right)  $.}. Thus, the sum $\sum\limits
_{\substack{\lambda\text{ is a partition;}\\\lambda\text{ is a permutation of
}\alpha}}\mathbf{x}^{\alpha}$ has only one addend. Therefore, this sum
simplifies as follows:%
\begin{equation}
\sum\limits_{\substack{\lambda\text{ is a partition;}\\\lambda\text{ is a permutation
of }\alpha}}\mathbf{x}^{\alpha}=\mathbf{x}^{\alpha}.
\label{sol.Lambda.triangular.a.3}%
\end{equation}
Thus,%
\begin{equation}
\mathbf{x}^{\alpha}=\underbrace{\sum\limits_{\substack{\lambda\text{ is a
partition;}\\\lambda\text{ is a permutation of }\alpha}}}_{=\sum\limits
_{\substack{\lambda\text{ is a partition;}\\\alpha\text{ is a permutation of
}\lambda}}=\sum\limits_{\substack{\lambda\text{ is a partition;}\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}=\sum\limits_{\substack{\lambda
\in\Par;\\\alpha\in\Symm_{\left(  \infty\right)
}\lambda}}}\mathbf{x}^{\alpha}=\sum\limits_{\substack{\lambda\in\Par%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
}. \label{sol.Lambda.triangular.a.5}%
\end{equation}


Now, forget that we fixed $\alpha$. We thus have proven that
(\ref{sol.Lambda.triangular.a.5}) holds for each $\alpha\in\operatorname*{WC}$.

Recall that each $\lambda\in\Par$ satisfies%
\begin{equation}
m_{\lambda}=\sum\limits_{\substack{\alpha\in\operatorname*{WC};\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha}
\label{sol.Lambda.triangular.a.mlambda=}%
\end{equation}
(by (\ref{monomial-symmetric-function-definition})).

Now,
\begin{align}
f  &  =\sum_{\alpha\in\operatorname*{WC}}c_{\alpha}\underbrace{{\xx%
}^{\alpha}}_{\substack{=\sum\limits_{\substack{\lambda\in\Par%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
}\\\text{(by (\ref{sol.Lambda.triangular.a.5}))}}}=\sum_{\alpha\in
\operatorname*{WC}}c_{\alpha}\sum\limits_{\substack{\lambda\in\Par%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
} \nonumber\\
&=\underbrace{\sum_{\alpha\in\operatorname*{WC}}\sum\limits_{\substack{\lambda
\in\Par;\\\alpha\in\Symm_{\left(  \infty\right)
}\lambda}}}_{=\sum_{\lambda\in\Par}\sum\limits_{\substack{\alpha
\in\operatorname*{WC};\\\alpha\in\Symm_{\left(  \infty\right)  }%
\lambda}}}\underbrace{c_{\alpha}}_{\substack{=c_{\lambda}\\\text{(by
(\ref{sol.Lambda.triangular.a.c=c}), applied to }\beta=\lambda\\\text{(since
}\alpha,\lambda\text{ lie in the same }\Symm_{\left(  \infty\right)
}\text{-orbit}\\\text{(since }\alpha\in\Symm_{\left(  \infty\right)
}\lambda\text{)))}}}\mathbf{x}^{\alpha}\nonumber\\
&  =\sum_{\lambda\in\Par}\sum\limits_{\substack{\alpha\in
\operatorname*{WC};\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda
}}c_{\lambda}\mathbf{x}^{\alpha}=\sum_{\lambda\in\Par}
c_{\lambda}\underbrace{\sum\limits_{\substack{\alpha\in\operatorname*{WC}%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
}}_{\substack{=m_{\lambda}\\\text{(by (\ref{sol.Lambda.triangular.a.mlambda=}%
))}}}=\sum_{\lambda\in\Par}c_{\lambda}m_{\lambda}.
\label{sol.Lambda.triangular.a.half}%
\end{align}


Now, let $\pi_{n}:\kk\left[  \left[  \xx\right]  \right]
\rightarrow\kk\left[  \left[  \xx\right]  \right]  $ be the
projection that maps each power series $g\in\kk\left[  \left[
\xx\right]  \right]  $ to its $n$-th homogeneous component. Then, the
definition of $\pi_{n}$ shows that the following holds:

\begin{itemize}
\item If $g\in\kk\left[  \left[  \xx\right]  \right]  $ is a
homogeneous power series of degree $n$, then
\begin{equation}
\pi_{n}\left(  g\right)  =g. \label{sol.Lambda.triangular.a.hg-1-1}%
\end{equation}


\item If $g\in\kk\left[  \left[  \xx\right]  \right]  $ is a
homogeneous power series of degree $\neq n$, then
\begin{equation}
\pi_{n}\left(  g\right)  =0. \label{sol.Lambda.triangular.a.hg-1-2}%
\end{equation}

\end{itemize}

We can now use this to compute $\pi_{n}\left(  m_{\lambda}\right)  $ for each
$\lambda\in\Par$.

First, let us notice that
\begin{equation}
m_{\lambda}\text{ is a homogeneous power series of degree }\left\vert
\lambda\right\vert \label{sol.Lambda.triangular.a.mlambda-hg}%
\end{equation}
for each $\lambda\in\Par$. (This is clear, because each of the
addends $\mathbf{x}^{\alpha}$ on the right hand side of
(\ref{sol.Lambda.triangular.a.mlambda=}) is a monomial of degree $\left\vert
\alpha\right\vert =\left\vert \lambda\right\vert $.)

Now, we conclude the following:

\begin{itemize}
\item If $\lambda\in\Par$ satisfies $\left\vert \lambda
\right\vert =n$, then%
\begin{equation}
\pi_{n}\left(  m_{\lambda}\right)  =m_{\lambda}
\label{sol.Lambda.triangular.a.hg-conc-1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.triangular.a.hg-conc-1}):} Let
$\lambda\in\Par$ be such that $\left\vert \lambda\right\vert
=n$. Then, (\ref{sol.Lambda.triangular.a.mlambda-hg}) shows that $m_{\lambda}$
is a homogeneous power series of degree $\left\vert \lambda\right\vert =n$.
Hence, (\ref{sol.Lambda.triangular.a.hg-1-1}) (applied to $g=m_{\lambda}$)
shows that $\pi_{n}\left(  m_{\lambda}\right)  =m_{\lambda}$. This proves
(\ref{sol.Lambda.triangular.a.hg-conc-1}).}.

\item If $\lambda\in\Par$ satisfies $\left\vert \lambda
\right\vert \neq n$, then%
\begin{equation}
\pi_{n}\left(  m_{\lambda}\right)  =0
\label{sol.Lambda.triangular.a.hg-conc-2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.triangular.a.hg-conc-2}):} Let
$\lambda\in\Par$ be such that $\left\vert \lambda\right\vert
\neq n$. Then, (\ref{sol.Lambda.triangular.a.mlambda-hg}) shows that
$m_{\lambda}$ is a homogeneous power series of degree $\left\vert
\lambda\right\vert \neq n$. Hence, (\ref{sol.Lambda.triangular.a.hg-1-2})
(applied to $g=m_{\lambda}$) shows that $\pi_{n}\left(  m_{\lambda}\right)
=0$. This proves (\ref{sol.Lambda.triangular.a.hg-conc-2}).}.
\end{itemize}

Finally, let us apply the map $\pi_{n}$ to both sides of the equality
(\ref{sol.Lambda.triangular.a.half}). We thus obtain%
\begin{align*}
\pi_{n}\left(  f\right)   &  =\pi_{n}\left(  \sum_{\lambda\in
\Par}c_{\lambda}m_{\lambda}\right)  =\sum_{\lambda
\in\Par}c_{\lambda}\pi_{n}\left(  m_{\lambda}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\pi_{n}\text{ respects infinite }\kk\text{-linear
combinations}\\
\text{(because }\pi_{n}\text{ is }\kk\text{-linear and continuous)}%
\end{array}
\right) \\
&  =\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}}_{\substack{=\sum_{\lambda\in\Par_{n}}\\
\text{(since }\left\{  \lambda\in\Par%
\ \mid\ \left\vert \lambda\right\vert =n\right\}  =\Par_{n}
\text{)}}}c_{\lambda}\underbrace{\pi_{n}\left(  m_{\lambda
}\right)  }_{\substack{=m_{\lambda}\\\text{(by
(\ref{sol.Lambda.triangular.a.hg-conc-1}))}}}+\sum\limits_{\substack{\lambda
\in\Par;\\\left\vert \lambda\right\vert \neq n}}c_{\lambda
}\underbrace{\pi_{n}\left(  m_{\lambda}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.Lambda.triangular.a.hg-conc-2}))}}}\\
&  =\sum_{\lambda\in\Par_{n}}c_{\lambda}m_{\lambda
}+\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert \neq n}}c_{\lambda}0}_{=0}=\sum_{\lambda\in
\Par_{n}}c_{\lambda}m_{\lambda}.
\end{align*}


But $f$ is a homogeneous power series of degree $n$ (since $f\in\Lambda_{n}$).
Thus, (\ref{sol.Lambda.triangular.a.hg-1-1}) (applied to $g=f$) yields
$\pi_{n}\left(  f\right)  =f$. Hence,%
\[
f=\pi_{n}\left(  f\right)
=\sum_{\lambda\in\Par_n}c_{\lambda}m_{\lambda}
=\sum_{\mu\in\Par_{n}}c_{\mu}m_{\mu}
\]
(here, we have renamed the summation index $\lambda$ as $\mu$). Comparing this
with%
\[
\sum_{\mu\in\Par_{n}}\underbrace{\left(  \left[
\mathbf{x}^{\mu}\right]  f\right)  }_{\substack{=c_{\mu}\\\text{(by
(\ref{sol.Lambda.triangular.a.coeff}) (applied to }\alpha=\mu\text{))}}%
}m_{\mu}=\sum_{\mu\in\Par_{n}}c_{\mu}m_{\mu},
\]
we obtain $f=\sum_{\mu\in\Par_{n}}\left(  \left[
\mathbf{x}^{\mu}\right]  f\right)  m_{\mu}$. This solves Exercise
\ref{exe.Lambda.triangular}(a).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(b).
\end{commentedout}

(b) Let $\lambda$ be a partition. Let $\mu$ be a weak composition. We must
prove that the number $K_{\lambda,\mu}$ is well-defined. In other words, we
must prove that there are only finitely many column-strict tableaux $T$ of
shape $\lambda$ having $\cont \left(  T\right)  =\mu$.

Let $F$ be the Ferrers diagram of $\lambda$ (as a set of cells). Thus, $F$ is
a finite subset of $\left\{  1,2,3,\ldots\right\}  ^{2}$.

Any column-strict tableau of shape $\lambda$ is an assignment of entries in
$\{1,2,3,\ldots\}$ to the cells of the Ferrers diagram of $\lambda$. In other
words, any column-strict tableau of shape $\lambda$ is map $F\rightarrow
\left\{  1,2,3,\ldots\right\}  $ (since $F$ is the set of all cells of the
Ferrers diagram of $\lambda$).

The sequence $\mu$ is a weak composition, and thus has a finite support. In
other words, the support of $\mu$ is finite. Let $W$ be this support. Thus,
$W$ is a finite set, and satisfies%
\begin{align}
W  &  =\left(  \text{the support of }\mu\right) \nonumber\\
&  =\left(  \text{the set of all positive integers }i\text{ for which }\mu
_{i}\neq0\right)  \label{sol.Lambda.triangular.b.W=}%
\end{align}
(by the definition of the support of $\mu$). Thus, $W\subset\left\{
1,2,3,\ldots\right\}  $.

It is well-known that if $X$, $Y$ and $Z$ are three sets such that $X\subset
Y$, then%
\[
X^{Z}\cong\left\{  f\in Y^{Z}\ \mid\ f\left(  Z\right)  \subset X\right\}
\ \ \ \ \ \ \ \ \ \ \text{as sets.}%
\]
Applying this to $X=W$, $Y=\left\{  1,2,3,\ldots\right\}  $ and $Z=F$, we
conclude that%
\begin{equation}
W^{F}\cong\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  \ \ \ \ \ \ \ \ \ \ \text{as sets.}
\label{sol.Lambda.triangular.b.as-sets}%
\end{equation}
But $F$ and $W$ are finite sets. Hence, $W^{F}$ is also a finite set. Thus,
$\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(  F\right)
\subset W\right\}  $ is a finite set (because of
(\ref{sol.Lambda.triangular.b.as-sets})).

Now, let $T$ be a column-strict tableau of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$. We shall prove that
$T\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  $.

Indeed, $T$ is a map $F\rightarrow\left\{  1,2,3,\ldots\right\}  $ (since any
column-strict tableau of shape $\lambda$ is a map $F\rightarrow\left\{
1,2,3,\ldots\right\}  $), hence an element of $\left\{  1,2,3,\ldots\right\}
^{F}$.

Furthermore, $T\left(  F\right)  \subset W$\ \ \ \ \footnote{\textit{Proof.}
Let $j\in T\left(  F\right)  $. Then, $j\in\left\{  1,2,3,\ldots\right\}  $.
Furthermore, there exists some $c\in F$ satisfying $j=T\left(  c\right)  $
(since $j\in T\left(  F\right)  $). Consider this $c$. From $j=T\left(
c\right)  $, we obtain $c\in T^{-1}\left(  j\right)  $. Hence, the set
$T^{-1}\left(  j\right)  $ contains at least one element (namely, $c$).
Therefore, $\left\vert T^{-1}\left(  j\right)  \right\vert \geq1$.
\par
But $\mu=\cont \left(  T\right)  =\left(  \left\vert
T^{-1}\left(  1\right)  \right\vert ,\left\vert T^{-1}\left(  2\right)
\right\vert ,\left\vert T^{-1}\left(  3\right)  \right\vert ,\ldots\right)  $
(by the definition of $\cont \left(  T\right)  $). Hence,
$\mu_{j}=\left\vert T^{-1}\left(  j\right)  \right\vert \geq1>0$. Hence,
$\mu_{j}\neq0$. Thus, $j$ belongs to the set of all positive integers $i$ for
which $\mu_{i}\neq0$. In other words, $j\in\left(  \text{the set of all
positive integers }i\text{ for which }\mu_{i}\neq0\right)  $. In light of
(\ref{sol.Lambda.triangular.b.W=}), this rewrites as $j\in W$.
\par
Now, forget that we fixed $j$. We thus have shown that $j\in W$ for each $j\in
T\left(  F\right)  $. In other words, $T\left(  F\right)  \subset W$.}. Hence,
$T$ is an $f\in\left\{  1,2,3,\ldots\right\}  ^{F}$ satisfying $f\left(
F\right)  \subset W$ (since $T$ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{F}$ and satisfies $T\left(  F\right)  \subset W$). In other words,%
\[
T\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(  F\right)
\subset W\right\}  .
\]


Now, forget that we fixed $T$. We thus have shown that every column-strict
tableau $T$ of shape $\lambda$ having $\cont \left(  T\right)
=\mu$ satisfies $T\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}%
\ \mid\ f\left(  F\right)  \subset W\right\}  $. In other words,%
\begin{align*}
&  \left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{
having }\cont \left(  T\right)  =\mu\right\}
 \subset\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  .
\end{align*}
Hence, $\left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{
having }\cont \left(  T\right)  =\mu\right\}  $ is a finite set
(since \newline
 $\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  $ is a finite set). In other words, there are
only finitely many column-strict tableaux $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$. This solves Exercise
\ref{exe.Lambda.triangular}(b).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(c).
\end{commentedout}

(c) Let $\lambda\in\Par_{n}$. The definition of
$s_{\lambda}$ yields%
\[
s_{\lambda}=\sum_{T}{\mathbf{x}}^{\cont \left(  T\right)  },
\]
where $T$ runs through all column-strict tableaux of shape $\lambda$. In other
words,%
\[
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda}}{\mathbf{x}}^{\cont \left(  T\right)  }.
\]


Now, every $\mu\in\Par_{n}$ satisfies
\begin{align}
&  \left[  \mathbf{x}^{\mu}\right]  \left(  \underbrace{s_{\lambda}}%
_{=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}{\mathbf{x}}^{\cont \left(  T\right)  }}\right)
\nonumber\\
&  =\left[  \mathbf{x}^{\mu}\right]  \left(  \sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda}}{\mathbf{x}}%
^{\cont \left(  T\right)  }\right)  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda}}\underbrace{\left[
\mathbf{x}^{\mu}\right]  \left(  {\mathbf{x}}^{\cont \left(
T\right)  }\right)  }_{\substack{=\delta_{\mu,\cont \left(
T\right)  }\\\text{(by (\ref{sol.Lambda.triangular.coeff-mon}) (applied to
}\alpha=\cont \left(  T\right)  \text{))}}}\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}\delta_{\mu,\cont \left(  T\right)  }=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
;\\\mu=\cont \left(  T\right)  }}\underbrace{\delta
_{\mu,\cont \left(  T\right)  }}_{\substack{=1\\\text{(since
}\mu=\cont \left(  T\right)  \text{)}}}+\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda;\\\mu\neq
\cont \left(  T\right)  }}\underbrace{\delta_{\mu
,\cont \left(  T\right)  }}_{\substack{=0\\\text{(since }\mu
\neq\cont \left(  T\right)  \text{)}}}\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda;\\\mu=\cont \left(  T\right)  }}1+\underbrace{\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
;\\\mu\neq\cont \left(  T\right)  }}0}_{=0}=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
;\\\mu=\cont \left(  T\right)  }}1\nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\mu=\cont \left(  T\right)  \right)
\cdot1\nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\mu=\cont \left(  T\right)  \right)
\nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\cont \left(  T\right)  =\mu\right)
\nonumber\\
&  =K_{\lambda,\mu}
\label{sol.Lambda.triangular.c.5}
\end{align}
(since $K_{\lambda,\mu}$ is defined as the number of all column-strict
tableaux $T$ of shape $\lambda$ having $\cont \left(  T\right)
=\mu$).

But $\lambda\in\Par_{n}$. Hence, $\left\vert
\lambda\right\vert =n$. Recall that the power series $s_{\lambda}$ is a
symmetric function (by Proposition \ref{Schur-functions-are-symmetric-prop}),
and is homogeneous of degree $\left\vert \lambda\right\vert $. Thus,
$s_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }=\Lambda_{n}$ (since
$\left\vert \lambda\right\vert =n$). Hence, Exercise
\ref{exe.Lambda.triangular}(a) (applied to $f=s_{\lambda}$) yields%
\[
s_{\lambda}=\sum_{\mu\in\Par_{n}}\underbrace{\left(
\left[  \mathbf{x}^{\mu}\right]  \left(  s_{\lambda}\right)  \right)
}_{\substack{=K_{\lambda,\mu}\\\text{(by (\ref{sol.Lambda.triangular.c.5}))}%
}}m_{\mu}=\sum_{\mu\in\Par_{n}}K_{\lambda,\mu}m_{\mu
}.
\]
This solves Exercise \ref{exe.Lambda.triangular}(c).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(d).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(d), let us show a lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.kostka}Let $n\in \NN$. Let $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$.
Let $T$ be a column-strict tableau of shape $\lambda$
satisfying $\cont \left(  T\right)  =\mu$. Let $F$ be the
Ferrers diagram of $\lambda$ (as a set of cells). For each positive integer
$i$, we let $F_{i}$ be the $i$-th row of $F$ (that is, the set of all cells of
$F$ that have the form $\left(  i,j\right)  $ for some $j\geq1$).

\begin{enumerate}
\item[(a)] We have $T\left(  i,j\right)  \geq i$ for each $\left(  i,j\right)
\in F$.

\item[(b)] We have $T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)
\cup\cdots\cup T^{-1}\left(  k\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup
F_{k}$ for each $k\in \NN$.

\item[(c)] We have $\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{k}\right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}$ for each $k\in \NN$.

\item[(d)] We have $\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(
2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \right\vert =\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$ for each $k\in \NN$.

\item[(e)] We have $\lambda\triangleright\mu$.

\item[(f)] If $\mu=\lambda$, then each $\left(  i,j\right)  \in F$ satisfies
$T\left(  i,j\right)  =i$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}.]Recall that $F$ is the
Ferrers diagram of $\lambda$. In other words, $F$ is the set of all pairs
$\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying
$j\leq\lambda_{i}$ (by the definition of a Ferrers diagram). In other words,
\[
F=\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}%
\ \mid\ j\leq\lambda_{i}\right\}  .
\]
Hence, every $\left(  p,q\right)  \in F$ satisfies%
\begin{equation}
\left(  p^{\prime},q\right)  \in F\ \ \ \ \ \ \ \ \ \ \text{for each
}p^{\prime}\in\left\{  1,2,\ldots,p\right\}
\label{pf.lem.sol.Lambda.triangular.kostka.go-up}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.kostka.go-up}):}
Let $\left(  p,q\right)  \in F$. Thus, $\left(  p,q\right)  \in F=\left\{
\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq
\lambda_{i}\right\}  $. In other words, $\left(  p,q\right)  $ is an element
of $\left\{  1,2,3,\ldots\right\}  ^{2}$ and satisfies $q\leq\lambda_{p}$.
\par
Now, let $p^{\prime}\in\left\{  1,2,\ldots,p\right\}  $. We must prove that
$\left(  p^{\prime},q\right)  \in F$.
\par
From $p^{\prime}\in\left\{  1,2,\ldots,p\right\}  \subset\left\{
1,2,3,\ldots\right\}  $ and $q\in\left\{  1,2,3,\ldots\right\}  $, we conclude
that $\left(  p^{\prime},q\right)  $ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{2}$. Also, $p^{\prime}\leq p$ (since $p^{\prime}\in\left\{
1,2,\ldots,p\right\}  $).
\par
But $\lambda$ is a partition. Hence, $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq\cdots$. Therefore, from $p^{\prime}\leq p$, we obtain
$\lambda_{p^{\prime}}\geq\lambda_{p}$. Hence, $q\leq\lambda_{p}\leq
\lambda_{p^{\prime}}$ (since $\lambda_{p^{\prime}}\geq\lambda_{p}$).
\par
Now, $\left(  p^{\prime},q\right)  $ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{2}$ and satisfies $q\leq\lambda_{p^{\prime}}$. In other words,
$\left(  p^{\prime},q\right)  \in\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $. In other words,
$\left(  p^{\prime},q\right)  \in F$ (since $F=\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $).
This completes the proof of (\ref{pf.lem.sol.Lambda.triangular.kostka.go-up}%
).}.

Any column-strict tableau of shape $\lambda$ is an assignment of entries in
$\{1,2,3,\ldots\}$ to the cells of the Ferrers diagram of $\lambda$. In other
words, any column-strict tableau of shape $\lambda$ is a map $F\rightarrow
\left\{  1,2,3,\ldots\right\}  $ (since $F$ is the Ferrers diagram of
$\lambda$). Hence, $T$ is a map $F\rightarrow\left\{  1,2,3,\ldots\right\}  $
(since $T$ is a column-strict tableau of shape $\lambda$).

Recall that $T$ is a column-strict tableau. Thus, the entries of $T$ are
strictly increasing top-to-bottom down columns (by the definition of a
column-strict tableau).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(a).
\end{commentedout}

(a) Let $\left(  i,j\right)  \in F$. We must prove that $T\left(  i,j\right)
\geq i$.

Assume the contrary. Thus, $T\left(  i,j\right)  <i$, so that $T\left(
i,j\right)  \leq i-1$ (since $T\left(  i,j\right)  $ and $i$ are integers).

We have $\left(  p^{\prime},j\right)  \in F$ for each $p^{\prime}\in\left\{
1,2,\ldots,i\right\}  $ (by (\ref{pf.lem.sol.Lambda.triangular.kostka.go-up})
(applied to $\left(  p,q\right)  =\left(  i,j\right)  $)). In other words, all
of the cells $\left(  1,j\right)  ,\left(  2,j\right)  ,\ldots,\left(
i,j\right)  $ belong to $F$. These cells therefore all lie in the $j$-th
column of $F$; more precisely, they are the first $i$ cells of the $j$-th
column of $F$. Hence, we have%
\begin{equation}
T\left(  1,j\right)  <T\left(  2,j\right)  <\cdots<T\left(  i,j\right)
\label{pf.lem.sol.Lambda.triangular.kostka.a.4}%
\end{equation}
(since the entries of $T$ are strictly increasing top-to-bottom down columns).
Hence, $T\left(  1,j\right)  <T\left(  2,j\right)  <\cdots<T\left(
i,j\right)  \leq i-1$. Thus, all of the $i$ numbers $T\left(  1,j\right)
,T\left(  2,j\right)  ,\ldots,T\left(  i,j\right)  $ are elements of the set
$\left\{  1,2,\ldots,i-1\right\}  $ (since these numbers all belong to
$\left\{  1,2,3,\ldots\right\}  $ and are $\leq i-1$). By the pigeonhole
principle, we thus conclude that two of these $i$ numbers are equal (since the
set $\left\{  1,2,\ldots,i-1\right\}  $ has size $i-1<i$). This contradicts
the fact that these $i$ numbers are distinct (because of
(\ref{pf.lem.sol.Lambda.triangular.kostka.a.4})).This contradiction proves
that our assumption was wrong; hence, we must have $T\left(  i,j\right)  \geq
i$. This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(a).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(b).
\end{commentedout}

(b) Let $k\in \NN$. Each $p\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
T^{-1}\left(  p\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}.
\label{pf.lem.sol.Lambda.triangular.kostka.b.1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.kostka.b.1}):} Let
$p\in\left\{  1,2,\ldots,k\right\}  $. Let $c\in T^{-1}\left(  p\right)  $.
Then, $c\in F$ (since $T$ is a map $F\rightarrow\left\{  1,2,3,\ldots\right\}
$). Also, $T\left(  c\right)  =p$ (since $c\in T^{-1}\left(  p\right)  $).

Write the cell $c$ in the form $c=\left(  i,j\right)  $ for some $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. Then, $\left(
i,j\right)  =c\in F$ and $T\left(  \underbrace{c}_{=\left(  i,j\right)
}\right)  =T\left(  i,j\right)  $, so that $T\left(  i,j\right)  =T\left(
c\right)  =p$. Hence, $p=T\left(  i,j\right)  \geq i$ (by Lemma
\ref{lem.sol.Lambda.triangular.kostka}(a)). Hence, $i\leq p\leq k$ (since
$p\in\left\{  1,2,\ldots,k\right\}  $), so that $i\in\left\{  1,2,\ldots
,k\right\}  $. Therefore, $F_{i}\subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}$.

But the cell $\left(  i,j\right)  $ belongs to the $i$-th row (since its first
coordinate is $i$) and also belongs to $F$ (since $\left(  i,j\right)  \in
F$). Hence, the cell $\left(  i,j\right)  $ belongs to $F_{i}$ (since $F_{i}$
is the $i$-th row of $F$). In other words, $\left(  i,j\right)  \in F_{i}$.
Hence, $c=\left(  i,j\right)  \in F_{i}\subset F_{1}\cup F_{2}\cup\cdots\cup
F_{k}$.

Now, forget that we fixed $c$. We thus have proven that $c\in F_{1}\cup
F_{2}\cup\cdots\cup F_{k}$ for each $c\in T^{-1}\left(  p\right)  $. In other
words, $T^{-1}\left(  p\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}$.
This proves (\ref{pf.lem.sol.Lambda.triangular.kostka.b.1}).]

Now,
\begin{align*}
T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)   &  =\bigcup_{p\in\left\{  1,2,\ldots,k\right\}
}\underbrace{T^{-1}\left(  p\right)  }_{\substack{\subset F_{1}\cup F_{2}%
\cup\cdots\cup F_{k}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.kostka.b.1}%
))}}}\\
&  \subset\bigcup_{p\in\left\{  1,2,\ldots,k\right\}  }\left(  F_{1}\cup
F_{2}\cup\cdots\cup F_{k}\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(b).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(c).
\end{commentedout}

(c) Every positive integer $i$ satisfies%
\begin{align}
\left\vert F_{i}\right\vert  &  =\left(  \text{the size of }F_{i}\right)
=\left(  \text{the size of the }i\text{-th row of }F\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F_{i}\text{ is the }i\text{-th row
of }F\right) \nonumber\\
&  =\left(  \text{the size of the }i\text{-th row of the Ferrers diagram of
}\lambda\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is the Ferrers diagram of
}\lambda\right) \nonumber\\
&  =\lambda_{i}. \label{pf.lem.sol.Lambda.triangular.kostka.c.2}%
\end{align}


Let $k\in \NN$. The sets $F_{1},F_{2},\ldots,F_{k}$ are disjoint (since
they are different rows of $F$). Hence, the size of their union equals the sum
of their sizes. In other words, $\left\vert F_{1}\cup F_{2}\cup\cdots\cup
F_{k}\right\vert =\left\vert F_{1}\right\vert +\left\vert F_{2}\right\vert
+\cdots+\left\vert F_{k}\right\vert $. Thus,%
\begin{align*}
\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{k}\right\vert  &  =\left\vert
F_{1}\right\vert +\left\vert F_{2}\right\vert +\cdots+\left\vert
F_{k}\right\vert =\sum_{i=1}^{k}\underbrace{\left\vert F_{i}\right\vert
}_{\substack{=\lambda_{i}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.kostka.c.2}))}}}=\sum_{i=1}^{k}\lambda
_{i}\\
&  =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(c).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(d).
\end{commentedout}

(d) We have $\mu=\cont \left(  T\right)  =\left(  \left\vert
T^{-1}\left(  1\right)  \right\vert ,\left\vert T^{-1}\left(  2\right)
\right\vert ,\left\vert T^{-1}\left(  3\right)  \right\vert ,\ldots\right)  $
(by the definition of $\cont \left(  T\right)  $). Hence,
\begin{equation}
\mu_{i}=\left\vert T^{-1}\left(  i\right)  \right\vert
\label{pf.lem.sol.Lambda.triangular.kostka.d.1}%
\end{equation}
for every positive integer $i$.

Let $k\in \NN$. The sets $T^{-1}\left(  1\right)  ,T^{-1}\left(
2\right)  ,\ldots,T^{-1}\left(  k\right)  $ are disjoint (since they are
different fibers of the map $T$). Hence, the size of their union equals the
sum of their sizes. In other words, $\left\vert T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \right\vert
=\left\vert T^{-1}\left(  1\right)  \right\vert +\left\vert T^{-1}\left(
2\right)  \right\vert +\cdots+\left\vert T^{-1}\left(  k\right)  \right\vert
$. Thus,%
\begin{align*}
\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)  \right\vert  &  =\left\vert T^{-1}\left(  1\right)
\right\vert +\left\vert T^{-1}\left(  2\right)  \right\vert +\cdots+\left\vert
T^{-1}\left(  k\right)  \right\vert =\sum_{i=1}^{k}\underbrace{\left\vert
T^{-1}\left(  i\right)  \right\vert }_{\substack{=\mu_{i}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.kostka.d.1}))}}}=\sum_{i=1}^{k}\mu_{i}\\
&  =\mu_{1}+\mu_{2}+\cdots+\mu_{k}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(d).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(e).
\end{commentedout}

(e) Let $k\in\left\{  1,2,\ldots,n\right\}  $. Then, Lemma
\ref{lem.sol.Lambda.triangular.kostka}(b) yields $T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \subset
F_{1}\cup F_{2}\cup\cdots\cup F_{k}$. Thus,%
\[
\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)  \right\vert \leq\left\vert F_{1}\cup F_{2}\cup
\cdots\cup F_{k}\right\vert .
\]
But Lemma \ref{lem.sol.Lambda.triangular.kostka}(d) yields $\left\vert
T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)  \right\vert =\mu_{1}+\mu_{2}+\cdots+\mu_{k}$. Hence,%
\begin{align*}
\mu_{1}+\mu_{2}+\cdots+\mu_{k}  &  =\left\vert T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \right\vert \\
&  \leq\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{k}\right\vert =\lambda
_{1}+\lambda_{2}+\cdots+\lambda_{k}%
\end{align*}
(by Lemma \ref{lem.sol.Lambda.triangular.kostka}(c)). In other words,
$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}$.

Now, forget that we fixed $k$. We thus have shown that%
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,n\right\}  .
\label{pf.lem.sol.Lambda.triangular.kostka.e.1}%
\end{equation}


In other words, $\lambda\triangleright\mu$ (by the definition of the dominance
order). This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(e).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(f).
\end{commentedout}

(f) Assume that $\mu=\lambda$. We must prove that each $\left(  i,j\right)
\in F$ satisfies $T\left(  i,j\right)  =i$.

Indeed, let $\left(  i,j\right)  \in F$ be arbitrary. We must prove that
$T\left(  i,j\right)  =i$.

Define two sets $A$ and $B$ by $A=T^{-1}\left(  1\right)  \cup T^{-1}\left(
2\right)  \cup\cdots\cup T^{-1}\left(  i\right)  $ and $B=F_{1}\cup F_{2}%
\cup\cdots\cup F_{i}$. Then,%
\begin{align*}
A  &  =T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  i\right) \\
&  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma \ref{lem.sol.Lambda.triangular.kostka}(b) (applied to
}k=i\text{)}\right) \\
&  =B.
\end{align*}
In other words, $A$ is a subset of $B$.

Applying Lemma \ref{lem.sol.Lambda.triangular.kostka}(c) to $k=i$, we find%
\begin{equation}
\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{i}\right\vert =\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{i}.
\label{pf.lem.sol.Lambda.triangular.kostka.f.1}%
\end{equation}


Moreover, from $A=T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)
\cup\cdots\cup T^{-1}\left(  i\right)  $, we obtain%
\begin{align*}
\left\vert A\right\vert  &  =\left\vert T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  i\right)  \right\vert
=\mu_{1}+\mu_{2}+\cdots+\mu_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.Lambda.triangular.kostka}(d) (applied to }k=i\text{)}\right) \\
&  =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mu=\lambda\right) \\
&  =\left\vert \underbrace{F_{1}\cup F_{2}\cup\cdots\cup F_{i}}_{=B}%
\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.kostka.f.1})}\right) \\
&  =\left\vert B\right\vert .
\end{align*}
Moreover, $B$ is a finite set (since $\left\vert B\right\vert =\lambda
_{1}+\lambda_{2}+\cdots+\lambda_{i}\in \NN$).

It is well-known that if $Y$ is a finite set, and if $X$ is a subset of $Y$
satisfying $\left\vert X\right\vert =\left\vert Y\right\vert $, then $X=Y$.
Applying this to $X=A$ and $Y=B$, we obtain $A=B$ (since $A$ is a subset of
$B$ and satisfies $\left\vert A\right\vert =\left\vert B\right\vert $).

Now, the cell $\left(  i,j\right)  $ belongs to the $i$-th row of $F$ (since
it belongs to $F$, and since its first coordinate is $i$). In other words,
$\left(  i,j\right)  \in F_{i}$ (since $F_{i}$ is the $i$-th row of $F$).
Hence,%
\begin{align*}
\left(  i,j\right)   &  \in F_{i}\subset F_{1}\cup F_{2}\cup\cdots\cup
F_{i}=B=A\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A=B\right) \\
&  =T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  i\right)  .
\end{align*}
In other words, $\left(  i,j\right)  \in T^{-1}\left(  p\right)  $ for some
$p\in\left\{  1,2,\ldots,i\right\}  $. Consider this $p$.

From $\left(  i,j\right)  \in T^{-1}\left(  p\right)  $, we obtain $T\left(
i,j\right)  =p$. Thus, $T\left(  i,j\right)  =p\leq i$ (since $p\in\left\{
1,2,\ldots,i\right\}  $). But Lemma \ref{lem.sol.Lambda.triangular.kostka}(a)
yields $T\left(  i,j\right)  \geq i$. Combining this with $T\left(
i,j\right)  \leq i$, we obtain $T\left(  i,j\right)  =i$.

Now, forget that we fixed $\left(  i,j\right)  $. We thus have proven that
each $\left(  i,j\right)  \in F$ satisfies $T\left(  i,j\right)  =i$. This
proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(f).
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(d).
\end{commentedout}

(d) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$ be two partitions that don't satisfy
$\lambda\triangleright\mu$. We must prove that $K_{\lambda,\mu}=0$.

Indeed, there exists no column-strict tableau $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, there exists a column-strict tableau $T$ of shape
$\lambda$ having $\cont \left(  T\right)  =\mu$. Consider this
$T$.
\par
Lemma \ref{lem.sol.Lambda.triangular.kostka}(e) yields $\lambda\triangleright
\mu$. This contradicts the fact that we don't have $\lambda\triangleright\mu$.
This contradiction proves that our assumption was wrong. Qed.}. Thus, the
number of all column-strict tableaux $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$ equals $0$. In other words,
$K_{\lambda,\mu}$ equals $0$ (since $K_{\lambda,\mu}$ is the number of all
column-strict tableaux $T$ of shape $\lambda$ having $\cont
\left(  T\right)  =\mu$). This solves Exercise \ref{exe.Lambda.triangular}(d).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(e).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(e), let us state another
simple lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.kostka1}Let $n\in \NN$. Let $\lambda
\in\Par_{n}$. Let $F$ be the Ferrers diagram of
$\lambda$ (as a set of cells). Let $T_{0}:F\rightarrow\left\{  1,2,3,\ldots
\right\}  $ be the map that sends each $\left(  i,j\right)  \in F$ to $i$.
Then, $T_{0}$ is a column-strict tableau of shape $\lambda$ and satisfies
$\cont \left(  T_{0}\right)  =\lambda$.
\end{lemma}

\begin{example}
Let $n=8$ and $\lambda=\left(  3,2,2,1\right)  \in\Par_{8}$.
Then, the column-strict tableau $T_{0}$ defined in Lemma
\ref{lem.sol.Lambda.triangular.kostka1} looks as follows:%
\[
T_{0}=%
\begin{matrix}
1 & 1 & 1\\
2 & 2 & \\
3 & 3 & \\
4 &  &
\end{matrix}
.
\]

\end{example}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.kostka1}.]Every $\left(
i,j\right)  \in F$ satisfies $i\in\left\{  1,2,3,\ldots\right\}  $ (since
$F\subset\left\{  1,2,3,\ldots\right\}  ^{2}$). Hence, the map $T_{0}$ is well-defined.

The map $T_{0}$ is a map from $F$ to $\left\{  1,2,3,\ldots\right\}  $. In
other words, $T_{0}$ is an assignment of entries in $\{1,2,3,\ldots\}$ to the
cells of the Ferrers diagram of $\lambda$ (since $F$ is the Ferrers diagram of
$\lambda$). We shall now show that $T_{0}$ is a column-strict tableau of shape
$\lambda$.

The definition of $T_{0}$ shows that for each $i\in\left\{  1,2,3,\ldots
\right\}  $, all entries in the $i$-th row of $T_{0}$ equal $i$. Therefore,
the entries of $T_{0}$ are weakly increasing left-to-right in rows (because
they are all equal in a given row) and strictly increasing top-to-bottom in
columns (since the topmost entry is a $1$, the next entry is a $2$, and so
on). Thus, $T_{0}$ is a column-strict tableau of shape $\lambda$ (by the
definition of a ``column-strict tableau of shape $\lambda$'').

It remains to prove that $\cont \left(  T_{0}\right)  =\lambda$.

The definition of $\cont \left(  T_{0}\right)  $ shows that
$\cont \left(  T_{0}\right)  =\left(  \left\vert \left(
T_{0}\right)  ^{-1}\left(  1\right)  \right\vert ,\left\vert \left(
T_{0}\right)  ^{-1}\left(  2\right)  \right\vert ,\left\vert \left(
T_{0}\right)  ^{-1}\left(  3\right)  \right\vert ,\ldots\right)  $. In other
words,%
\begin{equation}
\left(  \cont \left(  T_{0}\right)  \right)  _{i}=\left\vert
\left(  T_{0}\right)  ^{-1}\left(  i\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.
\label{pf.lem.sol.Lambda.triangular.kostka1.cont1}%
\end{equation}


Let $k$ be a positive integer. Then,
\begin{align*}
\left(  T_{0}\right)  ^{-1}\left(  k\right)   &  =\left\{  \left(  i,j\right)
\in F\ \mid\ \underbrace{T_{0}\left(  i,j\right)  }_{\substack{=i\\\text{(by
the definition of }T_{0}\text{)}}}=k\right\}  =\left\{  \left(  i,j\right)
\in F\ \mid\ \underbrace{i=k}_{\Longleftrightarrow\ \left(  \left(
i,j\right)  \text{ lies in the }k\text{-th row}\right)  }\right\} \\
&  =\left\{  \left(  i,j\right)  \in F\ \mid\ \left(  i,j\right)  \text{ lies
in the }k\text{-th row}\right\}  =\left(  \text{the set of all cells of
}F\text{ that lie in the }k\text{-th row}\right) \\
&  =\left(  \text{the }k\text{-th row of }F\right)  .
\end{align*}
Hence,%
\[
\left\vert \left(  T_{0}\right)  ^{-1}\left(  k\right)  \right\vert
=\left\vert \left(  \text{the }k\text{-th row of }F\right)  \right\vert
=\left(  \text{the size of the }k\text{-th row of }F\right)  =\lambda_{k}%
\]
(since $F$ is the Ferrers diagram of $\lambda$). But now,
(\ref{pf.lem.sol.Lambda.triangular.kostka1.cont1}) (applied to $i=k$) yields
$\left(  \cont \left(  T_{0}\right)  \right)  _{k}=\left\vert
\left(  T_{0}\right)  ^{-1}\left(  k\right)  \right\vert =\lambda_{k}$.

Now, forget that we fixed $k$. We thus have proven that $\left(
\cont \left(  T_{0}\right)  \right)  _{k}=\lambda_{k}$ for each
positive integer $k$. In other words, $\cont \left(
T_{0}\right)  =\lambda$. This completes the proof of Lemma
\ref{lem.sol.Lambda.triangular.kostka1}.
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(e).
\end{commentedout}

(e) Let $\lambda\in\Par_{n}$. We must prove that
$K_{\lambda,\lambda}=1$.

Define $F$ and $T_{0}$ as in Lemma \ref{lem.sol.Lambda.triangular.kostka1}.
Then, Lemma \ref{lem.sol.Lambda.triangular.kostka1} shows that $T_{0}$ is a
column-strict tableau of shape $\lambda$ and satisfies $\cont
\left(  T_{0}\right)  =\lambda$. Hence, there exists \textbf{at least one}
column-strict tableau $T$ of shape $\lambda$ having $\cont
\left(  T\right)  =\lambda$ (namely, $T=T_{0}$).

On the other hand, using Lemma \ref{lem.sol.Lambda.triangular.kostka}(f), it
is easy to see that every column-strict tableau $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\lambda$ must be equal to $T_{0}%
$\ \ \ \ \footnote{\textit{Proof.} Let $T$ be a column-strict tableau of shape
$\lambda$ having $\cont \left(  T\right)  =\lambda$. We must
prove that $T$ is equal to $T_{0}$.
\par
Lemma \ref{lem.sol.Lambda.triangular.kostka}(f) (applied to $\mu=\lambda$)
shows that each $\left(  i,j\right)  \in F$ satisfies $T\left(  i,j\right)
=i$ (since $\lambda=\lambda$). Thus, each $\left(  i,j\right)  \in F$
satisfies%
\[
T\left(  i,j\right)  =i=T_{0}\left(  i,j\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }T_{0}\left(  i,j\right)  \text{ is defined to be }i\right)  .
\]
\par
Recall that any column-strict tableau of shape $\lambda$ is a map
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $. Hence, $T$ and $T_{0}$ are maps
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $ (since $T$ and $T_{0}$ are
column-strict tableaux of shape $\lambda$). Therefore, we conclude that
$T=T_{0}$ (since each $\left(  i,j\right)  \in F$ satisfies $T\left(
i,j\right)  =T_{0}\left(  i,j\right)  $). In other words, $T$ is equal to
$T_{0}$. Qed.}. Hence, there exists \textbf{at most one} column-strict tableau
$T$ of shape $\lambda$ having $\cont \left(  T\right)  =\lambda$.

We know that $K_{\lambda,\lambda}$ is the number of all column-strict tableaux
$T$ of shape $\lambda$ having $\cont \left(  T\right)  =\lambda$
(by the definition of $K_{\lambda,\lambda}$). Since there exists
\textbf{exactly one} such tableau $T$ (because we have shown that there exists
\textbf{at least one} such tableau $T$, and we have also shown that there
exists \textbf{at most one} such tableau $T$), we thus conclude that
$K_{\lambda,\lambda}=1$. This solves Exercise \ref{exe.Lambda.triangular}(e).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(f).
\end{commentedout}

(f) Let $\lambda$ and $\mu$ be two partitions. We must prove that the number
$a_{\lambda,\mu}$ is well-defined. In other words, we must prove that there
are only finitely many $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$.

But this is easy: Any $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ is a map from the set
$\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{
1,2,\ldots,\ell\left(  \mu\right)  \right\}  $ to the set $\left\{
0,1\right\}  $. Thus,%
\begin{align*}
&  \left\{  \left\{  0,1\right\}  \text{-matrices of size }\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  \right\} \\
&  =\left\{  \text{maps from the set }\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  \text{ to the set }\left\{  0,1\right\}  \right\} \\
&  =\left\{  0,1\right\}  ^{\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  }%
\end{align*}
is a finite set (since both $\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  $ and
$\left\{  0,1\right\}  $ are finite sets). In other words, there are only
finitely many $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $. Hence, there are only
finitely many $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$. This solves Exercise \ref{exe.Lambda.triangular}(f).

[\textit{Remark:} We can prove a slightly stronger claim: Namely, there are
only finitely many matrices in $\mathbb{N}^{\ell\left(  \lambda\right)
\times\ell\left(  \mu\right)  }$ having row sums $\lambda$ and column sums
$\mu$.

Let us sketch the proof of this claim. Indeed, let $N=\left\vert
\lambda\right\vert $. Then, if $A$ is any matrix in $\mathbb{N}^{\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  }$ having row sums $\lambda$ and
column sums $\mu$, then the sum of all entries of $A$ must equal $\left\vert
\lambda\right\vert =N$, and therefore each entry of $A$ must be $\leq N$
(since a sum of nonnegative integers is always $\geq$ to each of its addends);
but this entails that each entry of $A$ belongs to the finite set $\left\{
0,1,\ldots,N\right\}  $, and therefore there are only finitely many choices
for each entry, which leads to only finitely many possible matrices $A$.]

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(g).
\end{commentedout}

(g) Exercise \ref{exe.Lambda.triangular}(g) is truly not a deep fact, but its
proof requires some bookkeeping. In order to make this bookkeeping more
palatable, we are going to introduce various auxiliary notations.

\begin{definition}
Let $q\in \NN$. Let $\overline{\mathbf{x}}_{q}$ denote the $q$-tuple
$\left(  x_{1},x_{2},\ldots,x_{q}\right)  $ of indeterminates. Let
$\kk\left[  \overline{\mathbf{x}}_{q}\right]  $ denote the polynomial
ring $\kk\left[  x_{1},x_{2},\ldots,x_{q}\right]  $. Let $\eta
_{q}:R\left(  \xx\right)  \rightarrow\kk\left[  \overline{\xx}_{q}
\right]  $ be the map that sends every power series $f\in
R\left(  \xx\right)  $ to the polynomial $f\left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  $. (This is well-defined, because Exercise
\ref{exe.Lambda.subs.fin} (applied to
$A=\kk\left[  \overline{\xx}_{q}\right]  $ and $k=q$) shows that
substituting $x_{1}%
,x_{2},\ldots,x_{q},0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in $f$ yields
an infinite sum in which all but finitely many addends are zero.)

The map $\eta_{q}$ is an evaluation homomorphism (in an appropriate
sense\footnote{i.e., it acts on a power series $f\in R\left(  \xx%
\right)  $ by substituting certain values for the indeterminates $x_{1}%
,x_{2},x_{3},\ldots$}); thus, it is a $\kk$-algebra homomorphism.

If $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  \in
\mathbb{N}^{q}$ is a $q$-tuple of nonnegative integers, then
$\overline{\xx}_{q}^{\beta}$ shall denote the monomial
$x_{1}^{\beta_{1}}%
x_{2}^{\beta_{2}}\cdots x_{q}^{\beta_{q}}$. This is a monomial in the
polynomial ring $\kk\left[  \overline{\mathbf{x}}_{q}\right]  $.

For every $f\in\kk\left[  \left[  \overline{\mathbf{x}}_{q}\right]
\right]  $ and $\beta\in\mathbb{N}^{q}$, we let $\left[  \overline{\xx%
}_{q}^{\beta}\right]  f$ denote the coefficient of the monomial
$\overline{\xx}_{q}^{\beta}$ in the power series $f$.
\end{definition}

Let us make the following simple observations:

\begin{itemize}
\item Every $q\in \NN$ and $i\in\left\{  1,2,\ldots,q\right\}  $ satisfy%
\begin{align}
\eta_{q}\left(  x_{i}\right)   &  =x_{i}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta_{q}\right) \nonumber\\
&  =x_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\in\left\{  1,2,\ldots
,q\right\}  \right)  . \label{pf.lem.sol.Lambda.triangular.etaq-coeff.eta1}%
\end{align}


\item Every $q\in \NN$ and $i\in\left\{  q+1,q+2,q+3,\ldots\right\}  $
satisfy%
\begin{align}
\eta_{q}\left(  x_{i}\right)   &  =x_{i}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta_{q}\right) \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\in\left\{  q+1,q+2,q+3,\ldots
\right\}  \right)  . \label{pf.lem.sol.Lambda.triangular.etaq-coeff.eta2}%
\end{align}


\item Any two $q$-tuples $\phi\in\mathbb{N}^{q}$ and $\psi\in\mathbb{N}^{q}$
satisfy%
\begin{equation}
\left[  \overline{\mathbf{x}}_{q}^{\phi}\right]  \left(  \overline{\xx%
}_{q}^{\psi}\right)  =\delta_{\phi,\psi}.
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.4}%
\end{equation}
(Indeed, $\overline{\mathbf{x}}_{q}^{\psi}$ and $\overline{\mathbf{x}}%
_{q}^{\phi}$ are two distinct monomials if $\phi\neq\psi$, and are two
identical monomials if $\phi=\psi$.)
\end{itemize}

\begin{lemma}
\label{lem.sol.Lambda.triangular.etaq-coeff}Let $q\in \NN$. Let $\beta$
be a weak composition. Assume that $\beta_{i}=0$ for every integer $i>q$. Let
$f\in R\left(  \xx\right)  $. Then,%
\[
\left[  \mathbf{x}^{\beta}\right]  f=\left[  \overline{\mathbf{x}}%
_{q}^{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(
\eta_{q}\left(  f\right)  \right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.etaq-coeff}.]Let us prove that
every weak composition $\alpha$ satisfies%
\begin{equation}
\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)
=\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(  \mathbf{x}^{\alpha
}\right)  \right)  . \label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}):}
Let $\alpha$ be a weak composition. We must prove the equality
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}).

We distinguish between two cases:

\textit{Case 1:} We have $\left(  \alpha_{i}=0\text{ for every integer
}i>q\right)  $.

\textit{Case 2:} We don't have $\left(  \alpha_{i}=0\text{ for every integer
}i>q\right)  $.

Let us first consider Case 1. In this case, we have
\begin{equation}
\left(  \alpha_{i}=0\text{ for every integer }i>q\right)  .
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1}%
\end{equation}


Thus,%
\begin{equation}
\prod_{i=q+1}^{\infty}\underbrace{x_{i}^{\alpha_{i}}}_{\substack{=x_{i}%
^{0}\\\text{(since }\alpha_{i}=0\text{ (by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1})))}}%
}=\prod_{i=q+1}^{\infty}\underbrace{x_{i}^{0}}_{=1}=1.
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1prod}%
\end{equation}
Now,%
\begin{equation}
\delta_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }=\delta_{\beta,\alpha}
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}):} We are in
one of the following two subcases:
\par
\textit{Subcase 1.1:} We have $\beta\neq\alpha$.
\par
\textit{Subcase 1.2:} We have $\beta=\alpha$.
\par
Let us first consider Subcase 1.1. In this subcase, we have $\beta\neq\alpha$.
In other words, $\alpha\neq\beta$. Hence, there exists some $k\in\left\{
1,2,3,\ldots\right\}  $ satisfying $\alpha_{k}\neq\beta_{k}$. Consider this
$k$.
\par
We claim that $k\leq q$. Indeed, assume the contrary (for the sake of
contradiction). Then, $k>q$. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1}) (applied to
$i=k$) yields $\alpha_{k}=0$. But let us recall that $\beta_{i}=0$ for every
integer $i>q$. Applying this to $i=k$, we find $\beta_{k}=0$. Hence,
$\alpha_{k}=0=\beta_{k}$. This contradicts $\alpha_{k}\neq\beta_{k}$.
\par
This contradiction completes the proof of $k\leq q$. Hence, $k\in\left\{
1,2,\ldots,q\right\}  $. Hence, there exists some $i\in\left\{  1,2,\ldots
,q\right\}  $ satisfying $\alpha_{i}\neq\beta_{i}$ (namely, $i=k$). Therefore,
$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  \neq\left(  \beta
_{1},\beta_{2},\ldots,\beta_{q}\right)  $. In other words, $\left(  \beta
_{1},\beta_{2},\ldots,\beta_{q}\right)  \neq\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{q}\right)  $. Thus, $\delta_{\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{q}\right)  }=0$. Comparing this with $\delta_{\beta,\alpha}=0$ (since
$\beta\neq\alpha$), we obtain $\delta_{\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{q}\right)  }=\delta_{\beta,\alpha}$. Thus,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) is proven in
Subcase 1.1.
\par
Let us now consider Subcase 1.2. In this subcase, we have $\beta=\alpha$.
Hence, $\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  =\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  $. Thus, $\delta_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{q}\right)  }=1$. Comparing this with $\delta_{\beta
,\alpha}=1$ (since $\beta=\alpha$), we obtain $\delta_{\left(  \beta_{1}%
,\beta_{2},\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{q}\right)  }=\delta_{\beta,\alpha}$. Thus,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) is proven in
Subcase 1.2.
\par
We have now proven
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) in each of
the two Subcases 1.1 and 1.2. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) always
holds.}.

Now, the definition of $\mathbf{x}^{\alpha}$ yields%
\[
\mathbf{x}^{\alpha}=\prod_{i\geq1}x_{i}^{\alpha_{i}}=\left(  \prod_{i=1}%
^{q}x_{i}^{\alpha_{i}}\right)  \underbrace{\left(  \prod_{i=q+1}^{\infty}%
x_{i}^{\alpha_{i}}\right)  }_{\substack{=1\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1prod}))}}%
}=\prod_{i=1}^{q}x_{i}^{\alpha_{i}}.
\]
Applying the map $\eta_{q}$ to both sides of this equality, we obtain%
\begin{align}
\eta_{q}\left(  \mathbf{x}^{\alpha}\right)   &  =\eta_{q}\left(  \prod
_{i=1}^{q}x_{i}^{\alpha_{i}}\right)  =\prod_{i=1}^{q}\left(  \underbrace{\eta
_{q}\left(  x_{i}\right)  }_{\substack{=x_{i}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta1}))}}}\right)  ^{\alpha_{i}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{q}\text{ is a }%
\kk\text{-algebra homomorphism}\right) \nonumber\\
&  =\prod_{i=1}^{q}x_{i}^{\alpha_{i}}=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{q}^{\alpha_{q}}=\overline{\mathbf{x}}_{q}^{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.3}%
\end{align}
(since $\overline{\mathbf{x}}_{q}^{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{q}\right)  }$ is defined to be $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{q}^{\alpha_{q}}$).

Now,%
\begin{align*}
&  \left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  }_{\substack{=\overline{\mathbf{x}}_{q}^{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.3}))}}}\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{q}\right)  }\right]  \left(  \overline{\mathbf{x}}_{q}^{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }\right)  =\delta_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{q}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.4}) (applied to
}\phi=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  \text{ and }%
\psi=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  \text{)}\right)
\\
&  =\delta_{\beta,\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2})}\right)  .
\end{align*}
Comparing this with%
\[
\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)
=\delta_{\beta,\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.coeff-mon}) (applied to }\mu=\beta\text{)}\right)
,
\]
we obtain $\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha
}\right)  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(  \xx%
^{\alpha}\right)  \right)  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) is proven in Case 1.

Let us now consider Case 2. In this case, we don't have $\left(  \alpha
_{i}=0\text{ for every integer }i>q\right)  $. In other words, there exists
some integer $i>q$ such that $\alpha_{i}\neq0$. Consider this $i$, and denote
it by $k$. Thus, $k$ is an integer such that $k>q$ and $\alpha_{k}\neq0$.

Recall that $\beta_{i}=0$ for every integer $i>q$. Applying this to $i=k$, we
obtain $\beta_{k}=0\neq\alpha_{k}$ (since $\alpha_{k}\neq0$). Thus, $\beta
\neq\alpha$, so that $\delta_{\beta,\alpha}=0$.

On the other hand, $\alpha_{k}\neq0$ and thus $\alpha_{k}>0$ (since
$\alpha_{k}\in \NN$). Hence, the monomial $\mathbf{x}^{\alpha}$ is
divisible by $x_{k}$. In other words, there exists a monomial $g\in
\kk\left[  \left[  \xx\right]  \right]  $ such that
$\mathbf{x}^{\alpha}=gx_{k}$. Consider this $g$. From $k>q$, we obtain
$k\in\left\{  q+1,q+2,q+3,\ldots\right\}  $. Thus,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta2}) (applied to $i=k$) yields
$\eta_{q}\left(  x_{k}\right)  =0$. Now,%
\begin{align*}
\eta_{q}\left(  \underbrace{\mathbf{x}^{\alpha}}_{=gx_{k}}\right)   &
=\eta_{q}\left(  gx_{k}\right)  =\eta_{q}\left(  g\right)  \underbrace{\eta
_{q}\left(  x_{k}\right)  }_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\eta_{q}\text{ is a }\kk\text{-algebra homomorphism}\right) \\
&  =0.
\end{align*}
Hence,%
\[
\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  }_{=0}\right)  =\left[  \overline{\mathbf{x}}%
_{q}^{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(
0\right)  =0.
\]
Comparing this with%
\begin{align*}
\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)   &
=\delta_{\beta,\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.coeff-mon}) (applied to }\mu=\beta\text{)}\right)
\\
&  =0,
\end{align*}
we obtain $\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha
}\right)  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(  \xx%
^{\alpha}\right)  \right)  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) is proven in Case 2.

We have now proven (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) in
each of the two Cases 1 and 2. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) always holds.]

\begin{commentedout}
Let us first notice that every polynomial $g\in\kk\left[
\overline{\mathbf{x}}_{q}\right]  $ satisfies%
\begin{equation}
g=\sum_{\alpha\in\mathbb{N}^{q}}\left[  \overline{\mathbf{x}}_{q}^{\alpha
}\right]  \left(  g\right)  \cdot\overline{\mathbf{x}}_{q}^{\alpha}
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.1}%
\end{equation}
(since the family of the coefficients of the polynomial $g$ is $\left(
\left[  \overline{\mathbf{x}}_{q}^{\alpha}\right]  \left(  g\right)  \right)
_{\alpha\in\mathbb{N}^{q}}$).
\end{commentedout}

Now, let us notice that every power series $g\in\kk\left[  \left[
\xx\right]  \right]  $ satisfies%
\begin{equation}
g=\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  g\right)  \cdot\mathbf{x}^{\alpha}
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.6}%
\end{equation}
(since the family of the coefficients of $g$ is $\left(  \left[
\mathbf{x}^{\alpha}\right]  \left(  g\right)  \right)  _{\alpha\in
\operatorname*{WC}}$). Applying this to $g=f$, we obtain%
\[
f=\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\mathbf{x}^{\alpha}.
\]
Substituting $x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots$ for the variables
$x_{1},x_{2},x_{3},\ldots$ in this equality, we obtain%
\begin{align}
f\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)   &  =\sum_{\alpha
\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)
\cdot\underbrace{\mathbf{x}^{\alpha}\left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  }_{\substack{=\eta_{q}\left(  \mathbf{x}^{\alpha
}\right)  \\\text{(since }\eta_{q}\left(  \mathbf{x}^{\alpha}\right)  \text{
is defined to be }\mathbf{x}^{\alpha}\left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  \text{)}}}\nonumber\\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\eta_{q}\left(  \mathbf{x}^{\alpha}\right)  .
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.9}%
\end{align}
Now, the definition of $\eta_{q}$ yields%
\begin{align*}
\eta_{q}\left(  f\right)   &  =f\left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right) \\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\eta_{q}\left(  \mathbf{x}^{\alpha}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.9})}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(  f\right)
}_{=\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\eta_{q}\left(  \mathbf{x}^{\alpha}\right)  }\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{q}\right)  }\right]  \left(  \sum_{\alpha\in\operatorname*{WC}%
}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)  \cdot\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  \right) \\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\underbrace{\left[  \overline{\mathbf{x}}_{q}^{\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(
\mathbf{x}^{\alpha}\right)  \right)  }_{\substack{=\left[  \mathbf{x}^{\beta
}\right]  \left(  \mathbf{x}^{\alpha}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}))}}}\\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\left[  \mathbf{x}^{\beta}\right]  \left(
\mathbf{x}^{\alpha}\right)  .
\end{align*}
Comparing this with%
\[
\left[  \mathbf{x}^{\beta}\right]  \underbrace{f}_{=\sum_{\alpha
\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)
\cdot\mathbf{x}^{\alpha}}=\left[  \mathbf{x}^{\beta}\right]  \left(
\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]  \left(
f\right)  \cdot\mathbf{x}^{\alpha}\right)  =\sum_{\alpha\in\operatorname*{WC}%
}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)  \cdot\left[
\mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)  ,
\]
we obtain $\left[  \mathbf{x}^{\beta}\right]  f=\left[  \overline{\xx%
}_{q}^{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(
\eta_{q}\left(  f\right)  \right)  $. This proves Lemma
\ref{lem.sol.Lambda.triangular.etaq-coeff}.
\end{proof}

We now need to introduce some more notations.

\begin{definition}
If $q\in \NN$, and if $\beta=\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  \in\mathbb{N}^{q}$ is a $q$-tuple of nonnegative integers,
then $\left\vert \beta\right\vert $ shall denote the sum $\beta_{1}+\beta
_{2}+\cdots+\beta_{q}\in \NN$.
\end{definition}

\begin{definition}
We shall use the so-called \dfn{Iverson bracket notation}: For every
assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $ denote the
integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
\end{definition}

Let us state a fundamental fact in combinatorics:

\begin{lemma}
\label{lem.sol.Lambda.triangular.IPK}Let $q\in \NN$. Let $m\in
\mathbb{N}$. Define a set $\mathfrak{I}$ by%
\[
\mathfrak{I}=\left\{  \left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m}\ \mid\ i_{1}<i_{2}<\cdots<i_{m}\right\}  .
\]
Let $\mathfrak{P}$ be the set of all $m$-element subsets of $\left\{
1,2,\ldots,q\right\}  $. Define a set $\mathfrak{K}$ by%
\[
\mathfrak{K}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  .
\]


\begin{enumerate}
\item[(a)] The map $\mathfrak{I}\rightarrow\mathfrak{P},\ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
$ is well-defined and bijective.

\item[(b)] The map $\mathfrak{P}\rightarrow\mathfrak{K},\ T\mapsto\left(
\left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]
\right)  $ is well-defined and bijective.

\item[(c)] There exists a bijection $\Psi:\mathfrak{I}\rightarrow\mathfrak{K}$
with the property that every $\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\mathfrak{I}$ satisfies $\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.IPK}.](a) The set
$\mathfrak{I}$ is the set of all strictly increasing lists of $m$ elements of
$\left\{  1,2,\ldots,q\right\}  $. Meanwhile, the set $\mathfrak{P}$ is the
set of all $m$-element subsets of $\left\{  1,2,\ldots,q\right\}  $. Hence,
there are well-known bijections between these two sets: The maps%
\[
\mathfrak{I}\rightarrow\mathfrak{P},\ \ \ \ \ \ \ \ \ \ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
\]
and%
\[
\mathfrak{P}\rightarrow\mathfrak{I},\ \ \ \ \ \ \ \ \ \ T\mapsto\left(
\text{the increasing list of }T\right)
\]
\footnote{Here, the \emph{increasing list} of a subset $T$ of $\left\{
1,2,\ldots,q\right\}  $ is defined to be the list of all elements of $T$ in
increasing order (with no repetitions).} are mutually inverse bijections. In
particular, the map $\mathfrak{I}\rightarrow\mathfrak{P},\ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
$ is well-defined and bijective. This proves Lemma
\ref{lem.sol.Lambda.triangular.IPK}(a).

(b) It is well-known that the map%
\[
\Phi:\left\{  \text{subsets }T\text{ of }\left\{  1,2,\ldots,q\right\}
\right\}  \rightarrow\left\{  0,1\right\}  ^{q},\ \ \ \ \ \ \ \ \ \ T\mapsto
\left(  \left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in
T\right]  \right)
\]
is a bijection\footnote{Indeed, its inverse is the map that sends any $\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  \in\left\{  0,1\right\}  ^{q}$ to
the subset $\left\{  i\in\left\{  1,2,\ldots,q\right\}  \ \mid\ \beta
_{i}=1\right\}  $ of $\left\{  1,2,\ldots,q\right\}  $.}. Furthermore, this
bijection $\Phi$ has the property that each subset $T$ of $\left\{
1,2,\ldots,q\right\}  $ satisfies $\left\vert \Phi\left(  T\right)
\right\vert =\left\vert T\right\vert $. Thus, in particular, a subset $T$ of
$\left\{  1,2,\ldots,q\right\}  $ satisfies $\left\vert \Phi\left(  T\right)
\right\vert =m$ if and only if it satisfies $\left\vert T\right\vert =m$.
Hence, $\Phi$ restricts to a bijection%
\begin{align}
\left\{  \text{subsets }T\text{ of }\left\{  1,2,\ldots,q\right\}
\ \mid\ \left\vert T\right\vert =m\right\}   &  \rightarrow\left\{  \beta
\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert \beta\right\vert =m\right\}
,\nonumber\\
T  &  \mapsto\left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)  .
\label{pf.lem.sol.Lambda.triangular.IPK.b.3}%
\end{align}
Thus, the map (\ref{pf.lem.sol.Lambda.triangular.IPK.b.3}) is well-defined and
bijective. Since $\left\{  \text{subsets }T\text{ of }\left\{  1,2,\ldots
,q\right\}  \ \mid\ \left\vert T\right\vert =m\right\}  =\mathfrak{P}$ and
$\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert \beta
\right\vert =m\right\}  =\mathfrak{K}$, this rewrites as follows: The map
$\mathfrak{P}\rightarrow\mathfrak{K},\ T\mapsto\left(  \left[  1\in T\right]
,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]  \right)  $ is
well-defined and bijective. This proves Lemma
\ref{lem.sol.Lambda.triangular.IPK}(b).

(c) Let $A$ be the map $\mathfrak{I}\rightarrow\mathfrak{P},\ \left(
j_{1},j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots
,j_{m}\right\}  $. Lemma \ref{lem.sol.Lambda.triangular.IPK}(a) shows that
this map $A$ is well-defined and bijective.

Let $B$ be the map $\mathfrak{P}\rightarrow\mathfrak{K},\ T\mapsto\left(
\left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]
\right)  $. Lemma \ref{lem.sol.Lambda.triangular.IPK}(b) shows that this map
$B$ is well-defined and bijective.

So the maps $B$ and $A$ are bijective. Hence, their composition $B\circ A$ is
also bijective. Thus, $B\circ A:\mathfrak{I}\rightarrow\mathfrak{K}$ is a
bijection. It has the property that every $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\mathfrak{I}$ satisfies $\overline{\mathbf{x}}_{q}^{\left(
B\circ A\right)  \left(  i_{1},i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}$. We must prove that
$\overline{\mathbf{x}}_{q}^{\left(  B\circ A\right)  \left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}$.
\par
From $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}$, we conclude
that $\left(  i_{1},i_{2},\ldots,i_{m}\right)  $ is an element of $\left\{
1,2,\ldots,q\right\}  ^{m}$ satisfying $i_{1}<i_{2}<\cdots<i_{m}$ (by the
definition of $\mathfrak{I}$).
\par
Define $T\in\mathfrak{P}$ by $T=A\left(  i_{1},i_{2},\ldots,i_{m}\right)  $.
Then, $T=A\left(  i_{1},i_{2},\ldots,i_{m}\right)  =\left\{  i_{1}%
,i_{2},\ldots,i_{m}\right\}  $ (by the definition of $A$). Hence, $\left(
i_{1},i_{2},\ldots,i_{m}\right)  $ is a list of all elements of $T$.
Furthermore, this list has no repetitions (since $i_{1},i_{2},\ldots,i_{m}$
are distinct (since $i_{1}<i_{2}<\cdots<i_{m}$)). Hence, $\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  $ is a list of all elements of $T$ with no
repetitions. Therefore,%
\[
\prod_{i\in T}x_{i}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}.
\]
Also, $T$ is a subset of $\left\{  1,2,\ldots,q\right\}  $ (since
$T\in\mathfrak{P}$). Now,%
\[
\left(  B\circ A\right)  \left(  i_{1},i_{2},\ldots,i_{m}\right)  =B\left(
\underbrace{A\left(  i_{1},i_{2},\ldots,i_{m}\right)  }_{=T}\right)  =B\left(
T\right)  =\left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)
\]
(by the definition of $B$). Hence,%
\begin{align*}
&  \overline{\mathbf{x}}_{q}^{\left(  B\circ A\right)  \left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }\\
&  =\overline{\mathbf{x}}_{q}^{\left(  \left[  1\in T\right]  ,\left[  2\in
T\right]  ,\ldots,\left[  q\in T\right]  \right)  }=x_{1}^{\left[  1\in
T\right]  }x_{2}^{\left[  2\in T\right]  }\cdots x_{q}^{\left[  q\in T\right]
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{\xx%
}_{q}^{\left(  \left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[
q\in T\right]  \right)  }\right) \\
&  =\prod_{i\in\left\{  1,2,\ldots,q\right\}  }x_{i}^{\left[  i\in T\right]
}=\left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}\underbrace{x_{i}^{\left[  i\in T\right]  }}_{\substack{=x_{i}%
^{1}\\\text{(since }\left[  i\in T\right]  =1\\\text{(since }i\in T\text{))}%
}}\right)  \left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}
;\\\text{not }i\in T}}\underbrace{x_{i}^{\left[  i\in T\right]  }%
}_{\substack{=x_{i}^{0}\\\text{(since }\left[  i\in T\right]  =0\\\text{(since
we don't have }i\in T\text{))}}}\right) \\
&  =\left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}\underbrace{x_{i}^{1}}_{=x_{i}}\right)  \left(  \prod\limits_{\substack{i\in
\left\{  1,2,\ldots,q\right\}  ;\\\text{not }i\in T}}\underbrace{x_{i}^{0}%
}_{=1}\right)  =\left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}
;\\i\in T}}x_{i}\right)  \underbrace{\left(  \prod\limits_{\substack{i\in\left\{
1,2,\ldots,q\right\}  ;\\\text{not }i\in T}}1\right)  }_{=1}\\
&  =\underbrace{\prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}}_{\substack{=\prod_{i\in T}\\\text{(since }T\text{ is a subset of
}\left\{  1,2,\ldots,q\right\}  \text{)}}}x_{i}=\prod_{i\in T}x_{i}=x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}.
\end{align*}
}. Hence, there exists a bijection $\Psi:\mathfrak{I}\rightarrow\mathfrak{K}$
with the property that every $\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\mathfrak{I}$ satisfies $\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}$ (namely,
$\Psi=B\circ A$). This proves Lemma \ref{lem.sol.Lambda.triangular.IPK}(c).
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.em}Let $q\in \NN$. Let $m\in\mathbb{N}%
$. Then,%
\[
\eta_{q}\left(  e_{m}\right)  =\sum\limits_{\substack{\beta\in\left\{  0,1\right\}
^{q};\\\left\vert \beta\right\vert =m}}\overline{\mathbf{x}}_{q}^{\beta}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.em}.]The definition of $e_{m}$
yields%
\[
e_{m}=\underbrace{\sum_{i_{1}<i_{2}<\cdots<i_{m}}}_{=\sum\limits_{\substack{\left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}
^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}%
=\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}.
\]
Substituting $x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots$ for the variables
$x_{1},x_{2},x_{3},\ldots$ in this equality, we obtain%
\begin{align}
& e_{m}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  \nonumber\\
& =\sum\limits
_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  . \label{pf.lem.sol.Lambda.triangular.em.1}%
\end{align}


Note that $\left\{  1,2,\ldots,q\right\}  ^{m}$ is a subset of $\left\{
1,2,3,\ldots\right\}  ^{m}$. Moreover, the following holds:

\begin{itemize}
\item If $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{m}$ is an $m$-tuple that does not satisfy $\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$, then
\begin{equation}
\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  =0
\label{pf.lem.sol.Lambda.triangular.em.2a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.em.2a}):} Let
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}
^{m}$ be an $m$-tuple that does not satisfy $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$.
\par
If every $k\in\left\{  1,2,\ldots,m\right\}  $ would satisfy $i_{k}\in\left\{
1,2,\ldots,q\right\}  $, then we would have $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$, which would contradict
the fact that we do not have $\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}$. Hence, not every $k\in\left\{
1,2,\ldots,m\right\}  $ satisfies $i_{k}\in\left\{  1,2,\ldots,q\right\}  $.
In other words, there exists some $k\in\left\{  1,2,\ldots,m\right\}  $
satisfying $i_{k}\notin\left\{  1,2,\ldots,q\right\}  $. Consider this $k$.
\par
The element $i_{k}$ belongs to $\left\{  1,2,3,\ldots\right\}  $ but not to
$\left\{  1,2,\ldots,q\right\}  $ (since $i_{k}\notin\left\{  1,2,\ldots
,q\right\}  $). Thus, $i_{k}\in\left\{  1,2,3,\ldots\right\}  \setminus
\left\{  1,2,\ldots,q\right\}  =\left\{  q+1,q+2,q+3,\ldots\right\}  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta2}) (applied to $i=i_{k}$)
yields $\eta_{q}\left(  x_{i_{k}}\right)  =0$.
\par
Now,
\begin{align*}
&  \underbrace{\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  }%
_{=\prod_{j\in\left\{  1,2,\ldots,m\right\}  }x_{i_{j}}}\left(  x_{1}%
,x_{2},\ldots,x_{q},0,0,0,\ldots\right) \\
&  =\left(  \prod_{j\in\left\{  1,2,\ldots,m\right\}  }x_{i_{j}}\right)
\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  =\prod_{j\in\left\{
1,2,\ldots,m\right\}  }\underbrace{x_{i_{j}}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  }_{\substack{=\eta_{q}\left(  x_{i_{j}}\right)
\\\text{(since }\eta_{q}\left(  x_{i_{j}}\right)  \text{ is defined to be
}x_{i_{j}}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  \text{)}}}\\
&  =\prod_{j\in\left\{  1,2,\ldots,m\right\}  }\eta_{q}\left(  x_{i_{j}%
}\right)  =\underbrace{\eta_{q}\left(  x_{i_{k}}\right)  }_{=0}\cdot
\prod\limits_{\substack{j\in\left\{  1,2,\ldots,m\right\}  ;\\j\neq k}}\eta
_{q}\left(  x_{i_{j}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have
split off the factor for }j=k\text{ from the product}\right) \\
&  =0\cdot\prod\limits_{\substack{j\in\left\{  1,2,\ldots,m\right\}  ;\\j\neq k}%
}\eta_{q}\left(  x_{i_{j}}\right)  =0.
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.em.2a}).}.

\item If $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots
,q\right\}  ^{m}$, then%
\begin{equation}
\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  =x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}
\label{pf.lem.sol.Lambda.triangular.em.2b}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.em.2b}):} Let
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}
^{m}$.
\par
Let $k\in\left\{  1,2,\ldots,m\right\}  $. From $\left(  i_{1},i_{2}%
,\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$, we obtain
$i_{k}\in\left\{  1,2,\ldots,q\right\}  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta1}) (applied to $i=i_{k}$)
yields $\eta_{q}\left(  x_{i_{k}}\right)  =x_{i_{k}}$.
\par
Let us forget that we fixed $k$. We thus have shown that $\eta_{q}\left(
x_{i_{k}}\right)  =x_{i_{k}}$ for each $k\in\left\{  1,2,\ldots,m\right\}  $.
Hence, $\prod_{k=1}^{m}\underbrace{\eta_{q}\left(  x_{i_{k}}\right)
}_{=x_{i_{k}}}=\prod_{k=1}^{m}x_{i_{k}}$. Now,
\begin{align*}
&  \underbrace{\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  }%
_{=\prod_{k=1}^{m}x_{i_{k}}}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots
\right) \\
&  =\left(  \prod_{k=1}^{m}x_{i_{k}}\right)  \left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  =\prod_{k=1}^{m}\underbrace{x_{i_{k}}\left(
x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  }_{\substack{=\eta_{q}\left(
x_{i_{k}}\right)  \\\text{(since }\eta_{q}\left(  x_{i_{k}}\right)  \text{ is
defined to be }x_{i_{k}}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)
\text{)}}}\\
&  =\prod_{k=1}^{m}\eta_{q}\left(  x_{i_{k}}\right)  =\prod_{k=1}^{m}x_{i_{k}%
}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}.
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.em.2b}).}.
\end{itemize}

Now, (\ref{pf.lem.sol.Lambda.triangular.em.1}) becomes%
\begin{align}
&  e_{m}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m};\\\left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}}%
}}_{\substack{=\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}%
}}\\\text{(since }\left\{  1,2,\ldots,q\right\}  ^{m}\text{ is a subset of
}\left\{  1,2,3,\ldots\right\}  ^{m}\text{)}}}\underbrace{\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  }_{\substack{=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}%
}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.em.2b}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}%
<\cdots<i_{m};\\\text{not }\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}}}\underbrace{\left(  x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots
\right)  }_{\substack{=0\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.em.2a}%
))}}}\nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}+\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}%
<\cdots<i_{m};\\\text{not }\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}}}0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}. \label{pf.lem.sol.Lambda.triangular.em.4}%
\end{align}


Now, let us define the sets $\mathfrak{I}$, $\mathfrak{P}$ and $\mathfrak{K}$
as in Lemma \ref{lem.sol.Lambda.triangular.IPK}. Then, Lemma
\ref{lem.sol.Lambda.triangular.IPK}(c) shows that there exists a bijection
$\Psi:\mathfrak{I}\rightarrow\mathfrak{K}$ with the property that every
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}$ satisfies
\begin{equation}
\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)
}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}.
\label{pf.lem.sol.Lambda.triangular.em.Psi}%
\end{equation}
Consider this $\Psi$.

Now, (\ref{pf.lem.sol.Lambda.triangular.em.4}) becomes%
\begin{align*}
&  e_{m}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right) \\
&  =\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}%
}}_{\substack{=\sum_{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}%
}\\\text{(since }\mathfrak{I}=\left\{  \left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}\ \mid\ i_{1}%
<i_{2}<\cdots<i_{m}\right\}  \text{)}}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}%
=\sum_{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}}%
\underbrace{x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}}_{\substack{=
\overline{\xx}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)  }\\
\text{(by (\ref{pf.lem.sol.Lambda.triangular.em.Psi}))}}}\\
&  =\sum_{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}}%
\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)
}=\underbrace{\sum_{\beta\in\mathfrak{K}}}_{\substack{=\sum\limits_{\substack{\beta
\in\left\{  0,1\right\}  ^{q};\\\left\vert \beta\right\vert =m}}\\\text{(since
}\mathfrak{K}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  \text{)}}}\overline{\mathbf{x}}_{q}^{\beta}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\beta\text{ for }\Psi\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \text{ in the sum,}\\
\text{since the map }\Psi:\mathfrak{I}\rightarrow\mathfrak{K}\text{ is a
bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\beta\in\left\{  0,1\right\}  ^{q};\\\left\vert
\beta\right\vert =m}}\overline{\mathbf{x}}_{q}^{\beta}.
\end{align*}
Now, the definition of $\eta_{q}$ yields%
\[
\eta_{q}\left(  e_{m}\right)  =e_{m}\left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  =\sum\limits_{\substack{\beta\in\left\{  0,1\right\}
^{q};\\\left\vert \beta\right\vert =m}}\overline{\mathbf{x}}_{q}^{\beta}.
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.em}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.g-set}Let $X$ and $Y$ be two sets. Let
$\phi:X\rightarrow Y$ be a bijection. Let $Z$ be a subset of $X$. Then, the
map $Z\rightarrow\phi\left(  Z\right)  ,\ A\mapsto\phi\left(  A\right)  $ is
well-defined and is a bijection.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.g-set}.]Lemma
\ref{lem.sol.Lambda.triangular.g-set} is a fundamental and trivial fact of
set theory.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.ela}Let $q\in \NN$. Let $p\in
\mathbb{N}$. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{p}\right)  \in\mathbb{N}^{p}$ be a $p$-tuple of nonnegative integers. Then,%
\[
\eta_{q}\left(  e_{\alpha_{1}}e_{\alpha_{2}}\cdots e_{\alpha_{p}}\right)
=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\alpha}}
\overline{\xx}_{q}^{\operatorname*{colsums}A}.
\]
Here, $\operatorname*{colsums}A$ denotes the column sums of $A$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.ela}.]For each matrix
$A\in\left\{  0,1\right\}  ^{p\times q}$ and each $i\in\left\{  1,2,\ldots
,p\right\}  $, we let $\operatorname*{row}\nolimits_{i}A$ denote the $i$-th
row of $A$. This $\operatorname{row}_{i}A$ is an element of $\left\{
0,1\right\}  ^{q}$.

Let $\Phi:\left\{  0,1\right\}  ^{p\times q}\rightarrow\left(  \left\{
0,1\right\}  ^{q}\right)  ^{p}$ be the map that sends each matrix
$A\in\left\{  0,1\right\}  ^{p\times q}$ to the list \newline
$\left(
\operatorname*{row}\nolimits_{1}A,\operatorname{row}_{2}A,\ldots
,\operatorname{row}_{p}A\right)  $ of all the rows of $A$. Then, the map
$\Phi$ is a bijection (because a matrix can be viewed as a list of rows).

For each matrix $A\in\left\{  0,1\right\}  ^{p\times q}$, we let
$\operatorname*{rowsums}A$ denote the row sums of $A$. (This is a $p$-tuple in
$\mathbb{N}^{p}$.) Furthermore, for each matrix $A\in\left\{  0,1\right\}
^{p\times q}$, we let $\operatorname*{colsums}A$ denote the column sums of
$A$. (This is a $q$-tuple in $\mathbb{N}^{q}$.)

Every matrix $A\in\left\{  0,1\right\}  ^{p\times q}$ satisfies%
\begin{equation}
\prod_{i=1}^{p}\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}%
=\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}A}
\label{pf.lem.sol.Lambda.triangular.ela.prodxA}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.prodxA}):}
Let $A\in\left\{  0,1\right\}  ^{p\times q}$ be a matrix.
\par
Write the $p\times q$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}$.
\par
Let $i\in\left\{  1,2,\ldots,p\right\}  $. The definition of
$\operatorname{row}_{i}A$ yields%
\[
\operatorname{row}_{i}A=\left(  \text{the }i\text{-th row of }A\right)
=\left(  a_{i,1},a_{1,2},\ldots,a_{i,q}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }A=\left(  a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq
q}\right)  .
\]
Hence,%
\begin{align*}
\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}  &  =\overline{\xx%
}_{q}^{\left(  a_{i,1},a_{1,2},\ldots,a_{i,q}\right)  }=x_{1}^{a_{i,1}}%
x_{2}^{a_{i,2}}\cdots x_{q}^{a_{i,q}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\overline{\mathbf{x}}_{q}^{\left(  a_{i,1},a_{1,2}%
,\ldots,a_{i,q}\right)  }\right) \\
&  =\prod_{j=1}^{q}x_{j}^{a_{i,j}}.
\end{align*}
\par
Now, forget that we fixed $i$. We thus have proven $\overline{\mathbf{x}}%
_{q}^{\operatorname{row}_{i}A}=\prod_{j=1}^{q}x_{j}^{a_{i,j}}$ for each
$i\in\left\{  1,2,\ldots,p\right\}  $. Hence,%
\begin{equation}
\prod_{i=1}^{p}\underbrace{\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}%
A}}_{=\prod_{j=1}^{q}x_{j}^{a_{i,j}}}=\underbrace{\prod_{i=1}^{p}\prod
_{j=1}^{q}}_{=\prod_{j=1}^{q}\prod_{i=1}^{p}}x_{j}^{a_{i,j}}=\prod_{j=1}%
^{q}\prod_{i=1}^{p}x_{j}^{a_{i,j}}.
\label{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.3}%
\end{equation}
\par
Now, for each $j\in\left\{  1,2,\ldots,q\right\}  $, we let $c_{j}$ be the sum
of all entries in the $j$-th column of $A$. Then, the definition of
$\operatorname*{colsums}A$ yields%
\[
\operatorname*{colsums}A=\left(  \text{the column sums of }A\right)  =\left(
c_{1},c_{2},\ldots,c_{q}\right)
\]
(by the definition of the column sums of $A$). Thus,
\begin{align}
\overline{\xx}_{q}^{\operatorname*{colsums}A}  &  =
\overline{\xx}_{q}^{\left(  c_{1},c_{2},\ldots,c_{q}\right)  }
=x_{1}^{c_{1}}x_{2}^{c_{2}}\cdots x_{q}^{c_{q}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\overline{\xx}_{q}^{\left(  c_{1},c_{2},\ldots
,c_{q}\right)  }\right) \nonumber\\
&  =\prod_{j=1}^{q}x_{j}^{c_{j}}.
\label{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.4}%
\end{align}
\par
But each $j\in\left\{  1,2,\ldots,q\right\}  $ satisfies%
\begin{align*}
c_{j}  &  =\left(  \text{the sum of all entries in }\underbrace{\text{the
}j\text{-th column of }A}_{\substack{=\left(  a_{1,j},a_{2,j},\ldots
,a_{p,j}\right)  ^{T}\\\text{(since }A=\left(  a_{i,j}\right)  _{1\leq i\leq
p,\ 1\leq j\leq q}\text{)}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }c_{j}\right) \\
&  =\left(  \text{the sum of all entries in }\left(  a_{1,j},a_{2,j}%
,\ldots,a_{p,j}\right)  ^{T}\right)  =a_{1,j}+a_{2,j}+\cdots+a_{p,j}%
=\sum_{i=1}^{p}a_{i,j},
\end{align*}
and thus $x_{j}^{c_{j}}=x_{j}^{\sum_{i=1}^{p}a_{i,j}}=\prod_{i=1}^{p}%
x_{j}^{a_{i,j}}$. Hence,
\[
\prod_{j=1}^{q}\underbrace{x_{j}^{c_{j}}}_{=\prod_{i=1}^{p}x_{j}^{a_{i,j}}%
}=\prod_{j=1}^{q}\prod_{i=1}^{p}x_{j}^{a_{i,j}}=\prod_{i=1}^{p}
\overline{\xx}_{q}^{\operatorname{row}_{i}A}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.3})}\right)  .
\]
Thus,
\[
\prod_{i=1}^{p}\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}=\prod
_{j=1}^{q}x_{j}^{c_{j}}=\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}%
A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.4})}\right)  .
\]
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.prodxA}).}.

Thus, every $\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in\left(
\left\{  0,1\right\}  ^{q}\right)  ^{p}$ satisfies%
\begin{equation}
\prod_{i=1}^{p}\overline{\mathbf{x}}_{q}^{\beta_{i}}=\overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \beta_{1},\beta
_{2},\ldots,\beta_{p}\right)  \right)  }
\label{pf.lem.sol.Lambda.triangular.ela.prodx}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.prodx}):}
Let $\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in\left(  \left\{
0,1\right\}  ^{q}\right)  ^{p}$. Define $A\in\left\{  0,1\right\}  ^{p\times
q}$ by $A=\Phi^{-1}\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  $.
Thus, $\Phi\left(  A\right)  =\left(  \beta_{1},\beta_{2},\ldots,\beta
_{p}\right)  $. Hence, $\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)
=\Phi\left(  A\right)  =\left(  \operatorname*{row}\nolimits_{1}%
A,\operatorname{row}_{2}A,\ldots,\operatorname{row}_{p}A\right)  $ (by the
definition of $\Phi$). In other words, $\beta_{i}=\operatorname*{row}%
\nolimits_{i}A$ for each $i\in\left\{  1,2,\ldots,p\right\}  $. Thus,
$\overline{\mathbf{x}}_{q}^{\beta_{i}}=\overline{\mathbf{x}}_{q}%
^{\operatorname{row}_{i}A}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $.
Hence,
\begin{align*}
\prod_{i=1}^{p}\underbrace{\overline{\mathbf{x}}_{q}^{\beta_{i}}}%
_{=\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}}  &  =\prod_{i=1}%
^{p}\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}=\overline{\xx%
}_{q}^{\operatorname*{colsums}A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodxA})}\right) \\
&  =\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(
\beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A=\Phi^{-1}\left(  \beta_{1}%
,\beta_{2},\ldots,\beta_{p}\right)  \right)  .
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.prodx}).}.

Furthermore, every matrix $A\in\left\{  0,1\right\}  ^{p\times q}$ satisfies%
\begin{equation}
\operatorname*{rowsums}A=\left(  \text{the row sums of }A\right)  =\left(
\left\vert \operatorname{row}_{1}A\right\vert ,\left\vert \operatorname{row}%
_{2}A\right\vert ,\ldots,\left\vert \operatorname{row}_{p}A\right\vert
\right)  \label{pf.lem.sol.Lambda.triangular.ela.rowsums}%
\end{equation}
(because for each $i\in\left\{  1,2,\ldots,p\right\}  $, the sum of the
entries of the $i$-th row of $A$ is precisely $\left\vert \operatorname*{row}%
\nolimits_{i}A\right\vert $).

Let $\mathfrak{M}_{\alpha}$ be the set of all matrices $A\in\left\{
0,1\right\}  ^{p\times q}$ satisfying $\operatorname*{rowsums}A=\alpha$. Thus,%
\begin{align}
\mathfrak{M}_{\alpha}  &  =\left\{  A\in\left\{  0,1\right\}  ^{p\times
q}\ \mid\ \operatorname*{rowsums}A=\alpha\right\}
\label{pf.lem.sol.Lambda.triangular.ela.Malpha=}\\
&  =\left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid\ A\text{ is a
}\left\{  0,1\right\}  \text{-matrix having row sums }\alpha\right\}
\label{pf.lem.sol.Lambda.triangular.ela.Malpha=2}\\
&  \subset\left\{  0,1\right\}  ^{p\times q}.\nonumber
\end{align}


For each $m\in \NN$, we define a subset $\mathfrak{K}_{m}$ of $\left\{
0,1\right\}  ^{q}$ by%
\[
\mathfrak{K}_{m}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  .
\]
This set $\mathfrak{K}_{m}$ is finite (since it is a subset of the finite set
$\left\{  0,1\right\}  ^{q}$).

We have $\mathfrak{K}_{\alpha_{i}}\subset\left\{  0,1\right\}  ^{q}$ for each
$i\in\left\{  1,2,\ldots,p\right\}  $. Hence, $\prod_{i=1}^{p}\mathfrak{K}%
_{\alpha_{i}}\subset\prod_{i=1}^{p}\left(  \left\{  0,1\right\}  ^{q}\right)
=\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}$. Thus, $\mathfrak{K}%
_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}%
_{\alpha_{p}}=\prod_{i=1}^{p}\mathfrak{K}_{\alpha_{i}}\subset\left(  \left\{
0,1\right\}  ^{q}\right)  ^{p}$.

It is easy to see that for each matrix $A\in\left\{  0,1\right\}  ^{p\times
q}$, we have the following logical equivalence:%
\begin{equation}
\left(  \Phi\left(  A\right)  \in\mathfrak{K}_{\alpha_{1}}\times
\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}\right)
\ \Longleftrightarrow\ \left(  \operatorname*{rowsums}A=\alpha\right)
\label{pf.lem.sol.Lambda.triangular.ela.Phi-rs}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}):}
Let $A\in\left\{  0,1\right\}  ^{p\times q}$ be a matrix.
\par
Let $i\in\left\{  1,2,\ldots,p\right\}  $. Recall that $\operatorname{row}%
_{i}A$ denotes the $i$-th row of $A$; this $i$-th row is an element of
$\left\{  0,1\right\}  ^{q}$. Thus, $\operatorname{row}_{i}A\in\left\{
0,1\right\}  ^{q}$. Now, we have the following chain of equivalences:%
\begin{align}
\left(  \operatorname{row}_{i}A\in\underbrace{\mathfrak{K}_{\alpha_{i}}%
}_{\substack{=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =\alpha_{i}\right\}  \\\text{(by the definition of
}\mathfrak{K}_{\alpha_{i}}\text{)}}}\right)  \  &  \Longleftrightarrow
\ \left(  \operatorname{row}_{i}A\in\left\{  \beta\in\left\{  0,1\right\}
^{q}\ \mid\ \left\vert \beta\right\vert =\alpha_{i}\right\}  \right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\operatorname{row}_{i}A\in\left\{
0,1\right\}  ^{q}}_{\substack{\text{This is always true}\\\text{(since we know
that }\operatorname{row}_{i}A\in\left\{  0,1\right\}  ^{q}\text{)}}}\text{ and
}\left\vert \operatorname{row}_{i}A\right\vert =\alpha_{i}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left\vert \operatorname{row}_{i}A\right\vert
=\alpha_{i}\right)  . \label{pf.lem.sol.Lambda.triangular.ela.Phi-rs.pf.1}%
\end{align}
\par
Now, forget that we fixed $i$. We thus have proven the equivalence
(\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs.pf.1}) for each $i\in\left\{
1,2,\ldots,p\right\}  $.
\par
The definition of $\Phi$ yields $\Phi\left(  A\right)  =\left(
\operatorname*{row}\nolimits_{1}A,\operatorname{row}_{2}A,\ldots
,\operatorname{row}_{p}A\right)  $. Now, we have the following chain of
logical equivalences:%
\begin{align*}
&  \ \left(  \underbrace{\Phi\left(  A\right)  }_{=\left(  \operatorname*{row}%
\nolimits_{1}A,\operatorname{row}_{2}A,\ldots,\operatorname{row}_{p}A\right)
}\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}\right) \\
&  \Longleftrightarrow\ \left(  \left(  \operatorname*{row}\nolimits_{1}%
A,\operatorname{row}_{2}A,\ldots,\operatorname{row}_{p}A\right)
\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\operatorname{row}_{i}%
A\in\mathfrak{K}_{\alpha_{i}}}_{\substack{\Longleftrightarrow\ \left(
\left\vert \operatorname{row}_{i}A\right\vert =\alpha_{i}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs.pf.1}))}}}\text{ for each }%
i\in\left\{  1,2,\ldots,p\right\}  \right) \\
&  \Longleftrightarrow\ \left(  \left\vert \operatorname{row}_{i}A\right\vert
=\alpha_{i}\text{ for each }i\in\left\{  1,2,\ldots,p\right\}  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \left\vert
\operatorname{row}_{1}A\right\vert ,\left\vert \operatorname{row}%
_{2}A\right\vert ,\ldots,\left\vert \operatorname{row}_{p}A\right\vert
\right)  }_{\substack{=\operatorname*{rowsums}A\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.rowsums}))}}}=\underbrace{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  }_{=\alpha}\right) \\
&  \Longleftrightarrow\ \left(  \operatorname*{rowsums}A=\alpha\right)  .
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}).}. Now, applying
the map $\Phi$ to the equality (\ref{pf.lem.sol.Lambda.triangular.ela.Malpha=}%
), we find%
\begin{align}
\Phi\left(  \mathfrak{M}_{\alpha}\right)   &  =\Phi\left(  \left\{
A\in\left\{  0,1\right\}  ^{p\times q}\ \mid\ \operatorname*{rowsums}%
A=\alpha\right\}  \right) \nonumber\\
&  =\left\{  \Phi\left(  A\right)  \ \mid\ A\in\left\{  0,1\right\}  ^{p\times
q};\ \underbrace{\operatorname*{rowsums}A=\alpha}%
_{\substack{\Longleftrightarrow\ \left(  \Phi\left(  A\right)  \in
\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}))}}}\right\} \nonumber\\
&  =\left\{  \Phi\left(  A\right)  \ \mid\ A\in\left\{  0,1\right\}  ^{p\times
q};\ \Phi\left(  A\right)  \in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}%
_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}\right\} \nonumber\\
&  =\left\{  \beta\ \mid\ \beta\in\left(  \left\{  0,1\right\}  ^{q}\right)
^{p};\ \beta\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}%
\times\cdots\times\mathfrak{K}_{\alpha_{p}}\right\} \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\beta\text{ for }\Phi\left(  A\right)
\text{, since the map}\\
\Phi:\left\{  0,1\right\}  ^{p\times q}\rightarrow\left(  \left\{
0,1\right\}  ^{q}\right)  ^{p}\text{ is a bijection}%
\end{array}
\right) \nonumber\\
&  =\left\{  \beta\in\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}%
\ \mid\ \beta\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}%
\times\cdots\times\mathfrak{K}_{\alpha_{p}}\right\}  =\mathfrak{K}_{\alpha
_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha
_{p}} \label{pf.lem.sol.Lambda.triangular.ela.3}%
\end{align}
(since $\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times
\cdots\times\mathfrak{K}_{\alpha_{p}}\subset\left(  \left\{  0,1\right\}
^{q}\right)  ^{p}$).

Now, $\mathfrak{M}_{\alpha}$ is a subset of $\left\{  0,1\right\}  ^{p\times
q}$ (since $\mathfrak{M}_{\alpha}\subset\left\{  0,1\right\}  ^{p\times q}$).
Hence, Lemma \ref{lem.sol.Lambda.triangular.g-set} (applied to $X=\left\{
0,1\right\}  ^{p\times q}$, $Y=\left(  \left\{  0,1\right\}  ^{q}\right)
^{p}$, $\phi=\Phi$ and $Z=\mathfrak{M}_{\alpha}$) yields that the map
$\mathfrak{M}_{\alpha}\rightarrow\Phi\left(  \mathfrak{M}_{\alpha}\right)
,\ A\mapsto\Phi\left(  A\right)  $ is well-defined and is a bijection.

Lemma \ref{lem.sol.Lambda.triangular.em} yields that%
\begin{equation}
\eta_{q}\left(  e_{m}\right)  =\underbrace{\sum\limits_{\substack{\beta\in\left\{
0,1\right\}  ^{q};\\\left\vert \beta\right\vert =m}}}_{\substack{=\sum
_{\beta\in\mathfrak{K}_{m}}\\\text{(since }\mathfrak{K}_{m}=\left\{  \beta
\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert \beta\right\vert =m\right\}
\text{)}}}\overline{\mathbf{x}}_{q}^{\beta}=\sum_{\beta\in\mathfrak{K}_{m}%
}\overline{\mathbf{x}}_{q}^{\beta} \label{pf.lem.sol.Lambda.triangular.ela.em}%
\end{equation}
for every $m\in \NN$.

But applying the map $\eta_{q}$ to the equality $e_{\alpha_{1}}e_{\alpha_{2}%
}\cdots e_{\alpha_{p}}=\prod_{i=1}^{p}e_{\alpha_{i}}$, we obtain%
\begin{align*}
&  \eta_{q}\left(  e_{\alpha_{1}}e_{\alpha_{2}}\cdots e_{\alpha_{p}}\right) \\
&  =\eta_{q}\left(  \prod_{i=1}^{p}e_{\alpha_{i}}\right)  =\prod_{i=1}%
^{p}\underbrace{\eta_{q}\left(  e_{\alpha_{i}}\right)  }_{\substack{=\sum
_{\beta\in\mathfrak{K}_{\alpha_{i}}}\overline{\mathbf{x}}_{q}^{\beta
}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.ela.em}) (applied to }%
m=\alpha_{i}\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{q}\text{
is a }\kk\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{p}\sum_{\beta\in\mathfrak{K}_{\alpha_{i}}}
\overline{\xx}_{q}^{\beta}=\sum_{\left(  \beta_{1},\beta_{2},\ldots,\beta
_{p}\right)  \in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}%
\times\cdots\times\mathfrak{K}_{\alpha_{p}}}\underbrace{\prod_{i=1}%
^{p}\overline{\mathbf{x}}_{q}^{\beta_{i}}}_{\substack{=\overline{\xx%
}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \beta_{1},\beta
_{2},\ldots,\beta_{p}\right)  \right)  }\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodx})}\\\text{(since }\left(
\beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in\mathfrak{K}_{\alpha_{1}%
}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}\subset\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\sum_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in
\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}}\overline{\mathbf{x}}_{q}%
^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{p}\right)  \right)  }=\sum_{C\in\mathfrak{K}_{\alpha_{1}}%
\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}%
}\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(
C\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \text{ as }C\right) \\
&  =\sum_{C\in\Phi\left(  \mathfrak{M}_{\alpha}\right)  }\overline{\xx%
}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  C\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{K}_{\alpha_{1}}%
\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}=\Phi\left(  \mathfrak{M}_{\alpha}\right)  \text{ (by
(\ref{pf.lem.sol.Lambda.triangular.ela.3}))}\right) \\
&  =\underbrace{\sum_{A\in\mathfrak{M}_{\alpha}}}_{\substack{=\sum\limits
_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\alpha}}\\\text{(because
of (\ref{pf.lem.sol.Lambda.triangular.ela.Malpha=2}))}}}\underbrace{
\overline{\xx}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \Phi
\left(  A\right)  \right)  \right)  }}_{\substack{=\overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}A}\\\text{(since }\Phi^{-1}\left(  \Phi\left(
A\right)  \right)  =A\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\Phi\left(  A\right)  \text{ for }C\text{ in
the sum, since}\\
\text{the map }\mathfrak{M}_{\alpha}\rightarrow\Phi\left(  \mathfrak{M}%
_{\alpha}\right)  ,\ A\mapsto\Phi\left(  A\right)  \text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a
}\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }\alpha
}}\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}A}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.ela}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.g-coeff}Let $\lambda\in\Par$
and $\mu\in\Par$. Then, $\left[  \mathbf{x}^{\mu}\right]
\left(  e_{\lambda}\right)  =a_{\lambda,\mu}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.g-coeff}.]Let $p=\ell\left(
\lambda\right)  $. Thus, $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{p}\right)  $ with $\lambda_{1}\geq\lambda_{2}\geq\cdots
\geq\lambda_{p}>0$. Hence, $e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots
e_{\lambda_{p}}$ (by the definition of $e_{\lambda}$). Hence, $e_{\lambda}%
\in\Lambda$ (since $e_{i}\in\Lambda$ for each $i\in \NN$).

Let $q=\ell\left(  \mu\right)  $. Thus, $\mu_{q+1}=\mu_{q+2}=\mu_{q+3}%
=\cdots=0$. Therefore, $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $
with $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{q}>0$.

For any matrix $A\in\left\{  0,1\right\}  ^{p\times q}$, we let
$\operatorname*{colsums}A$ denote the column sums of $A$.

The definition of $a_{\lambda,\mu}$ yields%
\begin{align}
&  a_{\lambda,\mu}\nonumber\\
&  =\left(  \text{the number of all }\left\{  0,1\right\}  \text{-matrices of
size }\underbrace{\ell\left(  \lambda\right)  }_{=p}\times\underbrace{\ell
\left(  \mu\right)  }_{=q}\text{ having row sums }\lambda\text{ and column
sums }\mu\right) \nonumber\\
&  =\left(  \text{the number of all }\left\{  0,1\right\}  \text{-matrices of
size }p\times q\text{ having row sums }\lambda\text{ and column sums }%
\mu\right) \nonumber\\
&  =\left(  \text{the number of all matrices }A\in\left\{  0,1\right\}
^{p\times q}\text{ having row sums }\lambda\text{ and column sums }\mu\right)
\nonumber\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\underbrace{\text{the column sums of }A\text{ are }\mu}%
_{\substack{\Longleftrightarrow\ \left(  \operatorname*{colsums}A=\mu\right)
\\\text{(since the column sums of }A\text{ is }\operatorname*{colsums}%
A\text{)}}}\right\}  \right\vert \nonumber\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\operatorname*{colsums}A=\mu\right\}  \right\vert .
\label{pf.lem.sol.Lambda.triangular.g-coeff.a1}%
\end{align}


But $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)
\in\mathbb{N}^{p}$ is a $p$-tuple of nonnegative integers. Hence, Lemma
\ref{lem.sol.Lambda.triangular.ela} (applied to $\lambda$ and $\lambda_{i}$
instead of $\alpha$ and $\alpha_{i}$) yields
\[
\eta_{q}\left(  e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}\right)
=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda}}
\overline{\xx}_{q}^{\operatorname*{colsums}A}.
\]


But $e_{\lambda}\in\Lambda\subset R\left(  \xx\right)  $. Also,
$\mu_{i}=0$ for every integer $i>q$ (since $q=\ell\left(  \mu\right)  $).
Hence, Lemma \ref{lem.sol.Lambda.triangular.etaq-coeff} (applied to $\beta
=\mu$ and $f=e_{\lambda}$) yields%
\begin{align*}
&  \left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  }\right]  \left(  \eta_{q}\left(  \underbrace{e_{\lambda}%
}_{=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}}\right)  \right)
=\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(  e_{\lambda_{1}%
}e_{\lambda_{2}}\cdots e_{\lambda_{p}}\right)  }_{=\sum\limits_{\substack{A\in
\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{  0,1\right\}
\text{-matrix}\\\text{having row sums }\lambda}}\overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}A}}\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  }\right]  \left(  \sum\limits_{\substack{A\in\left\{  0,1\right\}
^{p\times q}\text{ is a }\left\{  0,1\right\}  \text{-matrix}\\\text{having
row sums }\lambda}}\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}%
A}\right)  \\
&=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a
}\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda
}}\underbrace{\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu
_{2},\ldots,\mu_{q}\right)  }\right]  \left(  \overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}A}\right)  }_{\substack{=\delta_{\left(  \mu
_{1},\mu_{2},\ldots,\mu_{q}\right)  ,\operatorname*{colsums}A}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.4}) (applied to
}\phi=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  \\\text{and }%
\psi=\operatorname*{colsums}A\text{))}}}\\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a
}\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda
}}\underbrace{\delta_{\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
,\operatorname*{colsums}A}}_{\substack{=\delta_{\mu,\operatorname*{colsums}%
A}\\\text{(since }\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  =\mu\text{)}%
}}=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda}}\delta
_{\mu,\operatorname*{colsums}A}\\
&  =\underbrace{\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{
is a }\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }%
\lambda;\\\operatorname*{colsums}A=\mu}}}_{=\sum\limits_{\substack{A\in\left\{
0,1\right\}  ^{p\times q}\text{;}\\\text{the row sums of }A\text{ are }%
\lambda;\\\operatorname*{colsums}A=\mu}}}\underbrace{\delta_{\mu
,\operatorname*{colsums}A}}_{\substack{=1\\\text{(since }%
\operatorname*{colsums}A=\mu\text{)}}}+\sum\limits_{\substack{A\in\left\{
0,1\right\}  ^{p\times q}\text{ is a }\left\{  0,1\right\}  \text{-matrix}%
\\\text{having row sums }\lambda;\\\operatorname*{colsums}A\neq\mu
}}\underbrace{\delta_{\mu,\operatorname*{colsums}A}}%
_{\substack{=0\\\text{(since }\operatorname*{colsums}A\neq\mu\text{)}}}\\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{;}\\\text{the
row sums of }A\text{ are }\lambda;\\\operatorname*{colsums}A=\mu
}}1+\underbrace{\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{
is a }\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }%
\lambda;\\\operatorname*{colsums}A\neq\mu}}0}_{=0}=\sum\limits_{\substack{A\in
\left\{  0,1\right\}  ^{p\times q}\text{;}\\\text{the row sums of }A\text{ are
}\lambda;\\\operatorname*{colsums}A=\mu}}1\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\operatorname*{colsums}A=\mu\right\}  \right\vert \cdot1\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\operatorname*{colsums}A=\mu\right\}  \right\vert .
\end{align*}
Comparing this with (\ref{pf.lem.sol.Lambda.triangular.g-coeff.a1}), we obtain
$\left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right)  =a_{\lambda,\mu
}$. This proves Lemma \ref{lem.sol.Lambda.triangular.g-coeff}.
\end{proof}

Now, let $\lambda\in\Par_{n}$. Let $p=\ell\left(
\lambda\right)  $. We have $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{p}\right)  $ with $\lambda_{1}\geq\lambda_{2}\geq\cdots
\geq\lambda_{p}>0$ (since $p=\ell\left(  \lambda\right)  $). Hence,
$e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}$ (by the
definition of $e_{\lambda}$).

Moreover, $\lambda\in\Par_{n}$, so that $\left\vert
\lambda\right\vert =n$. Hence, $n=\left\vert \underbrace{\lambda}_{=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  }\right\vert =\left\vert
\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  \right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}$.

Let $i\in\left\{  1,2,\ldots,p\right\}  $. Then, $e_{\lambda_{i}}$ is a
homogeneous element of $\Lambda$ having degree $\lambda_{i}$ (because for each
$m\in \NN$, the element $e_{m}$ is a homogeneous element of $\Lambda$
having degree $m$).

Now, forget that we fixed $i$. We thus have shown that for each $i\in\left\{
1,2,\ldots,p\right\}  $, the element $e_{\lambda_{i}}$ is a homogeneous
element of $\Lambda$ having degree $\lambda_{i}$. In other words,
$e_{\lambda_{1}},e_{\lambda_{2}},\ldots,e_{\lambda_{p}}$ are homogeneous
elements of $\Lambda$ having degrees $\lambda_{1},\lambda_{2},\ldots
,\lambda_{p}$, respectively. Hence, the product $e_{\lambda_{1}}e_{\lambda
_{2}}\cdots e_{\lambda_{p}}$ of these elements is a homogeneous element of
$\Lambda$ having degree $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}$. In light
of $e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}$ and
$n=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}$, this rewrites as follows: The
element $e_{\lambda}$ is a homogeneous element of $\Lambda$ having degree $n$.
In other words, $e_{\lambda}\in\Lambda_{n}$. Thus, Exercise
\ref{exe.Lambda.triangular}(a) (applied to $f=e_{\lambda}$) yields%
\[
e_{\lambda}=\sum_{\mu\in\Par_{n}}\underbrace{\left(
\left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right)  \right)
}_{\substack{=a_{\lambda,\mu}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.g-coeff})}}}m_{\mu}=\sum_{\mu\in
\operatorname{Par}_{n}}a_{\lambda,\mu}m_{\mu}.
\]
This solves Exercise \ref{exe.Lambda.triangular}(g).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(h).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(h), let us show a few lemmas:

\begin{lemma}
\label{lem.sol.Lambda.triangular.ineq-eq}Let $p\in \NN$. Let
$a_{1},a_{2},\ldots,a_{p}$ be $p$ integers. Let $b_{1},b_{2},\ldots,b_{p}$ be
$p$ integers. Assume that%
\begin{equation}
a_{i}\geq b_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots
,p\right\}  . \label{eq.lem.sol.Lambda.triangular.ineq-eq.ass1}%
\end{equation}
Assume furthermore that $\sum_{i=1}^{p}a_{i}\leq\sum_{i=1}^{p}b_{i}$. Then,
$a_{i}=b_{i}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.ineq-eq}.]Let $j\in\left\{
1,2,\ldots,p\right\}  $. Then,
(\ref{eq.lem.sol.Lambda.triangular.ineq-eq.ass1}) (applied to $i=j$) shows
that $a_{j}\geq b_{j}$. But
\begin{align*}
\underbrace{\sum_{i=1}^{p}}_{=\sum_{i\in\left\{  1,2,\ldots,p\right\}  }%
}a_{i}  &  =\sum_{i\in\left\{  1,2,\ldots,p\right\}  }a_{i}=a_{j}%
+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq j}%
}\underbrace{a_{i}}_{\substack{\geq b_{i}\\\text{(by
(\ref{eq.lem.sol.Lambda.triangular.ineq-eq.ass1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=j\text{ from the sum}\right) \\
&  \geq a_{j}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq
j}}b_{i}.
\end{align*}
Hence,%
\begin{align*}
a_{j}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq j}}b_{i}  &
\leq\sum_{i=1}^{p}a_{i}\leq\underbrace{\sum_{i=1}^{p}}_{=\sum_{i\in\left\{
1,2,\ldots,p\right\}  }}b_{i}=\sum_{i\in\left\{  1,2,\ldots,p\right\}  }%
b_{i}=b_{j}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq
j}}b_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=j\text{ from the sum}\right)  .
\end{align*}
Subtracting $\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq
j}}b_{i}$ from both sides of this inequality, we obtain $a_{j}\leq b_{j}$.
Combining this with $a_{j}\geq b_{j}$, we obtain $a_{j}=b_{j}$.

Now, forget that we fixed $j$. We thus have shown that $a_{j}=b_{j}$ for each
$j\in\left\{  1,2,\ldots,p\right\}  $. Renaming the variable $j$ as $i$ in
this statement, we conclude that $a_{i}=b_{i}$ for each $i\in\left\{
1,2,\ldots,p\right\}  $. This proves Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.ineq-eq2}Let $k\in \NN$. Let
$a_{1},a_{2},\ldots,a_{k}$ be $k$ integers. Let $b_{1},b_{2},\ldots,b_{k}$ be
$k$ integers. Assume that%
\[
a_{j}\geq b_{j}\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,\ldots
,k\right\}  .
\]
Assume furthermore that $\sum_{j=1}^{k}a_{j}\leq\sum_{j=1}^{k}b_{j}$. Then,
$a_{j}=b_{j}$ for each $j\in\left\{  1,2,\ldots,k\right\}  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.ineq-eq2}.]Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq2} is obtained from Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq} upon renaming the variables $p$ and
$i$ as $k$ and $j$. Thus, Lemma \ref{lem.sol.Lambda.triangular.ineq-eq2}
follows from Lemma \ref{lem.sol.Lambda.triangular.ineq-eq}.
\end{proof}

The next few lemmas use the so-called \dfn{Iverson bracket notation}:

\begin{definition}
For every assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $
denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
\end{definition}

\begin{lemma}
\label{lem.sol.Lambda.triangular.sum-iver}Let $q\in \NN$ and
$r\in \NN$ be such that $r\leq q$. Then, $\sum_{j=1}^{q}\left[  j\leq
r\right]  =r$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.sum-iver}.]We have $0\leq r\leq
q$. Hence,%
\[
\sum_{j=1}^{q}\left[  j\leq r\right]  =\sum_{j=1}^{r}\underbrace{\left[  j\leq
r\right]  }_{\substack{=1\\\text{(since }j\leq r\text{)}}}+\sum_{j=r+1}%
^{q}\underbrace{\left[  j\leq r\right]  }_{\substack{=0\\\text{(since we don't
have }j\leq r\\\text{(since }j\geq r+1>r\text{))}}}=\sum_{j=1}^{r}%
1+\underbrace{\sum_{j=r+1}^{q}0}_{=0}=\sum_{j=1}^{r}1=r\cdot1=r.
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.sum-iver}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.01pqmat}Let $p\in \NN$ and
$q\in \NN$. Let $A=\left(  a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq
q}\in\left\{  0,1\right\}  ^{p\times q}$ be a $\left\{  0,1\right\}  $-matrix.
Let $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $ be the row
sums of $A$. Let $\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $ be the
column sums of $A$. Then:

\begin{enumerate}
\item[(a)] We have $\lambda_{i}=\sum_{j=1}^{q}a_{i,j}$ for each $i\in\left\{
1,2,\ldots,p\right\}  $.

\item[(b)] We have $\mu_{j}=\sum_{i=1}^{p}a_{i,j}$ for each $j\in\left\{
1,2,\ldots,q\right\}  $.

\item[(c)] We have $\sum_{i=1}^{p}\lambda_{i}=\sum_{j=1}^{q}\mu_{j}$.

\item[(d)] We have $\min\left\{  \lambda_{i},k\right\}  \geq\sum_{j=1}%
^{k}a_{i,j}$ for each $k\in\left\{  0,1,\ldots,q\right\}  $ and $i\in\left\{
1,2,\ldots,p\right\}  $.

\item[(e)] We have $\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  \geq
\sum_{j=1}^{k}\mu_{j}$ for each $k\in\left\{  0,1,\ldots,q\right\}  $.

\item[(f)] If $\left(  \sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}
=\sum_{j=1}^{k}\mu_{j}\text{ for each }k\in\left\{  1,2,\ldots,q\right\}
\right)  $, then $A=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq
i\leq p,\ 1\leq j\leq q}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.01pqmat}.]We have $\left(
a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq q}=A\in\left\{  0,1\right\}
^{p\times q}$. Thus, $a_{i,j}\in\left\{  0,1\right\}  $ for each $i\in\left\{
1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}  $. Hence, for
each $i\in\left\{  1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots
,q\right\}  $, we have%
\begin{equation}
a_{i,j}\geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a_{i,j}\in\left\{
0,1\right\}  \right)  \label{pf.lem.sol.Lambda.triangular.01pqmat.aij0}%
\end{equation}
and%
\begin{equation}
a_{i,j}\leq1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a_{i,j}\in\left\{
0,1\right\}  \right)  . \label{pf.lem.sol.Lambda.triangular.01pqmat.aij1}%
\end{equation}


(a) Recall that $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $
is the row sums of $A$. By the definition of ``row
sums'', this means that for each $i\in\left\{  1,2,\ldots
,p\right\}  $, the number $\lambda_{i}$ is the sum of all entries in the
$i$-th row of $A$. Thus, for each $i\in\left\{  1,2,\ldots,p\right\}  $, we
have
\begin{align*}
\lambda_{i} &  =\left(  \text{the sum of all entries in }\underbrace{\text{the
}i\text{-th row of }A}_{\substack{=\left(  a_{i,1},a_{i,2},\ldots
,a_{i,q}\right)  \\\text{(since }A=\left(  a_{i,j}\right)  _{1\leq i\leq
p,\ 1\leq j\leq q}\text{)}}}\right)  \\
&  =\left(  \text{the sum of all entries in }\left(  a_{i,1},a_{i,2}%
,\ldots,a_{i,q}\right)  \right)  =a_{i,1}+a_{i,2}+\cdots+a_{i,q}=\sum
_{j=1}^{q}a_{i,j}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a).

(b) The proof of Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) can easily
be adapted (mutatis mutandis) to yield a proof of Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(b).

(c) Comparing%
\[
\sum_{i=1}^{p}\underbrace{\lambda_{i}}_{\substack{=\sum_{j=1}^{q}%
a_{i,j}\\\text{(by Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a))}%
}}=\underbrace{\sum_{i=1}^{p}\sum_{j=1}^{q}}_{=\sum_{j=1}^{q}\sum_{i=1}^{p}%
}a_{i,j}=\sum_{j=1}^{q}\sum_{i=1}^{p}a_{i,j}%
\]
with%
\[
\sum_{j=1}^{q}\underbrace{\mu_{j}}_{\substack{=\sum_{i=1}^{p}a_{i,j}%
\\\text{(by Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(b))}}}=\sum
_{j=1}^{q}\sum_{i=1}^{p}a_{i,j},
\]
we obtain $\sum_{i=1}^{p}\lambda_{i}=\sum_{j=1}^{q}\mu_{j}$. This proves Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(c).

(d) Let $k\in\left\{  0,1,\ldots,q\right\}  $ and $i\in\left\{  1,2,\ldots
,p\right\}  $. Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) yields
$\lambda_{i}=\sum_{j=1}^{q}a_{i,j}$. But
\[
\sum_{j=1}^{k}\underbrace{a_{i,j}}_{\substack{\leq1\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij1}))}}}\leq\sum_{j=1}%
^{k}1=k\cdot1=k.
\]


Furthermore, $k\in\left\{  0,1,\ldots,q\right\}  $, so that $0\leq k\leq q$.
Hence,%
\[
\sum_{j=1}^{q}a_{i,j}=\sum_{j=1}^{k}a_{i,j}+\sum_{j=k+1}^{q}%
\underbrace{a_{i,j}}_{\substack{\geq0\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij0}))}}}\geq\sum_{j=1}^{k}%
a_{i,j}+\underbrace{\sum_{j=k+1}^{q}0}_{=0}=\sum_{j=1}^{k}a_{i,j},
\]
so that%
\[
\sum_{j=1}^{k}a_{i,j}\leq\sum_{j=1}^{q}a_{i,j}=\lambda_{i}.
\]


But let us recall the following basic fact about integers: If three integers
$\alpha,\beta,\gamma$ satisfy $\alpha\leq\beta$ and $\alpha\leq\gamma$, then
$\alpha\leq\min\left\{  \beta,\gamma\right\}  $. Applying this fact to
$\alpha=\sum_{j=1}^{k}a_{i,j}$, $\beta=\lambda_{i}$ and $\gamma=k$, we
conclude that $\sum_{j=1}^{k}a_{i,j}\leq\min\left\{  \lambda_{i},k\right\}  $
(since $\sum_{j=1}^{k}a_{i,j}\leq\lambda_{i}$ and $\sum_{j=1}^{k}a_{i,j}\leq
k$). In other words, $\min\left\{  \lambda_{i},k\right\}  \geq\sum_{j=1}%
^{k}a_{i,j}$. This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(d).

(e) Let $k\in\left\{  0,1,\ldots,q\right\}  $. Now,%
\begin{align*}
\sum_{i=1}^{p}\underbrace{\min\left\{  \lambda_{i},k\right\}  }%
_{\substack{\geq\sum_{j=1}^{k}a_{i,j}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(d))}}} &  \geq\underbrace{\sum
_{i=1}^{p}\sum_{j=1}^{k}}_{=\sum_{j=1}^{k}\sum_{i=1}^{p}}a_{i,j}=\sum
_{j=1}^{k}\underbrace{\sum_{i=1}^{p}a_{i,j}}_{\substack{=\mu_{j}\\\text{(by
Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(b))}}}
  =\sum_{j=1}^{k}\mu_{j}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(e).

(f) Assume that
\begin{equation}
\left(  \sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}%
\mu_{j}\text{ for each }k\in\left\{  1,2,\ldots,q\right\}  \right)  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.f.ass}%
\end{equation}
We want to show that $A=\left(  \left[  j\leq\lambda_{i}\right]  \right)
_{1\leq i\leq p,\ 1\leq j\leq q}$.

Let $k\in\left\{  1,2,\ldots,q\right\}  $. Thus, $k\in\left\{  1,2,\ldots
,q\right\}  \subset\left\{  0,1,\ldots,q\right\}  $. But
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.ass}) yields
\[
\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}%
\underbrace{\mu_{j}}_{\substack{=\sum_{i=1}^{p}a_{i,j}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(b))}}}=\underbrace{\sum_{j=1}^{k}%
\sum_{i=1}^{p}}_{=\sum_{i=1}^{p}\sum_{j=1}^{k}}a_{i,j}=\sum_{i=1}^{p}%
\sum_{j=1}^{k}a_{i,j}\leq\sum_{i=1}^{p}\sum_{j=1}^{k}a_{i,j}.
\]
Furthermore, we know that $\min\left\{  \lambda_{i},k\right\}  \geq\sum
_{j=1}^{k}a_{i,j}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $ (by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(d)). Hence, Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq} (applied to $\min\left\{  \lambda
_{i},k\right\}  $ and $\sum_{j=1}^{k}a_{i,j}$ instead of $a_{i}$ and $b_{i}$)
shows that%
\begin{equation}
\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}a_{i,j}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,p\right\}
.\label{pf.lem.sol.Lambda.triangular.01pqmat.f.3}%
\end{equation}


Now, forget that we fixed $k$. We thus have proven
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.3}) for each $k\in\left\{
1,2,\ldots,q\right\}  $.

Now, pick $i\in\left\{  1,2,\ldots,p\right\}  $. We shall show that%
\begin{equation}
a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  \ \ \ \ \ \ \ \ \ \ \text{for each
}k\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.f.4}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.4}):} Let
$k\in\left\{  1,2,\ldots,q\right\}  $. We must prove the inequality
$a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  $.

Indeed, we are in one of the following two cases:

\textit{Case 1:} We have $k\leq\lambda_{i}$.

\textit{Case 2:} We don't have $k\leq\lambda_{i}$.

Let us first consider Case 1. In this case, we have $k\leq\lambda_{i}$. Thus,
$\left[  k\leq\lambda_{i}\right]  =1$. Also, $k\geq1$ (since $k\in\left\{
1,2,\ldots,q\right\}  $), so that $k\in\left\{  1,2,\ldots,k\right\}  $. But
from $k\leq\lambda_{i}$, we also obtain $\min\left\{  \lambda_{i},k\right\}
=k$. Thus,%
\[
\sum_{j=1}^{k}1=k\cdot1=k\leq k=\min\left\{  \lambda_{i},k\right\}
=\sum_{j=1}^{k}a_{i,j}%
\]
(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.3})). Furthermore, $1\geq
a_{i,j}$ for each $j\in\left\{  1,2,\ldots,k\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,k\right\}  $.
Then, $j\in\left\{  1,2,\ldots,k\right\}  \subset\left\{  1,2,\ldots
,q\right\}  $ (since $k\in\left\{  1,2,\ldots,q\right\}  $). Hence,
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij1}) yields $a_{i,j}\leq1$. In
other words, $1\geq a_{i,j}$. Qed.}. Hence, Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq2} (applied to $1$ and $a_{i,j}$ instead
of $a_{j}$ and $b_{j}$) shows that $1=a_{i,j}$ for each $j\in\left\{
1,2,\ldots,k\right\}  $. Applying this to $j=k$, we obtain $1=a_{i,k}$ (since
$k\in\left\{  1,2,\ldots,k\right\}  $). Hence, $a_{i,k}=1=\left[  k\leq
\lambda_{i}\right]  $, so that $a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  $.
Thus, the inequality $a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  $ is proven
in Case 1.

Next, let us consider Case 2. In this case, we don't have $k\leq\lambda_{i}$.
Thus, $\left[  k\leq\lambda_{i}\right]  =0$. But
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij0}) (applied to $j=k$) yields
$a_{i,k}\geq0$. Thus, $a_{i,k}\geq0=\left[  k\leq\lambda_{i}\right]  $. Hence,
the inequality $a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  $ is proven in
Case 2.

We have now proven the inequality $a_{i,k}\geq\left[  k\leq\lambda_{i}\right]
$ in each of the two Cases 1 and 2. Thus, the inequality $a_{i,k}\geq\left[
k\leq\lambda_{i}\right]  $ always holds. This proves
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.4}).]

Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) yields $\lambda_{i}%
=\sum_{j=1}^{q}a_{i,j}$. But Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(d)
(applied to $k=q$) yields $\min\left\{  \lambda_{i},q\right\}  \geq\sum
_{j=1}^{q}a_{i,j}$. Hence, $\sum_{j=1}^{q}a_{i,j}\leq\min\left\{  \lambda
_{i},q\right\}  $, so that $\lambda_{i}=\sum_{j=1}^{q}a_{i,j}\leq\min\left\{
\lambda_{i},q\right\}  $. If we had $\lambda_{i}>q$, then we would have
$\min\left\{  \lambda_{i},q\right\}  =q<\lambda_{i}\leq\min\left\{
\lambda_{i},q\right\}  $, which would be absurd. Thus, we cannot have
$\lambda_{i}>q$. We therefore must have $\lambda_{i}\leq q$. Hence, Lemma
\ref{lem.sol.Lambda.triangular.sum-iver} (applied to $r=\lambda_{i}$) yields%
\[
\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  =\lambda_{i}=\sum_{j=1}%
^{q}a_{i,j}.
\]
Thus,%
\[
\sum_{j=1}^{q}a_{i,j}=\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  \leq
\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  .
\]
Also,
\[
a_{i,j}\geq\left[  j\leq\lambda_{i}\right]  \ \ \ \ \ \ \ \ \ \ \text{for each
}j\in\left\{  1,2,\ldots,q\right\}
\]
(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.4}), applied to $k=j$).
Hence, Lemma \ref{lem.sol.Lambda.triangular.ineq-eq2} (applied to $q$,
$a_{i,j}$ and $\left[  j\leq\lambda_{i}\right]  $ instead of $k$, $a_{j}$ and
$b_{j}$) shows that%
\begin{equation}
a_{i,j}=\left[  j\leq\lambda_{i}\right]  \ \ \ \ \ \ \ \ \ \ \text{for each
}j\in\left\{  1,2,\ldots,q\right\}
.\label{pf.lem.sol.Lambda.triangular.01pqmat.f.7}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.7}) for each $i\in\left\{
1,2,\ldots,p\right\}  $.

Now,
\[
A=\left(  \underbrace{a_{i,j}}_{\substack{=\left[  j\leq\lambda_{i}\right]
\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.7}))}}}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}=\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}.
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(f).
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.majo-mu}Let $n\in \NN$ and
$q\in \NN$. Let $\lambda\in\Par_{n}$ and $\mu
\in\Par_{n}$ be such that $\ell\left(  \mu\right)
\leq q$. Assume that%
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\]
Then, $\lambda\triangleright\mu$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.majo-mu}.]We have assumed that%
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}
.\label{pf.lem.sol.Lambda.triangular.majo-mu.have}%
\end{equation}
Hence,
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{q}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{q}\label{pf.lem.sol.Lambda.triangular.majo-mu.q}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}):} If
$q=0$, then the inequality (\ref{pf.lem.sol.Lambda.triangular.majo-mu.q})
holds because both of its sides equal $0$ (in fact, an empty sum is $0$).
Hence, for the rest of this proof, we WLOG assume that $q\neq0$. Thus, $q$ is
a positive integer (since $q\in \NN$). Thus, $q\in\left\{
1,2,\ldots,q\right\}  $. Therefore,
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.have}) (applied to $k=q$) yields
$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{q}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{q}%
$. This proves (\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}).}.

Both $\lambda$ and $\mu$ are partitions of $n$ (since $\lambda\in
\Par_{n}$ and $\mu\in\Par_{n}%
$). Thus, we have $\lambda\triangleright\mu$ if and only if
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\label{pf.lem.sol.Lambda.triangular.majo-mu.wish}%
\end{equation}
(by the definition of the dominance order).

Now, we are going to prove (\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}).

[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}):} Let
$k\in\left\{  1,2,\ldots,n\right\}  $. We must prove that $\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}$.

If $k\in\left\{  1,2,\ldots,q\right\}  $, then this follows immediately from
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.have}). Hence, for the rest of this
proof, we WLOG assume that $k\notin\left\{  1,2,\ldots,q\right\}  $. Combining
$k\in\left\{  1,2,\ldots,n\right\}  \subset\left\{  1,2,3,\ldots\right\}  $
with $k\notin\left\{  1,2,\ldots,q\right\}  $, we obtain $k\in\left\{
1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots,q\right\}  =\left\{
q+1,q+2,q+3,\ldots\right\}  $. Thus, $k\geq q+1>q$. Hence, $q<k$. Recall also
that $\ell\left(  \mu\right)  \leq q$, hence $q\geq\ell\left(  \mu\right)  $.
Now,%
\begin{align}
\mu_{1}+\mu_{2}+\cdots+\mu_{k} &  =\sum_{i=1}^{k}\mu_{i}=\sum_{i=1}^{q}\mu
_{i}+\sum_{i=q+1}^{k}\underbrace{\mu_{i}}_{\substack{=0\\\text{(since }i\geq
q+1>q\geq\ell\left(  \mu\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }0\leq q<k\right)  \nonumber\\
&  =\sum_{i=1}^{q}\mu_{i}+\underbrace{\sum_{i=q+1}^{k}0}_{=0}=\sum_{i=1}%
^{q}\mu_{i}=\mu_{1}+\mu_{2}+\cdots+\mu_{q}%
.\label{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.5}%
\end{align}


On the other hand,%
\begin{align*}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k} &  =\sum_{i=1}^{k}\lambda_{i}%
=\sum_{i=1}^{q}\lambda_{i}+\sum_{i=q+1}^{k}\underbrace{\lambda_{i}}_{\geq
0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq q<k\right)  \\
&  \geq\sum_{i=1}^{q}\lambda_{i}+\underbrace{\sum_{i=q+1}^{k}0}_{=0}%
=\sum_{i=1}^{q}\lambda_{i}=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{q}\\
&  \geq\mu_{1}+\mu_{2}+\cdots+\mu_{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.q})}\right)  \\
&  =\mu_{1}+\mu_{2}+\cdots+\mu_{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.5})}\right)  .
\end{align*}
This proves $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$. Thus, (\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish})
is proven.]

Recall that we have $\lambda\triangleright\mu$ if and only if
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}) holds. Thus, we have
$\lambda\triangleright\mu$ (since
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}) holds). This proves Lemma
\ref{lem.sol.Lambda.triangular.majo-mu}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.majo-lam}Let $n\in \NN$ and
$q\in \NN$. Let $\lambda\in\Par_{n}$ and $\mu
\in\Par_{n}$ be such that $\ell\left(  \lambda\right)
\leq q$. Assume that%
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\]
Then, $\lambda\triangleright\mu$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.majo-lam}.]We have assumed that%
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.majo-lam.have}%
\end{equation}


Both $\lambda$ and $\mu$ are partitions of $n$ (since $\lambda\in
\Par_{n}$ and $\mu\in\Par_{n}%
$). Thus, we have $\lambda\triangleright\mu$ if and only if
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\label{pf.lem.sol.Lambda.triangular.majo-lam.wish}%
\end{equation}
(by the definition of the dominance order).

Now, we are going to prove (\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}).

[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}):} Let
$k\in\left\{  1,2,\ldots,n\right\}  $. We must prove that $\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}$.

If $k\in\left\{  1,2,\ldots,q\right\}  $, then this follows immediately from
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.have}). Hence, for the rest of
this proof of $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$, we WLOG assume that $k\notin\left\{  1,2,\ldots
,q\right\}  $. Combining $k\in\left\{  1,2,\ldots,n\right\}  \subset\left\{
1,2,3,\ldots\right\}  $ with $k\notin\left\{  1,2,\ldots,q\right\}  $, we
obtain $k\in\left\{  1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots
,q\right\}  =\left\{  q+1,q+2,q+3,\ldots\right\}  $. Thus, $k\geq
q+1>q\geq\ell\left(  \lambda\right)  $ (since $\ell\left(  \lambda\right)
\leq q$). 

From $\mu\in\Par_{n}$, we obtain $\left\vert
\mu\right\vert =n$. Hence, $n=\left\vert \mu\right\vert =\mu_{1}+\mu_{2}%
+\mu_{3}+\cdots$ (by the definition of $\left\vert \mu\right\vert $).

But the definition of $\left\vert \lambda\right\vert $ yields
\begin{align*}
\left\vert \lambda\right\vert  &  =\lambda_{1}+\lambda_{2}+\lambda_{3}%
+\cdots=\sum_{i=1}^{\infty}\lambda_{i}=\sum_{i=1}^{k}\lambda_{i}+\sum
_{i=k+1}^{\infty}\underbrace{\lambda_{i}}_{\substack{=0\\\text{(since }i\geq
k+1>k>\ell\left(  \lambda\right)  \text{)}}}=\sum_{i=1}^{k}\lambda
_{i}+\underbrace{\sum_{i=k+1}^{\infty}0}_{=0}\\
&  =\sum_{i=1}^{k}\lambda_{i}=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}.
\end{align*}
Thus,%
\begin{align*}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k} &  =\left\vert \lambda\right\vert
=n\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda\in\Par_{n}\right)  \\
&  =\mu_{1}+\mu_{2}+\mu_{3}+\cdots=\sum_{i=1}^{\infty}\mu_{i}=\sum_{i=1}%
^{k}\mu_{i}+\sum_{i=k+1}^{\infty}\underbrace{\mu_{i}}_{\geq0}\\
&  \geq\sum_{i=1}^{k}\mu_{i}+\underbrace{\sum_{i=k+1}^{\infty}0}_{=0}%
=\sum_{i=1}^{k}\mu_{i}=\mu_{1}+\mu_{2}+\cdots+\mu_{k}.
\end{align*}
This proves $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$. Thus, (\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish})
is proven.]

Recall that we have $\lambda\triangleright\mu$ if and only if
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}) holds. Thus, we have
$\lambda\triangleright\mu$ (since
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}) holds). This proves Lemma
\ref{lem.sol.Lambda.triangular.majo-lam}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.01mat}Let $n\in \NN$, $p\in \NN$
and $q\in \NN$. Let $\lambda\in\Par_{n}$ and
$\mu\in\Par_{n}$. Let $A\in\left\{  0,1\right\}
^{p\times q}$ be a $\left\{  0,1\right\}  $-matrix having row sums $\lambda$
and column sums $\mu$. Then:

\begin{enumerate}
\item[(a)] We have $\lambda^{t}\triangleright\mu$.

\item[(b)] If $\mu=\lambda^{t}$, then $A=\left(  \left[  j\leq\lambda
_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.01mat}.]The row sums of $A$ is
$\lambda$ (since $A$ has row sums $\lambda$). Thus, $\lambda$ is a $p$-tuple
(since the row sums of $A$ is a $p$-tuple). Therefore, $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $. Thus, $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $ is the row sums of $A$
(since $\lambda$ is the row sums of $A$).

The column sums of $A$ is $\mu$ (since $A$ has column sums $\mu$). Thus, $\mu$
is a $q$-tuple (since the column sums of $A$ is a $q$-tuple). Therefore,
$\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $. Thus, $\left(  \mu
_{1},\mu_{2},\ldots,\mu_{q}\right)  $ is the column sums of $A$ (since $\mu$
is the column sums of $A$). From $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  $, we obtain $\ell\left(  \mu\right)  \leq q$.

We have $\left\vert \lambda^{t}\right\vert =\left\vert \lambda\right\vert $
(since the transpose of a partition always has the same size as the partition
itself). But $\left\vert \lambda\right\vert =n$ (since $\lambda\in
\Par_{n}$). Hence, $\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert =n$, so that $\lambda^{t}\in\Par_{n}$.

Write the $p\times q$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}$. Hence, Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(e) shows that we have%
\begin{equation}
\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  \geq\sum_{j=1}^{k}\mu_{j}
\label{pf.lem.sol.Lambda.triangular.01mat.1}%
\end{equation}
for each $k\in\left\{  0,1,\ldots,q\right\}  $.

But $\lambda\in\Par_{n}\subset\Par$.
Hence, Lemma \ref{lem.sol.Lambda.triangular.transpose1} (applied to
$\nu=\lambda$) shows that
\begin{equation}
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},k\right\}  \label{pf.lem.sol.Lambda.triangular.01mat.2}%
\end{equation}
for every $k\in \NN$. Hence, every $k\in\left\{  1,2,\ldots,q\right\}  $
satisfies%
\begin{align}
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k} &  =\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},k\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.01mat.2})}\right)  \nonumber\\
&  =\sum_{j=1}^{p}\min\left\{  \lambda_{j},k\right\}  +\sum_{j=p+1}^{\infty
}\underbrace{\min\left\{  \lambda_{j},k\right\}  }_{\substack{\geq
0}}\nonumber\\
&  \geq\sum_{j=1}^{p}\min\left\{  \lambda_{j},k\right\}  +\underbrace{\sum
_{j=p+1}^{\infty}0}_{=0}=\sum_{j=1}^{p}\min\left\{  \lambda_{j},k\right\}
\nonumber\\
&  =\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}
\label{pf.lem.sol.Lambda.triangular.01mat.4}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}j\text{ as }i\right)  \nonumber\\
&  \geq\sum_{j=1}^{k}\mu_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.01mat.1})}\right)  \nonumber\\
&  =\mu_{1}+\mu_{2}+\cdots+\mu_{k}.\nonumber
\end{align}
Hence, Lemma \ref{lem.sol.Lambda.triangular.majo-mu} (applied to $\lambda^{t}$
instead of $\lambda$) shows that $\lambda^{t}\triangleright\mu$. This proves
Lemma \ref{lem.sol.Lambda.triangular.01mat}(a).

(b) Assume that $\mu=\lambda^{t}$.

Let $k\in\left\{  1,2,\ldots,q\right\}  $. Then,
(\ref{pf.lem.sol.Lambda.triangular.01mat.4}) shows that
\[
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}\geq\sum_{i=1}^{p}\min\left\{
\lambda_{i},k\right\}  .
\]
Hence,
\[
\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  \leq\left(  \lambda
^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda
^{t}\right)  _{k}=\sum_{j=1}^{k}\left(  \underbrace{\lambda^{t}}_{=\mu
}\right)  _{j}=\sum_{j=1}^{k}\mu_{j}.
\]
Combining this with (\ref{pf.lem.sol.Lambda.triangular.01mat.1}), we obtain
$\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}\mu_{j}$.

Now, forget that we fixed $k$. We thus have proven that $\left(  \sum
_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}\mu_{j}\text{ for
each }k\in\left\{  1,2,\ldots,q\right\}  \right)  $. Hence, Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(f) shows that $A=\left(  \left[
j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$. This
proves Lemma \ref{lem.sol.Lambda.triangular.01mat}(b).
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.01matB}Let $p\in \NN$ and
$q\in \NN$. Let $\lambda$ be a partition satisfying $\ell\left(
\lambda\right)  \leq p$ and $\lambda_{1}\leq q$.

Let $B$ be the $p\times q$-matrix $\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$.

Then, $B$ is a $\left\{  0,1\right\}  $-matrix having row sums $\lambda$ and
column sums $\lambda^{t}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.01matB}.]We have $\left[
j\leq\lambda_{i}\right]  \in\left\{  0,1\right\}  $ for each $i\in\left\{
1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}  $ (since
$\left[  \mathcal{A}\right]  \in\left\{  0,1\right\}  $ for any logical
statement $\mathcal{A}$). Hence, $\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}\in\left\{  0,1\right\}  ^{p\times
q}$.

Thus, $B=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq
p,\ 1\leq j\leq q}\in\left\{  0,1\right\}  ^{p\times q}$. Therefore, $B$ is a
$\left\{  0,1\right\}  $-matrix.

From $\ell\left(  \lambda\right)  \leq p$, we obtain $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $.

Let $\left(  \widetilde{\lambda}_{1},\widetilde{\lambda}_{2},\ldots
,\widetilde{\lambda}_{p}\right)  $ be the row sums of $B$. Let $\left(
\widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{q}\right)  $
be the column sums of $B$.

Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) (applied to $B$,
$\widetilde{\lambda}_{i}$, $\widetilde{\mu}_{j}$ and $\left[  j\leq\lambda
_{i}\right]  $ instead of $A$, $\lambda_{i}$, $\mu_{j}$ and $a_{i,j}$) shows
that we have%
\begin{equation}
\widetilde{\lambda}_{i}=\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,p\right\}
\label{pf.lem.sol.Lambda.triangular.01matB.lati}%
\end{equation}
(since $B=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq
p,\ 1\leq j\leq q}$).

Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(b) (applied to $B$,
$\widetilde{\lambda}_{i}$, $\widetilde{\mu}_{j}$ and $\left[  j\leq\lambda
_{i}\right]  $ instead of $A$, $\lambda_{i}$, $\mu_{j}$ and $a_{i,j}$) shows
that we have%
\begin{equation}
\widetilde{\mu}_{j}=\sum_{i=1}^{p}\left[  j\leq\lambda_{i}\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,\ldots,q\right\}
\label{pf.lem.sol.Lambda.triangular.01matB.muti}%
\end{equation}
(since $B=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq
p,\ 1\leq j\leq q}$).

We have%
\begin{equation}
\lambda_{i}\leq q\label{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}%
\end{equation}
for each $i\in\left\{  1,2,3,\ldots\right\}  $\ \ \ \ \footnote{\textit{Proof
of (\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}):} Let $i\in\left\{
1,2,3,\ldots\right\}  $. Thus, $i\geq1$, so that $1\leq i$.
\par
The sequence $\lambda$ is a partition, and thus is weakly decreasing. In other
words, $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$. Hence,
$\lambda_{1}\geq\lambda_{i}$ (since $1\leq i$). Thus, $\lambda_{i}\leq
\lambda_{1}\leq q$. This proves
(\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}).}.

Let $i\in\left\{  1,2,\ldots,p\right\}  $. Then, $\lambda_{i}\leq q$ (by
(\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq})). Hence, Lemma
\ref{lem.sol.Lambda.triangular.sum-iver} (applied to $r=\lambda_{i}$) yields
$\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  =\lambda_{i}$. Hence,
(\ref{pf.lem.sol.Lambda.triangular.01matB.lati}) becomes $\widetilde{\lambda
}_{i}=\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  =\lambda_{i}$.

Now, forget that we fixed $i$. We thus have shown that $\widetilde{\lambda
}_{i}=\lambda_{i}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $. In other
words, $\left(  \widetilde{\lambda}_{1},\widetilde{\lambda}_{2},\ldots
,\widetilde{\lambda}_{p}\right)  =\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{p}\right)  =\lambda$ (since $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $). Thus, the row sums of $B$ is
$\lambda$ (since the row sums of $B$ is $\left(  \widetilde{\lambda}%
_{1},\widetilde{\lambda}_{2},\ldots,\widetilde{\lambda}_{p}\right)  $). In
other words, the matrix $B$ has row sums $\lambda$.

On the other hand, the definition (\ref{eq.partition.transpose}) of the
conjugate partition $\lambda^{t}$ of $\lambda$ shows that%
\begin{equation}
\left(  \lambda^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert
\label{pf.lem.sol.Lambda.triangular.01matB.4gen}%
\end{equation}
for every positive integer $i$.

Now, let $k\in\left\{  1,2,\ldots,q\right\}  $. Applying
(\ref{pf.lem.sol.Lambda.triangular.01matB.4gen}) to $i=k$, we obtain
\begin{equation}
\left(  \lambda^{t}\right)  _{k}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}  \right\vert
.\label{pf.lem.sol.Lambda.triangular.01matB.4}%
\end{equation}
But (\ref{pf.lem.sol.Lambda.triangular.01matB.muti}) (applied to $j=k$) shows
that%
\begin{align}
\widetilde{\mu}_{k} &  =\underbrace{\sum_{i=1}^{p}}_{=\sum_{i\in\left\{
1,2,\ldots,p\right\}  }}\left[  \underbrace{k\leq\lambda_{i}}%
_{\Longleftrightarrow\ \left(  \lambda_{i}\geq k\right)  }\right]  =\sum
_{i\in\left\{  1,2,\ldots,p\right\}  }\left[  \lambda_{i}\geq k\right]
\nonumber\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\\lambda_{i}\geq
k}}\underbrace{\left[  \lambda_{i}\geq k\right]  }_{\substack{=1\\\text{(since
}\lambda_{i}\geq k\text{)}}}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,p\right\}  ;\\\text{not }\lambda_{i}\geq k}}\underbrace{\left[  \lambda
_{i}\geq k\right]  }_{\substack{=0\\\text{(since we don't have }\lambda
_{i}\geq k\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\\lambda_{i}\geq
k}}1+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}
;\\\text{not }\lambda_{i}\geq k}}0}_{=0}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,p\right\}  ;\\\lambda_{i}\geq k}}1\nonumber\\
&  =\left\vert \left\{  i\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{i}\geq k\right\}  \right\vert \cdot1=\left\vert \left\{  i\in\left\{
1,2,\ldots,p\right\}  \ \mid\ \lambda_{i}\geq k\right\}  \right\vert
\nonumber\\
&  =\left\vert \left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{j}\geq k\right\}  \right\vert \label{pf.lem.sol.Lambda.triangular.01matB.5}%
\end{align}
(here, we have renamed the index $i$ as $j$). But
\begin{equation}
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}
=\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  \label{pf.lem.sol.Lambda.triangular.01matB.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.01matB.6}):} Let
$i\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  $. Thus, $i$ is an element of $\left\{  1,2,3,\ldots\right\}  $ and
satisfies $\lambda_{i}\geq k$.
\par
We have $\lambda_{i}\geq k>0$ (since $k\in\left\{  1,2,\ldots,q\right\}  $).
Hence, $\lambda_{i}\neq0$.
\par
Assume (for the sake of contradiction) that $i>p$. Hence, $i>p\geq\ell\left(
\lambda\right)  $ (since $\ell\left(  \lambda\right)  \leq p$). But each
integer $j>\ell\left(  \lambda\right)  $ satisfies $\lambda_{j}=0$ (by the
definition of $\ell\left(  \lambda\right)  $). Applying this to $j=i$, we
obtain $\lambda_{i}=0$ (since $i>\ell\left(  \lambda\right)  $). This
contradicts $\lambda_{i}\neq0$. This contradiction shows that our assumption
(that $i>p$) was wrong. Hence, we must have $i\leq p$. Thus, $i\in\left\{
1,2,\ldots,p\right\}  $.
\par
Now, forget that we fixed $i$. We thus have shown that each $i\in\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}  $
satisfies $i\in\left\{  1,2,\ldots,p\right\}  $. In other words, $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}
\subset\left\{  1,2,\ldots,p\right\}  $. Hence,%
\begin{align*}
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}
& =\left\{  1,2,\ldots,p\right\}  \cap\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq k\right\}  \\
& =\left\{  j\in\underbrace{\left\{  1,2,\ldots,p\right\}  \cap\left\{
1,2,3,\ldots\right\}  }_{=\left\{  1,2,\ldots,p\right\}  }\ \mid\ \lambda
_{j}\geq k\right\}  =\left\{  j\in\left\{  1,2,\ldots,p\right\}
\ \mid\ \lambda_{j}\geq k\right\}  .
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.01matB.6}).}. Hence,
(\ref{pf.lem.sol.Lambda.triangular.01matB.4}) becomes%
\[
\left(  \lambda^{t}\right)  _{k}=\left\vert \underbrace{\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}  }%
_{\substack{=\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{j}\geq k\right\}  \\\text{(by (\ref{pf.lem.sol.Lambda.triangular.01matB.6}%
))}}}\right\vert =\left\vert \left\{  j\in\left\{  1,2,\ldots,p\right\}
\ \mid\ \lambda_{j}\geq k\right\}  \right\vert =\widetilde{\mu}_{k}%
\]
(by (\ref{pf.lem.sol.Lambda.triangular.01matB.5})).

Now, forget that we fixed $k$. We thus have shown that $\left(  \lambda
^{t}\right)  _{k}=\widetilde{\mu}_{k}$ for each $k\in\left\{  1,2,\ldots
,q\right\}  $. In other words,%
\[
\left(  \left(  \lambda^{t}\right)  _{1},\left(  \lambda^{t}\right)
_{2},\ldots,\left(  \lambda^{t}\right)  _{q}\right)  =\left(  \widetilde{\mu
}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{q}\right)  .
\]


But $\left(  \lambda^{t}\right)  _{i}=0$ for each integer $i>q$%
\ \ \ \ \footnote{\textit{Proof.} Let $i>q$ be an integer. Then, $q<i$. But
each $j\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\lambda_{j}\leq q$ (by
(\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}), applied to $j$
instead of $i$). Hence, each $j\in\left\{  1,2,3,\ldots\right\}  $ satisfies
$\lambda_{j}<i$ (since $\lambda_{j}\leq q<i$). In other words, no
$j\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\lambda_{j}\geq i$. In other
words, $\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i\right\}  =\varnothing$. Hence,
(\ref{pf.lem.sol.Lambda.triangular.01matB.4gen}) becomes $\left(  \lambda
^{t}\right)  _{i}=\left\vert \underbrace{\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq i\right\}  }_{=\varnothing}\right\vert
=\left\vert \varnothing\right\vert =0$.}. In other words, $\left(  \lambda
^{t}\right)  _{q+1}=\left(  \lambda^{t}\right)  _{q+2}=\left(  \lambda
^{t}\right)  _{q+3}=\cdots=0$. Hence, $\lambda^{t}=\left(  \left(  \lambda
^{t}\right)  _{1},\left(  \lambda^{t}\right)  _{2},\ldots,\left(  \lambda
^{t}\right)  _{q}\right)  $ (since we omit trailing zeroes from a partition).
Thus,%
\[
\lambda^{t}=\left(  \left(  \lambda^{t}\right)  _{1},\left(  \lambda
^{t}\right)  _{2},\ldots,\left(  \lambda^{t}\right)  _{q}\right)  =\left(
\widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{q}\right)
=\left(  \text{the column sums of }B\right)
\]
(since $\left(  \widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu
}_{q}\right)  $ is the column sums of $B$). In other words, the matrix $B$ has
column sums $\lambda^{t}$.

Hence, we know that the matrix $B$ has row sums $\lambda$ and column sums
$\lambda^{t}$. Thus, $B$ is a $\left\{  0,1\right\}  $-matrix having row sums
$\lambda$ and column sums $\lambda^{t}$ (since $B$ is a $\left\{  0,1\right\}
$-matrix). This proves Lemma \ref{lem.sol.Lambda.triangular.01matB}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.lenlat}Let $\lambda$ be a partition. Then:

\begin{enumerate}
\item[(a)] We have $\left(  \lambda^{t}\right)  _{1}=\ell\left(
\lambda\right)  $.

\item[(b)] We have $\lambda_{1}=\ell\left(  \lambda^{t}\right)  $.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.lenlat}.](a) Write the
partition $\lambda$ in the form $\lambda=\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{k},0,0,0,\ldots\right)  $ with $\lambda_{1}\geq
\lambda_{2}\geq\cdots\geq\lambda_{k}>0$. Then, $\ell\left(  \lambda\right)
=k$ (by the definition of $\ell\left(  \lambda\right)  $). Thus,
we have%
\begin{equation}
\lambda_{i}=0\ \ \ \ \ \ \ \ \ \ \text{for each integer }i>\ell\left(
\lambda\right)  \label{pf.lem.sol.Lambda.triangular.lenlat.a.3}%
\end{equation}
(by the definition of $\ell\left(  \lambda\right)  $). Thus,%
\begin{equation}
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}
=\left\{  1,2,\ldots,k\right\}
\label{pf.lem.sol.Lambda.triangular.lenlat.a.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.lenlat.a.5}):}
Let $i\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}%
\geq1\right\}  $. Thus, $i$ is an element of $\left\{  1,2,3,\ldots\right\}  $
and satisfies $\lambda_{i}\geq1$. Thus, $\lambda_{i}\geq1>0$, so that
$\lambda_{i}\neq0$.
\par
If we had $i>\ell\left(  \lambda\right)  $, then we would have $\lambda_{i}=0$
(by (\ref{pf.lem.sol.Lambda.triangular.lenlat.a.3})), which would contradict
$\lambda_{i}\neq0$. Hence, we cannot have $i>\ell\left(  \lambda\right)  $.
Thus, we must have $i\leq\ell\left(  \lambda\right)  $. Hence, $i\leq
\ell\left(  \lambda\right)  =k$, so that $i\in\left\{  1,2,\ldots,k\right\}
$.
\par
Now, forget that we fixed $i$. We thus have shown that $i\in\left\{
1,2,\ldots,k\right\}  $ for each $i\in\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq1\right\}  $. In other words, $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}
\subset\left\{  1,2,\ldots,k\right\}  $.
\par
On the other hand, let $h\in\left\{  1,2,\ldots,k\right\}  $. Then,
$\lambda_{h}>0$ (since $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}%
>0$). Hence, $\lambda_{h}\geq1$ (since $\lambda_{h}$ is an integer). Hence,
$h$ is an element of $\left\{  1,2,3,\ldots\right\}  $ and satisfies
$\lambda_{h}\geq1$. In other words, $h\in\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq1\right\}  $.
\par
Now, forget that we fixed $h$. We thus have shown that $h\in\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}  $ for
each $h\in\left\{  1,2,\ldots,k\right\}  $. In other words, $\left\{
1,2,\ldots,k\right\}  \subset\left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq1\right\}  $. Combining this with $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}
\subset\left\{  1,2,\ldots,k\right\}  $, we obtain $\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}  =\left\{
1,2,\ldots,k\right\}  $. This proves
(\ref{pf.lem.sol.Lambda.triangular.lenlat.a.5}).}.

The definition (\ref{eq.partition.transpose}) of the conjugate partition
$\lambda^{t}$ of $\lambda$ shows that $\left(  \lambda^{t}\right)
_{i}=\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid
\ \lambda_{j}\geq i\right\}  \right\vert $ for each positive integer $i$.
Applying this to $i=1$, we obtain
\[
\left(  \lambda^{t}\right)  _{1}=\left\vert \underbrace{\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}  }_{\substack{=\left\{
1,2,\ldots,k\right\}  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.lenlat.a.5}))}}}\right\vert =\left\vert
\left\{  1,2,\ldots,k\right\}  \right\vert =k=\ell\left(  \lambda\right)  .
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.lenlat}(a).

(b) It is known that $\left(
\lambda^{t}\right)  ^{t}=\lambda$. But Lemma
\ref{lem.sol.Lambda.triangular.lenlat}(a) (applied to $\lambda^{t}$ instead of
$\lambda$) yields $\left(  \left(  \lambda^{t}\right)  ^{t}\right)  _{1}%
=\ell\left(  \lambda^{t}\right)  $. In light of $\left(  \lambda^{t}\right)
^{t}=\lambda$, this rewrites as $\lambda_{1}=\ell\left(  \lambda^{t}\right)
$. This proves Lemma \ref{lem.sol.Lambda.triangular.lenlat}(b).
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(h).
\end{commentedout}

(h) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$ be two partitions that don't satisfy
$\lambda^{t}\triangleright\mu$. We must prove that $a_{\lambda,\mu}=0$.

Indeed, there exists no $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$\ \ \ \ \footnote{\textit{Proof.} Let $A$ be a $\left\{
0,1\right\}  $-matrix of size $\ell\left(  \lambda\right)  \times\ell\left(
\mu\right)  $ having row sums $\lambda$ and column sums $\mu$. We shall derive
a contradiction.
\par
We have $A\in\left\{  0,1\right\}  ^{\ell\left(  \lambda\right)  \times
\ell\left(  \mu\right)  }$ (since $A$ is a $\left\{  0,1\right\}  $-matrix of
size $\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)  $). Thus,
Lemma \ref{lem.sol.Lambda.triangular.01mat}(a) shows that $\lambda
^{t}\triangleright\mu$. This contradicts the fact that we don't have
$\lambda^{t}\triangleright\mu$.
\par
Now, forget that we fixed $A$. We thus have found a contradiction for each
$\left\{  0,1\right\}  $-matrix $A$ of size $\ell\left(  \lambda\right)
\times\ell\left(  \mu\right)  $ having row sums $\lambda$ and column sums
$\mu$. Thus, there exists no $\left\{  0,1\right\}  $-matrix of size
$\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)  $ having row sums
$\lambda$ and column sums $\mu$. Qed.}. Thus, the number of all $\left\{
0,1\right\}  $-matrices of size $\ell\left(  \lambda\right)  \times\ell\left(
\mu\right)  $ having row sums $\lambda$ and column sums $\mu$ equals $0$. In
other words, $a_{\lambda,\mu}$ equals $0$ (since $a_{\lambda,\mu}$ is the
number of all $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$). In other words, $a_{\lambda,\mu}=0$. This solves Exercise
\ref{exe.Lambda.triangular}(h).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(i).
\end{commentedout}

(i) Let $\lambda\in\Par_{n}$. We shall prove that
$a_{\lambda,\lambda^{t}}=1$.

Let $p=\ell\left(  \lambda\right)  $ and $q=\lambda_{1}$. Then, $\ell\left(
\lambda\right)  =p\leq p$ and $\lambda_{1}=q\leq q$. Also, $q=\lambda_{1}%
=\ell\left(  \lambda^{t}\right)  $ (by Lemma
\ref{lem.sol.Lambda.triangular.lenlat}(b)).

Define the $p\times q$-matrix $B$ as in Lemma
\ref{lem.sol.Lambda.triangular.01matB}. Then, Lemma
\ref{lem.sol.Lambda.triangular.01matB} shows that $B$ is a $\left\{
0,1\right\}  $-matrix having row sums $\lambda$ and column sums $\lambda^{t}$.
Furthermore, $B$ is a $\left\{  0,1\right\}  $-matrix of size $p\times q$. In
other words, $B$ is a $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  $ (since $p=\ell\left(
\lambda\right)  $ and $q=\ell\left(  \lambda^{t}\right)  $). Hence, there
exists \textbf{at least one} $\left\{  0,1\right\}  $-matrix of size
$\ell\left(  \lambda\right)  \times\ell\left(  \lambda^{t}\right)  $ having
row sums $\lambda$ and column sums $\lambda^{t}$ (namely, $B$).

On the other hand, using Lemma \ref{lem.sol.Lambda.triangular.01mat}(b), it is
easy to see that every $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  $ having row sums
$\lambda$ and column sums $\lambda^{t}$ must be equal to $B$%
\ \ \ \ \footnote{\textit{Proof.} Let $A$ be a $\left\{  0,1\right\}  $-matrix
of size $\ell\left(  \lambda\right)  \times\ell\left(  \lambda^{t}\right)  $
having row sums $\lambda$ and column sums $\lambda^{t}$. We must prove that
$A=B$.
\par
We know that $A$ is a $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  $. In other words, $A$
is a $\left\{  0,1\right\}  $-matrix of size $p\times q$ (since $\ell\left(
\lambda\right)  =p$ and $\ell\left(  \lambda^{t}\right)  =q$). In other words,
$A\in\left\{  0,1\right\}  ^{p\times q}$.
\par
We have $\left\vert \lambda^{t}\right\vert =\left\vert \lambda\right\vert $
(since the transpose of a partition always has the same size as the partition
itself). But $\left\vert \lambda\right\vert =n$ (since $\lambda\in
\Par_{n}$). Hence, $\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert =n$, so that $\lambda^{t}\in\Par_{n}$.
Hence, Lemma \ref{lem.sol.Lambda.triangular.01mat}(b) (applied
to $\mu=\lambda^{t}$) shows that $A=\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$. But the definition of $B$ yields
$B=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq
j\leq q}$. Comparing this with $A=\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$, we obtain $A=B$. This proves
$A=B$, qed.}. Hence, there exists \textbf{at most one} $\left\{  0,1\right\}
$-matrix of size $\ell\left(  \lambda\right)  \times\ell\left(  \lambda
^{t}\right)  $ having row sums $\lambda$ and column sums $\lambda^{t}$.

We know that $a_{\lambda,\lambda^{t}}$ is the number of all $\left\{
0,1\right\}  $-matrices of size $\ell\left(  \lambda\right)  \times\ell\left(
\lambda^{t}\right)  $ having row sums $\lambda$ and column sums $\lambda^{t}$
(by the definition of $a_{\lambda,\lambda^{t}}$). Since there exists
\textbf{exactly one} such matrix (because we have shown that there exists
\textbf{at least one} such matrix, and we have also shown that there exists
\textbf{at most one} such matrix), we thus conclude that $a_{\lambda
,\lambda^{t}}=1$.

Now, forget that we fixed $\lambda$. We thus have shown that
\begin{equation}
a_{\lambda,\lambda^{t}}=1\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda
\in\Par_{n}. \label{sol.Lambda.triangular.i.almost}%
\end{equation}


Now, let $\lambda\in\Par_{n}$. We shall show that
$a_{\lambda^{t},\lambda}=1$.

It is known that $\left(
\lambda^{t}\right)  ^{t}=\lambda$. But $\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert $ (since the transpose of a partition always
has the same size as the partition itself). But $\left\vert \lambda\right\vert
=n$ (since $\lambda\in\Par_{n}$). Hence, $\left\vert
\lambda^{t}\right\vert =\left\vert \lambda\right\vert =n$, so that
$\lambda^{t}\in\Par_{n}$. Thus,
(\ref{sol.Lambda.triangular.i.almost}) (applied to $\lambda^{t}$ instead of
$\lambda$) yields $a_{\lambda^{t},\left(  \lambda^{t}\right)  ^{t}}=1$. In
light of $\left(  \lambda^{t}\right)  ^{t}=\lambda$, this rewrites as
$a_{\lambda^{t},\lambda}=1$. This solves Exercise \ref{exe.Lambda.triangular}(i).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(j).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(j), let us introduce some
terminology and prove some lemmas.

\begin{definition}
\label{def.sol.Lambda.triangular.j.1}Let $m\in \NN$. Then, $\left[
m\right]  $ shall denote the subset $\left\{  1,2,\ldots,m\right\}  $ of
$\left\{  1,2,3,\ldots\right\}  $. Notice that $\left\vert \left[  m\right]
\right\vert =m$ for each $m\in \NN$.
\end{definition}

\begin{definition}
\label{def.sol.Lambda.triangular.j.2}Let $p\in \NN$ and $q\in\mathbb{N}%
$. Let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]  $ be any map.
Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Then, we define a $q$-tuple $\varphi_{\ast
}\alpha\in\mathbb{N}^{q}$ by%
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)  .
\]
(Recall that if $U$ is any subset of $\left[  q\right]  $, then $\varphi
^{-1}U$ denotes the subset $\left\{  i\in\left[  p\right]  \ \mid
\ \varphi\left(  i\right)  \in U\right\}  $ of $\left[  p\right]  $.)
\end{definition}

\begin{example}
Let $p=5$ and $q=4$, and let $\varphi:\left[  p\right]  \rightarrow\left[
q\right]  $ be the map that sends $1,2,3,4,5$ to $1,4,4,2,2$, respectively.
Let $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha
_{5}\right)  \in\mathbb{N}^{5}$ be a $5$-tuple. Then, the $4$-tuple
$\varphi_{\ast}\alpha\in\mathbb{N}^{4}$ is
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  4\right\}  }\alpha_{i}\right)  =\left(  \alpha_{1},\alpha
_{4}+\alpha_{5},0,\alpha_{2}+\alpha_{3}\right)
\]
(since $\varphi^{-1}\left\{  1\right\}  =\left\{  1\right\}  $, $\varphi
^{-1}\left\{  2\right\}  =\left\{  4,5\right\}  $, $\varphi^{-1}\left\{
3\right\}  =\varnothing$ and $\varphi^{-1}\left\{  4\right\}  =\left\{
2,3\right\}  $).
\end{example}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l1}Let $p$, $q$ and $r$ be three elements
of $\NN$. Let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]  $
and $\psi:\left[  q\right]  \rightarrow\left[  r\right]  $ be any maps. Let
$\alpha\in\mathbb{N}^{p}$. Then, $\left(  \psi\circ\varphi\right)  _{\ast
}\alpha=\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l1}.]Write the
$p$-tuple $\alpha\in\mathbb{N}^{p}$ in the form $\alpha=\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $ for some elements $\alpha
_{1},\alpha_{2},\ldots,\alpha_{p}$ of $\NN$. Thus,%
\begin{equation}
\left(  \psi\circ\varphi\right)  _{\ast}\alpha=\left(  \sum_{i\in\left(
\psi\circ\varphi\right)  ^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  2\right\}  }\alpha
_{i},\ldots,\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{
r\right\}  }\alpha_{i}\right)  \label{pf.prop.Lambda.triangular.j.l1.LHS}%
\end{equation}
(by the definition of $\left(  \psi\circ\varphi\right)  _{\ast}\alpha$).

Write the $q$-tuple $\varphi_{\ast}\alpha\in\mathbb{N}^{q}$ in the form
$\varphi_{\ast}\alpha=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  $
for some elements $\beta_{1},\beta_{2},\ldots,\beta_{q}$ of $\NN$.
Then,%
\[
\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  =\varphi_{\ast}%
\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i}%
,\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots,\sum
_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)
\]
(by the definition of $\varphi_{\ast}\alpha$, because $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $). Thus,%
\begin{equation}
\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left[  q\right]  .
\label{pf.prop.Lambda.triangular.j.l1.1}%
\end{equation}


Write the $r$-tuple $\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)
\in\mathbb{N}^{r}$ in the form $\psi_{\ast}\left(  \varphi_{\ast}%
\alpha\right)  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{r}\right)  $ for
some elements $\gamma_{1},\gamma_{2},\ldots,\gamma_{r}$ of $\NN$. Then,%
\[
\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{r}\right)  =\psi_{\ast}\left(
\varphi_{\ast}\alpha\right)  =\left(  \sum_{i\in\psi^{-1}\left\{  1\right\}
}\beta_{i},\sum_{i\in\psi^{-1}\left\{  2\right\}  }\beta_{i},\ldots,\sum
_{i\in\psi^{-1}\left\{  r\right\}  }\beta_{i}\right)
\]
(by the definition of $\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)  $,
because $\varphi_{\ast}\alpha=\left(  \beta_{1},\beta_{2},\ldots,\beta
_{q}\right)  $). Thus, for each $m\in\left[  r\right]  $, we have%
\begin{align*}
\gamma_{m}  &  =\underbrace{\sum_{i\in\psi^{-1}\left\{  m\right\}  }%
}_{\substack{=\sum\limits_{\substack{i\in\left[  q\right]  ;\\\psi\left(  i\right)
=m}}}}\beta_{i}=\sum\limits_{\substack{i\in\left[  q\right]  ;\\\psi\left(  i\right)
=m}}\beta_{i}=\sum\limits_{\substack{j\in\left[  q\right]  ;\\\psi\left(  j\right)
=m}}\underbrace{\beta_{j}}_{\substack{=\sum_{i\in\varphi^{-1}\left\{
j\right\}  }\alpha_{i}\\\text{(by (\ref{pf.prop.Lambda.triangular.j.l1.1}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the}\\
\text{summation index }i\text{ as }j
\end{array}
\right) \\
&  =\sum\limits_{\substack{j\in\left[  q\right]  ;\\\psi\left(  j\right)
=m}}\underbrace{\sum_{i\in\varphi^{-1}\left\{  j\right\}  }}_{\substack{=\sum\limits
_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}}}}\alpha
_{i}=\underbrace{\sum\limits_{\substack{j\in\left[  q\right]  ;\\\psi\left(
j\right)  =m}}\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}}_{=\sum_{j\in\left[  q\right]  }\sum\limits_{\substack{i\in\left[
p\right]  ;\\\psi\left(  j\right)  =m;\\\varphi\left(  i\right)  =j}}}%
\alpha_{i}\\
&  =\sum_{j\in\left[  q\right]  }\underbrace{\sum\limits_{\substack{i\in\left[
p\right]  ;\\\psi\left(  j\right)  =m;\\\varphi\left(  i\right)  =j}%
}}_{\substack{_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j;\\\psi\left(  j\right)  =m}}=\sum\limits
_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j;\\\psi\left(
\varphi\left(  i\right)  \right)  =m}}}}\\\text{(because for any }i\in\left[
p\right]  \text{ satisfying }\varphi\left(  i\right)  =j\text{,}\\\text{the
condition }\left(  \psi\left(  j\right)  =m\right)  \text{ is equivalent to
}\left(  \psi\left(  \varphi\left(  i\right)  \right)  =m\right)
\\\text{(since }j=\varphi\left(  i\right)  \text{))}}}\alpha_{i}%
=\underbrace{\sum_{j\in\left[  q\right]  }\sum\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  =j;\\\psi\left(  \varphi\left(
i\right)  \right)  =m}}}_{\substack{=\sum_{i\in\left[  p\right]  }%
\sum\limits_{\substack{j\in\left[  q\right]  ;\\\varphi\left(  i\right)
=j;\\\psi\left(  \varphi\left(  i\right)  \right)  =m}}\\=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\psi\left(  \varphi\left(  i\right)  \right)  =m}%
}\sum\limits_{\substack{j\in\left[  q\right]  ;\\\varphi\left(  i\right)  =j}%
}}}\alpha_{i}\\
&  =\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(
\varphi\left(  i\right)  \right)  =m}}}_{=\sum\limits_{\substack{i\in\left[
p\right]  ;\\\left(  \psi\circ\varphi\right)  \left(  i\right)  =m}%
}=\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  m\right\}  }%
}\underbrace{\sum\limits_{\substack{j\in\left[  q\right]  ;\\\varphi\left(  i\right)
=j}}\alpha_{i}}_{\substack{=\alpha_{i}\\\text{(since there is a}\\\text{unique
}j\in\left[  q\right]  \text{ satisfying }\varphi\left(  i\right)  =j\text{)}%
}}=\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  m\right\}  }%
\alpha_{i}.
\end{align*}
In other words, we have%
\[
\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{r}\right)  =\left(  \sum
_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  1\right\}  }\alpha
_{i},\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  2\right\}
}\alpha_{i},\ldots,\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{
r\right\}  }\alpha_{i}\right)  .
\]
Comparing this with (\ref{pf.prop.Lambda.triangular.j.l1.LHS}), we obtain
\[
\left(  \psi\circ\varphi\right)  _{\ast}\alpha=\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{r}\right)  =\psi_{\ast}\left(  \varphi_{\ast}%
\alpha\right)
\]
(since $\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)  =\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{r}\right)  $). This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l1}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l9}Let $p\in \NN$. Let
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  \in
\mathbb{N}^{p}$. Let $\sigma\in\Symm_p$. Then:

\begin{enumerate}
\item[(a)] The $p$-tuple $\sigma_{\ast}\alpha$ is well-defined and satisfies
$\sigma_{\ast}\alpha=\left(  \alpha_{\sigma^{-1}\left(  1\right)  }%
,\alpha_{\sigma^{-1}\left(  2\right)  },\ldots,\alpha_{\sigma^{-1}\left(
p\right)  }\right)  $.

\item[(b)] We have $\sigma_{\ast}\alpha=\alpha$ if and only if $\left(
\alpha_{\sigma\left(  i\right)  }=\alpha_{i}\text{ for all }i\in\left[
p\right]  \right)  $.
\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l9}.]We have
$\sigma\in\Symm_p$. In other words, $\sigma$ is a permutation of
$\left[  p\right]  $ (since $\Symm_p$ is the set of all permutations
of $\left[  p\right]  $ (because $\left[  p\right]  =\left\{  1,2,\ldots
,p\right\}  $)). In other words, $\sigma$ is a bijection $\left[  p\right]
\rightarrow\left[  p\right]  $. Thus, the $p$-tuple $\sigma_{\ast}\alpha$ is
well-defined (since $\alpha\in\mathbb{N}^{p}$). The definition of
$\sigma_{\ast}\alpha$ yields%
\begin{equation}
\sigma_{\ast}\alpha=\left(  \sum_{i\in\sigma^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\sigma^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\sigma^{-1}\left\{  p\right\}  }\alpha_{i}\right)  .
\label{pf.prop.Lambda.triangular.j.l9.0}%
\end{equation}
But each $j\in\left[  p\right]  $ satisfies%
\begin{align*}
\sum_{i\in\sigma^{-1}\left\{  j\right\}  }\alpha_{i}  &  =\sum_{i\in\left\{
\sigma^{-1}\left(  j\right)  \right\}  }\alpha_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sigma^{-1}\left\{  j\right\}  =\left\{  \sigma^{-1}\left(
j\right)  \right\}  \text{ (because }\sigma\text{ is a bijection)}\right) \\
&  =\alpha_{\sigma^{-1}\left(  j\right)  }.
\end{align*}
In other words,%
\[
\left(  \sum_{i\in\sigma^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in
\sigma^{-1}\left\{  2\right\}  }\alpha_{i},\ldots,\sum_{i\in\sigma
^{-1}\left\{  p\right\}  }\alpha_{i}\right)  =\left(  \alpha_{\sigma
^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(  2\right)  },\ldots
,\alpha_{\sigma^{-1}\left(  p\right)  }\right)  .
\]
Hence, (\ref{pf.prop.Lambda.triangular.j.l9.0}) becomes
\[
\sigma_{\ast}\alpha=\left(  \sum_{i\in\sigma^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\sigma^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\sigma^{-1}\left\{  p\right\}  }\alpha_{i}\right)  =\left(
\alpha_{\sigma^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(  2\right)
},\ldots,\alpha_{\sigma^{-1}\left(  p\right)  }\right)  .
\]
This completes the proof of Proposition \ref{prop.sol.Lambda.triangular.j.l9}(a).

(b) We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \underbrace{\sigma_{\ast}\alpha}_{=\left(  \alpha_{\sigma
^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(  2\right)  },\ldots
,\alpha_{\sigma^{-1}\left(  p\right)  }\right)  }=\underbrace{\alpha
}_{=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  }\right) \\
&  \Longleftrightarrow\ \left(  \left(  \alpha_{\sigma^{-1}\left(  1\right)
},\alpha_{\sigma^{-1}\left(  2\right)  },\ldots,\alpha_{\sigma^{-1}\left(
p\right)  }\right)  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\right) \\
&  \Longleftrightarrow\ \left(  \alpha_{\sigma^{-1}\left(  j\right)  }%
=\alpha_{j}\text{ for all }j\in\left[  p\right]  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\alpha_{\sigma^{-1}\left(
\sigma\left(  i\right)  \right)  }}_{\substack{=\alpha_{i}\\\text{(since
}\sigma^{-1}\left(  \sigma\left(  i\right)  \right)  =i\text{)}}%
}=\alpha_{\sigma\left(  i\right)  }\text{ for all }i\in\left[  p\right]
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }\sigma\left(
i\right)  \text{ for }j\text{, since the map }\sigma:\left[  p\right]
\rightarrow\left[  p\right]  \text{ is a bijection}\right) \\
&  \Longleftrightarrow\ \left(  \alpha_{i}=\alpha_{\sigma\left(  i\right)
}\text{ for all }i\in\left[  p\right]  \right) \\
&  \Longleftrightarrow\ \left(  \alpha_{\sigma\left(  i\right)  }=\alpha
_{i}\text{ for all }i\in\left[  p\right]  \right)  .
\end{align*}
This proves Proposition \ref{prop.sol.Lambda.triangular.j.l9}(b).
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l2}Let $p\in \NN$. Let $\alpha
\in\mathbb{N}^{p}$. Then, $\left(  \id_{\left[
p\right]  }\right)  _{\ast}\alpha=\alpha$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l2}.]Clearly,
$\id_{\left[  p\right]  }\in\Symm_p$. Write
the $p$-tuple $\alpha\in\mathbb{N}^{p}$ in the form $\alpha=\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $ for some elements $\alpha
_{1},\alpha_{2},\ldots,\alpha_{p}$ of $\NN$. Thus, Proposition
\ref{prop.sol.Lambda.triangular.j.l9}(a) (applied to $\sigma
=\id_{\left[  p\right]  }$) shows that the $p$-tuple
$\left(  \id_{\left[  p\right]  }\right)  _{\ast
}\alpha$ is well-defined and satisfies $\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  _{\ast}\alpha=\left(  \alpha_{\left(
\id_{\left[  p\right]  }\right)  ^{-1}\left(
1\right)  },\alpha_{\left(  \id_{\left[  p\right]
}\right)  ^{-1}\left(  2\right)  },\ldots,\alpha_{\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left(  p\right)  }\right)  $.
Hence,%
\begin{align*}
\left(  \id_{\left[  p\right]  }\right)  _{\ast}\alpha
&  =\left(  \alpha_{\left(  \id_{\left[  p\right]
}\right)  ^{-1}\left(  1\right)  },\alpha_{\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left(  2\right)  },\ldots
,\alpha_{\left(  \id_{\left[  p\right]  }\right)
^{-1}\left(  p\right)  }\right) \\
&  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left(  i\right)  =i\text{ for all
}i\in\left[  p\right]  \right) \\
&  =\alpha.
\end{align*}
This proves Proposition \ref{prop.sol.Lambda.triangular.j.l2}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.j.l10}Let $p\in \NN$ and $q\in
\mathbb{N}$. Let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]  $ be
any map. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Let $\beta=\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  \in\mathbb{N}^{q}$ be a $q$-tuple such that
$\left(  \beta_{i}>0\text{ for each }i\in\left[  q\right]  \right)  $ and
$\beta=\varphi_{\ast}\alpha$. Then, the map $\varphi$ is surjective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.j.l10}.]We have%
\[
\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  =\beta=\varphi_{\ast
}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i}%
,\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots,\sum
_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)
\]
(by the definition of $\varphi_{\ast}\alpha$, since $\alpha=\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $). In other words,
\begin{equation}
\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left[  q\right]  .
\label{pf.lem.Lambda.triangular.j.l10.1}%
\end{equation}


Let $j\in\left[  q\right]  $. Assume (for the sake of contradiction) that
$\varphi^{-1}\left\{  j\right\}  =\varnothing$. Thus, $\sum_{i\in\varphi
^{-1}\left\{  j\right\}  }\alpha_{i}=\sum_{i\in\varnothing}\alpha_{i}=\left(
\text{empty sum}\right)  =0$. Thus, (\ref{pf.lem.Lambda.triangular.j.l10.1})
becomes $\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}=0$.
But recall that $\left(  \beta_{i}>0\text{ for each }i\in\left[  q\right]
\right)  $. Applying this to $i=j$, we obtain $\beta_{j}>0$. This contradicts
$\beta_{j}=0$.

This contradiction shows that our assumption (that $\varphi^{-1}\left\{
j\right\}  =\varnothing$) was wrong. Hence, we have $\varphi^{-1}\left\{
j\right\}  \neq\varnothing$. Hence, there exists some $k\in\varphi
^{-1}\left\{  j\right\}  $. Consider this $k$.

From $k\in\varphi^{-1}\left\{  j\right\}  $, we obtain $\varphi\left(
k\right)  \in\left\{  j\right\}  $, so that $\varphi\left(  k\right)  =j$.
Hence, $j=\varphi\left(  \underbrace{k}_{\in\varphi^{-1}\left\{  j\right\}
\subset\left[  p\right]  }\right)  \in\varphi\left(  \left[  p\right]
\right)  $.

Now, forget that we fixed $j$. We thus have shown that $j\in\varphi\left(
\left[  p\right]  \right)  $ for each $j\in\left[  q\right]  $. In other
words, $\left[  q\right]  \subset\varphi\left(  \left[  p\right]  \right)  $.
In other words, the map $\varphi$ is surjective. This proves Lemma
\ref{lem.sol.Lambda.triangular.j.l10}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.j.l3}Let $\left(  \beta_{1},\beta_{2}%
,\beta_{3},\ldots\right)  \in\mathbb{N}^{\infty}$ be such that $\beta_{1}%
\geq\beta_{2}\geq\beta_{3}\geq\cdots$. Let $k\in \NN$. Let $R$ be a
finite subset of $\left\{  1,2,3,\ldots\right\}  $ such that $\left\vert
R\right\vert \leq k$. Then,%
\[
\sum_{r\in R}\beta_{r}\leq\beta_{1}+\beta_{2}+\cdots+\beta_{k}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.j.l3}.]We have $\beta_{1}%
\geq\beta_{2}\geq\beta_{3}\geq\cdots$. In other words, if $u$ and $v$ are two
elements of $\left\{  1,2,3,\ldots\right\}  $ satisfying $u\leq v$, then%
\begin{equation}
\beta_{u}\geq\beta_{v}. \label{pf.lem.Lambda.triangular.j.l3.decr}%
\end{equation}


Let $\left(  r_{1},r_{2},\ldots,r_{p}\right)  $ be a list of all elements of
$R$ in increasing order (with no repetitions). Thus, $\sum_{r\in R}\beta
_{r}=\beta_{r_{1}}+\beta_{r_{2}}+\cdots+\beta_{r_{p}}$ and $R=\left\{
r_{1},r_{2},\ldots,r_{p}\right\}  $ and $\left\vert R\right\vert =p$ and
$r_{1}<r_{2}<\cdots<r_{p}$.

Hence, $p=\left\vert R\right\vert \leq k$, so that $0\leq p\leq k$.
Furthermore, $\left\{  r_{1},r_{2},\ldots,r_{p}\right\}  =R\subset\left\{
1,2,3,\ldots\right\}  $.

But each $j\in\left\{  1,2,\ldots,p-1\right\}  $ satisfies%
\begin{equation}
r_{j+1}-r_{j}\geq1 \label{pf.lem.Lambda.triangular.j.l3.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.Lambda.triangular.j.l3.1}):} Let
$j\in\left\{  1,2,\ldots,p-1\right\}  $. Then, $r_{j}<r_{j+1}$ (since
$r_{1}<r_{2}<\cdots<r_{p}$). Hence, $r_{j}\leq r_{j+1}-1$ (since $r_{j}$ and
$r_{j+1}$ are integers). In other words, $r_{j+1}-r_{j}\geq1$. This proves
(\ref{pf.lem.Lambda.triangular.j.l3.1}).}. Thus, each $i\in\left\{
1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
r_{i}\geq i \label{pf.lem.Lambda.triangular.j.l3.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.Lambda.triangular.j.l3.2}):} Let
$i\in\left\{  1,2,\ldots,p\right\}  $. Then, $\sum_{j=1}^{i-1}\left(
r_{j+1}-r_{j}\right)  =r_{i}-r_{1}$ (by the telescope principle). Hence,%
\[
r_{i}-r_{1}=\sum_{j=1}^{i-1}\underbrace{\left(  r_{j+1}-r_{j}\right)
}_{\substack{\geq1\\\text{(by (\ref{pf.lem.Lambda.triangular.j.l3.1}))}}%
}\geq\sum_{j=1}^{i-1}1=i-1.
\]
Hence, $r_{i}\geq\left(  i-1\right)  +r_{1}$. But $r_{1}\geq1$ (since
$r_{1}\in\left\{  r_{1},r_{2},\ldots,r_{p}\right\}  \subset\left\{
1,2,3,\ldots\right\}  $). Thus, $r_{i}\geq\left(  i-1\right)
+\underbrace{r_{1}}_{\geq1}\geq\left(  i-1\right)  +1=i$. This proves
(\ref{pf.lem.Lambda.triangular.j.l3.2}).}. Hence, each $i\in\left\{
1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
\beta_{r_{i}}\leq\beta_{i} \label{pf.lem.Lambda.triangular.j.l3.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.Lambda.triangular.j.l3.3}):} Let
$i\in\left\{  1,2,\ldots,p\right\}  $. From
(\ref{pf.lem.Lambda.triangular.j.l3.2}), we obtain $r_{i}\geq i$. Thus, $i\leq
r_{i}$. Hence, (\ref{pf.lem.Lambda.triangular.j.l3.decr}) (applied to $r_{i}$
and $i$ instead of $u$ and $v$) shows that $\beta_{i}\geq\beta_{r_{i}}$. In
other words, $\beta_{r_{i}}\leq\beta_{i}$. This proves
(\ref{pf.lem.Lambda.triangular.j.l3.3}).}.

Now,%
\begin{equation}
\sum_{r\in R}\beta_{r}=\beta_{r_{1}}+\beta_{r_{2}}+\cdots+\beta_{r_{p}}%
=\sum_{i=1}^{p}\underbrace{\beta_{r_{i}}}_{\substack{\leq\beta_{i}\\\text{(by
(\ref{pf.lem.Lambda.triangular.j.l3.3}))}}}\leq\sum_{i=1}^{p}\beta_{i}.
\label{pf.lem.Lambda.triangular.j.l3.6}%
\end{equation}


But
\begin{align*}
\beta_{1}+\beta_{2}+\cdots+\beta_{k}  &  =\sum_{i=1}^{k}\beta_{i}=\sum
_{i=1}^{p}\beta_{i}+\sum_{i=p+1}^{k}\underbrace{\beta_{i}}_{\substack{\geq
0\\\text{(since }\beta_{i}\in\mathbb{N}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the sum into two at
}i=p\text{ (since }0\leq p\leq k\text{)}\right) \\
&  \geq\sum_{i=1}^{p}\beta_{i}+\underbrace{\sum_{i=p+1}^{k}0}_{=0}=\sum
_{i=1}^{p}\beta_{i}\geq\sum_{r\in R}\beta_{r}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.Lambda.triangular.j.l3.6})}\right)  .
\end{align*}
In other words, $\sum_{r\in R}\beta_{r}\leq\beta_{1}+\beta_{2}+\cdots
+\beta_{k}$. This proves Lemma \ref{lem.sol.Lambda.triangular.j.l3}.
\end{proof}

We now introduce some notations:

\begin{definition}
\label{def.sol.Lambda.triangular.j.B}Let $p\in \NN$ and $q\in\mathbb{N}%
$. Let $\alpha\in\mathbb{N}^{p}$ and $\beta\in\mathbb{N}^{q}$. Then, we define
a set $\mathfrak{B}_{\alpha,\beta,p,q}$ by%
\[
\mathfrak{B}_{\alpha,\beta,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \beta=\varphi_{\ast}\alpha\right\}  .
\]
Note that this set $\mathfrak{B}_{\alpha,\beta,p,q}$ is a subset of $\left\{
\varphi:\left[  p\right]  \rightarrow\left[  q\right]  \right\}  =\left[
q\right]  ^{\left[  p\right]  }$, and therefore is a finite set (since
$\left[  q\right]  ^{\left[  p\right]  }$ is a finite set).
\end{definition}

\begin{example}
Let $p=3$, $q=4$, $\alpha=\left(  2,1,2\right)  $ and $\beta=\left(
3,0,2,0\right)  $. Then, $\mathfrak{B}_{\alpha,\beta,p,q}=\left\{  \varphi
_{1},\varphi_{2}\right\}  $, where $\varphi_{1}$ and $\varphi_{2}$ are the two
maps $\left[  3\right]  \rightarrow\left[  4\right]  $ defined by%
\begin{align*}
\varphi_{1}\left(  1\right)   &  =1,\ \ \ \ \ \ \ \ \ \ \varphi_{1}\left(
2\right)  =1,\ \ \ \ \ \ \ \ \ \ \varphi_{1}\left(  3\right)  =3;\\
\varphi_{2}\left(  1\right)   &  =3,\ \ \ \ \ \ \ \ \ \ \varphi_{2}\left(
2\right)  =1,\ \ \ \ \ \ \ \ \ \ \varphi_{2}\left(  3\right)  =1.
\end{align*}


Notice that $\mathfrak{B}_{\alpha,\beta,p,q}$ depends nontrivially on $p$ and
$q$. Indeed, recall that we are identifying any $k$-tuple $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  \in\mathbb{N}^{k}$ with the weak composition
$\left(  a_{1},a_{2},\ldots,a_{k},0,0,0,\ldots\right)  $; therefore, any two
tuples of nonnegative integers that differ only in trailing zeroes are equated
with each other (for example, $\left(  2,3\right)  $ is equated with $\left(
2,3,0\right)  $), because they are being identified with one and the same weak
composition. Thus, for example, the $3$-tuple $\alpha=\left(  2,1,2\right)  $
is equated with the $4$-tuple $\alpha^{\prime}=\left(  2,1,2,0\right)  $. But
the set $\mathfrak{B}_{\alpha,\beta,p,q}=\mathfrak{B}_{\alpha,\beta,3,4}$
cannot be equated with $\mathfrak{B}_{\alpha^{\prime},\beta,4,4}$; indeed, the
set $\mathfrak{B}_{\alpha^{\prime},\beta,4,4}$ has more than two elements (due
to the extra choice in picking the image of $4$ under the map $\varphi
\in\mathfrak{B}_{\alpha^{\prime},\beta,4,4}$), and so we have $\left\vert
\mathfrak{B}_{\alpha,\beta,3,4}\right\vert \neq\left\vert \mathfrak{B}%
_{\alpha^{\prime},\beta,4,4}\right\vert $.
\end{example}

\begin{remark}
We recall that every $k$-tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
\in\mathbb{N}^{k}$ is identified with the weak composition $\left(  a_{1}%
,a_{2},\ldots,a_{k},0,0,0,\ldots\right)  $. Thus, conversely, any weak
composition $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $
is identified with any $k$-tuple that is obtained from it by removing trailing
zeroes. For example, the weak composition $\left(  2,0,3,0,0,0,\ldots\right)
$ is identified with the $3$-tuple $\left(  2,0,3\right)  $, with the
$4$-tuple $\left(  2,0,3,0\right)  $, and so on.

These identifications have an important consequence: If $\lambda$ and $\mu$
are two weak compositions, then there exist nonnegative integers $p$ and $q$
for which $\lambda$ can be identified with a $p$-tuple (namely, with $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $) and $\mu$ can be
identified with a $q$-tuple (namely, with $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{q}\right)  $). Any two such integers $p$ and $q$ give rise to a
well-defined set $\mathfrak{B}_{\lambda,\mu,p,q}$, defined by regarding
$\lambda$ as a $p$-tuple and regarding $\mu$ as a $q$-tuple.
\end{remark}

\begin{definition}
\label{def.sol.Lambda.triangular.j.B'}Let $\lambda\in\Par$ and
$\mu\in\mathbb{N}^{\infty}$. Let $\ell=\ell\left(  \lambda\right)  $. Then, we
define a set $\mathfrak{B}_{\lambda,\mu}^{\prime}$ by%
\[
\mathfrak{B}_{\lambda,\mu}^{\prime}=\left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  \ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all
}j\in\left\{  1,2,3,\ldots\right\}  \right\}  .
\]

\end{definition}

\begin{definition}
Let $X$, $Y$ and $Z$ be three sets such that $X\subset Y$. Let $g:Z\rightarrow
X$ be any map. Then, we define a map $g\mid^{Y}:Z\rightarrow Y$ by%
\[
\left(  \left(  g\mid^{Y}\right)  \left(  z\right)  =g\left(  z\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }z\in Z\right)  .
\]
(This is well-defined, since each $z\in Z$ satisfies $g\left(  z\right)  \in
X\subset Y$.)

The map $g\mid^{Y}$
is identical to the map $g$ except for the fact that its target
is $Y$ rather than $X$.
\end{definition}

If $X$, $Y$ and $Z$ are three sets such that $X\subset Y$, then the
maps of the form $g \mid^Y$ (with $g$ being a map $Z \to X$)
are exactly those maps
$Z\rightarrow Y$ whose image is contained in $X$. More precisely, the
following lemma holds:

\begin{lemma}
\label{lem.sol.Lambda.triangular.j.l5l}Let $X$, $Y$ and $Z$ be three sets such
that $X\subset Y$. Then, the map%
\begin{align*}
X^{Z}  &  \rightarrow\left\{  f\in Y^{Z}\ \mid\ f\left(  Z\right)  \subset
X\right\}  ,\\
g  &  \mapsto g\mid^{Y}%
\end{align*}
is a bijection.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.j.l5l}.]Lemma
\ref{lem.sol.Lambda.triangular.j.l5l} is a fundamental fact about sets. We
thus omit its proof.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l5}Let $\lambda\in\Par$ and
$\mu\in\operatorname*{WC}$. Let $p=\ell\left(  \lambda\right)  $. From
$p=\ell\left(  \lambda\right)  $, we obtain $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $; thus, $\lambda$ is a $p$-tuple
in $\mathbb{N}^{p}$. Let $q\in \NN$ be such that $\mu=\left(  \mu
_{1},\mu_{2},\ldots,\mu_{q}\right)  $ (that is, $\mu_{i}=0$ for all integers
$i>q$). Thus, $\mu$ is a $q$-tuple in $\mathbb{N}^{q}$. Hence, the set
$\mathfrak{B}_{\lambda,\mu,p,q}$ is well-defined (since $\lambda\in
\mathbb{N}^{p}$ and $\mu\in\mathbb{N}^{q}$).

Now, $\mathfrak{B}_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,p,q}$
as sets.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l5}.]We have
$\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $. In other words,%
\begin{equation}
\mu_{j}=0\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  q+1,q+2,q+3,\ldots
\right\}  . \label{pf.prop.Lambda.triangular.j.l5.muj=0}%
\end{equation}


We have $p=\ell\left(  \lambda\right)  $. Hence, $\lambda_{1}\geq\lambda
_{2}\geq\cdots\geq\lambda_{p}>0$ (by the definition of $\ell\left(
\lambda\right)  $). Therefore,%
\begin{equation}
\lambda_{k}>0\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left[  p\right]  .
\label{pf.prop.Lambda.triangular.j.l5.lambdaneq0}%
\end{equation}


The definition of $\mathfrak{B}_{\lambda,\mu,p,q}$ shows that
\[
\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
\]
(where, of course, $\lambda$ and $\mu$ are regarded as the tuples $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  \in\mathbb{N}^{p}$ and
$\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  \in\mathbb{N}^{q}$). Thus,%
\begin{equation}
\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
=\left\{  \zeta:\left[  p\right]  \rightarrow\left[  q\right]  \ \mid
\ \mu=\zeta_{\ast}\lambda\right\}
\label{pf.prop.Lambda.triangular.j.l5.Blmpq=}%
\end{equation}
(here, we have renamed the index $\varphi$ as $\zeta$).

On the other hand, $p=\ell\left(  \lambda\right)  $. Hence, the definition of
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ yields
\begin{align}
\mathfrak{B}_{\lambda,\mu}^{\prime}  &  =\left\{  \varphi:\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all
}j\in\left\{  1,2,3,\ldots\right\}  \right\} \nonumber\\
&  =\left\{  \zeta:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
\ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\zeta\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right\}  \label{pf.prop.Lambda.triangular.j.l5.B'lm=}%
\end{align}
(here, we have renamed the index $\varphi$ as $\zeta$).

But $\left[  q\right]  =\left\{  1,2,\ldots,q\right\}  \subset\left\{
1,2,3,\ldots\right\}  $. Thus, Lemma \ref{lem.sol.Lambda.triangular.j.l5l}
(applied to $X=\left[  q\right]  $, $Y=\left\{  1,2,3,\ldots\right\}  $ and
$Z=\left[  p\right]  $) shows that the map
\begin{align*}
\left[  q\right]  ^{\left[  p\right]  }  &  \rightarrow\left\{  f\in\left\{
1,2,3,\ldots\right\}  ^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]
\right)  \subset\left[  q\right]  \right\}  ,\\
g  &  \mapsto g\mid^{\left\{  1,2,3,\ldots\right\}  }%
\end{align*}
is a bijection. Denote this bijection by $\Phi$.

We have $\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
\subset\left\{  \varphi:\left[  p\right]  \rightarrow\left[  q\right]
\right\}  =\left[  q\right]  ^{\left[  p\right]  }$. In other words,
$\mathfrak{B}_{\lambda,\mu,p,q}$ is a subset of $\left[  q\right]  ^{\left[
p\right]  }$. Thus, the image $\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  $ is well-defined.

The map $\Phi$ is bijective (since it is a bijection), and therefore injective.

If $U$ and $V$ are two finite sets, and if $T$ is a subset of $U$, and if
$\Psi:U\rightarrow V$ is an injective map, then $\Psi\left(  T\right)  \cong
T$ as sets.\footnote{This is a basic fact about sets.} Applying this to
$U=\left[  q\right]  ^{\left[  p\right]  }$, $V=\left\{  f\in\left\{
1,2,3,\ldots\right\}  ^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]
\right)  \subset\left[  q\right]  \right\}  $, $T=\mathfrak{B}_{\lambda
,\mu,p,q}$ and $\Psi=\Phi$, we conclude that%
\begin{equation}
\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  \cong\mathfrak{B}%
_{\lambda,\mu,p,q}\ \ \ \ \ \ \ \ \ \ \text{as sets}
\label{pf.prop.Lambda.triangular.j.l5.5}%
\end{equation}
(since $\Phi$ is injective).

On the other hand, let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]
$ be any map. Then,%
\begin{equation}
\mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  q+1,q+2,q+3,\ldots\right\}
\label{pf.prop.Lambda.triangular.j.l5.equiv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l5.equiv1}):} Let
$j\in\left\{  q+1,q+2,q+3,\ldots\right\}  $. Thus, $j\geq q+1$ and $\mu_{j}=0$
(by (\ref{pf.prop.Lambda.triangular.j.l5.muj=0})).
\par
Let $i\in\left[  p\right]  $ be such that $\varphi\left(  i\right)  =j$. Thus,
$j=\varphi\left(  i\right)  \in\left[  q\right]  =\left\{  1,2,\ldots
,q\right\}  $, so that $j\leq q$. This contradicts $j\geq q+1>q$.
\par
Now, forget that we fixed $i$. We thus have found a contradiction for each
$i\in\left[  p\right]  $ satisfying $\varphi\left(  i\right)  =j$. Hence,
there exists no $i\in\left[  p\right]  $ satisfying $\varphi\left(  i\right)
=j$. Thus, the sum $\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}$ is empty. Hence, $\sum\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}=\left(  \text{empty
sum}\right)  =0$. Comparing this with $\mu_{j}=0$, we obtain $\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}$. This proves (\ref{pf.prop.Lambda.triangular.j.l5.equiv1}).}.

Recall that $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{p}\right)  $. Hence, the definition of $\varphi_{\ast}\lambda$ shows that%
\[
\varphi_{\ast}\lambda=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}
}\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i}%
,\ldots,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\lambda_{i}\right)  .
\]
Thus, we have the following chain of logical equivalences:%
\begin{align}
&  \ \left(  \underbrace{\mu}_{=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
}=\underbrace{\varphi_{\ast}\lambda}_{=\left(  \sum_{i\in\varphi^{-1}\left\{
1\right\}  }\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda
_{i},\ldots,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\lambda_{i}\right)
}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\lambda_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i},\ldots,\sum_{i\in
\varphi^{-1}\left\{  q\right\}  }\lambda_{i}\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu_{j}=\underbrace{\sum_{i\in\varphi
^{-1}\left\{  j\right\}  }}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}}}\lambda_{i}\text{ for all }j\in\left\{
1,2,\ldots,q\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,\ldots,q\right\}  \right)  . \label{pf.prop.Lambda.triangular.j.l5.equiv2}%
\end{align}


Now, we have the following logical implication:%
\begin{equation}
\left(  \mu=\varphi_{\ast}\lambda\right)  \ \Longrightarrow\ \left(  \mu
_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)
\label{pf.prop.Lambda.triangular.j.l5.equiv3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}):}
Assume that $\left(  \mu=\varphi_{\ast}\lambda\right)  $ holds. We must show
that $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  $ holds.
\par
From the equivalence (\ref{pf.prop.Lambda.triangular.j.l5.equiv2}), we
conclude that
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,\ldots,q\right\}
\right)  \label{pf.prop.Lambda.triangular.j.l5.equiv3a.pf.1}%
\end{equation}
holds (because $\left(  \mu=\varphi_{\ast}\lambda\right)  $ holds).
\par
Now, let $j\in\left\{  1,2,3,\ldots\right\}  $. Then, we want to prove that
$\mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}$. If $j\in\left\{  q+1,q+2,q+3,\ldots\right\}  $, then this
follows immediately from (\ref{pf.prop.Lambda.triangular.j.l5.equiv1}). Hence,
for the rest of this proof, we WLOG assume that $j\notin\left\{
q+1,q+2,q+3,\ldots\right\}  $. Combining this with $j\in\left\{
1,2,3,\ldots\right\}  $, we obtain $j\in\left\{  1,2,3,\ldots\right\}
\setminus\left\{  q+1,q+2,q+3,\ldots\right\}  =\left\{  1,2,\ldots,q\right\}
$. Hence, from (\ref{pf.prop.Lambda.triangular.j.l5.equiv3a.pf.1}), we obtain
$\mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}$.
\par
Now, forget that we fixed $j$. We thus have shown that $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)  $
holds. This concludes the proof of the implication
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}).}.

Now, forget that we fixed $\varphi$. We thus have proven the implication
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}) for each map $\varphi:\left[
p\right]  \rightarrow\left[  q\right]  $.

Now, it is easy to see that $\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  \subset\mathfrak{B}_{\lambda,\mu}^{\prime}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\Phi\left(  \mathfrak{B}%
_{\lambda,\mu,p,q}\right)  $. Thus, $\psi\in\Phi\left(  \mathfrak{B}%
_{\lambda,\mu,p,q}\right)  \subset\left\{  f\in\left\{  1,2,3,\ldots\right\}
^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]  \right)
\subset\left[  q\right]  \right\}  $. In other words, $\psi$ is an element of
$\left\{  1,2,3,\ldots\right\}  ^{\left[  p\right]  }$ and satisfies
$\psi\left(  \left[  p\right]  \right)  \subset\left[  q\right]  $.
\par
But $\psi\in\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  $. In other
words, there exists some $\varphi\in\mathfrak{B}_{\lambda,\mu,p,q}$ such that
$\psi=\Phi\left(  \varphi\right)  $. Consider this $\varphi$. We have
$\psi=\Phi\left(  \varphi\right)  =\varphi\mid^{\left\{  1,2,3,\ldots\right\}
}$ (by the definition of $\Phi$).
\par
We have $\varphi\in\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \zeta:\left[
p\right]  \rightarrow\left[  q\right]  \ \mid\ \mu=\zeta_{\ast}\lambda
\right\}  $ (by (\ref{pf.prop.Lambda.triangular.j.l5.Blmpq=})). In other
words, $\varphi$ is a map $\left[  p\right]  \rightarrow\left[  q\right]  $
and satisfies $\mu=\varphi_{\ast}\lambda$.
\par
Each $i\in\left[  p\right]  $ satisfies%
\begin{equation}
\underbrace{\psi}_{=\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }}\left(
i\right)  =\left(  \varphi\mid^{\left\{  1,2,3,\ldots\right\}  }\right)
\left(  i\right)  =\varphi\left(  i\right)
\label{pf.prop.Lambda.triangular.j.l5.8.pf.0}%
\end{equation}
(by the definition of $\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }$).
\par
We have $\left(  \mu=\varphi_{\ast}\lambda\right)  $. Hence, from the
implication (\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}), we conclude that
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  . \label{pf.prop.Lambda.triangular.j.l5.8.pf.1}%
\end{equation}
Now, each $j\in\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align*}
\mu_{j}  &  =\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\psi\left(  i\right)  =j}}\\\text{(because every }i\in\left[  p\right]
\text{ satisfies }\varphi\left(  i\right)  =\psi\left(  i\right)  \\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l5.8.pf.0})))}}}\lambda_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.Lambda.triangular.j.l5.8.pf.1})}\right) \\
&  =\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}%
}\lambda_{i}.
\end{align*}
In other words, we have $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\psi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right)  $.
\par
Now, $\psi$ is a map $\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  $ (since $\psi$ is an element of $\left\{  1,2,3,\ldots\right\}
^{\left[  p\right]  }$) and satisfies $\left(  \mu_{j}=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\psi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\in\left\{  1,2,3,\ldots\right\}  \right)  $. In other words,%
\[
\psi\in\left\{  \zeta:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\zeta
\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots
\right\}  \right\}  .
\]
In light of (\ref{pf.prop.Lambda.triangular.j.l5.B'lm=}), this rewrites as
$\psi\in\mathfrak{B}_{\lambda,\mu}^{\prime}$.
\par
Now, forget that we fixed $\psi$. We thus have proven that $\psi
\in\mathfrak{B}_{\lambda,\mu}^{\prime}$ for each $\psi\in\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  $. In other words, $\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  \subset\mathfrak{B}_{\lambda,\mu
}^{\prime}$. Qed.} and $\mathfrak{B}_{\lambda,\mu}^{\prime}\subset\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$\psi\in\mathfrak{B}_{\lambda,\mu}^{\prime}$. Thus, $\psi\in\mathfrak{B}%
_{\lambda,\mu}^{\prime}=\left\{  \zeta:\left[  p\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\zeta\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right\}  $ (by
(\ref{pf.prop.Lambda.triangular.j.l5.B'lm=})). In other words, $\psi$ is a map
$\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ and satisfies
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)  .
\label{pf.prop.Lambda.triangular.j.l5.9.pf.2}%
\end{equation}
\par
Now, let $k\in\left[  p\right]  $. We shall show that $\psi\left(  k\right)
\in\left[  q\right]  $.
\par
Indeed, assume the contrary. Thus, $\psi\left(  k\right)  \notin\left[
q\right]  $. Set $j=\psi\left(  k\right)  $. Combining $j=\psi\left(
k\right)  \in\left\{  1,2,3,\ldots\right\}  $ with $j=\psi\left(  k\right)
\notin\left[  q\right]  =\left\{  1,2,\ldots,q\right\}  $, we obtain
$j\in\left\{  1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots,q\right\}
=\left\{  q+1,q+2,q+3,\ldots\right\}  $. Hence, $j\geq q+1$ and $\mu_{j}=0$
(by (\ref{pf.prop.Lambda.triangular.j.l5.muj=0})).
\par
But $k$ is an element of $\left[  p\right]  $ and satisfies $\psi\left(
k\right)  =j$ (since $j=\psi\left(  k\right)  $). Hence, the sum
$\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}%
}\lambda_{i}$ has an addend for $i=k$. If we split off this addend from this
sum, then we obtain%
\[
\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}}\lambda
_{i}=\lambda_{k}+\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(
i\right)  =j;\\i\neq k}}\underbrace{\lambda_{i}}_{\substack{\geq
0\\\text{(since }\lambda_{i}\in\mathbb{N}\text{)}}}\geq\lambda_{k}%
+\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)
=j;\\i\neq k}}0}_{=0}=\lambda_{k}>0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.Lambda.triangular.j.l5.lambdaneq0})}\right)  .
\]
Hence, (\ref{pf.prop.Lambda.triangular.j.l5.9.pf.2}) yields $\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}%
}\lambda_{i}>0$. This contradicts $\mu_{j}=0$.
\par
This contradiction completes our proof of $\psi\left(  k\right)  \in\left[
q\right]  $.
\par
Now, forget that we fixed $k$. We thus have proven that $\psi\left(  k\right)
\in\left[  q\right]  $ for each $k\in\left[  p\right]  $. In other words,
$\psi\left(  \left[  p\right]  \right)  \subset\left[  q\right]  $.
\par
Altogether, we know that $\psi$ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{\left[  p\right]  }$ (since $\psi$ is a map $\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  $) and satisfies $\psi\left(
\left[  p\right]  \right)  \subset\left[  q\right]  $. In other words,
$\psi\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{\left[  p\right]
}\ \mid\ f\left(  \left[  p\right]  \right)  \subset\left[  q\right]
\right\}  $. Hence, an element $\Phi^{-1}\left(  \psi\right)  \in\left[
q\right]  ^{\left[  p\right]  }$ is well-defined (since $\Phi$ is a bijection
$\left[  q\right]  ^{\left[  p\right]  }\rightarrow\left\{  f\in\left\{
1,2,3,\ldots\right\}  ^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]
\right)  \subset\left[  q\right]  \right\}  $). Let us denote this element
$\Phi^{-1}\left(  \psi\right)  $ by $\varphi$. Thus, $\varphi=\Phi^{-1}\left(
\psi\right)  \in\left[  q\right]  ^{\left[  p\right]  }$. In other words,
$\varphi$ is a map $\left[  p\right]  \rightarrow\left[  q\right]  $.
\par
From $\varphi=\Phi^{-1}\left(  \psi\right)  $, we obtain $\psi=\Phi\left(
\varphi\right)  =\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }$ (by the
definition of $\Phi$).
\par
Each $i\in\left[  p\right]  $ satisfies%
\begin{equation}
\underbrace{\psi}_{=\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }}\left(
i\right)  =\left(  \varphi\mid^{\left\{  1,2,3,\ldots\right\}  }\right)
\left(  i\right)  =\varphi\left(  i\right)
\label{pf.prop.Lambda.triangular.j.l5.9.pf.0}%
\end{equation}
(by the definition of $\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }$).
\par
Now, each $j\in\left\{  1,2,\ldots,q\right\}  $ satisfies%
\begin{align*}
\mu_{j}  &  =\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(
i\right)  =j}}}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\\\text{(because every }i\in\left[  p\right]
\text{ satisfies }\psi\left(  i\right)  =\varphi\left(  i\right)  \\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l5.9.pf.0})))}}}\lambda_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.Lambda.triangular.j.l5.9.pf.2})}\right) \\
&  =\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}.
\end{align*}
In other words, we have $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,\ldots,q\right\}  \right)  $. Hence, the equivalence
(\ref{pf.prop.Lambda.triangular.j.l5.equiv2}) shows that $\left(  \mu
=\varphi_{\ast}\lambda\right)  $.
\par
Altogether, we now know that $\varphi$ is a map $\left[  p\right]
\rightarrow\left[  q\right]  $ and satisfies $\mu=\varphi_{\ast}\lambda$. In
other words, $\varphi\in\left\{  \zeta:\left[  p\right]  \rightarrow\left[
q\right]  \ \mid\ \mu=\zeta_{\ast}\lambda\right\}  $. In light of
(\ref{pf.prop.Lambda.triangular.j.l5.Blmpq=}), this rewrites as $\varphi
\in\mathfrak{B}_{\lambda,\mu,p,q}$. Hence, $\psi=\Phi\left(
\underbrace{\varphi}_{\in\mathfrak{B}_{\lambda,\mu,p,q}}\right)  \in
\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  $.
\par
Now, forget that we fixed $\psi$. We thus have proven that $\psi\in\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  $ for each $\psi\in\mathfrak{B}%
_{\lambda,\mu}^{\prime}$. In other words, $\mathfrak{B}_{\lambda,\mu}^{\prime
}\subset\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  $. Qed.}. Combining
these two inclusions, we obtain $\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  =\mathfrak{B}_{\lambda,\mu}^{\prime}$. Hence,%
\[
\mathfrak{B}_{\lambda,\mu}^{\prime}=\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  \cong\mathfrak{B}_{\lambda,\mu,p,q}\ \ \ \ \ \ \ \ \ \ \text{as
sets}%
\]
(by (\ref{pf.prop.Lambda.triangular.j.l5.5})). This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l5}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l6}Let $n\in \NN$. Let $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$.
Assume that $\mathfrak{B}_{\lambda,\mu}^{\prime}%
\neq\varnothing$. Then, $\mu\triangleright\lambda$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l6}.]Let $\ell
=\ell\left(  \lambda\right)  $. We have $\mathfrak{B}_{\lambda,\mu}^{\prime
}\neq\varnothing$. In other words, there exists some $\psi\in\mathfrak{B}%
_{\lambda,\mu}^{\prime}$. Consider this $\psi$. We have%
\[
\psi=\mathfrak{B}_{\lambda,\mu}^{\prime}=\left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  \ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all
}j\in\left\{  1,2,3,\ldots\right\}  \right\}
\]
(by the definition of $\mathfrak{B}_{\lambda,\mu}^{\prime}$ (since $\ell
=\ell\left(  \lambda\right)  $)). In other words, $\psi$ is a map $\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ and satisfies%
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\psi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  . \label{pf.prop.Lambda.triangular.j.l6.psi-def}%
\end{equation}


Let $k\in\left\{  1,2,\ldots,\ell\right\}  $. Then, $1\leq k\leq\ell$. Hence,
$\left[  k\right]  $ is a subset of $\left[  \ell\right]  $.

Define a subset $R$ of $\left\{  1,2,3,\ldots\right\}  $ by $R=\psi\left(
\left[  k\right]  \right)  $. This set $R=\psi\left(  \left[  k\right]
\right)  $ is finite (since $\left[  k\right]  $ is finite).

Each $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}\lambda_{i}=\lambda_{i}
\label{pf.prop.Lambda.triangular.j.l6.stupid-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l6.stupid-sum}):}
Let $i\in\left\{  1,2,\ldots,k\right\}  $. Thus, $i\in\left\{  1,2,\ldots
,k\right\}  =\left[  k\right]  $. Hence, $\psi\left(  \underbrace{i}%
_{\in\left[  k\right]  }\right)  \in\psi\left(  \left[  k\right]  \right)
=R$. Thus, there exists a $j\in R$ satisfying $j=\psi\left(  i\right)  $
(namely, $j=\psi\left(  i\right)  $). This $j$ is furthermore unique (since
the condition $j=\psi\left(  i\right)  $ determines $j$ uniquely). Thus, there
exists \textbf{exactly one} $j\in R$ satisfying $\psi\left(  i\right)  =j$
(namely, $j=\psi\left(  i\right)  $). Hence, the sum $\sum\limits_{\substack{j\in
R;\\\psi\left(  i\right)  =j}}\lambda_{i}$ has exactly one addend (namely, the
addend for $j=\psi\left(  i\right)  $). Thus, this sum simplifies as follows:
$\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}\lambda_{i}=\lambda_{i}$.
This proves (\ref{pf.prop.Lambda.triangular.j.l6.stupid-sum}).}.

It is well-known that if $X$ and $Y$ are two sets, if $f:X\rightarrow Y$ is
any map, and if $T$ is a finite subset of $X$, then $\left\vert f\left(
T\right)  \right\vert \leq\left\vert T\right\vert $. Applying this to
$X=\left[  \ell\right]  $, $Y=\left\{  1,2,3,\ldots\right\}  $, $f=\psi$ and
$T=\left[  k\right]  $, we conclude that $\left\vert \psi\left(  \left[
k\right]  \right)  \right\vert \leq\left\vert \left[  k\right]  \right\vert
=k$. In light of $R=\psi\left(  \left[  k\right]  \right)  $, this rewrites as
$\left\vert R\right\vert \leq k$.

We know that $\mu$ is a partition; thus, $\mu_{1}\geq\mu_{2}\geq\mu_{3}%
\geq\cdots$. Moreover, $\mu\in\Par \subset\mathbb{N}^{\infty}$.
Hence, $\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  =\mu\in\mathbb{N}%
^{\infty}$. Thus, Lemma \ref{lem.sol.Lambda.triangular.j.l3} (applied to
$\beta_{i}=\mu_{i}$) yields $\sum_{r\in R}\mu_{r}\leq\mu_{1}+\mu_{2}%
+\cdots+\mu_{k}$. Hence,%
\begin{align*}
\mu_{1}+\mu_{2}+\cdots+\mu_{k}  &  \geq\sum_{r\in R}\mu_{r}\\
&  =\sum_{j\in R}\underbrace{\mu_{j}}_{\substack{=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\psi\left(  i\right)  =j}}\lambda_{i}\\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l6.psi-def}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have renamed the summation index }r\text{ as }j\right) \\
&  =\underbrace{\sum_{j\in R}\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\psi\left(  i\right)  =j}}}_{=\sum_{i\in\left[  \ell\right]  }%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}}\lambda_{i}%
=\underbrace{\sum_{i\in\left[  \ell\right]  }}_{=\sum_{i=1}^{\ell}}%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}\lambda_{i}\\
&  =\sum_{i=1}^{\ell}\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)
=j}}\lambda_{i}=\sum_{i=1}^{k}\underbrace{\sum\limits_{\substack{j\in R;\\\psi\left(
i\right)  =j}}\lambda_{i}}_{\substack{=\lambda_{i}\\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l6.stupid-sum}))}}}+\sum_{i=k+1}^{\ell}%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}\underbrace{\lambda_{i}%
}_{\substack{\geq0\\\text{(since }\lambda_{i}\in\mathbb{N}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the outer sum at
}i=k\text{, because }1\leq k\leq\ell\right) \\
&  \geq\sum_{i=1}^{k}\lambda_{i}+\underbrace{\sum_{i=k+1}^{\ell}%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}0}_{=0}=\sum_{i=1}%
^{k}\lambda_{i}=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}.
\end{align*}


Now, let us forget that we fixed $k$. We thus have shown that $\mu_{1}+\mu
_{2}+\cdots+\mu_{k}\geq\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}$ for each
$k\in\left\{  1,2,\ldots,\ell\right\}  $. Therefore, Lemma
\ref{lem.sol.Lambda.triangular.majo-mu} (applied to $\ell$, $\lambda$ and
$\mu$ instead of $q$, $\mu$ and $\lambda$) yields $\mu\triangleright\lambda$
(since $\ell\left(  \lambda\right)  =\ell\leq\ell$). This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l6}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l7}Let $\mu\in\Par$. Let
$k=\ell\left(  \mu\right)  $. Then:

\begin{enumerate}
\item[(a)] We have $\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in
\Symm_k\ \mid\ \mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for each
}i\in\left[  k\right]  \right\}  $.

\item[(b)] The set $\mathfrak{B}_{\mu,\mu,k,k}$ is a subgroup of
$\Symm_k$.
\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l7}.]We know that
$\Symm_k$ is the set of all permutations of $\left\{  1,2,\ldots
,k\right\}  $. In other words, $\Symm_k$ is the set of all
permutations of $\left[  k\right]  $ (since $\left[  k\right]  =\left\{
1,2,\ldots,k\right\}  $).

Recall that $\ell\left(  \mu\right)  =k$. Thus, $\mu=\left(  \mu_{1},\mu
_{2},\ldots,\mu_{k}\right)  \in\mathbb{N}^{k}$. Also, from $\ell\left(
\mu\right)  =k$, we obtain $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{k}>0$ (by the
definition of $\ell\left(  \mu\right)  $). Hence,%
\begin{equation}
\mu_{i}>0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left[  k\right]  .
\label{pf.prop.sol.Lambda.triangular.j.l7.gr0}%
\end{equation}


The definition of $\mathfrak{B}_{\mu,\mu,k,k}$ yields%
\begin{equation}
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[  k\right]  \rightarrow
\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  ,
\label{pf.prop.sol.Lambda.triangular.j.l7.gr1}%
\end{equation}
where $\mu$ is regarded as the $k$-tuple $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{k}\right)  \in\mathbb{N}^{k}$.

We have $\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$%
\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\mathfrak{B}_{\mu,\mu,k,k}$.
Thus, $\psi\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[  k\right]
\rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In
other words, $\psi$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $
and satisfies $\mu=\psi_{\ast}\mu$. Thus, $\psi_{\ast}\mu=\mu$.
\par
But Lemma \ref{lem.sol.Lambda.triangular.j.l10} (applied to $p=k$, $q=k$,
$\varphi=\psi$, $\alpha=\mu$, $\alpha_{i}=\mu_{i}$, $\beta=\mu$ and $\beta
_{i}=\mu_{i}$) shows that the map $\psi$ is surjective (because
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr0}) shows that $\left(  \mu
_{i}>0\text{ for each }i\in\left[  k\right]  \right)  $). Hence, $\psi$ is a
surjective map $\left[  k\right]  \rightarrow\left[  k\right]  $.
\par
Now, recall the following known fact about finite sets: If $T$ is a finite
set, then each surjective map $T\rightarrow T$ is bijective. Applying this
fact to $T=\left[  k\right]  $, we conclude that each surjective map $\left[
k\right]  \rightarrow\left[  k\right]  $ is bijective (since $\left[
k\right]  $ is a finite set). Thus, the map $\psi$ is bijective (since $\psi$
is a surjective map $\left[  k\right]  \rightarrow\left[  k\right]  $). Thus,
$\psi$ is a bijection $\left[  k\right]  \rightarrow\left[  k\right]  $. In
other words, $\psi$ is a permutation of $\left[  k\right]  $. In other words,
$\psi\in\Symm_k$ (since $\Symm_k$ is the set of all
permutations of $\left[  k\right]  $).
\par
Now, forget that we fixed $\psi$. We thus have shown that $\psi\in
\Symm_k$ for each $\psi\in\mathfrak{B}_{\mu,\mu,k,k}$. In other
words, $\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$. Qed.}. Hence, we
have the two inclusions $\mathfrak{B}_{\mu,\mu,k,k}\subset\left\{  \sigma
\in\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\mathfrak{B}_{\mu,\mu,k,k}$.
Thus, $\psi\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[  k\right]
\rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In
other words, $\psi$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $
and satisfies $\mu=\psi_{\ast}\mu$. Thus, $\psi_{\ast}\mu=\mu$. Also, $\psi
\in\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$. Hence, $\psi$ is an
element of $\Symm_k$ and satisfies $\psi_{\ast}\mu=\mu$. In other
words, $\psi\in\left\{  \sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu
=\mu\right\}  $.
\par
Now, forget that we fixed $\psi$. We thus have shown that $\psi\in\left\{
\sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $ for each
$\psi\in\mathfrak{B}_{\mu,\mu,k,k}$. In other words, $\mathfrak{B}_{\mu
,\mu,k,k}\subset\left\{  \sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu
=\mu\right\}  $. Qed.} and $\left\{  \sigma\in\Symm_k\ \mid
\ \sigma_{\ast}\mu=\mu\right\}  \subset\mathfrak{B}_{\mu,\mu,k,k}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\left\{  \sigma\in
\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $. Thus, $\psi$ is an
element of $\Symm_k$ and satisfies $\psi_{\ast}\mu=\mu$.
\par
We know that $\psi$ is an element of $\Symm_k$. In other words,
$\psi$ is a permutation of $\left[  k\right]  $ (since $\Symm_k$ is
the set of all permutations of $\left[  k\right]  $). In other words, $\psi$
is a bijection $\left[  k\right]  \rightarrow\left[  k\right]  $.
\par
Thus, $\psi$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\psi_{\ast}\mu$ (since $\psi_{\ast}\mu=\mu$). In other words,
$\psi\in\left\{  \varphi:\left[  k\right]  \rightarrow\left[  k\right]
\ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as $\psi
\in\mathfrak{B}_{\mu,\mu,k,k}$.
\par
Now, forget that we fixed $\psi$. We thus have proven that $\psi
\in\mathfrak{B}_{\mu,\mu,k,k}$ for each $\psi\in\left\{  \sigma\in
\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $. In other words,
$\left\{  \sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}
\subset\mathfrak{B}_{\mu,\mu,k,k}$. Qed.}. Combining these two inclusions, we
obtain
\begin{equation}
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in\Symm_k\ \mid
\ \sigma_{\ast}\mu=\mu\right\}  .
\label{pf.prop.sol.Lambda.triangular.j.l7.gr3}%
\end{equation}


But let $\sigma\in\Symm_k$ be arbitrary. Then, Proposition
\ref{prop.sol.Lambda.triangular.j.l9}(b) (applied to $p=k$, $\alpha=\mu$ and
$\alpha_{i}=\mu_{i}$) shows that we have $\sigma_{\ast}\mu=\mu$ if and only if
$\left(  \mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for all }i\in\left[
k\right]  \right)  $. In other words, we have the following logical
equivalence:%
\begin{equation}
\left(  \sigma_{\ast}\mu=\mu\right)  \Longleftrightarrow\left(  \mu
_{\sigma\left(  i\right)  }=\mu_{i}\text{ for all }i\in\left[  k\right]
\right)  . \label{pf.prop.sol.Lambda.triangular.j.l7.gr5}%
\end{equation}


Now, forget that we fixed $\sigma$. We thus have proven the equivalence
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr5}) for each $\sigma\in
\Symm_k$.

Now, the equality (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr3}) becomes%
\[
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in\Symm_k\ \mid
\ \underbrace{\sigma_{\ast}\mu=\mu}_{\substack{\Longleftrightarrow\ \left(
\mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for all }i\in\left[  k\right]
\right)  \\\text{(by (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr5}))}%
}}\right\}  =\left\{  \sigma\in\Symm_k\ \mid\ \mu_{\sigma\left(
i\right)  }=\mu_{i}\text{ for each }i\in\left[  k\right]  \right\}  .
\]
This proves Proposition \ref{prop.sol.Lambda.triangular.j.l7}(a).

(b) The neutral element of the group $\Symm_k$ is $\operatorname*{id}%
\nolimits_{\left\{  1,2,\ldots,k\right\}  }=\operatorname*{id}%
\nolimits_{\left[  k\right]  }$ (since $\left\{  1,2,\ldots,k\right\}
=\left[  k\right]  $).

The following four observations hold:

\begin{itemize}
\item The set $\mathfrak{B}_{\mu,\mu,k,k}$ is a subset of $\Symm_k$
(since $\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$).

\item If $\gamma$ and $\delta$ are two elements of $\mathfrak{B}_{\mu,\mu
,k,k}$, then $\gamma\circ\delta\in\mathfrak{B}_{\mu,\mu,k,k}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\gamma$ and $\delta$ be two elements of
$\mathfrak{B}_{\mu,\mu,k,k}$. We must prove that $\gamma\circ\delta
\in\mathfrak{B}_{\mu,\mu,k,k}$.
\par
We have $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1})). In other words,
$\gamma$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\gamma_{\ast}\mu$. The same argument (applied to $\delta$
instead of $\gamma$) shows that $\delta$ is a map $\left[  k\right]
\rightarrow\left[  k\right]  $ and satisfies $\mu=\delta_{\ast}\mu$. Now,
Proposition \ref{prop.sol.Lambda.triangular.j.l1} (applied to $p=k$, $q=k$,
$r=k$, $\psi=\gamma$, $\varphi=\delta$ and $\alpha=\mu$) yields $\left(
\gamma\circ\delta\right)  _{\ast}\mu=\gamma_{\ast}\left(  \underbrace{\delta
_{\ast}\mu}_{=\mu}\right)  =\gamma_{\ast}\mu=\mu$. In other words,
$\mu=\left(  \gamma\circ\delta\right)  _{\ast}\mu$. Hence, $\gamma\circ\delta$
is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ (since $\gamma$ and
$\delta$ are maps $\left[  k\right]  \rightarrow\left[  k\right]  $) and
satisfies $\mu=\left(  \gamma\circ\delta\right)  _{\ast}\mu$. In other words,
$\gamma\circ\delta\in\left\{  \varphi:\left[  k\right]  \rightarrow\left[
k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as $\gamma
\circ\delta\in\mathfrak{B}_{\mu,\mu,k,k}$. Qed.}.

\item We have $\id_{\left[  k\right]  }\in
\mathfrak{B}_{\mu,\mu,k,k}$\ \ \ \ \footnote{\textit{Proof.} Proposition
\ref{prop.sol.Lambda.triangular.j.l2} (applied to $p=k$ and $\alpha=\mu$)
yields $\left(  \id_{\left[  k\right]  }\right)
_{\ast}\mu=\mu$. Hence, $\mu=\left(  \id_{\left[
k\right]  }\right)  _{\ast}\mu$.
\par
Thus, $\id_{\left[  k\right]  }$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu=\left(
\id_{\left[  k\right]  }\right)  _{\ast}\mu$. In other
words, $\id_{\left[  k\right]  }\in\left\{
\varphi:\left[  k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu
=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as
$\id_{\left[  k\right]  }\in\mathfrak{B}_{\mu,\mu
,k,k}$. Qed.}.

\item If $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}$, then $\gamma^{-1}%
\in\mathfrak{B}_{\mu,\mu,k,k}$\ \ \ \ \footnote{\textit{Proof.} Let $\gamma
\in\mathfrak{B}_{\mu,\mu,k,k}$. We must prove that $\gamma^{-1}\in
\mathfrak{B}_{\mu,\mu,k,k}$.
\par
We have $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1})). In other words,
$\gamma$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\gamma_{\ast}\mu$. Also, $\gamma\in\mathfrak{B}_{\mu,\mu
,k,k}\subset\Symm_k$; therefore, $\gamma$ has an inverse $\gamma
^{-1}$ (since $\Symm_k$ is a group). Proposition
\ref{prop.sol.Lambda.triangular.j.l1} (applied to $p=k$, $q=k$, $r=k$,
$\psi=\gamma^{-1}$, $\varphi=\gamma$ and $\alpha=\mu$) yields $\left(
\gamma^{-1}\circ\gamma\right)  _{\ast}\mu=\left(  \gamma^{-1}\right)  _{\ast
}\left(  \underbrace{\gamma_{\ast}\mu}_{=\mu}\right)  =\left(  \gamma
^{-1}\right)  _{\ast}\mu$. Hence, $\left(  \gamma^{-1}\right)  _{\ast}%
\mu=\left(  \underbrace{\gamma^{-1}\circ\gamma}_{=\operatorname*{id}%
\nolimits_{\left[  k\right]  }}\right)  _{\ast}\mu=\left(  \operatorname*{id}%
\nolimits_{\left[  k\right]  }\right)  _{\ast}\mu$.
\par
But Proposition \ref{prop.sol.Lambda.triangular.j.l2} (applied to $p=k$ and
$\alpha=\mu$) yields $\left(  \id_{\left[  k\right]
}\right)  _{\ast}\mu=\mu$. Hence, $\mu=\left(  \operatorname*{id}%
\nolimits_{\left[  k\right]  }\right)  _{\ast}\mu$. Comparing this with
$\left(  \gamma^{-1}\right)  _{\ast}\mu=\left(  \operatorname*{id}%
\nolimits_{\left[  k\right]  }\right)  _{\ast}\mu$, we obtain $\mu=\left(
\gamma^{-1}\right)  _{\ast}\mu$. Hence, $\gamma^{-1}$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ (since $\gamma$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $) and satisfies $\mu=\left(
\gamma^{-1}\right)  _{\ast}\mu$. In other words, $\gamma^{-1}\in\left\{
\varphi:\left[  k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu
=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as $\gamma
^{-1}\in\mathfrak{B}_{\mu,\mu,k,k}$. Qed.}.
\end{itemize}

\noindent
Combining these four observations, we conclude that $\mathfrak{B}_{\mu
,\mu,k,k}$ is a subgroup of $\Symm_k$ (since the neutral element of
the group $\Symm_k$ is $\id_{\left[
k\right]  }$). This proves Proposition \ref{prop.sol.Lambda.triangular.j.l7}(b).
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l1}Let $p\in \NN$. Let
$\varphi:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ be any
map. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Then,%
\[
\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)
\]
is a weak composition.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l1}.]For each
$j\in\left\{  1,2,3,\ldots\right\}  $, the sum $\sum_{i\in\varphi^{-1}\left\{
j\right\}  }\alpha_{i}$ is a well-defined element of $\mathbb{N}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,3,\ldots\right\}  $.
Then, $\varphi^{-1}\left\{  j\right\}  $ is a subset of $\left[  p\right]  $,
and thus a finite set (since $\left[  p\right]  $ is a finite set). Hence, the
sum $\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}$ is well-defined.
Moreover, this sum belongs to $\NN$ (since each of its addends
$\alpha_{i}$ belongs to $\NN$). Thus, the sum $\sum_{i\in\varphi
^{-1}\left\{  j\right\}  }\alpha_{i}$ is a well-defined element of
$\NN$. Qed.}. Hence, the sequence%
\[
\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)
\]
is a well-defined element of $\mathbb{N}^{\infty}$. Denote this sequence by
$\beta$. We shall now show that this sequence $\beta$ is a weak composition.

Write the sequence $\beta$ in the form $\beta=\left(  \beta_{1},\beta
_{2},\beta_{3},\ldots\right)  $. Thus,%
\[
\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  =\beta=\left(  \sum
_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  3\right\}
}\alpha_{i},\ldots\right)  .
\]
In other words,
\begin{equation}
\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,3,\ldots\right\}  .
\label{pf.prop.Lambda.triangular.k.l1.betaj=}%
\end{equation}


Let $Z$ be the support of the sequence $\beta$. Then,%
\begin{align*}
Z  &  =\left(  \text{the support of the sequence }\beta\right) \\
&  =\left(  \text{the set of all positive integers }i\text{ for which }%
\beta_{i}\neq0\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the support of a
sequence}\right) \\
&  =\left\{  i\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \beta_{i}%
\neq0\right\}  .
\end{align*}


Every $j\in Z$ satisfies $j\in\varphi\left(  \left[  p\right]  \right)
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in Z$. We must show that
$j\in\varphi\left(  \left[  p\right]  \right)  $.
\par
We have $j\in Z=\left\{  i\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \beta
_{i}\neq0\right\}  $. In other words, $j$ is an element of $\left\{
1,2,3,\ldots\right\}  $ and satisfies $\beta_{j}\neq0$.
\par
Assume (for the sake of contradiction) that $\varphi^{-1}\left\{  j\right\}
=\varnothing$. Thus, $\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha
_{i}=\sum_{i\in\varnothing}\alpha_{i}=\left(  \text{empty sum}\right)  =0$.
Hence, (\ref{pf.prop.Lambda.triangular.k.l1.betaj=}) yields $\beta_{j}%
=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}=0$. This contradicts
$\beta_{j}\neq0$.
\par
This contradiction shows that our assumption (that $\varphi^{-1}\left\{
j\right\}  =\varnothing$) was false. Hence, we must have $\varphi^{-1}\left\{
j\right\}  \neq\varnothing$. In other words, there exists some $k\in
\varphi^{-1}\left\{  j\right\}  $. Consider this $k$.
\par
From $k\in\varphi^{-1}\left\{  j\right\}  $, we obtain $\varphi\left(
k\right)  \in\left\{  j\right\}  $. Hence, $\varphi\left(  k\right)  =j$.
Therefore, $j=\varphi\left(  \underbrace{k}_{\in\varphi^{-1}\left\{
j\right\}  \subset\left[  p\right]  }\right)  \in\varphi\left(  \left[
p\right]  \right)  $. This completes our proof.}. In other words, we have
$Z\subset\varphi\left(  \left[  p\right]  \right)  $. Hence, the set $Z$ is
finite (since the set $\varphi\left(  \left[  p\right]  \right)  $ is finite
(since the set $\left[  p\right]  $ is finite)). In other words, the support
of the sequence $\beta$ is finite (since $Z$ is the support of the sequence
$\beta$).

Hence, we know that $\beta$ is a sequence in $\mathbb{N}^{\infty}$ having
finite support. In other words, $\beta$ is a weak composition (since a weak
composition is defined as a sequence in $\mathbb{N}^{\infty}$ having finite
support). In other words,%
\[
\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)
\]
is a weak composition (since $\beta=\left(  \sum_{i\in\varphi^{-1}\left\{
1\right\}  }\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha
_{i},\sum_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)  $).
This proves Proposition \ref{prop.sol.Lambda.triangular.k.l1}.
\end{proof}

Proposition \ref{prop.sol.Lambda.triangular.k.l1} allows us to make the
following definition (which is similar to Definition
\ref{def.sol.Lambda.triangular.j.2}, but uses the infinite set $\left\{
1,2,3,\ldots\right\}  $ instead of $\left[  q\right]  $):

\begin{definition}
\label{def.sol.Lambda.triangular.k.2}Let $p\in \NN$. Let $\varphi
:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ be any map. Let
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  \in
\mathbb{N}^{p}$ be a $p$-tuple. Then, we define a weak composition
$\varphi_{\ast}\alpha\in\operatorname*{WC}$ by%
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)  .
\]
(This is indeed a weak composition, because of Proposition
\ref{prop.sol.Lambda.triangular.k.l1}.)
\end{definition}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l3}Let $p\in \NN$. Let
$\varphi:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ be any
map. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Then, $\mathbf{x}^{\varphi_{\ast}\alpha
}=x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)
}^{\alpha_{2}}\cdots x_{\varphi\left(  p\right)  }^{\alpha_{p}}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l3}.]The definition of
$\varphi_{\ast}\alpha$ yields
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)  .
\]
Hence, the definition of $\mathbf{x}^{\varphi_{\ast}\alpha}$ shows that%
\begin{align*}
\mathbf{x}^{\varphi_{\ast}\alpha}  &  =x_{1}^{\sum_{i\in\varphi^{-1}\left\{
1\right\}  }\alpha_{i}}x_{2}^{\sum_{i\in\varphi^{-1}\left\{  2\right\}
}\alpha_{i}}x_{3}^{\sum_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i}%
}\cdots=\underbrace{\prod_{j=1}^{\infty}}_{=\prod_{j\in\left\{  1,2,3,\ldots
\right\}  }}\underbrace{x_{j}^{\sum_{i\in\varphi^{-1}\left\{  j\right\}
}\alpha_{i}}}_{=\prod_{i\in\varphi^{-1}\left\{  j\right\}  }x_{j}^{\alpha_{i}%
}}\\
&  =\prod_{j\in\left\{  1,2,3,\ldots\right\}  }\underbrace{\prod_{i\in
\varphi^{-1}\left\{  j\right\}  }}_{\substack{=\prod\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  =j}}}}x_{j}^{\alpha_{i}}=\prod
_{j\in\left\{  1,2,3,\ldots\right\}  }\prod\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}x_{j}^{\alpha_{i}}.
\end{align*}
Comparing this with%
\begin{align*}
x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)
}^{\alpha_{2}}\cdots x_{\varphi\left(  p\right)  }^{\alpha_{p}}%
&=\underbrace{\prod_{i\in\left[  p\right]  }}_{\substack{=\prod_{j\in\left\{
1,2,3,\ldots\right\}  }\prod\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}\\\text{(since }\varphi\left(  i\right)  \in\left\{
1,2,3,\ldots\right\}  \text{ for each }i\in\left[  p\right]  \text{)}%
}}x_{\varphi\left(  i\right)  }^{\alpha_{i}}=\prod_{j\in\left\{
1,2,3,\ldots\right\}  }\prod\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}\underbrace{x_{\varphi\left(  i\right)  }^{\alpha_{i}}%
}_{\substack{=x_{j}^{\alpha_{i}}\\\text{(since }\varphi\left(  i\right)
=j\text{)}}}
\\
&=\prod_{j\in\left\{  1,2,3,\ldots\right\}  }\prod\limits_{\substack{i\in
\left[  p\right]  ;\\\varphi\left(  i\right)  =j}}x_{j}^{\alpha_{i}},
\end{align*}
we obtain $\mathbf{x}^{\varphi_{\ast}\alpha}=x_{\varphi\left(  1\right)
}^{\alpha_{1}}x_{\varphi\left(  2\right)  }^{\alpha_{2}}\cdots x_{\varphi
\left(  p\right)  }^{\alpha_{p}}$. This proves Proposition
\ref{prop.sol.Lambda.triangular.k.l3}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.k.l4auxbij}Let $\ell\in \NN$. Let $X$
be a set. Then, the map%
\[
X^{\left\{  1,2,\ldots,\ell\right\}  }\rightarrow X^{\ell}%
,\ \ \ \ \ \ \ \ \ \ \varphi\mapsto\left(  \varphi\left(  1\right)
,\varphi\left(  2\right)  ,\ldots,\varphi\left(  \ell\right)  \right)
\]
is a bijection.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.k.l4auxbij}.]It is well-known
that the $\ell$-tuples of elements of $X$ are in bijection with the maps
$\left\{  1,2,\ldots,\ell\right\}  \rightarrow X$\ \ \ \ \footnote{In fact,
depending on your definition of an ``$\ell$-tuple'', you might even
consider the $\ell$-tuples of
elements of $X$ to be \textbf{exactly} the maps $\left\{  1,2,\ldots
,\ell\right\}  \rightarrow X$.}. Lemma
\ref{lem.sol.Lambda.triangular.k.l4auxbij} is merely a way to precisely
formulate this bijection. Thus, we omit its proof.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l4}Let $\ell\in \NN$. Let
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\mathbb{N}^{\ell}$ be such that $\left(  \alpha_{i}>0\text{ for each }%
i\in\left[  \ell\right]  \right)  $. Then,%
\[
p_{\alpha_{1}}p_{\alpha_{2}}\cdots p_{\alpha_{\ell}}=\sum_{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  }\xx%
^{\varphi_{\ast}\alpha}%
\]
in $\kk\left[  \left[  \xx\right]  \right]  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l4}.]We assumed that
$\left(  \alpha_{i}>0\text{ for each }i\in\left[  \ell\right]  \right)  $.
Thus, each $i\in\left[  \ell\right]  $ satisfies $\alpha_{i}>0$ and therefore
\begin{align}
p_{\alpha_{i}}  &  =x_{1}^{\alpha_{i}}+x_{2}^{\alpha_{i}}+x_{3}^{\alpha_{i}%
}+\cdots\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }p_{\alpha_{i}%
}\right) \nonumber\\
&  =\sum_{j\in\left\{  1,2,3,\ldots\right\}  }x_{j}^{\alpha_{i}}.
\label{pf.prop.Lambda.triangular.k.l4.1}%
\end{align}


Lemma \ref{lem.sol.Lambda.triangular.k.l4auxbij} (applied to $X=\left\{
1,2,3,\ldots\right\}  $) shows that the map%
\[
\left\{  1,2,3,\ldots\right\}  ^{\left\{  1,2,\ldots,\ell\right\}
}\rightarrow\left\{  1,2,3,\ldots\right\}  ^{\ell},\ \ \ \ \ \ \ \ \ \ \varphi
\mapsto\left(  \varphi\left(  1\right)  ,\varphi\left(  2\right)
,\ldots,\varphi\left(  \ell\right)  \right)
\]
is a bijection. In view of $\left[  \ell\right]  =\left\{  1,2,\ldots
,\ell\right\}  $, this rewrites as follows: The map%
\begin{equation}
\left\{  1,2,3,\ldots\right\}  ^{\left[  \ell\right]  }\rightarrow\left\{
1,2,3,\ldots\right\}  ^{\ell},\ \ \ \ \ \ \ \ \ \ \varphi\mapsto\left(
\varphi\left(  1\right)  ,\varphi\left(  2\right)  ,\ldots,\varphi\left(
\ell\right)  \right)  \label{pf.prop.Lambda.triangular.k.l4.bij}%
\end{equation}
is a bijection.

Now,%
\begin{align*}
p_{\alpha_{1}}p_{\alpha_{2}}\cdots p_{\alpha_{\ell}}  &  =\prod_{i=1}^{\ell
}\underbrace{p_{\alpha_{i}}}_{\substack{=\sum_{j\in\left\{  1,2,3,\ldots
\right\}  }x_{j}^{\alpha_{i}}\\\text{(by
(\ref{pf.prop.Lambda.triangular.k.l4.1}))}}}=\prod_{i=1}^{\ell}\sum
_{j\in\left\{  1,2,3,\ldots\right\}  }x_{j}^{\alpha_{i}}\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{\ell}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{\ell}}\prod_{i=1}^{\ell}x_{j_{i}}^{\alpha_{i}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\underbrace{\sum_{\varphi\in\left\{  1,2,3,\ldots\right\}  ^{\left[
\ell\right]  }}}_{=\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  }}\underbrace{\prod_{i=1}^{\ell}x_{\varphi\left(
i\right)  }^{\alpha_{i}}}_{=x_{\varphi\left(  1\right)  }^{\alpha_{1}%
}x_{\varphi\left(  2\right)  }^{\alpha_{2}}\cdots x_{\varphi\left(
\ell\right)  }^{\alpha_{\ell}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  \varphi\left(  1\right)
,\varphi\left(  2\right)  ,\ldots,\varphi\left(  \ell\right)  \right)  \text{
for }\left(  j_{1},j_{2},\ldots,j_{\ell}\right) \\
\text{in the sum, since the map (\ref{pf.prop.Lambda.triangular.k.l4.bij}) is
a bijection}%
\end{array}
\right) \\
&  =\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  }x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(
2\right)  }^{\alpha_{2}}\cdots x_{\varphi\left(  \ell\right)  }^{\alpha_{\ell
}}.
\end{align*}
Comparing this with%
\[
\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
}\underbrace{\mathbf{x}^{\varphi_{\ast}\alpha}}_{\substack{=x_{\varphi\left(
1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)  }^{\alpha_{2}}\cdots
x_{\varphi\left(  \ell\right)  }^{\alpha_{\ell}}\\\text{(by Proposition
\ref{prop.sol.Lambda.triangular.k.l3} (applied to }p=\ell\text{))}}%
}=\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
}x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)
}^{\alpha_{2}}\cdots x_{\varphi\left(  \ell\right)  }^{\alpha_{\ell}},
\]
we obtain $p_{\alpha_{1}}p_{\alpha_{2}}\cdots p_{\alpha_{\ell}}=\sum
_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
}\mathbf{x}^{\varphi_{\ast}\alpha}$. This proves Proposition
\ref{prop.sol.Lambda.triangular.k.l4}.
\end{proof}

Finally, let us state a proposition that follows immediately from the
definition of a weak composition:

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l5}Let $\mu$ be a weak composition. Then,
there exists a $q\in \NN$ such that $\mu=\left(  \mu_{1},\mu_{2}%
,\ldots,\mu_{q}\right)  $.\ \ \ \ \footnote{Keep in mind that we are
identifying any $k$-tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
\in\mathbb{N}^{k}$ with the weak composition $\left(  a_{1},a_{2},\ldots
,a_{k},0,0,0,\ldots\right)  $. Thus, the $q$-tuple $\left(  \mu_{1},\mu
_{2},\ldots,\mu_{q}\right)  $ is identified with the weak composition $\left(
\mu_{1},\mu_{2},\ldots,\mu_{q},0,0,0,\ldots\right)  $.}
\end{proposition}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(j).
\end{commentedout}

(j) Let $\lambda$ be a partition. Let $\mu$ be a weak composition. Let
$\ell=\ell\left(  \lambda\right)  $. We must prove that the number
$b_{\lambda,\mu}$ is well-defined. In other words, we must prove that there
are only finitely many maps $\varphi:\left\{  1,2,\ldots,\ell\right\}
\rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,\ell\right\}  ;\\\varphi
(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $.

We have $\ell=\ell\left(  \lambda\right)  $; thus, $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $. Hence, $\lambda$ is
an $\ell$-tuple in $\mathbb{N}^{\ell}$.

We have $\ell=\ell\left(  \lambda\right)  $. Hence, the definition of
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ yields%
\begin{align}
\mathfrak{B}_{\lambda,\mu}^{\prime}  &  =\left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \underbrace{\mu_{j}%
=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
}_{\substack{\Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  \\\text{(since the }j\in\left\{  1,2,3,\ldots\right\}  \text{
are precisely the integers }j\geq1\text{)}}}\right\} \nonumber\\
&  =\left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}
\nonumber\\
&  =\left\{  \varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,\ell\right\}  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for
all }j\geq1\right\}  \label{sol.Lambda.triangular.B'lm=}%
\end{align}
(since $\left[  \ell\right]  =\left\{  1,2,\ldots,\ell\right\}  $).

Proposition \ref{prop.sol.Lambda.triangular.k.l5} shows that there exists some
$q\in \NN$ such that $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
$. Consider this $q$. Thus, $\mu$ is a $q$-tuple in $\mathbb{N}^{q}$ (since
$\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $). Hence, the set
$\mathfrak{B}_{\lambda,\mu,\ell,q}$ is well-defined (since $\lambda
\in\mathbb{N}^{\ell}$ and $\mu\in\mathbb{N}^{q}$). Also, $\mu\in
\operatorname*{WC}$ (since $\mu$ is a weak composition). Thus, Proposition
\ref{prop.sol.Lambda.triangular.j.l5} (applied to $p=\ell$) shows that
$\mathfrak{B}_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,\ell,q}$ as
sets. But the definition of $\mathfrak{B}_{\lambda,\mu,\ell,q}$ yields
\[
\mathfrak{B}_{\lambda,\mu,\ell,q}=\left\{  \varphi:\left[  \ell\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
\subset\left\{  \varphi:\left[  \ell\right]  \rightarrow\left[  q\right]
\right\}  =\left[  q\right]  ^{\left[  \ell\right]  }.
\]
Hence, $\mathfrak{B}_{\lambda,\mu,\ell,q}$ is a finite set (since $\left[
q\right]  ^{\left[  \ell\right]  }$ is a finite set). Therefore, the set
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ is a finite set as well (since
$\mathfrak{B}_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,\ell,q}$ as
sets). In view of (\ref{sol.Lambda.triangular.B'lm=}), this rewrites as
follows: The set%
\[
\left\{  \varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,\ell\right\}  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for
all }j\geq1\right\}
\]
is a finite set. In other words, there are only finitely many maps
$\varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow\left\{  1,2,3,\ldots
\right\}  $ satisfying $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,\ell\right\}  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }%
j\geq1\right)  $. This solves Exercise \ref{exe.Lambda.triangular}(j).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(k).
\end{commentedout}

(k) Let $\lambda\in\Par_{n}$. Let $\ell=\ell\left(
\lambda\right)  $. Thus, $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell}\right)  $ (by the definition of $\ell\left(
\lambda\right)  $). Thus, $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell}\right)  \in\mathbb{N}^{\ell}$. Therefore, for every map
$\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $, a
weak composition $\varphi_{\ast}\lambda\in\operatorname*{WC}$ is defined
(according to Definition \ref{def.sol.Lambda.triangular.k.2}, applied to
$p=\ell$, $\alpha=\lambda$ and $\alpha_{i}=\lambda_{i}$). The definition of
this weak composition $\varphi_{\ast}\lambda$ yields%
\begin{equation}
\varphi_{\ast}\lambda=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}
}\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\lambda_{i},\ldots\right)
\label{sol.Lambda.triangular.phi*lam=}%
\end{equation}
for any map $\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  $. Hence, for each map $\varphi:\left[  \ell\right]  \rightarrow
\left\{  1,2,3,\ldots\right\}  $, we have the following chain of logical
equivalences:%
\begin{align*}
&  \ \left(  \underbrace{\mu}_{=\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)
}=\underbrace{\varphi_{\ast}\lambda}_{\substack{=\left(  \sum_{i\in
\varphi^{-1}\left\{  1\right\}  }\lambda_{i},\sum_{i\in\varphi^{-1}\left\{
2\right\}  }\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  3\right\}  }\lambda
_{i},\ldots\right)  \\\text{(by (\ref{sol.Lambda.triangular.phi*lam=}))}%
}}\right) \\
&  \Longleftrightarrow\ \left(  \left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)
=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\lambda_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\lambda_{i},\ldots\right)  \right) \\
&  \Longleftrightarrow\ \left(  \mu_{j}=\underbrace{\sum\limits_{i\in
\varphi^{-1}\left\{  j\right\}  }}_{\substack{=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  =j}}}}\lambda_{i}\text{ for all
}j\geq1\right) \\
&  \Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  .
\end{align*}
Thus,%
\begin{align}
&  \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu=\varphi_{\ast}\lambda\right\} \nonumber\\
&  =\left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}  .
\label{sol.Lambda.triangular.equivs.sets}%
\end{align}


We have $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)
$ with $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{\ell}>0$ (since
$\ell=\ell\left(  \lambda\right)  $). Hence, $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ (by the definition of $p_{\lambda}$).

From $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{\ell}>0$, we also
obtain $\left(  \lambda_{i}>0\text{ for each }i\in\left[  \ell\right]
\right)  $. Hence, Proposition \ref{prop.sol.Lambda.triangular.k.l4} (applied
to $\alpha=\lambda$ and $\alpha_{i}=\lambda_{i}$) yields%
\[
p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}=\sum_{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  }\xx%
^{\varphi_{\ast}\lambda}%
\]
in $\kk\left[  \left[  \xx\right]  \right]  $.

Now, let $\mu\in\Par_{n}$. Thus, $\mu\in
\Par_{n}\subset\Par \subset
\operatorname*{WC}$. From $p_{\lambda}=p_{\lambda_{1}}p_{\lambda_{2}}\cdots
p_{\lambda_{\ell}}=\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  }\mathbf{x}^{\varphi_{\ast}\lambda}$, we obtain%
\begin{align}
&  \left[  \mathbf{x}^{\mu}\right]  \left(  p_{\lambda}\right) \nonumber\\
&  =\left[  \mathbf{x}^{\mu}\right]  \left(  \sum_{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  }\xx%
^{\varphi_{\ast}\lambda}\right)  =\sum_{\varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  }\underbrace{\left[  \xx%
^{\mu}\right]  \left(  \mathbf{x}^{\varphi_{\ast}\lambda}\right)
}_{\substack{=\delta_{\mu,\varphi_{\ast}\lambda}\\\text{(by
(\ref{sol.Lambda.triangular.coeff-mon}) (applied to }\alpha=\varphi_{\ast
}\lambda\text{))}}}\nonumber\\
&  =\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  }\delta_{\mu,\varphi_{\ast}\lambda}\nonumber\\
&  =\sum\limits_{\substack{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  ;\\\mu=\varphi_{\ast}\lambda}}\underbrace{\delta
_{\mu,\varphi_{\ast}\lambda}}_{\substack{=1\\\text{(since }\mu=\varphi_{\ast
}\lambda\text{)}}}+\sum\limits_{\substack{\varphi:\left[  \ell\right]  \rightarrow
\left\{  1,2,3,\ldots\right\}  ;\\\mu\neq\varphi_{\ast}\lambda}%
}\underbrace{\delta_{\mu,\varphi_{\ast}\lambda}}_{\substack{=0\\\text{(since
}\mu\neq\varphi_{\ast}\lambda\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  ;\\\mu=\varphi_{\ast}\lambda}}1+\underbrace{\sum\limits
_{\substack{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  ;\\\mu\neq\varphi_{\ast}\lambda}}0}_{=0}=\sum\limits_{\substack{\varphi
:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  ;\\\mu
=\varphi_{\ast}\lambda}}1\nonumber\\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}  \right\vert
\cdot1\nonumber\\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}  \right\vert
\nonumber\\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}
\right\vert \label{sol.Lambda.triangular.k.4}%
\end{align}
(by (\ref{sol.Lambda.triangular.equivs.sets})).

On the other hand, the definition of $b_{\lambda,\mu}$ shows that
$b_{\lambda,\mu}$ is the number of all maps $\varphi:\left\{  1,2,\ldots
,\ell\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(
\mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,\ell\right\}
;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $ (since $\ell
=\ell\left(  \lambda\right)  $). In light of $\left\{  1,2,\ldots
,\ell\right\}  =\left[  \ell\right]  $, this rewrites as follows:
$b_{\lambda,\mu}$ is the number of all maps $\varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi(i)=j}}\lambda
_{i}\text{ for all }j\geq1\right)  $. In other words,
\[
b_{\lambda,\mu}=\left\vert \left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum
\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi(i)=j}}\lambda_{i}\text{
for all }j\geq1\right\}  \right\vert .
\]
Comparing this with (\ref{sol.Lambda.triangular.k.4}), we obtain%
\begin{equation}
\left[  \mathbf{x}^{\mu}\right]  \left(  p_{\lambda}\right)  =b_{\lambda,\mu}.
\label{sol.Lambda.triangular.k.coeff=}%
\end{equation}


Now, forget that we fixed $\mu$. We thus have proven the equality
(\ref{sol.Lambda.triangular.k.coeff=}) for every $\mu\in\Par_{n}$.

But $\lambda\in\Par_{n}$, so that $\left\vert
\lambda\right\vert =n$. Hence, $n=\left\vert \underbrace{\lambda}_{=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  }\right\vert =\left\vert
\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  \right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}$.

Let $i\in\left\{  1,2,\ldots,\ell\right\}  $. Then, $p_{\lambda_{i}}$ is a
homogeneous element of $\Lambda$ having degree $\lambda_{i}$ (because for each
positive integer $m$, the element $p_{m}$ is a homogeneous element of
$\Lambda$ having degree $m$).

Now, forget that we fixed $i$. We thus have shown that for each $i\in\left\{
1,2,\ldots,\ell\right\}  $, the element $p_{\lambda_{i}}$ is a homogeneous
element of $\Lambda$ having degree $\lambda_{i}$. In other words,
$p_{\lambda_{1}},p_{\lambda_{2}},\ldots,p_{\lambda_{\ell}}$ are homogeneous
elements of $\Lambda$ having degrees $\lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell}$, respectively. Hence, the product $p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ of these elements is a homogeneous
element of $\Lambda$ having degree $\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{\ell}$. In light of $p_{\lambda}=p_{\lambda_{1}}p_{\lambda_{2}%
}\cdots p_{\lambda_{\ell}}$ and $n=\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{\ell}$, this rewrites as follows: The element $p_{\lambda}$ is a
homogeneous element of $\Lambda$ having degree $n$. In other words,
$p_{\lambda}\in\Lambda_{n}$. Thus, Exercise \ref{exe.Lambda.triangular}(a)
(applied to $f=p_{\lambda}$) yields%
\[
p_{\lambda}=\sum_{\mu\in\Par_{n}}\underbrace{\left(
\left[  \mathbf{x}^{\mu}\right]  \left(  p_{\lambda}\right)  \right)
}_{\substack{=b_{\lambda,\mu}\\\text{(by (\ref{sol.Lambda.triangular.k.coeff=}%
))}}}m_{\mu}=\sum_{\mu\in\operatorname{Par}_{n}}b_{\lambda,\mu}m_{\mu}.
\]
This solves Exercise \ref{exe.Lambda.triangular}(k).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(l).
\end{commentedout}

Before we come to the solution of Exercise \ref{exe.Lambda.triangular}(l), let
us state a simple lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.k.1}Let $\lambda$ be a partition. Let $\mu$
be a weak composition. Then:

\begin{enumerate}
\item[(a)] We have $b_{\lambda,\mu}=\left\vert \mathfrak{B}_{\lambda,\mu
}^{\prime}\right\vert $.

\item[(b)] Let $p=\ell\left(  \lambda\right)  $. Let $q\in \NN$ be such
that $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $ (that is, $\mu
_{i}=0$ for all integers $i>q$). Then, the set $\mathfrak{B}_{\lambda,\mu
,p,q}$ is well-defined and satisfies $b_{\lambda,\mu}=\left\vert
\mathfrak{B}_{\lambda,\mu,p,q}\right\vert $.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.k.1}.]We know that $\lambda$ is
a partition.

We have $p=\ell\left(  \lambda\right)  $. Hence, the definition of
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ yields%
\begin{align}
\mathfrak{B}_{\lambda,\mu}^{\prime}  &  =\left\{  \varphi:\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \underbrace{\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  }%
_{\substack{\Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  \\\text{(since the }j\in\left\{  1,2,3,\ldots\right\}  \text{
are precisely the integers }j\geq1\text{)}}}\right\} \nonumber\\
&  =\left\{  \varphi:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}  .
\label{pf.lem.sol.Lambda.triangular.k.1.1}%
\end{align}


On the other hand, the definition of $b_{\lambda,\mu}$ shows that
$b_{\lambda,\mu}$ is the number of all maps $\varphi:\left\{  1,2,\ldots
,p\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(
\mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}
;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $ (since
$p=p\left(  \lambda\right)  $). In view of $\left\{  1,2,\ldots,p\right\}
=\left[  p\right]  $, this rewrites as follows: $b_{\lambda,\mu}$ is the
number of all maps $\varphi:\left[  p\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  $ satisfying $\left(  \mu_{j}=\sum
\limits_{\substack{i\in\left[  p\right]  ;\\\varphi(i)=j}}\lambda_{i}\text{
for all }j\geq1\right)  $. In other words,%
\[
b_{\lambda,\mu}=\left\vert \underbrace{\left\{  \varphi:\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum
\limits_{\substack{i\in\left[  p\right]  ;\\\varphi(i)=j}}\lambda_{i}\text{
for all }j\geq1\right\}  }_{\substack{=\mathfrak{B}_{\lambda,\mu}^{\prime
}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.k.1.1}))}}}\right\vert
=\left\vert \mathfrak{B}_{\lambda,\mu}^{\prime}\right\vert .
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.k.1}(a).

(b) From $p=\ell\left(  \lambda\right)  $, we obtain $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $; thus, $\lambda$ is a
$p$-tuple in $\mathbb{N}^{p}$. Also, $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{q}\right)  \in\mathbb{N}^{q}$. Thus, $\mu$ is a $q$-tuple in
$\mathbb{N}^{q}$. Hence, the set $\mathfrak{B}_{\lambda,\mu,p,q}$ is
well-defined (since $\lambda\in\mathbb{N}^{p}$ and $\mu\in\mathbb{N}^{q}$).
Proposition \ref{prop.sol.Lambda.triangular.j.l5} yields that $\mathfrak{B}%
_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,p,q}$ as sets. But Lemma
\ref{lem.sol.Lambda.triangular.k.1}(a) shows that $b_{\lambda,\mu}=\left\vert
\mathfrak{B}_{\lambda,\mu}^{\prime}\right\vert =\left\vert \mathfrak{B}%
_{\lambda,\mu,p,q}\right\vert $ (since $\mathfrak{B}_{\lambda,\mu}^{\prime
}\cong\mathfrak{B}_{\lambda,\mu,p,q}$ as sets). Thus, Lemma
\ref{lem.sol.Lambda.triangular.k.1}(b) is proven.
\end{proof}

We now resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(l).
\end{commentedout}

(l) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$ be any partitions that don't satisfy
$\mu\triangleright\lambda$. We must prove that $b_{\lambda,\mu}=0$.

Indeed, assume the contrary. Thus, $b_{\lambda,\mu}\neq0$. But Lemma
\ref{lem.sol.Lambda.triangular.k.1}(a) yields $b_{\lambda,\mu}=\left\vert
\mathfrak{B}_{\lambda,\mu}^{\prime}\right\vert $ (since $\mu$ is a weak
composition (since $\mu$ is a partition)). Hence, $\left\vert \mathfrak{B}%
_{\lambda,\mu}^{\prime}\right\vert =b_{\lambda,\mu}\neq0$. In other words,
$\mathfrak{B}_{\lambda,\mu}^{\prime}\neq\varnothing$. Thus, Proposition
\ref{prop.sol.Lambda.triangular.j.l6} shows that $\mu\triangleright\lambda$.
This contradicts the fact that we don't have $\mu\triangleright\lambda$.

This contradiction proves that our assumption was wrong. Hence, $b_{\lambda
,\mu}=0$ is proven. This solves Exercise \ref{exe.Lambda.triangular}(l).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(m).
\end{commentedout}

(m) Let $\lambda\in\Par_{n}$. We must prove that
$b_{\lambda,\lambda}$ is a positive integer.

Let $k=\ell\left(  \lambda\right)  $. Thus, $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{k}\right)  $ (by the definition of
$\ell\left(  \lambda\right)  $). Clearly, $\lambda$ is a weak composition
(since $\lambda$ is a partition). Hence, Lemma
\ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $\mu=\lambda$, $p=k$ and
$q=k$) shows that the set $\mathfrak{B}_{\lambda,\lambda,k,k}$ is well-defined
and satisfies $b_{\lambda,\lambda}=\left\vert \mathfrak{B}_{\lambda
,\lambda,k,k}\right\vert $. But $\lambda\in\Par_{n}%
\subset\Par$. Hence, Proposition
\ref{prop.sol.Lambda.triangular.j.l7}(b) (applied to $\mu=\lambda$) shows that
$\mathfrak{B}_{\lambda,\lambda,k,k}$ is a subgroup of $\Symm_k$.
Thus, the set $\mathfrak{B}_{\lambda,\lambda,k,k}$ contains the neutral
element of $\Symm_k$. Thus, the set $\mathfrak{B}_{\lambda
,\lambda,k,k}$ contains at least one element, and therefore is nonempty. Thus,
$\left\vert \mathfrak{B}_{\lambda,\lambda,k,k}\right\vert >0$.

Hence, $b_{\lambda,\lambda}=\left\vert \mathfrak{B}_{\lambda,\lambda
,k,k}\right\vert >0$. Since $b_{\lambda,\lambda}$ is an integer, we can
therefore conclude that $b_{\lambda,\lambda}$ is a positive integer. This
solves Exercise \ref{exe.Lambda.triangular}(m).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(n).
\end{commentedout}

(n) Let $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{k}\right)  \in
\Par_{n}$ be a partition. Let $k=\ell\left(
\mu\right)  $. We must show that the integer $b_{\mu,\mu}$ is the size of the
subgroup of $\Symm_k$ consisting of all permutations $\sigma
\in\Symm_k$ having each $i$ satisfy $\mu_{\sigma(i)}=\mu_{i}$. In
particular, we must show that this subgroup is indeed a subgroup.

Clearly, $\mu$ is a weak composition (since $\mu$ is a partition). Hence,
Lemma \ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $\lambda=\mu$, $p=k$
and $q=k$) shows that the set $\mathfrak{B}_{\mu,\mu,k,k}$ is well-defined and
satisfies $b_{\mu,\mu}=\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert $.

We have $\mu\in\Par_{n}\subset\Par$.
Hence, Proposition \ref{prop.sol.Lambda.triangular.j.l7}(a) shows that
\begin{equation}
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in\Symm_k\ \mid
\ \mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for each }i\in\left[  k\right]
\right\}  . \label{sol.Lambda.triangular.n.1}%
\end{equation}
Furthermore, Proposition \ref{prop.sol.Lambda.triangular.j.l7}(b) shows that
$\mathfrak{B}_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$.

Now,%
\begin{align}
&  \left(  \text{the set of all permutations }\sigma\in\Symm_k\text{
having each }i\text{ satisfy }\mu_{\sigma\left(  i\right)  }=\mu_{i}\right)
\nonumber\\
&  =\left\{  \sigma\in\Symm_k\ \mid\ \text{each }i\text{ satisfies
}\mu_{\sigma\left(  i\right)  }=\mu_{i}\right\} \nonumber \\
&  =\left\{  \sigma
\in\Symm_k\ \mid\ \text{each }i\in\underbrace{\left\{  1,2,\ldots
,k\right\}  }_{=\left[  k\right]  }\text{ satisfies }\mu_{\sigma\left(
i\right)  }=\mu_{i}\right\} \nonumber\\
&  =\left\{  \sigma\in\Symm_k\ \mid\ \text{each }i\in\left[
k\right]  \text{ satisfies }\mu_{\sigma\left(  i\right)  }=\mu_{i}\right\}
\nonumber \\
&  =
\left\{  \sigma\in\Symm_k\ \mid\ \mu_{\sigma\left(  i\right)  }%
=\mu_{i}\text{ for each }i\in\left[  k\right]  \right\} \nonumber\\
&  =\mathfrak{B}_{\mu,\mu,k,k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.n.1})}\right)  . \label{sol.Lambda.triangular.n.3}%
\end{align}


But recall that $\mathfrak{B}_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$.
In view of (\ref{sol.Lambda.triangular.n.3}), this rewrites as follows:
The set of all permutations $\sigma\in\Symm_k$ having each $i$
satisfy $\mu_{\sigma\left(  i\right)  }=\mu_{i}$ is a subgroup of
$\Symm_k$. This subgroup thus is the subgroup of $\Symm_k$
consisting of all permutations $\sigma\in\Symm_k$ having each $i$
satisfy $\mu_{\sigma(i)}=\mu_{i}$. The size of this subgroup is clearly%
\[
\left\vert \underbrace{\left\{  \sigma\in\Symm_k\ \mid\ \text{each
}i\text{ satisfies }\mu_{\sigma\left(  i\right)  }=\mu_{i}\right\}
}_{=\mathfrak{B}_{\mu,\mu,k,k}}\right\vert =\left\vert \mathfrak{B}_{\mu
,\mu,k,k}\right\vert =b_{\mu,\mu}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}b_{\mu,\mu}=\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert \right)  .
\]
Thus, $b_{\mu,\mu}$ is the size of the subgroup of $\Symm_k$
consisting of all permutations $\sigma\in\Symm_k$ having each $i$
satisfy $\mu_{\sigma(i)}=\mu_{i}$. (In particular, we have shown that this
subgroup is indeed a subgroup.) This solves Exercise
\ref{exe.Lambda.triangular}(n).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(o).
\end{commentedout}

Before we come to the solution of Exercise \ref{exe.Lambda.triangular}(o), let
us prove some further auxiliary results.

\begin{proposition}
\label{prop.sol.Lambda.triangular.o.l1}Let $\mu\in\Par$. Let
$k=\ell\left(  \mu\right)  $. Proposition
\ref{prop.sol.Lambda.triangular.j.l7}(b) shows that the set $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$. Thus, $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a group.

Let $p\in \NN$. Let $\lambda\in\mathbb{N}^{p}$. Then, the set
$\mathfrak{B}_{\lambda,\mu,p,k}$ can be made into a left $\mathfrak{B}%
_{\mu,\mu,k,k}$-set (i.e., it can be equipped with an action of the group
$\mathfrak{B}_{\mu,\mu,k,k}$ from the left) in such a way that the group
$\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on $\mathfrak{B}_{\lambda,\mu,p,k}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.o.l1}.]For any
$\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$ and $\beta\in\mathfrak{B}_{\lambda
,\mu,p,k}$, we have $\alpha\circ\beta\in\mathfrak{B}_{\lambda,\mu,p,k}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$
and $\beta\in\mathfrak{B}_{\lambda,\mu,p,k}$. We must show that $\alpha
\circ\beta\in\mathfrak{B}_{\lambda,\mu,p,k}$.
\par
We have $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by the definition of $\mathfrak{B}_{\mu,\mu,k,k}$). In other words,
$\alpha$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\alpha_{\ast}\mu$.
\par
We have $\beta\in\mathfrak{B}_{\lambda,\mu,p,k}=\left\{  \varphi:\left[
p\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}%
\lambda\right\}  $ (by the definition of $\mathfrak{B}_{\lambda,\mu,p,k}$). In
other words, $\beta$ is a map $\left[  p\right]  \rightarrow\left[  k\right]
$ and satisfies $\mu=\beta_{\ast}\lambda$.
\par
Now, Proposition \ref{prop.sol.Lambda.triangular.j.l1} (applied to $k$, $k$,
$\beta$, $\alpha$ and $\lambda$ instead of $q$, $r$, $\varphi$, $\psi$ and
$\alpha$) yields $\left(  \alpha\circ\beta\right)  _{\ast}\lambda=\alpha
_{\ast}\left(  \underbrace{\beta_{\ast}\lambda}_{=\mu}\right)  =\alpha_{\ast
}\mu=\mu$. In other words, $\mu=\left(  \alpha\circ\beta\right)  _{\ast
}\lambda$. Hence, $\alpha\circ\beta$ is a map $\left[  p\right]
\rightarrow\left[  k\right]  $ (since $\alpha$ is a map $\left[  k\right]
\rightarrow\left[  k\right]  $, and since $\beta$ is a map $\left[  p\right]
\rightarrow\left[  k\right]  $), and satisfies $\mu=\left(  \alpha\circ
\beta\right)  _{\ast}\lambda$. In other words, $\alpha\circ\beta\in\left\{
\varphi:\left[  p\right]  \rightarrow\left[  k\right]  \ \mid\ \mu
=\varphi_{\ast}\lambda\right\}  $. This rewrites as $\alpha\circ\beta
\in\mathfrak{B}_{\lambda,\mu,p,k}$ (since $\mathfrak{B}_{\lambda,\mu
,p,k}=\left\{  \varphi:\left[  p\right]  \rightarrow\left[  k\right]
\ \mid\ \mu=\varphi_{\ast}\lambda\right\}  $). Qed.}. Thus, we can try to
define an action of the group $\mathfrak{B}_{\mu,\mu,k,k}$ on the set
$\mathfrak{B}_{\lambda,\mu,p,k}$ from the left by setting%
\begin{equation}
\left(  \alpha\beta=\alpha\circ\beta\ \ \ \ \ \ \ \ \ \ \text{for all }%
\alpha\in\mathfrak{B}_{\mu,\mu,k,k}\text{ and }\beta\in\mathfrak{B}%
_{\lambda,\mu,p,k}\right)  .
\label{pf.prop.sol.Lambda.triangular.o.l1.def-act}%
\end{equation}
In order to show that this definition actually defines an action of the group
$\mathfrak{B}_{\mu,\mu,k,k}$ on the set $\mathfrak{B}_{\lambda,\mu,p,k}$, we
need to prove the following two observations:

\begin{statement}
\textit{Observation 1:} We have $\id_{\left\{
1,2,\ldots,k\right\}  }\circ\gamma=\gamma$ for all $\gamma\in\mathfrak{B}%
_{\lambda,\mu,p,k}$.
\end{statement}

\begin{statement}
\textit{Observation 2:} We have $\left(  \alpha\circ\beta\right)  \circ
\gamma=\alpha\circ\left(  \beta\circ\gamma\right)  $ for all $\alpha
\in\mathfrak{B}_{\mu,\mu,k,k}$, $\beta\in\mathfrak{B}_{\mu,\mu,k,k}$ and
$\gamma\in\mathfrak{B}_{\lambda,\mu,p,k}$.
\end{statement}

However, both Observation 1 and Observation 2 are obvious. Thus, we have shown
that (\ref{pf.prop.sol.Lambda.triangular.o.l1.def-act}) actually defines an
action of the group $\mathfrak{B}_{\mu,\mu,k,k}$ on the set $\mathfrak{B}%
_{\lambda,\mu,p,k}$. Consider this action. Thus, the set $\mathfrak{B}%
_{\lambda,\mu,p,k}$ has been made into a left $\mathfrak{B}_{\mu,\mu,k,k}$-set.

We shall now prove that the group $\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on
$\mathfrak{B}_{\lambda,\mu,p,k}$. In order to do so, we must prove the
following observation:

\begin{statement}
\textit{Observation 3:} If $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$, $\beta
\in\mathfrak{B}_{\mu,\mu,k,k}$ and $\gamma\in\mathfrak{B}_{\lambda,\mu,p,k}$
are such that $\alpha\circ\gamma=\beta\circ\gamma$, then $\alpha=\beta$.
\end{statement}

[\textit{Proof of Observation 3:} Let $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$,
$\beta\in\mathfrak{B}_{\mu,\mu,k,k}$ and $\gamma\in\mathfrak{B}_{\lambda
,\mu,p,k}$ be such that $\alpha\circ\gamma=\beta\circ\gamma$. We must prove
that $\alpha=\beta$.

We have $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by the definition of $\mathfrak{B}_{\mu,\mu,k,k}$). In other words,
$\alpha$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\alpha_{\ast}\mu$. The same argument (applied to $\beta$
instead of $\alpha$) shows that $\beta$ is a map $\left[  k\right]
\rightarrow\left[  k\right]  $ and satisfies $\mu=\beta_{\ast}\mu$.

We have $\gamma\in\mathfrak{B}_{\lambda,\mu,p,k}=\left\{  \varphi:\left[
p\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}%
\lambda\right\}  $ (by the definition of $\mathfrak{B}_{\lambda,\mu,p,k}$). In
other words, $\gamma$ is a map $\left[  p\right]  \rightarrow\left[  k\right]
$ and satisfies $\mu=\gamma_{\ast}\lambda$.

Write the $p$-tuple $\lambda$ in the form $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{p}\right)  $.

Also, $\ell\left(  \mu\right)  =k$. Hence, $\mu=\left(  \mu_{1},\mu_{2}%
,\ldots,\mu_{k}\right)  \in\mathbb{N}^{k}$ and $\mu_{1}\geq\mu_{2}\geq
\cdots\geq\mu_{k}>0$. From $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{k}>0$, we
obtain $\left(  \mu_{i}>0\text{ for each }i\in\left[  k\right]  \right)  $.
Hence, Lemma \ref{lem.sol.Lambda.triangular.j.l10} (applied to $q=k$,
$\varphi=\gamma$, $\alpha=\lambda$, $\alpha_{i}=\lambda_{i}$, $\beta=\mu$ and
$\beta_{i}=\mu_{i}$) shows that the map $\gamma$ is surjective. Hence, we can
cancel $\gamma$ from the equality $\alpha\circ\gamma=\beta\circ\gamma$. Thus,
we obtain $\alpha=\beta$. This proves Observation 3.]

Observation 3 shows that the group $\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on
$\mathfrak{B}_{\lambda,\mu,p,k}$ (by the definition of ``acting freely'').
Thus, the set $\mathfrak{B}_{\lambda
,\mu,p,k}$ can be made into a left $\mathfrak{B}_{\mu,\mu,k,k}$-set in such a
way that the group $\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on $\mathfrak{B}%
_{\lambda,\mu,p,k}$ (namely, by using the above-defined left action of
$\mathfrak{B}_{\mu,\mu,k,k}$ on $\mathfrak{B}_{\lambda,\mu,p,k}$). This proves
Proposition \ref{prop.sol.Lambda.triangular.o.l1}.
\end{proof}

Let us now recall a basic fact from abstract algebra:

\begin{proposition}
\label{prop.sol.Lambda.triangular.o.l2}Let $G$ be a finite group. Let $X$ be a
finite left $G$-set. Assume that $G$ acts freely on $X$. Then, $\left\vert
G\right\vert \mid\left\vert X\right\vert $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.o.l2}.]This fact is
well-known, so let us merely sketch the proof: The $G$-set $X$ is a disjoint
union of orbits (since every $G$-set is a disjoint union of orbits). Thus, the
$G$-set $X$ is a disjoint union of finitely many orbits (since $X$ is finite).
In other words, we have $X=O_{1}\cup O_{2}\cup\cdots\cup O_{k}$ for some list
$\left(  O_{1},O_{2},\ldots,O_{k}\right)  $ of disjoint orbits of $G$ on $X$.
Consider this list $\left(  O_{1},O_{2},\ldots,O_{k}\right)  $.

Every $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies $\left\vert
O_{i}\right\vert =\left\vert G\right\vert $\ \ \ \ \footnote{\textit{Proof.}
Let $i\in\left\{  1,2,\ldots,k\right\}  $. Thus, $O_{i}$ is an orbit of $G$ on
$X$. In other words, $O_{i}=Gy$ for some $y\in X$. Consider this $y$.
\par
The elements $gy$ for $g\in G$ are all distinct (because if $gy=hy$ for two
elements $g,h\in G$, then we must have $g=h$ (because $G$ acts freely on
$X$)). Thus, the number of these elements is precisely $\left\vert
G\right\vert $. In other words, $\left\vert \left\{  gy\ \mid\ g\in G\right\}
\right\vert =\left\vert G\right\vert $. In view of $\left\{  gy\ \mid\ g\in
G\right\}  =Gy=O_{i}$, this rewrites as $\left\vert O_{i}\right\vert
=\left\vert G\right\vert $. Qed.}. Hence, $\sum_{i=1}^{k}%
\underbrace{\left\vert O_{i}\right\vert }_{=\left\vert G\right\vert }%
=\sum_{i=1}^{k}\left\vert G\right\vert =k\left\vert G\right\vert $.

From $X=O_{1}\cup O_{2}\cup\cdots\cup O_{k}$, we obtain
\begin{align*}
\left\vert X\right\vert  &  =\left\vert O_{1}\cup O_{2}\cup\cdots\cup
O_{k}\right\vert =\left\vert O_{1}\right\vert +\left\vert O_{2}\right\vert
+\cdots+\left\vert O_{k}\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since
the orbits }O_{1},O_{2},\ldots,O_{k}\text{ are disjoint}\right) \\
&  =\sum_{i=1}^{k}\left\vert O_{i}\right\vert =k\left\vert G\right\vert .
\end{align*}
Thus, $\left\vert G\right\vert \mid k\left\vert G\right\vert =\left\vert
X\right\vert $. This proves Proposition \ref{prop.sol.Lambda.triangular.o.l2}.
\end{proof}

We now resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(o).
\end{commentedout}

(o) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$. We must prove that $b_{\mu,\mu}\mid
b_{\lambda,\mu}$.

Let $p=\ell\left(  \lambda\right)  $. Thus, $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $ (by the definition of
$\ell\left(  \lambda\right)  $). Hence, $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{p}\right)  \in\mathbb{N}^{p}$.

We have $\mu\in\Par_{n}\subset\Par$.
Let $k=\ell\left(  \mu\right)  $. Proposition
\ref{prop.sol.Lambda.triangular.j.l7}(b) shows that the set $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$. Thus, $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a finite group (since $\Symm_k$ is a finite group).

But $\lambda$ is a partition (since $\lambda\in \Par_{n}\subset\Par$).
Also, $\mu$ is a partition (since
$\mu\in\Par$), thus a weak composition. Furthermore,
$k=\ell\left(  \mu\right)  $; thus, $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{k}\right)  $ (by the definition of $\ell\left(  \mu\right)  $). Hence,
Lemma \ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $q=k$) shows that the
set $\mathfrak{B}_{\lambda,\mu,p,k}$ is well-defined and satisfies
$b_{\lambda,\mu}=\left\vert \mathfrak{B}_{\lambda,\mu,p,k}\right\vert $.

Furthermore, Lemma \ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $k$, $k$
and $\mu$ instead of $p$, $q$ and $\lambda$) shows that the set $\mathfrak{B}%
_{\mu,\mu,k,k}$ is well-defined and satisfies $b_{\mu,\mu}=\left\vert
\mathfrak{B}_{\mu,\mu,k,k}\right\vert $.

The definition of $\mathfrak{B}_{\lambda,\mu,p,k}$ yields $\mathfrak{B}%
_{\lambda,\mu,p,k}=\left\{  \varphi:\left[  p\right]  \rightarrow\left[
k\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}  \subset\left\{
\varphi:\left[  p\right]  \rightarrow\left[  k\right]  \right\}  =\left[
k\right]  ^{\left[  p\right]  }$. Thus, $\mathfrak{B}_{\lambda,\mu,p,k}$ is a
finite set (since $\left[  k\right]  ^{\left[  p\right]  }$ is a finite set).

Proposition \ref{prop.sol.Lambda.triangular.o.l1} shows that the set
$\mathfrak{B}_{\lambda,\mu,p,k}$ can be made into a left $\mathfrak{B}%
_{\mu,\mu,k,k}$-set (i.e., it can be equipped with an action of the group
$\mathfrak{B}_{\mu,\mu,k,k}$ from the left) in such a way that the group
$\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on $\mathfrak{B}_{\lambda,\mu,p,k}$.
Consider this action. Proposition \ref{prop.sol.Lambda.triangular.o.l2}
(applied to $G=\mathfrak{B}_{\mu,\mu,k,k}$ and $X=\mathfrak{B}_{\lambda
,\mu,p,k}$) now yields $\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert
\mid\left\vert \mathfrak{B}_{\lambda,\mu,p,k}\right\vert $. Thus, $b_{\mu,\mu
}=\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert \mid\left\vert
\mathfrak{B}_{\lambda,\mu,p,k}\right\vert =b_{\lambda,\mu}$ (since
$b_{\lambda,\mu}=\left\vert \mathfrak{B}_{\lambda,\mu,p,k}\right\vert $). This
solves Exercise \ref{exe.Lambda.triangular}(o).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.triangular}}

\textit{Solution to Exercise \ref{exe.Lambda.triangular}.} Let us first
introduce some terminology.

\begin{definition}
If $\alpha \in \NN^\infty$ is any sequence, and if $i$ is any positive integer,
then $\alpha_i$ shall denote the $i$-th entry of $\alpha$.
(This generalizes the notation $\lambda_i$ for the $i$-th entry of a
partition $\lambda$.)
Thus, any sequence $\alpha \in \NN^\infty$ satisfies
$\alpha = \left(\alpha_1, \alpha_2, \alpha_3, \ldots\right)$.
\end{definition}

\begin{definition}
Let \dfn{$\operatorname*{WC}$} denote the set of all weak compositions. For
every $f\in\kk\left[  \left[  \xx\right]  \right]  $ and $\mu
\in\operatorname*{WC}$, we let $\left[  \mathbf{x}^{\mu}\right]  f$ denote the
coefficient of the monomial $\mathbf{x}^{\mu}$ in the power series $f$. (This
generalizes the notation $\left[  \mathbf{x}^{\mu}\right]  f$ introduced in
Exercise \ref{exe.Lambda.triangular}(a).)
\end{definition}

We observe the following obvious facts:

\begin{itemize}
\item For any $\alpha\in\operatorname*{WC}$ and $\mu\in\operatorname*{WC}$, we
have%
\begin{equation}
\left[  \mathbf{x}^{\mu}\right]  \left(  \mathbf{x}^{\alpha}\right)
=\delta_{\mu,\alpha}. \label{sol.Lambda.triangular.coeff-mon}%
\end{equation}


\item If a power series $f\in\kk\left[  \left[  \xx\right]
\right]  $ is written in the form $f=\sum_{\alpha\in\operatorname*{WC}%
}c_{\alpha}{\mathbf{x}}^{\alpha}$ for some family $\left(  c_{\alpha}\right)
_{\alpha\in\operatorname*{WC}}\in\kk^{\operatorname*{WC}}$ of scalars,
then
\begin{equation}
\text{every }\alpha\in\operatorname*{WC}\text{ satisfies }\left[
\mathbf{x}^{\alpha}\right]  f=c_{\alpha} \label{sol.Lambda.triangular.coeff-f}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.triangular.coeff-f}):} Let a power
series $f\in\kk\left[  \left[  \xx\right]  \right]  $ be written
in the form $f=\sum_{\alpha\in\operatorname*{WC}}c_{\alpha}{\xx%
}^{\alpha}$ for some family $\left(  c_{\alpha}\right)  _{\alpha
\in\operatorname*{WC}}\in\kk^{\operatorname*{WC}}$ of scalars. Then,
the family $\left(  c_{\alpha}\right)  _{\alpha\in\operatorname*{WC}}$ is a
family of scalars satisfying $f=\sum_{\alpha\in\operatorname*{WC}}c_{\alpha
}{\mathbf{x}}^{\alpha}$. In other words, this family $\left(  c_{\alpha
}\right)  _{\alpha\in\operatorname*{WC}}$ is the family of coefficients of the
power series $f$. More precisely, for every $\alpha\in\operatorname*{WC}$, the
coefficient of $\mathbf{x}^{\alpha}$ in $f$ is $c_{\alpha}$. In other words,
for every $\alpha\in\operatorname*{WC}$, we have $\left[  \mathbf{x}^{\alpha
}\right]  f=c_{\alpha}$. This proves (\ref{sol.Lambda.triangular.coeff-f}).}.
\end{itemize}

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(a).
\end{commentedout}

(a) The definition of $\Lambda$ yields%
\begin{align*}
\Lambda &  =\left\{  f=\sum_{\alpha\in\operatorname*{WC}}c_{\alpha}%
{\mathbf{x}}^{\alpha}\in R\left(  \xx\right)  \ \mid\ c_{\alpha
}=c_{\beta}\text{ if }\alpha,\beta\text{ lie in the same }
\Symm_{\left(  \infty\right)  }\text{-orbit}\right\} \\
&  =\left\{  \sum_{\alpha\in\operatorname*{WC}}c_{\alpha}{\mathbf{x}}^{\alpha
}\in R\left(  \xx\right)  \ \mid\ c_{\alpha}=c_{\beta}\text{ if }%
\alpha,\beta\text{ lie in the same }\Symm_{\left(  \infty\right)
}\text{-orbit}\right\}  .
\end{align*}


Now, let $f\in\Lambda_{n}$. Thus,%
\[
f\in\Lambda_{n}\subset\Lambda=\left\{  \sum_{\alpha\in\operatorname*{WC}%
}c_{\alpha}{\mathbf{x}}^{\alpha}\in R\left(  \xx\right)  \ \mid
\ c_{\alpha}=c_{\beta}\text{ if }\alpha,\beta\text{ lie in the same
}\Symm_{\left(  \infty\right)  }\text{-orbit}\right\}  .
\]
In other words, $f$ can be written in the form $f=\sum_{\alpha\in
\operatorname*{WC}}c_{\alpha}{\mathbf{x}}^{\alpha}$ for some family $\left(
c_{\alpha}\right)  _{\alpha\in\operatorname*{WC}}\in\kk%
^{\operatorname*{WC}}$ of scalars having the property that
\begin{equation}
\left(  c_{\alpha}=c_{\beta}\text{ if }\alpha,\beta\text{ lie in the same
}\Symm_{\left(  \infty\right)  }\text{-orbit}\right)  .
\label{sol.Lambda.triangular.a.c=c}%
\end{equation}
Consider this family $\left(  c_{\alpha}\right)  _{\alpha\in\operatorname*{WC}%
}$.

From (\ref{sol.Lambda.triangular.coeff-f}), we conclude that every $\alpha
\in\operatorname*{WC}$ satisfies%
\begin{equation}
\left[  \mathbf{x}^{\alpha}\right]  f=c_{\alpha}.
\label{sol.Lambda.triangular.a.coeff}%
\end{equation}


Now, fix $\alpha\in\operatorname*{WC}$. Thus, $\alpha$ is a weak composition
(since $\operatorname*{WC}$ is the set of all weak compositions). Hence, there
exists a \textbf{unique} partition $\lambda$ that is a permutation of $\alpha
$\ \ \ \ \footnote{Namely, this partition $\lambda$ is the result of sorting
the entries of $\alpha$ into decreasing order (or, more precisely: moving the
positive entries of $\alpha$ to the left of all zero entries, and then sorting
the former into decreasing order). For example, if $\alpha=\left(
0,2,5,0,3,1,0,0,1,0,0,\ldots\right)  $, then this partition $\lambda$ is
$\left(  5,3,2,1,1,0,0,\ldots\right)  $.}. Thus, the sum $\sum\limits
_{\substack{\lambda\text{ is a partition;}\\\lambda\text{ is a permutation of
}\alpha}}\mathbf{x}^{\alpha}$ has only one addend. Therefore, this sum
simplifies as follows:%
\begin{equation}
\sum\limits_{\substack{\lambda\text{ is a partition;}\\\lambda\text{ is a permutation
of }\alpha}}\mathbf{x}^{\alpha}=\mathbf{x}^{\alpha}.
\label{sol.Lambda.triangular.a.3}%
\end{equation}
But if $\lambda$ is any partition, then we have the following equivalence of
assertions:%
\[
\left(  \lambda\text{ is a permutation of }\alpha\right)
\ \Longleftrightarrow\ \left(  \lambda\in\Symm_{\left(  \infty\right)
}\alpha\right)  \ \Longleftrightarrow\ \left(  \Symm_{\left(
\infty\right)  }\lambda=\Symm_{\left(  \infty\right)  }\alpha\right)
\ \Longleftrightarrow\ \left(  \alpha\in\Symm_{\left(  \infty\right)
}\lambda\right)  .
\]
Hence, we have the following equality of summation signs:%
\[
\sum\limits_{\substack{\lambda\text{ is a partition;}\\\lambda\text{ is a permutation
of }\alpha}}=\sum\limits_{\substack{\lambda\text{ is a partition;}\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}=\sum\limits_{\substack{\lambda
\in\Par;\\\alpha\in\Symm_{\left(  \infty\right)
}\lambda}}
\]
(since the set of all partitions is $\Par$). Thus, the equality
(\ref{sol.Lambda.triangular.a.3}) rewrites as%
\begin{equation}
\sum\limits_{\substack{\lambda\in\Par;\\\alpha\in\Symm_{\left(
\infty\right)  }\lambda}}\mathbf{x}^{\alpha}=\mathbf{x}^{\alpha}.
\label{sol.Lambda.triangular.a.5}%
\end{equation}
But every $\alpha\in\operatorname*{WC}$ satisfying
$\alpha\in\Symm_{\left(  \infty\right)  }\lambda$ satisfies
\begin{equation}
c_{\alpha}=c_{\lambda} \label{sol.Lambda.triangular.a.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.triangular.a.6}):} Let $\alpha
\in\operatorname*{WC}$ be such that $\alpha\in\Symm_{\left(
\infty\right)  }\lambda$. Then, $\alpha$ and $\lambda$ lie in the same
$\Symm_{\left(  \infty\right)  }$-orbit (since $\alpha\in
\Symm_{\left(  \infty\right)  }\lambda$). Hence,
(\ref{sol.Lambda.triangular.a.c=c}) (applied to $\beta=\lambda$) yields
$c_{\alpha}=c_{\lambda}$. This proves (\ref{sol.Lambda.triangular.a.6}).}.

Now, forget that we fixed $\alpha$. We thus have proven that
(\ref{sol.Lambda.triangular.a.5}) holds for each $\alpha\in\operatorname*{WC}$.

Recall that each $\lambda\in\Par$ satisfies%
\begin{equation}
m_{\lambda}=\sum\limits_{\substack{\alpha\in\operatorname*{WC};\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha}
\label{sol.Lambda.triangular.a.mlambda=}%
\end{equation}
(by (\ref{monomial-symmetric-function-definition})).

Now,
\begin{align}
f  &  =\sum_{\alpha\in\operatorname*{WC}}c_{\alpha}\underbrace{{\xx%
}^{\alpha}}_{\substack{=\sum\limits_{\substack{\lambda\in\Par%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
}\\\text{(by (\ref{sol.Lambda.triangular.a.5}))}}}=\sum_{\alpha\in
\operatorname*{WC}}c_{\alpha}\sum\limits_{\substack{\lambda\in\Par%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
}=\underbrace{\sum_{\alpha\in\operatorname*{WC}}\sum\limits_{\substack{\lambda
\in\Par;\\\alpha\in\Symm_{\left(  \infty\right)
}\lambda}}}_{=\sum_{\lambda\in\Par}\sum\limits_{\substack{\alpha
\in\operatorname*{WC};\\\alpha\in\Symm_{\left(  \infty\right)  }%
\lambda}}}\underbrace{c_{\alpha}}_{\substack{=c_{\lambda}\\\text{(by
(\ref{sol.Lambda.triangular.a.6}))}}}\mathbf{x}^{\alpha}\nonumber\\
&  =\sum_{\lambda\in\Par}\sum\limits_{\substack{\alpha\in
\operatorname*{WC};\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda
}}c_{\lambda}\mathbf{x}^{\alpha}=\sum_{\lambda\in\Par}
c_{\lambda}\underbrace{\sum\limits_{\substack{\alpha\in\operatorname*{WC}%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha
}}_{\substack{=m_{\lambda}\\\text{(by (\ref{sol.Lambda.triangular.a.mlambda=}%
))}}}=\sum_{\lambda\in\Par}c_{\lambda}m_{\lambda}.
\label{sol.Lambda.triangular.a.half}%
\end{align}


Now, let $\pi_{n}:\kk\left[  \left[  \xx\right]  \right]
\rightarrow\kk\left[  \left[  \xx\right]  \right]  $ be the
projection that maps each power series $g\in\kk\left[  \left[
\xx\right]  \right]  $ to its $n$-th homogeneous component. Then, the
definition of $\pi_{n}$ shows that the following holds:

\begin{itemize}
\item If $g\in\kk\left[  \left[  \xx\right]  \right]  $ is a
homogeneous power series of degree $n$, then
\begin{equation}
\pi_{n}\left(  g\right)  =g. \label{sol.Lambda.triangular.a.hg-1-1}%
\end{equation}


\item If $g\in\kk\left[  \left[  \xx\right]  \right]  $ is a
homogeneous power series of degree $\neq n$, then
\begin{equation}
\pi_{n}\left(  g\right)  =0. \label{sol.Lambda.triangular.a.hg-1-2}%
\end{equation}

\end{itemize}

We can now use this to compute $\pi_{n}\left(  m_{\lambda}\right)  $ for each
$\lambda\in\Par$.

First, let us notice that
\begin{equation}
m_{\lambda}\text{ is a homogeneous power series of degree }\left\vert
\lambda\right\vert \label{sol.Lambda.triangular.a.mlambda-hg}%
\end{equation}
for each $\lambda\in\Par$\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.Lambda.triangular.a.mlambda-hg}):} Let $\lambda\in
\Par$. Let $\alpha\in\operatorname*{WC}$ be such that
$\alpha\in\Symm_{\left(  \infty\right)  }\lambda$.
\par
From $\alpha\in\Symm_{\left(  \infty\right)  }\lambda$, we conclude
that $\alpha$ is a permutation of $\lambda$. Hence, $\left\vert \alpha
\right\vert =\left\vert \lambda\right\vert $ (because the size of a weak
composition does not change when we permute it). But $\mathbf{x}^{\alpha}$ is
a monomial of degree $\left\vert \alpha\right\vert $. Hence, $\xx%
^{\alpha}$ is a monomial of degree $\left\vert \lambda\right\vert $ (since
$\left\vert \alpha\right\vert =\left\vert \lambda\right\vert $).
\par
Now, forget that we fixed $\alpha$. We thus have shown that $\xx%
^{\alpha}$ is a monomial of degree $\left\vert \lambda\right\vert $ for every
$\alpha\in\operatorname*{WC}$ satisfying $\alpha\in\Symm_{\left(
\infty\right)  }\lambda$. Hence, the sum $\sum\limits_{\substack{\lambda
\in\Par;\\\alpha\in\Symm_{\left(  \infty\right)
}\lambda}}\mathbf{x}^{\alpha}$ is a sum of monomials of degree $\left\vert
\lambda\right\vert $. Therefore, this sum $\sum\limits_{\substack{\lambda
\in\Par;\\\alpha\in\Symm_{\left(  \infty\right)
}\lambda}}\mathbf{x}^{\alpha}$ is a homogeneous power series of degree
$\left\vert \lambda\right\vert $. In other words, $m_{\lambda}$ is a
homogeneous power series of degree $\left\vert \lambda\right\vert $ (since
$m_{\lambda}=\sum\limits_{\substack{\alpha\in\operatorname*{WC};\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}\mathbf{x}^{\alpha}$ (by
(\ref{sol.Lambda.triangular.a.mlambda=}))). This proves
(\ref{sol.Lambda.triangular.a.mlambda-hg}).}.

Now, we conclude the following:

\begin{itemize}
\item If $\lambda\in\Par$ satisfies $\left\vert \lambda
\right\vert =n$, then%
\begin{equation}
\pi_{n}\left(  m_{\lambda}\right)  =m_{\lambda}
\label{sol.Lambda.triangular.a.hg-conc-1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.triangular.a.hg-conc-1}):} Let
$\lambda\in\Par$ be such that $\left\vert \lambda\right\vert
=n$. Then, (\ref{sol.Lambda.triangular.a.mlambda-hg}) shows that $m_{\lambda}$
is a homogeneous power series of degree $\left\vert \lambda\right\vert $. In
other words, $m_{\lambda}$ is a homogeneous power series of degree $n$ (since
$\left\vert \lambda\right\vert =n$). Hence,
(\ref{sol.Lambda.triangular.a.hg-1-1}) (applied to $g=m_{\lambda}$) shows that
$\pi_{n}\left(  m_{\lambda}\right)  =m_{\lambda}$. This proves
(\ref{sol.Lambda.triangular.a.hg-conc-1}).}.

\item If $\lambda\in\Par$ satisfies $\left\vert \lambda
\right\vert \neq n$, then%
\begin{equation}
\pi_{n}\left(  m_{\lambda}\right)  =0
\label{sol.Lambda.triangular.a.hg-conc-2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.triangular.a.hg-conc-2}):} Let
$\lambda\in\Par$ be such that $\left\vert \lambda\right\vert
\neq n$. Then, (\ref{sol.Lambda.triangular.a.mlambda-hg}) shows that
$m_{\lambda}$ is a homogeneous power series of degree $\left\vert
\lambda\right\vert $. Thus, $m_{\lambda}$ is a homogeneous power series of
degree $\neq n$ (since $\left\vert \lambda\right\vert \neq n$). Hence,
(\ref{sol.Lambda.triangular.a.hg-1-2}) (applied to $g=m_{\lambda}$) shows that
$\pi_{n}\left(  m_{\lambda}\right)  =0$. This proves
(\ref{sol.Lambda.triangular.a.hg-conc-2}).}.
\end{itemize}

Finally, let us apply the map $\pi_{n}$ to both sides of the equality
(\ref{sol.Lambda.triangular.a.half}). We thus obtain%
\begin{align*}
\pi_{n}\left(  f\right)   &  =\pi_{n}\left(  \sum_{\lambda\in
\Par}c_{\lambda}m_{\lambda}\right)  =\sum_{\lambda
\in\Par}c_{\lambda}\pi_{n}\left(  m_{\lambda}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\pi_{n}\text{ respects infinite }\kk\text{-linear
combinations}\\
\text{(because }\pi_{n}\text{ is }\kk\text{-linear and continuous)}%
\end{array}
\right) \\
&  =\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}}_{\substack{=\sum_{\lambda\in\Par_{n}}\\
\text{(since }\left\{  \lambda\in\Par%
\ \mid\ \left\vert \lambda\right\vert =n\right\}  =\Par_{n}
\text{)}}}c_{\lambda}\underbrace{\pi_{n}\left(  m_{\lambda
}\right)  }_{\substack{=m_{\lambda}\\\text{(by
(\ref{sol.Lambda.triangular.a.hg-conc-1}))}}}+\sum\limits_{\substack{\lambda
\in\Par;\\\left\vert \lambda\right\vert \neq n}}c_{\lambda
}\underbrace{\pi_{n}\left(  m_{\lambda}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.Lambda.triangular.a.hg-conc-2}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }\lambda\in\Par%
\text{ satisfies either }\left\vert \lambda\right\vert =n\text{ or }\left\vert
\lambda\right\vert \neq n\text{ (but not both)}\right) \\
&  =\sum_{\lambda\in\Par_{n}}c_{\lambda}m_{\lambda
}+\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert \neq n}}c_{\lambda}0}_{=0}=\sum_{\lambda\in
\Par_{n}}c_{\lambda}m_{\lambda}.
\end{align*}


But $f$ is a homogeneous power series of degree $n$ (since $f\in\Lambda_{n}$).
Thus, (\ref{sol.Lambda.triangular.a.hg-1-1}) (applied to $g=f$) yields
$\pi_{n}\left(  f\right)  =f$. Hence,%
\[
f=\pi_{n}\left(  f\right)  =\sum_{\lambda\in\Par_{n}%
}c_{\lambda}m_{\lambda}=\sum_{\mu\in\Par_{n}}c_{\mu
}m_{\mu}%
\]
(here, we have renamed the summation index $\lambda$ as $\mu$). Comparing this
with%
\[
\sum_{\mu\in\Par_{n}}\underbrace{\left(  \left[
\mathbf{x}^{\mu}\right]  f\right)  }_{\substack{=c_{\mu}\\\text{(by
(\ref{sol.Lambda.triangular.a.coeff}) (applied to }\alpha=\mu\text{))}}%
}m_{\mu}=\sum_{\mu\in\Par_{n}}c_{\mu}m_{\mu},
\]
we obtain $f=\sum_{\mu\in\Par_{n}}\left(  \left[
\mathbf{x}^{\mu}\right]  f\right)  m_{\mu}$. This solves Exercise
\ref{exe.Lambda.triangular}(a).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(b).
\end{commentedout}

(b) Let $\lambda$ be a partition. Let $\mu$ be a weak composition. We must
prove that the number $K_{\lambda,\mu}$ is well-defined. In other words, we
must prove that there are only finitely many column-strict tableaux $T$ of
shape $\lambda$ having $\cont \left(  T\right)  =\mu$.

Let $F$ be the Ferrers diagram of $\lambda$ (as a set of cells). Thus, $F$ is
a finite subset of $\left\{  1,2,3,\ldots\right\}  ^{2}$.

Recall that a column-strict tableau of shape $\lambda$ is an assignment of
entries in $\{1,2,3,\ldots\}$ to the cells of the Ferrers diagram of $\lambda$
(satisfying certain conditions). Hence, any column-strict tableau of shape
$\lambda$ is an assignment of entries in $\{1,2,3,\ldots\}$ to the cells of
the Ferrers diagram of $\lambda$. In other words, any column-strict tableau of
shape $\lambda$ is an assignment of entries in $\{1,2,3,\ldots\}$ to the
elements of $F$ (since the elements of $F$ are the cells of the Ferrers
diagram of $\lambda$ (since $F$ is the Ferrers diagram of $\lambda$)). In
other words, any column-strict tableau of shape $\lambda$ is a map
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $.

The sequence $\mu$ is a weak composition, and thus has a finite support. In
other words, the support of $\mu$ is finite. Let $W$ be this support. Thus,%
\begin{align}
W  &  =\left(  \text{the support of }\mu\right) \nonumber\\
&  =\left(  \text{the set of all positive integers }i\text{ for which }\mu
_{i}\neq0\right)  \label{sol.Lambda.triangular.b.W=}%
\end{align}
(since the support of $\mu$ is defined to be the set of all positive integers
$i$ for which $\mu_{i}\neq0$). Hence, $W$ is a set of positive integers. In
other words, $W\subset\left\{  1,2,3,\ldots\right\}  $.

The support of $\mu$ is finite. In other words, the set $W$ is finite (since
the set $W$ is the support of $\mu$).

It is well-known that if $X$, $Y$ and $Z$ are three sets such that $X\subset
Y$, then%
\[
X^{Z}\cong\left\{  f\in Y^{Z}\ \mid\ f\left(  Z\right)  \subset X\right\}
\ \ \ \ \ \ \ \ \ \ \text{as sets.}%
\]
Applying this to $X=W$, $Y=\left\{  1,2,3,\ldots\right\}  $ and $Z=F$, we
conclude that%
\begin{equation}
W^{F}\cong\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  \ \ \ \ \ \ \ \ \ \ \text{as sets.}
\label{sol.Lambda.triangular.b.as-sets}%
\end{equation}
But $F$ and $W$ are finite sets. Hence, $W^{F}$ is also a finite set. Thus,
$\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(  F\right)
\subset W\right\}  $ is a finite set (because of
(\ref{sol.Lambda.triangular.b.as-sets})).

Now, let $T$ be a column-strict tableau of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$. We shall prove that
$T\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  $.

Indeed, $T$ is a map $F\rightarrow\left\{  1,2,3,\ldots\right\}  $ (since any
column-strict tableau of shape $\lambda$ is a map $F\rightarrow\left\{
1,2,3,\ldots\right\}  $), hence an element of $\left\{  1,2,3,\ldots\right\}
^{F}$.

Furthermore, $T\left(  F\right)  \subset W$\ \ \ \ \footnote{\textit{Proof.}
Let $j\in T\left(  F\right)  $. Then, $j\in\left\{  1,2,3,\ldots\right\}  $.
Furthermore, there exists some $c\in F$ satisfying $j=T\left(  c\right)  $
(since $j\in T\left(  F\right)  $). Consider this $c$. From $j=T\left(
c\right)  $, we obtain $c\in T^{-1}\left(  j\right)  $. Hence, the set
$T^{-1}\left(  j\right)  $ contains at least one element (namely, $c$).
Therefore, $\left\vert T^{-1}\left(  j\right)  \right\vert \geq1$.
\par
But $\mu=\cont \left(  T\right)  =\left(  \left\vert
T^{-1}\left(  1\right)  \right\vert ,\left\vert T^{-1}\left(  2\right)
\right\vert ,\left\vert T^{-1}\left(  3\right)  \right\vert ,\ldots\right)  $
(by the definition of $\cont \left(  T\right)  $). Hence,
$\mu_{i}=\left\vert T^{-1}\left(  i\right)  \right\vert $ for every positive
integer $i$. Applying this to $i=j$, we obtain $\mu_{j}=\left\vert
T^{-1}\left(  j\right)  \right\vert \geq1>0$. Hence, $\mu_{j}\neq0$. Thus, $j$
belongs to the set of all positive integers $i$ for which $\mu_{i}\neq0$. In
other words, $j\in\left(  \text{the set of all positive integers }i\text{ for
which }\mu_{i}\neq0\right)  $. In light of (\ref{sol.Lambda.triangular.b.W=}),
this rewrites as $j\in W$.
\par
Now, forget that we fixed $j$. We thus have shown that $j\in W$ for each $j\in
T\left(  F\right)  $. In other words, $T\left(  F\right)  \subset W$.}. Hence,
$T$ is an $f\in\left\{  1,2,3,\ldots\right\}  ^{F}$ satisfying $f\left(
F\right)  \subset W$ (since $T$ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{F}$ and satisfies $T\left(  F\right)  \subset W$). In other words,%
\[
T\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(  F\right)
\subset W\right\}  .
\]


Now, forget that we fixed $T$. We thus have shown that every column-strict
tableau $T$ of shape $\lambda$ having $\cont \left(  T\right)
=\mu$ satisfies $T\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}%
\ \mid\ f\left(  F\right)  \subset W\right\}  $. In other words,%
\begin{align*}
&  \left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{
having }\cont \left(  T\right)  =\mu\right\}
 \subset\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  .
\end{align*}
Hence, $\left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{
having }\cont \left(  T\right)  =\mu\right\}  $ is a finite set
(since \newline
$\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{F}\ \mid\ f\left(
F\right)  \subset W\right\}  $ is a finite set). In other words, there are
only finitely many column-strict tableaux $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$. Thus, the number $K_{\lambda,\mu
}$ is well-defined (since this number is defined as the number of all
column-strict tableaux $T$ of shape $\lambda$ having $\cont
\left(  T\right)  =\mu$). This solves Exercise \ref{exe.Lambda.triangular}(b).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(c).
\end{commentedout}

(c) Let $\lambda\in\Par_{n}$. The definition of
$s_{\lambda}$ yields%
\[
s_{\lambda}=\sum_{T}{\mathbf{x}}^{\cont \left(  T\right)  },
\]
where $T$ runs through all column-strict tableaux of shape $\lambda$. In other
words,%
\[
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda}}{\mathbf{x}}^{\cont \left(  T\right)  }.
\]


Now, every $\mu\in\Par_{n}$ satisfies%
\begin{align}
&  K_{\lambda,\mu}\nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\cont \left(  T\right)  =\mu\right)
\label{sol.Lambda.triangular.c.3}%
\end{align}
(since $K_{\lambda,\mu}$ is the number of all column-strict tableaux $T$ of
shape $\lambda$ having $\cont \left(  T\right)  =\mu$). But every
$\mu\in\Par_{n}$ satisfies
\begin{align}
&  \left[  \mathbf{x}^{\mu}\right]  \left(  \underbrace{s_{\lambda}}%
_{=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}{\mathbf{x}}^{\cont \left(  T\right)  }}\right)
\nonumber\\
&  =\left[  \mathbf{x}^{\mu}\right]  \left(  \sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda}}{\mathbf{x}}%
^{\cont \left(  T\right)  }\right)  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda}}\underbrace{\left[
\mathbf{x}^{\mu}\right]  \left(  {\mathbf{x}}^{\cont \left(
T\right)  }\right)  }_{\substack{=\delta_{\mu,\cont \left(
T\right)  }\\\text{(by (\ref{sol.Lambda.triangular.coeff-mon}) (applied to
}\alpha=\cont \left(  T\right)  \text{))}}}\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}\delta_{\mu,\cont \left(  T\right)  }=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
;\\\mu=\cont \left(  T\right)  }}\underbrace{\delta
_{\mu,\cont \left(  T\right)  }}_{\substack{=1\\\text{(since
}\mu=\cont \left(  T\right)  \text{)}}}+\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda;\\\mu\neq
\cont \left(  T\right)  }}\underbrace{\delta_{\mu
,\cont \left(  T\right)  }}_{\substack{=0\\\text{(since }\mu
\neq\cont \left(  T\right)  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every column-strict tableau }T\text{ of shape }\lambda\text{
satisfies}\\
\text{either }\mu=\cont \left(  T\right)  \text{ or }\mu
\neq\cont \left(  T\right)  \text{ (but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda;\\\mu=\cont \left(  T\right)  }}1+\underbrace{\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
;\\\mu\neq\cont \left(  T\right)  }}0}_{=0}=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
;\\\mu=\cont \left(  T\right)  }}1\nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\mu=\cont \left(  T\right)  \right)
\cdot1\nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\underbrace{\mu=\cont \left(  T\right)
}_{\Longleftrightarrow\ \left(  \cont \left(  T\right)
=\mu\right)  }\right) \nonumber\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda\text{ having }\cont \left(  T\right)  =\mu\right)
\nonumber\\
&  =K_{\lambda,\mu}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.c.3})}\right)  . \label{sol.Lambda.triangular.c.5}%
\end{align}


But $\lambda\in\Par_{n}$. Hence, $\left\vert
\lambda\right\vert =n$. Recall that the power series $s_{\lambda}$ is a
symmetric function (by Proposition \ref{Schur-functions-are-symmetric-prop}),
and is homogeneous of degree $\left\vert \lambda\right\vert $. Thus,
$s_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }=\Lambda_{n}$ (since
$\left\vert \lambda\right\vert =n$). Hence, Exercise
\ref{exe.Lambda.triangular}(a) (applied to $f=s_{\lambda}$) yields%
\[
s_{\lambda}=\sum_{\mu\in\Par_{n}}\underbrace{\left(
\left[  \mathbf{x}^{\mu}\right]  \left(  s_{\lambda}\right)  \right)
}_{\substack{=K_{\lambda,\mu}\\\text{(by (\ref{sol.Lambda.triangular.c.5}))}%
}}m_{\mu}=\sum_{\mu\in\Par_{n}}K_{\lambda,\mu}m_{\mu
}.
\]
This solves Exercise \ref{exe.Lambda.triangular}(c).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(d).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(d), let us show a lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.kostka}Let $n\in \NN$. Let $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$.
Let $T$ be a column-strict tableau of shape $\lambda$
satisfying $\cont \left(  T\right)  =\mu$. Let $F$ be the
Ferrers diagram of $\lambda$ (as a set of cells). For each positive integer
$i$, we let $F_{i}$ be the $i$-th row of $F$ (that is, the set of all cells of
$F$ that have the form $\left(  i,j\right)  $ for some $j\geq1$).

\begin{enumerate}
\item[(a)] We have $T\left(  i,j\right)  \geq i$ for each $\left(  i,j\right)
\in F$.

\item[(b)] We have $T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)
\cup\cdots\cup T^{-1}\left(  k\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup
F_{k}$ for each $k\in \NN$.

\item[(c)] We have $\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{k}\right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}$ for each $k\in \NN$.

\item[(d)] We have $\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(
2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \right\vert =\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$ for each $k\in \NN$.

\item[(e)] We have $\lambda\triangleright\mu$.

\item[(f)] If $\mu=\lambda$, then each $\left(  i,j\right)  \in F$ satisfies
$T\left(  i,j\right)  =i$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}.]Recall that the
Ferrers diagram of $\lambda$ is the set of all pairs $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$ (by the
definition of the Ferrers diagram). In other words, $F$ is the set of all
pairs $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying
$j\leq\lambda_{i}$ (since $F$ is the Ferrers diagram of $\lambda$). In other
words,%
\[
F=\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}%
\ \mid\ j\leq\lambda_{i}\right\}  .
\]
Hence, every $\left(  p,q\right)  \in F$ satisfies%
\begin{equation}
\left(  p^{\prime},q\right)  \in F\ \ \ \ \ \ \ \ \ \ \text{for each
}p^{\prime}\in\left\{  1,2,\ldots,p\right\}
\label{pf.lem.sol.Lambda.triangular.kostka.go-up}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.kostka.go-up}):}
Let $\left(  p,q\right)  \in F$. Thus, $\left(  p,q\right)  \in F=\left\{
\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq
\lambda_{i}\right\}  $. In other words, $\left(  p,q\right)  $ is an $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq
\lambda_{i}$. In other words, $\left(  p,q\right)  $ is an element of
$\left\{  1,2,3,\ldots\right\}  ^{2}$ and satisfies $q\leq\lambda_{p}$.
\par
Now, let $p^{\prime}\in\left\{  1,2,\ldots,p\right\}  $. We must prove that
$\left(  p^{\prime},q\right)  \in F$.
\par
Recall that $\left(  p,q\right)  $ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{2}$. Thus, $p\in\left\{  1,2,3,\ldots\right\}  $ and $q\in\left\{
1,2,3,\ldots\right\}  $. Combining $p^{\prime}\in\left\{  1,2,\ldots
,p\right\}  \subset\left\{  1,2,3,\ldots\right\}  $ with $q\in\left\{
1,2,3,\ldots\right\}  $, we obtain $\left(  p^{\prime},q\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$. In other words, $\left(  p^{\prime},q\right)  $
is an element of $\left\{  1,2,3,\ldots\right\}  ^{2}$. Also, $p^{\prime}\leq
p$ (since $p^{\prime}\in\left\{  1,2,\ldots,p\right\}  $).
\par
But $\lambda$ is a partition. Hence, $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq\cdots$. In other words, $\lambda_{u}\geq\lambda_{v}$ for any
two elements $u$ and $v$ of $\left\{  1,2,3,\ldots\right\}  $ satisfying
$u\leq v$. We can apply this to $u=p^{\prime}$ and $v=p$ (since $p^{\prime
}\leq p$); thus we obtain $\lambda_{p^{\prime}}\geq\lambda_{p}$. Hence,
$q\leq\lambda_{p}\leq\lambda_{p^{\prime}}$ (since $\lambda_{p^{\prime}}%
\geq\lambda_{p}$).
\par
Now, $\left(  p^{\prime},q\right)  $ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{2}$ and satisfies $q\leq\lambda_{p^{\prime}}$. In other words,
$\left(  p^{\prime},q\right)  $ is an $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$. In other words,
$\left(  p^{\prime},q\right)  \in\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $. In other words,
$\left(  p^{\prime},q\right)  \in F$ (since $F=\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $).
This completes the proof of (\ref{pf.lem.sol.Lambda.triangular.kostka.go-up}%
).}.

Recall that a column-strict tableau of shape $\lambda$ is an assignment of
entries in $\{1,2,3,\ldots\}$ to the cells of the Ferrers diagram of $\lambda$
(satisfying certain conditions). Hence, any column-strict tableau of shape
$\lambda$ is an assignment of entries in $\{1,2,3,\ldots\}$ to the cells of
the Ferrers diagram of $\lambda$. In other words, any column-strict tableau of
shape $\lambda$ is an assignment of entries in $\{1,2,3,\ldots\}$ to the
elements of $F$ (since the elements of $F$ are the cells of the Ferrers
diagram of $\lambda$ (since $F$ is the Ferrers diagram of $\lambda$)). In
other words, any column-strict tableau of shape $\lambda$ is a map
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $. Hence, $T$ is a map
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $ (since $T$ is a column-strict
tableau of shape $\lambda$).

Recall that $T$ is a column-strict tableau. Thus, the entries of $T$ are
strictly increasing top-to-bottom down columns (by the definition of a
column-strict tableau).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(a).
\end{commentedout}

(a) Let $\left(  i,j\right)  \in F$. We must prove that $T\left(  i,j\right)
\geq i$.

From $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$, we obtain
$i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in\left\{  1,2,3,\ldots\right\}
$. From $i\in\left\{  1,2,3,\ldots\right\}  $, we conclude that both $1$ and
$i$ are elements of $\left\{  1,2,3,\ldots\right\}  $.

We have $\left(  p^{\prime},j\right)  \in F$ for each $p^{\prime}\in\left\{
1,2,\ldots,i\right\}  $ (by (\ref{pf.lem.sol.Lambda.triangular.kostka.go-up})
(applied to $\left(  p,q\right)  =\left(  i,j\right)  $)). In other words, all
of the cells $\left(  1,j\right)  ,\left(  2,j\right)  ,\ldots,\left(
i,j\right)  $ belong to $F$. These cells therefore all lie in the $j$-th
column of $F$; more precisely, they are the first $i$ cells of the $j$-th
column of $F$. Hence, we have%
\[
T\left(  1,j\right)  <T\left(  2,j\right)  <\cdots<T\left(  i,j\right)
\]
(since the entries of $T$ are strictly increasing top-to-bottom down columns).
In other words, every $u\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies%
\begin{equation}
T\left(  u,j\right)  <T\left(  u+1,j\right)  .
\label{pf.lem.sol.Lambda.triangular.kostka.a.1}%
\end{equation}


Thus, each $u\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies $T\left(
u,j\right)  -u\leq T\left(  u+1,j\right)  -\left(  u+1\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $u\in\left\{  1,2,\ldots,i-1\right\}  $.
From (\ref{pf.lem.sol.Lambda.triangular.kostka.a.1}), we obtain $T\left(
u,j\right)  <T\left(  u+1,j\right)  $. But both $T\left(  u,j\right)  $ and
$T\left(  u,j+1\right)  $ are elements of $\left\{  1,2,3,\ldots\right\}  $
and therefore integers. Thus, from $T\left(  u,j\right)  <T\left(
u+1,j\right)  $, we obtain $T\left(  u,j\right)  \leq T\left(  u+1,j\right)
-1$. Subtracting $u$ from both sides of this inequality, we obtain $T\left(
u,j\right)  -u\leq T\left(  u+1,j\right)  -1-u=T\left(  u+1,j\right)  -\left(
u+1\right)  $. Qed.}. In other words, we have%
\[
T\left(  1,j\right)  -1\leq T\left(  2,j\right)  -2\leq\cdots\leq T\left(
i,j\right)  -i.
\]
Thus, $T\left(  1,j\right)  -1\leq T\left(  i,j\right)  -i$. (Here, we are
using the fact that both $1$ and $i$ are elements of $\left\{  1,2,\ldots
,i\right\}  $.)

But $T\left(  1,j\right)  \in\left\{  1,2,3,\ldots\right\}  $, so that
$T\left(  1,j\right)  \geq1$. Hence, $T\left(  1,j\right)  -1\geq0$. Thus,
$0\leq T\left(  1,j\right)  -1\leq T\left(  i,j\right)  -i$. Therefore, $i\leq
T\left(  i,j\right)  $, so that $T\left(  i,j\right)  \geq i$. This proves
Lemma \ref{lem.sol.Lambda.triangular.kostka}(a).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(b).
\end{commentedout}

(b) Let $k\in \NN$. Each $p\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
T^{-1}\left(  p\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}.
\label{pf.lem.sol.Lambda.triangular.kostka.b.1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.kostka.b.1}):} Let
$p\in\left\{  1,2,\ldots,k\right\}  $. Let $c\in T^{-1}\left(  p\right)  $.
Then, $c\in F$ (since $T$ is a map $F\rightarrow\left\{  1,2,3,\ldots\right\}
$). Also, $T\left(  c\right)  =p$ (since $c\in T^{-1}\left(  p\right)  $).

Write the cell $c$ in the form $c=\left(  i,j\right)  $ for some $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. Then, $\left(
i,j\right)  =c\in F$ and $T\left(  \underbrace{c}_{=\left(  i,j\right)
}\right)  =T\left(  i,j\right)  $, so that $T\left(  i,j\right)  =T\left(
c\right)  =p$. Hence, $p=T\left(  i,j\right)  \geq i$ (by Lemma
\ref{lem.sol.Lambda.triangular.kostka}(a)). Hence, $i\leq p\leq k$ (since
$p\in\left\{  1,2,\ldots,k\right\}  $), so that $i\in\left\{  1,2,\ldots
,k\right\}  $ (since $i\in\left\{  1,2,3,\ldots\right\}  $ (since $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$)). Therefore, $F_{i}$ is a
term of the union $F_{1}\cup F_{2}\cup\cdots\cup F_{k}$. Thus, $F_{i}\subset
F_{1}\cup F_{2}\cup\cdots\cup F_{k}$.

But the cell $\left(  i,j\right)  $ belongs to the $i$-th row (since its first
coordinate is $i$) and also belongs to $F$ (since $\left(  i,j\right)  \in
F$). Hence, the cell $\left(  i,j\right)  $ belongs to the $i$-th row of $F$.
In other words, the cell $\left(  i,j\right)  $ belongs to $F_{i}$ (since
$F_{i}$ is the $i$-th row of $F$ (by the definition of $F_{i}$)). In other
words, $\left(  i,j\right)  \in F_{i}$. Hence, $c=\left(  i,j\right)  \in
F_{i}\subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}$.

Now, forget that we fixed $c$. We thus have proven that $c\in F_{1}\cup
F_{2}\cup\cdots\cup F_{k}$ for each $c\in T^{-1}\left(  p\right)  $. In other
words, $T^{-1}\left(  p\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}$.
This proves (\ref{pf.lem.sol.Lambda.triangular.kostka.b.1}).]

Now,
\begin{align*}
T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)   &  =\bigcup_{p\in\left\{  1,2,\ldots,k\right\}
}\underbrace{T^{-1}\left(  p\right)  }_{\substack{\subset F_{1}\cup F_{2}%
\cup\cdots\cup F_{k}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.kostka.b.1}%
))}}}\\
&  \subset\bigcup_{p\in\left\{  1,2,\ldots,k\right\}  }\left(  F_{1}\cup
F_{2}\cup\cdots\cup F_{k}\right)  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{k}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(b).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(c).
\end{commentedout}

(c) Every positive integer $i$ satisfies%
\begin{align}
\left\vert F_{i}\right\vert  &  =\left(  \text{the size of }F_{i}\right)
=\left(  \text{the size of the }i\text{-th row of }F\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F_{i}\text{ is the }i\text{-th row
of }F\text{ (by the definition of }F_{i}\text{)}\right) \nonumber\\
&  =\left(  \text{the size of the }i\text{-th row of the Ferrers diagram of
}\lambda\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is the Ferrers diagram of
}\lambda\right) \nonumber\\
&  =\lambda_{i}. \label{pf.lem.sol.Lambda.triangular.kostka.c.2}%
\end{align}


Let $k\in \NN$. If $u$ and $v$ are two distinct elements of $\left\{
1,2,\ldots,k\right\}  $, then the sets $F_{u}$ and $F_{v}$ are
disjoint\footnote{\textit{Proof.} Let $u$ and $v$ be two distinct elements of
$\left\{  1,2,\ldots,k\right\}  $. We must prove that the sets $F_{u}$ and
$F_{v}$ are disjoint.
\par
Indeed, assume the contrary. Thus, the sets $F_{u}$ and $F_{v}$ are not
disjoint. In other words, $F_{u}\cap F_{v}\neq\varnothing$. Hence, there
exists some $c\in F_{u}\cap F_{v}$. Consider this $c$.
\par
We have $c\in F_{u}\cap F_{v}\subset F_{u}\subset\left\{  1,2,3,\ldots
\right\}  ^{2}$. Thus, we can write $c$ in the form $c=\left(  i,j\right)  $
for some $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in\left\{
1,2,3,\ldots\right\}  $. Consider these $i$ and $j$.
\par
We have $\left(  i,j\right)  =c\in F_{u}\cap F_{v}\subset F_{u}$. In other
words, $\left(  i,j\right)  $ belongs to $F_{u}$. In other words, $\left(
i,j\right)  $ belongs to the $u$-th row of $F$ (since $F_{u}$ is the $u$-th
row of $F$ (by the definition of $F_{i}$)). Hence, $\left(  i,j\right)  $
belongs to the $u$-th row. In other words, the first coordinate of $\left(
i,j\right)  $ is $u$. In other words, $i$ is $u$ (since the first coordinate
of $\left(  i,j\right)  $ is clearly $i$). In other words, $i=u$.
\par
The same argument (applied to $v$ instead of $u$) shows that $i=v$. Thus,
$u=i=v$. This contradicts the fact that $u$ and $v$ are distinct. This
contradiction shows that our assumption was false. Thus, we have proven that
the sets $F_{u}$ and $F_{v}$ are disjoint. Qed.}. In other words, the sets
$F_{1},F_{2},\ldots,F_{k}$ are disjoint. Hence, the size of their union equals
the sum of their sizes. In other words, $\left\vert F_{1}\cup F_{2}\cup
\cdots\cup F_{k}\right\vert =\left\vert F_{1}\right\vert +\left\vert
F_{2}\right\vert +\cdots+\left\vert F_{k}\right\vert $. Thus,%
\begin{align*}
\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{k}\right\vert  &  =\left\vert
F_{1}\right\vert +\left\vert F_{2}\right\vert +\cdots+\left\vert
F_{k}\right\vert =\sum_{i=1}^{k}\underbrace{\left\vert F_{i}\right\vert
}_{\substack{=\lambda_{i}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.kostka.c.2}))}}}=\sum_{i=1}^{k}\lambda
_{i}\\
&  =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(c).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(d).
\end{commentedout}

(d) We have $\mu=\cont \left(  T\right)  =\left(  \left\vert
T^{-1}\left(  1\right)  \right\vert ,\left\vert T^{-1}\left(  2\right)
\right\vert ,\left\vert T^{-1}\left(  3\right)  \right\vert ,\ldots\right)  $
(by the definition of $\cont \left(  T\right)  $). Hence,
\begin{equation}
\mu_{i}=\left\vert T^{-1}\left(  i\right)  \right\vert
\label{pf.lem.sol.Lambda.triangular.kostka.d.1}%
\end{equation}
for every positive integer $i$.

Let $k\in \NN$. If $u$ and $v$ are two distinct elements of $\left\{
1,2,\ldots,k\right\}  $, then the sets $T^{-1}\left(  u\right)  $ and
$T^{-1}\left(  v\right)  $ are disjoint\footnote{\textit{Proof.} Let $u$ and
$v$ be two distinct elements of $\left\{  1,2,\ldots,k\right\}  $. We must
prove that the sets $T^{-1}\left(  u\right)  $ and $T^{-1}\left(  v\right)  $
are disjoint.
\par
Indeed, assume the contrary. Thus, the sets $T^{-1}\left(  u\right)  $ and
$T^{-1}\left(  v\right)  $ are not disjoint. In other words, $T^{-1}\left(
u\right)  $ and $T^{-1}\left(  v\right)  \neq\varnothing$. Hence, there exists
some $c\in T^{-1}\left(  u\right)  \cap T^{-1}\left(  v\right)  $. Consider
this $c$.
\par
We have $c\in T^{-1}\left(  u\right)  \cap T^{-1}\left(  v\right)  \subset
T^{-1}\left(  u\right)  $, so that $T\left(  c\right)  =u$. But we also have
$c\in T^{-1}\left(  u\right)  \cap T^{-1}\left(  v\right)  \subset
T^{-1}\left(  v\right)  $, so that $T\left(  c\right)  =v$. Thus, $u=T\left(
c\right)  =v$. This contradicts the fact that $u$ and $v$ are distinct. This
contradiction shows that our assumption was false. Thus, we have proven that
the sets $T^{-1}\left(  u\right)  $ and $T^{-1}\left(  v\right)  $ are
disjoint. Qed.}. In other words, the sets $T^{-1}\left(  1\right)
,T^{-1}\left(  2\right)  ,\ldots,T^{-1}\left(  k\right)  $ are disjoint.
Hence, the size of their union equals the sum of their sizes. In other words,
$\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup
\cdots\cup T^{-1}\left(  k\right)  \right\vert =\left\vert T^{-1}\left(
1\right)  \right\vert +\left\vert T^{-1}\left(  2\right)  \right\vert
+\cdots+\left\vert T^{-1}\left(  k\right)  \right\vert $. Thus,%
\begin{align*}
\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)  \right\vert  &  =\left\vert T^{-1}\left(  1\right)
\right\vert +\left\vert T^{-1}\left(  2\right)  \right\vert +\cdots+\left\vert
T^{-1}\left(  k\right)  \right\vert =\sum_{i=1}^{k}\underbrace{\left\vert
T^{-1}\left(  i\right)  \right\vert }_{\substack{=\mu_{i}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.kostka.d.1}))}}}=\sum_{i=1}^{k}\mu_{i}\\
&  =\mu_{1}+\mu_{2}+\cdots+\mu_{k}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(d).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(e).
\end{commentedout}

(e) Let $k\in\left\{  1,2,\ldots,n\right\}  $. Then, Lemma
\ref{lem.sol.Lambda.triangular.kostka}(b) yields $T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \subset
F_{1}\cup F_{2}\cup\cdots\cup F_{k}$. Thus,%
\[
\left\vert T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)  \right\vert \leq\left\vert F_{1}\cup F_{2}\cup
\cdots\cup F_{k}\right\vert .
\]
But Lemma \ref{lem.sol.Lambda.triangular.kostka}(d) yields $\left\vert
T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  k\right)  \right\vert =\mu_{1}+\mu_{2}+\cdots+\mu_{k}$. Hence,%
\begin{align*}
\mu_{1}+\mu_{2}+\cdots+\mu_{k}  &  =\left\vert T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  k\right)  \right\vert \\
&  \leq\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{k}\right\vert =\lambda
_{1}+\lambda_{2}+\cdots+\lambda_{k}%
\end{align*}
(by Lemma \ref{lem.sol.Lambda.triangular.kostka}(c)). In other words,
$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}$.

Now, forget that we fixed $k$. We thus have shown that%
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,n\right\}  .
\label{pf.lem.sol.Lambda.triangular.kostka.e.1}%
\end{equation}


But recall that we have $\lambda\triangleright\mu$ if and only if we have
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,n\right\}
\]
(by the definition of the dominance order). Hence, we have $\lambda
\triangleright\mu$ (since we have
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,n\right\}
\]
). This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(e).

\begin{commentedout}
Here begins the proof of Lemma \ref{lem.sol.Lambda.triangular.kostka}(f).
\end{commentedout}

(f) Assume that $\mu=\lambda$. We must prove that each $\left(  i,j\right)
\in F$ satisfies $T\left(  i,j\right)  =i$.

Indeed, let $\left(  i,j\right)  \in F$ be arbitrary. We must prove that
$T\left(  i,j\right)  =i$.

Indeed, from $\left(  i,j\right)  \in F\subset\left\{  1,2,3,\ldots\right\}
^{2}$, we obtain $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in\left\{
1,2,3,\ldots\right\}  $. Thus, $i\in\left\{  1,2,3,\ldots\right\}
\subset\mathbb{N}$.

Define two sets $A$ and $B$ by $A=T^{-1}\left(  1\right)  \cup T^{-1}\left(
2\right)  \cup\cdots\cup T^{-1}\left(  i\right)  $ and $B=F_{1}\cup F_{2}%
\cup\cdots\cup F_{i}$. Then,%
\begin{align*}
A  &  =T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  i\right) \\
&  \subset F_{1}\cup F_{2}\cup\cdots\cup F_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma \ref{lem.sol.Lambda.triangular.kostka}(b) (applied to
}k=i\text{)}\right) \\
&  =B.
\end{align*}
In other words, $A$ is a subset of $B$.

Applying Lemma \ref{lem.sol.Lambda.triangular.kostka}(c) to $k=i$, we find%
\begin{equation}
\left\vert F_{1}\cup F_{2}\cup\cdots\cup F_{i}\right\vert =\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{i}.
\label{pf.lem.sol.Lambda.triangular.kostka.f.1}%
\end{equation}


Moreover, from $A=T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)
\cup\cdots\cup T^{-1}\left(  i\right)  $, we obtain%
\begin{align*}
\left\vert A\right\vert  &  =\left\vert T^{-1}\left(  1\right)  \cup
T^{-1}\left(  2\right)  \cup\cdots\cup T^{-1}\left(  i\right)  \right\vert
=\mu_{1}+\mu_{2}+\cdots+\mu_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.Lambda.triangular.kostka}(d) (applied to }k=i\text{)}\right) \\
&  =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mu=\lambda\right) \\
&  =\left\vert \underbrace{F_{1}\cup F_{2}\cup\cdots\cup F_{i}}_{=B}%
\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.kostka.f.1})}\right) \\
&  =\left\vert B\right\vert .
\end{align*}
Moreover, $B$ is a finite set (since $\left\vert B\right\vert =\lambda
_{1}+\lambda_{2}+\cdots+\lambda_{i}\in \NN$).

It is well-known that if $Y$ is a finite set, and if $X$ is a subset of $Y$
satisfying $\left\vert X\right\vert =\left\vert Y\right\vert $, then $X=Y$.
Applying this to $X=A$ and $Y=B$, we obtain $A=B$ (since $A$ is a subset of
$B$ and satisfies $\left\vert A\right\vert =\left\vert B\right\vert $).

Now, $\left(  i,j\right)  \in F_{i}$\ \ \ \ \footnote{\textit{Proof.} The cell
$\left(  i,j\right)  $ belongs to the $i$-th row (since its first coordinate
is $i$). Hence, this cell $\left(  i,j\right)  $ belongs to the $i$-th row of
$F$ (since $\left(  i,j\right)  \in F$). In other words, this cell $\left(
i,j\right)  $ belongs to $F_{i}$ (since $F_{i}$ is the $i$-th row of $F$). In
other words, $\left(  i,j\right)  \in F_{i}$.}. But $i\in\left\{
1,2,3,\ldots\right\}  $; hence, $i\in\left\{  1,2,\ldots,i\right\}  $. Thus,
$F_{i}$ is a term of the union $F_{1}\cup F_{2}\cup\cdots\cup F_{i}$. Thus,
$F_{i}\subset F_{1}\cup F_{2}\cup\cdots\cup F_{i}$. Therefore,%
\begin{align*}
\left(  i,j\right)   &  \in F_{i}\subset F_{1}\cup F_{2}\cup\cdots\cup
F_{i}=B=A\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A=B\right) \\
&  =T^{-1}\left(  1\right)  \cup T^{-1}\left(  2\right)  \cup\cdots\cup
T^{-1}\left(  i\right)  .
\end{align*}
In other words, $\left(  i,j\right)  \in T^{-1}\left(  p\right)  $ for some
$p\in\left\{  1,2,\ldots,i\right\}  $. Consider this $p$.

From $\left(  i,j\right)  \in T^{-1}\left(  p\right)  $, we obtain $T\left(
i,j\right)  =p$. Thus, $T\left(  i,j\right)  =p\leq i$ (since $p\in\left\{
1,2,\ldots,i\right\}  $). But Lemma \ref{lem.sol.Lambda.triangular.kostka}(a)
yields $T\left(  i,j\right)  \geq i$. Combining this with $T\left(
i,j\right)  \leq i$, we obtain $T\left(  i,j\right)  =i$.

Now, forget that we fixed $\left(  i,j\right)  $. We thus have proven that
each $\left(  i,j\right)  \in F$ satisfies $T\left(  i,j\right)  =i$.

This proves Lemma \ref{lem.sol.Lambda.triangular.kostka}(f).
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(d).
\end{commentedout}

(d) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$ be two partitions that don't satisfy
$\lambda\triangleright\mu$. We must prove that $K_{\lambda,\mu}=0$.

Indeed, there exists no column-strict tableau $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, there exists a column-strict tableau $T$ of shape
$\lambda$ having $\cont \left(  T\right)  =\mu$. Consider this
$T$.
\par
Lemma \ref{lem.sol.Lambda.triangular.kostka}(e) yields $\lambda\triangleright
\mu$. This contradicts the fact that we don't have $\lambda\triangleright\mu$.
This contradiction proves that our assumption was wrong. Qed.}. Thus, the
number of all column-strict tableaux $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$ equals $0$. In other words,
$K_{\lambda,\mu}$ equals $0$ (since $K_{\lambda,\mu}$ is the number of all
column-strict tableaux $T$ of shape $\lambda$ having $\cont
\left(  T\right)  =\mu$). In other words, $K_{\lambda,\mu}=0$. This solves
Exercise \ref{exe.Lambda.triangular}(d).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(e).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(e), let us state another
simple lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.kostka1}Let $n\in \NN$. Let $\lambda
\in\Par_{n}$. Let $F$ be the Ferrers diagram of
$\lambda$ (as a set of cells). Let $T_{0}:F\rightarrow\left\{  1,2,3,\ldots
\right\}  $ be the map that sends each $\left(  i,j\right)  \in F$ to $i$.
Then, $T_{0}$ is a column-strict tableau of shape $\lambda$ and satisfies
$\cont \left(  T_{0}\right)  =\lambda$.
\end{lemma}

\begin{example}
Let $n=8$ and $\lambda=\left(  3,2,2,1\right)  \in\Par_{8}$.
Then, the column-strict tableau $T_{0}$ defined in Lemma
\ref{lem.sol.Lambda.triangular.kostka1} looks as follows:%
\[
T_{0}=%
\begin{matrix}
1 & 1 & 1\\
2 & 2 & \\
3 & 3 & \\
4 &  &
\end{matrix}
.
\]

\end{example}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.kostka1}.]We have
$F\subset\left\{  1,2,3,\ldots\right\}  ^{2}$. Thus, every $\left(
i,j\right)  \in F$ satisfies $i\in\left\{  1,2,3,\ldots\right\}
$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  i,j\right)  \in F$. Then,
$\left(  i,j\right)  \in F\subset\left\{  1,2,3,\ldots\right\}  ^{2}$. In
other words, $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in\left\{
1,2,3,\ldots\right\}  $. Hence, $i\in\left\{  1,2,3,\ldots\right\}  $, qed.}.
Hence, the map $T_{0}$ is well-defined.

The map $T_{0}$ is a map from $F$ to $\left\{  1,2,3,\ldots\right\}  $. In
other words, the map $T_{0}$ is an assignment of entries in $\{1,2,3,\ldots\}$
to the elements of $F$. In other words, the map $T_{0}$ is an assignment of
entries in $\{1,2,3,\ldots\}$ to the cells of the Ferrers diagram of $\lambda$
(since the elements of $F$ are the cells of the Ferrers diagram of $\lambda$
(since $F$ is the Ferrers diagram of $\lambda$)). We shall now show that
$T_{0}$ is a column-strict tableau of shape $\lambda$.

For each $j\in\left\{  1,2,3,\ldots\right\}  $, the entries of the $j$-th
column of $T_{0}$ are strictly increasing
top-to-bottom\footnote{\textit{Proof.} Let $j\in\left\{  1,2,3,\ldots\right\}
$. Then, for each $i\in\left\{  1,2,3,\ldots\right\}  $, the entry of $T_{0}$
in cell $\left(  i,j\right)  $ (if this cell belongs to $F$) equals $i$ (since
$T_{0}\left(  i,j\right)  =i$ (by the definition of $T_{0}$)). Hence, this
entry strictly increases when $i$ increases. In other words, the entries of
the $j$-th column of $T_{0}$ are strictly increasing top-to-bottom. Qed.}. In
other words, the entries of $T_{0}$ are strictly increasing top-to-bottom in columns.

For each $i\in\left\{  1,2,3,\ldots\right\}  $, the entries of the $i$-th row
of $T_{0}$ are weakly increasing left-to-right\footnote{\textit{Proof.} Let
$i\in\left\{  1,2,3,\ldots\right\}  $. Then, for each $j\in\left\{
1,2,3,\ldots\right\}  $, the entry of $T_{0}$ in cell $\left(  i,j\right)  $
(if this cell belongs to $F$) equals $i$ (since $T_{0}\left(  i,j\right)  =i$
(by the definition of $T_{0}$)). In other words, each entry in the $i$-th row
of $T_{0}$ equals $i$. Hence, the entries in the $i$-th row of $T_{0}$ are
weakly increasing left-to-right (because they all equal $i$). Qed.}. In other
words, the entries of $T_{0}$ are weakly increasing left-to-right in rows.

Altogether, we have shown that $T_{0}$ is an assignment of entries in
$\left\{  1,2,3,\ldots\right\}  $ to the cells of the Ferrers diagram of
$\lambda$, weakly increasing left-to-right in rows, and strictly increasing
top-to-bottom in columns. In other words, $T_{0}$ is a column-strict tableau
of shape $\lambda$ (by the definition of a ``column-strict
tableau of shape $\lambda$'').

It remains to prove that $\cont \left(  T_{0}\right)  =\lambda$.

The definition of $\cont \left(  T_{0}\right)  $ shows that
$\cont \left(  T_{0}\right)  =\left(  \left\vert \left(
T_{0}\right)  ^{-1}\left(  1\right)  \right\vert ,\left\vert \left(
T_{0}\right)  ^{-1}\left(  2\right)  \right\vert ,\left\vert \left(
T_{0}\right)  ^{-1}\left(  3\right)  \right\vert ,\ldots\right)  $. In other
words,%
\begin{equation}
\left(  \cont \left(  T_{0}\right)  \right)  _{i}=\left\vert
\left(  T_{0}\right)  ^{-1}\left(  i\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.
\label{pf.lem.sol.Lambda.triangular.kostka1.cont1}%
\end{equation}


Now, let $i$ be a positive integer. Let $F_{i}$ be the $i$-th row of $F$ (that
is, the set of all cells of $F$ that have the form $\left(  i,j\right)  $ for
some $j\geq1$). We have%
\begin{align}
\left\vert F_{i}\right\vert  &  =\left(  \text{the size of }F_{i}\right)
=\left(  \text{the size of the }i\text{-th row of }F\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F_{i}\text{ is the }i\text{-th row
of }F\right) \nonumber\\
&  =\left(  \text{the size of the }i\text{-th row of the Ferrers diagram of
}\lambda\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is the Ferrers diagram of
}\lambda\right) \nonumber\\
&  =\lambda_{i}. \label{pf.lem.sol.Lambda.triangular.kostka1.absFi=}%
\end{align}


But $F_{i}\subset\left(  T_{0}\right)  ^{-1}\left(  i\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $c\in F_{i}$. Thus, $c$ belongs to
$F_{i}$. In other words, $c$ belongs to the $i$-th row of $F$ (since $F_{i}$
is the $i$-th row of $F$). Thus, in particular, $c$ belongs to $F$. Hence,
$c\in F\subset\left\{  1,2,3,\ldots\right\}  ^{2}$. Thus, we can write $c$ in
the form $c=\left(  u,v\right)  $ for some $u\in\left\{  1,2,3,\ldots\right\}
$ and $v\in\left\{  1,2,3,\ldots\right\}  $. Consider these $u$ and $v$.
\par
The cell $c$ belongs to the $i$-th row of $F$. Thus, the cell $c$ lies in the
$i$-th row. In other words, the first coordinate of $c$ equals $i$. In other
words, $u$ equals $i$ (since the first coordinate of $c$ is $u$ (since
$c=\left(  u,v\right)  $)). In other words, $u=i$. But the definition of
$T_{0}$ yields $T_{0}\left(  u,v\right)  =u=i$. Since $\left(  u,v\right)
=c$, this rewrites as $T_{0}\left(  c\right)  =i$. Hence, $c\in\left(
T_{0}\right)  ^{-1}\left(  i\right)  $.
\par
Now, forget that we fixed $c$. We thus have shown that $c\in\left(
T_{0}\right)  ^{-1}\left(  i\right)  $ for each $c\in F_{i}$. In other words,
$F_{i}\subset\left(  T_{0}\right)  ^{-1}\left(  i\right)  $.} and $\left(
T_{0}\right)  ^{-1}\left(  i\right)  \subset F_{i}$%
\ \ \ \ \footnote{\textit{Proof.} Let $c\in\left(  T_{0}\right)  ^{-1}\left(
i\right)  $. Thus, $c\in F\subset\left\{  1,2,3,\ldots\right\}  ^{2}$. Thus,
we can write $c$ in the form $c=\left(  u,v\right)  $ for some $u\in\left\{
1,2,3,\ldots\right\}  $ and $v\in\left\{  1,2,3,\ldots\right\}  $. Consider
these $u$ and $v$.
\par
The definition of $T_{0}$ yields $T_{0}\left(  u,v\right)  =u$. Since $\left(
u,v\right)  =c$, this rewrites as $T_{0}\left(  c\right)  =u$. Hence,
$u=T_{0}\left(  c\right)  =i$ (since $c\in\left(  T_{0}\right)  ^{-1}\left(
i\right)  $).
\par
The first coordinate of $c$ equals $u$ (since $c=\left(  u,v\right)  $). In
other words, the cell $c$ lies in the $u$-th row. In other words, the cell $c$
lies in the $i$-th row (since $u=i$). Thus, the cell $c$ belongs to the $i$-th
row of $F$ (since $c\in F$). In other words, the cell $c$ belongs to $F_{i}$
(since $F_{i}$ is the $i$-th row of $F$). In other words, $c\in F_{i}$.
\par
Now, forget that we fixed $c$. We thus have shown that $c\in F_{i}$ for each
$c\in\left(  T_{0}\right)  ^{-1}\left(  i\right)  $. In other words, $\left(
T_{0}\right)  ^{-1}\left(  i\right)  \subset F_{i}$.}. Combining these two
inclusions, we obtain $F_{i}=\left(  T_{0}\right)  ^{-1}\left(  i\right)  $.
Hence, $\left\vert F_{i}\right\vert =\left\vert \left(  T_{0}\right)
^{-1}\left(  i\right)  \right\vert $. Comparing this with
(\ref{pf.lem.sol.Lambda.triangular.kostka1.absFi=}), we obtain $\left\vert
\left(  T_{0}\right)  ^{-1}\left(  i\right)  \right\vert =\lambda_{i}$. Now,
(\ref{pf.lem.sol.Lambda.triangular.kostka1.cont1}) yields $\left(
\cont \left(  T_{0}\right)  \right)  _{i}=\left\vert \left(
T_{0}\right)  ^{-1}\left(  i\right)  \right\vert =\lambda_{i}$.

Now, forget that we fixed $i$. We thus have proven that $\left(
\cont \left(  T_{0}\right)  \right)  _{i}=\lambda_{i}$ for each
positive integer $i$. In other words, $\cont \left(
T_{0}\right)  =\lambda$. This completes the proof of Lemma
\ref{lem.sol.Lambda.triangular.kostka1}.
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(e).
\end{commentedout}

(e) Let $\lambda\in\Par_{n}$. We must prove that
$K_{\lambda,\lambda}=1$.

Define $F$ and $T_{0}$ as in Lemma \ref{lem.sol.Lambda.triangular.kostka1}.
Then, Lemma \ref{lem.sol.Lambda.triangular.kostka1} shows that $T_{0}$ is a
column-strict tableau of shape $\lambda$ and satisfies $\cont
\left(  T_{0}\right)  =\lambda$. In other words, $T_{0}$ is a column-strict
tableau $T$ of shape $\lambda$ having $\cont \left(  T\right)
=\lambda$. In other words,%
\[
T_{0}\in\left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{
having }\cont \left(  T\right)  =\lambda\right\}  .
\]
Thus,%
\begin{equation}
\left\{  T_{0}\right\}  \subset\left\{  \text{column-strict tableaux }T\text{
of shape }\lambda\text{ having }\cont \left(  T\right)
=\lambda\right\}  . \label{sol.Lambda.triangular.e.1}%
\end{equation}


On the other hand, the definition of $T_{0}$ shows that each $\left(
i,j\right)  \in F$ satisfies%
\begin{equation}
T_{0}\left(  i,j\right)  =i. \label{sol.Lambda.triangular.e.2}%
\end{equation}


Using Lemma \ref{lem.sol.Lambda.triangular.kostka}(f), it is easy to see that
every column-strict tableau $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\lambda$ must belong to the set
$\left\{  T_{0}\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $T$ be a
column-strict tableau of shape $\lambda$ having $\cont \left(
T\right)  =\lambda$. We must prove that $T$ must belong to the set $\left\{
T_{0}\right\}  $.
\par
Lemma \ref{lem.sol.Lambda.triangular.kostka}(f) (applied to $\mu=\lambda$)
shows that each $\left(  i,j\right)  \in F$ satisfies $T\left(  i,j\right)
=i$ (since $\lambda=\lambda$). Thus, each $\left(  i,j\right)  \in F$
satisfies%
\[
T\left(  i,j\right)  =i=T_{0}\left(  i,j\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Lambda.triangular.e.2})}\right)  .
\]
\par
Recall that any column-strict tableau of shape $\lambda$ is a map
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $. Hence, $T$ and $T_{0}$ are maps
$F\rightarrow\left\{  1,2,3,\ldots\right\}  $ (since $T$ and $T_{0}$ are
column-strict tableaux of shape $\lambda$). Therefore, we conclude that
$T=T_{0}$ (since each $\left(  i,j\right)  \in F$ satisfies $T\left(
i,j\right)  =T_{0}\left(  i,j\right)  $). Hence, $T=T_{0}\in\left\{
T_{0}\right\}  $. In other words, $T$ must belong to the set $\left\{
T_{0}\right\}  $. Qed.}. In other words,%
\[
\left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{ having
}\cont \left(  T\right)  =\lambda\right\}  \subset\left\{
T_{0}\right\}  .
\]
Combining this with (\ref{sol.Lambda.triangular.e.1}), we obtain%
\[
\left\{  \text{column-strict tableaux }T\text{ of shape }\lambda\text{ having
}\cont \left(  T\right)  =\lambda\right\}  =\left\{
T_{0}\right\}  .
\]


We know that $K_{\lambda,\lambda}$ is the number of all column-strict tableaux
$T$ of shape $\lambda$ having $\cont \left(  T\right)  =\lambda$
(by the definition of $K_{\lambda,\lambda}$). Thus,%
\begin{align*}
K_{\lambda,\lambda}  &  =\left(  \text{the number of all column-strict
tableaux }T\text{ of shape }\lambda\text{ having }\cont \left(
T\right)  =\lambda\right) \\
&  =\left\vert \underbrace{\left\{  \text{column-strict tableaux }T\text{ of
shape }\lambda\text{ having }\cont \left(  T\right)
=\lambda\right\}  }_{=\left\{  T_{0}\right\}  }\right\vert =\left\vert
\left\{  T_{0}\right\}  \right\vert =1.
\end{align*}
This solves Exercise \ref{exe.Lambda.triangular}(e).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(f).
\end{commentedout}

(f) Let $\lambda$ and $\mu$ be two partitions. We must prove that the number
$a_{\lambda,\mu}$ is well-defined. In other words, we must prove that there
are only finitely many $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$.

But this is easy: Any $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ is a map from the set
$\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{
1,2,\ldots,\ell\left(  \mu\right)  \right\}  $ to the set $\left\{
0,1\right\}  $. Thus,%
\begin{align*}
&  \left\{  \left\{  0,1\right\}  \text{-matrices of size }\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  \right\} \\
&  =\left\{  \text{maps from the set }\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  \text{ to the set }\left\{  0,1\right\}  \right\} \\
&  =\left\{  0,1\right\}  ^{\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  }%
\end{align*}
is a finite set (since both $\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  $ and
$\left\{  0,1\right\}  $ are finite sets). In other words, there are only
finitely many $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $. Hence, there are only
finitely many $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$. Hence, the number $a_{\lambda,\mu}$ is well-defined (since
this number is defined as the number of all $\left\{  0,1\right\}  $-matrices
of size $\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)  $ having
row sums $\lambda$ and column sums $\mu$). This solves Exercise
\ref{exe.Lambda.triangular}(f).

[\textit{Remark:} We can prove a slightly stronger claim: Namely, there are
only finitely many matrices in $\mathbb{N}^{\ell\left(  \lambda\right)
\times\ell\left(  \mu\right)  }$ having row sums $\lambda$ and column sums
$\mu$.

Let us sketch the proof of this claim. Indeed, let $N=\left\vert
\lambda\right\vert $. Then, if $A$ is any matrix in $\mathbb{N}^{\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  }$ having row sums $\lambda$ and
column sums $\mu$, then the sum of all entries of $A$ must equal $\left\vert
\lambda\right\vert =N$, and therefore each entry of $A$ must be $\leq N$
(since a sum of nonnegative integers is always $\geq$ to each of its addends);
but this entails that each entry of $A$ belongs to the finite set $\left\{
0,1,\ldots,N\right\}  $, and therefore there are only finitely many choices
for each entry, which leads to only finitely many possible matrices $A$.]

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(g).
\end{commentedout}

(g) Exercise \ref{exe.Lambda.triangular}(g) is truly not a deep fact, but its
proof requires some bookkeeping. In order to make this bookkeeping more
palatable, we are going to introduce various auxiliary notations.

\begin{definition}
Let $q\in \NN$. Let $\overline{\mathbf{x}}_{q}$ denote the $q$-tuple
$\left(  x_{1},x_{2},\ldots,x_{q}\right)  $ of indeterminates. Let
$\kk\left[  \overline{\mathbf{x}}_{q}\right]  $ denote the polynomial
ring $\kk\left[  x_{1},x_{2},\ldots,x_{q}\right]  $. Let $\eta
_{q}:R\left(  \xx\right)  \rightarrow\kk\left[  \overline{\xx}_{q}\right]  $
be the map that sends every power series $f\in
R\left(  \xx\right)  $ to the polynomial $f\left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  $. (This is well-defined, because Exercise
\ref{exe.Lambda.subs.fin} (applied to $A=\kk\left[  \overline{\xx}_{q}
\right]  $ and $k=q$) shows that substituting $x_{1}%
,x_{2},\ldots,x_{q},0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in $f$ yields
an infinite sum in which all but finitely many addends are zero.)

The map $\eta_{q}$ is an evaluation homomorphism (in an appropriate
sense\footnote{i.e., it acts on a power series $f\in R\left(  \xx%
\right)  $ by substituting certain values for the indeterminates $x_{1}%
,x_{2},x_{3},\ldots$}); thus, it is a $\kk$-algebra homomorphism.

If $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  \in
\mathbb{N}^{q}$ is a $q$-tuple of nonnegative integers, then
$\overline{\xx}_{q}^{\beta}$ shall denote the monomial
$x_{1}^{\beta_{1}}%
x_{2}^{\beta_{2}}\cdots x_{q}^{\beta_{q}}$. This is a monomial in the
polynomial ring $\kk\left[  \overline{\xx}_{q}\right]  $.

For every $f\in\kk\left[  \left[  \overline{\mathbf{x}}_{q}\right]
\right]  $ and $\beta\in\mathbb{N}^{q}$, we let $\left[  \overline{\xx%
}_{q}^{\beta}\right]  f$ denote the coefficient of the monomial
$\overline{\xx}_{q}^{\beta}$ in the power series $f$.
\end{definition}

Let us make the following simple observations:

\begin{itemize}
\item Every $q\in \NN$ and $i\in\left\{  1,2,\ldots,q\right\}  $ satisfy%
\begin{align}
\eta_{q}\left(  x_{i}\right)   &  =x_{i}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta_{q}\right) \nonumber\\
&  =x_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\in\left\{  1,2,\ldots
,q\right\}  \right)  . \label{pf.lem.sol.Lambda.triangular.etaq-coeff.eta1}%
\end{align}


\item Every $q\in \NN$ and $i\in\left\{  q+1,q+2,q+3,\ldots\right\}  $
satisfy%
\begin{align}
\eta_{q}\left(  x_{i}\right)   &  =x_{i}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta_{q}\right) \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\in\left\{  q+1,q+2,q+3,\ldots
\right\}  \right)  . \label{pf.lem.sol.Lambda.triangular.etaq-coeff.eta2}%
\end{align}


\item Any two $q$-tuples $\phi\in\mathbb{N}^{q}$ and $\psi\in\mathbb{N}^{q}$
satisfy%
\begin{equation}
\left[  \overline{\mathbf{x}}_{q}^{\phi}\right]  \left(  \overline{\xx%
}_{q}^{\psi}\right)  =\delta_{\phi,\psi}.
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.4}%
\end{equation}
(Indeed, $\overline{\mathbf{x}}_{q}^{\psi}$ and $\overline{\mathbf{x}}%
_{q}^{\phi}$ are two distinct monomials if $\phi\neq\psi$, and are two
identical monomials if $\phi=\psi$.)
\end{itemize}

\begin{lemma}
\label{lem.sol.Lambda.triangular.etaq-coeff}Let $q\in \NN$. Let $\beta$
be a weak composition. Assume that $\beta_{i}=0$ for every integer $i>q$. Let
$f\in R\left(  \xx\right)  $. Then,%
\[
\left[  \mathbf{x}^{\beta}\right]  f=\left[  \overline{\mathbf{x}}%
_{q}^{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(
\eta_{q}\left(  f\right)  \right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.etaq-coeff}.]Let us prove that
every weak composition $\alpha$ satisfies%
\begin{equation}
\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)
=\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(  \mathbf{x}^{\alpha
}\right)  \right)  . \label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}):}
Let $\alpha$ be a weak composition. We must prove the equality
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}).

We distinguish between two cases:

\textit{Case 1:} We have $\left(  \alpha_{i}=0\text{ for every integer
}i>q\right)  $.

\textit{Case 2:} We don't have $\left(  \alpha_{i}=0\text{ for every integer
}i>q\right)  $.

Let us first consider Case 1. In this case, we have
\begin{equation}
\left(  \alpha_{i}=0\text{ for every integer }i>q\right)  .
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1}%
\end{equation}


Now,%
\begin{equation}
\delta_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }=\delta_{\beta,\alpha}
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}):} We are in
one of the following two subcases:
\par
\textit{Subcase 1.1:} We have $\alpha\neq\beta$.
\par
\textit{Subcase 1.2:} We have $\alpha=\beta$.
\par
Let us first consider Subcase 1.1. In this subcase, we have $\alpha\neq\beta$.
Hence, there exists some $k\in\left\{  1,2,3,\ldots\right\}  $ satisfying
$\alpha_{k}\neq\beta_{k}$. Consider this $k$.
\par
We claim that $k\in\left\{  1,2,\ldots,q\right\}  $. Indeed, assume the
contrary (for the sake of contradiction). Then, $k\notin\left\{
1,2,\ldots,q\right\}  $. Combining $k\in\left\{  1,2,3,\ldots\right\}  $ with
$k\notin\left\{  1,2,\ldots,q\right\}  $, we obtain $k\in\left\{
1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots,q\right\}  =\left\{
q+1,q+2,q+3,\ldots\right\}  $. Hence, $k\geq q+1>q$. Thus,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1}) (applied to
$i=k$) yields $\alpha_{k}=0$. But let us recall that $\beta_{i}=0$ for every
integer $i>q$. Applying this to $i=k$, we find $\beta_{k}=0$. Hence,
$\alpha_{k}=0=\beta_{k}$. This contradicts $\alpha_{k}\neq\beta_{k}$.
\par
This contradiction completes the proof of $k\in\left\{  1,2,\ldots,q\right\}
$. Hence, we now know that $k$ is an element of $\left\{  1,2,\ldots
,q\right\}  $ and satisfies $\alpha_{k}\neq\beta_{k}$. Thus, there exists some
$i\in\left\{  1,2,\ldots,q\right\}  $ satisfying $\alpha_{i}\neq\beta_{i}$
(namely, $i=k$). Therefore, $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{q}\right)  \neq\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  $. In
other words, $\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  \neq\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  $. Thus, $\delta_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{q}\right)  }=0$. Comparing this with $\delta_{\beta
,\alpha}=0$ (since $\beta\neq\alpha$ (since $\alpha\neq\beta$)), we obtain
$\delta_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }=\delta_{\beta,\alpha}$.
Thus, (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) is
proven in Subcase 1.1.
\par
Let us now consider Subcase 1.2. In this subcase, we have $\alpha=\beta$.
Hence, each $k\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\alpha_{k}%
=\beta_{k}$. Thus, each $k\in\left\{  1,2,\ldots,q\right\}  $ satisfies
$\alpha_{k}=\beta_{k}$. In other words, $\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{q}\right)  =\left(  \beta_{1},\beta_{2},\ldots,\beta
_{q}\right)  $. In other words, $\left(  \beta_{1},\beta_{2},\ldots,\beta
_{q}\right)  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  $. Thus,
$\delta_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }=1$. Comparing this with
$\delta_{\beta,\alpha}=1$ (since $\beta=\alpha$ (since $\alpha=\beta$)), we
obtain $\delta_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }=\delta_{\beta,\alpha}$.
Thus, (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) is
proven in Subcase 1.2.
\par
We have now proven
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) in each of
the two Subcases 1.1 and 1.2. Since these two Subcases cover the whole Case 1,
we thus conclude that
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2}) always
holds.}.

Now, the definition of $\mathbf{x}^{\alpha}$ yields%
\begin{align*}
\mathbf{x}^{\alpha}  &  =x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha
_{3}}\cdots=\prod_{i\geq1}x_{i}^{\alpha_{i}}=\left(  \prod_{i=1}^{q}%
x_{i}^{\alpha_{i}}\right)  \left(  \prod_{i=q+1}^{\infty}\underbrace{x_{i}%
^{\alpha_{i}}}_{\substack{=x_{i}^{0}\\\text{(since }\alpha_{i}=0\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.1})}%
\\\text{(because }i\geq q+1>q\text{)))}}}\right) \\
&  =\left(  \prod_{i=1}^{q}x_{i}^{\alpha_{i}}\right)  \left(  \prod
_{i=q+1}^{\infty}\underbrace{x_{i}^{0}}_{=1}\right)  =\left(  \prod_{i=1}%
^{q}x_{i}^{\alpha_{i}}\right)  \underbrace{\left(  \prod_{i=q+1}^{\infty
}1\right)  }_{=1}=\prod_{i=1}^{q}x_{i}^{\alpha_{i}}.
\end{align*}
Applying the map $\eta_{q}$ to both sides of this equality, we obtain%
\begin{align}
\eta_{q}\left(  \mathbf{x}^{\alpha}\right)   &  =\eta_{q}\left(  \prod
_{i=1}^{q}x_{i}^{\alpha_{i}}\right)  =\prod_{i=1}^{q}\left(  \underbrace{\eta
_{q}\left(  x_{i}\right)  }_{\substack{=x_{i}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta1}))}}}\right)  ^{\alpha_{i}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{q}\text{ is a }%
\kk\text{-algebra homomorphism}\right) \nonumber\\
&  =\prod_{i=1}^{q}x_{i}^{\alpha_{i}}=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{q}^{\alpha_{q}}=\overline{\mathbf{x}}_{q}^{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.3}%
\end{align}
(since $\overline{\mathbf{x}}_{q}^{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{q}\right)  }=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots x_{q}%
^{\alpha_{q}}$ (by the definition of $\overline{\mathbf{x}}_{q}^{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }$)).

Now,%
\begin{align*}
&  \left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  }_{\substack{=\overline{\mathbf{x}}_{q}^{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.3}))}}}\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{q}\right)  }\right]  \left(  \overline{\mathbf{x}}_{q}^{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  }\right)  =\delta_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  ,\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{q}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.4}) (applied to
}\phi=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  \text{ and }%
\psi=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{q}\right)  \text{)}\right)
\\
&  =\delta_{\beta,\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.2})}\right)  .
\end{align*}
Comparing this with%
\[
\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)
=\delta_{\beta,\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.coeff-mon}) (applied to }\mu=\beta\text{)}\right)
,
\]
we obtain $\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha
}\right)  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(  \xx%
^{\alpha}\right)  \right)  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) is proven in Case 1.

Let us now consider Case 2. In this case, we don't have $\left(  \alpha
_{i}=0\text{ for every integer }i>q\right)  $. In other words, not every
integer $i>q$ satisfies $\alpha_{i}=0$. In other words, there exists some
integer $i>q$ that does not satisfy $\alpha_{i}=0$. Consider this $i$, and
denote it by $k$. Thus, $k$ is an integer such that $k>q$ and such that $k$
does not satisfy $\alpha_{k}=0$.

We have $\alpha_{k}\neq0$ (since $k$ does not satisfy $\alpha_{k}=0$). Hence,
$\beta\neq\alpha$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus,
$\beta=\alpha$. Hence, $\beta_{k}=\alpha_{k}\neq0$. But recall that $\beta
_{i}=0$ for every integer $i>q$. Applying this to $i=k$, we obtain $\beta
_{k}=0$. This contradicts $\beta_{k}\neq0$. This contradiction shows that our
assumption was wrong. Qed.}. Thus, $\delta_{\beta,\alpha}=0$.

On the other hand, $\alpha_{k}\neq0$ and thus $\alpha_{k}>0$ (since
$\alpha_{k}\in \NN$). Hence, there exists a $g\in\kk\left[
\left[  \xx\right]  \right]  $ such that $\mathbf{x}^{\alpha}=gx_{k}%
$\ \ \ \ \footnote{\textit{Proof.} We have $k>q\geq0$ (since $q\in\mathbb{N}%
$); thus, $k$ is a positive integer. In other words, $k\in\left\{
1,2,3,\ldots\right\}  $. Now,
\begin{align*}
\mathbf{x}^{\alpha}  &  =x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha
_{3}}\cdots=\prod_{i\in\left\{  1,2,3,\ldots\right\}  }x_{i}^{\alpha_{i}%
}=\left(  \prod\limits_{\substack{i\in\left\{  1,2,3,\ldots\right\}  ;\\i\neq
k}}x_{i}^{\alpha_{i}}\right)  \underbrace{x_{k}^{\alpha_{k}}}%
_{\substack{=x_{k}^{\alpha_{k}-1}x_{k}\\\text{(since }\alpha_{k}>0\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the factor for }i=k\text{ from the product}\\
\text{(since }k\in\left\{  1,2,3,\ldots\right\}  \text{)}%
\end{array}
\right) \\
&  =\left(  \prod\limits_{\substack{i\in\left\{  1,2,3,\ldots\right\}  ;\\i\neq
k}}x_{i}^{\alpha_{i}}\right)  x_{k}^{\alpha_{k}-1}x_{k}.
\end{align*}
Hence, there exists a $g\in\kk\left[  \left[  \xx\right]
\right]  $ such that $\mathbf{x}^{\alpha}=gx_{k}$ (namely, $g=\left(
\prod\limits_{\substack{i\in\left\{  1,2,3,\ldots\right\}  ;\\i\neq k}}x_{i}%
^{\alpha_{i}}\right)  x_{k}^{\alpha_{k}-1}$). Qed.}. Consider this $g$. From
$k>q$, we obtain $k\in\left\{  q+1,q+2,q+3,\ldots\right\}  $. Thus,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta2}) (applied to $i=k$) yields
$\eta_{q}\left(  x_{k}\right)  =0$. Now,%
\begin{align*}
\eta_{q}\left(  \underbrace{\mathbf{x}^{\alpha}}_{=gx_{k}}\right)   &
=\eta_{q}\left(  gx_{k}\right)  =\eta_{q}\left(  g\right)  \underbrace{\eta
_{q}\left(  x_{k}\right)  }_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\eta_{q}\text{ is a }\kk\text{-algebra homomorphism}\right) \\
&  =0.
\end{align*}
Hence,%
\[
\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  }_{=0}\right)  =\left[  \overline{\mathbf{x}}%
_{q}^{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(
0\right)  =0.
\]
Comparing this with%
\begin{align*}
\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)   &
=\delta_{\beta,\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.coeff-mon}) (applied to }\mu=\beta\text{)}\right)
\\
&  =0,
\end{align*}
we obtain $\left[  \mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha
}\right)  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(  \xx%
^{\alpha}\right)  \right)  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) is proven in Case 2.

We have now proven (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}) in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
we thus conclude that (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase})
always holds.]

\begin{commentedout}
Let us first notice that every polynomial $g\in\kk\left[
\overline{\mathbf{x}}_{q}\right]  $ satisfies%
\begin{equation}
g=\sum_{\alpha\in\mathbb{N}^{q}}\left[  \overline{\mathbf{x}}_{q}^{\alpha
}\right]  \left(  g\right)  \cdot\overline{\mathbf{x}}_{q}^{\alpha}
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.1}%
\end{equation}
(since the family of the coefficients of the polynomial $g$ is $\left(
\left[  \overline{\mathbf{x}}_{q}^{\alpha}\right]  \left(  g\right)  \right)
_{\alpha\in\mathbb{N}^{q}}$).
\end{commentedout}

Now, let us notice that every power series $g\in\kk\left[  \left[
\xx\right]  \right]  $ satisfies%
\begin{equation}
g=\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  g\right)  \cdot\mathbf{x}^{\alpha}
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.6}%
\end{equation}
(since the family of the coefficients of $g$ is $\left(  \left[
\mathbf{x}^{\alpha}\right]  \left(  g\right)  \right)  _{\alpha\in
\operatorname*{WC}}$). Applying this to $g=f$, we obtain%
\[
f=\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\mathbf{x}^{\alpha}.
\]
Substituting $x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots$ for the variables
$x_{1},x_{2},x_{3},\ldots$ in this equality, we obtain%
\begin{align}
f\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)   &  =\left(
\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]  \left(
f\right)  \cdot\mathbf{x}^{\alpha}\right)  \left(  \left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  \right) \nonumber\\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\underbrace{\mathbf{x}^{\alpha}\left(  x_{1}%
,x_{2},\ldots,x_{q},0,0,0,\ldots\right)  }_{\substack{=\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  \\\text{(since }\eta_{q}\left(  \xx%
^{\alpha}\right)  =\mathbf{x}^{\alpha}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  \\\text{(by the definition of }\eta_{q}\text{))}%
}}\nonumber\\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\eta_{q}\left(  \mathbf{x}^{\alpha}\right)  .
\label{pf.lem.sol.Lambda.triangular.etaq-coeff.9}%
\end{align}
Now, the definition of $\eta_{q}$ yields%
\begin{align*}
\eta_{q}\left(  f\right)   &  =f\left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right) \\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\eta_{q}\left(  \mathbf{x}^{\alpha}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.9})}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(  f\right)
}_{=\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\eta_{q}\left(  \mathbf{x}^{\alpha}\right)  }\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{q}\right)  }\right]  \left(  \sum_{\alpha\in\operatorname*{WC}%
}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)  \cdot\eta_{q}\left(
\mathbf{x}^{\alpha}\right)  \right) \\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\underbrace{\left[  \overline{\mathbf{x}}_{q}^{\left(
\beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(  \eta_{q}\left(
\mathbf{x}^{\alpha}\right)  \right)  }_{\substack{=\left[  \mathbf{x}^{\beta
}\right]  \left(  \mathbf{x}^{\alpha}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase}))}}}\\
&  =\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]
\left(  f\right)  \cdot\left[  \mathbf{x}^{\beta}\right]  \left(
\mathbf{x}^{\alpha}\right)  .
\end{align*}
Comparing this with%
\[
\left[  \mathbf{x}^{\beta}\right]  \underbrace{f}_{=\sum_{\alpha
\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)
\cdot\mathbf{x}^{\alpha}}=\left[  \mathbf{x}^{\beta}\right]  \left(
\sum_{\alpha\in\operatorname*{WC}}\left[  \mathbf{x}^{\alpha}\right]  \left(
f\right)  \cdot\mathbf{x}^{\alpha}\right)  =\sum_{\alpha\in\operatorname*{WC}%
}\left[  \mathbf{x}^{\alpha}\right]  \left(  f\right)  \cdot\left[
\mathbf{x}^{\beta}\right]  \left(  \mathbf{x}^{\alpha}\right)  ,
\]
we obtain $\left[  \mathbf{x}^{\beta}\right]  f=\left[  \overline{\xx%
}_{q}^{\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right]  \left(
\eta_{q}\left(  f\right)  \right)  $. This proves Lemma
\ref{lem.sol.Lambda.triangular.etaq-coeff}.
\end{proof}

We now need to introduce some more notations.

\begin{definition}
If $q\in \NN$, and if $\beta=\left(  \beta_{1},\beta_{2},\ldots
,\beta_{q}\right)  \in\mathbb{N}^{q}$ is a $q$-tuple of nonnegative integers,
then $\left\vert \beta\right\vert $ shall denote the sum $\beta_{1}+\beta
_{2}+\cdots+\beta_{q}\in \NN$.
\end{definition}

\begin{definition}
We shall use the so-called \dfn{Iverson bracket notation}: For every
assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $ denote the
integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
\end{definition}

Let us state a fundamental fact in combinatorics:

\begin{lemma}
\label{lem.sol.Lambda.triangular.IPK}Let $q\in \NN$. Let $m\in
\mathbb{N}$. Define a set $\mathfrak{I}$ by%
\[
\mathfrak{I}=\left\{  \left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m}\ \mid\ i_{1}<i_{2}<\cdots<i_{m}\right\}  .
\]
Let $\mathfrak{P}$ be the set of all $m$-element subsets of $\left\{
1,2,\ldots,q\right\}  $. Define a set $\mathfrak{K}$ by%
\[
\mathfrak{K}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  .
\]


\begin{enumerate}
\item[(a)] The map $\mathfrak{I}\rightarrow\mathfrak{P},\ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
$ is well-defined and bijective.

\item[(b)] The map $\mathfrak{P}\rightarrow\mathfrak{K},\ T\mapsto\left(
\left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]
\right)  $ is well-defined and bijective.

\item[(c)] There exists a bijection $\Psi:\mathfrak{I}\rightarrow\mathfrak{K}$
with the property that every $\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\mathfrak{I}$ satisfies $\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.IPK}.]Each $\left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \in\mathfrak{I}$ satisfies $\left\{  j_{1}%
,j_{2},\ldots,j_{m}\right\}  \in\mathfrak{P}$\ \ \ \ \footnote{\textit{Proof.}
Let $\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\mathfrak{I}$. Thus,%
\[
\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\mathfrak{I}=\left\{  \left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}%
\ \mid\ i_{1}<i_{2}<\cdots<i_{m}\right\}  .
\]
In other words, $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is an $\left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$
satisfying $i_{1}<i_{2}<\cdots<i_{m}$. In other words, $\left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  $ is an element of $\left\{  1,2,\ldots,q\right\}
^{m}$ and satisfies $j_{1}<j_{2}<\cdots<j_{m}$.
\par
We know that $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is an element of
$\left\{  1,2,\ldots,q\right\}  ^{m}$. Thus, $j_{1},j_{2},\ldots,j_{m}$ are
elements of $\left\{  1,2,\ldots,q\right\}  $. Hence, $\left\{  j_{1}%
,j_{2},\ldots,j_{m}\right\}  $ is a subset of $\left\{  1,2,\ldots,q\right\}
$.
\par
Moreover, the $m$ integers $j_{1},j_{2},\ldots,j_{m}$ are distinct (since
$j_{1}<j_{2}<\cdots<j_{m}$). Thus, $\left\vert \left\{  j_{1},j_{2}%
,\ldots,j_{m}\right\}  \right\vert =m$. In other words, $\left\{  j_{1}%
,j_{2},\ldots,j_{m}\right\}  $ is an $m$-element set.
\par
Thus, we know that $\left\{  j_{1},j_{2},\ldots,j_{m}\right\}  $ is an
$m$-element subset of $\left\{  1,2,\ldots,q\right\}  $. In other words,
$\left\{  j_{1},j_{2},\ldots,j_{m}\right\}  $ belongs to the set of all
$m$-element subsets of $\left\{  1,2,\ldots,q\right\}  $. In other words,
$\left\{  j_{1},j_{2},\ldots,j_{m}\right\}  $ belongs to $\mathfrak{P}$ (since
$\mathfrak{P}$ is the set of all $m$-element subsets of $\left\{
1,2,\ldots,q\right\}  $). In other words, $\left\{  j_{1},j_{2},\ldots
,j_{m}\right\}  \in\mathfrak{P}$, qed.}. Hence, the map%
\[
\mathfrak{I}\rightarrow\mathfrak{P},\ \ \ \ \ \ \ \ \ \ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
\]
is well-defined. Let us denote this map by $A$.

Each $T\in\mathfrak{P}$ satisfies $\left(  \left[  1\in T\right]  ,\left[
2\in T\right]  ,\ldots,\left[  q\in T\right]  \right)  \in\mathfrak{K}%
$\ \ \ \ \footnote{\textit{Proof.} Let $T\in\mathfrak{P}$. Thus, $T$ belongs
to $\mathfrak{P}$. In other words, $T$ belongs to the set of all $m$-element
subsets of $\left\{  1,2,\ldots,q\right\}  $ (since $\mathfrak{P}$ is the set
of all $m$-element subsets of $\left\{  1,2,\ldots,q\right\}  $). In other
words, $T$ is an $m$-element subset of $\left\{  1,2,\ldots,q\right\}  $.
\par
We have $\left[  i\in T\right]  \in\left\{  0,1\right\}  $ for each
$i\in\left\{  1,2,\ldots,q\right\}  $ (because any assertion $\mathcal{A}$
satisfies $\left[  \mathcal{A}\right]  \in\left\{  0,1\right\}  $). Hence,
$\left(  \left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in
T\right]  \right)  \in\left\{  0,1\right\}  ^{q}\subset\mathbb{N}^{q}$. Thus,
the number $\left\vert \left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)  \right\vert $ is well-defined. The
definition of this number yields%
\begin{align*}
&  \left\vert \left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)  \right\vert \\
&  =\left[  1\in T\right]  +\left[  2\in T\right]  +\cdots+\left[  q\in
T\right]  =\sum_{i\in\left\{  1,2,\ldots,q\right\}  }\left[  i\in T\right] \\
&  =\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}}_{\substack{=\sum_{i\in T}\\\text{(since }T\text{ is a subset}\\\text{of
}\left\{  1,2,\ldots,q\right\}  \text{)}}}\underbrace{\left[  i\in T\right]
}_{\substack{=1\\\text{(since }i\in T\text{)}}}+\sum\limits_{\substack{i\in\left\{
1,2,\ldots,q\right\}  ;\\\text{not }i\in T}}\underbrace{\left[  i\in T\right]
}_{\substack{=0\\\text{(since we don't have }i\in T\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }i\in\left\{  1,2,\ldots,q\right\}  \text{ satisfies either
}i\in T\text{ or }\left(  \text{not }i\in T\right) \\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum_{i\in T}1+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,q\right\}  ;\\\text{not }i\in T}}0}_{=0}=\sum_{i\in T}1=\left\vert
T\right\vert \cdot1=\left\vert T\right\vert =m
\end{align*}
(since $T$ is an $m$-element set).
\par
Thus, $\left(  \left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[
q\in T\right]  \right)  $ is a $\beta\in\left\{  0,1\right\}  ^{q}$ satisfying
$\left\vert \beta\right\vert =m$ (since $\left(  \left[  1\in T\right]
,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]  \right)  \in\left\{
0,1\right\}  ^{q}$ and $\left\vert \left(  \left[  1\in T\right]  ,\left[
2\in T\right]  ,\ldots,\left[  q\in T\right]  \right)  \right\vert =m$). In
other words,%
\[
\left(  \left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in
T\right]  \right)  \in\left\{  \beta\in\left\{  0,1\right\}  ^{q}%
\ \mid\ \left\vert \beta\right\vert =m\right\}  .
\]
In other words, $\left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)  \in\mathfrak{K}$ (since $\mathfrak{K}%
=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert \beta
\right\vert =m\right\}  $). Qed.}. Hence, the map%
\[
\mathfrak{P}\rightarrow\mathfrak{K},\ \ \ \ \ \ \ \ \ \ T\mapsto\left(
\left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]
\right)
\]
is well-defined. Let us denote this map by $B$.

If $T$ is a finite set of integers, then the \emph{increasing list} of $T$
shall mean the list of all elements of $T$ in increasing order (with no
repetitions). It is well-known that the increasing list of $T$ always exists
and is unique, and has exactly $\left\vert T\right\vert $ entries. For
example, the increasing list of the set $\left\{  2,6,4\right\}  $ is $\left(
2,4,6\right)  $.

Let us first observe the following fact: If $j\in\mathfrak{I}$ is arbitrary,
then
\begin{equation}
j\text{ is the increasing list of the set }A\left(  j\right)
\label{pf.lem.sol.Lambda.triangular.IPK.inclist}%
\end{equation}
\footnote{\textit{Proof.} Let $j\in\mathfrak{J}$. We shall show that $j$ is
the increasing list of the set $A\left(  j\right)  $.
\par
We have
\begin{align*}
j  &  \in\mathfrak{I}=\left\{  \left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}\ \mid\ i_{1}<i_{2}<\cdots<i_{m}\right\}
\\
&  =\left\{  \left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m}\ \mid\ j_{1}<j_{2}<\cdots<j_{m}\right\}
\end{align*}
(here, we have renamed the index $\left(  i_{1},i_{2},\ldots,i_{m}\right)  $
as $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $). In other words, $j$ can be
written in the form $j=\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ for some
$\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\left\{  1,2,\ldots,q\right\}
^{m}$ satisfying $j_{1}<j_{2}<\cdots<j_{m}$.
Consider this $\left(  j_{1},j_{2},\ldots,j_{m}\right)$.
Applying the map $A$ to both
sides of the equality $j=\left(  j_{1},j_{2},\ldots,j_{m}\right)  $, we obtain
$A\left(  j\right)  =A\left(  j_{1},j_{2},\ldots,j_{m}\right)  =\left\{
j_{1},j_{2},\ldots,j_{m}\right\}  $ (by the definition of $A$).
\par
The list $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is a list of all elements
of the set $A\left(  j\right)  $ (since $A\left(  j\right)  =\left\{
j_{1},j_{2},\ldots,j_{m}\right\}  $) in increasing order (since $j_{1}%
<j_{2}<\cdots<j_{m}$) with no repetitions (since $j_{1}<j_{2}<\cdots<j_{m}$).
In other words, the list $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is the
increasing list of $A\left(  j\right)  $ (by the definition of
``increasing list''). In other words, the list
$j$ is the increasing list of $A\left(  j\right)  $ (since $j=\left(
j_{1},j_{2},\ldots,j_{m}\right)  $). This proves
(\ref{pf.lem.sol.Lambda.triangular.IPK.inclist}).}.

The map $A$ is injective\footnote{\textit{Proof.} Let $j$ and $k$ be two
elements of $\mathfrak{I}$ satisfying $A\left(  j\right)  =A\left(  k\right)
$. We shall show that $j=k$.
\par
From (\ref{pf.lem.sol.Lambda.triangular.IPK.inclist}), we conclude that $j$ is
the increasing list of the set $A\left(  j\right)  $. In other words,
$j=\left(  \text{the increasing list of }A\left(  j\right)  \right)  $. The
same argument (applied to $k$ instead of $j$) shows that $k=\left(
\text{the increasing list of }A\left(  k\right)  \right)  $. Hence,%
\[
j=\left(  \text{the increasing list of }\underbrace{A\left(  j\right)
}_{=A\left(  k\right)  }\right)  =\left(  \text{the increasing list of
}A\left(  k\right)  \right)  =k.
\]
\par
Now, forget that we fixed $j$ and $k$. We thus have shown that if $j$ and $k$
are two elements of $\mathfrak{I}$ satisfying $A\left(  j\right)  =A\left(
k\right)  $, then $j=k$. In other words, the map $A$ is injective.} and
surjective\footnote{\textit{Proof.} Let $T\in\mathfrak{P}$ be arbitrary. We
shall show that $T\in A\left(  \mathfrak{I}\right)  $.
\par
We know that $T$ belongs to $\mathfrak{P}$. In other words, $T$ belongs to the
set of all $m$-element subsets of $\left\{  1,2,\ldots,q\right\}  $ (since
$\mathfrak{P}$ is the set of all $m$-element subsets of $\left\{
1,2,\ldots,q\right\}  $). In other words, $T$ is an $m$-element subset of
$\left\{  1,2,\ldots,q\right\}  $. Thus, $\left\vert T\right\vert =m$ (since
$T$ is an $m$-element set).
\par
Let $j$ be the increasing list of $T$. Then, the list $j$ has exactly
$\left\vert T\right\vert $ entries. In other words, the list $j$ has exactly
$m$ entries (since $\left\vert T\right\vert =m$). Thus, we can write $j$ in
the form $j=\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ for some $m$ objects
$j_{1},j_{2},\ldots,j_{m}$. Consider these $j_{1},j_{2},\ldots,j_{m}$.
\par
The list $j$ is the increasing list of $T$. In other words, the list $\left(
j_{1},j_{2},\ldots,j_{m}\right)  $ is the increasing list of $T$ (since
$j=\left(  j_{1},j_{2},\ldots,j_{m}\right)  $). In other words, the list
$\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is the list of all elements of $T$
in increasing order (with no repetitions) (since the increasing list of $T$ is
the list of all elements of $T$ in increasing order (with no repetitions) (by
the definition of ``increasing list'')). Thus,
the list $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is a list of all elements
of $T$. In other words, $T=\left\{  j_{1},j_{2},\ldots,j_{m}\right\}  $.
Furthermore, the list $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is strictly
increasing (since it is the list of all elements of $T$ in increasing order
(with no repetitions)). In other words, $j_{1}<j_{2}<\cdots<j_{m}$.
\par
Each $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $j_{i}\in\left\{
j_{1},j_{2},\ldots,j_{m}\right\}  =T\subset\left\{  1,2,\ldots,q\right\}  $
(since $T$ is a subset of $\left\{  1,2,\ldots,q\right\}  $). Hence, $\left(
j_{1},j_{2},\ldots,j_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^m$.
\par
Now, the list $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ belongs to $\left\{
1,2,\ldots,q\right\}  ^m$ (since $\left(  j_{1},j_{2},\ldots,j_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^m$) and satisfies $j_{1}<j_{2}%
<\cdots<j_{m}$. In other words, $\left(  j_{1},j_{2},\ldots,j_{m}\right)  $ is
an $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}
^{m}$ satisfying $i_{1}<i_{2}<\cdots<i_{m}$. In other words,%
\[
\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\left\{  \left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}%
\ \mid\ i_{1}<i_{2}<\cdots<i_{m}\right\}  .
\]
This rewrites as $\left(  j_{1},j_{2},\ldots,j_{m}\right)  \in\mathfrak{I}$
(since $\mathfrak{I}=\left\{  \left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}\ \mid\ i_{1}<i_{2}<\cdots
<i_{m}\right\}  $).
\par
The definition of $A$ yields $A\left(  j_{1},j_{2},\ldots,j_{m}\right)
=\left\{  j_{1},j_{2},\ldots,j_{m}\right\}  =T$. Hence,
$T=A\underbrace{\left(  j_{1},j_{2},\ldots,j_{m}\right)  }_{\in\mathfrak{I}%
}\in A\left(  \mathfrak{I}\right)  $.
\par
Now, forget that we fixed $T$. We thus have proven that each $T\in
\mathfrak{P}$ satisfies $T\in A\left(  \mathfrak{I}\right)  $. In other words,
$\mathfrak{P}\subset A\left(  \mathfrak{I}\right)  $. In other words, the map
$A$ is surjective.}. Hence, the map $A$ is bijective. In other words, the map%
\[
\mathfrak{I}\rightarrow\mathfrak{P},\ \ \ \ \ \ \ \ \ \ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
\]
is bijective (since the map $A$ is the map
\[
\mathfrak{I}\rightarrow\mathfrak{P},\ \ \ \ \ \ \ \ \ \ \left(  j_{1}%
,j_{2},\ldots,j_{m}\right)  \mapsto\left\{  j_{1},j_{2},\ldots,j_{m}\right\}
\]
). This completes the proof of Lemma \ref{lem.sol.Lambda.triangular.IPK}(a)
(since we already know that this map is well-defined).

The map $B$ is injective\footnote{\textit{Proof.} Let $T_{1}$ and $T_{2}$ be
two elements of $\mathfrak{P}$ satisfying $B\left(  T_{1}\right)  =B\left(
T_{2}\right)  $. We shall prove that $T_{1}=T_{2}$.
\par
The definition of $B$ yields $B\left(  T_{1}\right)  =\left(  \left[  1\in
T_{1}\right]  ,\left[  2\in T_{1}\right]  ,\ldots,\left[  q\in T_{1}\right]
\right)  $. Hence,%
\[
\left(  \left[  1\in T_{1}\right]  ,\left[  2\in T_{1}\right]  ,\ldots,\left[
q\in T_{1}\right]  \right)  =B\left(  T_{1}\right)  =B\left(  T_{2}\right)
=\left(  \left[  1\in T_{2}\right]  ,\left[  2\in T_{2}\right]  ,\ldots
,\left[  q\in T_{2}\right]  \right)
\]
(by the definition of $B$). In other words, we have%
\begin{equation}
\left[  i\in T_{1}\right]  =\left[  i\in T_{2}\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.IPK.b.inj.1}%
\end{equation}
\par
Now, let $i\in T_{1}$. Thus, $\left[  i\in T_{1}\right]  =1$. But
(\ref{pf.lem.sol.Lambda.triangular.IPK.b.inj.1}) yields $\left[  i\in
T_{2}\right]  =\left[  i\in T_{1}\right]  =1\neq0$. Thus, we have $i\in T_{2}$
(because otherwise, we would have $\left(  \text{not }i\in T_{2}\right)  $ and
therefore $\left[  i\in T_{2}\right]  =0$, which would contradict $\left[
i\in T_{2}\right]  \neq0$).
\par
Now, forget that we fixed $i$. We thus have proven that $i\in T_{2}$ for each
$i\in T_{1}$. In other words, $T_{1}\subset T_{2}$. The same argument (applied
to $T_{2}$ and $T_{1}$ instead of $T_{1}$ and $T_{2}$) shows that
$T_{2}\subset T_{1}$. Combining $T_{1}\subset T_{2}$ with $T_{2}\subset T_{1}%
$, we obtain $T_{1}=T_{2}$.
\par
Now, forget that we fixed $T_{1}$ and $T_{2}$. We thus have shown that if
$T_{1}$ and $T_{2}$ are two elements of $\mathfrak{P}$ satisfying $B\left(
T_{1}\right)  =B\left(  T_{2}\right)  $, then $T_{1}=T_{2}$. In other words,
the map $B$ is injective.} and surjective\footnote{\textit{Proof.} Let
$\gamma\in\mathfrak{K}$. Thus, $\gamma\in\mathfrak{K}=\left\{  \beta
\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert \beta\right\vert =m\right\}
$. In other words, $\gamma$ is a $\beta\in\left\{  0,1\right\}  ^{q}$
satisfying $\left\vert \beta\right\vert =m$. In other words, $\gamma$ is an
element of $\left\{  0,1\right\}  ^{q}$ and satisfies $\left\vert
\gamma\right\vert =m$.
\par
We know that $\gamma$ is an element of $\left\{  0,1\right\}  ^{q}$. Thus, we
can write $\gamma$ in the form $\gamma=\left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{q}\right)  $, where $\gamma_{1},\gamma_{2},\ldots,\gamma_{q}$
are elements of $\left\{  0,1\right\}  $. Consider these $\gamma_{1}%
,\gamma_{2},\ldots,\gamma_{q}$.
Thus, $\left\vert \gamma\right\vert =\gamma_{1}+\gamma_{2}+\cdots+\gamma
_{q}$ (by the definition of $\left\vert \gamma\right\vert$).
\par
Each $i\in\left\{  1,2,\ldots,q\right\}  $ satisfying $\gamma_{i}\neq1$ must
satisfy%
\begin{equation}
\gamma_{i}=0. \label{pf.lem.sol.Lambda.triangular.IPK.b.surj.1}%
\end{equation}
\par
[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.1}):} Let
$i\in\left\{  1,2,\ldots,q\right\}  $ be such that $\gamma_{i}\neq1$. Then,
$\gamma_{i}\in\left\{  0,1\right\}  $ (since $\gamma_{1},\gamma_{2}%
,\ldots,\gamma_{q}$ are elements of $\left\{  0,1\right\}  $). Combining this
with $\gamma_{i}\neq1$, we obtain $\gamma_{i}\in\left\{  0,1\right\}
\setminus\left\{  1\right\}  =\left\{  0\right\}  $. Thus, $\gamma_{i}=0$.
This proves (\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.1}).]
\par
Define a subset $T$ of $\left\{  1,2,\ldots,q\right\}  $ by $T=\left\{
i\in\left\{  1,2,\ldots,q\right\}  \ \mid\ \gamma_{i}=1\right\}  $. Then,%
\begin{align*}
\sum_{i\in\left\{  1,2,\ldots,q\right\}  }\gamma_{i}  &  =\sum\limits_{\substack{i\in
\left\{  1,2,\ldots,q\right\}  ;\\\gamma_{i}=1}}\underbrace{\gamma_{i}}%
_{=1}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\\gamma_{i}\neq
1}}\underbrace{\gamma_{i}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }i\in\left\{  1,2,\ldots,q\right\}  \text{ satisfies either
}\gamma_{i}=1\text{ or }\gamma_{i}\neq1\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\\gamma_{i}%
=1}}1+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}
;\\\gamma_{i}\neq1}}0}_{=0}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}
;\\\gamma_{i}=1}}1\\
&  =\left\vert \underbrace{\left\{  i\in\left\{  1,2,\ldots,q\right\}
\ \mid\ \gamma_{i}=1\right\}  }_{=T}\right\vert \cdot1=\left\vert T\right\vert
\cdot1=\left\vert T\right\vert .
\end{align*}
But%
\begin{align*}
m  &  =\left\vert \gamma\right\vert =\gamma_{1}+\gamma_{2}+\cdots+\gamma
_{q} =\sum_{i\in\left\{  1,2,\ldots,q\right\}  }\gamma_{i}=\left\vert
T\right\vert .
\end{align*}
In other words, $\left\vert T\right\vert =m$. Thus, the set $T$ is an
$m$-element set. Hence, we know that $T$ is an $m$-element subset of $\left\{
1,2,\ldots,q\right\}  $. In other words, $T$ belongs to the set of all
$m$-element subsets of $\left\{  1,2,\ldots,q\right\}  $. In other words, $T$
belongs to $\mathfrak{P}$ (since $\mathfrak{P}$ is the set of all $m$-element
subsets of $\left\{  1,2,\ldots,q\right\}  $). In other words, $T\in
\mathfrak{P}$.
\par
We have $T=\left\{  i\in\left\{  1,2,\ldots,q\right\}  \ \mid\ \gamma
_{i}=1\right\}  $. Hence, for any $i\in\left\{  1,2,\ldots,q\right\}  $, we
have the following logical equivalence:%
\begin{equation}
\left(  i\in T\right)  \Longleftrightarrow\left(  \gamma_{i}=1\right)  .
\label{pf.lem.sol.Lambda.triangular.IPK.b.surj.3}%
\end{equation}
Each $i\in\left\{  1,2,\ldots,q\right\}  $ satisfies%
\begin{equation}
\left[  i\in T\right]  =\gamma_{i}.
\label{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}%
\end{equation}
\par
[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}):} Let
$i\in\left\{  1,2,\ldots,q\right\}  $. Then, $\gamma_{i}\in\left\{
0,1\right\}  $ (since $\gamma_{1},\gamma_{2},\ldots,\gamma_{q}$ are elements
of $\left\{  0,1\right\}  $). We are thus in one of the following two cases:
\par
\textit{Case 1:} We have $\gamma_{i}=0$.
\par
\textit{Case 2:} We have $\gamma_{i}=1$.
\par
Let us first consider Case 1. In this case, we have $\gamma_{i}=0$. Thus,
$\gamma_{i}\neq1$. In other words, we don't have $\gamma_{i}=1$. Hence, we
don't have $i\in T$ (because we have $i\in T$ if and only if $\gamma_{i}=1$
(by (\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.3}))). Thus, $\left[  i\in
T\right]  =0$. Comparing this with $\gamma_{i}=0$, we obtain $\left[  i\in
T\right]  =\gamma_{i}$. Hence,
(\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $\gamma_{i}=1$. Hence, $i\in
T$ (because we have $i\in T$ if and only if $\gamma_{i}=1$ (by
(\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.3}))). Thus, $\left[  i\in
T\right]  =1$. Comparing this with $\gamma_{i}=1$, we obtain $\left[  i\in
T\right]  =\gamma_{i}$. Hence,
(\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}) is proven in Case 2.
\par
We have now proven (\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}) in each
of the two Cases 1 and 2. Since these two Cases cover all possibilities, we
thus conclude that (\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}) always
holds.]
\par
The definition of $B$ yields
\begin{align*}
B\left(  T\right)   &  =\left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)  =\left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{q}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }i\in\left\{  1,2,\ldots
,q\right\}  \text{ satisfies }\left[  i\in T\right]  =\gamma_{i}\text{ (by
(\ref{pf.lem.sol.Lambda.triangular.IPK.b.surj.2}))}\right) \\
&  =\gamma.
\end{align*}
Hence, $\gamma=B\left(  \underbrace{T}_{\in\mathfrak{P}}\right)  \in B\left(
\mathfrak{P}\right)  $.
\par
Now, forget that we fixed $\gamma$. We thus have shown that each $\gamma
\in\mathfrak{K}$ satisfies $\gamma\in B\left(  \mathfrak{P}\right)  $. In
other words, $\mathfrak{K}\subset B\left(  \mathfrak{P}\right)  $. In other
words, the map $B$ is surjective.}. Hence, the map $B$ is bijective. In other
words, the map%
\[
\mathfrak{P}\rightarrow\mathfrak{K},\ \ \ \ \ \ \ \ \ \ T\mapsto\left(
\left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]
\right)
\]
is bijective (since the map $B$ is the map
\[
\mathfrak{P}\rightarrow\mathfrak{K},\ \ \ \ \ \ \ \ \ \ T\mapsto\left(
\left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[  q\in T\right]
\right)
\]
). This completes the proof of Lemma \ref{lem.sol.Lambda.triangular.IPK}(b)
(since we already know that this map is well-defined).

(c) We have shown that the maps $B$ and $A$ are bijective. Hence, their
composition $B\circ A$ is also bijective. Thus, $B\circ A:\mathfrak{I}%
\rightarrow\mathfrak{K}$ is a bijection. It has the property that every
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}$ satisfies
$\overline{\mathbf{x}}_{q}^{\left(  B\circ A\right)  \left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\mathfrak{I}$. We must prove that $\overline{\mathbf{x}}%
_{q}^{\left(  B\circ A\right)  \left(  i_{1},i_{2},\ldots,i_{m}\right)
}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}$.
\par
Define $T\in\mathfrak{P}$ by $T=A\left(  i_{1},i_{2},\ldots,i_{m}\right)  $.
Then, $T=A\left(  i_{1},i_{2},\ldots,i_{m}\right)  =\left\{  i_{1}%
,i_{2},\ldots,i_{m}\right\}  $ (by the definition of $A$).
\par
Applying (\ref{pf.lem.sol.Lambda.triangular.IPK.inclist}) to $j=\left(
i_{1},i_{2},\ldots,i_{m}\right)  $, we conclude that $\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  $ is the increasing list of the set $A\left(
i_{1},i_{2},\ldots,i_{m}\right)  $. In other words, $\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  $ is the increasing list of the set $T$ (since
$T=A\left(  i_{1},i_{2},\ldots,i_{m}\right)  $). In other words, $\left(
i_{1},i_{2},\ldots,i_{m}\right)  $ is the list of all elements of $T$ in
increasing order (with no repetitions) (because the increasing list of the set
$T$ is the list of all elements of $T$ in increasing order (with no
repetitions) (by the definition of ``increasing list'')).
Thus, in particular, $\left(  i_{1},i_{2}%
,\ldots,i_{m}\right)  $ is a list of all elements of $T$ with no repetitions.
Hence,%
\[
\prod_{i\in T}x_{i}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}.
\]
\par
We know that $T$ belongs to $\mathfrak{P}$. In other words, $T$ belongs to the
set of all $m$-element subsets of $\left\{  1,2,\ldots,q\right\}  $ (since
$\mathfrak{P}$ is the set of all $m$-element subsets of $\left\{
1,2,\ldots,q\right\}  $). In other words, $T$ is an $m$-element subset of
$\left\{  1,2,\ldots,q\right\}  $.
\par
Now,%
\[
\left(  B\circ A\right)  \left(  i_{1},i_{2},\ldots,i_{m}\right)  =B\left(
\underbrace{A\left(  i_{1},i_{2},\ldots,i_{m}\right)  }_{=T}\right)  =B\left(
T\right)  =\left(  \left[  1\in T\right]  ,\left[  2\in T\right]
,\ldots,\left[  q\in T\right]  \right)
\]
(by the definition of $B$). Hence,%
\begin{align*}
&  \overline{\mathbf{x}}_{q}^{\left(  B\circ A\right)  \left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  }\\
&  =\overline{\mathbf{x}}_{q}^{\left(  \left[  1\in T\right]  ,\left[  2\in
T\right]  ,\ldots,\left[  q\in T\right]  \right)  }=x_{1}^{\left[  1\in
T\right]  }x_{2}^{\left[  2\in T\right]  }\cdots x_{q}^{\left[  q\in T\right]
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{\xx%
}_{q}^{\left(  \left[  1\in T\right]  ,\left[  2\in T\right]  ,\ldots,\left[
q\in T\right]  \right)  }\right) \\
&  =\prod_{i\in\left\{  1,2,\ldots,q\right\}  }x_{i}^{\left[  i\in T\right]
}=\left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}\underbrace{x_{i}^{\left[  i\in T\right]  }}_{\substack{=x_{i}%
^{1}\\\text{(since }\left[  i\in T\right]  =1\\\text{(since }i\in T\text{))}%
}}\right)  \left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}
;\\\text{not }i\in T}}\underbrace{x_{i}^{\left[  i\in T\right]  }%
}_{\substack{=x_{i}^{0}\\\text{(since }\left[  i\in T\right]  =0\\\text{(since
we don't have }i\in T\text{))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }i\in\left\{  1,2,\ldots,q\right\}  \text{ satisfies either
}i\in T\text{ or }\left(  \text{not }i\in T\right) \\
\text{(but not both)}%
\end{array}
\right) \\
&  =\left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}\underbrace{x_{i}^{1}}_{=x_{i}}\right)  \left(  \prod\limits_{\substack{i\in
\left\{  1,2,\ldots,q\right\}  ;\\\text{not }i\in T}}\underbrace{x_{i}^{0}%
}_{=1}\right)  =\left(  \prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}
;\\i\in T}}x_{i}\right)  \underbrace{\left(  \prod\limits_{\substack{i\in\left\{
1,2,\ldots,q\right\}  ;\\\text{not }i\in T}}1\right)  }_{=1}\\
&  =\underbrace{\prod\limits_{\substack{i\in\left\{  1,2,\ldots,q\right\}  ;\\i\in
T}}}_{\substack{=\prod_{i\in T}\\\text{(since }T\text{ is a subset of
}\left\{  1,2,\ldots,q\right\}  \text{)}}}x_{i}=\prod_{i\in T}x_{i}=x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}.
\end{align*}
Qed.}. Hence, there exists a bijection $\Psi:\mathfrak{I}\rightarrow
\mathfrak{K}$ with the property that every $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\mathfrak{I}$ satisfies $\overline{\mathbf{x}}_{q}%
^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)  }=x_{i_{1}}x_{i_{2}}\cdots
x_{i_{m}}$ (namely, $\Psi=B\circ A$). This proves Lemma
\ref{lem.sol.Lambda.triangular.IPK}(c).
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.em}Let $q\in \NN$. Let $m\in\mathbb{N}%
$. Then,%
\[
\eta_{q}\left(  e_{m}\right)  =\sum\limits_{\substack{\beta\in\left\{  0,1\right\}
^{q};\\\left\vert \beta\right\vert =m}}\overline{\mathbf{x}}_{q}^{\beta}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.em}.]The definition of $e_{m}$
yields%
\begin{align*}
e_{m}  &  =\sum_{i_{1}<\cdots<i_{m}}x_{i_{1}}\cdots x_{i_{m}}=\underbrace{\sum
_{i_{1}<i_{2}<\cdots<i_{m}}}_{=\sum\limits_{\substack{\left(  i_{1},i_{2}%
,\ldots,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}%
<i_{2}<\cdots<i_{m}}}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}.
\end{align*}
Substituting $x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots$ for the variables
$x_{1},x_{2},x_{3},\ldots$ in this equality, we obtain%
\begin{align}
e_{m}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)   &  =\left(
\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)
\nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  . \label{pf.lem.sol.Lambda.triangular.em.1}%
\end{align}


Note that $\left\{  1,2,\ldots,q\right\}  ^{m}$ is a subset of $\left\{
1,2,3,\ldots\right\}  ^{m}$ (since $\left\{  1,2,\ldots,q\right\}  $ is a
subset of $\left\{  1,2,3,\ldots\right\}  $). Moreover, the following holds:

\begin{itemize}
\item If $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{m}$ is an $m$-tuple that does not satisfy $\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$, then
\begin{equation}
\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  =0
\label{pf.lem.sol.Lambda.triangular.em.2a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.em.2a}):} Let
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}
^{m}$ be an $m$-tuple that does not satisfy $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$.
\par
If every $k\in\left\{  1,2,\ldots,m\right\}  $ would satisfy $i_{k}\in\left\{
1,2,\ldots,q\right\}  $, then we would have $\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$, which would contradict
the fact that we do not have $\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}$. Hence, not every $k\in\left\{
1,2,\ldots,m\right\}  $ satisfies $i_{k}\in\left\{  1,2,\ldots,q\right\}  $.
In other words, there exists some $k\in\left\{  1,2,\ldots,m\right\}  $
satisfying $i_{k}\notin\left\{  1,2,\ldots,q\right\}  $. Consider this $k$.
\par
The elements $i_{1},i_{2},\ldots,i_{m}$ belong to $\left\{  1,2,3,\ldots
\right\}  $ (since $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m}$). Hence, in particular, $i_{k}\in\left\{
1,2,3,\ldots\right\}  $. Combining this with $i_{k}\notin\left\{
1,2,\ldots,q\right\}  $, we obtain $i_{k}\in\left\{  1,2,3,\ldots\right\}
\setminus\left\{  1,2,\ldots,q\right\}  =\left\{  q+1,q+2,q+3,\ldots\right\}
$. Hence, (\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta2}) (applied to
$i=i_{k}$) yields $\eta_{q}\left(  x_{i_{k}}\right)  =0$.
\par
Now,
\begin{align*}
&  \underbrace{\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  }%
_{=\prod_{j\in\left\{  1,2,\ldots,m\right\}  }x_{i_{j}}}\left(  x_{1}%
,x_{2},\ldots,x_{q},0,0,0,\ldots\right) \\
&  =\left(  \prod_{j\in\left\{  1,2,\ldots,m\right\}  }x_{i_{j}}\right)
\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  =\prod_{j\in\left\{
1,2,\ldots,m\right\}  }\underbrace{x_{i_{j}}\left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  }_{\substack{=\eta_{q}\left(  x_{i_{j}}\right)
\\\text{(since }\eta_{q}\left(  x_{i_{j}}\right)  =x_{i_{j}}\left(
x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  \\\text{(by the definition of
}\eta_{q}\text{))}}}\\
&  =\prod_{j\in\left\{  1,2,\ldots,m\right\}  }\eta_{q}\left(  x_{i_{j}%
}\right)  =\underbrace{\eta_{q}\left(  x_{i_{k}}\right)  }_{=0}\cdot
\prod\limits_{\substack{j\in\left\{  1,2,\ldots,m\right\}  ;\\j\neq k}}\eta
_{q}\left(  x_{i_{j}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the factor for }j=k\text{ from the product}\\
\text{(since }k\in\left\{  1,2,\ldots,m\right\}  \text{)}%
\end{array}
\right) \\
&  =0\cdot\prod\limits_{\substack{j\in\left\{  1,2,\ldots,m\right\}  ;\\j\neq k}%
}\eta_{q}\left(  x_{i_{j}}\right)  =0.
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.em.2a}).}.

\item If $\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots
,q\right\}  ^{m}$, then%
\begin{equation}
\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2}%
,\ldots,x_{q},0,0,0,\ldots\right)  =x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}
\label{pf.lem.sol.Lambda.triangular.em.2b}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.em.2b}):} Let
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}
^{m}$.
\par
Let $k\in\left\{  1,2,\ldots,m\right\}  $. The elements $i_{1},i_{2}%
,\ldots,i_{m}$ belong to $\left\{  1,2,\ldots,q\right\}  $ (since $\left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}$).
Hence, in particular, $i_{k}\in\left\{  1,2,\ldots,q\right\}  $. Hence,
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.eta1}) (applied to $i=i_{k}$)
yields $\eta_{q}\left(  x_{i_{k}}\right)  =x_{i_{k}}$.
\par
Let us forget that we fixed $k$. We thus have shown that $\eta_{q}\left(
x_{i_{k}}\right)  =x_{i_{k}}$ for each $k\in\left\{  1,2,\ldots,m\right\}  $.
Hence, $\prod_{k=1}^{m}\underbrace{\eta_{q}\left(  x_{i_{k}}\right)
}_{=x_{i_{k}}}=\prod_{k=1}^{m}x_{i_{k}}$. Now,
\begin{align*}
&  \underbrace{\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  }%
_{=\prod_{k=1}^{m}x_{i_{k}}}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots
\right) \\
&  =\left(  \prod_{k=1}^{m}x_{i_{k}}\right)  \left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  =\prod_{k=1}^{m}\underbrace{x_{i_{k}}\left(
x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  }_{\substack{=\eta_{q}\left(
x_{i_{k}}\right)  \\\text{(since }\eta_{q}\left(  x_{i_{k}}\right)  =x_{i_{k}%
}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right)  \\\text{(by the
definition of }\eta_{q}\text{))}}}\\
&  =\prod_{k=1}^{m}\eta_{q}\left(  x_{i_{k}}\right)  =\prod_{k=1}^{m}x_{i_{k}%
}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}.
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.em.2b}).}.
\end{itemize}

Now, (\ref{pf.lem.sol.Lambda.triangular.em.1}) becomes%
\begin{align}
&  e_{m}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m};\\\left(
i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}}%
}}_{\substack{=\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{m};\\\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}%
<\cdots<i_{m}}}\\=\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}%
}}\\\text{(since }\left\{  1,2,\ldots,q\right\}  ^{m}\text{ is a subset
of}\\\left\{  1,2,3,\ldots\right\}  ^{m}\text{)}}}\underbrace{\left(
x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots
,x_{q},0,0,0,\ldots\right)  }_{\substack{=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}%
}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.em.2b}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}%
<\cdots<i_{m};\\\text{not }\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}}}\underbrace{\left(  x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}\right)  \left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots
\right)  }_{\substack{=0\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.em.2a}%
))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{m}\text{ satisfies}\\
\text{either }\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m}\text{ or }\left(  \text{not }\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}\right) \\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}+\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{m};\\i_{1}<i_{2}%
<\cdots<i_{m};\\\text{not }\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m}}}0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\left\{
1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{m}}. \label{pf.lem.sol.Lambda.triangular.em.4}%
\end{align}


Now, let us define the sets $\mathfrak{I}$, $\mathfrak{P}$ and $\mathfrak{K}$
as in Lemma \ref{lem.sol.Lambda.triangular.IPK}. Then, Lemma
\ref{lem.sol.Lambda.triangular.IPK}(c) shows that there exists a bijection
$\Psi:\mathfrak{I}\rightarrow\mathfrak{K}$ with the property that every
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}$ satisfies
$\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)
}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}$. Consider this $\Psi$. Thus, every
$\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}$ satisfies%
\begin{equation}
\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)
}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}.
\label{pf.lem.sol.Lambda.triangular.em.Psi}%
\end{equation}


Now, (\ref{pf.lem.sol.Lambda.triangular.em.4}) becomes%
\begin{align*}
&  e_{m}\left(  x_{1},x_{2},\ldots,x_{q},0,0,0,\ldots\right) \\
&  =\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{m}\right)
\in\left\{  1,2,\ldots,q\right\}  ^{m};\\i_{1}<i_{2}<\cdots<i_{m}}%
}}_{\substack{=\sum_{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}%
}\\\text{(since }\mathfrak{I}=\left\{  \left(  i_{1},i_{2},\ldots
,i_{m}\right)  \in\left\{  1,2,\ldots,q\right\}  ^{m}\ \mid\ i_{1}%
<i_{2}<\cdots<i_{m}\right\}  \text{)}}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}%
=\sum_{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}}%
\underbrace{x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}}_{\substack{=
\overline{\xx}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)  }\\
\text{(by (\ref{pf.lem.sol.Lambda.triangular.em.Psi}))}}}\\
&  =\sum_{\left(  i_{1},i_{2},\ldots,i_{m}\right)  \in\mathfrak{I}}%
\overline{\mathbf{x}}_{q}^{\Psi\left(  i_{1},i_{2},\ldots,i_{m}\right)
}=\underbrace{\sum_{\beta\in\mathfrak{K}}}_{\substack{=\sum\limits_{\substack{\beta
\in\left\{  0,1\right\}  ^{q};\\\left\vert \beta\right\vert =m}}\\\text{(since
}\mathfrak{K}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  \text{)}}}\overline{\mathbf{x}}_{q}^{\beta}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\beta\text{ for }\Psi\left(  i_{1}%
,i_{2},\ldots,i_{m}\right)  \text{ in the sum,}\\
\text{since the map }\Psi:\mathfrak{I}\rightarrow\mathfrak{K}\text{ is a
bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\beta\in\left\{  0,1\right\}  ^{q};\\\left\vert
\beta\right\vert =m}}\overline{\mathbf{x}}_{q}^{\beta}.
\end{align*}
Now, the definition of $\eta_{q}$ yields%
\[
\eta_{q}\left(  e_{m}\right)  =e_{m}\left(  x_{1},x_{2},\ldots,x_{q}%
,0,0,0,\ldots\right)  =\sum\limits_{\substack{\beta\in\left\{  0,1\right\}
^{q};\\\left\vert \beta\right\vert =m}}\overline{\mathbf{x}}_{q}^{\beta}.
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.em}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.g-set}Let $X$ and $Y$ be two sets. Let
$\phi:X\rightarrow Y$ be a bijection. Let $Z$ be a subset of $X$. Then, the
map $Z\rightarrow\phi\left(  Z\right)  ,\ A\mapsto\phi\left(  A\right)  $ is
well-defined and is a bijection.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.g-set}.]Lemma
\ref{lem.sol.Lambda.triangular.g-set} is a fundamental and trivial fact from
set theory; we omit its proof.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.ela}Let $q\in \NN$. Let $p\in
\mathbb{N}$. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{p}\right)  \in\mathbb{N}^{p}$ be a $p$-tuple of nonnegative integers. Then,%
\[
\eta_{q}\left(  e_{\alpha_{1}}e_{\alpha_{2}}\cdots e_{\alpha_{p}}\right)
=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\alpha}}
\overline{\xx}_{q}^{\operatorname*{colsums}A}.
\]
Here, $\operatorname*{colsums}A$ denotes the column sums of $A$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.ela}.]For each matrix
$A\in\left\{  0,1\right\}  ^{p\times q}$ and each $i\in\left\{  1,2,\ldots
,p\right\}  $, we let $\operatorname*{row}\nolimits_{i}A$ denote the $i$-th
row of $A$. This $\operatorname{row}_{i}A$ is an element of $\left\{
0,1\right\}  ^{q}$.

Let $\Phi:\left\{  0,1\right\}  ^{p\times q}\rightarrow\left(  \left\{
0,1\right\}  ^{q}\right)  ^{p}$ be the map that sends each matrix
$A\in\left\{  0,1\right\}  ^{p\times q}$ to the list \newline
$\left(
\operatorname*{row}\nolimits_{1}A,\operatorname{row}_{2}A,\ldots
,\operatorname{row}_{p}A\right)  $ of all the rows of $A$. Then, the map
$\Phi$ is a bijection (because a matrix can be viewed as a list of rows).

For each matrix $A\in\left\{  0,1\right\}  ^{p\times q}$, we let
$\operatorname*{rowsums}A$ denote the row sums of $A$. (This is a $p$-tuple in
$\mathbb{N}^{p}$.) Furthermore, for each matrix $A\in\left\{  0,1\right\}
^{p\times q}$, we let $\operatorname*{colsums}A$ denote the column sums of
$A$. (This is a $q$-tuple in $\mathbb{N}^{q}$.)

Each matrix $A\in\left\{  0,1\right\}  ^{p\times q}$ satisfies%
\begin{equation}
\operatorname*{rowsums}A=\left(  \left\vert \operatorname{row}_{1}A\right\vert
,\left\vert \operatorname{row}_{2}A\right\vert ,\ldots,\left\vert
\operatorname{row}_{p}A\right\vert \right)
\label{pf.lem.sol.Lambda.triangular.ela.rowsums}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.rowsums}):}
Let $A\in\left\{  0,1\right\}  ^{p\times q}$ be a matrix.
\par
For each $i\in\left\{  1,2,\ldots,p\right\}  $, we let $r_{i}$ be the sum of
all entries in the $i$-th row of $A$. Then, the row sums of $A$ is the
$p$-tuple $\left(  r_{1},r_{2},\ldots,r_{p}\right)  $ (by the definition of
``row sums'').
\par
Now, let $i\in\left\{  1,2,\ldots,p\right\}  $. Recall that
$\operatorname{row}_{i}A$ denotes the $i$-th row of $A$; this $i$-th row is an
element of $\left\{  0,1\right\}  ^{q}$. Thus, $\operatorname{row}_{i}%
A\in\left\{  0,1\right\}  ^{q}$. Hence, we can write $\operatorname{row}%
_{i}A\in\left\{  0,1\right\}  ^{q}$ in the form $\operatorname{row}%
_{i}A=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  $ for some elements
$\beta_{1},\beta_{2},\ldots,\beta_{q}$ of $\left\{  0,1\right\}  $. Consider
these $\beta_{1},\beta_{2},\ldots,\beta_{q}$.
\par
We have $\operatorname{row}_{i}A=\left(  \beta_{1},\beta_{2},\ldots,\beta
_{q}\right)  $. Thus, the definition of $\left\vert \operatorname{row}%
_{i}A\right\vert $ yields $\left\vert \operatorname{row}_{i}A\right\vert
=\beta_{1}+\beta_{2}+\cdots+\beta_{q}$.
\par
But $r_{i}$ is the sum of all entries in the $i$-th row of $A$ (by the
definition of $r_{i}$). Hence,%
\begin{align*}
r_{i}  &  =\left(  \text{the sum of all entries in the }i\text{-th row of
}A\right) \\
&  =\left(  \text{the sum of all entries of }\underbrace{\text{the }i\text{-th
row of }A}_{\substack{=\operatorname{row}_{i}A\\\text{(since the }i\text{-th
row of }A\text{ is }\operatorname{row}_{i}A\text{)}}}\right) \\
&  =\left(  \text{the sum of all entries of }\underbrace{\operatorname{row}%
_{i}A}_{=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  }\right)
=\left(  \text{the sum of all entries of }\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{q}\right)  \right) \\
&  =\beta_{1}+\beta_{2}+\cdots+\beta_{q}=\left\vert \operatorname{row}%
_{i}A\right\vert .
\end{align*}
\par
Now, forget that we fixed $i$. We thus have shown that $r_{i}=\left\vert
\operatorname{row}_{i}A\right\vert $ for each $i\in\left\{  1,2,\ldots
,p\right\}  $. In other words, $\left(  r_{1},r_{2},\ldots,r_{p}\right)
=\left(  \left\vert \operatorname{row}_{1}A\right\vert ,\left\vert
\operatorname{row}_{2}A\right\vert ,\ldots,\left\vert \operatorname{row}%
_{p}A\right\vert \right)  $.
\par
Now, recall that $\operatorname*{rowsums}A$ is the row sums of $A$ (by the
definition of $\operatorname*{rowsums}A$). Hence,%
\begin{align*}
\operatorname*{rowsums}A  &  =\left(  \text{the row sums of }A\right)
=\left(  r_{1},r_{2},\ldots,r_{p}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the row sums of }A\text{ is the
}p\text{-tuple }\left(  r_{1},r_{2},\ldots,r_{p}\right)  \right) \\
&  =\left(  \left\vert \operatorname{row}_{1}A\right\vert ,\left\vert
\operatorname{row}_{2}A\right\vert ,\ldots,\left\vert \operatorname{row}%
_{p}A\right\vert \right)  .
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.rowsums}).}.

Furthermore, every matrix $A\in\left\{  0,1\right\}  ^{p\times q}$ satisfies%
\begin{equation}
\prod_{i=1}^{p}\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}%
=\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}A}
\label{pf.lem.sol.Lambda.triangular.ela.prodxA}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.prodxA}):}
Let $A\in\left\{  0,1\right\}  ^{p\times q}$ be a matrix.
\par
Write the $p\times q$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}$. Thus, each $i\in\left\{  1,2,\ldots
,p\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th row of }A\right)  =\left(  a_{i,1}%
,a_{1,2},\ldots,a_{i,q}\right)  .
\label{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.1}%
\end{equation}
Moreover, each $j\in\left\{  1,2,\ldots,q\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }j\text{-th column of }A\right)  =\left(
\begin{array}
[c]{c}%
a_{1,j}\\
a_{2,j}\\
\vdots\\
a_{p,j}%
\end{array}
\right)  \label{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.2}%
\end{equation}
(again because $A=\left(  a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$).
\par
Let $i\in\left\{  1,2,\ldots,p\right\}  $. Then, $\operatorname{row}_{i}A$ is
the $i$-th row of $A$ (by the definition of $\operatorname{row}_{i}A$). Thu,%
\[
\operatorname{row}_{i}A=\left(  \text{the }i\text{-th row of }A\right)
=\left(  a_{i,1},a_{1,2},\ldots,a_{i,q}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.1})}\right)  .
\]
Hence,%
\begin{align*}
\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}  &  =\overline{\xx%
}_{q}^{\left(  a_{i,1},a_{1,2},\ldots,a_{i,q}\right)  }=x_{1}^{a_{i,1}}%
x_{2}^{a_{i,2}}\cdots x_{q}^{a_{i,q}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\overline{\mathbf{x}}_{q}^{\left(  a_{i,1},a_{1,2}%
,\ldots,a_{i,q}\right)  }\right) \\
&  =\prod_{j=1}^{q}x_{j}^{a_{i,j}}.
\end{align*}
\par
Now, forget that we fixed $i$. We thus have proven $\overline{\mathbf{x}}%
_{q}^{\operatorname{row}_{i}A}=\prod_{j=1}^{q}x_{j}^{a_{i,j}}$ for each
$i\in\left\{  1,2,\ldots,p\right\}  $. Hence,%
\begin{equation}
\prod_{i=1}^{p}\underbrace{\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}%
A}}_{=\prod_{j=1}^{q}x_{j}^{a_{i,j}}}=\underbrace{\prod_{i=1}^{p}\prod
_{j=1}^{q}}_{=\prod_{j=1}^{q}\prod_{i=1}^{p}}x_{j}^{a_{i,j}}=\prod_{j=1}%
^{q}\prod_{i=1}^{p}x_{j}^{a_{i,j}}.
\label{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.3}%
\end{equation}
\par
Now, for each $j\in\left\{  1,2,\ldots,q\right\}  $, we let $c_{j}$ be the sum
of all entries in the $j$-th column of $A$. Then, the column sums of $A$ is
the $q$-tuple $\left(  c_{1},c_{2},\ldots,c_{q}\right)  $ (by the definition
of ``column sums''). In other words, $\left(
\text{the column sums of }A\right)  =\left(  c_{1},c_{2},\ldots,c_{q}\right)
$. But $\operatorname*{colsums}A$ is the column sums of $A$ (by the definition
of $\operatorname*{colsums}A$). Hence,%
\[
\operatorname*{colsums}A=\left(  \text{the column sums of }A\right)  =\left(
c_{1},c_{2},\ldots,c_{q}\right)  .
\]
Thus,
\begin{align}
\overline{\xx}_{q}^{\operatorname*{colsums}A}
&  = \overline{\xx}_{q}^{\left(  c_{1},c_{2},\ldots,c_{q}\right)  }
=x_{1}^{c_{1}}x_{2}^{c_{2}}\cdots x_{q}^{c_{q}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\overline{\mathbf{x}}_{q}^{\left(  c_{1},c_{2},\ldots
,c_{q}\right)  }\right) \nonumber\\
&  =\prod_{j=1}^{q}x_{j}^{c_{j}}.
\label{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.4}%
\end{align}
\par
Now, fix $j\in\left\{  1,2,\ldots,q\right\}  $. Then, $c_{j}$ is the sum of
all entries in the $j$-th column of $A$ (by the definition of $c_{j}$). Hence,%
\begin{align*}
c_{j}  &  =\left(  \text{the sum of all entries in }\underbrace{\text{the
}j\text{-th column of }A}_{\substack{=\left(
\begin{array}
[c]{c}%
a_{1,j}\\
a_{2,j}\\
\vdots\\
a_{p,j}%
\end{array}
\right)  \\\text{(by (\ref{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.2}))}%
}}\right)  =\left(  \text{the sum of all entries in }\left(
\begin{array}
[c]{c}%
a_{1,j}\\
a_{2,j}\\
\vdots\\
a_{p,j}%
\end{array}
\right)  \right) \\
&  =a_{1,j}+a_{2,j}+\cdots+a_{p,j}=\sum_{i=1}^{p}a_{i,j}.
\end{align*}
Hence, $x_{j}^{c_{j}}=x_{j}^{\sum_{i=1}^{p}a_{i,j}}=\prod_{i=1}^{p}%
x_{j}^{a_{i,j}}$.
\par
Now, forget that we fixed $j$. We thus have proven that $x_{j}^{c_{j}}%
=\prod_{i=1}^{p}x_{j}^{a_{i,j}}$ for each $j\in\left\{  1,2,\ldots,q\right\}
$. Hence,
\[
\prod_{j=1}^{q}\underbrace{x_{j}^{c_{j}}}_{=\prod_{i=1}^{p}x_{j}^{a_{i,j}}%
}=\prod_{j=1}^{q}\prod_{i=1}^{p}x_{j}^{a_{i,j}}=\prod_{i=1}^{p}
\overline{\xx}_{q}^{\operatorname{row}_{i}A}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.3})}\right)  .
\]
Thus,
\[
\prod_{i=1}^{p}\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}=\prod
_{j=1}^{q}x_{j}^{c_{j}}=\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}%
A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodxA.pf.4})}\right)  .
\]
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.prodxA}).}.

Thus, every $\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in\left(
\left\{  0,1\right\}  ^{q}\right)  ^{p}$ satisfies%
\begin{equation}
\prod_{i=1}^{p}\overline{\mathbf{x}}_{q}^{\beta_{i}}=\overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \beta_{1},\beta
_{2},\ldots,\beta_{p}\right)  \right)  }
\label{pf.lem.sol.Lambda.triangular.ela.prodx}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.prodx}):}
Let $\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in\left(  \left\{
0,1\right\}  ^{q}\right)  ^{p}$. Define $A\in\left\{  0,1\right\}  ^{p\times
q}$ by $A=\Phi^{-1}\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  $.
Thus, $\Phi\left(  A\right)  =\left(  \beta_{1},\beta_{2},\ldots,\beta
_{p}\right)  $ (since $A=\Phi^{-1}\left(  \beta_{1},\beta_{2},\ldots,\beta
_{p}\right)  $). Hence, $\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)
=\Phi\left(  A\right)  =\left(  \operatorname*{row}\nolimits_{1}%
A,\operatorname{row}_{2}A,\ldots,\operatorname{row}_{p}A\right)  $ (by the
definition of $\Phi$). In other words, $\beta_{i}=\operatorname*{row}%
\nolimits_{i}A$ for each $i\in\left\{  1,2,\ldots,p\right\}  $. Thus,
$\overline{\mathbf{x}}_{q}^{\beta_{i}}=\overline{\mathbf{x}}_{q}%
^{\operatorname{row}_{i}A}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $.
Hence,
\begin{align*}
\prod_{i=1}^{p}\underbrace{\overline{\mathbf{x}}_{q}^{\beta_{i}}}%
_{=\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}}  &  =\prod_{i=1}%
^{p}\overline{\mathbf{x}}_{q}^{\operatorname{row}_{i}A}=\overline{\xx%
}_{q}^{\operatorname*{colsums}A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodxA})}\right) \\
&  =\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(
\beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }A=\Phi^{-1}\left(  \beta_{1}%
,\beta_{2},\ldots,\beta_{p}\right)  \right)  .
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.prodx}).}.

Let $\mathfrak{M}_{\alpha}$ be the set of all matrices $A\in\left\{
0,1\right\}  ^{p\times q}$ satisfying $\operatorname*{rowsums}A=\alpha$. Thus,%
\begin{align*}
\mathfrak{M}_{\alpha}  &  =\left\{  A\in\left\{  0,1\right\}  ^{p\times
q}\ \mid\ \underbrace{\operatorname*{rowsums}A}_{=\left(  \text{the row sums
of }A\right)  }=\alpha\right\} \\
&  =\left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid\ \underbrace{\left(
\text{the row sums of }A\right)  =\alpha}_{\substack{\Longleftrightarrow
\ \left(  A\text{ has row sums }\alpha\right)  \\\Longleftrightarrow\ \left(
A\text{ is a }\left\{  0,1\right\}  \text{-matrix having row sums }%
\alpha\right)  \\\text{(since we know that }A\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{(because }A\in\left\{  0,1\right\}
^{p\times q}\text{))}}}\right\} \\
&  =\left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid\ A\text{ is a
}\left\{  0,1\right\}  \text{-matrix having row sums }\alpha\right\} \\
&  \subset\left\{  0,1\right\}  ^{p\times q}.
\end{align*}


For each $m\in \NN$, we define a subset $\mathfrak{K}_{m}$ of $\left\{
0,1\right\}  ^{q}$ by%
\[
\mathfrak{K}_{m}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  .
\]
This set $\mathfrak{K}_{m}$ is finite (since it is a subset of the finite set
$\left\{  0,1\right\}  ^{q}$).

We have $\mathfrak{K}_{\alpha_{i}}\subset\left\{  0,1\right\}  ^{q}$ for each
$i\in\left\{  1,2,\ldots,p\right\}  $ (since $\mathfrak{K}_{m}\subset\left\{
0,1\right\}  ^{q}$ for each $m\in \NN$ (by the definition of
$\mathfrak{K}_{m}$)). Hence, $\prod_{i=1}^{p}\mathfrak{K}_{\alpha_{i}}$ is a
subset of $\prod_{i=1}^{p}\left(  \left\{  0,1\right\}  ^{q}\right)  =\left(
\left\{  0,1\right\}  ^{q}\right)  ^{p}$. In other words, $\mathfrak{K}%
_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}%
_{\alpha_{p}}$ is a subset of $\left(  \left\{  0,1\right\}  ^{q}\right)
^{p}$ (since $\prod_{i=1}^{p}\mathfrak{K}_{\alpha_{i}}=\mathfrak{K}%
_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}%
_{\alpha_{p}}$). In other words, $\mathfrak{K}_{\alpha_{1}}\times
\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}%
\subset\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}$.

It is easy to see that the following holds: If $A\in\left\{  0,1\right\}
^{p\times q}$ is a matrix, then we have the following logical equivalence:%
\begin{equation}
\left(  \Phi\left(  A\right)  \in\mathfrak{K}_{\alpha_{1}}\times
\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}\right)
\ \Longleftrightarrow\ \left(  \operatorname*{rowsums}A=\alpha\right)
\label{pf.lem.sol.Lambda.triangular.ela.Phi-rs}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}):}
Let $A\in\left\{  0,1\right\}  ^{p\times q}$ be a matrix.
\par
Let $i\in\left\{  1,2,\ldots,p\right\}  $. Recall that $\operatorname{row}%
_{i}A$ denotes the $i$-th row of $A$; this $i$-th row is an element of
$\left\{  0,1\right\}  ^{q}$. Thus, $\operatorname{row}_{i}A\in\left\{
0,1\right\}  ^{q}$. Now, we have the following chain of equivalences:%
\begin{align}
\left(  \operatorname{row}_{i}A\in\underbrace{\mathfrak{K}_{\alpha_{i}}%
}_{\substack{=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =\alpha_{i}\right\}  \\\text{(by the definition of
}\mathfrak{K}_{\alpha_{i}}\text{)}}}\right)  \  &  \Longleftrightarrow
\ \left(  \operatorname{row}_{i}A\in\left\{  \beta\in\left\{  0,1\right\}
^{q}\ \mid\ \left\vert \beta\right\vert =\alpha_{i}\right\}  \right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\operatorname{row}_{i}A\in\left\{
0,1\right\}  ^{q}}_{\substack{\text{This is always true}\\\text{(since we know
that }\operatorname{row}_{i}A\in\left\{  0,1\right\}  ^{q}\text{)}}}\text{ and
}\left\vert \operatorname{row}_{i}A\right\vert =\alpha_{i}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left\vert \operatorname{row}_{i}A\right\vert
=\alpha_{i}\right)  . \label{pf.lem.sol.Lambda.triangular.ela.Phi-rs.pf.1}%
\end{align}
\par
Now, forget that we fixed $i$. We thus have proven the equivalence
(\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs.pf.1}) for each $i\in\left\{
1,2,\ldots,p\right\}  $.
\par
The definition of $\Phi$ yields $\Phi\left(  A\right)  =\left(
\operatorname*{row}\nolimits_{1}A,\operatorname{row}_{2}A,\ldots
,\operatorname{row}_{p}A\right)  $. Now, we have the following chain of
logical equivalences:%
\begin{align*}
&  \ \left(  \underbrace{\Phi\left(  A\right)  }_{=\left(  \operatorname*{row}%
\nolimits_{1}A,\operatorname{row}_{2}A,\ldots,\operatorname{row}_{p}A\right)
}\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}\right) \\
&  \Longleftrightarrow\ \left(  \left(  \operatorname*{row}\nolimits_{1}%
A,\operatorname{row}_{2}A,\ldots,\operatorname{row}_{p}A\right)
\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\operatorname{row}_{i}%
A\in\mathfrak{K}_{\alpha_{i}}}_{\substack{\Longleftrightarrow\ \left(
\left\vert \operatorname{row}_{i}A\right\vert =\alpha_{i}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs.pf.1}))}}}\text{ for each }%
i\in\left\{  1,2,\ldots,p\right\}  \right) \\
&  \Longleftrightarrow\ \left(  \left\vert \operatorname{row}_{i}A\right\vert
=\alpha_{i}\text{ for each }i\in\left\{  1,2,\ldots,p\right\}  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \left\vert
\operatorname{row}_{1}A\right\vert ,\left\vert \operatorname{row}%
_{2}A\right\vert ,\ldots,\left\vert \operatorname{row}_{p}A\right\vert
\right)  }_{\substack{=\operatorname*{rowsums}A\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.rowsums}))}}}=\underbrace{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  }_{=\alpha}\right) \\
&  \Longleftrightarrow\ \left(  \operatorname*{rowsums}A=\alpha\right)  .
\end{align*}
This proves (\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}).}.

Hence, we have $\Phi\left(  \mathfrak{M}_{\alpha}\right)  \subset
\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}$\ \ \ \ \footnote{\textit{Proof.} Let
$B\in\mathfrak{M}_{\alpha}$. Then, $B$ belongs to $\mathfrak{M}_{\alpha}$. In
other words, $B$ belongs to the set of all matrices $A\in\left\{  0,1\right\}
^{p\times q}$ satisfying $\operatorname*{rowsums}A=\alpha$ (since
$\mathfrak{M}_{\alpha}$ is the set of all matrices $A\in\left\{  0,1\right\}
^{p\times q}$ satisfying $\operatorname*{rowsums}A=\alpha$). In other words,
$B$ is a matrix $A\in\left\{  0,1\right\}  ^{p\times q}$ satisfying
$\operatorname*{rowsums}A=\alpha$. In other words, $B$ is a matrix in
$\left\{  0,1\right\}  ^{p\times q}$ and satisfies $\operatorname*{rowsums}%
B=\alpha$.
\par
The equivalence (\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}) (applied to
$A=B$) shows that $\left(  \Phi\left(  B\right)  \in\mathfrak{K}_{\alpha_{1}%
}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}\right)  \ \Longleftrightarrow\ \left(  \operatorname*{rowsums}%
B=\alpha\right)  $. Thus, $\Phi\left(  B\right)  \in\mathfrak{K}_{\alpha_{1}%
}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}$
holds (since $\operatorname*{rowsums}B=\alpha$ holds).
\par
Now, forget that we fixed $B$. We thus have shown that $\Phi\left(  B\right)
\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}$ for each $B\in\mathfrak{M}_{\alpha}$. In
other words, $\Phi\left(  \mathfrak{M}_{\alpha}\right)  \subset\mathfrak{K}%
_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}%
_{\alpha_{p}}$. Qed.} and $\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}%
_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}\subset\Phi\left(
\mathfrak{M}_{\alpha}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$C\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times
\cdots\times\mathfrak{K}_{\alpha_{p}}$. Then, $C\in\mathfrak{K}_{\alpha_{1}%
}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}\subset\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}$ (since $\mathfrak{K}%
_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}%
_{\alpha_{p}}$ is a subset of $\left(  \left\{  0,1\right\}  ^{q}\right)
^{p}$). Hence, $\Phi^{-1}\left(  C\right)  \in\left\{  0,1\right\}  ^{p\times
q}$ is well-defined (since $\Phi$ is a bijection $\left\{  0,1\right\}
^{p\times q}\rightarrow\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}$).
Define $B\in\left\{  0,1\right\}  ^{p\times q}$ by $B=\Phi^{-1}\left(
C\right)  $.
\par
From $B=\Phi^{-1}\left(  C\right)  $, we obtain $\Phi\left(  B\right)
=C\in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times
\cdots\times\mathfrak{K}_{\alpha_{p}}$.
\par
The equivalence (\ref{pf.lem.sol.Lambda.triangular.ela.Phi-rs}) (applied to
$A=B$) shows that $\left(  \Phi\left(  B\right)  \in\mathfrak{K}_{\alpha_{1}%
}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}\right)  \ \Longleftrightarrow\ \left(  \operatorname*{rowsums}%
B=\alpha\right)  $. Thus, $\operatorname*{rowsums}B=\alpha$ holds (since
$\Phi\left(  B\right)  \in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}%
_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}$ holds).
\par
Now, $B$ is a matrix $A\in\left\{  0,1\right\}  ^{p\times q}$ satisfying
$\operatorname*{rowsums}A=\alpha$ (since $B$ is a matrix in $\left\{
0,1\right\}  ^{p\times q}$ and satisfies $\operatorname*{rowsums}B=\alpha$).
In other words, $B$ belongs to the set of all matrices $A\in\left\{
0,1\right\}  ^{p\times q}$ satisfying $\operatorname*{rowsums}A=\alpha$. In
other words, $B$ belongs to $\mathfrak{M}_{\alpha}$ (since $\mathfrak{M}%
_{\alpha}$ is the set of all matrices $A\in\left\{  0,1\right\}  ^{p\times q}$
satisfying $\operatorname*{rowsums}A=\alpha$). In other words, $B\in
\mathfrak{M}_{\alpha}$. Now, $C=\Phi\left(  \underbrace{B}_{\in\mathfrak{M}%
_{\alpha}}\right)  \in\Phi\left(  \mathfrak{M}_{\alpha}\right)  $.
\par
Now, forget that we fixed $C$. We thus have shown that $C\in\Phi\left(
\mathfrak{M}_{\alpha}\right)  $ for each $C\in\mathfrak{K}_{\alpha_{1}}%
\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}$.
In other words, $\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}%
}\times\cdots\times\mathfrak{K}_{\alpha_{p}}\subset\Phi\left(  \mathfrak{M}%
_{\alpha}\right)  $. Qed.}. Combining these two inclusions, we obtain%
\begin{equation}
\Phi\left(  \mathfrak{M}_{\alpha}\right)  =\mathfrak{K}_{\alpha_{1}}%
\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}.
\label{pf.lem.sol.Lambda.triangular.ela.3}%
\end{equation}


Now, $\mathfrak{M}_{\alpha}$ is a subset of $\left\{  0,1\right\}  ^{p\times
q}$ (since $\mathfrak{M}_{\alpha}\subset\left\{  0,1\right\}  ^{p\times q}$).
Hence, Lemma \ref{lem.sol.Lambda.triangular.g-set} (applied to $X=\left\{
0,1\right\}  ^{p\times q}$, $Y=\left(  \left\{  0,1\right\}  ^{q}\right)
^{p}$, $\phi=\Phi$ and $Z=\mathfrak{M}_{\alpha}$) yields that the map
$\mathfrak{M}_{\alpha}\rightarrow\Phi\left(  \mathfrak{M}_{\alpha}\right)
,\ A\mapsto\Phi\left(  A\right)  $ is well-defined and is a bijection.

For each $m\in \NN$, we have%
\begin{align}
\eta_{q}\left(  e_{m}\right)   &  =\underbrace{\sum\limits_{\substack{\beta
\in\left\{  0,1\right\}  ^{q};\\\left\vert \beta\right\vert =m}}}%
_{\substack{=\sum_{\beta\in\mathfrak{K}_{m}}\\\text{(since }\mathfrak{K}%
_{m}=\left\{  \beta\in\left\{  0,1\right\}  ^{q}\ \mid\ \left\vert
\beta\right\vert =m\right\}  \\\text{(by the definition of }\mathfrak{K}%
_{m}\text{))}}}\overline{\mathbf{x}}_{q}^{\beta}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma \ref{lem.sol.Lambda.triangular.em}}\right) \nonumber\\
&  =\sum_{\beta\in\mathfrak{K}_{m}}\overline{\mathbf{x}}_{q}^{\beta}.
\label{pf.lem.sol.Lambda.triangular.ela.em}%
\end{align}


But%
\begin{align*}
&  \eta_{q}\left(  \underbrace{e_{\alpha_{1}}e_{\alpha_{2}}\cdots
e_{\alpha_{p}}}_{=\prod_{i=1}^{p}e_{\alpha_{i}}}\right) \\
&  =\eta_{q}\left(  \prod_{i=1}^{p}e_{\alpha_{i}}\right)  =\prod_{i=1}%
^{p}\underbrace{\eta_{q}\left(  e_{\alpha_{i}}\right)  }_{\substack{=\sum
_{\beta\in\mathfrak{K}_{\alpha_{i}}}\overline{\mathbf{x}}_{q}^{\beta
}\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.ela.em}) (applied to }%
m=\alpha_{i}\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta_{q}\text{
is a }\kk\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{p}\sum_{\beta\in\mathfrak{K}_{\alpha_{i}}}
\overline{\xx}_{q}^{\beta}=\sum_{\left(  \beta_{1},\beta_{2},\ldots,\beta
_{p}\right)  \in\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}%
\times\cdots\times\mathfrak{K}_{\alpha_{p}}}\underbrace{\prod_{i=1}%
^{p}\overline{\mathbf{x}}_{q}^{\beta_{i}}}_{\substack{=\overline{\xx%
}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \beta_{1},\beta
_{2},\ldots,\beta_{p}\right)  \right)  }\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.ela.prodx})}\\\text{(since }\left(
\beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in\mathfrak{K}_{\alpha_{1}%
}\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}\subset\left(  \left\{  0,1\right\}  ^{q}\right)  ^{p}\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\sum_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \in
\mathfrak{K}_{\alpha_{1}}\times\mathfrak{K}_{\alpha_{2}}\times\cdots
\times\mathfrak{K}_{\alpha_{p}}}\overline{\mathbf{x}}_{q}%
^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{p}\right)  \right)  }=\sum_{C\in\mathfrak{K}_{\alpha_{1}}%
\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}}%
}\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(
C\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\left(  \beta_{1},\beta_{2},\ldots,\beta_{p}\right)  \text{ as }C\right) \\
&  =\sum_{C\in\Phi\left(  \mathfrak{M}_{\alpha}\right)  }\overline{\xx%
}_{q}^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  C\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{K}_{\alpha_{1}}%
\times\mathfrak{K}_{\alpha_{2}}\times\cdots\times\mathfrak{K}_{\alpha_{p}%
}=\Phi\left(  \mathfrak{M}_{\alpha}\right)  \text{ (by
(\ref{pf.lem.sol.Lambda.triangular.ela.3}))}\right) \\
&  =\underbrace{\sum_{A\in\mathfrak{M}_{\alpha}}}_{\substack{=\sum\limits
_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\alpha}}\\\text{(since
}\mathfrak{M}_{\alpha}=\left\{  A\in\left\{  0,1\right\}  ^{p\times q}%
\ \mid\ A\text{ is a }\left\{  0,1\right\}  \text{-matrix having row sums
}\alpha\right\}  \text{)}}}\underbrace{\overline{\mathbf{x}}_{q}%
^{\operatorname*{colsums}\left(  \Phi^{-1}\left(  \Phi\left(  A\right)
\right)  \right)  }}_{\substack{=\overline{\mathbf{x}}_{q}%
^{\operatorname*{colsums}A}\\\text{(since }\Phi^{-1}\left(  \Phi\left(
A\right)  \right)  =A\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\Phi\left(  A\right)  \text{ for }C\text{ in
the sum, since}\\
\text{the map }\mathfrak{M}_{\alpha}\rightarrow\Phi\left(  \mathfrak{M}%
_{\alpha}\right)  ,\ A\mapsto\Phi\left(  A\right)  \text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a
}\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }\alpha
}}\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}A}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.ela}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.g-coeff}Let $\lambda\in\Par$
and $\mu\in\Par$. Then, $\left[  \mathbf{x}^{\mu}\right]
\left(  e_{\lambda}\right)  =a_{\lambda,\mu}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.g-coeff}.]Let $p=\ell\left(
\lambda\right)  $. Thus, $\lambda_{p+1}=\lambda_{p+2}=\lambda_{p+3}=\cdots=0$.
Therefore, $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)
$ (since we omit trailing zeroes from a partition). Thus, $\lambda$ can be
regarded as the $p$-tuple $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{p}\right)  $.

Let $q=\ell\left(  \mu\right)  $. Thus, $\mu_{q+1}=\mu_{q+2}=\mu_{q+3}%
=\cdots=0$. Therefore, $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $
(since we omit trailing zeroes from a partition). Thus, $\mu$ can be regarded
as the $q$-tuple $\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $.

For any matrix $A\in\left\{  0,1\right\}  ^{p\times q}$, we let
$\operatorname*{colsums}A$ denote the column sums of $A$. Thus, for any matrix
$A\in\left\{  0,1\right\}  ^{p\times q}$, we have%
\begin{equation}
\operatorname*{colsums}A=\left(  \text{the column sums of }A\right)  .
\label{pf.lem.sol.Lambda.triangular.g-coeff.colsumsA=}%
\end{equation}


We have defined $a_{\lambda,\mu}$ as the number of all $\left\{  0,1\right\}
$-matrices of size $\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)
$ having row sums $\lambda$ and column sums $\mu$. Thus,%
\begin{align}
&  a_{\lambda,\mu}\nonumber\\
&  =\left(  \text{the number of all }\left\{  0,1\right\}  \text{-matrices of
size }\underbrace{\ell\left(  \lambda\right)  }_{=p}\times\underbrace{\ell
\left(  \mu\right)  }_{=q}\text{ having row sums }\lambda\text{ and column
sums }\mu\right) \nonumber\\
&  =\left(  \text{the number of all }\left\{  0,1\right\}  \text{-matrices of
size }p\times q\text{ having row sums }\lambda\text{ and column sums }%
\mu\right) \nonumber\\
&  =\left(  \text{the number of all matrices }A\in\left\{  0,1\right\}
^{p\times q}\text{ having row sums }\lambda\text{ and column sums }\mu\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the }\left\{  0,1\right\}
\text{-matrices of size }p\times q\text{ are precisely the matrices }%
A\in\left\{  0,1\right\}  ^{p\times q}\right) \nonumber\\
&  =\left(  \text{the number of all matrices }A\in\left\{  0,1\right\}
^{p\times q}\text{ such that the row sums of }A\text{ are }\lambda
\text{,}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \text{and the column sums
of }A\text{ are }\mu\right) \nonumber\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\underbrace{\text{the column sums of }A\text{ are }\mu}_{\Longleftrightarrow
\ \left(  \left(  \text{the column sums of }A\right)  =\mu\right)  }\right\}
\right\vert \nonumber\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }\underbrace{\left(
\text{the column sums of }A\right)  }_{\substack{=\operatorname*{colsums}%
A\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.g-coeff.colsumsA=}))}}%
}=\mu\right\}  \right\vert \nonumber\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\operatorname*{colsums}A=\mu\right\}  \right\vert .
\label{pf.lem.sol.Lambda.triangular.g-coeff.a1}%
\end{align}


We have $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $
with $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{p}>0$ (since
$p=\ell\left(  \lambda\right)  $). Hence, $e_{\lambda}=e_{\lambda_{1}%
}e_{\lambda_{2}}\cdots e_{\lambda_{p}}$ (by the definition of $e_{\lambda}$).
Hence, $e_{\lambda}\in\Lambda$ (since $e_{i}\in\Lambda$ for each
$i\in \NN$).

But $e_{\lambda}\in\Lambda\subset R\left(  \xx\right)  $. Also,
$\mu_{i}=0$ for every integer $i>q$ (since $\mu_{q+1}=\mu_{q+2}=\mu
_{q+3}=\cdots=0$). Hence, Lemma \ref{lem.sol.Lambda.triangular.etaq-coeff}
(applied to $\beta=\mu$ and $f=e_{\lambda}$) yields%
\begin{align}
\left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right)   &  =\left[
\overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
}\right]  \left(  \eta_{q}\left(  \underbrace{e_{\lambda}}_{=e_{\lambda_{1}%
}e_{\lambda_{2}}\cdots e_{\lambda_{p}}}\right)  \right) \nonumber\\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  }\right]  \left(  \eta_{q}\left(  e_{\lambda_{1}}e_{\lambda_{2}%
}\cdots e_{\lambda_{p}}\right)  \right)  .
\label{pf.lem.sol.Lambda.triangular.g-coeff.4}%
\end{align}


But $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)
\in\mathbb{N}^{p}$ is a $p$-tuple of nonnegative integers. Hence, Lemma
\ref{lem.sol.Lambda.triangular.ela} (applied to $\lambda$ and $\lambda_{i}$
instead of $\alpha$ and $\alpha_{i}$) yields
\[
\eta_{q}\left(  e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}\right)
=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda}}
\overline{\xx}_{q}^{\operatorname*{colsums}A}.
\]
Hence, (\ref{pf.lem.sol.Lambda.triangular.g-coeff.4}) becomes%
\begin{align*}
&  \left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  }\right]  \left(  \underbrace{\eta_{q}\left(  e_{\lambda_{1}%
}e_{\lambda_{2}}\cdots e_{\lambda_{p}}\right)  }_{=\sum\limits_{\substack{A\in
\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{  0,1\right\}
\text{-matrix}\\\text{having row sums }\lambda}}\overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}A}}\right) \\
&  =\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  }\right]  \left(  \sum\limits_{\substack{A\in\left\{  0,1\right\}
^{p\times q}\text{ is a }\left\{  0,1\right\}  \text{-matrix}\\\text{having
row sums }\lambda}}\overline{\mathbf{x}}_{q}^{\operatorname*{colsums}A}\right)
\\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a
}\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda
}}\underbrace{\left[  \overline{\mathbf{x}}_{q}^{\left(  \mu_{1},\mu
_{2},\ldots,\mu_{q}\right)  }\right]  \left(  \overline{\mathbf{x}}%
_{q}^{\operatorname*{colsums}A}\right)  }_{\substack{=\delta_{\left(  \mu
_{1},\mu_{2},\ldots,\mu_{q}\right)  ,\operatorname*{colsums}A}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.etaq-coeff.monocase.pf.c1.4}) (applied to
}\phi=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  \\\text{and }%
\psi=\operatorname*{colsums}A\text{))}}}\\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a
}\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda
}}\underbrace{\delta_{\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
,\operatorname*{colsums}A}}_{\substack{=\delta_{\mu,\operatorname*{colsums}%
A}\\\text{(since }\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  =\mu\text{)}%
}}=\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{
0,1\right\}  \text{-matrix}\\\text{having row sums }\lambda}}\delta
_{\mu,\operatorname*{colsums}A}\\
&  =\underbrace{\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{
is a }\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }%
\lambda;\\\operatorname*{colsums}A=\mu}}}_{\substack{=\sum\limits_{\substack{A\in
\left\{  0,1\right\}  ^{p\times q}\text{ is a }\left\{  0,1\right\}
\text{-matrix;}\\\text{the row sums of }A\text{ are }\lambda
;\\\operatorname*{colsums}A=\mu}}\\=\sum\limits_{\substack{A\in\left\{  0,1\right\}
^{p\times q}\text{;}\\\text{the row sums of }A\text{ are }\lambda
;\\\operatorname*{colsums}A=\mu}}\\\text{(since every }A\in\left\{
0,1\right\}  ^{p\times q}\text{ is a }\left\{  0,1\right\}  \text{-matrix)}%
}}\underbrace{\delta_{\mu,\operatorname*{colsums}A}}%
_{\substack{=1\\\text{(since }\mu=\operatorname*{colsums}A\\\text{(since
}\operatorname*{colsums}A=\mu\text{))}}}+\sum\limits_{\substack{A\in\left\{
0,1\right\}  ^{p\times q}\text{ is a }\left\{  0,1\right\}  \text{-matrix}%
\\\text{having row sums }\lambda;\\\operatorname*{colsums}A\neq\mu
}}\underbrace{\delta_{\mu,\operatorname*{colsums}A}}%
_{\substack{=0\\\text{(since }\mu\neq\operatorname*{colsums}A\\\text{(since
}\operatorname*{colsums}A\neq\mu\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }A\in\left\{  0,1\right\}  ^{p\times q}\text{ satisfies
either }\operatorname*{colsums}A=\mu\text{ or }\operatorname*{colsums}A\neq
\mu\\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{;}\\\text{the
row sums of }A\text{ are }\lambda;\\\operatorname*{colsums}A=\mu
}}1+\underbrace{\sum\limits_{\substack{A\in\left\{  0,1\right\}  ^{p\times q}\text{
is a }\left\{  0,1\right\}  \text{-matrix}\\\text{having row sums }%
\lambda;\\\operatorname*{colsums}A\neq\mu}}0}_{=0}=\sum\limits_{\substack{A\in
\left\{  0,1\right\}  ^{p\times q}\text{;}\\\text{the row sums of }A\text{ are
}\lambda;\\\operatorname*{colsums}A=\mu}}1\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\operatorname*{colsums}A=\mu\right\}  \right\vert \cdot1\\
&  =\left\vert \left\{  A\in\left\{  0,1\right\}  ^{p\times q}\ \mid
\ \text{the row sums of }A\text{ are }\lambda\text{, and }%
\operatorname*{colsums}A=\mu\right\}  \right\vert .
\end{align*}
Comparing this with (\ref{pf.lem.sol.Lambda.triangular.g-coeff.a1}), we obtain
$\left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right)  =a_{\lambda,\mu
}$. This proves Lemma \ref{lem.sol.Lambda.triangular.g-coeff}.
\end{proof}

Now, let $\lambda\in\Par_{n}$. Let $p=\ell\left(
\lambda\right)  $. We have $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{p}\right)  $ with $\lambda_{1}\geq\lambda_{2}\geq\cdots
\geq\lambda_{p}>0$ (since $p=\ell\left(  \lambda\right)  $). Hence,
$e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}$ (by the
definition of $e_{\lambda}$).

Moreover, $\lambda\in\Par_{n}$, so that $\left\vert
\lambda\right\vert =n$. Hence, $n=\left\vert \underbrace{\lambda}_{=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  }\right\vert =\left\vert
\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  \right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}$.

Let $i\in\left\{  1,2,\ldots,p\right\}  $. Then, $e_{\lambda_{i}}$ is a
homogeneous element of $\Lambda$ having degree $\lambda_{i}$ (because for each
$m\in \NN$, the element $e_{m}$ is a homogeneous element of $\Lambda$
having degree $m$).

Now, forget that we fixed $i$. We thus have shown that for each $i\in\left\{
1,2,\ldots,p\right\}  $, the element $e_{\lambda_{i}}$ is a homogeneous
element of $\Lambda$ having degree $\lambda_{i}$. In other words,
$e_{\lambda_{1}},e_{\lambda_{2}},\ldots,e_{\lambda_{p}}$ are homogeneous
elements of $\Lambda$ having degrees $\lambda_{1},\lambda_{2},\ldots
,\lambda_{p}$, respectively. Hence, the product $e_{\lambda_{1}}e_{\lambda
_{2}}\cdots e_{\lambda_{p}}$ of these elements is a homogeneous element of
$\Lambda$ having degree $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}$. In light
of $e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{p}}$ and
$n=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{p}$, this rewrites as follows: The
element $e_{\lambda}$ is a homogeneous element of $\Lambda$ having degree $n$.
In other words, $e_{\lambda}\in\Lambda_{n}$. Thus, Exercise
\ref{exe.Lambda.triangular}(a) (applied to $f=e_{\lambda}$) yields%
\[
e_{\lambda}=\sum_{\mu\in\Par_{n}}\underbrace{\left(
\left[  \mathbf{x}^{\mu}\right]  \left(  e_{\lambda}\right)  \right)
}_{\substack{=a_{\lambda,\mu}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.g-coeff})}}}m_{\mu}=\sum_{\mu\in
\operatorname{Par}_{n}}a_{\lambda,\mu}m_{\mu}.
\]
This solves Exercise \ref{exe.Lambda.triangular}(g).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(h).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(h), let us show a few lemmas:

\begin{lemma}
\label{lem.sol.Lambda.triangular.transpose1}Let $\nu \in \Par$ and
$k \in \NN$. Then,
\begin{equation}
\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{  \nu_{j},k\right\}  .
\label{eq.lem.sol.Lambda.triangular.transpose1.claim}
\end{equation}
(Note that the right hand side of
(\ref{eq.lem.sol.Lambda.triangular.transpose1.claim}) is well-defined,
because
every sufficiently high $j$ satisfies $\min\left\{  \nu_{j},k\right\}  =0$
(since every sufficiently high $j$ satisfies $\nu_{j}=0$)).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.transpose1}.]Lemma
\ref{lem.sol.Lambda.triangular.transpose1} is precisely the same as
Proposition~\ref{prop.sol.partition.dominance.lemma},
and thus has already been proven.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.ineq-eq}Let $p\in \NN$. Let
$a_{1},a_{2},\ldots,a_{p}$ be $p$ integers. Let $b_{1},b_{2},\ldots,b_{p}$ be
$p$ integers. Assume that%
\begin{equation}
a_{i}\geq b_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots
,p\right\}  . \label{eq.lem.sol.Lambda.triangular.ineq-eq.ass1}%
\end{equation}
Assume furthermore that $\sum_{i=1}^{p}a_{i}\leq\sum_{i=1}^{p}b_{i}$. Then,
$a_{i}=b_{i}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.ineq-eq}.]Let $j\in\left\{
1,2,\ldots,p\right\}  $. Then,
(\ref{eq.lem.sol.Lambda.triangular.ineq-eq.ass1}) (applied to $i=j$) shows
that $a_{j}\geq b_{j}$. But
\begin{align*}
\underbrace{\sum_{i=1}^{p}}_{=\sum_{i\in\left\{  1,2,\ldots,p\right\}  }%
}a_{i}  &  =\sum_{i\in\left\{  1,2,\ldots,p\right\}  }a_{i}=a_{j}%
+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq j}%
}\underbrace{a_{i}}_{\substack{\geq b_{i}\\\text{(by
(\ref{eq.lem.sol.Lambda.triangular.ineq-eq.ass1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=j\\
\text{from the sum (since }j\in\left\{  1,2,\ldots,p\right\}  \text{)}%
\end{array}
\right) \\
&  \geq a_{j}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq
j}}b_{i}.
\end{align*}
Hence,%
\begin{align*}
a_{j}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq j}}b_{i}  &
\leq\sum_{i=1}^{p}a_{i}\leq\underbrace{\sum_{i=1}^{p}}_{=\sum_{i\in\left\{
1,2,\ldots,p\right\}  }}b_{i}=\sum_{i\in\left\{  1,2,\ldots,p\right\}  }%
b_{i}=b_{j}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq
j}}b_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=j\\
\text{from the sum (since }j\in\left\{  1,2,\ldots,p\right\}  \text{)}%
\end{array}
\right)  .
\end{align*}
Subtracting $\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\i\neq
j}}b_{i}$ from both sides of this inequality, we obtain $a_{j}\leq b_{j}$.
Combining this with $a_{j}\geq b_{j}$, we obtain $a_{j}=b_{j}$.

Now, forget that we fixed $j$. We thus have shown that $a_{j}=b_{j}$ for each
$j\in\left\{  1,2,\ldots,p\right\}  $. Renaming the variable $j$ as $i$ in
this statement, we conclude that $a_{i}=b_{i}$ for each $i\in\left\{
1,2,\ldots,p\right\}  $. This proves Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.ineq-eq2}Let $k\in \NN$. Let
$a_{1},a_{2},\ldots,a_{k}$ be $k$ integers. Let $b_{1},b_{2},\ldots,b_{k}$ be
$k$ integers. Assume that%
\[
a_{j}\geq b_{j}\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,\ldots
,k\right\}  .
\]
Assume furthermore that $\sum_{j=1}^{k}a_{j}\leq\sum_{j=1}^{k}b_{j}$. Then,
$a_{j}=b_{j}$ for each $j\in\left\{  1,2,\ldots,k\right\}  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.ineq-eq2}.]Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq2} is obtained from Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq} upon renaming the variables $p$ and
$i$ as $k$ and $j$. Thus, Lemma \ref{lem.sol.Lambda.triangular.ineq-eq2}
follows from Lemma \ref{lem.sol.Lambda.triangular.ineq-eq}.
\end{proof}

The next few lemmas use the so-called \dfn{Iverson bracket notation}:

\begin{definition}
For every assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $
denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
\end{definition}

\begin{lemma}
\label{lem.sol.Lambda.triangular.sum-iver}Let $q\in \NN$ and
$r\in \NN$ be such that $r\leq q$. Then, $\sum_{j=1}^{q}\left[  j\leq
r\right]  =r$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.sum-iver}.]We have $r\leq q$.
Hence, $\min\left\{  r,q\right\}  =r$. Therefore, $\left\{  j\in\left\{
1,2,\ldots,q\right\}  \ \mid\ j\leq r\right\}  =\left\{  1,2,\ldots,r\right\}
$\ \ \ \ \footnote{\textit{Proof.} Let $g\in\left\{  j\in\left\{
1,2,\ldots,q\right\}  \ \mid\ j\leq r\right\}  $. Thus, $g$ is a $j\in\left\{
1,2,\ldots,q\right\}  $ satisfying $j\leq r$. In other words, $g$ is an
element of $\left\{  1,2,\ldots,q\right\}  $ and satisfies $g\leq r$.
\par
From $g\in\left\{  1,2,\ldots,q\right\}  $, we conclude that $g$ is a positive
integer. Thus, $g\in\left\{  1,2,\ldots,r\right\}  $ (since $g\leq r$).
\par
Now, let us forget that we fixed $g$. We thus have shown that each
$g\in\left\{  j\in\left\{  1,2,\ldots,q\right\}  \ \mid\ j\leq r\right\}  $
satisfies $g\in\left\{  1,2,\ldots,r\right\}  $. In other words, $\left\{
j\in\left\{  1,2,\ldots,q\right\}  \ \mid\ j\leq r\right\}  \subset\left\{
1,2,\ldots,r\right\}  $.
\par
Now, let $h\in\left\{  1,2,\ldots,r\right\}  $. Thus, $h$ is a positive
integer satisfying $1\leq h\leq r$. From $h\leq r\leq q$, we conclude that
$h\in\left\{  1,2,\ldots,q\right\}  $ (since $h$ is a positive integer).
Hence, $h$ is an element of $\left\{  1,2,\ldots,q\right\}  $ and satisfies
$h\leq r$. In other words, $h$ is a $j\in\left\{  1,2,\ldots,q\right\}  $
satisfying $j\leq r$. In other words, $h\in\left\{  j\in\left\{
1,2,\ldots,q\right\}  \ \mid\ j\leq r\right\}  $.
\par
Now, forget that we fixed $h$. We thus have shown that each $h\in\left\{
1,2,\ldots,r\right\}  $ satisfies $h\in\left\{  j\in\left\{  1,2,\ldots
,q\right\}  \ \mid\ j\leq r\right\}  $. In other words, $\left\{
1,2,\ldots,r\right\}  \subset\left\{  j\in\left\{  1,2,\ldots,q\right\}
\ \mid\ j\leq r\right\}  $. Combining this with $\left\{  j\in\left\{
1,2,\ldots,q\right\}  \ \mid\ j\leq r\right\}  \subset\left\{  1,2,\ldots
,r\right\}  $, we obtain $\left\{  j\in\left\{  1,2,\ldots,q\right\}
\ \mid\ j\leq r\right\}  =\left\{  1,2,\ldots,r\right\}  $. Qed.}.

Now,%
\begin{align*}
\underbrace{\sum_{j=1}^{q}}_{=\sum_{j\in\left\{  1,2,\ldots,q\right\}  }%
}\left[  j\leq r\right]   &  =\sum_{j\in\left\{  1,2,\ldots,q\right\}
}\left[  j\leq r\right]  =\sum\limits_{\substack{j\in\left\{  1,2,\ldots,q\right\}
;\\j\leq r}}\underbrace{\left[  j\leq r\right]  }_{\substack{=1\\\text{(since
}j\leq r\text{)}}}+\sum\limits_{\substack{j\in\left\{  1,2,\ldots,q\right\}
;\\\text{not }j\leq r}}\underbrace{\left[  j\leq r\right]  }%
_{\substack{=0\\\text{(since we don't have }j\leq r\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }j\in\left\{  1,2,\ldots,q\right\}  \text{ satisfies either
}j\leq r\text{ or }\left(  \text{not }j\leq r\right) \\
\text{(but not both)}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{j\in\left\{  1,2,\ldots,q\right\}  ;\\j\leq r}%
}1+\underbrace{\sum\limits_{\substack{j\in\left\{  1,2,\ldots,q\right\}  ;\\\text{not
}j\leq r}}0}_{=0}=\sum\limits_{\substack{j\in\left\{  1,2,\ldots,q\right\}  ;\\j\leq
r}}1\\
&  =\left\vert \left\{  j\in\left\{  1,2,\ldots,q\right\}  \ \mid\ j\leq
r\right\}  \right\vert \cdot1\\
&  =\left\vert \underbrace{\left\{  j\in\left\{  1,2,\ldots,q\right\}
\ \mid\ j\leq r\right\}  }_{=\left\{  1,2,\ldots,r\right\}  }\right\vert
=\left\vert \left\{  1,2,\ldots,r\right\}  \right\vert =r.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.sum-iver}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.01pqmat}Let $p\in \NN$ and
$q\in \NN$. Let $A\in\left\{  0,1\right\}  ^{p\times q}$ be a $\left\{
0,1\right\}  $-matrix. Let $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{p}\right)  $ be the row sums of $A$. Let $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{q}\right)  $ be the column sums of $A$.

Write the $p\times q$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}$. Then:

\begin{enumerate}
\item[(a)] We have $\lambda_{i}=\sum_{j=1}^{q}a_{i,j}$ for each $i\in\left\{
1,2,\ldots,p\right\}  $.

\item[(b)] We have $\mu_{j}=\sum_{i=1}^{p}a_{i,j}$ for each $j\in\left\{
1,2,\ldots,q\right\}  $.

\item[(c)] We have $\sum_{i=1}^{p}\lambda_{i}=\sum_{j=1}^{q}\mu_{j}$.

\item[(d)] We have $\min\left\{  \lambda_{i},k\right\}  \geq\sum_{j=1}%
^{k}a_{i,j}$ for each $k\in\left\{  0,1,\ldots,q\right\}  $ and $i\in\left\{
1,2,\ldots,p\right\}  $.

\item[(e)] We have $\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  \geq
\sum_{j=1}^{k}\mu_{j}$ for each $k\in\left\{  0,1,\ldots,q\right\}  $.

\item[(f)] If $\left(  \sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}
=\sum_{j=1}^{k}\mu_{j}\text{ for each }k\in\left\{  1,2,\ldots,q\right\}
\right)  $, then $A=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq
i\leq p,\ 1\leq j\leq q}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.01pqmat}.]We have $\left(
a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq q}=A\in\left\{  0,1\right\}
^{p\times q}$. Thus, $a_{i,j}\in\left\{  0,1\right\}  $ for each $i\in\left\{
1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}  $. Hence, for
each $i\in\left\{  1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots
,q\right\}  $, we have%
\begin{equation}
a_{i,j}\geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a_{i,j}\in\left\{
0,1\right\}  \right)  \label{pf.lem.sol.Lambda.triangular.01pqmat.aij0}%
\end{equation}
and%
\begin{equation}
a_{i,j}\leq1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a_{i,j}\in\left\{
0,1\right\}  \right)  . \label{pf.lem.sol.Lambda.triangular.01pqmat.aij1}%
\end{equation}


(a) For every $i\in\left\{  1,2,\ldots,p\right\}  $, we let $r_{i}$ be the sum
of all entries in the $i$-th row of $A$. Then, the row sums of $A$ is the
$p$-tuple $\left(  r_{1},r_{2},\ldots,r_{p}\right)  $ (by the definition of
``row sums''). In other words, $\left(
\text{the row sums of }A\right)  =\left(  r_{1},r_{2},\ldots,r_{p}\right)  $.
But recall that $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $
is the row sums of $A$. Hence,%
\[
\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  =\left(  \text{the
row sums of }A\right)  =\left(  r_{1},r_{2},\ldots,r_{p}\right)  .
\]
In other words, each $i\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
\lambda_{i}=r_{i}. \label{pf.lem.sol.Lambda.triangular.01pqmat.a.1}%
\end{equation}


But $A=\left(  a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$. Hence, each
$i\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th row of }A\right)  =\left(  a_{i,1}%
,a_{i,2},\ldots,a_{i,q}\right)  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.a.2}%
\end{equation}


Let $i\in\left\{  1,2,\ldots,p\right\}  $. Then,
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.a.1}) yields%
\begin{align*}
\lambda_{i}  &  =r_{i}=\left(  \text{the sum of all entries in }%
\underbrace{\text{the }i\text{-th row of }A}_{\substack{=\left(
a_{i,1},a_{i,2},\ldots,a_{i,q}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.a.2}))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }r_{i}\text{ was defined as the sum
of all entries in the }i\text{-th row of }A\right) \\
&  =\left(  \text{the sum of all entries in }\left(  a_{i,1},a_{i,2}%
,\ldots,a_{i,q}\right)  \right) \\
&  =a_{i,1}+a_{i,2}+\cdots+a_{i,q}=\sum_{j=1}^{q}a_{i,j}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a).

(b) For every $j\in\left\{  1,2,\ldots,q\right\}  $, we let $c_{j}$ be the sum
of all entries in the $j$-th column of $A$. Then, the column sums of $A$ is
the $q$-tuple $\left(  c_{1},c_{2},\ldots,c_{q}\right)  $ (by the definition
of ``column sums''). In other words, $\left(
\text{the column sums of }A\right)  =\left(  c_{1},c_{2},\ldots,c_{q}\right)
$. But recall that $\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $ is the
column sums of $A$. Hence,%
\[
\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  =\left(  \text{the column sums
of }A\right)  =\left(  c_{1},c_{2},\ldots,c_{q}\right)  .
\]
In other words, each $j\in\left\{  1,2,\ldots,q\right\}  $ satisfies%
\begin{equation}
\mu_{j}=c_{j}. \label{pf.lem.sol.Lambda.triangular.01pqmat.b.1}%
\end{equation}


But $A=\left(  a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$. Hence, each
$j\in\left\{  1,2,\ldots,q\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }j\text{-th column of }A\right)  =\left(
\begin{array}
[c]{c}%
a_{1,j}\\
a_{2,j}\\
\vdots\\
a_{p,j}%
\end{array}
\right)  . \label{pf.lem.sol.Lambda.triangular.01pqmat.b.2}%
\end{equation}


Let $j\in\left\{  1,2,\ldots,q\right\}  $. Then,
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.b.1}) yields%
\begin{align*}
\mu_{j}  &  =c_{j}=\left(  \text{the sum of all entries in }%
\underbrace{\text{the }j\text{-th column of }A}_{\substack{=\left(
\begin{array}
[c]{c}%
a_{1,j}\\
a_{2,j}\\
\vdots\\
a_{p,j}%
\end{array}
\right)  \\\text{(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.b.2}))}%
}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }c_{j}\text{ was defined as the sum
of all entries in the }j\text{-th column of }A\right) \\
&  =\left(  \text{the sum of all entries in }\left(
\begin{array}
[c]{c}%
a_{1,j}\\
a_{2,j}\\
\vdots\\
a_{p,j}%
\end{array}
\right)  \right) \\
&  =a_{1,j}+a_{2,j}+\cdots+a_{p,j}=\sum_{i=1}^{p}a_{i,j}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(b).

(c) Comparing%
\[
\sum_{i=1}^{p}\underbrace{\lambda_{i}}_{\substack{=\sum_{j=1}^{q}%
a_{i,j}\\\text{(by Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a))}%
}}=\underbrace{\sum_{i=1}^{p}\sum_{j=1}^{q}}_{=\sum_{j=1}^{q}\sum_{i=1}^{p}%
}a_{i,j}=\sum_{j=1}^{q}\sum_{i=1}^{p}a_{i,j}%
\]
with%
\[
\sum_{j=1}^{q}\underbrace{\mu_{j}}_{\substack{=\sum_{i=1}^{p}a_{i,j}%
\\\text{(by Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(b))}}}=\sum
_{j=1}^{q}\sum_{i=1}^{p}a_{i,j},
\]
we obtain $\sum_{i=1}^{p}\lambda_{i}=\sum_{j=1}^{q}\mu_{j}$. This proves Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(c).

(d) Let $k\in\left\{  0,1,\ldots,q\right\}  $ and $i\in\left\{  1,2,\ldots
,p\right\}  $. Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) yields
$\lambda_{i}=\sum_{j=1}^{q}a_{i,j}$.

We have $\left\{  1,2,\ldots,k\right\}  \subset\left\{  1,2,\ldots,q\right\}
$ (since $k\leq q$ (since $k\in\left\{  0,1,\ldots,q\right\}  $)). Hence, each
$j\in\left\{  1,2,\ldots,k\right\}  $ satisfies $j\in\left\{  1,2,\ldots
,q\right\}  $. Thus, $a_{i,j}$ is well-defined for each $j\in\left\{
1,2,\ldots,k\right\}  $. Now,
\[
\sum_{j=1}^{k}\underbrace{a_{i,j}}_{\substack{\leq1\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij1}))}}}\leq\sum_{j=1}%
^{k}1=k\cdot1=k.
\]


Furthermore, $k\in\left\{  0,1,\ldots,q\right\}  $, so that $0\leq k\leq q$.
Hence,%
\[
\sum_{j=1}^{q}a_{i,j}=\sum_{j=1}^{k}a_{i,j}+\sum_{j=k+1}^{q}%
\underbrace{a_{i,j}}_{\substack{\geq0\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij0}))}}}\geq\sum_{j=1}^{k}%
a_{i,j}+\underbrace{\sum_{j=k+1}^{q}0}_{=0}=\sum_{j=1}^{k}a_{i,j},
\]
so that%
\[
\sum_{j=1}^{k}a_{i,j}\leq\sum_{j=1}^{q}a_{i,j}=\lambda_{i}.
\]


But let us recall the following basic fact about integers: If three integers
$\alpha,\beta,\gamma$ satisfy $\alpha\leq\beta$ and $\alpha\leq\gamma$, then
$\alpha\leq\min\left\{  \beta,\gamma\right\}  $. Applying this fact to
$\alpha=\sum_{j=1}^{k}a_{i,j}$, $\beta=\lambda_{i}$ and $\gamma=k$, we
conclude that $\sum_{j=1}^{k}a_{i,j}\leq\min\left\{  \lambda_{i},k\right\}  $
(since $\sum_{j=1}^{k}a_{i,j}\leq\lambda_{i}$ and $\sum_{j=1}^{k}a_{i,j}\leq
k$). In other words, $\min\left\{  \lambda_{i},k\right\}  \geq\sum_{j=1}%
^{k}a_{i,j}$. This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(d).

(e) Let $k\in\left\{  0,1,\ldots,q\right\}  $. Now,%
\begin{align*}
\sum_{i=1}^{p}\underbrace{\min\left\{  \lambda_{i},k\right\}  }%
_{\substack{\geq\sum_{j=1}^{k}a_{i,j}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(d))}}}  &  \geq\underbrace{\sum
_{i=1}^{p}\sum_{j=1}^{k}}_{=\sum_{j=1}^{k}\sum_{i=1}^{p}}a_{i,j}=\sum
_{j=1}^{k}\underbrace{\sum_{i=1}^{p}a_{i,j}}_{\substack{=\mu_{j}\\\text{(since
}\mu_{j}=\sum_{i=1}^{p}a_{i,j}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(b)))}}}\\
&  =\sum_{j=1}^{k}\mu_{j}.
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(e).

(f) Assume that
\begin{equation}
\left(  \sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}%
\mu_{j}\text{ for each }k\in\left\{  1,2,\ldots,q\right\}  \right)  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.f.ass}%
\end{equation}
We want to show that $A=\left(  \left[  j\leq\lambda_{i}\right]  \right)
_{1\leq i\leq p,\ 1\leq j\leq q}$.

Let $k\in\left\{  1,2,\ldots,q\right\}  $. Thus, $k\in\left\{  1,2,\ldots
,q\right\}  \subset\left\{  0,1,\ldots,q\right\}  $. But
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.ass}) yields
\begin{align*}
\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}   &  =\sum_{j=1}%
^{k}\underbrace{\mu_{j}}_{\substack{=\sum_{i=1}^{p}a_{i,j}\\\text{(by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(b))}}}=\underbrace{\sum_{j=1}^{k}%
\sum_{i=1}^{p}}_{=\sum_{i=1}^{p}\sum_{j=1}^{k}}a_{i,j}=\sum_{i=1}^{p}%
\sum_{j=1}^{k}a_{i,j}\\
&  \leq\sum_{i=1}^{p}\sum_{j=1}^{k}a_{i,j}.
\end{align*}
Furthermore, we know that $\min\left\{  \lambda_{i},k\right\}  \geq\sum
_{j=1}^{k}a_{i,j}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $ (by Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(d)). Hence, Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq} (applied to $\min\left\{  \lambda
_{i},k\right\}  $ and $\sum_{j=1}^{k}a_{i,j}$ instead of $a_{i}$ and $b_{i}$)
shows that%
\begin{equation}
\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}a_{i,j}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,p\right\}  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.f.3}%
\end{equation}


Now, forget that we fixed $k$. We thus have proven
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.3}) for each $k\in\left\{
1,2,\ldots,q\right\}  $.

Now, pick $i\in\left\{  1,2,\ldots,p\right\}  $. We shall show that%
\begin{equation}
a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  \ \ \ \ \ \ \ \ \ \ \text{for each
}k\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.f.4}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.4}):} Let
$k\in\left\{  1,2,\ldots,q\right\}  $. We must prove the inequality
$a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  $.

Indeed, we are in one of the following two cases:

\textit{Case 1:} We have $k\leq\lambda_{i}$.

\textit{Case 2:} We don't have $k\leq\lambda_{i}$.

Let us first consider Case 1. In this case, we have $k\leq\lambda_{i}$. Thus,
$\left[  k\leq\lambda_{i}\right]  =1$. Also, $k\geq1$ (since $k\in\left\{
1,2,\ldots,q\right\}  $). But from $k\leq\lambda_{i}$, we also obtain
$\min\left\{  \lambda_{i},k\right\}  =k$. Thus,%
\[
\sum_{j=1}^{k}1=k\cdot1=k\leq k=\min\left\{  \lambda_{i},k\right\}
=\sum_{j=1}^{k}a_{i,j}%
\]
(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.3})). Furthermore, $1\geq
a_{i,j}$ for each $j\in\left\{  1,2,\ldots,k\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,k\right\}  $.
Then, $j\in\left\{  1,2,\ldots,k\right\}  \subset\left\{  1,2,\ldots
,q\right\}  $ (since $k\leq q$ (since $k\in\left\{  1,2,\ldots,q\right\}  $)).
Hence, (\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij1}) yields $a_{i,j}\leq
1$. In other words, $1\geq a_{i,j}$. Qed.}. Hence, Lemma
\ref{lem.sol.Lambda.triangular.ineq-eq2} (applied to $1$ and $a_{i,j}$ instead
of $a_{j}$ and $b_{j}$) shows that $1=a_{i,j}$ for each $j\in\left\{
1,2,\ldots,k\right\}  $. Applying this to $j=k$, we obtain $1=a_{i,k}$ (since
$k\in\left\{  1,2,\ldots,k\right\}  $ (since $k\geq1$)). Hence, $a_{i,k}%
=1=\left[  k\leq\lambda_{i}\right]  $, so that $a_{i,k}\geq\left[
k\leq\lambda_{i}\right]  $. Thus, the inequality $a_{i,k}\geq\left[
k\leq\lambda_{i}\right]  $ is proven in Case 1.

Next, let us consider Case 2. In this case, we don't have $k\leq\lambda_{i}$.
Thus, $\left[  k\leq\lambda_{i}\right]  =0$. But
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.aij0}) (applied to $j=k$) yields
$a_{i,k}\geq0$. Thus, $a_{i,k}\geq0=\left[  k\leq\lambda_{i}\right]  $. Hence,
the inequality $a_{i,k}\geq\left[  k\leq\lambda_{i}\right]  $ is proven in
Case 2.

We have now proven the inequality $a_{i,k}\geq\left[  k\leq\lambda_{i}\right]
$ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, we thus conclude that the inequality $a_{i,k}\geq\left[
k\leq\lambda_{i}\right]  $ always holds. This proves
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.4}).]

Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) yields $\lambda_{i}%
=\sum_{j=1}^{q}a_{i,j}$. But Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(d)
(applied to $k=q$) yields $\min\left\{  \lambda_{i},q\right\}  \geq\sum
_{j=1}^{q}a_{i,j}$. Hence, $\sum_{j=1}^{q}a_{i,j}\leq\min\left\{  \lambda
_{i},q\right\}  $, so that $\lambda_{i}=\sum_{j=1}^{q}a_{i,j}\leq\min\left\{
\lambda_{i},q\right\}  $. If we had $\lambda_{i}>q$, then we would have
$\min\left\{  \lambda_{i},q\right\}  =q<\lambda_{i}\leq\min\left\{
\lambda_{i},q\right\}  $, which would be absurd. Thus, we cannot have
$\lambda_{i}>q$. We therefore must have $\lambda_{i}\leq q$. Hence, Lemma
\ref{lem.sol.Lambda.triangular.sum-iver} (applied to $r=\lambda_{i}$) yields%
\[
\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  =\lambda_{i}=\sum_{j=1}%
^{q}a_{i,j}.
\]
Thus,%
\[
\sum_{j=1}^{q}a_{i,j}=\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  \leq
\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  .
\]
Also,
\[
a_{i,j}\geq\left[  j\leq\lambda_{i}\right]  \ \ \ \ \ \ \ \ \ \ \text{for each
}j\in\left\{  1,2,\ldots,q\right\}
\]
(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.4}) (applied to $k=j$)).
Hence, Lemma \ref{lem.sol.Lambda.triangular.ineq-eq2} (applied to $q$,
$a_{i,j}$ and $\left[  j\leq\lambda_{i}\right]  $ instead of $k$, $a_{j}$ and
$b_{j}$) shows that%
\begin{equation}
a_{i,j}=\left[  j\leq\lambda_{i}\right]  \ \ \ \ \ \ \ \ \ \ \text{for each
}j\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.01pqmat.f.7}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven
(\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.7}) for each $i\in\left\{
1,2,\ldots,p\right\}  $.

Now,
\[
A=\left(  \underbrace{a_{i,j}}_{\substack{=\left[  j\leq\lambda_{i}\right]
\\\text{(by (\ref{pf.lem.sol.Lambda.triangular.01pqmat.f.7}))}}}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}=\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}.
\]
This proves Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(f).
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.majo-mu}Let $n\in \NN$ and
$q\in \NN$. Let $\lambda\in\Par_{n}$ and $\mu
\in\Par_{n}$ be such that $\ell\left(  \mu\right)
\leq q$. Assume that%
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\]
Then, $\lambda\triangleright\mu$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.majo-mu}.]We have assumed that%
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.majo-mu.have}%
\end{equation}
Hence,
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{q}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{q}
\label{pf.lem.sol.Lambda.triangular.majo-mu.q}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}):} We
are in one of the following two cases:
\par
\textit{Case 1:} We have $q=0$.
\par
\textit{Case 2:} We have $q\neq0$.
\par
Let us first consider Case 1. In this case, we have $q=0$. Hence, $\lambda
_{1}+\lambda_{2}+\cdots+\lambda_{q}=\lambda_{1}+\lambda_{2}+\cdots+\lambda
_{0}=\left(  \text{empty sum}\right)  =0$. Also, from $q=0$, we obtain
$\mu_{1}+\mu_{2}+\cdots+\mu_{q}=\mu_{1}+\mu_{2}+\cdots+\mu_{0}=\left(
\text{empty sum}\right)  =0$. Thus, $\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{q}=0\geq0=\mu_{1}+\mu_{2}+\cdots+\mu_{q}$. Thus,
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $q\neq0$. Hence, $q$ is a
positive integer (since $q\in \NN$). Thus, $q\in\left\{  1,2,\ldots
,q\right\}  $. Therefore, (\ref{pf.lem.sol.Lambda.triangular.majo-mu.have})
(applied to $k=q$) yields $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{q}\geq
\mu_{1}+\mu_{2}+\cdots+\mu_{q}$. Thus,
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}) is proven in Case 2.
\par
We have now proven (\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}) in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, we thus
conclude that (\ref{pf.lem.sol.Lambda.triangular.majo-mu.q}) always holds.}.

Both $\lambda$ and $\mu$ are partitions of $n$ (since $\lambda\in
\Par_{n}$ and $\mu\in\Par_{n}%
$). Thus, we have $\lambda\triangleright\mu$ if and only if
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\label{pf.lem.sol.Lambda.triangular.majo-mu.wish}%
\end{equation}
(by the definition of the dominance order).

Now, we are going to prove (\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}).

[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}):} Let
$k\in\left\{  1,2,\ldots,n\right\}  $. We must prove that $\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}$.

If $k\in\left\{  1,2,\ldots,q\right\}  $, then this follows immediately from
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.have}). Hence, for the rest of this
proof of $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}%
+\cdots+\mu_{k}$, we can WLOG assume that we don't have $k\in\left\{
1,2,\ldots,q\right\}  $. Assume this.

We have $k\notin\left\{  1,2,\ldots,q\right\}  $ (since we don't have
$k\in\left\{  1,2,\ldots,q\right\}  $). Combining $k\in\left\{  1,2,\ldots
,n\right\}  \subset\left\{  1,2,3,\ldots\right\}  $ with $k\notin\left\{
1,2,\ldots,q\right\}  $, we obtain $k\in\left\{  1,2,3,\ldots\right\}
\setminus\left\{  1,2,\ldots,q\right\}  =\left\{  q+1,q+2,q+3,\ldots\right\}
$. Thus, $k\geq q+1>q$. Hence, $q<k$. Now,%
\begin{equation}
\mu_{i}=0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  q+1,q+2,\ldots
,k\right\}  \label{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.3}):} Let $i\in\left\{
q+1,q+2,\ldots,k\right\}  $. Thus, $i\geq q+1>q\geq\ell\left(  \mu\right)  $
(since $\ell\left(  \mu\right)  \leq q$). But each integer $j>\ell\left(
\mu\right)  $ satisfies $\mu_{j}=0$ (by the definition of $\ell\left(
\mu\right)  $). Applying this to $j=i$, we obtain $\mu_{i}=0$ (since
$i>\ell\left(  \mu\right)  $). This proves
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.3}).}. Now,%
\begin{align}
\mu_{1}+\mu_{2}+\cdots+\mu_{k}  &  =\sum_{i=1}^{k}\mu_{i}=\sum_{i=1}^{q}%
\mu_{i}+\sum_{i=q+1}^{k}\underbrace{\mu_{i}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.3}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq q<k\right) \nonumber\\
&  =\sum_{i=1}^{q}\mu_{i}+\underbrace{\sum_{i=q+1}^{k}0}_{=0}=\sum_{i=1}%
^{q}\mu_{i}=\mu_{1}+\mu_{2}+\cdots+\mu_{q}.
\label{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.5}%
\end{align}


On the other hand,%
\begin{align*}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}  &  =\sum_{i=1}^{k}\lambda_{i}%
=\sum_{i=1}^{q}\lambda_{i}+\sum_{i=q+1}^{k}\underbrace{\lambda_{i}%
}_{\substack{\geq0\\\text{(since }\lambda_{i}\in\mathbb{N}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq q<k\right) \\
&  \geq\sum_{i=1}^{q}\lambda_{i}+\underbrace{\sum_{i=q+1}^{k}0}_{=0}%
=\sum_{i=1}^{q}\lambda_{i}=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{q}\\
&  \geq\mu_{1}+\mu_{2}+\cdots+\mu_{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.q})}\right) \\
&  =\mu_{1}+\mu_{2}+\cdots+\mu_{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish.pf.5})}\right)  .
\end{align*}
This proves $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$. Thus, (\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish})
is proven.]

Recall that we have $\lambda\triangleright\mu$ if and only if
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}) holds. Thus, we have
$\lambda\triangleright\mu$ (since
(\ref{pf.lem.sol.Lambda.triangular.majo-mu.wish}) holds). This proves Lemma
\ref{lem.sol.Lambda.triangular.majo-mu}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.majo-lam}Let $n\in \NN$ and
$q\in \NN$. Let $\lambda\in\Par_{n}$ and $\mu
\in\Par_{n}$ be such that $\ell\left(  \lambda\right)
\leq q$. Assume that%
\[
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\]
Then, $\lambda\triangleright\mu$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.majo-lam}.]We have assumed that%
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.sol.Lambda.triangular.majo-lam.have}%
\end{equation}


Both $\lambda$ and $\mu$ are partitions of $n$ (since $\lambda\in
\Par_{n}$ and $\mu\in\Par_{n}%
$). Thus, we have $\lambda\triangleright\mu$ if and only if
\begin{equation}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\label{pf.lem.sol.Lambda.triangular.majo-lam.wish}%
\end{equation}
(by the definition of the dominance order).

Now, we are going to prove (\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}).

[\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}):} Let
$k\in\left\{  1,2,\ldots,n\right\}  $. We must prove that $\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{k}$.

If $k\in\left\{  1,2,\ldots,q\right\}  $, then this follows immediately from
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.have}). Hence, for the rest of
this proof of $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$, we can WLOG assume that we don't have $k\in\left\{
1,2,\ldots,q\right\}  $. Assume this.

We have $k\notin\left\{  1,2,\ldots,q\right\}  $ (since we don't have
$k\in\left\{  1,2,\ldots,q\right\}  $). Combining $k\in\left\{  1,2,\ldots
,n\right\}  \subset\left\{  1,2,3,\ldots\right\}  $ with $k\notin\left\{
1,2,\ldots,q\right\}  $, we obtain $k\in\left\{  1,2,3,\ldots\right\}
\setminus\left\{  1,2,\ldots,q\right\}  =\left\{  q+1,q+2,q+3,\ldots\right\}
$. Thus, $k\geq q+1>q\geq\ell\left(  \lambda\right)  $ (since $\ell\left(
\lambda\right)  \leq q$). Thus, every integer $i>k$ satisfies
\begin{equation}
\lambda_{i}=0 \label{pf.lem.sol.Lambda.triangular.majo-lam.wish.pf.5}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish.pf.5}):} Let $i>k$ be an
integer. Then, $i>k>\ell\left(  \lambda\right)  $. But every integer
$j>\ell\left(  \lambda\right)  $ satisfies $\lambda_{j}=0$ (by the definition
of $\ell\left(  \lambda\right)  $). Applying this to $j=i$, we obtain
$\lambda_{i}=0$ (since $i>\ell\left(  \lambda\right)  $). This proves
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish.pf.5}).}.

From $\mu\in\Par_{n}$, we obtain $\left\vert
\mu\right\vert =n$. Hence, $n=\left\vert \mu\right\vert =\mu_{1}+\mu_{2}%
+\mu_{3}+\cdots$ (by the definition of $\left\vert \mu\right\vert $).

But the definition of $\left\vert \lambda\right\vert $ yields
\begin{align*}
\left\vert \lambda\right\vert  &  =\lambda_{1}+\lambda_{2}+\lambda_{3}%
+\cdots=\sum_{i=1}^{\infty}\lambda_{i}=\sum_{i=1}^{k}\lambda_{i}+\sum
_{i=k+1}^{\infty}\underbrace{\lambda_{i}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish.pf.5})}\\\text{(since }i\geq
k+1>k\text{))}}}=\sum_{i=1}^{k}\lambda_{i}+\underbrace{\sum_{i=k+1}^{\infty}%
0}_{=0}\\
&  =\sum_{i=1}^{k}\lambda_{i}=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}.
\end{align*}
Thus,%
\begin{align*}
\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}  &  =\left\vert \lambda\right\vert
=n\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda\in\Par_{n}\right) \\
&  =\mu_{1}+\mu_{2}+\mu_{3}+\cdots=\sum_{i=1}^{\infty}\mu_{i}=\sum_{i=1}%
^{k}\mu_{i}+\sum_{i=k+1}^{\infty}\underbrace{\mu_{i}}_{\substack{\geq
0\\\text{(since }\mu_{i}\in\mathbb{N}\text{)}}}\\
&  \geq\sum_{i=1}^{k}\mu_{i}+\underbrace{\sum_{i=k+1}^{\infty}0}_{=0}%
=\sum_{i=1}^{k}\mu_{i}=\mu_{1}+\mu_{2}+\cdots+\mu_{k}.
\end{align*}
This proves $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu
_{2}+\cdots+\mu_{k}$. Thus, (\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish})
is proven.]

Recall that we have $\lambda\triangleright\mu$ if and only if
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}) holds. Thus, we have
$\lambda\triangleright\mu$ (since
(\ref{pf.lem.sol.Lambda.triangular.majo-lam.wish}) holds). This proves Lemma
\ref{lem.sol.Lambda.triangular.majo-lam}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.01mat}Let $n\in \NN$, $p\in \NN$
and $q\in \NN$. Let $\lambda\in\Par_{n}$ and
$\mu\in\Par_{n}$. Let $A\in\left\{  0,1\right\}
^{p\times q}$ be a $\left\{  0,1\right\}  $-matrix having row sums $\lambda$
and column sums $\mu$. Then:

\begin{enumerate}
\item[(a)] We have $\lambda^{t}\triangleright\mu$.

\item[(b)] If $\mu=\lambda^{t}$, then $A=\left(  \left[  j\leq\lambda
_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.01mat}.]The row sums of $A$ is
$\lambda$ (since $A$ has row sums $\lambda$). In other words, $\lambda$ is the
row sums of $A$. Thus, $\lambda$ is a $p$-tuple (since the row sums of $A$ is
a $p$-tuple (by the definition of ``row sums'')).
Therefore, $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{p}\right)  $. Thus, $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{p}\right)  $ is the row sums of $A$ (since $\lambda$ is the row sums of $A$).

The column sums of $A$ is $\mu$ (since $A$ has column sums $\mu$). In other
words, $\mu$ is the column sums of $A$. Thus, $\mu$ is a $q$-tuple (since the
column sums of $A$ is a $q$-tuple (by the definition of ``column sums'')).
Therefore, $\mu=\left(  \mu_{1},\mu
_{2},\ldots,\mu_{q}\right)  $. Thus, $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{q}\right)  $ is the column sums of $A$ (since $\mu$ is the column sums of
$A$).

From $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $, we obtain
$\ell\left(  \mu\right)  \leq q$.

We have $\left\vert \lambda^{t}\right\vert =\left\vert \lambda\right\vert $
(since the transpose of a partition always has the same size as the partition
itself). But $\left\vert \lambda\right\vert =n$ (since $\lambda\in
\Par_{n}$). Hence, $\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert =n$, so that $\lambda^{t}\in\Par_{n}$.

Write the $p\times q$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq p,\ 1\leq j\leq q}$. Hence, Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(e) shows that we have%
\begin{equation}
\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  \geq\sum_{j=1}^{k}\mu_{j}
\label{pf.lem.sol.Lambda.triangular.01mat.1}%
\end{equation}
for each $k\in\left\{  0,1,\ldots,q\right\}  $.

But $\lambda\in\Par_{n}\subset\Par$.
Hence, Lemma \ref{lem.sol.Lambda.triangular.transpose1} (applied to
$\nu=\lambda$) shows that
\begin{equation}
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}=\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},k\right\}  \label{pf.lem.sol.Lambda.triangular.01mat.2}%
\end{equation}
for every $k\in \NN$. Hence, every $k\in\left\{  1,2,\ldots,q\right\}  $
satisfies%
\begin{align}
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}  &  =\sum_{j=1}^{\infty}\min\left\{
\lambda_{j},k\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.01mat.2}) (since }k\in\left\{
1,2,\ldots,q\right\}  \subset\mathbb{N}\text{)}\right) \nonumber\\
&  =\sum_{j=1}^{p}\min\left\{  \lambda_{j},k\right\}  +\sum_{j=p+1}^{\infty
}\underbrace{\min\left\{  \lambda_{j},k\right\}  }_{\substack{\geq
0\\\text{(since }\min\left\{  \lambda_{j},k\right\}  \in\mathbb{N}%
\\\text{(since }\lambda_{j}\in\mathbb{N}\text{ and }k\in\mathbb{N}\text{))}%
}}\nonumber\\
&  \geq\sum_{j=1}^{p}\min\left\{  \lambda_{j},k\right\}  +\underbrace{\sum
_{j=p+1}^{\infty}0}_{=0}=\sum_{j=1}^{p}\min\left\{  \lambda_{j},k\right\}
\nonumber\\
&  =\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}
\label{pf.lem.sol.Lambda.triangular.01mat.4}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}j\text{ as }i\right) \nonumber\\
&  \geq\sum_{j=1}^{k}\mu_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.triangular.01mat.1}) (since }k\in\left\{
1,2,\ldots,q\right\}  \subset\left\{  0,1,\ldots q\right\}  \text{)}\right)
\nonumber\\
&  =\mu_{1}+\mu_{2}+\cdots+\mu_{k}.\nonumber
\end{align}
Hence, Lemma \ref{lem.sol.Lambda.triangular.majo-mu} (applied to $\lambda^{t}$
instead of $\lambda$) shows that $\lambda^{t}\triangleright\mu$. This proves
Lemma \ref{lem.sol.Lambda.triangular.01mat}(a).

(b) Assume that $\mu=\lambda^{t}$.

Let $k\in\left\{  1,2,\ldots,q\right\}  $. Then,
(\ref{pf.lem.sol.Lambda.triangular.01mat.4}) shows that
\[
\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}\geq\sum_{i=1}^{p}\min\left\{
\lambda_{i},k\right\}  .
\]
Hence,
\[
\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  \leq\left(  \lambda
^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda
^{t}\right)  _{k}=\sum_{j=1}^{k}\left(  \underbrace{\lambda^{t}}_{=\mu
}\right)  _{j}=\sum_{j=1}^{k}\mu_{j}.
\]
Combining this with (\ref{pf.lem.sol.Lambda.triangular.01mat.1}) (which holds
because $k\in\left\{  1,2,\ldots,q\right\}  \subset\left\{  0,1,\ldots
q\right\}  $), we obtain $\sum_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}
=\sum_{j=1}^{k}\mu_{j}$.

Now, forget that we fixed $k$. We thus have proven that $\left(  \sum
_{i=1}^{p}\min\left\{  \lambda_{i},k\right\}  =\sum_{j=1}^{k}\mu_{j}\text{ for
each }k\in\left\{  1,2,\ldots,q\right\}  \right)  $. Hence, Lemma
\ref{lem.sol.Lambda.triangular.01pqmat}(f) shows that $A=\left(  \left[
j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$. This
proves Lemma \ref{lem.sol.Lambda.triangular.01mat}(b).
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.01matB}Let $p\in \NN$ and
$q\in \NN$. Let $\lambda$ be a partition satisfying $\ell\left(
\lambda\right)  \leq p$ and $\lambda_{1}\leq q$.

Let $B$ be the $p\times q$-matrix $\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}$.

Then, $B$ is a $\left\{  0,1\right\}  $-matrix having row sums $\lambda$ and
column sums $\lambda^{t}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.01matB}.]We have $\left[
j\leq\lambda_{i}\right]  \in\left\{  0,1\right\}  $ for each $i\in\left\{
1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}  $ (since
$\left[  \mathcal{A}\right]  \in\left\{  0,1\right\}  $ for any logical
statement $\mathcal{A}$). Hence, $\left(  \left[  j\leq\lambda_{i}\right]
\right)  _{1\leq i\leq p,\ 1\leq j\leq q}\in\left\{  0,1\right\}  ^{p\times
q}$.

Thus, $B=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq
p,\ 1\leq j\leq q}\in\left\{  0,1\right\}  ^{p\times q}$. Therefore, $B$ is a
$\left\{  0,1\right\}  $-matrix.

From $\ell\left(  \lambda\right)  \leq p$, we obtain $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $.

Let $\left(  \widetilde{\lambda}_{1},\widetilde{\lambda}_{2},\ldots
,\widetilde{\lambda}_{p}\right)  $ be the row sums of $B$. Let $\left(
\widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{q}\right)  $
be the column sums of $B$.

Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(a) (applied to $B$,
$\widetilde{\lambda}_{i}$, $\widetilde{\mu}_{j}$ and $\left[  j\leq\lambda
_{i}\right]  $ instead of $A$, $\lambda_{i}$, $\mu_{j}$ and $a_{i,j}$) shows
that we have%
\begin{equation}
\widetilde{\lambda}_{i}=\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots,p\right\}
\label{pf.lem.sol.Lambda.triangular.01matB.lati}%
\end{equation}
(since the $p\times q$-matrix $B$ is written in the form $B=\left(  \left[
j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$).

Lemma \ref{lem.sol.Lambda.triangular.01pqmat}(b) (applied to $B$,
$\widetilde{\lambda}_{i}$, $\widetilde{\mu}_{j}$ and $\left[  j\leq\lambda
_{i}\right]  $ instead of $A$, $\lambda_{i}$, $\mu_{j}$ and $a_{i,j}$) shows
that we have%
\begin{equation}
\widetilde{\mu}_{j}=\sum_{i=1}^{p}\left[  j\leq\lambda_{i}\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,\ldots,q\right\}
\label{pf.lem.sol.Lambda.triangular.01matB.muti}%
\end{equation}
(since the $p\times q$-matrix $B$ is written in the form $B=\left(  \left[
j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$).

We have%
\begin{equation}
\lambda_{i}\leq q \label{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}%
\end{equation}
for each $i\in\left\{  1,2,3,\ldots\right\}  $\ \ \ \ \footnote{\textit{Proof
of (\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}):} Let $i\in\left\{
1,2,3,\ldots\right\}  $. Thus, $i\geq1$, so that $1\leq i$.
\par
The sequence $\lambda$ is a partition, and thus is weakly decreasing. In other
words, $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\cdots$. In other words,
if $u$ and $v$ are two positive integers satisfying $u\leq v$, then
$\lambda_{u}\geq\lambda_{v}$. Applying this to $u=1$ and $v=i$, we obtain
$\lambda_{1}\geq\lambda_{i}$ (since $1\leq i$). Hence, $\lambda_{i}\leq
\lambda_{1}\leq q$. This proves
(\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}).}.

Let $i\in\left\{  1,2,\ldots,p\right\}  $. Then, $\lambda_{i}\leq
q$\ \ \ \ \footnote{\textit{Proof.} We have $i\in\left\{  1,2,\ldots
,p\right\}  \subset\left\{  1,2,3,\ldots\right\}  $. Thus, $\lambda_{i}\leq q$
(by (\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq})).}. Hence, Lemma
\ref{lem.sol.Lambda.triangular.sum-iver} (applied to $r=\lambda_{i}$) yields
$\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  =\lambda_{i}$. Hence,
(\ref{pf.lem.sol.Lambda.triangular.01matB.lati}) becomes $\widetilde{\lambda
}_{i}=\sum_{j=1}^{q}\left[  j\leq\lambda_{i}\right]  =\lambda_{i}$.

Now, forget that we fixed $i$. We thus have shown that $\widetilde{\lambda
}_{i}=\lambda_{i}$ for each $i\in\left\{  1,2,\ldots,p\right\}  $. In other
words, $\left(  \widetilde{\lambda}_{1},\widetilde{\lambda}_{2},\ldots
,\widetilde{\lambda}_{p}\right)  =\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{p}\right)  =\lambda$ (since $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $). Thus, the row sums of $B$ is
$\lambda$ (since the row sums of $B$ is $\left(  \widetilde{\lambda}%
_{1},\widetilde{\lambda}_{2},\ldots,\widetilde{\lambda}_{p}\right)  $). In
other words, the matrix $B$ has row sums $\lambda$.

On the other hand, let $k\in\left\{  1,2,\ldots,q\right\}  $. Applying
(\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}) to $i=k$, we
obtain
\begin{equation}
\left(  \lambda^{t}\right)  _{k}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}  \right\vert .
\label{pf.lem.sol.Lambda.triangular.01matB.4}%
\end{equation}
But (\ref{pf.lem.sol.Lambda.triangular.01matB.muti}) (applied to $j=k$) shows
that%
\begin{align}
\widetilde{\mu}_{k}  &  =\underbrace{\sum_{i=1}^{p}}_{=\sum_{i\in\left\{
1,2,\ldots,p\right\}  }}\left[  \underbrace{k\leq\lambda_{i}}%
_{\Longleftrightarrow\ \left(  \lambda_{i}\geq k\right)  }\right]  =\sum
_{i\in\left\{  1,2,\ldots,p\right\}  }\left[  \lambda_{i}\geq k\right]
\nonumber\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\\lambda_{i}\geq
k}}\underbrace{\left[  \lambda_{i}\geq k\right]  }_{\substack{=1\\\text{(since
}\lambda_{i}\geq k\text{)}}}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,p\right\}  ;\\\text{not }\lambda_{i}\geq k}}\underbrace{\left[  \lambda
_{i}\geq k\right]  }_{\substack{=0\\\text{(since we don't have }\lambda
_{i}\geq k\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }i\in\left\{  1,2,\ldots,p\right\}  \text{ satisfies either
}\lambda_{i}\geq k\text{ or }\left(  \text{not }\lambda_{i}\geq k\right) \\
\text{(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}  ;\\\lambda_{i}\geq
k}}1+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}
;\\\text{not }\lambda_{i}\geq k}}0}_{=0}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,p\right\}  ;\\\lambda_{i}\geq k}}1\nonumber\\
&  =\left\vert \left\{  i\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{i}\geq k\right\}  \right\vert \cdot1=\left\vert \left\{  i\in\left\{
1,2,\ldots,p\right\}  \ \mid\ \lambda_{i}\geq k\right\}  \right\vert
\nonumber\\
&  =\left\vert \left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{j}\geq k\right\}  \right\vert \label{pf.lem.sol.Lambda.triangular.01matB.5}%
\end{align}
(here, we have renamed the index $i$ as $j$). But
\begin{equation}
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}
=\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  \label{pf.lem.sol.Lambda.triangular.01matB.6}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.01matB.6}):} Let
$i\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  $. Thus, $i$ is a $j\in\left\{  1,2,3,\ldots\right\}  $ satisfying
$\lambda_{j}\geq k$. In other words, $i$ is an element of $\left\{
1,2,3,\ldots\right\}  $ and satisfies $\lambda_{i}\geq k$.
\par
We have $\lambda_{i}\geq k>0$ (since $k\in\left\{  1,2,\ldots,q\right\}  $).
Hence, $\lambda_{i}\neq0$.
\par
Assume (for the sake of contradiction) that $i\notin\left\{  1,2,\ldots
,p\right\}  $. Combining this with $i\in\left\{  1,2,3,\ldots\right\}  $
(since $i$ is an element of $\left\{  1,2,3,\ldots\right\}  $), we obtain
$i\in\left\{  1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots,p\right\}
=\left\{  p+1,p+2,p+3,\ldots\right\}  $. Hence, $i\geq p+1>p\geq\ell\left(
\lambda\right)  $ (since $\ell\left(  \lambda\right)  \leq p$). But each
integer $j>\ell\left(  \lambda\right)  $ satisfies $\lambda_{j}=0$ (by the
definition of $\ell\left(  \lambda\right)  $). Applying this to $j=i$, we
obtain $\lambda_{i}=0$ (since $i>\ell\left(  \lambda\right)  $). This
contradicts $\lambda_{i}\neq0$. This contradiction shows that our assumption
(that $i\notin\left\{  1,2,\ldots,p\right\}  $) was wrong. Hence, we cannot
have $i\notin\left\{  1,2,\ldots,p\right\}  $. In other words, we must have
$i\in\left\{  1,2,\ldots,p\right\}  $. Hence, $i$ is an element of $\left\{
1,2,\ldots,p\right\}  $. Thus, $i$ is a $j\in\left\{  1,2,\ldots,p\right\}  $
satisfying $\lambda_{j}\geq k$ (since $i$ is an element of $\left\{
1,2,\ldots,p\right\}  $ and satisfies $\lambda_{i}\geq k$). In other words,
$i\in\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  $.
\par
Now, forget that we fixed $i$. We thus have shown that $i\in\left\{
j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda_{j}\geq k\right\}  $ for
each $i\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  $. In other words,%
\[
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}
\subset\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  .
\]
Combining this inclusion with%
\[
\left\{  j\in\underbrace{\left\{  1,2,\ldots,p\right\}  }_{\subset\left\{
1,2,3,\ldots\right\}  }\ \mid\ \lambda_{j}\geq k\right\}  \subset\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}  ,
\]
we obtain $\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
k\right\}  =\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{j}\geq k\right\}  $. This proves
(\ref{pf.lem.sol.Lambda.triangular.01matB.6}).}. Hence,
(\ref{pf.lem.sol.Lambda.triangular.01matB.4}) becomes%
\[
\left(  \lambda^{t}\right)  _{k}=\left\vert \underbrace{\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq k\right\}  }%
_{\substack{=\left\{  j\in\left\{  1,2,\ldots,p\right\}  \ \mid\ \lambda
_{j}\geq k\right\}  \\\text{(by (\ref{pf.lem.sol.Lambda.triangular.01matB.6}%
))}}}\right\vert =\left\vert \left\{  j\in\left\{  1,2,\ldots,p\right\}
\ \mid\ \lambda_{j}\geq k\right\}  \right\vert =\widetilde{\mu}_{k}%
\]
(by (\ref{pf.lem.sol.Lambda.triangular.01matB.5})).

Now, forget that we fixed $k$. We thus have shown that $\left(  \lambda
^{t}\right)  _{k}=\widetilde{\mu}_{k}$ for each $k\in\left\{  1,2,\ldots
,q\right\}  $. In other words,%
\[
\left(  \left(  \lambda^{t}\right)  _{1},\left(  \lambda^{t}\right)
_{2},\ldots,\left(  \lambda^{t}\right)  _{q}\right)  =\left(  \widetilde{\mu
}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{q}\right)  .
\]


But $\left(  \lambda^{t}\right)  _{i}=0$ for each integer $i>q$%
\ \ \ \ \footnote{\textit{Proof.} Let $i>q$ be an integer. From
(\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}), we obtain
\[
\left(  \lambda^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert .
\]
\par
But let $h\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda
_{j}\geq i\right\}  $ be arbitrary. We shall derive a contradiction.
\par
We have $h\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda
_{j}\geq i\right\}  $. In other words, $h$ is a $j\in\left\{  1,2,3,\ldots
\right\}  $ satisfying $\lambda_{j}\geq i$. In other words, $h$ is an element
of $\left\{  1,2,3,\ldots\right\}  $ and satisfies $\lambda_{h}\geq i$. But
(\ref{pf.lem.sol.Lambda.triangular.01matB.lambdaileqq}) (applied to $i=h$)
yields $\lambda_{h}\leq q$. This contradicts $\lambda_{h}\geq i>q$.
\par
Now, forget that we fixed $h$. We thus have found a contradiction for each
$h\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq
i\right\}  $. Hence, there exists no $h\in\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq i\right\}  $. In other words, $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  $ is an
empty set. Thus, $\left\vert \left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq i\right\}  \right\vert =0$. Hence,%
\[
\left(  \lambda^{t}\right)  _{i}=\left\vert \left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert =0.
\]
Qed.}. In other words, $\left(  \lambda^{t}\right)  _{q+1}=\left(  \lambda
^{t}\right)  _{q+2}=\left(  \lambda^{t}\right)  _{q+3}=\cdots=0$. Hence,
$\lambda^{t}=\left(  \left(  \lambda^{t}\right)  _{1},\left(  \lambda
^{t}\right)  _{2},\ldots,\left(  \lambda^{t}\right)  _{q}\right)  $ (since we
omit trailing zeroes from a partition). Thus,%
\[
\lambda^{t}=\left(  \left(  \lambda^{t}\right)  _{1},\left(  \lambda
^{t}\right)  _{2},\ldots,\left(  \lambda^{t}\right)  _{q}\right)  =\left(
\widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu}_{q}\right)
=\left(  \text{the column sums of }B\right)
\]
(since $\left(  \widetilde{\mu}_{1},\widetilde{\mu}_{2},\ldots,\widetilde{\mu
}_{q}\right)  $ is the column sums of $B$). In other words, $\lambda^{t}$ is
the column sums of $B$. In other words, the matrix $B$ has column sums
$\lambda^{t}$.

Hence, we know that the matrix $B$ has row sums $\lambda$ and column sums
$\lambda^{t}$. Thus, $B$ is a $\left\{  0,1\right\}  $-matrix having row sums
$\lambda$ and column sums $\lambda^{t}$ (since $B$ is a $\left\{  0,1\right\}
$-matrix). This proves Lemma \ref{lem.sol.Lambda.triangular.01matB}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.lenlat}Let $\lambda$ be a partition. Then:

\begin{enumerate}
\item[(a)] We have $\left(  \lambda^{t}\right)  _{1}=\ell\left(
\lambda\right)  $.

\item[(b)] We have $\lambda_{1}=\ell\left(  \lambda^{t}\right)  $.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.lenlat}.](a) Write the
partition $\lambda$ in the form $\lambda=\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{k},0,0,0,\ldots\right)  $ with $\lambda_{1}\geq
\lambda_{2}\geq\cdots\geq\lambda_{k}>0$. Then, $\ell\left(  \lambda\right)
=k$ (by the definition of $\ell\left(  \lambda\right)  $). Moreover, we have%
\begin{equation}
\lambda_{i}>0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots
,k\right\}  \label{pf.lem.sol.Lambda.triangular.lenlat.a.1}%
\end{equation}
(since $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}>0$). Furthermore,
we have%
\begin{equation}
\lambda_{i}=0\ \ \ \ \ \ \ \ \ \ \text{for each integer }i>\ell\left(
\lambda\right)  \label{pf.lem.sol.Lambda.triangular.lenlat.a.3}%
\end{equation}
(by the definition of $\ell\left(  \lambda\right)  $). Thus,%
\begin{equation}
\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}
=\left\{  1,2,\ldots,k\right\}
\label{pf.lem.sol.Lambda.triangular.lenlat.a.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lambda.triangular.lenlat.a.5}):}
Let $i\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}%
\geq1\right\}  $. Thus, $i$ is a $j\in\left\{  1,2,3,\ldots\right\}  $
satisfying $\lambda_{j}\geq1$. In other words, $i$ is an element of $\left\{
1,2,3,\ldots\right\}  $ and satisfies $\lambda_{i}\geq1$. Thus, $\lambda
_{i}\geq1>0$, so that $\lambda_{i}\neq0$.
\par
If we had $i>\ell\left(  \lambda\right)  $, then we would have $\lambda_{i}=0$
(by (\ref{pf.lem.sol.Lambda.triangular.lenlat.a.3})), which would contradict
$\lambda_{i}\neq0$. Hence, we cannot have $i>\ell\left(  \lambda\right)  $.
Thus, we must have $i\leq\ell\left(  \lambda\right)  $. Hence, $i\leq
\ell\left(  \lambda\right)  =k$, so that $i\in\left\{  1,2,\ldots,k\right\}  $
(since $i$ is an element of $\left\{  1,2,3,\ldots\right\}  $).
\par
Now, forget that we fixed $i$. We thus have shown that $i\in\left\{
1,2,\ldots,k\right\}  $ for each $i\in\left\{  j\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \lambda_{j}\geq1\right\}  $. In other words, $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}
\subset\left\{  1,2,\ldots,k\right\}  $.
\par
On the other hand, let $h\in\left\{  1,2,\ldots,k\right\}  $. Then,
$\lambda_{h}>0$ (by (\ref{pf.lem.sol.Lambda.triangular.lenlat.a.1}) (applied
to $i=h$)). Hence, $\lambda_{h}\geq1$ (since $\lambda_{h}$ is an integer).
Also, $h\in\left\{  1,2,\ldots,k\right\}  \subset\left\{  1,2,3,\ldots
\right\}  $. Hence, $h$ is an element of $\left\{  1,2,3,\ldots\right\}  $ and
satisfies $\lambda_{h}\geq1$. In other words, $h$ is a $j\in\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{j}\geq1$. In other words,
$h\in\left\{  j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}%
\geq1\right\}  $.
\par
Now, forget that we fixed $h$. We thus have shown that $h\in\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}  $ for
each $h\in\left\{  1,2,\ldots,k\right\}  $. In other words, $\left\{
1,2,\ldots,k\right\}  \subset\left\{  j\in\left\{  1,2,3,\ldots\right\}
\ \mid\ \lambda_{j}\geq1\right\}  $. Combining this with $\left\{
j\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}
\subset\left\{  1,2,\ldots,k\right\}  $, we obtain $\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}  =\left\{
1,2,\ldots,k\right\}  $. This proves
(\ref{pf.lem.sol.Lambda.triangular.lenlat.a.5}).}.

Applying (\ref{eq.lem.sol.partition.dominance.lambdat-is-a-partition.1}) to
$i=1$, we obtain
\[
\left(  \lambda^{t}\right)  _{1}=\left\vert \underbrace{\left\{  j\in\left\{
1,2,3,\ldots\right\}  \ \mid\ \lambda_{j}\geq1\right\}  }_{\substack{=\left\{
1,2,\ldots,k\right\}  \\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.lenlat.a.5}))}}}\right\vert =\left\vert
\left\{  1,2,\ldots,k\right\}  \right\vert =k
\]
(since $k\in \NN$). Hence, $\left(  \lambda^{t}\right)  _{1}%
=k=\ell\left(  \lambda\right)  $. This proves Lemma
\ref{lem.sol.Lambda.triangular.lenlat}(a).

(b) Proposition \ref{prop.sol.partition.dominance.lambdatt} yields $\left(
\lambda^{t}\right)  ^{t}=\lambda$. But Lemma
\ref{lem.sol.Lambda.triangular.lenlat}(a) (applied to $\lambda^{t}$ instead of
$\lambda$) yields $\left(  \left(  \lambda^{t}\right)  ^{t}\right)  _{1}%
=\ell\left(  \lambda^{t}\right)  $. In light of $\left(  \lambda^{t}\right)
^{t}=\lambda$, this rewrites as $\lambda_{1}=\ell\left(  \lambda^{t}\right)
$. This proves Lemma \ref{lem.sol.Lambda.triangular.lenlat}(b).
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(h).
\end{commentedout}

(h) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$ be two partitions that don't satisfy
$\lambda^{t}\triangleright\mu$. We must prove that $a_{\lambda,\mu}=0$.

Indeed, there exists no $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$\ \ \ \ \footnote{\textit{Proof.} Let $A$ be a $\left\{
0,1\right\}  $-matrix of size $\ell\left(  \lambda\right)  \times\ell\left(
\mu\right)  $ having row sums $\lambda$ and column sums $\mu$. We shall derive
a contradiction.
\par
We have $A\in\left\{  0,1\right\}  ^{\ell\left(  \lambda\right)  \times
\ell\left(  \mu\right)  }$ (since $A$ is a $\left\{  0,1\right\}  $-matrix of
size $\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)  $). Thus,
Lemma \ref{lem.sol.Lambda.triangular.01mat}(a) shows that $\lambda
^{t}\triangleright\mu$. This contradicts the fact that we don't have
$\lambda^{t}\triangleright\mu$.
\par
Now, forget that we fixed $A$. We thus have found a contradiction for each
$\left\{  0,1\right\}  $-matrix $A$ of size $\ell\left(  \lambda\right)
\times\ell\left(  \mu\right)  $ having row sums $\lambda$ and column sums
$\mu$. Thus, there exists no $\left\{  0,1\right\}  $-matrix of size
$\ell\left(  \lambda\right)  \times\ell\left(  \mu\right)  $ having row sums
$\lambda$ and column sums $\mu$. Qed.}. Thus, the number of all $\left\{
0,1\right\}  $-matrices of size $\ell\left(  \lambda\right)  \times\ell\left(
\mu\right)  $ having row sums $\lambda$ and column sums $\mu$ equals $0$. In
other words, $a_{\lambda,\mu}$ equals $0$ (since $a_{\lambda,\mu}$ is the
number of all $\left\{  0,1\right\}  $-matrices of size $\ell\left(
\lambda\right)  \times\ell\left(  \mu\right)  $ having row sums $\lambda$ and
column sums $\mu$ (by the definition of $a_{\lambda,\mu}$)). In other words,
$a_{\lambda,\mu}=0$. This solves Exercise \ref{exe.Lambda.triangular}(h).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(i).
\end{commentedout}

(i) Let $\lambda\in\Par_{n}$. We shall prove that
$a_{\lambda,\lambda^{t}}=1$.

Let $p=\ell\left(  \lambda\right)  $ and $q=\lambda_{1}$. Then, $\ell\left(
\lambda\right)  =p\leq p$ and $\lambda_{1}=q\leq q$. Also, $q=\lambda_{1}%
=\ell\left(  \lambda^{t}\right)  $ (by Lemma
\ref{lem.sol.Lambda.triangular.lenlat}(b)).

Define the $p\times q$-matrix $B$ as in Lemma
\ref{lem.sol.Lambda.triangular.01matB}. Then, Lemma
\ref{lem.sol.Lambda.triangular.01matB} shows that $B$ is a $\left\{
0,1\right\}  $-matrix having row sums $\lambda$ and column sums $\lambda^{t}$.
Furthermore, $B$ is a $\left\{  0,1\right\}  $-matrix of size $p\times q$. In
other words, $B$ is a $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  $ (since $p=\ell\left(
\lambda\right)  $ and $q=\ell\left(  \lambda^{t}\right)  $). Altogether, we
thus know that $B$ is a $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  $ having row sums
$\lambda$ and column sums $\lambda^{t}$. In other words,%
\[
B\in\left\{  \left\{  0,1\right\}  \text{-matrices of size }\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  \text{ having row sums
}\lambda\text{ and column sums }\lambda^{t}\right\}  .
\]
Thus,%
\begin{equation}
\left\{  B\right\}  \subset\left\{  \left\{  0,1\right\}  \text{-matrices of
size }\ell\left(  \lambda\right)  \times\ell\left(  \lambda^{t}\right)  \text{
having row sums }\lambda\text{ and column sums }\lambda^{t}\right\}  .
\label{sol.Lambda.triangular.i.1}%
\end{equation}


Using Lemma \ref{lem.sol.Lambda.triangular.01mat}(b), it is easy to see that
every $\left\{  0,1\right\}  $-matrix of size $\ell\left(  \lambda\right)
\times\ell\left(  \lambda^{t}\right)  $ having row sums $\lambda$ and column
sums $\lambda^{t}$ must belong to the set $\left\{  B\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $A$ be a $\left\{  0,1\right\}  $-matrix
of size $\ell\left(  \lambda\right)  \times\ell\left(  \lambda^{t}\right)  $
having row sums $\lambda$ and column sums $\lambda^{t}$. We must prove that
$A$ must belong to the set $\left\{  B\right\}  $.
\par
We know that $A$ is a $\left\{  0,1\right\}  $-matrix of size $\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  $. In other words, $A$
is a $\left\{  0,1\right\}  $-matrix of size $p\times q$ (since $\ell\left(
\lambda\right)  =p$ and $\ell\left(  \lambda^{t}\right)  =q$). In other words,
$A\in\left\{  0,1\right\}  ^{p\times q}$.
\par
We have $\left\vert \lambda^{t}\right\vert =\left\vert \lambda\right\vert $
(since the transpose of a partition always has the same size as the partition
itself). But $\left\vert \lambda\right\vert =n$ (since $\lambda\in
\Par_{n}$). Hence, $\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert =n$, so that $\lambda^{t}\in\Par_{n}$.
Clearly, $\lambda^{t}=\lambda^{t}$. Hence, Lemma
\ref{lem.sol.Lambda.triangular.01mat}(b) (applied to $\mu=\lambda^{t}$) shows
that $A=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq
p,\ 1\leq j\leq q}$. But the definition of $B$ yields $B=\left(  \left[
j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq p,\ 1\leq j\leq q}$. Comparing
this with $A=\left(  \left[  j\leq\lambda_{i}\right]  \right)  _{1\leq i\leq
p,\ 1\leq j\leq q}$, we obtain $A=B$. Hence, $A=B\in\left\{  B\right\}  $. In
other words, $A$ must belong to the set $\left\{  B\right\}  $.
\par
Now, forget that we fixed $A$. We thus have shown that if $A$ is a $\left\{
0,1\right\}  $-matrix of size $\ell\left(  \lambda\right)  \times\ell\left(
\lambda^{t}\right)  $ having row sums $\lambda$ and column sums $\lambda^{t}$,
then $A$ must belong to the set $\left\{  B\right\}  $. In other words, every
$\left\{  0,1\right\}  $-matrix of size $\ell\left(  \lambda\right)
\times\ell\left(  \lambda^{t}\right)  $ having row sums $\lambda$ and column
sums $\lambda^{t}$ must belong to the set $\left\{  B\right\}  $. Qed.}. In
other words,%
\[
\left\{  \left\{  0,1\right\}  \text{-matrices of size }\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  \text{ having row sums
}\lambda\text{ and column sums }\lambda^{t}\right\}  \subset\left\{
B\right\}  .
\]
Combining this with (\ref{sol.Lambda.triangular.i.1}), we obtain%
\[
\left\{  \left\{  0,1\right\}  \text{-matrices of size }\ell\left(
\lambda\right)  \times\ell\left(  \lambda^{t}\right)  \text{ having row sums
}\lambda\text{ and column sums }\lambda^{t}\right\}  =\left\{  B\right\}  .
\]


We know that $a_{\lambda,\lambda^{t}}$ is the number of all $\left\{
0,1\right\}  $-matrices of size $\ell\left(  \lambda\right)  \times\ell\left(
\lambda^{t}\right)  $ having row sums $\lambda$ and column sums $\lambda^{t}$
(by the definition of $a_{\lambda,\lambda^{t}}$). Thus,%
\begin{align*}
&  a_{\lambda,\lambda^{t}}\\
&  =\left(  \text{the number of all }\left\{  0,1\right\}  \text{-matrices of
size }\ell\left(  \lambda\right)  \times\ell\left(  \lambda^{t}\right)  \text{
having row sums }\lambda\text{ and column sums }\lambda^{t}\right) \\
&  =\left\vert \underbrace{\left\{  \left\{  0,1\right\}  \text{-matrices of
size }\ell\left(  \lambda\right)  \times\ell\left(  \lambda^{t}\right)  \text{
having row sums }\lambda\text{ and column sums }\lambda^{t}\right\}
}_{=\left\{  B\right\}  }\right\vert =\left\vert \left\{  B\right\}
\right\vert =1.
\end{align*}


Now, forget that we fixed $\lambda$. We thus have shown that
\begin{equation}
a_{\lambda,\lambda^{t}}=1\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda
\in\Par_{n}. \label{sol.Lambda.triangular.i.almost}%
\end{equation}


Now, let $\lambda\in\Par_{n}$. We shall show that
$a_{\lambda^{t},\lambda}=1$.

Proposition \ref{prop.sol.partition.dominance.lambdatt} yields $\left(
\lambda^{t}\right)  ^{t}=\lambda$. But $\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert $ (since the transpose of a partition always
has the same size as the partition itself). But $\left\vert \lambda\right\vert
=n$ (since $\lambda\in\Par_{n}$). Hence, $\left\vert
\lambda^{t}\right\vert =\left\vert \lambda\right\vert =n$, so that
$\lambda^{t}\in\Par_{n}$. Thus,
(\ref{sol.Lambda.triangular.i.almost}) (applied to $\lambda^{t}$ instead of
$\lambda$) yields $a_{\lambda^{t},\left(  \lambda^{t}\right)  ^{t}}=1$. In
light of $\left(  \lambda^{t}\right)  ^{t}=\lambda$, this rewrites as
$a_{\lambda^{t},\lambda}=1$. This solves Exercise \ref{exe.Lambda.triangular}(i).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(j).
\end{commentedout}

Before we solve Exercise \ref{exe.Lambda.triangular}(j), let us introduce some
terminology and prove some lemmas.

\begin{definition}
\label{def.sol.Lambda.triangular.j.1}Let $m\in \NN$. Then, $\left[
m\right]  $ shall denote the subset $\left\{  1,2,\ldots,m\right\}  $ of
$\left\{  1,2,3,\ldots\right\}  $. Notice that $\left\vert \left[  m\right]
\right\vert =m$ for each $m\in \NN$.
\end{definition}

\begin{definition}
\label{def.sol.Lambda.triangular.j.2}Let $p\in \NN$ and $q\in\mathbb{N}%
$. Let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]  $ be any map.
Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Then, we define a $q$-tuple $\varphi_{\ast
}\alpha\in\mathbb{N}^{q}$ by%
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)  .
\]
(Recall that if $U$ is any subset of $\left[  q\right]  $, then $\varphi
^{-1}U$ denotes the subset $\left\{  i\in\left[  p\right]  \ \mid
\ \varphi\left(  i\right)  \in U\right\}  $ of $\left[  p\right]  $.)
\end{definition}

\begin{example}
Let $p=5$ and $q=4$, and let $\varphi:\left[  p\right]  \rightarrow\left[
q\right]  $ be the map that sends $1,2,3,4,5$ to $1,4,4,2,2$, respectively.
Let $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha
_{5}\right)  \in\mathbb{N}^{5}$ be a $5$-tuple. Then, the $4$-tuple
$\varphi_{\ast}\alpha\in\mathbb{N}^{4}$ is
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  4\right\}  }\alpha_{i}\right)  =\left(  \alpha_{1},\alpha
_{4}+\alpha_{5},0,\alpha_{2}+\alpha_{3}\right)
\]
(since $\varphi^{-1}\left\{  1\right\}  =\left\{  1\right\}  $, $\varphi
^{-1}\left\{  2\right\}  =\left\{  4,5\right\}  $, $\varphi^{-1}\left\{
3\right\}  =\varnothing$ and $\varphi^{-1}\left\{  4\right\}  =\left\{
2,3\right\}  $).
\end{example}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l1}Let $p$, $q$ and $r$ be three elements
of $\NN$. Let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]  $
and $\psi:\left[  q\right]  \rightarrow\left[  r\right]  $ be any maps. Let
$\alpha\in\mathbb{N}^{p}$. Then, $\left(  \psi\circ\varphi\right)  _{\ast
}\alpha=\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l1}.]Write the
$p$-tuple $\alpha\in\mathbb{N}^{p}$ in the form $\alpha=\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $ for some elements $\alpha
_{1},\alpha_{2},\ldots,\alpha_{p}$ of $\NN$. Thus,%
\begin{equation}
\left(  \psi\circ\varphi\right)  _{\ast}\alpha=\left(  \sum_{i\in\left(
\psi\circ\varphi\right)  ^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  2\right\}  }\alpha
_{i},\ldots,\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{
r\right\}  }\alpha_{i}\right)  \label{pf.prop.Lambda.triangular.j.l1.LHS}%
\end{equation}
(by the definition of $\left(  \psi\circ\varphi\right)  _{\ast}\alpha$).

Write the $q$-tuple $\varphi_{\ast}\alpha\in\mathbb{N}^{q}$ in the form
$\varphi_{\ast}\alpha=\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  $
for some elements $\beta_{1},\beta_{2},\ldots,\beta_{q}$ of $\NN$.
Then,%
\[
\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  =\varphi_{\ast}%
\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i}%
,\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots,\sum
_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)
\]
(by the definition of $\varphi_{\ast}\alpha$, because $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $). Thus,%
\begin{equation}
\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.prop.Lambda.triangular.j.l1.1}%
\end{equation}


Write the $r$-tuple $\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)
\in\mathbb{N}^{r}$ in the form $\psi_{\ast}\left(  \varphi_{\ast}%
\alpha\right)  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{r}\right)  $ for
some elements $\gamma_{1},\gamma_{2},\ldots,\gamma_{r}$ of $\NN$. Then,%
\[
\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{r}\right)  =\psi_{\ast}\left(
\varphi_{\ast}\alpha\right)  =\left(  \sum_{i\in\psi^{-1}\left\{  1\right\}
}\beta_{i},\sum_{i\in\psi^{-1}\left\{  2\right\}  }\beta_{i},\ldots,\sum
_{i\in\psi^{-1}\left\{  r\right\}  }\beta_{i}\right)
\]
(by the definition of $\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)  $,
because $\varphi_{\ast}\alpha=\left(  \beta_{1},\beta_{2},\ldots,\beta
_{q}\right)  $). Thus,%
\begin{equation}
\gamma_{m}=\sum_{i\in\psi^{-1}\left\{  m\right\}  }\beta_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }m\in\left\{  1,2,\ldots,r\right\}  .
\label{pf.prop.Lambda.triangular.j.l1.3}%
\end{equation}


If $i\in\left[  p\right]  $, $j\in\left[  q\right]  $ and $m\in\left[
r\right]  $ are arbitrary, then we have the following logical equivalence:%
\begin{equation}
\left(  \psi\left(  j\right)  =m\text{ and }\varphi\left(  i\right)
=j\right)  \ \Longleftrightarrow\ \left(  \left(  \psi\circ\varphi\right)
\left(  i\right)  =m\text{ and }j=\varphi\left(  i\right)  \right)
\label{pf.prop.Lambda.triangular.j.l1.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l1.5}):} Let
$i\in\left[  p\right]  $, $j\in\left[  q\right]  $ and $m\in\left[  r\right]
$. We shall first prove the implication $\left(  \psi\left(  j\right)
=m\text{ and }\varphi\left(  i\right)  =j\right)  \ \Longrightarrow\ \left(
\left(  \psi\circ\varphi\right)  \left(  i\right)  =m\text{ and }%
j=\varphi\left(  i\right)  \right)  $.
\par
Indeed, assume that $\left(  \psi\left(  j\right)  =m\text{ and }%
\varphi\left(  i\right)  =j\right)  $ holds. Thus, $\psi\left(  j\right)  =m$
and $\varphi\left(  i\right)  =j$. Hence, $\left(  \psi\circ\varphi\right)
\left(  i\right)  =\psi\left(  \underbrace{\varphi\left(  i\right)  }%
_{=j}\right)  =\psi\left(  j\right)  =m$ and $j=\varphi\left(  i\right)  $.
Hence, $\left(  \left(  \psi\circ\varphi\right)  \left(  i\right)  =m\text{
and }j=\varphi\left(  i\right)  \right)  $ holds.
\par
Now, forget our assumption that $\left(  \psi\left(  j\right)  =m\text{ and
}\varphi\left(  i\right)  =j\right)  $ holds. We thus have proven that
$\left(  \left(  \psi\circ\varphi\right)  \left(  i\right)  =m\text{ and
}j=\varphi\left(  i\right)  \right)  $ holds under the assumption that
$\left(  \psi\left(  j\right)  =m\text{ and }\varphi\left(  i\right)
=j\right)  $ holds. In other words, we have proven the implication%
\begin{equation}
\left(  \psi\left(  j\right)  =m\text{ and }\varphi\left(  i\right)
=j\right)  \ \Longrightarrow\ \left(  \left(  \psi\circ\varphi\right)  \left(
i\right)  =m\text{ and }j=\varphi\left(  i\right)  \right)  .
\label{pf.prop.Lambda.triangular.j.l1.5.pf.1}%
\end{equation}
\par
Now, let us assume that $\left(  \left(  \psi\circ\varphi\right)  \left(
i\right)  =m\text{ and }j=\varphi\left(  i\right)  \right)  $ holds. Thus,
$\left(  \psi\circ\varphi\right)  \left(  i\right)  =m$ and $j=\varphi\left(
i\right)  $. Hence, $m=\left(  \psi\circ\varphi\right)  \left(  i\right)
=\psi\left(  \underbrace{\varphi\left(  i\right)  }_{=j}\right)  =\psi\left(
j\right)  $, so that $\psi\left(  j\right)  =m$ and $\varphi\left(  i\right)
=j$. Hence, $\left(  \psi\left(  j\right)  =m\text{ and }\varphi\left(
i\right)  =j\right)  $ holds.
\par
Now, forget our assumption that $\left(  \left(  \psi\circ\varphi\right)
\left(  i\right)  =m\text{ and }j=\varphi\left(  i\right)  \right)  $ holds.
We thus have proven that $\left(  \psi\left(  j\right)  =m\text{ and }%
\varphi\left(  i\right)  =j\right)  $ holds under the assumption that $\left(
\left(  \psi\circ\varphi\right)  \left(  i\right)  =m\text{ and }%
j=\varphi\left(  i\right)  \right)  $ holds. In other words, we have proven
the implication%
\[
\left(  \left(  \psi\circ\varphi\right)  \left(  i\right)  =m\text{ and
}j=\varphi\left(  i\right)  \right)  \ \Longrightarrow\ \left(  \psi\left(
j\right)  =m\text{ and }\varphi\left(  i\right)  =j\right)  .
\]
Combining this implication with (\ref{pf.prop.Lambda.triangular.j.l1.5.pf.1}),
we conclude that the logical equivalence%
\[
\left(  \psi\left(  j\right)  =m\text{ and }\varphi\left(  i\right)
=j\right)  \ \Longleftrightarrow\ \left(  \left(  \psi\circ\varphi\right)
\left(  i\right)  =m\text{ and }j=\varphi\left(  i\right)  \right)
\]
holds. This proves (\ref{pf.prop.Lambda.triangular.j.l1.5}).}.

Now, each $m\in\left\{  1,2,\ldots,r\right\}  $ satisfies%
\[
\gamma_{m}=\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  m\right\}
}\alpha_{i}%
\]
\footnote{\textit{Proof.} Let $m\in\left\{  1,2,\ldots,r\right\}  $. Thus,
$m\in\left\{  1,2,\ldots,r\right\}  =\left[  r\right]  $. From
(\ref{pf.prop.Lambda.triangular.j.l1.3}), we obtain%
\begin{align*}
\gamma_{m}  &  =\underbrace{\sum_{i\in\psi^{-1}\left\{  m\right\}  }%
}_{\substack{=\sum\limits_{\substack{i\in\left[  q\right]  ;\\\psi\left(  i\right)
\in\left\{  m\right\}  }}=\sum\limits_{\substack{i\in\left[  q\right]  ;\\\psi\left(
i\right)  =m}}\\\text{(because for each }i\in\left[  q\right]  \text{,}%
\\\text{the condition }\left(  \psi\left(  i\right)  \in\left\{  m\right\}
\right)  \text{ is}\\\text{equivalent to }\left(  \psi\left(  i\right)
=m\right)  \text{)}}}\beta_{i}=\sum\limits_{\substack{i\in\left[  q\right]
;\\\psi\left(  i\right)  =m}}\beta_{i}=\sum\limits_{\substack{j\in\left[  q\right]
;\\\psi\left(  j\right)  =m}}\underbrace{\beta_{j}}_{\substack{=\sum
_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}\\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l1.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the}\\
\text{summation index }i\text{ as }j
\end{array}
\right) \\
&  =\sum\limits_{\substack{j\in\left[  q\right]  ;\\\psi\left(  j\right)
=m}}\underbrace{\sum_{i\in\varphi^{-1}\left\{  j\right\}  }}_{\substack{=\sum\limits
_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  \in\left\{
j\right\}  }}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\\\text{(because for each }i\in\left[  p\right]  \text{,}%
\\\text{the condition }\left(  \varphi\left(  i\right)  \in\left\{  j\right\}
\right)  \\\text{is equivalent to }\left(  \varphi\left(  i\right)  =j\right)
\text{)}}}\alpha_{i}=\underbrace{\sum\limits_{\substack{j\in\left[  q\right]
;\\\psi\left(  j\right)  =m}}\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}}_{=\sum_{j\in\left[  q\right]  }%
\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  j\right)  =m;\\\varphi
\left(  i\right)  =j}}}\alpha_{i}\\
&  =\sum_{j\in\left[  q\right]  }\underbrace{\sum\limits_{\substack{i\in\left[
p\right]  ;\\\psi\left(  j\right)  =m;\\\varphi\left(  i\right)  =j}%
}}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(  \psi
\circ\varphi\right)  \left(  i\right)  =m;\\j=\varphi\left(  i\right)
}}\\\text{(because for any given }i\in\left[  p\right]  \text{, the
condition}\\\left(  \psi\left(  j\right)  =m\text{ and }\varphi\left(
i\right)  =j\right)  \text{ holds if and only if the}\\\text{condition
}\left(  \left(  \psi\circ\varphi\right)  \left(  i\right)  =m\text{ and
}j=\varphi\left(  i\right)  \right)  \\\text{holds (because of
(\ref{pf.prop.Lambda.triangular.j.l1.5})))}}}\alpha_{i}=\underbrace{\sum
_{j\in\left[  q\right]  }\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(
\psi\circ\varphi\right)  \left(  i\right)  =m;\\j=\varphi\left(  i\right)  }%
}}_{\substack{=\sum_{i\in\left[  p\right]  }\sum\limits_{\substack{j\in\left[
q\right]  ;\\\left(  \psi\circ\varphi\right)  \left(  i\right)  =m;\\j=\varphi
\left(  i\right)  }}\\=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(
\psi\circ\varphi\right)  \left(  i\right)  =m}}\sum\limits_{\substack{j\in\left[
q\right]  ;\\j=\varphi\left(  i\right)  }}}}\alpha_{i}\\
&  =\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(  \psi\circ\varphi\right)
\left(  i\right)  =m}}\underbrace{\sum\limits_{\substack{j\in\left[  q\right]
;\\j=\varphi\left(  i\right)  }}\alpha_{i}}_{\substack{=\alpha_{i}%
\\\text{(since }\varphi\left(  i\right)  \in\left[  q\right]  \text{)}}%
}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(  \psi\circ\varphi\right)
\left(  i\right)  =m}}\alpha_{i}.
\end{align*}
Comparing this with%
\[
\underbrace{\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{
m\right\}  }}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(
\psi\circ\varphi\right)  \left(  i\right)  \in\left\{  m\right\}  }%
}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(  \psi\circ\varphi\right)
\left(  i\right)  =m}}\\\text{(because for each }i\in\left[  p\right]
\text{,}\\\text{the condition }\left(  \left(  \psi\circ\varphi\right)
\left(  i\right)  \in\left\{  m\right\}  \right)  \\\text{is equivalent to
}\left(  \left(  \psi\circ\varphi\right)  \left(  i\right)  =m\right)
\text{)}}}\alpha_{i}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\left(
\psi\circ\varphi\right)  \left(  i\right)  =m}}\alpha_{i},
\]
we obtain $\gamma_{m}=\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{
m\right\}  }\alpha_{i}$. Qed.}. In other words, we have%
\[
\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{r}\right)  =\left(  \sum
_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  1\right\}  }\alpha
_{i},\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{  2\right\}
}\alpha_{i},\ldots,\sum_{i\in\left(  \psi\circ\varphi\right)  ^{-1}\left\{
r\right\}  }\alpha_{i}\right)  .
\]
Comparing this with (\ref{pf.prop.Lambda.triangular.j.l1.LHS}), we obtain
\[
\left(  \psi\circ\varphi\right)  _{\ast}\alpha=\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{r}\right)  =\psi_{\ast}\left(  \varphi_{\ast}%
\alpha\right)
\]
(since $\psi_{\ast}\left(  \varphi_{\ast}\alpha\right)  =\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{r}\right)  $). This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l1}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l9}Let $p\in \NN$. Let
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  \in
\mathbb{N}^{p}$. Let $\sigma\in\Symm_p$. Then:

\begin{enumerate}
\item[(a)] The $p$-tuple $\sigma_{\ast}\alpha$ is well-defined and satisfies
$\sigma_{\ast}\alpha=\left(  \alpha_{\sigma^{-1}\left(  1\right)  }%
,\alpha_{\sigma^{-1}\left(  2\right)  },\ldots,\alpha_{\sigma^{-1}\left(
p\right)  }\right)  $.

\item[(b)] We have $\sigma_{\ast}\alpha=\alpha$ if and only if $\left(
\alpha_{\sigma\left(  i\right)  }=\alpha_{i}\text{ for all }i\in\left[
p\right]  \right)  $.
\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l9}.]We have
$\sigma\in\Symm_p$. In other words, $\sigma$ belongs to the set
$\Symm_p$. In other words, $\sigma$ belongs to the set of all
permutations of $\left\{  1,2,\ldots,p\right\}  $ (since $\Symm_p$ is
the set of all permutations of $\left\{  1,2,\ldots,p\right\}  $). In other
words, $\sigma$ is a permutation of $\left\{  1,2,\ldots,p\right\}  $. In
other words, $\sigma$ is a permutation of $\left[  p\right]  $ (since $\left[
p\right]  =\left\{  1,2,\ldots,p\right\}  $). In other words, $\sigma$ is a
bijection $\left[  p\right]  \rightarrow\left[  p\right]  $. In other words,
$\sigma$ is a bijective map $\left[  p\right]  \rightarrow\left[  p\right]  $.
Hence, the map $\sigma$ has an inverse $\sigma^{-1}$.

Recall that $\sigma$ is a map $\left[  p\right]  \rightarrow\left[  p\right]
$. Thus, the $p$-tuple $\sigma_{\ast}\alpha$ is well-defined (since $\alpha
\in\mathbb{N}^{p}$). The definition of $\sigma_{\ast}\alpha$ yields%
\begin{equation}
\sigma_{\ast}\alpha=\left(  \sum_{i\in\sigma^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\sigma^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\sigma^{-1}\left\{  p\right\}  }\alpha_{i}\right)  .
\label{pf.prop.Lambda.triangular.j.l9.0}%
\end{equation}
But each $j\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
\sum_{i\in\sigma^{-1}\left\{  j\right\}  }\alpha_{i}=\alpha_{\sigma
^{-1}\left(  j\right)  } \label{pf.prop.Lambda.triangular.j.l9.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l9.1}):} Let
$j\in\left\{  1,2,\ldots,p\right\}  $. Thus, $j\in\left\{  1,2,\ldots
,p\right\}  =\left[  p\right]  $. Hence, $\sigma^{-1}\left(  j\right)
\in\left[  p\right]  $ (since $\sigma^{-1}$ is a map $\left[  p\right]
\rightarrow\left[  p\right]  $). Now, the definition of $\sigma^{-1}\left\{
j\right\}  $ yields%
\[
\sigma^{-1}\left\{  j\right\}  =\left\{  k\in\left[  p\right]  \ \mid
\ \underbrace{\sigma\left(  k\right)  \in\left\{  j\right\}  }%
_{\substack{\Longleftrightarrow\ \left(  \sigma\left(  k\right)  =j\right)
\\\Longleftrightarrow\ \left(  k=\sigma^{-1}\left(  j\right)  \right)
}}\right\}  =\left\{  k\in\left[  p\right]  \ \mid\ k=\sigma^{-1}\left(
j\right)  \right\}  =\left\{  \sigma^{-1}\left(  j\right)  \right\}
\]
(since $\sigma^{-1}\left(  j\right)  \in\left[  p\right]  $). Hence,
$\sum_{i\in\sigma^{-1}\left\{  j\right\}  }\alpha_{i}=\sum_{i\in\left\{
\sigma^{-1}\left(  j\right)  \right\}  }\alpha_{i}=\alpha_{\sigma^{-1}\left(
j\right)  }$. This proves (\ref{pf.prop.Lambda.triangular.j.l9.1}).}. In other
words,%
\[
\left(  \sum_{i\in\sigma^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in
\sigma^{-1}\left\{  2\right\}  }\alpha_{i},\ldots,\sum_{i\in\sigma
^{-1}\left\{  p\right\}  }\alpha_{i}\right)  =\left(  \alpha_{\sigma
^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(  2\right)  },\ldots
,\alpha_{\sigma^{-1}\left(  p\right)  }\right)  .
\]
Hence, (\ref{pf.prop.Lambda.triangular.j.l9.0}) becomes
\[
\sigma_{\ast}\alpha=\left(  \sum_{i\in\sigma^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\sigma^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\sigma^{-1}\left\{  p\right\}  }\alpha_{i}\right)  =\left(
\alpha_{\sigma^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(  2\right)
},\ldots,\alpha_{\sigma^{-1}\left(  p\right)  }\right)  .
\]
Thus, we have shown that the $p$-tuple $\sigma_{\ast}\alpha$ is well-defined
and satisfies $\sigma_{\ast}\alpha=\left(  \alpha_{\sigma^{-1}\left(
1\right)  },\alpha_{\sigma^{-1}\left(  2\right)  },\ldots,\alpha_{\sigma
^{-1}\left(  p\right)  }\right)  $. This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l9}(a).

(b) We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \underbrace{\sigma_{\ast}\alpha}_{=\left(  \alpha_{\sigma
^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(  2\right)  },\ldots
,\alpha_{\sigma^{-1}\left(  p\right)  }\right)  }=\underbrace{\alpha
}_{=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  }\right) \\
&  \Longleftrightarrow\ \left(  \left(  \alpha_{\sigma^{-1}\left(  1\right)
},\alpha_{\sigma^{-1}\left(  2\right)  },\ldots,\alpha_{\sigma^{-1}\left(
p\right)  }\right)  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\right) \\
&  \Longleftrightarrow\ \left(  \alpha_{\sigma^{-1}\left(  j\right)  }%
=\alpha_{j}\text{ for all }j\in\left[  p\right]  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\alpha_{\sigma^{-1}\left(
\sigma\left(  i\right)  \right)  }}_{\substack{=\alpha_{i}\\\text{(since
}\sigma^{-1}\left(  \sigma\left(  i\right)  \right)  =i\text{)}}%
}=\alpha_{\sigma\left(  i\right)  }\text{ for all }i\in\left[  p\right]
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }\sigma\left(
i\right)  \text{ for }j\text{, since the map }\sigma:\left[  p\right]
\rightarrow\left[  p\right]  \text{ is a bijection}\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\alpha_{i}=\alpha_{\sigma\left(
i\right)  }}_{\Longleftrightarrow\ \left(  \alpha_{\sigma\left(  i\right)
}=\alpha_{i}\right)  }\text{ for all }i\in\left[  p\right]  \right) \\
&  \Longleftrightarrow\ \left(  \alpha_{\sigma\left(  i\right)  }=\alpha
_{i}\text{ for all }i\in\left[  p\right]  \right)  .
\end{align*}
In other words, we have $\sigma_{\ast}\alpha=\alpha$ if and only if $\left(
\alpha_{\sigma\left(  i\right)  }=\alpha_{i}\text{ for all }i\in\left[
p\right]  \right)  $. This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l9}(b).
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l2}Let $p\in \NN$. Let $\alpha
\in\mathbb{N}^{p}$. Then, $\left(  \id_{\left[
p\right]  }\right)  _{\ast}\alpha=\alpha$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l2}.]Write the
$p$-tuple $\alpha\in\mathbb{N}^{p}$ in the form $\alpha=\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $ for some elements $\alpha
_{1},\alpha_{2},\ldots,\alpha_{p}$ of $\NN$. Thus,%
\[
\left(  \id_{\left[  p\right]  }\right)  _{\ast}%
\alpha=\left(  \sum_{i\in\left(  \id_{\left[
p\right]  }\right)  ^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in\left(
\id_{\left[  p\right]  }\right)  ^{-1}\left\{
2\right\}  }\alpha_{i},\ldots,\sum_{i\in\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left\{  p\right\}  }\alpha
_{i}\right)
\]
(by the definition of $\left(  \id_{\left[  p\right]
}\right)  _{\ast}\alpha$).

But each $j\in\left\{  1,2,\ldots,p\right\}  $ satisfies $\sum_{i\in\left(
\id_{\left[  p\right]  }\right)  ^{-1}\left\{
j\right\}  }\alpha_{i}=\alpha_{j}$\ \ \ \ \footnote{\textit{Proof:} Let
$j\in\left\{  1,2,\ldots,p\right\}  $. Thus, $j\in\left\{  1,2,\ldots
,p\right\}  =\left[  p\right]  $. Hence, the definition of $\left(
\id_{\left[  p\right]  }\right)  ^{-1}\left\{
j\right\}  $ yields
\[
\left(  \id_{\left[  p\right]  }\right)  ^{-1}\left\{
j\right\}  =\left\{  u\in\left[  p\right]  \ \mid
\ \underbrace{\id_{\left[  p\right]  }\left(
u\right)  }_{=u}\in\left\{  j\right\}  \right\}  =\left\{  u\in\left[
p\right]  \ \mid\ \underbrace{u\in\left\{  j\right\}  }_{\Longleftrightarrow
\ \left(  u=j\right)  }\right\}  =\left\{  u\in\left[  p\right]
\ \mid\ u=j\right\}  =\left\{  j\right\}
\]
(since $j\in\left[  p\right]  $). Hence, $\sum_{i\in\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left\{  j\right\}  }\alpha
_{i}=\sum_{i\in\left\{  j\right\}  }\alpha_{i}=\alpha_{j}$. Qed.}. In other
words,%
\[
\left(  \sum_{i\in\left(  \id_{\left[  p\right]
}\right)  ^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in\left(
\id_{\left[  p\right]  }\right)  ^{-1}\left\{
2\right\}  }\alpha_{i},\ldots,\sum_{i\in\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left\{  p\right\}  }\alpha
_{i}\right)  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  .
\]
Hence,%
\[
\left(  \id_{\left[  p\right]  }\right)  _{\ast}%
\alpha=\left(  \sum_{i\in\left(  \id_{\left[
p\right]  }\right)  ^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in\left(
\id_{\left[  p\right]  }\right)  ^{-1}\left\{
2\right\}  }\alpha_{i},\ldots,\sum_{i\in\left(  \operatorname*{id}%
\nolimits_{\left[  p\right]  }\right)  ^{-1}\left\{  p\right\}  }\alpha
_{i}\right)  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
=\alpha.
\]
This proves Proposition \ref{prop.sol.Lambda.triangular.j.l2}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.j.l10}Let $p\in \NN$ and $q\in
\mathbb{N}$. Let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]  $ be
any map. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Let $\beta=\left(  \beta_{1},\beta
_{2},\ldots,\beta_{q}\right)  \in\mathbb{N}^{q}$ be a $q$-tuple such that
$\left(  \beta_{i}>0\text{ for each }i\in\left[  q\right]  \right)  $ and
$\beta=\varphi_{\ast}\alpha$. Then, the map $\varphi$ is surjective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.j.l10}.]We have $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  $. Hence, the definition of
$\varphi_{\ast}\alpha$ yields%
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots
,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)  .
\]
Thus,%
\[
\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  =\beta=\varphi_{\ast
}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i}%
,\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\ldots,\sum
_{i\in\varphi^{-1}\left\{  q\right\}  }\alpha_{i}\right)  .
\]
In other words,
\begin{equation}
\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,\ldots,q\right\}  .
\label{pf.lem.Lambda.triangular.j.l10.1}%
\end{equation}


Let $j\in\left[  q\right]  $. Assume (for the sake of contradiction) that
$\varphi^{-1}\left\{  j\right\}  =\varnothing$. Thus, $\sum_{i\in\varphi
^{-1}\left\{  j\right\}  }\alpha_{i}=\sum_{i\in\varnothing}\alpha_{i}=\left(
\text{empty sum}\right)  =0$.

But $j\in\left[  q\right]  =\left\{  1,2,\ldots,q\right\}  $. Hence,
(\ref{pf.lem.Lambda.triangular.j.l10.1}) becomes $\beta_{j}=\sum_{i\in
\varphi^{-1}\left\{  j\right\}  }\alpha_{i}=0$. But recall that $\left(
\beta_{i}>0\text{ for each }i\in\left[  q\right]  \right)  $. Applying this to
$i=j$, we obtain $\beta_{j}>0$. Thus, $\beta_{j}\neq0$. This contradicts
$\beta_{j}=0$.

This contradiction shows that our assumption (that $\varphi^{-1}\left\{
j\right\}  =\varnothing$) was wrong. Hence, we have $\varphi^{-1}\left\{
j\right\}  \neq\varnothing$. In other words, the set $\varphi^{-1}\left\{
j\right\}  $ is nonempty. Hence, there exists some $k\in\varphi^{-1}\left\{
j\right\}  $. Consider this $k$.

From $k\in\varphi^{-1}\left\{  j\right\}  $, we obtain $\varphi\left(
k\right)  \in\left\{  j\right\}  $, so that $\varphi\left(  k\right)  =j$.
Hence, $j=\varphi\left(  \underbrace{k}_{\in\varphi^{-1}\left\{  j\right\}
\subset\left[  p\right]  }\right)  \in\varphi\left(  \left[  p\right]
\right)  $.

Now, forget that we fixed $j$. We thus have shown that $j\in\varphi\left(
\left[  p\right]  \right)  $ for each $j\in\left[  q\right]  $. In other
words, $\left[  q\right]  \subset\varphi\left(  \left[  p\right]  \right)  $.
In other words, the map $\varphi$ is surjective. This proves Lemma
\ref{lem.sol.Lambda.triangular.j.l10}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.j.l3}Let $\left(  \beta_{1},\beta_{2}%
,\beta_{3},\ldots\right)  \in\mathbb{N}^{\infty}$ be such that $\beta_{1}%
\geq\beta_{2}\geq\beta_{3}\geq\cdots$. Let $k\in \NN$. Let $R$ be a
finite subset of $\left\{  1,2,3,\ldots\right\}  $ such that $\left\vert
R\right\vert \leq k$. Then,%
\[
\sum_{r\in R}\beta_{r}\leq\beta_{1}+\beta_{2}+\cdots+\beta_{k}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.j.l3}.]We have $\beta_{1}%
\geq\beta_{2}\geq\beta_{3}\geq\cdots$. In other words, if $u$ and $v$ are two
elements of $\left\{  1,2,3,\ldots\right\}  $ satisfying $u\leq v$, then%
\begin{equation}
\beta_{u}\geq\beta_{v}. \label{pf.lem.Lambda.triangular.j.l3.decr}%
\end{equation}


Let $\left(  r_{1},r_{2},\ldots,r_{p}\right)  $ be a list of all elements of
$R$ in increasing order (with no repetitions). Thus, $\sum_{r\in R}\beta
_{r}=\beta_{r_{1}}+\beta_{r_{2}}+\cdots+\beta_{r_{p}}$ and $R=\left\{
r_{1},r_{2},\ldots,r_{p}\right\}  $.

But $\left(  r_{1},r_{2},\ldots,r_{p}\right)  $ is a list of all elements of
$R$ in increasing order (with no repetitions). Thus, $p=\left\vert
R\right\vert $. Hence, $p=\left\vert R\right\vert \leq k$. Also, $p=\left\vert
R\right\vert \in \NN$, so that $p\geq0$. Hence, $0\leq p\leq k$.

We have $\left\{  r_{1},r_{2},\ldots,r_{p}\right\}  =R\subset\left\{
1,2,3,\ldots\right\}  $ (since $R$ is a subset of $\left\{  1,2,3,\ldots
\right\}  $).

Also, $r_{1}<r_{2}<\cdots<r_{p}$ (since $\left(  r_{1},r_{2},\ldots
,r_{p}\right)  $ is a list of all elements of $R$ in increasing order (with no
repetitions)). In other words, each $i\in\left\{  1,2,\ldots,p-1\right\}  $
satisfies%
\begin{equation}
r_{i}<r_{i+1}. \label{pf.lem.Lambda.triangular.j.l3.1}%
\end{equation}


Thus, each $i\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
r_{i}\geq i \label{pf.lem.Lambda.triangular.j.l3.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.Lambda.triangular.j.l3.2}):} We shall
prove (\ref{pf.lem.Lambda.triangular.j.l3.2}) by induction over $i$:
\par
\textit{Induction base:} Assume that $1\in\left\{  1,2,\ldots,p\right\}  $. We
want to show that $r_{1}\geq1$.
\par
From $1\in\left\{  1,2,\ldots,p\right\}  $, we obtain $r_{1}\in\left\{
r_{1},r_{2},\ldots,r_{p}\right\}  \subset\left\{  1,2,3,\ldots\right\}  $, so
that $r_{1}\geq1$.
\par
Now, forget our assumption that $1\in\left\{  1,2,\ldots,p\right\}  $. We thus
have proven that if $1\in\left\{  1,2,\ldots,p\right\}  $, then $r_{1}\geq1$.
In other words, (\ref{pf.lem.Lambda.triangular.j.l3.2}) holds for $i=1$. This
completes the induction base.
\par
\textit{Induction step:} Let $j\in\left\{  1,2,\ldots,p\right\}  $ be such
that $j>1$. Assume that (\ref{pf.lem.Lambda.triangular.j.l3.2}) holds for
$i=j-1$. We must then show that (\ref{pf.lem.Lambda.triangular.j.l3.2}) holds
for $i=j$.
\par
We have $j\neq1$ (since $j>1$). Combining $j\in\left\{  1,2,\ldots,p\right\}
$ with $j\neq1$, we obtain $j\in\left\{  1,2,\ldots,p\right\}  \setminus
\left\{  1\right\}  =\left\{  2,3,\ldots,p\right\}  $, so that $j-1\in\left\{
1,2,\ldots,p-1\right\}  \subset\left\{  1,2,\ldots,p\right\}  $.
\par
We have assumed that (\ref{pf.lem.Lambda.triangular.j.l3.2}) holds for
$i=j-1$. In other words, if $j-1\in\left\{  1,2,\ldots,p\right\}  $, then
$r_{j-1}\geq j-1$. Hence, $r_{j-1}\geq j-1$ (since $j-1\in\left\{
1,2,\ldots,p\right\}  $). But (\ref{pf.lem.Lambda.triangular.j.l3.1}) (applied
to $i=j-1$) yields $r_{j-1}<r_{\left(  j-1\right)  +1}=r_{j}$ (since $\left(
j-1\right)  +1=j$). Hence, $r_{j}>r_{j-1}\geq j-1$.
\par
But $j\in\left\{  1,2,\ldots,p\right\}  $, and thus $r_{j}\in\left\{
r_{1},r_{2},\ldots,r_{p}\right\}  \subset\left\{  1,2,3,\ldots\right\}  $.
Hence, $r_{j}$ is an integer. Thus, both $r_{j}$ and $j-1$ are integers.
Hence, from $r_{j}>j-1$, we obtain $r_{j}\geq\left(  j-1\right)  +1=j$. In
other words, (\ref{pf.lem.Lambda.triangular.j.l3.2}) holds for $i=j$. This
completes the induction step. Thus, the induction proof of
(\ref{pf.lem.Lambda.triangular.j.l3.2}) is complete.}. Hence, each
$i\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
\beta_{r_{i}}\leq\beta_{i} \label{pf.lem.Lambda.triangular.j.l3.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.Lambda.triangular.j.l3.3}):} Let
$i\in\left\{  1,2,\ldots,p\right\}  $. Hence, $r_{i}\in\left\{  r_{1}%
,r_{2},\ldots,r_{p}\right\}  \subset\left\{  1,2,3,\ldots\right\}  $. Thus,
$\beta_{r_{i}}$ is well-defined.
\par
Also, $i\in\left\{  1,2,\ldots,p\right\}  \subset\left\{  1,2,3,\ldots
\right\}  $. Thus, $\beta_{i}$ is well-defined.
\par
From (\ref{pf.lem.Lambda.triangular.j.l3.2}), we obtain $r_{i}\geq i$. Thus,
$i\leq r_{i}$. Hence, (\ref{pf.lem.Lambda.triangular.j.l3.decr}) (applied to
$r_{i}$ and $i$ instead of $u$ and $v$) shows that $\beta_{i}\geq\beta_{r_{i}%
}$. In other words, $\beta_{r_{i}}\leq\beta_{i}$. This proves
(\ref{pf.lem.Lambda.triangular.j.l3.3}).}.

Now,%
\begin{equation}
\sum_{r\in R}\beta_{r}=\beta_{r_{1}}+\beta_{r_{2}}+\cdots+\beta_{r_{p}}%
=\sum_{i=1}^{p}\underbrace{\beta_{r_{i}}}_{\substack{\leq\beta_{i}\\\text{(by
(\ref{pf.lem.Lambda.triangular.j.l3.3}))}}}\leq\sum_{i=1}^{p}\beta_{i}.
\label{pf.lem.Lambda.triangular.j.l3.6}%
\end{equation}


But
\begin{align*}
\beta_{1}+\beta_{2}+\cdots+\beta_{k}  &  =\sum_{i=1}^{k}\beta_{i}=\sum
_{i=1}^{p}\beta_{i}+\sum_{i=p+1}^{k}\underbrace{\beta_{i}}_{\substack{\geq
0\\\text{(since }\beta_{i}\in\mathbb{N}\\\text{(since }\left(  \beta_{1}%
,\beta_{2},\beta_{3},\ldots\right)  \in\mathbb{N}^{\infty}\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the sum into two at
}i=p\text{ (since }0\leq p\leq k\text{)}\right) \\
&  \geq\sum_{i=1}^{p}\beta_{i}+\underbrace{\sum_{i=p+1}^{k}0}_{=0}=\sum
_{i=1}^{p}\beta_{i}\geq\sum_{r\in R}\beta_{r}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.Lambda.triangular.j.l3.6})}\right)  .
\end{align*}
In other words, $\sum_{r\in R}\beta_{r}\leq\beta_{1}+\beta_{2}+\cdots
+\beta_{k}$. This proves Lemma \ref{lem.sol.Lambda.triangular.j.l3}.
\end{proof}

We now introduce some notations:

\begin{definition}
\label{def.sol.Lambda.triangular.j.B}Let $p\in \NN$ and $q\in\mathbb{N}%
$. Let $\alpha\in\mathbb{N}^{p}$ and $\beta\in\mathbb{N}^{q}$. Then, we define
a set $\mathfrak{B}_{\alpha,\beta,p,q}$ by%
\[
\mathfrak{B}_{\alpha,\beta,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \beta=\varphi_{\ast}\alpha\right\}  .
\]
Note that this set $\mathfrak{B}_{\alpha,\beta,p,q}$ is a subset of $\left\{
\varphi:\left[  p\right]  \rightarrow\left[  q\right]  \right\}  =\left[
q\right]  ^{\left[  p\right]  }$, and therefore is a finite set (since
$\left[  q\right]  ^{\left[  p\right]  }$ is a finite set).
\end{definition}

\begin{example}
Let $p=3$, $q=4$, $\alpha=\left(  2,1,2\right)  $ and $\beta=\left(
3,0,2,0\right)  $. Then, $\mathfrak{B}_{\alpha,\beta,p,q}=\left\{  \varphi
_{1},\varphi_{2}\right\}  $, where $\varphi_{1}$ and $\varphi_{2}$ are the two
maps $\left[  3\right]  \rightarrow\left[  4\right]  $ defined by%
\begin{align*}
\varphi_{1}\left(  1\right)   &  =1,\ \ \ \ \ \ \ \ \ \ \varphi_{1}\left(
2\right)  =1,\ \ \ \ \ \ \ \ \ \ \varphi_{1}\left(  3\right)  =3;\\
\varphi_{2}\left(  1\right)   &  =3,\ \ \ \ \ \ \ \ \ \ \varphi_{2}\left(
2\right)  =1,\ \ \ \ \ \ \ \ \ \ \varphi_{2}\left(  3\right)  =1.
\end{align*}


Notice that $\mathfrak{B}_{\alpha,\beta,p,q}$ depends nontrivially on $p$ and
$q$. Indeed, recall that we are identifying any $k$-tuple $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  \in\mathbb{N}^{k}$ with the weak composition
$\left(  a_{1},a_{2},\ldots,a_{k},0,0,0,\ldots\right)  $; therefore, any two
tuples of nonnegative integers that differ only in trailing zeroes are equated
with each other (for example, $\left(  2,3\right)  $ is equated with $\left(
2,3,0\right)  $), because they are being identified with one and the same weak
composition. Thus, for example, the $3$-tuple $\alpha=\left(  2,1,2\right)  $
is equated with the $4$-tuple $\alpha^{\prime}=\left(  2,1,2,0\right)  $. But
the set $\mathfrak{B}_{\alpha,\beta,p,q}=\mathfrak{B}_{\alpha,\beta,3,4}$
cannot be equated with $\mathfrak{B}_{\alpha^{\prime},\beta,4,4}$; indeed, the
set $\mathfrak{B}_{\alpha^{\prime},\beta,4,4}$ has more than two elements (due
to the extra choice in picking the image of $4$ under the map $\varphi
\in\mathfrak{B}_{\alpha^{\prime},\beta,4,4}$), and so we have $\left\vert
\mathfrak{B}_{\alpha,\beta,3,4}\right\vert \neq\left\vert \mathfrak{B}%
_{\alpha^{\prime},\beta,4,4}\right\vert $.
\end{example}

\begin{remark}
We recall that every $k$-tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
\in\mathbb{N}^{k}$ is identified with the weak composition $\left(  a_{1}%
,a_{2},\ldots,a_{k},0,0,0,\ldots\right)  $. Thus, conversely, any weak
composition $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $
is identified with any $k$-tuple that is obtained from it by removing trailing
zeroes. For example, the weak composition $\left(  2,0,3,0,0,0,\ldots\right)
$ is identified with the $3$-tuple $\left(  2,0,3\right)  $, with the
$4$-tuple $\left(  2,0,3,0\right)  $, and so on.

These identifications have an important consequence: If $\lambda$ and $\mu$
are two weak compositions, then there exist nonnegative integers $p$ and $q$
for which $\lambda$ can be identified with a $p$-tuple (namely, with $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $) and $\mu$ can be
identified with a $q$-tuple (namely, with $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{q}\right)  $). Any two such integers $p$ and $q$ give rise to a
well-defined set $\mathfrak{B}_{\lambda,\mu,p,q}$, defined by regarding
$\lambda$ as a $p$-tuple and regarding $\mu$ as a $q$-tuple.
\end{remark}

\begin{definition}
\label{def.sol.Lambda.triangular.j.B'}Let $\lambda\in\Par$ and
$\mu\in\mathbb{N}^{\infty}$. Let $\ell=\ell\left(  \lambda\right)  $. Then, we
define a set $\mathfrak{B}_{\lambda,\mu}^{\prime}$ by%
\[
\mathfrak{B}_{\lambda,\mu}^{\prime}=\left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  \ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all
}j\in\left\{  1,2,3,\ldots\right\}  \right\}  .
\]

\end{definition}

\begin{definition}
Let $X$, $Y$ and $Z$ be three sets such that $X\subset Y$. Let $g:Z\rightarrow
X$ be any map. Then, we define a map $g\mid^{Y}:Z\rightarrow Y$ by%
\[
\left(  \left(  g\mid^{Y}\right)  \left(  z\right)  =g\left(  z\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }z\in Z\right)  .
\]
(This is well-defined, since each $z\in Z$ satisfies $g\left(  z\right)  \in
X\subset Y$.)

The map $g\mid^{Y}$
is identical to the map $g$ except for the fact that its target
is $Y$ rather than $X$.
\end{definition}

If $X$, $Y$ and $Z$ are three sets such that $X\subset Y$, then the
maps of the form $g \mid^Y$ (with $g$ being a map $Z \to X$)
are exactly those maps
$Z\rightarrow Y$ whose image is contained in $X$. More precisely, the
following lemma holds:

\begin{lemma}
\label{lem.sol.Lambda.triangular.j.l5l}Let $X$, $Y$ and $Z$ be three sets such
that $X\subset Y$. Then, the map%
\begin{align*}
X^{Z}  &  \rightarrow\left\{  f\in Y^{Z}\ \mid\ f\left(  Z\right)  \subset
X\right\}  ,\\
g  &  \mapsto g\mid^{Y}%
\end{align*}
is a bijection.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.j.l5l}.]Lemma
\ref{lem.sol.Lambda.triangular.j.l5l} is a fundamental fact about sets. We
thus omit its proof.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l5}Let $\lambda\in\Par$ and
$\mu\in\operatorname*{WC}$. Let $p=\ell\left(  \lambda\right)  $. From
$p=\ell\left(  \lambda\right)  $, we obtain $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $; thus, $\lambda$ is a $p$-tuple
in $\mathbb{N}^{p}$. Let $q\in \NN$ be such that $\mu=\left(  \mu
_{1},\mu_{2},\ldots,\mu_{q}\right)  $ (that is, $\mu_{i}=0$ for all integers
$i>q$). Thus, $\mu$ is a $q$-tuple in $\mathbb{N}^{q}$. Hence, the set
$\mathfrak{B}_{\lambda,\mu,p,q}$ is well-defined (since $\lambda\in
\mathbb{N}^{p}$ and $\mu\in\mathbb{N}^{q}$).

Now, $\mathfrak{B}_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,p,q}$
as sets.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l5}.]We have
$\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $. Thus, $\mu_{q+1}%
=\mu_{q+2}=\mu_{q+3}=\cdots=0$. In other words,%
\begin{equation}
\mu_{j}=0\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  q+1,q+2,q+3,\ldots
\right\}  . \label{pf.prop.Lambda.triangular.j.l5.muj=0}%
\end{equation}


We have $p=\ell\left(  \lambda\right)  $. Hence,%
\begin{equation}
\lambda_{k}>0\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left[  p\right]
\label{pf.prop.Lambda.triangular.j.l5.lambdaneq0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l5.lambdaneq0}):}
Write the partition $\lambda$ in the form $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{h},0,0,0,\ldots\right)  $ with $\lambda_{1}%
\geq\lambda_{2}\geq\cdots\geq\lambda_{h}>0$. Then, $\ell\left(  \lambda
\right)  =h$ (by the definition of $\ell\left(  \lambda\right)  $). Hence,
$p=\ell\left(  \lambda\right)  =h$.
\par
Moreover, we have $\lambda_{k}>0$ for each $k\in\left\{  1,2,\ldots,h\right\}
$ (since $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{h}>0$). In other
words, $\lambda_{k}>0$ for each $k\in\left[  h\right]  $ (since $\left[
h\right]  =\left\{  1,2,\ldots,h\right\}  $). In other words, $\lambda_{k}>0$
for each $k\in\left[  p\right]  $ (since $p=h$). This proves
(\ref{pf.prop.Lambda.triangular.j.l5.lambdaneq0}).}.

The definition of $\mathfrak{B}_{\lambda,\mu,p,q}$ shows that
\[
\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
\]
(where, of course, $\lambda$ and $\mu$ are regarded as the tuples $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  \in\mathbb{N}^{p}$ and
$\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  \in\mathbb{N}^{q}$). Thus,%
\begin{equation}
\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
=\left\{  \zeta:\left[  p\right]  \rightarrow\left[  q\right]  \ \mid
\ \mu=\zeta_{\ast}\lambda\right\}
\label{pf.prop.Lambda.triangular.j.l5.Blmpq=}%
\end{equation}
(here, we have renamed the index $\varphi$ as $\zeta$).

On the other hand, $p=\ell\left(  \lambda\right)  $. Hence, the definition of
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ yields
\begin{align}
\mathfrak{B}_{\lambda,\mu}^{\prime}  &  =\left\{  \varphi:\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all
}j\in\left\{  1,2,3,\ldots\right\}  \right\} \nonumber\\
&  =\left\{  \zeta:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
\ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\zeta\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right\}  \label{pf.prop.Lambda.triangular.j.l5.B'lm=}%
\end{align}
(here, we have renamed the index $\varphi$ as $\zeta$).

But $\left[  q\right]  =\left\{  1,2,\ldots,q\right\}  \subset\left\{
1,2,3,\ldots\right\}  $. Thus, Lemma \ref{lem.sol.Lambda.triangular.j.l5l}
(applied to $X=\left[  q\right]  $, $Y=\left\{  1,2,3,\ldots\right\}  $ and
$Z=\left[  p\right]  $) shows that the map
\begin{align*}
\left[  q\right]  ^{\left[  p\right]  }  &  \rightarrow\left\{  f\in\left\{
1,2,3,\ldots\right\}  ^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]
\right)  \subset\left[  q\right]  \right\}  ,\\
g  &  \mapsto g\mid^{\left\{  1,2,3,\ldots\right\}  }%
\end{align*}
is a bijection. Denote this bijection by $\Phi$.

We have $\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \varphi:\left[  p\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
\subset\left\{  \varphi:\left[  p\right]  \rightarrow\left[  q\right]
\right\}  =\left[  q\right]  ^{\left[  p\right]  }$. In other words,
$\mathfrak{B}_{\lambda,\mu,p,q}$ is a subset of $\left[  q\right]  ^{\left[
p\right]  }$. Thus, the image $\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  $ is well-defined.

The map $\Phi$ is bijective (since it is a bijection), and therefore injective.

If $U$ and $V$ are two finite sets, and if $T$ is a subset of $U$, and if
$\Psi:U\rightarrow V$ is an injective map, then $\Psi\left(  T\right)  \cong
T$ as sets.\footnote{This is a basic fact about sets, and so we omit its
proof.} Applying this to $U=\left[  q\right]  ^{\left[  p\right]  }$,
$V=\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{\left[  p\right]  }%
\ \mid\ f\left(  \left[  p\right]  \right)  \subset\left[  q\right]  \right\}
$, $T=\mathfrak{B}_{\lambda,\mu,p,q}$ and $\Psi=\Phi$, we conclude that%
\begin{equation}
\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  \cong\mathfrak{B}%
_{\lambda,\mu,p,q}\ \ \ \ \ \ \ \ \ \ \text{as sets}
\label{pf.prop.Lambda.triangular.j.l5.5}%
\end{equation}
(since $\Phi$ is injective).

On the other hand, let $\varphi:\left[  p\right]  \rightarrow\left[  q\right]
$ be any map. Then,%
\begin{equation}
\mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  q+1,q+2,q+3,\ldots\right\}
\label{pf.prop.Lambda.triangular.j.l5.equiv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l5.equiv1}):} Let
$j\in\left\{  q+1,q+2,q+3,\ldots\right\}  $. Thus, $j\geq q+1$ and $\mu_{j}=0$
(by (\ref{pf.prop.Lambda.triangular.j.l5.muj=0})).
\par
Let $i\in\left[  p\right]  $ be such that $\varphi\left(  i\right)  =j$. Thus,
$j=\varphi\left(  i\right)  \in\left[  q\right]  $ (since $\varphi$ is a map
$\left[  p\right]  \rightarrow\left[  q\right]  $). Hence, $j\in\left[
q\right]  =\left\{  1,2,\ldots,q\right\}  $, so that $j\leq q$. This
contradicts $j\geq q+1>q$.
\par
Now, forget that we fixed $i$. We thus have found a contradiction for each
$i\in\left[  p\right]  $ satisfying $\varphi\left(  i\right)  =j$. Hence,
there exists no $i\in\left[  p\right]  $ satisfying $\varphi\left(  i\right)
=j$. Thus, the sum $\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}$ is empty. Hence, $\sum\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}=\left(  \text{empty
sum}\right)  =0$. Comparing this with $\mu_{j}=0$, we obtain $\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}$. This proves (\ref{pf.prop.Lambda.triangular.j.l5.equiv1}).}.

Recall that $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{p}\right)  $. Hence, the definition of $\varphi_{\ast}\lambda$ shows that%
\[
\varphi_{\ast}\lambda=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}
}\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i}%
,\ldots,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\lambda_{i}\right)  .
\]
Thus, we have the following chain of logical equivalences:%
\begin{align}
&  \ \left(  \underbrace{\mu}_{=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
}=\underbrace{\varphi_{\ast}\lambda}_{=\left(  \sum_{i\in\varphi^{-1}\left\{
1\right\}  }\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda
_{i},\ldots,\sum_{i\in\varphi^{-1}\left\{  q\right\}  }\lambda_{i}\right)
}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\lambda_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i},\ldots,\sum_{i\in
\varphi^{-1}\left\{  q\right\}  }\lambda_{i}\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu_{j}=\underbrace{\sum_{i\in\varphi
^{-1}\left\{  j\right\}  }}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  \in\left\{  j\right\}  }}=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\varphi\left(  i\right)  =j}}\\\text{(because for each
}i\in\left[  p\right]  \text{,}\\\text{the condition }\left(  \varphi\left(
i\right)  \in\left\{  j\right\}  \right)  \\\text{is equivalent to }\left(
\varphi\left(  i\right)  =j\right)  \text{)}}}\lambda_{i}\text{ for all }%
j\in\left\{  1,2,\ldots,q\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,\ldots,q\right\}  \right)  . \label{pf.prop.Lambda.triangular.j.l5.equiv2}%
\end{align}


Now, we can easily see that we have the following logical implications:%
\begin{equation}
\left(  \mu=\varphi_{\ast}\lambda\right)  \ \Longrightarrow\ \left(  \mu
_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)
\label{pf.prop.Lambda.triangular.j.l5.equiv3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}):}
Assume that $\left(  \mu=\varphi_{\ast}\lambda\right)  $ holds. We must show
that $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  $ holds.
\par
From the equivalence (\ref{pf.prop.Lambda.triangular.j.l5.equiv2}), we
conclude that
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,\ldots,q\right\}
\right)  \label{pf.prop.Lambda.triangular.j.l5.equiv3a.pf.1}%
\end{equation}
holds (because $\left(  \mu=\varphi_{\ast}\lambda\right)  $ holds).
\par
Now, let $j\in\left\{  1,2,3,\ldots\right\}  $. Then, we want to prove that
$\mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}$. If $j\in\left\{  q+1,q+2,q+3,\ldots\right\}  $, then this
follows immediately from (\ref{pf.prop.Lambda.triangular.j.l5.equiv1}). Hence,
for the rest of this proof, we can WLOG assume that we don't have
$j\in\left\{  q+1,q+2,q+3,\ldots\right\}  $. Assume this.
\par
We have $j\notin\left\{  q+1,q+2,q+3,\ldots\right\}  $ (since we don't have
$j\in\left\{  q+1,q+2,q+3,\ldots\right\}  $). Combining this with
$j\in\left\{  1,2,3,\ldots\right\}  $, we obtain $j\in\left\{  1,2,3,\ldots
\right\}  \setminus\left\{  q+1,q+2,q+3,\ldots\right\}  =\left\{
1,2,\ldots,q\right\}  $. Hence, from
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3a.pf.1}), we obtain $\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}$.
\par
Now, forget that we fixed $j$. We thus have shown that $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)  $
holds. This concludes the proof of the implication
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}).} and%
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  \ \Longrightarrow\ \left(  \mu=\varphi_{\ast}\lambda\right)
\label{pf.prop.Lambda.triangular.j.l5.equiv3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l5.equiv3b}):}
Assume that $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right)  $ holds. We must show that $\left(  \mu
=\varphi_{\ast}\lambda\right)  $ holds.
\par
We have assumed that
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  \label{pf.prop.Lambda.triangular.j.l5.equiv3b.pf.1}%
\end{equation}
holds.
\par
Now, let $j\in\left\{  1,2,\ldots,q\right\}  $. Then, $j\in\left\{
1,2,\ldots,q\right\}  \subset\left\{  1,2,3,\ldots\right\}  $. Hence, from
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3b.pf.1}), we obtain $\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}$.
\par
Now, forget that we fixed $j$. We thus have shown that $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}\text{ for all }j\in\left\{  1,2,\ldots,q\right\}  \right)  $
holds. Thus, from the equivalence (\ref{pf.prop.Lambda.triangular.j.l5.equiv2}%
), we obtain $\left(  \mu=\varphi_{\ast}\lambda\right)  $. This concludes the
proof of the implication (\ref{pf.prop.Lambda.triangular.j.l5.equiv3b}).}.
Combining these two implications (\ref{pf.prop.Lambda.triangular.j.l5.equiv3a}%
) and (\ref{pf.prop.Lambda.triangular.j.l5.equiv3b}), we obtain the logical
equivalence%
\begin{equation}
\left(  \mu=\varphi_{\ast}\lambda\right)  \ \Longleftrightarrow\ \left(
\mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)  .
\label{pf.prop.Lambda.triangular.j.l5.equiv3}%
\end{equation}


Now, forget that we fixed $\varphi$. We thus have proven the equivalence
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3}) for each map $\varphi:\left[
p\right]  \rightarrow\left[  q\right]  $.

Now, it is easy to see that $\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  \subset\mathfrak{B}_{\lambda,\mu}^{\prime}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\Phi\left(  \mathfrak{B}%
_{\lambda,\mu,p,q}\right)  $. Thus, $\psi\in\Phi\left(  \mathfrak{B}%
_{\lambda,\mu,p,q}\right)  \subset\left\{  f\in\left\{  1,2,3,\ldots\right\}
^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]  \right)
\subset\left[  q\right]  \right\}  $. In other words, $\psi$ is an
$f\in\left\{  1,2,3,\ldots\right\}  ^{\left[  p\right]  }$ satisfying
$f\left(  \left[  p\right]  \right)  \subset\left[  q\right]  $. In other
words, $\psi$ is an element of $\left\{  1,2,3,\ldots\right\}  ^{\left[
p\right]  }$ and satisfies $\psi\left(  \left[  p\right]  \right)
\subset\left[  q\right]  $.
\par
But $\psi\in\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  $. In other
words, there exists some $\varphi\in\mathfrak{B}_{\lambda,\mu,p,q}$ such that
$\psi=\Phi\left(  \varphi\right)  $. Consider this $\varphi$. We have
$\psi=\Phi\left(  \varphi\right)  =\varphi\mid^{\left\{  1,2,3,\ldots\right\}
}$ (by the definition of $\Phi$).
\par
We have $\varphi\in\mathfrak{B}_{\lambda,\mu,p,q}=\left\{  \zeta:\left[
p\right]  \rightarrow\left[  q\right]  \ \mid\ \mu=\zeta_{\ast}\lambda
\right\}  $ (by (\ref{pf.prop.Lambda.triangular.j.l5.Blmpq=})). In other
words, $\varphi$ is a map $\zeta:\left[  p\right]  \rightarrow\left[
q\right]  $ satisfying $\mu=\zeta_{\ast}\lambda$. In other words, $\varphi$ is
a map $\left[  p\right]  \rightarrow\left[  q\right]  $ and satisfies
$\mu=\varphi_{\ast}\lambda$.
\par
Each $i\in\left[  p\right]  $ satisfies%
\begin{equation}
\underbrace{\psi}_{=\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }}\left(
i\right)  =\left(  \varphi\mid^{\left\{  1,2,3,\ldots\right\}  }\right)
\left(  i\right)  =\varphi\left(  i\right)
\label{pf.prop.Lambda.triangular.j.l5.8.pf.0}%
\end{equation}
(by the definition of $\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }$).
\par
We have $\left(  \mu=\varphi_{\ast}\lambda\right)  $. Hence, from the
equivalence (\ref{pf.prop.Lambda.triangular.j.l5.equiv3}), we conclude that
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  . \label{pf.prop.Lambda.triangular.j.l5.8.pf.1}%
\end{equation}
Now, each $j\in\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align*}
\mu_{j}  &  =\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\psi\left(  i\right)  =j}}\\\text{(because every }i\in\left[  p\right]
\text{ satisfies }\varphi\left(  i\right)  =\psi\left(  i\right)  \\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l5.8.pf.0})))}}}\lambda_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.Lambda.triangular.j.l5.8.pf.1})}\right) \\
&  =\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}%
}\lambda_{i}.
\end{align*}
In other words, we have $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\psi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right)  $.
\par
Now, $\psi$ is a map $\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  $ (since $\psi$ is an element of $\left\{  1,2,3,\ldots\right\}
^{\left[  p\right]  }$) and satisfies $\left(  \mu_{j}=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\psi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\in\left\{  1,2,3,\ldots\right\}  \right)  $. In other words, $\psi$ is a map
$\zeta:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $
satisfying $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\zeta\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right)  $. In other words,%
\[
\psi\in\left\{  \zeta:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\zeta
\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots
\right\}  \right\}  .
\]
In light of (\ref{pf.prop.Lambda.triangular.j.l5.B'lm=}), this rewrites as
$\psi\in\mathfrak{B}_{\lambda,\mu}^{\prime}$.
\par
Now, forget that we fixed $\psi$. We thus have proven that $\psi
\in\mathfrak{B}_{\lambda,\mu}^{\prime}$ for each $\psi\in\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  $. In other words, $\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  \subset\mathfrak{B}_{\lambda,\mu
}^{\prime}$. Qed.} and $\mathfrak{B}_{\lambda,\mu}^{\prime}\subset\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$\psi\in\mathfrak{B}_{\lambda,\mu}^{\prime}$. Thus, $\psi\in\mathfrak{B}%
_{\lambda,\mu}^{\prime}=\left\{  \zeta:\left[  p\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\zeta\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right\}  $ (by
(\ref{pf.prop.Lambda.triangular.j.l5.B'lm=})). In other words, $\psi$ is a map
$\zeta:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $
satisfying $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\zeta\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right)  $. In other words, $\psi$ is a map $\left[
p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ and satisfies
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)  .
\label{pf.prop.Lambda.triangular.j.l5.9.pf.2}%
\end{equation}
\par
Now, let $k\in\left[  p\right]  $. We shall show that $\psi\left(  k\right)
\in\left[  q\right]  $.
\par
Indeed, assume the contrary. Thus, $\psi\left(  k\right)  \notin\left[
q\right]  $. Set $j=\psi\left(  k\right)  $. Combining $j=\psi\left(
k\right)  \in\left\{  1,2,3,\ldots\right\}  $ with $j=\psi\left(  k\right)
\notin\left[  q\right]  =\left\{  1,2,\ldots,q\right\}  $, we obtain
$j\in\left\{  1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots,q\right\}
=\left\{  q+1,q+2,q+3,\ldots\right\}  $. Hence, $j\geq q+1$ and $\mu_{j}=0$
(by (\ref{pf.prop.Lambda.triangular.j.l5.muj=0})).
\par
But $k$ is an element of $\left[  p\right]  $ and satisfies $\psi\left(
k\right)  =j$ (since $j=\psi\left(  k\right)  $). In other words, $k$ is an
$i\in\left[  p\right]  $ satisfying $\psi\left(  i\right)  =j$. Hence, the sum
$\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}%
}\lambda_{i}$ has an addend for $i=k$. If we split off this addend from this
sum, then we obtain%
\[
\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}}\lambda
_{i}=\lambda_{k}+\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(
i\right)  =j;\\i\neq k}}\underbrace{\lambda_{i}}_{\substack{\geq
0\\\text{(since }\lambda_{i}\in\mathbb{N}\text{)}}}\geq\lambda_{k}%
+\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)
=j;\\i\neq k}}0}_{=0}=\lambda_{k}>0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.Lambda.triangular.j.l5.lambdaneq0})}\right)  .
\]
Hence, (\ref{pf.prop.Lambda.triangular.j.l5.9.pf.2}) yields $\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(  i\right)  =j}%
}\lambda_{i}>0$. Thus, $\mu_{j}\neq0$. This contradicts $\mu_{j}=0$.
\par
This contradiction completes our proof of $\psi\left(  k\right)  \in\left[
q\right]  $.
\par
Now, forget that we fixed $k$. We thus have proven that $\psi\left(  k\right)
\in\left[  q\right]  $ for each $k\in\left[  p\right]  $. In other words,
$\psi\left(  \left[  p\right]  \right)  \subset\left[  q\right]  $.
\par
Altogether, we know that $\psi$ is an element of $\left\{  1,2,3,\ldots
\right\}  ^{\left[  p\right]  }$ (since $\psi$ is a map $\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  $) and satisfies $\psi\left(
\left[  p\right]  \right)  \subset\left[  q\right]  $. In other words, $\psi$
is an $f\in\left\{  1,2,3,\ldots\right\}  ^{\left[  p\right]  }$ satisfying
$f\left(  \left[  p\right]  \right)  \subset\left[  q\right]  $. In other
words, $\psi\in\left\{  f\in\left\{  1,2,3,\ldots\right\}  ^{\left[  p\right]
}\ \mid\ f\left(  \left[  p\right]  \right)  \subset\left[  q\right]
\right\}  $. Hence, an element $\Phi^{-1}\left(  \psi\right)  \in\left[
q\right]  ^{\left[  p\right]  }$ is well-defined (since $\Phi$ is a bijection
$\left[  q\right]  ^{\left[  p\right]  }\rightarrow\left\{  f\in\left\{
1,2,3,\ldots\right\}  ^{\left[  p\right]  }\ \mid\ f\left(  \left[  p\right]
\right)  \subset\left[  q\right]  \right\}  $). Let us denote this element
$\Phi^{-1}\left(  \psi\right)  $ by $\varphi$. Thus, $\varphi=\Phi^{-1}\left(
\psi\right)  \in\left[  q\right]  ^{\left[  p\right]  }$. In other words,
$\varphi$ is a map $\left[  p\right]  \rightarrow\left[  q\right]  $.
\par
From $\varphi=\Phi^{-1}\left(  \psi\right)  $, we obtain $\psi=\Phi\left(
\varphi\right)  =\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }$ (by the
definition of $\Phi$).
\par
Each $i\in\left[  p\right]  $ satisfies%
\begin{equation}
\underbrace{\psi}_{=\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }}\left(
i\right)  =\left(  \varphi\mid^{\left\{  1,2,3,\ldots\right\}  }\right)
\left(  i\right)  =\varphi\left(  i\right)
\label{pf.prop.Lambda.triangular.j.l5.9.pf.0}%
\end{equation}
(by the definition of $\varphi\mid^{\left\{  1,2,3,\ldots\right\}  }$).
\par
Now, each $j\in\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align*}
\mu_{j}  &  =\underbrace{\sum\limits_{\substack{i\in\left[  p\right]  ;\\\psi\left(
i\right)  =j}}}_{\substack{=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\\\text{(because every }i\in\left[  p\right]
\text{ satisfies }\psi\left(  i\right)  =\varphi\left(  i\right)  \\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l5.9.pf.0})))}}}\lambda_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.Lambda.triangular.j.l5.9.pf.2})}\right) \\
&  =\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}.
\end{align*}
In other words, we have $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{
1,2,3,\ldots\right\}  \right)  $. Hence, from the equivalence
(\ref{pf.prop.Lambda.triangular.j.l5.equiv3}), we conclude that $\left(
\mu=\varphi_{\ast}\lambda\right)  $.
\par
Altogether, we now know that $\varphi$ is a map $\left[  p\right]
\rightarrow\left[  q\right]  $ and satisfies $\mu=\varphi_{\ast}\lambda$. In
other words, $\varphi$ is a map $\zeta:\left[  p\right]  \rightarrow\left[
q\right]  $ satisfying $\mu=\zeta_{\ast}\lambda$. In other words, $\varphi
\in\left\{  \zeta:\left[  p\right]  \rightarrow\left[  q\right]  \ \mid
\ \mu=\zeta_{\ast}\lambda\right\}  $. In light of
(\ref{pf.prop.Lambda.triangular.j.l5.Blmpq=}), this rewrites as $\varphi
\in\mathfrak{B}_{\lambda,\mu,p,q}$. Hence, $\psi=\Phi\left(
\underbrace{\varphi}_{\in\mathfrak{B}_{\lambda,\mu,p,q}}\right)  \in
\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  $.
\par
Now, forget that we fixed $\psi$. We thus have proven that $\psi\in\Phi\left(
\mathfrak{B}_{\lambda,\mu,p,q}\right)  $ for each $\psi\in\mathfrak{B}%
_{\lambda,\mu}^{\prime}$. In other words, $\mathfrak{B}_{\lambda,\mu}^{\prime
}\subset\Phi\left(  \mathfrak{B}_{\lambda,\mu,p,q}\right)  $. Qed.}. Combining
these two inclusions, we obtain $\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  =\mathfrak{B}_{\lambda,\mu}^{\prime}$. Hence,%
\[
\mathfrak{B}_{\lambda,\mu}^{\prime}=\Phi\left(  \mathfrak{B}_{\lambda,\mu
,p,q}\right)  \cong\mathfrak{B}_{\lambda,\mu,p,q}\ \ \ \ \ \ \ \ \ \ \text{as
sets}%
\]
(by (\ref{pf.prop.Lambda.triangular.j.l5.5})). This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l5}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l6}Let $n\in \NN$. Let $\lambda
\in\Par_{n}$ and $\mu\in\Par_{n}$.
Assume that $\mathfrak{B}_{\lambda,\mu}^{\prime}%
\neq\varnothing$. Then, $\mu\triangleright\lambda$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l6}.]We have
$\lambda\in\Par_{n}\subset\Par$ and
$\mu\in\Par_{n}\subset\Par$.

Let $\ell=\ell\left(  \lambda\right)  $. We have $\mathfrak{B}_{\lambda,\mu
}^{\prime}\neq\varnothing$. In other words, the set $\mathfrak{B}_{\lambda
,\mu}^{\prime}$ contains at least one element. In other words, there exists
some $\psi\in\mathfrak{B}_{\lambda,\mu}^{\prime}$. Consider this $\psi$. We
have%
\[
\psi=\mathfrak{B}_{\lambda,\mu}^{\prime}=\left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  \ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all
}j\in\left\{  1,2,3,\ldots\right\}  \right\}
\]
(by the definition of $\mathfrak{B}_{\lambda,\mu}^{\prime}$ (since $\ell
=\ell\left(  \lambda\right)  $)). In other words, $\psi$ is a $\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(
\mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  \right)  $.
In other words, $\psi$ is a map $\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  $ and satisfies%
\begin{equation}
\left(  \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\psi\left(
i\right)  =j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
\right)  . \label{pf.prop.Lambda.triangular.j.l6.psi-def}%
\end{equation}


Let $k\in\left\{  1,2,\ldots,\ell\right\}  $. Then, $1\leq k\leq\ell$, so that
$\left[  k\right]  \subset\left[  \ell\right]  $. Hence, $\left[  k\right]  $
is a subset of $\left[  \ell\right]  $. Also, clearly, the set $\left[
k\right]  $ is finite.

Define a subset $R$ of $\left\{  1,2,3,\ldots\right\}  $ by $R=\psi\left(
\left[  k\right]  \right)  $. This set $R=\psi\left(  \left[  k\right]
\right)  $ is finite (since $\left[  k\right]  $ is finite).

Each $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}\lambda_{i}=\lambda_{i}
\label{pf.prop.Lambda.triangular.j.l6.stupid-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Lambda.triangular.j.l6.stupid-sum}):}
Let $i\in\left\{  1,2,\ldots,k\right\}  $. Thus, $i\in\left\{  1,2,\ldots
,k\right\}  =\left[  k\right]  $. Hence, $\psi\left(  \underbrace{i}%
_{\in\left[  k\right]  }\right)  \in\psi\left(  \left[  k\right]  \right)
=R$. Thus, there exists a $j\in R$ satisfying $j=\psi\left(  i\right)  $
(namely, $j=\psi\left(  i\right)  $). This $j$ is furthermore unique (since
the condition $j=\psi\left(  i\right)  $ determines $j$ uniquely). Thus, there
exists \textbf{exactly one} $j\in R$ satisfying $j=\psi\left(  i\right)  $
(namely, $j=\psi\left(  i\right)  $). Hence, the sum $\sum\limits_{\substack{j\in
R;\\j=\psi\left(  i\right)  }}\lambda_{i}$ has exactly one addend (namely, the
addend for $j=\psi\left(  i\right)  $). Thus, this sum simplifies as follows:
$\sum\limits_{\substack{j\in R;\\j=\psi\left(  i\right)  }}\lambda_{i}=\lambda_{i}$.
\par
Now,%
\[
\underbrace{\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}}%
_{\substack{=\sum\limits_{\substack{j\in R;\\j=\psi\left(  i\right)  }%
}\\\text{(because for any }j\in R\text{, the}\\\text{condition }\left(
\psi\left(  i\right)  =j\right)  \text{ is equivalent to}\\\text{the condition
}\left(  j=\psi\left(  i\right)  \right)  \text{)}}}\lambda_{i}=\sum\limits
_{\substack{j\in R;\\j=\psi\left(  i\right)  }}\lambda_{i}=\lambda_{i}.
\]
This proves (\ref{pf.prop.Lambda.triangular.j.l6.stupid-sum}).}.

It is well-known that if $X$ and $Y$ are two sets, if $f:X\rightarrow Y$ is
any map, and if $T$ is a finite subset of $X$, then $\left\vert f\left(
T\right)  \right\vert \leq\left\vert T\right\vert $. Applying this to
$X=\left[  \ell\right]  $, $Y=\left\{  1,2,3,\ldots\right\}  $, $f=\psi$ and
$T=\left[  k\right]  $, we conclude that $\left\vert \psi\left(  \left[
k\right]  \right)  \right\vert \leq\left\vert \left[  k\right]  \right\vert
=k$. In light of $R=\psi\left(  \left[  k\right]  \right)  $, this rewrites as
$\left\vert R\right\vert \leq k$.

We know that $\mu$ is a partition; thus, $\mu_{1}\geq\mu_{2}\geq\mu_{3}%
\geq\cdots$. Moreover, $\mu\in\Par \subset\mathbb{N}^{\infty}$.
Hence, $\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  =\mu\in\mathbb{N}%
^{\infty}$. Thus, Lemma \ref{lem.sol.Lambda.triangular.j.l3} (applied to
$\beta_{i}=\mu_{i}$) yields $\sum_{r\in R}\mu_{r}\leq\mu_{1}+\mu_{2}%
+\cdots+\mu_{k}$. Hence,%
\begin{align*}
\mu_{1}+\mu_{2}+\cdots+\mu_{k}  &  \geq\sum_{r\in R}\mu_{r}\\
&  =\sum_{j\in R}\underbrace{\mu_{j}}_{\substack{=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\psi\left(  i\right)  =j}}\lambda_{i}\\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l6.psi-def}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have renamed the summation index }r\text{ as }j\right) \\
&  =\underbrace{\sum_{j\in R}\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\psi\left(  i\right)  =j}}}_{=\sum_{i\in\left[  \ell\right]  }%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}}\lambda_{i}%
=\underbrace{\sum_{i\in\left[  \ell\right]  }}_{\substack{=\sum_{i\in\left\{
1,2,\ldots,\ell\right\}  }\\\text{(since }\left[  \ell\right]  =\left\{
1,2,\ldots,\ell\right\}  \text{)}}}\sum\limits_{\substack{j\in R;\\\psi\left(
i\right)  =j}}\lambda_{i}=\underbrace{\sum_{i\in\left\{  1,2,\ldots
,\ell\right\}  }}_{=\sum_{i=1}^{\ell}}\sum\limits_{\substack{j\in R;\\\psi\left(
i\right)  =j}}\lambda_{i}\\
&  =\sum_{i=1}^{\ell}\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)
=j}}\lambda_{i}=\sum_{i=1}^{k}\underbrace{\sum\limits_{\substack{j\in R;\\\psi\left(
i\right)  =j}}\lambda_{i}}_{\substack{=\lambda_{i}\\\text{(by
(\ref{pf.prop.Lambda.triangular.j.l6.stupid-sum}))}}}+\sum_{i=k+1}^{\ell}%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}\underbrace{\lambda_{i}%
}_{\substack{\geq0\\\text{(since }\lambda_{i}\in\mathbb{N}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split the outer sum at
}i=k\text{, because }1\leq k\leq\ell\right) \\
&  \geq\sum_{i=1}^{k}\lambda_{i}+\underbrace{\sum_{i=k+1}^{\ell}%
\sum\limits_{\substack{j\in R;\\\psi\left(  i\right)  =j}}0}_{=0}=\sum_{i=1}%
^{k}\lambda_{i}=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}.
\end{align*}


Now, let us forget that we fixed $k$. We thus have shown that $\mu_{1}+\mu
_{2}+\cdots+\mu_{k}\geq\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}$ for each
$k\in\left\{  1,2,\ldots,\ell\right\}  $. Therefore, Lemma
\ref{lem.sol.Lambda.triangular.majo-mu} (applied to $\ell$, $\lambda$ and
$\mu$ instead of $q$, $\mu$ and $\lambda$) yields $\mu\triangleright\lambda$
(since $\ell\left(  \lambda\right)  =\ell\leq\ell$). This proves Proposition
\ref{prop.sol.Lambda.triangular.j.l6}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.j.l7}Let $\mu\in\Par$. Let
$k=\ell\left(  \mu\right)  $. Then:

\begin{enumerate}
\item[(a)] We have $\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in
\Symm_k\ \mid\ \mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for each
}i\in\left[  k\right]  \right\}  $.

\item[(b)] The set $\mathfrak{B}_{\mu,\mu,k,k}$ is a subgroup of
$\Symm_k$.
\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.j.l7}.]Recall that
$\ell\left(  \mu\right)  =k$. Thus, $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{k}\right)  \in\mathbb{N}^{k}$. Also,%
\begin{equation}
\mu_{i}>0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left[  k\right]
\label{pf.prop.sol.Lambda.triangular.j.l7.gr0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr0}):}
Write the partition $\mu$ in the form $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{h},0,0,0,\ldots\right)  $ with $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu
_{h}>0$. Then, $\ell\left(  \mu\right)  =h$ (by the definition of $\ell\left(
\mu\right)  $). Hence, $k=\ell\left(  \mu\right)  =h$.
\par
Moreover, we have $\mu_{i}>0$ for each $i\in\left\{  1,2,\ldots,h\right\}  $
(since $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{h}>0$). In other words, $\mu
_{i}>0$ for each $i\in\left[  h\right]  $ (since $\left[  h\right]  =\left\{
1,2,\ldots,h\right\}  $). In other words, $\mu_{i}>0$ for each $i\in\left[
k\right]  $ (since $k=h$). This proves
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr0}).}.

The definition of $\mathfrak{B}_{\mu,\mu,k,k}$ yields%
\begin{equation}
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[  k\right]  \rightarrow
\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  ,
\label{pf.prop.sol.Lambda.triangular.j.l7.gr1}%
\end{equation}
where $\mu$ is regarded as the $k$-tuple $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{k}\right)  \in\mathbb{N}^{k}$.

We have $\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$%
\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\mathfrak{B}_{\mu,\mu,k,k}$.
Thus, $\psi\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[  k\right]
\rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In
other words, $\psi$ is a $\varphi:\left[  k\right]  \rightarrow\left[
k\right]  $ satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\psi$ is a
map $\left[  k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu
=\psi_{\ast}\mu$. Thus, $\psi_{\ast}\mu=\mu$.
\par
But Lemma \ref{lem.sol.Lambda.triangular.j.l10} (applied to $p=k$, $q=k$,
$\varphi=\psi$, $\alpha=\mu$, $\alpha_{i}=\mu_{i}$, $\beta=\mu$ and $\beta
_{i}=\mu_{i}$) shows that the map $\psi$ is surjective (because
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr0}) shows that $\left(  \mu
_{i}>0\text{ for each }i\in\left[  k\right]  \right)  $). Hence, $\psi$ is a
surjective map $\left[  k\right]  \rightarrow\left[  k\right]  $.
\par
Now, recall the following known fact about finite sets: If $T$ is a finite
set, then each surjective map $T\rightarrow T$ is bijective. Applying this
fact to $T=\left[  k\right]  $, we conclude that each surjective map $\left[
k\right]  \rightarrow\left[  k\right]  $ is bijective (since $\left[
k\right]  $ is a finite set). Thus, the map $\psi$ is bijective (since $\psi$
is a surjective map $\left[  k\right]  \rightarrow\left[  k\right]  $). Thus,
$\psi$ is a bijection $\left[  k\right]  \rightarrow\left[  k\right]  $. In
other words, $\psi$ is a permutation of $\left[  k\right]  $. In other words,
$\psi$ is a permutation of $\left\{  1,2,\ldots,k\right\}  $ (since $\left[
k\right]  =\left\{  1,2,\ldots,k\right\}  $). In other words, $\psi$ belongs
to the set of all permutations of $\left\{  1,2,\ldots,k\right\}  $. In other
words, $\psi$ belongs to $\Symm_k$ (since $\Symm_k$ is the
set of all permutations of $\left\{  1,2,\ldots,k\right\}  $). In other words,
$\psi\in\Symm_k$.
\par
Now, forget that we fixed $\psi$. We thus have shown that $\psi\in
\Symm_k$ for each $\psi\in\mathfrak{B}_{\mu,\mu,k,k}$. In other
words, $\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$. Qed.}. Hence, we
have the two inclusions $\mathfrak{B}_{\mu,\mu,k,k}\subset\left\{  \sigma
\in\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\mathfrak{B}_{\mu,\mu,k,k}$.
Thus, $\psi\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[  k\right]
\rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In
other words, $\psi$ is a $\varphi:\left[  k\right]  \rightarrow\left[
k\right]  $ satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\psi$ is a
map $\left[  k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu
=\psi_{\ast}\mu$. Thus, $\psi_{\ast}\mu=\mu$. Also, $\psi\in\mathfrak{B}%
_{\mu,\mu,k,k}\subset\Symm_k$. Hence, $\psi$ is an element of
$\Symm_k$ and satisfies $\psi_{\ast}\mu=\mu$. In other words, $\psi$
is a $\sigma\in\Symm_k$ satisfying $\sigma_{\ast}\mu=\mu$. In other
words, $\psi\in\left\{  \sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu
=\mu\right\}  $.
\par
Now, forget that we fixed $\psi$. We thus have shown that $\psi\in\left\{
\sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $ for each
$\psi\in\mathfrak{B}_{\mu,\mu,k,k}$. In other words, $\mathfrak{B}_{\mu
,\mu,k,k}\subset\left\{  \sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu
=\mu\right\}  $. Qed.} and $\left\{  \sigma\in\Symm_k\ \mid
\ \sigma_{\ast}\mu=\mu\right\}  \subset\mathfrak{B}_{\mu,\mu,k,k}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\psi\in\left\{  \sigma\in
\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $. Thus, $\psi$ is a
$\sigma\in\Symm_k$ satisfying $\sigma_{\ast}\mu=\mu$. In other words,
$\psi$ is an element of $\Symm_k$ and satisfies $\psi_{\ast}\mu=\mu$.
\par
We know that $\psi$ is an element of $\Symm_k$. In other words,
$\psi$ is an element of the set of all permutations of $\left\{
1,2,\ldots,k\right\}  $ (since $\Symm_k$ is the set of all
permutations of $\left\{  1,2,\ldots,k\right\}  $). In other words, $\psi$ is
a permutation of $\left\{  1,2,\ldots,k\right\}  $. In other words, $\psi$ is
a permutation of $\left[  k\right]  $ (since $\left[  k\right]  =\left\{
1,2,\ldots,k\right\}  $). In other words, $\psi$ is a bijection $\left[
k\right]  \rightarrow\left[  k\right]  $. In other words, $\psi$ is a
bijective map $\left[  k\right]  \rightarrow\left[  k\right]  $.
\par
Thus, $\psi$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\psi_{\ast}\mu$ (since $\psi_{\ast}\mu=\mu$). In other words,
$\psi$ is a $\varphi:\left[  k\right]  \rightarrow\left[  k\right]  $
satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\psi\in\left\{
\varphi:\left[  k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu
=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as $\psi
\in\mathfrak{B}_{\mu,\mu,k,k}$.
\par
Now, forget that we fixed $\psi$. We thus have proven that $\psi
\in\mathfrak{B}_{\mu,\mu,k,k}$ for each $\psi\in\left\{  \sigma\in
\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}  $. In other words,
$\left\{  \sigma\in\Symm_k\ \mid\ \sigma_{\ast}\mu=\mu\right\}
\subset\mathfrak{B}_{\mu,\mu,k,k}$. Qed.}. Combining these two inclusions, we
obtain
\begin{equation}
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in\Symm_k\ \mid
\ \sigma_{\ast}\mu=\mu\right\}  .
\label{pf.prop.sol.Lambda.triangular.j.l7.gr3}%
\end{equation}


But let $\sigma\in\Symm_k$ be arbitrary. Then, Proposition
\ref{prop.sol.Lambda.triangular.j.l9}(b) (applied to $p=k$, $\alpha=\mu$ and
$\alpha_{i}=\mu_{i}$) shows that we have $\sigma_{\ast}\mu=\mu$ if and only if
$\left(  \mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for all }i\in\left[
k\right]  \right)  $. In other words, we have the following logical
equivalence:%
\begin{equation}
\left(  \sigma_{\ast}\mu=\mu\right)  \Longleftrightarrow\left(  \mu
_{\sigma\left(  i\right)  }=\mu_{i}\text{ for all }i\in\left[  k\right]
\right)  . \label{pf.prop.sol.Lambda.triangular.j.l7.gr5}%
\end{equation}


Now, forget that we fixed $\sigma$. We thus have proven the equivalence
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr5}) for each $\sigma\in
\Symm_k$.

Now, the equality (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr3}) becomes%
\[
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in\Symm_k\ \mid
\ \underbrace{\sigma_{\ast}\mu=\mu}_{\substack{\Longleftrightarrow\ \left(
\mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for all }i\in\left[  k\right]
\right)  \\\text{(by (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr5}))}%
}}\right\}  =\left\{  \sigma\in\Symm_k\ \mid\ \mu_{\sigma\left(
i\right)  }=\mu_{i}\text{ for each }i\in\left[  k\right]  \right\}  .
\]
This proves Proposition \ref{prop.sol.Lambda.triangular.j.l7}(a).

(b) The neutral element of the group $\Symm_k$ is $\operatorname*{id}%
\nolimits_{\left\{  1,2,\ldots,k\right\}  }=\operatorname*{id}%
\nolimits_{\left[  k\right]  }$ (since $\left\{  1,2,\ldots,k\right\}
=\left[  k\right]  $).

The following four observations hold:

\begin{itemize}
\item The set $\mathfrak{B}_{\mu,\mu,k,k}$ is a subset of $\Symm_k$
(since $\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$).

\item If $\gamma$ and $\delta$ are two elements of $\mathfrak{B}_{\mu,\mu
,k,k}$, then $\gamma\circ\delta\in\mathfrak{B}_{\mu,\mu,k,k}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\gamma$ and $\delta$ be two elements of
$\mathfrak{B}_{\mu,\mu,k,k}$. We must prove that $\gamma\circ\delta
\in\mathfrak{B}_{\mu,\mu,k,k}$.
\par
We have $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1})). In other words,
$\gamma$ is a $\varphi:\left[  k\right]  \rightarrow\left[  k\right]  $
satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\gamma$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu=\gamma_{\ast}\mu$.
The same argument (applied to $\delta$ instead of $\gamma$) shows that
$\delta$ is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and
satisfies $\mu=\delta_{\ast}\mu$. Now, Proposition
\ref{prop.sol.Lambda.triangular.j.l1} (applied to $p=k$, $q=k$, $r=k$,
$\psi=\gamma$, $\varphi=\delta$ and $\alpha=\mu$) yields $\left(  \gamma
\circ\delta\right)  _{\ast}\mu=\gamma_{\ast}\left(  \underbrace{\delta_{\ast
}\mu}_{=\mu}\right)  =\gamma_{\ast}\mu=\mu$. In other words, $\mu=\left(
\gamma\circ\delta\right)  _{\ast}\mu$. Hence, $\gamma\circ\delta$ is a map
$\left[  k\right]  \rightarrow\left[  k\right]  $ (since $\gamma$ and $\delta$
are maps $\left[  k\right]  \rightarrow\left[  k\right]  $) and satisfies
$\mu=\left(  \gamma\circ\delta\right)  _{\ast}\mu$. In other words,
$\gamma\circ\delta$ is a $\varphi:\left[  k\right]  \rightarrow\left[
k\right]  $ satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\gamma
\circ\delta\in\left\{  \varphi:\left[  k\right]  \rightarrow\left[  k\right]
\ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as $\gamma
\circ\delta\in\mathfrak{B}_{\mu,\mu,k,k}$. Qed.}.

\item We have $\id_{\left[  k\right]  }\in
\mathfrak{B}_{\mu,\mu,k,k}$\ \ \ \ \footnote{\textit{Proof.} Proposition
\ref{prop.sol.Lambda.triangular.j.l2} (applied to $p=k$ and $\alpha=\mu$)
yields $\left(  \id_{\left[  k\right]  }\right)
_{\ast}\mu=\mu$. Hence, $\mu=\left(  \id_{\left[
k\right]  }\right)  _{\ast}\mu$.
\par
Thus, $\id_{\left[  k\right]  }$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu=\left(
\id_{\left[  k\right]  }\right)  _{\ast}\mu$. In other
words, $\id_{\left[  k\right]  }$ is a $\varphi
:\left[  k\right]  \rightarrow\left[  k\right]  $ satisfying $\mu
=\varphi_{\ast}\mu$. In other words, $\id_{\left[
k\right]  }\in\left\{  \varphi:\left[  k\right]  \rightarrow\left[  k\right]
\ \mid\ \mu=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as
$\id_{\left[  k\right]  }\in\mathfrak{B}_{\mu,\mu
,k,k}$. Qed.}.

\item If $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}$, then $\gamma^{-1}%
\in\mathfrak{B}_{\mu,\mu,k,k}$\ \ \ \ \footnote{\textit{Proof.} Let $\gamma
\in\mathfrak{B}_{\mu,\mu,k,k}$. We must prove that $\gamma^{-1}\in
\mathfrak{B}_{\mu,\mu,k,k}$.
\par
We have $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by (\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1})). In other words,
$\gamma$ is a $\varphi:\left[  k\right]  \rightarrow\left[  k\right]  $
satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\gamma$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu=\gamma_{\ast}\mu$.
Also, $\gamma\in\mathfrak{B}_{\mu,\mu,k,k}\subset\Symm_k$; therefore,
$\gamma$ has an inverse $\gamma^{-1}$ (since $\Symm_k$ is a group).
Proposition \ref{prop.sol.Lambda.triangular.j.l1} (applied to $p=k$, $q=k$,
$r=k$, $\psi=\gamma^{-1}$, $\varphi=\gamma$ and $\alpha=\mu$) yields $\left(
\gamma^{-1}\circ\gamma\right)  _{\ast}\mu=\left(  \gamma^{-1}\right)  _{\ast
}\left(  \underbrace{\gamma_{\ast}\mu}_{=\mu}\right)  =\left(  \gamma
^{-1}\right)  _{\ast}\mu$. Hence, $\left(  \gamma^{-1}\right)  _{\ast}%
\mu=\left(  \underbrace{\gamma^{-1}\circ\gamma}_{=\operatorname*{id}%
\nolimits_{\left[  k\right]  }}\right)  _{\ast}\mu=\left(  \operatorname*{id}%
\nolimits_{\left[  k\right]  }\right)  _{\ast}\mu$.
\par
But Proposition \ref{prop.sol.Lambda.triangular.j.l2} (applied to $p=k$ and
$\alpha=\mu$) yields $\left(  \id_{\left[  k\right]
}\right)  _{\ast}\mu=\mu$. Hence, $\mu=\left(  \operatorname*{id}%
\nolimits_{\left[  k\right]  }\right)  _{\ast}\mu$. Comparing this with
$\left(  \gamma^{-1}\right)  _{\ast}\mu=\left(  \operatorname*{id}%
\nolimits_{\left[  k\right]  }\right)  _{\ast}\mu$, we obtain $\mu=\left(
\gamma^{-1}\right)  _{\ast}\mu$. Hence, $\gamma^{-1}$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ (since $\gamma$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $) and satisfies $\mu=\left(
\gamma^{-1}\right)  _{\ast}\mu$. In other words, $\gamma^{-1}$ is a
$\varphi:\left[  k\right]  \rightarrow\left[  k\right]  $ satisfying
$\mu=\varphi_{\ast}\mu$. In other words, $\gamma^{-1}\in\left\{
\varphi:\left[  k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu
=\varphi_{\ast}\mu\right\}  $. In light of
(\ref{pf.prop.sol.Lambda.triangular.j.l7.gr1}), this rewrites as $\gamma
^{-1}\in\mathfrak{B}_{\mu,\mu,k,k}$. Qed.}.
\end{itemize}

\noindent
Combining these four observations, we conclude that $\mathfrak{B}_{\mu
,\mu,k,k}$ is a subgroup of $\Symm_k$ (since the neutral element of
the group $\Symm_k$ is $\id_{\left[
k\right]  }$). This proves Proposition \ref{prop.sol.Lambda.triangular.j.l7}(b).
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l1}Let $p\in \NN$. Let
$\varphi:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ be any
map. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Then,%
\[
\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)
\]
is a weak composition.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l1}.]For each
$j\in\left\{  1,2,3,\ldots\right\}  $, the sum $\sum_{i\in\varphi^{-1}\left\{
j\right\}  }\alpha_{i}$ is a well-defined element of $\mathbb{N}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,3,\ldots\right\}  $.
Then, $\varphi^{-1}\left\{  j\right\}  $ is a subset of $\left[  p\right]  $,
and thus a finite set (since $\left[  p\right]  $ is a finite set). Hence, the
sum $\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}$ is well-defined.
Moreover, this sum belongs to $\NN$ (since each of its addends
$\alpha_{i}$ belongs to $\NN$). Thus, the sum $\sum_{i\in\varphi
^{-1}\left\{  j\right\}  }\alpha_{i}$ is a well-defined element of
$\NN$. Qed.}. Hence, the sequence%
\[
\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)
\]
is a well-defined element of $\mathbb{N}^{\infty}$. Denote this sequence by
$\beta$. We shall now show that this sequence $\beta$ is a weak composition.

Write the sequence $\beta$ in the form $\beta=\left(  \beta_{1},\beta
_{2},\beta_{3},\ldots\right)  $. Thus,%
\[
\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  =\beta=\left(  \sum
_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  3\right\}
}\alpha_{i},\ldots\right)
\]
(by the definition of $\beta$). In other words,
\begin{equation}
\beta_{j}=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for each }j\in\left\{  1,2,3,\ldots\right\}  .
\label{pf.prop.Lambda.triangular.k.l1.betaj=}%
\end{equation}


We have $\beta=\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $. Thus,
the support of the sequence $\beta$ is the set of all positive integers $i$
for which $\beta_{i}\neq0$ (by the definition of ``support'').
In other words,%
\[
\left(  \text{the support of the sequence }\beta\right)  =\left(  \text{the
set of all positive integers }i\text{ for which }\beta_{i}\neq0\right)  .
\]


Let $Z$ be the support of the sequence $\beta$. Then,%
\begin{align*}
Z  &  =\left(  \text{the support of the sequence }\beta\right) \\
&  =\left(  \text{the set of all positive integers }i\text{ for which }%
\beta_{i}\neq0\right) \\
&  =\left\{  i\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \beta_{i}%
\neq0\right\}  .
\end{align*}


Every $j\in Z$ satisfies $j\in\varphi\left(  \left[  p\right]  \right)
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in Z$. We must show that
$j\in\varphi\left(  \left[  p\right]  \right)  $.
\par
We have $j\in Z=\left\{  i\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \beta
_{i}\neq0\right\}  $. In other words, $j$ is a $i\in\left\{  1,2,3,\ldots
\right\}  $ satisfying $\beta_{i}\neq0$. In other words, $j$ is an element of
$\left\{  1,2,3,\ldots\right\}  $ and satisfies $\beta_{j}\neq0$.
\par
Assume (for the sake of contradiction) that $\varphi^{-1}\left\{  j\right\}
=\varnothing$. Thus, $\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha
_{i}=\sum_{i\in\varnothing}\alpha_{i}=\left(  \text{empty sum}\right)  =0$.
Hence, (\ref{pf.prop.Lambda.triangular.k.l1.betaj=}) yields $\beta_{j}%
=\sum_{i\in\varphi^{-1}\left\{  j\right\}  }\alpha_{i}=0$. This contradicts
$\beta_{j}\neq0$.
\par
This contradiction shows that our assumption (that $\varphi^{-1}\left\{
j\right\}  =\varnothing$) was false. Hence, we cannot have $\varphi
^{-1}\left\{  j\right\}  =\varnothing$. Thus, we have $\varphi^{-1}\left\{
j\right\}  \neq\varnothing$. In other words, the set $\varphi^{-1}\left\{
j\right\}  $ is nonempty. Hence, there exists some $k\in\varphi^{-1}\left\{
j\right\}  $. Consider this $k$.
\par
From $k\in\varphi^{-1}\left\{  j\right\}  $, we obtain $\varphi\left(
k\right)  \in\left\{  j\right\}  $. Hence, $\varphi\left(  k\right)  =j$.
Therefore, $j=\varphi\left(  \underbrace{k}_{\in\varphi^{-1}\left\{
j\right\}  \subset\left[  p\right]  }\right)  \in\varphi\left(  \left[
p\right]  \right)  $. This completes our proof.}. In other words, we have
$Z\subset\varphi\left(  \left[  p\right]  \right)  $. In other words, $Z$ is a
subset of $\varphi\left(  \left[  p\right]  \right)  $.

The set $\left[  p\right]  $ is finite. Hence, its image $\varphi\left(
\left[  p\right]  \right)  $ is also finite. Thus, its subset $Z$ is also
finite (since $Z$ is a subset of $\varphi\left(  \left[  p\right]  \right)
$). In other words, the support of the sequence $\beta$ is finite (since $Z$
is the support of the sequence $\beta$).

Hence, we know that $\beta$ is a sequence in $\mathbb{N}^{\infty}$ having
finite support (since the support of the sequence $\beta$ is finite). In other
words, $\beta$ is a weak composition (since a weak composition is defined as a
sequence in $\mathbb{N}^{\infty}$ having finite support). In other words,%
\[
\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)
\]
is a weak composition (since $\beta=\left(  \sum_{i\in\varphi^{-1}\left\{
1\right\}  }\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha
_{i},\sum_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)  $
(by the definition of $\beta$)). This proves Proposition
\ref{prop.sol.Lambda.triangular.k.l1}.
\end{proof}

Proposition \ref{prop.sol.Lambda.triangular.k.l1} allows us to make the
following definition (which is similar to Definition
\ref{def.sol.Lambda.triangular.j.2}, but uses the infinite set $\left\{
1,2,3,\ldots\right\}  $ instead of $\left[  q\right]  $):

\begin{definition}
\label{def.sol.Lambda.triangular.k.2}Let $p\in \NN$. Let $\varphi
:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ be any map. Let
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)  \in
\mathbb{N}^{p}$ be a $p$-tuple. Then, we define a weak composition
$\varphi_{\ast}\alpha\in\operatorname*{WC}$ by%
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)  .
\]
(This is indeed a weak composition, because of Proposition
\ref{prop.sol.Lambda.triangular.k.l1}.)
\end{definition}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l3}Let $p\in \NN$. Let
$\varphi:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  $ be any
map. Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{p}\right)
\in\mathbb{N}^{p}$ be a $p$-tuple. Then, $\mathbf{x}^{\varphi_{\ast}\alpha
}=x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)
}^{\alpha_{2}}\cdots x_{\varphi\left(  p\right)  }^{\alpha_{p}}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l3}.]The definition of
$\varphi_{\ast}\alpha$ yields
\[
\varphi_{\ast}\alpha=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }%
\alpha_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\alpha_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i},\ldots\right)  .
\]
Hence, the definition of $\mathbf{x}^{\varphi_{\ast}\alpha}$ shows that%
\begin{align*}
\mathbf{x}^{\varphi_{\ast}\alpha}  &  =x_{1}^{\sum_{i\in\varphi^{-1}\left\{
1\right\}  }\alpha_{i}}x_{2}^{\sum_{i\in\varphi^{-1}\left\{  2\right\}
}\alpha_{i}}x_{3}^{\sum_{i\in\varphi^{-1}\left\{  3\right\}  }\alpha_{i}%
}\cdots=\underbrace{\prod_{j=1}^{\infty}}_{=\prod_{j\in\left\{  1,2,3,\ldots
\right\}  }}\underbrace{x_{j}^{\sum_{i\in\varphi^{-1}\left\{  j\right\}
}\alpha_{i}}}_{=\prod_{i\in\varphi^{-1}\left\{  j\right\}  }x_{j}^{\alpha_{i}%
}}\\
&  =\prod_{j\in\left\{  1,2,3,\ldots\right\}  }\underbrace{\prod_{i\in
\varphi^{-1}\left\{  j\right\}  }}_{\substack{=\prod\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  \in\left\{  j\right\}  }}=\prod\limits
_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\\\text{(because for each }i\in\left[  p\right]  \text{,}\\\text{the
condition }\left(  \varphi\left(  i\right)  \in\left\{  j\right\}  \right)
\\\text{is equivalent to }\left(  \varphi\left(  i\right)  =j\right)
\text{)}}}x_{j}^{\alpha_{i}}=\prod_{j\in\left\{  1,2,3,\ldots\right\}  }%
\prod\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}\underbrace{x_{j}^{\alpha_{i}}}_{\substack{=x_{\varphi\left(  i\right)
}^{\alpha_{i}}\\\text{(since }j=\varphi\left(  i\right)  \\\text{(since
}\varphi\left(  i\right)  =j\text{))}}}=\prod_{j\in\left\{  1,2,3,\ldots
\right\}  }\prod\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)
=j}}x_{\varphi\left(  i\right)  }^{\alpha_{i}}.
\end{align*}
Comparing this with%
\[
x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)
}^{\alpha_{2}}\cdots x_{\varphi\left(  p\right)  }^{\alpha_{p}}%
=\underbrace{\prod_{i\in\left\{  1,2,\ldots,p\right\}  }}_{\substack{=\prod
_{i\in\left[  p\right]  }\\\text{(since }\left\{  1,2,\ldots,p\right\}
=\left[  p\right]  \text{)}}}x_{\varphi\left(  i\right)  }^{\alpha_{i}%
}=\underbrace{\prod_{i\in\left[  p\right]  }}_{\substack{=\prod_{j\in\left\{
1,2,3,\ldots\right\}  }\prod\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}\\\text{(since }\varphi\left(  i\right)  \in\left\{
1,2,3,\ldots\right\}  \text{ for each }i\in\left[  p\right]  \text{)}%
}}x_{\varphi\left(  i\right)  }^{\alpha_{i}}=\prod_{j\in\left\{
1,2,3,\ldots\right\}  }\prod\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}x_{\varphi\left(  i\right)  }^{\alpha_{i}},
\]
we obtain $\mathbf{x}^{\varphi_{\ast}\alpha}=x_{\varphi\left(  1\right)
}^{\alpha_{1}}x_{\varphi\left(  2\right)  }^{\alpha_{2}}\cdots x_{\varphi
\left(  p\right)  }^{\alpha_{p}}$. This proves Proposition
\ref{prop.sol.Lambda.triangular.k.l3}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.triangular.k.l4auxbij}Let $\ell\in \NN$. Let $X$
be a set. Then, the map%
\[
X^{\left\{  1,2,\ldots,\ell\right\}  }\rightarrow X^{\ell}%
,\ \ \ \ \ \ \ \ \ \ \varphi\mapsto\left(  \varphi\left(  1\right)
,\varphi\left(  2\right)  ,\ldots,\varphi\left(  \ell\right)  \right)
\]
is a bijection.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.k.l4auxbij}.]It is well-known
that the $\ell$-tuples of elements of $X$ are in bijection with the maps
$\left\{  1,2,\ldots,\ell\right\}  \rightarrow X$\ \ \ \ \footnote{In fact,
depending on your definition of an ``$\ell$-tuple'', you might even
consider the $\ell$-tuples of
elements of $X$ to be \textbf{exactly} the maps $\left\{  1,2,\ldots
,\ell\right\}  \rightarrow X$.}. Lemma
\ref{lem.sol.Lambda.triangular.k.l4auxbij} is merely a way to precisely
formulate this bijection. Thus, we omit its proof.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l4}Let $\ell\in \NN$. Let
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\mathbb{N}^{\ell}$ be such that $\left(  \alpha_{i}>0\text{ for each }%
i\in\left[  \ell\right]  \right)  $. Then,%
\[
p_{\alpha_{1}}p_{\alpha_{2}}\cdots p_{\alpha_{\ell}}=\sum_{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  }\xx%
^{\varphi_{\ast}\alpha}%
\]
in $\kk\left[  \left[  \xx\right]  \right]  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l4}.]We have assumed
that%
\begin{equation}
\left(  \alpha_{i}>0\text{ for each }i\in\left[  \ell\right]  \right)  .
\label{pf.prop.Lambda.triangular.k.l4.pos}%
\end{equation}
Let $i\in\left\{  1,2,\ldots,\ell\right\}  $. Then, $i\in\left\{
1,2,\ldots,\ell\right\}  =\left[  \ell\right]  $. Hence, $\alpha_{i}>0$ (by
(\ref{pf.prop.Lambda.triangular.k.l4.pos})). Thus, the definition of
$p_{\alpha_{i}}$ yields
\begin{equation}
p_{\alpha_{i}}=x_{1}^{\alpha_{i}}+x_{2}^{\alpha_{i}}+x_{3}^{\alpha_{i}}%
+\cdots=\sum_{j\in\left\{  1,2,3,\ldots\right\}  }x_{j}^{\alpha_{i}}.
\label{pf.prop.Lambda.triangular.k.l4.1}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven
(\ref{pf.prop.Lambda.triangular.k.l4.1}) for each $i\in\left\{  1,2,\ldots
,\ell\right\}  $.

Lemma \ref{lem.sol.Lambda.triangular.k.l4auxbij} (applied to $X=\left\{
1,2,3,\ldots\right\}  $) shows that the map%
\[
\left\{  1,2,3,\ldots\right\}  ^{\left\{  1,2,\ldots,\ell\right\}
}\rightarrow\left\{  1,2,3,\ldots\right\}  ^{\ell},\ \ \ \ \ \ \ \ \ \ \varphi
\mapsto\left(  \varphi\left(  1\right)  ,\varphi\left(  2\right)
,\ldots,\varphi\left(  \ell\right)  \right)
\]
is a bijection. In view of $\left[  \ell\right]  =\left\{  1,2,\ldots
,\ell\right\}  $, this rewrites as follows: The map%
\begin{equation}
\left\{  1,2,3,\ldots\right\}  ^{\left[  \ell\right]  }\rightarrow\left\{
1,2,3,\ldots\right\}  ^{\ell},\ \ \ \ \ \ \ \ \ \ \varphi\mapsto\left(
\varphi\left(  1\right)  ,\varphi\left(  2\right)  ,\ldots,\varphi\left(
\ell\right)  \right)  \label{pf.prop.Lambda.triangular.k.l4.bij}%
\end{equation}
is a bijection.

Now,%
\begin{align*}
p_{\alpha_{1}}p_{\alpha_{2}}\cdots p_{\alpha_{\ell}}  &  =\prod_{i=1}^{\ell
}\underbrace{p_{\alpha_{i}}}_{\substack{=\sum_{j\in\left\{  1,2,3,\ldots
\right\}  }x_{j}^{\alpha_{i}}\\\text{(by
(\ref{pf.prop.Lambda.triangular.k.l4.1}))}}}=\prod_{i=1}^{\ell}\sum
_{j\in\left\{  1,2,3,\ldots\right\}  }x_{j}^{\alpha_{i}}\\
&  =\sum_{\left(  j_{1},j_{2},\ldots,j_{\ell}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{\ell}}\prod_{i=1}^{\ell}x_{j_{i}}^{\alpha_{i}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\underbrace{\sum_{\varphi\in\left\{  1,2,3,\ldots\right\}  ^{\left[
\ell\right]  }}}_{=\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  }}\underbrace{\prod_{i=1}^{\ell}x_{\varphi\left(
i\right)  }^{\alpha_{i}}}_{=x_{\varphi\left(  1\right)  }^{\alpha_{1}%
}x_{\varphi\left(  2\right)  }^{\alpha_{2}}\cdots x_{\varphi\left(
\ell\right)  }^{\alpha_{\ell}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  \varphi\left(  1\right)
,\varphi\left(  2\right)  ,\ldots,\varphi\left(  \ell\right)  \right)  \text{
for }\left(  j_{1},j_{2},\ldots,j_{\ell}\right) \\
\text{in the sum, since the map (\ref{pf.prop.Lambda.triangular.k.l4.bij}) is
a bijection}%
\end{array}
\right) \\
&  =\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  }x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(
2\right)  }^{\alpha_{2}}\cdots x_{\varphi\left(  \ell\right)  }^{\alpha_{\ell
}}.
\end{align*}
Comparing this with%
\[
\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
}\underbrace{\mathbf{x}^{\varphi_{\ast}\alpha}}_{\substack{=x_{\varphi\left(
1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)  }^{\alpha_{2}}\cdots
x_{\varphi\left(  \ell\right)  }^{\alpha_{\ell}}\\\text{(by Proposition
\ref{prop.sol.Lambda.triangular.k.l3} (applied to }p=\ell\text{))}}%
}=\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
}x_{\varphi\left(  1\right)  }^{\alpha_{1}}x_{\varphi\left(  2\right)
}^{\alpha_{2}}\cdots x_{\varphi\left(  \ell\right)  }^{\alpha_{\ell}},
\]
we obtain $p_{\alpha_{1}}p_{\alpha_{2}}\cdots p_{\alpha_{\ell}}=\sum
_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}
}\mathbf{x}^{\varphi_{\ast}\alpha}$. This proves Proposition
\ref{prop.sol.Lambda.triangular.k.l4}.
\end{proof}

\begin{proposition}
\label{prop.sol.Lambda.triangular.k.l5}Let $\mu$ be a weak composition. Then,
there exists a $q\in \NN$ such that $\mu=\left(  \mu_{1},\mu_{2}%
,\ldots,\mu_{q}\right)  $.\ \ \ \ \footnote{Keep in mind that we are
identifying any $k$-tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
\in\mathbb{N}^{k}$ with the weak composition $\left(  a_{1},a_{2},\ldots
,a_{k},0,0,0,\ldots\right)  $. Thus, the $q$-tuple $\left(  \mu_{1},\mu
_{2},\ldots,\mu_{q}\right)  $ is identified with the weak composition $\left(
\mu_{1},\mu_{2},\ldots,\mu_{q},0,0,0,\ldots\right)  $.}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.k.l5}.]We have
$\mu=\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  $. Hence, the support of
the sequence $\mu$ is the set of all positive integers $i$ for which $\mu
_{i}\neq0$ (by the definition of ``support''). In other words,%
\[
\left(  \text{the support of the sequence }\mu\right)  =\left(  \text{the set
of all positive integers }i\text{ for which }\mu_{i}\neq0\right)  .
\]


Let $Z$ be the support of the sequence $\mu$. Then,%
\begin{align*}
Z  &  =\left(  \text{the support of the sequence }\mu\right) \\
&  =\left(  \text{the set of all positive integers }i\text{ for which }\mu
_{i}\neq0\right) \\
&  =\left\{  i\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{i}\neq0\right\}
.
\end{align*}
Hence, $Z$ is a subset of $\left\{  1,2,3,\ldots\right\}  $.

The sequence $\mu$ is a weak composition, i.e., a sequence in
$\NN^{\infty}$ having finite support (because a ``weak composition''
is defined to mean a sequence in $\NN^{\infty}$ having finite support).
Thus, the support of the sequence $\mu$ is
finite. In other words, $Z$ is finite (since $Z$ is the support of the
sequence $\mu$). Hence, $Z$ is a finite subset of $\left\{  1,2,3,\ldots
\right\}  $. Therefore, the set $Z$ has an upper bound (since any finite
subset of $\left\{  1,2,3,\ldots\right\}  $ has an upper bound). In other
words, there exists some $k\in \NN$ such that $\left(  \text{every }z\in
Z\text{ satisfies }z\leq k\right)  $. Consider this $k$. Clearly, $\left\{
k+1,k+2,k+3,\ldots\right\}  \subset\left\{  1,2,3,\ldots\right\}  $ (since
$k\in \NN$).

Let $h\in\left\{  k+1,k+2,k+3,\ldots\right\}  $. Assume (for the sake of
contradiction) that $\mu_{h}\neq0$. Thus, $h$ is an element of $\left\{
1,2,3,\ldots\right\}  $ (since $h\in\left\{  k+1,k+2,k+3,\ldots\right\}
\subset\left\{  1,2,3,\ldots\right\}  $) and satisfies $\mu_{h}\neq0$. In
other words, $h$ is an $i\in\left\{  1,2,3,\ldots\right\}  $ satisfying
$\mu_{i}\neq0$. In other words, $h\in\left\{  i\in\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{i}\neq0\right\}  $. In other words, $h\in Z$ (since
$Z=\left\{  i\in\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{i}\neq0\right\}
$). But recall that $\left(  \text{every }z\in Z\text{ satisfies }z\leq
k\right)  $. Applying this to $z=h$, we conclude that $h\leq k$. On the other
hand, $h\in\left\{  k+1,k+2,k+3,\ldots\right\}  $, so that $h\geq k+1>k$. This
contradicts $h\leq k$. This contradiction shows that our assumption (that
$\mu_{h}\neq0$) was wrong. Hence, we cannot have $\mu_{h}\neq0$. In other
words, we have $\mu_{h}=0$.

Now, forget that we fixed $h$. We thus have proven that $\mu_{h}=0$ for each
$h\in\left\{  k+1,k+2,k+3,\ldots\right\}  $. In other words, $\mu_{k+1}%
=\mu_{k+2}=\mu_{k+3}=\cdots=0$. Now,%
\begin{align*}
\mu &  =\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  =\left(  \mu_{1}%
,\mu_{2},\ldots,\mu_{k},\mu_{k+1},\mu_{k+2},\mu_{k+3},\ldots\right) \\
&  =\left(  \mu_{1},\mu_{2},\ldots,\mu_{k},0,0,0,\ldots\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{k+1}=\mu_{k+2}=\mu_{k+3}%
=\cdots=0\right) \\
&  =\left(  \mu_{1},\mu_{2},\ldots,\mu_{k}\right)
\end{align*}
(since we are identifying any $k$-tuple $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  \in\mathbb{N}^{k}$ with the weak composition $\left(
a_{1},a_{2},\ldots,a_{k},0,0,0,\ldots\right)  $). Thus, there exists some
$q\in \NN$ such that $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
$ (namely, $q=k$). This proves Proposition
\ref{prop.sol.Lambda.triangular.k.l5}.
\end{proof}

Now, let us resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(j).
\end{commentedout}

(j) Let $\lambda$ be a partition. Let $\mu$ be a weak composition. Let
$\ell=\ell\left(  \lambda\right)  $. We must prove that the number
$b_{\lambda,\mu}$ is well-defined. In other words, we must prove that there
are only finitely many maps $\varphi:\left\{  1,2,\ldots,\ell\right\}
\rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(  \mu_{j}%
=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,\ell\right\}  ;\\\varphi
(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $.

We know that $\lambda$ is a partition. In other words, $\lambda\in
\Par$ (since $\Par$ is the set of all partitions).

We have $\ell=\ell\left(  \lambda\right)  $; thus, $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $. Hence, $\lambda$ is
an $\ell$-tuple in $\mathbb{N}^{\ell}$.

We have $\ell=\ell\left(  \lambda\right)  $. Hence, the definition of
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ yields%
\begin{align}
\mathfrak{B}_{\lambda,\mu}^{\prime}  &  =\left\{  \varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \underbrace{\mu_{j}%
=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi\left(  i\right)
=j}}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}
}_{\substack{\Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  \\\text{(since the }j\in\left\{  1,2,3,\ldots\right\}  \text{
are precisely the integers }j\geq1\text{)}}}\right\} \nonumber\\
&  =\left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}
\nonumber\\
&  =\left\{  \varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,\ell\right\}  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for
all }j\geq1\right\}  \label{sol.Lambda.triangular.B'lm=}%
\end{align}
(since $\left[  \ell\right]  =\left\{  1,2,\ldots,\ell\right\}  $).

We know that $\mu$ is a weak composition. In other words, $\mu\in
\operatorname*{WC}$ (since $\operatorname*{WC}$ is the set of all weak
compositions). Proposition \ref{prop.sol.Lambda.triangular.k.l5} shows that
there exists some $q\in \NN$ such that $\mu=\left(  \mu_{1},\mu
_{2},\ldots,\mu_{q}\right)  $. Consider this $q$. Thus, $\mu$ is a $q$-tuple
in $\mathbb{N}^{q}$ (since $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)
$). Hence, the set $\mathfrak{B}_{\lambda,\mu,\ell,q}$ is well-defined (since
$\lambda\in\mathbb{N}^{\ell}$ and $\mu\in\mathbb{N}^{q}$). Thus, Proposition
\ref{prop.sol.Lambda.triangular.j.l5} (applied to $p=\ell$) shows that
$\mathfrak{B}_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,\ell,q}$ as
sets. But the definition of $\mathfrak{B}_{\lambda,\mu,\ell,q}$ yields
\[
\mathfrak{B}_{\lambda,\mu,\ell,q}=\left\{  \varphi:\left[  \ell\right]
\rightarrow\left[  q\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}
\subset\left\{  \varphi:\left[  \ell\right]  \rightarrow\left[  q\right]
\right\}  =\left[  q\right]  ^{\left[  \ell\right]  }.
\]
Hence, $\mathfrak{B}_{\lambda,\mu,\ell,q}$ is a finite set (since $\left[
q\right]  ^{\left[  \ell\right]  }$ is a finite set). Therefore, the set
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ is a finite set as well (since
$\mathfrak{B}_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,\ell,q}$ as
sets). In view of (\ref{sol.Lambda.triangular.B'lm=}), this rewrites as
follows: The set%
\[
\left\{  \varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,\ell\right\}  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for
all }j\geq1\right\}
\]
is a finite set. In other words, there are only finitely many maps
$\varphi:\left\{  1,2,\ldots,\ell\right\}  \rightarrow\left\{  1,2,3,\ldots
\right\}  $ satisfying $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,\ell\right\}  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }%
j\geq1\right)  $. Thus, the number $b_{\lambda,\mu}$ is well-defined (since
this number is defined as the number of all maps $\varphi:\left\{
1,2,\ldots,\ell\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  $
satisfying $\left(  \mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,\ell\right\}  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $).
This solves Exercise \ref{exe.Lambda.triangular}(j).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(k).
\end{commentedout}

(k) Let $\lambda\in\Par_{n}$. Let $\ell=\ell\left(
\lambda\right)  $. Thus, $\lambda_{\ell+1}=\lambda_{\ell+2}=\lambda_{\ell
+3}=\cdots=0$. Therefore, $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell}\right)  $ (since we omit trailing zeroes from a
partition). Thus, $\lambda$ can be regarded as the $\ell$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $. Hence, $\lambda
=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  \in
\mathbb{N}^{\ell}$. Therefore, for every map $\varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  $, a weak composition $\varphi
_{\ast}\lambda\in\operatorname*{WC}$ is defined (according to Definition
\ref{def.sol.Lambda.triangular.k.2}, applied to $p=\ell$, $\alpha=\lambda$ and
$\alpha_{i}=\lambda_{i}$). The definition of this weak composition
$\varphi_{\ast}\lambda$ yields%
\begin{equation}
\varphi_{\ast}\lambda=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}
}\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i},\sum
_{i\in\varphi^{-1}\left\{  3\right\}  }\lambda_{i},\ldots\right)
\label{sol.Lambda.triangular.phi*lam=}%
\end{equation}
for any map $\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  $. Hence, for each map $\varphi:\left[  \ell\right]  \rightarrow
\left\{  1,2,3,\ldots\right\}  $, we have the following chain of logical
equivalences:%
\begin{align}
&  \ \left(  \underbrace{\mu}_{=\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)
}=\underbrace{\varphi_{\ast}\lambda}_{\substack{=\left(  \sum_{i\in
\varphi^{-1}\left\{  1\right\}  }\lambda_{i},\sum_{i\in\varphi^{-1}\left\{
2\right\}  }\lambda_{i},\sum_{i\in\varphi^{-1}\left\{  3\right\}  }\lambda
_{i},\ldots\right)  \\\text{(by (\ref{sol.Lambda.triangular.phi*lam=}))}%
}}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)
=\left(  \sum_{i\in\varphi^{-1}\left\{  1\right\}  }\lambda_{i},\sum
_{i\in\varphi^{-1}\left\{  2\right\}  }\lambda_{i},\sum_{i\in\varphi
^{-1}\left\{  3\right\}  }\lambda_{i},\ldots\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu_{j}=\underbrace{\sum\limits_{i\in
\varphi^{-1}\left\{  j\right\}  }}_{\substack{=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  \in\left\{  j\right\}  }%
}=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi\left(  i\right)
=j}}\\\text{(since for each }i\in\left[  \ell\right]  \text{,}\\\text{the
condition }\left(  \varphi\left(  i\right)  \in\left\{  j\right\}  \right)
\\\text{is equivalent to }\left(  \varphi\left(  i\right)  =j\right)
\text{)}}}\lambda_{i}\text{ for all }j\geq1\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  . \label{sol.Lambda.triangular.equivs}%
\end{align}


We have $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)
$ with $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{\ell}>0$ (since
$\ell=\ell\left(  \lambda\right)  $). Hence, $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ (by the definition of $p_{\lambda}%
$). Hence, $p_{\lambda}\in\Lambda$ (since $p_{i}\in\Lambda$ for each
$i\in \NN$), so that $p_{\lambda}\in\Lambda\subset R\left(
\xx\right)  $.

Also, $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)
\in\mathbb{N}^{\ell}$ and $\left(  \lambda_{i}>0\text{ for each }i\in\left[
\ell\right]  \right)  $\ \ \ \ \footnote{\textit{Proof.} We have $\lambda
_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{\ell}>0$. Hence, $\lambda_{i}>0$ for
each $i\in\left\{  1,2,\ldots,\ell\right\}  $. In other words, $\lambda_{i}>0$
for each $i\in\left[  \ell\right]  $ (since $\left[  \ell\right]  =\left\{
1,2,\ldots,\ell\right\}  $). Qed.}. Hence, Proposition
\ref{prop.sol.Lambda.triangular.k.l4} (applied to $\alpha=\lambda$ and
$\alpha_{i}=\lambda_{i}$) yields%
\[
p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}=\sum_{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  }\xx%
^{\varphi_{\ast}\lambda}%
\]
in $\kk\left[  \left[  \xx\right]  \right]  $.

Now, let $\mu\in\Par_{n}$. Thus, $\mu\in
\Par_{n}\subset\Par \subset \operatorname*{WC}$. Now,
\begin{align}
&  \left[  \mathbf{x}^{\mu}\right]  \left(  \underbrace{p_{\lambda}%
}_{\substack{=p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}%
\\=\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  }\mathbf{x}^{\varphi_{\ast}\lambda}}}\right) \nonumber\\
&  =\left[  \mathbf{x}^{\mu}\right]  \left(  \sum_{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  }\xx%
^{\varphi_{\ast}\lambda}\right)  =\sum_{\varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  }\underbrace{\left[  \xx%
^{\mu}\right]  \left(  \mathbf{x}^{\varphi_{\ast}\lambda}\right)
}_{\substack{=\delta_{\mu,\varphi_{\ast}\lambda}\\\text{(by
(\ref{sol.Lambda.triangular.coeff-mon}) (applied to }\alpha=\varphi_{\ast
}\lambda\text{))}}}\nonumber\\
&  =\sum_{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  }\delta_{\mu,\varphi_{\ast}\lambda}=\sum\limits_{\substack{\varphi:\left[
\ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  ;\\\mu=\varphi_{\ast
}\lambda}}\underbrace{\delta_{\mu,\varphi_{\ast}\lambda}}%
_{\substack{=1\\\text{(since }\mu=\varphi_{\ast}\lambda\text{)}}%
}+\sum\limits_{\substack{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  ;\\\mu\neq\varphi_{\ast}\lambda}}\underbrace{\delta
_{\mu,\varphi_{\ast}\lambda}}_{\substack{=0\\\text{(since }\mu\neq
\varphi_{\ast}\lambda\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each map }\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \text{ satisfies}\\
\text{either }\mu=\varphi_{\ast}\lambda\text{ or }\mu\neq\varphi_{\ast}%
\lambda\text{ (but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{\varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  ;\\\mu=\varphi_{\ast}\lambda}}1+\underbrace{\sum\limits
_{\substack{\varphi:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  ;\\\mu\neq\varphi_{\ast}\lambda}}0}_{=0}=\sum\limits_{\substack{\varphi
:\left[  \ell\right]  \rightarrow\left\{  1,2,3,\ldots\right\}  ;\\\mu
=\varphi_{\ast}\lambda}}1\nonumber\\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}  \right\vert
\cdot1\nonumber\\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \underbrace{\mu=\varphi_{\ast}\lambda
}_{\substack{\Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  \\\text{(by (\ref{sol.Lambda.triangular.equivs}))}}}\right\}
\right\vert \nonumber\\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]
;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}
\right\vert . \label{sol.Lambda.triangular.k.4}%
\end{align}


On the other hand, the definition of $b_{\lambda,\mu}$ shows that
$b_{\lambda,\mu}$ is the number of all maps $\varphi:\left\{  1,2,\ldots
,\ell\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(
\mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,\ell\right\}
;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $ (since $\ell
=\ell\left(  \lambda\right)  $). In other words,%
\begin{align*}
b_{\lambda,\mu}  &  =\left(  \text{the number of all maps }\varphi:\left\{
1,2,\ldots,\ell\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  \text{
satisfying }\left(  \mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,\ell\right\}  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)
\right) \\
&  =\left(  \text{the number of all maps }\varphi:\left[  \ell\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \text{ satisfying }\left(  \mu
_{j}=\sum\limits_{\substack{i\in\left[  \ell\right]  ;\\\varphi(i)=j}%
}\lambda_{i}\text{ for all }j\geq1\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\{  1,2,\ldots,\ell\right\}
=\left[  \ell\right]  \right) \\
&  =\left\vert \left\{  \varphi:\left[  \ell\right]  \rightarrow\left\{
1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[
\ell\right]  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right\}
\right\vert .
\end{align*}
Comparing this with (\ref{sol.Lambda.triangular.k.4}), we obtain%
\begin{equation}
\left[  \mathbf{x}^{\mu}\right]  \left(  p_{\lambda}\right)  =b_{\lambda,\mu}.
\label{sol.Lambda.triangular.k.coeff=}%
\end{equation}


Now, forget that we fixed $\mu$. We thus have proven the equality
(\ref{sol.Lambda.triangular.k.coeff=}) for every $\mu\in\Par_{n}$.

But $\lambda\in\Par_{n}$, so that $\left\vert
\lambda\right\vert =n$. Hence, $n=\left\vert \underbrace{\lambda}_{=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  }\right\vert =\left\vert
\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  \right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}$.

Let $i\in\left\{  1,2,\ldots,\ell\right\}  $. Then, $p_{\lambda_{i}}$ is a
homogeneous element of $\Lambda$ having degree $\lambda_{i}$ (because for each
positive integer $m$, the element $p_{m}$ is a homogeneous element of
$\Lambda$ having degree $m$).

Now, forget that we fixed $i$. We thus have shown that for each $i\in\left\{
1,2,\ldots,\ell\right\}  $, the element $p_{\lambda_{i}}$ is a homogeneous
element of $\Lambda$ having degree $\lambda_{i}$. In other words,
$p_{\lambda_{1}},p_{\lambda_{2}},\ldots,p_{\lambda_{\ell}}$ are homogeneous
elements of $\Lambda$ having degrees $\lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell}$, respectively. Hence, the product $p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ of these elements is a homogeneous
element of $\Lambda$ having degree $\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{\ell}$. In light of $p_{\lambda}=p_{\lambda_{1}}p_{\lambda_{2}%
}\cdots p_{\lambda_{\ell}}$ and $n=\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{\ell}$, this rewrites as follows: The element $p_{\lambda}$ is a
homogeneous element of $\Lambda$ having degree $n$. In other words,
$p_{\lambda}\in\Lambda_{n}$. Thus, Exercise \ref{exe.Lambda.triangular}(a)
(applied to $f=p_{\lambda}$) yields%
\[
p_{\lambda}=\sum_{\mu\in\Par_{n}}\underbrace{\left(
\left[  \mathbf{x}^{\mu}\right]  \left(  p_{\lambda}\right)  \right)
}_{\substack{=b_{\lambda,\mu}\\\text{(by (\ref{sol.Lambda.triangular.k.coeff=}%
))}}}m_{\mu}=\sum_{\mu\in\operatorname{Par}_{n}}b_{\lambda,\mu}m_{\mu}.
\]
This solves Exercise \ref{exe.Lambda.triangular}(k).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(l).
\end{commentedout}

Before we come to the solution of Exercise \ref{exe.Lambda.triangular}(l), let
us state a simple lemma:

\begin{lemma}
\label{lem.sol.Lambda.triangular.k.1}Let $\lambda$ be a partition. Let $\mu$
be a weak composition. Then:

\begin{enumerate}
\item[(a)] We have $b_{\lambda,\mu}=\left\vert \mathfrak{B}_{\lambda,\mu
}^{\prime}\right\vert $.

\item[(b)] Let $p=\ell\left(  \lambda\right)  $. Let $q\in \NN$ be such
that $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{q}\right)  $ (that is, $\mu
_{i}=0$ for all integers $i>q$). Then, the set $\mathfrak{B}_{\lambda,\mu
,p,q}$ is well-defined and satisfies $b_{\lambda,\mu}=\left\vert
\mathfrak{B}_{\lambda,\mu,p,q}\right\vert $.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.triangular.k.1}.]We know that $\lambda$ is
a partition. In other words, $\lambda \in \Par$ (since
$\Par$ is the set of all partitions).

We have $p=\ell\left(  \lambda\right)  $. Hence, the definition of
$\mathfrak{B}_{\lambda,\mu}^{\prime}$ yields%
\begin{align}
\mathfrak{B}_{\lambda,\mu}^{\prime}  &  =\left\{  \varphi:\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \ \mid\ \underbrace{\mu_{j}%
=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi\left(  i\right)  =j}%
}\lambda_{i}\text{ for all }j\in\left\{  1,2,3,\ldots\right\}  }%
_{\substack{\Longleftrightarrow\ \left(  \mu_{j}=\sum\limits_{\substack{i\in\left[
p\right]  ;\\\varphi\left(  i\right)  =j}}\lambda_{i}\text{ for all }%
j\geq1\right)  \\\text{(since the }j\in\left\{  1,2,3,\ldots\right\}  \text{
are precisely the integers }j\geq1\text{)}}}\right\} \nonumber\\
&  =\left\{  \varphi:\left[  p\right]  \rightarrow\left\{  1,2,3,\ldots
\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi
\left(  i\right)  =j}}\lambda_{i}\text{ for all }j\geq1\right\}  .
\label{pf.lem.sol.Lambda.triangular.k.1.1}%
\end{align}


On the other hand, the definition of $b_{\lambda,\mu}$ shows that
$b_{\lambda,\mu}$ is the number of all maps $\varphi:\left\{  1,2,\ldots
,p\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  $ satisfying $\left(
\mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,p\right\}
;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  $ (since
$p=p\left(  \lambda\right)  $). In other words,%
\begin{align*}
b_{\lambda,\mu}  &  =\left(  \text{the number of all maps }\varphi:\left\{
1,2,\ldots,p\right\}  \rightarrow\left\{  1,2,3,\ldots\right\}  \text{
satisfying }\left(  \mu_{j}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,p\right\}  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right)  \right)
\\
&  =\left(  \text{the number of all maps }\varphi:\left[  p\right]
\rightarrow\left\{  1,2,3,\ldots\right\}  \text{ satisfying }\left(  \mu
_{j}=\sum\limits_{\substack{i\in\left[  p\right]  ;\\\varphi(i)=j}}\lambda
_{i}\text{ for all }j\geq1\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\{  1,2,\ldots,p\right\}
=\left[  p\right]  \right) \\
&  =\left\vert \underbrace{\left\{  \varphi:\left[  p\right]  \rightarrow
\left\{  1,2,3,\ldots\right\}  \ \mid\ \mu_{j}=\sum\limits_{\substack{i\in
\left[  p\right]  ;\\\varphi(i)=j}}\lambda_{i}\text{ for all }j\geq1\right\}
}_{\substack{=\mathfrak{B}_{\lambda,\mu}^{\prime}\\\text{(by
(\ref{pf.lem.sol.Lambda.triangular.k.1.1}))}}}\right\vert =\left\vert
\mathfrak{B}_{\lambda,\mu}^{\prime}\right\vert .
\end{align*}
This proves Lemma \ref{lem.sol.Lambda.triangular.k.1}(a).

(b) From $p=\ell\left(  \lambda\right)  $, we obtain $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $; thus, $\lambda$ is a
$p$-tuple in $\mathbb{N}^{p}$. Also, $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{q}\right)  \in\mathbb{N}^{q}$. Thus, $\mu$ is a $q$-tuple in
$\mathbb{N}^{q}$. Hence, the set $\mathfrak{B}_{\lambda,\mu,p,q}$ is
well-defined (since $\lambda\in\mathbb{N}^{p}$ and $\mu\in\mathbb{N}^{q}$).
Proposition \ref{prop.sol.Lambda.triangular.j.l5} yields that $\mathfrak{B}%
_{\lambda,\mu}^{\prime}\cong\mathfrak{B}_{\lambda,\mu,p,q}$ as sets. But Lemma
\ref{lem.sol.Lambda.triangular.k.1}(a) shows that $b_{\lambda,\mu}=\left\vert
\mathfrak{B}_{\lambda,\mu}^{\prime}\right\vert =\left\vert \mathfrak{B}%
_{\lambda,\mu,p,q}\right\vert $ (since $\mathfrak{B}_{\lambda,\mu}^{\prime
}\cong\mathfrak{B}_{\lambda,\mu,p,q}$ as sets). Thus, Lemma
\ref{lem.sol.Lambda.triangular.k.1}(b) is proven.
\end{proof}

We now resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(l).
\end{commentedout}

(l) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$ be any partitions that don't satisfy
$\mu\triangleright\lambda$. We must prove that $b_{\lambda,\mu}=0$.

Indeed, assume the contrary. Thus, $b_{\lambda,\mu}\neq0$. But Lemma
\ref{lem.sol.Lambda.triangular.k.1}(a) yields $b_{\lambda,\mu}=\left\vert
\mathfrak{B}_{\lambda,\mu}^{\prime}\right\vert $ (since $\mu$ is a weak
composition (since $\mu$ is a partition)). Hence, $\left\vert \mathfrak{B}%
_{\lambda,\mu}^{\prime}\right\vert =b_{\lambda,\mu}\neq0$. In other words,
$\mathfrak{B}_{\lambda,\mu}^{\prime}\neq\varnothing$. Thus, Proposition
\ref{prop.sol.Lambda.triangular.j.l6} shows that $\mu\triangleright\lambda$.
This contradicts the fact that we don't have $\mu\triangleright\lambda$.

This contradiction proves that our assumption was wrong. Hence, $b_{\lambda
,\mu}=0$ is proven. This solves Exercise \ref{exe.Lambda.triangular}(l).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(m).
\end{commentedout}

(m) Let $\lambda\in\Par_{n}$. We must prove that
$b_{\lambda,\lambda}$ is a positive integer.

Let $k=\ell\left(  \lambda\right)  $. Thus, $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{k}\right)  $ (by the definition of
$\ell\left(  \lambda\right)  $). Clearly, $\lambda$ is a weak composition
(since $\lambda$ is a partition). Hence, Lemma
\ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $\mu=\lambda$, $p=k$ and
$q=k$) shows that the set $\mathfrak{B}_{\lambda,\lambda,k,k}$ is well-defined
and satisfies $b_{\lambda,\lambda}=\left\vert \mathfrak{B}_{\lambda
,\lambda,k,k}\right\vert $. But $\lambda\in\Par_{n}%
\subset\Par$. Hence, Proposition
\ref{prop.sol.Lambda.triangular.j.l7}(b) (applied to $\mu=\lambda$) shows that
$\mathfrak{B}_{\lambda,\lambda,k,k}$ is a subgroup of $\Symm_k$.
Thus, the set $\mathfrak{B}_{\lambda,\lambda,k,k}$ contains the neutral
element of $\Symm_k$. Thus, the set $\mathfrak{B}_{\lambda
,\lambda,k,k}$ contains at least one element, and therefore is nonempty. Thus,
$\left\vert \mathfrak{B}_{\lambda,\lambda,k,k}\right\vert >0$.

Hence, $b_{\lambda,\lambda}=\left\vert \mathfrak{B}_{\lambda,\lambda
,k,k}\right\vert >0$. Since $b_{\lambda,\lambda}$ is an integer, we can
therefore conclude that $b_{\lambda,\lambda}$ is a positive integer. This
solves Exercise \ref{exe.Lambda.triangular}(m).

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(n).
\end{commentedout}

(n) Let $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{k}\right)  \in
\Par_{n}$ be a partition. Let $k=\ell\left(
\mu\right)  $. We must show that the integer $b_{\mu,\mu}$ is the size of the
subgroup of $\Symm_k$ consisting of all permutations $\sigma
\in\Symm_k$ having each $i$ satisfy $\mu_{\sigma(i)}=\mu_{i}$. In
particular, we must show that this subgroup is indeed a subgroup.

Clearly, $\mu$ is a weak composition (since $\mu$ is a partition). Hence,
Lemma \ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $\lambda=\mu$, $p=k$
and $q=k$) shows that the set $\mathfrak{B}_{\mu,\mu,k,k}$ is well-defined and
satisfies $b_{\mu,\mu}=\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert $.

We have $\mu\in\Par_{n}\subset\Par$.
Hence, Proposition \ref{prop.sol.Lambda.triangular.j.l7}(a) shows that
\begin{equation}
\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \sigma\in\Symm_k\ \mid
\ \mu_{\sigma\left(  i\right)  }=\mu_{i}\text{ for each }i\in\left[  k\right]
\right\}  . \label{sol.Lambda.triangular.n.1}%
\end{equation}
Furthermore, Proposition \ref{prop.sol.Lambda.triangular.j.l7}(b) shows that
$\mathfrak{B}_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$.

Now,%
\begin{align}
&  \left(  \text{the set of all permutations }\sigma\in\Symm_k\text{
having each }i\text{ satisfy }\mu_{\sigma\left(  i\right)  }=\mu_{i}\right)
\nonumber\\
&  =\left\{  \sigma\in\Symm_k\ \mid\ \text{each }i\text{ satisfies
}\mu_{\sigma\left(  i\right)  }=\mu_{i}\right\}  \nonumber \\
&  =\left\{  \sigma
\in\Symm_k\ \mid\ \text{each }i\in\underbrace{\left\{  1,2,\ldots
,k\right\}  }_{=\left[  k\right]  }\text{ satisfies }\mu_{\sigma\left(
i\right)  }=\mu_{i}\right\} \nonumber\\
&  =\left\{  \sigma\in\Symm_k\ \mid\ \text{each }i\in\left[
k\right]  \text{ satisfies }\mu_{\sigma\left(  i\right)  }=\mu_{i}\right\}
\nonumber \\
&  =
\left\{  \sigma\in\Symm_k\ \mid\ \mu_{\sigma\left(  i\right)  }%
=\mu_{i}\text{ for each }i\in\left[  k\right]  \right\} \nonumber\\
&  =\mathfrak{B}_{\mu,\mu,k,k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.triangular.n.1})}\right)  . \label{sol.Lambda.triangular.n.3}%
\end{align}


But recall that $\mathfrak{B}_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$.
In view of (\ref{sol.Lambda.triangular.n.3}), this rewrites as follows:
The set of all permutations $\sigma\in\Symm_k$ having each $i$
satisfy $\mu_{\sigma\left(  i\right)  }=\mu_{i}$ is a subgroup of
$\Symm_k$. This subgroup thus is the subgroup of $\Symm_k$
consisting of all permutations $\sigma\in\Symm_k$ having each $i$
satisfy $\mu_{\sigma(i)}=\mu_{i}$. The size of this subgroup is clearly%
\[
\left\vert \underbrace{\left\{  \sigma\in\Symm_k\ \mid\ \text{each
}i\text{ satisfies }\mu_{\sigma\left(  i\right)  }=\mu_{i}\right\}
}_{=\mathfrak{B}_{\mu,\mu,k,k}}\right\vert =\left\vert \mathfrak{B}_{\mu
,\mu,k,k}\right\vert =b_{\mu,\mu}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}b_{\mu,\mu}=\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert \right)  .
\]
Thus, $b_{\mu,\mu}$ is the size of the subgroup of $\Symm_k$
consisting of all permutations $\sigma\in\Symm_k$ having each $i$
satisfy $\mu_{\sigma(i)}=\mu_{i}$. (In particular, we have shown that this
subgroup is indeed a subgroup.) This solves Exercise
\ref{exe.Lambda.triangular}(n).

\begin{commentedout}
Here begin preparations for the solution to Exercise
\ref{exe.Lambda.triangular}(o).
\end{commentedout}

Before we come to the solution of Exercise \ref{exe.Lambda.triangular}(o), let
us prove some further auxiliary results.

\begin{proposition}
\label{prop.sol.Lambda.triangular.o.l1}Let $\mu\in\Par$. Let
$k=\ell\left(  \mu\right)  $. Proposition
\ref{prop.sol.Lambda.triangular.j.l7}(b) shows that the set $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$. Thus, $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a group.

Let $p\in \NN$. Let $\lambda\in\mathbb{N}^{p}$. Then, the set
$\mathfrak{B}_{\lambda,\mu,p,k}$ can be made into a left $\mathfrak{B}%
_{\mu,\mu,k,k}$-set (i.e., it can be equipped with an action of the group
$\mathfrak{B}_{\mu,\mu,k,k}$ from the left) in such a way that the group
$\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on $\mathfrak{B}_{\lambda,\mu,p,k}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.o.l1}.]For any
$\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$ and $\beta\in\mathfrak{B}_{\lambda
,\mu,p,k}$, we have $\alpha\circ\beta\in\mathfrak{B}_{\lambda,\mu,p,k}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$
and $\beta\in\mathfrak{B}_{\lambda,\mu,p,k}$. We must show that $\alpha
\circ\beta\in\mathfrak{B}_{\lambda,\mu,p,k}$.
\par
We have $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by the definition of $\mathfrak{B}_{\mu,\mu,k,k}$). In other words,
$\alpha$ is a $\varphi:\left[  k\right]  \rightarrow\left[  k\right]  $
satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\alpha$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu=\alpha_{\ast}\mu$.
\par
We have $\beta\in\mathfrak{B}_{\lambda,\mu,p,k}=\left\{  \varphi:\left[
p\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}%
\lambda\right\}  $ (by the definition of $\mathfrak{B}_{\lambda,\mu,p,k}$). In
other words, $\beta$ is a $\varphi:\left[  p\right]  \rightarrow\left[
k\right]  $ satisfying $\mu=\varphi_{\ast}\lambda$. In other words, $\beta$ is
a map $\left[  p\right]  \rightarrow\left[  k\right]  $ and satisfies
$\mu=\beta_{\ast}\lambda$.
\par
Now, Proposition \ref{prop.sol.Lambda.triangular.j.l1} (applied to $k$, $k$,
$\beta$, $\alpha$ and $\lambda$ instead of $q$, $r$, $\varphi$, $\psi$ and
$\alpha$) yields $\left(  \alpha\circ\beta\right)  _{\ast}\lambda=\alpha
_{\ast}\left(  \underbrace{\beta_{\ast}\lambda}_{=\mu}\right)  =\alpha_{\ast
}\mu=\mu$. In other words, $\mu=\left(  \alpha\circ\beta\right)  _{\ast
}\lambda$. Hence, $\alpha\circ\beta$ is a map $\left[  p\right]
\rightarrow\left[  k\right]  $ (since $\alpha$ is a map $\left[  k\right]
\rightarrow\left[  k\right]  $, and since $\beta$ is a map $\left[  p\right]
\rightarrow\left[  k\right]  $), and satisfies $\mu=\left(  \alpha\circ
\beta\right)  _{\ast}\lambda$. In other words, $\alpha\circ\beta$ is a
$\varphi:\left[  p\right]  \rightarrow\left[  k\right]  $ satisfying
$\mu=\varphi_{\ast}\lambda$. In other words, $\alpha\circ\beta\in\left\{
\varphi:\left[  p\right]  \rightarrow\left[  k\right]  \ \mid\ \mu
=\varphi_{\ast}\lambda\right\}  $. This rewrites as $\alpha\circ\beta
\in\mathfrak{B}_{\lambda,\mu,p,k}$ (since $\mathfrak{B}_{\lambda,\mu
,p,k}=\left\{  \varphi:\left[  p\right]  \rightarrow\left[  k\right]
\ \mid\ \mu=\varphi_{\ast}\lambda\right\}  $). Qed.}. Thus, we can try to
define an action of the group $\mathfrak{B}_{\mu,\mu,k,k}$ on the set
$\mathfrak{B}_{\lambda,\mu,p,k}$ from the left by setting%
\begin{equation}
\left(  \alpha\beta=\alpha\circ\beta\ \ \ \ \ \ \ \ \ \ \text{for all }%
\alpha\in\mathfrak{B}_{\mu,\mu,k,k}\text{ and }\beta\in\mathfrak{B}%
_{\lambda,\mu,p,k}\right)  .
\label{pf.prop.sol.Lambda.triangular.o.l1.def-act}%
\end{equation}
In order to show that this definition actually defines an action of the group
$\mathfrak{B}_{\mu,\mu,k,k}$ on the set $\mathfrak{B}_{\lambda,\mu,p,k}$, we
need to prove the following two observations:

\begin{statement}
\textit{Observation 1:} We have $\id_{\left\{
1,2,\ldots,k\right\}  }\circ\gamma=\gamma$ for all $\gamma\in\mathfrak{B}%
_{\lambda,\mu,p,k}$.
\end{statement}

\begin{statement}
\textit{Observation 2:} We have $\left(  \alpha\circ\beta\right)  \circ
\gamma=\alpha\circ\left(  \beta\circ\gamma\right)  $ for all $\alpha
\in\mathfrak{B}_{\mu,\mu,k,k}$, $\beta\in\mathfrak{B}_{\mu,\mu,k,k}$ and
$\gamma\in\mathfrak{B}_{\lambda,\mu,p,k}$.
\end{statement}

However, both Observation 1 and Observation 2 are obvious. Thus, we have shown
that (\ref{pf.prop.sol.Lambda.triangular.o.l1.def-act}) actually defines an
action of the group $\mathfrak{B}_{\mu,\mu,k,k}$ on the set $\mathfrak{B}%
_{\lambda,\mu,p,k}$. Consider this action. Thus, the set $\mathfrak{B}%
_{\lambda,\mu,p,k}$ has been made into a left $\mathfrak{B}_{\mu,\mu,k,k}$-set.

We shall now prove that the group $\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on
$\mathfrak{B}_{\lambda,\mu,p,k}$. In order to do so, we must prove the
following observation:

\begin{statement}
\textit{Observation 3:} If $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$, $\beta
\in\mathfrak{B}_{\mu,\mu,k,k}$ and $\gamma\in\mathfrak{B}_{\lambda,\mu,p,k}$
are such that $\alpha\circ\gamma=\beta\circ\gamma$, then $\alpha=\beta$.
\end{statement}

[\textit{Proof of Observation 3:} Let $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}$,
$\beta\in\mathfrak{B}_{\mu,\mu,k,k}$ and $\gamma\in\mathfrak{B}_{\lambda
,\mu,p,k}$ be such that $\alpha\circ\gamma=\beta\circ\gamma$. We must prove
that $\alpha=\beta$.

We have $\alpha\in\mathfrak{B}_{\mu,\mu,k,k}=\left\{  \varphi:\left[
k\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}\mu\right\}
$ (by the definition of $\mathfrak{B}_{\mu,\mu,k,k}$). In other words,
$\alpha$ is a $\varphi:\left[  k\right]  \rightarrow\left[  k\right]  $
satisfying $\mu=\varphi_{\ast}\mu$. In other words, $\alpha$ is a map $\left[
k\right]  \rightarrow\left[  k\right]  $ and satisfies $\mu=\alpha_{\ast}\mu$.
The same argument (applied to $\beta$ instead of $\alpha$) shows that $\beta$
is a map $\left[  k\right]  \rightarrow\left[  k\right]  $ and satisfies
$\mu=\beta_{\ast}\mu$.

We have $\gamma\in\mathfrak{B}_{\lambda,\mu,p,k}=\left\{  \varphi:\left[
p\right]  \rightarrow\left[  k\right]  \ \mid\ \mu=\varphi_{\ast}%
\lambda\right\}  $ (by the definition of $\mathfrak{B}_{\lambda,\mu,p,k}$). In
other words, $\gamma$ is a $\varphi:\left[  p\right]  \rightarrow\left[
k\right]  $ satisfying $\mu=\varphi_{\ast}\lambda$. In other words, $\gamma$
is a map $\left[  p\right]  \rightarrow\left[  k\right]  $ and satisfies
$\mu=\gamma_{\ast}\lambda$.

Write the $p$-tuple $\lambda$ in the form $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{p}\right)  $.

Also, $\ell\left(  \mu\right)  =k$. Hence, $\mu=\left(  \mu_{1},\mu_{2}%
,\ldots,\mu_{k}\right)  \in\mathbb{N}^{k}$ and $\left(  \mu_{i}>0\text{ for
each }i\in\left[  k\right]  \right)  $\ \ \ \ \footnote{\textit{Proof.} We
have $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{k}>0$ (since $\ell\left(
\mu\right)  =k$). Hence, $\mu_{i}>0$ for each $i\in\left\{  1,2,\ldots
,k\right\}  $. In other words, $\mu_{i}>0$ for each $i\in\left[  k\right]  $
(since $\left[  k\right]  =\left\{  1,2,\ldots,k\right\}  $). Qed.}. Hence,
Lemma \ref{lem.sol.Lambda.triangular.j.l10} (applied to $q=k$, $\varphi
=\gamma$, $\alpha=\lambda$, $\alpha_{i}=\lambda_{i}$, $\beta=\mu$ and
$\beta_{i}=\mu_{i}$) shows that the map $\gamma$ is surjective.

Now, let $i\in\left[  k\right]  $. Hence, $i\in\left[  k\right]
=\gamma\left(  \left[  p\right]  \right)  $ (since the map $\gamma$ is
surjective). In other words, there exists some $j\in\left[  p\right]  $ such
that $i=\gamma\left(  j\right)  $. Consider this $j$. We have $\alpha\left(
\underbrace{i}_{=\gamma\left(  j\right)  }\right)  =\underbrace{\left(
\alpha\circ\gamma\right)  }_{=\beta\circ\gamma}\left(  j\right)  =\left(
\beta\circ\gamma\right)  \left(  j\right)  =\beta\left(  \underbrace{\gamma
\left(  j\right)  }_{=i}\right)  =\beta\left(  i\right)  $.

Now, forget that we fixed $i$. We thus have proven that $\alpha\left(
i\right)  =\beta\left(  i\right)  $ for each $i\in\left[  k\right]  $. In
other words, $\alpha=\beta$ (since both $\alpha$ and $\beta$ are maps $\left[
k\right]  \rightarrow\left[  k\right]  $). This proves Observation 3.]

Observation 3 shows that the group $\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on
$\mathfrak{B}_{\lambda,\mu,p,k}$ (by the definition of ``acting freely'').
Thus, the set $\mathfrak{B}_{\lambda
,\mu,p,k}$ can be made into a left $\mathfrak{B}_{\mu,\mu,k,k}$-set in such a
way that the group $\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on $\mathfrak{B}%
_{\lambda,\mu,p,k}$ (namely, by using the left action of $\mathfrak{B}%
_{\mu,\mu,k,k}$ on $\mathfrak{B}_{\lambda,\mu,p,k}$ that we defined above).
This proves Proposition \ref{prop.sol.Lambda.triangular.o.l1}.
\end{proof}

Let us now recall a basic fact from abstract algebra:

\begin{proposition}
\label{prop.sol.Lambda.triangular.o.l2}Let $G$ be a finite group. Let $X$ be a
finite left $G$-set. Assume that $G$ acts freely on $X$. Then, $\left\vert
G\right\vert \mid\left\vert X\right\vert $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lambda.triangular.o.l2}.]This fact is
well-known, so let us merely sketch the proof: The $G$-set $X$ is a disjoint
union of orbits (since every $G$-set is a disjoint union of orbits). Thus, the
$G$-set $X$ is a disjoint union of finitely many orbits (since $X$ is finite).
In other words, we have $X=O_{1}\cup O_{2}\cup\cdots\cup O_{k}$ for some list
$\left(  O_{1},O_{2},\ldots,O_{k}\right)  $ of disjoint orbits of $G$ on $X$.
Consider this list $\left(  O_{1},O_{2},\ldots,O_{k}\right)  $.

Every $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies $\left\vert
O_{i}\right\vert =\left\vert G\right\vert $\ \ \ \ \footnote{\textit{Proof.}
Let $i\in\left\{  1,2,\ldots,k\right\}  $. Thus, $O_{i}$ is an orbit of $G$ on
$X$. In other words, $O_{i}=Gy$ for some $y\in X$. Consider this $y$.
\par
The elements $gy$ for $g\in G$ are all distinct (because if $gy=hy$ for two
elements $g,h\in G$, then we must have $g=h$ (because $G$ acts freely on
$X$)). Thus, the number of these elements is precisely $\left\vert
G\right\vert $. In other words, $\left\vert \left\{  gy\ \mid\ g\in G\right\}
\right\vert =\left\vert G\right\vert $. In view of $\left\{  gy\ \mid\ g\in
G\right\}  =Gy=O_{i}$, this rewrites as $\left\vert O_{i}\right\vert
=\left\vert G\right\vert $. Qed.}. Hence, $\sum_{i=1}^{k}%
\underbrace{\left\vert O_{i}\right\vert }_{=\left\vert G\right\vert }%
=\sum_{i=1}^{k}\left\vert G\right\vert =k\left\vert G\right\vert $.

From $X=O_{1}\cup O_{2}\cup\cdots\cup O_{k}$, we obtain
\begin{align*}
\left\vert X\right\vert  &  =\left\vert O_{1}\cup O_{2}\cup\cdots\cup
O_{k}\right\vert =\left\vert O_{1}\right\vert +\left\vert O_{2}\right\vert
+\cdots+\left\vert O_{k}\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since
the orbits }O_{1},O_{2},\ldots,O_{k}\text{ are disjoint}\right) \\
&  =\sum_{i=1}^{k}\left\vert O_{i}\right\vert =k\left\vert G\right\vert .
\end{align*}
Thus, $\left\vert G\right\vert \mid k\left\vert G\right\vert =\left\vert
X\right\vert $. This proves Proposition \ref{prop.sol.Lambda.triangular.o.l2}.
\end{proof}

We now resume the solution of Exercise \ref{exe.Lambda.triangular}.

\begin{commentedout}
Here begins the solution to Exercise \ref{exe.Lambda.triangular}(o).
\end{commentedout}

(o) Let $\lambda\in\Par_{n}$ and $\mu\in
\Par_{n}$. We must prove that $b_{\mu,\mu}\mid
b_{\lambda,\mu}$.

Let $p=\ell\left(  \lambda\right)  $. Thus, $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{p}\right)  $ (by the definition of
$\ell\left(  \lambda\right)  $). Hence, $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{p}\right)  \in\mathbb{N}^{p}$.

We have $\mu\in\Par_{n}\subset\Par$.
Let $k=\ell\left(  \mu\right)  $. Proposition
\ref{prop.sol.Lambda.triangular.j.l7}(b) shows that the set $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a subgroup of $\Symm_k$. Thus, $\mathfrak{B}%
_{\mu,\mu,k,k}$ is a finite group (since $\Symm_k$ is a finite group).

But $\lambda$ is a partition (since $\lambda\in \Par_{n}\subset\Par$).
Also, $\mu$ is a partition (since
$\mu\in\Par$), thus a weak composition. Furthermore,
$k=\ell\left(  \mu\right)  $; thus, $\mu=\left(  \mu_{1},\mu_{2},\ldots
,\mu_{k}\right)  $ (by the definition of $\ell\left(  \mu\right)  $). Hence,
Lemma \ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $q=k$) shows that the
set $\mathfrak{B}_{\lambda,\mu,p,k}$ is well-defined and satisfies
$b_{\lambda,\mu}=\left\vert \mathfrak{B}_{\lambda,\mu,p,k}\right\vert $.

Furthermore, Lemma \ref{lem.sol.Lambda.triangular.k.1}(b) (applied to $k$, $k$
and $\mu$ instead of $p$, $q$ and $\lambda$) shows that the set $\mathfrak{B}%
_{\mu,\mu,k,k}$ is well-defined and satisfies $b_{\mu,\mu}=\left\vert
\mathfrak{B}_{\mu,\mu,k,k}\right\vert $.

The definition of $\mathfrak{B}_{\lambda,\mu,p,k}$ yields $\mathfrak{B}%
_{\lambda,\mu,p,k}=\left\{  \varphi:\left[  p\right]  \rightarrow\left[
k\right]  \ \mid\ \mu=\varphi_{\ast}\lambda\right\}  \subset\left\{
\varphi:\left[  p\right]  \rightarrow\left[  k\right]  \right\}  =\left[
k\right]  ^{\left[  p\right]  }$. Thus, $\mathfrak{B}_{\lambda,\mu,p,k}$ is a
finite set (since $\left[  k\right]  ^{\left[  p\right]  }$ is a finite set).

Proposition \ref{prop.sol.Lambda.triangular.o.l1} shows that the set
$\mathfrak{B}_{\lambda,\mu,p,k}$ can be made into a left $\mathfrak{B}%
_{\mu,\mu,k,k}$-set (i.e., it can be equipped with an action of the group
$\mathfrak{B}_{\mu,\mu,k,k}$ from the left) in such a way that the group
$\mathfrak{B}_{\mu,\mu,k,k}$ acts freely on $\mathfrak{B}_{\lambda,\mu,p,k}$.
Consider this action. Proposition \ref{prop.sol.Lambda.triangular.o.l2}
(applied to $G=\mathfrak{B}_{\mu,\mu,k,k}$ and $X=\mathfrak{B}_{\lambda
,\mu,p,k}$) now yields $\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert
\mid\left\vert \mathfrak{B}_{\lambda,\mu,p,k}\right\vert $. Thus, $b_{\mu,\mu
}=\left\vert \mathfrak{B}_{\mu,\mu,k,k}\right\vert \mid\left\vert
\mathfrak{B}_{\lambda,\mu,p,k}\right\vert =b_{\lambda,\mu}$ (since
$b_{\lambda,\mu}=\left\vert \mathfrak{B}_{\lambda,\mu,p,k}\right\vert $). This
solves Exercise \ref{exe.Lambda.triangular}(o).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.alggen-via-span}}

\textit{Solution to Exercise \ref{exe.alggen-via-span}.} Let us first consider
the polynomial ring $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ in
countably many indeterminates $x_{1},x_{2},x_{3},\ldots$.

Let $f:\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  \rightarrow A$ be
the $\kk$-algebra homomorphism that sends $x_{1},x_{2},x_{3},\ldots$ to
$v_{1},v_{2},v_{3},\ldots$, respectively. This is well-defined by the
universal property of the polynomial ring $\kk\left[  x_{1},x_{2}%
,x_{3},\ldots\right]  $ (since $A$ is commutative). The $\kk$-algebra
homomorphism $f$ is an instance of an evaluation homomorphism; it sends each
polynomial $P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ to
$P\left(  v_{1},v_{2},v_{3},\ldots\right)  \in A$. In other words,%
\begin{equation}
f\left(  P\right)  =P\left(  v_{1},v_{2},v_{3},\ldots\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }P\in\kk\left[  x_{1},x_{2}%
,x_{3},\ldots\right]  . \label{sol.alggen-via-span.fP=}%
\end{equation}
The map $f$ is a $\kk$-algebra homomorphism, and thus is $\kk$-linear.

For each partition $\lambda\in\Par$, we define a monomial
$\xx_{\lambda}\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $
by%
\[
\xx_{\lambda}=x_{\lambda_{1}}x_{\lambda_{2}}\cdots x_{\lambda
_{\ell\left(  \lambda\right)  }}.
\]
(This is well-defined, since $\lambda_{1},\lambda_{2},\ldots,\lambda
_{\ell\left(  \lambda\right)  }$ are positive integers whenever $\lambda
\in\Par$.)

If $\lambda$ is a partition and $i$ is a positive integer, then $m_{i}\left(
\lambda\right)  $ shall denote the multiplicity of $i$ in $\lambda$ (that is,
the number of parts of $\lambda$ equal to $i$).

Let \dfn{$\operatorname*{WC}$} denote the set of all weak compositions.

Every partition $\lambda$ can be uniquely written in the form $\lambda=\left(
1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for some weak composition $\left(
m_{1},m_{2},m_{3},\ldots\right)  $. Thus, the map
\begin{align*}
\mathbf{part}:\operatorname*{WC}  &  \rightarrow\Par,\\
\left(  m_{1},m_{2},m_{3},\ldots\right)   &  \mapsto\left(  1^{m_{1}}2^{m_{2}%
}3^{m_{3}}\cdots\right)
\end{align*}
is a bijection. The inverse of this bijection is the map%
\begin{align*}
\mathbf{mults}:\Par  &  \rightarrow\operatorname*{WC},\\
\lambda &  \mapsto\left(  m_{1}\left(  \lambda\right)  ,m_{2}\left(
\lambda\right)  ,m_{3}\left(  \lambda\right)  ,\ldots\right)  .
\end{align*}
Thus, this map $\mathbf{mults}$ is a bijection, too (since it is the inverse
of a bijection).

It is easy to see that
\begin{equation}
\xx_{\lambda}=\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)
}\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par%
\label{sol.alggen-via-span.xl=xl}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.alggen-via-span.xl=xl}):} Let $\lambda
\in\Par$. Then, $\lambda_{p}$ is a positive integer for each
$p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $.
\par
We have $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(
\lambda\right)  }\right)  $, and the entries $\lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell\left(  \lambda\right)  }$ of $\lambda$ are positive
integers, whereas each integer $p>\ell\left(  \lambda\right)  $ satisfies
$\lambda_{p}=0$. Thus, the parts of $\lambda$ are $\lambda
_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }$
(since a \emph{part}
of a partition means a nonzero entry of the partition). For each
$i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\begin{align*}
m_{i}\left(  \lambda\right)    & =\left(  \text{the multiplicity of }i\text{
in }\lambda\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}m_{i}\left(  \lambda\right)  \right)  \\
& =\left(  \text{the number of parts of }\lambda\text{ equal to }i\right)  \\
& =\left(  \text{the number of }p\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \text{ satisfying }\lambda_{p}=i\right)
\end{align*}
(since the parts of $\lambda$ are $\lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell\left(  \lambda\right)  }$). Hence, for each $i\in\left\{
1,2,3,\ldots\right\}  $, we have%
\begin{equation}
\left(  \text{the number of }p\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \text{ satisfying }\lambda_{p}=i\right)
=m_{i}\left(  \lambda\right)  .\label{sol.alggen-via-span.xl=xl.pf.mi}%
\end{equation}
\par
The definition of $\mathbf{mults}$ yields $\mathbf{mults}\left(
\lambda\right)  =\left(  m_{1}\left(  \lambda\right)  ,m_{2}\left(
\lambda\right)  ,m_{3}\left(  \lambda\right)  ,\ldots\right)  $. Thus, the
definition of $\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }$ yields%
\begin{equation}
\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }=x_{1}^{m_{1}\left(
\lambda\right)  }x_{2}^{m_{2}\left(  \lambda\right)  }x_{3}^{m_{3}\left(
\lambda\right)  }\cdots=\prod_{i=1}^{\infty}x_{i}^{m_{i}\left(  \lambda
\right)  }.\label{sol.alggen-via-span.xl=xl.pf.xml=}%
\end{equation}
\par
But the definition of $\xx_{\lambda}$ yields%
\begin{equation}
\xx_{\lambda}=x_{\lambda_{1}}x_{\lambda_{2}}\cdots x_{\lambda
_{\ell\left(  \lambda\right)  }}=\prod_{p\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  }x_{\lambda_{p}}=\prod_{i=1}^{\infty}\prod\limits
_{\substack{p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
;\\\lambda_{p}=i}}x_{\lambda_{p}}\nonumber
\end{equation}
(here, we have split the product according to the value of $\lambda_{p}$,
since $\lambda_{p}$ is a positive integer for each $p\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $). Hence,%
\[
\xx_{\lambda}=\prod_{i=1}^{\infty}\prod\limits_{\substack{p\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{p}=i}%
}\underbrace{x_{\lambda_{p}}}_{\substack{=x_{i}\\\text{(since }\lambda
_{p}=i\text{)}}}=\prod_{i=1}^{\infty}\underbrace{\prod\limits_{\substack{p\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{p}=i}}x_{i}%
}_{\substack{=x_{i}^{\left(  \text{the number of }p\in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  \text{ satisfying }\lambda
_{p}=i\right)  }=x_{i}^{m_{i}\left(  \lambda\right)  }\\\text{(by
(\ref{sol.alggen-via-span.xl=xl.pf.mi}))}}}=\prod_{i=1}^{\infty}x_{i}%
^{m_{i}\left(  \lambda\right)  }.
\]
Comparing this with (\ref{sol.alggen-via-span.xl=xl.pf.xml=}), we obtain
$\xx_{\lambda}=\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }$.
This proves (\ref{sol.alggen-via-span.xl=xl}).}. In other words, $\left(
\xx_{\lambda}\right)  _{\lambda\in\Par}=\left(
\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }\right)  _{\lambda
\in\Par}$.

The monomials in $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ have
the form $\mathbf{x}^{\alpha}$ for $\alpha\in\operatorname*{WC}$. It is
well-known that these monomials form a basis of the $\kk$-module
$\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ (since any polynomial
ring has a basis consisting of the monomials). In other words, the family
$\left(  \mathbf{x}^{\alpha}\right)  _{\alpha\in\operatorname*{WC}}$ is a
basis of the $\kk$-module $\kk\left[  x_{1},x_{2},x_{3}%
,\ldots\right]  $. The family $\left(  \mathbf{x}^{\mathbf{mults}\left(
\lambda\right)  }\right)  _{\lambda\in\Par}$ is a reindexing of
this basis $\left(  \mathbf{x}^{\alpha}\right)  _{\alpha\in\operatorname*{WC}%
}$ (since $\mathbf{mults}:\Par\rightarrow\operatorname*{WC}$ is
a bijection), and thus must also be a basis of the $\kk$-module
$\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ (since a reindexing of a
basis of a $\kk$-module is always a basis itself). In view of $\left(
\xx_{\lambda}\right)  _{\lambda\in\Par}=\left(
\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }\right)  _{\lambda
\in\Par}$, we can rewrite this as follows: The family $\left(
\xx_{\lambda}\right)  _{\lambda\in\Par}$ is a basis of
the $\kk$-module $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $.
Hence, this family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ is $\kk$-linearly independent and spans the
$\kk$-module $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $.

Let us now step to the solution of Exercise \ref{exe.alggen-via-span}:

(a) Let us use the following notation: If $U$ is any $\kk$-module, and
if $\left(  u_{i}\right)  _{i\in I}$ is any family of elements of $U$, then
$\left\langle u_{i}\ \mid\ i\in I\right\rangle _{\kk}$ shall denote the
$\kk$-submodule of $U$ spanned by this family $\left(  u_{i}\right)
_{i\in I}$.

We know that the family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ spans the $\kk$-module $\kk\left[
x_{1},x_{2},x_{3},\ldots\right]  $. In other words,
\[
\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  =\left\langle
\xx_{\lambda}\ \mid\ \lambda\in\Par\right\rangle
_{\kk}.
\]


We have%
\begin{equation}
f\left(  \xx_{\lambda}\right)  =v_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par
\label{sol.alggen-via-span.fxl=vl}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.alggen-via-span.fxl=vl}):} Let
$\lambda\in\Par$. Then, the definition of $\xx_{\lambda
}$ yields $\xx_{\lambda}=x_{\lambda_{1}}x_{\lambda_{2}}\cdots
x_{\lambda_{\ell\left(  \lambda\right)  }}$. Applying the map $f$ to both
sides of this equality, we obtain%
\begin{align*}
f\left(  \xx_{\lambda}\right)   &  =f\left(  x_{\lambda_{1}}%
x_{\lambda_{2}}\cdots x_{\lambda_{\ell\left(  \lambda\right)  }}\right)
=f\left(  x_{\lambda_{1}}\right)  f\left(  x_{\lambda_{2}}\right)  \cdots
f\left(  x_{\lambda_{\ell\left(  \lambda\right)  }}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }\kk%
\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{\ell\left(  \lambda\right)  }\underbrace{f\left(
x_{\lambda_{i}}\right)  }_{\substack{=v_{\lambda_{i}}\\\text{(by the
definition of }f\text{)}}}=\prod_{i=1}^{\ell\left(  \lambda\right)
}v_{\lambda_{i}}=v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(
\lambda\right)  }}.
\end{align*}
Comparing this with%
\[
v_{\lambda}=v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(
\lambda\right)  }}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}v_{\lambda}\right)  ,
\]
we obtain $f\left(  \xx_{\lambda}\right)  =v_{\lambda}$. This proves
(\ref{sol.alggen-via-span.fxl=vl}).}.

But $A$ is commutative; hence,%
\begin{align*}
&  \left(  \text{the }\kk\text{-subalgebra of }A\text{ generated by
}v_{1},v_{2},v_{3},\ldots\right)  \\
&  =\left\{  \underbrace{P\left(  v_{1},v_{2},v_{3},\ldots\right)
}_{\substack{=f\left(  P\right)  \\\text{(by (\ref{sol.alggen-via-span.fP=}%
))}}}\ \mid\ P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  \right\}
\\
&  =\left\{  f\left(  P\right)  \ \mid\ P\in\kk\left[  x_{1}%
,x_{2},x_{3},\ldots\right]  \right\}  =f\left(  \underbrace{\kk\left[
x_{1},x_{2},x_{3},\ldots\right]  }_{=\left\langle \xx_{\lambda}%
\ \mid\ \lambda\in\Par\right\rangle _{\kk}}\right)  \\
&  =f\left(  \left\langle \xx_{\lambda}\ \mid\ \lambda\in
\Par\right\rangle _{\kk}\right)  =\left\langle
\underbrace{f\left(  \xx_{\lambda}\right)  }_{\substack{=v_{\lambda
}\\\text{(by (\ref{sol.alggen-via-span.fxl=vl}))}}}\ \mid\ \lambda
\in\Par\right\rangle _{\kk}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }f\text{ is }\kk\text{-linear}\right)  \\
&  =\left\langle v_{\lambda}\ \mid\ \lambda\in\Par\right\rangle
_{\kk}=\left(  \text{the }\kk\text{-submodule of }A\text{
spanned by the family }\left(  v_{\lambda}\right)  _{\lambda\in
\Par}\right)  .
\end{align*}
This solves Exercise \ref{exe.alggen-via-span}(a).

(b) We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the elements }v_{1},v_{2},v_{3},\ldots\text{ generate the
}\kk\text{-algebra }A\right)  \\
&  \Longleftrightarrow\ \left(  \text{the }\kk\text{-subalgebra of
}A\text{ generated by }v_{1},v_{2},v_{3},\ldots\text{ is }A\right)  \\
&  \Longleftrightarrow\ \left(  \text{the }\kk\text{-submodule of
}A\text{ spanned by the family }\left(  v_{\lambda}\right)  _{\lambda
\in\Par}\text{ is }A\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Exercise \ref{exe.alggen-via-span}(a) shows that}\\
\text{the }\kk\text{-subalgebra of }A\text{ generated by }v_{1}%
,v_{2},v_{3},\ldots\\
\text{is the }\kk\text{-submodule of }A\text{ spanned by the family
}\left(  v_{\lambda}\right)  _{\lambda\in\Par}%
\end{array}
\right)  \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  v_{\lambda}\right)
_{\lambda\in\Par}\text{ spans the }\kk\text{-module
}A\right)  .
\end{align*}
This solves Exercise \ref{exe.alggen-via-span}(b).

(c) We shall prove the ``$\Longrightarrow$'' and the ``$\Longleftarrow$''
directions of Exercise \ref{exe.alggen-via-span}(c) separately:

$\Longrightarrow:$ Assume that the elements $v_{1},v_{2},v_{3},\ldots$ are
algebraically independent over $\kk$. We must show that the family
$\left(  v_{\lambda}\right)  _{\lambda\in\Par}$ is $\kk%
$-linearly independent.

Let $\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk^{\Par}$ be a family of scalars such that $\left(
\text{all but finitely many }\lambda\in\Par\text{ satisfy
}a_{\lambda}=0\right)  $ and $\sum_{\lambda\in\Par}a_{\lambda
}v_{\lambda}=0$. We shall show that $\left(  a_{\lambda}\right)  _{\lambda
\in\Par}=\left(  0\right)  _{\lambda\in\Par}$.

Define a polynomial $P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $
by $P=\sum_{\lambda\in\Par}a_{\lambda}\xx_{\lambda}$.
Then,
\begin{align*}
f\left(  \underbrace{P}_{=\sum_{\lambda\in\Par}a_{\lambda
}\xx_{\lambda}}\right)   &  =f\left(  \sum_{\lambda\in
\Par}a_{\lambda}\xx_{\lambda}\right)  =\sum_{\lambda
\in\Par}a_{\lambda}\underbrace{f\left(  \xx_{\lambda
}\right)  }_{\substack{=v_{\lambda}\\\text{(by
(\ref{sol.alggen-via-span.fxl=vl}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
the map }f\text{ is }\kk\text{-linear}\right)  \\
&  =\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}=0.
\end{align*}
Comparing this with (\ref{sol.alggen-via-span.fP=}), we obtain $P\left(
v_{1},v_{2},v_{3},\ldots\right)  =0$. This entails $P=0$, because the elements
$v_{1},v_{2},v_{3},\ldots$ are algebraically independent. Comparing this with
$P=\sum_{\lambda\in\Par}a_{\lambda}\xx_{\lambda}$, we
obtain $\sum_{\lambda\in\Par}a_{\lambda}\xx_{\lambda}%
=0$. But since the family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ is $\kk$-linearly independent, this entails
that $\left(  a_{\lambda}\right)  _{\lambda\in\Par}=\left(
0\right)  _{\lambda\in\Par}$.

Forget that we fixed $\left(  a_{\lambda}\right)  _{\lambda\in
\Par}$. We thus have proved that if $\left(  a_{\lambda
}\right)  _{\lambda\in\Par}\in\kk^{\Par}$
is a family of scalars such that $\left(  \text{all but finitely many }%
\lambda\in\Par\text{ satisfy }a_{\lambda}=0\right)  $ and
$\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}=0$, then $\left(
a_{\lambda}\right)  _{\lambda\in\Par}=\left(  0\right)
_{\lambda\in\Par}$. In other words, the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par}$ is $\kk$-linearly
independent. This proves the ``$\Longrightarrow$'' direction of
Exercise \ref{exe.alggen-via-span}(c).

$\Longleftarrow:$ Assume that the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ is $\kk$-linearly independent. We
shall show that the elements $v_{1},v_{2},v_{3},\ldots$ are algebraically
independent over $\kk$.

Indeed, let $P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ be a
polynomial that satisfies $P\left(  v_{1},v_{2},v_{3},\ldots\right)  =0$. We
shall show that $P=0$.

Recall that the family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ spans the $\kk$-module $\kk\left[
x_{1},x_{2},x_{3},\ldots\right]  $. Hence, we can write the polynomial
$P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ as a $\kk%
$-linear combination of this family. In other words, there exists a family
$\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk^{\Par}$ of scalars such that $\left(  \text{all but
finitely many }\lambda\in\Par\text{ satisfy }a_{\lambda
}=0\right)  $ and $P=\sum_{\lambda\in\Par}a_{\lambda}%
\xx_{\lambda}$. Consider this $\left(  a_{\lambda}\right)  _{\lambda
\in\Par}$.

Applying the map $f$ to both sides of the equality $P=\sum_{\lambda
\in\Par}a_{\lambda}\xx_{\lambda}$, we obtain%
\begin{align*}
f\left(  P\right)
&  =f\left(  \sum_{\lambda\in\Par}a_{\lambda}
\xx_{\lambda}\right)
= \sum_{\lambda\in\Par}a_{\lambda}
\underbrace{f\left(  \xx_{\lambda}\right)  }%
_{\substack{=v_{\lambda}\\\text{(by (\ref{sol.alggen-via-span.fxl=vl}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}.
\end{align*}
But (\ref{sol.alggen-via-span.fP=}) yields%
\[
f\left(  P\right)  =P\left(  v_{1},v_{2},v_{3},\ldots\right)  =0.
\]
Comparing these two equalities, we obtain
$\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}=0$.
This entails that $\left(  a_{\lambda}\right)
_{\lambda\in\Par}=\left(  0\right)  _{\lambda\in
\Par}$ (since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ is $\kk$-linearly independent). In
other words, $a_{\lambda}=0$ for each $\lambda\in\Par$. Hence,
$P=\sum_{\lambda\in\Par}\underbrace{a_{\lambda}}_{=0}%
\xx_{\lambda}=\sum_{\lambda\in\Par}0\xx_{\lambda
}=0$.

Now, forget that we fixed $P$. We thus have showed that every polynomial
$P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ satisfying $P\left(
v_{1},v_{2},v_{3},\ldots\right)  =0$ must satisfy $P=0$. In other words, the
elements $v_{1},v_{2},v_{3},\ldots$ are algebraically independent over
$\kk$. This proves the ``$\Longleftarrow$'' direction of
Exercise \ref{exe.alggen-via-span}(c).

Thus, both ``$\Longrightarrow$'' and
``$\Longleftarrow$'' directions of Exercise
\ref{exe.alggen-via-span}(c) are solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.alggen-via-span}}

\textit{Solution to Exercise \ref{exe.alggen-via-span}.} Let us first consider
the polynomial ring $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ in
countably many indeterminates $x_{1},x_{2},x_{3},\ldots$.

Let $f:\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  \rightarrow A$ be
the $\kk$-algebra homomorphism that sends $x_{1},x_{2},x_{3},\ldots$ to
$v_{1},v_{2},v_{3},\ldots$, respectively. This is well-defined by the
universal property of the polynomial ring $\kk\left[  x_{1},x_{2}%
,x_{3},\ldots\right]  $ (since $A$ is commutative). The $\kk$-algebra
homomorphism $f$ is an instance of an evaluation homomorphism; it sends each
polynomial $P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ to
$P\left(  v_{1},v_{2},v_{3},\ldots\right)  \in A$. In other words,%
\begin{equation}
f\left(  P\right)  =P\left(  v_{1},v_{2},v_{3},\ldots\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }P\in\kk\left[  x_{1},x_{2}%
,x_{3},\ldots\right]  . \label{sol.alggen-via-span.fP=}%
\end{equation}
The map $f$ is a $\kk$-algebra homomorphism, and thus is $\kk$-linear.

For each partition $\lambda\in\Par$, we define a monomial
$\xx_{\lambda}\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $
by%
\[
\xx_{\lambda}=x_{\lambda_{1}}x_{\lambda_{2}}\cdots x_{\lambda
_{\ell\left(  \lambda\right)  }}.
\]
(This is well-defined, since $\lambda_{1},\lambda_{2},\ldots,\lambda
_{\ell\left(  \lambda\right)  }$ are positive integers whenever $\lambda
\in\Par$.)

If $\lambda$ is a partition and $i$ is a positive integer, then $m_{i}\left(
\lambda\right)  $ shall denote the multiplicity of $i$ in $\lambda$ (that is,
the number of parts of $\lambda$ equal to $i$).

Let \dfn{$\operatorname*{WC}$} denote the set of all weak compositions.

Every partition $\lambda$ can be uniquely written in the form $\lambda=\left(
1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for some weak composition $\left(
m_{1},m_{2},m_{3},\ldots\right)  $. Thus, the map
\begin{align*}
\mathbf{part}:\operatorname*{WC}  &  \rightarrow\Par,\\
\left(  m_{1},m_{2},m_{3},\ldots\right)   &  \mapsto\left(  1^{m_{1}}2^{m_{2}%
}3^{m_{3}}\cdots\right)
\end{align*}
is a bijection. The inverse of this bijection is the map%
\begin{align*}
\mathbf{mults}:\Par  &  \rightarrow\operatorname*{WC},\\
\lambda &  \mapsto\left(  m_{1}\left(  \lambda\right)  ,m_{2}\left(
\lambda\right)  ,m_{3}\left(  \lambda\right)  ,\ldots\right)  .
\end{align*}
Thus, this map $\mathbf{mults}$ is a bijection, too (since it is the inverse
of a bijection).

It is easy to see that
\begin{equation}
\xx_{\lambda}=\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)
}\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par
\label{sol.alggen-via-span.xl=xl}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.alggen-via-span.xl=xl}):} Let $\lambda
\in\Par$. Recall that $\lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell\left(  \lambda\right)  }$ are positive integers. In other
words, $\lambda_{p}$ is a positive integer for each $p\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $. The definition of
$\mathbf{mults}$ yields $\mathbf{mults}\left(  \lambda\right)  =\left(
m_{1}\left(  \lambda\right)  ,m_{2}\left(  \lambda\right)  ,m_{3}\left(
\lambda\right)  ,\ldots\right)  $. Thus, the definition of $\xx%
^{\mathbf{mults}\left(  \lambda\right)  }$ yields%
\begin{equation}
\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }=x_{1}^{m_{1}\left(
\lambda\right)  }x_{2}^{m_{2}\left(  \lambda\right)  }x_{3}^{m_{3}\left(
\lambda\right)  }\cdots. \label{sol.alggen-via-span.xl=xl.pf.xml=}%
\end{equation}
\par
The definition of $\xx_{\lambda}$ yields%
\begin{align}
\xx_{\lambda}  &  =x_{\lambda_{1}}x_{\lambda_{2}}\cdots x_{\lambda
_{\ell\left(  \lambda\right)  }}=\underbrace{\prod_{p=1}^{\ell\left(
\lambda\right)  }}_{=\prod_{p\in\left\{  1,2,\ldots,\ell\left(  \lambda
\right)  \right\}  }}x_{\lambda_{p}}=\prod_{p\in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  }x_{\lambda_{p}}\nonumber\\
&  =\prod_{i=1}^{\infty}\prod\limits_{\substack{p\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  ;\\\lambda_{p}=i}}x_{\lambda_{p}}
\label{sol.alggen-via-span.xl=xl.pf.1}%
\end{align}
(here, we have split the product according to the value of $\lambda_{p}$,
since $\lambda_{p}$ is a positive integer for each $p\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $).
\par
Now, fix a positive integer $i$. Thus, $i>0$, so that $i\neq0$. But we have
$\lambda_{p}=0$ for each integer $p>\ell\left(  \lambda\right)  $ (by the
definition of $\ell\left(  \lambda\right)  $). Hence, for each integer
$p>\ell\left(  \lambda\right)  $, we have $\lambda_{p}=0\neq i$ (since
$i\neq0$). Thus, we obtain a contradiction for any $p\in\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{p}=i$ and $p>\ell\left(
\lambda\right)  $ (since the previous sentence yields $\lambda_{p}\neq i$,
which contradicts $\lambda_{p}=i$). Hence, there exists no $p\in\left\{
1,2,3,\ldots\right\}  $ satisfying $\lambda_{p}=i$ and $p>\ell\left(
\lambda\right)  $.
\par
Recall that $m_{i}\left(  \lambda\right)  $ is the number of parts of
$\lambda$ equal to $i$ (by the definition of $m_{i}\left(  \lambda\right)  $).
In other words,
\begin{align*}
m_{i}\left(  \lambda\right)   &  =\left(  \text{the number of parts of
}\lambda\text{ equal to }i\right) \\
&  =\left(  \text{the number of }p\in\left\{  1,2,3,\ldots\right\}  \text{
satisfying }\lambda_{p}=i\right) \\
&  =\left(  \text{the number of }p\in\left\{  1,2,3,\ldots\right\}  \text{
satisfying }\lambda_{p}=i\text{ and }p\leq\ell\left(  \lambda\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of }p\in\left\{
1,2,3,\ldots\right\}  \text{ satisfying }\lambda_{p}=i\text{ and }%
p>\ell\left(  \lambda\right)  \right)  }_{\substack{=0\\\text{(since there
exists no }p\in\left\{  1,2,3,\ldots\right\}  \text{ satisfying }\lambda
_{p}=i\text{ and }p>\ell\left(  \lambda\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since any }p\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies either
}p\leq\ell\left(  \lambda\right)  \text{ or }p>\ell\left(  \lambda\right) \\
\text{(but not both at the same time)}%
\end{array}
\right) \\
&  =\left(  \text{the number of }p\in\left\{  1,2,3,\ldots\right\}  \text{
satisfying }\lambda_{p}=i\text{ and }p\leq\ell\left(  \lambda\right)  \right)
\\
&  =\left(  \text{the number of }p\in\left\{  1,2,3,\ldots\right\}  \text{
satisfying }p\leq\ell\left(  \lambda\right)  \text{ and }\lambda_{p}=i\right)
\\
&  =\left(  \text{the number of }p\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \text{ satisfying }\lambda_{p}=i\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the }p\in\left\{  1,2,3,\ldots
\right\}  \text{ satisfying }p\leq\ell\left(  \lambda\right)  \text{ are
precisely the }p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\right)  .
\end{align*}
Hence,%
\[
x_{i}^{m_{i}\left(  \lambda\right)  }=x_{i}^{\left(  \text{the number of }%
p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \text{
satisfying }\lambda_{p}=i\right)  }.
\]
Comparing this with%
\[
\prod\limits_{\substack{p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
;\\\lambda_{p}=i}}\underbrace{x_{\lambda_{p}}}_{\substack{=x_{i}\\\text{(since
}\lambda_{p}=i\text{)}}}=\prod\limits_{\substack{p\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  ;\\\lambda_{p}=i}}x_{i}=x_{i}^{\left(  \text{the
number of }p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\text{ satisfying }\lambda_{p}=i\right)  },
\]
we obtain%
\begin{equation}
\prod\limits_{\substack{p\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
;\\\lambda_{p}=i}}x_{\lambda_{p}}=x_{i}^{m_{i}\left(  \lambda\right)  }.
\label{sol.alggen-via-span.xl=xl.pf.2}%
\end{equation}
\par
Now, forget that we fixed $i$. We thus have proved the equality
(\ref{sol.alggen-via-span.xl=xl.pf.2}) for each positive integer $i$. Now,
(\ref{sol.alggen-via-span.xl=xl.pf.1}) becomes%
\[
\xx_{\lambda}=\prod_{i=1}^{\infty}\underbrace{\prod\limits_{\substack{p\in
\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{p}%
=i}}x_{\lambda_{p}}}_{\substack{=x_{i}^{m_{i}\left(  \lambda\right)
}\\\text{(by (\ref{sol.alggen-via-span.xl=xl.pf.2}))}}}=\prod_{i=1}^{\infty
}x_{i}^{m_{i}\left(  \lambda\right)  }=x_{1}^{m_{1}\left(  \lambda\right)
}x_{2}^{m_{2}\left(  \lambda\right)  }x_{3}^{m_{3}\left(  \lambda\right)
}\cdots.
\]
Comparing this with (\ref{sol.alggen-via-span.xl=xl.pf.xml=}), we obtain
$\xx_{\lambda}=\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }$.
This proves (\ref{sol.alggen-via-span.xl=xl}).}. In other words, $\left(
\xx_{\lambda}\right)  _{\lambda\in\Par}=\left(
\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }\right)  _{\lambda
\in\Par}$.

The monomials in $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ have
the form $\mathbf{x}^{\alpha}$ for $\alpha\in\operatorname*{WC}$. It is
well-known that these monomials form a basis of the $\kk$-module
$\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ (since any polynomial
ring has a basis consisting of the monomials). In other words, the family
$\left(  \mathbf{x}^{\alpha}\right)  _{\alpha\in\operatorname*{WC}}$ is a
basis of the $\kk$-module $\kk\left[  x_{1},x_{2},x_{3}%
,\ldots\right]  $. The family $\left(  \mathbf{x}^{\mathbf{mults}\left(
\lambda\right)  }\right)  _{\lambda\in\Par}$ is a reindexing of
this basis $\left(  \mathbf{x}^{\alpha}\right)  _{\alpha\in\operatorname*{WC}%
}$ (since $\mathbf{mults}:\Par\rightarrow\operatorname*{WC}$ is
a bijection), and thus must also be a basis of the $\kk$-module
$\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ (since a reindexing of a
basis of a $\kk$-module is always a basis itself). In view of $\left(
\xx_{\lambda}\right)  _{\lambda\in\Par}=\left(
\mathbf{x}^{\mathbf{mults}\left(  \lambda\right)  }\right)  _{\lambda
\in\Par}$, we can rewrite this as follows: The family $\left(
\xx_{\lambda}\right)  _{\lambda\in\Par}$ is a basis of
the $\kk$-module $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $.
Hence, this family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ is $\kk$-linearly independent and spans the
$\kk$-module $\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $.

Let us now step to the solution of Exercise \ref{exe.alggen-via-span}:

(a) Let us use the following notation: If $U$ is any $\kk$-module, and
if $\left(  u_{i}\right)  _{i\in I}$ is any family of elements of $U$, then
$\left\langle u_{i}\ \mid\ i\in I\right\rangle _{\kk}$ shall denote the
$\kk$-submodule of $U$ spanned by this family $\left(  u_{i}\right)
_{i\in I}$.

We know that the family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ spans the $\kk$-module $\kk\left[
x_{1},x_{2},x_{3},\ldots\right]  $. In other words,
\[
\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  =\left\langle
\xx_{\lambda}\ \mid\ \lambda\in\Par\right\rangle
_{\kk}.
\]


We have%
\begin{equation}
f\left(  \xx_{\lambda}\right)  =v_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par
\label{sol.alggen-via-span.fxl=vl}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.alggen-via-span.fxl=vl}):} Let
$\lambda\in\Par$. Then, the definition of $\xx_{\lambda
}$ yields $\xx_{\lambda}=x_{\lambda_{1}}x_{\lambda_{2}}\cdots
x_{\lambda_{\ell\left(  \lambda\right)  }}$. Applying the map $f$ to both
sides of this equality, we obtain%
\begin{align*}
f\left(  \xx_{\lambda}\right)   &  =f\left(  x_{\lambda_{1}}%
x_{\lambda_{2}}\cdots x_{\lambda_{\ell\left(  \lambda\right)  }}\right)
=f\left(  x_{\lambda_{1}}\right)  f\left(  x_{\lambda_{2}}\right)  \cdots
f\left(  x_{\lambda_{\ell\left(  \lambda\right)  }}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }\kk%
\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{\ell\left(  \lambda\right)  }\underbrace{f\left(
x_{\lambda_{i}}\right)  }_{\substack{=v_{\lambda_{i}}\\\text{(by the
definition of }f\text{)}}}=\prod_{i=1}^{\ell\left(  \lambda\right)
}v_{\lambda_{i}}=v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(
\lambda\right)  }}.
\end{align*}
Comparing this with%
\[
v_{\lambda}=v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(
\lambda\right)  }}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}v_{\lambda}\right)  ,
\]
we obtain $f\left(  \xx_{\lambda}\right)  =v_{\lambda}$. This proves
(\ref{sol.alggen-via-span.fxl=vl}).}.

But $A$ is commutative; hence,%
\begin{align*}
&  \left(  \text{the }\kk\text{-subalgebra of }A\text{ generated by
}v_{1},v_{2},v_{3},\ldots\right) \\
&  =\left\{  \underbrace{P\left(  v_{1},v_{2},v_{3},\ldots\right)
}_{\substack{=f\left(  P\right)  \\\text{(by (\ref{sol.alggen-via-span.fP=}%
))}}}\ \mid\ P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  \right\}
\\
&  =\left\{  f\left(  P\right)  \ \mid\ P\in\kk\left[  x_{1}%
,x_{2},x_{3},\ldots\right]  \right\}  =f\left(  \underbrace{\kk\left[
x_{1},x_{2},x_{3},\ldots\right]  }_{=\left\langle \xx_{\lambda}%
\ \mid\ \lambda\in\Par\right\rangle _{\kk}}\right) \\
&  =f\left(  \left\langle \xx_{\lambda}\ \mid\ \lambda\in
\Par\right\rangle _{\kk}\right)  =\left\langle
\underbrace{f\left(  \xx_{\lambda}\right)  }_{\substack{=v_{\lambda
}\\\text{(by (\ref{sol.alggen-via-span.fxl=vl}))}}}\ \mid\ \lambda
\in\Par\right\rangle _{\kk}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }f\text{ is }\kk\text{-linear}\right) \\
&  =\left\langle v_{\lambda}\ \mid\ \lambda\in\Par\right\rangle
_{\kk}=\left(  \text{the }\kk\text{-submodule of }A\text{
spanned by the family }\left(  v_{\lambda}\right)  _{\lambda\in
\Par}\right)  .
\end{align*}
In other words, the $\kk$-subalgebra of $A$ generated by $v_{1}%
,v_{2},v_{3},\ldots$ is the $\kk$-submodule of $A$ spanned by the
family $\left(  v_{\lambda}\right)  _{\lambda\in\Par}$. This
solves Exercise \ref{exe.alggen-via-span}(a).

(b) We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the elements }v_{1},v_{2},v_{3},\ldots\text{ generate the
}\kk\text{-algebra }A\right) \\
&  \Longleftrightarrow\ \left(  \text{the }\kk\text{-subalgebra of
}A\text{ generated by }v_{1},v_{2},v_{3},\ldots\text{ is }A\right) \\
&  \Longleftrightarrow\ \left(  \text{the }\kk\text{-submodule of
}A\text{ spanned by the family }\left(  v_{\lambda}\right)  _{\lambda
\in\Par}\text{ is }A\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Exercise \ref{exe.alggen-via-span}(a) shows that}\\
\text{the }\kk\text{-subalgebra of }A\text{ generated by }v_{1}%
,v_{2},v_{3},\ldots\\
\text{is the }\kk\text{-submodule of }A\text{ spanned by the family
}\left(  v_{\lambda}\right)  _{\lambda\in\Par}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  v_{\lambda}\right)
_{\lambda\in\Par}\text{ spans the }\kk\text{-module
}A\right)  .
\end{align*}
In other words, the elements $v_{1},v_{2},v_{3},\ldots$ generate the
$\kk$-algebra $A$ if and only if the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par}$ spans the $\kk$-module $A$.
This solves Exercise \ref{exe.alggen-via-span}(b).

(c) We shall prove the ``$\Longrightarrow$'' and the
``$\Longleftarrow$'' directions of
Exercise \ref{exe.alggen-via-span}(c) separately:

$\Longrightarrow:$ Assume that the elements $v_{1},v_{2},v_{3},\ldots$ are
algebraically independent over $\kk$. We must show that the family
$\left(  v_{\lambda}\right)  _{\lambda\in\Par}$ is $\kk%
$-linearly independent.

Let $\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk^{\Par}$ be a family of scalars such that $\left(
\text{all but finitely many }\lambda\in\Par\text{ satisfy
}a_{\lambda}=0\right)  $ and $\sum_{\lambda\in\Par}a_{\lambda
}v_{\lambda}=0$. We shall show that $\left(  a_{\lambda}\right)  _{\lambda
\in\Par}=\left(  0\right)  _{\lambda\in\Par}$.

Define a polynomial $P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $
by $P=\sum_{\lambda\in\Par}a_{\lambda}\xx_{\lambda}$.
Then,
\begin{align*}
f\left(  \underbrace{P}_{=\sum_{\lambda\in\Par}a_{\lambda
}\xx_{\lambda}}\right)   &  =f\left(  \sum_{\lambda\in
\Par}a_{\lambda}\xx_{\lambda}\right)  =\sum_{\lambda
\in\Par}a_{\lambda}\underbrace{f\left(  \xx_{\lambda
}\right)  }_{\substack{=v_{\lambda}\\\text{(by
(\ref{sol.alggen-via-span.fxl=vl}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
the map }f\text{ is }\kk\text{-linear}\right) \\
&  =\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}=0.
\end{align*}
Comparing this with (\ref{sol.alggen-via-span.fP=}), we obtain $P\left(
v_{1},v_{2},v_{3},\ldots\right)  =0$.

Since the elements $v_{1},v_{2},v_{3},\ldots$ are algebraically independent,
we have $Q=0$ for any polynomial $Q\in\kk\left[  x_{1},x_{2}%
,x_{3},\ldots\right]  $ that satisfies $Q\left(  v_{1},v_{2},v_{3}%
,\ldots\right)  =0$. Applying this to $Q=P$, we obtain $P=0$ (since $P\left(
v_{1},v_{2},v_{3},\ldots\right)  =0$). Comparing this with $P=\sum_{\lambda
\in\Par}a_{\lambda}\xx_{\lambda}$, we obtain
$\sum_{\lambda\in\Par}a_{\lambda}\xx_{\lambda}=0$. But
since the family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ is $\kk$-linearly independent, this entails
that $\left(  a_{\lambda}\right)  _{\lambda\in\Par}=\left(
0\right)  _{\lambda\in\Par}$.

Forget that we fixed $\left(  a_{\lambda}\right)  _{\lambda\in
\Par}$. We thus have proved that if $\left(  a_{\lambda
}\right)  _{\lambda\in\Par}\in\kk^{\Par}$
is a family of scalars such that $\left(  \text{all but finitely many }%
\lambda\in\Par\text{ satisfy }a_{\lambda}=0\right)  $ and
$\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}=0$, then $\left(
a_{\lambda}\right)  _{\lambda\in\Par}=\left(  0\right)
_{\lambda\in\Par}$. In other words, the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par}$ is $\kk$-linearly
independent. This proves the ``$\Longrightarrow$'' direction of
Exercise \ref{exe.alggen-via-span}(c).

$\Longleftarrow:$ Assume that the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ is $\kk$-linearly independent. We
shall show that the elements $v_{1},v_{2},v_{3},\ldots$ are algebraically
independent over $\kk$.

Indeed, let $P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ be a
polynomial that satisfies $P\left(  v_{1},v_{2},v_{3},\ldots\right)  =0$. We
shall show that $P=0$.

Recall that the family $\left(  \xx_{\lambda}\right)  _{\lambda
\in\Par}$ spans the $\kk$-module $\kk\left[
x_{1},x_{2},x_{3},\ldots\right]  $. Hence, we can write the polynomial
$P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ as a $\kk%
$-linear combination of this family. In other words, there exists a family
$\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk^{\Par}$ of scalars such that $\left(  \text{all but
finitely many }\lambda\in\Par\text{ satisfy }a_{\lambda
}=0\right)  $ and $P=\sum_{\lambda\in\Par}a_{\lambda}%
\xx_{\lambda}$. Consider this $\left(  a_{\lambda}\right)  _{\lambda
\in\Par}$.

Applying the map $f$ to both sides of the equality $P=\sum_{\lambda
\in\Par}a_{\lambda}\xx_{\lambda}$, we obtain%
\begin{align*}
f\left(  P\right)
&  =f\left(  \sum_{\lambda\in\Par}a_{\lambda}
\xx_{\lambda}\right)
= \sum_{\lambda\in\Par}a_{\lambda}
\underbrace{f\left(  \xx_{\lambda}\right)  }%
_{\substack{=v_{\lambda}\\\text{(by (\ref{sol.alggen-via-span.fxl=vl}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}.
\end{align*}
But (\ref{sol.alggen-via-span.fP=}) yields%
\[
f\left(  P\right)  =P\left(  v_{1},v_{2},v_{3},\ldots\right)  =0.
\]
Comparing these two equalities, we obtain
$\sum_{\lambda\in\Par}a_{\lambda}v_{\lambda}=0$.
This entails that $\left(  a_{\lambda}\right)
_{\lambda\in\Par}=\left(  0\right)  _{\lambda\in
\Par}$ (since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ is $\kk$-linearly independent). In
other words, $a_{\lambda}=0$ for each $\lambda\in\Par$. Hence,
$P=\sum_{\lambda\in\Par}\underbrace{a_{\lambda}}_{=0}%
\xx_{\lambda}=\sum_{\lambda\in\Par}0\xx_{\lambda
}=0$.

Now, forget that we fixed $P$. We thus have showed that every polynomial
$P\in\kk\left[  x_{1},x_{2},x_{3},\ldots\right]  $ satisfying $P\left(
v_{1},v_{2},v_{3},\ldots\right)  =0$ must satisfy $P=0$. In other words, the
elements $v_{1},v_{2},v_{3},\ldots$ are algebraically independent over
$\kk$. This proves the ``$\Longleftarrow$'' direction of
Exercise \ref{exe.alggen-via-span}(c).

Thus, both ``$\Longrightarrow$'' and
``$\Longleftarrow$'' directions of Exercise
\ref{exe.alggen-via-span}(c) are solved.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.H(t)-and-E(t)}}

\textit{Solution to Exercise \ref{exe.H(t)-and-E(t)}.} Before we start solving
the exercise, let us recall our notion of a ``monomial''.
For us, a \emph{pure monomial} is (formally
speaking) just a symbol $\xx^{\alpha}$ indexed by a weak composition
$\alpha$. (Thus, pure monomials are combinatorial objects; in particular, they
have nothing to do with the ground ring $\kk$\ \ \ \ \footnote{In
particular, the pure monomials $\xx^{\alpha}$ for different weak
compositions $\alpha$ are distinct, although their images in the
polynomial ring $\kk\left[  \xx\right]  $ are equal when
$\kk=0$.}, and do not ``come with coefficients''.)

We let $\operatorname*{Mon}$ be the set of all pure monomials. We define the
product of two pure monomials $\xx^{\alpha}$ and $\xx^{\beta}$
by $\xx^{\alpha}\xx^{\beta}=\xx^{\alpha+\beta}$, where
$\alpha+\beta$ denotes the entrywise sum of the two weak compositions $\alpha$
and $\beta$ (in other words, if $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},\ldots\right)  $ and $\beta=\left(  \beta_{1},\beta_{2},\beta
_{3},\ldots\right)  $, then $\alpha+\beta=\left(  \alpha_{1}+\beta_{1}%
,\alpha_{2}+\beta_{2},\alpha_{3}+\beta_{3},\ldots\right)  $). This notion of
product makes the set $\operatorname*{Mon}$ of all pure monomials into a
monoid. The neutral element of this monoid $\operatorname*{Mon}$ is the pure
monomial $\xx^{\left(  0,0,0,\ldots\right)  }$, which we denote by $1$.
Whenever $i$ is a positive integer, we write $x_{i}$ for the pure monomial
$\xx^{t_{i}}$, where $t_{i}$ is the weak composition $\left(
\delta_{1,i},\delta_{2,i},\delta_{3,i},\ldots\right)  =\left(
\underbrace{0,0,\ldots,0}_{i-1\text{ zeroes}},1,0,0,0,\ldots\right)  $. This
notation allows us to rewrite any pure monomial $\xx^{\alpha}$ in the
form $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}}\cdots
=\prod_{i\geq1}x_{i}^{\alpha_{i}}$, where $\alpha$ is written in the form
$\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $.

So far we have only discussed pure monomials, but not the actual monomials we
encounter in polynomial rings or rings of power series. In practice, however,
most authors (including us) identify pure monomials with actual monomials. Let
us now introduce this identification.

For each weak composition $\alpha$, we identify the pure monomial
$\xx^{\alpha}\in\operatorname*{Mon}$ with the monomial $\xx^{\alpha}$
in the polynomial ring $\kk\left[  \xx\right]  $.
This identification is mostly harmless, because the map
\begin{align*}
\operatorname*{Mon}  &  \rightarrow\kk\left[  \xx\right]  ,\\
\xx^{\alpha}  &  \mapsto\xx^{\alpha}
\end{align*}
is a monoid homomorphism (from $\operatorname*{Mon}$ to the multiplicative
monoid of $\kk\left[  \xx\right]  $).\ \ \ \ \footnote{Also,
this map is injective unless $\kk=0$. However, we do not need the
injectivity of this map, because we will not derive any equalities between
pure monomials from equalities between monomials in $\kk\left[
\xx\right]  $.} Using this injectivity, we thus can consider every
monomial $\xx^{\alpha}\in\operatorname*{Mon}$ as an element of
$\kk\left[  \xx\right]  $, and thus also as an element of
$\kk\left[  \left[  \xx\right]  \right]  $. Whenever we write
sums of monomials (such as $\xx^{\alpha}+\xx^{\beta}$ or
$\sum_{\mathfrak{m}\in\operatorname*{Mon}}\mathfrak{m}$), we always mean these
sums to be computed in $\kk\left[  \xx\right]  $ or
$\kk\left[  \left[  \xx\right]  \right]  $.

For any pure monomial $\xx^{\alpha}$, we let $\deg\left(
\xx^{\alpha}\right)  $ denote the nonnegative integer $\alpha
_{1}+\alpha_{2}+\alpha_{3}+\cdots$. This integer is called the \emph{degree}
of the monomial $\xx^{\alpha}$.

Now, we notice that every pure monomial $\mathfrak{m}\in\operatorname*{Mon}$
can be uniquely represented in the form $\xx^{\alpha}$ for a weak
composition $\alpha$. Hence, we can substitute $\xx^{\alpha}$ for
$\mathfrak{m}$ in the sum $\sum_{\mathfrak{m}\in\operatorname*{Mon}%
}\mathfrak{m}t^{\deg\mathfrak{m}}$. We thus obtain%
\begin{align*}
\sum_{\mathfrak{m}\in\operatorname*{Mon}}\mathfrak{m}t^{\deg\mathfrak{m}}  &
=\sum\limits_{\substack{\alpha\text{ is a weak}\\\text{composition}}}
\xx^{\alpha}t^{\deg\left(  \xx^{\alpha}\right)  }=\sum\limits_{\substack{\left(
\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  \text{ is a weak}%
\\\text{composition}}}\underbrace{\xx^{\left(  \alpha_{1},\alpha
_{2},\alpha_{3},\ldots\right)  }}_{=\prod_{i\geq1}x_{i}^{\alpha_{i}}%
}\underbrace{t^{\deg\left(  \xx^{\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},\ldots\right)  }\right)  }}_{\substack{=t^{\alpha_{1}+\alpha
_{2}+\alpha_{3}+\cdots}\\\text{(since }\deg\left(  \xx^{\left(
\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  }\right)  =\alpha_{1}%
+\alpha_{2}+\alpha_{3}+\cdots\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\alpha\text{ as }\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\right) \\
&  =\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\text{ is a weak}\\\text{composition}}}
\left( \prod_{i\geq1}x_{i}^{\alpha_{i}} \right)
\underbrace{t^{\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots}}_{=t^{\sum_{i\geq
1}\alpha_{i}}=\prod_{i\geq1}t^{\alpha_{i}}}=\sum\limits_{\substack{\left(  \alpha
_{1},\alpha_{2},\alpha_{3},\ldots\right)  \text{ is a weak}%
\\\text{composition}}}\underbrace{\left(  \prod_{i\geq1}x_{i}^{\alpha_{i}%
}\right)  \left(  \prod_{i\geq1}t^{\alpha_{i}}\right)  }_{=\prod_{i\geq
1}\left(  x_{i}^{\alpha_{i}}t^{\alpha_{i}}\right)  }\\
&  =\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\text{ is a weak}\\\text{composition}}}\prod_{i\geq1}\underbrace{\left(
x_{i}^{\alpha_{i}}t^{\alpha_{i}}\right)  }_{=\left(  x_{i}t\right)
^{\alpha_{i}}}=\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  \text{ is a weak}\\\text{composition}}}\prod_{i\geq1}\left(
x_{i}t\right)  ^{\alpha_{i}}\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},k_{3},\ldots\right)  \text{ is a
weak}\\\text{composition}}}\prod_{i\geq1}\left(  x_{i}t\right)  ^{k_{i}}%
\end{align*}
(here, we renamed the summation index $\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},\ldots\right)  $ as $\left(  k_{1},k_{2},k_{3},\ldots\right)  $).
Compared with%
\begin{align*}
\underbrace{\prod_{i=1}^{\infty}}_{=\prod_{i\geq1}}\underbrace{\left(
1-x_{i}t\right)  ^{-1}}_{\substack{=\sum_{k\in\NN}\left(
x_{i}t\right)  ^{k}\\\text{(by the formula for the}\\\text{geometric series)}%
}}  &  =\prod_{i\geq1}\sum_{k\in\NN}\left(  x_{i}t\right)  ^{k}%
=\sum\limits_{\substack{\left(  k_{1},k_{2},k_{3},\ldots\right)  \text{ is a
weak}\\\text{composition}}}\prod_{i\geq1}\left(  x_{i}t\right)  ^{k_{i}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right)  ,
\end{align*}
this yields%
\begin{equation}
\sum_{\mathfrak{m}\in\operatorname*{Mon}}\mathfrak{m}t^{\deg\mathfrak{m}%
}=\prod_{i=1}^{\infty}\left(  1-x_{i}t\right)  ^{-1}.
\label{sol.H(t)-and-E(t).1}
\end{equation}


On the other hand, every pure monomial $\mathfrak{m}\in\operatorname*{Mon}$
can be uniquely represented in the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$
for some $n\in \NN$ and some weakly increasing $n$-tuple $\left(
i_{1},i_{2},\ldots,i_{n}\right)  $ of positive integers.\footnote{Indeed, this
$n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ can be found by writing
$\mathfrak{m}$ as a product of $x_{j}$'s, and sorting those $x_{j}$'s in
weakly increasing order.} Hence, we can substitute $x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}$ for $\mathfrak{m}$ in the sum $\sum_{\mathfrak{m}\in
\operatorname*{Mon}}\mathfrak{m}t^{\deg\mathfrak{m}}$. We thus obtain%
\begin{align*}
\sum_{\mathfrak{m}\in\operatorname*{Mon}}\mathfrak{m}t^{\deg\mathfrak{m}}  &
=\sum_{n\in\NN}\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  \text{ is a}\\\text{weakly increasing}\\n\text{-tuple of
positive integers}}}}_{=\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}}\left(
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right)  \underbrace{t^{\deg\left(
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right)  }}_{\substack{=t^n \\\text{(since
}\deg\left(  x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right)  =n\text{)}}}\\
&  =\sum_{n\in\NN}\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}\left(
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right)  t^n =\sum_{n\in\NN %
}\underbrace{\left(  \sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}\right)  }_{\substack{=h_{n}\\\text{(since
(\ref{def.hn}) yields}\\h_{n}=\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}%
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\text{)}}}t^n \\
&  =\sum_{n\in\NN}\underbrace{h_{n}}_{=h_{n}\left(  \xx\right)
}t^n =\sum_{n\geq0}h_{n}\left(  \xx\right)  t^n =\underbrace{h_{0}%
\left(  \xx\right)  }_{=h_{0}=1}+h_{1}\left(  \xx\right)
t+h_{2}\left(  \xx\right)  t^{2}+\cdots\\
&  =1+h_{1}\left(  \xx\right)  t+h_{2}\left(  \xx\right)
t^{2}+\cdots.
\end{align*}
Compared with (\ref{sol.H(t)-and-E(t).1}), this yields%
\[
\prod_{i=1}^{\infty}\left(  1-x_{i}t\right)  ^{-1}
= 1+h_{1}\left( \xx\right)  t+h_{2}\left( \xx\right)  t^{2}+\cdots=\sum_{n\geq0}%
h_{n}\left( \xx \right)  t^n .
\]
Thus, the first of the two identities that we need to prove is proven.

Next, let us define the notion of a \emph{squarefree monomial}. Indeed, let
us say that a pure monomial $\xx^{\alpha}\in\operatorname*{Mon}$ is
\emph{squarefree} if and only if every entry of the weak composition
$\alpha$ belongs to $\left\{  0,1\right\}  $. Thus, of course, every
squarefree monomial $\mathfrak{m}\in\operatorname*{Mon}$ can be uniquely
represented in the form $\xx^{\alpha}$ for a weak composition $\alpha$
whose every entry belongs to $\left\{  0,1\right\}  $. Hence, we can
substitute $\xx^{\alpha}$ for $\mathfrak{m}$ in the sum $\sum\limits
_{\substack{\mathfrak{m}\in\operatorname*{Mon};\\\mathfrak{m}\text{ is
squarefree}}}\mathfrak{m}t^{\deg\mathfrak{m}}$. We thus obtain%
\begin{align*}
&  \sum\limits_{\substack{\mathfrak{m}\in\operatorname*{Mon};\\\mathfrak{m}\text{ is
squarefree}}}\mathfrak{m}t^{\deg\mathfrak{m}}\\
&  =\underbrace{\sum\limits_{\substack{\alpha\text{ is a weak}\\\text{composition;}%
\\\text{every entry of }\alpha\\\text{belongs to }\left\{  0,1\right\}  }%
}}_{\substack{=\sum\limits_{\substack{\alpha\in\left\{  0,1\right\}  ^{\infty}\text{
is a weak}\\\text{composition}}}\\\text{(because if }\alpha\text{ is a
weak}\\\text{composition, then the statement}\\\left(  \text{every entry of
}\alpha\text{ belongs to }\left\{  0,1\right\}  \right)  \\\text{is equivalent
to }\left(  \alpha\in\left\{  0,1\right\}  ^{\infty}\right)  \text{)}%
}}\xx^{\alpha}t^{\deg\left(  \xx^{\alpha}\right)  }%
=\sum\limits_{\substack{\alpha\in\left\{  0,1\right\}  ^{\infty}\text{ is a
weak}\\\text{composition}}}\xx^{\alpha}t^{\deg\left(  \xx^{\alpha}
\right)  }\\
&  =\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\in\left\{  0,1\right\}  ^{\infty}\text{ is a}\\\text{weak composition}%
}}\underbrace{\xx^{\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  }}_{=\prod_{i\geq1}x_{i}^{\alpha_{i}}}\underbrace{t^{\deg
\left(  \xx^{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
}\right)  }}_{\substack{=t^{\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots
}\\\text{(since }\deg\left(  \xx^{\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},\ldots\right)  }\right)  =\alpha_{1}+\alpha_{2}+\alpha_{3}%
+\cdots\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\alpha\text{ as }\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\right)  \\
&  =\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\in\left\{  0,1\right\}  ^{\infty}\text{ is a}\\\text{weak composition}}%
}
\left( \prod_{i\geq1}x_{i}^{\alpha_{i}} \right)
\underbrace{t^{\alpha_{1}+\alpha_{2}%
+\alpha_{3}+\cdots}}_{=t^{\sum_{i\geq1}\alpha_{i}}=\prod_{i\geq1}t^{\alpha
_{i}}}=\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\in\left\{  0,1\right\}  ^{\infty}\text{ is a}\\\text{weak composition}%
}}\underbrace{\left(  \prod_{i\geq1}x_{i}^{\alpha_{i}}\right)  \left(
\prod_{i\geq1}t^{\alpha_{i}}\right)  }_{=\prod_{i\geq1}\left(  x_{i}%
^{\alpha_{i}}t^{\alpha_{i}}\right)  }\\
&  =\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\in\left\{  0,1\right\}  ^{\infty}\text{ is a}\\\text{weak composition}}%
}\prod_{i\geq1}\underbrace{\left(  x_{i}^{\alpha_{i}}t^{\alpha_{i}}\right)
}_{=\left(  x_{i}t\right)  ^{\alpha_{i}}}=\sum\limits_{\substack{\left(  \alpha
_{1},\alpha_{2},\alpha_{3},\ldots\right)  \in\left\{  0,1\right\}  ^{\infty
}\text{ is a}\\\text{weak composition}}}\prod_{i\geq1}\left(  x_{i}t\right)
^{\alpha_{i}}\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},k_{3},\ldots\right)  \in\left\{
0,1\right\}  ^{\infty}\text{ is a}\\\text{weak composition}}}\prod_{i\geq
1}\left(  x_{i}t\right)  ^{k_{i}}%
\end{align*}
(here, we renamed the summation index $\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},\ldots\right)  $ as $\left(  k_{1},k_{2},k_{3},\ldots\right)  $).
Compared with%
\begin{align*}
\underbrace{\prod_{i=1}^{\infty}}_{=\prod_{i\geq1}}\underbrace{\left(
1+x_{i}t\right)  }_{\substack{=\sum_{k\in\left\{  0,1\right\}  }\left(
x_{i}t\right)  ^{k}\\\text{(since }\sum_{k\in\left\{  0,1\right\}  }\left(
x_{i}t\right)  ^{k}\\=\left(  x_{i}t\right)  ^{0}+\left(  x_{i}t\right)
^{1}=1+x_{i}t\\\text{(since }\left(  x_{i}t\right)  ^{0}=1\text{ and }\left(
x_{i}t\right)  ^{1}=x_{i}t\text{))}}} &  =\prod_{i\geq1}\sum_{k\in\left\{
0,1\right\}  }\left(  x_{i}t\right)  ^{k}=\sum\limits_{\substack{\left(  k_{1}%
,k_{2},k_{3},\ldots\right)  \in\left\{  0,1\right\}  ^{\infty}
\text{ is a}\\
\text{weak composition}}}\prod_{i\geq1}\left(  x_{i}t\right)  ^{k_{i}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right)  ,
\end{align*}
this yields%
\begin{equation}
\sum\limits_{\substack{\mathfrak{m}\in\operatorname*{Mon};\\\mathfrak{m}\text{ is
squarefree}}}\mathfrak{m}t^{\deg\mathfrak{m}}=\prod_{i=1}^{\infty}\left(
1+x_{i}t\right)  .
\label{sol.H(t)-and-E(t).5}
\end{equation}


On the other hand, every squarefree monomial $\mathfrak{m}\in
\operatorname*{Mon}$ can be uniquely represented in the form $x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}$ for some $n\in \NN$ and some strictly
increasing $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ of positive
integers.\footnote{\textit{Proof.} This follows from the following
observations:
\par
\begin{itemize}
\item If $\mathfrak{m}\in\operatorname*{Mon}$ is a squarefree monomial, then
$\mathfrak{m}$ can be represented in the form $x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}$ for some $n\in \NN$ and some strictly increasing $n$-tuple
$\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ of positive integers. (Indeed, we
know already that every pure monomial $\mathfrak{m}\in\operatorname*{Mon}$ can
be uniquely represented in the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ for
some $n\in \NN$ and some weakly increasing $n$-tuple $\left(
i_{1},i_{2},\ldots,i_{n}\right)  $ of positive integers. If $\mathfrak{m}$ is
squarefree, then this $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $
must consist of $n$ \textbf{distinct} integers (because otherwise,
the product $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ would fail to be squarefree,
which would contradict the fact that $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}%
}=\mathfrak{m}$ is squarefree), and therefore is strictly increasing (because
it is weakly increasing). Hence, every squarefree monomial $\mathfrak{m}$ can
be be represented in the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ for some
$n\in \NN$ and some strictly increasing $n$-tuple $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  $ of positive integers.)
\par
\item If $\mathfrak{m}\in\operatorname*{Mon}$ is a squarefree monomial, then
the representation of $\mathfrak{m}$ in the form $x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}$ for some $n\in \NN$ and some strictly increasing $n$-tuple
$\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ of positive integers is unique.
(Indeed, even if we only require $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $
to be weakly increasing rather than strictly increasing, then the
representation is unique (because we know already that every pure monomial
$\mathfrak{m}\in\operatorname*{Mon}$ can be uniquely represented in the form
$x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ for some $n\in \NN$ and some weakly
increasing $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ of positive
integers).)
\par
\item Every product of the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ for some
$n\in \NN$ and some strictly increasing $n$-tuple $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  $ of positive integers is a squarefree monomial.
(This is obvious, because the elements $i_{1},i_{2},\ldots,i_{n}$ of a
strictly increasing $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ are
distinct.)
\end{itemize}
} Hence, we can substitute $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ for
$\mathfrak{m}$ in the sum $\sum\limits_{\substack{\mathfrak{m}\in\operatorname*{Mon}%
;\\\mathfrak{m}\text{ is squarefree}}}\mathfrak{m}t^{\deg\mathfrak{m}}$. We
thus obtain%
\begin{align*}
\sum\limits_{\substack{\mathfrak{m}\in\operatorname*{Mon};\\\mathfrak{m}\text{ is
squarefree}}}\mathfrak{m}t^{\deg\mathfrak{m}} &  =\sum_{n\in\NN %
}\underbrace{\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \text{
is a}\\\text{strictly increasing}\\n\text{-tuple of positive integers}}%
}}_{=\sum_{i_{1}<i_{2}<\cdots<i_{n}}}\left(  x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\right)  \underbrace{t^{\deg\left(  x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\right)  }}_{\substack{=t^n \\\text{(since }\deg\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}\right)  =n\text{)}}}\\
&  =\sum_{n\in\NN}\sum_{i_{1}<i_{2}<\cdots<i_{n}}\left(  x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}\right)  t^n =\sum_{n\in\NN}%
\underbrace{\left(  \sum_{i_{1}<i_{2}<\cdots<i_{n}}x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\right)  }_{\substack{=e_{n}\\\text{(since (\ref{def.en})
yields}\\e_{n}=\sum_{i_{1}<i_{2}<\cdots<i_{n}}x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\text{)}}}t^n \\
&  =\sum_{n\in\NN}\underbrace{e_{n}}_{=e_{n}\left(  \xx\right)
}t^n =\sum_{n\geq0}e_{n}\left(  \xx\right)  t^n =\underbrace{e_{0}%
\left(  \xx\right)  }_{=e_{0}=1}+e_{1}\left(  \xx\right)
t+e_{2}\left(  \xx\right)  t^{2}+\cdots\\
&  =1+e_{1}\left(  \xx\right)  t+e_{2}\left(  \xx\right)
t^{2}+\cdots.
\end{align*}
Compared with (\ref{sol.H(t)-and-E(t).5}), this yields%
\[
\prod_{i=1}^{\infty}\left(  1+x_{i}t\right)  =1+e_{1}\left(  \xx
\right)  t+e_{2}\left(  \xx\right)  t^{2}+\cdots=\sum_{n\geq0}%
e_{n}\left(  \xx\right)  t^n .
\]
This completes the solution to Exercise \ref{exe.H(t)-and-E(t)}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.schur.rotate180}}

\textit{Solution to Exercise~\ref{exe.schur.rotate180}.}
(a) This is straightforward. If $\lambda$ and $\mu$ are two partitions
such that $\mu \subseteq \lambda$, and if $\LLL$ is any total order on
the positive integers, then we say that an assignment $T$ of entries in
$\left\{1,2,3,\ldots\right\}$ to the cells of the Ferrers diagram of
$\lambda / \mu$ is an \dfn{$\LLL$-column-strict tableau} if it is
weakly $\LLL$-increasing left-to-right in rows, and strictly
$\LLL$-increasing top-to-bottom in columns. (This definition of
$\LLL$-column-strict tableaux clearly extends the definition given
in Remark~\ref{other-linear-orders-remark} for tableaux of shape
$\lambda$.) Now, the analogue of
Proposition~\ref{prop.other-linear-orders.slambda} is the following
statement:

\begin{proposition}
\label{prop.sol.schur.rotate180.analogue}
Let $\lambda$ and $\mu$ be two partitions such that
$\mu \subseteq \lambda$. Then, for any total order $\LLL$ on the
positive integers,
\[
s_{\lambda / \mu} = \sum_T \xx^{\cont(T)}
\]
as $T$ runs through all $\LLL$-column-strict tableaux of shape
$\lambda / \mu$.
\end{proposition}

The proof of Proposition~\ref{prop.sol.schur.rotate180.analogue}
is completely analogous to the proof of
Proposition~\ref{prop.other-linear-orders.slambda}.

(b) If $\alpha$ and $\beta$ are two partitions such that
$\beta \subseteq \alpha$, then let $Y\left(\alpha / \beta\right)$
denote the skew Ferrers diagram $\alpha / \beta$. This is a
finite set of cells in $\left\{ 1, 2, 3, ... \right\}^2$.

We know that the skew Ferrers diagram $\lambda' / \mu'$ can be
obtained from the skew Ferrers diagram $\lambda / \mu$ by a
$180^{\circ}$ rotation. In other words, there exists a
$180^{\circ}$ rotation $r$ such that
$r \left(Y\left(\lambda / \mu\right) \right)
= Y\left(\lambda' / \mu'\right)$. Consider this $r$.

Let $\LLL$ be the total order on the set of all positive integers
which is defined by $\cdots <_\LLL 3 <_\LLL 2 <_\LLL 1$ (in other
words, let $\LLL$ be the reverse of the usual total order on
the set of all positive integers). Recall the definition of
$\LLL$-column-strict tableaux that we gave in the solution of
part (a) of this exercise. According to this definition, an
$\LLL$-column-strict tableau of shape $\lambda' / \mu'$ is an
assignment of entries in $\left\{1,2,3,\ldots\right\}$ to the cells
of the Ferrers diagram of $\lambda' / \mu'$ which is
weakly $\LLL$-increasing left-to-right in rows, and strictly
$\LLL$-increasing top-to-bottom in columns. Since
``weakly $\LLL$-increasing'' is the same as ``weakly
decreasing'' (because $\LLL$ is the reverse of the usual total
order on the set of all positive integers), and since
``strictly $\LLL$-increasing'' is the same as ``strictly
decreasing'' (for the same reason), this rewrites as follows: An
$\LLL$-column-strict tableau of shape $\lambda' / \mu'$ is an
assignment of entries in $\left\{1,2,3,\ldots\right\}$ to the cells
of the Ferrers diagram of $\lambda' / \mu'$ which is
weakly decreasing left-to-right in rows, and strictly
decreasing top-to-bottom in columns. Hence, if $T$ is an
$\LLL$-column-strict tableau of shape $\lambda' / \mu'$, then
$T \circ r$ (this composition is well-defined\footnote{because
$T$ is an assignment of entries in $\left\{1,2,3,\ldots\right\}$
to the cells of the Ferrers diagram of $\lambda' / \mu'$, that is,
a map
$Y\left(\lambda' / \mu'\right) \to \left\{1,2,3,\ldots\right\}$,
whereas $r$ is a map sending $Y\left(\lambda / \mu\right)$ to
$Y\left(\lambda' / \mu'\right)$}) is an
assignment of entries in $\left\{1,2,3,\ldots\right\}$ to the cells
of the Ferrers diagram of $\lambda / \mu$ which is weakly
decreasing right-to-left in rows, and strictly decreasing
bottom-to-top in columns\footnote{because $r$ is a $180^{\circ}$
rotation, and thus interchanges ``left-to-right'' with
``right-to-left'' and interchanges ``top-to-bottom'' with
``bottom-to-top''}. In other words, if $T$ is an
$\LLL$-column-strict tableau of shape $\lambda' / \mu'$, then
$T \circ r$ is an assignment of entries in
$\left\{1,2,3,\ldots\right\}$ to the cells
of the Ferrers diagram of $\lambda / \mu$ which is weakly
increasing left-to-right in rows, and strictly increasing
top-to-bottom in columns. In other words, if $T$ is an
$\LLL$-column-strict tableau of shape $\lambda' / \mu'$, then
$T \circ r$ is a column-strict tableau (in the usual sense)
of shape $\lambda / \mu$. Hence, we have constructed a map
\[
\left\{\LLL\text{-column-strict tableaux of shape }\lambda' / \mu'\right\}
\to
\left\{\text{column-strict tableaux of shape }\lambda / \mu\right\},
\]
which sends every $T$ to $T \circ r$. This map is easily seen to
be a bijection. Therefore, we can substitute $T \circ r$ for $T$
in the sum $\sum\limits_{\substack{T\text{ is a column-strict}\\ 
\text{tableau of shape }\lambda / \mu}} \xx^{\cont(T)}$. We thus obtain
\begin{align}
\sum\limits_{\substack{T\text{ is a column-strict}\\ 
\text{tableau of shape }\lambda / \mu}} \xx^{\cont(T)}
&= \sum\limits_{\substack{T\text{ is an }\LLL\text{-column-strict}\\ 
\text{tableau of shape }\lambda' / \mu'}}
\underbrace{\xx^{\cont\left(T \circ r\right)}}_{
\substack{= \xx^{\cont(T)} \\ \text{(since the multiset of entries} \\
\text{of }T \circ r \text{ is the multiset} \\ \text{of entries
 of } T\text{)}}} \nonumber \\
&= \sum\limits_{\substack{T\text{ is an }\LLL\text{-column-strict}\\ 
\text{tableau of shape }\lambda' / \mu'}} \xx^{\cont(T)}.
\label{sol.schur.rotate180.schureq}
\end{align}

But Proposition~\ref{prop.sol.schur.rotate180.analogue}
(applied to $\lambda'$ and $\mu'$ instead of $\lambda$ and $\mu$)
yields
\[
s_{\lambda' / \mu'} = \sum_T \xx^{\cont(T)}
\]
as $T$ runs through all $\LLL$-column-strict tableaux of shape
$\lambda' / \mu'$. In other words,
\begin{equation}
\label{sol.schur.rotate180.schureq2}
s_{\lambda' / \mu'}
= \sum\limits_{\substack{T\text{ is an }\LLL\text{-column-strict}\\ 
\text{tableau of shape }\lambda' / \mu'}} \xx^{\cont(T)}
= \sum\limits_{\substack{T\text{ is a column-strict}\\ 
\text{tableau of shape }\lambda / \mu}} \xx^{\cont(T)}
\end{equation}
(by \eqref{sol.schur.rotate180.schureq}).

But the definition of $s_{\lambda / \mu}$ yields
$s_{\lambda / \mu} = \sum_T \xx^{\cont(T)}$,
where the sum ranges over all column-strict tableaux $T$ of shape
$\lambda / \mu$. In other words,
\[
s_{\lambda / \mu} =
\sum\limits_{\substack{T\text{ is a column-strict}\\ 
\text{tableau of shape }\lambda / \mu}} \xx^{\cont(T)}.
\]
Compared with \eqref{sol.schur.rotate180.schureq2}, this yields
$s_{\lambda / \mu} = s_{\lambda' / \mu'}$. This solves part (b) of
the exercise.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.schur.disconnected}}

\textit{Solution to Exercise~\ref{exe.schur.disconnected}.} If
$\varphi$ and $\psi$ are two partitions such that $\psi\subseteq\varphi$, then
let $Y\left(  \varphi/\psi\right)  $ denote the skew Ferrers diagram
$\varphi/\psi$ (this is a subset of $\left\{  1,2,3,...\right\}  ^{2}$).

Whenever $Z$ is a subset of $\ZZ^2$, we define a
\emph{column-strict $Z$-tableau} to be an assignment of entries in
$\left\{  1,2,3,...\right\}  $ to the elements of $Z$ which is weakly
increasing left-to-right in rows and strictly increasing top-to-bottom in
columns. It is clear that if $\varphi$ and $\psi$ are two partitions such that
$\psi\subseteq\varphi$, then a column-strict tableau of shape $\varphi/\psi$
is the same as a column-strict $Y\left(  \varphi/\psi\right)  $-tableau. We
define the notation $\cont \left(  T\right)  $ (and therefore,
$\mathbf{x}^{\cont \left(  T\right)  }$) for a column-strict
$Z$-tableau $T$ in the same way as it is defined for a column-strict tableau
of shape $\lambda/\mu$ (for some partitions $\lambda$ and $\mu$).

The following is now more or less obvious:

\begin{lemma}
\label{lem.sol.schur.disconnected.trans}Let $\varphi$ and $\psi$ be two
partitions such that $\psi\subseteq\varphi$. Let $Z$ be a subset of
$\ZZ^2$. Assume that the skew Ferrers diagram $\varphi/\psi$ can be
obtained from $Z$ by parallel translation. Then,%
\[
s_{\varphi/\psi}=\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}%
}}\mathbf{x}^{\cont \left(  T\right)  }.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.schur.disconnected.trans}.]Let $R$ be the
parallel translation which sends the set $Z$ to $Y\left(  \varphi/\psi\right)
$.

The definition of $s_{\varphi/\psi}$ yields $s_{\varphi/\psi}=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\varphi/\psi
}}\mathbf{x}^{\cont \left(  T\right)  }$. It remains to prove
that the right hand side of this equality equals $\sum\limits_{\substack{T\text{ is a
column-strict}\\Z\text{-tableau}}}\mathbf{x}^{\cont \left(
T\right)  }$. To achieve this, it is clearly enough to find a bijection%
\[
\Gamma:\left(  \text{the set of all column-strict tableaux of shape }%
\varphi/\psi\right)  \rightarrow\left(  \text{the set of all column-strict
}Z\text{-tableaux}\right)
\]
which satisfies
\[
\left(  \mathbf{x}^{\cont\left(  \Gamma\left(  T\right)
\right)  }=\mathbf{x}^{\cont\left(  T\right)  }
\ \ \ \ \ \ \ \ \ \ \text{for every column-strict tableau }
T \text{ of shape } \varphi/\psi\right)  .
\]


But this is very easy: The bijection $\Gamma$ sends every column-strict
tableau $T$ of shape $\varphi/\psi$ to the column-strict $Z$-tableau $T\circ
R$. (The notation $T\circ R$ makes sense because $T$, being a column-strict
tableau of shape $\varphi/\psi$, is an assignment of entries in $\left\{
1,2,3,...\right\}  $ to the cells of the skew Ferrers diagram $\varphi/\psi$,
that is, a map $Y\left(  \varphi/\psi\right)  \rightarrow\left\{
1,2,3,...\right\}  $. Visually speaking, $T\circ R$ is the result of moving
the tableau $T$ so that it takes up the cells of $Z$ rather than the cells of
$\varphi/\psi$.) Lemma \ref{lem.sol.schur.disconnected.trans} is proven.
\end{proof}

Now, let us return to the solution of the exercise. Clearly, the subsets
$F_{\operatorname{rows}\leq k}$ and $F_{\operatorname{rows}>k}$ of $F$ are
disjoint, and their union is $F$.

Lemma \ref{lem.sol.schur.disconnected.trans} (applied to $\alpha$, $\beta$ and
$F_{\operatorname{rows}\leq k}$ instead of $\varphi$, $\psi$ and $Z$) yields%
\begin{equation}
s_{\alpha/\beta}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }=\sum\limits_{\substack{P\text{ is a
column-strict}\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\xx%
^{\cont \left(  P\right)  }
\label{sol.schur.disconnected.factor1}%
\end{equation}
(here, we renamed the summation index $T$ as $P$). Also, Lemma
\ref{lem.sol.schur.disconnected.trans} (applied to $\gamma$, $\delta$ and
$F_{\operatorname{rows}>k}$ instead of $\varphi$, $\psi$ and $Z$) yields%
\begin{equation}
s_{\gamma/\delta}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\F_{\operatorname{rows}>k}\text{-tableau}}}\mathbf{x}^{\cont
\left(  T\right)  }=\sum\limits_{\substack{Q\text{ is a column-strict}%
\\F_{\operatorname{rows}>k}\text{-tableau}}}\mathbf{x}^{\cont
\left(  Q\right)  } \label{sol.schur.disconnected.factor2}%
\end{equation}
(here, we renamed the summation index $T$ as $Q$). Multiplying the identities
(\ref{sol.schur.disconnected.factor1}) and
(\ref{sol.schur.disconnected.factor2}), we obtain%
\begin{align}
s_{\alpha/\beta}s_{\gamma/\delta}  &  =\left(  \sum\limits_{\substack{P\text{ is a
column-strict}\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\xx%
^{\cont \left(  P\right)  }\right)  \left(  \sum\limits
_{\substack{Q\text{ is a column-strict}\\F_{\operatorname{rows}>k}%
\text{-tableau}}}\mathbf{x}^{\cont \left(  Q\right)  }\right)
\nonumber\\
&  =\sum\limits_{\substack{P\text{ is a column-strict}\\F_{\operatorname{rows}\leq
k}\text{-tableau}}}\sum\limits_{\substack{Q\text{ is a column-strict}%
\\F_{\operatorname{rows}>k}\text{-tableau}}}\mathbf{x}^{\cont
\left(  P\right)  }\mathbf{x}^{\cont \left(  Q\right)
}\nonumber\\
&  =\sum\limits_{\substack{\left(  P,Q\right)  \in\\\left(  \text{the set of all
column-strict }F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)
\\\times\left(  \text{the set of all column-strict }F_{\operatorname*{rows}%
>k}\text{-tableaux}\right)  }}\mathbf{x}^{\cont \left(
P\right)  }\mathbf{x}^{\cont \left(  Q\right)  }.
\label{sol.schur.disconnected.product}%
\end{align}
On the other hand, the definition of $s_{\lambda/\mu}$ yields%
\begin{equation}
s_{\lambda/\mu}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\mu}}\mathbf{x}^{\cont \left(  T\right)  }.
\label{sol.schur.disconnected.otherside}%
\end{equation}
Our goal is to prove that the left-hand side of
(\ref{sol.schur.disconnected.otherside}) equals the left-hand side of
(\ref{sol.schur.disconnected.product}). For this, it is clearly enough to show
that the right-hand side of (\ref{sol.schur.disconnected.otherside}) equals
the right-hand side of (\ref{sol.schur.disconnected.product}). But to achieve
this, it clearly suffices to exhibit a bijection%
\begin{align*}
\Phi &  :\left(  \text{the set of all column-strict tableaux of shape }%
\lambda/\mu\right) \\
&  \rightarrow\left(  \text{the set of all column-strict }%
F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)  \times\left(  \text{the
set of all column-strict }F_{\operatorname*{rows}>k}\text{-tableaux}\right)
\end{align*}
which has the property that
\begin{equation}
\text{if }\left(  P,Q\right)  =\Phi\left(  T\right)  \text{ for some
column-strict tableau }T\text{ of shape }\lambda/\mu\text{, then }%
\mathbf{x}^{\cont \left(  P\right)  }\xx%
^{\cont \left(  Q\right)  }=\mathbf{x}^{\cont
\left(  T\right)  }. \label{sol.schur.disconnected.contprod}%
\end{equation}
We claim that such a bijection $\Phi$ can be defined by%
\begin{equation}
\left(  \Phi\left(  T\right)  =\left(  T\mid_{F_{\operatorname*{rows}\leq k}%
},T\mid_{F_{\operatorname*{rows}>k}}\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every column-strict tableau }T\text{ of shape }\lambda/\mu\right)  .
\label{sol.schur.disconnected.bij}%
\end{equation}
Indeed, it is clear that we can define a map
\begin{align*}
\Phi &  :\left(  \text{the set of all column-strict tableaux of shape }%
\lambda/\mu\right) \\
&  \rightarrow\left(  \text{the set of all column-strict }%
F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)  \times\left(  \text{the
set of all column-strict }F_{\operatorname*{rows}>k}\text{-tableaux}\right)
\end{align*}
by (\ref{sol.schur.disconnected.bij}), and that this map $\Phi$ satisfies
(\ref{sol.schur.disconnected.contprod}). All that remains to be proven is that
this map $\Phi$ is a bijection. It is clear that $\Phi$ is injective, so we
only need to prove that $\Phi$ is surjective.

Let $\left(  P,Q\right)  \in\left(  \text{the set of all column-strict
}F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)  \times\left(
\text{the set of all column-strict }F_{\operatorname*{rows}>k}\text{-tableaux}%
\right)  $ be arbitrary. We are going to prove that $\left(  P,Q\right)  $
lies in the image of $\Phi$.

Define a map $T:F\rightarrow\left\{  1,2,3,...\right\}  $ by
setting
\[
\left(  T\left(  p\right)  = \begin{cases}
P\left(  p\right)  ,& \text{if }p\in F_{\operatorname*{rows} \leq k};\\
Q\left(  p\right)  ,& \text{if }p\in F_{\operatorname*{rows} >k}
\end{cases}
\qquad \text{ for all } p \in F \right) .
\]
This map $T$ is clearly well-defined (since the subsets
$F_{\operatorname{rows}\leq k}$ and $F_{\operatorname{rows}>k}$ of $F$ are
disjoint, and their union is $F$), and thus is an assignment of entries in
$\left\{  1,2,3,...\right\}  $ to the cells of the skew Ferrers diagram
$\lambda/\mu$ (since $F$ is the set of those cells). It furthermore satisfies
$T\mid_{F_{\operatorname*{rows}\leq k}}=P$ and $T\mid_{F_{\operatorname*{rows}%
>k}}=Q$. We shall now show that this assignment $T$ is a column-strict tableau
of shape $\lambda/\mu$.

This rests on the following observation:

\begin{statement}
\textit{Assertion A:} Let $c$ and $d$ be two cells lying in $F$. Assume that
the cells $c$ and $d$ either lie in one and the same row, or lie in one and
the same column. Then, either both $c$ and $d$ belong to
$F_{\operatorname*{rows}\leq k}$, or both $c$ and $d$ belong to
$F_{\operatorname*{rows}>k}$.
\end{statement}

Assertion A is an easy consequence of our assumption that $\mu_{k}\geq
\lambda_{k+1}$.\ \ \ \ \footnote{\textit{Proof of Assertion A:} Assume the
contrary. Then, one of the cells $c$ and $d$ belongs to
$F_{\operatorname*{rows}\leq k}$, whereas the other belongs to
$F_{\operatorname*{rows}>k}$. We WLOG assume that $c$ belongs to
$F_{\operatorname*{rows}\leq k}$, whereas $d$ belongs to
$F_{\operatorname*{rows}>k}$ (since otherwise, we can simply switch $c$ with
$d$).
\par
Write the cell $c$ in the form $\left(  a,b\right)  $, so that $c$ lies in row
$a$ and column $b$. Write the cell $d$ in the form $\left(  a^{\prime
},b^{\prime}\right)  $, so that $d$ lies in row $a^{\prime}$ and column
$b^{\prime}$.
\par
We have $\left(  a,b\right)  =c\in F_{\operatorname*{rows}\leq k}$, so that
$a\leq k$ (by the definition of $F_{\operatorname*{rows}\leq k}$). We have
$\left(  a^{\prime},b^{\prime}\right)  =d\in F_{\operatorname*{rows}>k}$, so
that $a^{\prime}>k$ (by the definition of $F_{\operatorname*{rows}>k}$).
\par
We have $a\leq k<a^{\prime}$ (since $a^{\prime}>k$), hence $a\neq a^{\prime}$.
Thus, the cells $c$ and $d$ lie in different rows (since the cell $c$ lies in
row $a$, whereas the cell $d$ lies in row $a^{\prime}$). As a consequence, the
cells $c$ and $d$ must lie in one and the same column (since we assumed that
the cells $c$ and $d$ either lie in one and the same row, or lie in one and
the same column). In other words, $b=b^{\prime}$ (since the cell $c$ lies in
column $b$, while the cell $d$ lies in column $b^{\prime}$).
\par
Now, $\left(  a,b\right)  =c\in F=Y\left(  \lambda/\mu\right)  $. By the
definition of $Y\left(  \lambda/\mu\right)  $, this shows that $\mu_{a}%
<b\leq\lambda_{a}$. Similarly, $\mu_{a^{\prime}}<b^{\prime}\leq\lambda
_{a^{\prime}}$.
\par
We have $a^{\prime}>k$, hence $a^{\prime}\geq k+1$ (since $a^{\prime}$ and $k$
are integers), thus $k+1\leq a^{\prime}$.
\par
Since $\mu$ is a partition, we have $\mu_{1}\geq\mu_{2}\geq\mu_{3}\geq...$.
Hence, $\mu_{a}\geq\mu_{k}$ (since $a\leq k$). Thus, $\mu_{a}\geq\mu_{k}%
\geq\lambda_{k+1}$.
\par
Since $\lambda$ is a partition, we have $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq...$. Thus, $\lambda_{k+1}\geq\lambda_{a^{\prime}}$ (since
$k+1\leq a^{\prime}$). Thus, $\mu_{a}\geq\lambda_{k+1}\geq\lambda_{a^{\prime}%
}$, so that $\lambda_{a^{\prime}}\leq\mu_{a}$ and thus $b^{\prime}\leq
\lambda_{a^{\prime}}\leq\mu_{a}<b=b^{\prime}$, which is absurd. This
contradiction proves that our assumption was wrong, and thus Assertion A is
proven.} Now, we want to prove that $T$ is a column-strict tableau of shape
$\lambda/\mu$. To do so, we need to check that $T$ is weakly increasing
left-to-right in rows, and strictly increasing top-to-bottom in columns. We
will only prove the latter part of this statement, as the former part is
proven analogously. So we are going to show that $T$ is strictly increasing
top-to-bottom in columns. In other words, we are going to show that if $c$ and
$d$ are two cells of $\lambda/\mu$ lying in one and the same column, with $d$
lying strictly further south than $c$, then $T\left(  c\right)  <T\left(
d\right)  $. Indeed, consider two such cells $c$ and $d$. Assertion A shows
that either both $c$ and $d$ belong to $F_{\operatorname*{rows}\leq k}$, or
both $c$ and $d$ belong to $F_{\operatorname*{rows}>k}$. Let us WLOG assume
that we are in the first of these two cases (the other case is exactly
analogous). Then, both $c$ and $d$ belong to $F_{\operatorname*{rows}\leq k}$,
so that we have $T\left(  c\right)  =\left(  T\mid_{F_{\operatorname*{rows}%
\leq k}}\right)  \left(  c\right)  $ and $T\left(  d\right)  =\left(
T\mid_{F_{\operatorname*{rows}\leq k}}\right)  \left(  d\right)  $. Since
$T\mid_{F_{\operatorname*{rows}\leq k}}=P$, these two equalities rewrite as
$T\left(  c\right)  =P\left(  c\right)  $ and $T\left(  d\right)  =P\left(
d\right)  $. But since $P$ is strictly increasing top-to-bottom in columns
(because $P$ is a column-strict tableau), we have $P\left(  c\right)
<P\left(  d\right)  $, and thus $T\left(  c\right)  =P\left(  c\right)
<P\left(  d\right)  =T\left(  d\right)  $. Thus, we have proven that $T\left(
c\right)  <T\left(  d\right)  $. This completes the proof that $T$ is a
column-strict tableau of shape $\lambda/\mu$. Hence, $\Phi\left(  T\right)  $
is well-defined, and the definition of $\Phi\left(  T\right)  $ shows that
$\Phi\left(  T\right)  =\left(  \underbrace{T\mid_{F_{\operatorname*{rows}\leq
k}}}_{=P},\underbrace{T\mid_{F_{\operatorname*{rows}>k}}}_{=Q}\right)
=\left(  P,Q\right)  $. Thus, $\left(  P,Q\right)  $ lies in the image of
$\Phi$.

Now, let us forget that we fixed $\left(  P,Q\right)  $. We thus have shown
that every \newline$\left(  P,Q\right)  \in\left(  \text{the set of all
column-strict }F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)
\times\left(  \text{the set of all column-strict }F_{\operatorname*{rows}%
>k}\text{-tableaux}\right)  $ lies in the image of $\Phi$. In other words, the
map $\Phi$ is surjective, which (as we know that $\Phi$ is injective) yields
that $\Phi$ is a bijection. As explained above, this completes the solution of
Exercise~\ref{exe.schur.disconnected}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.schur.disconnected}}

\textit{Solution to Exercise~\ref{exe.schur.disconnected}.} If
$\varphi$ and $\psi$ are two partitions such that $\psi\subseteq\varphi$, then
let $Y\left(  \varphi/\psi\right)  $ denote the skew Ferrers diagram
$\varphi/\psi$ (this is a subset of $\left\{  1,2,3,...\right\}  ^{2}$).

Recall that $F$ is the skew Ferrers diagram $\lambda/\mu$. Thus, the cells of
the skew Ferrers diagram $\lambda/\mu$ are exactly the elements of $F$, and we
have $F=Y\left(  \lambda/\mu\right)  $. Hence, $Y\left(  \lambda/\mu\right)
=F$.

Whenever $Z$ is a subset of $\ZZ^2$, we define a
\emph{column-strict $Z$-tableau} to be an assignment of entries in
$\left\{  1,2,3,...\right\}  $ to the elements of $Z$ which is weakly
increasing left-to-right in rows and strictly increasing top-to-bottom in
columns. It is clear that if $\varphi$ and $\psi$ are two partitions such that
$\psi\subseteq\varphi$, then a column-strict tableau of shape $\varphi/\psi$
is the same as a column-strict $Y\left(  \varphi/\psi\right)  $-tableau (since
the elements of $Y\left(  \varphi/\psi\right)  $ are precisely the cells of
the skew Ferrers diagram $\varphi/\psi$). We define the notation
$\cont \left(  T\right)  $ (and therefore, $\xx%
^{\cont \left(  T\right)  }$) for a column-strict $Z$-tableau
$T$ in the same way as it is defined for a column-strict tableau of shape
$\lambda/\mu$ (for some partitions $\lambda$ and $\mu$).

We shall now show two simple lemmas.

\begin{lemma}
\label{lem.sol.schur.disconnected.trans-gen}Let $Z$ and $W$ be two finite
subsets of $\ZZ^2$. Assume that the set $W$ can be obtained from $Z$
by parallel translation. Then,%
\[
\sum\limits_{\substack{T\text{ is a column-strict}\\W\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }=\sum\limits_{\substack{T\text{ is a
column-strict}\\Z\text{-tableau}}}\mathbf{x}^{\cont \left(
T\right)  }.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.schur.disconnected.trans-gen}.]The set $W$ can be
obtained from $Z$ by parallel translation. In other words, there exists a
parallel translation $R$ such that $W=R\left(  Z\right)  $. Consider this $R$.
Since $R$ is a parallel translation, the map $R$ is invertible, and its
inverse $R^{-1}$ is also a parallel translation. This parallel translation
$R^{-1}$ satisfies $Z=R^{-1}\left(  W\right)  $ (since $R\left(  Z\right)  =W$).

Let $T$ be a column-strict $W$-tableau. Then, $T$ is an assignment of entries
in $\left\{  1,2,3,\ldots\right\}  $ to the elements of $W$ which is weakly
increasing left-to-right in rows, and strictly increasing top-to-bottom in
columns (by the definition of a column-strict $W$-tableau). So $T$ is a map
$W\rightarrow\left\{  1,2,3,...\right\}  $ (since $T$ is an assignment of
entries in $\left\{  1,2,3,\ldots\right\}  $ to the elements of $W$), and thus
the composition $T\circ R$ is a well-defined map $Z\rightarrow\left\{
1,2,3,...\right\}  $ (since $R$ sends $Z$ to $R\left(  Z\right)  =W$). In
other words, $T\circ R$ is an assignment of entries in $\left\{
1,2,3,...\right\}  $ to the elements of $Z$. This assignment $T\circ R$ is
weakly increasing left-to-right in rows\footnote{\textit{Proof.} Let $c$ and
$d$ be two cells lying in one and the same row of $T\circ R$, with $d$ lying
strictly further east than $c$. We shall prove that $\left(  T\circ R\right)
\left(  c\right)  \leq\left(  T\circ R\right)  \left(  d\right)  $.
\par
Since $c$ is a cell of $T\circ R$, we have $c\in Z$ (since the domain of the
map $T\circ R$ is $Z$). Hence, $R\left(  c\right)  \in R\left(  Z\right)  =W$.
The same argument (but for $d$ instead of $c$) shows that $R\left(  d\right)
\in W$. Thus, the cells $R\left(  c\right)  $ and $R\left(  d\right)  $ belong
to $W$, and therefore are cells of $T$ (since the domain of the map $T$ is
$W$).
\par
Furthermore, the cells $c$ and $d$ lie in one and the same row, with $d$ lying
strictly further east than $c$. Since $R$ is a parallel translation (and
therefore preserves the property of lying in and one and the same row, as well
as the property of a point lying strictly further east than another), this
yields that the cells $R\left(  c\right)  $ and $R\left(  d\right)  $ lie in
one and the same row, with $R\left(  d\right)  $ lying strictly further east
than $R\left(  c\right)  $. Consequently, $T\left(  R\left(  c\right)
\right)  \leq T\left(  R\left(  d\right)  \right)  $ (since $T$ is weakly
increasing left-to-right in rows). Thus, $\left(  T\circ R\right)  \left(
c\right)  =T\left(  R\left(  c\right)  \right)  \leq T\left(  R\left(
d\right)  \right)  =\left(  T\circ R\right)  \left(  d\right)  $.
\par
Now, let us forget that we fixed $c$ and $d$. We thus have shown that if $c$
and $d$ are two cells lying in one and the same row of $T\circ R$, with $d$
lying strictly further east than $c$, then $\left(  T\circ R\right)  \left(
c\right)  \leq\left(  T\circ R\right)  \left(  d\right)  $. In other words,
the assignment $T\circ R$ is weakly increasing left-to-right in rows, qed.}
and strictly increasing top-to-bottom in columns\footnote{This is proven in
the same way as we have shown that $T\circ R$ is weakly increasing
left-to-right in rows, with the only difference that ``weakly'',
``left-to-right'', ``row'', ``east'' and ``$\leq$'' must be
replaced by ``strictly'', ``top-to-bottom'', ``column'',
``south'' and ``$<$'', respectively.}. Hence, $T\circ R$ is a column-strict
$Z$-tableau (by the definition of a column-strict $Z$-tableau).

Now, let us forget that we have fixed $T$. We thus have shown that if $T$ is a
column-strict $W$-tableau, then $T\circ R$ is a well-defined column-strict
$Z$-tableau. Hence, we can define a map%
\[
\Phi:\left(  \text{the set of all column-strict }W\text{-tableaux}\right)
\rightarrow\left(  \text{the set of all column-strict }Z\text{-tableaux}%
\right)
\]
by $\left(  \Phi\left(  T\right)  =T\circ R\text{ for all column-strict
}W\text{-tableaux }T\right)  $. Consider this map $\Phi$.

We can also define a map
\[
\Psi:\left(  \text{the set of all column-strict }Z\text{-tableaux}\right)
\rightarrow\left(  \text{the set of all column-strict }W\text{-tableaux}%
\right)
\]
by $\left(  \Psi\left(  T\right)  =T\circ R^{-1}\text{ for all column-strict
}Z\text{-tableaux }T\right)  $.\ \ \ \ \footnote{This can be proven in the
same way as we have proved that we can define a map%
\[
\Phi:\left(  \text{the set of all column-strict }W\text{-tableaux}\right)
\rightarrow\left(  \text{the set of all column-strict }Z\text{-tableaux}%
\right)
\]
by $\left(  \Phi\left(  T\right)  =T\circ R\text{ for all column-strict
}W\text{-tableaux }T\right)  $, with the only difference that we must replace
$W$, $Z$ and $R$ by $Z$, $W$ and $R^{-1}$ in our proof.} Consider this map
$\Psi$.

We have $\Phi\circ\Psi = \id$ (since every column-strict
$Z$-tableau $T$ satisfies
\begin{align*}
\left(  \Phi\circ\Psi\right)  \left(  T\right)   &  =\Phi\left(
\underbrace{\Psi\left(  T\right)  }_{=T\circ R^{-1}}\right)  =\Phi\left(
T\circ R^{-1}\right)  =\left(  T\circ R^{-1}\right)  \circ
R\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\right) \\
&  =T\circ\underbrace{R^{-1}\circ R}_{=\id}=T\circ
\id =T=\operatorname*{id}\left(  T\right)
\end{align*}
) and $\Psi\circ\Phi=\operatorname*{id}$ (for similar reasons). Thus, the maps
$\Phi$ and $\Psi$ are mutually inverse. Hence, $\Phi$ is a bijection. Hence,
we can substitute $T$ for $\Phi\left(  T\right)  $ in the sum $\sum\limits
_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }$. As a result, we obtain%
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }=\sum\limits_{\substack{T\text{ is a
column-strict}\\W\text{-tableau}}}\mathbf{x}^{\cont \left(
\Psi\left(  T\right)  \right)  }.
\label{pf.lem.sol.schur.disconnected.trans-gen.1}%
\end{equation}


Now, let us notice that every column-strict $W$-tableau $T$ satisfies%
\begin{equation}
\mathbf{x}^{\cont \left(  \Psi\left(  T\right)  \right)
}=\mathbf{x}^{\cont \left(  T\right)  }.
\label{pf.lem.sol.schur.disconnected.trans-gen.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.schur.disconnected.trans-gen.2}):}
Let $T$ be a column-strict $W$-tableau. Then, $\xx%
^{\cont \left(  T\right)  }=\prod_{i}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }$ (by the definition of $\xx%
^{\cont \left(  T\right)  }$) and $\xx%
^{\cont \left(  \Psi\left(  T\right)  \right)  }=\prod_{i}%
x_{i}^{\left\vert \left(  \Psi\left(  T\right)  \right)  ^{-1}\left(
i\right)  \right\vert }$ (by the definition of $\xx%
^{\cont \left(  \Psi\left(  T\right)  \right)  }$). Now, every
$i\in\left\{  1,2,3,...\right\}  $ satisfies
\[
\left(  \underbrace{\Psi\left(  T\right)  }_{\substack{=T\circ R^{-1}%
\\\text{(by the definition of }\Psi\left(  T\right)  \text{)}}}\right)
^{-1}\left(  i\right)  =\left(  T\circ R^{-1}\right)  ^{-1}\left(  i\right)
=\left(  R^{-1}\right)  ^{-1}\left(  T^{-1}\left(  i\right)  \right)  .
\]
Hence, every $i\in\left\{  1,2,3,...\right\}  $ satisfies
\begin{equation}
\left\vert \underbrace{\left(  \Psi\left(  T\right)  \right)  ^{-1}\left(
i\right)  }_{=\left(  R^{-1}\right)  ^{-1}\left(  T^{-1}\left(  i\right)
\right)  }\right\vert =\left\vert \left(  R^{-1}\right)  ^{-1}\left(
T^{-1}\left(  i\right)  \right)  \right\vert =\left\vert T^{-1}\left(
i\right)  \right\vert \label{pf.lem.sol.schur.disconnected.trans-gen.2.pf.2}%
\end{equation}
(since $R^{-1}$ is a bijection). Thus,
\[
\mathbf{x}^{\cont \left(  \Psi\left(  T\right)  \right)  }%
=\prod_{i}\underbrace{x_{i}^{\left\vert \left(  \Psi\left(  T\right)  \right)
^{-1}\left(  i\right)  \right\vert }}_{\substack{=x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\\\text{(by
(\ref{pf.lem.sol.schur.disconnected.trans-gen.2.pf.2}))}}}=\prod_{i}%
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }=\xx%
^{\cont \left(  T\right)  }.
\]
This proves (\ref{pf.lem.sol.schur.disconnected.trans-gen.2}).} Hence,
(\ref{pf.lem.sol.schur.disconnected.trans-gen.1}) becomes%
\[
\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }=\sum\limits_{\substack{T\text{ is a
column-strict}\\W\text{-tableau}}}\underbrace{\mathbf{x}^{\cont
\left(  \Psi\left(  T\right)  \right)  }}_{\substack{=\xx%
^{\cont \left(  T\right)  }\\\text{(by
(\ref{pf.lem.sol.schur.disconnected.trans-gen.2}))}}}=\sum\limits_{\substack{T\text{
is a column-strict}\\W\text{-tableau}}}\mathbf{x}^{\cont \left(
T\right)  }.
\]
This proves Lemma \ref{lem.sol.schur.disconnected.trans-gen}.
\end{proof}

\begin{lemma}
\label{lem.sol.schur.disconnected.trans}Let $\varphi$ and $\psi$ be two
partitions such that $\psi\subseteq\varphi$. Let $Z$ be a subset of
$\ZZ^2$. Assume that the skew Ferrers diagram $\varphi/\psi$ can be
obtained from $Z$ by parallel translation. Then,%
\[
s_{\varphi/\psi}=\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}%
}}\mathbf{x}^{\cont \left(  T\right)  }.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.schur.disconnected.trans}.]The skew Ferrers
diagram $\varphi/\psi$ can be obtained from $Z$ by parallel translation. In
other words, $Y\left(  \varphi/\psi\right)  $ can be obtained from $Z$ by
parallel translation (since $Y\left(  \varphi/\psi\right)  $ is the skew
Ferrers diagram $\varphi/\psi$). Hence, the sets $Y\left(  \varphi
/\psi\right)  $ and $Z$ have the same cardinality. Thus, the set $Z$ is finite
(since the set $Y\left(  \varphi/\psi\right)  $ is finite). Hence, Lemma
\ref{lem.sol.schur.disconnected.trans-gen} (applied to $W=Y\left(
\varphi/\psi\right)  $) yields%
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\Y\left(  \varphi/\psi\right)
\text{-tableau}}}\mathbf{x}^{\cont \left(  T\right)  }%
=\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }.
\label{pf.lem.sol.schur.disconnected.trans.0}%
\end{equation}


The definition of $s_{\varphi/\psi}$ yields $s_{\varphi/\psi}=\sum
_{T}\mathbf{x}^{\cont \left(  T\right)  }$, where the sum runs
over all column-strict tableaux of shape $\varphi/\psi$. In other words,%
\begin{equation}
s_{\varphi/\psi}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\varphi/\psi}}\mathbf{x}^{\cont \left(  T\right)  }.
\label{pf.lem.sol.schur.disconnected.trans.1}%
\end{equation}


But recall that a column-strict tableau of shape $\varphi/\psi$ is the same as
a column-strict $Y\left(  \varphi/\psi\right)  $-tableau. Hence, the summation
sign ``$\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\varphi/\psi}}$'' can be replaced
by ``$\sum\limits_{\substack{T\text{ is a column-strict}\\Y\left(
\varphi/\psi\right)  \text{-tableau}}}$'' on the right hand
side of (\ref{pf.lem.sol.schur.disconnected.trans.1}). We thus obtain%
\[
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\varphi/\psi}}\mathbf{x}^{\cont \left(  T\right)  }%
=\sum\limits_{\substack{T\text{ is a column-strict}\\Y\left(  \varphi/\psi\right)
\text{-tableau}}}\mathbf{x}^{\cont \left(  T\right)  }%
=\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.schur.disconnected.trans.0})}\right)  .
\]
Hence, (\ref{pf.lem.sol.schur.disconnected.trans.1}) becomes%
\[
s_{\varphi/\psi}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\varphi/\psi}}\mathbf{x}^{\cont \left(  T\right)  }%
=\sum\limits_{\substack{T\text{ is a column-strict}\\Z\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }.
\]
This proves Lemma \ref{lem.sol.schur.disconnected.trans}.
\end{proof}

Now, let us return to the solution of the exercise.

We can rewrite the definition of $F_{\operatorname*{rows}\leq k}$ as follows:%
\[
F_{\operatorname*{rows}\leq k}=\left\{  \left(  i,j\right)  \in F\ \mid\ i\leq
k\right\}  .
\]
We can rewrite the definition of $F_{\operatorname*{rows}>k}$ as follows:%
\[
F_{\operatorname*{rows}>k}=\left\{  \left(  i,j\right)  \in F\ \mid
\ i>k\right\}  .
\]


From the definitions of $F_{\operatorname*{rows}\leq k}$ and
$F_{\operatorname*{rows}>k}$, it follows that every element of $F$ lies in
exactly one of the two sets $F_{\operatorname*{rows}\leq k}$ and
$F_{\operatorname*{rows}>k}$\ \ \ \ \footnote{\textit{Proof.} Let $p$ be any
element of $F$. We shall show that $p$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$.
\par
Indeed, let us write $p$ in the form $p=\left(  u,v\right)  $ for some
$\left(  u,v\right)  \in\left\{  1,2,3,...\right\}  ^{2}$. We distinguish
between two cases:
\par
\textit{Case 1:} We have $u\leq k$.
\par
\textit{Case 2:} We have $u>k$.
\par
Let us first consider Case 1. In this case, we have $u\leq k$. Now, $p=\left(
u,v\right)  \in\left\{  \left(  i,j\right)  \in F\ \mid\ i\leq k\right\}  $
(since $\left(  u,v\right)  =p\in F$ and $u\leq k$). In other words, $p\in
F_{\operatorname*{rows}\leq k}$ (since $F_{\operatorname*{rows}\leq
k}=\left\{  \left(  i,j\right)  \in F\ \mid\ i\leq k\right\}  $). On the other
hand, $\left(  u,v\right)  \notin F_{\operatorname*{rows}>k}$ (because
otherwise, we would have $\left(  u,v\right)  \in F_{\operatorname*{rows}%
>k}=\left\{  \left(  i,j\right)  \in F\ \mid\ i>k\right\}  $, which would
yield $u>k$, which would contradict $u\leq k$). Thus, $p=\left(  u,v\right)
\notin F_{\operatorname*{rows}>k}$. Thus, we know that $p$ lies in the set
$F_{\operatorname*{rows}\leq k}$ (since $p\in F_{\operatorname*{rows}\leq k}$)
but not in the set $F_{\operatorname*{rows}>k}$ (since $p\notin
F_{\operatorname*{rows}>k}$). Hence, $p$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$.
\par
We thus have shown that $p$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$ in Case 1.
\par
Let us now consider Case 2. In this case, we have $u>k$. Now, $p=\left(
u,v\right)  \in\left\{  \left(  i,j\right)  \in F\ \mid\ i>k\right\}  $ (since
$\left(  u,v\right)  =p\in F$ and $u>k$). In other words, $p\in
F_{\operatorname*{rows}>k}$ (since $F_{\operatorname*{rows}>k}=\left\{
\left(  i,j\right)  \in F\ \mid\ i>k\right\}  $). On the other hand, $\left(
u,v\right)  \notin F_{\operatorname*{rows}\leq k}$ (because otherwise, we
would have $\left(  u,v\right)  \in F_{\operatorname*{rows}\leq k}=\left\{
\left(  i,j\right)  \in F\ \mid\ i\leq k\right\}  $, which would yield $u\leq
k$, which would contradict $u>k$). Thus, $p=\left(  u,v\right)  \notin
F_{\operatorname*{rows}\leq k}$. Thus, we know that $p$ lies in the set
$F_{\operatorname*{rows}>k}$ (since $p\in F_{\operatorname*{rows}>k}$) but not
in the set $F_{\operatorname*{rows}\leq k}$ (since $p\notin
F_{\operatorname*{rows}\leq k}$). Hence, $p$ lies in exactly one of the two
sets $F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$.
\par
We thus have shown that $p$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$ in Case 2.
\par
Now, in each of the two Cases 1 and 2, we have shown that $p$ lies in exactly
one of the two sets $F_{\operatorname*{rows}\leq k}$ and
$F_{\operatorname*{rows}>k}$. Since these two Cases cover all possibilities,
this yields that $p$ always lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$.
\par
Now, let us forget that we fixed $p$. Thus, we have shown that for every
element $p$ of $F$, the element $p$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$. In other
words, every element of $F$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$, qed.}.
Thus, the two subsets $F_{\operatorname*{rows}\leq k}$ and
$F_{\operatorname*{rows}>k}$ of $F$ are disjoint and their union is $F$. In
other words, $F_{\operatorname*{rows}\leq k}\cap F_{\operatorname*{rows}%
>k}=\varnothing$ and $F_{\operatorname*{rows}\leq k}\cup
F_{\operatorname*{rows}>k}=F$. Hence, $F=F_{\operatorname*{rows}\leq k}\cup
F_{\operatorname*{rows}>k}$, so that $\underbrace{F}_{=F_{\operatorname*{rows}%
\leq k}\cup F_{\operatorname*{rows}>k}}\setminus F_{\operatorname*{rows}\leq
k}=\left(  F_{\operatorname*{rows}\leq k}\cup F_{\operatorname*{rows}%
>k}\right)  \setminus F_{\operatorname*{rows}\leq k}=F_{\operatorname*{rows}%
>k}\setminus F_{\operatorname*{rows}\leq k}=F_{\operatorname*{rows}>k}$ (since
$F_{\operatorname*{rows}\leq k}\cap F_{\operatorname*{rows}>k}=\varnothing$)
and $\underbrace{F}_{=F_{\operatorname*{rows}\leq k}\cup
F_{\operatorname*{rows}>k}}\setminus F_{\operatorname*{rows}>k}=\left(
F_{\operatorname*{rows}\leq k}\cup F_{\operatorname*{rows}>k}\right)
\setminus F_{\operatorname*{rows}>k}=F_{\operatorname*{rows}\leq k}\setminus
F_{\operatorname*{rows}>k}=F_{\operatorname*{rows}\leq k}$ (since
$F_{\operatorname*{rows}\leq k}\cap F_{\operatorname*{rows}>k}=\varnothing$).

We can now make the following observation:

\begin{statement}
\textit{Assertion A:} Let $c$ and $d$ be two cells lying in $F$. Assume that
the cells $c$ and $d$ either lie in one and the same row, or lie in one and
the same column. Then, either both $c$ and $d$ belong to
$F_{\operatorname*{rows}\leq k}$, or both $c$ and $d$ belong to
$F_{\operatorname*{rows}>k}$.
\end{statement}

\textit{Proof of Assertion A:} Assume the contrary. Then, neither $\left(
\text{both }c\text{ and }d\text{ belong to }F_{\operatorname*{rows}\leq
k}\right)  $ nor \newline$\left(  \text{both }c\text{ and }d\text{ belong to
}F_{\operatorname*{rows}>k}\right)  $.

At least one of the cells $c$ and $d$ does not belong to
$F_{\operatorname*{rows}\leq k}$ (because we don't have $\left(  \text{both
}c\text{ and }d\text{ belong to }F_{\operatorname*{rows}\leq k}\right)  $). We
can WLOG assume that this cell is $d$ (since otherwise, we can simply switch
$c$ with $d$). Assume this. Thus, $d$ does not belong to
$F_{\operatorname*{rows}\leq k}$. In other words, $d\notin
F_{\operatorname*{rows}\leq k}$. Since $d\in F$ and $d\notin
F_{\operatorname*{rows}\leq k}$, we have $d\in F\setminus
F_{\operatorname*{rows}\leq k}=F_{\operatorname*{rows}>k}$.

If we had $c\in F_{\operatorname*{rows}>k}$, then both $c$ and $d$ would
belong to $F_{\operatorname*{rows}>k}$, which would contradict the fact that
we don't have $\left(  \text{both }c\text{ and }d\text{ belong to
}F_{\operatorname*{rows}>k}\right)  $. Hence, we cannot have $c\in
F_{\operatorname*{rows}>k}$. We thus have $c\notin F_{\operatorname*{rows}>k}%
$. Since $c\in F$ and $c\notin F_{\operatorname*{rows}>k}$, we have $c\in
F\setminus F_{\operatorname*{rows}>k}=F_{\operatorname*{rows}\leq k}$.

Write the cell $c$ in the form $\left(  a,b\right)  $, and write the cell $d$
in the form $\left(  a^{\prime},b^{\prime}\right)  $. The cell $c$ lies in row
$a$ (since $c=\left(  a,b\right)  $) and column $b$ (since $c=\left(
a,b\right)  $). The cell $d$ lies in row $a^{\prime}$ (since $d=\left(
a^{\prime},b^{\prime}\right)  $) and column $b^{\prime}$ (since $d=\left(
a^{\prime},b^{\prime}\right)  $).

We have $\left(  a,b\right)  =c\in F_{\operatorname*{rows}\leq k}=\left\{
\left(  i,j\right)  \in F\ \mid\ i\leq k\right\}  $, so that $a\leq k$. We
have $\left(  a^{\prime},b^{\prime}\right)  =d\in F_{\operatorname*{rows}%
>k}=\left\{  \left(  i,j\right)  \in F\ \mid\ i>k\right\}  $, so that
$a^{\prime}>k$.

We have $a\leq k<a^{\prime}$ (since $a^{\prime}>k$), hence $a\neq a^{\prime}$.
Thus, the cells $c$ and $d$ lie in different rows (since the cell $c$ lies in
row $a$, whereas the cell $d$ lies in row $a^{\prime}$). In other words, the
cells $c$ and $d$ cannot lie in one and the same row. As a consequence, the
cells $c$ and $d$ must lie in one and the same column (since we assumed that
the cells $c$ and $d$ either lie in one and the same row, or lie in one and
the same column). In other words, $b=b^{\prime}$ (since the cell $c$ lies in
column $b$, while the cell $d$ lies in column $b^{\prime}$).

Now, $\left(  a,b\right)  =c\in F=Y\left(  \lambda/\mu\right)  =\left\{
\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{i}%
<j\leq\lambda_{i}\right\}  $ (by the definition of $Y\left(  \lambda
/\mu\right)  $). Thus, this shows that $\mu_{a}<b\leq\lambda_{a}$. The same
argument (but applied to $a^{\prime}$, $b^{\prime}$ and $d$ instead of $a$,
$b$ and $c$) yields $\mu_{a^{\prime}}<b^{\prime}\leq\lambda_{a^{\prime}}$.

We have $a^{\prime}>k$, hence $a^{\prime}\geq k+1$ (since $a^{\prime}$ and $k$
are integers), thus $k+1\leq a^{\prime}$.

Since $\mu$ is a partition, we have $\mu_{1}\geq\mu_{2}\geq\mu_{3}\geq...$. In
other words, $\mu_{u}\geq\mu_{v}$ for any two positive integers $u$ and $v$
satisfying $u\leq v$. Applying this to $u=a$ and $v=k$, we obtain $\mu_{a}%
\geq\mu_{k}$ (since $a\leq k$). Thus, $\mu_{a}\geq\mu_{k}\geq\lambda_{k+1}$.

Since $\lambda$ is a partition, we have $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq...$. In other words, $\lambda_{u}\geq\lambda_{v}$ for any two
positive integers $u$ and $v$ satisfying $u\leq v$. Applying this to $u=k+1$
and $v=a^{\prime}$, we obtain $\lambda_{k+1}\geq\lambda_{a^{\prime}}$ (since
$k+1\leq a^{\prime}$). Thus, $\mu_{a}\geq\lambda_{k+1}\geq\lambda_{a^{\prime}%
}$, so that $\lambda_{a^{\prime}}\leq\mu_{a}$ and thus $b^{\prime}\leq
\lambda_{a^{\prime}}\leq\mu_{a}<b=b^{\prime}$, which is absurd. This
contradiction proves that our assumption was wrong, and thus Assertion A is proven.

Now, we define a map
\begin{align*}
\Phi &  :\left(  \text{the set of all column-strict tableaux of shape }%
\lambda/\mu\right) \\
&  \rightarrow\left(  \text{the set of all column-strict }%
F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)  \times\left(  \text{the
set of all column-strict }F_{\operatorname*{rows}>k}\text{-tableaux}\right)
\end{align*}
as follows: Let $T$ be a column-strict tableau of shape $\lambda/\mu$. Then,
$T$ is an assignment of entries in $\left\{  1,2,3,...\right\}  $ to the cells
of the skew Ferrers diagram $\lambda/\mu$ which is weakly increasing
left-to-right in rows, and strictly increasing top-to-bottom in columns. In
other words, $T$ is an assignment of entries in $\left\{  1,2,3,...\right\}  $
to the elements of $F$ (since the cells of the skew Ferrers diagram
$\lambda/\mu$ are exactly the elements of $F$). Hence, $T\mid
_{F_{\operatorname*{rows}\leq k}}$ is an assignment of entries in $\left\{
1,2,3,...\right\}  $ to the elements of $F_{\operatorname*{rows}\leq k}$
(since $F_{\operatorname*{rows}\leq k}$ is a subset of $F$) which is weakly
increasing left-to-right in rows (since $T$ is weakly increasing left-to-right
in rows), and strictly increasing top-to-bottom in columns (since $T$ is
strictly increasing top-to-bottom in columns). In other words, $T\mid
_{F_{\operatorname*{rows}\leq k}}$ is a column-strict $F_{\operatorname*{rows}%
\leq k}$-tableau. Similarly, $T\mid_{F_{\operatorname*{rows}>k}}$ is a
column-strict $F_{\operatorname*{rows}>k}$-tableau. Now, we define
$\Phi\left(  T\right)  $ to be $\left(  T\mid_{F_{\operatorname*{rows}\leq k}%
},T\mid_{F_{\operatorname*{rows}>k}}\right)  $.

We have thus defined a map $\Phi$. We now notice that the map $\Phi$ is
injective\footnote{\textit{Proof.} Let $T$ and $S$ be two column-strict
tableaux of shape $\lambda/\mu$ such that $\Phi\left(  T\right)  =\Phi\left(
S\right)  $. We are going to show that $T=S$.
\par
Let $c$ be a cell of the skew Ferrers diagram $\lambda/\mu$. We shall show
that $T\left(  c\right)  =S\left(  c\right)  $.
\par
We have $\Phi\left(  T\right)  =\left(  T\mid_{F_{\operatorname*{rows}\leq k}%
},T\mid_{F_{\operatorname*{rows}>k}}\right)  $ (by the definition of
$\Phi\left(  T\right)  $) and $\Phi\left(  S\right)  =\left(  S\mid
_{F_{\operatorname*{rows}\leq k}},S\mid_{F_{\operatorname*{rows}>k}}\right)  $
(by the definition of $\Phi\left(  S\right)  $). Hence, $\left(
T\mid_{F_{\operatorname*{rows}\leq k}},T\mid_{F_{\operatorname*{rows}>k}%
}\right)  =\Phi\left(  T\right)  =\Phi\left(  S\right)  =\left(
S\mid_{F_{\operatorname*{rows}\leq k}},S\mid_{F_{\operatorname*{rows}>k}%
}\right)  $, and therefore $T\mid_{F_{\operatorname*{rows}\leq k}}%
=S\mid_{F_{\operatorname*{rows}\leq k}}$ and $T\mid_{F_{\operatorname*{rows}%
>k}}=S\mid_{F_{\operatorname*{rows}>k}}$.
\par
We know that $c$ is a cell of the skew Ferrers diagram $\lambda/\mu$. In other
words, $c$ is an element of $F$ (since the cells of the skew Ferrers diagram
$\lambda/\mu$ are exactly the elements of $F$). Hence, $c$ lies in exactly one
of the two sets $F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}%
>k}$ (since every element of $F$ lies in exactly one of the two sets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$). Hence, we
must be in one of the following two cases:
\par
\textit{Case 1:} The element $c$ lies in $F_{\operatorname*{rows}\leq k}$.
\par
\textit{Case 2:} The element $c$ lies in $F_{\operatorname*{rows}>k}$.
\par
Let us consider Case 1 first. In this case, the element $c$ lies in
$F_{\operatorname*{rows}\leq k}$. Thus, the elements $\left(  T\mid
_{F_{\operatorname*{rows}\leq k}}\right)  \left(  c\right)  $ and $\left(
S\mid_{F_{\operatorname*{rows}\leq k}}\right)  \left(  c\right)  $ are
well-defined. Clearly, $\left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
\left(  c\right)  =T\left(  c\right)  $, so that%
\[
T\left(  c\right)  =\underbrace{\left(  T\mid_{F_{\operatorname*{rows}\leq k}%
}\right)  }_{=S\mid_{F_{\operatorname*{rows}\leq k}}}\left(  c\right)
=\left(  S\mid_{F_{\operatorname*{rows}\leq k}}\right)  \left(  c\right)
=S\left(  c\right)  .
\]
Thus, we have proven that $T\left(  c\right)  =S\left(  c\right)  $ in Case 1.
\par
The same argument (but with $F_{\operatorname*{rows}\leq k}$ replaced by
$F_{\operatorname*{rows}>k}$) shows that $T\left(  c\right)  =S\left(
c\right)  $ in Case 2. We have now proven that $T\left(  c\right)  =S\left(
c\right)  $ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this yields that $T\left(  c\right)  =S\left(  c\right)  $
always holds.
\par
Now, forget that we fixed $c$. We thus have proven that $T\left(  c\right)
=S\left(  c\right)  $ for every cell $c$ of the skew Ferrers diagram
$\lambda/\mu$. In other words, $T=S$.
\par
Now, forget that we fixed $T$ and $S$. We thus have shown that if $T$ and $S$
are two column-strict tableaux of shape $\lambda/\mu$ such that $\Phi\left(
T\right)  =\Phi\left(  S\right)  $, then $T=S$. In other words, the map $\Phi$
is injective, qed.}. Let us now show that this map $\Phi$ is surjective.

In fact, let
\[
U\in\left(  \text{the set of all column-strict }F_{\operatorname*{rows}\leq
k}\text{-tableaux}\right)  \times\left(  \text{the set of all column-strict
}F_{\operatorname*{rows}>k}\text{-tableaux}\right)
\]
be arbitrary. Then, we can write $U$ in the form $U=\left(  P,Q\right)  $,
where $P$ is a column-strict $F_{\operatorname*{rows}\leq k}$-tableau, and
where $Q$ is a column-strict $F_{\operatorname*{rows}>k}$-tableau. Consider
these $P$ and $Q$.

Recall that $P$ is a column-strict $F_{\operatorname*{rows}\leq k}$-tableau.
In other words, $P$ is an assignment of entries in $\left\{
1,2,3,...\right\}  $ to the elements of $F_{\operatorname*{rows}\leq k}$ which
is weakly increasing left-to-right in rows and strictly increasing
top-to-bottom in columns (by the definition of a column-strict
$F_{\operatorname*{rows}\leq k}$-tableau). Similarly, $Q$ is an assignment of
entries in $\left\{  1,2,3,...\right\}  $ to the elements of
$F_{\operatorname*{rows}>k}$ which is weakly increasing left-to-right in rows
and strictly increasing top-to-bottom in columns (by the definition of a
column-strict $F_{\operatorname*{rows}>k}$-tableau).

We now can define a map $T:F\rightarrow\left\{  1,2,3,...\right\}  $ by
setting
\[
\left(  T\left(  p\right)  = \begin{cases}
P\left(  p\right)  ,& \text{if }p\in F_{\operatorname*{rows} \leq k};\\
Q\left(  p\right)  ,& \text{if }p\in F_{\operatorname*{rows} >k}
\end{cases}
\qquad \text{ for all } p \in F \right)
\]
\footnote{This is well-defined, because the two subsets
$F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$ of $F$ are
disjoint and their union is $F$.}. Consider this map $T$. This map $T$ is an
assignment of entries in $\left\{  1,2,3,...\right\}  $ to the elements of
$F$. In other words, this map $T$ is an assignment of entries in $\left\{
1,2,3,...\right\}  $ to the cells of the skew Ferrers diagram $\lambda/\mu$
(since the cells of the skew Ferrers diagram $\lambda/\mu$ are exactly the
elements of $F$). We shall now prove the following two assertions:

\begin{statement}
\textit{Assertion B:} The map $T$ is weakly increasing left-to-right in rows.
\end{statement}

\begin{statement}
\textit{Assertion C:} The map $T$ is strictly increasing top-to-bottom in columns.
\end{statement}

\textit{Proof of Assertion B:} Let $c$ and $d$ be two cells lying in one and
the same row of $T$, with $d$ lying strictly further east than $c$. We shall
prove that $T\left(  c\right)  \leq T\left(  d\right)  $.

The cells $c$ and $d$ are cells of $T$, and thus lie in $F$ (since the domain
of the map $T$ is $F$). We know that the cells $c$ and $d$ either lie in one
and the same row, or lie in one and the same column (because we know that the
cells $c$ and $d$ lie in one and the same row). Hence, Assertion A yields that
either both $c$ and $d$ belong to $F_{\operatorname*{rows}\leq k}$, or both
$c$ and $d$ belong to $F_{\operatorname*{rows}>k}$. We must therefore be in
one of the following two cases:

\textit{Case 1:} Both $c$ and $d$ belong to $F_{\operatorname*{rows}\leq k}$.

\textit{Case 2:} Both $c$ and $d$ belong to $F_{\operatorname*{rows}>k}$.

Let us consider Case 1 first. In this case, both $c$ and $d$ belong to
$F_{\operatorname*{rows}\leq k}$. Hence, $c\in F_{\operatorname*{rows}\leq k}$
and $d\in F_{\operatorname*{rows}\leq k}$. Thus, $c$ and $d$ are cells of $P$
(since the domain of the map $P$ is $F_{\operatorname*{rows}\leq k}$).
Therefore, $c$ and $d$ are two cells lying in one and the same row of $P$,
with $d$ lying strictly further east than $c$. Consequently, $P\left(
c\right)  \leq P\left(  d\right)  $ (since the map $P$ is weakly increasing
left-to-right in rows). Now, the definition of $T\left(  c\right)  $ yields
$T\left(  c\right)  = \begin{cases}
P\left(  c\right)  ,& \text{if }c\in F_{\operatorname*{rows} \leq k};\\
Q\left(  c\right)  ,& \text{if }c\in F_{\operatorname*{rows} >k}
\end{cases} \quad  =P\left(  c\right)  $
(since $c\in F_{\operatorname*{rows}\leq k}$).
Also, the definition of $T\left(  d\right)  $ yields $T\left(  d\right)
= \begin{cases}
P\left(  d\right)  ,& \text{if }d\in F_{\operatorname*{rows} \leq k};\\
Q\left(  d\right)  ,& \text{if }d\in F_{\operatorname*{rows} >k}%
\end{cases} \quad
=P\left(  d\right)  $ (since $d\in F_{\operatorname*{rows}\leq k}$),
so that $P\left(  d\right)  =T\left(  d\right)  $. Now, $T\left(  c\right)
=P\left(  c\right)  \leq P\left(  d\right)  =T\left(  d\right)  $. Thus,
$T\left(  c\right)  \leq T\left(  d\right)  $ is proven in Case 1.

The same argument (but with $P$ and $F_{\operatorname*{rows}\leq k}$ replaced
by $Q$ and $F_{\operatorname*{rows}>k}$) shows that $T\left(  c\right)  \leq
T\left(  d\right)  $ holds in Case 2.

We thus have proven that $T\left(  c\right)  \leq T\left(  d\right)  $ in each
of the two Cases 1 and 2. Since these two Cases cover all possibilities, we
thus conclude that $T\left(  c\right)  \leq T\left(  d\right)  $ always holds.

Now, let us forget that we fixed $c$ and $d$. We thus have shown that if $c$
and $d$ are two cells lying in one and the same row of $T$, with $d$ lying
strictly further east than $c$, then we have $T\left(  c\right)  \leq T\left(
d\right)  $. In other words, the map $T$ is weakly increasing left-to-right in
rows. This proves Assertion B.

The \textit{proof of Assertion C} can be obtained from the above proof of
Assertion B by replacing ``weakly'',
``left-to-right'', ``row'', ``east'' and ``$\leq$'' by ``strictly'',
``top-to-bottom'', ``column'', ``south'' and ``$<$'', respectively.

Now, both Assertions B and C are proven. It follows that $T$ is an assignment
of entries in $\left\{  1,2,3,...\right\}  $ to the cells of the skew Ferrers
diagram $\lambda/\mu$ which is weakly increasing left-to-right in rows, and
strictly increasing top-to-bottom in columns. In other words, $T$ is a
column-strict tableau of shape $\lambda/\mu$ (according to the definition of a
column-strict tableau of shape $\lambda/\mu$). We shall now prove that
$\Phi\left(  T\right)  =U$.

Indeed, it is easy to see that $T\mid_{F_{\operatorname*{rows}\leq k}}%
=P$\ \ \ \ \footnote{This is because every $c\in F_{\operatorname*{rows}\leq
k}$ satisfies%
\begin{align*}
\left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)  \left(  c\right)   &
=T\left(  c\right)  = \begin{cases}
P\left(  c\right)  ,& \text{if }c\in F_{\operatorname*{rows} \leq k};\\
Q\left(  c\right)  ,& \text{if }c\in F_{\operatorname*{rows} >k}
\end{cases}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }T\left(
c\right)  \right) \\
&  =P\left(  c\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }c\in
F_{\operatorname*{rows}\leq k}\right)  .
\end{align*}
} and $T\mid_{F_{\operatorname*{rows}>k}}=Q$\ \ \ \ \footnote{This can be
proven in the same way as we have proven $T\mid_{F_{\operatorname*{rows}\leq
k}}=P$, with the only difference that $F_{\operatorname*{rows}\leq k}$ and $P$
must be replaced by $F_{\operatorname*{rows}>k}$ and $Q$.}. Now, the
definition of $\Phi\left(  T\right)  $ yields $\Phi\left(  T\right)  =\left(
\underbrace{T\mid_{F_{\operatorname*{rows}\leq k}}}_{=P},\underbrace{T\mid
_{F_{\operatorname*{rows}>k}}}_{=Q}\right)  =\left(  P,Q\right)  =U$ (since
$U=\left(  P,Q\right)  $). Thus, $U=\Phi\left(  T\right)  $ lies in the image
of $\Phi$.

Now, let us forget that we fixed $U$. We thus have proven that every%
\[
U\in\left(  \text{the set of all column-strict }F_{\operatorname*{rows}\leq
k}\text{-tableaux}\right)  \times\left(  \text{the set of all column-strict
}F_{\operatorname*{rows}>k}\text{-tableaux}\right)
\]
lies in the image of $\Phi$. Hence, the map $\Phi$ is surjective. Combined
with the fact that $\Phi$ is injective, this yields that $\Phi$ is bijective.
Thus, the map $\Phi$ is a bijection%
\begin{align*}
&  \left(  \text{the set of all column-strict tableaux of shape }\lambda
/\mu\right) \\
&  \rightarrow\left(  \text{the set of all column-strict }%
F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)  \times\left(  \text{the
set of all column-strict }F_{\operatorname*{rows}>k}\text{-tableaux}\right)  .
\end{align*}


Next, let us define a monomial $\mathbf{x}^{\cont \left(
U\right)  }$ for every $U\in\left(  \text{the set of all column-strict
}F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)  \times\left(
\text{the set of all column-strict }F_{\operatorname*{rows}>k}\text{-tableaux}%
\right)  $ as follows: We set $\mathbf{x}^{\cont \left(
U\right)  }=\mathbf{x}^{\cont \left(  P\right)  }\xx%
^{\cont \left(  Q\right)  }$, where $U$ is written in the form
$U=\left(  P,Q\right)  $ for $P$ is a column-strict $F_{\operatorname*{rows}%
\leq k}$-tableau, and where $Q$ is a column-strict $F_{\operatorname*{rows}%
>k}$-tableau.

We can now easily check that%
\begin{equation}
\mathbf{x}^{\cont \left(  \Phi\left(  T\right)  \right)
}=\mathbf{x}^{\cont \left(  T\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every column-strict tableau }T\text{ of shape
}\lambda/\mu. \label{sol.schur.disconnected.cont=cont}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.schur.disconnected.cont=cont}):} Let $T$
be a column-strict tableau of shape $\lambda/\mu$. By the definition of
$\Phi\left(  T\right)  $, we have $\Phi\left(  T\right)  =\left(
T\mid_{F_{\operatorname*{rows}\leq k}},T\mid_{F_{\operatorname*{rows}>k}%
}\right)  $. Hence, the definition of $\mathbf{x}^{\cont \left(
\Phi\left(  T\right)  \right)  }$ yields
\begin{align}
\mathbf{x}^{\cont \left(  \Phi\left(  T\right)  \right)  }  &
=\underbrace{\mathbf{x}^{\cont \left(  T\mid
_{F_{\operatorname*{rows}\leq k}}\right)  }}_{\substack{=\prod_{i}%
x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
^{-1}\left(  i\right)  \right\vert }\\\text{(by the definition of }%
\mathbf{x}^{\cont \left(  T\mid_{F_{\operatorname*{rows}\leq k}%
}\right)  }\text{)}}}\underbrace{\mathbf{x}^{\cont \left(
T\mid_{F_{\operatorname*{rows}>k}}\right)  }}_{\substack{=\prod_{i}%
x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}>k}}\right)
^{-1}\left(  i\right)  \right\vert }\\\text{(by the definition of }%
\mathbf{x}^{\cont \left(  T\mid_{F_{\operatorname*{rows}>k}%
}\right)  }\text{)}}}\nonumber\\
&  =\left(  \prod_{i}x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}%
\leq k}}\right)  ^{-1}\left(  i\right)  \right\vert }\right)  \left(
\prod_{i}x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}>k}}\right)
^{-1}\left(  i\right)  \right\vert }\right)  =\prod_{i}\underbrace{x_{i}%
^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
^{-1}\left(  i\right)  \right\vert }x_{i}^{\left\vert \left(  T\mid
_{F_{\operatorname*{rows}>k}}\right)  ^{-1}\left(  i\right)  \right\vert }%
}_{=x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
^{-1}\left(  i\right)  \right\vert +\left\vert \left(  T\mid
_{F_{\operatorname*{rows}>k}}\right)  ^{-1}\left(  i\right)  \right\vert }%
}\nonumber\\
&  =\prod_{i}x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}\leq k}%
}\right)  ^{-1}\left(  i\right)  \right\vert +\left\vert \left(
T\mid_{F_{\operatorname*{rows}>k}}\right)  ^{-1}\left(  i\right)  \right\vert
}. \label{sol.schur.disconnected.cont=cont.pf.1}%
\end{align}
\par
Now, fix any $i\in\left\{  1,2,3,...\right\}  $. Then, recall that the two
subsets $F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}>k}$ of
$F$ are disjoint and their union is $F$. In other words,
$F_{\operatorname*{rows}\leq k}\cap F_{\operatorname*{rows}>k}=\varnothing$
and $F_{\operatorname*{rows}\leq k}\cup F_{\operatorname*{rows}>k}=F$. Thus,
the two sets $\left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
^{-1}\left(  i\right)  $ and $\left(  T\mid_{F_{\operatorname*{rows}>k}%
}\right)  ^{-1}\left(  i\right)  $ are disjoint (since
\begin{align*}
&  \underbrace{\left(  \left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
^{-1}\left(  i\right)  \right)  }_{=F_{\operatorname*{rows}\leq k}\cap
T^{-1}\left(  i\right)  }\cap\underbrace{\left(  \left(  T\mid
_{F_{\operatorname*{rows}>k}}\right)  ^{-1}\left(  i\right)  \right)
}_{=F_{\operatorname*{rows}>k}\cap T^{-1}\left(  i\right)  }\\
&  =F_{\operatorname*{rows}\leq k}\cap T^{-1}\left(  i\right)  \cap
F_{\operatorname*{rows}>k}\cap T^{-1}\left(  i\right)
=\underbrace{F_{\operatorname*{rows}\leq k}\cap F_{\operatorname*{rows}>k}%
}_{=\varnothing}\cap T^{-1}\left(  i\right)  \cap T^{-1}\left(  i\right)
=\varnothing\cap T^{-1}\left(  i\right)  \cap T^{-1}\left(  i\right)
=\varnothing
\end{align*}
) and their union is $T^{-1}\left(  i\right)  $ (since%
\begin{align*}
&  \underbrace{\left(  \left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)
^{-1}\left(  i\right)  \right)  }_{=F_{\operatorname*{rows}\leq k}\cap
T^{-1}\left(  i\right)  }\cup\underbrace{\left(  \left(  T\mid
_{F_{\operatorname*{rows}>k}}\right)  ^{-1}\left(  i\right)  \right)
}_{=F_{\operatorname*{rows}>k}\cap T^{-1}\left(  i\right)  }\\
&  =\left(  F_{\operatorname*{rows}\leq k}\cap T^{-1}\left(  i\right)
\right)  \cup\left(  F_{\operatorname*{rows}>k}\cap T^{-1}\left(  i\right)
\right)  =\left(  \underbrace{F_{\operatorname*{rows}\leq k}\cup
F_{\operatorname*{rows}>k}}_{=F}\right)  \cap T^{-1}\left(  i\right)  =F\cap
T^{-1}\left(  i\right)  =T^{-1}\left(  i\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }T^{-1}\left(  i\right)  \subset F\right)
\end{align*}
). Hence, the sum of the cardinalities of these two subsets is the cardinality
of $T^{-1}\left(  i\right)  $. In other words,
\begin{equation}
\left\vert \left(  T\mid_{F_{\operatorname*{rows}\leq k}}\right)  ^{-1}\left(
i\right)  \right\vert +\left\vert \left(  T\mid_{F_{\operatorname*{rows}>k}%
}\right)  ^{-1}\left(  i\right)  \right\vert =\left\vert T^{-1}\left(
i\right)  \right\vert . \label{sol.schur.disconnected.cont=cont.pf.3}%
\end{equation}
\par
Now, forget that we fixed $i$. We thus have shown that every $i\in\left\{
1,2,3,...\right\}  $ satisfies (\ref{sol.schur.disconnected.cont=cont.pf.3}).
Now, (\ref{sol.schur.disconnected.cont=cont.pf.1}) becomes%
\begin{equation}
\mathbf{x}^{\cont \left(  \Phi\left(  T\right)  \right)  }%
=\prod_{i}\underbrace{x_{i}^{\left\vert \left(  T\mid_{F_{\operatorname*{rows}%
\leq k}}\right)  ^{-1}\left(  i\right)  \right\vert +\left\vert \left(
T\mid_{F_{\operatorname*{rows}>k}}\right)  ^{-1}\left(  i\right)  \right\vert
}}_{\substack{=x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert
}\\\text{(by (\ref{sol.schur.disconnected.cont=cont.pf.3}))}}}=\prod_{i}%
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }=\xx%
^{\cont \left(  T\right)  }\nonumber
\end{equation}
(since $\mathbf{x}^{\cont \left(  T\right)  }=\prod_{i}%
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }$ (by the definition of
$\mathbf{x}^{\cont \left(  T\right)  }$)). This proves
(\ref{sol.schur.disconnected.cont=cont}).}

Now, Lemma \ref{lem.sol.schur.disconnected.trans} (applied to $\alpha$,
$\beta$ and $F_{\operatorname{rows}\leq k}$ instead of $\varphi$, $\psi$ and
$Z$) yields%
\begin{equation}
s_{\alpha/\beta}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\xx%
^{\cont \left(  T\right)  }=\sum\limits_{\substack{P\text{ is a
column-strict}\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\xx%
^{\cont \left(  P\right)  }
\label{sol.schur.disconnected.factor1}%
\end{equation}
(here, we renamed the summation index $T$ as $P$). Also, Lemma
\ref{lem.sol.schur.disconnected.trans} (applied to $\gamma$, $\delta$ and
$F_{\operatorname{rows}>k}$ instead of $\varphi$, $\psi$ and $Z$) yields%
\begin{equation}
s_{\gamma/\delta}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\F_{\operatorname{rows}>k}\text{-tableau}}}\mathbf{x}^{\cont
\left(  T\right)  }=\sum\limits_{\substack{Q\text{ is a column-strict}%
\\F_{\operatorname{rows}>k}\text{-tableau}}}\mathbf{x}^{\cont
\left(  Q\right)  } \label{sol.schur.disconnected.factor2}%
\end{equation}
(here, we renamed the summation index $T$ as $Q$). Multiplying the identities
(\ref{sol.schur.disconnected.factor1}) and
(\ref{sol.schur.disconnected.factor2}), we obtain%
\begin{align}
s_{\alpha/\beta}s_{\gamma/\delta}  &  =\left(  \sum\limits_{\substack{P\text{ is a
column-strict}\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\xx%
^{\cont \left(  P\right)  }\right)  \left(  \sum\limits
_{\substack{Q\text{ is a column-strict}\\F_{\operatorname{rows}>k}%
\text{-tableau}}}\mathbf{x}^{\cont \left(  Q\right)  }\right)
\nonumber\\
&  =\underbrace{\sum\limits_{\substack{P\text{ is a column-strict}%
\\F_{\operatorname{rows}\leq k}\text{-tableau}}}\sum\limits_{\substack{Q\text{ is a
column-strict}\\F_{\operatorname{rows}>k}\text{-tableau}}}}_{=\sum\limits
_{\substack{\left(  P,Q\right)  \in\\\left(  \text{the set of all
column-strict }F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)
\\\times\left(  \text{the set of all column-strict }F_{\operatorname*{rows}%
>k}\text{-tableaux}\right)  }}}\underbrace{\mathbf{x}^{\cont
\left(  P\right)  }\mathbf{x}^{\cont \left(  Q\right)  }%
}_{\substack{=\mathbf{x}^{\cont \left(  \left(  P,Q\right)
\right)  }\\\text{(since }\mathbf{x}^{\cont \left(  \left(
P,Q\right)  \right)  }=\mathbf{x}^{\cont \left(  P\right)
}\mathbf{x}^{\cont \left(  Q\right)  }\\\text{(by the definition
of }\mathbf{x}^{\cont \left(  \left(  P,Q\right)  \right)
}\text{))}}}\nonumber\\
&  =\sum\limits_{\substack{\left(  P,Q\right)  \in\\\left(  \text{the set of all
column-strict }F_{\operatorname*{rows}\leq k}\text{-tableaux}\right)
\\\times\left(  \text{the set of all column-strict }F_{\operatorname*{rows}%
>k}\text{-tableaux}\right)  }}\mathbf{x}^{\cont \left(  \left(
P,Q\right)  \right)  }\nonumber\\
&  =\underbrace{\sum\limits_{\substack{T\in\\\left(  \text{the set of all
column-strict tableaux of shape }\lambda/\mu\right)  }}}_{=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda/\mu
}}}\mathbf{x}^{\cont \left(  \Phi\left(  T\right)  \right)
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\Phi\left(  T\right)  \text{ for }\left(
P,Q\right)  \text{ in the sum}\\
\text{(since the map }\Phi\text{ is a bijection)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu}}\underbrace{\mathbf{x}^{\cont \left(  \Phi\left(
T\right)  \right)  }}_{\substack{=\mathbf{x}^{\cont \left(
T\right)  }\\\text{(by (\ref{sol.schur.disconnected.cont=cont}))}}%
}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\mu}}\mathbf{x}^{\cont \left(  T\right)  }.
\label{sol.schur.disconnected.almostthere}%
\end{align}


But the definition of $s_{\lambda/\mu}$ yields $s_{\lambda/\mu}=\sum
_{T}\mathbf{x}^{\cont \left(  T\right)  }$, where the sum runs
over all column-strict tableaux of shape $\lambda/\mu$. In other words,%
\[
s_{\lambda/\mu}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\mu}}\mathbf{x}^{\cont \left(  T\right)  }.
\]
Compared with (\ref{sol.schur.disconnected.almostthere}), this yields
$s_{\lambda/\mu}=s_{\alpha/\beta}s_{\gamma/\delta}$. This solves
Exercise~\ref{exe.schur.disconnected}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.cocommutative}}

\textit{Solution to Exercise~\ref{exe.Lambda.cocommutative}.} As usual,
let $T$ denote the twist map $\Lambda\otimes\Lambda\rightarrow\Lambda
\otimes\Lambda$ (that is, the $\kk$-linear map sending every $c\otimes
d\in\Lambda\otimes\Lambda$ to $d\otimes c$). By the definition of
``cocommutative'', we know that the Hopf
algebra $\Lambda$ is cocommutative if and only if the diagram
\begin{equation}
\xymatrix{
\Lambda\otimes\Lambda\ar[rr]^{T} & & \Lambda\otimes\Lambda\\
& \Lambda\ar[ru]_{\Delta} \ar[lu]^{\Delta} &
}
\label{sol.Lambda.cocommutative.diagram}
\end{equation}
commutes. Hence, in order to solve Exercise~\ref{exe.Lambda.cocommutative}(a),
it is enough to check that the diagram
\eqref{sol.Lambda.cocommutative.diagram} commutes.

The set $\{h_{n}\}_{n=1,2,\ldots}$ generates the $\kk$-algebra
$\Lambda$ (due to Proposition~\ref{symm-antipode-and-h-basis}). In other
words, the set $\left\{  h_{1},h_{2},h_{3},...\right\}  $ is a generating set
of the $\kk$-algebra $\Lambda$.

By the axioms of a bialgebra, the comultiplication $\Delta$ of $\Lambda$ is a
$\kk$-algebra homomorphism (since $\Lambda$ is a bialgebra). Hence,
$T\circ\Delta$ also is a $\kk$-algebra homomorphism (since $T$ and
$\Delta$ are $\kk$-algebra homomorphisms). Every positive integer $n$
satisfies
\begin{align*}
\left(  T\circ\Delta\right)  \left(  h_{n}\right)    & =T\left(
\underbrace{\Delta h_{n}}_{\substack{=\sum_{i+j=n}h_{i}\otimes h_{j}
\\\text{(by Proposition \ref{symm-comultiplication-formulas-prop}(iii))}}}
\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left( \begin{array}{cc}
\text{here, we are using the notation }
\sum_{i+j=n}h_{i}\otimes h_{j} \\
\text{ in the same way as explained in
Proposition~\ref{symm-comultiplication-formulas-prop}}
\end{array} \right)  \\
& =T\left(  \sum_{i+j=n}h_{i}\otimes h_{j}\right)  =\sum_{i+j=n}h_{j}\otimes
h_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the twist map
}T\right)  \\
& =\underbrace{\sum_{j+i=n}}_{=\sum_{i+j=n}}h_{i}\otimes h_{j}
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  i,j\right)  \text{ as }\left(  j,i\right)  \right)  \\
& =\sum_{i+j=n}h_{i}\otimes h_{j}=\Delta h_{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{symm-comultiplication-formulas-prop}(iii)}\right)
\\
& =\Delta\left(  h_{n}\right)  .
\end{align*}
In other words, for every positive integer $n$, the two maps $T\circ\Delta$
and $\Delta$ are equal to each other on the element $h_n$. In other words,
the two maps $T\circ\Delta$ and $\Delta$ are equal to each other on the set
$\left\{  h_{1},h_{2},h_{3},...\right\}  $. Hence, the two maps $T\circ\Delta$
and $\Delta$ are equal to each other on a generating set of the $\kk$-algebra
$\Lambda$ (since the set $\left\{  h_{1},h_{2},h_{3},...\right\}  $
is a generating set of the $\kk$-algebra $\Lambda$). Since these two
maps $T\circ\Delta$ and $\Delta$ are $\kk$-algebra homomorphisms, this
shows that the two maps $T\circ\Delta$ and $\Delta$ must be identical (because
if two $\kk$-algebra homomorphisms with the same domain and the same
target are equal to each other on a generating set of their domain, then these
two homomorphisms must be identical). In other words, $T\circ\Delta=\Delta$.
Hence, the diagram \eqref{sol.Lambda.cocommutative.diagram} commutes. As we
know, this shows that the Hopf algebra $\Lambda$ is cocommutative. This solves
Exercise~\ref{exe.Lambda.cocommutative}(a).

(b) Let $\lambda$ and $\nu$ be two partitions. We have shown above that
$T\circ\Delta=\Delta$. Hence, $\Delta=T\circ\Delta$. Applying both sides of
this equality to $s_{\lambda/\nu}$, we obtain
\begin{align*}
\Delta s_{\lambda/\nu}  & =\left(  T\circ\Delta\right)  \left(  s_{\lambda
/\nu}\right)  =T\left(  \underbrace{\Delta s_{\lambda/\nu}}_{\substack{=\sum\limits
_{\substack{\mu\in\Par :\\\nu\subseteq\mu\subseteq\lambda}
}s_{\mu/\nu}\otimes s_{\lambda/\mu}\\\text{(by
Proposition \ref{symm-comultiplication-formulas-prop}(v))}}}\right) 
=T\left( \sum\limits_{\substack{\mu\in\Par :\\\nu\subseteq\mu\subseteq\lambda}}
s_{\mu/\nu}\otimes s_{\lambda/\mu}\right)  \\
& =\sum\limits_{\substack{\mu\in\Par :\\\nu\subseteq\mu\subseteq\lambda
}}s_{\lambda/\mu}\otimes s_{\mu/\nu}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of the twist map }T\right)  .
\end{align*}
This solves Exercise~\ref{exe.Lambda.cocommutative}(b).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.schur.finite}}

\textit{Solution to Exercise~\ref{exe.schur.finite}.} (a) Whenever
$\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)  $ is a weak
composition satisfying $\left(  \alpha_{i}=0\text{ for every }i>n\right)  $,
the monomial $\xx^\alpha$ is a monomial in $\kk\left[x_1,x_2,...,x_n\right]$.
This will be often used in the following.
%let us write $\left(  x_{1},x_{2},\ldots,x_{n}\right)  ^{\alpha}$ for the
%monomial $\prod_{i=1}^{n}x_{i}^{\alpha_{i}}$. (This is the same monomial as
%$\mathbf{x}^{\alpha}$, but in $\kk \left[  x_{1},x_{2},...,x_{n}\right]
%$ instead of $\kk \left[  \left[  \xx\right]  \right]  $.) This
%generalizes the notation $\left(  x_{1},x_{2},\ldots,x_{n}\right)
%^{\cont\left(  T\right)  }$ introduced in the exercise.

Recall that $s_{\lambda/\mu}$ is defined as $\sum\limits_{\substack{T\text{ is
a column-strict}\\\text{tableau of shape }\lambda/\mu}}\xx%
^{\cont\left(  T\right)  }$. Now, $s_{\lambda/\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right)  $ is the result of substituting $x_{1}%
,x_{2},...,x_{n},0,0,0,...$ for $x_{1},x_{2},x_{3},...$ in $s_{\lambda/\mu}$.
This substitution has the following effect on any given monomial
$\mathbf{x}^{\alpha}$:

\begin{itemize}
\item if none of the indeterminates $x_{n+1},x_{n+2},x_{n+3},...$ occur in
this monomial $\mathbf{x}^{\alpha}$, then the monomial $\mathbf{x}^{\alpha}$
stays fixed;

\item otherwise, the monomial $\mathbf{x}^{\alpha}$ goes to $0$.
\end{itemize}

\noindent
Hence, the effect of this substitution on the power series $s_{\lambda/\mu
}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu}}\mathbf{x}^{\cont\left(  T\right)  }$ is that:

\begin{itemize}
\item every addend in the sum $\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu}}\xx%
^{ \cont \left(  T\right)  }$ for which none of the
indeterminates $x_{n+1},x_{n+2},x_{n+3},...$ occur in the monomial
$\mathbf{x}^{\cont\left(  T\right)  }$ stays fixed;

\item all other addends go to $0$.
\end{itemize}

\noindent
The result of the substitution is therefore $\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu\text{;}\\\text{none of the
indeterminates }x_{n+1},x_{n+2},x_{n+3},...\\\text{occur in the monomial
}\mathbf{x}^{ \cont \left(  T\right)  }}}
\xx^{ \cont \left(  T\right)  }$. We thus have
\[
s_{\lambda/\mu}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
/\mu\text{;}\\\text{none of the indeterminates }x_{n+1},x_{n+2},x_{n+3}%
,...\\\text{occur in the monomial }\mathbf{x}^{ \cont \left(
T\right)  }}}\xx ^{ \cont \left(  T\right)  }.
\]
But this rewrites as
\[
s_{\lambda/\mu}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
/\mu\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,n\right\}  }} \xx^{ \cont \left(  T\right)  }
\]
(because for a column-strict tableau $T$, saying that none of the
indeterminates $x_{n+1},x_{n+2},x_{n+3},...$ occur in the monomial
$\mathbf{x}^{ \cont \left(  T\right)  }$ is equivalent to saying
that all entries of $T$ belong to $\left\{  1,2,...,n\right\}  $). This solves
Exercise \ref{exe.schur.finite}(a).

(b) Let $\lambda$ be a partition having more than $n$ parts. We have to prove
that $s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =0$. Since%
\[
\underbrace{s_{\lambda}}_{=s_{\lambda/\varnothing}}\left(  x_{1},x_{2}%
,\ldots,x_{n}\right)  =s_{\lambda/\varnothing}\left(  x_{1},x_{2},\ldots
,x_{n}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\varnothing\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,n\right\}  }}
\xx^{ \cont \left(  T\right)  }
\]
(by Exercise \ref{exe.schur.finite}(a), applied to $\mu=\varnothing$), this
goal will clearly be achieved if we can show that the sum $\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,n\right\}  }}\xx^{ \cont \left(  T\right)  }$
is empty, i.e., that there exists no
column-strict tableau $T$ of shape $\lambda/\varnothing$ such that all entries
of $T$ belong to $\left\{  1,2,...,n\right\}  $.

Assume the contrary. Thus, there exists a column-strict tableau $T$ of shape
$\lambda/\varnothing$ such that all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $. This tableau has more than $n$ rows (since the partition
$\lambda$ has more than $n$ parts), and thus the first column of this tableau
must have more than $n$ entries. These entries must be strictly increasing
top-to-bottom (since the entries of a column-strict tableau strictly increase
top-to-bottom along columns) and hence be distinct, but at the same
time (like all entries of $T$) they must belong to $\left\{
1,2,...,n\right\}  $. So we have found more than $n$ entries which are
distinct and belong to $\left\{  1,2,...,n\right\}  $. This
contradicts the fact that the set $\left\{  1,2,...,n\right\}  $ does not have
more than $n$ distinct elements. This contradiction concludes our
proof, and Exercise \ref{exe.schur.finite}(b) is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.schur.finite}}

\textit{Solution to Exercise~\ref{exe.schur.finite}.} (a) Let us
consider both $\kk$-algebras $\kk \left[ \left[ \xx \right] \right]$ and
$\kk \left[ \left[ x_{1},x_{2},...,x_{n} \right] \right]$
as topological $\kk$-algebras, endowed with the
usual topology\footnote{The usual topology on a ring $U$ of formal power series
over $\kk$ is the direct-product topology obtained by viewing the set
$U$ as a direct product of many copies of $\kk$ (this is done by
identifying every power series with the family of its coefficients), each of
which is endowed with the discrete topology. In this topology, a sequence (or,
more generally, a net) $\left( f_i \right) _i$ of power series converges
to a power series $f$ if and only if for every monomial $\mathfrak{m}$, the
sequence (resp. net)
$\left( \text{the coefficient of } \mathfrak{m} \text{ in } f_i \right) _i$
converges to
$\left( \text{the coefficient of } \mathfrak{m} \text{ in } f\right)$
with respect to the discrete topology.
(The notion of a \emph{net} is a generalization of the notion of a
sequence; it is useful in topology. See \cite{Vermeeren-nets} for an
introduction to it.)}.
Then, there exists a unique continuous $\kk$-algebra homomorphism
$\Gamma: \kk \left[  \left[ \xx \right]  \right]  \rightarrow
\kk \left[  \left[  x_{1},x_{2},...,x_{n}\right]  \right]  $ satisfying
\begin{equation}
\left(  \Gamma\left(  x_{i}\right)  = \begin{cases}
x_{i},& \text{if }i\leq n;\\
0,& \text{if }i>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,3,...\right\}
\right)  .
\label{sol.schur.finite.a.Gamma.2}
\end{equation}
\footnote{This follows from the universal property of the power series ring
$\kk \left[  \left[ \xx \right]  \right]  $.} Consider this
continuous $\kk$-algebra homomorphism $\Gamma$.

Every $i\in\left\{  1,2,...,n\right\}  $ satisfies
\begin{align}
\Gamma\left(  x_{i}\right)   &  = \begin{cases}
x_{i},& \text{if }i\leq n;\\
0,& \text{if }i>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.2}}\right)  \nonumber\\
&  =x_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq n\text{ (since }
i\in\left\{  1,2,...,n\right\}  \text{)}\right)
.
\label{sol.schur.finite.a.Gamma.2a}
\end{align}
Every $i\in\left\{  n+1,n+2,n+3,...\right\}  $ satisfies
\begin{align}
\Gamma\left(  x_{i}\right)   &  = \begin{cases}
x_{i},& \text{if }i\leq n;\\
0,& \text{if }i>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.2}}\right)  \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>n\text{ (since }i\in\left\{
n+1,n+2,n+3,...\right\}  \text{)}\right)  .
\label{sol.schur.finite.a.Gamma.2b}
\end{align}


It is easy to see that
\begin{equation}
\Gamma\left(  f\right)  =f\left(  x_{1},x_{2},...,x_{n}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in R\left(  \xx \right)  .
\label{sol.schur.finite.a.Gamma.=subst}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.schur.finite.a.Gamma.=subst}:} Let $f\in
R\left(  \xx \right)  $. Then, we can write $f$ in the form
$f=\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)
\\\text{is a weak composition}}}\lambda_{\left(  \alpha_{1},\alpha_{2}
,\alpha_{3},...\right)  }\xx ^{\left(  \alpha_{1},\alpha_{2},\alpha
_{3},...\right)  }$, where $\left(  \lambda_{\left(  \alpha_{1},\alpha
_{2},\alpha_{3},...\right)  }\right)  _{\left(  \alpha_{1},\alpha_{2}
,\alpha_{3},...\right)  \text{ is a weak composition}}$ is a family of
elements of $\kk$ indexed by the weak compositions. Consider this
family $\left(  \lambda_{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)
}\right)  _{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)  \text{ is a
weak composition}}$.
\par
Now,
\[
f=\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)
\\\text{is a weak composition}}}\lambda_{\left(  \alpha_{1},\alpha_{2}
,\alpha_{3},...\right)  }\underbrace{\xx ^{\left(  \alpha_{1},\alpha
_{2},\alpha_{3},...\right)  }}_{=\prod_{i=1}^{\infty}x_{i}^{\alpha_{i}}}
=\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)
\\\text{is a weak composition}}}\lambda_{\left(  \alpha_{1},\alpha_{2}
,\alpha_{3},...\right)  }\prod_{i=1}^{\infty}x_{i}^{\alpha_{i}}.
\]
Applying the map $\Gamma$ to both sides of this equality, we obtain
\begin{equation}
\Gamma\left(  f\right)  =\Gamma\left(  \sum\limits_{\substack{\left(
\alpha_{1},\alpha_{2},\alpha_{3},...\right)  \\
\text{is a weak composition}}}
\lambda_{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)  }\prod
_{i=1}^{\infty}x_{i}^{\alpha_{i}}\right)  =\sum\limits_{\substack{\left(
\alpha_{1},\alpha_{2},\alpha_{3},...\right)  \\\text{is a weak composition}%
}}\lambda_{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)  }\prod
_{i=1}^{\infty}\left(  \Gamma\left(  x_{i}\right)  \right)  ^{\alpha_{i}}
\label{sol.schur.finite.a.Gamma.=subst.pf.1}%
\end{equation}
(since $\Gamma$ is a continuous $\kk$-algebra homomorphism).
\par
However, for every $i\in\left\{  1,2,3,...\right\}  $, we have
\begin{align*}
&  \left(  \text{the }i\text{-th entry of the sequence }\left(  x_{1}%
,x_{2},...,x_{n},0,0,0,...\right)  \right) \\
&  = \begin{cases}
x_{i},& \text{if }i\leq n;\\
0,& \text{if }i>n
\end{cases} \quad
=\Gamma\left(  x_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.2}}\right) \\
&  =\left(  \text{the }i\text{-th entry of the sequence }\left(  \Gamma\left(
x_{1}\right)  ,\Gamma\left(  x_{2}\right)  ,\Gamma\left(  x_{3}\right)
,...\right)  \right)  .
\end{align*}
Thus, the sequences $\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)  $ and
$\left(  \Gamma\left(  x_{1}\right)  ,\Gamma\left(  x_{2}\right)
,\Gamma\left(  x_{3}\right)  ,...\right)  $ are equal to each other in every
term. Thus,%
\[
\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)  =\left(  \Gamma\left(
x_{1}\right)  ,\Gamma\left(  x_{2}\right)  ,\Gamma\left(  x_{3}\right)
,...\right)  .
\]
\par
On the other hand, the definition of $f\left(  x_{1},x_{2},...,x_{n}\right)  $
yields that $f\left(  x_{1},x_{2},...,x_{n}\right)  $ is the result of
substituting $x_{1},x_{2},...,x_{n},0,0,0,...$ for $x_{1},x_{2},x_{3},\ldots$
in $f$. In other words, $f\left(  x_{1},x_{2},...,x_{n}\right)  $ is the
result of substituting $\Gamma\left(  x_{1}\right)  ,\Gamma\left(
x_{2}\right)  ,\Gamma\left(  x_{3}\right)  ,...$ for $x_{1},x_{2},x_{3}%
,\ldots$ in $f$ (since $\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)
=\left(  \Gamma\left(  x_{1}\right)  ,\Gamma\left(  x_{2}\right)
,\Gamma\left(  x_{3}\right)  ,...\right)  $). In other words,%
\begin{align*}
&  f\left(  x_{1},x_{2},...,x_{n}\right) \\
&  =\left(  \text{the result of substituting }\Gamma\left(  x_{1}\right)
,\Gamma\left(  x_{2}\right)  ,\Gamma\left(  x_{3}\right)  ,...\text{
for}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  x_{1},x_{2},x_{3},...\text{
in }\underbrace{f}_{=\sum\limits_{\substack{\left(  \alpha_{1},\alpha
_{2},\alpha_{3},...\right)  \\\text{is a weak composition}}}\lambda_{\left(
\alpha_{1},\alpha_{2},\alpha_{3},...\right)  }\prod_{i=1}^{\infty}%
x_{i}^{\alpha_{i}}}\right) \\
&  =\left(  \text{the result of substituting }\Gamma\left(  x_{1}\right)
,\Gamma\left(  x_{2}\right)  ,\Gamma\left(  x_{3}\right)  ,...\text{
for}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  x_{1},x_{2},x_{3},...\text{
in }\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,...\right)  \\\text{is a weak composition}}}\lambda_{\left(  \alpha
_{1},\alpha_{2},\alpha_{3},...\right)  }\prod_{i=1}^{\infty}x_{i}^{\alpha_{i}%
}\right) \\
&  =\sum\limits_{\substack{\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,...\right)  \\\text{is a weak composition}}}\lambda_{\left(  \alpha
_{1},\alpha_{2},\alpha_{3},...\right)  }\prod_{i=1}^{\infty}\left(
\Gamma\left(  x_{i}\right)  \right)  ^{\alpha_{i}}=\Gamma\left(  f\right)
\end{align*}
(by \eqref{sol.schur.finite.a.Gamma.=subst.pf.1}). This proves
\eqref{sol.schur.finite.a.Gamma.=subst}.}

Now, let $\lambda$ and $\mu$ be two partitions. Recall the definition of the
Schur function $s_{\lambda/\mu}$. It yields $s_{\lambda/\mu}=\sum
_{T}{\xx }^{\cont \left(  T\right)  }$, where $T$ runs
through all column-strict tableaux of shape $\lambda/\mu$. In other words,
\begin{equation}
s_{\lambda/\mu}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda/\mu}}\underbrace{{\xx }%
^{\cont \left(  T\right)  }}_{=\prod_{i\geq1}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }}=\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu}}\prod_{i\geq1}%
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\label{sol.schur.finite.a.slambda.y}%
\end{equation}


We now notice that if $T$ is any column-strict tableau such that all entries
of $T$ belong to $\left\{  1,2,...,n\right\}  $, then
\begin{equation}
\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }.
\label{sol.schur.finite.a.Gamma.T.good}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.schur.finite.a.Gamma.T.good}:} Let $T$
be a column-strict tableau such that all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $.
\par
Let $i\in\left\{  n+1,n+2,n+3,...\right\}  $ be arbitrary. Then, $i>n$. Let
$c\in T^{-1}\left(  i\right)  $ be arbitrary. Then,
\[
c\in T^{-1}\left(  i\right)  =\left(  \text{the set of all cells of the shape
of }T\text{ which are filled with the entry }i\text{ in the tableau }T\right)
\]
(since $T^{-1}\left(  i\right)  $ is the set of all cells of the shape of $T$
which are filled with the entry $i$ in the tableau $T$). In other words, $c$
is a cell of the shape of $T$ which is filled with the entry $i$ in the
tableau $T$. Hence, the entry of the tableau $T$ in cell $c$ is $i$. Thus, $i$
is the entry of the tableau $T$ in cell $i$. In particular, this shows that
$i$ is an entry of $T$. Since all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $, this yields that $i$ belongs to $\left\{
1,2,...,n\right\}  $. Hence, $i\leq n$. This contradicts $i>n$.
\par
Now, forget that we fixed $c$. We thus have obtained a contradiction for every
$c\in T^{-1}\left(  i\right)  $. As a consequence, we see that there exists no
$c\in T^{-1}\left(  i\right)  $. In other words, $T^{-1}\left(  i\right)
=\varnothing$, so that $\left\vert \underbrace{T^{-1}\left(  i\right)
}_{=\varnothing}\right\vert =\left\vert \varnothing\right\vert =0$ and thus
$x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }=x_{i}^{0}=1$.
\par
Now, forget that we fixed $i$. We now have proven that $x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }=1$ for every $i\in\left\{
n+1,n+2,n+3,...\right\}  $. Hence, $\prod_{i=n+1}^{\infty}\underbrace{x_{i}%
^{\left\vert T^{-1}\left(  i\right)  \right\vert }}_{=1}=\prod_{i=n+1}%
^{\infty}1=1$. But%
\[
\underbrace{\prod_{i\geq1}}_{=\prod_{i=1}^{\infty}}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }=\prod_{i=1}^{\infty}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }=\prod_{i=1}^{n}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\cdot\underbrace{\prod_{i=n+1}^{\infty
}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }}_{=1}=\prod_{i=1}%
^{n}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\]
Applying the map $\Gamma$ to both sides of this equality, we obtain%
\begin{align*}
\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)   &  =\Gamma\left(  \prod_{i=1}^{n}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\right)  =\prod_{i=1}^{n}\left(
\underbrace{\Gamma\left(  x_{i}\right)  }_{\substack{=x_{i}\\\text{(by
\eqref{sol.schur.finite.a.Gamma.2a})}}}\right)  ^{\left\vert T^{-1}\left(
i\right)  \right\vert }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{
is a }\kk \text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\end{align*}
This proves \eqref{sol.schur.finite.a.Gamma.T.good}.} On the other hand, if
$T$ is any column-strict tableau such that not all entries of $T$ belong to
$\left\{  1,2,...,n\right\}  $, then
\begin{equation}
\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =0.
\label{sol.schur.finite.a.Gamma.T.bad}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.schur.finite.a.Gamma.T.bad}:} Let $T$ be
a column-strict tableau such that not all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $.
\par
We know that not all entries of $T$ belong to $\left\{ 1,2,...,n \right\}$.
In other words, there exists an entry of $T$ which does not belong to
$\left\{  1,2,...,n\right\}  $. Let $j$ be this entry. Thus, $j$ is an entry
of $T$ which does not belong to $\left\{  1,2,...,n\right\}  $.
\par
We have $j\in\left\{  1,2,3,...\right\}  $ (since $j$ is an entry of $T$) but
$j\notin\left\{  1,2,...,n\right\}  $ (since $j$ does not belong to $\left\{
1,2,...,n\right\}  $). In other words, $j\in\left\{  1,2,3,...\right\}
\setminus\left\{  1,2,...,n\right\}  =\left\{  n+1,n+2,n+3,...\right\}  $.
Thus, \eqref{sol.schur.finite.a.Gamma.2b} (applied to $i=j$) yields
$\Gamma\left(  x_{j}\right)  =0$.
\par
Now, recall that $j$ is an entry of $T$. In other words, there exists a cell
$c$ of the shape of $T$ such that $j$ is the entry of the tableau $T$ in cell
$c$. Consider this cell $c$.
\par
We know that
\begin{equation}
T^{-1}\left(  j\right)  =\left(  \text{the set of all cells of the shape of
}T\text{ which are filled with the entry }j\text{ in the tableau }T\right)
\label{sol.schur.finite.a.Gamma.T.bad.pf.1}
\end{equation}
(since $T^{-1}\left(  j\right)  $ is the set of all cells of the shape of $T$
which are filled with the entry $j$ in the tableau $T$). But $j$ is the entry
of the tableau $T$ in cell $c$. In other words, the cell $c$ is filled with
the entry $j$ in the tableau $T$. Also, $c$ is a cell of the shape of $T$.
Hence, $c$ is a cell of the shape of $T$ which is filled with the entry $j$ in
the tableau $T$. In other words,
\begin{align*}
c  &  \in\left(  \text{the set of all cells of the shape of }T\text{ which are
filled with the entry }j\text{ in the tableau }T\right) \\
&  =T^{-1}\left(  j\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.T.bad.pf.1}}\right)  .
\end{align*}
Hence, the set $T^{-1}\left(  j\right)  $ contains at least one element
(namely, $c$), and thus is nonempty. Therefore, $\left\vert T^{-1}\left(
j\right)  \right\vert >0$. But since $\Gamma\left(  x_{j}\right)  =0$, we have
$\left(  \underbrace{\Gamma\left(  x_{j}\right)  }_{=0}\right)  ^{\left\vert
T^{-1}\left(  j\right)  \right\vert }=0^{\left\vert T^{-1}\left(  j\right)
\right\vert }=0$ (since $\left\vert T^{-1}\left(  j\right)  \right\vert >0$).
\par
But $j\in\left\{  1,2,3,...\right\}  $, so that
\[
\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }
=x_{j}^{\left\vert T^{-1}\left(  j\right)  \right\vert }\cdot\prod\limits
_{\substack{i\geq1;\\i\neq j}}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }.
\]
Applying the map $\Gamma$ to both sides of this equality, we obtain%
\begin{align*}
\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)   &  =\Gamma\left(  x_{j}^{\left\vert T^{-1}\left(
j\right)  \right\vert }\cdot\prod\limits_{\substack{i\geq1;\\i\neq j}}x_{i}%
^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right) \\
&  =\underbrace{\left(  \Gamma\left(  x_{j}\right)  \right)  ^{\left\vert
T^{-1}\left(  j\right)  \right\vert }}_{=0}\cdot\Gamma\left(  \prod\limits
_{\substack{i\geq1;\\i\neq j}}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is
a }\kk \text{-algebra homomorphism}\right) \\
&  =0.
\end{align*}
This proves \eqref{sol.schur.finite.a.Gamma.T.bad}.}

Now, applying the map $\Gamma$ to both sides of the identity
\eqref{sol.schur.finite.a.slambda.y}, we obtain
\begin{align}
\Gamma\left(  s_{\lambda/\mu}\right)   &  =\Gamma\left(  \sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu}}\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =\sum\limits_{\substack{T\text{ is a column-strict}
\\\text{tableau of shape }\lambda/\mu}}\Gamma\left(  \prod_{i\geq1}
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is a continuous
}\kk \text{-algebra homomorphism}\right) \nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\underbrace{\Gamma\left(  \prod_{i\geq1}
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)  }
_{\substack{=\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\\\text{(by \eqref{sol.schur.finite.a.Gamma.T.good})}}}
+\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu\text{;}\\\text{not all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\underbrace{\Gamma\left(  \prod_{i\geq1}
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)  }
_{\substack{=0\\\text{(by \eqref{sol.schur.finite.a.Gamma.T.bad})}}}
\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }+\underbrace{\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu\text{;}\\\text{not all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,...,n\right\}  }}0}%
_{=0}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\mu\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }.
\label{sol.schur.finite.a.Gamma.T.Gamma}
\end{align}
Compared with%
\[
\Gamma\left(  s_{\lambda/\mu}\right)  =s_{\lambda/\mu}\left(  x_{1}%
,x_{2},...,x_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.=subst}, applied to }f=s_{\lambda/\mu}\right)
,
\]
this yields%
\begin{align*}
s_{\lambda/\mu}\left(  x_{1},x_{2},...,x_{n}\right)   &  =\sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\underbrace{\prod_{i=1}^{n}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }}_{\substack{=\left(  x_{1}%
,x_{2},...,x_{n}\right)  ^{\cont \left(  T\right)
}\\\text{(because }\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }\\=\prod_{i=1}^{n}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\text{)}}}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }.
\end{align*}
This solves Exercise~\ref{exe.schur.finite}(a).

(b) Let $\lambda$ be a partition having more than $n$ parts. Then,
$\ell\left(  \lambda\right)  >n$ (since $\lambda$ has more than $n$ parts).
Thus, $\ell\left(  \lambda\right)  \geq n+1$ (since $\ell\left(
\lambda\right)  $ and $n$ are integers), so that $\lambda_{n+1}\neq0$. Thus,
$\lambda_{n+1}>0$ (since $\lambda_{n+1}\geq0$). Thus, $\lambda_{n+1}\geq1$
(since $\lambda_{n+1}$ is an integer).

We know that $\lambda$ is a partition. Hence, $\lambda_{1}\geq\lambda_{2}%
\geq\lambda_{3}\geq...$. In other words, any two positive integers $u$ and $v$
satisfying $u\leq v$ satisfy $\lambda_{u}\geq\lambda_{v}$.

Let $T$ be any column-strict tableau of shape $\lambda$. We are going to show
that not all entries of $T$ belong to $\left\{  1,2,...,n\right\}  $.

In fact, assume the contrary. Thus, all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $.

We know that $T$ is a column-strict tableau of shape $\lambda$. In other
words, $T$ is an assignment of entries of $\left\{  1,2,3,...\right\}  $ to
the cells of the Ferrers diagram of $\lambda$, weakly increasing left-to-right
in rows, and strictly increasing top-to-bottom in columns (because this is how
a column-strict tableau of shape $\lambda$ was defined).

Let $Y\left(  \lambda\right)  $ denote the Ferrers diagram of $\lambda$. Then,%
\begin{equation}
Y\left(  \lambda\right)  =\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ q\leq\lambda_{p}\right\}
\label{sol.schur.finite.a.Gamma.T.verybad.pf.1}%
\end{equation}
(because this is how $Y\left(  \lambda\right)  $ is defined). Also, the cells
of the Ferrers diagram of $\lambda$ are the elements of $Y\left(
\lambda\right)  $ (since $Y\left(  \lambda\right)  $ is the Ferrers diagram of
$\lambda$).

We know that $T$ is an assignment of entries of $\left\{  1,2,3,...\right\}  $
to the cells of the Ferrers diagram of $\lambda$. In other words, $T$ is an
assignment of entries of $\left\{  1,2,3,...\right\}  $ to the elements of
$Y\left(  \lambda\right)  $ (since the cells of the Ferrers diagram of
$\lambda$ are the elements of $Y\left(  \lambda\right)  $). In other words,
$T$ is a map $Y\left(  \lambda\right)  \rightarrow\left\{  1,2,3,...\right\}
$. We have%
\begin{equation}
T\left(  \left(  p,q\right)  \right)  <T\left(  \left(  p^{\prime},q\right)
\right)  \ \ \ \ \ \ \ \ \ \ \text{for any integers }p\text{, }p^{\prime
}\text{ and }q\text{ satisfying }\left(  p,q\right)  \in Y\left(
\lambda\right)  \text{, }\left(  p^{\prime},q\right)  \in Y\left(
\lambda\right)  \text{ and }p<p^{\prime}
\label{sol.schur.finite.a.Gamma.T.verybad.pf.2}%
\end{equation}
(because $T$ is strictly increasing top-to-bottom in columns).

Now, let $i\in\left\{  1,2,...,n+1\right\}  $ be arbitrary. Then, $i\leq n+1$
and $\left(  i,1\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ (since
$i\in\left\{  1,2,...,n+1\right\}  \subset\left\{  1,2,3,...\right\}  $ and
$1\in\left\{  1,2,3,...\right\}  $). Recall that any two positive integers $u$
and $v$ satisfying $u\leq v$ satisfy $\lambda_{u}\geq\lambda_{v}$. Applying
this to $u=i$ and $v=n+1$, we obtain $\lambda_{i}\geq\lambda_{n+1}\geq1$ and
thus $1\leq\lambda_{i}$. Now, $\left(  i,1\right)  $ is an element of the set
$\left\{  1,2,3,...\right\}  ^{2}$ satisfying $1\leq\lambda_{i}$. In other
words, $\left(  i,1\right)  $ is an element $\left(  p,q\right)  $ of the set
$\left\{  1,2,3,...\right\}  ^{2}$ satisfying $q\leq\lambda_{p}$. In other
words,%
\[
\left(  i,1\right)  \in\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ q\leq\lambda_{p}\right\}  =Y\left(
\lambda\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.T.verybad.pf.1}}\right)  .
\]


Now, forget that we fixed $i$. We thus have shown that
\begin{equation}
\left(  i,1\right)  \in Y\left(  \lambda\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  1,2,...,n+1\right\}  .
\label{sol.schur.finite.a.Gamma.T.verybad.pf.3}%
\end{equation}
Hence, $T\left(  \left(  i,1\right)  \right)  $ is well-defined for every
$i\in\left\{  1,2,...,n+1\right\}  $ (because $T$ is a map $Y\left(
\lambda\right)  \rightarrow\left\{  1,2,3,...\right\}  $). Applied to $i=n+1$,
this yields that $T\left(  \left(  n+1,1\right)  \right)  $ is well-defined.
Now, $T\left(  \left(  n+1,1\right)  \right)  $ is the entry of the tableau
$T$ in the cell $\left(  n+1,1\right)  $. Thus, in particular, $T\left(
\left(  n+1,1\right)  \right)  $ is an entry of $T$, and therefore belongs to
$\left\{  1,2,...,n\right\}  $ (since all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $). Hence, $T\left(  \left(  n+1,1\right)  \right)  \leq n$.

Let now $i\in\left\{  1,2,...,n\right\}  $. Then, $i\in\left\{
1,2,...,n\right\}  \subset\left\{  1,2,...,n+1\right\}  $, so that $\left(
i,1\right)  \in Y\left(  \lambda\right)  $ (by
\eqref{sol.schur.finite.a.Gamma.T.verybad.pf.3}). Also, $i\in\left\{
1,2,...,n\right\}  $ shows that $i+1\in\left\{  2,3,...,n+1\right\}
\subset\left\{  1,2,...,n+1\right\}  $, so that $\left(  i+1,1\right)  \in
Y\left(  \lambda\right)  $ (by
\eqref{sol.schur.finite.a.Gamma.T.verybad.pf.3}, applied to $i+1$ instead
of $i$). Also, $i<i+1$. Hence,
\eqref{sol.schur.finite.a.Gamma.T.verybad.pf.2} (applied to $p=i$, $p^{\prime
}=i+1$ and $q=1$) yields $T\left(  \left(  i,1\right)  \right)  <T\left(
\left(  i+1,1\right)  \right)  $. But $T\left(  \left(  i,1\right)  \right)  $
and $T\left(  \left(  i+1,1\right)  \right)  $ are elements of $\left\{
1,2,3,...\right\}  $ (since $T$ is a map $Y\left(  \lambda\right)
\rightarrow\left\{  1,2,3,...\right\}  $) and thus integers. Hence, from
$T\left(  \left(  i,1\right)  \right)  <T\left(  \left(  i+1,1\right)
\right)  $, we obtain $T\left(  \left(  i,1\right)  \right)  \leq T\left(
\left(  i+1,1\right)  \right)  -1$. Thus, $T\left(  \left(  i+1,1\right)
\right)  -1\geq T\left(  \left(  i,1\right)  \right)  $, so that $T\left(
\left(  i+1,1\right)  \right)  -T\left(  \left(  i,1\right)  \right)  \geq1$.

Now, forget that we fixed $i$. We thus have proven that every $i\in\left\{
1,2,...,n\right\}  $ satisfies%
\[
T\left(  \left(  i+1,1\right)  \right)  -T\left(  \left(  i,1\right)  \right)
\geq1.
\]
Summing these inequalities over all $i\in\left\{  1,2,...,n\right\}  $, we
obtain%
\[
\sum_{i\in\left\{  1,2,...,n\right\}  }\left(  T\left(  \left(  i+1,1\right)
\right)  -T\left(  \left(  i,1\right)  \right)  \right)  \geq\sum
_{i\in\left\{  1,2,...,n\right\}  }1.
\]
Hence,%
\begin{align*}
\sum_{i\in\left\{  1,2,...,n\right\}  }1  &  \leq\underbrace{\sum
_{i\in\left\{  1,2,...,n\right\}  }}_{=\sum_{i=1}^{n}}\left(  T\left(  \left(
i+1,1\right)  \right)  -T\left(  \left(  i,1\right)  \right)  \right) \\
&  =\sum_{i=1}^{n}\left(  T\left(  \left(  i+1,1\right)  \right)  -T\left(
\left(  i,1\right)  \right)  \right)  =\underbrace{\sum_{i=1}^{n}T\left(
\left(  i+1,1\right)  \right)  }_{\substack{=\sum_{i=2}^{n+1}T\left(  \left(
i,1\right)  \right)  \\\text{(here, we substituted }i\text{ for }i+1\\\text{in
the sum)}}}-\sum_{i=1}^{n}T\left(  \left(  i,1\right)  \right) \\
&  =\underbrace{\sum_{i=2}^{n+1}T\left(  \left(  i,1\right)  \right)  }%
_{=\sum_{i=2}^{n}T\left(  \left(  i,1\right)  \right)  +T\left(  \left(
n+1,1\right)  \right)  }-\underbrace{\sum_{i=1}^{n}T\left(  \left(
i,1\right)  \right)  }_{=T\left(  \left(  1,1\right)  \right)  +\sum_{i=2}%
^{n}T\left(  \left(  i,1\right)  \right)  }\\
&  =\left(  \sum_{i=2}^{n}T\left(  \left(  i,1\right)  \right)  +T\left(
\left(  n+1,1\right)  \right)  \right)  -\left(  T\left(  \left(  1,1\right)
\right)  +\sum_{i=2}^{n}T\left(  \left(  i,1\right)  \right)  \right) \\
&  =\underbrace{T\left(  \left(  n+1,1\right)  \right)  }_{\substack{\leq
n}}-\underbrace{T\left(  \left(  1,1\right)  \right)  }_{\substack{\geq
1\\\text{(since }T\left(  \left(  1,1\right)  \right)  \in\left\{
1,2,3,...\right\}  \\\text{(because }T\text{ is a map }Y\left(  \lambda
\right)  \rightarrow\left\{  1,2,3,...\right\}  \text{))}}}\leq n-1<n,
\end{align*}
which contradicts $\sum_{i\in\left\{  1,2,...,n\right\} } 1
= \underbrace{\left\vert \left\{  1,2,...,n\right\}  \right\vert }_{=n}
\cdot 1 = n\cdot 1 = n$.
This contradiction shows that our assumption was wrong.
Hence, not all entries of $T$ belong to $\left\{  1,2,...,n\right\}  $. Thus,
$\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =0$ (according to \eqref{sol.schur.finite.a.Gamma.T.bad}).

Now, forget that we fixed $T$. We thus have proven that
\begin{equation}
\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =0 \label{sol.schur.finite.b.Gamma.T.bad-lambda}%
\end{equation}
for every column-strict tableau $T$ of shape $\lambda$.

Now, the definition of $s_{\lambda}$ yields $s_{\lambda}=\sum_{T}{\xx %
}^{\cont \left(  T\right)  }$, where $T$ runs through all
column-strict tableaux of shape $\lambda$. In other words,%
\[
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau
of shape }\lambda}}\underbrace{\xx ^{\cont \left(
T\right)  }}_{=\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda}}\prod_{i\geq1}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }.
\]
Applying the map $\Gamma$ to both sides of this equality, we obtain%
\begin{align*}
\Gamma\left(  s_{\lambda}\right)   &  =\Gamma\left(  \sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert
}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau
of shape }\lambda}}\underbrace{\Gamma\left(  \prod_{i\geq1}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\right)  }_{\substack{=0\\\text{(by
\eqref{sol.schur.finite.b.Gamma.T.bad-lambda})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is a continuous
}\kk \text{-algebra homomorphism}\right) \\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}0=0.
\end{align*}
Compared with%
\[
\Gamma\left(  s_{\lambda}\right)  =s_{\lambda}\left(  x_{1},x_{2}%
,...,x_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.schur.finite.a.Gamma.=subst}, applied to }f=s_{\lambda/\mu}\right)
,
\]
this yields $s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  =0$. This solves
Exercise~\ref{exe.schur.finite}(b).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.omega-and-S}}

\textit{Solution to Exercise \ref{exe.Lambda.omega-and-S}.}

\begin{proof}
[Proof of Proposition \ref{prop.Lambda.omega-and-S}.]The map $\omega$ is
precisely the $\kk$-algebra homomorphism $\omega$ defined in the proof
of Proposition~\ref{symm-antipode-and-h-basis}. Thus, in particular, $\omega$
is a $\kk$-algebra homomorphism; hence, $\omega\left(  1\right)  =1$
and $\omega\left(  0\right)  =0$.

(a) Let $n\in\mathbb{Z}$. We must prove that $\omega\left(  e_{n}\right)
=h_{n}$.

If $n$ is a positive integer, then this follows immediately from the
definition of $\omega$. Thus, for the rest of the proof of $\omega\left(
e_{n}\right)  =h_{n}$, we can WLOG assume that $n$ is not a positive integer.
Assume this.

\begin{vershort}
From $e_{0}=1$, we obtain $\omega\left(  e_{0}\right)  =\omega\left(
1\right)  =1=h_{0}$ (since $h_{0}=1$). Hence, $\omega\left(  e_{n}\right)
=h_{n}$ holds for $n=0$. Thus, for the rest of the proof of $\omega\left(
e_{n}\right)  =h_{n}$, we can WLOG assume that we don't have $n=0$. Assume this.
\end{vershort}

\begin{verlong}
We have $n\in{\mathbb{Z}}$, but we don't have $n\in\left\{  1,2,3,...\right\}
$ (since $n$ is not a positive integer). Hence, we have $n\in{\mathbb{Z}%
}\setminus\left\{  1,2,3,...\right\}  =\left\{  0,-1,-2,-3,...\right\}  $.

We have $\omega\left(  \underbrace{e_{0}}_{=1}\right)  =\omega\left(
1\right)  =1$. Compared with $h_{0}=1$, this yields $\omega\left(
e_{0}\right)  =h_{0}$. In other words, $\omega\left(  e_{n}\right)  =h_{n}$
holds for $n=0$. Hence, if $n=0$, then $\omega\left(  e_{n}\right)  =h_{n}$ is
true. Thus, for the rest of the proof of $\omega\left(  e_{n}\right)  =h_{n}$,
we can WLOG assume that we don't have $n=0$. Assume this.
\end{verlong}

\begin{vershort}
Now, $n$ is a negative integer (since $n$ is not a positive integer, and since
we don't have $n=0$). This yields $h_{n}=0$ and $e_{n}=0$. Now, $\omega\left(
\underbrace{e_{n}}_{=0}\right)  =\omega\left(  0\right)  =0=h_{n}$. This
proves $\omega\left(  e_{n}\right)  =h_{n}$. This proves Proposition
\ref{prop.Lambda.omega-and-S}(a).
\end{vershort}

\begin{verlong}
We have $n\in\left\{  0,-1,-2,-3,...\right\}  $, but we don't have $n=0$.
Hence, we have $n\in\left\{  0,-1,-2,-3,...\right\}  \setminus\left\{
0\right\}  =\left\{  -1,-2,-3,...\right\}  $. Thus, $n<0$. This yields
$h_{n}=0$ and $e_{n}=0$. Now, $\omega\left(  \underbrace{e_{n}}_{=0}\right)
=\omega\left(  0\right)  =0=h_{n}$ (since $h_{n}=0$). This proves
$\omega\left(  e_{n}\right)  =h_{n}$. This proves Proposition
\ref{prop.Lambda.omega-and-S}(a).
\end{verlong}

\begin{vershort}
(b) The proof of Proposition \ref{prop.Lambda.omega-and-S}(b) is completely
analogous to the proof of Proposition \ref{prop.Lambda.omega-and-S}(a) given
above, except that $e_{n}$ and $h_{n}$ trade places (and that we need to use
\eqref{the-fundamental-involution-he} instead of the definition of $\omega$).
\end{vershort}

\begin{verlong}
Let $n\in\mathbb{Z}$. We must prove that $\omega\left(  h_{n}\right)  =e_{n}$.

If $n$ is a positive integer, then this follows immediately from
\eqref{the-fundamental-involution-he}. Thus, for the rest of the proof of
$\omega\left(  h_{n}\right)  =e_{n}$, we can WLOG assume that $n$ is not a
positive integer. Assume this.

We have $n\in{\mathbb{Z}}$, but we don't have $n\in\left\{  1,2,3,...\right\}
$ (since $n$ is not a positive integer). Hence, we have $n\in{\mathbb{Z}%
}\setminus\left\{  1,2,3,...\right\}  =\left\{  0,-1,-2,-3,...\right\}  $.

We have $\omega\left(  \underbrace{h_{0}}_{=1}\right)  =\omega\left(
1\right)  =1$. Compared with $e_{0}=1$, this yields $\omega\left(
h_{0}\right)  =e_{0}$. In other words, $\omega\left(  h_{n}\right)  =e_{n}$
holds for $n=0$. Hence, if $n=0$, then $\omega\left(  h_{n}\right)  =e_{n}$ is
true. Thus, for the rest of the proof of $\omega\left(  h_{n}\right)  =e_{n}$,
we can WLOG assume that we don't have $n=0$. Assume this.

We have $n\in\left\{  0,-1,-2,-3,...\right\}  $, but we don't have $n=0$.
Hence, we have $n\in\left\{  0,-1,-2,-3,...\right\}  \setminus\left\{
0\right\}  =\left\{  -1,-2,-3,...\right\}  $. Thus, $n<0$. This yields
$e_{n}=0$ and $h_{n}=0$. Now, $\omega\left(  \underbrace{h_{n}}_{=0}\right)
=\omega\left(  0\right)  =0=e_{n}$ (since $e_{n}=0$). This proves
$\omega\left(  h_{n}\right)  =e_{n}$. This proves Proposition
\ref{prop.Lambda.omega-and-S}(b).
\end{verlong}

(d) In the proof of Proposition~\ref{symm-antipode-and-h-basis}, we have shown
that $\omega$ is an involution and a $\kk$-algebra automorphism of
$\Lambda$. This proves Proposition \ref{prop.Lambda.omega-and-S}(d).

(e) Proposition \ref{antipodes-are-antiendomorphisms} (applied to $A=\Lambda$)
shows that the antipode $S$ of $\Lambda$ is a $\kk$-algebra
anti-endomorphism. In other words, $S$ is a $\kk$-algebra
anti-homomorphism from $\Lambda$ to $\Lambda$. But Exercise
\ref{exe.comm-cocomm.anti}(a) (applied to $A=\Lambda$ and $B=\Lambda$) shows
that the $\kk$-algebra anti-homomorphisms from $\Lambda$ to $\Lambda$
are the same as the $\kk$-algebra homomorphisms from $\Lambda$ to
$\Lambda$ (since the $\kk$-algebra $\Lambda$ is commutative). Hence,
$S$ is a $\kk$-algebra homomorphism from $\Lambda$ to $\Lambda$ (since
$S$ is a $\kk$-algebra anti-homomorphism from $\Lambda$ to $\Lambda$).

Recall that $\Lambda$ is a graded $\kk$-bialgebra. Let $q=-1\in
\kk$. Consider the $\kk$-linear map $D_{q}:\Lambda
\rightarrow\Lambda$ constructed in Exercise \ref{exe.graded.Dq} (applied to
$A=\Lambda$). From Exercise \ref{exe.graded.Dq} (applied to $A=\Lambda$), we
know that this map $D_{q}$ is a $\kk$-bialgebra homomorphism; thus, in
particular, $D_{q}$ is a $\kk$-algebra homomorphism. The definition of
$D_{q}$ shows that%
\begin{equation}
D_{q}\left(  a\right)  =q^{n}a\ \ \ \ \ \ \ \ \ \ \text{for each }%
n\in\mathbb{N}\text{ and each }a\in\Lambda_{n}.
\label{pf.prop.Lambda.omega-and-S.e.Dqa1}%
\end{equation}


Now, the map $\omega\circ D_{q}:\Lambda\rightarrow\Lambda$ is a
$\kk$-algebra homomorphism (since it is the composition of the two
$\kk$-algebra homomorphisms $\omega$ and $D_{q}$).

Let $n$ be a positive integer. Then, $\omega\left(  e_{n}\right)  =h_{n}$ (by
the definition of $\omega$). But $e_{n}\in\Lambda_{n}$; hence,
(\ref{pf.prop.Lambda.omega-and-S.e.Dqa1}) (applied to $a=e_{n}$) yields
$D_{q}\left(  e_{n}\right)  =q^{n}e_{n}=\left(  -1\right)  ^{n}e_{n}$ (since
$q=-1$). Now,%
\begin{align*}
\left(  \omega\circ D_{q}\right)  \left(  e_{n}\right)   &  =\omega\left(
\underbrace{D_{q}\left(  e_{n}\right)  }_{=\left(  -1\right)  ^{n}e_{n}%
}\right)  =\omega\left(  \left(  -1\right)  ^{n}e_{n}\right)  =\left(
-1\right)  ^{n}\underbrace{\omega\left(  e_{n}\right)  }_{=h_{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\omega\text{ is }%
\kk\text{-linear}\right) \\
&  =\left(  -1\right)  ^{n}h_{n}.
\end{align*}
Comparing this with%
\[
S\left(  e_{n}\right)  =\left(  -1\right)  ^{n}h_{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{symm-antipode-and-h-basis}(ii)}\right)  ,
\]
we obtain $\left(  \omega\circ D_{q}\right)  \left(  e_{n}\right)  =S\left(
e_{n}\right)  $.

Now, forget that we fixed $n$. We thus have shown that $\left(  \omega\circ
D_{q}\right)  \left(  e_{n}\right)  =S\left(  e_{n}\right)  $ for each
positive integer $n$. In other words, the two $\kk$-algebra
homomorphisms $\omega\circ D_{q}$ and $S$ (from $\Lambda$ to $\Lambda$) agree
on each element of the family $\left\{  e_{n}\right\}  _{n\geq1}$ of elements
of $\Lambda$.

But Proposition \ref{symm-antipode-and-h-basis} tells us that the family
$\left\{  e_{n}\right\}  _{n\geq1}$ generates the $\kk$-algebra
$\Lambda$. Thus, if two $\kk$-algebra homomorphisms from $\Lambda$
agree on each element of this family $\left\{  e_{n}\right\}  _{n\geq1}$, then
these two homomorphisms must be equal. Hence, the two $\kk$-algebra
homomorphisms $\omega\circ D_{q}$ and $S$ must be equal (since they agree on
each element of this family $\left\{  e_{n}\right\}  _{n\geq1}$). In other
words, we have $\omega\circ D_{q}=S$.

Now, let $n\in\mathbb{N}$ and let $f\in\Lambda_{n}$. We must show that
$S\left(  f\right)  =\left(  -1\right)  ^{n}\omega\left(  f\right)  $. And
indeed, we have $\underbrace{\left(  \omega\circ D_{q}\right)  }_{=S}\left(
f\right)  =S\left(  f\right)  $, so that%
\begin{align*}
S\left(  f\right)   &  =\left(  \omega\circ D_{q}\right)  \left(  f\right)
=\omega\left(  \underbrace{D_{q}\left(  f\right)  }_{\substack{=q^{n}%
f\\\text{(by (\ref{pf.prop.Lambda.omega-and-S.e.Dqa1}), applied to
}a=f\text{)}}}\right)  =\omega\left(  q^{n}f\right) \\
&  =q^{n}\omega\left(  f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }\omega\text{ is } \kk\text{-linear}\right) \\
&  =\left(  -1\right)  ^{n}\omega\left(  f\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }q=-1\right)  .
\end{align*}
This proves Proposition \ref{prop.Lambda.omega-and-S}(e).

\begin{vershort}
(c) Let $n$ be a positive integer. Proposition~\ref{symm-antipode-and-h-basis}%
(i) yields $S\left(  p_{n}\right)  =-p_{n}$. But
\eqref{antipode-versus-omega-in-Sym} (applied to $f=p_{n}$) yields $S\left(
p_{n}\right)  =\left(  -1\right)  ^{n}\omega\left(  p_{n}\right)  $ (since
$p_{n}\in\Lambda_{n}$). Comparing these two equalities yields $\left(
-1\right)  ^{n}\omega\left(  p_{n}\right)  =-p_{n}$, and this quickly rewrites
as $\omega\left(  p_{n}\right)  =\left(  -1\right)  ^{n-1}p_{n}$. This proves
Proposition \ref{prop.Lambda.omega-and-S}(c).
\end{vershort}

\begin{verlong}
(c) Let $n$ be a positive integer. Proposition~\ref{symm-antipode-and-h-basis}%
(i) yields $S\left(  p_{n}\right)  =-p_{n}$. But
\eqref{antipode-versus-omega-in-Sym} (applied to $f=p_{n}$) yields $S\left(
p_{n}\right)  =\left(  -1\right)  ^{n}\omega\left(  p_{n}\right)  $ (since
$p_{n}\in\Lambda_{n}$). Hence, $\left(  -1\right)  ^{n}\omega\left(
p_{n}\right)  =S\left(  p_{n}\right)  =-p_{n}$. Multiplying both sides of this
equality by $\left(  -1\right)  ^{n}$, we obtain
\[
\left(  -1\right)  ^{n}\left(  -1\right)  ^{n}\omega\left(  p_{n}\right)
=\left(  -1\right)  ^{n}\left(  -p_{n}\right)  =\underbrace{-\left(
-1\right)  ^{n}}_{=\left(  -1\right)  ^{n-1}}p_{n}=\left(  -1\right)
^{n-1}p_{n}.
\]
Comparing this with $\underbrace{\left(  -1\right)  ^{n}\left(  -1\right)
^{n}}_{\substack{=\left(  \left(  -1\right)  ^{n}\right)  ^{2}=\left(
-1\right)  ^{2n}=1\\\text{(since }2n\text{ is even)}}}\omega\left(
p_{n}\right)  =\omega\left(  p_{n}\right)  $, we obtain $\omega\left(
p_{n}\right)  =\left(  -1\right)  ^{n-1}p_{n}$. This proves Proposition
\ref{prop.Lambda.omega-and-S}(c).
\end{verlong}

\begin{vershort}
(f) To show that $\omega$ is a coalgebra homomorphism, it suffices to check
that $\left(  \omega\otimes\omega\right)  \circ\Delta=\Delta\circ\omega$ and
$\epsilon=\epsilon\circ\omega$.

Let us first prove that $\left(  \omega\otimes\omega\right)  \circ
\Delta=\Delta\circ\omega$. Indeed, both sides of this equality being
$\kk$-algebra homomorphisms, it only needs to be checked on the algebra
generators $h_{n}$ of $\Lambda$ (indeed, we know from Proposition
\ref{symm-antipode-and-h-basis} that the family $\left\{  h_{n}\right\}
_{n\geq1}$ generates the $\kk$-algebra $\Lambda$). On these generators
this is easy to check: Comparing
\begin{align*}
\left(  \left(  \omega\otimes\omega\right)  \circ\Delta\right)  \left(
h_{n}\right)    & =\left(  \omega\otimes\omega\right)  \left(  \Delta\left(
h_{n}\right)  \right)  =\left(  \omega\otimes\omega\right)  \left(  \sum
_{i+j=n}h_{i}\otimes h_{j}\right)
\ \ \ \ \ \ \ \ \ \ %
\left(\text{by Proposition \ref{symm-comultiplication-formulas-prop}(iii)}\right) \\
& =\sum_{i+j=n}\underbrace{\omega\left(  h_{i}\right)  }_{\substack{=e_{i}%
\\\text{(by Proposition \ref{prop.Lambda.omega-and-S}(b))}}}\otimes
\underbrace{\omega\left(  h_{j}\right)  }_{\substack{=e_{j}\\\text{(by
Proposition \ref{prop.Lambda.omega-and-S}(b))}}}
=\sum_{i+j=n}e_{i}\otimes e_{j}%
\end{align*}
and
\[
\left(  \Delta\circ\omega\right)  \left(  h_{n}\right)  =\Delta\left(
\underbrace{\omega\left(  h_{n}\right)  }_{\substack{=e_{n}\\\text{(by
Proposition \ref{prop.Lambda.omega-and-S}(b))}}}\right)  =\Delta\left(
e_{n}\right)  = \sum_{i+j=n}e_{i}\otimes e_{j}
\ \ \ \ \ \ \ \ \ \ %
\left(\text{by Proposition \ref{symm-comultiplication-formulas-prop}(ii)}\right)
\]
shows that $\left(  \left(  \omega\otimes\omega\right)  \circ\Delta\right)
\left(  h_{n}\right)  =\left(  \Delta\circ\omega\right)  \left(  h_{n}\right)
$ for all $n\geq1$. So we obtain $\left(  \omega\otimes\omega\right)
\circ\Delta=\Delta\circ\omega$. The equality $\epsilon=\epsilon\circ\omega$ is
even easier to check. Thus, $\omega$ is an algebra and coalgebra morphism,
thus a bialgebra morphism, thus a Hopf morphism by
Corollary~\ref{cor.bialg-mor-is-Hopf}. Since $\omega$ is invertible (by part
(d)), we conclude that $\omega$ is a Hopf algebra automorphism. This proves
Proposition \ref{prop.Lambda.omega-and-S}(f).
\end{vershort}

\begin{verlong}
(f) We already know that $\omega$ is a $\kk$-algebra automorphism of
$\Lambda$ (by Proposition \ref{prop.Lambda.omega-and-S}(d)). Thus, $\omega$ is
invertible. Let us next show that $\omega:\Lambda\rightarrow\Lambda$ is a
$\kk$-coalgebra homomorphism.

Indeed, in order to show this, we shall check that the equalities%
\begin{equation}
\left(  \omega\otimes\omega\right)  \circ\Delta=\Delta\circ\omega
\label{pf.prop.Lambda.omega-and-S.f.1}%
\end{equation}
and%
\begin{equation}
\epsilon=\epsilon\circ\omega\label{pf.prop.Lambda.omega-and-S.f.2}%
\end{equation}
hold.

[\textit{Proof of \eqref{pf.prop.Lambda.omega-and-S.f.1}:} The map
$\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ is a $\kk$-algebra
homomorphism (since $\Lambda$ is a $\kk$-bialgebra). The map
$\omega\otimes\omega:\Lambda\otimes\Lambda\rightarrow\Lambda\otimes\Lambda$ is
a $\kk$-algebra homomorphism (since $\omega$ is a $\kk$-algebra
homomorphism). Hence, the maps $\left(  \omega\otimes\omega\right)
\circ\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ and $\Delta\circ
\omega:\Lambda\rightarrow\Lambda\otimes\Lambda$ are $\kk$-algebra
homomorphisms (since they are obtained by composing the maps $\omega
\otimes\omega$, $\Delta$ and $\omega$, all of which are $\kk$-algebra homomorphisms).

Let $n$ be a positive integer. We shall show that $\left(  \left(
\omega\otimes\omega\right)  \circ\Delta\right)  \left(  h_{n}\right)  =\left(
\Delta\circ\omega\right)  \left(  h_{n}\right)  $.

Indeed, Proposition \ref{symm-comultiplication-formulas-prop}(ii) yields%
\[
\Delta\left(  e_{n}\right)  =\sum_{i+j=n}e_{i}\otimes e_{j}%
\]
(where the summation sign \textquotedblleft$\sum_{i+j=n}$\textquotedblright%
\ stands for \textquotedblleft$\sum_{\substack{\left(  i,j\right)
\in\mathbb{N}^{2};\\i+j=n}}$\textquotedblright). Likewise, Proposition
\ref{symm-comultiplication-formulas-prop}(iii) yields%
\[
\Delta\left(  h_{n}\right)  =\sum_{i+j=n}h_{i}\otimes h_{j}.
\]
Now,
\begin{align*}
\left(  \left(  \omega\otimes\omega\right)  \circ\Delta\right)  \left(
h_{n}\right)    & =\left(  \omega\otimes\omega\right)  \left(
\underbrace{\Delta\left(  h_{n}\right)  }_{=\sum_{i+j=n}h_{i}\otimes h_{j}%
}\right)  =\left(  \omega\otimes\omega\right)  \left(  \sum_{i+j=n}%
h_{i}\otimes h_{j}\right)  \\
& =\sum_{i+j=n}\underbrace{\omega\left(  h_{i}\right)  }_{\substack{=e_{i}%
\\\text{(by Proposition \ref{prop.Lambda.omega-and-S}(b),}\\\text{applied to
}i\\\text{instead of }n\text{)}}}\otimes\underbrace{\omega\left(
h_{j}\right)  }_{\substack{=e_{j}\\\text{(by Proposition
\ref{prop.Lambda.omega-and-S}(b),}\\\text{applied to }j\\\text{instead of
}n\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\omega\otimes\omega\right)  \\
& =\sum_{i+j=n}e_{i}\otimes e_{j}.
\end{align*}
Comparing this with
\[
\left(  \Delta\circ\omega\right)  \left(  h_{n}\right)  =\Delta\left(
\underbrace{\omega\left(  h_{n}\right)  }_{\substack{=e_{n}\\\text{(by
Proposition \ref{prop.Lambda.omega-and-S}(b))}}}\right)  =\Delta\left(
e_{n}\right)  =\sum_{i+j=n}e_{i}\otimes e_{j},
\]
we obtain $\left(  \left(  \omega\otimes\omega\right)  \circ\Delta\right)
\left(  h_{n}\right)  =\left(  \Delta\circ\omega\right)  \left(  h_{n}\right)
$.

Now, forget that we fixed $n$. We thus have shown that $\left(  \left(
\omega\otimes\omega\right)  \circ\Delta\right)  \left(  h_{n}\right)  =\left(
\Delta\circ\omega\right)  \left(  h_{n}\right)  $ for each positive integer
$n$. In other words, the two $\kk$-algebra homomorphisms $\left(
\omega\otimes\omega\right)  \circ\Delta$ and $\Delta\circ\omega$ (from
$\Lambda$ to $\Lambda\otimes\Lambda$) agree on each element of the family
$\left\{  h_{n}\right\}  _{n\geq1}$ of elements of $\Lambda$.

But Proposition \ref{symm-antipode-and-h-basis} tells us that the family
$\left\{  h_{n}\right\}  _{n\geq1}$ generates the $\kk$-algebra
$\Lambda$. Thus, if two $\kk$-algebra homomorphisms from $\Lambda$
agree on each element of this family $\left\{  h_{n}\right\}  _{n\geq1}$, then
these two homomorphisms must be equal. Hence, the two $\kk$-algebra
homomorphisms $\left(  \omega\otimes\omega\right)  \circ\Delta$ and
$\Delta\circ\omega$ must be equal (since they agree on each element of this
family $\left\{  h_{n}\right\}  _{n\geq1}$). In other words, we have $\left(
\omega\otimes\omega\right)  \circ\Delta=\Delta\circ\omega$.\ This proves \eqref{pf.prop.Lambda.omega-and-S.f.1}.]

[\textit{Proof of \eqref{pf.prop.Lambda.omega-and-S.f.2}:} The map
$\epsilon:\Lambda\rightarrow \kk$ is a $\kk$-algebra homomorphism
(since $\Lambda$ is a $\kk$-bialgebra). The map $\epsilon\circ
\omega:\Lambda\rightarrow \kk$ is a $\kk$-algebra homomorphism
(since it is obtained by composing the maps $\epsilon$ and $\omega$, both of
which are $\kk$-algebra homomorphisms).

Let $n$ be a positive integer. We shall show that $\epsilon\left(
h_{n}\right)  =\left(  \epsilon\circ\omega\right)  \left(  h_{n}\right)  $.

Indeed, the counit $\epsilon$ of $\Lambda$ annihilates the graded component
$\Lambda_{n}$ (since $n$ is positive). In other words, $\epsilon\left(
\Lambda_{n}\right)  =0$. But $e_{n}\in\Lambda_{n}$, so that $\epsilon\left(
e_{n}\right)  \in\epsilon\left(  \Lambda_{n}\right)  =0$, and thus
$\epsilon\left(  e_{n}\right)  =0$. The same argument (applied to $h_{n}$
instead of $e_{n}$) yields $\epsilon\left(  h_{n}\right)  =0$. Comparing this
with
\[
\left(  \epsilon\circ\omega\right)  \left(  h_{n}\right)  =\epsilon\left(
\underbrace{\omega\left(  h_{n}\right)  }_{\substack{=e_{n}\\\text{(by
Proposition \ref{prop.Lambda.omega-and-S}(b))}}}\right)  =\epsilon\left(
e_{n}\right)  =0,
\]
we obtain $\epsilon\left(  h_{n}\right)  =\left(  \epsilon\circ\omega\right)
\left(  h_{n}\right)  $.

Now, forget that we fixed $n$. We thus have shown that $\epsilon\left(
h_{n}\right)  =\left(  \epsilon\circ\omega\right)  \left(  h_{n}\right)  $ for
each positive integer $n$. In other words, the two $\kk$-algebra
homomorphisms $\epsilon$ and $\epsilon\circ\omega$ (from $\Lambda$ to
$\kk$) agree on each element of the family $\left\{  h_{n}\right\}
_{n\geq1}$ of elements of $\Lambda$.

But Proposition \ref{symm-antipode-and-h-basis} tells us that the family
$\left\{  h_{n}\right\}  _{n\geq1}$ generates the $\kk$-algebra
$\Lambda$. Thus, if two $\kk$-algebra homomorphisms from $\Lambda$
agree on each element of this family $\left\{  h_{n}\right\}  _{n\geq1}$, then
these two homomorphisms must be equal. Hence, the two $\kk$-algebra
homomorphisms $\epsilon$ and $\epsilon\circ\omega$ must be equal (since they
agree on each element of this family $\left\{  h_{n}\right\}  _{n\geq1}$). In
other words, we have $\epsilon=\epsilon\circ\omega$.\ This proves \eqref{pf.prop.Lambda.omega-and-S.f.2}.]

We have now proved both identities \eqref{pf.prop.Lambda.omega-and-S.f.1} and
\eqref{pf.prop.Lambda.omega-and-S.f.2}. These identities show that
$\omega:\Lambda\rightarrow\Lambda$ is a $\kk$-coalgebra homomorphism
(by the definition of a \textquotedblleft$\kk$-coalgebra
homomorphism\textquotedblright, since we already know that $\omega$ is
$\kk$-linear).

Thus, we know that $\omega$ is a $\kk$-algebra homomorphism and a
$\kk$-coalgebra homomorphism. In other words, $\omega$ is a
$\kk$-bialgebra homomorphism. Hence, Corollary
\ref{cor.bialg-mor-is-Hopf} (applied to $H_{1}=\Lambda$, $H_{2}=\Lambda$,
$S_{1}=S$ and $S_{2}=S$) shows that $\omega$ is a Hopf morphism. Since
$\omega$ is invertible, we thus conclude that $\omega$ is a Hopf algebra
isomorphism. Thus, $\omega$ is a Hopf algebra automorphism of $\Lambda$. This
proves Proposition \ref{prop.Lambda.omega-and-S}(f).
\end{verlong}

\begin{vershort}
(g) Part (g) of Proposition \ref{prop.Lambda.omega-and-S} can be proved in the
same way as we proved part (f) above, with the following three differences:

\begin{itemize}
\item We need to know that $S$ is a $\kk$-algebra homomorphism. (This
was shown during the proof of Proposition \ref{prop.Lambda.omega-and-S}(e) above.)

\item We need to know that $S$ is invertible. (This can be concluded from
Corollary \ref{commutative-implies-involutive-antipode-cor}, or from the fact
that $\omega$ is invertible using Proposition \ref{prop.Lambda.omega-and-S}(e).)

\item Instead of the formula $\omega\left(  h_{n}\right)  =e_{n}$ we now need
to use the formula $S\left(  h_{n}\right)  =\left(  -1\right)  ^{n}e_{n}$
(which is Proposition \ref{symm-antipode-and-h-basis}(iii)), and thus we incur
some signs in the computation.
\end{itemize}
\end{vershort}

\begin{verlong}
(g) Recall that $S$ is a $\kk$-algebra homomorphism from $\Lambda$ to
$\Lambda$. (Indeed, this was shown during the proof of Proposition
\ref{prop.Lambda.omega-and-S}(e) above.) Moreover, $\Lambda$ is commutative;
thus, Corollary \ref{commutative-implies-involutive-antipode-cor} (applied to
$A=\Lambda$) shows that the antipode $S$ of $\Lambda$ is an involution: that
is, $S^{2}=\operatorname*{id}\nolimits_{\Lambda}$. This yields that the map
$S$ is invertible.

Let us next show that $S:\Lambda\rightarrow\Lambda$ is a
$\kk$-coalgebra homomorphism.

Indeed, in order to show this, we shall check that the equalities%
\begin{equation}
\left(  S\otimes S\right)  \circ\Delta=\Delta\circ
S\label{pf.prop.Lambda.omega-and-S.g.1}%
\end{equation}
and%
\begin{equation}
\epsilon=\epsilon\circ S\label{pf.prop.Lambda.omega-and-S.g.2}%
\end{equation}
hold.

[\textit{Proof of \eqref{pf.prop.Lambda.omega-and-S.g.1}:} The map
$\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ is a $\kk$-algebra
homomorphism (since $\Lambda$ is a $\kk$-bialgebra). The map $S\otimes
S:\Lambda\otimes\Lambda\rightarrow\Lambda\otimes\Lambda$ is a
$\kk$-algebra homomorphism (since $S$ is a $\kk$-algebra homomorphism).
Hence, the maps $\left(  S\otimes S\right)  \circ\Delta:\Lambda\rightarrow
\Lambda\otimes\Lambda$ and $\Delta\circ S:\Lambda\rightarrow\Lambda
\otimes\Lambda$ are $\kk$-algebra homomorphisms (since they are
obtained by composing the maps $S\otimes S$, $\Delta$ and $S$, all of which
are $\kk$-algebra homomorphisms).

Let $n$ be a positive integer. We shall show that $\left(  \left(  S\otimes
S\right)  \circ\Delta\right)  \left(  h_{n}\right)  =\left(  \Delta\circ
S\right)  \left(  h_{n}\right)  $.

Indeed, Proposition \ref{symm-comultiplication-formulas-prop}(ii) yields%
\[
\Delta\left(  e_{n}\right)  =\sum_{i+j=n}e_{i}\otimes e_{j}%
\]
(where the summation sign \textquotedblleft$\sum_{i+j=n}$\textquotedblright%
\ stands for \textquotedblleft$\sum_{\substack{\left(  i,j\right)
\in\mathbb{N}^{2};\\i+j=n}}$\textquotedblright). Likewise, Proposition
\ref{symm-comultiplication-formulas-prop}(iii) yields%
\[
\Delta\left(  h_{n}\right)  =\sum_{i+j=n}h_{i}\otimes h_{j}.
\]
Now,
\begin{align*}
\left(  \left(  S\otimes S\right)  \circ\Delta\right)  \left(  h_{n}\right)
& =\left(  S\otimes S\right)  \left(  \underbrace{\Delta\left(  h_{n}\right)
}_{=\sum_{i+j=n}h_{i}\otimes h_{j}}\right)  =\left(  S\otimes S\right)
\left(  \sum_{i+j=n}h_{i}\otimes h_{j}\right)  \\
& =\sum_{i+j=n}\underbrace{S\left(  h_{i}\right)  }_{\substack{=\left(
-1\right)  ^{i}e_{i}\\\text{(by Proposition \ref{symm-antipode-and-h-basis}%
(iii),}\\\text{applied to }i\\\text{instead of }n\text{)}}}\otimes
\underbrace{S\left(  h_{j}\right)  }_{\substack{=\left(  -1\right)  ^{j}%
e_{j}\\\text{(by Proposition \ref{symm-antipode-and-h-basis}(iii),}%
\\\text{applied to }j\\\text{instead of }n\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }S\otimes
S\right)  \\
& =\sum_{i+j=n}\underbrace{\left(  -1\right)  ^{i}e_{i}\otimes\left(
-1\right)  ^{j}e_{j}}_{=\left(  -1\right)  ^{i}\left(  -1\right)  ^{j}%
e_{i}\otimes e_{j}}=\sum_{i+j=n}\underbrace{\left(  -1\right)  ^{i}\left(
-1\right)  ^{j}}_{\substack{=\left(  -1\right)  ^{i+j}=\left(  -1\right)
^{n}\\\text{(since }i+j=n\text{)}}}e_{i}\otimes e_{j}=\sum_{i+j=n}\left(
-1\right)  ^{n}e_{i}\otimes e_{j}.
\end{align*}
Comparing this with
\begin{align*}
\left(  \Delta\circ S\right)  \left(  h_{n}\right)    & =\Delta\left(
\underbrace{S\left(  h_{n}\right)  }_{\substack{=\left(  -1\right)  ^{n}%
e_{n}\\\text{(by Proposition \ref{symm-antipode-and-h-basis}(iii))}}}\right)
=\Delta\left(  \left(  -1\right)  ^{n}e_{n}\right)  \\
& =\left(  -1\right)  ^{n}\underbrace{\Delta\left(  e_{n}\right)  }%
_{=\sum_{i+j=n}e_{i}\otimes e_{j}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }\Delta\text{ is } \kk\text{-linear}\right)  \\
& =\left(  -1\right)  ^{n}\sum_{i+j=n}e_{i}\otimes e_{j}=\sum_{i+j=n}\left(
-1\right)  ^{n}e_{i}\otimes e_{j},
\end{align*}
we obtain $\left(  \left(  S\otimes S\right)  \circ\Delta\right)  \left(
h_{n}\right)  =\left(  \Delta\circ S\right)  \left(  h_{n}\right)  $.

Now, forget that we fixed $n$. We thus have shown that $\left(  \left(
S\otimes S\right)  \circ\Delta\right)  \left(  h_{n}\right)  =\left(
\Delta\circ S\right)  \left(  h_{n}\right)  $ for each positive integer $n$.
In other words, the two $\kk$-algebra homomorphisms $\left(  S\otimes
S\right)  \circ\Delta$ and $\Delta\circ S$ (from $\Lambda$ to $\Lambda
\otimes\Lambda$) agree on each element of the family $\left\{  h_{n}\right\}
_{n\geq1}$ of elements of $\Lambda$.

But Proposition \ref{symm-antipode-and-h-basis} tells us that the family
$\left\{  h_{n}\right\}  _{n\geq1}$ generates the $\kk$-algebra
$\Lambda$. Thus, if two $\kk$-algebra homomorphisms from $\Lambda$
agree on each element of this family $\left\{  h_{n}\right\}  _{n\geq1}$, then
these two homomorphisms must be equal. Hence, the two $\kk$-algebra
homomorphisms $\left(  S\otimes S\right)  \circ\Delta$ and $\Delta\circ S$
must be equal (since they agree on each element of this family $\left\{
h_{n}\right\}  _{n\geq1}$). In other words, we have $\left(  S\otimes
S\right)  \circ\Delta=\Delta\circ S$.\ This proves \eqref{pf.prop.Lambda.omega-and-S.g.1}.]

[\textit{Proof of \eqref{pf.prop.Lambda.omega-and-S.g.2}:} The map
$\epsilon:\Lambda\rightarrow \kk$ is a $\kk$-algebra homomorphism
(since $\Lambda$ is a $\kk$-bialgebra). The map $\epsilon\circ
S:\Lambda\rightarrow \kk$ is a $\kk$-algebra homomorphism (since
it is obtained by composing the maps $\epsilon$ and $S$, both of which are
$\kk$-algebra homomorphisms).

Let $n$ be a positive integer. We shall show that $\epsilon\left(
h_{n}\right)  =\left(  \epsilon\circ S\right)  \left(  h_{n}\right)  $.

Indeed, the counit $\epsilon$ of $\Lambda$ annihilates the graded component
$\Lambda_{n}$ (since $n$ is positive). In other words, $\epsilon\left(
\Lambda_{n}\right)  =0$. But $e_{n}\in\Lambda_{n}$, so that $\epsilon\left(
e_{n}\right)  \in\epsilon\left(  \Lambda_{n}\right)  =0$, and thus
$\epsilon\left(  e_{n}\right)  =0$. The same argument (applied to $h_{n}$
instead of $e_{n}$) yields $\epsilon\left(  h_{n}\right)  =0$. Comparing this
with
\begin{align*}
\left(  \epsilon\circ S\right)  \left(  h_{n}\right)    & =\epsilon\left(
\underbrace{S\left(  h_{n}\right)  }_{\substack{=\left(  -1\right)  ^{n}%
e_{n}\\\text{(by Proposition \ref{symm-antipode-and-h-basis}(iii))}}}\right)
=\epsilon\left(  \left(  -1\right)  ^{n}e_{n}\right)  \\
& =\left(  -1\right)  ^{n}\underbrace{\epsilon\left(  e_{n}\right)  }%
_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon\text{ is
} \kk\text{-linear}\right)  \\
& =0,
\end{align*}
we obtain $\epsilon\left(  h_{n}\right)  =\left(  \epsilon\circ S\right)
\left(  h_{n}\right)  $.

Now, forget that we fixed $n$. We thus have shown that $\epsilon\left(
h_{n}\right)  =\left(  \epsilon\circ S\right)  \left(  h_{n}\right)  $ for
each positive integer $n$. In other words, the two $\kk$-algebra
homomorphisms $\epsilon$ and $\epsilon\circ S$ (from $\Lambda$ to $\kk$)
agree on each element of the family $\left\{  h_{n}\right\}  _{n\geq1}$ of
elements of $\Lambda$.

But Proposition \ref{symm-antipode-and-h-basis} tells us that the family
$\left\{  h_{n}\right\}  _{n\geq1}$ generates the $\kk$-algebra
$\Lambda$. Thus, if two $\kk$-algebra homomorphisms from $\Lambda$
agree on each element of this family $\left\{  h_{n}\right\}  _{n\geq1}$, then
these two homomorphisms must be equal. Hence, the two $\kk$-algebra
homomorphisms $\epsilon$ and $\epsilon\circ S$ must be equal (since they agree
on each element of this family $\left\{  h_{n}\right\}  _{n\geq1}$). In other
words, we have $\epsilon=\epsilon\circ S$.\ This proves \eqref{pf.prop.Lambda.omega-and-S.g.2}.]

We have now proved both identities \eqref{pf.prop.Lambda.omega-and-S.g.1} and
\eqref{pf.prop.Lambda.omega-and-S.g.2}. These identities show that
$S:\Lambda\rightarrow\Lambda$ is a $\kk$-coalgebra homomorphism (by the
definition of a \textquotedblleft$\kk$-coalgebra
homomorphism\textquotedblright, since we already know that $S$ is $\kk$-linear).

Thus, we know that $S$ is a $\kk$-algebra homomorphism and a
$\kk$-coalgebra homomorphism. In other words, $S$ is a
$\kk$-bialgebra homomorphism. Hence, Corollary \ref{cor.bialg-mor-is-Hopf}
(applied to $H_{1}=\Lambda$, $H_{2}=\Lambda$, $S_{1}=S$ and $S_{2}=S$) shows
that $S$ is a Hopf morphism. Since $S$ is invertible, we thus conclude that
$S$ is a Hopf algebra isomorphism. Thus, $S$ is a Hopf algebra automorphism of
$\Lambda$. This proves Proposition \ref{prop.Lambda.omega-and-S}(g).
\end{verlong}

[\textit{Remark:} Let us sketch alternative approaches to proving parts (f)
and (g) of Proposition \ref{prop.Lambda.omega-and-S}:

\textit{Alternative proof of Proposition \ref{prop.Lambda.omega-and-S}(g)
(sketched):} We have already seen (during the proof of Proposition
\ref{prop.Lambda.omega-and-S}(e) above) that $S$ is a $\kk$-algebra
homomorphism. Corollary \ref{commutative-implies-involutive-antipode-cor}
(applied to $S = \Lambda$)
yields that $S$ is an involution; thus, $S$ is invertible. Moreover, the Hopf
algebra $\Lambda$ is cocommutative (by Exercise \ref{exe.Lambda.cocommutative}%
(a)). But Exercise \ref{exe.coalg-anti} (applied to $A=\Lambda$) yields that
the antipode $S$ of $\Lambda$ is a $\kk$-coalgebra anti-endomorphism of
$\Lambda$. In other words, $S$ is a $\kk$-coalgebra anti-homomorphism
from $\Lambda$ to $\Lambda$. But Exercise \ref{exe.comm-cocomm.anti}(b)
(applied to $A=\Lambda$ and $B=\Lambda$) shows that the $\kk$-coalgebra
anti-homomorphisms from $\Lambda$ to $\Lambda$ are the same as the
$\kk$-coalgebra homomorphisms from $\Lambda$ to $\Lambda$ (since the
$\kk$-coalgebra $\Lambda$ is cocommutative). Hence, $S$ is a
$\kk$-coalgebra homomorphism from $\Lambda$ to $\Lambda$ (since $S$ is
a $\kk$-coalgebra anti-homomorphism from $\Lambda$ to $\Lambda$). We
now know that $S$ is a $\kk$-algebra homomorphism and a
$\kk$-coalgebra homomorphism at the same time. Hence, $S$ is a
$\kk$-bialgebra homomorphism, therefore a Hopf morphism (by
Corollary~\ref{cor.bialg-mor-is-Hopf}), and thus a Hopf algebra automorphism
(since $S$ is invertible). This proves Proposition
\ref{prop.Lambda.omega-and-S}(g) again.

\textit{Alternative proof of Proposition \ref{prop.Lambda.omega-and-S}(f)
(sketched):} Proposition \ref{prop.Lambda.omega-and-S}(g) shows that $S$ is a
Hopf algebra automorphism. Define $q$ and $D_{q}$ as in the proof of
Proposition \ref{prop.Lambda.omega-and-S}(e) above. It is easy to see that
$D_{q}$ is an involution; thus, $D_{q}$ is invertible. But $D_{q}$ is a
$\kk$-bialgebra homomorphism (as we have seen in the proof of
Proposition \ref{prop.Lambda.omega-and-S}(e) above), therefore a Hopf morphism
(by Corollary~\ref{cor.bialg-mor-is-Hopf}), and thus a Hopf algebra
automorphism (since $D_{q}$ is invertible). Now, recall that $\omega\circ
D_{q}=S$ (as we have seen in the proof of Proposition
\ref{prop.Lambda.omega-and-S}(e) above). Hence, $\omega=S\circ\left(
D_{q}\right)  ^{-1}$ (since $D_{q}$ is invertible). This shows that $\omega$
is a Hopf algebra automorphism (since both $S$ and $D_{q}$ are Hopf algebra
automorphisms). This proves Proposition \ref{prop.Lambda.omega-and-S}(f) again.]

(h) Let $\lambda$ be a partition.

\begin{vershort}
We can easily obtain \eqref{eq.exe.Lambda.omegahlam.p}
by multiplicativity:

[\textit{Proof of \eqref{eq.exe.Lambda.omegahlam.p}:}
Let $\ell$ be the length $\ell\left(  \lambda\right)  $ of the partition
$\lambda$. Then, $p_{\lambda}$ is defined as $p_{\lambda_{1}}p_{\lambda_{2}%
}\cdots p_{\lambda_{\ell}}$, and thus we have
\begin{align*}
\omega\left(  p_{\lambda}\right)   &  =\omega\left(  p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}\right)  =\omega\left(  p_{\lambda
_{1}}\right)  \omega\left(  p_{\lambda_{2}}\right)  \cdots\omega\left(
p_{\lambda_{\ell}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\omega\text{ is a } \kk\text{-algebra homomorphism}\right) \\
&  =\left(  -1\right)  ^{\lambda_{1}-1}p_{\lambda_{1}}\cdot\left(  -1\right)
^{\lambda_{2}-1}p_{\lambda_{2}}\cdots\left(  -1\right)
^{\lambda_{\ell}-1}p_{\lambda_{\ell}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{prop.Lambda.omega-and-S}(c)}\right) \\
&  =\underbrace{\left(  -1\right)  ^{\left(  \lambda_{1}+\lambda_{2}%
+\cdots+\lambda_{\ell}\right)  -\ell}}_{\substack{=\left(  -1\right)
^{\left\vert \lambda\right\vert -\ell\left(  \lambda\right)  }\\\text{(since
}\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}=\left\vert \lambda\right\vert
\\\text{and }\ell=\ell\left(  \lambda\right)  \text{)}}}\underbrace{p_{\lambda
_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}}_{=p_{\lambda}}=\left(
-1\right)  ^{\left\vert \lambda\right\vert -\ell\left(  \lambda\right)
}p_{\lambda}.
\end{align*}
This proves \eqref{eq.exe.Lambda.omegahlam.p}.]

A similar argument (but using Proposition~\ref{prop.Lambda.omega-and-S}(b) instead of
Proposition~\ref{prop.Lambda.omega-and-S}(c)) proves \eqref{eq.exe.Lambda.omegahlam.h}.
Finally, a similar argument (but using Proposition~\ref{prop.Lambda.omega-and-S}(a)
instead of Proposition~\ref{prop.Lambda.omega-and-S}(c)) proves
\eqref{eq.exe.Lambda.omegahlam.e}.
\end{vershort}

\begin{verlong}
Write the partition $\lambda$ in the form $\lambda
=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $, where
$\ell=\ell\left(  \lambda\right)  $. Then, the definition of $p_{\lambda}$
yields%
\begin{equation}
p_{\lambda}=p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}.
\label{sol.Lambda.omegahlam.plam=}%
\end{equation}
Applying the map $\omega$ to both sides of this equality, we obtain
$\omega\left(  p_{\lambda}\right)  =\omega\left(  p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}\right)  =\omega\left(  p_{\lambda
_{1}}\right)  \omega\left(  p_{\lambda_{2}}\right)  \cdots\omega\left(
p_{\lambda_{\ell}}\right)  $ (since $\omega$ is a $\kk$-algebra
homomorphism). Thus,
\[
\omega\left(  p_{\lambda}\right)  =\omega\left(  p_{\lambda_{1}}\right)
\omega\left(  p_{\lambda_{2}}\right)  \cdots\omega\left(  p_{\lambda_{\ell}%
}\right)  =\prod_{i=1}^{\ell}\omega\left(  p_{\lambda_{i}}\right)  .
\]


But $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $.
Hence, the definition of $\left\vert \lambda\right\vert $ yields $\left\vert
\lambda\right\vert =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}=\sum
_{i=1}^{\ell}\lambda_{i}$, so that $\sum_{i=1}^{\ell}\lambda_{i}=\left\vert
\lambda\right\vert $ and
\[
\sum_{i=1}^{\ell}\left(  \lambda_{i}-1\right)  =\underbrace{\sum_{i=1}^{\ell
}\lambda_{i}}_{=\left\vert \lambda\right\vert }-\underbrace{\sum_{i=1}^{\ell
}1}_{=\ell\cdot1=\ell=\ell\left(  \lambda\right)  }=\left\vert \lambda
\right\vert -\ell\left(  \lambda\right)  .
\]


For every $i\in\left\{  1,2,\ldots,\ell\right\}  $, we have $i\leq\ell
=\ell\left(  \lambda\right)  $. Hence, for every $i\in\left\{  1,2,\ldots
,\ell\right\}  $, the number $\lambda_{i}$ is a positive integer. Hence, for
every $i\in\left\{  1,2,\ldots,\ell\right\}  $, we have the equality
$\omega\left(  p_{\lambda_{i}}\right)  =\left(  -1\right)  ^{\lambda_{i}%
-1}p_{\lambda_{i}}$ (by Proposition~\ref{prop.Lambda.omega-and-S}(c),
applied to $n=\lambda_{i}$). Therefore,%
\begin{align*}
\prod_{i=1}^{\ell}\underbrace{\omega\left(  p_{\lambda_{i}}\right)
}_{=\left(  -1\right)  ^{\lambda_{i}-1}e_{\lambda_{i}}} &  =\prod_{i=1}^{\ell
}\left(  \left(  -1\right)  ^{\lambda_{i}-1}e_{\lambda_{i}}\right)
=\underbrace{\left(  \prod_{i=1}^{\ell}\left(  -1\right)  ^{\lambda_{i}%
-1}\right)  }_{\substack{=\left(  -1\right)  ^{\sum_{i=1}^{\ell}\left(
\lambda_{i}-1\right)  }\\=\left(  -1\right)  ^{\left\vert \lambda\right\vert
-\ell\left(  \lambda\right)  }\\\text{(since }\sum_{i=1}^{\ell}\left(
\lambda_{i}-1\right)  =\left\vert \lambda\right\vert -\ell\left(
\lambda\right)  \text{)}}}\underbrace{\prod_{i=1}^{\ell}p_{\lambda_{i}}%
}_{\substack{=p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}%
}=p_{\lambda}\\\text{(by \eqref{sol.Lambda.omegahlam.plam=})}}}\\
&  =\left(  -1\right)  ^{\left\vert \lambda\right\vert -\ell\left(
\lambda\right)  }p_{\lambda}.
\end{align*}
Hence,%
\[
\omega\left(  p_{\lambda}\right)  =\prod_{i=1}^{\ell}\omega\left(
p_{\lambda_{i}}\right)  =\left(  -1\right)  ^{\left\vert \lambda\right\vert
-\ell\left(  \lambda\right)  }p_{\lambda}.
\]
This proves \eqref{eq.exe.Lambda.omegahlam.p}.

Recall that we have written $\lambda$ in the form $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $, where $\ell=\ell\left(
\lambda\right)  $. Then,
\begin{equation}
h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }h_{\lambda}\right)
\label{sol.Lambda.omegahlam.hlam=}%
\end{equation}
and%
\begin{equation}
e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{\ell}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }e_{\lambda}\right)  .
\label{sol.Lambda.omegahlam.elam=}%
\end{equation}
Applying the map $\omega$ to both sides of the equality
\eqref{sol.Lambda.omegahlam.hlam=}, we obtain $\omega\left(  h_{\lambda}\right)
=\omega\left(  h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}\right)
=\omega\left(  h_{\lambda_{1}}\right)  \omega\left(  h_{\lambda_{2}}\right)
\cdots\omega\left(  h_{\lambda_{\ell}}\right)  $ (since $\omega$ is a
$\kk$-algebra homomorphism). Thus,
\[
\omega\left(  h_{\lambda}\right)  =\omega\left(  h_{\lambda_{1}}\right)
\omega\left(  h_{\lambda_{2}}\right)  \cdots\omega\left(  h_{\lambda_{\ell}%
}\right)  =\prod_{i=1}^{\ell}\omega\left(  h_{\lambda_{i}}\right)  .
\]


For every $i\in\left\{  1,2,\ldots,\ell\right\}  $, the number $\lambda_{i}$
is a positive integer (as we have already proved). Hence, for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $, we have the equality $\omega\left(
h_{\lambda_{i}}\right)  =e_{\lambda_{i}}$ (by
Proposition~\ref{prop.Lambda.omega-and-S}(b), applied to $n=\lambda_{i}$).
Hence,
\[
\prod_{i=1}^{\ell}\underbrace{\omega\left(  h_{\lambda_{i}}\right)
}_{=e_{\lambda_{i}}}=\prod_{i=1}^{\ell}e_{\lambda_{i}}=e_{\lambda_{1}%
}e_{\lambda_{2}}\cdots e_{\lambda_{\ell}}=e_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.Lambda.omegahlam.elam=}}\right)  .
\]
Thus,%
\[
\omega\left(  h_{\lambda}\right)  =\prod_{i=1}^{\ell}\omega\left(
h_{\lambda_{i}}\right)  =e_{\lambda}.
\]
This proves \eqref{eq.exe.Lambda.omegahlam.h}.

Finally, applying the map $\omega$ to both sides of the equality
\eqref{sol.Lambda.omegahlam.elam=}, we obtain $\omega\left(  e_{\lambda}\right)
=\omega\left(  e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{\ell}}\right)
=\omega\left(  e_{\lambda_{1}}\right)  \omega\left(  e_{\lambda_{2}}\right)
\cdots\omega\left(  e_{\lambda_{\ell}}\right)  $ (since $\omega$ is a
$\kk$-algebra homomorphism). Thus,
\[
\omega\left(  e_{\lambda}\right)  =\omega\left(  e_{\lambda_{1}}\right)
\omega\left(  e_{\lambda_{2}}\right)  \cdots\omega\left(  e_{\lambda_{\ell}%
}\right)  =\prod_{i=1}^{\ell}\omega\left(  e_{\lambda_{i}}\right)  .
\]


For every $i\in\left\{  1,2,\ldots,\ell\right\}  $, the number $\lambda_{i}$
is a positive integer (as we have already proved). Hence, for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $, we have the equality $\omega\left(
e_{\lambda_{i}}\right)  =h_{\lambda_{i}}$ (by
Proposition~\ref{prop.Lambda.omega-and-S}(a), applied to $n=\lambda_{i}$).
Hence,
\[
\prod_{i=1}^{\ell}\underbrace{\omega\left(  e_{\lambda_{i}}\right)
}_{=h_{\lambda_{i}}}=\prod_{i=1}^{\ell}h_{\lambda_{i}}=h_{\lambda_{1}%
}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}=h_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.Lambda.omegahlam.hlam=}}\right)  .
\]
Thus,%
\[
\omega\left(  e_{\lambda}\right)  =\prod_{i=1}^{\ell}\omega\left(
e_{\lambda_{i}}\right)  =h_{\lambda}.
\]
This proves \eqref{eq.exe.Lambda.omegahlam.e}.
\end{verlong}

We have now proved all three equalities \eqref{eq.exe.Lambda.omegahlam.h},
\eqref{eq.exe.Lambda.omegahlam.e} and \eqref{eq.exe.Lambda.omegahlam.p}.
Thus, Proposition~\ref{prop.Lambda.omega-and-S}(h) is proved.

(i) Proposition \ref{prop.Lambda.omega-and-S}(d) yields that the map $\omega$
is a $\kk$-algebra automorphism of $\Lambda$ and an involution. Thus,
$\omega$ is an involution. In other words, $\omega$ is invertible, and the
inverse of $\omega$ is $\omega$ itself.

We know that $\Lambda$ is a connected graded bialgebra. Thus, Proposition
\ref{graded-connected-bialgebras-have-antipodes} (applied to $A=\Lambda$)
yields that $\Lambda$ has a unique antipode $S$, which is a graded map
$\Lambda\overset{S}{\longrightarrow}\Lambda$, endowing it with a Hopf
structure. Thus, in particular, the antipode $S$ of $\Lambda$ is a graded map.
In other words,
\begin{equation}
S\left(  \Lambda_n \right)  \subset\Lambda_n \ \ \ \ \ \ \ \ \ \ \text{for
each }n\in\NN .\label{pf.prop.Lambda.omega-and-S.j.1}%
\end{equation}


Now, let $n\in\NN $. Let $f\in\Lambda_n $. Then, Proposition
\ref{prop.Lambda.omega-and-S}(e) yields $S\left(  f\right)  =\left(
-1\right)  ^n \omega\left(  f\right)  $. Hence,%
\[
\left(  -1\right)  ^n \underbrace{S\left(  f\right)  }_{=\left(  -1\right)
^n \omega\left(  f\right)  }=\underbrace{\left(  -1\right)  ^n \left(
-1\right)  ^n }_{\substack{=\left(  -1\right)  ^{n+n}=\left(  -1\right)
^{2n}=1\\\text{(since }2n\text{ is even)}}}\omega\left(  f\right)
=\omega\left(  f\right)  ,
\]
so that $\omega\left(  f\right)  =\left(  -1\right)  ^n S\left(
\underbrace{f}_{\in\Lambda_n }\right)  \in\left(  -1\right)  ^n %
\underbrace{S\left(  \Lambda_n \right)  }_{\substack{\subset\Lambda
_n \\\text{(by \eqref{pf.prop.Lambda.omega-and-S.j.1})}}}\subset\left(
-1\right)  ^n \Lambda_n \subset\Lambda_n $ (since $\Lambda_n $ is a
$\kk$-module).

Forget that we fixed $f$. We thus have proved that $\omega\left(  f\right)
\in\Lambda_n $ for each $f\in\Lambda_n $. In other words, $\omega\left(
\Lambda_n \right)  \subset\Lambda_n $.

Forget that we fixed $n$. We thus have shown that $\omega\left(  \Lambda
_n \right)  \subset\Lambda_n $ for each $n\in\NN $. Hence, the
$\kk$-linear map $\omega:\Lambda\rightarrow\Lambda$ is graded. In other
words, the inverse of $\omega$ is graded (since the inverse of $\omega$ is
$\omega$ itself).

Now, we know that the $\kk$-linear map $\omega$ is graded, and its
inverse is also graded. In other words, the $\kk$-linear map $\omega$
is an isomorphism of graded $\kk$-modules. This proves Proposition
\ref{prop.Lambda.omega-and-S}(i).

(j) From \eqref{eq.exe.Lambda.omegahlam.e}, we know that $\omega\left(
e_{\lambda}\right)  =h_{\lambda}$ for each partition $\lambda$. In other
words, $\omega\left(  e_{\lambda}\right)  =h_{\lambda}$ for each $\lambda
\in \Par$. In other words, $\left(  \omega\left(  e_{\lambda
}\right)  \right)  _{\lambda\in\Par }=\left(  h_{\lambda
}\right)  _{\lambda\in\Par }$.

Proposition \ref{many-symmetric-function-bases-prop} says (among other things)
that the family $\left\{  e_{\lambda}\right\}  $, as $\lambda$ runs through
all partitions, is a graded basis of the graded $\kk$-module $\Lambda$.
In other words, the family $\left(  e_{\lambda}\right)  _{\lambda
\in\Par }$ is a graded basis of the graded $\kk$-module
$\Lambda$. On the other hand, Proposition \ref{prop.Lambda.omega-and-S}(i)
shows that the map $\omega$ is an isomorphism of graded $\kk$-modules.

Now, recall the following well-known fact: If $V$ and $W$ are two graded
$\kk$-modules, and if $f:V\rightarrow W$ is an isomorphism of graded
$\kk$-modules, then $f$ sends any graded basis of $V$ to a graded basis
of $W$. In other words, if $V$ and $W$ are two graded $\kk$-modules,
and if $f:V\rightarrow W$ is an isomorphism of graded $\kk$-modules,
and if $\left(  v_{i}\right)  _{i\in I}$ is a graded basis of $V$, then
$\left(  f\left(  v_{i}\right)  \right)  _{i\in I}$ is a graded basis of $W$.

We can apply this fact to $V=\Lambda$, $W=\Lambda$, $f=\omega$,
$I=\Par$ and $\left(  v_{i}\right)  _{i\in I}=\left(
e_{\lambda}\right)  _{\lambda\in\Par }$ (since we know that
$\omega:\Lambda\rightarrow\Lambda$ is an isomorphism of graded $\kk %
$-modules, and since we know that the family $\left(  e_{\lambda}\right)
_{\lambda\in\Par }$ is a graded basis of $\Lambda$). We thus
conclude that $\left(  \omega\left(  e_{\lambda}\right)  \right)  _{\lambda
\in\Par }$ is a graded basis of $\Lambda$. In other words, the
family $\left(  h_{\lambda}\right)  _{\lambda\in\Par }$ is a
graded basis of $\Lambda$ (since $\left(  \omega\left(  e_{\lambda}\right)
\right)  _{\lambda\in\Par }=\left(  h_{\lambda}\right)
_{\lambda\in\Par }$). This proves Proposition
\ref{prop.Lambda.omega-and-S}(j).
\end{proof}

We have now proved Proposition \ref{prop.Lambda.omega-and-S}; thus, Exercise
\ref{exe.Lambda.omega-and-S} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.infinite-indep}}
\textit{Solution to Exercise \ref{exe.Lambda.infinite-indep}.} Recall that for
each partition $\lambda$,%
\begin{equation}
\text{the element }q_{\lambda}\in\Lambda\text{ is homogeneous of degree
}\left\vert \lambda\right\vert \text{.} \label{sol.Lambda.infinite-indep.hg}%
\end{equation}


(a) Let $\left(  a_{\lambda}\right)  _{\lambda\in \Par }%
\in\kk^{ \Par }$ and $\left(  b_{\lambda}\right)
_{\lambda\in \Par }\in\kk^{ \Par }$ be two
families satisfying (\ref{eq.exe.Lambda.infinite-indep.a.ass}) in
$\kk\left[  \left[  \xx\right]  \right]  $. We must prove that
$\left(  a_{\lambda}\right)  _{\lambda\in \Par }=\left(
b_{\lambda}\right)  _{\lambda\in \Par }$.

Fix $n\in \NN$. Consider the $\kk$-linear map $\pi_{n}%
:\kk\left[  \left[  {\mathbf{x}}\right]  \right]  \rightarrow
\kk\left[  \left[  {\mathbf{x}}\right]  \right]  $ that sends each
power series $f\in\kk\left[  \left[  {\mathbf{x}}\right]  \right]  $ to
its homogeneous component of degree $n$. Thus, $\pi_{n}$ has the following properties:

\begin{itemize}
\item If $f\in\kk\left[  \left[  \xx\right]  \right]  $ is a
power series that is homogeneous of degree $n$, then%
\begin{equation}
\pi_{n}\left(  f\right)  =f. \label{sol.Lambda.infinite-indep.a.hg1}%
\end{equation}


\item If $f\in\kk\left[  \left[  \xx\right]  \right]  $ is a
power series that is homogeneous of degree $\neq n$, then%
\begin{equation}
\pi_{n}\left(  f\right)  =0. \label{sol.Lambda.infinite-indep.a.hg2}%
\end{equation}


\item The map $\pi_{n}$ is $\kk$-linear and continuous (with respect to
the topology on $\kk\left[  \left[  \xx\right]  \right]  $).
\end{itemize}

Hence, each $\lambda\in \Par $ satisfying $\left\vert
\lambda\right\vert =n$ satisfies%
\begin{equation}
\pi_{n}\left(  q_{\lambda}\left(  \xx\right)  \right)  =q_{\lambda
}\left(  \xx\right)  \label{sol.Lambda.infinite-indep.a.pin1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.a.pin1}):} Let
$\lambda\in \Par $ be such that $\left\vert \lambda\right\vert
=n$. From (\ref{sol.Lambda.infinite-indep.hg}), we know that the element
$q_{\lambda}\in\Lambda$ is homogeneous of degree $\left\vert \lambda
\right\vert $. Thus, the power series $q_{\lambda}\left(  \xx\right)  $
is homogeneous of degree $\left\vert \lambda\right\vert =n$. Hence,
(\ref{sol.Lambda.infinite-indep.a.hg1}) (applied to $f=q_{\lambda}\left(
\xx\right)  $) yields $\pi_{n}\left(  q_{\lambda}\left(  \xx%
\right)  \right)  =q_{\lambda}\left(  \xx\right)  $. This proves
(\ref{sol.Lambda.infinite-indep.a.pin1}).}. Furthermore, each $\lambda
\in \Par $ satisfying $\left\vert \lambda\right\vert \neq n$
satisfies%
\begin{equation}
\pi_{n}\left(  q_{\lambda}\left(  \xx\right)  \right)
=0\label{sol.Lambda.infinite-indep.a.pin2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.a.pin2}):} Let
$\lambda\in \Par $ be such that $\left\vert \lambda\right\vert
\neq n$. From (\ref{sol.Lambda.infinite-indep.hg}), we know that the element
$q_{\lambda}\in\Lambda$ is homogeneous of degree $\left\vert \lambda
\right\vert $. Thus, the power series $q_{\lambda}\left(  \xx\right)  $
is homogeneous of degree $\left\vert \lambda\right\vert \neq n$. Hence,
(\ref{sol.Lambda.infinite-indep.a.hg2}) (applied to $f=q_{\lambda}\left(
\xx\right)  $) yields $\pi_{n}\left(  q_{\lambda}\left(  \xx%
\right)  \right)  =0$. This proves (\ref{sol.Lambda.infinite-indep.a.pin2}).}.

But the map $\pi_{n}$ is $\kk$-linear and continuous. Thus, it respects
infinite sums. Hence,
\begin{align}
\pi_{n}\left(  \sum_{\lambda\in \Par }a_{\lambda}q_{\lambda
}\left(  \xx\right)  \right)   &  =\sum_{\lambda\in \Par}
a_{\lambda}\pi_{n}\left(  q_{\lambda}\left(  \xx\right)  \right)
=\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert \lambda\right\vert
=n}}a_{\lambda}\underbrace{\pi_{\lambda}\left(  q_{\lambda}\left(
\xx\right)  \right)  }_{\substack{=q_{\lambda}\left(  \xx%
\right)  \\\text{(by (\ref{sol.Lambda.infinite-indep.a.pin1}))}}%
}+\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert \lambda
\right\vert \neq n}}a_{\lambda}\underbrace{\pi_{\lambda}\left(  q_{\lambda
}\left(  \xx\right)  \right)  }_{\substack{=0\\\text{(by
(\ref{sol.Lambda.infinite-indep.a.pin2}))}}}\nonumber\\
&  =\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}q_{\lambda}\left(  \xx\right)
+\underbrace{\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert
\lambda\right\vert \neq n}}a_{\lambda}0}_{=0}=\sum\limits_{\substack{\lambda
\in \Par ;\\\left\vert \lambda\right\vert =n}}a_{\lambda
}q_{\lambda}\left(  \xx\right)  . \label{sol.Lambda.infinite-indep.4}%
\end{align}
The same argument (applied to the family $\left(  b_{\lambda}\right)
_{\lambda\in \Par }$ instead of $\left(  a_{\lambda}\right)
_{\lambda\in \Par }$) yields%
\begin{equation}
\pi_{n}\left(  \sum_{\lambda\in \Par }b_{\lambda}q_{\lambda
}\left(  \xx\right)  \right)  =\sum\limits_{\substack{\lambda\in
 \Par ;\\\left\vert \lambda\right\vert =n}}b_{\lambda}q_{\lambda
}\left(  \xx\right)  . \label{sol.Lambda.infinite-indep.5}%
\end{equation}


Now, applying the map $\pi_{n}$ to both sides of the equality
(\ref{eq.exe.Lambda.infinite-indep.a.ass}), we obtain%
\[
\pi_{n}\left(  \sum_{\lambda\in \Par }a_{\lambda}q_{\lambda
}\left(  \xx\right)  \right)  =\pi_{n}\left(  \sum_{\lambda
\in \Par }b_{\lambda}q_{\lambda}\left(  \xx\right)
\right)  =\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert
\lambda\right\vert =n}}b_{\lambda}q_{\lambda}\left(  \xx\right)  .
\]
Comparing this with (\ref{sol.Lambda.infinite-indep.4}), we obtain%
\begin{equation}
\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert \lambda\right\vert
=n}}a_{\lambda}q_{\lambda}\left(  \xx\right)  =\sum\limits_{\substack{\lambda
\in \Par ;\\\left\vert \lambda\right\vert =n}}b_{\lambda
}q_{\lambda}\left(  \xx\right)  . \label{sol.Lambda.infinite-indep.7}%
\end{equation}
Notice that both sums appearing in this equality are finite (since there are
only finitely many $\lambda\in \Par $ satisfying $\left\vert
\lambda\right\vert =n$). Since the family $\left(  q_{\lambda}\left(
\xx\right)  \right)  _{\lambda\in \Par ;\ \left\vert
\lambda\right\vert =n}$ is $\kk$-linearly
independent\footnote{\textit{Proof.} The family $\left(
\underbrace{q_{\lambda}\left(  \xx\right)  }_{=q_{\lambda}}\right)
_{\lambda\in \Par }=\left(  q_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\kk$-module $\Lambda$, and
thus is $\kk$-linearly independent. Hence, the family $\left(
q_{\lambda}\left(  \xx\right)  \right)  _{\lambda\in \Par %
;\ \left\vert \lambda\right\vert =n}$ is $\kk$-linearly independent as
well (since it is a subfamily of this family $\left(  q_{\lambda}\left(
\xx\right)  \right)  _{\lambda\in \Par }$).}, we can thus
conclude from (\ref{sol.Lambda.infinite-indep.7}) that%
\begin{equation}
a_{\lambda}=b_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda
\in \Par \text{ satisfying }\left\vert \lambda\right\vert =n.
\label{sol.Lambda.infinite-indep.9}%
\end{equation}


Now, forget that we fixed $n$. We thus have proven
(\ref{sol.Lambda.infinite-indep.9}) for each $n\in \NN$.

Now, let $\lambda\in \Par $ be arbitrary. Then, $\left\vert
\lambda\right\vert \in \NN$. Hence, (\ref{sol.Lambda.infinite-indep.9})
(applied to $n=\left\vert \lambda\right\vert $) yields $a_{\lambda}%
=b_{\lambda}$.

Now, forget that we fixed $\lambda$. We thus have proven that $a_{\lambda
}=b_{\lambda}$ for each $\lambda\in \Par $. In other words,
$\left(  a_{\lambda}\right)  _{\lambda\in \Par }=\left(
b_{\lambda}\right)  _{\lambda\in \Par }$. This solves Exercise
\ref{exe.Lambda.infinite-indep} (a).

(c) We first go afield. Recall that $\left(  m_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\kk$-module $\Lambda$.
Hence, the family \newline$\left(  m_{\mu}\otimes m_{\nu}\otimes m_{\lambda
}\right)  _{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ is a
basis of the $\kk$-module $\Lambda\otimes\Lambda\otimes\Lambda$.

On the other hand, $\left(  q_{\lambda}\right)  _{\lambda\in
 \Par }$ is a basis of the $\kk$-module $\Lambda$. Hence,
the family $\left(  q_{\mu}\otimes q_{\nu}\otimes q_{\lambda}\right)
_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ is a basis of
the $\kk$-module $\Lambda\otimes\Lambda\otimes\Lambda$.

We shall now show that the family $\left(  m_{\mu}\left(  {\mathbf{x}}\right)
m_{\nu}\left(  {\mathbf{y}}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)
\right)  _{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ of
elements of $\kk\left[  \left[  {\mathbf{x}},{\mathbf{y}},{\mathbf{z}%
}\right]  \right]  $ is $\kk$-linearly independent.

If $\mathfrak{m}$ is a monomial, and if $f$ is a power series, then we let
$\left[  \mathfrak{m}\right]  f$ denote the coefficient of $\mathfrak{m}$ in
$f$.

Any $\tau\in \Par $ and $\lambda\in \Par $ satisfy%
\begin{equation}
\left[  \mathbf{x}^{\tau}\right]  \left(  m_{\lambda}\left(  \xx%
\right)  \right)  =\delta_{\tau,\lambda}
\label{sol.Lambda.infinite-indep.c.kronecker1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.c.kronecker1}):}
Let $\tau\in \Par $ and $\lambda\in \Par $. The set
$\Symm_{\left(  \infty\right)  }\lambda$ clearly contains $\lambda$,
since $\lambda=\underbrace{\operatorname*{id}}_{\in\Symm_{\left(
\infty\right)  }}\cdot\lambda\in\Symm_{\left(  \infty\right)  }\lambda
$. Moreover, each $\alpha\in\Symm_{\left(  \infty\right)  }\lambda$
satisfying $\alpha\neq\lambda$ must satisfy%
\begin{equation}
\delta_{\tau,\alpha}=0. \label{sol.Lambda.infinite-indep.c.kronecker1.pf.2}%
\end{equation}
\par
[\textit{Proof of (\ref{sol.Lambda.infinite-indep.c.kronecker1.pf.2}):} Let
$\alpha\in\Symm_{\left(  \infty\right)  }\lambda$ be such that
$\alpha\neq\lambda$.
\par
Assume (for the sake of contradiction) that $\tau=\alpha$. Then, $\alpha
=\tau\in \Par $. Hence, the sequence $\alpha$ is nonincreasing.
But the sequence $\lambda$ is also nonincreasing (since $\lambda
\in \Par $).
\par
From $\alpha\in\Symm_{\left(  \infty\right)  }\lambda$, we conclude
that $\alpha$ is a rearrangement of the partition $\lambda$. But $\lambda$ is
also a rearrangement of $\lambda$. However, there is only one nonincreasing
rearrangement of $\lambda$. In other words, if $\mu$ and $\nu$ are two
nonincreasing rearrangements of $\lambda$, then $\mu=\nu$. Applying this to
$\mu=\alpha$ and $\nu=\lambda$, we conclude that $\alpha=\lambda$ (since both
$\alpha$ and $\lambda$ are nonincreasing rearrangements of $\lambda$). This
contradicts $\alpha\neq\lambda$.
\par
This contradiction shows that our assumption (that $\tau=\alpha$) was false.
Hence, we have $\tau\neq\alpha$. Thus, $\delta_{\tau,\alpha}=0$. This proves
(\ref{sol.Lambda.infinite-indep.c.kronecker1.pf.2}).]
\par
We have $m_{\lambda}\left(  \xx\right)  =m_{\lambda}=\sum_{\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}{\mathbf{x}}^{\alpha}$ (by
(\ref{monomial-symmetric-function-definition})). Thus,%
\begin{align*}
\left[  \mathbf{x}^{\tau}\right]  \left(  \underbrace{m_{\lambda}\left(
\xx\right)  }_{=\sum_{\alpha\in\Symm_{\left(  \infty\right)
}\lambda}{\mathbf{x}}^{\alpha}}\right)   &  =\left[  \mathbf{x}^{\tau}\right]
\left(  \sum_{\alpha\in\Symm_{\left(  \infty\right)  }\lambda
}{\mathbf{x}}^{\alpha}\right)  =\sum_{\alpha\in\Symm_{\left(
\infty\right)  }\lambda}\underbrace{\left[  \mathbf{x}^{\tau}\right]  \left(
{\mathbf{x}}^{\alpha}\right)  }_{=\delta_{\tau,\alpha}}=\sum_{\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}\delta_{\tau,\alpha}\\
&  =\delta_{\tau,\lambda}+\sum\limits_{\substack{\alpha\in\Symm_{\left(
\infty\right)  }\lambda;\\\alpha\neq\lambda}}\underbrace{\delta_{\tau,\alpha}%
}_{\substack{=0\\\text{(by (\ref{sol.Lambda.infinite-indep.c.kronecker1.pf.2}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }\alpha=\lambda\text{ from}\\
\text{the sum (since }\lambda\in\Symm_{\left(  \infty\right)  }%
\lambda\text{)}%
\end{array}
\right) \\
&  =\delta_{\tau,\lambda}+\underbrace{\sum\limits_{\substack{\alpha\in
\Symm_{\left(  \infty\right)  }\lambda;\\\alpha\neq\lambda}}0}_{=0}=\delta
_{\tau,\lambda}.
\end{align*}
This proves (\ref{sol.Lambda.infinite-indep.c.kronecker1}).}.

For any three partitions $\alpha,\beta,\gamma\in \Par $ and any
three partitions $\mu,\nu,\lambda\in \Par $, we have
\begin{equation}
\left[  {\mathbf{x}}^{\alpha}{\mathbf{y}}^{\beta}{\mathbf{z}}^{\gamma}\right]
\left(  m_{\mu}\left(  {\mathbf{x}}\right)  m_{\nu}\left(  {\mathbf{y}%
}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)  \right)  =\delta_{\left(
\alpha,\beta,\gamma\right)  ,\left(  \mu,\nu,\lambda\right)  }
\label{sol.Lambda.infinite-indep.c.xyz1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.c.xyz1}):} Let
$\alpha,\beta,\gamma\in \Par $ be three partitions. Let $\mu
,\nu,\lambda\in \Par $ be three partitions. Then,
(\ref{sol.Lambda.infinite-indep.c.kronecker1}) (applied to $\alpha$ and $\mu$
instead of $\tau$ and $\lambda$) yields $\left[  {\mathbf{x}}^{\alpha}\right]
\left(  m_{\mu}\left(  {\mathbf{x}}\right)  \right)  =\delta_{\alpha,\mu}$.
Also, (\ref{sol.Lambda.infinite-indep.c.kronecker1}) (applied to $\beta$ and
$\nu$ instead of $\tau$ and $\lambda$) yields $\left[  {\mathbf{x}}^{\beta
}\right]  \left(  m_{\nu}\left(  {\mathbf{x}}\right)  \right)  =\delta
_{\beta,\nu}$. Renaming the indeterminates $\xx$ as $\mathbf{y}$ in
this fact, we obtain $\left[  {\mathbf{y}}^{\beta}\right]  \left(  m_{\nu
}\left(  {\mathbf{y}}\right)  \right)  =\delta_{\beta,\nu}$. Finally,
(\ref{sol.Lambda.infinite-indep.c.kronecker1}) (applied to $\gamma$ and
$\lambda$ instead of $\tau$ and $\lambda$) yields $\left[  \mathbf{x}^{\gamma
}\right]  \left(  m_{\lambda}\left(  \xx\right)  \right)
=\delta_{\gamma,\lambda}$. Renaming the indeterminates $\xx$ as
$\mathbf{z}$ in this fact, we obtain $\left[  {\mathbf{z}}^{\gamma}\right]
\left(  m_{\lambda}\left(  {\mathbf{z}}\right)  \right)  =\delta
_{\gamma,\lambda}$. Now,%
\begin{align*}
\left[  {\mathbf{x}}^{\alpha}{\mathbf{y}}^{\beta}{\mathbf{z}}^{\gamma
}\right]  \left(  m_{\mu}\left(  {\mathbf{x}}\right)  m_{\nu}\left(
{\mathbf{y}}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)  \right)
&  =\underbrace{\left(  \left[  {\mathbf{x}}^{\alpha}\right]  \left(  m_{\mu
}\left(  {\mathbf{x}}\right)  \right)  \right)  }_{=\delta_{\alpha,\mu}}%
\cdot\underbrace{\left(  \left[  \mathbf{y}^{\beta}\right]  \left(  m_{\nu
}\left(  \mathbf{y}\right)  \right)  \right)  }_{=\delta_{\beta,\nu}}%
\cdot\underbrace{\left(  \left[  {\mathbf{z}}^{\gamma}\right]  \left(
m_{\lambda}\left(  {\mathbf{z}}\right)  \right)  \right)  }_{=\delta
_{\gamma,\lambda}}\\
&  =\delta_{\alpha,\mu}\cdot\delta_{\beta,\nu}\cdot\delta_{\gamma,\lambda
}=\delta_{\left(  \alpha,\beta,\gamma\right)  ,\left(  \mu,\nu,\lambda\right)
}.
\end{align*}
This proves (\ref{sol.Lambda.infinite-indep.c.xyz1}).}.

Hence, if $\left(  c_{\mu,\nu,\lambda}\right)  _{\left(  \mu,\nu
,\lambda\right)  \in \Par^3}\in\kk^{ \Par^3 }$
is a family of elements of $\kk$ satisfying%
\begin{equation}
\sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}c_{\mu
,\nu,\lambda}m_{\mu}\left(  {\mathbf{x}}\right)  m_{\nu}\left(  {\mathbf{y}%
}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)  =0,
\label{sol.Lambda.infinite-indep.c.lindass}%
\end{equation}
then $\left(  c_{\mu,\nu,\lambda}\right)  _{\left(  \mu,\nu,\lambda\right)
\in\operatorname{Par}^{3}}=\left(  0\right)  _{\left(  \mu,\nu,\lambda\right)
\in\operatorname{Par}^{3}}$\ \ \ \ \footnote{\textit{Proof.} Let $\left(
c_{\mu,\nu,\lambda}\right)  _{\left(  \mu,\nu,\lambda\right)  \in
\operatorname{Par}^{3}}\in\kk^{  \Par^3 }$ be a
family of elements of $\kk$ satisfying
(\ref{sol.Lambda.infinite-indep.c.lindass}). We must show that
$\left(  c_{\mu
,\nu,\lambda}\right)  _{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3}}=\left(  0\right)  _{\left(  \mu,\nu,\lambda\right)  \in
\operatorname{Par}^{3}}$.
\par
Fix any $\left(  \alpha,\beta,\gamma\right)  \in \Par^3$. Then,%
\begin{align*}
&  \left[  {\mathbf{x}}^{\alpha}{\mathbf{y}}^{\beta}{\mathbf{z}}^{\gamma
}\right]  \left(  \sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3}}c_{\mu,\nu,\lambda}m_{\mu}\left(  {\mathbf{x}}\right)  m_{\nu}\left(
{\mathbf{y}}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)  \right) \\
&  =\sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}c_{\mu
,\nu,\lambda}\underbrace{\left[  {\mathbf{x}}^{\alpha}{\mathbf{y}}^{\beta
}{\mathbf{z}}^{\gamma}\right]  \left(  m_{\mu}\left(  {\mathbf{x}}\right)
m_{\nu}\left(  {\mathbf{y}}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)
\right)  }_{\substack{=\delta_{\left(  \alpha,\beta,\gamma\right)  ,\left(
\mu,\nu,\lambda\right)  }\\\text{(by (\ref{sol.Lambda.infinite-indep.c.xyz1}%
))}}}\\
&  =\sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}c_{\mu
,\nu,\lambda}\delta_{\left(  \alpha,\beta,\gamma\right)  ,\left(  \mu
,\nu,\lambda\right)  }\\
&  =c_{\alpha,\beta,\gamma}\underbrace{\delta_{\left(  \alpha,\beta
,\gamma\right)  ,\left(  \alpha,\beta,\gamma\right)  }}_{=1}+\sum\limits
_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left(  \mu,\nu,\lambda\right)  \neq\left(  \alpha,\beta,\gamma\right)
}}c_{\mu,\nu,\lambda}\underbrace{\delta_{\left(  \alpha,\beta,\gamma\right)
,\left(  \mu,\nu,\lambda\right)  }}_{\substack{=0\\\text{(since }\left(
\alpha,\beta,\gamma\right)  \neq\left(  \mu,\nu,\lambda\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}\left(  \mu,\nu,\lambda\right)  =\left(  \alpha,\beta,\gamma\right)  \text{
from the sum}\right) \\
&  =c_{\alpha,\beta,\gamma}+\underbrace{\sum\limits_{\substack{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3};\\\left(  \mu,\nu
,\lambda\right)  \neq\left(  \alpha,\beta,\gamma\right)  }}c_{\mu,\nu,\lambda
}0}_{=0}=c_{\alpha,\beta,\gamma},
\end{align*}
so that%
\[
c_{\alpha,\beta,\gamma}=\left[  {\mathbf{x}}^{\alpha}{\mathbf{y}}^{\beta
}{\mathbf{z}}^{\gamma}\right]  \left(  \underbrace{\sum_{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3}}c_{\mu,\nu,\lambda}m_{\mu
}\left(  {\mathbf{x}}\right)  m_{\nu}\left(  {\mathbf{y}}\right)  m_{\lambda
}\left(  {\mathbf{z}}\right)  }_{=0}\right)  =\left[  {\mathbf{x}}^{\alpha
}{\mathbf{y}}^{\beta}{\mathbf{z}}^{\gamma}\right]  0=0.
\]
\par
Now, forget that we fixed $\left(  \alpha,\beta,\gamma\right)  $. We thus have
shown that $c_{\alpha,\beta,\gamma}=0$ for each $\left(  \alpha,\beta
,\gamma\right)  \in  \Par^3 $. In other words, $\left(
c_{\alpha,\beta,\gamma}\right)  _{\left(  \alpha,\beta,\gamma\right)
\in\operatorname{Par}^{3}}=\left(  0\right)  _{\left(  \alpha,\beta
,\gamma\right)  \in\operatorname{Par}^{3}}$. Renaming the index $\left(
\alpha,\beta,\gamma\right)  $ as $\left(  \mu,\nu,\lambda\right)  $ in this
equality, we obtain $\left(  c_{\mu,\nu,\lambda}\right)  _{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3}}=\left(  0\right)  _{\left(
\mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$. Qed.}. Thus, the family
$\left(  m_{\mu}\left(  {\mathbf{x}}\right)  m_{\nu}\left(  {\mathbf{y}%
}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)  \right)  _{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ of elements of $\kk%
\left[  \left[  {\mathbf{x}},{\mathbf{y}},{\mathbf{z}}\right]  \right]  $ is
$\kk$-linearly independent.

Hence, the $\kk$-linear map
\[
\Lambda\otimes\Lambda\otimes\Lambda\rightarrow\kk\left[  \left[
{\mathbf{x}},{\mathbf{y}},{\mathbf{z}}\right]  \right]  ,\qquad f\otimes
g\otimes h\mapsto f\left(  {\mathbf{x}}\right)  g\left(  {\mathbf{y}}\right)
h\left(  {\mathbf{z}}\right)
\]
maps the basis $\left(  m_{\mu}\otimes m_{\nu}\otimes m_{\lambda}\right)
_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ of
$\Lambda\otimes\Lambda\otimes\Lambda$ to the linearly independent family
\newline$\left(  m_{\mu}\left(  {\mathbf{x}}\right)  m_{\nu}\left(
{\mathbf{y}}\right)  m_{\lambda}\left(  {\mathbf{z}}\right)  \right)
_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$. Consequently,
this map is injective\footnote{because any $\kk$-linear map that maps a
basis of its domain to a linearly independent family must be injective}.
Therefore, the family \newline$\left(  q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)
\right)  _{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ of
elements of $\kk\left[  \left[  {\mathbf{x}},{\mathbf{y}},{\mathbf{z}%
}\right]  \right]  $ is also linearly independent (because it is the image of
the basis $\left(  q_{\mu}\otimes q_{\nu}\otimes q_{\lambda}\right)  _{\left(
\mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ of $\Lambda\otimes
\Lambda\otimes\Lambda$ under this injective map\footnote{Here, we are using
the fact that the image of a basis under an injective $\kk$-linear map
is always linearly independent.}). If the sums appearing in
(\ref{eq.exe.Lambda.infinite-indep.c.ass}) were finite, then this observation
would already yield Exercise \ref{exe.Lambda.infinite-indep} (c). However,
these sums are infinite (and linear independence makes no claims about
infinite sums being $0$), so the solution of Exercise
\ref{exe.Lambda.infinite-indep} (c) takes some more work:

Let $\left(  a_{\lambda,\mu,\nu}\right)  _{\left(  \mu,\nu,\lambda\right)
\in\operatorname{Par}^{3}}\in\kk^{\operatorname{Par}^{3}}$ and $\left(
b_{\lambda,\mu,\nu}\right)  _{\left(  \mu,\nu,\lambda\right)  \in
\operatorname{Par}^{3}}\in\kk^{\operatorname{Par}^{3}}$ be two families
satisfying (\ref{eq.exe.Lambda.infinite-indep.c.ass}) in $\kk\left[
\left[  \xx,\mathbf{y},\mathbf{z}\right]  \right]  $. We must prove
that $\left(  a_{\lambda,\mu,\nu}\right)  _{\left(  \mu,\nu,\lambda\right)
\in\operatorname{Par}^{3}}=\left(  b_{\lambda,\mu,\nu}\right)  _{\left(
\mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$.

For each $\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}$,
\begin{equation}
\text{the power series }q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \text{ is
homogeneous of degree }\left\vert \mu\right\vert +\left\vert \nu\right\vert
+\left\vert \lambda\right\vert \label{sol.Lambda.infinite-indep.c.hg}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.c.hg}):} Let
$\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}$.
\par
The power series $q_{\lambda}\left(  \xx\right)  =q_{\lambda}$ is
homogeneous of degree $\left\vert \lambda\right\vert $ (by
(\ref{sol.Lambda.infinite-indep.hg})). Renaming the indeterminates
$\xx$ as $\mathbf{z}$ in this fact, we conclude that the power series
$q_{\lambda}\left(  \mathbf{z}\right)  $ is homogeneous of degree $\left\vert
\lambda\right\vert $. Similarly, the power series $q_{\mu}\left(
\xx\right)  $ is homogeneous of degree $\left\vert \mu\right\vert $.
Similarly, the power series $q_{\nu}\left(  \mathbf{y}\right)  $ is
homogeneous of degree $\left\vert \nu\right\vert $.
\par
Now, we know that the three power series $q_{\mu}\left(  {\mathbf{x}}\right)
,q_{\nu}\left(  {\mathbf{y}}\right)  ,q_{\lambda}\left(  {\mathbf{z}}\right)
$ are homogeneous of degrees $\left\vert \mu\right\vert ,\left\vert
\nu\right\vert ,\left\vert \lambda\right\vert $, respectively. Hence, their
product $q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}%
}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  $ is homogeneous of degree
$\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert $. This proves (\ref{sol.Lambda.infinite-indep.c.hg}).}.

Fix $n\in \NN$. Note that there are only finitely many $\left(
\mu,\nu,\lambda\right)  \in  \Par^3 $ satisfying
$\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n$\ \ \ \ \footnote{\textit{Proof.} If $\left(  \mu
,\nu,\lambda\right)  \in  \Par^3 $ satisfies
$\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n$, then each of the three partitions $\mu,\nu,\lambda$
has to have size $\leq n$ (because $\left\vert \mu\right\vert \leq\left\vert
\mu\right\vert +\left\vert \nu\right\vert +\left\vert \lambda\right\vert =n$,
and similarly $\left\vert \nu\right\vert \leq n$ and $\left\vert
\lambda\right\vert \leq n$). This leaves only finitely many possibilities for
each of these partitions $\mu,\nu,\lambda$. Thus, there are only finitely many
$\left(  \mu,\nu,\lambda\right)  \in  \Par^3 $
satisfying $\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n$.}.

Consider the $\kk$-linear map $\pi_{n}:\kk\left[  \left[
{\mathbf{x}},{\mathbf{y}},{\mathbf{z}}\right]  \right]  \rightarrow
\kk\left[  \left[  {\mathbf{x}},{\mathbf{y}},{\mathbf{z}}\right]
\right]  $ that sends each power series $f\in\kk\left[  \left[
{\mathbf{x}},\mathbf{y},\mathbf{z}\right]  \right]  $ to its homogeneous
component of degree $n$. Thus, $\pi_{n}$ has the following properties:

\begin{itemize}
\item If $f\in\kk\left[  \left[  {\mathbf{x}},\mathbf{y},\mathbf{z}%
\right]  \right]  $ is a power series that is homogeneous of degree $n$, then%
\begin{equation}
\pi_{n}\left(  f\right)  =f. \label{sol.Lambda.infinite-indep.c.hg1}%
\end{equation}


\item If $f\in\kk\left[  \left[  {\mathbf{x}},\mathbf{y},\mathbf{z}%
\right]  \right]  $ is a power series that is homogeneous of degree $\neq n$,
then%
\begin{equation}
\pi_{n}\left(  f\right)  =0. \label{sol.Lambda.infinite-indep.c.hg2}%
\end{equation}


\item The map $\pi_{n}$ is $\kk$-linear and continuous (with respect to
the topology on $\kk\left[  \left[  {\mathbf{x}},\mathbf{y}%
,\mathbf{z}\right]  \right]  $).
\end{itemize}

Hence, each $\left(  \mu,\nu,\lambda\right)  \in \Par^3$
satisfying $\left\vert \mu\right\vert +\left\vert
\nu\right\vert +\left\vert \lambda\right\vert =n$ satisfies%
\begin{equation}
\pi_{n}\left(  q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}%
}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  =q_{\mu}\left(
{\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(
{\mathbf{z}}\right)  \label{sol.Lambda.infinite-indep.c.pin1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.c.pin1}):} Let
$\left(  \mu,\nu,\lambda\right)  \in  \Par^3 $ be such
that $\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n$. From (\ref{sol.Lambda.infinite-indep.c.hg}), we know
that the power series $q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  $ is homogeneous
of degree $\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n$. Hence, (\ref{sol.Lambda.infinite-indep.c.hg1})
(applied to $f=q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}%
}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  $) yields $\pi_{n}\left(
q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)
q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  =q_{\mu}\left(  {\xx%
}\right)  q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}%
}\right)  $. This proves (\ref{sol.Lambda.infinite-indep.c.pin1}).}.
Furthermore, each $\left(  \mu,\nu,\lambda\right)  \in \Par^3$
satisfying $\left\vert \mu\right\vert +\left\vert
\nu\right\vert +\left\vert \lambda\right\vert \neq n$ satisfies%
\begin{equation}
\pi_{n}\left(  q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}%
}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  =0
\label{sol.Lambda.infinite-indep.c.pin2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.infinite-indep.c.pin2}):} Let
$\left(  \mu,\nu,\lambda\right)  \in  \Par^3 $ be such
that $\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert \neq n$. From (\ref{sol.Lambda.infinite-indep.c.hg}), we
know that the power series $q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  $ is homogeneous
of degree $\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert \neq n$. Hence, (\ref{sol.Lambda.infinite-indep.c.hg2})
(applied to $f=q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}%
}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  $) yields $\pi_{n}\left(
q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)
q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  =0$. This proves
(\ref{sol.Lambda.infinite-indep.c.pin2}).}.

But the map $\pi_{n}$ is $\kk$-linear and continuous. Thus, it respects
infinite sums. Hence,%
\begin{align}
&  \pi_{n}\left(  \sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3}}a_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right)
\nonumber\\
&  =\sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}%
a_{\lambda,\mu,\nu}\pi_{n}\left(  q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu
}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right)
\nonumber\\
&  =\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}a_{\lambda,\mu,\nu}\underbrace{\pi_{n}\left(  q_{\mu
}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda
}\left(  {\mathbf{z}}\right)  \right)  }_{\substack{=q_{\mu}\left(
{\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(
{\mathbf{z}}\right)  \\\text{(by (\ref{sol.Lambda.infinite-indep.c.pin1}))}%
}}+\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert \neq n}}a_{\lambda,\mu,\nu}\underbrace{\pi_{n}\left(
q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)
q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  }_{\substack{=0\\\text{(by
(\ref{sol.Lambda.infinite-indep.c.pin2}))}}}\nonumber\\
&  =\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}a_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)
+\underbrace{\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in
\operatorname{Par}^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert
+\left\vert \lambda\right\vert \neq n}}a_{\lambda,\mu,\nu}0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}a_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  .
\label{sol.Lambda.infinite-indep.c.4}%
\end{align}
The same argument (applied to the family $\left(  b_{\lambda,\mu,\nu}\right)
_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$ instead of
$\left(  a_{\lambda,\mu,\nu}\right)  _{\left(  \mu,\nu,\lambda\right)
\in\operatorname{Par}^{3}}$) yields%
\begin{equation}
\pi_{n}\left(  \sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3}}b_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right)
=\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}b_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  .
\label{sol.Lambda.infinite-indep.c.5}%
\end{equation}


Now, applying the map $\pi_{n}$ to both sides of the equality
(\ref{eq.exe.Lambda.infinite-indep.c.ass}), we obtain%
\begin{align*}
&  \pi_{n}\left(  \sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3}}a_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right) \\
&  =\pi_{n}\left(  \sum_{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3}}b_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(
{\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  \right)
=\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}b_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  .
\end{align*}
Comparing this with (\ref{sol.Lambda.infinite-indep.c.4}), we obtain%
\begin{equation}
\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}a_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)
=\sum\limits_{\substack{\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}%
^{3};\\\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n}}b_{\lambda,\mu,\nu}q_{\mu}\left(  {\mathbf{x}}\right)
q_{\nu}\left(  {\mathbf{y}}\right)  q_{\lambda}\left(  {\mathbf{z}}\right)  .
\label{sol.Lambda.infinite-indep.c.7}%
\end{equation}
Notice that both sums appearing in this equality are finite (since there are
only finitely many $\left(  \mu,\nu,\lambda\right)  \in \Par^3$
satisfying $\left\vert \mu\right\vert +\left\vert
\nu\right\vert +\left\vert \lambda\right\vert =n$). Since the family $\left(
q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)
q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  _{\left(  \mu,\nu
,\lambda\right)  \in\operatorname{Par}^{3};\ \left\vert \mu\right\vert
+\left\vert \nu\right\vert +\left\vert \lambda\right\vert =n}$
is $\kk$-linearly independent\footnote{\textit{Proof.}
We know that the family $\left(
q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)
q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  _{\left(  \mu,\nu
,\lambda\right)  \in\operatorname{Par}^{3}}$ is $\kk$-linearly
independent. Hence, the family
$\left(
q_{\mu}\left(  {\mathbf{x}}\right)  q_{\nu}\left(  {\mathbf{y}}\right)
q_{\lambda}\left(  {\mathbf{z}}\right)  \right)  _{\left(  \mu,\nu
,\lambda\right)  \in\operatorname{Par}^{3};\ \left\vert \mu\right\vert
+\left\vert \nu\right\vert +\left\vert \lambda\right\vert =n}$
(being a subfamily of it) must also be $\kk$-linearly
independent.}, we can thus conclude from
(\ref{sol.Lambda.infinite-indep.c.7})
that%
\begin{equation}
a_{\lambda,\mu,\nu}=b_{\lambda,\mu,\nu}\ \ \ \ \ \ \ \ \ \ \text{for each
}\left(  \mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}\text{ satisfying
}\left\vert \mu\right\vert +\left\vert \nu\right\vert +\left\vert
\lambda\right\vert =n. \label{sol.Lambda.infinite-indep.c.9}%
\end{equation}


Now, forget that we fixed $n$. We thus have proven
(\ref{sol.Lambda.infinite-indep.c.9}) for each $n\in \NN$.

Now, let $\left(  \mu,\nu,\lambda\right)  \in  \Par^3 $
be arbitrary. Then, $\left\vert \mu\right\vert +\left\vert \nu\right\vert
+\left\vert \lambda\right\vert \in \NN$. Hence,
(\ref{sol.Lambda.infinite-indep.c.9}) (applied to $n=\left\vert \mu\right\vert
+\left\vert \nu\right\vert +\left\vert \lambda\right\vert $) yields
$a_{\lambda,\mu,\nu}=b_{\lambda,\mu,\nu}$.

Now, forget that we fixed $\left(  \mu,\nu,\lambda\right)  $. We thus have
proven that $a_{\lambda,\mu,\nu}=b_{\lambda,\mu,\nu}$ for each $\left(
\mu,\nu,\lambda\right)  \in\operatorname{Par}^{3}$. In other words, $\left(
a_{\lambda,\mu,\nu}\right)  _{\left(  \mu,\nu,\lambda\right)  \in
\operatorname{Par}^{3}}=\left(  b_{\lambda,\mu,\nu}\right)  _{\left(  \mu
,\nu,\lambda\right)  \in\operatorname{Par}^{3}}$. This solves Exercise
\ref{exe.Lambda.infinite-indep} (c).

(b) The solution to Exercise \ref{exe.Lambda.infinite-indep} (b) is analogous
to that of Exercise \ref{exe.Lambda.infinite-indep} (c) (the difference being
that there are now just two families $\xx$ and $\mathbf{y}$ of
indeterminates, rather than three families $\xx$, $\mathbf{y}$ and
$\mathbf{z}$).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.bialg-c}}

\textit{Solution to Exercise~\ref{exe.Lambda.bialg-c}.} Recall that
$\left(  s_{\kappa}\right)  _{\kappa\in\operatorname{Par}}$ is a basis of the
$\kk$-module $\Lambda$. Hence, $\left(  s_{\kappa}\otimes s_{\lambda
}\right)  _{\left(  \kappa,\lambda\right)  \in\operatorname{Par}%
\times\operatorname{Par}}$ is a basis of the $\kk$-module
$\Lambda\otimes\Lambda$. We will refer to this basis as the ``Schur basis''
of $\Lambda\otimes\Lambda$.

Now, the first diagram in (\ref{bialgebra-diagrams}) commutes when $A$ is set
to $\Lambda$ (since $\Lambda$ is a bialgebra). In other words,%
\begin{equation}
\Delta\circ m=\left(  m\otimes m\right)  \circ\left(   \id %
\otimes T\otimes \id \right)  \circ\left(  \Delta\otimes
\Delta\right)  :\Lambda\otimes\Lambda\rightarrow\Lambda\otimes\Lambda
.
\label{sol.Lambda.bialg-c.axiom}
\end{equation}
Fix four partitions $\varphi$, $\psi$, $\kappa$ and $\lambda$.
Applying both sides of the equality
(\ref{sol.Lambda.bialg-c.axiom}) to $s_{\varphi}\otimes s_{\psi}$, we obtain%
\[
\left(  \Delta\circ m\right)  \left(  s_{\varphi}\otimes s_{\psi}\right)
=\left(  \left(  m\otimes m\right)  \circ\left(   \id \otimes
T\otimes \id \right)  \circ\left(  \Delta\otimes\Delta\right)
\right)  \left(  s_{\varphi}\otimes s_{\psi}\right)  .
\]
Comparing the coefficients before $s_{\kappa}\otimes s_{\lambda}$ in this
equality, we obtain
\[
\sum_{\rho\in\operatorname{Par}}c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi
}^{\rho}
=
\sum_{\left(\alpha , \beta , \gamma , \delta\right) \in \Par^4}
c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}^{\varphi
}c_{\gamma,\delta}^{\psi}
\]
(after a straightforward computation using \eqref{symm-structure-constants.1},
\eqref{symm-structure-constants.2}
and Corollary \ref{self-duality-corollary}). This solves the exercise.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.bialg-c}}

\textit{Solution to Exercise~\ref{exe.Lambda.bialg-c}.} Recall that
$\left(  s_{\kappa}\right)  _{\kappa\in\Par}$ is a basis of the
$\kk$-module $\Lambda$. Hence, $\left(  s_{\kappa}\otimes s_{\lambda
}\right)  _{\left(  \kappa,\lambda\right)  \in\Par %
\times\Par}$ is a basis of the $\kk$-module
$\Lambda\otimes\Lambda$. In particular, the family $\left(  s_{\kappa}\otimes
s_{\lambda}\right)  _{\left(  \kappa,\lambda\right)  \in\Par %
\times\Par}$ is $\kk$-linearly independent. Thus, if
$\left(  g_{\left(  \kappa,\lambda\right)  }\right)  _{\left(  \kappa
,\lambda\right)  \in\Par \times\Par}\in
\kk^{\Par \times\Par}$ is a family of
elements of $\kk$ such that \newline
$\left(  \text{all but finitely many
}\left(  \kappa,\lambda\right)  \in\Par \times
\Par \text{ satisfy }g_{\left(  \kappa,\lambda\right)
}=0\right)  $ and $\sum_{\left(  \kappa,\lambda\right)  \in\Par %
\times\Par}g_{\left(  \kappa,\lambda\right)  }s_{\kappa}\otimes
s_{\lambda}=0$, then%
\begin{equation}
\left(  g_{\left(  \kappa,\lambda\right)  }\right)  _{\left(  \kappa
,\lambda\right)  \in\Par \times\Par}=\left(
0\right)  _{\left(  \kappa,\lambda\right)  \in\Par %
\times\Par}.\label{sol.Lambda.bialg-c.linind}%
\end{equation}


But recall that every partition $\lambda$ satisfies
\begin{align}
\Delta\left(  s_{\lambda}\right)   &  =\underbrace{\sum_{\left(  \mu
,\nu\right)  \in\Par \times\Par}}_{=\sum_{\mu
\in\Par}\sum_{\nu\in\Par}}%
\underbrace{\hat{c}_{\mu,\nu}^{\lambda}}_{\substack{=c_{\mu,\nu}^{\lambda
}\\\text{(by Corollary \ref{self-duality-corollary})}}}
s_{\mu}\otimes s_{\nu}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{symm-structure-constants.2}}
\right)  \nonumber\\
&  =\sum_{\mu\in\Par}\sum_{\nu\in\Par}c_{\mu
,\nu}^{\lambda}s_{\mu}\otimes s_{\nu}\label{sol.Lambda.bialg-c.copro}\\
&  =\sum_{\alpha\in\Par}\sum_{\beta\in\Par %
}c_{\alpha,\beta}^{\lambda}s_{\alpha}\otimes s_{\beta}%
\label{sol.Lambda.bialg-c.copro2}%
\end{align}
(here, we renamed the summation indices $\mu$ and $\nu$ as $\alpha$ and
$\beta$, respectively). Also, every partitions $\mu$ and $\nu$ satisfy%
\begin{align}
s_{\mu}s_{\nu} &  =\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda
}s_{\lambda}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{symm-structure-constants.1}}
\right)
\label{sol.Lambda.bialg-c.pro3}\\
&  =\sum_{\kappa\in\Par}c_{\mu,\nu}^{\kappa}s_{\kappa
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\lambda\text{ as }\kappa\right) .
\label{sol.Lambda.bialg-c.pro}
\end{align}


Now, let $\varphi$ and $\psi$ be two partitions. We need to prove that%
\[
\sum_{\rho\in\Par }c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi
}^{\rho}=\sum_{\alpha\in\Par }\sum_{\beta\in\Par %
}\sum_{\gamma\in\Par }\sum_{\delta\in\Par }
c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}^{\varphi
}c_{\gamma,\delta}^{\psi}%
\]
holds for any two partitions $\kappa$ and $\lambda$.

Since $\Lambda$ is a $\kk$-bialgebra, we know that the comultiplication
$\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ of $\Lambda$ is a
$\kk$-algebra homomorphism. Hence,
\begin{align*}
\Delta\left(  s_{\varphi}s_{\psi}\right)   &  =\underbrace{\Delta\left(
s_{\varphi}\right)  }_{\substack{=\sum_{\alpha\in\Par}
\sum_{\beta\in\Par}c_{\alpha,\beta}^{\varphi}s_{\alpha}\otimes
s_{\beta}\\\text{(by \eqref{sol.Lambda.bialg-c.copro2}, applied}\\\text{to
}\varphi\text{ instead of }\lambda\text{)}}}\underbrace{\Delta\left(  s_{\psi
}\right)  }_{\substack{=\sum_{\mu\in\Par}\sum_{\nu
\in\Par}c_{\mu,\nu}^{\psi}s_{\mu}\otimes s_{\nu}\\\text{(by
\eqref{sol.Lambda.bialg-c.copro}, applied}\\\text{to }\psi\text{ instead of
}\lambda\text{)}}}\\
&  =\left(  \sum_{\alpha\in\Par}\sum_{\beta\in
\Par}c_{\alpha,\beta}^{\varphi}s_{\alpha}\otimes s_{\beta
}\right)  \left(  \sum_{\mu\in\Par}\sum_{\nu\in
\Par}c_{\mu,\nu}^{\psi}s_{\mu}\otimes s_{\nu}\right) \\
&=\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\mu\in\Par}\sum_{\nu\in\Par}%
c_{\alpha,\beta}^{\varphi}c_{\mu,\nu}^{\psi}\underbrace{\left(  s_{\alpha
}\otimes s_{\beta}\right)  \left(  s_{\mu}\otimes s_{\nu}\right)
}_{=s_{\alpha}s_{\mu}\otimes s_{\beta}s_{\nu}}\\
&  =\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\mu\in\Par}\sum_{\nu\in\Par}%
c_{\alpha,\beta}^{\varphi}c_{\mu,\nu}^{\psi}s_{\alpha}s_{\mu}\otimes s_{\beta
}s_{\nu}\\
&  =\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\beta}^{\varphi}c_{\gamma,\delta}^{\psi}\underbrace{s_{\alpha
}s_{\gamma}}_{\substack{=\sum_{\kappa\in\Par}c_{\alpha,\gamma
}^{\kappa}s_{\kappa}\\\text{(by \eqref{sol.Lambda.bialg-c.pro}, applied}
\\\text{to }\alpha\text{ and }\gamma\text{ instead of }\mu\text{ and }
\nu\text{)}}}\otimes\underbrace{s_{\beta}s_{\delta}}_{\substack{=\sum
_{\lambda\in\Par}c_{\beta,\delta}^{\lambda}s_{\lambda
}\\\text{(by \eqref{sol.Lambda.bialg-c.pro3}, applied}\\\text{to }\beta\text{
and }\delta\text{ instead of }\mu\text{ and }\nu\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation indices
}\mu\text{ and }\nu\text{ as }\gamma\text{ and }\delta\text{, respectively}%
\right)  \\
&  =\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\beta}^{\varphi}c_{\gamma,\delta}^{\psi}\underbrace{\left(
\sum_{\kappa\in\Par}c_{\alpha,\gamma}^{\kappa}s_{\kappa
}\right)  \otimes\left(  \sum_{\lambda\in\Par}c_{\beta,\delta
}^{\lambda}s_{\lambda}\right)  }_{\substack{=\sum_{\kappa\in
\Par}\sum_{\lambda\in\Par}c_{\alpha,\gamma
}^{\kappa}c_{\beta,\delta}^{\lambda}s_{\kappa}\otimes s_{\lambda
}\\\text{(since taking tensor products is }\kk \text{-bilinear)}}}\\
&  =\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\beta}^{\varphi}c_{\gamma,\delta}^{\psi}\sum_{\kappa\in
\Par}\sum_{\lambda\in\Par}c_{\alpha,\gamma
}^{\kappa}c_{\beta,\delta}^{\lambda}s_{\kappa}\otimes s_{\lambda}\\
&  =\sum_{\kappa\in\Par}\sum_{\lambda\in\Par %
}\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}%
^{\varphi}c_{\gamma,\delta}^{\psi}s_{\kappa}\otimes s_{\lambda}\\
&  =\sum_{\kappa\in\Par}\sum_{\lambda\in\Par %
}\left(  \sum_{\alpha\in\Par}\sum_{\beta\in\Par %
}\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}%
^{\varphi}c_{\gamma,\delta}^{\psi}\right)  s_{\kappa}\otimes s_{\lambda}.
\end{align*}
Compared with
\begin{align*}
& \Delta\left(  \underbrace{s_{\varphi}s_{\psi}}_{\substack{=\sum_{\kappa
\in\Par}c_{\varphi,\psi}^{\kappa}s_{\kappa}\\\text{(by
\eqref{sol.Lambda.bialg-c.pro}, applied to }\varphi\text{ and }\psi
\\\text{instead of }\mu\text{ and }\nu\text{)}}}\right)  \\
& =\Delta\left(  \sum_{\kappa\in\Par}c_{\varphi,\psi}^{\kappa
}s_{\kappa}\right)  =\sum_{\kappa\in\Par}c_{\varphi,\psi
}^{\kappa}\Delta\left(  s_{\kappa}\right)  =\sum_{\rho\in\Par %
}c_{\varphi,\psi}^{\rho}\Delta\left(  s_{\rho}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index }%
\kappa\text{ as }\rho\right)  \\
& =\sum_{\rho\in\Par}c_{\varphi,\psi}^{\rho}\underbrace{\Delta
\left(  s_{\rho}\right)  }_{\substack{=\sum_{\alpha\in\Par}%
\sum_{\beta\in\Par}c_{\alpha,\beta}^{\rho}s_{\alpha}\otimes
s_{\beta}\\\text{(by \eqref{sol.Lambda.bialg-c.copro2}, applied to}
\\\rho\text{ instead of }\lambda\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\Delta\text{ is }\kk \text{-linear}\right)  \\
& =\sum_{\rho\in\Par}c_{\varphi,\psi}^{\rho}\sum_{\alpha
\in\Par}\sum_{\beta\in\Par}c_{\alpha,\beta
}^{\rho}s_{\alpha}\otimes s_{\beta}=\sum_{\alpha\in\Par}%
\sum_{\beta\in\Par}\sum_{\rho\in\Par}%
c_{\alpha,\beta}^{\rho}c_{\varphi,\psi}^{\rho}s_{\alpha}\otimes s_{\beta}\\
& =\sum_{\kappa\in\Par}\sum_{\lambda\in\Par}%
\sum_{\rho\in\Par}c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi
}^{\rho}s_{\kappa}\otimes s_{\lambda}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation indices
}\alpha\text{ and }\beta\text{ as }\kappa\text{ and }\lambda\text{,
respectively}\right)  \\
& =\sum_{\kappa\in\Par}\sum_{\lambda\in\Par %
}\left(  \sum_{\rho\in\Par}c_{\kappa,\lambda}^{\rho}%
c_{\varphi,\psi}^{\rho}\right)  s_{\kappa}\otimes s_{\lambda},
\end{align*}
this yields%
\begin{align}
& \sum_{\kappa\in\Par}\sum_{\lambda\in\Par %
}\left(  \sum_{\alpha\in\Par}\sum_{\beta\in\Par %
}\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}%
^{\varphi}c_{\gamma,\delta}^{\psi}\right)  s_{\kappa}\otimes s_{\lambda
}\nonumber\\
& =\sum_{\kappa\in\Par}\sum_{\lambda\in\Par %
}\left(  \sum_{\rho\in\Par}c_{\kappa,\lambda}^{\rho}%
c_{\varphi,\psi}^{\rho}\right)  s_{\kappa}\otimes s_{\lambda}%
.\label{sol.Lambda.bialg-c.almost}%
\end{align}


Now, define a family $\left(  g_{\left(  \kappa,\lambda\right)  }\right)
_{\left(  \kappa,\lambda\right)  \in\Par \times \Par}
\in\kk^{\Par \times \Par}$ of elements of $\kk$ by
\begin{equation}
\left(  g_{\left(  \kappa,\lambda\right)  }=\sum_{\rho\in\Par %
}c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi}^{\rho}-\sum_{\alpha\in
\Par }\sum_{\beta\in\Par }\sum_{\gamma
\in\Par }\sum_{\delta\in\Par }c_{\alpha,\gamma
}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}^{\varphi}c_{\gamma
,\delta}^{\psi}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  \kappa
,\lambda\right)  \in\Par \times\Par \right)
.\label{sol.Lambda.bialg-c.g}%
\end{equation}
Then, subtracting $\sum_{\kappa\in\Par}\sum_{\lambda
\in\Par}\left(  \sum_{\alpha\in\Par}\sum
_{\beta\in\Par}\sum_{\gamma\in\Par}\sum
_{\delta\in\Par}c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta
}^{\lambda}c_{\alpha,\beta}^{\varphi}c_{\gamma,\delta}^{\psi}\right)
s_{\kappa}\otimes s_{\lambda}$ from the identity
\eqref{sol.Lambda.bialg-c.almost}, we obtain
\begin{align*}
0  & =\sum_{\kappa\in\Par}\sum_{\lambda\in\Par}
\left(  \sum_{\rho\in\Par}c_{\kappa,\lambda}^{\rho}
c_{\varphi,\psi}^{\rho}\right)  s_{\kappa}\otimes s_{\lambda}-\sum_{\kappa
\in\Par}\sum_{\lambda\in\Par}\left(
\sum_{\alpha\in\Par}\sum_{\beta\in\Par}
\sum_{\gamma\in\Par}\sum_{\delta\in\Par}
c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}
c_{\alpha,\beta}^{\varphi}c_{\gamma,\delta}^{\psi}\right)
s_{\kappa}\otimes s_{\lambda}\\
& =\underbrace{\sum_{\kappa\in\Par}\sum_{\lambda\in
\Par}}_{=\sum_{\left(  \kappa,\lambda\right)  \in
\Par \times\Par}}\underbrace{\left(  \sum
_{\rho\in\Par}c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi}^{\rho
}-\sum_{\alpha\in\Par}\sum_{\beta\in\Par}%
\sum_{\gamma\in\Par}\sum_{\delta\in\Par %
}c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}%
^{\varphi}c_{\gamma,\delta}^{\psi}\right)  }_{\substack{=g_{\left(
\kappa,\lambda\right)  }\\\text{(by \eqref{sol.Lambda.bialg-c.g})}}}
s_{\kappa}\otimes s_{\lambda}\\
& =\sum_{\left(  \kappa,\lambda\right)  \in\Par \times
\Par}g_{\left(  \kappa,\lambda\right)  }s_{\kappa}\otimes
s_{\lambda}.
\end{align*}
Thus, $\sum_{\left(  \kappa,\lambda\right)  \in\Par %
\times\Par}g_{\left(  \kappa,\lambda\right)  }s_{\kappa}\otimes
s_{\lambda}=0$. In particular, the sum $\sum_{\left(  \kappa,\lambda\right)
\in\Par \times\Par}g_{\left(  \kappa
,\lambda\right)  }$ is well-defined (without taking limits in any other than
the discrete topology), so that we see that all but finitely many $\left(
\kappa,\lambda\right)  \in\Par \times\Par $
satisfy $g_{\left(  \kappa,\lambda\right)  }=0$. Hence,
\eqref{sol.Lambda.bialg-c.linind} yields $\left(  g_{\left(  \kappa
,\lambda\right)  }\right)  _{\left(  \kappa,\lambda\right)  \in
\Par \times\Par}=\left(  0\right)  _{\left(
\kappa,\lambda\right)  \in\Par \times\Par}$. In
other words,
\begin{equation}
g_{\left(  \kappa,\lambda\right)  }=0\ \ \ \ \ \ \ \ \ \ 
\text{for every }\left(  \kappa,\lambda\right)
 \in \Par \times \Par .
\label{sol.Lambda.bialg-c.almost2}
\end{equation}


Now, let $\kappa$ and $\lambda$ be two partitions. Then, $\kappa
\in\Par $ and $\lambda\in\Par $, so that $\left(
\kappa,\lambda\right)  \in\Par \times\Par $.
Hence, \eqref{sol.Lambda.bialg-c.almost2} yields $g_{\left(  \kappa
,\lambda\right)  }=0$. Compared with
\[
g_{\left(  \kappa,\lambda\right)  }=\sum_{\rho\in\Par }%
c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi}^{\rho}-\sum_{\alpha\in
\Par }\sum_{\beta\in\Par }\sum_{\gamma
\in\Par }\sum_{\delta\in\Par }c_{\alpha,\gamma
}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}^{\varphi}c_{\gamma
,\delta}^{\psi},
\]
this yields
\[
\sum_{\rho\in\Par }c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi
}^{\rho}-\sum_{\alpha\in\Par }\sum_{\beta\in\Par %
}\sum_{\gamma\in\Par }\sum_{\delta\in\Par }%
c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}^{\varphi
}c_{\gamma,\delta}^{\psi}=0.
\]
In other words,
\[
\sum_{\rho\in\Par }c_{\kappa,\lambda}^{\rho}c_{\varphi,\psi
}^{\rho}=\sum_{\alpha\in\Par }\sum_{\beta\in\Par %
}\sum_{\gamma\in\Par }\sum_{\delta\in\Par }%
c_{\alpha,\gamma}^{\kappa}c_{\beta,\delta}^{\lambda}c_{\alpha,\beta}^{\varphi
}c_{\gamma,\delta}^{\psi}.
\]
This solves Exercise~\ref{exe.Lambda.bialg-c}.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cauchy.skew}}

\textit{Solution to Exercise~\ref{exe.cauchy.skew}.} Before we step to
the solution of this problem, let us make some general observations.

\begin{itemize}
\item Every two partitions $\lambda$ and $\mu$ satisfy $s_{\lambda/\mu}%
=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum ranges over all
partitions $\nu$ (according to Remark \ref{third-L-R-interpretation-remark}).
In other words, every two partitions $\lambda$ and $\mu$ satisfy
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in\operatorname{Par}}c_{\mu,\nu}^{\lambda}s_{\nu}.
\label{sol.cauchy.skew.slash}%
\end{equation}


\item On the other hand,
\eqref{symm-structure-constants.1} yields
\begin{equation}
s_{\mu}s_{\nu}=\sum_{\lambda\in\operatorname{Par}}c_{\mu,\nu}^{\lambda
}s_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }\mu\text{ and
}\nu. \label{sol.cauchy.skew.prod}%
\end{equation}

\end{itemize}

(a) Let $\mu$ be a partition. We have
\begin{equation}
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\mathbf{x}}\right)
\underbrace{s_{\lambda/\mu}\left(  {\mathbf{y}}\right)  }_{\substack{=\sum
_{\nu\in\operatorname{Par}}c_{\mu,\nu}^{\lambda}s_{\nu}\left(  {\mathbf{y}%
}\right)  \\\text{(by \eqref{sol.cauchy.skew.slash}, evaluated}\\\text{at the
variable set }\mathbf{y}\text{)}}}=\sum_{\lambda\in\operatorname{Par}%
}s_{\lambda}\left(  {\xx}\right)  \sum_{\nu\in\operatorname{Par}%
}c_{\mu,\nu}^{\lambda}s_{\nu}\left(  {\mathbf{y}}\right)  .
\label{sol.cauchy.skew.a.1}%
\end{equation}
On the other hand, \eqref{Cauchy-identity} yields
\[
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda
\in\operatorname{Par}}s_{\lambda}\left(  {\xx}\right)  s_{\lambda
}\left(  {\mathbf{y}}\right)  =\sum_{\nu\in\operatorname{Par}}s_{\nu}\left(
{\xx}\right)  s_{\nu}\left(  {\mathbf{y}}\right)
\]
(here, we renamed the summation index $\lambda$ as $\nu$). Multiplying this
equality with $s_{\mu}\left(  {\mathbf{x}}\right)  $, we obtain
\begin{align*}
s_{\mu}\left(  {\xx}\right)  \cdot\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}  &  =s_{\mu}\left(  {\mathbf{x}}\right)  \cdot
\sum_{\nu\in\operatorname{Par}}s_{\nu}\left(  {\xx}\right)  s_{\nu
}\left(  {\mathbf{y}}\right)  =\sum_{\nu\in\operatorname{Par}}%
\underbrace{s_{\mu}\left(  {\mathbf{x}}\right)  s_{\nu}\left(  {\xx}
\right)  }_{\substack{=\sum_{\lambda\in\operatorname{Par}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  {\mathbf{x}}\right)  \\\text{(by
\eqref{sol.cauchy.skew.prod})}}}s_{\nu}\left(  {\mathbf{y}}\right) \\
&  =\sum_{\nu\in\operatorname{Par}}\left(  \sum_{\lambda\in\operatorname{Par}%
}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  {\mathbf{x}}\right)  \right)  s_{\nu
}\left(  {\mathbf{y}}\right)  =\sum_{\lambda\in\operatorname{Par}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  {\xx}\right)  \sum_{\nu\in
\operatorname{Par}}s_{\nu}\left(  {\mathbf{y}}\right) \\
&  =\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\xx}
\right)  \sum_{\nu\in\operatorname{Par}}c_{\mu,\nu}^{\lambda}s_{\nu}\left(
{\mathbf{y}}\right)  .
\end{align*}
Compared with \eqref{sol.cauchy.skew.a.1}, this yields
\[
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\mathbf{x}}\right)
s_{\lambda/\mu}\left(  {\mathbf{y}}\right)  =s_{\mu}\left(  {\xx%
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\]
This solves Exercise~\ref{exe.cauchy.skew}(a).

(b) Let $\kk \left[  \left[  {\mathbf{x}},{\mathbf{y}},{\mathbf{z}%
},\mathbf{w}\right]  \right]  $ denote the ring
\[
 \kk \left[  \left[  x_{1},x_{2},x_{3},...,y_{1},y_{2},y_{3}%
,...,z_{1},z_{2},z_{3},...,w_{1},w_{2},w_{3},...\right]  \right]  .
\]
This ring clearly contains $\kk \left[  \left[  {\mathbf{x}}%
,{\mathbf{y}}\right]  \right]  $ as a subring. We will use the obvious
abbreviations for variable sets: $\xx=\left(  x_{1},x_{2}%
,x_{3},...\right)  $, $\left(  {\mathbf{x}},{\mathbf{z}}\right)  =\left(
x_{1},x_{2},x_{3},...,z_{1},z_{2},z_{3},...\right)  $, etc.

Every partition $\lambda$ satisfies%
\begin{align*}
s_{\lambda}\left(  {\mathbf{y}},{\mathbf{x}}\right)   &  =s_{\lambda}\left(
{\mathbf{x}},{\mathbf{y}}\right)  =\sum_{\mu\subseteq\lambda}s_{\mu}\left(
{\xx}\right)  s_{\lambda/\mu}\left(  {\mathbf{y}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{eq.Lambda.s.xy}}\right) \\
&  =\sum_{\mu\in\operatorname{Par}}s_{\mu}\left(  {\xx}\right)
s_{\lambda/\mu}\left(  {\mathbf{y}}\right)  -\sum\limits_{\substack{\mu
\in\operatorname{Par};\\\mu\nsubseteq\lambda}}s_{\mu}\left(  {\xx}
\right)  \underbrace{s_{\lambda/\mu}\left(  {\mathbf{y}}\right)
}_{\substack{=0\\\text{(since }\mu\nsubseteq\lambda\text{)}}}=\sum_{\mu
\in\operatorname{Par}}s_{\mu}\left(  {\xx}\right)  s_{\lambda/\mu
}\left(  {\mathbf{y}}\right)  -\underbrace{\sum\limits_{\substack{\mu\in
\operatorname{Par};\\\mu\nsubseteq\lambda}}s_{\mu}\left(  {\xx}
\right)  0}_{=0}\\
&  =\sum_{\mu\in\operatorname{Par}}s_{\mu}\left(  {\xx}\right)
s_{\lambda/\mu}\left(  {\mathbf{y}}\right)  .
\end{align*}
Applying this equality to the variables ${\mathbf{z}}$ and ${\mathbf{x}}$
instead of ${\mathbf{x}}$ and ${\mathbf{y}}$, we obtain%
\begin{equation}
s_{\lambda}\left(  {\mathbf{x}},{\mathbf{z}}\right)  =\sum_{\mu\in
\operatorname{Par}}s_{\mu}\left(  {\zz}\right)  s_{\lambda/\mu
}\left(  {\xx}\right)  . \label{sol.cauchy.skew.b.1.1}%
\end{equation}


Applying this equality to the variables ${\mathbf{y}}$ and $\mathbf{w}$
instead of ${\mathbf{x}}$ and ${\mathbf{z}} $, we obtain%
\begin{equation}
s_{\lambda}\left(  {\mathbf{y}} ,\mathbf{w}\right)  =\sum_{\mu\in
\operatorname{Par}}s_{\mu}\left(  {\mathbf{w}}\right)  s_{\lambda/\mu}\left(
{ {\mathbf{y}} }\right)  =\sum_{\nu\in\operatorname{Par}}s_{\nu}\left(
\mathbf{w}\right)  s_{\lambda/\nu}\left(  {\mathbf{y}} \right)
\label{sol.cauchy.skew.b.1.2}%
\end{equation}
(here, we renamed the summation index $\mu$ as $\nu$) for any partition
$\lambda$.

Now,%
\begin{align}
&  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\prod
_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty
}\left(  1-z_{i}y_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty}\left(  1-z_{i}%
w_{j}\right)  ^{-1}\nonumber\\
&  =\prod\limits_{\substack{a\in\left(  {\mathbf{x}} , {\mathbf{z}} \right)
;\\b\in\left(  {\mathbf{y}} ,\mathbf{w}\right)  }}\left(  1-ab\right)
^{-1}=\sum_{\lambda\in\operatorname{Par}}\underbrace{s_{\lambda}\left(
{\mathbf{x}} , {\mathbf{z}} \right)  }_{\substack{=\sum_{\mu\in
\operatorname{Par}}s_{\mu}\left(  {\zz}\right)  s_{\lambda/\mu
}\left(  {\xx}\right)  \\\text{(by \eqref{sol.cauchy.skew.b.1.1})}%
}}\underbrace{s_{\lambda}\left(  {\mathbf{y}} ,\mathbf{w}\right)
}_{\substack{=\sum_{\nu\in\operatorname{Par}} s_{\nu}\left(  \mathbf{w}
\right)  s_{\lambda/\nu}\left(  {\mathbf{y}} \right)  \\\text{(by
\eqref{sol.cauchy.skew.b.1.2})}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{Cauchy-identity}, applied to
the variable sets }\left(  {\mathbf{x}},{\mathbf{z}} \right)  \text{ and
}\left(  {\mathbf{y}},\mathbf{w}\right)  \text{ instead of }{\mathbf{x}}
\text{ and } {\mathbf{y}} \right) \nonumber\\
&  =\sum_{\lambda\in\operatorname{Par}}\left(  \sum_{\mu\in\operatorname{Par}%
}s_{\mu}\left(  {\zz}\right)  s_{\lambda/\mu}\left(  {{\mathbf{x}}
}\right)  \right)  \left(  \sum_{\nu\in\operatorname{Par}}s_{\nu}\left(
{\mathbf{w}}\right)  s_{\lambda/\nu}\left(  {\mathbf{y}} \right)  \right)
\nonumber\\
&  =\sum_{\mu\in\operatorname{Par}}\sum_{\nu\in\operatorname{Par}}s_{\mu
}\left(  {\mathbf{z}} \right)  s_{\nu}\left(  \mathbf{w}\right)  \left(
\sum_{\lambda\in\operatorname{Par}}s_{\lambda/\mu}\left(  {{\mathbf{x}}
}\right)  s_{\lambda/\nu}\left(  {\mathbf{y}} \right)  \right)  .
\label{sol.cauchy.skew.b.1.6}%
\end{align}


On the other hand, every partition $\kappa$ satisfies
\begin{equation}
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\xx}\right)
s_{\lambda/\kappa}\left(  {\mathbf{y}}\right)  =s_{\kappa}\left(
{\xx}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)
^{-1}\label{sol.cauchy.skew.b.1.9}%
\end{equation}
(by Exercise~\ref{exe.cauchy.skew}(a), applied to $\kappa$ instead of $\mu$).
Applying this to the variable sets $\mathbf{w}$ and ${\mathbf{x}}$ instead of
${\mathbf{x}}$ and ${\mathbf{y}}$, we obtain%
\begin{align}
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\mathbf{w}}\right)
s_{\lambda/\kappa}\left(  {\xx}\right)   &  =s_{\kappa}\left(
{\mathbf{w}}\right)  \cdot\underbrace{\prod_{i,j=1}^{\infty}\left(
1-w_{i}x_{j}\right)  ^{-1}}_{\substack{=\prod_{i,j=1}^{\infty}\left(
1-x_{j}w_{i}\right)  ^{-1}\\=\prod_{j,i=1}^{\infty}\left(  1-x_{i}%
w_{j}\right)  ^{-1}\\\text{(here, we renamed the index }\left(  i,j\right)
\text{ as }\left(  j,i\right)  \text{)}}}\nonumber\\
&  =s_{\kappa}\left(  \mathbf{w}\right)  \cdot\underbrace{\prod_{j,i=1}%
^{\infty}}_{=\prod_{i,j=1}^{\infty}}\left(  1-x_{i}w_{j}\right)
^{-1}=s_{\kappa}\left(  \mathbf{w}\right)  \cdot\prod_{i,j=1}^{\infty}\left(
1-x_{i}w_{j}\right)  ^{-1}\label{sol.cauchy.skew.b.1.10}%
\end{align}
for every partition $\kappa$. But applying both sides of the identity
\eqref{sol.cauchy.skew.b.1.9} to the variable set ${\mathbf{z}}$ instead of
${\mathbf{x}}$, and renaming the summation index $\lambda$ as $\mu$ on the
left hand side of this equality, we obtain%
\begin{equation}
\sum_{\mu\in\operatorname{Par}}s_{\mu}\left(  {\zz}\right)
s_{\mu/\kappa}\left(  {\mathbf{y}}\right)  =s_{\kappa}\left(  {{\mathbf{z}}%
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-z_{i}y_{j}\right)
^{-1}.\label{sol.cauchy.skew.b.1.11}%
\end{equation}


Now, applying \eqref{Cauchy-identity} to the variable sets ${\mathbf{z}}$ and
$\mathbf{w}$ instead of ${\mathbf{x}}$ and ${\mathbf{y}}$, we obtain%
\[
\prod_{i,j=1}^{\infty}\left(  1-z_{i}w_{j}\right)  ^{-1}=\sum_{\lambda
\in\operatorname{Par}}s_{\lambda}\left(  {\mathbf{z}}\right)  s_{\lambda
}\left(  \mathbf{w}\right)  =\sum_{\kappa\in\operatorname{Par}}s_{\kappa
}\left(  {\mathbf{z}}\right)  s_{\kappa}\left(  \mathbf{w}\right)  .
\]
Hence,%
\begin{align*}
&  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\prod
_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty
}\left(  1-z_{i}y_{j}\right)  ^{-1}\underbrace{\prod_{i,j=1}^{\infty}\left(
1-z_{i}w_{j}\right)  ^{-1}}_{=\sum_{\kappa\in\operatorname{Par}}s_{\kappa
}\left(  {\mathbf{z}}\right)  s_{\kappa}\left(  \mathbf{w}\right)  }\\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\prod
_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty
}\left(  1-z_{i}y_{j}\right)  ^{-1}\cdot\sum_{\kappa\in\operatorname{Par}%
}s_{\kappa}\left(  {\mathbf{z}}\right)  s_{\kappa}\left(  \mathbf{w}\right)
\\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\cdot\sum
_{\kappa\in\operatorname{Par}}\underbrace{\left(  s_{\kappa}\left(
\mathbf{w}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)
^{-1}\right)  }_{\substack{=\sum_{\lambda\in\operatorname{Par}}s_{\lambda
}\left(  {\mathbf{w}}\right)  s_{\lambda/\kappa}\left(  {\xx}\right)
\\\text{(by \eqref{sol.cauchy.skew.b.1.10})}}}\underbrace{\left(  s_{\kappa
}\left(  {\mathbf{z}}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-z_{i}%
y_{j}\right)  ^{-1}\right)  }_{\substack{=\sum_{\mu\in\operatorname{Par}%
}s_{\mu}\left(  {\zz}\right)  s_{\mu/\kappa}\left(  {{\mathbf{y}}%
}\right)  \\\text{(by \eqref{sol.cauchy.skew.b.1.11})}}}\\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\cdot\sum
_{\kappa\in\operatorname{Par}}\left(  \sum_{\lambda\in\operatorname{Par}%
}s_{\lambda}\left(  {\mathbf{w}}\right)  s_{\lambda/\kappa}\left(
{\xx}\right)  \right)  \left(  \sum_{\mu\in\operatorname{Par}}s_{\mu
}\left(  {\zz}\right)  s_{\mu/\kappa}\left(  {\mathbf{y}}\right)
\right)  \\
&  =\sum_{\mu\in\operatorname{Par}}\sum_{\lambda\in\operatorname{Par}}s_{\mu
}\left(  {\zz}\right)  s_{\lambda}\left(  {\mathbf{w}}\right)
\left(  \sum_{\kappa\in\operatorname{Par}}s_{\lambda/\kappa}\left(
{\xx}\right)  s_{\mu/\kappa}\left(  {\mathbf{y}}\right)  \right)
\cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\\
&  =\sum_{\mu\in\operatorname{Par}}\sum_{\nu\in\operatorname{Par}}s_{\mu
}\left(  {\zz}\right)  s_{\nu}\left(  {\mathbf{w}}\right)  \left(
\sum_{\rho\in\operatorname{Par}}s_{\nu/\rho}\left(  {\xx}\right)
s_{\mu/\rho}\left(  {\mathbf{y}}\right)  \right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}%
\end{align*}
(here, we renamed the summation indices $\lambda$ and $\kappa$ as $\nu$ and
$\rho$, respectively). Comparing this with \eqref{sol.cauchy.skew.b.1.6}, we
obtain%
\begin{align*}
&  \sum_{\mu\in\operatorname{Par}}\sum_{\nu\in\operatorname{Par}}s_{\mu
}\left(  {\mathbf{z}}\right)  s_{\nu}\left(  \mathbf{w}\right)  \left(
\sum_{\lambda\in\operatorname{Par}}s_{\lambda/\mu}\left(  {\xx}
\right)  s_{\lambda/\nu}\left(  {\mathbf{y}}\right)  \right)  \\
&  =\sum_{\mu\in\operatorname{Par}}\sum_{\nu\in\operatorname{Par}}s_{\mu
}\left(  {\zz}\right)  s_{\nu}\left(  {\mathbf{w}}\right)  \left(
\sum_{\rho\in\operatorname{Par}}s_{\nu/\rho}\left(  {\xx}\right)
s_{\mu/\rho}\left(  {\mathbf{y}}\right)  \right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
We can regard this as an identity in the ring $\left(   \kk \left[
\left[ \xx , \yy \right]  \right]  \right)  \left[  \left[
\zz , \mathbf{w}\right]  \right]  $ of formal power series in the
variables $\left(  {\mathbf{z}},\mathbf{w}\right)  =\left(  z_{1},z_{2}%
,z_{3},...,w_{1},w_{2},w_{3},...\right)  $ over the ring $\kk \left[
\left[  {\mathbf{x}},{\mathbf{y}}\right]  \right]  $. Extracting the
coefficients in front of $s_{\alpha}\left(  {\mathbf{z}}\right)  s_{\beta
}\left(  \mathbf{w}\right)  $ in this identity%
\footnote{This notion of ``extracting the coefficients'' relies on the fact
that if two families $\left(  a_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)
\in\operatorname{Par}^{2}}\in
\kk\left[\left[\xx, \yy\right]\right]^{\operatorname{Par}^{2}}$ and $\left(
b_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}%
\in
\kk\left[\left[\xx, \yy\right]\right]^{\operatorname{Par}^{2}}$ satisfy
\[
\sum_{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}a_{\mu,\nu}
s_\mu (\zz) s_\nu (\mathbf{w})
=\sum_{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}b_{\mu,\nu}
s_\mu (\zz) s_\nu (\mathbf{w})
\]
in $\left(\kk\left[\left[\xx, \yy\right]\right]\right)
\left[\left[\zz, \mathbf{w}\right]\right]$, then
$\left(  a_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}%
^{2}}=\left(  b_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in
\operatorname{Par}^{2}}$.
This fact is a consequence of
Exercise~\ref{exe.Lambda.infinite-indep}(b),
applied to $q_\lambda = s_\lambda$
(with the base ring $\kk$ replaced by
$\kk\left[\left[\xx, \yy\right]\right]$, and with the families
$\xx$ and $\yy$ renamed as $\zz$ and $\mathbf{w}$).},
% (this is well-defined because
% the family $\left(  s_{\mu}\left(  {\zz}\right)  s_{\nu}\left(
% {\mathbf{w}}\right)  \right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}%
% \times\operatorname{Par}}\in\left(  \left(   \kk \left[  \left[
% {\mathbf{x}},{\mathbf{y}}\right]  \right]  \right)  \left[  \left[
% {\mathbf{z}},\mathbf{w}\right]  \right]  \right)  ^{\operatorname{Par}%
% \times\operatorname{Par}}$ is $\kk \left[  \left[  {\xx%
% },{\mathbf{y}}\right]  \right]  $-linearly independent),
we obtain%
\[
\sum_{\lambda\in\operatorname{Par}}s_{\lambda/\alpha}\left(  {\xx}
\right)  s_{\lambda/\beta}\left(  {\mathbf{y}}\right)  =\left(  \sum_{\rho
\in\operatorname{Par}}s_{\beta/\rho}\left(  {\xx}\right)
s_{\alpha/\rho}\left(  {\mathbf{y}}\right)  \right)  \cdot\prod_{i,j=1}%
^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\]
This solves Exercise~\ref{exe.cauchy.skew}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cauchy.skew}}

\textit{Solution to Exercise~\ref{exe.cauchy.skew}.} Before we step to
the solution of this problem, let us make some general observations.

\begin{itemize}
\item Every two partitions $\lambda$ and $\mu$ satisfy $s_{\lambda/\mu}%
=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum ranges over all
partitions $\nu$ (according to Remark \ref{third-L-R-interpretation-remark}).
In other words, every two partitions $\lambda$ and $\mu$ satisfy
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu}.
\label{sol.cauchy.skew.slash}
\end{equation}
Applying both sides of this equality to the variable set $\yy = \left(
y_{1},y_{2},y_{3},...\right)  $ in the power series ring $\kk \left[
\left[ \yy \right] \right]  \subset \kk \left[  \left[
\xx , \yy \right]  \right]  $, we obtain%
\begin{equation}
s_{\lambda/\mu}\left(   \yy \right)  =\left(  \sum_{\nu\in
\Par}c_{\mu,\nu}^{\lambda}s_{\nu}\right)  \left(
 \yy \right)  =\sum_{\nu\in\Par}c_{\mu,\nu}^{\lambda
}s_{\nu}\left(   \yy \right)
\label{sol.cauchy.skew.slash.y}
\end{equation}
for every two partitions $\lambda$ and $\mu$. On the other hand, applying both
sides of the equality \eqref{sol.cauchy.skew.slash} to the variable set
$\xx =\left(  x_{1},x_{2},x_{3},...\right)  $ in the power series ring
$\kk \left[ \left[ \xx \right] \right]  \subset
\kk \left[ \left[ \xx , \yy \right] \right]$, we obtain
\begin{equation}
s_{\lambda/\mu}\left(  \xx \right)  =\left(  \sum_{\nu\in
\Par}c_{\mu,\nu}^{\lambda}s_{\nu}\right)  \left(
\xx \right)  =\sum_{\nu\in\Par}c_{\mu,\nu}^{\lambda
}s_{\nu}\left(  \xx \right)
\label{sol.cauchy.skew.slash.x}
\end{equation}
for every two partitions $\lambda$ and $\mu$.

\item On the other hand,
\eqref{symm-structure-constants.1} yields
\begin{equation}
s_{\mu} s_{\nu}
= \sum_{\lambda\in\Par} c_{\mu,\nu}^{\lambda} s_{\lambda}
\ \ \ \ \ \ \ \ \ \ 
\text{for any two partitions }\mu\text{ and }\nu.
\label{sol.cauchy.skew.prod}
\end{equation}
Applying both sides of this equality to the variable set $\xx =\left(
x_{1},x_{2},x_{3},...\right)  $ in the power series ring $\kk \left[
\left[ \xx \right]  \right]  \subset \kk \left[  \left[
\xx , \yy \right]  \right]  $, we obtain%
\begin{equation}
\left(  s_{\mu}s_{\nu}\right)  \left(  \xx \right)  =\left(
\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda}s_{\lambda}\right)
\left(  \xx \right)  =\sum_{\lambda\in\Par}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  \xx \right)
\label{sol.cauchy.skew.prod.x}
\end{equation}
for any two partitions $\mu$ and $\nu$.

Also, \eqref{sol.cauchy.skew.prod} yields
\begin{equation}
s_{\mu}s_{\nu}=\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda
}s_{\lambda}=\sum_{\varphi\in\Par}c_{\mu,\nu}^{\varphi
}s_{\varphi} \label{sol.cauchy.skew.prod.2}%
\end{equation}
(here, we renamed the summation index $\lambda$ as $\varphi$) for any two
partitions $\mu$ and $\nu$. Applying both sides of this equality to the
variable set $ \yy =\left(  y_{1},y_{2},y_{3},...\right)$ in the power
series ring $\kk \left[ \left[ \yy \right] \right]
\subset \kk \left[ \left[ \xx , \yy \right] \right]$, we
obtain
\begin{equation}
\left(  s_{\mu}s_{\nu}\right)  \left(   \yy \right)  =\left(
\sum_{\varphi\in\Par}c_{\mu,\nu}^{\varphi}s_{\varphi}\right)
\left(   \yy \right)  =\sum_{\varphi\in\Par}c_{\mu,\nu
}^{\varphi}s_{\varphi}\left(   \yy \right)  =\sum_{\psi\in
\Par}c_{\mu,\nu}^{\psi}s_{\psi}\left(   \yy \right)
\label{sol.cauchy.skew.prod.y}%
\end{equation}
(here, we renamed the summation index $\varphi$ as $\psi$) for any two
partitions $\mu$ and $\nu$.

\item Any four partitions $\alpha$, $\nu$, $\beta$ and $\psi$ satisfy%
\begin{equation}
\sum_{\pi\in\Par }c_{\alpha,\nu}^{\pi}c_{\beta,\psi}^{\pi}%
=\sum_{\rho\in\Par }\sum_{\gamma\in\Par }%
\sum_{\delta\in\Par }\sum_{\lambda\in\Par }%
c_{\rho,\delta}^{\alpha}c_{\gamma,\lambda}^{\nu}c_{\rho,\gamma}^{\beta
}c_{\delta,\lambda}^{\psi}. \label{sol.cauchy.skew.bialgebra}%
\end{equation}
(This follows from Exercise~\ref{exe.Lambda.bialg-c} upon renaming $\kappa$,
$\lambda$, $\varphi$, $\psi$, $\rho$, $\alpha$, $\beta$, $\gamma$, and
$\delta$ as $\alpha$, $\nu$, $\beta$, $\psi$, $\pi$, $\rho$, $\gamma$,
$\delta$, and $\lambda$, respectively.) Notice that both sides of
\eqref{sol.cauchy.skew.bialgebra} are well-defined (i.e., the sums appearing
on them converge with respect to the discrete topology).
\end{itemize}

(a) Let $\mu$ be a partition. We have
\begin{equation}
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
\underbrace{s_{\lambda/\mu}\left( \yy \right)  }_{\substack{=\sum
_{\nu\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu}\left(  \yy
\right)  \\\text{(by \eqref{sol.cauchy.skew.slash.y})}}}=\sum_{\lambda
\in\Par }s_{\lambda}\left(  {\xx }\right)  \sum_{\nu
\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu}\left( \yy \right) .
\label{sol.cauchy.skew.a.1}
\end{equation}
On the other hand, \eqref{Cauchy-identity} yields
\[
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda
\in\Par }s_{\lambda}\left(  {\xx }\right)  s_{\lambda
}\left(   \yy \right)  =\sum_{\nu\in\Par }s_{\nu}\left(
{\xx }\right)  s_{\nu}\left(   \yy \right)
\]
(here, we renamed the summation index $\lambda$ as $\nu$). Multiplying this
equality with $s_{\mu}\left(  \xx \right)  $, we obtain
\begin{align*}
s_{\mu}\left(  {\xx }\right)  \cdot\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}  &  =s_{\mu}\left(  \xx \right)  \cdot
\sum_{\nu\in\Par }s_{\nu}\left(  {\xx }\right)  s_{\nu
}\left(   \yy \right)  =\sum_{\nu\in\Par }\underbrace{s_{\mu
}\left(  \xx \right)  s_{\nu}\left(  {\xx }\right)  }
_{\substack{=\left(  s_{\mu}s_{\nu}\right)  \left(  \xx \right)
=\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
\xx \right)  \\\text{(by \eqref{sol.cauchy.skew.prod.x})}}}s_{\nu
}\left(   \yy \right) \\
&  =\sum_{\nu\in\Par }\left(  \sum_{\lambda\in\Par}
c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  \xx \right)  \right)  s_{\nu
}\left(   \yy \right)  =\sum_{\lambda\in\Par }c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  {\xx }\right)  \sum_{\nu\in
\Par}s_{\nu}\left(   \yy \right) \\
&=\sum_{\lambda
\in\Par }s_{\lambda}\left(  {\xx }\right)  \sum_{\nu
\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu}\left(   \yy \right)
.
\end{align*}
Compared with \eqref{sol.cauchy.skew.a.1}, this yields
\[
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda/\mu}\left( \yy \right)
= s_{\mu} \left( \xx \right) \cdot
  \prod_{i,j=1}^{\infty} \left( 1 - x_i y_j \right) ^{-1}.
\]
This solves Exercise~\ref{exe.cauchy.skew}(a).

(b) We shall give two solutions of Exercise~\ref{exe.cauchy.skew}(b). The
first solution will follow the hint and work in the power series ring
$\kk \left[  \left[  x_{1},x_{2},x_{3},...,y_{1},y_{2},y_{3}
,...,z_{1},z_{2},z_{3},...,w_{1},w_{2},w_{3},...\right]  \right]  $; the
second one will work in $\kk \left[ \left[ \xx , \yy \right] \right]$
only and make use of Exercise~\ref{exe.Lambda.bialg-c}.
\footnote{The two solutions are actually one and the same argument in
different disguises.}

\textit{First solution of Exercise~\ref{exe.cauchy.skew}(b):} Let
$\kk \left[  \left[ \xx, \yy, \zz, \mathbf{w} \right]
\right]$ denote the ring
\[
\kk \left[  \left[  x_{1},x_{2},x_{3},...,y_{1},y_{2},y_{3}%
,...,z_{1},z_{2},z_{3},...,w_{1},w_{2},w_{3},...\right]  \right]  .
\]
This ring clearly contains $\kk \left[ \left[ \xx , \yy \right] \right]$
as a subring. Recall that $\xx$ denotes
the variable set $\left(  x_{1},x_{2},x_{3},...\right)  $, and $\yy$
denotes the variable set $\left(  y_{1},y_{2},y_{3},...\right)  $. We
furthermore denote the variable set $\left(  z_{1},z_{2},z_{3},...\right)  $
by $ \zz $, and the variable set $\left(  w_{1},w_{2},w_{3},...\right)  $
by $\mathbf{w}$. Also, we denote by $\left(  \xx , \zz \right)  $
the variable set $\left(  x_{1},x_{2},x_{3},...,z_{1},z_{2},z_{3},...\right)
$, and by $\left(   \yy ,\mathbf{w}\right)  $ the variable set $\left(
y_{1},y_{2},y_{3},...,w_{1},w_{2},w_{3},...\right)  $. Finally, we denote by
$\left(  \xx , \yy \right)  $ the variable set $\left(  x_{1}%
,x_{2},x_{3},...,y_{1},y_{2},y_{3},...\right)  $.

Now, let us notice that every partition $\lambda$ satisfies%
\begin{align*}
s_{\lambda}\left(   \yy ,\xx \right)   &  =s_{\lambda}\left(
\xx , \yy \right)  =\sum_{\mu\subseteq\lambda}s_{\mu}\left(
{\xx }\right)  s_{\lambda/\mu}\left( \yy \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{eq.Lambda.s.xy}}\right) \\
&  =\sum_{\mu\in\Par}s_{\mu}\left(  {\xx }\right)
s_{\lambda/\mu}\left( \yy \right)  -\sum\limits_{\substack{\mu
\in\Par ;\\\mu\nsubseteq\lambda}}s_{\mu}\left(  {\xx %
}\right)  \underbrace{s_{\lambda/\mu}\left( \yy \right)
}_{\substack{=0\\\text{(since }\mu\nsubseteq\lambda\text{)}}}=\sum_{\mu
\in\Par}s_{\mu}\left(  {\xx }\right)  s_{\lambda/\mu
}\left( \yy \right)  -\underbrace{\sum\limits_{\substack{\mu\in
\Par ;\\\mu\nsubseteq\lambda}}s_{\mu}\left(  {\xx %
}\right)  0}_{=0}\\
&  =\sum_{\mu\in\Par}s_{\mu}\left(  {\xx }\right)
s_{\lambda/\mu}\left( \yy \right)  .
\end{align*}
Applying this equality to the variables $ \zz $ and $\xx$ instead
of $\xx$ and $\yy$, we obtain%
\begin{equation}
s_{\lambda}\left(  \xx , \zz \right)  =\sum_{\mu\in
\Par}s_{\mu}\left(  { \zz }\right)  s_{\lambda/\mu}\left(
{\xx }\right)  . \label{sol.cauchy.skew.b.1.1}%
\end{equation}


Applying this equality to the variables $\yy$ and $\mathbf{w}$ instead
of $\xx$ and $ \zz $, we obtain%
\begin{equation}
s_{\lambda}\left(   \yy ,\mathbf{w}\right)  =\sum_{\mu\in
\Par}s_{\mu}\left(  {\mathbf{w}}\right)  s_{\lambda/\mu}\left(
{ \yy }\right)  =\sum_{\nu\in\Par}s_{\nu}\left(
\mathbf{w}\right)  s_{\lambda/\nu}\left(   \yy \right)
\label{sol.cauchy.skew.b.1.2}%
\end{equation}
(here, we renamed the summation index $\mu$ as $\nu$) for any partition
$\lambda$.

Now,%
\begin{align}
&  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\prod
_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty
}\left(  1-z_{i}y_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty}\left(  1-z_{i}%
w_{j}\right)  ^{-1}\nonumber\\
&  =\prod\limits_{\substack{a\in\left(  \xx , \zz \right)  ;\\b\in\left(
 \yy ,\mathbf{w}\right)  }}\left(  1-ab\right)  ^{-1}=\sum_{\lambda
\in\Par}\underbrace{s_{\lambda}\left(  \xx %
, \zz \right)  }_{\substack{=\sum_{\mu\in\Par}s_{\mu
}\left(  { \zz }\right)  s_{\lambda/\mu}\left(  {\xx }\right)
\\\text{(by \eqref{sol.cauchy.skew.b.1.1})}}}\underbrace{s_{\lambda}\left(
 \yy ,\mathbf{w}\right)  }_{\substack{=\sum_{\nu\in\Par}
s_{\nu}\left( \mathbf{w} \right)  s_{\lambda/\nu}\left( \yy \right)
\\\text{(by \eqref{sol.cauchy.skew.b.1.2})}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{Cauchy-identity}, applied to
the variable sets }\left( \xx,\zz \right)  \text{ and }\left(
\yy,\mathbf{w}\right)  \text{ instead of }\xx \text{ and
} \yy \right) \nonumber\\
&  =\sum_{\lambda\in\Par}\left(  \sum_{\mu\in
\Par}s_{\mu}\left(  { \zz }\right)  s_{\lambda/\mu}\left(
{\xx }\right)  \right)  \left(  \sum_{\nu\in\Par}s_{\nu
}\left(  {\mathbf{w}}\right)  s_{\lambda/\nu}\left(   \yy \right)
\right) \nonumber\\
&  =\sum_{\mu\in\Par}\sum_{\nu\in\Par}s_{\mu
}\left(   \zz \right)  s_{\nu}\left(  \mathbf{w}\right)  \left(
\sum_{\lambda\in\Par}s_{\lambda/\mu}\left(  {\xx %
}\right)  s_{\lambda/\nu}\left(   \yy \right)  \right)  .
\label{sol.cauchy.skew.b.1.6}
\end{align}


On the other hand, every partition $\kappa$ satisfies
\begin{equation}
\sum_{\lambda\in\Par }s_{\lambda}\left(  {\xx }\right)
s_{\lambda/\kappa}\left( \yy \right)  =s_{\kappa}\left(  {\xx %
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)
^{-1}
\label{sol.cauchy.skew.b.1.9}
\end{equation}
(by Exercise~\ref{exe.cauchy.skew}(a), applied to $\kappa$ instead of $\mu$).
Applying this to the variable sets $\mathbf{w}$ and $\xx$ instead of
$\xx$ and $\yy$, we obtain%
\begin{align}
\sum_{\lambda\in\Par }s_{\lambda}\left(  {\mathbf{w}}\right)
s_{\lambda/\kappa}\left(  {\xx }\right)    & =s_{\kappa}\left(
{\mathbf{w}}\right)  \cdot\underbrace{\prod_{i,j=1}^{\infty}\left(
1-w_{i}x_{j}\right)  ^{-1}}_{\substack{=\prod_{i,j=1}^{\infty}\left(
1-x_{j}w_{i}\right)  ^{-1}\\=\prod_{j,i=1}^{\infty}\left(  1-x_{i}%
w_{j}\right)  ^{-1}\\\text{(here, we renamed the index }\left(  i,j\right)
\text{ as }\left(  j,i\right)  \text{)}}}\nonumber\\
& =s_{\kappa}\left(  \mathbf{w}\right)  \cdot\underbrace{\prod_{j,i=1}%
^{\infty}}_{=\prod_{i,j=1}^{\infty}}\left(  1-x_{i}w_{j}\right)
^{-1}=s_{\kappa}\left(  \mathbf{w}\right)  \cdot\prod_{i,j=1}^{\infty}\left(
1-x_{i}w_{j}\right)  ^{-1}\label{sol.cauchy.skew.b.1.10}%
\end{align}
for every partition $\kappa$. But applying both sides of the identity
\eqref{sol.cauchy.skew.b.1.9} to the variable set $ \zz $ instead of
$\xx$, we obtain
\[
\sum_{\lambda\in\Par }s_{\lambda}\left(  { \zz }\right)
s_{\lambda/\kappa}\left( \yy \right)  =s_{\kappa}\left(  { \zz %
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-z_{i}y_{j}\right)  ^{-1}.
\]
Renaming the summation index $\lambda$ as $\mu$ on the left hand side of this
equality, we obtain%
\begin{equation}
\sum_{\mu\in\Par }s_{\mu}\left(  { \zz }\right)
s_{\mu/\kappa}\left( \yy \right)  =s_{\kappa}\left(  { \zz %
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-z_{i}y_{j}\right)
^{-1}.\label{sol.cauchy.skew.b.1.11}%
\end{equation}


Now, applying \eqref{Cauchy-identity} to the variable sets $ \zz $ and
$\mathbf{w}$ instead of $\xx$ and $\yy$, we obtain%
\[
\prod_{i,j=1}^{\infty}\left(  1-z_{i}w_{j}\right)  ^{-1}=\sum_{\lambda
\in\Par}s_{\lambda}\left(   \zz \right)  s_{\lambda
}\left(  \mathbf{w}\right)  .
\]
Hence,%
\begin{align*}
&  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\prod
_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty
}\left(  1-z_{i}y_{j}\right)  ^{-1}\underbrace{\prod_{i,j=1}^{\infty}\left(
1-z_{i}w_{j}\right)  ^{-1}}_{=\sum_{\lambda\in\Par}s_{\lambda
}\left(   \zz \right)  s_{\lambda}\left(  \mathbf{w}\right)  }\\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\prod
_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\prod_{i,j=1}^{\infty
}\left(  1-z_{i}y_{j}\right)  ^{-1}\cdot\sum_{\lambda\in\Par %
}s_{\lambda}\left(   \zz \right)  s_{\lambda}\left(  \mathbf{w}\right)
\\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\cdot\sum
_{\lambda\in\Par}\left(  s_{\lambda}\left(  \mathbf{w}\right)
\cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)  ^{-1}\right)  \left(
s_{\lambda}\left(   \zz \right)  \cdot\prod_{i,j=1}^{\infty}\left(
1-z_{i}y_{j}\right)  ^{-1}\right)  \\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\cdot\sum
_{\kappa\in\Par}\underbrace{\left(  s_{\kappa}\left(
\mathbf{w}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}w_{j}\right)
^{-1}\right)  }_{\substack{=\sum_{\lambda\in\Par }s_{\lambda
}\left(  {\mathbf{w}}\right)  s_{\lambda/\kappa}\left(  {\xx }\right)
\\\text{(by \eqref{sol.cauchy.skew.b.1.10})}}}\underbrace{\left(  s_{\kappa
}\left(   \zz \right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-z_{i}%
y_{j}\right)  ^{-1}\right)  }_{\substack{=\sum_{\mu\in\Par}
s_{\mu}\left(  { \zz }\right)  s_{\mu/\kappa}\left(  { \yy %
}\right)  \\\text{(by \eqref{sol.cauchy.skew.b.1.11})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\lambda\text{ as }\kappa\right)  \\
&  =\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\cdot\sum
_{\kappa\in\Par}\left(  \sum_{\lambda\in\Par %
}s_{\lambda}\left(  {\mathbf{w}}\right)  s_{\lambda/\kappa}\left(
{\xx }\right)  \right)  \left(  \sum_{\mu\in\Par }s_{\mu
}\left(  { \zz }\right)  s_{\mu/\kappa}\left( \yy \right)
\right)  \\
&  =\sum_{\mu\in\Par }\sum_{\lambda\in\Par }s_{\mu
}\left(  { \zz }\right)  s_{\lambda}\left(  {\mathbf{w}}\right)  \left(
\sum_{\kappa\in\Par }s_{\lambda/\kappa}\left(  {\xx %
}\right)  s_{\mu/\kappa}\left( \yy \right)  \right)  \cdot
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\\
&  =\sum_{\mu\in\Par }\sum_{\nu\in\Par }s_{\mu
}\left(  { \zz }\right)  s_{\nu}\left(  {\mathbf{w}}\right)  \left(
\sum_{\rho\in\Par }s_{\nu/\rho}\left(  {\xx }\right)
s_{\mu/\rho}\left( \yy \right)  \right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}%
\end{align*}
(here, we renamed the summation indices $\lambda$ and $\kappa$ as $\nu$ and
$\rho$, respectively). Comparing this with \eqref{sol.cauchy.skew.b.1.6}, we
obtain%
\begin{align*}
& \sum_{\mu\in\Par}\sum_{\nu\in\Par}s_{\mu
}\left(   \zz \right)  s_{\nu}\left(  \mathbf{w}\right)  \left(
\sum_{\lambda\in\Par}s_{\lambda/\mu}\left(  {\xx %
}\right)  s_{\lambda/\nu}\left(   \yy \right)  \right)  \\
& =\sum_{\mu\in\Par }\sum_{\nu\in\Par }s_{\mu
}\left(  { \zz }\right)  s_{\nu}\left(  {\mathbf{w}}\right)  \left(
\sum_{\rho\in\Par }s_{\nu/\rho}\left(  {\xx }\right)
s_{\mu/\rho}\left( \yy \right)  \right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
We can regard this as an identity in the ring $\left(   \kk \left[
\left[ \xx , \yy \right]  \right]  \right)  \left[  \left[
\zz , \mathbf{w}\right]  \right]  $ of formal power series in the
variables $\left(  {\mathbf{z}},\mathbf{w}\right)  =\left(  z_{1},z_{2}%
,z_{3},...,w_{1},w_{2},w_{3},...\right)  $ over the ring $\kk \left[
\left[  {\mathbf{x}},{\mathbf{y}}\right]  \right]  $. Extracting the
coefficients in front of $s_{\alpha}\left(  {\mathbf{z}}\right)  s_{\beta
}\left(  \mathbf{w}\right)  $ in this identity%
\footnote{This notion of ``extracting the coefficients'' relies on the fact
that if two families $\left(  a_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)
\in\operatorname{Par}^{2}}\in
\kk\left[\left[\xx, \yy\right]\right]^{\operatorname{Par}^{2}}$ and $\left(
b_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}%
\in
\kk\left[\left[\xx, \yy\right]\right]^{\operatorname{Par}^{2}}$ satisfy
\[
\sum_{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}a_{\mu,\nu}
s_\mu (\zz) s_\nu (\mathbf{w})
=\sum_{\left(  \mu,\nu\right)  \in\operatorname{Par}^{2}}b_{\mu,\nu}
s_\mu (\zz) s_\nu (\mathbf{w})
\]
in $\left(\kk\left[\left[\xx, \yy\right]\right]\right)
\left[\left[\zz, \mathbf{w}\right]\right]$, then
$\left(  a_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}%
^{2}}=\left(  b_{\mu,\nu}\right)  _{\left(  \mu,\nu\right)  \in
\operatorname{Par}^{2}}$.
This fact is a consequence of
Exercise~\ref{exe.Lambda.infinite-indep}(b),
applied to $q_\lambda = s_\lambda$
(with the base ring $\kk$ replaced by
$\kk\left[\left[\xx, \yy\right]\right]$, and with the families
$\xx$ and $\yy$ renamed as $\zz$ and $\mathbf{w}$).},
% (this is well-defined because
% the family $\left(  s_{\mu}\left(  {\zz}\right)  s_{\nu}\left(
% {\mathbf{w}}\right)  \right)  _{\left(  \mu,\nu\right)  \in\operatorname{Par}%
% \times\operatorname{Par}}\in\left(  \left(   \kk \left[  \left[
% {\mathbf{x}},{\mathbf{y}}\right]  \right]  \right)  \left[  \left[
% {\mathbf{z}},\mathbf{w}\right]  \right]  \right)  ^{\operatorname{Par}%
% \times\operatorname{Par}}$ is $\kk \left[  \left[  {\xx%
% },{\mathbf{y}}\right]  \right]  $-linearly independent),
we obtain%
\[
\sum_{\lambda\in\operatorname{Par}}s_{\lambda/\alpha}\left(  {\xx}
\right)  s_{\lambda/\beta}\left(  {\mathbf{y}}\right)  =\left(  \sum_{\rho
\in\operatorname{Par}}s_{\beta/\rho}\left(  {\xx}\right)
s_{\alpha/\rho}\left(  {\mathbf{y}}\right)  \right)  \cdot\prod_{i,j=1}%
^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\]
This solves Exercise~\ref{exe.cauchy.skew}(b).

\textit{Second solution of Exercise~\ref{exe.cauchy.skew}(b):} We have
$s_{\lambda/\alpha}=\sum_{\nu\in\Par}c_{\alpha,\nu}^{\lambda
}s_{\nu}$ for every partition $\lambda$ (by \eqref{sol.cauchy.skew.slash},
applied to $\mu=\alpha$). Applying both sides of this equality to the variable
set $\xx =\left(  x_{1},x_{2},x_{3},...\right)  $ in the power series
ring $\kk \left[  \left[ \xx \right]  \right]  \subset
\kk \left[  \left[  \xx , \yy \right]  \right]  $, we obtain%
\begin{equation}
s_{\lambda/\alpha}\left(  \xx \right)  =\left(  \sum_{\nu\in
\Par}c_{\alpha,\nu}^{\lambda}s_{\nu}\right)  \left(
\xx \right)  =\sum_{\nu\in\Par}c_{\alpha,\nu}^{\lambda
}s_{\nu}\left(  \xx \right)
\label{sol.cauchy.skew.b.2.1}
\end{equation}
for every partition $\lambda$.

But we have $s_{\lambda/\beta}=\sum_{\nu\in\Par}c_{\beta,\nu
}^{\lambda}s_{\nu}$ for any partition $\lambda$ (by
\eqref{sol.cauchy.skew.slash}, applied to $\mu=\beta$). Applying both sides of
this equality to the variable set
$\yy =\left(  y_{1},y_{2},y_{3},...\right)  $ in the
power series ring $\kk \left[ \left[ \yy \right] \right] \subset
\kk \left[ \left[ \xx , \yy \right] \right]$, we obtain
\begin{equation}
s_{\lambda/\beta}\left(   \yy \right)  =\left(  \sum_{\nu\in
\Par}c_{\beta,\nu}^{\lambda}s_{\nu}\right)  \left(
 \yy \right)  =\sum_{\nu\in\Par}c_{\beta,\nu}^{\lambda
}s_{\nu}\left( \yy \right) = \sum_{\psi\in\Par}
c_{\beta,\psi}^{\lambda}s_{\psi}\left(   \yy \right)
\label{sol.cauchy.skew.b.2.2}
\end{equation}
(here, we renamed the summation index $\nu$ as $\psi$) for every partition
$\lambda$. Now,
\begin{align}
&  \sum_{\lambda\in\Par }\underbrace{s_{\lambda/\alpha}\left(
{\xx }\right)  }_{\substack{=\sum_{\nu\in\Par}%
c_{\alpha,\nu}^{\lambda}s_{\nu}\left(  \xx \right)  \\\text{(by
\eqref{sol.cauchy.skew.b.2.1})}}}\underbrace{s_{\lambda/\beta}\left(
{ \yy }\right)  }_{\substack{=\sum_{\psi\in\Par}%
c_{\beta,\psi}^{\lambda}s_{\psi}\left(   \yy \right)  \\\text{(by
\eqref{sol.cauchy.skew.b.2.2})}}}\nonumber\\
&  =\sum_{\lambda\in\Par }\left(  \sum_{\nu\in\Par %
}c_{\alpha,\nu}^{\lambda}s_{\nu}\left(  \xx \right)  \right)  \left(
\sum_{\psi\in\Par}c_{\beta,\psi}^{\lambda}s_{\psi}\left(
 \yy \right)  \right)  =\underbrace{\sum_{\lambda\in\Par %
}\sum_{\nu\in\Par}\sum_{\psi\in\Par}}%
_{=\sum_{\nu\in\Par}\sum_{\psi\in\Par}%
\sum_{\lambda\in\Par }}c_{\alpha,\nu}^{\lambda}c_{\beta,\psi
}^{\lambda}s_{\nu}\left(  \xx \right)  s_{\psi}\left(   \yy \right)
\nonumber\\
&  =\sum_{\nu\in\Par}\sum_{\psi\in\Par %
}\underbrace{\sum_{\lambda\in\Par }c_{\alpha,\nu}^{\lambda
}c_{\beta,\psi}^{\lambda}}_{\substack{=\sum_{\pi\in\Par %
}c_{\alpha,\nu}^{\pi}c_{\beta,\psi}^{\pi}\\\text{(here, we replaced the
summation index }\lambda\text{ as }\pi\text{)}}}s_{\nu}\left(  \xx %
\right)  s_{\psi}\left(   \yy \right) \nonumber\\
&  =\sum_{\nu\in\Par}\sum_{\psi\in\Par %
}\underbrace{\sum_{\pi\in\Par }c_{\alpha,\nu}^{\pi}c_{\beta,\psi
}^{\pi}}_{\substack{=\sum_{\rho\in\Par }\sum_{\gamma
\in\Par }\sum_{\delta\in\Par }\sum_{\lambda
\in\Par }c_{\rho,\delta}^{\alpha}c_{\gamma,\lambda}^{\nu}%
c_{\rho,\gamma}^{\beta}c_{\delta,\lambda}^{\psi}\\\text{(by
\eqref{sol.cauchy.skew.bialgebra})}}}s_{\nu}\left(  \xx \right)
s_{\psi}\left(   \yy \right) \nonumber\\
&  =\sum_{\nu\in\Par}\sum_{\psi\in\Par}\left(
\sum_{\rho\in\Par }\sum_{\gamma\in\Par }\sum
_{\delta\in\Par }\sum_{\lambda\in\Par }%
c_{\rho,\delta}^{\alpha}c_{\gamma,\lambda}^{\nu}c_{\rho,\gamma}^{\beta
}c_{\delta,\lambda}^{\psi}\right)  s_{\nu}\left(  \xx \right)  s_{\psi
}\left(   \yy \right) \nonumber\\
&  =\sum_{\rho\in\Par }\sum_{\gamma\in\Par }%
\sum_{\delta\in\Par }\sum_{\lambda\in\Par %
}\underbrace{\sum_{\nu\in\Par}\sum_{\psi\in\Par %
}c_{\rho,\delta}^{\alpha}c_{\gamma,\lambda}^{\nu}c_{\rho,\gamma}^{\beta
}c_{\delta,\lambda}^{\psi}s_{\nu}\left(  \xx \right)  s_{\psi}\left(
 \yy \right)  }_{=c_{\rho,\delta}^{\alpha}c_{\rho,\gamma}^{\beta}\left(
\sum_{\nu\in\Par}c_{\gamma,\lambda}^{\nu}s_{\nu}\left(
\xx \right)  \right)  \left(  \sum_{\psi\in\Par}%
c_{\delta,\lambda}^{\psi}s_{\psi}\left(   \yy \right)  \right)
}\nonumber\\
&  =\sum_{\rho\in\Par }\sum_{\gamma\in\Par }%
\sum_{\delta\in\Par }\sum_{\lambda\in\Par }%
c_{\rho,\delta}^{\alpha}c_{\rho,\gamma}^{\beta}\left(  \sum_{\nu
\in\Par}c_{\gamma,\lambda}^{\nu}s_{\nu}\left(  \xx %
\right)  \right)  \left(  \sum_{\psi\in\Par}c_{\delta,\lambda
}^{\psi}s_{\psi}\left(   \yy \right)  \right)  .
\label{sol.cauchy.skew.b.2.4}%
\end{align}


But any two partitions $\delta$ and $\lambda$ satisfy%
\begin{equation}
\left(  s_{\delta}s_{\lambda}\right)  \left(   \yy \right)  =\sum
_{\psi\in\Par}c_{\delta,\lambda}^{\psi}s_{\psi}\left(
 \yy \right)
\label{sol.cauchy.skew.b.2.5}
\end{equation}
(by \eqref{sol.cauchy.skew.prod.y}, applied to $\delta$ and $\lambda$ instead
of $\mu$ and $\nu$). Furthermore, any two partitions $\gamma$ and $\lambda$
satisfy
\begin{align*}
s_{\gamma}s_{\lambda}  &  =\sum_{\varphi\in\Par}c_{\gamma
,\lambda}^{\varphi}s_{\varphi}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.skew.prod.2}, applied to }\gamma\text{ and }\lambda\text{
instead of }\mu\text{ and }\nu\right) \\
&  =\sum_{\nu\in\Par}c_{\gamma,\lambda}^{\nu}s_{\nu
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\varphi\text{ as }\nu\right)  .
\end{align*}
Applying both sides of this equality to the variable set $\xx =\left(
x_{1},x_{2},x_{3},...\right)  $ in the power series ring $\kk \left[
\left[ \xx \right]  \right]  \subset \kk \left[  \left[
\xx , \yy \right]  \right]  $, we obtain
\begin{equation}
\left(  s_{\gamma}s_{\lambda}\right)  \left(  \xx \right)  =\left(
\sum_{\nu\in\Par}c_{\gamma,\lambda}^{\nu}s_{\nu}\right)
\left(  \xx \right)  =\sum_{\nu\in\Par}c_{\gamma,\lambda
}^{\nu}s_{\nu}\left(  \xx \right)  . \label{sol.cauchy.skew.b.2.6}%
\end{equation}


Now, \eqref{sol.cauchy.skew.b.2.4} becomes
\begin{align}
&  \sum_{\lambda\in\Par }s_{\lambda/\alpha}\left( \xx \right)
 s_{\lambda/\beta}\left( \yy \right) \nonumber\\
&  =\sum_{\rho\in\Par }\sum_{\gamma\in\Par }
\sum_{\delta\in\Par }\sum_{\lambda\in\Par }
c_{\rho,\delta}^{\alpha}c_{\rho,\gamma}^{\beta}\underbrace{\left(  \sum
_{\nu\in\Par}c_{\gamma,\lambda}^{\nu}s_{\nu}\left(
\xx \right)  \right)  }_{\substack{=\left(  s_{\gamma}s_{\lambda
}\right)  \left(  \xx \right)  \\
\text{(by \eqref{sol.cauchy.skew.b.2.6})}}}
\underbrace{\left(  \sum_{\psi\in\Par}c_{\delta,\lambda
}^{\psi}s_{\psi}\left(   \yy \right)  \right)  }_{\substack{=\left(
s_{\delta}s_{\lambda}\right)  \left(   \yy \right)  \\\text{(by
\eqref{sol.cauchy.skew.b.2.5})}}}\nonumber\\
&  =\sum_{\rho\in\Par }\sum_{\gamma\in\Par }
\sum_{\delta\in\Par }\sum_{\lambda\in\Par }
c_{\rho,\delta}^{\alpha}c_{\rho,\gamma}^{\beta}\underbrace{\left(  s_{\gamma
}s_{\lambda}\right)  \left(  \xx \right)  }_{=s_{\gamma}\left(
\xx \right)  \cdot s_{\lambda}\left(  \xx \right)  }%
\cdot\underbrace{\left(  s_{\delta}s_{\lambda}\right)  \left(   \yy %
\right)  }_{=s_{\delta}\left(   \yy \right)  \cdot s_{\lambda}\left(
 \yy \right)  }\nonumber\\
&  =\sum_{\rho\in\Par }\sum_{\gamma\in\Par }%
\sum_{\delta\in\Par }\sum_{\lambda\in\Par }%
c_{\rho,\delta}^{\alpha}c_{\rho,\gamma}^{\beta}s_{\gamma}\left(
\xx \right)  \cdot s_{\lambda}\left(  \xx \right)  \cdot
s_{\delta}\left(   \yy \right)  \cdot s_{\lambda}\left(   \yy %
\right) \nonumber\\
&  =\sum_{\rho\in\Par }\left(  \sum_{\gamma\in\Par %
}c_{\rho,\gamma}^{\beta}s_{\gamma}\left(  \xx \right)  \right)  \left(
\sum_{\delta\in\Par }c_{\rho,\delta}^{\alpha}s_{\delta}\left(
 \yy \right)  \right)  \cdot\left(  \sum_{\lambda\in\Par %
}s_{\lambda}\left(  \xx \right)  s_{\lambda}\left(   \yy \right)
\right)  . \label{sol.cauchy.skew.b.2.9}%
\end{align}


Now, every $\rho\in\Par $ satisfies%
\begin{align}
s_{\alpha/\rho}\left(   \yy \right)   &  =\sum_{\nu\in\Par}
c_{\rho,\nu}^{\alpha}s_{\nu}\left(   \yy \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.skew.slash.y}, applied
to }\alpha\text{ and }\rho\text{ instead of }\lambda\text{ and }\mu\right)
\nonumber\\
&  =\sum_{\delta\in\Par }c_{\rho,\delta}^{\alpha}s_{\delta
}\left(   \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
the summation index }\nu\text{ as }\delta\right)  .
\label{sol.cauchy.skew.b.2.11}
\end{align}
Also, every $\rho\in\Par$ satisfies
\begin{align}
s_{\beta/\rho}\left(  \xx \right)   &  =\sum_{\nu\in\Par}
c_{\rho,\nu}^{\beta}s_{\nu}\left(  \xx \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.skew.slash.x}, applied
to }\beta\text{ and }\rho\text{ instead of }\lambda\text{ and }\mu\right)
\nonumber\\
&  =\sum_{\gamma\in\Par}c_{\rho,\gamma}^{\beta}s_{\gamma
}\left(  \xx \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
the summation index }\nu\text{ as }\gamma\right)  .
\label{sol.cauchy.skew.b.2.12}
\end{align}
Now, \eqref{sol.cauchy.skew.b.2.9} becomes
\begin{align*}
&  \sum_{\lambda\in\Par }s_{\lambda/\alpha}\left( \xx \right)
 s_{\lambda/\beta}\left( \yy \right) \\
&  =\sum_{\rho\in\Par }\underbrace{\left(  \sum_{\gamma
\in\Par }c_{\rho,\gamma}^{\beta}s_{\gamma}\left(  \xx %
\right)  \right)  }_{\substack{=s_{\beta/\rho}\left(  \xx \right)
\\\text{(by \eqref{sol.cauchy.skew.b.2.12})}}}
\underbrace{\left(  \sum
_{\delta\in\Par }c_{\rho,\delta}^{\alpha}s_{\delta}\left(
 \yy \right)  \right)  }_{\substack{=s_{\alpha/\rho}\left(
 \yy \right)  \\\text{(by \eqref{sol.cauchy.skew.b.2.11})}}}
\cdot\underbrace{\left(  \sum_{\lambda\in\Par }s_{\lambda
}\left(  \xx \right)  s_{\lambda}\left(   \yy \right)  \right)
}_{\substack{=\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)
^{-1}\\\text{(by \eqref{Cauchy-identity})}}}\\
&  =\sum_{\rho\in\Par }s_{\beta/\rho}\left(  {\xx }\right)
s_{\alpha/\rho}\left( \yy \right)  \cdot\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}\\
&  =\left(  \sum_{\rho\in\Par }s_{\beta/\rho}\left(  {\xx %
}\right)  s_{\alpha/\rho}\left( \yy \right)  \right)  \cdot
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
This solves Exercise~\ref{exe.cauchy.skew}(b) again.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.hall.gr}}

\textit{Solution to Exercise~\ref{exe.Lambda.hall.gr}.} We know that
$\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ is a basis of
the $\kk$-module $\Lambda$, but we can also say something more
specific: For every $n\in \NN$, the family $\left(  s_{\lambda}\right)
_{\lambda\in \Par_n}$ is a basis of the $\kk$-module $\Lambda_n$%
\ \ \ \ \footnote{\textit{Proof.} We have $s_{\lambda} \in
\Lambda_{\left\vert \lambda\right\vert} = \Lambda_n$ for every
$\lambda\in \Par_n$. Thus,
$\left(  s_{\lambda}  \right)  _{\lambda\in \Par_n}$ is a family of
elements of $\Lambda_{n}$. This family is $\kk$-linearly independent
(because it is part of the basis $\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ of $\Lambda$). It remains to prove that this family
spans the $\kk$-module $\Lambda_{n}$.
\par
Let $f\in\Lambda_{n}$. We can write $f$ in the form $f=\sum_{\lambda
\in \Par }a_{\lambda}s_{\lambda}$ for some family $\left(
a_{\lambda}\right)  _{\lambda\in \Par }\in \kk %
^{ \Par }$ of scalars. Consider this $\left(  a_{\lambda}\right)
_{\lambda\in \Par }$ and notice that
\[
f=\sum_{\lambda\in \Par }a_{\lambda}s_{\lambda}=\sum_{\lambda
\in \Par_{n}}a_{\lambda}s_{\lambda}+\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda\notin \Par_{n}}}
a_{\lambda}s_{\lambda},
\]
so that $f-\sum_{\lambda\in \Par_{n}}a_{\lambda
}s_{\lambda}=\sum\limits_{\substack{\lambda\in \Par ;\\\lambda
\notin \Par_{n}}}a_{\lambda}s_{\lambda}$. The left hand
side of this latter equality is homogeneous of degree $n$ (since $f$ and all
the $s_{\lambda}$ with $\lambda\in \Par_{n}$ are
homogeneous of degree $n$), while the right hand side is a sum of homogeneous
elements of degrees different from $n$. So the only way these two sides can be
equal is if they both are $0$. In particular, this shows that the left hand
side is $0$. In other words, $f-\sum_{\lambda\in \Par_{n}}
a_{\lambda}s_{\lambda}=0$, so that $f=\sum_{\lambda
\in \Par_{n}}a_{\lambda}s_{\lambda}$. Hence, $f$ is a
$\kk$-linear combination of the $s_{\lambda}$ for $\lambda
\in \Par_{n}$.
\par
Since we have proven this for every $f\in\Lambda_{n}$, we thus conclude that
the family $\left(  s_{\lambda}\right)  _{\lambda\in \Par_{n}}$
spans the $\kk$-module $\Lambda_{n}$, qed.}.

The basis $\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ of
$\Lambda$ is orthonormal with respect to the Hall inner product. In other
words,
\begin{equation}
\left(  s_{\lambda},s_{\mu}\right)  =\delta_{\lambda,\mu}%
\ \ \ \ \ \ \ \ \ \ \text{for any partitions }\lambda\text{ and }\mu.
\label{sol.Lambda.hall.gr.5}
\end{equation}


(a) Let $n$ and $m$ be two distinct nonnegative integers. Let $f\in\Lambda
_{n}$ and $g\in\Lambda_{m}$.

We need to prove that $\left(  f,g\right)  =0$. Since this equality is
$\kk$-linear in $f$, we can WLOG assume that $f$ is an element of the
basis $\left(  s_{\lambda}\right)  _{\lambda\in \Par_{n}}$ of
$\Lambda_{n}$. In other words, we can WLOG assume that
$f=s_{\lambda}$ for some $\lambda\in \Par_{n}$. Assume
this, and similarly assume that $g=s_{\mu}$ for some $\mu\in
 \Par_{m}$. These two partitions $\lambda$ and $\mu$
must be distinct (because their sizes differ: $\left\vert \lambda\right\vert
=n\neq m=\left\vert \mu\right\vert $), and so they satisfy $\delta
_{\lambda,\mu}=0$. Now,%
\begin{align*}
\left(  \underbrace{f}_{=s_{\lambda}},\underbrace{g}_{=s_{\mu}}\right)   &
=\left(  s_{\lambda},s_{\mu}\right)  =\delta_{\lambda,\mu}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Lambda.hall.gr.5})}\right) \\
&  =0.
\end{align*}


This solves Exercise~\ref{exe.Lambda.hall.gr}(a).

(b) Let $n\in{ \NN }$ and $f\in\Lambda_{n}$.

We need to prove that $\left(  h_{n},f\right)  =f\left(  1\right)  $. Since
this equality is $\kk$-linear in $f$, we can WLOG assume that $f$ is an
element of the basis $\left(  s_{\lambda}\right)  _{\lambda\in
 \Par_{n}}$ of $\Lambda_{n}$. In other words, we can
WLOG assume that $f=s_{\lambda}$ for some $\lambda\in \Par_{n}$. Assume this.

We must be in one of the two cases:

\textit{Case 1:} We have $\ell\left(  \lambda\right)  \leq1$.

\textit{Case 2:} We have $\ell\left(  \lambda\right)  >1$.

Let us consider Case 1 first. In this case, $\ell\left(  \lambda\right)
\leq1$. Hence, $\lambda=\left(  n\right)  $ (since $\lambda\in
 \Par_{n}$), so that $s_{\lambda}=s_{\left(  n\right)
}$. Hence, $f=s_{\lambda}=s_{\left(  n\right)  }=h_{n}$ and thus $f\left(
1\right)  =h_{n}\left(  1\right)  =1$. Compared with%
\begin{align*}
\left(  \underbrace{h_{n}}_{=s_{\left(  n\right)  }},\underbrace{f}%
_{=s_{\left(  n\right)  }}\right)    & =\left(  s_{\left(  n\right)
},s_{\left(  n\right)  }\right)  =\delta_{\left(  n\right)  ,\left(  n\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Lambda.hall.gr.5})}\right)
\\
& =1,
\end{align*}
this yields $\left(  h_{n},f\right)  =f\left(  1\right)  $. Hence, $\left(
h_{n},f\right)  =f\left(  1\right)  $ is proven in Case 1.

Let us now consider Case 2. In this case, $\ell\left(  \lambda\right)  >1$, so
the partition $\lambda$ has more than $1$ part. Exercise
\ref{exe.schur.finite}(b) (applied to $1$ instead of $n$) thus yields
$s_{\lambda}\left(  x_{1}\right)  =0$. But $s_{\lambda}\left(  1\right)  $ can
be seen as the result of substituting $1$ for $x_{1}$ in $s_{\lambda}\left(
x_{1}\right)  $, and therefore must be $0$ as well (since $s_{\lambda}\left(
x_{1}\right)  =0$). Thus, we have $s_{\lambda}\left( 1 \right)  =0$. Since
$f=s_{\lambda}$, we now have $f\left(  1\right)  =s_{\lambda}\left(  1\right)
=0$. Compared with%
\begin{align*}
\left(  \underbrace{h_{n}}_{=s_{\left(  n\right)  }},\underbrace{f}%
_{=s_{\lambda}}\right)    & =\left(  s_{\left(  n\right)  },s_{\lambda
}\right)  =\delta_{\left(  n\right)  ,\lambda}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Lambda.hall.gr.5})}\right)  \\
& =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  n\right)  \neq
\lambda\text{ (because }\ell\left(  \left(  n\right)  \right)  =1<\ell\left(
\lambda\right) \text{)} \right)  ,
\end{align*}
this yields $\left(  h_{n},f\right)  =f\left(  1\right)  $. Hence, $\left(
h_{n},f\right)  =f\left(  1\right)  $ is proven in Case 2.

Now, $\left(  h_{n},f\right)  =f\left(  1\right)  $ is proven in both Cases,
which solves Exercise~\ref{exe.Lambda.hall.gr}(b).

(c) We know that the Hall inner product has an orthonormal basis
(namely, $\left(  s_{\lambda}\right)  _{\lambda\in \Par }$).
Thus, the Hall inner product is symmetric (since any bilinear form that
has an orthonormal basis must be symmetric).
In other words, $\left(f, g\right) = \left(g, f\right)$ for all
$f \in \Lambda$ and $g \in \Lambda$.
This solves Exercise~\ref{exe.Lambda.hall.gr}(c).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.hall.gr}}

\textit{Solution to Exercise~\ref{exe.Lambda.hall.gr}.} For every
$N \in \NN$, let $\pi_N$ denote the projection from $\Lambda$ onto the
$N$-th graded component $\Lambda_N$ of $\Lambda$. These projections have the
following properties:
\begin{equation}
\pi_N \left(  h\right)  =h\ \ \ \ \ \ \ \ \ \ \text{for every } N\in
\NN \text{ and } h \in \Lambda_N,
\label{sol.Lambda.hall.gr.1}
\end{equation}
whereas
\begin{equation}
\pi_N \left(  h\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every } M\in
\NN \text{ and } N \in \NN \text{ and } h \in \Lambda_M
\text{ satisfying }M \neq N.
\label{sol.Lambda.hall.gr.2}
\end{equation}


(a) Let $n$ and $m$ be two distinct nonnegative integers. Let
$f \in \Lambda_n$ and $g \in \Lambda_m$.

We have $f \in \Lambda_n \subset \Lambda$. Hence, we can write $f$ in the form
$f=\sum_{\lambda\in\Par}a_{\lambda}s_{\lambda}$ for some family
$\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk ^{\Par}$ of elements of $\kk$ such that all
but finitely many $\lambda\in\Par$ satisfy $a_{\lambda}=0$
(since the family $\left(  s_{\lambda}\right)  _{\lambda\in\Par}$
is a $\kk$-module basis of $\Lambda$). Consider this family $\left(
a_{\lambda}\right)  _{\lambda\in\Par}$.

We have $g\in\Lambda_{m}\subset\Lambda$. Hence, we can write $g$ in the form
$g=\sum_{\lambda\in\Par}b_{\lambda}s_{\lambda}$ for some family
$\left(  b_{\lambda}\right)  _{\lambda\in\Par}\in
\kk ^{\Par}$ of elements of $\kk$ such that all
but finitely many $\lambda\in\Par$ satisfy $b_{\lambda}=0$
(since the family $\left(  s_{\lambda}\right)  _{\lambda\in\Par}$
is a $\kk$-module basis of $\Lambda$). Consider this family $\left(
b_{\lambda}\right)  _{\lambda\in\Par}$.

We have $f\in\Lambda_{n}$, so that $\pi_{n}\left(  f\right)  =f$ (by
\eqref{sol.Lambda.hall.gr.1}, applied to $N=n$ and $h=f$). Thus,
\begin{align}
f  &  =\pi_{n}\left(  \underbrace{f}_{=\sum_{\lambda\in\Par}
a_{\lambda}s_{\lambda}}\right)  =\pi_{n}\left(  \sum_{\lambda\in
\Par}a_{\lambda}s_{\lambda}\right)  =\sum_{\lambda
\in\Par}a_{\lambda}\pi_{n}\left(  s_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi_{n}\text{ is }
\kk\text{-linear}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}\underbrace{\pi_{n}\left(  s_{\lambda}\right)
}_{\substack{=s_{\lambda}\\\text{(by \eqref{sol.Lambda.hall.gr.1} (applied to
}N=n\\\text{and }h=s_{\lambda}\text{) (since }s_{\lambda}\in\Lambda
_{\left\vert \lambda\right\vert }=\Lambda_{n}\text{ (since }\left\vert
\lambda\right\vert =n\text{)))}}}+\sum\limits_{\substack{\lambda\in
\Par ;\\\left\vert \lambda\right\vert \neq n}}a_{\lambda
}\underbrace{\pi_{n}\left(  s_{\lambda}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.Lambda.hall.gr.2} (applied to }M=\left\vert \lambda\right\vert
\text{, }N=n\\\text{and }h=s_{\lambda}\text{) (since }s_{\lambda}\in
\Lambda_{\left\vert \lambda\right\vert }\text{ and }\left\vert \lambda
\right\vert \neq n\text{))}}}\nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}s_{\lambda}+\underbrace{\sum\limits_{\substack{\lambda
\in\Par ;\\\left\vert \lambda\right\vert \neq n}}a_{\lambda}
0}_{=0}=\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert
\lambda\right\vert =n}}a_{\lambda}s_{\lambda}.
\label{sol.Lambda.hall.gr.a.1}
\end{align}


We have $g\in\Lambda_{m}$, so that $\pi_{m}\left(  g\right)  =g$ (by
\eqref{sol.Lambda.hall.gr.2}, applied to $N=m$ and $h=g$). Thus,
\begin{align*}
g  &  =\pi_{m}\left(  \underbrace{g}_{=\sum_{\lambda\in\Par}
b_{\lambda}s_{\lambda}}\right)  =\pi_{m}\left(  \sum_{\lambda\in
\Par}b_{\lambda}s_{\lambda}\right)  =\sum_{\lambda
\in\Par}b_{\lambda}\pi_{m}\left(  s_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi_{m}\text{ is }
\kk\text{-linear}\right) \\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =m}}b_{\lambda}\underbrace{\pi_{m}\left(  s_{\lambda}\right)
}_{\substack{=s_{\lambda}\\\text{(by \eqref{sol.Lambda.hall.gr.1} (applied to
}N=m\\\text{and }h=s_{\lambda}\text{) (since }s_{\lambda}\in\Lambda
_{\left\vert \lambda\right\vert }=\Lambda_{m}\text{ (since }\left\vert
\lambda\right\vert =m\text{)))}}}+\sum\limits_{\substack{\lambda\in
\Par ;\\\left\vert \lambda\right\vert \neq m}}b_{\lambda
}\underbrace{\pi_{m}\left(  s_{\lambda}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.Lambda.hall.gr.2} (applied to }M=\left\vert \lambda\right\vert
\text{, }N=m\\\text{and }h=s_{\lambda}\text{) (since }s_{\lambda}\in
\Lambda_{\left\vert \lambda\right\vert }\text{ and }\left\vert \lambda
\right\vert \neq m\text{))}}}\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =m}}b_{\lambda}s_{\lambda}+\underbrace{\sum\limits_{\substack{\lambda
\in\Par ;\\\left\vert \lambda\right\vert \neq m}}b_{\lambda}
0}_{=0}=\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert
\lambda\right\vert =m}}b_{\lambda}s_{\lambda}=\sum\limits_{\substack{\nu
\in\Par ;\\\left\vert \nu\right\vert =m}}b_{\nu}s_{\nu}
\end{align*}
(here, we have renamed the summation index $\lambda$ as $\nu$).

But we know (from the definition of the Hall inner product) that
\begin{equation}
\left(  s_{\lambda},s_{\nu}\right)  =\delta_{\lambda,\nu}
\label{sol.Lambda.hall.gr.ss}
\end{equation}
for any two partitions $\lambda$ and $\nu$. Hence, any two partitions
$\lambda\in\Par$ and $\nu\in\Par$ satisfying
$\left\vert \lambda\right\vert =n$ and $\left\vert \nu\right\vert =m$ must
satisfy
\begin{equation}
\left(  s_{\lambda},s_{\nu}\right)  =\delta_{\lambda,\nu}
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda\neq\nu\text{ (because
}\left\vert \lambda\right\vert =n\neq m=\left\vert \nu\right\vert
\text{)}\right)  .
\label{sol.Lambda.hall.gr.5}
\end{equation}
Now,
\begin{align*}
\left(  \underbrace{f}_{=\sum\limits_{\substack{\lambda\in\Par ;
\\\left\vert \lambda\right\vert =n}}a_{\lambda}s_{\lambda}},\underbrace{g}
_{=\sum\limits_{\substack{\nu\in\Par ;\\\left\vert \nu\right\vert
=m}}b_{\nu}s_{\nu}}\right)   &  =\left(  \sum\limits_{\substack{\lambda
\in\Par ;\\\left\vert \lambda\right\vert =n}}a_{\lambda
}s_{\lambda},\sum\limits_{\substack{\nu\in\Par ;\\\left\vert
\nu\right\vert =m}}b_{\nu}s_{\nu}\right)  =\sum\limits_{\substack{\lambda
\in\Par ;\\\left\vert \lambda\right\vert =n}}a_{\lambda}
\sum\limits_{\substack{\nu\in\Par ;\\\left\vert \nu\right\vert
=m}}b_{\nu}\underbrace{\left(  s_{\lambda},s_{\nu}\right)  }
_{\substack{=0\\\text{(by \eqref{sol.Lambda.hall.gr.5})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk \text{-bilinear}\right) \\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}\sum\limits_{\substack{\nu\in\Par ;
\\\left\vert \nu\right\vert =m}}b_{\nu}0=0.
\end{align*}


This solves Exercise~\ref{exe.Lambda.hall.gr}(a).

(b) Let $n\in\NN$ and $f\in\Lambda_n$.

We have $f\in\Lambda_{n}\subset\Lambda$. Hence, we can write $f$ in the form
$f=\sum_{\lambda\in\Par}a_{\lambda}s_{\lambda}$ for some family
$\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk ^{\Par}$ of elements of $\kk$ such that all
but finitely many $\lambda\in\Par$ satisfy $a_{\lambda}=0$
(since the family $\left(  s_{\lambda}\right)  _{\lambda\in\Par}$ is a
$\kk$-module basis of $\Lambda$). Consider this family $\left(
a_{\lambda}\right)  _{\lambda\in\Par}$. Then,
\eqref{sol.Lambda.hall.gr.a.1} holds (this can be proven just as in the
solution of Exercise~\ref{exe.Lambda.hall.gr}(a)).

But consider the partition $\left(  n\right)  $ (note that this is $\left(
0\right)  =\varnothing$ when $n=0$). It is well-known that $s_{\left(
n\right)  }=h_{n}$, so that $h_{n}=s_{\left(  n\right)  }$. Now,
\begin{align}
\left(  \underbrace{h_{n}}_{=s_{\left(  n\right)  }},\underbrace{f}
_{\substack{=\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert
\lambda\right\vert =n}}a_{\lambda}s_{\lambda}\\\text{(by
\eqref{sol.Lambda.hall.gr.a.1})}}}\right)   &  =\left(  s_{\left(  n\right)
},\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}s_{\lambda}\right)  =\sum\limits_{\substack{\lambda
\in\Par ;\\\left\vert \lambda\right\vert =n}}a_{\lambda
}\underbrace{\left(  s_{\left(  n\right)  },s_{\lambda}\right)  }
_{\substack{=\delta_{\left(  n\right)  ,\lambda}\\\text{(by
\eqref{sol.Lambda.hall.gr.ss}, applied to}\\\left(  n\right)  \text{ and
}\lambda\text{ instead of }\lambda\text{ and }\nu\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk \text{-bilinear}\right)  \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}\delta_{\left(  n\right)  ,\lambda}=\sum\limits
_{\substack{\lambda\in\Par ;\\\left\vert \lambda\right\vert
=n;\\\lambda\neq\left(  n\right)  }}a_{\lambda}\underbrace{\delta_{\left(
n\right)  ,\lambda}}_{\substack{=0\\\text{(since }\left(  n\right)
\neq\lambda\\\text{(since }\lambda\neq\left(  n\right)  \text{))}}}+a_{\left(
n\right)  }\underbrace{\delta_{\left(  n\right)  ,\left(  n\right)  }}
_{\substack{=1\\\text{(since }\left(  n\right)  =\left(  n\right) 
\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  n\right)  \text{ is a
partition }\lambda\in\Par \text{ satisfying }\left\vert
\lambda\right\vert =n\right)  \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert
\lambda\right\vert =n;\\\lambda\neq\left(  n\right)  }}a_{\lambda}0}
_{=0}+a_{\left(  n\right)  }1=a_{\left(  n\right)  }1=a_{\left(  n\right)
}.
\label{sol.Lambda.hall.gr.b.1}
\end{align}


On the other hand, every partition $\lambda$ satisfies $s_{\lambda}=\sum
_{T}{\xx }^{\cont (T)}$, where $T$ runs through all
column-strict tableaux of shape $\lambda$. In other words, every partition
$\lambda$ satisfies
\begin{equation}
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda}}\underbrace{\xx ^{\cont\left(  T\right)
}}_{\substack{=\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\\=x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert }
\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }}}
=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }
\lambda}}\left(  x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert }
\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)
.
\label{sol.Lambda.hall.gr.b.2}
\end{equation}
Now, \eqref{sol.Lambda.hall.gr.a.1} yields
\begin{align}
f  &  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert
\lambda\right\vert =n}}a_{\lambda}\underbrace{s_{\lambda}}_{\substack{=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
}}\left(  x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert }\prod
_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)
\\\text{(by \eqref{sol.Lambda.hall.gr.b.2})}}}\nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}\sum\limits_{\substack{T\text{ is a column-strict}
\\\text{tableau of shape }\lambda}}\left(  x_{1}^{\left\vert T^{-1}\left(
1\right)  \right\vert }\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  .
\label{sol.Lambda.hall.gr.b.4}
\end{align}


But $f\left(  1\right)$ is the result of substituting $1,0,0,0,...$ for
$x_{1},x_{2},x_{3},...$ in $f$ (because this is how $f\left(  1\right)  $ was
defined). In other words,
\begin{align}
&  f\left(  1\right)  \nonumber\\
&  =\left(  \text{the result of substituting }1,0,0,0,...\text{ for }
x_{1},x_{2},x_{3},...\right.  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{ in }\underbrace{f}_{\substack{=\sum\limits
_{\substack{\lambda\in\Par ;\\\left\vert \lambda\right\vert
=n}}a_{\lambda}\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda}}\left(  x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert
}\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)
\\\text{(by \eqref{sol.Lambda.hall.gr.b.4})}}}\right)  \nonumber\\
&  =\left(  \text{the result of substituting }1,0,0,0,...\text{ for }
x_{1},x_{2},x_{3},...\right.  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{ in }\sum\limits_{\substack{\lambda
\in\Par ;\\\left\vert \lambda\right\vert =n}}a_{\lambda}
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
}}\left(  x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert }\prod
_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)
\right)  \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}\sum\limits_{\substack{T\text{ is a column-strict}
\\\text{tableau of shape }\lambda}}\left(  \underbrace{1^{\left\vert
T^{-1}\left(  1\right)  \right\vert }}_{=1}\underbrace{\prod_{i\geq
2}0^{\left\vert T^{-1}\left(  i\right)  \right\vert }}_{=0^{\left\vert
T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)
\right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert +...}}\right)
\nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n}}a_{\lambda}\sum\limits_{\substack{T\text{ is a column-strict}
\\\text{tableau of shape }\lambda}}0^{\left\vert T^{-1}\left(  2\right)
\right\vert +\left\vert T^{-1}\left(  3\right)  \right\vert +\left\vert
T^{-1}\left(  4\right)  \right\vert +...}.
\label{sol.Lambda.hall.gr.b.6}
\end{align}
However, every partition $\lambda\in\Par$ such
that $\left\vert \lambda\right\vert =n$ and $\lambda\neq\left(  n\right)  $
satisfies
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
}}0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(
3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert
+...}=0.\label{sol.Lambda.hall.gr.b.simpl1}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.Lambda.hall.gr.b.simpl1}:} Let
$\lambda\in\Par$ be a partition such that $\left\vert
\lambda\right\vert =n$ and $\lambda\neq\left(  n\right)  $. Let $T$ be a
column-strict tableau of shape $\lambda$.
\par
We notice that $T$ is a column-strict tableau of shape $\lambda$, thus a map
from the Ferrers diagram of $\lambda$ to the set $\left\{  1,2,3,...\right\}
$.
\par
We have $n=\left\vert \lambda\right\vert =\lambda_{1}+\lambda_{2}+\lambda
_{3}+...=\lambda_{1}+\left(  \lambda_{2}+\lambda_{3}+\lambda_{4}+...\right)
$.
\par
We have $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq...\geq0$ (since
$\lambda$ is a partition). Thus, $\lambda_{2}\geq\lambda_{3}\geq\lambda
_{4}\geq...\geq0$. In particular, $\lambda_{2}\geq\lambda_{i}$ for every
$i\in\left\{  2,3,4,...\right\}  $.
\par
Assume (for the sake of contradiction) that $\lambda_{2}=0$. Then,
$0=\lambda_{2}\geq\lambda_{i}$ for every $i\in\left\{  2,3,4,...\right\}  $.
Hence, every $i\in\left\{  2,3,4,...\right\}  $ satisfies $\lambda_{i}=0$
(because every $i\in\left\{  2,3,4,...\right\}  $ satisfies $0=\lambda_{2}
\geq\lambda_{i}$ and $\lambda_{i}\geq0$ at the same time). In other words,
$\left(  \lambda_{2},\lambda_{3},\lambda_{4},...\right)  =\left(
0,0,0,...\right)  $. Taking sums on both sides of this equality, we obtain
$\lambda_{2}+\lambda_{3}+\lambda_{4}+...=0+0+0+...=0$. Now, $n=\lambda
_{1}+\underbrace{\left(  \lambda_{2}+\lambda_{3}+\lambda_{4}+...\right)
}_{=0}=\lambda_{1}$, so that $\lambda_{1}=n$. But
\begin{align*}
\lambda &  =\left(  \lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}
,...\right)  =\left(  \underbrace{\lambda_{1}}_{=n},0,0,0,...\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \lambda_{2},\lambda
_{3},\lambda_{4},...\right)  =\left(  0,0,0,...\right)  \right)  \\
&  =\left(  n,0,0,0,...\right)  =\left(  n\right)  ,
\end{align*}
which contradicts $\lambda\neq\left(  n\right)  $. This contradiction shows
that our assumption (that $\lambda_{2}=0$) was wrong. Hence, we cannot have
$\lambda_{2}=0$. We thus must have $\lambda_{2}\neq0$, so that $\lambda_{2}
>0$. Consequently, the cell $\left(  2,1\right)  $ belongs to the Ferrers
diagram of $\lambda$. Thus, $T\left(  \left(  2,1\right)  \right)  $ is a
well-defined element of $\left\{  1,2,3,...\right\}  $ (since $T$ is a map
from the Ferrers diagram of $\lambda$ to the set $\left\{  1,2,3,...\right\}
$). Moreover, since $\lambda$ is a partition, we have $\lambda_{1}\geq
\lambda_{2}>0$, and thus the cell $\left(  1,1\right)  $ belongs to the
Ferrers diagram of $\lambda$. Thus, $T\left(  \left(  1,1\right)  \right)  $
is a well-defined element of $\left\{  1,2,3,...\right\}  $ (since $T$ is a
map from the Ferrers diagram of $\lambda$ to the set $\left\{
1,2,3,...\right\}  $). Hence, $T\left(  \left(  1,1\right)  \right)  \geq1$.
\par
The tableau $T$ is column-strict, and thus strictly increasing top-to-bottom
in columns. In other words, if $\left(  p,q\right)  $ and $\left(  p^{\prime
},q\right)  $ are two cells in the Ferrers diagram of $\lambda$ satisfying
$p<p^{\prime}$, then $T\left(  \left(  p,q\right)  \right)  <T\left(  \left(
p^{\prime},q\right)  \right)  $. Applying this to $p=1$, $q=1$ and $p^{\prime
}=2$, we obtain $T\left(  \left(  1,1\right)  \right)  <T\left(  \left(
2,1\right)  \right)  $ (since $1<2$). Thus, $T\left(  \left(  2,1\right)
\right)  >T\left(  \left(  1,1\right)  \right)  \geq1$, so that $T\left(
\left(  2,1\right)  \right)  \neq1$. Combining $T\left(  \left(  2,1\right)
\right)  \in\left\{  1,2,3,...\right\}  $ with $T\left(  \left(  2,1\right)
\right)  \neq1$, we obtain $T\left(  \left(  2,1\right)  \right)  \in\left\{
1,2,3,...\right\}  \setminus\left\{  1\right\}  =\left\{  2,3,4,...\right\}
$. Hence, $\left(  2,1\right)  =T^{-1}\left(  T\left(  \left(  2,1\right)
\right)  \right)  \in T^{-1}\left(  T\left(  \left(  2,1\right)  \right)
\right)  $. Thus, the set $T^{-1}\left(  T\left(  \left(  2,1\right)  \right)
\right)  $ is nonempty, so that we have $\left\vert T^{-1}\left(  T\left(
\left(  2,1\right)  \right)  \right)  \right\vert >0$. Altogether, we have
$T\left(  \left(  2,1\right)  \right)  \in\left\{  2,3,4,...\right\}  $ and
$\left\vert T^{-1}\left(  T\left(  \left(  2,1\right)  \right)  \right)
\right\vert >0$. Therefore, there exists some $i\in\left\{  2,3,4,...\right\}
$ satisfying $\left\vert T^{-1}\left(  i\right)  \right\vert >0$ (namely,
$i=T\left(  \left(  2,1\right)  \right)  $). In other words, the sum
$\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(
3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert +...$
has at least one positive summand. Since all summands of this sum are
nonnegative, this shows that this sum is positive, i.e., we have $\left\vert
T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)
\right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert +...>0$. Thus,
$0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(
3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert +...}
=0$.
\par
Now, forget that we fixed $T$. We thus have proven that every column-strict
tableau $T$ of shape $\lambda$ satisfies $0^{\left\vert T^{-1}\left(
2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)  \right\vert
+\left\vert T^{-1}\left(  4\right)  \right\vert +...}=0$. Hence,
\[
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
}}\underbrace{0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert
T^{-1}\left(  3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)
\right\vert +...}}_{=0}=\sum\limits_{\substack{T\text{ is a column-strict}
\\\text{tableau of shape }\lambda}}0=0.
\]
This proves \eqref{sol.Lambda.hall.gr.b.simpl1}.} Also,
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\left(
n\right)  }}0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert
T^{-1}\left(  3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)
\right\vert +...}=1.\label{sol.Lambda.hall.gr.b.simpl2}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.Lambda.hall.gr.b.simpl2}:} The Ferrers
diagram of the partition $\left(  n\right)  $ is the set $\left\{  \left(
1,1\right)  ,\left(  1,2\right)  ,...,\left(  1,n\right)  \right\}  =\left\{
\left(  1,i\right)  \ \mid\ i\in\left\{  1,2,...,n\right\}  \right\}  $. Let
$T_{0}$ denote the map from this Ferrers diagram to the set $\left\{
1,2,3,...\right\}  $ which is defined by
\begin{equation}
\left(  T_{0}\left(  x\right)  =1\ \ \ \ \ \ \ \ \ \ \text{for every }
x\in\left(  \text{the Ferrers diagram of }\left(  n\right)  \right)  \right)
.
\label{sol.Lambda.hall.gr.b.simpl2.1}
\end{equation}
Then, $T_{0}$ is an assignment of entries of $\left\{  1,2,3,...\right\}  $ to
the cells of the Ferrers diagram for $\left(  n\right)  $, weakly increasing
left-to-right in rows (because all values of $T_{0}$ are $1$), and strictly
increasing top-to-bottom in columns (because every column of the Ferrers
diagram for $\left(  n\right)  $ contains at most one cell). In other words,
$T_{0}$ is a column-strict tableau of shape $\left(  n\right)  $.
\par
Also, every $i\in\left\{  2,3,4,...\right\}  $ satisfies
\begin{align*}
T_{0}^{-1}\left(  i\right)   &  =\left\{  x\in\left(  \text{the Ferrers
diagram of }\left(  n\right)  \right)  \ \mid\ \underbrace{T_{0}\left(
x\right)  }
_{\substack{=1\\\text{(by \eqref{sol.Lambda.hall.gr.b.simpl1})}}}
=i\right\}  \\
&  =\left\{  x\in\left(  \text{the Ferrers diagram of }\left(  n\right)
\right)  \ \mid\ \underbrace{1=i}_{\substack{\text{this is false}
\\\text{(because }i\neq1\\\text{(since }i\in\left\{  2,3,4,...\right\}
\text{))}}}\right\}  \\
&  =\left\{  x\in\left(  \text{the Ferrers diagram of }\left(  n\right)
\right)  \ \mid\ \text{false}\right\}  =\varnothing.
\end{align*}
Hence, every $i\in\left\{  2,3,4,...\right\}  $ satisfies $\left\vert
\underbrace{T_{0}^{-1}\left(  i\right)  }_{=\varnothing}\right\vert
=\left\vert \varnothing\right\vert =0$. Summing up these equalities over all
$i\in\left\{  2,3,4,...\right\}  $, we obtain $\left\vert T_{0}^{-1}\left(
2\right)  \right\vert +\left\vert T_{0}^{-1}\left(  3\right)  \right\vert
+\left\vert T_{0}^{-1}\left(  4\right)  \right\vert +...=0+0+0+...=0$. Hence,
\begin{equation}
0^{\left\vert T_{0}^{-1}\left(  2\right)  \right\vert 
+\left\vert T_{0}^{-1}\left(  3\right)  \right\vert 
+\left\vert T_{0}^{-1}\left(  4\right)
\right\vert +...}=0^{0}=1.
\label{sol.Lambda.hall.gr.b.simpl2.1a}
\end{equation}
\par
Now, let $T$ be a column-strict tableau of shape $\left(  n\right)  $ such
that $T\neq T_{0}$. We shall prove that $0^{\left\vert T^{-1}\left(  2\right)
\right\vert +\left\vert T^{-1}\left(  3\right)  \right\vert +\left\vert
T^{-1}\left(  4\right)  \right\vert +...}=0$.
\par
There exists at least one $x\in\left(  \text{the Ferrers diagram of }\left(
n\right)  \right)  $ such that $T\left(  x\right)  \neq1$ (because otherwise,
every $x\in\left(  \text{the Ferrers diagram of }\left(  n\right)  \right)  $
would satisfy $T\left(  x\right)  =1$, and thus every $x\in\left(  \text{the
Ferrers diagram of }\left(  n\right)  \right)  $ would satisfy $T\left(
x\right) = 1 = T_0 \left( x \right)$ (by
\eqref{sol.Lambda.hall.gr.b.simpl2.1}), and thus we would have
$T=T_0$, which would contradict $T\neq T_0$).
Consider this $x$.
\par
We notice that $T$ is a column-strict tableau of shape $\left(  n\right)  $,
thus a map from the Ferrers diagram of $\left(  n\right)  $ to the set
$\left\{  1,2,3,...\right\}  $. Hence, $T\left(  x\right)  \in\left\{
1,2,3,...\right\}  $. Combining this with $T\left(  x\right)  \neq1$, we
obtain $T\left(  x\right)  \in\left\{  1,2,3,...\right\}  \setminus\left\{
1\right\}  =\left\{  2,3,4,...\right\}  $. Moreover, $T^{-1}\left(  T\left(
x\right)  \right)  $ is a nonempty set (since $T^{-1}\left(  T\left(
x\right)  \right)  $ contains the element $x$), so that $\left\vert
T^{-1}\left(  T\left(  x\right)  \right)  \right\vert >0$. We now have
$T\left(  x\right) \in \left\{ 2,3,4,... \right\}$ and
$\left\vert T^{-1}\left(  T\left(  x\right) \right)  \right\vert >0$.
Thus, there exists an $i\in\left\{
2,3,4,...\right\}$ such that $\left\vert T^{-1}\left( i \right) \right\vert
>0$ (namely, $i=T\left(x\right)$). In other words, the sum $\left\vert
T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)
\right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert +...$ has at least
one positive summand. Since all summands of this sum are nonnegative, this
shows that this sum is positive, i.e., we have $\left\vert T^{-1}\left(
2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)  \right\vert
+\left\vert T^{-1}\left(  4\right)  \right\vert +...>0$. Thus, $0^{\left\vert
T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)
\right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert +...}=0$.
\par
Now, forget that we fixed $T$. We thus have proven that every column-strict
tableau $T$ of shape $\left(  n\right)  $ such that $T\neq T_{0}$ satisfies
\begin{equation}
0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(
3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert
+...}=0.
\label{sol.Lambda.hall.gr.b.simpl2.2}
\end{equation}
But recall that $T_0$ is a column-strict tableau of shape
$\left(n\right)$. Thus,
\begin{align*}
&  \sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\left(  n\right)  }}0^{\left\vert T^{-1}\left(  2\right)  \right\vert
+\left\vert T^{-1}\left(  3\right)  \right\vert +\left\vert T^{-1}\left(
4\right)  \right\vert +...}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\left(  n\right)  ;\\T\neq T_{0}}}\underbrace{0^{\left\vert T^{-1}\left(
2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)  \right\vert
+\left\vert T^{-1}\left(  4\right)  \right\vert +...}}
_{\substack{=0\\\text{(by \eqref{sol.Lambda.hall.gr.b.simpl2.2})}}}
+\underbrace{0^{\left\vert T_{0}^{-1}\left(  2\right)  \right\vert
+\left\vert T_{0}^{-1}\left(  3\right)  \right\vert 
+\left\vert T_{0}^{-1}\left(  4\right)  \right\vert +...}}
_{\substack{=1\\\text{(by
\eqref{sol.Lambda.hall.gr.b.simpl2.1a})}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(\text{here, we have split off the addend for }
T = T_0 \text{ from the sum}\right) \\
&  =\underbrace{\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\left(  n\right)  ;\\T\neq T_{0}}}0}_{=0}+1=1.
\end{align*}
This proves \eqref{sol.Lambda.hall.gr.b.simpl2}.}

Now,
\eqref{sol.Lambda.hall.gr.b.6} becomes
\begin{align*}
f\left(  1\right)    & =\sum\limits_{\substack{\lambda\in\Par ;
\\\left\vert \lambda\right\vert =n}}a_{\lambda}\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda}}0^{\left\vert T^{-1}\left(
2\right)  \right\vert +\left\vert T^{-1}\left(  3\right)  \right\vert
+\left\vert T^{-1}\left(  4\right)  \right\vert +...}\\
& =\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert \lambda
\right\vert =n;\\\lambda\neq\left(  n\right)  }}a_{\lambda}\underbrace{\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
}}0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert T^{-1}\left(
3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)  \right\vert
+...}}_{\substack{=0\\\text{(by \eqref{sol.Lambda.hall.gr.b.simpl1})}}}\\
& \ \ \ \ \ \ \ \ \ \ +a_{\left(  n\right)  }\underbrace{\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\left(
n\right)  }}0^{\left\vert T^{-1}\left(  2\right)  \right\vert +\left\vert
T^{-1}\left(  3\right)  \right\vert +\left\vert T^{-1}\left(  4\right)
\right\vert +...}}_{\substack{=1\\
\text{(by \eqref{sol.Lambda.hall.gr.b.simpl2})}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  n\right)  \text{ is an
element }\lambda\in\Par \text{ satisfying }\left\vert
\lambda\right\vert =n\right)  \\
& =\underbrace{\sum\limits_{\substack{\lambda\in\Par ;\\\left\vert
\lambda\right\vert =n;\\\lambda\neq\left(  n\right)  }}a_{\lambda}0}
_{=0}+a_{\left(  n\right)  }1=a_{\left(  n\right)  }1=a_{\left(  n\right)
}=\left(  h_{n},f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.Lambda.hall.gr.b.1}}\right)  .
\end{align*}
In other words, $\left(  h_{n},f\right)  =f\left(  1\right)  $. This solves
Exercise~\ref{exe.Lambda.hall.gr}(b).

(c) We know (from the definition of the Hall inner product) that
\begin{equation}
\left(  s_{\lambda},s_{\nu}\right)  =\delta_{\lambda,\nu}%
\label{sol.Lambda.hall.gr.c.ss}%
\end{equation}
for any two partitions $\lambda$ and $\nu$. In other words, the basis $\left(
s_{\lambda}\right)  _{\lambda\in\Par}\in\Lambda
^{\Par}$ is orthonormal with respect to the Hall inner product.
Hence, the Hall inner product has an orthonormal basis. This readily yields
that we have $\left(  f,g\right)  =\left(  g,f\right)  $ for all $f\in\Lambda$
and $g\in\Lambda$\ \ \ \ \footnote{Here is the \textit{proof} in detail:
\par
Let $f\in\Lambda$ and $g\in\Lambda$.
\par
We have $f\in\Lambda$. Hence, we can write $f$ in the form $f=\sum_{\lambda
\in\Par}a_{\lambda}s_{\lambda}$ for some family $\left(
a_{\lambda}\right)  _{\lambda\in\Par}\in\kk%
^{\Par}$ of elements of $\kk$ such that all but finitely
many $\lambda\in\Par$ satisfy $a_{\lambda}=0$ (since the family
$\left(  s_{\lambda}\right)  _{\lambda\in\Par}$ is a
$\kk$-module basis of $\Lambda$). Consider this family $\left(
a_{\lambda}\right)  _{\lambda\in\Par}$. Clearly, $f=\sum
_{\lambda\in\Par}a_{\lambda}s_{\lambda}=\sum_{\nu\in
\Par}a_{\nu}s_{\nu}$ (here, we have renamed the summation index
$\lambda$ as $\nu$).
\par
We have $g\in\Lambda$. Hence, we can write $g$ in the form $g=\sum_{\lambda
\in\Par}b_{\lambda}s_{\lambda}$ for some family $\left(
b_{\lambda}\right)  _{\lambda\in\Par}\in\kk%
^{\Par}$ of elements of $\kk$ such that all but finitely
many $\lambda\in\Par$ satisfy $b_{\lambda}=0$ (since the family
$\left(  s_{\lambda}\right)  _{\lambda\in\Par}$ is a
$\kk$-module basis of $\Lambda$). Consider this family $\left(
b_{\lambda}\right)  _{\lambda\in\Par}$. Clearly, $g=\sum
_{\lambda\in\Par}b_{\lambda}s_{\lambda}=\sum_{\nu\in
\Par}b_{\nu}s_{\nu}$ (here, we have renamed the summation index
$\lambda$ as $\nu$).
\par
Comparing%
\begin{align*}
\left(  \underbrace{f}_{=\sum_{\lambda\in\Par}a_{\lambda
}s_{\lambda}},\underbrace{g}_{=\sum_{\nu\in\Par}b_{\nu}s_{\nu}%
}\right)    & =\left(  \sum_{\lambda\in\Par}a_{\lambda
}s_{\lambda},\sum_{\nu\in\Par}b_{\nu}s_{\nu}\right)
=\sum_{\lambda\in\Par}a_{\lambda}\underbrace{\left(  s_{\lambda
},\sum_{\nu\in\Par}b_{\nu}s_{\nu}\right)  }_{\substack{=\sum
_{\nu\in\Par}b_{\nu}\left(  s_{\lambda},s_{\nu}\right)
\\\text{(since the Hall inner product is }\kk\text{-bilinear)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk\text{-bilinear}\right)  \\
& =\sum_{\lambda\in\Par}a_{\lambda}\sum_{\nu\in
\Par}b_{\nu}\left(  s_{\lambda},s_{\nu}\right)  =\sum
_{\lambda\in\Par}\sum_{\nu\in\Par}a_{\lambda
}b_{\nu}\underbrace{\left(  s_{\lambda},s_{\nu}\right)  }_{\substack{=\delta
_{\lambda,\nu}\\\text{(by \eqref{sol.Lambda.hall.gr.c.ss})}}}\\
& =\sum_{\lambda\in\Par}\underbrace{\sum_{\nu\in
\Par}a_{\lambda}b_{\nu}\delta_{\lambda,\nu}}%
_{\substack{=a_{\lambda}b_{\lambda}\delta_{\lambda,\lambda}+\sum
\limits_{\substack{\nu\in\Par;\\\nu\neq\lambda}}a_{\lambda
}b_{\nu}\delta_{\lambda,\nu}\\\text{(here, we have split off the addend for
}\nu=\lambda\text{ from the sum)}}}\\
& =\sum_{\lambda\in\Par}\left(  a_{\lambda}b_{\lambda
}\underbrace{\delta_{\lambda,\lambda}}_{\substack{=1\\\text{(since }%
\lambda=\lambda\text{)}}}+\sum\limits_{\substack{\nu\in\Par
;\\\nu\neq\lambda}}a_{\lambda}b_{\nu}\underbrace{\delta_{\lambda,\nu}%
}_{\substack{=0\\\text{(since }\lambda\neq\nu\\\text{(because }\nu\neq
\lambda\text{))}}}\right)  =\sum_{\lambda\in\Par}\left(
a_{\lambda}b_{\lambda}+\underbrace{\sum\limits_{\substack{\nu\in
\Par;\\\nu\neq\lambda}}a_{\lambda}b_{\nu}0}_{=0}\right)  \\
& =\sum_{\lambda\in\Par}a_{\lambda}b_{\lambda}%
\end{align*}
with%
\begin{align*}
\left(  \underbrace{g}_{=\sum_{\lambda\in\Par}b_{\lambda
}s_{\lambda}},\underbrace{f}_{=\sum_{\nu\in\Par}a_{\nu}s_{\nu}%
}\right)    & =\left(  \sum_{\lambda\in\Par}b_{\lambda
}s_{\lambda},\sum_{\nu\in\Par}a_{\nu}s_{\nu}\right)
=\sum_{\lambda\in\Par}b_{\lambda}\underbrace{\left(  s_{\lambda
},\sum_{\nu\in\Par}a_{\nu}s_{\nu}\right)  }_{\substack{=\sum
_{\nu\in\Par}a_{\nu}\left(  s_{\lambda},s_{\nu}\right)
\\\text{(since the Hall inner product is }\kk\text{-bilinear)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk\text{-bilinear}\right)  \\
& =\sum_{\lambda\in\Par}b_{\lambda}\sum_{\nu\in
\Par}a_{\nu}\left(  s_{\lambda},s_{\nu}\right)  =\sum
_{\lambda\in\Par}\sum_{\nu\in\Par}%
\underbrace{b_{\lambda}a_{\nu}}_{=a_{\nu}b_{\lambda}}\underbrace{\left(
s_{\lambda},s_{\nu}\right)  }_{\substack{=\delta_{\lambda,\nu}\\\text{(by
\eqref{sol.Lambda.hall.gr.c.ss})}}}\\
& =\sum_{\lambda\in\Par}\underbrace{\sum_{\nu\in
\Par}a_{\nu}b_{\lambda}\delta_{\lambda,\nu}}%
_{\substack{=a_{\lambda}b_{\lambda}\delta_{\lambda,\lambda}+\sum
\limits_{\substack{\nu\in\Par;\\\nu\neq\lambda}}a_{\nu
}b_{\lambda}\delta_{\lambda,\nu}\\\text{(here, we have split off the addend
for }\nu=\lambda\text{ from the sum)}}}\\
& =\sum_{\lambda\in\Par}\left(  a_{\lambda}b_{\lambda
}\underbrace{\delta_{\lambda,\lambda}}_{\substack{=1\\\text{(since }%
\lambda=\lambda\text{)}}}+\sum\limits_{\substack{\nu\in\Par;
\\\nu\neq\lambda}}a_{\nu}b_{\lambda}\underbrace{\delta_{\lambda,\nu}%
}_{\substack{=0\\\text{(since }\lambda\neq\nu\\\text{(because }\nu\neq
\lambda\text{))}}}\right)  =\sum_{\lambda\in\Par}\left(
a_{\lambda}b_{\lambda}+\underbrace{\sum\limits_{\substack{\nu\in
\Par;\\\nu\neq\lambda}}a_{\nu}b_{\lambda}0}_{=0}\right)  \\
& =\sum_{\lambda\in\Par}a_{\lambda}b_{\lambda},
\end{align*}
we obtain $\left(  f,g\right)  =\left(  g,f\right)  $. Qed.}. In other words,
the Hall inner product is symmetric. This solves Exercise
\ref{exe.Lambda.hall.gr}(c).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.linalg.sur-bij}}

\textit{Solution to Exercise~\ref{exe.linalg.sur-bij}.}
(a) This is a well-known fact; for a proof, see Theorem~5.3 in Keith Conrad,
\textit{Universal Identities I},
\url{http://www.math.uconn.edu/~kconrad/blurbs/}.
Another proof (more complicated, but with the advantage of proving
a more general result) can be found in
\cite[proof of Corollary 0.2]{Grinberg-orzech}.

(b) Consider the endomorphism of the $\kk$-module
$A$ defined by sending $\gamma_{i}$ to $\beta_{i}$ for every $i\in
I$. This endomorphism is well-defined (since $\left(  \gamma_{i}\right)
_{i\in I}$ is a basis of $A$) and surjective (since the family $\left(
\beta_{i}\right)  _{i\in I}$ spans $A$), therefore is a $\kk$-module
isomorphism (according to Exercise~\ref{exe.linalg.sur-bij}(a)).
As a consequence, it must send the basis $\left(  \gamma_{i}\right)  _{i\in
I}$ of $A$ to a basis of $A$ (because a $\kk$-module isomorphism sends
any basis to a basis). Since it sends the basis $\left(  \gamma_{i}\right)
_{i\in I}$ to $\left(  \beta_{i}\right)  _{i\in I}$, this yields that $\left(
\beta_{i}\right)  _{i\in I}$ must be a basis of $A$. This solves
Exercise~\ref{exe.linalg.sur-bij}(b).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.basis-vlam}}

\textit{Solution to Exercise \ref{exe.Lambda.basis-vlam}.} Fix $n\in
{\NN}$. Recall that $v_{\lambda}\in\Lambda_{\left\vert \lambda
\right\vert }$ for each partition $\lambda$.
Hence, for each $\lambda\in\Par_{n}$, we have
$v_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert
}=\Lambda_{n}$ (since $\left\vert \lambda\right\vert =n$ (because $\lambda
\in\Par_{n}$)). Thus, $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ is a family of elements of $\Lambda_{n}$.

But we have assumed that the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par}$ spans the $\kk$-module $\Lambda$. Since
everything is graded, this yields that the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ spans the $\kk$-module
$\Lambda_{n}$\ \ \ \ \footnote{\textit{Proof.} Here is this argument in more
detail:
\par
Let $f\in\Lambda_{n}$. We shall prove that $f$ is a $\kk$-linear
combination of the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par_{n}}$.
\par
We have $f\in\Lambda_{n}\subset\Lambda$. Thus, $f$ is a $\kk$-linear
combination of the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par}$ (since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ spans the $\kk$-module $\Lambda$). In
other words, there exists a family $\left(  c_{\lambda}\right)  _{\lambda
\in\Par}\in \kk^{\Par}$ such that $\left(
\text{all but finitely many }\lambda\in\Par\text{ satisfy
}c_{\lambda}=0\right)  $ and $f=\sum_{\lambda\in\Par}c_{\lambda
}v_{\lambda}$. Consider this $\left(  c_{\lambda}\right)  _{\lambda
\in\Par}$.
\par
Let $\pi_{n}:\Lambda\rightarrow\Lambda_{n}$ be the projection from the graded
$\kk$-module $\Lambda$ onto its $n$-th graded component. Then, $\pi
_{n}$ is a $\kk$-linear map with the properties that%
\begin{equation}
\pi_{n}\left(  g\right)  =g\ \ \ \ \ \ \ \ \ \ \text{for each }g\in\Lambda
_{n},\label{sol.Lambda.basis-vlam.fn1.pf.1}%
\end{equation}
and%
\begin{equation}
\pi_{n}\left(  g\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for each }g\in\Lambda
_{m}\text{ for each }m\in{\NN}\text{ satisfying }m\neq
n.\label{sol.Lambda.basis-vlam.fn1.pf.2}%
\end{equation}
\par
If $\lambda\in\Par$ satisfies $\left\vert \lambda\right\vert
=n$, then $v_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }=\Lambda_{n}$
(since $\left\vert \lambda\right\vert =n$) and thus%
\begin{equation}
\pi_{n}\left(  v_{\lambda}\right)  =v_{\lambda}%
\label{sol.Lambda.basis-vlam.fn1.pf.1use}%
\end{equation}
(by (\ref{sol.Lambda.basis-vlam.fn1.pf.1}), applied to $g=v_{\lambda}$).
\par
If $\lambda\in\Par$ satisfies $\left\vert \lambda\right\vert
\neq n$, then $v_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }$ and
thus%
\begin{equation}
\pi_{n}\left(  v_{\lambda}\right)  =0\label{sol.Lambda.basis-vlam.fn1.pf.2use}%
\end{equation}
(by (\ref{sol.Lambda.basis-vlam.fn1.pf.2}), applied to $g=v_{\lambda}$ and
$m=\left\vert \lambda\right\vert $).
\par
Applying (\ref{sol.Lambda.basis-vlam.fn1.pf.1}) to $g=f$, we obtain $\pi
_{n}\left(  f\right)  =f$, so that%
\begin{align*}
f &  =\pi_{n}\left(  \underbrace{f}_{=\sum_{\lambda\in\Par%
}c_{\lambda}v_{\lambda}}\right)  =\pi_{n}\left(  \sum_{\lambda\in
\Par}c_{\lambda}v_{\lambda}\right)  =\sum_{\lambda
\in\Par}c_{\lambda}\pi_{n}\left(  v_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{n}\text{ is a } \kk%
\text{-linear map}\right)  \\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert =n}}c_{\lambda}\underbrace{\pi_{n}\left(  v_{\lambda}\right)
}_{\substack{=v_{\lambda}\\\text{(by (\ref{sol.Lambda.basis-vlam.fn1.pf.1use}%
))}}}+\sum\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert \neq n}}c_{\lambda}\underbrace{\pi_{n}\left(  v_{\lambda}\right)
}_{\substack{=0\\\text{(by (\ref{sol.Lambda.basis-vlam.fn1.pf.2use}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }\lambda\in\Par%
\text{ satisfies either }\left\vert \lambda\right\vert =n\text{ or }\left\vert
\lambda\right\vert \neq n\text{ (but not both)}\right)  \\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert =n}}c_{\lambda}v_{\lambda}+\underbrace{\sum\limits_{\substack{\lambda
\in\Par;\\\left\vert \lambda\right\vert \neq n}}c_{\lambda}%
0}_{=0}=\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}}_{=\sum_{\lambda\in\Par_{n}}}c_{\lambda
}v_{\lambda}=\sum_{\lambda\in\Par_{n}}c_{\lambda}v_{\lambda}.
\end{align*}
This shows that $f$ is a $\kk$-linear combination of the family
$\left(  v_{\lambda}\right)  _{\lambda\in\Par_{n}}$.
\par
Now, forget that we fixed $f$. We thus have proven that every $f\in\Lambda
_{n}$ is a $\kk$-linear combination of the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par_{n}}$. Therefore, the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par_{n}}$ spans the
$\kk$-module $\Lambda_{n}$.}.

Recall that $v_{\lambda}\in\Lambda_{n}$ for each $\lambda\in
\Par_{n}$. Hence, we can define the $\kk$-linear map
\begin{align*}
\beta:\Lambda_{n} &  \rightarrow\Lambda_{n},\\
m_{\lambda} &  \mapsto v_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }%
\lambda\in\Par_{n}%
\end{align*}
(since $\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a
basis of the $\kk$-module $\Lambda_{n}$). Consider this map $\beta$.
This map $\beta$ is surjective (since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ spans the $\kk$-module
$\Lambda_{n}$)\ \ \ \ \footnote{\textit{Proof.} In slightly more detail:
The map $\beta$ is $\kk$-linear; thus, its image
$\beta\left(\Lambda_n\right)$ is a $\kk$-submodule of $\Lambda_n$.
This $\kk$-submodule $\beta\left(\Lambda_n\right)$ clearly contains
$v_\lambda$ for each $\lambda \in \Par_n$
(since the definition of $\beta$ yields $v_\lambda = \beta\left(m_\lambda\right)$).
Hence, $\beta\left(\Lambda_n\right)$ is a $\kk$-submodule of
$\Lambda_n$ that contains $v_\lambda$ for each $\lambda \in \Par_n$.
But the only such $\kk$-submodule is the whole module $\Lambda_n$
(since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ spans the $\kk$-module
$\Lambda_{n}$).
Thus, we conclude that $\beta\left(\Lambda_n\right)$ must be
the whole module $\Lambda_n$.
In other words, $\beta$ is surjective.}.

But $\Lambda_{n}$ is a finitely generated $\kk$-module (since it has a
finite basis $\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}%
$). The map $\beta$ is an endomorphism of this $\kk$-module (since
$\beta:\Lambda_{n}\rightarrow\Lambda_{n}$ is $\kk$-linear), and is
surjective. Thus, Exercise~\ref{exe.linalg.sur-bij}(a) (applied to this
endomorphism) yields that $\beta$ is a $\kk$-module isomorphism.

Now, the map $\beta$ satisfies $\beta\left(  m_{\lambda}\right)  =v_{\lambda}$
for each $\lambda\in\Par_{n}$ (by the definition of
$\beta$). In other words, $\left(  \beta\left(  m_{\lambda}\right)  \right)
_{\lambda\in\Par_{n}}=\left(  v_{\lambda}\right)  _{\lambda
\in\Par_{n}}$.

But the family $\left(  m_{\lambda}\right)  _{\lambda\in\Par%
_{n}}$ is a basis of the $\kk$-module $\Lambda_{n}$. Hence, the image
of this family $\left(  m_{\lambda}\right)  _{\lambda\in\Par%
_{n}}$ under $\beta$ is also a basis of the $\kk$-module $\Lambda_{n}$
(since $\beta$ is a $\kk$-module isomorphism). In other words, the
family $\left(  v_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a
basis of the $\kk$-module $\Lambda_{n}$ (since the image of the family
$\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}$ under
$\beta$ is $\left(  \beta\left(  m_{\lambda}\right)  \right)  _{\lambda
\in\Par_{n}}=\left(  v_{\lambda}\right)  _{\lambda
\in\Par_{n}}$).

Forget that we fixed $n$. We thus have proved that $\left(  v_{\lambda
}\right)  _{\lambda\in\Par_{n}}$ is a basis of the
$\kk$-module $\Lambda_{n}$ for each $n\in\NN$.
Hence, the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par}$ is a graded basis of the
$\kk$-module $\bigoplus_{n\in\NN}\Lambda_{n}$ (since
$\Par_{0},\Par_{1},\Par_{2},\ldots$ is a partition of the set
$\Par$). In other words, the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par}$ is a graded basis of the
$\kk$-module $\Lambda$ (since $\Lambda=\bigoplus_{n\in\NN}
\Lambda_{n}$). This solves Exercise \ref{exe.Lambda.basis-vlam}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.basis-vlam}}

\textit{Solution to Exercise \ref{exe.Lambda.basis-vlam}.} Fix $n\in
{\NN}$. Recall that $v_{\lambda}\in\Lambda_{\left\vert \lambda
\right\vert }$ for each partition $\lambda$. In other words, $v_{\lambda}%
\in\Lambda_{\left\vert \lambda\right\vert }$ for each $\lambda\in
\Par$. Hence, for each $\lambda\in\Par_{n}$, we have
$v_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert
}=\Lambda_{n}$ (since $\left\vert \lambda\right\vert =n$ (because $\lambda
\in\Par_{n}$)). Thus, $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ is a family of elements of $\Lambda_{n}$.

But we have assumed that the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par}$ spans the $\kk$-module $\Lambda$. Since
everything is graded, this yields that the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ spans the $\kk$-module
$\Lambda_{n}$\ \ \ \ \footnote{\textit{Proof.} Here is this argument in more
detail:
\par
Let $f\in\Lambda_{n}$. We shall prove that $f$ is a $\kk$-linear
combination of the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par_{n}}$.
\par
We have $f\in\Lambda_{n}\subset\Lambda$. Thus, $f$ is a $\kk$-linear
combination of the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par}$ (since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ spans the $\kk$-module $\Lambda$). In
other words, there exists a family $\left(  c_{\lambda}\right)  _{\lambda
\in\Par}\in \kk^{\Par}$ such that $\left(
\text{all but finitely many }\lambda\in\Par\text{ satisfy
}c_{\lambda}=0\right)  $ and $f=\sum_{\lambda\in\Par}c_{\lambda
}v_{\lambda}$. Consider this $\left(  c_{\lambda}\right)  _{\lambda
\in\Par}$.
\par
Let $\pi_{n}:\Lambda\rightarrow\Lambda_{n}$ be the projection from the graded
$\kk$-module $\Lambda$ onto its $n$-th graded component. Then, $\pi
_{n}$ is a $\kk$-linear map with the properties that%
\begin{equation}
\pi_{n}\left(  g\right)  =g\ \ \ \ \ \ \ \ \ \ \text{for each }g\in\Lambda
_{n},\label{sol.Lambda.basis-vlam.fn1.pf.1}%
\end{equation}
and%
\begin{equation}
\pi_{n}\left(  g\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for each }g\in\Lambda
_{m}\text{ for each }m\in{\NN}\text{ satisfying }m\neq
n.\label{sol.Lambda.basis-vlam.fn1.pf.2}%
\end{equation}
\par
If $\lambda\in\Par$ satisfies $\left\vert \lambda\right\vert
=n$, then $v_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }=\Lambda_{n}$
(since $\left\vert \lambda\right\vert =n$) and thus%
\begin{equation}
\pi_{n}\left(  v_{\lambda}\right)  =v_{\lambda}%
\label{sol.Lambda.basis-vlam.fn1.pf.1use}%
\end{equation}
(by (\ref{sol.Lambda.basis-vlam.fn1.pf.1}), applied to $g=v_{\lambda}$).
\par
If $\lambda\in\Par$ satisfies $\left\vert \lambda\right\vert
\neq n$, then $v_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }$ and
thus%
\begin{equation}
\pi_{n}\left(  v_{\lambda}\right)  =0\label{sol.Lambda.basis-vlam.fn1.pf.2use}%
\end{equation}
(by (\ref{sol.Lambda.basis-vlam.fn1.pf.2}), applied to $g=v_{\lambda}$ and
$m=\left\vert \lambda\right\vert $).
\par
Applying (\ref{sol.Lambda.basis-vlam.fn1.pf.1}) to $g=f$, we obtain $\pi
_{n}\left(  f\right)  =f$, so that%
\begin{align*}
f &  =\pi_{n}\left(  \underbrace{f}_{=\sum_{\lambda\in\Par%
}c_{\lambda}v_{\lambda}}\right)  =\pi_{n}\left(  \sum_{\lambda\in
\Par}c_{\lambda}v_{\lambda}\right)  =\sum_{\lambda
\in\Par}c_{\lambda}\pi_{n}\left(  v_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{n}\text{ is a } \kk%
\text{-linear map}\right)  \\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert =n}}c_{\lambda}\underbrace{\pi_{n}\left(  v_{\lambda}\right)
}_{\substack{=v_{\lambda}\\\text{(by (\ref{sol.Lambda.basis-vlam.fn1.pf.1use}%
))}}}+\sum\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert \neq n}}c_{\lambda}\underbrace{\pi_{n}\left(  v_{\lambda}\right)
}_{\substack{=0\\\text{(by (\ref{sol.Lambda.basis-vlam.fn1.pf.2use}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }\lambda\in\Par%
\text{ satisfies either }\left\vert \lambda\right\vert =n\text{ or }\left\vert
\lambda\right\vert \neq n\text{ (but not both)}\right)  \\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert =n}}c_{\lambda}v_{\lambda}+\underbrace{\sum\limits_{\substack{\lambda
\in\Par;\\\left\vert \lambda\right\vert \neq n}}c_{\lambda}%
0}_{=0}=\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}}_{=\sum_{\lambda\in\Par_{n}}}c_{\lambda
}v_{\lambda}=\sum_{\lambda\in\Par_{n}}c_{\lambda}v_{\lambda}.
\end{align*}
This shows that $f$ is a $\kk$-linear combination of the family
$\left(  v_{\lambda}\right)  _{\lambda\in\Par_{n}}$.
\par
Now, forget that we fixed $f$. We thus have proven that every $f\in\Lambda
_{n}$ is a $\kk$-linear combination of the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par_{n}}$. Therefore, the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par_{n}}$ spans the
$\kk$-module $\Lambda_{n}$.}.

Recall that $v_{\lambda}\in\Lambda_{n}$ for each $\lambda\in
\Par_{n}$. Hence, we can define the $\kk$-linear map
\begin{align*}
\beta:\Lambda_{n} &  \rightarrow\Lambda_{n},\\
m_{\lambda} &  \mapsto v_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }%
\lambda\in\Par_{n}%
\end{align*}
(since $\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a
basis of the $\kk$-module $\Lambda_{n}$). Consider this map $\beta$.
This map $\beta$ is surjective (since the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par_{n}}$ spans the $\kk$-module
$\Lambda_{n}$)\ \ \ \ \footnote{\textit{Proof.} Let $f\in\Lambda_{n}$. Then,
$f$ is a $\kk$-linear combination of the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par_{n}}$ (since the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par_{n}}$ spans the
$\kk$-module $\Lambda_{n}$). In other words, we can write $f$ in the
form $f=\sum_{\lambda\in\Par_{n}}a_{\lambda}%
v_{\lambda}$ for some family $\left(  a_{\lambda}\right)  _{\lambda
\in\Par_{n}}\in \kk^{\Par_n}$ of scalars.
Consider this family $\left(  a_{\lambda}\right)
_{\lambda\in\Par_{n}}$. As we know, $m_{\lambda}%
\in\Lambda_{n}$ for each $\lambda\in\Par_{n}$ (since
$\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a basis
of the $\kk$-module $\Lambda_{n}$); thus, $\sum_{\lambda\in
\Par_{n}}a_{\lambda}\underbrace{m_{\lambda}}%
_{\in\Lambda_{n}}\in\Lambda_{n}$ (since $\Lambda_{n}$ is a
$\kk$-module). Hence,
$\beta\left(  \sum_{\lambda\in\Par_n}a_{\lambda}m_{\lambda}\right)  $
is well-defined. Moreover, the
map $\beta$ is $\kk$-linear, so that we have%
\[
\beta\left(  \sum_{\lambda\in\Par_{n}}a_{\lambda
}m_{\lambda}\right)  =\sum_{\lambda\in\Par_{n}%
}a_{\lambda}\underbrace{\beta\left(  m_{\lambda}\right)  }%
_{\substack{=v_{\lambda}\\\text{(by the definition of }\beta\text{)}}%
}=\sum_{\lambda\in\Par_{n}}a_{\lambda}v_{\lambda}.
\]
Comparing this with $f=\sum_{\lambda\in\Par_{n}%
}a_{\lambda}v_{\lambda}$, we obtain $f=\beta\left(  \underbrace{\sum
_{\lambda\in\Par_{n}}a_{\lambda}m_{\lambda}}%
_{\in\Lambda_{n}}\right)  \in\beta\left(  \Lambda_{n}\right)  $.
\par
Forget that we fixed $f$. We thus have proved that $f\in\beta\left(
\Lambda_{n}\right)  $ for every $f\in\Lambda_{n}$. In other words,
$\Lambda_{n}\subset\beta\left(  \Lambda_{n}\right)  $. In other words, the map
$\beta$ is surjective.}.

But $\Lambda_{n}$ is a finitely generated $\kk$-module (since it has a
finite basis $\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}%
$). The map $\beta$ is an endomorphism of this $\kk$-module (since
$\beta:\Lambda_{n}\rightarrow\Lambda_{n}$ is $\kk$-linear), and is
surjective. Thus, Exercise~\ref{exe.linalg.sur-bij}(a) (applied to this
endomorphism) yields that $\beta$ is a $\kk$-module isomorphism.

Now, the map $\beta$ satisfies $\beta\left(  m_{\lambda}\right)  =v_{\lambda}$
for each $\lambda\in\Par_{n}$ (by the definition of
$\beta$). In other words, $\left(  \beta\left(  m_{\lambda}\right)  \right)
_{\lambda\in\Par_{n}}=\left(  v_{\lambda}\right)  _{\lambda
\in\Par_{n}}$.

But the family $\left(  m_{\lambda}\right)  _{\lambda\in\Par%
_{n}}$ is a basis of the $\kk$-module $\Lambda_{n}$. Hence, the image
of this family $\left(  m_{\lambda}\right)  _{\lambda\in\Par%
_{n}}$ under $\beta$ is also a basis of the $\kk$-module $\Lambda_{n}$
(since $\beta$ is a $\kk$-module isomorphism). In other words, the
family $\left(  v_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a
basis of the $\kk$-module $\Lambda_{n}$ (since the image of the family
$\left(  m_{\lambda}\right)  _{\lambda\in\Par_{n}}$ under
$\beta$ is $\left(  \beta\left(  m_{\lambda}\right)  \right)  _{\lambda
\in\Par_{n}}=\left(  v_{\lambda}\right)  _{\lambda
\in\Par_{n}}$).

Forget that we fixed $n$. We thus have proved that $\left(  v_{\lambda
}\right)  _{\lambda\in\Par_{n}}$ is a basis of the
$\kk$-module $\Lambda_{n}$ for each $n\in\NN$.
Hence, the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par}$ is a graded basis of the
$\kk$-module $\bigoplus_{n\in\NN}\Lambda_{n}$ (since
$\Par_{0},\Par_{1},\Par_{2},\ldots$ is a partition of the set
$\Par$). In other words, the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par}$ is a graded basis of the
$\kk$-module $\Lambda$ (since $\Lambda=\bigoplus_{n\in\NN}
\Lambda_{n}$). This solves Exercise \ref{exe.Lambda.basis-vlam}.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.bases}}

\textit{Solution to Exercise~\ref{exe.Lambda.bases}.} (a) This can be
solved by following the proof of Corollary~\ref{dual-bases-h-m-p}, but in
doing so one has to be careful about how one obtains the invertibility of $A$:
it is no longer a consequence of $A$ being a transition matrix between two
bases (because we do not know in advance that $\left(  u_{\lambda}\right)
_{\lambda\in\Par}$ is a basis). Instead, one has to argue as in
the footnote: The matrices $A$ and $B^{t}$ are block-diagonal, with each
diagonal block corresponding to the partitions of size $n$ for a given
$n\in\NN$\ \ \ \ \footnote{It is here that we are using the assumption that
$u_{\lambda}$ and $v_{\lambda}$ are homogeneous of degree $\left\vert
\lambda\right\vert $.}. In particular, they are block-diagonal matrices with
each block being a square matrix of finite size. It is known that if such a
matrix is right-invertible, then it is left-invertible\footnote{In fact, it is
clearly enough to prove this statement for square matrices of finite size
(because block-diagonal matrices can be inverted block-by-block). However,
for square matrices, this follows from the fact that a surjective endomorphism
of a finitely-generated $\kk$-module is an isomorphism
(Exercise~\ref{exe.linalg.sur-bij}(a)).}; therefore, since $A$ is
right-invertible (because $AB^{t}=I$), we conclude that $A$ is invertible.

(b) We know that for every partition $\lambda$, the symmetric functions
$h_{\lambda}$ and $m_{\lambda}$ are two homogeneous elements of $\Lambda$,
both of degree $\left\vert \lambda\right\vert $. We also know (from
Proposition~\ref{prop-Two-other-Cauchy-expansions}) that
\[
\prod_{i,j=1}^{\infty}(1-x_{i}y_{j})^{-1}=\sum_{\lambda\in\Par}
h_{\lambda}(\xx )m_{\lambda}(\yy ).
\]
Compared with \eqref{Cauchy-identity}, this yields $\sum_{\lambda
\in\Par}s_{\lambda}\left( \xx \right)  s_{\lambda
}\left( \yy \right)  =\sum_{\lambda\in\Par}
h_{\lambda}\left( \xx \right)  m_{\lambda}\left(  \yy \right)  $.
Thus, Exercise~\ref{exe.Lambda.bases}(a) (applied to $u_{\lambda
}=h_{\lambda}$ and $v_{\lambda}=m_{\lambda}$) yields that $\left(  h_{\lambda
}\right)  _{\lambda\in\Par}$ and $\left(  m_{\lambda}\right)
_{\lambda\in\Par}$ are $\kk$-bases of $\Lambda$, and
actually are dual bases with respect to the Hall inner product on $\Lambda$.
In particular, $\left(  h_{\lambda}\right)  _{\lambda\in\Par}$
is a $\kk$-basis of $\Lambda$. This solves
Exercise~\ref{exe.Lambda.bases}(b).

\textit{Remark.} We can use Exercise~\ref{exe.Lambda.bases}(b) to prove that
$\left(  e_{\lambda}\right)  _{\lambda\in\Par}$ is a $\kk$-basis of
$\Lambda$ as well (in a different way than we have done in the
proof of Proposition~\ref{many-symmetric-function-bases-prop}). In fact, let
us sketch this proof. We are going to proceed similarly to the above proof of
Proposition~\ref{symm-antipode-and-h-basis}, but with the roles of the
families $\left( e_n\right) _{n \geq 1}$ and
$\left( h_n\right) _{n \geq 1}$ switched.

We know (from Exercise~\ref{exe.Lambda.bases}(b)) that $\left(  h_{\lambda
}\right)  _{\lambda\in\Par}$ is a $\kk$-basis of
$\Lambda$. In other words, the family $\left(  h_{n}\right)  _{n\geq1}$ is
algebraically independent and generates the $\kk$-algebra $\Lambda$
(by parts (b) and (c) of
Exercise~\ref{exe.alggen-via-span}, applied to $v_n = h_n$
and $v_\lambda = h_\lambda$).
Thus, we can define a $\kk$-algebra homomorphism $\omega^{\prime
}:\Lambda\rightarrow\Lambda$ by setting
\begin{equation}
\omega^{\prime}\left(  h_{n}\right)  =e_{n} \qquad
\text{for every }n\geq 1.
\label{sol.Lambda.bases.c.omega'}
\end{equation}
\footnote{This $\omega^{\prime}$ is, of course, precisely the $\omega$
constructed in the proof of Proposition~\ref{symm-antipode-and-h-basis}. But
we cannot use the $\omega$ constructed in the proof of
Proposition~\ref{symm-antipode-and-h-basis} at this point, because its
construction made use of the fact that
$\left( e_{\lambda}\right) _{\lambda\in \Par}$
is a $\kk$-basis of $\Lambda$ (and we want to
prove this fact).} The identical form of the two recursions
\eqref{e-h-recursions.2} and \eqref{e-h-recursions.3}
shows that this $\omega^{\prime}$ also satisfies
\begin{align}
\omega^{\prime}\left(e_n\right) = h_n \qquad \text{ for each }
n \geq 0 .
\label{sol.Lambda.bases.c.omega'-eh}
\end{align}
\footnote{The details of this proof proceed just like the proof
of \eqref{the-fundamental-involution-he}, except for the fact
that the $e_n$ have traded places with the $h_n$.}
Combining this with \eqref{sol.Lambda.bases.c.omega'}, we
conclude that
$\left(\omega^{\prime}\circ\omega^{\prime}\right) \left(h_n\right)
= h_n$ for each $n \geq 1$.
Therefore, the two $\kk$-algebra homomorphisms
$\omega^{\prime} \circ \omega^{\prime} : \Lambda \to \Lambda$
and $\id : \Lambda \to \Lambda$
agree on each element of the generating set $\{h_n\}$ of $\Lambda$.
Hence, they are equal, i.e., we have
$\omega^{\prime} \circ \omega^{\prime} = \id$.
Hence, $\omega^{\prime}$ is an involution, and therefore an
automorphism of the $\kk$-algebra $\Lambda$. In particular,
$\omega^{\prime}$ is a $\kk$-module isomorphism $\Lambda\rightarrow
\Lambda$. By multiplicativity and \eqref{sol.Lambda.bases.c.omega'}, we
obtain
\begin{equation}
\omega^{\prime}\left(  h_{\lambda}\right)  =e_{\lambda}
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda
.
\label{sol.Lambda.bases.c.1}
\end{equation}
Thus, the image of the basis $\left(  h_{\lambda}\right)  _{\lambda
\in\Par}$ of $\Lambda$ under the $\kk$-module isomorphism
$\omega^{\prime}$ is the family $\left(  e_{\lambda}\right)  _{\lambda
\in\Par}$. Therefore, this latter family $\left(  e_{\lambda
}\right)  _{\lambda\in\Par}$ must be a basis of $\Lambda$ (being
the image of a basis under a $\kk$-module isomorphism).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.bases}}

\textit{Solution to Exercise \ref{exe.Lambda.bases}.} (a) Define
transition matrices $A=\left(  a_{\nu,\lambda}\right)  $ and $B=\left(
b_{\nu,\lambda}\right)  $ as in the proof of Corollary \ref{dual-bases-h-m-p}.
We then have
\begin{equation}
u_{\lambda}=\sum_{\nu\in\Par }a_{\nu,\lambda}s_{\nu
}\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda,
\label{sol.Lambda.bases.a.u}
\end{equation}
and
\begin{equation}
v_{\lambda}=\sum_{\nu\in\Par }b_{\nu,\lambda}s_{\nu
}\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.Lambda.bases.a.v}
\end{equation}


Notice that
\begin{equation}
\text{any two partitions }\nu\text{ and }\lambda\text{ with }\left\vert
\nu\right\vert \neq\left\vert \lambda\right\vert \text{ satisfy }
a_{\nu,\lambda}=0 \label{sol.Lambda.bases.a.graded}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.Lambda.bases.a.graded}:}
Let $\nu$ and $\lambda$ be two partitions with
$\left\vert\nu\right\vert \neq\left\vert \lambda\right\vert$. We know that
$u_{\lambda}$ is homogeneous of degree $\left\vert \lambda\right\vert$.
Thus, the expansion of $u_{\lambda}$ in the basis $\left(  s_{\mu
}\right)_{\mu\in\Par}$ contains only those $s_{\mu}$ with
nonzero coefficient which satisfy $\left\vert \mu\right\vert =\left\vert
\lambda\right\vert $. In other words, if a partition $\kappa$ satisfies
$\left\vert\kappa\right\vert \neq\left\vert \lambda\right\vert$, then the
coefficient of $s_\kappa$ in the expansion of $u_{\lambda}$ in the basis
$\left(  s_{\mu}\right)_{\mu\in\Par}$ is zero. Applying this to
$\kappa = \nu$, we conclude that the coefficient of $s_\nu$ in the
expansion of $u_{\lambda}$ in the basis
$\left(  s_{\mu}\right)_{\mu\in\Par}$ is zero. But since the coefficient of
$s_\nu$ in the expansion of $u_{\lambda}$ in the basis
$\left(  s_{\mu}\right)_{\mu\in\Par}$ is $a_{\nu, \lambda}$ (because
this expansion is given by \eqref{sol.Lambda.bases.a.u}), this rewrites as
follows: The element $a_{\nu, \lambda}$ is zero. In other words,
$a_{\nu, \lambda} = 0$. This proves \eqref{sol.Lambda.bases.a.graded}.}

For every partition $\lambda$, we have
\begin{equation}
v_{\lambda}=\sum_{\nu\in\Par }b_{\nu,\lambda}s_{\nu}=\sum
_{\rho\in\Par }b_{\rho,\lambda}s_{\rho}
\label{sol.Lambda.bases.a.v.2}
\end{equation}
(here, we renamed the summation index $\nu$ as $\rho$).

We can prove (as in the proof of Corollary \ref{dual-bases-h-m-p}) that
$\sum_{\lambda}a_{\nu,\lambda}b_{\rho,\lambda}=\delta_{\nu,\rho}$ for any two
partitions $\nu$ and $\rho$, where the sum ranges over all partitions
$\lambda$. In other words,
\begin{equation}
\sum_{\lambda\in\Par }a_{\nu,\lambda}b_{\rho,\lambda}
=\delta_{\nu,\rho}\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }\nu\text{
and }\rho. \label{sol.Lambda.bases.a.orth}
\end{equation}


Now, fix $n\in\NN$. Define a $\kk$-linear map $\Phi:\Lambda
_{n}\rightarrow\Lambda_{n}$ by setting
\[
\Phi\left(  s_{\nu}\right)  =\sum_{\lambda\in\Par_n}
a_{\nu,\lambda}s_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }\nu
\in\Par_{n}.
\]
\footnote{This is well-defined since $\left(  s_{\nu}\right)  _{\nu
\in\Par_{n}}$ is a $\kk$-basis of $\Lambda_{n}$.}

Also, recall that $v_{\lambda}$ is homogeneous of degree $\left\vert
\lambda\right\vert $ for every partition $\lambda$. Thus, $v_{\lambda}
\in\Lambda_{n}$ for every $\lambda\in\Par_{n}$. Hence,
we can define a $\kk$-linear map $\Psi:\Lambda_{n}\rightarrow
\Lambda_{n}$ by setting
\[
\Psi\left(  s_{\lambda}\right)  =v_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for
every }\lambda\in\Par_{n}\text{.}
\]
\footnote{This is well-defined since $\left(  s_{\lambda}\right)  _{\lambda
\in\Par_{n}}$ is a $\kk$-basis of $\Lambda_{n}$.}
Consider this map $\Psi$.

Fix a partition $\nu\in\Par_{n}$. Then, $\left\vert
\nu\right\vert =n$. Now, $a_{\nu,\lambda}=0$ for every $\lambda\in
\Par $ satisfying $\lambda\notin\Par_n$\ \ \ \ \footnote{\textit{Proof.}
Let $\lambda\in \Par$ satisfy $\lambda\notin\Par_{n}$.
Then, $\left\vert \lambda\right\vert \neq n=\left\vert \nu\right\vert $.
Hence, (\ref{sol.Lambda.bases.a.graded}) yields $a_{\nu,\lambda}=0$, qed.}.
Hence, $\sum\limits_{\substack{\lambda\in\Par ;\\\lambda\notin\Par_{n}}}
\underbrace{a_{\nu,\lambda}}_{=0}v_{\lambda}
=\sum\limits_{\substack{\lambda\in\Par ;\\\lambda\notin\Par_{n}}}0v_{\lambda}=0$
(in particular, this shows
that the sum $\sum\limits_{\substack{\lambda\in\Par ;\\\lambda
\notin\Par_{n}}}a_{\nu,\lambda}v_{\lambda}$ is
well-defined). Now,
\begin{equation}
\sum_{\lambda\in\Par }a_{\nu,\lambda}v_{\lambda}=\sum\limits
_{\substack{\lambda\in\Par ;\\\lambda\in\Par_{n}}}
a_{\nu,\lambda}v_{\lambda}+\underbrace{\sum\limits_{\substack{\lambda
\in\Par ;\\\lambda\notin\Par_{n}}}
a_{\nu,\lambda}v_{\lambda}}_{=0}=\underbrace{\sum\limits_{\substack{\lambda
\in\Par ;\\\lambda\in\Par_{n}}}}
_{\substack{=\sum_{\lambda\in\Par_{n}}\\\text{(since
}\Par_{n}\subset \Par \text{)}}}
a_{\nu,\lambda}v_{\lambda}=\sum_{\lambda\in\Par_{n}}
a_{\nu,\lambda}v_{\lambda} \label{sol.Lambda.bases.a.5}
\end{equation}
(in particular, this shows that the sum $\sum_{\lambda\in\Par}
a_{\nu,\lambda}v_{\lambda}$ is well-defined). But
\begin{align*}
\left(  \Psi\circ\Phi\right)  \left(  s_{\nu}\right)   &  =\Psi\left(
\underbrace{\Phi\left(  s_{\nu}\right)  }_{=\sum_{\lambda\in
\Par_{n}}a_{\nu,\lambda}s_{\lambda}}\right)
=\Psi\left(  \sum_{\lambda\in\Par_{n}}a_{\nu,\lambda
}s_{\lambda}\right) \\
&  =\sum_{\lambda\in\Par_{n}}a_{\nu,\lambda
}\underbrace{\Psi\left(  s_{\lambda}\right)  }_{=v_{\lambda}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Psi
\text{ is }\kk \text{-linear}\right) \\
&  =\sum_{\lambda\in\Par_{n}}a_{\nu,\lambda}
v_{\lambda}=\sum_{\lambda\in\Par }a_{\nu,\lambda}
\underbrace{v_{\lambda}}_{\substack{=\sum_{\rho\in\Par }
b_{\rho,\lambda}s_{\rho}
\\\text{(by (\ref{sol.Lambda.bases.a.v.2}))}}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Lambda.bases.a.5})}\right) \\
&  =\sum_{\lambda\in\Par }a_{\nu,\lambda}\sum_{\rho
\in\Par }b_{\rho,\lambda}s_{\rho}=\underbrace{\sum_{\lambda
\in\Par }\sum_{\rho\in\Par }}_{=\sum_{\rho
\in\Par }\sum_{\lambda\in\Par }}a_{\nu,\lambda}
b_{\rho,\lambda}s_{\rho}=\sum_{\rho\in\Par }\underbrace{\sum
_{\lambda\in\Par }a_{\nu,\lambda}b_{\rho,\lambda}}
_{\substack{=\delta_{\nu,\rho}\\
\text{(by (\ref{sol.Lambda.bases.a.orth}))}}}s_{\rho}\\
&  =\sum_{\rho\in\Par }\delta_{\nu,\rho}s_{\rho}
=\underbrace{\delta_{\nu,\nu}}_{\substack{=1\\
\text{(since }\nu=\nu\text{)}}}
s_{\nu}+\sum\limits_{\substack{\rho\in\Par ;\\\rho\neq\nu
}}\underbrace{\delta_{\nu,\rho}}_{\substack{=0\\\text{(since }\nu\neq
\rho\\\text{(since }\rho\neq\nu\text{))}}}s_{\rho}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\nu\in\Par \right) \\
&  =s_{\nu}+\underbrace{\sum\limits_{\substack{\rho\in\Par ;\\\rho
\neq\nu}}0s_{\rho}}_{=0}=s_{\nu}=\id_{\Lambda_{n}}
\left(  s_{\nu}\right)  .
\end{align*}


Now, forget that we fixed $\nu$. We thus have proven that $\left(  \Psi
\circ\Phi\right)  \left(  s_{\nu}\right) 
 =\id_{\Lambda_{n}}\left(  s_{\nu}\right)  $ for every $\nu\in
\Par_{n}$. In other words, the two maps $\Psi\circ
\Phi$ and $\id_{\Lambda_{n}}$ are equal to each other
on every element of the basis $\left(  s_{\nu}\right)  _{\nu\in
\Par_n}$ of the $\kk$-module $\Lambda_{n}$.
Since these two maps $\Psi\circ\Phi$ and $\id_{\Lambda
_{n}}$ are both $\kk$-linear, this yields that these two maps
$\Psi\circ\Phi$ and $\id_{\Lambda_{n}}$ must be
identical (because if two $\kk$-linear maps are equal to each other on
every element of a given basis of their domain, then they must be identical).
In other words, $\Psi\circ\Phi=\id_{\Lambda_{n}}$.
Hence, the map $\Psi$ is right-invertible, and thus surjective. Since $\Psi$
is an endomorphism of the finitely generated $\kk$-module $\Lambda_{n}$,
we can therefore apply Exercise \ref{exe.linalg.sur-bij}(a) to this
endomorphism $\Psi$. We thus conclude that $\Psi$ is a $\kk$-module
isomorphism. Hence, $\Psi$ must map the basis $\left(  s_{\lambda}\right)
_{\lambda\in\Par_{n}}$ of the $\kk$-module
$\Lambda_{n}$ to a basis of the $\kk$-module $\Lambda_{n}$ (since any
$\kk$-module isomorphism maps any basis of its domain to a basis of its
target). In other words, the image of the basis $\left(  s_{\lambda}\right)
_{\lambda\in\Par_{n}}$ under $\Psi$ must be a basis of
the $\kk$-module $\Lambda_{n}$. Since the image of the basis $\left(
s_{\lambda}\right)  _{\lambda\in\Par_{n}}$ under
$\Psi$ is the family $\left(  v_{\lambda}\right)  _{\lambda\in
\Par_{n}}$ (because $\Psi\left(  s_{\lambda}\right)
=v_{\lambda}$ for every $\lambda\in\Par_{n}$), this
rewrites as follows:
\begin{equation}
\text{The family }\left(  v_{\lambda}\right)  _{\lambda\in\Par_{n}}
\text{ is a basis of the }\kk \text{-module }\Lambda_{n}.
\label{sol.Lambda.bases.a.vbasis}
\end{equation}


Now, recall that we assumed that
\begin{equation}
\sum_{\lambda\in\Par}s_{\lambda}\left(  {\xx }\right)
s_{\lambda}\left(  {\yy }\right)  =\sum_{\lambda\in\Par}
u_{\lambda}\left(  {\xx }\right)  v_{\lambda}\left(  {\yy}
\right)  . \label{sol.Lambda.bases.a.ass}
\end{equation}


But there is a continuous\footnote{``Continuous'' here means
``continuous with
respect to the pointwise topology on $\kk \left[  \left[  \xx 
,\yy \right]  \right]  $''.} endomorphism $\rho$ of the
$\kk$-algebra $\kk \left[  \left[  \xx ,\yy \right]
\right]  $ which satisfies
\begin{align*}
\rho\left(  x_{i}\right)   &  =y_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }
i\geq1,\ \ \ \ \ \ \ \ \ \ \text{and}\\
\rho\left(  y_{i}\right)   &  =x_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }
i\geq1.
\end{align*}
Consider this $\rho$. Since the map $\rho$ is a continuous endomorphism of a
topological $\kk$-algebra, this map $\rho$ commutes with the application of any
power series. Hence,
\begin{equation}
\rho\left(  f\left(  \xx \right)  \right)  =f\left(  \yy \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in\Lambda\label{sol.Lambda.bases.a.rho.1}
\end{equation}
and
\begin{equation}
\rho\left(  f\left(  \yy \right)  \right)  =f\left(  \xx \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in\Lambda. \label{sol.Lambda.bases.a.rho.2}
\end{equation}
Since $\rho$ is a continuous $\kk$-algebra homomorphism, we have
\begin{align}
\rho\left(  \sum_{\lambda\in\Par}s_{\lambda}\left(  {\xx }
\right)  s_{\lambda}\left(  {\yy }\right)  \right)   &  =\sum
_{\lambda\in\Par}\underbrace{\rho\left(  s_{\lambda}\left(
{\xx }\right)  \right)  }_{\substack{=s_{\lambda}\left(  \yy 
\right)  \\\text{(by (\ref{sol.Lambda.bases.a.rho.1}), applied to }f=s_{\lambda
}\text{)}}}\underbrace{\rho\left(  s_{\lambda}\left(  {\yy }\right)
\right)  }_{\substack{=s_{\lambda}\left(  \xx \right)  \\\text{(by
(\ref{sol.Lambda.bases.a.rho.2}), applied to }f=s_{\lambda}\text{)}}}\nonumber\\
&  =\sum_{\lambda\in\Par}s_{\lambda}\left(  \yy \right)
s_{\lambda}\left(  \xx \right)  =\sum_{\lambda\in\Par}
s_{\lambda}\left(  {\xx }\right)  s_{\lambda}\left(  {\yy }
\right)
\label{sol.Lambda.bases.a.rho.3}
\end{align}
and
\begin{align}
\rho\left(  \sum_{\lambda\in\Par}u_{\lambda}\left(  {\xx }
\right)  v_{\lambda}\left(  {\yy }\right)  \right)   &  =\sum
_{\lambda\in\Par}\underbrace{\rho\left(  u_{\lambda}\left(
{\xx }\right)  \right)  }_{\substack{=u_{\lambda}\left(  \yy 
\right)  \\\text{(by (\ref{sol.Lambda.bases.a.rho.1}), applied to }f=u_{\lambda
}\text{)}}}\underbrace{\rho\left(  v_{\lambda}\left(  {\yy }\right)
\right)  }_{\substack{=v_{\lambda}\left(  \xx \right)  \\\text{(by
(\ref{sol.Lambda.bases.a.rho.2}), applied to }f=v_{\lambda}\text{)}}}\nonumber\\
&  =\sum_{\lambda\in\Par}u_{\lambda}\left(  \yy \right)
v_{\lambda}\left(  \xx \right)  =\sum_{\lambda\in\Par}
v_{\lambda}\left(  \xx \right)  u_{\lambda}\left(  \yy \right)  .
\label{sol.Lambda.bases.a.rho.4}
\end{align}
Now, (\ref{sol.Lambda.bases.a.rho.3}) yields
\begin{align*}
\sum_{\lambda\in\Par}s_{\lambda}\left(  {\xx }\right)
s_{\lambda}\left(  {\yy }\right)   &  =\rho\left(  \underbrace{\sum
_{\lambda\in\Par}s_{\lambda}\left(  {\xx }\right)
s_{\lambda}\left(  {\yy }\right)  }_{\substack{=\sum_{\lambda
\in\Par}u_{\lambda}\left(  {\xx }\right)  v_{\lambda
}\left(  {\yy }\right)  \\
\text{(by (\ref{sol.Lambda.bases.a.ass}))}}}\right) 
 =\rho\left(  \sum_{\lambda\in\Par}u_{\lambda}\left(
{\xx }\right)  v_{\lambda}\left(  {\yy }\right)  \right) \\
&  =\sum_{\lambda\in\Par}v_{\lambda}\left(  \xx \right)
u_{\lambda}\left(  \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.bases.a.rho.4})}\right)  .
\end{align*}
Hence, the assumptions of Exercise \ref{exe.Lambda.bases}(a) are still
satisfied if we switch $u_{\lambda}$ with $v_{\lambda}$ for every $\lambda
\in\Par $. As a consequence, we can apply
(\ref{sol.Lambda.bases.a.vbasis}) to $v_{\lambda}$ and $u_{\lambda}$ instead
of $u_{\lambda}$ and $v_{\lambda}$. As a result, we obtain:
\[
\text{The family }\left(  u_{\lambda}\right)  _{\lambda\in\Par_{n}}
\text{ is a basis of the }\kk \text{-module }\Lambda_{n}.
\]


We thus know that $\left(  u_{\lambda}\right)  _{\lambda\in\Par_{n}}$
and $\left(  v_{\lambda}\right)  _{\lambda\in \Par_{n}}$ are bases of
the $\kk$-module $\Lambda_{n}$.

Now, forget that we fixed $n$. We thus know that $\left(  u_{\lambda}\right)
_{\lambda\in\Par_{n}}$ and $\left(  v_{\lambda
}\right)  _{\lambda\in\Par_{n}}$ are bases of the
$\kk$-module $\Lambda_{n}$ for every $n\in\NN$. Now, $\left(
u_{\lambda}\right)  _{\lambda\in\Par }$ is a basis of the
$\kk$-module $\Lambda$ (since $\left(  u_{\lambda}\right)  _{\lambda
\in\Par_{n}}$ is a basis of the $\kk$-module
$\Lambda_{n}$ for every $n\in\NN$). Also, $\left(  v_{\lambda}\right)
_{\lambda\in\Par }$ is a basis of the $\kk$-module
$\Lambda$ (since $\left(  v_{\lambda}\right) _{\lambda\in\Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$ for every
$n\in\NN$). It remains to prove that these two bases $\left(
u_{\lambda}\right)  _{\lambda\in\Par }$ and $\left(  v_{\lambda
}\right)  _{\lambda\in\Par }$ are dual bases with respect to the
Hall inner product on $\Lambda$. This can be proven just as in our proof of
Corollary \ref{dual-bases-h-m-p} above. This completes the solution of
Exercise \ref{exe.Lambda.bases}(a).

(b) We know that for every partition $\lambda$, the symmetric functions
$h_{\lambda}$ and $m_{\lambda}$ are two homogeneous elements of $\Lambda$,
both of degree $\left\vert \lambda\right\vert $. We also know (from
Proposition \ref{prop-Two-other-Cauchy-expansions}) that
\[
\prod_{i,j=1}^{\infty}(1-x_{i}y_{j})^{-1}=\sum_{\lambda\in\Par}
h_{\lambda}({\xx })m_{\lambda}(\yy ).
\]
Compared with (\ref{Cauchy-identity}), this yields $\sum_{\lambda
\in\Par}s_{\lambda}\left(  {\xx }\right)  s_{\lambda
}\left(  {\yy }\right)  =\sum_{\lambda\in\Par}h_{\lambda
}\left(  {\xx }\right)  m_{\lambda}\left(  {\yy }\right)  $. Thus,
Exercise \ref{exe.Lambda.bases}(a) (applied to $u_{\lambda}=h_{\lambda}$ and
$v_{\lambda}=m_{\lambda}$) yields that $\left(  h_{\lambda}\right)
_{\lambda\in\Par}$ and $\left(  m_{\lambda}\right)  _{\lambda
\in\Par}$ are $\kk$-bases of $\Lambda$, and actually are
dual bases with respect to the Hall inner product on $\Lambda$. In particular,
$\left(  h_{\lambda}\right)  _{\lambda\in\Par}$ is a
$\kk$-basis of $\Lambda$. This solves Exercise \ref{exe.Lambda.bases}(b).

\textit{Remark.} We can use Exercise \ref{exe.Lambda.bases}(b) to prove that
$\left(  e_{\lambda}\right)  _{\lambda\in\Par}$ is a
$\kk$-basis of $\Lambda$ as well (in a different way than we have done
in the proof of Proposition \ref{many-symmetric-function-bases-prop}). In
fact, let us sketch this proof. We are going to proceed similarly to the
above proof of Proposition
\ref{symm-antipode-and-h-basis}, but with the roles of the families $\left(
e_{n}\right)  _{n\geq1}$ and $\left(  h_{n}\right)  _{n\geq1}$ switched.

We know (from Exercise \ref{exe.Lambda.bases}(b)) that $\left(  h_{\lambda
}\right)  _{\lambda\in\Par}$ is a $\kk$-basis of
$\Lambda$. In other words, the family $\left(  h_{n}\right)  _{n\geq1}$ is
algebraically independent and generates the $\kk$-algebra $\Lambda$
(by parts (b) and (c) of
Exercise~\ref{exe.alggen-via-span}, applied to $v_n = h_n$
and $v_\lambda = h_\lambda$).
Thus, we can define a $\kk$-algebra homomorphism $\omega^{\prime
}:\Lambda\rightarrow\Lambda$ by setting
\begin{equation}
\omega^{\prime}\left(  h_{n}\right)  =e_{n} \qquad
\text{for every }n\geq 1.
\label{sol.Lambda.bases.c.omega'}
\end{equation}
\footnote{This $\omega^{\prime}$ is, of course, precisely the $\omega$
constructed in the proof of Proposition~\ref{symm-antipode-and-h-basis}. But
we cannot use the $\omega$ constructed in the proof of
Proposition~\ref{symm-antipode-and-h-basis} at this point, because its
construction made use of the fact that
$\left( e_{\lambda}\right) _{\lambda\in \Par}$
is a $\kk$-basis of $\Lambda$ (and we want to
prove this fact).} The identical form of the two recursions
\eqref{e-h-recursions.2} and \eqref{e-h-recursions.3}
shows that this $\omega^{\prime}$ also satisfies
\begin{align}
\omega^{\prime}\left(e_n\right) = h_n \qquad \text{ for each }
n \geq 0 .
\label{sol.Lambda.bases.c.omega'-eh}
\end{align}
\footnote{The details of this proof proceed just like the proof
of \eqref{the-fundamental-involution-he}, except for the fact
that the $e_n$ have traded places with the $h_n$.}
Combining this with \eqref{sol.Lambda.bases.c.omega'}, we
conclude that
$\left(\omega^{\prime}\circ\omega^{\prime}\right) \left(h_n\right)
= h_n$ for each $n \geq 1$.
Therefore, the two $\kk$-algebra homomorphisms
$\omega^{\prime} \circ \omega^{\prime} : \Lambda \to \Lambda$
and $\id : \Lambda \to \Lambda$
agree on each element of the generating set $\{h_n\}$ of $\Lambda$.
Hence, they are equal, i.e., we have
$\omega^{\prime} \circ \omega^{\prime} = \id$.
Hence, $\omega^{\prime}$ is an involution, and therefore an
automorphism of the $\kk$-algebra $\Lambda$. In particular,
$\omega^{\prime}$ is a $\kk$-module isomorphism $\Lambda\rightarrow
\Lambda$. But clearly,
\begin{equation}
\omega^{\prime}\left(  h_{\lambda}\right)  =e_{\lambda}
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda
.\label{sol.Lambda.bases.c.1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.bases.c.1}):} Let $\lambda$ be a
partition. Write this partition $\lambda$ in the form $\lambda=\left(
\lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)  $ for $\ell=\ell\left(
\lambda\right)  $. Then, $\lambda_{i}\geq1$ for every $i\in\left\{
1,2,...,\ell\right\}  $. Hence, $\omega^{\prime}\left(  h_{\lambda_{i}
}\right)  =e_{\lambda_{i}}$ for every $i\in\left\{  1,2,...,\ell\right\}  $
(by (\ref{sol.Lambda.bases.c.omega'}), applied to $n=\lambda_{i}$).
Multiplying these equalities for all $i\in\left\{  1,2,...,\ell\right\}  $, we
obtain
\[
\omega^{\prime}\left(  h_{\lambda_{1}}\right)  \omega^{\prime}\left(
h_{\lambda_{2}}\right)  ...\omega^{\prime}\left(  h_{\lambda_{\ell}}\right)
=e_{\lambda_{1}}e_{\lambda_{2}}...e_{\lambda_{\ell}}.
\]
\par
But recall again that $\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda
_{\ell}\right)  $. Hence,
$h_{\lambda} = h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{\ell}}$
(by the definition of $h_{\lambda}$) and $e_{\lambda
}=e_{\lambda_{1}}e_{\lambda_{2}}...e_{\lambda_{\ell}}$ (by the definition of
$e_{\lambda}$). Now,
\begin{align*}
\omega^{\prime}\left(  \underbrace{h_{\lambda}}_{=h_{\lambda_{1}}
h_{\lambda_{2}}...h_{\lambda_{\ell}}}\right)    & =\omega^{\prime}\left(
h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{\ell}}\right)  =\omega^{\prime
}\left(  h_{\lambda_{1}}\right)  \omega^{\prime}\left(  h_{\lambda_{2}}
\right)  ...\omega^{\prime}\left(  h_{\lambda_{\ell}}\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega^{\prime}\text{ is a
}\kk \text{-algebra homomorphism}\right)  \\
& =e_{\lambda_{1}}e_{\lambda_{2}}...e_{\lambda_{\ell}}=e_{\lambda
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }e_{\lambda}=e_{\lambda_{1}}
e_{\lambda_{2}}...e_{\lambda_{\ell}}\right)  .
\end{align*}
This proves (\ref{sol.Lambda.bases.c.1}).}

Now, recall that the map $\omega^{\prime}$ is a $\kk$-module
isomorphism $\Lambda\rightarrow\Lambda$. Hence, this map $\omega^{\prime}$
must map the basis $\left(  h_{\lambda}\right)  _{\lambda\in
\Par }$ of the $\kk$-module $\Lambda$ to a basis of the
$\kk$-module $\Lambda$ (since any $\kk$-module isomorphism maps
any basis of its domain to a basis of its target). In other words, the image
of the basis $\left(  h_{\lambda}\right)  _{\lambda\in\Par }$
under $\omega^{\prime}$ must be a basis of the $\kk$-module $\Lambda$.
Since the image of the basis $\left(  h_{\lambda}\right)  _{\lambda
\in\Par }$ under $\omega^{\prime}$ is the family $\left(
e_{\lambda}\right)  _{\lambda\in\Par }$ (because of
(\ref{sol.Lambda.bases.c.1})), this rewrites as follows: The family $\left(
e_{\lambda}\right)  _{\lambda\in\Par }$ is a basis of the
$\kk$-module $\Lambda$. Thus, we have found a new proof of the fact
that $\left(  e_{\lambda}\right)  _{\lambda\in\Par }$ is a basis
of the $\kk$-module $\Lambda$.

(Of course, the map $\omega^\prime$ that we constructed in this proof
is exactly the map $\omega$ defined in the proof of Proposition
\ref{symm-antipode-and-h-basis}, since it satisfies
\eqref{sol.Lambda.bases.c.1}.)
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.E(t)-H(t)-P(t)-relation}}

\textit{Solution to Exercise~\ref{exe.E(t)-H(t)-P(t)-relation}.} When
$\QQ$ is a subring of $\kk$, the statement of
Exercise~\ref{exe.E(t)-H(t)-P(t)-relation} has been proven in
\eqref{E(t)-H(t)-P(t)-relation}. However, there is no immediate way to reuse
the proof of \eqref{E(t)-H(t)-P(t)-relation} in the general case (because this
proof made use of logarithms, and these are only defined when $\QQ$ is
a subring of $\kk$). Nevertheless, it is possible to imitate that proof
by defining a notion of ``logarithmic derivative'' even in the absence
of a logarithm. Let us elaborate on this.

Let $A$ be any commutative ring. Whenever $Q\in A\left[  \left[  t\right]
\right]  $ is a formal power series with constant term $1$, we define the
\dfn{logarithmic derivative}\index{$\operatorname*{lder}Q$}
of $Q$ to be the power series $\dfrac
{Q^{\prime}}{Q}\in A\left[  \left[  t\right]  \right]$ (where
$Q^\prime$ denotes the derivative of $Q$, as usual). We denote this
logarithmic derivative by $\operatorname*{lder}Q$. When $\QQ$ is a
subring of $A$, we have $\operatorname*{lder}Q=\dfrac{d}{dt}\left(  \log
Q\right)  $, but the concept of $\operatorname*{lder}Q$ is defined even when
$\log$ is not. Some authors find it instructive to write $\dfrac{d}{dt}\left(
\log Q\right)  $ for $\operatorname*{lder}Q$, but we prefer not to do so,
since this might tempt us to write things which make no sense.

It is easy to see that the map
\begin{align*}
\left\{  Q\in A\left[  \left[  t\right]  \right]  \ \mid\ Q\text{ has constant
term }1\right\}    & \rightarrow A\left[  \left[  t\right]  \right]  ,\\
Q  & \mapsto\operatorname*{lder}Q
\end{align*}
is continuous (where we equip $A\left[  \left[  t\right]  \right]  $ with the
usual topology -- i.e., the product topology obtained by regarding the set
$A \left[ \left[ t \right] \right]$ as a direct product of infinitely many
copies of $A$). Moreover, whenever $I$ is a set and
$\left( Q_i \right) _{i \in I}
\in\left(  A\left[  \left[  t\right]  \right]  \right)  ^{I}$ is a family of
formal power series with constant term $1$ such that the product $\prod_{i\in
I}Q_{i}$ converges (with respect to our topology on $A\left[  \left[
t\right]  \right]  $), then
\begin{equation}
\operatorname*{lder}\left(  \prod_{i\in I}Q_{i}\right)  =\sum_{i\in
I}\operatorname*{lder}\left(  Q_{i}\right)
.
\label{sol.E(t)-H(t)-P(t)-relation.prodsum}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.E(t)-H(t)-P(t)-relation.prodsum}:} Let
$I$ be a set, and let $\left(  Q_{i}\right)  _{i\in I}\in\left(  A\left[
\left[  t\right]  \right]  \right)  ^{I}$ be a family of formal power series
with constant term $1$ such that the product $\prod_{i\in I}Q_{i}$ converges
(with respect to our topology on $A\left[  \left[  t\right]  \right]  $). We
need to prove \eqref{sol.E(t)-H(t)-P(t)-relation.prodsum}. Since the map
\begin{align*}
\left\{  Q\in A\left[  \left[  t\right]  \right]  \ \mid\ Q\text{ has constant
term }1\right\}    & \rightarrow A\left[  \left[  t\right]  \right]  ,\\
Q  & \mapsto\operatorname*{lder}Q
\end{align*}
is continuous (and since infinite products are limits of finite products, and
infinite sums are limits of finite sums), we can WLOG assume that the set $I$
is finite. Assuming this, we recall that the product rule for differentiating
a product of several power series yields%
\[
\left(  \prod_{i\in I}Q_{i}\right)  ^{\prime}=\sum_{j\in I}Q_{j}^{\prime
}\left(  \prod_{i\in I\setminus\left\{  j\right\}  }Q_{i}\right)  .
\]
Now, by the definition of $\operatorname*{lder}\left(  \prod_{i\in I}%
Q_{i}\right)  $, we have
\begin{align*}
\operatorname*{lder}\left(  \prod_{i\in I}Q_{i}\right)    & =\dfrac{\left(
\prod_{i\in I}Q_{i}\right)  ^{\prime}}{\prod_{i\in I}Q_{i}}=\dfrac{1}%
{\prod_{i\in I}Q_{i}}\underbrace{\left(  \prod_{i\in I}Q_{i}\right)  ^{\prime
}}_{=\sum_{j\in I}Q_{j}^{\prime}\left(  \prod_{i\in I\setminus\left\{
j\right\}  }Q_{i}\right)  }=\dfrac{1}{\prod_{i\in I}Q_{i}}\sum_{j\in I}%
Q_{j}^{\prime}\left(  \prod_{i\in I\setminus\left\{  j\right\}  }Q_{i}\right)
\\
& =\sum_{j\in I}Q_{j}^{\prime}\cdot\dfrac{\prod_{i\in I\setminus\left\{
j\right\}  }Q_{i}}{\prod_{i\in I}Q_{i}}=\sum_{j\in I}Q_{j}^{\prime}%
\cdot\underbrace{\dfrac{\prod_{i\in I\setminus\left\{  j\right\}  }Q_{i}%
}{Q_{j}\prod_{i\in I\setminus\left\{  j\right\}  }Q_{i}}}_{=\dfrac{1}{Q_{j}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\prod_{i\in I}Q_{i}=Q_{j}\prod_{i\in
I\setminus\left\{  j\right\}  }Q_{i}\text{ for every }j\in I\right)  \\
& =\sum_{j\in I}Q_{j}^{\prime}\cdot\dfrac{1}{Q_{j}}=\sum_{j\in I}
\underbrace{\dfrac{Q_{j}^{\prime}}{Q_{j}}}_{\substack{=\operatorname*{lder}
\left(  Q_{j}\right)  \\\text{(since }\operatorname*{lder}\left(
Q_{j}\right)  \text{ is}\\\text{defined as }\dfrac{Q_{j}^{\prime}}{Q_{j}}
\text{)}}}=\sum_{j\in I}\operatorname*{lder}\left(  Q_{j}\right)  =\sum_{i\in
I}\operatorname*{lder}\left(  Q_{i}\right)  .
\end{align*}
This proves \eqref{sol.E(t)-H(t)-P(t)-relation.prodsum}.} This equality can be
used as a substitute for the famous property of the logarithm to take products
into sums in situations where the logarithm is not defined. Furthermore, we
have
\begin{equation}
\operatorname*{lder}\left(  Q^{-1}\right)
 = -\operatorname*{lder} Q
\label{sol.E(t)-H(t)-P(t)-relation.inv}
\end{equation}
whenever $Q\in A\left[  \left[  t\right]  \right]  $ is a formal power series
with constant term $1$ (where $Q^{-1}$ means the multiplicative inverse of
$Q$)\ \ \ \ \footnote{\textit{Proof of
\eqref{sol.E(t)-H(t)-P(t)-relation.inv}:} Let
$Q\in A\left[  \left[  t\right]  \right]  $ be a formal power series
with constant term $1$. Then, the product rule for the derivative of a product
of two power series yields $\left(  QQ^{-1}\right)  ^{\prime}=Q^{\prime}%
Q^{-1}+Q\left(  Q^{-1}\right)  ^{\prime}$, so that $Q^{\prime}Q^{-1}+Q\left(
Q^{-1}\right)  ^{\prime}=\left(  \underbrace{QQ^{-1}}_{=1}\right)  ^{\prime
}=1^{\prime}=0$ and thus $Q\left(  Q^{-1}\right)  ^{\prime}
= -Q^{\prime}Q^{-1}$, so that
$\dfrac{\left(  Q^{-1}\right)  ^{\prime}}{Q^{-1}}=-\dfrac
{Q^{\prime}}{Q}$. But the definition of $\operatorname*{lder}\left(
Q^{-1}\right)  $ yields $\operatorname*{lder}\left(  Q^{-1}\right)
=\dfrac{\left(  Q^{-1}\right)  ^{\prime}}{Q^{-1}}=-\underbrace{\dfrac
{Q^{\prime}}{Q}}_{=\operatorname*{lder}Q}=-\operatorname*{lder}Q$. This proves
\eqref{sol.E(t)-H(t)-P(t)-relation.inv}.}.

Now, set $A=\Lambda$. We have
\[
H\left(  t\right)  =\prod_{i=1}^{\infty}\left(  1-x_{i}t\right)  ^{-1}%
=\prod_{i\in\left\{  1,2,3,...\right\}  }\left(  1-x_{i}t\right)  ^{-1},
\]
so that
\begin{align*}
\operatorname*{lder}\left(  H\left(  t\right)  \right)    &
=\operatorname*{lder}\left(  \prod_{i\in\left\{  1,2,3,...\right\}  }\left(
1-x_{i}t\right)  ^{-1}\right)  =\sum_{i\in\left\{  1,2,3,...\right\}
}\underbrace{\operatorname*{lder}\left(  \left(  1-x_{i}t\right)
^{-1}\right)  }_{\substack{=-\operatorname*{lder}\left(  1-x_{i}t\right)
\\\text{(by \eqref{sol.E(t)-H(t)-P(t)-relation.inv}, applied to }
Q=1-x_{i}t\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.E(t)-H(t)-P(t)-relation.prodsum}, applied to }I=\left\{
1,2,3,...\right\}  \text{ and }Q_{i}=\left(  1-x_{i}t\right)  ^{-1}\right)
\\
& =\sum_{i\in\left\{  1,2,3,...\right\}  }\left(
-\underbrace{\operatorname*{lder}\left(  1-x_{i}t\right)  }_{\substack{=\dfrac
{\left(  1-x_{i}t\right)  ^{\prime}}{1-x_{i}t}\\\text{(by the definition of
}\operatorname*{lder}\left(  1-x_{i}t\right)  \text{)}}}\right)  =\sum
_{i\in\left\{  1,2,3,...\right\}  }\underbrace{\left(  -\dfrac{\left(
1-x_{i}t\right)  ^{\prime}}{1-x_{i}t}\right)  }_{\substack{=-\dfrac{-x_{i}%
}{1-x_{i}}t\\\text{(since }\left(  1-x_{i}t\right)  ^{\prime}=-x_{i}\text{)}%
}}\\
& =\sum_{i\in\left\{  1,2,3,...\right\}  }\underbrace{\left(  -\dfrac{-x_{i}%
}{1-x_{i}t}\right)  }_{=x_{i}\cdot\dfrac{1}{1-x_{i}t}}=\sum_{i\in\left\{
1,2,3,...\right\}  }x_{i}\cdot\underbrace{\dfrac{1}{1-x_{i}t}}_{=\sum_{m\geq
0}\left(  x_{i}t\right)  ^{m}=\sum_{m\geq0}x_{i}^{m}t^{m}}\\
& =\sum_{i\in\left\{  1,2,3,...\right\}  }x_{i}\cdot\sum_{m\geq0}x_{i}%
^{m}t^{m}=\sum_{m\geq0}\left(  \sum_{i\in\left\{  1,2,3,...\right\}
}\underbrace{x_{i}\cdot x_{i}^{m}}_{=x_{i}^{m+1}}\right)  t^{m} \\
&=\sum_{m\geq
0}\underbrace{\left(  \sum_{i\in\left\{  1,2,3,...\right\}  }x_{i}%
^{m+1}\right)  }_{=p_{m+1}}t^{m}=\sum_{m\geq0}p_{m+1}t^{m}.
\end{align*}
Thus,
\begin{align*}
\sum_{m\geq0}p_{m+1}t^{m}  & =\operatorname*{lder}\left(  H\left(  t\right)
\right)  =\dfrac{\left(  H\left(  t\right)  \right)  ^{\prime}}{H\left(
t\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{lder}\left(  H\left(  t\right)  \right)  \right)  \\
& =\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }.
\end{align*}
This solves the exercise.

\textit{Remark:} Another solution of
Exercise~\ref{exe.E(t)-H(t)-P(t)-relation} proceeds by noticing that
the statement of this exercise can be rewritten in the form
$\left(\sum_{m \geq 0} p_{m+1} t^m\right) \cdot H\left(t\right)
= H^\prime \left(t\right)$,
which (by comparison of coefficients) is equivalent to saying that
\begin{equation}
\label{sol.E(t)-H(t)-P(t)-relation.encoded}
\text{every } m \in \NN \text{ satisfies }
\sum_{i=0}^m p_{i+1} h_{m-i} = \left(m+1\right) h_{m+1}.
\end{equation}
But it is clear that once \eqref{sol.E(t)-H(t)-P(t)-relation.encoded}
is proven for $\kk = \ZZ$, it immediately follows that
\eqref{sol.E(t)-H(t)-P(t)-relation.encoded} also holds for all $\kk$
(since $\Lambda_\kk = \kk \otimes_\ZZ \Lambda_\ZZ$). Proving
\eqref{sol.E(t)-H(t)-P(t)-relation.encoded} for $\kk = \ZZ$, in turn,
boils down to proving \eqref{sol.E(t)-H(t)-P(t)-relation.encoded}
for $\kk = \QQ$, which we already know how to do (from the proof
of \eqref{E(t)-H(t)-P(t)-relation}).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.alggen-ind}}

\textit{Solution to Exercise \ref{exe.Lambda.alggen-ind}.} For every partition
$\lambda$, define an element $v_{\lambda}\in\Lambda$ by $v_{\lambda
}=v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(  \lambda\right)
}}$. (This is well-defined, since $\lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell\left(  \lambda\right)  }$ are positive integers whenever
$\lambda$ is a partition.) Thus, Exercise \ref{exe.alggen-via-span}(b)
(applied to $A=\Lambda$) shows that the elements $v_{1},v_{2},v_{3},\ldots$
generate the $\kk$-algebra $\Lambda$ if and only if the family $\left(
v_{\lambda}\right)  _{\lambda\in\Par}$ spans the $\kk%
$-module $\Lambda$. Hence, the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ spans the $\kk$-module $\Lambda$
(since the elements $v_{1},v_{2},v_{3},\ldots$ generate the $\kk%
$-algebra $\Lambda$).

Furthermore, $v_{\lambda}$ is an element of $\Lambda_{\left\vert
\lambda\right\vert }$ for each partition $\lambda$%
\ \ \ \ \footnote{\textit{Proof.} Let $\lambda$ be a partition. Thus,
$\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(
\lambda\right)  }\right)  $, so that $\left\vert \lambda\right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell\left(  \lambda\right)  }$.
\par
Let $i\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $. Then,
the number $\lambda_{i}$ is a positive integer (by the definition of
$\ell\left(  \lambda\right)  $). But recall that $v_{n}\in\Lambda_{n}$ for
each positive integer $n$. Applying this to $n=\lambda_{i}$, we obtain
$v_{\lambda_{i}}\in\Lambda_{\lambda_{i}}$ (since $\lambda_{i}$ is a positive
integer). In other words, the symmetric function $v_{\lambda_{i}}$ is
homogeneous of degree $\lambda_{i}$.
\par
Forget that we fixed $i$. We thus have showed that the symmetric function
$v_{\lambda_{i}}$ is homogeneous of degree $\lambda_{i}$ for each
$i\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $. Hence, the
product $v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(
\lambda\right)  }}$ of these $\ell\left(  \lambda\right)  $ symmetric
functions is homogeneous of degree%
\begin{align*}
&  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell\left(  \lambda\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Lambda\text{ is a graded
}\kk\text{-algebra}\right) \\
&  =\left\vert \lambda\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left\vert \lambda\right\vert =\lambda_{1}+\lambda_{2}+\cdots+\lambda
_{\ell\left(  \lambda\right)  }\right)  .
\end{align*}
In other words, $v_{\lambda_{1}}v_{\lambda_{2}}\cdots v_{\lambda_{\ell\left(
\lambda\right)  }}\in\Lambda_{\left\vert \lambda\right\vert }$. But our
definition of $v_{\lambda}$ yields $v_{\lambda}=v_{\lambda_{1}}v_{\lambda_{2}%
}\cdots v_{\lambda_{\ell\left(  \lambda\right)  }}\in\Lambda_{\left\vert
\lambda\right\vert }$. In other words, $v_{\lambda}$ is an element of
$\Lambda_{\left\vert \lambda\right\vert }$. Qed.}. Hence, Exercise
\ref{exe.Lambda.basis-vlam} yields that the family $\left(  v_{\lambda
}\right)  _{\lambda\in\Par}$ is a graded basis of the graded
$\kk$-module $\Lambda$ (where the indexing set $\Par$ is
partitioned into $\Par_{0},\Par_{1},\Par_{2},\ldots$). This solves
Exercise \ref{exe.Lambda.alggen-ind}(b).

(a) The family $\left(  v_{\lambda}\right)  _{\lambda\in\Par}$
is a basis of the $\kk$-module $\Lambda$ (since it is a graded basis of
the graded $\kk$-module $\Lambda$). Hence, in particular, this family
$\left(  v_{\lambda}\right)  _{\lambda\in\Par}$ is $\kk%
$-linearly independent.

But Exercise \ref{exe.alggen-via-span}(c) (applied to $A=\Lambda$) yields that
the elements $v_{1},v_{2},v_{3},\ldots$ are algebraically independent over
$\kk$ if and only if the family $\left(  v_{\lambda}\right)
_{\lambda\in\Par}$ is $\kk$-linearly independent. Hence,
the elements $v_{1},v_{2},v_{3},\ldots$ are algebraically independent over
$\kk$ (since the family $\left(  v_{\lambda}\right)  _{\lambda
\in\Par}$ is $\kk$-linearly independent). This solves
Exercise \ref{exe.Lambda.alggen-ind}(a).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.alggen-crit1}}

\textit{Solution to Exercise \ref{exe.Lambda.alggen-crit1}.} Let $V$ denote
the $\kk$-subalgebra of $\Lambda$ generated by $v_{1},v_{2}%
,v_{3},\ldots$. Thus, $V$ is a $\kk$-submodule of $\Lambda$.

We shall now show that%
\begin{equation}
h_{n}\in V\ \ \ \ \ \ \ \ \ \ \text{for each positive integer }n.
\label{sol.Lambda.alggen-crit1.1}%
\end{equation}


[\textit{Proof of (\ref{sol.Lambda.alggen-crit1.1}):} We shall prove
(\ref{sol.Lambda.alggen-crit1.1}) by strong induction on $n$:

\textit{Induction step:} Let $m$ be a positive integer. Assume that
(\ref{sol.Lambda.alggen-crit1.1}) holds for all $n<m$. We must prove that
(\ref{sol.Lambda.alggen-crit1.1}) holds for $n=m$. In other words, we must
prove that $h_{m}\in V$.

We have assumed that (\ref{sol.Lambda.alggen-crit1.1}) holds for all $n<m$. In
other words, we have%
\begin{equation}
h_{n}\in V\ \ \ \ \ \ \ \ \ \ \text{for each positive integer }n<m.
\label{sol.Lambda.alggen-crit1.1.pf.IH}%
\end{equation}


We have assumed that the element $a_{\left(  n\right)  }\in\kk$ is
invertible for each positive integer $n$. Applying this to $n=m$, we conclude
that the element $a_{\left(  m\right)  }\in\kk$ is invertible. Hence,
its inverse $a_{\left(  m\right)  }^{-1}\in\kk$ is well-defined.

But we have also assumed that each positive integer $n$ satisfies $v_{n}%
=\sum_{\lambda\in\Par_{n}}a_{\lambda}h_{\lambda}$.
Applying this to $n=m$, we find%
\[
v_{m}=\sum_{\lambda\in\Par_{m}}a_{\lambda}h_{\lambda
}=a_{\left(  m\right)  }h_{\left(  m\right)  }+\sum\limits_{\substack{\lambda
\in\Par_{m};\\\lambda\neq\left(  m\right)
}}a_{\lambda}h_{\lambda}%
\]
(here, we have split off the addend for $\lambda=\left(  m\right)  $ from the
sum, since $\left(  m\right)  \in\Par_{m}$). Hence,%
\begin{equation}
a_{\left(  m\right)  }h_{\left(  m\right)  }=v_{m}-\sum\limits_{\substack{\lambda
\in\Par_{m};\\\lambda\neq\left(  m\right)
}}a_{\lambda}h_{\lambda}.\label{sol.Lambda.alggen-crit1.1.pf.3}%
\end{equation}
But each $\lambda\in\Par_{m}$ satisfying $\lambda
\neq\left(  m\right)  $ satisfies $a_{\lambda}h_{\lambda}\in V$%
\ \ \ \ \footnote{\textit{Proof.} Let $\lambda\in\Par_{m}$ satisfy
$\lambda\neq\left(  m\right)  $. We must show that
$a_{\lambda}h_{\lambda}\in V$.
\par
Write the partition $\lambda$ in the form $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{\ell}\right)  $ with $\ell=\ell\left(
\lambda\right)  $. Then, the definition of $h_{\lambda}$ yields $h_{\lambda
}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}$.
\par
From $\lambda\in\Par_{m}$, we conclude that
$m=\left\vert \lambda\right\vert =\lambda_{1}+\lambda_{2}+\cdots+\lambda
_{\ell}$ (since $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell
}\right)  $).
\par
Assume (for the sake of contradiction) that $\ell=1$. Then, we have
$m=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}=\lambda_{1}$ (since $\ell
=1$), so that $\lambda_{1}=m$. Moreover, from $\ell=1$, we obtain $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  =\left(  \lambda
_{1}\right)  =\left(  m\right)  $ (since $\lambda_{1}=m$), and therefore
$\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)
=\left(  m\right)  $. This contradicts $\lambda\neq\left(  m\right)  $.
\par
This contradiction shows that our assumption was wrong. Hence, we must have
$\ell\neq1$.
\par
Now, let $i\in\left\{  1,2,\ldots,\ell\right\}  $ be arbitrary. Then,
$i\leq\ell=\ell\left(  \lambda\right)  $. Hence, $\lambda_{i}$ is a positive
integer.
\par
Also, from $i\in\left\{  1,2,\ldots,\ell\right\}  $, we obtain $1\leq
i\leq\ell$, so that $\ell\geq1$. Combining this with $\ell\neq1$, we obtain
$\ell>1$, so that $\ell\geq2$. Hence, $2\leq\ell=\ell\left(  \lambda\right)
$, so that $\lambda_{2}>0$ (by the definition of $\ell\left(  \lambda\right)
$).
\par
But $\lambda$ is a partition; thus, $\lambda_{1}\geq\lambda_{2}\geq\cdots
\geq\lambda_{\ell}$. Hence, $\lambda_{1}\geq\lambda_{i}$ (since $1\leq
i\leq\ell$), so that $\lambda_{i}\leq\lambda_{1}$. Also,%
\[
m=\left\vert \lambda\right\vert =\lambda_{1}+\lambda_{2}+\lambda_{3}%
+\cdots=\lambda_{1}+\underbrace{\lambda_{2}}_{>0}+\underbrace{\left(
\lambda_{3}+\lambda_{4}+\lambda_{5}+\cdots\right)  }_{\geq0}>\lambda_{1}.
\]
In other words, $\lambda_{1}<m$. Thus, $\lambda_{i}\leq\lambda_{1}<m$. Hence,
(\ref{sol.Lambda.alggen-crit1.1.pf.IH}) (applied to $n=\lambda_{i}$) yields
$h_{\lambda_{i}}\in V$ (since $\lambda_{i}$ is a positive integer).
\par
Forget that we fixed $i$. We thus have shown that $h_{\lambda_{i}}\in V$ for
each $i\in\left\{  1,2,\ldots,\ell\right\}  $. In other words, $h_{\lambda
_{1}},h_{\lambda_{2}},\ldots,h_{\lambda_{\ell}}$ are elements of $V$. Thus,
$h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}$ is a product of
elements of $V$, and therefore must itself be an element of $V$ (since $V$ is
a $\kk$-subalgebra of $\Lambda$). In other words, $h_{\lambda_{1}%
}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}\in V$. Now, $h_{\lambda}%
=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}\in V$. Hence,
$a_{\lambda}\underbrace{h_{\lambda}}_{\in V}\in a_{\lambda}V\subset V$ (since
$V$ is a $\kk$-module). Qed.}. Hence,%
\[
\sum\limits_{\substack{\lambda\in\Par_{m};\\\lambda
\neq\left(  m\right)  }}\underbrace{a_{\lambda}h_{\lambda}}_{\in V}\in
\sum\limits_{\substack{\lambda\in\Par_{m};\\\lambda
\neq\left(  m\right)  }}V\subset V\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}V\text{ is a }\kk\text{-module}\right)  .
\]


Furthermore, $V$ contains $v_{1},v_{2},v_{3},\ldots$ (since $V$ is the
$\kk$-subalgebra of $\Lambda$ generated by $v_{1},v_{2},v_{3},\ldots$).
Thus, in particular, $V$ contains $v_{m}$ (since $m$ is a positive integer).
In other words, $v_{m}\in V$.

Now, (\ref{sol.Lambda.alggen-crit1.1.pf.3}) becomes%
\[
a_{\left(  m\right)  }h_{\left(  m\right)  }=\underbrace{v_{m}}_{\in
V}-\underbrace{\sum\limits_{\substack{\lambda\in\Par_{m}%
;\\\lambda\neq\left(  m\right)  }}a_{\lambda}h_{\lambda}}_{\in V}\in
V-V\subset V\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V\text{ is a }%
\kk\text{-module}\right)  .
\]
Now, since $a_{\left(  m\right)  }^{-1}$ is well-defined, we have%
\[
h_{\left(  m\right)  }=a_{\left(  m\right)  }^{-1}\underbrace{a_{\left(
m\right)  }h_{\left(  m\right)  }}_{\in V}\in a_{\left(  m\right)  }%
^{-1}V\subset V\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V\text{ is a
}\kk\text{-module}\right)  .
\]
In view of $h_{\left(  m\right)  }=h_{m}$, this rewrites as $h_{m}\in V$. In
other words, (\ref{sol.Lambda.alggen-crit1.1}) holds for $n=m$. This completes
the induction step. Thus, (\ref{sol.Lambda.alggen-crit1.1}) is proved by
strong induction.]

Let us now draw some conclusions. We can restate
(\ref{sol.Lambda.alggen-crit1.1}) as follows: The elements $h_{1},h_{2}%
,h_{3},\ldots$ all belong to $V$. In other words, $V$ contains $h_{1}%
,h_{2},h_{3},\ldots$.

Proposition \ref{symm-antipode-and-h-basis} shows that the family $\left(
h_{n}\right)  _{n=1,2,\ldots}$ generates the $\kk$-algebra
$\Lambda_{\kk}$. In other words, the family $\left(  h_{n}\right)
_{n=1,2,\ldots}$ generates the $\kk$-algebra $\Lambda$ (since
$\Lambda_{\kk}=\Lambda$). In other words, the elements $h_{1}%
,h_{2},h_{3},\ldots$ generate the $\kk$-algebra $\Lambda$. Thus, the
only $\kk$-subalgebra of $\Lambda$ that contains these elements
$h_{1},h_{2},h_{3},\ldots$ is the whole algebra $\Lambda$. In other words,%
\[
\left(  \text{if }B\text{ is any }\kk\text{-subalgebra of }%
\Lambda\text{ that contains }h_{1},h_{2},h_{3},\ldots\text{, then }%
B=\Lambda\right)  .
\]
We can apply this to $B=V$ (since $V$ is a $\kk$-subalgebra of
$\Lambda$ that contains $h_{1},h_{2},h_{3},\ldots$), and thus conclude that
$V=\Lambda$. In other words, the $\kk$-subalgebra of $\Lambda$
generated by $v_{1},v_{2},v_{3},\ldots$ is $\Lambda$ (since $V$ was defined to
be the $\kk$-subalgebra of $\Lambda$ generated by $v_{1},v_{2}%
,v_{3},\ldots$). In other words, the elements $v_{1},v_{2},v_{3},\ldots$
generate the $\kk$-algebra $\Lambda$. Moreover, we have $v_{n}%
\in\Lambda_{n}$ for each positive integer $n$\ \ \ \ \footnote{\textit{Proof.}
We know that $h_{n}\in\Lambda_{n}$ for each positive integer $n$. Furthermore,
we know that the family $\left(  h_{n}\right)  _{n=1,2,\ldots}$ generates the
$\kk$-algebra $\Lambda$; in other words, $h_{1},h_{2},h_{3},\ldots$
generate the $\kk$-algebra $\Lambda$. Finally, for every partition
$\lambda$, the symmetric function $h_{\lambda}\in\Lambda$ is defined by
$h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell\left(
\lambda\right)  }}$. Hence, Exercise \ref{exe.Lambda.alggen-ind}(b) (applied
to $h_{n}$ and $h_{\lambda}$ instead of $v_{n}$ and $v_{\lambda}$) shows that
the family $\left(  h_{\lambda}\right)  _{\lambda\in\Par}$ is a
graded basis of the graded $\kk$-module $\Lambda$. In other words, for
each $n\in \NN$,
\begin{equation}
\text{the family }\left(  h_{\lambda}\right)  _{\lambda\in\Par_{n}}
\text{ is a basis of the }\kk\text{-module }\Lambda
_{n}.\label{sol.Lambda.alggen-crit1.1.pf.fn4.1}%
\end{equation}
\par
Now, let $n$ be a positive integer. Then,
(\ref{sol.Lambda.alggen-crit1.1.pf.fn4.1}) shows that the family $\left(
h_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a basis
of the $\kk$-module $\Lambda_{n}$. Hence,
\begin{equation}
h_{\lambda}\in\Lambda_{n}\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda
\in\Par_{n}.\label{sol.Lambda.alggen-crit1.1.pf.fn4.2}%
\end{equation}
But by assumption, we have%
\[
v_{n}=\sum_{\lambda\in\Par_{n}}a_{\lambda
}\underbrace{h_{\lambda}}_{\substack{\in\Lambda_{n}\\\text{(by
(\ref{sol.Lambda.alggen-crit1.1.pf.fn4.2}))}}}\in\sum_{\lambda\in
\Par_{n}}a_{\lambda}\Lambda_{n}\subset\Lambda
_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Lambda_{n}\text{ is a
}\kk\text{-module}\right)  .
\]
Qed.}. Thus, Exercise \ref{exe.Lambda.alggen-ind}(a) shows that $v_{1}%
,v_{2},v_{3},\ldots$ are algebraically independent over $\kk$. This
completes the solution to Exercise \ref{exe.Lambda.alggen-crit1}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.alggen-crit1}}

\textit{Solution to Exercise \ref{exe.Lambda.alggen-crit1}.} Let $V$ denote
the $\kk$-subalgebra of $\Lambda$ generated by $v_{1},v_{2}%
,v_{3},\ldots$. Thus, $V$ is a $\kk$-submodule of $\Lambda$.

We shall now show that%
\begin{equation}
h_{n}\in V\ \ \ \ \ \ \ \ \ \ \text{for each positive integer }n.
\label{sol.Lambda.alggen-crit1.1}%
\end{equation}


[\textit{Proof of (\ref{sol.Lambda.alggen-crit1.1}):} We shall prove
(\ref{sol.Lambda.alggen-crit1.1}) by strong induction on $n$:

\textit{Induction step:} Let $m$ be a positive integer. Assume that
(\ref{sol.Lambda.alggen-crit1.1}) holds for all $n<m$. We must prove that
(\ref{sol.Lambda.alggen-crit1.1}) holds for $n=m$. In other words, we must
prove that $h_{m}\in V$.

We have assumed that (\ref{sol.Lambda.alggen-crit1.1}) holds for all $n<m$. In
other words, we have%
\begin{equation}
h_{n}\in V\ \ \ \ \ \ \ \ \ \ \text{for each positive integer }n<m.
\label{sol.Lambda.alggen-crit1.1.pf.IH}%
\end{equation}


We know that $m$ is a positive integer. Hence, $\left(  m\right)  $ is a
partition of $m$. In other words, $\left(  m\right)  \in\Par_{m}$.

We have assumed that the element $a_{\left(  n\right)  }\in\kk$ is
invertible for each positive integer $n$. Applying this to $n=m$, we conclude
that the element $a_{\left(  m\right)  }\in\kk$ is invertible. Hence,
its inverse $a_{\left(  m\right)  }^{-1}\in\kk$ is well-defined.

But we have also assumed that each positive integer $n$ satisfies $v_{n}%
=\sum_{\lambda\in\Par_{n}}a_{\lambda}h_{\lambda}$.
Applying this to $n=m$, we find%
\[
v_{m}=\sum_{\lambda\in\Par_{m}}a_{\lambda}h_{\lambda
}=a_{\left(  m\right)  }h_{\left(  m\right)  }+\sum\limits_{\substack{\lambda
\in\Par_{m};\\\lambda\neq\left(  m\right)
}}a_{\lambda}h_{\lambda}%
\]
(here, we have split off the addend for $\lambda=\left(  m\right)  $ from the
sum, since $\left(  m\right)  \in\Par_{m}$). Hence,%
\begin{equation}
a_{\left(  m\right)  }h_{\left(  m\right)  }=v_{m}-\sum\limits_{\substack{\lambda
\in\Par_{m};\\\lambda\neq\left(  m\right)
}}a_{\lambda}h_{\lambda}. \label{sol.Lambda.alggen-crit1.1.pf.3}%
\end{equation}
But each $\lambda\in\Par_{m}$ satisfying $\lambda
\neq\left(  m\right)  $ satisfies $a_{\lambda}h_{\lambda}\in V$%
\ \ \ \ \footnote{\textit{Proof.} Let $\lambda\in\Par_{m}$ satisfy
$\lambda\neq\left(  m\right)  $. We must show that
$a_{\lambda}h_{\lambda}\in V$.
\par
Clearly, $\lambda$ is a partition (since $\lambda\in\Par_{m}$). Write
this partition $\lambda$ in the form $\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)  $ with $\ell=\ell\left(
\lambda\right)  $. Then, the definition of $h_{\lambda}$ yields $h_{\lambda
}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}$.
\par
But from $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell
}\right)  $, we obtain $\left\vert \lambda\right\vert =\lambda_{1}+\lambda
_{2}+\cdots+\lambda_{\ell}$. But $\left\vert \lambda\right\vert =m$ (since
$\lambda\in\Par_{m}$), so that $m=\left\vert
\lambda\right\vert =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}$.
\par
Assume (for the sake of contradiction) that $\ell=1$. Thus, $\lambda
_{1}+\lambda_{2}+\cdots+\lambda_{\ell}=\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{1}=\lambda_{1}$, so that $m=\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{\ell}=\lambda_{1}$. Thus, $\lambda_{1}=m$. Moreover, from $\ell=1$,
we obtain $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right)
=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{1}\right)  =\left(
\lambda_{1}\right)  =\left(  m\right)  $ (since $\lambda_{1}=m$), and
therefore $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell
}\right)  =\left(  m\right)  $. This contradicts $\lambda\neq\left(  m\right)
$.
\par
This contradiction shows that our assumption was wrong. Hence, we cannot have
$\ell=1$. In other words, $\ell\neq1$.
\par
Now, let $i\in\left\{  1,2,\ldots,\ell\right\}  $ be arbitrary. Then,
$i\leq\ell=\ell\left(  \lambda\right)  $. Hence, $\lambda_{i}$ is a positive
integer (by the definition of $\ell\left(  \lambda\right)  $).
\par
Also, from $i\in\left\{  1,2,\ldots,\ell\right\}  $, we obtain $1\leq
i\leq\ell$, so that $\ell\geq1$. Combining this with $\ell\neq1$, we obtain
$\ell>1$, so that $\ell\geq2$ (since $\ell$ is an integer). Hence, $2\leq
\ell=\ell\left(  \lambda\right)  $, so that $\lambda_{2}>0$ (by the definition
of $\ell\left(  \lambda\right)  $).
\par
But $\lambda$ is a partition; thus, $\lambda_{1}\geq\lambda_{2}\geq\cdots
\geq\lambda_{\ell}$. Hence, $\lambda_{1}\geq\lambda_{i}$ (since $1\leq
i\leq\ell$), so that $\lambda_{i}\leq\lambda_{1}$. Also,%
\begin{align*}
m  &  =\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}=\left(  \lambda
_{1}+\lambda_{2}\right)  +\underbrace{\left(  \lambda_{3}+\lambda_{4}%
+\cdots+\lambda_{\ell}\right)  }_{\substack{\geq0\\\text{(since }\lambda
_{3},\lambda_{4},\ldots,\lambda_{\ell}\text{ are nonnegative)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\geq2\right) \\
&  \geq\lambda_{1}+\underbrace{\lambda_{2}}_{>0}>\lambda_{1}.
\end{align*}
In other words, $\lambda_{1}<m$. Thus, $\lambda_{i}\leq\lambda_{1}<m$. Hence,
(\ref{sol.Lambda.alggen-crit1.1.pf.IH}) (applied to $n=\lambda_{i}$) yields
$h_{\lambda_{i}}\in V$ (since $\lambda_{i}$ is a positive integer).
\par
Forget that we fixed $i$. We thus have shown that $h_{\lambda_{i}}\in V$ for
each $i\in\left\{  1,2,\ldots,\ell\right\}  $. In other words, $h_{\lambda
_{1}},h_{\lambda_{2}},\ldots,h_{\lambda_{\ell}}$ are elements of $V$. Thus,
$h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}$ is a product of
elements of $V$. But any product of elements of $V$ must itself belong to $V$
(since $V$ is a $\kk$-subalgebra of $\Lambda$). Hence, $h_{\lambda_{1}%
}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}$ must belong to $V$ (since
$h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}$ is a product of
elements of $V$). In other words, $h_{\lambda_{1}}h_{\lambda_{2}}\cdots
h_{\lambda_{\ell}}\in V$. Now, $h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}%
}\cdots h_{\lambda_{\ell}}\in V$. Hence, $a_{\lambda}\underbrace{h_{\lambda}%
}_{\in V}\in a_{\lambda}V\subset V$ (since $V$ is a $\kk$-module).
Qed.}. Hence,%
\[
\sum\limits_{\substack{\lambda\in\Par_{m};\\\lambda
\neq\left(  m\right)  }}\underbrace{a_{\lambda}h_{\lambda}}_{\in V}\in
\sum\limits_{\substack{\lambda\in\Par_{m};\\\lambda
\neq\left(  m\right)  }}V\subset V\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}V\text{ is a }\kk\text{-module}\right)  .
\]


Furthermore, $V$ contains $v_{1},v_{2},v_{3},\ldots$ (since $V$ is the
$\kk$-subalgebra of $\Lambda$ generated by $v_{1},v_{2},v_{3},\ldots$).
Thus, in particular, $V$ contains $v_{m}$ (since $m$ is a positive integer).
In other words, $v_{m}\in V$.

Now, (\ref{sol.Lambda.alggen-crit1.1.pf.3}) becomes%
\[
a_{\left(  m\right)  }h_{\left(  m\right)  }=\underbrace{v_{m}}_{\in
V}-\underbrace{\sum\limits_{\substack{\lambda\in\Par_{m}%
;\\\lambda\neq\left(  m\right)  }}a_{\lambda}h_{\lambda}}_{\in V}\in
V-V\subset V\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V\text{ is a }%
\kk\text{-module}\right)  .
\]
Now, since $a_{\left(  m\right)  }^{-1}$ is well-defined, we have%
\[
h_{\left(  m\right)  }=a_{\left(  m\right)  }^{-1}\underbrace{a_{\left(
m\right)  }h_{\left(  m\right)  }}_{\in V}\in a_{\left(  m\right)  }%
^{-1}V\subset V\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V\text{ is a
}\kk\text{-module}\right)  .
\]
In view of $h_{\left(  m\right)  }=h_{m}$, this rewrites as $h_{m}\in V$. In
other words, (\ref{sol.Lambda.alggen-crit1.1}) holds for $n=m$. This completes
the induction step. Thus, (\ref{sol.Lambda.alggen-crit1.1}) is proved by
strong induction.]

Let us now draw some conclusions. We can restate
(\ref{sol.Lambda.alggen-crit1.1}) as follows: The elements $h_{1},h_{2}%
,h_{3},\ldots$ all belong to $V$. In other words, $V$ contains $h_{1}%
,h_{2},h_{3},\ldots$.

Proposition \ref{symm-antipode-and-h-basis} shows that the family $\left(
h_{n}\right)  _{n=1,2,\ldots}$ generates the $\kk$-algebra
$\Lambda_{\kk}$. In other words, the family $\left(  h_{n}\right)
_{n=1,2,\ldots}$ generates the $\kk$-algebra $\Lambda$ (since
$\Lambda_{\kk}=\Lambda$). In other words, the $\kk$-algebra
$\Lambda$ is generated by $h_{1},h_{2},h_{3},\ldots$. In other words, the
$\kk$-subalgebra of $\Lambda$ generated by $h_{1},h_{2},h_{3},\ldots$
is $\Lambda$ itself.

But the $\kk$-subalgebra of $\Lambda$ generated by $h_{1},h_{2}%
,h_{3},\ldots$ is the smallest $\kk$-subalgebra of $\Lambda$ that
contains $h_{1},h_{2},h_{3},\ldots$. In other words, $\Lambda$ is the smallest
$\kk$-subalgebra of $\Lambda$ that contains $h_{1},h_{2},h_{3},\ldots$
(because we know that the $\kk$-subalgebra of $\Lambda$ generated by
$h_{1},h_{2},h_{3},\ldots$ is $\Lambda$ itself). Thus, in particular,
$\Lambda$ is the \textbf{smallest} such $\kk$-subalgebra. Hence,%
\[
\left(  \text{if }B\text{ is any }\kk\text{-subalgebra of }%
\Lambda\text{ that contains }h_{1},h_{2},h_{3},\ldots\text{, then }%
\Lambda\subset B\right)  .
\]
We can apply this to $B=V$ (since $V$ is a $\kk$-subalgebra of
$\Lambda$ that contains $h_{1},h_{2},h_{3},\ldots$), and thus conclude that
$\Lambda\subset V$. Combining this with $V\subset\Lambda$ (which is obvious),
we obtain $V=\Lambda$. In other words, $V$ is $\Lambda$. In other words, the
$\kk$-subalgebra of $\Lambda$ generated by $v_{1},v_{2},v_{3},\ldots$
is $\Lambda$ (since $V$ was defined to be the $\kk$-subalgebra of
$\Lambda$ generated by $v_{1},v_{2},v_{3},\ldots$). In other words, the
elements $v_{1},v_{2},v_{3},\ldots$ generate the $\kk$-algebra
$\Lambda$. Moreover, we have $v_{n}\in\Lambda_{n}$ for each positive integer
$n$\ \ \ \ \footnote{\textit{Proof.} We know that $h_{n}\in\Lambda_{n}$ for
each positive integer $n$. Furthermore, we know that the family $\left(
h_{n}\right)  _{n=1,2,\ldots}$ generates the $\kk$-algebra $\Lambda$;
in other words, $h_{1},h_{2},h_{3},\ldots$ generate the $\kk$-algebra
$\Lambda$. Finally, for every partition $\lambda$, the symmetric function
$h_{\lambda}\in\Lambda$ is defined by $h_{\lambda}=h_{\lambda_{1}}%
h_{\lambda_{2}}\cdots h_{\lambda_{\ell\left(  \lambda\right)  }}$. Hence,
Exercise \ref{exe.Lambda.alggen-ind}(b) (applied to $h_{n}$ and $h_{\lambda}$
instead of $v_{n}$ and $v_{\lambda}$) shows that the family $\left(
h_{\lambda}\right)  _{\lambda\in\Par}$ is a graded basis of the
graded $\kk$-module $\Lambda$. In other words, for each $n\in
\mathbb{N}$,
\begin{equation}
\text{the family }\left(  h_{\lambda}\right)  _{\lambda\in\Par_{n}}
\text{ is a basis of the }\kk\text{-module }\Lambda_{n}.
\label{sol.Lambda.alggen-crit1.1.pf.fn4.1}%
\end{equation}
\par
Now, let $n$ be a positive integer. Then,
(\ref{sol.Lambda.alggen-crit1.1.pf.fn4.1}) shows that the family $\left(
h_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a basis
of the $\kk$-module $\Lambda_{n}$. Hence,
\begin{equation}
h_{\lambda}\in\Lambda_{n}\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda
\in\Par_{n}.
\label{sol.Lambda.alggen-crit1.1.pf.fn4.2}%
\end{equation}
But by assumption, we have%
\[
v_{n}=\sum_{\lambda\in\Par_{n}}a_{\lambda
}\underbrace{h_{\lambda}}_{\substack{\in\Lambda_{n}\\\text{(by
(\ref{sol.Lambda.alggen-crit1.1.pf.fn4.2}))}}}\in\sum_{\lambda\in
\Par_{n}}a_{\lambda}\Lambda_{n}\subset\Lambda
_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Lambda_{n}\text{ is a
}\kk\text{-module}\right)  .
\]
Qed.}. Thus, Exercise \ref{exe.Lambda.alggen-ind}(a) shows that $v_{1}%
,v_{2},v_{3},\ldots$ are algebraically independent over $\kk$. This
completes the solution to Exercise \ref{exe.Lambda.alggen-crit1}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.alggen-crit1hall}}

\textit{Solution to Exercise \ref{exe.Lambda.alggen-crit1hall}.} Corollary
\ref{dual-bases-h-m-p}(a) says that the bases $\left(  h_{\lambda}\right)
_{\lambda\in\Par}$ and $\left(  m_{\lambda}\right)
_{\lambda\in\Par}$ of $\Lambda$ are dual with respect to the
Hall inner product on $\Lambda$. In other words,%
\begin{equation}
\left(  h_{\lambda},m_{\mu}\right)  =\delta_{\lambda,\mu}%
\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par\text{ and
}\mu\in\Par. \label{sol.Lambda.alggen-crit1hall.dual}%
\end{equation}


Proposition \ref{symm-antipode-and-h-basis} shows that the family $\left(
h_{n}\right)  _{n=1,2,\ldots}$ generates the $\kk$-algebra
$\Lambda_{\kk}$. In other words, the family $\left(  h_{n}\right)
_{n=1,2,\ldots}$ generates the $\kk$-algebra $\Lambda$ (since
$\Lambda_{\kk}=\Lambda$). In other words, $h_{1},h_{2},h_{3},\ldots$
generate the $\kk$-algebra $\Lambda$. Furthermore, $h_{n}\in\Lambda
_{n}$ for each positive integer $n$. Finally, for every partition $\lambda$,
the symmetric function $h_{\lambda}\in\Lambda$ is defined by $h_{\lambda
}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell\left(  \lambda\right)
}}$. Hence, Exercise \ref{exe.Lambda.alggen-ind}(b) (applied to $h_{n}$ and
$h_{\lambda}$ instead of $v_{n}$ and $v_{\lambda}$) shows that the family
$\left(  h_{\lambda}\right)  _{\lambda\in\Par}$ is a graded
basis of the graded $\kk$-module $\Lambda$. In other words, for each
$n\in \NN$,
\begin{equation}
\text{the family }\left(  h_{\lambda}\right)  _{\lambda\in\Par_{n}}
\text{ is a basis of the }\kk\text{-module }\Lambda_{n}.
\label{sol.Lambda.alggen-crit1hall.1}%
\end{equation}


Now, let $n$ be a positive integer. Then, $v_{n}\in\Lambda_{n}$ (by
assumption) and $\left(  n\right)  \in\Par_{n}$ (since
$\left(  n\right)  $ is a partition of $n$). But
(\ref{sol.Lambda.alggen-crit1hall.1}) shows that the family $\left(
h_{\lambda}\right)  _{\lambda\in\Par_{n}}$ is a basis
of the $\kk$-module $\Lambda_{n}$. Hence, this family spans
$\Lambda_{n}$. Thus, $v_{n}$ is a $\kk$-linear combination of the
family $\left(  h_{\lambda}\right)  _{\lambda\in\Par_{n}}$
(since $v_{n}\in\Lambda_{n}$). In other words, there exists
some family $\left(  b_{\lambda}\right)  _{\lambda\in\Par_{n}}
\in\kk^{\Par_{n}}$ of scalars
such that $v_{n}=\sum_{\lambda\in\Par_{n}}b_{\lambda
}h_{\lambda}$. Consider this family $\left(  b_{\lambda}\right)  _{\lambda
\in\Par_{n}}$.

Now, recall that the Hall inner product has an orthonormal basis (namely,
$\left(  s_{\lambda}\right)  _{\lambda\in\Par}$), and thus is
symmetric. Therefore, every $\mu\in\Par_{n}$ satisfies%
\begin{align*}
\left(  m_{\mu},v_{n}\right)   &  =\left(  \underbrace{v_{n}}_{=\sum
_{\lambda\in\Par_{n}}b_{\lambda}h_{\lambda}},m_{\mu
}\right)  =\left(  \sum_{\lambda\in\Par_{n}}%
b_{\lambda}h_{\lambda},m_{\mu}\right)  =\sum_{\lambda\in\Par_{n}}
b_{\lambda}\underbrace{\left(  h_{\lambda},m_{\mu}\right)
}_{\substack{=\delta_{\lambda,\mu}\\\text{(by
(\ref{sol.Lambda.alggen-crit1hall.dual}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk\text{-bilinear}\right)  \\
&  =\sum_{\lambda\in\Par_{n}}b_{\lambda}%
\delta_{\lambda,\mu}=b_{\mu}\underbrace{\delta_{\mu,\mu}}%
_{\substack{=1\\\text{(since }\mu=\mu\text{)}}}+\sum\limits_{\substack{\lambda
\in\Par_{n};\\\lambda\neq\mu}}b_{\lambda
}\underbrace{\delta_{\lambda,\mu}}_{\substack{=0\\\text{(since }\lambda\neq
\mu\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}\lambda=\mu\text{ from the sum, since }\mu\in\Par_{n}\right)  \\
&  =b_{\mu}+\underbrace{\sum\limits_{\substack{\lambda\in\Par_{n};\\
\lambda\neq\mu}}b_{\lambda}0}_{=0}=b_{\mu}.
\end{align*}
Renaming $\mu$ as $\lambda$ in this statement, we obtain the following: Every
$\lambda\in\Par_{n}$ satisfies $\left(  m_{\lambda
},v_{n}\right)  =b_{\lambda}$. Hence, every $\lambda\in\Par_{n}$
satisfies%
\begin{equation}
b_{\lambda}=\left(  m_{\lambda},v_{n}\right)  =\left(  m_{\lambda
},v_{\left\vert \lambda\right\vert }\right)
\label{sol.Lambda.alggen-crit1hall.2}%
\end{equation}
(since $n=\left\vert \lambda\right\vert $ (because $\lambda\in
\Par_{n}$ and thus $\left\vert \lambda\right\vert
=n$)). Now,%
\[
v_{n}=\sum_{\lambda\in\Par_{n}}\underbrace{b_{\lambda
}}_{\substack{=\left(  m_{\lambda},v_{\left\vert \lambda\right\vert }\right)
\\\text{(by (\ref{sol.Lambda.alggen-crit1hall.2}))}}}h_{\lambda}=\sum
_{\lambda\in\Par_{n}}\left(  m_{\lambda},v_{\left\vert
\lambda\right\vert }\right)  h_{\lambda}.
\]


Forget that we fixed $n$. Thus, we have showed that each positive integer $n$
satisfies $v_{n}=\sum_{\lambda\in\Par_{n}}\left(
m_{\lambda},v_{\left\vert \lambda\right\vert }\right)  h_{\lambda}$. Moreover,
the element $\left(  m_{\left(  n\right)  },v_{\left\vert \left(  n\right)
\right\vert }\right)  \in\kk$ is invertible for each positive integer
$n$\ \ \ \ \footnote{\textit{Proof.} We have assumed that $\left(  p_{n}%
,v_{n}\right)  \in\kk$ is invertible for each positive integer $n$. In
other words, $\left(  m_{\left(  n\right)  },v_{\left\vert \left(  n\right)
\right\vert }\right)  \in\kk$ is invertible for each positive integer
$n$ (since each positive integer $n$ satisfies $m_{\left(  n\right)  }=p_{n}$
and $\left\vert \left(  n\right)  \right\vert =n$).}. Hence, Exercise
\ref{exe.Lambda.alggen-crit1} (applied to $a_{\lambda}=\left(  m_{\lambda
},v_{\left\vert \lambda\right\vert }\right)  $) yields that the elements
$v_{1},v_{2},v_{3},\ldots$ generate the $\kk$-algebra $\Lambda$ and are
algebraically independent over $\kk$. This solves Exercise
\ref{exe.Lambda.alggen-crit1hall}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.hall-coefficient}}

\textit{Solution to Exercise \ref{exe.Lambda.hall-coefficient}.} Corollary
\ref{dual-bases-h-m-p}(a) yields that $\left(  h_{\lambda}\right)
_{\lambda\in\operatorname{Par}}$ and $\left(  m_{\lambda}\right)  _{\lambda
\in\operatorname{Par}}$ are dual bases with respect to the Hall inner product
on $\Lambda$.

Recall the following fundamental fact from linear algebra: If $\kk$ is
a commutative ring, if $A$ is a $\kk$-module, if $\left(  \cdot
,\cdot\right)  :A\times A\rightarrow \kk$ is a symmetric $\kk$-bilinear
form on $A$, and if $\left(  u_{\lambda}\right)  _{\lambda\in L}$
and $\left(  v_{\lambda}\right)  _{\lambda\in L}$ are two $\kk$-bases
of $A$ which are dual to each other with respect to the form $\left(
\cdot,\cdot\right)  $ (where $L$ is some indexing set), then every $a\in A$
satisfies
\[
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}.
\]
We can apply this fact to $A=\Lambda$, $L=\operatorname{Par}$, $\left(
u_{\lambda}\right)  _{\lambda\in L}=\left(  h_{\lambda}\right)  _{\lambda
\in\operatorname{Par}}$ and $\left(  v_{\lambda}\right)  _{\lambda\in
L}=\left(  m_{\lambda}\right)  _{\lambda\in\operatorname{Par}}$ (since the
bases $\left(  h_{\lambda}\right)  _{\lambda\in\operatorname{Par}}$ and
$\left(  m_{\lambda}\right)  _{\lambda\in\operatorname{Par}}$ of $\Lambda$ are
dual to each other with respect to the Hall inner product $\left(  \cdot
,\cdot\right)  $). As a result, we obtain that every $a\in\Lambda$ satisfies%
\begin{equation}
a=\sum_{\lambda\in\operatorname{Par}}\left(  h_{\lambda},a\right)  m_{\lambda
}. \label{sol.Lambda.hall-coefficient.a=sum}%
\end{equation}

\begin{vershort}
Now, let $\operatorname{WC}$ denote the set of all weak compositions.
Thus, $\beta \in \operatorname{WC}$.
Recall (from Section \ref{subsect.Sym-section.def}) the finitary
symmetric group $\Symm_{\left(  \infty\right)  }$ and its action on
the set $\operatorname{WC}$ of all weak compositions. Applying the equality
(\ref{sol.Lambda.hall-coefficient.a=sum}) to $a=f$, we obtain%
\begin{align*}
f  &  =\sum\limits_{\lambda\in\operatorname{Par}}\left(  h_{\lambda},f\right)
\underbrace{m_{\lambda}}_{\substack{=\sum\limits_{\alpha\in
\Symm_{\left(  \infty\right)  }\lambda}\mathbf{x}^{\alpha}\\
\text{(by \eqref{monomial-symmetric-function-definition})}}}
=\sum\limits_{\lambda\in\operatorname{Par}}
\left(  h_{\lambda},f\right)
\sum\limits_{\alpha\in\Symm_{\left(  \infty\right)  }\lambda}
\mathbf{x}^{\alpha}
\\
&  =\underbrace{\sum\limits_{\lambda\in\operatorname{Par}}\ \ \sum
\limits_{\alpha\in\Symm_{\left(  \infty\right)  }\lambda}
}_{=\sum\limits_{\alpha\in
\operatorname*{WC}}\ \ \sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}}\left(
h_{\lambda},f\right)  \mathbf{x}^{\alpha}=\sum\limits_{\alpha\in
\operatorname*{WC}}\ \ \sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\left(  h_{\lambda
},f\right)  \mathbf{x}^{\alpha}.
\end{align*}
Hence,
\begin{align}
\left(  \text{the coefficient of }\mathbf{x}^{\beta}\text{ in }f\right)   &
=\left(  \text{the coefficient of }\mathbf{x}^{\beta}\text{ in }%
\sum\limits_{\alpha\in\operatorname*{WC}}
\sum\limits_{\substack{\lambda\in\operatorname{Par};\\
\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}
\left(  h_{\lambda},f\right) \mathbf{x}^{\alpha}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\\\beta\in
\Symm_{\left(  \infty\right)  }\lambda}}\left(  h_{\lambda},f\right)  .
\label{sol.Lambda.hall-coefficient.3}
\end{align}
\end{vershort}

\begin{verlong}
Now, let $\operatorname{WC}$ denote the set of all weak compositions.
Thus, $\beta \in \operatorname{WC}$ (since $\beta$ is a weak composition).
Recall (from Section \ref{subsect.Sym-section.def}) the finitary
symmetric group $\Symm_{\left(  \infty\right)  }$ and its action on
$\operatorname*{WC}$. Applying the equality
(\ref{sol.Lambda.hall-coefficient.a=sum}) to $a=f$, we obtain%
\begin{align*}
f  &  =\sum\limits_{\lambda\in\operatorname{Par}}\left(  h_{\lambda},f\right)
\underbrace{m_{\lambda}}_{\substack{=\sum\limits_{\alpha\in
\Symm_{\left(  \infty\right)  }\lambda}\mathbf{x}^{\alpha}\\
\text{(by \eqref{monomial-symmetric-function-definition})}}}
=\sum\limits_{\lambda
\in\operatorname{Par}}\left(  h_{\lambda},f\right)  \underbrace{\sum
\limits_{\alpha\in\Symm_{\left(  \infty\right)  }\lambda}%
}_{\substack{=\sum\limits_{\substack{\alpha\in\operatorname*{WC};\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}\\\text{(since }%
\Symm_{\left(  \infty\right)  }\lambda\subset\operatorname*{WC}%
\text{)}}}\mathbf{x}^{\alpha}=\sum\limits_{\lambda\in\operatorname{Par}%
}\left(  h_{\lambda},f\right)  \sum\limits_{\substack{\alpha\in
\operatorname*{WC};\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda
}}\mathbf{x}^{\alpha}\\
&  =\underbrace{\sum\limits_{\lambda\in\operatorname{Par}}\ \ \sum
\limits_{\substack{\alpha\in\operatorname*{WC};\\\alpha\in
\Symm_{\left(  \infty\right)  }\lambda}}}_{=\sum\limits_{\alpha\in
\operatorname*{WC}}\ \ \sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}}\left(
h_{\lambda},f\right)  \mathbf{x}^{\alpha}=\sum\limits_{\alpha\in
\operatorname*{WC}}\ \ \sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\\\alpha\in\Symm_{\left(  \infty\right)  }\lambda}}\left(  h_{\lambda
},f\right)  \mathbf{x}^{\alpha}=\sum\limits_{\alpha\in\operatorname*{WC}%
}\left(  \sum\limits_{\substack{\lambda\in\operatorname{Par};\\\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}}\left(  h_{\lambda
},f\right)  \right)  \mathbf{x}^{\alpha}.
\end{align*}
Hence,
\begin{align}
\left(  \text{the coefficient of }\mathbf{x}^{\beta}\text{ in }f\right)   &
=\left(  \text{the coefficient of }\mathbf{x}^{\beta}\text{ in }%
\sum\limits_{\alpha\in\operatorname*{WC}}\left(  \sum
\limits_{\substack{\lambda\in\operatorname{Par};\\\alpha\in
\Symm_{\left(  \infty\right)  }\lambda}}
\left(  h_{\lambda},f\right)  \right)
\mathbf{x}^{\alpha}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\\\beta\in
\Symm_{\left(  \infty\right)  }\lambda}}\left(  h_{\lambda},f\right)  .
\label{sol.Lambda.hall-coefficient.3}
\end{align}
\end{verlong}

Recall that $\mu$ is the partition consisting of the nonzero entries of
$\beta$ (sorted in decreasing order). Hence, the partition $\mu$ has the same
nonzero entries as the weak composition $\beta$ (just possibly in a different
order). Thus, the partition $\beta$ can be obtained by rearranging the entries
of the weak composition $\mu$. More precisely, we can transform $\mu$ into
$\beta$ by rearranging finitely many entries of $\mu$\ \ \ \ \footnote{Indeed,
there is an $i\in\left\{  1,2,3,\ldots\right\}  $ such that $\beta_{i+1}%
=\beta_{i+2}=\beta_{i+3}=\cdots=0$ and $\mu_{i+1}=\mu_{i+2}=\mu_{i+3}%
=\cdots=0$ (since $\beta$ and $\mu$ are weak compositions). We only need to
rearrange the first $i$ entries in order to transform $\mu$ into $\beta$.}. In
other words, there exists a permutation $\sigma\in\Symm_{\left(
\infty\right)  }$ such that $\beta=\sigma\left(  \mu\right)  $ (because
rearranging finitely many entries of a weak composition is tantamount to
applying a permutation in $\Symm_{\left(  \infty\right)  }$ to it). In
other words, $\beta\in\Symm_{\left(  \infty\right)  }\mu$. Hence, $\mu$
is a $\lambda\in\operatorname*{Par}$ such that $\beta\in\Symm_{\left(
\infty\right)  }\lambda$.

On the other hand, recall that every weak composition $\alpha$ lies in the
$\Symm_{(\infty)}$-orbit of a unique partition $\lambda$. In other
words, for every weak composition $\alpha$, there is a unique partition
$\lambda$ such that $\alpha\in\Symm_{\left(  \infty\right)  }\lambda$.
Applying this to $\alpha=\beta$, we conclude that there is a unique partition
$\lambda$ such that $\beta\in\Symm_{\left(  \infty\right)  }\lambda$.
In other words, there is a unique $\lambda\in\operatorname*{Par}$ such that
$\beta\in\Symm_{\left(  \infty\right)  }\lambda$. This unique $\lambda$
must be $\mu$ (since $\mu$ is a $\lambda\in\operatorname*{Par}$ such that
$\beta\in\Symm_{\left(  \infty\right)  }\lambda$).

So we know that there is a unique $\lambda\in\operatorname*{Par}$ such that
$\beta\in\Symm_{\left(  \infty\right)  }\lambda$, and this unique
$\lambda$ is $\mu$. Hence, the sum $\sum\limits_{\substack{\lambda
\in\operatorname{Par};\\\beta\in\Symm_{\left(  \infty\right)  }\lambda
}}\left(  h_{\lambda},f\right)  $ has exactly one addend, namely the addend
for $\lambda=\mu$. Thus, this sum simplifies as follows: $\sum
\limits_{\substack{\lambda\in\operatorname{Par};\\\beta\in
\Symm_{\left(  \infty\right)  }\lambda}}
\left(  h_{\lambda},f\right)  =\left(  h_{\mu},f\right)  $.

Hence, \eqref{sol.Lambda.hall-coefficient.3} becomes%
\[
\left(  \text{the coefficient of }\mathbf{x}^{\beta}\text{ in }f\right)
=\sum\limits_{\substack{\lambda\in\operatorname{Par};\\\beta\in
\Symm_{\left(  \infty\right)  }\lambda}}
\left(  h_{\lambda},f\right)  =\left(  h_{\mu},f\right)  .
\]
Combining this with
\[
\left(  f,h_{\mu}\right)  =\left(  h_{\mu},f\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.Lambda.hall.gr}(c),
applied to }g=h_{\mu}\right)  ,
\]
we obtain%
\[
\left(  f,h_{\mu}\right)  =\left(  h_{\mu},f\right)  =\left(  \text{the
coefficient of }\mathbf{x}^{\beta}\text{ in }f\right)  .
\]
This solves Exercise \ref{exe.Lambda.hall-coefficient}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lambda.h-e-through-p}}

\textit{Solution to Exercise \ref{exe.Lambda.h-e-through-p}.} Let us first
prove three simple claims:

\begin{statement}
\textit{Claim 1:} For every $\lambda\in\operatorname*{Par}$, we have
$p_{\lambda}\in\Lambda_{\left\vert \lambda\right\vert }$.
\end{statement}

[\textit{Proof of Claim 1:} Let $n=\left\vert \lambda\right\vert $. Thus,
$\lambda\in\operatorname*{Par}\nolimits_{n}$ (since $\lambda\in
\operatorname*{Par}$). Hence, we can show (just as in the solution to
Exercise~\ref{exe.Lambda.triangular}(k)) that $p_{\lambda}\in\Lambda_{n}$. In
view of $n=\left\vert \lambda\right\vert $, this rewrites as $p_{\lambda}%
\in\Lambda_{\left\vert \lambda\right\vert }$. This proves Claim 1.]

\begin{statement}
\textit{Claim 2:} Let $n\in\mathbb{N}$ and $\lambda\in\operatorname*{Par}%
\nolimits_{n}$. Then, $\left(  p_{\lambda},h_{n}\right)  =1$.
\end{statement}

[\textit{Proof of Claim 2:} We have $\left\vert \lambda\right\vert =n$ (since
$\lambda\in\operatorname*{Par}\nolimits_{n}$). Claim 1 yields $p_{\lambda}%
\in\Lambda_{\left\vert \lambda\right\vert }=\Lambda_{n}$ (since $\left\vert
\lambda\right\vert =n$). Hence, Exercise \ref{exe.Lambda.hall.gr}(b) (applied
to $f=p_{\lambda}$) yields $\left(  h_{n},p_{\lambda}\right)  =p_{\lambda
}\left(  1\right)  $ (where $p_{\lambda}\left(  1\right)  $ is defined as in
Exercise~\ref{exe.Lambda.subs.fin}).

Recall that $p_{\lambda}\left(  1\right)  $ is defined to be the result of
substituting $1,0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in $p_{\lambda}$.
In other words, $p_{\lambda}\left(  1\right)  =p_{\lambda}\left(
1,0,0,0,\ldots\right)  $.

\begin{vershort}
Now, write the partition $\lambda$ as $\lambda=\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{\ell}\right)  $, where $\ell=\ell\left(  \lambda\right)
$. Then, the definition of $p_{\lambda}$ yields $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}=\prod_{i=1}^{\ell}p_{\lambda_{i}}$.
Substituting $1,0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ on both sides of
this equality, we obtain%
\begin{align*}
p_{\lambda}\left(  1,0,0,0,\ldots\right)    & =\prod_{i=1}^{\ell
}\underbrace{p_{\lambda_{i}}\left(  1,0,0,0,\ldots\right)  }%
_{\substack{=1^{\lambda_{i}}+0^{\lambda_{i}}+0^{\lambda_{i}}+0^{\lambda_{i}%
}+\cdots\\\text{(since }p_{\lambda_{i}}=x_{1}^{\lambda_{i}}+x_{2}^{\lambda
_{i}}+x_{3}^{\lambda_{i}}+\cdots\text{)}}}=\prod_{i=1}^{\ell}%
\underbrace{\left(  1^{\lambda_{i}}+0^{\lambda_{i}}+0^{\lambda_{i}}%
+0^{\lambda_{i}}+\cdots\right)  }_{\substack{=1+0+0+0+\cdots\\\text{(since
}1^{\lambda_{i}}=1\text{ and }0^{\lambda_{i}}=0\\\text{(because }\lambda
_{i}\text{ is positive (since }i\leq\ell=\ell\left(  \lambda\right)
\text{)))}}}\\
& =\prod_{i=1}^{\ell}\underbrace{\left(  1+0+0+0+\cdots\right)  }_{=1}=1.
\end{align*}
Summarizing what we have proved so far, we obtain%
\[
\left(  h_{n},p_{\lambda}\right)  =p_{\lambda}\left(  1\right)  =p_{\lambda
}\left(  1,0,0,0,\ldots\right)  =1.
\]

\end{vershort}

\begin{verlong}
Now, write the partition $\lambda$ as $\lambda=\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{\ell}\right)  $, where $\ell=\ell\left(  \lambda\right)
$. Then, the definition of $p_{\lambda}$ yields $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}=\prod_{i=1}^{\ell}p_{\lambda_{i}}$.
Substituting $1,0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ on both sides of
this equality, we obtain%
\[
p_{\lambda}\left(  1,0,0,0,\ldots\right)  =\left(  \prod_{i=1}^{\ell
}p_{\lambda_{i}}\right)  \left(  1,0,0,0,\ldots\right)  =\prod_{i=1}^{\ell
}p_{\lambda_{i}}\left(  1,0,0,0,\ldots\right)  .
\]


But each $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies
\begin{equation}
p_{\lambda_{i}}\left(  1,0,0,0,\ldots\right)
=1.\label{sol.Lambda.h-e-through-p.C2.pf.1}%
\end{equation}


[\textit{Proof of \eqref{sol.Lambda.h-e-through-p.C2.pf.1}:} Let $i\in\left\{
1,2,\ldots,\ell\right\}  $. Thus, $i\leq\ell=\ell\left(  \lambda\right)  $, so
that the entry $\lambda_{i}$ of $\lambda$ is positive. Hence, the definition
of $p_{\lambda_{i}}$ yields $p_{\lambda_{i}}=x_{1}^{\lambda_{i}}%
+x_{2}^{\lambda_{i}}+x_{3}^{\lambda_{i}}+\cdots$. Substituting $1,0,0,0,\ldots
$ for $x_{1},x_{2},x_{3},\ldots$ on both sides of this equality, we obtain%
\[
p_{\lambda_{i}}\left(  1,0,0,0,\ldots\right)  =\underbrace{1^{\lambda_{i}}%
}_{=1}+\underbrace{0^{\lambda_{i}}+0^{\lambda_{i}}+0^{\lambda_{i}}+\cdots
}_{\substack{=0+0+0+\cdots\\\text{(since }0^{\lambda_{i}}=0\text{ (because
}\lambda_{i}\text{ is positive))}}}=1+\underbrace{0+0+0+\cdots}_{=0}=1.
\]
This proves \eqref{sol.Lambda.h-e-through-p.C2.pf.1}.]

Now,%
\[
\left(  h_{n},p_{\lambda}\right)  =p_{\lambda}\left(  1\right)  =p_{\lambda
}\left(  1,0,0,0,\ldots\right)  =\prod_{i=1}^{\ell}\underbrace{p_{\lambda_{i}%
}\left(  1,0,0,0,\ldots\right)  }_{\substack{=1\\\text{(by
\eqref{sol.Lambda.h-e-through-p.C2.pf.1})}}}=\prod_{i=1}^{\ell}1=1.
\]

\end{verlong}

\noindent But Exercise \ref{exe.Lambda.hall.gr}(c) (applied to $f=p_{\lambda}$
and $g=h_{n}$) yields $\left(  p_{\lambda},h_{n}\right)  =\left(
h_{n},p_{\lambda}\right)  =1$. This proves Claim 2.]

\begin{statement}
\textit{Claim 3:} Let $n\in\mathbb{N}$ and $\lambda\in\operatorname*{Par}$ be
such that $\lambda\notin\operatorname*{Par}\nolimits_{n}$. Then, $\left(
p_{\lambda},h_{n}\right)  =0$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 3:} We have $\left\vert \lambda\right\vert \neq n$
(since $\lambda\in\operatorname*{Par}$ but $\lambda\notin\operatorname*{Par}%
\nolimits_{n}$). Therefore, $\left\vert \lambda\right\vert $ and $n$ are two
distinct nonnegative integers. Moreover, $p_{\lambda}\in\Lambda_{\left\vert
\lambda\right\vert }$ (by Claim 1) and $h_{n}\in\Lambda_{n}$. Hence, Exercise
\ref{exe.Lambda.hall.gr}(a) (applied to $\left\vert \lambda\right\vert $, $n$,
$p_{\lambda}$ and $h_{n}$ instead of $n$, $m$, $f$ and $g$) yields $\left(
p_{\lambda},h_{n}\right)  =0$. This proves Claim 3.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 3:} If we had $\left\vert \lambda\right\vert =n$, then
we would have $\lambda\in\operatorname*{Par}\nolimits_{n}$ (since $\lambda
\in\operatorname*{Par}$), which would contradict $\lambda\notin%
\operatorname*{Par}\nolimits_{n}$. Hence, we cannot have $\left\vert
\lambda\right\vert =n$. Thus, we have $\left\vert \lambda\right\vert \neq n$.
Therefore, $\left\vert \lambda\right\vert $ and $n$ are two distinct
nonnegative integers. Moreover, $p_{\lambda}\in\Lambda_{\left\vert
\lambda\right\vert }$ (by Claim 1) and $h_{n}\in\Lambda_{n}$. Hence, Exercise
\ref{exe.Lambda.hall.gr}(a) (applied to $\left\vert \lambda\right\vert $, $n$,
$p_{\lambda}$ and $h_{n}$ instead of $n$, $m$, $f$ and $g$) yields $\left(
p_{\lambda},h_{n}\right)  =0$. This proves Claim 3.]
\end{verlong}

% \begin{statement}
% \textit{Claim 4:} Let $n\in\mathbb{N}$. Then, $\omega\left(  h_{n}\right)
% =e_{n}$.
% \end{statement}

% [\textit{Proof of Claim 4:} If $n$ is a positive integer, then this follows
% immediately from Proposition~\ref{prop.Lambda.omega-and-S}(b). Hence, for the rest of
% this proof of Claim 4, we WLOG assume that $n$ is not a positive integer. In
% other words, $n\notin\left\{  1,2,3,\ldots\right\}  $ (since $\left\{
% 1,2,3,\ldots\right\}  $ is the set of all positive integers).

% Combining $n\in\mathbb{N}$ with $n\notin\left\{  1,2,3,\ldots\right\}  $, we
% obtain $n\in\mathbb{N}\setminus\left\{  1,2,3,\ldots\right\}  =\left\{
% 0\right\}  $. Thus, $n=0$. Hence, $h_{n}=h_{0}=1$ and $e_{n}=e_{0}=1$.
% Applying the map $\omega$ to both sides of the equality $h_{n}=1$, we obtain
% $\omega\left(  h_{n}\right)  =\omega\left(  1\right)  =1$ (since $\omega$ is a
% $\kk$-algebra homomorphism). Comparing this with $e_{n}=1$, we obtain
% $\omega\left(  h_{n}\right)  =e_{n}$. This proves Claim 4.]

Corollary \ref{dual-bases-h-m-p}(b) yields that $\left(  p_{\lambda}\right)
_{\lambda\in\operatorname{Par}}$ and $\left(  z_{\lambda}^{-1}p_{\lambda
}\right)  _{\lambda\in\operatorname{Par}}$ are dual bases with respect to the
Hall inner product on $\Lambda$. Recall the following fundamental fact from
linear algebra: If $\kk$ is a commutative ring, if $A$ is a
$\kk$-module, if $\left(  \cdot,\cdot\right)  :A\times A\rightarrow
\kk$ is a symmetric $\kk$-bilinear form on $A$, and if $\left(
u_{\lambda}\right)  _{\lambda\in L}$ and $\left(  v_{\lambda}\right)
_{\lambda\in L}$ are two $\kk$-bases of $A$ which are dual to each
other with respect to the form $\left(  \cdot,\cdot\right)  $ (where $L$ is
some indexing set), then every $a\in A$ satisfies
\[
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}.
\]
We can apply this fact to $A=\Lambda$, $L=\operatorname{Par}$, $\left(
u_{\lambda}\right)  _{\lambda\in L}=\left(  p_{\lambda}\right)  _{\lambda
\in\operatorname{Par}}$ and $\left(  v_{\lambda}\right)  _{\lambda\in
L}=\left(  z_{\lambda}^{-1}p_{\lambda}\right)  _{\lambda\in\operatorname{Par}%
}$ (since the bases $\left(  p_{\lambda}\right)  _{\lambda\in
\operatorname{Par}}$ and $\left(  z_{\lambda}^{-1}p_{\lambda}\right)
_{\lambda\in\operatorname{Par}}$ of $\Lambda$ are dual to each other with
respect to the Hall inner product $\left(  \cdot,\cdot\right)  $). As a
result, we obtain that every $a\in\Lambda$ satisfies%
\begin{equation}
a=\sum_{\lambda\in\operatorname{Par}}\left(  p_{\lambda},a\right)  z_{\lambda
}^{-1}p_{\lambda}. \label{sol.Lambda.h-e-through-p.a=sum}%
\end{equation}


Now, let $n\in\mathbb{N}$. Applying \eqref{sol.Lambda.h-e-through-p.a=sum} to
$a=h_{n}$, we find%
\begin{align*}
h_{n}  &  =\sum_{\lambda\in\operatorname{Par}}\left(  p_{\lambda}%
,h_{n}\right)  z_{\lambda}^{-1}p_{\lambda}=\underbrace{\sum_{\substack{\lambda
\in\operatorname{Par};\\\lambda\in\operatorname*{Par}\nolimits_{n}}%
}}_{\substack{=\sum_{\lambda\in\operatorname*{Par}\nolimits_{n}}\\\text{(since
}\operatorname*{Par}\nolimits_{n}\subset\operatorname*{Par}\text{)}%
}}\underbrace{\left(  p_{\lambda},h_{n}\right)  }_{\substack{=1\\\text{(by
Claim 2)}}}z_{\lambda}^{-1}p_{\lambda}+\sum_{\substack{\lambda\in
\operatorname{Par};\\\lambda\notin\operatorname*{Par}\nolimits_{n}%
}}\underbrace{\left(  p_{\lambda},h_{n}\right)  }_{\substack{=0\\\text{(by
Claim 3)}}}z_{\lambda}^{-1}p_{\lambda}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }\lambda\in\operatorname*{Par}\text{ satisfies either
}\lambda\in\operatorname*{Par}\nolimits_{n}\text{ or }\lambda\notin%
\operatorname*{Par}\nolimits_{n}\\
\text{(but not both at the same time)}%
\end{array}
\right) \\
&  =\sum_{\lambda\in\operatorname*{Par}\nolimits_{n}}z_{\lambda}%
^{-1}p_{\lambda}+\underbrace{\sum_{\substack{\lambda\in\operatorname{Par}%
;\\\lambda\notin\operatorname*{Par}\nolimits_{n}}}0z_{\lambda}^{-1}p_{\lambda
}}_{=0}=\sum_{\lambda\in\operatorname*{Par}\nolimits_{n}}z_{\lambda}%
^{-1}p_{\lambda}.
\end{align*}
This proves \eqref{eq.exe.Lambda.h-e-through-p.h}.

Recall the map $\omega : \Lambda \to \Lambda$ introduced in
Definition~\ref{def.Lambda.omega}.
Applying this map $\omega$ to both sides of the
equality \eqref{eq.exe.Lambda.h-e-through-p.h}, we obtain
\begin{align*}
\omega\left(  h_{n}\right)   &  =\omega\left(  \sum_{\lambda\in
\operatorname*{Par}\nolimits_{n}}z_{\lambda}^{-1}p_{\lambda}\right)
=\sum_{\lambda\in\operatorname*{Par}\nolimits_{n}}z_{\lambda}^{-1}%
\underbrace{\omega\left(  p_{\lambda}\right)  }_{\substack{=\left(  -1\right)
^{\left\vert \lambda\right\vert -\ell\left(  \lambda\right)  }p_{\lambda
}\\\text{(by \eqref{eq.exe.Lambda.omegahlam.p})}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\omega\text{ is } \kk\text{-linear}\right) \\
&  =\sum_{\lambda\in\operatorname*{Par}\nolimits_{n}}\underbrace{z_{\lambda
}^{-1}\left(  -1\right)  ^{\left\vert \lambda\right\vert -\ell\left(
\lambda\right)  }}_{=\left(  -1\right)  ^{\left\vert \lambda\right\vert
-\ell\left(  \lambda\right)  }z_{\lambda}^{-1}}p_{\lambda}=\sum_{\lambda
\in\operatorname*{Par}\nolimits_{n}}\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }z_{\lambda}^{-1}p_{\lambda}.
\end{align*}
But Proposition~\ref{prop.Lambda.omega-and-S}(b) yields
$\omega\left(  h_{n}\right)  =e_{n}$. Comparing these two
equalities, we obtain%
\[
e_{n}=\sum_{\lambda\in\operatorname*{Par}\nolimits_{n}}\left(  -1\right)
^{\left\vert \lambda\right\vert -\ell\left(  \lambda\right)  }z_{\lambda}%
^{-1}p_{\lambda}.
\]
This proves \eqref{eq.exe.Lambda.h-e-through-p.e}. Thus, the solution to
Exercise \ref{exe.Lambda.h-e-through-p} is complete.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.horistrip.postnikov}}

\textit{Solution to Exercise~\ref{exe.horistrip.postnikov}.} We recall
that the cells of $\lambda/\mu$ are the cells $\left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ satisfying $\mu_{p}<q\leq\lambda_{p}$.

(a) In order to solve Exercise~\ref{exe.horistrip.postnikov}(a),
we need to prove the following two claims:
\begin{equation}
\left(  \text{if }\lambda/\mu\text{ is a horizontal strip, then every }%
i\in\left\{  1,2,3,...\right\}  \text{ satisfies }\mu_{i}\geq\lambda
_{i+1}\right)  \label{sol.horistrip.postnikov.goal1}%
\end{equation}
and
\begin{equation}
\left(  \text{if every }i\in\left\{  1,2,3,...\right\}  \text{ satisfies }%
\mu_{i}\geq\lambda_{i+1}\text{, then }\lambda/\mu\text{ is a horizontal
strip}\right)  .\label{sol.horistrip.postnikov.goal2}%
\end{equation}


\textit{Proof of \eqref{sol.horistrip.postnikov.goal1}:} Assume that
$\lambda/\mu$ is a horizontal strip. In other words, no two cells of
$\lambda/\mu$ lie in the same column.

Let $i\in\left\{  1,2,3,...\right\}  $. Assume (for the sake of contradiction)
that $\mu_{i}<\lambda_{i+1}$. Hence, $\lambda_{i+1}>\mu_{i}\geq0$. Also,
$\mu\subseteq\lambda$ yields $\mu_{i}\leq\lambda_{i}$. Hence, $\mu_{i+1}%
\leq\mu_{i}<\lambda_{i+1}$. Now, $\left(  i,\lambda_{i+1}\right)  $ is a cell
of $\lambda/\mu$ (since $\mu_{i}<\lambda_{i+1}\leq\lambda_{i}$). Also,
$\left(  i+1,\lambda_{i+1}\right)  $ is a cell of $\lambda/\mu$ (since
$\mu_{i+1}<\lambda_{i+1}\leq\lambda_{i+1}$). Thus, $\left(  i,\lambda
_{i+1}\right)  $ and $\left(  i+1,\lambda_{i+1}\right)  $ are two cells of
$\lambda/\mu$ lying in the same column. This contradicts the fact that no two
cells of $\lambda/\mu$ lie in the same column. This contradiction shows that
our assumption (that $\mu_{i}<\lambda_{i+1}$) was wrong. Hence, $\mu_{i}%
\geq\lambda_{i+1}$. This proves \eqref{sol.horistrip.postnikov.goal1}.

\textit{Proof of \eqref{sol.horistrip.postnikov.goal2}:} Assume that every
$i\in\left\{  1,2,3,...\right\}  $ satisfies $\mu_{i}\geq\lambda_{i+1}$.

Now, let us (for the sake of contradiction) assume that there exist two
distinct cells of $\lambda/\mu$ which lie in the same column. Let $c$ and $d$
be two such cells. Thus, $c$ and $d$ are two distinct cells of $\lambda/\mu$
which lie in the same column.

Write $c$ and $d$ in the forms $c=\left(  p_{c},q_{c}\right)  $ and $d=\left(
p_{d},q_{d}\right)  $ for some positive integers $p_{c},q_{c},p_{d},q_{d}$. We
have $\mu_{p_{c}}<q_{c}\leq\lambda_{p_{c}}$ (since $\left(  p_{c}%
,q_{c}\right)  =c$ is a cell of $\lambda/\mu$) and $\mu_{p_{d}}<q_{d}%
\leq\lambda_{p_{d}}$ (similarly). The cells $c$ and $d$ lie in the same
column; in other words, $q_{c}=q_{d}$ (because the cell $c=\left(  p_{c}%
,q_{c}\right)  $ lies in column $q_{c}$, and the cell $d=\left(  p_{d}%
,q_{d}\right)  $ lies in column $q_{d}$).

Our situation so far is symmetric with respect to interchanging $c$ with $d$.
Hence, we can WLOG assume that $p_{c}\leq p_{d}$ (because otherwise, we can
switch $c$ with $d$). Assume this.

If we had $p_{c}=p_{d}$, then we would have $c=\left(  \underbrace{p_{c}%
}_{=p_{d}},\underbrace{q_{c}}_{=q_{d}}\right)  =\left(  p_{d},q_{d}\right)
=d$, which would contradict the fact that $c$ and $d$ are distinct. Hence, we
cannot have $p_{c}=p_{d}$. Thus, we have $p_{c}\neq p_{d}$. Combined with
$p_{c}\leq p_{d}$, this yields $p_{c}<p_{d}$. Thus, $p_{c}\leq p_{d}-1$ (since
$p_{c}$ and $p_{d}$ are integers), so that $p_{c}+1\leq p_{d}$.

Since $\lambda$ is a partition, we have $\lambda_{u}\geq\lambda_{v}$ for any
positive integers $u$ and $v$ satisfying $u\leq v$. Applying this to
$u=p_{c}+1$ and $v=p_{d}$, we obtain $\lambda_{p_{c}+1}\geq\lambda_{p_{d}}$
(since $p_{c}+1\leq p_{d}$). But recall that every $i\in\left\{
1,2,3,...\right\}  $ satisfies $\mu_{i}\geq\lambda_{i+1}$. Applying this to
$i=p_{c}$, we obtain $\mu_{p_{c}}\geq\lambda_{p_{c}+1}$. Now, recall that
$\mu_{p_{c}}<q_{c}$, so that $q_{c}>\mu_{p_{c}}\geq\lambda_{p_{c}+1}%
\geq\lambda_{p_{d}}\geq q_{d}$ (since $q_{d}\leq\lambda_{p_{d}}$). This
contradicts $q_{c}=q_{d}$. This contradiction shows that our assumption (that
there exist two distinct cells of $\lambda/\mu$ which lie in the same column)
was wrong. Hence, no two cells of $\lambda/\mu$ lie in the same column. In
other words, $\lambda/\mu$ is a horizontal strip. This proves \eqref{sol.horistrip.postnikov.goal2}.

Now, both \eqref{sol.horistrip.postnikov.goal1} and
\eqref{sol.horistrip.postnikov.goal2} are proven. Thus,
Exercise~\ref{exe.horistrip.postnikov}(a) is solved.

(b) We have the following equivalence of statements:
\begin{align*}
& \ \left(  \lambda/\mu\text{ is a vertical strip}\right)  \\
& \Longleftrightarrow\ \left(  \text{no two cells of }\lambda/\mu\text{ lie in
the same row}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of a
``vertical strip''}\right)  \\
& \Longleftrightarrow\ \left(  \text{for every }i\in\left\{  1,2,3,\ldots
\right\}  \text{, no two cells of }\lambda/\mu\text{ lie in row }i\right)  \\
& \Longleftrightarrow\ \left(  \text{for every }i\in\left\{  1,2,3,\ldots
\right\}  \text{, }\underbrace{\text{the number of cells of }\lambda/\mu\text{
in row }i}_{=\lambda_{i}-\mu_{i}}\text{ is }\leq1\right)  \\
& \Longleftrightarrow\ \left(  \text{for every }i\in\left\{  1,2,3,\ldots
\right\}  \text{, we have }\lambda_{i}-\mu_{i}\leq1\right)  \\
& \Longleftrightarrow\ \left(  \text{for every }i\in\left\{  1,2,3,\ldots
\right\}  \text{, we have }\lambda_{i}\leq\mu_{i}+1\right)  .
\end{align*}
This solves Exercise~\ref{exe.horistrip.postnikov}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.horistrip.postnikov}}

\textit{Solution to Exercise~\ref{exe.horistrip.postnikov}.} In the
following, for every partition $\nu$, let $Y\left(  \nu\right)  $ denote the
Ferrers diagram of $\nu$. Then, for every partition $\nu$, we have
\begin{equation}
Y\left(  \nu\right)  =\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ q\leq\nu_{p}\right\}
,
\label{sol.horistrip.postnikov.Y}
\end{equation}
where $\nu$ is written in the form $\left(  \nu_{1},\nu_{2},\nu_{3}
,...\right)  $. By the definition of the skew diagram of $\lambda/\mu$, we
have
\begin{align}
& \left(  \text{the set of cells of }\lambda/\mu\right)  \nonumber\\
& =\underbrace{\left(  \text{the set of cells of }\lambda\right)
}_{\substack{=Y\left(  \lambda\right)  \\=\left\{  \left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ q\leq\lambda_{p}\right\}
\\\text{(by \eqref{sol.horistrip.postnikov.Y}, applied to }\nu=\lambda
\text{)}}}\setminus\underbrace{\left(  \text{the set of cells of }\mu\right)
}_{\substack{=Y\left(  \mu\right)  \\=\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ q\leq\mu_{p}\right\}  \\\text{(by
\eqref{sol.horistrip.postnikov.Y}, applied to }\nu=\mu\text{)}}}\nonumber\\
& =\left\{  \left(  p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}
\ \mid\ q\leq\lambda_{p}\right\}  \setminus\left\{  \left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ q\leq\mu_{p}\right\}  \nonumber\\
& =\left\{  \left(  p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}
\ \mid\ \underbrace{q\leq\lambda_{p}\text{ but not }q\leq\mu_{p}}_{\text{this
is equivalent to }\left(  q\leq\lambda_{p}\text{ and }q>\mu_{p}\right)
}\right\}  \nonumber\\
& =\left\{  \left(  p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}
\ \mid\ \underbrace{q\leq\lambda_{p}\text{ and }q>\mu_{p}}_{\text{this is
equivalent to }\mu_{p}<q\leq\lambda_{p}}\right\}  \nonumber\\
& =\left\{  \left(  p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}\ \mid
\ \mu_{p}<q\leq\lambda_{p}\right\}  .\label{sol.horistrip.postnikov.cells}
\end{align}


(a) In order to solve Exercise~\ref{exe.horistrip.postnikov}(a),
we need to prove the following two claims:
\begin{equation}
\left(  \text{if }\lambda/\mu\text{ is a horizontal strip, then every }
i\in\left\{  1,2,3,...\right\}  \text{ satisfies }\mu_{i}\geq\lambda
_{i+1}\right)  \label{sol.horistrip.postnikov.goal1}
\end{equation}
and
\begin{equation}
\left(  \text{if every }i\in\left\{  1,2,3,...\right\}  \text{ satisfies }
\mu_{i}\geq\lambda_{i+1}\text{, then }\lambda/\mu\text{ is a horizontal
strip}\right)  .
\label{sol.horistrip.postnikov.goal2}
\end{equation}


\textit{Proof of \eqref{sol.horistrip.postnikov.goal1}:} Assume that
$\lambda/\mu$ is a horizontal strip. In other words, no two cells of
$\lambda/\mu$ lie in the same column (because $\lambda/\mu$ is a horizontal
strip if and only if no two cells of $\lambda/\mu$ lie in the same column (by
the definition of a ``horizontal strip'')).

Let $i\in\left\{  1,2,3,...\right\}  $. Assume (for the sake of contradiction)
that $\mu_{i}<\lambda_{i+1}$. Then, $\lambda_{i+1}>\mu_{i}\geq0$, so that
$\lambda_{i+1}\in\left\{  1,2,3,...\right\}  $ and thus $\left(
i,\lambda_{i+1}\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ (since
$i\in\left\{  1,2,3,...\right\}  $ and $\lambda_{i+1}\in\left\{
1,2,3,...\right\}  $) and $\left(  i+1,\lambda_{i+1}\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ (since $i+1\in\left\{  1,2,3,...\right\}  $ and
$\lambda_{i+1}\in\left\{  1,2,3,...\right\}  $).

We know that $\lambda$ is a partition, so that $\lambda_{1}\geq\lambda_{2}
\geq\lambda_{3}\geq...$. Thus, $\lambda_{i}\geq\lambda_{i+1}$, so that
$\lambda_{i+1}\leq\lambda_{i}$. Hence, $\mu_{i}<\lambda_{i+1}\leq\lambda_{i}$.
Now, $\left(  i,\lambda_{i+1}\right)  $ is an element $\left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}$ satisfying $\mu_{p}<q\leq\lambda_{p}$
(because $\left(  i,\lambda_{i+1}\right)  \in\left\{  1,2,3,...\right\}  ^{2}$
and $\mu_{i}<\lambda_{i+1}\leq\lambda_{i}$). In other words,
\begin{align*}
\left(  i,\lambda_{i+1}\right)    & \in\left\{  \left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq\lambda_{p}\right\}
\\
& =\left(  \text{the set of cells of }\lambda/\mu\right)
\ \ \ \ \ \ \ \ \ \ 
\left(  \text{by \eqref{sol.horistrip.postnikov.cells}}\right)  .
\end{align*}
In other words, $\left(  i,\lambda_{i+1}\right)  $ is a cell of $\lambda/\mu$.

Also, $\mu$ is a partition, so that $\mu_{1}\geq\mu_{2}\geq\mu_{3}\geq...$.
Thus, $\mu_{i}\geq\mu_{i+1}$, so that $\mu_{i+1}\leq\mu_{i}<\lambda_{i+1}$.
Hence, $\mu_{i+1}<\lambda_{i+1}\leq\lambda_{i+1}$. Now, $\left(
i+1,\lambda_{i+1}\right)  $ is an element $\left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ satisfying $\mu_{p}<q\leq\lambda_{p}$ (because
$\left(  i+1,\lambda_{i+1}\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ and
$\mu_{i+1}<\lambda_{i+1}\leq\lambda_{i+1}$). In other words,
\begin{align*}
\left(  i+1,\lambda_{i+1}\right)    & \in\left\{  \left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq\lambda_{p}\right\}
\\
& =\left(  \text{the set of cells of }\lambda/\mu\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.horistrip.postnikov.cells}}
\right)  .
\end{align*}
In other words, $\left(  i+1,\lambda_{i+1}\right)  $ is a cell of $\lambda
/\mu$.

Now, we know that $\left(  i,\lambda_{i+1}\right)  $ and $\left(
i+1,\lambda_{i+1}\right)  $ are two cells of $\lambda/\mu$. These two cells
are distinct (since $i\neq i+1$) but lie in the same column (since
$\lambda_{i+1}=\lambda_{i+1}$). Hence, there exist two distinct cells of
$\lambda/\mu$ lying in the same column (namely, the two cells $\left(
i,\lambda_{i+1}\right)  $ and $\left(  i+1,\lambda_{i+1}\right)  $). This
contradicts the fact that no two cells of $\lambda/\mu$ lie in the same
column. This contradiction shows that our assumption (that $\mu_{i}
<\lambda_{i+1}$) was wrong. Hence, we cannot have $\mu_{i}<\lambda_{i+1}$.
Thus, we must have $\mu_{i}\geq\lambda_{i+1}$. Now, let us forget that we
fixed $i$. We thus have shown that every $i\in\left\{  1,2,3,...\right\}  $
satisfies $\mu_{i}\geq\lambda_{i+1}$. This proves
\eqref{sol.horistrip.postnikov.goal1}.

\textit{Proof of \eqref{sol.horistrip.postnikov.goal2}:} Assume that every
$i\in\left\{  1,2,3,...\right\}  $ satisfies $\mu_{i}\geq\lambda_{i+1}$.

Now, let us (for the sake of contradiction) assume that there exist two
distinct cells of $\lambda/\mu$ which lie in the same column. Let $c$ and $d$
be two such cells. Thus, $c$ and $d$ are two distinct cells of $\lambda/\mu$
which lie in the same column.

We know that $c$ is a cell of $\lambda/\mu$; thus,
\[
c\in\left(  \text{the set of cells of }\lambda/\mu\right)  =\left\{  \left(
p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq
\lambda_{p}\right\}  .
\]
In other words, there exists some $\left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ such that $\mu_{p}<q\leq\lambda_{p}$ and $c=\left(
p,q\right)  $. Denote this $\left(  p,q\right)  $ by $\left(  p_{c},
q_{c}\right)  $. Thus, $\left(  p_{c},q_{c}\right)  $ is an element of
$\left\{  1,2,3,...\right\}  ^{2}$ such that $\mu_{p_{c}}<q_{c}\leq
\lambda_{p_{c}}$ and $c=\left(  p_{c},q_{c}\right)  $. The column in which the
cell $c$ lies is $q_{c}$ (since $c=\left(  p_{c},q_{c}\right)  $).

We know that $d$ is a cell of $\lambda/\mu$; thus,
\[
d\in\left(  \text{the set of cells of }\lambda/\mu\right)  =\left\{  \left(
p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq
\lambda_{p}\right\}  .
\]
In other words, there exists some $\left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ such that $\mu_{p}<q\leq\lambda_{p}$ and $d=\left(
p,q\right)  $. Denote this $\left(  p,q\right)  $ by $\left(  p_{d},
q_{d}\right)  $. Thus, $\left(  p_{d},q_{d}\right)  $ is an element of
$\left\{  1,2,3,...\right\}  ^{2}$ such that $\mu_{p_{d}}<q_{d}\leq
\lambda_{p_{d}}$ and $d=\left(  p_{d},q_{d}\right)  $. The column in which the
cell $d$ lies is $q_{d}$ (since $d=\left(  p_{d},q_{d}\right)  $).

We know that the cells $c$ and $d$ lie in the same column. Thus, the column in
which the cell $c$ lies is the column in which the cell $d$ lies. In other
words, $q_{c}$ is $q_{d}$ (because the column in which the cell $c$ lies is
$q_{c}$, while the column in which the cell $d$ lies is $q_{d}$). Thus,
$q_{c}=q_{d}$.

Our situation so far is symmetric with respect to interchanging $c$ with $d$.
Hence, we can WLOG assume that $p_{c}\leq p_{d}$ (because otherwise, we can
switch $c$ with $d$). Assume this.

If we had $p_{c}=p_{d}$, then we would have
$c=\left(  \underbrace{p_{c}}_{=p_{d}},\underbrace{q_{c}}_{=q_{d}}\right) 
 =\left(  p_{d},q_{d}\right) =d$,
which would contradict the fact that $c$ and $d$ are distinct. Hence, we
cannot have $p_{c}=p_{d}$. Thus, we have $p_{c}\neq p_{d}$. Combined with
$p_{c}\leq p_{d}$, this yields $p_{c}<p_{d}$. Thus, $p_{c}\leq p_{d}-1$ (since
$p_{c}$ and $p_{d}$ are integers), so that $p_{c}+1\leq p_{d}$.

Since $\lambda$ is a partition, we have $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq...$. In other words, $\lambda_{u}\geq\lambda_{v}$ for any
positive integers $u$ and $v$ satisfying $u\leq v$. Applying this to
$u=p_{c}+1$ and $v=p_{d}$, we obtain $\lambda_{p_{c}+1}\geq\lambda_{p_{d}}$
(since $p_{c}+1\leq p_{d}$). But recall that every $i\in\left\{
1,2,3,...\right\}  $ satisfies $\mu_{i}\geq\lambda_{i+1}$. Applying this to
$i=p_{c}$, we obtain $\mu_{p_{c}}\geq\lambda_{p_{c}+1}$. Now, recall that
$\mu_{p_{c}}<q_{c}$, so that $q_{c}>\mu_{p_{c}}\geq\lambda_{p_{c}+1}
\geq\lambda_{p_{d}}\geq q_{d}$ (since $q_{d}\leq\lambda_{p_{d}}$). This
contradicts $q_{c}=q_{d}$. This contradiction shows that our assumption (that
there exist two distinct cells of $\lambda/\mu$ which lie in the same column)
was wrong. Hence, there exist no two distinct cells of $\lambda/\mu$ which lie
in the same column. In other words, no two cells of $\lambda/\mu$ lie in the
same column. In other words, $\lambda/\mu$ is a horizontal strip (because
$\lambda/\mu$ is a horizontal strip if and only if no two cells of
$\lambda/\mu$ lie in the same column (by the definition of a
``horizontal strip'')). This proves
\eqref{sol.horistrip.postnikov.goal2}.

Now, both \eqref{sol.horistrip.postnikov.goal1} and
\eqref{sol.horistrip.postnikov.goal2} are proven. Thus,
Exercise~\ref{exe.horistrip.postnikov}(a) is solved.

(b) In order to solve Exercise~\ref{exe.horistrip.postnikov}(b), we need to prove
the following two claims:
\begin{equation}
\left(  \text{if }\lambda/\mu\text{ is a vertical strip, then every }%
i\in\left\{  1,2,3,...\right\}  \text{ satisfies }\lambda_{i}\leq\mu
_{i}+1\right)  \label{sol.horistrip.postnikov.b.goal1}%
\end{equation}
and
\begin{equation}
\left(  \text{if every }i\in\left\{  1,2,3,...\right\}  \text{ satisfies
}\lambda_{i}\leq\mu_{i}+1\text{, then }\lambda/\mu\text{ is a vertical
strip}\right)  . \label{sol.horistrip.postnikov.b.goal2}%
\end{equation}


\textit{Proof of \eqref{sol.horistrip.postnikov.b.goal1}:} Assume that
$\lambda/\mu$ is a vertical strip. In other words, no two cells of
$\lambda/\mu$ lie in the same row (because $\lambda/\mu$ is a vertical strip
if and only if no two cells of $\lambda/\mu$ lie in the same row (by the
definition of a ``vertical strip'')). Let
$i\in\left\{  1,2,3,...\right\}  $. Assume (for the sake of contradiction)
that $\lambda_{i}>\mu_{i}+1$. Then, $\lambda_{i}\geq\left(  \mu_{i}+1\right)
+1$ (since $\lambda_{i}$ and $\mu_{i}+1$ are integers). Hence, $\lambda
_{i}\geq\left(  \mu_{i}+1\right)  +1=\mu_{i}+2$, so that $\mu_{i}+2\leq
\lambda_{i}$ and thus $\mu_{i}+1\leq\mu_{i}+2\leq\lambda_{i}$. Also,
$\underbrace{\mu_{i}}_{\geq0}+1\geq1$, so that $\mu_{i}+1\in\left\{
1,2,3,...\right\}  $ and thus $\left(  i,\mu_{i}+1\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ (since $i\in\left\{  1,2,3,...\right\}  $ and
$\mu_{i}+1\in\left\{  1,2,3,...\right\}  $). Also, $\mu_{i}+2\geq\mu_{i}%
+1\geq1$, so that $\mu_{i}+2\in\left\{  1,2,3,...\right\}  $ and thus $\left(
i,\mu_{i}+2\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ (since $i\in\left\{
1,2,3,...\right\}  $ and $\mu_{i}+2\in\left\{  1,2,3,...\right\}  $).

Now, $\left(  i,\mu_{i}+1\right)  $ is an element $\left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}$ satisfying $\mu_{p}<q\leq\lambda_{p}$
(because $\left(  i,\mu_{i}+1\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ and
$\mu_{i}<\mu_{i}+1\leq\lambda_{i}$). In other words,%
\begin{align*}
\left(  i,\mu_{i}+1\right)   &  \in\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq\lambda_{p}\right\}  \\
&  =\left(  \text{the set of cells of }\lambda/\mu\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.horistrip.postnikov.cells}}%
\right)  .
\end{align*}
In other words, $\left(  i,\mu_{i}+1\right)  $ is a cell of $\lambda/\mu$.

Also, $\left(  i,\mu_{i}+2\right)  $ is an element $\left(  p,q\right)
\in\left\{  1,2,3,...\right\}  ^{2}$ satisfying $\mu_{p}<q\leq\lambda_{p}$
(because $\left(  i,\mu_{i}+2\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ and
$\mu_{i}<\mu_{i}+2\leq\lambda_{i}$). In other words,%
\begin{align*}
\left(  i,\mu_{i}+2\right)   &  \in\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq\lambda_{p}\right\}  \\
&  =\left(  \text{the set of cells of }\lambda/\mu\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.horistrip.postnikov.cells}}%
\right)  .
\end{align*}
In other words, $\left(  i,\mu_{i}+2\right)  $ is a cell of $\lambda/\mu$.

Now, we know that $\left(  i,\mu_{i}+1\right)  $ and $\left(  i,\mu
_{i}+2\right)  $ are two cells of $\lambda/\mu$. These two cells are distinct
(since $\mu_{i}+1\neq\mu_{i}+2$) but lie in the same row (since $i=i$). Hence,
there exist two distinct cells of $\lambda/\mu$ lying in the same row (namely,
the two cells $\left(  i,\mu_{i}+1\right)  $ and $\left(  i,\mu_{i}+2\right)
$). This contradicts the fact that no two cells of $\lambda/\mu$ lie in the
same row. This contradiction shows that our assumption (that $\lambda_{i}%
>\mu_{i}+1$) was wrong. Hence, we cannot have $\lambda_{i}>\mu_{i}+1$. Thus,
we must have $\lambda_{i}\leq\mu_{i}+1$. Now, let us forget that we fixed $i$.
We thus have shown that every $i\in\left\{  1,2,3,...\right\}  $ satisfies
$\lambda_{i}\leq\mu_{i}+1$. This proves \eqref{sol.horistrip.postnikov.b.goal1}.

\textit{Proof of \eqref{sol.horistrip.postnikov.b.goal2}:} Assume that every
$i\in\left\{  1,2,3,...\right\}  $ satisfies $\lambda_{i}\leq\mu_{i}+1$.

Now, let us (for the sake of contradiction) assume that there exist two
distinct cells of $\lambda/\mu$ which lie in the same row. Let $c$ and $d$ be
two such cells. Thus, $c$ and $d$ are two distinct cells of $\lambda/\mu$
which lie in the same row.

We know that $c$ is a cell of $\lambda/\mu$; thus,
\[
c\in\left(  \text{the set of cells of }\lambda/\mu\right)  =\left\{  \left(
p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq
\lambda_{p}\right\}  .
\]
In other words, there exists some $\left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ such that $\mu_{p}<q\leq\lambda_{p}$ and $c=\left(
p,q\right)  $. Denote this $\left(  p,q\right)  $ by $\left(  p_{c}%
,q_{c}\right)  $. Thus, $\left(  p_{c},q_{c}\right)  $ is an element of
$\left\{  1,2,3,...\right\}  ^{2}$ such that $\mu_{p_{c}}<q_{c}\leq
\lambda_{p_{c}}$ and $c=\left(  p_{c},q_{c}\right)  $. The row in which the
cell $c$ lies is $p_{c}$ (since $c=\left(  p_{c},q_{c}\right)  $).

We know that $d$ is a cell of $\lambda/\mu$; thus,
\[
d\in\left(  \text{the set of cells of }\lambda/\mu\right)  =\left\{  \left(
p,q\right)  \in\left\{  1,2,3,...\right\}  ^{2}\ \mid\ \mu_{p}<q\leq
\lambda_{p}\right\}  .
\]
In other words, there exists some $\left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ such that $\mu_{p}<q\leq\lambda_{p}$ and $d=\left(
p,q\right)  $. Denote this $\left(  p,q\right)  $ by $\left(  p_{d}%
,q_{d}\right)  $. Thus, $\left(  p_{d},q_{d}\right)  $ is an element of
$\left\{  1,2,3,...\right\}  ^{2}$ such that $\mu_{p_{d}}<q_{d}\leq
\lambda_{p_{d}}$ and $d=\left(  p_{d},q_{d}\right)  $. The row in which the
cell $d$ lies is $p_{d}$ (since $d=\left(  p_{d},q_{d}\right)  $).

We know that the cells $c$ and $d$ lie in the same row. Thus, the row in which
the cell $c$ lies is the row in which the cell $d$ lies. In other words,
$p_{c}$ is $p_{d}$ (because the row in which the cell $c$ lies is $p_{c}$,
while the row in which the cell $d$ lies is $p_{d}$). Thus, $p_{c}=p_{d}$.
Hence, $\lambda_{p_{c}}=\lambda_{p_{d}}$.

Our situation so far is symmetric with respect to interchanging $c$ with $d$.
Hence, we can WLOG assume that $q_{c}\leq q_{d}$ (because otherwise, we can
switch $c$ with $d$). Assume this.

If we had $q_{c}=q_{d}$, then we would have $c=\left(  \underbrace{p_{c}%
}_{=p_{d}},\underbrace{q_{c}}_{=q_{d}}\right)  =\left(  p_{d},q_{d}\right)
=d$, which would contradict the fact that $c$ and $d$ are distinct. Hence, we
cannot have $q_{c}=q_{d}$. Thus, we have $q_{c}\neq q_{d}$. Combined with
$q_{c}\leq q_{d}$, this yields $q_{c}<q_{d}$. Thus, $q_{c}\leq q_{d}-1$ (since
$q_{c}$ and $q_{d}$ are integers), so that $q_{c}+1\leq q_{d}$.

Now, recall that every $i\in\left\{  1,2,3,...\right\}  $ satisfies
$\lambda_{i}\leq\mu_{i}+1$. Applying this to $i=p_{c}$, we obtain
$\lambda_{p_{c}}\leq\underbrace{\mu_{p_{c}}}_{<q_{c}}+1<q_{c}+1\leq q_{d}%
\leq\lambda_{p_{d}}$. But this contradicts $\lambda_{p_{c}}=\lambda_{p_{d}}$.
This contradiction shows that our assumption (that there exist two distinct
cells of $\lambda/\mu$ which lie in the same row) was wrong. Hence, there
exist no two distinct cells of $\lambda/\mu$ which lie in the same row. In
other words, no two cells of $\lambda/\mu$ lie in the same row. In other
words, $\lambda/\mu$ is a vertical strip (because $\lambda/\mu$ is a vertical
strip if and only if no two cells of $\lambda/\mu$ lie in the same row (by the
definition of a ``vertical strip'')). This
proves \eqref{sol.horistrip.postnikov.b.goal2}.

Now, both \eqref{sol.horistrip.postnikov.b.goal1} and
\eqref{sol.horistrip.postnikov.b.goal2} are proven. Thus,
Exercise~\ref{exe.horistrip.postnikov}(b) is solved.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.h2}}

\textit{Solution to Exercise~\ref{exe.pieri.h2}.} (a) We use the
notation $f\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ defined in
Exercise~\ref{exe.Lambda.subs.fin} whenever $a_{1},a_{2},\ldots,a_{k}$ are
elements of a commutative $\kk$-algebra $A$ and $f\in R\left(
{\mathbf{x}}\right)  $. In particular, $f\left(  1\right)  $ is a well-defined
element of $\kk$ for every $f\in R\left(  {\mathbf{x}}\right)  $. Every
$f\in R\left(  {\mathbf{x}}\right)  $ satisfies
\begin{equation}
f\left(  1\right)  =\left(  \text{the result of substituting }%
1,0,0,0,...\text{ for }x_{1},x_{2},x_{3},...\text{ in }f\right)  .
\label{sol.pieri.h2.a.f1}%
\end{equation}


We have $s_{\lambda/\mu}=\sum_{T}{\xx}^{ \cont (T)}$,
where $T$ runs through all column-strict tableaux of shape $\lambda/\mu$. In
other words, $s_{\lambda/\mu}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda/\mu}}{\mathbf{x}}^{ \cont %
\left(  T\right)  }$. Hence,%
\begin{align}
s_{\lambda/\mu}\left(  1\right)   &  =\left(  \sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu}}{\xx%
}^{ \cont \left(  T\right)  }\right)  \left(  1\right) \nonumber\\
&  =\left(  \text{the result of substituting }1,0,0,0,...\text{ for }%
x_{1},x_{2},x_{3},...\text{ in }\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda/\mu}}{\mathbf{x}}^{ \cont %
\left(  T\right)  }\right)  \label{sol.pieri.h2.a.f2}%
\end{align}
(by (\ref{sol.pieri.h2.a.f1})).

But the substitution $1,0,0,0,...$ for $x_{1},x_{2},x_{3},...$ has the
following effect on any given monomial $\mathbf{x}^{\alpha}$:

\begin{itemize}
\item if none of the indeterminates $x_{2},x_{3},x_{4},...$ occur in this
monomial $\mathbf{x}^{\alpha}$, then the monomial $\mathbf{x}^{\alpha}$ goes
to $1$;

\item otherwise, the monomial $\mathbf{x}^{\alpha}$ goes to $0$.
\end{itemize}

\noindent
Hence, applying this substitution to $\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu}}{\xx}
^{ \cont (T)}$ yields a sum of $1$'s over all column-strict
tableaux $T$ of shape $\lambda/\mu$ having the property that none of the
indeterminates $x_{2},x_{3},x_{4},...$ occur in this monomial ${\xx}
^{ \cont (T)}$. In other words,%
\begin{align*}
&  \left(  \text{the result of substituting }1,0,0,0,...\text{ for }%
x_{1},x_{2},x_{3},...\text{ in }\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda/\mu}}{\xx}^{ \cont %
(T)}\right) \\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau of shape }\lambda
/\mu;\\\text{none of the indeterminates }x_{2},x_{3},x_{4},...\\\text{occur in
the monomial }\mathbf{x}^{ \cont \left(  T\right)  }}}1.
\end{align*}
Therefore, (\ref{sol.pieri.h2.a.f2}) rewrites as
\[
s_{\lambda/\mu}\left(  1\right)  =\sum\limits_{\substack{T\text{ is a column-strict
tableau of shape }\lambda/\mu;\\\text{none of the indeterminates }x_{2}%
,x_{3},x_{4},...\\\text{occur in the monomial }\xx%
^{ \cont \left(  T\right)  }}}1.
\]
This rewrites as%
\begin{equation}
s_{\lambda/\mu}\left(  1\right)  =\sum\limits_{\substack{T\text{ is a column-strict
tableau of shape }\lambda/\mu;\\\text{all entries of }T\text{ are }1}}1
\label{sol.pieri.h2.a.f3}%
\end{equation}
(because for a column-strict tableau $T$, saying that none of the
indeterminates $x_{2},x_{3},x_{4},...$ occur in the monomial $\xx%
^{ \cont \left(  T\right)  }$ is equivalent to saying that all
entries of $T$ are $1$).

In order to solve Exercise~\ref{exe.pieri.h2}(a), we need to prove the
following two statements:
\begin{equation}
\left(  \text{if }\lambda/\mu\text{ is a horizontal }n\text{-strip, then
}\left(  h_{n},s_{\lambda/\mu}\right)  =1\right)  \label{sol.pieri.h2.a.goal1}%
\end{equation}
and%
\begin{equation}
\left(  \text{if }\lambda/\mu\text{ is not a horizontal }n\text{-strip, then
}\left(  h_{n},s_{\lambda/\mu}\right)  =0\right)  .
\label{sol.pieri.h2.a.goal2}%
\end{equation}


\textit{Proof of \eqref{sol.pieri.h2.a.goal1}:} Assume that $\lambda/\mu$ is a
horizontal $n$-strip. Thus, in particular, we have $\left\vert \lambda
/\mu\right\vert =n$. Hence, $s_{\lambda/\mu}\in\Lambda_{\left\vert \lambda
/\mu\right\vert }=\Lambda_{n}$ (since $\left\vert \lambda/\mu\right\vert =n$).
Thus, Exercise \ref{exe.Lambda.hall.gr}(b) (applied to $f=s_{\lambda/\mu}$)
yields
\begin{equation}
\left(  h_{n},s_{\lambda/\mu}\right)  =s_{\lambda/\mu}\left(  1\right)
=\sum\limits_{\substack{T\text{ is a column-strict tableau of shape }\lambda
/\mu;\\\text{all entries of }T\text{ are }1}}1\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.pieri.h2.a.f3})}\right)  . \label{sol.pieri.h2.a.goal1.1}%
\end{equation}


Recall that $\lambda/\mu$ is a horizontal $n$-strip, therefore a horizontal
strip. In other words, no two cells of $\lambda/\mu$ lie in the same column.
Hence, if we fill every cell of the Ferrers diagram of $\lambda/\mu$ with a
$1$, we obtain a column-strict tableau $T$ of shape $\lambda/\mu$ having the
property that all entries of $T$ are $1$. Therefore, such a tableau exists. It
is also clearly unique (because requiring that all entries be $1$ does not
leave any freedom in choosing the entries). Therefore, there exists exactly
one such tableau. This shows that the sum on the right hand side of
(\ref{sol.pieri.h2.a.goal1.1}) has exactly one addend, and therefore equals
$1$ (since the addend is $1$). Hence, (\ref{sol.pieri.h2.a.goal1.1}) rewrites
as $\left(  h_{n},s_{\lambda/\mu}\right)  =1$, and thus
(\ref{sol.pieri.h2.a.goal1}) is proven.

\textit{Proof of \eqref{sol.pieri.h2.a.goal2}:} Assume that $\lambda/\mu$ is
not a horizontal $n$-strip. We need to show that $\left(  h_{n},s_{\lambda
/\mu}\right)  =0$.

If $\left\vert \lambda/\mu\right\vert \neq n$, then
Exercise~\ref{exe.Lambda.hall.gr}(a) (applied to $m=\left\vert \lambda
/\mu\right\vert $, $f=h_{n}$ and $g=s_{\lambda/\mu}$) yields $\left(
h_{n},s_{\lambda/\mu}\right)  =0$ (since $h_{n}\in\Lambda_{n}$ and
$s_{\lambda/\mu}\in\Lambda_{\left\vert \lambda/\mu\right\vert }$). Thus,
$\left(  h_{n},s_{\lambda/\mu}\right)  =0$ is proven if $\left\vert
\lambda/\mu\right\vert \neq n$. Hence, for the rest of the proof of $\left(
h_{n},s_{\lambda/\mu}\right)  =0$, we WLOG assume that we don't have
$\left\vert \lambda/\mu\right\vert \neq n$. Thus, $\left\vert \lambda
/\mu\right\vert =n$. We can thus prove (\ref{sol.pieri.h2.a.goal1.1}) just as
we did before (in the proof of \eqref{sol.pieri.h2.a.goal1}).

But if $\lambda/\mu$ were a horizontal strip, then $\lambda/\mu$ would be a
horizontal $n$-strip (since $\left\vert \lambda/\mu\right\vert =n$), which
would contradict our assumption that $\lambda/\mu$ is not a horizontal
$n$-strip. Hence, $\lambda/\mu$ cannot be a horizontal strip. As a
consequence, there must be two cells of $\lambda/\mu$ which lie in the same
column. If $T$ is a column-strict tableau of shape $\lambda/\mu$, then these
two cells must be filled with two different entries in $T$ (because the
entries of a column-strict tableau are strictly increasing top-to-bottom down
columns, and hence all entries in any given column must be distinct),
which is impossible if all entries of $T$ are to be $1$. Therefore, there
exists no column-strict tableau $T$ of shape $\lambda/\mu$ such that all
entries of $T$ are $1$. Hence, the sum on the right hand side of
(\ref{sol.pieri.h2.a.goal1.1}) is empty, and therefore equals $0$. So
(\ref{sol.pieri.h2.a.goal1.1}) rewrites as $\left(  h_{n},s_{\lambda/\mu
}\right)  =0$, and thus (\ref{sol.pieri.h2.a.goal2}) is proven.

Now that both (\ref{sol.pieri.h2.a.goal1}) and (\ref{sol.pieri.h2.a.goal2})
are proven, Exercise~\ref{exe.pieri.h2}(a) is solved.

(b) Notice first that%
\begin{equation}
\left(  h_{n},s_{\lambda/\mu}\right)  =c_{\mu,\left(  n\right)  }^{\lambda
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in{ \NN }\text{, }\lambda
\in\operatorname{Par}\text{ and }\mu\in\operatorname{Par}.
\label{sol.pieri.h2.b.scalp_n}%
\end{equation}
\footnote{\textit{Proof.} Let $n\in \NN$, $\lambda\in\operatorname{Par}$
and $\mu\in\operatorname{Par}$. From Remark
\ref{third-L-R-interpretation-remark}, we know that $s_{\lambda/\mu}=\sum
_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum is over all $\nu
\in\operatorname{Par}$. Thus, for every given partition $\tau$, the $s_{\tau}%
$-coordinate of $s_{\lambda/\mu}$ in the basis $\left(  s_{\nu}\right)
_{\nu\in \Par }$ of $\Lambda$ equals $c_{\mu,\tau}^{\lambda}$.
But since $\left(  s_{\nu}\right)  _{\nu\in \Par }$ is an
orthonormal basis of the $\kk$-module $\Lambda$ with respect to the
Hall inner product, this $s_{\tau}$-coordinate also equals $\left(  s_{\tau
},s_{\lambda/\mu}\right)  $. Comparing these two expressions for this
$s_{\tau}$-coordinate, we obtain $c_{\mu,\tau}^{\lambda}=\left(  s_{\tau
},s_{\lambda/\mu}\right)  $. Applying this to $\tau=\left(  n\right)  $, we
obtain $c_{\mu,\left(  n\right)  }^{\lambda}=\left(  \underbrace{s_{\left(
n\right)  }}_{=h_{n}},s_{\lambda/\mu}\right)  =\left(  h_{n},s_{\lambda/\mu
}\right)  $, which proves (\ref{sol.pieri.h2.b.scalp_n}).}

Recall that any two partitions $\mu$ and $\nu$ satisfy
\begin{equation}
s_{\mu}s_{\nu}=\sum_{\lambda\in\operatorname{Par}}c_{\mu,\nu}^{\lambda
}s_{\lambda}=\sum_{\tau\in\operatorname{Par}}c_{\mu,\nu}^{\tau}s_{\tau}
\label{sol.pieri.h2.b.LR}%
\end{equation}
(here, we renamed the summation index $\lambda$ as $\tau$).

Let $n\in{ \NN }$. Let $\lambda$ be a partition. We have $s_{\left(
n\right)  }=h_{n}$, so that $h_{n}=s_{\left(  n\right)  }$. Now,
\begin{align*}
s_{\lambda}\underbrace{h_{n}}_{=s_{\left(  n\right)  }} &  =s_{\lambda
}s_{\left(  n\right)  }=\sum_{\tau\in\operatorname{Par}}\underbrace{c_{\lambda
,\left(  n\right)  }^{\tau}}_{\substack{=\left(  h_{n},s_{\tau/\lambda
}\right)  \\\text{(because \eqref{sol.pieri.h2.b.scalp_n} (applied to }%
\tau\\\text{and }\lambda\text{ instead of }\lambda\text{ and }\mu\text{)
yields}\\\left(  h_{n},s_{\tau/\lambda}\right)  =c_{\lambda,\left(  n\right)
}^{\tau}\text{)}}}s_{\tau}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.pieri.h2.b.LR}, applied to }\mu=\lambda\text{ and }\nu=\left(
n\right)  \right)  \\
&  =\sum_{\tau\in\operatorname{Par}}\left(  h_{n},s_{\tau/\lambda}\right)
s_{\tau}=\sum\limits_{\substack{\tau\in\operatorname{Par};\\\lambda\subseteq\tau
}}\left(  h_{n},s_{\tau/\lambda}\right)  s_{\tau}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have ridden the sum of all its addends in which }%
\lambda\nsubseteq\tau;\\
\text{these addends were zero (because if }\lambda\nsubseteq\tau\text{, then
}\left(  h_{n},\underbrace{s_{\tau/\lambda}}_{=0}\right)  s_{\tau}=0\text{)}%
\end{array}
\right)  \\
&  =\sum\limits_{\substack{\tau\in\operatorname{Par};\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is a horizontal }n\text{-strip}}}\underbrace{\left(
h_{n},s_{\tau/\lambda}\right)  }_{\substack{=1\\\text{(by
\eqref{sol.pieri.h2.a.goal1}, applied}\\\text{to }\tau\text{ and }%
\lambda\text{ instead of }\lambda\text{ and }\mu\text{)}}}s_{\tau}%
+\sum\limits_{\substack{\tau\in\operatorname{Par};\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is not a horizontal }n\text{-strip}}}\underbrace{\left(
h_{n},s_{\tau/\lambda}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.pieri.h2.a.goal2}, applied}\\\text{to }\tau\text{ and }%
\lambda\text{ instead of }\lambda\text{ and }\mu\text{)}}}s_{\tau}\\
&  =\sum\limits_{\substack{\tau\in\operatorname{Par};\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is a horizontal }n\text{-strip}}}s_{\tau}%
+\underbrace{\sum\limits_{\substack{\tau\in\operatorname{Par};\\\lambda\subseteq
\tau;\\\tau/\lambda\text{ is not a horizontal }n\text{-strip}}}0s_{\tau}}%
_{=0}\\
&  =\sum\limits_{\substack{\tau\in\operatorname{Par};\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is a horizontal }n\text{-strip}}}s_{\tau}=\sum\limits
_{\substack{\lambda^{+}\in\operatorname{Par};\\\lambda\subseteq\lambda
^{+};\\\lambda^{+}/\lambda\text{ is a horizontal }n\text{-strip}}%
}s_{\lambda^{+}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\tau\text{ as }\lambda^{+}\right)  \\
&  =\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}.
\end{align*}
This proves \eqref{eq.pieri.h}. Thus, Exercise~\ref{exe.pieri.h2}(b) is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.h2}}

\textit{Solution to Exercise~\ref{exe.pieri.h2}.} (a) We use the
notation $f\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ defined in
Exercise~\ref{exe.Lambda.subs.fin} whenever
$a_{1},a_{2},\ldots,a_{k}$ are elements of
a commutative $\kk$-algebra $A$ and $f\in R\left(\xx\right)$. In particular,
$f\left(  1\right)  $ is a well-defined element of $\kk$ for every
$f\in R\left(\xx\right)$. Every $f\in R\left(\xx\right)$ satisfies
\begin{equation}
f\left(  1\right)  =\left(  \text{the result of substituting }
1,0,0,0,...\text{ for }x_{1},x_{2},x_{3},...\text{ in }f\right)  .
\label{sol.pieri.h2.a.f1}
\end{equation}


We have $s_{\lambda/\mu}=\sum_{T}{\xx}^{\cont (T)}$, where
$T$ runs through all column-strict tableaux of shape $\lambda/\mu$. In other
words,
\[
s_{\lambda/\mu}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\mu}}\underbrace{\xx ^{\cont\left(
T\right)  }}_{\substack{=\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }\\=x_{1}^{\left\vert T^{-1}\left(  1\right)
\right\vert }\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }}}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\mu}}x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert
}\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\]
But if $\left\vert \lambda/\mu\right\vert = n$, then $s_{\lambda/\mu}\in
\Lambda_{\left\vert \lambda/\mu\right\vert} = \Lambda_n$ (since $\left\vert
\lambda/\mu\right\vert =n$). Hence, if $\left\vert \lambda/\mu\right\vert =n$,
then
\begin{align}
& \left(  h_n, s_{\lambda/\mu}\right) \nonumber \\
&  =s_{\lambda/\mu}\left(  1\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.Lambda.hall.gr}(b)
(applied to }f=s_{\lambda/\mu}\text{)}\right) \nonumber\\
&  =\left(  \text{the result of substituting }1,0,0,0,...\text{ for }
x_{1},x_{2},x_{3},...\text{ in }s_{\lambda/\mu}\right) \nonumber \\
& \ \ \ \ \ \ \ \ \ \ \left(
 \text{by \eqref{sol.pieri.h2.a.f1} (applied to }f=s_{\lambda/\mu}
 \text{)}\right) \nonumber\\
&  =\left(  \text{the result of substituting }1,0,0,0,...\text{ for }
x_{1},x_{2},x_{3},...\text{ in }\sum\limits_{\substack{T\text{ is a column-strict}
\\\text{tableau of shape }\lambda/\mu}}x_{1}^{\left\vert T^{-1}\left(
1\right)  \right\vert }\prod_{i\geq2}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }s_{\lambda/\mu}=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda/\mu
}}x_{1}^{\left\vert T^{-1}\left(  1\right)  \right\vert }\prod_{i\geq2}
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right) \nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu}}
\underbrace{1^{\left\vert T^{-1}\left(  1\right)  \right\vert }}_{=1}
\underbrace{\prod_{i\geq2}0^{\left\vert T^{-1}\left(  i\right)
\right\vert }}_{\substack{=0^{\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)
\right\vert }=\left[  \sum_{i\geq2}\left\vert T^{-1}\left(  i\right)
\right\vert =0\right]  \\\text{(since }0^{u}=\left[  u=0\right]  \text{ for
every }u\in\NN \text{)}}}\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu}}\left[  \sum_{i\geq2}\left\vert T^{-1}\left(  i\right)
\right\vert =0\right]  .
\label{sol.pieri.h2.a.hns1}
\end{align}


Now, let $T_{0}$ be the map from the set of cells of $\lambda/\mu$ to the set
$\left\{  1,2,3,...\right\}  $ which is defined by
\begin{equation}
\left(  T_{0}\left(  x\right)  =1\ \ \ \ \ \ \ \ \ \ \text{for every }
x\in\left(  \text{the set of cells of }\lambda/\mu\right)  \right)  .
\label{sol.pieri.h2.a.T0.def}
\end{equation}
Thus, all values of the map $T_{0}$ are $1$. This map $T_{0}$ might and might
not be a column-strict tableau of shape $\lambda/\mu$ (in fact, this depends
on whether or not $\lambda/\mu$ is a horizontal strip).

We have
\begin{equation}
\left[  \sum_{i\geq2}\left\vert T_{0}^{-1}\left(  i\right)  \right\vert
=0\right]  =1.
\label{sol.pieri.h2.a.T0.res}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.pieri.h2.a.T0.res}:} Every $i\in\left\{
2,3,4,...\right\}  $ satisfies
\begin{align*}
T_{0}^{-1}\left(  i\right)   &  =\left\{  x\in\left(  \text{the set of cells
of }\lambda/\mu\right)  \ \mid\ \underbrace{T_{0}\left(  x\right)
}_{\substack{=1\\\text{(by \eqref{sol.pieri.h2.a.T0.def})}}}=i\right\} \\
&  =\left\{  x\in\left(  \text{the set of cells of }\lambda/\mu\right)
\ \mid\ \underbrace{1=i}_{\substack{\text{this is false}\\\text{(because
}i\neq1\\\text{(since }i\in\left\{  2,3,4,...\right\}  \text{))}}}\right\} \\
&  =\left\{  x\in\left(  \text{the set of cells of }\lambda/\mu\right)
\ \mid\ \text{false}\right\}  =\varnothing.
\end{align*}
Hence, every $i\in\left\{  2,3,4,...\right\}  $ satisfies $\left\vert
\underbrace{T_{0}^{-1}\left(  i\right)  }_{=\varnothing}\right\vert
=\left\vert \varnothing\right\vert =0$. Hence, $\sum_{i\geq2}
\underbrace{\left\vert T_{0}^{-1}\left(  i\right)  \right\vert }_{=0}
=\sum_{i\geq2}0=0$, so that $\left[  \sum_{i\geq2}\left\vert T_{0}^{-1}\left(
i\right)  \right\vert =0\right]  =1$, qed.} Also,
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }
\lambda/\mu;\\T\neq T_{0}}}\left[  \sum_{i\geq2}\left\vert T^{-1}\left(
i\right)  \right\vert =0\right]  =0.
\label{sol.pieri.h2.a.T.res}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.pieri.h2.a.T.res}:} Let $T$ be a
column-strict tableau of shape $\lambda/\mu$ such that $T\neq T_{0}$. We shall
prove that $\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert \neq
0$.
\par
There exists at least one $x\in\left(  \text{the set of cells of }\lambda
/\mu\right)  $ such that $T\left(  x\right)  \neq1$ (because otherwise, every
$x\in\left(  \text{the set of cells of }\lambda/\mu\right)  $ would satisfy
$T\left(  x\right)  =1$, and thus every $x\in\left(  \text{the set of cells of
}\lambda/\mu\right)  $ would satisfy $T\left(  x\right)  =1=T_{0}\left(
x\right)  $ (by \eqref{sol.pieri.h2.a.T0.def}), and thus we would have
$T=T_{0}$, which would contradict $T\neq T_{0}$). Consider this $x$.
\par
Recall that $T$ is a column-strict tableau of shape $\lambda/\mu$, therefore a
map from the set of cells of $\lambda/\mu$ to the set $\left\{
1,2,3,...\right\}  $. Hence, $T\left(  x\right)  \in\left\{
1,2,3,...\right\}  $. Combining this with $T\left(  x\right)  \neq1$, we
obtain $T\left(  x\right)  \in\left\{  1,2,3,...\right\}  \setminus\left\{
1\right\}  =\left\{  2,3,4,...\right\}  $. Moreover, $T^{-1}\left(  T\left(
x\right)  \right)  $ is a nonempty set (since $T^{-1}\left(  T\left(
x\right)  \right)  $ contains the element $x$), so that $\left\vert
T^{-1}\left(  T\left(  x\right)  \right)  \right\vert >0$. We now have
$T\left(  x\right)  \in\left\{  2,3,4,...\right\}  $ and $\left\vert
T^{-1}\left(  T\left(  x\right)  \right)  \right\vert >0$. Thus, there exists
an $i\in\left\{  2,3,4,...\right\}  $ such that $\left\vert T^{-1}\left(
i\right)  \right\vert >0$ (namely, $i=T\left(  x\right)  $). In other words,
the sum $\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert $ has at
least one positive summand. Since all summands of this sum are nonnegative,
this shows that this sum is positive, i.e., we have $\sum_{i\geq2}\left\vert
T^{-1}\left(  i\right)  \right\vert >0$. Thus, $\sum_{i\geq2}\left\vert
T^{-1}\left(  i\right)  \right\vert \neq0$. Hence, the assertion $\sum
_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert =0$ is false. Thus,
$\left[  \sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert
=0\right]  =0$.
\par
Now, forget that we fixed $T$. We thus have proven that every column-strict
tableau $T$ of shape $\lambda/\mu$ such that $T\neq T_{0}$ satisfies $\left[
\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert =0\right]  =0$.
Thus,
\[
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }
\lambda/\mu;\\T\neq T_{0}}}\underbrace{\left[  \sum_{i\geq2}\left\vert
T^{-1}\left(  i\right)  \right\vert =0\right]  }_{=0}=\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda/\mu;\\T\neq T_{0}}}0=0,
\]
qed.} Now, if $\left\vert \lambda/\mu\right\vert =n$, then
\begin{align}
\left(  h_{n},s_{\lambda/\mu}\right)   &  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu}}\left[  \sum_{i\geq
2}\left\vert T^{-1}\left(  i\right)  \right\vert =0\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.pieri.h2.a.hns1}}\right)
\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu;\\T=T_{0}}}\left[  \sum_{i\geq2}\left\vert T^{-1}\left(
i\right)  \right\vert =0\right]  +\underbrace{\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu;\\T\neq T_{0}}}\left[
\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert =0\right]
}_{\substack{=0\\\text{(by \eqref{sol.pieri.h2.a.T.res})}}}\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda/\mu;\\T=T_{0}}}\left[  \sum_{i\geq2}\left\vert T^{-1}\left(
i\right)  \right\vert =0\right]  .
\label{sol.pieri.h2.a.hns2}
\end{align}


In order to solve Exercise~\ref{exe.pieri.h2}(a), we need to prove the
following two statements:
\begin{equation}
\left(  \text{if }\lambda/\mu
\text{ is a horizontal }n\text{-strip, then }
\left(  h_{n},s_{\lambda/\mu}\right)  =1\right)
\label{sol.pieri.h2.a.goal1}
\end{equation}
and%
\begin{equation}
\left(  \text{if }\lambda/\mu
\text{ is not a horizontal }n\text{-strip, then }
\left(  h_{n},s_{\lambda/\mu}\right)  =0\right)  .
\label{sol.pieri.h2.a.goal2}
\end{equation}


\textit{Proof of \eqref{sol.pieri.h2.a.goal1}:} Assume that $\lambda/\mu$ is a
horizontal $n$-strip. Thus, in particular, we have $\left\vert \lambda
/\mu\right\vert =n$. Hence, \eqref{sol.pieri.h2.a.hns2} can be applied.

Recall that $\lambda/\mu$ is a horizontal $n$-strip, therefore a horizontal
strip. In other words, no two cells of $\lambda/\mu$ lie in the same column
(because $\lambda/\mu$ is a horizontal strip if and only if no two cells of
$\lambda/\mu$ lie in the same column (by the definition of a
``horizontal strip'')). Thus, every column of the Ferrers
diagram of $\lambda/\mu$ contains at most one cell.

Now, $T_{0}$ is an assignment of entries of $\left\{  1,2,3,...\right\}  $ to
the cells of $\lambda/\mu$ (because $T_{0}$ is a map from the set of cells of
$\lambda/\mu$ to the set $\left\{  1,2,3,...\right\}  $), weakly increasing
left-to-right in rows (because all values of $T_{0}$ are $1$), and strictly
increasing top-to-bottom in columns (because every column of the Ferrers
diagram of $\lambda/\mu$ contains at most one cell). In other words, $T_{0}$
is a column-strict tableau of shape $\lambda/\mu$. Now,
\eqref{sol.pieri.h2.a.hns2} becomes
\[
\left(  h_{n},s_{\lambda/\mu}\right)  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu;\\T=T_{0}}}\left[
\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert =0\right]
=\left[  \sum_{i\geq2}\left\vert T_{0}^{-1}\left(  i\right)  \right\vert
=0\right]  =1
\]
(by \eqref{sol.pieri.h2.a.T0.res}). This proves \eqref{sol.pieri.h2.a.goal1}.

\textit{Proof of \eqref{sol.pieri.h2.a.goal2}:} Assume that $\lambda/\mu$ is
not a horizontal $n$-strip. We need to show that $\left(  h_{n},s_{\lambda
/\mu}\right)  =0$.

If $\left\vert \lambda/\mu\right\vert \neq n$, then
Exercise~\ref{exe.Lambda.hall.gr}(a) (applied to $m=\left\vert \lambda
/\mu\right\vert $, $f=h_{n}$ and $g=s_{\lambda/\mu}$) yields $\left(
h_{n},s_{\lambda/\mu}\right)  =0$ (since $h_{n}\in\Lambda_{n}$ and
$s_{\lambda/\mu}\in\Lambda_{\left\vert \lambda/\mu\right\vert }$). Thus,
$\left(  h_{n},s_{\lambda/\mu}\right)  =0$ is proven if $\left\vert
\lambda/\mu\right\vert \neq n$. Hence, for the rest of the proof of $\left(
h_{n},s_{\lambda/\mu}\right)  =0$, we can WLOG assume that we don't have
$\left\vert \lambda/\mu\right\vert \neq n$. Assume this.

We don't have $\left\vert \lambda/\mu\right\vert \neq n$. Hence, we have
$\left\vert \lambda/\mu\right\vert =n$. Thus, \eqref{sol.pieri.h2.a.hns2}
can be applied.

Now, $T_{0}$ is not a column-strict tableau of shape $\lambda/\mu
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $T_{0}$ is a
column-strict tableau of shape $\lambda/\mu$. In other words, $T_{0}$ is an
assignment of entries of $\left\{  1,2,3,...\right\}  $ to the cells of
$\lambda/\mu$, weakly increasing left-to-right in rows, and strictly
increasing top-to-bottom in columns. In particular, the map $T_{0}$ cannot
take any value twice in the same column (since $T_{0}$ is strictly increasing
top-to-bottom in columns).
\par
Let $c$ and $d$ be two distinct cells of $\lambda/\mu$ which lie in the same
column. Recall that the map $T_{0}$ cannot take any value twice in the same
column. Thus, the map $T_{0}$ cannot take the same value at $c$ and at $d$
(because $c$ and $d$ are two distinct cells of $\lambda/\mu$ which lie in the
same column). In other words, $T_{0}\left(  c\right)  \neq T_{0}\left(
d\right)  $. But \eqref{sol.pieri.h2.a.T0.def} (applied to $x=c$) yields
$T_{0}\left(  c\right)  =1$, and furthermore, \eqref{sol.pieri.h2.a.T0.def}
(applied to $x=d$) yields $T_{0}\left(  d\right)  =1$. Now, $T_{0}\left(
c\right)  \neq T_{0}\left(  d\right)  =1$ contradicts $T_{0}\left(  c\right)
=1$.
\par
Now, forget that we fixed $c$ and $d$. We thus have obtained a contradiction
whenever $c$ and $d$ are two distinct cells of $\lambda/\mu$ which lie in the
same column. Hence, there are no two distinct cells $c$ and $d$ of
$\lambda/\mu$ which lie in the same column. In other words, no two cells of
$\lambda/\mu$ lie in the same column. In other words, $\lambda/\mu$ is a
horizontal strip (because $\lambda/\mu$ is a horizontal strip if and only if
no two cells of $\lambda/\mu$ lie in the same column (by the definition of a
``horizontal strip'')). Since $\lambda/\mu$
has size $n$ (because $\left\vert \lambda/\mu\right\vert =n$), this yields
that $\lambda/\mu$ is a horizontal $n$-strip; but this contradicts the fact
that $\lambda/\mu$ is not a horizontal $n$-strip. This contradiction shows
that our assumption was wrong, qed.}. Hence, there exists no column-strict
tableau $T$ of shape $\lambda/\mu$ such that $T=T_{0}$. Hence, the sum
$\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }
\lambda/\mu;\\T=T_{0}}}\left[  \sum_{i\geq2}\left\vert T^{-1}\left(  i\right)
\right\vert =0\right]  $ is empty, and thus we have $\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda/\mu;\\T=T_{0}}}\left[
\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert =0\right]  =0$.
Now, \eqref{sol.pieri.h2.a.hns2} becomes
\[
\left(  h_{n},s_{\lambda/\mu}\right)  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda/\mu;\\T=T_{0}}}\left[
\sum_{i\geq2}\left\vert T^{-1}\left(  i\right)  \right\vert =0\right]  =0.
\]
This proves \eqref{sol.pieri.h2.a.goal2}.

Now, both \eqref{sol.pieri.h2.a.goal1} and \eqref{sol.pieri.h2.a.goal2} are
proven. This completes the solution of Exercise~\ref{exe.pieri.h2}(a).

(b) Recall that $\left(  s_{\lambda}\right)  _{\lambda\in\Par}$
is an orthonormal basis of the $\kk$-module $\Lambda$ with respect to
the Hall inner product. Thus,
\begin{equation}
\left(  s_{\lambda},s_{\nu}\right)  =\delta_{\lambda,\nu}
\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par \text{ and
}\nu\in\Par.
\label{sol.pieri.h2.b.orthonormal}
\end{equation}

\begin{comment}
% The following part of the solution turned out to be useless:

Recall the following basic fact from linear algebra: If $A$ is a
$\kk$-module with a given inner product $\left(  \cdot,\cdot\right)  $, and if
$\left(  \gamma_{i}\right)  _{i\in I}$ is an orthonormal basis of $A$ with
respect to this inner product, then
\[
a=\sum_{i\in I}\left(  a,\gamma_{i}\right)  \cdot\gamma_{i}
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{.}
\]
Applying this to $A=\Lambda$, $I=\Par$ and $\left(  \gamma
_{i}\right)  _{i\in I}=\left(  s_{\lambda}\right)  _{\lambda\in
\Par}$, we obtain
\begin{equation}
a=\sum_{\lambda\in\Par}\left(  a,s_{\lambda}\right)  \cdot
s_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\Lambda\text{.}
\label{sol.pieri.h2.b.orthobasis}
\end{equation}
Thus, every $a\in\Lambda$ satisfies
\[
a=\sum_{\lambda\in\Par}\left(  a,s_{\lambda}\right)  \cdot
s_{\lambda}=\sum_{\nu\in\Par}\left(  a,s_{\nu}\right)  \cdot
s_{\nu}
\]
(here, we renamed the summation index $\lambda$ as $\nu$).
\end{comment}

Now, it is easy to see that
\begin{equation}
\left(  s_{\tau},s_{\lambda/\mu}\right)  =c_{\mu,\tau}^{\lambda}
\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\Par \text{, }
\mu\in\Par \text{ and }\tau\in\Par 
.
\label{sol.pieri.h2.b.scalp}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.pieri.h2.b.scalp}:} Let $\lambda
\in\Par$, $\mu\in\Par$ and $\tau\in
\Par$. From Remark \ref{third-L-R-interpretation-remark}, we
know that $s_{\lambda/\mu}=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the
sum is over all $\nu\in\Par$. Thus,
\begin{align*}
\left(  s_{\tau},\underbrace{s_{\lambda/\mu}}
_{=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}}\right)
& =\left(  s_{\tau},\sum_{\nu}c_{\mu,\nu
}^{\lambda}s_{\nu}\right)  =\sum_{\nu}c_{\mu,\nu}^{\lambda}\underbrace{\left(
s_{\tau},s_{\nu}\right)  }_{\substack{=\delta_{\tau,\nu}\\\text{(by
\eqref{sol.pieri.h2.b.orthonormal}, applied}\\\text{to }\nu\text{ and }
\tau\text{ instead of }\lambda\text{ and }\nu\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk \text{-bilinear}\right)  \\
& =\sum_{\nu}c_{\mu,\nu}^{\lambda}\underbrace{\delta_{\tau,\nu}}_{=\delta
_{\nu,\tau}}=\sum_{\nu}c_{\mu,\nu}^{\lambda}\delta_{\nu,\tau}=\sum\limits
_{\substack{\nu;\\\nu\neq\tau}}c_{\mu,\nu}^{\lambda}\underbrace{\delta
_{\nu,\tau}}_{\substack{=0\\\text{(since }\nu\neq\tau\text{)}}}+c_{\mu,\tau
}^{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau\in\Par 
\right)  \\
& =\underbrace{\sum\limits_{\substack{\nu;\\\nu\neq\tau}}c_{\mu,\nu}^{\lambda}0}
_{=0}+c_{\mu,\tau}^{\lambda}=c_{\mu,\tau}^{\lambda}.
\end{align*}
This proves \eqref{sol.pieri.h2.b.scalp}.} As a consequence,
\begin{equation}
\left(  h_{n},s_{\lambda/\mu}\right)  =c_{\mu,\left(  n\right)  }^{\lambda
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\NN \text{, }\lambda
\in\Par \text{ and }\mu\in\Par 
.
\label{sol.pieri.h2.b.scalp_n}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.pieri.h2.b.scalp_n}:} Let $n\in
\NN$, $\lambda\in\Par$ and $\mu\in\Par$.
We have $s_{\left( n\right) } = h_n$, so that $h_n = s_{\left( n\right)}$.
Thus,
\[
\left(  \underbrace{h_n}_{=s_{\left(  n\right)  }},s_{\lambda/\mu}\right)
=\left(  s_{\left(  n\right)  },s_{\lambda/\mu}\right)  =c_{\mu,\left(
n\right)  }^{\lambda}
\]
(by \eqref{sol.pieri.h2.b.scalp}, applied to $\tau=\left(  n\right)  $). This
proves \eqref{sol.pieri.h2.b.scalp_n}.}

Recall that any two partitions $\mu$ and $\nu$ satisfy
\begin{align}
s_{\mu}s_{\nu}  & =\sum_{\lambda}c_{\mu,\nu}^{\lambda}s_{\lambda
}\ \ \ \ \ \ \ \ \ \ \left(  \text{where the sum is over all partitions}
\right)  \nonumber\\
& =\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda}s_{\lambda}
=\sum_{\tau\in\Par}c_{\mu,\nu}^{\tau}s_{\tau}
\label{sol.pieri.h2.b.LR}
\end{align}
(here, we renamed the summation index $\lambda$ as $\tau$).

Let $n \in \NN$. Let $\lambda$ be a partition. We have
$s_{\left(  n\right)  }=h_{n}$, so that $h_{n}=s_{\left(  n\right)  }$. Now,
\begin{align*}
s_{\lambda}\underbrace{h_{n}}_{=s_{\left(  n\right)  }}  & =s_{\lambda
}s_{\left(  n\right)  }=\sum_{\tau\in\Par}
\underbrace{c_{\lambda,\left(  n\right)  }^{\tau}}_{\substack{=\left(
h_{n},s_{\tau/\lambda}\right)  \\
\text{(because \eqref{sol.pieri.h2.b.scalp_n} (applied to }\tau\\
\text{and }\lambda\text{ instead of }\lambda\text{ and }\mu
\text{) yields}\\\left(  h_{n},s_{\tau/\lambda}\right)  =c_{\lambda,\left(
n\right)  }^{\tau}\text{)}}}s_{\tau}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.pieri.h2.b.LR}, applied to }\mu=\lambda\text{ and }\nu=\left(
n\right)  \right)  \\
& =\sum_{\tau\in\Par}\left(  h_{n},s_{\tau/\lambda}\right)
s_{\tau}=\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq\tau
}}\left(  h_{n},s_{\tau/\lambda}\right)  s_{\tau}+\sum\limits_{\substack{\tau
\in\Par ;\\\lambda\nsubseteq\tau}}\left(  h_{n}%
,\underbrace{s_{\tau/\lambda}}_{\substack{=0\\\text{(since }\lambda
\nsubseteq\tau\text{)}}}\right)  s_{\tau}\\
& =\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq\tau}}\left(
h_{n},s_{\tau/\lambda}\right)  s_{\tau}+\sum\limits_{\substack{\tau\in
\Par ;\\\lambda\nsubseteq\tau}}\underbrace{\left(
h_{n},0\right)  }_{=0}s_{\tau}=\sum\limits_{\substack{\tau\in\Par
;\\\lambda\subseteq\tau}}\left(  h_{n},s_{\tau/\lambda}\right)  s_{\tau
}+\underbrace{\sum\limits_{\substack{\tau\in\Par ;\\\lambda
\nsubseteq\tau}}0s_{\tau}}_{=0}\\
& =\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is a horizontal }n\text{-strip}}}\underbrace{\left(
h_{n},s_{\tau/\lambda}\right)  }_{\substack{=1\\\text{(by
\eqref{sol.pieri.h2.a.goal1}, applied}\\\text{to }\tau\text{ and }%
\lambda\text{ instead of }\lambda\text{ and }\mu\text{)}}}s_{\tau}%
+\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is not a horizontal }n\text{-strip}}}\underbrace{\left(
h_{n},s_{\tau/\lambda}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.pieri.h2.a.goal2}, applied}\\\text{to }\tau\text{ and }%
\lambda\text{ instead of }\lambda\text{ and }\mu\text{)}}}s_{\tau}\\
& =\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is a horizontal }n\text{-strip}}}s_{\tau}%
+\underbrace{\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq
\tau;\\\tau/\lambda\text{ is not a horizontal }n\text{-strip}}}0s_{\tau}}%
_{=0}\\
& =\sum\limits_{\substack{\tau\in\Par ;\\\lambda\subseteq\tau
;\\\tau/\lambda\text{ is a horizontal }n\text{-strip}}}s_{\tau}=\sum\limits
_{\substack{\lambda^{+}\in\Par ;\\\lambda\subseteq\lambda
^{+};\\\lambda^{+}/\lambda\text{ is a horizontal }n\text{-strip}}%
}s_{\lambda^{+}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index }%
\tau\text{ as }\lambda^{+}\right)  \\
& =\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}.
\end{align*}
This proves \eqref{eq.pieri.h}. Thus, Exercise~\ref{exe.pieri.h2}(b) is solved.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.RSK}}

\textit{Solution to Exercise~\ref{exe.pieri.RSK}.}
Let us use the notations introduced in the proof of
Theorem~\ref{thm.cauchy-identity}. In particular, we use the words
``letter'' and ``positive integer'' as synonyms.

We shall use the following lemma:

\begin{lemma}
\label{lem.sol.pieri.RSK.rbl}Let $P$ be a column-strict tableau, and let $j$
and $j^{\prime}$ be two letters. Applying RS-insertion to the tableau $P$ and
the letter $j$ yields a new column-strict tableau $P^{\prime}$ and a corner
cell $c$. Applying RS-insertion to the tableau $P^{\prime}$ and the letter
$j^{\prime}$ yields a new column-strict tableau $P^{\prime\prime}$ and a
corner cell $c^{\prime}$.

\begin{enumerate}
\item[(a)] Assume that $j\leq j^{\prime}$. Then, the cell $c^{\prime}$ is in
the same row as the cell $c$ or in a row further up; it is also in a column
further right than $c$.

\item[(b)] Assume instead that $j>j^{\prime}$. Then, the cell $c^{\prime}$ is
in a row further down than the cell $c$; it is also in the same column as $c$
or in a column further left.
\end{enumerate}
\end{lemma}

Lemma \ref{lem.sol.pieri.RSK.rbl} is part of the Row bumping lemma that
appeared in our proof of Theorem~\ref{thm.cauchy-identity}.

We shall first concentrate on proving \eqref{eq.pieri.h}.

\begin{proof}
[Alternative proof of \eqref{eq.pieri.h}.]Let $\lambda$ be a partition, and
let $n\in \NN$. The definition of $s_{\lambda}$ yields%
\[
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda}}{\mathbf{x}}^{\cont \left(  T\right)  }.
\]
The definition of $h_{n}$ yields
\begin{equation}
h_{n}=\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}. \label{sol.exe.pieri.RSK.h.1}%
\end{equation}
Multiplying these two identities, we obtain
\begin{align}
s_{\lambda}h_{n}  &  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda}}{\mathbf{x}}^{\cont \left(
T\right)  }\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\nonumber\\
&  =\sum_{\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\in\mathbf{A}}{\mathbf{x}}^{\cont \left(  T\right)  }x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}, \label{sol.exe.pieri.RSK.h.lhs}%
\end{align}
where $\mathbf{A}$ is the set of all pairs $\left(  T,\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \right)  $ of a column-strict tableau $T$ of shape
$\lambda$ and an $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ of
positive integers satisfying $i_{1}\leq i_{2}\leq\cdots\leq i_{n}$. Consider
this set $\mathbf{A}$.

On the other hand, every partition $\lambda^{+}$ satisfies
\begin{align}
s_{\lambda^{+}}  &  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda^{+}}}{\mathbf{x}}^{\cont
\left(  T\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}s_{\lambda^{+}}\right) \nonumber\\
&  =\sum\limits_{\substack{S\text{ is a column-strict}\\\text{tableau of shape
}\lambda^{+}}}{\mathbf{x}}^{\cont \left(  S\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}T\text{ as }S\right)  . \label{sol.exe.pieri.RSK.h.slp}%
\end{align}
Hence,%
\begin{align}
\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }\\\text{horizontal
}n\text{-strip}}}\underbrace{s_{\lambda^{+}}}_{\substack{=\sum\limits
_{\substack{S\text{ is a column-strict}\\\text{tableau of shape }\lambda^{+}%
}}{\mathbf{x}}^{\cont \left(  S\right)  }\\\text{(by
(\ref{sol.exe.pieri.RSK.h.slp}))}}}  &  =\sum\limits_{\substack{\lambda^{+}%
:\lambda^{+}/\lambda\text{ is a }\\\text{horizontal }n\text{-strip}}%
}\sum\limits_{\substack{S\text{ is a column-strict}\\\text{tableau of shape }%
\lambda^{+}}}{\mathbf{x}}^{\cont \left(  S\right)  }\nonumber\\
&  =\sum_{\left(  \lambda^{+},S\right)  \in\mathbf{B}}{\xx%
}^{\cont \left(  S\right)  }, \label{sol.exe.pieri.RSK.h.rhs}%
\end{align}
where $\mathbf{B}$ is the set of all pairs $\left(  \lambda^{+},S\right)  $ of
a partition $\lambda^{+}$ and a column-strict tableau $S$ of shape
$\lambda^{+}$ such that $\lambda^{+}/\lambda$ is a horizontal $n$-strip.
Consider this set $\mathbf{B}$.

We shall now prove that there exists a bijection $\mathbf{i}:\mathbf{A}%
\rightarrow\mathbf{B}$ which has the property that%
\begin{equation}
{\mathbf{x}}^{\cont \left(  T\right)  }x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}={\mathbf{x}}^{\cont \left(  S\right)  }
\label{sol.exe.pieri.RSK.h.prop}%
\end{equation}
whenever some $\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\in\mathbf{A}$ and $\left(  \lambda^{+},S\right)  \in\mathbf{B}$ satisfy
$\mathbf{i}\left(  \left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\right)  =\left(  \lambda^{+},S\right)  $. Once this will be proven, it will
immediately follow that $\sum_{\left(  T,\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \right)  \in\mathbf{A}}{\mathbf{x}}^{\cont \left(
T\right)  }x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}=\sum_{\left(  \lambda
^{+},S\right)  \in\mathbf{B}}{\mathbf{x}}^{\cont \left(
S\right)  }$, and therefore (\ref{sol.exe.pieri.RSK.h.lhs}) will become%
\[
s_{\lambda}h_{n}=\sum_{\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)
\right)  \in\mathbf{A}}{\mathbf{x}}^{\cont \left(  T\right)
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}=\sum_{\left(  \lambda^{+},S\right)
\in\mathbf{B}}{\mathbf{x}}^{\cont \left(  S\right)  }%
=\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.exe.pieri.RSK.h.rhs})}\right)  ,
\]
and thus \eqref{eq.pieri.h} will be proven. Hence, in order to complete the
proof of \eqref{eq.pieri.h}, it is enough to prove that there exists a
bijection $\mathbf{i}:\mathbf{A}\rightarrow\mathbf{B}$ which has the property
(\ref{sol.exe.pieri.RSK.h.prop}).

We construct such a bijection $\mathbf{i}:\mathbf{A}\rightarrow\mathbf{B}$
explicitly. Namely, for every $\left(  T,\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \right)  \in\mathbf{A}$, we define $\mathbf{i}\left(  \left(
T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  $ as follows:
Construct a sequence $\left(  T_{0},T_{1},\ldots,T_{n}\right)  $ of
column-strict tableaux recursively: We set $T_{0}=T$. For every $k\in\left\{
1,2,\ldots,n\right\}  $, if $T_{k-1}$ is already defined, we let $T_{k}$ be
the column-strict tableau obtained by applying RS-insertion to the tableau
$T_{k-1}$ and the letter $i_{k}$. (This RS-insertion also returns a corner
cell, but we do not care about it.) Thus, a sequence $\left(  T_{0}%
,T_{1},\ldots,T_{n}\right)  $ is defined. We now set $S=T_{n}$, and let
$\lambda^{+}$ be the shape of $S$. It is easy to see that $\lambda^{+}%
/\lambda$ is a horizontal $n$-strip\footnote{\textit{Proof.} Notice that
$i_{1}\leq i_{2}\leq\cdots\leq i_{n}$ (since $\left(  T,\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \right)  \in\mathbf{A}$). The tableau $S$ has been
obtained from $T$ by applying RS-insertion $n$ times, using the letters
$i_{1},i_{2},\ldots,i_{n}$ in this order. Each time that we have applied
RS-insertion, the shape of our tableau has grown by a new cell. According to
Lemma \ref{lem.sol.pieri.RSK.rbl}(a), each of these cells (except for the
first one) lies in a column further right than the previous one (because the
letters $i_{1},i_{2},\ldots,i_{n}$ that we inserted satisfy $i_{1}\leq
i_{2}\leq\cdots\leq i_{n}$). Therefore, no two of these cells lie in the same
column. Since these cells are precisely the cells of $\lambda^{+}/\lambda$
(because $\lambda^{+}$ is the shape of $S$, while $\lambda$ is the shape of
$T$), this means that no two cells of $\lambda^{+}/\lambda$ lie in the same
column. In other words, $\lambda^{+}/\lambda$ is a horizontal strip. Since
$\lambda^{+}/\lambda$ has precisely $n$ cells (because we have applied
RS-insertion exactly $n$ times, gaining precisely one cell every time), this
yields that $\lambda^{+}/\lambda$ is a horizontal $n$-strip, qed.}. Thus,
$\left(  \lambda^{+},S\right)  \in\mathbf{B}$. Now, set $\mathbf{i}\left(
\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  =\left(
\lambda^{+},S\right)  $. We have therefore defined a map $\mathbf{i}%
:\mathbf{A}\rightarrow\mathbf{B}$. It remains to prove that this map
$\mathbf{i}$ is a bijection and satisfies (\ref{sol.exe.pieri.RSK.h.prop}).

Proving that the map $\mathbf{i}$ satisfies (\ref{sol.exe.pieri.RSK.h.prop})
is easy\footnote{\textit{Proof.} Assume that some $\left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  \in\mathbf{A}$ and $\left(
\lambda^{+},S\right)  \in\mathbf{B}$ satisfy $\mathbf{i}\left(  \left(
T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  =\left(
\lambda^{+},S\right)  $. We need to show that (\ref{sol.exe.pieri.RSK.h.prop})
holds.
\par
According to the definition of $\mathbf{i}\left(  \left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  $, the tableau $S$ is
obtained by successively applying RS-insertion to the tableau $T$ using the
letters $i_{1},i_{2},\ldots,i_{n}$. But whenever a tableau $V$ results from
applying RS-insertion to a column-strict tableau $U$ and a letter $j$, the
multiset of entries of $V$ is obtained from the multiset of entries of $U$ by
tossing in the letter $j$. Thus, the multiset of entries of $S$ is obtained
from the multiset of entries of $T$ by tossing in the $n$ letters $i_{1}%
,i_{2},\ldots,i_{n}$ (because $S$ is obtained by successively applying
RS-insertion to the tableau $T$ using the letters $i_{1},i_{2},\ldots,i_{n}$).
Hence, ${\mathbf{x}}^{\cont \left(  S\right)  }={\xx%
}^{\cont \left(  T\right)  }x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$.
Thus, (\ref{sol.exe.pieri.RSK.h.prop}) is proven.}. It remains to show that
$\mathbf{i}$ is a bijection. We will achieve this by constructing an inverse map.

Indeed, let us define a map $\mathbf{r}:\mathbf{B}\rightarrow\mathbf{A}$. For
every $\left(  \lambda^{+},S\right)  \in\mathbf{B}$, we define $\mathbf{r}%
\left(  \left(  \lambda^{+},S\right)  \right)  $ as follows: We know that
$\lambda^{+}/\lambda$ is a horizontal $n$-strip (since $\left(  \lambda
^{+},S\right)  \in\mathbf{B}$). We can thus uniquely label the $n$ cells of
$\lambda^{+}/\lambda$ by $c_{1},c_{2},\ldots,c_{n}$ from right to left.
Consider these cells $c_{1},c_{2},\ldots,c_{n}$. Construct a sequence $\left(
S_{0},S_{1},\ldots,S_{n}\right)  $ of column-strict tableaux and a sequence
$\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ of positive integers recursively:
We set $S_{0}=S$. For every $k\in\left\{  1,2,\ldots,n\right\}  $, if
$S_{k-1}$ is already defined, we apply reverse bumping to the tableau
$S_{k-1}$ and its corner cell $c_{k}$. We denote the resulting tableau by
$S_{k}$, and the resulting letter by $j_{k}$.\ \ \ \ \footnote{In order to
verify that this definition makes sense, we need to check that $c_k$ is a
corner cell of $S_{k-1}$ for each $k \in \left\{1, 2, \ldots, n\right\}$.
Let us sketch a proof of this fact.
\par
In fact, we shall show a stronger claim:
\par
\begin{statement}
\textit{Claim CC:} For each
$k \in \left\{1, 2, \ldots, n\right\}$, the shape of the tableau $S_{k-1}$
is obtained from $\lambda^{+}$ by removing the corner cells $c_1, c_2,
\ldots, c_{k-1}$, and the cell $c_k$ is a corner cell of $S_{k-1}$.
\end{statement}
\par
\textit{Proof.} We use induction over $k$.
\par
\textit{Induction base:} The shape of the
tableau $S_0$ is $\lambda^{+}$ (since $S_0 = S$ has shape $\lambda^{+}$),
which is clearly the shape obtained from $\lambda^{+}$ by removing the
corner cells $c_1, c_2, \ldots, c_0$.
Moreover, the cell $c_1$ is the rightmost cell of $\lambda^{+}/\lambda$
(since we labelled the cells of $\lambda^{+}/\lambda$ by $c_1, c_2, \ldots,
c_n$ from right to left), and thus has no cells of $\lambda^{+}$ to its
right in its row;
but it also has no cells of $\lambda^{+}$ below it in its column (since any
such cell would have to belong to $\lambda^{+}/\lambda$, but
$\lambda^{+}/\lambda$ is a horizontal strip and therefore cannot have two
cells in the same column). Thus, $c_1$ is a corner cell of $S = S_0$.
Therefore, Claim CC is proven for $k = 1$. This completes the induction
base.
\par
\textit{Induction step:} Fix some
$K \in \left\{1, 2, \ldots, n-1\right\}$. Assume that Claim CC is proven
for $k = K$. We need to show that Claim CC holds for $k = K+1$.
\par
We know that Claim CC is proven for $k = K$. In other words, the shape of
the tableau $S_{K-1}$ is obtained from $\lambda^{+}$ by removing the
corner cells $c_1, c_2, \ldots, c_{K-1}$, and the cell $c_K$ is a corner
cell of $S_{K-1}$. Now, the shape of the tableau $S_K$ is obtained from
the shape of $S_{K-1}$ by removing the corner cell $c_K$ (because $S_K$
is obtained by applying reverse bumping to the tableau $S_{K-1}$ and its
corner cell $c_K$), and thus is obtained from $\lambda^{+}$ by removing
the corner cells $c_1, c_2, \ldots, c_K$ (since the shape of
the tableau $S_{K-1}$ is obtained from $\lambda^{+}$ by removing the
corner cells $c_1, c_2, \ldots, c_{K-1}$).
Hence, $c_1, c_2, \ldots, c_K$ are not cells of $S_K$, but
$c_{K+1}$ is a cell of $S_K$ (since $c_{K+1}$ is a cell of
$\lambda^{+} / \lambda$, therefore is a cell of $\lambda^{+}$,
and is not among the $K$ cells $c_1, c_2, \ldots, c_K$
that were removed).
We shall now show that the cell $c_{K+1}$ is a corner cell of $S_K$.
\par
Indeed, assume the contrary.
Thus, $c_{K+1}$ is a cell of $S_K$, but not a corner cell.
Hence, there exists a cell $d$ of $S_K$ that is either the bottom
neighbor or the right neighbor of $c_{K+1}$ (that is, it either lies
in the same column as $c_{K+1}$ but one step further down, or lies in
the same row as $c_{K+1}$ but one step further right).
Consider this cell $d$.
\par
Recall that the cells of $\lambda^{+}/\lambda$ have been labelled
$c_1, c_2, \ldots, c_n$ from right to left. Therefore, the cell $c_{K+1}$
belongs to $\lambda^{+}/\lambda$.
Hence, $c_{K+1}$ is not a cell of $\lambda$.
Therefore, $d$ is not a cell of $\lambda$ either
(since $d$ is either the bottom neighbor or the right neighbor of $c_{K+1}$).
But $d$ is a cell of $S_K$, thus a cell of $\lambda^{+}$
(since the shape of the tableau $S_K$ is obtained from $\lambda^{+}$
by removing some cells).
Hence, the cell $d$ must also belong to $\lambda^{+}/\lambda$
(since $d$ is not a cell of $\lambda$).
If $d$ was the bottom neighbor of $c_{K+1}$, we would thus conclude
that $c_{K+1}$ and $d$ are two cells of $\lambda^{+}/\lambda$ that
lie in the same column; but this is impossible
(since $\lambda^{+} / \lambda$ is a horizontal strip and thus cannot have
two cells in the same column).
Hence, $d$ cannot be the bottom neighbor of $c_{K+1}$.
Thus, $d$ must be the right neighbor of $c_{K+1}$ (since $d$ is
either the bottom neighbor or the right neighbor of $c_{K+1}$).
Thus, $d$ lies further right than $c_{K+1}$.
But all cells of $\lambda^{+}/\lambda$
that lie further right than $c_{K+1}$ are $c_1, c_2, \ldots, c_K$
(since the cells of $\lambda^{+}/\lambda$ have been labelled
$c_1, c_2, \ldots, c_n$ from right to left).
Hence, any cell of $\lambda^{+}/\lambda$ that lies further right
than $c_{K+1}$ must be one of $c_1, c_2, \ldots, c_K$.
Thus, $d$ must be one of $c_1, c_2, \ldots, c_K$
(since $d$ is a cell of $\lambda^{+}/\lambda$ that lies further right
than $c_{K+1}$).
Hence, $d$ cannot be a cell of $S_K$
(since $c_1, c_2, \ldots, c_K$ are not cells of $S_K$).
This contradicts the fact that $d$ is a cell of $S_K$.
\par
This contradiction shows that our assumption was wrong.
Hence, the cell $c_{K+1}$ is a corner cell of $S_K$.
\par
Thus we have shown that the shape of the tableau $S_K$
is obtained from $\lambda^{+}$ by removing the corner cells $c_1, c_2,
\ldots, c_K$, and the cell $c_{K+1}$ is a corner cell of $S_K$.
This proves Claim CC for $k = K+1$. Thus, the induction step is
finished. Claim CC is thus proven. We hence conclude that our
definition makes sense.}
Thus, two sequences $\left(
S_{0},S_{1},\ldots,S_{n}\right)  $ and $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $ are defined. We now set $T=S_{n}$ and $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  =\left(  j_{n},j_{n-1},\ldots,j_{1}\right)  $. So
the tableau $T$ is obtained from $S$ by successively applying reverse bumping
using the corner cells $c_{1},c_{2},\ldots,c_{n}$ (in this order), and
$j_{1},j_{2},\ldots,j_{n}$ are the letters that are obtained from the
reverse-bumping procedure. Since reverse bumping is the inverse map to
RS-insertion, this yields that we can obtain $S$ back from $T$ by successively
applying RS-insertion using the letters $j_{n},j_{n-1},\ldots,j_{1}$ (that is,
the letters $i_{1},i_{2},\ldots,i_{n}$), and that these successive
RS-insertion steps recover the corner cells $c_{n},c_{n-1},\ldots,c_{1}$ in
this order. Now, it is easy to see that the tableau $T$ has shape $\lambda$
(because in passing from $S$ to $T$, we lost the corner cells $c_{n}%
,c_{n-1},\ldots,c_{1}$, which are exactly the cells of $\lambda^{+}/\lambda$)
and the letters $i_{1},i_{2},\ldots,i_{n}$ satisfy $i_{1}\leq i_{2}\leq
\cdots\leq i_{n}$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then,
we don't have $i_{1}\leq i_{2}\leq\cdots\leq i_{n}$. In other words, we don't
have $j_{n}\leq j_{n-1}\leq\cdots\leq j_{1}$ (since $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  =\left(  j_{n},j_{n-1},\ldots,j_{1}\right)  $).
Hence, there exists a $k\in\left\{  2,3,\ldots,n\right\}  $ such that
$j_{k}>j_{k-1}$. Consider this $k$. Recall that the tableau $S_{k}$ and the
letter $j_{k}$ were obtained by applying reverse bumping to the tableau
$S_{k-1}$ and its corner cell $c_{k}$, while the tableau $S_{k-1}$ and the
letter $j_{k-1}$ were obtained by applying reverse bumping to the tableau
$S_{k-2}$ and its corner cell $c_{k-1}$. Since reverse bumping is the inverse
map to RS-insertion, this entails that conversely, the tableau $S_{k-1}$ and
its corner cell $c_{k}$ are obtained by applying RS-insertion to the tableau
$S_{k}$ and the letter $j_{k}$, and the tableau $S_{k-2}$ and its corner cell
$c_{k-1}$ are obtained by applying RS-insertion to the tableau $S_{k-1}$ and
the letter $j_{k-1}$. Hence, Lemma \ref{lem.sol.pieri.RSK.rbl}(b) (applied to
$S_{k}$, $j_{k}$, $j_{k-1}$, $S_{k-1}$, $c_{k}$, $S_{k-2}$ and $c_{k-1}$
instead of $P$, $j$, $j^{\prime}$, $P^{\prime}$, $c$, $P^{\prime\prime}$ and
$c^{\prime}$) yields that the cell $c_{k-1}$ is in the same column as $c_{k}$
or in a column further left. But this contradicts the fact that $c_{k-1}$ lies
in a column further right than $c_{k}$ (since the cells of $\lambda
^{+}/\lambda$ were labelled by $c_{1},c_{2},\ldots,c_{n}$ \textbf{from right
to left}, and lie in different columns). This contradiction completes our
proof.}. Hence, $\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\in\mathbf{A}$. We set $\mathbf{r}\left(  \left(  \lambda^{+},S\right)
\right)  =\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)  $. The
map $\mathbf{r}:\mathbf{B}\rightarrow\mathbf{A}$ is thus defined. It is now
easy to prove that the maps $\mathbf{i}$ and $\mathbf{r}$ are mutually inverse
(since RS-insertion and reverse bumping are inverse maps), and thus
$\mathbf{i}$ is a bijection. Thus, there exists a bijection $\mathbf{i}%
:\mathbf{A}\rightarrow\mathbf{B}$ which has the property
(\ref{sol.exe.pieri.RSK.h.prop}). This completes our proof of \eqref{eq.pieri.h}.
\end{proof}

The proof of \eqref{eq.pieri.e} is almost entirely analogous. We give it for
the sake of completeness (but most of it is copypasted material from the proof above).

\begin{proof}
[Alternative proof of \eqref{eq.pieri.e}.]Let $\lambda$ be a partition, and
let $n\in \NN$. The definition of $s_{\lambda}$ yields%
\[
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda}}{\mathbf{x}}^{\cont \left(  T\right)  }.
\]
The definition of $h_{n}$ yields
\begin{equation}
e_{n}=\sum_{i_{1}<i_{2}<\cdots<i_{n}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}%
=\sum_{i_{n}>i_{n-1}>\cdots>i_{1}}x_{i_{n}}x_{i_{n-1}}\cdots x_{i_{1}}%
=\sum_{i_{1}>i_{2}>\cdots>i_{n}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}
\label{sol.exe.pieri.RSK.e.1}%
\end{equation}
(here, we substituted $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ for $\left(
i_{n},i_{n-1},\ldots,i_{1}\right)  $ in the sum). Multiplying these two
identities, we obtain
\begin{align}
s_{\lambda}e_{n}  &  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda}}{\mathbf{x}}^{\cont \left(
T\right)  }\sum_{i_{1}>i_{2}>\cdots>i_{n}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}%
}\nonumber\\
&  =\sum_{\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\in\mathbf{A}}{\mathbf{x}}^{\cont \left(  T\right)  }x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}, \label{sol.exe.pieri.RSK.e.lhs}%
\end{align}
where $\mathbf{A}$ is the set of all pairs $\left(  T,\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  \right)  $ of a column-strict tableau $T$ of shape
$\lambda$ and an $n$-tuple $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ of
positive integers satisfying $i_{1}>i_{2}>\cdots>i_{n}$. Consider this set
$\mathbf{A}$.

On the other hand, every partition $\lambda^{+}$ satisfies
\begin{align}
s_{\lambda^{+}}  &  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda^{+}}}{\mathbf{x}}^{\cont
\left(  T\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}s_{\lambda^{+}}\right) \nonumber\\
&  =\sum\limits_{\substack{S\text{ is a column-strict}\\\text{tableau of shape
}\lambda^{+}}}{\mathbf{x}}^{\cont \left(  S\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}T\text{ as }S\right)  . \label{sol.exe.pieri.RSK.e.slp}%
\end{align}
Hence,%
\begin{align}
\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }\\\text{vertical
}n\text{-strip}}}\underbrace{s_{\lambda^{+}}}_{\substack{=\sum\limits
_{\substack{S\text{ is a column-strict}\\\text{tableau of shape }\lambda^{+}%
}}{\mathbf{x}}^{\cont \left(  S\right)  }\\\text{(by
(\ref{sol.exe.pieri.RSK.e.slp}))}}}  &  =\sum\limits_{\substack{\lambda^{+}%
:\lambda^{+}/\lambda\text{ is a }\\\text{vertical }n\text{-strip}}%
}\sum\limits_{\substack{S\text{ is a column-strict}\\\text{tableau of shape }%
\lambda^{+}}}{\mathbf{x}}^{\cont \left(  S\right)  }\nonumber\\
&  =\sum_{\left(  \lambda^{+},S\right)  \in\mathbf{B}}{\xx%
}^{\cont \left(  S\right)  }, \label{sol.exe.pieri.RSK.e.rhs}%
\end{align}
where $\mathbf{B}$ is the set of all pairs $\left(  \lambda^{+},S\right)  $ of
a partition $\lambda^{+}$ and a column-strict tableau $S$ of shape
$\lambda^{+}$ such that $\lambda^{+}/\lambda$ is a vertical $n$-strip.
Consider this set $\mathbf{B}$.

We shall now prove that there exists a bijection $\mathbf{i}:\mathbf{A}%
\rightarrow\mathbf{B}$ which has the property that%
\begin{equation}
{\mathbf{x}}^{\cont \left(  T\right)  }x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}={\mathbf{x}}^{\cont \left(  S\right)  }
\label{sol.exe.pieri.RSK.e.prop}%
\end{equation}
whenever some $\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\in\mathbf{A}$ and $\left(  \lambda^{+},S\right)  \in\mathbf{B}$ satisfy
$\mathbf{i}\left(  \left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)
\right)  =\left(  \lambda^{+},S\right)  $. Once this will be proven, it will
immediately follow that $\sum_{\left(  T,\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \right)  \in\mathbf{A}}{\mathbf{x}}^{\cont \left(
T\right)  }x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}=\sum_{\left(  \lambda
^{+},S\right)  \in\mathbf{B}}{\mathbf{x}}^{\cont \left(
S\right)  }$, and therefore (\ref{sol.exe.pieri.RSK.e.lhs}) will become%
\[
s_{\lambda}e_{n}=\sum_{\left(  T,\left(  i_{1},i_{2},\ldots,i_{n}\right)
\right)  \in\mathbf{A}}{\mathbf{x}}^{\cont \left(  T\right)
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}=\sum_{\left(  \lambda^{+},S\right)
\in\mathbf{B}}{\mathbf{x}}^{\cont \left(  S\right)  }%
=\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }\\\text{vertical
}n\text{-strip}}}s_{\lambda^{+}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.pieri.RSK.e.rhs})}\right)  ,
\]
and thus \eqref{eq.pieri.e} will be proven. Hence, in order to complete the
proof of \eqref{eq.pieri.e}, it is enough to prove that there exists a
bijection $\mathbf{i}:\mathbf{A}\rightarrow\mathbf{B}$ which has the property
(\ref{sol.exe.pieri.RSK.e.prop}).

We construct such a bijection $\mathbf{i}:\mathbf{A}\rightarrow\mathbf{B}$
explicitly. Namely, for every $\left(  T,\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \right)  \in\mathbf{A}$, we define $\mathbf{i}\left(  \left(
T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  $ as follows:
Construct a sequence $\left(  T_{0},T_{1},\ldots,T_{n}\right)  $ of
column-strict tableaux recursively: We set $T_{0}=T$. For every $k\in\left\{
1,2,\ldots,n\right\}  $, if $T_{k-1}$ is already defined, we let $T_{k}$ be
the column-strict tableau obtained by applying RS-insertion to the tableau
$T_{k-1}$ and the letter $i_{k}$. (This RS-insertion also returns a corner
cell, but we do not care about it.) Thus, a sequence $\left(  T_{0}%
,T_{1},\ldots,T_{n}\right)  $ is defined. We now set $S=T_{n}$, and let
$\lambda^{+}$ be the shape of $S$. It is easy to see that $\lambda^{+}%
/\lambda$ is a vertical $n$-strip\footnote{\textit{Proof.} Notice that
$i_{1}>i_{2}>\cdots>i_{n}$ (since $\left(  T,\left(  i_{1},i_{2},\ldots
,i_{n}\right)  \right)  \in\mathbf{A}$). The tableau $S$ has been obtained
from $T$ by applying RS-insertion $n$ times, using the letters $i_{1}%
,i_{2},\ldots,i_{n}$ in this order. Each time that we have applied
RS-insertion, the shape of our tableau has grown by a new cell. According to
Lemma \ref{lem.sol.pieri.RSK.rbl}(b), each of these cells (except for the
first one) lies in a row further down than the previous one (because the
letters $i_{1},i_{2},\ldots,i_{n}$ that we inserted satisfy $i_{1}%
>i_{2}>\cdots>i_{n}$). Therefore, no two of these cells lie in the same row.
Since these cells are precisely the cells of $\lambda^{+}/\lambda$ (because
$\lambda^{+}$ is the shape of $S$, while $\lambda$ is the shape of $T$), this
means that no two cells of $\lambda^{+}/\lambda$ lie in the same row. In other
words, $\lambda^{+}/\lambda$ is a vertical strip. Since $\lambda^{+}/\lambda$
has precisely $n$ cells (because we have applied RS-insertion exactly $n$
times, gaining precisely one cell every time), this yields that $\lambda
^{+}/\lambda$ is a vertical $n$-strip, qed.}. Thus, $\left(  \lambda
^{+},S\right)  \in\mathbf{B}$. Now, set $\mathbf{i}\left(  \left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  =\left(  \lambda
^{+},S\right)  $. We have therefore defined a map $\mathbf{i}:\mathbf{A}%
\rightarrow\mathbf{B}$. It remains to prove that this map $\mathbf{i}$ is a
bijection and satisfies (\ref{sol.exe.pieri.RSK.e.prop}).

Proving that the map $\mathbf{i}$ satisfies (\ref{sol.exe.pieri.RSK.e.prop})
is easy\footnote{\textit{Proof.} Assume that some $\left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  \in\mathbf{A}$ and $\left(
\lambda^{+},S\right)  \in\mathbf{B}$ satisfy $\mathbf{i}\left(  \left(
T,\left(  i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  =\left(
\lambda^{+},S\right)  $. We need to show that (\ref{sol.exe.pieri.RSK.e.prop})
holds.
\par
According to the definition of $\mathbf{i}\left(  \left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  \right)  $, the tableau $S$ is
obtained by successively applying RS-insertion to the tableau $T$ using the
letters $i_{1},i_{2},\ldots,i_{n}$. But whenever a tableau $V$ results from
applying RS-insertion to a column-strict tableau $U$ and a letter $j$, the
multiset of entries of $V$ is obtained from the multiset of entries of $U$ by
tossing in the letter $j$. Thus, the multiset of entries of $S$ is obtained
from the multiset of entries of $T$ by tossing in the $n$ letters $i_{1}%
,i_{2},\ldots,i_{n}$ (because $S$ is obtained by successively applying
RS-insertion to the tableau $T$ using the letters $i_{1},i_{2},\ldots,i_{n}$).
Hence, ${\mathbf{x}}^{\cont \left(  S\right)  }={\xx%
}^{\cont \left(  T\right)  }x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$.
Thus, (\ref{sol.exe.pieri.RSK.e.prop}) is proven.}. It remains to show that
$\mathbf{i}$ is a bijection. We will achieve this by constructing an inverse map.

\begin{vershort}
Indeed, let us define a map $\mathbf{r}:\mathbf{B}\rightarrow\mathbf{A}$. For
every $\left(  \lambda^{+},S\right)  \in\mathbf{B}$, we define $\mathbf{r}%
\left(  \left(  \lambda^{+},S\right)  \right)  $ as follows: We know that
$\lambda^{+}/\lambda$ is a vertical $n$-strip (since $\left(  \lambda
^{+},S\right)  \in\mathbf{B}$). We can thus uniquely label the $n$ cells of
$\lambda^{+}/\lambda$ by $c_{1},c_{2},\ldots,c_{n}$ from bottom to top.
Consider these cells $c_{1},c_{2},\ldots,c_{n}$. Construct a sequence $\left(
S_{0},S_{1},\ldots,S_{n}\right)  $ of column-strict tableaux and a sequence
$\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ of positive integers recursively:
We set $S_{0}=S$. For every $k\in\left\{  1,2,\ldots,n\right\}  $, if
$S_{k-1}$ is already defined, we apply reverse bumping to the tableau
$S_{k-1}$ and its corner cell $c_{k}$. We denote the resulting tableau by
$S_{k}$, and the resulting letter by $j_{k}$.\ \ \ \ \footnote{One again
has to check that this is well-defined. The proof is very similar to the
analogous argument in our proof of \eqref{eq.pieri.h}, and is left to the
reader.}
\end{vershort}
%
\begin{verlong}
Indeed, let us define a map $\mathbf{r}:\mathbf{B}\rightarrow\mathbf{A}$. For
every $\left(  \lambda^{+},S\right)  \in\mathbf{B}$, we define $\mathbf{r}%
\left(  \left(  \lambda^{+},S\right)  \right)  $ as follows: We know that
$\lambda^{+}/\lambda$ is a vertical $n$-strip (since $\left(  \lambda
^{+},S\right)  \in\mathbf{B}$). We can thus uniquely label the $n$ cells of
$\lambda^{+}/\lambda$ by $c_{1},c_{2},\ldots,c_{n}$ from bottom to top.
Consider these cells $c_{1},c_{2},\ldots,c_{n}$. Construct a sequence $\left(
S_{0},S_{1},\ldots,S_{n}\right)  $ of column-strict tableaux and a sequence
$\left(  j_{1},j_{2},\ldots,j_{n}\right)  $ of positive integers recursively:
We set $S_{0}=S$. For every $k\in\left\{  1,2,\ldots,n\right\}  $, if
$S_{k-1}$ is already defined, we apply reverse bumping to the tableau
$S_{k-1}$ and its corner cell $c_{k}$. We denote the resulting tableau by
$S_{k}$, and the resulting letter by $j_{k}$.\ \ \ \ \footnote{In order to
verify that this definition makes sense, we need to check that $c_k$ is a
corner cell of $S_{k-1}$ for each $k \in \left\{1, 2, \ldots, n\right\}$.
Let us sketch a proof of this fact.
\par
In fact, we shall show a stronger claim:
\par
\begin{statement}
\textit{Claim CC:} For each
$k \in \left\{1, 2, \ldots, n\right\}$, the shape of the tableau $S_{k-1}$
is obtained from $\lambda^{+}$ by removing the corner cells $c_1, c_2,
\ldots, c_{k-1}$, and the cell $c_k$ is a corner cell of $S_{k-1}$.
\end{statement}
\par
\textit{Proof.} We use induction over $k$.
\par
\textit{Induction base:} The shape of the
tableau $S_0$ is $\lambda^{+}$ (since $S_0 = S$ has shape $\lambda^{+}$),
which is clearly the shape obtained from $\lambda^{+}$ by removing the
corner cells $c_1, c_2, \ldots, c_0$.
Moreover, the cell $c_1$ is the bottommost cell of $\lambda^{+}/\lambda$
(since we labelled the cells of $\lambda^{+}/\lambda$ by $c_1, c_2, \ldots,
c_n$ from bottom to top), and thus has no cells of $\lambda^{+}$ below
it in its column;
but it also has no cells of $\lambda^{+}$ to its right in its row
(since any such cell would have to belong to $\lambda^{+}/\lambda$, but
$\lambda^{+}/\lambda$ is a vertical strip and therefore cannot have two
cells in the same row). Thus, $c_1$ is a corner cell of $S = S_0$.
Therefore, Claim CC is proven for $k = 1$. This completes the
induction base.
\par
\textit{Induction step:} Fix some
$K \in \left\{1, 2, \ldots, n-1\right\}$. Assume that Claim CC is proven
for $k = K$. We need to show that Claim CC holds for $k = K+1$.
\par
We know that Claim CC is proven for $k = K$. In other words, the shape of
the tableau $S_{K-1}$ is obtained from $\lambda^{+}$ by removing the
corner cells $c_1, c_2, \ldots, c_{K-1}$, and the cell $c_K$ is a corner
cell of $S_{K-1}$. Now, the shape of the tableau $S_K$ is obtained from
the shape of $S_{K-1}$ by removing the corner cell $c_K$ (because $S_K$
is obtained by applying reverse bumping to the tableau $S_{K-1}$ and its
corner cell $c_K$), and thus is obtained from $\lambda^{+}$ by removing
the corner cells $c_1, c_2, \ldots, c_K$ (since the shape of
the tableau $S_{K-1}$ is obtained from $\lambda^{+}$ by removing the
corner cells $c_1, c_2, \ldots, c_{K-1}$).
Hence, $c_1, c_2, \ldots, c_K$ are not cells of $S_K$, but
$c_{K+1}$ is a cell of $S_K$ (since $c_{K+1}$ is a cell of
$\lambda^{+} / \lambda$, therefore is a cell of $\lambda^{+}$,
and is not among the $K$ cells $c_1, c_2, \ldots, c_K$
that were removed).
We shall now show that the cell $c_{K+1}$ is a corner cell of $S_K$.
\par
Indeed, assume the contrary.
Thus, $c_{K+1}$ is a cell of $S_K$, but not a corner cell.
Hence, there exists a cell $d$ of $S_K$ that is either the bottom
neighbor or the right neighbor of $c_{K+1}$ (that is, it either lies
in the same column as $c_{K+1}$ but one step further down, or lies in
the same row as $c_{K+1}$ but one step further right).
Consider this cell $d$.
\par
Recall that the cells of $\lambda^{+}/\lambda$ have been labelled
$c_1, c_2, \ldots, c_n$ from bottom to top. Therefore, the cell $c_{K+1}$
belongs to $\lambda^{+}/\lambda$.
Hence, $c_{K+1}$ is not a cell of $\lambda$.
Therefore, $d$ is not a cell of $\lambda$ either
(since $d$ is either the bottom neighbor or the right neighbor of $c_{K+1}$).
But $d$ is a cell of $S_K$, thus a cell of $\lambda^{+}$
(since the shape of the tableau $S_K$ is obtained from $\lambda^{+}$
by removing some cells).
Hence, the cell $d$ must also belong to $\lambda^{+}/\lambda$
(since $d$ is not a cell of $\lambda$).
If $d$ was the right neighbor of $c_{K+1}$, we would thus conclude
that $c_{K+1}$ and $d$ are two cells of $\lambda^{+}/\lambda$ that
lie in the same row; but this is impossible
(since $\lambda^{+} / \lambda$ is a vertical strip and thus cannot have
two cells in the same row).
Hence, $d$ cannot be the right neighbor of $c_{K+1}$.
Thus, $d$ must be the bottom neighbor of $c_{K+1}$ (since $d$ is
either the bottom neighbor or the right neighbor of $c_{K+1}$).
Thus, $d$ lies further below than $c_{K+1}$.
But all cells of $\lambda^{+}/\lambda$
that lie further below than $c_{K+1}$ are $c_1, c_2, \ldots, c_K$
(since the cells of $\lambda^{+}/\lambda$ have been labelled
$c_1, c_2, \ldots, c_n$ from bottom to top).
Hence, any cell of $\lambda^{+}/\lambda$ that lies further below
than $c_{K+1}$ must be one of $c_1, c_2, \ldots, c_K$.
Thus, $d$ must be one of $c_1, c_2, \ldots, c_K$
(since $d$ is a cell of $\lambda^{+}/\lambda$ that lies further below
than $c_{K+1}$).
Hence, $d$ cannot be a cell of $S_K$
(since $c_1, c_2, \ldots, c_K$ are not cells of $S_K$).
This contradicts the fact that $d$ is a cell of $S_K$.
\par
This contradiction shows that our assumption was wrong.
Hence, the cell $c_{K+1}$ is a corner cell of $S_K$.
\par
Thus we have shown that the shape of the tableau $S_K$
is obtained from $\lambda^{+}$ by removing the corner cells $c_1, c_2,
\ldots, c_K$, and the cell $c_{K+1}$ is a corner cell of $S_K$.
This proves Claim CC for $k = K+1$. Thus, the induction step is
finished. Claim CC is thus proven. We hence conclude that our
definition makes sense.}
\end{verlong}
Thus, two sequences $\left(
S_{0},S_{1},\ldots,S_{n}\right)  $ and $\left(  j_{1},j_{2},\ldots
,j_{n}\right)  $ are defined. We now set $T=S_{n}$ and $\left(  i_{1}%
,i_{2},\ldots,i_{n}\right)  =\left(  j_{n},j_{n-1},\ldots,j_{1}\right)  $. So
the tableau $T$ is obtained from $S$ by successively applying reverse bumping
using the corner cells $c_{1},c_{2},\ldots,c_{n}$ (in this order), and
$j_{1},j_{2},\ldots,j_{n}$ are the letters that are obtained from the
reverse-bumping procedure. Since reverse bumping is the inverse map to
RS-insertion, this yields that we can obtain $S$ back from $T$ by successively
applying RS-insertion using the letters $j_{n},j_{n-1},\ldots,j_{1}$ (that is,
the letters $i_{1},i_{2},\ldots,i_{n}$), and that these successive
RS-insertion steps recover the corner cells $c_{n},c_{n-1},\ldots,c_{1}$ in
this order. Now, it is easy to see that the tableau $T$ has shape $\lambda$
(because in passing from $S$ to $T$, we lost the corner cells $c_{n}%
,c_{n-1},\ldots,c_{1}$, which are exactly the cells of $\lambda^{+}/\lambda$)
and the letters $i_{1},i_{2},\ldots,i_{n}$ satisfy $i_{1}>i_{2}>\cdots>i_{n}%
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, we don't have
$i_{1}>i_{2}>\cdots>i_{n}$. In other words, we don't have $j_{n}%
>j_{n-1}>\cdots>j_{1}$ (since $\left(  i_{1},i_{2},\ldots,i_{n}\right)
=\left(  j_{n},j_{n-1},\ldots,j_{1}\right)  $). Hence, there exists a
$k\in\left\{  2,3,\ldots,n\right\}  $ such that $j_{k}\leq j_{k-1}$. Consider
this $k$. Recall that the tableau $S_{k}$ and the letter $j_{k}$ were obtained
by applying reverse bumping to the tableau $S_{k-1}$ and its corner cell
$c_{k}$, while the tableau $S_{k-1}$ and the letter $j_{k-1}$ were obtained by
applying reverse bumping to the tableau $S_{k-2}$ and its corner cell
$c_{k-1}$. Since reverse bumping is the inverse map to RS-insertion, this
entails that conversely, the tableau $S_{k-1}$ and its corner cell $c_{k}$ are
obtained by applying RS-insertion to the tableau $S_{k}$ and the letter
$j_{k}$, and the tableau $S_{k-2}$ and its corner cell $c_{k-1}$ are obtained
by applying RS-insertion to the tableau $S_{k-1}$ and the letter $j_{k-1}$.
Hence, Lemma \ref{lem.sol.pieri.RSK.rbl}(a) (applied to $S_{k}$, $j_{k}$,
$j_{k-1}$, $S_{k-1}$, $c_{k}$, $S_{k-2}$ and $c_{k-1}$ instead of $P$, $j$,
$j^{\prime}$, $P^{\prime}$, $c$, $P^{\prime\prime}$ and $c^{\prime}$) yields
that the cell $c_{k-1}$ is in the same row as $c_{k}$ or in a row further up.
But this contradicts the fact that $c_{k-1}$ lies in a row further down than
$c_{k}$ (since the cells of $\lambda^{+}/\lambda$ were labelled by
$c_{1},c_{2},\ldots,c_{n}$ \textbf{from bottom to top}, and lie in different
rows). This contradiction completes our proof.}. Hence, $\left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  \in\mathbf{A}$. We set
$\mathbf{r}\left(  \left(  \lambda^{+},S\right)  \right)  =\left(  T,\left(
i_{1},i_{2},\ldots,i_{n}\right)  \right)  $. The map $\mathbf{r}%
:\mathbf{B}\rightarrow\mathbf{A}$ is thus defined. It is now easy to prove
that the maps $\mathbf{i}$ and $\mathbf{r}$ are mutually inverse (since
RS-insertion and reverse bumping are inverse maps), and thus $\mathbf{i}$ is a
bijection. Thus, there exists a bijection $\mathbf{i}:\mathbf{A}%
\rightarrow\mathbf{B}$ which has the property (\ref{sol.exe.pieri.RSK.e.prop}%
). This completes our proof of \eqref{eq.pieri.e}.
\end{proof}

Now, both \eqref{eq.pieri.h} and \eqref{eq.pieri.e} are proven. Hence,
Theorem~\ref{Pieri-rules} is proven again.
%\end{solutions}




\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.alt-e}}

\textit{Solution to Exercise~\ref{exe.pieri.alt-e}.} We do \textbf{not} fix
$\lambda$ and $n$. (In fact, we shall use the letters $\lambda$ and $n$ for
other things than what they mean in (\ref{eq.pieri.e}).)

We first prove a very simple fact:

\begin{lemma}
\label{lem.sol.pieri.alt-e.projlim}Let $f\in R\left(  \xx\right)  $.

\begin{enumerate}
\item[(a)] If $n\in \NN$ and a weak composition $\alpha=\left(
\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $ are such that \newline%
$\left(  \alpha_{i}=0\text{ for every positive integer }i>n\right)  $, then
\begin{align*}
&  \left(  \text{the coefficient of }\mathbf{x}^{\alpha}\text{ in }f\right) \\
&  =\left(  \text{the coefficient of }x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{n}^{\alpha_{n}}\text{ in }f\left(  x_{1},x_{2},\ldots,x_{n}\right)
\right)  .
\end{align*}
(See Exercise \ref{exe.Lambda.subs.fin} for the definition of $f\left(
x_{1},x_{2},\ldots,x_{n}\right)  $.)

\item[(b)] Let $\mathbf{n}\in \NN$. Assume that $f\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  =0$ for every $n\in \NN$ satisfying
$n\geq\mathbf{n}$. Then, $f=0$.
\end{enumerate}
\end{lemma}

This lemma is elementary and very easy, but let us nevertheless give a proof.

\begin{proof}
[Proof of Lemma \ref{lem.sol.pieri.alt-e.projlim}.](a) Whenever $\mathfrak{m}$
is a monomial and $p$ is a power series, we shall let $\operatorname*{Coeff}%
\nolimits_{\mathfrak{m}}p$ denote the coefficient of $\mathfrak{m}$ in $p$.

Let $n\in \NN$ and a weak composition $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\alpha_{3},\ldots\right)  $ be such that $\left(  \alpha
_{i}=0\text{ for every positive integer }i>n\right)  $. We clearly have%
\[
f=\sum\limits_{\substack{\beta\text{ is a}\\\text{weak composition}}}\left(
\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)  \xx%
^{\beta}.
\]
But
$f\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ is the result of substituting
$x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in $f$
(because this is how $f\left(  x_{1},x_{2},\ldots,x_{n}\right)  $
is defined).
Hence,%
\begin{align}
&  f\left(  x_{1},x_{2},\ldots,x_{n}\right) \nonumber\\
&  =\left(  \text{the result of substituting }x_{1},x_{2},\ldots
,x_{n},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in
}\underbrace{f}_{=\sum\limits_{\substack{\beta\text{ is a}\\\text{weak composition}%
}}\left(  \operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)
\mathbf{x}^{\beta}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }x_{1},x_{2},\ldots
,x_{n},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in }%
\sum\limits_{\substack{\beta\text{ is a}\\\text{weak composition}}}\left(
\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)  \xx%
^{\beta}\right) \nonumber\\
&  =\sum\limits_{\substack{\beta\text{ is a}\\\text{weak composition}}}\left(
\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)  \left(  \text{the
result of substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{ for }%
x_{1},x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\beta}\right)  .
\label{pf.lem.sol.pieri.alt-e.projlim.1}%
\end{align}


Now, let us say that a weak composition $\left(  \beta_{1},\beta_{2},\beta
_{3},\ldots\right)  $ is \emph{$n$-finished} if $\left(  \beta_{i}=0\text{
for every positive integer }i>n\right)  $. Let $\mathbf{F}$ denote the set of
all $n$-finished weak compositions. Let us first notice that if $\beta$ is a
weak composition which is \textbf{not} $n$-finished, then%
\begin{equation}
\left(  \text{the result of substituting }x_{1},x_{2},\ldots,x_{n}%
,0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\beta
}\right)  =0. \label{pf.lem.sol.pieri.alt-e.projlim.2a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.projlim.2a}):} Let
$\beta$ be a weak composition which is \textbf{not} $n$-finished. Write
$\beta$ in the form $\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $.
We know that the weak composition $\beta$ is not $n$-finished. In other words,
the weak composition $\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $
is not $n$-finished (since $\beta=\left(  \beta_{1},\beta_{2},\beta_{3}%
,\ldots\right)  $). In other words, we do not have $\left(  \beta_{i}=0\text{
for every positive integer }i>n\right)  $ (because the weak composition
$\left(  \beta
_{1},\beta_{2},\beta_{3},\ldots\right)  $ is $n$-finished if and only if
$\left(  \beta_{i}=0\text{ for every positive integer }i>n\right)  $
(according to the definition of ``$n$-finished'')).
In other words, there exists some positive
integer $i>n$ which does not satisfy $\beta_{i}=0$. Consider this $i$.
\par
We do not have $\beta_{i}=0$. Thus, $\beta_{i}>0$ (since $\beta_{i}%
\in \NN$), so that $0^{\beta_{i}}=0$. Now, $i$ is a positive integer
such that $i>n$. Hence, $i\geq n+1$ (since $i$ and $n$ are integers). Hence,
$0^{\beta_{i}}$ is a factor in the product $\prod_{j=n+1}^{\infty}0^{\beta
_{j}}$. This factor is $0$ (since $0^{\beta_{i}}=0$). Therefore, the product
$\prod_{j=n+1}^{\infty}0^{\beta_{j}}$ has one of its factors equal to $0$. As
a consequence, the whole product must be $0$. In other words, $\prod
_{j=n+1}^{\infty}0^{\beta_{j}}=0$.
\par
We have $\beta=\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $ and thus%
\begin{align*}
\mathbf{x}^{\beta}  &  =\mathbf{x}^{\left(  \beta_{1},\beta_{2},\beta
_{3},\ldots\right)  }=\prod_{j\geq1}x_{j}^{\beta_{j}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{x}^{\left(
\beta_{1},\beta_{2},\beta_{3},\ldots\right)  }\right) \\
&  =\left(  \prod_{j=1}^{n}x_{j}^{\beta_{j}}\right)  \left(  \prod
_{j=n+1}^{\infty}x_{j}^{\beta_{j}}\right)  .
\end{align*}
Thus,%
\begin{align*}
&  \left(  \text{the result of substituting }x_{1},x_{2},\ldots,x_{n}%
,0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in }%
\underbrace{\mathbf{x}^{\beta}}_{=\left(  \prod_{j=1}^{n}x_{j}^{\beta_{j}%
}\right)  \left(  \prod_{j=n+1}^{\infty}x_{j}^{\beta_{j}}\right)  }\right) \\
&  =\left(  \text{the result of substituting }x_{1},x_{2},\ldots
,x_{n},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in }\left(
\prod_{j=1}^{n}x_{j}^{\beta_{j}}\right)  \left(  \prod_{j=n+1}^{\infty}%
x_{j}^{\beta_{j}}\right)  \right) \\
&  =\left(  \prod_{j=1}^{n}x_{j}^{\beta_{j}}\right)  \underbrace{\left(
\prod_{j=n+1}^{\infty}0^{\beta_{j}}\right)  }_{=0}=0.
\end{align*}
This proves (\ref{pf.lem.sol.pieri.alt-e.projlim.2a}).} On the other hand, for
every $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \in
\NN^n$, we have $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n},0,0,0,\ldots\right)  \in\mathbf{F}$\ \ \ \ \footnote{\textit{Proof.} Let
$\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \in\NN^n$.
Let $\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  =\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  $. Then, $\left(
\beta_{1},\beta_{2},\beta_{3},\ldots\right)  $ is a sequence of nonnegative
integers. Also, every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies
\[
\beta_{i}=\begin{cases}
\gamma_{i},&\text{if }i\leq n;\\
0,&\text{if }i>n
\end{cases}
\]
(since $\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  =\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  $). Thus, every
$i\in\left\{  1,2,3,\ldots\right\}  $ satisfying $i>n$ satisfies $\beta
_{i}=
\begin{cases}
\gamma_{i},&\text{if }i\leq n;\\
0,&\text{if }i>n
\end{cases} \quad
=0$ (since $i>n$). Hence, all but finitely many $i\in\left\{
1,2,3,\ldots\right\}  $ satisfy $\beta_{i}=0$ (since all but finitely many
$i\in\left\{  1,2,3,\ldots\right\}  $ satisfy $i>n$). Thus, $\left(  \beta
_{1},\beta_{2},\beta_{3},\ldots\right)  $ is a weak composition.
\par
Every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfying $i>n$ satisfies
$\beta_{i}=0$. In other words, every positive integer $i$ satisfying $i>n$
satisfies $\beta_{i}=0$. In other words, $\beta_{i}=0$ for every positive
integer $i>n$. In other words, the weak composition $\left(  \beta_{1}%
,\beta_{2},\beta_{3},\ldots\right)  $ is $n$-finished (since the weak
composition $\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $ is
$n$-finished if and only if $\left(  \beta_{i}=0\text{ for every positive
integer }i>n\right)  $ (according to the definition of
``$n$-finished'')). In other words, $\left(  \beta_{1},\beta
_{2},\beta_{3},\ldots\right)  \in\mathbf{F}$ (because $\left(  \beta_{1}%
,\beta_{2},\beta_{3},\ldots\right)  \in\mathbf{F}$ if and only if the weak
composition $\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $ is
$n$-finished (since $\mathbf{F}$ is the set of all $n$-finished weak
compositions)). In other words, $\left(  \gamma_{1},\gamma_{2},\ldots
,\gamma_{n},0,0,0,\ldots\right)  \in\mathbf{F}$ (since $\left(  \beta
_{1},\beta_{2},\beta_{3},\ldots\right)  =\left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{n},0,0,0,\ldots\right)  $), qed.}. Hence, we can define a map
$\mathcal{T}:\NN^{n}\rightarrow\mathbf{F}$ by setting%
\[
\left(  \mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n}\right)  \right)  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n},0,0,0,\ldots\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \in\NN^{n}\right)  .
\]
Consider this map $\mathcal{T}$. Also, define a map $\mathcal{S}%
:\mathbf{F}\rightarrow\NN^n$ by setting
\[
\left(  \mathcal{S}\left(  \left(  \gamma_{1},\gamma_{2},\gamma_{3}%
,\ldots\right)  \right)  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \gamma_{1}%
,\gamma_{2},\gamma_{3},\ldots\right)  \in\mathbf{F}\right)  .
\]
Then, $\mathcal{T}\circ\mathcal{S}=\operatorname*{id}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\gamma\in\mathbf{F}$. Then, $\gamma$ is
an element of $F$. In other words, $\gamma$ is an $n$-finished weak
composition (since $\mathbf{F}$ is the set of all $n$-finished weak
compositions).
\par
Let us write $\gamma$ in the form $\left(  \gamma_{1},\gamma_{2},\gamma
_{3},\ldots\right)  $. Recall that the weak composition $\gamma$ is
$n$-finished. In other words, the weak composition $\left(  \gamma_{1}%
,\gamma_{2},\gamma_{3},\ldots\right)  $ is $n$-finished (since $\gamma=\left(
\gamma_{1},\gamma_{2},\gamma_{3},\ldots\right)  $). In other words,
$\gamma_{i}=0$ for every positive integer $i>n$ (since the weak composition
$\left(  \gamma_{1},\gamma_{2},\gamma_{3},\ldots\right)  $ is $n$-finished if
and only if $\left(  \gamma_{i}=0\text{ for every positive integer
}i>n\right)  $ (according to the definition of ``$n$-finished'')).
In other words, $\left(  \gamma_{n+1}%
,\gamma_{n+2},\gamma_{n+3},\ldots\right)  =\left(  0,0,0,\ldots\right)  $.
Now,%
\begin{align*}
\left(  \mathcal{T}\circ\mathcal{S}\right)  \left(  \gamma\right)   &
=\mathcal{T}\left(  \mathcal{S}\left(  \underbrace{\gamma}_{=\left(
\gamma_{1},\gamma_{2},\gamma_{3},\ldots\right)  }\right)  \right)
=\mathcal{T}\left(  \underbrace{\mathcal{S}\left(  \left(  \gamma_{1}%
,\gamma_{2},\gamma_{3},\ldots\right)  \right)  }_{\substack{=\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \\\text{(by the definition of
}\mathcal{S}\text{)}}}\right) \\
&  =\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)
.
\end{align*}
Compared to%
\begin{align*}
\operatorname*{id}\left(  \gamma\right)   &  =\gamma=\left(  \gamma_{1}%
,\gamma_{2},\gamma_{3},\ldots\right)  =\left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{n},\gamma_{n+1},\gamma_{n+2},\gamma_{n+3},\ldots\right) \\
&  =\left(  \text{the concatenation of the sequence }\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{n}\right)  \text{ with the sequence }%
\underbrace{\left(  \gamma_{n+1},\gamma_{n+2},\gamma_{n+3},\ldots\right)
}_{=\left(  0,0,0,\ldots\right)  }\right) \\
&  =\left(  \text{the concatenation of the sequence }\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{n}\right)  \text{ with the sequence }\left(
0,0,0,\ldots\right)  \right) \\
&  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  ,
\end{align*}
this yields $\left(  \mathcal{T}\circ\mathcal{S}\right)  \left(
\gamma\right)  =\operatorname*{id}\left(  \gamma\right)  $.
\par
Let us now forget that we fixed $\gamma$. We thus have proven that $\left(
\mathcal{T}\circ\mathcal{S}\right)  \left(  \gamma\right)  =\operatorname*{id}%
\left(  \gamma\right)  $ for every $\gamma\in\mathbf{F}$. In other words,
$\mathcal{T}\circ\mathcal{S}=\operatorname*{id}$, qed.} and $\mathcal{S}%
\circ\mathcal{T}=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} Let
$\gamma\in\NN^n$. Let us write $\gamma$ in the form $\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  $. Then,%
\begin{align*}
\left(  \mathcal{S}\circ\mathcal{T}\right)  \left(  \gamma\right)   &
=\mathcal{S}\left(  \mathcal{T}\left(  \underbrace{\gamma}_{=\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  }\right)  \right)
=\mathcal{S}\left(  \underbrace{\mathcal{T}\left(  \left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{n}\right)  \right)  }_{\substack{=\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  \\\text{(by the
definition of }\mathcal{T}\text{)}}}\right)  =\mathcal{S}\left(  \left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  \right) \\
&  =\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathcal{S}\right) \\
&  =\gamma\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma=\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \right) \\
&  =\operatorname*{id}\left(  \gamma\right)  .
\end{align*}
\par
Let us now forget that we fixed $\gamma$. We thus have proven that $\left(
\mathcal{S}\circ\mathcal{T}\right)  \left(  \gamma\right)  =\operatorname*{id}%
\left(  \gamma\right)  $ for every $\gamma\in\NN^n$. In other words,
$\mathcal{S}\circ\mathcal{T}=\operatorname*{id}$, qed.}. In other words, the
maps $\mathcal{S}$ and $\mathcal{T}$ are mutually inverse. Hence, the map
$\mathcal{T}$ is a bijection. Also, every $\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{n}\right)  \in\NN^n$ satisfies%
\begin{align}
&  \left(  \text{the result of substituting }x_{1},x_{2},\ldots,x_{n}%
,0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in }\xx%
^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  }\right) \nonumber\\
&  =x_{1}^{\gamma_{1}}x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}.
\label{pf.lem.sol.pieri.alt-e.projlim.2b}%
\end{align}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.projlim.2b}):} Let
$\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \in\NN^n$.
\par
Let us write the weak composition $\mathcal{T}\left(  \left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \right)  $ in the form $\left(
\beta_{1},\beta_{2},\beta_{3},\ldots\right)  $. Then,%
\[
\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  =\mathcal{T}\left(
\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  \right)  =\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)
\]
(by the definition of $\mathcal{T}$). Hence, for every positive integer $j$,
we have%
\begin{align}
\beta_{j}  &  =\left(  \text{the }j\text{-th entry of the sequence
}\underbrace{\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  }_{=\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  }\right)
\nonumber\\
&  =\left(  \text{the }j\text{-th entry of the sequence }\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{n},0,0,0,\ldots\right)  \right) \nonumber\\
&  =\begin{cases}
\gamma_{j},&\text{if }j\leq n;\\
0,&\text{if }j>n
\end{cases} \quad
. \label{pf.lem.sol.pieri.alt-e.projlim.2b.pf.1}
\end{align}
Hence, every $j\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
\beta_{j}  &  =
\begin{cases}
\gamma_{j},&\text{if }j\leq n;\\
0,&\text{if }j>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.pieri.alt-e.projlim.2b.pf.1})}\right) \\
&  =\gamma_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j\leq n\right)  .
\end{align*}
Hence, $\prod_{j=1}^{n}\underbrace{x_{j}^{\beta_{j}}}_{\substack{=x_{j}%
^{\gamma_{j}}\\\text{(since }\beta_{j}=\gamma_{j}\text{)}}}=\prod_{j=1}%
^{n}x_{j}^{\gamma_{j}}$. Also, every $j\in\left\{  n+1,n+2,n+3,\ldots\right\}
$ satisfies%
\begin{align*}
\beta_{j}  &  =
\begin{cases}
\gamma_{j},&\text{if }j\leq n;\\
0,&\text{if }j>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.pieri.alt-e.projlim.2b.pf.1})}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j>n\right)  .
\end{align*}
Hence, $\prod_{j=n+1}^{\infty}\underbrace{x_{j}^{\beta_{j}}}_{\substack{=x_{j}%
^{0}\\\text{(since }\beta_{j}=0\text{)}}}=\prod_{j=n+1}^{\infty}%
\underbrace{x_{j}^{0}}_{=1}=\prod_{j=n+1}^{\infty}1=1$.
\par
We have $\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n}\right)  \right)  =\left(  \beta_{1},\beta_{2},\beta_{3},\ldots\right)  $
and thus%
\begin{align*}
\mathbf{x}^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n}\right)  \right)  }  &  =\mathbf{x}^{\left(  \beta_{1},\beta_{2},\beta
_{3},\ldots\right)  }=\prod_{j\geq1}x_{j}^{\beta_{j}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{x}^{\left(
\beta_{1},\beta_{2},\beta_{3},\ldots\right)  }\right) \\
&  =\underbrace{\left(  \prod_{j=1}^{n}x_{j}^{\beta_{j}}\right)  }%
_{=\prod_{j=1}^{n}x_{j}^{\gamma_{j}}}\underbrace{\left(  \prod_{j=n+1}%
^{\infty}x_{j}^{\beta_{j}}\right)  }_{=1}=\left(  \prod_{j=1}^{n}x_{j}%
^{\gamma_{j}}\right)  1=\prod_{j=1}^{n}x_{j}^{\gamma_{j}}=x_{1}^{\gamma_{1}%
}x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}.
\end{align*}
This proves (\ref{pf.lem.sol.pieri.alt-e.projlim.2b}).}

Now, (\ref{pf.lem.sol.pieri.alt-e.projlim.1}) becomes%
\begin{align*}
&  f\left(  x_{1},x_{2},\ldots,x_{n}\right) \\
&  =\sum\limits_{\substack{\beta\text{ is a}\\\text{weak composition}}}\left(
\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)  \left(  \text{the
result of substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{ for }%
x_{1},x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\beta}\right) \\
&  =\underbrace{\sum\limits_{\substack{\beta\text{ is a}\\\text{weak composition;}%
\\\beta\text{ is }n\text{-finished}}}}_{\substack{=\sum\limits_{\substack{\beta\text{
is an }n\text{-finished}\\\text{weak composition}}}\\=\sum_{\beta\in
\mathbf{F}}\\\text{(since }\mathbf{F}\text{ is the set of all}%
\\n\text{-finished weak compositions)}}}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\beta}}f\right)  \left(  \text{the result of
substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{ for }x_{1}%
,x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\beta}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\beta\text{ is a}\\\text{weak
composition;}\\\beta\text{ is not }n\text{-finished}}}\left(
\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)
\underbrace{\left(  \text{the result of substituting }x_{1},x_{2},\ldots
,x_{n},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ in }%
\mathbf{x}^{\beta}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.projlim.2a}))}}}\\
&  =\sum_{\beta\in\mathbf{F}}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\beta}}f\right)  \left(  \text{the result of
substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{ for }x_{1}%
,x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\beta}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{\beta\text{ is
a}\\\text{weak composition;}\\\beta\text{ is not }n\text{-finished}}}\left(
\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\beta}}f\right)  0}_{=0}\\
&  =\sum_{\beta\in\mathbf{F}}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\beta}}f\right)  \left(  \text{the result of
substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{ for }x_{1}%
,x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\beta}\right) \\
&  =\sum_{\gamma\in\NN^n}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\mathcal{T}\left(  \gamma\right)  }}f\right)  \left(
\text{the result of substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{
for }x_{1},x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\mathcal{T}\left(
\gamma\right)  }\right)
\end{align*}
(here, we have substituted $\mathcal{T}\left(  \gamma\right)  $ for $\beta$ in
the sum, since the map $\mathcal{T}:\NN^{n}\rightarrow\mathbf{F}$ is a
bijection). Thus,%
\begin{align*}
&  f\left(  x_{1},x_{2},\ldots,x_{n}\right) \\
&  =\sum_{\gamma\in\NN^n}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\mathcal{T}\left(  \gamma\right)  }}f\right)  \left(
\text{the result of substituting }x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{
for }x_{1},x_{2},x_{3},\ldots\text{ in }\mathbf{x}^{\mathcal{T}\left(
\gamma\right)  }\right) \\
&  =\sum_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\in\NN^n}\left(  \operatorname*{Coeff}\nolimits_{\xx%
^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  }}f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \underbrace{\left(  \text{the result of substituting
}x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3}%
,\ldots\text{ in }\mathbf{x}^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{n}\right)  \right)  }\right)  }_{\substack{=x_{1}%
^{\gamma_{1}}x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.projlim.2b}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\gamma\text{ as }\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right) \\
&  =\sum_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\in\NN^n}\left(  \operatorname*{Coeff}\nolimits_{\xx%
^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  }}f\right)  x_{1}^{\gamma_{1}}x_{2}^{\gamma_{2}}\cdots x_{n}%
^{\gamma_{n}}.
\end{align*}
Therefore,%
\begin{align}
&  \left(  \text{the coefficient of }x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{n}^{\alpha_{n}}\text{ in }\underbrace{f\left(  x_{1},x_{2}%
,\ldots,x_{n}\right)  }_{=\sum_{\left(  \gamma_{1},\gamma_{2},\ldots
,\gamma_{n}\right)  \in\NN^n}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{n}\right)  \right)  }}f\right)  x_{1}^{\gamma_{1}}%
x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}}\right) \nonumber\\
&  =\left(  \text{the coefficient of }x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{n}^{\alpha_{n}}\text{ in }\sum_{\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{n}\right)  \in\NN^n}\left(  \operatorname*{Coeff}%
\nolimits_{\mathbf{x}^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{n}\right)  \right)  }}f\right)  x_{1}^{\gamma_{1}}%
x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}\right) \nonumber\\
&  =\sum_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\in\NN^n}\left(  \operatorname*{Coeff}\nolimits_{\xx%
^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  }}f\right)  \left(  \text{the coefficient of }x_{1}^{\alpha_{1}}%
x_{2}^{\alpha_{2}}\cdots x_{n}^{\alpha_{n}}\text{ in }x_{1}^{\gamma_{1}}%
x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}\right) \nonumber\\
&  =\left(  \operatorname*{Coeff}\nolimits_{\mathbf{x}^{\mathcal{T}\left(
\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \right)  }}f\right)
\underbrace{\left(  \text{the coefficient of }x_{1}^{\alpha_{1}}x_{2}%
^{\alpha_{2}}\cdots x_{n}^{\alpha_{n}}\text{ in }x_{1}^{\alpha_{1}}%
x_{2}^{\alpha_{2}}\cdots x_{n}^{\alpha_{n}}\right)  }_{=1}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{n}\right)  \in\NN^n;\\\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{n}\right)  \neq\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{n}\right)  }}\left(  \operatorname*{Coeff}\nolimits_{\xx%
^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  }}f\right)  \underbrace{\left(  \text{the coefficient of }%
x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots x_{n}^{\alpha_{n}}\text{ in }%
x_{1}^{\gamma_{1}}x_{2}^{\gamma_{2}}\cdots x_{n}^{\gamma_{n}}\right)
}_{\substack{=0\\\text{(since }\left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n}\right)  \neq\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  =\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \text{ from the sum}\right)
\nonumber\\
&  =\left(  \operatorname*{Coeff}\nolimits_{\mathbf{x}^{\mathcal{T}\left(
\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \right)  }}f\right)
+\underbrace{\sum\limits_{\substack{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma
_{n}\right)  \in\NN^n;\\\left(  \gamma_{1},\gamma_{2},\ldots
,\gamma_{n}\right)  \neq\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }}\left(  \operatorname*{Coeff}\nolimits_{\xx%
^{\mathcal{T}\left(  \left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)
\right)  }}f\right)  0}_{=0}\nonumber\\
&  =\operatorname*{Coeff}\nolimits_{\mathbf{x}^{\mathcal{T}\left(  \left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \right)  }}f=\left(  \text{the
coefficient of }\mathbf{x}^{\mathcal{T}\left(  \left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  \right)  }\text{ in }f\right)
\label{pf.lem.sol.pieri.alt-e.projlim.4}%
\end{align}
But $\left(  \alpha_{i}=0\text{ for every positive integer }i>n\right)  $. In
other words, the weak composition $\left(  \alpha_{1},\alpha_{2},\alpha
_{3},\ldots\right)  $ is $n$-finished (since the weak composition $\left(
\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $ is $n$-finished if and only
if $\left(  \alpha_{i}=0\text{ for every positive integer }i>n\right)  $
(according to the definition of ``$n$-finished'')). In other words,
$\left(  \alpha_{1},\alpha
_{2},\alpha_{3},\ldots\right)  $ belongs to the set of all $n$-finished weak
compositions. In other words, $\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  $ belongs to $\mathbf{F}$ (since $\mathbf{F}$ is the set of
all $n$-finished weak compositions). Since $\mathcal{T}\circ\mathcal{S}%
=\operatorname*{id}$, we thus have $\left(  \mathcal{T}\circ\mathcal{S}%
\right)  \left(  \left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
\right)  =\operatorname*{id}\left(  \left(  \alpha_{1},\alpha_{2},\alpha
_{3},\ldots\right)  \right)  =\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  $, so that%
\[
\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  =\left(
\mathcal{T}\circ\mathcal{S}\right)  \left(  \left(  \alpha_{1},\alpha
_{2},\alpha_{3},\ldots\right)  \right)  =\mathcal{T}\left(
\underbrace{\mathcal{S}\left(  \left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  \right)  }_{\substack{=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  \\\text{(by the definition of }\mathcal{S}\text{)}%
}}\right)  =\mathcal{T}\left(  \left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \right)  .
\]
Thus, $\mathcal{T}\left(  \left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \right)  =\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)
=\alpha$, so that $\mathbf{x}^{\mathcal{T}\left(  \left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{n}\right)  \right)  }=\mathbf{x}^{\alpha}$. Hence,
(\ref{pf.lem.sol.pieri.alt-e.projlim.4}) becomes%
\begin{align*}
&  \left(  \text{the coefficient of }x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{n}^{\alpha_{n}}\text{ in }f\left(  x_{1},x_{2},\ldots,x_{n}\right)
\right) \\
&  =\left(  \text{the coefficient of }\underbrace{\mathbf{x}^{\mathcal{T}%
\left(  \left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \right)  }%
}_{=\mathbf{x}^{\alpha}}\text{ in }f\right)  =\left(  \text{the coefficient of
}\mathbf{x}^{\alpha}\text{ in }f\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.pieri.alt-e.projlim}(a).

(b) Let $\alpha$ be a weak composition. Let us write $\alpha$ in the form
$\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $. Then, $\left(
\alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  =\alpha$ is a weak
composition. In other words, all but finitely many positive integers $i$
satisfy $\alpha_{i}=0$. In other words, there exists a finite subset
$\mathbf{S}$ of $\left\{  1,2,3,\ldots\right\}  $ such that%
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \setminus\mathbf{S}\text{
satisfies }\alpha_{i}=0. \label{pf.lem.sol.pieri.alt-e.projlim.b.1}%
\end{equation}
Consider this $\mathbf{S}$. Both $\mathbf{S}$ and $\left\{  \mathbf{n}%
\right\}  $ are finite subsets of $ \NN$. Thus, their union
$\mathbf{S}\cup\left\{  \mathbf{n}\right\}  $ also is a finite subset of
$ \NN$.

Every finite subset of $ \NN$ has an upper bound. Applying this to the
finite subset $\mathbf{S}\cup\left\{  \mathbf{n}\right\}  $ of $ \NN$,
we conclude that the set $\mathbf{S}\cup\left\{  \mathbf{n}\right\}  $ has an
upper bound. Let $m$ be such a upper bound. Then, $m\geq u$ for every
$u\in\mathbf{S}\cup\left\{  \mathbf{n}\right\}  $. Applying this to
$u=\mathbf{n}$, we obtain $m\geq\mathbf{n}$ (since $\mathbf{n}\in\left\{
\mathbf{n}\right\}  \subset\mathbf{S}\cup\left\{  \mathbf{n}\right\}  $).

We assumed that $f\left(  x_{1},x_{2},\ldots,x_{n}\right)  =0$ for every
$n\in \NN$ satisfying $n\geq\mathbf{n}$. Applying this to $n=m$, we
obtain $f\left(  x_{1},x_{2},\ldots,x_{m}\right)  =0$ (since $m\geq\mathbf{n}$).

Let $i$ be any positive integer such that $i>m$. We assume (for the sake of
contradiction) that $i\in\mathbf{S}$. Recall that $m\geq u$ for every
$u\in\mathbf{S}\cup\left\{  \mathbf{n}\right\}  $. Applying this to $u=i$, we
obtain $m\geq i$ (since $i\in\mathbf{S}\subset\mathbf{S}\cup\left\{
\mathbf{n}\right\}  $). This contradicts $i>m$. This contradiction proves that
our assumption (that $i\in\mathbf{S}$) was wrong. Hence, we cannot have
$i\in\mathbf{S}$. Thus, we have $i\notin\mathbf{S}$. Therefore, $i\in\left\{
1,2,3,\ldots\right\}  \setminus\mathbf{S}$ (since $i\in\left\{  1,2,3,\ldots
\right\}  $). Therefore, $\alpha_{i}=0$ (according to
(\ref{pf.lem.sol.pieri.alt-e.projlim.b.1})).

Let us now forget that we fixed $i$. We thus have shown that $\alpha_{i}=0$
for every positive integer $i>m$. Hence, Lemma
\ref{lem.sol.pieri.alt-e.projlim}(a) (applied to $n=m$) yields
\begin{align*}
&  \left(  \text{the coefficient of }\mathbf{x}^{\alpha}\text{ in }f\right) \\
&  =\left(  \text{the coefficient of }x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}%
}\cdots x_{m}^{\alpha_{m}}\text{ in }\underbrace{f\left(  x_{1},x_{2}%
,\ldots,x_{m}\right)  }_{=0}\right)  =\left(  \text{the coefficient of }%
x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots x_{m}^{\alpha_{m}}\text{ in
}0\right)  =0.
\end{align*}


Let us now forget that we fixed $\alpha$. We thus have shown that $\left(
\text{the coefficient of }\mathbf{x}^{\alpha}\text{ in }f\right)  =0$ for
every weak composition $\alpha$. In other words, all coefficients of $f$ equal
$0$. Thus, $f$ equals $0$. In other words, $f=0$. This proves Lemma
\ref{lem.sol.pieri.alt-e.projlim}(b).
\end{proof}

We shall now show the following lemma, which is at the heart of our solution
of Exercise \ref{exe.pieri.alt-e}:

\begin{lemma}
\label{lem.sol.pieri.alt-e.1xs}Let $n\in \NN$. Let $\mu$ be a partition
such that $\ell\left(  \mu\right)  \leq n$. Then,%
\[
\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  s_{\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right)  =\sum\limits_{\substack{\lambda\in
 \Par ;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{
is a vertical strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
t^{\left\vert \lambda/\mu\right\vert }%
\]
in the polynomial ring $\left(  \kk\left[  x_{1},x_{2},\ldots
,x_{n}\right]  \right)  \left[  t\right]  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.pieri.alt-e.1xs}.]We regard the partition $\mu$
as an $n$-tuple in $\NN^n$ (this is legitimate since $\ell\left(
\mu\right)  \leq n$). If $\alpha\in\NN^n$ is an $n$-tuple, and $i$ is
an element of $\left\{  1,2,\ldots,n\right\}  $, then $\alpha_{i}$ shall
denote the $i$-th entry of $\alpha$. Thus, every $\alpha\in\NN^n$
satisfies $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $.
Notice that $\left\{  0,1\right\}  ^{n}$ is a subset of $\NN^n$, and
thus the notation $\alpha_{i}$ (meaning the $i$-th entry of $\alpha$) is also
defined for every $\alpha\in\left\{  0,1\right\}  ^{n}$ and $i\in\left\{
1,2,\ldots,n\right\}  $.

We define a componentwise addition on $\NN^n$; that is, $\alpha
+\beta=\left(  \alpha_{1}+\beta_{1},\alpha_{2}+\beta_{2},\ldots,\alpha
_{n}+\beta_{n}\right)  $ for any two $n$-tuples $\alpha$ and $\beta$ in
$\NN^n$. In other words, $\left(  \alpha+\beta\right)  _{i}%
=\alpha_{i}+\beta_{i}$ for any $\alpha\in\NN^n$, any $\beta
\in\NN^n$ and any $i\in\left\{  1,2,\ldots,n\right\}  $.

For every $n$-tuple $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in{\NN}^{n}$, we define the alternant $a_{\alpha}%
\in\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ as in
Definition~\ref{def.alternant}. (But other than this, we are not adapting the
notations of Section~\ref{bialternant-section}.)

Let $\rho$ be the $n$-tuple $\left(  n-1,n-2,...,2,1,0\right)  \in
\NN^n$. We can regard $\rho$ as a weak composition by padding it with zeroes
at the end (i.e., identifying $\rho$ with the weak composition \newline%
$\left(  n-1,n-2,...,2,1,0,0,0,0,...\right)  $). We have $\left(  \rho
_{1},\rho_{2},...,\rho_{n}\right)  =\rho=\left(  n-1,n-2,...,2,1,0\right)  $.
Thus,%
\begin{equation}
\rho_{i}=n-i\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,...,n\right\}  . \label{pf.lem.sol.pieri.alt-e.1xs.rhoi}%
\end{equation}


If $\alpha\in\NN^n$ is an $n$-tuple satisfying $\alpha_{1}\geq
\alpha_{2}\geq\cdots\geq\alpha_{n}$, then%
\begin{equation}
\text{the }n\text{-tuple }\alpha+\rho\text{ contains no two equal entries}
\label{pf.lem.sol.pieri.alt-e.1xs.alphaprho}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.1xs.alphaprho}):} Let
$\alpha\in\NN^n$ be an $n$-tuple satisfying $\alpha_{1}\geq\alpha
_{2}\geq\cdots\geq\alpha_{n}$. Let $j\in\left\{  1,2,\ldots,n-1\right\}  $.
Then, $\alpha_{j}\geq\alpha_{j+1}$ (since $\alpha_{1}\geq\alpha_{2}\geq
\cdots\geq\alpha_{n}$). But (\ref{pf.lem.sol.pieri.alt-e.1xs.rhoi}) (applied
to $i=j$) yields $\rho_{j}=n-j$. Also, (\ref{pf.lem.sol.pieri.alt-e.1xs.rhoi})
(applied to $i=j+1$) yields $\rho_{j+1}=n-\left(  j+1\right)  $. Thus,
$\underbrace{\rho_{j+1}}_{=n-\left(  j+1\right)  }+1=n-\left(  j+1\right)
+1=n-j=\rho_{j}$ (since $\rho_{j}=n-j$), so that $\rho_{j}=\rho_{j+1}%
+1>\rho_{j+1}$.
\par
But $\left(  \alpha+\rho\right)  _{j}=\underbrace{\alpha_{j}}_{\geq
\alpha_{j+1}}+\underbrace{\rho_{j}}_{>\rho_{j+1}}>\alpha_{j+1}+\rho
_{j+1}=\left(  \alpha+\rho\right)  _{j+1}$.
\par
Let us now forget that we fixed $j$. We thus have shown that $\left(
\alpha+\rho\right)  _{j}>\left(  \alpha+\rho\right)  _{j+1}$ for every
$j\in\left\{  1,2,\ldots,n-1\right\}  $. Hence, $\left(  \alpha+\rho\right)
_{1}>\left(  \alpha+\rho\right)  _{2}>\cdots>\left(  \alpha+\rho\right)  _{n}%
$. Therefore, the numbers $\left(  \alpha+\rho\right)  _{1}$, $\left(
\alpha+\rho\right)  _{2}$, $\ldots$, $\left(  \alpha+\rho\right)  _{n}$ are
distinct. In other words, no two of the numbers $\left(  \alpha
+\rho\right)  _{1}$, $\left(  \alpha+\rho\right)  _{2}$, $\ldots$, $\left(
\alpha+\rho\right)  _{n}$ are equal. In other words, no two of the entries of
the $n$-tuple $\alpha+\rho$ are equal (since the numbers $\left(  \alpha
+\rho\right)  _{1}$, $\left(  \alpha+\rho\right)  _{2}$, $\ldots$, $\left(
\alpha+\rho\right)  _{n}$ are the entries of the $n$-tuple $\alpha+\rho$). In
other words, the $n$-tuple $\alpha+\rho$ contains no two equal entries. This
proves (\ref{pf.lem.sol.pieri.alt-e.1xs.alphaprho}).}.

Recall that
\begin{equation}
a_{\alpha}=\det\left[
\begin{matrix}
x_{1}^{\alpha_{1}} & \cdots & x_{1}^{\alpha_{n}}\\
x_{2}^{\alpha_{1}} & \cdots & x_{2}^{\alpha_{n}}\\
\vdots & \ddots & \vdots\\
x_{n}^{\alpha_{1}} & \cdots & x_{n}^{\alpha_{n}}%
\end{matrix}
\right]  =\det\left(  \left(  x_{i}^{\alpha_{j}}\right)  _{i,j=1,2,...,n}%
\right)  \label{pf.lem.sol.pieri.alt-e.1xs.a}%
\end{equation}
for every $\alpha=\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)
\in{\NN}^{n}$. Also, recall that%
\begin{equation}
a_{\alpha}=0 \label{pf.lem.sol.pieri.alt-e.1xs.a=0}%
\end{equation}
for every $n$-tuple $\alpha\in\NN^n$ which contains at least two
equal entries.

Recall that every commutative ring $A$, every $m\in \NN$ and every
matrix $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in A^{m\times m}$
satisfy%
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\sum_{\sigma\in\Symm_m}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{m}\alpha_{i,\sigma\left(  i\right)  }.
\label{pf.lem.sol.pieri.alt-e.1xs.det}%
\end{equation}
(This is simply the explicit formula for the determinant of a matrix as a sum
over permutations.) Now, if $\gamma\in\NN^n$ is arbitrary, then
(\ref{pf.lem.sol.pieri.alt-e.1xs.a}) (applied to $\alpha=\gamma+\rho$) yields%
\begin{equation}
a_{\gamma+\rho}=\det\left(  \left(  x_{i}^{\left(  \gamma+\rho\right)  _{j}%
}\right)  _{i,j=1,2,...,n}\right)  =\sum_{\sigma\in\Symm_n}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{n}x_{i}^{\left(  \gamma+\rho\right)
_{\sigma\left(  i\right)  }} \label{pf.lem.sol.pieri.alt-e.1xs.aga}%
\end{equation}
(by (\ref{pf.lem.sol.pieri.alt-e.1xs.det}), applied to $A=\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $, $m=n$ and $\alpha_{i,j}=x_{i}^{\left(
\gamma+\rho\right)  _{j}}$). Applying this to $\gamma=\mu$, we obtain%
\begin{equation}
a_{\mu+\rho}=\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}%
\prod_{i=1}^{n}x_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }}
.
\label{pf.lem.sol.pieri.alt-e.1xs.amu}
\end{equation}


On the other hand, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
$\sum_{\eta\in\left\{  0,1\right\}  }x_{i}^{\eta}t^{\eta}=\underbrace{x_{i}%
^{0}}_{=1}\underbrace{t^{0}}_{=1}+\underbrace{x_{i}^{1}}_{=x_{i}%
}\underbrace{t^{1}}_{=t}=1+x_{i}t$. Thus, every $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfies%
\begin{equation}
1+x_{i}t=\sum_{\eta\in\left\{  0,1\right\}  }x_{i}^{\eta}t^{\eta}.
\label{pf.lem.sol.pieri.alt-e.1xs.1xit}%
\end{equation}


Now, every $\sigma\in\Symm_n$ satisfies%
\begin{equation}
\prod_{i=1}^{n}\left(  1+x_{i}t\right)  =\sum_{\left(  \eta_{1},\eta
_{2},\ldots,\eta_{n}\right)  \in\left\{  0,1\right\}  ^{n}}\left(  \prod
_{i=1}^{n}x_{i}^{\eta_{\sigma\left(  i\right)  }}\right)  t^{\eta_{1}+\eta
_{2}+\cdots+\eta_{n}}\label{pf.lem.sol.pieri.alt-e.1xs.everysi}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.1xs.everysi}):} Let
$\sigma\in\Symm_n$ be arbitrary. Then, $\sigma$ is a permutation of
$\left\{  1,2,\ldots,n\right\}  $, thus a bijection. Hence, its inverse
$\sigma^{-1}$ is a bijection. Now,
\begin{align*}
\underbrace{\prod_{i=1}^{n}}_{=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
}\left(  1+x_{i}t\right)   &  =\prod_{i\in\left\{  1,2,\ldots,n\right\}
}\left(  1+x_{i}t\right)  =\underbrace{\prod_{i\in\left\{  1,2,\ldots
,n\right\}  }}_{=\prod_{i=1}^{n}}\underbrace{\left(  1+x_{\sigma^{-1}\left(
i\right)  }t\right)  }_{\substack{=\sum_{\eta\in\left\{  0,1\right\}
}x_{\sigma^{-1}\left(  i\right)  }^{\eta}t^{\eta}\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.1xs.1xit}), applied to }\sigma^{-1}\left(
i\right)  \\\text{instead of }i\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\sigma^{-1}\left(  i\right)  \text{ for
}i\text{ in the product,}\\
\text{since the map }\sigma^{-1}:\left\{  1,2,\ldots,n\right\}  \rightarrow
\left\{  1,2,\ldots,n\right\}  \text{ is a bijection}%
\end{array}
\right)  \\
&  =\prod_{i=1}^{n}\left(  \sum_{\eta\in\left\{  0,1\right\}  }x_{\sigma
^{-1}\left(  i\right)  }^{\eta}t^{\eta}\right)  =\sum_{\left(  \eta_{1}%
,\eta_{2},\ldots,\eta_{n}\right)  \in\left\{  0,1\right\}  ^{n}}%
\underbrace{\prod_{i=1}^{n}\left(  x_{\sigma^{-1}\left(  i\right)  }^{\eta
_{i}}t^{\eta_{i}}\right)  }_{=\left(  \prod_{i=1}^{n}x_{\sigma^{-1}\left(
i\right)  }^{\eta_{i}}\right)  \left(  \prod_{i=1}^{n}t^{\eta_{i}}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right)  \\
&  =\sum_{\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{
0,1\right\}  ^{n}}\left(  \underbrace{\prod_{i=1}^{n}}_{=\prod_{i\in\left\{
1,2,\ldots,n\right\}  }}x_{\sigma^{-1}\left(  i\right)  }^{\eta_{i}}\right)
\underbrace{\left(  \prod_{i=1}^{n}t^{\eta_{i}}\right)  }_{=t^{\sum_{i=1}%
^{n}\eta_{i}}=t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}}\\
&  =\sum_{\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{
0,1\right\}  ^{n}}\underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots
,n\right\}  }x_{\sigma^{-1}\left(  i\right)  }^{\eta_{i}}\right)
}_{\substack{=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{\sigma^{-1}\left(
\sigma\left(  i\right)  \right)  }^{\eta_{\sigma\left(  i\right)  }%
}\\\text{(here, we have substituted }\sigma\left(  i\right)  \text{ for
}i\\\text{in the product, since the map}\\\sigma:\left\{  1,2,\ldots
,n\right\}  \rightarrow\left\{  1,2,\ldots,n\right\}  \text{ is a bijection)}%
}}t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}\\
&  =\sum_{\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{
0,1\right\}  ^{n}}\left(  \underbrace{\prod_{i\in\left\{  1,2,\ldots
,n\right\}  }}_{=\prod_{i=1}^{n}}\underbrace{x_{\sigma^{-1}\left(
\sigma\left(  i\right)  \right)  }^{\eta_{\sigma\left(  i\right)  }}%
}_{\substack{=x_{i}^{\eta_{\sigma\left(  i\right)  }}\\\text{(since }%
\sigma^{-1}\left(  \sigma\left(  i\right)  \right)  =i\text{)}}}\right)
t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}\\
&  =\sum_{\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{
0,1\right\}  ^{n}}\left(  \prod_{i=1}^{n}x_{i}^{\eta_{\sigma\left(  i\right)
}}\right)  t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}.
\end{align*}
This proves (\ref{pf.lem.sol.pieri.alt-e.1xs.everysi}).}.

Now, multiplying the equality (\ref{pf.lem.sol.pieri.alt-e.1xs.amu}) with
$\prod_{i=1}^{n}\left(  1+x_{i}t\right)  $, we obtain%
\begin{align}
\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \cdot a_{\mu+\rho}
&  =\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \cdot\sum
_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{n}%
x_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }}\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}%
\underbrace{\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)
}_{\substack{=\sum_{\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)
\in\left\{  0,1\right\}  ^{n}}\left(  \prod_{i=1}^{n}x_{i}^{\eta
_{\sigma\left(  i\right)  }}\right)  t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}%
}\\\text{(by (\ref{pf.lem.sol.pieri.alt-e.1xs.everysi}))}}}\prod_{i=1}%
^{n}x_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }}\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\left(
\sum_{\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{
0,1\right\}  ^{n}}\left(  \prod_{i=1}^{n}x_{i}^{\eta_{\sigma\left(  i\right)
}}\right)  t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}\right)  \prod_{i=1}^{n}%
x_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }}\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\left(
\eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{  0,1\right\}  ^{n}%
}\underbrace{\left(  \prod_{i=1}^{n}x_{i}^{\eta_{\sigma\left(  i\right)  }%
}\right)  \left(  \prod_{i=1}^{n}x_{i}^{\left(  \mu+\rho\right)
_{\sigma\left(  i\right)  }}\right)  }_{=\prod_{i=1}^{n}\left(  x_{i}%
^{\eta_{\sigma\left(  i\right)  }}x_{i}^{\left(  \mu+\rho\right)
_{\sigma\left(  i\right)  }}\right)  }t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}%
}\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\left(
\eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{  0,1\right\}  ^{n}%
}\left(  \prod_{i=1}^{n}\underbrace{\left(  x_{i}^{\eta_{\sigma\left(
i\right)  }}x_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }%
}\right)  }_{=x_{i}^{\eta_{\sigma\left(  i\right)  }+\left(  \mu+\rho\right)
_{\sigma\left(  i\right)  }}}\right)  t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}%
}\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\left(
\eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \in\left\{  0,1\right\}  ^{n}%
}\left(  \prod_{i=1}^{n}x_{i}^{\eta_{\sigma\left(  i\right)  }+\left(
\mu+\rho\right)  _{\sigma\left(  i\right)  }}\right)  t^{\eta_{1}+\eta
_{2}+\cdots+\eta_{n}}\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\eta
\in\left\{  0,1\right\}  ^{n}}\left(  \prod_{i=1}^{n}\underbrace{x_{i}%
^{\eta_{\sigma\left(  i\right)  }+\left(  \mu+\rho\right)  _{\sigma\left(
i\right)  }}}_{\substack{=x_{i}^{\left(  \eta+\mu+\rho\right)  _{\sigma\left(
i\right)  }}\\\text{(since }\eta_{\sigma\left(  i\right)  }+\left(  \mu
+\rho\right)  _{\sigma\left(  i\right)  }=\left(  \eta+\mu+\rho\right)
_{\sigma\left(  i\right)  }\\\text{(since }\left(  \eta+\mu+\rho\right)
_{\sigma\left(  i\right)  }=\left(  \eta+\left(  \mu+\rho\right)  \right)
_{\sigma\left(  i\right)  }\\=\eta_{\sigma\left(  i\right)  }+\left(  \mu
+\rho\right)  _{\sigma\left(  i\right)  }\\\text{(since the addition on
}\NN^{n}\text{ is componentwise)))}}}\right)  t^{\eta_{1}+\eta
_{2}+\cdots+\eta_{n}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\left(  \eta_{1},\eta_{2},\ldots,\eta_{n}\right)  \text{ as }\eta\right)
\nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\eta
\in\left\{  0,1\right\}  ^{n}}\left(  \prod_{i=1}^{n}x_{i}^{\left(  \eta
+\mu+\rho\right)  _{\sigma\left(  i\right)  }}\right)  t^{\eta_{1}+\eta
_{2}+\cdots+\eta_{n}}\nonumber\\
&  =\sum_{\eta\in\left\{  0,1\right\}  ^{n}}\underbrace{\left(  \sum
_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{n}%
x_{i}^{\left(  \eta+\mu+\rho\right)  _{\sigma\left(  i\right)  }}\right)
}_{\substack{=a_{\eta+\mu+\rho}\\\text{(since }a_{\eta+\mu+\rho}=\sum
_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{n}%
x_{i}^{\left(  \eta+\mu+\rho\right)  _{\sigma\left(  i\right)  }}\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.1xs.aga}), applied to }\gamma=\eta+\mu\text{))}%
}}t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}\nonumber\\
&  =\sum_{\eta\in\left\{  0,1\right\}  ^{n}}a_{\eta+\mu+\rho}t^{\eta_{1}%
+\eta_{2}+\cdots+\eta_{n}}. \label{pf.lem.sol.pieri.alt-e.1xs.step1}%
\end{align}


Now, let $\mathbf{F}$ denote the set of all $\eta\in\left\{  0,1\right\}
^{n}$ such that the $n$-tuple $\eta+\mu+\rho$ contains no two equal entries.
Then, (\ref{pf.lem.sol.pieri.alt-e.1xs.step1}) becomes%
\begin{align}
&  \left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \cdot a_{\mu+\rho
}\nonumber\\
&  =\sum_{\eta\in\left\{  0,1\right\}  ^{n}}a_{\eta+\mu+\rho}t^{\eta_{1}%
+\eta_{2}+\cdots+\eta_{n}}\nonumber\\
&  =\underbrace{\sum\limits_{\substack{\eta\in\left\{  0,1\right\}  ^{n};\\\text{the
}n\text{-tuple }\eta+\mu+\rho\text{ contains}\\\text{no two equal entries}}%
}}_{\substack{=\sum_{\eta\in\mathbf{F}}\\\text{(since }\mathbf{F}\text{ is the
set of all }\eta\in\left\{  0,1\right\}  ^{n}\\\text{such that the
}n\text{-tuple }\eta+\mu+\rho\\\text{contains no two equal entries)}}%
}a_{\eta+\mu+\rho}t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}+\sum\limits_{\substack{\eta
\in\left\{  0,1\right\}  ^{n};\\\text{the }n\text{-tuple }\eta+\mu+\rho\text{
contains}\\\text{two equal entries}}}\underbrace{a_{\eta+\mu+\rho}%
}_{\substack{=0\\\text{(by (\ref{pf.lem.sol.pieri.alt-e.1xs.a=0}), applied
to}\\\alpha=\eta+\mu+\rho\text{)}}}t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for every }\eta\in\left\{  0,1\right\}  ^{n}\text{, either the
}n\text{-tuple }\eta+\mu+\rho\\
\text{contains no two equal entries, or the }n\text{-tuple }\eta+\mu+\rho\\
\text{contains two equal entries (but not both of these at the same time)}%
\end{array}
\right) \nonumber\\
&  =\sum_{\eta\in\mathbf{F}}a_{\eta+\mu+\rho}t^{\eta_{1}+\eta_{2}+\cdots
+\eta_{n}}+\underbrace{\sum\limits_{\substack{\eta\in\left\{  0,1\right\}
^{n};\\\text{the }n\text{-tuple }\eta+\mu+\rho\text{ contains}\\\text{two
equal entries}}}0t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}}_{=0}=\sum_{\eta
\in\mathbf{F}}a_{\eta+\mu+\rho}t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}.
\label{pf.lem.sol.pieri.alt-e.1xs.step4}%
\end{align}


On the other hand, let $\mathbf{G}$ denote the set of all $\lambda
\in \Par $ such that $\ell\left(  \lambda\right)  \leq n$ and
such that $\lambda/\mu$ is a vertical strip. For every $\eta\in\mathbf{F}$, we
have $\eta+\mu\in\mathbf{G}$ (where we identify $n$-tuples $\alpha
\in\NN^n$ satisfying $\alpha_{1}\geq\alpha_{2}\geq\cdots\geq
\alpha_{n}$ with partitions of length $\leq n$%
)\ \ \ \ \footnote{\textit{Proof.} Let $\gamma\in\mathbf{F}$. Then, $\gamma$
is an element of $\mathbf{F}$. In other words, $\gamma$ is an $n$-tuple
$\eta\in\left\{  0,1\right\}  ^{n}$ such that the $n$-tuple $\eta+\mu+\rho$
contains no two equal entries (since $\mathbf{F}$ is the set of all $\eta
\in\left\{  0,1\right\}  ^{n}$ such that the $n$-tuple $\eta+\mu+\rho$
contains no two equal entries). In other words, $\gamma$ is an $n$-tuple in
$\left\{  0,1\right\}  ^{n}$ and has the property that the $n$-tuple
$\gamma+\mu+\rho$ contains no two equal entries.
\par
Now, let $j\in\left\{  1,2,\ldots,n-1\right\}  $. We shall now show that
$\left(  \eta+\mu\right)  _{j}\geq\left(  \eta+\mu\right)  _{j+1}$.
\par
In fact, let us (for the sake of contradiction) assume the contrary. Then,
$\left(  \eta+\mu\right)  _{j}<\left(  \eta+\mu\right)  _{j+1}$. Hence,
$\left(  \eta+\mu\right)  _{j}\leq\left(  \eta+\mu\right)  _{j+1}-1$ (since
$\left(  \eta+\mu\right)  _{j}$ and $\left(  \eta+\mu\right)  _{j+1}$ are
integers).
\par
We have $\mu_{1}\geq\mu_{2}\geq\mu_{3}\geq\cdots$ (since $\mu$ is a
partition). Thus, $\mu_{i}\geq\mu_{i+1}$ for every $i\in\left\{
1,2,3,\ldots\right\}  $. Applying this to $i=j$, we obtain $\mu_{j}\geq
\mu_{j+1}$.
\par
We have $\eta_{j}\in\left\{  0,1\right\}  $ (since $\eta\in\left\{
0,1\right\}  ^{n}$), thus $\eta_{j}\geq0$. Also, we have $\eta_{j+1}%
\in\left\{  0,1\right\}  $ (since $\eta\in\left\{  0,1\right\}  ^{n}$), so
that $\eta_{j+1}\leq1$. Thus, $1\geq\eta_{j+1}$. Now,%
\begin{align*}
\left(  \eta+\mu\right)  _{j}  &  =\underbrace{\eta_{j}}_{\geq0=1-1}%
+\underbrace{\mu_{j}}_{\geq\mu_{j+1}}\geq\underbrace{1}_{\geq\eta_{j+1}}%
-1+\mu_{j+1}\geq\eta_{j+1}-1+\mu_{j+1}=\underbrace{\left(  \eta_{j+1}%
+\mu_{j+1}\right)  }_{=\left(  \eta+\mu\right)  _{j+1}}-1
=\left(  \eta+\mu\right)  _{j+1}-1.
\end{align*}
Combining this with $\left(  \eta+\mu\right)  _{j}\leq\left(  \eta+\mu\right)
_{j+1}-1$, we obtain $\left(  \eta+\mu\right)  _{j}=\left(  \eta+\mu\right)
_{j+1}-1$.
\par
On the other hand, (\ref{pf.lem.sol.pieri.alt-e.1xs.rhoi}) (applied to $i=j$)
yields $\rho_{j}=n-j$. Also, (\ref{pf.lem.sol.pieri.alt-e.1xs.rhoi}) (applied
to $i=j+1$) yields $\rho_{j+1}=n-\left(  j+1\right)  $, so that
$\underbrace{\rho_{j+1}}_{=n-\left(  j+1\right)  }+1=n-\left(  j+1\right)
+1=n-j=\rho_{j}$ (since $\rho_{j}=n-j$), and thus $\rho_{j}=\rho_{j+1}+1$.
Now,%
\begin{align*}
\left(  \eta+\mu+\rho\right)  _{j}  &  =\left(  \left(  \eta+\mu\right)
+\rho\right)  _{j}=\underbrace{\left(  \eta+\mu\right)  _{j}}_{=\left(
\eta+\mu\right)  _{j+1}-1}+\underbrace{\rho_{j}}_{=\rho_{j+1}+1}=\left(
\left(  \eta+\mu\right)  _{j+1}-1\right)  +\left(  \rho_{j+1}+1\right) \\
&  =\left(  \eta+\mu\right)  _{j+1}+\rho_{j+1}=\left(  \left(  \eta
+\mu\right)  +\rho\right)  _{j+1}=\left(  \eta+\mu+\rho\right)  _{j+1}.
\end{align*}
In other words, the $j$-th entry of the $n$-tuple $\eta+\mu+\rho$ equals the
$\left(  j+1\right)  $-th entry of this $n$-tuple. Thus, the $n$-tuple
$\eta+\mu+\rho$ contains two equal entries (namely, its $j$-th entry and its
$\left(  j+1\right)  $-th entry). This contradicts the fact that the $n$-tuple
$\eta+\mu+\rho$ contains no two equal entries. This contradiction shows that
our assumption was wrong. Hence, $\left(  \eta+\mu\right)  _{j}\geq\left(
\eta+\mu\right)  _{j+1}$ is proven.
\par
Now, let us forget that we fixed $j$. We thus have shown that $\left(
\eta+\mu\right)  _{j}\geq\left(  \eta+\mu\right)  _{j+1}$ for every
$j\in\left\{  1,2,\ldots,n-1\right\}  $. Hence,%
\[
\left(  \eta+\mu\right)  _{1}\geq\left(  \eta+\mu\right)  _{2}\geq\cdots
\geq\left(  \eta+\mu\right)  _{n}.
\]
Also, clearly, $\eta+\mu\in\NN^n$ (since $\eta\in\left\{
0,1\right\}  ^{n}\subset\NN^n$ and $\mu\in\NN^n$).
\par
Now, let us recall that every $\alpha\in\NN^n$ satisfying $\alpha
_{1}\geq\alpha_{2}\geq\cdots\geq\alpha_{n}$ is a partition having length $\leq
n$. Applying this to $\alpha=\eta+\mu$, we thus obtain that $\eta+\mu$ is a
partition having length $\leq n$. In other words, $\eta+\mu$ is a partition
and satisfies $\ell\left(  \eta+\mu\right)  \leq n$. We have $\eta+\mu
\in \Par $ (since $\eta+\mu$ is a partition).
\par
We shall now show that $\mu\subseteq\eta+\mu$.
\par
Fix $i\in\left\{  1,2,3,\ldots\right\}  $. We shall show that $\mu_{i}%
\leq\left(  \eta+\mu\right)  _{i}$.
\par
Indeed, let us first assume that $i>n$. Then, $n\geq\ell\left(  \mu\right)  $
(since $\ell\left(  \mu\right)  \leq n$) and thus $i>n\geq\ell\left(
\mu\right)  $, so that $\mu_{i}=0$. Hence, $\mu_{i}=0\leq\left(  \eta
+\mu\right)  _{i}$. Hence, $\mu_{i}\leq\left(  \eta+\mu\right)  _{i}$ is
proven under the assumption that $i>n$. Thus, for the rest of the proof of
$\mu_{i}\leq\left(  \eta+\mu\right)  _{i}$, we can WLOG assume that we don't
have $i>n$. Assume this.
\par
We don't have $i>n$. Thus, we have $i\leq n$, so that $i\in\left\{
1,2,\ldots,n\right\}  $. Thus, $\eta_{i}\in\left\{  0,1\right\}  $ (since
$\eta\in\left\{  0,1\right\}  ^{n}$) and therefore $\eta_{i}\geq0$. Now,
$\left(  \eta+\mu\right)  _{i}=\underbrace{\eta_{i}}_{\geq0}+\mu_{i}\geq
\mu_{i}$, so that $\mu_{i}\leq\left(  \eta+\mu\right)  _{i}$. This completes
our proof of $\mu_{i}\leq\left(  \eta+\mu\right)  _{i}$.
\par
Now, let us forget that we fixed $i$. We thus have shown that $\mu_{i}%
\leq\left(  \eta+\mu\right)  _{i}$ for every $i\in\left\{  1,2,3,\ldots
\right\}  $. In other words, $\mu\subseteq\eta+\mu$ (since $\mu\subseteq
\eta+\mu$ if and only if we have $\mu_{i}\leq\left(  \eta+\mu\right)  _{i}$
for every $i\in\left\{  1,2,3,\ldots\right\}  $ (according to the definition
of the relation $\subseteq$ on $ \Par $)).
\par
Now, we shall show that $\left(  \eta+\mu\right)  /\mu$ is a vertical strip.
In fact, let $i\in\left\{  1,2,3,\ldots\right\}  $. We shall show that
$\left(  \eta+\mu\right)  _{i}\leq\mu_{i}+1$.
\par
Indeed, let us first assume that $i>n$. Then, $n\geq\ell\left(  \eta
+\mu\right)  $ (since $\ell\left(  \eta+\mu\right)  \leq n$), and thus
$i>n\geq\ell\left(  \eta+\mu\right)  $. Therefore, $\left(  \eta+\mu\right)
_{i}=0$, so that $\left(  \eta+\mu\right)  _{i}=0\leq\mu_{i}\leq\mu_{i}+1$.
Hence, $\left(  \eta+\mu\right)  _{i}\leq\mu_{i}+1$ is proven under the
assumption that $i>n$. Thus, for the rest of the proof of $\left(  \eta
+\mu\right)  _{i}\leq\mu_{i}+1$, we can WLOG assume that we don't have $i>n$.
Assume this.
\par
We don't have $i>n$. Thus, we have $i\leq n$, so that $i\in\left\{
1,2,\ldots,n\right\}  $. Thus, $\eta_{i}\in\left\{  0,1\right\}  $ (since
$\eta\in\left\{  0,1\right\}  ^{n}$) and therefore $\eta_{i}\leq1$. Now,
$\left(  \eta+\mu\right)  _{i}=\underbrace{\eta_{i}}_{\leq1}+\mu_{i}\leq
1+\mu_{i}=\mu_{i}+1$. This completes our proof of $\left(  \eta+\mu\right)
_{i}\leq\mu_{i}+1$.
\par
Now, let us forget that we fixed $i$. We thus have shown that every
$i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\left(  \eta+\mu\right)
_{i}\leq\mu_{i}+1$. In other words, $\left(  \eta+\mu\right)  /\mu$ is a
vertical strip (since $\left(  \eta+\mu\right)  /\mu$ is a vertical strip if
and only if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\left(
\eta+\mu\right)  _{i}\leq\mu_{i}+1$ (according to Exercise
\ref{exe.horistrip.postnikov}(b), applied to $\lambda=\eta+\mu$)).
\par
Altogether, we thus have shown that $\eta+\mu$ is an element of
$ \Par $ with the properties that $\ell\left(  \eta+\mu\right)
\leq n$ and that $\left(  \eta+\mu\right)  /\mu$ is a vertical strip. In other
words, $\eta+\mu$ is a $\lambda\in \Par $ such that $\ell\left(
\lambda\right)  \leq n$ and such that $\lambda/\mu$ is a vertical strip. In
other words, $\eta+\mu$ is an element of $\mathbf{G}$ (since $\mathbf{G}$ is
the set of all $\lambda\in \Par $ such that $\ell\left(
\lambda\right)  \leq n$ and such that $\lambda/\mu$ is a vertical strip). In
other words, $\eta+\mu\in\mathbf{G}$, qed.}. Thus, we can define a map
$\Phi:\mathbf{F}\rightarrow\mathbf{G}$ by%
\[
\left(  \Phi\left(  \eta\right)  =\eta+\mu\ \ \ \ \ \ \ \ \ \ \text{for every
}\eta\in\mathbf{F}\right)  .
\]
Consider this map $\Phi$. This map $\Phi$ is
surjective\footnote{\textit{Proof.} Let $\nu\in\mathbf{G}$. Then, $\nu$ is an
element of $\mathbf{G}$. In other words, $\nu$ is a $\lambda\in
 \Par $ such that $\ell\left(  \lambda\right)  \leq n$ and such
that $\lambda/\mu$ is a vertical strip (since $\mathbf{G}$ is the set of all
$\lambda\in \Par $ such that $\ell\left(  \lambda\right)  \leq n$
and such that $\lambda/\mu$ is a vertical strip). In other words, $\nu$ is an
element of $ \Par $ and has the properties that $\ell\left(
\nu\right)  \leq n$ and that $\nu/\mu$ is a vertical strip.
\par
We know that $\nu$ is a partition (since $\nu\in \Par $) having
length $\leq n$ (since $\ell\left(  \nu\right)  \leq n$). Every partition
$\lambda$ having length $\leq n$ can be regarded as an $n$-tuple $\lambda
\in\NN^n$ satisfying $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq
\lambda_{n}$. Applying this to the partition $\nu$, we conclude that $\nu$ can
be regarded as an $n$-tuple $\nu\in\NN^n$ satisfying $\nu_{1}\geq
\nu_{2}\geq\cdots\geq\nu_{n}$. Let us regard $\nu$ in this way. Thus, $\nu$ is
an $n$-tuple in $\NN^n$ satisfying $\nu_{1}\geq\nu_{2}\geq\cdots
\geq\nu_{n}$.
\par
We have $\mu\subseteq\nu$ (since $\nu/\mu$ is a vertical strip). In other
words,%
\begin{equation}
\mu_{i}\leq\nu_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,3,\ldots\right\}  \label{pf.lem.sol.pieri.alt-e.1xs.surj1}%
\end{equation}
(since $\mu\subseteq\nu$ if and only if we have $\mu_{i}\leq\nu_{i}$ for every
$i\in\left\{  1,2,3,\ldots\right\}  $ (according to the definition of the
relation $\subseteq$ on $ \Par $)). Also, $\nu/\mu$ is a vertical
strip. In other words,%
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies }\nu_{i}%
\leq\mu_{i}+1 \label{pf.lem.sol.pieri.alt-e.1xs.surj2}%
\end{equation}
(because $\nu/\mu$ is a vertical strip if and only if every $i\in\left\{
1,2,3,\ldots\right\}  $ satisfies $\nu_{i}\leq\mu_{i}+1$ (according to
Exercise \ref{exe.horistrip.postnikov}(b), applied to $\lambda=\nu$)).
\par
Now, let $\gamma\in \ZZ^{n}$ denote the $n$-tuple $\left(  \nu_{1}%
-\mu_{1},\nu_{2}-\mu_{2},\ldots,\nu_{n}-\mu_{n}\right)  $. Then, for every
$i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align}
\gamma_{i}  &  =\nu_{i}-\underbrace{\mu_{i}}_{\substack{\leq\nu_{i}\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.1xs.surj1}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\gamma=\left(  \nu_{1}-\mu_{1},\nu_{2}-\mu_{2},\ldots,\nu_{n}%
-\mu_{n}\right)  \right) \label{pf.lem.sol.pieri.alt-e.1xs.surjetai}\\
&  \geq\nu_{i}-\nu_{i}=0. \label{pf.lem.sol.pieri.alt-e.1xs.surj3}%
\end{align}
Also, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{equation}
\gamma_{i}=\underbrace{\nu_{i}}_{\substack{\leq\mu_{i}+1\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.1xs.surj2}))}}}-\mu_{i}\leq\left(  \mu
_{i}+1\right)  -\mu_{i}=1. \label{pf.lem.sol.pieri.alt-e.1xs.surj4}%
\end{equation}
\par
Now, let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, $\gamma_{i}$ is an
integer satisfying $\gamma_{i}\geq0$ (by
(\ref{pf.lem.sol.pieri.alt-e.1xs.surj3})) and $\gamma_{i}\leq1$ (by
(\ref{pf.lem.sol.pieri.alt-e.1xs.surj2})). In other words, $\gamma_{i}$ is an
element of $\left\{  0,1,\ldots,1\right\}  =\left\{  0,1\right\}  $. Now, let
us forget that we have fixed $i$. We thus have proven that $\gamma_{i}%
\in\left\{  0,1\right\}  $ for every $i\in\left\{  1,2,\ldots,n\right\}  $. In
other words, $\gamma\in\left\{  0,1\right\}  ^{n}$.
\par
Now, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\[
\left(  \gamma+\mu\right)  _{i}=\underbrace{\gamma_{i}}_{\substack{=\nu
_{i}-\mu_{i}\\\text{(by (\ref{pf.lem.sol.pieri.alt-e.1xs.surjetai}))}}%
}+\mu_{i}=\left(  \nu_{i}-\mu_{i}\right)  +\mu_{i}=\nu_{i}.
\]
In other words, corresponding entries of the $n$-tuples $\gamma+\mu$ and $\nu$
are equal to each other. Hence, the $n$-tuples $\gamma+\mu$ and $\nu$ are
equal. In other words, $\gamma+\mu=\nu$.
\par
But $\nu$ is an $n$-tuple in $\NN^n$ satisfying $\nu_{1}\geq\nu
_{2}\geq\cdots\geq\nu_{n}$. Hence, the $n$-tuple $\nu+\rho$ contains no two
equal entries (according to (\ref{pf.lem.sol.pieri.alt-e.1xs.alphaprho}),
applied to $\alpha=\nu+\rho$). In other words, the $n$-tuple $\gamma+\mu+\rho$
contains no two equal entries (since $\gamma+\mu=\nu$).
\par
So we know that $\gamma$ is an element of $\left\{  0,1\right\}  ^{n}$ having
the property that the $n$-tuple $\gamma+\mu+\rho$ contains no two equal
entries. In other words, $\gamma$ is an $\eta\in\left\{  0,1\right\}  ^{n}$
such that the $n$-tuple $\eta+\mu+\rho$ contains no two equal entries. In
other words, $\gamma$ is an element of $\mathbf{F}$ (since $\mathbf{F}$ is the
set of all $\eta\in\left\{  0,1\right\}  ^{n}$ such that the $n$-tuple
$\eta+\mu+\rho$ contains no two equal entries). In other words, $\gamma
\in\mathbf{F}$. Hence, the definition of $\Phi$ yields $\Phi\left(
\gamma\right)  =\gamma+\mu=\nu$. Thus, $\nu=\Phi\left(  \underbrace{\gamma
}_{\in\mathbf{F}}\right)  \in\Phi\left(  \mathbf{F}\right)  $.
\par
Let us now forget that we fixed $\nu$. We thus have proven that $\nu\in
\Phi\left(  \mathbf{F}\right)  $ for every $\nu\in\mathbf{G}$. In other words,
$\mathbf{G}\subset\Phi\left(  \mathbf{F}\right)  $. In other words, the map
$\Phi$ is surjective, qed.} and injective\footnote{\textit{Proof.} Let $\eta$
and $\eta^{\prime}$ be two elements of $\mathbf{F}$ such that $\Phi\left(
\eta\right)  =\Phi\left(  \eta^{\prime}\right)  $.
\par
We have $\Phi\left(  \eta\right)  =\eta+\mu$ (by the definition of $\Phi$) and
$\Phi\left(  \eta^{\prime}\right)  =\eta^{\prime}+\mu$ (by the definition of
$\Phi$). From $\Phi\left(  \eta\right)  =\eta+\mu$, we obtain $\eta+\mu
=\Phi\left(  \eta\right)  =\Phi\left(  \eta^{\prime}\right)  =\eta^{\prime
}+\mu$.
\par
Now, let $i\in\left\{  1,2,\ldots,n\right\}  $. We have $\left(  \eta
+\mu\right)  _{i}=\eta_{i}+\mu_{i}$. Compared with $\left(  \underbrace{\eta
+\mu}_{=\eta^{\prime}+\mu}\right)  _{i}=\left(  \eta^{\prime}+\mu\right)
_{i}=\left(  \eta^{\prime}\right)  _{i}+\mu_{i}$, this yields $\eta_{i}%
+\mu_{i}=\left(  \eta^{\prime}\right)  _{i}+\mu_{i}$. Subtracting $\mu_{i}$
from both sides of this equation, we obtain $\eta_{i}=\left(  \eta^{\prime
}\right)  _{i}$.
\par
Let us now forget that we fixed $i$. We thus have shown that $\eta_{i}=\left(
\eta^{\prime}\right)  _{i}$ for every $i\in\left\{  1,2,\ldots,n\right\}  $.
In other words, corresponding entries of the $n$-tuples $\eta$ and
$\eta^{\prime}$ are equal. Thus, the $n$-tuples $\eta$ and $\eta^{\prime}$ are
equal. That is, $\eta=\eta^{\prime}$.
\par
Now, let us forget that we fixed $\eta$ and $\eta^{\prime}$. We thus have
proven that if $\eta$ and $\eta^{\prime}$ are two elements of $\mathbf{F}$
such that $\Phi\left(  \eta\right)  =\Phi\left(  \eta^{\prime}\right)  $, then
$\eta=\eta^{\prime}$. In other words, the map $\Phi$ is injective, qed.}.
Hence, the map $\Phi$ is bijective. In other words, $\Phi$ is a bijection.
Also,%
\begin{equation}
\left\vert \Phi\left(  \eta\right)  /\mu\right\vert =\eta_{1}+\eta_{2}%
+\cdots+\eta_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }\eta\in\mathbf{F}.
\label{pf.lem.sol.pieri.alt-e.1xs.Phisize}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.1xs.Phisize}):} Let
$\eta\in\mathbf{F}$. Then, $\Phi\left(  \eta\right)  $ is an element of
$\mathbf{G}$. In other words, $\Phi\left(  \eta\right)  $ is a $\lambda
\in \Par $ such that $\ell\left(  \lambda\right)  \leq n$ and
such that $\lambda/\mu$ is a vertical strip (since $\mathbf{G}$ is the set of
all $\lambda\in \Par $ such that $\ell\left(  \lambda\right)
\leq n$ and such that $\lambda/\mu$ is a vertical strip). In other words,
$\Phi\left(  \eta\right)  $ is an element of $ \Par $ and has the
properties that $\ell\left(  \Phi\left(  \eta\right)  \right)  \leq n$ and
such that $\Phi\left(  \eta\right)  /\mu$ is a vertical strip. Thus, in
particular, the skew shape $\Phi\left(  \eta\right)  /\mu$ is well-defined.
\par
We have $n\geq\ell\left(  \Phi\left(  \eta\right)  \right)  $ (since
$\ell\left(  \Phi\left(  \eta\right)  \right)  \leq n$) and $n\geq\ell\left(
\mu\right)  $ (since $\ell\left(  \mu\right)  \leq n$). Now,%
\begin{align*}
\left\vert \Phi\left(  \eta\right)  /\mu\right\vert  &
=\underbrace{\left\vert \Phi\left(  \eta\right)  \right\vert }%
_{\substack{=\sum_{i=1}^{\infty}\left(  \Phi\left(  \eta\right)  \right)
_{i}\\=\sum_{i=1}^{n}\left(  \Phi\left(  \eta\right)  \right)  _{i}%
+\sum_{i=n+1}^{\infty}\left(  \Phi\left(  \eta\right)  \right)  _{i}%
}}-\underbrace{\left\vert \mu\right\vert }_{\substack{=\sum_{i=1}^{\infty}%
\mu_{i}\\=\sum_{i=1}^{n}\mu_{i}+\sum_{i=n+1}^{\infty}\mu_{i}}}\\
&  =\left(  \sum_{i=1}^{n}\underbrace{\left(  \Phi\left(  \eta\right)
\right)  _{i}}_{\substack{=\left(  \eta+\mu\right)  _{i}\\\text{(since }%
\Phi\left(  \eta\right)  =\eta+\mu\\\text{(by the definition of }\Phi
\text{))}}}+\sum_{i=n+1}^{\infty}\underbrace{\left(  \Phi\left(  \eta\right)
\right)  _{i}}_{\substack{=0\\\text{(since }i>n\geq\ell\left(  \Phi\left(
\eta\right)  \right)  \text{)}}}\right)  -\left(  \sum_{i=1}^{n}\mu_{i}%
+\sum_{i=n+1}^{\infty}\underbrace{\mu_{i}}_{\substack{=0\\\text{(since
}i>n\geq\ell\left(  \mu\right)  \text{)}}}\right) \\
&  =\left(  \sum_{i=1}^{n}\left(  \eta+\mu\right)  _{i}+\underbrace{\sum
_{i=n+1}^{\infty}0}_{=0}\right)  -\left(  \sum_{i=1}^{n}\mu_{i}%
+\underbrace{\sum_{i=n+1}^{\infty}0}_{=0}\right)  =\sum_{i=1}^{n}%
\underbrace{\left(  \eta+\mu\right)  _{i}}_{\substack{=\eta_{i}+\mu
_{i}\\\text{(by the definition}\\\text{of addition on }\NN^{n}\text{)}%
}}-\sum_{i=1}^{n}\mu_{i}\\
&  =\underbrace{\sum_{i=1}^{n}\left(  \eta_{i}+\mu_{i}\right)  }_{=\sum
_{i=1}^{n}\eta_{i}+\sum_{i=1}^{n}\mu_{i}}-\sum_{i=1}^{n}\mu_{i}=\left(
\sum_{i=1}^{n}\eta_{i}+\sum_{i=1}^{n}\mu_{i}\right)  -\sum_{i=1}^{n}\mu
_{i}=\sum_{i=1}^{n}\eta_{i}\\
&  =\eta_{1}+\eta_{2}+\cdots+\eta_{n}.
\end{align*}
This proves (\ref{pf.lem.sol.pieri.alt-e.1xs.Phisize}).} Now,
(\ref{pf.lem.sol.pieri.alt-e.1xs.step4}) becomes%
\begin{align}
\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \cdot a_{\mu+\rho}
&  =\sum_{\eta\in\mathbf{F}}\underbrace{a_{\eta+\mu+\rho}}_{\substack{=a_{\Phi
\left(  \eta\right)  +\rho}\\\text{(since }\eta+\mu=\Phi\left(  \eta\right)
\\\text{(since }\Phi\left(  \eta\right)  =\eta+\mu\\\text{(by the definition
of }\Phi\text{)))}}}\underbrace{t^{\eta_{1}+\eta_{2}+\cdots+\eta_{n}}%
}_{\substack{=t^{\left\vert \Phi\left(  \eta\right)  /\mu\right\vert
}\\\text{(since }\eta_{1}+\eta_{2}+\cdots+\eta_{n}=\left\vert \Phi\left(
\eta\right)  /\mu\right\vert \\\text{(by
\eqref{pf.lem.sol.pieri.alt-e.1xs.Phisize}))}}}
=\sum_{\eta\in\mathbf{F}}%
a_{\Phi\left(  \eta\right)  +\rho}t^{\left\vert \Phi\left(  \eta\right)
/\mu\right\vert }\nonumber\\
&  =\underbrace{\sum_{\lambda\in\mathbf{G}}}_{\substack{=\sum\limits
_{\substack{\lambda\in \Par ;\\\ell\left(  \lambda\right)  \leq
n;\\\lambda/\mu\text{ is a vertical strip}}}\\\text{(since }\mathbf{G}\text{
is the set of all }\lambda\in \Par \\\text{such that }\ell\left(
\lambda\right)  \leq n\text{ and such that}\\\lambda/\mu\text{ is a vertical
strip)}}}a_{\lambda+\rho}t^{\left\vert \lambda/\mu\right\vert }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }\lambda\text{
for }\Phi\left(  \eta\right)  \text{, since the map }\Phi:\mathbf{F}%
\rightarrow\mathbf{G}\text{ is a bijection}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(  \lambda
\right)  \leq n;\\\lambda/\mu\text{ is a vertical strip}}}a_{\lambda+\rho
}t^{\left\vert \lambda/\mu\right\vert }.
\label{pf.lem.sol.pieri.alt-e.1xs.step6}%
\end{align}


From Corollary \ref{cor.stembridge.slambda}, we obtain
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\dfrac{a_{\lambda+\rho}%
}{a_{\rho}}\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda\text{
satisfying }\ell\left(  \lambda\right)  \leq n.
\label{pf.lem.sol.pieri.alt-e.1xs.slam}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.1xs.slam}):} Recall
that Corollary \ref{cor.stembridge.slambda} used the notations and conventions
of Section \ref{bialternant-section}. In particular, ``partition'' means
``partition having at most $n$ parts'' in
Corollary \ref{cor.stembridge.slambda}. Thus,
Corollary \ref{cor.stembridge.slambda} can only be applied to a partition
$\lambda$ if $\lambda$ has at most $n$ parts.
\par
Let $\lambda$ be a partition satisfying $\ell\left(  \lambda\right)  \leq n$.
Then, the partition $\lambda$ has at most $n$ parts (since $\ell\left(
\lambda\right)  \leq n$). Thus, Corollary \ref{cor.stembridge.slambda} can be
applied to $\lambda$. As a result, we obtain $s_{\lambda}\left(
\xx\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$, where the notations
of Section \ref{bialternant-section} are used. Translating this result into
our notations, we obtain $s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
=\dfrac{a_{\lambda+\rho}}{a_{\rho}}$ (since what was denoted by $\xx$
in Corollary \ref{cor.stembridge.slambda} is what we now call $\left(
x_{1},x_{2},\ldots,x_{n}\right)  $). This proves
(\ref{pf.lem.sol.pieri.alt-e.1xs.slam}).} Notice that this equality makes
sense, since division by $a_{\rho}$ is unambiguous in the ring $\kk%
\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ (this was proven in Corollary
\ref{cor.stembridge.slambda} as well). Now,
\begin{align*}
&  \left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \underbrace{s_{\mu
}\left(  x_{1},x_{2},\ldots,x_{n}\right)  }_{\substack{=\dfrac{a_{\mu+\rho}%
}{a_{\rho}}\\\text{(by (\ref{pf.lem.sol.pieri.alt-e.1xs.slam}), applied to
}\lambda=\mu\text{)}}}\\
&  =\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \dfrac
{a_{\mu+\rho}}{a_{\rho}}=\dfrac{\left(  \prod_{i=1}^{n}\left(  1+x_{i}%
t\right)  \right)  \cdot a_{\mu+\rho}}{a_{\rho}}\\
&  =\dfrac{\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(
\lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical strip}}%
}a_{\lambda+\rho}t^{\left\vert \lambda/\mu\right\vert }}{a_{\rho}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{this follows by dividing the equality
(\ref{pf.lem.sol.pieri.alt-e.1xs.step6}) by }a_{\rho}\right) \\
&  =\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(  \lambda
\right)  \leq n;\\\lambda/\mu\text{ is a vertical strip}}}\underbrace{\dfrac
{a_{\lambda+\rho}}{a_{\rho}}}_{\substack{=s_{\lambda}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  \\\text{(according to
(\ref{pf.lem.sol.pieri.alt-e.1xs.slam}))}}}t^{\left\vert \lambda
/\mu\right\vert }=\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(
\lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical strip}}}s_{\lambda
}\left(  x_{1},x_{2},\ldots,x_{n}\right)  t^{\left\vert \lambda/\mu\right\vert
}.
\end{align*}
This proves Lemma \ref{lem.sol.pieri.alt-e.1xs}.
\end{proof}

Next, we notice the following simple fact:

\begin{lemma}
\label{lem.sol.pieri.alt-e.finite-e}For every $n\in \NN$, we have%
\begin{equation}
\prod_{i=1}^{n}\left(  1+x_{i}t\right)  =\sum_{k\geq0}e_{k}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  t^{k}
\label{sol.pieri.alt-e.es.e-generating-function}%
\end{equation}
in the polynomial ring $\left(  \kk\left[  x_{1},x_{2},\ldots
,x_{n}\right]  \right)  \left[  t\right]  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.pieri.alt-e.finite-e}.]From
(\ref{e-generating-function-no-HE}), we know that
$\prod_{i=1}^{\infty}\left(  1+x_{i}t\right)
=\sum_{n\geq0}e_{n}\left(  \xx\right)  t^{n}
=\sum_{k\geq0}e_{k}\left(  \xx\right)  t^{k}$ (here, we renamed the
summation index $n$ as $k$).

Now, fix $n \in \NN$.
Let $\Phi$ denote the map $R\left(  \xx\right)  \rightarrow
\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ which sends every $f\in
R\left(  \xx\right)  $ to the result of substituting $x_{1}%
,x_{2},\ldots,x_{n},0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in $f$. Then,
$\Phi$ is a substitution homomorphism, and thus a $\kk$-algebra
homomorphism. Also,%
\begin{equation}
\Phi\left(  f\right)  =f\left(  x_{1},x_{2},\ldots,x_{n}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in R\left(  \xx\right)  .
\label{pf.lem.sol.pieri.alt-e.finite-e.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.finite-e.1}):} Let
$f\in R\left(  \xx\right)  $. Then, $\Phi\left(  f\right)  $ is the
result of substituting $x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for
$x_{1},x_{2},x_{3},\ldots$ in $f$ (according to the definition of $\Phi$). In
other words, $\Phi\left(  f\right)  $ is $f\left(  x_{1},x_{2},\ldots
,x_{n}\right)  $ (since $f\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ is the
result of substituting $x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for
$x_{1},x_{2},x_{3},\ldots$ in $f$ (according to the definition of $f\left(
x_{1},x_{2},\ldots,x_{n}\right)  $)). In other words, $\Phi\left(  f\right)
=f\left(  x_{1},x_{2},\ldots,x_{n}\right)  $. This proves
(\ref{pf.lem.sol.pieri.alt-e.finite-e.1}).} Also,%
\begin{equation}
\Phi\left(  x_{i}\right)  =x_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  1,2,\ldots,n\right\}  \label{pf.lem.sol.pieri.alt-e.finite-e.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.finite-e.2}):} Let
$i\in\left\{  1,2,\ldots,n\right\}  $. Then, $\Phi\left(  x_{i}\right)  $ is
the result of substituting $x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for
$x_{1},x_{2},x_{3},\ldots$ in $x_{i}$ (according to the definition of $\Phi$).
In other words, $\Phi\left(  x_{i}\right)  $ is $x_{i}$ (since the result of
substituting $x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for $x_{1},x_{2}%
,x_{3},\ldots$ in $x_{i}$ is clearly $x_{i}$ (because $i\in\left\{
1,2,\ldots,n\right\}  $)). In other words, $\Phi\left(  x_{i}\right)  =x_{i}$.
This proves (\ref{pf.lem.sol.pieri.alt-e.finite-e.2}).} Moreover,%
\begin{equation}
\Phi\left(  x_{i}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
n+1,n+2,n+3,\ldots\right\}  \label{pf.lem.sol.pieri.alt-e.finite-e.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.pieri.alt-e.finite-e.3}):} Let
$i\in\left\{  n+1,n+2,n+3,\ldots\right\}  $. Then, $\Phi\left(  x_{i}\right)
$ is the result of substituting $x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for
$x_{1},x_{2},x_{3},\ldots$ in $x_{i}$ (according to the definition of $\Phi$).
In other words, $\Phi\left(  x_{i}\right)  $ is $0$ (since the result of
substituting $x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots$ for $x_{1},x_{2}%
,x_{3},\ldots$ in $x_{i}$ is clearly $0$ (because $i\in\left\{
n+1,n+2,n+3,\ldots\right\}  $)). In other words, $\Phi\left(  x_{i}\right)
=0$. This proves (\ref{pf.lem.sol.pieri.alt-e.finite-e.3}).}

Recall that if $A$ and $B$ are two commutative $\kk$-algebras, and if
$\Psi:A\rightarrow B$ is a $\kk$-algebra homomorphism, then a
continuous\footnote{This notion of ``continuous'' refers to the
usual topologies on $A\left[
\left[  t\right]  \right]  $ and $B\left[  \left[  t\right]  \right]  $.}
$\kk$-algebra homomorphism $\Psi\left[  \left[  t\right]  \right]
:A\left[  \left[  t\right]  \right]  \rightarrow B\left[  \left[  t\right]
\right]  $ between the rings $A\left[  \left[  t\right]  \right]  $ and
$B\left[  \left[  t\right]  \right]  $ is defined by%
\[
\left(  \left(  \Psi\left[\left[  t\right]\right]  \right)
\left(  \sum_{i\geq0}a_{i}%
t^{i}\right)  =\sum_{i\geq0}\Psi\left(  a_{i}\right)  t^{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{i}\right)  _{i\geq0}\in
A^{\NN}\right)  .
\]
\footnote{The commutativity of $A$ and $B$ is not really necessary for this
statement to hold, but we require it because the commutative case of this
statement is better known.}

Applying this to $A=R\left(  \xx\right)  $, $B=\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $ and $\Psi=\Phi$, we conclude that the
$\kk$-algebra homomorphism $\Phi:R\left(  \xx\right)
\rightarrow\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ gives rise to
a continuous $\kk$-algebra homomorphism $\Phi\left[  \left[  t\right]
\right]  :\left(  R\left(  \xx\right)  \right)  \left[  \left[
t\right]  \right]  \rightarrow\left(  \kk\left[  x_{1},x_{2}%
,\ldots,x_{n}\right]  \right)  \left[  \left[  t\right]  \right]  $. Applying
this homomorphism $\Phi\left[  \left[  t\right]  \right]  $ to both sides of
the identity $\prod_{i=1}^{\infty}\left(  1+x_{i}t\right)  =\sum_{k\geq0}%
e_{k}\left(  \xx\right)  t^{k}$, we obtain%
\begin{align*}
\left(  \Phi\left[  \left[  t\right]  \right]  \right)  \left(  \prod
_{i=1}^{\infty}\left(  1+x_{i}t\right)  \right)   &  =\left(  \Phi\left[
\left[  t\right]  \right]  \right)  \left(  \sum_{k\geq0}e_{k}\left(
\xx\right)  t^{k}\right)  =\sum_{k\geq0}\underbrace{\Phi\left(
e_{k}\right)  }_{\substack{=e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)
\\\text{(by (\ref{pf.lem.sol.pieri.alt-e.finite-e.1}), applied to }%
f=e_{k}\text{)}}}t^{k}\\
&  =\sum_{k\geq0}e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)  t^{k},
\end{align*}
so that%
\begin{align*}
\sum_{k\geq0}e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)  t^{k}  &  =\left(
\Phi\left[  \left[  t\right]  \right]  \right)  \left(  \prod_{i=1}^{\infty
}\left(  1+x_{i}t\right)  \right)  =\prod_{i=1}^{\infty}\underbrace{\left(
\Phi\left[  \left[  t\right]  \right]  \right)  \left(  1+x_{i}t\right)
}_{\substack{=\left(  \Phi\left[  \left[  t\right]  \right]  \right)  \left(
1\right)  +\left(  \Phi\left[  \left[  t\right]  \right]  \right)  \left(
x_{i}\right)  \cdot\left(  \Phi\left[  \left[  t\right]  \right]  \right)
\left(  t\right)  \\\text{(since }\left(  \Phi\left[  \left[  t\right]
\right]  \right)  \text{ is a }\kk\text{-algebra homomorphism)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\Phi\left[  \left[  t\right]  \right]  \text{ preserves
convergent infinite products}\\
\text{(because }\Phi\left[  \left[  t\right]  \right]  \text{ is a continuous
}\kk\text{-algebra homomorphism)}%
\end{array}
\right) \\
&  =\prod_{i=1}^{\infty}\left(  \underbrace{\left(  \Phi\left[  \left[
t\right]  \right]  \right)  \left(  1\right)  }_{\substack{=1\\\text{(by the
definition of }\Phi\left[  \left[  t\right]  \right]  \text{)}}%
}+\underbrace{\left(  \Phi\left[  \left[  t\right]  \right]  \right)  \left(
x_{i}\right)  }_{\substack{=\Phi\left(  x_{i}\right)  \\\text{(by the
definition of }\Phi\left[  \left[  t\right]  \right]  \text{)}}}\cdot
\underbrace{\left(  \Phi\left[  \left[  t\right]  \right]  \right)  \left(
t\right)  }_{\substack{=t\\\text{(by the definition of }\Phi\left[  \left[
t\right]  \right]  \text{)}}}\right) \\
&  =\prod_{i=1}^{\infty}\left(  1+\Phi\left(  x_{i}\right)  \cdot t\right)
=\prod_{i=1}^{\infty}\left(  1+\Phi\left(  x_{i}\right)  t\right) \\
&  =\left(  \prod_{i=1}^{n}\left(  1+\underbrace{\Phi\left(  x_{i}\right)
}_{\substack{=x_{i}\\\text{(by (\ref{pf.lem.sol.pieri.alt-e.finite-e.2}))}%
}}t\right)  \right)  \cdot\left(  \prod_{i=n+1}^{\infty}\left(
1+\underbrace{\Phi\left(  x_{i}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.pieri.alt-e.finite-e.3}))}}}t\right)  \right) \\
&  =\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \cdot\left(
\prod_{i=n+1}^{\infty}\underbrace{\left(  1+0t\right)  }_{=1}\right)  =\left(
\prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  \cdot\underbrace{\left(
\prod_{i=n+1}^{\infty}1\right)  }_{=1}=\prod_{i=1}^{n}\left(  1+x_{i}t\right)
.
\end{align*}
This proves Lemma \ref{lem.sol.pieri.alt-e.finite-e}.
\end{proof}

As a corollary of Lemma \ref{lem.sol.pieri.alt-e.1xs}, we now obtain:

\begin{corollary}
\label{cor.sol.pieri.alt-e.es}Let $n\in \NN$. Let $\mu$ be a partition
such that $\ell\left(  \mu\right)  \leq n$. Let $N\in \NN$. Then,%
\[
e_{N}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  =\sum\limits_{\substack{\lambda\in \Par %
;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical
}N\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
\]
in the polynomial ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.sol.pieri.alt-e.es}.]In the polynomial ring
$\left(  \kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  \right)  \left[
t\right]  $, we have%
\begin{align*}
&  \left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  s_{\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right) \\
&  =\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(  \lambda
\right)  \leq n;\\\lambda/\mu\text{ is a vertical strip}}}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{n}\right)  t^{\left\vert \lambda/\mu\right\vert
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.sol.pieri.alt-e.1xs}%
}\right) \\
&  =\sum_{k\in\NN}\underbrace{\sum\limits_{\substack{\lambda\in
 \Par ;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{
is a vertical strip;}\\\left\vert \lambda/\mu\right\vert =k}}}%
_{\substack{=\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(
\lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical }k\text{-strip}%
}}\\\text{(because for a partition }\lambda\in \Par %
\text{,}\\\text{the condition}\\\left(  \lambda/\mu\text{ is a vertical strip
and }\left\vert \lambda/\mu\right\vert =k\right)  \\\text{is equivalent to the
condition}\\\left(  \lambda/\mu\text{ is a vertical }k\text{-strip}\right)
\text{)}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
\underbrace{t^{\left\vert \lambda/\mu\right\vert }}_{\substack{=t^{k}%
\\\text{(since }\left\vert \lambda/\mu\right\vert =k\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert \lambda/\mu\right\vert
\in\NN \text{ whenever }\lambda\in \Par \text{ has the
property that }\lambda/\mu\text{ is a vertical strip}\right) \\
&  =\sum_{k\in\NN}\sum\limits_{\substack{\lambda\in \Par %
;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical
}k\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
t^{k}=\sum_{k\in\NN}\left(  \sum\limits_{\substack{\lambda\in
 \Par ;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{
is a vertical }k\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{n}\right)  \right)  t^{k}.
\end{align*}
Thus,
\begin{align}
&  \left(  \text{the coefficient of }t^{N}\text{ in the polynomial
}\underbrace{\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  s_{\mu
}\left(  x_{1},x_{2},\ldots,x_{n}\right)  }_{=\sum_{k\in\NN}\left(
\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(  \lambda\right)
\leq n;\\\lambda/\mu\text{ is a vertical }k\text{-strip}}}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{n}\right)  \right)  t^{k}}\right) \nonumber\\
&  =\left(  \text{the coefficient of }t^{N}\text{ in the polynomial }%
\sum_{k\in\NN}\left(  \sum\limits_{\substack{\lambda\in \Par %
;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical
}k\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \right)
t^{k}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(  \lambda
\right)  \leq n;\\\lambda/\mu\text{ is a vertical }N\text{-strip}}}s_{\lambda
}\left(  x_{1},x_{2},\ldots,x_{n}\right)  .
\label{pf.cor.sol.pieri.alt-e.es.1}%
\end{align}


On the other hand,%
\begin{align*}
&  \underbrace{\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)
}_{\substack{=\sum_{k\geq0}e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)
t^{k}\\\text{(by Lemma \ref{lem.sol.pieri.alt-e.finite-e})}}}s_{\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right) \\
&  =\left(  \sum_{k\geq0}e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)
t^{k}\right)  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =\sum_{k\geq
0}e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right)  t^{k}.
\end{align*}
Hence,%
\begin{align*}
&  \left(  \text{the coefficient of }t^{N}\text{ in the polynomial
}\underbrace{\left(  \prod_{i=1}^{n}\left(  1+x_{i}t\right)  \right)  s_{\mu
}\left(  x_{1},x_{2},\ldots,x_{n}\right)  }_{=\sum_{k\geq0}e_{k}\left(
x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(  x_{1},x_{2},\ldots
,x_{n}\right)  t^{k}}\right) \\
&  =\left(  \text{the coefficient of }t^{N}\text{ in the polynomial }%
\sum_{k\geq0}e_{k}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu
}\left(  x_{1},x_{2},\ldots,x_{n}\right)  t^{k}\right) \\
&  =e_{N}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right)  .
\end{align*}
Compared with (\ref{pf.cor.sol.pieri.alt-e.es.1}), this yields%
\[
e_{N}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  =\sum\limits_{\substack{\lambda\in \Par %
;\\\ell\left(  \lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical
}N\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  .
\]
This proves Corollary \ref{cor.sol.pieri.alt-e.es}.
\end{proof}

We can rewrite Corollary \ref{cor.sol.pieri.alt-e.es} one step further:

\begin{corollary}
\label{cor.sol.pieri.alt-e.es.2}Let $n\in \NN$. Let $\mu$ be a partition
such that $\ell\left(  \mu\right)  \leq n$. Let $N\in \NN$. Then,%
\[
e_{N}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  =\sum\limits_{\substack{\lambda\in \Par %
;\\\lambda/\mu\text{ is a vertical }N\text{-strip}}}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{n}\right)
\]
in the polynomial ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.sol.pieri.alt-e.es.2}.]For any $\lambda
\in \Par $ satisfying $\ell\left(  \lambda\right)  >n$, we have
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =0
\label{pf.cor.sol.pieri.alt-e.es.2.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.sol.pieri.alt-e.es.2.1}):} Let
$\lambda\in \Par $ be such that $\ell\left(  \lambda\right)  >n$.
Then, $\ell\left(  \lambda\right)  >n$. In other words, the number of parts of
the partition $\lambda$ is $>n$ (since $\ell\left(  \lambda\right)  $ is the
number of parts of the partition $\lambda$). In other words, the partition
$\lambda$ has more than $n$ parts. Thus, Exercise \ref{exe.schur.finite}(b)
shows that $s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =0$. This
proves (\ref{pf.cor.sol.pieri.alt-e.es.2.1}).}. Now, every $\lambda
\in \Par $ satisfies either $\ell\left(  \lambda\right)  \leq n$
or $\ell\left(  \lambda\right)  >n$ (but not both). Hence,%
\begin{align*}
&  \sum\limits_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a
vertical }N\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
\\
&  =\underbrace{\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(
\lambda\right)  \leq n;\\\lambda/\mu\text{ is a vertical }N\text{-strip}%
}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  }_{\substack{=e_{N}%
\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  \\\text{(by Corollary \ref{cor.sol.pieri.alt-e.es}%
)}}}+\sum\limits_{\substack{\lambda\in \Par ;\\\ell\left(
\lambda\right)  >n;\\\lambda/\mu\text{ is a vertical }N\text{-strip}%
}}\underbrace{s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
}_{\substack{=0\\\text{(by (\ref{pf.cor.sol.pieri.alt-e.es.2.1}))}}}\\
&  =e_{N}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(
x_{1},x_{2},\ldots,x_{n}\right)  +\underbrace{\sum\limits_{\substack{\lambda
\in \Par ;\\\ell\left(  \lambda\right)  >n;\\\lambda/\mu\text{ is
a vertical }N\text{-strip}}}0}_{=0}=e_{N}\left(  x_{1},x_{2},\ldots
,x_{n}\right)  \cdot s_{\mu}\left(  x_{1},x_{2},\ldots,x_{n}\right)  .
\end{align*}
This proves Corollary \ref{cor.sol.pieri.alt-e.es.2}.
\end{proof}

Now, let us finally complete the solution of Exercise \ref{exe.pieri.alt-e}.
Fix a partition $\mu$ and an $m\in \NN$. We shall prove that%
\[
s_{\mu}e_{m}=\sum\limits_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{
is a vertical }m\text{-strip}}}s_{\lambda}.
\]


Let us first notice that every $\lambda\in \Par $ for which
$\lambda/\mu$ is a vertical $m$-strip must be an element of
$\Par_{\left\vert \mu\right\vert +m}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\lambda\in \Par $ be such
that $\lambda/\mu$ is a vertical $m$-strip. We know that $\lambda/\mu$ is a
vertical $m$-strip. In other words, $\lambda/\mu$ is a vertical strip and
satisfies $\left\vert \lambda/\mu\right\vert =m$ (because $\lambda/\mu$ is a
vertical $m$-strip if and only if $\lambda/\mu$ is a vertical strip and
satisfies $\left\vert \lambda/\mu\right\vert =m$ (according to the definition
of a ``vertical $m$-strip'')). Now, we have
$\mu\subseteq\lambda$ (since $\lambda/\mu$ is a vertical trip) and thus
$\left\vert \lambda/\mu\right\vert =\left\vert \lambda\right\vert -\left\vert
\mu\right\vert $, so that $\left\vert \lambda\right\vert -\left\vert
\mu\right\vert =\left\vert \lambda/\mu\right\vert =m$. Therefore, $\left\vert
\lambda\right\vert =\left\vert \mu\right\vert +m$. In other words, $\lambda
\in\Par_{\left\vert \mu\right\vert +m}$ (since
$\lambda\in \Par $), qed.}. Thus, there are only finitely many
$\lambda\in \Par $ for which $\lambda/\mu$ is a vertical
$m$-strip (since there are only finitely many elements of
$\Par_{\left\vert \mu\right\vert +m}$). Hence, the sum $\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a vertical
}m\text{-strip}}}s_{\lambda}$ is a finite sum, and thus is a well-defined
element of $\Lambda$.

Define an $f\in\Lambda$ by%
\begin{equation}
f=s_{\mu}e_{m}-\sum\limits_{\substack{\lambda\in \Par ;\\\lambda
/\mu\text{ is a vertical }m\text{-strip}}}s_{\lambda}.
\label{sol.pieri.alt-e.es.last}%
\end{equation}
We shall prove that $f=0$.

Indeed, every $n\in \NN$ satisfying $n\geq\ell\left(  \mu\right)  $
satisfies%
\begin{align*}
f\left(  x_{1},x_{2},\ldots,x_{n}\right)   &  =\left(  s_{\mu}e_{m}%
-\sum\limits_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a
vertical }m\text{-strip}}}s_{\lambda}\right)  \left(  x_{1},x_{2},\ldots
,x_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f=s_{\mu}e_{m}-\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a vertical
}m\text{-strip}}}s_{\lambda}\right) \\
&  =\underbrace{s_{\mu}\left(  x_{1},x_{2},\ldots,x_{n}\right)  \cdot
e_{m}\left(  x_{1},x_{2},\ldots,x_{n}\right)  }_{\substack{=e_{m}\left(
x_{1},x_{2},\ldots,x_{n}\right)  \cdot s_{\mu}\left(  x_{1},x_{2},\ldots
,x_{n}\right)  \\=\sum\limits_{\substack{\lambda\in \Par ;\\\lambda
/\mu\text{ is a vertical }m\text{-strip}}}s_{\lambda}\left(  x_{1}%
,x_{2},\ldots,x_{n}\right)  \\\text{(by Corollary
\ref{cor.sol.pieri.alt-e.es.2}, applied to }N=m\text{)}}}-\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a vertical
}m\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right) \\
&  =\sum\limits_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a
vertical }m\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
-\sum\limits_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a
vertical }m\text{-strip}}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)
=0.
\end{align*}
Hence, Lemma \ref{lem.sol.pieri.alt-e.projlim}(b) (applied to $\mathbf{n}%
=\ell\left(  \mu\right)  $) yields $f=0$. Compared with
(\ref{sol.pieri.alt-e.es.last}), this yields $s_{\mu}e_{m}-\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a vertical
}m\text{-strip}}}s_{\lambda}=0$. In other words, $s_{\mu}e_{m}=\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{ is a vertical
}m\text{-strip}}}s_{\lambda}$. We thus have%
\begin{equation}
s_{\mu}e_{m}=\sum\limits_{\substack{\lambda\in \Par ;\\\lambda/\mu\text{
is a vertical }m\text{-strip}}}s_{\lambda}=\sum\limits_{\substack{\lambda^{+}%
\in \Par ;\\\lambda^{+}/\mu\text{ is a vertical }m\text{-strip}%
}}s_{\lambda^{+}} \label{sol.pieri.alt-e.es.verylast}%
\end{equation}
(here, we have renamed the summation index $\lambda$ as $\lambda^{+}$).

Let us now forget that we fixed $\mu$ and $m$. We thus have shown that
(\ref{sol.pieri.alt-e.es.verylast}) holds for every partition $\mu$ and every
$m\in \NN$. Thus, for every partition $\lambda$ and any $n\in
{\NN}$, we have%
\[
s_{\lambda}e_{n}=\sum\limits_{\substack{\lambda^{+}\in \Par %
;\\\lambda^{+}/\lambda\text{ is a vertical }n\text{-strip}}}s_{\lambda^{+}}%
\]
(by (\ref{sol.pieri.alt-e.es.verylast}), applied to $\mu=\lambda$ and $m=n$).
Thus, (\ref{eq.pieri.e}) is proven. Exercise~\ref{exe.pieri.alt-e} is solved.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cauchy.cauchy}}

\textit{Solution to Exercise~\ref{exe.cauchy.cauchy}.} Before we start
solving any specific part of this exercise, let us state some general
properties of determinants (and prove some of them):

\begin{itemize}
\item Every $m\in{ \NN }$ and every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ satisfy%
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\sum_{\sigma\in\Symm_m}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{m}\alpha_{i,\sigma\left(  i\right)  }. \label{sol.cauchy.cauchy.det.3}%
\end{equation}
(This is simply the explicit formula for the determinant of a matrix as a sum
over permutations.)

\item If a positive integer $m$ and a matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ are such that every $j\in\left\{
1,2,...,m-1\right\}  $ satisfies $\alpha_{m,j}=0$, then%
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\alpha_{m,m}\cdot\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m-1}%
\right)  . \label{sol.cauchy.cauchy.det.7}%
\end{equation}
\footnote{In fact, the condition that every $j\in\left\{  1,2,...,m-1\right\}
$ satisfies $\alpha_{m,j}=0$ means that all entries of the $m$-th row of the
$m\times m$-matrix $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}$ are zeroes
apart from (possibly) the last entry. Hence, applying Laplace expansion to the
determinant $\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
$ of this matrix yields a sum of products, all of which are zero apart from%
\[
\alpha_{m,m}\cdot\underbrace{\left(  \text{the }\left(  m,m\right)  \text{-th
cofactor of the matrix }\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
}_{=\left(  -1\right)  ^{m+m}\cdot\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m-1}\right)  =\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m-1}\right)  }=\alpha_{m,m}\cdot\det\left(  \left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m-1}\right)  .
\]
}

\item Every positive integer $m$ and every matrix $\left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\in A^{m\times m}$ satisfy%
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}\right)
_{i,j=1,2,...,m-1}\right)  =\alpha_{m,m}^{m-2}\cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  \label{sol.cauchy.cauchy.det.17}%
\end{equation}
if $\alpha_{m,m}$ is an invertible element of $A$.

\textit{Proof of \eqref{sol.cauchy.cauchy.det.17}:} Let $m$ be a positive
integer. Let $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in A^{m\times m}$
be a matrix such that $\alpha_{m,m}$ is an invertible element of $A$.

For every $\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^2$, define an
element $\beta_{i,j}$ of $A$ by
\begin{equation}
\beta_{i,j}= \begin{cases}
\alpha_{m,m},& \text{if }i=j;\\
-\alpha_{m,j},& \text{if }i=m\text{ and }j\neq m
\end{cases} \quad .
\label{sol.cauchy.cauchy.det.17.pf.beta}
\end{equation}


Then, the matrix $\left(  \beta_{i,j}\right)  _{i,j=1,2,...,m}$ is
lower-triangular. Since the determinant of a lower-triangular matrix equals
the product of its diagonal entries, we thus have
\begin{equation}
\det\left(  \left(  \beta_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\prod
_{i=1}^{m}\underbrace{\beta_{i,i}}_{\substack{=\alpha_{m,m}\\\text{(this
is}\\\text{easily seen)}}}=\prod_{i=1}^{m}\alpha_{m,m}=\alpha_{m,m}^{m}.
\label{sol.cauchy.cauchy.det.17.pf.beta.det2}%
\end{equation}


But since the determinant of a product of two square matrices equals the
product of their determinants, we have
\begin{align*}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\cdot\left(
\beta_{i,j}\right)  _{i,j=1,2,...,m}\right)   &  =\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  \cdot\underbrace{\det\left(
\left(  \beta_{i,j}\right)  _{i,j=1,2,...,m}\right)  }_{\substack{=\alpha
_{m,m}^{m}\\\text{(by \eqref{sol.cauchy.cauchy.det.17.pf.beta.det2})}}}\\
&  =\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
\cdot\alpha_{m,m}^{m}.
\end{align*}


Since $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\cdot\left(  \beta
_{i,j}\right)  _{i,j=1,2,...,m}=\left(  \sum_{k=1}^{m}\alpha_{i,k}\beta
_{k,j}\right)  _{i,j=1,2,...,m}$, this rewrites as%
\begin{equation}
\det\left(  \left(  \sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}\right)
_{i,j=1,2,...,m}\right)  =\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  \cdot\alpha_{m,m}^{m}.
\label{sol.cauchy.cauchy.det.17.pf.prod-det1}%
\end{equation}


Now, for every $\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}$,
define an element $\gamma_{i,j}$ of $A$ by
\begin{equation}
\gamma_{i,j}=\sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}.
\label{sol.cauchy.cauchy.det.17.pf.gamma}
\end{equation}
Then,
\begin{equation}
\det\left(  \left(  \gamma_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\det\left(
\left(  \sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}\right)  _{i,j=1,2,...,m}\right)
=\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  \cdot
\alpha_{m,m}^{m}
\label{sol.cauchy.cauchy.det.17.pf.gamma-det.1}
\end{equation}
(by \eqref{sol.cauchy.cauchy.det.17.pf.prod-det1}).

However, for every $\left(  i,j\right)  $, we can simplify the expression for
$\gamma_{i,j}$ given by \eqref{sol.cauchy.cauchy.det.17.pf.gamma} by plugging
in the definition of $\beta_{k,j}$ (which guarantees that no more than two of
the terms of the sum will be nonzero). We obtain%
\begin{equation}
\gamma_{i,j}= \begin{cases}
\alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j},
& \text{if }i\neq m\text{ and }j\neq m;\\
0, & \text{if }i=m\text{ and }j\neq m;\\
\alpha_{i,m}\alpha_{m,m}, & \text{if }j=m
\end{cases} \quad .
\label{sol.cauchy.cauchy.det.17.pf.gamma.explicit}%
\end{equation}
In particular, this shows that every $j\in\left\{  1,2,...,m-1\right\}  $
satisfies $\gamma_{m,j}=0$. Hence, \eqref{sol.cauchy.cauchy.det.7} (applied to
$\gamma_{i,j}$ instead of $\alpha_{i,j}$) yields
\begin{align*}
\det\left(  \left(  \gamma_{i,j}\right)  _{i,j=1,2,...,m}\right)   &
=\underbrace{\gamma_{m,m}}_{\substack{=\alpha_{m,m}\alpha_{m,m}\\\text{(by
\eqref{sol.cauchy.cauchy.det.17.pf.gamma.explicit}, since }m=m\text{)}}%
}\cdot\det\left(  \left(  \underbrace{\gamma_{i,j}}_{\substack{=\alpha
_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}\\\text{(by
\eqref{sol.cauchy.cauchy.det.17.pf.gamma.explicit}, since }i\neq m\text{ and
}j\neq m\text{)}}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\alpha_{m,m}\alpha_{m,m}\cdot\det\left(  \left(  \alpha_{i,j}\alpha
_{m,m}-\alpha_{i,m}\alpha_{m,j}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\alpha_{m,m}^{2}\cdot\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}%
-\alpha_{i,m}\alpha_{m,j}\right)  _{i,j=1,2,...,m-1}\right)  .
\end{align*}
Compared with \eqref{sol.cauchy.cauchy.det.17.pf.gamma-det.1}, this yields%
\[
\alpha_{m,m}^{2}\cdot\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}-\alpha
_{i,m}\alpha_{m,j}\right)  _{i,j=1,2,...,m-1}\right)  =\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  \cdot\alpha_{m,m}^{m}.
\]
We can divide both sides of this equality by $\alpha_{m,m}^{2}$ (since
$\alpha_{m,m}$ is invertible in $A$), and thus obtain%
\[
\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}\right)
_{i,j=1,2,...,m-1}\right)  =\alpha_{m,m}^{m-2}\cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\]
This proves \eqref{sol.cauchy.cauchy.det.17}.

[\textit{Remark:} The equality \eqref{sol.cauchy.cauchy.det.17} holds even
without requiring that $\alpha_{m,m}$ be invertible, if we have $m\geq2$ (of
course, if $m$ is not $\geq2$, then the $\alpha_{m,m}^{m-2}$ on the right hand
side of \eqref{sol.cauchy.cauchy.det.17} does not make sense unless
$\alpha_{m,m}$ is invertible). There are several ways to see why this is so.
One of these ways proceeds as follows: First of all, one should notice that
our above proof of \eqref{sol.cauchy.cauchy.det.17} works without requiring
that $\alpha_{m,m}$ be invertible, as long as $\alpha_{m,m}$ is a
non-zero-divisor\footnote{A \emph{non-zero-divisor} in a commutative ring
$B$ means an element $b\in B$ such that every element $c\in B$ satisfying
$bc=0$ must satisfy $c=0$.} in $A$ and we have $m\geq2$. However, for any
fixed $m\geq2$, the equality \eqref{sol.cauchy.cauchy.det.17} is a polynomial
identity in the elements $\alpha_{i,j}$ of $A$; thus, it suffices to prove it
when $\alpha_{i,j}$ are distinct indeterminates $X_{i,j}$ in the polynomial
ring $\ZZ \left[  X_{i,j}\ \mid\ \left(  i,j\right)  \in\left\{
1,2,...,m\right\}  ^{2}\right]  $. But in this case, $\alpha_{m,m}$ is clearly
a non-zero-divisor, and so our above proof applies. (An alternative approach
would be to replace $\alpha_{m,m}$ by $X+\alpha_{m,m}$ in the polynomial ring
$A\left[  X\right]  $; again, $X+\alpha_{m,m}$ is a non-zero-divisor even if
$\alpha_{m,m}$ is not.)]
\end{itemize}

Now, rather than solve parts (a) and (b) of the exercise separately, we are
going to prove a result from which both of these parts will easily follow:

\begin{proposition}
\label{prop.sol.cauchy.cauchy}
Let $A$ be a commutative ring. Let
$n \in \NN$. For every $i\in\left\{  1,2,...,n\right\}  $, let $a_{i}$,
$b_{i}$, $c_{i}$ and $d_{i}$ be four elements of $A$. Assume that $a_{i}%
d_{j}-b_{i}c_{j}$ is an invertible element of $A$ for every $i\in\left\{
1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $. Then,%
\[
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots
,n}\right)  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }%
\]

\end{proposition}

\textit{Proof of Proposition \ref{prop.sol.cauchy.cauchy}.} We prove this by
induction over $n$. The base case ($n=0$) is obvious, as it claims an equality
between the determinant of a $0\times0$-matrix (defined to be $1$) and the
ratio of two empty products (thus $\dfrac{1}{1}=1$). For the induction step,
we fix some positive integer $m$, and we set out to prove the equality%
\begin{equation}
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m}%
\right)  =\dfrac{\prod_{1\leq j<i\leq m}\left(  \left(  a_{i}b_{j}-a_{j}%
b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }{\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(  a_{i}d_{j}%
-b_{i}c_{j}\right)  }, \label{sol.cauchy.cauchy.a.indgoal}%
\end{equation}
assuming that we already know
\begin{equation}
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots
,m-1}\right)  =\dfrac{\prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  } \label{sol.cauchy.cauchy.a.indhyp}%
\end{equation}
to be true.

Now, we know that $\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}$ is an invertible element
of $A$ (because it is the inverse of $a_{m}d_{m}-b_{m}c_{m}$). Thus,
\eqref{sol.cauchy.cauchy.det.17} (applied to $\alpha_{i,j}=\dfrac{1}{a_{i}%
d_{j}-b_{i}c_{j}}$) yields
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}%
{a_{m}d_{m}-b_{m}c_{m}}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}%
{a_{m}d_{j}-b_{m}c_{j}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}\cdot\det\left(
\left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m}\right)  .
\end{align*}
Solving this for $\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)
_{i,j=1,2,...,m}\right)  $, we obtain%
\begin{align}
&  \det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)
_{i,j=1,2,...,m}\right) \nonumber\\
&  =\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\det\left(  \left(
\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}%
-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)
_{i,j=1,2,...,m-1}\right)  . \label{sol.cauchy.cauchy.a.1}%
\end{align}


But straightforward computations show that every $\left(  i,j\right)
\in\left\{  1,2,...,m\right\}  ^{2}$ satisfy%
\begin{align*}
&  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}%
}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\\
&  =\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\left(  a_{m}d_{j}-b_{m}c_{j}\right)  }\cdot\dfrac{a_{m}b_{i}-a_{i}b_{m}%
}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}.
\end{align*}
Hence, the matrix%
\[
\left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}%
}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}%
}\right)  _{i,j=1,2,...,m-1}%
\]
can be rewritten as%
\[
\left(  \dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\left(  a_{m}d_{j}-b_{m}c_{j}\right)  }\cdot\dfrac{a_{m}b_{i}-a_{i}b_{m}%
}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)
_{i,j=1,2,...,m-1}.
\]
This means that this matrix can be obtained from the matrix $\left(  \dfrac
{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m-1}$ by multiplying every
row with $\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}$, where $i$ is
the index of this row, and then multiplying every column with $\dfrac
{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)  \left(
a_{m}d_{j}-b_{m}c_{j}\right)  }$, where $j$ is the index of the column. As a
consequence, the determinant of this matrix is%
\[
\left(  \prod_{j=1}^{m-1}
\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\left(  a_{m}d_{j}-b_{m}c_{j}\right)  }  \right)
\cdot\left(  \prod_{i=1}^{m-1}\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}%
-b_{i}c_{m}}\right)  \cdot\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}%
}\right)  _{i,j=1,2,...,m-1}\right)
\]
(because when a row of a matrix is multiplied by a scalar, the determinant of
the matrix gets multiplied by the same scalar, and the same rule holds for
columns). Hence, we have shown that%
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}%
{a_{m}d_{m}-b_{m}c_{m}}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}%
{a_{m}d_{j}-b_{m}c_{j}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\left(  \prod_{j=1}^{m-1}
\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\left(  a_{m}d_{j}-b_{m}c_{j}\right)  } \right)
\cdot\left(  \prod_{i=1}^{m-1}\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}%
-b_{i}c_{m}}\right)  \cdot\underbrace{\det\left(  \left(  \dfrac{1}{a_{i}%
d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m-1}\right)  }_{\substack{=\dfrac
{\prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}b_{j}-a_{j}b_{i}\right)
\left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}\\\text{(by \eqref{sol.cauchy.cauchy.a.indhyp})}}}\\
&  =\left(  \prod_{j=1}^{m-1}
\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\left(  a_{m}d_{j}-b_{m}c_{j}\right)  }  \right)
\cdot\left(  \prod_{i=1}^{m-1}\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}%
-b_{i}c_{m}}\right)  \cdot\dfrac{\prod_{1\leq j<i\leq m-1}\left(  \left(
a_{i}b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)
}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }.
\end{align*}
We can now plug this into (\ref{sol.cauchy.cauchy.a.1}) and make
straightforward simplifications (splitting apart and pulling together
products), and obtain (\ref{sol.cauchy.cauchy.a.indgoal}). Thus, the induction
step is complete, and Proposition \ref{prop.sol.cauchy.cauchy} is proven.

Now, let us solve the exercise.

(a) \textit{First solution of Exercise~\ref{exe.cauchy.cauchy}(a):}
Exercise~\ref{exe.cauchy.cauchy}(a) follows from Proposition
\ref{prop.sol.cauchy.cauchy} (applied to $a_{i}$, $1$, $b_{i}$ and $1$ instead
of $a_{i}$, $b_{i}$, $c_{i}$ and $d_{i}$).

\textit{Second solution of Exercise~\ref{exe.cauchy.cauchy}(a) (sketched):}
The statement of Exercise~\ref{exe.cauchy.cauchy}(a) is an identity between
two rational functions in the variables $a_{1},a_{2},...,a_{n},b_{1}%
,b_{2},...,b_{n}$, and is easily seen to be equivalent to a polynomial
identity in these variables (by multiplying both sides through with the common
denominator $\prod_{\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}%
}\left(  a_{i}-b_{j}\right)  $). It is well-known that such identities need
only be checked on complex numbers to ensure that they hold for any elements
of any ring. So we only need to prove the statement of Exercise
\ref{exe.cauchy.cauchy}(a) in the case when $A=\CC$. But the statement
is well-known in this case (see, e.g., \cite[Lemma 5.15.3 (Lemma 4.48 in the
arXiv version)]{Etingof-et-al} and various other sources for the proof).

(b) \textit{First solution of Exercise~\ref{exe.cauchy.cauchy}(b):}
Exercise~\ref{exe.cauchy.cauchy}(b) follows from Proposition
\ref{prop.sol.cauchy.cauchy} (applied to $1$, $a_{i}$, $b_{i}$ and $1$ instead
of $a_{i}$, $b_{i}$, $c_{i}$ and $d_{i}$).

\textit{Second solution of Exercise~\ref{exe.cauchy.cauchy}(b):} Just as in
the Second solution of Exercise~\ref{exe.cauchy.cauchy}(a), we can see that it
is enough to solve Exercise~\ref{exe.cauchy.cauchy}(b) in the case when
$A=\CC$. But in this case, the statement of this exercise is
well-known, and proven, e.g., in \cite[Corollary 5.15.4 (Corollary 4.49 in
the arXiv version)]{Etingof-et-al} and
\cite[Cauchy's lemma, p. 18]{CrawleyBoevey}.

\textit{Remark:} One could also derive the statement of Exercise
\ref{exe.cauchy.cauchy}(b) from Exercise~\ref{exe.cauchy.cauchy}(a) by
applying the latter to $1/a_{i}$ instead of $a_{i}$, after first WLOG assuming
that the $a_{i}$ are invertible (but one needs to justify this WLOG assumption).

(c) \textit{Alternative proof of Theorem \ref{thm.cauchy-identity}.} We are
going to show that, for every $n\in{ \NN }$, we have%
\begin{equation}
\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}=\sum
\limits_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  x_{1},x_{2}%
,...,x_{n}\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)
\label{sol.cauchy.cauchy.c.2}%
\end{equation}
in the ring $\kk \left[  \left[  x_{1},x_{2},...,x_{n},y_{1}%
,y_{2},...,y_{n}\right]  \right]  $. Once this is proven, Theorem
\ref{thm.cauchy-identity} will easily follow.\footnote{\textit{Proof.} Assume
that (\ref{sol.cauchy.cauchy.c.2}) is proven for all $n\in \NN$. We need
to show that
\[
\prod\limits_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}=\sum
\limits_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  \xx\right)
s_{\lambda}\left(  \mathbf{y}\right)
\]
in the ring $\kk \left[  \left[  {\mathbf{x}},{\mathbf{y}}\right]
\right]  = \kk \left[  \left[  x_{1},x_{2},x_{3},...,y_{1},y_{2}%
,y_{3},...\right]  \right]  $. In order to do so, it is clearly enough to
prove that for any two weak compositions $\alpha$ and $\beta$, the coefficient
of $\mathbf{x}^{\alpha}\mathbf{y}^{\beta}$ in $\prod\limits_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}$ equals the coefficient of $\xx%
^{\alpha}\mathbf{y}^{\beta}$ in $\sum\limits_{\lambda\in\operatorname{Par}%
}s_{\lambda}\left(  \xx\right)  s_{\lambda}\left(  \mathbf{y}\right)
$. So fix two weak compositions $\alpha$ and $\beta$. Write $\alpha$ and
$\beta$ as $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)  $ and
$\beta=\left(  \beta_{1},\beta_{2},\beta_{3},...\right)  $. Choose some
$n\in \NN$ such that every integer $m>n$ satisfies $\alpha_{m}=\beta
_{m}=0$. (Such an $m$ clearly exists, since $\alpha$ and $\beta$ are finitely
supported.) Then, the coefficient of $\mathbf{x}^{\alpha}\mathbf{y}^{\beta}$
in $\prod\limits_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}$ equals
the coefficient of $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}%
}y_{1}^{\beta_{1}}y_{2}^{\beta_{2}}...y_{n}^{\beta_{n}}$ in $\prod
\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}$, whereas the
coefficient of $\mathbf{x}^{\alpha}\mathbf{y}^{\beta}$ in $\sum
\limits_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  \xx\right)
s_{\lambda}\left(  \mathbf{y}\right)  $ equals the coefficient of
$x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}}y_{1}^{\beta_{1}%
}y_{2}^{\beta_{2}}...y_{n}^{\beta_{n}}$ in $\sum\limits_{\lambda
\in\operatorname{Par}}s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  $. Since the coefficients of
$x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{n}^{\alpha_{n}}y_{1}^{\beta_{1}%
}y_{2}^{\beta_{2}}...y_{n}^{\beta_{n}}$ in $\prod\limits_{i,j=1}^{n}\left(
1-x_{i}y_{j}\right)  ^{-1}$ and in $\sum\limits_{\lambda\in\operatorname{Par}%
}s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  s_{\lambda}\left(
y_{1},y_{2},...,y_{n}\right)  $ are equal (because of
(\ref{sol.cauchy.cauchy.c.2})), this shows that the coefficients of
$\mathbf{x}^{\alpha}\mathbf{y}^{\beta}$ in $\prod\limits_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}$ and in $\sum\limits_{\lambda
\in\operatorname{Par}}s_{\lambda}\left(  \xx\right)  s_{\lambda}\left(
\mathbf{y}\right)  $ are equal; but this is exactly what we need to prove.
Thus, Theorem \ref{thm.cauchy-identity} follows if
(\ref{sol.cauchy.cauchy.c.2}) is proven.}

So let $n\in{ \NN }$. For every $i\in\left\{  1,2,...,n\right\}  $ and
$j\in\left\{  1,2,...,n\right\}  $, the element $1-x_{i}y_{j}$ of the ring
\newline$\kk \left[  \left[  x_{1},x_{2},...,x_{n},y_{1},y_{2}%
,...,y_{n}\right]  \right]  $ is invertible. Hence, Exercise
\ref{exe.cauchy.cauchy}(b) (applied to $A= \kk \left[  \left[
x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}\right]  \right]  $, $a_{i}=x_{i}$
and $b_{j}=y_{j}$) yields%
\begin{align}
\det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
&  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  x_{i}-x_{j}\right)  \left(
y_{i}-y_{j}\right)  \right)  }{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}}\left(  1-x_{i}y_{j}\right)  }=\dfrac{\prod_{1\leq
i<j\leq n}\left(  \left(  x_{i}-x_{j}\right)  \left(  y_{i}-y_{j}\right)
\right)  }{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  1-x_{i}y_{j}\right)  }\nonumber\\
&  =\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)  \right)
\cdot\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}-y_{j}\right)
\right)  \cdot\prod\limits_{i,j=1}^{n}\dfrac{1}{1-x_{i}y_{j}}
\label{sol.cauchy.cauchy.c.cauchy}%
\end{align}
in the ring $\kk \left[  \left[  x_{1},x_{2},...,x_{n},y_{1}%
,y_{2},...,y_{n}\right]  \right]  $.

We will use the notations of Definition \ref{def.alternant} and Proposition
\ref{prop.alternants.space} (but we do not require $\kk$ to be
$\ZZ$ or a field of characteristic not equal to $2$ as was done in
Proposition~\ref{prop.alternants.space}). We have $a_{\rho} = \prod
\limits_{1\leq i<j\leq n} \left(  x_{i} - x_{j} \right)  $ (as proven in the
proof of Proposition~\ref{prop.alternants.space}). Thus, $a_{\rho}\left(
x_{1},x_{2},...,x_{n}\right)  =\prod\limits_{1\leq i<j\leq n}\left(
x_{i}-x_{j}\right)  $ and $a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
=\prod\limits_{1\leq i<j\leq n}\left(  y_{i}-y_{j}\right)  $. Thus,
\eqref{sol.cauchy.cauchy.c.cauchy} becomes
\begin{align}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
\nonumber\\
&  =\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}%
-x_{j}\right)  \right)  }_{=a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)
}\cdot\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}%
-y_{j}\right)  \right)  }_{=a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}\cdot\prod\limits_{i,j=1}^{n}\underbrace{\dfrac{1}{1-x_{i}y_{j}}}_{=\left(
1-x_{i}y_{j}\right)  ^{-1}}\nonumber\\
&  =a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho}\left(
y_{1},y_{2},...,y_{n}\right)  \cdot\prod\limits_{i,j=1}^{n}\left(
1-x_{i}y_{j}\right)  ^{-1}. \label{sol.cauchy.cauchy.c.cauchy2}%
\end{align}


On the other hand, \eqref{sol.cauchy.cauchy.det.3} (applied to $m=n$ and
$\alpha_{i,j}=\dfrac{1}{1-x_{i}y_{j}}$) yields
\begin{align}
\det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod
\limits_{i=1}^{n}\underbrace{\dfrac{1}{1-x_{i}y_{\sigma\left(  i\right)  }}%
}_{\substack{=\sum\limits_{k\in{ \NN }}\left(  x_{i}y_{\sigma\left(
i\right)  }\right)  ^{k}\\\text{(by the formula for the geometric series)}%
}}\nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
\limits_{i=1}^{n}\sum\limits_{k\in{ \NN }}\left(  x_{i}y_{\sigma\left(
i\right)  }\right)  ^{k}}_{\substack{=\sum\limits_{\left(  k_{1}%
,k_{2},...,k_{n}\right)  \in{ \NN }^{n}}\left(  x_{1}y_{\sigma\left(
1\right)  }\right)  ^{k_{1}}\left(  x_{2}y_{\sigma\left(  2\right)  }\right)
^{k_{2}}...\left(  x_{n}y_{\sigma\left(  n\right)  }\right)  ^{k_{n}%
}\\\text{(by the product rule)}}}\nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}%
}\underbrace{\left(  x_{1}y_{\sigma\left(  1\right)  }\right)  ^{k_{1}}\left(
x_{2}y_{\sigma\left(  2\right)  }\right)  ^{k_{2}}...\left(  x_{n}%
y_{\sigma\left(  n\right)  }\right)  ^{k_{n}}}_{\substack{=\left(
x_{1}^{k_{1}}y_{\sigma\left(  1\right)  }^{k_{1}}\right)  \left(  x_{2}%
^{k_{2}}y_{\sigma\left(  2\right)  }^{k_{2}}\right)  ...\left(  x_{n}^{k_{n}%
}y_{\sigma\left(  n\right)  }^{k_{n}}\right)  \\=\left(  x_{1}^{k_{1}}%
x_{2}^{k_{2}}...x_{n}^{k_{n}}\right)  \left(  y_{\sigma\left(  1\right)
}^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(  n\right)
}^{k_{n}}\right)  }}\nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}}\left(
x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\right)  \left(  y_{\sigma\left(
1\right)  }^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(
n\right)  }^{k_{n}}\right) \nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}%
}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\sum\limits_{\sigma\in S_{n}%
}\left(  -1\right)  ^{\sigma}y_{\sigma\left(  1\right)  }^{k_{1}}%
y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(  n\right)  }^{k_{n}}.
\label{sol.cauchy.cauchy.c.6}%
\end{align}
But every $\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}$
satisfies
\begin{align}
\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)   &
=\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
\limits_{i=1}^{n}y_{\sigma\left(  i\right)  }^{k_{i}}}_{=y_{\sigma\left(
1\right)  }^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(
n\right)  }^{k_{n}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.cauchy.det.3}, applied to }m=n\text{ and }\alpha_{i,j}%
=y_{j}^{k_{i}}\right) \nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}y_{\sigma\left(
1\right)  }^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(
n\right)  }^{k_{n}}. \label{sol.cauchy.cauchy.c.ydet}%
\end{align}
Hence, \eqref{sol.cauchy.cauchy.c.6} becomes
\begin{align}
\det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}%
}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\underbrace{\sum\limits_{\sigma\in
S_{n}}\left(  -1\right)  ^{\sigma}y_{\sigma\left(  1\right)  }^{k_{1}%
}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(  n\right)  }^{k_{n}}%
}_{\substack{=\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  \\\text{(by \eqref{sol.cauchy.cauchy.c.ydet})}}}\nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}%
}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{,
}...\text{, }k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}%
}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  . \label{sol.cauchy.cauchy.c.8}%
\end{align}
(Here, we have removed from our sum all addends in which the integers $k_{1}$,
$k_{2}$, $...$, $k_{n}$ are not distinct. This did not change the
value of the sum, because all these addends are zero\footnote{In fact, if
$\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}$ is such that the
integers $k_{1}$, $k_{2}$, $...$, $k_{n}$ are \textbf{not} distinct,
then the matrix $\left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}$ has two equal
rows, which causes its determinant $\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  $ to be $0$, and thus the addend corresponding to
this $\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}$ is%
\[
x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\underbrace{\det\left(  \left(
y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  }_{=0}=0.
\]
}.)

On the other hand, every $\left(  k_{1},k_{2},...,k_{n}\right)  \in
{ \NN }^{n}$ satisfies
\begin{align*}
\det\left(  \left(  x_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)   &
=\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
\limits_{i=1}^{n}x_{\sigma\left(  i\right)  }^{k_{i}}}_{=x_{\sigma\left(
1\right)  }^{k_{1}}x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(
n\right)  }^{k_{n}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.cauchy.det.3}, applied to }m=n\text{ and }\alpha_{i,j}%
=x_{j}^{k_{i}}\right)  \\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}x_{\sigma\left(
1\right)  }^{k_{1}}x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(
n\right)  }^{k_{n}}.
\end{align*}
Thus,
\begin{align*}
&  \sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
{ \NN }^{n};\\k_{1}>k_{2}>...>k_{n}}}\underbrace{\det\left(  \left(
x_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  }_{=\sum\limits_{\sigma\in
S_{n}}\left(  -1\right)  ^{\sigma}x_{\sigma\left(  1\right)  }^{k_{1}%
}x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(  n\right)  }^{k_{n}}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  \\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\k_{1}>k_{2}>...>k_{n}}}\sum\limits_{\sigma\in S_{n}%
}\left(  -1\right)  ^{\sigma}x_{\sigma\left(  1\right)  }^{k_{1}}%
x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(  n\right)  }^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  \\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\sum
\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }%
^{n};\\k_{1}>k_{2}>...>k_{n}}}x_{\sigma\left(  1\right)  }^{k_{1}}%
x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(  n\right)  }^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)
}_{\substack{=\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)
}>...>k_{\sigma\left(  n\right)  }}}x_{\sigma\left(  1\right)  }%
^{k_{\sigma\left(  1\right)  }}x_{\sigma\left(  2\right)  }^{k_{\sigma\left(
2\right)  }}...x_{\sigma\left(  n\right)  }^{k_{\sigma\left(  n\right)  }}%
\det\left(  \left(  y_{j}^{k_{\sigma\left(  i\right)  }}\right)
_{i,j=1,2,...,n}\right)  \\\text{(here, we substituted }\left(  k_{\sigma
\left(  1\right)  },k_{\sigma\left(  2\right)  },...,k_{\sigma\left(
n\right)  }\right)  \text{ for }\left(  k_{1},k_{2},...,k_{n}\right)  \text{
in the sum}\\\text{(this is allowed, since }\sigma\text{ is a permutation))}%
}}\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }%
^{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }%
>...>k_{\sigma\left(  n\right)  }}}\underbrace{x_{\sigma\left(  1\right)
}^{k_{\sigma\left(  1\right)  }}x_{\sigma\left(  2\right)  }^{k_{\sigma\left(
2\right)  }}...x_{\sigma\left(  n\right)  }^{k_{\sigma\left(  n\right)  }}%
}_{\substack{=x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\\\text{(since }%
\sigma\text{ is a permutation)}}}\underbrace{\det\left(  \left(
y_{j}^{k_{\sigma\left(  i\right)  }}\right)  _{i,j=1,2,...,n}\right)
}_{\substack{=\left(  -1\right)  ^{\sigma}\det\left(  \left(  y_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right)  \\\text{(since permuting the rows of a
matrix}\\\text{multiplies its determinant by the sign }\\\text{of the
permutation)}}}\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }%
^{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }%
>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}%
}\left(  -1\right)  ^{\sigma}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  \\
&  =\sum\limits_{\sigma\in S_{n}}\underbrace{\left(  -1\right)  ^{\sigma
}\left(  -1\right)  ^{\sigma}}_{=\left(  \left(  -1\right)  ^{\sigma}\right)
^{2}=1}\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)
}>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  \\
&  =\underbrace{\sum\limits_{\sigma\in S_{n}}\sum\limits_{\substack{\left(
k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}}%
_{=\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}}%
\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}}x_{1}%
^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  \\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in{ \NN }^{n}%
}\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  \\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{,
}...\text{, }k_{n}\\\text{are distinct}}}\underbrace{\sum
\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)  }%
>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  }_{\substack{=x_{1}^{k_{1}}x_{2}^{k_{2}}%
...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  \\\text{(because there exists exactly one }\sigma\in S_{n}\text{ such
that }k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }>...>k_{\sigma
\left(  n\right)  }\\\text{(since the integers }k_{1}\text{, }k_{2}\text{,
}...\text{, }k_{n}\text{ are distinct, and thus}\\\text{there is
exactly one permutation which arranges these integers}\\\text{in decreasing
order))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  k_{1},k_{2}%
,...,k_{n}\right)  \in{ \NN }^{n};\\\text{the integers }k_{1}\text{,
}k_{2}\text{, }...\text{, }k_{n}\\\text{are not distinct}}%
}\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\underbrace{\det\left(  \left(  y_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=0\\\text{(because the matrix
}\left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\\\text{has two equal rows
(since the integers}\\\text{ }k_{1}\text{, }k_{2}\text{, }...\text{, }%
k_{n}\text{ are not distinct,}\\\text{i. e., there are two equal
among them)) }}}
\end{align*}%
\begin{align*}
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{,
}...\text{, }k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}%
}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{\left(  k_{1}%
,k_{2},...,k_{n}\right)  \in{ \NN }^{n};\\\text{the integers }%
k_{1}\text{, }k_{2}\text{, }...\text{, }k_{n}\\\text{are not
distinct}}}\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}0}_{=0}\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{,
}...\text{, }k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}%
}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  .
\end{align*}
Compared with \eqref{sol.cauchy.cauchy.c.8}, this yields
\begin{align}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}%
\right)  \nonumber\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n};\\k_{1}>k_{2}>...>k_{n}}}\det\left(  \left(  x_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right)  \det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  \lambda_{1},\lambda
_{2},...,\lambda_{n}\right)  \in{ \NN }^{n};\\\lambda_{1}\geq\lambda
_{2}\geq...\geq\lambda_{n}}}}_{=\sum\limits_{\substack{\lambda=\left(
\lambda_{1},\lambda_{2},...,\lambda_{n}\right)  \in{ \NN }^{n}\\\text{is
a partition with at most }n\text{ parts}}}}\det\left(  \left(  x_{j}%
^{\lambda_{i}+n-i}\right)  _{i,j=1,2,...,n}\right)  \det\left(  \left(
y_{j}^{\lambda_{i}+n-i}\right)  _{i,j=1,2,...,n}\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  \lambda_{1}+n-1,\lambda_{2}%
+n-2,...,\lambda_{n}+n-n\right)  \text{ for }\left(  k_{1},k_{2}%
,...,k_{n}\right)  \\
\text{in the sum (since the map}\\
\left\{  \left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)
\in{ \NN }^{n}\ \mid\ \lambda_{1}\geq\lambda_{2}\geq...\geq\lambda
_{n}\right\}  \rightarrow\left\{  \left(  k_{1},k_{2},...,k_{n}\right)
\in{ \NN }^{n}\ \mid\ k_{1}>k_{2}>...>k_{n}\right\}  ,\\
\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  \mapsto\left(
\lambda_{1}+n-1,\lambda_{2}+n-2,...,\lambda_{n}+n-n\right)  \\
\text{is a bijection)}%
\end{array}
\right)  \nonumber\\
&  =\sum\limits_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{n}\right)  \in{ \NN }^{n}\\\text{is a partition with at
most }n\text{ parts}}}\det\left(  \left(  x_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  \det\left(  \left(  y_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  .\label{sol.cauchy.cauchy.c.10}%
\end{align}


But whenever $\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)
$ is a partition with at most $n$ parts, we have
\begin{align*}
\underbrace{\lambda}_{=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)
}+\underbrace{\rho}_{=\left(  n-1,n-2,...,0\right)  } &  =\left(  \lambda
_{1},\lambda_{2},...,\lambda_{n}\right)  +\left(  n-1,n-2,...,0\right)  \\
&  =\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,...,\lambda_{n}+n-n\right)
\end{align*}
and thus%
\begin{align*}
a_{\lambda+\rho} &  =a_{\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,...,\lambda
_{n}+n-n\right)  }=\det\left(  \underbrace{\left(  x_{i}^{\lambda_{j}%
+n-j}\right)  _{i,j=1,2,...,n}}_{=\left(  \left(  x_{j}^{\lambda_{i}%
+n-i}\right)  _{i,j=1,2,...,n}\right)  ^{T}}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }a_{\left(
\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }\right)  \\
&  =\det\left(  \left(  \left(  x_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  ^{T}\right)  =\det\left(  \left(  x_{j}^{\lambda
_{i}+n-i}\right)  _{i,j=1,2,...,n}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the determinant of a matrix is
preserved under transposition}\right)  .
\end{align*}
Hence, we get the two equalities $a_{\lambda+\rho}\left(  x_{1},x_{2}%
,...,x_{n}\right)  =\det\left(  \left(  x_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  $ and $a_{\lambda+\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  =\det\left(  \left(  y_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  $ for any partition $\lambda=\left(  \lambda
_{1},\lambda_{2},...,\lambda_{n}\right)  $ with at most $n$ parts. Thus,
\eqref{sol.cauchy.cauchy.c.10} becomes
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}%
\right)  \\
&  =\sum\limits_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{n}\right)  \in{ \NN }^{n}\\\text{is a partition with at
most }n\text{ parts}}}\underbrace{\det\left(  \left(  x_{j}^{\lambda_{i}%
+n-i}\right)  _{i,j=1,2,...,n}\right)  }_{=a_{\lambda+\rho}\left(  x_{1}%
,x_{2},...,x_{n}\right)  }\underbrace{\det\left(  \left(  y_{j}^{\lambda
_{i}+n-i}\right)  _{i,j=1,2,...,n}\right)  }_{=a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  }\\
&  =\sum\limits_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{n}\right)  \in{ \NN }^{n}\\\text{is a partition with at
most }n\text{ parts}}}a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)
a_{\lambda+\rho}\left(  y_{1},y_{2},...,y_{n}\right)  \\
&  =\sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  .
\end{align*}
Compared to \eqref{sol.cauchy.cauchy.c.cauchy}, this yields
\begin{align}
&  \sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  \nonumber\\
&  =\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}%
-x_{j}\right)  \right)  }_{=a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)
}\cdot\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}%
-y_{j}\right)  \right)  }_{=a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}\cdot\prod\limits_{i,j=1}^{n}\underbrace{\dfrac{1}{1-x_{i}y_{j}}}_{=\left(
1-x_{i}y_{j}\right)  ^{-1}}\nonumber\\
&  =a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho}\left(
y_{1},y_{2},...,y_{n}\right)  \cdot\prod\limits_{i,j=1}^{n}\left(
1-x_{i}y_{j}\right)  ^{-1}.\label{sol.cauchy.cauchy.c.13}%
\end{align}


Now, Corollary \ref{cor.stembridge.slambda} says that $s_{\lambda}\left(
x_{1},x_{2},...,x_{n}\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$ for every
partition $\lambda$ with at most $n$ parts. Thus, $s_{\lambda}\left(
x_{1},x_{2},...,x_{n}\right)  =\dfrac{a_{\lambda+\rho}\left(  x_{1}%
,x_{2},...,x_{n}\right)  }{a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  }$
and $s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  =\dfrac{a_{\lambda+\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  }$ for every partition $\lambda$ with at most $n$ parts.
Hence,%
\begin{align}
&  \sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}\underbrace{s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  }_{=\dfrac
{a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  }{a_{\rho}\left(
x_{1},x_{2},...,x_{n}\right)  }}\underbrace{s_{\lambda}\left(  y_{1}%
,y_{2},...,y_{n}\right)  }_{=\dfrac{a_{\lambda+\rho}\left(  y_{1}%
,y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  }%
}\nonumber\\
&  =\sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}\dfrac{a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  }{a_{\rho
}\left(  x_{1},x_{2},...,x_{n}\right)  }\cdot\dfrac{a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}\nonumber\\
&  =\dfrac{1}{a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }\underbrace{\sum\limits_{\lambda\text{
is a partition with at most }n\text{ parts}}a_{\lambda+\rho}\left(
x_{1},x_{2},...,x_{n}\right)  a_{\lambda+\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  }_{\substack{=a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)
\cdot a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  \cdot\prod\limits_{i,j=1}%
^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}\\\text{(by
\eqref{sol.cauchy.cauchy.c.13})}}}\nonumber\\
&  =\dfrac{1}{a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }\cdot a_{\rho}\left(  x_{1}%
,x_{2},...,x_{n}\right)  \cdot a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
\cdot\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}\nonumber\\
&  =\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\label{sol.cauchy.cauchy.c.16}%
\end{align}


We are almost done. Let us now notice that every partition $\lambda$ with more
than $n$ parts satisfies
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
=0\label{sol.cauchy.cauchy.c.overflow}%
\end{equation}
(according to Exercise~\ref{exe.schur.finite}(b)). Thus, in the sum
$\sum\limits_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  x_{1}%
,x_{2},...,x_{n}\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  $,
all addends corresponding to partitions $\lambda$ having more than $n$ parts
are $0$. We can therefore remove these addends from the sum. Hence,%
\begin{align*}
&  \sum\limits_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  x_{1}%
,x_{2},...,x_{n}\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  \\
&  =\sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  s_{\lambda}\left(
y_{1},y_{2},...,y_{n}\right)  \\
&  =\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.cauchy.c.16}}\right)  .
\end{align*}
Thus, we have proven that \eqref{sol.cauchy.cauchy.c.2} holds for every
$n\in{ \NN }$. As we explained above, this yields that Theorem
\ref{thm.cauchy-identity} is true.

\textit{Remark:} In our above solution, we solved Exercise
\ref{exe.cauchy.cauchy}(b) first, and then used it to prove Theorem
\ref{thm.cauchy-identity}. It is also possible (more or less by treading the
above proof backwards) to conversely derive the statement of Exercise
\ref{exe.cauchy.cauchy}(b) from Theorem \ref{thm.cauchy-identity} instead
(though Exercise~\ref{exe.cauchy.cauchy}(b) is usually considered a more
elementary fact than Theorem \ref{thm.cauchy-identity}).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cauchy.cauchy}}

\textit{Solution to Exercise~\ref{exe.cauchy.cauchy}.} Before we start
solving any specific part of this exercise, let us state some general
properties of determinants (and prove some of them):

\begin{itemize}
\item The determinant of the transpose of any square matrix $M$ equals the
determinant of $M$. In other words,%
\begin{equation}
\det\left(  M^{T}\right)  =\det M\ \ \ \ \ \ \ \ \ \ \text{for any }%
m\in\NN \text{ and any }M\in A^{m\times m}.
\label{sol.cauchy.cauchy.det.0}%
\end{equation}
Hence, every $m\in\NN$ and every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ satisfy
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\det\left(
\left(  \alpha_{j,i}\right)  _{i,j=1,2,...,m}\right)  .
\label{sol.cauchy.cauchy.det.1}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.1}:} Let
$m\in\NN$. Let $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in
A^{m\times m}$ be a matrix. Then, \eqref{sol.cauchy.cauchy.det.0} (applied to
$M=\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}$) yields $\det\left(  \left(
\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  ^{T}\right)
=\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  $. But the
definition of the transpose matrix $\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  ^{T}$ yields $\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  ^{T}=\left(  \alpha_{j,i}\right)  _{i,j=1,2,...,m}$.
Hence, $\det\left(  \underbrace{\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  ^{T}}_{=\left(  \alpha_{j,i}\right)  _{i,j=1,2,...,m}%
}\right)  =\det\left(  \left(  \alpha_{j,i}\right)  _{i,j=1,2,...,m}\right)
$. Compared with $\det\left(  \left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  ^{T}\right)  =\det\left(  \left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\right)  $, this yields $\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\det\left(  \left(
\alpha_{j,i}\right)  _{i,j=1,2,...,m}\right)  $. This proves
\eqref{sol.cauchy.cauchy.det.1}.}

\item Every $m\in\NN$ and every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ satisfy%
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\sum_{\sigma\in\Symm_{m}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{m}\alpha_{i,\sigma\left(  i\right)  }. \label{sol.cauchy.cauchy.det.3}%
\end{equation}
(This is simply the explicit formula for the determinant of a matrix as a sum
over permutations.)

\item Every $m\in\NN$, every matrix $M\in A^{m\times m}$ and every
$i\in\left\{  1,2,...,m\right\}  $ satisfy
\begin{equation}
\det M=\sum_{k=1}^{m}\left(  \text{the }\left(  i,k\right)  \text{-th entry of
the matrix }M\right)  \cdot\left(  \text{the }\left(  i,k\right)  \text{-th
cofactor of the matrix }M\right)  . \label{sol.cauchy.cauchy.det.5}%
\end{equation}
(This is the well-known Laplace expansion for the determinant of $M$ in its
$i$-th row.) Let us recall how a cofactor of a matrix is defined: If $m$ is a
nonnegative integer, if $M\in A^{m\times m}$ is a matrix, and if $i$ and $j$
are two elements of $\left\{  1,2,...,m\right\}  $, then the $\left(
i,j\right)  $-th cofactor of the matrix $M$ is defined as $\left(  -1\right)
^{i+j}$ times the determinant of the matrix obtained by removing the $i$-th
row and the $j$-th column from the matrix $M$. Writing this down as an
equality, we obtain the following: If $m$ is a nonnegative integer, if $M\in
A^{m\times m}$ is a matrix, and if $i$ and $j$ are two elements of $\left\{
1,2,...,m\right\}  $, then%
\begin{align}
&  \left(  \text{the }\left(  i,j\right)  \text{-th cofactor of the matrix
}M\right) \nonumber\\
&  =\left(  -1\right)  ^{i+j}\cdot\det\left(  \text{the matrix obtained by
removing the }i\text{-th row and the }j\text{-th column from the matrix
}M\right)  . \label{sol.cauchy.cauchy.det.cofactor}%
\end{align}


Now, it is clear that every positive integer $m$ and every $\left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\in A^{m\times m}$ satisfy%
\begin{equation}
\left(  \text{the }\left(  m,m\right)  \text{-th cofactor of the matrix
}\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m-1}\right)  .
\label{sol.cauchy.cauchy.det.6}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.6}:} Let $m$ be a
positive integer. Let $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in
A^{m\times m}$. The matrix obtained by removing the $m$-th row and the $m$-th
column from the matrix $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}$ is
obviously $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m-1}$. In other words,%
\begin{align*}
& \left(  \text{the matrix obtained by removing the }m\text{-th row and the
}m\text{-th column from the matrix }\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right) \\
&= \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m-1}.
\end{align*}
Now, \eqref{sol.cauchy.cauchy.det.cofactor} (applied to $i=m$, $j=m$ and
$M=\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}$) yields
\begin{align*}
&  \left(  \text{the }\left(  m,m\right)  \text{-th cofactor of the matrix
}\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right) \\
&  =\left(  -1\right)  ^{m+m}\cdot\det\underbrace{\left(  \text{the matrix
obtained by removing the }m\text{-th row and the }m\text{-th column from the
matrix }\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  }_{=\left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m-1}}\\
&  =\underbrace{\left(  -1\right)  ^{m+m}}_{\substack{=1\\\text{(since
}m+m=2m\text{ is even)}}}\cdot\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m-1}\right)  =\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m-1}\right)  .
\end{align*}
This proves \eqref{sol.cauchy.cauchy.det.6}.}

\item If a positive integer $m$ and a matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ are such that every $j\in\left\{
1,2,...,m-1\right\}  $ satisfies $\alpha_{m,j}=0$, then%
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\alpha_{m,m}\cdot\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m-1}%
\right)  . \label{sol.cauchy.cauchy.det.7}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.7}:} Let a positive
integer $m$ and a matrix $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in
A^{m\times m}$ be such that
\begin{equation}
\text{every }j\in\left\{  1,2,...,m-1\right\}  \text{ satisfies }\alpha
_{m,j}=0. \label{sol.cauchy.cauchy.det.7.pf.1}%
\end{equation}
Applying \eqref{sol.cauchy.cauchy.det.5} to $M=\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}$ and $i=m$, we obtain
\begin{align*}
&  \det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right) \\
&  =\sum_{k=1}^{m}\underbrace{\left(  \text{the }\left(  m,k\right)  \text{-th
entry of the matrix }\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
}_{=\alpha_{m,k}}\cdot\left(  \text{the }\left(  m,k\right)  \text{-th
cofactor of the matrix }\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
\\
&  =\sum_{k=1}^{m}\alpha_{m,k}\cdot\left(  \text{the }\left(  m,k\right)
\text{-th cofactor of the matrix }\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right) \\
&  =\sum_{k=1}^{m-1}\underbrace{\alpha_{m,k}}_{\substack{=0\\\text{(by
\eqref{sol.cauchy.cauchy.det.7.pf.1}, applied to }j=k\text{)}}}\cdot\left(
\text{the }\left(  m,k\right)  \text{-th cofactor of the matrix }\left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\alpha_{m,m}\cdot\underbrace{\left(  \text{the
}\left(  m,m\right)  \text{-th cofactor of the matrix }\left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\right)  }_{\substack{=\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m-1}\right)  \\\text{(by
\eqref{sol.cauchy.cauchy.det.6})}}}\\
&  =\underbrace{\sum_{k=1}^{m-1}0\cdot\left(  \text{the }\left(  m,k\right)
\text{-th cofactor of the matrix }\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  }_{=0}+\alpha_{m,m}\cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\alpha_{m,m}\cdot\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m-1}\right)  .
\end{align*}
This proves \eqref{sol.cauchy.cauchy.det.7}.}

\item Every $m\in\NN$ and any two matrices $M\in A^{m\times m}$ and
$N\in A^{m\times m}$ satisfy
\begin{equation}
\det\left(  MN\right)  =\det M\cdot\det N.
\label{sol.cauchy.cauchy.det.9}
\end{equation}
(This is simply the well-known fact that the determinant of the product of two
square matrices equals the product of their determinants.)

\item If an $m\in\NN$ and a matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ are such that
\begin{equation}
\text{every }\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}\text{
satisfying }i<j\text{ satisfy }\alpha_{i,j}=0,
\label{sol.cauchy.cauchy.det.11.cond}
\end{equation}
then
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\prod
_{i=1}^{m}\alpha_{i,i} .
\label{sol.cauchy.cauchy.det.11}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.11}:} Of course,
\eqref{sol.cauchy.cauchy.det.11} is just a restatement of the well-known fact
that the determinant of a lower-triangular square matrix equals the product of
its diagonal entries (in fact, the condition
\eqref{sol.cauchy.cauchy.det.11.cond} means precisely that the matrix $\left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}$ is lower-triangular). Nevertheless, let
us give a proof of \eqref{sol.cauchy.cauchy.det.11} here.
\par
Let us first prove a simple property of permutations: If a permutation
$\sigma\in\Symm_{m}$ satisfies $\sigma\neq\id$, then
\begin{equation}
\text{there exists an }i\in\left\{  1,2,...,m\right\}  \text{ such that
}i<\sigma\left(  i\right)  . \label{sol.cauchy.cauchy.det.11.pf.perm}%
\end{equation}
\par
\textit{Proof of \eqref{sol.cauchy.cauchy.det.11.pf.perm}:} Let $\sigma
\in\Symm_{m}$ be a permutation satisfying $\sigma\neq\id$.
We need to prove \eqref{sol.cauchy.cauchy.det.11.pf.perm}.
\par
If every $k\in\left\{  1,2,...,m\right\}  $ would satisfy $\sigma\left(
k\right)  =\id \left(  k\right)  $, then we would have
$\sigma=\id$, which would contradict $\sigma\neq
\id$. Hence, not every $k\in\left\{  1,2,...,m\right\}  $
satisfies $\sigma\left(  k\right)  =\id \left(  k\right)  $. In
other words, there exists at least one $k\in\left\{  1,2,...,m\right\}  $
which does not satisfy $\sigma\left(  k\right)  =\id \left(
k\right)  $. Consider this $k$. We know that $k$ does not satisfy
$\sigma\left(  k\right)  =\id \left(  k\right)  $. In other
words, $\sigma\left(  k\right)  \neq\id \left(  k\right)  $.
Thus, $\sigma\left(  k\right)  \neq\id \left(  k\right)  =k$.
Thus, $k$ is an element of $\left\{  1,2,...,m\right\}  $ satisfying
$\sigma\left(  k\right)  \neq k$. In other words, $k$ is an element $i$ of
$\left\{  1,2,...,m\right\}  $ satisfying $\sigma\left(  i\right)  \neq i$. In
other words, $k\in\left\{  i\in\left\{  1,2,...,m\right\}  \ \mid
\ \sigma\left(  i\right)  \neq i\right\}  $.
\par
Now, let $Z$ be the set $\left\{  i\in\left\{  1,2,...,m\right\}
\ \mid\ \sigma\left(  i\right)  \neq i\right\}  $. This set $Z$ is a finite
set of integers (because it is a subset of $\left\{  1,2,...,m\right\}  $).
Also, $k\in\left\{  i\in\left\{  1,2,...,m\right\}  \ \mid\ \sigma\left(
i\right)  \neq i\right\}  =Z$ (because $Z=\left\{  i\in\left\{
1,2,...,m\right\}  \ \mid\ \sigma\left(  i\right)  \neq i\right\}  $), so that
the set $Z$ contains at least one element (namely, $k$); thus, $Z$ is
nonempty. Hence, $Z$ is a nonempty finite set of integers. Thus, the set $Z$
has a minimum (since every nonempty finite set of integers has a minimum).
Denote this minimum by $z$.
\par
The element $z$ is the minimum of $Z$, and therefore must belong to $Z$. We
thus have $z\in Z=\left\{  i\in\left\{  1,2,...,m\right\}  \ \mid
\ \sigma\left(  i\right)  \neq i\right\}  $. In other words, $z$ is an element
$i$ of $\left\{  1,2,...,m\right\}  $ satisfying $\sigma\left(  i\right)  \neq
i$. In other words, $z$ is an element of $\left\{  1,2,...,m\right\}  $
satisfying $\sigma\left(  z\right)  \neq z$. Also, $z$ is the minimum of $Z$,
and thus $\leq$ to every element of $Z$. In other words, $z\leq w$ for every
$w\in Z$.
\par
Now, assume (for the sake of contradiction) that $\sigma\left(  z\right)  \leq
z$. Combining this with $\sigma\left(  z\right)  \neq z$, we obtain
$\sigma\left(  z\right)  <z$. But $\sigma$ is a permutation (since $\sigma
\in\Symm_{m}$), hence a bijective map, hence an injective map. If we
had $\sigma\left(  \sigma\left(  z\right)  \right)  =\sigma\left(  z\right)
$, then we would have $\sigma\left(  z\right)  =z$ (because $\sigma$ is
injective), which would contradict $\sigma\left(  z\right)  \neq z$. Hence, we
cannot have $\sigma\left(  \sigma\left(  z\right)  \right)  =\sigma\left(
z\right)  $. In other words, we must have $\sigma\left(  \sigma\left(
z\right)  \right)  \neq\sigma\left(  z\right)  $.
\par
So we know that $\sigma\left(  z\right)  $ is an element of $\left\{
1,2,...,m\right\}  $ satisfying $\sigma\left(  \sigma\left(  z\right)
\right)  \neq\sigma\left(  z\right)  $. In other words, $\sigma\left(
z\right)  $ is an element $i$ of $\left\{  1,2,...,m\right\}  $ satisfying
$\sigma\left(  i\right)  \neq i$. In other words, $\sigma\left(  z\right)
\in\left\{  i\in\left\{  1,2,...,m\right\}  \ \mid\ \sigma\left(  i\right)
\neq i\right\}  =Z$. Now, recall that $z\leq w$ for every $w\in Z$. Applying
this to $w=\sigma\left(  z\right)  $, we obtain $z\leq\sigma\left(  z\right)
$ (because $\sigma\left(  z\right)  \in Z$). This contradicts $\sigma\left(
z\right)  <z$. This contradiction shows that our assumption (that
$\sigma\left(  z\right)  \leq z$) was wrong. Hence, we cannot have
$\sigma\left(  z\right)  \leq z$. Thus, we must have $z<\sigma\left(
z\right)  $. Thus, there exists an $i\in\left\{  1,2,...,m\right\}  $ such
that $i<\sigma\left(  i\right)  $ (namely, $i=z$). This proves
\eqref{sol.cauchy.cauchy.det.11.pf.perm}.
\par
We can now easily see that
\begin{equation}
\prod_{i=1}^{m}\alpha_{i,\sigma\left(  i\right)  }
=0\ \ \ \ \ \ \ \ \ \ \text{for every }\sigma\in\Symm_{m}\text{
satisfying }\sigma\neq\id .
\label{sol.cauchy.cauchy.det.11.pf.0}
\end{equation}
\par
\textit{Proof of \eqref{sol.cauchy.cauchy.det.11.pf.0}:} Let $\sigma
\in\Symm_{m}$ be such that $\sigma\neq\id$. Thus,
$\sigma$ is a permutation satisfying $\sigma\neq\id$. Hence,
there exists an $i\in\left\{  1,2,...,m\right\}  $ such that $i<\sigma\left(
i\right)  $ (according to \eqref{sol.cauchy.cauchy.det.11.pf.perm}). Let $k$
be such an $i$. Thus, $k\in\left\{  1,2,...,m\right\}  $ and $k<\sigma\left(
k\right)  $. Applying \eqref{sol.cauchy.cauchy.det.11.cond} to $i=k$ and
$j=\sigma\left(  k\right)  $, we obtain $\alpha_{k,\sigma\left(  k\right)
}=0$ (since $k<\sigma\left(  k\right)  $). But
\begin{align*}
\underbrace{\prod_{i=1}^{m}}_{=\prod_{i\in\left\{  1,2,...,m\right\}  }}%
\alpha_{i,\sigma\left(  i\right)  }  &  =\prod_{i\in\left\{
1,2,...,m\right\}  }\alpha_{i,\sigma\left(  i\right)  }=\underbrace{\alpha
_{k,\sigma\left(  k\right)  }}_{=0}\cdot\prod\limits_{\substack{i\in\left\{
1,2,...,m\right\}  ;\\i\neq k}}\alpha_{i,\sigma\left(  i\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\in\left\{  1,2,...,m\right\}
\right) \\
&  =0\cdot\prod\limits_{\substack{i\in\left\{  1,2,...,m\right\}  ;\\i\neq k}%
}\alpha_{i,\sigma\left(  i\right)  }=0,
\end{align*}
so that \eqref{sol.cauchy.cauchy.det.11.pf.0} is proven.
\par
Now, \eqref{sol.cauchy.cauchy.det.3} yields
\begin{align*}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)   &
=\sum_{\sigma\in\Symm_{m}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{m}\alpha_{i,\sigma\left(  i\right)  }=\underbrace{\sum\limits_{\substack{\sigma
\in\Symm_{m};\\\sigma=\id }}\left(  -1\right)  ^{\sigma
}\prod_{i=1}^{m}\alpha_{i,\sigma\left(  i\right)  }}_{\substack{=\left(
-1\right)  ^{\id }\prod_{i=1}^{m}\alpha_{i,\id %
\left(  i\right)  }\\\text{(since }\id \in\Symm_m
\text{)}}}+\sum\limits_{\substack{\sigma\in\Symm_{m};\\\sigma
\neq\id }}\left(  -1\right)  ^{\sigma}\underbrace{\prod
_{i=1}^{m}\alpha_{i,\sigma\left(  i\right)  }}_{\substack{=0\\\text{(by
\eqref{sol.cauchy.cauchy.det.11.pf.0})}}}\\
&  =\left(  -1\right)  ^{\id }\prod_{i=1}^{m}\alpha
_{i,\id \left(  i\right)  }+\underbrace{\sum\limits_{\substack{\sigma
\in\Symm_{m};\\\sigma\neq\id }}\left(  -1\right)
^{\sigma}0}_{=0}=\underbrace{\left(  -1\right)  ^{\id }}%
_{=1}\prod_{i=1}^{m}\underbrace{\alpha_{i,\id \left(  i\right)
}}_{\substack{=\alpha_{i,i}\\\text{(since }\id \left(  i\right)
=i\text{)}}}=\prod_{i=1}^{m}\alpha_{i,i}.
\end{align*}
This proves \eqref{sol.cauchy.cauchy.det.11}.}

\item Every $m\in\NN$, every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ and every family $\left(  \gamma
_{i}\right)  _{i\in\left\{  1,2,...,m\right\}  }\in A^{\left\{
1,2,...,m\right\}  }$ of elements of $A$ satisfy
\begin{equation}
\det\left(  \left(  \gamma_{i}\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\left(  \prod_{i=1}^{m}\gamma_{i}\right)  \cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\label{sol.cauchy.cauchy.det.13}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.13}:} Let
$m\in\NN$. Let $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in
A^{m\times m}$ be a matrix. Let $\left(  \gamma_{i}\right)  _{i\in\left\{
1,2,...,m\right\}  }\in A^{\left\{  1,2,...,m\right\}  }$ be a family of
elements of $A$. Applying \eqref{sol.cauchy.cauchy.det.3} to $\gamma_{i}%
\alpha_{i,j}$ instead of $\alpha_{i,j}$, we obtain
\begin{align*}
\det\left(  \left(  \gamma_{i}\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
&  =\sum_{\sigma\in\Symm_{m}}\left(  -1\right)  ^{\sigma}
\underbrace{\prod_{i=1}^{m}\left(  \gamma_{i}\alpha_{i,\sigma\left(  i\right)
}\right)  }_{=\left(  \prod_{i=1}^{m}\gamma_{i}\right)  \cdot\left(
\prod_{i=1}^{m}\alpha_{i,\sigma\left(  i\right)  }\right)  }=\sum_{\sigma
\in\Symm_{m}}\left(  -1\right)  ^{\sigma}\left(  \prod_{i=1}^{m}
\gamma_{i}\right)  \cdot\left(  \prod_{i=1}^{m}\alpha_{i,\sigma\left(
i\right)  }\right) \\
&  =\left(  \prod_{i=1}^{m}\gamma_{i}\right)  \cdot\underbrace{\sum_{\sigma
\in\Symm_{m}}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{m}\alpha
_{i,\sigma\left(  i\right)  }}_{\substack{=\det\left(  \left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\right)  \\\text{(by
\eqref{sol.cauchy.cauchy.det.3})}}}=\left(  \prod_{i=1}^{m}\gamma_{i}\right)
\cdot\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\end{align*}
This proves \eqref{sol.cauchy.cauchy.det.13}.}

\item Every $m\in\NN$, every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ and every family $\left(  \delta
_{j}\right)  _{j\in\left\{  1,2,...,m\right\}  }\in A^{\left\{
1,2,...,m\right\}  }$ of elements of $A$ satisfy
\begin{equation}
\det\left(  \left(  \delta_{j}\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\left(  \prod_{j=1}^{m}\delta_{j}\right)  \cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\label{sol.cauchy.cauchy.det.15}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.15}:} Let
$m\in\NN$. Let $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in
A^{m\times m}$ be a matrix. Let $\left(  \delta_{j}\right)  _{j\in\left\{
1,2,...,m\right\}  }\in A^{\left\{  1,2,...,m\right\}  }$ be a family of
elements of $A$. We have%
\begin{align*}
\det\left(  \left(  \delta_{j}\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
&  =\det\left(  \left(  \delta_{i}\alpha_{j,i}\right)  _{i,j=1,2,...,m}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.cauchy.det.1},
applied to }\delta_{j}\alpha_{i,j}\text{ instead of }\alpha_{i,j}\right) \\
&  =\underbrace{\left(  \prod_{i=1}^{m}\delta_{i}\right)  }_{\substack{=\prod
_{j=1}^{m}\delta_{j}\\\text{(here, we renamed the index}\\i\text{ as }j\text{
in the product)}}}\cdot\underbrace{\det\left(  \left(  \alpha_{j,i}\right)
_{i,j=1,2,...,m}\right)  }_{\substack{=\det\left(  \left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\right)  \\\text{(by
\eqref{sol.cauchy.cauchy.det.1})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.cauchy.det.13},
applied to }\alpha_{j,i}\text{ and }\delta_{i}\text{ instead of }\alpha
_{i,j}\text{ and }\gamma_{i}\right) \\
&  =\left(  \prod_{j=1}^{m}\delta_{j}\right)  \cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\end{align*}
This proves \eqref{sol.cauchy.cauchy.det.15}.}

\item Every $m\in\NN$, every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ and every matrix $\left(  \beta
_{i,j}\right)  _{i,j=1,2,...,m}\in A^{m\times m}$ satisfy%
\begin{equation}
\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\cdot\left(  \beta_{i,j}\right)
_{i,j=1,2,...,m}=\left(  \sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}\right)
_{i,j=1,2,...,m}. \label{sol.cauchy.cauchy.det.AB}%
\end{equation}
This is plainly the definition of the product of two matrices (in the
particular case when the two matrices are square).

\item Every positive integer $m$ and every matrix $\left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\in A^{m\times m}$ satisfy
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}\right)
_{i,j=1,2,...,m-1}\right)  =\alpha_{m,m}^{m-2}\cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
\label{sol.cauchy.cauchy.det.17}
\end{equation}
if $\alpha_{m,m}$ is an invertible element of $A$.

\textit{Proof of \eqref{sol.cauchy.cauchy.det.17}:} Let $m$ be a positive
integer. Let $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\in A^{m\times m}$
be a matrix such that $\alpha_{m,m}$ is an invertible element of $A$.

In the following, whenever $\mathcal{A}$ is any assertion, we will use the
notation $\left[  \mathcal{A}\right]  $ for the truth value of $\mathcal{A}$
(that is, the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$).

For every $\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}$, define an
element $\beta_{i,j}$ of $A$ by
\begin{equation}
\beta_{i,j}=\left[  i=j\right]  \alpha_{m,m}-\left[  i=m\text{ and }j\neq
m\right]  \alpha_{m,j}.
\label{sol.cauchy.cauchy.det.17.pf.beta}
\end{equation}


Then, it is easy to see that
\begin{equation}
\text{every }\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}\text{
satisfying }i<j\text{ satisfy }\beta_{i,j}=0.
\label{sol.cauchy.cauchy.det.17.pf.beta.triang}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.17.pf.beta.triang}:}
Let $\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}$ be such that
$i<j$. Then, $i\neq j$ (since $i<j$), and thus we don't have $i=j$. Hence,
$\left[  i=j\right]  =0$. Also, $j\in\left\{  1,2,...,m\right\}  $ (since
$\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}$) and thus $j\leq m$,
so that $i<j\leq m$ and thus $i\neq m$. Hence, we don't have $i=m$. Thus, we
don't have $\left(  i=m\text{ and }j\neq m\right)  $ either. In other words,
$\left[  i=m\text{ and }j\neq m\right]  =0$. Now, $\beta_{i,j}%
=\underbrace{\left[  i=j\right]  }_{=0}\alpha_{m,m}-\underbrace{\left[
i=m\text{ and }j\neq m\right]  }_{=0}\alpha_{m,j}=\underbrace{0\alpha_{m,m}%
}_{=0}-\underbrace{0\alpha_{m,j}}_{=0}=0$. This proves
\eqref{sol.cauchy.cauchy.det.17.pf.beta.triang}.} Hence,
\eqref{sol.cauchy.cauchy.det.11} (applied to $\beta_{i,j}$ instead of
$\alpha_{i,j}$) yields
\begin{equation}
\det\left(  \left(  \beta_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\prod
_{i=1}^{m}\beta_{i,i}.
\label{sol.cauchy.cauchy.det.17.pf.beta.det1}
\end{equation}


But
\begin{equation}
\text{every }i\in\left\{  1,2,...,m\right\}  \text{ satisfies }\beta
_{i,i}=\alpha_{m,m}.
\label{sol.cauchy.cauchy.det.17.pf.beta.diag}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.17.pf.beta.diag}:} Let
$i\in\left\{  1,2,...,m\right\}  $. Then, we don't have $\left(  i=m\text{ and
}i\neq m\right)  $ (because the two statements $i=m$ and $i\neq m$ clearly
contradict each other). In other words, $\left[  i=m\text{ and }i\neq
m\right]  =0$. Now, \eqref{sol.cauchy.cauchy.det.17.pf.beta} (applied to
$j=i$) yields%
\[
\beta_{i,i}=\underbrace{\left[  i=i\right]  }_{\substack{=1\\\text{(since
}i=i\text{ is true)}}}\alpha_{m,m}-\underbrace{\left[  i=m\text{ and }i\neq
m\right]  }_{=0}\alpha_{m,i}=\underbrace{1\alpha_{m,m}}_{=\alpha_{m,m}%
}-\underbrace{0\alpha_{m,i}}_{=0}=\alpha_{m,m}.
\]
This proves \eqref{sol.cauchy.cauchy.det.17.pf.beta.diag}.} Thus,
\eqref{sol.cauchy.cauchy.det.17.pf.beta.det1} becomes
\begin{equation}
\det\left(  \left(  \beta_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\prod
_{i=1}^{m}\underbrace{\beta_{i,i}}_{\substack{=\alpha_{m,m}\\\text{(by
\eqref{sol.cauchy.cauchy.det.17.pf.beta.diag})}}}=\prod_{i=1}^{m}\alpha
_{m,m}=\alpha_{m,m}^{m}.
\label{sol.cauchy.cauchy.det.17.pf.beta.det2}
\end{equation}


Now, \eqref{sol.cauchy.cauchy.det.9} (applied to $M=\left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}$ and $N=\left(  \beta_{i,j}\right)
_{i,j=1,2,...,m}$) yields
\begin{align*}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\cdot\left(
\beta_{i,j}\right)  _{i,j=1,2,...,m}\right)   &  =\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  \cdot\underbrace{\det\left(
\left(  \beta_{i,j}\right)  _{i,j=1,2,...,m}\right)  }_{\substack{=\alpha
_{m,m}^{m}\\\text{(by \eqref{sol.cauchy.cauchy.det.17.pf.beta.det2})}}}\\
&  =\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)
\cdot\alpha_{m,m}^{m}=\alpha_{m,m}^{m}\cdot\det\left(  \left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\end{align*}


Since $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\cdot\left(  \beta
_{i,j}\right)  _{i,j=1,2,...,m}=\left(  \sum_{k=1}^{m}\alpha_{i,k}\beta
_{k,j}\right)  _{i,j=1,2,...,m}$ (by \eqref{sol.cauchy.cauchy.det.AB}), this
rewrites as
\begin{equation}
\det\left(  \left(  \sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}\right)
_{i,j=1,2,...,m}\right)  =\alpha_{m,m}^{m}\cdot\det\left(  \left(
\alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\label{sol.cauchy.cauchy.det.17.pf.prod-det1}
\end{equation}


Now, for every $\left(  i,j\right)  \in\left\{  1,2,...,m\right\}  ^{2}$,
define an element $\gamma_{i,j}$ of $A$ by
\begin{equation}
\gamma_{i,j}=\sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}.
\label{sol.cauchy.cauchy.det.17.pf.gamma}
\end{equation}
Then,
\begin{equation}
\det\left(  \left(  \underbrace{\gamma_{i,j}}_{=\sum_{k=1}^{m}\alpha
_{i,k}\beta_{k,j}}\right)  _{i,j=1,2,...,m}\right)  =\det\left(  \left(
\sum_{k=1}^{m}\alpha_{i,k}\beta_{k,j}\right)  _{i,j=1,2,...,m}\right)
=\alpha_{m,m}^{m}\cdot\det\left(  \left(  \alpha_{i,j}\right)
_{i,j=1,2,...,m}\right)  \label{sol.cauchy.cauchy.det.17.pf.gamma-det.1}%
\end{equation}
(by \eqref{sol.cauchy.cauchy.det.17.pf.prod-det1}).

But it is easy to see that
\begin{equation}
\text{every }j\in\left\{  1,2,...,m-1\right\}  \text{ satisfies }\gamma
_{m,j}=0
\label{sol.cauchy.cauchy.det.17.pf.prod-rowm}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.17.pf.prod-rowm}:} Let
$j\in\left\{  1,2,...,m-1\right\}  $. Then, $j\leq m-1<m$, so that $j\neq m$.
Now, \eqref{sol.cauchy.cauchy.det.17.pf.gamma} (applied to $i=m$) yields%
\begin{align}
&  \gamma_{m,j}=\underbrace{\sum_{k=1}^{m}}_{=\sum_{k\in\left\{
1,2,...,m\right\}  }}\alpha_{m,k}\underbrace{\beta_{k,j}}_{\substack{=\left[
k=j\right]  \alpha_{m,m}-\left[  k=m\text{ and }j\neq m\right]  \alpha
_{m,j}\\\text{(by \eqref{sol.cauchy.cauchy.det.17.pf.beta}, applied to
}i=k\text{)}}}\nonumber\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\underbrace{\alpha_{m,k}\left(
\left[  k=j\right]  \alpha_{m,m}-\left[  k=m\text{ and }j\neq m\right]
\alpha_{m,j}\right)  }_{=\left[  k=j\right]  \alpha_{m,k}\alpha_{m,m}-\left[
k=m\text{ and }j\neq m\right]  \alpha_{m,k}\alpha_{m,j}}\nonumber\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\left(  \left[  k=j\right]
\alpha_{m,k}\alpha_{m,m}-\left[  k=m\text{ and }j\neq m\right]  \alpha
_{m,k}\alpha_{m,j}\right) \nonumber\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=j\right]  \alpha
_{m,k}\alpha_{m,m}-\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=m\text{
and }j\neq m\right]  \alpha_{m,k}\alpha_{m,j}.
\label{sol.cauchy.cauchy.det.17.pf.prod-rowm.pf.1}
\end{align}
But
\begin{align*}
\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=j\right]  \alpha_{m,k}%
\alpha_{m,m}  &  =\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}
;\\k=j}}\left[  k=j\right]  \alpha_{m,k}\alpha_{m,m}}_{\substack{=\left[
j=j\right]  \alpha_{m,j}\alpha_{m,m}\\\text{(since }j\in\left\{
1,2,...,m\right\}  \text{)}}}+\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}
;\\k\neq j}}\underbrace{\left[  k=j\right]  }_{\substack{=0\\\text{(since
}k=j\text{ does not hold}\\\text{(since }k\neq j\text{))}}}\alpha_{m,k}%
\alpha_{m,m}\\
&  =\left[  j=j\right]  \alpha_{m,j}\alpha_{m,m}+\underbrace{\sum\limits
_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq j}}0\alpha_{m,k}%
\alpha_{m,m}}_{=0}=\underbrace{\left[  j=j\right]  }%
_{\substack{=1\\\text{(since }j=j\text{ holds)}}}\alpha_{m,j}\alpha
_{m,m}=\alpha_{m,j}\alpha_{m,m}
\end{align*}
and%
\begin{align*}
&  \sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=m\text{ and }j\neq
m\right]  \alpha_{m,k}\alpha_{m,j}\\
&  =\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k=m}%
}\left[  k=m\text{ and }j\neq m\right]  \alpha_{m,k}\alpha_{m,j}%
}_{\substack{=\left[  m=m\text{ and }j\neq m\right]  \alpha_{m,m}\alpha
_{m,j}\\\text{(since }m\in\left\{  1,2,...,m\right\}  \text{)}}}+\sum\limits
_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq m}}\underbrace{\left[
k=m\text{ and }j\neq m\right]  }_{\substack{=0\\\text{(since }\left(
k=m\text{ and }j\neq m\right)  \text{ does not hold}\\\text{(since }k=m\text{
does not hold}\\\text{(since }k\neq m\text{)))}}}\alpha_{m,k}\alpha_{m,j}\\
&  =\left[  m=m\text{ and }j\neq m\right]  \alpha_{m,m}\alpha_{m,j}%
+\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq
m}}0\alpha_{m,k}\alpha_{m,j}}_{=0}=\underbrace{\left[  m=m\text{ and }j\neq
m\right]  }_{\substack{=1\\\text{(since }\left(  m=m\text{ and }j\neq
m\right)  \text{ holds}\\\text{(since }m=m\text{ and }j\neq m\text{))}%
}}\underbrace{\alpha_{m,m}\alpha_{m,j}}_{=\alpha_{m,j}\alpha_{m,m}}%
=\alpha_{m,j}\alpha_{m,m}.
\end{align*}
\par
Hence, \eqref{sol.cauchy.cauchy.det.17.pf.prod-rowm.pf.1} becomes
\[
\gamma_{m,j}=\underbrace{\sum_{k\in\left\{  1,2,...,m\right\}  }\left[
k=j\right]  \alpha_{m,k}\alpha_{m,m}}_{=\alpha_{m,j}\alpha_{m,m}%
}-\underbrace{\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=m\text{ and
}j\neq m\right]  \alpha_{m,k}\alpha_{m,j}}_{=\alpha_{m,j}\alpha_{m,m}}%
=\alpha_{m,j}\alpha_{m,m}-\alpha_{m,j}\alpha_{m,m}=0.
\]
This proves \eqref{sol.cauchy.cauchy.det.17.pf.prod-rowm}.}. Hence,
\eqref{sol.cauchy.cauchy.det.7} (applied to $\gamma_{i,j}$ instead of
$\alpha_{i,j}$) yields
\begin{equation}
\det\left(  \left(  \gamma_{i,j}\right)  _{i,j=1,2,...,m}\right)
=\gamma_{m,m}\cdot\det\left(  \left(  \gamma_{i,j}\right)  _{i,j=1,2,...,m-1}%
\right)  .
\label{sol.cauchy.cauchy.det.17.pf.gamma-det.2}
\end{equation}


Now, it is easy to notice that every $\left(  i,j\right)  \in\left\{
1,2,...,m-1\right\}  ^{2}$ satisfy
\begin{equation}
\gamma_{i,j}=\alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}.
\label{sol.cauchy.cauchy.det.17.pf.gamma.ij}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.17.pf.gamma.ij}:} Let
$\left(  i,j\right)  \in\left\{  1,2,...,m-1\right\}  ^{2}$. Since $\left(
i,j\right)  \in\left\{  1,2,...,m-1\right\}  ^{2}$, we have $i\in\left\{
1,2,...,m-1\right\}  $ and $j\in\left\{  1,2,...,m-1\right\}  $.
\par
Since $j\in\left\{  1,2,...,m-1\right\}  $, we have $j\leq m-1<m$, so that
$j\neq m$. Now, \eqref{sol.cauchy.cauchy.det.17.pf.gamma} yields
\begin{align}
&  \gamma_{i,j}=\underbrace{\sum_{k=1}^{m}}_{=\sum_{k\in\left\{
1,2,...,m\right\}  }}\alpha_{i,k}\underbrace{\beta_{k,j}}_{\substack{=\left[
k=j\right]  \alpha_{m,m}-\left[  k=m\text{ and }j\neq m\right]  \alpha
_{m,j}\\\text{(by \eqref{sol.cauchy.cauchy.det.17.pf.beta}, applied to
}k\text{ instead of }i\text{)}}}\nonumber\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\underbrace{\alpha_{i,k}\left(
\left[  k=j\right]  \alpha_{m,m}-\left[  k=m\text{ and }j\neq m\right]
\alpha_{m,j}\right)  }_{=\left[  k=j\right]  \alpha_{i,k}\alpha_{m,m}-\left[
k=m\text{ and }j\neq m\right]  \alpha_{i,k}\alpha_{m,j}}\nonumber\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\left(  \left[  k=j\right]
\alpha_{i,k}\alpha_{m,m}-\left[  k=m\text{ and }j\neq m\right]  \alpha
_{i,k}\alpha_{m,j}\right) \nonumber\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=j\right]  \alpha
_{i,k}\alpha_{m,m}-\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=m\text{
and }j\neq m\right]  \alpha_{i,k}\alpha_{m,j}.
\label{sol.cauchy.cauchy.det.17.pf.gamma.ij.pf.1}
\end{align}
But
\begin{align*}
\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=j\right]  \alpha_{i,k}%
\alpha_{m,m}  &  =\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}
;\\k=j}}\left[  k=j\right]  \alpha_{i,k}\alpha_{m,m}}_{\substack{=\left[
j=j\right]  \alpha_{i,j}\alpha_{m,m}\\\text{(since }j\in\left\{
1,2,...,m\right\}  \text{)}}}+\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}
;\\k\neq j}}\underbrace{\left[  k=j\right]  }_{\substack{=0\\\text{(since
}k=j\text{ does not hold}\\\text{(since }k\neq j\text{))}}}\alpha_{i,k}%
\alpha_{m,m}\\
&  =\left[  j=j\right]  \alpha_{i,j}\alpha_{m,m}+\underbrace{\sum\limits
_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq j}}0\alpha_{i,k}%
\alpha_{m,m}}_{=0}=\underbrace{\left[  j=j\right]  }%
_{\substack{=1\\\text{(since }j=j\text{ holds)}}}\alpha_{i,j}\alpha
_{m,m}=\alpha_{i,j}\alpha_{m,m}
\end{align*}
and
\begin{align*}
&  \sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=m\text{ and }j\neq
m\right]  \alpha_{i,k}\alpha_{m,j}\\
&  =\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k=m}%
}\left[  k=m\text{ and }j\neq m\right]  \alpha_{i,k}\alpha_{m,j}%
}_{\substack{=\left[  m=m\text{ and }j\neq m\right]  \alpha_{i,m}\alpha
_{m,j}\\\text{(since }m\in\left\{  1,2,...,m\right\}  \text{)}}}+\sum\limits
_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq m}}\underbrace{\left[
k=m\text{ and }j\neq m\right]  }_{\substack{=0\\\text{(since }\left(
k=m\text{ and }j\neq m\right)  \text{ does not hold}\\\text{(since }k=m\text{
does not hold}\\\text{(since }k\neq m\text{)))}}}\alpha_{i,k}\alpha_{m,j}\\
&  =\left[  m=m\text{ and }j\neq m\right]  \alpha_{i,m}\alpha_{m,j}%
+\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq
m}}0\alpha_{i,k}\alpha_{m,j}}_{=0}=\underbrace{\left[  m=m\text{ and }j\neq
m\right]  }_{\substack{=1\\\text{(since }\left(  m=m\text{ and }j\neq
m\right)  \text{ holds}\\\text{(since }m=m\text{ and }j\neq m\text{))}}%
}\alpha_{i,m}\alpha_{m,j}=\alpha_{i,m}\alpha_{m,j}.
\end{align*}
\par
Hence, \eqref{sol.cauchy.cauchy.det.17.pf.gamma.ij.pf.1} becomes
\[
\gamma_{i,j}=\underbrace{\sum_{k\in\left\{  1,2,...,m\right\}  }\left[
k=j\right]  \alpha_{m,k}\alpha_{m,m}}_{=\alpha_{i,j}\alpha_{m,m}}
-\underbrace{\sum_{k\in\left\{  1,2,...,m\right\}  }\left[  k=m\text{ and
}j\neq m\right]  \alpha_{m,k}\alpha_{m,j}}_{=\alpha_{i,m}\alpha_{m,j}}
=\alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}.
\]
This proves \eqref{sol.cauchy.cauchy.det.17.pf.gamma.ij}.} Furthermore,
\begin{equation}
\gamma_{m,m}=\alpha_{m,m}^{2}.
\label{sol.cauchy.cauchy.det.17.pf.gamma.mm}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.det.17.pf.gamma.mm}:} The
equality \eqref{sol.cauchy.cauchy.det.17.pf.gamma} (applied to $i=m$ and
$j=m$) yields
\begin{align*}
\gamma_{m,m}  &  =\underbrace{\sum_{k=1}^{m}}_{=\sum_{k\in\left\{
1,2,...,m\right\}  }}\alpha_{m,k}\underbrace{\beta_{k,m}}_{\substack{=\left[
k=m\right]  \alpha_{m,m}-\left[  k=m\text{ and }m\neq m\right]  \alpha
_{m,m}\\\text{(by \eqref{sol.cauchy.cauchy.det.17.pf.beta}, applied to
}i=k\text{ and }j=m\text{)}}}\\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\alpha_{m,k}\left(  \left[
k=m\right]  \alpha_{m,m}-\underbrace{\left[  k=m\text{ and }m\neq m\right]
}_{\substack{=0\\\text{(since }\left(  k=m\text{ and }m\neq m\right)  \text{
is false}\\\text{(because }m\neq m\text{ is false))}}}\alpha_{m,m}\right) \\
&  =\sum_{k\in\left\{  1,2,...,m\right\}  }\alpha_{m,k}\left(  \left[
k=m\right]  \alpha_{m,m}-\underbrace{0\alpha_{m,m}}_{=0}\right)  =\sum
_{k\in\left\{  1,2,...,m\right\}  }\alpha_{m,k}\left[  k=m\right]
\alpha_{m,m}\\
&  =\underbrace{\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k=m}%
}\alpha_{m,k}\left[  k=m\right]  \alpha_{m,m}}_{\substack{=\alpha_{m,m}\left[
m=m\right]  \alpha_{m,m}\\\text{(because }m\in\left\{  1,2,...,m\right\}
\text{)}}}+\sum\limits_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq m}%
}\alpha_{m,k}\underbrace{\left[  k=m\right]  }_{\substack{=0\\\text{(since
}k=m\text{ is false}\\\text{(because }k\neq m\text{))}}}\alpha_{m,m}\\
&  =\alpha_{m,m}\left[  m=m\right]  \alpha_{m,m}+\underbrace{\sum\limits
_{\substack{k\in\left\{  1,2,...,m\right\}  ;\\k\neq m}}\alpha_{m,k}%
0\alpha_{m,m}}_{=0}=\alpha_{m,m}\underbrace{\left[  m=m\right]  }%
_{\substack{=1\\\text{(since }m=m\text{ is true)}}}\alpha_{m,m}=\alpha
_{m,m}\alpha_{m,m}=\alpha_{m,m}^{2}.
\end{align*}
This proves \eqref{sol.cauchy.cauchy.det.17.pf.gamma.mm}.} Now,
\eqref{sol.cauchy.cauchy.det.17.pf.gamma-det.2} becomes
\begin{align*}
\det\left(  \left(  \gamma_{i,j}\right)  _{i,j=1,2,...,m}\right)   &
=\underbrace{\gamma_{m,m}}_{\substack{=\alpha_{m,m}^{2}\\\text{(by
\eqref{sol.cauchy.cauchy.det.17.pf.gamma.mm})}}}\cdot\det\left(  \left(
\underbrace{\gamma_{i,j}}_{\substack{=\alpha_{i,j}\alpha_{m,m}-\alpha
_{i,m}\alpha_{m,j}\\
\text{(by \eqref{sol.cauchy.cauchy.det.17.pf.gamma.ij})}}}
\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\alpha_{m,m}^{2}\cdot\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}
-\alpha_{i,m}\alpha_{m,j}\right)  _{i,j=1,2,...,m-1}\right)  .
\end{align*}
Compared with \eqref{sol.cauchy.cauchy.det.17.pf.gamma-det.1}, this yields%
\[
\alpha_{m,m}^{2}\cdot\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}-\alpha
_{i,m}\alpha_{m,j}\right)  _{i,j=1,2,...,m-1}\right)  =\alpha_{m,m}^{m}%
\cdot\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\]
We can divide both sides of this equality by $\alpha_{m,m}^{2}$ (since
$\alpha_{m,m}^{2}$ is invertible in $A$ (because $\alpha_{m,m}$ is invertible
in $A$)), and thus obtain%
\[
\det\left(  \left(  \alpha_{i,j}\alpha_{m,m}-\alpha_{i,m}\alpha_{m,j}\right)
_{i,j=1,2,...,m-1}\right)  =\underbrace{\dfrac{1}{\alpha_{m,m}^{2}}%
\alpha_{m,m}^{m}}_{=\alpha_{m,m}^{m-2}}\cdot\det\left(  \left(  \alpha
_{i,j}\right)  _{i,j=1,2,...,m}\right)  =\alpha_{m,m}^{m-2}\cdot\det\left(
\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,m}\right)  .
\]
This proves \eqref{sol.cauchy.cauchy.det.17}.

[\textit{Remark:} The equality \eqref{sol.cauchy.cauchy.det.17} holds even
without requiring that $\alpha_{m,m}$ be invertible, if we have $m\geq2$ (of
course, if $m$ is not $\geq2$, then the $\alpha_{m,m}^{m-2}$ on the right hand
side of \eqref{sol.cauchy.cauchy.det.17} does not make sense unless
$\alpha_{m,m}$ is invertible). There are several ways to see why this is so.
One of these ways proceeds as follows: First of all, one should notice that
our above proof of \eqref{sol.cauchy.cauchy.det.17} works without requiring
that $\alpha_{m,m}$ be invertible, as long as $\alpha_{m,m}$ is a
non-zero-divisor\footnote{A \emph{non-zero-divisor} in a commutative ring
$B$ means an element $b\in B$ such that every element $c\in B$ satisfying
$bc=0$ must satisfy $c=0$.} in $A$ and we have $m\geq2$. However, for any
fixed $m\geq2$, the equality \eqref{sol.cauchy.cauchy.det.17} is a polynomial
identity in the elements $\alpha_{i,j}$ of $A$; thus, it suffices to prove it
when $\alpha_{i,j}$ are distinct indeterminates $X_{i,j}$ in the polynomial
ring $\ZZ \left[  X_{i,j}\ \mid\ \left(  i,j\right)  \in\left\{
1,2,...,m\right\}  ^{2}\right]  $. But in this case, $\alpha_{m,m}$ is clearly
a non-zero-divisor, and so our above proof applies. (An alternative approach
would be to replace $\alpha_{m,m}$ by $X+\alpha_{m,m}$ in the polynomial ring
$A\left[  X\right]  $; again, $X+\alpha_{m,m}$ is a non-zero-divisor even if
$\alpha_{m,m}$ is not.)]
\end{itemize}

Now, rather than solve parts (a) and (b) of the exercise separately, we are
going to prove a result from which both of these parts will easily follow.
Namely, we will show the following statement:

\begin{proposition}
\label{prop.sol.cauchy.cauchy}
Let $A$ be a commutative ring. Let
$n \in \NN$. For every $i\in\left\{  1,2,...,n\right\}  $, let $a_{i}$,
$b_{i}$, $c_{i}$ and $d_{i}$ be four elements of $A$. Assume that $a_{i}
d_{j}-b_{i}c_{j}$ is an invertible element of $A$ for every $i\in\left\{
1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $. Then,
\[
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots
,n}\right)  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}b_{j}
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }
\]

\end{proposition}

\textit{Proof of Proposition \ref{prop.sol.cauchy.cauchy}.} We are going to
show that
\begin{equation}
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots
,k}\right)  =\dfrac{\prod_{1\leq j<i\leq k}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,k\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  } \label{sol.cauchy.cauchy.a.goal}%
\end{equation}
for every $k\in\left\{  0,1,...,n\right\}  $.

\textit{Proof of \eqref{sol.cauchy.cauchy.a.goal}:} We will prove
\eqref{sol.cauchy.cauchy.a.goal} by induction over $k$:

\textit{Induction base:} It is easy to see that
\eqref{sol.cauchy.cauchy.a.goal} holds if $k=0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $k=0$. Then, $\left(  \dfrac
{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots,k}=\left(  \dfrac{1}%
{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots,0}$ is a $0\times0$-matrix,
and thus has determinant $1$ (since every $0\times0$-matrix has determinant
$1$). In other words, $\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}%
}\right)  _{i,j=1,2,\ldots,k}\right)  =1$.
\par
But then there exist no integers $i$ and $j$ satisfying $1\leq j<i\leq k$
(because any such integers would have to satisfy $1\leq j<i\leq k=0<1$, which
is absurd). Thus, $\prod_{1\leq j<i\leq k}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  $ is an
empty product. Hence, $\prod_{1\leq j<i\leq k}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  =\left(
\text{empty product}\right)  =1$. Also, since $k=0$, we have $\left\{
1,2,\ldots,k\right\}  =\left\{  1,2,...,0\right\}  =\varnothing$, so that
$\left\{  1,2,\ldots,k\right\}  ^{2}=\varnothing^{2}=\varnothing$. Thus,
$\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,k\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  $ is an empty product. Hence, $\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,k\right\}  ^{2}}\left(  a_{i}d_{j}%
-b_{i}c_{j}\right)  =\left(  \text{empty product}\right)  =1$.
\par
Now, dividing the equality $\prod_{1\leq j<i\leq k}\left(  \left(  a_{i}%
b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  =1$ by
the equality $\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,k\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  =1$, we obtain%
\[
\dfrac{\prod_{1\leq j<i\leq k}\left(  \left(  a_{i}b_{j}-a_{j}b_{i}\right)
\left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,k\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}=\dfrac{1}{1}=1.
\]
Comparing this with $\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}%
}\right)  _{i,j=1,2,\ldots,k}\right)  =1$, we obtain $\det\left(  \left(
\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots,k}\right)
=\dfrac{\prod_{1\leq j<i\leq k}\left(  \left(  a_{i}b_{j}-a_{j}b_{i}\right)
\left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,k\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}$. Thus, \eqref{sol.cauchy.cauchy.a.goal} holds if $k=0$, qed.}. Hence, the
induction base is complete.

\textit{Induction step:} Let $m\in\left\{  1,2,...,n\right\}  $. Assume that
\eqref{sol.cauchy.cauchy.a.goal} is proven for $k=m-1$. We need to show that
\eqref{sol.cauchy.cauchy.a.goal} holds for $k=m$.

We know that \eqref{sol.cauchy.cauchy.a.goal} is proven for $k=m-1$. Thus,%
\begin{equation}
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots
,m-1}\right)  =\dfrac{\prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  } .
\label{sol.cauchy.cauchy.a.indhyp}
\end{equation}


Now, we know that $\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}$ is an invertible element
of $A$ (because it is the inverse of $a_{m}d_{m}-b_{m}c_{m}$). Thus,
\eqref{sol.cauchy.cauchy.det.17} (applied to $\alpha_{i,j}=\dfrac{1}{a_{i}
d_{j}-b_{i}c_{j}}$) yields
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}%
{a_{m}d_{m}-b_{m}c_{m}}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}%
{a_{m}d_{j}-b_{m}c_{j}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}\cdot\det\left(
\left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m}\right)  .
\end{align*}
Multiplying both sides of this equality with $\left(  a_{m}d_{m}-b_{m}
c_{m}\right)  ^{m-2}$, we obtain
\begin{align}
&  \left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\det\left(  \left(
\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}
-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)
_{i,j=1,2,...,m-1}\right) \nonumber\\
&  =\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\underbrace{\left(
\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}}_{=\dfrac{1}{\left(
a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}}}\cdot\det\left(  \left(  \dfrac{1}%
{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m}\right) \nonumber\\
&  =\underbrace{\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\dfrac
{1}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}}}_{=1}\cdot\det\left(
\left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m}\right)
\nonumber\\
&  =\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)
_{i,j=1,2,...,m}\right)  .
\label{sol.cauchy.cauchy.a.1}
\end{align}


But every $\left(  i,j\right)  \in\left\{  1,2,...,m-1\right\}  ^{2}$ satisfy
\begin{align*}
&  \underbrace{\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}{a_{m}d_{m}%
-b_{m}c_{m}}}_{=\dfrac{1}{\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \left(
a_{m}d_{m}-b_{m}c_{m}\right)  }}-\underbrace{\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}%
}\cdot\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}}_{=\dfrac{1}{\left(  a_{i}d_{m}%
-b_{i}c_{m}\right)  \left(  a_{m}d_{j}-b_{m}c_{j}\right)  }}\\
&  =\dfrac{1}{\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \left(  a_{m}d_{m}%
-b_{m}c_{m}\right)  }-\dfrac{1}{\left(  a_{i}d_{m}-b_{i}c_{m}\right)  \left(
a_{m}d_{j}-b_{m}c_{j}\right)  }\\
&  =\dfrac{\left(  a_{i}d_{m}-b_{i}c_{m}\right)  \left(  a_{m}d_{j}-b_{m}%
c_{j}\right)  -\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \left(  a_{m}d_{m}%
-b_{m}c_{m}\right)  }{\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \left(  a_{m}%
d_{m}-b_{m}c_{m}\right)  \left(  a_{i}d_{m}-b_{i}c_{m}\right)  \left(
a_{m}d_{j}-b_{m}c_{j}\right)  }\\
&  =\dfrac{\left(  a_{m}b_{i}-a_{i}b_{m}\right)  \left(  c_{j}d_{m}-c_{m}%
d_{j}\right)  }{\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \left(  a_{m}d_{m}%
-b_{m}c_{m}\right)  \left(  a_{i}d_{m}-b_{i}c_{m}\right)  \left(  a_{m}%
d_{j}-b_{m}c_{j}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\underbrace{\left(  a_{i}d_{m}-b_{i}c_{m}\right)  \left(
a_{m}d_{j}-b_{m}c_{j}\right)  }_{=a_{i}d_{m}a_{m}d_{j}-a_{i}d_{m}b_{m}%
c_{j}-b_{i}c_{m}a_{m}d_{j}+b_{i}c_{m}b_{m}c_{j}}-\underbrace{\left(
a_{i}d_{j}-b_{i}c_{j}\right)  \left(  a_{m}d_{m}-b_{m}c_{m}\right)  }%
_{=a_{i}d_{j}a_{m}d_{m}-a_{i}d_{j}b_{m}c_{m}-b_{i}c_{j}a_{m}d_{m}+b_{i}%
c_{j}b_{m}c_{m}}\\
=\left(  \underbrace{a_{i}d_{m}a_{m}d_{j}}_{=a_{i}d_{j}a_{m}d_{m}%
}-\underbrace{a_{i}d_{m}b_{m}c_{j}}_{=a_{i}b_{m}c_{j}d_{m}}-\underbrace{b_{i}%
c_{m}a_{m}d_{j}}_{=a_{m}b_{i}c_{m}d_{j}}+\underbrace{b_{i}c_{m}b_{m}c_{j}%
}_{=b_{i}c_{j}b_{m}c_{m}}\right)  -\left(  a_{i}d_{j}a_{m}d_{m}%
-\underbrace{a_{i}d_{j}b_{m}c_{m}}_{=a_{i}b_{m}c_{m}d_{j}}-\underbrace{b_{i}%
c_{j}a_{m}d_{m}}_{=a_{m}b_{i}c_{j}d_{m}}+b_{i}c_{j}b_{m}c_{m}\right) \\
=\left(  a_{i}d_{j}a_{m}d_{m}-a_{i}b_{m}c_{j}d_{m}-a_{m}b_{i}c_{m}d_{j}%
+b_{i}c_{j}b_{m}c_{m}\right)  -\left(  a_{i}d_{j}a_{m}d_{m}-a_{i}b_{m}%
c_{m}d_{j}-a_{m}b_{i}c_{j}d_{m}+b_{i}c_{j}b_{m}c_{m}\right) \\
=a_{m}b_{i}c_{j}d_{m}-a_{m}b_{i}c_{m}d_{j}-a_{i}b_{m}c_{j}d_{m}+a_{i}%
b_{m}c_{m}d_{j}=\left(  a_{m}b_{i}-a_{i}b_{m}\right)  \left(  c_{j}d_{m}%
-c_{m}d_{j}\right)
\end{array}
\right) \\
&  =\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\left(  a_{m}d_{j}-b_{m}c_{j}\right)  }\cdot\dfrac{a_{m}b_{i}-a_{i}b_{m}%
}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}.
\end{align*}
Hence,
\begin{align*}
&  \det\left(  \left(  \underbrace{\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot
\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot
\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}}_{=\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(
a_{m}d_{m}-b_{m}c_{m}\right)  \left(  a_{m}d_{j}-b_{m}c_{j}\right)  }%
\cdot\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}%
{a_{i}d_{j}-b_{i}c_{j}}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\det\left(  \left(  \dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}%
-b_{m}c_{m}\right)  \left(  a_{m}d_{j}-b_{m}c_{j}\right)  }\cdot\dfrac
{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{i}d_{j}%
-b_{i}c_{j}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\left(  \prod_{j=1}^{m-1}\underbrace{\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(
a_{m}d_{m}-b_{m}c_{m}\right)  \left(  a_{m}d_{j}-b_{m}c_{j}\right)  }%
}_{=\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\cdot\dfrac{c_{j}d_{m}-c_{m}d_{j}}%
{a_{m}d_{j}-b_{m}c_{j}}}\right)  \cdot\underbrace{\det\left(  \left(
\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{i}%
d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m-1}\right)  }_{\substack{=\left(
\prod_{i=1}^{m-1}\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\right)
\cdot\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)
_{i,j=1,2,...,m-1}\right)  \\\text{(by \eqref{sol.cauchy.cauchy.det.13},
applied to }m-1\text{, }\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\\\text{and }%
\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\text{ instead of
}m\text{, }\alpha_{i,j}\text{ and }\gamma_{i}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by \eqref{sol.cauchy.cauchy.det.15}, applied to }m-1\text{, }
\dfrac{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{i}
d_{j}-b_{i}c_{j}}\text{ and}\\
\dfrac{c_{j}d_{m}-c_{m}d_{j}}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)  \left(
a_{m}d_{j}-b_{m}c_{j}\right)  }\text{ instead of }m\text{, }\alpha_{i,j}\text{
and }\delta_{j}%
\end{array}
\right) \\
&  =\underbrace{\left(  \prod_{j=1}^{m-1}\left(  \dfrac{1}{a_{m}d_{m}%
-b_{m}c_{m}}\cdot\dfrac{c_{j}d_{m}-c_{m}d_{j}}{a_{m}d_{j}-b_{m}c_{j}}\right)
\right)  }_{=\left(  \prod_{j=1}^{m-1}\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)
\cdot\left(  \prod_{j=1}^{m-1}\dfrac{c_{j}d_{m}-c_{m}d_{j}}{a_{m}d_{j}%
-b_{m}c_{j}}\right)  }\cdot\left(  \prod_{i=1}^{m-1}\dfrac{a_{m}b_{i}%
-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}\right)  \cdot\underbrace{\det\left(
\left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m-1}\right)
}_{\substack{=\dfrac{\prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }\\
\text{(by \eqref{sol.cauchy.cauchy.a.indhyp})}}}\\
&  =\underbrace{\left(  \prod_{j=1}^{m-1}\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}%
}\right)  }_{=\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-1}}%
\cdot\left(  \prod_{j=1}^{m-1}\underbrace{\dfrac{c_{j}d_{m}-c_{m}d_{j}}%
{a_{m}d_{j}-b_{m}c_{j}}}_{=\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\cdot\left(
c_{j}d_{m}-c_{m}d_{j}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \prod_{i=1}^{m-1}\underbrace{\dfrac
{a_{m}b_{i}-a_{i}b_{m}}{a_{i}d_{m}-b_{i}c_{m}}}_{=\dfrac{1}{a_{i}d_{m}%
-b_{i}c_{m}}\cdot\left(  a_{m}b_{i}-a_{i}b_{m}\right)  }\right)
\cdot\underbrace{\dfrac{\prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}%
b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)
}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }}_{=\dfrac{1}{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}\cdot\left(  \prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}b_{j}-a_{j}%
b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  \right)  }%
\end{align*}%
\begin{align}
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-1}\cdot
\underbrace{\left(  \prod_{j=1}^{m-1}\left(  \dfrac{1}{a_{m}d_{j}-b_{m}c_{j}%
}\cdot\left(  c_{j}d_{m}-c_{m}d_{j}\right)  \right)  \right)  }_{=\left(
\prod_{j=1}^{m-1}\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)  \cdot\left(
\prod_{j=1}^{m-1}\left(  c_{j}d_{m}-c_{m}d_{j}\right)  \right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \prod_{i=1}^{m-1}\left(
\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\left(  a_{m}b_{i}-a_{i}b_{m}\right)
\right)  \right)  }_{=\left(  \prod_{i=1}^{m-1}\dfrac{1}{a_{i}d_{m}-b_{i}%
c_{m}}\right)  \cdot\left(  \prod_{i=1}^{m-1}\left(  a_{m}b_{i}-a_{i}%
b_{m}\right)  \right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }%
\cdot\underbrace{\left(  \prod_{1\leq j<i\leq m-1}\left(  \left(  a_{i}%
b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)
\right)  }_{=\left(  \prod_{1\leq j<i\leq m-1}\left(  a_{i}b_{j}-a_{j}%
b_{i}\right)  \right)  \left(  \prod_{1\leq j<i\leq m-1}\left(  c_{j}%
d_{i}-c_{i}d_{j}\right)  \right)  }\nonumber\\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-1}\cdot\left(
\prod_{j=1}^{m-1}\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)  \cdot\left(
\prod_{j=1}^{m-1}\left(  c_{j}d_{m}-c_{m}d_{j}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \prod_{i=1}^{m-1}\dfrac{1}{a_{i}%
d_{m}-b_{i}c_{m}}\right)  \cdot\left(  \prod_{i=1}^{m-1}\left(  a_{m}%
b_{i}-a_{i}b_{m}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }%
\cdot\left(  \prod_{1\leq j<i\leq m-1}\left(  a_{i}b_{j}-a_{j}b_{i}\right)
\right)  \left(  \prod_{1\leq j<i\leq m-1}\left(  c_{j}d_{i}-c_{i}%
d_{j}\right)  \right) \nonumber\\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-1}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \prod_{j=1}^{m-1}\dfrac{1}{a_{m}%
d_{j}-b_{m}c_{j}}\right)  \cdot\left(  \prod_{i=1}^{m-1}\dfrac{1}{a_{i}%
d_{m}-b_{i}c_{m}}\right)  \cdot\dfrac{1}{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \prod_{i=1}^{m-1}\left(  a_{m}b_{i}%
-a_{i}b_{m}\right)  \right)  \cdot\left(  \prod_{1\leq j<i\leq m-1}\left(
a_{i}b_{j}-a_{j}b_{i}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \prod_{j=1}^{m-1}\left(  c_{j}d_{m}%
-c_{m}d_{j}\right)  \right)  \cdot\left(  \prod_{1\leq j<i\leq m-1}\left(
c_{j}d_{i}-c_{i}d_{j}\right)  \right)  .
\label{sol.cauchy.cauchy.a.3}
\end{align}


But notice that
\begin{align*}
&  \underbrace{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}
^{2}}}_{=\prod_{i\in\left\{  1,2,...,m\right\}  }\prod_{j\in\left\{
1,2,...,m\right\}  }}\left(  a_{i}d_{j}-b_{i}c_{j}\right) \\
&  =\underbrace{\prod_{i\in\left\{  1,2,...,m\right\}  }}_{=\prod_{i=1}^{m}%
}\underbrace{\prod_{j\in\left\{  1,2,...,m\right\}  }}_{=\prod_{j=1}^{m}%
}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  =\prod_{i=1}^{m}\underbrace{\prod
_{j=1}^{m}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }_{\substack{=\left(
\prod_{j=1}^{m-1}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \right)  \cdot\left(
a_{i}d_{m}-b_{i}c_{m}\right)  \\\text{(here, we have split off the}\\j=m\text{
term from the product)}}}\\
&  =\prod_{i=1}^{m}\left(  \left(  \prod_{j=1}^{m-1}\left(  a_{i}d_{j}%
-b_{i}c_{j}\right)  \right)  \cdot\left(  a_{i}d_{m}-b_{i}c_{m}\right)
\right) \\
&  =\underbrace{\left(  \prod_{i=1}^{m}\prod_{j=1}^{m-1}\left(  a_{i}%
d_{j}-b_{i}c_{j}\right)  \right)  }_{\substack{=\left(  \prod_{i=1}^{m-1}%
\prod_{j=1}^{m-1}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \right)  \cdot\left(
\prod_{j=1}^{m-1}\left(  a_{m}d_{j}-b_{m}c_{j}\right)  \right)  \\\text{(here,
we have split off the }i=m\text{ term from the outer product)}}}\cdot
\underbrace{\left(  \prod_{i=1}^{m}\left(  a_{i}d_{m}-b_{i}c_{m}\right)
\right)  }_{\substack{=\left(  \prod_{i=1}^{m-1}\left(  a_{i}d_{m}-b_{i}%
c_{m}\right)  \right)  \cdot\left(  a_{m}d_{m}-b_{m}c_{m}\right)
\\\text{(here, we have split off the }i=m\text{ term}\\\text{from the
product)}}}\\
&  =\left(  \underbrace{\prod_{i=1}^{m-1}}_{=\prod_{i\in\left\{
1,2,...,m-1\right\}  }}\underbrace{\prod_{j=1}^{m-1}}_{=\prod_{j\in\left\{
1,2,...,m-1\right\}  }}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \right)
\cdot\left(  \prod_{j=1}^{m-1}\left(  a_{m}d_{j}-b_{m}c_{j}\right)  \right)
\cdot\left(  \prod_{i=1}^{m-1}\left(  a_{i}d_{m}-b_{i}c_{m}\right)  \right)
\cdot\left(  a_{m}d_{m}-b_{m}c_{m}\right) \\
&  =\left(  \underbrace{\prod_{i\in\left\{  1,2,...,m-1\right\}  }\prod
_{j\in\left\{  1,2,...,m-1\right\}  }}_{=\prod_{\left(  i,j\right)
\in\left\{  1,2,...,m-1\right\}  ^{2}}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
\right)  \cdot\left(  \prod_{j=1}^{m-1}\left(  a_{m}d_{j}-b_{m}c_{j}\right)
\right)  \cdot\left(  \prod_{i=1}^{m-1}\left(  a_{i}d_{m}-b_{i}c_{m}\right)
\right)  \cdot\left(  a_{m}d_{m}-b_{m}c_{m}\right) \\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,...,m-1\right\}  ^{2}%
}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \right)  \cdot\left(  \prod_{j=1}%
^{m-1}\left(  a_{m}d_{j}-b_{m}c_{j}\right)  \right)  \cdot\left(  \prod
_{i=1}^{m-1}\left(  a_{i}d_{m}-b_{i}c_{m}\right)  \right)  \cdot\left(
a_{m}d_{m}-b_{m}c_{m}\right)  .
\end{align*}
Taking the (multiplicative) inverse of both sides of this equality, we obtain%
\begin{align*}
&  \dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\\
&  =\dfrac{1}{\left(  \prod_{\left(  i,j\right)  \in\left\{
1,2,...,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  \right)
\cdot\left(  \prod_{j=1}^{m-1}\left(  a_{m}d_{j}-b_{m}c_{j}\right)  \right)
\cdot\left(  \prod_{i=1}^{m-1}\left(  a_{i}d_{m}-b_{i}c_{m}\right)  \right)
\cdot\left(  a_{m}d_{m}-b_{m}c_{m}\right)  }\\
&  =\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,...,m-1\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\cdot\underbrace{\dfrac{1}%
{\prod_{j=1}^{m-1}\left(  a_{m}d_{j}-b_{m}c_{j}\right)  }}_{=\prod_{j=1}%
^{m-1}\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}}\cdot\underbrace{\dfrac{1}{\prod
_{i=1}^{m-1}\left(  a_{i}d_{m}-b_{i}c_{m}\right)  }}_{=\prod_{i=1}^{m-1}%
\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}}\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\\
&  =\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,...,m-1\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\cdot\left(  \prod_{j=1}%
^{m-1}\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)  \cdot\left(  \prod_{i=1}%
^{m-1}\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\right)  \cdot\dfrac{1}{a_{m}d_{m}%
-b_{m}c_{m}}\\
&  =\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\cdot\left(  \prod_{j=1}^{m-1}\dfrac
{1}{a_{m}d_{j}-b_{m}c_{j}}\right)  \cdot\left(  \prod_{i=1}^{m-1}\dfrac
{1}{a_{i}d_{m}-b_{i}c_{m}}\right)  \cdot\dfrac{1}{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}.
\end{align*}
Thus,
\begin{align}
&  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\cdot\left(  \prod_{j=1}^{m-1}\dfrac
{1}{a_{m}d_{j}-b_{m}c_{j}}\right)  \cdot\left(  \prod_{i=1}^{m-1}\dfrac
{1}{a_{i}d_{m}-b_{i}c_{m}}\right)  \cdot\dfrac{1}{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}\nonumber\\
&  =\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }. \label{sol.cauchy.cauchy.a.5a}%
\end{align}
Multiplying both sides of this equality by $a_{m}d_{m}-b_{m}c_{m}$, we obtain%
\begin{align}
&  \left(  \prod_{j=1}^{m-1}\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)
\cdot\left(  \prod_{i=1}^{m-1}\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\right)
\cdot\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m-1\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\nonumber\\
&  =\left(  a_{m}d_{m}-b_{m}c_{m}\right)  \cdot\dfrac{1}{\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(  a_{i}d_{j}%
-b_{i}c_{j}\right)  }.
\label{sol.cauchy.cauchy.a.5aa}
\end{align}


On the other hand,
\begin{align}
\underbrace{\prod_{1\leq j<i\leq m}}_{=\prod_{i=1}^{m}\prod_{j=1}^{i-1}%
}\left(  a_{i}b_{j}-a_{j}b_{i}\right)   &  =\prod_{i=1}^{m}\prod_{j=1}%
^{i-1}\left(  a_{i}b_{j}-a_{j}b_{i}\right)  =\left(  \underbrace{\prod
_{i=1}^{m-1}\prod_{j=1}^{i-1}}_{=\prod_{1\leq j<i\leq m-1}}\left(  a_{i}%
b_{j}-a_{j}b_{i}\right)  \right)  \cdot\prod_{j=1}^{m-1}\left(  a_{m}%
b_{j}-a_{j}b_{m}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the }i=m\text{
term from the outer product}\right) \nonumber\\
&  =\left(  \prod_{1\leq j<i\leq m-1}\left(  a_{i}b_{j}-a_{j}b_{i}\right)
\right)  \cdot\underbrace{\left(  \prod_{j=1}^{m-1}\left(  a_{m}b_{j}%
-a_{j}b_{m}\right)  \right)  }_{\substack{=\prod_{i=1}^{m-1}\left(  a_{m}%
b_{i}-a_{i}b_{m}\right)  \\\text{(here, we renamed the index }j\\\text{as
}i\text{ in the product)}}}\nonumber\\
&  =\left(  \prod_{1\leq j<i\leq m-1}\left(  a_{i}b_{j}-a_{j}b_{i}\right)
\right)  \cdot\left(  \prod_{i=1}^{m-1}\left(  a_{m}b_{i}-a_{i}b_{m}\right)
\right) \nonumber\\
&  =\left(  \prod_{i=1}^{m-1}\left(  a_{m}b_{i}-a_{i}b_{m}\right)  \right)
\cdot\left(  \prod_{1\leq j<i\leq m-1}\left(  a_{i}b_{j}-a_{j}b_{i}\right)
\right)  . \label{sol.cauchy.cauchy.a.5b}%
\end{align}


Also,%
\begin{align}
\underbrace{\prod_{1\leq j<i\leq m}}_{=\prod_{i=1}^{m}\prod_{j=1}^{i-1}%
}\left(  c_{j}d_{i}-c_{i}d_{j}\right)   &  =\prod_{i=1}^{m}\prod_{j=1}%
^{i-1}\left(  c_{j}d_{i}-c_{i}d_{j}\right)  =\left(  \underbrace{\prod
_{i=1}^{m-1}\prod_{j=1}^{i-1}}_{=\prod_{1\leq j<i\leq m-1}}\left(  c_{j}%
d_{i}-c_{i}d_{j}\right)  \right)  \cdot\prod_{j=1}^{m-1}\left(  c_{j}%
d_{m}-c_{m}d_{j}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the }i=m\text{
term from the outer product}\right) \nonumber\\
&  =\left(  \prod_{1\leq j<i\leq m-1}\left(  c_{j}d_{i}-c_{i}d_{j}\right)
\right)  \cdot\prod_{j=1}^{m-1}\left(  c_{j}d_{m}-c_{m}d_{j}\right)
\nonumber\\
&  =\left(  \prod_{j=1}^{m-1}\left(  c_{j}d_{m}-c_{m}d_{j}\right)  \right)
\cdot\left(  \prod_{1\leq j<i\leq m-1}\left(  c_{j}d_{i}-c_{i}d_{j}\right)
\right)  .
\label{sol.cauchy.cauchy.a.5c}
\end{align}
Now, \eqref{sol.cauchy.cauchy.a.3} becomes
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}%
{a_{m}d_{m}-b_{m}c_{m}}-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}%
{a_{m}d_{j}-b_{m}c_{j}}\right)  _{i,j=1,2,...,m-1}\right) \\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-1}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \prod_{j=1}^{m-1}\dfrac
{1}{a_{m}d_{j}-b_{m}c_{j}}\right)  \cdot\left(  \prod_{i=1}^{m-1}\dfrac
{1}{a_{i}d_{m}-b_{i}c_{m}}\right)  \cdot\dfrac{1}{\prod_{\left(  i,j\right)
\in\left\{  1,2,\ldots,m-1\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)
}}_{\substack{=\left(  a_{m}d_{m}-b_{m}c_{m}\right)  \cdot\dfrac{1}%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }\\
\text{(by \eqref{sol.cauchy.cauchy.a.5aa})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \prod_{i=1}^{m-1}\left(
a_{m}b_{i}-a_{i}b_{m}\right)  \right)  \cdot\left(  \prod_{1\leq j<i\leq
m-1}\left(  a_{i}b_{j}-a_{j}b_{i}\right)  \right)  }_{\substack{=\prod_{1\leq
j<i\leq m}\left(  a_{i}b_{j}-a_{j}b_{i}\right)  \\\text{(by
\eqref{sol.cauchy.cauchy.a.5b})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \prod_{j=1}^{m-1}\left(
c_{j}d_{m}-c_{m}d_{j}\right)  \right)  \cdot\left(  \prod_{1\leq j<i\leq
m-1}\left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }_{\substack{=\prod_{1\leq
j<i\leq m}\left(  c_{j}d_{i}-c_{i}d_{j}\right)  \\\text{(by
\eqref{sol.cauchy.cauchy.a.5c})}}}\\
&  =\underbrace{\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-1}%
}_{\substack{=\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{\left(
m-2\right)  +1}\\=\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}%
\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}}}\cdot\left(  a_{m}d_{m}-b_{m}%
c_{m}\right)  \cdot\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,m\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \prod_{1\leq j<i\leq m}\left(
a_{i}b_{j}-a_{j}b_{i}\right)  \right)  \cdot\left(  \prod_{1\leq j<i\leq
m}\left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }_{=\prod_{1\leq j<i\leq
m}\left(  \left(  a_{i}b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}%
d_{j}\right)  \right)  }\\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}\cdot
\underbrace{\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\cdot\left(  a_{m}d_{m}-b_{m}%
c_{m}\right)  }_{=1}\cdot\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,m\right\}  ^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\prod_{1\leq j<i\leq m}\left(  \left(  a_{i}%
b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right) \\
&  =\left(  \dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}\cdot\dfrac
{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }\cdot\prod_{1\leq j<i\leq m}\left(  \left(
a_{i}b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  .
\end{align*}
Multiplying both sides of this equality by $\left(  a_{m}d_{m}-b_{m}%
c_{m}\right)  ^{m-2}$, we obtain%
\begin{align*}
&  \left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\det\left(  \left(
\dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\cdot\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}%
-\dfrac{1}{a_{i}d_{m}-b_{i}c_{m}}\cdot\dfrac{1}{a_{m}d_{j}-b_{m}c_{j}}\right)
_{i,j=1,2,...,m-1}\right) \\
&  =\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\underbrace{\left(
\dfrac{1}{a_{m}d_{m}-b_{m}c_{m}}\right)  ^{m-2}}_{=\dfrac{1}{\left(
a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}}}\cdot\dfrac{1}{\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(  a_{i}d_{j}%
-b_{i}c_{j}\right)  }\cdot\prod_{1\leq j<i\leq m}\left(  \left(  a_{i}%
b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right) \\
&  =\underbrace{\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}\cdot\dfrac
{1}{\left(  a_{m}d_{m}-b_{m}c_{m}\right)  ^{m-2}}}_{=1}\cdot\dfrac{1}%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }\cdot\prod_{1\leq j<i\leq m}\left(  \left(
a_{i}b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right) \\
&  =\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}
^{2}}\left(  a_{i}d_{j}-b_{i}c_{j}\right)  }\cdot\prod_{1\leq j<i\leq
m}\left(  \left(  a_{i}b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}%
d_{j}\right)  \right)  =\dfrac{\prod_{1\leq j<i\leq m}\left(  \left(
a_{i}b_{j}-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)
}{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }.
\end{align*}
Compared with \eqref{sol.cauchy.cauchy.a.1}, this yields
\[
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,...,m}%
\right)  =\dfrac{\prod_{1\leq j<i\leq m}\left(  \left(  a_{i}b_{j}-a_{j}%
b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }{\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}}\left(  a_{i}d_{j}%
-b_{i}c_{j}\right)  }.
\]
In other words, \eqref{sol.cauchy.cauchy.a.goal} holds for $k=m$. This
completes the induction step. The induction proof of
\eqref{sol.cauchy.cauchy.a.goal} is thus complete.

So we have proven \eqref{sol.cauchy.cauchy.a.goal}. We thus can apply
\eqref{sol.cauchy.cauchy.a.goal} to $k=n$, and obtain
\[
\det\left(  \left(  \dfrac{1}{a_{i}d_{j}-b_{i}c_{j}}\right)  _{i,j=1,2,\ldots
,n}\right)  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}b_{j}%
-a_{j}b_{i}\right)  \left(  c_{j}d_{i}-c_{i}d_{j}\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(
a_{i}d_{j}-b_{i}c_{j}\right)  }.
\]
This proves Proposition \ref{prop.sol.cauchy.cauchy}.

Now, let us solve the exercise.

(a) \textit{First solution of Exercise~\ref{exe.cauchy.cauchy}(a):} Assume
that $a_{i}-b_{j}$ is an invertible element of $A$ for every $i\in\left\{
1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $.

We know that $a_{i}-b_{j}$ is an invertible element of $A$ for every
$i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $. In
other words, $a_{i}\cdot1-1\cdot b_{j}$ is an invertible element of $A$ for
every $i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}
$ (because $\underbrace{a_{i}\cdot1}_{=a_{i}}-\underbrace{1\cdot b_{j}%
}_{=b_{j}}=a_{i}-b_{j}$ for every $i\in\left\{  1,2,...,n\right\}  $ and
$j\in\left\{  1,2,...,n\right\}  $). Hence, Proposition
\ref{prop.sol.cauchy.cauchy} (applied to $a_{i}$, $1$, $b_{i}$ and $1$ instead
of $a_{i}$, $b_{i}$, $c_{i}$ and $d_{i}$) yields%
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{a_{i}\cdot1-1\cdot b_{j}}\right)
_{i,j=1,2,\ldots,n}\right) \\
&  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}\cdot1-a_{j}%
\cdot1\right)  \left(  b_{j}\cdot1-b_{i}\cdot1\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(
a_{i}\cdot1-1\cdot b_{j}\right)  }\\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  \underbrace{a_{i}\cdot1}_{=a_{i}}-\underbrace{1\cdot b_{j}}_{=b_{j}%
}\right)  \right)  ^{-1}\cdot\prod_{1\leq j<i\leq n}\left(  \left(
\underbrace{a_{i}\cdot1}_{=a_{i}}-\underbrace{a_{j}\cdot1}_{=a_{j}}\right)
\left(  \underbrace{b_{j}\cdot1}_{=b_{j}}-\underbrace{b_{i}\cdot1}_{=b_{i}%
}\right)  \right) \\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  a_{i}-b_{j}\right)  \right)  ^{-1}\cdot\prod_{1\leq j<i\leq n}\left(
\left(  a_{i}-a_{j}\right)  \left(  b_{j}-b_{i}\right)  \right)  =\dfrac
{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}-a_{j}\right)  \left(
b_{j}-b_{i}\right)  \right)  }{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}}\left(  a_{i}-b_{j}\right)  }.
\end{align*}
This rewrites as $\det\left(  \left(  \dfrac{1}{a_{i}-b_{j}}\right)
_{i,j=1,2,\ldots,n}\right)  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(
a_{i}-a_{j}\right)  \left(  b_{j}-b_{i}\right)  \right)  }{\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(  a_{i}-b_{j}\right)
}$ (because every $i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{
1,2,...,n\right\}  $ satisfy $\underbrace{a_{i}\cdot1}_{=a_{i}}%
-\underbrace{1\cdot b_{j}}_{=b_{j}}=a_{i}-b_{j}$). This solves Exercise
\ref{exe.cauchy.cauchy}(a).

\textit{Second solution of Exercise~\ref{exe.cauchy.cauchy}(a):} Theorem 7
from \cite{Grinberg-19-9} states the following:\footnote{We have renamed the
elements denoted by $a_{1}$, $a_{2}$, $...$, $a_{m}$ in \cite[Theorem
7]{Grinberg-19-9} as $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{m}$, and we
have renamed the elements denoted by $b_{1}$, $b_{2}$, $...$, $b_{m}$ in
\cite[Theorem 7]{Grinberg-19-9} as $\beta_{1}$, $\beta_{2}$, $...$, $\beta
_{m}$, because they are not a-priori related to our elements $a_{1}$, $a_{2}$,
$...$, $a_{n}$ and $b_{1}$, $b_{2}$, $...$, $b_{n}$. We have also changed the
notations from \cite{Grinberg-19-9} to the ones that we are using.}

\begin{proposition}
\label{prop.sol.cauchy.cauchy.t7}Let $R$ be a commutative ring. Let
$m\in\NN$. Let $\alpha_{1}$, $\alpha_{2}$, $...$, $\alpha_{m}$ be $m$
elements of $R$. Let $\beta_{1}$, $\beta_{2}$, $...$, $\beta_{m}$ be $m$
elements of $R$. Then,%
\[
\det\left(  \left(  \prod_{\ell\in\left\{  1,2,...,m\right\}  \setminus
\left\{  i\right\}  }\left(  \alpha_{j}-\beta_{\ell}\right)  \right)
_{i,j=1,2,...,m}\right)  =\prod_{1\leq j<i\leq m}\left(  \left(  \alpha
_{i}-\alpha_{j}\right)  \left(  \beta_{j}-\beta_{i}\right)  \right)  .
\]

\end{proposition}

Now, assume that $a_{i}-b_{j}$ is an invertible element of $A$ for every
$i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $. We
can apply Proposition \ref{prop.sol.cauchy.cauchy.t7} to $m=n$, $\alpha
_{i}=a_{i}$ and $\beta_{i}=b_{i}$. As a result, we obtain%
\begin{equation}
\det\left(  \left(  \prod_{\ell\in\left\{  1,2,...,n\right\}  \setminus
\left\{  i\right\}  }\left(  a_{j}-b_{\ell}\right)  \right)  _{i,j=1,2,...,n}%
\right)  =\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}-a_{j}\right)  \left(
b_{j}-b_{i}\right)  \right)  . \label{sol.cauchy.cauchy.a2.use}%
\end{equation}
But every $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}$ satisfies%
\begin{equation}
\prod_{\ell\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}
}\left(  a_{j}-b_{\ell}\right)  =\left(  \prod_{\ell\in\left\{
1,2,...,n\right\}  }\left(  a_{j}-b_{\ell}\right)  \right)  \cdot\dfrac
{1}{a_{j}-b_{i}} \label{sol.cauchy.cauchy.a2.1}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.a2.1}:} Let $\left(
i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}$. Then,
\[
\prod_{\ell\in\left\{  1,2,...,n\right\}  }\left(  a_{j}-b_{\ell}\right)
=\left(  a_{j}-b_{i}\right)  \cdot\prod_{\ell\in\left\{  1,2,...,n\right\}
\setminus\left\{  i\right\}  }\left(  a_{j}-b_{\ell}\right)
\]
(here, we have split off the $\ell=i$ factor from the product $\prod_{\ell
\in\left\{  1,2,...,n\right\}  }\left(  a_{j}-b_{\ell}\right)  $, since
$i\in\left\{  1,2,...,n\right\}  $). Dividing both sides of this equality by
$a_{j}-b_{i}$, we obtain%
\[
\dfrac{1}{a_{j}-b_{i}}\prod_{\ell\in\left\{  1,2,...,n\right\}  }\left(
a_{j}-b_{\ell}\right)  =\prod_{\ell\in\left\{  1,2,...,n\right\}
\setminus\left\{  i\right\}  }\left(  a_{j}-b_{\ell}\right)  .
\]
Thus,%
\[
\prod_{\ell\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}
}\left(  a_{j}-b_{\ell}\right)  =\dfrac{1}{a_{j}-b_{i}}\prod_{\ell\in\left\{
1,2,...,n\right\}  }\left(  a_{j}-b_{\ell}\right)  =\left(  \prod_{\ell
\in\left\{  1,2,...,n\right\}  }\left(  a_{j}-b_{\ell}\right)  \right)
\cdot\dfrac{1}{a_{j}-b_{i}}.
\]
This proves \eqref{sol.cauchy.cauchy.a2.1}.}.

Thus,
\begin{align*}
&  \det\left(  \left(  \underbrace{\prod_{\ell\in\left\{  1,2,...,n\right\}
\setminus\left\{  i\right\}  }\left(  a_{j}-b_{\ell}\right)  }%
_{\substack{=\left(  \prod_{\ell\in\left\{  1,2,...,n\right\}  }\left(
a_{j}-b_{\ell}\right)  \right)  \cdot\dfrac{1}{a_{j}-b_{i}}\\\text{(by
\eqref{sol.cauchy.cauchy.a2.1})}}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\det\left(  \left(  \left(  \prod_{\ell\in\left\{  1,2,...,n\right\}
}\left(  a_{j}-b_{\ell}\right)  \right)  \cdot\dfrac{1}{a_{j}-b_{i}}\right)
_{i,j=1,2,...,n}\right) \\
&  =\underbrace{\left(  \prod_{j=1}^{n}\prod_{\ell\in\left\{
1,2,...,n\right\}  }\left(  a_{j}-b_{\ell}\right)  \right)  }%
_{\substack{=\prod_{i=1}^{n}\prod_{\ell\in\left\{  1,2,...,n\right\}  }\left(
a_{i}-b_{\ell}\right)  \\\text{(here, we renamed the index }j\text{ as
}i\\\text{in the outer product)}}}\cdot\underbrace{\det\left(  \left(
\dfrac{1}{a_{j}-b_{i}}\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=\det
\left(  \left(  \dfrac{1}{a_{i}-b_{j}}\right)  _{i,j=1,2,...,n}\right)
\\\text{(by \eqref{sol.cauchy.cauchy.det.1}, applied}\\\text{to }m=n\text{ and
}\alpha_{i,j}=\dfrac{1}{a_{j}-b_{i}}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.cauchy.det.15},
applied to }m=n\text{, }\alpha_{i,j}=\dfrac{1}{a_{j}-b_{i}}\text{ and }%
\delta_{j}=\prod_{\ell\in\left\{  1,2,...,n\right\}  }\left(  a_{j}-b_{\ell
}\right)  \right) \\
&  =\left(  \underbrace{\prod_{i=1}^{n}}_{=\prod_{i\in\left\{
1,2,...,n\right\}  }}\underbrace{\prod_{\ell\in\left\{  1,2,...,n\right\}
}\left(  a_{i}-b_{\ell}\right)  }_{\substack{=\prod_{j\in\left\{
1,2,...,n\right\}  }\left(  a_{i}-b_{j}\right)  \\\text{(here, we renamed the
index }\ell\\\text{as }j\text{ in the product)}}}\right)  \cdot\det\left(
\left(  \dfrac{1}{a_{i}-b_{j}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\left(  \underbrace{\prod_{i\in\left\{  1,2,...,n\right\}  }\prod
_{j\in\left\{  1,2,...,n\right\}  }}_{=\prod_{\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}}}\left(  a_{i}-b_{j}\right)  \right)  \cdot\det\left(
\left(  \dfrac{1}{a_{i}-b_{j}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}%
}\left(  a_{i}-b_{j}\right)  \right)  \cdot\det\left(  \left(  \dfrac{1}%
{a_{i}-b_{j}}\right)  _{i,j=1,2,...,n}\right)  ,
\end{align*}
so that%
\begin{align*}
&  \left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}%
}\left(  a_{i}-b_{j}\right)  \right)  \cdot\det\left(  \left(  \dfrac{1}%
{a_{i}-b_{j}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\det\left(  \left(  \prod_{\ell\in\left\{  1,2,...,n\right\}
\setminus\left\{  i\right\}  }\left(  a_{j}-b_{\ell}\right)  \right)
_{i,j=1,2,...,n}\right)  =\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}%
-a_{j}\right)  \left(  b_{j}-b_{i}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.cauchy.a2.use}}\right)
.
\end{align*}
Dividing both sides of this equality by $\prod_{\left(  i,j\right)
\in\left\{  1,2,...,n\right\}  ^{2}}\left(  a_{i}-b_{j}\right)  $, we obtain%
\[
\det\left(  \left(  \dfrac{1}{a_{i}-b_{j}}\right)  _{i,j=1,2,...,n}\right)
=\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}-a_{j}\right)  \left(
b_{j}-b_{i}\right)  \right)  }{\prod_{\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}}\left(  a_{i}-b_{j}\right)  }.
\]
Thus, Exercise~\ref{exe.cauchy.cauchy}(a) is solved once again.

\textit{Third solution of Exercise~\ref{exe.cauchy.cauchy}(a) (sketched):} The
statement of Exercise~\ref{exe.cauchy.cauchy}(a) is an identity between two
rational functions in the variables $a_{1},a_{2},...,a_{n},b_{1}
,b_{2},...,b_{n}$, and is easily seen to be equivalent to a polynomial
identity in these variables (by multiplying both sides through with the common
denominator $\prod_{\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}%
}\left(  a_{i}-b_{j}\right)  $). It is well-known that such identities need
only be checked on complex numbers to ensure that they hold for any elements
of any ring. So we only need to prove the statement of Exercise
\ref{exe.cauchy.cauchy}(a) in the case when $A=\CC$. But the statement
is well-known in this case (see, e.g., \cite[Lemma 5.15.3 (Lemma 4.48 in the
arXiv version)]{Etingof-et-al} and various other sources for the proof).

(b) \textit{First solution of Exercise~\ref{exe.cauchy.cauchy}(b):} Assume
that $1-a_{i}b_{j}$ is an invertible element of $A$ for every $i\in\left\{
1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $.

We know that $1-a_{i}b_{j}$ is an invertible element of $A$ for every
$i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $. In
other words, $1\cdot1-a_{i}b_{j}$ is an invertible element of $A$ for every
$i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{  1,2,...,n\right\}  $
(since $1\cdot1=1$). Hence, Proposition \ref{prop.sol.cauchy.cauchy} (applied
to $1$, $a_{i}$, $b_{i}$ and $1$ instead of $a_{i}$, $b_{i}$, $c_{i}$ and
$d_{i}$) yields%
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{1\cdot1-a_{i}\cdot b_{j}}\right)
_{i,j=1,2,\ldots,n}\right) \\
&  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  1\cdot a_{j}-1\cdot
a_{i}\right)  \left(  b_{j}\cdot1-b_{i}\cdot1\right)  \right)  }%
{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(
1\cdot1-a_{i}b_{j}\right)  }\\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  \underbrace{1\cdot1}_{=1}-a_{i}b_{j}\right)  \right)  ^{-1}\cdot
\prod_{1\leq j<i\leq n}\left(  \left(  \underbrace{1\cdot a_{j}}_{=a_{j}%
}-\underbrace{1\cdot a_{i}}_{=a_{i}}\right)  \left(  \underbrace{b_{j}\cdot
1}_{=b_{j}}-\underbrace{b_{i}\cdot1}_{=b_{i}}\right)  \right) \\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  1-a_{i}b_{j}\right)  \right)  ^{-1}\cdot\prod_{1\leq j<i\leq
n}\left(  \underbrace{\left(  a_{j}-a_{i}\right)  }_{=-\left(  a_{i}%
-a_{j}\right)  }\underbrace{\left(  b_{j}-b_{i}\right)  }_{=-\left(
b_{i}-b_{j}\right)  }\right) \\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  1-a_{i}b_{j}\right)  \right)  ^{-1}\cdot\prod_{1\leq j<i\leq
n}\left(  \underbrace{\left(  -\left(  a_{i}-a_{j}\right)  \right)  \left(
-\left(  b_{i}-b_{j}\right)  \right)  }_{=\left(  a_{i}-a_{j}\right)  \left(
b_{i}-b_{j}\right)  }\right) \\
&  =\left(  \prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
}\left(  1-a_{i}b_{j}\right)  \right)  ^{-1}\cdot\prod_{1\leq j<i\leq
n}\left(  \left(  a_{i}-a_{j}\right)  \left(  b_{i}-b_{j}\right)  \right)
=\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  a_{i}-a_{j}\right)  \left(
b_{i}-b_{j}\right)  \right)  }{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}}\left(  1-a_{i}b_{j}\right)  }.
\end{align*}
This rewrites as $\det\left(  \left(  \dfrac{1}{1-a_{i}b_{j}}\right)
_{i,j=1,2,\ldots,n}\right)  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(
a_{i}-a_{j}\right)  \left(  b_{i}-b_{j}\right)  \right)  }{\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\left(  1-a_{i}%
b_{j}\right)  }$ (because $1\cdot1=1$). This solves Exercise
\ref{exe.cauchy.cauchy}(b).

\textit{Second solution of Exercise~\ref{exe.cauchy.cauchy}(b):} Just as in
the Third solution of Exercise~\ref{exe.cauchy.cauchy}(a), we can see that it
is enough to solve Exercise~\ref{exe.cauchy.cauchy}(b) in the case when
$A=\CC$. But in this case, the statement of this exercise is
well-known, and proven, e.g., in \cite[Corollary 5.15.4 (Corollary 4.49 in
the arXiv version)]{Etingof-et-al} and
\cite[Cauchy's lemma, p. 18]{CrawleyBoevey}.

\textit{Remark:} One could also derive the statement of Exercise
\ref{exe.cauchy.cauchy}(b) from Exercise~\ref{exe.cauchy.cauchy}(a) by
applying the latter to $1/a_{i}$ instead of $a_{i}$, after first WLOG assuming
that the $a_{i}$ are invertible (but one needs to justify this WLOG assumption).

(c) \textit{Alternative proof of Theorem \ref{thm.cauchy-identity}.} We are
going to show that, for every $n\in\NN$, we have%
\begin{equation}
\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}=\sum
\limits_{\lambda\in \Par }s_{\lambda}\left(  x_{1},x_{2}%
,...,x_{n}\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)
\label{sol.cauchy.cauchy.c.2}%
\end{equation}
in the ring $\kk \left[  \left[  x_{1},x_{2},...,x_{n},y_{1}%
,y_{2},...,y_{n}\right]  \right]  $. Once this is proven, Theorem
\ref{thm.cauchy-identity} will easily follow.\footnote{\textit{Proof.} Assume
that we have proven that \eqref{sol.cauchy.cauchy.c.2} holds for every
$n\in\NN$.
\par
Let $\kk \left[  \left[  \xx , \yy \right]  \right]  $ denote
the ring of power series $\kk \left[  \left[  x_{1},x_{2},x_{3}%
,...,y_{1},y_{2},y_{3},...\right]  \right]  $. For every $n\in\NN$, we
embed the ring $\kk \left[  \left[  x_{1},x_{2},...,x_{n},y_{1}%
,y_{2},...,y_{n}\right]  \right]  $ in this ring $\kk \left[  \left[
x_{1},x_{2},x_{3},...,y_{1},y_{2},y_{3},...\right]  \right]  =\kk %
\left[  \left[  \xx , \yy \right]  \right]  $.
\par
Let us regard the identity \eqref{Cauchy-identity} as an equality in this ring
$\kk \left[  \left[  \xx , \yy \right]  \right]  $. For every
monomial $\xx ^{\mathbf{i}} \yy ^{\mathbf{j}}$ in the variables
$\left(  \xx , \yy \right)  =\left(  x_{1},x_{2},x_{3}%
,...,y_{1},y_{2},y_{3},...\right)  $, we define the \emph{skew degree} of
$\xx ^{\mathbf{i}} \yy ^{\mathbf{j}}$ as the sum $\left(
1i_{1}+2i_{2}+3i_{3}+...\right)  +\left(  1j_{1}+2j_{2}+3j_{3}+...\right)  $,
where the weak compositions $\mathbf{i}$ and $\mathbf{j}$ are written in the
forms $\mathbf{i}=\left(  i_{1},i_{2},i_{3},...\right)  $ and $\mathbf{j}%
=\left(  j_{1},j_{2},j_{3},...\right)  $. For every $n\in\NN$, let
$Q_{n}$ be the subset%
\[
\left\{  f\in\kk \left[  \left[  \xx , \yy \right]  \right]
\ \mid\ \text{the coefficient of }f\text{ before any monomial of skew degree
}<n\text{ equals }0\right\}
\]
of $\kk \left[  \left[  \xx , \yy \right]  \right]  $. This
$Q_{n}$ is an ideal of $\kk \left[  \left[  \xx , \yy %
\right]  \right]  $ for every $n\in\NN$.
\par
Notice that $Q_{0} \subset Q_{1} \subset Q_{2} \subset...$.
\par
We define a topology on $\kk \left[  \left[  \xx , \yy %
\right]  \right]  $ by proclaiming that a subset $U$ of $\kk \left[
\left[  \xx , \yy \right]  \right]  $ be open if and only if for
every $f\in U$ there exists an $n\in\NN$ satisfying $f+Q_{n}\subset U$.
It is easy to see that this topology makes $\kk \left[  \left[
\xx , \yy \right]  \right]  $ into a complete and Hausdorff
topological ring. For every $n\in\NN$, the ideal $Q_{n}$ is closed in
this topology. (Moreover, the polynomial ring $\kk \left[
\xx , \yy \right]  :=\kk \left[  x_{1},x_{2},x_{3}%
,...,y_{1},y_{2},y_{3},...\right]  $ is dense in $\kk \left[  \left[
\xx , \yy \right]  \right]  $. We will not use this, however.)
\par
Every $n\in\NN$ satisfies
\begin{align*}
\prod\limits_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}
&= \prod\limits_{i=1}^{\infty} \prod\limits_{j=1}^{\infty} \left(  1-x_{i}y_{j}\right)  ^{-1}
= \left(\prod\limits_{i=1}^{n} \prod\limits_{j=1}^{\infty} \left(  1-x_{i}y_{j}\right)  ^{-1}\right)
 \underbrace{\left(
\prod\limits_{i=n+1}^{\infty}\prod\limits_{j=1}^{\infty}\left(  1-x_{i}%
y_{j}\right)  ^{-1}\right)  }_{\substack{\equiv1\operatorname{mod}%
Q_{n}\\\text{(since every }i\in\left\{  n+1,n+2,n+3,...\right\}  \text{ and
}j\in\left\{  1,2,3,,...\right\}  \\\text{satisfy }x_{i}\in Q_{n}\text{, thus
}x_{i}\equiv0\operatorname{mod}Q_{n}\\\text{and thus }1-x_{i}y_{j}%
\equiv1\operatorname{mod}Q_{n}\text{)}}} \\
&\equiv \prod\limits_{i=1}^{n} \underbrace{
 \prod\limits_{j=1}^{\infty} \left(  1-x_{i}y_{j}\right)  ^{-1} }
_{= \left( \prod\limits_{j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
 \cdot \left( \prod\limits_{j=n+1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
} \\
&= \prod\limits_{i=1}^n \left(
 \left( \prod\limits_{j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
\cdot\underbrace{\left(  \prod\limits_{j=n+1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)  }_{\substack{\equiv
1\operatorname{mod}Q_{n}\\\text{(since every }
j\in\left\{  n+1,n+2,n+3,...\right\}  \\\text{satisfies }y_{j}\in
Q_{n}\text{, thus }y_{j}\equiv0\operatorname{mod}Q_{n}\\\text{and thus
}1-x_{i}y_{j}\equiv1\operatorname{mod}Q_{n}\text{)}}}\right) \\
&  \equiv \prod\limits_{i=1}^n \prod\limits_{j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}
= \prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}
\operatorname{mod}Q_{n}.
\end{align*}
Thus,
\[
\prod\limits_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}=\lim
\limits_{n\rightarrow\infty}\prod\limits_{i,j=1}^{n}\left(  1-x_{i}%
y_{j}\right)  ^{-1}%
\]
(where limits are taken with respect to our topology).
\par
Furthermore, every $n\in\NN$ and $\lambda \in \Par$
satisfy
\begin{align*}
s_{\lambda}\left(  \xx \right)   &  =s_{\lambda}\left(  x_{1}
,x_{2},x_{3},...\right)  =s_{\lambda}\left(  x_{1},x_{2},...,x_{n}
,x_{n+1},x_{n+2},x_{n+3},...\right)
\equiv s_{\lambda}\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in\left\{
n+1,n+2,n+3,...\right\}  \text{ satisfies }x_{i}\in Q_{n}\text{, thus }
x_{i}\equiv0\operatorname{mod}Q_{n}\right) \\
&  =s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
\operatorname{mod} Q_{n}
\end{align*}
and similarly $s_{\lambda}\left(   \yy \right)  \equiv s_{\lambda}\left(
y_{1},y_{2},...,y_{n}\right)  \operatorname{mod}Q_{n}$. Hence, every
$n\in\NN$ satisfies
\[
\sum\limits_{\lambda\in \Par }\underbrace{s_{\lambda}\left(
\xx \right)  }_{\equiv s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
\operatorname{mod} Q_n}
\underbrace{s_{\lambda}\left(   \yy \right)  }_{\equiv
s_{\lambda}\left( y_{1},y_{2},...,y_{n}\right) \operatorname{mod} Q_n}
\equiv\sum\limits_{\lambda\in
 \Par }s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  \operatorname{mod}Q_{n}.
\]
Thus,
\[
\sum\limits_{\lambda\in \Par} s_{\lambda}\left( \xx \right)
 s_{\lambda}\left( \yy \right)
= \lim\limits_{n\rightarrow \infty}
 \sum\limits_{\lambda\in \Par}
 s_{\lambda}\left( x_{1},x_{2},...,x_{n} \right)
 s_{\lambda}\left( y_{1},y_{2},...,y_{n} \right)
.
\]
\par
Now,
\begin{align*}
\prod\limits_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}  &
=\lim\limits_{n\rightarrow\infty}\underbrace{\prod\limits_{i,j=1}^{n}\left(
1-x_{i}y_{j}\right)  ^{-1}}_{\substack{=\sum\limits_{\lambda\in
 \Par }s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  \\\text{(by
\eqref{sol.cauchy.cauchy.c.2} (since we assumed that
\eqref{sol.cauchy.cauchy.c.2}}\\\text{is already proven))}}}\\
&  =\lim\limits_{n\rightarrow\infty}\sum\limits_{\lambda\in \Par %
}s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  s_{\lambda}\left(
y_{1},y_{2},...,y_{n}\right)  =\sum\limits_{\lambda\in \Par %
}s_{\lambda}\left(  \xx \right)  s_{\lambda}\left(   \yy \right)  ,
\end{align*}
so that Theorem \ref{thm.cauchy-identity} is proven.
\par
(Alternatively, we could have derived Theorem \ref{thm.cauchy-identity} from
\eqref{sol.cauchy.cauchy.c.2} by comparison of coefficients, choosing for
every coefficient an appropriate $n\in\NN$ to apply
\eqref{sol.cauchy.cauchy.c.2} to. But this would be exactly the above proof,
just written in a less slick way.)}

So let $n\in\NN$.

For every $i\in\left\{  1,2,...,n\right\}  $ and $j\in\left\{
1,2,...,n\right\}  $, the element $1-x_{i}y_{j}$ of the ring $\kk %
\left[  \left[  x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}\right]  \right]  $
is invertible (because this element is a power series with constant term $1$,
and every such power series is invertible). Hence, Exercise
\ref{exe.cauchy.cauchy}(b) (applied to $A=\kk \left[  \left[
x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}\right]  \right]  $, $a_{i}=x_{i}$
and $b_{j}=y_{j}$) yields%
\begin{align}
\det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
&  =\dfrac{\prod_{1\leq j<i\leq n}\left(  \left(  x_{i}-x_{j}\right)  \left(
y_{i}-y_{j}\right)  \right)  }{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}}\left(  1-x_{i}y_{j}\right)  }\nonumber\\
&  =\underbrace{\dfrac{1}{\prod_{\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}}\left(  1-x_{i}y_{j}\right)  }}_{=\prod_{\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}\dfrac{1}{1-x_{i}y_{j}}%
}\cdot\prod_{1\leq j<i\leq n}\left(  \underbrace{\left(  x_{i}-x_{j}\right)
}_{=-\left(  x_{j}-x_{i}\right)  }\underbrace{\left(  y_{i}-y_{j}\right)
}_{=-\left(  y_{j}-y_{i}\right)  }\right) \nonumber\\
&  =\left(  \underbrace{\prod_{\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}}}_{=\prod\limits_{i,j=1}^{n}}\dfrac{1}{1-x_{i}y_{j}}\right)
\cdot\prod_{1\leq j<i\leq n}\left(  \underbrace{\left(  -\left(  x_{j}%
-x_{i}\right)  \right)  \left(  -\left(  y_{j}-y_{i}\right)  \right)
}_{=\left(  x_{j}-x_{i}\right)  \left(  y_{j}-y_{i}\right)  }\right)
\nonumber\\
&  =\left(  \prod\limits_{i,j=1}^{n}\dfrac{1}{1-x_{i}y_{j}}\right)
\cdot\underbrace{\prod_{1\leq j<i\leq n}\left(  \left(  x_{j}-x_{i}\right)
\left(  y_{j}-y_{i}\right)  \right)  }_{\substack{=\prod_{1\leq i<j\leq
n}\left(  \left(  x_{i}-x_{j}\right)  \left(  y_{i}-y_{j}\right)  \right)
\\\text{(here, we renamed the index }\left(  j,i\right)  \\\text{as }\left(
i,j\right)  \text{ in the product)}}}\nonumber\\
&  =\left(  \prod\limits_{i,j=1}^{n}\dfrac{1}{1-x_{i}y_{j}}\right)
\cdot\underbrace{\prod_{1\leq i<j\leq n}\left(  \left(  x_{i}-x_{j}\right)
\left(  y_{i}-y_{j}\right)  \right)  }_{=\left(  \prod\limits_{1\leq i<j\leq
n}\left(  x_{i}-x_{j}\right)  \right)  \cdot\left(  \prod\limits_{1\leq
i<j\leq n}\left(  y_{i}-y_{j}\right)  \right)  }\nonumber\\
&  =\left(  \prod\limits_{i,j=1}^{n}\dfrac{1}{1-x_{i}y_{j}}\right)
\cdot\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)
\right)  \cdot\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}%
-y_{j}\right)  \right) \nonumber\\
&  =\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)  \right)
\cdot\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}-y_{j}\right)
\right)  \cdot\prod\limits_{i,j=1}^{n}\dfrac{1}{1-x_{i}y_{j}}
\label{sol.cauchy.cauchy.c.cauchy}
\end{align}
in the ring
$\kk \left[ \left[ x_1, x_2, ..., x_n, y_1, y_2, ..., y_n \right] \right]$.

We will use the notations of Definition~\ref{def.alternant} and
Proposition~\ref{prop.alternants.space} (but we do not require $\kk$ to
be $\ZZ$ or a field of characteristic not equal to $2$ as was done in
Proposition~\ref{prop.alternants.space}).
We have
$a_{\rho} = \prod\limits_{1\leq i<j\leq n} \left( x_i - x_j \right)$
(as proven in the proof of Proposition~\ref{prop.alternants.space}).
Thus, $a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right) 
 =\prod\limits_{1\leq i<j\leq n}\left(  x_{i}-x_{j}\right)$
and $a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  =\prod\limits_{1\leq
i<j\leq n}\left(  y_{i}-y_{j}\right)  $. Thus,
\eqref{sol.cauchy.cauchy.c.cauchy} becomes
\begin{align}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
\nonumber\\
&  =\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}%
-x_{j}\right)  \right)  }_{=a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)
}\cdot\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}%
-y_{j}\right)  \right)  }_{=a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}\cdot\prod\limits_{i,j=1}^{n}\underbrace{\dfrac{1}{1-x_{i}y_{j}}}_{=\left(
1-x_{i}y_{j}\right)  ^{-1}}\nonumber\\
&  =a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho}\left(
y_{1},y_{2},...,y_{n}\right)  \cdot\prod\limits_{i,j=1}^{n}\left(
1-x_{i}y_{j}\right)  ^{-1}.
\label{sol.cauchy.cauchy.c.cauchy2}
\end{align}


On the other hand, \eqref{sol.cauchy.cauchy.det.3} (applied to $m=n$ and
$\alpha_{i,j}=\dfrac{1}{1-x_{i}y_{j}}$) yields
\begin{align}
\det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod
\limits_{i=1}^{n}\underbrace{\dfrac{1}{1-x_{i}y_{\sigma\left(  i\right)  }}%
}_{\substack{=\sum\limits_{k\in\NN}\left(  x_{i}y_{\sigma\left(
i\right)  }\right)  ^{k}\\\text{(by the formula for the geometric series)}%
}}\nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
\limits_{i=1}^{n}\sum\limits_{k\in\NN}\left(  x_{i}y_{\sigma\left(
i\right)  }\right)  ^{k}}_{\substack{=\sum\limits_{\left(  k_{1}%
,k_{2},...,k_{n}\right)  \in\NN ^{n}}\left(  x_{1}y_{\sigma\left(
1\right)  }\right)  ^{k_{1}}\left(  x_{2}y_{\sigma\left(  2\right)  }\right)
^{k_{2}}...\left(  x_{n}y_{\sigma\left(  n\right)  }\right)  ^{k_{n}%
}\\\text{(by the product rule)}}}\nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}\left(
x_{1}y_{\sigma\left(  1\right)  }\right)  ^{k_{1}}\left(  x_{2}y_{\sigma
\left(  2\right)  }\right)  ^{k_{2}}...\left(  x_{n}y_{\sigma\left(  n\right)
}\right)  ^{k_{n}}\nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}
\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\left(
x_{1}y_{\sigma\left(  1\right)  }\right)  ^{k_{1}}\left(  x_{2}y_{\sigma
\left(  2\right)  }\right)  ^{k_{2}}...\left(  x_{n}y_{\sigma\left(  n\right)
}\right)  ^{k_{n}}}_{\substack{=\left(  x_{1}^{k_{1}}y_{\sigma\left(
1\right)  }^{k_{1}}\right)  \left(  x_{2}^{k_{2}}y_{\sigma\left(  2\right)
}^{k_{2}}\right)  ...\left(  x_{n}^{k_{n}}y_{\sigma\left(  n\right)  }^{k_{n}%
}\right)  \\=\left(  x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\right)
\left(  y_{\sigma\left(  1\right)  }^{k_{1}}y_{\sigma\left(  2\right)
}^{k_{2}}...y_{\sigma\left(  n\right)  }^{k_{n}}\right)  }}\nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}%
\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\left(  x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\right)  \left(  y_{\sigma\left(  1\right)
}^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(  n\right)
}^{k_{n}}\right) \nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}
x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\sum\limits_{\sigma\in S_{n}}\left(
-1\right)  ^{\sigma}y_{\sigma\left(  1\right)  }^{k_{1}}y_{\sigma\left(
2\right)  }^{k_{2}}...y_{\sigma\left(  n\right)  }^{k_{n}}.
\label{sol.cauchy.cauchy.c.6}
\end{align}
But every $\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}$ satisfies
\begin{align}
\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)   &
=\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
\limits_{i=1}^{n}y_{\sigma\left(  i\right)  }^{k_{i}}}_{=y_{\sigma\left(
1\right)  }^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(
n\right)  }^{k_{n}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.cauchy.det.3}, applied to }m=n\text{ and }\alpha_{i,j}%
=y_{j}^{k_{i}}\right) \nonumber\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}y_{\sigma\left(
1\right)  }^{k_{1}}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(
n\right)  }^{k_{n}}.
\label{sol.cauchy.cauchy.c.ydet}
\end{align}
Hence, \eqref{sol.cauchy.cauchy.c.6} becomes
\begin{align}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
\nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}
x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\underbrace{\sum\limits_{\sigma\in
S_{n}}\left(  -1\right)  ^{\sigma}y_{\sigma\left(  1\right)  }^{k_{1}%
}y_{\sigma\left(  2\right)  }^{k_{2}}...y_{\sigma\left(  n\right)  }^{k_{n}}%
}_{\substack{=\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  \\\text{(by \eqref{sol.cauchy.cauchy.c.ydet})}}}\nonumber\\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}
x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}}%
...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  k_{1},k_{2}%
,...,k_{n}\right)  \in\NN ^{n};\\\text{the integers }k_{1}\text{, }%
k_{2}\text{, }...\text{, }k_{n}\\\text{are not distinct}}%
}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\underbrace{\det\left(  \left(
y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=0\\\text{(because
the matrix }\left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\\\text{has two
equal rows (since the integers}\\k_{1}\text{, }k_{2}\text{, }...\text{, }%
k_{n}\text{ are not distinct,}\\\text{i. e., there are two equal
among them)) }}}\nonumber\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}}%
...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  +\underbrace{\sum\limits_{\substack{\left(  k_{1},k_{2}%
,...,k_{n}\right)  \in\NN ^{n};\\\text{the integers }k_{1}\text{, }%
k_{2}\text{, }...\text{, }k_{n}\\\text{are not distinct}}%
}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}}%
...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  . \label{sol.cauchy.cauchy.c.8}%
\end{align}


On the other hand, every $\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n}$ satisfies
\begin{align*}
\det\left(  \left(  x_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)   &
=\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
\limits_{i=1}^{n}x_{\sigma\left(  i\right)  }^{k_{i}}}_{=x_{\sigma\left(
1\right)  }^{k_{1}}x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(
n\right)  }^{k_{n}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.cauchy.det.3}, applied to } m = n \text{ and }\alpha
_{i,j}=x_{j}^{k_{i}}\right) \\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}x_{\sigma\left(
1\right)  }^{k_{1}}x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(
n\right)  }^{k_{n}}.
\end{align*}
Thus,
\begin{align*}
&  \sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\k_{1}>k_{2}>...>k_{n}}}\underbrace{\det\left(  \left(
x_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)  }_{=\sum\limits_{\sigma\in
S_{n}}\left(  -1\right)  ^{\sigma}x_{\sigma\left(  1\right)  }^{k_{1}%
}x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(  n\right)  }^{k_{n}}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\k_{1}>k_{2}>...>k_{n}}}\sum\limits_{\sigma\in S_{n}}\left(
-1\right)  ^{\sigma}x_{\sigma\left(  1\right)  }^{k_{1}}x_{\sigma\left(
2\right)  }^{k_{2}}...x_{\sigma\left(  n\right)  }^{k_{n}}\det\left(  \left(
y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{\sum
\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n};
\\k_{1}>k_{2}>...>k_{n}}}x_{\sigma\left(  1\right)  }^{k_{1}}%
x_{\sigma\left(  2\right)  }^{k_{2}}...x_{\sigma\left(  n\right)  }^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right)
}_{\substack{=\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in\NN ^{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)
}>...>k_{\sigma\left(  n\right)  }}}x_{\sigma\left(  1\right)  }%
^{k_{\sigma\left(  1\right)  }}x_{\sigma\left(  2\right)  }^{k_{\sigma\left(
2\right)  }}...x_{\sigma\left(  n\right)  }^{k_{\sigma\left(  n\right)  }}%
\det\left(  \left(  y_{j}^{k_{\sigma\left(  i\right)  }}\right)
_{i,j=1,2,...,n}\right)  \\\text{(here, we substituted }\left(  k_{\sigma
\left(  1\right)  },k_{\sigma\left(  2\right)  },...,k_{\sigma\left(
n\right)  }\right)  \text{ for }\left(  k_{1},k_{2},...,k_{n}\right)  \text{
in the sum}\\\text{(this is allowed, since }\sigma\text{ is a permutation))}%
}}\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n};
\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }%
>...>k_{\sigma\left(  n\right)  }}}\underbrace{x_{\sigma\left(  1\right)
}^{k_{\sigma\left(  1\right)  }}x_{\sigma\left(  2\right)  }^{k_{\sigma\left(
2\right)  }}...x_{\sigma\left(  n\right)  }^{k_{\sigma\left(  n\right)  }}%
}_{\substack{=x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\\\text{(since }%
\sigma\text{ is a permutation)}}}\underbrace{\det\left(  \left(
y_{j}^{k_{\sigma\left(  i\right)  }}\right)  _{i,j=1,2,...,n}\right)
}_{\substack{=\left(  -1\right)  ^{\sigma}\det\left(  \left(  y_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right)  \\\text{(since permuting the rows of a
matrix}\\\text{multiplies its determinant by the sign }\\\text{of the
permutation)}}}\\
&  =\sum\limits_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum
\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n};
\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }%
>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}%
}\left(  -1\right)  ^{\sigma}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right) \\
&  =\sum\limits_{\sigma\in S_{n}}\underbrace{\left(  -1\right)  ^{\sigma
}\left(  -1\right)  ^{\sigma}}_{=\left(  \left(  -1\right)  ^{\sigma}\right)
^{2}=1}\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)
\in\NN ^{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)
}>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right) \\
&  =\underbrace{\sum\limits_{\sigma\in S_{n}}\sum\limits_{\substack{\left(
k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}}%
_{=\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}%
\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}}x_{1}%
^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right) \\
&  =\sum\limits_{\left(  k_{1},k_{2},...,k_{n}\right)  \in\NN ^{n}}%
\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right) \\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}\sum\limits_{\substack{\sigma\in
S_{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }%
>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  k_{1},k_{2}%
,...,k_{n}\right)  \in\NN ^{n};\\\text{the integers }k_{1}\text{, }%
k_{2}\text{, }...\text{, }k_{n}\\\text{are not distinct}}%
}\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\underbrace{\det\left(  \left(  y_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=0\\\text{(because the matrix
}\left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\\\text{has two equal rows
(since the integers}\\\text{ }k_{1}\text{, }k_{2}\text{, }...\text{, }%
k_{n}\text{ are not distinct,}\\\text{i. e., there are two equal
among them)) }}}
\end{align*}%
\begin{align*}
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}\sum\limits_{\substack{\sigma\in
S_{n};\\k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }%
>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}}x_{2}^{k_{2}}...x_{n}^{k_{n}%
}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{\left(  k_{1}%
,k_{2},...,k_{n}\right)  \in\NN ^{n};\\\text{the integers }k_{1}\text{,
}k_{2}\text{, }...\text{, }k_{n}\\\text{are not distinct}}%
}\sum\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)
}>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}0}_{=0}\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}\underbrace{\sum
\limits_{\substack{\sigma\in S_{n};\\k_{\sigma\left(  1\right)  }%
>k_{\sigma\left(  2\right)  }>...>k_{\sigma\left(  n\right)  }}}x_{1}^{k_{1}%
}x_{2}^{k_{2}}...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right)  }_{\substack{=x_{1}^{k_{1}}x_{2}^{k_{2}}%
...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  \\\text{(because there exists exactly one }\sigma\in S_{n}\text{ such
that }k_{\sigma\left(  1\right)  }>k_{\sigma\left(  2\right)  }>...>k_{\sigma
\left(  n\right)  }\\\text{(since the integers }k_{1}\text{, }k_{2}\text{,
}...\text{, }k_{n}\text{ are distinct, and thus}\\\text{there is
exactly one permutation which arranges these integers}\\\text{in decreasing
order))}}}\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\\text{the integers }k_{1}\text{, }k_{2}\text{, }...\text{,
}k_{n}\\\text{are distinct}}}x_{1}^{k_{1}}x_{2}^{k_{2}}%
...x_{n}^{k_{n}}\det\left(  \left(  y_{j}^{k_{i}}\right)  _{i,j=1,2,...,n}%
\right)  .
\end{align*}
Compared with \eqref{sol.cauchy.cauchy.c.8}, this yields
\begin{align}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
\nonumber\\
&  =\sum\limits_{\substack{\left(  k_{1},k_{2},...,k_{n}\right)  \in
\NN ^{n};\\k_{1}>k_{2}>...>k_{n}}}\det\left(  \left(  x_{j}^{k_{i}%
}\right)  _{i,j=1,2,...,n}\right)  \det\left(  \left(  y_{j}^{k_{i}}\right)
_{i,j=1,2,...,n}\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  \lambda_{1},\lambda
_{2},...,\lambda_{n}\right)  \in\NN ^{n};\\\lambda_{1}\geq\lambda
_{2}\geq...\geq\lambda_{n}}}}_{\substack{=\sum\limits_{\substack{\left(
\lambda_{1},\lambda_{2},...,\lambda_{n}\right)  \in\NN ^{n}\\\text{is a
partition with at most }n\text{ parts}}}\\=\sum\limits_{\substack{\lambda
=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  \in\NN ^{n}
\\\text{is a partition with at most }n\text{ parts}}}}}\det\left(  \left(
x_{j}^{\lambda_{i}+n-i}\right)  _{i,j=1,2,...,n}\right)  \det\left(  \left(
y_{j}^{\lambda_{i}+n-i}\right)  _{i,j=1,2,...,n}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  \lambda_{1}+n-1,\lambda_{2}%
+n-2,...,\lambda_{n}+n-n\right)  \text{ for }\left(  k_{1},k_{2}%
,...,k_{n}\right) \\
\text{in the sum (since the map}\\
\left\{  \left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  \in
\NN ^{n}\ \mid\ \lambda_{1}\geq\lambda_{2}\geq...\geq\lambda
_{n}\right\}  \rightarrow\left\{  \left(  k_{1},k_{2},...,k_{n}\right)
\in\NN ^{n}\ \mid\ k_{1}>k_{2}>...>k_{n}\right\}  ,\\
\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  \mapsto\left(
\lambda_{1}+n-1,\lambda_{2}+n-2,...,\lambda_{n}+n-n\right) \\
\text{is a bijection)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{n}\right)  \in\NN ^{n}\\\text{is a partition with at most
}n\text{ parts}}}\det\left(  \left(  x_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  \det\left(  \left(  y_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  . \label{sol.cauchy.cauchy.c.10}%
\end{align}


But whenever $\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)
$ is a partition with at most $n$ parts, we have
\begin{align*}
\underbrace{\lambda}_{=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)
}+\underbrace{\rho}_{=\left(  n-1,n-2,...,0\right)  }  &  =\left(  \lambda
_{1},\lambda_{2},...,\lambda_{n}\right)  +\left(  n-1,n-2,...,0\right) \\
&  =\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,...,\lambda_{n}+n-n\right)
\end{align*}
and thus%
\begin{align*}
a_{\lambda+\rho}  &  =a_{\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,...,\lambda
_{n}+n-n\right)  }=\det\left(  \underbrace{\left(  x_{i}^{\lambda_{j}%
+n-j}\right)  _{i,j=1,2,...,n}}_{=\left(  \left(  x_{j}^{\lambda_{i}%
+n-i}\right)  _{i,j=1,2,...,n}\right)  ^{T}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)
\in\NN ^{n}\text{ satisfies }\\
a_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }=\det\left(  \left(
x_{i}^{\alpha_{j}}\right)  _{i,j=1,2,...,n}\right)  \text{ (by the definition
of }a_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }\text{)}%
\end{array}
\right) \\
&  =\det\left(  \left(  \left(  x_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  ^{T}\right)  =\det\left(  \left(  x_{j}^{\lambda
_{i}+n-i}\right)  _{i,j=1,2,...,n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the determinant of a matrix is
preserved under transposition}\right)  .
\end{align*}
Hence, we get the two equalities $a_{\lambda+\rho}\left(  x_{1},x_{2}%
,...,x_{n}\right)  =\det\left(  \left(  x_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  $ and $a_{\lambda+\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  =\det\left(  \left(  y_{j}^{\lambda_{i}+n-i}\right)
_{i,j=1,2,...,n}\right)  $ for any partition $\lambda=\left(  \lambda
_{1},\lambda_{2},...,\lambda_{n}\right)  $ with at most $n$ parts. Thus,
\eqref{sol.cauchy.cauchy.c.10} becomes
\begin{align*}
&  \det\left(  \left(  \dfrac{1}{1-x_{i}y_{j}}\right)  _{i,j=1,2,...,n}\right)
\\
&  =\sum\limits_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{n}\right)  \in\NN ^{n}\\\text{is a partition with at most
}n\text{ parts}}}\underbrace{\det\left(  \left(  x_{j}^{\lambda_{i}%
+n-i}\right)  _{i,j=1,2,...,n}\right)  }_{=a_{\lambda+\rho}\left(  x_{1}%
,x_{2},...,x_{n}\right)  }\underbrace{\det\left(  \left(  y_{j}^{\lambda
_{i}+n-i}\right)  _{i,j=1,2,...,n}\right)  }_{=a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  }\\
&  =\sum\limits_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2}%
,...,\lambda_{n}\right)  \in\NN ^{n}\\\text{is a partition with at most
}n\text{ parts}}}a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)
a_{\lambda+\rho}\left(  y_{1},y_{2},...,y_{n}\right) \\
&  =\sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  .
\end{align*}
Compared to \eqref{sol.cauchy.cauchy.c.cauchy}, this yields
\begin{align}
&  \sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right) \nonumber\\
&  =\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  x_{i}%
-x_{j}\right)  \right)  }_{=a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)
}\cdot\underbrace{\left(  \prod\limits_{1\leq i<j\leq n}\left(  y_{i}%
-y_{j}\right)  \right)  }_{=a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}\cdot\prod\limits_{i,j=1}^{n}\underbrace{\dfrac{1}{1-x_{i}y_{j}}}_{=\left(
1-x_{i}y_{j}\right)  ^{-1}}\nonumber\\
&  =a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho}\left(
y_{1},y_{2},...,y_{n}\right)  \cdot\prod\limits_{i,j=1}^{n}\left(
1-x_{i}y_{j}\right)  ^{-1}. \label{sol.cauchy.cauchy.c.13}%
\end{align}


Now, Corollary \ref{cor.stembridge.slambda} says that $s_{\lambda}\left(
x_{1},x_{2},...,x_{n}\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$ for every
partition $\lambda$ with at most $n$ parts. Thus, $s_{\lambda}\left(
x_{1},x_{2},...,x_{n}\right)  =\dfrac{a_{\lambda+\rho}\left(  x_{1}%
,x_{2},...,x_{n}\right)  }{a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  }$
and $s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  =\dfrac{a_{\lambda+\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  }$ for every partition $\lambda$ with at most $n$ parts.
Hence,%
\begin{align}
&  \sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}\underbrace{s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  }_{=\dfrac
{a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  }{a_{\rho}\left(
x_{1},x_{2},...,x_{n}\right)  }}\underbrace{s_{\lambda}\left(  y_{1}%
,y_{2},...,y_{n}\right)  }_{=\dfrac{a_{\lambda+\rho}\left(  y_{1}%
,y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  }%
}\nonumber\\
&  =\sum\limits_{\lambda\text{ is a partition with at most }n\text{ parts}%
}\dfrac{a_{\lambda+\rho}\left(  x_{1},x_{2},...,x_{n}\right)  }{a_{\rho
}\left(  x_{1},x_{2},...,x_{n}\right)  }\cdot\dfrac{a_{\lambda+\rho}\left(
y_{1},y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}\nonumber\\
&  =\dfrac{1}{a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }\underbrace{\sum\limits_{\lambda\text{
is a partition with at most }n\text{ parts}}a_{\lambda+\rho}\left(
x_{1},x_{2},...,x_{n}\right)  a_{\lambda+\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  }_{\substack{=a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)
\cdot a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  \cdot\prod\limits_{i,j=1}%
^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}\\\text{(by
\eqref{sol.cauchy.cauchy.c.13})}}}\nonumber\\
&  =\dfrac{1}{a_{\rho}\left(  x_{1},x_{2},...,x_{n}\right)  \cdot a_{\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }\cdot a_{\rho}\left(  x_{1}%
,x_{2},...,x_{n}\right)  \cdot a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)
\cdot\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}\nonumber\\
&  =\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\label{sol.cauchy.cauchy.c.16}%
\end{align}


We are almost done. Let us now notice that every partition $\lambda$ with more
than $n$ parts satisfies
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)
=0
\label{sol.cauchy.cauchy.c.overflow}
\end{equation}
(according to Exercise~\ref{exe.schur.finite}(b)).
%% The following argument has been commented out since it is old and not
%% as detailed as the one in the solution to Exercise~\ref{exe.schur.finite}(b):
%\ \ \ \ \footnote{\textit{Proof of \eqref{sol.cauchy.cauchy.c.overflow}:} Let
%$\lambda$ be a partition with more than $n$ parts. Then, the shape of
%$\lambda$ is a Ferrers diagram with $>n$ rows. Hence, the cell $\left(
%n+1,1\right)  $ (this is the cell in the $\left(  n+1\right)  $-th row and
%$1$-st column) belongs to the shape of $\lambda$.
%\par
%Let $T$ be a column-strict tableau of shape $\lambda$. Then, for every
%$i\in\NN$ and $j\in\NN$, the entry in cell $\left(  i,j\right)
%$ of $T$ must be $\geq i$ (if this entry exists). Applied to $i=n+1$ and
%$j=1$, this yields: The entry in cell $\left(  n+1,1\right)  $ of $T$ must be
%$\geq n+1$ (because we know that this entry exists). Let $u$ be this entry.
%Thus, $u\geq n+1$.
%\par
%Denote the monomial $\prod\limits_{\substack{i\in\left\{  1,2,3,...\right\}
%;\\i\neq u}}x_{i}^{\left(  \cont T\right)  _{i}}$ by $Q$.
%\par
%By the definition of $\cont T$, the number $\left(
%\cont T\right)  _{u}$ is the number of entries of $T$ equal to
%$u$. Since there exists at least one entry of $T$ equal to $u$ (because we
%defined $u$ as an entry of $T$), we thus have $\left(  \cont 
%T\right)  _{u}\geq1$. Thus,
%\[
%\xx ^{\cont T}=\prod\limits_{i\in\left\{
%1,2,3,...\right\}  }x_{i}^{\left(  \cont T\right)  _{i}}%
%=x_{u}^{\left(  \cont T\right)  _{u}}\underbrace{\prod
%\limits_{\substack{i\in\left\{  1,2,3,...\right\}  ;\\i\neq u}}x_{i}^{\left(
%\cont T\right)  _{i}}}_{=Q}=x_{u}^{\left(  \cont %
%T\right)  _{u}}Q.
%\]
%Evaluating this equation at $\xx =\left(  x_{1},x_{2},...,x_{n}%
%,0,0,0,...\right)  $, we obtain
%\begin{align*}
%\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)  ^{\cont T} &
%=\underbrace{0^{\left(  \cont T\right)  _{u}}}%
%_{\substack{=0\\\text{(since }\left(  \cont T\right)  _{u}
%\geq1\text{)}}}Q\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)  \\
%&  \ \ \ \ \ \ \ \ \ \ \left(
%\begin{array}
%[c]{c}
%\text{because evaluation at }\xx =\left(  x_{1},x_{2},...,x_{n}%
%,0,0,0,...\right)  \\
%\text{maps }x_{u}\text{ to }0\text{ (because }u\geq n+1\text{)}%
%\end{array}
%\right)  \\
%&  =0\cdot Q\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)  =0.
%\end{align*}
%\par
%Now, forget that we fixed $T$. We thus have proven that $\left(  x_{1}%
%,x_{2},...,x_{n},0,0,0,...\right)  ^{\cont T}=0$ for each
%column-strict tableau of shape $\lambda$.
%\par
%Now, by the definition of $s_{\lambda}$, we have
%\[
%s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict tableau}%
%\\\text{of shape }\lambda}}\xx ^{\cont T},
%\]
%so that
%\[
%s_{\lambda}\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)  =\sum
%\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
%}\lambda}}\underbrace{\left(  x_{1},x_{2},...,x_{n},0,0,0,...\right)
%^{\cont T}}_{=0}=\sum\limits_{\substack{T\text{ is a
%column-strict tableau}\\\text{of shape }\lambda}}0=0,
%\]
%and thus \eqref{sol.cauchy.cauchy.c.overflow} is proven.}.
Thus,
\begin{align*}
& \underbrace{\sum\limits_{\lambda\in \Par }}_{=\sum
\limits_{\lambda\text{ is a partition}}}s_{\lambda}\left(  x_{1}%
,x_{2},...,x_{n}\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right) \\
&  =\sum\limits_{\lambda\text{ is a partition}}s_{\lambda}\left(  x_{1}
,x_{2},...,x_{n}\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  \\
&  =\underbrace{\sum\limits_{\lambda\text{ is a partition with at most
}n\text{ parts}}s_{\lambda}\left(  x_{1},x_{2},...,x_{n}\right)  s_{\lambda
}\left(  y_{1},y_{2},...,y_{n}\right)  }
_{\substack{=\prod\limits_{i,j=1}^{n}
\left(  1-x_{i}y_{j}\right)  ^{-1}\\
\text{(by \eqref{sol.cauchy.cauchy.c.16})}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\lambda\text{ is a partition with more
than }n\text{ parts}}\underbrace{s_{\lambda}\left(  x_{1},x_{2},...,x_{n}
\right)  s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  }
_{\substack{=0\\\text{(by \eqref{sol.cauchy.cauchy.c.overflow})}}}\\
&  =\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}
+\underbrace{\sum\limits_{\lambda\text{ is a partition with more than }n\text{
parts}}0}_{=0}=\prod\limits_{i,j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
Thus, we have proven that \eqref{sol.cauchy.cauchy.c.2} holds for every
$n\in\NN$. As we explained above, this yields that Theorem
\ref{thm.cauchy-identity} is true.

\textit{Remark:} In our above solution, we solved Exercise
\ref{exe.cauchy.cauchy}(b) first, and then used it to prove Theorem
\ref{thm.cauchy-identity}. It is also possible (more or less by treading the
above proof backwards) to conversely derive the statement of Exercise
\ref{exe.cauchy.cauchy}(b) from Theorem \ref{thm.cauchy-identity} instead
(though Exercise~\ref{exe.cauchy.cauchy}(b) is usually considered a more
elementary fact than Theorem \ref{thm.cauchy-identity}).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.jacobi-trudi.len2.0}}

\textit{Solution to Exercise~\ref{exe.jacobi-trudi.len2.0}.}
Let us first check that we have
\begin{equation}
h_{u}h_{v}=\sum_{i=0}^{v}s_{\left(  u+i,v-i\right)  }
\label{sol.jacobi-trudi.len2.0.huhv}
\end{equation}
for any two nonnegative integers $u$ and $v$ satisfying $u\geq v$.

\textit{Proof of \eqref{sol.jacobi-trudi.len2.0.huhv}:} Let $u$ and $v$ be
nonnegative integers satisfying $u\geq v$. We WLOG assume that $u>0$
(otherwise, $u=0$, and $u\geq v$ forces $v=0$, so that
\eqref{sol.jacobi-trudi.len2.0.huhv} can be checked immediately). We have
$h_u = s_{\left( u\right)}$ and thus
\begin{equation}
h_u h_v = s_{\left( u\right)} h_v = \sum\limits_{\substack{\lambda^{+}
\ :\ \lambda^{+}/\left( u\right) \text{ is a}\\\text{horizontal
}v\text{-strip}}}s_{\lambda^{+}}
\label{sol.jacobi-trudi.len2.0.1}
\end{equation}
(by \eqref{eq.pieri.h}, applied to $n=v$ and $\lambda=\left( u\right)$).

Now, fix a partition $\lambda^{+}$ such that $\lambda^{+}/\left(  u\right)  $
is a horizontal $v$-strip. Then, the skew diagram $\lambda^{+}/\left(
u\right)  $ has at most one cell in column $1$ (since $\lambda^{+}/\left(
u\right)  $ is a horizontal $v$-strip and thus contains no two cells in the
same column). Hence, the partition $\lambda^{+}$ has at most $2$ rows (because
otherwise, the skew diagram $\lambda^{+}/\left(  u\right)  $ would contain at
least two cells in column $1$, which would contradict the fact that the skew
diagram $\lambda^{+}/\left(  u\right)  $ has at most one cell in column $1$).
Thus, the partition $\lambda^{+}$ has the form $\left(  p,q\right)  $ for two
nonnegative integers $p$ and $q$ satisfying $p\geq q$. Consider these two
integers $p$ and $q$. Since $\lambda^{+}/\left(  u\right)  $ is a horizontal
$v$-strip, we have $\left\vert \lambda^{+}/\left(  u\right)  \right\vert =v$
and thus $v=\left\vert \lambda^{+}/\left(  u\right)  \right\vert =\left\vert
\underbrace{\lambda^{+}}_{=\left(  p,q\right)  }\right\vert
-\underbrace{\left\vert \left(  u\right)  \right\vert }_{=u}
=\underbrace{\left\vert \left(  p,q\right)  \right\vert }_{=p+q}-u=p+q-u$.
Moreover, $\lambda^{+}/\left(  u\right)  $ is a horizontal $v$-strip, so that
$\left(  u\right)  \subseteq\lambda^{+}=\left(  p,q\right)  $. Thus, $u\leq
p$. Hence, there exists an $i\in\NN$ such that $p=u+i$. Consider this $i$.
Now, $v=\underbrace{p}_{=u+i}+q-u=u+i+q-u=q+i$, so that $q=v-i$ and thus
$v-i=q\geq0$, so that $i\leq v$ and thus $i\in\left\{  0,1,...,v\right\}  $
(since $i\in\NN$). Hence, $\lambda^{+}=\left(  \underbrace{p}
_{=u+i},\underbrace{q}_{=v-i}\right)  =\left(  u+i,v-i\right)  $. We have thus
shown that the partition $\lambda^{+}$ has the form $\lambda^{+}=\left(
u+i,v-i\right)  $ for some $i\in\left\{  0,1,...,v\right\}  $.

Now forget that we fixed $\lambda^{+}$. We thus have proven that every
partition $\lambda^{+}$ such that $\lambda^{+}/\left(  u\right)  $ is a
horizontal $v$-strip has the form $\lambda^{+}=\left(  u+i,v-i\right)  $ for
some $i\in\left\{  0,1,...,v\right\}$. Conversely, it is clear that for
every $i\in\left\{  0,1,...,v\right\}$, the weak composition $\left(
u+i,v-i\right)  $ is a partition $\lambda^{+}$ such that $\lambda^{+}/\left(
u\right)  $ is a horizontal $v$-strip. Combining the previous two sentences,
we conclude that the partitions $\lambda^{+}$ such that $\lambda^{+}/\left(
u\right)  $ is a horizontal $v$-strip are precisely the weak compositions of
the form $\left(  u+i,v-i\right)  $ for $i\in\left\{  0,1,...,v\right\}  $.
Therefore, $\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\left(
u\right)  \text{ is a}\\\text{horizontal }v\text{-strip}}}s_{\lambda^{+}}
=\sum\limits_{i\in\left\{  0,1,...,v\right\}  }s_{\left(  u+i,v-i\right)
}=\sum\limits_{i=0}^{v}s_{\left(  u+i,v-i\right)  }$. Now,
\eqref{sol.jacobi-trudi.len2.0.1} becomes
\[
h_{u}h_{v}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\left(
u\right)  \text{ is a}\\\text{horizontal }v\text{-strip}}}s_{\lambda^{+}}
=\sum\limits_{i=0}^{v}s_{\left(  u+i,v-i\right)  }.
\]
This proves \eqref{sol.jacobi-trudi.len2.0.huhv}.

Now, fix two integers $a$ and $b$ satisfying $a\geq b\geq0$. We need to prove
that $s_{\left( a,b\right) } = h_{a}h_{b}-h_{a+1}h_{b-1}$.

If $b=0$, then proving $s_{\left(  a,b\right)  }=h_{a}h_{b}-h_{a+1}h_{b-1}$ is
very easy\footnote{In fact, assume that $b=0$. Then,
\[
s_{\left(  a,b\right)  }=s_{\left(  a,0\right)  }=s_{\left(  a\right)  }=h_{a}
\]
and
\[
h_{a}\underbrace{h_{b}}_{=h_{0}=1}-h_{a+1}\underbrace{h_{b-1}}_{=h_{-1}
=0}=h_{a}1-h_{a+1}0=h_{a}.
\]
Hence, $s_{\left(  a,b\right)  }=h_{a}=h_{a}h_{b}-h_{a+1}h_{b-1}$, qed.}.
Hence, for the rest of the proof of $s_{\left(  a,b\right)  }=h_{a}
h_{b}-h_{a+1}h_{b-1}$, we assume WLOG that we don't have $b=0$. Thus, $b\geq
1$, so that $b-1\geq0$. Clearly, $a\geq b$ shows that $a+1\geq b+1\geq b-1$,
so that we can apply \eqref{sol.jacobi-trudi.len2.0.huhv} to $u=a+1$ and
$v=b-1$. Now,
\begin{align*}
& \underbrace{h_{a}h_{b}}_{\substack{=\sum_{i=0}^{b}s_{\left(  a+i,b-i\right)
}\\\text{(by \eqref{sol.jacobi-trudi.len2.0.huhv}, applied}\\\text{to
}u=a\text{ and }v=b\text{)}}}-\underbrace{h_{a+1}h_{b-1}}_{\substack{=\sum
_{i=0}^{b-1}s_{\left(  a+1+i,b-1-i\right)  }\\\text{(by
\eqref{sol.jacobi-trudi.len2.0.huhv}, applied}\\\text{to }u=a+1\text{ and }
v=b-1\text{)}}}\\
& =\underbrace{\sum_{i=0}^{b}s_{\left(  a+i,b-i\right)  }}_{=s_{\left(
a+0,b-0\right)  }+\sum_{i=1}^{b}s_{\left(  a+i,b-i\right)  }}-\sum_{i=0}
^{b-1}\underbrace{s_{\left(  a+1+i,b-1-i\right)  }}_{=s_{\left(  a+\left(
i+1\right)  ,b-\left(  i+1\right)  \right)  }}=s_{\left(  a+0,b-0\right)
}+\sum_{i=1}^{b}s_{\left(  a+i,b-i\right)  }-\sum_{i=0}^{b-1}s_{\left(
a+\left(  i+1\right)  ,b-\left(  i+1\right)  \right)  }\\
& =s_{\left(  a+0,b-0\right)  }+\sum_{i=1}^{b}s_{\left(  a+i,b-i\right)
}-\sum_{i=1}^{b}s_{\left(  a+i,b-i\right)  }\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }i\text{ for
}i+1\text{ in the second sum}\right)  \\
& =s_{\left(  a+0,b-0\right)  }=s_{\left(  a,b\right)  }.
\end{align*}
We thus have proven $s_{\left( a,b\right)} = h_a h_b - h_{a+1} h_{b-1}$.
The exercise is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.cauchy}}

\textit{Solution to Exercise~\ref{exe.pieri.cauchy}.} (a) We shall
prove the statement of Exercise~\ref{exe.pieri.cauchy}(a) by induction over
the length $\ell\left(  \mu\right)  $ of $\mu$.

The \textit{induction base} (i.e., the case $\ell\left(  \mu\right)  = 0$) is
trivial. For the \textit{induction step}, we fix a positive integer $L$ and
assume (as the induction hypothesis) that Exercise~\ref{exe.pieri.cauchy}(a)
has been solved for all $\mu$ satisfying $\ell\left(  \mu\right)  = L-1$. We
now have to solve Exercise~\ref{exe.pieri.cauchy}(a) for every partition $\mu$
satisfying $\ell\left(  \mu\right)  = L$.

So let $\mu$ be a partition satisfying $\ell\left(  \mu\right)  =L$. Write
$\mu$ in the form $\left(  \mu_{1},\mu_{2},\ldots,\mu_{L}\right)  $. Let
$\overline{\mu}$ be the partition $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{L-1}\right)  $ (this is well-defined since $L$ is positive). Then,
$\ell\left(  \overline{\mu}\right)  =L-1$, and hence (by the induction
hypothesis) we can apply Exercise~\ref{exe.pieri.cauchy}(a) to $\overline{\mu
}$ instead of $\mu$. As a result, we obtain $h_{\overline{\mu}}=\sum_{\lambda
}K_{\lambda,\overline{\mu}}s_{\lambda}$, where the sum ranges over all
partitions $\lambda$.

But the definition of $h_{\mu}$ yields $h_{\mu}=h_{\mu_{1}}h_{\mu_{2}}\ldots
h_{\mu_{L}}$; similarly, $h_{\overline{\mu}}=h_{\mu_{1}}h_{\mu_{2}}\ldots
h_{\mu_{L-1}}$. Hence,
\begin{align}
h_{\mu} &  =h_{\mu_{1}}h_{\mu_{2}}\ldots h_{\mu_{L}}
= \underbrace{h_{\mu_{1}}h_{\mu_{2}}\ldots h_{\mu_{L-1}}}
  _{=h_{\overline{\mu}}=\sum_{\lambda}K_{\lambda,\overline{\mu}}s_{\lambda}}
  h_{\mu_{L}}
= \left(  \sum_{\lambda}
  K_{\lambda,\overline{\mu}}s_{\lambda}\right)  h_{\mu_{L}}
\nonumber
\\
&= \sum_{\lambda}K_{\lambda,\overline{\mu}}
  \underbrace{s_{\lambda}h_{\mu_{L}}}_{\substack{
  =\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a
  horizontal}\\ \mu_{L}\text{-strip}}}
  s_{\lambda^{+}}\\\text{(by \eqref{eq.pieri.h}, applied to }n=\mu_{L}
  \text{)}}}
=\sum_{\lambda}K_{\lambda,\overline{\mu}}\sum\limits_{\substack{\lambda^{+}\ :
  \ \lambda^{+}/\lambda\text{ is a horizontal}\\ \mu_{L}\text{-strip}}}
  s_{\lambda^{+}}
\nonumber\\
&= \underbrace{\sum_{\lambda}\sum\limits_{\substack{\lambda^{+}\ :\ \lambda
  ^{+}/\lambda\text{ is a horizontal}\\ \mu_{L}\text{-strip}}}}
  _{=\sum_{\lambda^{+}}\sum\limits_{\substack{\lambda\ :\ \lambda^{+}/\lambda
  \text{ is a horizontal}\\ \mu_{L}\text{-strip}}}}
  K_{\lambda,\overline{\mu}} s_{\lambda^{+}}
=\sum_{\lambda^{+}}\sum\limits_{\substack{\lambda\ :\ \lambda^{+}/\lambda
  \text{ is a horizontal}\\ \mu_{L}\text{-strip}}}
  K_{\lambda,\overline{\mu}}s_{\lambda^{+}}
\nonumber\\
&= \sum_{\lambda}\sum\limits_{\substack{\lambda^{-}\ :\ \lambda/\lambda^{-}
  \text{ is a horizontal}\\ \mu_{L}\text{-strip}}}
  K_{\lambda^{-},\overline{\mu}} s_{\lambda}
\label{sol.pieri.cauchy.a.hmu}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation
  indices }\lambda^{+}\text{ and }\lambda\text{ as }\lambda
  \text{ and }\lambda^{-}\right)  ,
\nonumber
\end{align}
where all summation indices are supposed to be partitions.

Now, let us fix a partition $\lambda$. We shall show that
\begin{equation}
\sum\limits_{\substack{\lambda^{-}\ :\ \lambda/\lambda^{-}\text{ is a horizontal}
\\ \mu_{L}\text{-strip}}}K_{\lambda^{-},\overline{\mu}}=K_{\lambda,\mu}.
\label{sol.pieri.cauchy.a.1}
\end{equation}


\textit{Proof of \eqref{sol.pieri.cauchy.a.1}:} For every partition
$\lambda^{-}$, the number $K_{\lambda^{-},\overline{\mu}}$ is the number of
column-strict tableaux $S$ of shape $\lambda^{-}$ having $\cont
\left( S\right)  =\overline{\mu}$. Hence, the sum $\sum
\limits_{\substack{\lambda^{-}\ :\ \lambda/\lambda^{-}\text{ is a
horizontal}\\ \mu_{L}\text{-strip}}}K_{\lambda^{-},\overline{\mu}}$ is the
number of all pairs $\left(  \lambda^{-},S\right)  $, where:

\begin{itemize}
\item $\lambda^{-}$ is a partition such that $\lambda/\lambda^{-}$ is a
horizontal $\mu_{L}$-strip;

\item $S$ is a column-strict tableau of shape $\lambda^{-}$ having
$\cont \left(  S\right)  =\overline{\mu}$.
\end{itemize}

\noindent
We will refer to such pairs $\left(  \lambda^{-},S\right)  $ as
\emph{$\left(  \lambda,\mu\right)  $-last-step pairs}. So the sum
$\sum\limits_{\substack{\lambda^{-}\ :\ \lambda/\lambda^{-}\text{ is a
horizontal}\\ \mu_{L}\text{-strip}}}K_{\lambda^{-},\overline{\mu}}$ is the
number of $\left(  \lambda,\mu\right)  $-last-step pairs. On the other hand,
$K_{\lambda,\mu}$ is the number of all column-strict tableaux $T$ of shape
$\lambda$ having $\cont \left(  T\right)  =\mu$. We now will
construct a bijection between the $\left(  \lambda,\mu\right)  $-last-step
pairs and the column-strict tableaux $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$.

Indeed, let us first define a map $\Phi$ from the set of all $\left(
\lambda,\mu\right)  $-last-step pairs to the set of all column-strict tableaux
$T$ of shape $\lambda$ having $\cont \left(  T\right)  =\mu$.
Namely, let $\left(  \lambda^{-},S\right)  $ be a $\left(  \lambda,\mu\right)
$-last-step pair. Then, $\lambda^{-}$ is a partition such that $\lambda
/\lambda^{-}$ is a horizontal $\mu_{L}$-strip, whereas $S$ is a
column-strict tableau of shape $\lambda^{-}$ having $\cont 
\left(  S\right)  =\overline{\mu}$. In particular, all entries of $S$ are $<L$
(since $\cont \left(  S\right)  =\overline{\mu}=\left(  \mu
_{1},\mu_{2},\ldots,\mu_{L-1}\right)  $). Now, we can extend the column-strict
tableau $S$ of shape $\lambda^{-}$ to a tableau of shape $\lambda$ by filling
the number $L$ into all cells of the skew shape $\lambda/\lambda^{-}$. The
resulting tableau is a column-strict tableau $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$ (indeed, its column-strictness
follows from the fact that $\lambda/\lambda^{-}$ is a horizontal
$\mu_{L}$-strip whereas all entries of $S$ are $<L$;
and the property $\cont\left( T\right)  =\mu$ follows from the facts that
$\cont \left( S\right)  =\overline{\mu}$ and
$\left\vert \lambda/\lambda^{-}\right\vert = \mu_{L}$). We define
$\Phi\left(  \lambda^{-},S\right)  $ to be this
tableau. Thus we have defined a map $\Phi$.

Conversely, let us define a map $\Psi$ from the set of all column-strict
tableaux $T$ of shape $\lambda$ having $\cont \left(  T\right)
=\mu$ to the set of all $\left(  \lambda,\mu\right)  $-last-step pairs.
Namely, let $T$ be a column-strict tableau of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$. Then, all entries of $T$ are
$\leq L$ (since $\cont \left(  T\right)  =\mu=\left(  \mu
_{1},\mu_{2},\ldots,\mu_{L}\right)  $), and the cells containing the entries
$L$ form a horizontal strip (since $T$ is column-strict). Hence, if we remove
all entries $L$ from $T$ (along with their cells), the result will be a
column-strict tableau $S$ of some shape $\lambda^{-}$ such that $\lambda^{-}$
is a partition (because all entries of $T$ were $\leq L$, so the entries we
removed were maximal), $\lambda/\lambda^{-}$ is a horizontal
$\mu_{L}$-strip (since we removed a total of $\mu_{L}$ entries and
they formed a horizontal strip), and $\cont\left( S\right) =\overline{\mu}$.
Consider these $S$ and $\lambda^{-}$, and define $\Psi\left(  T\right)  $ to
be the pair $\left(  \lambda^{-},S\right)  $; it is clear that this
$\Psi\left(  T\right)  $ is a $\left(  \lambda,\mu\right)  $-last-step pair.
Hence, we have defined a map $\Psi$.

It is fairly obvious that the maps $\Phi$ and $\Psi$ are mutually inverse.
Hence, they are bijections; in particular, $\Phi$ is a bijection. Thus, we
have a bijection between the set of all $\left(  \lambda,\mu\right)
$-last-step pairs and the set of all column-strict tableaux $T$ of shape
$\lambda$ having $\cont \left(  T\right)  =\mu$. As a
consequence, the number of all $\left(  \lambda,\mu\right)  $-last-step pairs
equals the number of all column-strict tableaux $T$ of shape $\lambda$ having
$\cont \left(  T\right)  =\mu$. In other words, $\sum
\limits_{\substack{\lambda^{-}\ :\ \lambda/\lambda^{-}\text{ is a
horizontal}\\ \mu_{L}\text{-strip}}}K_{\lambda^{-},\overline{\mu}}$ equals
$K_{\lambda,\mu}$ (because $\sum\limits_{\substack{\lambda^{-}\ :\ \lambda
/\lambda^{-}\text{ is a horizontal}\\ \mu_{L}\text{-strip}}}K_{\lambda
^{-},\overline{\mu}}$ is the number of $\left(  \lambda,\mu\right)
$-last-step pairs, whereas  $K_{\lambda,\mu}$ is the number of all
column-strict tableaux $T$ of shape $\lambda$ having
$\cont\left(  T\right)  =\mu$). This proves \eqref{sol.pieri.cauchy.a.1}.

Now, \eqref{sol.pieri.cauchy.a.hmu} becomes
\[
h_{\mu}=\sum_{\lambda}\underbrace{\sum\limits_{\substack{\lambda^{-}\ :\ \lambda
/\lambda^{-}\text{ is a horizontal}\\ \mu_{L}\text{-strip}}}K_{\lambda
^{-},\overline{\mu}}}_{\substack{=K_{\lambda,\mu}\\\text{(by
\eqref{sol.pieri.cauchy.a.1})}}}s_{\lambda}=\sum_{\lambda}K_{\lambda,\mu
}s_{\lambda}.
\]
Hence, Exercise~\ref{exe.pieri.cauchy}(a) is solved for our partition $\mu$.
Thus, Exercise~\ref{exe.pieri.cauchy}(a) is solved for every partition $\mu$
satisfying $\ell\left(  \mu\right)  =L$. This completes the induction step,
and thus Exercise~\ref{exe.pieri.cauchy}(a) is solved by induction.

(b) Let us work in the power series ring $\kk \left[  \left[
\xx ,\yy \right]  \right]  :=\kk \left[  \left[
x_{1},x_{2},\ldots,y_{1},y_{2},\ldots\right]  \right]  $. Every partition
$\mu$ satisfies $h_{\mu}\left(  \xx \right)  =\sum_{\lambda}
K_{\lambda,\mu}s_{\lambda}\left(  \xx \right)  $, where the sum ranges
over all partitions $\lambda$ (because Exercise~\ref{exe.pieri.cauchy}(a)
yields $h_{\mu}=\sum_{\lambda}K_{\lambda,\mu}s_{\lambda}$ in $\Lambda$). In
other words, every partition $\mu$ satisfies $h_{\mu}\left(  \xx 
\right)  =\sum_{\lambda\in\Par} K_{\lambda,\mu}s_{\lambda
}\left(  \xx \right)  $. On the other hand, every partition $\lambda$
satisfies $s_{\lambda}=\sum_{\mu}K_{\lambda,\mu}m_{\mu}$, where the sum ranges
over all partitions $\mu$ (this was shown in the proof of
Proposition~\ref{many-symmetric-function-bases-prop}). In other words, every
partition $\lambda$ satisfies
$s_{\lambda}=\sum_{\mu\in\Par }K_{\lambda,\mu}m_{\mu}$, so that
\begin{equation}
s_{\lambda}\left(  \yy \right)  =\sum_{\mu\in\Par }
K_{\lambda,\mu}m_{\mu}\left(  \yy \right)
.\label{sol.pieri.cauchy.b.1}
\end{equation}
Now,
\begin{align*}
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}  & =\sum_{\lambda
\in\Par}h_{\lambda}\left( \xx \right)  m_{\lambda
}\left( \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{Two-other-Cauchy-expansions}}\right)  \\
& =\sum_{\mu\in\Par }\underbrace{h_{\mu}\left( \xx \right) }
_{=\sum_{\lambda\in\Par} K_{\lambda,\mu}s_{\lambda
}\left(  \xx \right)  }m_{\mu}\left( \yy \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\lambda\text{ as }\mu\right)  \\
& =\underbrace{\sum_{\mu\in \Par}\sum_{\lambda\in \Par}}
_{=\sum_{\lambda\in \Par}\sum_{\mu\in \Par}} K_{\lambda,\mu}
s_{\lambda}\left(  \xx \right) m_{\mu}\left( \yy \right)  \\
& =\sum_{\lambda\in\Par }\sum_{\mu\in\Par }
K_{\lambda,\mu}s_{\lambda}\left(  \xx \right)  m_{\mu}\left(
\yy \right)  =\sum_{\lambda\in\Par }s_{\lambda}\left(
\xx \right)  \underbrace{\sum_{\mu\in\Par }K_{\lambda,\mu}
m_{\mu}\left( \yy \right)  }_{\substack{=s_{\lambda}\left(
\yy \right)  \\\text{(by \eqref{sol.pieri.cauchy.b.1})}}}\\
& =\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  .
\end{align*}
This proves Theorem~\ref{thm.cauchy-identity}. Thus,
Exercise~\ref{exe.pieri.cauchy}(b) is solved.

(c) We start out just as in the proof of
Proposition~\ref{many-symmetric-function-bases-prop}: We
fix an $n \in \NN$, and we
restrict our attention to a given homogeneous component
$\Lambda_n$ and partitions of $n$. Regard the set $\Par_n$ as
a poset with smaller-or-equal relation $\triangleright$.
We will check that the family
$\left(h_\lambda\right)_{\lambda\in\Par_n}$
expands unitriangularly\footnote{See
Definition~\ref{def.STmat.expansion-tria}(c) for what this means.}
in the basis
$\left(s_\lambda\right)_{\lambda\in\Par_n}$.

If $\lambda$ and $\mu$ are two partitions satisfying
$\left|\lambda\right| \neq \left|\mu\right|$,
then
\begin{equation}
K_{\lambda, \mu} = 0
\label{sol.pieri.cauchy.c.graded-1}
\end{equation}
(because $K_{\lambda, \mu}$ counts the column-strict tableaux
$T$ of shape $\lambda$ having $\cont\left(T\right) = \mu$; but
no such tableaux exist when
$\left|\lambda\right| \neq \left|\mu\right|$).

In Exercise~\ref{exe.pieri.cauchy}(a), we have shown that
$h_\mu = \sum_{\lambda \in \Par} K_{\lambda, \mu} s_\lambda$
for every $\mu \in \Par$.
Thus, for every $\mu \in \Par_n$, we have
\begin{align}
h_\mu = \sum_{\lambda \in \Par} K_{\lambda, \mu} s_\lambda
&= \underbrace{\sum\limits_{\substack{\lambda \in \Par ; \\ \left|\lambda\right| = n}}}_{= \sum_{\lambda \in \Par_n}}
     K_{\lambda, \mu} s_\lambda
  + \sum\limits_{\substack{\lambda \in \Par ; \\ \left|\lambda\right| \neq n}}
     \underbrace{K_{\lambda, \mu}}_{\substack{
        = 0 \\ \text{(by \eqref{sol.pieri.cauchy.c.graded-1}, since }
                \left|\lambda\right| \neq n = \left|\mu\right|
                \text{)}}}
     s_\lambda
\nonumber \\
&= \sum_{\lambda \in \Par_n} K_{\lambda, \mu} s_\lambda
  + \underbrace{\sum\limits_{\substack{\lambda \in \Par ; \\ \left|\lambda\right| \neq n}} 0 s_\lambda}_{= 0}
= \sum_{\lambda \in \Par_n} K_{\lambda, \mu} s_\lambda
.
\label{sol.pieri.cauchy.c.graded-2}
\end{align}

In the proof of Proposition~\ref{many-symmetric-function-bases-prop},
we showed that any two partitions $\lambda$ and $\mu$ in $\Par_n$
satisfy $K_{\lambda, \mu} = 0$ unless $\lambda \triangleright
\mu$. Hence, the $\Par_n \times \Par_n$-matrix
$\left(K_{\lambda, \mu}\right)_{\left(\mu, \lambda\right) \in \Par_n \times \Par_n}$
is triangular\footnote{See Definition~\ref{def.STmat.tria} for the
meaning of this.}. This matrix is furthermore unitriangular
(since $K_{\lambda, \lambda} = 1$ for every partition $\lambda$,
as shown in the proof of
Proposition~\ref{many-symmetric-function-bases-prop}), and therefore
invertibly triangular.
But the family 
$\left(h_\lambda\right)_{\lambda\in\Par_n}$
expands in the family
$\left(s_\lambda\right)_{\lambda\in\Par_n}$
through this matrix
$\left(K_{\lambda, \mu}\right)_{\left(\mu, \lambda\right) \in \Par_n \times \Par_n}$
(because of \eqref{sol.pieri.cauchy.c.graded-2}). Therefore,
the family 
$\left(h_\lambda\right)_{\lambda\in\Par_n}$
expands invertibly triangularly in the family
$\left(s_\lambda\right)_{\lambda\in\Par_n}$ (since our
matrix is invertibly triangular).
Hence,
Corollary~\ref{cor.STmat.expansion-tria-inv}(e) (applied
to $\Lambda_n$, $\Par_n$, $\left( h_\lambda \right)_{\lambda \in \Par_n}$
and $\left( s_\lambda \right)_{\lambda \in \Par_n}$ instead of
$M$, $S$, $\left(e_s\right)_{s \in S}$ and $\left(f_s\right)_{s \in S}$)
shows that the family
$\left( h_\lambda \right)_{\lambda \in \Par_n}$
is a basis of the $\kk$-module $\Lambda_n$ if and only if
the family $\left( s_\lambda \right)_{\lambda \in \Par_n}$
is a basis of the $\kk$-module $\Lambda_n$. Therefore,
the family
$\left( h_\lambda \right)_{\lambda \in \Par_n}$
is a basis of the $\kk$-module $\Lambda_n$ (since
the family $\left( s_\lambda \right)_{\lambda \in \Par_n}$
is a basis of the $\kk$-module $\Lambda_n$).
Since this has been proven for all
$n \in \NN$, we can combine this to conclude that the family 
$\left(h_\lambda\right)_{\lambda\in\Par}$ is a graded basis of
the graded $\kk$-module
$\Lambda$. This solves Exercise~\ref{exe.pieri.cauchy}(c).

\textit{Remark.} Just as in the Remark after the solution of
Exercise~\ref{exe.Lambda.bases}(b),
we can use our solution of Exercise~\ref{exe.pieri.cauchy}(c) to
further prove that $\left(e_\lambda\right)_{\lambda\in\Par}$ is a
$\kk$-basis of $\Lambda$ (in a different way than we have done in
the proof of Proposition~\ref{many-symmetric-function-bases-prop}).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.omega}}

\textit{Solution to Exercise~\ref{exe.pieri.omega}.} Let us
first observe that
\begin{equation}
\mathfrak{Z}\left(  h_{n}\right) = e_{n}\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }n. \label{sol.pieri.omega.hn}
\end{equation}
(In fact, $h_n = s_{\left(n\right)}$, and so the map $\mathfrak{Z}$
maps $h_n$ to $s_{\left(n\right)^t} = s_{\left(1^n\right)} = e_n$.)

(a) Fix $n\in\NN$. We need to prove that
\begin{equation}
\mathfrak{Z}\left(  fh_{n}\right)  =\mathfrak{Z}\left(  f\right)
\cdot\mathfrak{Z}\left(  h_{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}f\in\Lambda. \label{sol.pieri.omega.a.goal}
\end{equation}
Since this equality is linear in $f$, it is clearly enough to prove it
when $f$ is of the form $s_\lambda$ for some $\lambda \in \Par$
(because
$\left(  s_{\lambda}\right)  _{\lambda\in\Par }$ is a
$\kk$-basis of $\Lambda$). In other words, it is enough to show that
every $\lambda\in\Par $ satisfies
\begin{equation}
\mathfrak{Z}\left(  s_{\lambda}h_{n}\right)  =\mathfrak{Z}\left(  s_{\lambda
}\right)  \cdot\mathfrak{Z}\left(  h_{n}\right)  .
\label{sol.pieri.omega.a.1}
\end{equation}
Thus, we will focus on proving \eqref{sol.pieri.omega.a.1} now.

We WLOG assume that $n \neq 0$, since otherwise \eqref{sol.pieri.omega.a.1}
is obvious. Let $\lambda \in \Par$. The equality \eqref{eq.pieri.h} yields
\[
s_{\lambda}h_{n}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is
a}\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}.
\]
Applying the map $\mathfrak{Z}$ to both sides of this equality, we obtain
\begin{equation}
\mathfrak{Z}\left(  s_{\lambda}h_{n}\right)   =\mathfrak{Z}\left(
\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}
\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}\right)  =\sum\limits
_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}\\\text{horizontal
}n\text{-strip}}}\underbrace{\mathfrak{Z}\left(  s_{\lambda^{+}}\right)
}_{=s_{\left(  \lambda^{+}\right)  ^{t}}}
=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is
a}\\\text{horizontal }n\text{-strip}}}s_{\left(  \lambda^{+}\right)  ^{t}}.
\label{sol.pieri.omega.a.1.pf.1}
\end{equation}


However, for any given partition $\lambda^{+}$, the assertion that $\lambda
^{+}/\lambda$ be a horizontal $n$-strip is equivalent to the assertion that
$\left(  \lambda^{+}\right)  ^{t}/\lambda^{t}$ be a vertical
$n$-strip\footnote{This is because the notion of a vertical $n$-strip is
obtained
from the notion of a horizontal $n$-strip by interchanging the roles of rows
and columns, and the operation which sends a partition $\mu$ to its transpose
partition $\mu^{t}$ interchanges the roles of rows and columns as well.}.
Hence, we can replace the summation sign ``$\sum\limits
_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}\\\text{horizontal
}n\text{-strip}}}$'' in \eqref{sol.pieri.omega.a.1.pf.1} by a
``$\sum\limits_{\substack{\lambda^{+}\ :\ \left(  \lambda^{+}\right)
^{t}/\lambda^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}$'' sign. Thus,
we obtain
\begin{align*}
\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}
\\\text{horizontal }n\text{-strip}}}s_{\left(  \lambda^{+}\right)  ^{t}}  &
=\sum\limits_{\substack{\lambda^{+}\ :\ \left(  \lambda^{+}\right)  ^{t}/\lambda
^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}s_{\left(  \lambda
^{+}\right)  ^{t}}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda
^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }\lambda^{+}\text{ for }\left(  \lambda^{+}\right)
^{t}\text{ in the sum, because}\\
\text{the map }\Par\rightarrow\Par,\ \mu\mapsto\mu^{t}\text{ is a
bijection}
\end{array}
\right)  .
\end{align*}
Thus, \eqref{sol.pieri.omega.a.1.pf.1} becomes
\begin{equation}
\mathfrak{Z}\left(  s_{\lambda}h_{n}\right)  =\sum\limits_{\substack{\lambda
^{+}\ :\ \lambda^{+}/\lambda\text{ is a}\\\text{horizontal }n\text{-strip}}}
s_{\left(  \lambda^{+}\right)  ^{t}}=\sum\limits_{\substack{\lambda^{+}
\ :\ \lambda^{+}/\lambda^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}
s_{\lambda^{+}}. \label{sol.pieri.omega.a.1.pf.2}
\end{equation}


On the other hand, the equality \eqref{eq.pieri.e} yields
\[
s_{\lambda}e_{n}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is
a}\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}.
\]
Applying this to $\lambda^{t}$ instead of $\lambda$, we obtain
\[
s_{\lambda^{t}}e_{n}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda
^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}.
\]


Now,
\[
\underbrace{\mathfrak{Z}\left(  s_{\lambda}\right)  }_{=s_{\lambda
^{t}}}\cdot\underbrace{\mathfrak{Z}
\left(  h_{n}\right)  }_{=e_{n}}=s_{\lambda^{t}}e_{n}=\sum\limits_{\substack{\lambda
^{+}\ :\ \lambda^{+}/\lambda^{t}\text{ is a}\\
\text{vertical }n\text{-strip}}}s_{\lambda^{+}}.
\]
Compared with \eqref{sol.pieri.omega.a.1.pf.2}, this yields $\mathfrak{Z}
\left(  s_{\lambda}h_{n}\right)  =\mathfrak{Z}\left(  s_{\lambda}\right)
\cdot\mathfrak{Z}\left(  h_{n}\right)  $. This proves
\eqref{sol.pieri.omega.a.1}. This concludes the solution of
Exercise~\ref{exe.pieri.omega}(a).

(b) We shall show that
\begin{equation}
\mathfrak{Z}\left(  h_{\lambda}\right)  =\omega\left(  h_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.pieri.omega.b.goal}
\end{equation}


\textit{Proof of \eqref{sol.pieri.omega.b.goal}:} We will prove
\eqref{sol.pieri.omega.b.goal} by induction over $\ell\left(  \lambda\right)
$. The \textit{induction base} is the case $\ell\left(  \lambda\right)  =0$,
which is utterly obvious. For the \textit{induction step}, we fix a positive
integer $L$ and assume (as the induction hypothesis) that the equality
\eqref{sol.pieri.omega.b.goal} holds if $\ell\left(  \lambda\right)  =L-1$. We
need to prove that the equality \eqref{sol.pieri.omega.b.goal} holds if
$\ell\left(  \lambda\right)  =L$.

Let $\lambda$ be a partition satisfying $\ell\left(  \lambda\right)  =L$.
Write $\lambda$ in the form $\lambda=\left(  \lambda_{1},\lambda
_{2},...,\lambda_{L}\right)  $ with all of $\lambda_{1}$, $\lambda_{2}$,
$...$, $\lambda_{L}$ being positive integers. Let $\overline{\lambda}$ be the
partition $\left(  \lambda_{1},\lambda_{2},...,\lambda_{L-1}\right)  $ (this
is obviously well-defined); then, $\ell\left(  \overline{\lambda}\right)
=L-1$.

Recall that $e_n = \omega\left(h_n\right)$ for every $n \geq 1$
(by Proposition~\ref{prop.Lambda.omega-and-S}(b)).

Recall that $\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{L}\right)
$. Hence, by the definition of $h_{\lambda}$, we have
\[
h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{L}}
=\underbrace{h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{L-1}}}
_{\substack{=h_{\overline{\lambda}}\\\text{(since }\overline{\lambda}=\left(
\lambda_{1},\lambda_{2},...,\lambda_{L-1}\right)  \text{)}}}h_{\lambda_{L}}
=h_{\overline{\lambda}}h_{\lambda_{L}}.
\]
Applying the map $\mathfrak{Z}$ to both sides of this equality, we obtain
\begin{align*}
\mathfrak{Z}\left(  h_{\lambda}\right)    & =\mathfrak{Z}\left(
h_{\overline{\lambda}}h_{\lambda_{L}}\right)  =\underbrace{\mathfrak{Z}\left(
h_{\overline{\lambda}}\right)  }_{\substack{=\omega\left(
h_{\overline{\lambda}}\right)  \\
\text{(by the induction}\\\text{hypothesis,}\\\text{since
}\ell\left(  \overline{\lambda}\right)  =L-1\text{)}}}\cdot
\underbrace{\mathfrak{Z}\left(  h_{\lambda_{L}}\right)  }
_{\substack{=e_{\lambda_{L}}\\\text{(by \eqref{sol.pieri.omega.hn})}}} \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.pieri.omega}(a),
applied to }h_{\overline{\lambda}}\text{ and }\lambda_{L}\text{ instead of
}f\text{ and }n\right)  \\
& =\omega\left(  h_{\overline{\lambda}}\right)  \cdot\underbrace{e_{\lambda
_{L}}}_{\substack{=\omega\left(  h_{\lambda_{L}}\right)  \\\text{(since }
e_{n}=\omega\left(  h_{n}\right)  \\\text{for every }n\geq1\text{)}}}
=\omega\left(  h_{\overline{\lambda}}\right)  \cdot\omega\left(
h_{\lambda_{L}}\right)  =\omega\left(  \underbrace{h_{\overline{\lambda}}
h_{\lambda_{L}}}_{=h_{\lambda}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\omega\text{ is a }\kk \text{-algebra morphism}\right)  \\
& =\omega\left(  h_{\lambda}\right)  .
\end{align*}
In other words, the equality \eqref{sol.pieri.omega.b.goal} holds if
$\ell\left(  \lambda\right)  =L$. This completes the induction step. The
induction proof of \eqref{sol.pieri.omega.b.goal} is therefore complete.

The equality \eqref{sol.pieri.omega.b.goal} shows that the $\kk$-linear
maps $\mathfrak{Z}:\Lambda\rightarrow\Lambda$ and $\omega:\Lambda
\rightarrow\Lambda$ are equal to each other on every element of the basis
$\left( h_{\lambda} \right) _{\lambda\in\Par}$ of the
$\kk$-module $\Lambda$. Hence, $\mathfrak{Z}=\omega$. This solves
Exercise~\ref{exe.pieri.omega}(b).

(c) Fix two partitions $\mu$ and $\nu$. Then,
\eqref{symm-structure-constants.1} yields
\begin{equation}
s_{\mu}s_{\nu}
=\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda}s_{\lambda}.
\label{sol.pieri.omega.c.1}
\end{equation}
Applying this to $\mu^{t}$ and $\nu^{t}$ instead of $\mu$ and $\nu$, we
obtain
\begin{equation}
s_{\mu^{t}}s_{\nu^{t}}=\sum_{\lambda\in\Par}c_{\mu^{t},\nu^{t}}^{\lambda}
s_{\lambda}.
\label{sol.pieri.omega.c.2}
\end{equation}


Applying the map $\mathfrak{Z}$ to the identity \eqref{sol.pieri.omega.c.1},
we obtain
\begin{align}
\mathfrak{Z}\left(  s_{\mu}s_{\nu}\right)   &  =\mathfrak{Z}\left(
\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda}s_{\lambda}\right)
=\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda}
\underbrace{\mathfrak{Z}\left(  s_{\lambda}\right)  }_{=s_{\lambda^{t}}}
\nonumber\\
&  =\sum_{\lambda\in\Par}\underbrace{c_{\mu,\nu}^{\lambda}}
_{\substack{=c_{\mu,\nu}^{\left(  \lambda^{t}\right)  ^{t}}\\\text{(since
}\lambda=\left(  \lambda^{t}\right)  ^{t}\text{)}}}s_{\lambda^{t}}
=\sum_{\lambda\in\Par}c_{\mu,\nu}^{\left(  \lambda^{t}\right)
^{t}}s_{\lambda^{t}}=\sum_{\lambda\in\Par}c_{\mu,\nu}^{\lambda^{t}}
s_{\lambda}
\label{sol.pieri.omega.c.4}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }\lambda\text{ for }\lambda^{t}\text{ in the sum,
since}\\
\text{the map }\Par\rightarrow\Par,\ \lambda\mapsto\lambda^{t}\text{ is
a bijection}
\end{array}
\right)  .\nonumber
\end{align}
But recall that $\omega$ is a $\kk$-algebra homomorphism. Since
$\mathfrak{Z}=\omega$ (by Exercise~\ref{exe.pieri.omega}(b)), this shows that
$\mathfrak{Z}$ is a $\kk$-algebra homomorphism. Hence,
\[
\mathfrak{Z}\left(  s_{\mu}s_{\nu}\right)  =\underbrace{\mathfrak{Z}\left(
s_{\mu}\right)  }_{=s_{\mu^{t}}}\cdot\underbrace{\mathfrak{Z}\left(  s_{\nu
}\right)  }_{=s_{\nu^{t}}}=s_{\mu^{t}}s_{\nu^{t}}=\sum_{\lambda\in
\Par}c_{\mu^{t},\nu^{t}}^{\lambda}s_{\lambda}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.pieri.omega.c.2}}\right)  .
\]
Compared with \eqref{sol.pieri.omega.c.4}, this yields $\sum_{\lambda
\in\Par}c_{\mu,\nu}^{\lambda^{t}}s_{\lambda}=\sum_{\lambda
\in\Par}c_{\mu^{t},\nu^{t}}^{\lambda}s_{\lambda}$. Since
$\left(  s_{\lambda}\right)  _{\lambda\in\Par}$ is a basis of
the $\kk$-module $\Lambda$, we can compare coefficients in this
equality, and obtain
\begin{equation}
c_{\mu,\nu}^{\lambda^{t}}=c_{\mu^{t},\nu^{t}}^{\lambda}
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\Par .
\label{sol.pieri.omega.c.6}
\end{equation}
We can substitute $\lambda^{t}$ for $\lambda$ in this result, and conclude
that $c_{\mu,\nu}^{\left(  \lambda^{t}\right)  ^{t}}
=c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$ for every $\lambda\in\Par$. Since $\left(
\lambda^{t}\right)  ^{t}=\lambda$, this rewrites as $c_{\mu,\nu}^{\lambda
}=c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$. This solves
Exercise~\ref{exe.pieri.omega}(c).

(d) Let $\mu$ and $\lambda$ be two partitions such that $\mu\subseteq\lambda$.
In Remark~\ref{third-L-R-interpretation-remark}, it has been shown that
$s_{\lambda/\mu}=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum ranges
over all partitions $\nu$. In other words,
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu
}.
\label{sol.pieri.omega.d.1}
\end{equation}
Applying this to $\lambda^{t}$ and $\mu^{t}$ instead of $\lambda$ and $\mu$,
we obtain
\begin{equation}
s_{\lambda^{t}/\mu^{t}}=\sum_{\nu\in\Par}c_{\mu^{t},\nu
}^{\lambda^{t}}s_{\nu}.\label{sol.pieri.omega.d.2}
\end{equation}
Applying the map $\mathfrak{Z}$ to both sides of the identity
\eqref{sol.pieri.omega.d.1}, we obtain
\begin{align*}
\mathfrak{Z}\left(  s_{\lambda/\mu}\right)   &  =\mathfrak{Z}\left(  \sum
_{\nu\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu}\right)  =\sum_{\nu
\in\Par}\underbrace{c_{\mu,\nu}^{\lambda}}_{\substack{=c_{\mu
^{t},\nu^{t}}^{\lambda^{t}}\\
\text{(by Exercise~\ref{exe.pieri.omega}(c))}}}
\underbrace{\mathfrak{Z}\left(  s_{\nu}\right)  }_{=s_{\nu^{t}}}=\sum
_{\nu\in\Par}c_{\mu^{t},\nu^{t}}^{\lambda^{t}}s_{\nu^{t}}
=\sum_{\nu\in\Par}c_{\mu^{t},\nu}^{\lambda^{t}}s_{\nu}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }\nu\text{ for }\nu^{t}\text{ in the sum, since}\\
\text{the map }\Par\rightarrow\Par,\ \nu\mapsto\nu^{t}\text{ is a
bijection}
\end{array}
\right)  \\
&  =s_{\lambda^{t}/\mu^{t}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.pieri.omega.d.2}}\right)  .
\end{align*}
Since $\mathfrak{Z}=\omega$ (by Exercise~\ref{exe.pieri.omega}(b)), this
rewrites as $\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}$.
This proves the first identity of \eqref{skew-Schur-antipodes}.

It is clear that the skew Schur function $s_{\lambda/\mu}$ is homogeneous of
degree $\left\vert \lambda/\mu\right\vert$. In other words, $s_{\lambda/\mu}
\in\Lambda_{\left\vert \lambda/\mu\right\vert}$. Hence,
\eqref{antipode-versus-omega-in-Sym} (applied to $f=s_{\lambda/\mu}$ and
$n=\left\vert \lambda/\mu\right\vert $) yields
\[
S\left(  s_{\lambda/\mu}\right)  =\left(  -1\right)  ^{\left\vert \lambda
/\mu\right\vert }\underbrace{\omega\left(  s_{\lambda/\mu}\right)
}_{=s_{\lambda^{t}/\mu^{t}}}=\left(  -1\right)  ^{\left\vert \lambda
/\mu\right\vert }s_{\lambda^{t}/\mu^{t}}.
\]
This proves the second identity of \eqref{skew-Schur-antipodes}. The proof of
\eqref{skew-Schur-antipodes} is thus complete.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pieri.omega}}

\textit{Solution to Exercise \ref{exe.pieri.omega}.} We have defined
$\mathfrak{Z}$ as the $\kk$-linear map $\Lambda\rightarrow\Lambda$
which sends $s_{\lambda}$ to $s_{\lambda^{t}}$ for every partition $\lambda$.
Thus,
\begin{equation}
\mathfrak{Z}\left(  s_{\lambda}\right)  =s_{\lambda^{t}}
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.pieri.omega.def}
\end{equation}
Hence,
\begin{equation}
\mathfrak{Z}\left(  h_{n}\right)  =e_{n}\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }n. \label{sol.pieri.omega.hn}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.pieri.omega.hn}):} Let $n$ be a positive
integer. We have $h_{n}=s_{\left(  n\right)  }$, and thus
\begin{align*}
\mathfrak{Z}\left(  \underbrace{h_{n}}_{=s_{\left(  n\right)  }}\right)   &
=\mathfrak{Z}\left(  s_{\left(  n\right)  }\right)  =s_{\left(  n\right)
^{t}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.pieri.omega.def}), applied
to }\lambda=\left(  n\right)  \right) \\
&  =s_{\left(  1^{n}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
n\right)  ^{t}=\left(  1^{n}\right)  \right) \\
&  =e_{n}.
\end{align*}
This proves (\ref{sol.pieri.omega.hn}).}

(a) Fix $n\in\NN$. It is clearly enough to prove that
\begin{equation}
\mathfrak{Z}\left(  fh_{n}\right)  =\mathfrak{Z}\left(  f\right)
\cdot\mathfrak{Z}\left(  h_{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}f\in\Lambda. \label{sol.pieri.omega.a.goal}
\end{equation}


\textit{Proof of (\ref{sol.pieri.omega.a.goal}):} Let $f\in\Lambda$ be
arbitrary. We can write $f$ in the form
$f=\sum_{\lambda\in\Par}a_{\lambda}s_{\lambda}$ for some family
$\left(  a_{\lambda}\right)_{\lambda\in\Par }\in\kk ^{\Par }$ of
elements of $\kk$ such that $\left(  \text{all but finitely many
}\lambda\in\Par \text{ satisfy }a_{\lambda}=0\right)  $ (because
$\left(  s_{\lambda}\right)  _{\lambda\in\Par }$ is a
$\kk$-basis of $\Lambda$). Consider this family $\left(  a_{\lambda
}\right)  _{\lambda\in\Par }\in\kk ^{\Par }$.

If $n=0$, then (\ref{sol.pieri.omega.a.goal}) is true\footnote{\textit{Proof.}
Assume that $n=0$. Then, $h_{n}=h_{0}=1$, so that $\mathfrak{Z}\left(
f\underbrace{h_{n}}_{=1}\right)  =\mathfrak{Z}\left(  f\right)  $. But
$h_{n}=1=s_{\varnothing}$, so that
$\mathfrak{Z}\left(  \underbrace{h_{n}}_{=s_{\varnothing}}\right)
 =\mathfrak{Z}\left(  s_{\varnothing}\right) =s_{\varnothing^{t}}$
(by (\ref{sol.pieri.omega.def}), applied to
$\lambda=\varnothing$). Since $\varnothing^{t}=\varnothing$, we have
$s_{\varnothing^{t}}=s_{\varnothing}=1$, so that $\mathfrak{Z}\left(
h_{n}\right)  =s_{\varnothing^{t}}=1$ and thus $\mathfrak{Z}\left(  f\right)
\cdot\underbrace{\mathfrak{Z}\left(  h_{n}\right)  }_{=1}=\mathfrak{Z}\left(
f\right)  $. Compared with $\mathfrak{Z}\left(  fh_{n}\right)  =\mathfrak{Z}
\left(  f\right)  $, this yields $\mathfrak{Z}\left(  fh_{n}\right)
=\mathfrak{Z}\left(  f\right)  \cdot\mathfrak{Z}\left(  h_{n}\right)  $. Thus,
(\ref{sol.pieri.omega.a.goal}) is proven under the assumption that $n=0$,
qed.}. Hence, for the rest of the proof of (\ref{sol.pieri.omega.a.goal}), we
can WLOG assume that we don't have $n=0$. Assume this. We don't have $n=0$.
Thus, $n$ is a positive integer. Hence, (\ref{sol.pieri.omega.hn}) yields that
$\mathfrak{Z}\left(  h_{n}\right)  =e_{n}$.

We shall now show that every $\lambda\in\Par $ satisfies
\begin{equation}
\mathfrak{Z}\left(  s_{\lambda}h_{n}\right)  =\mathfrak{Z}\left(  s_{\lambda
}\right)  \cdot\mathfrak{Z}\left(  h_{n}\right)  . \label{sol.pieri.omega.a.1}
\end{equation}


\textit{Proof of (\ref{sol.pieri.omega.a.1}):} Let $\lambda\in
\Par $. The equality \eqref{eq.pieri.h} yields
\[
s_{\lambda}h_{n}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is
a}\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}.
\]
Applying the map $\mathfrak{Z}$ to both sides of this equality, we obtain
\begin{align}
\mathfrak{Z}\left(  s_{\lambda}h_{n}\right)   &  =\mathfrak{Z}\left(
\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}
\\\text{horizontal }n\text{-strip}}}s_{\lambda^{+}}\right)  =\sum\limits
_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}\\\text{horizontal
}n\text{-strip}}}\underbrace{\mathfrak{Z}\left(  s_{\lambda^{+}}\right)
}_{\substack{=s_{\left(  \lambda^{+}\right)  ^{t}}\\\text{(by
(\ref{sol.pieri.omega.def}), applied}\\\text{to }\lambda^{+}\text{ instead of
}\lambda\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\mathfrak{Z}\text{ is }\kk \text{-linear}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is
a}\\\text{horizontal }n\text{-strip}}}s_{\left(  \lambda^{+}\right)  ^{t}}.
\label{sol.pieri.omega.a.1.pf.1}
\end{align}


However, for any given partition $\lambda^{+}$, the assertion that $\lambda
^{+}/\lambda$ be a horizontal $n$-strip is equivalent to the assertion that
$\left(  \lambda^{+}\right)  ^{t}/\lambda^{t}$ be a vertical
$n$-strip\footnote{This is because the notion of a vertical $n$-strip is
obtained
from the notion of a horizontal $n$-strip by interchanging the roles of rows
and columns, and the operation which sends a partition $\mu$ to its transpose
partition $\mu^{t}$ interchanges the roles of rows and columns as well.}.
Hence, we can replace the summation sign ``$\sum\limits
_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}\\\text{horizontal
}n\text{-strip}}}$'' in (\ref{sol.pieri.omega.a.1.pf.1}) by a
``$\sum\limits_{\substack{\lambda^{+}\ :\ \left(  \lambda^{+}\right)
^{t}/\lambda^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}$'' sign. Thus,
we obtain
\begin{align*}
\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is a}
\\\text{horizontal }n\text{-strip}}}s_{\left(  \lambda^{+}\right)  ^{t}}  &
=\sum\limits_{\substack{\lambda^{+}\ :\ \left(  \lambda^{+}\right)  ^{t}/\lambda
^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}s_{\left(  \lambda
^{+}\right)  ^{t}}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda
^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }\lambda^{+}\text{ for }\left(  \lambda^{+}\right)
^{t}\text{ in the sum, because}\\
\text{the map }\Par\rightarrow\Par,\ \mu\mapsto\mu^{t}\text{ is a
bijection}
\end{array}
\right)  .
\end{align*}
Thus, (\ref{sol.pieri.omega.a.1.pf.1}) becomes
\begin{equation}
\mathfrak{Z}\left(  s_{\lambda}h_{n}\right)  =\sum\limits_{\substack{\lambda
^{+}\ :\ \lambda^{+}/\lambda\text{ is a}\\\text{horizontal }n\text{-strip}}}
s_{\left(  \lambda^{+}\right)  ^{t}}=\sum\limits_{\substack{\lambda^{+}
\ :\ \lambda^{+}/\lambda^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}
s_{\lambda^{+}}. \label{sol.pieri.omega.a.1.pf.2}
\end{equation}


On the other hand, the equality \eqref{eq.pieri.e} yields
\[
s_{\lambda}e_{n}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda\text{ is
a}\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}.
\]
Applying this to $\lambda^{t}$ instead of $\lambda$, we obtain
\[
s_{\lambda^{t}}e_{n}=\sum\limits_{\substack{\lambda^{+}\ :\ \lambda^{+}/\lambda
^{t}\text{ is a}\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}.
\]


Now,
\[
\underbrace{\mathfrak{Z}\left(  s_{\lambda}\right)  }_{\substack{=s_{\lambda
^{t}}\\\text{(by (\ref{sol.pieri.omega.def}))}}}\cdot\underbrace{\mathfrak{Z}
\left(  h_{n}\right)  }_{=e_{n}}=s_{\lambda^{t}}e_{n}=\sum\limits_{\substack{\lambda
^{+}\ :\ \lambda^{+}/\lambda^{t}\text{ is a}\\
\text{vertical }n\text{-strip}}}s_{\lambda^{+}}.
\]
Compared with (\ref{sol.pieri.omega.a.1.pf.2}), this yields $\mathfrak{Z}
\left(  s_{\lambda}h_{n}\right)  =\mathfrak{Z}\left(  s_{\lambda}\right)
\cdot\mathfrak{Z}\left(  h_{n}\right)  $. This proves
(\ref{sol.pieri.omega.a.1}).

Now, recall that $f=\sum_{\lambda\in\Par }a_{\lambda}s_{\lambda
}$. Hence, $fh_{n}=\left(  \sum_{\lambda\in\Par }a_{\lambda
}s_{\lambda}\right)  h_{n}=\sum_{\lambda\in\Par }a_{\lambda
}s_{\lambda}h_{n}$. Applying the map $\mathfrak{Z}$ to both sides of this
equality, we obtain
\begin{align*}
\mathfrak{Z}\left(  fh_{n}\right)   &  =\mathfrak{Z}\left(  \sum_{\lambda
\in\Par }a_{\lambda}s_{\lambda}h_{n}\right)  =\sum_{\lambda
\in\Par }a_{\lambda}\underbrace{\mathfrak{Z}\left(  s_{\lambda
}h_{n}\right)  }_{\substack{=\mathfrak{Z}\left(  s_{\lambda}\right)
\cdot\mathfrak{Z}\left(  h_{n}\right)  \\
\text{(by (\ref{sol.pieri.omega.a.1}))}}}\ \ \ \ \ \ \ \ \ \ 
\left(  \text{since the map }\mathfrak{Z}
\text{ is }\kk \text{-linear}\right) \\
&  =\sum_{\lambda\in\Par }a_{\lambda}\mathfrak{Z}\left(
s_{\lambda}\right)  \cdot\mathfrak{Z}\left(  h_{n}\right)
=\underbrace{\left(  \sum_{\lambda\in\Par }a_{\lambda
}\mathfrak{Z}\left(  s_{\lambda}\right)  \right)  }_{\substack{=\mathfrak{Z}
\left(  \sum_{\lambda\in\Par }a_{\lambda}s_{\lambda}\right)
\\\text{(since the map }\mathfrak{Z}\text{ is }\kk \text{-linear)}}}
\cdot\mathfrak{Z}\left(  h_{n}\right) \\
&  =\mathfrak{Z}\left(  \underbrace{\sum_{\lambda\in\Par}
a_{\lambda}s_{\lambda}}_{\substack{=f\\\text{(since }f=\sum_{\lambda
\in\Par }a_{\lambda}s_{\lambda}\text{)}}}\right)  \cdot
\mathfrak{Z}\left(  h_{n}\right)  =\mathfrak{Z}\left(  f\right)
\cdot\mathfrak{Z}\left(  h_{n}\right)  .
\end{align*}
This proves (\ref{sol.pieri.omega.a.goal}). Hence, Exercise
\ref{exe.pieri.omega}(a) is solved.

(b) We shall show that
\begin{equation}
\mathfrak{Z}\left(  h_{\lambda}\right)  =\omega\left(  h_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.pieri.omega.b.goal}
\end{equation}


\textit{Proof of (\ref{sol.pieri.omega.b.goal}):} We will prove
(\ref{sol.pieri.omega.b.goal}) by induction over $\ell\left(  \lambda\right)
$:

\textit{Induction base:} The equality (\ref{sol.pieri.omega.b.goal}) clearly
holds if $\ell\left(  \lambda\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Let
$\lambda$ be a partition such that $\ell\left(  \lambda\right)  =0$. Then,
$\lambda$ is the empty partition, i.e., we have $\lambda=\varnothing$. Hence,
$h_{\lambda}=h_{\varnothing}=1$, so that $\omega\left(  \underbrace{h_{\lambda
}}_{=1}\right)  =\omega\left(  1\right)  =1$ (since $\omega$ is a
$\kk$-algebra homomorphism). But $h_{\lambda}=1=s_{\varnothing}$, so that
$\mathfrak{Z}\left(  \underbrace{h_{\lambda}}_{=s_{\varnothing}}\right)
=\mathfrak{Z}\left(  s_{\varnothing}\right)  =s_{\varnothing^{t}}$ (by
(\ref{sol.pieri.omega.def}), applied to $\varnothing$ instead of $\lambda$).
Since $\varnothing^{t}=\varnothing$, we have $s_{\varnothing^{t}}
=s_{\varnothing}$. Thus, $\mathfrak{Z}\left(  h_{\lambda}\right)
=s_{\varnothing^{t}}=s_{\varnothing}=1=\omega\left(  h_{\lambda}\right)  $
(since $\omega\left(  h_{\lambda}\right)  =1$). Thus,
(\ref{sol.pieri.omega.b.goal}) is proven under the assumption that
$\ell\left(  \lambda\right)  =0$, qed.}. This completes the induction base.

\textit{Induction step:} Let $L$ be a positive integer. Assume that the
equality (\ref{sol.pieri.omega.b.goal}) holds if $\ell\left(  \lambda\right)
=L-1$. We need to prove that the equality (\ref{sol.pieri.omega.b.goal}) holds
if $\ell\left(  \lambda\right)  =L$.

We have assumed that the equality (\ref{sol.pieri.omega.b.goal}) holds if
$\ell\left(  \lambda\right)  =L-1$. In other words,
\begin{equation}
\mathfrak{Z}\left(  h_{\lambda}\right)  =\omega\left(  h_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda\text{ satisfying }
\ell\left(  \lambda\right)  =L-1. \label{sol.pieri.omega.b.goal.pf.1}
\end{equation}


Now, let $\lambda$ be a partition satisfying $\ell\left(  \lambda\right)  =L$.
Then, we can write $\lambda$ in the form $\lambda=\left(  \lambda_{1},
\lambda_{2},...,\lambda_{L}\right)  $ with all of $\lambda_{1}$, $\lambda
_{2}$, $...$, $\lambda_{L}$ being positive integers. Consider these
$\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{L}$. Notice that $\lambda_{L}$
is well-defined since $L\geq1$ (because $L$ is a positive integer). Thus,
$\lambda_{L}$ is a positive integer.

Since $\left(  \lambda_{1},\lambda_{2},...,\lambda_{L}\right)  =\lambda$ is a
partition, we have $\lambda_{1}\geq\lambda_{2}\geq...\geq\lambda_{L}$. Hence,
$\lambda_{1}\geq\lambda_{2}\geq...\geq\lambda_{L-1}$. Thus, $\left(
\lambda_{1},\lambda_{2},...,\lambda_{L-1}\right)  $ is a partition. Denote
this partition by $\overline{\lambda}$.

By the definition of $h_{\overline{\lambda}}$, we have
$h_{\overline{\lambda}}=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{L-1}}$
(since $\overline{\lambda}
=\left(  \lambda_{1},\lambda_{2},...,\lambda_{L-1}\right)  $). By the
definition of $e_{\overline{\lambda}}$, we have $e_{\overline{\lambda}}
=e_{\lambda_{1}}e_{\lambda_{2}}...e_{\lambda_{L-1}}$ (since
$\overline{\lambda}
=\left(  \lambda_{1},\lambda_{2},...,\lambda_{L-1}\right)  $).

By the definition of $h_{\lambda}$, we have
\begin{align*}
h_{\lambda}  &  =h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{L}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda=\left(  \lambda_{1},
\lambda_{2},...,\lambda_{L}\right)  \right) \\
&  =\underbrace{h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{L-1}}}
_{\substack{=h_{\overline{\lambda}}\\\text{(since }h_{\overline{\lambda}}
=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{L-1}}\text{)}}}h_{\lambda_{L}}
=h_{\overline{\lambda}}h_{\lambda_{L}}.
\end{align*}
Applying the map $\mathfrak{Z}$ to both sides of this equality, we obtain
\[
\mathfrak{Z}\left(  h_{\lambda}\right)
= \mathfrak{Z}\left(  h_{\overline{\lambda}}h_{\lambda_{L}}\right)  .
\]


But $h_{\overline{\lambda}}\in\Lambda$, whereas $\lambda_{L}\in\NN$
(since $\lambda_{L}$ is a positive integer). Hence, Exercise
\ref{exe.pieri.omega}(a) (applied to $h_{\overline{\lambda}}$ and $\lambda_{L}$
instead of $f$ and $n$) yields
$\mathfrak{Z}\left(  h_{\overline {\lambda}}h_{\lambda_{L}}\right)
 =\mathfrak{Z}\left(  h_{\overline{\lambda}}\right)
  \cdot\mathfrak{Z}\left(  h_{\lambda_{L}}\right)  $. Furthermore,
$\lambda_{L}$ is a positive integer, so that $\mathfrak{Z}\left(
h_{\lambda_{L}}\right)  =e_{\lambda_{L}}$ (by (\ref{sol.pieri.omega.hn}),
applied to $\lambda_{L}$ instead of $n$).

On the other hand, recall that $\omega\left(  h_{n}\right)  =e_{n}$ for every
positive integer $n$ (by Proposition~\ref{prop.Lambda.omega-and-S}(b)).
Applying this to $n=\lambda_{L}$, we conclude that
$\omega\left(  h_{\lambda_{L}}\right)  =e_{\lambda_{L}}$ (since $\lambda_{L}$
is a positive integer). Thus, $e_{\lambda_{L}}=\omega\left(  h_{\lambda_{L}}
\right)  $, so that $\mathfrak{Z}\left(  h_{\lambda_{L}}\right)
=e_{\lambda_{L}}=\omega\left(  h_{\lambda_{L}}\right)  $.

But $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{L}$ are positive integers.
Hence, $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{L-1}$ are positive
integers. Thus, the partition $\left(  \lambda_{1},\lambda_{2},...,\lambda
_{L-1}\right)  $ has length $L-1$. In other words, the partition
$\overline{\lambda}$ has length $L-1$ (since $\overline{\lambda}=\left(
\lambda_{1},\lambda_{2},...,\lambda_{L-1}\right)  $). In other words,
$\ell\left(  \overline{\lambda}\right)  =L-1$. Therefore,
(\ref{sol.pieri.omega.b.goal.pf.1}) (applied to $\overline{\lambda}$ instead
of $\lambda$) yields $\mathfrak{Z}\left(  h_{\overline{\lambda}}\right)
=\omega\left(  h_{\overline{\lambda}}\right)  $. Now,
\begin{align*}
\mathfrak{Z}\left(  h_{\lambda}\right)   &  =\mathfrak{Z}\left(
h_{\overline{\lambda}}h_{\lambda_{L}}\right)  =\underbrace{\mathfrak{Z}\left(
h_{\overline{\lambda}}\right)  }_{=\omega\left(  h_{\overline{\lambda}}
\right)  }\cdot\underbrace{\mathfrak{Z}\left(  h_{\lambda_{L}}\right)
}_{=\omega\left(  h_{\lambda_{L}}\right)  }=\omega\left(
h_{\overline{\lambda}}\right)  \cdot\omega\left(  h_{\lambda_{L}}\right) \\
&  =\omega\left(  \underbrace{h_{\overline{\lambda}}h_{\lambda_{L}}}
_{\substack{=h_{\lambda}\\\text{(since }h_{\lambda}=h_{\overline{\lambda}}
h_{\lambda_{L}}\text{)}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\omega\text{ is a }\kk \text{-algebra homomorphism}\right) \\
&  =\omega\left(  h_{\lambda}\right)  .
\end{align*}


Now, forget that we fixed $\lambda$. We thus have proven that $\mathfrak{Z}
\left(  h_{\lambda}\right)  =\omega\left(  h_{\lambda}\right)  $ for every
partition $\lambda$ satisfying $\ell\left(  \lambda\right)  =L$. In other
words, the equality (\ref{sol.pieri.omega.b.goal}) holds if $\ell\left(
\lambda\right)  =L$. This completes the induction step. The induction proof of
(\ref{sol.pieri.omega.b.goal}) is therefore complete.

Now, let us consider the two $\kk$-linear maps $\mathfrak{Z}
:\Lambda\rightarrow\Lambda$ and $\omega:\Lambda\rightarrow\Lambda$. These two
$\kk$-linear maps $\mathfrak{Z}$ and $\omega$ are equal to each other
on every element of the basis $\left( h_{\lambda} \right) _{\lambda
\in\Par }$ of the $\kk$-module $\Lambda$ (because every
$\lambda\in\Par $ satisfies $\mathfrak{Z}\left(  h_{\lambda
}\right)  =\omega\left(  h_{\lambda}\right)  $ (according to
(\ref{sol.pieri.omega.b.goal}))). Hence, these two $\kk$-linear maps
$\mathfrak{Z}$ and $\omega$ must be identical (because it is known that if two
$\kk$-linear maps are equal to each other on every element of a given
basis of their domain, then they must be identical). In other words,
$\mathfrak{Z}=\omega$. This solves Exercise \ref{exe.pieri.omega}(b).

(c) Fix two partitions $\mu$ and $\nu$. The equality
\eqref{symm-structure-constants.1} yields $s_{\mu}s_{\nu}=\sum_{\lambda}
c_{\mu,\nu}^{\lambda}s_{\lambda}$, where the sum ranges over all partitions
$\lambda$. In other words,
\begin{equation}
s_{\mu}s_{\nu}=\sum_{\lambda\in\Par }c_{\mu,\nu}^{\lambda
}s_{\lambda}.
\label{sol.pieri.omega.c.1}
\end{equation}
Applying this to $\mu^{t}$ and $\nu^{t}$ instead of $\mu$ and $\nu$, we obtain
\begin{equation}
s_{\mu^{t}}s_{\nu^{t}}=\sum_{\lambda\in\Par }c_{\mu^{t},\nu^{t}}^{\lambda}
s_{\lambda}.
\label{sol.pieri.omega.c.2}
\end{equation}


Applying the map $\mathfrak{Z}$ to the identity (\ref{sol.pieri.omega.c.1}),
we obtain
\begin{align}
\mathfrak{Z}\left(  s_{\mu}s_{\nu}\right)   &  =\mathfrak{Z}\left(
\sum_{\lambda\in\Par }c_{\mu,\nu}^{\lambda}s_{\lambda}\right)
=\sum_{\lambda\in\Par }c_{\mu,\nu}^{\lambda}
\underbrace{\mathfrak{Z}\left(  s_{\lambda}\right)  }_{\substack{=s_{\lambda
^{t}}\\\text{(by (\ref{sol.pieri.omega.def}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\mathfrak{Z}\text{ is }\kk \text{-linear}\right)
\nonumber\\
&  =\sum_{\lambda\in\Par }\underbrace{c_{\mu,\nu}^{\lambda}}
_{\substack{=c_{\mu,\nu}^{\left(  \lambda^{t}\right)  ^{t}}\\\text{(since
}\lambda=\left(  \lambda^{t}\right)  ^{t}\text{)}}}s_{\lambda^{t}}
=\sum_{\lambda\in\Par }c_{\mu,\nu}^{\left(  \lambda^{t}\right)
^{t}}s_{\lambda^{t}}=\sum_{\lambda\in\Par }c_{\mu,\nu}^{\lambda^{t}}
s_{\lambda}\label{sol.pieri.omega.c.4}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }\lambda\text{ for }\lambda^{t}\text{ in the sum,
since}\\
\text{the map }\Par\rightarrow\Par,\ \lambda\mapsto\lambda^{t}\text{ is
a bijection}
\end{array}
\right)  .\nonumber
\end{align}
But recall that the map $\omega$ is a $\kk$-algebra homomorphism. Since
$\mathfrak{Z}=\omega$ (by Exercise \ref{exe.pieri.omega}(b)), this rewrites as
follows: The map $\mathfrak{Z}$ is a $\kk$-algebra homomorphism. Hence,
\begin{align*}
\mathfrak{Z}\left(  s_{\mu}s_{\nu}\right)   &  =\underbrace{\mathfrak{Z}
\left(  s_{\mu}\right)  }_{\substack{=s_{\mu^{t}}\\\text{(by
(\ref{sol.pieri.omega.def}), applied to }\lambda=\nu\text{)}}}\cdot
\underbrace{\mathfrak{Z}\left(  s_{\nu}\right)  }_{\substack{=s_{\nu^{t}}
\\\text{(by (\ref{sol.pieri.omega.def}), applied to }\lambda=\nu\text{)}}}\\
&  =s_{\mu^{t}}s_{\nu^{t}}=\sum_{\lambda\in\Par }
c_{\mu^{t},\nu^{t}}^{\lambda}s_{\lambda}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.pieri.omega.c.2})}\right)  .
\end{align*}
Subtracting this equality from (\ref{sol.pieri.omega.c.4}), we obtain
\[
\mathfrak{Z}\left(  s_{\mu}s_{\nu}\right)  -\mathfrak{Z}\left(  s_{\mu}s_{\nu
}\right)  =\sum_{\lambda\in\Par }c_{\mu,\nu}^{\lambda^{t}}s_{\lambda}
-\sum_{\lambda\in\Par }c_{\mu^{t},\nu^{t}}^{\lambda
}s_{\lambda}=\sum_{\lambda\in\Par }\left(  c_{\mu,\nu}^{\lambda^{t}}
-c_{\mu^{t},\nu^{t}}^{\lambda}\right)  s_{\lambda}.
\]
Hence,
\[
\sum_{\lambda\in\Par }\left(  c_{\mu,\nu}^{\lambda^{t}}
-c_{\mu^{t},\nu^{t}}^{\lambda}\right)  s_{\lambda}=\mathfrak{Z}\left(  s_{\mu
}s_{\nu}\right)  -\mathfrak{Z}\left(  s_{\mu}s_{\nu}\right)  =0.
\]
In particular, the sum $\sum_{\lambda\in\Par }\left(  c_{\mu
,\nu}^{\lambda^{t}}-c_{\mu^{t},\nu^{t}}^{\lambda}\right)  s_{\lambda}$ is
well-defined; hence, all but finitely many $\lambda\in\Par $
satisfy $c_{\mu,\nu}^{\lambda^{t}}-c_{\mu^{t},\nu^{t}}^{\lambda}=0$.

But recall that $\left(  s_{\lambda}\right)  _{\lambda\in\Par }$
is a basis of the $\kk$-module $\Lambda$, and thus a linearly
independent family. Hence, if $\left(  \alpha_{\lambda}\right)  _{\lambda
\in\Par }\in\kk ^{\Par }$ is a family of
elements of $\kk$ such that $\left(  \text{all but finitely many
}\lambda\in\Par \text{ satisfy }\alpha_{\lambda}=0\right)  $ and
$\sum_{\lambda\in\Par }\alpha_{\lambda}s_{\lambda}=0$, then
$\left(  \alpha_{\lambda}\right)  _{\lambda\in\Par }=\left(
0\right)  _{\lambda\in\Par }$. We can apply this to $\left(
\alpha_{\lambda}\right)  _{\lambda\in\Par }=\left(  c_{\mu,\nu
}^{\lambda^{t}}-c_{\mu^{t},\nu^{t}}^{\lambda}\right)  _{\lambda\in
\Par }$ (since we know that $\sum_{\lambda\in\Par }
\left(  c_{\mu,\nu}^{\lambda^{t}}-c_{\mu^{t},\nu^{t}}^{\lambda}\right)
s_{\lambda}=0$, and since we know that all but finitely many $\lambda
\in\Par $ satisfy
$c_{\mu,\nu}^{\lambda^{t}}-c_{\mu^{t},\nu^{t}}^{\lambda}=0$). As a
result, we obtain $\left(  c_{\mu,\nu}^{\lambda^{t}}
-c_{\mu^{t},\nu^{t}}^{\lambda}\right)  _{\lambda\in\Par }
=\left(  0\right)  _{\lambda\in\Par }$. In other words, every
$\lambda\in\Par $ satisfies $c_{\mu,\nu}^{\lambda^{t}}
-c_{\mu^{t},\nu^{t}}^{\lambda}=0$. In other words,
\begin{equation}
\text{every }\lambda\in\Par \text{ satisfies }c_{\mu,\nu
}^{\lambda^{t}}=c_{\mu^{t},\nu^{t}}^{\lambda}. \label{sol.pieri.omega.c.6}
\end{equation}


Now, let $\lambda$ be a partition. Applying (\ref{sol.pieri.omega.c.6}) to
$\lambda^{t}$ instead of $\lambda$, we obtain $c_{\mu,\nu}^{\left(
\lambda^{t}\right)  ^{t}}=c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$. Since $\left(
\lambda^{t}\right)  ^{t}=\lambda$, this rewrites as $c_{\mu,\nu}^{\lambda
}=c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$. This solves
Exercise~\ref{exe.pieri.omega}(c).

(d) Let $\mu$ and $\lambda$ be two partitions such that $\mu\subseteq\lambda$.
In Remark \ref{third-L-R-interpretation-remark}, it has been shown that
$s_{\lambda/\mu}=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum ranges
over all partitions $\nu$. In other words,
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in\Par}c_{\mu,\nu}^{\lambda}s_{\nu} .
\label{sol.pieri.omega.d.1}
\end{equation}
Applying this to $\lambda^{t}$ and $\mu^{t}$ instead of $\lambda$ and $\mu$,
we obtain
\begin{equation}
s_{\lambda^{t}/\mu^{t}}=\sum_{\nu\in\Par}c_{\mu^{t},\nu
}^{\lambda^{t}}s_{\nu}
\label{sol.pieri.omega.d.2}
\end{equation}
(since $\mu^{t}\subseteq\lambda^{t}$ (because $\mu\subseteq\lambda$)).
Applying the map $\mathfrak{Z}$ to both sides of the identity
(\ref{sol.pieri.omega.d.1}), we obtain
\begin{align*}
\mathfrak{Z}\left(  s_{\lambda/\mu}\right)    & =\mathfrak{Z}\left(  \sum
_{\nu\in\Par }c_{\mu,\nu}^{\lambda}s_{\nu}\right)  =\sum_{\nu
\in\Par }c_{\mu,\nu}^{\lambda}\underbrace{\mathfrak{Z}\left(
s_{\nu}\right)  }_{\substack{=s_{\nu^{t}}\\\text{(by
(\ref{sol.pieri.omega.def}), applied to }\nu\\\text{instead of }
\lambda\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }
\mathfrak{Z}\text{ is }\kk \text{-linear}\right)  \\
& =\sum_{\nu\in\Par }\underbrace{c_{\mu,\nu}^{\lambda}}
_{\substack{=c_{\mu^{t},\nu^{t}}^{\lambda^{t}}\\\text{(by Exercise
\ref{exe.pieri.omega}(c))}}}s_{\nu^{t}}=\sum_{\nu\in\Par }
c_{\mu^{t},\nu^{t}}^{\lambda^{t}}s_{\nu^{t}}=\sum_{\nu\in\Par }
c_{\mu^{t},\nu}^{\lambda^{t}}s_{\nu}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }\nu\text{ for }\nu^{t}\text{ in the sum,
since}\\
\text{the map }\Par\rightarrow\Par,\ \nu\mapsto\nu^{t}\text{ is
a bijection}
\end{array}
\right)  \\
& =s_{\lambda^{t}/\mu^{t}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.pieri.omega.d.2})}\right)  .
\end{align*}
Since $\mathfrak{Z}=\omega$ (by Exercise \ref{exe.pieri.omega}(b)), this
rewrites as $\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}$.
This proves the first identity of \eqref{skew-Schur-antipodes}.

It is clear that the skew Schur function $s_{\lambda/\mu}$ is homogeneous of
degree $\left\vert \lambda/\mu\right\vert $. In other words, $s_{\lambda/\mu
}\in\Lambda_{\left\vert \lambda/\mu\right\vert }$. Hence,
(\ref{antipode-versus-omega-in-Sym}) (applied to $f=s_{\lambda/\mu}$ and
$n=\left\vert \lambda/\mu\right\vert $) yields
\[
S\left(  s_{\lambda/\mu}\right)  =\left(  -1\right)  ^{\left\vert \lambda
/\mu\right\vert }\underbrace{\omega\left(  s_{\lambda/\mu}\right)
}_{=s_{\lambda^{t}/\mu^{t}}}=\left(  -1\right)  ^{\left\vert \lambda
/\mu\right\vert }s_{\lambda^{t}/\mu^{t}}.
\]
This proves the second identity of \eqref{skew-Schur-antipodes}. The proof of
\eqref{skew-Schur-antipodes} is thus complete.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cauchy.dual}}

\textit{Solution to Exercise~\ref{exe.cauchy.dual}.} (a)

\textit{First solution to Exercise~\ref{exe.cauchy.dual}(a):}
Let us first
show that $\prod_{i,j=1}^{\infty}\left(  1+x_{i}y_{j}\right)  =\sum
_{\lambda\in\Par  }e_{\lambda}\left( \xx \right)
m_{\lambda}\left(  \yy \right)  $.

In fact, in the proof of Proposition~\ref{prop-Two-other-Cauchy-expansions},
we showed that $\prod_{j=1}^{\infty}\sum_{n\geq0} h_{n} \left( \xx
\right)  y_{j}^{n}=\sum_{\lambda\in\Par } h_{\lambda}\left(
\xx \right)  m_{\lambda}\left(  \yy \right)  $. The same
argument (with all appearances of the letter ``$h$''
replaced by the letter ``$e$'') shows that
\begin{equation}
\prod_{j=1}^{\infty} \sum_{n\geq0} e_{n} \left( \xx \right)
y_{j}^{n} =\sum_{\lambda\in\Par } e_{\lambda} \left(
\xx \right)  m_{\lambda} \left(  \yy \right)  .
\label{sol.cauchy.dual.a.1}
\end{equation}


But \eqref{e-generating-function-no-HE} yields
\begin{equation}
\prod_{i=1}^{\infty}\left(  1+x_{i}t\right)  =\sum_{n\geq0}e_{n}\left(
\xx \right)  t^{n}
\label{sol.cauchy.dual.a.e}
\end{equation}
in the ring $\left(  \kk \left[  \left[ \xx \right]  \right]
\right)  \left[  \left[  t\right]  \right]  $. For every $j\in\left\{
1,2,3,...\right\}  $, we have
\begin{equation}
\prod_{i=1}^{\infty}\left(  1+x_{i}y_{j}\right)  =\sum_{n\geq0}e_{n}\left(
\xx \right)  y_{j}^{n}
\label{sol.cauchy.dual.a.e2}
\end{equation}
in the ring $\left(  \kk \left[  \left[ \xx \right]  \right]
\right)  \left[  \left[  \yy \right]  \right]  $ (in fact, this
results from \eqref{sol.cauchy.dual.a.e} by substituting $y_{j}$ for $t$).
Thus, \eqref{sol.cauchy.dual.a.e2} holds in $\kk  \left[  \left[
\xx , \yy \right]  \right]  $ (since $\left(   \kk 
\left[  \left[ \xx \right]  \right]  \right)  \left[  \left[
\yy \right]  \right]  = \kk \left[  \left[  \xx ,
\yy \right]  \right]  $ as rings). Now,
\begin{align}
\underbrace{\prod_{i,j=1}^{\infty}}_{=\prod_{j=1}^{\infty}\prod_{i=1}^{\infty
}}\left(  1+x_{i}y_{j}\right)   &  =\prod_{j=1}^{\infty}\underbrace{\prod
_{i=1}^{\infty}\left(  1+x_{i}y_{j}\right)  }_{\substack{=\sum_{n\geq0}%
e_{n}\left( \xx \right)  y_{j}^{n}\\\text{(by
\eqref{sol.cauchy.dual.a.e2})}}}=\prod_{j=1}^{\infty}\sum_{n\geq0}e_{n}\left(
\xx \right)  y_{j}^{n}\nonumber\\
&  =\sum_{\lambda\in\Par  }e_{\lambda}\left(  \xx
\right)  m_{\lambda}\left(  \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{sol.cauchy.dual.a.1}}\right)  .
\label{sol.cauchy.dual.a.part1done}
\end{align}


Next, we shall show that $\sum_{\lambda\in\Par  }s_{\lambda
}\left( \xx \right)  s_{\lambda^{t}}\left(  \yy \right)
=\sum_{\lambda\in\Par  }e_{\lambda}\left(  \xx
\right)  m_{\lambda}\left(  \yy \right)  $.

In fact, consider the $\kk$-algebra $\left(   \kk \left[  \left[
\xx\right]  \right]  \right)  \left[  \left[  \yy\right]
\right]  = \kk \left[  \left[  \xx,\yy\right]  \right]
$. This $\kk$-algebra $\left(   \kk \left[  \left[ \xx
\right]  \right]  \right)  \left[  \left[  \yy\right]  \right]  $ is
a $\kk \left[  \left[  \yy\right]  \right]  $-algebra, and
contains $\Lambda\left[  \left[  \yy\right]  \right]  $ as a
$\kk \left[  \left[  \yy\right]  \right]  $-subalgebra.

The equality \eqref{Cauchy-identity} yields
\begin{equation}
\sum_{\lambda\in\Par  }s_{\lambda}\left( \xx \right)
s_{\lambda}\left(  \yy \right)  =\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda\in\Par  }h_{\lambda
}\left( \xx \right)  m_{\lambda}\left(  \yy \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{Two-other-Cauchy-expansions}}%
\right)  . \label{sol.cauchy.dual.a.5}%
\end{equation}
This is an equality in $\Lambda\left[  \left[  \yy \right]  \right]
$ (because $s_{\lambda}\left( \xx \right)  $ and $h_{\lambda}\left(
\xx \right)  $ belong to $\Lambda$ for every $\lambda\in
\Par $).

Recall that $\omega:\Lambda\rightarrow\Lambda$ is a $\kk$-algebra
homomorphism. It thus induces a $\kk \left[  \left[ \yy \right]
\right]  $-algebra homomorphism $\omega\left[  \left[ \yy \right]
\right]  :\Lambda\left[  \left[ \yy \right]  \right]  \rightarrow
\Lambda\left[  \left[ \yy \right]  \right]  $ which sends every
$q\in\Lambda$ to $\omega\left(  q\right)  $, and is continuous with respect to
the usual topology\footnote{The usual topology on a power series ring
$Z\left[  \left[  \yy\right]  \right]  $ (where $Z$ is a commutative
ring) is the direct-product topology obtained by viewing the set $Z\left[
\left[  \yy\right]  \right]  $ as a direct product of many copies of
$Z$ (this is done by identifying every power series with the family of its
coefficients), each of which is endowed with the discrete topology. In this
topology, a sequence (or, more generally, a net) $\left(  f_{i}\right)  _{i}$
of power series converges to a power series $f$ if and only if for every
monomial $\mathfrak{m}$, the sequence (resp. net) $\left(  \text{the
coefficient of }\mathfrak{m}\text{ in }f_{i}\right)  _{i}$ converges to
$\left(  \text{the coefficient of }\mathfrak{m}\text{ in }f\right)  $ with
respect to the discrete topology.
(The notion of a \emph{net} is a generalization of the notion of a
sequence; it is useful in topology. See \cite{Vermeeren-nets} for an
introduction to it.)} on $\Lambda\left[  \left[  \yy
\right]  \right]  $. Applying this homomorphism $\omega\left[  \left[
\yy \right]  \right]  $ to both sides of (\ref{sol.cauchy.dual.a.5}), we
obtain
\begin{equation}
\sum_{\lambda\in\Par }\omega\left(  s_{\lambda}\left(
\xx \right)  \right)  s_{\lambda}\left( \yy \right)
=\sum_{\lambda\in\Par }\omega\left(  h_{\lambda}\left(
\xx \right)  \right)  m_{\lambda}\left( \yy \right)
\label{sol.cauchy.dual.a.5-omega}%
\end{equation}
(because $\omega\left[  \left[ \yy \right]  \right]  $, being a
$\kk \left[  \left[ \yy \right]  \right]  $-algebra homomorphism,
leaves the $s_{\lambda}\left( \yy \right)  $ and $m_{\lambda}\left(
\yy \right)  $ terms unchanged, while the $s_{\lambda}\left(
\xx \right)  $ and $h_{\lambda}\left( \xx \right)  $
terms are elements of $\Lambda$ and thus are transformed as by $\omega$).

But we know that
\begin{equation}
\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}%
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.cauchy.dual.a.omega.s2}
\end{equation}
(This follows from the first equality in \eqref{skew-Schur-antipodes} by
setting $\mu=\varnothing$.) Using (\ref{eq.exe.Lambda.omegahlam.h}) and
(\ref{sol.cauchy.dual.a.omega.s2}), we can rewrite
(\ref{sol.cauchy.dual.a.5-omega}) as
\[
\sum_{\lambda\in\Par }s_{\lambda^{t}}\left(  \xx
\right)  s_{\lambda}\left( \yy \right)  =\sum_{\lambda
\in\Par }e_{\lambda}\left( \xx \right)  m_{\lambda
}\left( \yy \right)  .
\]
Hence,
\begin{align}
\sum_{\lambda\in\Par }e_{\lambda}\left( \xx \right)
m_{\lambda}\left( \yy \right)   &  =\sum_{\lambda\in
\Par }s_{\lambda^{t}}\left( \xx \right)  s_{\lambda
}\left( \yy \right)  =\sum_{\lambda\in\Par %
}\underbrace{s_{\left(  \lambda^{t}\right)  ^{t}}\left(  \xx
\right)  }_{\substack{=s_{\lambda}\left(  \xx \right)  \\\text{(since
}\left(  \lambda^{t}\right)  ^{t}=\lambda\text{)}}}s_{\lambda^{t}}\left(
\yy\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\lambda^{t}\text{ for }\lambda\text{ in the sum,
since the map}\\
\Par \rightarrow\Par ,\ \lambda\mapsto\lambda
^{t}\text{ is a bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{\lambda\in\Par }s_{\lambda}\left(  \xx
\right)  s_{\lambda^{t}}\left( \yy \right)  .
\label{sol.cauchy.dual.a.omega.leftside}
\end{align}
Combined with \eqref{sol.cauchy.dual.a.part1done}, this yields
\begin{equation}
\prod_{i,j=1}^{\infty}\left(  1+x_{i}y_{j}\right)  =\sum_{\lambda
\in\Par }e_{\lambda}\left( \xx \right)  m_{\lambda
}\left( \yy \right)  =\sum_{\lambda\in\Par }s_{\lambda
}\left( \xx \right)  s_{\lambda^{t}}\left( \yy \right)
.
\label{sol.cauchy.dual.a.omega.end}
\end{equation}
This solves Exercise~\ref{exe.cauchy.dual}(a).

\textit{Second solution to Exercise~\ref{exe.cauchy.dual}(a):}
We can prove \eqref{sol.cauchy.dual.a.part1done} as in the First
solution to Exercise~\ref{exe.cauchy.dual}(a). Thus, in order to
solve Exercise~\ref{exe.cauchy.dual}(a), it remains to verify
\begin{equation}
\prod_{i,j=1}^{\infty} \left(1 + x_i y_j\right)
= \sum_{\lambda \in \Par} s_\lambda\left(\xx\right) s_{\lambda^t}\left(\yy\right) .
\label{sol.cauchy.dual.a.secondproof.goal}
\end{equation}

Instead of proving \eqref{sol.cauchy.dual.a.secondproof.goal}, we will
verify the identity
\begin{equation}
\prod_{i,j=1}^{\infty} \left(1 + t x_i y_j\right)
= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_\lambda\left(\xx\right) s_{\lambda^t}\left(\yy\right)
\label{sol.cauchy.dual.a.secondproof.goal2}
\end{equation}
in the ring $R\left(\xx,\yy\right)\left[\left[t\right]\right]$. This
identity will clearly yield \eqref{sol.cauchy.dual.a.secondproof.goal}
(by substituting $t = 1$), and thus conclude the solution of
Exercise~\ref{exe.cauchy.dual}(a).

We shall prove \eqref{sol.cauchy.dual.a.secondproof.goal2} in a
similar way to
how we proved \eqref{Cauchy-identity-with-t}, but using a variation
on the RSK correspondence. This is not in itself a new idea; indeed,
this is how the equivalent identity
\eqref{sol.cauchy.dual.a.secondproof.goal}
is proven in \cite[Theorem 7.14.3]{Stanley}, in
\cite[\S 7]{Knuth-RSK} and in \cite[Theorem 4.8.6]{Sagan}. However,
instead of using the dual RSK algorithm (also known as the
RSK$^\ast$ algorithm; see \cite[\S 7.14]{Stanley},
\cite[\S 5]{Knuth-RSK} and \cite[Theorem 4.8.5]{Sagan} for it)
like these proofs do, we introduce a variation of the RSK algorithm
that relies on the same RS-insertion operation but changes the
order in which the biletters are processed. (We will reprove
\eqref{sol.cauchy.dual.a.secondproof.goal2} using the dual RSK
algorithm in the Third solution further below.)

For a given partition $\lambda$, let us define a \dfn{row-strict
tableau} of shape $\lambda$ to be an assignment $T$ of entries in
$\left\{1,2,3,\ldots\right\}$ to the cells of the Ferrers diagram
for $\lambda$ which is strictly increasing left-to-right in rows, and
weakly increasing top-to-bottom in columns. It is clear that if
$\lambda$ is a partition, then the row-strict tableaux of shape
$\lambda$ are in 1-to-1 correspondence with the column-strict
tableaux of shape $\lambda^t$, and the correspondence is given
by transposing the tableau (i.e., taking whatever entry was assigned
to a cell $(i,j)$ in the input tableau, and reassigning it to the
cell $(j,i)$ in the output tableau). Hence, for every partition
$\lambda$, we have
\begin{align}
\sum\limits_{\substack{Q \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(Q\right)}
&= \sum\limits_{\substack{T \text{ is a column-strict} \\ \text{tableau of
shape } \lambda^t}} \xx^{\cont\left(T\right)}
\nonumber \\
&= s_{\lambda^t} \qquad \qquad
\left(\text{since this is how }s_{\lambda^t}\text{ is defined}\right)
\label{sol.cauchy.dual.a.secondproof.slat.x}
\end{align}
(where $\xx^{\cont\left(Q\right)}$ is defined for a row-strict
tableau $Q$ in the same way as it is defined for a column-strict
tableau $Q$). Substituting $\yy$ for $\xx$ in this equality,
we obtain
\begin{equation}
\sum\limits_{\substack{Q \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \yy^{\cont\left(Q\right)}
= s_{\lambda^t} \left(\yy\right) .
\label{sol.cauchy.dual.a.secondproof.slat}
\end{equation}
We also have
\begin{equation}
\sum\limits_{\substack{P \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(P\right)}
= s_{\lambda} \left(\xx\right)
\label{sol.cauchy.dual.a.secondproof.sla}
\end{equation}
(because this is how $s_{\lambda} = s_{\lambda} \left(\xx\right)$
is defined) for every partition $\lambda$.

A \dfn{tableau-cotableau pair} will mean a pair
$\left(P, Q\right)$ such that $P$ is a column-strict
tableau, $Q$ is a row-strict tableau, and $P$ and $Q$
both have shape $\lambda$ for one and the same partition
$\lambda$. Multiplying the identities
\eqref{sol.cauchy.dual.a.secondproof.sla} and
\eqref{sol.cauchy.dual.a.secondproof.slat} and multiplying
the result with $t^{\left|\lambda\right|}$, we obtain
\[
t^{\left|\lambda\right|}
\sum\limits_{\substack{
\left(P, Q\right) \text{ is a pair with} \\
P \text{ being a column-strict tableau} \\
\text{of shape } \lambda \text{, and } Q \text{ being} \\
\text{a row-strict tableau of shape }\lambda
}} \xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= t^{\left|\lambda\right|}
s_{\lambda} \left(\xx\right) s_{\lambda^t} \left(\yy\right)
\]
for every partition $\lambda$. Summing this equality over
all partitions $\lambda$, we obtain
\begin{equation}
\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda} \left(\xx\right) s_{\lambda^t} \left(\yy\right) ,
\label{sol.cauchy.dual.a.secondproof.rhs}
\end{equation}
where the sum on the left hand side is over all
tableau-cotableau pairs $\left(P, Q\right)$ and where
$\lambda$ denotes the common shape of $P$ and $Q$.

We shall use all notations introduced in the proof of
Theorem~\ref{thm.cauchy-identity}. Define the
\dfn{antilexicographic order} $\leq_{alex}$ to be the total order
on the set of all biletters which is given by
\[
\binom{i_1}{j_1} \leq_{alex} \binom{i_2}{j_2}
\quad \Longleftrightarrow \quad
\left( \text{we have } i_1 \leq i_2 \text{, and if }
i_1 = i_2 \text{, then } j_1 \geq j_2 \right).
\]
We denote by $<_{alex}$ the (strict) smaller relation of this
order. A \dfn{strict cobiword} will mean an array
$\binom{\ii}{\jj} = \binom{i_1\ldots i_\ell}{j_1\ldots j_\ell}$
in which the biletters satisfy
$\binom{i_1}{j_1} <_{alex} \cdots <_{alex} \binom{i_\ell}{j_\ell}$
(that is, the biletters are distinct and ordered).
Strict cobiwords are clearly in 1-to-1 correspondence
with sets (not multisets!) of biletters.
Now, the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is
$\prod_{i,j=1}^\infty \left(1 + t x_i y_j\right)
= \prod_{i,j=1}^\infty \left(1 + t x_j y_i\right)$
(here, we substituted $\left(j,i\right)$ for the
index $\left(i,j\right)$ in the product), and thus can be
rewritten as the sum of
$t^\ell \left(x_{j_1} y_{i_1}\right) \left(x_{j_2} y_{i_2}\right)
\cdots \left(x_{j_\ell} y_{i_\ell}\right)$
over all sets
$\left\{\binom{i_1}{j_1}, \ldots, \binom{i_\ell}{j_\ell}\right\}$
of biletters. Thus,
the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^\ell \xx^{\cont(\jj)} \yy^{\cont(\ii)}$ over all
strict cobiwords $\binom{\ii}{\jj}$, where $\ell$ stands
for the number of biletters in the strict cobiword. Meanwhile, we
know that the right hand side of
\eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}$
over all tableau-cotableau pairs $\left(P, Q\right)$
(because of
\eqref{sol.cauchy.dual.a.secondproof.rhs}). Thus,
in order to prove \eqref{sol.cauchy.dual.a.secondproof.goal2},
we only need to construct a bijection between the
strict cobiwords $\binom{\ii}{\jj}$ and the
tableau-cotableau pairs $\left(P,Q\right)$, which has
the property that
\begin{align*}
\cont\left(\ii\right) &= \cont\left(Q\right); \\
\cont\left(\jj\right) &= \cont\left(P\right).
\end{align*}
\footnote{Such a bijection will then automatically satisfy
$\left|\lambda\right| = \ell$, where $\lambda$ is the (common)
shape of $P$ and $Q$, and where $\ell$ is the length of the
strict cobiword $\binom{\ii}{\jj}$. This is because
$\left|\lambda\right| = \left|\cont\left(Q\right)\right|$
and $\ell = \left|\cont\left(\ii\right)\right|$.}

This bijection is the \dfn{coRSK algorithm}, which we
shall now define.\footnote{This algorithm appears in
Fulton's \cite[\S A.4]{Fulton} as construction (1d).}

Let $\binom{\ii}{\jj}$ be a strict cobiword.
Starting with the pair $(P_0,Q_0)=(\varnothing,\varnothing)$ and
$m=0$, the algorithm applies the following steps
(see Example~\ref{sol.cauchy.dual.a.secondproof.coRSK-example} below):
\begin{itemize}
\item If $i_{m+1}$ does not exist (that is, $m$ is the length of
$\ii$), stop.
\item Apply RS-insertion to the column-strict tableau $P_m$
and the letter $j_{m+1}$ (the bottom letter of
$\binom{i_{m+1}}{j_{m+1}}$). Let $P_{m+1}$ be the resulting
column-strict tableau, and let $c_{m+1}$ be the resulting
corner cell.
\item Create $Q_{m+1}$ from $Q_m$ by adding the top letter
$i_{m+1}$ of $\binom{i_{m+1}}{j_{m+1}}$ to $Q_m$ in the cell
$c_{m+1}$ (which, as we recall, is the extra corner cell of
$P_{m+1}$ not present in $P_m$).
\item Set $m$ to $m+1$.
\end{itemize}
After all of the biletters have been thus processed,
the result of the coRSK algorithm
is $(P_\ell,Q_\ell)=:(P,Q)$.

\begin{example}
\label{sol.cauchy.dual.a.secondproof.coRSK-example}
The term in the expansion of the left side of
\eqref{sol.cauchy.dual.a.secondproof.goal} corresponding to
\[
(x_4 y_1)(x_2 y_1)(x_1 y_2)(x_3 y_4)(x_1 y_4)(x_2 y_5)
\]
is the strict cobiword
$\binom{\ii}{\jj}=\binom{112445}{421312}$, and the coRSK
algorithm applied to this cobiword proceeds
as follows:
\[
\begin{array}{rcl|rcl}
P_0&= &\varnothing & Q_0 &=& \varnothing \\
 & & & & &\\
P_1&= &\begin{matrix} 4 \end{matrix}
 & Q_1 &= &\begin{matrix} 1 \end{matrix} \\
 & & & & &\\
P_2&= &\begin{matrix} 2 \\ 4 \end{matrix}
 & Q_2 &= &\begin{matrix} 1 \\ 1  \end{matrix} \\
 & & & & &\\
P_3&= &\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}
 & Q_3 &= &\begin{matrix} 1 \\ 1 \\ 3 \end{matrix} \\
 & & & & &\\
P_4&= &\begin{matrix} 1 & 3 \\ 2 \\ 4 \end{matrix}
 & Q_4 &= &\begin{matrix} 1 & 4 \\ 1 \\ 2 \end{matrix} \\
 & & & & &\\
P_5&= &\begin{matrix} 1 & 1 \\ 2 & 3 \\ 4 \end{matrix}
 & Q_5 &= &\begin{matrix} 1 & 4 \\ 1 & 4 \\ 2 \end{matrix} \\
 & & & & &\\
P:=P_6&= &\begin{matrix} 1 & 1 & 2 \\ 2 & 3 \\ 4 \end{matrix}
 & Q:=Q_6 &= &\begin{matrix} 1 & 4 & 5 \\ 1 & 4 \\ 2 \end{matrix} \\
 & & & & &\\
\end{array}
\]
\end{example}

It is clear that $P_m$ remains a column-strict tableau of some
Ferrers shape throughout the execution of the coRSK algorithm, and
that $Q_m$ remains a filling of the same shape as $P_m$ which is
(at least) weakly increasing left-to-right along rows and weakly
increasing top-to-bottom in columns. But we can also see that
$Q_m$ is strictly increasing left-to-right along rows\footnote{Indeed,
this follows from the observation that when one has a string of equal
letters $i_m=i_{m+1}=\cdots=i_{m+r}$ on top of the strict cobiword,
then the bottom letters bumped in are
$j_m > j_{m+1} > \cdots > j_{m+r}$, and therefore (as a consequence
of the last claim of part (b) of the row bumping lemma) the new cells
form a \emph{vertical strip}, that is, no two of these cells lie
in the same row. Actually, more can be said: Each of these new cells
(except for the first one) is in a row further down than the
previous one. We will use this stronger fact further below.}, so
that $Q_m$ is a row-strict tableau. Thus,
the result $\left(P, Q\right)$ of the coRSK algorithm
is a tableau-cotableau pair.

To see that the coRSK algorithm is a bijection, we show how to
recover $\binom{\ii}{\jj}$ from $(P,Q)$.  This is done by
reverse bumping in the same way as for the usual RSK algorithm,
with the only difference that now $Q_m$ is obtained by removing
the bottommost (rather than the rightmost)
occurrence of the letter $i_{m+1}$ from $Q_{m+1}$.
\footnote{It necessarily has to be the bottommost occurrence, since
(according to the previous footnote) the cell into which
$i_{m+1}$ was filled at the step from $Q_m$ to $Q_{m+1}$ lies further
down than any existing cell of $Q_m$ containing the letter $i_{m+1}$.}

Finally, to see that the coRSK map is surjective, one needs to show
that the reverse bumping procedure can be applied to any
tableau-cotableau pair $(P,Q)$, and will result in a strict
cobiword $\binom{\ii}{\jj}$.
We leave this verification to the reader.\footnote{It is easy to
see that repeatedly applying reverse bumping to $(P,Q)$ will
result in a sequence
$\binom{i_\ell}{j_\ell}, \binom{i_{\ell-1}}{j_{\ell-1}},
\ldots, \binom{i_1}{j_1}$ of biletters such that applying the
coRSK algorithm to
$\binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$ gives back
$(P,Q)$. The question is why we have
$\binom{i_1}{j_1} <_{alex} \cdots <_{alex} \binom{i_\ell}{j_\ell}$.
Since the chain of inequalities $i_1 \leq i_2 \leq \cdots \leq i_\ell$
is clear from the choice of entry to reverse-bump, it only remains
to show that for every string $i_m=i_{m+1}=\cdots=i_{m+r}$
of equal top letters, the corresponding bottom letters
strictly decrease (that is,
$j_m > j_{m+1} > \cdots > j_{m+r}$). One way to see
this is the following:
\par
Assume the contrary; i.e., assume that
the bottom letters corresponding
to some string $i_m=i_{m+1}=\cdots=i_{m+r}$ of equal top
letters do not strictly decrease.
Thus, $j_{m+p} \leq j_{m+p+1}$ for some
$p \in \left\{0, 1, \ldots, r-1\right\}$.
Consider this $p$.
\par
Let us consider the cells containing the equal letters
$i_m=i_{m+1}=\cdots=i_{m+r}$ in the tableau $Q_{m+r}$.
Label these cells as $c_m, c_{m+1}, \ldots, c_{m+r}$
from top to bottom (noticing that no two of them lie
in the same row, since $Q_{m+r}$ is row-strict).
By the definition of reverse bumping, the first entry
to be reverse bumped from $P_{m+r}$ is the entry in
position $c_{m+r}$ (since this is the bottommost
occurrence of the letter $i_{m+r}$ in $Q_{m+r}$); then, the next entry
to be reverse bumped is the one in position $c_{m+r-1}$,
etc., moving further and further up.
Thus, for each $q \in \left\{0, 1, \ldots, r\right\}$,
the tableau $P_{m+q-1}$ is obtained from $P_{m+q}$ by
reverse bumping the entry in position $c_{m+q}$.
Hence, conversely, the tableau $P_{m+q}$ is obtained
from $P_{m+q-1}$ by RS-inserting the entry $j_{m+q}$,
which creates the corner cell $c_{m+q}$.
\par
But recall that $j_{m+p} \leq j_{m+p+1}$.
Hence, part (a) of the row bumping lemma (applied to
$P_{m+p-1}$, $j_{m+p}$, $j_{m+p+1}$, $P_{m+p}$,
$c_{m+p}$, $P_{m+p+1}$ and $c_{m+p+1}$
instead of $P$, $j$, $j'$, $P'$, $c$, $P''$ and $c'$)
shows that the cell $c_{m+p+1}$ is in the same
row as the cell $c_{m+p}$ or in a row further
up.
But this contradicts the fact that the cell $c_{m+p+1}$ is
in a row further down than the cell $c_{m+p}$ (since
we have labeled our cells as
$c_m, c_{m+1}, \ldots, c_{m+r}$ from top to bottom,
and no two of them lied in the same row).
This contradiction completes our proof.}

So the coRSK map is a bijection having the required properties.
As we have said, this proves
\eqref{sol.cauchy.dual.a.secondproof.goal2}.
This solves Exercise~\ref{exe.cauchy.dual}(a) again.

[\textit{Remark:} By combining the two solutions of
Exercise~\ref{exe.cauchy.dual}(a) given above, one can obtain a
new proof of the equality \eqref{sol.cauchy.dual.a.omega.s2}
(a proof which does not use \eqref{skew-Schur-antipodes}).
Indeed, using \eqref{eq.exe.Lambda.omegahlam.h}, we can rewrite
\eqref{sol.cauchy.dual.a.5-omega} as
\[
\sum_{\lambda \in \Par} \omega\left(s_\lambda \left(\xx\right)\right)
s_\lambda \left(\yy\right)
= \sum_{\lambda \in \Par} e_\lambda\left(\xx\right)
m_\lambda\left(\yy\right) .
\]
Compared with \eqref{sol.cauchy.dual.a.part1done}, this yields
\begin{align*}
& \sum_{\lambda \in \Par} \omega\left(s_\lambda \left(\xx\right)\right)
s_\lambda \left(\yy\right) \\
&= \prod_{i,j=1}^{\infty} \left(1 + x_i y_j\right)
= \prod_{i,j=1}^{\infty} \left(1 + x_j y_i\right)
\qquad \qquad \left(\text{here, we substituted } \left(j,i\right)
\text{ for }\left(i,j\right)\text{ in the sum}\right) \\
&= \prod_{i,j=1}^{\infty} \left(1 + y_i x_j\right)
= \sum_{\lambda \in \Par} s_\lambda\left(\yy\right) s_{\lambda^t}\left(\xx\right)
\qquad \qquad \left(
\text{by \eqref{sol.cauchy.dual.a.secondproof.goal}, with }
\xx \text{ and } \yy \text{ substituted for } \yy \text{ and }
\xx \right) \\
&= \sum_{\lambda \in \Par} s_{\lambda^t}\left(\xx\right) s_\lambda\left(\yy\right) .
\end{align*}
Since the $s_\lambda\left(\yy\right)$ for $\lambda \in \Par$ are linearly
independent over $\kk\left[\left[\xx\right]\right]$, we can compare
coefficients before $s_\lambda\left(\yy\right)$ in this equality.
As a result, we obtain $\omega\left(s_\lambda \left(\xx\right)\right)
= s_{\lambda^t}\left(\xx\right)$ for every $\lambda \in \Par$. Thus,
\eqref{sol.cauchy.dual.a.omega.s2} is proven again.]

\textit{Third solution to Exercise~\ref{exe.cauchy.dual}(a):}
In order to solve Exercise~\ref{exe.cauchy.dual}(a), it suffices to
verify the identity \eqref{sol.cauchy.dual.a.secondproof.goal2}.
(This can be proven by the same argument as in the Second solution to
Exercise~\ref{exe.cauchy.dual}(a).)

We shall now verify the identity
\eqref{sol.cauchy.dual.a.secondproof.goal2} using the so-called
\emph{dual RSK algorithm} (also known as
\emph{RSK$^\ast$ algorithm})\footnote{We will define this algorithm
further below. It also frequently appears in literature: see, e.g.,
\cite[\S 7.14]{Stanley}, \cite[\S 5]{Knuth-RSK},
\cite[Theorem 4.8.5]{Sagan} and (with different conventions)
\cite[\S A.4.3, Prop. 3]{Fulton}.}. This will be fairly similar to the
Second solution to Exercise~\ref{exe.cauchy.dual}(a) given above, but
not identical to it; in particular, the dual RSK algorithm that we will
introduce below will (unlike the coRSK algorithm from the Second
solution) not rely on the same row bumping operation as the usual
RSK algorithm, but on a somewhat modified version of it.

For a given partition $\lambda$, let us define a \dfn{row-strict
tableau} of shape $\lambda$ to be an assignment $T$ of entries in
$\left\{1,2,3,\ldots\right\}$ to the cells of the Ferrers diagram
for $\lambda$ which is strictly increasing left-to-right in rows, and
weakly increasing top-to-bottom in columns. The equality
\eqref{sol.cauchy.dual.a.secondproof.slat.x} holds\footnote{Indeed,
it can be proven as in the
Second solution to Exercise~\ref{exe.cauchy.dual}(a).}; thus, we have
\begin{align}
\sum\limits_{\substack{P \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(P\right)}
&= \sum\limits_{\substack{Q \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(Q\right)}
= s_{\lambda^t}
\qquad \left(\text{by \eqref{sol.cauchy.dual.a.secondproof.slat.x}}
\right) \nonumber \\
&= s_{\lambda^t} \left(\xx\right) .
\label{sol.cauchy.dual.a.thirdproof.slat}
\end{align}
We also have
\[
\sum\limits_{\substack{Q \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(Q\right)}
= \sum\limits_{\substack{P \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(P\right)}
= s_{\lambda} \left(\xx\right)
\]
(because this is how $s_{\lambda} = s_{\lambda} \left(\xx\right)$
is defined) for every partition $\lambda$. Substituting $\yy$ for
$\xx$ in this equality, we obtain
\begin{equation}
\sum\limits_{\substack{Q \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \yy^{\cont\left(Q\right)}
= s_{\lambda} \left(\yy\right) .
\label{sol.cauchy.dual.a.thirdproof.sla}
\end{equation}

A \dfn{cotableau-tableau pair} will mean a pair
$\left(P, Q\right)$ such that $P$ is a row-strict
tableau, $Q$ is a column-strict tableau, and $P$ and $Q$
both have shape $\lambda$ for one and the same partition
$\lambda$. Multiplying the identities
\eqref{sol.cauchy.dual.a.thirdproof.slat} and
\eqref{sol.cauchy.dual.a.thirdproof.sla} and multiplying
the result with $t^{\left|\lambda\right|}$, we obtain
\[
t^{\left|\lambda\right|}
\sum\limits_{\substack{
\left(P, Q\right) \text{ is a pair with} \\
P \text{ being a row-strict tableau} \\
\text{of shape } \lambda \text{, and } Q \text{ being} \\
\text{a column-strict tableau of shape }\lambda
}} \xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= t^{\left|\lambda\right|}
s_{\lambda^t} \left(\xx\right) s_{\lambda} \left(\yy\right)
\]
for every partition $\lambda$. Summing this equality over
all partitions $\lambda$, we obtain
\[
\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda^t} \left(\xx\right) s_{\lambda} \left(\yy\right) ,
\]
where the sum on the left hand side is over all
cotableau-tableau pairs $\left(P, Q\right)$ and where
$\lambda$ denotes the common shape of $P$ and $Q$. This becomes
\begin{align}
\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
&= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda^t} \left(\xx\right) s_{\lambda} \left(\yy\right)
= \sum_{\lambda \in \Par}
\underbrace{t^{\left|\lambda^t\right|}}_{\substack{
  = t^{\left|\lambda\right|} \\
  \text{(since } \left|\lambda^t\right| = \left|\lambda\right| \text{)}}}
\underbrace{s_{\left(\lambda^t\right)^t} \left(\xx\right)}_{
 \substack{=s_{\lambda} \left(\xx\right) \\ \text{(since }
 \left(\lambda^t\right)^t = \lambda \text{)}}}
s_{\lambda^t} \left(\yy\right)
\nonumber \\
&\qquad \left(
\begin{array}[c]{c}
\text{here, we substituted } \lambda^t \text{ for }\lambda
\text{ in the sum, since the map}\\
\Par \rightarrow\Par ,\ \lambda \mapsto \lambda^t
\text{ is a bijection}
\end{array}
\right)
\nonumber \\
&= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda} \left(\xx\right)
s_{\lambda^t} \left(\yy\right) .
\label{sol.cauchy.dual.a.thirdproof.rhs}
\end{align}

We shall use all notations introduced in the proof of
Theorem~\ref{thm.cauchy-identity}. A \dfn{strict biword} will mean
an array
$\binom{\ii}{\jj} = \binom{i_1\ldots i_\ell}{j_1\ldots j_\ell}$
in which the biletters satisfy
$\binom{i_1}{j_1} <_{lex} \cdots <_{lex} \binom{i_\ell}{j_\ell}$
(that is, the biletters are distinct and ordered with respect to the
lexicographic order).
Strict biwords are clearly in 1-to-1 correspondence
with sets (not multisets!) of biletters.
Now, the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is
$\prod_{i,j=1}^\infty \left(1 + t x_i y_j\right)
= \prod_{i,j=1}^\infty \left(1 + t x_j y_i\right)$
(here, we substituted $\left(j,i\right)$ for the
index $\left(i,j\right)$ in the product), and thus can be
rewritten as the sum of
$t^\ell \left(x_{j_1} y_{i_1}\right) \left(x_{j_2} y_{i_2}\right)
\cdots \left(x_{j_\ell} y_{i_\ell}\right)$
over all sets
$\left\{\binom{i_1}{j_1}, \ldots, \binom{i_\ell}{j_\ell}\right\}$
of biletters. Thus,
the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^\ell \xx^{\cont(\jj)} \yy^{\cont(\ii)}$ over all
strict biwords $\binom{\ii}{\jj}$, where $\ell$ stands
for the number of biletters in the biword. Meanwhile, we
know that the right hand side of
\eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}$
over all cotableau-tableau pairs $\left(P, Q\right)$
(because of
\eqref{sol.cauchy.dual.a.thirdproof.rhs}). Thus,
in order to prove \eqref{sol.cauchy.dual.a.secondproof.goal2},
we only need to construct a bijection between the
strict biwords $\binom{\ii}{\jj}$ and the
cotableau-tableau pairs $\left(P,Q\right)$, which has
the property that
\begin{align*}
\cont\left(\ii\right) &= \cont\left(Q\right); \\
\cont\left(\jj\right) &= \cont\left(P\right).
\end{align*}
\footnote{Such a bijection will then automatically satisfy
$\left|\lambda\right| = \ell$, where $\lambda$ is the (common)
shape of $P$ and $Q$, and where $\ell$ is the length of the
strict biword $\binom{\ii}{\jj}$. This is because
$\left|\lambda\right| = \left|\cont\left(Q\right)\right|$
and $\ell = \left|\cont\left(\ii\right)\right|$.}

This bijection is the \dfn{dual RSK algorithm}, which we
shall define below.

First, we shall introduce a simpler operation which we call
\dfn{dual RS-insertion} (and which is similar to RS-insertion, but
not identical with it\footnote{We will leave many of its properties
unproven, because their proofs are analogous (or at least very similar)
to the proofs of the corresponding properties of RS-insertion, and thus
can be easily reconstructed by the reader.}).

Dual RS-insertion takes as input a row-strict tableau $P$ and a letter
$j$, and returns a new row-strict tableau $P'$ along with a corner
cell $c$ of $P'$, which is constructed as follows:
Start out by setting $P' = P$. The letter $j$ tries to insert itself
into the first row of $P'$ by either bumping out the leftmost letter
in the first row \textbf{larger or equal to} $j$, or else placing
itself at the right end of the row if
no such letter (larger or equal to $j$) exists.
If a letter was bumped from the
first row, this letter follows
the same rules to insert itself into the second row, and so on.
This series of bumps must eventually come to an
end.  At the end of the bumping,
the tableau $P'$ created has an extra corner cell not present in $P$.
If we call this corner cell $c$, then $P'$ (in its final form) and $c$
are what the dual RS-insertion operation returns. One says that $P'$ is the
result of \dfn{dually inserting} $j$ into the tableau $P$. It is
straightforward
to see that this resulting filling $P'$ is a row-strict
tableau\footnote{Indeed, the reader can check that $P'$ remains a
row-strict tableau throughout the algorithm that defines dual
RS-insertion.}.

\begin{example}
To give an example
of this operation, let us dually insert the letter $j = 3$ into the
row-strict tableau
$\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 4 & \\ 3 & 4 & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix}$ (we are showing all
intermediate states of $P'$; the underlined letter is always
the one that is going to be bumped out at the next step):
\begin{align*}
\begin{matrix} 1 & 2 & \underline{3} & 5 \\ 1 & 2 & 4 & \\ 3 & 4 & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix}
\quad & \overset{\longmapsto}{\substack{\text{insert 3;} \\ \text{bump out 3}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & \underline{4} & \\ 3 & 4 & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 3;} \\ \text{bump out 4}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & \underline{4} & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix} \\
\quad & \overset{\longmapsto}{\substack{\text{insert 4;} \\ \text{bump out 4}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & 4 & 7 & \\
3 & \underline{6} & & \\ 4 & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 4;} \\ \text{bump out 6}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & 4 & 7 & \\
3 & 4 & & \\ 4 & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 6;} \\ \text{done}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & 4 & 7 & \\
3 & 4 & & \\ 4 & 6 & & \end{matrix} .
\end{align*}
The last tableau in this sequence is the row-strict
tableau that is returned. The corner cell that is returned
is the second cell of the fifth row (the one containing $6$).
\end{example}

Dual RS-insertion will be used as a step in the dual RSK algorithm; the
construction will rely on a simple fact known as the
\emph{dual row bumping lemma}. Let us first define the notion of
a \emph{dual bumping path} (or \dfn{dual bumping route}):
If $P$ is a row-strict tableau, and $j$ is a letter, then
some letters are inserted into some cells when dual RS-insertion
is applied to $P$ and $j$. The sequence of these cells (in
the order in which they see letters inserted into them)\footnote{In
particular, this includes those cells whose entries did not change under
the insertion (because the entry inserted was the same as the entry
they contained before the insertion).} is called
the \dfn{dual bumping path} for $P$ and $j$. This dual bumping path
always ends with the corner cell $c$ which is returned by
dual RS-insertion. As an example, when
$j=3$ is dually inserted into the tableau $P$
shown below, the result $P'$ is shown with all entries on
the dual bumping path underlined:
\[
P=\begin{matrix}
1&2&3&4 \\
1&2&4&6 \\
2&4&5&6 \\
2&4& & \\
3& & &
\end{matrix}
\qquad
\underset{j=3}{\overset{\text{dually insert}}{\longmapsto}}
\qquad
P'=\begin{matrix}
1&2&\underline{3}&4 \\
1&2&\underline{3}&6 \\
2&\underline{4}&5&6 \\
2&\underline{4}& & \\
3&\underline{4}& &
\end{matrix}
\]
A first simple observation about dual bumping paths is that
dual bumping paths trend weakly left (just as bumping paths for regular
RS-insertion do). A subtler property of bumping paths is the following
\dfn{dual row bumping lemma}\footnote{This lemma is equivalent to
the ``column bumping lemma'' in Fulton \cite[p. 187]{Fulton}.}:

\begin{statement}
\textbf{Dual row bumping lemma:} Let $P$ be a row-strict tableau,
and let $j$ and $j'$ be two letters. Applying dual RS-insertion to the
tableau $P$ and the letter $j$ yields a new row-strict tableau
$P'$ and a corner cell $c$. Applying dual RS-insertion to the tableau
$P'$ and the letter $j'$ yields a new row-strict tableau $P''$
and a corner cell $c'$.
\begin{enumerate}
\item[(a)] Assume that $j < j'$. Then, the dual bumping path
for $P'$ and $j'$
stays strictly to the right, within each row, of the dual bumping path
for $P$ and $j$.
The cell $c'$ (in which the dual bumping path for $P'$ and $j'$ ends) is
in the same row as the cell $c$ (in which the dual bumping path for $P$
and $j$ ends) or in a row further up; it is also in a column further
right than $c$.
\item[(b)] Assume instead that $j \geq j'$. Then, the dual bumping path
for $P'$ and $j'$
stays weakly to the left, within each row, of the dual bumping path
for $P$ and $j$.
The cell $c'$ (in which the dual bumping path for $P'$ and $j'$ ends) is
in a row further down than the cell $c$ (in which the dual bumping
path for $P$ and $j$ ends); it is also in the same column as $c$ or
in a column further left.
\end{enumerate}
\end{statement}
This lemma can be easily proven by induction over the
row (similarly to the usual row bumping lemma).

We will now define the dual RSK algorithm.
Let $\binom{\ii}{\jj}$ be a strict biword.
Starting with the pair $(P_0,Q_0)=(\varnothing,\varnothing)$ and
$m=0$, the algorithm applies the following steps
(see Example~\ref{sol.cauchy.dual.a.thirdproof.dualRSK-example} below):
\begin{itemize}
\item If $i_{m+1}$ does not exist (that is, $m$ is the length of
$\ii$), stop.
\item Apply dual RS-insertion to the row-strict tableau $P_m$
and the letter $j_{m+1}$ (the bottom letter of
$\binom{i_{m+1}}{j_{m+1}}$). Let $P_{m+1}$ be the resulting
row-strict tableau, and let $c_{m+1}$ be the resulting
corner cell.
\item Create $Q_{m+1}$ from $Q_m$ by adding the top letter
$i_{m+1}$ of $\binom{i_{m+1}}{j_{m+1}}$ to $Q_m$ in the cell
$c_{m+1}$ (which, as we recall, is the extra corner cell of
$P_{m+1}$ not present in $P_m$).
\item Set $m$ to $m+1$.
\end{itemize}
After all of the biletters have been thus processed,
the result of the dual RSK algorithm
is $(P_\ell,Q_\ell)=:(P,Q)$.

\begin{example}
\label{sol.cauchy.dual.a.thirdproof.dualRSK-example}
The term in the expansion of the left side of
\eqref{sol.cauchy.dual.a.secondproof.goal} corresponding to
\[
(x_2 y_1)(x_4 y_1)(x_1 y_2)(x_1 y_4)(x_3 y_4)(x_2 y_5)
\]
is the strict biword
$\binom{\ii}{\jj}=\binom{112445}{241132}$, and the dual RSK
algorithm applied to this biword proceeds
as follows:
\[
\begin{array}{rcl|rcl}
P_0&= &\varnothing & Q_0 &=& \varnothing \\
 & & & & &\\
P_1&= &\begin{matrix} 2 \end{matrix}
 & Q_1 &= &\begin{matrix} 1 \end{matrix} \\
 & & & & &\\
P_2&= &\begin{matrix} 2 & 4 \end{matrix}
 & Q_2 &= &\begin{matrix} 1 & 1 \end{matrix} \\
 & & & & &\\
P_3&= &\begin{matrix} 1 & 4 \\ 2 \end{matrix}
 & Q_3 &= &\begin{matrix} 1 & 1 \\ 2 \end{matrix} \\
 & & & & &\\
P_4&= &\begin{matrix} 1 & 4 \\ 1 \\ 2 \end{matrix}
 & Q_4 &= &\begin{matrix} 1 & 1 \\ 2 \\ 4 \end{matrix} \\
 & & & & &\\
P_5&= &\begin{matrix} 1 & 3 \\ 1 & 4 \\ 2 \end{matrix}
 & Q_5 &= &\begin{matrix} 1 & 1 \\ 2 & 4 \\ 4 \end{matrix} \\
 & & & & &\\
P:=P_6&= &\begin{matrix} 1 & 2 \\ 1 & 3 \\ 2 & 4 \end{matrix}
 & Q:=Q_6 &= &\begin{matrix} 1 & 1 \\ 2 & 4 \\ 4 & 5 \end{matrix} \\
 & & & & &\\
\end{array}
\]
\end{example}

It is clear that $P_m$ remains a row-strict tableau of some
Ferrers shape throughout the execution of the dual RSK algorithm, and
that $Q_m$ remains a filling of the same shape as $P_m$ which is
(at least) weakly increasing left-to-right along rows and weakly
increasing top-to-bottom in columns. But we can also see that
$Q_m$ is strictly increasing top-to-bottom along columns\footnote{Indeed,
this follows from the observation that when one has a string of equal
letters $i_m=i_{m+1}=\cdots=i_{m+r}$ on top of the strict biword,
then the bottom letters bumped in are
$j_m < j_{m+1} < \cdots < j_{m+r}$, and therefore (as a consequence
of the second-to-last claim of part (a) of the dual row bumping lemma)
the new
cells form a \emph{horizontal strip}, that is, no two of these cells
lie in the same column. Actually, more can be said: Each of these new
cells (except for the first one) is in a column further right than
the previous one. We will use this stronger fact further below.},
so that $Q_m$ is a column-strict tableau. Thus,
the result $\left(P, Q\right)$ of the dual RSK algorithm
is a cotableau-tableau pair.

To see that the dual RSK algorithm is a bijection, we show how to
recover $\binom{\ii}{\jj}$ from $(P,Q)$.  This is done by
\dfn{dually reverse bumping} from $(P_{m+1},Q_{m+1})$ to recover
both the biletter $\binom{i_{m+1}}{j_{m+1}}$ and the tableaux
$(P_m,Q_m)$, as follows.
Firstly, $i_{m+1}$ is the maximum entry of $Q_{m+1}$, and
$Q_m$ is obtained by removing
the rightmost occurrence of this letter $i_{m+1}$ from $Q_{m+1}$.
\footnote{It necessarily has to be the rightmost occurrence, since
(according to the previous footnote) the cell into which
$i_{m+1}$ was filled at the step from $Q_m$ to $Q_{m+1}$ lies further
right than any existing cell of $Q_m$ containing the letter $i_{m+1}$.}
To produce $P_m$ and $j_{m+1}$, find the position of the
rightmost occurrence of $i_{m+1}$ in $Q_{m+1}$,
and start \emph{dually reverse bumping} in $P_{m+1}$ from the entry in
this same position, where dually reverse bumping an
entry means inserting it into one row higher by having it bump
out the rightmost entry which is smaller or equal to
it.\footnote{Let us give a few more details on this
``dually reverse bumping'' procedure.
Dually reverse bumping (also known as \dfn{dual RS-deletion} or
\dfn{reverse dual RS-insertion}) is an operation which takes a
row-strict tableau $P'$ and a corner cell $c$ of $P'$,
and constructs a row-strict tableau $P$ and a letter $j$
such that dual RS-insertion for $P$ and $j$ yields $P'$ and $c$.
It starts by setting $P = P'$, and removing the entry in the
cell $c$ from $P$. This removed entry is then denoted by $k$,
and is inserted into the row of $P$ above $c$, bumping out the
rightmost entry which is smaller or equal to $k$. The letter which is
bumped out -- say, $\ell$ --, in turn, is inserted into the row
above it, bumping out the rightmost entry which is smaller or equal
to $\ell$. This procedure continues in the same way until an entry
is bumped out of the first row (which will eventually happen).
The dually reverse bumping operation returns the
resulting tableau $P$ and the entry which is bumped out of the
first row.
\par
It is straightforward to check that the dually reverse bumping
operation is well-defined (i.e., $P$ does stay a row-strict
tableau throughout the procedure) and is the inverse of the
dual RS-insertion operation. (In fact, these two operations undo each
other step by step.)}
The entry bumped out of the
first row is $j_{m+1}$, and the resulting tableau is $P_{m}$.

Finally, to see that the dual RSK map is surjective, one needs to show
that the dually reverse bumping procedure can be applied to any
cotableau-tableau pair $(P,Q)$, and will result in a strict
biword $\binom{\ii}{\jj}$.
We leave this verification to the reader.\footnote{It is easy to
see that repeatedly applying dually reverse bumping to $(P,Q)$ will
result in a sequence
$\binom{i_\ell}{j_\ell}, \binom{i_{\ell-1}}{j_{\ell-1}},
\ldots, \binom{i_1}{j_1}$ of biletters such that applying the
dual RSK algorithm to
$\binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$ gives back
$(P,Q)$. The question is why we have
$\binom{i_1}{j_1} <_{lex} \cdots <_{lex} \binom{i_\ell}{j_\ell}$.
Since the chain of inequalities $i_1 \leq i_2 \leq \cdots \leq i_\ell$
is clear from the choice of entry to dually reverse-bump, it only
remains to show that for every string $i_m=i_{m+1}=\cdots=i_{m+r}$
of equal top letters, the corresponding bottom letters
strictly increase (that is,
$j_m < j_{m+1} < \cdots < j_{m+r}$). One way to see
this is the following:
\par
Assume the contrary; i.e., assume that
the bottom letters corresponding
to some string $i_m=i_{m+1}=\cdots=i_{m+r}$ of equal top
letters do not strictly increase.
Thus, $j_{m+p} \geq j_{m+p+1}$ for some
$p \in \left\{0, 1, \ldots, r-1\right\}$.
Consider this $p$.
\par
Let us consider the cells containing the equal letters
$i_m=i_{m+1}=\cdots=i_{m+r}$ in the tableau $Q_{m+r}$.
Label these cells as $c_m, c_{m+1}, \ldots, c_{m+r}$
from left to right (noticing that no two of them lie
in the same column, since $Q_{m+r}$ is column-strict).
By the definition of dually reverse bumping, the first entry
to be dually reverse bumped from $P_{m+r}$ is the entry in
position $c_{m+r}$ (since this is the rightmost
occurrence of the letter $i_{m+r}$ in $Q_{m+r}$); then, the next entry
to be dually reverse bumped is the one in position $c_{m+r-1}$,
etc., moving further and further left.
Thus, for each $q \in \left\{0, 1, \ldots, r\right\}$,
the tableau $P_{m+q-1}$ is obtained from $P_{m+q}$ by
dually reverse bumping the entry in position $c_{m+q}$.
Hence, conversely, the tableau $P_{m+q}$ is obtained
from $P_{m+q-1}$ by dually RS-inserting the entry $j_{m+q}$,
which creates the corner cell $c_{m+q}$.
\par
But recall that $j_{m+p} \geq j_{m+p+1}$.
Hence, part (b) of the dual row bumping lemma (applied to
$P_{m+p-1}$, $j_{m+p}$, $j_{m+p+1}$, $P_{m+p}$,
$c_{m+p}$, $P_{m+p+1}$ and $c_{m+p+1}$
instead of $P$, $j$, $j'$, $P'$, $c$, $P''$ and $c'$)
shows that the cell $c_{m+p+1}$ is in the same
column as the cell $c_{m+p}$ or in a column further
left.
But this contradicts the fact that the cell $c_{m+p+1}$ is
in a column further right than the cell $c_{m+p}$ (since
we have labeled our cells as
$c_m, c_{m+1}, \ldots, c_{m+r}$ from left to right,
and no two of them lied in the same column).
This contradiction completes our proof.}

So the dual RSK map is a bijection having the required properties.
As we have said, this proves
\eqref{sol.cauchy.dual.a.secondproof.goal2}.
This solves Exercise~\ref{exe.cauchy.dual}(a) again.

(b) Let us consider the map $\omega\left[  \left[  \yy\right]
\right]  :\Lambda\left[  \left[  \yy\right]  \right]  \rightarrow
\Lambda\left[  \left[  \yy\right]  \right]  $ defined as in the
solution of Exercise \ref{exe.cauchy.dual}(a).

Now, the equality \eqref{Cauchy-identity} yields
\begin{equation}
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  =\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda\in\Par }z_{\lambda}%
^{-1}p_{\lambda}\left( \xx \right)  p_{\lambda}\left(
\yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{Two-other-Cauchy-expansions}}\right)  .\label{sol.cauchy.dual.b.5}%
\end{equation}
This is an equality in $\Lambda\left[  \left[  \yy\right]  \right]  $
(because $s_{\lambda}\left( \xx \right)  $ and $p_{\lambda}\left(
\xx \right)  $ belong to $\Lambda$ for every $\lambda\in
\Par $). Hence, we can apply the map $\omega\left[  \left[
\yy\right]  \right]  :\Lambda\left[  \left[  \yy\right]
\right]  \rightarrow\Lambda\left[  \left[  \yy\right]  \right]  $ to
both sides of this equality. As a result, we obtain%
\[
\sum_{\lambda\in\Par }\omega\left(  s_{\lambda}\left(
\xx \right)  \right)  s_{\lambda}\left( \yy \right)
=\sum_{\lambda\in\Par }z_{\lambda}^{-1}\omega\left(  p_{\lambda
}\left( \xx \right)  \right)  p_{\lambda}\left(  \yy
\right)
\]
(because $\omega\left[  \left[ \yy \right]  \right]  $, being a
$\kk \left[  \left[ \yy \right]  \right]  $-algebra homomorphism,
leaves the $s_{\lambda}\left( \yy \right)  $ and $p_{\lambda}\left(
\yy \right)  $ terms unchanged, while the $s_{\lambda}\left(
\xx \right)  $ and $p_{\lambda}\left( \xx \right)  $
terms are elements of $\Lambda$ and thus are transformed as by $\omega$). Due
to \eqref{eq.exe.Lambda.omegahlam.p} and \eqref{sol.cauchy.dual.a.omega.s2},
this rewrites as%
\[
\sum_{\lambda\in\Par }s_{\lambda^{t}}\left( \xx
\right)  s_{\lambda}\left( \yy \right)  =\sum_{\lambda
\in\Par }z_{\lambda}^{-1}\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }p_{\lambda}\left(
\xx \right)  p_{\lambda}\left( \yy \right)  .
\]
Since $\sum_{\lambda\in\Par }s_{\lambda^{t}}\left( \xx
\right)  s_{\lambda}\left( \yy \right)  =\sum_{\lambda
\in\Par }s_{\lambda}\left( \xx \right)  s_{\lambda
^{t}}\left( \yy \right)  $ (this can be proven just as in
(\ref{sol.cauchy.dual.a.omega.leftside})), this rewrites as%
\[
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda^{t}}\left( \yy \right)  =\sum_{\lambda\in
\Par }z_{\lambda}^{-1}\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }p_{\lambda}\left(
\xx \right)  p_{\lambda}\left( \yy \right)  =\sum_{\lambda
\in\Par }\left(  -1\right)  ^{\left\vert \lambda\right\vert
-\ell\left(  \lambda\right)  }z_{\lambda}^{-1}p_{\lambda}\left( \xx
\right)  p_{\lambda}\left( \yy \right)  .
\]
Hence,
\[
\sum_{\lambda\in\Par }\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }z_{\lambda}^{-1}p_{\lambda
}\left( \xx \right)  p_{\lambda}\left( \yy \right)
=\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda^{t}}\left( \yy \right)  =\prod_{i,j=1}^{\infty}\left(
1+x_{i}y_{j}\right)
\]
(by \eqref{sol.cauchy.dual.a.omega.end}). This solves
Exercise~\ref{exe.cauchy.dual}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cauchy.dual}}

\textit{Solution to Exercise~\ref{exe.cauchy.dual}.} (a)

\textit{First solution to Exercise~\ref{exe.cauchy.dual}(a):}
Let us first
show that $\prod_{i,j=1}^{\infty}\left(  1+x_{i}y_{j}\right)  =\sum
_{\lambda\in\Par }e_{\lambda}\left( \xx \right)
m_{\lambda}\left( \yy \right)  $.

In fact, in the proof of Proposition~\ref{prop-Two-other-Cauchy-expansions},
we showed that $\prod_{j=1}^{\infty}\sum_{n\geq0} h_n \left( \xx
\right)  y_{j}^{n}=\sum_{\lambda\in\Par} h_{\lambda}\left( \xx
\right)  m_{\lambda}\left( \yy \right)  $. The same
argument (with all appearances of the letter ``$h$''
replaced by the letter ``$e$'') shows that
\begin{equation}
\prod_{j=1}^{\infty} \sum_{n\geq 0} e_n \left( \xx \right) y_{j}^{n}
=\sum_{\lambda\in\Par} e_{\lambda} \left( \xx \right)
 m_{\lambda} \left( \yy \right)  .
\label{sol.cauchy.dual.a.1}
\end{equation}


But \eqref{e-generating-function-no-HE} yields
\begin{equation}
\prod_{i=1}^{\infty}\left(  1+x_{i}t\right)  =\sum_{n\geq0}e_{n}\left(
\xx \right)  t^{n}
\label{sol.cauchy.dual.a.e}
\end{equation}
in the ring $\left( \kk \left[ \left[ \xx \right] \right]
\right)  \left[  \left[  t\right]  \right]  $. For every $j\in\left\{
1,2,3,...\right\}  $, we have
\begin{equation}
\prod_{i=1}^{\infty}\left(  1+x_{i}y_{j}\right)  =\sum_{n\geq0}e_{n}\left(
\xx \right)  y_{j}^{n}
\label{sol.cauchy.dual.a.e2}
\end{equation}
in the ring $\left( \kk \left[  \left[ \xx \right]  \right]
\right)  \left[  \left[   \yy \right]  \right]  $ (in fact, this results
from \eqref{sol.cauchy.dual.a.e} by substituting $y_{j}$ for $t$). Thus,
\eqref{sol.cauchy.dual.a.e2} holds in $\kk \left[  \left[
\xx , \yy \right]  \right]  $ (since $\left( \kk \left[
\left[ \xx \right]  \right]  \right)  \left[  \left[   \yy 
\right]  \right]  = \kk \left[  \left[  \xx , \yy \right]
\right]  $ as rings). Now,
\begin{align}
\underbrace{\prod_{i,j=1}^{\infty}}_{=\prod_{j=1}^{\infty}\prod_{i=1}^{\infty
}}\left(  1+x_{i}y_{j}\right)   &  =\prod_{j=1}^{\infty}\underbrace{\prod
_{i=1}^{\infty}\left(  1+x_{i}y_{j}\right)  }_{\substack{=\sum_{n\geq0}%
e_{n}\left(  \xx \right)  y_{j}^{n}\\\text{(by
\eqref{sol.cauchy.dual.a.e2})}}}=\prod_{j=1}^{\infty}\sum_{n\geq0}e_{n}\left(
\xx \right)  y_{j}^{n}\nonumber\\
&  =\sum_{\lambda\in\Par }e_{\lambda}\left( \xx \right)
m_{\lambda}\left( \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.dual.a.1}}\right)  .
\label{sol.cauchy.dual.a.part1done}
\end{align}


Next, we shall show that $\sum_{\lambda\in\Par }s_{\lambda
}\left( \xx \right)  s_{\lambda^{t}}\left( \yy \right)
=\sum_{\lambda\in\Par }e_{\lambda}\left( \xx \right)
m_{\lambda}\left( \yy \right)  $.

In fact, consider the $\kk$-algebra $\left( \kk \left[  \left[
\xx \right]  \right]  \right)  \left[  \left[   \yy \right]
\right]  = \kk \left[  \left[  \xx , \yy \right]  \right]  $.
This $\kk$-algebra $\left( \kk \left[  \left[  \xx %
\right]  \right]  \right)  \left[  \left[   \yy \right]  \right]  $ is a
$\kk \left[  \left[   \yy \right]  \right]  $-algebra because
$\underbrace{\kk}_{\subset \kk \left[  \left[ \xx \right]
\right]  }\left[  \left[   \yy \right]  \right]  \subset\left(
\kk \left[  \left[ \xx \right]  \right]  \right)  \left[  \left[
 \yy \right]  \right]  $. Furthermore, $\Lambda\left[  \left[
 \yy \right]  \right]  $ is a $\kk \left[  \left[   \yy %
\right]  \right]  $-subalgebra of $\left( \kk \left[  \left[
\xx \right]  \right]  \right)  \left[  \left[   \yy \right]
\right]  $ (since $\underbrace{\kk}_{\subset\Lambda}\left[  \left[
 \yy \right]  \right]  \subset\underbrace{\Lambda}_{\subset
\kk \left[  \left[ \xx \right]  \right]  }\left[  \left[
 \yy \right]  \right]  \subset\left(  \kk \left[  \left[
\xx \right]  \right]  \right)  \left[  \left[   \yy \right]
\right]  $).

The equality \eqref{Cauchy-identity} yields
\begin{equation}
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  =\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda\in\Par }h_{\lambda
}\left( \xx \right)  m_{\lambda}\left( \yy \right)
\ \ \ \ \ \ \ \ \ \ \left(
 \text{by \eqref{Two-other-Cauchy-expansions}}\right)  .
\label{sol.cauchy.dual.a.5}
\end{equation}
This is an equality in $\Lambda\left[  \left[   \yy \right]  \right]  $
(because $s_{\lambda}\left( \xx \right)  $ and $h_{\lambda}\left(
\xx \right)$ belong to $\Lambda$ for every $\lambda\in \Par$).

Recall that $\omega:\Lambda\rightarrow\Lambda$ is a $\kk$-algebra
homomorphism. Furthermore,
\begin{equation}
\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.cauchy.dual.a.omega.s2}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.cauchy.dual.a.omega.s2}:} Let $\lambda$
be a partition. From the first equation in \eqref{skew-Schur-antipodes}, we
obtain $\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}$ for
every partition $\mu$ satisfying $\mu\subseteq\lambda$. Applying this to
$\mu=\varnothing$, we see that $\omega\left(  s_{\lambda/\varnothing}\right)
=s_{\lambda^{t}/\varnothing^{t}}$ (since $\varnothing\subseteq\mu$). Since
$s_{\lambda/\varnothing}=s_{\lambda}$ and $s_{\lambda^{t}/\varnothing^{t}%
}=s_{\lambda^{t}/\varnothing}=s_{\lambda^{t}}$, this rewrites as
$\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}$. This proves
\eqref{sol.cauchy.dual.a.omega.s2}.}

If $Q$ and $R$ are two $\kk$-algebras, then every $\kk$-algebra
homomorphism $\varphi:Q\rightarrow R$ canonically gives rise to a continuous
$\kk \left[  \left[   \yy \right]  \right]  $-algebra
homomorphism\footnote{Here, the word ``continuous'' refers to the
usual topologies on the rings
$Q\left[  \left[   \yy \right]  \right]  $ and $R\left[  \left[
 \yy \right]  \right]  $. Let us recall how these topologies are defined:
The usual topology on a power series ring $Z\left[ \left[ \yy \right]
\right]$ (where $Z$ is a commutative ring) is the direct-product topology
obtained by viewing the set $Z\left[ \left[ \yy \right]
\right]$ as a direct product of many copies of $Z$ (this
is done by identifying every power series with the family of its
coefficients), each of which is endowed with the discrete topology. In this
topology, a sequence (or, more generally, a net) $\left(  f_{i}\right)  _{i}$
of power series converges to a power series $f$ if and only if for every
monomial $\mathfrak{m}$, the sequence (resp. net) $\left(  \text{the
coefficient of }\mathfrak{m}\text{ in }f_{i}\right)  _{i}$ converges to
$\left(  \text{the coefficient of }\mathfrak{m}\text{ in }f\right)  $ with
respect to the discrete topology.
(The notion of a \emph{net} is a generalization of the notion of a
sequence; it is useful in topology. See \cite{Vermeeren-nets} for an
introduction to it.)} $\varphi\left[  \left[   \yy \right]
\right]  :Q\left[  \left[   \yy \right]  \right]  \rightarrow R\left[
\left[   \yy \right]  \right]  $ which satisfies%
\begin{equation}
\left(  \varphi\left[  \left[   \yy \right]  \right]  \right)  \left(
q\right)  =\varphi\left(  q\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }q\in
Q. \label{sol.cauchy.dual.a.phi}%
\end{equation}
\footnote{Explicitly, this map $\varphi\left[  \left[   \yy \right]
\right]  :Q\left[  \left[   \yy \right]  \right]  \rightarrow R\left[
\left[   \yy \right]  \right]  $ is defined by%
\[
\left(  \varphi\left[  \left[   \yy \right]  \right]  \right)  \left(
\sum_{\alpha\text{ is a weak composition}}q_{\alpha} \yy ^{\alpha
}\right)  =\sum_{\alpha\text{ is a weak composition}}\varphi\left(  q_{\alpha
}\right)   \yy ^{\alpha}%
\]
for any family $\left(  q_{\alpha}\right)  _{\alpha\text{ is a weak
composition}}$ of elements of $Q$.} Applying this to $Q=\Lambda$, $R=\Lambda$
and $\varphi=\omega$, we obtain a continuous $\kk \left[  \left[
 \yy \right]  \right]  $-algebra homomorphism $\omega\left[  \left[
 \yy \right]  \right]  :\Lambda\left[  \left[   \yy \right]
\right]  \rightarrow\Lambda\left[  \left[   \yy \right]  \right]  $ which
satisfies
\begin{equation}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
q\right)  =\omega\left(  q\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
q\in\Lambda.
\label{sol.cauchy.dual.a.omega}
\end{equation}
We also have
\begin{equation}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
p\right)  =p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\kk \left[  \left[
 \yy \right]  \right]
\label{sol.cauchy.dual.a.omega2}
\end{equation}
(since $\omega\left[  \left[   \yy \right]  \right]  $ is a
$\kk \left[  \left[   \yy \right]  \right]  $-algebra homomorphism).

Now, for every partition $\lambda$, we have $h_{\lambda}\left(  \xx %
\right)  \in\Lambda$. Hence, for every partition $\lambda$, we have%
\begin{align}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
h_{\lambda}\left( \xx \right)  \right)   &  =\omega\left(
\underbrace{h_{\lambda}\left(  \xx \right)  }_{=h_{\lambda}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.dual.a.omega}, applied
to }q=h_{\lambda}\left(  \xx \right)  \right) \nonumber\\
&  =\omega\left(  h_{\lambda}\right)  =e_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{eq.exe.Lambda.omegahlam.h}}\right) \nonumber\\
&  =e_{\lambda}\left(  \xx \right)  .
\label{sol.cauchy.dual.a.omega.h4}
\end{align}
Also, for every partition $\lambda$, we have $s_{\lambda}\left(
\xx \right)  \in\Lambda$. Thus, for every partition $\lambda$, we have%
\begin{align}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
s_{\lambda}\left( \xx \right)  \right)   &  =\omega\left(
\underbrace{s_{\lambda}\left(  \xx \right)  }_{=s_{\lambda}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.dual.a.omega}, applied
to }q=s_{\lambda}\left(  \xx \right)  \right) \nonumber\\
&  =\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}
\ \ \ \ \ \ \ \ \ \ \left(
 \text{by \eqref{sol.cauchy.dual.a.omega.s2}}\right) \nonumber\\
&  =s_{\lambda^{t}}\left(  \xx \right)  .
\label{sol.cauchy.dual.a.omega.s4}
\end{align}
Furthermore, for every partition $\lambda$, we have $m_{\lambda}\left(
 \yy \right)  \in\kk \left[  \left[   \yy \right]  \right]  $.
Hence, for every partition $\lambda$, we have
\begin{equation}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
m_{\lambda}\left(   \yy \right)  \right)  =m_{\lambda}\left(
 \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.dual.a.omega2}, applied to }p=m_{\lambda}\left(
 \yy \right)  \right)  .
\label{sol.cauchy.dual.a.omega.m5}
\end{equation}
Furthermore, for every partition $\lambda$, we have $s_{\lambda}\left(
 \yy \right)  \in\kk \left[  \left[   \yy \right]  \right]  $.
Hence, for every partition $\lambda$, we have%
\begin{equation}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
s_{\lambda}\left(   \yy \right)  \right)  =s_{\lambda}\left(
 \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.dual.a.omega2}, applied to }p=s_{\lambda}\left(
 \yy \right)  \right)  .
\label{sol.cauchy.dual.a.omega.s5}
\end{equation}


Recall that \eqref{sol.cauchy.dual.a.5} is an equality in $\Lambda\left[
\left[   \yy \right]  \right]  $. Hence, we can apply the map
$\omega\left[  \left[   \yy \right]  \right]  :\Lambda\left[  \left[
 \yy \right]  \right]  \rightarrow\Lambda\left[  \left[   \yy
\right]  \right]  $ to both sides of this equality. As a result, we obtain
\begin{align*}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  \right)   &  =\left(  \omega\left[
\left[   \yy \right]  \right]  \right)  \left(  \sum_{\lambda
\in\Par }h_{\lambda}\left( \xx \right)  m_{\lambda
}\left( \yy \right)  \right)  =\sum_{\lambda\in\Par }
\underbrace{\left(  \omega\left[  \left[   \yy \right]  \right]
\right)  \left(  h_{\lambda}\left( \xx \right)  \right)
}_{\substack{=e_{\lambda}\left(  \xx \right)  \\\text{(by
\eqref{sol.cauchy.dual.a.omega.h4})}}}\cdot\underbrace{\left(  \omega\left[
\left[   \yy \right]  \right]  \right)  \left(  m_{\lambda}\left(
 \yy \right)  \right)  }_{\substack{=m_{\lambda}\left( \yy
\right)  \\\text{(by \eqref{sol.cauchy.dual.a.omega.m5})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\left[  \left[
 \yy \right]  \right]  \text{ is a continuous }\kk \left[  \left[
 \yy \right]  \right]  \text{-algebra homomorphism}\right)  \\
&  =\sum_{\lambda\in\Par }e_{\lambda}\left(  \xx \right)
m_{\lambda}\left(   \yy \right)  .
\end{align*}
Compared with%
\begin{align}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  \right)   &  =\sum_{\lambda
\in\Par }\underbrace{\left(  \omega\left[  \left[   \yy %
\right]  \right]  \right)  \left(  s_{\lambda}\left( \xx \right)
\right)  }_{\substack{=s_{\lambda^{t}}\left(  \xx \right)  \\\text{(by
\eqref{sol.cauchy.dual.a.omega.s4})}}}\cdot\underbrace{\left(  \omega\left[
\left[   \yy \right]  \right]  \right)  \left(  s_{\lambda}\left(
{ \yy }\right)  \right)  }_{\substack{=s_{\lambda}\left(   \yy %
\right)  \\\text{(by \eqref{sol.cauchy.dual.a.omega.s5})}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\left[  \left[
 \yy \right]  \right]  \text{ is a continuous }\kk \left[  \left[
 \yy \right]  \right]  \text{-algebra homomorphism}\right)  \nonumber\\
&  =\sum_{\lambda\in\Par }s_{\lambda^{t}}\left(  \xx
\right)  s_{\lambda}\left( \yy \right)  =\sum_{\lambda
\in\Par }\underbrace{s_{\left(  \lambda^{t}\right)  ^{t}}\left(
\xx \right)  }_{\substack{=s_{\lambda}\left(  \xx \right)
\\\text{(since }\left(  \lambda^{t}\right)  ^{t}=\lambda\text{)}}%
}s_{\lambda^{t}}\left( \yy \right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\lambda^{t}\text{ for }\lambda\text{ in the sum,
since the map}\\
\Par \rightarrow\Par ,\ \lambda\mapsto\lambda
^{t}\text{ is a bijection}%
\end{array}
\right)  \nonumber\\
&  =\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda^{t}}\left( \yy \right)
,\label{sol.cauchy.dual.a.omega.leftside}%
\end{align}
this yields $\sum_{\lambda\in\Par }e_{\lambda}\left(
\xx \right)  m_{\lambda}\left(   \yy \right)  =\sum_{\lambda
\in\Par }s_{\lambda}\left( \xx \right)  s_{\lambda^{t}%
}\left( \yy \right)  $. Combined with
\eqref{sol.cauchy.dual.a.part1done}, this yields
\begin{equation}
\prod_{i,j=1}^{\infty}\left(  1+x_{i}y_{j}\right)  =\sum_{\lambda
\in\Par }e_{\lambda}\left( \xx \right)  m_{\lambda
}\left( \yy \right)  =\sum_{\lambda\in\Par }s_{\lambda
}\left( \xx \right)  s_{\lambda^{t}}\left( \yy \right)
.
\label{sol.cauchy.dual.a.omega.end}
\end{equation}
This solves Exercise~\ref{exe.cauchy.dual}(a).

\textit{Second solution to Exercise~\ref{exe.cauchy.dual}(a):}
We can prove \eqref{sol.cauchy.dual.a.part1done} as in the First
solution to Exercise~\ref{exe.cauchy.dual}(a). Thus, in order to
solve Exercise~\ref{exe.cauchy.dual}(a), it remains to verify
\begin{equation}
\prod_{i,j=1}^{\infty} \left(1 + x_i y_j\right)
= \sum_{\lambda \in \Par} s_\lambda\left(\xx\right) s_{\lambda^t}\left(\yy\right) .
\label{sol.cauchy.dual.a.secondproof.goal}
\end{equation}

Instead of proving \eqref{sol.cauchy.dual.a.secondproof.goal}, we will
verify the identity
\begin{equation}
\prod_{i,j=1}^{\infty} \left(1 + t x_i y_j\right)
= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_\lambda\left(\xx\right) s_{\lambda^t}\left(\yy\right)
\label{sol.cauchy.dual.a.secondproof.goal2}
\end{equation}
in the ring $R\left(\xx,\yy\right)\left[\left[t\right]\right]$. This
identity will clearly yield \eqref{sol.cauchy.dual.a.secondproof.goal}
(by substituting $t = 1$), and thus conclude the solution of
Exercise~\ref{exe.cauchy.dual}(a).

We shall prove \eqref{sol.cauchy.dual.a.secondproof.goal2} in a
similar way to
how we proved \eqref{Cauchy-identity-with-t}, but using a variation
on the RSK correspondence. This is not in itself a new idea; indeed,
this is how the equivalent identity
\eqref{sol.cauchy.dual.a.secondproof.goal}
is proven in \cite[Theorem 7.14.3]{Stanley}, in
\cite[\S 7]{Knuth-RSK} and in \cite[Theorem 4.8.6]{Sagan}. However,
instead of using the dual RSK algorithm (also known as the
RSK$^\ast$ algorithm; see \cite[\S 7.14]{Stanley},
\cite[\S 5]{Knuth-RSK} and \cite[Theorem 4.8.5]{Sagan} for it)
like these proofs do, we introduce a variation of the RSK algorithm
that relies on the same RS-insertion operation but changes the
order in which the biletters are processed. (We will reprove
\eqref{sol.cauchy.dual.a.secondproof.goal2} using the dual RSK
algorithm in the Third solution further below.)

For a given partition $\lambda$, let us define a \dfn{row-strict
tableau} of shape $\lambda$ to be an assignment $T$ of entries in
$\left\{1,2,3,\ldots\right\}$ to the cells of the Ferrers diagram
for $\lambda$ which is strictly increasing left-to-right in rows, and
weakly increasing top-to-bottom in columns. It is clear that if
$\lambda$ is a partition, then the row-strict tableaux of shape
$\lambda$ are in 1-to-1 correspondence with the column-strict
tableaux of shape $\lambda^t$, and the correspondence is given
by transposing the tableau (i.e., taking whatever entry was assigned
to a cell $(i,j)$ in the input tableau, and reassigning it to the
cell $(j,i)$ in the output tableau). Hence, for every partition
$\lambda$, we have
\begin{align}
\sum\limits_{\substack{Q \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(Q\right)}
&= \sum\limits_{\substack{T \text{ is a column-strict} \\ \text{tableau of
shape } \lambda^t}} \xx^{\cont\left(T\right)}
\nonumber \\
&= s_{\lambda^t} \qquad \qquad
\left(\text{since this is how }s_{\lambda^t}\text{ is defined}\right)
\label{sol.cauchy.dual.a.secondproof.slat.x}
\end{align}
(where $\xx^{\cont\left(Q\right)}$ is defined for a row-strict
tableau $Q$ in the same way as it is defined for a column-strict
tableau $Q$). Substituting $\yy$ for $\xx$ in this equality,
we obtain
\begin{equation}
\sum\limits_{\substack{Q \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \yy^{\cont\left(Q\right)}
= s_{\lambda^t} \left(\yy\right) .
\label{sol.cauchy.dual.a.secondproof.slat}
\end{equation}
We also have
\begin{equation}
\sum\limits_{\substack{P \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(P\right)}
= s_{\lambda} \left(\xx\right)
\label{sol.cauchy.dual.a.secondproof.sla}
\end{equation}
(because this is how $s_{\lambda} = s_{\lambda} \left(\xx\right)$
is defined) for every partition $\lambda$.

A \dfn{tableau-cotableau pair} will mean a pair
$\left(P, Q\right)$ such that $P$ is a column-strict
tableau, $Q$ is a row-strict tableau, and $P$ and $Q$
both have shape $\lambda$ for one and the same partition
$\lambda$. Multiplying the identities
\eqref{sol.cauchy.dual.a.secondproof.sla} and
\eqref{sol.cauchy.dual.a.secondproof.slat} and multiplying
the result with $t^{\left|\lambda\right|}$, we obtain
\[
t^{\left|\lambda\right|}
\sum\limits_{\substack{
\left(P, Q\right) \text{ is a pair with} \\
P \text{ being a column-strict tableau} \\
\text{of shape } \lambda \text{, and } Q \text{ being} \\
\text{a row-strict tableau of shape }\lambda
}} \xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= t^{\left|\lambda\right|}
s_{\lambda} \left(\xx\right) s_{\lambda^t} \left(\yy\right)
\]
for every partition $\lambda$. Summing this equality over
all partitions $\lambda$, we obtain
\begin{equation}
\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda} \left(\xx\right) s_{\lambda^t} \left(\yy\right) ,
\label{sol.cauchy.dual.a.secondproof.rhs}
\end{equation}
where the sum on the left hand side is over all
tableau-cotableau pairs $\left(P, Q\right)$ and where
$\lambda$ denotes the common shape of $P$ and $Q$.

We shall use all notations introduced in the proof of
Theorem~\ref{thm.cauchy-identity}. Define the
\dfn{antilexicographic order} $\leq_{alex}$ to be the total order
on the set of all biletters which is given by
\[
\binom{i_1}{j_1} \leq_{alex} \binom{i_2}{j_2}
\quad \Longleftrightarrow \quad
\left( \text{we have } i_1 \leq i_2 \text{, and if }
i_1 = i_2 \text{, then } j_1 \geq j_2 \right).
\]
We denote by $<_{alex}$ the (strict) smaller relation of this
order. A \dfn{strict cobiword} will mean an array
$\binom{\ii}{\jj} = \binom{i_1\ldots i_\ell}{j_1\ldots j_\ell}$
in which the biletters satisfy
$\binom{i_1}{j_1} <_{alex} \cdots <_{alex} \binom{i_\ell}{j_\ell}$
(that is, the biletters are distinct and ordered).
Strict cobiwords are clearly in 1-to-1 correspondence
with sets (not multisets!) of biletters.
Now, the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is
$\prod_{i,j=1}^\infty \left(1 + t x_i y_j\right)
= \prod_{i,j=1}^\infty \left(1 + t x_j y_i\right)$
(here, we substituted $\left(j,i\right)$ for the
index $\left(i,j\right)$ in the product), and thus can be
rewritten as the sum of
$t^\ell \left(x_{j_1} y_{i_1}\right) \left(x_{j_2} y_{i_2}\right)
\cdots \left(x_{j_\ell} y_{i_\ell}\right)$
over all sets
$\left\{\binom{i_1}{j_1}, \ldots, \binom{i_\ell}{j_\ell}\right\}$
of biletters. Thus,
the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^\ell \xx^{\cont(\jj)} \yy^{\cont(\ii)}$ over all
strict cobiwords $\binom{\ii}{\jj}$, where $\ell$ stands
for the number of biletters in the strict cobiword. Meanwhile, we
know that the right hand side of
\eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}$
over all tableau-cotableau pairs $\left(P, Q\right)$
(because of
\eqref{sol.cauchy.dual.a.secondproof.rhs}). Thus,
in order to prove \eqref{sol.cauchy.dual.a.secondproof.goal2},
we only need to construct a bijection between the
strict cobiwords $\binom{\ii}{\jj}$ and the
tableau-cotableau pairs $\left(P,Q\right)$, which has
the property that
\begin{align*}
\cont\left(\ii\right) &= \cont\left(Q\right); \\
\cont\left(\jj\right) &= \cont\left(P\right).
\end{align*}
\footnote{Such a bijection will then automatically satisfy
$\left|\lambda\right| = \ell$, where $\lambda$ is the (common)
shape of $P$ and $Q$, and where $\ell$ is the length of the
strict cobiword $\binom{\ii}{\jj}$. This is because
$\left|\lambda\right| = \left|\cont\left(Q\right)\right|$
and $\ell = \left|\cont\left(\ii\right)\right|$.}

This bijection is the \dfn{coRSK algorithm}, which we
shall now define.\footnote{This algorithm appears in
Fulton's \cite[\S A.4]{Fulton} as construction (1d).}

Let $\binom{\ii}{\jj}$ be a strict cobiword.
Starting with the pair $(P_0,Q_0)=(\varnothing,\varnothing)$ and
$m=0$, the algorithm applies the following steps
(see Example~\ref{sol.cauchy.dual.a.secondproof.coRSK-example} below):
\begin{itemize}
\item If $i_{m+1}$ does not exist (that is, $m$ is the length of
$\ii$), stop.
\item Apply RS-insertion to the column-strict tableau $P_m$
and the letter $j_{m+1}$ (the bottom letter of
$\binom{i_{m+1}}{j_{m+1}}$). Let $P_{m+1}$ be the resulting
column-strict tableau, and let $c_{m+1}$ be the resulting
corner cell.
\item Create $Q_{m+1}$ from $Q_m$ by adding the top letter
$i_{m+1}$ of $\binom{i_{m+1}}{j_{m+1}}$ to $Q_m$ in the cell
$c_{m+1}$ (which, as we recall, is the extra corner cell of
$P_{m+1}$ not present in $P_m$).
\item Set $m$ to $m+1$.
\end{itemize}
After all of the biletters have been thus processed,
the result of the coRSK algorithm
is $(P_\ell,Q_\ell)=:(P,Q)$.

\begin{example}
\label{sol.cauchy.dual.a.secondproof.coRSK-example}
The term in the expansion of the left side of
\eqref{sol.cauchy.dual.a.secondproof.goal} corresponding to
\[
(x_4 y_1)(x_2 y_1)(x_1 y_2)(x_3 y_4)(x_1 y_4)(x_2 y_5)
\]
is the strict cobiword
$\binom{\ii}{\jj}=\binom{112445}{421312}$, and the coRSK
algorithm applied to this cobiword proceeds
as follows:
\[
\begin{array}{rcl|rcl}
P_0&= &\varnothing & Q_0 &=& \varnothing \\
 & & & & &\\
P_1&= &\begin{matrix} 4 \end{matrix}
 & Q_1 &= &\begin{matrix} 1 \end{matrix} \\
 & & & & &\\
P_2&= &\begin{matrix} 2 \\ 4 \end{matrix}
 & Q_2 &= &\begin{matrix} 1 \\ 1  \end{matrix} \\
 & & & & &\\
P_3&= &\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}
 & Q_3 &= &\begin{matrix} 1 \\ 1 \\ 3 \end{matrix} \\
 & & & & &\\
P_4&= &\begin{matrix} 1 & 3 \\ 2 \\ 4 \end{matrix}
 & Q_4 &= &\begin{matrix} 1 & 4 \\ 1 \\ 2 \end{matrix} \\
 & & & & &\\
P_5&= &\begin{matrix} 1 & 1 \\ 2 & 3 \\ 4 \end{matrix}
 & Q_5 &= &\begin{matrix} 1 & 4 \\ 1 & 4 \\ 2 \end{matrix} \\
 & & & & &\\
P:=P_6&= &\begin{matrix} 1 & 1 & 2 \\ 2 & 3 \\ 4 \end{matrix}
 & Q:=Q_6 &= &\begin{matrix} 1 & 4 & 5 \\ 1 & 4 \\ 2 \end{matrix} \\
 & & & & &\\
\end{array}
\]
\end{example}

It is clear that $P_m$ remains a column-strict tableau of some
Ferrers shape throughout the execution of the coRSK algorithm, and
that $Q_m$ remains a filling of the same shape as $P_m$ which is
(at least) weakly increasing left-to-right along rows and weakly
increasing top-to-bottom in columns. But we can also see that
$Q_m$ is strictly increasing left-to-right along rows\footnote{Indeed,
this follows from the observation that when one has a string of equal
letters $i_m=i_{m+1}=\cdots=i_{m+r}$ on top of the strict cobiword,
then the bottom letters bumped in are
$j_m > j_{m+1} > \cdots > j_{m+r}$, and therefore (as a consequence
of the last claim of part (b) of the row bumping lemma) the new cells
form a \emph{vertical strip}, that is, no two of these cells lie
in the same row. Actually, more can be said: Each of these new cells
(except for the first one) is in a row further down than the
previous one. We will use this stronger fact further below.}, so
that $Q_m$ is a row-strict tableau. Thus,
the result $\left(P, Q\right)$ of the coRSK algorithm
is a tableau-cotableau pair.

To see that the coRSK algorithm is a bijection, we show how to
recover $\binom{\ii}{\jj}$ from $(P,Q)$.  This is done by
reverse bumping in the same way as for the usual RSK algorithm,
with the only difference that now $Q_m$ is obtained by removing
the bottommost (rather than the rightmost)
occurrence of the letter $i_{m+1}$ from $Q_{m+1}$.
\footnote{It necessarily has to be the bottommost occurrence, since
(according to the previous footnote) the cell into which
$i_{m+1}$ was filled at the step from $Q_m$ to $Q_{m+1}$ lies further
down than any existing cell of $Q_m$ containing the letter $i_{m+1}$.}

Finally, to see that the coRSK map is surjective, one needs to show
that the reverse bumping procedure can be applied to any
tableau-cotableau pair $(P,Q)$, and will result in a strict
cobiword $\binom{\ii}{\jj}$.
We leave this verification to the reader.\footnote{It is easy to
see that repeatedly applying reverse bumping to $(P,Q)$ will
result in a sequence
$\binom{i_\ell}{j_\ell}, \binom{i_{\ell-1}}{j_{\ell-1}},
\ldots, \binom{i_1}{j_1}$ of biletters such that applying the
coRSK algorithm to
$\binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$ gives back
$(P,Q)$. The question is why we have
$\binom{i_1}{j_1} <_{alex} \cdots <_{alex} \binom{i_\ell}{j_\ell}$.
Since the chain of inequalities $i_1 \leq i_2 \leq \cdots \leq i_\ell$
is clear from the choice of entry to reverse-bump, it only remains
to show that for every string $i_m=i_{m+1}=\cdots=i_{m+r}$
of equal top letters, the corresponding bottom letters
strictly decrease (that is,
$j_m > j_{m+1} > \cdots > j_{m+r}$). One way to see
this is the following:
\par
Assume the contrary; i.e., assume that
the bottom letters corresponding
to some string $i_m=i_{m+1}=\cdots=i_{m+r}$ of equal top
letters do not strictly decrease.
Thus, $j_{m+p} \leq j_{m+p+1}$ for some
$p \in \left\{0, 1, \ldots, r-1\right\}$.
Consider this $p$.
\par
Let us consider the cells containing the equal letters
$i_m=i_{m+1}=\cdots=i_{m+r}$ in the tableau $Q_{m+r}$.
Label these cells as $c_m, c_{m+1}, \ldots, c_{m+r}$
from top to bottom (noticing that no two of them lie
in the same row, since $Q_{m+r}$ is row-strict).
By the definition of reverse bumping, the first entry
to be reverse bumped from $P_{m+r}$ is the entry in
position $c_{m+r}$ (since this is the bottommost
occurrence of the letter $i_{m+r}$ in $Q_{m+r}$); then, the next entry
to be reverse bumped is the one in position $c_{m+r-1}$,
etc., moving further and further up.
Thus, for each $q \in \left\{0, 1, \ldots, r\right\}$,
the tableau $P_{m+q-1}$ is obtained from $P_{m+q}$ by
reverse bumping the entry in position $c_{m+q}$.
Hence, conversely, the tableau $P_{m+q}$ is obtained
from $P_{m+q-1}$ by RS-inserting the entry $j_{m+q}$,
which creates the corner cell $c_{m+q}$.
\par
But recall that $j_{m+p} \leq j_{m+p+1}$.
Hence, part (a) of the row bumping lemma (applied to
$P_{m+p-1}$, $j_{m+p}$, $j_{m+p+1}$, $P_{m+p}$,
$c_{m+p}$, $P_{m+p+1}$ and $c_{m+p+1}$
instead of $P$, $j$, $j'$, $P'$, $c$, $P''$ and $c'$)
shows that the cell $c_{m+p+1}$ is in the same
row as the cell $c_{m+p}$ or in a row further
up.
But this contradicts the fact that the cell $c_{m+p+1}$ is
in a row further down than the cell $c_{m+p}$ (since
we have labeled our cells as
$c_m, c_{m+1}, \ldots, c_{m+r}$ from top to bottom,
and no two of them lied in the same row).
This contradiction completes our proof.}

So the coRSK map is a bijection having the required properties.
As we have said, this proves
\eqref{sol.cauchy.dual.a.secondproof.goal2}.
This solves Exercise~\ref{exe.cauchy.dual}(a) again.

[\textit{Remark:} By combining the two solutions of
Exercise~\ref{exe.cauchy.dual}(a) given above, one can obtain a
new proof of the equality \eqref{sol.cauchy.dual.a.omega.s2}
(a proof which does not use \eqref{skew-Schur-antipodes}).
Indeed, comparing
\begin{align*}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  \right)
&=\sum_{\lambda\in\Par }e_{\lambda}\left(  \xx \right)
m_{\lambda}\left(   \yy \right)
= \prod_{i,j=1}^\infty \left(1 + x_i y_j\right)
\qquad \qquad
\left(\text{by \eqref{sol.cauchy.dual.a.part1done}}\right) \\
&= \sum_{\lambda \in \Par} s_\lambda\left(\xx\right) s_{\lambda^t}\left(\yy\right)
\qquad \qquad
\left(\text{by \eqref{sol.cauchy.dual.a.secondproof.goal}}\right) \\
&= \sum_{\lambda \in\Par } s_{\lambda^{t}}\left( \xx \right)
\underbrace{s_{\left( \lambda^{t}\right)  ^{t}}\left(
\yy \right)  }_{\substack{=s_{\lambda}\left(  \yy \right)
\\\text{(since }\left(  \lambda^{t}\right)  ^{t}=\lambda\text{)}}}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}[c]{c}
\text{here, we substituted }\lambda^{t}\text{ for }\lambda\text{ in the sum,
since the map}\\
\Par \rightarrow\Par ,\ \lambda\mapsto\lambda
^{t}\text{ is a bijection}
\end{array}
\right) \\
&  =\sum_{\lambda\in\Par } s_{\lambda^t}\left( \xx \right)
s_\lambda \left( \yy \right)
\end{align*}
with
\begin{align*}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  \right)   &  =\sum_{\lambda
\in\Par }\underbrace{\left(  \omega\left[  \left[   \yy %
\right]  \right]  \right)  \left(  s_{\lambda}\left( \xx \right)
\right)  }_{= \omega\left(s_\lambda\left(  \xx \right)\right)}
\cdot\underbrace{\left(  \omega\left[
\left[   \yy \right]  \right]  \right)  \left(  s_{\lambda}\left(
{ \yy }\right)  \right)  }_{\substack{=s_{\lambda}\left(   \yy %
\right)  \\\text{(by \eqref{sol.cauchy.dual.a.omega.s5})}}} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\left[  \left[
 \yy \right]  \right]  \text{ is a continuous }\kk \left[  \left[
 \yy \right]  \right]  \text{-algebra homomorphism}\right) \\
&  =\sum_{\lambda\in\Par }
\omega\left(s_\lambda\left(  \xx \right)\right)
s_{\lambda}\left( \yy \right) ,
\end{align*}
we obtain
$\sum_{\lambda \in \Par} \omega\left(s_\lambda \left(\xx\right)\right)
s_\lambda\left(\yy\right)
= \sum_{\lambda \in \Par} s_{\lambda^t}\left(\xx\right) s_\lambda\left(\yy\right)
$.
Since the $s_\lambda\left(\yy\right)$ for $\lambda \in \Par$ are linearly
independent over $\kk\left[\left[\xx\right]\right]$, we can compare
coefficients before $s_\lambda\left(\yy\right)$ in this equality.
As a result, we obtain $\omega\left(s_\lambda \left(\xx\right)\right)
= s_{\lambda^t}\left(\xx\right)$ for every $\lambda \in \Par$. Thus,
\eqref{sol.cauchy.dual.a.omega.s2} is proven again.]

\textit{Third solution to Exercise~\ref{exe.cauchy.dual}(a):}
In order to solve Exercise~\ref{exe.cauchy.dual}(a), it suffices to
verify the identity \eqref{sol.cauchy.dual.a.secondproof.goal2}.
(This can be proven by the same argument as in the Second solution to
Exercise~\ref{exe.cauchy.dual}(a).)

We shall now verify the identity
\eqref{sol.cauchy.dual.a.secondproof.goal2} using the so-called
\emph{dual RSK algorithm} (also known as
\dfn{RSK$^\ast$ algorithm})\footnote{We will define this algorithm
further below. It also frequently appears in literature: see, e.g.,
\cite[\S 7.14]{Stanley}, \cite[\S 5]{Knuth-RSK},
\cite[Theorem 4.8.5]{Sagan} and (with different conventions)
\cite[\S A.4.3, Prop. 3]{Fulton}.}. This will be fairly similar to the
Second solution to Exercise~\ref{exe.cauchy.dual}(a) given above, but
not identical to it; in particular, the dual RSK algorithm that we will
introduce below will (unlike the coRSK algorithm from the Second
solution) not rely on the same row bumping operation as the usual
RSK algorithm, but on a somewhat modified version of it.

For a given partition $\lambda$, let us define a \dfn{row-strict
tableau} of shape $\lambda$ to be an assignment $T$ of entries in
$\left\{1,2,3,\ldots\right\}$ to the cells of the Ferrers diagram
for $\lambda$ which is strictly increasing left-to-right in rows, and
weakly increasing top-to-bottom in columns. The equality
\eqref{sol.cauchy.dual.a.secondproof.slat.x} holds\footnote{Indeed,
it can be proven as in the
Second solution to Exercise~\ref{exe.cauchy.dual}(a).}; thus, we have
\begin{align}
\sum\limits_{\substack{P \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(P\right)}
&= \sum\limits_{\substack{Q \text{ is a row-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(Q\right)}
= s_{\lambda^t}
\qquad \left(\text{by \eqref{sol.cauchy.dual.a.secondproof.slat.x}}
\right) \nonumber \\
&= s_{\lambda^t} \left(\xx\right) .
\label{sol.cauchy.dual.a.thirdproof.slat}
\end{align}
We also have
\[
\sum\limits_{\substack{Q \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(Q\right)}
= \sum\limits_{\substack{P \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \xx^{\cont\left(P\right)}
= s_{\lambda} \left(\xx\right)
\]
(because this is how $s_{\lambda} = s_{\lambda} \left(\xx\right)$
is defined) for every partition $\lambda$. Substituting $\yy$ for
$\xx$ in this equality, we obtain
\begin{equation}
\sum\limits_{\substack{Q \text{ is a column-strict} \\ \text{tableau of
shape } \lambda}} \yy^{\cont\left(Q\right)}
= s_{\lambda} \left(\yy\right) .
\label{sol.cauchy.dual.a.thirdproof.sla}
\end{equation}

A \dfn{cotableau-tableau pair} will mean a pair
$\left(P, Q\right)$ such that $P$ is a row-strict
tableau, $Q$ is a column-strict tableau, and $P$ and $Q$
both have shape $\lambda$ for one and the same partition
$\lambda$. Multiplying the identities
\eqref{sol.cauchy.dual.a.thirdproof.slat} and
\eqref{sol.cauchy.dual.a.thirdproof.sla} and multiplying
the result with $t^{\left|\lambda\right|}$, we obtain
\[
t^{\left|\lambda\right|}
\sum\limits_{\substack{
\left(P, Q\right) \text{ is a pair with} \\
P \text{ being a row-strict tableau} \\
\text{of shape } \lambda \text{, and } Q \text{ being} \\
\text{a column-strict tableau of shape }\lambda
}} \xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= t^{\left|\lambda\right|}
s_{\lambda^t} \left(\xx\right) s_{\lambda} \left(\yy\right)
\]
for every partition $\lambda$. Summing this equality over
all partitions $\lambda$, we obtain
\[
\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda^t} \left(\xx\right) s_{\lambda} \left(\yy\right) ,
\]
where the sum on the left hand side is over all
cotableau-tableau pairs $\left(P, Q\right)$ and where
$\lambda$ denotes the common shape of $P$ and $Q$. This becomes
\begin{align}
\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}
&= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda^t} \left(\xx\right) s_{\lambda} \left(\yy\right)
= \sum_{\lambda \in \Par}
\underbrace{t^{\left|\lambda^t\right|}}_{\substack{
  = t^{\left|\lambda\right|} \\
  \text{(since } \left|\lambda^t\right| = \left|\lambda\right| \text{)}}}
\underbrace{s_{\left(\lambda^t\right)^t} \left(\xx\right)}_{
 \substack{=s_{\lambda} \left(\xx\right) \\ \text{(since }
 \left(\lambda^t\right)^t = \lambda \text{)}}}
s_{\lambda^t} \left(\yy\right)
\nonumber \\
&\qquad \left(
\begin{array}[c]{c}
\text{here, we substituted } \lambda^t \text{ for }\lambda
\text{ in the sum, since the map}\\
\Par \rightarrow\Par ,\ \lambda \mapsto \lambda^t
\text{ is a bijection}
\end{array}
\right)
\nonumber \\
&= \sum_{\lambda \in \Par} t^{\left|\lambda\right|}
s_{\lambda} \left(\xx\right)
s_{\lambda^t} \left(\yy\right) .
\label{sol.cauchy.dual.a.thirdproof.rhs}
\end{align}

We shall use all notations introduced in the proof of
Theorem~\ref{thm.cauchy-identity}. A \dfn{strict biword} will mean
an array
$\binom{\ii}{\jj} = \binom{i_1\ldots i_\ell}{j_1\ldots j_\ell}$
in which the biletters satisfy
$\binom{i_1}{j_1} <_{lex} \cdots <_{lex} \binom{i_\ell}{j_\ell}$
(that is, the biletters are distinct and ordered with respect to the
lexicographic order).
Strict biwords are clearly in 1-to-1 correspondence
with sets (not multisets!) of biletters.
Now, the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is
$\prod_{i,j=1}^\infty \left(1 + t x_i y_j\right)
= \prod_{i,j=1}^\infty \left(1 + t x_j y_i\right)$
(here, we substituted $\left(j,i\right)$ for the
index $\left(i,j\right)$ in the product), and thus can be
rewritten as the sum of
$t^\ell \left(x_{j_1} y_{i_1}\right) \left(x_{j_2} y_{i_2}\right)
\cdots \left(x_{j_\ell} y_{i_\ell}\right)$
over all sets
$\left\{\binom{i_1}{j_1}, \ldots, \binom{i_\ell}{j_\ell}\right\}$
of biletters. Thus,
the left hand side
of \eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^\ell \xx^{\cont(\jj)} \yy^{\cont(\ii)}$ over all
strict biwords $\binom{\ii}{\jj}$, where $\ell$ stands
for the number of biletters in the biword. Meanwhile, we
know that the right hand side of
\eqref{sol.cauchy.dual.a.secondproof.goal2} is the sum
$\sum t^{\left|\lambda\right|}
\xx^{\cont\left(P\right)} \yy^{\cont\left(Q\right)}$
over all cotableau-tableau pairs $\left(P, Q\right)$
(because of
\eqref{sol.cauchy.dual.a.thirdproof.rhs}). Thus,
in order to prove \eqref{sol.cauchy.dual.a.secondproof.goal2},
we only need to construct a bijection between the
strict biwords $\binom{\ii}{\jj}$ and the
cotableau-tableau pairs $\left(P,Q\right)$, which has
the property that
\begin{align*}
\cont\left(\ii\right) &= \cont\left(Q\right); \\
\cont\left(\jj\right) &= \cont\left(P\right).
\end{align*}
\footnote{Such a bijection will then automatically satisfy
$\left|\lambda\right| = \ell$, where $\lambda$ is the (common)
shape of $P$ and $Q$, and where $\ell$ is the length of the
strict biword $\binom{\ii}{\jj}$. This is because
$\left|\lambda\right| = \left|\cont\left(Q\right)\right|$
and $\ell = \left|\cont\left(\ii\right)\right|$.}

This bijection is the \dfn{dual RSK algorithm}, which we
shall define below.

First, we shall introduce a simpler operation which we call
\dfn{dual RS-insertion} (and which is similar to RS-insertion, but
not identical with it\footnote{We will leave many of its properties
unproven, because their proofs are analogous (or at least very similar)
to the proofs of the corresponding properties of RS-insertion, and thus
can be easily reconstructed by the reader.}).

Dual RS-insertion takes as input a row-strict tableau $P$ and a letter
$j$, and returns a new row-strict tableau $P'$ along with a corner
cell $c$ of $P'$, which is constructed as follows:
Start out by setting $P' = P$. The letter $j$ tries to insert itself
into the first row of $P'$ by either bumping out the leftmost letter
in the first row \textbf{larger or equal to} $j$, or else placing
itself at the right end of the row if
no such letter (larger or equal to $j$) exists.
If a letter was bumped from the
first row, this letter follows
the same rules to insert itself into the second row, and so on.
This series of bumps must eventually come to an
end.  At the end of the bumping,
the tableau $P'$ created has an extra corner cell not present in $P$.
If we call this corner cell $c$, then $P'$ (in its final form) and $c$
are what the dual RS-insertion operation returns. One says that $P'$ is the
result of \dfn{dually inserting} $j$ into the tableau $P$. It is
straightforward
to see that this resulting filling $P'$ is a row-strict
tableau\footnote{Indeed, the reader can check that $P'$ remains a
row-strict tableau throughout the algorithm that defines dual
RS-insertion.}.

\begin{example}
To give an example
of this operation, let us dually insert the letter $j = 3$ into the
row-strict tableau
$\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 4 & \\ 3 & 4 & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix}$ (we are showing all
intermediate states of $P'$; the underlined letter is always
the one that is going to be bumped out at the next step):
\begin{align*}
\begin{matrix} 1 & 2 & \underline{3} & 5 \\ 1 & 2 & 4 & \\ 3 & 4 & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix}
\quad & \overset{\longmapsto}{\substack{\text{insert 3;} \\ \text{bump out 3}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & \underline{4} & \\ 3 & 4 & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 3;} \\ \text{bump out 4}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & \underline{4} & 7 & \\
3 & 6 & & \\ 4 & & & \end{matrix} \\
\quad & \overset{\longmapsto}{\substack{\text{insert 4;} \\ \text{bump out 4}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & 4 & 7 & \\
3 & \underline{6} & & \\ 4 & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 4;} \\ \text{bump out 6}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & 4 & 7 & \\
3 & 4 & & \\ 4 & & & \end{matrix}
\quad \overset{\longmapsto}{\substack{\text{insert 6;} \\ \text{done}}} \quad
\begin{matrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & \\ 3 & 4 & 7 & \\
3 & 4 & & \\ 4 & 6 & & \end{matrix} .
\end{align*}
The last tableau in this sequence is the row-strict
tableau that is returned. The corner cell that is returned
is the second cell of the fifth row (the one containing $6$).
\end{example}

Dual RS-insertion will be used as a step in the dual RSK algorithm; the
construction will rely on a simple fact known as the
\emph{dual row bumping lemma}. Let us first define the notion of
a \emph{dual bumping path} (or \dfn{dual bumping route}):
If $P$ is a row-strict tableau, and $j$ is a letter, then
some letters are inserted into some cells when dual RS-insertion
is applied to $P$ and $j$. The sequence of these cells (in
the order in which they see letters inserted into them)\footnote{In
particular, this includes those cells whose entries did not change under
the insertion (because the entry inserted was the same as the entry
they contained before the insertion).} is called
the \dfn{dual bumping path} for $P$ and $j$. This dual bumping path
always ends with the corner cell $c$ which is returned by
dual RS-insertion. As an example, when
$j=3$ is dually inserted into the tableau $P$
shown below, the result $P'$ is shown with all entries on
the dual bumping path underlined:
\[
P=\begin{matrix}
1&2&3&4 \\
1&2&4&6 \\
2&4&5&6 \\
2&4& & \\
3& & &
\end{matrix}
\qquad
\underset{j=3}{\overset{\text{dually insert}}{\longmapsto}}
\qquad
P'=\begin{matrix}
1&2&\underline{3}&4 \\
1&2&\underline{3}&6 \\
2&\underline{4}&5&6 \\
2&\underline{4}& & \\
3&\underline{4}& &
\end{matrix}
\]
A first simple observation about dual bumping paths is that
dual bumping paths trend weakly left (just as bumping paths for regular
RS-insertion do). A subtler property of bumping paths is the following
\dfn{dual row bumping lemma}\footnote{This lemma is equivalent to
the ``column bumping lemma'' in Fulton \cite[p. 187]{Fulton}.}:

\begin{statement}
\textbf{Dual row bumping lemma:} Let $P$ be a row-strict tableau,
and let $j$ and $j'$ be two letters. Applying dual RS-insertion to the
tableau $P$ and the letter $j$ yields a new row-strict tableau
$P'$ and a corner cell $c$. Applying dual RS-insertion to the tableau
$P'$ and the letter $j'$ yields a new row-strict tableau $P''$
and a corner cell $c'$.
\begin{enumerate}
\item[(a)] Assume that $j < j'$. Then, the dual bumping path
for $P'$ and $j'$
stays strictly to the right, within each row, of the dual bumping path
for $P$ and $j$.
The cell $c'$ (in which the dual bumping path for $P'$ and $j'$ ends) is
in the same row as the cell $c$ (in which the dual bumping path for $P$
and $j$ ends) or in a row further up; it is also in a column further
right than $c$.
\item[(b)] Assume instead that $j \geq j'$. Then, the dual bumping path
for $P'$ and $j'$
stays weakly to the left, within each row, of the dual bumping path
for $P$ and $j$.
The cell $c'$ (in which the dual bumping path for $P'$ and $j'$ ends) is
in a row further down than the cell $c$ (in which the dual bumping
path for $P$ and $j$ ends); it is also in the same column as $c$ or
in a column further left.
\end{enumerate}
\end{statement}
This lemma can be easily proven by induction over the
row (similarly to the usual row bumping lemma).

We will now define the dual RSK algorithm.
Let $\binom{\ii}{\jj}$ be a strict biword.
Starting with the pair $(P_0,Q_0)=(\varnothing,\varnothing)$ and
$m=0$, the algorithm applies the following steps
(see Example~\ref{sol.cauchy.dual.a.thirdproof.dualRSK-example} below):
\begin{itemize}
\item If $i_{m+1}$ does not exist (that is, $m$ is the length of
$\ii$), stop.
\item Apply dual RS-insertion to the row-strict tableau $P_m$
and the letter $j_{m+1}$ (the bottom letter of
$\binom{i_{m+1}}{j_{m+1}}$). Let $P_{m+1}$ be the resulting
row-strict tableau, and let $c_{m+1}$ be the resulting
corner cell.
\item Create $Q_{m+1}$ from $Q_m$ by adding the top letter
$i_{m+1}$ of $\binom{i_{m+1}}{j_{m+1}}$ to $Q_m$ in the cell
$c_{m+1}$ (which, as we recall, is the extra corner cell of
$P_{m+1}$ not present in $P_m$).
\item Set $m$ to $m+1$.
\end{itemize}
After all of the biletters have been thus processed,
the result of the dual RSK algorithm
is $(P_\ell,Q_\ell)=:(P,Q)$.

\begin{example}
\label{sol.cauchy.dual.a.thirdproof.dualRSK-example}
The term in the expansion of the left side of
\eqref{sol.cauchy.dual.a.secondproof.goal} corresponding to
\[
(x_2 y_1)(x_4 y_1)(x_1 y_2)(x_1 y_4)(x_3 y_4)(x_2 y_5)
\]
is the strict biword
$\binom{\ii}{\jj}=\binom{112445}{241132}$, and the dual RSK
algorithm applied to this biword proceeds
as follows:
\[
\begin{array}{rcl|rcl}
P_0&= &\varnothing & Q_0 &=& \varnothing \\
 & & & & &\\
P_1&= &\begin{matrix} 2 \end{matrix}
 & Q_1 &= &\begin{matrix} 1 \end{matrix} \\
 & & & & &\\
P_2&= &\begin{matrix} 2 & 4 \end{matrix}
 & Q_2 &= &\begin{matrix} 1 & 1 \end{matrix} \\
 & & & & &\\
P_3&= &\begin{matrix} 1 & 4 \\ 2 \end{matrix}
 & Q_3 &= &\begin{matrix} 1 & 1 \\ 2 \end{matrix} \\
 & & & & &\\
P_4&= &\begin{matrix} 1 & 4 \\ 1 \\ 2 \end{matrix}
 & Q_4 &= &\begin{matrix} 1 & 1 \\ 2 \\ 4 \end{matrix} \\
 & & & & &\\
P_5&= &\begin{matrix} 1 & 3 \\ 1 & 4 \\ 2 \end{matrix}
 & Q_5 &= &\begin{matrix} 1 & 1 \\ 2 & 4 \\ 4 \end{matrix} \\
 & & & & &\\
P:=P_6&= &\begin{matrix} 1 & 2 \\ 1 & 3 \\ 2 & 4 \end{matrix}
 & Q:=Q_6 &= &\begin{matrix} 1 & 1 \\ 2 & 4 \\ 4 & 5 \end{matrix} \\
 & & & & &\\
\end{array}
\]
\end{example}

It is clear that $P_m$ remains a row-strict tableau of some
Ferrers shape throughout the execution of the dual RSK algorithm, and
that $Q_m$ remains a filling of the same shape as $P_m$ which is
(at least) weakly increasing left-to-right along rows and weakly
increasing top-to-bottom in columns. But we can also see that
$Q_m$ is strictly increasing top-to-bottom along columns\footnote{Indeed,
this follows from the observation that when one has a string of equal
letters $i_m=i_{m+1}=\cdots=i_{m+r}$ on top of the strict biword,
then the bottom letters bumped in are
$j_m < j_{m+1} < \cdots < j_{m+r}$, and therefore (as a consequence
of the second-to-last claim of part (a) of the dual row bumping lemma)
the new
cells form a \emph{horizontal strip}, that is, no two of these cells
lie in the same column. Actually, more can be said: Each of these new
cells (except for the first one) is in a column further right than
the previous one. We will use this stronger fact further below.},
so that $Q_m$ is a column-strict tableau. Thus,
the result $\left(P, Q\right)$ of the dual RSK algorithm
is a cotableau-tableau pair.

To see that the dual RSK algorithm is a bijection, we show how to
recover $\binom{\ii}{\jj}$ from $(P,Q)$.  This is done by
\dfn{dually reverse bumping} from $(P_{m+1},Q_{m+1})$ to recover
both the biletter $\binom{i_{m+1}}{j_{m+1}}$ and the tableaux
$(P_m,Q_m)$, as follows.
Firstly, $i_{m+1}$ is the maximum entry of $Q_{m+1}$, and
$Q_m$ is obtained by removing
the rightmost occurrence of this letter $i_{m+1}$ from $Q_{m+1}$.
\footnote{It necessarily has to be the rightmost occurrence, since
(according to the previous footnote) the cell into which
$i_{m+1}$ was filled at the step from $Q_m$ to $Q_{m+1}$ lies further
right than any existing cell of $Q_m$ containing the letter $i_{m+1}$.}
To produce $P_m$ and $j_{m+1}$, find the position of the
rightmost occurrence of $i_{m+1}$ in $Q_{m+1}$,
and start \dfn{dually reverse bumping} in $P_{m+1}$ from the entry in
this same position, where dually reverse bumping an
entry means inserting it into one row higher by having it bump
out the rightmost entry which is smaller or equal to
it.\footnote{Let us give a few more details on this
``dually reverse bumping'' procedure.
Dually reverse bumping (also known as \dfn{dual RS-deletion} or
\dfn{reverse dual RS-insertion}) is an operation which takes a
row-strict tableau $P'$ and a corner cell $c$ of $P'$,
and constructs a row-strict tableau $P$ and a letter $j$
such that dual RS-insertion for $P$ and $j$ yields $P'$ and $c$.
It starts by setting $P = P'$, and removing the entry in the
cell $c$ from $P$. This removed entry is then denoted by $k$,
and is inserted into the row of $P$ above $c$, bumping out the
rightmost entry which is smaller or equal to $k$. The letter which is
bumped out -- say, $\ell$ --, in turn, is inserted into the row
above it, bumping out the rightmost entry which is smaller or equal
to $\ell$. This procedure continues in the same way until an entry
is bumped out of the first row (which will eventually happen).
The dually reverse bumping operation returns the
resulting tableau $P$ and the entry which is bumped out of the
first row.
\par
It is straightforward to check that the dually reverse bumping
operation is well-defined (i.e., $P$ does stay a row-strict
tableau throughout the procedure) and is the inverse of the
dual RS-insertion operation. (In fact, these two operations undo each
other step by step.)}
The entry bumped out of the
first row is $j_{m+1}$, and the resulting tableau is $P_{m}$.

Finally, to see that the dual RSK map is surjective, one needs to show
that the dually reverse bumping procedure can be applied to any
cotableau-tableau pair $(P,Q)$, and will result in a strict
biword $\binom{\ii}{\jj}$.
We leave this verification to the reader.\footnote{It is easy to
see that repeatedly applying dually reverse bumping to $(P,Q)$ will
result in a sequence
$\binom{i_\ell}{j_\ell}, \binom{i_{\ell-1}}{j_{\ell-1}},
\ldots, \binom{i_1}{j_1}$ of biletters such that applying the
dual RSK algorithm to
$\binom{i_1 \cdots i_\ell}{j_1 \cdots j_\ell}$ gives back
$(P,Q)$. The question is why we have
$\binom{i_1}{j_1} <_{lex} \cdots <_{lex} \binom{i_\ell}{j_\ell}$.
Since the chain of inequalities $i_1 \leq i_2 \leq \cdots \leq i_\ell$
is clear from the choice of entry to dually reverse-bump, it only
remains to show that for every string $i_m=i_{m+1}=\cdots=i_{m+r}$
of equal top letters, the corresponding bottom letters
strictly increase (that is,
$j_m < j_{m+1} < \cdots < j_{m+r}$). One way to see
this is the following:
\par
Assume the contrary; i.e., assume that
the bottom letters corresponding
to some string $i_m=i_{m+1}=\cdots=i_{m+r}$ of equal top
letters do not strictly increase.
Thus, $j_{m+p} \geq j_{m+p+1}$ for some
$p \in \left\{0, 1, \ldots, r-1\right\}$.
Consider this $p$.
\par
Let us consider the cells containing the equal letters
$i_m=i_{m+1}=\cdots=i_{m+r}$ in the tableau $Q_{m+r}$.
Label these cells as $c_m, c_{m+1}, \ldots, c_{m+r}$
from left to right (noticing that no two of them lie
in the same column, since $Q_{m+r}$ is column-strict).
By the definition of dually reverse bumping, the first entry
to be dually reverse bumped from $P_{m+r}$ is the entry in
position $c_{m+r}$ (since this is the rightmost
occurrence of the letter $i_{m+r}$ in $Q_{m+r}$); then, the next entry
to be dually reverse bumped is the one in position $c_{m+r-1}$,
etc., moving further and further left.
Thus, for each $q \in \left\{0, 1, \ldots, r\right\}$,
the tableau $P_{m+q-1}$ is obtained from $P_{m+q}$ by
dually reverse bumping the entry in position $c_{m+q}$.
Hence, conversely, the tableau $P_{m+q}$ is obtained
from $P_{m+q-1}$ by dually RS-inserting the entry $j_{m+q}$,
which creates the corner cell $c_{m+q}$.
\par
But recall that $j_{m+p} \geq j_{m+p+1}$.
Hence, part (b) of the dual row bumping lemma (applied to
$P_{m+p-1}$, $j_{m+p}$, $j_{m+p+1}$, $P_{m+p}$,
$c_{m+p}$, $P_{m+p+1}$ and $c_{m+p+1}$
instead of $P$, $j$, $j'$, $P'$, $c$, $P''$ and $c'$)
shows that the cell $c_{m+p+1}$ is in the same
column as the cell $c_{m+p}$ or in a column further
left.
But this contradicts the fact that the cell $c_{m+p+1}$ is
in a column further right than the cell $c_{m+p}$ (since
we have labeled our cells as
$c_m, c_{m+1}, \ldots, c_{m+r}$ from left to right,
and no two of them lied in the same column).
This contradiction completes our proof.}

So the dual RSK map is a bijection having the required properties.
As we have said, this proves
\eqref{sol.cauchy.dual.a.secondproof.goal2}.
This solves Exercise~\ref{exe.cauchy.dual}(a) again.

(b) Let us consider the map $\omega\left[  \left[   \yy \right]  \right]
:\Lambda\left[  \left[   \yy \right]  \right]  \rightarrow\Lambda\left[
\left[   \yy \right]  \right]  $ defined as in the solution of Exercise
\ref{exe.cauchy.dual}(a).

For every partition $\lambda$, we have $p_{\lambda}\left(  \xx \right)
\in\Lambda$. Thus, for every partition $\lambda$, we have%
\begin{align}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
p_{\lambda}\left( \xx \right)  \right)   &  =\omega\left(
\underbrace{p_{\lambda}\left(  \xx \right)  }_{=p_{\lambda}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.cauchy.dual.a.omega}, applied
to }q=p_{\lambda}\left(  \xx \right)  \right)  \nonumber\\
&  =\omega\left(  p_{\lambda}\right)  =\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }
\underbrace{p_{\lambda}}_{=p_{\lambda}\left(  \xx \right)  }
\ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{eq.exe.Lambda.omegahlam.p}}\right)  \nonumber\\
&  =\left(  -1\right)  ^{\left\vert \lambda\right\vert -\ell\left(
\lambda\right)  }p_{\lambda}\left(  \xx \right) .
\label{sol.cauchy.dual.a.omega.p4}
\end{align}
Furthermore, for every partition $\lambda$, we have $p_{\lambda}\left(
 \yy \right)  \in\kk \left[  \left[   \yy \right]  \right]  $.
Hence, for every partition $\lambda$, we have
\begin{equation}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
p_{\lambda}\left(   \yy \right)  \right)  =p_{\lambda}\left(
 \yy \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.cauchy.dual.a.omega2}, applied to }p=p_{\lambda}\left(
 \yy \right)  \right)  .
\label{sol.cauchy.dual.a.omega.p5}
\end{equation}


Now, the equality \eqref{Cauchy-identity} yields
\begin{equation}
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  =\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda\in\Par }z_{\lambda}%
^{-1}p_{\lambda}\left( \xx \right)  p_{\lambda}\left(  { \yy %
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{Two-other-Cauchy-expansions}}\right)  .
\label{sol.cauchy.dual.b.5}
\end{equation}
This is an equality in $\Lambda\left[  \left[   \yy \right]  \right]  $
(because $s_{\lambda}\left( \xx \right)  $ and $p_{\lambda}\left(
\xx \right)  $ belong to $\Lambda$ for every $\lambda\in
\Par $). Hence, we can apply the map $\omega\left[  \left[
 \yy \right]  \right]  :\Lambda\left[  \left[   \yy \right]
\right]  \rightarrow\Lambda\left[  \left[   \yy \right]  \right]  $ to
both sides of this equality. As a result, we obtain
\begin{align*}
\left(  \omega\left[  \left[   \yy \right]  \right]  \right)  \left(
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda}\left( \yy \right)  \right)   &  =\left(  \omega\left[
\left[   \yy \right]  \right]  \right)  \left(  \sum_{\lambda
\in\Par }z_{\lambda}^{-1}p_{\lambda}\left( \xx \right)
p_{\lambda}\left( \yy \right)  \right)  =\sum_{\lambda\in
\Par }z_{\lambda}^{-1}\underbrace{\left(  \omega\left[  \left[
 \yy \right]  \right]  \right)  \left(  p_{\lambda}\left(  \xx
\right)  \right)  }_{\substack{=\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }p_{\lambda}\left(
\xx \right)  \\\text{(by \eqref{sol.cauchy.dual.a.omega.p4})}}}
\cdot\underbrace{\left(  \omega\left[  \left[   \yy \right]  \right]
\right)  \left(  p_{\lambda}\left( \yy \right)  \right)
}_{\substack{=p_{\lambda}\left(   \yy \right)  \\\text{(by
\eqref{sol.cauchy.dual.a.omega.p5})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\left[  \left[
 \yy \right]  \right]  \text{ is a continuous }\kk \left[  \left[
 \yy \right]  \right]  \text{-algebra homomorphism}\right)  \\
&  =\sum_{\lambda\in\Par }z_{\lambda}^{-1}\left(  -1\right)
^{\left\vert \lambda\right\vert -\ell\left(  \lambda\right)  }p_{\lambda
}\left(  \xx \right)  p_{\lambda}\left(   \yy \right)
=\sum_{\lambda\in\Par }\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }z_{\lambda}^{-1}p_{\lambda
}\left( \xx \right)  p_{\lambda}\left( \yy \right)  .
\end{align*}
Compared with \eqref{sol.cauchy.dual.a.omega.leftside}, this yields
\[
\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda^{t}}\left( \yy \right)  =\sum_{\lambda\in
\Par }\left(  -1\right)  ^{\left\vert \lambda\right\vert
-\ell\left(  \lambda\right) }z_{\lambda}^{-1}
p_{\lambda}\left( \xx \right) p_{\lambda}\left( \yy \right)  .
\]
Hence,
\[
\sum_{\lambda\in\Par }\left(  -1\right)  ^{\left\vert
\lambda\right\vert -\ell\left(  \lambda\right)  }z_{\lambda}^{-1}p_{\lambda
}\left( \xx \right)  p_{\lambda}\left( \yy \right)
=\sum_{\lambda\in\Par }s_{\lambda}\left( \xx \right)
s_{\lambda^{t}}\left( \yy \right)  =\prod_{i,j=1}^{\infty}\left(
1+x_{i}y_{j}\right)
\]
(by \eqref{sol.cauchy.dual.a.omega.end}). This solves
Exercise~\ref{exe.cauchy.dual}(b).
%\end{solutions}

\end{verlong}



\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.jacobi-trudi}}

\textit{Solution to Exercise~\ref{exe.jacobi-trudi}.} Before we begin
the actual solution, let us recall a basic fact from linear algebra, namely
the explicit formula for the determinant of a matrix as a sum over
permutations: Any matrix $\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,\ell}$
over a commutative ring satisfies
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,\ell}\right)
=\sum_{\sigma\in\Symm_{\ell}}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{\ell}\alpha_{i,\sigma\left(  i\right)  }.
\label{sol.jacobi-trudi.b.detexp}
\end{equation}


(a) Fix two partitions $\lambda$ and $\mu$. We need to show that
$\widetilde{s}_{\lambda/\mu}^{[k]}$ does not depend on $k$ as long as
$k\geq\ell\left(  \lambda\right)  $ and $k\geq\ell\left(  \mu\right)  $. For
this, it is clearly enough to prove that if $k\in\NN$ satisfies
$k\geq\ell\left(  \lambda\right)  $ and $k\geq\ell\left(  \mu\right)  $, then
$\widetilde{s}_{\lambda/\mu}^{[k]}=\widetilde{s}_{\lambda/\mu}^{[k+1]}$. So
let us prove this.

Fix a $k\in\NN$ satisfying $k\geq\ell\left(  \lambda\right)  $ and
$k\geq\ell\left(  \mu\right)  $. We have
$\widetilde{s}_{\lambda/\mu}^{[k+1]}
= \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,k+1}\right)  $ (by the definition of
$\widetilde{s}_{\lambda/\mu}^{[k+1]}$)
and $\widetilde{s}_{\lambda/\mu}^{[k]}=\det\left(
\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)  $ (by
the definition of $\widetilde{s}_{\lambda/\mu}^{[k]}$).

Notice that $k+1>k\geq\ell\left(  \lambda\right)  $, so that
$\lambda_{k+1}=0$. Similarly, $\mu_{k+1}=0$.

Expanding the determinant $\det\left(  \left(  h_{\lambda_{i}-\mu_{j}
-i+j}\right)  _{i,j=1,2,\ldots,k+1}\right)  $ with respect to its $\left(
k+1\right)  $-st row, we obtain
\begin{align}
&  \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,k+1}\right) \nonumber\\
&  =\sum_{r=1}^{k+1}h_{\lambda_{k+1}-\mu_{r}-\left(  k+1\right)  +r}
\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \cdot
\left(  \text{the }\left(  k+1,r\right)  \text{-th cofactor of the matrix
}\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k+1}\right) .
\label{sol.jacobi-trudi.a.1}
\end{align}
For every $r\in\left\{  1,2,...,k\right\}  $, we have $h_{\lambda_{k+1}
-\mu_{r}-\left(  k+1\right)  +r}=0$\ \ \ \ \footnote{\textit{Proof.} Let
$r\in\left\{  1,2,...,k\right\}  $. Then, $r\leq k$. Hence,
$\underbrace{\lambda_{k+1}}_{=0}-\mu_{r}-\left(  k+1\right) 
+\underbrace{r}_{\leq k}
\leq -\mu_{r}-\left(  k+1\right)  +k = -\mu_{r}-1
< -\underbrace{\mu_{r}}_{\geq0} \leq 0$,
so that $h_{\lambda_{k+1}-\mu_{r}-\left(  k+1\right)  +r}=0$,
qed.}. Hence, all terms of the sum on the right hand side of
\eqref{sol.jacobi-trudi.a.1} vanish except possibly the $r=k+1$ term. Thus,
\eqref{sol.jacobi-trudi.a.1} simplifies to
\begin{align*}
&  \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,k+1}\right) \\
&  =\underbrace{h_{\lambda_{k+1}-\mu_{k+1}-\left(  k+1\right)  +\left(
k+1\right)  }}_{\substack{=h_{0}\\\text{(since }\lambda_{k+1}-\mu
_{k+1}-\left(  k+1\right)  +\left(  k+1\right)
 =\underbrace{\lambda_{k+1}}_{=0}
-\underbrace{\mu_{k+1}}_{=0}=0\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \text{the }\left(
k+1,k+1\right)  \text{-th cofactor of the matrix }\left(  h_{\lambda_{i}
-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k+1}\right)  }_{=\left(  -1\right)
^{\left(  k+1\right)  +\left(  k+1\right)  }\det\left(  \left(  h_{\lambda
_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)  }\\
&  =\underbrace{h_{0}}_{=1}\cdot\underbrace{\left(  -1\right)  ^{\left(
k+1\right)  +\left(  k+1\right)  }}_{=1}\underbrace{\det\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)  }
_{=\widetilde{s}_{\lambda/\mu}^{[k]}}=\widetilde{s}_{\lambda/\mu}^{[k]}.
\end{align*}
Hence,
\[
\widetilde{s}_{\lambda/\mu}^{[k+1]}
= \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,k+1}\right)
= \widetilde{s}_{\lambda/\mu}^{[k]} .
\]
Thus, $\widetilde{s}_{\lambda/\mu}^{[k]}=\widetilde{s}_{\lambda/\mu}^{[k+1]}$
is proven. Exercise~\ref{exe.jacobi-trudi}(a) is solved.

(b) Let us first show a few more fundamental statements. First, we claim the
following: If $k\in\NN$ and if $\left(  a_{1},a_{2},...,a_{k}\right)
\in\ZZ ^{k}$ and $\left(  b_{1},b_{2},...,b_{k}\right)  \in
\ZZ ^{k}$ are two $k$-tuples of integers which satisfy $a_{1}%
>a_{2}>...>a_{k}$ and $b_{1}>b_{2}>...>b_{k}$ but do \textbf{not} satisfy
$\left(  a_{u}\geq b_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}
\right)  $, then
\begin{equation}
\det\left(  \left(  h_{a_{i}-b_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  =0.
\label{sol.jacobi-trudi.b.lem}%
\end{equation}


[\textit{Proof of \eqref{sol.jacobi-trudi.b.lem}:} Let $k\in\NN$. Let
$\left(  a_{1},a_{2},...,a_{k}\right)  \in\ZZ ^{k}$ and $\left(
b_{1},b_{2},...,b_{k}\right)  \in\ZZ ^{k}$ be two $k$-tuples of integers
which satisfy $a_{1}>a_{2}>...>a_{k}$ and $b_{1}>b_{2}>...>b_{k}$ but do
\textbf{not} satisfy $\left(  a_{u}\geq b_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  \right)  $. Then, there exists an $u\in\left\{
1,2,...,k\right\}  $ such that we don't have $a_{u}\geq b_{u}$ (because we do
\textbf{not} have \newline $\left(  a_{u}\geq b_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  \right)  $). Consider this $u$. We don't have $a_{u}\geq
b_{u}$; hence, we have $a_{u}<b_{u}$. Applying
\eqref{sol.jacobi-trudi.b.detexp} to $\ell=k$ and $\alpha_{i,j}=h_{a_{i}%
-b_{j}}$, we obtain%
\begin{equation}
\det\left(  \left(  h_{a_{i}-b_{j}}\right)  _{i,j=1,2,\ldots,k}\right)
=\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{k}h_{a_{i}-b_{\sigma\left(  i\right)  }}.
\label{sol.jacobi-trudi.b.lem.pf.2}%
\end{equation}


Now, let $\sigma\in\Symm_{k}$ be arbitrary. The two subsets $\left\{
u,u+1,...,k\right\}  $ and $\sigma^{-1}\left(  \left\{  1,2,...,u\right\}
\right)  $ of $\left\{ 1, 2, ..., k \right\}$
have $k-u+1$ and $u$ elements, respectively, and therefore cannot
be disjoint (since otherwise, they would have altogether $\left(
k-u+1\right)  +u=k+1$ elements, which is more than the cardinality of the set
$\left\{  1,2,...,k\right\}  $ which contains them both). Thus, there exists a
$v\in\left\{  1,2,...,k\right\}  $ which belongs both to $\left\{
u,u+1,...,k\right\}  $ and to $\sigma^{-1}\left(  \left\{  1,2,...,u\right\}
\right)  $. Consider this $v$. Since $v\in\left\{  u,u+1,...,k\right\}  $, we
have $v\geq u$ and thus $a_{v}\leq a_{u}$ (since $a_{1}>a_{2}>...>a_{k}$).
Since $v\in\sigma^{-1}\left(  \left\{  1,2,...,u\right\}  \right)  $, we have
$\sigma\left(  v\right)  \in\left\{  1,2,...,u\right\}  $, thus $\sigma\left(
v\right)  \leq u$ and therefore $b_{\sigma\left(  v\right)  }\geq b_{u}$
(since $b_{1}>b_{2}>...>b_{k}$). Now, $\underbrace{a_{v}}_{\leq a_{u}<b_{u}%
}-\underbrace{b_{\sigma\left(  v\right)  }}_{\geq b_{u}}<b_{u}-b_{u}=0$, so
that $h_{a_{v}-b_{\sigma\left(  v\right)  }}=0$. Hence, the product
$\prod_{i=1}^{k}h_{a_{i}-b_{\sigma\left(  i\right)  }}$ has at least one
factor equal to $0$ (namely, the $i=v$ factor), and thus must be $0$.

Now, forget that we fixed $\sigma$. We thus have shown that $\prod_{i=1}%
^{k}h_{a_{i}-b_{\sigma\left(  i\right)  }}=0$ for every $\sigma\in
\Symm_{k}$. Now, \eqref{sol.jacobi-trudi.b.lem.pf.2} becomes
\[
\det\left(  \left(  h_{a_{i}-b_{j}}\right)  _{i,j=1,2,\ldots,k}\right)
=\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\underbrace{\prod
_{i=1}^{k}h_{a_{i}-b_{\sigma\left(  i\right)  }}}_{=0}=\sum_{\sigma
\in\Symm_{k}}\left(  -1\right)  ^{\sigma}0=0.
\]
This proves \eqref{sol.jacobi-trudi.b.lem}.]

Next, let us notice that if $k\in\NN$ and if $\kappa$ and $\tau$ are
any two partitions satisfying $\ell\left(  \kappa\right)  \leq k$ and
$\ell\left(  \tau\right)  \leq k$, then
\begin{equation}
\widetilde{s}_{\kappa/\tau}=\det\left(  \left(  h_{\left(  \kappa
_{i}-i\right)  -\left(  \tau_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  .
\label{sol.jacobi-trudi.b.pf.kappatau}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.b.pf.kappatau}:} If
$k\in\NN$ and if $\kappa$ and $\tau$ are any two partitions satisfying
$\ell\left(  \kappa\right)  \leq k$ and $\ell\left(  \tau\right)  \leq k$,
then
\begin{align*}
\widetilde{s}_{\kappa/\tau}  &  =\widetilde{s}_{\kappa/\tau}^{[k]}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{s}%
_{\kappa/\tau}\right) \\
&  =\det\left(  \left(  \underbrace{h_{\kappa_{i}-\tau_{j}-i+j}}_{=h_{\left(
\kappa_{i}-i\right)  -\left(  \tau_{j}-j\right)  }}\right)  _{i,j=1,2,\ldots
,k}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{s}_{\kappa/\tau}^{[k]}\right) \\
&  =\det\left(  \left(  h_{\left(  \kappa_{i}-i\right)  -\left(  \tau
_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  .
\end{align*}
This proves \eqref{sol.jacobi-trudi.b.pf.kappatau}.}

Now let us step to the solution of Exercise~\ref{exe.jacobi-trudi}(b).

Let $\lambda$ and $\mu$ be two partitions which do not satisfy $\mu
\subseteq\lambda$. Then, there exists a $u\in\left\{  1,2,3,...\right\}  $
such that $\lambda_{u}<\mu_{u}$ (because otherwise, every $u\in\left\{
1,2,3,...\right\}  $ would satisfy $\lambda_{u}\geq\mu_{u}$, so that we would
have $\mu\subseteq\lambda$, which contradicts the fact that $\lambda$ and
$\mu$ do not satisfy $\mu\subseteq\lambda$). Consider this $u$. We have
$0\leq\lambda_{u}<\mu_{u}$ and thus $u\leq\ell\left(  \mu\right)  $.

Now, fix any $k\in\NN$ such that $k\geq\ell\left(  \lambda\right)  $
and $k\geq\ell\left(  \mu\right)  $ (such a $k$ clearly exists). Then,
$k\geq\ell\left(  \mu\right)  \geq u$, so that $u\in\left\{
1,2,...,k\right\}  $. Also, $\lambda_{u}-u<\mu_{u}-u$ (since $\lambda_{u}%
<\mu_{u}$). Now, forget that we fixed $u$. We thus have found a $u\in\left\{
1,2,...,k\right\}  $ such that $\lambda_{u}-u<\mu_{u}-u$. Hence, we do
\textbf{not} have $\left(  \lambda_{u}-u\geq\mu_{u}-u\text{ for all }%
u\in\left\{  1,2,...,k\right\}  \right)  $.

Applying \eqref{sol.jacobi-trudi.b.pf.kappatau} to $\kappa=\lambda$ and
$\tau=\mu$, we obtain
\begin{equation}
\widetilde{s}_{\lambda/\mu}=\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -\left(  \mu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  \label{sol.jacobi-trudi.b.pf.4}%
\end{equation}
(since $\ell\left(  \lambda\right)  \leq k$ and $\ell\left(  \mu\right)  \leq
k$).

But $\lambda_{1}\geq\lambda_{2}\geq...\geq\lambda_{k}$ (since $\lambda$ is a
partition). Adding this chain of inequalities to the obvious chain of strict
inequalities $-1>-2>...>-k$, we obtain $\lambda_{1}-1>\lambda_{2}%
-2>...>\lambda_{k}-k$. Similarly, $\mu_{1}-1>\mu_{2}-2>...>\mu_{k}-k$. We can
thus apply \eqref{sol.jacobi-trudi.b.lem} to $a_{i}=\lambda_{i}-i$ and
$b_{i}=\mu_{i}-i$. As a result, we obtain $\det\left(  \left(  h_{\left(
\lambda_{i}-i\right)  -\left(  \mu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  =0$. Thus, \eqref{sol.jacobi-trudi.b.pf.4} becomes
$\widetilde{s}_{\lambda/\mu}
=\det\left(  \left(  h_{\left(  \lambda_{i}-i\right)  -\left(
\mu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  =0$. This solves
Exercise~\ref{exe.jacobi-trudi}(b).

(c) Notice first that every $n\in\NN$ satisfies
\begin{equation}
\Delta h_n = \sum_{p\in\ZZ} h_{n-p}\otimes h_{p}.
\label{sol.jacobi-trudi.c.Delh}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.c.Delh}:} Let
$n\in\NN$. From
Proposition~\ref{symm-comultiplication-formulas-prop}(iii), we have
\[
\Delta h_{n}=\sum_{i+j=n}h_{i}\otimes h_{j}=\sum_{p=0}^{n}h_{n-p}\otimes
h_{p}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }\left(
n-p,p\right)  \text{ for the summation index }\left(  i,j\right)  \right)  .
\]
Compared with
\begin{align*}
\sum_{p\in\ZZ }h_{n-p}\otimes h_{p}  &  =\sum_{p=-\infty}^{-1}%
h_{n-p}\otimes\underbrace{h_{p}}_{\substack{=0\\\text{(since }p<0\text{)}%
}}+\sum_{p=0}^{n}h_{n-p}\otimes h_{p}+\sum_{p=n+1}^{\infty}\underbrace{h_{n-p}%
}_{\substack{=0\\\text{(since }p>n\text{ and thus}\\n-p<0\text{)}}}\otimes
h_{p}\\
&  =\underbrace{\sum_{p=-\infty}^{-1}h_{n-p}\otimes0}_{=0}+\sum_{p=0}%
^{n}h_{n-p}\otimes h_{p}+\underbrace{\sum_{p=n+1}^{\infty}0\otimes h_{p}}%
_{=0}=\sum_{p=0}^{n}h_{n-p}\otimes h_{p},
\end{align*}
this yields $\Delta h_{n}=\sum_{p\in\ZZ }h_{n-p}\otimes h_{p}$. Thus,
\eqref{sol.jacobi-trudi.c.Delh} is proven.}

Let $\lambda$ and $\nu$ be two partitions. Fix any $k\in\NN$ such that
$k\geq\ell\left(  \lambda\right)  $ and $k\geq\ell\left(  \nu\right)  $ (such
a $k$ clearly exists). Define two $k$-tuples $\left(  \overline{\lambda}%
_{1},\overline{\lambda}_{2},...,\overline{\lambda}_{k}\right)  \in
\ZZ ^{k}$ and $\left(  \overline{\nu}_{1},\overline{\nu}_{2}%
,...,\overline{\nu}_{k}\right)  \in\ZZ ^{k}$ by setting%
\[
\overline{\lambda}_{u}=\lambda_{u}-u\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \overline{\nu}_{u}=\nu_{u}-u\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in\left\{  1,2,...,k\right\}  \text{.}%
\]
Then, $\overline{\lambda}_{1}>\overline{\lambda}_{2}>...>\overline{\lambda
}_{k}$ (since $\lambda$ is a partition) and $\overline{\nu}_{1}>\overline{\nu
}_{2}>...>\overline{\nu}_{k}$ (similarly). Any $i\in\left\{
1,2,...,k\right\}  $ and $j\in\left\{  1,2,...,k\right\}  $ satisfy%
\[
\underbrace{\overline{\lambda}_{i}}_{=\lambda_{i}-i}
-\underbrace{\overline{\nu}_{j}}_{=\nu_{j}-j}
=\left(  \lambda_{i}-i\right)  -\left(  \nu_{j}-j\right)
=\lambda_{i}-\nu_{j}-i+j,
\]
so that%
\begin{equation}
h_{\overline{\lambda}_{i}-\overline{\nu}_{j}}=h_{\lambda_{i}-\nu_{j}-i+j}.
\label{sol.jacobi-trudi.c.dash}%
\end{equation}


By the definition of $\widetilde{s}_{\lambda/\nu}$, we have
\begin{align*}
\widetilde{s}_{\lambda/\nu}  &  =\widetilde{s}_{\lambda/\nu}^{[k]}=\det\left(
\left(  \underbrace{h_{\lambda_{i}-\nu_{j}-i+j}}_{\substack{=
h_{\overline{\lambda}_{i}-\overline{\nu}_{j}}\\
\text{(by \eqref{sol.jacobi-trudi.c.dash})}}}
\right)  _{i,j=1,2,\ldots,k}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\widetilde{s}_{\lambda/\nu}^{[k]}\right) \\
&  =\det\left(  \left(  h_{\overline{\lambda}_{i}-\overline{\nu}_{j}}\right)
_{i,j=1,2,\ldots,k}\right)  =\sum_{\sigma\in\Symm_{k}}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-\overline{\nu
}_{\sigma\left(  i\right)  }}
\end{align*}
(by \eqref{sol.jacobi-trudi.b.detexp}, applied to $\ell=k$ and $\alpha
_{i,j}=h_{\overline{\lambda}_{i}-\overline{\nu}_{j}}$). Applying $\Delta$ to
both sides of this equality, we obtain
\begin{align}
\Delta\widetilde{s}_{\lambda/\nu}  &  =\Delta\left(  \sum_{\sigma
\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{k}
h_{\overline{\lambda}_{i}-\overline{\nu}_{\sigma\left(  i\right)  }}
\right)  =\sum
_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{k}\underbrace{\Delta\left(  h_{\overline{\lambda}_{i}-\overline{\nu}%
_{\sigma\left(  i\right)  }}\right)  }_{\substack{=\sum_{p\in\ZZ}
h_{\overline{\lambda}_{i}-\overline{\nu}_{\sigma\left(  i\right)  }-p}\otimes
h_{p}\\\text{(by \eqref{sol.jacobi-trudi.c.Delh})}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a }\kk %
\text{-algebra homomorphism}\right) \nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{k}\underbrace{\sum_{p\in\ZZ }h_{\overline{\lambda}_{i}%
-\overline{\nu}_{\sigma\left(  i\right)  }-p}\otimes h_{p}}_{\substack{=\sum
_{p\in\ZZ }h_{\overline{\lambda}_{i}-p}\otimes h_{p-\overline{\nu
}_{\sigma\left(  i\right)  }}\\\text{(here, we substituted }p-\overline{\nu
}_{\sigma\left(  i\right)  }\\\text{for }p\text{ in the sum)}}}\nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}%
\underbrace{\prod_{i=1}^{k}\sum_{p\in\ZZ }h_{\overline{\lambda}_{i}%
-p}\otimes h_{p-\overline{\nu}_{\sigma\left(  i\right)  }}}_{\substack{=\sum
_{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}}\prod_{i=1}%
^{k}\left(  h_{\overline{\lambda}_{i}-p_{i}}\otimes h_{p_{i}-\overline{\nu
}_{\sigma\left(  i\right)  }}\right)  \\\text{(by the product rule)}%
}}\nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\sum_{\left(
p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}}\underbrace{\prod_{i=1}%
^{k}\left(  h_{\overline{\lambda}_{i}-p_{i}}\otimes h_{p_{i}-\overline{\nu
}_{\sigma\left(  i\right)  }}\right)  }_{=\left(  \prod_{i=1}^{k}%
h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes\left(  \prod_{i=1}%
^{k}h_{p_{i}-\overline{\nu}_{\sigma\left(  i\right)  }}\right)  }\nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\sum_{\left(
p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}}\left(  \prod_{i=1}%
^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes\left(  \prod_{i=1}%
^{k}h_{p_{i}-\overline{\nu}_{\sigma\left(  i\right)  }}\right) \nonumber\\
&  =\sum_{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes
\underbrace{\left(  \sum_{\sigma\in\Symm_{k}}\left(  -1\right)
^{\sigma}\prod_{i=1}^{k}h_{p_{i}-\overline{\nu}_{\sigma\left(  i\right)  }%
}\right)  }_{\substack{=\det\left(  \left(  h_{p_{i}-\overline{\nu}_{j}%
}\right)  _{i,j=1,2,\ldots,k}\right)  \\\text{(since
\eqref{sol.jacobi-trudi.b.detexp} (applied to }\ell=k\text{ and }\alpha
_{i,j}=h_{p_{i}-\overline{\nu}_{j}}\text{) yields}\\\det\left(  \left(
h_{p_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  =\sum
_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{k}h_{p_{i}-\overline{\nu}_{\sigma\left(  i\right)  }}\text{)}}}\nonumber\\
&  =\sum_{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes\det\left(
\left(  h_{p_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)
\label{sol.jacobi-trudi.c.10}
\end{align}
\begin{align}
&  =\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\\text{some two of the }p_{1},p_{2},...,p_{k}\text{ are equal}}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes
\underbrace{\det\left(  \left(  h_{p_{i}-\overline{\nu}_{j}}\right)
_{i,j=1,2,\ldots,k}\right)  }_{\substack{=0\\\text{(because some two of the
}p_{1},p_{2},...,p_{k}\\\text{ are equal, and thus the matrix }\\
\left( h_{p_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\text{ has two} \\
\text{equal rows, which yields that the} \\
\text{determinant of this matrix is }0\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)
\in\ZZ ^{k};\\\text{no two of the }p_{1},p_{2},...,p_{k}\text{ are
equal}}}\left(  \prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)
\otimes\det\left(  \left(  h_{p_{i}-\overline{\nu}_{j}}\right)
_{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)
\in\ZZ ^{k};\\\text{some two of the }p_{1},p_{2},...,p_{k}\text{ are
equal}}}\left(  \prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)
\otimes0}_{=0}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)
\in\ZZ ^{k};\\\text{no two of the }p_{1},p_{2},...,p_{k}\text{ are
equal}}}\left(  \prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)
\otimes\det\left(  \left(  h_{p_{i}-\overline{\nu}_{j}}\right)
_{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\\text{no two of the }p_{1},p_{2},...,p_{k}\text{ are equal}}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes\det\left(
\left(  h_{p_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  .
\label{sol.jacobi-trudi.c.11}%
\end{align}


But every $k$-tuple $\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}$
such that no two of the $p_{1},p_{2},...,p_{k}$ are equal can be written
\textbf{uniquely} in the form $\left(  q_{\sigma\left(  1\right)  }%
,q_{\sigma\left(  2\right)  },...,q_{\sigma\left(  k\right)  }\right)  $ for
some permutation $\sigma\in\Symm_{k}$ and some $k$-tuple $\left(
q_{1},q_{2},...,q_{k}\right)  \in\ZZ ^{k}$ satisfying $q_{1}%
>q_{2}>...>q_{k}$. Conversely, of course, for any permutation $\sigma
\in\Symm_{k}$ and any $k$-tuple $\left(  q_{1},q_{2},...,q_{k}\right)
\in\ZZ ^{k}$ satisfying $q_{1}>q_{2}>...>q_{k}$, the $k$-tuple $\left(
q_{\sigma\left(  1\right)  },q_{\sigma\left(  2\right)  },...,q_{\sigma\left(
k\right)  }\right)  $ has the property that no two of the $q_{\sigma\left(
1\right)  },q_{\sigma\left(  2\right)  },...,q_{\sigma\left(  k\right)  }$ are
equal. Thus, we can replace the summation sign ``%
$\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\\text{no two of the }p_{1},p_{2},...,p_{k}\text{ are equal}}%
}$'' in \eqref{sol.jacobi-trudi.c.11} by ``%
$\sum_{\sigma\in\Symm_{k}}\sum\limits_{\substack{\left(  q_{1},q_{2}%
,...,q_{k}\right)  \in\ZZ ^{k};\\q_{1}>q_{2}>...>q_{k}}}$'',
while at the same time substituting $\left(
q_{\sigma\left(  1\right)  },q_{\sigma\left(  2\right)  },...,q_{\sigma\left(
k\right)  }\right)  $ for $\left(  p_{1},p_{2},...,p_{k}\right)  $. As a
result, we obtain%
\begin{align}
&  \sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\\text{no two of the }p_{1},p_{2},...,p_{k}\text{ are equal}}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes\det\left(
\left(  h_{p_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)
\nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\sum\limits_{\substack{\left(  q_{1}%
,q_{2},...,q_{k}\right)  \in\ZZ ^{k};\\q_{1}>q_{2}>...>q_{k}}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-q_{\sigma\left(  i\right)  }}\right)
\otimes\underbrace{\det\left(  \left(  h_{q_{\sigma\left(  i\right)
}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  }%
_{\substack{=\left(  -1\right)  ^{\sigma}\det\left(  \left(  h_{q_{i}%
-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \\\text{(because
permuting the rows of}\\\text{a matrix multiplies its determinant}\\\text{by
the sign of the permutation used)}}}\nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\sum\limits_{\substack{\left(  q_{1}%
,q_{2},...,q_{k}\right)  \in\ZZ ^{k};\\q_{1}>q_{2}>...>q_{k}}}\left(
\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-q_{\sigma\left(  i\right)  }}\right)
\otimes\left(  -1\right)  ^{\sigma}\det\left(  \left(  h_{q_{i}-\overline{\nu
}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}%
\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k}}}\left(  \prod_{i=1}^{k}h_{\overline{\lambda}%
_{i}-q_{\sigma\left(  i\right)  }}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k}}}\underbrace{\left(  \sum_{\sigma\in
\Symm_{k}}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{k}
h_{\overline{\lambda}_{i}-q_{\sigma\left(  i\right)  }}\right)
}_{\substack{=\det\left(
\left(  h_{\overline{\lambda}_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)
\\\text{(since \eqref{sol.jacobi-trudi.b.detexp} (applied to }\ell=k\text{
and}\\\alpha_{i,j}=h_{\overline{\lambda}_{i}-q_{j}}\text{) yields}%
\\\det\left(  \left(  h_{\overline{\lambda}_{i}-q_{j}}\right)
_{i,j=1,2,\ldots,k}\right)  =\sum_{\sigma\in\Symm_{k}}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{k}h_{\overline{\lambda}_{i}-q_{\sigma\left(
i\right)  }}\text{)}}}\otimes\det\left(  \left(  h_{q_{i}-\overline{\nu}_{j}%
}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k}}}\det\left(  \left(  h_{\overline{\lambda}%
_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  .
\label{sol.jacobi-trudi.c.21}%
\end{align}


Now, consider any summand of the sum on the right hand side of
\eqref{sol.jacobi-trudi.c.21}. If this summand corresponds to a $\left(
q_{1},q_{2},...,q_{k}\right)  \in\ZZ ^{k}$ which does \textbf{not}
satisfy $\left(  \overline{\lambda}_{u}\geq q_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  \right)  $, then this summand must vanish\footnote{In fact,
in this case, we have
$\det\left(  \left(  h_{\overline{\lambda}_{i}-q_{j}}\right) 
_{i,j=1,2,\ldots,k}\right)  =0$ (by \eqref{sol.jacobi-trudi.b.lem},
applied to $a_{i}=\overline{\lambda}_{i}$ and $b_{i}=q_{i}$), and thus the
whole summand is%
\[
\underbrace{\det\left(  \left(  h_{\overline{\lambda}_{i}-q_{j}}\right)
_{i,j=1,2,\ldots,k}\right)  }_{=0}\otimes\det\left(  \left(  h_{q_{i}%
-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  =0.
\]
}. Hence, all such summands can be removed from the sum without changing the
value of the sum. In other words,%
\begin{align}
&  \sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k}}}\det\left(  \left(  h_{\overline{\lambda}%
_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k};\\\overline{\lambda}_{u}\geq q_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\overline{\lambda
}_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  .
\label{sol.jacobi-trudi.c.23}%
\end{align}


Furthermore, consider any summand of the sum on the right hand side of
\eqref{sol.jacobi-trudi.c.23}. If this summand corresponds to a $\left(
q_{1},q_{2},...,q_{k}\right)  \in\ZZ ^{k}$ which does \textbf{not}
satisfy $\left(  q_{u}\geq\overline{\nu}_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  \right)  $, then this summand must vanish\footnote{In fact,
in this case, we have $\det\left(  \left(  h_{q_{i}-\overline{\nu}_{j}}
\right)  _{i,j=1,2,\ldots,k}\right)  =0$ (by \eqref{sol.jacobi-trudi.b.lem},
applied to $a_{i}=q_{i}$ and $b_{i}=\overline{\nu}_{i}$), and thus the whole
summand is%
\[
\det\left(  \left(  h_{\overline{\lambda}_{i}-q_{j}}\right)  _{i,j=1,2,\ldots
,k}\right)  \otimes\underbrace{\det\left(  \left(  h_{q_{i}-\overline{\nu}%
_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  }_{=0}=0.
\]
}. Hence, all such summands can be removed from the sum without changing the
value of the sum. In other words,%
\begin{align}
&  \sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k};\\\overline{\lambda}_{u}\geq q_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\overline{\lambda
}_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k};\\\overline{\lambda}_{u}\geq q_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\q_{u}\geq\overline{\nu}_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\overline{\lambda
}_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k};\\\lambda_{u}-u\geq q_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  ;\\q_{u}\geq\nu_{u}-u\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(
\left(  h_{q_{i}-\left(  \nu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  \label{sol.jacobi-trudi.c.27}%
\end{align}
(here we have used the fact that $\overline{\lambda}_{u}=\lambda_{u}-u$ and
$\overline{\nu}_{u}=\nu_{u}-u$ for all $u\in\left\{  1,2,...,k\right\}  $).

Now, \eqref{sol.jacobi-trudi.c.11} becomes
\begin{align}
\Delta\widetilde{s}_{\lambda/\nu}  &  =\sum\limits_{\substack{\left(  p_{1}%
,p_{2},...,p_{k}\right)  \in\ZZ ^{k};\\\text{no two of the }p_{1}%
,p_{2},...,p_{k}\text{ are equal}}}\left(  \prod_{i=1}^{k}
h_{\overline{\lambda}_{i}-p_{i}}\right)  \otimes
\det\left(  \left(  h_{p_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}
\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ %
^{k};\\q_{1}>q_{2}>...>q_{k}}}\det\left(  \left(  h_{\overline{\lambda}%
_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.jacobi-trudi.c.21}}\right)
\nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ ^{k};
\\q_{1}>q_{2}>...>q_{k};\\\overline{\lambda}_{u}\geq q_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\overline{\lambda
}_{i}-q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(
h_{q_{i}-\overline{\nu}_{j}}\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.jacobi-trudi.c.23}}\right)
\nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{k}\right)  \in\ZZ ^{k};
\\q_{1}>q_{2}>...>q_{k};\\\lambda_{u}-u\geq q_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  ;\\q_{u}\geq\nu_{u}-u\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -q_{j}}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(
\left(  h_{q_{i}-\left(  \nu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.jacobi-trudi.c.27}}\right)
\nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)
\in\ZZ ^{k};\\p_{1}-1>p_{2}-2>...>p_{k}-k;\\\lambda_{u}-u\geq
p_{u}-u\text{ for all }u\in\left\{  1,2,...,k\right\}  ;\\p_{u}-u\geq\nu
_{u}-u\text{ for all }u\in\left\{  1,2,...,k\right\}  }}}_{\substack{=\sum\limits
_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}%
;\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}-u\geq p_{u}-u\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\p_{u}-u\geq\nu_{u}-u\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\\\text{(because the condition}%
\\p_{1}-1>p_{2}-2>...>p_{k}-k\text{ on a tuple}\\\left(  p_{1},p_{2}%
,...,p_{k}\right)  \in\ZZ ^{k}\text{ of integers}\\\text{is equivalent
to}\\p_{1}\geq p_{2}\geq...\geq p_{k}\text{)}}}\det\left(  \left(  h_{\left(
\lambda_{i}-i\right)  -\left(  p_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  \otimes\det\left(  \left(  h_{\left(  p_{i}-i\right)  -\left(
\nu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }\left(
p_{1}-1,p_{2}-2,...,p_{k}-k\right)  \text{ for }\left(  q_{1},q_{2}%
,...,q_{k}\right)  \text{ in the sum}\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)
\in\ZZ ^{k};\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}-u\geq
p_{u}-u\text{ for all }u\in\left\{  1,2,...,k\right\}  ;\\p_{u}-u\geq\nu
_{u}-u\text{ for all }u\in\left\{  1,2,...,k\right\}  }}}_{\substack{=\sum\limits
_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ ^{k}%
;\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}\geq p_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\\\text{(because the condition }\lambda
_{u}-u\geq p_{u}-u\\\text{is equivalent to }\lambda_{u}\geq p_{u}\text{, and
the condition}\\p_{u}-u\geq\nu_{u}-u\text{ is equivalent to }p_{u}\geq\nu
_{u}\text{)}}}\det\left(  \left(  h_{\left(  \lambda_{i}-i\right)  -\left(
p_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(
\left(  h_{\left(  p_{i}-i\right)  -\left(  \nu_{j}-j\right)  }\right)
_{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}\geq p_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -\left(  p_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)
\otimes\det\left(  \left(  h_{\left(  p_{i}-i\right)  -\left(  \nu
_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  .
\label{sol.jacobi-trudi.c.47}
\end{align}


But every $\left(  p_{1},p_{2},...,p_{k}\right)  \in \NN^k$ belongs to
$\ZZ ^{k}$ (obviously), whereas every $\left(  p_{1},p_{2}
,...,p_{k}\right)  \in\ZZ ^{k}$ which satisfies $\left(  p_{u}\geq
\nu_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}  \right)  $ belongs to
$\NN^k$ (because every such $\left(  p_{1},p_{2},...,p_{k}\right)
\in\ZZ ^{k}$ satisfies $p_{u}\geq\nu_{u}\geq0$ for all $u\in\left\{
1,2,...,k\right\}  $). Hence, the summation sign ``%
$\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}\geq p_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}$'' on the right hand side
of \eqref{sol.jacobi-trudi.c.47} can be replaced by ``%
$\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in \NN^{k};
\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}\geq p_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}$''. As a result, we obtain%
\begin{align}
&  \sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \in\ZZ %
^{k};\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}\geq p_{u}\text{ for all
}u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -\left(  p_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)
\otimes\det\left(  \left(  h_{\left(  p_{i}-i\right)  -\left(  \nu
_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)
\in\NN^{k};\\p_{1}\geq p_{2}\geq...\geq p_{k};\\\lambda_{u}\geq
p_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{
for all }u\in\left\{  1,2,...,k\right\}  }}}_{\substack{=\sum\limits
_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \text{ is a partition of
length }\leq k;\\\lambda_{u}\geq p_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  }}\\\text{(since the }k\text{-tuples }\left(  p_{1}%
,p_{2},...,p_{k}\right)  \in\NN^{k}\text{ satisfying}\\p_{1}\geq
p_{2}\geq...\geq p_{k}\text{ are precisely the partitions of length }\leq
k\text{)}}}\det\left(  \left(  h_{\left(  \lambda_{i}-i\right)  -\left(
p_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(
\left(  h_{\left(  p_{i}-i\right)  -\left(  \nu_{j}-j\right)  }\right)
_{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  p_{1},p_{2},...,p_{k}\right)  \text{ is a
partition of length }\leq k;\\\lambda_{u}\geq p_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  ;\\p_{u}\geq\nu_{u}\text{ for all }%
u\in\left\{  1,2,...,k\right\}  }}\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -\left(  p_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)
\otimes\det\left(  \left(  h_{\left(  p_{i}-i\right)  -\left(  \nu
_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\mu\text{ is a partition of length }\leq
k;\\\lambda_{u}\geq\mu_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}
;\\\mu_{u}\geq\nu_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}  }%
}}_{\substack{=\sum\limits_{\substack{\mu\text{ is a partition of length }\leq
k;\\\mu\subseteq\lambda;\\\nu\subseteq\mu}}\\\text{(because the condition}%
\\\left(  \lambda_{u}\geq\mu_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  \right)  \\\text{on a partition }\mu\text{ of length }\leq
k\\\text{is equivalent to }\mu\subseteq\lambda\text{, and the}%
\\\text{condition }\left(  \mu_{u}\geq\nu_{u}\text{ for all }u\in\left\{
1,2,...,k\right\}  \right)  \\\text{on a partition }\mu\text{ of length }\leq
k\\\text{is equivalent to }\nu\subseteq\mu\text{)}}}\underbrace{\det\left(
\left(  h_{\left(  \lambda_{i}-i\right)  -\left(  \mu_{j}-j\right)  }\right)
_{i,j=1,2,\ldots,k}\right)  }_{\substack{=\widetilde{s}_{\lambda/\mu
}\\\text{(since \eqref{sol.jacobi-trudi.b.pf.kappatau} (applied to}%
\\\kappa=\lambda\text{ and }\tau=\mu\text{) yields}\\\widetilde{s}%
_{\lambda/\mu}=\det\left(  \left(  h_{\left(  \lambda_{i}-i\right)  -\left(
\mu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  \text{)}}%
}\otimes\underbrace{\det\left(  \left(  h_{\left(  \mu_{i}-i\right)  -\left(
\nu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  }%
_{\substack{=\widetilde{s}_{\mu/\nu}\\\text{(since
\eqref{sol.jacobi-trudi.b.pf.kappatau} (applied to}\\\kappa=\mu\text{ and
}\tau=\nu\text{) yields}\\\widetilde{s}_{\mu/\nu}=\det\left(  \left(
h_{\left(  \mu_{i}-i\right)  -\left(  \nu_{j}-j\right)  }\right)
_{i,j=1,2,\ldots,k}\right)  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  p_{1},p_{2},...,p_{k}\right)  \text{ as }\mu\right) \nonumber\\
&  =\sum\limits_{\substack{\mu\text{ is a partition of length }\leq k;\\\mu
\subseteq\lambda;\\\nu\subseteq\mu}}\widetilde{s}_{\lambda/\mu}\otimes
\widetilde{s}_{\mu/\nu}. \label{sol.jacobi-trudi.c.59}%
\end{align}
But every partition $\mu$ belongs to $\Par $, whereas every
partition $\mu$ satisfying $\mu\subseteq\lambda$ must be a partition of length
$\leq k$ (because if $\mu$ is a partition satisfying $\mu\subseteq\lambda$,
then $\ell\left(  \mu\right)  \leq\ell\left(  \lambda\right)  \leq k$). Hence,
the summation sign ``$\sum\limits_{\substack{\mu\text{ is a partition
of length }\leq k;\\\mu\subseteq\lambda;\\\nu\subseteq\mu}}$'' on the
right hand side of \eqref{sol.jacobi-trudi.c.59} can be replaced by
``$\sum\limits_{\substack{\mu\in\Par :\\\mu
\subseteq\lambda;\\\nu\subseteq\mu}}$''. Thus, we obtain%
\begin{equation}
\sum\limits_{\substack{\mu\text{ is a partition of length }\leq k;\\\mu
\subseteq\lambda;\\\nu\subseteq\mu}}\widetilde{s}_{\lambda/\mu}\otimes
\widetilde{s}_{\mu/\nu}
=\sum\limits_{\substack{\mu\in\Par :\\\mu
\subseteq\lambda;\\\nu\subseteq\mu}}\widetilde{s}_{\lambda/\mu}\otimes
\widetilde{s}_{\mu/\nu}
=\sum\limits_{\substack{\mu\in\Par %
:\\\nu\subseteq\mu\subseteq\lambda}}\widetilde{s}_{\lambda/\mu}\otimes
\widetilde{s}_{\mu/\nu}.
\label{sol.jacobi-trudi.c.68}
\end{equation}


Now, \eqref{sol.jacobi-trudi.c.47} becomes
\begin{align*}
\Delta\widetilde{s}_{\lambda/\nu}  &  =\sum\limits_{\substack{\left(  p_{1}%
,p_{2},...,p_{k}\right)  \in\ZZ ^{k};\\p_{1}\geq p_{2}\geq...\geq
p_{k};\\\lambda_{u}\geq p_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}
;\\p_{u}\geq\nu_{u}\text{ for all }u\in\left\{  1,2,...,k\right\}  }%
}\det\left(  \left(  h_{\left(  \lambda_{i}-i\right)  -\left(  p_{j}-j\right)
}\right)  _{i,j=1,2,\ldots,k}\right)  \otimes\det\left(  \left(  h_{\left(
p_{i}-i\right)  -\left(  \nu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right) \\
&  =\sum\limits_{\substack{\mu\text{ is a partition of length }\leq k;\\\mu
\subseteq\lambda;\\\nu\subseteq\mu}}\widetilde{s}_{\lambda/\mu}\otimes
\widetilde{s}_{\mu/\nu}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.jacobi-trudi.c.59}}\right) \\
&  =\sum\limits_{\substack{\mu\in\Par :\\\nu\subseteq\mu\subseteq
\lambda}}\widetilde{s}_{\lambda/\mu}\otimes\widetilde{s}_{\mu/\nu
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.jacobi-trudi.c.68}}\right)  .
\end{align*}
This solves Exercise~\ref{exe.jacobi-trudi}(c).

(d) Let us first observe the following simple fact:
If $k \in \NN$ and $m \in \NN$, and if $a_1, a_2, \ldots, a_k$ are $k$
integers satisfying $\sum_{j=1}^k a_j = m$,
then
\begin{equation}
\prod_{i=1}^k h_{a_i} \in \Lambda_m .
\label{sol.jacobi-trudi.d.sum-has-degree}
\end{equation}

[\textit{Proof of \eqref{sol.jacobi-trudi.d.sum-has-degree}:}
Let $k \in \NN$ and $m \in \NN$. Let $a_1, a_2, \ldots, a_k$ be $k$
integers satisfying $\sum_{j=1}^k a_j = m$.
We must prove that $\prod_{i=1}^k h_{a_i} \in \Lambda_m$.

Assume first that at least one $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$a_j < 0$. Consider this $j$. Then, $h_{a_j}=0$ (since $a_j < 0$).
The product $\prod_{i=1}^{k} h_{a_i}$ thus has a factor equal to zero
(namely, the factor for $i=j$), and therefore equals zero itself.
In other words, $\prod_{i=1}^{k} h_{a_i} = 0$.
Hence, $\prod_{i=1}^{k} h_{a_i} = 0 \in \Lambda_m$.

Now, forget our assumption that at least one
$j\in\left\{  1,2,\ldots,k\right\}$ satisfies $a_j < 0$.
We thus have shown that $\prod_{i=1}^k h_{a_i} \in \Lambda_m$
if at least one $j\in\left\{  1,2,\ldots,k\right\}  $
satisfies $a_j < 0$.
Hence, for the rest of the proof of
$\prod_{i=1}^{k} h_{a_i} \in \Lambda_m$,
we can WLOG assume that it is not true that at
least one $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$a_j < 0$. Assume this.

It is not true that at least one $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$a_j < 0$. In other words, no $j\in\left\{ 1,2,\ldots,k\right\}  $
satisfies $a_j < 0$.
In other words, every $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$a_j \geq 0$.
Thus, every $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$a_j \in \NN$.
Hence, every $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$h_{a_j} \in \Lambda_{a_j}$.
Multiplying these relations for all $j\in\left\{  1,2,\ldots,k\right\}  $,
we obtain
$h_{a_1} h_{a_2} \cdots h_{a_k}
\subset \Lambda_{a_1} \Lambda_{a_2} \cdots \Lambda_{a_k}$.

But $\Lambda$ is a graded $\kk$-algebra. Hence, every $\left(
b_{1},b_{2},\ldots,b_{k}\right)  \in\NN^{k}$ satisfies
$\Lambda_{b_1} \Lambda_{b_2} \cdots \Lambda_{b_k}
\subset \Lambda_{b_1 + b_2 + \cdots + b_k}$.
Applying this to $b_j = a_j$, we obtain
$\Lambda_{a_1} \Lambda_{a_2} \cdots \Lambda_{a_k}
\subset \Lambda_{a_1 + a_2 + \cdots + a_k} = \Lambda_m$
(since $a_1 + a_2 + \cdots + a_k = \sum_{j=1}^k a_j = m$).
Now,
\begin{align*}
\prod_{i=1}^k h_{a_i}
&= h_{a_1} h_{a_2} \cdots h_{a_k}
\subset \Lambda_{a_1} \Lambda_{a_2} \cdots \Lambda_{a_k}
\subset \Lambda_m .
\end{align*}
This proves \eqref{sol.jacobi-trudi.d.sum-has-degree}.]

Now,
let $\mu$ and $\lambda$ be two partitions such that $\mu\subseteq\lambda$.

Fix any $k\in\NN$ such that $k\geq\ell\left(  \lambda\right)  $ and
$k\geq\ell\left(  \mu\right)  $ (such a $k$ clearly exists). Then,
\eqref{sol.jacobi-trudi.b.pf.kappatau} (applied to $\kappa=\lambda$ and
$\tau=\mu$) yields%
\[
\widetilde{s}_{\lambda/\mu}=\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -\left(  \mu_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  =\sum_{\sigma\in\Symm_{k}}\left(  -1\right)  ^{\sigma}%
\prod_{i=1}^{k}h_{\left(  \lambda_{i}-i\right)  -\left(  \mu_{\sigma\left(
i\right)  }-\sigma\left(  i\right)  \right)  }%
\]
(by \eqref{sol.jacobi-trudi.b.detexp}, applied to $\ell=k$ and $\alpha
_{i,j}=h_{\left(  \lambda_{i}-i\right)  -\left(  \mu_{j}-j\right)  }$).

Notice that $k\geq\ell\left(  \lambda\right)  $, so that $\lambda=\left(
\lambda_{1},\lambda_{2},...,\lambda_{k}\right)  $ and thus $\left\vert
\lambda\right\vert =\lambda_{1}+\lambda_{2}+...+\lambda_{k}=\sum_{j\in\left\{
1,2,...,k\right\}  }\lambda_{j}$. Similarly, $\left\vert \mu\right\vert
=\sum_{j\in\left\{  1,2,...,k\right\}  }\mu_{j}$. Now,
\begin{equation}
\left\vert \lambda/\mu\right\vert =\underbrace{\left\vert \lambda\right\vert
}_{=\sum_{j\in\left\{  1,2,...,k\right\}  }\lambda_{j}}-\underbrace{\left\vert
\mu\right\vert }_{=\sum_{j\in\left\{  1,2,...,k\right\}  }\mu_{j}}=\sum
_{j\in\left\{  1,2,...,k\right\}  }\lambda_{j}-\sum_{j\in\left\{
1,2,...,k\right\}  }\mu_{j}.
\label{sol.jacobi-trudi.d.pf.triv}
\end{equation}


We shall now show that $\prod_{i=1}^{k}h_{\left(  \lambda_{i}-i\right)
-\left(  \mu_{\sigma\left(  i\right)  }-\sigma\left(  i\right)  \right)  }%
\in\Lambda_{\left\vert \lambda/\mu\right\vert }$ for every $\sigma
\in\Symm_{k}$.

In fact, let $\sigma\in\Symm_{k}$. We need to prove that $\prod
_{i=1}^{k}h_{\left(  \lambda_{i}-i\right)  -\left(  \mu_{\sigma\left(
i\right)  }-\sigma\left(  i\right)  \right)  }\in\Lambda_{\left\vert
\lambda/\mu\right\vert }$.

Notice that the map $\sigma:\left\{  1,2,...,k\right\}  \rightarrow\left\{
1,2,...,k\right\}  $ is a permutation (since $\sigma\in\Symm_{k}$),
thus a bijection.

We have
\begin{align*}
& \underbrace{\sum_{j=1}^{k}}_{=\sum_{j\in\left\{  1,2,...,k\right\}  }}\left(
\left(  \lambda_{j}-j\right)  -\left(  \mu_{\sigma\left(  j\right)  }%
-\sigma\left(  j\right)  \right)  \right) \\
&  =\sum_{j\in\left\{
1,2,...,k\right\}  }\left(  \left(  \lambda_{j}-j\right)  -\left(  \mu
_{\sigma\left(  j\right)  }-\sigma\left(  j\right)  \right)  \right) \\
&  =\sum_{j\in\left\{  1,2,...,k\right\}  }\left(  \lambda_{j}-j\right)
-\underbrace{\sum_{j\in\left\{  1,2,...,k\right\}  }\left(  \mu_{\sigma\left(
j\right)  }-\sigma\left(  j\right)  \right)  }_{\substack{=\sum_{j\in\left\{
1,2,...,k\right\}  }\left(  \mu_{j}-j\right)  \\\text{(here, we substituted
}j\text{ for }\sigma\left(  j\right)  \text{ in the sum,}\\\text{since the map
}\sigma:\left\{  1,2,...,k\right\}  \rightarrow\left\{  1,2,...,k\right\}
\text{ is a bijection)}}}\\
&  =\sum_{j\in\left\{  1,2,...,k\right\}  }\left(  \lambda_{j}-j\right)
-\sum_{j\in\left\{  1,2,...,k\right\}  }\left(  \mu_{j}-j\right) \\
&  =\left(  \sum_{j\in\left\{  1,2,...,k\right\}  }\lambda_{j}-\sum
_{j\in\left\{  1,2,...,k\right\}  }j\right)  -\left(  \sum_{j\in\left\{
1,2,...,k\right\}  }\mu_{j}-\sum_{j\in\left\{  1,2,...,k\right\}  }j\right) \\
&  =\sum_{j\in\left\{  1,2,...,k\right\}  }\lambda_{j}-\sum_{j\in\left\{
1,2,...,k\right\}  }\mu_{j}=\left\vert \lambda/\mu\right\vert
\ \ \ \ \ \ \ \ \ \ \left(
 \text{by \eqref{sol.jacobi-trudi.d.pf.triv}}\right) .
\end{align*}
Hence, \eqref{sol.jacobi-trudi.d.sum-has-degree}
(applied to
$a_j = \left(  \lambda_{j}-j\right)  -\left(  \mu_{\sigma\left(  j\right)  }
-\sigma\left(  j\right)  \right)$
and $m = \left| \lambda / \mu \right|$)
yields
\begin{align*}
\prod_{i=1}^{k}h_{\left(  \lambda_{i}-i\right)  -\left(  \mu_{\sigma\left(
i\right)  }-\sigma\left(  i\right)  \right)  }
\in \Lambda_{\left\vert \lambda/\mu\right\vert } .
\end{align*}

Now, forget that we fixed $\sigma$. We thus have shown that every $\sigma
\in\Symm_{k}$ satisfies%
\begin{equation}
\prod_{i=1}^{k}h_{\left(  \lambda_{i}-i\right)  -\left(  \mu_{\sigma\left(
i\right)  }-\sigma\left(  i\right)  \right)  }\in\Lambda_{\left\vert
\lambda/\mu\right\vert }. \label{sol.jacobi-trudi.e.pf.astep5}%
\end{equation}
Now,%
\[
\widetilde{s}_{\lambda/\mu}=\sum_{\sigma\in\Symm_{k}}\left(  -1\right)
^{\sigma}\underbrace{\prod_{i=1}^{k}h_{\left(  \lambda_{i}-i\right)  -\left(
\mu_{\sigma\left(  i\right)  }-\sigma\left(  i\right)  \right)  }%
}_{\substack{\in\Lambda_{\left\vert \lambda/\mu\right\vert }\\\text{(by
\eqref{sol.jacobi-trudi.e.pf.astep5})}}}\in\sum_{\sigma\in\Symm_{k}}
\left(  -1\right)  ^{\sigma}\Lambda_{\left\vert \lambda/\mu\right\vert
}\subset\Lambda_{\left\vert \lambda/\mu\right\vert }
\]
(since $\Lambda_{\left\vert \lambda/\mu\right\vert }$ is a $\kk$-module).
This solves Exercise~\ref{exe.jacobi-trudi}(d).

(e) In the following, whenever $\mathcal{A}$ is an assertion, we will use the
notation $\left[  \mathcal{A}\right]  $ for the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.

We use the notation $f\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ defined in
Exercise~\ref{exe.Lambda.subs.fin} whenever $a_{1},a_{2},\ldots,a_{k}$ are
elements of a commutative $\kk$-algebra $A$ and $f\in R\left(\xx\right)$. In
particular, $f\left(  1\right)  $ is a well-defined element of $\kk$
for every $f\in R\left(\xx\right)$.

Let $\operatorname{ev}_{1}$ be the map $\Lambda\rightarrow
\kk$ which sends every $f\in\Lambda$ to $f\left(  1\right)
\in\kk$. Every $f\in\Lambda$ satisfies
\begin{align}
\operatorname{ev}_{1}\left(  f\right)   &  =f\left(  1\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }
\operatorname{ev}_{1}\right) \nonumber\\
&  =\left(  \text{the result of substituting }1,0,0,0,...\text{ for }
x_{1},x_{2},x_{3},...\text{ in }f\right) \label{sol.jacobi-trudi.e.pf.ev1}
\end{align}
(since $f\left(  1\right)$ was defined as the result of substituting
$1,0,0,0,...$ for $x_{1},x_{2},x_{3},...$ in $f$).

It is easy to see that $\operatorname{ev}_{1}$ is a
$\kk$-algebra homomorphism\footnote{\textit{Proof.} Every
$f\in\Lambda$ and $g\in\Lambda$ satisfy
\begin{align*}
\operatorname{ev}_{1}\left(  fg\right)   &  =\left(  \text{the
result of substituting }1,0,0,0,...\text{ for }x_{1},x_{2},x_{3},...\text{ in
}fg\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.jacobi-trudi.e.pf.ev1},
applied to }fg\text{ instead of }f\right) \\
&  =\underbrace{\left(  \text{the result of substituting }1,0,0,0,...\text{
for }x_{1},x_{2},x_{3},...\text{ in }f\right)  }%
_{\substack{=\operatorname{ev}_{1}\left(  f\right)  \\\text{(by
\eqref{sol.jacobi-trudi.e.pf.ev1})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\underbrace{\left(  \text{the result of
substituting }1,0,0,0,...\text{ for }x_{1},x_{2},x_{3},...\text{ in }g\right)
}_{\substack{=\operatorname{ev}_{1}\left(  g\right)  \\\text{(by
\eqref{sol.jacobi-trudi.e.pf.ev1}, applied to }g\text{ instead of }f\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since substitution of variables in power
series preserves products}\right) \\
&  =\operatorname{ev}_{1}\left(  f\right)  \cdot\operatorname{ev}_{1}
\left(  g\right)  .
\end{align*}
Similarly, one can show that every $f\in\Lambda$ and $g\in\Lambda$ satisfy
$\operatorname{ev}_{1}\left(  f+g\right)  =\operatorname{ev}_{1}
\left(  f\right)  +\operatorname{ev}_{1}\left(
g\right)  $, and that every $\zeta\in\kk$ and $f\in\Lambda$ satisfy
$\operatorname{ev}_{1}\left(  \zeta f\right)  =\zeta
\operatorname{ev}_{1}\left(  f\right)  $. Thus, $\operatorname{ev}_{1}$
is a $\kk$-linear map. Since every $f\in\Lambda$ and
$g\in\Lambda$ satisfy $\operatorname{ev}_{1}\left(  fg\right)
=\operatorname{ev}_{1}\left(  f\right)  \cdot
\operatorname{ev}_{1}\left(  g\right)  $, and since we can easily see
that $\operatorname{ev}_{1}\left(  1\right)  =1$, we thus conclude that
$\operatorname{ev}_{1}$ is a $\kk$-algebra homomorphism,
qed.}.

Every $n \in \ZZ$ satisfies
\begin{equation}
\operatorname{ev}_{1}\left(  h_{n}\right)  = \left[  n\geq0\right]  .
\label{sol.jacobi-trudi.e.pf.ev1h}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.e.pf.ev1h}:} Let $n\in\ZZ$. If
$n<0$, then \eqref{sol.jacobi-trudi.e.pf.ev1h} is true (because if $n<0$,
then
\begin{align*}
\operatorname{ev}_{1}\left(  \underbrace{h_{n}}_{=0}\right)   &
=\operatorname{ev}_{1}\left(  0\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname{ev}_{1}%
\text{ is a }\kk \text{-algebra homomorphism}\right)  \\
&  =\left[  n\geq0\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[
n\geq0\right]  =0\text{ (since }n\geq0\text{ is false (since }n<0\text{))}%
\right)
\end{align*}
). Hence, for the rest of the proof of \eqref{sol.jacobi-trudi.e.pf.ev1h}, we
can WLOG assume that we don't have $n<0$. Assume this. We don't have $n<0$;
hence, we have $n\geq0$. Thus, $s_{\left(  n\right)  /\varnothing}=s_{\left(
n\right)  }=h_{n}$.

The partitions $\left(  n\right)  $ and $\varnothing$ satisfy $\varnothing
\subseteq\left(  n\right)  $. Hence, Exercise~\ref{exe.pieri.h2}(a) (applied
to $\left(  n\right)  $ and $\varnothing$ instead of $\lambda$ and $\mu$)
yields that $\left(  h_{n},s_{\left(  n\right)  /\varnothing}\right)  $ equals
$1$ if $\left(  n\right)  /\varnothing$ is a horizontal $n$-strip, and equals
$0$ otherwise. Since $\left(  n\right)  /\varnothing$ is a horizontal
$n$-strip, this shows that $\left(  h_{n},s_{\left(  n\right)  /\varnothing
}\right)  $ equals $1$. Thus, $\left(  h_{n},s_{\left(  n\right)
/\varnothing}\right)  =1$. Since $s_{\left(  n\right)  /\varnothing}=h_{n}$,
this rewrites as $\left(  h_{n},h_{n}\right)  =1$.

But Exercise~\ref{exe.Lambda.hall.gr}(b) (applied to $h_{n}$ instead of $f$)
yields $\left(  h_{n},h_{n}\right)  =h_{n}\left(  1\right)  $, so that
$h_{n}\left(  1\right)  =\left(  h_{n},h_{n}\right)  =1$.

Now, the definition of the map $\operatorname{ev}_{1}$ yields
$\operatorname{ev}_{1}\left(  h_{n}\right)  =h_{n}\left(  1\right)
=1=\left[  n\geq0\right]  $ (since $\left[  n\geq0\right]  =1$ (since $n\geq0$
holds)). This proves \eqref{sol.jacobi-trudi.e.pf.ev1h}.}

Now, let $\lambda$ and $\mu$ be two partitions. Let $n\in\NN$. We need
to prove that $\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\left(
h_{n},s_{\lambda/\mu}\right)  $.

Fix any $k\in\NN$ such that $k\geq\ell\left(  \lambda\right)  $ and
$k\geq\ell\left(  \mu\right)  $ (such a $k$ clearly exists). Then,%
\begin{equation}
\widetilde{s}_{\lambda/\mu}=\widetilde{s}_{\lambda/\mu}^{[k]}=\det\left(
\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{s}%
_{\lambda/\mu}^{[k]}\right)  . \label{sol.jacobi-trudi.e.pf.slm}%
\end{equation}


If we don't have $\mu\subseteq\lambda$, then both $\widetilde{s}_{\lambda/\mu
}$ and $s_{\lambda/\mu}$ are $0$ (in fact, for $\widetilde{s}_{\lambda/\mu}$
this follows from Exercise~\ref{exe.jacobi-trudi}(b), and for $s_{\lambda/\mu
}$ this follows from the definition of $s_{\lambda/\mu}$). Hence, if we don't
have $\mu\subseteq\lambda$, then our claim $\left(  h_{n},\widetilde{s}%
_{\lambda/\mu}\right)  =\left(  h_{n},s_{\lambda/\mu}\right)  $ is obvious.
Thus, for the rest of the proof of $\left(  h_{n},\widetilde{s}_{\lambda/\mu
}\right)  =\left(  h_{n},s_{\lambda/\mu}\right)  $, we assume WLOG that we do
have $\mu\subseteq\lambda$.

Recall that $\widetilde{s}_{\lambda/\mu}\in\Lambda_{\left\vert \lambda
/\mu\right\vert }$ (by Exercise~\ref{exe.jacobi-trudi}(d)) and $s_{\lambda
/\mu}\in\Lambda_{\left\vert \lambda/\mu\right\vert }$.

Assume first that $\left\vert \lambda/\mu\right\vert \neq n$. Then,
Exercise~\ref{exe.Lambda.hall.gr}(a) (applied to $m=\left\vert \lambda
/\mu\right\vert $, $f=h_{n}$ and $g=\widetilde{s}_{\lambda/\mu}$) yields
$\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =0$ (since $h_{n}\in
\Lambda_{n}$ and $\widetilde{s}_{\lambda/\mu}\in\Lambda_{\left\vert
\lambda/\mu\right\vert }$). Also, Exercise~\ref{exe.Lambda.hall.gr}(a)
(applied to $m=\left\vert \lambda/\mu\right\vert $, $f=h_{n}$ and
$g=s_{\lambda/\mu}$) yields $\left(  h_{n},s_{\lambda/\mu}\right)  =0$ (since
$h_{n}\in\Lambda_{n}$ and $s_{\lambda/\mu}\in\Lambda_{\left\vert \lambda
/\mu\right\vert }$). Thus, $\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)
=0=\left(  h_{n},s_{\lambda/\mu}\right)  $. Now, forget that we assumed that
$\left\vert \lambda/\mu\right\vert \neq n$. We have thus proven our claim
$\left( h_n, \widetilde{s}_{\lambda/\mu} \right) = \left( h_n,
s_{\lambda/\mu} \right)$ in the case when $\left\vert \lambda/\mu\right\vert
\neq n$. Hence, for the rest of our proof, we WLOG assume that we don't have
$\left\vert \lambda/\mu\right\vert \neq n$. Hence, we have $\left\vert
\lambda/\mu\right\vert =n$. Now, $\widetilde{s}_{\lambda/\mu}\in
\Lambda_{\left\vert \lambda/\mu\right\vert }=\Lambda_{n}$ (since $\left\vert
\lambda/\mu\right\vert =n$), and therefore Exercise~\ref{exe.Lambda.hall.gr}%
(b) (applied to $\widetilde{s}_{\lambda/\mu}$ instead of $f$) yields $\left(
h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\widetilde{s}_{\lambda/\mu}\left(
1\right)  $.

Recall that $\operatorname{ev}_{1}$ is a $\kk$-algebra
homomorphism. As a consequence, the map $\operatorname{ev}_{1}$
commutes with taking determinants; that is, every matrix $A\in\Lambda
^{\ell\times\ell}$ with $\ell\in\NN$ satisfies
$\operatorname{ev}_{1}\left(  \det A\right) =
\det\left(  \operatorname{ev}_{1}^{\ell\times\ell}A\right)$,
where $\operatorname{ev}_{1}^{\ell\times\ell}A$ means the result of
applying $\operatorname{ev}_{1}$ to every entry of the matrix $A$.
Applying this to $\ell=k$ and
$A = \left( h_{\lambda_i-\mu_j-i+j} \right) _{i,j=1,2,\ldots,k}$,
we obtain
\begin{align}
\operatorname{ev}_{1}\left(  \det\left(  \left(  h_{\lambda_{i}
-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)  \right)   &  =\det\left(
\underbrace{\operatorname{ev}_{1}^{k\times k}\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)  }_{=\left(
\operatorname{ev}_{1}\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
\right)  _{i,j=1,2,\ldots,k}}\right)  \nonumber\\
&  =\det\left(  \left(  \underbrace{\operatorname{ev}_{1}\left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  }_{\substack{=\left[  \lambda_{i}-\mu
_{j}-i+j\geq0\right]  \\\text{(by \eqref{sol.jacobi-trudi.e.pf.ev1h},
applied}\\\text{to }\lambda_{i}-\mu_{j}-i+j\text{ instead of }n\text{)}%
}}\right)  _{i,j=1,2,\ldots,k}\right)  \nonumber\\
&  =\det\left(  \left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)
_{i,j=1,2,\ldots,k}\right)  .
\label{sol.jacobi-trudi.e.pf.3}
\end{align}
Now, applying the map $\operatorname{ev}_{1}$ to both sides of the
equality \eqref{sol.jacobi-trudi.e.pf.slm}, we obtain%
\[
\operatorname{ev}_{1}\left(  \widetilde{s}_{\lambda/\mu}\right)
=\operatorname{ev}_{1}\left(  \det\left(  \left(  h_{\lambda_{i}%
-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,k}\right)  \right)  =\det\left(
\left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots
,k}\right)
\]
(by \eqref{sol.jacobi-trudi.e.pf.3}). Compared with
\begin{align*}
\operatorname{ev}_{1}\left(  \widetilde{s}_{\lambda/\mu}\right)   &
=\widetilde{s}_{\lambda/\mu}\left(  1\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\operatorname{ev}_{1}\right)  \\
&  =\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  h_{n},\widetilde{s}%
_{\lambda/\mu}\right)  =\widetilde{s}_{\lambda/\mu}\left(  1\right)  \right)
,
\end{align*}
this yields%
\begin{equation}
\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\det\left(  \left(  \left[
\lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}\right)
.\label{sol.jacobi-trudi.e.pf.5}%
\end{equation}


Now, let us distinguish between two cases:

\textit{Case 1:} The shape $\lambda/\mu$ is a horizontal $n$-strip.

\textit{Case 2:} The shape $\lambda/\mu$ is not a horizontal $n$-strip.

Let us consider Case 1 first. In this case, the shape $\lambda/\mu$ is a
horizontal $n$-strip. Hence,
\begin{equation}
\text{every }i\in\left\{  1,2,3,...\right\}  \text{ satisfies }\mu_{i}%
\geq\lambda_{i+1} \label{sol.jacobi-trudi.e.pf.c1.0}%
\end{equation}
(because Exercise~\ref{exe.horistrip.postnikov}(a)
shows that $\lambda/\mu$ is a
horizontal strip if and only if every $i\in\left\{  1,2,3,\ldots\right\}  $
satisfies $\mu_{i}\geq\lambda_{i+1}$).

Now, every $\left(  i,j\right)  \in\left\{  1,2,...,k\right\}  ^{2}$
satisfying $i>j$ satisfies%
\[
\left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  =0.
\]
\footnote{\textit{Proof:} Recall that $\lambda$ is a partition, and thus
$\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq...$.
\par
Now, let $\left(  i,j\right)  \in\left\{  1,2,...,k\right\}  ^{2}$ be such
that $i>j$. Then, $i\geq j+1$ (since $i$ and $j$ are integers). In other
words, $j+1\leq i$. Now, $\lambda_{u}\geq\lambda_{v}$ for any two positive
integers $u$ and $v$ satisfying $u\leq v$ (since $\lambda_{1}\geq\lambda
_{2}\geq\lambda_{3}\geq...$). Applying this to $u=j+1$ and $v=i$, we obtain
$\lambda_{j+1}\geq\lambda_{i}$. But \eqref{sol.jacobi-trudi.e.pf.c1.0}
(applied to $j$ instead of $i$) yields $\mu_{j}\geq\lambda_{j+1}\geq
\lambda_{i}$. Thus, $\lambda_{i}-\underbrace{\mu_{j}}_{\geq\lambda_{i}%
}-\underbrace{i}_{>j}+j<\lambda_{i}-\lambda_{i}-j+j=0$. Thus, $\lambda_{i}%
-\mu_{j}-i+j\geq0$ is false. Hence, $\left[  \lambda_{i}-\mu_{j}%
-i+j\geq0\right]  =0$, qed.} Hence, the matrix $\left(  \left[  \lambda
_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}$ is
upper-triangular. Since the determinant of an upper-triangular matrix equals
the product of its diagonal entries, we thus conclude that the determinant of
this matrix can be written as%
\begin{align*}
\det\left(  \left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)
_{i,j=1,2,\ldots,k}\right)   &  =\prod_{i=1}^{k}\left[  \underbrace{\lambda
_{i}-\mu_{i}-i+i}_{=\lambda_{i}-\mu_{i}}\geq0\right]  =\prod_{i=1}%
^{k}\underbrace{\left[  \lambda_{i}-\mu_{i}\geq0\right]  }%
_{\substack{=1\\\text{(because }\lambda_{i}-\mu_{i}\geq0\text{ is
true}\\\text{(since }\lambda_{i}\geq\mu_{i}\text{ (because }\mu\subseteq
\lambda\text{)))}}}\\
&  =\prod_{i=1}^{k}1=1.
\end{align*}
Now, \eqref{sol.jacobi-trudi.e.pf.5} becomes
\[
\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\det\left(  \left(  \left[
\lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}\right)  =1.
\]


But on the other hand, Exercise~\ref{exe.pieri.h2}(a) yields that $\left(
h_{n},s_{\lambda/\mu}\right)  $ equals $1$ if $\lambda/\mu$ is a horizontal
$n$-strip, and equals $0$ otherwise. Thus, $\left(  h_{n},s_{\lambda/\mu
}\right)  $ equals $1$ (since $\lambda/\mu$ is a horizontal $n$-strip). In
other words, $\left(  h_{n},s_{\lambda/\mu}\right)  =1$. Compared with
$\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =1$, this yields $\left(
h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\left(  h_{n},s_{\lambda/\mu
}\right)  $. Hence, $\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)
=\left(  h_{n},s_{\lambda/\mu}\right)  $ is proven in Case 1.

Let us now handle Case 2. In this case, the shape $\lambda/\mu$ is not a
horizontal $n$-strip. Hence, it is not true that every $i\in\left\{
1,2,3,...\right\}  $ satisfies $\mu_{i}\geq\lambda_{i+1}$ (because Exercise
\ref{exe.horistrip.postnikov}(a)
shows that $\lambda/\mu$ is a horizontal strip
if and only if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu
_{i}\geq\lambda_{i+1}$). In other words, there exists an $i\in\left\{
1,2,3,...\right\}  $ such that we don't have $\mu_{i}\geq\lambda_{i+1}$.
Therefore, there exist the \textbf{smallest }$i\in\left\{  1,2,3,...\right\}
$ such that we don't have $\mu_{i}\geq\lambda_{i+1}$. Denote this $i$ by $w$.
Thus, $w$ is the \textbf{smallest }$i\in\left\{  1,2,3,...\right\}  $ such
that we don't have $\mu_{i}\geq\lambda_{i+1}$. In particular, $w$ is an
$i\in\left\{  1,2,3,...\right\}  $ such that we don't have $\mu_{i}\geq
\lambda_{i+1}$. That is, $w\in\left\{  1,2,3,...\right\}  $ is such that we
don't have $\mu_{w}\geq\lambda_{w+1}$.

We have $\mu_{w}<\lambda_{w+1}$ (since we don't have $\mu_{w}\geq\lambda
_{w+1}$). Hence, $\lambda_{w+1}>\mu_{w}\geq0$, so that $w+1\leq\ell\left(
\lambda\right)  \leq k$. Hence, both $w$ and $w+1$ belong to the set $\left\{
1,2,...,k\right\}  $.

Recall that $w$ is the \textbf{smallest }$i\in\left\{  1,2,3,...\right\}  $
such that we don't have $\mu_{i}\geq\lambda_{i+1}$. Hence,%
\begin{equation}
\text{if }i\in\left\{  1,2,3,...\right\}  \text{ is such that we don't have
}\mu_{i}\geq\lambda_{i+1}\text{, then }i\geq w.
\label{sol.jacobi-trudi.e.pf.c2.0}%
\end{equation}


It is now easy to see that%
\begin{equation}
\text{if }p\text{ and }q\text{ are two integers satisfying }p\geq q\text{ and
}\left[  p\geq0\right]  \neq\left[  q\geq0\right]  \text{, then }p\geq0>q.
\label{sol.jacobi-trudi.e.pf.c2.lem}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.e.pf.c2.lem}:} Let $p$ and
$q$ be two integers satisfying $p\geq q$ and $\left[  p\geq0\right]
\neq\left[  q\geq0\right]  $. We have $q\leq p$ (since $p\geq q$). If we had
$p<0$, then we would have
\begin{align*}
\left[  p\geq0\right]   &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
p\geq0\text{ is false (since we have }p<0\text{)}\right) \\
&  =\left[  q\geq0\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
q\geq0\text{ is false (because we have }q\leq p<0\text{)}\right)  ,
\end{align*}
which would contradict $\left[  p\geq0\right]  \neq\left[  q\geq0\right]  $.
Hence, we cannot have $p<0$. We thus must have $p\geq0$. If we had $q\geq0$,
then we would have%
\begin{align*}
\left[  p\geq0\right]   &  =1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
p\geq0\text{ is true (since we have }p\geq q\geq0\text{)}\right) \\
&  =\left[  q\geq0\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
q\geq0\text{ is true (because we have }q\geq0\text{)}\right)  ,
\end{align*}
which would contradict $\left[  p\geq0\right]  \neq\left[  q\geq0\right]  $.
Hence, we cannot have $q\geq0$. We thus must have $q<0$, so that $0>q$. Thus,
$p\geq0>q$. This proves \eqref{sol.jacobi-trudi.e.pf.c2.lem}.}

Now, we have
\begin{equation}
\left[  \lambda_{w}-\mu_{j}-w+j\geq0\right]  =\left[  \lambda_{w+1}-\mu
_{j}-\left(  w+1\right)  +j\geq0\right]  \label{sol.jacobi-trudi.e.pf.c2.1}%
\end{equation}
for every $j\in\left\{  1,2,...,k\right\}  $\ \ \ \ \footnote{\textit{Proof of
\eqref{sol.jacobi-trudi.e.pf.c2.1}:} Assume the contrary. Then, we don't have
$\left[  \lambda_{w}-\mu_{j}-w+j\geq0\right]  =\left[  \lambda_{w+1}-\mu
_{j}-\left(  w+1\right)  +j\geq0\right]  $ for every $j\in\left\{
1,2,...,k\right\}  $. In other words, there exists some $j\in\left\{
1,2,...,k\right\}  $ such that we don't have $\left[  \lambda_{w}-\mu
_{j}-w+j\geq0\right]  =\left[  \lambda_{w+1}-\mu_{j}-\left(  w+1\right)
+j\geq0\right]  $. Consider this $j$.
\par
We know that $\lambda$ is a partition; thus, $\lambda_{1}\geq\lambda_{2}%
\geq\lambda_{3}\geq...$. Hence, $\lambda_{u}\geq\lambda_{u+1}$ for every
$u\in\left\{  1,2,3,...\right\}  $. Applying this to $u=w$, we obtain
$\lambda_{w}\geq\lambda_{w+1}$. Now,%
\[
\underbrace{\lambda_{w}}_{\geq\lambda_{w+1}}-\mu_{j}-\underbrace{w}_{\leq
w+1}+j\geq\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j.
\]
Also, $\left[  \lambda_{w}-\mu_{j}-w+j\geq0\right]  \neq\left[  \lambda
_{w+1}-\mu_{j}-\left(  w+1\right)  +j\geq0\right]  $ (because we don't have
$\left[  \lambda_{w}-\mu_{j}-w+j\geq0\right]  =\left[  \lambda_{w+1}-\mu
_{j}-\left(  w+1\right)  +j\geq0\right]  $). Hence, we can apply
\eqref{sol.jacobi-trudi.e.pf.c2.lem} to $p=\lambda_{w}-\mu_{j}-w+j$ and
$q=\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j$. As a result, we obtain%
\[
\lambda_{w}-\mu_{j}-w+j\geq0>\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j.
\]
\par
Let us first assume (for the sake of contradiction) that $j\geq w$. Thus,
$w\leq j$. We have $\mu_{1}\geq\mu_{2}\geq\mu_{3}\geq...$ (since $\mu$ is a
partition). Hence, any positive integers $p$ and $q$ satisfying $p\leq q$
satisfy $\mu_{p}\geq\mu_{q}$. Applying this to $p=w$ and $q=j$, we obtain
$\mu_{w}\geq\mu_{j}$ (since $w\leq j$). Now, $\lambda_{w+1}>\mu_{w}\geq\mu
_{j}$, so that%
\[
\underbrace{\lambda_{w+1}}_{>\mu_{j}}-\mu_{j}-\left(  \underbrace{w}_{\leq
j}+1\right)  +j>\mu_{j}-\mu_{j}-\left(  j+1\right)  +j=-1.
\]
Hence, $\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j\geq-1+1$ (since
$\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j$ and $-1$ are integers). Hence,
$\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j\geq-1+1=0$, which contradicts
$0>\lambda_{w+1}-\mu_{j}-\left(  w+1\right)  +j$. This contradiction shows
that our assumption (that $j\geq w$) was wrong. Hence, we cannot have $j\geq
w$. We thus must have $j<w$. Hence, $j\leq w-1$ (since $j$ and $w$ are
integers), so that $j+1\leq w$.
\par
We have $\lambda_{w}-\mu_{j}-w+j\geq0$. Adding $\mu_{j}+w-j$ to this
inequality, we obtain $\lambda_{w}\geq\mu_{j}+w-\underbrace{j}_{<w}>\mu
_{j}+w-w=\mu_{j}$.
\par
But $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq...$. Hence, any positive
integers $p$ and $q$ satisfying $p\leq q$ satisfy $\lambda_{p}\geq\lambda_{q}%
$. Applying this to $p=j+1$ and $q=w$, we obtain $\lambda_{j+1}\geq\lambda
_{w}$ (since $j+1\leq w$). Thus, $\lambda_{j+1}\geq\lambda_{w}>\mu_{j}$. In
other words, we don't have $\mu_{j}\geq\lambda_{j+1}$. Hence,
\eqref{sol.jacobi-trudi.e.pf.c2.0} (applied to $i=j$) yields $j\geq w$. This
contradicts $j<w$. This contradiction shows that our assumption was wrong.
Hence, \eqref{sol.jacobi-trudi.e.pf.c2.1} is proven.}. Thus, for every
$j\in\left\{  1,2,...,k\right\}  $, we have
\begin{align*}
&  \left(  \text{the }j\text{-th entry of the }w\text{-th row of the matrix
}\left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)
_{i,j=1,2,\ldots,k}\right) \\
&  =\left[  \lambda_{w}-\mu_{j}-w+j\geq0\right]  =\left[  \lambda_{w+1}%
-\mu_{j}-\left(  w+1\right)  +j\geq0\right] \\
&  =\left(  \text{the }j\text{-th entry of the }\left(  w+1\right)  \text{-th
row of the matrix }\left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]
\right)  _{i,j=1,2,\ldots,k}\right)
\end{align*}
\footnote{Here, we are using the fact that the $w$-th row and the $\left(
w+1\right)  $-th row of the matrix $\left(  \left[  \lambda_{i}-\mu
_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}$ are well-defined. (This is
true because both $w$ and $w+1$ belong to the set $\left\{  1,2,...,k\right\}
$.)}. In other words, the $w$-th row of the matrix $\left(  \left[
\lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}$ equals the
$\left(  w+1\right)  $-th row of the matrix $\left(  \left[  \lambda_{i}%
-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}$. Hence, the matrix
$\left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)
_{i,j=1,2,\ldots,k}$ has two equal rows. Since every matrix which has two
equal rows must have determinant $0$, this yields that the matrix $\left(
\left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}$ has
determinant $0$. In other words,%
\[
\det\left(  \left(  \left[  \lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)
_{i,j=1,2,\ldots,k}\right)  =0.
\]
Now, \eqref{sol.jacobi-trudi.e.pf.5} becomes
\[
\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\det\left(  \left(  \left[
\lambda_{i}-\mu_{j}-i+j\geq0\right]  \right)  _{i,j=1,2,\ldots,k}\right)  =0.
\]


But on the other hand, Exercise~\ref{exe.pieri.h2}(a) yields that $\left(
h_{n},s_{\lambda/\mu}\right)  $ equals $1$ if $\lambda/\mu$ is a horizontal
$n$-strip, and equals $0$ otherwise. Thus, $\left(  h_{n},s_{\lambda/\mu
}\right)  $ equals $0$ (since $\lambda/\mu$ is not a horizontal $n$-strip). In
other words, $\left(  h_{n},s_{\lambda/\mu}\right)  =0$. Compared with
$\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  =0$, this yields $\left(
h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\left(  h_{n},s_{\lambda/\mu
}\right)  $. Hence, $\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)
=\left(  h_{n},s_{\lambda/\mu}\right)  $ is proven in Case 2.

We have thus proven $\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)
=\left(  h_{n},s_{\lambda/\mu}\right)  $ in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this yields that $\left(
h_{n},\widetilde{s}_{\lambda/\mu}\right)  =\left(  h_{n},s_{\lambda/\mu
}\right)  $ always holds. This solves Exercise~\ref{exe.jacobi-trudi}(e).

(f) Let $x$ be a primitive element of $\Lambda$ such that every $n\in
\NN$ satisfies $\left(  h_{n},x\right)  =0$. We need to prove that
$x=0$.

Recall the following fundamental fact from linear algebra: If $A$ is a
$\kk$-module, if $\left(  \cdot,\cdot\right) : A\times A\rightarrow
\kk$ is a symmetric $\kk$-bilinear form on $A$, and if $\left(
u_{\lambda}\right) _{\lambda\in L}$ and $\left( v_{\lambda}\right)
_{\lambda\in L}$ are two $\kk$-bases of $A$ which are dual to each
other with respect to the form $\left(  \cdot,\cdot\right)  $ (where $L$ is
some indexing set), then every $a\in A$ satisfies
\begin{equation}
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}.
\label{sol.jacobi-trudi.f.pf.general-form}
\end{equation}
We can apply this fact to $A=\Lambda$, $L=\Par $, $\left(
u_{\lambda}\right)  _{\lambda\in L}=\left(  h_{\lambda}\right)  _{\lambda
\in\Par}$, $\left(  v_{\lambda}\right)  _{\lambda\in L}=\left(
m_{\lambda}\right)  _{\lambda\in\Par}$ and $a=x$ (because the
bases $\left(  h_{\lambda}\right)  _{\lambda\in\Par}$ and
$\left(  m_{\lambda}\right)  _{\lambda\in\Par}$ of $\Lambda$
are dual to each other with respect to the Hall inner product $\left(
\cdot,\cdot\right)  $). As a result, we obtain%
\begin{equation}
x=\sum_{\lambda\in\Par}\left(  h_{\lambda},x\right)
m_{\lambda}. \label{sol.jacobi-trudi.f.pf.1}%
\end{equation}


Now, we will show that $\left(  h_{\lambda},x\right)  =0$ for every
$\lambda\in\Par $.

\textit{Proof.} Let $\lambda\in\Par $. We need to prove that
$\left(  h_{\lambda},x\right)  =0$.

Write $\lambda$ in the form $\lambda=\left(  \lambda_{1},\lambda
_{2},...,\lambda_{\ell}\right)  $, where $\ell=\ell\left(  \lambda\right)  $.
Then, all of $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{\ell}$ are
positive integers, and the definition of $h_{\lambda}$ yields $h_{\lambda
}=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{\ell}}$.

If $\ell=0$, then it is easy to see that $\left(  h_{\lambda},x\right)
=0$\ \ \ \ \footnote{\textit{Proof.} Assume that $\ell=0$. Hence,
\begin{align*}
\lambda &  =\left(  \lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)
=\left(  \text{empty tuple}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\ell=0\right) \\
&  =\varnothing.
\end{align*}
Thus, $h_{\lambda}=h_{\varnothing}=1=h_{0}$. Now, recall that every
$n\in\NN$ satisfies $\left(  h_{n},x\right)  =0$. Applying this to
$n=0$, we obtain $\left(  h_{0},x\right)  =0$. Hence, $\left(
\underbrace{h_{\lambda}}_{=h_{0}},x\right)  =\left(  h_{0},x\right)  =0$,
qed.}. Hence, for the rest of the proof of $\left(  h_{\lambda},x\right)  =0$,
we can WLOG assume that we don't have $\ell=0$. Assume this. We don't have
$\ell=0$. Thus, $\ell\neq0$, so that $\ell\geq1$ (since $\ell\in\NN$)
and thus $\ell\in\left\{  1,2,3,...\right\}  $. Thus, $\lambda_{1}$ is a
positive integer (since all of $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{\ell}$ are positive integers). Hence, $\lambda_{1}\neq0$.

If $\ell=1$, then it is easy to see that $\left(  h_{\lambda},x\right)
=0$\ \ \ \ \footnote{\textit{Proof.} Assume that $\ell=1$. Hence,
$\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)  =\left(
\lambda_{1}\right)  $ (since $\ell=1$). Thus, $h_{\lambda}=h_{\left(
\lambda_{1}\right)  }=h_{\lambda_{1}}$. Now, recall that every $n\in
\NN$ satisfies $\left(  h_{n},x\right)  =0$. Applying this to
$n=\lambda_{1}$, we obtain $\left(  h_{\lambda_{1}},x\right)  =0$. Hence,
$\left(  \underbrace{h_{\lambda}}_{=h_{\lambda_{1}}},x\right)  =\left(
h_{\lambda_{1}},x\right)  =0$, qed.}. Hence, for the rest of the proof of
$\left(  h_{\lambda},x\right)  =0$, we can WLOG assume that we don't have
$\ell=1$. Assume this. We don't have $\ell=1$. Thus, $\ell\neq1$, so that
$\ell\geq2$ (since $\ell\in\left\{  1,2,3,...\right\}  $). Thus, the sum
$\lambda_{2}+\lambda_{3}+...+\lambda_{\ell}$ is nonempty. Therefore,
$\lambda_{2}+\lambda_{3}+...+\lambda_{\ell}$ is a nonempty sum of positive
integers (since $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{\ell}$ are
positive integers), hence a positive integer itself. Thus, $\lambda
_{2}+\lambda_{3}+...+\lambda_{\ell}\neq0$.

Now, define two elements $u$ and $v$ of $\Lambda$ by $u=h_{\lambda_{1}}$ and
$v=h_{\lambda_{2}}h_{\lambda_{3}}...h_{\lambda_{\ell}}$ (these are both
well-defined since $\ell\geq1$). Then,%
\[
\underbrace{u}_{=h_{\lambda_{1}}}\underbrace{v}_{=h_{\lambda_{2}}%
h_{\lambda_{3}}...h_{\lambda_{\ell}}}=h_{1}\left(  h_{\lambda_{2}}%
h_{\lambda_{3}}...h_{\lambda_{\ell}}\right)  =h_{\lambda_{1}}h_{\lambda_{2}%
}...h_{\lambda_{\ell}}=h_{\lambda}.
\]


We have%
\[
v=\underbrace{h_{\lambda_{2}}}_{\in\Lambda_{\lambda_{2}}}%
\underbrace{h_{\lambda_{3}}}_{\in\Lambda_{\lambda_{3}}}%
...\underbrace{h_{\lambda_{\ell}}}_{\in\Lambda_{\lambda_{\ell}}}\in
\Lambda_{\lambda_{2}}\Lambda_{\lambda_{3}}...\Lambda_{\lambda_{\ell}}%
\subset\Lambda_{\lambda_{2}+\lambda_{3}+...+\lambda_{\ell}}.
\]
Hence, Exercise~\ref{exe.Lambda.hall.gr}(a) (applied to $\lambda_{2}%
+\lambda_{3}+...+\lambda_{\ell}$, $0$, $v$ and $1$ instead of $n$, $m$, $f$
and $g$) yields $\left(  v,1\right)  =0$ (since $\lambda_{2}+\lambda
_{3}+...+\lambda_{\ell}\neq0$).

Also, $u=h_{\lambda_{1}}\in\Lambda_{\lambda_{1}}$. Hence, Exercise
\ref{exe.Lambda.hall.gr}(a) (applied to $\lambda_{1}$, $0$, $u$ and $1$
instead of $n$, $m$, $f$ and $g$) yields $\left(  u,1\right)  =0$ (since
$\lambda_{1}\neq0$).

Now, $x$ is primitive, so that $\Delta_{\Lambda}\left(  x\right)  =1\otimes
x+x\otimes1$.

But let $\Phi$ denote the isomorphism $\Lambda\rightarrow\Lambda^{o}$ induced
by the Hall inner product. This isomorphism $\Phi$ is an isomorphism of Hopf
algebras (according to Corollary \ref{cor.Lambda.selfdual}), hence a
$\kk$-algebra homomorphism. Thus,
\[
\Phi\left(  uv\right)  =\Phi\left(  u\right)  \cdot\Phi\left(  v\right)
=m_{\Lambda^{o}}\left(  \Phi\left(  u\right)  \otimes\Phi\left(  v\right)
\right)  .
\]
We can identify $\Lambda^{o}\otimes\Lambda^{o}$ with $\left(  \Lambda
\otimes\Lambda\right)  ^{o}$ (since $\Lambda$ is of finite type), and then the
definition of the $\kk$-algebra structure on $\Lambda^{o}$ yields
$m_{\Lambda^{o}}=\Delta_{\Lambda}^{\ast}$. Thus,%
\[
\Phi\left(  uv\right)  =\underbrace{m_{\Lambda^{o}}}_{=\Delta_{\Lambda}^{\ast
}}\left(  \Phi\left(  u\right)  \otimes\Phi\left(  v\right)  \right)
=\Delta_{\Lambda}^{\ast}\left(  \Phi\left(  u\right)  \otimes\Phi\left(
v\right)  \right)  =\left(  \Phi\left(  u\right)  \otimes\Phi\left(  v\right)
\right)  \circ\Delta_{\Lambda}%
\]
(by the definition of the map $\Delta_{\Lambda}^{\ast}$).

But the definition of $\Phi$ yields%
\begin{equation}
\left(  \Phi\left(  p\right)  \right)  \left(  q\right)  =\left(  p,q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\Lambda\text{ and }q\in\Lambda.
\label{sol.jacobi-trudi.f.pf.2}%
\end{equation}
Applying this to $p=uv$ and $q=x$, we obtain $\left(  \Phi\left(  uv\right)
\right)  \left(  x\right)  =\left(  \underbrace{uv}_{=h_{\lambda}},x\right)
=\left(  h_{\lambda},x\right)  $. Hence,%
\begin{align*}
\left(  h_{\lambda},x\right)   &  =\left(  \underbrace{\Phi\left(  uv\right)
}_{=\left(  \Phi\left(  u\right)  \otimes\Phi\left(  v\right)  \right)
\circ\Delta_{\Lambda}}\right)  \left(  x\right)  =\left(  \left(  \Phi\left(
u\right)  \otimes\Phi\left(  v\right)  \right)  \circ\Delta_{\Lambda}\right)
\left(  x\right) \\
&  =\left(  \Phi\left(  u\right)  \otimes\Phi\left(  v\right)  \right)
\left(  \underbrace{\Delta_{\Lambda}\left(  x\right)  }_{=1\otimes
x+x\otimes1}\right)  =\left(  \Phi\left(  u\right)  \otimes\Phi\left(
v\right)  \right)  \left(  1\otimes x+x\otimes1\right) \\
&  =\underbrace{\left(  \Phi\left(  u\right)  \right)  \left(  1\right)
}_{\substack{=\left(  u,1\right)  \\\text{(by \eqref{sol.jacobi-trudi.f.pf.2},
applied}\\\text{to }p=u\text{ and }q=1\text{)}}}\otimes\left(  \Phi\left(
v\right)  \right)  \left(  x\right)  +\left(  \Phi\left(  u\right)  \right)
\left(  x\right)  \otimes\underbrace{\left(  \Phi\left(  v\right)  \right)
\left(  1\right)  }_{\substack{=\left(  v,1\right)  \\\text{(by
\eqref{sol.jacobi-trudi.f.pf.2}, applied}\\\text{to }p=v\text{ and
}q=1\text{)}}}\\
&  =\underbrace{\left(  u,1\right)  }_{=0}\otimes\left(  \Phi\left(  v\right)
\right)  \left(  x\right)  +\left(  \Phi\left(  u\right)  \right)  \left(
x\right)  \otimes\underbrace{\left(  v,1\right)  }_{=0}=\underbrace{0\otimes
\left(  \Phi\left(  v\right)  \right)  \left(  x\right)  }_{=0}%
+\underbrace{\left(  \Phi\left(  u\right)  \right)  \left(  x\right)
\otimes0}_{=0}=0.
\end{align*}
Thus, $\left(  h_{\lambda},x\right)  =0$ is proven.

Now, forget that we fixed $\lambda\in\Par $. We thus have shown
that $\left(  h_{\lambda},x\right)  =0$ for every $\lambda\in
\Par $. Hence, \eqref{sol.jacobi-trudi.f.pf.1} becomes
$x=\sum_{\lambda\in\Par}\underbrace{\left(  h_{\lambda
},x\right)  }_{=0}m_{\lambda}=\sum_{\lambda\in\Par}0m_{\lambda
}=0$. Exercise~\ref{exe.jacobi-trudi}(f) is thus proven.

(g) First, let us notice that
\begin{equation}
s_{\lambda/\lambda}=1\ \ \ \ \ \ \ \ \ \ \text{for every partition }
\lambda .
\label{sol.jacobi-trudi.g.s1}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.g.s1}:} Let
$\lambda$ be a partition. The definition of $s_{\lambda/\lambda}$ yields
$s_{\lambda/\lambda}=\sum_{T}{\xx }^{\cont \left(
T\right)  }$, where the sum ranges over all column-strict tableaux $T$ of
shape $\lambda/\lambda$. In other words,
\[
s_{\lambda/\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau
of shape }\lambda/\lambda}}{\xx }^{\cont \left(  T\right)
}
\]
\par
But the shape $\lambda/\lambda$ has no cells. Hence, there exists exactly one
column-strict tableau $T$ of shape $\lambda/\lambda$, namely the unique map
from the empty set to $\left\{  1,2,3,...\right\}  $. Denote this tableau $T$
by $T_{0}$. So $T_{0}$ is the unique column-strict tableau $T$ of shape
$\lambda/\lambda$. Hence, the sum $\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda/\lambda}}{\xx }^{\cont %
\left(  T\right)  }$ has exactly one addend, namely the addend for $T=T_{0}$.
Consequently,%
\[
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\lambda}}{\xx }^{\cont \left(  T\right)
}=\xx ^{\cont\left(  T_{0}\right)  }=1
\]
(since $\cont\left(  T_{0}\right)  =\left(  0,0,0,...\right)  $
(because $T_{0}$ is a map from the empty set)). Thus, $s_{\lambda/\lambda
}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\lambda}}{\xx }^{\cont \left(  T\right)  }=1$,
qed.}. Furthermore,
\begin{equation}
\widetilde{s}_{\lambda/\lambda}=1\ \ \ \ \ \ \ \ \ \ \text{for every partition
}\lambda.\label{sol.jacobi-trudi.g.s2}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.g.s2}:} Let $\lambda$ be a
partition. Fix any $k\in\NN$ such that $k\geq\ell\left(  \lambda
\right)  $ (such a $k$ clearly exists). Then,
\eqref{sol.jacobi-trudi.b.pf.kappatau} (applied to $\kappa=\lambda$ and
$\tau=\lambda$) yields%
\[
\widetilde{s}_{\lambda/\lambda}=\det\left(  \left(  h_{\left(  \lambda
_{i}-i\right)  -\left(  \lambda_{j}-j\right)  }\right)  _{i,j=1,2,\ldots
,k}\right)  .
\]
\par
Now, let $\left(  i,j\right)  \in\left\{  1,2,...,k\right\}  ^{2}$ such that
$i>j$. Then, $j<i$, so that $j\leq i$. Recall that $\lambda$ is a partition,
so that $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq...$. Hence, any two
positive integers $p$ and $q$ satisfying $p\leq q$ satisfy $\lambda_{p}%
\geq\lambda_{q}$. Applying this to $p=j$ and $q=i$, we obtain $\lambda_{j}%
\geq\lambda_{i}$ (since $j\leq i$). Now, $\left(  \lambda_{i}-\underbrace{i}%
_{>j}\right)  -\left(  \underbrace{\lambda_{j}}_{\geq\lambda_{i}}-j\right)
<\left(  \lambda_{i}-j\right)  -\left(  \lambda_{i}-j\right)  =0$, so that
$h_{\left(  \lambda_{i}-i\right)  -\left(  \lambda_{j}-j\right)  }=0$.
\par
Now, forget that we fixed $\left(  i,j\right)  $. We thus have proven that
$h_{\left(  \lambda_{i}-i\right)  -\left(  \lambda_{j}-j\right)  }=0$ for
every $\left(  i,j\right)  \in\left\{  1,2,...,k\right\}  ^{2}$ such that
$i>j$. The matrix $\left(  h_{\left(  \lambda_{i}-i\right)  -\left(
\lambda_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}$ is therefore
upper-triangular. Since the determinant of any upper-triangular matrix equals
the product of its diagonal entries, this yields that the determinant of the
matrix $\left(  h_{\left(  \lambda_{i}-i\right)  -\left(  \lambda
_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}$ equals the product of its
diagonal entries. In other words,%
\[
\det\left(  \left(  h_{\left(  \lambda_{i}-i\right)  -\left(  \lambda
_{j}-j\right)  }\right)  _{i,j=1,2,\ldots,k}\right)  =\prod_{i=1}%
^{k}\underbrace{h_{\left(  \lambda_{i}-i\right)  -\left(  \lambda
_{i}-i\right)  }}_{=h_{0}=1}=\prod_{i=1}^{k}1=1.
\]
Hence, $\widetilde{s}_{\lambda/\lambda}=\det\left(  \left(  h_{\left(
\lambda_{i}-i\right)  -\left(  \lambda_{j}-j\right)  }\right)
_{i,j=1,2,\ldots,k}\right)  =1$, qed.}. We also notice that%
\begin{equation}
\mu=\lambda\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }\lambda\text{ and
}\mu\text{ satisfying }\mu\subseteq\lambda\text{ and }\left\vert \lambda
/\mu\right\vert =0.
\label{sol.jacobi-trudi.g.s3}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.g.s3}:} Let $\lambda$ and
$\mu$ be two partitions satisfying $\mu\subseteq\lambda$ and $\left\vert
\lambda/\mu\right\vert =0$.
\par
Since $\mu\subseteq\lambda$, we have
\[
\left\vert \lambda/\mu\right\vert =\underbrace{\left\vert \lambda\right\vert
}_{=\lambda_{1}+\lambda_{2}+\lambda_{3}+...=\sum_{k\geq1}\lambda_{k}%
}-\underbrace{\left\vert \mu\right\vert }_{=\mu_{1}+\mu_{2}+\mu_{3}%
+...=\sum_{k\geq1}\mu_{k}}=\sum_{k\geq1}\lambda_{k}-\sum_{k\geq1}\mu_{k}%
=\sum_{k\geq1}\left(  \lambda_{k}-\mu_{k}\right)  .
\]
Hence, $\sum_{k\geq1}\left(  \lambda_{k}-\mu_{k}\right)  =\left\vert
\lambda/\mu\right\vert =0$. But $\lambda_{k}\geq\mu_{k}$ for every $k\geq1$
(since $\mu\subseteq\lambda$). In other words, $\lambda_{k}-\mu_{k}\geq0$ for
every $k\geq1$. Now, let $i\in\left\{  1,2,3,...\right\}  $. Then,
\begin{align*}
0  & =\sum_{k\geq1}\left(  \lambda_{k}-\mu_{k}\right)  =\left(  \lambda
_{i}-\mu_{i}\right)  +\sum\limits_{\substack{k\geq1;\\k\neq i}}\underbrace{\left(
\lambda_{k}-\mu_{k}\right)  }_{\geq0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}i\in\left\{  1,2,3,...\right\}  \right)  \\
& \geq\left(  \lambda_{i}-\mu_{i}\right)  +\underbrace{\sum\limits_{\substack{k\geq
1;\\k\neq i}}0}_{=0}=\lambda_{i}-\mu_{i},
\end{align*}
so that $\mu_{i}\geq\lambda_{i}$. Combined with $\lambda_{i}\geq\mu_{i}$ (this
holds because $\mu\subseteq\lambda$), this yields $\mu_{i}=\lambda_{i}$. Now,
forget that we fixed $i$. We thus have shown that $\mu_{i}=\lambda_{i}$ for
every $i\in\left\{  1,2,3,...\right\}  $. In other words, $\mu=\lambda$. This
proves \eqref{sol.jacobi-trudi.g.s3}.}

We shall now prove that
\begin{equation}
\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}\ \ \ \ \ \ \ \ \ \ \text{for any
two partitions }\lambda\text{ and }\mu\text{ satisfying }\mu\subseteq
\lambda.
\label{sol.jacobi-trudi.g.claim}
\end{equation}


\textit{Proof of \eqref{sol.jacobi-trudi.g.claim}:} We will prove
\eqref{sol.jacobi-trudi.g.claim} by strong induction over the nonnegative
integer $\left\vert \lambda/\mu\right\vert $.

\textit{Induction step:}\footnote{A strong induction needs no induction base.}
Let $N\in\NN$. Assume that \eqref{sol.jacobi-trudi.g.claim} is proven
in the case when $\left\vert \lambda/\mu\right\vert <N$. We need to show that
\eqref{sol.jacobi-trudi.g.claim} holds in the case when $\left\vert
\lambda/\mu\right\vert =N$.

We have assumed that \eqref{sol.jacobi-trudi.g.claim} is proven in the case
when $\left\vert \lambda/\mu\right\vert <N$. Thus,
\begin{equation}
\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}\ \ \ \ \ \ \ \ \ \ \text{for any
two partitions }\lambda\text{ and }\mu\text{ satisfying }\mu\subseteq
\lambda\text{ and }\left\vert \lambda/\mu\right\vert  <N.
\label{sol.jacobi-trudi.g.indhyp}
\end{equation}


Now, let $\lambda$ and $\mu$ be two partitions satisfying $\mu\subseteq
\lambda$ and $\left\vert \lambda/\mu\right\vert =N$. We are going to prove
that $\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}$.

If $\mu=\lambda$, then it is easy to see that $\widetilde{s}_{\lambda/\mu
}=s_{\lambda/\mu}$\ \ \ \ \footnote{\textit{Proof.} Assume that $\mu=\lambda$.
Then, $\widetilde{s}_{\lambda/\mu}=\widetilde{s}_{\lambda/\lambda}=1$ (by
\eqref{sol.jacobi-trudi.g.s1}). But since $\mu=\lambda$, we also have
$s_{\lambda/\mu}=s_{\lambda/\lambda}=1$ (by \eqref{sol.jacobi-trudi.g.s2}).
Hence, $\widetilde{s}_{\lambda/\mu}=1=s_{\lambda/\mu}$, qed.}. Hence, for the
rest of the proof of $\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}$, we can
WLOG assume that we don't have $\mu=\lambda$. Assume this.

We don't have $\mu=\lambda$. Hence, $\mu\neq\lambda$. Therefore, $\mu$ and
$\lambda$ are two \textbf{distinct} partitions $\tau$ satisfying $\mu
\subseteq\tau\subseteq\lambda$ (since obviously, $\mu$ and $\lambda$ satisfy
$\mu\subseteq\mu\subseteq\lambda$ and $\mu\subseteq\lambda\subseteq\lambda$).

Every partition $\nu$ satisfies $\Delta\widetilde{s}_{\lambda/\nu}
=\sum\limits_{\substack{\tau\in\Par :\\\nu\subseteq\tau\subseteq\lambda
}}\widetilde{s}_{\lambda/\tau}\otimes\widetilde{s}_{\tau/\nu}$ (this simply
the statement of Exercise~\ref{exe.jacobi-trudi}(c), with the index $\mu$
replaced by $\tau$). Applying this to $\nu=\mu$, we obtain
\begin{equation}
\Delta\widetilde{s}_{\lambda/\mu}=\sum\limits_{\substack{\tau\in\Par
:\\\mu\subseteq\tau\subseteq\lambda}}\widetilde{s}_{\lambda/\tau}
\otimes\widetilde{s}_{\tau/\mu}.
\label{sol.jacobi-trudi.g.eq1}
\end{equation}


Every partition $\nu$ satisfies $\Delta s_{\lambda/\nu}=\sum\limits_{\substack{\tau
\in\Par :\\\nu\subseteq\tau\subseteq\lambda}}s_{\lambda/\tau
}\otimes s_{\tau/\nu}$ (this simply the statement of Exercise
\ref{exe.Lambda.cocommutative}(b), with the index $\mu$ replaced by $\tau$).
Applying this to $\nu=\mu$, we obtain%
\begin{equation}
\Delta s_{\lambda/\mu}=\sum\limits_{\substack{\tau\in\Par %
:\\\mu\subseteq\tau\subseteq\lambda}}s_{\lambda/\tau}\otimes s_{\tau/\mu
}.\label{sol.jacobi-trudi.g.eq2}%
\end{equation}


Subtracting the equality \eqref{sol.jacobi-trudi.g.eq1} from the equality
\eqref{sol.jacobi-trudi.g.eq2}, we obtain
\begin{align}
\Delta\widetilde{s}_{\lambda/\mu}-\Delta s_{\lambda/\mu}  & =\sum\limits
_{\substack{\tau\in\Par :\\\mu\subseteq\tau\subseteq\lambda
}}\widetilde{s}_{\lambda/\tau}\otimes\widetilde{s}_{\tau/\mu}-\sum\limits
_{\substack{\tau\in\Par :\\\mu\subseteq\tau\subseteq\lambda
}}s_{\lambda/\tau}\otimes s_{\tau/\mu}=\sum\limits_{\substack{\tau\in
\Par :\\\mu\subseteq\tau\subseteq\lambda}}\left(  \widetilde{s}%
_{\lambda/\tau}\otimes\widetilde{s}_{\tau/\mu}-s_{\lambda/\tau}\otimes
s_{\tau/\mu}\right)  \nonumber\\
& =\left(  \widetilde{s}_{\lambda/\mu}\otimes\widetilde{s}_{\mu/\mu
}-s_{\lambda/\mu}\otimes s_{\mu/\mu}\right)  +\left(  \widetilde{s}%
_{\lambda/\lambda}\otimes\widetilde{s}_{\lambda/\mu}-s_{\lambda/\lambda
}\otimes s_{\lambda/\mu}\right)  \nonumber\\
& \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\tau\in\Par %
:\\\mu\subseteq\tau\subseteq\lambda;\\\tau\neq\mu;\ \tau\neq\lambda}}\left(
\widetilde{s}_{\lambda/\tau}\otimes\widetilde{s}_{\tau/\mu}-s_{\lambda/\tau
}\otimes s_{\tau/\mu}\right)
\label{sol.jacobi-trudi.g.eq3}
\end{align}
(here, we have split the sum using the fact that $\mu$ and $\lambda$ are two
\textbf{distinct} partitions $\tau$ satisfying $\mu\subseteq\tau
\subseteq\lambda$).

Now, let $\tau\in\Par$ be such that $\mu\subseteq\tau
\subseteq\lambda$ and $\tau\neq\mu$ and $\tau\neq\lambda$. We are going to
show that $\widetilde{s}_{\lambda/\tau}\otimes\widetilde{s}_{\tau/\mu
}-s_{\lambda/\tau}\otimes s_{\tau/\mu}=0$.

Indeed, we have $\mu\subseteq\lambda$ and therefore $\left\vert \lambda
/\mu\right\vert =\left\vert \lambda\right\vert -\left\vert \mu\right\vert
=\underbrace{\left(  \left\vert \lambda\right\vert -\left\vert \tau\right\vert
\right)  }_{\substack{=\left\vert \lambda/\tau\right\vert \\\text{(since }%
\tau\subseteq\lambda\text{)}}}+\underbrace{\left(  \left\vert \tau\right\vert
-\left\vert \mu\right\vert \right)  }_{\substack{=\left\vert \tau
/\mu\right\vert \\\text{(since }\mu\subseteq\tau\text{)}}}=\left\vert
\lambda/\tau\right\vert +\left\vert \tau/\mu\right\vert $. Furthermore, we
have $\left\vert \lambda/\tau\right\vert >0$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Then, we don't have $\left\vert \lambda/\tau\right\vert
>0$. Hence, we have $\left\vert \lambda/\tau\right\vert \leq0$. Combined with
$\left\vert \lambda/\tau\right\vert \geq0$ (since $\left\vert \lambda
/\tau\right\vert $ is the number of cells of $\lambda/\tau$), this yields
$\left\vert \lambda/\tau\right\vert =0$. Therefore,
\eqref{sol.jacobi-trudi.g.s3} (applied to $\tau$ instead of $\mu$) yields
$\tau=\lambda$. This contradicts $\tau\neq\lambda$. This contradiction shows
that our assumption was wrong, qed.}. Now, $\left\vert \lambda/\mu\right\vert
=\underbrace{\left\vert \lambda/\tau\right\vert }_{>0}+\left\vert \tau
/\mu\right\vert >\left\vert \tau/\mu\right\vert $, so that $\left\vert
\tau/\mu\right\vert <\left\vert \lambda/\mu\right\vert =N$. Hence,
\eqref{sol.jacobi-trudi.g.indhyp} (applied to $\tau$ instead of $\lambda$)
yields $\widetilde{s}_{\tau/\mu}=s_{\tau/\mu}$.

Also, $\left\vert \tau/\mu\right\vert >0$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Then, we don't have $\left\vert \tau/\mu\right\vert >0$.
Hence, we have $\left\vert \tau/\mu\right\vert \leq0$. Combined with
$\left\vert \tau/\mu\right\vert \geq0$ (since $\left\vert \tau/\mu\right\vert
$ is the number of cells of $\tau/\mu$), this yields $\left\vert \tau
/\mu\right\vert =0$. Therefore, \eqref{sol.jacobi-trudi.g.s3} (applied to
$\tau$ instead of $\lambda$) yields $\mu=\tau$, so that $\tau=\mu$ This
contradicts $\tau\neq\mu$. This contradiction shows that our assumption was
wrong, qed.}. Now, $\left\vert \lambda/\mu\right\vert =\left\vert \lambda
/\tau\right\vert +\underbrace{\left\vert \tau/\mu\right\vert }_{>0}>\left\vert
\lambda/\tau\right\vert $, so that $\left\vert \lambda/\tau\right\vert
<\left\vert \lambda/\mu\right\vert =N$. Hence,
\eqref{sol.jacobi-trudi.g.indhyp} (applied to $\tau$ instead of $\mu$) yields
$\widetilde{s}_{\lambda/\tau}=s_{\lambda/\tau}$. Now,
\[
\underbrace{\widetilde{s}_{\lambda/\tau}}_{=s_{\lambda/\tau}}\otimes
\underbrace{\widetilde{s}_{\tau/\mu}}_{=s_{\tau/\mu}}-s_{\lambda/\tau}\otimes
s_{\tau/\mu}=s_{\lambda/\tau}\otimes s_{\tau/\mu}-s_{\lambda/\tau}\otimes
s_{\tau/\mu}=0.
\]


Now, forget that we fixed $\tau$. We thus have proven that
\begin{equation}
\widetilde{s}_{\lambda/\tau}\otimes\widetilde{s}_{\tau/\mu}-s_{\lambda/\tau
}\otimes s_{\tau/\mu}=0\ \ \ \ \ \ \ \ \ \ \text{for every }\tau
\in\Par \text{ such that }\mu\subseteq\tau\subseteq\lambda\text{
and }\tau\neq\mu\text{ and }\tau\neq\lambda .
\label{sol.jacobi-trudi.g.5}
\end{equation}
Now, \eqref{sol.jacobi-trudi.g.eq3} becomes
\begin{align*}
& \Delta\widetilde{s}_{\lambda/\mu}-\Delta s_{\lambda/\mu}  \\
& =\left(
\widetilde{s}_{\lambda/\mu}\otimes\underbrace{\widetilde{s}_{\mu/\mu}%
}_{\substack{=1\\\text{(by \eqref{sol.jacobi-trudi.g.s2}, applied}\\\text{to
}\mu\text{ instead of }\lambda\text{)}}}-s_{\lambda/\mu}\otimes
\underbrace{s_{\mu/\mu}}_{\substack{=1\\
\text{(by \eqref{sol.jacobi-trudi.g.s1}, applied}\\
\text{to }\mu\text{ instead of }\lambda\text{)}}}\right)
+\left(  \underbrace{\widetilde{s}_{\lambda/\lambda}}_{\substack{=1\\\text{(by
\eqref{sol.jacobi-trudi.g.s2})}}}\otimes\widetilde{s}_{\lambda/\mu
}-\underbrace{s_{\lambda/\lambda}}_{\substack{=1\\\text{(by
\eqref{sol.jacobi-trudi.g.s1})}}}\otimes s_{\lambda/\mu}\right)  \\
& \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\tau\in\Par
:\\\mu\subseteq\tau\subseteq\lambda;\\\tau\neq\mu;\ \tau\neq\lambda
}}\underbrace{\left(  \widetilde{s}_{\lambda/\tau}\otimes\widetilde{s}%
_{\tau/\mu}-s_{\lambda/\tau}\otimes s_{\tau/\mu}\right)  }%
_{\substack{=0\\\text{(by \eqref{sol.jacobi-trudi.g.5})}}}\\
& =\underbrace{\left(  \widetilde{s}_{\lambda/\mu}\otimes1-s_{\lambda/\mu
}\otimes1\right)  }_{=\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu
}\right)  \otimes1}+\underbrace{\left(  1\otimes\widetilde{s}_{\lambda/\mu
}-1\otimes s_{\lambda/\mu}\right)  }_{=1\otimes\left(  \widetilde{s}%
_{\lambda/\mu}-s_{\lambda/\mu}\right)  }+\underbrace{\sum\limits_{\substack{\tau
\in\Par :\\\mu\subseteq\tau\subseteq\lambda;\\\tau\neq\mu
;\ \tau\neq\lambda}}0}_{=0}\\
& =\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  \otimes
1+1\otimes\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)
+0=\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  \otimes
1+1\otimes\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  \\
& =1\otimes\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)
+\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  \otimes1.
\end{align*}
Since $\Delta\widetilde{s}_{\lambda/\mu}-\Delta s_{\lambda/\mu}=\Delta\left(
\widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  $, this rewrites as
$\Delta\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)
=1\otimes\left(  \widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  +\left(
\widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)  \otimes1$. In other words,
the element $\widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}$ of $\Lambda$ is primitive.

But every $n\in\NN$ satisfies%
\begin{align*}
\left(  h_{n},\widetilde{s}_{\lambda/\mu}-s_{\lambda/\mu}\right)    &
=\underbrace{\left(  h_{n},\widetilde{s}_{\lambda/\mu}\right)  }%
_{\substack{=\left(  h_{n},s_{\lambda/\mu}\right)  \\\text{(by Exercise
\ref{exe.jacobi-trudi}(e))}}}-\left(  h_{n},s_{\lambda/\mu}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is }
\kk\text{-bilinear}\right)  \\
& =\left(  h_{n},s_{\lambda/\mu}\right)  -\left(  h_{n},s_{\lambda/\mu
}\right)  =0.
\end{align*}
Hence, Exercise~\ref{exe.jacobi-trudi}(f) (applied to $x=\widetilde{s}%
_{\lambda/\mu}-s_{\lambda/\mu}$) yields $\widetilde{s}_{\lambda/\mu
}-s_{\lambda/\mu}=0$. In other words,
$\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}$.

Now, forget that we fixed $\lambda$ and $\mu$. We thus have proven
$\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}$ for any two partitions $\lambda$
and $\mu$ satisfying $\mu\subseteq\lambda$ and $\left\vert \lambda
/\mu\right\vert =N$. In other words, \eqref{sol.jacobi-trudi.g.claim} holds in
the case when $\left\vert \lambda/\mu\right\vert =N$. This completes the
induction step. The induction proof of \eqref{sol.jacobi-trudi.g.claim} is
thus complete. In other words, Exercise~\ref{exe.jacobi-trudi}(g) is solved
(since the statement of Exercise~\ref{exe.jacobi-trudi}(g) is exactly
\eqref{sol.jacobi-trudi.g.claim}).

(h) \textit{Proof of Theorem~\ref{Jacobi-Trudi-formulae}.} Let $\lambda$ and
$\mu$ be two partitions, and let $\ell\in\NN$ be such that $\ell\left(
\lambda\right)  \leq\ell$ and $\ell\left(  \mu\right)  \leq\ell$. We need to
prove the equalities \eqref{eq.jacobi-trudi.h} and \eqref{eq.jacobi-trudi.e}.

Let us first prove the equality \eqref{eq.jacobi-trudi.h}. Recall that (by the
definition of $\widetilde{s}_{\lambda/\mu}$) we have $\widetilde{s}%
_{\lambda/\mu}=\widetilde{s}_{\lambda/\mu}^{[k]}$, where $k$ is any element of
${\NN }$ satisfying $k\geq\ell\left(  \lambda\right)  $ and $k\geq
\ell\left(  \mu\right)  $. Applying this to $k=\ell$, we obtain $\widetilde{s}%
_{\lambda/\mu}=\widetilde{s}_{\lambda/\mu}^{[\ell]}$ (since $\ell\geq
\ell\left(  \lambda\right)  $ and $\ell\geq\ell\left(  \mu\right)  $). Thus,%
\[
\widetilde{s}_{\lambda/\mu}=\widetilde{s}_{\lambda/\mu}^{[\ell]}=\det\left(
\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{s}%
_{\lambda/\mu}^{[\ell]}\right)  .
\]


If we don't have $\mu\subseteq\lambda$, then the equality
\eqref{eq.jacobi-trudi.h} is easily seen to hold.\footnote{\textit{Proof.}
Assume that we don't have $\mu\subseteq\lambda$. Then, $\widetilde{s}%
_{\lambda/\mu}=0$ (by Exercise~\ref{exe.jacobi-trudi}(b)). On the other hand,
$s_{\lambda/\mu}=0$ (since we don't have $\mu\subseteq\lambda$), so that%
\[
s_{\lambda/\mu}=0=\widetilde{s}_{\lambda/\mu}=\det\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  .
\]
Thus, \eqref{eq.jacobi-trudi.h} holds, qed.} Hence, for the rest of this proof
of \eqref{eq.jacobi-trudi.h}, we can WLOG assume that we do have $\mu
\subseteq\lambda$.

Assume this. We do have $\mu\subseteq\lambda$. Hence, Exercise
\ref{exe.jacobi-trudi}(g) yields $\widetilde{s}_{\lambda/\mu}=s_{\lambda/\mu}%
$, and thus we have $s_{\lambda/\mu}=\widetilde{s}_{\lambda/\mu}=\det\left(
\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  $.
This completes the proof of \eqref{eq.jacobi-trudi.h}.

Now, it remains to prove \eqref{eq.jacobi-trudi.e}.

Every $m \in \ZZ$ satisfies
\begin{equation}
\omega\left( h_m \right) = e_m.
\label{sol.jacobi-trudi.h.wh}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.h.wh}:} Let $m\in
\ZZ$. Then, Proposition~\ref{prop.Lambda.omega-and-S}(b) (applied to $m$
instead of $n$) yields $\omega\left(  h_{m}\right)  =e_{m}$.
In other words, \eqref{sol.jacobi-trudi.h.wh} is proven.}

We know that $\omega$ is a $\kk$-algebra homomorphism. As a
consequence, the map $\omega$ commutes with taking determinants; that is,
every matrix $A\in\Lambda^{k\times k}$ with $k\in\NN$ satisfies
$\omega\left(  \det A\right)  =\det\left(  \omega^{k\times k}A\right)  $,
where $\omega^{k\times k}A$ means the result of applying $\omega$ to every
entry of the matrix $A$. Applying this to $k=\ell$ and $A=\left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}$, we obtain%
\begin{align}
\omega\left(  \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  \right)   &  =\det\left(  \underbrace{\omega
^{\ell\times\ell}\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  }_{=\left(  \omega\left(  h_{\lambda_{i}%
-\mu_{j}-i+j}\right)  \right)  _{i,j=1,2,\ldots,\ell}}\right)  \nonumber\\
&  =\det\left(  \left(  \underbrace{\omega\left(  h_{\lambda_{i}-\mu_{j}%
-i+j}\right)  }_{\substack{=e_{\lambda_{i}-\mu_{j}-i+j}\\\text{(by
\eqref{sol.jacobi-trudi.h.wh}, applied}\\\text{to }m=\lambda_{i}-\mu
_{j}-i+j\text{)}}}\right)  _{i,j=1,2,\ldots,\ell}\right)  \nonumber\\
&  =\det\left(  \left(  e_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,\ell}\right)  .
\label{sol.jacobi-trudi.h.4}
\end{align}
But \eqref{eq.jacobi-trudi.h} yields $s_{\lambda/\mu}=\det\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  $. Applying
the map $\omega$ to both sides of this equality, we obtain
\[
\omega\left(  s_{\lambda/\mu}\right)  =\omega\left(  \det\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  \right)
=\det\left(  \left(  e_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,\ell}\right)
\]
(by \eqref{sol.jacobi-trudi.h.4}).

But using \eqref{skew-Schur-antipodes}, it is easy to see that
$\omega\left(  s_{\lambda/\mu}\right) =s_{\lambda^{t}/\mu^{t}}$ \ \ \ \ 
\footnote{\textit{Proof.} If $\mu \subseteq \lambda$, then
$\omega\left(  s_{\lambda/\mu}\right) =s_{\lambda^{t}/\mu^{t}}$ follows
immediately from \eqref{skew-Schur-antipodes}. Hence, for the rest of
this proof of
$\omega\left(  s_{\lambda/\mu}\right) =s_{\lambda^{t}/\mu^{t}}$, we can
WLOG assume that we don't have $\mu \subseteq \lambda$. Assume this.
We don't have $\mu \subseteq \lambda$. Hence, we don't have
$\mu^t \subseteq \lambda^t$ either. Thus,
$s_{\lambda^{t}/\mu^{t}} = 0$, so that $0 = s_{\lambda^{t}/\mu^{t}}$.
But we also have $s_{\lambda/\mu} = 0$ (since we don't have
$\mu \subseteq \lambda$), so that
$\omega\left(\underbrace{s_{\lambda/\mu}}_{=0}\right) = \omega\left(0\right)
= 0$ (since $\omega$ is $\kk$-linear), thus
$\omega\left(  s_{\lambda/\mu}\right) = 0 = s_{\lambda^{t}/\mu^{t}}$, qed.}.
(See, e.g., Exercise~\ref{exe.pieri.omega}(c) for
the proof of \eqref{skew-Schur-antipodes}.) Comparing the equalities
$\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}$ and
$\omega\left(  s_{\lambda/\mu}\right)  =\det\left(  \left(  e_{\lambda_{i}
-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  $, we obtain
$s_{\lambda^{t}/\mu^{t}}=\det\left(  \left(  e_{\lambda_{i}-\mu_{j}
-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  $. This proves
\eqref{eq.jacobi-trudi.e}. Thus, the proof of Theorem
\ref{Jacobi-Trudi-formulae} is complete. Exercise~\ref{exe.jacobi-trudi}(h) is
thus proven.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.jacobi-trudi.macdonald}}

\textit{Solution to Exercise~\ref{exe.jacobi-trudi.macdonald}.}
\textit{Proof of Theorem~\ref{Jacobi-Trudi-formulae}.}

Let $n\in \NN$. Let $\mu$ be a partition having at most $n$ parts.
Exercise \ref{exe.cauchy.skew}(a) yields%
\[
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\mathbf{x}}\right)
s_{\lambda/\mu}\left(  {\mathbf{y}}\right)  =s_{\mu}\left(  {\xx%
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}
\]
in the ring $\kk \left[  \left[  {\mathbf{x}},{\mathbf{y}}\right]
\right]  = \kk \left[  \left[  x_{1},x_{2},x_{3},...,y_{1},y_{2}%
,y_{3},...\right]  \right]  $. Switching the roles of the variables
$\xx$ and $\mathbf{y}$ in this equality, we obtain%
\[
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  {\mathbf{y}}\right)
s_{\lambda/\mu}\left(  {\mathbf{x}}\right)  =s_{\mu}\left(  {\mathbf{y}%
}\right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-y_{i}x_{j}\right)  ^{-1}.
\]
We can now substitute $\left(  y_{1},y_{2},...,y_{n},0,0,0,...\right)  $ for
$\left(  y_{1},y_{2},y_{3},...\right)  $ on both sides of this, and obtain the
equality%
\[
\sum_{\lambda\in\operatorname{Par}}s_{\lambda}\left(  y_{1},y_{2}%
,...,y_{n}\right)  s_{\lambda/\mu}\left(  {\mathbf{x}}\right)  =s_{\mu}\left(
y_{1},y_{2},...,y_{n}\right)  \cdot\prod_{i=1}^{n}\prod_{j=1}^{\infty}\left(
1-y_{i}x_{j}\right)  ^{-1}%
\]
in the subring $\left(   \kk \left[  \left[  \xx\right]  \right]
\right)  \left[  \left[  y_{1},y_{2},...,y_{n}\right]  \right]  $ of
$\kk \left[  \left[  {\mathbf{x}},{\mathbf{y}}\right]  \right]  $
(notice that the $\prod_{i,j=1}^{\infty}\left(  1-y_{i}x_{j}\right)  ^{-1}$ on
the right hand side became $\prod_{i=1}^{n}\prod_{j=1}^{\infty}\left(
1-y_{i}x_{j}\right)  ^{-1}$ because all factors $1-y_{i}x_{j}$ with $i>n$ got
sent to $1-0x_{j}=1$). In the sum on the left hand side of this equality, all
addends corresponding to partitions $\lambda$ having more than $n$ parts are
$0$ (because Exercise \ref{exe.schur.finite}(b) yields that all such $\lambda$
satisfy $s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{n}\right)  =0$, hence
$s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  =0$). Thus, we can remove
all these addends, and the equality thus becomes%
\begin{equation}
\sum\limits_{\substack{\lambda\in\operatorname{Par};\\\lambda\text{ has at most
}n\text{ parts}}}s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)
s_{\lambda/\mu}\left(  {\mathbf{x}}\right)  =s_{\mu}\left(  y_{1}%
,y_{2},...,y_{n}\right)  \cdot\prod_{i=1}^{n}\prod_{j=1}^{\infty}\left(
1-y_{i}x_{j}\right)  ^{-1}.
\label{sol.jacobi-trudi.macdonald.switched-and-substituted}%
\end{equation}


Let $\rho$ be the $n$-tuple $\left(  n-1,n-2,...,2,1,0\right)  \in{ \NN %
}^{n}$. We can regard $\rho$ as a weak composition by padding it with zeroes
at the end (i.e., identifying $\rho$ with the weak composition $\left(
n-1,n-2,...,2,1,0,0,0,0,...\right)  $).

For every $n$-tuple $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in{ \NN }^{n}$, we define the alternant $a_{\alpha}%
\in \kk \left[  x_{1},x_{2},\ldots,x_{n}\right]  $ as in
Definition~\ref{def.alternant}. (But other than this, we are not adapting the
notations of Section~\ref{bialternant-section}.)

Now, Corollary \ref{cor.stembridge.slambda} states that $s_{\lambda}\left(
x_{1},x_{2},...,x_{n}\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$ in
$\kk \left[  x_{1},x_{2},...,x_{n}\right]  $ whenever $\lambda$ is a
partition having at most $n$ parts. Applied to the variables $y_{1}%
,y_{2},...,y_{n}$ instead of $x_{1},x_{2},...,x_{n}$, this yields that
$s_{\lambda}\left(  y_{1},y_{2},...,y_{n}\right)  =\dfrac{a_{\lambda+\rho
}\left(  y_{1},y_{2},...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  }$ whenever $\lambda$ is a partition having at most $n$
parts. This equality, applied to $\lambda=\mu$, yields $s_{\mu}\left(
y_{1},y_{2},...,y_{n}\right)  =\dfrac{a_{\mu+\rho}\left(  y_{1},y_{2}%
,...,y_{n}\right)  }{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  }$.
Substituting the last two equalities into
(\ref{sol.jacobi-trudi.macdonald.switched-and-substituted}), we obtain%
\[
\sum\limits_{\substack{\lambda\in\operatorname{Par};\\\lambda\text{ has at most
}n\text{ parts}}}\dfrac{a_{\lambda+\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  }s_{\lambda/\mu}\left(
{\mathbf{x}}\right)  =\dfrac{a_{\mu+\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}{a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  }\cdot\prod_{i=1}^{n}%
\prod_{j=1}^{\infty}\left(  1-y_{i}x_{j}\right)  ^{-1}.
\]
Multiplied with $a_{\rho}\left(  y_{1},y_{2},...,y_{n}\right)  $, this becomes%
\[
\sum\limits_{\substack{\lambda\in\operatorname{Par};\\\lambda\text{ has at most
}n\text{ parts}}}a_{\lambda+\rho}\left(  y_{1},y_{2},...,y_{n}\right)
s_{\lambda/\mu}\left(  {\mathbf{x}}\right)  =a_{\mu+\rho}\left(  y_{1}%
,y_{2},...,y_{n}\right)  \cdot\prod_{i=1}^{n}\prod_{j=1}^{\infty}\left(
1-y_{i}x_{j}\right)  ^{-1}.
\]
Renaming $\lambda$ as $\nu$ on the left hand side of this equality, we obtain%
\begin{equation}
\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts}}}a_{\nu+\rho}\left(  y_{1},y_{2},...,y_{n}\right)  s_{\nu/\mu}\left(
{\mathbf{x}}\right)  =a_{\mu+\rho}\left(  y_{1},y_{2},...,y_{n}\right)
\cdot\prod_{i=1}^{n}\prod_{j=1}^{\infty}\left(  1-y_{i}x_{j}\right)  ^{-1}.
\label{sol.jacobi-trudi.macdonald.alternant-form}%
\end{equation}


Now, let $\lambda$ be a partition having at most $n$ parts. We want to find
the coefficient of $y_{1}^{\lambda_{1}+n-1}y_{2}^{\lambda_{2}+n-2}%
...y_{n}^{\lambda_{n}+n-n}$ on the left and the right hand sides of
(\ref{sol.jacobi-trudi.macdonald.alternant-form}). Here, we regard
(\ref{sol.jacobi-trudi.macdonald.alternant-form}) as an equality in the ring
$\left(   \kk \left[  \left[  \xx\right]  \right]  \right)
\left[  \left[  y_{1},y_{2},...,y_{n}\right]  \right]  $, so that we consider
the variables $x_{1},x_{2},x_{3},...$ as constants, and thus (for example) the
coefficient of $y_{1}$ in $\left(  1+x_{1}\right)  \left(  1+y_{1}\right)  $
is $1+x_{1}$ rather than $1$.

We first notice that%
\begin{equation}
\prod_{i=1}^{n}\prod_{j=1}^{\infty}\left(  1-y_{i}x_{j}\right)  ^{-1}%
=\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)
\label{sol.jacobi-trudi.macdonald.enter-h}%
\end{equation}
\footnote{\textit{Proof:} The equality \eqref{h-generating-function-no-HE}
(with the indices $i$ and $n$ renamed as $j$ and $q$) yields
\begin{equation}
\prod_{j=1}^{\infty}\left(  1-x_{j}t\right)  ^{-1}=1+h_{1}\left(  {\xx%
}\right)  t+h_{2}\left(  {\mathbf{x}}\right)  t^{2}+...=\sum_{q\geq0}%
h_{q}\left(  {\mathbf{x}}\right)  t^{q}\ \ \ \ \ \ \ \ \ \ \text{in }\left(
 \kk \left[  \left[  \xx\right]  \right]  \right)  \left[  \left[
t\right]  \right]  . \label{sol.jacobi-trudi.macdonald.enter-h.pf.1}%
\end{equation}
\par
For every $i\in\left\{  1,2,...,n\right\}  $, we have
\[
\prod_{j=1}^{\infty}\left(  1-y_{i}x_{j}\right)  ^{-1}=\sum_{q\geq0}%
h_{q}\left(  {\mathbf{x}}\right)  y_{i}^{q}%
\]
(this follows by substituting $y_{i}$ for $t$ on both sides of
\eqref{sol.jacobi-trudi.macdonald.enter-h.pf.1}). Thus,
\begin{align*}
\prod_{i=1}^{n}\underbrace{\prod_{j=1}^{\infty}\left(  1-y_{i}x_{j}\right)
^{-1}}_{=\sum_{q\geq0}h_{q}\left(  {\mathbf{x}}\right)  y_{i}^{q}}  &
=\prod_{i=1}^{n}\sum_{q\geq0}h_{q}\left(  {\mathbf{x}}\right)  y_{i}^{q}%
=\prod_{j=1}^{n}\sum_{q\geq0}h_{q}\left(  {\mathbf{x}}\right)  y_{j}^{q}\\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}%
}\underbrace{\prod_{j=1}^{n}\left(  h_{q_{j}}\left(  {\mathbf{x}}\right)
\cdot y_{j}^{q_{j}}\right)  }_{=\left(  \prod_{j=1}^{n}h_{q_{j}}\left(
{\mathbf{x}}\right)  \right)  \cdot\left(  \prod_{j=1}^{n}y_{j}^{q_{j}%
}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  ,
\end{align*}
qed.}.

Let us recall a basic fact from linear algebra, namely the explicit formula
for the determinant of a matrix as a sum over permutations: Any matrix
$\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,\ell}$ over a commutative ring
satisfies
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,\ell}\right)
=\sum_{\sigma\in\Symm_{\ell}}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{\ell}\alpha_{i,\sigma\left(  i\right)  }.
\label{sol.jacobi-trudi.macdonald.detexp}%
\end{equation}
Applying this to $\ell=n$ and $\alpha_{i,j}=y_{i}^{\left(  \mu+\rho\right)
_{j}}$, we obtain
\begin{equation}
\det\left(  \left(  y_{i}^{\left(  \mu+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)  =\sum_{\sigma\in\Symm_n}\left(  -1\right)
^{\sigma}\prod_{i=1}^{n}y_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(
i\right)  }}=\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}%
\prod_{j=1}^{n}y_{j}^{\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}
\label{sol.jacobi-trudi.macdonald.detexp.used1}%
\end{equation}
(here, we renamed the index $i$ as $j$ in the product). Now, the right hand
side of (\ref{sol.jacobi-trudi.macdonald.alternant-form}) becomes%
\begin{align*}
&  \underbrace{a_{\mu+\rho}\left(  y_{1},y_{2},...,y_{n}\right)
}_{\substack{=\det\left(  \left(  y_{i}^{\left(  \mu+\rho\right)  _{j}%
}\right)  _{i,j=1,2,...,n}\right)  \\\text{(by the definition of the alternant
}a_{\mu+\rho}\text{)}}}\cdot\underbrace{\prod_{i=1}^{n}\prod_{j=1}^{\infty
}\left(  1-y_{i}x_{j}\right)  ^{-1}}_{\substack{=\sum_{\left(  q_{1}%
,q_{2},...,q_{n}\right)  \in{ \NN }^{n}}\left(  \prod_{j=1}^{n}h_{q_{j}%
}\left(  {\mathbf{x}}\right)  \right)  \cdot\left(  \prod_{j=1}^{n}%
y_{j}^{q_{j}}\right)  \\\text{(by (\ref{sol.jacobi-trudi.macdonald.enter-h}%
))}}}\\
&  =\underbrace{\det\left(  \left(  y_{i}^{\left(  \mu+\rho\right)  _{j}%
}\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=\sum_{\sigma\in
\Symm_n}\left(  -1\right)  ^{\sigma}\prod_{j=1}^{n}y_{j}^{\left(
\mu+\rho\right)  _{\sigma\left(  j\right)  }}\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.detexp.used1}))}}}\cdot\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  \right)  \cdot\left(  \prod
_{j=1}^{n}y_{j}^{q_{j}}\right) \\
&  =\left(  \sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}%
\prod_{j=1}^{n}y_{j}^{\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }%
}\right)  \cdot\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }%
^{n}}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  \right)
\cdot\left(  \prod_{j=1}^{n}y_{j}^{q_{j}}\right) \\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  \right)  \cdot\underbrace{\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \left(  \prod_{j=1}^{n}y_{j}^{\left(
\mu+\rho\right)  _{\sigma\left(  j\right)  }}\right)  }_{=\prod_{j=1}%
^{n}\left(  y_{j}^{q_{j}}y_{j}^{\left(  \mu+\rho\right)  _{\sigma\left(
j\right)  }}\right)  =\prod_{j=1}^{n}y_{j}^{q_{j}+\left(  \mu+\rho\right)
_{\sigma\left(  j\right)  }}}\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  \right)  \cdot\left(  \prod
_{j=1}^{n}y_{j}^{q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }%
}\right)  .
\end{align*}
Hence, the coefficient of $y_{1}^{\lambda_{1}+n-1}y_{2}^{\lambda_{2}%
+n-2}...y_{n}^{\lambda_{n}+n-n}$ on the right hand side of
(\ref{sol.jacobi-trudi.macdonald.alternant-form}) equals%
\begin{equation}
\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }^{n}%
;\\q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}\prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  .
\label{sol.jacobi-trudi.macdonald.right1}%
\end{equation}
However, for every $\sigma\in\Symm_n$, it is easy to see that
\begin{equation}
\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }%
^{n};\\q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}\prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  =\prod_{j=1}^{n}h_{\lambda
_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}\left(
{\mathbf{x}}\right)  . \label{sol.jacobi-trudi.macdonald.right2}%
\end{equation}
\footnote{\textit{Proof.} Let $\sigma\in\Symm_n$. It is clear that
the sum
\begin{equation}
\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }%
^{n};\\q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}\prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)
\label{sol.jacobi-trudi.macdonald.right2.pf.0}%
\end{equation}
has at most one addend: namely, the one corresponding to the $n$-tuple
$\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \ZZ }^{n}$ defined by%
\begin{equation}
q_{j}=\lambda_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(  j\right)
}\text{ for every }j\in\left\{  1,2,...,n\right\}  .
\label{sol.jacobi-trudi.macdonald.right2.pf.1}%
\end{equation}
If this $n$-tuple $\left(  q_{1},q_{2},...,q_{n}\right)  $ belongs to
$\NN ^{n}$, then the sum (\ref{sol.jacobi-trudi.macdonald.right2.pf.0})
does have this summand. Thus, if the $n$-tuple $\left(  q_{1},q_{2}%
,...,q_{n}\right)  $ belongs to $\NN ^{n}$, we have%
\[
\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in{ \NN }%
^{n};\\q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}\prod_{j=1}%
^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  =\prod_{j=1}^{n}h_{\lambda
_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}\left(
{\mathbf{x}}\right) .
\]
Thus, if the $n$-tuple $\left(  q_{1},q_{2},...,q_{n}\right)  $ defined by
(\ref{sol.jacobi-trudi.macdonald.right2.pf.1}) belongs to $\NN ^{n}$,
the equality (\ref{sol.jacobi-trudi.macdonald.right2}) is proven. It remains
to consider the case when the $n$-tuple $\left(  q_{1},q_{2},...,q_{n}\right)
$ defined by (\ref{sol.jacobi-trudi.macdonald.right2.pf.1}) does not belong to
$\NN ^{n}$. In this case, the sum
(\ref{sol.jacobi-trudi.macdonald.right2.pf.0}) is empty (because the only
addend it can have corresponds to the $n$-tuple $\left(  q_{1},q_{2}%
,...,q_{n}\right)  $ defined by (\ref{sol.jacobi-trudi.macdonald.right2.pf.1}%
), but this $n$-tuple does not belong to $\NN ^{n}$ and therefore does
not appear in the sum), hence equals $0$. But since the $n$-tuple $\left(
q_{1},q_{2},...,q_{n}\right)  $ defined by
(\ref{sol.jacobi-trudi.macdonald.right2.pf.1}) does not belong to
$\NN ^{n}$, there must be a $j\in\left\{  1,2,...,n\right\}  $
satisfying $\lambda_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(  j\right)
}\notin \NN$. For this $j$, we have $h_{\lambda_{j}+n-j-\left(  \mu
+\rho\right)  _{\sigma\left(  j\right)  }}=0$. Hence, the product $\prod
_{j=1}^{n}h_{\lambda_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(
j\right)  }}\left(  {\mathbf{x}}\right)  $ has a factor equal to $0$;
consequently, the product $\prod_{j=1}^{n}h_{\lambda_{j}+n-j-\left(  \mu
+\rho\right)  _{\sigma\left(  j\right)  }}\left(  {\mathbf{x}}\right)  $ must
be $0$. Thus, both the sum (\ref{sol.jacobi-trudi.macdonald.right2.pf.0}) and
the product $\prod_{j=1}^{n}h_{\lambda_{j}+n-j-\left(  \mu+\rho\right)
_{\sigma\left(  j\right)  }}\left(  {\mathbf{x}}\right)  $ are $0$, and thus
(\ref{sol.jacobi-trudi.macdonald.right2}) is proven.} Hence, the coefficient
of $y_{1}^{\lambda_{1}+n-1}y_{2}^{\lambda_{2}+n-2}...y_{n}^{\lambda_{n}+n-n}$
on the right hand side of (\ref{sol.jacobi-trudi.macdonald.alternant-form})
equals%
\begin{align}
&  \sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}%
\underbrace{\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)
\in{ \NN }^{n};\\q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)
}=\lambda_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }%
}\prod_{j=1}^{n}h_{q_{j}}\left(  {\mathbf{x}}\right)  }_{\substack{=\prod
_{j=1}^{n}h_{\lambda_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(
j\right)  }}\left(  {\mathbf{x}}\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.right2}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.jacobi-trudi.macdonald.right1})}\right) \nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod
_{j=1}^{n}\underbrace{h_{\lambda_{j}+n-j-\left(  \mu+\rho\right)
_{\sigma\left(  j\right)  }}}_{\substack{=h_{\lambda_{j}-\mu_{\sigma\left(
j\right)  }-j+\sigma\left(  j\right)  }\\\text{(since it is easy to see
that}\\\lambda_{j}+n-j-\left(  \mu+\rho\right)  _{\sigma\left(  j\right)
}=\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(  j\right)
\text{)}}}\left(  {\mathbf{x}}\right)  \nonumber\\
&=\sum_{\sigma\in\Symm_n%
}\left(  -1\right)  ^{\sigma}\prod_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(
j\right)  }-j+\sigma\left(  j\right)  }\left(  {\mathbf{x}}\right) \nonumber\\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{n}h_{\lambda_{i}-\mu_{\sigma\left(  i\right)  }-i+\sigma\left(
i\right)  }\left(  {\mathbf{x}}\right)  =\det\left(  \left(  h_{\lambda
_{i}-\mu_{j}-i+j}\left(  \xx\right)  \right)  _{i,j=1,2,\ldots
,n}\right)  \label{sol.jacobi-trudi.macdonald.right}%
\end{align}
(because applying (\ref{sol.jacobi-trudi.macdonald.detexp}) to $\ell=n$ and
$\alpha_{i,j}=h_{\lambda_{i}-\mu_{j}-i+j}\left(  \xx\right)  $ yields
$\det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\left(  \xx\right)
\right)  _{i,j=1,2,\ldots,n}\right)  =\sum_{\sigma\in\Symm_n}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{n}h_{\lambda_{i}-\mu_{\sigma\left(  i\right)
}-i+\sigma\left(  i\right)  }\left(  {\mathbf{x}}\right)  $).

On the other hand, let us study the left hand side of
(\ref{sol.jacobi-trudi.macdonald.alternant-form}). Every partition $\nu$
having at most $n$ parts satisfies%
\begin{align*}
a_{\nu+\rho}\left(  y_{1},y_{2},...,y_{n}\right)   &  =\det\left(  \left(
y_{i}^{\left(  \nu+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the alternant }%
a_{\nu+\rho}\right) \\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{n}y_{i}^{\left(  \nu+\rho\right)  _{\sigma\left(  i\right)  }%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.jacobi-trudi.macdonald.detexp}%
), applied to }\ell=n\text{ and }\alpha_{i,j}=y_{i}^{\left(  \nu+\rho\right)
_{j}}\right)  .
\end{align*}
Hence, the left hand side of (\ref{sol.jacobi-trudi.macdonald.alternant-form})
is%
\begin{align*}
&  \sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts}}}\underbrace{a_{\nu+\rho}\left(  y_{1},y_{2},...,y_{n}\right)  }%
_{=\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}y_{i}^{\left(  \nu+\rho\right)  _{\sigma\left(  i\right)  }}}s_{\nu/\mu
}\left(  {\mathbf{x}}\right) \\
&  =\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts}}}\left(  \sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma
}\prod_{i=1}^{n}y_{i}^{\left(  \nu+\rho\right)  _{\sigma\left(  i\right)  }%
}\right)  s_{\nu/\mu}\left(  {\mathbf{x}}\right) \\
&  =\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}%
\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts}}}\left(  \prod_{i=1}^{n}y_{i}^{\left(  \nu+\rho\right)  _{\sigma\left(
i\right)  }}\right)  s_{\nu/\mu}\left(  {\mathbf{x}}\right)  .
\end{align*}
Thus, the coefficient of $y_{1}^{\lambda_{1}+n-1}y_{2}^{\lambda_{2}%
+n-2}...y_{n}^{\lambda_{n}+n-n}$ on the left hand side of
(\ref{sol.jacobi-trudi.macdonald.alternant-form}) equals%
\begin{equation}
\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts;}\\\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}s_{\nu/\mu}\left(
{\mathbf{x}}\right)  . \label{sol.jacobi-trudi.macdonald.left1}%
\end{equation}
Now let us simplify this. First, we claim that every permutation $\sigma
\in\Symm_n$ satisfying $\sigma\neq \id$ satisfies%
\begin{equation}
\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts;}\\\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}s_{\nu/\mu}\left(
{\mathbf{x}}\right)  =0. \label{sol.jacobi-trudi.macdonald.left2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.left2}):} Let
$\sigma\in\Symm_n$ be a permutation satisfying $\sigma\neq
 \id$. We need to prove (\ref{sol.jacobi-trudi.macdonald.left2}%
). Of course, it is enough to show that the sum on the left hand side of
(\ref{sol.jacobi-trudi.macdonald.left2}) is empty, i.e., that there exists no
$\nu\in \Par $ such that $\nu$ has at most $n$ parts and
satisfies $\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j$ for every $j\in\left\{  1,2,...,n\right\}  $. Assume the contrary.
Then, there exists a $\nu\in \Par $ such that $\nu$ has at most
$n$ parts and satisfies $\left(  \nu+\rho\right)  _{\sigma\left(  j\right)
}=\lambda_{j}+n-j$ for every $j\in\left\{  1,2,...,n\right\}  $. Consider this
$\nu$. Since $\nu$ is a partition, $\nu+\rho$ is a strict partition, i.e., we
have $\left(  \nu+\rho\right)  _{1}>\left(  \nu+\rho\right)  _{2}>...>\left(
\nu+\rho\right)  _{n}$. In other words, the entries of the $n$-tuple $\nu
+\rho$ are in strictly decreasing order. Since $\sigma\neq \id$,
the permutation $\sigma$ must mess up the order of the entries of $\nu+\rho$;
thus, we \textbf{cannot have }$\left(  \nu+\rho\right)  _{\sigma\left(
1\right)  }>\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  }>...>\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }$. Since $\left(  \nu+\rho\right)
_{\sigma\left(  j\right)  }=\lambda_{j}+n-j$ for every $j\in\left\{
1,2,...,n\right\}  $, this rewrites as follows: We \textbf{cannot} have
$\lambda_{1}+n-1>\lambda_{2}+n-2>...>\lambda_{n}+n-n$.
\par
But since $\lambda$ is a partition, we have $\lambda_{1}\geq\lambda_{2}%
\geq...\geq\lambda_{n}$, thus $\lambda_{1}+n-1>\lambda_{2}+n-2>...>\lambda
_{n}+n-n$. This contradicts the fact that we cannot have $\lambda
_{1}+n-1>\lambda_{2}+n-2>...>\lambda_{n}+n-n$. This contradiction finishes the
proof.} Thus, in the outer sum in (\ref{sol.jacobi-trudi.macdonald.left1}),
all addends which correspond to permutations $\sigma\in\Symm_n$
satisfying $\sigma\neq \id$ are $0$. We can therefore remove all
these addends, leaving only the addend corresponding to $\sigma
= \id$. Thus, the sum simplifies as follows:%
\begin{align*}
\sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts;}\\\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}s_{\nu/\mu}\left(
{\mathbf{x}}\right)   &  =\underbrace{\left(  -1\right)  ^{ \id %
}}_{=1}\underbrace{\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at
most }n\text{ parts;}\\\left(  \nu+\rho\right)  _{ \id \left(
j\right)  }=\lambda_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}
}}}_{\substack{=\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at
most }n\text{ parts;}\\\nu_{j}+n-j=\lambda_{j}+n-j\text{ for every }%
j\in\left\{  1,2,...,n\right\}  }}\\\text{(since }\left(  \nu+\rho\right)
_{ \id \left(  j\right)  }=\left(  \nu+\rho\right)  _{j}=\nu
_{j}+n-j\text{ for all }j\text{)}}}s_{\nu/\mu}\left(  {\mathbf{x}}\right) \\
&  =\sum\limits_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts;}\\\nu_{j}+n-j=\lambda_{j}+n-j\text{ for every }j\in\left\{
1,2,...,n\right\}  }}s_{\nu/\mu}\left(  \xx\right)  =s_{\lambda/\mu
}\left(  \xx\right)
\end{align*}
(because there is only one $\nu\in \Par $ such that $\nu$ has at
most $n$ parts and satisfies $\nu_{j}+n-j=\lambda_{j}+n-j$ for every
$j\in\left\{  1,2,...,n\right\}  $; namely, this $\nu$ is $\lambda$). Hence,
the coefficient of $y_{1}^{\lambda_{1}+n-1}y_{2}^{\lambda_{2}+n-2}%
...y_{n}^{\lambda_{n}+n-n}$ on the left hand side of
(\ref{sol.jacobi-trudi.macdonald.alternant-form}) equals%
\begin{align}
&  \sum_{\sigma\in\Symm_n}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\operatorname{Par};\\\nu\text{ has at most }n\text{
parts;}\\\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }=\lambda
_{j}+n-j\text{ for every }j\in\left\{  1,2,...,n\right\}  }}s_{\nu/\mu}\left(
{\mathbf{x}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.jacobi-trudi.macdonald.left1})}\right) \nonumber\\
&  =s_{\lambda/\mu}\left(  \xx\right)  .
\label{sol.jacobi-trudi.macdonald.left}%
\end{align}
But the coefficients of $y_{1}^{\lambda_{1}+n-1}y_{2}^{\lambda_{2}%
+n-2}...y_{n}^{\lambda_{n}+n-n}$ on the left hand side of
(\ref{sol.jacobi-trudi.macdonald.alternant-form}) and on the right hand side
of (\ref{sol.jacobi-trudi.macdonald.alternant-form}) must be equal. Since the
former coefficient is $s_{\lambda/\mu}\left(  \xx\right)  $ (by
(\ref{sol.jacobi-trudi.macdonald.left})), and the latter coefficient is
$\det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\left(  \xx\right)
\right)  _{i,j=1,2,\ldots,n}\right)  $ (by
(\ref{sol.jacobi-trudi.macdonald.right})), this shows that%
\[
s_{\lambda/\mu}\left(  \xx\right)  =\det\left(  \left(  h_{\lambda
_{i}-\mu_{j}-i+j}\left(  \xx\right)  \right)  _{i,j=1,2,\ldots
,n}\right)  .
\]
In other words, $s_{\lambda/\mu}=\det\left(  \left(  h_{\lambda_{i}-\mu
_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  $.

Now, forget that we fixed $n$, $\lambda$ and $\mu$. We thus have proven that
if $n\in \NN$, and if $\lambda$ and $\mu$ are two partitions having at
most $n$ parts (each), then $s_{\lambda/\mu}=\det\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  $. Renaming
$n$ as $\ell$ in this claim, we obtain: If $\ell\in \NN$, and if
$\lambda$ and $\mu$ are two partitions having at most $\ell$ parts (each),
then%
\begin{equation}
s_{\lambda/\mu}=\det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  . \label{sol.jacobi-trudi.macdonald.h.done}%
\end{equation}
This proves (\ref{eq.jacobi-trudi.h}).

Now it remains to prove (\ref{eq.jacobi-trudi.e}). Let $\ell\in \NN$,
and let $\lambda$ and $\mu$ be two partitions having at most $\ell$ parts (each).

Let us first notice that every $m\in \ZZ$ satisfies
\begin{equation}
\omega\left(  h_{m}\right)  =e_{m}. \label{sol.jacobi-trudi.macdonald.h.wh}%
\end{equation}
(Indeed, this follows from Proposition~\ref{prop.Lambda.omega-and-S}(b),
applied to $m$ instead of $n$.) But using \eqref{skew-Schur-antipodes}, it
is easy to see that%
\begin{equation}
\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}.
\label{sol.jacobi-trudi.macdonald.omega.h}%
\end{equation}
\footnote{In fact, (\ref{sol.jacobi-trudi.macdonald.omega.h}) follows
immediately from \eqref{skew-Schur-antipodes} in the case when $\mu
\subseteq\lambda$; but otherwise it follows from $s_{\lambda/\mu}=0$ and
$s_{\lambda^{t}/\mu^{t}}=0$.} Thus,%
\begin{align*}
s_{\lambda^{t}/\mu^{t}}  &  =\omega\left(  s_{\lambda/\mu}\right)
=\omega\left(  \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.jacobi-trudi.macdonald.h.done})}\right) \\
&  =\det\left(  \left(  \underbrace{\omega\left(  h_{\lambda_{i}-\mu_{j}%
-i+j}\right)  }_{\substack{=e_{\lambda_{i}-\mu_{j}-i+j}\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.h.wh}))}}}\right)  _{i,j=1,2,\ldots,\ell
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\omega\text{ is a } \kk \text{-algebra homomorphism, and
thus}\\
\text{commutes with taking determinants}%
\end{array}
\right) \\
&  =\det\left(  \left(  e_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,\ell}\right)  .
\end{align*}


This proves (\ref{eq.jacobi-trudi.e}). Therefore, the proof of Theorem
\ref{Jacobi-Trudi-formulae} is complete.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.jacobi-trudi.macdonald}}

\textit{Solution to Exercise~\ref{exe.jacobi-trudi.macdonald}.}
\textit{Proof of Theorem~\ref{Jacobi-Trudi-formulae}.}

\underline{\textit{Step 1: Introducing notations.}}

Let $\mathbf{p}$ be the $\kk$-algebra $\kk \left[  \left[
\xx \right]  \right]  =\kk \left[  \left[  x_{1},x_{2}%
,x_{3},...\right]  \right]  $. Thus, $\kk \left[  \left[  \xx %
\right]  \right]  =\mathbf{p}$. Clearly, $x_{i}\in\kk \left[  \left[
\xx \right]  \right]  =\mathbf{p}$ for every $i\in\left\{
1,2,3,...\right\}  $.

We have a canonical $\kk \left[  \left[ \xx \right]  \right]
$-algebra isomorphism from $\kk \left[  \left[  \xx %
, \yy \right]  \right]  =\kk \left[  \left[  x_{1},x_{2}%
,x_{3},...,y_{1},y_{2},y_{3},...\right]  \right]  $ to $\left(  \kk %
\left[  \left[ \xx \right]  \right]  \right)  \left[  \left[
 \yy \right]  \right]  =\left(  \kk \left[  \left[  x_{1}%
,x_{2},x_{3},...\right]  \right]  \right)  \left[  \left[  y_{1},y_{2}%
,y_{3},...\right]  \right]  $. We shall regard this $\kk \left[  \left[
\xx \right]  \right]  $-algebra isomorphism as the identity. Thus,
$\kk \left[  \left[  \xx , \yy \right]  \right]
=\underbrace{\left(  \kk \left[  \left[ \xx \right]  \right]
\right)  }_{=\mathbf{p}}\left[  \left[   \yy \right]  \right]
=\mathbf{p}\left[  \left[   \yy \right]  \right]  $.

Let $n\in\NN$. Also, let us consider the $\mathbf{p}$-subalgebra
$\mathbf{p}\left[  \left[  y_{1},y_{2},...,y_{n}\right]  \right]  $ of
$\mathbf{p}\left[  \left[   \yy \right]  \right]  $.

Let $\rho$ be the $n$-tuple $\left(  n-1,n-2,...,2,1,0\right)  \in
\NN^n$. We can regard $\rho$ as a weak composition by padding it with
zeroes at the end (i.e., identifying $\rho$ with the weak composition $\left(
n-1,n-2,...,2,1,0,0,0,0,...\right)  $). As usual, write $\rho$ in the form
$\rho=\left(  \rho_{1},\rho_{2},...,\rho_{n}\right)  $; then, $\rho$ satisfies%
\begin{equation}
\rho_{i}=n-i\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,...,n\right\}  .
\label{sol.jacobi-trudi.macdonald.preludium.rhoi}
\end{equation}

For every $n$-tuple $\alpha = \left(\alpha_1, \alpha_2, \ldots, \alpha_n\right)
\in \NN^n$, we define the alternant $a_\alpha \in \kk\left[x_1, x_2, \ldots,
x_n\right]$ as in Definition~\ref{def.alternant}. (But other than this, we are
not adapting the notations of Section~\ref{bialternant-section}.)


\underline{\textit{Step 2: The alternant identity.}}

We start out by bringing Corollary~\ref{cor.stembridge.slambda} into a form
that will be most useful to us. Literally, Corollary
\ref{cor.stembridge.slambda} states that for any partition $\lambda$, we have
$s_{\lambda}({\xx })=\dfrac{a_{\lambda+\rho}}{a_{\rho}}$ in
$\kk \left[  x_{1},x_{2},...,x_{n}\right]  $ (this equality entails, in
particular, that $a_{\lambda+\rho}$ is uniquely divisible by $a_{\rho}$ in
$\kk \left[  x_{1},x_{2},...,x_{n}\right]  $). However, this statement
uses the conventions of Section~\ref{bialternant-section}; thus,

\begin{itemize}
\item ``any partition $\lambda$'' means
``any partition $\lambda$ having at most $n$ parts''
in this statement;

\item the letter ``$\xx$'' means the
finite variable set $\left(  x_{1},x_{2},...,x_{n}\right)  $ (rather than the
countable variable set $\left(  x_{1},x_{2},x_{3},...\right)  $ that it
usually denotes) in this statement;

\item the term ``$s_{\lambda}\left(  \xx \right) $'' means
the expression $\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape } \lambda {\text;}\\\text{all entries of }T\text{ belong}%
\\\text{to }\left\{  1,2,...,n\right\}  }}\left(  x_{1},x_{2},...,x_{n}%
\right)  ^{\cont \left(  T\right)  }$ (rather than \newline%
$\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda%
}}{\xx }^{\cont \left(  T\right)  }$ as it means usually)
in this statement, where $\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }$ is defined as $\prod_{i=1}^{n}%
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }$ for any column-strict
tableau $T$ having the property that all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $.
\end{itemize}

Since we are not using these conventions in this solution, let us reformulate
this statement without using these conventions. We obtain the following: For
any partition $\lambda$ having at most $n$ parts, we have%
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,...,n\right\}  }}\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }=\dfrac{a_{\lambda+\rho}}{a_{\rho}%
}\ \ \ \ \ \ \ \ \ \ \text{in }\kk \left[  x_{1},x_{2},...,x_{n}\right]
. \label{sol.jacobi-trudi.macdonald.preludium.1}%
\end{equation}
Hence, for any partition $\lambda$ having at most $n$ parts, we have%
\begin{equation}
a_{\rho}\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau
of shape }\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }=a_{\lambda+\rho}%
\ \ \ \ \ \ \ \ \ \ \text{in }\kk \left[  x_{1},x_{2},...,x_{n}\right]
. \label{sol.jacobi-trudi.macdonald.preludium.2}%
\end{equation}


Let now $\lambda$ be a partition having at most $n$ parts. Then,
$\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  $ (since
$\lambda$ has at most $n$ parts).

Recall that
\begin{equation}
a_{\alpha}=\det\left[
\begin{matrix}
x_{1}^{\alpha_{1}} & \cdots & x_{1}^{\alpha_{n}}\\
x_{2}^{\alpha_{1}} & \cdots & x_{2}^{\alpha_{n}}\\
\vdots & \ddots & \vdots\\
x_{n}^{\alpha_{1}} & \cdots & x_{n}^{\alpha_{n}}%
\end{matrix}
\right]  =\det\left(  \left(  x_{i}^{\alpha_{j}}\right)  _{i,j=1,2,...,n}%
\right)  \label{sol.jacobi-trudi.macdonald.preludium.a}%
\end{equation}
for every $\alpha=\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)
\in\NN^n$. Applying this to $\alpha=\lambda+\rho$, we obtain%
\[
a_{\lambda+\rho}=\det\left(  \left(  x_{i}^{\left(  \lambda+\rho\right)  _{j}%
}\right)  _{i,j=1,2,...,n}\right)  .
\]
On the other hand, applying \eqref{sol.jacobi-trudi.macdonald.preludium.a} to
$\alpha=\rho$, we obtain
\[
a_{\rho}=\det\left(  \left(  x_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
.
\]
Multiplying both sides of this equality with $\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda\text{;}\\\text{all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,...,n\right\}  }}\left(
x_{1},x_{2},...,x_{n}\right)  ^{\cont \left(  T\right)  }$, we
obtain%
\begin{align*}
a_{\rho}\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau
of shape }\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }  &  =\det\left(  \left(  x_{i}%
^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)  \cdot\sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,...,n\right\}  }}\underbrace{\left(  x_{1},x_{2},...,x_{n}\right)
^{\cont \left(  T\right)  }}_{\substack{=\prod_{i=1}^{n}%
x_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\\\text{(by the
definition of}\\\left(  x_{1},x_{2},...,x_{n}\right)  ^{\cont %
\left(  T\right)  }\text{)}}}\\
&  =\det\left(  \left(  x_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }.
\end{align*}
Thus,%
\begin{align*}
&  \det\left(  \left(  x_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}x_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }\\
&  =a_{\rho}\cdot\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,...,n\right\}  }}\left(  x_{1},x_{2}%
,...,x_{n}\right)  ^{\cont \left(  T\right)  }=a_{\lambda+\rho
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.jacobi-trudi.macdonald.preludium.2}}\right) \\
&  =\det\left(  \left(  x_{i}^{\left(  \lambda+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)  .
\end{align*}
This is an identity in the polynomial ring $\kk \left[  x_{1}%
,x_{2},...,x_{n}\right]  $. Substituting $y_{1},y_{2},...,y_{n}$ for
$x_{1},x_{2},...,x_{n}$ in this identity, we obtain%
\begin{align}
&  \det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }\nonumber\\
&  =\det\left(  \left(  y_{i}^{\left(  \lambda+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)  \label{sol.jacobi-trudi.macdonald.preludium.y-id}%
\end{align}
in the polynomial ring $\kk \left[  y_{1},y_{2},...,y_{n}\right]  $.
Notice that $\kk \left[  y_{1},y_{2},...,y_{n}\right]  \subset
\underbrace{\kk }_{\subset\mathbf{p}}\left[  \left[  y_{1}%
,y_{2},...,y_{n}\right]  \right]  \subset\mathbf{p}\left[  \left[  y_{1}%
,y_{2},...,y_{n}\right]  \right]  $, so we can regard
\eqref{sol.jacobi-trudi.macdonald.preludium.y-id} as an identity in
$\mathbf{p}\left[  \left[  y_{1},y_{2},...,y_{n}\right]  \right]  $ as well.

Now, forget that we fixed $\lambda$. We thus have proven that if $\lambda$ is
a partition having at most $n$ parts, then
\eqref{sol.jacobi-trudi.macdonald.preludium.y-id} holds.

\underline{\textit{Step 3: The morphism $\Gamma:\mathbf{p}\left[  \left[
 \yy \right]  \right]  \rightarrow\mathbf{p}\left[  \left[  y_{1}%
,y_{2},...,y_{n}\right]  \right]  $.}}

Now, let us consider both $\mathbf{p}$-algebras $\mathbf{p}\left[  \left[
 \yy \right]  \right]  $ and $\mathbf{p}\left[  \left[  y_{1}%
,y_{2},...,y_{n}\right]  \right]  $ as topological $\mathbf{p}$-algebras,
endowed with the usual topology\footnote{The usual topology on a ring $U$ of
formal power series over $\mathbf{p}$ is the direct-product topology obtained
by viewing the set $U$ as a direct product of many copies of $\mathbf{p}$ (this
is done by identifying every power series with the family of its
coefficients), each of which is endowed with the discrete topology. In this
topology, a sequence (or, more generally, a net) $\left(  f_{i}\right)  _{i}$
of power series converges to a power series $f$ if and only if for every
monomial $\mathfrak{m}$, the sequence (resp. net) $\left(  \text{the
coefficient of }\mathfrak{m}\text{ in }f_{i}\right)  _{i}$ converges to
$\left(  \text{the coefficient of }\mathfrak{m}\text{ in }f\right)  $ with
respect to the discrete topology.
\par
Notice that we are using $\mathbf{p}$ as the base ring here! Even though
$\mathbf{p}\left[  \left[   \yy \right]  \right]  =\kk \left[
\left[  \xx , \yy \right]  \right]  $ as $\mathbf{p}$-algebras, the
usual topology on the $\mathbf{p}$-algebra $\mathbf{p}\left[  \left[
 \yy \right]  \right]  $ is not the same as the usual topology on the
$\kk$-algebra $\kk \left[  \left[  \xx , \yy \right]
\right]  $.}. Then, there exists a unique continuous $\mathbf{p}$-algebra
homomorphism $\Gamma:\mathbf{p}\left[  \left[   \yy \right]  \right]
\rightarrow\mathbf{p}\left[  \left[  y_{1},y_{2},...,y_{n}\right]  \right]  $
satisfying%
\begin{equation}
\left(  \Gamma\left(  y_{i}\right)  =
\begin{cases}
y_{i},&\text{if }i\leq n;\\
0,&\text{if }i>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,3,...\right\}
\right)  . \label{sol.jacobi-trudi.macdonald.Gamma.2}%
\end{equation}
\footnote{This follows from the universal property of the power series ring
$\mathbf{p}\left[  \left[   \yy \right]  \right]  $.} Consider this
continuous $\mathbf{p}$-algebra homomorphism $\Gamma$.

Every $i\in\left\{  1,2,...,n\right\}  $ satisfies%
\begin{align}
\Gamma\left(  y_{i}\right)   &  =
\begin{cases}
y_{i},&\text{if }i\leq n;\\
0,&\text{if }i>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.jacobi-trudi.macdonald.Gamma.2}}\right) \nonumber\\
&  =y_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq n\text{ (since }%
i\in\left\{  1,2,...,n\right\}  \text{)}\right)  .
\label{sol.jacobi-trudi.macdonald.Gamma.2a}
\end{align}
Every $i\in\left\{  n+1,n+2,n+3,...\right\}  $ satisfies%
\begin{align}
\Gamma\left(  y_{i}\right)   &  =
\begin{cases}
y_{i},&\text{if }i\leq n;\\
0,&\text{if }i>n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.jacobi-trudi.macdonald.Gamma.2}}\right) \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>n\text{ (since }i\in\left\{
n+1,n+2,n+3,...\right\}  \text{)}\right)  .
\label{sol.jacobi-trudi.macdonald.Gamma.2b}
\end{align}
Notice that
\begin{equation}
\Gamma\left(  p\right)  =p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbf{p}
\label{sol.jacobi-trudi.macdonald.Gamma.p}%
\end{equation}
(since $\Gamma$ is a $\mathbf{p}$-algebra homomorphism). Thus,%
\begin{equation}
\Gamma\left(  s_{\lambda/\mu}\left(  \xx \right)  \right)
=s_{\lambda/\mu}\left(  \xx \right)  \ \ \ \ \ \ \ \ \ \ \text{for any
partitions }\lambda\text{ and }\mu
\label{sol.jacobi-trudi.macdonald.Gamma.p.schur}
\end{equation}
(by \eqref{sol.jacobi-trudi.macdonald.Gamma.p} (applied to $p=s_{\lambda/\mu
}\left(  \xx \right)  $) (since $s_{\lambda/\mu}\left(  \xx %
\right)  \in\kk \left[  \left[ \xx \right]  \right]  =\mathbf{p}$)).

It is now easy to see that
\begin{equation}
\Gamma\left(  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
=\prod_{i=1}^{\infty}\prod_{j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\label{sol.jacobi-trudi.macdonald.Gamma.prod}%
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.macdonald.Gamma.prod}:} We
know that $\Gamma$ is a continuous $\mathbf{p}$-algebra homomorphism. Hence,
\begin{align*}
\Gamma\left(  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
&  =\underbrace{\prod_{i,j=1}^{\infty}}_{=\prod_{i=1}^{\infty}\prod
_{j=1}^{\infty}}\left(  1-\underbrace{\Gamma\left(  x_{i}\right)
}_{\substack{=x_{i}\\\text{(by \eqref{sol.jacobi-trudi.macdonald.Gamma.p},
applied to }p=x_{i}\text{)}}}\Gamma\left(  y_{j}\right)  \right)  ^{-1}\\
&  =\prod_{i=1}^{\infty}\underbrace{\prod_{j=1}^{\infty}\left(  1-x_{i}%
\Gamma\left(  y_{j}\right)  \right)  ^{-1}}_{=\left(
\prod_{j=1}^{n}\left( 1-x_{i}\Gamma\left(  y_{j}\right)  \right)  ^{-1}
\right)
\cdot
\left(
\prod_{j=n+1}^{\infty}\left(  1-x_{i}\Gamma\left(  y_{j}\right)  \right)  ^{-1}
\right)
}\\
&  =\prod_{i=1}^{\infty}\left(
\left(
\prod_{j=1}^{n}\left(  1-x_{i}
\underbrace{\Gamma\left(  y_{j}\right)  }_{\substack{=y_{j}\\\text{(by
\eqref{sol.jacobi-trudi.macdonald.Gamma.2a}, applied}\\\text{to }i=j\text{)}%
}}\right)  ^{-1}
\right)
\cdot
\left(
\prod_{j=n+1}^{\infty}\left(  1-x_{i}\underbrace{\Gamma
\left(  y_{j}\right)  }_{\substack{=0\\\text{(by
\eqref{sol.jacobi-trudi.macdonald.Gamma.2b}, applied}\\\text{to }i=j\text{)}%
}}\right)  ^{-1}
\right)
\right) \\
&  =\prod_{i=1}^{\infty}\left(
\left( \prod_{j=1}^{n}\left(  1-x_{i}y_{j}\right) ^{-1} \right)
\cdot
\left( \prod_{j=n+1}^{\infty}\left(  \underbrace{1-x_{i}0}_{=1-0=1}\right)
^{-1} \right)
\right) \\
&  =
\prod_{i=1}^{\infty}\left(
\left( \prod_{j=1}^{n}\left(  1-x_{i}y_{j}\right) ^{-1} \right)
\cdot
\left( \prod_{j=n+1}^{\infty}\underbrace{1^{-1}}_{=1} \right)
\right)
=
\prod_{i=1}^{\infty}\left(
\left( \prod_{j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1} \right)
\cdot
\left( \underbrace{\prod_{j=n+1}^{\infty}1}_{=1} \right)
\right) \\
&=
\prod_{i=1}^{\infty}\prod_{j=1}^{n}\left(  1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
This proves \eqref{sol.jacobi-trudi.macdonald.Gamma.prod}.} We can rewrite
this further, obtaining
\begin{equation}
\Gamma\left(  \prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
=\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  .
\label{sol.jacobi-trudi.macdonald.Gamma.prodh}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.macdonald.Gamma.prodh}:}
The equality \eqref{h-generating-function-no-HE}
(with the summation index $n$ renamed as $q$) yields
\begin{equation}
\prod_{i=1}^{\infty} \left( 1 - x_i t \right) ^{-1}
= 1 + h_1 \left( \xx \right) t + h_2 \left( \xx \right) t^2 + ...
= \sum_{q \geq 0} h_q \left( \xx \right) t^q
\ \ \ \ \ \ \ \ \ \ \text{in }
\mathbf{p} \left[ \left[ t\right] \right] .
\label{sol.jacobi-trudi.macdonald.Gamma.prodh.pf.1}
\end{equation}
\par
For every $j\in\left\{  1,2,...,n\right\}$, we have
\begin{equation}
\prod_{i=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}
= \sum_{q \geq 0} h_q \left( \xx \right) y_j^q
\label{sol.jacobi-trudi.macdonald.Gamma.prodh.pf.2}
\end{equation}
(this follows by substituting $y_{j}$ for $t$ on both sides of
\eqref{sol.jacobi-trudi.macdonald.Gamma.prodh.pf.1}). Now,
\eqref{sol.jacobi-trudi.macdonald.Gamma.prod} becomes
\begin{align*}
\Gamma \left( 
 \prod_{i,j=1}^{\infty} \left( 1 - x_i y_j \right) ^{-1}
\right)
& = \underbrace{\prod_{i=1}^{\infty} \prod_{j=1}^{n}}_{= \prod_{j=1}^{n}
\prod_{i=1}^{\infty}} \left( 1 - x_i y_j \right) ^{-1}
= \prod_{j=1}^{n}
\underbrace{\prod_{i=1}^{\infty} \left( 1 - x_i y_j \right) ^{-1}}
_{\substack{= \sum_{q \geq 0} h_q \left( \xx \right) y_j^q \\
  \text{(by \eqref{sol.jacobi-trudi.macdonald.Gamma.prodh.pf.2})}}}\\
& = \prod_{j=1}^{n} \sum_{q \geq 0} h_q \left( \xx \right) y_j^q
= \sum_{\left( q_1, q_2, ..., q_n \right) \in \NN^n}
\underbrace{\prod_{j=1}^{n} \left( h_{q_j} \left( \xx \right) \cdot
y_j^{q_j} \right) }_{= \left( \prod_{j=1}^{n} h_{q_j} \left(
\xx \right) \right) \cdot\left(  \prod_{j=1}^{n}y_{j}^{q_{j}}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  .
\end{align*}
This proves \eqref{sol.jacobi-trudi.macdonald.Gamma.prodh}.}

Now, fix a partition $\lambda$ again. Recall the definition of the Schur
function $s_{\lambda}$. It yields
$s_{\lambda} = \sum_{T}{\xx }^{\cont \left( T\right)}$,
where $T$ runs through all column-strict tableaux of shape $\lambda$.
In other words,
\[
s_{\lambda}=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau
of shape }\lambda}}\underbrace{{\xx }^{\cont \left(
T\right)  }}_{=\prod_{i\geq1}x_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }}=\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda}}\prod_{i\geq1}x_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }.
\]
This is an identity in the power series ring $\kk \left[  \left[
\xx \right]  \right]  =\kk \left[  \left[  x_{1},x_{2}%
,x_{3},...\right]  \right]  $. Substituting $y_{1},y_{2},y_{3},...$ for
$x_{1},x_{2},x_{3},...$ in this identity, we obtain%
\begin{equation}
s_{\lambda}\left(   \yy \right)  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda}}\prod_{i\geq1}y_{i}%
^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\label{sol.jacobi-trudi.macdonald.slambda.y}%
\end{equation}
This is an identity in the power series ring $\kk \left[  \left[
 \yy \right]  \right]  =\kk \left[  \left[  y_{1},y_{2}%
,y_{3},...\right]  \right]  $, and thus can also be viewed as an identity in
the power series ring $\mathbf{p}\left[  \left[   \yy \right]  \right]
=\mathbf{p}\left[  \left[  y_{1},y_{2},y_{3},...\right]  \right]  $ (since
$\underbrace{\kk }_{\subset\mathbf{p}}\left[  \left[   \yy \right]
\right]  \subset\mathbf{p}\left[  \left[   \yy \right]  \right]  $).

Now, forget that we fixed $\lambda$. We thus have shown that every partition
$\lambda$ satisfies \eqref{sol.jacobi-trudi.macdonald.slambda.y}.

We now notice that if $T$ is any column-strict tableau such that all entries
of $T$ belong to $\left\{  1,2,...,n\right\}  $, then%
\begin{equation}
\Gamma\left(  \prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }.
\label{sol.jacobi-trudi.macdonald.Gamma.T.good}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.macdonald.Gamma.T.good}:}
Let $T$ be a column-strict tableau such that all entries of $T$ belong to
$\left\{  1,2,...,n\right\}  $.
\par
Let $i\in\left\{  n+1,n+2,n+3,...\right\}  $ be arbitrary. Then, $i>n$. Let
$c\in T^{-1}\left(  i\right)  $ be arbitrary. Then,
\[
c\in T^{-1}\left(  i\right)  =\left(  \text{the set of all cells of the shape
of }T\text{ which are filled with the entry }i\text{ in the tableau }T\right)
\]
(since $T^{-1}\left(  i\right)  $ is the set of all cells of the shape of $T$
which are filled with the entry $i$ in the tableau $T$). In other words, $c$
is a cell of the shape of $T$ which is filled with the entry $i$ in the
tableau $T$. Hence, the entry of the tableau $T$ in cell $c$ is $i$. Thus, $i$
is the entry of the tableau $T$ in cell $i$. In particular, this shows that
$i$ is an entry of $T$. Since all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $, this yields that $i$ belongs to $\left\{
1,2,...,n\right\}  $. Hence, $i\leq n$. This contradicts $i>n$.
\par
Now, forget that we fixed $c$. We thus have obtained a contradiction for every
$c\in T^{-1}\left(  i\right)  $. As a consequence, we see that there exists no
$c\in T^{-1}\left(  i\right)  $. In other words, $T^{-1}\left(  i\right)
=\varnothing$, so that $\left\vert \underbrace{T^{-1}\left(  i\right)
}_{=\varnothing}\right\vert =\left\vert \varnothing\right\vert =0$ and thus
$y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }=y_{i}^{0}=1$.
\par
Now, forget that we fixed $i$. We now have proven that $y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }=1$ for every $i\in\left\{
n+1,n+2,n+3,...\right\}  $. Hence, $\prod_{i=n+1}^{\infty}\underbrace{y_{i}%
^{\left\vert T^{-1}\left(  i\right)  \right\vert }}_{=1}=\prod_{i=n+1}%
^{\infty}1=1$. But%
\[
\underbrace{\prod_{i\geq1}}_{=\prod_{i=1}^{\infty}}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }=\prod_{i=1}^{\infty}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }=\prod_{i=1}^{n}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\cdot\underbrace{\prod_{i=n+1}^{\infty
}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }}_{=1}=\prod_{i=1}%
^{n}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\]
Applying the map $\Gamma$ to both sides of this equality, we obtain%
\begin{align*}
\Gamma\left(  \prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)   &  =\Gamma\left(  \prod_{i=1}^{n}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\right)  =\prod_{i=1}^{n}\left(
\underbrace{\Gamma\left(  y_{i}\right)  }_{\substack{=y_{i}\\\text{(by
\eqref{sol.jacobi-trudi.macdonald.Gamma.2a})}}}\right)  ^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\Gamma\text{ is a }\mathbf{p}\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }.
\end{align*}
This proves \eqref{sol.jacobi-trudi.macdonald.Gamma.T.good}.} On the other
hand, if $T$ is any column-strict tableau such that not all entries of $T$
belong to $\left\{  1,2,...,n\right\}  $, then%
\begin{equation}
\Gamma\left(  \prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =0.
\label{sol.jacobi-trudi.macdonald.Gamma.T.bad}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.jacobi-trudi.macdonald.Gamma.T.bad}:}
Let $T$ be a column-strict tableau such that not all entries of $T$ belong to
$\left\{  1,2,...,n\right\}  $.
\par
We know that not all entries of $T$ belong to $\left\{  1,2,...,n\right\}  $.
In other words, there exists an entry of $T$ which does not belong to
$\left\{  1,2,...,n\right\}  $. Let $j$ be this entry. Thus, $j$ is an entry
of $T$ which does not belong to $\left\{  1,2,...,n\right\}  $.
\par
We have $j\in\left\{  1,2,3,...\right\}  $ (since $j$ is an entry of $T$) but
$j\notin\left\{  1,2,...,n\right\}  $ (since $j$ does not belong to $\left\{
1,2,...,n\right\}  $). In other words, $j\in\left\{  1,2,3,...\right\}
\setminus\left\{  1,2,...,n\right\}  =\left\{  n+1,n+2,n+3,...\right\}  $.
Thus, \eqref{sol.jacobi-trudi.macdonald.Gamma.2b} (applied to $i=j$) yields
$\Gamma\left(  y_{j}\right)  =0$.
\par
Now, recall that $j$ is an entry of $T$. In other words, there exists a cell
$c$ of the shape of $T$ such that $j$ is the entry of the tableau $T$ in cell
$c$. Consider this cell $c$.
\par
We know that
\begin{equation}
T^{-1}\left(  j\right)  =\left(  \text{the set of all cells of the shape of
}T\text{ which are filled with the entry }j\text{ in the tableau }T\right)
\label{sol.jacobi-trudi.macdonald.Gamma.T.bad.pf.1}
\end{equation}
(since $T^{-1}\left(  j\right)  $ is the set of all cells of the shape of $T$
which are filled with the entry $j$ in the tableau $T$). But $j$ is the entry
of the tableau $T$ in cell $c$. In other words, the cell $c$ is filled with
the entry $j$ in the tableau $T$. Also, $c$ is a cell of the shape of $T$.
Hence, $c$ is a cell of the shape of $T$ which is filled with the entry $j$ in
the tableau $T$. In other words,
\begin{align*}
c  &  \in\left(  \text{the set of all cells of the shape of }T\text{ which are
filled with the entry }j\text{ in the tableau }T\right) \\
&  =T^{-1}\left(  j\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.jacobi-trudi.macdonald.Gamma.T.bad.pf.1}}\right)  .
\end{align*}
Hence, the set $T^{-1}\left(  j\right)  $ contains at least one element
(namely, $c$), and thus is nonempty. Therefore, $\left\vert T^{-1}\left(
j\right)  \right\vert >0$. But since $\Gamma\left(  y_{j}\right)  =0$, we have
$\left(  \underbrace{\Gamma\left(  y_{j}\right)  }_{=0}\right)  ^{\left\vert
T^{-1}\left(  j\right)  \right\vert }=0^{\left\vert T^{-1}\left(  j\right)
\right\vert }=0$ (since $\left\vert T^{-1}\left(  j\right)  \right\vert >0$).
\par
But $j\in\left\{  1,2,3,...\right\}  $, so that%
\[
\prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }%
=y_{j}^{\left\vert T^{-1}\left(  j\right)  \right\vert }\cdot\prod\limits
_{\substack{i\geq1;\\i\neq j}}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }.
\]
Applying the map $\Gamma$ to both sides of this equality, we obtain%
\begin{align*}
\Gamma\left(  \prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)   &  =\Gamma\left(  y_{j}^{\left\vert T^{-1}\left(
j\right)  \right\vert }\cdot\prod\limits_{\substack{i\geq1;\\i\neq j}}y_{i}%
^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right) \\
&  =\underbrace{\left(  \Gamma\left(  y_{j}\right)  \right)  ^{\left\vert
T^{-1}\left(  j\right)  \right\vert }}_{=0}\cdot\Gamma\left(  \prod\limits
_{\substack{i\geq1;\\i\neq j}}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is
a }\mathbf{p}\text{-algebra homomorphism}\right) \\
&  =0.
\end{align*}
This proves \eqref{sol.jacobi-trudi.macdonald.Gamma.T.bad}.} We can use these
results to show the following: If $\lambda$ is a partition, then%
\begin{equation}
\Gamma\left(  s_{\lambda}\left(   \yy \right)  \right)  =\sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,...,n\right\}  }}\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }. \label{sol.jacobi-trudi.macdonald.Gamma.T.verygood}%
\end{equation}
\footnote{\textit{Proof of
\eqref{sol.jacobi-trudi.macdonald.Gamma.T.verygood}:} Let $\lambda$ be
a partition. Then, applying the map $\Gamma$ to both
sides of the identity \eqref{sol.jacobi-trudi.macdonald.slambda.y}, we obtain
\begin{align*}
\Gamma\left(  s_{\lambda}\left(   \yy \right)  \right)   &
=\Gamma\left(  \sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda}}\prod_{i\geq1}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\right)  =\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda}}\Gamma\left(
\prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is a continuous
}\mathbf{p}\text{-algebra homomorphism}\right) \\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,...,n\right\}  }}\underbrace{\Gamma\left(  \prod_{i\geq1}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\right)  }_{\substack{=\prod_{i=1}%
^{n}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\\\text{(by
\eqref{sol.jacobi-trudi.macdonald.Gamma.T.good})}}}+\sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{not all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\underbrace{\Gamma\left(  \prod_{i\geq1}%
y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)  }%
_{\substack{=0\\\text{(by \eqref{sol.jacobi-trudi.macdonald.Gamma.T.bad})}}}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,...,n\right\}  }}\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }+\underbrace{\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda\text{;}\\\text{not all entries of }T\text{
belong}\\\text{to }\left\{  1,2,...,n\right\}  }}0}_{=0}=\sum
\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,...,n\right\}  }}\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }.
\end{align*}
This proves (\ref{sol.jacobi-trudi.macdonald.Gamma.T.verygood}).} On the other
hand, if $\lambda$ is any partition having more than $n$ parts\footnote{Recall
that a \emph{part} of a partition means a nonzero entry of the partition.},
then
\begin{equation}
\Gamma\left(  s_{\lambda}\left(   \yy \right)  \right)  =0.
\label{sol.jacobi-trudi.macdonald.Gamma.T.verybad}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad}%
):} Let $\lambda$ be a partition having more than $n$ parts. Then,
$\ell\left(  \lambda\right)  >n$ (since $\lambda$ has more than $n$ parts).
Thus, $\ell\left(  \lambda\right)  \geq n+1$ (since $\ell\left(
\lambda\right)  $ and $n$ are integers), so that $\lambda_{n+1}\neq0$. Thus,
$\lambda_{n+1}>0$ (since $\lambda_{n+1}\geq0$). Thus, $\lambda_{n+1}\geq1$
(since $\lambda_{n+1}$ is an integer).
\par
We know that $\lambda$ is a partition. Hence, $\lambda_{1}\geq\lambda_{2}%
\geq\lambda_{3}\geq...$. In other words, any two positive integers $u$ and $v$
satisfying $u\leq v$ satisfy $\lambda_{u}\geq\lambda_{v}$.
\par
Let $T$ be any column-strict tableau of shape $\lambda$. We are going to show
that not all entries of $T$ belong to $\left\{  1,2,...,n\right\}  $.
\par
In fact, assume the contrary. Thus, all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $.
\par
We know that $T$ is a column-strict tableau of shape $\lambda$. In other
words, $T$ is an assignment of entries of $\left\{  1,2,3,...\right\}  $ to
the cells of the Ferrers diagram of $\lambda$, weakly increasing left-to-right
in rows, and strictly increasing top-to-bottom in columns (because this is how
a column-strict tableau of shape $\lambda$ was defined).
\par
Let $Y\left(  \lambda\right)  $ denote the Ferrers diagram of $\lambda$. Then,%
\begin{equation}
Y\left(  \lambda\right)  =\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ q\leq\lambda_{p}\right\}
\label{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.1}%
\end{equation}
(because this is how $Y\left(  \lambda\right)  $ is defined). Also, the cells
of the Ferrers diagram of $\lambda$ are the elements of $Y\left(
\lambda\right)  $ (since $Y\left(  \lambda\right)  $ is the Ferrers diagram of
$\lambda$).
\par
We know that $T$ is an assignment of entries of $\left\{  1,2,3,...\right\}  $
to the cells of the Ferrers diagram of $\lambda$. In other words, $T$ is an
assignment of entries of $\left\{  1,2,3,...\right\}  $ to the elements of
$Y\left(  \lambda\right)  $ (since the cells of the Ferrers diagram of
$\lambda$ are the elements of $Y\left(  \lambda\right)  $). In other words,
$T$ is a map $Y\left(  \lambda\right)  \rightarrow\left\{  1,2,3,...\right\}
$. We have%
\begin{equation}
T\left(  \left(  p,q\right)  \right)  <T\left(  \left(  p^{\prime},q\right)
\right)  \ \ \ \ \ \ \ \ \ \ \text{for any integers }p\text{, }p^{\prime
}\text{ and }q\text{ satisfying }\left(  p,q\right)  \in Y\left(
\lambda\right)  \text{, }\left(  p^{\prime},q\right)  \in Y\left(
\lambda\right)  \text{ and }p<p^{\prime}
\label{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.2}%
\end{equation}
(because $T$ is strictly increasing top-to-bottom in columns).
\par
Now, let $i\in\left\{  1,2,...,n+1\right\}  $ be arbitrary. Then, $i\leq n+1$
and $\left(  i,1\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ (since
$i\in\left\{  1,2,...,n+1\right\}  \subset\left\{  1,2,3,...\right\}  $ and
$1\in\left\{  1,2,3,...\right\}  $). Recall that any two positive integers $u$
and $v$ satisfying $u\leq v$ satisfy $\lambda_{u}\geq\lambda_{v}$. Applying
this to $u=i$ and $v=n+1$, we obtain $\lambda_{i}\geq\lambda_{n+1}\geq1$ and
thus $1\leq\lambda_{i}$. Now, $\left(  i,1\right)  $ is an element of the set
$\left\{  1,2,3,...\right\}  ^{2}$ satisfying $1\leq\lambda_{i}$. In other
words, $\left(  i,1\right)  $ is an element $\left(  p,q\right)  $ of the set
$\left\{  1,2,3,...\right\}  ^{2}$ satisfying $q\leq\lambda_{p}$. In other
words,%
\[
\left(  i,1\right)  \in\left\{  \left(  p,q\right)  \in\left\{
1,2,3,...\right\}  ^{2}\ \mid\ q\leq\lambda_{p}\right\}  =Y\left(
\lambda\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.1})}\right)  .
\]
\par
Now, forget that we fixed $i$. We thus have shown that
\begin{equation}
\left(  i,1\right)  \in Y\left(  \lambda\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  1,2,...,n+1\right\}  .
\label{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.3}%
\end{equation}
Hence, $T\left(  \left(  i,1\right)  \right)  $ is well-defined for every
$i\in\left\{  1,2,...,n+1\right\}  $ (because $T$ is a map $Y\left(
\lambda\right)  \rightarrow\left\{  1,2,3,...\right\}  $). Applied to $i=n+1$,
this yields that $T\left(  \left(  n+1,1\right)  \right)  $ is well-defined.
Now, $T\left(  \left(  n+1,1\right)  \right)  $ is the entry of the tableau
$T$ in the cell $\left(  n+1,1\right)  $. Thus, in particular, $T\left(
\left(  n+1,1\right)  \right)  $ is an entry of $T$, and therefore belongs to
$\left\{  1,2,...,n\right\}  $ (since all entries of $T$ belong to $\left\{
1,2,...,n\right\}  $). Hence, $T\left(  \left(  n+1,1\right)  \right)  \leq
n$.
\par
Let now $i\in\left\{  1,2,...,n\right\}  $. Then, $i\in\left\{
1,2,...,n\right\}  \subset\left\{  1,2,...,n+1\right\}  $, so that $\left(
i,1\right)  \in Y\left(  \lambda\right)  $ (by
(\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.3})). Also, $i\in\left\{
1,2,...,n\right\}  $ shows that $i+1\in\left\{  2,3,...,n+1\right\}
\subset\left\{  1,2,...,n+1\right\}  $, so that $\left(  i+1,1\right)  \in
Y\left(  \lambda\right)  $ (by
(\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.3}), applied to $i+1$
instead of $i$). Also, $i<i+1$. Hence,
(\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad.pf.2}) (applied to $p=i$,
$p^{\prime}=i+1$ and $q=1$) yields $T\left(  \left(  i,1\right)  \right)
<T\left(  \left(  i+1,1\right)  \right)  $. But $T\left(  \left(  i,1\right)
\right)  $ and $T\left(  \left(  i+1,1\right)  \right)  $ are elements of
$\left\{  1,2,3,...\right\}  $ (since $T$ is a map $Y\left(  \lambda\right)
\rightarrow\left\{  1,2,3,...\right\}  $) and thus integers. Hence, from
$T\left(  \left(  i,1\right)  \right)  <T\left(  \left(  i+1,1\right)
\right)  $, we obtain $T\left(  \left(  i,1\right)  \right)  \leq T\left(
\left(  i+1,1\right)  \right)  -1$. Thus, $T\left(  \left(  i+1,1\right)
\right)  -1\geq T\left(  \left(  i,1\right)  \right)  $, so that $T\left(
\left(  i+1,1\right)  \right)  -T\left(  \left(  i,1\right)  \right)  \geq1$.
\par
Now, forget that we fixed $i$. We thus have proven that every $i\in\left\{
1,2,...,n\right\}  $ satisfies%
\[
T\left(  \left(  i+1,1\right)  \right)  -T\left(  \left(  i,1\right)  \right)
\geq1.
\]
Summing these inequalities over all $i\in\left\{  1,2,...,n\right\}  $, we
obtain%
\[
\sum_{i\in\left\{  1,2,...,n\right\}  }\left(  T\left(  \left(  i+1,1\right)
\right)  -T\left(  \left(  i,1\right)  \right)  \right)  \geq\sum
_{i\in\left\{  1,2,...,n\right\}  }1.
\]
Hence,%
\begin{align*}
\sum_{i\in\left\{  1,2,...,n\right\}  }1  &  \leq\underbrace{\sum
_{i\in\left\{  1,2,...,n\right\}  }}_{=\sum_{i=1}^{n}}\left(  T\left(  \left(
i+1,1\right)  \right)  -T\left(  \left(  i,1\right)  \right)  \right) \\
&  =\sum_{i=1}^{n}\left(  T\left(  \left(  i+1,1\right)  \right)  -T\left(
\left(  i,1\right)  \right)  \right)  =\underbrace{\sum_{i=1}^{n}T\left(
\left(  i+1,1\right)  \right)  }_{\substack{=\sum_{i=2}^{n+1}T\left(  \left(
i,1\right)  \right)  \\\text{(here, we substituted }i\text{ for }i+1\\\text{in
the sum)}}}-\sum_{i=1}^{n}T\left(  \left(  i,1\right)  \right) \\
&  =\underbrace{\sum_{i=2}^{n+1}T\left(  \left(  i,1\right)  \right)  }%
_{=\sum_{i=2}^{n}T\left(  \left(  i,1\right)  \right)  +T\left(  \left(
n+1,1\right)  \right)  }-\underbrace{\sum_{i=1}^{n}T\left(  \left(
i,1\right)  \right)  }_{=T\left(  \left(  1,1\right)  \right)  +\sum_{i=2}%
^{n}T\left(  \left(  i,1\right)  \right)  }\\
&  =\left(  \sum_{i=2}^{n}T\left(  \left(  i,1\right)  \right)  +T\left(
\left(  n+1,1\right)  \right)  \right)  -\left(  T\left(  \left(  1,1\right)
\right)  +\sum_{i=2}^{n}T\left(  \left(  i,1\right)  \right)  \right) \\
&  =\underbrace{T\left(  \left(  n+1,1\right)  \right)  }_{\substack{\leq
n}}-\underbrace{T\left(  \left(  1,1\right)  \right)  }_{\substack{\geq
1\\\text{(since }T\left(  \left(  1,1\right)  \right)  \in\left\{
1,2,3,...\right\}  \\\text{(because }T\text{ is a map }Y\left(  \lambda
\right)  \rightarrow\left\{  1,2,3,...\right\}  \text{))}}}\leq n-1<n,
\end{align*}
which contradicts $\sum_{i\in\left\{  1,2,...,n\right\}  }%
1=\underbrace{\left\vert \left\{  1,2,...,n\right\}  \right\vert }_{=n}%
\cdot1=n\cdot1=n$. This contradiction shows that our assumption was wrong.
Hence, not all entries of $T$ belong to $\left\{  1,2,...,n\right\}  $. Thus,
$\Gamma\left(  \prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)
\right\vert }\right)  =0$ (according to
(\ref{sol.jacobi-trudi.macdonald.Gamma.T.bad})).
\par
Now, forget that we fixed $T$. We thus have proven that $\Gamma\left(
\prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\right)
=0$ for every column-strict tableau $T$ of shape $\lambda$. Now, applying the
map $\Gamma$ to both sides of the identity
(\ref{sol.jacobi-trudi.macdonald.slambda.y}), we obtain%
\begin{align*}
\Gamma\left(  s_{\lambda}\left(   \yy \right)  \right)   &
=\Gamma\left(  \sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda}}\prod_{i\geq1}y_{i}^{\left\vert
T^{-1}\left(  i\right)  \right\vert }\right)  =\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda}}\underbrace{\Gamma
\left(  \prod_{i\geq1}y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert
}\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is a continuous
}\mathbf{p}\text{-algebra homomorphism}\right) \\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda}}0=0.
\end{align*}
This proves (\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad}).}

\underline{\textit{Step 4: Adapting the skew Cauchy identity.}}

Now, fix a partition $\mu$ having at most $n$ parts. Exercise
\ref{exe.cauchy.skew}(a) yields
\begin{align*}
\sum_{\lambda\in\Par }s_{\lambda}\left(  {\xx }\right)
s_{\lambda/\mu}\left( \yy \right)   &  =\underbrace{s_{\mu}\left(
{\xx }\right)  }_{=s_{\mu}\left(  x_{1},x_{2},x_{3},...\right)  }%
\cdot\underbrace{\prod_{i,j=1}^{\infty}}_{=\prod_{j,i=1}^{\infty}}\left(
1-\underbrace{x_{i}y_{j}}_{=y_{j}x_{i}}\right)  ^{-1}\\
&  =s_{\mu}\left(  x_{1},x_{2},x_{3},...\right)  \cdot\prod_{j,i=1}^{\infty
}\left(  1-y_{j}x_{i}\right)  ^{-1}\\
&  =s_{\mu}\left(  x_{1},x_{2},x_{3},...\right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-y_{i}x_{j}\right)  ^{-1}%
\end{align*}
(here, we renamed the index $\left(  j,i\right)  $ as $\left(  i,j\right)  $
in the product) in the ring $\kk \left[  \left[  \xx %
, \yy \right]  \right]  $. Thus,%
\begin{align*}
s_{\mu}\left(  x_{1},x_{2},x_{3},...\right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-y_{i}x_{j}\right)  ^{-1}  &  =\sum_{\lambda\in\Par %
}\underbrace{s_{\lambda}\left(  {\xx }\right)  }_{=s_{\lambda}\left(
x_{1},x_{2},x_{3},...\right)  }\underbrace{s_{\lambda/\mu}\left(  { \yy %
}\right)  }_{=s_{\lambda/\mu}\left(  y_{1},y_{2},y_{3},...\right)  }\\
&  =\sum_{\lambda\in\Par }s_{\lambda}\left(  x_{1},x_{2}%
,x_{3},...\right)  s_{\lambda/\mu}\left(  y_{1},y_{2},y_{3},...\right)  .
\end{align*}
Substituting $y_{1},y_{2},y_{3},...,x_{1},x_{2},x_{3},...$ for $x_{1}%
,x_{2},x_{3},...,y_{1},y_{2},y_{3},...$ in this identity, we obtain%
\[
s_{\mu}\left(  y_{1},y_{2},y_{3},...\right)  \cdot\prod_{i,j=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}=\sum_{\lambda\in\Par %
}\underbrace{s_{\lambda}\left(  y_{1},y_{2},y_{3},...\right)  }_{=s_{\lambda
}\left( \yy \right)  }\underbrace{s_{\lambda/\mu}\left(  x_{1}%
,x_{2},x_{3},...\right)  }_{=s_{\lambda/\mu}\left(  {\xx }\right)
}=\sum_{\lambda\in\Par }s_{\lambda}\left( \yy \right)
s_{\lambda/\mu}\left(  {\xx }\right)  .
\]
Thus,%
\[
\sum_{\lambda\in\Par }s_{\lambda}\left( \yy \right)
s_{\lambda/\mu}\left(  {\xx }\right)  =\underbrace{s_{\mu}\left(
y_{1},y_{2},y_{3},...\right)  }_{=s_{\mu}\left(   \yy \right)  }%
\cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}=s_{\mu}\left(
 \yy \right)  \cdot\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)
^{-1}.
\]
This is an identity in the ring $\kk \left[  \left[  {\xx %
},{ \yy }\right]  \right]  =\mathbf{p}\left[  \left[   \yy \right]
\right]  $, and thus we can apply the map $\Gamma$ to both sides of this
identity. As a result, we obtain%
\begin{align*}
&  \Gamma\left(  \sum_{\lambda\in\Par }s_{\lambda}\left(
 \yy \right)  s_{\lambda/\mu}\left(  {\xx }\right)  \right) \\
&  =\Gamma\left(  s_{\mu}\left( \yy \right)  \cdot\prod
_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)  =\Gamma\left(
s_{\mu}\left( \yy \right)  \right)  \cdot\underbrace{\Gamma\left(
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}\right)
}_{\substack{=\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^{n}%
}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)
\cdot\left(  \prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.Gamma.prodh}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is a }\mathbf{p}%
\text{-algebra homomorphism}\right) \\
&  =\Gamma\left(  s_{\mu}\left( \yy \right)  \right)  \cdot
\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \Gamma\left(  \sum_{\lambda\in\Par }s_{\lambda}\left(
 \yy \right)  s_{\lambda/\mu}\left(  {\xx }\right)  \right) \\
&  =\sum_{\lambda\in\Par }\Gamma\left(  s_{\lambda}\left(
{ \yy }\right)  \right)  \underbrace{\Gamma\left(  s_{\lambda/\mu}\left(
{\xx }\right)  \right)  }_{\substack{=s_{\lambda/\mu}\left(
\xx \right)  \\\text{(by (\ref{sol.jacobi-trudi.macdonald.Gamma.p.schur}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Gamma\text{ is a continuous
}\mathbf{p}\text{-algebra homomorphism}\right) \\
&  =\sum_{\lambda\in\Par }\Gamma\left(  s_{\lambda}\left(
{ \yy }\right)  \right)  s_{\lambda/\mu}\left(  \xx \right) \\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition
}\lambda\text{ has}\\\text{at most }n\text{ parts}}}\Gamma\left(  s_{\lambda
}\left( \yy \right)  \right)  s_{\lambda/\mu}\left(  \xx %
\right)  +\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition
}\lambda\text{ has}\\\text{more than }n\text{ parts}}}\underbrace{\Gamma
\left(  s_{\lambda}\left( \yy \right)  \right)  }%
_{\substack{=0\\\text{(by (\ref{sol.jacobi-trudi.macdonald.Gamma.T.verybad}%
))}}}s_{\lambda/\mu}\left(  \xx \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because every partition }\lambda\text{
either has at most }n\text{ parts or has more than }n\text{ parts, but not
both at once}\right) \\
&  =\Gamma\left(  s_{\lambda}\left( \yy \right)  \right)
s_{\lambda/\mu}\left(  \xx \right)  +\underbrace{\sum\limits_{\substack{\lambda
\in\Par ;\\\text{the partition }\lambda\text{ has}\\\text{more
than }n\text{ parts}}}0s_{\lambda/\mu}\left(  \xx \right)  }_{=0}%
=\Gamma\left(  s_{\lambda}\left( \yy \right)  \right)  s_{\lambda
/\mu}\left(  \xx \right)  ,
\end{align*}
this yields%
\begin{equation}
\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition }%
\lambda\text{ has}\\\text{at most }n\text{ parts}}}\Gamma\left(  s_{\lambda
}\left( \yy \right)  \right)  s_{\lambda/\mu}\left(  \xx %
\right)  =\Gamma\left(  s_{\mu}\left(   \yy \right)  \right)  \cdot
\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  .
\label{sol.jacobi-trudi.macdonald.5}
\end{equation}


\underline{\textit{Step 5: Rewriting the (adapted) skew Cauchy identity using
alternants.}}

Now, every partition $\lambda$ having at most $n$ parts satisfies%
\begin{align}
&  \det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\underbrace{\Gamma\left(  s_{\lambda}\left( \yy \right)
\right)  }_{\substack{=\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}%
y_{i}^{\left\vert T^{-1}\left(  i\right)  \right\vert }\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.Gamma.T.verygood}))}}}\nonumber\\
&  =\det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,...,n\right\}  }}\prod_{i=1}^{n}y_{i}^{\left\vert T^{-1}\left(
i\right)  \right\vert }=\det\left(  \left(  y_{i}^{\left(  \lambda
+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)
\label{sol.jacobi-trudi.macdonald.6}%
\end{align}
(by (\ref{sol.jacobi-trudi.macdonald.preludium.y-id})). Applying this to
$\lambda=\mu$, we obtain%
\begin{equation}
\det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\Gamma\left(  s_{\mu}\left( \yy \right)  \right)  =\det\left(
\left(  y_{i}^{\left(  \mu+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)
. \label{sol.jacobi-trudi.macdonald.6.mu}%
\end{equation}
Now, multiplying both sides of the identity
(\ref{sol.jacobi-trudi.macdonald.5}) with $\det\left(  \left(  y_{i}^{\rho
_{j}}\right)  _{i,j=1,2,...,n}\right)  $, we obtain%
\begin{align*}
&  \det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition
}\lambda\text{ has}\\\text{at most }n\text{ parts}}}\Gamma\left(  s_{\lambda
}\left( \yy \right)  \right)  s_{\lambda/\mu}\left(  \xx %
\right) \\
&  =\underbrace{\det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}%
\right)  \cdot\Gamma\left(  s_{\mu}\left(   \yy \right)  \right)
}_{\substack{=\det\left(  \left(  y_{i}^{\left(  \mu+\rho\right)  _{j}%
}\right)  _{i,j=1,2,...,n}\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.6.mu}))}}}\cdot\sum_{\left(  q_{1}%
,q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}^{n}h_{q_{j}%
}\left(  \xx \right)  \right)  \cdot\left(  \prod_{j=1}^{n}y_{j}^{q_{j}%
}\right) \\
&  =\det\left(  \left(  y_{i}^{\left(  \mu+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)  \cdot\sum_{\left(  q_{1},q_{2},...,q_{n}\right)
\in\NN^n}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)
\right)  \cdot\left(  \prod_{j=1}^{n}y_{j}^{q_{j}}\right) \\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \det\left(  \left(  y_{i}^{\left(
\mu+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)  .
\end{align*}
Thus,%
\begin{align}
&  \sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \det\left(  \left(  y_{i}^{\left(
\mu+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right) \nonumber\\
&  =\det\left(  \left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)
\cdot\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition
}\lambda\text{ has}\\\text{at most }n\text{ parts}}}\Gamma\left(  s_{\lambda
}\left( \yy \right)  \right)  s_{\lambda/\mu}\left(  \xx %
\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition
}\lambda\text{ has}\\\text{at most }n\text{ parts}}}\underbrace{\det\left(
\left(  y_{i}^{\rho_{j}}\right)  _{i,j=1,2,...,n}\right)  \cdot\Gamma\left(
s_{\lambda}\left( \yy \right)  \right)  }_{\substack{=\det\left(
\left(  y_{i}^{\left(  \lambda+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}
\right)  \\\text{(by (\ref{sol.jacobi-trudi.macdonald.6}))}}}
s_{\lambda/\mu} \left( \xx \right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par ;\\\text{the partition
}\lambda\text{ has}\\\text{at most }n\text{ parts}}}\det\left(  \left(
y_{i}^{\left(  \lambda+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)
s_{\lambda/\mu}\left(  \xx \right)  =\sum\limits_{\substack{\lambda
\in\Par ;\\\text{the partition }\lambda\text{ has}\\\text{at most
}n\text{ parts}}}s_{\lambda/\mu}\left(  \xx \right)  \det\left(  \left(
y_{i}^{\left(  \lambda+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)
\nonumber\\
&  =\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\det\left(  \left(  y_{i}^{\left(  \nu+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)
\label{sol.jacobi-trudi.macdonald.8}
\end{align}
(here, we renamed the summation index $\lambda$ as $\nu$).

\underline{\textit{Step 6: Expanding the determinants.}}

Now, let us recall a basic fact from linear algebra, namely the explicit
formula for the determinant of a matrix as a sum over permutations: Any matrix
$\left(  \alpha_{i,j}\right)  _{i,j=1,2,...,\ell}$ over a commutative ring
satisfies
\begin{equation}
\det\left(  \left(  \alpha_{i,j}\right)  _{i,j=1,2,...,\ell}\right)
=\sum_{\sigma\in\Symm_{\ell}}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{\ell}\alpha_{i,\sigma\left(  i\right)  }.
\label{sol.jacobi-trudi.macdonald.detexp}
\end{equation}
Applying (\ref{sol.jacobi-trudi.macdonald.detexp}) to $\ell=n$ and
$\alpha_{i,j}=y_{i}^{\left(  \mu+\rho\right)  _{j}}$, we obtain
\[
\det\left(  \left(  y_{i}^{\left(  \mu+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)
^{\sigma}\prod_{i=1}^{n}y_{i}^{\left(  \mu+\rho\right)  _{\sigma\left(
i\right)  }}=\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}
\prod_{j=1}^{n}y_{j}^{\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}
\]
(here, we renamed the index $i$ as $j$ in the product). Thus,
\begin{align*}
&  \sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \underbrace{\det\left(  \left(
y_{i}^{\left(  \mu+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right)
}_{=\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\prod
_{j=1}^{n}y_{j}^{\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}}\\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \sum_{\sigma\in\Symm_{n}}\left(
-1\right)  ^{\sigma}\prod_{j=1}^{n}y_{j}^{\left(  \mu+\rho\right)
_{\sigma\left(  j\right)  }}\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\underbrace{\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \left(  \prod_{j=1}^{n}y_{j}^{\left(
\mu+\rho\right)  _{\sigma\left(  j\right)  }}\right)  }_{=\prod_{j=1}%
^{n}\left(  y_{j}^{q_{j}}y_{j}^{\left(  \mu+\rho\right)  _{\sigma\left(
j\right)  }}\right)  }\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(  \prod_{j=1}%
^{n}\underbrace{\left(  y_{j}^{q_{j}}y_{j}^{\left(  \mu+\rho\right)
_{\sigma\left(  j\right)  }}\right)  }_{=y_{j}^{q_{j}+\left(  \mu+\rho\right)
_{\sigma\left(  j\right)  }}}\right) \\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left( 
 \prod_{j=1}^{n}
 y_{j}^{q_{j} + \left( \mu+\rho \right) _{\sigma\left( j \right)}}
\right)
.
\end{align*}
Hence,
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(  \prod_{j=1}%
^{n}y_{j}^{q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}\right)
\nonumber\\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\left(
\prod_{j=1}^{n}y_{j}^{q_{j}}\right)  \det\left(  \left(  y_{i}^{\left(
\mu+\rho\right)  _{j}}\right)  _{i,j=1,2,...,n}\right) \nonumber\\
&  =\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\underbrace{\det\left(  \left(  y_{i}^{\left(  \nu+\rho\right)  _{j}}\right)
_{i,j=1,2,...,n}\right)  }_{\substack{=\sum_{\sigma\in\Symm_{n}}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{n}y_{i}^{\left(  \nu+\rho\right)
_{\sigma\left(  i\right)  }}\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.detexp}), applied to}\\\ell=n\text{ and
}\alpha_{i,j}=y_{i}^{\left(  \nu+\rho\right)  _{j}}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.jacobi-trudi.macdonald.8}%
)}\right) \nonumber\\
&  =\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}y_{i}^{\left(  \nu+\rho\right)  _{\sigma\left(  i\right)  }}=\sum
_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum\limits_{\substack{\nu
\in\Par ;\\\text{the partition }\nu\text{ has}\\\text{at most
}n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)  \prod_{i=1}^{n}%
y_{i}^{\left(  \nu+\rho\right)  _{\sigma\left(  i\right)  }}\nonumber\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}%
\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\prod_{j=1}^{n}y_{j}^{\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }}
\label{sol.jacobi-trudi.macdonald.12}%
\end{align}
(here, we renamed the index $i$ as $j$ in the product).

\underline{\textit{Step 7: Extracting coefficients.}}

Now, for every $\left(  r_{1},r_{2},...,r_{n}\right)  \in\NN^n$,
define a map
$\operatorname{coeff}_{\left(  r_{1},r_{2},...,r_{n}\right)  }
 : \mathbf{p}\left[ \left[ \yy \right] \right] \rightarrow
   \mathbf{p}$
by
\[
\operatorname{coeff}_{\left(  r_{1},r_{2},...,r_{n}\right)  }\left(
p\right)  =\left(  \text{the coefficient of the monomial }y_{1}^{r_{1}}%
y_{2}^{r_{2}}...y_{n}^{r_{n}}\text{ in }p\right)
\]
(where, of course, we regard $p$ as a power series over $\mathbf{p}$, not as a
power series over $\kk$, so that (for example) the coefficient of the
monomial $y_{1}$ in $1+x_{1}+y_{1}+x_{1}y_{1}$ is $1+x_{1}$ rather than $1$).
This map $\operatorname{coeff}_{\left(  r_{1},r_{2},...,r_{n}%
\right)  }$ is simply the map projecting every power series $p\in
\mathbf{p}\left[  \left[   \yy \right]  \right]  $ onto its coefficient
before the monomial $y_{1}^{r_{1}}y_{2}^{r_{2}}...y_{n}^{r_{n}}$; thus, this
map $\operatorname{coeff}_{\left(  r_{1},r_{2},...,r_{n}\right)  }$
is a continuous $\mathbf{p}$-linear map (for every $\left(  r_{1}%
,r_{2},...,r_{n}\right)  \in\NN^n$). Moreover, for any $\left(
r_{1},r_{2},...,r_{n}\right)  \in\NN^n$ and $\left(  t_{1}%
,t_{2},...,t_{n}\right)  \in\NN^n$, we have%
\begin{align*}
\operatorname{coeff}_{\left(  r_{1},r_{2},...,r_{n}\right)  }\left(
\underbrace{\prod_{j=1}^{n}y_{j}^{t_{j}}}_{=y_{1}^{t_{1}}y_{2}^{t_{2}}%
...y_{n}^{t_{n}}}\right)   &  =\operatorname{coeff}_{\left(
r_{1},r_{2},...,r_{n}\right)  }\left(  y_{1}^{t_{1}}y_{2}^{t_{2}}%
...y_{n}^{t_{n}}\right) \\
&  =\left(  \text{the coefficient of the monomial }y_{1}^{r_{1}}y_{2}^{r_{2}%
}...y_{n}^{r_{n}}\text{ in }y_{1}^{t_{1}}y_{2}^{t_{2}}...y_{n}^{t_{n}}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map
}\operatorname{coeff}_{\left(  r_{1},r_{2},...,r_{n}\right)
}\right) \\
&  =\delta_{\left(  r_{1},r_{2},...,r_{n}\right)  ,\left(  t_{1}%
,t_{2},...,t_{n}\right)  }.
\end{align*}
Renaming $\left(  r_{1},r_{2},...,r_{n}\right)  $ as $r$ in this statement, we
conclude the following: For any $r\in\NN^n$ and $\left(  t_{1}%
,t_{2},...,t_{n}\right)  \in\NN^n$, we have%
\begin{equation}
\operatorname{coeff}_{r}\left(  \prod_{j=1}^{n}y_{j}^{t_{j}}\right)
=\delta_{r,\left(  t_{1},t_{2},...,t_{n}\right)  }.
\label{sol.jacobi-trudi.macdonald.coeff}
\end{equation}


Now, let $\lambda$ be a partition having at most $n$ parts. Applying the map
$\operatorname{coeff}_{\lambda+\rho}$ to both sides of the identity
(\ref{sol.jacobi-trudi.macdonald.12}), we obtain%
\begin{align*}
&  \operatorname{coeff}_{\lambda+\rho}\left(  \sum_{\sigma
\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(  q_{1}%
,q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}^{n}h_{q_{j}%
}\left(  \xx \right)  \right)  \cdot\left(  \prod_{j=1}^{n}y_{j}%
^{q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}\right)  \right)
\\
&  =\operatorname{coeff}_{\lambda+\rho}\left(  \sum_{\sigma
\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum\limits_{\substack{\nu
\in\Par ;\\\text{the partition }\nu\text{ has}\\\text{at most
}n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)  \prod_{j=1}^{n}%
y_{j}^{\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }}\right) \\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}%
\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\underbrace{\operatorname{coeff}_{\lambda+\rho}\left(  \prod
_{j=1}^{n}y_{j}^{\left(  \nu+\rho\right)  _{\sigma\left(  j\right)  }}\right)
}_{\substack{=\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)
_{\sigma\left(  1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)
},...,\left(  \nu+\rho\right)  _{\sigma\left(  n\right)  }\right)
}\\\text{(by (\ref{sol.jacobi-trudi.macdonald.coeff}), applied to}%
\\r=\lambda+\rho\text{ and }\left(  t_{1},t_{2},...,t_{n}\right)  =\left(
\left(  \nu+\rho\right)  _{\sigma\left(  1\right)  },\left(  \nu+\rho\right)
_{\sigma\left(  2\right)  },...,\left(  \nu+\rho\right)  _{\sigma\left(
n\right)  }\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\operatorname{coeff}_{\lambda+\rho}\text{ is a continuous }
\mathbf{p}\text{-linear map,
and since }s_{\nu/\mu}\left(  \xx \right)  \in\kk \left[  \left[
\xx \right]  \right]  =\mathbf{p}\text{ for every }\nu\in
\Par \right) \\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}%
\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }.
\end{align*}
Therefore,%
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }\nonumber\\
&  =\operatorname{coeff}_{\lambda+\rho}\left(  \sum_{\sigma
\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(  q_{1}%
,q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}^{n}h_{q_{j}%
}\left(  \xx \right)  \right)  \cdot\left(  \prod_{j=1}^{n}y_{j}%
^{q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }}\right)  \right)
\nonumber\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot
\underbrace{\operatorname{coeff}_{\lambda+\rho}\left(  \prod
_{j=1}^{n}y_{j}^{q_{j}+\left(  \mu+\rho\right)  _{\sigma\left(  j\right)  }%
}\right)  }_{\substack{=\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }\\\text{(by (\ref{sol.jacobi-trudi.macdonald.coeff}),
applied to}\\r=\lambda+\rho\text{ and }\left(  t_{1},t_{2},...,t_{n}\right)
=\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  }%
,q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(
\mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname{coeff}_{\lambda+\rho}\text{ is a
continuous }\mathbf{p}\text{-linear map, and since}\\
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \in\kk \left[
\left[ \xx \right]  \right]  =\mathbf{p}\text{ for every }\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^{n}%
\end{array}
\right) \nonumber\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum_{\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(  \prod_{j=1}%
^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot\delta_{\lambda
+\rho,\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  }%
,q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(
\mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }.
\label{sol.jacobi-trudi.macdonald.15}%
\end{align}


Our next goal is to simplify the sums on both sides of this equality.

\underline{\textit{Step 8: Simplifying the left-hand side of
(\ref{sol.jacobi-trudi.macdonald.15}).}}

First of all, we will show the following auxiliary statement: If $\nu$ is a
partition having at most $n$ parts, and $p$ and $q$ are two elements of
$\left\{  1,2,...,n\right\}  $ satisfying $p<q$, then%
\begin{equation}
\left(  \nu+\rho\right)  _{p}>\left(  \nu+\rho\right)  _{q}.
\label{sol.jacobi-trudi.macdonald.simpl.strongmon}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.simpl.strongmon}%
):} Let $\nu$ be a partition having at most $n$ parts. Then, $\nu_{1}\geq
\nu_{2}\geq\nu_{3}\geq...$ (since $\nu$ is a partition). Hence, any two
positive integers $p$ and $q$ satisfying $p<q$ must satisfy $\nu_{p}\geq
\nu_{q}$.
\par
Now, let $p$ and $q$ be two elements of $\left\{  1,2,...,n\right\}  $
satisfying $p<q$. As we know, we then have $\nu_{p}\geq\nu_{q}$ (since $p<q$).
Now, $\rho_{p}=n-p$ (by (\ref{sol.jacobi-trudi.macdonald.preludium.rhoi}),
applied to $i=p$) and $\rho_{q}=n-q$ (for similar reasons). Thus,%
\[
\left(  \nu+\rho\right)  _{p}=\nu_{p}+\underbrace{\rho_{p}}_{=n-p}%
=\underbrace{\nu_{p}}_{\geq\nu_{q}}+n-\underbrace{p}_{<q}>\nu_{q}%
+\underbrace{n-q}_{\substack{=\rho_{q}\\\text{(since }\rho_{q}=n-q\text{)}%
}}=\nu_{q}+\rho_{q}=\left(  \nu+\rho\right)  _{q},
\]
so that (\ref{sol.jacobi-trudi.macdonald.simpl.strongmon}) is proven.}

Now, let us show that whenever $\nu$ is a partition having at most $n$ parts,
and $\sigma\in\Symm_{n}$ is a permutation such that $\sigma
\neq\id$, then%
\begin{equation}
\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=0.
\label{sol.jacobi-trudi.macdonald.simpl1}%
\end{equation}


\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.simpl1}):} Let $\nu$ be a
partition having at most $n$ parts. Let $\sigma\in\Symm_{n}$ be a
permutation such that $\sigma\neq\id$. We need to prove that
$\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=0$.

Indeed, assume (for the sake of contradiction) that $\lambda+\rho=\left(
\left(  \nu+\rho\right)  _{\sigma\left(  1\right)  },\left(  \nu+\rho\right)
_{\sigma\left(  2\right)  },...,\left(  \nu+\rho\right)  _{\sigma\left(
n\right)  }\right)  $. Thus,%
\[
\left(  \left(  \nu+\rho\right)  _{\sigma\left(  1\right)  },\left(  \nu
+\rho\right)  _{\sigma\left(  2\right)  },...,\left(  \nu+\rho\right)
_{\sigma\left(  n\right)  }\right)  =\lambda+\rho=\left(  \left(  \lambda
+\rho\right)  _{1},\left(  \lambda+\rho\right)  _{2},...,\left(  \lambda
+\rho\right)  _{n}\right)  .
\]
In other words,
\begin{equation}
\text{every }i\in\left\{  1,2,...,n\right\}  \text{ satisfies }\left(
\nu+\rho\right)  _{\sigma\left(  i\right)  }=\left(  \lambda+\rho\right)
_{i}. \label{sol.jacobi-trudi.macdonald.simpl1.pf.1}%
\end{equation}


Recall that $\sigma\neq\id$. Thus, it is easy to see that there
exists some $i\in\left\{  1,2,...,n-1\right\}  $ such that $\sigma\left(
i\right)  >\sigma\left(  i+1\right)  $\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Thus, there exists no $i\in\left\{  1,2,...,n-1\right\}  $ such
that $\sigma\left(  i\right)  >\sigma\left(  i+1\right)  $. In other words,
every $i\in\left\{  1,2,...,n-1\right\}  $ satisfies $\left(  \text{not
}\sigma\left(  i\right)  >\sigma\left(  i+1\right)  \right)  $. In other
words,
\begin{equation}
\text{every }i\in\left\{  1,2,...,n-1\right\}  \text{ satisfies }\sigma\left(
i\right)  \leq\sigma\left(  i+1\right)  .
\label{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.1}%
\end{equation}
every $i\in\left\{  1,2,...,n-1\right\}  $ satisfies $\sigma\left(  i\right)
\leq\sigma\left(  i+1\right)  $.
\par
We know that $\sigma$ is a permutation. Thus, $\sigma$ is a bijection. We thus
see that $\sigma$ is bijective, and therefore injective.
\par
We shall now show that
\begin{equation}
\sigma\left(  i\right)  <\sigma\left(  j\right)  \ \ \ \ \ \ \ \ \ \ \text{for
any two elements }i\text{ and }j\text{ of }\left\{  1,2,...,n\right\}  \text{
satisfying }i<j. \label{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.2}%
\end{equation}
\par
\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.2}):} Let $i$
and $j$ be two elements of $\left\{  1,2,...,n\right\}  $ satisfying $i<j$.
Then, every $k\in\left\{  i,i+1,...,j-1\right\}  $ satisfies $k\in\left\{
1,2,...,n-1\right\}  $ (because every $k\in\left\{  i,i+1,...,j-1\right\}  $
satisfies $k\geq i\geq1$ and $k\leq\underbrace{j}_{\leq n}-1\leq n-1$). Hence,
every $k\in\left\{  i,i+1,...,j-1\right\}  $ satisfies $\sigma\left(
k\right)  \leq\sigma\left(  k+1\right)  $ (by
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.1})). Thus,%
\[
\sum_{k=i}^{j-1}\left(  \sigma\left(  k+1\right)  -\underbrace{\sigma\left(
k\right)  }_{\leq\sigma\left(  k+1\right)  }\right)  \geq\sum_{k=i}%
^{j-1}\underbrace{\left(  \sigma\left(  k+1\right)  -\sigma\left(  k+1\right)
\right)  }_{=0}=\sum_{k=i}^{j-1}0=0.
\]
Hence,%
\begin{align*}
0  &  \leq\sum_{k=i}^{j-1}\left(  \sigma\left(  k+1\right)  -\sigma\left(
k\right)  \right)  =\underbrace{\sum_{k=i}^{j-1}\sigma\left(  k+1\right)
}_{\substack{=\sum_{k=i+1}^{j}\sigma\left(  k\right)  \\\text{(here, we
substituted }k\text{ for }k+1\\\text{in the sum)}}}-\sum_{k=i}^{j-1}%
\sigma\left(  k\right)  =\underbrace{\sum_{k=i+1}^{j}\sigma\left(  k\right)
}_{=\sum_{k=i+1}^{j-1}\sigma\left(  k\right)  +\sigma\left(  j\right)
}-\underbrace{\sum_{k=i}^{j-1}\sigma\left(  k\right)  }_{=\sigma\left(
i\right)  +\sum_{k=i+1}^{j-1}\sigma\left(  k\right)  }\\
&  =\left(  \sum_{k=i+1}^{j-1}\sigma\left(  k\right)  +\sigma\left(  j\right)
\right)  -\left(  \sigma\left(  i\right)  +\sum_{k=i+1}^{j-1}\sigma\left(
k\right)  \right)  =\sigma\left(  j\right)  -\sigma\left(  i\right)  .
\end{align*}
Adding $\sigma\left(  i\right)  $ to this inequality, we obtain $\sigma\left(
i\right)  \leq\sigma\left(  j\right)  $.
\par
But if we had $\sigma\left(  i\right)  =\sigma\left(  j\right)  $, then we
would have $i=j$ (because $\sigma$ is injective), which would contradict
$i<j$. Hence, we cannot have $\sigma\left(  i\right)  =\sigma\left(  j\right)
$. Thus, we have $\sigma\left(  i\right)  \neq\sigma\left(  j\right)  $.
Combined with $\sigma\left(  i\right)  \leq\sigma\left(  j\right)  $, this
yields $\sigma\left(  i\right)  <\sigma\left(  j\right)  $. This completes the
proof of (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.2}).
\par
Let us now return to proving that there exists some $i\in\left\{
1,2,...,n-1\right\}  $ such that $\sigma\left(  i\right)  >\sigma\left(
i+1\right)  $. Let us introduce the following notation: If $\mathcal{A}$ is
any assertion, then we will denote by $\left[  \mathcal{A}\right]  $ the truth
value of $\mathcal{A}$ (that is, the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$). We now claim that
\begin{equation}
\left[  \sigma\left(  i\right)  <\sigma\left(  j\right)  \right]  =\left[
i<j\right]  \ \ \ \ \ \ \ \ \ \ \text{for any two elements }i\text{ and
}j\text{ of }\left\{  1,2,...,n\right\}  .
\label{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}
\end{equation}
\par
\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}):} Let $i$
and $j$ be two elements of $\left\{  1,2,...,n\right\}  $. We need to prove
that (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}) holds.
\par
If $i<j$, then $\sigma\left(  i\right)  <\sigma\left(  j\right)  $ (by
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.2})). Hence, if $i<j$, then
\begin{align*}
\left[  \sigma\left(  i\right)  <\sigma\left(  j\right)  \right]   &
=1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma\left(  i\right)
<\sigma\left(  j\right)  \right) \\
&  =\left[  i<j\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[
i<j\right]  =1\text{ (since }i<j\text{)}\right)  .
\end{align*}
Hence, we have proven (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}) in
the case when $i<j$. Thus, for the rest of the proof of
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}), we can WLOG assume that
we don't have $i<j$. Assume this.
\par
We have assumed that we don't have $i<j$. Hence, we have $i\geq j$.
\par
If $i=j$, then%
\begin{align*}
\left[  \sigma\left(  \underbrace{i}_{=j}\right)  <\sigma\left(  j\right)
\right]   &  =\left[  \sigma\left(  j\right)  <\sigma\left(  j\right)
\right]  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since we don't have }%
\sigma\left(  j\right)  <\sigma\left(  j\right)  \right) \\
&  =\left[  \underbrace{j}_{\substack{=i\\\text{(since }i=j\text{)}%
}}<j\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[  j<j\right]
=0\text{ (since we don't have }j<j\text{)}\right) \\
&  =\left[  i<j\right]  .
\end{align*}
Hence, we have proven (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}) in
the case when $i=j$. Thus, for the rest of the proof of
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}), we can WLOG assume that
we don't have $i=j$. Assume this.
\par
We have assumed that we don't have $i=j$. Hence, we have $i\neq j$. Combined
with $i\geq j$, this yields $i>j$, so that $j<i$. Thus,
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.2}) (applied to $j$ and $i$
instead of $i$ and $j$) yields $\sigma\left(  j\right)  <\sigma\left(
i\right)  $. On the other hand, we have $i>j$, so that $i\geq j$, and thus we
don't have $i<j$. Hence, $\left[  i<j\right]  =0$.
\par
But we have $\sigma\left(  j\right)  <\sigma\left(  i\right)  $, thus
$\sigma\left(  i\right)  >\sigma\left(  j\right)  $, hence $\sigma\left(
i\right)  \geq\sigma\left(  j\right)  $. Therefore, we don't have
$\sigma\left(  i\right)  <\sigma\left(  j\right)  $. Thus,%
\[
\left[  \sigma\left(  i\right)  <\sigma\left(  j\right)  \right]  =0=\left[
i<j\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[  i<j\right]
=0\right)  .
\]
This proves (\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}).
\par
Let us again return to proving that there exists some $i\in\left\{
1,2,...,n-1\right\}  $ such that $\sigma\left(  i\right)  >\sigma\left(
i+1\right)  $. Every $j\in\left\{  1,2,...,n\right\}  $ satisfies%
\begin{align}
\sum_{i\in\left\{  1,2,...,n\right\}  }\left[  i<j\right]   &  =\sum\limits
_{\substack{i\in\left\{  1,2,...,n\right\}  ;\\i<j}}\underbrace{\left[
i<j\right]  }_{\substack{=1\\\text{(since we have }i<j\text{)}}}+\sum\limits
_{\substack{i\in\left\{  1,2,...,n\right\}  ;\\\text{we don't have }%
i<j}}\underbrace{\left[  i<j\right]  }_{\substack{=0\\\text{(since we don't
have }i<j\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,...,n\right\}  ;\\i<j}}1+\underbrace{\sum\limits
_{\substack{i\in\left\{  1,2,...,n\right\}  ;\\\text{we don't have }i<j}%
}0}_{=0}=\sum\limits_{\substack{i\in\left\{  1,2,...,n\right\}  ;\\i<j}}1\nonumber\\
&  =\left\vert \underbrace{\left\{  i\in\left\{  1,2,...,n\right\}
\ \mid\ i<j\right\}  }_{=\left\{  1,2,...,j-1\right\}  }\right\vert
\cdot1\nonumber\\
&  =\underbrace{\left\vert \left\{  1,2,...,j-1\right\}  \right\vert }%
_{=j-1}\cdot1=\left(  j-1\right)  \cdot1=j-1.
\label{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.4}%
\end{align}
\par
But now, let us fix $j\in\left\{  1,2,...,n\right\}  $. Then,
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.4}) (applied to $\sigma\left(
j\right)  $ instead of $j$) yields
\[
\sum_{i\in\left\{  1,2,...,n\right\}  }\left[  i<\sigma\left(  j\right)
\right]  =\sigma\left(  j\right)  -1.
\]
Compared with%
\begin{align*}
\sum_{i\in\left\{  1,2,...,n\right\}  }\left[  i<\sigma\left(  j\right)
\right]   &  =\sum_{i\in\left\{  1,2,...,n\right\}  }\underbrace{\left[
\sigma\left(  i\right)  <\sigma\left(  j\right)  \right]  }%
_{\substack{=\left[  i<j\right]  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.3}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\sigma\left(  i\right)  \text{ for }i\text{ in the
sum,}\\
\text{since }\sigma:\left\{  1,2,...,n\right\}  \rightarrow\left\{
1,2,...,n\right\}  \text{ is a bijection}%
\end{array}
\right) \\
&  =\sum_{i\in\left\{  1,2,...,n\right\}  }\left[  i<j\right]
=j-1\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1.pf.4})}\right)  ,
\end{align*}
this yields $\sigma\left(  j\right)  -1=j-1$. Hence, $\sigma\left(  j\right)
= j = \id \left(  j\right)  $.
\par
Now, let us forget that we fixed $j$. We thus have shown that $\sigma\left(
j\right)  = \id \left(  j\right)  $ for every $j\in\left\{
1,2,...,n\right\}  $. In other words, $\sigma=\id$. This
contradicts $\sigma\neq\id$. This contradiction completes our
proof, qed.}. Consider this $i$. We have $i\in\left\{  1,2,...,n-1\right\}
\subset\left\{  1,2,...,n\right\}  $. Also, from $i\in\left\{
1,2,...,n-1\right\}  $, we obtain $i+1\in\left\{  2,3,...,n\right\}
\subset\left\{  1,2,...,n\right\}  $. Applying
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1}) to $i+1$ instead of $i$, we
therefore obtain $\left(  \nu+\rho\right)  _{\sigma\left(  i+1\right)
}=\left(  \lambda+\rho\right)  _{i+1}$.

We have $\sigma\left(  i\right)  >\sigma\left(  i+1\right)  $, so that
$\sigma\left(  i+1\right)  <\sigma\left(  i\right)  $. Hence, $\left(
\nu+\rho\right)  _{\sigma\left(  i+1\right)  }>\left(  \nu+\rho\right)
_{\sigma\left(  i\right)  }$ (by
(\ref{sol.jacobi-trudi.macdonald.simpl.strongmon}), applied to $p=\sigma
\left(  i+1\right)  $ and $q=\sigma\left(  i\right)  $). Thus, $\left(
\nu+\rho\right)  _{\sigma\left(  i\right)  }<\left(  \nu+\rho\right)
_{\sigma\left(  i+1\right)  }$. Now,
(\ref{sol.jacobi-trudi.macdonald.simpl1.pf.1}) yields $\left(  \lambda
+\rho\right)  _{i}=\left(  \nu+\rho\right)  _{\sigma\left(  i\right)
}<\left(  \nu+\rho\right)  _{\sigma\left(  i+1\right)  }=\left(  \lambda
+\rho\right)  _{i+1}$.

But we can apply (\ref{sol.jacobi-trudi.macdonald.simpl.strongmon}) to
$\lambda$, $i$ and $i+1$ instead of $\lambda$, $p$ and $q$ (because $i<i+1$).
As a result, we obtain $\left(  \lambda+\rho\right)  _{i}>\left(  \lambda
+\rho\right)  _{i+1}$. This contradicts $\left(  \lambda+\rho\right)
_{i}<\left(  \lambda+\rho\right)  _{i+1}$. This contradiction proves that our
assertion (that $\lambda+\rho=\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  $) was wrong. Hence, we
don't have $\lambda+\rho=\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  $. Thus, we have
$\lambda+\rho\neq\left(  \left(  \nu+\rho\right)  _{\sigma\left(  1\right)
},\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(  \nu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  $. Consequently,
$\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=0$. This proves
(\ref{sol.jacobi-trudi.macdonald.simpl1}).

Now, the left-hand side of (\ref{sol.jacobi-trudi.macdonald.15}) rewrites as%
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }\nonumber\\
&  =\underbrace{\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma
= \id}}\left(  -1\right)  ^{\sigma}\sum\limits_{\substack{\nu
\in\Par ;\\\text{the partition }\nu\text{ has}\\\text{at most
}n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)  \cdot\delta
_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(  1\right)
},\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(  \nu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  }}_{\substack{=\left(
-1\right)  ^{\id}\sum\limits_{\substack{\nu\in\Par %
;\\\text{the partition }\nu\text{ has}\\\text{at most }n\text{ parts}}%
}s_{\nu/\mu}\left(  \xx \right)  \cdot\delta_{\lambda+\rho,\left(
\left(  \nu+\rho\right)  _{\id \left(  1\right)  },\left(
\nu+\rho\right)  _{\id \left(  2\right)  },...,\left(  \nu
+\rho\right)  _{\id \left(  n\right)  }\right)  }\\\text{(since
}\id \in\Symm_{n}\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma
\neq\id}}\left(  -1\right)  ^{\sigma}\sum\limits_{\substack{\nu
\in\Par ;\\\text{the partition }\nu\text{ has}\\\text{at most
}n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)  \cdot\underbrace{\delta
_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(  1\right)
},\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(  \nu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  }}_{\substack{=0\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl1}))}}}\nonumber\\
&  =\underbrace{\left(  -1\right)  ^{\id}}_{=1}\sum\limits
_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\underbrace{\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)
_{\id \left(  1\right)  },\left(  \nu+\rho\right)
_{\id \left(  2\right)  },...,\left(  \nu+\rho\right)
_{\id \left(  n\right)  }\right)  }}_{\substack{=\delta
_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{1},\left(  \nu+\rho\right)
_{2},...,\left(  \nu+\rho\right)  _{n}\right)  }\\\text{(because
} \id \left(  1\right)  =1\text{, }\id \left(
2\right)  =2\text{, }...\text{, }\id \left(  n\right)
=n\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma
\neq\id}}\left(  -1\right)  ^{\sigma}\underbrace{\sum\limits
_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\underbrace{\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)
_{1},\left(  \nu+\rho\right)  _{2},...,\left(  \nu+\rho\right)  _{n}\right)
}}_{\substack{=\delta_{\lambda+\rho,\nu+\rho}\\\text{(since }\left(  \left(
\nu+\rho\right)  _{1},\left(  \nu+\rho\right)  _{2},...,\left(  \nu
+\rho\right)  _{n}\right)  =\nu+\rho\text{)}}}+\underbrace{\sum\limits
_{\substack{\sigma\in\Symm_{n};\\\sigma\neq\id}}\left(
-1\right)  ^{\sigma}0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\delta_{\lambda+\rho,\nu+\rho}+0=\sum\limits_{\substack{\nu\in\Par %
;\\\text{the partition }\nu\text{ has}\\\text{at most }n\text{ parts}}%
}s_{\nu/\mu}\left(  \xx \right)  \cdot\delta_{\lambda+\rho,\nu+\rho
}\nonumber\\
&  =\underbrace{\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition
}\nu\text{ has}\\\text{at most }n\text{ parts;}\\\nu=\lambda}}s_{\nu/\mu
}\left(  \xx \right)  \cdot\delta_{\lambda+\rho,\nu+\rho}}%
_{\substack{=s_{\lambda/\mu}\left(  \xx \right)  \cdot\delta
_{\lambda+\rho,\lambda+\rho}\\\text{(since }\lambda\text{ is an element of
}\Par \\\text{such that }\lambda\text{ has at most }n\text{
parts)}}}+\sum\limits_{\substack{\nu\in\Par ;\\\text{the partition }%
\nu\text{ has}\\\text{at most }n\text{ parts;}\\\nu\neq\lambda}}s_{\nu/\mu
}\left(  \xx \right)  \cdot\underbrace{\delta_{\lambda+\rho,\nu+\rho}%
}_{\substack{=0\\\text{(since }\lambda+\rho\neq\nu+\rho\\\text{(since }%
\lambda\neq\nu\text{ (since }\nu\neq\lambda\text{)))}}}\nonumber\\
&  =s_{\lambda/\mu}\left(  \xx \right)  \cdot\delta_{\lambda
+\rho,\lambda+\rho}+\underbrace{\sum\limits_{\substack{\nu\in\Par %
;\\\text{the partition }\nu\text{ has}\\\text{at most }n\text{ parts;}%
\\\nu\neq\lambda}}s_{\nu/\mu}\left(  \xx \right)  \cdot0}_{=0}%
=s_{\lambda/\mu}\left(  \xx \right)  \cdot\underbrace{\delta
_{\lambda+\rho,\lambda+\rho}}_{\substack{=1\\\text{(since }\lambda
+\rho=\lambda+\rho\text{)}}}=s_{\lambda/\mu}\left(  \xx \right)  .
\label{sol.jacobi-trudi.macdonald.simpl.lhs}%
\end{align}


\underline{\textit{Step 9: Simplifying the right-hand side of
(\ref{sol.jacobi-trudi.macdonald.15}).}}

Now, we shall simplify the right-hand side of
(\ref{sol.jacobi-trudi.macdonald.15}). In order to do so, we will show that%
\begin{equation}
\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot
\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(
1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)
},...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)  }\right)
}=\prod_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(
j\right)  }\left(  \xx \right)
\label{sol.jacobi-trudi.macdonald.simpl2}%
\end{equation}
for every $\sigma\in\Symm_{n}$.

\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.simpl2}):} We first notice
that every $i\in\left\{  1,2,...,n\right\}  $ satisfies $\lambda_{i}%
-\mu_{\sigma\left(  i\right)  }-i+\sigma\left(  i\right)  \in\ZZ$.
Hence, we can define an $n$-tuple $\left(  r_{1},r_{2},...,r_{n}\right)
\in\ZZ^n$ by
\begin{equation}
\left(  r_{i}=\lambda_{i}-\mu_{\sigma\left(  i\right)  }-i+\sigma\left(
i\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,n\right\}
\right)  . \label{sol.jacobi-trudi.macdonald.simpl2.pf.ri}%
\end{equation}
Consider this $n$-tuple $\left(  r_{1},r_{2},...,r_{n}\right)  \in
\ZZ^n$.

We are going to show that every $n$-tuple $\left(  q_{1},q_{2},...,q_{n}%
\right)  \in\NN^n$ satisfying $\left(  q_{1},q_{2},...,q_{n}\right)
\neq\left(  r_{1},r_{2},...,r_{n}\right)  $ satisfies%
\begin{equation}
\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(
1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)
},...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=0.
\label{sol.jacobi-trudi.macdonald.simpl2.pf.1}%
\end{equation}


\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.simpl2.pf.1}):} Let $\left(
q_{1},q_{2},...,q_{n}\right)  \in\NN^n$ be an $n$-tuple satisfying
$\left(  q_{1},q_{2},...,q_{n}\right)  \neq\left(  r_{1},r_{2},...,r_{n}%
\right)  $. Assume (for the sake of contradiction) that $\lambda+\rho=\left(
q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(
\mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  $. Then,%
\begin{align*}
&  \left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  }%
,q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(
\mu+\rho\right)  _{\sigma\left(  n\right)  }\right) \\
&  =\lambda+\rho=\left(  \left(  \lambda+\rho\right)  _{1},\left(
\lambda+\rho\right)  _{2},...,\left(  \lambda+\rho\right)  _{n}\right)  .
\end{align*}
In other words, every $i\in\left\{  1,2,...,n\right\}  $ satisfies
$q_{i}+\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }=\left(
\lambda+\rho\right)  _{i}$. Hence, every $i\in\left\{  1,2,...,n\right\}  $
satisfies%
\begin{align*}
q_{i}  &  =\underbrace{\left(  \lambda+\rho\right)  _{i}}_{=\lambda_{i}%
+\rho_{i}}-\underbrace{\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }%
}_{=\mu_{\sigma\left(  i\right)  }+\rho_{\sigma\left(  i\right)  }}=\left(
\lambda_{i}+\underbrace{\rho_{i}}_{\substack{=n-i\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.preludium.rhoi}))}}}\right)  -\left(
\mu_{\sigma\left(  i\right)  }+\underbrace{\rho_{\sigma\left(  i\right)  }%
}_{\substack{=n-\sigma\left(  i\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.preludium.rhoi}), applied to}\\\sigma\left(
i\right)  \text{ instead of }i\text{)}}}\right) \\
&  =\left(  \lambda_{i}+\left(  n-i\right)  \right)  -\left(  \mu_{i}+\left(
n-\sigma\left(  i\right)  \right)  \right)  =\lambda_{i}-\mu_{\sigma\left(
i\right)  }-i+\sigma\left(  i\right)  =r_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.jacobi-trudi.macdonald.simpl2.pf.ri})}\right)  .
\end{align*}
In other words, $\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1}%
,r_{2},...,r_{n}\right)  $, which contradicts $\left(  q_{1},q_{2}%
,...,q_{n}\right)  \neq\left(  r_{1},r_{2},...,r_{n}\right)  $. This
contradiction shows that our assumption (that $\lambda+\rho=\left(
q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(
\mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  $) was wrong. Hence, we
don't have $\lambda+\rho=\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma
\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)
},...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  $. In
other words, we must have $\lambda+\rho\neq\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  $. Hence, \newline$\delta_{\lambda+\rho,\left(
q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(
\mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=0$. This proves
(\ref{sol.jacobi-trudi.macdonald.simpl2.pf.1}).

Let us now return to the proof of (\ref{sol.jacobi-trudi.macdonald.simpl2}).
Every $\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n$ satisfies
either $\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2}%
,...,r_{n}\right)  $ or $\left(  q_{1},q_{2},...,q_{n}\right)  \neq\left(
r_{1},r_{2},...,r_{n}\right)  $ but not both. Hence, we can split the sum on
the left-hand side of (\ref{sol.jacobi-trudi.macdonald.simpl2}) as follows:%
\begin{align}
&  \sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot
\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(
1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)
},...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)  }\right)
}\nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n ;
\\\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2}%
,...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx %
\right)  \right)  \cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)
\in\NN^n;\\\left(  q_{1},q_{2},...,q_{n}\right)  \neq\left(
r_{1},r_{2},...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(
\xx \right)  \right)  \cdot\underbrace{\delta_{\lambda+\rho,\left(
q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(
\mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu
+\rho\right)  _{\sigma\left(  n\right)  }\right)  }}_{\substack{=0\\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl2.pf.1}))}}}\nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN %
^{n};\\\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2}%
,...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx %
\right)  \right)  \cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{\left(  q_{1}%
,q_{2},...,q_{n}\right)  \in\NN^n;\\\left(  q_{1},q_{2}%
,...,q_{n}\right)  \neq\left(  r_{1},r_{2},...,r_{n}\right)  }}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot0}%
_{=0}\nonumber\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN %
^{n};\\\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2}%
,...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx %
\right)  \right)  \cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }. \label{sol.jacobi-trudi.macdonald.simpl2.pf.4}%
\end{align}


Now, we distinguish between two cases:

\textit{Case 1:} Every $i\in\left\{  1,2,...,n\right\}  $ satisfies $r_{i}%
\in\NN$.

\textit{Case 2:} Not every $i\in\left\{  1,2,...,n\right\}  $ satisfies
$r_{i}\in\NN$.

Let us consider Case 1 first. In this case, every $i\in\left\{
1,2,...,n\right\}  $ satisfies $r_{i}\in\NN$. Hence, $\left(
r_{1},r_{2},...,r_{n}\right)  \in\NN^n$.

Every $i\in\left\{  1,2,...,n\right\}  $ satisfies
\begin{align*}
\underbrace{r_{i}}_{\substack{=\lambda_{i}-\mu_{\sigma\left(  i\right)
}-i+\sigma\left(  i\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl2.pf.ri}))}}}+\underbrace{\left(
\mu+\rho\right)  _{\sigma\left(  i\right)  }}_{=\mu_{\sigma\left(  i\right)
}+\rho_{\sigma\left(  i\right)  }}-\rho_{i}  &  =\left(  \lambda_{i}%
-\mu_{\sigma\left(  i\right)  }-i+\sigma\left(  i\right)  \right)
+\mu_{\sigma\left(  i\right)  }+\underbrace{\rho_{\sigma\left(  i\right)  }%
}_{\substack{=n-\sigma\left(  i\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.preludium.rhoi}), applied to}\\\sigma\left(
i\right)  \text{ instead of }i\text{)}}}-\underbrace{\rho_{i}}%
_{\substack{=n-i\\\text{(by (\ref{sol.jacobi-trudi.macdonald.preludium.rhoi}%
))}}}\\
&  =\left(  \lambda_{i}-\mu_{\sigma\left(  i\right)  }-i+\sigma\left(
i\right)  \right)  +\mu_{\sigma\left(  i\right)  }+\left(  n-\sigma\left(
i\right)  \right)  -\left(  n-i\right)  =\lambda_{i}.
\end{align*}
In other words, every $i\in\left\{  1,2,...,n\right\}  $ satisfies
$\lambda_{i}=r_{i}+\left(  \mu+\rho\right)  _{\sigma\left(  i\right)  }%
-\rho_{i}$. Hence, every $i\in\left\{  1,2,...,n\right\}  $ satisfies
\[
\left(  \lambda+\rho\right)  _{i}=\underbrace{\lambda_{i}}_{=r_{i}+\left(
\mu+\rho\right)  _{\sigma\left(  i\right)  }-\rho_{i}}+\rho_{i}=r_{i}+\left(
\mu+\rho\right)  _{\sigma\left(  i\right)  }-\rho_{i}+\rho_{i}=r_{i}+\left(
\mu+\rho\right)  _{\sigma\left(  i\right)  }.
\]
In other words,%
\[
\left(  \left(  \lambda+\rho\right)  _{1},\left(  \lambda+\rho\right)
_{2},...,\left(  \lambda+\rho\right)  _{n}\right)  =\left(  r_{1}+\left(
\mu+\rho\right)  _{\sigma\left(  1\right)  },r_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,r_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  .
\]
Compared with $\left(  \left(  \lambda+\rho\right)  _{1},\left(  \lambda
+\rho\right)  _{2},...,\left(  \lambda+\rho\right)  _{n}\right)  =\lambda
+\rho$, this yields%
\[
\lambda+\rho=\left(  r_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)
},r_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)  },...,r_{n}+\left(
\mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  .
\]


Now, recall that $\left(  r_{1},r_{2},...,r_{n}\right)  \in\NN^n$.
Hence, the sum \newline$\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)
\in\NN^n;\\\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1}%
,r_{2},...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(
\xx \right)  \right)  \cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(
\mu+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }$ has precisely one addend (namely, the addend for
$\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2},...,r_{n}\right)
$), and we obtain%
\begin{align*}
&  \sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN %
^{n};\\\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2}%
,...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx %
\right)  \right)  \cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }\\
&  =\left(  \prod_{j=1}^{n}h_{r_{j}}\left(  \xx \right)  \right)
\cdot\underbrace{\delta_{\lambda+\rho,\left(  r_{1}+\left(  \mu+\rho\right)
_{\sigma\left(  1\right)  },r_{2}+\left(  \mu+\rho\right)  _{\sigma\left(
2\right)  },...,r_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)
}\right)  }}_{\substack{=1\\\text{(since }\lambda+\rho=\left(  r_{1}+\left(
\mu+\rho\right)  _{\sigma\left(  1\right)  },r_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,r_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  \text{)}}}\\
&  =\prod_{j=1}^{n}\underbrace{h_{r_{j}}\left(  \xx \right)
}_{\substack{=h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(
j\right)  }\left(  \xx \right)  \\\text{(since }r_{j}=\lambda_{j}%
-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(  j\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl2.pf.ri}), applied to }i=j\text{))}%
}}=\prod_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma
\left(  j\right)  }\left(  \xx \right)  .
\end{align*}
Thus, (\ref{sol.jacobi-trudi.macdonald.simpl2.pf.4}) becomes%
\begin{align*}
&  \sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot
\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(
1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)
},...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }\\
&  =\sum\limits_{\substack{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN %
^{n};\\\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2}%
,...,r_{n}\right)  }}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx %
\right)  \right)  \cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu
+\rho\right)  _{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)
_{\sigma\left(  2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(
n\right)  }\right)  }=\prod_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(
j\right)  }-j+\sigma\left(  j\right)  }\left(  \xx \right)  .
\end{align*}
Hence, (\ref{sol.jacobi-trudi.macdonald.simpl2}) is proven in Case 1.

Let us now consider Case 2. In this case, not every $i\in\left\{
1,2,...,n\right\}  $ satisfies $r_{i}\in\NN$. Hence, there exists an
$i\in\left\{  1,2,...,n\right\}  $ which does not satisfy $r_{i}\in\NN %
$. Choose such an $i$ and denote it by $k$. Thus, $k$ is an $i\in\left\{
1,2,...,n\right\}  $ which does not satisfy $r_{i}\in\NN$. Thus,
$k\in\left\{  1,2,...,n\right\}  $ and we don't have $r_{k}\in\NN$.

We have $r_{k}=\lambda_{k}-\mu_{\sigma\left(  k\right)  }-k+\sigma\left(
k\right)  \in\ZZ$ but not $r_{k}\in\NN$. Hence, $r_{k}%
\in\ZZ \setminus\NN =\left\{  -1,-2,-3,...\right\}  $, so that
$r_{k}<0$ and thus $h_{r_{k}}=0$. Now,%

\begin{align}
\underbrace{\prod_{j=1}^{n}}_{=\prod_{j\in\left\{  1,2,...,n\right\}  }%
}h_{r_{j}}\left(  \xx \right)   &  =\prod_{j\in\left\{
1,2,...,n\right\}  }h_{r_{j}}\left(  \xx \right)  =\underbrace{h_{r_{k}%
}\left(  \xx \right)  }_{=h_{r_{k}}=0}\cdot\prod\limits_{\substack{j\in\left\{
1,2,...,n\right\}  ;\\j\neq k}}h_{r_{j}}\left(  \xx \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\in\left\{  1,2,...,n\right\}
\right) \nonumber\\
&  =0. \label{sol.jacobi-trudi.macdonald.simpl2.pf.7}%
\end{align}


If we had $\left(  r_{1},r_{2},...,r_{n}\right)  \in\NN^n$, then
every $i\in\left\{  1,2,...,n\right\}  $ would satisfy $r_{i}\in\NN$,
and this would contradict the fact that not every $i\in\left\{
1,2,...,n\right\}  $ satisfies $r_{i}\in\NN$. Hence, we cannot have
$\left(  r_{1},r_{2},...,r_{n}\right)  \in\NN^n$.

Now, let $\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n$ be
arbitrary. If we had $\left(  q_{1},q_{2},...,q_{n}\right)  =\left(
r_{1},r_{2},...,r_{n}\right)  $, then we would have $\left(  r_{1}%
,r_{2},...,r_{n}\right)  =\left(  q_{1},q_{2},...,q_{n}\right)  \in
\NN^n$, which would contradict the fact that we cannot have $\left(
r_{1},r_{2},...,r_{n}\right)  \in\NN^n$. Hence, we cannot have
$\left(  q_{1},q_{2},...,q_{n}\right)  =\left(  r_{1},r_{2},...,r_{n}\right)
$. Thus, we must have $\left(  q_{1},q_{2},...,q_{n}\right)  \neq\left(
r_{1},r_{2},...,r_{n}\right)  $. Consequently,
(\ref{sol.jacobi-trudi.macdonald.simpl2.pf.1}) yields $\delta_{\lambda
+\rho,\left(  q_{1}+\left(  \mu+\rho\right)  _{\sigma\left(  1\right)  }%
,q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(  2\right)  },...,q_{n}+\left(
\mu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=0$.

Now, forget that we fixed $\left(  q_{1},q_{2},...,q_{n}\right)  $. We thus
have shown that every $\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN %
^{n}$ satisfies $\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)
_{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(
2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)
}\right)  }=0$. Thus,
\begin{align*}
&  \sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot
\underbrace{\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)
_{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(
2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)
}\right)  }}_{=0}\\
&  =\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^n}\left(
\prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)  \cdot
0=0=\prod_{j=1}^{n}\underbrace{h_{r_{j}}\left(  \xx \right)
}_{\substack{=h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(
j\right)  }\left(  \xx \right)  \\\text{(since }r_{j}=\lambda_{j}%
-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(  j\right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl2.pf.ri}), applied to }i=j\text{))}%
}}=\prod_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma
\left(  j\right)  }\left(  \xx \right)  .
\end{align*}
Hence, (\ref{sol.jacobi-trudi.macdonald.simpl2}) is proven in Case 2.

We have now proven (\ref{sol.jacobi-trudi.macdonald.simpl2}) in each of the
two Cases 1 and 2. Since these two Cases cover all possibilities, this yields
that (\ref{sol.jacobi-trudi.macdonald.simpl2}) always holds. We thus are
finished proving (\ref{sol.jacobi-trudi.macdonald.simpl2}).

\underline{\textit{Step 10: Taming (\ref{sol.jacobi-trudi.macdonald.15}).}}

The equality (\ref{sol.jacobi-trudi.macdonald.15}) becomes%
\begin{align*}
&  \sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}%
\underbrace{\sum_{\left(  q_{1},q_{2},...,q_{n}\right)  \in\NN^{n}%
}\left(  \prod_{j=1}^{n}h_{q_{j}}\left(  \xx \right)  \right)
\cdot\delta_{\lambda+\rho,\left(  q_{1}+\left(  \mu+\rho\right)
_{\sigma\left(  1\right)  },q_{2}+\left(  \mu+\rho\right)  _{\sigma\left(
2\right)  },...,q_{n}+\left(  \mu+\rho\right)  _{\sigma\left(  n\right)
}\right)  }}_{\substack{=\prod_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(
j\right)  }-j+\sigma\left(  j\right)  }\left(  \xx \right)  \\\text{(by
(\ref{sol.jacobi-trudi.macdonald.simpl2}))}}}\\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\prod
_{j=1}^{n}\underbrace{h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }%
-j+\sigma\left(  j\right)  }\left(  \xx \right)  }_{=h_{\lambda_{j}%
-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(  j\right)  }}=\sum_{\sigma
\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\prod_{j=1}^{n}h_{\lambda
_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(  j\right)  }.
\end{align*}
Compared with%
\begin{align*}
\det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)   &  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma
}\prod_{i=1}^{n}h_{\lambda_{i}-\mu_{\sigma\left(  i\right)  }-i+\sigma\left(
i\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.jacobi-trudi.macdonald.detexp}), applied to }\ell=n\text{ and
}\alpha_{i,j}=h_{\lambda_{i}-\mu_{j}-i+j}\right)  \\
&  =\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\prod
_{j=1}^{n}h_{\lambda_{j}-\mu_{\sigma\left(  j\right)  }-j+\sigma\left(
j\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }i\text{ as
}j\text{ in the product}\right)  ,
\end{align*}
this yields%
\[
\sum_{\sigma\in\Symm_{n}}\left(  -1\right)  ^{\sigma}\sum\limits
_{\substack{\nu\in\Par ;\\\text{the partition }\nu\text{
has}\\\text{at most }n\text{ parts}}}s_{\nu/\mu}\left(  \xx \right)
\cdot\delta_{\lambda+\rho,\left(  \left(  \nu+\rho\right)  _{\sigma\left(
1\right)  },\left(  \nu+\rho\right)  _{\sigma\left(  2\right)  },...,\left(
\nu+\rho\right)  _{\sigma\left(  n\right)  }\right)  }=\det\left(  \left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  .
\]
Comparing this with (\ref{sol.jacobi-trudi.macdonald.simpl.lhs}), this yields
$s_{\lambda/\mu}\left(  \xx \right)  =\det\left(  \left(  h_{\lambda
_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  $. Since $s_{\lambda
/\mu}\left(  \xx \right)  =s_{\lambda/\mu}$, this rewrites as%
\begin{equation}
s_{\lambda/\mu}=\det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  .\label{sol.jacobi-trudi.macdonald.res.h}%
\end{equation}


\underline{\textit{Step 11: Proving (\ref{eq.jacobi-trudi.h}).}}

Now, let us forget that we fixed $\lambda$, $\mu$ and $n$. We thus have shown
that if $n\in\NN$, and if $\lambda$ and $\mu$ are two partitions having
at most $n$ parts (each), then (\ref{sol.jacobi-trudi.macdonald.res.h}) holds.

Now, let $\lambda$ and $\mu$ be two partitions, and let $\ell\in\NN$ be
such that $\ell\left(  \lambda\right)  \leq\ell$ and $\ell\left(  \mu\right)
\leq\ell$. Then, the partition $\lambda$ has at most $\ell$ parts (since
$\ell\left(  \lambda\right)  \leq\ell$), and the partition $\mu$ has at most
$\ell$ parts (since $\ell\left(  \mu\right)  \leq\ell$). Thus, we can apply
(\ref{sol.jacobi-trudi.macdonald.res.h}) to $n=\ell$. As a result, we obtain%
\begin{equation}
s_{\lambda/\mu}=\det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  . \label{sol.jacobi-trudi.macdonald.h.done}%
\end{equation}
This proves (\ref{eq.jacobi-trudi.h}).

\underline{\textit{Step 12: Proving (\ref{eq.jacobi-trudi.e}).}}

Now it remains to prove (\ref{eq.jacobi-trudi.e}).

Let us first notice that every $m \in \ZZ$ satisfies
\begin{equation}
\omega\left(  h_{m}\right)  =e_{m}. \label{sol.jacobi-trudi.macdonald.h.wh}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.jacobi-trudi.macdonald.h.wh}):} Let
$m \in \ZZ$. Then, Proposition~\ref{prop.Lambda.omega-and-S}(b) (applied to $m$
instead of $n$) yields $\omega\left(  h_{m}\right)  =e_{m}$.
In other words, \eqref{sol.jacobi-trudi.macdonald.h.wh} is proven.}

Recall that the determinant of a matrix is a polynomial in the entries of the
matrix. Hence, determinants commute with algebra homomorphisms. In other
words, if $\mathbf{r}$ and $\mathbf{q}$ are two commutative $\kk %
$-algebras, and if $k\in\NN$ is arbitrary, and if $M\in\mathbf{r}%
^{k\times k}$ is a $k\times k$-matrix over $\mathbf{r}$, and if $\varphi
:\mathbf{r}\rightarrow\mathbf{s}$ is a $\kk$-algebra homomorphism, then%
\begin{equation}
\varphi\left(  \det M\right)  =\det\left(  \varphi^{k\times k}\left(
M\right)  \right)  . \label{sol.jacobi-trudi.macdonald.det-functor}%
\end{equation}
Here, $\varphi^{k\times k}\left(  M\right)  $ denotes the matrix obtained from
$M$ by applying the map $\varphi$ to every entry of $M$; in other words, if
$M$ is written in the form $M=\left(  b_{i,j}\right)  _{i,j=1,2,...,k}$ for
some $b_{i,j}\in\mathbf{r}$, then $\varphi^{k\times k}\left(  M\right)  $ is
defined as the matrix $\left(  \varphi\left(  b_{i,j}\right)  \right)
_{i,j=1,2,...,k}\in\mathbf{s}^{k\times k}$.

But using \eqref{skew-Schur-antipodes}, it is easy to see that%
\begin{equation}
\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}
\label{sol.jacobi-trudi.macdonald.omega.h}
\end{equation}
\footnote{\textit{Proof.} If $\mu\subseteq\lambda$, then $\omega\left(
s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}$ follows immediately from the
first identity in \eqref{skew-Schur-antipodes}. Hence, for the rest of this
proof of $\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}$, we
can WLOG assume that we don't have $\mu\subseteq\lambda$. Assume this. We
don't have $\mu\subseteq\lambda$. Hence, we don't have $\mu^{t}\subseteq
\lambda^{t}$ either. Thus, $s_{\lambda^{t}/\mu^{t}}=0$, so that $0=s_{\lambda
^{t}/\mu^{t}}$. But we also have $s_{\lambda/\mu}=0$ (since we don't have
$\mu\subseteq\lambda$), so that $\omega\left(  \underbrace{s_{\lambda/\mu}%
}_{=0}\right)  =\omega\left(  0\right)  =0$ (since $\omega$ is $\kk %
$-linear), thus $\omega\left(  s_{\lambda/\mu}\right)  =0=s_{\lambda^{t}%
/\mu^{t}}$, qed.}.

We know that $\omega$ is a $\kk$-algebra homomorphism. As a
consequence, we can apply (\ref{sol.jacobi-trudi.macdonald.det-functor}) to
$\mathbf{r}=\Lambda$, $\mathbf{s}=\Lambda$, $k=\ell$, $\varphi=\omega$ and
$M=\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}$. As a
result, we obtain%
\begin{equation}
\omega\left(  \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  \right)  =\det\left(  \omega^{\ell\times \ell}\left(
\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)
\right)  . \label{sol.jacobi-trudi.macdonald.e.1}%
\end{equation}


Now, the matrix $\left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots
,\ell}$ is obviously written in the form \newline$\left(  h_{\lambda_{i}%
-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}=\left(  h_{\lambda_{i}-\mu
_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}$. Hence, by the definition of the
matrix $\omega^{\ell\times \ell}\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  $, we have
\[
\omega^{\ell\times \ell}\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  =\left(  \underbrace{\omega\left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  }_{\substack{=e_{\lambda_{i}-\mu_{j}%
-i+j}\\\text{(by (\ref{sol.jacobi-trudi.macdonald.h.wh}), applied}\\\text{to
}m=\lambda_{i}-\mu_{j}-i+j\text{)}}}\right)  _{i,j=1,2,\ldots,\ell}=\left(
e_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}.
\]
Thus, (\ref{sol.jacobi-trudi.macdonald.e.1}) becomes
\[
\omega\left(  \det\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  \right)  =\det\left(  \underbrace{\omega
^{\ell\times \ell}\left(  \left(  h_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}\right)  }_{=\left(  e_{\lambda_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,\ell}}\right)  =\det\left(  \left(  e_{\lambda_{i}-\mu
_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  .
\]
Compared with%
\[
\omega\left(  \underbrace{\det\left(  \left(  h_{\lambda_{i}-\mu_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  }_{\substack{=s_{\lambda/\mu
}\\\text{(by (\ref{sol.jacobi-trudi.macdonald.h.done}))}}}\right)
=\omega\left(  s_{\lambda/\mu}\right)  =s_{\lambda^{t}/\mu^{t}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.jacobi-trudi.macdonald.omega.h}%
)}\right)  ,
\]
this yields $s_{\lambda^{t}/\mu^{t}}=\det\left(  \left(  e_{\lambda_{i}%
-\mu_{j}-i+j}\right)  _{i,j=1,2,\ldots,\ell}\right)  $. This proves
(\ref{eq.jacobi-trudi.e}). Therefore, the proof of Theorem
\ref{Jacobi-Trudi-formulae} is complete.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.hall-e-f}}

\textit{Solution to Exercise \ref{exe.hall-e-f}.} (a) The second identity of
\eqref{skew-Schur-antipodes} shows that%
\begin{equation}
S\left(  s_{\lambda/\mu}\right)  =\left(  -1\right)  ^{\left\vert \lambda
/\mu\right\vert }s_{\lambda^{t}/\mu^{t}} \label{sol.hall-e-f.a.1}%
\end{equation}
whenever $\lambda$ and $\mu$ are partitions satisfying $\mu\subseteq\lambda$.
Hence,%
\begin{equation}
S\left(  s_{\lambda}\right)  =\left(  -1\right)  ^{\left\vert \lambda
\right\vert }s_{\lambda^{t}}\ \ \ \ \ \ \ \ \ \ \text{for any partition
}\lambda. \label{sol.hall-e-f.a.2}%
\end{equation}


[\textit{Proof of (\ref{sol.hall-e-f.a.2}):} Let $\lambda$ be any partition.
Then, the empty partition $\varnothing$ clearly satisfies $\varnothing
\subseteq\lambda$. Hence, (\ref{sol.hall-e-f.a.1}) (applied to $\mu
=\varnothing$) yields $S\left(  s_{\lambda/\varnothing}\right)  =\left(
-1\right)  ^{\left\vert \lambda/\varnothing\right\vert }s_{\lambda
^{t}/\varnothing^{t}}$. In view of $s_{\lambda/\varnothing}=s_{\lambda}$ and
$\varnothing^{t}=\varnothing$, this rewrites as $S\left(  s_{\lambda}\right)
=\left(  -1\right)  ^{\left\vert \lambda/\varnothing\right\vert }%
s_{\lambda^{t}/\varnothing}$. In view of $\left\vert \lambda/\varnothing
\right\vert =\left\vert \lambda\right\vert $ and $s_{\lambda^{t}/\varnothing
}=s_{\lambda^{t}}$, this rewrites as $S\left(  s_{\lambda}\right)  =\left(
-1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda^{t}}$. This proves
(\ref{sol.hall-e-f.a.2}).]

On the other hand, the definition of the Hall inner product $\left(
\cdot,\cdot\right)  $ shows that%
\begin{equation}
\left(  s_{\lambda},s_{\nu}\right)  =\delta_{\lambda,\nu}%
\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }\lambda\text{ and }\nu.
\label{sol.hall-e-f.a.orthonormal-s}%
\end{equation}


Now, let $f\in\Lambda$ and $g\in\Lambda$.

Proposition \ref{many-symmetric-function-bases-prop} shows that the family
$\left(  s_{\lambda}\right)  _{\lambda\in\Par}$ is a basis of
the $\kk$-module $\Lambda$. Hence, $f$ can be written in the form
$f=\sum_{\lambda\in\Par}a_{\lambda}s_{\lambda}$ for some family
$\left(  a_{\lambda}\right)  _{\lambda\in\Par}\in
\kk^{\Par}$ of elements of $\kk$ such that all
but finitely many $\lambda\in\Par$ satisfy $a_{\lambda}=0$.
Consider this family $\left(  a_{\lambda}\right)  _{\lambda\in
\Par}$.

Proposition \ref{many-symmetric-function-bases-prop} shows that the family
$\left(  s_{\lambda}\right)  _{\lambda\in\Par}$ is a basis of
the $\kk$-module $\Lambda$. Hence, $g$ can be written in the form
$g=\sum_{\lambda\in\Par}b_{\lambda}s_{\lambda}$ for some family
$\left(  b_{\lambda}\right)  _{\lambda\in\Par}\in
\kk^{\Par}$ of elements of $\kk$ such that all
but finitely many $\lambda\in\Par$ satisfy $b_{\lambda}=0$.
Consider this family $\left(  b_{\lambda}\right)  _{\lambda\in
\Par}$.

The map $\Par\rightarrow\Par,\ \lambda \mapsto\lambda^{t}$
is a bijection. (Indeed, this map is inverse to itself,
since each partition $\lambda$ satisfies $\left(  \lambda^{t}\right)
^{t}=\lambda$).

Applying the map $S$ to both sides of the equality $f=\sum_{\lambda
\in\Par}a_{\lambda}s_{\lambda}$, we obtain%
\begin{align*}
S\left(  f\right)   &  =S\left(  \sum_{\lambda\in\Par%
}a_{\lambda}s_{\lambda}\right)  =\sum_{\lambda\in\Par%
}a_{\lambda}\underbrace{S\left(  s_{\lambda}\right)  }_{\substack{=\left(
-1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda^{t}}\\\text{(by
(\ref{sol.hall-e-f.a.2}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}S\text{ is }\kk\text{-linear}\right) \\
&  =\sum_{\lambda\in\Par}a_{\lambda}\left(  -1\right)
^{\left\vert \lambda\right\vert }s_{\lambda^{t}}=\sum_{\lambda\in
\Par}a_{\lambda^{t}}\underbrace{\left(  -1\right)  ^{\left\vert
\lambda^{t}\right\vert }}_{\substack{=\left(  -1\right)  ^{\left\vert
\lambda\right\vert }\\\text{(since }\left\vert \lambda^{t}\right\vert
=\left\vert \lambda\right\vert \text{)}}}\underbrace{s_{\left(  \lambda
^{t}\right)  ^{t}}}_{\substack{=s_{\lambda}\\\text{(since }\left(  \lambda
^{t}\right)  ^{t}=\lambda\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\lambda^{t}\text{ for }\lambda\text{ in the
sum, since}\\
\text{the map }\Par\rightarrow\Par%
,\ \lambda\mapsto\lambda^{t}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum_{\lambda\in\Par}a_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }s_{\lambda}.
\end{align*}
The same argument (applied to $g$ and $b_{\lambda}$ instead of $f$ and
$a_{\lambda}$) yields%
\begin{align*}
S\left(  g\right)   &  =\sum_{\lambda\in\Par}b_{\lambda^{t}%
}\left(  -1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g=\sum_{\lambda\in\Par}
b_{\lambda}s_{\lambda}\right) \\
&  =\sum_{\mu\in\Par}b_{\mu^{t}}\left(  -1\right)  ^{\left\vert
\mu\right\vert }s_{\mu}%
\end{align*}
(here, we have renamed the summation index $\lambda$ as $\mu$ in the sum).
Also,%
\[
g=\sum_{\lambda\in\Par}b_{\lambda}s_{\lambda}=\sum_{\mu
\in\Par}b_{\mu}s_{\mu}%
\]
(here, we have renamed the summation index $\lambda$ as $\mu$ in the sum).

Now,%
\begin{align*}
&  \left(  \underbrace{S\left(  f\right)  }_{=\sum_{\lambda\in
\Par}a_{\lambda^{t}}\left(  -1\right)  ^{\left\vert
\lambda\right\vert }s_{\lambda}},\underbrace{S\left(  g\right)  }_{=\sum
_{\mu\in\Par}b_{\mu^{t}}\left(  -1\right)  ^{\left\vert
\mu\right\vert }s_{\mu}}\right)  \\
&  =\left(  \sum_{\lambda\in\Par}a_{\lambda^{t}}\left(
-1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda},\sum_{\mu
\in\Par}b_{\mu^{t}}\left(  -1\right)  ^{\left\vert
\mu\right\vert }s_{\mu}\right)  \\
&  =\sum_{\lambda\in\Par}a_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }\sum_{\mu\in\Par}b_{\mu^{t}%
}\left(  -1\right)  ^{\left\vert \mu\right\vert }\underbrace{\left(
s_{\lambda},s_{\mu}\right)  }_{\substack{=\delta_{\lambda,\mu}\\\text{(by
(\ref{sol.hall-e-f.a.orthonormal-s}),}\\\text{applied to }\nu=\mu\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product }\left(
\cdot,\cdot\right)  \text{ is }\kk\text{-bilinear}\right)  \\
&  =\sum_{\lambda\in\Par}a_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }\underbrace{\sum_{\mu\in\Par%
}b_{\mu^{t}}\left(  -1\right)  ^{\left\vert \mu\right\vert }\delta
_{\lambda,\mu}}_{\substack{=b_{\lambda^{t}}\left(  -1\right)  ^{\left\vert
\lambda\right\vert }\delta_{\lambda,\lambda}+\sum\limits_{\substack{\mu
\in\Par;\\\mu\neq\lambda}}b_{\mu^{t}}\left(  -1\right)
^{\left\vert \mu\right\vert }\delta_{\lambda,\mu}\\\text{(here, we have split
off the addend for }\mu=\lambda\text{ from the sum)}}}\\
&  =\sum_{\lambda\in\Par}a_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }\left(  b_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }\underbrace{\delta_{\lambda,\lambda}}%
_{=1}+\sum\limits_{\substack{\mu\in\Par;\\\mu\neq\lambda}}b_{\mu^{t}%
}\left(  -1\right)  ^{\left\vert \mu\right\vert }\underbrace{\delta
_{\lambda,\mu}}_{\substack{=0\\\text{(since }\lambda\neq\mu\\\text{(since }%
\mu\neq\lambda\text{))}}}\right)  \\
&  =\sum_{\lambda\in\Par}a_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }\left(  b_{\lambda^{t}}\left(  -1\right)
^{\left\vert \lambda\right\vert }+\underbrace{\sum\limits_{\substack{\mu
\in\Par;\\\mu\neq\lambda}}b_{\mu^{t}}\left(  -1\right)
^{\left\vert \mu\right\vert }0}_{=0}\right)  \\
&  =\sum_{\lambda\in\Par}a_{\lambda^{t}}\underbrace{\left(
-1\right)  ^{\left\vert \lambda\right\vert }b_{\lambda^{t}}}_{=b_{\lambda^{t}%
}\left(  -1\right)  ^{\left\vert \lambda\right\vert }}\left(  -1\right)
^{\left\vert \lambda\right\vert }=\sum_{\lambda\in\Par%
}a_{\lambda^{t}}b_{\lambda^{t}}\underbrace{\left(  -1\right)  ^{\left\vert
\lambda\right\vert }\left(  -1\right)  ^{\left\vert \lambda\right\vert }%
}_{\substack{=\left(  -1\right)  ^{\left\vert \lambda\right\vert +\left\vert
\lambda\right\vert }=1\\\text{(since }\left\vert \lambda\right\vert
+\left\vert \lambda\right\vert =2\left\vert \lambda\right\vert \text{ is
even)}}}=\sum_{\lambda\in\Par}a_{\lambda^{t}}b_{\lambda^{t}%
}=\sum_{\lambda\in\Par}a_{\lambda}b_{\lambda}%
\end{align*}
(here, we have substituted $\lambda^{t}$ for $\lambda$ in the sum, since the
map $\Par\rightarrow\Par,\ \lambda\mapsto
\lambda^{t}$ is a bijection). Comparing this with%
\begin{align*}
&  \left(  \underbrace{f}_{=\sum_{\lambda\in\Par}a_{\lambda
}s_{\lambda}},\underbrace{g}_{=\sum_{\mu\in\Par}b_{\mu}s_{\mu}%
}\right)  \\
&  =\left(  \sum_{\lambda\in\Par}a_{\lambda}s_{\lambda}%
,\sum_{\mu\in\Par}b_{\mu}s_{\mu}\right)  =\sum_{\lambda
\in\Par}a_{\lambda}\sum_{\mu\in\Par}b_{\mu
}\underbrace{\left(  s_{\lambda},s_{\mu}\right)  }_{\substack{=\delta
_{\lambda,\mu}\\\text{(by (\ref{sol.hall-e-f.a.orthonormal-s}),}%
\\\text{applied to }\nu=\mu\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product }\left(
\cdot,\cdot\right)  \text{ is }\kk\text{-bilinear}\right)  \\
&  =\sum_{\lambda\in\Par}a_{\lambda}\underbrace{\sum_{\mu
\in\Par}b_{\mu}\delta_{\lambda,\mu}}_{\substack{=b_{\lambda
}\delta_{\lambda,\lambda}+\sum\limits_{\substack{\mu\in\Par;\\\mu
\neq\lambda}}b_{\mu}\delta_{\lambda,\mu}\\\text{(here, we have split off the
addend for }\mu=\lambda\text{ from the sum)}}}\\
&  =\sum_{\lambda\in\Par}a_{\lambda}\left(  b_{\lambda
}\underbrace{\delta_{\lambda,\lambda}}_{=1}+\sum\limits_{\substack{\mu\in
\Par;\\\mu\neq\lambda}}b_{\mu}\underbrace{\delta_{\lambda,\mu}%
}_{\substack{=0\\\text{(since }\lambda\neq\mu\\\text{(since }\mu\neq
\lambda\text{))}}}\right)  =\sum_{\lambda\in\Par}a_{\lambda
}\left(  b_{\lambda}+\underbrace{\sum\limits_{\substack{\mu\in\Par%
;\\\mu\neq\lambda}}b_{\mu}0}_{=0}\right)  =\sum_{\lambda\in\Par%
}a_{\lambda}b_{\lambda},
\end{align*}
we obtain $\left(  S\left(  f\right)  ,S\left(  g\right)  \right)  =\left(
f,g\right)  $. This solves Exercise \ref{exe.hall-e-f}(a).

(b) Let $n \in \NN$ and $f \in \Lambda_n$.
We know that the Hopf algebra $\Lambda$ is graded; thus, its antipode
$S$ is a graded $\kk$-linear map. Hence, $S\left(\Lambda_n\right)
\subset \Lambda_n$.
Now, from $f \in \Lambda_n$, we obtain
$S\left(f\right) \in S\left(\Lambda_n\right)
\subset \Lambda_n$.
Hence, Exercise~\ref{exe.Lambda.hall.gr}(b) (applied to $S\left(f\right)$
instead of $f$) yields
$\left(h_n, S\left(f\right)\right)
= \left(S\left(f\right)\right)\left(1\right)$.

But Proposition \ref{symm-antipode-and-h-basis}(ii) yields $S\left(
e_{n}\right)  =\left(  -1\right)  ^{n}h_{n}$. Hence,
\begin{align*}
&  \left(  \underbrace{S\left(  e_{n}\right)  }_{=\left(  -1\right)  ^{n}%
h_{n}},S\left(  f\right)  \right) \\
&  =\left(  \left(  -1\right)  ^{n}h_{n},S\left(  f\right)  \right)  =\left(
-1\right)  ^{n}\cdot\underbrace{\left(  h_{n},S\left(  f\right)  \right)
}_{=\left(  S\left(  f\right)  \right)  \left(  1\right)}
\ \ \ \ \ \ \ \ \ \ \left(
\text{since the Hall inner product }\left(  \cdot,\cdot\right)  \text{ is
}\kk\text{-bilinear}\right) \\
&  =\left(  -1\right)  ^{n}\cdot\left(  S\left(  f\right)  \right)  \left(
1\right)  .
\end{align*}
But Exercise \ref{exe.hall-e-f}%
(a) (applied to $e_{n}$ and $f$ instead of $f$ and $g$) yields $\left(
S\left(  e_{n}\right)  ,S\left(  f\right)  \right)  =\left(  e_{n},f\right)
$.
Comparing these two equalities, we obtain
$\left(e_n, f\right) =\left(  -1\right)  ^{n}\cdot\left(  S\left(  f\right)  \right)  \left(
1\right)$. 
This solves Exercise \ref{exe.hall-e-f}(b).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.coPieri-rules}}

\textit{Solution to Exercise~\ref{exe.Lambda.coPieri-rules}.}
Let us first prove two lemmas:

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.s-determines}Let $f\in\Lambda$ and
$g\in\Lambda$. Assume that
\begin{equation}
\left(  s_{\lambda},f\right)  =\left(  s_{\lambda},g\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }\lambda\in \Par .
\label{eq.lem.sol.Lambda.staircase-skew-2.s-determines.1}
\end{equation}
Then, $f=g$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.s-determines}.]The basis
$\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ of $\Lambda$ is
orthonormal with respect to the Hall inner product $\left(  \cdot
,\cdot\right)  $ (by Definition \ref{def.hall-inner-product}). In other words,
the two (identical) bases $\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ and $\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ of $\Lambda$ are dual to each other with respect to
the Hall inner product $\left(  \cdot,\cdot\right)  $.

Recall the following fundamental fact from linear algebra: If $\kk$ is
a commutative ring, if $A$ is a $\kk$-module, if $\left(  \cdot
,\cdot\right)  :A\times A\rightarrow\kk$ is a symmetric $\kk%
$-bilinear form on $A$, and if $\left(  u_{\lambda}\right)  _{\lambda\in L}$
and $\left(  v_{\lambda}\right)  _{\lambda\in L}$ are two $\kk$-bases
of $A$ which are dual to each other with respect to the form $\left(
\cdot,\cdot\right)  $ (where $L$ is some indexing set), then every $a\in A$
satisfies
\begin{equation}
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}.
\label{pf.lem.sol.Lambda.staircase-skew-2.s-determines.1}%
\end{equation}
We can apply this fact to $A=\Lambda$, $L=\operatorname{Par}$, $\left(
u_{\lambda}\right)  _{\lambda\in L}=\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ and $\left(  v_{\lambda}\right)  _{\lambda\in
L}=\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ (since the
bases $\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ and
$\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ of $\Lambda$
are dual to each other with respect to the Hall inner product $\left(
\cdot,\cdot\right)  $). We thus conclude that every $a\in\Lambda$ satisfies%
\begin{equation}
a=\sum_{\lambda\in \Par }\left(  s_{\lambda},a\right)
s_{\lambda}.
\label{pf.lem.sol.Lambda.staircase-skew-2.s-determines.2}
\end{equation}
Applying this to $a=f$, we obtain%
\[
f=\sum_{\lambda\in \Par }\underbrace{\left(  s_{\lambda
},f\right)  }_{\substack{=\left(  s_{\lambda},g\right)  \\\text{(by
(\ref{eq.lem.sol.Lambda.staircase-skew-2.s-determines.1}))}}}s_{\lambda}%
=\sum_{\lambda\in \Par }\left(  s_{\lambda},g\right)  s_{\lambda
}.
\]
Comparing this with%
\[
g=\sum_{\lambda\in \Par }\left(  s_{\lambda},g\right)
s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.s-determines.2}), applied to
}a=g\right)  ,
\]
we obtain $f=g$. This proves Lemma
\ref{lem.sol.Lambda.staircase-skew-2.s-determines}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.coPieri}Let $\gamma\in
 \Par $ and $k\in \NN$.

\begin{enumerate}
\item[(a)] We have%
\begin{equation}
h_{k}^{\perp}s_{\gamma}=\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}s_{\nu}.
\label{eq.lem.sol.Lambda.staircase-skew-2.coPieri.h}
\end{equation}


\item[(b)] We have%
\begin{equation}
e_{k}^{\perp}s_{\gamma}=\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}.
\label{eq.lem.sol.Lambda.staircase-skew-2.coPieri.e}
\end{equation}

\end{enumerate}
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}.]We know that
every $f\in\Lambda$, $g\in\Lambda$ and $a\in\Lambda$ satisfy
\begin{equation}
\left(  g,f^{\perp}\left(  a\right)  \right)  =\left(  fg,a\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.adjt}
\end{equation}
Indeed, this follows from Proposition \ref{skewing-properties-prop}(i)
(applied to $A=\Lambda$), after we make the standard identification of
$\Lambda^o$ with $\Lambda$ via the Hall inner product.

Recall that the basis $\left(  s_{\lambda}\right)  _{\lambda\in
 \Par }$ of $\Lambda$ is orthonormal with respect to the Hall
inner product $\left(  \cdot,\cdot\right)  $ (by Definition
\ref{def.hall-inner-product}). In other words,
\begin{equation}
\left(  s_{\alpha},s_{\beta}\right)  =\delta_{\alpha,\beta}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \alpha,\beta\right)
\in \Par \times \Par .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}
\end{equation}


(b) We claim that%
\begin{equation}
\left(  s_{\lambda},e_{k}^{\perp}s_{\gamma}\right)  =\left(  s_{\lambda}%
,\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}%
\end{equation}
for every $\lambda\in \Par $.

\textit{Proof of (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}):}
Let $\lambda\in \Par $. We have%
\begin{equation}
e_{k}s_{\lambda}=s_{\lambda}e_{k}=\sum\limits_{\substack{\lambda^{+}:\lambda
^{+}/\lambda\text{ is a }\\\text{vertical }k\text{-strip}}}s_{\lambda^{+}}
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-e}%
\end{equation}
(by (\ref{eq.pieri.e}), applied to $n=k$). But
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.adjt}) (applied to $f=e_{k}$,
$a=s_{\gamma}$ and $g=s_{\lambda}$) yields%
\begin{align}
\left(  s_{\lambda},e_{k}^{\perp}s_{\gamma}\right)   &  =\left(
e_{k}s_{\lambda},s_{\gamma}\right)  =\left(  \sum\limits_{\substack{\lambda
^{+}:\lambda^{+}/\lambda\text{ is a }\\\text{vertical }k\text{-strip}%
}}s_{\lambda^{+}},s_{\gamma}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-e})}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{vertical }k\text{-strip}}}\underbrace{\left(  s_{\lambda^{+}%
},s_{\gamma}\right)  }_{\substack{=\delta_{\lambda^{+},\gamma}\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}), applied}%
\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda^{+},\gamma\right)
\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk\text{-bilinear}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{vertical }k\text{-strip}}}\delta_{\lambda^{+},\gamma}=%
\begin{cases}
1, & \text{if }\gamma/\lambda\text{ is a vertical }k\text{-strip};\\
0, & \text{otherwise}%
\end{cases} \quad
. \label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3}
\end{align}


On the other hand, the Hall inner product is $\kk$-bilinear. Thus,%
\begin{align*}
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)   &
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}\underbrace{\left(  s_{\lambda},s_{\nu
}\right)  }_{\substack{=\delta_{\lambda,\nu}\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}), applied}%
\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda,\nu\right)
\text{)}}}\\
&  =\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}\delta_{\lambda,\nu}=%
\begin{cases}
1, & \text{if }\gamma/\lambda\text{ is a vertical }k\text{-strip};\\
0, & \text{otherwise}%
\end{cases} \quad
.
\end{align*}
Comparing this with (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3}), we
obtain%
\[
\left(  s_{\lambda},e_{k}^{\perp}s_{\gamma}\right)  =\left(  s_{\lambda}%
,\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)  .
\]
Thus, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}) is proven.

Now, we have proven (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal})
for all $\lambda\in \Par $. Hence, Lemma
\ref{lem.sol.Lambda.staircase-skew-2.s-determines} (applied to $f=e_{k}%
^{\perp}s_{\gamma}$ and $g=\sum\limits_{\substack{\nu\in \Par %
;\\\gamma/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}$) yields
$e_{k}^{\perp}s_{\gamma}=\sum\limits_{\substack{\nu\in \Par %
;\\\gamma/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}$. This
proves Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}(b).

(a) We claim that
\begin{equation}
\left(  s_{\lambda},h_{k}^{\perp}s_{\gamma}\right)  =\left(  s_{\lambda}%
,\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}
\end{equation}
for every $\lambda\in \Par $.

\textit{Proof of (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}):}
Let $\lambda\in \Par $. Then,%
\begin{equation}
h_{k}s_{\lambda}=s_{\lambda}h_{k}=\sum\limits_{\substack{\lambda^{+}:\lambda
^{+}/\lambda\text{ is a }\\\text{horizontal }k\text{-strip}}}s_{\lambda^{+}%
}\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-h}%
\end{equation}
(by (\ref{eq.pieri.h}), applied to $n=k$). But
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.adjt}) (applied to $f=h_{k}$,
$a=s_{\gamma}$ and $g=s_{\lambda}$) yields%
\begin{align}
\left(  s_{\lambda},h_{k}^{\perp}s_{\gamma}\right)   &  =\left(
h_{k}s_{\lambda},s_{\gamma}\right)  =\left(  \sum\limits_{\substack{\lambda
^{+}:\lambda^{+}/\lambda\text{ is a }\\\text{horizontal }k\text{-strip}%
}}s_{\lambda^{+}},s_{\gamma}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-h})}\right)  \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{horizontal }k\text{-strip}}}\underbrace{\left(  s_{\lambda^{+}%
},s_{\gamma}\right)  }_{\substack{=\delta_{\lambda^{+},\gamma}\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}), applied}%
\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda^{+},\gamma\right)
\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
}\kk\text{-bilinear}\right)  \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a }%
\\\text{horizontal }k\text{-strip}}}\delta_{\lambda^{+},\gamma}=%
\begin{cases}
1, & \text{if }\gamma/\lambda\text{ is a horizontal }k\text{-strip};\\
0, & \text{otherwise}%
\end{cases} \quad
.\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3}
\end{align}


On the other hand, the Hall inner product is $\kk$-bilinear. Thus,%
\begin{align*}
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)   &
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}\underbrace{\left(  s_{\lambda},s_{\nu
}\right)  }_{\substack{=\delta_{\lambda,\nu}\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}), applied}%
\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda,\nu\right)
\text{)}}}\\
&  =\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}\delta_{\lambda,\nu}=%
\begin{cases}
1, & \text{if }\gamma/\lambda\text{ is a horizontal }k\text{-strip};\\
0, & \text{otherwise}%
\end{cases} \quad
.
\end{align*}
Comparing this with (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3}), we
obtain%
\[
\left(  s_{\lambda},h_{k}^{\perp}s_{\gamma}\right)  =\left(  s_{\lambda}%
,\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)  .
\]
Thus, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}) is proven.

Now, we have proven (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal})
for all $\lambda\in \Par $. Hence, Lemma
\ref{lem.sol.Lambda.staircase-skew-2.s-determines} (applied to $f=h_{k}%
^{\perp}s_{\gamma}$ and $g=\sum\limits_{\substack{\nu\in \Par %
;\\\gamma/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}s_{\nu}$) yields
$h_{k}^{\perp}\left(  s_{\gamma}\right)  =\sum\limits_{\substack{\nu\in
 \Par ;\\\gamma/\nu\text{ is a}\\\text{horizontal }%
k\text{-strip}}}s_{\nu}$. This proves Lemma
\ref{lem.sol.Lambda.staircase-skew-2.coPieri}(a).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}.]There exist
only finitely many $\nu\in \Par $ satisfying $\nu\subseteq\gamma
$. Hence, there exist only finitely many $\nu\in \Par $ for which
$\gamma/\nu$ is a horizontal $k$-strip (since all such $\nu$ must satisfy
$\nu\subseteq\gamma$). Thus, the sum $\sum\limits_{\substack{\nu\in
 \Par ;\\\gamma/\nu\text{ is a}\\\text{horizontal }%
k\text{-strip}}}s_{\nu}$ on the right hand side of
(\ref{eq.lem.sol.Lambda.staircase-skew-2.coPieri.h}) is well-defined.
Similarly, the sum $\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}$ on the right hand
side of (\ref{eq.lem.sol.Lambda.staircase-skew-2.coPieri.e}) is well-defined.

We know that every $f\in\Lambda$, $g\in\Lambda$ and $a\in\Lambda$ satisfy%
\begin{equation}
\left(  g,f^{\perp}\left(  a\right)  \right)  =\left(  fg,a\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.adjt}%
\end{equation}
Indeed, this follows from Proposition \ref{skewing-properties-prop}(i)
(applied to $A=\Lambda$), after we make the standard identification of
$\Lambda^o$ with $\Lambda$ via the Hall inner product.

Recall that the basis $\left(  s_{\lambda}\right)  _{\lambda\in
 \Par }$ of $\Lambda$ is orthonormal with respect to the Hall
inner product $\left(  \cdot,\cdot\right)  $ (by Definition
\ref{def.hall-inner-product}). In other words,%
\begin{equation}
\left(  s_{\alpha},s_{\beta}\right)  =\delta_{\alpha,\beta}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \alpha,\beta\right)
\in \Par \times \Par .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}%
\end{equation}


(b) We claim that%
\begin{equation}
\left(  s_{\lambda},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(
s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}%
\end{equation}
for every $\lambda\in \Par $.

\textit{Proof of (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}):}
Let $\lambda\in \Par $. The equality (\ref{eq.pieri.e}) (applied
to $n=k$) yields%
\begin{equation}
s_{\lambda}e_{k}=\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a
}\\\text{vertical }k\text{-strip}}}s_{\lambda^{+}}=\sum\limits_{\substack{\lambda
^{+}\in \Par ;\\\lambda^{+}/\lambda\text{ is a }\\\text{vertical
}k\text{-strip}}}s_{\lambda^{+}}.
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-e}%
\end{equation}
But (\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.adjt}) (applied to
$f=e_{k}$, $a=s_{\gamma}$ and $g=s_{\lambda}$) yields%
\begin{align}
\left(  s_{\lambda},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)   &
=\left(  \underbrace{e_{k}s_{\lambda}}_{\substack{=s_{\lambda}e_{k}%
\\=\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}%
/\lambda\text{ is a }\\\text{vertical }k\text{-strip}}}s_{\lambda^{+}%
}\\\text{(by (\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-e}))}%
}},s_{\gamma}\right)  =\left(  \sum\limits_{\substack{\lambda^{+}\in
 \Par ;\\\lambda^{+}/\lambda\text{ is a }\\\text{vertical
}k\text{-strip}}}s_{\lambda^{+}},s_{\gamma}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}%
/\lambda\text{ is a }\\\text{vertical }k\text{-strip}}}\underbrace{\left(
s_{\lambda^{+}},s_{\gamma}\right)  }_{\substack{=\delta_{\lambda^{+},\gamma
}\\\text{(by (\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}),
applied}\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda^{+}%
,\gamma\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall
inner product is }\kk\text{-bilinear}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}%
/\lambda\text{ is a }\\\text{vertical }k\text{-strip}}}\delta_{\lambda
^{+},\gamma}. \label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3}%
\end{align}


On the other hand, the Hall inner product is $\kk$-bilinear. Thus,%
\begin{align}
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)   &
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}\underbrace{\left(  s_{\lambda},s_{\nu
}\right)  }_{\substack{=\delta_{\lambda,\nu}\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}), applied}%
\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda,\nu\right)
\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}\delta_{\lambda,\nu}.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.4}%
\end{align}


Now, we must be in one of the following two cases:

\textit{Case 1:} The skew shape $\gamma/\lambda$ is a vertical $k$-strip.

\textit{Case 2:} The skew shape $\gamma/\lambda$ is not a vertical $k$-strip.

(The case when the skew shape $\gamma/\lambda$ is not well-defined (i.e., when
$\lambda\not \subseteq \gamma$) is counted towards Case 2.)

Let us first consider Case 1. In this case, the skew shape $\gamma/\lambda$ is
a vertical $k$-strip. Hence, the sum $\sum\limits_{\substack{\lambda^{+}%
\in \Par ;\\\lambda^{+}/\lambda\text{ is a }\\\text{vertical
}k\text{-strip}}}\delta_{\lambda^{+},\gamma}$ has an addend for $\lambda
^{+}=\gamma$. Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{
is a }\\\text{vertical }k\text{-strip}}}\delta_{\lambda^{+},\gamma
}=\underbrace{\delta_{\gamma,\gamma}}_{\substack{=1\\\text{(since }%
\gamma=\gamma\text{)}}}+\sum\limits_{\substack{\lambda^{+}\in \Par %
;\\\lambda^{+}/\lambda\text{ is a }\\\text{vertical }k\text{-strip;}%
\\\lambda^{+}\neq\gamma}}\underbrace{\delta_{\lambda^{+},\gamma}%
}_{\substack{=0\\\text{(since }\lambda^{+}\neq\gamma\text{)}}%
}=1+\underbrace{\sum\limits_{\substack{\lambda^{+}\in \Par %
;\\\lambda^{+}/\lambda\text{ is a }\\\text{vertical }k\text{-strip;}%
\\\lambda^{+}\neq\gamma}}0}_{=0}=1.
\]
Hence, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3}) becomes%
\begin{equation}
\left(  s_{\lambda},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)
=\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{
is a }\\\text{vertical }k\text{-strip}}}\delta_{\lambda^{+},\gamma}=1.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3.c1}%
\end{equation}
On the other hand, the skew shape $\gamma/\lambda$ is a vertical $k$-strip.
Hence, the sum $\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{
is a}\\\text{vertical }k\text{-strip}}}\delta_{\lambda,\nu}$ has an addend for
$\nu=\lambda$. Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}%
\\\text{vertical }k\text{-strip}}}\delta_{\lambda,\nu}=\underbrace{\delta
_{\lambda,\lambda}}_{\substack{=1\\\text{(since }\lambda=\lambda\text{)}%
}}+\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip;}\\\nu\neq\lambda}}\underbrace{\delta
_{\lambda,\nu}}_{\substack{=0\\\text{(since }\lambda\neq\nu\\\text{(since }%
\nu\neq\lambda\text{))}}}=1+\underbrace{\sum\limits_{\substack{\nu\in
 \Par ;\\\gamma/\nu\text{ is a}\\\text{vertical }k\text{-strip;}%
\\\nu\neq\lambda}}0}_{=0}=1.
\]
Hence, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.4}) becomes%
\begin{equation}
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}\delta_{\lambda,\nu}=1.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.4.c1}%
\end{equation}
Comparing this with (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3.c1}),
we obtain%
\[
\left(  s_{\lambda},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(
s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)  .
\]
Thus, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}) is proven in
Case 1.

Let us now consider Case 2. In this case, the skew shape $\gamma/\lambda$ is
not a vertical $k$-strip. Hence, if $\lambda^{+}\in \Par $ has
the property that $\lambda^{+}/\lambda$ is a vertical $k$-strip, then%
\begin{equation}
\delta_{\lambda^{+},\gamma}=0
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.1}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.1}):} Let $\lambda
^{+}\in \Par $ have the property that $\lambda^{+}/\lambda$ is a
vertical $k$-strip. Assume (for the sake of contradiction) that $\lambda
^{+}=\gamma$. Then, $\lambda^{+}/\lambda$ is a vertical $k$-strip. In other
words, $\gamma/\lambda$ is a vertical $k$-strip (since $\lambda^{+}=\gamma$).
This contradicts the fact that $\gamma/\lambda$ is not a vertical $k$-strip.
\par
This contradiction shows that our assumption (that $\lambda^{+}=\gamma$) was
false. Thus, $\lambda^{+}\neq\gamma$. Hence, $\delta_{\lambda^{+},\gamma}=0$.
This proves (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.1}).}. Now,
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.3}) becomes%
\begin{equation}
\left(  s_{\lambda},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)
=\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{
is a }\\\text{vertical }k\text{-strip}}}\underbrace{\delta_{\lambda^{+}%
,\gamma}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.1}))}}}=\sum\limits
_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{ is a
}\\\text{vertical }k\text{-strip}}}0=0.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.2}%
\end{equation}
On the other hand, if $\nu\in \Par $ has the property that
$\gamma/\nu$ is a vertical $k$-strip, then
\begin{equation}
\delta_{\lambda,\nu}=0
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.3}):} Let $\nu
\in \Par $ have the property that $\gamma/\nu$ is a vertical
$k$-strip. Assume (for the sake of contradiction) that $\lambda=\nu$. Then,
$\gamma/\nu$ is a vertical $k$-strip. In other words, $\gamma/\lambda$ is a
vertical $k$-strip (since $\lambda=\nu$). This contradicts the fact that
$\gamma/\lambda$ is not a vertical $k$-strip.
\par
This contradiction shows that our assumption (that $\lambda=\nu$) was false.
Thus, $\lambda\neq\nu$. Hence, $\delta_{\lambda,\nu}=0$. This proves
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.3}).}. Now,
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.4}) becomes%
\[
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}\underbrace{\delta_{\lambda,\nu}%
}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.3}))}}}=\sum\limits
_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}\\\text{vertical
}k\text{-strip}}}0=0.
\]
Comparing this with (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.c2.2}),
we obtain%
\[
\left(  s_{\lambda},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(
s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)  .
\]
Thus, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}) is proven in
Case 2.

We have now proven (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal})
in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this shows that
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal}) always holds.

Now, we have proven (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.b.goal})
for all $\lambda\in \Par $. In other words, $\left(  s_{\lambda
},e_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(  s_{\lambda}%
,\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{vertical }k\text{-strip}}}s_{\nu}\right)  $ for each $\lambda
\in \Par $. Thus, Lemma
\ref{lem.sol.Lambda.staircase-skew-2.s-determines} (applied to $f=e_{k}%
^{\perp}\left(  s_{\gamma}\right)  $ and $g=\sum\limits_{\substack{\nu\in
 \Par ;\\\gamma/\nu\text{ is a}\\\text{vertical }k\text{-strip}%
}}s_{\nu}$) yields $e_{k}^{\perp}\left(  s_{\gamma}\right)  =\sum\limits
_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}\\\text{vertical
}k\text{-strip}}}s_{\nu}$. Thus,%
\[
e_{k}^{\perp}s_{\gamma}=e_{k}^{\perp}\left(  s_{\gamma}\right)  =\sum\limits
_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}\\\text{vertical
}k\text{-strip}}}s_{\nu}.
\]
This proves Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}(b).

(a) We claim that
\begin{equation}
\left(  s_{\lambda},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(
s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}
\end{equation}
for every $\lambda\in \Par $.

\textit{Proof of (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}):}
Let $\lambda\in \Par $. The equality (\ref{eq.pieri.h}) (applied
to $n=k$) yields%
\begin{equation}
s_{\lambda}h_{k}=\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\lambda\text{ is a
}\\\text{horizontal }k\text{-strip}}}s_{\lambda^{+}}=\sum\limits_{\substack{\lambda
^{+}\in \Par ;\\\lambda^{+}/\lambda\text{ is a }%
\\\text{horizontal }k\text{-strip}}}s_{\lambda^{+}}.
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-h}%
\end{equation}
But (\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.adjt}) (applied to
$f=h_{k}$, $a=s_{\gamma}$ and $g=s_{\lambda}$) yields%
\begin{align}
\left(  s_{\lambda},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)   &
=\left(  \underbrace{h_{k}s_{\lambda}}_{\substack{=s_{\lambda}h_{k}%
\\=\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}%
/\lambda\text{ is a }\\\text{horizontal }k\text{-strip}}}s_{\lambda^{+}%
}\\\text{(by (\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.pieri-h}))}%
}},s_{\gamma}\right)  =\left(  \sum\limits_{\substack{\lambda^{+}\in
 \Par ;\\\lambda^{+}/\lambda\text{ is a }\\\text{horizontal
}k\text{-strip}}}s_{\lambda^{+}},s_{\gamma}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}%
/\lambda\text{ is a }\\\text{horizontal }k\text{-strip}}}\underbrace{\left(
s_{\lambda^{+}},s_{\gamma}\right)  }_{\substack{=\delta_{\lambda^{+},\gamma
}\\\text{(by (\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}),
applied}\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda^{+}%
,\gamma\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall
inner product is }\kk\text{-bilinear}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}%
/\lambda\text{ is a }\\\text{horizontal }k\text{-strip}}}\delta_{\lambda
^{+},\gamma}. \label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3}%
\end{align}


On the other hand, the Hall inner product is $\kk$-bilinear. Thus,%
\begin{align}
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)   &
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}\underbrace{\left(  s_{\lambda},s_{\nu
}\right)  }_{\substack{=\delta_{\lambda,\nu}\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.orthonormal}), applied}%
\\\text{to }\left(  \alpha,\beta\right)  =\left(  \lambda,\nu\right)
\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}\delta_{\lambda,\nu}.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.4}%
\end{align}


Now, we must be in one of the following two cases:

\textit{Case 1:} The skew shape $\gamma/\lambda$ is a horizontal $k$-strip.

\textit{Case 2:} The skew shape $\gamma/\lambda$ is not a horizontal $k$-strip.

(The case when the skew shape $\gamma/\lambda$ is not well-defined (i.e., when
$\lambda\not \subseteq \gamma$) is counted towards Case 2.)

Let us first consider Case 1. In this case, the skew shape $\gamma/\lambda$ is
a horizontal $k$-strip. Hence, the sum $\sum\limits_{\substack{\lambda^{+}%
\in \Par ;\\\lambda^{+}/\lambda\text{ is a }\\\text{horizontal
}k\text{-strip}}}\delta_{\lambda^{+},\gamma}$ has an addend for $\lambda
^{+}=\gamma$. Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{
is a }\\\text{horizontal }k\text{-strip}}}\delta_{\lambda^{+},\gamma
}=\underbrace{\delta_{\gamma,\gamma}}_{\substack{=1\\\text{(since }%
\gamma=\gamma\text{)}}}+\sum\limits_{\substack{\lambda^{+}\in \Par %
;\\\lambda^{+}/\lambda\text{ is a }\\\text{horizontal }k\text{-strip;}%
\\\lambda^{+}\neq\gamma}}\underbrace{\delta_{\lambda^{+},\gamma}%
}_{\substack{=0\\\text{(since }\lambda^{+}\neq\gamma\text{)}}%
}=1+\underbrace{\sum\limits_{\substack{\lambda^{+}\in \Par %
;\\\lambda^{+}/\lambda\text{ is a }\\\text{horizontal }k\text{-strip;}%
\\\lambda^{+}\neq\gamma}}0}_{=0}=1.
\]
Hence, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3}) becomes%
\begin{equation}
\left(  s_{\lambda},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)
=\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{
is a }\\\text{horizontal }k\text{-strip}}}\delta_{\lambda^{+},\gamma}=1.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3.c1}%
\end{equation}
On the other hand, the skew shape $\gamma/\lambda$ is a horizontal $k$-strip.
Hence, the sum $\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{
is a}\\\text{horizontal }k\text{-strip}}}\delta_{\lambda,\nu}$ has an addend
for $\nu=\lambda$. Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}%
\\\text{horizontal }k\text{-strip}}}\delta_{\lambda,\nu}=\underbrace{\delta
_{\lambda,\lambda}}_{\substack{=1\\\text{(since }\lambda=\lambda\text{)}%
}}+\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip;}\\\nu\neq\lambda}}\underbrace{\delta
_{\lambda,\nu}}_{\substack{=0\\\text{(since }\lambda\neq\nu\\\text{(since }%
\nu\neq\lambda\text{))}}}=1+\underbrace{\sum\limits_{\substack{\nu\in
 \Par ;\\\gamma/\nu\text{ is a}\\\text{horizontal }%
k\text{-strip;}\\\nu\neq\lambda}}0}_{=0}=1.
\]
Hence, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.4}) becomes%
\begin{equation}
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}\delta_{\lambda,\nu}=1.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.4.c1}%
\end{equation}
Comparing this with (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3.c1}),
we obtain%
\[
\left(  s_{\lambda},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(
s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)  .
\]
Thus, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}) is proven in
Case 1.

Let us now consider Case 2. In this case, the skew shape $\gamma/\lambda$ is
not a horizontal $k$-strip. Hence, if $\lambda^{+}\in \Par $ has
the property that $\lambda^{+}/\lambda$ is a horizontal $k$-strip, then%
\begin{equation}
\delta_{\lambda^{+},\gamma}=0
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.1}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.1}):} Let $\lambda
^{+}\in \Par $ have the property that $\lambda^{+}/\lambda$ is a
horizontal $k$-strip. Assume (for the sake of contradiction) that $\lambda
^{+}=\gamma$. Then, $\lambda^{+}/\lambda$ is a horizontal $k$-strip. In other
words, $\gamma/\lambda$ is a horizontal $k$-strip (since $\lambda^{+}=\gamma
$). This contradicts the fact that $\gamma/\lambda$ is not a horizontal
$k$-strip.
\par
This contradiction shows that our assumption (that $\lambda^{+}=\gamma$) was
false. Thus, $\lambda^{+}\neq\gamma$. Hence, $\delta_{\lambda^{+},\gamma}=0$.
This proves (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.1}).}. Now,
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.3}) becomes%
\begin{equation}
\left(  s_{\lambda},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)
=\sum\limits_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{
is a }\\\text{horizontal }k\text{-strip}}}\underbrace{\delta_{\lambda
^{+},\gamma}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.1}))}}}=\sum\limits
_{\substack{\lambda^{+}\in \Par ;\\\lambda^{+}/\lambda\text{ is a
}\\\text{horizontal }k\text{-strip}}}0=0.
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.2}%
\end{equation}
On the other hand, if $\nu\in \Par $ has the property that
$\gamma/\nu$ is a horizontal $k$-strip, then
\begin{equation}
\delta_{\lambda,\nu}=0
\label{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.3}):} Let $\nu
\in \Par $ have the property that $\gamma/\nu$ is a horizontal
$k$-strip. Assume (for the sake of contradiction) that $\lambda=\nu$. Then,
$\gamma/\nu$ is a horizontal $k$-strip. In other words, $\gamma/\lambda$ is a
horizontal $k$-strip (since $\lambda=\nu$). This contradicts the fact that
$\gamma/\lambda$ is not a horizontal $k$-strip.
\par
This contradiction shows that our assumption (that $\lambda=\nu$) was false.
Thus, $\lambda\neq\nu$. Hence, $\delta_{\lambda,\nu}=0$. This proves
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.3}).}. Now,
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.4}) becomes%
\[
\left(  s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma
/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}\underbrace{\delta_{\lambda,\nu}%
}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.3}))}}}=\sum\limits
_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}%
\\\text{horizontal }k\text{-strip}}}0=0.
\]
Comparing this with (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.c2.2}),
we obtain%
\[
\left(  s_{\lambda},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(
s_{\lambda},\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)  .
\]
Thus, (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}) is proven in
Case 2.

We have now proven (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal})
in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this shows that
(\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal}) always holds.

Now, we have proven (\ref{pf.lem.sol.Lambda.staircase-skew-2.coPieri.a.goal})
for all $\lambda\in \Par $. In other words, $\left(  s_{\lambda
},h_{k}^{\perp}\left(  s_{\gamma}\right)  \right)  =\left(  s_{\lambda}%
,\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}\right)  $ for each $\lambda
\in \Par $. Thus, Lemma
\ref{lem.sol.Lambda.staircase-skew-2.s-determines} (applied to $f=h_{k}%
^{\perp}\left(  s_{\gamma}\right)  $ and $g=\sum\limits_{\substack{\nu\in
 \Par ;\\\gamma/\nu\text{ is a}\\\text{horizontal }%
k\text{-strip}}}s_{\nu}$) yields $h_{k}^{\perp}\left(  s_{\gamma}\right)
=\sum\limits_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is
a}\\\text{horizontal }k\text{-strip}}}s_{\nu}$. Thus,%
\[
h_{k}^{\perp}s_{\gamma}=h_{k}^{\perp}\left(  s_{\gamma}\right)  =\sum\limits
_{\substack{\nu\in \Par ;\\\gamma/\nu\text{ is a}%
\\\text{horizontal }k\text{-strip}}}s_{\nu}.
\]
This proves Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}(a).
\end{proof}
\end{verlong}

Now, proving Proposition~\ref{prop.coPieri-rules} is just a matter
of renaming symbols in Lemma~\ref{lem.sol.Lambda.staircase-skew-2.coPieri}:

\begin{proof}[Proof of Proposition~\ref{prop.coPieri-rules}.]
Let $\lambda$ be a partition. Let $n \in \NN$.
Then, Lemma~\ref{lem.sol.Lambda.staircase-skew-2.coPieri}(a)
(applied to $\gamma = \lambda$ and $k = n$) yields
\begin{align*}
h_n^\perp s_\lambda
= \sum\limits_{\substack{\nu\in \Par ;\\\lambda
/\nu\text{ is a}\\\text{horizontal }n\text{-strip}}} s_\nu
= \sum\limits_{\substack{\lambda^- \in \Par; \\ \lambda/\lambda^- \text{ is a }\\\text{horizontal }n\text{-strip}}}  s_{\lambda^-}
\end{align*}
(here, we have renamed the summation index $\nu$ as $\lambda^-$).
This is precisely the equality \eqref{eq.copieri.h}.
Thus, \eqref{eq.copieri.h} is proven.

Lemma~\ref{lem.sol.Lambda.staircase-skew-2.coPieri}(b)
(applied to $\gamma = \lambda$ and $k = n$) yields
\begin{align*}
e_n^\perp s_\lambda
= \sum\limits_{\substack{\nu\in \Par ;\\\lambda
/\nu\text{ is a}\\\text{vertical }n\text{-strip}}} s_\nu
= \sum\limits_{\substack{\lambda^- \in \Par; \\ \lambda/\lambda^- \text{ is a }\\\text{vertical }n\text{-strip}}}  s_{\lambda^-}
\end{align*}
(here, we have renamed the summation index $\nu$ as $\lambda^-$).
This is precisely the equality \eqref{eq.copieri.e}.
Thus, \eqref{eq.copieri.e} is proven.
This completes the proof of Proposition~\ref{prop.coPieri-rules}.
\end{proof}

Thus, Exercise~\ref{exe.Lambda.coPieri-rules} is solved.

%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.aguiar.appli-omega}}

\textit{Solution to Exercise~\ref{exe.aguiar.appli-omega}.}

Let $\left(\Lambda_{\QQ}\right)_n$ denote the $n$-th graded component of the $\QQ$-vector space $\Lambda_{\QQ}$. We are going to work in $\Lambda_{\QQ}$ in this solution, making use of the fact that both $\left(s_\lambda\right)_{\lambda\in\Par_n}$ and $\left(p_\lambda\right)_{\lambda\in\Par_n}$ are bases of the $\QQ$-vector space $\left(\Lambda_{\QQ}\right)_n$ (due to Proposition~\ref{many-symmetric-function-bases-prop}). Of course, notions such as comultiplication, the antipode, skewing etc. are defined in $\Lambda_{\QQ}$ just in the same way as they have been defined in $\Lambda$, and their properties are proven analogously.

Our solutions for both parts of the exercise rely on the fact that the trace of an endomorphism of a finite-dimensional vector space can be computed using any basis of the vector space. Specifically, we will be applying this fact to certain endomorphisms of $\left(\Lambda_{\QQ}\right)_n$, and as bases we will use $\left(p_\lambda\right)_{\lambda\in \Par_n}$ and $\left(s_\lambda\right)_{\lambda\in \Par_n}$.

(a) The antipode $S$ of $\Lambda_\QQ$ is a graded map, and thus
it restricts to an endomorphism of the $\QQ$-vector space $\left(
\Lambda _\QQ\right) _n$. Denote this endomorphism by $S_{n}$.
We want to compute $\trace \left( S_n \right) $. (This is well-defined
since $\left( \Lambda_\QQ\right) _n$ is a finite-dimensional
$\QQ$-vector space.)

From \eqref{skew-Schur-antipodes}, we see that $S\left( s_{\lambda /\mu}
\right) =\left( -1\right) ^{\left\vert \lambda /\mu \right\vert }s_{\lambda
^{t}/\mu ^{t}}$ for any partitions $\lambda$ and $\mu$ satisfying
$\mu \subseteq \lambda$. In particular,
$S\left( s_{\lambda }\right) =\left( -1\right) ^{\left\vert \lambda
\right\vert }s_{\lambda ^{t}}$ for any partition $\lambda $. For any
partition $\lambda \in \Par_{n}$, we now have $S_{n}\left( s_{\lambda
}\right) =S\left( s_{\lambda }\right) =\left( -1\right) ^{\left\vert \lambda
\right\vert }s_{\lambda ^{t}}=\left( -1\right) ^{n}s_{\lambda ^{t}}$. Thus,
the matrix which represents the endomorphism $S_{n}$ of
$\left( \Lambda_\QQ\right) _{n}$ with respect to the basis $\left( s_{\lambda
}\right) _{\lambda \in \Par_{n}}$ of $\left( \Lambda_\QQ\right) _{n}$ is
the matrix whose $\left( \lambda ,\mu \right) $-th entry
(for any $\lambda \in \Par_n$ and $\mu \in \Par_n$) is
$\left( -1\right) ^{n}$ whenever $\mu =\lambda ^{t}$, and
otherwise $0$. But the trace $\trace \left( S_{n}\right) $ of $S_{n}$
(by its definition) is the sum of the diagonal entries of this matrix;
hence, this trace equals
\begin{equation*}
\trace \left( S_{n}\right) =\sum\limits_{\substack{ \lambda \in \Par_n;
 \\ \lambda =\lambda ^{t}}}\left( -1\right) ^{n}=\left(
-1\right) ^{n}\underbrace{\left( \text{number of all }\lambda \in 
\Par_{n}\text{ satisfying }\lambda =\lambda ^{t}\right) }_{=c\left(
n\right) }=\left( -1\right) ^{n}c\left( n\right) .
\end{equation*}

On the other hand, every partition $\lambda$ satisfies $S\left( p_{\lambda
}\right) =\left( -1\right) ^{\ell \left( \lambda \right) }p_{\lambda }$
(this follows from Proposition~\ref{symm-antipode-and-h-basis}(i), after
recalling that $S$ is an algebra morphism and that $p_{\left( \lambda
_{1},\lambda _{2},...,\lambda _{\ell }\right) }=p_{\lambda _{1}}p_{\lambda
_{2}}...p_{\lambda _{\ell }}$). Hence, the matrix which represents the
endomorphism $S_{n}$ of $\left( \Lambda_\QQ\right) _{n}$ with
respect to the basis $\left( p_{\lambda }\right) _{\lambda \in \Par 
_{n}}$ of $\left( \Lambda_\QQ\right) _{n}$ is a diagonal matrix,
whose $\lambda $-th diagonal entry (for any $\lambda \in \Par_n$)
is $\left( -1\right) ^{\ell \left( \lambda \right) }$. But
the trace $\trace \left( S_{n}\right) $ of $S_{n}$ (by its definition)
is the sum of the diagonal entries of this matrix; hence, this trace equals
\begin{align*}
\trace \left( S_{n}\right)
&= \sum_{\lambda \in \Par_{n}}
\left( -1\right) ^{\ell \left( \lambda \right) }
=\sum_{k=0}^{n}
\underbrace{\sum\limits_{\substack{ \lambda \in \Par_{n}; \\ \ell
\left( \lambda \right) =k}}\left( -1\right) ^{k}}_{=\left( -1\right)
^{k}\left( \text{number of all }\lambda \in \Par_{n}\text{
satisfying }\ell \left( \lambda \right) =k\right) } \\
& \ \ \ \ \ \ \ \ \ \
\left( \text{since }\ell \left( \lambda \right) \in \left\{
0,1,...,n\right\} \text{ for every }\lambda \in \Par_{n}\right)  \\
&= \sum_{k=0}^{n}\left( -1\right) ^{k}\underbrace{\left( \text{number of all 
}\lambda \in \Par_{n}\text{ satisfying }\ell \left(
\lambda \right) =k\right) }_{=p\left( n,k\right) }=\sum_{k=0}^{n}\left(
-1\right) ^{k}p\left( n,k\right) .
\end{align*}
Compared with $\trace \left( S_{n}\right) =\left( -1\right) ^{n}c\left(
n\right) $, this yields $\left( -1\right) ^{n}c\left( n\right)
=\sum_{k=0}^{n}\left( -1\right) ^{k}p\left( n,k\right) $. This solves part
(a) of the exercise.

(b) Consider the map $s_{1}s_{1}^{\perp }:\Lambda \rightarrow \Lambda $
which sends every $f\in \Lambda $ to $s_{1}s_{1}^{\perp }f$. This map
$s_{1}s_{1}^{\perp }$ is graded (because the map $s_{1}^{\perp }:\Lambda
\rightarrow \Lambda $ lowers the degree of any homogeneous element by $1$,
while the map $s_{1}$ raises it back by $1$) and thus restricts to an
endomorphism of the $\QQ$-vector space
$\left( \Lambda_\QQ\right) _{n}$. Denote this endomorphism by $P_{n}$.
We want to compute $\trace \left( P_{n}\right) $. (This is well-defined
since $\left(\Lambda_\QQ\right) _{n}$ is a finite-dimensional
$\QQ$-vector space.)

In the following, if $\mathcal{A}$ is any statement, then
$\left[ \mathcal{A}\right] $ will denote the truth value of $\mathcal{A}$
(that is, $1$ if $\mathcal{A}$ holds, and $0$ if it doesn't).

We know that the basis
$\left( s_{\lambda }\right) _{\lambda \in \Par_{n}}$ of
$\left( \Lambda_\QQ\right) _{n}$ is orthonormal with
respect to the Hall inner product (restricted to
$\left( \Lambda_\QQ\right) _{n}$). Hence, for any
$\lambda \in \Par_{n}$, we have
\begin{align}
&\phantom{=}\left( \text{the }s_{\lambda }\text{-coordinate of }P_{n}s_{\lambda }\text{
with respect to this basis}\right)   \notag \\
&=\left( \underbrace{P_{n}s_{\lambda }}_{=s_{1}s_{1}^{\perp }s_{\lambda
}},s_{\lambda }\right) =\left( s_{1}s_{1}^{\perp }s_{\lambda },s_{\lambda
}\right) =\left( s_{1}^{\perp }s_{\lambda },s_{1}^{\perp }s_{\lambda
}\right)   \notag \\
&=\sum_{\mu \in \Par_{n-1}}\left( \underbrace{\left(
s_{1}^{\perp }s_{\lambda },s_{\mu }\right) }_{=\left( s_{\lambda
},s_{1}s_{\mu }\right) }\right) ^{2}  \notag \\
&\ \ \ \ \ \ \ \ \ \ \left( 
\begin{array}{c}
\text{since }s_{1}^{\perp }s_{\lambda }\in \left( \Lambda_\QQ\right)
 _{n-1}\text{, and since the basis }\left( s_{\mu }\right) _{\mu \in 
\Par_{n-1}} \\ 
\text{of }\left( \Lambda_\QQ\right) _{n-1}\text{ is orthonormal
with respect to the Hall} \\ 
\text{inner product (restricted to }\left( \Lambda_\QQ\right)
_{n-1}\text{)}
\end{array}
\right)   \notag \\
&=\sum_{\mu \in \Par_{n-1}}\left( s_{\lambda
},s_{1}s_{\mu }\right) ^{2}.  \label{sol.aguiar.appli-omega.b.3}
\end{align}
Now, every $\mu \in \Par_{n-1}$ satisfies
\begin{equation*}
\underbrace{s_{1}}_{=h_{1}}s_{\mu }=h_{1}s_{\mu }
\overset{\eqref{eq.pieri.h}}{=}
\sum\limits_{\substack{ \mu ^{+}\
:\ \mu ^{+}/\mu \text{ is a} \\ \text{horizontal }1\text{-strip}}}s_{\mu
^{+}}=\sum_{\mu ^{+}\ :\ \left\vert \mu ^{+}/\mu \right\vert =1}s_{\mu
^{+}}=\sum\limits_{\substack{ \mu ^{+}\in \Par_{n}; \\ \mu
\subseteq \mu ^{+}}}s_{\mu ^{+}}.
\end{equation*}
Hence, every $\lambda \in \Par_n$ and $\mu \in \Par_{n-1}$ satisfy
\begin{align*}
\left( s_{\lambda },s_{1}s_{\mu }\right)  &=\left( s_{\lambda },\sum\limits
_{\substack{ \mu ^{+}\in \Par_{n}; \\ \mu \subseteq \mu
^{+}}}s_{\mu ^{+}}\right) =\sum\limits_{\substack{ \mu ^{+}\in \Par_{n};
 \\ \mu \subseteq \mu ^{+}}}\underbrace{\left( s_{\lambda
},s_{\mu ^{+}}\right) }_{\substack{ =\left[ \lambda =\mu ^{+}\right]  \\ 
\text{(since the Schur functions} \\ \text{are orthonormal with respect} \\ 
\text{to the Hall inner product)}}}=\sum\limits_{\substack{ \mu ^{+}\in \Par_{n};
 \\ \mu \subseteq \mu ^{+}}}\left[ \lambda =\mu ^{+}\right] 
\\
&=\left[ \mu \subseteq \lambda \right] .
\end{align*}

Now, consider again the basis $\left( s_{\lambda }\right) _{\lambda \in 
\Par_{n}}$ of $\left( \Lambda_\QQ\right) _{n}$. For every
$\lambda \in \Par_{n}$, we have
\begin{align}
&\phantom{=}\left( \text{the }s_{\lambda }\text{-coordinate of }P_{n}s_{\lambda }\text{
with respect to this basis}\right)   \notag \\
&=\sum_{\mu \in \Par_{n-1}}\left( \underbrace{\left(
s_{\lambda },s_{1}s_{\mu }\right) }_{=\left[ \mu \subseteq \lambda \right]
}\right) ^{2}\ \ \ \ \ \ \ \ \ \ \left( \text{by
 \eqref{sol.aguiar.appli-omega.b.3}}\right)   \notag \\
&=\sum_{\mu \in \Par_{n-1}}\underbrace{\left[ \mu
\subseteq \lambda \right] ^{2}}_{=\left[ \mu \subseteq \lambda \right]
}=\sum_{\mu \in \Par_{n-1}}\left[ \mu \subseteq \lambda \right]   \notag \\
&=\left( \text{the number of all }\mu \in \Par_{n-1}\text{
such that }\mu \subseteq \lambda \right)   \notag \\
&=\left( \text{the number of all ways to remove a single cell from }\lambda 
\text{ to obtain a partition}\right)   \notag \\
&=\left( \text{the number of all corners of the Ferrers diagram of }\lambda
\right)   \notag \\
&= C\left( \lambda \right) .
\label{sol.aguiar.appli-omega.b.5}
\end{align}
Now, consider the matrix which represents the endomorphism $P_{n}$ of
$\left( \Lambda_\QQ\right) _{n}$ with respect to the basis
$\left( s_{\lambda }\right) _{\lambda \in \Par_{n}}$ of $\left( \Lambda
_\QQ\right) _{n}$. For any $\lambda \in \Par_{n}
$ and $\mu \in \Par_{n}$, the $\left( \lambda ,\mu \right) $-th
entry of this matrix is
\begin{equation*}
\left( \text{the }s_{\lambda }\text{-coordinate of }P_{n}s_{\mu }\text{ with
respect to this basis}\right) .
\end{equation*}
The trace $\trace \left( P_n\right) $ of $P_{n}$ (by its definition)
is the sum of the diagonal entries of this matrix; hence, this trace equals
\[
\trace \left( P_{n}\right)  = \sum_{\lambda \in \Par_{n}}
\underbrace{\left( \text{the }s_{\lambda }\text{-coordinate of 
}P_{n}s_{\lambda }\text{ with respect to this basis}\right) }_{\substack{ 
=C\left( \lambda \right)  \\ \text{(by \eqref{sol.aguiar.appli-omega.b.5})}}}
= \sum_{\lambda \in \Par_{n}}C\left( \lambda \right) .
\]

On the other hand, it is easy to see that
\begin{equation}
s_{1}^{\perp }\left( p_{n}\right) =\left[ n=1\right] \ \ \ \ \ \ \ \ \ \ 
\text{for every positive integer }n  \label{sol.aguiar.appli-omega.b.6}
\end{equation}
(in fact, recall that $p_{n}$ is primitive, so that $\Delta \left(
p_{n}\right) =p_{n}\otimes 1+1\otimes p_{n}$ and thus $s_{1}^{\perp }\left(
p_{n}\right) =\underbrace{\left( s_{1},p_{n}\right) }_{=\left[ n=1\right] }1+
\underbrace{\left( s_{1},1\right) }_{=0}p_{n}=\left[ n=1\right] $). But the
map $s_{1}^{\perp }:\Lambda \rightarrow \Lambda $ is a derivation (by
Proposition~\ref{skewing-properties-prop}(iv), since $s_{1}$ is primitive).
Hence, every partition
$\lambda = \left( \lambda_{1},\lambda_{2},...,\lambda _{\ell }\right)$
with $\ell =\ell \left( \lambda \right)$ satisfies
\begin{align}
s_{1}^{\perp }\left( p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda _{\ell
}}\right)  &= \sum_{k=1}^{\ell}p_{\lambda _{1}}p_{\lambda
_{2}}...p_{\lambda _{k-1}}\underbrace{s_{1}^{\perp }\left( p_{\lambda
_{k}}\right) }_{\substack{ =\left[ \lambda _{k}=1\right]  \\ \text{(by
\eqref{sol.aguiar.appli-omega.b.6})}}}p_{\lambda _{k+1}}p_{\lambda
_{k+2}}...p_{\lambda _{\ell }}  \notag \\
&= \sum_{k=1}^{\ell }p_{\lambda_{1}}p_{\lambda_{2}}...p_{\lambda_{k-1}}
\left[ \lambda_{k}=1\right]
p_{\lambda_{k+1}}p_{\lambda_{k+2}}...p_{\lambda_{\ell}}
\label{sol.aguiar.appli-omega.b.8}
\end{align}
and 
\begin{align*}
P_n p_{\lambda}
&= s_{1}s_{1}^{\perp}
\underbrace{p_{\lambda}}
_{=p_{\lambda_1}p_{\lambda_2}...p_{\lambda_{\ell}}}
=\underbrace{s_{1}}_{=p_{1}}
\underbrace{s_{1}^{\perp}
\left( p_{\lambda_1}p_{\lambda_2}...p_{\lambda_{\ell}}\right)}
_{\substack{ =\sum_{k=1}^{\ell}
p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda _{k-1}}
\left[ \lambda _{k}=1 \right]
p_{\lambda _{k+1}}p_{\lambda _{k+2}}...p_{\lambda _{\ell }} \\
\text{(by \eqref{sol.aguiar.appli-omega.b.8})}}} \\
&= p_{1}\sum_{k=1}^{\ell }p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda
_{k-1}}\left[ \lambda _{k}=1\right] p_{\lambda _{k+1}}p_{\lambda
_{k+2}}...p_{\lambda _{\ell }} \\
&= \sum_{k=1}^{\ell }p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda _{k-1}}
\underbrace{\left[ \lambda _{k}=1\right] p_{1}}_{\substack{ =\left[ \lambda
_{k}=1\right] p_{\lambda _{k}} \\ \text{(because if }\lambda _{k}\neq 1
\text{, then both sides} \\ \text{of this are }0\text{, and otherwise they} \\ 
\text{are clearly equal)}}}p_{\lambda _{k+1}}p_{\lambda _{k+2}}...p_{\lambda
_{\ell }} \\
&= \sum_{k=1}^{\ell }p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda _{k-1}}
\left[ \lambda _{k}=1\right] p_{\lambda _{k}}p_{\lambda _{k+1}}p_{\lambda
_{k+2}}...p_{\lambda _{\ell }}=\sum_{k=1}^{\ell }\left[ \lambda _{k}=1\right]
p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda _{\ell }} \\
&= \underbrace{\left( \sum_{k=1}^{\ell }\left[ \lambda _{k}=1\right] \right) 
}_{\substack{ =\left( \text{the number of }k\in \left\{ 1,2,...,\ell
\right\} \text{ such that }\lambda _{k}=1\right)  \\ =\left( \text{the
number of parts of }\lambda \text{ equal to }1\right)  \\ =\mu _{1}\left(
\lambda \right) }}\underbrace{p_{\lambda _{1}}p_{\lambda _{2}}...p_{\lambda
_{\ell }}}_{=p_{\lambda }} \\
&= \mu _{1}\left( \lambda \right) p_{\lambda }.
\end{align*}
Hence, the matrix which represents the endomorphism $P_{n}$ of $\left(
\Lambda_\QQ\right) _{n}$ with respect to the basis $\left(
p_{\lambda }\right) _{\lambda \in \Par_{n}}$ of
$\left( \Lambda_\QQ\right) _{n}$ is a diagonal matrix, whose
$\lambda $-th diagonal entry (for any $\lambda \in \Par_{n}$)
is $\mu _{1}\left( \lambda \right) $. But the trace
$\trace \left( P_{n}\right) $ of $P_{n}$ (by its definition) is the sum
of the diagonal entries of this matrix; hence, this trace equals
\begin{equation*}
\trace \left( P_{n}\right) =\sum_{\lambda \in \Par_{n}}
\mu _{1}\left( \lambda \right) .
\end{equation*}
Compared with $\trace \left( P_{n}\right) =\sum_{\lambda \in \Par_n}
C\left( \lambda \right) $, this yields
$\sum_{\lambda \in \Par_{n}}C\left( \lambda \right)
=\sum_{\lambda \in \Par_{n}}\mu _{1}\left( \lambda \right) 
$. This solves part (b) of the exercise.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{Darij's-skewing-exercise}}

\textit{Solution to Exercise~\ref{Darij's-skewing-exercise}.}
We assume WLOG that $\kk = \ZZ$, since it is enough to prove what we want
over $\ZZ$.

(a) This is exactly the equality \eqref{eq.exe.Lambda.omegahlam.p},
and has already been proved. Here is a slightly different way to
prove it:

Use the primitivity of $p_n$ and Proposition~\ref{antipode-on-primitives}
to obtain $S\left(p_n\right) = -p_n$ for every $n \geq 1$.

But the antipode $S$ of $\Lambda$ is an
algebra anti-endomorphism (by Proposition~\ref{antipodes-are-antiendomorphisms}),
therefore an algebra endomorphism (by Exercise~\ref{exe.comm-cocomm.anti}(a),
since $\Lambda$ is commutative).
Hence, from the fact that $S\left(p_n\right) = -p_n$ for every $n \geq 1$,
we can deduce by multiplicativity that
$S\left(p_\lambda\right) = \left(-1\right)^{\ell\left(\lambda\right)} p_\lambda$
for every partition $\lambda$. Now recall \eqref{antipode-versus-omega-in-Sym},
and the claim of part (a) follows.

(b) For this part of the exercise, we shall work in $\Lambda_\RR$ and prove
that the endomorphism $\omega$ of $\Lambda_\RR$ is an isometry. This will
clearly yield the analogous statement over $\ZZ$.

Recall that $\left\{\dfrac{p_\lambda}{\sqrt{z_\lambda}} \right\}$
is an orthonormal basis of $\Lambda_\RR$ (by
Corollary~\ref{dual-bases-h-m-p}(c)). Hence,
in order to prove that the endomorphism $\omega$ of $\Lambda_\RR$ is an
isometry, it is enough to show that
\[
\left( \omega\left(\dfrac{p_\lambda}{\sqrt{z_\lambda}}\right),
       \omega\left(\dfrac{p_\mu}{\sqrt{z_\mu}}\right) \right)
= \left( \dfrac{p_\lambda}{\sqrt{z_\lambda}},
         \dfrac{p_\mu}{\sqrt{z_\mu}} \right)
\]
for any two partitions $\lambda$ and $\mu$. But this follows from part
(a) and the fact that the basis
$\left\{\dfrac{p_\lambda}{\sqrt{z_\lambda}} \right\}$ is orthonormal. Hence,
we have shown that the endomorphism $\omega$ of $\Lambda_\RR$ is an isometry.
The same holds therefore for the endomorphism $\omega$ of $\Lambda$, and thus
part (b) of the exercise is solved.

(c) Part (c) of the exercise is precisely the statement of
Proposition~\ref{prop.Lambda.omega-and-S}(f), which was proved
in the solution to Exercise~\ref{exe.Lambda.omega-and-S}.
Thus, we need not prove it again.

(d) follows from (b) and (c): Indeed, $\omega$ is a coalgebra morphism
(by part (c) of the exercise). Now, Definition~\ref{def.skewing} yields
\[
\left(\omega\left(a\right)\right)^\perp \omega\left(b\right)
= \sum_{\left(\omega\left(b\right)\right)}
 \left(\omega\left(a\right), \left(\omega\left(b\right)\right)_1\right)
 \left(\omega\left(b\right)\right)_2
= \sum_{\left(b\right)}
 \left(\omega\left(a\right), \omega\left(b_1\right)\right)
 \omega\left(b_2\right)
\]
(since $\omega$ is a coalgebra morphism and thus we have
$\sum_{\left(\omega\left(b\right)\right)}
 \left(\omega\left(b\right)\right)_1 \otimes \left(\omega\left(b\right)\right)_2
 = \sum_{\left(b\right)} \omega\left(b_1\right) \otimes \omega\left(b_2\right)$).
Since $\omega$ is an isometry, this further simplifies to
\[
\left(\omega\left(a\right)\right)^\perp \omega\left(b\right)
= \sum_{\left(b\right)}
 \left(a, b_1\right) \omega\left(b_2\right)
= \omega\left(\underbrace{\sum_{\left(b\right)}
 \left(a, b_1\right) b_2}_{=a^\perp b}\right)
= \omega\left(a^\perp b\right),
\]
whence part (d) of the exercise is solved.

(e) and (f) follow from the fact that
$s_{\mu}^{\perp}\left(  s_{\lambda}\right)  =s_{\lambda / \mu}$
(applied, respectively, to $\mu = \left(1^\ell\right)$ and to
$\mu = \left(\lambda_1\right)$), and the
fact that a parallel translation of a skew Ferrers shape doesn't change the
corresponding skew Schur function.

(g) It should be clear that $S\left(  s_{\lambda / \mu}\right)  =\left(
-1\right)  ^{\left\vert \lambda / \mu\right\vert }s_{\lambda^{t} /
\mu^{t}}$ follows from $\omega\left(  s_{\lambda / \mu}\right)
=s_{\lambda^{t} / \mu^{t}}$, and this, in turn, follows from (d)
if one knows that $\omega\left( s_{\lambda} \right) = s_{\lambda^{t}}$ for
every partition $\lambda$.

So it remains to prove that
$\omega\left( s_{\lambda} \right) = s_{\lambda^{t}}$ for every partition
$\lambda$. To do so, we use induction over
$\left\vert \lambda\right\vert $, and fix $\lambda$. Part (b) yields that
$\left(\omega\left(s_{\lambda}\right), \omega\left(s_{\lambda}\right)\right)
 = 1$. Since $\omega \left( s_{\lambda} \right) \in\Lambda$, this shows
that
$\omega\left( s_{\lambda} \right) = \pm s_{\nu}$ for some partition $\nu$
(since a length-$1$ integral vector cannot have more than one nonzero
coordinate). Hence, $\omega\left(  s_{\nu}\right)  =\pm
s_{\lambda}$ (since $\omega$ is an involution).
Writing $\lambda$ as $\left(  \lambda_{1},\lambda_{2}
,\ldots,\lambda_{\ell}\right)  $ with all of $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{\ell}$ being positive, and writing $\nu$ as $\left(  \nu_{1},\nu
_{2},\nu_{3},\ldots\right)  $, we easily see (using (d) and
(f)) that $\omega\left(  e_{\nu_{1}}^{\perp}s_{\lambda}\right)  =\pm
h_{\nu_{1}}^{\perp}\left(  s_{\nu}\right)  =\pm s_{\left(  \nu_{2},\nu_{3}
,\nu_{4},\ldots\right)  }\neq0$, so that $e_{\nu_{1}}^{\perp}s_{\lambda}\neq0$,
and therefore $\ell\geq\nu_{1}$. On the other hand, (d) and (e)
yield $\omega\left(  h_{\ell}^{\perp}s_{\nu}\right)  =\pm
e_{\ell}^{\perp}\left(  s_{\lambda}\right)  =
\pm s_{\left(  \lambda_{1}-1,\lambda_{2}-1,\ldots,\lambda_{\ell}-1\right) }
\neq0$, thus $h_{\ell}^{\perp} s_{\nu}\neq0$, so that $\ell\leq\nu_{1}$.
Combined with $\ell\geq \nu_{1}$, this yields $\ell=\nu_{1}$. Hence,
$h_{\ell}^{\perp}s_{\nu}
=h_{\nu_{1}}^{\perp}s_{\nu_{1}}=s_{\left(  \nu_{2},\nu_{3},\nu_{4},\ldots\right)
}$ (by (f)), so that
\[
\omega\left(  h_{\ell}^{\perp}s_{\nu}\right)  =\omega\left(  s_{\left(
\nu_{2},\nu_{3},\nu_{4},\ldots\right)  }\right)  =s_{\left(  \nu_{2},\nu_{3}
,\nu_{4},\ldots\right)  ^{t}}
\]
by the induction assumption. Compared with
\[
\omega\left(  h_{\ell}^{\perp}s_{\nu}\right)  =\pm e_{\ell}^{\perp}\left(
s_{\lambda}\right)  =\pm s_{\left(  \lambda_{1}-1,\lambda_{2}-1,\ldots,\lambda
_{\ell}-1\right)  }
\]
(a consequence of (d) and (e)), this yields $s_{\left(
\nu_{2},\nu_{3},\nu_{4},\ldots\right)  ^{t}}=\pm s_{\left(  \lambda_{1}
-1,\lambda_{2}-1,\ldots,\lambda_{\ell}-1\right)  }$, from which it directly
follows that $\pm=+$ and $\left(  \nu_{2},\nu_{3},\nu_{4},\ldots\right)
^{t}=\left(  \lambda_{1}-1,\lambda_{2}-1,\ldots,\lambda_{\ell}-1\right)  $. This
quickly yields $\nu=\lambda^{t}$, so that $\omega\left(  s_{\lambda}\right)
=\pm s_{\nu}$ becomes $\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}$
(the $\pm$ in $\omega\left(  s_{\lambda}\right)  =\pm s_{\nu}$ is the same as
in $s_{\left(  \nu_{2},\nu_{3},\nu_{4},\ldots\right)  ^{t}}=\pm s_{\left(
\lambda_{1}-1,\lambda_{2}-1,\ldots,\lambda_{\ell}-1\right)  }$), so we are done.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.skew-p-by-e}}

\textit{Solution to Exercise \ref{exe.skew-p-by-e}.} Proposition
\ref{symm-comultiplication-formulas-prop}(i) yields $\Delta p_{n}=1\otimes
p_{n}+p_{n}\otimes1$. Comparing this with
$\Delta p_{n} = \sum_{\left(p_n\right)} \left(  p_{n}\right)_{1}
\otimes\left(  p_{n}\right)  _{2}$ (here, we are using the Sweedler
notation), we obtain%
\begin{equation}
\sum_{\left(p_n\right)}
\left(  p_{n}\right)  _{1}\otimes\left(  p_{n}\right)  _{2}
= 1\otimes p_{n}+p_{n}\otimes1.
\label{sol.skew-p-by-e.Delp}
\end{equation}


(a) Proposition \ref{symm-antipode-and-h-basis}(i) yields $S\left(
p_{n}\right)  =-p_{n}$.

But $p_{n}\in\Lambda_{n}$. Hence, Exercise \ref{exe.hall-e-f}(b) (applied to
$f=p_{n}$) yields
\begin{align*}
\left(  e_{n},p_{n}\right)    & =\left(  -1\right)  ^{n}\cdot
\underbrace{\left(  S\left(  p_{n}\right)  \right)  }_{=-p_{n}}\left(
1\right)  =\left(  -1\right)  ^{n}\cdot\underbrace{\left(  -p_{n}\right)
\left(  1\right)  }_{=-p_{n}\left(  1\right)  }\\
& =\left(  -1\right)  ^{n}\cdot\left(  -p_{n}\left(  1\right)  \right)
=\underbrace{-\left(  -1\right)  ^{n}}_{\substack{=\left(  -1\right)
^{n+1}=\left(  -1\right)  ^{n-1}\\\text{(since }n+1\equiv
n-1\operatorname{mod}2\text{)}}}\cdot\underbrace{p_{n}\left(  1\right)
}_{\substack{=1^{n}\\\text{(by the definition of }p_{n}\text{)}}}\\
& =\left(  -1\right)  ^{n-1}\cdot\underbrace{1^{n}}_{=1}=\left(  -1\right)
^{n-1}.
\end{align*}
This solves Exercise \ref{exe.skew-p-by-e}(a).

(b) Let $m \in \NN$ satisfy $m \neq n$. Then, $e_m \in \Lambda_m$ and
$p_n \in \Lambda_n$. But $m$ and $n$ are two distinct nonnegative integers
(since $m \neq n$). Hence, Exercise \ref{exe.Lambda.hall.gr}(a) (applied
to $m$, $n$, $e_m$ and $p_n$ instead of $n$, $m$, $f$ and $g$) yields
$\left( e_m, p_n \right) = 0$.
This solves Exercise \ref{exe.skew-p-by-e}(b).

(c) We have $n \neq 0$ (since $n$ is positive). Thus, $n$ and $0$ are two
distinct nonnegative integers. Hence, Exercise \ref{exe.Lambda.hall.gr}(a)
(applied to $m = 0$, $f = e_n$ and $g=1$) yields
$\left( e_n, 1 \right) = 0$
(since $e_n \in \Lambda_n$ and $1 \in \Lambda_0$).

By the definition of $e_{n}^{\perp}p_{n}$, we have
\begin{align*}
e_{n}^{\perp}p_{n}  & =\sum_{\left(  p_{n}\right)  }\left(  e_{n},\left(
p_{n}\right)  _{1}\right)  \cdot\left(  p_{n}\right)  _{2}=\underbrace{\left(
e_{n},1\right)  }_{=0}\cdot p_{n}+\underbrace{\left(  e_{n},p_{n}\right)
}_{\substack{=\left(  -1\right)  ^{n-1}\\\text{(by Exercise
\ref{exe.skew-p-by-e}(a))}}}\cdot1\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since } \sum_{\left(p_n\right)}
\left(  p_{n}\right)  _{1} \otimes\left(  p_{n}\right)  _{2}
=1\otimes p_{n}+p_{n}\otimes1\text{ (by
(\ref{sol.skew-p-by-e.Delp}))}\right)  \\
& =\underbrace{0\cdot p_{n}}_{=0}+\left(  -1\right)  ^{n-1}=\left(  -1\right)
^{n-1}.
\end{align*}
This solves Exercise \ref{exe.skew-p-by-e}(c).

(d) Let $m$ be a positive integer satisfying $m\neq n$. Then, Exercise
\ref{exe.skew-p-by-e}(b) yields $\left(  e_{m},p_{n}\right)  =0$.

We have $m \neq 0$ (since $m$ is positive). Thus, $m$ and $0$ are two
distinct nonnegative integers. Hence, Exercise \ref{exe.Lambda.hall.gr}(a)
(applied to $m$, $0$, $e_m$ and $1$ instead of $n$, $m$, $f$ and $g$)
yields $\left( e_m, 1 \right) = 0$
(since $e_m \in \Lambda_m$ and $1 \in \Lambda_0$).

By the definition of $e_{m}^{\perp}p_{n}$, we have
\begin{align*}
e_{m}^{\perp}p_{n}  & =\sum_{\left(  p_{n}\right)  }\left(  e_{m},\left(
p_{n}\right)  _{1}\right)  \cdot\left(  p_{n}\right)  _{2}=\underbrace{\left(
e_{m},1\right)  }_{=0}\cdot p_{n}+\underbrace{\left(  e_{m},p_{n}\right)
}_{=0}\cdot1\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since } \sum_{\left(p_n\right)}
\left(  p_{n}\right)  _{1}\otimes\left(  p_{n}\right)  _{2}
=1\otimes p_{n}+p_{n}\otimes1\text{ (by
(\ref{sol.skew-p-by-e.Delp}))}\right)  \\
& =\underbrace{0\cdot p_{n}}_{=0}+\underbrace{0\cdot1}_{=0}=0.
\end{align*}
This solves Exercise \ref{exe.skew-p-by-e}(d).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.bernstein}}

\textit{Solution to Exercise \ref{exe.bernstein}.} (a) The claim that
the sum $\sum_{i\in \NN }\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}f$
is convergent is very easy to see: Any given $f\in\Lambda=\bigoplus
_{n\in \NN }\Lambda_{n}$ lives in a finite direct sum $\bigoplus
_{n=0}^{m}\Lambda_{n}\subset\Lambda$; if we take $i\in \NN$ higher than
$m$, then $e_{i}^{\perp}f=0$ for degree reasons\footnote{In fact, the map
$e_{i}^{\perp}$ lowers degree by $i$, and thus annihilates $\bigoplus
_{n=0}^{m}\Lambda_{n}$ when $i>m$.} and therefore $\left(  -1\right)
^{i}h_{m+i}e_{i}^{\perp}f=0$.

The claim that the map $\mathbf{B}_{m}$ is $\kk$-linear is obvious.
Exercise \ref{exe.bernstein}(a) is solved.

(c) Exercise \ref{exe.bernstein}(c) makes three claims. Let us first prove the
first of them: the identity (\ref{eq.exe.bernstein.c.1}).

\textit{Proof of (\ref{eq.exe.bernstein.c.1}):} Let $\lambda=\left(
\lambda_{1},\lambda_{2},\lambda_{3},...\right)  $ be a partition having at
most $n$ parts. Then,%
\begin{align*}
s_{\lambda}  &  =s_{\lambda/\varnothing}%
\overset{\text{(\ref{eq.jacobi-trudi.h})}}{=}\det\left(  \left(
h_{\lambda_{i}-\varnothing_{j}-i+j}\right)  _{i,j=1,2,...,n}\right)
=\det\left(  \left(  h_{\lambda_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varnothing_{j}=0\right) \\
&  =\overline{s}_{\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{s}_{\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  }\right)  .
\end{align*}
This proves (\ref{eq.exe.bernstein.c.1}).

Next, we will show that for every $n$-tuple $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$, the symmetric function
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
either is $0$ or equals $\pm s_{\nu}$ for some partition $\nu$ having at most
$n$ parts.

\textit{Proof.} Let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
\in{ \ZZ }^{n}$ be any $n$-tuple. The definition of
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  .
\]
If the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are
not distinct, then the matrix $\left(  h_{\alpha_{i}-i+j}\right)
_{i,j=1,2,...,n}$ has two equal rows and thus its determinant is $0$, so that
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  =0$
in this case. So the claim is proven in this case, and this is a case we do
not need to address anymore. Thus, assume WLOG that the $n$ integers
$\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are distinct.
Hence, there exists a (unique) permutation $\tau\in\Symm_n$
satisfying $\alpha_{\tau\left(  1\right)  }-\tau\left(  1\right)
>\alpha_{\tau\left(  2\right)  }-\tau\left(  2\right)  >...>\alpha
_{\tau\left(  n\right)  }-\tau\left(  n\right)  $. Consider this $\tau$.

Define an $n$-tuple $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)
\in \ZZ ^{n}$ by setting%
\[
\gamma_{i}=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)
+i\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,n\right\}  .
\]
Then, it is easy to see that $\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}%
$.\ \ \ \ \footnote{To prove this, it is enough to show that $\gamma_{i}%
\geq\gamma_{i+1}$ for every $i\in\left\{  1,2,...,n-1\right\}  $. But this
simplifies to $\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)
\geq\alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)  +1$ (upon adding
$i$), and this follows from $\alpha_{\tau\left(  i\right)  }-\tau\left(
i\right)  >\alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)  $.}
Moreover, every $i\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma
_{i}-i=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  $ (by the
definition of $\gamma_{i}$), and so the matrix $\left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}$ is obtained from the matrix $\left(
h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}$ by permuting its rows according
to the permutation $\tau$. Since permuting the rows of a matrix multiplies the
determinant of the matrix by the sign of the permutation, we thus see%
\[
\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\alpha_{i}%
-i+j}\right)  _{i,j=1,2,...,n}\right)  }_{=\overline{s}_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }}=\left(  -1\right)  ^{\tau
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  },
\]
so that%
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  . \label{sol.bernstein.c.1.permuted}%
\end{equation}
If $\gamma_{n}<0$, then the matrix $\left(  h_{\gamma_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}$ on the right hand side of this equality has its $n$-th
row consist of zeroes only, which implies that the determinant of this matrix
is $0$, and thus (\ref{sol.bernstein.c.1.permuted}) becomes
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=0$.
Hence, we
are done in the case $\gamma_{n}<0$. So assume WLOG that we don't have
$\gamma_{n}<0$. Thus, $\gamma_{n}\geq0$. Combined with $\gamma_{1}\geq
\gamma_{2}\geq...\geq\gamma_{n}$, this yields that $\left(  \gamma_{1}%
,\gamma_{2},...,\gamma_{n}\right)  $ is a partition. This partition satisfies%
\[
s_{\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  }=s_{\left(
\gamma_{1},\gamma_{2},...,\gamma_{n}\right)  /\varnothing}%
\overset{\text{(\ref{eq.jacobi-trudi.h})}}{=}\det\left(  \left(  h_{\gamma
_{i}-\varnothing_{j}-i+j}\right)  _{i,j=1,2,...,n}\right)  =\det\left(
\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(since $\varnothing_{j}=0$ for all $j$). Thus,
(\ref{sol.bernstein.c.1.permuted}) becomes%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=s_{\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n}\right)  }}=\left(  -1\right)  ^{\tau}s_{\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n}\right)  },
\]
and therefore the symmetric function $\overline{s}_{\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{n}\right)  }$ equals $\pm s_{\nu}$ for some
partition $\nu$ having at most $n$ parts (namely, the partition $\nu$ is
$\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  $, and the $\pm$ sign is
$\left(  -1\right)  ^{\tau}$).

We thus have shown that for every $n$-tuple $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$, the symmetric function
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
either is $0$ or equals $\pm s_{\nu}$ for some partition $\nu$ having at most
$n$ parts.

In order to complete the solution of Exercise \ref{exe.bernstein}(c), it now
remains to prove (\ref{eq.exe.bernstein.c.3}).

\textit{Proof of (\ref{eq.exe.bernstein.c.3}):} Let $\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$ and $\left(
\beta_{1},\beta_{2},\ldots,\beta_{n}\right)  \in{ \NN }^{n}$ be two
$n$-tuples. We need to prove that (\ref{eq.exe.bernstein.c.3}) holds.

The definition of $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{n}\right)  }$ yields
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  .
\]
If the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are
not distinct, then (\ref{eq.exe.bernstein.c.3}) holds (because in
this case, the matrices $\left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}$
and $\left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}$ have
two equal rows each, so their determinants vanish, rendering both sides of
(\ref{eq.exe.bernstein.c.3}) equal to zero\footnote{The left hand side is
affected because of $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{n}\right)  }=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)
_{i,j=1,2,...,n}\right)  $.}). Thus, we WLOG assume that the $n$ integers
$\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are distinct.
Thus, there exists a (unique) permutation $\tau\in\Symm_n$ satisfying
$\alpha_{\tau\left(  1\right)  }-\tau\left(  1\right)  >\alpha_{\tau\left(
2\right)  }-\tau\left(  2\right)  >...>\alpha_{\tau\left(  n\right)  }%
-\tau\left(  n\right)  $. Consider this $\tau$, and define an $n$-tuple
$\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \in \ZZ ^{n}$ by
setting%
\[
\gamma_{i}=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)
+i\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,n\right\}  .
\]
Then, it is easy to see that $\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$.
Moreover, the matrix $\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}$
is obtained from the matrix $\left(  h_{\alpha_{i}-i+j}\right)
_{i,j=1,2,...,n}$ by permuting its rows according to the permutation $\tau$
(because every $i\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma
_{i}-i=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  $). This leads to
(\ref{sol.bernstein.c.1.permuted}) again (as in the previous proof), so that%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=\overline{s}_{\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{n}\right)  }}=\left(  -1\right)  ^{\tau
}\overline{s}_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  }.
\]
This equality, along with%
\[
\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}%
-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(which, again, follows from the fact that the matrix on the right hand side is
obtained from the matrix on the left hand side by permuting the rows according
to $\tau$), shows that in order to prove (\ref{eq.exe.bernstein.c.3}), it is
enough to show that
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  }%
=\det\left(  \left(  h_{\gamma_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  .
\]
But this is the same equality as (\ref{eq.exe.bernstein.c.3}), except with
$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $ replaced by
$\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  $. The advantage of
$\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{n}\right)  $ over $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $ is that we know that
$\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$ (while the analogous chain of
inequalities $\alpha_{1}\geq\alpha_{2}\geq...\geq\alpha_{n}$ does not
necessarily hold). So, we can \textbf{WLOG assume that }$\alpha_{1}\geq
\alpha_{2}\geq...\geq\alpha_{n}$ (because otherwise, we can replace $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $ by $\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{n}\right)  $). Assume this.

If $\alpha_{n}<0$, then both matrices $\left(  h_{\alpha_{i}-i+j}\right)
_{i,j=1,2,...,n}$ and $\left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}$ have their $n$-th row consisting of only zeroes, and so
their determinants both vanish, which shows that both sides of the equality
(\ref{eq.exe.bernstein.c.3}) are zero\footnote{For the left hand side, this is
because of $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}%
\right)  $.}. So this is a case in which (\ref{eq.exe.bernstein.c.3})
trivially holds. We thus assume WLOG that we are not in this case; hence, we
don't have $\alpha_{n}<0$, so we have $\alpha_{n}\geq0$. Combined with
$\alpha_{1}\geq\alpha_{2}\geq...\geq\alpha_{n}$, this yields that $\left(
\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  $ is a partition.

We have%
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }%
=\det\left(  \left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  .
\]
If the $n$ integers $\beta_{1}-1$, $\beta_{2}-2$, $...$, $\beta_{n}-n$ are not
distinct, then (\ref{eq.exe.bernstein.c.3}) holds (because in this
case, the matrix $\left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,...,n}$ has two
equal rows while the matrix $\left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}$ has two equal columns; thus, both of these matrices have
determinant $0$, so that both sides of (\ref{eq.exe.bernstein.c.3}) equal to
zero\footnote{This time, the reason why this causes the left hand side to be
zero is the identity $\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{n}\right)  }=\det\left(  \left(  h_{\beta_{i}-i+j}\right)
_{i,j=1,2,...,n}\right)  $.}). Thus, we WLOG assume that the $n$ integers
$\beta_{1}-1$, $\beta_{2}-2$, $...$, $\beta_{n}-n$ are distinct.
Thus, there exists a (unique) permutation $\zeta\in\Symm_n$
satisfying $\beta_{\zeta\left(  1\right)  }-\zeta\left(  1\right)
>\beta_{\zeta\left(  2\right)  }-\zeta\left(  2\right)  >...>\beta
_{\zeta\left(  n\right)  }-\zeta\left(  n\right)  $. Consider this $\zeta$.

Define an $n$-tuple $\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)
\in \ZZ ^{n}$ by setting%
\[
\delta_{j}=\beta_{\zeta\left(  j\right)  }-\zeta\left(  j\right)
+j\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{  1,2,...,n\right\}  .
\]
Then, it is easy to see that $\delta_{1}\geq\delta_{2}\geq...\geq\delta
_{n}\geq0$\ \ \ \ \footnote{The $\delta_{1}\geq\delta_{2}\geq...\geq\delta
_{n}$ part here is proven like the similar inequalities $\gamma_{1}\geq
\gamma_{2}\geq...\geq\gamma_{n}$ shown above; but the $\delta_{n}\geq0$ part
might require explanation. It stems from the fact that $\delta_{n}%
=\beta_{\zeta\left(  n\right)  }-\underbrace{\zeta\left(  n\right)  }_{\leq
n}+n\geq\beta_{\zeta\left(  n\right)  }-n+n=\beta_{\zeta\left(  n\right)
}\geq0$ (because we have assumed $\left(  \beta_{1},\beta_{2},...,\beta
_{n}\right)  \in \NN ^{n}$).}. Thus, $\left(  \delta_{1},\delta
_{2},...,\delta_{n}\right)  $ is a partition.

The matrix $\left(  h_{\delta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}$ is
obtained from the matrix $\left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,...,n}$
by permuting its rows according to the permutation $\zeta$ (since the
definition of $\delta_{i}$ yields $\delta_{i}=\beta_{\zeta\left(  i\right)
}-\zeta\left(  i\right)  +i$, thus $\delta_{i}-i=\beta_{\zeta\left(  i\right)
}-\zeta\left(  i\right)  $ for every $i\in\left\{  1,2,...,n\right\}  $). This
leads to $\det\left(  \left(  h_{\delta_{i}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\zeta}\underbrace{\det\left(  \left(
h_{\beta_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  }_{=\overline{s}_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }}=\left(  -1\right)  ^{\zeta
}\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }$ and
thus
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }=\left(
-1\right)  ^{\zeta}\underbrace{\det\left(  \left(  h_{\delta_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  }_{=\overline{s}_{\left(  \delta_{1},\delta
_{2},...,\delta_{n}\right)  }}=\left(  -1\right)  ^{\zeta}
\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  }.
\]
This equality, and the equality%
\[
\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\zeta}\det\left(  \left(  h_{\alpha
_{i}-\delta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(this time because the matrix on the right hand side is obtained from that on
the left hand side by permuting its \textbf{columns} according to $\zeta$)
show that in order to prove (\ref{eq.exe.bernstein.c.3}), it is enough to show
that
\[
\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-\delta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  .
\]
This is, of course, the same equality as (\ref{eq.exe.bernstein.c.3}), except
with $\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  $ replaced by
$\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  $. The advantage of
$\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  $ over $\left(
\beta_{1},\beta_{2},\ldots,\beta_{n}\right)  $ is that we know that $\left(
\delta_{1},\delta_{2},...,\delta_{n}\right)  $ is a partition. So, we can
\textbf{WLOG assume that }$\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  $
is a partition (because otherwise, we can replace $\left(  \beta_{1},\beta
_{2},\ldots,\beta_{n}\right)  $ by $\left(  \delta_{1},\delta_{2}%
,...,\delta_{n}\right)  $). Assume this.

Through a series of WLOG assumptions, we have now ensured that both $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $ and $\left(  \beta_{1}%
,\beta_{2},\ldots,\beta_{n}\right)  $ are partitions. Thus,%
\begin{align*}
s_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  } &  =s_{\left(
\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  /\varnothing}%
\overset{\text{(\ref{eq.jacobi-trudi.h})}}{=}\det\left(  \left(  h_{\alpha
_{i}-\varnothing_{j}-i+j}\right)  _{i,j=1,2,...,n}\right)  \\
&  =\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varnothing_{j}=0\right)  \\
&  =\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  },
\end{align*}
so that $\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)
}=s_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }$. Similarly
$\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}=s_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }$. Hence,%
\begin{align*}
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }  &
=s_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp}s_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=s_{\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{n}\right)  /\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{n}\right)  }\\
& \overset{\text{(\ref{eq.jacobi-trudi.h})}}{=}\det\left(  \left(
h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  .
\end{align*}
This proves (\ref{eq.exe.bernstein.c.3}) and thus completes the proof of
Exercise \ref{exe.bernstein}(c).

(d) In the following, we use the so-called \dfn{Iverson bracket notation}:
For every assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $
denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
(This integer is called the \dfn{truth value} of $\mathcal{A}$.)

Before we come to the solution of Exercise \ref{exe.bernstein}(d), let us
recall a general fact about determinants:

\begin{itemize}
\item Every commutative ring $A$, every positive integer $N$ and every matrix
$\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$ satisfy%
\begin{equation}
\det\left(  \left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right)  =\sum
_{k=1}^{N}\left(  -1\right)  ^{k-1}\beta_{1,k}\det\left(  \left(
\beta_{i+1,j+\left[  j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\label{sol.bernstein.d.laplace1r}%
\end{equation}
(This is just one possible way to write the Laplace expansion formula for the
expansion of the determinant of a matrix with respect to its first row.)
\end{itemize}

Let us now solve Exercise \ref{exe.bernstein}(d). Let $n \in \NN$, let
$m \in \ZZ$ and let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in{ \ZZ }^{n}$. We must prove (\ref{eq.exe.bernstein.d}).

Define an $\left(  n+1\right)  $-tuple $\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n+1}\right)  \in \ZZ ^{n+1}$ by $\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n+1}\right)  =\left(  m,\alpha_{1},\alpha
_{2},...,\alpha_{n}\right)  $. Then, $\gamma_{1}=m$, whereas every
$i\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma_{i+1}=\alpha_{i}$. Since
$\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  =\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n+1}\right)  $, we have%
\begin{align}
\overline{s}_{\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  } &
=\overline{s}_{\left(  \gamma_{1},\gamma_{2},...,\gamma_{n+1}\right)  }%
=\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n+1}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{s}_{\left(
\gamma_{1},\gamma_{2},...,\gamma_{n+1}\right)  }\right)  \nonumber\\
&  =\sum_{k=1}^{n+1}\left(  -1\right)  ^{k-1}h_{\gamma_{1}-1+k}\det\left(
\left(  h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.bernstein.d.laplace1r}),
applied to }A=\Lambda\text{, }N=n+1\text{ and }\beta_{i,j}=h_{\gamma_{i}%
-i+j}\right)  \nonumber\\
&  =\sum_{k=0}^{n}\underbrace{\left(  -1\right)  ^{\left(  k+1\right)  -1}%
}_{=\left(  -1\right)  ^{k}}\underbrace{h_{\gamma_{1}-1+\left(  k+1\right)  }%
}_{\substack{=h_{m+k}\\\text{(since }\gamma_{1}-1+\left(  k+1\right)
=\gamma_{1}+k=m+k\\\text{(because }\gamma_{1}=m\text{))}}}\det\left(  \left(
\underbrace{h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq
k+1\right]  \right)  }}_{\substack{=h_{\alpha_{i}-\left(  i+1\right)  +\left(
j+\left[  j\geq k+1\right]  \right)  }\\\text{(since }\gamma_{i+1}=\alpha
_{i}\text{)}}}\right)  _{i,j=1,2,...,n}\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k+1\text{ for
}k\text{ in the sum}\right)  \nonumber\\
&  =\sum_{k=0}^{n}\left(  -1\right)  ^{k}h_{m+k}\det\left(  \left(
h_{\alpha_{i}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)
.\label{sol.bernstein.d.rhs.first-formula}%
\end{align}


Now, let us check that
\begin{equation}
\det\left(  \left(  h_{\alpha_{i}-\left(  i+1\right)  +\left(  j+\left[  j\geq
k+1\right]  \right)  }\right)  _{i,j=1,2,...,n}\right)  =e_{k}^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,...,n\right\}
.\label{sol.bernstein.d.lhs.small-terms}%
\end{equation}


\textit{Proof of (\ref{sol.bernstein.d.lhs.small-terms}):} Let $k\in\left\{
0,1,...,n\right\}  $. The partition $\left(  1^{k}\right)  $ has length $k\leq
n$, and thus can be identified with the $n$-tuple $\left(
\underbrace{1,1,...,1}_{k\text{ times}},\underbrace{0,0,...,0}_{n-k\text{
times}}\right)  =\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  $, where
we set $\beta_{j}=
\begin{cases}
1, & \text{if }j\leq k;\\
0, & \text{if }j>k
\end{cases} \quad
=\left[  j\leq k\right]  =1-\left[  j\geq k+1\right]  $ for every
$j\in\left\{  1,2,...,n\right\}  $. Thus, $s_{\left(  1^{k}\right)
}=s_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }
=\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }$ (by
(\ref{eq.exe.bernstein.c.1}), applied to $\lambda=\left(  \beta_{1},\beta
_{2},\ldots,\beta_{n}\right)  $). Hence, $e_{k}=s_{\left(  1^{k}\right)
}=\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }$, so
that%
\begin{align*}
e_{k}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }  & =\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta
_{n}\right)  }^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  }=\det\left(  \left(  h_{\alpha_{i}-\beta_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.exe.bernstein.c.3})}\right)  \\
& =\det\left(  \left(  h_{\alpha_{i}-\left(  1-\left[  j\geq k+1\right]
\right)  -i+j}\right)  _{i,j=1,2,\ldots,n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\beta_{j}=1-\left[  j\geq k+1\right]  \text{ for every }%
j\in\left\{  1,2,...,n\right\}  \right)  \\
& =\det\left(  \left(  h_{\alpha_{i}-\left(  i+1\right)  +\left(  j+\left[
j\geq k+1\right]  \right)  }\right)  _{i,j=1,2,...,n}\right)
\end{align*}
(since $\alpha_{i}-\left(  1-\left[  j\geq k+1\right]  \right)
-i+j=\alpha_{i}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)$ for all $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}$).
This proves (\ref{sol.bernstein.d.lhs.small-terms}).

Now, (\ref{sol.bernstein.d.rhs.first-formula}) becomes%
\begin{align}
\overline{s}_{\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  } &
=\sum_{k=0}^{n}\left(  -1\right)  ^{k}h_{m+k}\underbrace{\det\left(  \left(
h_{\alpha_i - \left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=e_{k}^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}\\\text{(by (\ref{sol.bernstein.d.lhs.small-terms}))}}}\nonumber\\
&  =\sum_{k=0}^{n}\left(  -1\right)  ^{k}h_{m+k}e_{k}^{\perp}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }
=\sum_{i=0}%
^{n}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
\label{sol.bernstein.d.rhs.proven}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}k\text{ as }i\right)  .\nonumber
\end{align}
But in order to solve Exercise \ref{exe.bernstein}(d), we need to prove a very
similar yet different formula:
\begin{equation}
\overline{s}_{\left(  m,\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}%
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}.\label{sol.bernstein.d.rhs.to-prove}%
\end{equation}
These two formulas differ in a minor detail: The sum on the right hand side of
(\ref{sol.bernstein.d.rhs.to-prove}) runs over all $i\in \NN$, whereas
the sum on the right hand side of (\ref{sol.bernstein.d.rhs.proven}) only runs
over $i\in\left\{  0,1,...,n\right\}  $. If we can show that these two sums
are equal, then it will follow that the equality
(\ref{sol.bernstein.d.rhs.to-prove}) that we are proving and the equality
(\ref{sol.bernstein.d.rhs.proven}) that we have proven are equivalent, and so
the former equality must hold, and Exercise \ref{exe.bernstein}(d) will be solved.

So we need to prove that the sum on the right hand side of
(\ref{sol.bernstein.d.rhs.to-prove}) and the sum on the right hand side of
(\ref{sol.bernstein.d.rhs.proven}) are equal. To achieve this, it clearly
suffices to show that all addends in which these sums differ are $0$. But
these addends are the $\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
for $i\in \NN$ satisfying $i>n$. So we need to prove that $\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }=0$ for all $i\in \NN$ satisfying $i>n$.
Let us do this now.

Let $i\in \NN$ be such that $i>n$. We need to show that $\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }=0$.

The second statement of Exercise \ref{exe.bernstein}(c) says that the
symmetric function $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{n}\right)  }$ either is $0$ or equals $\pm s_{\nu}$ for some
partition $\nu$ having at most $n$ parts. If it is $0$, then $\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}\underbrace{\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }}_{=0}=0$ is obvious, and so
we are done. Thus, we can WLOG assume that we are in the other case, i.e., the
function $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }$ equals $\pm s_{\nu}$ for some partition $\nu$ having at most
$n$ parts. Assume that we are in this case, and consider this $\nu$. Since
$\nu$ has at most $n$ parts, we have $\ell\left(  \nu\right)  \leq n$. Now,
$\ell\left(  \left(  1^{i}\right)  \right)  =i>n\geq\ell\left(  \nu\right)  $,
and therefore $\left(  1^{i}\right)  \nsubseteq\nu$. But $e_{i}=s_{\left(
1^{i}\right)  }$ and thus $e_{i}^{\perp}s_{\nu}=s_{\left(  1^{i}\right)
}^{\perp}s_{\nu}=s_{\nu/\left(  1^{i}\right)  }=0$ (because $\left(
1^{i}\right)  \nsubseteq\nu$). Now,%
\[
\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}\underbrace{\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }}_{=\pm s_{\nu}}=\pm\left(
-1\right)  ^{i}h_{m+i}\underbrace{e_{i}^{\perp}s_{\nu}}_{=0}=0,
\]
which concludes our proof. Exercise \ref{exe.bernstein}(d) is thus solved.

(b) Let $n=\ell\left(  \lambda\right)  $. Then, $\lambda=\left(  \lambda
_{1},\lambda_{2},...,\lambda_{n}\right)  $, and the equality
(\ref{eq.exe.bernstein.c.1}) yields $s_{\lambda}=\overline{s}_{\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  }$.

Also, $\ell\left(  \left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)
\right)  =n+1$, and thus (\ref{eq.exe.bernstein.c.1}) (applied to $n+1$ and
$\left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)  $ instead of $n$
and $\lambda=\left(  \lambda_{1},\lambda_{2},\lambda_{3},...\right)  $) yields%
\[
s_{\left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)  }
= \overline{s}_{\left(  m,\lambda_{1},\lambda_{2},...,\lambda_{n}\right)  }.
\]
Compared with%
\begin{align*}
\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp
}\underbrace{s_{\lambda}}_{=\overline{s}_{\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{n}\right)  }} &  =\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{n}\right)  }\\
&  =\overline{s}_{\left(  m,\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.bernstein.d}), applied to
}\alpha_{j}=\lambda_{j}\right)  ,
\end{align*}
this yields $\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}%
^{\perp}s_{\lambda}=s_{\left(  m,\lambda_{1},\lambda_{2},\lambda
_{3},...\right)  }$. This solves Exercise \ref{exe.bernstein}(b).

(e) This follows by induction over $n$ from Exercise~\ref{exe.bernstein}(d).
The induction base (the $n=0$ case) is trivial. For the induction step, we
need to prove that
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n+1}\right)  }=\left(
\mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots\circ
\mathbf{B}_{\alpha_{n+1}}\right)  \left(  1\right)  ,
\]
assuming that
\begin{equation}
\overline{s}_{\left(  \alpha_{2},\alpha_{3},\ldots,\alpha_{n+1}\right)
}=\left(  \mathbf{B}_{\alpha_{2}}\circ\mathbf{B}_{\alpha_{3}}\circ\cdots
\circ\mathbf{B}_{\alpha_{n+1}}\right)  \left(  1\right)
.\label{sol.bernstein.e.ih}%
\end{equation}
But this is fairly straightforward: Applying the map $\mathbf{B}_{\alpha_{1}}$
to (\ref{sol.bernstein.e.ih}), we obtain%
\[
\mathbf{B}_{\alpha_{1}}\left(  \overline{s}_{\left(  \alpha_{2},\alpha
_{3},\ldots,\alpha_{n+1}\right)  }\right)  =\mathbf{B}_{\alpha_{1}}\left(
\left(  \mathbf{B}_{\alpha_{2}}\circ\mathbf{B}_{\alpha_{3}}\circ\cdots
\circ\mathbf{B}_{\alpha_{n+1}}\right)  \left(  1\right)  \right)  =\left(
\mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots\circ
\mathbf{B}_{\alpha_{n+1}}\right)  \left(  1\right)  .
\]
Compared with%
\begin{align*}
\mathbf{B}_{\alpha_{1}}\left(  \overline{s}_{\left(  \alpha_{2},\alpha
_{3},\ldots,\alpha_{n+1}\right)  }\right)   &  =\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{\alpha_{1}+i}e_{i}^{\perp}\overline{s}_{\left(  \alpha
_{2},\alpha_{3},\ldots,\alpha_{n+1}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the map }\mathbf{B}_{\alpha_{1}}\right)  \\
&  =\overline{s}_{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...,\alpha
_{n+1}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Exercise \ref{exe.bernstein}(d), applied to }\alpha_{1}\text{ and}\\
\left(  \alpha_{2},\alpha_{3},\ldots,\alpha_{n+1}\right)  \text{ instead of
}m\text{ and }\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
\end{array}
\right)  \\
&  =\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n+1}\right)  },
\end{align*}
this yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n+1}\right)  }=\left(
\mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots\circ
\mathbf{B}_{\alpha_{n+1}}\right)  \left(  1\right)  ,
\]
which completes the induction step and thus the proof.

(f) Let $m\in \ZZ$. Let $n$ be a positive integer. We have $e_{0}=1$ and
thus $e_{0}^{\perp}=1^{\perp}=\operatorname*{id}$ (by Proposition
\ref{skewing-properties-prop}(iii), applied to $A=\Lambda$). Hence,
$e_{0}^{\perp}p_{n}=\operatorname*{id}\left(  p_{n}\right)  =p_{n}$.

If $i$ is a positive integer satisfying $i\neq n$, then%
\begin{equation}
e_{i}^{\perp}p_{n}=0\label{sol.bernstein.f.eiperppn=0}%
\end{equation}
(by Exercise \ref{exe.skew-p-by-e}(d), applied to $m=i$).

The definition of $\mathbf{B}_{m}$ yields%
\begin{align*}
\mathbf{B}_{m}\left(  p_{n}\right)    & =\sum_{i\in{\NN}}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}\\
& =\underbrace{\left(  -1\right)  ^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}%
}\underbrace{e_{0}^{\perp}p_{n}}_{=p_{n}}+\sum_{i>0}\left(  -1\right)
^{i}h_{m+i}e_{i}^{\perp}p_{n}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=0\text{ from the sum}\right)  \\
& =h_{m}p_{n}+\sum_{i>0}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}.
\end{align*}
In view of%
\begin{align*}
& \sum_{i>0}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}\\
& =\left(  -1\right)  ^{n}h_{m+n}\underbrace{e_{n}^{\perp}p_{n}}%
_{\substack{=\left(  -1\right)  ^{n-1}\\\text{(by Exercise
\ref{exe.skew-p-by-e}(c))}}}+\sum\limits_{\substack{i>0;\\i\neq n}}\left(  -1\right)
^{i}h_{m+i}\underbrace{e_{i}^{\perp}p_{n}}_{\substack{=0\\\text{(by
(\ref{sol.bernstein.f.eiperppn=0}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=n\text{ from the sum,}\\
\text{since }n\text{ is a positive integer}%
\end{array}
\right)  \\
& =\left(  -1\right)  ^{n}h_{m+n}\left(  -1\right)  ^{n-1}+\underbrace{\sum\limits
_{\substack{i>0;\\i\neq n}}\left(  -1\right)  ^{i}h_{m+i}0}_{=0}=\left(
-1\right)  ^{n}h_{m+n}\left(  -1\right)  ^{n-1}=\underbrace{\left(  -1\right)
^{n}\left(  -1\right)  ^{n-1}}_{\substack{=\left(  -1\right)  ^{n+\left(
n-1\right)  }=-1\\\text{(since }n+\left(  n-1\right)  =2n-1\\\text{is odd)}%
}}h_{m+n}=-h_{m+n},
\end{align*}
this becomes
\[
\mathbf{B}_{m}\left(  p_{n}\right)  =h_{m}p_{n}+\underbrace{\sum_{i>0}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}}_{=-h_{m+n}}=h_{m}p_{n}+\left(
-h_{m+n}\right)  =h_{m}p_{n}-h_{m+n}.
\]
This solves Exercise~\ref{exe.bernstein}(f).

\textit{Remark:} Our solution to Exercise~\ref{exe.bernstein}(d) was modelled
after the rough sketch of a solution to Exercise~\ref{exe.bernstein}(b) given
in \cite[\S 4.20]{Zelevinsky}; but it involved many technicalities which are
not necessary if one is only interested in a solution to
Exercise~\ref{exe.bernstein}(b). (Specifically, if one only wants to solve
Exercise~\ref{exe.bernstein}(b), one can avoid the use of
Exercise~\ref{exe.bernstein}(c).)

We solved Exercise~\ref{exe.bernstein}(e) using Exercise~\ref{exe.bernstein}%
(d), but of course one can just as well turn this around and solve
Exercise~\ref{exe.bernstein}(d) using Exercise~\ref{exe.bernstein}(e) if one
has an independent solution to Exercise~\ref{exe.bernstein}(e). Such an
independent solution can be extracted from \cite[Corollary 3.30]%
{BergBergeronSaliolaSerranoZabrocki-immac1} (using the realization that the
immaculate creation operators $\mathbb{B}_{m}$ of
\cite{BergBergeronSaliolaSerranoZabrocki-immac1} are lifts of our Bernstein
operators $\mathbf{B}_{m}$ to $\Nsym$).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.bernstein}}

\textit{Solution to Exercise \ref{exe.bernstein}.} Before we start
solving this exercise, we notice the following fact:%
\begin{equation}
s_{\mu}^{\perp}s_{\lambda}=s_{\lambda/\mu}\ \ \ \ \ \ \ \ \ \ \text{for any
two partitions }\lambda\text{ and }\mu. \label{sol.bernstein.skewing.general}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.skewing.general}):} Recall
first that the basis $\left(  s_{\lambda}\right)  _{\lambda\in
 \Par }$ of $\Lambda$ is orthonormal with respect to the Hall
inner product. Hence,%
\begin{equation}
\left(  s_{\lambda},s_{\mu}\right)  =\delta_{\lambda,\mu}%
\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }\lambda\text{ and }\mu.
\label{sol.bernstein.orthonormal}%
\end{equation}
\par
On the other hand, every partition $\lambda$ satisfies%
\begin{align*}
\sum_{\mu\in \Par }s_{\mu}\otimes s_{\lambda/\mu}  &
=\underbrace{\sum\limits_{\substack{\mu\in \Par ;\\\mu\subseteq\lambda
}}s_{\mu}\otimes s_{\lambda/\mu}}_{\substack{=\sum_{\mu\subseteq\lambda}%
s_{\mu}\otimes s_{\lambda/\mu}=\Delta s_{\lambda}\\\text{(by Proposition
\ref{symm-comultiplication-formulas-prop}(iv))}}}+\sum\limits_{\substack{\mu
\in \Par ;\\\mu\nsubseteq\lambda}}s_{\mu}\otimes
\underbrace{s_{\lambda/\mu}}_{\substack{=0\\\text{(since }\mu\nsubseteq
\lambda\text{)}}}\\
&  =\Delta s_{\lambda}+\underbrace{\sum\limits_{\substack{\mu\in \Par %
;\\\mu\nsubseteq\lambda}}s_{\mu}\otimes0}_{=0}=\Delta s_{\lambda}%
=\sum_{\left(  \lambda\right)  }\left(  s_{\lambda}\right)  _{\left(
1\right)  }\otimes\left(  s_{\lambda}\right)  _{\left(  2\right)  }%
\end{align*}
(where we use the Sweedler notation). Hence,%
\begin{equation}
\sum_{\left(  \lambda\right)  }\left(  s_{\lambda}\right)  _{\left(  1\right)
}\otimes\left(  s_{\lambda}\right)  _{\left(  2\right)  }=\sum_{\mu
\in \Par }s_{\mu}\otimes s_{\lambda/\mu}=\sum_{\nu\in
 \Par }s_{\nu}\otimes s_{\lambda/\nu}
\label{sol.bernstein.skewing.general.pf.2}%
\end{equation}
(here, we renamed the summation index $\mu$ as $\nu$).
\par
Now, let $\lambda$ and $\mu$ be two partitions. By the definition of $s_{\mu
}^{\perp}s_{\lambda}$, we have%
\begin{align*}
s_{\mu}^{\perp}s_{\lambda}  &  =\sum_{\left(  s_{\lambda}\right)  }\left(
s_{\mu},\left(  s_{\lambda}\right)  _{1}\right)  \left(  s_{\lambda}\right)
_{2}=\sum_{\nu\in \Par }\underbrace{\left(  s_{\mu},s_{\nu
}\right)  }_{\substack{=\delta_{\mu,\nu}\\\text{(by
(\ref{sol.bernstein.orthonormal}), applied to}\\\mu\text{ and }\nu\text{
instead of }\lambda\text{ and }\mu\text{)}}}s_{\lambda/\nu}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{\left(  \lambda\right)
}\left(  s_{\lambda}\right)  _{\left(  1\right)  }\otimes\left(  s_{\lambda
}\right)  _{\left(  2\right)  }=\sum_{\nu\in \Par }s_{\nu}\otimes
s_{\lambda/\nu}\text{ (by (\ref{sol.bernstein.skewing.general.pf.2}))}\right)
\\
&  =\sum_{\nu\in \Par }\delta_{\mu,\nu}s_{\lambda/\nu
}=\underbrace{\sum\limits_{\substack{\nu\in \Par ;\\\nu=\mu}}\delta
_{\mu,\nu}s_{\lambda/\nu}}_{\substack{=\delta_{\mu,\mu}s_{\lambda/\mu
}\\\text{(since }\mu\in \Par \text{)}}}+\sum\limits_{\substack{\nu
\in \Par ;\\\nu\neq\mu}}\underbrace{\delta_{\mu,\nu}%
}_{\substack{=0\\\text{(since }\mu\neq\nu\\\text{(since }\nu\neq\mu\text{))}%
}}s_{\lambda/\nu}=\underbrace{\delta_{\mu,\mu}}_{\substack{=1\\\text{(since
}\mu=\mu\text{)}}}s_{\lambda/\mu}+\underbrace{\sum\limits_{\substack{\nu
\in \Par ;\\\nu\neq\mu}}0s_{\lambda/\nu}}_{=0}=s_{\lambda/\mu}.
\end{align*}
This proves (\ref{sol.bernstein.skewing.general}).}

Now, it is easy to see that
\begin{equation}
e_{i}^{\perp}s_{\lambda}=s_{\lambda/\left(  1^{i}\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for any partition }\lambda\text{ and every }%
i\in \NN . \label{sol.bernstein.skewing.e}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.skewing.e}):} Let $\lambda$ be
any partition, and let $i\in \NN$. Then, $s_{\left(  1^{i}\right)
}=e_{i}$, so that $e_{i}=s_{\left(  1^{i}\right)  }$ and thus $e_{i}^{\perp
}=s_{\left(  1^{i}\right)  }^{\perp}$. Hence, $\underbrace{e_{i}^{\perp}%
}_{=s_{\left(  1^{i}\right)  }^{\perp}}s_{\lambda}=s_{\left(  1^{i}\right)
}^{\perp}s_{\lambda}=s_{\lambda/\left(  1^{i}\right)  }$ (by
(\ref{sol.bernstein.skewing.general}), applied to $\mu=\left(  1^{i}\right)
$). This proves (\ref{sol.bernstein.skewing.e}).} Also,%
\begin{equation}
e_{i}^{\perp}s_{\lambda}=0\ \ \ \ \ \ \ \ \ \ \text{for any partition }%
\lambda\text{ and every }i\in \NN \text{ satisfying }i>\ell\left(
\lambda\right)  . \label{sol.bernstein.skewing.e0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.skewing.e0}):} Let $\lambda$ be
any partition, and let $i\in \NN$ be such that $i>\ell\left(
\lambda\right)  $. Then, $s_{\left(  1^{i}\right)  }=e_{i}$, so that
$e_{i}=s_{\left(  1^{i}\right)  }$ and thus $e_{i}^{\perp}=s_{\left(
1^{i}\right)  }^{\perp}$.
\par
But let us write the partition $\lambda$ in the form $\lambda=\left(
\lambda_{1},\lambda_{2},\lambda_{3},...\right)  $, and let us write the
partition $\left(  1^{i}\right)  $ in the form $\left(  1^{i}\right)  =\left(
\nu_{1},\nu_{2},\nu_{3},...\right)  $. Then, $\nu_{i}$ is the $i$-th entry of
the partition $\left(  1^{i}\right)  $ (since $\left(  \nu_{1},\nu_{2},\nu
_{3},...\right)  =\left(  1^{i}\right)  $), and therefore equals $1$ (because
the $i$-th entry of the partition $\left(  1^{i}\right)  $ equals $0$). In
other words, $\nu_{i}=1$. Meanwhile, $\lambda_{i}$ is the $i$-th entry of the
partition $\lambda$ (since $\left(  \lambda_{1},\lambda_{2},\lambda
_{3},...\right)  =\lambda$), and thus equals $0$ (since the $i$-th entry of
the partition $\lambda$ equals $0$ (since $i>\ell\left(  \lambda\right)  $)).
In other words, $\lambda_{i}=0$.
\par
Now, assume (for the sake of contradiction) that $\left(  1^{i}\right)
\subseteq\lambda$. Thus, $\left(  \nu_{1},\nu_{2},\nu_{3},...\right)  =\left(
1^{i}\right)  \subseteq\lambda=\left(  \lambda_{1},\lambda_{2},\lambda
_{3},...\right)  $. Hence, every positive integer $j$ satisfies $\nu_{j}%
\leq\lambda_{j}$ (because (according to the definition of the relation
$\subseteq$) we have $\left(  \nu_{1},\nu_{2},\nu_{3},...\right)
\subseteq\left(  \lambda_{1},\lambda_{2},\lambda_{3},...\right)  $ if and only
if every positive integer $j$ satisfies $\nu_{j}\leq\lambda_{j}$). Applied to
$j=i$, this yields $\nu_{i}\leq\lambda_{i}$. But this contradicts $\nu
_{i}=1>0=\lambda_{i}$. This contradiction shows that our assumption (that
$\left(  1^{i}\right)  \subseteq\lambda$) was false. Hence, we cannot have
$\left(  1^{i}\right)  \subseteq\lambda$. Thus, we have $\left(  1^{i}\right)
\nsubseteq\lambda$.
\par
Now, it is known that $s_{\mu}^{\perp}s_{\lambda}=0$ for every partition $\mu$
satisfying $\mu\nsubseteq\lambda$. Applying this to $\mu=\left(  1^{i}\right)
$, we obtain $s_{\left(  1^{i}\right)  }^{\perp}s_{\lambda}=0$. Hence,
$\underbrace{e_{i}^{\perp}}_{=s_{\left(  1^{i}\right)  }^{\perp}}s_{\lambda
}=s_{\left(  1^{i}\right)  }^{\perp}s_{\lambda}=0$. This proves
(\ref{sol.bernstein.skewing.e0}).}

(a) We need to prove the following two assertions:

\begin{statement}
\textit{Assertion A.1:} For every $f\in\Lambda$, the infinite sum $\sum
_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}f$ is convergent
in the discrete topology.
\end{statement}

\begin{statement}
\textit{Assertion A.2:} The map $\mathbf{B}_{m}$ is $\kk$-linear.
\end{statement}

\textit{Proof of Assertion A.1:} Let $f\in\Lambda$. Then, we can write $f$ in
the form $f=\sum_{\lambda\in \Par }a_{\lambda}s_{\lambda}$ for
some family $\left(  a_{\lambda}\right)  _{\lambda\in \Par }%
\in \kk ^{ \Par }$ of elements of $\kk$ having the
property that $\left(  \text{all but finitely many }\lambda\in
 \Par \text{ satisfy }a_{\lambda}=0\right)  $ (because $\left(
s_{\lambda}\right)  _{\lambda\in \Par }$ is a basis of the
$\kk$-module $\Lambda$). Consider this family $\left(  a_{\lambda
}\right)  _{\lambda\in \Par }$.

We know that all but finitely many $\lambda\in \Par $ satisfy
$a_{\lambda}=0$. In other words, there exists a finite set $I\subset
 \Par $ such that all $\lambda\in \Par \setminus I$
satisfy $a_{\lambda}=0$. Consider this $I$. The set $I$ is finite, and thus
the set $\left\{  \ell\left(  \lambda\right)  \ \mid\ \lambda\in I\right\}  $
is finite as well. Hence, $\left\{  \ell\left(  \lambda\right)  \ \mid
\ \lambda\in I\right\}  $ is a finite set of integers. Therefore, the set
$\left\{  \ell\left(  \lambda\right)  \ \mid\ \lambda\in I\right\}  $ has an
upper bound (since every finite set of integers has an upper bound). Let $N$
be such an upper bound.

We know that $N$ is an upper bound of the set $\left\{  \ell\left(
\lambda\right)  \ \mid\ \lambda\in I\right\}  $. In other words, every element
of the set $\left\{  \ell\left(  \lambda\right)  \ \mid\ \lambda\in I\right\}
$ is $\leq N$. Since $\left\{  \ell\left(  \lambda\right)  \ \mid\ \lambda\in
I\right\}  =\left\{  \ell\left(  \mu\right)  \ \mid\ \mu\in I\right\}  $
(here, we renamed the index $\lambda$ as $\mu$), this rewrites as follows:
Every element of the set $\left\{  \ell\left(  \mu\right)  \ \mid\ \mu\in
I\right\}  $ is $\leq N$.

We know that all $\lambda\in \Par \setminus I$ satisfy
$a_{\lambda}=0$. Hence, $\sum_{\lambda\in \Par \setminus
I}\underbrace{a_{\lambda}}_{=0}s_{\lambda}=\sum_{\lambda\in \Par %
\setminus I}0s_{\lambda}=0$.

We have%
\begin{equation}
f=\sum_{\lambda\in \Par }a_{\lambda}s_{\lambda}=\underbrace{\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda\in I}}}_{\substack{=\sum
_{\lambda\in I}\\\text{(since }I\subset \Par \text{)}}%
}a_{\lambda}s_{\lambda}+\underbrace{\sum\limits_{\substack{\lambda\in
 \Par ;\\\lambda\notin I}}}_{=\sum_{\lambda\in \Par %
\setminus I}}a_{\lambda}s_{\lambda}=\sum_{\lambda\in I}a_{\lambda}s_{\lambda
}+\underbrace{\sum_{\lambda\in \Par \setminus I}a_{\lambda
}s_{\lambda}}_{=0}=\sum_{\lambda\in I}a_{\lambda}s_{\lambda}.
\label{sol.bernstein.a.1.pf.1}%
\end{equation}


But now, let $i\in \NN$ be such that $i>N$.

Let $\lambda\in I$ be arbitrary. Then, $\ell\left(  \lambda\right)  $ is an
element of the set $\left\{  \ell\left(  \mu\right)  \ \mid\ \mu\in I\right\}
$ (since $\ell\left(  \lambda\right)  $ has the form $\ell\left(  \mu\right)
$ for some $\mu\in I$ (namely, for $\mu=\lambda$)). Therefore, $\ell\left(
\lambda\right)  $ is $\leq N$ (because every element of the set $\left\{
\ell\left(  \mu\right)  \ \mid\ \mu\in I\right\}  $ is $\leq N$). In other
words, $\ell\left(  \lambda\right)  \leq N$. Hence, $N\geq\ell\left(
\lambda\right)  $ and thus $i>N\geq\ell\left(  \lambda\right)  $. Therefore,
(\ref{sol.bernstein.skewing.e0}) yields $e_{i}^{\perp}s_{\lambda}=0$.

Now, forget that we fixed $\lambda$. We thus have proven that every
$\lambda\in I$ satisfies $e_{i}^{\perp}s_{\lambda}=0$. Hence, $\sum
_{\lambda\in I}a_{\lambda}\underbrace{e_{i}^{\perp}s_{\lambda}}_{=0}%
=\sum_{\lambda\in I}a_{\lambda}0=0$. Now, applying the map $e_{i}^{\perp}$ to
both sides of the equality (\ref{sol.bernstein.a.1.pf.1}), we obtain%
\begin{align*}
e_{i}^{\perp}f  &  =e_{i}^{\perp}\left(  \sum_{\lambda\in I}a_{\lambda
}s_{\lambda}\right)  =\sum_{\lambda\in I}a_{\lambda}e_{i}^{\perp}s_{\lambda
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }e_{i}^{\perp}\text{ is
} \kk \text{-linear}\right) \\
&  =0.
\end{align*}


Hence, $\left(  -1\right)  ^{i}h_{m+i}\underbrace{e_{i}^{\perp}f}_{=0}=0$.

Now, forget that we fixed $i$. We thus have shown that $\left(  -1\right)
^{i}h_{m+i}e_{i}^{\perp}f=0$ for every $i\in \NN$ such that $i>N$. In
other words, the $i$-th addend of the infinite sum $\sum_{i\in{ \NN }%
}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}f$ is $0$ for every $i\in
 \NN$ such that $i>N$. Thus, all but finitely many addends of the
infinite sum $\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}%
e_{i}^{\perp}f$ are $0$ (because all but finitely many $i\in \NN$
satisfy $i>N$). Therefore, the infinite sum $\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}f$ is convergent in the discrete topology.
This proves Assertion A.1.

\textit{Proof of Assertion A.2.} Recall that every $f\in\Lambda$ satisfies%
\begin{equation}
\mathbf{B}_{m}\left(  f\right)  =\sum_{i\in{ \NN }}\left(  -1\right)
^{i}h_{m+i}e_{i}^{\perp}f. \label{sol.bernstein.a.2.pf.def}%
\end{equation}
Renaming $f$ as $g$ in this equality, we obtain the following: Every
$g\in\Lambda$ satisfies%
\begin{equation}
\mathbf{B}_{m}\left(  g\right)  =\sum_{i\in{ \NN }}\left(  -1\right)
^{i}h_{m+i}e_{i}^{\perp}g. \label{sol.bernstein.a.2.pf.defg}%
\end{equation}
By the definition of $\mathbf{B}_{m}$, we have%
\begin{equation}
\mathbf{B}_{m}\left(  0\right)  =\sum_{i\in{ \NN }}\left(  -1\right)
^{i}h_{m+i}\underbrace{e_{i}^{\perp}0}_{\substack{=0\\\text{(since the map
}e_{i}^{\perp}\\\text{is } \kk \text{-linear)}}}=\sum_{i\in{ \NN }%
}\left(  -1\right)  ^{i}h_{m+i}0=0. \label{sol.bernstein.a.2.pf.1}%
\end{equation}


Also, every $\lambda\in \kk$ and $f\in\Lambda$ satisfy%
\begin{align}
\mathbf{B}_{m}\left(  \lambda f\right)   &  =\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{m+i}\underbrace{e_{i}^{\perp}\left(  \lambda f\right)
}_{\substack{=\lambda e_{i}^{\perp}f\\\text{(since the map }e_{i}^{\perp
}\\\text{is } \kk \text{-linear)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\mathbf{B}_{m}\right) \nonumber\\
&  =\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}\lambda e_{i}^{\perp
}f=\lambda\underbrace{\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}%
h_{m+i}e_{i}^{\perp}f}_{\substack{=\mathbf{B}_{m}\left(  f\right)  \\\text{(by
(\ref{sol.bernstein.a.2.pf.def}))}}}=\lambda\mathbf{B}_{m}\left(  f\right)  .
\label{sol.bernstein.a.2.pf.2}%
\end{align}


Finally, every $f\in\Lambda$ and $g\in\Lambda$ satisfy%
\begin{align*}
\mathbf{B}_{m}\left(  f+g\right)   &  =\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{m+i}\underbrace{e_{i}^{\perp}\left(  f+g\right)
}_{\substack{=e_{i}^{\perp}f+e_{i}^{\perp}g\\\text{(since the map }%
e_{i}^{\perp}\\\text{is } \kk \text{-linear)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{B}_{m}\right)
\\
&  =\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}\left(  e_{i}^{\perp
}f+e_{i}^{\perp}g\right)  =\underbrace{\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}f}_{\substack{=\mathbf{B}_{m}\left(
f\right)  \\\text{(by (\ref{sol.bernstein.a.2.pf.def}))}}}+\underbrace{\sum
_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}g}%
_{\substack{=\mathbf{B}_{m}\left(  g\right)  \\\text{(by
(\ref{sol.bernstein.a.2.pf.defg}))}}}\\
&  =\mathbf{B}_{m}\left(  f\right)  +\mathbf{B}_{m}\left(  g\right)  .
\end{align*}
Combining this with the identities (\ref{sol.bernstein.a.2.pf.1}) and
(\ref{sol.bernstein.a.2.pf.2}), we conclude that the map $\mathbf{B}_{m}$ is
$\kk$-linear. This proves Assertion A.2.

Thus, both Assertions A.1 and A.2 are proven. This completes the solution of
Exercise \ref{exe.bernstein}(a).

(c) Before we start solving this part of the exercise, let us recall two
general facts about determinants:

\begin{itemize}
\item If $A$ is any commutative ring, and if $N\in \NN$, and if $\left(
\beta_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$ is a square matrix, and
if $\sigma\in\Symm_N$ is a permutation, then%
\begin{equation}
\det\left(  \left(  \beta_{\sigma\left(  i\right)  ,j}\right)
_{i,j=1,2,...,N}\right)  =\left(  -1\right)  ^{\sigma}\det\left(  \left(
\beta_{i,j}\right)  _{i,j=1,2,...,N}\right)  .
\label{sol.bernstein.c.det.perm-rows}%
\end{equation}
(This is just a restatement of the well-known fact that if we permute the rows
of a square matrix by some permutation $\sigma$, then the determinant of this
matrix gets multiplied by $\left(  -1\right)  ^{\sigma}$.)

\item If $A$ is any commutative ring, and if $N\in \NN$, and if $\left(
\beta_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$ is a square matrix, and
if $\sigma\in\Symm_N$ is a permutation, then%
\begin{equation}
\det\left(  \left(  \beta_{i,\sigma\left(  j\right)  }\right)
_{i,j=1,2,...,N}\right)  =\left(  -1\right)  ^{\sigma}\det\left(  \left(
\beta_{i,j}\right)  _{i,j=1,2,...,N}\right)  .
\label{sol.bernstein.c.det.perm-cols}%
\end{equation}
(This is just a restatement of the well-known fact that if we permute the
columns of a square matrix by some permutation $\sigma$, then the determinant
of this matrix gets multiplied by $\left(  -1\right)  ^{\sigma}$.)
\end{itemize}

Now let us come to the solution of Exercise \ref{exe.bernstein}(c). Let us
first prove (\ref{eq.exe.bernstein.c.1}):

\textit{Proof of (\ref{eq.exe.bernstein.c.1}):} Let $\lambda$ be a partition
having at most $n$ parts. Write $\lambda$ in the form $\lambda=\left(
\lambda_{1},\lambda_{2},\lambda_{3},...\right)  $.

The partition $\lambda$ has at most $n$ parts. In other words, the number of
parts of $\lambda$ is $\leq n$. In other words, $\ell\left(  \lambda\right)  $
is $\leq n$ (since $\ell\left(  \lambda\right)  $ is the number of parts of
$\lambda$). Thus, $\ell\left(  \lambda\right)  \leq n$. Also, $\ell\left(
\varnothing\right)  =0\leq n$. Write the partition $\varnothing$ in the form
$\varnothing=\left(  \varnothing_{1},\varnothing_{2},\varnothing
_{3},...\right)  $. Then, $\varnothing_{j}=0$ for every positive integer $j$.
Now, (\ref{eq.jacobi-trudi.h}) (applied to $\mu=\varnothing$ and $\ell=n$)
yields%
\begin{align*}
s_{\lambda/\varnothing}  &  =\det\left(  \left(  \underbrace{h_{\lambda
_{i}-\varnothing_{j}-i+j}}_{\substack{=h_{\lambda_{i}-0-i+j}\\\text{(since
}\varnothing_{j}=0\text{)}}}\right)  _{i,j=1,2,...,n}\right)  =\det\left(
\left(  \underbrace{h_{\lambda_{i}-0-i+j}}_{=h_{\lambda_{i}-i+j}}\right)
_{i,j=1,2,...,n}\right) \\
&  =\det\left(  \left(  h_{\lambda_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)
=\overline{s}_{\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  }%
\end{align*}
(since $\overline{s}_{\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{n}\right)  }=\det\left(  \left(  h_{\lambda_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  $ (by the definition of $\overline{s}_{\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  }$)).

Hence,
$ s_\lambda = s_{\lambda/\varnothing}
= \overline{s}_{\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  }$.
This proves (\ref{eq.exe.bernstein.c.1}).

Next, we are going to prove the following four statements:

\begin{statement}
\textit{Assertion C.1:} For every $n$-tuple $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$, we have%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
\in\left\{  0\right\}  \cup\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid
\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in \Par %
;\ \ell\left(  \nu\right)  \leq n\right\}  .
\]
\end{statement}

\begin{statement}
\textit{Assertion C.2:} Let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in \ZZ^n$ be an $n$-tuple such that the $n$ integers
$\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are distinct.
Let $\tau\in\Symm_n$ be a permutation such that $\alpha_{\tau\left(
1\right)  }-\tau\left(  1\right)  \geq\alpha_{\tau\left(  2\right)  }%
-\tau\left(  2\right)  \geq...\geq\alpha_{\tau\left(  n\right)  }-\tau\left(
n\right)  $. Let $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)
\in \ZZ ^{n}$ be an $n$-tuple such that every $p\in\left\{
1,2,...,n\right\}  $ satisfies $\gamma_{p}=\alpha_{\tau\left(  p\right)
}-\tau\left(  p\right)  +p$. Then, $\gamma_{1}\geq\gamma_{2}\geq...\geq
\gamma_{n}$.
\end{statement}

\begin{statement}
\textit{Assertion C.3:} Let $f: \ZZ \times \ZZ \rightarrow
 \ZZ$ be any map. Let $\left(  \mathfrak{a}_{1},\mathfrak{a}%
_{2},...,\mathfrak{a}_{n}\right)  \in \ZZ ^{n}$ and $\left(
\mathfrak{b}_{1},\mathfrak{b}_{2},...,\mathfrak{b}_{n}\right)  \in
 \ZZ ^{n}$ be two $n$-tuples of integers. Assume that the $n$ integers
$\mathfrak{a}_{1}$, $\mathfrak{a}_{2}$, $...$, $\mathfrak{a}_{n}$ are not
distinct. Then, $\det\left(  \left(  h_{f\left(  \mathfrak{a}%
_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$.
\end{statement}

\begin{statement}
\textit{Assertion C.4:} Let $f: \ZZ \times \ZZ \rightarrow
 \ZZ$ be any map. Let $\left(  \mathfrak{a}_{1},\mathfrak{a}%
_{2},...,\mathfrak{a}_{n}\right)  \in \ZZ ^{n}$ and $\left(
\mathfrak{b}_{1},\mathfrak{b}_{2},...,\mathfrak{b}_{n}\right)  \in
 \ZZ ^{n}$ be two $n$-tuples of integers. Assume that the $n$ integers
$\mathfrak{b}_{1}$, $\mathfrak{b}_{2}$, $...$, $\mathfrak{b}_{n}$ are not
distinct. Then, $\det\left(  \left(  h_{f\left(  \mathfrak{a}%
_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$.
\end{statement}

Note that Assertion C.1 is more or less one of the claims of Exercise
\ref{exe.bernstein}(c), whereas Assertions C.2, C.3 and C.4 are auxiliary
results that will be used in proving Assertion C.1 but will also come in handy
later. We will first prove Assertion C.3, then Assertion C.4, then Assertion
C.2 and then Assertion C.1.

\textit{Proof of Assertion C.3:} We have assumed that the $n$ integers
$\mathfrak{a}_{1}$, $\mathfrak{a}_{2}$, $...$, $\mathfrak{a}_{n}$ are not
distinct. Thus, two of these integers are equal. In other words,
there exist two distinct elements $u$ and $v$ of $\left\{  1,2,...,n\right\}
$ such that $\mathfrak{a}_{u}=\mathfrak{a}_{v}$. Consider these $u$ and $v$.

For every $p\in\left\{  1,2,...,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }p\text{-th entry of the }u\text{-th row of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right) \\
&  =\left(  \text{the }\left(  u,p\right)  \text{-th entry of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right) \\
&  =h_{f\left(  \mathfrak{a}_{u},\mathfrak{b}_{p}\right)  }=h_{f\left(
\mathfrak{a}_{v},\mathfrak{b}_{p}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathfrak{a}_{u}=\mathfrak{a}_{v}\right) \\
&  =\left(  \text{the }\left(  v,p\right)  \text{-th entry of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \text{the }\left(
v,p\right)  \text{-th entry of the matrix }\left(  h_{f\left(  \mathfrak{a}%
_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,\ldots,n}\right)
=h_{f\left(  \mathfrak{a}_{v},\mathfrak{b}_{p}\right)  }\right) \\
&  =\left(  \text{the }p\text{-th entry of the }v\text{-th row of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right)  .
\end{align*}
Hence,%
\[
\left(  \text{the }u\text{-th row of the matrix }\left(  h_{f\left(
\mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  \text{the }v\text{-th row of the matrix }\left(
h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right)  .
\]
Since $u\neq v$, this yields that the matrix $\left(  h_{f\left(
\mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,\ldots,n}$ has
two equal rows (namely, the $u$-th row and the $v$-th row). Thus, the matrix
$\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}$ must have zero determinant (because any square matrix
which has two equal rows must have zero determinant). In other words,
$\det\left(  \left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)
}\right)  _{i,j=1,2,\ldots,n}\right)  =0$. This proves Assertion C.3.

\textit{Proof of Assertion C.4:} We have assumed that the $n$ integers
$\mathfrak{b}_{1}$, $\mathfrak{b}_{2}$, $...$, $\mathfrak{b}_{n}$ are not
distinct. Thus, two of these integers are equal. In other words,
there exist two distinct elements $u$ and $v$ of $\left\{  1,2,...,n\right\}
$ such that $\mathfrak{b}_{u}=\mathfrak{b}_{v}$. Consider these $u$ and $v$.

For every $p\in\left\{  1,2,...,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }p\text{-th entry of the }u\text{-th column of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right) \\
&  =\left(  \text{the }\left(  p,u\right)  \text{-th entry of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right) \\
&  =h_{f\left(  \mathfrak{a}_{p},\mathfrak{b}_{u}\right)  }=h_{f\left(
\mathfrak{a}_{p},\mathfrak{b}_{v}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathfrak{v}_{u}=\mathfrak{v}_{v}\right) \\
&  =\left(  \text{the }\left(  p,v\right)  \text{-th entry of the matrix
}\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \text{the }\left(
p,v\right)  \text{-th entry of the matrix }\left(  h_{f\left(  \mathfrak{a}%
_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,\ldots,n}\right)
=h_{f\left(  \mathfrak{a}_{p},\mathfrak{b}_{v}\right)  }\right) \\
&  =\left(  \text{the }p\text{-th entry of the }v\text{-th column of the
matrix }\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)
}\right)  _{i,j=1,2,\ldots,n}\right)  .
\end{align*}
Hence,%
\[
\left(  \text{the }u\text{-th column of the matrix }\left(  h_{f\left(
\mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  \text{the }v\text{-th column of the matrix }\left(
h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)
_{i,j=1,2,\ldots,n}\right)  .
\]
Since $u\neq v$, this yields that the matrix $\left(  h_{f\left(
\mathfrak{a}_{i},\mathfrak{b}_{j}\right)  }\right)  _{i,j=1,2,\ldots,n}$ has
two equal columns (namely, the $u$-th column and the $v$-th column). Thus, the
matrix $\left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)
}\right)  _{i,j=1,2,\ldots,n}$ must have zero determinant (because any square
matrix which has two equal column must have zero determinant). In other words,
$\det\left(  \left(  h_{f\left(  \mathfrak{a}_{i},\mathfrak{b}_{j}\right)
}\right)  _{i,j=1,2,\ldots,n}\right)  =0$. This proves Assertion C.4.

\textit{Proof of Assertion C.2:} Let $i\in\left\{  1,2,...,n-1\right\}  $.
Then, we are going to prove that $\gamma_{i}\geq\gamma_{i+1}$.

Indeed, we have $\alpha_{\tau\left(  1\right)  }-\tau\left(  1\right)
\geq\alpha_{\tau\left(  2\right)  }-\tau\left(  2\right)  \geq...\geq
\alpha_{\tau\left(  n\right)  }-\tau\left(  n\right)  $ and therefore
$\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  \geq\alpha_{\tau\left(
i+1\right)  }-\tau\left(  i+1\right)  $.

But $\tau$ is a permutation, and therefore bijective, and thus injective.
Hence, if we had $\tau\left(  i\right)  =\tau\left(  i+1\right)  $, then we
would have $i=i+1$, which is absurd. Thus, we cannot have $\tau\left(
i\right)  =\tau\left(  i+1\right)  $. We thus have $\tau\left(  i\right)
\neq\tau\left(  i+1\right)  $. The elements $\tau\left(  i\right)  $ and
$\tau\left(  i+1\right)  $ of $\left\{  1,2,...,n\right\}  $ are therefore distinct.

But the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are
distinct. In other words, for any two distinct elements $u$ and $v$
of $\left\{  1,2,...,n\right\}  $, we have $\alpha_{u}-u\neq\alpha_{v}-v$.
Applying this to $u=\tau\left(  i\right)  $ and $v=\tau\left(  i+1\right)  $,
we obtain $\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  \neq
\alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)  $. Combined with
$\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  \geq\alpha_{\tau\left(
i+1\right)  }-\tau\left(  i+1\right)  $, this yields $\alpha_{\tau\left(
i\right)  }-\tau\left(  i\right)  >\alpha_{\tau\left(  i+1\right)  }%
-\tau\left(  i+1\right)  $. Since $\alpha_{\tau\left(  i\right)  }-\tau\left(
i\right)  $ and $\alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)  $
are integers, this yields $\alpha_{\tau\left(  i\right)  }-\tau\left(
i\right)  \geq\alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)  +1$.

Now, every $p\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma_{p}%
=\alpha_{\tau\left(  p\right)  }-\tau\left(  p\right)  +p$. Applying this to
$p=i+1$, we obtain $\gamma_{i+1}=\alpha_{\tau\left(  i+1\right)  }-\tau\left(
i+1\right)  +\left(  i+1\right)  =\left(  \alpha_{\tau\left(  i+1\right)
}-\tau\left(  i+1\right)  +1\right)  +i$. Hence, $\left(  \alpha_{\tau\left(
i+1\right)  }-\tau\left(  i+1\right)  +1\right)  +i=\gamma_{i+1}$.

But recall that every $p\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma
_{p}=\alpha_{\tau\left(  p\right)  }-\tau\left(  p\right)  +p$. Applying this
to $p=i$, we obtain%
\[
\gamma_{i}=\underbrace{\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)
}_{\geq\alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)  +1}%
+i\geq\left(  \alpha_{\tau\left(  i+1\right)  }-\tau\left(  i+1\right)
+1\right)  +i=\gamma_{i+1}.
\]


Now, let us forget that we fixed $i$. We thus have proven that $\gamma_{i}%
\geq\gamma_{i+1}$ for every $i\in\left\{  1,2,...,n-1\right\}  $. In other
words, $\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$. This proves Assertion C.2.

\textit{Proof of Assertion C.1:} Let $\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$ be any $n$-tuple. We need to
prove that%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
\in\left\{  0\right\}  \cup\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid
\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in \Par %
;\ \ell\left(  \nu\right)  \leq n\right\}  .
\]


It is easy to see that if the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$,
$...$, $\alpha_{n}-n$ are not distinct, then $\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that the $n$ integers $\alpha_{1}-1$,
$\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not distinct. Define a map
$f: \ZZ \times \ZZ \rightarrow \ZZ$ by%
\[
\left(  f\left(  u,v\right)  =u+v\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
u,v\right)  \in \ZZ \times \ZZ \right)  .
\]
Then, Assertion C.3 (applied to $\mathfrak{a}_{i}=\alpha_{i}-i$ and
$\mathfrak{b}_{j}=j$) yields $\det\left(  \left(  h_{f\left(  \alpha
_{i}-i,j\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$ (since the $n$
integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not
distinct). Since every $\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}$ satisfy $f\left(  \alpha_{i}-i,j\right)  =\alpha
_{i}-i+j$ (by the definition of $f$), this rewrites as $\det\left(  \left(
h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  =0$. Now,
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }
=\det\left(
\left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =0$, qed.}.
Hence, if the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha
_{n}-n$ are not distinct, then
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=0\in\left\{  0\right\}  \subset\left\{  0\right\}  \cup\left\{
\mathfrak{s}\cdot s_{\nu}\ \mid\ \mathfrak{s}\in\left\{  -1,1\right\}
;\ \nu\in \Par ;\ \ell\left(  \nu\right)  \leq n\right\}  .
\]
Thus, if the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha
_{n}-n$ are not distinct, then Assertion C.1 is proven. Hence, for
the rest of the proof of Assertion C.1, we can WLOG assume that it is
\textbf{not} true that the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$,
$\alpha_{n}-n$ are not distinct. Assume this.

We assumed that it is \textbf{not} true that the $n$ integers $\alpha_{1}-1$,
$\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not distinct. In other
words, the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$
are distinct.

For every $n$-tuple $\left(  \eta_{1},\eta_{2},...,\eta_{n}\right)  $ of
integers, there exists a permutation $\tau\in\Symm_n$ satisfying
$\eta_{\tau\left(  1\right)  }\geq\eta_{\tau\left(  2\right)  }\geq...\geq
\eta_{\tau\left(  n\right)  }$\ \ \ \ \footnote{This is just a restatement of
the known fact that any $n$-tuple of integers can be sorted in decreasing
order.}. Applied to $\eta_{i}=\alpha_{i}-i$, this yields the following: There
exists a permutation $\tau\in\Symm_n$ satisfying $\alpha_{\tau\left(
1\right)  }-\tau\left(  1\right)  \geq\alpha_{\tau\left(  2\right)  }%
-\tau\left(  2\right)  \geq...\geq\alpha_{\tau\left(  n\right)  }-\tau\left(
n\right)  $. Consider this permutation $\tau$.

Define an $n$-tuple $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)
\in \ZZ ^{n}$ by%
\[
\left(  \gamma_{p}=\alpha_{\tau\left(  p\right)  }-\tau\left(  p\right)
+p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\left\{  1,2,...,n\right\}
\right)  .
\]
Then, $\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$%
\ \ \ \ \footnote{\textit{Proof.} This follows from Assertion C.2 (because
every $p\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma_{p}=\alpha
_{\tau\left(  p\right)  }-\tau\left(  p\right)  +p$).}. Moreover,%
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  . \label{sol.bernstein.c.1.permuted}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.c.1.permuted}):} The sign of
every permutation belongs to $\left\{  1,-1\right\}  $. Hence, $\left(
-1\right)  ^{\tau}\in\left\{  1,-1\right\}  $. Consequently, $\left(  \left(
-1\right)  ^{\tau}\right)  ^{2}=1$.
\par
Now, every $i\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma_{i}%
=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  +i$ (by the definition
of $\gamma_{i}$). Hence, every $i\in\left\{  1,2,...,n\right\}  $ satisfies
$\gamma_{i}-i=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  $. Thus,%
\[
\det\left(  \left(  \underbrace{h_{\gamma_{i}-i+j}}_{\substack{=h_{\alpha
_{\tau\left(  i\right)  }-\tau\left(  i\right)  +j}\\\text{(since }\gamma
_{i}-i=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  \text{)}%
}}\right)  _{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\alpha
_{\tau\left(  i\right)  }-\tau\left(  i\right)  +j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\alpha_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(by (\ref{sol.bernstein.c.det.perm-rows}), applied to $A=\Lambda$, $N=n$,
$\beta_{i,j}=h_{\alpha_{i}-i+j}$ and $\sigma=\tau$). Multiplying both sides of
this equality by $\left(  -1\right)  ^{\tau}$, we obtain%
\begin{align*}
\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)   &  =\underbrace{\left(  -1\right)  ^{\tau}\left(
-1\right)  ^{\tau}}_{=\left(  \left(  -1\right)  ^{\tau}\right)  ^{2}=1}%
\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right) \\
&  =\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\overline{s}_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=\det\left(  \left(  h_{\alpha
_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  \right)  .
\end{align*}
This proves (\ref{sol.bernstein.c.1.permuted}).} It is easy to see that if
$\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \notin \NN ^{n}$,
then $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=0$\ \ \ \ \footnote{\textit{Proof.} Assume that $\left(  \gamma_{1}%
,\gamma_{2},...,\gamma_{n}\right)  \notin \NN ^{n}$. Then, there exists
at least one $p\in\left\{  1,2,...,n\right\}  $ which does not satisfy
$\gamma_{p}\in \NN$ (because otherwise, every $p\in\left\{
1,2,...,n\right\}  $ would satisfy $\gamma_{p}\in \NN$, and therefore we
would have $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \in
 \NN ^{n}$, which would contradict $\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n}\right)  \notin \NN ^{n}$). Consider this $p$. We know
that $p$ satisfies $\gamma_{p}\in \ZZ$ (since $\left(  \gamma_{1}%
,\gamma_{2},...,\gamma_{n}\right)  \in \ZZ ^{n}$) but not $\gamma_{p}%
\in \NN$. Hence, $\gamma_{p}\in \ZZ \setminus \NN =\left\{
-1,-2,-3,...\right\}  $, so that $\gamma_{p}<0$.
\par
Now, $p\in\left\{  1,2,...,n\right\}  $ yields $1\leq p\leq n$, so that
$n\geq1$ and thus $n\in\left\{  1,2,...,n\right\}  $. Hence, $\tau\left(
n\right)  $ and $\gamma_{n}$ are well-defined. Now, we have $\gamma_{1}%
\geq\gamma_{2}\geq...\geq\gamma_{n}$. Thus, any two elements $u$ and $v$ of
$\left\{  1,2,...,n\right\}  $ satisfying $u\leq v$ satisfy $\gamma_{u}%
\geq\gamma_{v}$. Applying this to $u=p$ and $v=n$, we obtain $\gamma_{p}%
\geq\gamma_{n}$. Thus, $\gamma_{n}\leq\gamma_{p}<0$.
\par
Now, for every $q\in\left\{  1,2,...,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }q\text{-th entry of the }n\text{-th row of the matrix
}\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right) \\
&  =h_{\gamma_{n}-n+q}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\underbrace{\gamma_{n}}_{<0}-n+\underbrace{q}_{\substack{\leq n\\\text{(since
}q\in\left\{  1,2,...,n\right\}  \text{)}}}<0-n+n=0\right) \\
&  =\left(  \text{the }p\text{-th entry of the row vector }0\text{ of length
}n\right)  .
\end{align*}
Thus,%
\[
\left(  \text{the }n\text{-th row of the matrix }\left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =\left(  \text{the row vector
}0\text{ of length }n\right)  .
\]
In other words, the $n$-th row of the matrix $\left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}$ is the zero vector. Hence, the matrix
$\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}$ has a row which is
the zero vector (namely, its $n$-th row). As a consequence, this matrix
$\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}$ must have zero
determinant (because it is well-known that if a square matrix has a row which
is the zero vector, then this square matrix must have zero determinant). In
other words, $\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =0$. Now, (\ref{sol.bernstein.c.1.permuted}) yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=0}=0,
\]
qed.}. Hence, if $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)
\notin \NN ^{n}$, then
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=0\in\left\{  0\right\}  \subset\left\{  0\right\}  \cup\left\{
\mathfrak{s}\cdot s_{\nu}\ \mid\ \mathfrak{s}\in\left\{  -1,1\right\}
;\ \nu\in \Par ;\ \ell\left(  \nu\right)  \leq n\right\}  .
\]
Thus, if $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \notin%
 \NN ^{n}$, then Assertion C.1 is proven. Hence, for the rest of the
proof of Assertion C.1, we can WLOG assume that we don't have $\left(
\gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \notin \NN ^{n}$. Assume this.

We assumed that we don't have $\left(  \gamma_{1},\gamma_{2},...,\gamma
_{n}\right)  \notin \NN ^{n}$. Thus, we do have $\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n}\right)  \in \NN ^{n}$. Combined with
$\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$, this yields that $\left(
\gamma_{1},\gamma_{2},...,\gamma_{n}\right)  $ is a partition having at most
$n$ parts. Denote this partition $\left(  \gamma_{1},\gamma_{2},...,\gamma
_{n}\right)  $ as $\kappa$. Write $\kappa$ in the form $\left(  \kappa
_{1},\kappa_{2},\kappa_{3},...\right)  $. Then, $\kappa=\left(  \kappa
_{1},\kappa_{2},...,\kappa_{n}\right)  $ (since $\kappa=\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n}\right)  $ is a partition having at most $n$
parts). Thus, $\left(  \kappa_{1},\kappa_{2},...,\kappa_{n}\right)
=\kappa=\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  $. But
(\ref{eq.exe.bernstein.c.1}) (applied to $\kappa$ instead of $\lambda$) yields%
\begin{align*}
s_{\kappa}  &  =\overline{s}_{\left(  \kappa_{1},\kappa_{2},...,\kappa
_{n}\right)  }=\overline{s}_{\left(  \gamma_{1},\gamma_{2},...,\gamma
_{n}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \kappa
_{1},\kappa_{2},...,\kappa_{n}\right)  =\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n}\right)  \right) \\
&  =\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{s}_{\left(
\gamma_{1},\gamma_{2},...,\gamma_{n}\right)  }\right)  ,
\end{align*}
so that $\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =s_{\kappa}$. Now, (\ref{sol.bernstein.c.1.permuted}) becomes%
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=s_{\kappa}}=\left(  -1\right)
^{\tau}s_{\kappa}. \label{sol.bernstein.c.1.almost}%
\end{equation}


But $\left(  -1\right)  ^{\tau}\in\left\{  -1,1\right\}  $ and $\kappa
\in \Par $ (since $\kappa$ is a partition) and $\ell\left(
\kappa\right)  \leq n$\ \ \ \ \footnote{\textit{Proof.} The partition $\kappa$
has at most $n$ parts. In other words, the number of parts of the partition
$\kappa$ is at most $n$. In other words, $\ell\left(  \kappa\right)  $ is at
most $n$ (since $\ell\left(  \kappa\right)  $ is the number of parts of the
partition $\kappa$). In other words, $\ell\left(  \kappa\right)  \leq n$,
qed.}. Hence, (\ref{sol.bernstein.c.1.almost}) shows that
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
can be written
in the form $\mathfrak{s}\cdot s_{\nu}$ for an $s\in\left\{  -1,1\right\}  $
and a $\nu\in \Par $ satisfying $\ell\left(  \nu\right)  \leq n$
(namely, for $s=\left(  -1\right)  ^{\tau}$ and $\nu=\kappa$). In other words,%
\begin{align*}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }  &
\in\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid\ \mathfrak{s}\in\left\{
-1,1\right\}  ;\ \nu\in \Par ;\ \ell\left(  \nu\right)  \leq
n\right\} \\
&  \subset\left\{  0\right\}  \cup\left\{  \mathfrak{s}\cdot s_{\nu}%
\ \mid\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in \Par %
;\ \ell\left(  \nu\right)  \leq n\right\}  .
\end{align*}
This proves Assertion C.1.

Now, we can readily see that for every $n$-tuple $\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$, the symmetric
function $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }$ either is $0$ or equals $\pm s_{\nu}$ for some partition $\nu$
having at most $n$ parts\footnote{\textit{Proof.} Let $\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$ be an
$n$-tuple. We must prove that $\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }$ either is $0$ or equals $\pm s_{\nu}$ for
some partition $\nu$ having at most $n$ parts.
\par
Assertion C.1 yields%
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
\in\left\{  0\right\}  \cup\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid
\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in \Par %
;\ \ell\left(  \nu\right)  \leq n\right\}  . \label{sol.bernstein.c.2.pf}%
\end{equation}
In other words, either $\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  }\in\left\{  0\right\}  $ or $\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=\left\{  \mathfrak{s}\cdot
s_{\nu}\ \mid\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in
 \Par ;\ \ell\left(  \nu\right)  \leq n\right\}  $.
\par
If $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
is $0$, then it is clear that $\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }$ either is $0$ or equals $\pm s_{\nu}$ for
some partition $\nu$ having at most $n$ parts. Hence, if
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
is $0$, then
our proof is finished. Thus, for the rest of this proof, we can WLOG assume
that $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
is not $0$. Assume this.
\par
We assumed that $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }$ is not $0$. In other words, $\overline{s}_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }\neq0$. Combining this with
(\ref{sol.bernstein.c.2.pf}), we obtain%
\begin{align*}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }  &
\in\left(  \left\{  0\right\}  \cup\left\{  \mathfrak{s}\cdot s_{\nu}%
\ \mid\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in \Par %
;\ \ell\left(  \nu\right)  \leq n\right\}  \right)  \setminus\left\{
0\right\} \\
&  \subset\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid\ \mathfrak{s}\in\left\{
-1,1\right\}  ;\ \nu\in \Par ;\ \ell\left(  \nu\right)  \leq
n\right\}  .
\end{align*}
In other words, $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }$ has the form $\mathfrak{s}\cdot s_{\nu}$ for some
$\mathfrak{s}\in\left\{  -1,1\right\}  $ and some $\nu\in \Par $
satisfying $\ell\left(  \nu\right)  \leq n$. Consider this $\mathfrak{s}$ and
this $\nu$. Clearly, $\nu$ is a partition. The number $\ell\left(  \nu\right)
$ is $\leq n$. In other words, the number of parts of $\nu$ is $\leq n$ (since
the number of parts of $\nu$ is $\ell\left(  \nu\right)  $). In other words,
the partition $\nu$ has at most $n$ parts. We have $\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=\mathfrak{s}\cdot s_{\nu
}=\pm s_{\nu}$ (since $\mathfrak{s}\in\left\{  -1,1\right\}  $).
\par
Now, forget that we defined $\nu$. We thus have shown that
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
equals $\pm
s_{\nu}$ for some partition $\nu$ having at most $n$ parts. Hence,
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$
either is $0$ or equals $\pm s_{\nu}$ for some partition $\nu$ having at most
$n$ parts, qed.}. Thus, we have proven two of the statements of Exercise
\ref{exe.bernstein}(c) (namely, the statement made in the preceding sentence,
and (\ref{eq.exe.bernstein.c.1})). In order to complete the solution of
Exercise \ref{exe.bernstein}(c), it now remains to prove
(\ref{eq.exe.bernstein.c.3}). Before we prove (\ref{eq.exe.bernstein.c.3}) in
full generality, let us show the following assertion (which is a particular
case of (\ref{eq.exe.bernstein.c.3})):

\begin{statement}
\textit{Assertion C.5:} Let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in{ \ZZ }^{n}$ be an $n$-tuple, and let $\left(  \beta
_{1},\beta_{2},\ldots,\beta_{n}\right)  $ be a partition having at most $n$
parts. Then, (\ref{eq.exe.bernstein.c.3}) holds.
\end{statement}

\textit{Proof of Assertion C.5:} We need to prove that
(\ref{eq.exe.bernstein.c.3}) holds.

It is easy to see that if the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$,
$...$, $\alpha_{n}-n$ are not distinct, then
(\ref{eq.exe.bernstein.c.3}) holds\footnote{\textit{Proof.} Assume that the
$n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not
distinct.
\par
Define a function $f: \ZZ \times \ZZ \rightarrow \ZZ$ by%
\[
\left(  f\left(  u,v\right)  =u+v\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
u,v\right)  \in \ZZ \times \ZZ \right)  .
\]
Then, Assertion C.3 (applied to $\mathfrak{a}_{i}=\alpha_{i}-i$ and
$\mathfrak{b}_{j}=j$) yields $\det\left(  \left(  h_{f\left(  \alpha
_{i}-i,j\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$ (since the $n$
integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not
distinct). Since every $\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}$ satisfy $f\left(  \alpha_{i}-i,j\right)  =\alpha
_{i}-i+j$ (by the definition of $f$), this rewrites as $\det\left(  \left(
h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,...,n}\right)  =0$. Thus,
$\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }
=\det\left(
\left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =0$. Hence,%
\begin{equation}
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\underbrace{\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }}_{=0}=\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{n}\right)  }^{\perp}0=0 \label{sol.bernstein.c.3.case1.1}%
\end{equation}
(since $\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}^{\perp}$ is a $\kk$-linear map).
\par
But Assertion C.3 (applied to $\mathfrak{a}_{i}=\alpha_{i}-i$ and
$\mathfrak{b}_{j}=j-\beta_{j}$) yields $\det\left(  \left(  h_{f\left(
\alpha_{i}-i,j-\beta_{j}\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$ (since
the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not
distinct). Since every $\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}$ satisfy%
\begin{align*}
f\left(  \alpha_{i}-i,j-\beta_{j}\right)   &  =\alpha_{i}-i+j-\beta
_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f\right) \\
&  =\alpha_{i}-\beta_{j}-i+j,
\end{align*}
this rewrites as $\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,...,n}\right)  =0$. Compared with (\ref{sol.bernstein.c.3.case1.1}),
this yields%
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  .
\]
Thus, (\ref{eq.exe.bernstein.c.3}) holds, qed.}. Thus, for the rest of the
proof of (\ref{eq.exe.bernstein.c.3}), we can WLOG assume that it is
\textbf{not} true that the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$,
$\alpha_{n}-n$ are not distinct. Assume this.

We assumed that it is \textbf{not} true that the $n$ integers $\alpha_{1}-1$,
$\alpha_{2}-2$, $...$, $\alpha_{n}-n$ are not distinct. In other
words, the $n$ integers $\alpha_{1}-1$, $\alpha_{2}-2$, $...$, $\alpha_{n}-n$
are distinct.

For every $n$-tuple $\left(  \eta_{1},\eta_{2},...,\eta_{n}\right)  $ of
integers, there exists a permutation $\tau\in\Symm_n$ satisfying
$\eta_{\tau\left(  1\right)  }\geq\eta_{\tau\left(  2\right)  }\geq...\geq
\eta_{\tau\left(  n\right)  }$\ \ \ \ \footnote{This is just a restatement of
the known fact that any $n$-tuple of integers can be sorted in decreasing
order.}. Applied to $\eta_{i}=\alpha_{i}-i$, this yields the following: There
exists a permutation $\tau\in\Symm_n$ satisfying $\alpha_{\tau\left(
1\right)  }-\tau\left(  1\right)  \geq\alpha_{\tau\left(  2\right)  }%
-\tau\left(  2\right)  \geq...\geq\alpha_{\tau\left(  n\right)  }-\tau\left(
n\right)  $. Consider this permutation $\tau$.

Define an $n$-tuple $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)
\in \ZZ ^{n}$ by%
\[
\left(  \gamma_{p}=\alpha_{\tau\left(  p\right)  }-\tau\left(  p\right)
+p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\left\{  1,2,...,n\right\}
\right)  .
\]
Then, $\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$%
\ \ \ \ \footnote{\textit{Proof.} This follows from Assertion C.2 (because
every $p\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma_{p}=\alpha
_{\tau\left(  p\right)  }-\tau\left(  p\right)  +p$).}.

Moreover,%
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  . \label{sol.bernstein.c.5.permuted}%
\end{equation}
\footnote{This is proven in the same way as (\ref{sol.bernstein.c.1.permuted})
was proven in the proof of Assertion C.1.} Furthermore,
\begin{equation}
\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}%
-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\label{sol.bernstein.c.5.permuted-}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.c.5.permuted-}):} The sign of
every permutation belongs to $\left\{  1,-1\right\}  $. Hence, $\left(
-1\right)  ^{\tau}\in\left\{  1,-1\right\}  $. Consequently, $\left(  \left(
-1\right)  ^{\tau}\right)  ^{2}=1$.
\par
Now, every $i\in\left\{  1,2,...,n\right\}  $ satisfies $\gamma_{i}%
=\alpha_{\tau\left(  i\right)  }-\tau\left(  i\right)  +i$ (by the definition
of $\gamma_{i}$). Hence, every $i\in\left\{  1,2,...,n\right\}  $ satisfies%
\[
\underbrace{\gamma_{i}}_{=\alpha_{\tau\left(  i\right)  }-\tau\left(
i\right)  +i}-\beta_{j}-i+j=\left(  \alpha_{\tau\left(  i\right)  }%
-\tau\left(  i\right)  +i\right)  -\beta_{j}-i+j=\alpha_{\tau\left(  i\right)
}-\beta_{j}-\tau\left(  i\right)  +j.
\]
Thus,%
\[
\det\left(  \left(  \underbrace{h_{\gamma_{i}-\beta_{j}-i+j}}%
_{\substack{=h_{\alpha_{\tau\left(  i\right)  }-\beta_{j}-\tau\left(
i\right)  +j}\\\text{(since }\gamma_{i}-\beta_{j}-i+j=\alpha_{\tau\left(
i\right)  }-\beta_{j}-\tau\left(  i\right)  +j\text{)}}}\right)
_{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\alpha_{\tau\left(
i\right)  }-\beta_{j}-\tau\left(  i\right)  +j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\alpha_{i}%
-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(by (\ref{sol.bernstein.c.det.perm-rows}), applied to $A=\Lambda$, $N=n$,
$\beta_{i,j}=h_{\alpha_{i}-\beta_{j}-i+j}$ and $\sigma=\tau$). Multiplying
both sides of this equality by $\left(  -1\right)  ^{\tau}$, we obtain%
\[
\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}-\beta_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =\underbrace{\left(  -1\right)
^{\tau}\left(  -1\right)  ^{\tau}}_{=\left(  \left(  -1\right)  ^{\tau
}\right)  ^{2}=1}\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\alpha_{i}-\beta_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  .
\]
This proves (\ref{sol.bernstein.c.5.permuted-}).}. It is now easy to see that
if $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \notin \NN ^{n}%
$, then (\ref{eq.exe.bernstein.c.3}) holds\ \ \ \ \footnote{\textit{Proof.}
Assume that $\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)
\notin \NN ^{n}$. Then, there exists at least one $p\in\left\{
1,2,...,n\right\}  $ which does not satisfy $\gamma_{p}\in \NN$ (because
otherwise, every $p\in\left\{  1,2,...,n\right\}  $ would satisfy $\gamma
_{p}\in \NN$, and therefore we would have $\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n}\right)  \in \NN ^{n}$, which would contradict
$\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \notin \NN ^{n}$).
Consider this $p$. We know that $p$ satisfies $\gamma_{p}\in \ZZ$ (since
$\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \in \ZZ ^{n}$) but
not $\gamma_{p}\in \NN$. Hence, $\gamma_{p}\in \ZZ \setminus
 \NN =\left\{  -1,-2,-3,...\right\}  $, so that $\gamma_{p}<0$.
\par
Now, $p\in\left\{  1,2,...,n\right\}  $ yields $1\leq p\leq n$, so that
$n\geq1$ and thus $n\in\left\{  1,2,...,n\right\}  $. Hence, $\tau\left(
n\right)  $ and $\gamma_{n}$ are well-defined. Now, we have $\gamma_{1}%
\geq\gamma_{2}\geq...\geq\gamma_{n}$. Thus, any two elements $u$ and $v$ of
$\left\{  1,2,...,n\right\}  $ satisfying $u\leq v$ satisfy $\gamma_{u}%
\geq\gamma_{v}$. Applying this to $u=p$ and $v=n$, we obtain $\gamma_{p}%
\geq\gamma_{n}$. Thus, $\gamma_{n}\leq\gamma_{p}<0$.
\par
Now, for every $q\in\left\{  1,2,...,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }q\text{-th entry of the }n\text{-th row of the matrix
}\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right) \\
&  =h_{\gamma_{n}-n+q}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\underbrace{\gamma_{n}}_{<0}-n+\underbrace{q}_{\substack{\leq n\\\text{(since
}q\in\left\{  1,2,...,n\right\}  \text{)}}}<0-n+n=0\right) \\
&  =\left(  \text{the }p\text{-th entry of the row vector }0\text{ of length
}n\right)  .
\end{align*}
Thus,%
\[
\left(  \text{the }n\text{-th row of the matrix }\left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =\left(  \text{the row vector
}0\text{ of length }n\right)  .
\]
In other words, the $n$-th row of the matrix $\left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}$ is the zero vector. Hence, the matrix
$\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}$ has a row which is
the zero vector (namely, its $n$-th row). As a consequence, this matrix
$\left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}$ must have zero
determinant (because it is well-known that if a square matrix has a row which
is the zero vector, then this square matrix must have zero determinant). In
other words, $\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =0$. Now, (\ref{sol.bernstein.c.5.permuted}) yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\gamma_{i}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=0}=0.
\]
Thus,%
\begin{equation}
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\underbrace{\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }}_{=0}=\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{n}\right)  }^{\perp}0=0 \label{sol.bernstein.c.3.case3.1}%
\end{equation}
(since the map $\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta
_{n}\right)  }^{\perp}$ is $\kk$-linear).
\par
On the other hand, for every $q\in\left\{  1,2,...,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }q\text{-th entry of the }n\text{-th row of the matrix
}\left(  h_{\gamma_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right) \\
&  =h_{\gamma_{n}-\beta_{q}-n+q}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\underbrace{\gamma_{n}}_{<0}-\underbrace{\beta_{q}}_{\substack{\geq
0\\\text{(since }\beta_{q}\in \NN \\\text{(since }\left(  \beta_{1}%
,\beta_{2},...,\beta_{q}\right)  \\\text{is a partition))}}}-n+\underbrace{q}%
_{\substack{\leq n\\\text{(since }q\in\left\{  1,2,...,n\right\}  \text{)}%
}}<0-0-n+n=0\right) \\
&  =\left(  \text{the }p\text{-th entry of the row vector }0\text{ of length
}n\right)  .
\end{align*}
Thus,%
\[
\left(  \text{the }n\text{-th row of the matrix }\left(  h_{\gamma_{i}%
-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =\left(  \text{the row
vector }0\text{ of length }n\right)  .
\]
In other words, the $n$-th row of the matrix $\left(  h_{\gamma_{i}-\beta
_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}$ is the zero vector. Hence, the matrix
$\left(  h_{\gamma_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}$ has a row
which is the zero vector (namely, its $n$-th row). As a consequence, this
matrix $\left(  h_{\gamma_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}$ must
have zero determinant (because it is well-known that if a square matrix has a
row which is the zero vector, then this square matrix must have zero
determinant). In other words, $\det\left(  \left(  h_{\gamma_{i}-\beta
_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =0$. Now,
(\ref{sol.bernstein.c.5.permuted-}) becomes%
\[
\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(
h_{\gamma_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=0}=0.
\]
Compared with (\ref{sol.bernstein.c.3.case3.1}), this yields%
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  .
\]
Thus, (\ref{eq.exe.bernstein.c.3}) holds, qed.}. Hence, for the rest of the
proof of (\ref{eq.exe.bernstein.c.3}), we can WLOG assume that we don't have
$\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  \notin \NN ^{n}$.
Assume this.

We assumed that we don't have $\left(  \gamma_{1},\gamma_{2},...,\gamma
_{n}\right)  \notin \NN ^{n}$. Thus, we do have $\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n}\right)  \in \NN ^{n}$. Combined with
$\gamma_{1}\geq\gamma_{2}\geq...\geq\gamma_{n}$, this yields that $\left(
\gamma_{1},\gamma_{2},...,\gamma_{n}\right)  $ is a partition having at most
$n$ parts. Denote this partition $\left(  \gamma_{1},\gamma_{2},...,\gamma
_{n}\right)  $ as $\kappa$. Write $\kappa$ in the form $\left(  \kappa
_{1},\kappa_{2},\kappa_{3},...\right)  $. Then, $\kappa=\left(  \kappa
_{1},\kappa_{2},...,\kappa_{n}\right)  $ (since $\kappa=\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n}\right)  $ is a partition having at most $n$
parts). Thus, $\left(  \kappa_{1},\kappa_{2},...,\kappa_{n}\right)
=\kappa=\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  $. But
(\ref{eq.exe.bernstein.c.1}) (applied to $\kappa$ instead of $\lambda$) yields%
\begin{align*}
s_{\kappa}  &  =\overline{s}_{\left(  \kappa_{1},\kappa_{2},...,\kappa
_{n}\right)  }=\overline{s}_{\left(  \gamma_{1},\gamma_{2},...,\gamma
_{n}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \kappa
_{1},\kappa_{2},...,\kappa_{n}\right)  =\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n}\right)  \right) \\
&  =\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{s}_{\left(
\gamma_{1},\gamma_{2},...,\gamma_{n}\right)  }\right)  ,
\end{align*}
so that $\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =s_{\kappa}$. Now, (\ref{sol.bernstein.c.5.permuted}) becomes
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\left(  -1\right)  ^{\tau}\underbrace{\det\left(  \left(  h_{\gamma_{i}
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  }_{=s_{\kappa}}=\left(  -1\right)
^{\tau}s_{\kappa}.
\label{sol.bernstein.c.5.salpha-as-det}
\end{equation}


On the other hand, $\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  $ is a
partition having at most $n$ parts. Denote this partition $\left(  \beta
_{1},\beta_{2},...,\beta_{n}\right)  $ as $\lambda$. Write $\lambda$ in the
form $\left(  \lambda_{1},\lambda_{2},\lambda_{3},...\right)  $. Then,
$\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  $ (since
$\lambda=\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  $ is a partition
having at most $n$ parts). Thus, $\left(  \lambda_{1},\lambda_{2}
,...,\lambda_{n}\right)  =\lambda=\left(  \beta_{1},\beta_{2},...,\beta
_{n}\right)  $. But (\ref{eq.exe.bernstein.c.1}) yields
\[
s_{\lambda}=\overline{s}_{\left(  \lambda_{1},\lambda_{2},...,\lambda
_{n}\right)  }=\overline{s}_{\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \lambda_{1},\lambda
_{2},...,\lambda_{n}\right)  =\left(  \beta_{1},\beta_{2},...,\beta
_{n}\right)  \right)  .
\]


But $\ell\left(  \lambda\right)  \leq n$\ \ \ \ \footnote{\textit{Proof.} We
know that $\lambda=\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  $ is a
partition having at most $n$ parts. In other words, the number of parts of
$\lambda$ is at most $n$. Since the number of parts of $\lambda$ is
$\ell\left(  \lambda\right)  $, this rewrites as follows: $\ell\left(
\lambda\right)  $ is at most $n$. In other words, $\ell\left(  \lambda\right)
\leq n$, qed.} and $\ell\left(  \kappa\right)  \leq n$%
\ \ \ \ \footnote{\textit{Proof.} We know that $\kappa=\left(  \gamma
_{1},\gamma_{2},...,\gamma_{n}\right)  $ is a partition having at most $n$
parts. In other words, the number of parts of $\kappa$ is at most $n$. Since
the number of parts of $\kappa$ is $\ell\left(  \kappa\right)  $, this
rewrites as follows: $\ell\left(  \kappa\right)  $ is at most $n$. In other
words, $\ell\left(  \kappa\right)  \leq n$, qed.}. Therefore,
(\ref{eq.jacobi-trudi.h}) (applied to $\kappa$, $\lambda$ and $n$ instead of
$\lambda$, $\mu$ and $\ell$) yields%
\[
s_{\kappa/\lambda}=\det\left(  \left(  h_{\kappa_{i}-\lambda_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  .
\]
Since every $i\in\left\{  1,2,...,n\right\}  $ satisfies $\kappa_{i}
=\gamma_{i}$ (because $\left(  \kappa_{1},\kappa_{2},...,\kappa_{n}\right)
=\left(  \gamma_{1},\gamma_{2},...,\gamma_{n}\right)  $), this rewrites as
\[
s_{\kappa/\lambda}=\det\left(  \left(  h_{\gamma_{i}-\lambda_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  .
\]
Since every $j\in\left\{  1,2,...,n\right\}  $ satisfies $\lambda_{j}
=\beta_{j}$ (because $\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)
=\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  $), this rewrites as
\begin{equation}
s_{\kappa/\lambda}=\det\left(  \left(  h_{\gamma_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  .
\label{sol.bernstein.c.5.almost}
\end{equation}


But (\ref{sol.bernstein.skewing.general}) (applied to $\kappa$ and $\lambda$
instead of $\lambda$ and $\mu$) yields $s_{\lambda}^{\perp}s_{\kappa
}=s_{\kappa/\lambda}$. Thus,
\[
s_{\kappa/\lambda}=\underbrace{s_{\lambda}^{\perp}}_{\substack{=
\overline{s}_{\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  }^{\perp}
\\\text{(since
}s_{\lambda}=\overline{s}_{\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)
}\text{)}}}s_{\kappa}=\overline{s}_{\left(  \beta_{1},\beta_{2},...,\beta
_{n}\right)  }^{\perp}s_{\kappa}.
\]
Compared with (\ref{sol.bernstein.c.5.almost}), this yields
\begin{equation}
\overline{s}_{\left(  \beta_{1},\beta_{2},...,\beta_{n}\right)  }^{\perp
}s_{\kappa}=\det\left(  \left(  h_{\gamma_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  .
\label{sol.bernstein.c.3.kappa-skewed-by-beta}
\end{equation}
Now,
\begin{align*}
&  \overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }%
^{\perp}\underbrace{\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{n}\right)  }}_{\substack{=\left(  -1\right)  ^{\tau}s_{\kappa
}\\\text{(by (\ref{sol.bernstein.c.5.salpha-as-det}))}}}\\
&  =\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}^{\perp}\left(  \left(  -1\right)  ^{\tau}s_{\kappa}\right)  =\left(
-1\right)  ^{\tau}\underbrace{\overline{s}_{\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{n}\right)  }^{\perp}s_{\kappa}}_{\substack{=\det\left(  \left(
h_{\gamma_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  \\\text{(by
(\ref{sol.bernstein.c.3.kappa-skewed-by-beta}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
the map }\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}^{\perp}\text{ is } \kk \text{-linear}\right) \\
&  =\left(  -1\right)  ^{\tau}\det\left(  \left(  h_{\gamma_{i}-\beta_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\alpha
_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.bernstein.c.5.permuted-}%
)}\right)  .
\end{align*}
In other words, (\ref{eq.exe.bernstein.c.3}) holds. Thus, Assertion C.5 is proven.

Now, we can finally come to the proof of (\ref{eq.exe.bernstein.c.3}) in full generality:

\textit{Proof of (\ref{eq.exe.bernstein.c.3}):} Let $\left(  \alpha_{1}
,\alpha_{2},\ldots,\alpha_{n}\right)  \in{ \ZZ }^{n}$ and $\left(
\beta_{1},\beta_{2},\ldots,\beta_{n}\right)  \in{ \NN }^{n}$ be two
$n$-tuples. We need to prove that (\ref{eq.exe.bernstein.c.3}) holds.

It is easy to see that if the $n$ integers $\beta_{1}-1$, $\beta_{2}-2$,
$...$, $\beta_{n}-n$ are not distinct, then
(\ref{eq.exe.bernstein.c.3}) holds\footnote{\textit{Proof.} Assume that the
$n$ integers $\beta_{1}-1$, $\beta_{2}-2$, $...$, $\beta_{n}-n$ are not
distinct.
\par
Define a function $f: \ZZ \times \ZZ \rightarrow \ZZ$ by
\[
\left(  f\left(  u,v\right)  =u-v\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
u,v\right)  \in \ZZ \times \ZZ \right)  .
\]
Then, Assertion C.3 (applied to $\mathfrak{a}_{i}=\beta_{i}-i$ and
$\mathfrak{b}_{j}=-j$) yields $\det\left(  \left(  h_{f\left(  \beta
_{i}-i,-j\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$ (since the $n$
integers $\beta_{1}-1$, $\beta_{2}-2$, $...$, $\beta_{n}-n$ are not
distinct). Since every $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}
^{2}$ satisfy%
\begin{align*}
f\left(  \beta_{i}-i,-j\right)   &  =\beta_{i}-i-\left(  -j\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f\right) \\
&  =\beta_{i}-i+j,
\end{align*}
this rewrites as $\det\left(  \left(  h_{\beta_{i}-i+j}\right)
_{i,j=1,2,...,n}\right)  =0$. But the definition of $\overline{s}_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }$ yields $\overline{s}_{\left(
\beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }=\det\left(  \left(
h_{\beta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =0$. Hence,
$\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}=0^{\perp}=0$. Now,%
\begin{equation}
\underbrace{\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}^{\perp}}_{=0^{\perp}}\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  }=0^{\perp}\overline{s}_{\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{n}\right)  }=0. \label{sol.bernstein.c.3.case2.1}%
\end{equation}
\par
But Assertion C.4 (applied to $\mathfrak{a}_{i}=\alpha_{i}-i$ and
$\mathfrak{b}_{j}=\beta_{j}-j$) yields $\det\left(  \left(  h_{f\left(
\alpha_{i}-i,\beta_{j}-j\right)  }\right)  _{i,j=1,2,...,n}\right)  =0$ (since
the $n$ integers $\beta_{1}-1$, $\beta_{2}-2$, $...$, $\beta_{n}-n$ are not
distinct). Since every $\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}$ satisfy%
\begin{align*}
f\left(  \alpha_{i}-i,\beta_{j}-j\right)   &  =\left(  \alpha_{i}-i\right)
-\left(  \beta_{j}-j\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }f\right) \\
&  =\alpha_{i}-\beta_{j}-i+j,
\end{align*}
this rewrites as $\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,...,n}\right)  =0$. Compared with (\ref{sol.bernstein.c.3.case2.1}),
this yields%
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,...,n}%
\right)  .
\]
Thus, (\ref{eq.exe.bernstein.c.3}) holds, qed.}. Thus, for the rest of the
proof of (\ref{eq.exe.bernstein.c.3}), we can WLOG assume that it is
\textbf{not} true that the $n$ integers $\beta_{1}-1$, $\beta_{2}-2$, $...$,
$\beta_{n}-n$ are not distinct. Assume this.

We assumed that it is \textbf{not} true that the $n$ integers $\beta_{1}-1$,
$\beta_{2}-2$, $...$, $\beta_{n}-n$ are not distinct. In other words,
the $n$ integers $\beta_{1}-1$, $\beta_{2}-2$, $...$, $\beta_{n}-n$ are
distinct.

For every $n$-tuple $\left(  \eta_{1},\eta_{2},...,\eta_{n}\right)  $ of
integers, there exists a permutation $\zeta\in\Symm_n$ satisfying
$\eta_{\zeta\left(  1\right)  }\geq\eta_{\zeta\left(  2\right)  }\geq
...\geq\eta_{\zeta\left(  n\right)  }$\ \ \ \ \footnote{This is just a
restatement of the known fact that any $n$-tuple of integers can be sorted in
decreasing order.}. Applied to $\eta_{j}=\beta_{j}-j$, this yields the
following: There exists a permutation $\zeta\in\Symm_n$ satisfying
$\beta_{\zeta\left(  1\right)  }-\zeta\left(  1\right)  \geq\beta
_{\zeta\left(  2\right)  }-\zeta\left(  2\right)  \geq...\geq\beta
_{\zeta\left(  n\right)  }-\zeta\left(  n\right)  $. Consider this permutation
$\zeta$.

Define an $n$-tuple $\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)
\in \ZZ ^{n}$ by%
\[
\left(  \delta_{p}=\beta_{\zeta\left(  p\right)  }-\zeta\left(  p\right)
+p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\left\{  1,2,...,n\right\}
\right)  .
\]
Then, $\delta_{1}\geq\delta_{2}\geq...\geq\delta_{n}$%
\ \ \ \ \footnote{\textit{Proof.} This follows from Assertion C.2, applied to
$\beta_{j}$, $\zeta$ and $\delta_{j}$ instead of $\alpha_{j}$, $\tau$ and
$\gamma_{j}$ (because every $p\in\left\{  1,2,...,n\right\}  $ satisfies
$\delta_{p}=\beta_{\zeta\left(  p\right)  }-\zeta\left(  p\right)  +p$).}.
Also, $\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  \in \NN ^{n}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,...,n\right\}  $. We
are going to prove that $\delta_{j}\in \NN$.
\par
We have $1\leq j\leq n$ (since $j\in\left\{  1,2,...,n\right\}  $). Thus,
$n\geq1$, so that $n\in\left\{  1,2,...,n\right\}  $. Hence, $\delta_{n}$ and
$\zeta\left(  n\right)  $ are well-defined.
\par
But we have $\delta_{1}\geq\delta_{2}\geq...\geq\delta_{n}$. Hence, any two
elements $u$ and $v$ of $\left\{  1,2,...,n\right\}  $ satisfying $u\leq v$
satisfy $\delta_{u}\geq\delta_{v}$. Applying this to $u=j$ and $v=n$, we
obtain $\delta_{j}\geq\delta_{n}$. But the definition of $\delta_{n}$ yields
$\delta_{n}=\beta_{\zeta\left(  n\right)  }-\zeta\left(  n\right)  +n$. Since
$\zeta\left(  n\right)  \in\left\{  1,2,...,n\right\}  $ (because $\zeta
\in\Symm_n$), we have $\zeta\left(  n\right)  \leq n$, so that
$\delta_{n}=\beta_{\zeta\left(  n\right)  }-\underbrace{\zeta\left(  n\right)
}_{\leq n}+n\geq\beta_{\zeta\left(  n\right)  }-n+n=\beta_{\zeta\left(
n\right)  }$.
\par
But $\beta_{\zeta\left(  n\right)  }\in \NN$ (since $\left(  \beta
_{1},\beta_{2},\ldots,\beta_{n}\right)  \in{ \NN }^{n}$), and thus
$\beta_{\zeta\left(  n\right)  }\geq0$. Hence, $\delta_{j}\geq\delta_{n}%
\geq\beta_{\zeta\left(  n\right)  }\geq0$, so that $\delta_{j}\in \NN$.
\par
Now, forget that we fixed $j$. We thus have proven that $\delta_{j}%
\in \NN$ for every $j\in\left\{  1,2,...,n\right\}  $. In other words,
$\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  \in \NN ^{n}$,
qed.}. Combined with $\delta_{1}\geq\delta_{2}\geq...\geq\delta_{n}$, this
yields that $\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  $ is a
partition having at most $n$ parts. Hence, Assertion C.5 (applied to $\left(
\delta_{1},\delta_{2},...,\delta_{n}\right)  $ instead of $\left(  \beta
_{1},\beta_{2},\ldots,\beta_{n}\right)  $) yields that
(\ref{eq.exe.bernstein.c.3}) holds with $\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{n}\right)  $ replaced by $\left(  \delta_{1},\delta
_{2},...,\delta_{n}\right)  $. In other words, we have
\begin{equation}
\overline{s}_{\left(  \delta_{1},\delta_{2},\ldots,\delta_{n}\right)  }%
^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\det\left(  \left(  h_{\alpha_{i}-\delta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  .
\label{sol.bernstein.c.3.alpha-skewed-by-delta}
\end{equation}


We have
\begin{equation}
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }=\left(
-1\right)  ^{\zeta}\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta
_{n}\right)  }.
\label{sol.bernstein.c.3.permuted}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.c.3.permuted}):} The definition
of $\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }$
yields $\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}=\det\left(  \left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  $.
\par
The sign of every permutation belongs to $\left\{  1,-1\right\}  $. Hence,
$\left(  -1\right)  ^{\zeta}\in\left\{  1,-1\right\}  $. Consequently,
$\left(  \left(  -1\right)  ^{\zeta}\right)  ^{2}=1$.
\par
Every $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}$ satisfies
\[
\underbrace{\delta_{i}}_{\substack{=\beta_{\zeta\left(  i\right)  }
-\zeta\left(  i\right)  +i\\\text{(by the definition of }\delta_{i}\text{)}%
}}-i+j=\beta_{\zeta\left(  i\right)  }-\zeta\left(  i\right)  +i-i+j=\beta
_{\zeta\left(  i\right)  }-\zeta\left(  i\right)  +j.
\]
\par
The definition of $\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta
_{n}\right)  }$ yields
\[
\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)  }
=\det\left(  \left(  \underbrace{h_{\delta_{i}-i+j}}_{\substack{=h_{\beta
_{\zeta\left(  i\right)  }-\zeta\left(  i\right)  +j}\\\text{(since }%
\delta_{i}-i+j=\beta_{\zeta\left(  i\right)  }-\zeta\left(  i\right)
+j\text{)}}}\right)  _{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(
h_{\beta_{\zeta\left(  i\right)  }-\zeta\left(  i\right)  +j}\right)
_{i,j=1,2,\ldots,n}\right)  =\left(  -1\right)  ^{\zeta}\det\left(  \left(
h_{\beta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(by (\ref{sol.bernstein.c.det.perm-rows}), applied to $A=\Lambda$, $N=n$,
$\beta_{i,j}=h_{\beta_{i}-i+j}$ and $\sigma=\zeta$). Multiplying both sides of
this equality by $\left(  -1\right)  ^{\zeta}$, we obtain
\[
\left(  -1\right)  ^{\zeta}\overline{s}_{\left(  \delta_{1},\delta
_{2},...,\delta_{n}\right)  }=\underbrace{\left(  -1\right)  ^{\zeta}\left(
-1\right)  ^{\zeta}}_{=\left(  \left(  -1\right)  ^{\zeta}\right)  ^{2}=1}
\det\left(  \left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
=\det\left(  \left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
=\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }
\]
(since $\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}=\det\left(  \left(  h_{\beta_{i}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
$). This proves (\ref{sol.bernstein.c.3.permuted}).} Thus,%
\begin{equation}
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}=\left(  \left(  -1\right)  ^{\zeta}\overline{s}_{\left(  \delta_{1}%
,\delta_{2},...,\delta_{n}\right)  }\right)  ^{\perp}=\left(  -1\right)
^{\zeta}\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta_{n}\right)
}^{\perp}
\label{sol.bernstein.c.3.permuted.perp}
\end{equation}
(since the map $\Lambda\rightarrow\operatorname*{End}\Lambda$ sending every
$f\in\Lambda$ to $f^{\perp}$ is $\kk$-linear). Thus,
\begin{align}
\underbrace{\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)
}^{\perp}}_{=\left(  -1\right)  ^{\zeta}\overline{s}_{\left(  \delta
_{1},\delta_{2},...,\delta_{n}\right)  }^{\perp}}\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }  &  =\left(  -1\right)
^{\zeta}\underbrace{\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta
_{n}\right)  }^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2}
,\ldots,\alpha_{n}\right)  }}_{\substack{=\det\left(  \left(  h_{\alpha
_{i}-\delta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  \\\text{(by
(\ref{sol.bernstein.c.3.alpha-skewed-by-delta}))}}}\nonumber\\
&  =\left(  -1\right)  ^{\zeta}\det\left(  \left(  h_{\alpha_{i}-\delta
_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  .
\label{sol.bernstein.c.3.alpha-skewed-by-beta}
\end{align}


Furthermore,
\begin{equation}
\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\zeta}\det\left(  \left(  h_{\alpha
_{i}-\delta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\label{sol.bernstein.c.3.permuted-}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.c.3.permuted-}):} The sign of
every permutation belongs to $\left\{  1,-1\right\}  $. Hence, $\left(
-1\right)  ^{\zeta}\in\left\{  1,-1\right\}  $. Consequently, $\left(  \left(
-1\right)  ^{\zeta}\right)  ^{2}=1$.
\par
Every $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}$ satisfies%
\[
\alpha_{i}-\underbrace{\delta_{j}}_{\substack{=\beta_{\zeta\left(  j\right)
}-\zeta\left(  j\right)  +j\\\text{(by the definition of }\delta_{j}\text{)}%
}}-i+j=\alpha_{i}-\left(  \beta_{\zeta\left(  j\right)  }-\zeta\left(
j\right)  +j\right)  -i+j=\alpha_{i}-\beta_{\zeta\left(  j\right)  }%
-i+\zeta\left(  j\right)  .
\]
\par
The definition of $\overline{s}_{\left(  \delta_{1},\delta_{2},...,\delta
_{n}\right)  }$ yields%
\[
\det\left(  \left(  \underbrace{h_{\alpha_{i}-\delta_{j}-i+j}}%
_{\substack{=h_{\alpha_{i}-\beta_{\zeta\left(  j\right)  }-i+\zeta\left(
j\right)  }\\\text{(since }\alpha_{i}-\delta_{j}-i+j\\=\alpha_{i}-\beta
_{\zeta\left(  j\right)  }-i+\zeta\left(  j\right)  \text{)}}}\right)
_{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\alpha_{i}-\beta
_{\zeta\left(  j\right)  }-i+\zeta\left(  j\right)  }\right)  _{i,j=1,2,\ldots
,n}\right)  =\left(  -1\right)  ^{\zeta}\det\left(  \left(  h_{\alpha
_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\]
(by (\ref{sol.bernstein.c.det.perm-cols}), applied to $A=\Lambda$, $N=n$,
$\beta_{i,j}=h_{\alpha_{i}-\beta_{j}-i+j}$ and $\sigma=\zeta$). Multiplying
both sides of this equality by $\left(  -1\right)  ^{\zeta}$, we obtain%
\[
\left(  -1\right)  ^{\zeta}\det\left(  \left(  h_{\alpha_{i}-\delta_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  =\underbrace{\left(  -1\right)
^{\zeta}\left(  -1\right)  ^{\zeta}}_{=\left(  \left(  -1\right)  ^{\zeta
}\right)  ^{2}=1}\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\alpha_{i}-\beta_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)  .
\]
This proves (\ref{sol.bernstein.c.3.permuted-}).}. Comparing this with
(\ref{sol.bernstein.c.3.alpha-skewed-by-beta}), we obtain%
\[
\overline{s}_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{n}\right)  }^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-\beta_{j}-i+j}\right)  _{i,j=1,2,\ldots
,n}\right)  .
\]
Thus, (\ref{eq.exe.bernstein.c.3}) is proven. This completes the proof of
Exercise \ref{exe.bernstein}(c).

(d) In the following, we use the so-called \dfn{Iverson bracket notation}:
For every assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $
denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
(This integer is called the \dfn{truth value} of $\mathcal{A}$.)

Before we come to the solution of Exercise \ref{exe.bernstein}(d), let us
recall some general facts about determinants:

\begin{itemize}
\item Every commutative ring $A$, every $N\in{ \NN }$, every matrix
$\mathbf{M}\in A^{N\times N}$ and every $i\in\left\{  1,2,...,N\right\}  $
satisfy
\begin{equation}
\det\mathbf{M}=\sum_{k=1}^{N}\left(  \text{the }\left(  i,k\right)  \text{-th
entry of the matrix }\mathbf{M}\right)  \cdot\left(  \text{the }\left(
i,k\right)  \text{-th cofactor of the matrix }\mathbf{M}\right)  .
\label{sol.bernstein.d.laplace}%
\end{equation}
(This is the well-known Laplace expansion for the determinant of $\mathbf{M}$
in its $i$-th row.)

\item Every commutative ring $A$, every $N\in{ \NN }$, every matrix
$\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$, every
$p\in\left\{  1,2,...,N\right\}  $ and every $q\in\left\{  1,2,...,N\right\}
$ satisfy
\begin{align}
&  \left(  \text{the }\left(  p,q\right)  \text{-th cofactor of the matrix
}\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \nonumber\\
&  =\left(  -1\right)  ^{p+q}\det\left(  \left(  \beta_{i+\left[  i\geq
p\right]  ,j+\left[  j\geq q\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\label{sol.bernstein.d.cofactor.def}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.bernstein.d.cofactor.def}):} Let $A$ be a
commutative ring. Let $N\in \NN$. Let $\left(  \beta_{i,j}\right)
_{i,j=1,2,...,N}\in A^{N\times N}$ be a matrix. Let $p\in\left\{
1,2,...,N\right\}  $ and every $q\in\left\{  1,2,...,N\right\}  $. Denote the
matrix $\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}$ by $B$. Obviously, every
$u\in\left\{  1,2,...,N\right\}  $ and $v\in\left\{  1,2,...,N\right\}  $
satisfy%
\begin{align}
\left(  \text{the }\left(  u,v\right)  \text{-th entry of the matrix
}\underbrace{B}_{=\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}}\right)   &
=\left(  \text{the }\left(  u,v\right)  \text{-th entry of the matrix }\left(
\beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \nonumber\\
&  =\beta_{u,v}. \label{sol.bernstein.d.cofactor.def.pf.Bentry}%
\end{align}
\par
The $\left(  p,q\right)  $-th cofactor of the matrix $B$ is $\left(
-1\right)  ^{p+q}$ times the determinant of the matrix obtained by removing
row $p$ and column $q$ from the matrix $B$ (because this is how this $\left(
p,q\right)  $-th cofactor is defined). In other words,
\begin{align}
&  \left(  \text{the }\left(  p,q\right)  \text{-th cofactor of the matrix
}B\right) \nonumber\\
&  =\left(  -1\right)  ^{p+q}\det\left(  \text{the matrix obtained by removing
row }p\text{ and column }q\text{ from the matrix }B\right)  .
\label{sol.bernstein.d.cofactor.def.pf.1}%
\end{align}
But let $C$ denote the matrix obtained by removing column $q$ from the matrix
$B$. Thus,%
\begin{align}
C  &  =\left(  \text{the matrix obtained by removing column }q\text{ from the
matrix }B\right) \label{sol.bernstein.d.cofactor.def.pf.2}\\
&  =\left(  \text{the }N\times\left(  N-1\right)  \text{-matrix whose columns
are the columns }1\text{, }2\text{, }...\text{, }q-1\text{, }q+1\text{,
}q+2\text{, }...\text{, }N\text{ of the matrix }B\right)  .
\label{sol.bernstein.d.cofactor.def.pf.3}%
\end{align}
\par
Define an $\left(  N-1\right)  $-tuple $\left(  \mathbf{c}_{1},\mathbf{c}%
_{2},...,\mathbf{c}_{N-1}\right)  \in \ZZ ^{N-1}$ by%
\[
\left(  \mathbf{c}_{j}=j+\left[  j\geq q\right]  \ \ \ \ \ \ \ \ \ \ \text{for
every }j\in\left\{  1,2,...,N-1\right\}  \right)  .
\]
\par
Then, every $j\in\left\{  1,2,...,q-1\right\}  $ satisfies%
\[
\mathbf{c}_{j}=j+\underbrace{\left[  j\geq q\right]  }%
_{\substack{=0\\\text{(since }j\geq q\text{ is false}\\\text{(since }j<q\text{
(since}\\j\in\left\{  1,2,...,q-1\right\}  \text{)))}}}=j.
\]
Hence, $\left(  \mathbf{c}_{1},\mathbf{c}_{2},...,\mathbf{c}_{q-1}\right)
=\left(  1,2,...,q-1\right)  $. Also, every $j\in\left\{
q,q+1,...,N-1\right\}  $ satisfies%
\[
\mathbf{c}_{j}=j+\underbrace{\left[  j\geq q\right]  }%
_{\substack{=1\\\text{(since }j\geq q\text{ is true}\\\text{(since}%
\\j\in\left\{  q,q+1,...,N-1\right\}  \text{))}}}=j+1.
\]
Hence, $\left(  \mathbf{c}_{q},\mathbf{c}_{q+1},...,\mathbf{c}_{N-1}\right)
=\left(  q+1,\left(  q+1\right)  +1,...,\left(  N-1\right)  +1\right)
=\left(  q+1,q+2,...,N\right)  $. Now,%
\begin{align*}
\left(  \mathbf{c}_{1},\mathbf{c}_{2},...,\mathbf{c}_{N-1}\right)   &
=\left(  \text{the concatenation of the two tuples }\underbrace{\left(
\mathbf{c}_{1},\mathbf{c}_{2},...,\mathbf{c}_{q-1}\right)  }_{=\left(
1,2,...,q-1\right)  }\text{ and }\underbrace{\left(  \mathbf{c}_{q}%
,\mathbf{c}_{q+1},...,\mathbf{c}_{N-1}\right)  }_{=\left(
q+1,q+2,...,N\right)  }\right) \\
&  =\left(  \text{the concatenation of the two tuples }\left(
1,2,...,q-1\right)  \text{ and }\left(  q+1,q+2,...,N\right)  \right) \\
&  =\left(  1,2,...,q-1,q+1,q+2,...,N\right)  .
\end{align*}
Thus, $\left(  1,2,...,q-1,q+1,q+2,...,N\right)  =\left(  \mathbf{c}%
_{1},\mathbf{c}_{2},...,\mathbf{c}_{N-1}\right)  $. Therefore,%
\begin{align}
&  \left(  \text{the columns }\underbrace{1\text{, }2\text{, }...\text{,
}q-1\text{, }q+1\text{, }q+2\text{, }...\text{, }N}_{=\left(  \mathbf{c}%
_{1},\mathbf{c}_{2},...,\mathbf{c}_{N-1}\right)  }\text{ of the matrix
}B\right) \nonumber\\
&  =\left(  \text{the columns }\mathbf{c}_{1}\text{, }\mathbf{c}_{2}\text{,
}...\text{, }\mathbf{c}_{N-1}\text{ of the matrix }B\right)  .
\label{sol.bernstein.d.cofactor.def.pf.4}%
\end{align}
Now, (\ref{sol.bernstein.d.cofactor.def.pf.3}) becomes%
\begin{align*}
C  &  =\left(  \text{the }N\times\left(  N-1\right)  \text{-matrix whose
columns are }\underbrace{\text{the columns }1\text{, }2\text{, }...\text{,
}q-1\text{, }q+1\text{, }q+2\text{, }...\text{, }N\text{ of the matrix }%
B}_{\substack{=\left(  \text{the columns }\mathbf{c}_{1}\text{, }%
\mathbf{c}_{2}\text{, }...\text{, }\mathbf{c}_{N-1}\text{ of the matrix
}B\right)  \\\text{(by (\ref{sol.bernstein.d.cofactor.def.pf.4}))}}}\right) \\
&  =\left(  \text{the }N\times\left(  N-1\right)  \text{-matrix whose columns
are the columns }\mathbf{c}_{1}\text{, }\mathbf{c}_{2}\text{, }...\text{,
}\mathbf{c}_{N-1}\text{ of the matrix }B\right)  .
\end{align*}
Thus, the columns of the matrix $C$ are the columns $\mathbf{c}_{1}$,
$\mathbf{c}_{2}$, $...$, $\mathbf{c}_{N-1}$ of the matrix $B$. In other words,
for every $v\in\left\{  1,2,...,N-1\right\}  $, we have%
\begin{equation}
\left(  \text{the column }v\text{ of the matrix }C\right)  =\left(  \text{the
column }\mathbf{c}_{v}\text{ of the matrix }B\right)  .
\label{sol.bernstein.d.cofactor.def.pf.6}%
\end{equation}
Thus, every $\left(  u,v\right)  \in\left\{  1,2,...,N-1\right\}
\times\left\{  1,2,...,N\right\}  $ satisfies%
\begin{align}
&  \left(  \text{the }\left(  u,v\right)  \text{-th entry of the matrix
}C\right) \nonumber\\
&  =\left(  \text{the }u\text{-th entry of }\underbrace{\text{the column
}v\text{ of the matrix }C}_{\substack{=\left(  \text{the column }%
\mathbf{c}_{v}\text{ of the matrix }B\right)  \\\text{(by
(\ref{sol.bernstein.d.cofactor.def.pf.6}))}}}\right) \nonumber\\
&  =\left(  \text{the }u\text{-th entry of the column }\mathbf{c}_{v}\text{ of
the matrix }B\right) \nonumber\\
&  =\left(  \text{the }\left(  u,\mathbf{c}_{v}\right)  \text{-th entry of the
matrix }B\right)  =\beta_{u,\mathbf{c}_{v}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.bernstein.d.cofactor.def.pf.Bentry}), applied to
}\mathbf{c}_{v}\text{ instead of }v\right)  .
\label{sol.bernstein.d.cofactor.def.pf.Centry}%
\end{align}
\par
But now let, let $D$ denote the matrix obtained by removing row $p$ and column
$q$ from the matrix $B$. Then,
\begin{align}
D  &  =\left(  \text{the matrix obtained by removing row }p\text{ and column
}q\text{ from the matrix }B\right)  \label{sol.bernstein.d.cofactor.def.pf.12}%
\\
&  =\left(  \text{the matrix obtained by removing row }p\text{ from
}\underbrace{\text{the matrix obtained by removing column }q\text{ from the
matrix }B}_{\substack{=C\\\text{(by (\ref{sol.bernstein.d.cofactor.def.pf.2}%
))}}}\right) \nonumber\\
&  =\left(  \text{the matrix obtained by removing row }p\text{ from }C\right)
\nonumber\\
&  =\left(  \text{the }\left(  N-1\right)  \times\left(  N-1\right)
\text{-matrix whose rows are the rows }1\text{, }2\text{, }...\text{,
}p-1\text{, }p+1\text{, }p+2\text{, }...\text{, }N\text{ of the matrix
}C\right)  . \label{sol.bernstein.d.cofactor.def.pf.13}%
\end{align}
\par
Define an $\left(  N-1\right)  $-tuple $\left(  \mathbf{r}_{1},\mathbf{r}%
_{2},...,\mathbf{r}_{N-1}\right)  \in \ZZ ^{N-1}$ by%
\[
\left(  \mathbf{r}_{i}=i+\left[  i\geq p\right]  \ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  1,2,...,N-1\right\}  \right)  .
\]
\par
Then, every $i\in\left\{  1,2,...,p-1\right\}  $ satisfies%
\[
\mathbf{r}_{i}=i+\underbrace{\left[  i\geq p\right]  }%
_{\substack{=0\\\text{(since }i\geq p\text{ is false}\\\text{(since }i<p\text{
(since}\\i\in\left\{  1,2,...,p-1\right\}  \text{)))}}}=i.
\]
Hence, $\left(  \mathbf{r}_{1},\mathbf{r}_{2},...,\mathbf{r}_{p-1}\right)
=\left(  1,2,...,p-1\right)  $. Also, every $i\in\left\{
p,p+1,...,N-1\right\}  $ satisfies%
\[
\mathbf{r}_{i}=i+\underbrace{\left[  i\geq p\right]  }%
_{\substack{=1\\\text{(since }i\geq p\text{ is true}\\\text{(since}%
\\i\in\left\{  p,p+1,...,N-1\right\}  \text{))}}}=i+1.
\]
Hence, $\left(  \mathbf{r}_{p},\mathbf{r}_{p+1},...,\mathbf{r}_{N-1}\right)
=\left(  p+1,\left(  p+1\right)  +1,...,\left(  N-1\right)  +1\right)
=\left(  p+1,p+2,...,N\right)  $. Now,%
\begin{align*}
\left(  \mathbf{r}_{1},\mathbf{r}_{2},...,\mathbf{r}_{N-1}\right)   &
=\left(  \text{the concatenation of the two tuples }\underbrace{\left(
\mathbf{r}_{1},\mathbf{r}_{2},...,\mathbf{r}_{p-1}\right)  }_{=\left(
1,2,...,p-1\right)  }\text{ and }\underbrace{\left(  \mathbf{r}_{p}%
,\mathbf{r}_{p+1},...,\mathbf{r}_{N-1}\right)  }_{=\left(
p+1,p+2,...,N\right)  }\right) \\
&  =\left(  \text{the concatenation of the two tuples }\left(
1,2,...,p-1\right)  \text{ and }\left(  p+1,p+2,...,N\right)  \right) \\
&  =\left(  1,2,...,p-1,p+1,p+2,...,N\right)  .
\end{align*}
Thus, $\left(  1,2,...,p-1,p+1,p+2,...,N\right)  =\left(  \mathbf{r}%
_{1},\mathbf{r}_{2},...,\mathbf{r}_{N-1}\right)  $. Therefore,%
\begin{align}
&  \left(  \text{the rows }\underbrace{1\text{, }2\text{, }...\text{,
}p-1\text{, }p+1\text{, }p+2\text{, }...\text{, }N}_{=\left(  \mathbf{r}%
_{1},\mathbf{r}_{2},...,\mathbf{r}_{N-1}\right)  }\text{ of the matrix
}C\right) \nonumber\\
&  =\left(  \text{the rows }\mathbf{r}_{1}\text{, }\mathbf{r}_{2}\text{,
}...\text{, }\mathbf{r}_{N-1}\text{ of the matrix }C\right)  .
\label{sol.bernstein.d.cofactor.def.pf.14}%
\end{align}
Now, (\ref{sol.bernstein.d.cofactor.def.pf.13}) becomes%
\begin{align*}
D  &  =\left(  \text{the }\left(  N-1\right)  \times\left(  N-1\right)
\text{-matrix whose rows are }\underbrace{\text{the rows }1\text{, }2\text{,
}...\text{, }p-1\text{, }p+1\text{, }p+2\text{, }...\text{, }N\text{ of the
matrix }C}_{\substack{=\left(  \text{the rows }\mathbf{r}_{1}\text{,
}\mathbf{r}_{2}\text{, }...\text{, }\mathbf{r}_{N-1}\text{ of the matrix
}C\right)  \\\text{(by (\ref{sol.bernstein.d.cofactor.def.pf.14}))}}}\right)
\\
&  =\left(  \text{the }\left(  N-1\right)  \times\left(  N-1\right)
\text{-matrix whose rows are the rows }\mathbf{r}_{1}\text{, }\mathbf{r}%
_{2}\text{, }...\text{, }\mathbf{r}_{N-1}\text{ of the matrix }C\right) .
\end{align*}
Thus, the rows of the matrix $D$ are the rows $\mathbf{r}_{1}$, $\mathbf{r}%
_{2}$, $...$, $\mathbf{r}_{N-1}$ of the matrix $C$. In other words, for every
$u\in\left\{  1,2,...,N-1\right\}  $, we have%
\begin{equation}
\left(  \text{the row }u\text{ of the matrix }D\right)  =\left(  \text{the row
}\mathbf{r}_{u}\text{ of the matrix }C\right)  .
\label{sol.bernstein.d.cofactor.def.pf.16}%
\end{equation}
Thus, every $\left(  u,v\right)  \in\left\{  1,2,...,N-1\right\}
\times\left\{  1,2,...,N-1\right\}  $ satisfies
\begin{align}
&  \left(  \text{the }\left(  u,v\right)  \text{-th entry of the matrix
}D\right) \nonumber\\
&  =\left(  \text{the }v\text{-th entry of }\underbrace{\text{the row }u\text{
of the matrix }D}_{\substack{=\left(  \text{the row }\mathbf{r}_{u}\text{ of
the matrix }C\right)  \\\text{(by (\ref{sol.bernstein.d.cofactor.def.pf.16}%
))}}}\right) \nonumber\\
&  =\left(  \text{the }v\text{-th entry of the row }\mathbf{r}_{u}\text{ of
the matrix }C\right) \nonumber\\
&  =\left(  \text{the }\left(  \mathbf{r}_{u},v\right)  \text{-th entry of the
matrix }C\right)  =\beta_{\mathbf{r}_{u},\mathbf{c}_{v}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.bernstein.d.cofactor.def.pf.Centry}), applied to }\mathbf{r}%
_{u}\text{ instead of }u\right) \nonumber\\
&  =\beta_{u+\left[  u\geq p\right]  ,v+\left[  v\geq q\right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{r}_{u}=i_{u}+\left[  u\geq
p\right]  \text{ (by the definition of }\mathbf{r}_{u}\text{) and }%
\mathbf{c}_{v}=v+\left[  v\geq q\right]  \text{ (by the definition of
}\mathbf{c}_{v}\text{)}\right)  .
\label{sol.bernstein.d.cofactor.def.pf.Dentry}%
\end{align}
\par
Now, $D$ is an $\left(  N-1\right)  \times\left(  N-1\right)  $-matrix. Thus,%
\begin{align*}
D  &  =\left(  \underbrace{\left(  \text{the }\left(  i,j\right)  \text{-th
entry of the matrix }D\right)  }_{\substack{=\beta_{i+\left[  i\geq p\right]
,j+\left[  j\geq q\right]  }\\\text{(by
(\ref{sol.bernstein.d.cofactor.def.pf.Dentry}), applied to }\left(
u,v\right)  =\left(  i,j\right)  \text{)}}}\right)  _{i,j=1,2,...,N-1}\\
&  =\left(  \beta_{i+\left[  i\geq p\right]  ,j+\left[  j\geq q\right]
}\right)  _{i,j=1,2,...,N-1}.
\end{align*}
Compared with (\ref{sol.bernstein.d.cofactor.def.pf.12}), this yields%
\begin{align}
&  \left(  \text{the matrix obtained by removing row }p\text{ and column
}q\text{ from the matrix }B\right) \nonumber\\
&  =\left(  \beta_{i+\left[  i\geq p\right]  ,j+\left[  j\geq q\right]
}\right)  _{i,j=1,2,...,N-1}. \label{sol.bernstein.d.cofactor.def.pf.22}%
\end{align}
\par
Now, (\ref{sol.bernstein.d.cofactor.def.pf.1}) becomes%
\begin{align*}
&  \left(  \text{the }\left(  p,q\right)  \text{-th cofactor of the matrix
}B\right) \\
&  =\left(  -1\right)  ^{p+q}\det\underbrace{\left(  \text{the matrix obtained
by removing row }p\text{ and column }q\text{ from the matrix }B\right)
}_{\substack{=\left(  \beta_{i+\left[  i\geq p\right]  ,j+\left[  j\geq
q\right]  }\right)  _{i,j=1,2,...,N-1}\\\text{(by
(\ref{sol.bernstein.d.cofactor.def.pf.22}))}}}\\
&  =\left(  -1\right)  ^{p+q}\det\left(  \left(  \beta_{i+\left[  i\geq
p\right]  ,j+\left[  j\geq q\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\end{align*}
Since $B=\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}$, this rewrites as
\begin{align*}
&  \left(  \text{the }\left(  p,q\right)  \text{-th cofactor of the matrix
}\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \\
&  =\left(  -1\right)  ^{p+q}\det\left(  \left(  \beta_{i+\left[  i\geq
p\right]  ,j+\left[  j\geq q\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\end{align*}
This proves (\ref{sol.bernstein.d.cofactor.def}).}

\item Every commutative ring $A$, every positive integer $N$, every matrix
$\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$ and every
$k\in\left\{  1,2,...,N\right\}  $ satisfy
\begin{align}
&  \left(  \text{the }\left(  1,k\right)  \text{-th cofactor of the matrix
}\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \nonumber\\
&  =\left(  -1\right)  ^{k-1}\det\left(  \left(  \beta_{i+1,j+\left[  j\geq
k\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\label{sol.bernstein.d.cofactor.1}%
\end{align}
\footnote{\textit{Proof of (\ref{sol.bernstein.d.cofactor.1}):} Since $N$ is a
positive integer, we have $1\in\left\{  1,2,...,N\right\}  $. Thus, we can
apply (\ref{sol.bernstein.d.cofactor.def}) to $p=1$ and $q=k$. As a result, we
obtain%
\begin{align}
&  \left(  \text{the }\left(  1,k\right)  \text{-th cofactor of the matrix
}\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \nonumber\\
&  =\left(  -1\right)  ^{1+k}\det\left(  \left(  \beta_{i+\left[
i\geq1\right]  ,j+\left[  j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}\right)
. \label{sol.bernstein.d.cofactor.1.pf.1}%
\end{align}
But every $i\in\left\{  1,2,...,N-1\right\}  $ satisfies $\left[
i\geq1\right]  =1$ (since $i\geq1$ is true (since $i\in\left\{
1,2,...,N-1\right\}  $)). Hence, $\left(  \underbrace{\beta_{i+\left[
i\geq1\right]  ,j+\left[  j\geq k\right]  }}_{\substack{=\beta_{i+1,j+\left[
j\geq k\right]  }\\\text{(since }\left[  i\geq1\right]  =1\text{)}}}\right)
_{i,j=1,2,...,N-1}=\left(  \beta_{i+1,j+\left[  j\geq k\right]  }\right)
_{i,j=1,2,...,N-1}$. Furthermore, $1+k=2+\left(  k-1\right)  $, so that
$\left(  -1\right)  ^{1+k}=\left(  -1\right)  ^{2+\left(  k-1\right)
}=\underbrace{\left(  -1\right)  ^{2}}_{=1}\left(  -1\right)  ^{k-1}=\left(
-1\right)  ^{k-1}$. Thus, (\ref{sol.bernstein.d.cofactor.1.pf.1}) becomes%
\begin{align*}
&  \left(  \text{the }\left(  1,k\right)  \text{-th cofactor of the matrix
}\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \\
&  =\underbrace{\left(  -1\right)  ^{1+k}}_{=\left(  -1\right)  ^{k-1}}%
\det\left(  \underbrace{\left(  \beta_{i+\left[  i\geq1\right]  ,j+\left[
j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}}_{=\left(  \beta_{i+1,j+\left[
j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}}\right)  =\left(  -1\right)
^{k-1}\det\left(  \left(  \beta_{i+1,j+\left[  j\geq k\right]  }\right)
_{i,j=1,2,...,N-1}\right)  .
\end{align*}
This proves (\ref{sol.bernstein.d.cofactor.1}).}

\item Every commutative ring $A$, every positive integer $N$ and every matrix
$\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$ satisfy%
\begin{equation}
\det\left(  \left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right)  =\sum
_{k=1}^{N}\left(  -1\right)  ^{k-1}\beta_{1,k}\det\left(  \left(
\beta_{i+1,j+\left[  j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\label{sol.bernstein.d.laplace1r}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.d.laplace1r}):} Let $A$ be a
commutative ring. Let $N$ be a positive integer. Let $\left(  \beta
_{i,j}\right)  _{i,j=1,2,...,N}\in A^{N\times N}$ be a matrix. Since $N$ is a
positive integer, we have $1\in\left\{  1,2,...,N\right\}  $. Thus,
(\ref{sol.bernstein.d.laplace}) (applied to $\mathbf{M}=\left(  \beta
_{i,j}\right)  _{i,j=1,2,...,N}$ and $i=1$) yields%
\begin{align*}
&  \det\left(  \left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right) \\
&  =\sum_{k=1}^{N}\underbrace{\left(  \text{the }\left(  1,k\right)  \text{-th
entry of the matrix }\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}\right)
}_{=\beta_{1,k}}\cdot\underbrace{\left(  \text{the }\left(  1,k\right)
\text{-th cofactor of the matrix }\left(  \beta_{i,j}\right)  _{i,j=1,2,...,N}%
\right)  }_{\substack{=\left(  -1\right)  ^{k-1}\det\left(  \left(
\beta_{i+1,j+\left[  j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}\right)
\\\text{(by (\ref{sol.bernstein.d.cofactor.1}))}}}\\
&  =\sum_{k=1}^{N}\beta_{1,k}\cdot\left(  -1\right)  ^{k-1}\det\left(  \left(
\beta_{i+1,j+\left[  j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}\right) \\
&  =\sum_{k=1}^{N}\left(  -1\right)  ^{k-1}\beta_{1,k}\det\left(  \left(
\beta_{i+1,j+\left[  j\geq k\right]  }\right)  _{i,j=1,2,...,N-1}\right)  .
\end{align*}
This proves (\ref{sol.bernstein.d.laplace1r}).}
\end{itemize}

Let us now solve Exercise \ref{exe.bernstein}(d). Let $n \in \NN$, let
$m \in \ZZ$ and let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  \in{ \ZZ }^{n}$. We must prove (\ref{eq.exe.bernstein.d}).

First, let us notice that%
\begin{equation}
e_{i}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha
_{n}\right)  }=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in \NN \text{
satisfying }i>n. \label{sol.bernstein.d.zeroing}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.d.zeroing}):} Let
$i\in \NN$ be such that $i>n$. We need to prove that $e_{i}^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }=0$.
\par
Let us recall Assertion C.1 that we proved during our solution of Exercise
\ref{exe.bernstein}(c). This Assertion yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }%
\in\left\{  0\right\}  \cup\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid
\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu\in \Par %
;\ \ell\left(  \nu\right)  \leq n\right\}  .
\]
Thus, the element $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{n}\right)  }$ belongs to (at least) one of the sets $\left\{
0\right\}  $ and $\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid\ \mathfrak{s}%
\in\left\{  -1,1\right\}  ;\ \nu\in \Par ;\ \ell\left(
\nu\right)  \leq n\right\}  $. In other words, we must be in one of the
following two cases:
\par
\textit{Case 1:} The element $\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }$ belongs to the set $\left\{  0\right\}  $.
\par
\textit{Case 2:} The element $\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }$ belongs to the set $\left\{  \mathfrak{s}%
\cdot s_{\nu}\ \mid\ \mathfrak{s}\in\left\{  -1,1\right\}  ;\ \nu
\in \Par ;\ \ell\left(  \nu\right)  \leq n\right\}  $.
\par
Let us consider Case 1 first. In this case, the element $\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$ belongs to the set
$\left\{  0\right\}  $. Thus, $\overline{s}_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{n}\right)  }=0$. Hence, $e_{i}^{\perp}%
\underbrace{\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }}_{=0}=e_{i}^{\perp}0=0$ (since the map $e_{i}^{\perp}$ is
$\kk$-linear). Thus, $e_{i}^{\perp}\overline{s}_{\left(  \alpha
_{1},\alpha_{2},...,\alpha_{n}\right)  }=0$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, the element $\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$ belongs to the set
$\left\{  \mathfrak{s}\cdot s_{\nu}\ \mid\ \mathfrak{s}\in\left\{
-1,1\right\}  ;\ \nu\in \Par ;\ \ell\left(  \nu\right)  \leq
n\right\}  $. In other words, the element $\overline{s}_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$ can be written in the form
$\mathfrak{s}\cdot s_{\nu}$ for some $\mathfrak{s}\in\left\{  -1,1\right\}  $
and $\nu\in \Par $ satisfying $\ell\left(  \nu\right)  \leq n$.
Consider this $\mathfrak{s}$ and this $\nu$. We have $\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }=\mathfrak{s}\cdot s_{\nu}$.
But we have $\ell\left(  \nu\right)  \leq n$, thus $n\geq\ell\left(
\nu\right)  $, and therefore $i>n\geq\ell\left(  \nu\right)  $. Hence,
(\ref{sol.bernstein.skewing.e0}) (applied to $\nu$ instead of $\lambda$)
yields $e_{i}^{\perp}s_{\nu}=0$. Now,%
\begin{align*}
e_{i}^{\perp}\underbrace{\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  }}_{=\mathfrak{s}\cdot s_{\nu}}  &  =e_{i}^{\perp
}\left(  \mathfrak{s}\cdot s_{\nu}\right)  =\mathfrak{s}\cdot\underbrace{e_{i}%
^{\perp}s_{\nu}}_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
e_{i}^{\perp}\text{ is } \kk \text{-linear (and since }\mathfrak{s}%
\in\left\{  -1,1\right\}  \subset \kk \text{ is a scalar)}\right) \\
&  =0.
\end{align*}
Thus, $e_{i}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha
_{n}\right)  }=0$ is proven in Case 2.
\par
Hence, $e_{i}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,...,\alpha_{n}\right)  }=0$ is proven in both Cases 1 and 2. Since these two
Cases cover all possibilities, this yields that $e_{i}^{\perp}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }=0$
always holds.
This completes the proof of (\ref{sol.bernstein.d.zeroing}).}

Define an $\left(  n+1\right)  $-tuple $\left(  \gamma_{1},\gamma
_{2},...,\gamma_{n+1}\right)  \in \ZZ ^{n+1}$ by%
\begin{equation}
\left(  \gamma_{i}=
\begin{cases}
m,&\text{if }i=1;\\
\alpha_{i-1},&\text{if }i>1
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,...,n+1\right\}  \right)  . \label{sol.bernstein.d.gamma}%
\end{equation}
Then, every $i\in\left\{  1,2,...,n\right\}  $ satisfies%
\begin{align}
\gamma_{i+1}  &  =
\begin{cases}
m,&\text{if }i+1=1;\\
\alpha_{\left(  i+1\right)  -1},&\text{if }i+1>1
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\gamma
_{i+1}\right) \nonumber\\
&  =\alpha_{\left(  i+1\right)  -1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}i+1>1\text{ (because }i>0\text{ (since }i\in\left\{  1,2,...,n\right\}
\text{))}\right) \nonumber\\
&  =\alpha_{i}. \label{sol.bernstein.d.gamma-future}%
\end{align}
Thus, $\left(  \gamma_{1+1},\gamma_{2+1},...,\gamma_{n+1}\right)  =\left(
\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  $. Compared with the obvious
equality $\left(  \gamma_{1+1},\gamma_{2+1},...,\gamma_{n+1}\right)  =\left(
\gamma_{2},\gamma_{3},...,\gamma_{n+1}\right)  $, this yields $\left(
\gamma_{2},\gamma_{3},...,\gamma_{n+1}\right)  =\left(  \alpha_{1},\alpha
_{2},...,\alpha_{n}\right)  $.

On the other hand, the definition of $\gamma_{1}$ yields%
\[
\gamma_{1}=
\begin{cases}
m,&\text{if }1=1;\\
\alpha_{1-1},&\text{if }1>1
\end{cases}
\quad
=m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1=1\right)  .
\]
Now,
\begin{align*}
\left(  \gamma_{1},\gamma_{2},...,\gamma_{n+1}\right)   &  =\left(  \text{the
concatenation of the tuples }\left(  \underbrace{\gamma_{1}}_{=m}\right)
\text{ and }\underbrace{\left(  \gamma_{2},\gamma_{3},...,\gamma_{n+1}\right)
}_{=\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }\right) \\
&  =\left(  \text{the concatenation of the tuples }\left(  m\right)  \text{
and }\left(  \alpha_{1},\alpha_{2},...,\alpha_{n}\right)  \right) \\
&  =\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  .
\end{align*}


Thus, $\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  =\left(
\gamma_{1},\gamma_{2},...,\gamma_{n+1}\right)  $, so that%
\begin{align}
\overline{s}_{\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  } &
=\overline{s}_{\left(  \gamma_{1},\gamma_{2},...,\gamma_{n+1}\right)  }%
=\det\left(  \left(  h_{\gamma_{i}-i+j}\right)  _{i,j=1,2,\ldots,n+1}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\overline{s}_{\left(
\gamma_{1},\gamma_{2},...,\gamma_{n+1}\right)  }\right)  \nonumber\\
&  =\sum_{k=1}^{n+1}\left(  -1\right)  ^{k-1}h_{\gamma_{1}-1+k}\det\left(
\left(  h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k\right]
\right)  }\right)  _{i,j=1,2,...,\left(  n+1\right)  -1}\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.bernstein.d.laplace1r}),
applied to }A=\Lambda\text{, }N=n+1\text{ and }\beta_{i,j}=h_{\gamma_{i}%
-i+j}\right)  \nonumber\\
&  =\sum_{k=1}^{n+1}\left(  -1\right)  ^{k-1}h_{\gamma_{1}-1+k}\det\left(
\left(  h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  n+1\right)  -1=1\right)  \nonumber\\
&  =\sum_{k=0}^{n}\underbrace{\left(  -1\right)  ^{\left(  k+1\right)  -1}%
}_{=\left(  -1\right)  ^{k}}\underbrace{h_{\gamma_{1}-1+\left(  k+1\right)  }%
}_{\substack{=h_{m+k}\\\text{(since }\gamma_{1}-1+\left(  k+1\right)
=\gamma_{1}+k=m+k\\\text{(because }\gamma_{1}=m\text{))}}}\det\left(  \left(
h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k+1\text{ for
}k\text{ in the sum}\right)  \nonumber\\
&  =\sum_{k=0}^{n}\left(  -1\right)  ^{k}h_{m+k}\det\left(  \left(
h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)
.\label{sol.bernstein.d.rhs.first-formula}%
\end{align}


Let us notice a triviality: Any two integers $j$ and $k$ satisfy
\begin{equation}
\begin{cases}
1,&\text{if }j\leq k;\\
0,&\text{if }j>k
\end{cases}
\quad
=1-\left[  j\geq k+1\right]  .
\label{sol.bernstein.d.triv}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.d.triv}):} Let $j$ and $k$ be
two integers. We must prove (\ref{sol.bernstein.d.triv}).
\par
We distinguish between two cases:
\par
\textit{Case 1:} We have $j\leq k$.
\par
\textit{Case 2:} We have $j>k$.
\par
Let us consider Case 1 first. In this case, we have $j\leq k$. Thus, $j\leq
k<k+1$. Hence, the statement $j\geq k+1$ is false. Thus, $\left[  j\geq
k+1\right]  =0$. Now,%
\begin{align*}
\begin{cases}
1,&\text{if }j\leq k;\\
0,&\text{if }j>k
\end{cases}
\quad
& =1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j\geq k\right)  \\
& =1-\left[  j\geq k+1\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}1-\underbrace{\left[  j\geq k+1\right]  }_{=0}=1\right)  .
\end{align*}
Hence, (\ref{sol.bernstein.d.triv}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $j>k$. Thus, $j\geq k+1$
(since $j$ and $k$ are integers). Hence, $\left[  j\geq k+1\right]  =1$. Now,%
\begin{align*}
\begin{cases}
1,&\text{if }j\leq k;\\
0,&\text{if }j>k
\end{cases}
\quad
& =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j>k\right)  \\
& =1-\left[  j\geq k+1\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}1-\underbrace{\left[  j\geq k+1\right]  }_{=1}=1-1=0\right)  .
\end{align*}
Hence, (\ref{sol.bernstein.d.triv}) is proven in Case 2.
\par
Now, (\ref{sol.bernstein.d.triv}) is proven in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this yields that
(\ref{sol.bernstein.d.triv}) always holds. This completes the proof of
(\ref{sol.bernstein.d.triv}).}

Now, it is not hard to check that%
\begin{equation}
e_{k}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }=\det\left(  \left(  h_{\gamma_{i+1}-\left(  i+1\right)
+\left(  j+\left[  j\geq k+1\right]  \right)  }\right)  _{i,j=1,2,...,n}%
\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,...,n\right\}
\label{sol.bernstein.d.lhs.small-terms}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.bernstein.d.lhs.small-terms}):} Let
$k\in\left\{  0,1,...,n\right\}  $. Then, $k\leq n$.
\par
Let $\mu$ denote the partition $\left(  1^{k}\right)  $. Then, $\mu$ is a
partition having $k$ parts. Hence, $\mu$ is a partition having at most $n$
parts (since $k\leq n$). Since $\mu=\left(  1^{k}\right)  $, we have $s_{\mu
}=s_{\left(  1^{k}\right)  }=e_{k}$.
\par
Write the partition $\mu$ in the form $\mu=\left(  \mu_{1},\mu_{2},\mu
_{3},...\right)  $. Then, every positive integer $j$ satisfies%
\begin{align}
\mu_{j} &  =\left(  \text{the }j\text{-th entry of the partition
}\underbrace{\left(  \mu_{1},\mu_{2},\mu_{3},...\right)  }_{=\mu=\left(
1^{k}\right)  =\left(  \underbrace{1,1,...,1}_{k\text{ times}}\right)
}\right)  \nonumber\\
&  =\left(  \text{the }j\text{-th entry of the partition }\left(
\underbrace{1,1,...,1}_{k\text{ times}}\right)  \right)  =
\begin{cases}
1,&\text{if }j\leq k;\\
0,&\text{if }j>k
\end{cases}
\quad
=1-\left[  j\geq k+1\right]
\label{sol.bernstein.d.lhs.small-terms.pf.1}
\end{align}
(by (\ref{sol.bernstein.d.triv})).
\par
But $\mu$ is a partition having at most $n$ parts. Thus,
(\ref{eq.exe.bernstein.c.1}) (applied to $\mu$ instead of $\lambda$) yields
$s_{\mu}=\overline{s}_{\left(  \mu_{1},\mu_{2},...,\mu_{n}\right)  }$.
Compared with $s_{\mu}=e_{k}$, this yields $e_{k}=\overline{s}_{\left(
\mu_{1},\mu_{2},...,\mu_{n}\right)  }$. Thus, $e_{k}^{\perp}
= \overline{s}_{\left(  \mu_{1},\mu_{2},...,\mu_{n}\right)  }^{\perp}$.
Therefore,%
\begin{equation}
\underbrace{e_{k}^{\perp}}_{=\overline{s}_{\left(  \mu_{1},\mu_{2},...,\mu
_{n}\right)  }^{\perp}}\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  }=\overline{s}_{\left(  \mu_{1},\mu_{2},...,\mu
_{n}\right)  }^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  }=\det\left(  \left(  h_{\alpha_{i}-\mu_{j}%
-i+j}\right)  _{i,j=1,2,\ldots,n}\right)
\label{sol.bernstein.d.lhs.small-terms.pf.3}%
\end{equation}
(by (\ref{eq.exe.bernstein.c.3}), applied to $\beta_{i}=\mu_{i}$).
\par
Now, every $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}  ^{2}$ satisfies%
\begin{align*}
\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]  \right)
& =\underbrace{\gamma_{i+1}}_{\substack{=\alpha_{i}\\\text{(by
(\ref{sol.bernstein.d.gamma-future}))}}}-\underbrace{\left(  1-\left[  j\geq
k+1\right]  \right)  }_{\substack{=\mu_{j}\\\text{(by
(\ref{sol.bernstein.d.lhs.small-terms.pf.1}))}}}-i+j\\
& =\alpha_{i}-\mu_{j}-i+j.
\end{align*}
In other words, every $\left(  i,j\right)  \in\left\{  1,2,...,n\right\}
^{2}$ satisfies $\alpha_{i}-\mu_{j}-i+j=\gamma_{i+1}-\left(  i+1\right)
+\left(  j+\left[  j\geq k+1\right]  \right)  $. Hence,%
\[
\det\left(  \left(  \underbrace{h_{\alpha_{i}-\mu_{j}-i+j}}%
_{\substack{=h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq
k+1\right]  \right)  }\\\text{(since }\alpha_{i}-\mu_{j}-i+j\\=\gamma
_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]  \right)
\text{)}}}\right)  _{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(
h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)  .
\]
Now, (\ref{sol.bernstein.d.lhs.small-terms.pf.3}) becomes%
\[
e_{k}^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  }=\det\left(  \left(  h_{\alpha_{i}-\mu_{j}-i+j}\right)
_{i,j=1,2,\ldots,n}\right)  =\det\left(  \left(  h_{\gamma_{i+1}-\left(
i+1\right)  +\left(  j+\left[  j\geq k+1\right]  \right)  }\right)
_{i,j=1,2,...,n}\right)  .
\]
This proves (\ref{sol.bernstein.d.lhs.small-terms}).}. Now,
(\ref{sol.bernstein.d.rhs.first-formula}) becomes%
\begin{align*}
\overline{s}_{\left(  m,\alpha_{1},\alpha_{2},...,\alpha_{n}\right)  }  &
=\sum_{k=0}^{n}\left(  -1\right)  ^{k}h_{m+k}\underbrace{\det\left(  \left(
h_{\gamma_{i+1}-\left(  i+1\right)  +\left(  j+\left[  j\geq k+1\right]
\right)  }\right)  _{i,j=1,2,...,n}\right)  }_{\substack{=e_{k}^{\perp
}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}\\\text{(by (\ref{sol.bernstein.d.lhs.small-terms}))}}}\\
& =\sum_{k=0}^{n}\left(  -1\right)  ^{k}h_{m+k}e_{k}^{\perp}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }
=\sum_{i=0}%
^{n}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}k\text{ as }i\right)  .
\end{align*}
Compared with%
\begin{align*}
& \sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}%
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }\\
& =\underbrace{\sum\limits_{\substack{i\in{ \NN };\\i\leq n}}}_{=\sum_{i=0}^{n}%
}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }+\sum\limits_{\substack{i\in{ \NN %
};\\i>n}}\left(  -1\right)  ^{i}h_{m+i}\underbrace{e_{i}^{\perp}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }}%
_{\substack{=0\\\text{(by (\ref{sol.bernstein.d.zeroing}))}}}\\
& =\sum_{i=0}^{n}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  },
\end{align*}
this yields $\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}%
^{\perp}\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
}=\overline{s}_{\left(  m,\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  }$.
This solves Exercise \ref{exe.bernstein}(d).

(b) Let $n=\ell\left(  \lambda\right)  $. Then, the partition $\lambda$ has
length $\ell\left(  \lambda\right)  =n$. Thus, $\lambda=\left(  \lambda
_{1},\lambda_{2},...,\lambda_{n}\right)  $ (since $\lambda=\left(  \lambda
_{1},\lambda_{2},\lambda_{3},\ldots\right)  $). 

The number of parts of the partition $\lambda$ is $\ell\left(  \lambda\right)
=n$. In other words, the partition $\lambda$ has exactly $n$ parts. Hence, the
partition $\lambda$ has at most $n$ parts. Thus,
(\ref{eq.exe.bernstein.c.1}) yields $s_{\lambda}=\overline{s}_{\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  }$.

But as partitions, $\left(  \lambda_{1},\lambda_{2},\lambda_{3},...\right)
=\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n}\right)  =\left(
\lambda_{1},\lambda_{2},...,\lambda_{n},0,0,0,...\right)  $. Hence, $\left(
\lambda_{1},\lambda_{2},\lambda_{3},...\right)  =\left(  \lambda_{1}%
,\lambda_{2},...,\lambda_{n},0,0,0,...\right)  $ as sequences. Now,%
\begin{align*}
\left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)    & =\left(
\text{the concatenation of the sequences }\left(  m\right)  \text{ and
}\underbrace{\left(  \lambda_{1},\lambda_{2},\lambda_{3},...\right)
}_{=\left(  \lambda_{1},\lambda_{2},...,\lambda_{n},0,0,0,...\right)
}\right)  \\
& =\left(  \text{the concatenation of the sequences }\left(  m\right)  \text{
and }\left(  \lambda_{1},\lambda_{2},...,\lambda_{n},0,0,0,...\right)
\right)  \\
& =\left(  m,\lambda_{1},\lambda_{2},...,\lambda_{n},0,0,0,...\right)  .
\end{align*}
Thus, as partitions,%
\[
\left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)  =\left(
m,\lambda_{1},\lambda_{2},...,\lambda_{n},0,0,0,...\right)  =\left(
m,\lambda_{1},\lambda_{2},...,\lambda_{n}\right)  .
\]
Therefore, the partition $\left(  m,\lambda_{1},\lambda_{2},\lambda
_{3},...\right)  $ has at most $n+1$ parts (since the partition $\left(
m,\lambda_{1},\lambda_{2},...,\lambda_{n}\right)  $ clearly has at most $n+1$
parts). Hence, (\ref{eq.exe.bernstein.c.1}) (applied to $n+1$ and
$\left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)  $ instead of $n$
and $\lambda=\left(  \lambda_{1},\lambda_{2},\lambda_{3},...\right)  $) yields%
\[
s_{\left(  m,\lambda_{1},\lambda_{2},\lambda_{3},...\right)  }
= \overline{s}_{\left(  m,\lambda_{1},\lambda_{2},...,\lambda_{n}\right)  }.
\]


Compared with%
\begin{align*}
\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp
}\underbrace{s_{\lambda}}_{=\overline{s}_{\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{n}\right)  }}  & =\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}\overline{s}_{\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{n}\right)  }\\
& =\overline{s}_{\left(  m,\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.bernstein.d}), applied to
}\alpha_{j}=\lambda_{j}\right)  ,
\end{align*}
this yields $\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{m+i}e_{i}%
^{\perp}s_{\lambda}=s_{\left(  m,\lambda_{1},\lambda_{2},\lambda
_{3},...\right)  }$. This solves Exercise \ref{exe.bernstein}(b).

(e) We shall solve Exercise~\ref{exe.bernstein}(e) by induction over $n$:

\textit{Induction base:} The statement of Exercise \ref{exe.bernstein}(e) is
clearly valid for $n=0$\ \ \ \ \footnote{\textit{Proof.} We need to prove that
the statement of Exercise \ref{exe.bernstein}(e) is valid for $n=0$. In other
words, we need to prove that for every $0$-tuple $\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{0}\right)  \in{ \ZZ }^{0}$, we have
\begin{equation}
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{0}\right)
}=\left(  \mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots
\circ\mathbf{B}_{\alpha_{0}}\right)  \left(  1\right)
.\label{sol.bernstein.e.indbase}%
\end{equation}
So let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{0}\right)  \in
{ \ZZ }^{0}$ be a $0$-tuple. Of course, $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{0}\right)  =\left(  \text{empty }0\text{-tuple}\right)  $.
The matrix $\left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,0}$ is a
$0\times0$-matrix, and thus is the empty $0\times0$-matrix (since the only
$0\times0$-matrix is the empty $0\times0$-matrix), and hence has determinant
$1$. In other words, $\det\left(  \left(  h_{\alpha_{i}-i+j}\right)
_{i,j=1,2,\ldots,0}\right)  =1$.
\par
The definition of $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{0}\right)  }$ yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{0}\right)  }%
=\det\left(  \left(  h_{\alpha_{i}-i+j}\right)  _{i,j=1,2,\ldots,0}\right)
=1.
\]
Compared with%
\[
\underbrace{\left(  \mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}%
\circ\cdots\circ\mathbf{B}_{\alpha_{0}}\right)  }_{\substack{=\left(
\text{empty composition of maps}\right)  \\= \id }}\left(
1\right)  = \id \left(  1\right)  =1,
\]
this yields $\overline{s}_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{0}\right)  }=\left(  \mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}%
}\circ\cdots\circ\mathbf{B}_{\alpha_{0}}\right)  \left(  1\right)  $. This
completes the proof of (\ref{sol.bernstein.e.indbase}). Thus, the statement of
Exercise \ref{exe.bernstein}(e) is valid for $n=0$, qed.}. Hence, the
induction base is complete.

\textit{Induction step:} Let $N\in \NN$. Assume that the statement of
Exercise \ref{exe.bernstein}(e) is valid for $n=N$. We need to prove that the
statement of Exercise \ref{exe.bernstein}(e) is valid for $n=N+1$.

Let $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{N+1}\right)  \in{ \ZZ %
}^{N+1}$ be an $\left(  N+1\right)  $-tuple. We can apply Exercise
\ref{exe.bernstein}(e) to $N$ and $\left(  \alpha_{2},\alpha_{3}%
,...,\alpha_{N+1}\right)  $ instead of $n$ and $\left(  \alpha_{1},\alpha
_{2},...,\alpha_{n}\right)  $ (since we have assumed that the statement of
Exercise \ref{exe.bernstein}(e) is valid for $n=N$). As a consequence, we
obtain%
\[
\overline{s}_{\left(  \alpha_{2},\alpha_{3},\ldots,\alpha_{N+1}\right)
}=\left(  \mathbf{B}_{\alpha_{2}}\circ\mathbf{B}_{\alpha_{3}}\circ\cdots
\circ\mathbf{B}_{\alpha_{N+1}}\right)  \left(  1\right)  .
\]


Applying the map $\mathbf{B}_{\alpha_{1}}$ to both sides of this identity, we
obtain%
\begin{align*}
\mathbf{B}_{\alpha_{1}}\left(  \overline{s}_{\left(  \alpha_{2},\alpha
_{3},\ldots,\alpha_{N+1}\right)  }\right)    & =\mathbf{B}_{\alpha_{1}}\left(
\left(  \mathbf{B}_{\alpha_{2}}\circ\mathbf{B}_{\alpha_{3}}\circ\cdots
\circ\mathbf{B}_{\alpha_{N+1}}\right)  \left(  1\right)  \right)
=\underbrace{\left(  \mathbf{B}_{\alpha_{1}}\circ\left(  \mathbf{B}%
_{\alpha_{2}}\circ\mathbf{B}_{\alpha_{3}}\circ\cdots\circ\mathbf{B}%
_{\alpha_{N+1}}\right)  \right)  }_{=\mathbf{B}_{\alpha_{1}}\circ
\mathbf{B}_{\alpha_{2}}\circ\cdots\circ\mathbf{B}_{\alpha_{N+1}}}\left(
1\right)  \\
& =\left(  \mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots
\circ\mathbf{B}_{\alpha_{N+1}}\right)  \left(  1\right)  .
\end{align*}
Compared with%
\begin{align*}
\mathbf{B}_{\alpha_{1}}\left(  \overline{s}_{\left(  \alpha_{2},\alpha
_{3},\ldots,\alpha_{N+1}\right)  }\right)    & =\sum_{i\in{ \NN }}\left(
-1\right)  ^{i}h_{\alpha_{1}+i}e_{i}^{\perp}\overline{s}_{\left(  \alpha
_{2},\alpha_{3},\ldots,\alpha_{N+1}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the map }\mathbf{B}_{\alpha_{1}}\right)  \\
& =\overline{s}_{\left(  \alpha_{1},\alpha_{2},\alpha_{3},...,\alpha
_{N+1}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Exercise \ref{exe.bernstein}(d), applied to }N\text{, }\alpha
_{1}\text{ and}\\
\left(  \alpha_{2},\alpha_{3},\ldots,\alpha_{N+1}\right)  \text{ instead of
}n\text{, }m\text{ and }\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)
\end{array}
\right)  \\
& =\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{N+1}\right)  },
\end{align*}
this yields%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{N+1}\right)  }=\left(
\mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots\circ
\mathbf{B}_{\alpha_{N+1}}\right)  \left(  1\right)  .
\]


Now, forget that we fixed $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{N+1}\right)  $. We thus have shown that for every $\left(  N+1\right)
$-tuple $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{N+1}\right)
\in{ \ZZ }^{N+1}$, we have%
\[
\overline{s}_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{N+1}\right)  }=\left(
\mathbf{B}_{\alpha_{1}}\circ\mathbf{B}_{\alpha_{2}}\circ\cdots\circ
\mathbf{B}_{\alpha_{N+1}}\right)  \left(  1\right)  .
\]
In other words, the statement of Exercise \ref{exe.bernstein}(e) is valid for
$n=N+1$. This completes the induction step. We thus have solved Exercise
\ref{exe.bernstein}(e) by induction.

(f) Let $m\in \ZZ$. Let $n$ be a positive integer. We have $e_{0}=1$ and
thus $e_{0}^{\perp}=1^{\perp}=\operatorname*{id}$ (by Proposition
\ref{skewing-properties-prop}(iii), applied to $A=\Lambda$). Hence,
$e_{0}^{\perp}p_{n}=\operatorname*{id}\left(  p_{n}\right)  =p_{n}$.

If $i$ is a positive integer satisfying $i\neq n$, then%
\begin{equation}
e_{i}^{\perp}p_{n}=0\label{sol.bernstein.f.eiperppn=0}%
\end{equation}
(by Exercise \ref{exe.skew-p-by-e}(d), applied to $m=i$).

The definition of $\mathbf{B}_{m}$ yields%
\begin{align*}
\mathbf{B}_{m}\left(  p_{n}\right)    & =\sum_{i\in{\NN}}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}\\
& =\underbrace{\left(  -1\right)  ^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}%
}\underbrace{e_{0}^{\perp}p_{n}}_{=p_{n}}+\sum_{i>0}\left(  -1\right)
^{i}h_{m+i}e_{i}^{\perp}p_{n}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=0\text{ from the sum}\right)  \\
& =h_{m}p_{n}+\sum_{i>0}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}.
\end{align*}
In view of%
\begin{align*}
& \sum_{i>0}\left(  -1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}\\
& =\left(  -1\right)  ^{n}h_{m+n}\underbrace{e_{n}^{\perp}p_{n}}%
_{\substack{=\left(  -1\right)  ^{n-1}\\\text{(by Exercise
\ref{exe.skew-p-by-e}(c))}}}+\sum\limits_{\substack{i>0;\\i\neq n}}\left(  -1\right)
^{i}h_{m+i}\underbrace{e_{i}^{\perp}p_{n}}_{\substack{=0\\\text{(by
(\ref{sol.bernstein.f.eiperppn=0}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=n\text{ from the sum,}\\
\text{since }n\text{ is a positive integer}%
\end{array}
\right)  \\
& =\left(  -1\right)  ^{n}h_{m+n}\left(  -1\right)  ^{n-1}+\underbrace{\sum\limits
_{\substack{i>0;\\i\neq n}}\left(  -1\right)  ^{i}h_{m+i}0}_{=0}=\left(
-1\right)  ^{n}h_{m+n}\left(  -1\right)  ^{n-1}=\underbrace{\left(  -1\right)
^{n}\left(  -1\right)  ^{n-1}}_{\substack{=\left(  -1\right)  ^{n+\left(
n-1\right)  }=-1\\\text{(since }n+\left(  n-1\right)  =2n-1\\\text{is odd)}%
}}h_{m+n}=-h_{m+n},
\end{align*}
this becomes
\[
\mathbf{B}_{m}\left(  p_{n}\right)  =h_{m}p_{n}+\underbrace{\sum_{i>0}\left(
-1\right)  ^{i}h_{m+i}e_{i}^{\perp}p_{n}}_{=-h_{m+n}}=h_{m}p_{n}+\left(
-h_{m+n}\right)  =h_{m}p_{n}-h_{m+n}.
\]
This solves Exercise~\ref{exe.bernstein}(f).

\textit{Remark:} Our solution to Exercise~\ref{exe.bernstein}(d) was modelled
after the rough sketch of a solution to Exercise~\ref{exe.bernstein}(b) given
in \cite[\S 4.20]{Zelevinsky}; but it involved many technicalities which
are not necessary if one is only interested in a solution to
Exercise~\ref{exe.bernstein}(b). (Specifically, if one only wants to solve
Exercise~\ref{exe.bernstein}(b), one can avoid the use of
Exercise~\ref{exe.bernstein}(c), and one can replace the proof of
\eqref{sol.bernstein.d.zeroing} by a simple application of
\eqref{sol.bernstein.skewing.general}.)

We solved Exercise~\ref{exe.bernstein}(e) using
Exercise~\ref{exe.bernstein}(d), but of course one can just as well
turn this around and solve Exercise~\ref{exe.bernstein}(d) using
Exercise~\ref{exe.bernstein}(e) if one has an independent solution to
Exercise~\ref{exe.bernstein}(e). Such an independent solution can be
extracted from
\cite[Corollary 3.30]{BergBergeronSaliolaSerranoZabrocki-immac1}
(using the realization that the immaculate creation operators
$\mathbb{B}_m$ of \cite{BergBergeronSaliolaSerranoZabrocki-immac1}
are lifts of our Bernstein operators $\mathbf{B}_m$ to $\Nsym$).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.witt}}

\textit{Solution to Exercise~\ref{exe.witt}.}
(a) We shall prove a more general result:
\begin{statement}
\textit{Claim 1:} Let $A$ be \textbf{any}
commutative ring. Let $f\in A\left[  \left[  t\right]
\right]  $ be a power series with constant term $1$.
Then, there is a unique family $\left(
x_{n}\right)  _{n\geq 1}$ of elements of $A$ such that
\begin{equation}
f=\prod_{n=1}^{\infty}\left(  1-x_{n}t^{n}\right)  ^{-1}. \label{sol.witt.1}
\end{equation}
\end{statement}
[\textit{Proof of Claim 1:}
This family is constructed recursively: If $x_1, x_2, \ldots, x_{k-1}$ have
been determined (for some $k\geq 1$), then $x_k$ is obtained by comparing
coefficients before $t^k$ in the equation \eqref{sol.witt.1} (the coefficient
before $t^k$ on the left hand side is a known constant,
whereas the coefficient
before $t^k$ on the right hand side can be written in the form
$x_k + \left(  \text{some polynomial in }x_{1},x_{2},...,x_{k-1} \right)  $,
and thus the equality of these coefficients gives a linear equation in $x_k$
which can be uniquely solved for $x_k$). The family
$\left(x_n\right)_{n \geq 1}$ thus constructed satisfies \eqref{sol.witt.1}
(because the constant terms on both sides of \eqref{sol.witt.1} are $1$,
whereas for every $k \geq 1$, the coefficients before $t^k$ on both sides of
\eqref{sol.witt.1} are equal due to the construction of $x_k$).
Moreover, it is the only family that satisfies \eqref{sol.witt.1}
(since its construction was dictated by \eqref{sol.witt.1}).
Thus, Claim 1 is proven.]

Exercise~\ref{exe.witt}(a) follows by applying Claim 1 to $A = \Lambda$
and $f = H\left(t\right)$.

(b) Again, this generalizes: Let us say that a power series $f\in
A\left[  \left[  t\right]  \right]  $ over a graded commutative ring $A$ is
\emph{equigraded}\index{equigraded power series}
if, for every $n \in \NN$, the coefficient of $f$
before $t^{n}$ is homogeneous of degree $n$. Then, if $f\in A\left[  \left[
t\right]  \right]  $ is an equigraded power series with constant term $1$,
then the unique family
$\left(  x_{n}\right)  _{n\geq 1}$ satisfying \eqref{sol.witt.1} has the
property that $x_{n}$ is homogeneous of degree $n$ for every positive $n$.
This is rather easy to see by induction.

(c) By the definition of the $w_{n}$, we have $H\left(  t\right)
=\prod_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}$. Expanding and
comparing coefficients yields precisely $\sum_{\lambda\in\Par_n}w_{\lambda
}=h_{n}$.\ \ \ \ \footnote{Here are some more details of this argument.
From \eqref{h-generating-function},
we have $H\left(  t\right)  =\sum_{n\geq0}h_{n}\left( \xx \right)
t^{n}=\sum_{n\geq0}h_{n}t^{n}$. Thus,%
\begin{align}
\sum_{n\geq0}h_{n}t^{n}  & =H\left(  t\right)  =\prod_{n=1}^{\infty
}\underbrace{\left(  1-w_{n}t^{n}\right)  ^{-1}}_{=\sum_{m\in \NN %
}\left(  w_{n}t^{n}\right)  ^{m}}=\prod_{n=1}^{\infty}\sum_{m\in \NN %
}\underbrace{\left(  w_{n}t^{n}\right)  ^{m}}_{=w_{n}^{m}t^{nm}}\nonumber\\
& =\prod_{n=1}^{\infty}\sum_{m\in \NN }w_{n}^{m}t^{nm}=\sum\limits
_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\underbrace{\prod_{n=1}^{\infty}w_{n}^{m_{n}}t^{nm_{n}}%
}_{=\left(  \prod_{n=1}^{\infty}w_{n}^{m_{n}}\right)  t^{1m_{1}+2m_{2}%
+3m_{3}+\cdots}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right)
\nonumber\\
& =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\left(  \prod_{n=1}^{\infty}w_{n}^{m_{n}}\right)  t^{1m_{1}%
+2m_{2}+3m_{3}+\cdots}.\label{sol.witt.c.some-more-details.1}%
\end{align}
But every partition $\lambda$ can be uniquely written in the form
$\lambda=\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for some weak
composition $\left(  m_{1},m_{2},m_{3},\ldots\right)  $. Thus, we can
substitute $\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for $\lambda$
in the sum $\sum_{\lambda\in\Par}w_{\lambda}t^{\left\vert
\lambda\right\vert }$. As a result, we obtain%
\begin{align*}
\sum_{\lambda\in\Par}w_{\lambda}t^{\left\vert \lambda
\right\vert }  & =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)
\\\text{weak composition}}}\underbrace{w_{\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}%
}\cdots\right)  }}_{\substack{=w_{1}^{m_{1}}w_{2}^{m_{2}}w_{3}^{m_{3}}%
\cdots\\\text{(by the definition of }w_{\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}%
}\cdots\right)  }\text{)}}}\underbrace{t^{\left\vert \left(  1^{m_{1}}%
2^{m_{2}}3^{m_{3}}\cdots\right)  \right\vert }}_{\substack{=t^{1m_{1}%
+2m_{2}+3m_{3}+\cdots}\\\text{(since }\left\vert \left(  1^{m_{1}}2^{m_{2}%
}3^{m_{3}}\cdots\right)  \right\vert =1m_{1}+2m_{2}+3m_{3}+\cdots\text{)}}}\\
& =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\underbrace{\left(  w_{1}^{m_{1}}w_{2}^{m_{2}}w_{3}^{m_{3}%
}\cdots\right)  }_{=\prod_{n=1}^{\infty}w_{n}^{m_{n}}}t^{1m_{1}+2m_{2}%
+3m_{3}+\cdots}
 =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\left(  \prod_{n=1}^{\infty}w_{n}^{m_{n}}\right)  t^{1m_{1}%
+2m_{2}+3m_{3}+\cdots}.
\end{align*}
Compared with (\ref{sol.witt.c.some-more-details.1}), this yields%
\[
\sum_{n\geq0}h_{n}t^{n}=\sum_{\lambda\in\Par}w_{\lambda
}t^{\left\vert \lambda\right\vert }=\sum_{n\geq0}\underbrace{\sum\limits
_{\substack{\lambda\in\Par;\\\left\vert \lambda\right\vert
=n}}}_{=\sum_{\lambda\in\Par_{n}}}w_{\lambda
}\underbrace{t^{\left\vert \lambda\right\vert }}_{\substack{=t^{n}%
\\\text{(since }\left\vert \lambda\right\vert =n\text{)}}}=\sum_{n\geq0}%
\sum_{\lambda\in\Par_{n}}w_{\lambda}t^{n}.
\]
Comparing coefficients before $t^{n}$ in this equality of power series, we
conclude that $h_{n}=\sum_{\lambda\in\Par_{n}}w_{\lambda}$
for every $n\in\NN$, qed.}

(d) \textit{First solution to Exercise~\ref{exe.witt}(d) (sketched).} If
$\lambda^{\left(  1\right)  },\lambda^{\left(  2\right)  },\ldots
,\lambda^{\left(  \ell\right)  }$ are finitely many partitions, then we let
$\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup\cdots
\cup\lambda^{\left(  \ell\right)  }$ denote the partition obtained by sorting
all the parts of $\lambda^{\left(  1\right)  },\lambda^{\left(  2\right)
},...,\lambda^{\left(  \ell\right)  }$ in decreasing order. For instance,
$\left(  3,2,1\right)  \cup\left(  4,2\right)  \cup\left(  5,1\right)
=\left(  5,4,3,2,2,1,1\right)  $. Clearly, if $\lambda^{\left(  1\right)
},\lambda^{\left(  2\right)  },...,\lambda^{\left(  \ell\right)  }$ are
finitely many partitions, then%
\begin{equation}
w_{\lambda^{\left(  1\right)  }}w_{\lambda^{\left(  2\right)  }}\cdots
w_{\lambda^{\left(  \ell\right)  }}=w_{\lambda^{\left(  1\right)  }\cup
\lambda^{\left(  2\right)  }\cup\cdots\cup\lambda^{\left(  \ell\right)  }%
}\label{sol.witt.d.sol1.prod}%
\end{equation}
(due to the definition of $w_{\lambda}$ for a partition $\lambda$).

From (c), we know that $h_{n}=\sum_{\lambda\in\operatorname{Par}_{n}%
}w_{\lambda}$ for every $n\in \NN$. In other words,
\begin{equation}
h_{n}=\sum_{\lambda\vdash n}w_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in{\NN},\label{sol.witt.d.sol1.hn=}%
\end{equation}
where we are using the notation $\lambda\vdash n$ for $\lambda\in
\operatorname{Par}_{n}$.

Every partition $\mu=\left(  \mu_{1},\mu_{2},...,\mu_{\ell}\right)  $ with
$\ell=\ell\left(  \mu\right)  $ satisfies
\begin{align}
h_{\mu}  & =h_{\mu_{1}}h_{\mu_{2}}\cdots h_{\mu_{\ell}}=\prod_{i=1}^{\ell
}\underbrace{h_{\mu_{i}}}_{\substack{=\sum_{\lambda\vdash\left\vert \mu
_{i}\right\vert }w_{\lambda}\\\text{(by (\ref{sol.witt.d.sol1.hn=}), applied
to }n=\mu_{i}\text{)}}}\nonumber\\
& =\prod_{i=1}^{\ell}\left(  \sum_{\lambda\vdash\left\vert \mu_{i}\right\vert
}w_{\lambda}\right)  =\sum\limits_{\substack{\left(  \lambda^{\left(  1\right)
},\lambda^{\left(  2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)
\in\operatorname{Par}^{\ell};\\\lambda^{\left(  i\right)  }\vdash\mu_{i}\text{
for every }i}}\underbrace{w_{\lambda^{\left(  1\right)  }}w_{\lambda^{\left(
2\right)  }}\cdots w_{\lambda^{\left(  \ell\right)  }}}_{\substack{=w_{\lambda
^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup\cdots\cup
\lambda^{\left(  \ell\right)  }}\\\text{(by (\ref{sol.witt.d.sol1.prod}))}%
}}\nonumber\\
& =\sum\limits_{\substack{\left(  \lambda^{\left(  1\right)  },\lambda^{\left(
2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)  \in
\operatorname{Par}^{\ell};\\\lambda^{\left(  i\right)  }\vdash\mu_{i}\text{
for every }i}}w_{\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)
}\cup\cdots\cup\lambda^{\left(  \ell\right)  }}.\label{sol.witt.d.sol1.hmu=}%
\end{align}


Now, let us fix $n\in \NN$. We shall prove that $\left(  w_{\lambda
}\right)  _{\lambda\in  \Par_{n}}$ is a basis of the
$\kk$-module $\Lambda_{n}$. First of all, we know that this family
$\left(  w_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a family of elements of $\Lambda_{n}$ (since for each $\lambda\in
  \Par_{n}$, the symmetric function $w_{\lambda}$ is
homogeneous of degree $\left\vert \lambda\right\vert =n$).

Now, we define a binary relation $\underset{\operatorname*{ref}}{\leq}$ on the
set $\Par_{n}$ as follows: If $\lambda\in
  \Par_{n}$ and $\mu\in  \Par_{n}%
$, then we let $\lambda\underset{\operatorname*{ref}}{\leq}\mu$ if and only if
there exists a tuple $\left(  \lambda^{\left(  1\right)  },\lambda^{\left(
2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)  \in
\operatorname{Par}^{\ell}$ satisfying $\ell=\ell\left(  \mu\right)  $,
$\left(  \lambda^{\left(  i\right)  }\vdash\mu_{i}\text{ for every }i\right)
$ and $\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup
\cdots\cup\lambda^{\left(  \ell\right)  }=\lambda$. The intuitive meaning
behind this is the following: We have $\lambda\underset{\operatorname*{ref}%
}{\leq}\mu$ if we can obtain the partition $\lambda$ by splitting each part
$\mu_{i}$ of $\mu$ into several smaller parts\footnote{We allow the
``several smaller parts'' to be one single
part (namely, $\mu_{i}$).} (which are positive integers summing up to $\mu
_{i}$) and sorting the resulting list into decreasing order. For instance,
$\left(  5,3,2,2,2,1\right)  \underset{\operatorname*{ref}}{\leq}\left(
6,5,4\right)  $ (because the tuple $\left(  \lambda^{\left(  1\right)
},\lambda^{\left(  2\right)  },\lambda^{\left(  3\right)  }\right)  =\left(
\left(  3,2,1\right)  ,\left(  5\right)  ,\left(  2,2\right)  \right)  $
satisfies $\left(  3,2,1\right)  \vdash6$, $\left(  5\right)  \vdash5$ and
$\left(  2,2\right)  \vdash4$ and $\left(  3,2,1\right)  \cup\left(  5\right)
\cup\left(  2,2\right)  =\left(  5,3,2,2,2,1\right)  $).

It is easy to see that the relation $\underset{\operatorname*{ref}}{\leq}$ is
transitive, antisymmetric and reflexive\footnote{In proving these properties
(specifically, antisymmetry), it helps to observe the
following fact: If $\lambda\in  \Par_{n}$ and $\mu
\in  \Par_{n}$ satisfy $\lambda
\underset{\operatorname*{ref}}{\leq}\mu$, then either $\lambda=\mu$ or
$\ell\left(  \lambda\right)  >\ell\left(  \mu\right)  $.}. Hence,
$\underset{\operatorname*{ref}}{\leq}$ is the smaller-or-equal relation of a
partial order on the set $\Par_{n}$. Consider
$\Par_{n}$ as a poset, equipped with this partial
order. (This partial order is called the
\emph{refinement order}\index{refinement order on partitions} on
partitions\footnote{Caution: This order is \textbf{not} a restriction of the
refinement order on compositions.}.)

Now, for any two partitions $\lambda$ and $\mu$, let $b_{\mu,\lambda}$ denote
the number of tuples $\left(  \lambda^{\left(  1\right)  },\lambda^{\left(
2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)  \in
\operatorname{Par}^{\ell}$ satisfying $\ell=\ell\left(  \mu\right)  $,
$\left(  \lambda^{\left(  i\right)  }\vdash\mu_{i}\text{ for every }i\right)
$ and $\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup
\cdots\cup\lambda^{\left(  \ell\right)  }=\lambda$. Then, the formula
(\ref{sol.witt.d.sol1.hmu=}) rewrites as%
\[
h_{\mu}=\sum_{\lambda\in \Par }b_{\mu,\lambda}w_{\lambda}.
\]
Thus, for every $\mu\in  \Par_{n}$, we have%
\begin{align*}
h_{\mu}  & =\sum_{\lambda\in \Par }b_{\mu,\lambda}w_{\lambda
}=\underbrace{\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert
\lambda\right\vert =n}}}_{=\sum_{\lambda\in  \Par_{n}}%
}b_{\mu,\lambda}w_{\lambda}+\sum\limits_{\substack{\lambda\in \Par %
;\\\left\vert \lambda\right\vert \neq n}}\underbrace{b_{\mu,\lambda}%
}_{\substack{=0\\\text{(this easily follows}\\\text{from }\left\vert
\lambda\right\vert \neq n=\left\vert \mu\right\vert \text{)}}}w_{\lambda}\\
& =\sum_{\lambda\in  \Par_{n}}b_{\mu,\lambda}w_{\lambda
}+\underbrace{\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert
\lambda\right\vert \neq n}}0w_{\lambda}}_{=0}=\sum_{\lambda\in
  \Par_{n}}b_{\mu,\lambda}w_{\lambda}.
\end{align*}
Hence, the family $\left(  h_{\lambda}\right)  _{\lambda\in \Par_n}$
expands\footnote{Here, we are using the terminology of Section
\ref{sect.STmat}.} in the family $\left(  w_{\lambda}\right)  _{\lambda
\in  \Par_{n}}$ through the matrix $\left(
b_{\mu,\lambda}\right)  _{\left(  \mu,\lambda\right)  \in
\Par_{n}\times  \Par_{n}}$. But this matrix
$\left(  b_{\mu,\lambda}\right)  _{\left(  \mu,\lambda\right)  \in
  \Par_{n}\times  \Par_{n}}$ is
easily seen to be unitriangular (indeed, $b_{\mu,\lambda}=0$ for any $\left(
\mu,\lambda\right)  \in  \Par_{n}$ which do not satisfy
$\lambda\underset{\operatorname*{ref}}{\leq}\mu$, and furthermore, every
$\lambda\in  \Par_{n}$ satisfies $b_{\lambda,\lambda}%
=1$) and therefore invertibly triangular. Hence, the family $\left(
h_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ expands
invertibly triangularly in the family $\left(  w_{\lambda}\right)
_{\lambda\in  \Par_{n}}$. Corollary
\ref{cor.STmat.expansion-tria-inv}(e) (applied to $\Lambda_{n}$,
$\Par_{n}$, $\left(  h_{\lambda}\right)  _{\lambda
\in  \Par_{n}}$ and $\left(  w_{\lambda}\right)
_{\lambda\in  \Par_{n}}$) thus shows that the family
$\left(  h_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$ if and only if the family
$\left(  w_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$. Hence, the family $\left(
w_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is a basis
of the $\kk$-module $\Lambda_{n}$ (since we know that the family
$\left(  h_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$).

Now, forget that we fixed $n$. We thus have shown that, for every
$n\in \NN$, the family $\left(  w_{\lambda}\right)  _{\lambda
\in  \Par_{n}}$ is a basis of the $\kk$-module
$\Lambda_{n}$. Hence, the disjoint union of the families $\left(  w_{\lambda
}\right)  _{\lambda\in  \Par_{n}}$ over all
$n\in \NN$ is a basis of the direct sum $\bigoplus_{n\in\NN}
\Lambda_{n}$. Since the former disjoint union is the family $\left(
w_{\lambda}\right)  _{\lambda\in \Par }$, whereas the latter
direct sum is $\bigoplus_{n\in\NN}\Lambda_{n}=\Lambda$, this result
rewrites as follows: The family $\left(  w_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\kk$-module $\Lambda$. This
solves Exercise~\ref{exe.witt}(d).

[\textit{Remark:} We could have slightly simplified this argument by using a
coarser partial order instead of $\underset{\operatorname*{ref}}{\leq}$.
Namely, we can define a binary relation $\underset{\operatorname*{len}}{\leq}$
on the set $\Par_{n}$ as follows: If $\lambda
\in \Par_{n}$ and $\mu\in \Par_{n}$,
then we let $\lambda\underset{\operatorname*{len}}{\leq}\mu$
if and only if either $\ell\left(  \lambda\right)  >\ell\left(  \mu\right)  $
or $\lambda=\mu$. Then, clearly, $\underset{\operatorname*{len}}{\leq}$ is the
smaller-or-equal relation of a partial order on the set $\Par_{n}$.
If we consider $\Par_{n}$ as a poset
equipped with this partial order, then the matrix $\left(  b_{\mu,\lambda
}\right)  _{\left(  \mu,\lambda\right)  \in  \Par_{n}%
\times  \Par_{n}}$ is still unitriangular, and thus our
argument above still works, but we save ourselves the trouble of proving that
$\underset{\operatorname*{ref}}{\leq}$ is a partial order.]

\textit{Second solution to Exercise~\ref{exe.witt}(d) (sketched).} From part
(c), we see that $h_{n}$ can be written as a polynomial in the $w_{1}%
,w_{2},w_{3},\ldots$ for each $n\in \NN$. Therefore, $w_{1},w_{2}%
,w_{3},\ldots$ generate $\Lambda$ as a $\kk$-algebra (because
$h_{1},h_{2},h_{3},\ldots$ generate $\Lambda$ as a $\kk$-algebra). In
other words, the family $\left(  w_{\lambda}\right)  _{\lambda\in
\Par}$ spans the $\kk$-module $\Lambda$%
\ \ \ \ \footnote{We have used Exercise~\ref{exe.alggen-via-span}(b)
(applied to $w_m$ and $w_\lambda$ instead of $v_m$ and $v_\lambda$)
here. Alternatively, this can also be seen using
(\ref{sol.witt.d.sol1.hmu=}).}.
Recall that $w_{\lambda}\in\Lambda_{\left\vert
\lambda\right\vert }$ for each $\lambda\in \Par $. In other words,
$w_{\lambda}$ is an element of $\Lambda_{\left\vert
\lambda\right\vert }$ for each partition $\lambda$.
Hence, Exercise~\ref{exe.Lambda.basis-vlam}
(applied to $v_\lambda = w_\lambda$) yields that
the family $\left(w_\lambda\right)_{\lambda \in \Par}$ is a graded
basis of the graded $\kk$-module $\Lambda$.
Thus, in particular, this family
$\left(w_\lambda\right)_{\lambda \in \Par}$ is a
basis of the $\kk$-module $\Lambda$.
This solves Exercise~\ref{exe.witt}(d) again.

(e) By the definition of the $w_{n}$, we have $H\left(  t\right)
=\prod_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}$. Hence,\footnote{We
will be using logarithms here, so prima facie our argument only works when
the base ring $\kk$ is a $\QQ$-algebra. However, it is easy to see that
our argument can easily be adapted to work in the general case as well.
For example, one can argue that even if the power series $\log t$ is not
defined if $\kk$ is not a $\QQ$-algebra, the notion of the logarithmic
derivative $\frac{d}{dt}\left(\log Q\right)$ is well-defined for every
$\kk$ whenever $Q \in \kk\left[\left[t\right]\right]$ is a power series
with constant term $1$ (for example, one could use the formula
$\frac{d}{dt}\left(\log Q\right)
= \frac{Q^\prime\left(t\right)}{Q\left(t\right)}$ as the \emph{definition}
of the logarithmic derivative) and still has the familiar property of
turning products into sums in this generality. See the solution to
Exercise~\ref{exe.E(t)-H(t)-P(t)-relation} for details about this.}
\begin{align*}
\dfrac{d}{dt}\left(  \log H\left(  t\right)  \right)   &  =\dfrac{d}
{dt}\left(  \log\prod_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}\right)
=\dfrac{d}{dt}\sum_{n=1}^{\infty}\log\left(  \left(  1-w_{n}t^{n}\right)
^{-1}\right) \\
&  =\sum_{n=1}^{\infty}\underbrace{\dfrac{d}{dt}\log\left(  \left(
1-w_{n}t^{n}\right)  ^{-1}\right)  }_{=\dfrac{nw_{n}t^{n-1}}{1-w_{n}t^{n}}
}=\sum_{n=1}^{\infty}\dfrac{nw_{n}t^{n-1}}{1-w_{n}t^{n}}.
\end{align*}
Compared with
\[
\dfrac{d}{dt}\left(  \log H\left(  t\right)  \right) 
= \dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }
= \sum_{m \geq 0} p_{m+1} t^m
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.E(t)-H(t)-P(t)-relation}}\right)  ,
\]
this yields
\[
\sum_{m\geq0}p_{m+1}t^{m}
= \sum_{n=1}^{\infty}\dfrac{nw_{n}t^{n-1}}{1-w_{n}t^{n}}.
\]
Multiplying this by $t$, we obtain
\begin{align*}
\sum_{m\geq0}p_{m+1}t^{m+1}  &  =\sum_{n=1}^{\infty}\underbrace{\dfrac
{nw_{n}t^{n}}{1-w_{n}t^{n}}}_{=n\sum_{k=1}^{\infty}w_{n}^{k}t^{nk}}=\sum
_{n=1}^{\infty}\sum_{k=1}^{\infty}nw_{n}^{k}t^{nk}\\
&  =\sum_{m=1}^{\infty}\sum_{d\mid m}dw_{d}^{m/d}t^{m}=\sum_{n=1}^{\infty}
\sum_{d\mid n}dw_{d}^{n/d}t^{n},
\end{align*}
so that $\sum_{n=1}^{\infty}\sum_{d\mid n}dw_{d}^{n/d}t^{n}=\sum_{m\geq 0}
p_{m+1}t^{m+1}=\sum_{n\geq1}p_{n}t^{n}$. Comparing coefficients yields the
claim of part (e).

(f) This is done in \cite{Doran}. For the sake of completeness:
First, let $n\geq k\geq2$. Then, every $i\in\left\{  2,3,...,k-1\right\}  $
satisfies
$f_{i,i} = \sum\limits_{\substack{\lambda\in\Par_i, \\ \min\lambda\geq i}}
w_{\lambda} = w_{\left( i\right) } = w_i$. Thus,
\begin{align*}
\sum_{i=2}^{k-1}\underbrace{f_{i,i}}_{=w_{i}}\underbrace{f_{n-i,i}}
_{=\sum\limits_{\substack{\lambda\in\Par_{n-i},\\\min\lambda\geq i}}w_{\lambda}}  &
=\sum_{i=2}^{k-1}\underbrace{w_{i}\sum\limits_{\substack{\lambda\in\Par_{n-i},\\\min
\lambda\geq i}}w_{\lambda}}
_{=\sum\limits_{\substack{\lambda\in\Par_{n-i},\\\min\lambda\geq i}}w_{\lambda} w_i
=\sum\limits_{\substack{\lambda\in\Par_n,\\\min\lambda=i}}w_{\lambda}}=\sum_{i=2}^{k-1}\sum\limits_{\substack{\lambda\in
\Par_n,\\\min\lambda=i}}w_{\lambda}\\
&  =\sum\limits_{\substack{\lambda\in\Par_n,\\2\leq\min\lambda<k}}w_{\lambda
}=\underbrace{\sum_{\lambda\in\Par_n}w_{\lambda}}_{\substack{=h_{n}\\\text{(by
part (c))}}}-\underbrace{\sum\limits_{\substack{\lambda\in\Par_n,\\\min
\lambda=1}}w_{\lambda}}_{=w_{1}\sum_{\lambda\in\Par_{n-1}}w_{\lambda}
}-\underbrace{\sum\limits_{\substack{\lambda\in\Par_n,\\\min\lambda\geq k}}w_{\lambda
}}_{=f_{n,k}}\\
&  =h_{n}-\underbrace{w_{1}}_{=h_{1}}\underbrace{\sum_{\lambda\in
\Par_{n-1}}w_{\lambda}}_{\substack{=h_{n-1}\\\text{(by part (c))}}
}-f_{n,k}\\
&  =\underbrace{h_{n}}_{=s_{\left(  n\right)  }}-h_{1}\underbrace{h_{n-1}
}_{=s_{\left(  n-1\right)  }}-f_{n,k}\\
&  =s_{\left(  n\right)  }-\underbrace{h_{1}s_{\left(  n-1\right)  }
}_{\substack{=s_{\left(  n\right)  }+s_{\left(  n-1,1\right)  }\\\text{(by the
Pieri rule)}}}-f_{n,k}\\
&  =-s_{\left(  n-1,1\right)  }-f_{n,k};
\end{align*}
in other words,
\[
-f_{n,k}=s_{\left(  n-1,1\right)  }+\sum_{i=2}^{k-1}f_{i,i}f_{n-i,i}.
\]
From this, we can conclude inductively that $-f_{n,k}$ is a sum of Schur
functions for every $n\in \NN$ and $k\geq 2$ (in fact, the trivial cases
with $n<2$ have to be taken as induction base). Since
$f_{n,n}
= \sum\limits_{\substack{ \lambda\in\Par_n, \\ \min\lambda\geq n }} w_{\lambda}
=w_{\left(n\right)  } = w_n$, this yields that $-w_{n}$ is a sum of
Schur functions for every $n\geq 2$.

(g) Every partition $\lambda$ can be uniquely written in the form
$\lambda=\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for some weak
composition $\left(  m_{1},m_{2},m_{3},\ldots\right)  $. Thus,
\begin{align*}
\sum_{\lambda\in\Par}w_{\lambda}\left(  \xx\right)
r_{\lambda}\left(  \yy\right)   &  =\sum\limits_{\substack{\left(  m_{1}
,m_{2},m_{3},\ldots\right)  \\\text{weak composition}}}\underbrace{w_{\left(
1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  }\left(  \xx\right)
}_{\substack{=\prod_{i\geq1}\left(  w_{i}\left(  \xx\right)  \right)
^{m_{i}}\\\text{(by the definition of }w_{\lambda}\text{)}}
}\underbrace{r_{\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  }\left(
\yy\right)  }_{\substack{=\prod_{i\geq1}h_{m_{i}}\left(  y_{1}
^{i},y_{2}^{i},y_{3}^{i},\ldots\right)  \\\text{(by the definition of
}r_{\lambda}\text{)}}}\\
&  =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\left(  \prod_{i\geq1}\left(  w_{i}\left(  \xx\right)
\right)  ^{m_{i}}\right)  \left(  \prod_{i\geq1}h_{m_{i}}\left(  y_{1}
^{i},y_{2}^{i},y_{3}^{i},\ldots\right)  \right) \\
&  =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\prod_{i\geq1} \left(  w_{i}\left(  \xx
\right)  \right)  ^{m_{i}}h_{m_{i}}\left(  y_{1}^{i},y_{2}^{i},y_{3}
^{i},\ldots\right)\\
&  =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\prod_{i\geq1} h_{m_{i}}\left(  y_{1}^{i},y_{2}^{i},y_{3}
^{i},\ldots\right) \left(  w_{i}\left(  \xx\right)  \right)  ^{m_{i}} \\
&  =\prod_{i\geq1}\underbrace{\sum_{m\in\NN}h_{m}\left(  y_{1}^{i}
,y_{2}^{i},y_{3}^{i},\ldots\right) \left(w_i\left(\xx\right)\right)^m }
_{\substack{
=\prod_{j\geq 1}\left(  1-y_{j}^{i}w_{i}\left(  \xx\right)  \right)  ^{-1}
\\ \text{(by \eqref{h-generating-function-no-HE}, upon substitution of }
\left(y_1,y_2,y_3,\ldots\right)\text{ and }w_i\left(\xx\right)\text{ for }
\xx\text{ and }t\text{)}}}\\
&  =\prod_{i\geq1}\prod_{j\geq1}\left(  1-y_{j}^{i}w_{i}\left(  \xx
\right)  \right)  ^{-1}=\prod_{j\geq1}\underbrace{\prod_{i\geq1}\left(
1-y_{j}^{i}w_{i}\left(  \xx\right)  \right)  ^{-1}}_{\substack{=\prod
_{n\geq1}\left(  1-y_{j}^{n}w_{n}\left(  \xx\right)  \right)
^{-1}\\=\prod_{n=1}^{\infty}\left(  1-w_{n}\left(  \xx\right)
y_{j}^{n}\right)  ^{-1}\\=H\left(  y_{j}\right)  \\\text{(since }\prod
_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}=H\left(  t\right)  \text{)}
}}\\
&  =\prod_{j\geq1}\underbrace{H\left(  y_{j}\right)  }_{=\prod_{i=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}}=\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
This proves (g).

(h) For every partition $\lambda$, both symmetric functions $w_\lambda$
and $r_\lambda$ are homogeneous of degree $\left|\lambda\right|$. (In fact,
for $w_\lambda$ this follows from Exercise~\ref{exe.witt}(c), whereas for
$r_\lambda$ this is easily derived from the definition.)

From Exercise~\ref{exe.witt}(g), we obtain
$\sum_{\lambda \in \Par} w_\lambda\left(\xx\right)
r_\lambda\left(\yy\right)
= \prod_{i,j=1}^{\infty} \left(1-x_iy_j\right)^{-1}
=\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy)$
(by \eqref{Cauchy-identity}).
Thus, we can apply
Exercise~\ref{exe.Lambda.bases}(a) to $u_\lambda = w_\lambda$ and
$v_\lambda = r_\lambda$. As a result, we obtain that
$\left( w_\lambda\right)_{\lambda\in\Par}$ and
$\left( r_\lambda\right)_{\lambda\in\Par}$ are
$\kk$-bases of $\Lambda$, and actually are dual bases with respect
to the Hall inner product on $\Lambda$. Thus we have solved
Exercise~\ref{exe.witt}(h), but also given another proof of
Exercise~\ref{exe.witt}(d) in the process (because we have shown
once again that $\left( w_\lambda\right)_{\lambda\in\Par}$ is a
$\kk$-basis of $\Lambda$).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.witt}}

\textit{Solution to Exercise~\ref{exe.witt}.}
(a) We shall prove a more general result:
\begin{statement}
\textit{Claim 1:} Let $A$ be \textbf{any}
commutative ring. Let $f\in A\left[  \left[  t\right]
\right]  $ be a power series with constant term $1$.
Then, there is a unique family $\left(
x_{n}\right)  _{n\geq 1}$ of elements of $A$ such that
\begin{equation}
f=\prod_{n=1}^{\infty}\left(  1-x_{n}t^{n}\right)  ^{-1}. \label{sol.witt.1}
\end{equation}
\end{statement}
[\textit{Proof of Claim 1:}
This family is constructed recursively: If $x_1, x_2, \ldots, x_{k-1}$ have
been determined (for some $k\geq 1$), then $x_k$ is obtained by comparing
coefficients before $t^k$ in the equation \eqref{sol.witt.1} (the coefficient
before $t^k$ on the left hand side is a known constant,
whereas the coefficient
before $t^k$ on the right hand side can be written in the form
$x_k + \left(  \text{some polynomial in }x_{1},x_{2},...,x_{k-1} \right)  $,
and thus the equality of these coefficients gives a linear equation in $x_k$
which can be uniquely solved for $x_k$). The family
$\left(x_n\right)_{n \geq 1}$ thus constructed satisfies \eqref{sol.witt.1}
(because the constant terms on both sides of \eqref{sol.witt.1} are $1$,
whereas for every $k \geq 1$, the coefficients before $t^k$ on both sides of
\eqref{sol.witt.1} are equal due to the construction of $x_k$).
Moreover, it is the only family that satisfies \eqref{sol.witt.1}
(since its construction was dictated by \eqref{sol.witt.1}).
Thus, Claim 1 is proven.]

Exercise~\ref{exe.witt}(a) follows by applying Claim 1 to $A = \Lambda$
and $f = H\left(t\right)$.

(b) Again, this generalizes: Let us say that a power series $f\in
A\left[  \left[  t\right]  \right]  $ over a graded commutative ring $A$ is
\emph{equigraded}\index{equigraded power series}
if, for every $n \in \NN$, the coefficient of $f$
before $t^{n}$ is homogeneous of degree $n$. Then, if $f\in A\left[  \left[
t\right]  \right]  $ is an equigraded power series with constant term $1$,
then the unique family
$\left(  x_{n}\right)  _{n\geq 1}$ satisfying \eqref{sol.witt.1} has the
property that $x_{n}$ is homogeneous of degree $n$ for every positive $n$.
This is rather easy to see by induction.

(c) By the definition of the $w_{n}$, we have $H\left(  t\right)
=\prod_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}$. Expanding and
comparing coefficients yields precisely $\sum_{\lambda\in\Par_n}w_{\lambda
}=h_{n}$.\ \ \ \ \footnote{Here are some more details of this argument.
From \eqref{h-generating-function},
we have $H\left(  t\right)  =\sum_{n\geq0}h_{n}\left(  {\mathbf{x}}\right)
t^{n}=\sum_{n\geq0}h_{n}t^{n}$. Thus,%
\begin{align}
\sum_{n\geq0}h_{n}t^{n}  & =H\left(  t\right)  =\prod_{n=1}^{\infty
}\underbrace{\left(  1-w_{n}t^{n}\right)  ^{-1}}_{=\sum_{m\in \NN %
}\left(  w_{n}t^{n}\right)  ^{m}}=\prod_{n=1}^{\infty}\sum_{m\in \NN %
}\underbrace{\left(  w_{n}t^{n}\right)  ^{m}}_{=w_{n}^{m}t^{nm}}\nonumber\\
& =\prod_{n=1}^{\infty}\sum_{m\in \NN }w_{n}^{m}t^{nm}=\sum\limits
_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\underbrace{\prod_{n=1}^{\infty}w_{n}^{m_{n}}t^{nm_{n}}%
}_{=\left(  \prod_{n=1}^{\infty}w_{n}^{m_{n}}\right)  t^{1m_{1}+2m_{2}%
+3m_{3}+\cdots}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right)
\nonumber\\
& =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\left(  \prod_{n=1}^{\infty}w_{n}^{m_{n}}\right)  t^{1m_{1}%
+2m_{2}+3m_{3}+\cdots}.\label{sol.witt.c.some-more-details.1}%
\end{align}
But every partition $\lambda$ can be uniquely written in the form
$\lambda=\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for some weak
composition $\left(  m_{1},m_{2},m_{3},\ldots\right)  $. Thus, we can
substitute $\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for $\lambda$
in the sum $\sum_{\lambda\in\Par}w_{\lambda}t^{\left\vert
\lambda\right\vert }$. As a result, we obtain%
\begin{align*}
\sum_{\lambda\in\Par}w_{\lambda}t^{\left\vert \lambda
\right\vert }  & =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)
\\\text{weak composition}}}\underbrace{w_{\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}%
}\cdots\right)  }}_{\substack{=w_{1}^{m_{1}}w_{2}^{m_{2}}w_{3}^{m_{3}}%
\cdots\\\text{(by the definition of }w_{\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}%
}\cdots\right)  }\text{)}}}\underbrace{t^{\left\vert \left(  1^{m_{1}}%
2^{m_{2}}3^{m_{3}}\cdots\right)  \right\vert }}_{\substack{=t^{1m_{1}%
+2m_{2}+3m_{3}+\cdots}\\\text{(since }\left\vert \left(  1^{m_{1}}2^{m_{2}%
}3^{m_{3}}\cdots\right)  \right\vert =1m_{1}+2m_{2}+3m_{3}+\cdots\text{)}}}\\
& =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\underbrace{\left(  w_{1}^{m_{1}}w_{2}^{m_{2}}w_{3}^{m_{3}%
}\cdots\right)  }_{=\prod_{n=1}^{\infty}w_{n}^{m_{n}}}t^{1m_{1}+2m_{2}%
+3m_{3}+\cdots}
 =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\left(  \prod_{n=1}^{\infty}w_{n}^{m_{n}}\right)  t^{1m_{1}%
+2m_{2}+3m_{3}+\cdots}.
\end{align*}
Compared with (\ref{sol.witt.c.some-more-details.1}), this yields%
\[
\sum_{n\geq0}h_{n}t^{n}=\sum_{\lambda\in\Par}w_{\lambda
}t^{\left\vert \lambda\right\vert }=\sum_{n\geq0}\underbrace{\sum\limits
_{\substack{\lambda\in\Par;\\\left\vert \lambda\right\vert
=n}}}_{=\sum_{\lambda\in\Par_{n}}}w_{\lambda
}\underbrace{t^{\left\vert \lambda\right\vert }}_{\substack{=t^{n}%
\\\text{(since }\left\vert \lambda\right\vert =n\text{)}}}=\sum_{n\geq0}%
\sum_{\lambda\in\Par_{n}}w_{\lambda}t^{n}.
\]
Comparing coefficients before $t^{n}$ in this equality of power series, we
conclude that $h_{n}=\sum_{\lambda\in\Par_{n}}w_{\lambda}$
for every $n\in\NN$, qed.}

(d) \textit{First solution to Exercise~\ref{exe.witt}(d) (sketched).} If
$\lambda^{\left(  1\right)  },\lambda^{\left(  2\right)  },\ldots
,\lambda^{\left(  \ell\right)  }$ are finitely many partitions, then we let
$\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup\cdots
\cup\lambda^{\left(  \ell\right)  }$ denote the partition obtained by sorting
all the parts of $\lambda^{\left(  1\right)  },\lambda^{\left(  2\right)
},...,\lambda^{\left(  \ell\right)  }$ in decreasing order. For instance,
$\left(  3,2,1\right)  \cup\left(  4,2\right)  \cup\left(  5,1\right)
=\left(  5,4,3,2,2,1,1\right)  $. Clearly, if $\lambda^{\left(  1\right)
},\lambda^{\left(  2\right)  },...,\lambda^{\left(  \ell\right)  }$ are
finitely many partitions, then%
\begin{equation}
w_{\lambda^{\left(  1\right)  }}w_{\lambda^{\left(  2\right)  }}\cdots
w_{\lambda^{\left(  \ell\right)  }}=w_{\lambda^{\left(  1\right)  }\cup
\lambda^{\left(  2\right)  }\cup\cdots\cup\lambda^{\left(  \ell\right)  }%
}\label{sol.witt.d.sol1.prod}%
\end{equation}
(due to the definition of $w_{\lambda}$ for a partition $\lambda$).

From (c), we know that $h_{n}=\sum_{\lambda\in\operatorname{Par}_{n}%
}w_{\lambda}$ for every $n\in \NN$. In other words,
\begin{equation}
h_{n}=\sum_{\lambda\vdash n}w_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in{\NN},\label{sol.witt.d.sol1.hn=}%
\end{equation}
where we are using the notation $\lambda\vdash n$ for $\lambda\in
\operatorname{Par}_{n}$.

Every partition $\mu=\left(  \mu_{1},\mu_{2},...,\mu_{\ell}\right)  $ with
$\ell=\ell\left(  \mu\right)  $ satisfies
\begin{align}
h_{\mu}  & =h_{\mu_{1}}h_{\mu_{2}}\cdots h_{\mu_{\ell}}=\prod_{i=1}^{\ell
}\underbrace{h_{\mu_{i}}}_{\substack{=\sum_{\lambda\vdash\left\vert \mu
_{i}\right\vert }w_{\lambda}\\\text{(by (\ref{sol.witt.d.sol1.hn=}), applied
to }n=\mu_{i}\text{)}}}\nonumber\\
& =\prod_{i=1}^{\ell}\left(  \sum_{\lambda\vdash\left\vert \mu_{i}\right\vert
}w_{\lambda}\right)  =\sum\limits_{\substack{\left(  \lambda^{\left(  1\right)
},\lambda^{\left(  2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)
\in\operatorname{Par}^{\ell};\\\lambda^{\left(  i\right)  }\vdash\mu_{i}\text{
for every }i}}\underbrace{w_{\lambda^{\left(  1\right)  }}w_{\lambda^{\left(
2\right)  }}\cdots w_{\lambda^{\left(  \ell\right)  }}}_{\substack{=w_{\lambda
^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup\cdots\cup
\lambda^{\left(  \ell\right)  }}\\\text{(by (\ref{sol.witt.d.sol1.prod}))}%
}}\nonumber\\
& =\sum\limits_{\substack{\left(  \lambda^{\left(  1\right)  },\lambda^{\left(
2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)  \in
\operatorname{Par}^{\ell};\\\lambda^{\left(  i\right)  }\vdash\mu_{i}\text{
for every }i}}w_{\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)
}\cup\cdots\cup\lambda^{\left(  \ell\right)  }}.\label{sol.witt.d.sol1.hmu=}%
\end{align}


Now, let us fix $n\in \NN$. We shall prove that $\left(  w_{\lambda
}\right)  _{\lambda\in  \Par_{n}}$ is a basis of the
$\kk$-module $\Lambda_{n}$. First of all, we know that this family
$\left(  w_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a family of elements of $\Lambda_{n}$ (since for each $\lambda\in
  \Par_{n}$, the symmetric function $w_{\lambda}$ is
homogeneous of degree $\left\vert \lambda\right\vert =n$).

Now, we define a binary relation $\underset{\operatorname*{ref}}{\leq}$ on the
set $\Par_{n}$ as follows: If $\lambda\in
  \Par_{n}$ and $\mu\in  \Par_{n}%
$, then we let $\lambda\underset{\operatorname*{ref}}{\leq}\mu$ if and only if
there exists a tuple $\left(  \lambda^{\left(  1\right)  },\lambda^{\left(
2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)  \in
\operatorname{Par}^{\ell}$ satisfying $\ell=\ell\left(  \mu\right)  $,
$\left(  \lambda^{\left(  i\right)  }\vdash\mu_{i}\text{ for every }i\right)
$ and $\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup
\cdots\cup\lambda^{\left(  \ell\right)  }=\lambda$. The intuitive meaning
behind this is the following: We have $\lambda\underset{\operatorname*{ref}%
}{\leq}\mu$ if we can obtain the partition $\lambda$ by splitting each part
$\mu_{i}$ of $\mu$ into several smaller parts\footnote{We allow the
``several smaller parts'' to be one single
part (namely, $\mu_{i}$).} (which are positive integers summing up to $\mu
_{i}$) and sorting the resulting list into decreasing order. For instance,
$\left(  5,3,2,2,2,1\right)  \underset{\operatorname*{ref}}{\leq}\left(
6,5,4\right)  $ (because the tuple $\left(  \lambda^{\left(  1\right)
},\lambda^{\left(  2\right)  },\lambda^{\left(  3\right)  }\right)  =\left(
\left(  3,2,1\right)  ,\left(  5\right)  ,\left(  2,2\right)  \right)  $
satisfies $\left(  3,2,1\right)  \vdash6$, $\left(  5\right)  \vdash5$ and
$\left(  2,2\right)  \vdash4$ and $\left(  3,2,1\right)  \cup\left(  5\right)
\cup\left(  2,2\right)  =\left(  5,3,2,2,2,1\right)  $).

It is easy to see that the relation $\underset{\operatorname*{ref}}{\leq}$ is
transitive, antisymmetric and reflexive\footnote{In proving these properties
(specifically, antisymmetry), it helps to observe the
following fact: If $\lambda\in  \Par_{n}$ and $\mu
\in  \Par_{n}$ satisfy $\lambda
\underset{\operatorname*{ref}}{\leq}\mu$, then either $\lambda=\mu$ or
$\ell\left(  \lambda\right)  >\ell\left(  \mu\right)  $.}. Hence,
$\underset{\operatorname*{ref}}{\leq}$ is the smaller-or-equal relation of a
partial order on the set $\Par_{n}$. Consider
$\Par_{n}$ as a poset, equipped with this partial
order. (This partial order is called the
\emph{refinement order}\index{refinement order on partitions} on
partitions\footnote{Caution: This order is \textbf{not} a restriction of the
refinement order on compositions.}.)

Now, for any two partitions $\lambda$ and $\mu$, let $b_{\mu,\lambda}$ denote
the number of tuples $\left(  \lambda^{\left(  1\right)  },\lambda^{\left(
2\right)  },\ldots,\lambda^{\left(  \ell\right)  }\right)  \in
\operatorname{Par}^{\ell}$ satisfying $\ell=\ell\left(  \mu\right)  $,
$\left(  \lambda^{\left(  i\right)  }\vdash\mu_{i}\text{ for every }i\right)
$ and $\lambda^{\left(  1\right)  }\cup\lambda^{\left(  2\right)  }\cup
\cdots\cup\lambda^{\left(  \ell\right)  }=\lambda$. Then, the formula
(\ref{sol.witt.d.sol1.hmu=}) rewrites as%
\[
h_{\mu}=\sum_{\lambda\in \Par }b_{\mu,\lambda}w_{\lambda}.
\]
Thus, for every $\mu\in  \Par_{n}$, we have%
\begin{align*}
h_{\mu}  & =\sum_{\lambda\in \Par }b_{\mu,\lambda}w_{\lambda
}=\underbrace{\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert
\lambda\right\vert =n}}}_{=\sum_{\lambda\in  \Par_{n}}%
}b_{\mu,\lambda}w_{\lambda}+\sum\limits_{\substack{\lambda\in \Par %
;\\\left\vert \lambda\right\vert \neq n}}\underbrace{b_{\mu,\lambda}%
}_{\substack{=0\\\text{(this easily follows}\\\text{from }\left\vert
\lambda\right\vert \neq n=\left\vert \mu\right\vert \text{)}}}w_{\lambda}\\
& =\sum_{\lambda\in  \Par_{n}}b_{\mu,\lambda}w_{\lambda
}+\underbrace{\sum\limits_{\substack{\lambda\in \Par ;\\\left\vert
\lambda\right\vert \neq n}}0w_{\lambda}}_{=0}=\sum_{\lambda\in
  \Par_{n}}b_{\mu,\lambda}w_{\lambda}.
\end{align*}
Hence, the family $\left(  h_{\lambda}\right)  _{\lambda\in \Par_n}$
expands\footnote{Here, we are using the terminology of Section
\ref{sect.STmat}.} in the family $\left(  w_{\lambda}\right)  _{\lambda
\in  \Par_{n}}$ through the matrix $\left(
b_{\mu,\lambda}\right)  _{\left(  \mu,\lambda\right)  \in
\Par_{n}\times  \Par_{n}}$. But this matrix
$\left(  b_{\mu,\lambda}\right)  _{\left(  \mu,\lambda\right)  \in
  \Par_{n}\times  \Par_{n}}$ is
easily seen to be unitriangular (indeed, $b_{\mu,\lambda}=0$ for any $\left(
\mu,\lambda\right)  \in  \Par_{n}$ which do not satisfy
$\lambda\underset{\operatorname*{ref}}{\leq}\mu$, and furthermore, every
$\lambda\in  \Par_{n}$ satisfies $b_{\lambda,\lambda}%
=1$) and therefore invertibly triangular. Hence, the family $\left(
h_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ expands
invertibly triangularly in the family $\left(  w_{\lambda}\right)
_{\lambda\in  \Par_{n}}$. Corollary
\ref{cor.STmat.expansion-tria-inv}(e) (applied to $\Lambda_{n}$,
$\Par_{n}$, $\left(  h_{\lambda}\right)  _{\lambda
\in  \Par_{n}}$ and $\left(  w_{\lambda}\right)
_{\lambda\in  \Par_{n}}$) thus shows that the family
$\left(  h_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$ if and only if the family
$\left(  w_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$. Hence, the family $\left(
w_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is a basis
of the $\kk$-module $\Lambda_{n}$ (since we know that the family
$\left(  h_{\lambda}\right)  _{\lambda\in  \Par_{n}}$ is
a basis of the $\kk$-module $\Lambda_{n}$).

Now, forget that we fixed $n$. We thus have shown that, for every
$n\in \NN$, the family $\left(  w_{\lambda}\right)  _{\lambda
\in  \Par_{n}}$ is a basis of the $\kk$-module
$\Lambda_{n}$. Hence, the disjoint union of the families $\left(  w_{\lambda
}\right)  _{\lambda\in  \Par_{n}}$ over all
$n\in \NN$ is a basis of the direct sum $\bigoplus_{n\in\NN %
}\Lambda_{n}$. Since the former disjoint union is the family $\left(
w_{\lambda}\right)  _{\lambda\in \Par }$, whereas the latter
direct sum is $\bigoplus_{n\in\NN}\Lambda_{n}=\Lambda$, this result
rewrites as follows: The family $\left(  w_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\kk$-module $\Lambda$. This
solves Exercise~\ref{exe.witt}(d).

[\textit{Remark:} We could have slightly simplified this argument by using a
coarser partial order instead of $\underset{\operatorname*{ref}}{\leq}$.
Namely, we can define a binary relation $\underset{\operatorname*{len}}{\leq}$
on the set $\Par_{n}$ as follows: If $\lambda
\in \Par_{n}$ and $\mu\in \Par_{n}$,
then we let $\lambda\underset{\operatorname*{len}}{\leq}\mu$
if and only if either $\ell\left(  \lambda\right)  >\ell\left(  \mu\right)  $
or $\lambda=\mu$. Then, clearly, $\underset{\operatorname*{len}}{\leq}$ is the
smaller-or-equal relation of a partial order on the set $\Par_{n}$.
If we consider $\Par_{n}$ as a poset
equipped with this partial order, then the matrix $\left(  b_{\mu,\lambda
}\right)  _{\left(  \mu,\lambda\right)  \in  \Par_{n}%
\times  \Par_{n}}$ is still unitriangular, and thus our
argument above still works, but we save ourselves the trouble of proving that
$\underset{\operatorname*{ref}}{\leq}$ is a partial order.]

\textit{Second solution to Exercise~\ref{exe.witt}(d) (sketched).} From part
(c), we see that $h_{n}$ can be written as a polynomial in the $w_{1}%
,w_{2},w_{3},\ldots$ for each $n\in \NN$. Therefore, $w_{1},w_{2}%
,w_{3},\ldots$ generate $\Lambda$ as a $\kk$-algebra (because
$h_{1},h_{2},h_{3},\ldots$ generate $\Lambda$ as a $\kk$-algebra). In
other words, the family $\left(  w_{\lambda}\right)  _{\lambda\in
\Par}$ spans the $\kk$-module $\Lambda$%
\ \ \ \ \footnote{We have used Exercise~\ref{exe.alggen-via-span}(b)
(applied to $w_m$ and $w_\lambda$ instead of $v_m$ and $v_\lambda$)
here. Alternatively, this can also be seen using
(\ref{sol.witt.d.sol1.hmu=}).}.
Recall that $w_{\lambda}\in\Lambda_{\left\vert
\lambda\right\vert }$ for each $\lambda\in \Par $. In other words,
$w_{\lambda}$ is an element of $\Lambda_{\left\vert
\lambda\right\vert }$ for each partition $\lambda$.
Hence, Exercise~\ref{exe.Lambda.basis-vlam}
(applied to $v_\lambda = w_\lambda$) yields that
the family $\left(w_\lambda\right)_{\lambda \in \Par}$ is a graded
basis of the graded $\kk$-module $\Lambda$.
Thus, in particular, this family
$\left(w_\lambda\right)_{\lambda \in \Par}$ is a
basis of the $\kk$-module $\Lambda$.
This solves Exercise~\ref{exe.witt}(d) again.

(e) By the definition of the $w_{n}$, we have $H\left(  t\right)
=\prod_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}$. Hence,\footnote{We
will be using logarithms here, so prima facie our argument only works when
the base ring $\kk$ is a $\QQ$-algebra. However, it is easy to see that
our argument can easily be adapted to work in the general case as well.
For example, one can argue that even if the power series $\log t$ is not
defined if $\kk$ is not a $\QQ$-algebra, the notion of the logarithmic
derivative $\frac{d}{dt}\left(\log Q\right)$ is well-defined for every
$\kk$ whenever $Q \in \kk\left[\left[t\right]\right]$ is a power series
with constant term $1$ (for example, one could use the formula
$\frac{d}{dt}\left(\log Q\right)
= \frac{Q^\prime\left(t\right)}{Q\left(t\right)}$ as the \emph{definition}
of the logarithmic derivative) and still has the familiar property of
turning products into sums in this generality. See the solution to
Exercise~\ref{exe.E(t)-H(t)-P(t)-relation} for details about this.}
\begin{align*}
\dfrac{d}{dt}\left(  \log H\left(  t\right)  \right)   &  =\dfrac{d}
{dt}\left(  \log\prod_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}\right)
=\dfrac{d}{dt}\sum_{n=1}^{\infty}\log\left(  \left(  1-w_{n}t^{n}\right)
^{-1}\right) \\
&  =\sum_{n=1}^{\infty}\underbrace{\dfrac{d}{dt}\log\left(  \left(
1-w_{n}t^{n}\right)  ^{-1}\right)  }_{=\dfrac{nw_{n}t^{n-1}}{1-w_{n}t^{n}}
}=\sum_{n=1}^{\infty}\dfrac{nw_{n}t^{n-1}}{1-w_{n}t^{n}}.
\end{align*}
Compared with
\[
\dfrac{d}{dt}\left(  \log H\left(  t\right)  \right) 
= \dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }
= \sum_{m \geq 0} p_{m+1} t^m
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.E(t)-H(t)-P(t)-relation}}\right)  ,
\]
this yields
\[
\sum_{m\geq0}p_{m+1}t^{m}
= \sum_{n=1}^{\infty}\dfrac{nw_{n}t^{n-1}}{1-w_{n}t^{n}}.
\]
Multiplying this by $t$, we obtain
\begin{align*}
\sum_{m\geq0}p_{m+1}t^{m+1}  &  =\sum_{n=1}^{\infty}\underbrace{\dfrac
{nw_{n}t^{n}}{1-w_{n}t^{n}}}_{=n\sum_{k=1}^{\infty}w_{n}^{k}t^{nk}}=\sum
_{n=1}^{\infty}\sum_{k=1}^{\infty}nw_{n}^{k}t^{nk}\\
&  =\sum_{m=1}^{\infty}\sum_{d\mid m}dw_{d}^{m/d}t^{m}=\sum_{n=1}^{\infty}
\sum_{d\mid n}dw_{d}^{n/d}t^{n},
\end{align*}
so that $\sum_{n=1}^{\infty}\sum_{d\mid n}dw_{d}^{n/d}t^{n}=\sum_{m\geq 0}
p_{m+1}t^{m+1}=\sum_{n\geq1}p_{n}t^{n}$. Comparing coefficients yields the
claim of part (e).

(f) This is done in \cite{Doran}. For the sake of completeness:
First, let $n\geq k\geq2$. Then, every $i\in\left\{  2,3,...,k-1\right\}  $
satisfies
$f_{i,i} = \sum\limits_{\substack{\lambda\in\Par_i, \\ \min\lambda\geq i}}
w_{\lambda} = w_{\left( i\right) } = w_i$. Thus,
\begin{align*}
\sum_{i=2}^{k-1}\underbrace{f_{i,i}}_{=w_{i}}\underbrace{f_{n-i,i}}
_{=\sum\limits_{\substack{\lambda\in\Par_{n-i},\\\min\lambda\geq i}}w_{\lambda}}  &
=\sum_{i=2}^{k-1}\underbrace{w_{i}\sum\limits_{\substack{\lambda\in\Par_{n-i},\\\min
\lambda\geq i}}w_{\lambda}}
_{=\sum\limits_{\substack{\lambda\in\Par_{n-i},\\\min\lambda\geq i}}w_{\lambda} w_i
=\sum\limits_{\substack{\lambda\in\Par_n,\\\min\lambda=i}}w_{\lambda}}=\sum_{i=2}^{k-1}\sum\limits_{\substack{\lambda\in
\Par_n,\\\min\lambda=i}}w_{\lambda}\\
&  =\sum\limits_{\substack{\lambda\in\Par_n,\\2\leq\min\lambda<k}}w_{\lambda
}=\underbrace{\sum_{\lambda\in\Par_n}w_{\lambda}}_{\substack{=h_{n}\\\text{(by
part (c))}}}-\underbrace{\sum\limits_{\substack{\lambda\in\Par_n,\\\min
\lambda=1}}w_{\lambda}}_{=w_{1}\sum_{\lambda\in\Par_{n-1}}w_{\lambda}
}-\underbrace{\sum\limits_{\substack{\lambda\in\Par_n,\\\min\lambda\geq k}}w_{\lambda
}}_{=f_{n,k}}\\
&  =h_{n}-\underbrace{w_{1}}_{=h_{1}}\underbrace{\sum_{\lambda\in
\Par_{n-1}}w_{\lambda}}_{\substack{=h_{n-1}\\\text{(by part (c))}}
}-f_{n,k}\\
&  =\underbrace{h_{n}}_{=s_{\left(  n\right)  }}-h_{1}\underbrace{h_{n-1}
}_{=s_{\left(  n-1\right)  }}-f_{n,k}\\
&  =s_{\left(  n\right)  }-\underbrace{h_{1}s_{\left(  n-1\right)  }
}_{\substack{=s_{\left(  n\right)  }+s_{\left(  n-1,1\right)  }\\\text{(by the
Pieri rule)}}}-f_{n,k}\\
&  =-s_{\left(  n-1,1\right)  }-f_{n,k};
\end{align*}
in other words,
\[
-f_{n,k}=s_{\left(  n-1,1\right)  }+\sum_{i=2}^{k-1}f_{i,i}f_{n-i,i}.
\]
From this, we can conclude inductively that $-f_{n,k}$ is a sum of Schur
functions for every $n\in \NN$ and $k\geq 2$ (in fact, the trivial cases
with $n<2$ have to be taken as induction base). Since
$f_{n,n}
= \sum\limits_{\substack{ \lambda\in\Par_n, \\ \min\lambda\geq n }} w_{\lambda}
=w_{\left(n\right)  } = w_n$, this yields that $-w_{n}$ is a sum of
Schur functions for every $n\geq 2$.

(g) Every partition $\lambda$ can be uniquely written in the form
$\lambda=\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  $ for some weak
composition $\left(  m_{1},m_{2},m_{3},\ldots\right)  $. Thus,
\begin{align*}
\sum_{\lambda\in\Par}w_{\lambda}\left(  \xx\right)
r_{\lambda}\left(  \yy\right)   &  =\sum\limits_{\substack{\left(  m_{1}
,m_{2},m_{3},\ldots\right)  \\\text{weak composition}}}\underbrace{w_{\left(
1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  }\left(  \xx\right)
}_{\substack{=\prod_{i\geq1}\left(  w_{i}\left(  \xx\right)  \right)
^{m_{i}}\\\text{(by the definition of }w_{\lambda}\text{)}}
}\underbrace{r_{\left(  1^{m_{1}}2^{m_{2}}3^{m_{3}}\cdots\right)  }\left(
\yy\right)  }_{\substack{=\prod_{i\geq1}h_{m_{i}}\left(  y_{1}
^{i},y_{2}^{i},y_{3}^{i},\ldots\right)  \\\text{(by the definition of
}r_{\lambda}\text{)}}}\\
&  =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\left(  \prod_{i\geq1}\left(  w_{i}\left(  \xx\right)
\right)  ^{m_{i}}\right)  \left(  \prod_{i\geq1}h_{m_{i}}\left(  y_{1}
^{i},y_{2}^{i},y_{3}^{i},\ldots\right)  \right) \\
&  =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\prod_{i\geq1} \left(  w_{i}\left(  \xx
\right)  \right)  ^{m_{i}}h_{m_{i}}\left(  y_{1}^{i},y_{2}^{i},y_{3}
^{i},\ldots\right)\\
&  =\sum\limits_{\substack{\left(  m_{1},m_{2},m_{3},\ldots\right)  \\\text{weak
composition}}}\prod_{i\geq1} h_{m_{i}}\left(  y_{1}^{i},y_{2}^{i},y_{3}
^{i},\ldots\right) \left(  w_{i}\left(  \xx\right)  \right)  ^{m_{i}} \\
&  =\prod_{i\geq1}\underbrace{\sum_{m\in\NN}h_{m}\left(  y_{1}^{i}
,y_{2}^{i},y_{3}^{i},\ldots\right) \left(w_i\left(\xx\right)\right)^m }
_{\substack{
=\prod_{j\geq 1}\left(  1-y_{j}^{i}w_{i}\left(  \xx\right)  \right)  ^{-1}
\\ \text{(by \eqref{h-generating-function-no-HE}, upon substitution of }
\left(y_1,y_2,y_3,\ldots\right)\text{ and }w_i\left(\xx\right)\text{ for }
\xx\text{ and }t\text{)}}}\\
&  =\prod_{i\geq1}\prod_{j\geq1}\left(  1-y_{j}^{i}w_{i}\left(  \xx
\right)  \right)  ^{-1}=\prod_{j\geq1}\underbrace{\prod_{i\geq1}\left(
1-y_{j}^{i}w_{i}\left(  \xx\right)  \right)  ^{-1}}_{\substack{=\prod
_{n\geq1}\left(  1-y_{j}^{n}w_{n}\left(  \xx\right)  \right)
^{-1}\\=\prod_{n=1}^{\infty}\left(  1-w_{n}\left(  \xx\right)
y_{j}^{n}\right)  ^{-1}\\=H\left(  y_{j}\right)  \\\text{(since }\prod
_{n=1}^{\infty}\left(  1-w_{n}t^{n}\right)  ^{-1}=H\left(  t\right)  \text{)}
}}\\
&  =\prod_{j\geq1}\underbrace{H\left(  y_{j}\right)  }_{=\prod_{i=1}^{\infty
}\left(  1-x_{i}y_{j}\right)  ^{-1}}=\prod_{i,j=1}^{\infty}\left(
1-x_{i}y_{j}\right)  ^{-1}.
\end{align*}
This proves (g).

(h) For every partition $\lambda$, both symmetric functions $w_\lambda$
and $r_\lambda$ are homogeneous of degree $\left|\lambda\right|$. (In fact,
for $w_\lambda$ this follows from Exercise~\ref{exe.witt}(c), whereas for
$r_\lambda$ this is easily derived from the definition.)

From Exercise~\ref{exe.witt}(g), we obtain
$\sum_{\lambda \in \Par} w_\lambda\left(\xx\right)
r_\lambda\left(\yy\right)
= \prod_{i,j=1}^{\infty} \left(1-x_iy_j\right)^{-1}
=\sum_{\lambda \in \Par} s_{\lambda}(\xx) s_\lambda(\yy)$
(by \eqref{Cauchy-identity}).
Thus, we can apply
Exercise~\ref{exe.Lambda.bases}(a) to $u_\lambda = w_\lambda$ and
$v_\lambda = r_\lambda$. As a result, we obtain that
$\left( w_\lambda\right)_{\lambda\in\Par}$ and
$\left( r_\lambda\right)_{\lambda\in\Par}$ are
$\kk$-bases of $\Lambda$, and actually are dual bases with respect
to the Hall inner product on $\Lambda$. Thus we have solved
Exercise~\ref{exe.witt}(h), but also given another proof of
Exercise~\ref{exe.witt}(d) in the process (because we have shown
once again that $\left( w_\lambda\right)_{\lambda\in\Par}$ is a
$\kk$-basis of $\Lambda$).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.maps-on-pn}}

\textit{Solution to Exercise~\ref{exe.Lambda.maps-on-pn}.} (a) Let
$f\in\Lambda$.

Recall the following fundamental fact from linear algebra: If $\kk$ is
a commutative ring, if $A$ is a $\kk$-module, if $\left(  \cdot
,\cdot\right)  :A\times A\rightarrow \kk$ is a symmetric $\kk %
$-bilinear form on $A$, and if $\left(  u_{\lambda}\right)  _{\lambda\in L}$
and $\left(  v_{\lambda}\right)  _{\lambda\in L}$ are two $\kk$-bases
of $A$ which are dual to each other with respect to the form $\left(
\cdot,\cdot\right)  $ (where $L$ is some indexing set), then every $a\in A$
satisfies
\begin{equation}
\label{sol.exe.Lambda.maps-on-pn.a.bases}
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}.
\end{equation}
We can apply this fact to $\kk = \QQ $, $A=\Lambda_{ \QQ }$,
$L= \Par $, $\left(  u_{\lambda}\right)  _{\lambda\in L}=\left(
p_{\lambda}\right)  _{\lambda\in \Par }$, $\left(  v_{\lambda
}\right)  _{\lambda\in L}=\left(  z_{\lambda}^{-1}p_{\lambda}\right)
_{\lambda\in \Par }$ and $a=f$ (because the bases $\left(
p_{\lambda}\right)  _{\lambda\in \Par }$ and $\left(  z_{\lambda
}^{-1}p_{\lambda}\right)  _{\lambda\in \Par }$ of $\Lambda
_{ \QQ }$ are dual to each other with respect to the Hall inner product
$\left(  \cdot,\cdot\right)  $, as Corollary \ref{dual-bases-h-m-p}(b) shows).
As a result, we obtain
\[
f=\sum_{\lambda\in \Par }\left(  p_{\lambda},f\right)  z_{\lambda
}^{-1}p_{\lambda}.
\]
Hence,
\begin{align*}
Z\left(  f\right)   &  =Z\left(  \sum_{\lambda\in \Par }\left(
p_{\lambda},f\right)  z_{\lambda}^{-1}p_{\lambda}\right)  =\sum_{\lambda
\in \Par }\left(  p_{\lambda},f\right)  z_{\lambda}^{-1}
\underbrace{Z\left(  p_{\lambda}\right)  }_{=z_{\lambda}p_{\lambda}}
=\sum_{\lambda\in \Par }\left(  p_{\lambda},f\right)  z_{\lambda
}^{-1}z_{\lambda}p_{\lambda}\\
&  =\sum_{\lambda\in \Par }\underbrace{\left(  p_{\lambda
},f\right)  }_{\substack{\in \ZZ \\\text{(since }f\in\Lambda\text{ and
}p_{\lambda}\in\Lambda\text{)}}}p_{\lambda}\in\sum_{\lambda\in
 \Par } \ZZ p_{\lambda}\subset\Lambda.
\end{align*}
Since we have proven this for every $f\in\Lambda$, we thus have shown that
$Z\left(  \Lambda\right)  \subset\Lambda$. This solves
Exercise~\ref{exe.Lambda.maps-on-pn}(a).

(b) \textit{First solution of Exercise~\ref{exe.Lambda.maps-on-pn}(b):}
Consider two variable sets $ \xx =\left(  x_{1},x_{2},x_{3},...\right)  $
and $ \yy =\left(  y_{1},y_{2},y_{3},...\right)  $. Let $ \xx \yy $
denote the variable set
\begin{align*}
\left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}}  &  =\left(  x_{1}y_{1},x_{1}y_{2},x_{1}%
y_{3},...,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  x_{2}y_{1},x_{2}y_{2},x_{2}y_{3},...,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  x_{3}y_{1},x_{3}y_{2},x_{3}y_{3},...,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  ...\right)  .
\end{align*}


Now, we claim that for every $f\in\Lambda_{ \QQ }$,
\begin{equation}
\text{there is a well-defined element }f\left(   \xx \yy \right)  :=f\left(
\left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}}\right)  \text{ of } \QQ \left[  \left[
 \xx , \yy \right]  \right]  .
\label{sol.Lambda.maps-on-pn.b.sol1.1}%
\end{equation}


This claim (\ref{sol.Lambda.maps-on-pn.b.sol1.1}) is not obvious! For example,
there is generally no well-defined element \newline $f\left(  \left(  x_{i}%
+y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}%
}\right)  $, because e.g. in the case of $f=e_{1}$ we would have $e_{1}\left(
\left(  x_{i}+y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}}\right)  =\sum_{\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}}\left(  x_{i}+y_{j}\right)  $, which is a sum
containing infinitely many $x_{1}$'s. So there is some subtlety which allows
us to make sense of $f\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,3,...\right\}  ^{2}}\right)  $ but not of $f\left(  \left(
x_{i}+y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}%
}\right)  $.

Here is a sketch of a \textit{proof of (\ref{sol.Lambda.maps-on-pn.b.sol1.1}%
):} Consider a new variable set $\mathbf{s}:=\left(  s_{i,j}\right)  _{\left(
i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}}$ whose variables are indexed
by \textbf{pairs} of positive integers. This variable set $\mathbf{s}$ is
still countably infinite, and so there is an isomorphism $\Lambda_{ \QQ %
}=\Lambda_{ \QQ }\left(   \xx \right)  \rightarrow\Lambda
_{ \QQ }\left(  \mathbf{s}\right)  $. On the other hand, we can consider
the ring $ \QQ \left[  \left[  \mathbf{s}\right]  \right]  $ of formal
power series in the variables from the set $\mathbf{s}$, and then we have
$\Lambda_{ \QQ }\left(  \mathbf{s}\right)  \subset \QQ \left[
\left[  \mathbf{s}\right]  \right]  $. It is easy to see that $g\left(
\left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}}\right)  $ is a well-defined element of $ \QQ %
\left[  \left[ \xx , \yy \right]  \right]  $ for every
$g\in \QQ \left[  \left[  \mathbf{s}\right]  \right]  $%
\ \ \ \ \footnote{\textit{Proof.} Let $g\in \QQ \left[  \left[
\mathbf{s}\right]  \right]  $. When $\left(  x_{i}y_{j}\right)  _{\left(
i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}}$ is substituted for $\left(
s_{i,j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}}$ in
the power series $g\in \QQ \left[  \left[  \mathbf{s}\right]  \right]  $,
every monomial in the variables $\mathbf{s}$ turns into a monomial in the
variables $\left(   \xx , \yy \right)  $, and \textbf{any given
monomial in }$\left(   \xx , \yy \right)  $ \textbf{can only be
obtained (this way) from finitely many monomials in }$\mathbf{s}$ (indeed, a
given monomial $ \xx ^{\alpha} \yy ^{\beta}$ in $\left(
 \xx , \yy \right)  $ can only be obtained from monomials
$\prod_{i,j=1,2,3,...}s_{i,j}^{\gamma_{i,j}}$ whose exponents $\gamma_{i,j}$
satisfy the equations%
\begin{align*}
\sum_{i=1}^{\infty}\gamma_{i,j}  &  =\beta_{j}\ \ \ \ \ \ \ \ \ \ \text{for
all }j\in\left\{  1,2,3,...\right\}  ;\\
\sum_{j=1}^{\infty}\gamma_{i,j}  &  =\alpha_{i}\ \ \ \ \ \ \ \ \ \ \text{for
all }i\in\left\{  1,2,3,...\right\}  ;
\end{align*}
but it is easy to see that these equations leave only finitely many
possibilities for the monomial $\prod_{i,j=1,2,3,...}s_{i,j}^{\gamma_{i,j}}$).
Hence, the substitution yields an infinite sum of monomials in which every
monomial occurs \textbf{only finitely often}; therefore, this sum converges.
This shows that $g\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,3,...\right\}  ^{2}}\right)  $ is a well-defined element of
$ \QQ \left[  \left[ \xx , \yy \right]  \right]  $.}. Hence,
for every $f\in\Lambda_{ \QQ }$, there is a well-defined element
$f\left(   \xx \yy \right)  :=f\left(  \left(  x_{i}y_{j}\right)  _{\left(
i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}}\right)  $ of $ \QQ %
\left[  \left[ \xx , \yy \right]  \right]  $ (because we can
regard $f$ as an element of $\Lambda_{ \QQ }\left(  \mathbf{s}\right)
\subset \QQ \left[  \left[  \mathbf{s}\right]  \right]  $ by means of the
isomorphism $\Lambda_{ \QQ }=\Lambda_{ \QQ }\left(   \xx %
\right)  \rightarrow\Lambda_{ \QQ }\left(  \mathbf{s}\right)  $). This
proves (\ref{sol.Lambda.maps-on-pn.b.sol1.1}).

Due to (\ref{sol.Lambda.maps-on-pn.b.sol1.1}), we can define a map%
\begin{align*}
\widetilde{\Delta}_{\times}:\Lambda_{ \QQ }  &  \rightarrow
 \QQ \left[  \left[ \xx , \yy \right]  \right]  ,\\
f  &  \mapsto f\left(   \xx \yy \right)  =f\left(  \left(  x_{i}%
y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}%
}\right)  .
\end{align*}
This map $\widetilde{\Delta}_{\times}$ is an evaluation map (in an appropriate
sense), and thus a $\QQ$-algebra homomorphism. It is also clear that
$\widetilde{\Delta}_{\times}\left(  \Lambda\right)  \subset \ZZ \left[
\left[ \xx , \yy \right]  \right]  $ (where $ \ZZ \left[
\left[ \xx , \yy \right]  \right]  $ is regarded as a subring of
$ \QQ \left[  \left[ \xx , \yy \right]  \right]  $ in the
obvious way).

Now, recall the $\QQ$-algebra isomorphism $\Lambda_{ \QQ }%
\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow R_{ \QQ }\left(
{ \xx },{ \yy }\right)  ^{\Symm_{(\infty)}\times
\Symm_{(\infty)}}$ constructed in
(\ref{symmetric-functions-in-two-sets-iso}) (applied to $\kk %
= \QQ $). This entails a $\QQ$-algebra injection $\Lambda
_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow
 \QQ \left[  \left[ \xx , \yy \right]  \right]  $ (since
$R_{ \QQ }\left(  { \xx },{ \yy }\right) ^{
\Symm_{(\infty)}\times\Symm_{(\infty)}}\subset R_{ \QQ }\left(
{ \xx },{ \yy }\right)  \subset \QQ \left[  \left[
 \xx , \yy \right]  \right]  $). Denote this injection by $\iota$.

We shall now show that
\begin{equation}
\iota\circ\Delta_{\times}=\widetilde{\Delta}_{\times}.
\label{sol.Lambda.maps-on-pn.b.sol1.2}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.b.sol1.2}):} Indeed,
(\ref{sol.Lambda.maps-on-pn.b.sol1.2}) is an equality between $ \QQ %
$-algebra homomorphisms (since $\widetilde{\Delta}_{\times}$, $\iota$ and
$\Delta_{\times}$ are $\QQ$-algebra homomorphisms), and thus, in order
to prove it, we only need to check that it holds on a generating set of the
$\QQ$-algebra $\Lambda_{ \QQ }$. We do this on the generating set
$\left(  p_{n}\right)  _{n\geq1}$, by noticing that every $n\geq1$ satisfies%
\begin{align*}
\left(  \iota\circ\Delta_{\times}\right)  \left(  p_{n}\right)   &
=\iota\left(  \underbrace{\Delta_{\times}\left(  p_{n}\right)  }%
_{=p_{n}\otimes p_{n}}\right)  =\iota\left(  p_{n}\otimes p_{n}\right)
=\underbrace{p_{n}\left(   \xx \right)  }_{=\sum_{i\geq1}x_{i}^{n}%
}\underbrace{p_{n}\left(   \yy \right)  }_{=\sum_{j\geq1}y_{j}^{n}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\iota\right) \\
&  =\left(  \sum_{i\geq1}x_{i}^{n}\right)  \left(  \sum_{j\geq1}y_{j}%
^{n}\right)  =\underbrace{\sum_{i\geq1}\sum_{j\geq1}}_{=\sum_{\left(
i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}}}\underbrace{x_{i}^{n}%
y_{j}^{n}}_{=\left(  x_{i}y_{j}\right)  ^{n}}=\sum_{\left(  i,j\right)
\in\left\{  1,2,3,...\right\}  ^{2}}\left(  x_{i}y_{j}\right)  ^{n}\\
&  =p_{n}\left(  \underbrace{\left(  x_{i}y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,3,...\right\}  ^{2}}}_{= \xx \yy }\right)  =p_{n}\left(
 \xx \yy \right)  =\widetilde{\Delta}_{\times}\left(  p_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{\Delta}_{\times}\left(
p_{n}\right)  =p_{n}\left(   \xx \yy \right)  \text{ (by the definition of
}\widetilde{\Delta}_{\times}\left(  p_{n}\right)  \text{)}\right)  .
\end{align*}
So (\ref{sol.Lambda.maps-on-pn.b.sol1.2}) is proven.

From (\ref{sol.Lambda.maps-on-pn.b.sol1.2}), we obtain $\underbrace{\left(
\iota\circ\Delta_{\times}\right)  }_{=\widetilde{\Delta}_{\times}}\left(
\Lambda\right)  =\widetilde{\Delta}_{\times}\left(  \Lambda\right)
\subset \ZZ \left[  \left[ \xx , \yy \right]  \right]  $, so
that $\iota\left(  \Delta_{\times}\left(  \Lambda\right)  \right)  =\left(
\iota\circ\Delta_{\times}\right)  \left(  \Lambda\right)  \subset
 \ZZ \left[  \left[ \xx , \yy \right]  \right]  $ and thus
$\Delta_{\times}\left(  \Lambda\right)  \subset\iota^{-1}\left(
 \ZZ \left[  \left[ \xx , \yy \right]  \right]  \right)  $.

But it so happens that $\iota^{-1}\left(   \ZZ \left[  \left[
 \xx , \yy \right]  \right]  \right)  =\Lambda\otimes_{ \ZZ %
}\Lambda$\ \ \ \ \footnote{\textit{Proof.} It is clear that $\iota\left(
\Lambda\otimes_{ \ZZ }\Lambda\right)  \subset \ZZ \left[  \left[
 \xx , \yy \right]  \right]  $, so that $\Lambda\otimes_{ \ZZ %
}\Lambda\subset\iota^{-1}\left(   \ZZ \left[  \left[   \xx %
, \yy \right]  \right]  \right)  $. We are now going to prove the reverse
inclusion $\iota^{-1}\left(   \ZZ \left[  \left[   \xx %
, \yy \right]  \right]  \right)  \subset\Lambda\otimes_{ \ZZ %
}\Lambda$.
\par
Indeed, let $p\in\iota^{-1}\left(   \ZZ \left[  \left[   \xx %
, \yy \right]  \right]  \right)  $. Then, $p$ is an element of
$\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }$ satisfying
$\iota\left(  p\right)  \in \ZZ \left[  \left[   \xx , \yy %
\right]  \right]  $. All coefficients of the power series $\iota\left(
p\right)  $ are integers (since $\iota\left(  p\right)  \in \ZZ \left[
\left[ \xx , \yy \right]  \right]  $).
\par
We can write $p$ in the form $p=\sum_{\left(  \lambda,\mu\right)
\in \Par \times \Par }\rho_{\lambda,\mu}m_{\lambda
}\otimes m_{\mu}$ with $\rho_{\lambda,\mu}$ being elements of $\QQ$
(because $\left(  m_{\lambda}\otimes m_{\mu}\right)  _{\left(  \lambda
,\mu\right)  \in \Par \times \Par }$ is a
$\QQ$-basis of $\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda
_{ \QQ }$ (since $\left(  m_{\lambda}\right)  _{\lambda\in
 \Par }$ is a $\QQ$-basis of $\Lambda_{ \QQ }$)).
Consider these elements $\rho_{\lambda,\mu}$. Since $p=\sum_{\left(
\lambda,\mu\right)  \in \Par \times \Par }%
\rho_{\lambda,\mu}m_{\lambda}\otimes m_{\mu}$, we have $\iota\left(  p\right)
=\sum_{\left(  \lambda,\mu\right)  \in \Par \times
 \Par }\rho_{\lambda,\mu}m_{\lambda}\left(   \xx \right)
m_{\mu}\left(   \yy \right)  $ (by the definition of $\iota$). Therefore,
for every $\left(  \alpha,\beta\right)  \in \Par \times
 \Par $, the rational number $\rho_{\alpha,\beta}$ is the
coefficient of the power series $\iota\left(  p\right)  $ before the monomial
$ \xx ^{\alpha} \yy ^{\beta}$ (since there is clearly only one term
in the sum $\sum_{\left(  \lambda,\mu\right)  \in \Par %
\times \Par }\rho_{\lambda,\mu}m_{\lambda}\left(   \xx %
\right)  m_{\mu}\left(   \yy \right)  $ contributing to this coefficient,
namely the term for $\left(  \lambda,\mu\right)  =\left(  \alpha,\beta\right)
$, and this term contributes $\rho_{\alpha,\beta}$). In particular, this
number $\rho_{\alpha,\beta}$ must be an integer (since all coefficients of the
power series $\iota\left(  p\right)  $ are integers). So we have shown that
$\rho_{\alpha,\beta}$ is an integer for every $\left(  \alpha,\beta\right)
\in \Par \times \Par $. In other words,
$\rho_{\lambda,\mu}$ is an integer for every $\left(  \lambda,\mu\right)
\in \Par \times \Par $. Now, $p=\sum_{\left(
\lambda,\mu\right)  \in \Par \times \Par %
}\underbrace{\rho_{\lambda,\mu}}_{\text{an integer}}m_{\lambda}\otimes m_{\mu
}\in\Lambda\otimes_{ \ZZ }\Lambda$.
\par
Forget now that we fixed $p$. We thus have shown that $p\in\Lambda
\otimes_{ \ZZ }\Lambda$ for every $p\in\iota^{-1}\left(   \ZZ %
\left[  \left[ \xx , \yy \right]  \right]  \right)  $. In other
words, $\iota^{-1}\left(   \ZZ \left[  \left[   \xx , \yy %
\right]  \right]  \right)  \subset\Lambda\otimes_{ \ZZ }\Lambda$.
Combined with $\Lambda\otimes_{ \ZZ }\Lambda\subset\iota^{-1}\left(
 \ZZ \left[  \left[ \xx , \yy \right]  \right]  \right)  $,
this yields $\iota^{-1}\left(   \ZZ \left[  \left[   \xx %
, \yy \right]  \right]  \right)  =\Lambda\otimes_{ \ZZ }\Lambda$,
qed.}. Hence, $\Delta_{\times}\left(  \Lambda\right)  \subset\iota^{-1}\left(
 \ZZ \left[  \left[ \xx , \yy \right]  \right]  \right)
=\Lambda\otimes_{ \ZZ }\Lambda$. This solves Exercise
\ref{exe.Lambda.maps-on-pn}(b).

\textit{Second solution of Exercise~\ref{exe.Lambda.maps-on-pn}(b):} The
following solution of Exercise~\ref{exe.Lambda.maps-on-pn}(b) is a variation
on the First one given above. It avoids the use of
(\ref{sol.Lambda.maps-on-pn.b.sol1.1}) in favor of working with finite
variable sets. (This has the advantage that we no longer need to bother about
technicalities; but more importantly, this approach will be very useful in
solving Exercise~\ref{exe.Lambda.maps-on-pn}(f) later on.)

Let $N\in \NN$ be arbitrary. We define a $\QQ$-linear map
\begin{align*}
\mathcal{E}_{N}:\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }
&  \rightarrow \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2}%
,...,y_{N}\right]  ,\\
f\otimes g  &  \mapsto f\left(  x_{1},x_{2},...,x_{N}\right)  g\left(
y_{1},y_{2},...,y_{N}\right)  .
\end{align*}
This is well-defined because $f\left(  x_{1},x_{2},...,x_{N}\right)
\in \QQ \left[  x_{1},x_{2},...,x_{N}\right]  $ and $g\left(  y_{1}%
,y_{2},...,y_{N}\right)  \in \QQ \left[  y_{1},y_{2},...,y_{N}\right]  $
are well-defined polynomials for every $f\in\Lambda_{ \QQ }$ and
$g\in\Lambda_{ \QQ }$ (by Exercise~\ref{exe.Lambda.subs.fin}) and depend
linearly on $f$ and $g$, respectively. It is easy to see that the map
$\mathcal{E}_{N}$ is a $\QQ$-algebra homomorphism.

Next, we define a map%
\begin{align*}
\mathcal{K}_{N}:\Lambda_{ \QQ }  &  \rightarrow \QQ \left[
x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  ,\\
f  &  \mapsto f\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,...,N\right\}  ^{2}}\right)  .
\end{align*}
Here, $f\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,...,N\right\}  ^{2}}\right)  $ is defined as follows: Let $\left(
u_{1},u_{2},...,u_{N^{2}}\right)  $ be a list of all $N^{2}$ elements of the
family $\left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,...,N\right\}  ^{2}}$ in any arbitrary order, and set $f\left(  \left(
x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,...,N\right\}  ^{2}%
}\right)  =f\left(  u_{1},u_{2},...,u_{N^{2}}\right)  $. (The result does not
depend on the order chosen, because $f$ is symmetric.)

Again, $\mathcal{K}_{N}$ is a $\QQ$-algebra homomorphism (since
$\mathcal{K}_{N}$ is an evaluation map in an appropriate sense).

We now claim that%
\begin{equation}
\mathcal{E}_{N}\circ\Delta_{\times}=\mathcal{K}_{N}.
\label{sol.Lambda.maps-on-pn.b.sol2.2}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.b.sol2.2}):} The equality
(\ref{sol.Lambda.maps-on-pn.b.sol2.2}) is an equality between $ \QQ %
$-algebra homomorphisms (since $\mathcal{K}_{N}$, $\mathcal{E}_{N}$ and
$\Delta_{\times}$ are $\QQ$-algebra homomorphisms), and thus, in order
to prove it, we only need to check that it holds on a generating set of the
$\QQ$-algebra $\Lambda_{ \QQ }$. We do this on the generating set
$\left(  p_{n}\right)  _{n\geq1}$, by noticing that every $n\geq1$ satisfies%
\begin{align*}
\left(  \mathcal{E}_{N}\circ\Delta_{\times}\right)  \left(  p_{n}\right)   &
=\mathcal{E}_{N}\left(  \underbrace{\Delta_{\times}\left(  p_{n}\right)
}_{=p_{n}\otimes p_{n}}\right)  =\mathcal{E}_{N}\left(  p_{n}\otimes
p_{n}\right) \\
&  =\underbrace{p_{n}\left(  x_{1},x_{2},...,x_{N}\right)  }_{=\sum_{i=1}%
^{N}x_{i}^{n}}\underbrace{p_{n}\left(  y_{1},y_{2},...,y_{N}\right)  }%
_{=\sum_{j=1}^{N}y_{j}^{n}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }\mathcal{E}_{N}\right) \\
&  =\left(  \sum_{i=1}^{N}x_{i}^{n}\right)  \left(  \sum_{j=1}^{N}y_{j}%
^{n}\right)  =\underbrace{\sum_{i=1}^{N}\sum_{j=1}^{N}}_{=\sum_{\left(
i,j\right)  \in\left\{  1,2,...,N\right\}  ^{2}}}\underbrace{x_{i}^{n}%
y_{j}^{n}}_{=\left(  x_{i}y_{j}\right)  ^{n}}=\sum_{\left(  i,j\right)
\in\left\{  1,2,...,N\right\}  ^{2}}\left(  x_{i}y_{j}\right)  ^{n}\\
&  =p_{n}\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,...,N\right\}  ^{2}}\right)  =\mathcal{K}_{N}\left(  p_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{K}_{N}\left(
p_{n}\right)  =p_{n}\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,...,N\right\}  ^{2}}\right)  \text{ (by the definition of
}\mathcal{K}_{N}\left(  p_{n}\right)  \text{)}\right)  .
\end{align*}
This proves (\ref{sol.Lambda.maps-on-pn.b.sol2.2}).

Now,
\[
\mathcal{E}_{N}\left(  \Delta_{\times}\left(  \Lambda\right)  \right)
=\left(  \underbrace{\mathcal{E}_{N}\circ\Delta_{\times}}%
_{\substack{=\mathcal{K}_{N}\\\text{(by (\ref{sol.Lambda.maps-on-pn.b.sol2.2}%
))}}}\right)  \left(  \Lambda\right)  =\mathcal{K}_{N}\left(  \Lambda\right)
\subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
\]
(the latter inclusion is evident from the definition of $\mathcal{K}_{N}$).
Hence, every $p\in\Delta_{\times}\left(  \Lambda\right)  $ satisfies%
\begin{equation}
\mathcal{E}_{N}\left(  p\right)  \in\mathcal{E}_{N}\left(  \Delta_{\times
}\left(  \Lambda\right)  \right)  = \ZZ \left[  x_{1},x_{2}%
,...,x_{N},y_{1},y_{2},...,y_{N}\right]  .
\label{sol.Lambda.maps-on-pn.b.sol2.3}%
\end{equation}


Now, forget that we fixed $N$. We thus have defined a $\QQ$-algebra
homomorphism $\mathcal{E}_{N}:\Lambda_{ \QQ }\otimes_{ \QQ }%
\Lambda_{ \QQ }\rightarrow \QQ \left[  x_{1},x_{2},...,x_{N}%
,y_{1},y_{2},...,y_{N}\right]  $ for every $N\in \NN$, and we have shown
that every $p\in\Delta_{\times}\left(  \Lambda\right)  $ satisfies
(\ref{sol.Lambda.maps-on-pn.b.sol2.3}) for every $N\in \NN$.

Now, we are going to show that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if some }p\in\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda
_{ \QQ }\text{ satisfies}\\
\left(  \mathcal{E}_{N}\left(  p\right)  \in \ZZ \left[  x_{1}%
,x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  \text{ for every }%
N\in \NN \right)  \text{,}\\
\text{then }p\in\Lambda\otimes_{ \ZZ }\Lambda
\end{array}
\right)  \label{sol.Lambda.maps-on-pn.b.sol2.reduction}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.b.sol2.reduction}):} Let
$p\in\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }$ be such
that
\begin{equation}
\left(  \mathcal{E}_{N}\left(  p\right)  \in \ZZ \left[  x_{1}%
,x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  \text{ for every }%
N\in \NN \right)  . \label{sol.Lambda.maps-on-pn.b.sol2.reduction.pf.1}%
\end{equation}


We can write $p$ in the form $p=\sum_{\left(  \lambda,\mu\right)
\in \Par \times \Par }\rho_{\lambda,\mu}m_{\lambda
}\otimes m_{\mu}$ with $\rho_{\lambda,\mu}$ being elements of $\QQ$
(because $\left(  m_{\lambda}\otimes m_{\mu}\right)  _{\left(  \lambda
,\mu\right)  \in \Par \times \Par }$ is a
$\QQ$-basis of $\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda
_{ \QQ }$ (since $\left(  m_{\lambda}\right)  _{\lambda\in
 \Par }$ is a $\QQ$-basis of $\Lambda_{ \QQ }$)).
Consider these elements $\rho_{\lambda,\mu}$.

Now, let $\left(  \alpha,\beta\right)  \in \Par \times
 \Par $ be arbitrary. Choose some $N\in \NN$ satisfying
$N\geq\ell\left(  \alpha\right)  $ and $N\geq\ell\left(  \beta\right)  $ (such
an $N$ clearly exists). Then, $\alpha=\left(  \alpha_{1},\alpha_{2}%
,...,\alpha_{N}\right)  $ and $\beta=\left(  \beta_{1},\beta_{2},...,\beta
_{N}\right)  $. Since $p=\sum_{\left(  \lambda,\mu\right)  \in
 \Par \times \Par }\rho_{\lambda,\mu}m_{\lambda
}\otimes m_{\mu}$, we have
\begin{equation}
\mathcal{E}_{N}\left(  p\right)  =\sum_{\left(  \lambda,\mu\right)
\in \Par \times \Par }\rho_{\lambda,\mu}m_{\lambda
}\left(  x_{1},x_{2},...,x_{N}\right)  m_{\mu}\left(  y_{1},y_{2}%
,...,y_{N}\right)  . \label{sol.Lambda.maps-on-pn.b.sol2.reduction.pf.2}%
\end{equation}
The only addend on the right hand side of this equality which has a nonzero
coefficient before $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{N}^{\alpha_{N}%
}y_{1}^{\beta_{1}}y_{2}^{\beta_{2}}...y_{N}^{\beta_{N}}$ is the addend for
$\left(  \lambda,\mu\right)  =\left(  \alpha,\beta\right)  $ (since $\alpha$
and $\beta$ are partitions, so the only partition $\lambda$ such that the
monomial $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{N}^{\alpha_{N}}$ appears
in $m_{\lambda}\left(  x_{1},x_{2},...,x_{N}\right)  $ is $\alpha$, and the
only partition $\mu$ such that the monomial $y_{1}^{\beta_{1}}y_{2}^{\beta
_{2}}...y_{N}^{\beta_{N}}$ appears in $m_{\mu}\left(  y_{1},y_{2}%
,...,y_{N}\right)  $ is $\beta$). This coefficient is $\rho_{\alpha,\beta}$.
Hence, (\ref{sol.Lambda.maps-on-pn.b.sol2.reduction.pf.2}) shows that the
coefficient before $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}...x_{N}^{\alpha_{N}%
}y_{1}^{\beta_{1}}y_{2}^{\beta_{2}}...y_{N}^{\beta_{N}}$ in the power series
$\mathcal{E}_{N}\left(  p\right)  $ is $\rho_{\alpha,\beta}$. But this
coefficient must be an integer (in fact,
(\ref{sol.Lambda.maps-on-pn.b.sol2.reduction.pf.1}) shows that every
coefficient of the power series $\mathcal{E}_{N}\left(  p\right)  $ is an
integer). Thus, $\rho_{\alpha,\beta}$ is an integer.

So we have shown that $\rho_{\alpha,\beta}$ is an integer for every $\left(
\alpha,\beta\right)  \in \Par \times \Par $. In
other words, $\rho_{\lambda,\mu}$ is an integer for every $\left(  \lambda
,\mu\right)  \in \Par \times \Par $. Now,
$p=\sum_{\left(  \lambda,\mu\right)  \in \Par \times
 \Par }\underbrace{\rho_{\lambda,\mu}}_{\text{an integer}%
}m_{\lambda}\otimes m_{\mu}\in\Lambda\otimes_{ \ZZ }\Lambda$. This proves
(\ref{sol.Lambda.maps-on-pn.b.sol2.reduction}).

Now, we are almost done. Let $p \in\Delta_{\times}\left(  \Lambda\right)  $.
We know that $\mathcal{E}_{N}\left(  p\right)  \in \ZZ \left[
x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  $ for every $N\in
 \NN$ (according to (\ref{sol.Lambda.maps-on-pn.b.sol2.3})). Hence,
(\ref{sol.Lambda.maps-on-pn.b.sol2.reduction}) yields that $p\in\Lambda
\otimes_{ \ZZ }\Lambda$. Since we have proven this for every $p
\in\Delta_{\times}\left(  \Lambda\right)  $, we thus obtain $\Delta_{\times
}\left(  \Lambda\right)  \subset\Lambda\otimes_{ \ZZ }\Lambda$. This
solves Exercise~\ref{exe.Lambda.maps-on-pn}(b).

\textit{Third solution of Exercise~\ref{exe.Lambda.maps-on-pn}(b):} Here is
another solution of Exercise~\ref{exe.Lambda.maps-on-pn}(b), which entirely
gets by without using substitutions. We are going to prove that%
\begin{equation}
\Delta_{\times}\left(  h_{n}\right)  =\sum_{\lambda\vdash n}s_{\lambda}\otimes
s_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \NN .
\label{sol.Lambda.maps-on-pn.b.sol3.goal}
\end{equation}
\footnote{The notation ``$\lambda \vdash n$'' here is a synonym for
``$\lambda \in \Par_n$''.}
Once this is proven, it will follow that $\Delta_{\times}\left(  h_{n}\right)
\in\Lambda\otimes_{ \ZZ }\Lambda$ for every $n\in \NN$, and
therefore
\begin{align*}
\Delta_{\times}\left(  \underbrace{h_{\lambda}}_{=h_{\lambda_{1}}%
h_{\lambda_{2}}h_{\lambda_{3}}...}\right)   &  =\Delta_{\times}\left(
h_{\lambda_{1}}h_{\lambda_{2}}h_{\lambda_{3}}...\right) \\
&  =\underbrace{\Delta_{\times}\left(  h_{\lambda_{1}}\right)  }_{\in
\Lambda\otimes_{ \ZZ }\Lambda}\cdot\underbrace{\Delta_{\times}\left(
h_{\lambda_{2}}\right)  }_{\in\Lambda\otimes_{ \ZZ }\Lambda}%
\cdot\underbrace{\Delta_{\times}\left(  h_{\lambda_{3}}\right)  }_{\in
\Lambda\otimes_{ \ZZ }\Lambda}\cdot...\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\Delta_{\times}\text{ is a } \QQ \text{-algebra
homomorphism}\right) \\
&  \in\left(  \Lambda\otimes_{ \ZZ }\Lambda\right)  \cdot\left(
\Lambda\otimes_{ \ZZ }\Lambda\right)  \cdot\left(  \Lambda\otimes
_{ \ZZ }\Lambda\right)  \cdot...\\
&  \subset\Lambda\otimes_{ \ZZ }\Lambda
\end{align*}
for every partition $\lambda$; and this will immediately yield that
$\Delta_{\times}\left(  \Lambda\right)  \subset\Lambda\otimes_{ \ZZ %
}\Lambda$ (because $\Delta_{\times}$ is $\ZZ$-linear, and because
$\left(  h_{\lambda}\right)  _{\lambda\in \Par }$ is a basis of
the $\ZZ$-module $\Lambda$), which will solve Exercise
\ref{exe.Lambda.maps-on-pn}(b). Hence, in order to solve Exercise
\ref{exe.Lambda.maps-on-pn}(b), it is enough to prove
(\ref{sol.Lambda.maps-on-pn.b.sol3.goal}).

We need to prove (\ref{sol.Lambda.maps-on-pn.b.sol3.goal}). In order to
do so, it is clearly enough to show that%
\begin{equation}
\sum_{n\geq0}\Delta_{\times}\left(  h_{n}\right)  t^{n}=\sum_{n\geq0}\left(
\sum_{\lambda\vdash n}s_{\lambda}\otimes s_{\lambda}\right)  t^{n}%
\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \Lambda_{\QQ}\otimes
_{\QQ}\Lambda_{\QQ}\right)  \left[  \left[  t\right]
\right]  \label{sol.Lambda.maps-on-pn.b.sol3.goal2}%
\end{equation}
(because comparing coefficients in (\ref{sol.Lambda.maps-on-pn.b.sol3.goal2})
immediately yields (\ref{sol.Lambda.maps-on-pn.b.sol3.goal})).

Recall that for any two $\QQ$-algebras $\mathcal{A}$ and $\mathcal{B}$,
every $\QQ$-algebra homomorphism $\varphi:\mathcal{A}\rightarrow
\mathcal{B}$ induces a continuous\footnote{The word ``continuous''
refers to the usual topologies on the power
series rings $\mathcal{A}\left[  \left[  t\right]  \right]  $ and
$\mathcal{B}\left[  \left[  t\right]  \right]  $.} $ \QQ \left[  \left[
t\right]  \right]  $-algebra homomorphism $\varphi\left[  \left[  t\right]
\right]  :\mathcal{A}\left[  \left[  t\right]  \right]  \rightarrow
\mathcal{B}\left[  \left[  t\right]  \right]  $ which is given by%
\[
\left(  \varphi\left[  \left[  t\right]  \right]  \right)  \left(  \sum
_{k\geq0}a_{k}t^{k}\right)  =\sum_{k\geq0}\varphi\left(  a_{k}\right)
t^{k}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{k}\right)  _{k\geq0}%
\in\mathcal{A}^{ \NN }.
\]
In particular, the $\QQ$-algebra homomorphism $\Delta_{\times}%
:\Lambda_{\QQ}\rightarrow\Lambda_{\QQ}\otimes_{{ \QQ }%
}\Lambda_{\QQ}$ induces a $ \QQ \left[  \left[  t\right]
\right]  $-algebra homomorphism $\Delta_{\times}\left[  \left[  t\right]
\right]  :\Lambda_{\QQ}\left[  \left[  t\right]  \right]
\rightarrow\left(  \Lambda_{\QQ}\otimes_{\QQ}\Lambda
_{\QQ}\right)  \left[  \left[  t\right]  \right]  $. Recall the power
series $H\left(  t\right)  \in\Lambda\left[  \left[  t\right]  \right]  $
defined in (\ref{h-generating-function}); it satisfies $H\left(  t\right)
=\sum_{n\geq0}h_{n}t^{n}$. Applying the map $\Delta_{\times}\left[  \left[
t\right]  \right]  $ to both sides of this equality, we obtain
\begin{equation}
\left(  \Delta_{\times}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \right)  =\left(  \Delta_{\times}\left[  \left[  t\right]
\right]  \right)  \left(  \sum_{n\geq0}h_{n}t^{n}\right)  =\sum_{n\geq0}%
\Delta_{\times}\left(  h_{n}\right)  t^{n}
\label{sol.Lambda.maps-on-pn.b.sol3.goal2.lhs}%
\end{equation}
(by the definition of $\Delta_{\times}\left[  \left[  t\right]  \right]  $).

On the other hand, recall the $\QQ$-algebra isomorphism $\Lambda
_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow
R_{ \QQ }\left(  { \xx },{ \yy }\right)  ^{
\Symm_{(\infty)}\times\Symm_{(\infty)}}$ constructed in
(\ref{symmetric-functions-in-two-sets-iso}) (applied to $\kk %
= \QQ $). This entails a $\QQ$-algebra injection $\Lambda
_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow
 \QQ \left[  \left[ \xx , \yy \right]  \right]  $ (since
$R_{ \QQ }\left(  { \xx },{ \yy }\right)  ^{
\Symm_{(\infty)}\times\Symm_{(\infty)}}\subset R_{ \QQ }\left(
{ \xx },{ \yy }\right)  \subset \QQ \left[  \left[
 \xx , \yy \right]  \right]  $). Denote this injection by $\iota$.
Then, the $\QQ$-algebra homomorphism $\iota:\Lambda_{ \QQ }%
\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow \QQ \left[  \left[
 \xx , \yy \right]  \right]  $ induces a $ \QQ \left[  \left[
t\right]  \right]  $-algebra homomorphism $\iota\left[  \left[  t\right]
\right]  :\left(  \Lambda_{\QQ}\otimes_{\QQ}\Lambda
_{\QQ}\right)  \left[  \left[  t\right]  \right]  \rightarrow
 \QQ \left[  \left[ \xx , \yy \right]  \right]  \left[
\left[  t\right]  \right]  $. This homomorphism $\iota\left[  \left[
t\right]  \right]  $ is injective (since $\iota$ is injective).

We need to prove (\ref{sol.Lambda.maps-on-pn.b.sol3.goal2}). For this, it is
enough to prove%
\begin{equation}
\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(  \sum_{n\geq
0}\Delta_{\times}\left(  h_{n}\right)  t^{n}\right)  =\left(  \iota\left[
\left[  t\right]  \right]  \right)  \left(  \sum_{n\geq0}\left(  \sum
_{\lambda\vdash n}s_{\lambda}\otimes s_{\lambda}\right)  t^{n}\right)
\label{sol.Lambda.maps-on-pn.b.sol3.goal3}%
\end{equation}
(because the injectivity of $\iota\left[  \left[  t\right]  \right]  $ ensures
that (\ref{sol.Lambda.maps-on-pn.b.sol3.goal3}) implies
(\ref{sol.Lambda.maps-on-pn.b.sol3.goal2})).

In $ \QQ \left[  \left[ \xx , \yy \right]  \right]  \left[
\left[  t\right]  \right]  $, we have%
\begin{align}
\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(  \sum_{n\geq
0}\left(  \sum_{\lambda\vdash n}s_{\lambda}\otimes s_{\lambda}\right)
t^{n}\right)   &  =\sum_{n\geq0}\underbrace{\iota\left(  \sum_{\lambda\vdash
n}s_{\lambda}\otimes s_{\lambda}\right)  }_{\substack{=\sum_{\lambda\vdash
n}s_{\lambda}\left(   \xx \right)  s_{\lambda}\left(   \yy \right)
\\\text{(by the definition of }\iota\text{)}}}t^{n}\nonumber\\
&  =\sum_{n\geq0}\sum_{\lambda\vdash n}s_{\lambda}\left(   \xx \right)
s_{\lambda}\left(   \yy \right)  t^{n}=\sum_{\lambda\in
 \Par }s_{\lambda}\left(   \xx \right)
\underbrace{s_{\lambda}\left(   \yy \right)  t^{\left\vert \lambda
\right\vert }}_{\substack{=s_{\lambda}\left(  ty_{1},ty_{2},ty_{3},...\right)
\\\text{(since }s_{\lambda}\text{ is homogeneous of degree }\left\vert
\lambda\right\vert \text{)}}}\nonumber\\
&  =\sum_{\lambda\in \Par }s_{\lambda}\left(   \xx \right)
s_{\lambda}\left(  ty_{1},ty_{2},ty_{3},...\right)  =\prod_{i,j=1}^{\infty
}\left(  1-x_{i}\cdot ty_{j}\right)  ^{-1}%
\label{sol.Lambda.maps-on-pn.b.sol3.3}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since (\ref{Cauchy-identity}) (applied to }\left(  ty_{1},ty_{2}%
,ty_{3},...\right)  \text{ instead of } \yy \text{)}\\
\text{yields }\prod_{i,j=1}^{\infty}\left(  1-x_{i}\cdot ty_{j}\right)
^{-1}=\sum_{\lambda\in \Par }s_{\lambda}\left(   \xx %
\right)  s_{\lambda}\left(  ty_{1},ty_{2},ty_{3},...\right)
\end{array}
\right)  .\nonumber
\end{align}


Meanwhile, exponentiating both sides of the equality (\ref{H(t)-logarithm})
yields
\[
H\left(  t\right)  =\exp\left(  \sum_{m=1}^{\infty}\frac{1}{m}p_{m}\left(
 \xx \right)  t^{m}\right)  ,
\]
so that%
\begin{align*}
\left(  \Delta_{\times}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \right)   &  =\left(  \Delta_{\times}\left[  \left[
t\right]  \right]  \right)  \left(  \exp\left(  \sum_{m=1}^{\infty}\frac{1}%
{m}p_{m}\left(   \xx \right)  t^{m}\right)  \right)  =\exp\left(
\underbrace{\left(  \Delta_{\times}\left[  \left[  t\right]  \right]  \right)
\left(  \sum_{m=1}^{\infty}\frac{1}{m}p_{m}\left(   \xx \right)
t^{m}\right)  }_{=\sum_{m=1}^{\infty}\Delta_{\times}\left(  \frac{1}{m}%
p_{m}\left(   \xx \right)  \right)  t^{m}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\Delta_{\times}\left[  \left[  t\right]  \right]  \text{ is a
continuous } \QQ \text{-algebra homomorphism,}\\
\text{and thus commutes with }\exp
\end{array}
\right) \\
&  =\exp\left(  \sum_{m=1}^{\infty}\underbrace{\Delta_{\times}\left(  \frac
{1}{m}p_{m}\left(   \xx \right)  \right)  }_{=\dfrac{1}{m}\Delta_{\times
}\left(  p_{m}\left(   \xx \right)  \right)  }t^{m}\right)  =\exp\left(
\sum_{m=1}^{\infty}\frac{1}{m}\underbrace{\Delta_{\times}\left(  p_{m}\left(
 \xx \right)  \right)  }_{\substack{=\Delta_{\times}\left(  p_{m}\right)
=p_{m}\otimes p_{m}\\\text{(by the definition of }\Delta_{\times}\text{)}%
}}t^{m}\right) \\
&  =\exp\left(  \sum_{m=1}^{\infty}\frac{1}{m}p_{m}\otimes p_{m}t^{m}\right)
.
\end{align*}
Compared with (\ref{sol.Lambda.maps-on-pn.b.sol3.goal2.lhs}), this yields%
\[
\sum_{n\geq0}\Delta_{\times}\left(  h_{n}\right)  t^{n}=\exp\left(  \sum
_{m=1}^{\infty}\frac{1}{m}p_{m}\otimes p_{m}t^{m}\right)  .
\]
Applying the map $\iota\left[  \left[  t\right]  \right]  $ to both sides of
this equality, we obtain%
\begin{align}
\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(  \sum_{n\geq
0}\Delta_{\times}\left(  h_{n}\right)  t^{n}\right)   &  =\left(  \iota\left[
\left[  t\right]  \right]  \right)  \left(  \exp\left(  \sum_{m=1}^{\infty
}\frac{1}{m}p_{m}\otimes p_{m}t^{m}\right)  \right)  =\exp\left(
\underbrace{\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(
\sum_{m=1}^{\infty}\frac{1}{m}p_{m}\otimes p_{m}t^{m}\right)  }_{=\sum
_{m=1}^{\infty}\iota\left(  \frac{1}{m}p_{m}\otimes p_{m}\right)  t^{m}%
}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\iota\left[  \left[  t\right]  \right]  \text{ is a continuous
} \QQ \text{-algebra homomorphism,}\\
\text{and thus commutes with }\exp
\end{array}
\right) \nonumber\\
&  =\exp\left(  \sum_{m=1}^{\infty}\underbrace{\iota\left(  \frac{1}{m}%
p_{m}\otimes p_{m}\right)  }_{\substack{=\dfrac{1}{m}p_{m}\left(
 \xx \right)  p_{m}\left(   \yy \right)  \\\text{(by the definition
of }\iota\text{)}}}t^{m}\right)  =\exp\left(  \sum_{m=1}^{\infty}\dfrac{1}%
{m}p_{m}\left(   \xx \right)  \underbrace{p_{m}\left(   \yy \right)
t^{m}}_{\substack{=p_{m}\left(  ty_{1},ty_{2},ty_{3},...\right)
\\\text{(since }p_{m}\text{ is homogeneous of}\\\text{degree }m\text{)}%
}}\right) \nonumber\\
&  =\exp\left(  \sum_{m=1}^{\infty}\dfrac{1}{m}p_{m}\left(   \xx \right)
p_{m}\left(  ty_{1},ty_{2},ty_{3},...\right)  \right)  .
\label{sol.Lambda.maps-on-pn.b.sol3.5}%
\end{align}


But exponentiating both sides of the equality
(\ref{eq.cauchy-via-simple-power-sums}) yields%
\[
\prod_{i,j=1}^{\infty}\left(  1-x_{i}y_{j}\right)  ^{-1}=\exp\left(
\sum_{m=1}^{\infty}\dfrac{1}{m}p_{m}\left(   \xx \right)  p_{m}\left(
 \yy \right)  \right)  .
\]
Substituting $\left(  ty_{1},ty_{2},ty_{3},...\right)  $ for $\yy$ in
this equality, we obtain%
\[
\prod_{i,j=1}^{\infty}\left(  1-x_{i}\cdot ty_{j}\right)  ^{-1}=\exp\left(
\sum_{m=1}^{\infty}\dfrac{1}{m}p_{m}\left(   \xx \right)  p_{m}\left(
ty_{1},ty_{2},ty_{3},...\right)  \right)  .
\]
Compared with (\ref{sol.Lambda.maps-on-pn.b.sol3.5}), this yields
\[
\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(  \sum_{n\geq
0}\Delta_{\times}\left(  h_{n}\right)  t^{n}\right)  =\prod_{i,j=1}^{\infty
}\left(  1-x_{i}\cdot ty_{j}\right)  ^{-1}.
\]
Compared with (\ref{sol.Lambda.maps-on-pn.b.sol3.3}), this yields
\[
\left(  \iota\left[  \left[  t\right]  \right]  \right)  \left(  \sum_{n\geq
0}\Delta_{\times}\left(  h_{n}\right)  t^{n}\right)  =\left(  \iota\left[
\left[  t\right]  \right]  \right)  \left(  \sum_{n\geq0}\left(  \sum
_{\lambda\vdash n}s_{\lambda}\otimes s_{\lambda}\right)  t^{n}\right)  .
\]
This proves (\ref{sol.Lambda.maps-on-pn.b.sol3.goal3}). Thus, Exercise
\ref{exe.Lambda.maps-on-pn}(b) is solved.

\textit{Remark:} The three solutions we gave for Exercise
\ref{exe.Lambda.maps-on-pn}(b) are not that different. The Second solution is
a variation on the First solution which trades the use of a substitution of
infinitely many variables (with the technical troubles that come along with
it) for the inconvenience of having to consider a
``sufficiently high $N\in \NN$''. The Third solution looks
like a different beast, but its main idea -- the equality
(\ref{sol.Lambda.maps-on-pn.b.sol3.goal}) -- is really just an afterthought of
the First solution. Indeed, knowing the equality
(\ref{sol.Lambda.maps-on-pn.b.sol1.2}) in the First solution, we can easily
prove (\ref{sol.Lambda.maps-on-pn.b.sol3.goal}) as follows: Using the
notations of the First solution, we have%
\begin{align*}
\sum_{n\geq0}h_{n}\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,3,...\right\}  ^{2}}\right)  t^{n}  &  =\prod_{i,j=1}^{\infty
}\left(  1-tx_{i}y_{j}\right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{h-generating-function-no-HE}), evaluated on the variable set }
\xx \yy \right) \\
&  =\sum_{\lambda\in \Par }t^{\left\vert \lambda\right\vert
}s_{\lambda}\left(  { \xx }\right)  s_{\lambda}\left(   \yy \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{Cauchy-identity-with-t})}\right) \\
&  =\sum_{n\in \NN }\left(  \sum_{\lambda\vdash n}s_{\lambda}\left(
 \xx \right)  s_{\lambda}\left(   \yy \right)  \right)  t^{n}%
\end{align*}
in $ \QQ \left[  \left[ \xx , \yy \right]  \right]  \left[
\left[  t\right]  \right]  $. Comparing coefficients before $t^{n}$ in this
equality, we obtain%
\[
h_{n}\left(  \left(  x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}}\right)  =\sum_{\lambda\vdash n}s_{\lambda}\left(
 \xx \right)  s_{\lambda}\left(   \yy \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \NN .
\]
Thus, for every $n\in \NN$, we have%
\begin{align*}
\iota\left(  \Delta_{\times}\left(  h_{n}\right)  \right)   &
=\underbrace{\left(  \iota\circ\Delta_{\times}\right)  }%
_{\substack{=\widetilde{\Delta}_{\times}\\\text{(by
(\ref{sol.Lambda.maps-on-pn.b.sol1.2}))}}}\left(  h_{n}\right)
=\widetilde{\Delta}_{\times}\left(  h_{n}\right)  =h_{n}\left(  \left(
x_{i}y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}%
}\right)  =\sum_{\lambda\vdash n}s_{\lambda}\left(   \xx \right)
s_{\lambda}\left(   \yy \right) \\
&  =\iota\left(  \sum_{\lambda\vdash n}s_{\lambda}\otimes s_{\lambda}\right)
,
\end{align*}
which (by the injectivity of $\iota$) yields $\Delta_{\times}\left(
h_{n}\right)  =\sum_{\lambda\vdash n}s_{\lambda}\otimes s_{\lambda}$. Thus,
(\ref{sol.Lambda.maps-on-pn.b.sol3.goal}) is proven again.

In a similar vein, one can show that%
\[
\Delta_{\times}\left(  e_{n}\right)  =\sum_{\lambda\vdash n}s_{\lambda}\otimes
s_{\lambda^{t}}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \NN .
\]
(One would need to use the dual Cauchy identity, i.e., Exercise
\ref{exe.cauchy.dual}(a), instead of (\ref{Cauchy-identity-with-t}) this time.)

(c) We are going to show that
\begin{equation}
\epsilon_{r}\left(  h_{n}\right)  =\left(  -1\right)  ^{n}\dbinom{-r}%
{n}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \NN .
\label{sol.Lambda.maps-on-pn.c.goal}%
\end{equation}
This will yield the statement of Exercise~\ref{exe.Lambda.maps-on-pn}(c) in
the same way as (\ref{sol.Lambda.maps-on-pn.b.sol3.goal}) yielded the
statement of Exercise~\ref{exe.Lambda.maps-on-pn}(b) in the Third solution of
Exercise~\ref{exe.Lambda.maps-on-pn}(b) above. Hence, we only need to prove
(\ref{sol.Lambda.maps-on-pn.c.goal}) in order to be done with Exercise
\ref{exe.Lambda.maps-on-pn}(c).

Recall that for any two $\QQ$-algebras $\mathcal{A}$ and $\mathcal{B}$,
every $\QQ$-algebra homomorphism $\varphi:\mathcal{A}\rightarrow
\mathcal{B}$ induces a continuous\footnote{The word ``continuous''
refers to the usual topologies on the power
series rings $\mathcal{A}\left[  \left[  t\right]  \right]  $ and
$\mathcal{B}\left[  \left[  t\right]  \right]  $.} $ \QQ \left[  \left[
t\right]  \right]  $-algebra homomorphism $\varphi\left[  \left[  t\right]
\right]  :\mathcal{A}\left[  \left[  t\right]  \right]  \rightarrow
\mathcal{B}\left[  \left[  t\right]  \right]  $ which is given by%
\[
\left(  \varphi\left[  \left[  t\right]  \right]  \right)  \left(  \sum
_{k\geq0}a_{k}t^{k}\right)  =\sum_{k\geq0}\varphi\left(  a_{k}\right)
t^{k}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{k}\right)  _{k\geq0}%
\in\mathcal{A}^{ \NN }.
\]
Hence, the $\QQ$-algebra homomorphism $\epsilon_{r}:\Lambda
_{ \QQ }\rightarrow \QQ $ induces a continuous $ \QQ \left[
\left[  t\right]  \right]  $-algebra homomorphism $\epsilon_{r}\left[  \left[
t\right]  \right]  :\Lambda_{ \QQ }\left[  \left[  t\right]  \right]
\rightarrow \QQ \left[  \left[  t\right]  \right]  $.

Recall the power series $H\left(  t\right)  \in\Lambda\left[  \left[
t\right]  \right]  $ defined in (\ref{h-generating-function}); it satisfies
$H\left(  t\right)  =\sum_{n\geq0}h_{n}t^{n}$. Applying the homomorphism
$\epsilon_{r}\left[  \left[  t\right]  \right]  $ to both sides of this
equality, we obtain%
\begin{equation}
\left(  \epsilon_{r}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \right)  =\left(  \epsilon_{r}\left[  \left[  t\right]
\right]  \right)  \left(  \sum_{n\geq0}h_{n}t^{n}\right)  =\sum_{n\geq
0}\epsilon_{r}\left(  h_{n}\right)  t^{n}. \label{sol.Lambda.maps-on-pn.c.1}%
\end{equation}
On the other hand, exponentiating both sides of the equality
(\ref{H(t)-logarithm}) yields
\[
H\left(  t\right)  =\exp\left(  \sum_{m=1}^{\infty}\frac{1}{m}p_{m}\left(
 \xx \right)  t^{m}\right)  .
\]
Applying the map $\epsilon_{r}\left[  \left[  t\right]  \right]  $ to both
sides of this equality, we obtain%
\begin{align*}
\left(  \epsilon_{r}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \right)   &  =\left(  \epsilon_{r}\left[  \left[  t\right]
\right]  \right)  \left(  \exp\left(  \sum_{m=1}^{\infty}\frac{1}{m}%
p_{m}\left(   \xx \right)  t^{m}\right)  \right)  =\exp\left(
\underbrace{\left(  \epsilon_{r}\left[  \left[  t\right]  \right]  \right)
\left(  \sum_{m=1}^{\infty}\frac{1}{m}p_{m}\left(   \xx \right)
t^{m}\right)  }_{=\sum_{m=1}^{\infty}\epsilon_{r}\left(  \frac{1}{m}%
p_{m}\left(   \xx \right)  \right)  t^{m}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\epsilon_{r}\left[  \left[  t\right]  \right]  \text{ is a
continuous } \QQ \text{-algebra homomorphism,}\\
\text{and thus commutes with }\exp
\end{array}
\right) \\
&  =\exp\left(  \sum_{m=1}^{\infty}\underbrace{\epsilon_{r}\left(  \frac{1}%
{m}p_{m}\left(   \xx \right)  \right)  }_{=\epsilon_{r}\left(  \dfrac
{1}{m}p_{m}\right)  =\dfrac{1}{m}\epsilon_{r}\left(  p_{m}\right)  }%
t^{m}\right)  =\exp\left(  \sum_{m=1}^{\infty}\dfrac{1}{m}\underbrace{\epsilon
_{r}\left(  p_{m}\right)  }_{\substack{=r\\\text{(by the definition of
}\epsilon_{r}\text{)}}}t^{m}\right) \\
&  =\exp\left(  \sum_{m=1}^{\infty}\dfrac{1}{m}rt^{m}\right)  =\exp\left(
r\cdot\underbrace{\sum_{m=1}^{\infty}\dfrac{1}{m}t^{m}}_{=-\log\left(
1-t\right)  }\right)  =\exp\left(  r\cdot\left(  -\log\left(  1-t\right)
\right)  \right) \\
&  =\left(  \underbrace{\exp\left(  -\log\left(  1-t\right)  \right)
}_{=\left(  1-t\right)  ^{-1}}\right)  ^{r}=\left(  \left(  1-t\right)
^{-1}\right)  ^{r}=\left(  1-t\right)  ^{-r}=\sum_{n\geq0}\left(  -1\right)
^{n}\dbinom{-r}{n}t^{n}%
\end{align*}
(by Newton's binomial formula). Comparing this with
(\ref{sol.Lambda.maps-on-pn.c.1}), we obtain%
\[
\sum_{n\geq0}\epsilon_{r}\left(  h_{n}\right)  t^{n}=\sum_{n\geq0}\left(
-1\right)  ^{n}\dbinom{-r}{n}t^{n}.
\]
Comparing coefficients in this equality of power series, we see that
$\epsilon_{r}\left(  h_{n}\right)  =\left(  -1\right)  ^{n}\dbinom{-r}{n}$ for
every $n\in \NN$. Thus, (\ref{sol.Lambda.maps-on-pn.c.goal}) is proven,
and so Exercise~\ref{exe.Lambda.maps-on-pn}(c) is solved.

(d) Consider the $\QQ$-algebra homomorphism $\epsilon_{r}$ defined in
Exercise~\ref{exe.Lambda.maps-on-pn}(c), and the $\QQ$-algebra
homomorphism $\Delta_{\times}$ defined in Exercise~\ref{exe.Lambda.maps-on-pn}%
(b). We know that both of these maps $\epsilon_{r}$ and $\Delta_{\times}$ are
$\QQ$-algebra homomorphisms. In particular, $ \id %
:\Lambda_{ \QQ }\rightarrow\Lambda_{ \QQ }$ and $\epsilon
_{r}:\Lambda_{ \QQ }\rightarrow \QQ $ are $\QQ$-algebra
homomorphisms. Thus, $ \id \otimes\epsilon_{r}:\Lambda
_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow
\Lambda_{ \QQ }\otimes_{ \QQ } \QQ $ is a $\QQ$-algebra
homomorphism (by Exercise~\ref{exe.tensorprod.morphism}(a)). Also, let
$\operatorname*{can}:\Lambda_{ \QQ }\otimes_{ \QQ } \QQ %
\rightarrow\Lambda_{ \QQ }$ be the canonical $\QQ$-vector space
isomorphism sending every $f\otimes\alpha\in\Lambda_{ \QQ }%
\otimes_{ \QQ } \QQ $ to $\alpha f\in\Lambda_{ \QQ }$.

Now, we claim that%
\begin{equation}
\operatorname*{can}\circ\left(   \id \otimes\epsilon_{r}\right)
\circ\Delta_{\times}=\mathbf{i}_{r}. \label{sol.Lambda.maps-on-pn.d.goal}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.d.goal}):} The equality
(\ref{sol.Lambda.maps-on-pn.d.goal}) is an equality between $ \QQ %
$-algebra homomorphisms (since $\operatorname*{can}$, $ \id %
\otimes\epsilon_{r}$, $\Delta_{\times}$ and $\mathbf{i}_{r}$ are $ \QQ %
$-algebra homomorphisms).
Consequently, in order to prove it, we only need to
check that it holds on a generating set of the $\QQ$-algebra
$\Lambda_{ \QQ }$. We do this on the generating set $\left(
p_{n}\right)  _{n\geq1}$, by noticing that every $n\geq1$ satisfies%
\begin{align*}
 \left(  \operatorname*{can}\circ\left(   \id \otimes
\epsilon_{r}\right)  \circ\Delta_{\times}\right)  \left(  p_{n}\right)
&  =\operatorname*{can}\left(  \left(   \id \otimes\epsilon
_{r}\right)  \left(  \underbrace{\Delta_{\times}\left(  p_{n}\right)
}_{=p_{n}\otimes p_{n}}\right)  \right)  =\operatorname*{can}\left(
\underbrace{\left(   \id \otimes\epsilon_{r}\right)  \left(
p_{n}\otimes p_{n}\right)  }_{=p_{n}\otimes\epsilon_{r}\left(  p_{n}\right)
}\right) \\
&  =\operatorname*{can}\left(  p_{n}\otimes\epsilon_{r}\left(  p_{n}\right)
\right)  =\underbrace{\epsilon_{r}\left(  p_{n}\right)  }_{=r}p_{n}%
=rp_{n}=\mathbf{i}_{r}\left(  p_{n}\right)  .
\end{align*}
This proves (\ref{sol.Lambda.maps-on-pn.d.goal}).

Now, $\underbrace{\left(  \operatorname*{can}\circ\left(   \id %
\otimes\epsilon_{r}\right)  \circ\Delta_{\times}\right)  }_{=\mathbf{i}_{r}%
}\left(  \Lambda\right)  =\mathbf{i}_{r}\left(  \Lambda\right)  $, so that%
\begin{align*}
\mathbf{i}_{r}\left(  \Lambda\right)   &  =\left(  \operatorname*{can}%
\circ\left(   \id \otimes\epsilon_{r}\right)  \circ\Delta
_{\times}\right)  \left(  \Lambda\right)  =\operatorname*{can}\left(  \left(
 \id \otimes\epsilon_{r}\right)  \underbrace{\left(
\Delta_{\times}\left(  \Lambda\right)  \right)  }_{\substack{\subset
\Lambda\otimes_{ \ZZ }\Lambda\\\text{(by Exercise
\ref{exe.Lambda.maps-on-pn}(b))}}}\right)  \subset\operatorname*{can}\left(
\underbrace{\left(   \id \otimes\epsilon_{r}\right)  \left(
\Lambda\otimes_{ \ZZ }\Lambda\right)  }_{=\Lambda\otimes_\ZZ\epsilon
_{r}\left(  \Lambda\right)  }\right) \\
&  =\operatorname*{can}\left(  \Lambda\otimes_\ZZ\underbrace{\epsilon_{r}\left(
\Lambda\right)  }_{\substack{\subset \ZZ \\\text{(by Exercise
\ref{exe.Lambda.maps-on-pn}(c))}}}\right) \subset \operatorname*{can}\left(
\Lambda\otimes \ZZ \right)  = \ZZ \cdot\Lambda
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{can}%
\right) \\
&  =\Lambda.
\end{align*}
This solves Exercise~\ref{exe.Lambda.maps-on-pn}(d).

(e) Consider the $\QQ$-algebra homomorphism $\Delta_{\times}$ defined
in Exercise~\ref{exe.Lambda.maps-on-pn}(b). We have $\Delta_{\times}\left(
p_{n}\right)  =p_{n}\otimes p_{n}$ for every positive integer $n$. Now, it is
easy to see that
\begin{equation}
\Delta_{\times}\left(  p_{\lambda}\right)  =p_{\lambda}\otimes p_{\lambda
}\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.Lambda.maps-on-pn.e.lem}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.maps-on-pn.e.lem}):} Let $\lambda$
be a partition. Write $\lambda$ in the form $\lambda=\left(  \lambda
_{1},\lambda_{2},...,\lambda_{\ell}\right)  $ with $\ell=\ell\left(
\lambda\right)  $. Then, the definition of $p_{\lambda}$ yields $p_{\lambda
}=p_{\lambda_{1}}p_{\lambda_{2}}...p_{\lambda_{\ell}}$. Applying the map
$\Delta_{\times}$ to both sides of this equality, we obtain%
\begin{align*}
\Delta_{\times}\left(  p_{\lambda}\right)   &  =\Delta_{\times}\left(
p_{\lambda_{1}}p_{\lambda_{2}}...p_{\lambda_{\ell}}\right)  =\Delta_{\times
}\left(  p_{\lambda_{1}}\right)  \cdot\Delta_{\times}\left(  p_{\lambda_{2}%
}\right)  \cdot...\cdot\Delta_{\times}\left(  p_{\lambda_{\ell}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{\times}\text{ is a
} \QQ \text{-algebra homomorphism}\right) \\
&  =\left(  p_{\lambda_{1}}\otimes p_{\lambda_{1}}\right)  \cdot\left(
p_{\lambda_{2}}\otimes p_{\lambda_{2}}\right)  \cdot...\cdot\left(
p_{\lambda_{\ell}}\otimes p_{\lambda_{\ell}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta_{\times}\left(  p_{n}\right)
=p_{n}\otimes p_{n}\text{ for every positive integer }n\right) \\
&  =\underbrace{\left(  p_{\lambda_{1}}p_{\lambda_{2}}...p_{\lambda_{\ell}%
}\right)  }_{=p_{\lambda}}\otimes\underbrace{\left(  p_{\lambda_{1}}%
p_{\lambda_{2}}...p_{\lambda_{\ell}}\right)  }_{=p_{\lambda}}=p_{\lambda
}\otimes p_{\lambda},
\end{align*}
which proves (\ref{sol.Lambda.maps-on-pn.e.lem}).}

Now, consider the multiplication map $m_{\Lambda_{ \QQ }}:\Lambda
_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow
\Lambda_{ \QQ }$ of the $\QQ$-algebra $\Lambda_{ \QQ }$. We
claim that
\[
\operatorname*{Sq}=m_{\Lambda_{ \QQ }}\circ\Delta_{\times}.
\]
Indeed, every partition $\lambda$ satisfies
\[
\operatorname*{Sq}\left(  p_{\lambda}\right)  =p_{\lambda}^{2}=p_{\lambda
}p_{\lambda}=m_{\Lambda_{ \QQ }}\left(  \underbrace{p_{\lambda}\otimes
p_{\lambda}}_{\substack{=\Delta_{\times}\left(  p_{\lambda}\right)
\\\text{(by (\ref{sol.Lambda.maps-on-pn.e.lem}))}}}\right)  =m_{\Lambda
_{ \QQ }}\left(  \Delta_{\times}\left(  p_{\lambda}\right)  \right)
=\left(  m_{\Lambda_{ \QQ }}\circ\Delta_{\times}\right)  \left(
p_{\lambda}\right)  .
\]
Since $\left( p_\lambda \right) _{\lambda \in \Par}$ is a
$\QQ$-basis of $\Lambda_\QQ$ (and since $\operatorname*{Sq}$ and
$m_{\Lambda_\QQ} \circ \Delta_{\times}$ are $\QQ$-linear maps),
this shows that $\operatorname*{Sq} = m_{\Lambda_\QQ} \circ
\Delta_{\times}$. Hence,
\begin{align*}
\operatorname*{Sq}\left(  \Lambda\right)   &  =\left(  m_{\Lambda_{ \QQ %
}}\circ\Delta_{\times}\right)  \left(  \Lambda\right)  =m_{\Lambda
_{ \QQ }}\underbrace{\left(  \Delta_{\times}\left(  \Lambda\right)
\right)  }_{\substack{\subset\Lambda\otimes_{ \ZZ }\Lambda\\\text{(by
Exercise \ref{exe.Lambda.maps-on-pn}(b))}}}\subset m_{\Lambda_{ \QQ }%
}\left(  \Lambda\otimes_{ \ZZ }\Lambda\right) \\
&  =\Lambda\cdot\Lambda\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}m_{\Lambda_{ \QQ }}\right) \\
&  =\Lambda.
\end{align*}
This solves Exercise~\ref{exe.Lambda.maps-on-pn}(e).

(f) For every $N\in \NN$, we define a $\QQ$-algebra homomorphism
$\mathcal{E}_{N}:\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ %
}\rightarrow \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}%
\right]  $ as it was done in our Second solution of Exercise
\ref{exe.Lambda.maps-on-pn}(b).

Recall the definition of the maps $\mathbf{i}_{r}$ in Exercise
\ref{exe.Lambda.maps-on-pn}(d). We are first going to show that every
$a\in \ZZ$ and $b\in \ZZ$ satisfy%
\begin{equation}
\Delta_{a}=\Delta_{b}\star\left(  \Delta_{\Lambda_{ \QQ }}\circ
\mathbf{i}_{a-b}\right)  \ \ \ \ \ \ \ \ \ \ \text{in }\Hom
\left(  \Lambda_{ \QQ },\Lambda_{ \QQ }\otimes_{ \QQ }%
\Lambda_{ \QQ }\right)  \label{sol.Lambda.maps-on-pn.f.a-b}%
\end{equation}
(where $\Delta_{\Lambda_{ \QQ }}:\Lambda_{ \QQ }\rightarrow
\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda_{ \QQ }$ is the usual
comultiplication of $\Lambda_{ \QQ }$).

\textit{Proof of (\ref{sol.Lambda.maps-on-pn.f.a-b}):} Let $a\in \ZZ$
and $b\in \ZZ$. We know that $\Delta_{a}$ and $\Delta_{b}$ are
$\QQ$-algebra homomorphisms. Also, $\Delta_{\Lambda_{ \QQ }}$ is a
$\QQ$-algebra homomorphism (by the axioms of a bialgebra, which we know
are satisfied for $\Lambda_{ \QQ }$), and $\mathbf{i}_{a-b}$ is a
$\QQ$-algebra homomorphism. Hence, the composition $\Delta
_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}$ is a $\QQ$-algebra
homomorphism. Thus, the convolution $\Delta_{b}\star\left(  \Delta
_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)  $ is a $ \QQ %
$-algebra homomorphism (by Exercise~\ref{exe.alghoms.comm}(a), applied to
$\QQ$, $\Lambda_{ \QQ }$, $\Lambda_{ \QQ }\otimes
_{ \QQ }\Lambda_{ \QQ }$, $\Delta_{b}$ and $\Delta_{\Lambda
_{ \QQ }}\circ\mathbf{i}_{a-b}$ instead of $\kk$, $H$, $A$, $f$
and $g$). Hence, (\ref{sol.Lambda.maps-on-pn.f.a-b}) is an equality between
$\QQ$-algebra homomorphisms. Therefore, in order to prove it, we only
need to check that it holds on a generating set of the $\QQ$-algebra
$\Lambda_{ \QQ }$. We do this on the generating set $\left(
p_{n}\right)  _{n\geq1}$, by noticing that every $n\geq1$ satisfies%
\begin{align*}
&  \underbrace{\left(  \Delta_{b}\star\left(  \Delta_{\Lambda_{ \QQ }%
}\circ\mathbf{i}_{a-b}\right)  \right)  }_{\substack{=m_{\Lambda_{ \QQ }%
}\circ\left(  \Delta_{b}\otimes\left(  \Delta_{\Lambda_{ \QQ }}%
\circ\mathbf{i}_{a-b}\right)  \right)  \circ\Delta_{\Lambda_{ \QQ }%
}\\\text{(by the definition of convolution)}}}\left(  p_{n}\right) \\
&  =\left(  m_{\Lambda_{ \QQ }}\circ\left(  \Delta_{b}\otimes\left(
\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)  \right)
\circ\Delta_{\Lambda_{ \QQ }}\right)  \left(  p_{n}\right) \\
&  =m_{\Lambda_{ \QQ }}\left(  \left(  \Delta_{b}\otimes\left(
\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)  \right)  \left(
\underbrace{\Delta_{\Lambda_{ \QQ }}\left(  p_{n}\right)  }%
_{\substack{=1\otimes p_{n}+p_{n}\otimes1\\\text{(since }p_{n}\text{ is
primitive)}}}\right)  \right) \\
&  =m_{\Lambda_{ \QQ }}\left(  \underbrace{\left(  \Delta_{b}%
\otimes\left(  \Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)
\right)  \left(  1\otimes p_{n}+p_{n}\otimes1\right)  }_{=\Delta_{b}\left(
1\right)  \otimes\left(  \Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}%
_{a-b}\right)  \left(  p_{n}\right)  +\Delta_{b}\left(  p_{n}\right)
\otimes\left(  \Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)
\left(  1\right)  }\right) \\
&  =m_{\Lambda_{ \QQ }}\left(  \Delta_{b}\left(  1\right)  \otimes\left(
\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)  \left(
p_{n}\right)  +\Delta_{b}\left(  p_{n}\right)  \otimes\left(  \Delta
_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)  \left(  1\right)  \right)
\\
&  =\underbrace{\Delta_{b}\left(  1\right)  }_{=1\otimes1}\cdot
\underbrace{\left(  \Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\right)
\left(  p_{n}\right)  }_{=\Delta_{\Lambda_{ \QQ }}\left(  \mathbf{i}%
_{a-b}\left(  p_{n}\right)  \right)  }+\Delta_{b}\left(  p_{n}\right)
\cdot\underbrace{\left(  \Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}%
_{a-b}\right)  \left(  1\right)  }_{\substack{=1\otimes1\\\text{(since }%
\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{a-b}\text{ is a } \QQ %
\text{-algebra}\\\text{homomorphism)}}}\\
&  =\left(  1\otimes1\right)  \cdot\left(  \Delta_{\Lambda_{ \QQ }%
}\left(  \mathbf{i}_{a-b}\left(  p_{n}\right)  \right)  \right)  +\Delta
_{b}\left(  p_{n}\right)  \cdot\left(  1\otimes1\right)  =\Delta
_{\Lambda_{ \QQ }}\left(  \underbrace{\mathbf{i}_{a-b}\left(
p_{n}\right)  }_{\substack{=\left(  a-b\right)  p_{n}\\\text{(by the
definition of }\mathbf{i}_{a-b}\text{)}}}\right)  +\underbrace{\Delta
_{b}\left(  p_{n}\right)  }_{\substack{=\sum_{i=1}^{n-1}\dbinom{n}{i}%
p_{i}\otimes p_{n-i}+b\otimes p_{n}+p_{n}\otimes b\\\text{(by the definition
of }\Delta_{b}\text{)}}}\\
&  =\Delta_{\Lambda_{ \QQ }}\left(  \left(  a-b\right)  p_{n}\right)
+\sum_{i=1}^{n-1}\dbinom{n}{i}p_{i}\otimes p_{n-i}+b\otimes p_{n}+p_{n}\otimes
b\\
&  =\left(  a-b\right)  \underbrace{\Delta_{\Lambda_{ \QQ }}\left(
p_{n}\right)  }_{\substack{=1\otimes p_{n}+p_{n}\otimes1\\\text{(since }%
p_{n}\text{ is primitive)}}}+\sum_{i=1}^{n-1}\dbinom{n}{i}p_{i}\otimes
p_{n-i}+b\otimes p_{n}+p_{n}\otimes b\\
&  =\left(  a-b\right)  \left(  1\otimes p_{n}+p_{n}\otimes1\right)
+\sum_{i=1}^{n-1}\dbinom{n}{i}p_{i}\otimes p_{n-i}+b\otimes p_{n}+p_{n}\otimes
b\\
&  =\sum_{i=1}^{n-1}\dbinom{n}{i}p_{i}\otimes p_{n-i}+\underbrace{b\otimes
p_{n}+p_{n}\otimes b+\left(  a-b\right)  \left(  1\otimes p_{n}+p_{n}%
\otimes1\right)  }_{=a\otimes p_{n}+p_{n}\otimes a}\\
&  =\sum_{i=1}^{n-1}\dbinom{n}{i}p_{i}\otimes p_{n-i}+a\otimes p_{n}%
+p_{n}\otimes a=\Delta_{a}\left(  p_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the definition of }\Delta_{a}\text{
yields }\Delta_{a}\left(  p_{n}\right)  =\sum_{i=1}^{n-1}\dbinom{n}{i}%
p_{i}\otimes p_{n-i}+a\otimes p_{n}+p_{n}\otimes a\right)  .
\end{align*}
This proves (\ref{sol.Lambda.maps-on-pn.f.a-b}).

Before we move on, let us record a simple fact about convolution of maps.
Namely, if $\kk$ is a commutative ring, and $C$ is a $\kk %
$-coalgebra, and $A$ and $A^{\prime}$ are two $\kk$-algebras, and $f$
and $g$ are two $\kk$-linear maps $C\rightarrow A$, and $\alpha
:A\rightarrow A^{\prime}$ is a $\kk$-algebra homomorphism, then%
\begin{equation}
\alpha\circ\left(  f\star g\right)  =\left(  \alpha\circ f\right)
\star\left(  \alpha\circ g\right)  .
\label{sol.Lambda.maps-on-pn.f.banal-lem}
\end{equation}
This is merely the particular case of
\eqref{pre-and-post-composition-in-convolution} when $C^{\prime}=C$ and
$\gamma= \id$, and so requires no proof anymore.

Now, let $N\in \NN$. Our next goal is to show that%
\begin{equation}
\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  \Lambda\right)
\subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
.\label{sol.Lambda.maps-on-pn.f.goal-NN}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.f.goal-NN}):} Let us define a map%
\begin{align*}
\mathcal{L}_{N}:\Lambda_{ \QQ } &  \rightarrow \QQ \left[
x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  ,\\
f &  \mapsto f\left(  \left(  x_{i}+y_{j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,...,N\right\}  ^{2}}\right)  .
\end{align*}
Here, $f\left(  \left(  x_{i}+y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,...,N\right\}  ^{2}}\right)  $ is defined as follows: Let $\left(
u_{1},u_{2},...,u_{N^{2}}\right)  $ be a list of all $N^{2}$ elements of the
family $\left(  x_{i}+y_{j}\right)  _{\left(  i,j\right)  \in\left\{
1,2,...,N\right\}  ^{2}}$ in any arbitrary order, and set $f\left(  \left(
x_{i}+y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,...,N\right\}  ^{2}%
}\right)  =f\left(  u_{1},u_{2},...,u_{N^{2}}\right)  $. (The result does not
depend on the order chosen, because $f$ is symmetric.)

The map $\mathcal{L}_{N}$ is a $\QQ$-algebra homomorphism (since
$\mathcal{L}_{N}$ is an evaluation map in an appropriate sense).

For every positive integer $n$, we have%
\begin{align*}
\Delta_{N}\left(  p_{n}\right)    & =\sum_{i=1}^{n-1}\binom{n}{i}p_{i}\otimes
p_{n-i}+N\otimes p_{n}+p_{n}\otimes N\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\Delta_{N}\left(  p_{n}\right)  \right)  \\
& =\sum_{k=1}^{n-1}\binom{n}{k}p_{k}\otimes p_{n-k}+N\otimes p_{n}%
+p_{n}\otimes N
\end{align*}
and thus%
\begin{align}
&  \left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  p_{n}\right)
\nonumber\\
&  =\mathcal{E}_{N}\left(  \underbrace{\Delta_{N}\left(  p_{n}\right)
}_{=\sum_{k=1}^{n-1}\binom{n}{k}p_{k}\otimes p_{n-k}+N\otimes p_{n}%
+p_{n}\otimes N}\right)  =\mathcal{E}_{N}\left(  \sum_{k=1}^{n-1}\binom{n}%
{k}p_{k}\otimes p_{n-k}+N\otimes p_{n}+p_{n}\otimes N\right)  \nonumber\\
&  =\sum_{k=1}^{n-1}\binom{n}{k}\underbrace{p_{k}\left(  x_{1},x_{2}%
,...,x_{N}\right)  }_{=\sum_{i=1}^{N}x_{i}^{k}}\underbrace{p_{n-k}\left(
y_{1},y_{2},...,y_{N}\right)  }_{=\sum_{j=1}^{N}y_{j}^{n-k}}%
+N\underbrace{p_{n}\left(  y_{1},y_{2},...,y_{N}\right)  }_{=\sum_{j=1}%
^{N}y_{j}^{n}}+\underbrace{p_{n}\left(  x_{1},x_{2},...,x_{N}\right)  }%
_{=\sum_{i=1}^{N}x_{i}^{n}}N \nonumber \\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathcal{E}_{N}\right)  \nonumber\\
&  =\sum_{k=1}^{n-1}\binom{n}{k}\left(  \sum_{i=1}^{N}x_{i}^{k}\right)
\left(  \sum_{j=1}^{N}y_{j}^{n-k}\right)  +N\left(  \sum_{j=1}^{N}y_{j}%
^{n}\right)  +\left(  \sum_{i=1}^{N}x_{i}^{n}\right)
N,\label{sol.Lambda.maps-on-pn.f.goal-NN.pf.1}%
\end{align}
while at the same time%
\begin{align}
\mathcal{L}_{N}\left(  p_{n}\right)    & =p_{n}\left(  \left(  x_{i}%
+y_{j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,...,N\right\}  ^{2}%
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathcal{L}_{N}\left(  p_{n}\right)  \right)  \nonumber\\
& =\sum_{\left(  i,j\right)  \in\left\{  1,2,...,N\right\}  ^{2}%
}\underbrace{\left(  x_{i}+y_{j}\right)  ^{n}}_{=\sum_{k=0}^{n}\dbinom{n}%
{k}x_{i}^{k}y_{j}^{n-k}}=\sum_{\left(  i,j\right)  \in\left\{
1,2,...,N\right\}  ^{2}}\sum_{k=0}^{n}\dbinom{n}{k}x_{i}^{k}y_{j}%
^{n-k}\nonumber\\
& =\sum_{k=0}^{n}\dbinom{n}{k}\underbrace{\sum_{\left(  i,j\right)
\in\left\{  1,2,...,N\right\}  ^{2}}x_{i}^{k}y_{j}^{n-k}}_{=\left(  \sum
_{i=1}^{N}x_{i}^{k}\right)  \left(  \sum_{j=1}^{N}y_{j}^{n-k}\right)  }%
=\sum_{k=0}^{n}\dbinom{n}{k}\left(  \sum_{i=1}^{N}x_{i}^{k}\right)  \left(
\sum_{j=1}^{N}y_{j}^{n-k}\right)  \nonumber\\
& =\sum_{k=1}^{n-1}\binom{n}{k}\left(  \sum_{i=1}^{N}x_{i}^{k}\right)  \left(
\sum_{j=1}^{N}y_{j}^{n-k}\right)  +\underbrace{\dbinom{n}{0}}_{=1}%
\underbrace{\left(  \sum_{i=1}^{N}x_{i}^{0}\right)  }_{=N}\left(  \sum
_{j=1}^{N}y_{j}^{n}\right)  +\underbrace{\dbinom{n}{n}}_{=1}\left(  \sum
_{i=1}^{N}x_{i}^{n}\right)  \underbrace{\left(  \sum_{j=1}^{N}y_{j}%
^{0}\right)  }_{=N}\nonumber\\
& =\sum_{k=1}^{n-1}\binom{n}{k}\left(  \sum_{i=1}^{N}x_{i}^{k}\right)  \left(
\sum_{j=1}^{N}y_{j}^{n-k}\right)  +N\left(  \sum_{j=1}^{N}y_{j}^{n}\right)
+\left(  \sum_{i=1}^{N}x_{i}^{n}\right)
N.\label{sol.Lambda.maps-on-pn.f.goal-NN.pf.2}%
\end{align}
Comparing (\ref{sol.Lambda.maps-on-pn.f.goal-NN.pf.1}) with
(\ref{sol.Lambda.maps-on-pn.f.goal-NN.pf.2}) reveals that $\left(
\mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  p_{n}\right)  =\mathcal{L}%
_{N}\left(  p_{n}\right)  $ for every positive integer $n$. In other words,
the two maps $\mathcal{E}_{N}\circ\Delta_{N}$ and $\mathcal{L}_{N}$ are equal
to each other on the generating set $\left(  p_{n}\right)  _{n\geq1}$ of the
$\QQ$-algebra $\Lambda_{ \QQ }$. Since these two maps
$\mathcal{E}_{N}\circ\Delta_{N}$ and $\mathcal{L}_{N}$ are $ \QQ %
$-algebra homomorphisms (since $\mathcal{E}_{N}$, $\Delta_{N}$ and
$\mathcal{L}_{N}$ are $\QQ$-algebra homomorphisms), this yields that
these two maps must be identical, i.e., we have $\mathcal{E}_{N}\circ
\Delta_{N}=\mathcal{L}_{N}$. Hence,%
\[
\underbrace{\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  }_{=\mathcal{L}%
_{N}}\left(  \Lambda\right)  =\mathcal{L}_{N}\left(  \Lambda\right)
\subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
\]
(because the definition of $\mathcal{L}_{N}$ immediately shows that
$\mathcal{L}_{N}\left(  f\right)  \in \ZZ \left[  x_{1},x_{2}%
,...,x_{N},y_{1},y_{2},...,y_{N}\right]  $ for every $f\in\Lambda$). This
proves (\ref{sol.Lambda.maps-on-pn.f.goal-NN}).

Next, we are going to show that%
\begin{equation}
\left(  \mathcal{E}_{N}\circ\Delta_{r}\right)  \left(  \Lambda\right)
\subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
.\label{sol.Lambda.maps-on-pn.f.goal-Nr}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.f.goal-Nr}):} Applying
(\ref{sol.Lambda.maps-on-pn.f.a-b}) to $a=r$ and $b=N$, we obtain%
\[
\Delta_{r}=\Delta_{N}\star\left(  \Delta_{\Lambda_{ \QQ }}\circ
\mathbf{i}_{r-N}\right)  .
\]
Thus,%
\[
\mathcal{E}_{N}\circ\underbrace{\Delta_{r}}_{=\Delta_{N}\star\left(
\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{r-N}\right)  }=\mathcal{E}%
_{N}\circ\left(  \Delta_{N}\star\left(  \Delta_{\Lambda_{ \QQ }}%
\circ\mathbf{i}_{r-N}\right)  \right)  =\left(  \mathcal{E}_{N}\circ\Delta
_{N}\right)  \star\left(  \mathcal{E}_{N}\circ\Delta_{\Lambda_{ \QQ }%
}\circ\mathbf{i}_{r-N}\right)
\]
(by (\ref{sol.Lambda.maps-on-pn.f.banal-lem}), applied to $\QQ$,
$\Lambda_{ \QQ }$, $\Lambda_{ \QQ }\otimes_{ \QQ }%
\Lambda_{ \QQ }$, $ \QQ \left[  x_{1},x_{2},...,x_{N},y_{1}%
,y_{2},...,y_{N}\right]  $, $\Delta_{N}$, $\Delta_{\Lambda_{ \QQ }}%
\circ\mathbf{i}_{r-N}$ and $\mathcal{E}_{N}$ instead of $\kk$, $C$,
$A$, $A^{\prime}$, $f$, $g$ and $\alpha$). Thus,%
\[
\mathcal{E}_{N}\circ\Delta_{r}=\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)
\star\left(  \mathcal{E}_{N}\circ\Delta_{\Lambda_{ \QQ }}\circ
\mathbf{i}_{r-N}\right)  =m_{ \QQ \left[  x_{1},x_{2},...,x_{N}%
,y_{1},y_{2},...,y_{N}\right]  }\circ\left(  \left(  \mathcal{E}_{N}%
\circ\Delta_{N}\right)  \otimes\left(  \mathcal{E}_{N}\circ\Delta
_{\Lambda_{ \QQ }}\circ\mathbf{i}_{r-N}\right)  \right)  \circ
\Delta_{\Lambda_{ \QQ }}%
\]
(by the definition of convolution), so that%
\begin{align*}
& \underbrace{\left(  \mathcal{E}_{N}\circ\Delta_{r}\right)  }_{=m_{ \QQ %
\left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  }\circ\left(
\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \otimes\left(  \mathcal{E}%
_{N}\circ\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{r-N}\right)  \right)
\circ\Delta_{\Lambda_{ \QQ }}}\left(  \Lambda\right)  \\
& =\left(  m_{ \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2}%
,...,y_{N}\right]  }\circ\left(  \left(  \mathcal{E}_{N}\circ\Delta
_{N}\right)  \otimes\left(  \mathcal{E}_{N}\circ\Delta_{\Lambda_{ \QQ }%
}\circ\mathbf{i}_{r-N}\right)  \right)  \circ\Delta_{\Lambda_{ \QQ }%
}\right)  \left(  \Lambda\right)  \\
& =m_{ \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
}\left(  \left(  \left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \otimes\left(
\mathcal{E}_{N}\circ\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{r-N}\right)
\right)  \underbrace{\left(  \Delta_{\Lambda_{ \QQ }}\left(
\Lambda\right)  \right)  }_{=\Delta_{\Lambda}\left(  \Lambda\right)
\subset\Lambda\otimes_{ \ZZ }\Lambda}\right)  \\
& \subset m_{ \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2}%
,...,y_{N}\right]  }\left(  \underbrace{\left(  \left(  \mathcal{E}_{N}%
\circ\Delta_{N}\right)  \otimes\left(  \mathcal{E}_{N}\circ\Delta
_{\Lambda_{ \QQ }}\circ\mathbf{i}_{r-N}\right)  \right)  \left(
\Lambda\otimes_{ \ZZ }\Lambda\right)  }_{=\left(  \mathcal{E}_{N}%
\circ\Delta_{N}\right)  \left(  \Lambda\right)  \otimes_\ZZ \left(  \mathcal{E}%
_{N}\circ\Delta_{\Lambda_{ \QQ }}\circ\mathbf{i}_{r-N}\right)  \left(
\Lambda\right)  }\right)  \\
& =m_{ \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
}\left(  \left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  \Lambda
\right)  \otimes_\ZZ\left(  \mathcal{E}_{N}\circ\Delta_{\Lambda_{ \QQ }}%
\circ\mathbf{i}_{r-N}\right)  \left(  \Lambda\right)  \right)  \\
& =\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  \Lambda\right)
\cdot\underbrace{\left(  \mathcal{E}_{N}\circ\Delta_{\Lambda_{ \QQ }%
}\circ\mathbf{i}_{r-N}\right)  \left(  \Lambda\right)  }_{=\mathcal{E}%
_{N}\left(  \Delta_{\Lambda_{ \QQ }}\left(  \mathbf{i}_{r-N}\left(
\Lambda\right)  \right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}m_{ \QQ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
}\text{ is the multiplication map}\right)  \\
& =\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  \Lambda\right)
\cdot\mathcal{E}_{N}\left(  \Delta_{\Lambda_{ \QQ }}\left(
\underbrace{\mathbf{i}_{r-N}\left(  \Lambda\right)  }_{\substack{\subset
\Lambda\\\text{(by Exercise \ref{exe.Lambda.maps-on-pn}(d),}\\\text{applied to
}r-N\text{ instead of }r\text{)}}}\right)  \right)  \\
& \subset\underbrace{\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(
\Lambda\right)  }_{\substack{\subset \ZZ \left[  x_{1},x_{2}%
,...,x_{N},y_{1},y_{2},...,y_{N}\right]  \\\text{(by
(\ref{sol.Lambda.maps-on-pn.f.goal-NN}))}}}\cdot\mathcal{E}_{N}%
\underbrace{\left(  \Delta_{\Lambda_{ \QQ }}\left(  \Lambda\right)
\right)  }_{=\Delta_{\Lambda}\left(  \Lambda\right)  \subset\Lambda
\otimes_{ \ZZ }\Lambda}\\
& \subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
\cdot\underbrace{\mathcal{E}_{N}\left(  \Lambda\otimes_{ \ZZ }%
\Lambda\right)  }_{\substack{\subset \ZZ \left[  x_{1},x_{2}%
,...,x_{N},y_{1},y_{2},...,y_{N}\right]  \\\text{(because the definition of
}\mathcal{E}_{N}\text{ immediately yields}\\\mathcal{E}_{N}\left(  f\otimes
g\right)  \in \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2}%
,...,y_{N}\right]  \text{ for any }f\in\Lambda\text{ and }g\in\Lambda\text{)}%
}}\\
& \subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
\cdot \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
= \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]  .
\end{align*}
This proves (\ref{sol.Lambda.maps-on-pn.f.goal-Nr}).

Now, forget that we fixed $N$. We thus have proven
(\ref{sol.Lambda.maps-on-pn.f.goal-Nr}) to hold for every $N\in \NN$.

Now, let $p\in\Delta_{r}\left(  \Lambda\right)  $. Then,%
\[
\mathcal{E}_{N}\left(  \underbrace{p}_{\in\Delta_{r}\left(  \Lambda\right)
}\right)  \in\mathcal{E}_{N}\left(  \Delta_{r}\left(  \Lambda\right)  \right)
=\left(  \mathcal{E}_{N}\circ\Delta_{r}\right)  \left(  \Lambda\right)
\in \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1},y_{2},...,y_{N}\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Lambda.maps-on-pn.f.goal-Nr}%
)}\right)
\]
for every $N\in \NN$. Hence,
(\ref{sol.Lambda.maps-on-pn.b.sol2.reduction}) yields that $p\in\Lambda
\otimes_{ \ZZ }\Lambda$. Since we have shown this for every $p\in
\Delta_{r}\left(  \Lambda\right)  $, we thus conclude that $\Delta_{r}\left(
\Lambda\right)  \subset\Lambda\otimes_{ \ZZ }\Lambda$. This solves
Exercise~\ref{exe.Lambda.maps-on-pn}(f).

\textit{Remark.} Here is a rough sketch of an alternative way to conclude this
solution of Exercise \ref{exe.Lambda.maps-on-pn}(f) after proving
(\ref{sol.Lambda.maps-on-pn.f.goal-NN}). This is closer to Richard Stanley's
suggested solution than the above.

The family $\left(  m_{\lambda}\otimes m_{\mu}\right)  _{\left(  \lambda
,\mu\right)  \in \Par \times \Par }$ is a
$\QQ$-basis of $\Lambda_{ \QQ }\otimes_{ \QQ }\Lambda
_{ \QQ }$ (since $\left(  m_{\lambda}\right)  _{\lambda\in
 \Par }$ is a $\QQ$-basis of $\Lambda_{ \QQ }$). Let
us refer to this basis as the \emph{monomial basis} of $\Lambda_{ \QQ %
}\otimes_{ \QQ }\Lambda_{ \QQ }$. Then, $\Lambda\otimes_{ \ZZ %
}\Lambda$ is the subset of $\Lambda_{ \QQ }\otimes_{ \QQ }%
\Lambda_{ \QQ }$ consisting of all elements whose coordinates with
respect to the monomial basis all are integers.

We want to prove that $\Delta_{r}\left(  \Lambda\right)  \subset\Lambda
\otimes_{ \ZZ }\Lambda$ for every $r\in \ZZ$. In other words, we
want to prove that $\Delta_{r}\left(  f\right)  \in\Lambda\otimes_{ \ZZ %
}\Lambda$ for every $r\in \ZZ$ and $f\in\Lambda$. Let us fix
$f\in\Lambda$, but not fix $r$. We need to show that  $\Delta_{r}\left(
f\right)  \in\Lambda\otimes_{ \ZZ }\Lambda$ for every $r\in \ZZ$;
in other words, we need to show that for every $\left(  \alpha,\beta\right)
\in \Par $, the $m_{\alpha}\otimes m_{\beta}$-coordinate of
$\Delta_{r}\left(  f\right)  $ with respect to the monomial basis is an
integer for every $r\in \ZZ$.

So let us fix $\left(  \alpha,\beta\right)  \in \Par $. It is not
hard to see that the $m_{\alpha}\otimes m_{\beta}$-coordinate of $\Delta
_{r}\left(  f\right)  $ with respect to the monomial basis is a polynomial in
$r$ with rational coefficients.\footnote{This can be proven by noticing that
it holds whenever $f=p_{n}$ for some $n\geq1$ (by inspection of the definition
of $\Delta_{r}$), and if it holds for two given values of $f$ then it holds
for any of their $\QQ$-linear combinations and also for their product.}
We want to prove that this polynomial is integer-valued (i.e., all its values
at integer inputs are integers\footnote{This is a weaker statement than saying
that it has integer coefficients. (Actually, it does not in general have
integer coefficients.)}). To do so, it suffices to show that its values are
integers \textbf{at all sufficiently high} $r\in \ZZ$ (because for a
polynomial with rational coefficients, the non-integer values appear
periodically\footnote{To see this, just work modulo the common denominator of
the coefficients of the polynomial.}, and therefore if no non-integer values
appear from a given integer onwards, then the polynomial is integer-valued).
In other words, it suffices to show that the $m_{\alpha}\otimes m_{\beta}%
$-coordinate of $\Delta_{r}\left(  f\right)  $ with respect to the monomial
basis is an integer for all \textbf{sufficiently high }$r\in \ZZ$.

Let us prove this now. Our interpretation of ``sufficiently high'' will
be that $r\geq\ell\left(  \alpha\right)  $ and
$r\geq\ell\left(  \beta\right)  $. So what we need to prove is that the
$m_{\alpha}\otimes m_{\beta}$-coordinate of $\Delta_{r}\left(  f\right)  $
with respect to the monomial basis is an integer whenever $r$ is an integer
satisfying $r\geq\ell\left(  \alpha\right)  $ and $r\geq\ell\left(
\beta\right)  $.

Consider such an $r$. Set $N=r$ and $p=\Delta_{r}\left(  f\right)  $. Then,
$N=r\geq\ell\left(  \alpha\right)  $ and $N=r\geq\ell\left(  \beta\right)  $
and $\Delta_{N}\left(  f\right)  =\Delta_{r}\left(  f\right)  =p$. Then,%
\[
\mathcal{E}_{N}\left(  \underbrace{p}_{=\Delta_{N}\left(  f\right)  }\right)
=\mathcal{E}_{N}\left(  \Delta_{N}\left(  f\right)  \right)  =\left(
\mathcal{E}_{N}\circ\Delta_{N}\right)  \left(  \underbrace{f}_{\in\Lambda
}\right)  \in\left(  \mathcal{E}_{N}\circ\Delta_{N}\right)  \left(
\Lambda\right)  \subset \ZZ \left[  x_{1},x_{2},...,x_{N},y_{1}%
,y_{2},...,y_{N}\right]
\]
(by (\ref{sol.Lambda.maps-on-pn.f.goal-NN})). Therefore, arguing precisely as
in the proof of (\ref{sol.Lambda.maps-on-pn.b.sol2.reduction}) (but using our
specific $\alpha$ and $\beta$ rather than arbitrary $\alpha$ and $\beta$ as in
that proof), we can show that $\rho_{\alpha,\beta}$ is an integer, where $p$
is written in the form $p=\sum_{\left(  \lambda,\mu\right)  \in
 \Par \times \Par }\rho_{\lambda,\mu}m_{\lambda
}\otimes m_{\mu}$ with $\rho_{\lambda,\mu}$ being elements of $\QQ$.
But $\rho_{\alpha,\beta}$ is precisely the  $m_{\alpha}\otimes m_{\beta}%
$-coordinate of $\Delta_{r}\left(  f\right)  $ with respect to the monomial
basis. Thus, we have shown that the  $m_{\alpha}\otimes m_{\beta}$-coordinate
of $\Delta_{r}\left(  f\right)  $ with respect to the monomial basis is an
integer (for $r\in \ZZ$ satisfying $r\geq\ell\left(  \alpha\right)  $
and $r\geq\ell\left(  \beta\right)  $). As we know, this completes our solution
of Exercise~\ref{exe.Lambda.maps-on-pn}(f) again.

(g) Let us first show that the $\QQ$-module $\Lambda_{ \QQ }$,
endowed with the comultiplication $\Delta_{\times}$ and the counit
$\epsilon_{1}$, becomes a $\QQ$-coalgebra. In order to do so, we must
verifying that the diagrams (\ref{coassociativity-diagram}) and
(\ref{counit-diagram}), with $C$, $\Delta$ and $\epsilon$ replaced by
$\Lambda_{ \QQ }$, $\Delta_{\times}$ and $\epsilon_{1}$, commute. We will
only do this for the diagram (\ref{coassociativity-diagram}), while leaving
the diagram (\ref{counit-diagram}) to the reader.

So we must check that the diagram (\ref{coassociativity-diagram}), with $C$,
$\Delta$ and $\epsilon$ replaced by $\Lambda_{ \QQ }$, $\Delta_{\times}$
and $\epsilon_{1}$, commutes. In other words, we must prove the identity%
\begin{equation}
\left(  \Delta_{\times}\otimes \id \right)  \circ\Delta_{\times
}=\left(   \id \otimes\Delta_{\times}\right)  \circ\Delta
_{\times}.\label{sol.Lambda.maps-on-pn.g.delta-eq}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.d.goal}):} The equality
(\ref{sol.Lambda.maps-on-pn.g.delta-eq}) is an equality between $ \QQ %
$-algebra homomorphisms (since $\Delta_{\times}$, $ \id %
\otimes\Delta_{\times}$ and $\Delta_{\times}\otimes \id$ are
$\QQ$-algebra homomorphisms\footnote{Here, we are using Exercise
\ref{exe.tensorprod.morphism}(a) to see that $ \id \otimes
\Delta_{\times}$ and $\Delta_{\times}\otimes \id$ are
$\QQ$-algebra homomorphisms.}). Consequently, in order to prove it, we
only need to check that it holds on a generating set of the $ \QQ %
$-algebra $\Lambda_{ \QQ }$. But every $n\geq1$ satisfies%
\[
\left(  \left(  \Delta_{\times}\otimes \id \right)  \circ
\Delta_{\times}\right)  \left(  p_{n}\right)  =\left(  \Delta_{\times}%
\otimes \id \right)  \left(  \underbrace{\Delta_{\times}\left(
p_{n}\right)  }_{=p_{n}\otimes p_{n}}\right)  =\left(  \Delta_{\times}%
\otimes \id \right)  \left(  p_{n}\otimes p_{n}\right)
=\underbrace{\Delta_{\times}\left(  p_{n}\right)  }_{=p_{n}\otimes p_{n}%
}\otimes p_{n}=p_{n}\otimes p_{n}\otimes p_{n}%
\]
and similarly $\left(  \left(   \id \otimes\Delta_{\times
}\right)  \circ\Delta_{\times}\right)  \left(  p_{n}\right)  =p_{n}\otimes
p_{n}\otimes p_{n}$. Hence, every $n\geq1$ satisfies $\left(  \left(
\Delta_{\times}\otimes \id \right)  \circ\Delta_{\times}\right)
\left(  p_{n}\right)  =p_{n}\otimes p_{n}\otimes p_{n}=\left(  \left(
 \id \otimes\Delta_{\times}\right)  \circ\Delta_{\times}\right)
\left(  p_{n}\right)  $. Thus, we have checked that the equality
(\ref{sol.Lambda.maps-on-pn.g.delta-eq}) holds on a generating set of the
$\QQ$-algebra $\Lambda_{ \QQ }$ (namely, on the generating set
$\left(  p_{n}\right)  _{n\geq1}$). As a consequence, the proof of
(\ref{sol.Lambda.maps-on-pn.g.delta-eq}) is complete.

We have thus shown that the $\QQ$-module $\Lambda_{ \QQ }$,
endowed with the comultiplication $\Delta_{\times}$ and the counit
$\epsilon_{1}$, becomes a $\QQ$-coalgebra. This $\QQ$-coalgebra
becomes a $\QQ$-bialgebra when combined with the existing $ \QQ %
$-algebra structure on $\Lambda_{ \QQ }$ (this is because $\Delta
_{\times}$ and $\epsilon_{1}$ are $\QQ$-algebra homomorphisms), and
this $\QQ$-bialgebra is cocommutative (this follows from the equality
$T\circ\Delta_{\times}=\Delta_{\times}$, where $T:\Lambda_{ \QQ }%
\otimes_{ \QQ }\Lambda_{ \QQ }\rightarrow\Lambda_{ \QQ }%
\otimes_{ \QQ }\Lambda_{ \QQ }$ is the twist map; and this equality
can be proven in the same way as we have showed
(\ref{sol.Lambda.maps-on-pn.g.delta-eq})). This solves Exercise
\ref{exe.Lambda.maps-on-pn}(g).

(h) Corollary \ref{dual-bases-h-m-p}(b) yields that $\left(  p_{\lambda
}\right)  _{\lambda\in \Par }$ and $\left(  z_{\lambda}%
^{-1}p_{\lambda}\right)  _{\lambda\in \Par }$ are dual bases with
respect to the Hall inner product on $\Lambda$. In other words,%
\[
\left(  p_{\lambda},z_{\mu}^{-1}p_{\mu}\right)  =\delta_{\lambda,\mu
}\ \ \ \ \ \ \ \ \ \ \text{for any partitions }\lambda\text{ and }\mu.
\]
Hence,%
\begin{equation}
\left(  p_{\lambda},\underbrace{p_{\mu}}_{=z_{\mu}z_{\mu}^{-1}p_{\mu}}\right)
=\left(  p_{\lambda},z_{\mu}z_{\mu}^{-1}p_{\mu}\right)  =z_{\mu}%
\underbrace{\left(  p_{\lambda},z_{\mu}^{-1}p_{\mu}\right)  }_{=\delta
_{\lambda,\mu}}=z_{\mu}\delta_{\lambda,\mu}%
\label{sol.Lambda.maps-on-pn.h.dualbas}%
\end{equation}
for any partitions $\lambda$ and $\mu$.

The Hall inner product $\left(  \cdot,\cdot\right)  :\Lambda_{{ \QQ }%
}\times\Lambda_{\QQ}\rightarrow{ \QQ }$ is a bilinear form on
$\Lambda_{ \QQ }$. Hence, according to Definition~\ref{def.self-duality}%
(b) (below), this inner product induces a $\QQ$-bilinear form $\left(
\cdot,\cdot\right)  _{\Lambda_{\QQ}\otimes_{\QQ}%
\Lambda_{\QQ}}:\left(  \Lambda_{\QQ}\otimes_{{ \QQ }%
}\Lambda_{\QQ}\right)  \times\left(  \Lambda_{\QQ}%
\otimes_{\QQ}\Lambda_{\QQ}\right)  \rightarrow{ \QQ }$.
Similarly, the Hall inner product $\left(  \cdot,\cdot\right)  :\Lambda
\times\Lambda\rightarrow \ZZ$ induces a $\ZZ$-bilinear form
$\left(  \cdot,\cdot\right)  _{\Lambda\otimes_{\ZZ}\Lambda}:\left(
\Lambda\otimes_{\ZZ}\Lambda\right)  \times\left(  \Lambda
\otimes_{\ZZ}\Lambda\right)  \rightarrow \ZZ$, and this latter
$\ZZ$-bilinear form is clearly the restriction of the $ \QQ %
$-bilinear form $\left(  \cdot,\cdot\right)  _{\Lambda_{\QQ}%
\otimes_{\QQ}\Lambda_{\QQ}}$ to $\Lambda\otimes
_{\ZZ}\Lambda$.

We shall now show that
\begin{equation}
\left(  a\ast b,c\right)  =\left(  a\otimes b,\Delta_{\times}\left(  c\right)
\right)  _{\Lambda_{\QQ}\otimes_{\QQ}\Lambda_{\QQ}%
}\qquad\qquad\text{ for all }a\in\Lambda_{\QQ}\text{, }b\in
\Lambda_{\QQ}\text{ and }c\in\Lambda_{\QQ}%
.\label{sol.Lambda.maps-on-pn.h.selfadj}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.maps-on-pn.h.selfadj}):} Let $a\in
\Lambda_{\QQ}$, $b\in\Lambda_{\QQ}$ and $c\in\Lambda
_{\QQ}$. The equality (\ref{sol.Lambda.maps-on-pn.h.selfadj}) is
clearly $\QQ$-linear in each of $a$, $b$ and $c$. Hence, in proving
this equality, we can WLOG assume that $a$, $b$ and $c$ are elements of the
basis $\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ of the
$\QQ$-module $\Lambda_{ \QQ }$. Assume this. Thus, $a=p_{\lambda}%
$, $b=p_{\mu}$ and $c=p_{\nu}$ for some partitions $\lambda$, $\mu$ and $\nu$;
consider these partitions.

We have $\delta_{\lambda,\mu}z_{\lambda} = \delta_{\lambda,\mu}z_{\mu}$
(in fact, the two sides of this equality are equal when $\lambda = \mu$,
and both vanish otherwise). Hence,
$\underbrace{a}_{=p_{\lambda}}\ast\underbrace{b}_{=p_{\mu}}%
=p_{\lambda}\ast p_{\mu}
=\underbrace{\delta_{\lambda,\mu} z_{\lambda}}_{=\delta_{\lambda,\mu}z_{\mu}}
p_{\lambda}
= \delta_{\lambda, \mu} z_\mu p_\lambda$, so that%
\[
\left(  \underbrace{a\ast b}_{=\delta_{\lambda,\mu}z_{\mu}p_{\lambda}%
},\underbrace{c}_{=p_{\nu}}\right)  =\left(  \delta_{\lambda,\mu}z_{\mu
}p_{\lambda},p_{\nu}\right)  =\delta_{\lambda,\mu}z_{\lambda}%
\underbrace{\left(  p_{\lambda},p_{\nu}\right)  }_{\substack{=z_{\nu}%
\delta_{\lambda,\nu}\\\text{(by (\ref{sol.Lambda.maps-on-pn.h.dualbas}),
applied}\\\text{to }\nu\text{ instead of }\mu\text{)}}}=\delta_{\lambda,\mu
}z_{\lambda}z_{\nu}\delta_{\lambda,\nu}.
\]
On the other hand, $c=p_{\nu}$, so that
\[
\Delta_{\times}\left(  c\right)  =\Delta_{\times}\left(  p_{\nu}\right)
=p_{\nu}\otimes p_{\nu}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.maps-on-pn.e.lem}), applied to }\nu\text{ instead of }%
\lambda\right)  ,
\]
and thus%
\begin{align*}
\left(  \underbrace{a}_{=p_{\lambda}}\otimes\underbrace{b}_{=p_{\mu}%
},\underbrace{\Delta_{\times}\left(  c\right)  }_{=p_{\nu}\otimes p_{\nu}%
}\right)  _{\Lambda_{\QQ}\otimes_{\QQ}\Lambda_{{ \QQ }%
}}  & =\left(  p_{\lambda}\otimes p_{\mu},p_{\nu}\otimes p_{\nu}\right)
_{\Lambda_{\QQ}\otimes_{\QQ}\Lambda_{\QQ}%
}=\underbrace{\left(  p_{\lambda},p_{\nu}\right)  }_{\substack{=z_{\nu}%
\delta_{\lambda,\nu}\\\text{(by (\ref{sol.Lambda.maps-on-pn.h.dualbas}),
applied}\\\text{to }\nu\text{ instead of }\mu\text{)}}}\underbrace{\left(
p_{\mu},p_{\nu}\right)  }_{\substack{=z_{\nu}\delta_{\mu,\nu}\\\text{(by
(\ref{sol.Lambda.maps-on-pn.h.dualbas}), applied}\\\text{to }\mu\text{ and
}\nu\text{ instead of }\lambda\text{ and }\mu\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the bilinear form
}\left(  \cdot,\cdot\right)  _{\Lambda_{\QQ}\otimes_{{ \QQ }%
}\Lambda_{\QQ}}\right)  \\
& =z_{\nu}\delta_{\lambda,\nu}z_{\nu}\delta_{\mu,\nu}.
\end{align*}
The equality in question, (\ref{sol.Lambda.maps-on-pn.h.selfadj}), thus
rewrites as $\delta_{\lambda,\mu}z_{\lambda}z_{\nu}\delta_{\lambda,\nu}%
=z_{\nu}\delta_{\lambda,\nu}z_{\nu}\delta_{\mu,\nu}$ (because $\left(  a\ast
b,c\right)  =\delta_{\lambda,\mu}z_{\lambda}z_{\nu}\delta_{\lambda,\nu}$ and
$\left(  a\otimes b,\Delta_{\times}\left(  c\right)  \right)  _{\Lambda
_{\QQ}\otimes_{\QQ}\Lambda_{\QQ}}=z_{\nu}%
\delta_{\lambda,\nu}z_{\nu}\delta_{\mu,\nu}$). But the latter equality is
obvious (because both of its sides are $z_{\nu}^{2}$ if $\lambda=\mu=\nu$, and
vanish otherwise). Hence, (\ref{sol.Lambda.maps-on-pn.h.selfadj}) must hold as well.

Now that (\ref{sol.Lambda.maps-on-pn.h.selfadj}) is proven, let $f\in\Lambda$
and $g\in\Lambda$ be arbitrary. We can apply
(\ref{sol.exe.Lambda.maps-on-pn.a.bases}) to $\kk = \QQ $,
$A=\Lambda_{ \QQ }$, $L= \Par $, $\left(  u_{\lambda}\right)
_{\lambda\in L}=\left(  s_{\lambda}\right)  _{\lambda\in \Par }$,
$\left(  v_{\lambda}\right)  _{\lambda\in L}=\left(  s_{\lambda}\right)
_{\lambda\in \Par }$ and $a=f\ast g$ (because the basis $\left(
s_{\lambda}\right)  _{\lambda\in \Par }$ of $\Lambda_{ \QQ }$
is orthonormal with respect to the Hall inner product $\left(  \cdot
,\cdot\right)  $, and thus dual to itself with respect to this product). As a
result, we obtain%
\[
f\ast g=\sum_{\lambda\in \Par }\underbrace{\left(  s_{\lambda
},f\ast g\right)  }_{\substack{=\left(  f\ast g,s_{\lambda}\right)
\\\text{(since the Hall inner product}\\\text{is symmetric)}}}s_{\lambda}%
=\sum_{\lambda\in \Par }\underbrace{\left(  f\ast g,s_{\lambda
}\right)  }_{\substack{=\left(  f\otimes g,\Delta_{\times}\left(  s_{\lambda
}\right)  \right)  _{\Lambda_{\QQ}\otimes_{\QQ}%
\Lambda_{\QQ}}\\\text{(by (\ref{sol.Lambda.maps-on-pn.h.selfadj}))}%
}}s_{\lambda}=\sum_{\lambda\in \Par }\left(  f\otimes
g,\Delta_{\times}\left(  s_{\lambda}\right)  \right)  _{\Lambda_{ \QQ }
\otimes_{\QQ}\Lambda_{\QQ}}s_{\lambda}.
\]
But every $\lambda\in \Par $ satisfies $\Delta_{\times}\left(
\underbrace{s_{\lambda}}_{\in\Lambda}\right)  \in\Delta_{\times}\left(
\Lambda\right)  \subset\Lambda\otimes_{\ZZ}\Lambda$ (by Exercise
\ref{exe.Lambda.maps-on-pn}(b)) and thus
\[
\left(  f\otimes g,\Delta_{\times}\left(  s_{\lambda}\right)  \right)
_{\Lambda_{\QQ}\otimes_{\QQ}\Lambda_{\QQ}}=\left(
f\otimes g,\Delta_{\times}\left(  s_{\lambda}\right)  \right)  _{\Lambda
\otimes_{ \ZZ }\Lambda}\in\ZZ .
\]
Hence,
\[
f\ast g=\sum_{\lambda\in \Par }\underbrace{\left(  f\otimes
g,\Delta_{\times}\left(  s_{\lambda}\right)  \right)  _{\Lambda_{ \QQ }
\otimes_{\QQ}\Lambda_{\QQ}}}_{\in \ZZ }s_{\lambda}%
\in\sum_{\lambda\in \Par } \ZZ s_{\lambda}\subset\Lambda.
\]
This solves Exercise~\ref{exe.Lambda.maps-on-pn}(h).

(i) Define a map $U : \Lambda_\QQ \to \QQ$ by
\[
U\left(f\right) = f\left(1\right)
\qquad \text{for all } f \in \Lambda_\QQ .
\]
Notice that $U\left(f\right) = f\left(1\right)$ is the result of
substituting $1, 0, 0, 0, \ldots$ for $x_1, x_2, x_3, \ldots$ in $f$.
Hence, $U$ is a $\QQ$-algebra homomorphism.

We shall now show that $\epsilon_1 = U$.

For every integer $n \geq 1$, we have
\begin{align*}
U\left(p_n\right)
&= p_n\left(1\right)
\qquad \left(\text{by the definition of } U \right) \\
&= \left(\text{the result of substituting }
1, 0, 0, 0, \ldots \text{ for } x_1, x_2, x_3, \ldots \text{ in } p_n \right) \\
& \qquad \left(\text{by the definition of } p_n\left(1\right) \right) \\
&= \left(\text{the result of substituting }
1, 0, 0, 0, \ldots \text{ for } x_1, x_2, x_3, \ldots \text{ in }
x_1^n+x_2^n+x_3^n+x_4^n+\cdots \right) \\
& \qquad \left(\text{since } p_n = x_1^n+x_2^n+x_3^n+x_4^n+\cdots \right) \\
&= 1^n + 0^n + 0^n + 0^n + \cdots = 1 + 0 + 0 + 0 + \cdots = 1 \\
&= \epsilon_1\left(p_n\right)
\qquad \left(\text{since } \epsilon_1\left(p_n\right) =1
\text{ (by the definition of } \epsilon_1 \text{)} \right) .
\end{align*}
In other words, the two $\QQ$-algebra homomorphisms $U$ and
$\epsilon_1$ are equal to each other on each element of the family
$\left(p_n\right)_{n\geq 1}$. But since this family
$\left(p_n\right)_{n\geq 1}$ is a generating set of the
$\QQ$-algebra $\Lambda_\QQ$, this yields that the two
$\QQ$-algebra homomorphisms $U$ and $\epsilon_1$ must be identical
(because if two $\QQ$-algebra homomorphisms are equal to each other on
each element of a generating set of their domain, then they must be
identical). That is, $\epsilon_1 = U$. Hence, every $f \in \Lambda_\QQ$
satisfies
$\epsilon_1\left(f\right) = U\left(f\right) = f\left(1\right)$
(by the definition of $U$). This solves
Exercise~\ref{exe.Lambda.maps-on-pn}(i).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.witt.ghost-equiv}}

\textit{Solution to Exercise~\ref{exe.witt.ghost-equiv}.} We are going
to be brief; more detailed proofs for everything except of the (very easy)
equivalence $\mathcal{D}\Longleftrightarrow\mathcal{J}$ can be found at
\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/witt5f.pdf}
(along with generalizations and additional
equivalences in the case of $A = \ZZ$).

As suggested by the hint, we first prove some elementary facts of number theory:

\begin{itemize}
\item Every positive integer $n$ satisfies%
\begin{equation}
\sum_{d\mid n}\phi\left(  d\right)  =n. \label{sol.exe.witt.ghost-equiv.phi*1}%
\end{equation}


\textit{Proof:} Let $n$ be a positive integer. For every positive divisor $d$ of $n$,
there is a bijection%
\begin{align*}
\left\{  i\in\left\{  1,2,\ldots,n\right\}  \ \mid\ \gcd\left(  i,n\right)
=d\right\}   &  \rightarrow\left\{  j\in\left\{  1,2,\ldots,\dfrac{n}{d}\right\}
\ \mid\ j\text{ is coprime to }\dfrac{n}{d}\right\}  ,\\
i  &  \mapsto\dfrac{i}{d}.
\end{align*}
Hence, for every positive divisor $d$ of $n$, we have%
\begin{align}
\left\vert \left\{  i\in\left\{  1,2,\ldots,n\right\}  \ \mid\ \gcd\left(
i,n\right)  =d\right\}  \right\vert  &  =\left\vert \left\{  j\in\left\{
1,2,\ldots,\dfrac{n}{d}\right\}  \ \mid\ j\text{ is coprime to }\dfrac{n}%
{d}\right\}  \right\vert \nonumber\\
&  =\phi\left(  \dfrac{n}{d}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{because
this is how }\phi\left(  \dfrac{n}{d}\right)  \text{ was defined}\right)  .
\label{sol.exe.witt.ghost-equiv.phi*1.pf.num}%
\end{align}
Now,
\[
n=\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =\sum_{d\mid
n}\underbrace{\left\vert \left\{  i\in\left\{  1,2,\ldots,n\right\}  \ \mid
\ \gcd\left(  i,n\right)  =d\right\}  \right\vert }_{=\phi\left(  \dfrac{n}%
{d}\right)  }=\sum_{d\mid n}\phi\left(  \dfrac{n}{d}\right)  =\sum_{d\mid
n}\phi\left(  d\right)
\]
(here, we have substituted $d$ for $\dfrac{n}{d}$ in the sum). This proves
(\ref{sol.exe.witt.ghost-equiv.phi*1}).

\item Every positive integer $n$ satisfies%
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  =\delta_{n,1}.
\label{sol.exe.witt.ghost-equiv.mu*1}%
\end{equation}


\textit{Proof:} Let $n$ be a positive integer. Let $n=p_{1}^{a_{1}}%
p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ be the prime factorization of $n$, with all
of $a_{1},a_{2},\ldots,a_{k}$ being positive integers (and with
$p_1, p_2, \ldots, p_k$ being distinct primes). Then, the
\textbf{squarefree} positive divisors of $n$ all have the form $\prod_{i\in I}p_{i}$
for some subset $I$ of $\left\{  1,2,\ldots,k\right\}  $. More precisely,
there exists a bijection%
\begin{align}
\left\{  I\subset\left\{  1,2,\ldots,k\right\}  \right\}   &  \rightarrow
\left(  \text{the set of all squarefree positive divisors of }n\right)  ,\nonumber\\
I  &  \mapsto\prod_{i\in I}p_{i}. \label{sol.exe.witt.ghost-equiv.mu*1.pf.bij}%
\end{align}
Now, each positive divisor $d$ of $n$ is either squarefree or not.
Hence,
\begin{align*}
\sum_{d\mid n}\mu\left(  d\right)   &  =\sum\limits_{\substack{d\mid n;\\d\text{ is
squarefree}}}\mu\left(  d\right)  +\sum\limits_{\substack{d\mid n;\\d\text{ is not
squarefree}}}\underbrace{\mu\left(  d\right)  }_{\substack{=0\\\text{(by the
definition}\\\text{of }\mu\text{)}}}=\sum\limits_{\substack{d\mid n;\\d\text{ is
squarefree}}}\mu\left(  d\right) \\
&  =\sum_{I\subset\left\{  1,2,\ldots,k\right\}  }\underbrace{\mu\left(
\prod_{i\in I}p_{i}\right)  }_{\substack{=\left(  -1\right)  ^{\left\vert
I\right\vert }\\\text{(since }\prod_{i\in I}p_{i}\text{ is squarefree}%
\\\text{and has }\left\vert I\right\vert \text{ prime factors)}}} \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\prod_{i\in I}p_{i}\text{ for }d\\
\text{due to the bijection (\ref{sol.exe.witt.ghost-equiv.mu*1.pf.bij})}%
\end{array}
\right) \\
&  =\sum_{I\subset\left\{  1,2,\ldots,k\right\}  }
\left(  -1\right)^{\left\vert I\right\vert }
= \begin{cases}
1,& \text{if }\left\{  1,2,\ldots,k\right\}  =\varnothing;\\
0,& \text{otherwise}
\end{cases}
\quad
= \begin{cases}
1,& \text{if }k=0;\\
0,& \text{otherwise}
\end{cases} \\
& = \begin{cases}
1,& \text{if }n=1;\\
0,& \text{otherwise}
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }k\text{ is the number of distinct prime factors of }n\text{,}\\
\text{and thus we have }k=0\text{ if and only if }n=1
\end{array}
\right) \\
&  =\delta_{n,1},
\end{align*}
which proves (\ref{sol.exe.witt.ghost-equiv.mu*1}).

\item Every positive integer $n$ satisfies
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  \dfrac{n}{d}=\phi\left(  n\right)  .
\label{sol.exe.witt.ghost-equiv.mu*n}
\end{equation}


\textit{Proof:} This is an elementary fact, but we have a hammer at our
disposal, and this looks conspicuously like a nail. Let us define a
$\ZZ$-coalgebra. Namely, let $\mathbf{T}$ be the free $\ZZ$-module
with basis $\left( t_n \right)  _{n\geq 1}$. We define a
$\ZZ$-coalgebra structure on $\mathbf{T}$ by setting
\[
\Delta\left(  t_{n}\right)  =\sum\limits_{\substack{d,e\in\left\{  1,2,3,\ldots
\right\}  ;\\de=n}}t_{d}\otimes t_{e}=\sum_{d\mid n}t_{d}\otimes
t_{n/d}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \epsilon\left(
t_{n}\right)  =\delta_{n,1}
\]
for every $n\in\left\{  1,2,3,\ldots\right\}  $. It is straightforward to
check that this makes $\mathbf{T}$ into a cocommutative coalgebra. Hence,
$\left( \Hom_\ZZ \left( \mathbf{T} , \ZZ \right) , \star \right)$
is a $\ZZ$-algebra with unity $\epsilon$.\ \ \ \ \footnote{Number
theorists will recognize this $\ZZ$-algebra as an isomorphic
version of the so-called
\dfn{algebra of formal Dirichlet series over $\ZZ$}\index{Dirichlet series}.
The isomorphism from
$\left( \Hom_\ZZ \left( \mathbf{T} , \ZZ \right) , \star \right)$
to the algebra of formal Dirichlet series over $\ZZ$
takes an element
$f \in \left( \Hom_\ZZ \left( \mathbf{T} , \ZZ \right) , \star \right)$
to the formal Dirichlet series
$\sum_{n = 1}^\infty f \left( n \right) n^{-s}$.}
This algebra
$\left(   \Hom_{ \ZZ }\left(  \mathbf{T}, \ZZ \right)  ,\star\right)  $
is
commutative\footnote{This follows from
Exercise~\ref{exe.commutative.convolution} (applied to $\ZZ$,
$\mathbf{T}$ and $\ZZ$ instead of $\kk$, $C$ and $A$).}. We
define four elements of $ \Hom_{ \ZZ }\left(
\mathbf{T}, \ZZ \right)  $:

\begin{itemize}
\item a $\ZZ$-linear map $\widetilde{\phi}:\mathbf{T}\rightarrow
 \ZZ$ which sends $t_{n}$ to $\phi\left(  n\right)  $ for every $n\geq1$;

\item a $\ZZ$-linear map $\widetilde{\mu}:\mathbf{T}\rightarrow
 \ZZ$ which sends $t_{n}$ to $\mu\left(  n\right)  $ for every $n\geq1$;

\item a $\ZZ$-linear map $\widetilde{ \id }:\mathbf{T}%
\rightarrow \ZZ$ which sends $t_{n}$ to $n$ for every $n\geq1$;

\item a $\ZZ$-linear map $\widetilde{\mathbf{1}}:\mathbf{T}%
\rightarrow \ZZ$ which sends $t_{n}$ to $1$ for every $n\geq1$.
\end{itemize}

Then, the identity that we want to prove -- i.e., the identity
(\ref{sol.exe.witt.ghost-equiv.mu*n}) -- is equivalent to the claim that
$\widetilde{\mu}\star\widetilde{ \id }=\widetilde{\phi}%
$.\ \ \ \ \footnote{This is because the left hand side of
(\ref{sol.exe.witt.ghost-equiv.mu*n}) equals $\sum_{d\mid n}\widetilde{\mu
}\left(  t_{d}\right)  \widetilde{ \id }\left(  t_{n/d}\right)
=m\left(  \left(  \widetilde{\mu}\otimes\widetilde{ \id }\right)
\underbrace{\left(  \sum_{d\mid n}t_{d}\otimes t_{n/d}\right)  }%
_{=\Delta\left(  t_{n}\right)  }\right)  =m\left(  \left(  \widetilde{\mu
}\otimes\widetilde{ \id }\right)  \left(  \Delta\left(
t_{n}\right)  \right)  \right)  =\left(  \widetilde{\mu}\star
\widetilde{ \id }\right)  \left(  t_{n}\right)  $, whereas the
right hand side equals $\widetilde{\phi}\left(  t_{n}\right)  $.} Similarly,
the (already proven) identity (\ref{sol.exe.witt.ghost-equiv.phi*1}) is
equivalent to $\widetilde{\phi}\star\widetilde{\mathbf{1}}%
=\widetilde{ \id }$, and the (already proven) identity
(\ref{sol.exe.witt.ghost-equiv.mu*1}) is equivalent to $\widetilde{\mu}%
\star\widetilde{\mathbf{1}}=\epsilon$. Thus,
\begin{align*}
\widetilde{\mu}\star\underbrace{\widetilde{ \id }}%
_{=\widetilde{\phi}\star\widetilde{\mathbf{1}}}  &  =\widetilde{\mu}%
\star\widetilde{\phi}\star\widetilde{\mathbf{1}}=\underbrace{\widetilde{\mu
}\star\widetilde{\mathbf{1}}}_{=\epsilon}\star\widetilde{\phi}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \Hom_{ \ZZ }
\left(  \mathbf{T}, \ZZ \right)  ,\star\right)
\text{ is commutative}\right) \\
&  =\epsilon\star\widetilde{\phi}=\widetilde{\phi}.
\end{align*}
As we know, this is equivalent to (\ref{sol.exe.witt.ghost-equiv.mu*n}), so
that (\ref{sol.exe.witt.ghost-equiv.mu*n}) is proven.

It was not really necessary to phrase this argument in terms of coalgebras;
this was only done to illustrate a use of the latter. Our proof can just as
well be rewritten as a manipulation of sums, or (as a compromise between
concreteness and structure) it can be paraphrased by using the
\dfn{Dirichlet convolution}, which is the operation taking two maps
$f,g:\left\{  1,2,3,\ldots\right\}  \rightarrow \ZZ$ to a third map
$h:\left\{  1,2,3,\ldots\right\}  \rightarrow \ZZ$ defined by $h\left(
n\right)  =\sum_{d\mid n}f\left(  d\right)  g\left(  \dfrac{n}{d}\right)  $.
Of course, this Dirichlet convolution is the same as our convolution $\star$
on $ \Hom_{ \ZZ }\left(  \mathbf{T}, \ZZ %
\right)  $, with the only difference that $\ZZ$-linear maps
$\mathbf{T}\rightarrow \ZZ$ are replaced by arbitrary maps $\left\{
1,2,3,\ldots\right\}  \rightarrow \ZZ$.

\item Every positive integer $n$ satisfies
\begin{equation}
\sum_{d\mid n}d\mu\left(  d\right)  \phi\left(  \dfrac{n}{d}\right)
=\mu\left(  n\right)  .
\label{sol.exe.witt.ghost-equiv.nmu*phi}
\end{equation}


\textit{Proof:} We use the setup we prepared in the proof of
(\ref{sol.exe.witt.ghost-equiv.mu*n}). Additionally, we define $\widetilde{\mu
^{\prime}}:\mathbf{T}\rightarrow \ZZ$ as the $\ZZ$-linear map
which sends $t_{n}$ to $n\mu\left(  n\right)  $ for every $n\geq1$. Then,
$\widetilde{\mu^{\prime}}\star\widetilde{ \id }=\epsilon$,
because every positive integer $n$ satisfies%
\[
\left(  \widetilde{\mu^{\prime}}\star\widetilde{ \id }\right)
\left(  t_{n}\right)  =\sum_{d\mid n}d\mu\left(  d\right)  \dfrac{n}{d}%
=\sum_{d\mid n}n\mu\left(  d\right)  =n\underbrace{\sum_{d\mid n}\mu\left(
d\right)  }_{\substack{=\delta_{n,1}\\\text{(by
(\ref{sol.exe.witt.ghost-equiv.mu*1}))}}}=n\delta_{n,1}=\delta_{n,1}%
=\epsilon\left(  t_{n}\right)  .
\]
But recall that $\widetilde{\phi}=\widetilde{\mu}\star
\widetilde{ \id }=\widetilde{ \id }\star
\widetilde{\mu}$ (since $\left(   \Hom_{ \ZZ }
\left(  \mathbf{T}, \ZZ \right)  ,\star\right)  $ is commutative), and
thus
\[
\widetilde{\mu^{\prime}}\star\underbrace{\widetilde{\phi}}%
_{=\widetilde{ \id }\star\widetilde{\mu}}%
=\underbrace{\widetilde{\mu^{\prime}}\star\widetilde{ \id }%
}_{=\epsilon}\star\widetilde{\mu}=\epsilon\star\widetilde{\mu}=\widetilde{\mu
}.
\]
This is easily seen to be equivalent to
(\ref{sol.exe.witt.ghost-equiv.nmu*phi}), and so
(\ref{sol.exe.witt.ghost-equiv.nmu*phi}) is proven.

\item If $k$ is a positive integer, and if $p\in \NN$, $a\in A$ and
$b\in A$ are such that $a\equiv b\operatorname{mod}p^{k}A$, then
\begin{equation}
a^{p^{\ell}}\equiv b^{p^{\ell}}\operatorname{mod}p^{k+\ell}%
A\ \ \ \ \ \ \ \ \ \ \text{for every }\ell\in \NN .
\label{sol.exe.witt.ghost-equiv.hensel}
\end{equation}


\textit{Proof:} Let $k$ be a positive integer. Let $p\in \NN$, $a\in A$
and $b\in A$ be such that $a\equiv b\operatorname{mod}p^{k}A$. We need to
prove (\ref{sol.exe.witt.ghost-equiv.hensel}). It is clearly enough to show
that $a^{p}\equiv b^{p}\operatorname{mod}p^{k+1}A$, because then
(\ref{sol.exe.witt.ghost-equiv.hensel}) will follow by induction over $\ell$.
But we have $a\equiv b\operatorname{mod}pA$ (since $a\equiv
b\operatorname{mod}p^{k}A$ and since $k$ is positive) and therefore
\begin{align*}
a^{p-1}+a^{p-2}b+\cdots+b^{p-1}
&\equiv b^{p-1}+b^{p-2}b+\cdots+b^{p-1} \\
&=\underbrace{b^{p-1}+b^{p-1}+\cdots+b^{p-1}}_{p\text{ terms}}=pb^{p-1}%
\equiv0\operatorname{mod}pA,
\end{align*}
so that $a^{p-1}+a^{p-2}b+\cdots+b^{p-1}\in pA$. Thus,
\[
a^{p}-b^{p}=\underbrace{\left(  a-b\right)  }_{\substack{\in p^{k}%
A\\\text{(since }a\equiv b\operatorname{mod}p^{k}A\text{)}}%
}\underbrace{\left(  a^{p-1}+a^{p-2}b+\cdots+b^{p-1}\right)  }_{\in pA}
\in\left(  p^{k}A\right)  \left(  pA\right)  =p^{k+1}A,
\]
so that $a^{p}\equiv b^{p}\operatorname{mod}p^{k+1}A$. This proves
(\ref{sol.exe.witt.ghost-equiv.hensel}).

\item Here is a slightly more useful corollary of
(\ref{sol.exe.witt.ghost-equiv.hensel}): If a prime number $p$ and two
elements $a$ and $b$ of $A$ are such that $a\equiv b\operatorname{mod}pA$,
then
\begin{equation}
a^{N}\equiv b^{N}\operatorname{mod}p^{v_{p}\left(  N\right)  +1}%
A\ \ \ \ \ \ \ \ \ \ \text{for every }N\in\left\{  1,2,3,\ldots\right\}  .
\label{sol.exe.witt.ghost-equiv.hensel-v}
\end{equation}


\textit{Proof:} Let $p$ be a prime number, and let $a$ and $b$ be two elements
of $A$ such that $a\equiv b\operatorname{mod}pA$. Let $N\in\left\{
1,2,3,\ldots\right\}  $. Write $N$ in the form $N=p^{v_{p}\left(  N\right)
}M$ for some positive integer $M$. Then,
(\ref{sol.exe.witt.ghost-equiv.hensel}) (applied to $\ell=v_{p}\left(
N\right)  $ and $k=1$) yields $a^{p^{v_{p}\left(  N\right)  }}\equiv
b^{p^{v_{p}\left(  N\right)  }}\operatorname{mod}p^{v_{p}\left(  N\right)
+1}A$, and because of $N=p^{v_{p}\left(  N\right)  }M$ we have%
\begin{align*}
a^{N}  &  =a^{p^{v_{p}\left(  N\right)  }M}=\left(  a^{p^{v_{p}\left(
N\right)  }}\right)  ^{M}\equiv\left(  b^{p^{v_{p}\left(  N\right)  }}\right)
^{M}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{p^{v_{p}\left(  N\right)  }%
}\equiv b^{p^{v_{p}\left(  N\right)  }}\operatorname{mod}p^{v_{p}\left(
N\right)  +1}A\right) \\
&  =b^{p^{v_{p}\left(  N\right)  }M}=b^{N}\operatorname{mod}p^{v_{p}\left(
N\right)  +1}A\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p^{v_{p}\left(
N\right)  }M=N\right)  .
\end{align*}
This proves (\ref{sol.exe.witt.ghost-equiv.hensel-v}).
\end{itemize}

We shall also use a fact from commutative algebra -- namely, one of the
versions of the Chinese Remainder Theorem of ring theory:

\begin{theorem}
\label{thm.sol.exe.witt.ghost-equiv.CRT}
Let $A$ be a commutative ring. Let $\mathbf{S}$ be a finite set. For
every $s \in \mathbf{S}$, let $I_s$ be an ideal of $A$. Assume that
the ideals $I_s$ of $A$ are \emph{comaximal}\index{comaximal ideals}%
\footnote{Some authors
use the word ``coprime'' instead of ``comaximal'' here.}; this means
that every two distinct elements $s$ and $t$ of $\mathbf{S}$ satisfy
$I_s + I_t = A$.
Then:
\begin{enumerate}
\item[(a)] We have
\[
\bigcap\limits_{s \in \mathbf{S}} I_s
= \prod\limits_{s \in \mathbf{S}} I_s .
\]
\item[(b)] The canonical ring homomorphism
\[
A / \left(\bigcap\limits_{s \in \mathbf{S}} I_s\right) \to
\prod\limits_{s \in \mathbf{S}} \left(A / I_s\right), \qquad
a + \bigcap\limits_{s \in \mathbf{S}} I_s \mapsto
\left(a + I_s\right)_{s \in \mathbf{S}}
\]
is well-defined and a ring isomorphism.
\end{enumerate}
\end{theorem}

See \url{http://stacks.math.columbia.edu/tag/00DT} (Lemma 10.14.4 (1)
in the Stacks Project, as of 4 April 2020) or many other sources
for a proof of Theorem \ref{thm.sol.exe.witt.ghost-equiv.CRT}.
We shall only use part (a) of this theorem.

As a consequence of
Theorem \ref{thm.sol.exe.witt.ghost-equiv.CRT}(a), we have the
following:

\begin{itemize}
\item If $n$ is a positive integer and if $A$ is a commutative ring,
then
\begin{equation}
\label{sol.exe.witt.ghost-equiv.CRT.primes}
\bigcap\limits_{\substack{p\text{ is a prime}\\\text{factor of }n}}
p^{v_p \left( n \right)} A
= \prod\limits_{\substack{p\text{ is a prime}\\\text{factor of }n}}
\left( p^{v_p \left( n \right)} A \right) .
\end{equation}
Indeed, this follows from
Theorem \ref{thm.sol.exe.witt.ghost-equiv.CRT}(a), since the ideals
$p^{v_p\left(n\right)} A$ of $A$ (for varying $p$)
are comaximal (because for any
two distinct primes $p$ and $q$, we can find integers $x$ and $y$
satisfying $p^{v_p\left(n\right)} x + q^{v_q\left(n\right)} y = 1$,
and therefore we have
$p^{v_p\left(n\right)} A + q^{v_q\left(n\right)} A = A$).
\end{itemize}

Now, we can step to proving the actual equivalences.

\textit{Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{J}%
$:} Assume that Assertion $\mathcal{D}$ holds. That is, there exists a
family $\left(  \alpha_{n}\right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots
\right\}  }$ of elements of $A$ such that every positive integer $n$ satisfies
$b_{n}=\sum_{d\mid n}d\alpha_{d}^{n/d}$. Consider this family $\left(
\alpha_{n}\right)  _{n\geq1}$. Consider also the family $\left(  w_{n}\right)
_{n\geq1}\in\Lambda_{ \ZZ }^{\left\{  1,2,3,\ldots\right\}  }$ defined in
Exercise \ref{exe.witt}(a). This family $\left(  w_{n}\right)  _{n\geq1}$ is
an algebraically independent generating set of $\Lambda_\ZZ$ (indeed, this is a
restatement of Exercise \ref{exe.witt}(d)). Hence, there exists a unique
$\ZZ$-algebra homomorphism $f:\Lambda_\ZZ\rightarrow A$ which satisfies%
\[
f\left(  w_{n}\right)  =\alpha_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\left\{  1,2,3,\ldots\right\}  .
\]
Consider this $f$. We have $p_{n}=\sum_{d\mid n}dw_{d}^{n/d}$ for every
positive integer $n$ (by Exercise \ref{exe.witt}(e)). Thus, for every positive
integer $n$, we have%
\begin{align*}
f\left(  p_{n}\right)   &  =f\left(  \sum_{d\mid n}dw_{d}^{n/d}\right)
=\sum_{d\mid n}d\left(  \underbrace{f\left(  w_{d}\right)  }_{=\alpha_d}
\right)  ^{n/d}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a
} \ZZ \text{-algebra homomorphism}\right) \\
&  =\sum_{d\mid n}d\alpha_{d}^{n/d}=b_{n}.
\end{align*}
Hence, there exists a ring homomorphism $\Lambda_{ \ZZ }\rightarrow A$
which, for every positive integer $n$, sends $p_{n}$ to $b_{n}$ (namely, $f$).
That is, Assertion $\mathcal{J}$ holds, and the implication $\mathcal{D}%
\Longrightarrow\mathcal{J}$ is proven.

\textit{Proof of the implication $\mathcal{J}\Longrightarrow\mathcal{D}%
$:} Assume that Assertion $\mathcal{J}$ holds. That is, there exists a
ring homomorphism $\Lambda_{ \ZZ }\rightarrow A$ which, for every
positive integer $n$, sends $p_{n}$ to $b_{n}$. Let $f$ be such a
homomorphism. Consider the family $\left(  w_{n}\right)  _{n\geq1}\in
\Lambda_{ \ZZ }^{\left\{  1,2,3,\ldots\right\}  }$ defined in Exercise
\ref{exe.witt}(a). For every positive integer $n$, we have $p_{n}=\sum_{d\mid
n}dw_{d}^{n/d}$ (by Exercise \ref{exe.witt}(e)) and thus%
\[
f\left(  p_{n}\right)  =f\left(  \sum_{d\mid n}dw_{d}^{n/d}\right)
=\sum_{d\mid n}d\left(  f\left(  w_{d}\right)  \right)  ^{n/d}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a ring homomorphism}%
\right)  .
\]
Since $f\left(  p_{n}\right)  =b_{n}$ (by the definition of $f$), this
rewrites as $b_{n}=\sum_{d\mid n}d\left(  f\left(  w_{d}\right)  \right)
^{n/d}$. Thus, there exists a family $\left(  \alpha_{n}\right)  _{n\geq1}\in
A^{\left\{  1,2,3,\ldots\right\}  }$ of elements of $A$ such that every
positive integer $n$ satisfies $b_{n}=\sum_{d\mid n}d\alpha_{d}^{n/d}$
(namely, such a family can be defined by $\alpha_{n}=f\left(  w_{n}\right)
$). Assertion $\mathcal{D}$ thus holds. We have now proven the implication
$\mathcal{J}\Longrightarrow\mathcal{D}$.

\textit{Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{C}%
$:} Assume that Assertion $\mathcal{D}$ holds. That is, there exists a
family $\left(  \alpha_{n}\right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots
\right\}  }$ of elements of $A$ such that every positive integer $n$ satisfies
$b_{n}=\sum_{d\mid n}d\alpha_{d}^{n/d}$. Consider this family $\left(
\alpha_{n}\right)  _{n\geq1}$.

We need to prove that Assertion $\mathcal{C}$ holds, i.e., that we have%
\begin{equation}
\varphi_{p}\left(  b_{n/p}\right)  \equiv b_{n}\operatorname{mod}%
p^{v_{p}\left(  n\right)  }A \label{sol.exe.witt.ghost-equiv.DC.goal}%
\end{equation}
for every positive integer $n$ and every prime factor $p$ of $n$. So let us
fix a positive integer $n$ and a prime factor $p$ of $n$. We need to prove
(\ref{sol.exe.witt.ghost-equiv.DC.goal}).

The definition of the family $\left(  \alpha_{n}\right)  _{n\geq1}$ shows that
$b_{n/p}=\sum_{d\mid n/p}d\alpha_{d}^{\left(  n/p\right)  /d}$, so that%
\[
\varphi_{p}\left(  b_{n/p}\right)  =\varphi_{p}\left(  \sum_{d\mid n/p}%
d\alpha_{d}^{\left(  n/p\right)  /d}\right)  =\sum_{d\mid n/p}d\left(
\varphi_{p}\left(  \alpha_{d}\right)  \right)  ^{\left(  n/p\right)
/d}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi_{p}\text{ is a ring
endomorphism}\right)  .
\]
On the other hand,
\begin{equation}
\sum_{d\mid n}d\alpha_{d}^{n/d}=\sum_{d\mid n/p}d\alpha_{d}^{n/d}%
+\sum\limits_{\substack{d\mid n;\\d\nmid n/p}}\underbrace{d}_{\substack{\equiv
0\operatorname{mod}p^{v_{p}\left(  n\right)  }A\\\text{(since }d\mid n\text{
and }d\nmid n/p\\\text{yield }p^{v_{p}\left(  n\right)  }\mid d\text{)}%
}}\alpha_{d}^{n/d}\equiv\sum_{d\mid n/p}d\alpha_{d}^{n/d}\operatorname{mod}%
p^{v_{p}\left(  n\right)  }A. \label{sol.exe.witt.ghost-equiv.DC.1}%
\end{equation}
Thus, if we succeed to prove that
\begin{equation}
d\left(  \varphi_{p}\left(  \alpha_{d}\right)  \right)  ^{\left(  n/p\right)
/d}\equiv d\alpha_{d}^{n/d}\operatorname{mod}p^{v_{p}\left(  n\right)
}A\ \ \ \ \ \ \ \ \ \ \text{for every }d\mid n/p,
\label{sol.exe.witt.ghost-equiv.DC.goal2}
\end{equation}
then we will obtain%
\begin{align*}
\varphi_{p}\left(  b_{n/p}\right)   &  =\sum_{d\mid n/p}\underbrace{d\left(
\varphi_{p}\left(  \alpha_{d}\right)  \right)  ^{\left(  n/p\right)  /d}%
}_{\equiv d\alpha_{d}^{n/d}\operatorname{mod}p^{v_{p}\left(  n\right)  }%
A}\equiv\sum_{d\mid n/p}d\alpha_{d}^{n/d}\equiv\sum_{d\mid n}d\alpha_{d}%
^{n/d}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.witt.ghost-equiv.DC.1})}\right) \\
&  =b_{n}\operatorname{mod}p^{v_{p}\left(  n\right)  }%
A\ \ \ \ \ \ \ \ \ \ \left(  \text{since }b_{n}=\sum_{d\mid n}d\alpha
_{d}^{n/d}\right)  ,
\end{align*}
and thus our goal (proving (\ref{sol.exe.witt.ghost-equiv.DC.goal})) will be
achieved. Hence, it remains to prove (\ref{sol.exe.witt.ghost-equiv.DC.goal2}).

So let $d$ be any positive divisor of $n/p$. Then, $\varphi_{p}\left(
\alpha_{d}\right)  \equiv\alpha_{d}^{p}\operatorname{mod}pA$ (because of the
axiom $\varphi_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}pA$ for
every $a\in A$). Thus, (\ref{sol.exe.witt.ghost-equiv.hensel-v}) (applied to
$a=\varphi_{p}\left(  \alpha_{d}\right)  $, $b=\alpha_{p}^{d}$ and $N=\left(
n/p\right)  /d$) yields $\left(  \varphi_{p}\left(  \alpha_{d}\right)
\right)  ^{\left(  n/p\right)  /d}\equiv\left(  \alpha_{d}^{p}\right)
^{\left(  n/p\right)  /d}\operatorname{mod}p^{v_{p}\left(  \left(  n/p\right)
/d\right)  +1}A$. Since $\left(  \alpha_{d}^{p}\right)  ^{\left(  n/p\right)
/d}=\alpha_{d}^{n/d}$ and $v_{p}\left(  \left(  n/p\right)  /d\right)
+1=v_{p}\left(  n/d\right)  $, this rewrites as $\left(  \varphi_{p}\left(
\alpha_{d}\right)  \right)  ^{\left(  n/p\right)  /d}\equiv\alpha_{d}%
^{n/d}\operatorname{mod}p^{v_{p}\left(  n/d\right)  }A$. Multiplying this by $d$
results in $d\left(  \varphi_{p}\left(  \alpha_{d}\right)  \right)  ^{\left(
n/p\right)  /d}\equiv d\alpha_{d}^{n/d}\operatorname{mod}dp^{v_{p}\left(
n/d\right)  }A$. Since $dp^{v_{p}\left(  n/d\right)  }$ is divisible by
$p^{v_{p}\left(  n\right)  }$, this yields
(\ref{sol.exe.witt.ghost-equiv.DC.goal2}). This completes the proof of
(\ref{sol.exe.witt.ghost-equiv.DC.goal2}), and thus also that of the
implication $\mathcal{D}\Longrightarrow\mathcal{C}$.

\textit{Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{D}%
$:} Assume that Assertion $\mathcal{C}$ holds. Thus, for every
positive integer $n$ and every prime factor $p$ of $n$, we have%
\begin{equation}
\varphi_{p}\left(  b_{n/p}\right)  \equiv b_{n}\operatorname{mod}%
p^{v_{p}\left(  n\right)  }A. \label{sol.exe.witt.ghost-equiv.CD.C}%
\end{equation}
We now need to prove that Assertion $\mathcal{D}$ holds as well. In other
words, we need to show that there exists a family $\left(  \alpha_{n}\right)
_{n\geq1}\in A^{\left\{  1,2,3,\ldots\right\}  }$ of elements of $A$ such that
every positive integer $n$ satisfies $b_{n}=\sum_{d\mid n}d\alpha_{d}^{n/d}$.
In other words (renaming $n$ as $m$), we need to show that there exists a
family $\left(  \alpha_{m}\right)  _{m\geq1}\in A^{\left\{  1,2,3,\ldots
\right\}  }$ of elements of $A$ such that every positive integer $m$ satisfies
$b_{m}=\sum_{d\mid m}d\alpha_{d}^{m/d}$.

We construct this family $\left(  \alpha_{m}\right)  _{m\geq1}$ recursively.
So we fix some $n\geq1$, and assume that $\alpha_{m}$ is already constructed
for every positive integer $m<n$ in such a way that
\begin{equation}
\text{the equality }b_{m}=\sum_{d\mid m}d\alpha_{d}^{m/d}\text{ is satisfied
for every positive integer }m<n. \label{sol.exe.witt.ghost-equiv.CD.indass}%
\end{equation}
We now need to construct an $\alpha_{n}\in A$ such that $b_{m}=\sum_{d\mid
m}d\alpha_{d}^{m/d}$ is satisfied for $m=n$. In other words, we need to
construct an $\alpha_{n}\in A$ satisfying $b_{n}=\sum_{d\mid n}d\alpha
_{d}^{n/d}$.

Let us first choose $\alpha_{n}$ \textbf{arbitrarily} (with the intention to
tweak it later). Let $p$ be any prime factor of $n$. Then, applying
(\ref{sol.exe.witt.ghost-equiv.CD.indass}) to $m=n/p$, we obtain $b_{n/p}%
=\sum_{d\mid n/p}d\alpha_{d}^{\left(  n/p\right)  /d}$. This allows us to
prove that $\varphi_{p}\left(  b_{n/p}\right)  \equiv\sum_{d\mid n}d\alpha
_{d}^{n/d}\operatorname{mod}p^{v_{p}\left(  n\right)  }A$ holds (just as in
the proof of the implication $\mathcal{D}\Longrightarrow\mathcal{C}$).
Compared with (\ref{sol.exe.witt.ghost-equiv.CD.C}), this yields $b_{n}%
\equiv\sum_{d\mid n}d\alpha_{d}^{n/d}\operatorname{mod}p^{v_{p}\left(
n\right)  }A$. That is, $b_{n}-\sum_{d\mid n}d\alpha_{d}^{n/d}\in
p^{v_{p}\left(  n\right)  }A$.

Now, let us forget that we fixed $p$. We thus have shown that (with our
arbitrary choice of $\alpha_{n}$) we have $b_{n}-\sum_{d\mid n}d\alpha
_{d}^{n/d}\in p^{v_{p}\left(  n\right)  }A$ for every prime factor $p$ of $n$.
As a consequence,
\begin{align*}
b_{n}-\sum_{d\mid n}d\alpha_{d}^{n/d}  &  \in\bigcap\limits_{\substack{p\text{ is a
prime}\\\text{factor of }n}}p^{v_{p}\left(  n\right)  }A=\prod\limits
_{\substack{p\text{ is a prime}\\\text{factor of }n}}\left(  p^{v_{p}\left(
n\right)  }A\right)
\ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{sol.exe.witt.ghost-equiv.CRT.primes}}\right) \\
&  =\underbrace{\left(  \prod\limits_{\substack{p\text{ is a prime}\\\text{factor of
}n}}p^{v_{p}\left(  n\right)  }\right)  }_{=n}A=nA.
\end{align*}
In other words, there exists a $\gamma\in A$ such that $b_{n}-\sum_{d\mid
n}d\alpha_{d}^{n/d}=n\gamma$. Consider this $\gamma$. Now, if we replace
$\alpha_{n}$ by $\alpha_{n}+\gamma$, then the sum $\sum_{d\mid n}d\alpha
_{d}^{n/d}$ increases by $n\gamma=b_{n}-\sum_{d\mid n}d\alpha_{d}^{n/d}$, and
therefore becomes precisely $b_{n}$. Hence, by replacing $\alpha_{n}$ by
$\alpha_{n}+\gamma$, we achieve that $b_{n}=\sum_{d\mid n}d\alpha_{d}^{n/d}$
holds. Thus, we have found the $\alpha_{n}$ we were searching for, and the
recursive construction of the family $\left(  \alpha_{m}\right)  _{m\geq1}$
has proceeded by one more step. The proof of the implication $\mathcal{C}%
\Longrightarrow\mathcal{D}$ is thus complete.

\textit{Proof of the implication $\mathcal{E}\Longrightarrow\mathcal{C}%
$:} The proof of the implication $\mathcal{E}\Longrightarrow
\mathcal{C}$ proceeds exactly as our above proof of the implication
$\mathcal{D}\Longrightarrow\mathcal{C}$, with the following changes:

\begin{itemize}
\item Every appearance of $\alpha_{i}$ for some $i\geq1$ must be replaced by
the corresponding $\beta_{i}$.

\item Every time an element of $A$ was taken to the $k$-th power (for some
$k\in\left\{  1,2,3,\ldots\right\}  $) in our proof of the implication
$\mathcal{D}\Longrightarrow\mathcal{C}$, it needs now to be subjected to the
ring endomorphism $\varphi_{k}$ instead. So, for example, $\alpha_{d}^{n/d}$
is replaced by $\varphi_{n/d}\left(  \beta_{d}\right)  $ everywhere (remember
that $\alpha_{d}$ becomes $\beta_{d}$). Note that this concerns only elements
of $A$. We don't replace the power $p^{v_{p}\left(  n\right)  }$ by anything.

\item The equality%
\[
\varphi_{p}\left(  \sum_{d\mid n/p}d\alpha_{d}^{\left(  n/p\right)
/d}\right)  =\sum_{d\mid n/p}d\left(  \varphi_{p}\left(  \alpha_{d}\right)
\right)  ^{\left(  n/p\right)  /d}%
\]
is replaced by%
\[
\varphi_{p}\left(  \sum_{d\mid n/p}d\varphi_{\left(  n/p\right)  /d}\left(
\beta_{d}\right)  \right)  =\sum_{d\mid n/p}d\varphi_{\left(  n/p\right)
/d}\left(  \varphi_{p}\left(  \beta_{d}\right)  \right)  ,
\]
whose proof uses the $\ZZ$-linearity of $\varphi_{p}$ and the fact that
$\varphi_{p}\circ\varphi_{\left(  n/p\right)  /d}=\varphi_{n/d}=\varphi
_{\left(  n/p\right)  /d}\circ\varphi_{p}$.

\item The proof of (\ref{sol.exe.witt.ghost-equiv.DC.goal2}) needs to be
replaced by a proof of the congruence%
\begin{equation}
d\varphi_{\left(  n/p\right)  /d}\left(  \varphi_{p}\left(  \beta_{d}\right)
\right)  \equiv d\varphi_{n/d}\left(  \beta_{d}\right)  \operatorname{mod}%
p^{v_{p}\left(  n\right)  }A\ \ \ \ \ \ \ \ \ \ \text{for every }d\mid n/p.
\label{sol.exe.witt.ghost-equiv.EC.goal2}%
\end{equation}
Fortunately, the latter congruence is obvious, since $\varphi_{\left(
n/p\right)  /d}\circ\varphi_{p}=\varphi_{n/d}$.
\end{itemize}

\textit{Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{E}%
$:} The proof of the implication $\mathcal{C}\Longrightarrow
\mathcal{E}$ can be obtained from the proof of the implication $\mathcal{C}%
\Longrightarrow\mathcal{D}$ using the same changes that were made to transform
the proof of the implication $\mathcal{D}\Longrightarrow\mathcal{C}$ into a
proof of the implication $\mathcal{E}\Longrightarrow\mathcal{C}$. The only new
``idea'' is to use the fact that $\varphi
_{1}= \id$ (in showing that if we replace $\beta_{n}$ by
$\beta_{n}+\gamma$, then the sum $\sum_{d\mid n}d\varphi_{n/d}\left(
\beta_{d}\right)  $ increases by $n\gamma$).

\textit{Proof of the implication $\mathcal{E}\Longrightarrow\mathcal{F}%
$:} Assume that Assertion $\mathcal{E}$ holds. That is, there exists a
family $\left(  \beta_{n}\right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots
\right\}  }$ of elements of $A$ such that every positive integer $n$ satisfies%
\begin{equation}
b_{n}=\sum_{d\mid n}d\varphi_{n/d}\left(  \beta_{d}\right)  .
\label{sol.exe.witt.ghost-equiv.EF.1}%
\end{equation}
Consider this family $\left(  \beta_{n}\right)  _{n\geq1}$. We need to prove
that Assertion $\mathcal{F}$ holds, i.e., that every positive integer $n$
satisfies%
\[
\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)  \in nA.
\]
We fix a positive integer $n$. Then, every positive divisor $d$ of $n$
satisfies $b_{n/d}=\sum_{e\mid n/d}e\varphi_{\left(  n/d\right)  /e}\left(
\beta_{e}\right)  $ (by (\ref{sol.exe.witt.ghost-equiv.EF.1}), applied to $n/d$
instead of $n$, and with the summation index $d$ renamed as $e$). Hence,%
\begin{align*}
\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  \underbrace{b_{n/d}%
}_{=\sum_{e\mid n/d}e\varphi_{\left(  n/d\right)  /e}\left(  \beta_{e}\right)
}\right)   &  =\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(
\sum_{e\mid n/d}e\varphi_{\left(  n/d\right)  /e}\left(  \beta_{e}\right)
\right) \\
&  =\sum_{d\mid n}\mu\left(  d\right)  \sum_{e\mid n/d}e\varphi_{d}\left(
\varphi_{\left(  n/d\right)  /e}\left(  \beta_{e}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi_{d}\text{ is linear}\right)
\\
&  =\underbrace{\sum_{d\mid n}\sum_{e\mid n/d}}_{=\sum_{e\mid n}\sum_{d\mid
n/e}}\mu\left(  d\right)  e\underbrace{\varphi_{d}\left(  \varphi_{\left(
n/d\right)  /e}\left(  \beta_{e}\right)  \right)  }_{\substack{=\left(
\varphi_{d}\circ\varphi_{\left(  n/d\right)  /e}\right)  \left(  \beta
_{e}\right)  \\=\varphi_{n/e}\left(  \beta_{e}\right)  }}\\
&  =\sum_{e\mid n}\underbrace{\sum_{d\mid n/e}\mu\left(  d\right)
}_{\substack{=\delta_{n/e,1}\\\text{(by (\ref{sol.exe.witt.ghost-equiv.mu*1}),
applied}\\\text{to }n/e\text{ instead of }n\text{)}}}e\varphi_{n/e}\left(
\beta_{e}\right)
=\sum_{e\mid n}\underbrace{\delta_{n/e,1}}_{=\delta_{e,n}}e\varphi
_{n/e}\left(  \beta_{e}\right) \\
&= \sum_{e\mid n} \delta_{e,n} e\varphi_{n/e}\left(  \beta_{e}\right)
= n\underbrace{\varphi_{n/n}}_{=\varphi
_{1}= \id }\left(  \beta_{n}\right)  =n\beta_{n}\in nA.
\end{align*}
Thus, Assertion $\mathcal{F}$ holds, so that we have proven the implication
$\mathcal{E}\Longrightarrow\mathcal{F}$.

\textit{Proof of the implication $\mathcal{F}\Longrightarrow\mathcal{E}%
$:} Assume that Assertion $\mathcal{F}$ holds. That is, every positive
integer $n$ satisfies%
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)  \in nA.
\label{sol.exe.witt.ghost-equiv.FE.1}%
\end{equation}


Now we need to prove that Assertion $\mathcal{E}$ holds, i.e., that there
exists a family $\left(  \beta_{n}\right)  _{n\geq1}\in A^{\left\{
1,2,3,\ldots\right\}  }$ of elements of $A$ such that every positive integer
$n$ satisfies%
\begin{equation}
b_{n}=\sum_{d\mid n}d\varphi_{n/d}\left(  \beta_{d}\right)  .
\label{sol.exe.witt.ghost-equiv.FE.goal}%
\end{equation}
We shall construct such a family $\left(  \beta_{n}\right)  _{n\geq1}$
recursively. That is, we fix some $N\in\left\{  1,2,3,\ldots\right\}  $, and
we assume that we already have constructed a $\beta_{n}\in A$ for every
positive integer $n<N$ in such a way that
(\ref{sol.exe.witt.ghost-equiv.FE.goal}) is satisfied for every positive
integer $n<N$. We now need to find a $\beta_{N}\in A$ such that
(\ref{sol.exe.witt.ghost-equiv.FE.goal}) is satisfied for $n=N$ as well.

From (\ref{sol.exe.witt.ghost-equiv.FE.1}) (applied to $n=N$), we have
$\sum_{d\mid N}\mu\left(  d\right)  \varphi_{d}\left(  b_{N/d}\right)  \in
nA$. Thus, there exists a $t\in A$ such that $\sum_{d\mid N}\mu\left(
d\right)  \varphi_{d}\left(  b_{N/d}\right)  =Nt$. Set Consider this $t$. Set
$\beta_{N}=t$. We have%
\begin{align}
Nt  &  =\sum_{d\mid N}\mu\left(  d\right)  \varphi_{d}\left(  b_{N/d}\right)
=\sum_{e\mid N}\mu\left(  e\right)  \varphi_{e}\left(  b_{N/e}\right)
=\sum\limits_{\substack{e\mid N;\\e>1}}\mu\left(  e\right)  \varphi_{e}\left(
\underbrace{b_{N/e}}_{\substack{=\sum_{d\mid N/e}d\varphi_{\left(  N/e\right)
/d}\left(  \beta_{d}\right)  \\\text{(by
(\ref{sol.exe.witt.ghost-equiv.FE.goal}), applied}\\\text{to }n=N/e\text{,
because }N/e<n\text{)}}}\right)  +\underbrace{\mu\left(  1\right)  }%
_{=1}\underbrace{\varphi_{1}}_{= \id }\left(  \underbrace{b_{N/1}%
}_{=b_{N}}\right) \nonumber\\
&  =\sum\limits_{\substack{e\mid N;\\e>1}}\mu\left(  e\right)  \varphi_{e}\left(
\sum_{d\mid N/e}d\varphi_{\left(  N/e\right)  /d}\left(  \beta_{d}\right)
\right)  +b_{N}\nonumber\\
&  =\sum\limits_{\substack{e\mid N;\\e>1}}\mu\left(  e\right)  \sum_{d\mid
N/e}d\underbrace{\varphi_{e}\left(  \varphi_{\left(  N/e\right)  /d}\left(
\beta_{d}\right)  \right)  }_{\substack{=\left(  \varphi_{e}\circ
\varphi_{\left(  N/e\right)  /d}\right)  \left(  \beta_{d}\right)
\\=\varphi_{N/d}\left(  \beta_{d}\right)  }}+b_{N}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\varphi_{e}\text{ is linear}\right) \nonumber\\
&  =\sum\limits_{\substack{e\mid N;\\e>1}}\mu\left(  e\right)  \sum_{d\mid
N/e}d\varphi_{N/d}\left(  \beta_{d}\right)  +b_{N}=\underbrace{\sum\limits
_{\substack{e\mid N;\\e>1}}\sum_{d\mid N/e}}_{=\sum\limits_{\substack{d\mid
N;\\d<N}}\sum\limits_{\substack{e\mid N/d;\\e>1}}}\mu\left(  e\right)  d\varphi
_{N/d}\left(  \beta_{d}\right)  +b_{N}\nonumber\\
&  =\sum\limits_{\substack{d\mid N;\\d<N}}\underbrace{\sum\limits_{\substack{e\mid
N/d;\\e>1}}\mu\left(  e\right)  }_{\substack{=\sum_{e\mid N/d}\mu\left(
e\right)  -\mu\left(  1\right)  \\\text{(since }1|N/d\text{)}}}d\varphi
_{N/d}\left(  \beta_{d}\right)  +b_{N}=\sum\limits_{\substack{d\mid N;\\d<N}}\left(
\underbrace{\sum_{e\mid N/d}\mu\left(  e\right)  }_{\substack{=\delta
_{N/d,1}\\\text{(by (\ref{sol.exe.witt.ghost-equiv.mu*1}), applied}\\\text{to
}N/d\text{ instead of }n\text{)}}}-\underbrace{\mu\left(  1\right)  }%
_{=1}\right)  d\varphi_{N/d}\left(  \beta_{d}\right)  +b_{N}\nonumber\\
&  =\sum\limits_{\substack{d\mid N;\\d<N}}\left(  \underbrace{\delta_{N/d,1}%
}_{\substack{=0\\\text{(since }N/d>1\text{)}}}-1\right)  d\varphi_{N/d}\left(
\beta_{d}\right)  +b_{N}=\sum\limits_{\substack{d\mid N;\\d<N}}\left(  -1\right)
d\varphi_{N/d}\left(  \beta_{d}\right)  +b_{N} \nonumber \\
&= -\sum\limits_{\substack{d\mid
N;\\d<N}}d\varphi_{N/d}\left(  \beta_{d}\right)  +b_{N}.
\label{sol.exe.witt.ghost-equiv.FE.4}
\end{align}
Now,%
\begin{align*}
\sum_{d\mid N}d\varphi_{N/d}\left(  \beta_{d}\right)   &  =\sum\limits
_{\substack{d\mid N;\\d<N}}d\varphi_{N/d}\left(  \beta_{d}\right)
+N\underbrace{\varphi_{N/N}}_{=\varphi_{1}= \id }\left(
\underbrace{\beta_{N}}_{=t}\right)  =\sum\limits_{\substack{d\mid N;\\d<N}%
}d\varphi_{N/d}\left(  \beta_{d}\right)  +\underbrace{Nt}_{\substack{=-\sum\limits
_{\substack{d\mid N;\\d<N}}d\varphi_{N/d}\left(  \beta_{d}\right)
+b_{N}\\\text{(by (\ref{sol.exe.witt.ghost-equiv.FE.4}))}}}\\
&  =\sum\limits_{\substack{d\mid N;\\d<N}}d\varphi_{N/d}\left(  \beta_{d}\right)
+\left(  -\sum\limits_{\substack{d\mid N;\\d<N}}d\varphi_{N/d}\left(  \beta
_{d}\right)  +b_{N}\right)  =b_{N}.
\end{align*}
Thus, (\ref{sol.exe.witt.ghost-equiv.FE.goal}) is satisfied for $n=N$. We have
thus completed a step of our recursive construction of the family $\left(
\beta_{n}\right)  _{n\geq1}$; this family therefore exists, and the
implication $\mathcal{F}\Longrightarrow\mathcal{E}$ is proven.

\textit{Proof of the implication $\mathcal{F}\Longrightarrow\mathcal{G}%
$:} Assume that Assertion $\mathcal{F}$ holds. That is, every positive
integer $n$ satisfies%
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)  \in nA.
\label{sol.exe.witt.ghost-equiv.FG.1}%
\end{equation}


On the other hand, every positive integer $e$ satisfies%
\begin{equation}
\sum_{d\mid e}\mu\left(  d\right)  \dfrac{e}{d}=\phi\left(  e\right)
\label{sol.exe.witt.ghost-equiv.FG.mu*1}%
\end{equation}
(by (\ref{sol.exe.witt.ghost-equiv.mu*1}), applied to $n=e$). Now, every
positive integer $n$ satisfies%
\begin{align*}
&  \sum_{d\mid n}\phi\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)
=\sum_{e\mid n}\underbrace{\phi\left(  e\right)  }_{\substack{=\sum_{d\mid
e}\mu\left(  d\right)  \dfrac{e}{d}\\\text{(by
(\ref{sol.exe.witt.ghost-equiv.FG.mu*1}))}}}\varphi_{e}\left(  b_{n/e}\right)
\\
&  =\underbrace{\sum_{e\mid n}\sum_{d\mid e}}_{=\sum_{d\mid n}\sum\limits
_{\substack{e\mid n;\\d\mid e}}}\mu\left(  d\right)  \dfrac{e}{d}\varphi
_{e}\left(  \underbrace{b_{n/e}}_{=b_{\left(  n/d\right)  /\left(  e/d\right)
}}\right)  =\sum_{d\mid n}\sum\limits_{\substack{e\mid n;\\d\mid e}}\mu\left(
d\right)  \dfrac{e}{d}\varphi_{e}\left(  b_{\left(  n/d\right)  /\left(
e/d\right)  }\right) \\
&= \underbrace{\sum_{d\mid n}\sum_{e\mid n/d}}%
_{=\sum_{e\mid n}\sum_{d\mid n/e}}\mu\left(  d\right)  e\underbrace{\varphi
_{ed}}_{=\varphi_{e}\circ\varphi_{d}}\left(  \underbrace{
b_{\left(  n/d\right)  /e}}_{=b_{\left(n/e\right)/d}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }ed\text{ for
}e\text{ in the second sum}\right) \\
&  =\sum_{e\mid n}\sum_{d\mid n/e}\mu\left(  d\right)  e\left(  \varphi
_{e}\circ\varphi_{d}\right)  \left(  b_{\left(  n/e\right)  /d}\right)
=\sum_{e\mid n}\sum_{d\mid n/e}\mu\left(  d\right)  e\varphi_{e}\left(
\varphi_{d}\left(  b_{\left(  n/e\right)  /d}\right)  \right) \\
&  =\sum_{e\mid n}e\sum_{d\mid n/e}\mu\left(  d\right)  \varphi_{e}\left(
\varphi_{d}\left(  b_{\left(  n/e\right)  /d}\right)  \right)  =\sum_{e\mid
n}e\varphi_{e}\underbrace{\left(  \sum_{d\mid n/e}\mu\left(  d\right)
\varphi_{d}\left(  b_{\left(  n/e\right)  /d}\right)  \right)  }%
_{\substack{\in\left(  n/e\right)  A\\\text{(by
(\ref{sol.exe.witt.ghost-equiv.FG.1}), applied}\\\text{to }n/e\text{ instead
of }n\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi_{e}\text{ is
} \ZZ \text{-linear}\right) \\
&  \in\sum_{e\mid n}e\underbrace{\varphi_{e}\left(  \left(  n/e\right)
A\right)  }_{\substack{\subset\left(  n/e\right)  A\\\text{(since }\varphi
_{e}\text{ is } \ZZ \text{-linear)}}}\subset\sum_{e\mid n}%
\underbrace{e\left(  n/e\right)  }_{=n}A=\sum_{e\mid n}nA\subset nA.
\end{align*}
Thus, Assertion $\mathcal{G}$ holds. We have thus proven the implication
$\mathcal{F}\Longrightarrow\mathcal{G}$.

\textit{Proof of the implication $\mathcal{G}\Longrightarrow\mathcal{F}%
$:} Assume that Assertion $\mathcal{G}$ holds. That is, every positive
integer $n$ satisfies%
\begin{equation}
\sum_{d\mid n}\phi\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)  \in nA.
\label{sol.exe.witt.ghost-equiv.GF.1}%
\end{equation}


On the other hand, every positive integer $e$ satisfies $\sum_{d\mid e}%
d\mu\left(  d\right)  \phi\left(  \dfrac{e}{d}\right)  =\mu\left(  e\right)  $
(by (\ref{sol.exe.witt.ghost-equiv.nmu*phi}), applied to $n=e$). In other
words, every positive integer $e$ satisfies%
\begin{equation}
\mu\left(  e\right)  =\sum_{d\mid e}d\mu\left(  d\right)  \phi\left(
\dfrac{e}{d}\right)  =\sum_{d\mid e}\dfrac{e}{d}\mu\left(  \dfrac{e}%
{d}\right)  \phi\left(  d\right)  \label{sol.exe.witt.ghost-equiv.GF.nmu*phi}%
\end{equation}
(here, we have substituted $e/d$ for $d$ in the sum). Now, every positive
integer $n$ satisfies%
\begin{align*}
&  \sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  b_{n/d}\right)
=\sum_{e\mid n}\underbrace{\mu\left(  e\right)  }_{\substack{=\sum_{d\mid
e}\dfrac{e}{d}\mu\left(  \dfrac{e}{d}\right)  \phi\left(  d\right)
\\\text{(by (\ref{sol.exe.witt.ghost-equiv.GF.nmu*phi}))}}}\varphi_{e}\left(
b_{n/e}\right) \\
&  =\underbrace{\sum_{e\mid n}\sum_{d\mid e}}_{=\sum_{d\mid n}\sum\limits
_{\substack{e\mid n;\\d\mid e}}}\dfrac{e}{d}\mu\left(  \dfrac{e}{d}\right)
\phi\left(  d\right)  \varphi_{e}\left(  \underbrace{b_{n/e}}_{=b_{\left(
n/d\right)  /\left(  e/d\right)  }}\right) \\
&  =\sum_{d\mid n}\sum\limits_{\substack{e\mid n;\\d\mid e}}\dfrac{e}{d}\mu\left(
\dfrac{e}{d}\right)  \phi\left(  d\right)  \varphi_{e}\left(  b_{\left(
n/d\right)  /\left(  e/d\right)  }\right)  =\underbrace{\sum_{d\mid n}%
\sum_{e\mid n/d}}_{=\sum_{e\mid n}\sum_{d\mid n/e}}e\mu\left(  e\right)
\phi\left(  d\right)  \underbrace{\varphi_{ed}}_{=\varphi_{e}\circ\varphi_{d}%
}\left( \underbrace{b_{\left(  n/d\right)  /e}}_{=b_{\left(n/e\right)/d}}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }ed\text{ for
}e\text{ in the second sum}\right) \\
&  =\sum_{e\mid n}\sum_{d\mid n/e}e\mu\left(  e\right)  \phi\left(  d\right)
\left(  \varphi_{e}\circ\varphi_{d}\right)  \left(  b_{\left(  n/e\right)
/d}\right)  =\sum_{e\mid n}\sum_{d\mid n/e}e\mu\left(  e\right)  \phi\left(
d\right)  \varphi_{e}\left(  \varphi_{d}\left(  b_{\left(  n/e\right)
/d}\right)  \right) \\
&  =\sum_{e\mid n}e\mu\left(  e\right)  \sum_{d\mid n/e}\phi\left(  d\right)
\varphi_{e}\left(  \varphi_{d}\left(  b_{\left(  n/e\right)  /d}\right)
\right)  =\sum_{e\mid n}e\mu\left(  e\right)  \varphi_{e}\underbrace{\left(
\sum_{d\mid n/e}\phi\left(  d\right)  \varphi_{d}\left(  b_{\left(
n/e\right)  /d}\right)  \right)  }_{\substack{\in\left(  n/e\right)
A\\\text{(by (\ref{sol.exe.witt.ghost-equiv.GF.1}), applied}\\\text{to
}n/e\text{ instead of }n\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\varphi_{e}\text{ is } \ZZ \text{-linear}\right) \\
&  \in\sum_{e\mid n}e\mu\left(  e\right)  \underbrace{\varphi_{e}\left(
\left(  n/e\right)  A\right)  }_{\substack{\subset\left(  n/e\right)
A\\\text{(since }\varphi_{e}\text{ is } \ZZ \text{-linear)}}}\subset
\sum_{e\mid n}\underbrace{e\mu\left(  e\right)  \left(  n/e\right)  }%
_{=n\mu\left(  e\right)  }A=\sum_{e\mid n}n\mu\left(  e\right)  A\subset nA.
\end{align*}
Thus, Assertion $\mathcal{F}$ holds. We have thus proven the implication
$\mathcal{G}\Longrightarrow\mathcal{F}$.

\textit{Proof of the equivalence $\mathcal{G}\Longleftrightarrow\mathcal{H}%
$:} For every positive integer $n$, we have%
\begin{align*}
&  \sum_{i=1}^{n}\varphi_{n/\gcd\left(  i,n\right)  }\left(  b_{\gcd\left(
i,n\right)  }\right)  =\sum_{d\mid n}\underbrace{\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\\gcd\left(  i,n\right)  =d}}\varphi_{n/d}\left(
b_{d}\right)  }_{\substack{=\phi\left(  \dfrac{n}{d}\right)  \varphi
_{n/d}\left(  b_{d}\right)  \\\text{(by
(\ref{sol.exe.witt.ghost-equiv.phi*1.pf.num}))}}}\\
&  =\sum_{d\mid n}\phi\left(  \dfrac{n}{d}\right)  \varphi_{n/d}\left(
b_{d}\right)  =\sum_{d\mid n}\phi\left(  d\right)  \varphi_{d}\left(
b_{n/d}\right)
\end{align*}
(here, we have substituted $n/d$ for $d$ in the sum). This makes it clear that
Assertions $\mathcal{G}$ and $\mathcal{H}$ are equivalent.

The implications and equivalences that we have proven, combined, yield the
equivalence of all seven assertions $\mathcal{C}$, $\mathcal{D}$,
$\mathcal{E}$, $\mathcal{F}$, $\mathcal{G}$, $\mathcal{H}$ and $\mathcal{J}$.
This solves Exercise \ref{exe.witt.ghost-equiv}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.witt.ghost-exa}}

\textit{Solution to Exercise \ref{exe.witt.ghost-exa}.} We know that
the seven assertions $\mathcal{C}$, $\mathcal{D}$, $\mathcal{E}$,
$\mathcal{F}$, $\mathcal{G}$, $\mathcal{H}$ and $\mathcal{J}$ are equivalent;
hence, for each of our families, it suffices to prove one of these assertions.
We choose the assertion $\mathcal{C}$, as it is the easiest to prove.

\begin{itemize}
\item \textit{Proof of Assertion $\mathcal{C}$ for the family }$\left(
b_{n}\right)  _{n\geq1}=\left(  q^{n}\right)  _{n\geq1}$\textit{, where }%
$q$\textit{ is a given integer:} Let $q$ be an integer. We need to prove
Assertion $\mathcal{C}$ for the family $\left(  b_{n}\right)  _{n\geq
1}=\left(  q^{n}\right)  _{n\geq1}$. This means proving that for every
positive integer $n$ and every prime factor $p$ of $n$, we have%
\begin{equation}
\varphi_{p}\left(  q^{n/p}\right)  \equiv q^{n}\operatorname{mod}%
p^{v_{p}\left(  n\right)  } \ZZ . \label{sol.witt.ghost-exa.fam1.goal}%
\end{equation}
So let us prove this. Let $n$ be a positive integer, and let $p$ be a prime
factor of $n$. Fermat's little theorem yields $q^{p}\equiv q\operatorname{mod}%
p \ZZ$. Hence, (\ref{sol.exe.witt.ghost-equiv.hensel-v}) (applied to
$A= \ZZ$, $a=q^{p}$, $b=q$ and $N=n/p$) yields $\left(  q^{p}\right)
^{n/p}\equiv q^{n/p}\operatorname{mod}p^{v_{p}\left(  n/p\right)
+1} \ZZ$. Since $\left(  q^{p}\right)  ^{n/p}=q^{n}$ and $v_{p}\left(
n/p\right)  +1=v_{p}\left(  n\right)  $, this rewrites as $q^{n}\equiv
q^{n/p}\operatorname{mod}p^{v_{p}\left(  n\right)  } \ZZ$. Now,
$\varphi_{p}= \id$, so that $\varphi_{p}\left(  q^{n/p}\right)
=q^{n/p}\equiv q^{n}\operatorname{mod}p^{v_{p}\left(  n\right)  } \ZZ$.
Thus, (\ref{sol.witt.ghost-exa.fam1.goal}) is proven, and we are done with the
family $\left(  b_{n}\right)  _{n\geq1}=\left(  q^{n}\right)  _{n\geq1}$.

\item \textit{Proof of Assertion $\mathcal{C}$ for the family }$\left(
b_{n}\right)  _{n\geq1}=\left(  q\right)  _{n\geq1}$\textit{, where }%
$q$\textit{ is a given integer:} Let $q$ be an integer. We need to prove
Assertion $\mathcal{C}$ for the family $\left(  b_{n}\right)  _{n\geq
1}=\left(  q\right)  _{n\geq1}$. This means proving that for every positive
integer $n$ and every prime factor $p$ of $n$, we have%
\[
\varphi_{p}\left(  q\right)  \equiv q\operatorname{mod}p^{v_{p}\left(
n\right)  } \ZZ .
\]
But this is obvious, since $\varphi_{p}= \id$. Thus, the family
$\left(  b_{n}\right)  _{n\geq1}=\left(  q\right)  _{n\geq1}$ satisfies
Assertion $\mathcal{C}$.

\item \textit{Proof of Assertion $\mathcal{C}$ for the family }$\left(
b_{n}\right)  _{n\geq1}=\left(  \dbinom{qn}{rn}\right)  _{n\geq1}$\textit{,
where }$r\in\QQ$\textit{ and }$q\in \ZZ$\textit{ are given:} Let
$r\in\QQ$ and $q\in \ZZ$. We need to prove Assertion
$\mathcal{C}$ for the family $\left(  b_{n}\right)  _{n\geq1}=\left(
\dbinom{qn}{rn}\right)  _{n\geq1}$. This means proving that for every positive
integer $n$ and every prime factor $p$ of $n$, we have%
\begin{equation}
\varphi_{p}\left(  \dbinom{qn/p}{rn/p}\right)  \equiv\dbinom{qn}%
{rn}\operatorname{mod}p^{v_{p}\left(  n\right)  } \ZZ .
\label{sol.witt.ghost-exa.fam3.goal}%
\end{equation}
So let us prove this. Let $n$ be a positive integer, and $p$ be a prime factor
of $n$. We need to prove (\ref{sol.witt.ghost-exa.fam3.goal}). In other words,
we need to prove that%
\begin{equation}
\dbinom{qn/p}{rn/p}\equiv\dbinom{qn}{rn}\operatorname{mod}p^{v_{p}\left(
n\right)  } \ZZ  \label{sol.witt.ghost-exa.fam3.goal2}%
\end{equation}
(since $\varphi_{p}= \id$). We WLOG assume that $rn\in
 \ZZ$ (since otherwise, both sides of this congruence
(\ref{sol.witt.ghost-exa.fam3.goal2}) are $0$).

It is well-known that $\left(  1+X\right)  ^{p}\equiv1+X^{p}\operatorname{mod}%
p \ZZ \left[  X\right]  $ in the polynomial ring $ \ZZ \left[
X\right]  $. Hence, $\left(  \left(  1+X\right)  ^{p}\right)  ^{n/p}%
\equiv\left(  1+X^{p}\right)  ^{n/p}\operatorname{mod}p^{v_{p}\left(
n/p\right)  +1} \ZZ \left[  X\right]  $ (by
(\ref{sol.exe.witt.ghost-equiv.hensel-v}), applied to $ \ZZ \left[
X\right]  $, $\left(  1+X\right)  ^{p}$, $1+X^{p}$ and $n/p$ instead of $A$,
$a$, $b$ and $N$). Since $\left(  \left(  1+X\right)  ^{p}\right)
^{n/p}=\left(  1+X\right)  ^{n}$ and $v_{p}\left(  n/p\right)  +1=v_{p}\left(
n\right)  $, this rewrites as%
\[
\left(  1+X\right)  ^{n}\equiv\left(  1+X^{p}\right)  ^{n/p}\operatorname{mod}%
p^{v_{p}\left(  n\right)  } \ZZ \left[  X\right]  .
\]
Hence,
\[
\left(  1+X\right)  ^{n}\equiv\left(  1+X^{p}\right)  ^{n/p}\operatorname{mod}%
p^{v_{p}\left(  n\right)  } \ZZ \left[ \left[ X \right] \right]
\]
(since $\ZZ \left[ X \right]$ is a subring of
$\ZZ \left[\left[ X \right]\right]$).
We can take both sides of this congruence to the $q$-th power\footnote{This
works even if $q$ is negative, since $\left(  1+X\right)  ^{n}$ and $\left(
1+X^{p}\right)  ^{n/p}$ are invertible in
$\ZZ \left[ \left[ X \right] \right]
/ \left( p^{v_p \left( n \right)}
    \ZZ \left[ \left[ X \right] \right] \right)$.},
and thus obtain
\[
\left(  \left(  1+X\right)  ^{n}\right)  ^{q}\equiv\left(  \left(
1+X^{p}\right)  ^{n/p}\right)  ^{q}\operatorname{mod}p^{v_{p}\left(  n\right)
} \ZZ \left[ \left[ X \right] \right]  .
\]
In other words,
\[
\left(  1+X\right)  ^{qn}\equiv\left(  1+X^{p}\right)  ^{qn/p}
\operatorname{mod}
p^{v_{p}\left(  n\right)  } \ZZ \left[ \left[ X \right] \right]  .
\]
Comparing the coefficients before $X^{rn}$ on both sides of this congruence,
we obtain%
\[
\dbinom{qn}{rn}\equiv\dbinom{qn/p}{rn/p}\operatorname{mod}p^{v_{p}\left(
n\right)  } \ZZ .
\]
This proves (\ref{sol.witt.ghost-exa.fam3.goal2}) and thus
(\ref{sol.witt.ghost-exa.fam3.goal}). Hence, we are done with the family
$\left(  b_{n}\right)  _{n\geq1}=\left(  \dbinom{qn}{rn}\right)  _{n\geq1}$.

[\textit{Remark:} There is an alternative, combinatorial approach to proving
(\ref{sol.witt.ghost-exa.fam3.goal2}) when $q$ is nonnegative. This approach
proceeds by counting the $\left(  rn\right)  $-element subsets of the set
$ \ZZ /\left(  qn\right)  $. On the one hand, the number of such subsets
is clearly $\dbinom{qn}{rn}$. On the other hand, these subsets fall into two classes:

\begin{itemize}
\item the subsets which are invariant under the permutation%
\begin{align*}
 \ZZ /\left(  qn\right)   &  \rightarrow \ZZ /\left(  qn\right)  ,\\
i  &  \mapsto i+qn/p
\end{align*}
of $ \ZZ /\left(  qn\right)  $;

\item the subsets which are not invariant under this permutation.
\end{itemize}

It is easy to see that the number of all subsets in the first class is
$\dbinom{qn/p}{rn/p}$ (indeed, the intersection of such a subset with
$\left\{  0,1,\ldots,qn/p-1\right\}  \subset \ZZ /\left(  qn\right)  $ must
have $rn/p$ elements, and uniquely determines the whole subset by
``replication''), whereas the number of all
subsets in the second class is divisible by $p^{v_{p}\left(  n\right)  }$
(because the permutation $\ZZ /\left(  qn\right)  \rightarrow
 \ZZ /\left(  qn\right)  ,\ i\mapsto i+1$ acts on these subsets, and thus
splits them into orbits, each of which has size divisible by $p^{v_{p}\left(
n\right)  }$\ \ \ \ \footnote{Here is why:
Let $O$ be an orbit under the action of the permutation
$\ZZ /\left(  qn\right)  \rightarrow
\ZZ /\left(  qn\right)  ,\ i\mapsto i+1$ on the subsets in the
second class. We must prove that $O$ has size divisible by
$p^{v_p\left(n\right)}$.
\par
Note that $qn \neq 0$, since otherwise there would be no subsets in
the second class.
\par
Let $\xi$ be the permutation
\begin{align*}
 \ZZ /\left(  qn\right)   &  \rightarrow \ZZ /\left(  qn\right)  ,\\
i  &  \mapsto i+1
\end{align*}
of $ \ZZ /\left(  qn\right)  $.
Then, $O$ is an orbit under the action of $\xi$ on the subsets in
the second class.
Note that $\xi^{qn} = \id$, since $\xi$ is a cyclic permutation
of a $qn$-element set.
\par
Fix an element $L \in O$.
Thus, $L$ is a subset in the second class.
In other words, $L$ is a subset of $\ZZ / \left(qn\right)$ that is not
invariant under the permutation
\begin{align*}
 \ZZ /\left(  qn\right)   &  \rightarrow \ZZ /\left(  qn\right)  ,\\
i  &  \mapsto i+qn/p
\end{align*}
of $ \ZZ /\left(  qn\right)  $.
Since this permutation is $\xi^{qn/p}$, we can restate this
as follows:
$L$ is a subset of $\ZZ / \left(qn\right)$ that is not
invariant under the action of the permutation $\xi^{qn/p}$
on the subsets of $\ZZ / \left(qn\right)$.
That is, $\xi^{qn/p}\left(L\right) \neq L$.
But $\xi^{qn}\left(L\right) = L$ (since $\xi^{qn} = \id$).
Now, $O$ is the orbit of $L$ under the action of $\xi$.
Hence, the size of this orbit $O$ is a divisor of $qn$
(since $\xi^{qn}\left(L\right) = L$) but not a divisor
of $qn/p$ (since $\xi^{qn/p}\left(L\right) \neq L$).
Thus, this size must be divisible by $p^{v_p\left(qn\right)}$
(because a divisor of $qn$ that is not divisor of $qn/p$
must necessarily be divisible by $p^{v_p\left(qn\right)}$).
Hence, it is also divisible by $p^{v_p\left(n\right)}$
(since $v_p\left(qn\right) = \underbrace{v_p\left(q\right)}_{\geq 0} + v_p\left(n\right)
\geq v_p\left(n\right)$
and therefore $p^{v_p\left(n\right)} \mid p^{v_p\left(qn\right)}$).
Qed.}). Hence, the number of all subsets in both classes together is
$\equiv\dbinom{qn/p}{rn/p}\operatorname{mod}p^{v_{p}\left(  n\right)
} \ZZ$. Comparing these two answers, we obtain $\dbinom{qn}{rn}%
\equiv\dbinom{qn/p}{rn/p}\operatorname{mod}p^{v_{p}\left(  n\right)
} \ZZ$.

The downside of this nice approach is that it requires a modification in the
case when $q$ is negative. Here, the \dfn{upper negation formula}
$\dbinom{-a}{k}=\left(  -1\right)  ^{k}\dbinom{a+k-1}{k}$ needs to be used,
along with the ``stars-and-bars'' formula
$\dbinom{a+k-1}{k}$ for the number of $k$-element multisets whose elements
belong to $\left\{  1,2,\ldots,a\right\}  $. We leave the details to the reader.]

\item \textit{Proof of Assertion $\mathcal{C}$ for the family }$\left(
b_{n}\right)  _{n\geq1}=\left(  \dbinom{qn-1}{rn-1}\right)  _{n\geq1}%
$\textit{, where }$r\in \ZZ$\textit{ and }$q\in \ZZ$\textit{ are
given:} Let $r\in \ZZ$ and $q\in \ZZ$. We need to prove Assertion
$\mathcal{C}$ for the family $\left(  b_{n}\right)  _{n\geq1}=\left(
\dbinom{qn-1}{rn-1}\right)  _{n\geq1}$. This means proving that for every
positive integer $n$ and every prime factor $p$ of $n$, we have%
\begin{equation}
\varphi_{p}\left(  \dbinom{qn/p-1}{rn/p-1}\right)  \equiv\dbinom{qn-1}%
{rn-1}\operatorname{mod}p^{v_{p}\left(  n\right)  } \ZZ .
\label{sol.witt.ghost-exa.fam4.goal}%
\end{equation}
So let us prove this. Let $n$ be a positive integer, and $p$ be a prime factor
of $n$. We need to prove (\ref{sol.witt.ghost-exa.fam4.goal}). In other words,
we need to prove that%
\begin{equation}
\dbinom{qn/p-1}{rn/p-1}\equiv\dbinom{qn-1}{rn-1}\operatorname{mod}%
p^{v_{p}\left(  n\right)  } \ZZ  \label{sol.witt.ghost-exa.fam4.goal2}%
\end{equation}
(since $\varphi_{p}= \id$). By the recurrence of the binomial
coefficients, we have $\dbinom{qn/p-1}{rn/p-1}=\dbinom{qn/p}{rn/p}%
-\dbinom{qn/p-1}{rn/p}$ and $\dbinom{qn-1}{rn-1}=\dbinom{qn}{rn}-\dbinom
{qn-1}{rn}$. Hence, we can obtain the desired congruence
(\ref{sol.witt.ghost-exa.fam4.goal2}) by subtracting the congruence%
\begin{equation}
\dbinom{qn/p-1}{rn/p}\equiv\dbinom{qn-1}{rn}\operatorname{mod}p^{v_{p}\left(
n\right)  } \ZZ  \label{sol.witt.ghost-exa.fam4.goal3}%
\end{equation}
from the congruence (\ref{sol.witt.ghost-exa.fam3.goal2}). It therefore
remains to prove the congruence (\ref{sol.witt.ghost-exa.fam4.goal3}) (since
(\ref{sol.witt.ghost-exa.fam3.goal2}) has already been proven).

We recall the (easy-to-check) formula $\dbinom{a-1}{k}=\left(  -1\right)
^{k}\dbinom{k-a}{k}$ for every $a\in \ZZ$ and $k\in \NN$. This
formula allows us to rewrite both sides of
(\ref{sol.witt.ghost-exa.fam4.goal3}), and thus
(\ref{sol.witt.ghost-exa.fam4.goal3}) becomes%
\begin{equation}
\left(  -1\right)  ^{rn/p}\dbinom{rn/p-qn/p}{rn/p}\equiv\left(  -1\right)
^{rn}\dbinom{rn-qn}{rn}\operatorname{mod}p^{v_{p}\left(  n\right)  }%
 \ZZ . \label{sol.witt.ghost-exa.fam4.goal4}%
\end{equation}
So it remains to prove (\ref{sol.witt.ghost-exa.fam4.goal4}). We have $\left(
-1\right)  ^{rn/p}\equiv\left(  -1\right)  ^{rn}\operatorname{mod}%
p^{v_{p}\left(  n\right)  } \ZZ$\ \ \ \ \footnote{\textit{Proof.} The
only situation in which this is not obvious is when one of the integers $rn/p$
and $rn$ is even and the other is odd. This means, of course, that $rn/p$ is
odd and $rn$ is even (since $rn/p\mid rn$). Hence, in this situation, we have
$p=2$ and $v_{p}\left(  n\right)  \leq1$, and therefore $p^{v_{p}\left(
n\right)  }=2$, so that $\left(  -1\right)  ^{rn/p}\equiv\left(  -1\right)
^{rn}\operatorname{mod}p^{v_{p}\left(  n\right)  } \ZZ$ follows
immediately from the fact that any two integer powers of $-1$ are congruent to
each other modulo $2 \ZZ$.}. Thus, we can cancel the $\left(  -1\right)
^{rn/p}$ on the left hand side of (\ref{sol.witt.ghost-exa.fam4.goal4})
against the $\left(  -1\right)  ^{rn}$ on the right hand side, and therefore
(\ref{sol.witt.ghost-exa.fam4.goal4}) takes the equivalent form%
\[
\dbinom{rn/p-qn/p}{rn/p}\equiv\dbinom{rn-qn}{rn}\operatorname{mod}%
p^{v_{p}\left(  n\right)  } \ZZ .
\]
This further rewrites as%
\[
\dbinom{\left(  r-q\right)  n/p}{rn/p}\equiv\dbinom{\left(  r-q\right)  n}%
{rn}\operatorname{mod}p^{v_{p}\left(  n\right)  } \ZZ .
\]
But this follows from (\ref{sol.witt.ghost-exa.fam3.goal2}), applied to $r-q$
instead of $q$. Hence, (\ref{sol.witt.ghost-exa.fam4.goal4}) is proven, and
consequently Assertion $\mathcal{C}$ holds for the family $\left(
b_{n}\right)  _{n\geq1}=\left(  \dbinom{qn-1}{rn-1}\right)  _{n\geq1}$.
\end{itemize}

Exercise \ref{exe.witt.ghost-exa} is thus solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.witt.ghost-app}}

\textit{Solution to Exercise \ref{exe.witt.ghost-app}.} (a) This is
obvious since $\mathbf{f}_{n}$ is an evaluation homomorphism (in an
appropriate sense).\footnote{Or, in a more down-to-earth fashion, this is
obvious because (for example) multiplying two power series and then replacing
all variables in the product by their $n$-th powers gives the same result as
first replacing all variables by their $n$-th powers and then multiplying the
resulting two power series.}

(b) Let $n$ and $m$ be positive integers. Then,
\begin{align*}
\left(  \mathbf{f}_{n}\circ\mathbf{f}_{m}\right)  \left(  a\right)   &
=\mathbf{f}_{n}\left(  \underbrace{\mathbf{f}_{m}\left(  a\right)
}_{=a\left(  x_{1}^{m},x_{2}^{m},x_{3}^{m},\ldots\right)  }\right)
=\mathbf{f}_{n}\left(a\left(  x_{1}^{m},x_{2}^{m},x_{3}^{m},\ldots\right)
\right)  =\left(
a\left(  x_{1}^{m},x_{2}^{m},x_{3}^{m},\ldots\right)  \right)  \left(
x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right) \\
&  = a \left( \left(x_1^n\right)^m , \left(x_2^n\right)^m ,
\left(x_3^n\right)^m , \ldots \right)
= a\left(  x_{1}^{nm},x_{2}^{nm},x_{3}^{nm},\ldots\right)
= \mathbf{f}_{nm}\left(  a\right)
\end{align*}
for all $a\in \Lambda$. Thus,
$\mathbf{f}_{n}\circ\mathbf{f}_{m}=\mathbf{f}_{nm}$,
so that Exercise~\ref{exe.witt.ghost-app}(b) is solved.

(c) For every $a\in \Lambda$, we have
$\mathbf{f}_{1}\left(  a\right)  =a\left(
x_{1}^{1},x_{2}^{1},x_{3}^{1},\ldots\right)  =a$. Thus,
$\mathbf{f}_{1}= \id$, so that Exercise~\ref{exe.witt.ghost-app}(c) is solved.

(d) There are several ways to solve Exercise \ref{exe.witt.ghost-app}(d):

\begin{itemize}
\item One can check $\Delta\circ\mathbf{f}_{n}=\left(  \mathbf{f}_{n}%
\otimes\mathbf{f}_{n}\right)  \circ\Delta$ rather directly on the basis
$\left(  m_{\lambda}\right)  _{\lambda\in \Par }$ of $\Lambda$,
using (\ref{comultiplication-of-monomial-symmetric-function}) and the
definition of $\mathbf{f}_{n}$.

\item One can check $\Delta\circ\mathbf{f}_{n}=\left(  \mathbf{f}_{n}%
\otimes\mathbf{f}_{n}\right)  \circ\Delta$ very easily on the elements
$p_{1},p_{2},p_{3},\ldots$ of $\Lambda$, and then ``pretend'' that
these elements generate $\Lambda$ (in fact,
they do generate $\Lambda$ when $\kk = \QQ$, so this solves
Exercise \ref{exe.witt.ghost-app}(d) in the case of $\kk = \QQ$;
but this easily entails that Exercise \ref{exe.witt.ghost-app}(d) holds in the
case of $\kk = \ZZ$\ \ \ \ \footnote{Getting from $\kk
= \QQ$ to $\kk = \ZZ$ requires a standard functoriality
argument. See Step 2 of the solution of
Exercise~\ref{exe.Lambda.verschiebung}(g) further below for an example of
such an argument.}, and this, in turn, can
be used to solve Exercise \ref{exe.witt.ghost-app}(d) for arbitrary
$\kk$\ \ \ \ \footnote{Getting from $\kk = \ZZ$ to
arbitrary $\kk$ requires a standard functoriality argument. See Step 3
of the solution of Exercise \ref{exe.Lambda.verschiebung}(g) further below for
an example of such an argument.}).

\item One can derive Exercise \ref{exe.witt.ghost-app}(d) from Exercise
\ref{exe.Lambda.verschiebung}(e) using the self-duality of $\Lambda$ and
Exercise \ref{exe.Lambda.verschiebung}(f). This is somewhat complicated by the
fact that $\mathbf{v}_{n}$ and $\mathbf{f}_{n}$ (generally)
are not graded maps\footnote{They nevertheless have a property close
to being graded. Namely, let us say that a $\kk$-linear map $\varphi$ between
two graded $\kk$-modules $V = \bigoplus_{n \geq 0} V_n$ and
$W = \bigoplus_{n \geq 0} W_n$ \emph{scales the degree by $q$}
(where $q$ is some fixed rational number) if it has the property that
$\varphi \left(V_n\right) \subset W_{qn}$ for all $n \in \NN$ (where
$W_{qn}$ is understood to be $0$ when $qn$ is not an integer). Then,
$\mathbf{v}_n$ scales the degree by $1/n$, whereas $\mathbf{f}_n$ scales
the degree by $n$. When a $\kk$-linear map $\varphi$ between two graded
$\kk$-modules $V$ and $W$ scales the degree by a nonzero rational number
$q$, the adjoint map $\varphi^\ast : W^\ast \to V^\ast$ restricts to a
$\kk$-linear map $W^o \to V^o$, which scales the degree by $1/q$; thus
a large part of the theory of adjoint maps which is usually formulated
for graded maps can be carried over to the case of maps scaling the
degree by $q$.}, but with the correct arguments, one can completely
avoid any use of gradedness\footnote{Here is a sketch of how these
correct arguments look like.
\par
The Hall inner product $\left(\cdot, \cdot\right)_\Lambda$ on
$\Lambda$ gives rise to a symmetric bilinear form
$\left(\cdot, \cdot\right)_{\Lambda \otimes \Lambda}$ on
$\Lambda \otimes \Lambda$ (according to
Definition~\ref{primitives-split-off-prop}(b)). It is easy to see
that every three elements $a$, $b$ and $c$ of
$\Lambda$ satisfy
\begin{equation}
\left(a, m\left(b\otimes c\right)\right)_{\Lambda}
= \left(\Delta\left(a\right), b \otimes c\right)_{\Lambda \otimes \Lambda} .
\label{sol.witt.ghost-app.d.duality-trick.abc}
\end{equation}
\par
[\textit{Proof of \eqref{sol.witt.ghost-app.d.duality-trick.abc}:}
Let $a$, $b$ and $c$ be three elements of $\Lambda$.
We need to prove the equality
\eqref{sol.witt.ghost-app.d.duality-trick.abc}. Since this equality
is $\kk$-linear in each of $a$, $b$ and $c$, we can WLOG assume that
each of $a$, $b$ and $c$ belongs to the basis
$\left(s_\lambda\right)_{\lambda \in \Par}$ of the $\kk$-module
$\Lambda$. Assume this. Then, there exist three elements $\alpha$,
$\beta$ and $\gamma$ of $\Par$ such that $a = s_\alpha$,
$b = s_\beta$ and $c = s_\gamma$. Consider these $\alpha$, $\beta$ and
$\gamma$. Now, let us use the notations of
Corollary~\ref{self-duality-corollary}.
Applying Corollary~\ref{self-duality-corollary} to $\lambda = \alpha$,
$\mu = \beta$ and $\nu = \gamma$ yields
$c^\alpha_{\beta,\gamma} = \hat{c}^\alpha_{\beta,\gamma}$.
But the definition of the map $m$ yields
\[
m\left(b\otimes c\right) = \underbrace{b}_{=s_\beta} \underbrace{c}_{=s_\gamma}
= s_\beta s_\gamma = \sum_{\lambda \in \Par} c^\lambda_{\beta,\gamma} s_\lambda
\]
(by \eqref{symm-structure-constants.1}, applied
to $\mu = \beta$ and $\nu = \gamma$). Hence,
\begin{align*}
\left(\underbrace{a}_{=s_\alpha}, \underbrace{m\left(b\otimes c\right)}_{= \sum_{\lambda \in \Par} c^\lambda_{\beta,\gamma} s_\lambda}\right)_{\Lambda}
&= \left(s_\alpha, \sum_{\lambda \in \Par} c^\lambda_{\beta,\gamma} s_\lambda\right)_{\Lambda}
= \sum_{\lambda \in \Par} c^\lambda_{\beta,\gamma} \underbrace{\left(s_\alpha, s_\lambda\right)_{\Lambda}}_{=\delta_{\alpha,\lambda}}
= \sum_{\lambda \in \Par} c^\lambda_{\beta,\gamma} \delta_{\alpha,\lambda}
= c^\alpha_{\beta,\gamma} = \hat{c}^\alpha_{\beta,\gamma}.
\end{align*}
On the other hand, applying the map $\Delta$ to both sides of the
equality $a = s_\alpha$, we obtain
\[
\Delta a = \Delta s_\alpha = \sum_{\mu \in \Par ;\  \nu \in \Par} \hat{c}^\alpha_{\mu,\nu} s_\mu \otimes s_\nu
\]
(by \eqref{symm-structure-constants.2}, applied
to $\lambda = \alpha$). Hence,
\begin{align*}
& \left(\underbrace{\Delta a}_{=\sum_{\mu \in \Par ;\  \nu \in \Par} \hat{c}^\alpha_{\mu,\nu} s_\mu \otimes s_\nu}, \underbrace{b}_{=s_\beta} \otimes \underbrace{c}_{=s_\gamma}\right)_{\Lambda \otimes \Lambda}
= \left(\sum_{\mu \in \Par ;\  \nu \in \Par} \hat{c}^\alpha_{\mu,\nu} s_\mu \otimes s_\nu, s_\beta \otimes s_\gamma\right)_{\Lambda \otimes \Lambda} \\
&= \sum_{\mu \in \Par ;\  \nu \in \Par} \hat{c}^\alpha_{\mu,\nu} \underbrace{\left(s_\mu \otimes s_\nu, s_\beta \otimes s_\gamma\right)_{\Lambda \otimes \Lambda}}_{=\left(s_\mu, s_\beta\right)_{\Lambda} \left(s_\nu, s_\gamma\right)_{\Lambda}}
= \sum_{\mu \in \Par ;\  \nu \in \Par} \hat{c}^\alpha_{\mu,\nu} \underbrace{\left(s_\mu, s_\beta\right)_{\Lambda}}_{=\delta_{\mu, \beta}} \underbrace{\left(s_\nu, s_\gamma\right)_{\Lambda}}_{=\delta_{\nu, \gamma}} \\
&= \sum_{\mu \in \Par ;\  \nu \in \Par} \hat{c}^\alpha_{\mu,\nu} \delta_{\mu, \beta} \delta_{\nu, \gamma}
= \sum_{\mu \in \Par} \delta_{\mu, \beta} \underbrace{\sum_{\nu \in \Par} \hat{c}^\alpha_{\mu,\nu} \delta_{\nu, \gamma}}_{= \hat{c}^\alpha_{\mu,\gamma}}
= \sum_{\mu \in \Par} \delta_{\mu, \beta} \hat{c}^\alpha_{\mu,\gamma} = \hat{c}^\alpha_{\beta,\gamma}.
\end{align*}
Compared with $\left(a, m\left(b\otimes c\right)\right)_{\Lambda}
= \hat{c}^\alpha_{\beta,\gamma}$, this yields
$\left(a, m\left(b\otimes c\right)\right)_{\Lambda}
= \left(\Delta\left(a\right), b \otimes c\right)_{\Lambda \otimes \Lambda}$.
Thus, \eqref{sol.witt.ghost-app.d.duality-trick.abc} is proven.]
\par
Now, we can show that every $a \in \Lambda$ and $B \in \Lambda
\otimes \Lambda$ satisfy
\begin{equation}
\left(a, m\left(B\right)\right)_{\Lambda}
= \left(\Delta\left(a\right), B\right)_{\Lambda \otimes \Lambda} .
\label{sol.witt.ghost-app.d.duality-trick.aB}
\end{equation}
\par
[\textit{Proof of \eqref{sol.witt.ghost-app.d.duality-trick.aB}:}
Let $a \in \Lambda$ and $B \in \Lambda \otimes \Lambda$.
We need to prove the equality
\eqref{sol.witt.ghost-app.d.duality-trick.aB}. Since this equality
is $\kk$-linear in $B$, we can WLOG assume that $B$ is a pure
tensor (because the pure tensors span the $\kk$-module $\Lambda
\otimes \Lambda$). Assume this. Then, $B = b \otimes c$ for two
elements $b$ and $c$ of $\Lambda$. Consider these $b$ and $c$.
Then,
\begin{align*}
\left(a, m\left(\underbrace{B}_{=b \otimes c}\right)\right)_{\Lambda}
&= \left(a, m\left(b \otimes c\right)\right)_{\Lambda}
= \left(\Delta\left(a\right), \underbrace{b \otimes c}_{=B}\right)_{\Lambda \otimes \Lambda}
\qquad \left(\text{by \eqref{sol.witt.ghost-app.d.duality-trick.abc}}\right) \\
&= \left(\Delta\left(a\right), B\right)_{\Lambda \otimes \Lambda} .
\end{align*}
This proves \eqref{sol.witt.ghost-app.d.duality-trick.aB}.]
\par
Exercise \ref{exe.Lambda.verschiebung}(f) yields that the maps $\mathbf{f}%
_{n}:\Lambda\rightarrow\Lambda$ and $\mathbf{v}_{n}:\Lambda\rightarrow\Lambda$
are adjoint with respect to the Hall inner product on $\Lambda$. Thus,%
\begin{equation}
\left(  \mathbf{f}_{n}a,b\right)  _{\Lambda}=\left(  a,\mathbf{v}_{n}b\right)
_{\Lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\Lambda\text{ and }%
b\in\Lambda.\label{sol.witt.ghost-app.d.duality-trick.fv}%
\end{equation}


From this, it is easy to see that
\begin{equation}
\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  A,B\right)
_{\Lambda\otimes\Lambda}=\left(  A,\left(  \mathbf{v}_{n}\otimes\mathbf{v}%
_{n}\right)  B\right)  _{\Lambda\otimes\Lambda}\ \ \ \ \ \ \ \ \ \ \text{for
every }A\in\Lambda\otimes\Lambda\text{ and }B\in\Lambda\otimes\Lambda
.\label{sol.witt.ghost-app.d.duality-trick.ffvv}%
\end{equation}


[\textit{Proof of (\ref{sol.witt.ghost-app.d.duality-trick.ffvv}):} Let
$A\in\Lambda\otimes\Lambda$ and $B\in\Lambda\otimes\Lambda$. We have to prove
the equality (\ref{sol.witt.ghost-app.d.duality-trick.ffvv}). Since this
equality is $\kk$-linear in each of $A$ and $B$, we can WLOG assume
that $A$ and $B$ are pure tensors (since pure tensors span $\Lambda
\otimes\Lambda$ as a $\kk$-module). Assume this. Then, we can write $A$
and $B$ in the forms $A=a_{1}\otimes a_{2}$ and $B=b_{1}\otimes b_{2}$ for
some $a_{1},a_{2}\in\Lambda$ and $b_{1},b_{2}\in\Lambda$. Consider these
$a_{1},a_{2}$ and $b_{1},b_{2}$. Now,%
\begin{align*}
\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  \underbrace{A}%
_{=a_{1}\otimes a_{2}},\underbrace{B}_{=b_{1}\otimes b_{2}}\right)
_{\Lambda\otimes\Lambda}  & =\left(  \underbrace{\left(  \mathbf{f}_{n}%
\otimes\mathbf{f}_{n}\right)  \left(  a_{1}\otimes a_{2}\right)  }_{=\left(
\mathbf{f}_{n}a_{1}\right)  \otimes\left(  \mathbf{f}_{n}a_{2}\right)  }%
,b_{1}\otimes b_{2}\right)  _{\Lambda\otimes\Lambda}=\left(  \left(
\mathbf{f}_{n}a_{1}\right)  \otimes\left(  \mathbf{f}_{n}a_{2}\right)
,b_{1}\otimes b_{2}\right)  _{\Lambda\otimes\Lambda}\\
& =\underbrace{\left(  \mathbf{f}_{n}a_{1},b_{1}\right)  _{\Lambda}%
}_{\substack{=\left(  a_{1},\mathbf{v}_{n}b_{1}\right)  _{\Lambda}\\\text{(by
(\ref{sol.witt.ghost-app.d.duality-trick.fv}), applied to}\\a=a_{1}\text{ and
}b=b_{1}\text{)}}}\underbrace{\left(  \mathbf{f}_{n}a_{2},b_{2}\right)
_{\Lambda}}_{\substack{=\left(  a_{2},\mathbf{v}_{n}b_{2}\right)  _{\Lambda
}\\\text{(by (\ref{sol.witt.ghost-app.d.duality-trick.fv}), applied
to}\\a=a_{2}\text{ and }b=b_{2}\text{)}}}\\
& =\left(  a_{1},\mathbf{v}_{n}b_{1}\right)  _{\Lambda}\left(  a_{2}%
,\mathbf{v}_{n}b_{2}\right)  _{\Lambda}.
\end{align*}
Compared with%
\begin{align*}
\left(  \underbrace{A}_{=a_{1}\otimes a_{2}},\left(  \mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\right)  \underbrace{B}_{=b_{1}\otimes b_{2}}\right)
_{\Lambda\otimes\Lambda}  & =\left(  a_{1}\otimes a_{2},\underbrace{\left(
\mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(  b_{1}\otimes b_{2}\right)
}_{=\mathbf{v}_{n}\left(  b_{1}\right)  \otimes\mathbf{v}_{n}\left(
b_{2}\right)  }\right)  _{\Lambda\otimes\Lambda}\\
& =\left(  a_{1}\otimes a_{2},\mathbf{v}_{n}\left(  b_{1}\right)
\otimes\mathbf{v}_{n}\left(  b_{2}\right)  \right)  _{\Lambda\otimes\Lambda
}=\left(  a_{1},\mathbf{v}_{n}b_{1}\right)  _{\Lambda}\left(  a_{2}%
,\mathbf{v}_{n}b_{2}\right)  _{\Lambda},
\end{align*}
this yields $\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)
A,B\right)  _{\Lambda\otimes\Lambda}=\left(  A,\left(  \mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\right)  B\right)  _{\Lambda\otimes\Lambda}$. This proves
(\ref{sol.witt.ghost-app.d.duality-trick.ffvv}).]
\par
Let us now show that
$\left(\mathbf{f}_n \otimes \mathbf{f}_n\right) \circ \Delta
= \Delta \circ \mathbf{f}_n$ (this is one of the axioms that need to
be checked in order to show that $\mathbf{f}_n$ is a Hopf algebra
homomorphism).
\par
Indeed, $\mathbf{v}_n$ is a $\kk$-algebra homomorphism. Hence,
$m \circ \left(\mathbf{v}_n \otimes \mathbf{v}_n\right)
= \mathbf{v}_n \circ m$.
\par
Now, let $a \in \Lambda$. Every $B \in \Lambda \otimes \Lambda$
satisfies
\begin{align}
\left(  \underbrace{\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}%
_{n}\right)  \circ\Delta\right)  \left(  a\right)  }_{=\left(  \mathbf{f}%
_{n}\otimes\mathbf{f}_{n}\right)  \left(  \Delta a\right)  },B\right)
_{\Lambda\otimes\Lambda}
& =\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  \left(  \Delta
a\right)  ,B\right)  _{\Lambda\otimes\Lambda}=\left(  \Delta a,\left(
\mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  B\right)  _{\Lambda\otimes\Lambda
}
\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.witt.ghost-app.d.duality-trick.ffvv}), applied to }A=\Delta
a\right)
\nonumber\\
& =\left(  \Delta a,\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)
B\right)  _{\Lambda\otimes\Lambda}=\left(  a,\underbrace{m\left(  \left(
\mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  B\right)  }_{=\left(
m\circ\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \right)  \left(
B\right)  }\right)  _{\Lambda}
\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because }\left(  a,m\left(  \left(  \mathbf{v}_{n}\otimes\mathbf{v}%
_{n}\right)  B\right)  \right)  _{\Lambda}=\left(  \Delta a,\left(
\mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  B\right)  _{\Lambda\otimes\Lambda
} \\
\text{(by (\ref{sol.witt.ghost-app.d.duality-trick.aB}), applied to }\left(
\mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  B\text{ instead of }B\text{)}%
\end{array}
\right)
\nonumber\\
& =\left(  a,\underbrace{\left(  m\circ\left(  \mathbf{v}_{n}\otimes
\mathbf{v}_{n}\right)  \right)  }_{=\mathbf{v}_{n}\circ m}\left(  B\right)
\right)  _{\Lambda}=\left(  a,\underbrace{\left(  \mathbf{v}_{n}\circ
m\right)  \left(  B\right)  }_{=\mathbf{v}_{n}\left(  mB\right)  }\right)
_{\Lambda}=\left(  a,\mathbf{v}_{n}\left(  mB\right)  \right)  _{\Lambda
}=\left(  \mathbf{f}_{n}a,mB\right)  _{\Lambda}
\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because }\left(  \mathbf{f}_{n}a,mB\right)  _{\Lambda}=\left(
a,\mathbf{v}_{n}\left(  mB\right)  \right)  _{\Lambda}\\
\text{(by (\ref{sol.witt.ghost-app.d.duality-trick.fv}), applied to }mB\text{
instead of }b\text{)}%
\end{array}
\right)
\nonumber\\
& =\left(  \underbrace{\Delta\left(  \mathbf{f}_{n}a\right)  }_{=\left(
\Delta\circ\mathbf{f}_{n}\right)  \left(  a\right)  },B\right)  _{\Lambda
\otimes\Lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.witt.ghost-app.d.duality-trick.aB}), applied to }\mathbf{f}%
_{n}a\text{ instead of }a\right)
\nonumber\\
& =\left(  \left(  \Delta\circ\mathbf{f}_{n}\right)  \left(  a\right)
,B\right)  _{\Lambda\otimes\Lambda}.
\label{sol.witt.ghost-app.d.duality-trick.sample}
\end{align}
\par
Now, the bilinear form
$\left(\cdot, \cdot\right)_{\Lambda \otimes \Lambda}$ is nondegenerate
(in fact, $\left(s_\mu \otimes s_\nu\right)_{\left(\mu, \nu\right)
\in \Par \times \Par}$ is an orthonormal basis with respect to this
bilinear form). Hence, if $U$ and $V$ are two elements of
$\Lambda \otimes \Lambda$ such that every $B \in
\Lambda \otimes \Lambda$ satisfies
$\left(U, B\right)_{\Lambda \otimes \Lambda}
= \left(V, B\right)_{\Lambda \otimes \Lambda}$, then $U = V$. Applying
this to
$U = \left(\left(\mathbf{f}_n \otimes \mathbf{f}_n\right) \circ \Delta\right) a$
and $V = \left(\Delta \circ \mathbf{f}_n\right) a$, we obtain
$\left(\left(\mathbf{f}_n \otimes \mathbf{f}_n\right) \circ \Delta\right) a
= \left(\Delta \circ \mathbf{f}_n\right) a$ (because of
\eqref{sol.witt.ghost-app.d.duality-trick.sample}). Since we have proven
this for every $a \in \Lambda$, this yields that
$\left(\mathbf{f}_n \otimes \mathbf{f}_n\right) \circ \Delta
= \Delta \circ \mathbf{f}_n$. Thus,
$\left(\mathbf{f}_n \otimes \mathbf{f}_n\right) \circ \Delta
= \Delta \circ \mathbf{f}_n$ is proven.}.
\end{itemize}

Probably the following solution of Exercise \ref{exe.witt.ghost-app}(d) is the shortest:

Fix $n\in\left\{  1,2,3,\ldots\right\}  $. We need to prove that
$\mathbf{f}_{n}:\Lambda\rightarrow\Lambda$ is a Hopf algebra homomorphism. We
already know that $\mathbf{f}_{n}$ is a $\kk$-algebra homomorphism.
Therefore, if we can show that $\mathbf{f}_{n}$ is a $\kk$-coalgebra
homomorphism, then it will immediately follow that $\mathbf{f}_{n}$ is a
$\kk$-bialgebra homomorphism and thus a Hopf algebra homomorphism (due
to Corollary~\ref{cor.bialg-mor-is-Hopf}). Therefore, it
remains to show that $\mathbf{f}_{n}$ is a $\kk$-coalgebra
homomorphism. To do so, we need to check that $\epsilon\circ\mathbf{f}%
_{n}=\epsilon$ and $\Delta\circ\mathbf{f}_{n}=\left(  \mathbf{f}_{n}%
\otimes\mathbf{f}_{n}\right)  \circ\Delta$. We shall only prove $\Delta
\circ\mathbf{f}_{n}=\left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)
\circ\Delta$, while the easy proof of $\epsilon\circ\mathbf{f}_{n}=\epsilon$
is left to the reader.

Let us first notice that $\Lambda$ is a bialgebra, and therefore $\Delta$ is a
$\kk$-algebra homomorphism (by the axioms of a bialgebra). Also,
$\mathbf{f}_{n}$ is a $\kk$-algebra homomorphism. Thus, $\Delta
\circ\mathbf{f}_{n}$ and $\left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)
\circ\Delta$ are $\kk$-algebra homomorphisms.

For every $m\in \NN$, let us define an element $\widetilde{h}_{m}%
\in \kk \left[  \left[  \xx\right]  \right]  $ by $\widetilde{h}%
_{m}=\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{m}}x_{i_{1}}^{n}x_{i_{2}}%
^{n}\cdots x_{i_{m}}^{n}$. Then, it is easy to see that%
\begin{equation}
\mathbf{f}_{n}\left(  h_{m}\right)  =\widetilde{h}_{m}%
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in \NN %
.\label{sol.witt.ghost-app.d.fnhm}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.witt.ghost-app.d.fnhm}):} Let
$m\in \NN$. Then, (\ref{def.hn}) (applied to $m$ instead of $n$) yields
$h_{m}=\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{m}}x_{i_{1}}x_{i_{2}}\cdots
x_{i_{m}}$. Thus,%
\[
h_{m}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)  =\left(  \sum
_{i_{1}\leq i_{2}\leq\cdots\leq i_{m}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}%
}\right)  \left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)  =\sum
_{i_{1}\leq i_{2}\leq\cdots\leq i_{m}}x_{i_{1}}^{n}x_{i_{2}}^{n}\cdots
x_{i_{m}}^{n}.
\]
But the definition of $\mathbf{f}_{n}$ yields $\mathbf{f}_{n}\left(
h_{m}\right)  =h_{m}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)
=\sum_{i_{1}\leq i_{2}\leq\cdots\leq i_{m}}x_{i_{1}}^{n}x_{i_{2}}^{n}\cdots
x_{i_{m}}^{n}$. This proves (\ref{sol.witt.ghost-app.d.fnhm}).} As a
consequence, $\widetilde{h}_{m}=\mathbf{f}_{n}\left(  h_{m}\right)  \in
\Lambda$ for every $m\in \NN$. (Of course, this was obvious anyway.)

Fix a positive integer $r$. Proposition
\ref{symm-comultiplication-formulas-prop}(iii) (applied to $r$ instead of $n$)
states that $\Delta h_{r}=\sum_{i+j=r}h_{i}\otimes h_{j}$. The proof of this
proposition (which proceeded by observing that $h_{r}\left(  \xx%
,\mathbf{y}\right)  =\sum_{i+j=r}h_{i}\left(  \xx\right)  h_{j}\left(
\mathbf{y}\right)  $ in $\kk \left[  \left[  \xx,\mathbf{y}%
\right]  \right]  $) can be easily modified to obtain a proof of the equality
$\Delta\widetilde{h}_{r}=\sum_{i+j=r}\widetilde{h}_{i}\otimes\widetilde{h}%
_{j}$. Thus, we have%
\begin{equation}
\Delta\widetilde{h}_{r}=\sum_{i+j=r}\widetilde{h}_{i}\otimes\widetilde{h}%
_{j}.\label{sol.witt.ghost-app.d.1}%
\end{equation}


Now,
\[
\left(  \Delta\circ\mathbf{f}_{n}\right)  \left(  h_{r}\right)  =\Delta\left(
\underbrace{\mathbf{f}_{n}\left(  h_{r}\right)  }_{\substack{=\widetilde{h}%
_{r}\\\text{(by (\ref{sol.witt.ghost-app.d.fnhm}), applied to }m=r\text{)}%
}}\right)  =\Delta\widetilde{h}_{r}=\sum_{i+j=r}\widetilde{h}_{i}%
\otimes\widetilde{h}_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.witt.ghost-app.d.1})}\right)  .
\]
Compared with%
\begin{align*}
\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  \circ
\Delta\right)  \left(  h_{r}\right)    & =\left(  \mathbf{f}_{n}%
\otimes\mathbf{f}_{n}\right)  \left(  \underbrace{\Delta h_{r}}_{=\sum
_{i+j=r}h_{i}\otimes h_{j}}\right)  =\left(  \mathbf{f}_{n}\otimes
\mathbf{f}_{n}\right)  \left(  \sum_{i+j=r}h_{i}\otimes h_{j}\right)  \\
& =\sum_{i+j=r}\underbrace{\mathbf{f}_{n}\left(  h_{i}\right)  }%
_{\substack{=\widetilde{h}_{i}\\\text{(by (\ref{sol.witt.ghost-app.d.fnhm}),
applied to }m=i\text{)}}}\otimes\underbrace{\mathbf{f}_{n}\left(
h_{j}\right)  }_{\substack{=\widetilde{h}_{j}\\\text{(by
(\ref{sol.witt.ghost-app.d.fnhm}), applied to }m=j\text{)}}}=\sum
_{i+j=r}\widetilde{h}_{i}\otimes\widetilde{h}_{j},
\end{align*}
this yields $\left(  \Delta\circ\mathbf{f}_{n}\right)  \left(  h_{r}\right)
=\left(  \left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  \circ
\Delta\right)  \left(  h_{r}\right)  $.

Now, let us forget that we fixed $r$. We thus have proven that%
\begin{equation}
\left(  \Delta\circ\mathbf{f}_{n}\right)  \left(  h_{r}\right)  =\left(
\left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  \circ\Delta\right)
\left(  h_{r}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every positive integer
}r\text{.}\label{sol.witt.ghost-app.d.main}%
\end{equation}


Now, recall that the family $\left(  h_{r}\right)  _{r\geq1}$ generates the
$\kk$-algebra $\Lambda$ (according to Proposition
\ref{symm-antipode-and-h-basis}). In other words, $\left(  h_{r}\right)
_{r\geq1}$ is a generating set of the $\kk$-algebra $\Lambda$. The two
$\kk$-algebra homomorphisms $\Delta\circ\mathbf{f}_{n}$ and $\left(
\mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)  \circ\Delta$ are equal to each
other on this generating set (according to (\ref{sol.witt.ghost-app.d.main})),
and therefore must be identical (because if two $\kk$-algebra
homomorphisms from the same domain are equal to each other on a generating set
of their domain, then these two homomorphisms must be identical). In other
words, $\Delta\circ\mathbf{f}_{n}=\left(  \mathbf{f}_{n}\otimes\mathbf{f}%
_{n}\right)  \circ\Delta$. This completes the solution of Exercise
\ref{exe.witt.ghost-app}(d).

(e) Let $m\in \NN$. From (\ref{h-generating-function-no-HE}), we have
$\prod_{i=1}^{\infty}\left(  1-x_{i}t\right)  ^{-1}=\sum_{n\geq0}h_{n}\left(
\xx \right)  t^{n}$ in the ring $\Lambda\left[  \left[  t\right]
\right]  $. Substituting $x_{i}^{2}$ for $x_{i}$ and $t^{2}$ for $t$ in this
equality, we obtain%
\begin{equation}
\prod_{i=1}^{\infty}\left(  1-x_{i}^{2}t^{2}\right)  ^{-1}=\sum_{n\geq0}%
h_{n}\left(  x_{1}^{2},x_{2}^{2},x_{3}^{2},\ldots\right)  \left(
t^{2}\right)  ^{n}.\label{sol.witt.ghost-app.e.1}%
\end{equation}


But%
\begin{equation}
\prod_{i=1}^{\infty}\left(  1-x_{i}t\right)  ^{-1}=\sum_{n\geq0}%
\underbrace{h_{n}\left(  \xx\right)  }_{=h_{n}}t^{n}=\sum_{n\geq0}%
h_{n}t^{n}.\label{sol.witt.ghost-app.e.2}%
\end{equation}
Substituting $-t$ for $t$ in this equality, we obtain $\prod_{i=1}^{\infty
}\left(  1-x_{i}\left(  -t\right)  \right)  ^{-1}=\sum_{n\geq0}h_{n}\left(
-t\right)  ^{n}=\sum_{n\geq0}\left(  -1\right)  ^{n}h_{n}t^{n}$. Thus,%
\begin{equation}
\sum_{n\geq0}\left(  -1\right)  ^{n}h_{n}t^{n}=\prod_{i=1}^{\infty}\left(
\underbrace{1-x_{i}\left(  -t\right)  }_{=1+x_{i}t}\right)  ^{-1}=\prod
_{i=1}^{\infty}\left(  1+x_{i}t\right)  ^{-1}.\label{sol.witt.ghost-app.e.3}%
\end{equation}


Now,
\begin{align*}
\sum_{n\geq0}\underbrace{\mathbf{f}_{2}\left(  h_{n}\right)  }%
_{\substack{=h_{n}\left(  x_{1}^{2},x_{2}^{2},x_{3}^{2},\ldots\right)
\\\text{(by the definition of }\mathbf{f}_{2}\text{)}}}\underbrace{t^{2n}%
}_{=\left(  t^{2}\right)  ^{n}}  & =\sum_{n\geq0}h_{n}\left(  x_{1}^{2}%
,x_{2}^{2},x_{3}^{2},\ldots\right)  \left(  t^{2}\right)  ^{n}=\prod
_{i=1}^{\infty}\left(  \underbrace{1-x_{i}^{2}t^{2}}_{=\left(  1-x_{i}%
t\right)  \left(  1+x_{i}t\right)  }\right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.witt.ghost-app.e.1})}\right)  \\
& =\prod_{i=1}^{\infty}\left(  \left(  1-x_{i}t\right)  \left(  1+x_{i}%
t\right)  \right)  ^{-1}=\underbrace{\left(  \prod_{i=1}^{\infty}\left(
1-x_{i}t\right)  ^{-1}\right)  }_{\substack{=\sum_{n\geq0}h_{n}t^{n}%
\\\text{(by (\ref{sol.witt.ghost-app.e.2}))}}}\underbrace{\left(  \prod
_{i=1}^{\infty}\left(  1+x_{i}t\right)  ^{-1}\right)  }_{\substack{=\sum
_{n\geq0}\left(  -1\right)  ^{n}h_{n}t^{n}\\\text{(by
(\ref{sol.witt.ghost-app.e.3}))}}}\\
& =\left(  \sum_{n\geq0}h_{n}t^{n}\right)  \left(  \sum_{n\geq0}\left(
-1\right)  ^{n}h_{n}t^{n}\right)  =\sum_{n\geq0}\left(  \sum_{i=0}^{n}%
h_{i}\cdot\left(  -1\right)  ^{n-i}h_{n-i}\right)  t^{n}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the product of two
formal power series}\right)  .
\end{align*}
Comparing coefficients before $t^{2m}$ on both sides of this equality, we
obtain
\[
\mathbf{f}_{2}\left(  h_{m}\right)  =\sum_{i=0}^{2m}h_{i}\cdot
\underbrace{\left(  -1\right)  ^{2m-i}}_{\substack{=\left(  -1\right)
^{i}\\\text{(since }2m-i\equiv i\operatorname{mod}2\text{)}}}h_{2m-i}%
=\sum_{i=0}^{2m}h_{i}\cdot\left(  -1\right)  ^{i}h_{2m-i}=\sum_{i=0}%
^{2m}\left(  -1\right)  ^{i}h_{i}h_{2m-i}.
\]
Exercise~\ref{exe.witt.ghost-app}(e) is thus solved.

(f) Let $p$ be a prime number, and let $a\in\Lambda$. We need to prove that
$\mathbf{f}_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}p\Lambda$.

Indeed, let us first check that $\mathbf{f}_{p}\left(  a\right)  \equiv
a^{p}\operatorname{mod}p \ZZ \left[  \left[  \xx\right]  \right]
$. Since $\mathbf{f}_{p}\left(  a\right)  =a\left(  x_{1}^{p},x_{2}^{p}%
,x_{3}^{p},\ldots\right)  $, this is equivalent to showing that $a\left(
x_{1}^{p},x_{2}^{p},x_{3}^{p},\ldots\right)  \equiv a^{p}\operatorname{mod}%
p \ZZ \left[  \left[  \xx\right]  \right]  $. This, in turn, is
equivalent to proving that $\overline{a}\left(  x_{1}^{p},x_{2}^{p},x_{3}%
^{p},\ldots\right)  =\overline{a}^{p}$, where $\overline{a}$ denotes the
projection of the power series $a\in \ZZ \left[  \left[  \xx%
\right]  \right]  $ onto the ring $\left(   \ZZ /p \ZZ \right)
\left[  \left[  \xx\right]  \right]  $ (by reducing every coefficient
modulo $p$). So let us prove $\overline{a}\left(  x_{1}^{p},x_{2}^{p}%
,x_{3}^{p},\ldots\right)  =\overline{a}^{p}$ now.

Write the power series $\overline{a}$ in the form $\overline{a}=\sum_{\beta
}\kappa_{\beta}\mathbf{x}^{\beta}$, where the sum ranges over all weak
compositions $\beta$, and where $\kappa_{\beta}$ is an element of
$ \ZZ /p \ZZ$ for every weak composition $\beta$. Taking both sides
of this equality to the $p$-th power, we obtain%
\begin{equation}
\overline{a}^{p}=\left(  \sum_{\beta}\kappa_{\beta}\mathbf{x}^{\beta}\right)
^{p}. \label{sol.witt.ghost-app.1}%
\end{equation}


In the commutative ring $\left(   \ZZ /p \ZZ \right)  \left[
\left[  \xx\right]  \right]  $, we have $p\cdot1_{\left(
 \ZZ /p \ZZ \right)  \left[  \left[  \xx\right]  \right]
}=0$. Thus, taking the $p$-th power is a ring endomorphism of $\left(
 \ZZ /p \ZZ \right)  \left[  \left[  \xx\right]  \right]  $.
This ring endomorphism is moreover continuous (with respect to the usual
topology on $\left(   \ZZ /p \ZZ \right)  \left[  \left[
\xx\right]  \right]  $) and $\left(   \ZZ /p \ZZ \right)
$-linear (by virtue of being a ring endomorphism). Thus, this endomorphism
respects infinite $\left(   \ZZ /p \ZZ \right)  $-linear
combinations; hence,%
\begin{equation}
\left(  \sum_{\beta}\kappa_{\beta}\mathbf{x}^{\beta}\right)  ^{p}=\sum_{\beta
}\kappa_{\beta}\left(  \mathbf{x}^{\beta}\right)  ^{p}.
\label{sol.witt.ghost-app.2}%
\end{equation}
But $\overline{a}=\sum_{\beta}\kappa_{\beta}\mathbf{x}^{\beta}$, so that
$\overline{a}\left(  x_{1}^{p},x_{2}^{p},x_{3}^{p},\ldots\right)  =\sum
_{\beta}\kappa_{\beta}\left(  \mathbf{x}^{\beta}\right)  ^{p}$ (because
replacing all variables $x_{1},x_{2},x_{3},\ldots$ by their $p$-th powers
transforms every monomial $\mathbf{x}^{\beta}$ into $\left(  \xx%
^{\beta}\right)  ^{p}$). Thus,%
\begin{align*}
\overline{a}\left(  x_{1}^{p},x_{2}^{p},x_{3}^{p},\ldots\right)   &
=\sum_{\beta}\kappa_{\beta}\left(  \mathbf{x}^{\beta}\right)  ^{p}=\left(
\sum_{\beta}\kappa_{\beta}\mathbf{x}^{\beta}\right)  ^{p}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.witt.ghost-app.2})}\right) \\
&  =\overline{a}^{p}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.witt.ghost-app.1})}\right)  .
\end{align*}
We thus have proven that $\overline{a}\left(  x_{1}^{p},x_{2}^{p},x_{3}%
^{p},\ldots\right)  =\overline{a}^{p}$. As explained above, this yields
$\mathbf{f}_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}p \ZZ %
\left[  \left[  \xx\right]  \right]  $. In other words, the power
series $\dfrac{\mathbf{f}_{p}\left(  a\right)  -a^{p}}{p}$ (this is, a priori,
an element of $\QQ \left[  \left[  \xx\right]  \right]  $)
belongs to $ \ZZ \left[  \left[  \xx\right]  \right]  $. Since
this power series is also of bounded degree and symmetric (because so are
$\mathbf{f}_{p}\left(  a\right)  $ and $a^{p}$), it follows that it lies in
$\Lambda$. So we have $\dfrac{\mathbf{f}_{p}\left(  a\right)  -a^{p}}{p}%
\in\Lambda$, thus $\mathbf{f}_{p}\left(  a\right)  -a^{p}\in p\Lambda$ and
hence $\mathbf{f}_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}p\Lambda
$. This solves Exercise \ref{exe.witt.ghost-app}(f).

(g) Set $\kk = \ZZ$. Thus, the sign $\otimes$ will mean
$\otimes_{ \ZZ }$ in the remainder of this solution. Also, $\Lambda
=\Lambda_{ \ZZ }$. We define the notation $v_{p}\left(  n\right)  $ as in
Exercise \ref{exe.witt.ghost-equiv}.

Let us introduce a notion from commutative algebra:

In the following, a \dfn{special $\Psi$-ring} will mean a pair
$\left(  A,\left(  \varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}
}\right)  $, where $A$ is a commutative ring and $\left(  \varphi_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }$ is a family of ring endomorphisms
$\varphi_{n}:A\rightarrow A$ of $A$ satisfying the following properties:

\begin{itemize}
\item We have $\varphi_{n}\circ\varphi_{m}=\varphi_{nm}$ for any two positive
integers $n$ and $m$.

\item We have $\varphi_{1}= \id$.

\item We have $\varphi_{p}\left(  a\right)  \equiv a^{p}\operatorname{mod}pA$
for every $a\in A$ and every prime number $p$.
\end{itemize}

The tensor product of two special $\Psi$-rings $\left(  A,\left(  \varphi
_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ and $\left(
B,\left(  \psi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ is
defined to be the pair $\left(  A\otimes B,\left(  \varphi_{n}\otimes\psi
_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $. This pair
$\left(  A\otimes B,\left(  \varphi_{n}\otimes\psi_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }\right)  $ is a special $\Psi$-ring
again\footnote{\textit{Proof.} We need to prove the following five statements:
\par
\begin{statement}
\textit{Statement 1:} The ring $A\otimes B$ is a commutative ring.
\end{statement}
\par
\begin{statement}
\textit{Statement 2:} The family $\left(  \varphi_{n}\otimes\psi_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }$ is a family of ring endomorphisms
$\varphi_{n}\otimes\psi_{n}:A\otimes B\rightarrow A\otimes B$ of $A \otimes B$.
\end{statement}
\par
\begin{statement}
\textit{Statement 3:} We have $\left(  \varphi_{n}\otimes\psi_{n}\right)
\circ\left(  \varphi_{m}\otimes\psi_{n}\right)  =\varphi_{nm}\otimes\psi_{nm}$
for any two positive integers $n$ and $m$.
\end{statement}
\par
\begin{statement}
\textit{Statement 4:} We have $\varphi_{1}\otimes\psi_{1}= \id$.
\end{statement}
\par
\begin{statement}
\textit{Statement 5:} We have $\left(  \varphi_{p}\otimes\psi_{p}\right)
\left(  a\right)  \equiv a^{p}\operatorname{mod}p\left(  A\otimes B\right)  $
for every $a\in A\otimes B$ and every prime number $p$.
\end{statement}
\par
\textit{Proof of Statement 1:} This is obvious.
\par
\textit{Proof of Statement 2:} The family $\left(  \varphi_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }$ is a family of ring endomorphisms of
$A$ (since $\left(  A,\left(  \varphi_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }\right)  $ is a special $\Psi$-ring). Thus,
$\varphi_{n}$ is a ring endomorphism of $A$ for every positive integer $n$.
Similarly, $\psi_{n}$ is a ring endomorphism of $B$ for every positive integer
$n$. Thus, $\varphi_{n}\otimes\psi_{n}$ is a ring endomorphism of $A\otimes B$
for every positive integer $n$. In other words, Statement 2 holds.
\par
\textit{Proof of Statement 3:} Let $n$ and $m$ be positive integers. Then,
$\varphi_{n}\circ\varphi_{m}=\varphi_{nm}$ (since $\left(  A,\left(
\varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ is a
special $\Psi$-ring) and $\psi_{n}\circ\psi_{m}=\psi_{nm}$ (similarly). Now,
$\left(  \varphi_{n}\otimes\psi_{n}\right)  \circ\left(  \varphi_{m}%
\otimes\psi_{n}\right)  =\underbrace{\left(  \varphi_{n}\circ\varphi
_{m}\right)  }_{=\varphi_{nm}}\otimes\underbrace{\left(  \psi_{n}\circ\psi
_{m}\right)  }_{=\psi_{nm}}=\varphi_{nm}\otimes\psi_{nm}$, and thus Statement
3 is proven.
\par
\textit{Proof of Statement 4:} Since $\left(  A,\left(  \varphi_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ is a special $\Psi$-ring, we
have $\varphi_{1}= \id$. Similarly, $\psi_{1}= \id %
$. Thus, $\varphi_{1}\otimes\psi_{1}= \id \otimes
 \id = \id$. This proves Statement 4.
\par
\textit{Proof of Statement 5:} Fix a prime number $p$. For every commutative
ring $R$, we introduce three pieces of notation:
\par
\begin{itemize}
\item We let $\overline{R}$ denote the commutative ring $R/pR$.
\par
\item We let $\pi_{R}$ denote the canonical projection $R\rightarrow R/pR$.
\par
\item We let $\operatorname{pow}_{R}$ denote the map $R\rightarrow
R$ which sends every $r\in R$ to $r^{p}$. This is not a linear map in general,
but when $p\cdot1_{R}=0$, the map $\operatorname{pow}_{R}$ is a ring
endomorphism of $R$.
\end{itemize}
\par
Now, $\left(  A,\left(  \varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots
\right\}  }\right)  $ is a special $\Psi$-ring. Hence, $\varphi_{p}\left(
a\right)  \equiv a^{p}\operatorname{mod}pA$ for every $a\in A$. In other
words, $\pi_{A}\left(  \varphi_{p}\left(  a\right)  \right)  =\pi_{A}\left(
\underbrace{a^{p}}_{=\operatorname{pow}_{A}a}\right)  =\pi
_{A}\left(  \operatorname{pow}_{A}a\right)  $ for every $a\in A$. In
other words, $\pi_{A}\circ\varphi_{p}=\pi_{A}\circ\operatorname{pow}_{A}$.
But $\pi_{A}\circ\operatorname{pow}_{A}%
=\operatorname{pow}_{\overline{A}}\circ\pi_{A}$ (this is just saying
that taking the $p$-th power commutes with the projection $\pi_{A}$). Hence,%
\[
\pi_{A}\circ\varphi_{p}=\pi_{A}\circ\operatorname{pow}_{A}%
=\operatorname{pow}_{\overline{A}}\circ\pi_{A}.
\]
Similarly,%
\[
\pi_{B}\circ\psi_{p}=\pi_{B}\circ\operatorname{pow}_{B}%
=\operatorname{pow}_{\overline{B}}\circ\pi_{B}.
\]
\par
For every commutative ring $R$, the ring $\overline{R}$ satisfies
$p\cdot1_{\overline{R}}=0$, and thus $\operatorname{pow}_{\overline{R}}$
is a ring endomorphism of $\overline{R}$. Hence,
$\operatorname{pow}_{\overline{A}}$, $\operatorname{pow}_{\overline{B}}$ and
$\operatorname{pow}_{\overline{A\otimes B}}$ are ring endomorphisms,
thus $\ZZ$-algebra endomorphisms; consequently, the tensor product
$\operatorname{pow}_{\overline{A}}\otimes\operatorname{pow}_{\overline{B}}$
makes sense. Moreover, we have $\operatorname{pow}_{\overline{A}}
\otimes\operatorname{pow}_{\overline{B}}
=\operatorname{pow}_{\overline{A\otimes B}}$ (because both
$\operatorname{pow}_{\overline{A}}\otimes\operatorname{pow}_{\overline{B}}$
and $\operatorname{pow}_{\overline{A\otimes B}}$ are
ring endomorphisms of $\overline{A}\otimes\overline{B}$
sending pure tensors $\alpha\otimes\beta$ to $\alpha^{p}\otimes\beta
^{p}=\left(  \alpha\otimes\beta\right)  ^{p}$, and this characterizes them
uniquely).
\par
There is a canonical isomorphism $\overline{A}\otimes\overline{B}%
\rightarrow\overline{A\otimes B}$ (because tensoring is right-exact). We
identify $\overline{A\otimes B}$ with $\overline{A}\otimes\overline{B}$ along
this isomorphism. Then, $\pi_{A\otimes B}=\pi_{A}\otimes\pi_{B}$. Thus,%
\begin{align*}
\underbrace{\pi_{A\otimes B}}_{=\pi_{A}\otimes\pi_{B}}\circ\left(  \varphi
_{p}\otimes\psi_{p}\right)   &  =\left(  \pi_{A}\otimes\pi_{B}\right)
\circ\left(  \varphi_{p}\otimes\psi_{p}\right)  =\underbrace{\left(  \pi
_{A}\circ\varphi_{p}\right)  }_{=\operatorname{pow}_{\overline{A}%
}\circ\pi_{A}}\otimes\underbrace{\left(  \pi_{B}\circ\psi_{p}\right)
}_{=\operatorname{pow}_{\overline{B}}\circ\pi_{B}}\\
&  =\left(  \operatorname{pow}_{\overline{A}}\circ\pi_{A}\right)
\otimes\left(  \operatorname{pow}_{\overline{B}}\circ\pi_{B}\right)
=\underbrace{\left(  \operatorname{pow}_{\overline{A}}%
\otimes\operatorname{pow}_{\overline{B}}\right)  }%
_{=\operatorname{pow}_{\overline{A\otimes B}}}\circ
\underbrace{\left(  \pi_{A}\otimes\pi_{B}\right)  }_{=\pi_{A\otimes B}}\\
&  =\operatorname{pow}_{\overline{A\otimes B}}\circ\pi_{A\otimes
B}=\pi_{A\otimes B}\circ\operatorname{pow}_{A\otimes B}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since taking the }p\text{-th power
commutes with the projection }\pi_{A\otimes B}\right)  .
\end{align*}
Hence, every $a\in A\otimes B$ satisfies $\pi_{A\otimes B}\left(  \left(
\varphi_{p}\otimes\psi_{p}\right)  \left(  a\right)  \right)  =\pi_{A\otimes
B}\left(  \operatorname{pow}_{A\otimes B}\left(  a\right)  \right)
$. In other words, every $a\in A\otimes B$ satisfies $\left(  \varphi
_{p}\otimes\psi_{p}\right)  \left(  a\right)  \equiv
\operatorname{pow}_{A\otimes B}\left(  a\right) 
= a^{p}\operatorname{mod}p\left( A\otimes B\right)  $.
Thus, Statement 5 is proven.
\par
It thus follows that $\left(  A\otimes B,\left(  \varphi_{n}\otimes\psi
_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ is a special
$\Psi$-ring, qed.}.

Here are three examples of special $\Psi$-rings which we will need:

\begin{itemize}
\item The pair $\left(   \ZZ ,\left(   \id \right)
_{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ is a special $\Psi$-ring.
(The proof of this relies on Fermat's little theorem.)

\item The pair $\left(  \Lambda,\left(  \mathbf{f}_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }\right)  $ is a special $\Psi$-ring. (This follows from
parts (a), (b), (c) and (f) of Exercise~\ref{exe.witt.ghost-app}.)

\item The pair $\left(  \Lambda\otimes\Lambda,\left(  \mathbf{f}_{n}%
\otimes\mathbf{f}_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $
is a special $\Psi$-ring. (Indeed, this pair is the tensor product of the
special $\Psi$-ring $\left(  \Lambda,\left(  \mathbf{f}_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ with itself.)
\end{itemize}

Now, we will establish a pattern which we will follow in our new solutions to
parts (b), (c), (d), (e) and (f) of Exercise \ref{exe.Lambda.maps-on-pn}.
Namely, let $\left(  A,\left(  \varphi_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }\right)  $ be a special $\Psi$-ring such that the
$\ZZ$-module $A$ is free. Then, $A$ canonically injects into
$A\otimes \QQ$. Identify $A$ with a subring of $A\otimes \QQ$
using this injection. Also, consider $\Lambda=\Lambda_{ \ZZ }$ as a
subring of $\Lambda_{\QQ}$ as in Exercise \ref{exe.Lambda.maps-on-pn}.
Let $f:\Lambda_{\QQ}\rightarrow A\otimes \QQ$ be a $\QQ$-algebra
homomorphism such that every $n\in\left\{  1,2,3,\ldots\right\}  $
satisfies $f\left(  p_{n}\right)  \in A$. We can then ask for a criterion for
$f\left(  \Lambda\right)  \subset A$. Using Exercise
\ref{exe.witt.ghost-equiv}, we can obtain such an answer:
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{We have }f\left(  \Lambda\right)  \subset A\text{ if every positive
integer }n\text{ and every}\\
\text{prime factor }p\text{ of }n\text{ satisfy }\varphi_{p}\left(  f\left(
p_{n/p}\right)  \right)  \equiv f\left(  p_{n}\right)  \operatorname{mod}%
p^{v_{p}\left(  n\right)  }A
\end{array}
\right)  .\label{sol.witt.ghost-app.crit}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.witt.ghost-app.crit}):} Exercise
\ref{exe.witt.ghost-equiv} can be applied to the family $\left(  b_{n}\right)
_{n\geq1}=\left(  f\left(  p_{n}\right)  \right)  _{n\geq1}$ (indeed, the
conditions of Exercise \ref{exe.witt.ghost-equiv} are satisfied because
$\left(  A,\left(  \varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}
}\right)  $ is a special $\Psi$-ring). As a result, we see that the Assertions
$\mathcal{C}$, $\mathcal{D}$, $\mathcal{E}$, $\mathcal{F}$, $\mathcal{G}$,
$\mathcal{H}$ and $\mathcal{J}$ for $\left(  b_{n}\right)  _{n\geq1}=\left(
f\left(  p_{n}\right)  \right)  _{n\geq1}$ are equivalent.
\par
Assume that every positive integer $n$ and every prime factor $p$ of $n$
satisfy $\varphi_{p}\left(  f\left(  p_{n/p}\right)  \right)  \equiv f\left(
p_{n}\right)  \operatorname{mod}p^{v_{p}\left(  n\right)  }A$. Then, the
family $\left(  f\left(  p_{n}\right)  \right)  _{n\geq1}\in A^{\left\{
1,2,3,\ldots\right\}  }$ satisfies the Assertion $\mathcal{C}$ of Exercise
\ref{exe.witt.ghost-equiv}. But since Assertion $\mathcal{C}$ is equivalent to
Assertion $\mathcal{J}$, this yields that the family $\left(  f\left(
p_{n}\right)  \right)  _{n\geq1}\in A^{\left\{  1,2,3,\ldots\right\}  }$
satisfies the Assertion $\mathcal{J}$ as well. In other words, there exists a
ring homomorphism $\Lambda_{ \ZZ }\rightarrow A$ which, for every
positive integer $n$, sends $p_{n}$ to $f\left(  p_{n}\right)  $. Let $g$ be
such a ring homomorphism. Then, $g\left(  p_{n}\right)  =f\left(
p_{n}\right)  $ for every positive integer $n$.
\par
The $\ZZ$-algebra homomorphism $g:\Lambda_{ \ZZ }\rightarrow A$
can be extended to a $\QQ$-algebra homomorphism
$\Lambda_{\ZZ}\otimes \QQ \rightarrow A\otimes \QQ$ (by base change). Since
$\Lambda_{\ZZ}\otimes \QQ \cong\Lambda_{\QQ}$ canonically,
we can regard this latter $\QQ$-algebra homomorphism as a $\QQ$-algebra
homomorphism $\Lambda_{\QQ}\rightarrow A\otimes \QQ$.
Let us denote this $\QQ$-algebra homomorphism $\Lambda_{\QQ}
\rightarrow A\otimes \QQ$ by $\widetilde{g}$. Thus, $\widetilde{g}%
\mid_{\Lambda}=g$.
\par
Now, both $f$ and $\widetilde{g}$ are $\QQ$-algebra homomorphisms
$\Lambda_{\QQ}\rightarrow A\otimes \QQ$. These homomorphisms $f$
and $\widetilde{g}$ are equal to each other on the elements $p_{1},p_{2}%
,p_{3},\ldots$ of $\Lambda_{\QQ}$ (because for every positive integer
$n$, we have $\widetilde{g}\left(  p_{n}\right)  =\underbrace{\left(
\widetilde{g}\mid_{\Lambda}\right)  }_{=g}\left(  p_{n}\right)  =g\left(
p_{n}\right)  =f\left(  p_{n}\right)  $). Since the elements $p_{1}%
,p_{2},p_{3},\ldots$ generate the $\QQ$-algebra $\Lambda_\QQ$, this
forces said homomorphisms $f$ and $\widetilde{g}$ to be identical. That is, we
have $f=\widetilde{g}$. Hence, $f\left(  \Lambda\right)  =\widetilde{g}\left(
\Lambda\right)  =\underbrace{\left(  \widetilde{g}\mid_{\Lambda}\right)
}_{=g}\left(  \Lambda\right)  =g\left(  \Lambda\right)  \subset A$ (since the
target of $g$ is $A$). This proves (\ref{sol.witt.ghost-app.crit}).} (The
``if'' here can be extended to ``if and only if'', but we do not need the
``only if'' part.)

Let us now give alternative solutions to parts (b), (c), (d), (e) and (f) of
Exercise \ref{exe.Lambda.maps-on-pn}:

\textit{Alternative solution to part (b) of Exercise
\ref{exe.Lambda.maps-on-pn}:} The $\ZZ$-module $\Lambda\otimes\Lambda$
is free, and thus canonically injects into $\left(  \Lambda\otimes
\Lambda\right)  \otimes \QQ$. We use this identification to regard
$\Lambda\otimes\Lambda$ as a subring of $\left(  \Lambda\otimes\Lambda\right)
\otimes \QQ$. We also identify
$\Lambda_{\QQ} \otimes_{\QQ} \Lambda_{\QQ} $ with $\left(  \Lambda\otimes
\Lambda\right)  \otimes \QQ$. Notice that every $n\in\left\{
1,2,3,\ldots\right\}  $ satisfies $\Delta_{\times}\left(  p_{n}\right)
=p_{n}\otimes p_{n}\in\Lambda\otimes\Lambda$.

We need to prove that $\Delta_{\times}\left(  \Lambda\right)  \subset
\Lambda\otimes_{ \ZZ }\Lambda$. In other words, we need to prove that
$\Delta_{\times}\left(  \Lambda\right)  \subset\Lambda\otimes\Lambda$. This
will follow from (\ref{sol.witt.ghost-app.crit}) (applied to $\left(
A,\left(  \varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)
=\left(  \Lambda\otimes\Lambda,\left(  \mathbf{f}_{n}\otimes\mathbf{f}%
_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ and
$f=\Delta_{\times}$) once we have showed that every positive integer $n$ and
every prime factor $p$ of $n$ satisfy
\begin{equation}
\left(  \mathbf{f}_{p}\otimes\mathbf{f}_{p}\right)  \left(  \Delta_{\times
}\left(  p_{n/p}\right)  \right)  \equiv\Delta_{\times}\left(  p_{n}\right)
\operatorname{mod}p^{v_{p}\left(  n\right)  }\left(  \Lambda\otimes
\Lambda\right)  .\label{sol.witt.ghost-app.g.altsol-b.goal}%
\end{equation}
Thus, it remains to prove (\ref{sol.witt.ghost-app.g.altsol-b.goal}).

Let $n$ be a positive integer, and let $p$ be a prime factor of $n$. The
definition of $\mathbf{f}_{p}\left(  p_{n/p}\right)  $ yields
\begin{align}
\mathbf{f}_{p}\left(  p_{n/p}\right)
&=p_{n/p}\left(  x_{1}^{p},x_{2}%
^{p},x_{3}^{p},\ldots\right)  =\left(  x_{1}^{n/p}\right)  ^{p}+\left(
x_{2}^{n/p}\right)  ^{p}+\left(  x_{3}^{n/p}\right)  ^{p}+\cdots
\nonumber \\
&=x_{1}^{n}+x_{2}^{n}+x_{3}^{n}+\cdots=p_{n}.
\label{sol.witt.ghost-app.g.altsol-b.fp}
\end{align}
Now,%
\begin{align*}
\left(  \mathbf{f}_{p}\otimes\mathbf{f}_{p}\right)  \left(  \underbrace{\Delta
_{\times}\left(  p_{n/p}\right)  }_{\substack{=p_{n/p}\otimes p_{n/p}%
\\\text{(by the definition of }\Delta_{\times}\text{)}}}\right)    & =\left(
\mathbf{f}_{p}\otimes\mathbf{f}_{p}\right)  \left(  p_{n/p}\otimes
p_{n/p}\right)  =\underbrace{\mathbf{f}_{p}\left(  p_{n/p}\right)  }_{=p_{n}%
}\otimes\underbrace{\mathbf{f}_{p}\left(  p_{n/p}\right)  }_{=p_{n}}\\
& =p_{n}\otimes p_{n}=\Delta_{\times}\left(  p_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Delta_{\times
}\right)  .
\end{align*}
Hence, (\ref{sol.witt.ghost-app.g.altsol-b.goal}) holds. Thus, Exercise
\ref{exe.Lambda.maps-on-pn}(b) is solved again.

\textit{Alternative solution to part (c) of Exercise
\ref{exe.Lambda.maps-on-pn}:} We identify $\ZZ \otimes \QQ$ with
$\QQ$. Every $n\in\left\{  1,2,3,\ldots\right\}  $ satisfies
$\epsilon_{r}\left(  p_{n}\right)  =r\in \ZZ$.

We need to prove that $\epsilon_{r}\left(  \Lambda\right)  \subset{ \ZZ %
}$. This will follow from (\ref{sol.witt.ghost-app.crit}) (applied to $\left(
A,\left(  \varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)
=\left(   \ZZ ,\left(   \id \right)  _{n\in\left\{
1,2,3,\ldots\right\}  }\right)  $ and $f=\epsilon_{r}$) once we have showed
that every positive integer $n$ and every prime factor $p$ of $n$ satisfy%
\[
 \id \left(  \epsilon_{r}\left(  p_{n/p}\right)  \right)
\equiv\epsilon_{r}\left(  p_{n}\right)  \operatorname{mod}p^{v_{p}\left(
n\right)  } \ZZ .
\]
But this congruence follows immediately from%
\begin{align*}
 \id \left(  \epsilon_{r}\left(  p_{n/p}\right)  \right)    &
=\epsilon_{r}\left(  p_{n/p}\right)  =r\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\epsilon_{r}\right)  \\
& =\epsilon_{r}\left(  p_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\epsilon_{r}\right)  .
\end{align*}
Thus, Exercise \ref{exe.Lambda.maps-on-pn}(c) is solved again.

\textit{Alternative solution to part (d) of Exercise
\ref{exe.Lambda.maps-on-pn}:} The $\ZZ$-module $\Lambda$ is free, and
thus canonically injects into $\Lambda\otimes \QQ$. We use this
identification to regard $\Lambda$ as a subring of $\Lambda\otimes \QQ$.
We also identify $\Lambda_{\QQ}$ with $\Lambda\otimes \QQ$.
Notice that every $n\in\left\{  1,2,3,\ldots\right\}  $ satisfies
$\mathbf{i}_{r}\left(  p_{n}\right)  =rp_{n}\in\Lambda$.

We need to prove that $\mathbf{i}_{r}\left(  \Lambda\right)  \subset\Lambda$.
This will follow from (\ref{sol.witt.ghost-app.crit}) (applied to $\left(
A,\left(  \varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)
=\left(  \Lambda,\left(  \mathbf{f}_{n}\right)  _{n\in\left\{  1,2,3,\ldots
\right\}  }\right)  $ and $f=\mathbf{i}_{r}$) once we have showed that every
positive integer $n$ and every prime factor $p$ of $n$ satisfy%
\[
\mathbf{f}_{p}\left(  \mathbf{i}_{r}\left(  p_{n/p}\right)  \right)
\equiv\mathbf{i}_{r}\left(  p_{n}\right)  \operatorname{mod}p^{v_{p}\left(
n\right)  }\Lambda.
\]
But this congruence follows from%
\[
\mathbf{f}_{p}\left(  \underbrace{\mathbf{i}_{r}\left(  p_{n/p}\right)
}_{\substack{=rp_{n/p}\\\text{(by the definition of }\mathbf{i}_{r}\text{)}%
}}\right)  =\mathbf{f}_{p}\left(  rp_{n/p}\right)  =r\underbrace{\mathbf{f}%
_{p}\left(  p_{n/p}\right)  }_{\substack{=p_{n}\\\text{(by
(\ref{sol.witt.ghost-app.g.altsol-b.fp}))}}}=rp_{n}=\mathbf{i}_{r}\left(
p_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{i}_{r}\right)  .
\]
This solves Exercise \ref{exe.Lambda.maps-on-pn}(d) again.

\textit{Alternative solution to part (e) of Exercise
\ref{exe.Lambda.maps-on-pn}:} The $\ZZ$-module $\Lambda$ is free, and
thus canonically injects into $\Lambda\otimes \QQ$. We use this
identification to regard $\Lambda$ as a subring of $\Lambda\otimes \QQ$.
We also identify $\Lambda_{\QQ}$ with $\Lambda\otimes \QQ$.

It is easy to see that the map $\operatorname*{Sq}$ is a $\QQ$-algebra
homomorphism\footnote{\textit{Proof.} We need to check that
$\operatorname*{Sq}\left(  ab\right)  =\left(  \operatorname*{Sq}a\right)
\left(  \operatorname*{Sq}b\right)  $ for any $a\in\Lambda_{\QQ}$ and
$b\in\Lambda_{\QQ}$. Since $\operatorname*{Sq}$ is $\QQ$-linear,
this only needs to be checked on a basis of the $\QQ$-module
$\Lambda_{\QQ}$. For this we use the basis $\left(  p_{\lambda}\right)
_{\lambda\in\operatorname{Par}}$ of $\Lambda_{\QQ}$. Checking the
identity $\operatorname*{Sq}\left(  ab\right)  =\left(  \operatorname*{Sq}%
a\right)  \left(  \operatorname*{Sq}b\right)  $ on this basis amounts to
proving that $\operatorname*{Sq}\left(  p_{\lambda}p_{\mu}\right)  =\left(
\operatorname*{Sq}\left(  p_{\lambda}\right)  \right)  \left(
\operatorname*{Sq}\left(  p_{\mu}\right)  \right)  $ for any two partitions
$\lambda$ and $\mu$. So let $\lambda$ and $\mu$ be two partitions. It is clear
that there exists a partition $\nu$ such that $p_{\lambda}p_{\mu}=p_{\nu}$
(indeed, this $\nu$ is the partition obtained by sorting the list $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  },\mu
_{1},\mu_{2},\ldots,\mu_{\ell\left(  \mu\right)  }\right)  $ in decreasing
order). Consider this $\nu$. We have%
\begin{align*}
\operatorname*{Sq}\left(  \underbrace{p_{\lambda}p_{\mu}}_{=p_{\nu}}\right)
& =\operatorname*{Sq}\left(  p_{\nu}\right)  =p_{\nu}^{2}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Sq}%
\right)  \\
& =\left(  p_{\lambda}p_{\mu}\right)  ^{2}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }p_{\nu}=p_{\lambda}p_{\mu}\right)  \\
& =\underbrace{p_{\lambda}^{2}}_{\substack{=\operatorname*{Sq}\left(
p_{\lambda}\right)  \\\text{(by the definition of }\operatorname*{Sq}\text{)}%
}}\underbrace{p_{\mu}^{2}}_{\substack{=\operatorname*{Sq}\left(  p_{\mu
}\right)  \\\text{(by the definition of }\operatorname*{Sq}\text{)}}}=\left(
\operatorname*{Sq}\left(  p_{\lambda}\right)  \right)  \left(
\operatorname*{Sq}\left(  p_{\mu}\right)  \right)  ,
\end{align*}
which is what we wanted to prove. Thus we have checked that
$\operatorname*{Sq}\left(  ab\right)  =\left(  \operatorname*{Sq}a\right)
\left(  \operatorname*{Sq}b\right)  $ for any $a\in\Lambda_{\QQ}$ and
$b\in\Lambda_{\QQ}$. Hence, $\operatorname*{Sq}$ is a $\QQ$-algebra
homomorphism (since $\operatorname*{Sq}\left(  1\right)  =1$),
qed.}. Every $n\in\left\{  1,2,3,\ldots\right\}  $ satisfies
$\operatorname*{Sq}\left(  p_{n}\right)  =p_{n}^{2}\in\Lambda$.

We need to prove that $\operatorname*{Sq}\left(  \Lambda\right)
\subset\Lambda$. This will follow from (\ref{sol.witt.ghost-app.crit})
(applied to $\left(  A,\left(  \varphi_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }\right)  =\left(  \Lambda,\left(  \mathbf{f}%
_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ and
$f=\operatorname*{Sq}$) once we have showed that every positive integer $n$
and every prime factor $p$ of $n$ satisfy%
\[
\mathbf{f}_{p}\left(  \operatorname*{Sq}\left(  p_{n/p}\right)  \right)
\equiv\operatorname*{Sq}\left(  p_{n}\right)  \operatorname{mod}%
p^{v_{p}\left(  n\right)  }\Lambda.
\]
But this congruence follows from%
\begin{align*}
\mathbf{f}_{p}\left(  \underbrace{\operatorname*{Sq}\left(  p_{n/p}\right)
}_{\substack{=p_{n/p}^{2}\\\text{(by the definition of }\operatorname*{Sq}%
\text{)}}}\right)    & =\mathbf{f}_{p}\left(  p_{n/p}^{2}\right)  =\left(
\underbrace{\mathbf{f}_{p}\left(  p_{n/p}\right)  }_{\substack{=p_{n}%
\\\text{(by (\ref{sol.witt.ghost-app.g.altsol-b.fp}))}}}\right)
^{2}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{f}_{p}\text{ is a ring
homomorphism}\right)  \\
& =p_{n}^{2}=\operatorname*{Sq}\left(  p_{n}\right)  .
\end{align*}
This solves Exercise \ref{exe.Lambda.maps-on-pn}(e) again.

\textit{Alternative solution to part (f) of Exercise
\ref{exe.Lambda.maps-on-pn}:} The $\ZZ$-module $\Lambda\otimes\Lambda$
is free, and thus canonically injects into $\left(  \Lambda\otimes
\Lambda\right)  \otimes \QQ$. We use this identification to regard
$\Lambda\otimes\Lambda$ as a subring of $\left(  \Lambda\otimes\Lambda\right)
\otimes \QQ$. We also identify
$\Lambda_{\QQ} \otimes_{\QQ} \Lambda_{\QQ} $ with $\left(  \Lambda\otimes
\Lambda\right)  \otimes \QQ$. Notice that every $n\in\left\{
1,2,3,\ldots\right\}  $ satisfies $\Delta_{r}\left(  p_{n}\right)  =\sum
_{i=1}^{n-1}\binom{n}{i}p_{i}\otimes p_{n-i}+r\otimes p_{n}+p_{n}\otimes
r\in\Lambda\otimes\Lambda$.

We need to prove that $\Delta_{r}\left(  \Lambda\right)  \subset\Lambda
\otimes_{ \ZZ }\Lambda$. In other words, we need to prove that
$\Delta_{r}\left(  \Lambda\right)  \subset\Lambda\otimes\Lambda$. This will
follow from (\ref{sol.witt.ghost-app.crit}) (applied to $\left(  A,\left(
\varphi_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }\right)  =\left(
\Lambda\otimes\Lambda,\left(  \mathbf{f}_{n}\otimes\mathbf{f}_{n}\right)
_{n\in\left\{  1,2,3,\ldots\right\}  }\right)  $ and $f=\Delta_{r}$) once we
have showed that every positive integer $n$ and every prime factor $p$ of $n$
satisfy%
\begin{equation}
\left(  \mathbf{f}_{p}\otimes\mathbf{f}_{p}\right)  \left(  \Delta_{r}\left(
p_{n/p}\right)  \right)  \equiv\Delta_{r}\left(  p_{n}\right)
\operatorname{mod}p^{v_{p}\left(  n\right)  }\left(  \Lambda\otimes
\Lambda\right)  .\label{sol.witt.ghost-app.g.altsol-f.goal}%
\end{equation}
Thus, it remains to prove (\ref{sol.witt.ghost-app.g.altsol-f.goal}).

Let $n$ be a positive integer, and let $p$ be a prime factor of $n$. For the
sake of brevity, we denote $r$ by $p_{0}$. Then,%
\begin{align}
\Delta_{r}\left(  p_{n}\right)    & =\sum_{i=1}^{n-1}\binom{n}{i}p_{i}\otimes
p_{n-i}+\underbrace{r}_{=p_{0}}\otimes p_{n}+p_{n}\otimes\underbrace{r}%
_{=p_{0}}\nonumber\\
& =\sum_{i=1}^{n-1}\binom{n}{i}p_{i}\otimes p_{n-i}+p_{0}\otimes p_{n}%
+p_{n}\otimes p_{0}=\sum_{i=0}^{n}\binom{n}{i}p_{i}\otimes p_{n-i}%
.\label{sol.witt.ghost-app.g.altsol-f.shorter-form}%
\end{align}
Applying this to $n/p$ instead of $n$, we obtain%
\begin{equation}
\Delta_{r}\left(  p_{n/p}\right)  =\sum_{i=0}^{n/p}\binom{n/p}{i}p_{i}\otimes
p_{n/p-i}.\label{sol.witt.ghost-app.g.altsol-f.shorter-form.n/p}%
\end{equation}


Now, we will need two elementary congruences for binomial coefficients:

\begin{itemize}
\item For any $i\in \NN$ satisfying $p\mid i$, we have%
\begin{equation}
\dbinom{n/p}{i/p}\equiv\dbinom{n}{i}\operatorname{mod}p^{v_{p}\left(
n\right)  }.\label{sol.witt.ghost-app.g.altsol-f.bincong1}%
\end{equation}
(This follows from (\ref{sol.witt.ghost-exa.fam3.goal2}), applied to $q=1$ and
$r=i/n$.)

\item For any $i\in \NN$ satisfying $p\nmid i$, we have%
\begin{equation}
0\equiv\dbinom{n}{i}\operatorname{mod}p^{v_{p}\left(  n\right)  }%
.\label{sol.witt.ghost-app.g.altsol-f.bincong2}%
\end{equation}
(This follows from (\ref{sol.witt.ghost-exa.fam3.goal2}), applied to $q=1$ and
$r=i/n$, keeping in mind that $\dbinom{a}{b}=0$ if $b\notin \NN$.)
\end{itemize}

Now, applying the map $\mathbf{f}_{p}\otimes\mathbf{f}_{p}$ to both sides of
the equality (\ref{sol.witt.ghost-app.g.altsol-f.shorter-form.n/p}), we obtain%
\begin{align*}
\left(  \mathbf{f}_{p}\otimes\mathbf{f}_{p}\right)  \left(  \Delta_{r}\left(
p_{n/p}\right)  \right)    & =\left(  \mathbf{f}_{p}\otimes\mathbf{f}%
_{p}\right)  \left(  \sum_{i=0}^{n/p}\binom{n/p}{i}p_{i}\otimes p_{n/p-i}%
\right)  \\
& =\sum_{i=0}^{n/p}\binom{n/p}{i}\underbrace{\mathbf{f}_{p}\left(
p_{i}\right)  }_{\substack{=\mathbf{f}_{p}\left(  p_{pi/p}\right)
=p_{pi}\\\text{(by (\ref{sol.witt.ghost-app.g.altsol-b.fp}), applied
to}\\pi\text{ instead of }n\text{)}}}\otimes\underbrace{\mathbf{f}_{p}\left(
p_{n/p-i}\right)  }_{\substack{=\mathbf{f}_{p}\left(  p_{\left(  n-pi\right)
/p}\right)  =p_{n-pi}\\\text{(by (\ref{sol.witt.ghost-app.g.altsol-b.fp}),
applied to}\\n-pi\text{ instead of }n\text{)}}}\\
& =\sum_{i=0}^{n/p}\binom{n/p}{i}p_{pi}\otimes p_{n-pi}=\sum_{i\in\left\{
0,1,\ldots,n/p\right\}  }\binom{n/p}{i}p_{pi}\otimes p_{n-pi}=\sum\limits
_{\substack{i\in\left\{  0,1,\ldots,n\right\}  ;\\p\mid i}}\binom{n/p}%
{i/p}p_{i}\otimes p_{n-i}%
\end{align*}
(here, we have substituted $i/p$ for $i$ in the sum). Comparing this with%
\begin{align*}
\Delta_{r}\left(  p_{n}\right)   &  =\sum_{i=0}^{n}\binom{n}{i}p_{i}\otimes
p_{n-i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.witt.ghost-app.g.altsol-f.shorter-form})}\right)  \\
&  =\sum\limits_{\substack{i\in\left\{  0,1,\ldots,n\right\}  ;\\p\mid
i}}\underbrace{\dbinom{n}{i}}_{\substack{\equiv\dbinom{n/p}{i/p}%
\operatorname{mod}p^{v_{p}\left(  n\right)  }\left(  \Lambda\otimes
\Lambda\right)  \\\text{(by (\ref{sol.witt.ghost-app.g.altsol-f.bincong1}))}%
}}p_{i}\otimes p_{n-i}+\sum\limits_{\substack{i\in\left\{  0,1,\ldots,n\right\}
;\\p\nmid i}}\underbrace{\dbinom{n}{i}}_{\substack{\equiv0\operatorname{mod}%
p^{v_{p}\left(  n\right)  }\left(  \Lambda\otimes\Lambda\right)  \\\text{(by
(\ref{sol.witt.ghost-app.g.altsol-f.bincong2}))}}}p_{i}\otimes p_{n-i}\\
&  \equiv\sum\limits_{\substack{i\in\left\{  0,1,\ldots,n\right\}  ;\\p\mid
i}}\dbinom{n/p}{i/p}p_{i}\otimes p_{n-i}+\underbrace{\sum
\limits_{\substack{i\in\left\{  0,1,\ldots,n\right\}  ;\\p\nmid i}}0p_{i}\otimes
p_{n-i}}_{=0}\\
&  =\sum\limits_{\substack{i\in\left\{  0,1,\ldots,n\right\}  ;\\p\mid i}%
}\dbinom{n/p}{i/p}p_{i}\otimes p_{n-i}\operatorname{mod}p^{v_{p}\left(
n\right)  }\left(  \Lambda\otimes\Lambda\right)  ,
\end{align*}
we obtain%
\[
\left(  \mathbf{f}_{p}\otimes\mathbf{f}_{p}\right)  \left(  \Delta_{r}\left(
p_{n/p}\right)  \right)  \equiv\Delta_{r}\left(  p_{n}\right)
\operatorname{mod}p^{v_{p}\left(  n\right)  }\left(  \Lambda\otimes
\Lambda\right)  .
\]
Hence, (\ref{sol.witt.ghost-app.g.altsol-f.goal}) holds. Thus, Exercise
\ref{exe.Lambda.maps-on-pn}(f) is solved again.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.verschiebung}}

\textit{Solution to Exercise~\ref{exe.Lambda.verschiebung}.} Let us
first notice that every positive integer $n$ satisfies
\begin{equation}
\mathbf{v}_{n}\left(  h_{m}\right)  = \begin{cases}
h_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in \NN 
\label{sol.Lambda.verschiebung.vnhm}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.vnhm}):}
Let $n$ be a positive integer. Let $m \in \NN$. We need to prove that
(\ref{sol.Lambda.verschiebung.vnhm}) holds.
\par
If $m\neq0$, then $m$ is a positive integer. Hence, if $m\neq0$, then
(\ref{sol.Lambda.verschiebung.vnhm}) follows immediately from the definition
of $\mathbf{v}_{n}\left(  h_{m}\right)  $. Thus, for the rest of this proof of
(\ref{sol.Lambda.verschiebung.vnhm}), we can WLOG assume that we don't have
$m\neq0$. Assume this.
\par
We don't have $m\neq0$. Thus, we have $m=0$, so that $h_{m}=h_{0}=1$ and
therefore $\mathbf{v}_{n}\left(  \underbrace{h_{m}}_{=1}\right)
=\mathbf{v}_{n}\left(  1\right)  =1$ (since $\mathbf{v}_{n}$ is a
$\kk$-algebra homomorphism).
\par
Also, $ \begin{cases}
h_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad  =h_{m/n}$ (since $n\mid0=m$). Now,%
\begin{align*}
\mathbf{v}_{n}\left(  h_{m}\right)   &  =1=h_{0}=h_{m/n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h_{m/n}=h_{0}\text{ (since
}\underbrace{m}_{=0}/n=0/n=0\text{)}\right) \\
&  =  \begin{cases}
h_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad .
\end{align*}
This proves (\ref{sol.Lambda.verschiebung.vnhm}).}.

(b) Let $n$ be a positive integer. Recall that a ring of formal power series
$R\left[  \left[  t\right]  \right]  $ over a commutative ring $R$ has a
canonical topology which makes it into a topological $R$-algebra. Thus, in
particular, $\Lambda\left[  \left[  t\right]  \right]  $ becomes a topological
$\Lambda$-algebra. We shall be considering this topology when we speak of continuity.

The $\kk$-algebra homomorphism $\mathbf{v}_{n}:\Lambda\rightarrow
\Lambda$ induces a continuous $\kk$-algebra homomorphism $\mathbf{v}%
_{n}\left[  \left[  t\right]  \right]  :\Lambda\left[  \left[  t\right]
\right]  \rightarrow\Lambda\left[  \left[  t\right]  \right]  $ given by%
\[
\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\in \NN }a_{i}t^{i}\right)  =\sum_{i\in \NN }\mathbf{v}%
_{n}\left(  a_{i}\right)  t^{i}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
a_{i}\right)  _{i\in \NN }\in\Lambda^{ \NN }.
\]
Consider the power series $H\left(  t\right)  \in\Lambda\left[  \left[
t\right]  \right]  $ and $E\left(  t\right)  \in\Lambda\left[  \left[
t\right]  \right]  $ defined in the proof of Proposition
\ref{symm-antipode-and-h-basis}. We have
\[
H\left(  t\right)  =1+h_{1}\left(  \xx\right)  t+h_{2}\left(
{\mathbf{x}}\right)  t^{2}+\cdots=\sum_{i\geq0}h_{i}\left(  \xx\right)
t^{i}=\sum_{i\geq0}h_{i}t^{i}%
\]
and similarly $E\left(  t\right)  =\sum_{i\geq0}e_{i}t^{i}$.

Substituting $t^{n}$ for $t$ in the equality $H\left(  t\right)  =\sum
_{i\geq0}h_{i}t^{i}$, we obtain $H\left(  t^{n}\right)  =\sum_{i\geq0}%
h_{i}\left(  t^{n}\right)  ^{i}$. Substituting $-t$ for $t$ in the equality
$E\left(  t\right)  =\sum_{i\geq0}e_{i}t^{i}$, we obtain $E\left(  -t\right)
=\sum_{i\geq0}e_{i}\underbrace{\left(  -t\right)  ^{i}}_{=\left(  -1\right)
^{i}t^{i}}=\sum_{i\geq0}\left(  -1\right)  ^{i}e_{i}t^{i}$. Substituting
$-t^{n}$ for $t$ in the equality $E\left(  t\right)  =\sum_{i\geq0}e_{i}t^{i}%
$, we obtain $E\left(  -t^{n}\right)  =\sum_{i\geq0}e_{i}\left(
-t^{n}\right)  ^{i}$.

Applying the map $\mathbf{v}_{n}\left[  \left[  t\right]  \right]  $ to both
sides of the equality $H\left(  t\right)  =\sum_{i\geq0}h_{i}t^{i}$, we obtain%
\begin{align}
&  \left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \right) \nonumber\\
&  =\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\geq0}h_{i}t^{i}\right)  =\sum_{i\geq0}\underbrace{\mathbf{v}%
_{n}\left(  h_{i}\right)  }_{\substack{=
\begin{cases}
h_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases}
\\\text{(by (\ref{sol.Lambda.verschiebung.vnhm}), applied to
}m=i\text{)}}}t^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right) \nonumber\\
&  =\sum_{i\geq0}
\begin{cases}
h_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases} \cdot
t^{i}=\sum\limits_{\substack{i\geq0;\\n\mid i}}\underbrace{
\begin{cases}
h_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases}
}_{\substack{=h_{i/n}\\\text{(since }n\mid i\text{)}}}
\cdot t^{i}%
+\sum\limits_{\substack{i\geq0;\\n\nmid i}}\underbrace{
\begin{cases}
h_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases}}_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}h_{i/n}t^{i}+\underbrace{\sum\limits
_{\substack{i\geq0;\\n\nmid i}}0t^{i}}_{=0}=\sum\limits_{\substack{i\geq0;\\n\mid
i}}h_{i/n}t^{i}=\sum_{i\geq0}h_{i}\underbrace{t^{ni}}_{=\left(  t^{n}\right)
^{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }ni\text{ for
}i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\geq0}h_{i}\left(  t^{n}\right)  ^{i}=H\left(  t^{n}\right)  .
\label{sol.Lambda.verschiebung.a.H}%
\end{align}
But (\ref{E(t)-H(t)-relation}) yields $1=E\left(  -t\right)  H\left(
t\right)  $, so that $E\left(  -t\right)  =\dfrac{1}{H\left(  t\right)  }$.
Applying the map $\mathbf{v}_{n}\left[  \left[  t\right]  \right]  $ to both
sides of this equality, we obtain%
\begin{align}
\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
E\left(  -t\right)  \right)   &  =\left(  \mathbf{v}_{n}\left[  \left[
t\right]  \right]  \right)  \left(  \dfrac{1}{H\left(  t\right)  }\right)
=\dfrac{1}{\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)
\left(  H\left(  t\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{ is a } \kk %
\text{-algebra homomorphism}\right) \nonumber\\
&  =\dfrac{1}{H\left(  t^{n}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.verschiebung.a.H})}\right) \nonumber\\
&  =E\left(  -t^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because }E\left(  -t^{n}\right)  =\dfrac{1}{H\left(  t^{n}\right)
}\text{ (this follows by substituting }t^{n}\\
\text{for }t\text{ in the equality }E\left(  -t\right)  =\dfrac{1}{H\left(
t\right)  }\text{)}%
\end{array}
\right) \nonumber\\
&  =\sum_{i\geq0}e_{i}\underbrace{\left(  -t^{n}\right)  ^{i}}_{=\left(
-1\right)  ^{i}t^{ni}}=\sum_{i\geq0}\left(  -1\right)  ^{i}e_{i}t^{ni}.
\label{sol.Lambda.verschiebung.a.E1}%
\end{align}


Now,%
\begin{align}
&  \sum_{i\geq0}
\begin{cases}
\left(  -1\right)  ^{i/n}e_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases} \cdot
t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}\underbrace{
\begin{cases}
\left(  -1\right)  ^{i/n}e_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases}
}_{\substack{=\left(  -1\right)  ^{i/n}e_{i/n}\\\text{(since }n\mid
i\text{)}}} \cdot t^{i}
+\sum\limits_{\substack{i\geq0;\\n\nmid i}}\underbrace{
\begin{cases}
\left(  -1\right)  ^{i/n}e_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}\left(  -1\right)  ^{i/n}e_{i/n}%
t^{i}+\underbrace{\sum\limits_{\substack{i\geq0;\\n\nmid i}}0t^{i}}_{=0}%
=\sum\limits_{\substack{i\geq0;\\n\mid i}}\left(  -1\right)  ^{i/n}e_{i/n}%
t^{i}\nonumber\\
&  =\sum_{i\geq0}\left(  -1\right)  ^{i}e_{i}t^{ni}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we substituted }ni\text{ for }i\text{ in the sum}\right)
\nonumber\\
&  =\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
E\left(  -t\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.verschiebung.a.E1})}\right)  .
\label{sol.Lambda.verschiebung.a.E2}%
\end{align}


But applying the map $\mathbf{v}_{n}\left[  \left[  t\right]  \right]  $ to
both sides of the equality $E\left(  -t\right)  =\sum_{i\geq0}\left(
-1\right)  ^{i}e_{i}t^{i}$, we obtain%
\begin{align*}
\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
E\left(  -t\right)  \right)   &  =\left(  \mathbf{v}_{n}\left[  \left[
t\right]  \right]  \right)  \left(  \sum_{i\geq0}\left(  -1\right)  ^{i}%
e_{i}t^{i}\right)  =\sum_{i\geq0}\underbrace{\mathbf{v}_{n}\left(  \left(
-1\right)  ^{i}e_{i}\right)  }_{\substack{=\left(  -1\right)  ^{i}%
\mathbf{v}_{n}\left(  e_{i}\right)  \\\text{(since }\mathbf{v}_{n}\text{ is
} \kk \text{-linear)}}}t^{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right) \\
&  =\sum_{i\geq0}\left(  -1\right)  ^{i}\mathbf{v}_{n}\left(  e_{i}\right)
t^{i}.
\end{align*}
Hence,%
\begin{equation}
\sum_{i\geq0}\left(  -1\right)  ^{i}\mathbf{v}_{n}\left(  e_{i}\right)
t^{i}=\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)
\left(  E\left(  -t\right)  \right)  =\sum_{i\geq0}
\begin{cases}
\left(  -1\right)  ^{i/n}e_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases} \cdot
t^{i} \label{sol.Lambda.verschiebung.a.E3}%
\end{equation}
(by (\ref{sol.Lambda.verschiebung.a.E2})).

Now, let $m$ be a positive integer. Comparing coefficients before $t^{m}$ in
the equality (\ref{sol.Lambda.verschiebung.a.E3}), we obtain%
\[
\left(  -1\right)  ^{m}\mathbf{v}_{n}\left(  e_{m}\right)  =
\begin{cases}
\left(  -1\right)  ^{m/n}e_{m/n},&\text{if }n\mid m;\\
0,&\text{if }n\nmid m
\end{cases} \quad
.
\]
Dividing this by $\left(  -1\right)  ^{m}$, we obtain%
\begin{align*}
\mathbf{v}_{n}\left(  e_{m}\right)   &  =\dfrac{1}{\left(  -1\right)  ^{m}%
}
\begin{cases}
\left(  -1\right)  ^{m/n}e_{m/n},&\text{if }n\mid m;\\
0,&\text{if }n\nmid m
\end{cases} \quad
=
\begin{cases}
\dfrac{1}{\left(  -1\right)  ^{m}}\left(  -1\right)  ^{m/n}e_{m/n}%
,&\text{if }n\mid m;\\
0,&\text{if }n\nmid m
\end{cases} \\
&  =
\begin{cases}
\left(  -1\right)  ^{m-m/n} e_{m/n},&\text{if }n\mid m;\\
0,&\text{if }n\nmid m
\end{cases}
\end{align*}
(because if $n\mid m$, then $\dfrac{1}{\left(  -1\right)  ^{m}}\left(
-1\right)  ^{m/n}=\left(  -1\right)  ^{m/n-m}=\left(  -1\right)  ^{m-m/n}$).
This solves Exercise \ref{exe.Lambda.verschiebung}(b).

(a) Let $n$ be a positive integer. We define the continuous $\kk %
$-algebra homomorphism $\mathbf{v}_{n}\left[  \left[  t\right]  \right]
:\Lambda\left[  \left[  t\right]  \right]  \rightarrow\left[  \left[
t\right]  \right]  $ as in the solution of Exercise
\ref{exe.Lambda.verschiebung}(a).

Consider the power series $H\left(  t\right)  \in\Lambda\left[  \left[
t\right]  \right]  $ defined in the proof of Proposition
\ref{symm-antipode-and-h-basis}. We have $H\left(  t\right)  =\sum_{i\geq
0}h_{i}t^{i}$ (this can be proven just as in the solution of Exercise
\ref{exe.Lambda.verschiebung}(a)), so that $H^{\prime}\left(  t\right)
=\sum_{i\geq1}ih_{i}t^{i-1}$ (by the definition of the derivative of a power
series). We can also see that the equality (\ref{sol.Lambda.verschiebung.a.H})
holds (this can be proven just as in the solution of Exercise
\ref{exe.Lambda.verschiebung}(a)). The power series $H\left(  t\right)  $ is
invertible (since its constant term is $h_{0}=1$).

Exercise \ref{exe.E(t)-H(t)-P(t)-relation} yields $\sum_{m\geq0}p_{m+1}%
t^{m}=\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }$, so that%
\[
\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }=\sum_{m\geq0}%
p_{m+1}t^{m}=\sum_{i\geq1}p_{i}t^{i-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{here,
we substituted }i\text{ for }m+1\text{ in the sum}\right)  .
\]
Multiplying this equality with $t$, we obtain%
\[
t\cdot\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }=t\cdot
\sum_{i\geq1}p_{i}t^{i-1}=\sum_{i\geq1}p_{i}\underbrace{tt^{i-1}}_{=t^{i}%
}=\sum_{i\geq1}p_{i}t^{i}.
\]
Multiplying this equality with $H\left(  t\right)  $, we obtain%
\[
t\cdot H^{\prime}\left(  t\right)  =H\left(  t\right)  \cdot\sum_{i\geq1}%
p_{i}t^{i}.
\]
Hence,%
\begin{equation}
H\left(  t\right)  \cdot\sum_{i\geq1}p_{i}t^{i}=t\cdot\underbrace{H^{\prime
}\left(  t\right)  }_{=\sum_{i\geq1}ih_{i}t^{i-1}}=t\cdot\sum_{i\geq1}%
ih_{i}t^{i-1}=\sum_{i\geq1}ih_{i}\underbrace{tt^{i-1}}_{=t^{i}}=\sum_{i\geq
1}ih_{i}t^{i}. \label{sol.Lambda.verschiebung.b.4}%
\end{equation}
Substituting $t^{n}$ for $t$ in this equality, we obtain%
\begin{equation}
H\left(  t^{n}\right)  \cdot\sum_{i\geq1}p_{i}\left(  t^{n}\right)  ^{i}%
=\sum_{i\geq0}ih_{i}\left(  t^{n}\right)  ^{i}.
\label{sol.Lambda.verschiebung.b.5}%
\end{equation}


Applying the map $\mathbf{v}_{n}\left[  \left[  t\right]  \right]  $ to both
sides of the equality (\ref{sol.Lambda.verschiebung.b.4}), we obtain
\begin{align*}
&  \left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \cdot\sum_{i\geq1}p_{i}t^{i}\right) \\
&  =\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\geq1}ih_{i}t^{i}\right)  =\sum_{i\geq1}i\underbrace{\left(
\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
h_{i}\right)  }_{\substack{=\mathbf{v}_{n}\left(  h_{i}\right)  \\\text{(by
the definition of }\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{)}%
}}\underbrace{\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]
\right)  \left(  t^{i}\right)  }_{\substack{=t^{i}\\\text{(by the definition
of }\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{)}}%
} \\
& \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{ is a
continuous }\kk\text{-algebra homomorphism}
\right) \\
&  =\sum_{i\geq1}i\underbrace{\mathbf{v}_{n}\left(  h_{i}\right)
}_{\substack{=
\begin{cases}
h_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases}
\\\text{(by the definition of }\mathbf{v}_{n}\text{)}}}t^{i}\\
&  =\sum_{i\geq1}i
\begin{cases}
h_{i/n},&\text{if }n\mid i;\\
0,&\text{if }n\nmid i
\end{cases} \cdot
t^{i}=\sum\limits_{\substack{i\geq1;\\n\mid i}}i\underbrace{ \begin{cases}
h_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} }_{\substack{=h_{i/n}\\\text{(since }n\mid i\text{)}}}
\cdot t^{i}%
+\sum\limits_{\substack{i\geq1;\\n\nmid i}}i\underbrace{ \begin{cases}
h_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} }_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\\
&  =\sum\limits_{\substack{i\geq1;\\n\mid i}}ih_{i/n}t^{i}+\underbrace{\sum\limits
_{\substack{i\geq1;\\n\nmid i}}0t^{i}}_{=0}=\sum\limits_{\substack{i\geq1;\\n\mid
i}}ih_{i/n}t^{i}=\sum\limits_{\substack{i\geq0;\\n\mid i}}ih_{i/n}t^{i} \\
& \ \ \ \ \ \ \ \ \ \ \left(\text{since }
\sum\limits_{\substack{i\geq0;\\n\mid i}}ih_{i/n}t^{i}
= \sum\limits_{\substack{i\geq1;\\n\mid i}}ih_{i/n}t^{i} + \underbrace{0h_{0/n}t^0}_{=0}
= \sum\limits_{\substack{i\geq1;\\n\mid i}}ih_{i/n}t^{i} \right)
\\
&= \sum_{i\geq0}nih_{i}\underbrace{t^{ni}}_{=\left(
t^{n}\right)  ^{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted
}ni\text{ for }i\text{ in the sum}\right) \\
&  =\sum_{i\geq0}nih_{i}\left(  t^{n}\right)  ^{i}=n\underbrace{\sum_{i\geq
0}ih_{i}\left(  t^{n}\right)  ^{i}}_{\substack{=H\left(  t^{n}\right)
\cdot\sum_{i\geq1}p_{i}\left(  t^{n}\right)  ^{i}\\\text{(by
\eqref{sol.Lambda.verschiebung.b.5})}}}=n\cdot H\left(  t^{n}\right)
\cdot\sum_{i\geq1}p_{i}\left(  t^{n}\right)  ^{i}=H\left(  t^{n}\right)  \cdot
n\cdot\sum_{i\geq1}p_{i}\left(  t^{n}\right)  ^{i}.
\end{align*}
Compared with%
\begin{align*}
&  \left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
H\left(  t\right)  \cdot\sum_{i\geq1}p_{i}t^{i}\right) \\
&  =\underbrace{\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]
\right)  \left(  H\left(  t\right)  \right)  }_{\substack{=H\left(
t^{n}\right)  \\\text{(by (\ref{sol.Lambda.verschiebung.a.H}))}}}\cdot\left(
\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)  \left(  \sum
_{i\geq1}p_{i}t^{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{ is a } \kk %
\text{-algebra homomorphism}\right) \\
&  =H\left(  t^{n}\right)  \cdot\left(  \mathbf{v}_{n}\left[  \left[
t\right]  \right]  \right)  \left(  \sum_{i\geq1}p_{i}t^{i}\right)  ,
\end{align*}
this becomes%
\[
H\left(  t^{n}\right)  \cdot n\cdot\sum_{i\geq1}p_{i}\left(  t^{n}\right)
^{i}=H\left(  t^{n}\right)  \cdot\left(  \mathbf{v}_{n}\left[  \left[
t\right]  \right]  \right)  \left(  \sum_{i\geq1}p_{i}t^{i}\right)  .
\]
We can divide both sides of this equality by $H\left(  t^{n}\right)  $ (since
$H\left(  t^{n}\right)  $ is invertible\footnote{\textit{Proof.} We know that
the power series $H\left(  t\right)  $ is invertible. That is, there exists a
power series $A\left(  t\right)  \in\Lambda\left[  \left[  t\right]  \right]
$ such that $A\left(  t\right)  \cdot H\left(  t\right)  =1$. Consider this
$A$. Substituting $t^{n}$ for $t$ in the equality $A\left(  t\right)  \cdot
H\left(  t\right)  =1$, we obtain $A\left(  t^{n}\right)  \cdot H\left(
t^{n}\right)  =1$. Hence, the power series $H\left(  t^{n}\right)  $ is
invertible (and its inverse is $A\left(  t^{n}\right)  $), qed.}). As a
result, we obtain%
\begin{align}
n\cdot\sum_{i\geq1}p_{i}\left(  t^{n}\right)  ^{i}  &  =\left(  \mathbf{v}%
_{n}\left[  \left[  t\right]  \right]  \right)  \left(  \sum_{i\geq1}%
p_{i}t^{i}\right)  =\sum_{i\geq1}\underbrace{\left(  \mathbf{v}_{n}\left[
\left[  t\right]  \right]  \right)  \left(  p_{i}\right)  }%
_{\substack{=\mathbf{v}_{n}\left(  p_{i}\right)  \\\text{(by the definition of
}\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{)}}%
}\underbrace{\left(  \mathbf{v}_{n}\left[  \left[  t\right]  \right]  \right)
\left(  t^{i}\right)  }_{\substack{=t^{i}\\\text{(by the definition of
}\mathbf{v}_{n}\left[  \left[  t\right]  \right]  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{v}_{n}\left[  \left[
t\right]  \right]  \text{ is a continuous } \kk \text{-algebra
homomorphism}\right) \nonumber\\
&  =\sum_{i\geq1}\mathbf{v}_{n}\left(  p_{i}\right)  t^{i}.
\label{sol.Lambda.verschiebung.b.P}%
\end{align}
On the other hand,%
\begin{align}
&  \sum_{i\geq1} \begin{cases}
np_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} \cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq1;\\n\mid i}}\underbrace{ \begin{cases}
np_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}  }_{\substack{=np_{i/n}\\\text{(since }n\mid i\text{)}}}
\cdot t^{i}%
+\sum\limits_{\substack{i\geq1;\\n\nmid i}}\underbrace{ \begin{cases}
np_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}  }_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq1;\\n\mid i}}np_{i/n}t^{i}+\underbrace{\sum\limits
_{\substack{i\geq1;\\n\nmid i}}0t^{i}}_{=0}=\sum\limits_{\substack{i\geq1;\\n\mid
i}}np_{i/n}t^{i}\nonumber\\
&  =\sum_{i\geq1}np_{i}\underbrace{t^{ni}}_{=\left(
t^{n}\right)  ^{i}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted
}ni\text{ for }i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\geq1}np_{i}\left(  t^{n}\right)  ^{i}=n\cdot\sum_{i\geq1}%
p_{i}\left(  t^{n}\right)  ^{i}\nonumber\\
&  =\sum_{i\geq1}\mathbf{v}_{n}\left(
p_{i}\right)  t^{i}\qquad\qquad\left(  \text{by
\eqref{sol.Lambda.verschiebung.b.P}}\right)  .
\label{sol.Lambda.verschiebung.b.almost}%
\end{align}
Now, let $m$ be a positive integer. Comparing coefficients before $t^{m}$ in
the equality (\ref{sol.Lambda.verschiebung.b.almost}), we obtain%
\[
\begin{cases}
np_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad
=\mathbf{v}_{n}\left(  p_{m}\right)  .
\]
This solves Exercise \ref{exe.Lambda.verschiebung}(a).

(c) Fix two positive integers $n$ and $m$. We need to prove that
$\mathbf{v}_{n}\circ\mathbf{v}_{m}=\mathbf{v}_{nm}$. But both $\mathbf{v}%
_{n}\circ\mathbf{v}_{m}$ and $\mathbf{v}_{nm}$ are $\kk$-algebra
homomorphisms (since $\mathbf{v}_{n}$, $\mathbf{v}_{m}$ and $\mathbf{v}_{nm}$
are $\kk$-algebra homomorphisms).

Let us now show that%
\begin{equation}
\left(  \mathbf{v}_{n}\circ\mathbf{v}_{m}\right)  \left(  h_{r}\right)
=\mathbf{v}_{nm}\left(  h_{r}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }r\text{.}
\label{sol.Lambda.verschiebung.c.main}
\end{equation}


\textit{Proof of (\ref{sol.Lambda.verschiebung.c.main}):} Let $r$ be a
positive integer. If $m\nmid r$, then (\ref{sol.Lambda.verschiebung.c.main})
holds\footnote{\textit{Proof.} Assume that $m\nmid r$. Then, $nm\nmid r$
(because otherwise, we would have $nm\mid r$, and therefore $m\mid nm\mid r$,
which would contradict $m\nmid r$).
\par
By the definition of $\mathbf{v}_{m}$, we have $\mathbf{v}_{m}\left(
h_{r}\right)  = \begin{cases}
h_{r/m},& \text{if }m\mid r;\\
0,& \text{if }m\nmid r
\end{cases}  \quad =0$ (since $m\nmid r$).
By the definition of $\mathbf{v}_{nm}$, we
have $\mathbf{v}_{nm}\left(  h_{r}\right)  = \begin{cases}
h_{r/\left(  nm\right)  },& \text{if }nm\mid r;\\
0,& \text{if }nm\nmid r
\end{cases}  \quad =0$ (since $nm\nmid r$).
Now, $\left(  \mathbf{v}_{n}\circ
\mathbf{v}_{m}\right)  \left(  h_{r}\right)  =\mathbf{v}_{n}\left(
\underbrace{\mathbf{v}_{m}\left(  h_{r}\right)  }_{=0}\right)  =\mathbf{v}%
_{n}\left(  0\right)  =0$ (since $\mathbf{v}_{n}$ is $\kk$-linear).
Compared with $\mathbf{v}_{nm}\left(  h_{r}\right)  =0$, this yields $\left(
\mathbf{v}_{n}\circ\mathbf{v}_{m}\right)  \left(  h_{r}\right)  =\mathbf{v}%
_{nm}\left(  h_{r}\right)  $. Hence, (\ref{sol.Lambda.verschiebung.c.main})
holds is proven under the assumption that $m\nmid r$. Qed.}. Thus, for the
rest of this proof, we can WLOG assume that we don't have $m\nmid r$. Assume this.

We have $m\mid r$ (since we don't have $m\nmid r$). Thus, $r/m$ is a positive
integer. By the definition of $\mathbf{v}_{m}$, we have $\mathbf{v}_{m}\left(
h_{r}\right)  = \begin{cases}
h_{r/m},& \text{if }m\mid r;\\
0,& \text{if }m\nmid r
\end{cases}  \quad  =h_{r/m}$ (since $m\mid r$). Now,
\begin{align*}
\left(  \mathbf{v}_{n}\circ\mathbf{v}_{m}\right)  \left(  h_{r}\right)   &
=\mathbf{v}_{n}\left(  \underbrace{\mathbf{v}_{m}\left(  h_{r}\right)
}_{=h_{r/m}}\right)  =\mathbf{v}_{n}\left(  h_{r/m}\right) \\
&=
\begin{cases}
h_{\left(  r/m\right)  /n},& \text{if }n\mid r/m;\\
0,& \text{if }n\nmid r/m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of }
\mathbf{v}_n\right) \\
&=
\begin{cases}
h_{\left(  r/m\right)  /n},& \text{if }nm\mid r;\\
0,& \text{if }nm\nmid r
\end{cases} \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the condition }n\mid r/m\text{ is equivalent to the condition
}nm\mid r\text{,}\\
\text{and since the condition }n\nmid r/m\text{ is equivalent to the condition
}nm\nmid r
\end{array}
\right) \\
&  = \begin{cases}
h_{r/\left(  nm\right)  },& \text{if }nm\mid r;\\
0,& \text{if }nm\nmid r
\end{cases}   \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  r/m\right)
/n=r/\left(  nm\right)  \right)  .
\end{align*}
Compared with%
\[
\mathbf{v}_{nm}\left(  h_{r}\right)  = \begin{cases}
h_{r/\left(  nm\right)  },& \text{if }nm\mid r;\\
0,& \text{if }nm\nmid r
\end{cases}   \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{v}_{nm}\right)  ,
\]
this yields $\left(  \mathbf{v}_{n}\circ\mathbf{v}_{m}\right)  \left(
h_{r}\right)  =\mathbf{v}_{nm}\left(  h_{r}\right)  $. This proves
(\ref{sol.Lambda.verschiebung.c.main}).

Now, recall that the family $\left(  h_{r}\right)  _{r\geq1}$ generates the
$\kk$-algebra $\Lambda$ (according to Proposition
\ref{symm-antipode-and-h-basis}). In other words, $\left(  h_{r}\right)
_{r\geq1}$ is a generating set of the $\kk$-algebra $\Lambda$. The two
$\kk$-algebra homomorphisms $\mathbf{v}_{n}\circ\mathbf{v}_{m}$ and
$\mathbf{v}_{nm}$ are equal to each other on this generating set (according to
(\ref{sol.Lambda.verschiebung.c.main})), and therefore must be identical
(because if two $\kk$-algebra homomorphisms from the same domain are
equal to each other on a generating set of their domain, then these two
homomorphisms must be identical). In other words, $\mathbf{v}_{n}%
\circ\mathbf{v}_{m}=\mathbf{v}_{nm}$. This solves Exercise
\ref{exe.Lambda.verschiebung}(c).

(d) For every positive integer $r$, we have
\begin{align}
\mathbf{v}_1\left(h_r\right)
&= \begin{cases}
h_{r/1},& \text{if }1\mid r;\\
0,& \text{if }1\nmid r
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of } \mathbf{v}_1\right) \nonumber \\
&= h_{r/1} \ \ \ \ \ \ \ \ \ \ \left(\text{since } 1 \mid r\right) \nonumber \\
&= h_r = \id\left(h_r\right).
\label{sol.Lambda.verschiebung.newd.main}
\end{align}

Now, recall that the family $\left( h_r \right) _{r \geq 1}$ generates the
$\kk$-algebra $\Lambda$ (according to
Proposition~\ref{symm-antipode-and-h-basis}). In other words,
$\left( h_r \right) _{r \geq 1}$ is a generating set of the $\kk$-algebra
$\Lambda$. The two
$\kk$-algebra homomorphisms $\mathbf{v}_1$ and
$\id$ are equal to each other on this generating set (according to
(\ref{sol.Lambda.verschiebung.newd.main})), and therefore must be identical
(because if two $\kk$-algebra homomorphisms from the same domain are
equal to each other on a generating set of their domain, then these two
homomorphisms must be identical). In other words, $\mathbf{v}_{1}
= \id$. This solves Exercise~\ref{exe.Lambda.verschiebung}(d).

(e) Fix a positive integer $n$. We need to show that $\mathbf{v}_{n}%
:\Lambda\rightarrow\Lambda$ is a Hopf algebra homomorphism.

We know that $\mathbf{v}_{n}:\Lambda\rightarrow\Lambda$ is a $\kk %
$-algebra homomorphism. Thus, $\mathbf{v}_{n}\otimes\mathbf{v}_{n}%
:\Lambda\otimes\Lambda\rightarrow\Lambda\otimes\Lambda$ is a $\kk %
$-algebra homomorphism. Also, $\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$
is a $\kk$-algebra homomorphism (due to the axioms of a bialgebra,
since $\Lambda$ is a bialgebra). Hence, $\Delta\circ\mathbf{v}_{n}$ and
$\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ\Delta$ are
$\kk$-algebra homomorphisms.

For every $q\in \NN$, we have%
\begin{equation}
\Delta\left(  h_{q}\right)  =\sum_{i\in\left\{  0,1,\ldots,q\right\}  }%
h_{i}\otimes h_{q-i}. \label{sol.Lambda.verschiebung.d.Delh}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.d.Delh}):} Let
$q\in \NN$. Then, Proposition \ref{symm-comultiplication-formulas-prop}%
(iii) (applied to $q$ instead of $n$) yields%
\[
\Delta\left(  h_{q}\right)  =\sum_{i+j=q}h_{i}\otimes h_{j}=\sum_{i\in\left\{
0,1,\ldots,q\right\}  }h_{i}\otimes h_{q-i}%
\]
(here, we have substituted $\left(  i,q-i\right)  $ for $\left(  i,j\right)  $
in the sum), qed.}. Furthermore, for every $q\in \NN$, we have
\begin{equation}
\mathbf{v}_{n}\left(  h_{nq}\right)  =h_{q}
\label{sol.Lambda.verschiebung.d.vhnq}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.d.vhnq}):} Let
$q\in \NN$. Applying (\ref{sol.Lambda.verschiebung.vnhm}) to $nq$
instead of $m$, we obtain
\begin{align*}
\mathbf{v}_{n}\left(  h_{nq}\right)   &  =
\begin{cases}
h_{\left(  nq\right)  /n},& \text{if }n\mid nq;\\
0,& \text{if }n\nmid nq
\end{cases}  \quad
=h_{\left(  nq\right)  /n}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\mid nq\right) \\
&  =h_{q},
\end{align*}
qed.}.

Now, we shall prove that%
\begin{equation}
\left(  \Delta\circ\mathbf{v}_{n}\right)  \left(  h_{r}\right)  =\left(
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ\Delta\right)
\left(  h_{r}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every positive integer
}r\text{.}
\label{sol.Lambda.verschiebung.d.main}
\end{equation}


\textit{Proof of (\ref{sol.Lambda.verschiebung.d.main}):} Let $r$ be a
positive integer. We need to prove (\ref{sol.Lambda.verschiebung.d.main}).

Let us first notice that%
\begin{equation}
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes
h_{r-i}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
0,1,\ldots,r\right\}  \text{ which does not satisfy }\left(  n\mid i\text{ and
}n\mid r-i\right)  \label{sol.Lambda.verschiebung.d.main.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.d.main.pf.1}):} Let
$i\in\left\{  0,1,\ldots,r\right\}  $ be such that we don't have $\left(  n\mid
i\text{ and }n\mid r-i\right)  $. If $n\nmid i$, then
\begin{align*}
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes
h_{r-i}\right)   &  =\underbrace{\mathbf{v}_{n}\left(  h_{i}\right)
}_{\substack{= \begin{cases}
h_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} \\\text{(by (\ref{sol.Lambda.verschiebung.vnhm}), applied to
}m=i\text{)}}}\otimes\mathbf{v}_{n}\left(  h_{r-i}\right)
=\underbrace{ \begin{cases}
h_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} }_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\quad \otimes \mathbf{v}_{n}\left(  h_{r-i}\right) \\
&  =0\otimes\mathbf{v}_{n}\left(  h_{r-i}\right)  =0.
\end{align*}
Hence, if $n\nmid i$, then (\ref{sol.Lambda.verschiebung.d.main.pf.1}) is
proven. Thus, for the rest of this proof of
(\ref{sol.Lambda.verschiebung.d.main.pf.1}), we can WLOG assume that we don't
have $n\nmid i$. Assume this.
\par
We have $n\mid i$ (since we don't have $n\nmid i$). Hence, we don't have
$n\mid r-i$ (because otherwise, we would have $\left(  n\mid i\text{ and
}n\mid r-i\right)  $, which would contradict the fact that we don't have
$\left(  n\mid i\text{ and }n\mid r-i\right)  $). In other words, we have
$n\nmid r-i$. Now,%
\begin{align*}
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes
h_{r-i}\right)   &  =\mathbf{v}_{n}\left(  h_{i}\right)  \otimes
\underbrace{\mathbf{v}_{n}\left(  h_{r-i}\right)  }_{\substack{=
\begin{cases}
h_{\left(  r-i\right)  /n},& \text{if }n\mid r-i;\\
0,& \text{if }n\nmid r-i
\end{cases}
\\\text{(by (\ref{sol.Lambda.verschiebung.vnhm}), applied to
}m=r-i\text{)}}}=\mathbf{v}_{n}\left(  h_{i}\right)  \otimes
\underbrace{
\begin{cases}
h_{\left(  r-i\right)  /n},& \text{if }n\mid r-i;\\
0,& \text{if }n\nmid r-i
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid r-i\text{)}}}\\
&  =\mathbf{v}_{n}\left(  h_{i}\right)  \otimes0=0.
\end{align*}
This proves (\ref{sol.Lambda.verschiebung.d.main.pf.1}).}. In particular,%
\begin{equation}
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes
h_{r-i}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
0,1,\ldots,r\right\}  \text{ satisfying }n\nmid i
\label{sol.Lambda.verschiebung.d.main.pf.1a}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.d.main.pf.1a}):} Let
$i\in\left\{  0,1,\ldots,r\right\}  $ be such that $n\nmid i$. Then, we don't
have $n\mid i$. Hence, we don't have $\left(  n\mid i\text{ and }n\mid
r-i\right)  $. Thus, (\ref{sol.Lambda.verschiebung.d.main.pf.1a}) follows from
(\ref{sol.Lambda.verschiebung.d.main.pf.1}), qed.}.

Let us now assume that $n\nmid r$. Then, (\ref{sol.Lambda.verschiebung.vnhm})
(applied to $m=r$) yields $\mathbf{v}_{n}\left(  h_{r}\right)  =
\begin{cases}
h_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}  \quad =0$ (since $n\nmid r$), so that
$\left(  \Delta\circ\mathbf{v}%
_{n}\right)  \left(  h_{r}\right)  =\Delta\left(  \underbrace{\mathbf{v}%
_{n}\left(  h_{r}\right)  }_{=0}\right)  =\Delta\left(  0\right)  =0$. Also,%
\begin{equation}
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes
h_{r-i}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
0,1,\ldots,r\right\}  \label{sol.Lambda.verschiebung.d.main.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.d.main.pf.2}):} Let
$i\in\left\{  0,1,\ldots,r\right\}  $.
\par
Assume (for the sake of contradiction) that $\left(  n\mid i\text{ and }n\mid
r-i\right)  $. Then, $i\equiv0\operatorname{mod}n$ (since $n\mid i$) and
$r\equiv i\operatorname{mod}n$ (since $n\mid r-i$). Hence, $r\equiv
i\equiv0\operatorname{mod}n$, so that $n\mid r$. This contradicts $n\nmid r$.
This contradiction shows that our assumption (that $\left(  n\mid i\text{ and
}n\mid r-i\right)  $) was wrong. Hence, we do not have $\left(  n\mid i\text{
and }n\mid r-i\right)  $. Thus, (\ref{sol.Lambda.verschiebung.d.main.pf.2})
follows from (\ref{sol.Lambda.verschiebung.d.main.pf.1}), qed.}. Now,
\begin{align*}
\left(  \left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ
\Delta\right)  \left(  h_{r}\right)   &  =\left(  \mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\right)  \left(  \underbrace{\Delta\left(  h_{r}\right)
}_{\substack{=\sum_{i\in\left\{  0,1,\ldots,r\right\}  }h_{i}\otimes
h_{r-i}\\\text{(by (\ref{sol.Lambda.verschiebung.d.Delh}), applied to
}q=r\text{)}}}\right)  =\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)
\left(  \sum_{i\in\left\{  0,1,\ldots,r\right\}  }h_{i}\otimes h_{r-i}\right) \\
&  =\sum_{i\in\left\{  0,1,\ldots,r\right\}  }\underbrace{\left(  \mathbf{v}%
_{n}\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes h_{r-i}\right)
}_{\substack{=0\\\text{(by (\ref{sol.Lambda.verschiebung.d.main.pf.2}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\text{ is } \kk \text{-linear}\right) \\
&  =\sum_{i\in\left\{  0,1,\ldots,r\right\}  }0=0.
\end{align*}
Compared to $\left(  \Delta\circ\mathbf{v}_{n}\right)  \left(  h_{r}\right)
=0$, this yields $\left(  \Delta\circ\mathbf{v}_{n}\right)  \left(
h_{r}\right)  =\left(  \left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)
\circ\Delta\right)  \left(  h_{r}\right)  $. In other words,
(\ref{sol.Lambda.verschiebung.d.main}) holds.

Now, let us forget that we assumed that $n\nmid r$. We thus have proven that
(\ref{sol.Lambda.verschiebung.d.main}) holds under the assumption that $n\nmid
r$. Hence, for the rest of the proof of (\ref{sol.Lambda.verschiebung.d.main}%
), we can WLOG assume that we don't have $n\nmid r$. Assume this.

We have $n\mid r$ (since we don't have $n\nmid r$). Hence, $r/n$ is a positive
integer. Denote this positive integer $r/n$ by $s$. Then, $s=r/n$, so that
$r=ns$. The equality (\ref{sol.Lambda.verschiebung.vnhm}) (applied to $m=r$)
yields $\mathbf{v}_{n}\left(  h_{r}\right)  =
\begin{cases}
h_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}  \quad
=h_{r/n}$ (since $n\mid r$), and we have%
\begin{equation}
\left(  \Delta\circ\mathbf{v}_{n}\right)  \left(  h_{r}\right)  =\Delta\left(
\underbrace{\mathbf{v}_{n}\left(  h_{r}\right)  }_{\substack{=h_{r/n}%
=h_{s}\\\text{(since }r/n=s\text{)}}}\right)  =\Delta\left(  h_{s}\right)
=\sum_{i\in\left\{  0,1,\ldots,s\right\}  }h_{i}\otimes h_{s-i}
\label{sol.Lambda.verschiebung.d.main.pf.4}%
\end{equation}
(by (\ref{sol.Lambda.verschiebung.d.Delh}), applied to $q=s$).

On the other hand,%
\begin{align*}
\left(  \left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ
\Delta\right)  \left(  h_{r}\right)   &  =\left(  \mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\right)  \left(  \underbrace{\Delta\left(  h_{r}\right)
}_{\substack{=\sum_{i\in\left\{  0,1,\ldots,r\right\}  }h_{i}\otimes
h_{r-i}\\\text{(by (\ref{sol.Lambda.verschiebung.d.Delh}), applied to
}q=r\text{)}}}\right)  =\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)
\left(  \sum_{i\in\left\{  0,1,\ldots,r\right\}  }h_{i}\otimes h_{r-i}\right) \\
&  =\sum_{i\in\left\{  0,1,\ldots,r\right\}  }\left(  \mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes h_{r-i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathbf{v}_{n}\otimes
\mathbf{v}_{n}\text{ is } \kk \text{-linear}\right) \\
&  =\sum\limits_{\substack{i\in\left\{  0,1,\ldots,r\right\}  ;\\n\mid i}%
}\underbrace{\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \left(
h_{i}\otimes h_{r-i}\right)  }_{=\mathbf{v}_{n}\left(  h_{i}\right)
\otimes\mathbf{v}_{n}\left(  h_{r-i}\right)  }+\sum\limits_{\substack{i\in\left\{
0,1,\ldots,r\right\}  ;\\n\nmid i}}\underbrace{\left(  \mathbf{v}_{n}%
\otimes\mathbf{v}_{n}\right)  \left(  h_{i}\otimes h_{r-i}\right)
}_{\substack{=0\\\text{(by (\ref{sol.Lambda.verschiebung.d.main.pf.1a}))}}}\\
&  =\sum\limits_{\substack{i\in\left\{  0,1,\ldots,r\right\}  ;\\n\mid i}}\mathbf{v}%
_{n}\left(  h_{i}\right)  \otimes\mathbf{v}_{n}\left(  h_{r-i}\right)
+\underbrace{\sum\limits_{\substack{i\in\left\{  0,1,\ldots,r\right\}  ;\\n\nmid i}%
}0}_{=0}=\sum\limits_{\substack{i\in\left\{  0,1,\ldots,r\right\}  ;\\n\mid
i}}\mathbf{v}_{n}\left(  h_{i}\right)  \otimes\mathbf{v}_{n}\left(
h_{r-i}\right) \\
&  =\underbrace{\sum_{i\in\left\{  0,1,\ldots,r/n\right\}  }}_{\substack{=\sum
_{i\in\left\{  0,1,\ldots,s\right\}  }\\\text{(since }r/n=s\text{)}}%
}\mathbf{v}_{n}\left(  h_{ni}\right)  \otimes\mathbf{v}_{n}\left(
\underbrace{h_{r-ni}}_{\substack{=h_{n\left(  s-i\right)  }\\\text{(since
}r-ni=n\left(  s-i\right)  \\\text{(because }r=ns=ni+n\left(  s-i\right)
\text{))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }ni\text{ for
}i\text{ in the sum}\right) \\
&  =\sum_{i\in\left\{  0,1,\ldots,s\right\}  }\underbrace{\mathbf{v}_{n}\left(
h_{ni}\right)  }_{\substack{=h_{i}\\\text{(by
(\ref{sol.Lambda.verschiebung.d.vhnq}), applied}\\\text{to }q=i\text{)}%
}}\otimes\underbrace{\mathbf{v}_{n}\left(  h_{n\left(  s-i\right)  }\right)
}_{\substack{=h_{s-i}\\\text{(by (\ref{sol.Lambda.verschiebung.d.vhnq}),
applied}\\\text{to }q=s-i\text{)}}}=\sum_{i\in\left\{  0,1,\ldots,s\right\}
}h_{i}\otimes h_{s-i}.
\end{align*}
Compared with (\ref{sol.Lambda.verschiebung.d.main.pf.4}), this yields
$\left(  \Delta\circ\mathbf{v}_{n}\right)  \left(  h_{r}\right)  =\left(
\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ\Delta\right)
\left(  h_{r}\right)  $. Thus, (\ref{sol.Lambda.verschiebung.d.main}) is proven.

Now, recall that the family $\left(  h_{r}\right)  _{r\geq1}$ generates the
$\kk$-algebra $\Lambda$ (according to Proposition
\ref{symm-antipode-and-h-basis}). In other words, $\left(  h_{r}\right)
_{r\geq1}$ is a generating set of the $\kk$-algebra $\Lambda$. The two
$\kk$-algebra homomorphisms $\Delta\circ\mathbf{v}_{n}$ and $\left(
\mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ\Delta$ are equal to each
other on this generating set (according to
(\ref{sol.Lambda.verschiebung.d.main})), and therefore must be identical
(because if two $\kk$-algebra homomorphisms from the same domain are
equal to each other on a generating set of their domain, then these two
homomorphisms must be identical). In other words, $\Delta\circ\mathbf{v}%
_{n}=\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)  \circ\Delta$.

One can similarly check that $\epsilon\circ\mathbf{v}_{n}=\epsilon$. We can
now conclude that the map $\mathbf{v}_{n}$ is a $\kk$-coalgebra
homomorphism (since it is $\kk$-linear and satisfies $\Delta
\circ\mathbf{v}_{n}=\left(  \mathbf{v}_{n}\otimes\mathbf{v}_{n}\right)
\circ\Delta$ and $\epsilon\circ\mathbf{v}_{n}=\epsilon$), therefore a
$\kk$-bialgebra homomorphism (since it also is a $\kk$-algebra
homomorphism), and therefore a Hopf algebra homomorphism (by
Corollary~\ref{cor.bialg-mor-is-Hopf}). This solves Exercise
\ref{exe.Lambda.verschiebung}(e).

(f) Recall first that $\left( h_{\lambda} \right) _{\lambda \in \Par}$
and $\left( m_{\lambda} \right) _{\lambda \in \Par}$ are mutually dual
bases with respect to the Hall inner product on $\Lambda$ (according
to Corollary \ref{dual-bases-h-m-p}(a)). Thus,
\begin{equation}
\left(  m_{\lambda},h_{\mu}\right)  =\delta_{\lambda,\mu}
\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }\lambda\text{ and }\mu.
\label{sol.Lambda.verschiebung.e.duality}
\end{equation}


Let us introduce a notation: For every weak composition $\lambda$ and every
positive integer $s$, let $\lambda\left\{  s\right\}  $ denote the weak
composition $\left(  s\lambda_{1},s\lambda_{2},s\lambda_{3},\ldots\right)  $,
where $\lambda$ is written in the form $\left(  \lambda_{1},\lambda
_{2},\lambda_{3},\ldots\right)  $. Notice that if $\lambda$ is a partition and
$s$ is a positive integer, then $\lambda\left\{  s\right\}  $ is a partition
as well, and satisfies%
\begin{equation}
\ell\left(  \lambda\left\{  s\right\}  \right)  =\ell\left(  \lambda\right)  .
\label{sol.Lambda.verschiebung.e.llambdas}%
\end{equation}


We have%
\begin{equation}
\mathbf{f}_{n}m_{\lambda}=m_{\lambda\left\{  n\right\}  }%
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda
\label{sol.Lambda.verschiebung.e.f}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.e.f}):} Let $\lambda$
be a partition. Write $\lambda$ in the form $\lambda=\left(  \lambda
_{1},\lambda_{2},\lambda_{3},\ldots\right)  $. By the definition of
$m_{\lambda}$, we have $m_{\lambda}=\sum_{\alpha\in\Symm_{\left(
\infty\right)  }\lambda}{\mathbf{x}}^{\alpha}$. But the definition of
$\mathbf{f}_{n}$ yields%
\begin{equation}
\mathbf{f}_{n}m_{\lambda}
= \underbrace{m_{\lambda}}_{= \sum_{\alpha\in\Symm_{\left(
\infty\right)  }\lambda}{\mathbf{x}}^{\alpha}}
\left(  x_{1}^{n},x_{2}^{n},x_{3}%
^{n},\ldots\right)  =\left(  \sum_{\alpha\in\Symm_{\left(
\infty\right)  }\lambda}{\mathbf{x}}^{\alpha}\right)  \left(  x_{1}^{n}%
,x_{2}^{n},x_{3}^{n},\ldots\right)  =\sum_{\alpha\in\Symm_{\left(
\infty\right)  }\lambda}{\mathbf{x}}^{\alpha}\left(  x_{1}^{n},x_{2}^{n}%
,x_{3}^{n},\ldots\right)  . \label{sol.Lambda.verschiebung.e.f.pf.1}%
\end{equation}
\par
But every weak composition $\alpha$ satisfies%
\begin{equation}
{\mathbf{x}}^{\alpha}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)
=\mathbf{x}^{\alpha\left\{  n\right\}  }.
\label{sol.Lambda.verschiebung.e.f.pf.2}%
\end{equation}
\par
(\textit{Proof of (\ref{sol.Lambda.verschiebung.e.f.pf.2}):} Let $\alpha$ be a
weak composition. Write $\alpha$ as $\left(  \alpha_{1},\alpha_{2},\alpha
_{3},\ldots\right)  $. Then, $\alpha\left\{  n\right\}  =\left(  n\alpha
_{1},n\alpha_{2},n\alpha_{3},\ldots\right)  $ (by the definition of
$\alpha\left\{  n\right\}  $). Thus, $\mathbf{x}^{\alpha\left\{  n\right\}
}=x_{1}^{n\alpha_{1}}x_{2}^{n\alpha_{2}}x_{3}^{n\alpha_{3}}\cdots$ (by the
definition of $\mathbf{x}^{\alpha\left\{  n\right\}  }$). On the other hand,
$\mathbf{x}^{\alpha}=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}%
}\cdots$ (by the definition of $\mathbf{x}^{\alpha}$), so that
\[
\underbrace{{\mathbf{x}}^{\alpha}}_{=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}%
x_{3}^{\alpha_{3}}\cdots}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)
=\left(  x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}}\cdots\right)
\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)  =\left(  x_{1}%
^{n}\right)  ^{\alpha_{1}}\left(  x_{2}^{n}\right)  ^{\alpha_{2}}\left(
x_{3}^{n}\right)  ^{\alpha_{3}}\cdots=x_{1}^{n\alpha_{1}}x_{2}^{n\alpha_{2}%
}x_{3}^{n\alpha_{3}}\cdots=\mathbf{x}^{\alpha\left\{  n\right\}  }.
\]
Thus, (\ref{sol.Lambda.verschiebung.e.f.pf.2}) is proven.)
\par
Now, (\ref{sol.Lambda.verschiebung.e.f.pf.1}) becomes%
\begin{equation}
\mathbf{f}_{n}m_{\lambda}=\sum_{\alpha\in\Symm_{\left(  \infty\right)
}\lambda}\underbrace{{\mathbf{x}}^{\alpha}\left(  x_{1}^{n},x_{2}^{n}%
,x_{3}^{n},\ldots\right)  }_{\substack{=\mathbf{x}^{\alpha\left\{  n\right\}
}\\\text{(by (\ref{sol.Lambda.verschiebung.e.f.pf.2}))}}}=\sum_{\alpha
\in\Symm_{\left(  \infty\right)  }\lambda}\mathbf{x}^{\alpha\left\{
n\right\}  }. \label{sol.Lambda.verschiebung.e.f.pf.1a}%
\end{equation}
\par
On the other hand, for every weak composition $\alpha$ and every permutation
$\sigma\in\Symm_{\left(  \infty\right)  }$, we have%
\begin{equation}
\sigma\left(  \alpha\left\{  n\right\}  \right)  =\left(  \sigma\alpha\right)
\left\{  n\right\}  . \label{sol.Lambda.verschiebung.e.f.pf.3}%
\end{equation}
\par
(\textit{Proof of (\ref{sol.Lambda.verschiebung.e.f.pf.3}):} Let $\alpha$ be a
weak composition. Let $\sigma\in\Symm_{\left(  \infty\right)  }$. Write
$\alpha$ as $\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $. Then,
$\alpha\left\{  n\right\}  =\left(  n\alpha_{1},n\alpha_{2},n\alpha_{3}%
,\ldots\right)  $ (by the definition of $\alpha\left\{  n\right\}  $), so that
$\sigma\left(  \alpha\left\{  n\right\}  \right)  =\left(  n\alpha
_{\sigma^{-1}\left(  1\right)  },n\alpha_{\sigma^{-1}\left(  2\right)
},n\alpha_{\sigma^{-1}\left(  3\right)  },\ldots\right)  $. But $\sigma
\alpha=\left(  \alpha_{\sigma^{-1}\left(  1\right)  },\alpha_{\sigma
^{-1}\left(  2\right)  },\alpha_{\sigma^{-1}\left(  3\right)  },\ldots\right)
$ (since $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right)  $)
and thus $\left(  \sigma\alpha\right)  \left\{  n\right\}  =\left(
n\alpha_{\sigma^{-1}\left(  1\right)  },n\alpha_{\sigma^{-1}\left(  2\right)
},n\alpha_{\sigma^{-1}\left(  3\right)  },\ldots\right)  $ (by the definition of
$\left(  \sigma\alpha\right)  \left\{  n\right\}  $). Compared with
$\sigma\left(  \alpha\left\{  n\right\}  \right)  =\left(  n\alpha
_{\sigma^{-1}\left(  1\right)  },n\alpha_{\sigma^{-1}\left(  2\right)
},n\alpha_{\sigma^{-1}\left(  3\right)  },\ldots\right)  $, this yields
$\sigma\left(  \alpha\left\{  n\right\}  \right)  =\left(  \sigma
\alpha\right)  \left\{  n\right\}  $. This proves
(\ref{sol.Lambda.verschiebung.e.f.pf.3}).)
\par
Now,%
\begin{equation}
\alpha\left\{  n\right\}  \in\Symm_{\left(  \infty\right)  }\left(
\lambda\left\{  n\right\}  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}\alpha\in\Symm_{\left(  \infty\right)  }\lambda.
\label{sol.Lambda.verschiebung.e.f.pf.4}%
\end{equation}
\par
(\textit{Proof of (\ref{sol.Lambda.verschiebung.e.f.pf.4}):} Let $\alpha
\in\Symm_{\left(  \infty\right)  }\lambda$. Then, there exists a
$\sigma\in\Symm_{\left(  \infty\right)  }$ such that $\alpha
=\sigma\lambda$. Consider this $\sigma$. Then,
(\ref{sol.Lambda.verschiebung.e.f.pf.3}) (applied to $\lambda$ instead of
$\alpha$) yields that $\sigma\left(  \lambda\left\{  n\right\}  \right)
=\left(  \sigma\lambda\right)  \left\{  n\right\}  $. Compared with
$\underbrace{\alpha}_{=\sigma\lambda}\left\{  n\right\}  =\left(
\sigma\lambda\right)  \left\{  n\right\}  $, this yields $\alpha\left\{
n\right\}  =\underbrace{\sigma}_{\in\Symm_{\left(  \infty\right)  }%
}\left(  \lambda\left\{  n\right\}  \right)  \in\Symm_{\left(
\infty\right)  }\left(  \lambda\left\{  n\right\}  \right)  $. Thus,
(\ref{sol.Lambda.verschiebung.e.f.pf.4}) is proven.)
\par
Also,%
\begin{equation}
\text{every element of }\Symm_{\left(  \infty\right)  }\left(
\lambda\left\{  n\right\}  \right)  \text{ has the form }\alpha\left\{
n\right\}  \text{ for some }\alpha\in\Symm_{\left(  \infty\right)
}\lambda. \label{sol.Lambda.verschiebung.e.f.pf.5}%
\end{equation}
\par
(\textit{Proof of (\ref{sol.Lambda.verschiebung.e.f.pf.5}):} Let $\beta$ be an
element of $\Symm_{\left(  \infty\right)  }\left(  \lambda\left\{
n\right\}  \right)  $. Then, there exists a $\sigma\in\Symm_{\left(
\infty\right)  }$ such that $\beta=\sigma\left(  \lambda\left\{  n\right\}
\right)  $. Consider this $\sigma$. Now, $\underbrace{\sigma}_{\in
\Symm_{\left(  \infty\right)  }}\lambda\in\Symm_{\left(
\infty\right)  }\lambda$. Also, (\ref{sol.Lambda.verschiebung.e.f.pf.3})
(applied to $\lambda$ instead of $\alpha$) yields that $\sigma\left(
\lambda\left\{  n\right\}  \right)  =\left(  \sigma\lambda\right)  \left\{
n\right\}  $. Hence, $\beta=\sigma\left(  \lambda\left\{  n\right\}  \right)
=\left(  \sigma\lambda\right)  \left\{  n\right\}  $. Therefore, $\beta$ has
the form $\alpha\left\{  n\right\}  $ for some $\alpha\in\Symm_{\left(
\infty\right)  }\lambda$ (namely, for $\alpha=\sigma\lambda$). Let us now
forget that we fixed $\beta$. We thus have shown that every element $\beta$ of
$\Symm_{\left(  \infty\right)  }\left(  \lambda\left\{  n\right\}
\right)  $ has the form $\alpha\left\{  n\right\}  $ for some $\alpha
\in\Symm_{\left(  \infty\right)  }\lambda$. In other words, we have
proven (\ref{sol.Lambda.verschiebung.e.f.pf.5}).)
\par
Now, the map%
\begin{align*}
\Symm_{\left(  \infty\right)  }\lambda &  \rightarrow
\Symm_{\left(  \infty\right)  }\left(  \lambda\left\{  n\right\}  \right)  ,\\
\alpha &  \mapsto\alpha\left\{  n\right\}
\end{align*}
is well-defined (according to (\ref{sol.Lambda.verschiebung.e.f.pf.4})). This
map is injective (because any weak composition $\alpha$ can be uniquely
reconstructed from $\alpha\left\{  n\right\}  $) and surjective (due to
(\ref{sol.Lambda.verschiebung.e.f.pf.5})); therefore, this map is bijective.
\par
Now, (\ref{sol.Lambda.verschiebung.e.f.pf.1a}) becomes%
\[
\mathbf{f}_{n}m_{\lambda}=\sum_{\alpha\in\Symm_{\left(  \infty\right)
}\lambda}\mathbf{x}^{\alpha\left\{  n\right\}  }=\sum_{\alpha\in
\Symm_{\left(  \infty\right)  }\left(  \lambda\left\{  n\right\}
\right)  }\mathbf{x}^{\alpha}%
\]
(here, we substituted $\alpha$ for $\alpha\left\{  n\right\}  $ in the sum,
since the map
\begin{align*}
\Symm_{\left(  \infty\right)  }\lambda &  \rightarrow
\Symm_{\left(  \infty\right)  }\left(  \lambda\left\{  n\right\}  \right)  ,\\
\alpha &  \mapsto\alpha\left\{  n\right\}
\end{align*}
is bijective). Compared with%
\[
m_{\lambda\left\{  n\right\}  }=\sum_{\alpha\in\Symm_{\left(
\infty\right)  }\left(  \lambda\left\{  n\right\}  \right)  }{\xx%
}^{\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
m_{\lambda\left\{  n\right\}  }\right)  ,
\]
this yields $\mathbf{f}_{n}m_{\lambda}=m_{\lambda\left\{  n\right\}  }$. Thus,
(\ref{sol.Lambda.verschiebung.e.f}) is proven.}.

We need to prove that the maps $\mathbf{f}_{n}:\Lambda\rightarrow\Lambda$ and
$\mathbf{v}_{n}:\Lambda\rightarrow\Lambda$ are adjoint with respect to the
Hall inner product on $\Lambda$. In other words, we need to prove that%
\begin{equation}
\left(  \mathbf{f}_{n}a,b\right)  =\left(  a,\mathbf{v}_{n}b\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in\Lambda\text{ and }b\in\Lambda.
\label{sol.Lambda.verschiebung.e.goal}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.verschiebung.e.goal}):} Fix $a\in\Lambda$
and $b\in\Lambda$.

Recall that $\left(  m_{\lambda}\right)  _{\lambda\in \Par }$ is
a basis of the $\kk$-module $\Lambda$. Hence, in proving
(\ref{sol.Lambda.verschiebung.e.goal}), we can WLOG assume that $a$ is an
element of this basis $\left(  m_{\lambda}\right)  _{\lambda\in
 \Par }$ (because the equality
(\ref{sol.Lambda.verschiebung.e.goal}) is $\kk$-linear in $a$). Assume
this. Thus, $a$ is an element of the basis $\left(  m_{\lambda}\right)
_{\lambda\in \Par }$. In other words, there exists a $\mu
\in\Par$ such that $a=m_{\mu}$. Consider this $\mu$. We have
$\mathbf{f}_{n}\underbrace{a}_{=m_{\mu}}=\mathbf{f}_{n}m_{\mu}=m_{\mu\left\{
n\right\}  }$ (by (\ref{sol.Lambda.verschiebung.e.f}), applied to $\lambda
=\mu$).

Recall that $\left(  h_{\lambda}\right)  _{\lambda\in \Par }$ is
a basis of the $\kk$-module $\Lambda$. Hence, in proving
(\ref{sol.Lambda.verschiebung.e.goal}), we can WLOG assume that $b$ is an
element of this basis $\left(  h_{\lambda}\right)  _{\lambda\in
 \Par }$ (because the equality
(\ref{sol.Lambda.verschiebung.e.goal}) is $\kk$-linear in $b$). Assume
this. Thus, $b$ is an element of the basis $\left(  h_{\lambda}\right)
_{\lambda\in \Par }$. In other words, there exists a $\nu
\in\Par$ such that $b=h_{\nu}$. Consider this $\nu$. We have
\begin{equation}
\left(  \underbrace{\mathbf{f}_{n}a}_{=m_{\mu\left\{  n\right\}  }%
},\underbrace{b}_{=h_{\nu}}\right)  =\left(  m_{\mu\left\{  n\right\}
},h_{\nu}\right)  =\delta_{\mu\left\{  n\right\}  ,\nu}
\label{sol.Lambda.verschiebung.e.lhs}
\end{equation}
(by (\ref{sol.Lambda.verschiebung.e.duality}), applied to $\mu\left\{
n\right\}  $ and $\nu$ instead of $\lambda$ and $\mu$).

Let us write the partition $\nu$ in the form $\left(  \nu_{1},\nu_{2},\nu
_{3},\ldots\right)  $. Then, $\nu=\left(  \nu_{1},\nu_{2},\ldots,\nu
_{\ell\left(  \nu\right)  }\right)  $, so that $h_{\nu}=h_{\nu_{1}}h_{\nu_{2}%
}\cdots h_{\nu_{\ell\left(  \nu\right)  }}$ (by the definition of $h_{\nu}$).

Let us first assume that%
\begin{equation}
\left(  \text{there exists an }i\in\left\{  1,2,3,\ldots\right\}  \text{ such
that }n\nmid\nu_{i}\right)  .
\label{sol.Lambda.verschiebung.e.stupid-assumption}%
\end{equation}
Then, $\mathbf{v}_{n}\left(  h_{\nu}\right)  =0$%
\ \ \ \ \footnote{\textit{Proof.} There exists an $i\in\left\{
1,2,3,\ldots\right\}  $ such that $n\nmid\nu_{i}$. Consider this $i$. We have
$n\nmid\nu_{i}$, thus $\nu_{i}\neq0$, and therefore $\nu_{i}$ is a positive
integer. Hence, $i\leq\ell\left(  \nu\right)  $, so that $i\in\left\{
1,2,\ldots,\ell\left(  \nu\right)  \right\}  $. Also, the definition of
$\mathbf{v}_{n}$ yields $\mathbf{v}_{n}\left(  h_{\nu_{i}}\right)  =
\begin{cases}
h_{\nu_{i}/n},& \text{if }n\mid\nu_{i};\\
0,& \text{if }n\nmid\nu_{i}%
\end{cases}  \quad =0$ (since $n\nmid\nu_{i}$). But
\[
h_{\nu}=h_{\nu_{1}}h_{\nu_{2}}\cdots h_{\nu_{\ell\left(  \nu\right)  }%
}=\left(  h_{\nu_{1}}h_{\nu_{2}}\cdots h_{\nu_{i-1}}\right)  h_{\nu_{i}%
}\left(  h_{\nu_{i+1}}h_{\nu_{i+2}}\cdots h_{\nu_{\ell\left(  \nu\right)  }%
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\in\left\{  1,2,\ldots
,\ell\left(  \nu\right)  \right\}  \right)  .
\]
Applying the map $\mathbf{v}_{n}$ to both sides of this equality, we obtain%
\begin{align*}
\mathbf{v}_{n}\left(  h_{\nu}\right)   &  =\mathbf{v}_{n}\left(  \left(
h_{\nu_{1}}h_{\nu_{2}}\cdots h_{\nu_{i-1}}\right)  h_{\nu_{i}}\left(
h_{\nu_{i+1}}h_{\nu_{i+2}}\cdots h_{\nu_{\ell\left(  \nu\right)  }}\right)
\right) \\
&  =\mathbf{v}_{n}\left(  h_{\nu_{1}}h_{\nu_{2}}\cdots h_{\nu_{i-1}}\right)
\cdot\underbrace{\mathbf{v}_{n}\left(  h_{\nu_{i}}\right)  }_{=0}%
\cdot\mathbf{v}_{n}\left(  h_{\nu_{i+1}}h_{\nu_{i+2}}\cdots h_{\nu
_{\ell\left(  \nu\right)  }}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{v}_{n}\text{ is a } \kk \text{-algebra homomorphism}\right) \\
&  =0,
\end{align*}
qed.}. Also, $\mu\left\{  n\right\}  \neq\nu$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, $\mu\left\{  n\right\}  =\nu$. Let us write the
partition $\mu$ in the form $\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  $.
Then, $\mu\left\{  n\right\}  =\left(  n\mu_{1},n\mu_{2},n\mu_{3}%
,\ldots\right)  $ (by the definition of $\mu\left\{  n\right\}  $). Hence,
$\left(  \nu_{1},\nu_{2},\nu_{3},\ldots\right)  =\nu=\mu\left\{  n\right\}
=\left(  n\mu_{1},n\mu_{2},n\mu_{3},\ldots\right)  $. Thus, every positive
integer $j$ satisfies $\nu_{j}=n\mu_{j}$.
\par
But (\ref{sol.Lambda.verschiebung.e.stupid-assumption}) yields that there
exists an $i\in\left\{  1,2,3,\ldots\right\}  $ such that $n\nmid\nu_{i}$.
Consider this $i$. Now, recall that every positive integer $j$ satisfies
$\nu_{j}=n\mu_{j}$. Applied to $j=i$, this yields $\nu_{i}=n\mu_{i}$, so that
$n\mid\nu_{i}$. This contradicts $n\nmid\nu_{i}$. This contradiction proves
that our assumption was wrong, qed.}, so that $\delta_{\mu\left\{  n\right\}
,\nu}=0$. Thus, (\ref{sol.Lambda.verschiebung.e.lhs}) becomes $\left(
\mathbf{f}_{n}a,b\right)  =\delta_{\mu\left\{  n\right\}  ,\nu}=0$. Compared
with $\left(  a,\mathbf{v}_{n}\underbrace{b}_{=h_{\nu}}\right)  =\left(
a,\underbrace{\mathbf{v}_{n}\left(  h_{\nu}\right)  }_{=0}\right)  =\left(
a,0\right)  =0$ (since the Hall inner product is $\kk$-bilinear), this
yields $\left(  \mathbf{f}_{n}a,b\right)  =\left(  a,\mathbf{v}_{n}b\right)
$. Thus, (\ref{sol.Lambda.verschiebung.e.goal}) holds.

Now, let us forget that we assumed that
(\ref{sol.Lambda.verschiebung.e.stupid-assumption}) holds. We thus have proven
(\ref{sol.Lambda.verschiebung.e.goal}) under the assumption that
(\ref{sol.Lambda.verschiebung.e.stupid-assumption}) holds. Hence, for the rest
of our proof of (\ref{sol.Lambda.verschiebung.e.goal}), we can WLOG assume
that (\ref{sol.Lambda.verschiebung.e.stupid-assumption}) does not hold. Assume this.

We have assumed that (\ref{sol.Lambda.verschiebung.e.stupid-assumption}) does
not hold. In other words, there exists no $i\in\left\{  1,2,3,\ldots\right\}
$ such that $n\nmid\nu_{i}$. In other words,%
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies }n\mid\nu
_{i}. \label{sol.Lambda.verschiebung.e.anti}%
\end{equation}
Thus, $\nu_{i}/n$ is a nonnegative integer for every $i\in\left\{
1,2,3,\ldots\right\}  $. These nonnegative integers satisfy $\nu_{1}/n\geq
\nu_{2}/n\geq\nu_{3}/n\geq\cdots$ (since $\nu_{1}\geq\nu_{2}\geq\nu_{3}%
\geq\cdots$ (since $\left(  \nu_{1},\nu_{2},\nu_{3},\ldots\right)  =\nu$ is a
partition)) and \newline
$\left(  \nu_{i}/n=0\text{ for all sufficiently high
}i\right)  $ (since $\left(  \nu_{i}=0\text{ for all sufficiently high
}i\right)  $ (since $\left(  \nu_{1},\nu_{2},\nu_{3},\ldots\right)  =\nu$ is a
partition)). Hence, $\left(  \nu_{1}/n,\nu_{2}/n,\nu_{3}/n,\ldots\right)  $ is
a partition. Let us denote this partition by $\kappa$. We have $\kappa\left\{
n\right\}  =\nu$\ \ \ \ \footnote{\textit{Proof.} We have $\kappa=\left(
\nu_{1}/n,\nu_{2}/n,\nu_{3}/n,\ldots\right)  $. Hence, the definition of
$\kappa\left\{  n\right\}  $ yields $\kappa\left\{  n\right\}  =\left(
n\left(  \nu_{1}/n\right)  ,n\left(  \nu_{2}/n\right)  ,n\left(  \nu
_{3}/n\right)  ,\ldots\right)  =\left(  \nu_{1},\nu_{2},\nu_{3},\ldots\right)
=\nu$, qed.} and thus $\ell\left(  \kappa\right)  =\ell\left(  \nu\right)
$\ \ \ \ \footnote{\textit{Proof.} Applying
(\ref{sol.Lambda.verschiebung.e.llambdas}) to $\kappa$ and $n$ instead of
$\lambda$ and $s$, we obtain $\ell\left(  \kappa\left\{  n\right\}  \right)
=\ell\left(  \kappa\right)  $. Since $\kappa\left\{  n\right\}  =\nu$, this
rewrites as $\ell\left(  \nu\right)  =\ell\left(  \kappa\right)  $, qed.}. We
have $\mathbf{v}_{n}\left(  h_{\nu}\right)  =h_{\kappa}$%
\ \ \ \ \footnote{\textit{Proof.} We have $\kappa=\left(  \nu_{1}/n,\nu
_{2}/n,\nu_{3}/n,\ldots\right)  $ and therefore $\kappa=\left(  \nu_{1}%
/n,\nu_{2}/n,\ldots,\nu_{\ell\left(  \kappa\right)  }/n\right)  $. Since
$\ell\left(  \kappa\right)  =\ell\left(  \nu\right)  $, this rewrites as
$\kappa=\left(  \nu_{1}/n,\nu_{2}/n,\ldots,\nu_{\ell\left(  \nu\right)
}/n\right)  $. Hence, the definition of $h_{\kappa}$ yields $h_{\kappa}%
=h_{\nu_{1}/n}h_{\nu_{2}/n}\cdots h_{\nu_{\ell\left(  \nu\right)  }/n}%
=\prod_{i=1}^{\ell\left(  \nu\right)  }h_{\nu_{i}/n}$.
\par
On the other hand, $\nu=\left(  \nu_{1},\nu_{2},\ldots,\nu_{\ell\left(
\nu\right)  }\right)  $, so that the definition of $h_{\nu}$ yields $h_{\nu
}=h_{\nu_{1}}h_{\nu_{2}}\cdots h_{\nu_{\ell\left(  \nu\right)  }}=\prod
_{i=1}^{\ell\left(  \nu\right)  }h_{\nu_{i}}$. Applying the map $\mathbf{v}%
_{n}$ to both sides of this equality, we obtain%
\begin{align*}
\mathbf{v}_{n}\left(  h_{\nu}\right)   &  =\mathbf{v}_{n}\left(  \prod
_{i=1}^{\ell\left(  \nu\right)  }h_{\nu_{i}}\right)  =\prod_{i=1}^{\ell\left(
\nu\right)  }\underbrace{\mathbf{v}_{n}\left(  h_{\nu_{i}}\right)
}_{\substack{=
\begin{cases}
h_{\nu_{i}/n},& \text{if }n\mid\nu_{i};\\
0,& \text{if }n\nmid\nu_{i}%
\end{cases}
\\\text{(by (\ref{sol.Lambda.verschiebung.vnhm}), applied to }%
m=\nu_{i}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{v}%
_{n}\text{ is a } \kk \text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{\ell\left(  \nu\right)  }\underbrace{
\begin{cases}
h_{\nu_{i}/n},& \text{if }n\mid\nu_{i};\\
0,& \text{if }n\nmid\nu_{i}%
\end{cases}}_{\substack{=h_{\nu_{i}/n}\\\text{(since }n\mid\nu_{i}\text{ (by
(\ref{sol.Lambda.verschiebung.e.anti})))}}}
\quad =\prod_{i=1}^{\ell\left(
\nu\right)  }h_{\nu_{i}/n}=h_{\kappa}%
\end{align*}
(since $h_{\kappa}=\prod_{i=1}^{\ell\left(  \nu\right)  }h_{\nu_{i}/n}$),
qed.} and thus%
\begin{equation}
\left(  \underbrace{a}_{=m_{\mu}},\mathbf{v}_{n}\underbrace{b}_{=h_{\nu}%
}\right)  =\left(  m_{\mu},\underbrace{\mathbf{v}_{n}\left(  h_{\nu}\right)
}_{=h_{\kappa}}\right)  =\left(  m_{\mu},h_{\kappa}\right)  =\delta
_{\mu,\kappa} \label{sol.Lambda.verschiebung.e.rhs}%
\end{equation}
(by (\ref{sol.Lambda.verschiebung.e.duality}), applied to $\mu$ and $\kappa$
instead of $\lambda$ and $\mu$).

Now, let us write the partition $\mu$ in the form $\left(  \mu_{1},\mu_{2}%
,\mu_{3},\ldots\right)  $. Then, $\mu\left\{  n\right\}  =\left(  n\mu
_{1},n\mu_{2},n\mu_{3},\ldots\right)  $ (by the definition of $\mu\left\{
n\right\}  $).

Now, we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \mu\left\{  n\right\}  =\nu\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  n\mu_{1},n\mu_{2},n\mu_{3}%
,\ldots\right)  =\left(  \nu_{1},\nu_{2},\nu_{3},\ldots\right)  \right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu\left\{  n\right\}  =\left(
n\mu_{1},n\mu_{2},n\mu_{3},\ldots\right)  \text{ and }\nu=\left(  \nu_{1}%
,\nu_{2},\nu_{3},\ldots\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }i\in\left\{  1,2,3,\ldots\right\}
\text{ satisfies }n\mu_{i}=\nu_{i}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{every }i\in\left\{  1,2,3,\ldots\right\}
\text{ satisfies }\mu_{i}=\nu_{i}/n\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)
=\left(  \nu_{1}/n,\nu_{2}/n,\nu_{3}/n,\ldots\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \mu=\kappa\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  =\mu\text{ and
}\left(  \nu_{1}/n,\nu_{2}/n,\nu_{3}/n,\ldots\right)  =\kappa\right)  .
\label{sol.Lambda.verschiebung.e.equiv}%
\end{align}
But (\ref{sol.Lambda.verschiebung.e.lhs}) becomes%
\begin{align*}
\left(  \mathbf{f}_{n}a,b\right)   &  =\delta_{\mu\left\{  n\right\}  ,\nu
}=
\begin{cases}
1,& \text{if }\mu\left\{  n\right\}  =\nu;\\
0,& \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\delta
_{\mu\left\{  n\right\}  ,\nu}\right) \\
&  =
\begin{cases}
1,& \text{if }\mu=\kappa;\\
0,& \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu\left\{  n\right\}
=\nu\text{ is equivalent to }\mu=\kappa\text{ (by
(\ref{sol.Lambda.verschiebung.e.equiv}))}\right) \\
&  =\delta_{\mu,\kappa}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\delta
_{\mu,\kappa}=
\begin{cases}
1,& \text{if }\mu=\kappa;\\
0,& \text{otherwise}%
\end{cases}
\text{ (by the definition of }\delta_{\mu,\kappa}\text{)}\right) \\
&  =\left(  a,\mathbf{v}_{n}b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.verschiebung.e.rhs})}\right)  .
\end{align*}
Thus, (\ref{sol.Lambda.verschiebung.e.goal}) is proven. As we know, this
completes the solution of Exercise \ref{exe.Lambda.verschiebung}(f).

(g) Our solution to Exercise \ref{exe.Lambda.verschiebung}(g) shall proceed in
three steps:

\begin{itemize}
\item \textit{Step 1:} proving that Exercise \ref{exe.Lambda.verschiebung}(g)
holds if $\kk = \QQ$.

\item \textit{Step 2:} proving that Exercise \ref{exe.Lambda.verschiebung}(g)
holds if $\kk = \ZZ$.

\item \textit{Step 3:} proving that Exercise \ref{exe.Lambda.verschiebung}(g)
holds in the general case.
\end{itemize}

Let us now get to the details of these three steps:

\underline{\textit{Step 1:}} We shall prove that Exercise
\ref{exe.Lambda.verschiebung}(g) holds if $\kk = \QQ$.

Indeed, assume that $\kk = \QQ$. We know that $\mathbf{f}_{n}$ is
a $\kk$-algebra homomorphism (due to Exercise \ref{exe.witt.ghost-app}%
(a)), and that $\mathbf{v}_{n}$ is a $\kk$-algebra homomorphism. Hence,
$\mathbf{v}_{n}\circ\mathbf{f}_{n}$ is a $\kk$-algebra homomorphism.
Also, $\underbrace{\operatorname{id}_{\Lambda}\star
\operatorname{id}_{\Lambda}\star\cdots\star
\id_{\Lambda}}_{n\text{ times}}$ is a $\kk$-algebra homomorphism
(due to Exercise \ref{exe.alghoms.comm}(b), applied to $H=\Lambda$,
$A=\Lambda$, $k=n$ and $f_{i}=\operatorname{id}_{\Lambda}$). In
other words, $\operatorname{id}_{\Lambda}^{\star n}$ is a
$\kk$-algebra homomorphism.

Let $r$ be a positive integer. We shall now show that $\left(  \mathbf{v}%
_{n}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)  =
\id_{\Lambda}^{\star n}\left(  p_{r}\right)  $.

Indeed, it is easy to see that
\begin{equation}
\operatorname{id}_{\Lambda}^{\star a}\left(  p_{r}\right)
=ap_{r}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in \NN 
\label{sol.Lambda.verschiebung.f.step1.idstar}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.f.step1.idstar}):} We
shall prove (\ref{sol.Lambda.verschiebung.f.step1.idstar}) by induction over
$a$:
\par
\textit{Induction base:} We have
$\epsilon \left(p_r\right) = 0$ (since $r$ is positive).
Now,
$\underbrace{ \id_{\Lambda}^{\star0} }_{= u \circ \epsilon}
   \left( p_r \right)
= \left(u \circ \epsilon\right) \left( p_r \right)
= u \left( \underbrace{\epsilon \left(p_r\right)}_{= 0} \right)
= u \left( 0 \right) = 0$, so that $
\id_{\Lambda}^{\star0}\left(  p_{r}\right)  =0=0p_{r}$. In other words,
(\ref{sol.Lambda.verschiebung.f.step1.idstar}) holds for $a=0$. This completes
the induction base.
\par
\textit{Induction step:} Let $A$ be a positive integer. Assume that
(\ref{sol.Lambda.verschiebung.f.step1.idstar}) holds for $a=A-1$. We now need
to show that (\ref{sol.Lambda.verschiebung.f.step1.idstar}) holds for $a=A$.
\par
We know that (\ref{sol.Lambda.verschiebung.f.step1.idstar}) holds for $a=A-1$.
In other words, $\operatorname{id}_{\Lambda}^{\star\left(
A-1\right)  }\left(  p_{r}\right)  =\left(  A-1\right)  p_{r}$. Now, recall
that $\Delta p_{r}=1\otimes p_{r}+p_{r}\otimes1$ (by Proposition
\ref{symm-comultiplication-formulas-prop}(i), applied to $r$ instead of $n$).
On the other hand, $\operatorname{id}_{\Lambda}^{\star\left(
A-1\right)  }$ is a $\kk$-algebra homomorphism (in fact,
this is shown in the same way as we have shown that
$\operatorname{id}_{\Lambda}^{\star n}$ is a
$\kk$-algebra homomorphism), so that we have
$\operatorname{id}_{\Lambda}^{\star\left(  A-1\right)  }\left(
1\right)  =1$. Now, $\operatorname{id}_{\Lambda}^{\star
A}=\operatorname{id}_{\Lambda}^{\star\left(  A-1\right)  }%
\star\operatorname{id}_{\Lambda}=m\circ\left(  
\id_{\Lambda}^{\star\left(  A-1\right)  }\otimes
\id_{\Lambda}\right)  \circ\Delta$ (by the definition of convolution).
Applying both sides of this equality to $p_{r}$, we obtain%
\begin{align*}
\operatorname{id}_{\Lambda}^{\star A}\left(  p_{r}\right)   &
=\left(  m\circ\left(  \operatorname{id}_{\Lambda}^{\star\left(
A-1\right)  }\otimes\operatorname{id}_{\Lambda}\right)  \circ
\Delta\right)  \left(  p_{r}\right)  =m\left(  \left(  
\id_{\Lambda}^{\star\left(  A-1\right)  }\otimes
\id_{\Lambda}\right)  \underbrace{\left(  \Delta p_{r}\right)
}_{=1\otimes p_{r}+p_{r}\otimes1}\right) \\
&  =m\left(  \underbrace{\left(  \operatorname{id}_{\Lambda}%
^{\star\left(  A-1\right)  }\otimes\operatorname{id}_{\Lambda
}\right)  \left(  1\otimes p_{r}+p_{r}\otimes1\right)  }_{=
\id_{\Lambda}^{\star\left(  A-1\right)  }\left(  1\right)
\otimes\operatorname{id}_{\Lambda}\left(  p_{r}\right)
+\operatorname{id}_{\Lambda}^{\star\left(  A-1\right)  }\left(
p_{r}\right)  \otimes\operatorname{id}_{\Lambda}\left(  1\right)
}\right) \\
&  =m\left(  \operatorname{id}_{\Lambda}^{\star\left(  A-1\right)
}\left(  1\right)  \otimes\operatorname{id}_{\Lambda}\left(
p_{r}\right)  +\operatorname{id}_{\Lambda}^{\star\left(  A-1\right)
}\left(  p_{r}\right)  \otimes\operatorname{id}_{\Lambda}\left(
1\right)  \right) \\
&  =\underbrace{\operatorname{id}_{\Lambda}^{\star\left(
A-1\right)  }\left(  1\right)  }_{\substack{=1}}\cdot
\underbrace{\operatorname{id}_{\Lambda}\left(  p_{r}\right)
}_{=p_{r}}+\underbrace{\operatorname{id}_{\Lambda}^{\star\left(
A-1\right)  }\left(  p_{r}\right)  }_{=\left(  A-1\right)  p_{r}}%
\cdot\underbrace{\operatorname{id}_{\Lambda}\left(  1\right)  }%
_{=1}=p_{r}+\left(  A-1\right)  p_{r}=Ap_{r}.
\end{align*}
Thus, (\ref{sol.Lambda.verschiebung.f.step1.idstar}) holds for $a=A$. This
completes the induction step, so that
(\ref{sol.Lambda.verschiebung.f.step1.idstar}) is proven.}. Applied to $a=n$,
this yields $\operatorname{id}_{\Lambda}^{\star n}\left(
p_{r}\right)  =np_{r}$. But the definition of $\mathbf{f}_{n}\left(
p_{r}\right)  $ yields
\begin{align}
\mathbf{f}_{n}\left(  p_{r}\right)
&= p_{r}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)
  \nonumber \\
&= \left(x_1^n\right)^r + \left(x_2^n\right)^r
  + \left(x_3^n\right)^r + \cdots
  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
  p_r = x_1^r + x_2^r + x_3^r + \cdots \right)
  \nonumber \\
&  =x_{1}^{rn}+x_{2}^{rn}+x_{3}^{rn}+\ldots = p_{rn}.
\label{sol.Lambda.verschiebung.f.step1.fnpr}
\end{align}
Hence,%
\begin{align*}
\left(  \mathbf{v}_{n}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)   &
=\mathbf{v}_{n}\left(  \underbrace{\mathbf{f}_{n}\left(  p_{r}\right)
}_{=p_{rn}}\right)  =\mathbf{v}_{n}\left(  p_{rn}\right) \\
&=
\begin{cases}
np_{rn/n},& \text{if }n\mid rn;\\
0,& \text{if }n\nmid rn
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.Lambda.verschiebung}(a), applied to }m=rn\right) \\
&  =np_{rn/n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\mid rn\right) \\
&  =np_{r}=\operatorname{id}_{\Lambda}^{\star n}\left(
p_{r}\right)
\end{align*}
(since $\operatorname{id}_{\Lambda}^{\star n}\left(  p_{r}\right)
=np_{r}$).

Now, let us forget that we fixed $r$. We thus have proven that%
\begin{equation}
\left(  \mathbf{v}_{n}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)
=\operatorname{id}_{\Lambda}^{\star n}\left(  p_{r}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }r\text{.}
\label{sol.Lambda.verschiebung.f.step1.main}%
\end{equation}


We have assumed that $\kk = \QQ$.
Thus, $\QQ$ is a subring of $\kk$.
Hence, the elements $p_{1}%
,p_{2},p_{3},\ldots$ of $\Lambda$ generate the $\kk$-algebra $\Lambda$
(due to Proposition \ref{symm-antipode-and-h-basis}). But $\mathbf{v}_{n}%
\circ\mathbf{f}_{n}$ and $\operatorname{id}_{\Lambda}^{\star n}$ are
two $\kk$-algebra homomorphisms from $\Lambda$. These two homomorphisms
$\mathbf{v}_{n}\circ\mathbf{f}_{n}$ and $\operatorname{id}_{\Lambda
}^{\star n}$ are equal to each other on each of the elements $p_{1}%
,p_{2},p_{3},\ldots$ (due to (\ref{sol.Lambda.verschiebung.f.step1.main})),
and therefore are identical (because if two $\kk$-algebra homomorphisms
with one and the same domain are equal to each other on a generating set of
the domain, then these homomorphisms must be identical). In other words,
$\mathbf{v}_{n}\circ\mathbf{f}_{n}=\operatorname{id}_{\Lambda
}^{\star n}$. Thus, Exercise \ref{exe.Lambda.verschiebung}(g) is solved under
the assumption that $\kk = \QQ$. Our Step 1 is complete.

\underline{\textit{Step 2:}} We shall prove that Exercise
\ref{exe.Lambda.verschiebung}(g) holds if $\kk = \ZZ$.

Indeed, let us first consider the general case, without any assumptions on
$\kk$.

We shall now formalize the intuitive concept that the constructions relevant
to Exercise \ref{exe.Lambda.verschiebung}(g) (the Hopf algebra $\Lambda$, its
elements $m_{\lambda}$, $p_{\lambda}$, $e_{\lambda}$ etc., and the maps
$\mathbf{v}_{n}$, $\mathbf{f}_{n}$ and $\operatorname{id}_{\Lambda
}^{\star n}$) are ``functorial'' with respect
to the base ring $\kk$. This is chiefly a matter of introducing notations:

\begin{itemize}
\item We denote the $\kk$-algebra $\Lambda=\Lambda_{ \kk }$ by
$\Lambda^{\left[   \kk \right]  }$. (This is just a minor change of
notation that serves to make it more similar to the notations $\mathbf{f}%
_{n}^{\left[   \kk \right]  }$, $\mathbf{v}_{n}^{\left[   \kk %
\right]  }$ and $m_{\lambda}^{\left[   \kk \right]  }$ further below.)

\item If $\mathbf{m}$ and $\mathbf{n}$ are two commutative rings, and
$\varphi:\mathbf{m}\rightarrow\mathbf{n}$ is a ring homomorphism, then
$\varphi$ canonically induces a ring homomorphism $\Lambda^{\left[
\mathbf{m}\right]  }\rightarrow\Lambda^{\left[  \mathbf{n}\right]  }%
$.\ \ \ \ \footnote{In fact, this ring homomorphism $\Lambda^{\left[
\mathbf{m}\right]  }\rightarrow\Lambda^{\left[  \mathbf{n}\right]  }$ can be
defined by taking the ring homomorphism $\varphi\left[  \left[  \xx%
\right]  \right]  :\mathbf{m}\left[  \left[  \xx\right]  \right]
\rightarrow\mathbf{n}\left[  \left[  \xx\right]  \right]  $ canonically
induced by $\varphi$ and restricting it to the subring $\Lambda^{\left[
\mathbf{m}\right]  }$ of $\mathbf{m}\left[  \left[  \xx\right]
\right]  $.} We denote this latter homomorphism by $\Lambda^{\left[
\varphi\right]  }$.

Notice that while $\Lambda^{\left[   \kk \right]  }$ denotes a ring,
$\Lambda^{\left[  \varphi\right]  }$ denotes a map. This might look confusing,
but it makes sense from the viewpoint of category theory: There is a functor
from the category of commutative rings to itself, which sends every
commutative ring $\kk$ to $\Lambda^{\left[   \kk \right]  }$, and
every morphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$ of commutative rings
to $\Lambda^{\left[  \varphi\right]  }$. Unsurprisingly, this functor is
denoted by $\Lambda$.

\item A ring homomorphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$ also
induces a ring homomorphism $\Lambda^{\left[  \mathbf{m}\right]  }%
\otimes_{\mathbf{m}}\Lambda^{\left[  \mathbf{m}\right]  }\rightarrow
\Lambda^{\left[  \mathbf{n}\right]  }\otimes_{\mathbf{n}}\Lambda^{\left[
\mathbf{n}\right]  }$\ \ \ \ \footnote{There are several ways to define this
homomorphism. One of them is to define it as the $\mathbf{m}$-module
homomorphism $\Lambda^{\left[  \mathbf{m}\right]  }\otimes_{\mathbf{m}}%
\Lambda^{\left[  \mathbf{m}\right]  }\rightarrow\Lambda^{\left[
\mathbf{n}\right]  }\otimes_{\mathbf{n}}\Lambda^{\left[  \mathbf{n}\right]  }$
canonically induced by the $\mathbf{m}$-bilinear map
\begin{align*}
\Lambda^{\left[  \mathbf{m}\right]  }\times\Lambda^{\left[  \mathbf{m}\right]
}  &  \rightarrow\Lambda^{\left[  \mathbf{n}\right]  }\otimes_{\mathbf{n}%
}\Lambda^{\left[  \mathbf{n}\right]  },\\
\left(  a,b\right)   &  \mapsto\left(  \Lambda^{\left[  \varphi\right]
}\left(  a\right)  \right)  \otimes_{\mathbf{n}}\left(  \Lambda^{\left[
\varphi\right]  }\left(  b\right)  \right)  .
\end{align*}
(Here, $\Lambda^{\left[  \mathbf{n}\right]  }\otimes_{\mathbf{n}}%
\Lambda^{\left[  \mathbf{n}\right]  }$ is an $\mathbf{m}$-module because the
ring homomorphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$ makes $\mathbf{n}$
into an $\mathbf{m}$-algebra.)}. This makes $\Lambda\otimes\Lambda$ into a
functor of $\kk$.

\item For every partition $\lambda$, we shall denote the monomial symmetric
function $m_{\lambda}\in\Lambda$ by $m_{\lambda}^{\left[   \kk \right]
}$. This notation makes the dependency of $m_{\lambda}$ on the base ring
$\kk$ more explicit, and thus allows us to talk about these elements
defined over several different base rings at the same time (without the danger
of confusing them). Of course, the element $m_{\lambda}$ does not
``really'' depend on $\kk$, in the
sense that it is defined in the same way for every $\kk$. This entails
that for any two commutative rings $\mathbf{m}$ and $\mathbf{n}$ and any ring
homomorphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$, we have%
\begin{equation}
m_{\lambda}^{\left[  \mathbf{n}\right]  }=\Lambda^{\left[  \varphi\right]
}\left(  m_{\lambda}^{\left[  \mathbf{m}\right]  }\right)
\label{sol.Lambda.verschiebung.f.step2.func-m}%
\end{equation}
for every partition $\lambda$.

\item We shall denote the homomorphism $\mathbf{f}_{n}:\Lambda\rightarrow
\Lambda$ by $\mathbf{f}_{n}^{\left[   \kk \right]  }$. This notation
makes the dependency of $\mathbf{f}_{n}$ on the base ring $\kk$ more explicit.

The definition of $\mathbf{f}_{n}$ was functorial in $\kk$. Thus, for
any two commutative rings $\mathbf{m}$ and $\mathbf{n}$ and any ring
homomorphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$, we have%
\begin{equation}
\mathbf{f}_{n}^{\left[  \mathbf{n}\right]  }\circ\Lambda^{\left[
\varphi\right]  }=\Lambda^{\left[  \varphi\right]  }\circ\mathbf{f}%
_{n}^{\left[  \mathbf{m}\right]  }
\label{sol.Lambda.verschiebung.f.step2.func-f}%
\end{equation}
as maps from $\Lambda^{\left[  \mathbf{m}\right]  }$ to $\Lambda^{\left[
\mathbf{n}\right]  }$.

\item Similarly, we shall denote the homomorphism $\mathbf{v}_{n}%
:\Lambda\rightarrow\Lambda$ by $\mathbf{v}_{n}^{\left[   \kk \right]  }$.
This homomorphism is again defined in a way that is functorial in $\kk %
$. Thus, for any two commutative rings $\mathbf{m}$ and $\mathbf{n}$ and any
ring homomorphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$, we have%
\begin{equation}
\mathbf{v}_{n}^{\left[  \mathbf{n}\right]  }\circ\Lambda^{\left[
\varphi\right]  }=\Lambda^{\left[  \varphi\right]  }\circ\mathbf{v}%
_{n}^{\left[  \mathbf{m}\right]  }
\label{sol.Lambda.verschiebung.f.step2.func-v}%
\end{equation}
as maps from $\Lambda^{\left[  \mathbf{m}\right]  }$ to $\Lambda^{\left[
\mathbf{n}\right]  }$.

\item Notice that the Hopf algebra structure on $\Lambda$ is functorial in
$\kk$.\ \ \ \ \footnote{To make sense of this statement, we need to
recall that $\Lambda\otimes\Lambda$ has been made into a functor of
$\kk$.} As a consequence, for any two commutative rings $\mathbf{m}$
and $\mathbf{n}$ and any ring homomorphism $\varphi:\mathbf{m}\rightarrow
\mathbf{n}$, we have
\begin{equation}
\operatorname{id}_{\Lambda_{\mathbf{n}}}^{\star n}\circ
\Lambda^{\left[  \varphi\right]  }=\Lambda^{\left[  \varphi\right]  }%
\circ\operatorname{id}_{\Lambda_{\mathbf{m}}}^{\star n}
\label{sol.Lambda.verschiebung.f.step2.func-idn}%
\end{equation}
as maps from $\Lambda^{\left[  \mathbf{m}\right]  }$ to $\Lambda^{\left[
\mathbf{n}\right]  }$.

\item For every commutative ring $\kk$, there exists a canonical (and
unique) ring homomorphism $\ZZ \rightarrow \kk$. Denote this
homomorphism by $\rho_{ \kk }$.
\end{itemize}

In Step 1, we have shown that Exercise \ref{exe.Lambda.verschiebung}(g) holds
if $\kk = \QQ$. In other words, we have $\mathbf{v}_{n}%
\circ\mathbf{f}_{n}=\id_{\Lambda}^{\star n}$ if
$\kk = \QQ$. In other words,%
\begin{equation}
\mathbf{v}_{n}^{\left[  \QQ \right]  }\circ\mathbf{f}_{n}^{\left[
\QQ \right]  }=\operatorname{id}_{\Lambda_{\QQ}}^{\star
n}. \label{sol.Lambda.verschiebung.f.step2.step1}%
\end{equation}


Recall that $\left(  m_{\lambda}^{\left[   \kk \right]  }\right)
_{\lambda\in \Par }=\left(  m_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\kk$-module $\Lambda
_{ \kk }$ for every commutative ring $\kk$. Applying this to
$\kk = \ZZ$, we conclude that $\left(  m_{\lambda}^{\left[
 \ZZ \right]  }\right)  _{\lambda\in \Par }$ is a basis of
the $\ZZ$-module $\Lambda_{ \ZZ }$.

The ring homomorphism $\rho_{\QQ}: \ZZ \rightarrow \QQ$ is
just the canonical inclusion of $\ZZ$ into $\QQ$, and thus is
injective. Hence, the map $\Lambda^{\left[  \rho_{\QQ}\right]  }$ is
injective (because $\Lambda^{\left[  \varphi\right]  }$ is injective for every
injective ring homomorphism $\varphi$).

Now, let us recall that our goal (in Step 2) is to show that Exercise
\ref{exe.Lambda.verschiebung}(g) holds if $\kk = \ZZ$. In other
words, our goal is to show that%
\[
\mathbf{v}_{n}^{\left[   \ZZ \right]  }\circ\mathbf{f}_{n}^{\left[
 \ZZ \right]  }=\operatorname{id}_{\Lambda_{ \ZZ }}^{\star
n}.
\]
This is an equality between $\ZZ$-module homomorphisms (indeed, it is
clear that both $\mathbf{v}_{n}^{\left[   \ZZ \right]  }\circ
\mathbf{f}_{n}^{\left[   \ZZ \right]  }$ and $
\id_{\Lambda_{ \ZZ }}^{\star n}$ are $\ZZ$-module
homomorphisms). Hence, in order to prove it, it is sufficient to verify it on
the basis $\left(  m_{\lambda}^{\left[   \ZZ \right]  }\right)
_{\lambda\in \Par }$ of the $\ZZ$-module $\Lambda
_{ \ZZ }$. In other words, it is sufficient to prove that%
\begin{equation}
\left(  \mathbf{v}_{n}^{\left[   \ZZ \right]  }\circ\mathbf{f}%
_{n}^{\left[   \ZZ \right]  }\right)  \left(  m_{\lambda}^{\left[
 \ZZ \right]  }\right)  =\operatorname{id}_{\Lambda
_{ \ZZ }}^{\star n}\left(  m_{\lambda}^{\left[   \ZZ \right]
}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in \Par .
\label{sol.Lambda.verschiebung.f.step2.goal}%
\end{equation}


So let us prove (\ref{sol.Lambda.verschiebung.f.step2.goal}) now:

\textit{Proof of (\ref{sol.Lambda.verschiebung.f.step2.goal}):} Let
$\lambda\in \Par $. We have%
\begin{align*}
&  \Lambda^{\left[  \rho_{\QQ}\right]  }\left(  \left(  \mathbf{v}%
_{n}^{\left[   \ZZ \right]  }\circ\mathbf{f}_{n}^{\left[   \ZZ %
\right]  }\right)  \left(  m_{\lambda}^{\left[   \ZZ \right]  }\right)
\right) \\
&  =\left(  \underbrace{\Lambda^{\left[  \rho_{\QQ}\right]  }%
\circ\mathbf{v}_{n}^{\left[   \ZZ \right]  }}_{\substack{=\mathbf{v}%
_{n}^{\left[  \QQ \right]  }\circ\Lambda^{\left[  \rho_{\QQ}
\right]  }\\\text{(because of }\mathbf{v}_{n}^{\left[  \QQ\right]
}\circ\Lambda^{\left[  \rho_{\QQ}\right]  }=\Lambda^{\left[
\rho_{\QQ}\right]  }\circ\mathbf{v}_{n}^{\left[   \ZZ \right]
}\\\text{(by (\ref{sol.Lambda.verschiebung.f.step2.func-v}), applied}%
\\\text{to }\mathbf{m=} \ZZ \text{, }\mathbf{n}= \QQ \text{ and
}\varphi=\rho_{\QQ}\text{))}}}\circ\mathbf{f}_{n}^{\left[
 \ZZ \right]  }\right)  \left(  m_{\lambda}^{\left[   \ZZ \right]
}\right)  =\left(  \mathbf{v}_{n}^{\left[  \QQ \right]  }\circ
\underbrace{\Lambda^{\left[  \rho_{\QQ}\right]  }\circ\mathbf{f}%
_{n}^{\left[   \ZZ \right]  }}_{\substack{=\mathbf{f}_{n}^{\left[
\QQ \right]  }\circ\Lambda^{\left[  \rho_{\QQ}\right]
}\\\text{(because of }\mathbf{f}_{n}^{\left[  \QQ \right]  }\circ
\Lambda^{\left[  \rho_{\QQ}\right]  }=\Lambda^{\left[  \rho
_{\QQ}\right]  }\circ\mathbf{f}_{n}^{\left[   \ZZ \right]
}\\\text{(by (\ref{sol.Lambda.verschiebung.f.step2.func-f}), applied}%
\\\text{to }\mathbf{m=} \ZZ \text{, }\mathbf{n}= \QQ \text{ and
}\varphi=\rho_{\QQ}\text{))}}}\right)  \left(  m_{\lambda}^{\left[
 \ZZ \right]  }\right) \\
&  =\left(  \underbrace{\mathbf{v}_{n}^{\left[  \QQ \right]  }%
\circ\mathbf{f}_{n}^{\left[  \QQ \right]  }}%
_{\substack{=\operatorname{id}_{\Lambda_{\QQ}}^{\star
n}\\\text{(by (\ref{sol.Lambda.verschiebung.f.step2.step1}))}}}\circ
\Lambda^{\left[  \rho_{\QQ}\right]  }\right)  \left(  m_{\lambda
}^{\left[   \ZZ \right]  }\right)  =\underbrace{\left(
\operatorname{id}_{\Lambda_{\QQ}}^{\star n}\circ
\Lambda^{\left[  \rho_{\QQ}\right]  }\right)  }_{\substack{=\Lambda
^{\left[  \rho_{\QQ}\right]  }\circ\operatorname{id}_{\Lambda
_{\ZZ}}^{\star n}\\\text{(by
(\ref{sol.Lambda.verschiebung.f.step2.func-idn}), applied}\\\text{to
}\mathbf{m=} \ZZ \text{, }\mathbf{n}= \QQ \text{ and }\varphi
=\rho_{\QQ}\text{)}}}\left(  m_{\lambda}^{\left[   \ZZ \right]
}\right) \\
&  =\left(  \Lambda^{\left[  \rho_{\QQ}\right]  }\circ
\operatorname{id}_{\Lambda_{ \ZZ }}^{\star n}\right)  \left(
m_{\lambda}^{\left[   \ZZ \right]  }\right)  =\Lambda^{\left[
\rho_{\QQ}\right]  }\left(  \operatorname{id}_{\Lambda
_{ \ZZ }}^{\star n}\left(  m_{\lambda}^{\left[   \ZZ \right]
}\right)  \right)  .
\end{align*}
Since the map $\Lambda^{\left[  \rho_{\QQ}\right]  }$ is injective,
this yields $\left(  \mathbf{v}_{n}^{\left[   \ZZ \right]  }%
\circ\mathbf{f}_{n}^{\left[   \ZZ \right]  }\right)  \left(  m_{\lambda
}^{\left[   \ZZ \right]  }\right)  =\operatorname{id}_{\Lambda
_{ \ZZ }}^{\star n}\left(  m_{\lambda}^{\left[   \ZZ \right]
}\right)  $. Thus, (\ref{sol.Lambda.verschiebung.f.step2.goal}) is proven.

As we said, proving (\ref{sol.Lambda.verschiebung.f.step2.goal}) is sufficient
to showing that $\mathbf{v}_{n}^{\left[   \ZZ \right]  }\circ
\mathbf{f}_{n}^{\left[   \ZZ \right]  }=
\id_{\Lambda_{ \ZZ }}^{\star n}$. Hence, $\mathbf{v}_{n}^{\left[
 \ZZ \right]  }\circ\mathbf{f}_{n}^{\left[   \ZZ \right]
}=\operatorname{id}_{\Lambda_{ \ZZ }}^{\star n}$ is shown
(since (\ref{sol.Lambda.verschiebung.f.step2.goal}) is proven). In other
words, Exercise \ref{exe.Lambda.verschiebung}(g) holds if $\kk %
= \ZZ$. This completes Step 2.

\underline{\textit{Step 3:}} We shall now prove that Exercise
\ref{exe.Lambda.verschiebung}(g) holds in the general case.

Let us use all the notations that we introduced in Step 2.

In Step 2, we have shown that Exercise \ref{exe.Lambda.verschiebung}(g) holds
if $\kk = \ZZ$. In other words, we have $\mathbf{v}_{n}%
\circ\mathbf{f}_{n}=\operatorname{id}_{\Lambda}^{\star n}$ if
$\kk = \ZZ$. In other words,%
\begin{equation}
\mathbf{v}_{n}^{\left[   \ZZ \right]  }\circ\mathbf{f}_{n}^{\left[
 \ZZ \right]  }=\operatorname{id}_{\Lambda_{ \ZZ }}^{\star
n}. \label{sol.Lambda.verschiebung.f.step3.step2}%
\end{equation}


Recall that $\left(  m_{\lambda}\right)  _{\lambda\in \Par }$ is
a basis of the $\kk$-module $\Lambda_{ \kk }$.

Now, let us recall that our goal (in Step 3) is to show that Exercise
\ref{exe.Lambda.verschiebung}(g) holds. In other words, our goal is to show
that%
\[
\mathbf{v}_{n}\circ\mathbf{f}_{n}=\id^{\star n}.
\]
This is an equality between $\kk$-module homomorphisms (indeed, it is
clear that both $\mathbf{v}_{n}\circ\mathbf{f}_{n}$ and $\id^{\star n}$
are $\kk$-module homomorphisms). Hence, in order
to prove it, it is sufficient to verify it on the basis $\left(  m_{\lambda
}\right)  _{\lambda\in \Par }$ of the $\kk$-module
$\Lambda_{ \kk }$. In other words, it is sufficient to prove that%
\begin{equation}
\left(  \mathbf{v}_{n}\circ\mathbf{f}_{n}\right)  \left(  m_{\lambda}\right)
=\id^{\star n}\left(  m_{\lambda}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in \Par .
\label{sol.Lambda.verschiebung.f.step3.goal}%
\end{equation}


So let us prove (\ref{sol.Lambda.verschiebung.f.step3.goal}) now:

\textit{Proof of (\ref{sol.Lambda.verschiebung.f.step3.goal}):} Let
$\lambda\in \Par $. Then,%
\begin{equation}
m_{\lambda}=m_{\lambda}^{\left[   \kk \right]  }=\Lambda^{\left[
\rho_{ \kk }\right]  }\left(  m_{\lambda}^{\left[   \ZZ \right]
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.verschiebung.f.step2.func-m}) (applied to }\mathbf{m}%
= \ZZ \text{, }\mathbf{n}= \kk \text{ and }\varphi=\rho_{ \kk %
}\text{)}\right)  . \label{sol.Lambda.verschiebung.f.step3.goal.pf.1}%
\end{equation}
Hence,%
\begin{align*}
\left(  \underbrace{\mathbf{v}_{n}}_{=\mathbf{v}_{n}^{\left[   \kk %
\right]  }}\circ\underbrace{\mathbf{f}_{n}}_{=\mathbf{f}_{n}^{\left[
 \kk \right]  }}\right)  \left(  \underbrace{m_{\lambda}}_{=\Lambda
^{\left[  \rho_{ \kk }\right]  }\left(  m_{\lambda}^{\left[
 \ZZ \right]  }\right)  }\right)   &  =\left(  \mathbf{v}_{n}^{\left[
 \kk \right]  }\circ\mathbf{f}_{n}^{\left[   \kk \right]  }\right)
\left(  \Lambda^{\left[  \rho_{ \kk }\right]  }\left(  m_{\lambda
}^{\left[   \ZZ \right]  }\right)  \right)  =\left(  \mathbf{v}%
_{n}^{\left[   \kk \right]  }\circ\underbrace{\mathbf{f}_{n}^{\left[
 \kk \right]  }\circ\Lambda^{\left[  \rho_{ \kk }\right]  }%
}_{\substack{=\Lambda^{\left[  \rho_{ \kk }\right]  }\circ\mathbf{f}%
_{n}^{\left[   \ZZ \right]  }\\\text{(by
(\ref{sol.Lambda.verschiebung.f.step2.func-f}), applied}\\\text{to
}\mathbf{m=} \ZZ \text{, }\mathbf{n}= \kk \text{ and }\varphi
=\rho_{ \kk }\text{)}}}\right)  \left(  m_{\lambda}^{\left[
 \ZZ \right]  }\right) \\
&  =\left(  \underbrace{\mathbf{v}_{n}^{\left[   \kk \right]  }%
\circ\Lambda^{\left[  \rho_{ \kk }\right]  }}_{\substack{=\Lambda
^{\left[  \rho_{ \kk }\right]  }\circ\mathbf{v}_{n}^{\left[
 \ZZ \right]  }\\\text{(by (\ref{sol.Lambda.verschiebung.f.step2.func-v}%
), applied}\\\text{to }\mathbf{m=} \ZZ \text{, }\mathbf{n}= \kk %
\text{ and }\varphi=\rho_{ \kk }\text{)}}}\circ\mathbf{f}_{n}^{\left[
 \ZZ \right]  }\right)  \left(  m_{\lambda}^{\left[   \ZZ \right]
}\right) \\
&  =\left(  \Lambda^{\left[  \rho_{ \kk }\right]  }\circ
\underbrace{\mathbf{v}_{n}^{\left[   \ZZ \right]  }\circ\mathbf{f}%
_{n}^{\left[   \ZZ \right]  }}_{\substack{=
\id_{\Lambda_{ \ZZ }}^{\star n}\\\text{(by
(\ref{sol.Lambda.verschiebung.f.step3.step2}))}}}\right)  \left(  m_{\lambda
}^{\left[   \ZZ \right]  }\right)  =\underbrace{\left(  \Lambda^{\left[
\rho_{ \kk }\right]  }\circ\operatorname{id}_{\Lambda
_{ \ZZ }}^{\star n}\right)  }_{\substack{=
\id_{\Lambda_{ \kk }}^{\star n}\circ\Lambda^{\left[  \rho
_{ \kk }\right]  }\\\text{(because of }
\id_{\Lambda_{ \kk }}^{\star n}\circ\Lambda^{\left[  \rho
_{ \kk }\right]  }=\Lambda^{\left[  \rho_{ \kk }\right]  }%
\circ\operatorname{id}_{\Lambda_{ \ZZ }}^{\star n}\\\text{(by
(\ref{sol.Lambda.verschiebung.f.step2.func-idn}), applied}\\\text{to
}\mathbf{m=} \ZZ \text{, }\mathbf{n}= \kk \text{ and }\varphi
=\rho_{ \kk }\text{))}}}\left(  m_{\lambda}^{\left[   \ZZ \right]
}\right) \\
&  =\left(  \operatorname{id}_{\Lambda_{ \kk }}^{\star n}
\circ\Lambda^{\left[  \rho_{ \kk }\right]  }\right)  \left(  m_{\lambda
}^{\left[   \ZZ \right]  }\right)  =\underbrace{
\id_{\Lambda_{ \kk }}^{\star n}}_{=\id^{\star n}}
\left(  \underbrace{\Lambda^{\left[  \rho_{ \kk }\right]  }
\left(  m_{\lambda}^{\left[   \ZZ \right]  }\right)
}_{\substack{=m_{\lambda}\\\text{(by
(\ref{sol.Lambda.verschiebung.f.step3.goal.pf.1}))}}}\right)
=\id^{\star n}\left(  m_{\lambda}\right)  .
\end{align*}
Thus, (\ref{sol.Lambda.verschiebung.f.step3.goal}) is proven.

As we said, proving (\ref{sol.Lambda.verschiebung.f.step3.goal}) is sufficient
to showing that $\mathbf{v}_{n}\circ\mathbf{f}_{n} = \id^{\star n}$.
Hence, $\mathbf{v}_{n}\circ\mathbf{f}_{n} = \id^{\star n}$ is shown (since
(\ref{sol.Lambda.verschiebung.f.step3.goal}) is proven). In other words,
Exercise \ref{exe.Lambda.verschiebung}(g) holds in the general case. This
completes Step 3 and, with it, the solution of Exercise
\ref{exe.Lambda.verschiebung}(g).

(h) The structure of our solution of Exercise \ref{exe.Lambda.verschiebung}(h)
is similar to that of our solution of
Exercise~\ref{exe.Lambda.verschiebung}(g) above. It proceeds in three steps:

\begin{itemize}
\item \textit{Step 1:} proving that Exercise \ref{exe.Lambda.verschiebung}(h)
holds if $\kk = \QQ$.

\item \textit{Step 2:} proving that Exercise \ref{exe.Lambda.verschiebung}(h)
holds if $\kk = \ZZ$.

\item \textit{Step 3:} proving that Exercise \ref{exe.Lambda.verschiebung}(h)
holds in the general case.
\end{itemize}

The details of Steps 2 and 3 are very similar to the details of the
corresponding steps in the solution of Exercise \ref{exe.Lambda.verschiebung}%
(g), and so we will not dwell on these details. However, we need to give the
details of Step 1:

\underline{\textit{Step 1:}} We shall prove that Exercise
\ref{exe.Lambda.verschiebung}(h) holds if $\kk = \QQ$.

Indeed, assume that $\kk = \QQ$. We know that $\mathbf{f}_{n}$ is
a $\kk$-algebra homomorphism (due to Exercise \ref{exe.witt.ghost-app}%
(a)), and that $\mathbf{v}_{m}$ is a $\kk$-algebra homomorphism. Hence,
$\mathbf{f}_{n}\circ\mathbf{v}_{m}$ and $\mathbf{v}_{m}\circ\mathbf{f}_{n}$
are $\kk$-algebra homomorphisms.

Let $r$ be a positive integer. We shall show that $\left(  \mathbf{f}_{n}%
\circ\mathbf{v}_{m}\right)  \left(  p_{r}\right)  =\left(  \mathbf{v}_{m}%
\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)  $.

Indeed, we first notice that%
\begin{equation}
\mathbf{f}_{n}\left(  p_{r}\right)  =p_{rn}.
\label{sol.Lambda.verschiebung.g.step1.fnpr}%
\end{equation}
(This can be proven just as in (\ref{sol.Lambda.verschiebung.f.step1.fnpr}).)

Now, let us first assume that $m\nmid r$. Then, Exercise
\ref{exe.Lambda.verschiebung}(a) (applied to $m$ and $r$ instead of $n$ and
$m$) yields%
\[
\mathbf{v}_{m}\left(  p_{r}\right)  =
\begin{cases}
mp_{r/m},& \text{if }m\mid r;\\
0,& \text{if }m\nmid r
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m\nmid r\right)  ,
\]
so that $\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(
p_{r}\right)  =\mathbf{f}_{n}\left(  \underbrace{\mathbf{v}_{m}\left(
p_{r}\right)  }_{=0}\right)  =\mathbf{f}_{n}\left(  0\right)  =0$ (since the
map $\mathbf{f}_{n}$ is $\kk$-linear). On the other hand, if we had
$m\mid rn$, then we would have $m\mid r$ (since $m$ is coprime to $n$), which
would contradict $m\nmid r$. Hence, we cannot have $m\mid rn$. Thus, we have
$m\nmid rn$. Now,%
\begin{align*}
\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)   &
=\mathbf{v}_{m}\left(  \underbrace{\mathbf{f}_{n}\left(  p_{r}\right)
}_{\substack{=p_{rn}\\\text{(by (\ref{sol.Lambda.verschiebung.g.step1.fnpr}%
))}}}\right)  =\mathbf{v}_{m}\left(  p_{rn}\right)  =
\begin{cases}
mp_{rn/m},& \text{if }m\mid rn;\\
0,& \text{if }m\nmid rn
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.Lambda.verschiebung}%
(a), applied to }m\text{ and }rn\text{ instead of }n\text{ and }m\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m\nmid rn\right)  .
\end{align*}
Compared with $\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(
p_{r}\right)  =0$, this yields $\left(  \mathbf{f}_{n}\circ\mathbf{v}%
_{m}\right)  \left(  p_{r}\right)  =\left(  \mathbf{v}_{m}\circ\mathbf{f}%
_{n}\right)  \left(  p_{r}\right)  $.

Now, let us forget our assumption that $m\nmid r$. Hence, we have proven that
$\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(  p_{r}\right)
=\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)  $
under the assumption that $m\nmid r$. As a consequence, for the rest of the
proof of $\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(
p_{r}\right)  =\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)  \left(
p_{r}\right)  $, we can WLOG assume that we don't have $m\nmid r$. Assume this.

We have $m\mid r$ (since we don't have $m\nmid r$). Now, Exercise
\ref{exe.Lambda.verschiebung}(a) (applied to $m$ and $r$ instead of $n$ and
$m$) yields%
\[
\mathbf{v}_{m}\left(  p_{r}\right)  =
\begin{cases}
mp_{r/m},& \text{if }m\mid r;\\
0,& \text{if }m\nmid r
\end{cases}  \quad
=mp_{r/m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m\mid r\right)  ,
\]
so that%
\begin{align*}
\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(  p_{r}\right)   &
=\mathbf{f}_{n}\left(  \underbrace{\mathbf{v}_{m}\left(  pr\right)
}_{=mp_{r/m}}\right)  =\mathbf{f}_{n}\left(  mp_{r/m}\right)
=m\underbrace{\mathbf{f}_{n}\left(  p_{r/m}\right)  }_{\substack{=p_{\left(
r/m\right)  n}\\\text{(by (\ref{sol.Lambda.verschiebung.g.step1.fnpr}),
applied to}\\r/m\text{ instead of }r\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\mathbf{f}_{n}\text{ is } \kk \text{-linear}\right)
\\
&  =mp_{\left(  r/m\right)  n}=mp_{rn/m}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  r/m\right)  n=rn/m\right)  .
\end{align*}
Compared with%
\begin{align*}
\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)   &
=\mathbf{v}_{m}\left(  \underbrace{\mathbf{f}_{n}\left(  p_{r}\right)
}_{\substack{=p_{rn}\\\text{(by (\ref{sol.Lambda.verschiebung.g.step1.fnpr}%
))}}}\right)  =\mathbf{v}_{m}\left(  p_{rn}\right)  =
\begin{cases}
mp_{rn/m},& \text{if }m\mid rn;\\
0,& \text{if }m\nmid rn
\end{cases} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.Lambda.verschiebung}%
(a), applied to }m\text{ and }rn\text{ instead of }n\text{ and }m\right) \\
&  =mp_{rn/m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m\mid rn\text{ (since
}m\mid r\mid rn\text{)}\right)  ,
\end{align*}
this yields $\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(
p_{r}\right)  =\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)  \left(
p_{r}\right)  $. Hence, $\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)
\left(  p_{r}\right)  =\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)
\left(  p_{r}\right)  $ is proven.

Now, let us forget that we fixed $r$. We thus have proven that%
\begin{equation}
\left(  \mathbf{f}_{n}\circ\mathbf{v}_{m}\right)  \left(  p_{r}\right)
=\left(  \mathbf{v}_{m}\circ\mathbf{f}_{n}\right)  \left(  p_{r}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }r\text{.}
\label{sol.Lambda.verschiebung.g.step1.main}%
\end{equation}


We have assumed that $\kk = \QQ$.
Thus, $\QQ$ is a subring of $\kk$.
Hence, the elements $p_{1}%
,p_{2},p_{3},\ldots$ of $\Lambda$ generate the $\kk$-algebra $\Lambda$
(due to Proposition \ref{symm-antipode-and-h-basis}). But $\mathbf{f}_{n}%
\circ\mathbf{v}_{m}$ and $\mathbf{v}_{m}\circ\mathbf{f}_{n}$ are two
$\kk$-algebra homomorphisms from $\Lambda$. These two homomorphisms
$\mathbf{f}_{n}\circ\mathbf{v}_{m}$ and $\mathbf{v}_{m}\circ\mathbf{f}_{n}$
are equal to each other on each of the elements $p_{1},p_{2},p_{3},\ldots$
(due to (\ref{sol.Lambda.verschiebung.g.step1.main})), and therefore are
identical (because if two $\kk$-algebra homomorphisms with one and the
same domain are equal to each other on a generating set of the domain, then
these homomorphisms must be identical). In other words, $\mathbf{f}_{n}%
\circ\mathbf{v}_{m}=\mathbf{v}_{m}\circ\mathbf{f}_{n}$. Thus, Exercise
\ref{exe.Lambda.verschiebung}(h) is solved under the assumption that
$\kk = \QQ$. Our Step 1 is complete.

As already mentioned, Steps 2 and Steps 3 are very similar to the
corresponding steps in our above solution of Exercise
\ref{exe.Lambda.verschiebung}(g). Thus, we forego showing these steps.
Exercise \ref{exe.Lambda.verschiebung}(h) is thus solved.

(i) Our solution of Exercise \ref{exe.Lambda.verschiebung}(i) will be somewhat
similar to that of Exercise \ref{exe.Lambda.verschiebung}(g), but simpler. We
will only need two steps:

\begin{itemize}
\item \textit{Step 1:} proving that Exercise \ref{exe.Lambda.verschiebung}(i)
holds if $\kk = \ZZ$.

\item \textit{Step 2:} proving that Exercise \ref{exe.Lambda.verschiebung}(i)
holds in the general case.
\end{itemize}

Let us go through the details of each step:

\underline{\textit{Step 1:}} We shall prove that Exercise
\ref{exe.Lambda.verschiebung}(i) holds if $\kk = \ZZ$.

Indeed, assume that $\kk = \ZZ$. Every positive integer $r$
satisfies%
\begin{equation}
p_{r}=\sum_{d\mid r}dw_{d}^{r/d} \label{sol.Lambda.verschiebung.h.step1.witte}%
\end{equation}
(according to Exercise \ref{exe.witt}(e), applied to $r$ instead of $n$).

Now, for every positive integer $m$, we define an element $\widetilde{w}_{m}$
of $\Lambda$ by
\begin{equation}
\widetilde{w}_{m}=
\begin{cases}
w_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad .
\label{sol.Lambda.verschiebung.h.step1.wtilde}
\end{equation}


It is now easy to see that every positive integer $r$ satisfies%
\begin{equation}
\mathbf{v}_{n}\left(  p_{r}\right)  =\sum_{d\mid r}d\widetilde{w}_{d}^{r/d}.
\label{sol.Lambda.verschiebung.h.step1.witttilde}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.h.step1.witttilde}):}
Let $r$ be a positive integer. Let us first assume that $n\nmid r$. Then,
every positive divisor $d$ of $r$ satisfies%
\begin{align}
\widetilde{w}_{d}  &  =
\begin{cases}
w_{d/n},& \text{if }n\mid d;\\
0,& \text{if }n\nmid d
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\widetilde{w}_{d}\right) \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left( \begin{array}{c}
\text{since }n\nmid d\text{ (because
otherwise, we would have }n\mid d \\
\text{ and thus }n\mid d\mid r\text{, which
would contradict }n\nmid r\text{)}
\end{array}
\right)  .
\label{sol.Lambda.verschiebung.h.step1.witttilde.pf.1}
\end{align}
But Exercise \ref{exe.Lambda.verschiebung}(a) (applied to $r$ instead of $m$)
yields
\[
\mathbf{v}_{n}\left(  p_{r}\right)  =
\begin{cases}
np_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\nmid r\right)  .
\]
Compared with
\[
\sum_{d\mid r}d\widetilde{w}_{d}^{r/d}=\sum_{d\mid r}d\left(
\underbrace{\widetilde{w}_{d}}_{\substack{=0\\\text{(by
(\ref{sol.Lambda.verschiebung.h.step1.witttilde.pf.1}))}}}\right)  ^{r/d}%
=\sum_{d\mid r}d\underbrace{0^{r/d}}_{\substack{=0\\\text{(since
}r/d>0\text{)}}}=\sum_{d\mid r}d0=0,
\]
this yields $\mathbf{v}_{n}\left(  p_{r}\right)  =\sum_{d\mid r}%
d\widetilde{w}_{d}^{r/d}$. Thus,
(\ref{sol.Lambda.verschiebung.h.step1.witttilde}) holds.
\par
Let us now forget that we assumed that $n\nmid r$. We thus have proven
(\ref{sol.Lambda.verschiebung.h.step1.witttilde}) under the assumption that
$n\nmid r$. Hence, for the rest of the proof of
(\ref{sol.Lambda.verschiebung.h.step1.witttilde}), we can WLOG assume that we
don't have $n\nmid r$. Assume this.
\par
We have $n\mid r$ (since we don't have $n\nmid r$). Hence, $r/n$ is a positive
integer. Exercise \ref{exe.Lambda.verschiebung}(a) (applied to $r$ instead of
$m$) yields%
\begin{align*}
\mathbf{v}_{n}\left(  p_{r}\right)   &  =
\begin{cases}
np_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}  \quad
=n\underbrace{p_{r/n}}_{\substack{=\sum_{d\mid r/n}dw_{d}^{\left(
r/n\right)  /d}\\\text{(by (\ref{sol.Lambda.verschiebung.h.step1.witte}),
applied to}\\r/n\text{ instead of }r\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }n\mid r\right) \\
&  =n\sum_{d\mid r/n}dw_{d}^{\left(  r/n\right)  /d}=\sum_{d\mid
r/n}nd\underbrace{w_{d}^{\left(  r/n\right)  /d}}_{\substack{=w_{d}^{r/\left(
nd\right)  }\\\text{(since }\left(  r/n\right)  /d=r/\left(  nd\right)
\text{)}}}=\sum_{d\mid r/n}ndw_{d}^{r/\left(  nd\right)  }.
\end{align*}
Compared with%
\begin{align*}
\sum_{d\mid r}d\widetilde{w}_{d}^{r/d}  &  =\sum_{d\mid r}d\left(
\underbrace{\widetilde{w}_{d}}_{\substack{=
\begin{cases}
w_{d/n},& \text{if }n\mid d;\\
0,& \text{if }n\nmid d
\end{cases}
\\\text{(by the definition of }\widetilde{w}_{d}\text{)}}}\right)
^{r/d}=\sum_{d\mid r}d\left(
\begin{cases}
w_{d/n},& \text{if }n\mid d;\\
0,& \text{if }n\nmid d
\end{cases}
\right)  ^{r/d}\\
&  =\sum\limits_{\substack{d\mid r;\\n\mid d}}d\left(  \underbrace{
\begin{cases}
w_{d/n},& \text{if }n\mid d;\\
0,& \text{if }n\nmid d
\end{cases}
}_{\substack{=w_{d/n}\\\text{(since }n\mid d\text{)}}}\right)
^{r/d}+\sum\limits_{\substack{d\mid r;\\n\nmid d}}d\left(  \underbrace{
\begin{cases}
w_{d/n},& \text{if }n\mid d;\\
0,& \text{if }n\nmid d
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid d\text{)}}}\right)  ^{r/d}\\
&  =\sum\limits_{\substack{d\mid r;\\n\mid d}}dw_{d/n}^{r/d}+\sum\limits_{\substack{d\mid
r;\\n\nmid d}}d\underbrace{0^{r/d}}_{\substack{=0\\\text{(since }%
r/d>0\text{)}}}=\sum\limits_{\substack{d\mid r;\\n\mid d}}dw_{d/n}^{r/d}%
+\underbrace{\sum\limits_{\substack{d\mid r;\\n\nmid d}}d0}_{=0}=\sum\limits
_{\substack{d\mid r;\\n\mid d}}dw_{d/n}^{r/d}\\
&  =\sum_{d\mid r/n}nd\underbrace{w_{nd/n}^{r/\left(  nd\right)  }%
}_{\substack{=w_{d}^{r/\left(  nd\right)  }\\\text{(since }nd/n=d\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }nd\text{ for
}d\text{ in the sum}\right) \\
&  =\sum_{d\mid r/n}ndw_{d}^{r/\left(  nd\right)  },
\end{align*}
this yields $\mathbf{v}_{n}\left(  p_{r}\right)  =\sum_{d\mid r}%
d\widetilde{w}_{d}^{r/d}$. Thus,
(\ref{sol.Lambda.verschiebung.h.step1.witttilde}) is proven.}

We shall now prove that every positive integer $m$ satisfies%
\begin{equation}
\mathbf{v}_{n}\left(  w_{m}\right)  =\widetilde{w}_{m}.
\label{sol.Lambda.verschiebung.h.step1.main}%
\end{equation}


\textit{Proof of (\ref{sol.Lambda.verschiebung.h.step1.main}):} We proceed by
strong induction over $m$:

\textit{Induction step:} Fix a positive integer $M$. Assume that
(\ref{sol.Lambda.verschiebung.h.step1.main}) holds for every positive integer
$m<M$. We now must show that (\ref{sol.Lambda.verschiebung.h.step1.main})
holds for $m=M$.

The $\kk$-module $\Lambda$ is free, and thus torsionfree. Hence,
\begin{equation}
\text{every element }a\text{ of }\Lambda\text{ satisfying }Ma=0\text{
satisfies }a=0 \label{sol.Lambda.verschiebung.h.step1.main.pf.torfree}%
\end{equation}
(since $\kk = \ZZ$).

We have assumed that (\ref{sol.Lambda.verschiebung.h.step1.main}) holds for
every positive integer $m<M$. In other words,%
\begin{equation}
\mathbf{v}_{n}\left(  w_{m}\right)  =\widetilde{w}_{m}%
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }m<M\text{.}
\label{sol.Lambda.verschiebung.h.step1.main.pf.indhyp}%
\end{equation}


Applying (\ref{sol.Lambda.verschiebung.h.step1.witte}) to $r=M$, we obtain
$p_{M}=\sum_{d\mid M}dw_{d}^{M/d}$. Applying the map $\mathbf{v}_{n}$ to both
sides of this equality, we obtain%
\begin{align*}
\mathbf{v}_{n}\left(  p_{M}\right)   &  =\mathbf{v}_{n}\left(  \sum_{d\mid
M}dw_{d}^{M/d}\right)  =\sum_{d\mid M}d\left(  \mathbf{v}_{n}\left(
w_{d}\right)  \right)  ^{M/d}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{v}_{n}\text{ is a } \kk \text{-algebra homomorphism}\right) \\
&  =M\underbrace{\left(  \mathbf{v}_{n}\left(  w_{M}\right)  \right)  ^{M/M}%
}_{=\left(  \mathbf{v}_{n}\left(  w_{M}\right)  \right)  ^{1}=\mathbf{v}%
_{n}\left(  w_{M}\right)  }+\underbrace{\sum\limits_{\substack{d\mid M;\\d\neq M}%
}}_{=\sum\limits_{\substack{d\mid M;\\d<M}}}d\left(  \mathbf{v}_{n}\left(
w_{d}\right)  \right)  ^{M/d}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}d=M\text{ from the sum}\right) \\
&  =M\mathbf{v}_{n}\left(  w_{M}\right)  +\sum\limits_{\substack{d\mid M;\\d<M}%
}d\left(  \underbrace{\mathbf{v}_{n}\left(  w_{d}\right)  }%
_{\substack{=\widetilde{w}_{d}\\\text{(by
(\ref{sol.Lambda.verschiebung.h.step1.main.pf.indhyp}), applied}\\\text{to
}m=d\text{)}}}\right)  ^{M/d}
= M\mathbf{v}_{n}\left(  w_{M}\right)  +\sum\limits_{\substack{d\mid M;\\d<M}%
}d\widetilde{w}_{d}^{M/d}.
\end{align*}
Compared with
\begin{align*}
\mathbf{v}_{n}\left(  p_{M}\right)   &  =\sum_{d\mid M}d\widetilde{w}%
_{d}^{M/d}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.verschiebung.h.step1.witttilde}), applied to }r=M\right) \\
&  =M\underbrace{\widetilde{w}_{M}^{M/M}}_{=\widetilde{w}_{M}^{1}%
=\widetilde{w}_{M}}+\underbrace{\sum\limits_{\substack{d\mid M;\\d\neq M}}}%
_{=\sum\limits_{\substack{d\mid M;\\d<M}}}d\widetilde{w}_{d}^{M/d}
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}d=M\text{ from the sum}\right) \\
&  =M\widetilde{w}_{M}+\sum\limits_{\substack{d\mid M;\\d<M}}d\widetilde{w}_{d}%
^{M/d},
\end{align*}
this yields
\[
M\mathbf{v}_{n}\left(  w_{M}\right)  +\sum\limits_{\substack{d\mid M;\\d<M}%
}d\widetilde{w}_{d}^{M/d}=M\widetilde{w}_{M}+\sum\limits_{\substack{d\mid
M;\\d<M}}d\widetilde{w}_{d}^{M/d}.
\]
Subtracting $\sum\limits_{\substack{d\mid M;\\d<M}}d\widetilde{w}_{d}^{M/d}$ from
both sides of this equality, we obtain $M\mathbf{v}_{n}\left(  w_{M}\right)
=M\widetilde{w}_{M}$. Hence,
\[
M\left(  \mathbf{v}_{n}\left(  w_{M}\right)
-\widetilde{w}_{M}\right)  =\underbrace{M\mathbf{v}_{n}\left(  w_{M}\right)
}_{=M\widetilde{w}_{M}}-M\widetilde{w}_{M}=M\widetilde{w}_{M}-M\widetilde{w}%
_{M}=0 .
\]
Thus, (\ref{sol.Lambda.verschiebung.h.step1.main.pf.torfree})
(applied to $a=\mathbf{v}_{n}\left(  w_{M}\right)  -\widetilde{w}_{M}$) yields
$\mathbf{v}_{n}\left(  w_{M}\right)  -\widetilde{w}_{M}=0$, so that
$\mathbf{v}_{n}\left(  w_{M}\right)  =\widetilde{w}_{M}$. In other words,
(\ref{sol.Lambda.verschiebung.h.step1.main}) holds for $m=M$. This completes
the induction step. Thus, (\ref{sol.Lambda.verschiebung.h.step1.main}) is proven.

Now, every positive integer $m$ satisfies%
\begin{align*}
\mathbf{v}_{n}\left(  w_{m}\right)   &  =\widetilde{w}_{m}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.verschiebung.h.step1.main})}\right) \\
&  =
\begin{cases}
w_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\widetilde{w}_{m}\right)  .
\end{align*}
Thus, Exercise \ref{exe.Lambda.verschiebung}(i) is solved in the case when
$\kk = \ZZ$. In other words, Step 1 is finished.

\underline{\textit{Step 2:}} We shall now show that Exercise
\ref{exe.Lambda.verschiebung}(i) holds in the general case.

Indeed, let us use all the notations that we introduced in Step 2 of the
solution of Exercise \ref{exe.Lambda.verschiebung}(g). Let us furthermore
introduce one more notation:

\begin{itemize}
\item For every positive integer $m$, we shall denote the element $w_{m}$ of
$\Lambda$ by $w_{m}^{\left[   \kk \right]  }$. This notation makes the
dependency of $w_{m}$ on the base ring $\kk$ more explicit. Of course,
the element $w_{m}$ does not ``really'' depend on $\kk$, in the
sense that it is defined in the same way for
every $\kk$. This entails that for any two commutative rings
$\mathbf{m}$ and $\mathbf{n}$ and any ring homomorphism $\varphi
:\mathbf{m}\rightarrow\mathbf{n}$, we have%
\begin{equation}
w_{m}^{\left[  \mathbf{n}\right]  }=\Lambda^{\left[  \varphi\right]  }\left(
w_{m}^{\left[  \mathbf{m}\right]  }\right)
\label{sol.Lambda.verschiebung.h.step2.func-w}%
\end{equation}
for every positive integer $m$.
\end{itemize}

Now, fix a positive integer $m$. In Step 1, we have shown that Exercise
\ref{exe.Lambda.verschiebung}(i) holds if $\kk = \ZZ$. In other
words, we have%
\[
\mathbf{v}_{n}\left(  w_{m}\right)  =
\begin{cases}
w_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{if } \kk = \ZZ \text{.}
\]
In other words,
\begin{equation}
\mathbf{v}_{n}^{\left[   \ZZ \right]  }\left(  w_{m}^{\left[
 \ZZ \right]  }\right)  =
\begin{cases}
w_{m/n}^{\left[   \ZZ \right]  },& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad
. \label{sol.Lambda.verschiebung.h.step2.step1}
\end{equation}


Now,%
\begin{equation}
w_{m}=w_{m}^{\left[   \kk \right]  }=\Lambda^{\left[  \rho_{ \kk %
}\right]  }\left(  w_{m}^{\left[   \ZZ \right]  }\right)
\label{sol.Lambda.verschiebung.h.step2.wm}%
\end{equation}
(by (\ref{sol.Lambda.verschiebung.h.step2.func-w}) (applied to $\mathbf{m}%
= \ZZ$, $\mathbf{n}= \kk$ and $\varphi=\rho_{ \kk }$)). Thus,%
\begin{align}
\underbrace{\mathbf{v}_{n}}_{=\mathbf{v}_{n}^{\left[   \kk \right]  }%
}\left(  \underbrace{w_{m}}_{=\Lambda^{\left[  \rho_{ \kk }\right]
}\left(  w_{m}^{\left[   \ZZ \right]  }\right)  }\right)   &
=\mathbf{v}_{n}^{\left[   \kk \right]  }\left(  \Lambda^{\left[
\rho_{ \kk }\right]  }\left(  w_{m}^{\left[   \ZZ \right]  }\right)
\right)  =\underbrace{\left(  \mathbf{v}_{n}^{\left[   \kk \right]
}\circ\Lambda^{\left[  \rho_{ \kk }\right]  }\right)  }%
_{\substack{=\Lambda^{\left[  \rho_{ \kk }\right]  }\circ\mathbf{v}%
_{n}^{\left[   \ZZ \right]  }\\\text{(by
(\ref{sol.Lambda.verschiebung.f.step2.func-v}), applied to}\\\mathbf{m}%
= \ZZ \text{, }\mathbf{n}= \kk \text{ and }\varphi=\rho_{ \kk %
}\text{)}}}\left(  w_{m}^{\left[   \ZZ \right]  }\right) \nonumber\\
&  =\left(  \Lambda^{\left[  \rho_{ \kk }\right]  }\circ\mathbf{v}%
_{n}^{\left[   \ZZ \right]  }\right)  \left(  w_{m}^{\left[
 \ZZ \right]  }\right)  =\Lambda^{\left[  \rho_{ \kk }\right]
}\left(  \underbrace{\mathbf{v}_{n}^{\left[   \ZZ \right]  }\left(
w_{m}^{\left[   \ZZ \right]  }\right)  }_{\substack{=
\begin{cases}
w_{m/n}^{\left[   \ZZ \right]  },& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\\\text{(by (\ref{sol.Lambda.verschiebung.h.step2.step1}))}}}\right)
\nonumber\\
&  =\Lambda^{\left[  \rho_{ \kk }\right]  }\left(
\begin{cases}
w_{m/n}^{\left[   \ZZ \right]  },& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\right)  =
\begin{cases}
\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  w_{m/n}^{\left[
 \ZZ \right]  }\right)  ,& \text{if }n\mid m;\\
\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  0\right)
,& \text{if }n\nmid m
\end{cases}
\nonumber\\
&  =
\begin{cases}
\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  w_{m/n}^{\left[
 \ZZ \right]  }\right)  ,& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\label{sol.Lambda.verschiebung.h.step2.5}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  0\right)
=0\text{ (because }\Lambda^{\left[  \rho_{ \kk }\right]  }\text{ is a
ring homomorphism)}\\
\text{in the case when }n\nmid m
\end{array}
\right)  .  \nonumber
\end{align}


On the other hand, we have
\begin{equation}
\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  w_{m/n}^{\left[
 \ZZ \right]  }\right)  =w_{m/n}\ \ \ \ \ \ \ \ \ \ \text{in the case
when }n\mid m \label{sol.Lambda.verschiebung.h.step2.wm2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.verschiebung.h.step2.wm2}):}
Assume that $n\mid m$. Then, $m/n$ is a positive integer. Hence,
$w_{m/n}=\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  w_{m/n}^{\left[
 \ZZ \right]  }\right)  $ (in fact,
this follows from the same argument that was
used to prove (\ref{sol.Lambda.verschiebung.h.step2.wm}), but with $m$
replaced by $m/n$). In other words, $\Lambda^{\left[  \rho_{ \kk %
}\right]  }\left(  w_{m/n}^{\left[   \ZZ \right]  }\right)  =w_{m/n}$.
This proves (\ref{sol.Lambda.verschiebung.h.step2.wm2}).}. Now,
(\ref{sol.Lambda.verschiebung.h.step2.5}) becomes%
\begin{align*}
\mathbf{v}_{n}\left(  w_{m}\right)   &  =
\begin{cases}
\Lambda^{\left[  \rho_{ \kk }\right]  }\left(  w_{m/n}^{\left[
 \ZZ \right]  }\right)  ,& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}  \quad
=
\begin{cases}
w_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\Lambda^{\left[  \rho_{ \kk }\right]  }\left(
w_{m/n}^{\left[   \ZZ \right]  }\right)  =w_{m/n}\text{ (according to
(\ref{sol.Lambda.verschiebung.h.step2.wm2}))}\\
\text{in the case when }n\mid m
\end{array}
\right)  .
\end{align*}
Thus, Exercise \ref{exe.Lambda.verschiebung}(i) holds. This completes Step 2,
and thus Exercise \ref{exe.Lambda.verschiebung}(i) is solved.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.shareshian-wachs.path}}

\textit{Solution to Exercise \ref{exe.shareshian-wachs.path}.} (b) Let
us first notice that
\begin{equation}
X_{n,d,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
\ \ \ \ \ \ \ \ \ \ \text{for all }d\in{ \NN } \text{ and }%
s \in \NN .
\label{sol.shareshian-wachs.path.b.def}%
\end{equation}
(This is just the definition of $X_{n,d,s}$, written as a formula.)

Let us first show that if $n$ is a positive integer, then any $d\in \NN$
and $s\in \NN$ satisfy
\begin{align}
&  \left(  d+1\right)  X_{n,d+1,s}+\left(  s+1\right)  X_{n,d,s+1}+\left(
n-1-d-s\right)  X_{n,d,s}\nonumber\\
&  =\sum_{i=1}^{n-1}\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}.
\label{sol.shareshian-wachs.path.b.rec}
\end{align}


\textit{Proof of (\ref{sol.shareshian-wachs.path.b.rec}):} Let $n$ be a
positive integer. Let $d \in \NN$ and $s \in \NN$. Let
$i\in\left\{  1,2,...,n-1\right\}  $ be arbitrary. We make some more definitions:

\begin{itemize}
\item Define a power series $\mathcal{D}_{i}\in \kk  \left[  \left[ 
 \xx \right]  \right]  $ as the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d+1$, $\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s$ and $w_{i}>w_{i+1}$.

\item Define a power series $\mathcal{S}_{i}\in \kk  \left[  \left[ 
 \xx \right]  \right]  $ as the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$, $\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s+1$ and $w_{i}=w_{i+1}$.

\item Define a power series $\mathcal{A}_{i}\in \kk  \left[  \left[ 
 \xx \right]  \right]  $ as the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$, $\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s$ and $w_{i}<w_{i+1}$.
\end{itemize}

These three power series $\mathcal{D}_{i}$, $\mathcal{S}_{i}$ and
$\mathcal{A}_{i}$ are not necessarily symmetric (but will nevertheless come
useful). Let us notice that the definitions of $\mathcal{D}_{i}$,
$\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ can be rewritten as follows:

\begin{itemize}
\item The power series $\mathcal{D}_{i}\in \kk \left[  \left[ 
 \xx \right]  \right]  $ is the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{i}>w_{i+1}$.

\item The power series $\mathcal{S}_{i}\in \kk \left[  \left[ 
 \xx \right]  \right]  $ is the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{i}=w_{i+1}$.

\item The power series $\mathcal{A}_{i}\in \kk \left[ \left[
 \xx \right]  \right]  $ is the sum of the monomials $x_{w_{1}}
 x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{i}<w_{i+1}$.
\end{itemize}

These reformulations make it obvious that the sum $\mathcal{D}_{i}%
+\mathcal{S}_{i}+\mathcal{A}_{i}$ is precisely the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$ and $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$. In other words,
\begin{equation}
\mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}_{i}=\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}
;\\\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d;\ \left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.b.rec.pf.1}
\end{equation}


But the sum on the right hand side of
(\ref{sol.shareshian-wachs.path.b.rec.pf.1}) can be rewritten in terms of more
familiar sums. In fact, the $n$-tuples $w\in\left\{  1,2,3,\ldots\right\}
^{n}$ are in bijection with the pairs $\left(  u,v\right)  $ consisting of an
$i$-tuple $u\in\left\{  1,2,3,\ldots\right\}  ^{i}$ and an $\left(
n-i\right)  $-tuple $v\in\left\{  1,2,3,\ldots\right\}  ^{n-i}$. This
bijection sends an $n$-tuple $\left(  w_{1},w_{2},...,w_{n}\right)  $ to the
pair $\left(  \left(  w_{1},w_{2},...,w_{i}\right)  ,\left(  w_{i+1}%
,w_{i+2},...,w_{n}\right)  \right)  $, and has the property that $\left\vert
\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =\left\vert \operatorname*{Des}\left(  u\right)
\right\vert +\left\vert \operatorname*{Des}\left(  v\right)  \right\vert
$\ \ \ \ \footnote{In fact, $\left(  \operatorname*{Des}\left(  w\right)
\right)  \setminus\left\{  i\right\}  $ is the union of the set
$\operatorname*{Des}\left(  u\right)  $ with the set $\operatorname*{Des}
\left(  v\right)  $ shifted by $i$ (that is, the set $\left\{  p+i\ \mid
\ p\in\operatorname*{Des}\left(  v\right)  \right\}  $). This is a disjoint
union, and thus we find
$\left\vert
\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =\left\vert \operatorname*{Des}\left(  u\right)
\right\vert +\left\vert \operatorname*{Des}\left(  v\right)  \right\vert
$ (since the set $\operatorname*{Des} \left(  v\right)  $ shifted by $i$
has cardinality $\left\vert \operatorname*{Des}\left( v \right)
\right\vert$).
} and $\left\vert
\left(  \operatorname*{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =\left\vert \operatorname*{Stag}\left(  u\right)
\right\vert +\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert $
\ \ \ \ \footnote{for similar reasons}. Hence,
\begin{align}
&  \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\underbrace{\sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\v=\left(  v_{1},v_{2},...,v_{n-i}%
\right)  \in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert =d;\\\left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert =s}}}_{=\sum_{e=0}^{d}%
\sum_{t=0}^{s}\sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert =e;\ \left\vert \operatorname*{Stag}\left(  u\right)
\right\vert =t}}\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)
\in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(
v\right)  \right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}\nonumber\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\sum\limits_{\substack{u=\left(  u_{1},u_{2}%
,\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}\sum\limits
_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}\nonumber\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\left(  \sum\limits_{\substack{u=\left(  u_{1}%
,u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\right)  \left(  \sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}%
\right)  \in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert =d-e;\ \left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert =s-t}}x_{v_{1}}x_{v_{2}%
}\cdots x_{v_{n-i}}\right)  . \label{sol.shareshian-wachs.path.b.rec.pf.2}%
\end{align}
However, for every $e\in\left\{  0,1,...,d\right\}  $ and $t\in\left\{
0,1,...,s\right\}  $, the equality (\ref{sol.shareshian-wachs.path.b.def})
(applied to $i$, $e$ and $t$ instead of $n$, $d$ and $s$) yields
\begin{equation}
X_{i,e,t}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =e;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =t}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{i}}=\sum\limits_{\substack{u=\left(
u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i}%
;\\\left\vert \operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}.
\label{sol.shareshian-wachs.path.b.rec.pf.3a}
\end{equation}
Also, for every $e\in\left\{  0,1,...,d\right\}  $ and $t\in\left\{
0,1,...,s\right\}  $, the equality (\ref{sol.shareshian-wachs.path.b.def})
(applied to $n-i$, $d-e$ and $s-t$ instead of $n$, $d$ and $s$) yields
\begin{equation}
X_{n-i,d-e,s-t}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n-i}\right)
\in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d-e;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s-t}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n-i}}=\sum\limits
_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}.
\label{sol.shareshian-wachs.path.b.rec.pf.3b}
\end{equation}
Now, (\ref{sol.shareshian-wachs.path.b.rec.pf.2}) becomes
\begin{align*}
&  \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\underbrace{\left(  \sum\limits_{\substack{u=\left(
u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i}%
;\\\left\vert \operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\right)  }_{\substack{=X_{i,e,t}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.3a}))}}}\underbrace{\left(
\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}\right)  }%
_{\substack{=X_{n-i,d-e,s-t}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.3b}))}}}\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}.
\end{align*}
Hence, (\ref{sol.shareshian-wachs.path.b.rec.pf.1}) becomes%
\begin{align}
\mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}_{i}
&= \sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}%
;\\\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d;\ \left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}} \nonumber\\
&= \sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}.
\label{sol.shareshian-wachs.path.b.rec.pf.4}
\end{align}


Now, let us forget that we fixed $i$. We thus have defined $\mathcal{D}_{i}$,
$\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ and proven the equality
(\ref{sol.shareshian-wachs.path.b.rec.pf.4}) for all $i\in\left\{
1,2,...,n-1\right\}  $.

Now, let us take a closer look at the sums $\sum_{i=1}^{n-1}\mathcal{D}_{i}$,
$\sum_{i=1}^{n-1}\mathcal{S}_{i}$ and $\sum_{i=1}^{n-1}\mathcal{A}_{i}$:

\begin{itemize}
\item The definition of $\mathcal{D}_{i}$ can be rewritten as follows:%
\[
\mathcal{D}_{i}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\ w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,n-1\right\}  .
\]
Summing up these equations over all $i\in\left\{  1,2,...,n-1\right\}  $, we
obtain%
\begin{align}
\sum_{i=1}^{n-1}\mathcal{D}_{i}  &  =\sum_{i=1}^{n-1}\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}%
;\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\ w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,...,n-1\right\}
;\\w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}}_{\substack{=\left\vert
\left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}>w_{i+1}\right\}
\right\vert x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}\underbrace{\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}>w_{i+1}\right\}  \right\vert }_{=\left\vert \operatorname*{Des}%
w\right\vert =d+1}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\left(  d+1\right)  \underbrace{\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d+1;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}}_{\substack{=X_{n,d+1,s}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.def}), applied to }d+1\text{ instead of
}d\text{)}}}\nonumber\\
&  =\left(  d+1\right)  X_{n,d+1,s}.
\label{sol.shareshian-wachs.path.b.rec.pf.7a}%
\end{align}


\item Similarly, we can see that%
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{S}_{i}=\left(  s+1\right)  X_{n,d,s+1}.
\label{sol.shareshian-wachs.path.b.rec.pf.7b}%
\end{equation}


\item Similarly, we can see that
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{A}_{i}=\left(  n-1-d-s\right)  X_{n,d,s}.
\label{sol.shareshian-wachs.path.b.rec.pf.7c}%
\end{equation}
\footnote{In proving this, we have to observe that $\left\vert \left\{
i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}<w_{i+1}\right\}  \right\vert
=n-1-\left\vert \operatorname*{Des}w\right\vert -\left\vert
\operatorname*{Stag}w\right\vert $ for every $w=\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}$. This is clear from
realizing that the sets $\left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}<w_{i+1}\right\}  $, $\operatorname*{Des}w$ and
$\operatorname*{Stag}w$ are disjoint and their union is $\left\{
1,2,...,n-1\right\}  $.}
\end{itemize}

Now, summing the equality (\ref{sol.shareshian-wachs.path.b.rec.pf.4}) over
all $i\in\left\{  1,2,...,n-1\right\}  $ yields%
\[
\sum_{i=1}^{n-1}\left(  \mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}%
_{i}\right)  =\sum_{i=1}^{n-1}\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}%
X_{n-i,d-e,s-t}.
\]
Hence,%
\begin{align*}
&  \sum_{i=1}^{n-1}\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}\\
&  =\sum_{i=1}^{n-1}\left(  \mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}%
_{i}\right)  =\underbrace{\sum_{i=1}^{n-1}\mathcal{D}_{i}}_{\substack{=\left(
d+1\right)  X_{n,d+1,s}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.7a}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{S}_{i}}_{\substack{=\left(  s+1\right)  X_{n,d,s+1}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.7b}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{A}_{i}}_{\substack{=\left(  n-1-d-s\right)  X_{n,d,s}%
\\\text{(by (\ref{sol.shareshian-wachs.path.b.rec.pf.7c}))}}}\\
&  =\left(  d+1\right)  X_{n,d+1,s}+\left(  s+1\right)  X_{n,d,s+1}+\left(
n-1-d-s\right)  X_{n,d,s}.
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.b.rec}).

In solving Exercise~\ref{exe.shareshian-wachs.path}(b), we shall use
the following trick to simplify our life.
Recall that there is a canonical ring homomorphism $\varphi
: \ZZ \rightarrow \kk$. This homomorphism gives rise to a ring
homomorphism $\varphi\left[  \left[  \xx\right]  \right]
: \ZZ \left[  \left[  \xx\right]  \right]  \rightarrow
 \kk \left[  \left[  \xx\right]  \right]  $, and this latter
homomorphism $\varphi\left[  \left[  \xx\right]  \right]  $ sends
$\Lambda_{\ZZ}$ to $\Lambda_{\kk}$;
that is, we have $\left(  \varphi\left[  \left[
\mathbf{x}  \right]  \right]  \right)  \left(  
\Lambda_{\ZZ} \right)  \subset \Lambda_{\kk}$.
Moreover, it is clear that (for any nonnegative integers $d$ and $s$)
the ring homomorphism
$\varphi\left[  \left[  \xx\right]  \right]  $ sends the element
$X_{n,d,s}$ of $ \ZZ \left[  \left[
\xx\right]  \right]  $ to the element $X_{n,d,s}$ of
$\kk \left[  \left[  \xx\right]  \right]  $
(because the definition of $X_{n,d,s}$ is
functorial in the base ring $\kk$). Therefore, if we can prove
(for given nonnegative integers $d$ and $s$) that the
element $X_{n,d,s}$ of $ \ZZ \left[
\left[  \xx\right]  \right]  $ belongs to $\Lambda_{\ZZ}$,
then it will automatically follow that the element
$X_{n,d,s}$ of $\kk \left[  \left[  \xx\right]  \right]  $
belongs to $\left(  \varphi\left[  \left[  \xx\right]  \right]
\right)  \left(  \Lambda_{\ZZ} \right)
\subset \Lambda_{\kk}$; this will complete the
solution to Exercise~\ref{exe.shareshian-wachs.path}(b).
Hence, in order to solve Exercise~\ref{exe.shareshian-wachs.path}(b),
it only remains to prove (for any nonnegative integers $d$ and $s$)
that the element $X_{n,d,s}$ of $ \ZZ \left[
\left[  \xx\right]  \right]  $ belongs to $
\Lambda_{\ZZ}$. In other words, it only remains to solve
Exercise~\ref{exe.shareshian-wachs.path}(b) in the case of
$\kk = \ZZ$. Hence, in solving
Exercise~\ref{exe.shareshian-wachs.path}(b),
we can WLOG assume that $\kk = \ZZ$. Assume this.

Now, we are going to solve Exercise \ref{exe.shareshian-wachs.path}(b) by
strong induction over $n+d$. So (for the induction step) we need to show that
$X_{n,d,s}\in\Lambda$, and we can assume (as the induction hypothesis) that
$X_{n^{\prime},d^{\prime},s^{\prime}}\in\Lambda$ is already known to hold for
any nonnegative integers $n^{\prime}$, $d^{\prime}$ and $s^{\prime}$
satisfying $n^{\prime}+d^{\prime}<n+d$.

We must be in one of the following two cases:

\textit{Case 1:} We have $d>0$.

\textit{Case 2:} We have $d=0$.

Let us first consider Case 1. In this case, we have $d>0$. Hence,
$d-1 \in \NN$. We also WLOG assume that $n$ is positive (otherwise,
$X_{n,d,s}$ is a constant and thus lies in $\Lambda$ for sure). Applying
(\ref{sol.shareshian-wachs.path.b.rec}) to $d-1$ instead of $d$, we obtain
\[
dX_{n,d,s}+\left(  s+1\right)  X_{n,d-1,s+1}+\left(  n-d-s\right)
X_{n,d-1,s}=\sum_{i=1}^{n-1}\sum_{e=0}^{d-1}\sum_{t=0}^{s}X_{i,e,t}
X_{n-i,d-1-e,s-t}.
\]
Thus,
\begin{equation}
dX_{n,d,s}=\sum_{i=1}^{n-1}\sum_{e=0}^{d-1}\sum_{t=0}^{s}X_{i,e,t}%
X_{n-i,d-1-e,s-t}-\left(  s+1\right)  X_{n,d-1,s+1}+\left(  n-d-s\right)
X_{n,d-1,s}.
\label{sol.shareshian-wachs.path.b.5}
\end{equation}
Each of the power series $X_{i,e,t}$, $X_{n-i,d-1-e,s-t}$, $X_{n,d-1,s+1}$ and
$X_{n,d-1,s}$ on the right hand side of this equality is already known to lie
in $\Lambda$ (by the induction hypothesis). Hence, the right hand side of
(\ref{sol.shareshian-wachs.path.b.5}) lies in $\Lambda$ (since $\Lambda$ is a
ring), and thus (\ref{sol.shareshian-wachs.path.b.5}) shows that
$dX_{n,d,s}\in\Lambda$. Since $d>0$, this yields $X_{n,d,s}\in\Lambda$
(because if a power series $Q \in{ \ZZ }\left[  \left[  {\xx%
}\right]  \right]  $ satisfies $dQ \in\Lambda$, then $Q$ must belong to
$\Lambda$ itself\footnote{It is here that we are using our assumption that
$\kk  = { \ZZ }$.}). Hence, the induction step is complete in Case 1.

Let us now consider Case 2. In this case, we have $d=0$. Hence, $X_{n,d,s} =
X_{n,0,s}$.

We shall now show that
\begin{equation}
X_{n,0,s} = \sum\limits_{\substack{\lambda\in\operatorname{Par}_{n} ;\\\ell\left(
\lambda\right)  =n-s}} m_{\lambda} .
\label{sol.shareshian-wachs.path.b.8}
\end{equation}


\textit{Proof of (\ref{sol.shareshian-wachs.path.b.8}):} Summing up the
equality (\ref{monomial-symmetric-function-definition}) over all $\lambda
\in\operatorname{Par}_{n}$ satisfying $\ell\left(  \lambda\right)  =n-s$, we
obtain%
\begin{equation}
\sum\limits_{\substack{\lambda\in\operatorname{Par}_{n};\\\ell\left(  \lambda\right)
=n-s}}m_{\lambda}=\sum\limits_{\substack{\lambda\in\operatorname{Par}_{n}%
;\\\ell\left(  \lambda\right)  =n-s}}\sum_{\alpha\in\Symm_{\left(
\infty\right)  }\lambda}{\mathbf{x}}^{\alpha}.
\label{sol.shareshian-wachs.path.b.8.pf.1}
\end{equation}


On the other hand, the equality (\ref{sol.shareshian-wachs.path.b.def})
(applied to $0$ instead of $d$) yields%
\begin{equation}
X_{n,0,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =0;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.b.8.pf.3}
\end{equation}
The condition $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =0$
on an $n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ is equivalent to $w_{1}\leq w_{2}\leq...\leq w_{n}$,
and therefore (\ref{sol.shareshian-wachs.path.b.8.pf.3}) rewrites as%
\begin{equation}
X_{n,0,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\w_{1}\leq w_{2}\leq...\leq
w_{n};\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.b.8.pf.4}
\end{equation}
For an $n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfying $w_{1}\leq w_{2}\leq...\leq w_{n}$, the
condition $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$ is
equivalent to the condition that $\left\vert \left\{  w_{1},w_{2}
,...,w_{n}\right\}  \right\vert =n-s$. Hence,
(\ref{sol.shareshian-wachs.path.b.8.pf.4}) rewrites as%
\begin{equation}
X_{n,0,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\w_{1}\leq w_{2}\leq...\leq
w_{n};\ \left\vert \left\{  w_{1},w_{2},...,w_{n}\right\}  \right\vert
=n-s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.b.8.pf.5}
\end{equation}


For any fixed weak composition $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},...\right)  $, the monomial ${\mathbf{x}}^{\alpha}$ occurs
\textbf{at most once} on the right hand side of
(\ref{sol.shareshian-wachs.path.b.8.pf.5}) (because there is at most one way
to write $\mathbf{x}^{\alpha}$ in the form $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}$ for a $w = \left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfying $w_{1}\leq w_{2}\leq...\leq w_{n}$). We
can easily tell whether it occurs or not by looking at the size $\left\vert
\alpha\right\vert $ of $\alpha$ and the number of positive integers $i$
satisfying $\alpha_{i}\neq0$: Namely, the monomial $\mathbf{x}^{\alpha}$ for a
fixed weak composition $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha
_{3},...\right)  $ occurs on the right hand side of
(\ref{sol.shareshian-wachs.path.b.8.pf.5}) if and only if it satisfies the
following two properties:

\begin{itemize}
\item We have $\left\vert \alpha\right\vert =n$.

\item There are precisely $n-s$ positive integers $i$ satisfying $\alpha
_{i}\neq0$.
\end{itemize}

Thus, the monomials satisfying these two properties occur exactly once on the
right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5}), while all other
monomials don't occur there at all. But the same conclusion can be reached for
the right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.1}). Hence, for
any fixed weak composition $\alpha=\left(  \alpha_{1},\alpha_{2},\alpha
_{3},...\right)  $, the monomial $\mathbf{x}^{\alpha}$ occurs on the right
hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5}) precisely as often as
it occurs on the right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.1}%
). Hence, the right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5})
equals the right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.1}).
Therefore, the left hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5})
also equals the left hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.1}).
In other words, $X_{n,0,s}=\sum\limits_{\substack{\lambda\in\operatorname{Par}%
_{n};\\\ell\left(  \lambda\right)  =n-s}}m_{\lambda}$. This proves
(\ref{sol.shareshian-wachs.path.b.8}).

Now, (\ref{sol.shareshian-wachs.path.b.8}) yields $X_{n,0,s}=\sum\limits
_{\substack{\lambda\in\operatorname{Par}_{n};\\\ell\left(  \lambda\right)
=n-s}}\underbrace{m_{\lambda}}_{\in\Lambda}\in\Lambda$, so that $X_{n,d,s}%
=X_{n,0,s}\in\Lambda$. This completes the induction step in Case 2.

Thus, the induction step is complete in both Cases 1 and 2. This finally
completes the induction step. Exercise \ref{exe.shareshian-wachs.path}(b) is
thus solved.

(a) \textit{First solution of Exercise \ref{exe.shareshian-wachs.path}(a):}
Recall the power series $X_{n,d,s}$ defined in Exercise
\ref{exe.shareshian-wachs.path}(b) for any $d\in{ \NN } $ and
$s\in{ \NN } $. Exercise \ref{exe.shareshian-wachs.path}(b) yields that
$X_{n,k,0}\in\Lambda$. But an $n$-tuple $w$ is Smirnov if and only if its
stagnation set $\operatorname*{Stag}\left(  w\right)  $ is empty, i.e., if and
only if $\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =0$.
Hence, $X_{n,k,0}$ is the sum of the monomials $x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}$ over all Smirnov $n$-tuples $w\in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=k$. This shows that $X_{n,k,0}=X_{n,k}$. Thus, $X_{n,k}=X_{n,k,0}\in\Lambda$,
so that Exercise \ref{exe.shareshian-wachs.path}(a) is solved.

\textit{Second solution of Exercise \ref{exe.shareshian-wachs.path}(a):} Here
is an alternative solution to Exercise \ref{exe.shareshian-wachs.path}(a),
which avoids using part (b). It is an application of \cite[proof of Theorem
4.5]{ShareshianWachs-2014} to our special setting.

We proceed similarly to the proof of Proposition
\ref{Schur-functions-are-symmetric-prop}: It suffices to show that for every
positive integer $p$, the power series $X_{n,k}$ is invariant under swapping
the variables $x_{p}$ and $x_{p+1}$. So let us fix a positive integer $p$.

Let $S_{n,k}$ denote the set of all Smirnov $n$-tuples $w\in\left\{
1,2,3,\ldots\right\}  ^{n}$ satisfying $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =k$. Then, the definition of $X_{n,k}$ rewrites as
follows:%
\begin{equation}
X_{n,k}=\sum_{w=\left(  w_{1},w_{2},...,w_{n}\right)  \in S_{n,k}}x_{w_{1}%
}x_{w_{2}}...x_{w_{n}}. \label{sol.shareshian-wachs.path.a.1}%
\end{equation}


For every $w\in S_{n,k}$, let ${\mathbf{x}} _{w}$ denote the monomial
$x_{w_{1}}x_{w_{2}}\ldots x_{w_{n}}$, where $w$ is written in the form
$w=\left(  w_{1},w_{2},...,w_{n}\right)  $. Then,
(\ref{sol.shareshian-wachs.path.a.1}) rewrites as%
\begin{equation}
X_{n,k}=\sum_{w\in S_{n,k}} {\mathbf{x}} _{w}.
\label{sol.shareshian-wachs.path.a.2}%
\end{equation}


We need to show that this power series $X_{n,k}$ is invariant under swapping
the variables $x_{p}$ and $x_{p+1}$. In order to do so, it is clearly enough
to provide an involution $J:S_{n,k}\rightarrow S_{n,k}$ which has the property
that for every $w\in S_{n,k}$, the monomial ${\mathbf{x}} _{J\left(  w\right)
}$ is obtained from the monomial ${\mathbf{x}} _{w}$ by swapping the variables
$x_{p}$ and $x_{p+1}$.

To define the involution $J$, we introduce some notations. If $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}$ is any
$n$-tuple, then a nonempty interval $I$ of $\left\{  1,2,...,n\right\}  $ will
be called a \emph{$\left(  p,p+1\right)  $-interval} of $w$ if every $i\in
I$ satisfies $w_{i}\in\left\{  p,p+1\right\}  $. A
\emph{$\left(  p,p+1\right)$-run} of $w$ will mean a
$\left(  p,p+1\right)  $-interval of $w$
maximal with respect to inclusion. For instance, if $n=9$, $w=\left(
3,1,2,5,4,2,3,2,4\right)  $ and $p=2$, then the $\left(  p,p+1\right)
$-intervals of $w$ are $\left\{  1\right\}  $, $\left\{  3\right\}  $,
$\left\{  6\right\}  $, $\left\{  7\right\}  $, $\left\{  8\right\}  $,
$\left\{  6,7\right\}  $, $\left\{  7,8\right\}  $ and $\left\{
6,7,8\right\}  $ (since the letters of $w$ belonging to $\left\{
p,p+1\right\}  $ are the $1$-st, $3$-rd, the $6$-th, the $7$-th and the $8$-th
letter), while only $\left\{  1\right\}  $, $\left\{  3\right\}  $ and
$\left\{  6,7,8\right\}  $ are $\left(  p,p+1\right)  $-runs of $w$.

An interval of $\left\{  1,2,...,n\right\}  $ is said to be \emph{even} if
it has even size, and \emph{odd} if it has odd size.

It is easy to see that if $w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ is any Smirnov $n$-tuple, and $\left\{
a+1,a+2,...,b\right\}  $ is any $\left(  p,p+1\right)  $-interval of $w$, then
the $\left(  b-a\right)  $-tuple $\left(  w_{a+1},w_{a+2},...,w_{b}\right)  $
alternates between $p$'s and $\left(  p+1\right)  $'s, that is, has one of the
forms%
\[
\left(  p,p+1,p,p+1,p,...,p+1\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  p+1,p,p+1,p,p+1,...,p\right)
\]
(if $b-a$ is even) or one of the forms%
\[
\left(  p,p+1,p,p+1,p,...,p\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  p+1,p,p+1,p,p+1,...,p+1\right)
\]
(if $b-a$ is odd).

Now, we define a map $J:S_{n,k}\rightarrow S_{n,k}$ as follows: Let $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in S_{n,k}$. We know that $w$ is Smirnov and
satisfies $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =k$. It
is clear that the $\left(  p,p+1\right)  $-runs of $w$ are disjoint
and even separated from each other by at least $1$ (this means that if
$\alpha$ and $\beta$ are elements of two distinct $\left(  p,p+1\right)
$-runs of $w$, then $\left\vert \alpha-\beta\right\vert >1$). For every
$i\in\left\{  1,2,...,n\right\}  $, define a positive integer $w_{i}^{\prime}$
as follows:

\begin{itemize}
\item If $i$ belongs to an odd $\left(  p,p+1\right)  $-run of $w$, then set
$w_{i}^{\prime}=
\begin{cases}
p+1, & \text{if }w_{i}=p;\\
p, & \text{if }w_{i}=p+1
\end{cases} \quad$.

\item Otherwise, set $w_{i}^{\prime}=w_{i}$.
\end{itemize}

Thus, we have defined an $n$-tuple $\left(  w_{1}^{\prime},w_{2}^{\prime
},...,w_{n}^{\prime}\right)  $ of positive integers.\footnote{Informally
speaking, $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime}\right)  $
is simply obtained by changing those $p$'s and $p+1$'s in $w$ whose positions
belong to odd $\left(  p,p+1\right)  $-runs of $w$ into $p+1$'s and $p$'s,
respectively, while leaving all other entries of $w$ intact. For instance, in
our above example of $n=9$, $w=\left(  3,1,2,5,4,2,3,2,4\right)  $ and $p=2$,
we would have $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime
}\right)  =\left(  2,1,3,5,4,3,2,3,4\right)  $. (All letters $2$ and $3$ have
been changed here because all $\left(  p,p+1\right)  $-runs of this $w$ were
odd.) For another example, if $n=5$, $w=\left(  2,3,1,5,2\right)  $ and $p=2$,
then $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime}\right)
=\left(  2,3,1,5,3\right)  $ (the first two letters are unchanged since the
$\left(  p,p+1\right)  $-run $\left\{  1,2\right\}  $ is even).} Denote this
$n$-tuple $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime}\right)  $
by $w^{\prime}$. The reader can easily check that this new $n$-tuple
$w^{\prime}$ is again Smirnov and satisfies $\left\vert \operatorname{Des}%
\left(  w^{\prime}\right)  \right\vert =k$. In other words, $w^{\prime}\in
S_{n,k}$. Now, set $J\left(  w\right)  =w^{\prime}$. We have thus defined a
map $J:S_{n,k}\rightarrow S_{n,k}$. It is easy to see that, for every $w\in
S_{n,k}$, the $\left(  p,p+1\right)  $-runs of $J\left(  w\right)  $ are
exactly the $\left(  p,p+1\right)  $-runs of $w$, and applying the map $J$ to
$J\left(  w\right)  $ precisely reverts the changes made by the map $J$ to
$w$. In other words, $J\circ J= \id$, so that the map $J$ is an
involution. Finally, it is straightforward to see that for every $w\in
S_{n,k}$, we have the following facts:

\begin{enumerate}
\item The number of entries equal to $p+1$ in $J\left(  w\right)  $ equals the
number of entries equal to $p$ in $w$.

\item The number of entries equal to $p$ in $J\left(  w\right)  $ equals the
number of entries equal to $p+1$ in $w$.

\item For every $j\in\left\{  1,2,3,...\right\}  \setminus\left\{
p,p+1\right\}  $, the number of entries equal to $j$ in $J\left(  w\right)  $
equals the number of entries equal to $j$ in $w$.
\end{enumerate}

\footnote{The idea is that the map $J$ switches the number of $p$'s with the
number of $\left(  p+1\right)  $'s in any odd $\left(  p,p+1\right)  $-run,
while the numbers in an even $\left(  p,p+1\right)  $-run are already equal to
begin with.} These three statements, combined, show that for every $w\in
S_{n,k}$, the monomial ${\mathbf{x}}_{J\left(  w\right)  }$ is obtained from
the monomial ${\mathbf{x}}_{w}$ by swapping the variables $x_{p}$ and
$x_{p+1}$. This concludes our solution of Exercise
\ref{exe.shareshian-wachs.path}(a).

\textit{Remark:} We could have given an alternative solution to
Exercise~\ref{exe.shareshian-wachs.path}(b) that would still rely on
\eqref{sol.shareshian-wachs.path.b.rec}, but proceed by induction on $n+s$
(rather than $n+d$) and handle the case $s=0$ separately (rather than the case
$d=0$ as we did). In the case $s=0$, the assertion of
Exercise~\ref{exe.shareshian-wachs.path}(b) follows from
Exercise~\ref{exe.shareshian-wachs.path}(a). (But this only works combined
with a solution to Exercise~\ref{exe.shareshian-wachs.path}(a) that does not
rely on Exercise~\ref{exe.shareshian-wachs.path}(b).)

(c) For every $d\in \NN$, $s\in \NN$ and $i\in\left\{
1,2,...,n-1\right\}  $, we define the power series $\mathcal{D}_{i}$,
$\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ as in the solution to
Exercise~\ref{exe.shareshian-wachs.path}(b) above.

Let us first show that if $n$ is a positive integer, then any $d\in
{ \NN }$ and $s\in{ \NN }$ satisfy%
\begin{equation}
\left(  d+1\right)  U_{n,d+1,s}+\left(  s+1\right)  U_{n,d,s+1}+\left(
n-1-d-s\right)  U_{n,d,s}=\left(  d+1\right)  X_{n,d+1,s}.
\label{sol.shareshian-wachs.path.c.rec}%
\end{equation}


\textit{Proof of (\ref{sol.shareshian-wachs.path.c.rec}):} Let $n$ be a
positive integer. Let $d\in{ \NN }$ and $s\in{ \NN }$. Let
$i\in\left\{  1,2,...,n-1\right\}  $ be arbitrary.

Before we make any new definitions, let us recall a statement that we have
shown during our proof of (\ref{sol.shareshian-wachs.path.b.rec}) (back when
we were solving Exercise \ref{exe.shareshian-wachs.path}(b)): The power series
$\mathcal{D}_{i}\in \kk \left[  \left[  {\xx}\right]  \right]  $
is the sum of the monomials $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all
$n$-tuples $w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{n}$ satisfying $\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d$, $\left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s$ and $w_{i}>w_{i+1}$. Written as a formula, this
yields%
\[
\mathcal{D}_{i}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  i\right\}  \right\vert
=d;\ \left\vert \left(  \operatorname*{Stag}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =s;\\w_{i}>w_{i+1}}}x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}.
\]
Applying this to $n-i$ instead of $i$, we obtain%
\begin{equation}
\mathcal{D}_{n-i}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=d;\ \left\vert \left(  \operatorname*{Stag}\left(  w\right)  \right)
\setminus\left\{  n-i\right\}  \right\vert =s;\\w_{n-i}>w_{n-i+1}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}. \label{sol.shareshian-wachs.path.c.oldrec}%
\end{equation}


Now, let us make some more definitions:

\begin{itemize}
\item Define a power series $\mathcal{D}_{i}^{\prime}\in \kk \left[
\left[  {\xx}\right]  \right]  $ as the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d+1$,
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$, $w_{1}%
<w_{n}$ and $w_{i}>w_{i+1}$.

\item Define a power series $\mathcal{S}_{i}^{\prime}\in \kk \left[
\left[  {\xx}\right]  \right]  $ as the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$,
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s+1$,
$w_{1}<w_{n}$ and $w_{i}=w_{i+1}$.

\item Define a power series $\mathcal{A}_{i}^{\prime}\in \kk \left[
\left[  {\xx}\right]  \right]  $ as the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$,
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$, $w_{1}%
<w_{n}$ and $w_{i}<w_{i+1}$.
\end{itemize}

These three power series $\mathcal{D}_{i}^{\prime}$, $\mathcal{S}_{i}^{\prime
}$ and $\mathcal{A}_{i}^{\prime}$ were defined in obvious analogy to the power
series $\mathcal{D}_{i}$, $\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ defined in
our solution to Exercise \ref{exe.shareshian-wachs.path}(b) above. In the same
way as we have proved (\ref{sol.shareshian-wachs.path.b.rec.pf.1}) back there,
we can see that%
\begin{equation}
\mathcal{D}_{i}^{\prime}+\mathcal{S}_{i}^{\prime}+\mathcal{A}_{i}^{\prime
}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.c.rec.pf.1}%
\end{equation}


But the sum on the right hand side of
(\ref{sol.shareshian-wachs.path.c.rec.pf.1}) can be rewritten in terms of more
familiar sums. In fact, the $n$-tuples $w=\left(  w_{1},w_{2},...,w_{n}%
\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ are in bijection with the
pairs $\left(  u,v\right)  $ consisting of an $i$-tuple $u=\left(  u_{1}%
,u_{2},...,u_{i}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{i}$ and an
$\left(  n-i\right)  $-tuple $v=\left(  v_{1},v_{2},...,v_{n-i}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n-i}$. This bijection sends an $n$-tuple
$\left(  w_{1},w_{2},...,w_{n}\right)  $ to the pair $\left(  \left(
w_{1},w_{2},...,w_{i}\right)  ,\left(  w_{i+1},w_{i+2},...,w_{n}\right)
\right)  $, and has the property that $\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $\ \ \ \ \footnote{In fact,
$\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
i\right\}  $ is the union of the set $\operatorname*{Des}\left(  u\right)  $
with the set $\operatorname*{Des}\left(  v\right)  $ shifted by $i$ (that is,
the set $\left\{  p+i\ \mid\ p\in\operatorname*{Des}\left(  v\right)
\right\}  $). This is a disjoint union, and thus we find that
$\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $
(since the set $\operatorname*{Des} \left(  v\right)  $ shifted by $i$
has cardinality $\left\vert \operatorname*{Des}\left( v \right)
\right\vert$).
}, $\left\vert \left(  \operatorname*{Stag}\left(  w\right)
\right)  \setminus\left\{  i\right\}  \right\vert =\left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert $ \ \ \ \ \footnote{for
similar reasons}, $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}$, $w_{1}=u_{1}$ and
$w_{n}=v_{n-i}$. Hence,%
\begin{align}
&  \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
} \nonumber \\
&=\underbrace{\sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\v=\left(  v_{1},v_{2},...,v_{n-i}%
\right)  \in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert =d;\\\left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}<v_{n-i}}}}%
_{=\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}}\underbrace{x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}%
}\cdots x_{v_{n-i}}}_{=x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}x_{u_{1}}x_{u_{2}%
}\cdots x_{u_{i}}}\nonumber\\
&  =\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}. \label{sol.shareshian-wachs.path.c.rec.pf.15}%
\end{align}


On the other hand, the pairs $\left(  v,u\right)  $ consisting of an $\left(
n-i\right)  $-tuple $v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n-i}$ and an $i$-tuple $u=\left(  u_{1},u_{2}%
,...,u_{i}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{i}$ are in bijection
with the $n$-tuples $w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}$. This bijection sends a pair $\left(  v,u\right)
$ with $v=\left(  v_{1},v_{2},...,v_{n-i}\right)  $ and $u=\left(  u_{1}%
,u_{2},...,u_{i}\right)  $ to the $n$-tuple $\left(  v_{1},v_{2}%
,...,v_{n-i},u_{1},u_{2},...,u_{i}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$, and has the property that $\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $\ \ \ \ \footnote{In fact,
$\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
n-i\right\}  $ is the union of the set $\operatorname*{Des}\left(  v\right)  $
with the set $\operatorname*{Des}\left(  u\right)  $ shifted by $n-i$ (that
is, the set $\left\{  p+\left(  n-i\right)  \ \mid\ p\in\operatorname*{Des}%
\left(  u\right)  \right\}  $). This is a disjoint union; thus, we find
$\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $
(since the set $\operatorname*{Des} \left( u \right)  $ shifted by $n-i$
has cardinality $\left\vert \operatorname*{Des}\left( u \right)
\right\vert$).
}, $\left\vert \left(  \operatorname*{Stag}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert $ \ \ \ \ \footnote{for
similar reasons}, $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}$, $w_{n-i+1}=u_{1}$ and
$w_{n-i}=v_{n-i}$. We can use this bijection to transform the sum on the right
hand side of (\ref{sol.shareshian-wachs.path.c.rec.pf.15}), and obtain
\begin{align}
&  \sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}} \nonumber \\
&=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=d;\ \left\vert \left(  \operatorname*{Stag}\left(  w\right)  \right)
\setminus\left\{  n-i\right\}  \right\vert =s;\\w_{n-i+1}<w_{n-i}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}\left(
w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname*{Stag}\left(  w\right)  \right)  \setminus\left\{
n-i\right\}  \right\vert =s;\\w_{n-i}>w_{n-i+1}}}x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}=\mathcal{D}_{n-i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.oldrec})}\right)  .
\label{sol.shareshian-wachs.path.c.rec.pf.17}%
\end{align}
Now, (\ref{sol.shareshian-wachs.path.c.rec.pf.1}) becomes%
\begin{align}
\mathcal{D}_{i}^{\prime}+\mathcal{S}_{i}^{\prime}+\mathcal{A}_{i}^{\prime}  &
=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\nonumber\\
&  =\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.rec.pf.15})}\right) \nonumber\\
&  =\mathcal{D}_{n-i}. \label{sol.shareshian-wachs.path.c.rec.pf.4}%
\end{align}


Now, let us forget that we fixed $i$. We thus have defined $\mathcal{D}%
_{i}^{\prime}$, $\mathcal{S}_{i}^{\prime}$ and $\mathcal{A}_{i}^{\prime}$ and
proven the equality (\ref{sol.shareshian-wachs.path.c.rec.pf.4}) for all
$i\in\left\{  1,2,...,n-1\right\}  $.

Now, let us take a closer look at the sums $\sum_{i=1}^{n-1}\mathcal{D}%
_{i}^{\prime}$, $\sum_{i=1}^{n-1}\mathcal{S}_{i}^{\prime}$ and $\sum
_{i=1}^{n-1}\mathcal{A}_{i}^{\prime}$:

\begin{itemize}
\item Similarly to how we proved (\ref{sol.shareshian-wachs.path.b.rec.pf.7a}%
), we can show that%
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{D}_{i}^{\prime}=\left(  d+1\right)  U_{n,d+1,s}.
\label{sol.shareshian-wachs.path.c.rec.pf.7a}%
\end{equation}


\item Similarly to how we proved (\ref{sol.shareshian-wachs.path.b.rec.pf.7b}%
), we can see that%
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{S}_{i}^{\prime}=\left(  s+1\right)  U_{n,d,s+1}.
\label{sol.shareshian-wachs.path.c.rec.pf.7b}%
\end{equation}


\item Similarly to how we proved (\ref{sol.shareshian-wachs.path.b.rec.pf.7c}%
), we can see that
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{A}_{i}^{\prime}=\left(  n-1-d-s\right)  U_{n,d,s}.
\label{sol.shareshian-wachs.path.c.rec.pf.7c}%
\end{equation}

\end{itemize}

Now, summing the equality (\ref{sol.shareshian-wachs.path.c.rec.pf.4}) over
all $i\in\left\{  1,2,...,n-1\right\}  $ yields%
\[
\sum_{i=1}^{n-1}\left(  \mathcal{D}_{i}^{\prime}+\mathcal{S}_{i}^{\prime
}+\mathcal{A}_{i}^{\prime}\right)  =\sum_{i=1}^{n-1}\mathcal{D}_{n-i}
 =\sum_{i=1}^{n-1}\mathcal{D}_{i} =\left(
d+1\right)  X_{n,d+1,s}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.b.rec.pf.7a})}\right)  .
\]
Hence,%
\begin{align*}
\left(  d+1\right)  X_{n,d+1,s}  &  =\sum_{i=1}^{n-1}\left(  \mathcal{D}%
^{\prime}_{i}+\mathcal{S}^{\prime}_{i}+\mathcal{A}^{\prime}_{i} \right)  =
\underbrace{\sum_{i=1}^{n-1}\mathcal{D}^{\prime}_{i}}_{\substack{=\left(
d+1\right)  U_{n,d+1,s}\\\text{(by
(\ref{sol.shareshian-wachs.path.c.rec.pf.7a}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{S}^{\prime}_{i}}_{\substack{=\left(  s+1\right)  U_{n,d,s+1}%
\\\text{(by (\ref{sol.shareshian-wachs.path.c.rec.pf.7b}))}}}+\underbrace{\sum
_{i=1}^{n-1}\mathcal{A}^{\prime}_{i}}_{\substack{=\left(  n-1-d-s\right)
U_{n,d,s}\\\text{(by (\ref{sol.shareshian-wachs.path.c.rec.pf.7c}))}}}\\
&  =\left(  d+1\right)  U_{n,d+1,s}+\left(  s+1\right)  U_{n,d,s+1}+\left(
n-1-d-s\right)  U_{n,d,s}.
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.c.rec}).

In solving Exercise \ref{exe.shareshian-wachs.path}(c), we WLOG assume that
$\kk = \ZZ$ (for the same reason why we could assume
$\kk = \ZZ$ in solving Exercise \ref{exe.shareshian-wachs.path}(b)).

Now, we are going to prove $U_{n,d,s}\in\Lambda$ by strong induction over
$n+d$. So (for the induction step) we need to show that $U_{n,d,s}\in\Lambda$,
and we can assume (as the induction hypothesis) that $U_{n^{\prime},d^{\prime
},s^{\prime}}\in\Lambda$ is already known to hold for any nonnegative integers
$n^{\prime}$, $d^{\prime}$ and $s^{\prime}$ satisfying $n^{\prime}+d^{\prime
}<n+d$.

We must be in one of the following two cases:

\textit{Case 1:} We have $d>0$.

\textit{Case 2:} We have $d=0$.

In Case 1, we can proceed in the same way as in the corresponding case of the
solution of Exercise \ref{exe.shareshian-wachs.path}(b) (with the only
difference that we now have to use $X_{n,d,s}\in\Lambda$, but this follows
from the already solved Exercise \ref{exe.shareshian-wachs.path}(b)).

Let us now consider Case 2. In this case, we have $d=0$. We now distinguish
between two subcases:

\textit{Subcase 2.1:} We have $s=n-1$.

\textit{Subcase 2.2:} We have $s\neq n-1$.

Let us consider Subcase 2.1 first. In this subcase, we have $s=n-1$. We will
show that $U_{n,d,s}=0$ in this subcase.

Indeed, let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}$ be such that $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$, $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$ and $w_{1}<w_{n}$. Then, $\operatorname{Stag}\left(  w\right)
$ is a subset of $\left\{  1,2,...,n-1\right\}  $ whose cardinality is
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s=n-1=\left\vert
\left\{  1,2,...,n-1\right\}  \right\vert $. Obviously, the only such subset
is $\left\{  1,2,...,n-1\right\}  $ itself, and so $\operatorname*{Stag}%
\left(  w\right)  $ must be $\left\{  1,2,...,n-1\right\}  $. Hence, every
$j\in\left\{  1,2,...,n-1\right\}  $ satisfies $j\in\left\{
1,2,...,n-1\right\}  =\operatorname*{Stag}\left(  w\right)  =\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  :w_{i}=w_{i+1}\right\}  $ and thus
$w_{j}=w_{j+1}$. In other words, $w_{1}=w_{2}=...=w_{n}$. This contradicts
$w_{1}<w_{n}$.

Now, forget that we fixed $w$. We thus have obtained a contradiction for every
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d$, $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$ and
$w_{1}<w_{n}$. Therefore, there exists no such $w$. Hence, the sum on the
right hand side of (\ref{eq.exe.shareshian-wachs.path.c.U}) is empty and thus
equals $0$. Thus, (\ref{eq.exe.shareshian-wachs.path.c.U}) rewrites as
$U_{n,d,s}=0\in\Lambda$. Hence, the induction step is complete in Subcase 2.1.

Let us now consider Subcase 2.2. In this case, $s\neq n-1$. We will show that
$U_{n,d,s}=X_{n,d,s}$ in this subcase.

Indeed, let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}$ be such that $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$ and $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$. We will show that $w_{1}<w_{n}$.

We have $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d=0$, so
that the set $\operatorname{Des}\left(  w\right)  $ is empty. In other words,
every $j\in\left\{  1,2,...,n-1\right\}  $ satisfies $w_{j}\leq w_{j+1}$.
Hence, we have the chain of inequalities $w_{1}\leq w_{2}\leq...\leq w_{n}$.
At least one inequality in this chain must be strict (because otherwise, we
would have $w_{1}=w_{2}=...=w_{n}$, thus $w_{j}=w_{j+1}$ for every
$j\in\left\{  1,2,...,n-1\right\}  $, thus $j\in\operatorname*{Stag}\left(
w\right)  $ for every $j\in\left\{  1,2,...,n-1\right\}  $, which would lead
to $\operatorname*{Stag}\left(  w\right)  =\left\{  1,2,...,n-1\right\}  $ and
thus $\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =\left\vert
\left\{  1,2,...,n-1\right\}  \right\vert =n-1$, in contradiction to
$\operatorname*{Stag}\left(  w\right)  =s\neq n-1$), and thus we have
$w_{1}<w_{n}$.

Now, let us forget that we fixed $w$. We thus have proven that every
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d$ and $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$
automatically satisfies $w_{1}<w_{n}$. Hence, the condition $w_{1}<w_{n}$
under the summation sign ``$\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}$'' is redundant (i.e., can be removed
without changing the range of the summation). Thus,%
\[
\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}=X_{n,d,s}%
\]
(by (\ref{sol.shareshian-wachs.path.b.def})). Now,
(\ref{eq.exe.shareshian-wachs.path.c.U}) becomes%
\[
U_{n,d,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=X_{n,d,s}%
\in\Lambda
\]
(by Exercise \ref{exe.shareshian-wachs.path}(b)). Hence, the induction step is
complete in Subcase 2.2.

The induction step is thus complete in Case 1 and in each of the two Subcases
2.1 and 2.2. These are all cases, and so the induction step is finally
complete. We have thus proven that
\begin{equation}
U_{n,d,s}\in\Lambda\ \ \ \ \ \ \ \ \ \ \text{for all positive integers
}n\text{, all }d \in \NN \text{ and all }s \in \NN .
\label{sol.shareshian-wachs.path.c.Udone}
\end{equation}


In order to complete the solution of Exercise~\ref{exe.shareshian-wachs.path}(c),
we still need to prove that $V_{n, d, s}$ and $W_{n, d, s}$
belong to $\Lambda$ for all positive integers $n$, all
$d \in \NN$ and all $s \in \NN$.

Let $n$ be a positive integer, let $d \in \NN$ and let $s \in \NN$.
The equality (\ref{eq.exe.shareshian-wachs.path.c.W}) becomes%
\begin{align*}
W_{n,d,s} &  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}>w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}} \\
&=\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  \left(
w_{n},w_{n-1},\ldots,w_{1}\right)  \right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  \left(  w_{n},w_{n-1},\ldots,w_{1}\right)  \right)
\right\vert =s;\\w_{n}>w_{1}}}\underbrace{x_{w_{n}}x_{w_{n-1}}\cdots x_{w_{1}%
}}_{=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }\left(
w_{n},w_{n-1},\ldots,w_{1}\right)  \text{ for }w=\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \text{ in the sum}\right)  \\
&  =\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  \left(
w_{n},w_{n-1},\ldots,w_{1}\right)  \right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  \left(  w_{n},w_{n-1},\ldots,w_{1}\right)  \right)
\right\vert =s;\\w_{n}>w_{1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  \left(
w_{n},w_{n-1},\ldots,w_{1}\right)  \right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  \left(  w_{n},w_{n-1},\ldots,w_{1}\right)  \right)
\right\vert =s;\\w_{n}>w_{1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\n-1-\left\vert \operatorname{Des}\left(
w\right)  \right\vert -\left\vert \operatorname*{Stag}\left(  w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{n}>w_{1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}\text{ satisfies}\\
\left\vert \operatorname{Des}\left(  \left(  w_{n},w_{n-1},\ldots
,w_{1}\right)  \right)  \right\vert =n-1-\left\vert \operatorname{Des}\left(
w\right)  \right\vert -\left\vert \operatorname*{Stag}\left(  w\right)
\right\vert \\
\text{and }\left\vert \operatorname{Stag}\left(  \left(  w_{n},w_{n-1}%
,\ldots,w_{1}\right)  \right)  \right\vert =\left\vert \operatorname{Stag}%
\left(  w\right)  \right\vert \\
\text{(this is easily shown by observing that the set }\left\{
1,2,...,n-1\right\}  \\
\text{is the union of its three disjoint subsets}\\
\left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}<w_{i+1}\right\}
\text{, }\operatorname*{Des}w\text{ and }\operatorname*{Stag}w\text{)}%
\end{array}
\right)  \\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =n-1-d-s;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for a }w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n}\text{ satisfying}\\
\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s\text{, the
assertions }n-1-\left\vert \operatorname{Des}\left(  w\right)  \right\vert
-\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =d\\
\text{and }w_{n}>w_{1}\text{ are equivalent to }\left\vert \operatorname{Des}%
\left(  w\right)  \right\vert =n-1-d-s\text{ and }w_{1}<w_{n}%
\end{array}
\right)  \\
&  =U_{n,n-1-d-s,s}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.exe.shareshian-wachs.path.c.U}), applied to }n-1-d-s\text{ instead of
}d\right)  \\
&  \in\Lambda\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.Udone}), applied to }n-1-d-s\text{ instead
of }d\right)  .
\end{align*}
Finally, (\ref{sol.shareshian-wachs.path.b.def}) becomes%
\begin{align*}
X_{n,d,s} &  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  =\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
}_{\substack{=U_{n,d,s}\\\text{(by (\ref{eq.exe.shareshian-wachs.path.c.U}))}%
}}+\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}=w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
}_{\substack{=V_{n,d,s}\\\text{(by (\ref{eq.exe.shareshian-wachs.path.c.V}))}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}%
;\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}>w_{n}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}}_{\substack{=W_{n,d,s}\\\text{(by
(\ref{eq.exe.shareshian-wachs.path.c.W}))}}}\\
&  =U_{n,d,s}+V_{n,d,s}+W_{n,d,s},
\end{align*}
so that $V_{n,d,s}=\underbrace{X_{n,d,s}}_{\substack{\in\Lambda\\\text{(by
Exercise \ref{exe.shareshian-wachs.path}(b))}}}-\underbrace{U_{n,d,s}}%
_{\in\Lambda}-\underbrace{W_{n,d,s}}_{\in\Lambda}\in\Lambda-\Lambda
-\Lambda\subset\Lambda$.

We have thus shown that $V_{n,d,s}\in\Lambda$ and $W_{n,d,s}\in\Lambda$.
Combined with $U_{n,d,s}\in\Lambda$ (this follows from
(\ref{sol.shareshian-wachs.path.c.Udone})), this completes the solution of
Exercise~\ref{exe.shareshian-wachs.path}(c).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.shareshian-wachs.path}}

\textit{Solution to Exercise \ref{exe.shareshian-wachs.path}.} (b) Let
us first notice that
\begin{equation}
X_{n,d,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
\ \ \ \ \ \ \ \ \ \ \text{for all }d\in \NN \text{ and }s\in \NN .
\label{sol.shareshian-wachs.path.b.def}%
\end{equation}
(This is just the definition of $X_{n,d,s}$, written as a formula.)

Let us first show that if $n$ is a positive integer, then any $d\in \NN$
and $s\in \NN$ satisfy
\begin{align}
&  \left(  d+1\right)  X_{n,d+1,s}+\left(  s+1\right)  X_{n,d,s+1}+\left(
n-1-d-s\right)  X_{n,d,s}\nonumber\\
&  =\sum_{i=1}^{n-1}\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}.
\label{sol.shareshian-wachs.path.b.rec}
\end{align}


\textit{Proof of (\ref{sol.shareshian-wachs.path.b.rec}):} Let $n$ be a
positive integer. Let $d \in \NN$ and $s \in \NN$. Let
$i\in\left\{  1,2,...,n-1\right\}  $ be arbitrary. We make some more definitions:

\begin{itemize}
\item Define a power series $\mathcal{D}_{i}\in \kk \left[  \left[
 \xx \right]  \right]  $ as the sum of the monomials $x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d+1$, $\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s$ and $w_{i}>w_{i+1}$.

\item Define a power series $\mathcal{S}_{i}\in \kk \left[  \left[
 \xx \right]  \right]  $ as the sum of the monomials $x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$, $\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s+1$ and $w_{i}=w_{i+1}$.

\item Define a power series $\mathcal{A}_{i}\in \kk \left[  \left[
 \xx \right]  \right]  $ as the sum of the monomials $x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$, $\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s$ and $w_{i}<w_{i+1}$.
\end{itemize}

These three power series $\mathcal{D}_{i}$, $\mathcal{S}_{i}$ and
$\mathcal{A}_{i}$ are not necessarily symmetric (but will nevertheless come
useful). Let us notice that the definitions of $\mathcal{D}_{i}$,
$\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ can be rewritten as follows:

\begin{itemize}
\item The power series $\mathcal{D}_{i}\in \kk \left[  \left[
 \xx \right]  \right]  $ is the sum of the monomials $x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{i}>w_{i+1}$.

\item The power series $\mathcal{S}_{i}\in \kk \left[  \left[
 \xx \right]  \right]  $ is the sum of the monomials $x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{i}=w_{i+1}$.

\item The power series $\mathcal{A}_{i}\in \kk \left[ \left[
 \xx \right]  \right]  $ is the sum of the monomials $x_{w_{1}}
 x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{i}<w_{i+1}$.
\end{itemize}

These reformulations make it obvious that the sum $\mathcal{D}_{i}%
+\mathcal{S}_{i}+\mathcal{A}_{i}$ is precisely the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$ and $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$. In other words,
\begin{equation}
\mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}_{i}=\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}
;\\\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d;\ \left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.b.rec.pf.1}
\end{equation}


But the sum on the right hand side of
(\ref{sol.shareshian-wachs.path.b.rec.pf.1}) can be rewritten in terms of more
familiar sums. In fact, the $n$-tuples $w\in\left\{  1,2,3,\ldots\right\}
^{n}$ are in bijection with the pairs $\left(  u,v\right)  $ consisting of an
$i$-tuple $u\in\left\{  1,2,3,\ldots\right\}  ^{i}$ and an $\left(
n-i\right)  $-tuple $v\in\left\{  1,2,3,\ldots\right\}  ^{n-i}$. This
bijection sends an $n$-tuple $\left(  w_{1},w_{2},...,w_{n}\right)  $ to the
pair $\left(  \left(  w_{1},w_{2},...,w_{i}\right)  ,\left(  w_{i+1}%
,w_{i+2},...,w_{n}\right)  \right)  $, and has the property that $\left\vert
\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =\left\vert \operatorname*{Des}\left(  u\right)
\right\vert +\left\vert \operatorname*{Des}\left(  v\right)  \right\vert
$\ \ \ \ \footnote{In fact, $\left(  \operatorname*{Des}\left(  w\right)
\right)  \setminus\left\{  i\right\}  $ is the union of the set
$\operatorname*{Des}\left(  u\right)  $ with the set $\operatorname*{Des}
\left(  v\right)  $ shifted by $i$ (that is, the set $\left\{  p+i\ \mid
\ p\in\operatorname*{Des}\left(  v\right)  \right\}  $). This is a disjoint
union, and thus we find
$\left\vert
\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =\left\vert \operatorname*{Des}\left(  u\right)
\right\vert +\left\vert \operatorname*{Des}\left(  v\right)  \right\vert
$ (since the set $\operatorname*{Des} \left(  v\right)  $ shifted by $i$
has cardinality $\left\vert \operatorname*{Des}\left( v \right)
\right\vert$).
} and $\left\vert
\left(  \operatorname*{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =\left\vert \operatorname*{Stag}\left(  u\right)
\right\vert +\left\vert \operatorname*{Stag}\left(  v\right)
 \right\vert$ \ \ \ \ \footnote{for similar reasons}
and $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}
= x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}$.
Hence,
\begin{align}
&  \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{
1,2,3,...\right\}  ^{i};\\v=\left(  v_{1},v_{2},...,v_{n-i}\right)
\in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s}}x_{u_{1}%
}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}\nonumber\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\underbrace{\sum\limits_{\substack{u=\left(
u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i}%
;\\v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{  1,2,3,...\right\}
^{n-i};\\\left\vert \operatorname*{Des}\left(  u\right)  \right\vert
=e;\ \left\vert \operatorname*{Des}\left(  v\right)  \right\vert
=d-e;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
=t;\ \left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s-t}%
}}_{=\sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{
1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(  u\right)
\right\vert =e;\ \left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
=t}}\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}\nonumber\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\underbrace{\sum\limits_{\substack{u=\left(
u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i}%
;\\\left\vert \operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}\sum\limits
_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}}_{=\left(  \sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots
,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\right)  \left(  \sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}%
\right)  \in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert =d-e;\ \left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert =s-t}}x_{v_{1}}x_{v_{2}%
}\cdots x_{v_{n-i}}\right)  }\nonumber\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\left(  \sum\limits_{\substack{u=\left(  u_{1}%
,u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\right)  \left(  \sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}%
\right)  \in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert =d-e;\ \left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert =s-t}}x_{v_{1}}x_{v_{2}%
}\cdots x_{v_{n-i}}\right)  . \label{sol.shareshian-wachs.path.b.rec.pf.2}%
\end{align}
However, for every $e\in\left\{  0,1,...,d\right\}  $ and $t\in\left\{
0,1,...,s\right\}  $, the equality (\ref{sol.shareshian-wachs.path.b.def})
(applied to $i$, $e$ and $t$ instead of $n$, $d$ and $s$) yields
\begin{equation}
X_{i,e,t}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =e;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =t}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{i}}=\sum\limits_{\substack{u=\left(
u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i}%
;\\\left\vert \operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}.
\label{sol.shareshian-wachs.path.b.rec.pf.3a}
\end{equation}
Also, for every $e\in\left\{  0,1,...,d\right\}  $ and $t\in\left\{
0,1,...,s\right\}  $, the equality (\ref{sol.shareshian-wachs.path.b.def})
(applied to $n-i$, $d-e$ and $s-t$ instead of $n$, $d$ and $s$) yields
\begin{equation}
X_{n-i,d-e,s-t}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n-i}\right)
\in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d-e;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s-t}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n-i}}=\sum\limits
_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}.
\label{sol.shareshian-wachs.path.b.rec.pf.3b}
\end{equation}
Now, (\ref{sol.shareshian-wachs.path.b.rec.pf.2}) becomes
\begin{align*}
&  \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}\underbrace{\left(  \sum\limits_{\substack{u=\left(
u_{1},u_{2},\ldots,u_{i}\right)  \in\left\{  1,2,3,...\right\}  ^{i}%
;\\\left\vert \operatorname*{Des}\left(  u\right)  \right\vert =e;\ \left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert =t}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\right)  }_{\substack{=X_{i,e,t}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.3a}))}}}\underbrace{\left(
\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d-e;\ \left\vert \operatorname*{Stag}\left(  v\right)
\right\vert =s-t}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}\right)  }%
_{\substack{=X_{n-i,d-e,s-t}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.3b}))}}}\\
&  =\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}.
\end{align*}
Hence, (\ref{sol.shareshian-wachs.path.b.rec.pf.1}) becomes%
\begin{align}
\mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}_{i}
&= \sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}%
;\\\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d;\ \left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}} \nonumber\\
&= \sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}.
\label{sol.shareshian-wachs.path.b.rec.pf.4}
\end{align}


Now, let us forget that we fixed $i$. We thus have defined $\mathcal{D}_{i}$,
$\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ and proven the equality
(\ref{sol.shareshian-wachs.path.b.rec.pf.4}) for all $i\in\left\{
1,2,...,n-1\right\}  $.

Now, let us take a closer look at the sums $\sum_{i=1}^{n-1}\mathcal{D}_{i}$,
$\sum_{i=1}^{n-1}\mathcal{S}_{i}$ and $\sum_{i=1}^{n-1}\mathcal{A}_{i}$:

\begin{itemize}
\item If an $n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfies $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d+1$, then%
\begin{equation}
\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
>w_{i+1}\right\}  \right\vert =d+1
\label{sol.shareshian-wachs.path.b.rec.pf.6a}%
\end{equation}
(because $\left\vert \underbrace{\left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}>w_{i+1}\right\}  }_{=\operatorname*{Des}\left(  w\right)
}\right\vert =\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d+1$). Now, the definition of $\mathcal{D}_{i}$ can be rewritten as follows:%
\[
\mathcal{D}_{i}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\ w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,n-1\right\}  .
\]
Summing up these equations over all $i\in\left\{  1,2,...,n-1\right\}  $, we
obtain%
\begin{align}
\sum_{i=1}^{n-1}\mathcal{D}_{i}  &  =\underbrace{\sum_{i=1}^{n-1}%
\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\ w_{i}>w_{i+1}}}}_{=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d+1;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s}}\sum\limits_{\substack{i\in
\left\{  1,2,...,n-1\right\}  ;\\w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,...,n-1\right\}
;\\w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}}_{\substack{=\left\vert
\left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}>w_{i+1}\right\}
\right\vert x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\=\left(  d+1\right)  x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.6a}))}}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}\left(  d+1\right)  x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\left(  d+1\right)  \underbrace{\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d+1;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}}_{\substack{=X_{n,d+1,s}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.def}), applied to }d+1\text{ instead of
}d\text{)}}}\nonumber\\
&  =\left(  d+1\right)  X_{n,d+1,s}.
\label{sol.shareshian-wachs.path.b.rec.pf.7a}%
\end{align}


\item Similarly, we can see that%
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{S}_{i}=\left(  s+1\right)  X_{n,d,s+1}.
\label{sol.shareshian-wachs.path.b.rec.pf.7b}%
\end{equation}
(In proving this, we have to replace
(\ref{sol.shareshian-wachs.path.b.rec.pf.6a}) by the following fact: If an
$n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfies $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$ and $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s+1$, then
\[
\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
=w_{i+1}\right\}  \right\vert =s+1.
\]
This is proven in the same way as (\ref{sol.shareshian-wachs.path.b.rec.pf.6a}).)

\item Similarly, we can see that
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{A}_{i}=\left(  n-1-d-s\right)  X_{n,d,s}.
\label{sol.shareshian-wachs.path.b.rec.pf.7c}%
\end{equation}
(In proving this, we have to replace
(\ref{sol.shareshian-wachs.path.b.rec.pf.6a}) by the following fact: If an
$n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfies $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$ and $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$, then%
\begin{equation}
\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
<w_{i+1}\right\}  \right\vert =n-1-d-s.
\label{sol.shareshian-wachs.path.b.rec.pf.6c}%
\end{equation}
The proof of (\ref{sol.shareshian-wachs.path.b.rec.pf.6c}) is
easy\footnote{\textit{Proof of (\ref{sol.shareshian-wachs.path.b.rec.pf.6c}):}
Let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}
^{n}$ be an $n$-tuple satisfying $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$ and $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$. Every $i\in\left\{  1,2,...,n-1\right\}  $ satisfies exactly
one of the three statements $w_{i}<w_{i+1}$, $w_{i}=w_{i+1}$ and
$w_{i}>w_{i+1}$. Hence, the sets $\left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}<w_{i+1}\right\}  $, $\left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}=w_{i+1}\right\}  $ and $\left\{  i\in\left\{
1,2,...,n-1\right\}  \ \mid\ w_{i}>w_{i+1}\right\}  $ are disjoint and their
union is $\left\{  1,2,...,n-1\right\}  $. Thus,%
\begin{align*}
&  \left\vert \left\{  1,2,...,n-1\right\}  \right\vert \\
&  =\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid
\ w_{i}<w_{i+1}\right\}  \right\vert +\left\vert \underbrace{\left\{
i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}=w_{i+1}\right\}
}_{=\operatorname*{Stag}\left(  w\right)  }\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ +\left\vert \underbrace{\left\{  i\in\left\{
1,2,...,n-1\right\}  \ \mid\ w_{i}>w_{i+1}\right\}  }_{=\operatorname*{Des}%
\left(  w\right)  }\right\vert \\
&  =\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid
\ w_{i}<w_{i+1}\right\}  \right\vert +\underbrace{\left\vert
\operatorname*{Stag}\left(  w\right)  \right\vert }_{=s}%
+\underbrace{\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
}_{=d}\\
&  =\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid
\ w_{i}<w_{i+1}\right\}  \right\vert +s+d,
\end{align*}
so that%
\[
\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
<w_{i+1}\right\}  \right\vert +s+d=\left\vert \left\{  1,2,...,n-1\right\}
\right\vert =n-1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n>0\right)
\]
and thus $\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid
\ w_{i}<w_{i+1}\right\}  \right\vert =n-1-s-d=n-1-d-s$. This proves
(\ref{sol.shareshian-wachs.path.b.rec.pf.6c}).}.)
\end{itemize}

Now, summing the equality (\ref{sol.shareshian-wachs.path.b.rec.pf.4}) over
all $i\in\left\{  1,2,...,n-1\right\}  $ yields%
\[
\sum_{i=1}^{n-1}\left(  \mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}%
_{i}\right)  =\sum_{i=1}^{n-1}\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}%
X_{n-i,d-e,s-t}.
\]
Hence,%
\begin{align*}
&  \sum_{i=1}^{n-1}\sum_{e=0}^{d}\sum_{t=0}^{s}X_{i,e,t}X_{n-i,d-e,s-t}\\
&  =\sum_{i=1}^{n-1}\left(  \mathcal{D}_{i}+\mathcal{S}_{i}+\mathcal{A}%
_{i}\right)  =\underbrace{\sum_{i=1}^{n-1}\mathcal{D}_{i}}_{\substack{=\left(
d+1\right)  X_{n,d+1,s}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.7a}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{S}_{i}}_{\substack{=\left(  s+1\right)  X_{n,d,s+1}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.7b}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{A}_{i}}_{\substack{=\left(  n-1-d-s\right)  X_{n,d,s}%
\\\text{(by (\ref{sol.shareshian-wachs.path.b.rec.pf.7c}))}}}\\
&  =\left(  d+1\right)  X_{n,d+1,s}+\left(  s+1\right)  X_{n,d,s+1}+\left(
n-1-d-s\right)  X_{n,d,s}.
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.b.rec}).

In solving Exercise~\ref{exe.shareshian-wachs.path}(b), we shall use
the following trick to simplify our life.
Recall that there is a canonical ring homomorphism $\varphi
: \ZZ \rightarrow \kk$. This homomorphism gives rise to a ring
homomorphism $\varphi\left[  \left[  \xx\right]  \right]
: \ZZ \left[  \left[  \xx\right]  \right]  \rightarrow
 \kk \left[  \left[  \xx\right]  \right]  $, and this latter
homomorphism $\varphi\left[  \left[  \xx\right]  \right]  $ sends
$\Lambda_{\ZZ}$ to $\Lambda_{\kk}$;
that is, we have $\left(  \varphi\left[  \left[
\mathbf{x}  \right]  \right]  \right)  \left(  
\Lambda_{\ZZ} \right)  \subset \Lambda_{\kk}$.
Moreover, it is clear that (for any nonnegative integers $d$ and $s$)
the ring homomorphism
$\varphi\left[  \left[  \xx\right]  \right]  $ sends the element
$X_{n,d,s}$ of $ \ZZ \left[  \left[
\xx\right]  \right]  $ to the element $X_{n,d,s}$ of
$\kk \left[  \left[  \xx\right]  \right]  $
(because the definition of $X_{n,d,s}$ is
functorial in the base ring $\kk$). Therefore, if we can prove
(for given nonnegative integers $d$ and $s$) that the
element $X_{n,d,s}$ of $ \ZZ \left[
\left[  \xx\right]  \right]  $ belongs to $\Lambda_{\ZZ}$,
then it will automatically follow that the element
$X_{n,d,s}$ of $\kk \left[  \left[  \xx\right]  \right]  $
belongs to $\left(  \varphi\left[  \left[  \xx\right]  \right]
\right)  \left(  \Lambda_{\ZZ} \right)
\subset \Lambda_{\kk}$; this will complete the
solution to Exercise~\ref{exe.shareshian-wachs.path}(b).
Hence, in order to solve Exercise~\ref{exe.shareshian-wachs.path}(b),
it only remains to prove (for any nonnegative integers $d$ and $s$)
that the element $X_{n,d,s}$ of $ \ZZ \left[
\left[  \xx\right]  \right]  $ belongs to $
\Lambda_{\ZZ}$. In other words, it only remains to solve
Exercise~\ref{exe.shareshian-wachs.path}(b) in the case of
$\kk = \ZZ$. Hence, in solving
Exercise~\ref{exe.shareshian-wachs.path}(b),
we can WLOG assume that $\kk = \ZZ$. Assume this.

% OLD VERSION OF THE ABOVE PARAGRAPH:
% In solving Exercise~\ref{exe.shareshian-wachs.path}(b), we WLOG assume that
% $\kk = \ZZ$ (because there is a canonical ring homomorphism
% $\ZZ \left[ \left[ \xx \right] \right]  \rightarrow
%  \kk \left[ \left[ \xx \right] \right]$ for any
% commutative ring $\kk$, and because the power series $X_{n,d,s}$
% defined over $\kk$ is clearly the image of the power series $X_{n,d,s}$
% defined over $\ZZ$ under this homomorphism, whereas the image of the
% ring $\Lambda_\ZZ$ under this homomorphism is a subring of
% $\Lambda_\kk$).

Now, we are going to solve Exercise \ref{exe.shareshian-wachs.path}(b) by
strong induction over $n+d$. So (for the induction step) we need to show that
$X_{n,d,s}\in\Lambda$, and we can assume (as the induction hypothesis) that
$X_{n^{\prime},d^{\prime},s^{\prime}}\in\Lambda$ is already known to hold for
any nonnegative integers $n^{\prime}$, $d^{\prime}$ and $s^{\prime}$
satisfying $n^{\prime}+d^{\prime}<n+d$.

We must be in one of the following two cases:

\textit{Case 1:} We have $d>0$.

\textit{Case 2:} We have $d=0$.

Let us first consider Case 1. In this case, we have $d>0$. Hence,
$d-1\in \NN$. We also WLOG assume that $n$ is positive (otherwise,
$X_{n,d,s}$ is a constant and thus lies in $\Lambda$ for sure). Applying
(\ref{sol.shareshian-wachs.path.b.rec}) to $d-1$ instead of $d$, we obtain%
\[
dX_{n,d,s}+\left(  s+1\right)  X_{n,d-1,s+1}+\left(  n-d-s\right)
X_{n,d-1,s}=\sum_{i=1}^{n-1}\sum_{e=0}^{d-1}\sum_{t=0}^{s}X_{i,e,t}%
X_{n-i,d-1-e,s-t}.
\]
Thus,%
\begin{equation}
dX_{n,d,s}=\sum_{i=1}^{n-1}\sum_{e=0}^{d-1}\sum_{t=0}^{s}X_{i,e,t}%
X_{n-i,d-1-e,s-t}-\left(  s+1\right)  X_{n,d-1,s+1}+\left(  n-d-s\right)
X_{n,d-1,s}.
\label{sol.shareshian-wachs.path.b.5}
\end{equation}
Each of the power series $X_{i,e,t}$, $X_{n-i,d-1-e,s-t}$, $X_{n,d-1,s+1}$ and
$X_{n,d-1,s}$ on the right hand side of this equality is already known to lie
in $\Lambda$ (by the induction hypothesis). Hence, the right hand side of
(\ref{sol.shareshian-wachs.path.b.5}) lies in $\Lambda$ (since $\Lambda$ is a
ring), and thus (\ref{sol.shareshian-wachs.path.b.5}) shows that
$dX_{n,d,s}\in\Lambda$. Since $d>0$, this yields $X_{n,d,s}\in\Lambda$
(because if a power series $Q \in \ZZ\left[\left[\xx\right]\right]$
satisfies $dQ \in \Lambda$, then $Q$ must belong to $\Lambda$
itself\footnote{It is
here that we are using our assumption that $\kk = \ZZ$.}). Hence,
the induction step is complete in Case 1.

Let us now consider Case 2. In this case, we have $d=0$. Hence,
$X_{n,d,s} = X_{n,0,s}$.

We shall now show that
\begin{equation}
X_{n,0,s} = \sum\limits_{\substack{\lambda \in \Par_n ;\\
\ell\left(  \lambda\right)  =n-s}} m_{\lambda} .
\label{sol.shareshian-wachs.path.b.8}
\end{equation}


\textit{Proof of (\ref{sol.shareshian-wachs.path.b.8}):} Summing up the
equality (\ref{monomial-symmetric-function-definition}) over all $\lambda
\in\Par_{n}$ satisfying $\ell\left(  \lambda\right)
=n-s$, we obtain%
\begin{equation}
\sum\limits_{\substack{\lambda\in\Par_{n};\\\ell\left(
\lambda\right)  =n-s}}m_{\lambda}=\sum\limits_{\substack{\lambda\in
\Par_{n};\\\ell\left(  \lambda\right)  =n-s}}
\sum_{\alpha\in\Symm_{\left(  \infty\right)  }\lambda}
 \xx ^{\alpha}.
\label{sol.shareshian-wachs.path.b.8.pf.1}
\end{equation}
For any fixed weak composition $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\alpha_{3},...\right)  $, the monomial $\xx^{\alpha}$ occurs on the
right hand side of this equality if and only if it satisfies the following two properties:

\begin{itemize}
\item We have $\left\vert \alpha\right\vert =n$.

\item There are precisely $n-s$ positive integers $i$ satisfying $\alpha
_{i}\neq0$.
\end{itemize}

\noindent
Moreover, every such monomial occurs only once on this right hand side. Hence,
this right hand side can be rewritten as
\[
\sum\limits_{\substack{\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)
\text{ is}\\\text{a weak composition;}\\\left\vert \alpha\right\vert
=n;\\\text{there are precisely }n-s\\\text{positive integers }%
i\\\text{satisfying }\alpha_{i}\neq0}}{ \xx }^{\alpha}.
\]
Hence, (\ref{sol.shareshian-wachs.path.b.8.pf.1}) rewrites as%
\begin{equation}
\sum\limits_{\substack{\lambda\in\Par_{n};\\\ell\left(
\lambda\right)  =n-s}}m_{\lambda}=\sum\limits_{\substack{\alpha=\left(  \alpha
_{1},\alpha_{2},\alpha_{3},...\right)  \text{ is}\\\text{a weak composition;}%
\\\text{there are precisely }n-s\\\text{positive integers }i\\\text{satisfying
}\alpha_{i}\neq0}}{ \xx }^{\alpha}%
.\label{sol.shareshian-wachs.path.b.8.pf.2}%
\end{equation}


On the other hand, (\ref{sol.shareshian-wachs.path.b.def}) (applied to $0$
instead of $d$) yields%
\begin{equation}
X_{n,0,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =0;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
.\label{sol.shareshian-wachs.path.b.8.pf.3}%
\end{equation}
The condition $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =0$
on an $n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ is equivalent to $w_{1}\leq w_{2}\leq...\leq w_{n}$,
and therefore (\ref{sol.shareshian-wachs.path.b.8.pf.3}) rewrites as%
\begin{equation}
X_{n,0,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\w_{1}\leq w_{2}\leq...\leq
w_{n};\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}} .
\label{sol.shareshian-wachs.path.b.8.pf.4}
\end{equation}
For an $n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfying $w_{1}\leq w_{2}\leq...\leq w_{n}$, the
condition $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$ is
equivalent to the condition that $\left\vert \left\{  w_{1},w_{2}%
,...,w_{n}\right\}  \right\vert =n-s$\ \ \ \ \footnote{\textit{Proof.} Let
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}
^{n}$ be an $n$-tuple satisfying $w_{1}\leq w_{2}\leq...\leq w_{n}$. Then,
$\left\vert \left\{  w_{1},w_{2},...,w_{n}\right\}  \right\vert $ is the
number of \textbf{distinct} entries of the sequence $\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  $. Since this sequence is nondecreasing (because
$w_{1}\leq w_{2}\leq...\leq w_{n}$), this number equals $1$ plus the number of
times this sequence strictly increases, i.e., the number of $i\in\left\{
1,2,...,n-1\right\}  $ satisfying $w_{i}<w_{i+1}$. In other words,%
\[
\left\vert \left\{  w_{1},w_{2},...,w_{n}\right\}  \right\vert =1+\left\vert
\left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}<w_{i+1}\right\}
\right\vert .
\]
Hence,%
\[
\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
<w_{i+1}\right\}  \right\vert =\left\vert \left\{  w_{1},w_{2},...,w_{n}%
\right\}  \right\vert -1.
\]
But every $i\in\left\{  1,2,...,n-1\right\}  $ satisfies $w_{i}\leq w_{i+1}$
(since $w_{1}\leq w_{2}\leq...\leq w_{n}$). Thus, every $i\in\left\{
1,2,...,n-1\right\}  $ satisfies exactly one of the statements $w_{i}=w_{i+1}$
and $w_{i}<w_{i+1}$. Therefore, $\left\{  1,2,...,n-1\right\}  $ is the union
of the two disjoint subsets $\left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}<w_{i+1}\right\}  $ and $\left\{  i\in\left\{
1,2,...,n-1\right\}  \ \mid\ w_{i}=w_{i+1}\right\}  $. Hence,%
\begin{align*}
\left\vert \left\{  1,2,...,n-1\right\}  \right\vert  & =\left\vert \left\{
i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}<w_{i+1}\right\}  \right\vert
+\left\vert \underbrace{\left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}<w_{i+1}\right\}  }_{=\operatorname*{Stag}\left(  w\right)
}\right\vert \\
& =\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
<w_{i+1}\right\}  \right\vert +\left\vert \operatorname*{Stag}\left(
w\right)  \right\vert ,
\end{align*}
so that
\begin{align*}
\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert  &
=\underbrace{\left\vert \left\{  1,2,...,n-1\right\}  \right\vert }%
_{=n-1}-\underbrace{\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}<w_{i+1}\right\}  \right\vert }_{=\left\vert \left\{  w_{1}%
,w_{2},...,w_{n}\right\}  \right\vert -1}\\
& =n-1-\left(  \left\vert \left\{  w_{1},w_{2},...,w_{n}\right\}  \right\vert
-1\right)  =n-\left\vert \left\{  w_{1},w_{2},...,w_{n}\right\}  \right\vert .
\end{align*}
Thus, the condition $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$ is equivalent to the condition that $\left\vert \left\{
w_{1},w_{2},...,w_{n}\right\}  \right\vert =n-s$, qed.} Hence,
(\ref{sol.shareshian-wachs.path.b.8.pf.4}) rewrites as%
\begin{equation}
X_{n,0,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\w_{1}\leq w_{2}\leq...\leq
w_{n};\ \left\vert \left\{  w_{1},w_{2},...,w_{n}\right\}  \right\vert
=n-s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
.\label{sol.shareshian-wachs.path.b.8.pf.5}%
\end{equation}


Now, for any fixed weak composition $\alpha=\left(  \alpha_{1},\alpha
_{2},\alpha_{3},...\right)  $, the monomial $\xx^{\alpha}$ can be
written in the form $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ for an $n$-tuple
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}
^{n}$ satisfying $w_{1}\leq w_{2}\leq...\leq w_{n}$ if and only if this
monomial satisfies $\left\vert \alpha\right\vert =n$. If it does satisfy
$\left\vert \alpha\right\vert =n$, then it can be written in this form in
precisely one way. Hence, the only monomials $\xx^{\alpha}$ occurring on
the right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5}) are
monomials corresponding to weak compositions $\alpha$ satisfying $\left\vert
\alpha\right\vert =n$, and these monomials occur at most once on the right
hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5}). We can actually tell
which monomials do occur on this right hand side: Namely, the monomial
$\xx^{\alpha}$ corresponding to a fixed weak composition $\alpha
=\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)  $ occurs on the right
hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5}) if and only if it
satisfies the following two properties:

\begin{itemize}
\item We have $\left\vert \alpha\right\vert =n$.

\item There are precisely $n-s$ positive integers $i$ satisfying $\alpha
_{i}\neq0$. (This is caused by the condition $\left\vert \left\{  w_{1}%
,w_{2},...,w_{n}\right\}  \right\vert =n-s$ under the summation sign in
(\ref{sol.shareshian-wachs.path.b.8.pf.5}). Indeed, if a monomial
$\xx^{\alpha}$ is written in the form $x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}$ for an $n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n}$, then $\left\vert \left\{  w_{1}%
,w_{2},...,w_{n}\right\}  \right\vert $ is the number of all positive integers
$i$ satisfying $\alpha_{i}\neq0$ (because $w_{1}$, $w_{2}$, $...$, $w_{n}$
\textbf{are} all the positive integers $i$ satisfying $\alpha_{i}\neq0$).)
\end{itemize}

\noindent
Moreover, every monomial which satisfies these two properties occurs only once
on the right hand side of (\ref{sol.shareshian-wachs.path.b.8.pf.5}) (as we
already know). Hence, this right hand side can be rewritten as
\[
\sum\limits_{\substack{\alpha=\left(  \alpha_{1},\alpha_{2},\alpha_{3},...\right)
\text{ is}\\\text{a weak composition;}\\\left\vert \alpha\right\vert
=n;\\\text{there are precisely }n-s\\\text{positive integers }%
i\\\text{satisfying }\alpha_{i}\neq0}}{ \xx }^{\alpha}.
\]
Thus, (\ref{sol.shareshian-wachs.path.b.8.pf.5}) rewrites as%
\[
X_{n,0,s}=\sum\limits_{\substack{\alpha=\left(  \alpha_{1},\alpha_{2},\alpha
_{3},...\right)  \text{ is}\\\text{a weak composition;}\\\left\vert
\alpha\right\vert =n;\\\text{there are precisely }n-s\\\text{positive integers
}i\\\text{satisfying }\alpha_{i}\neq0}}{ \xx }^{\alpha}.
\]
Compared with (\ref{sol.shareshian-wachs.path.b.8.pf.2}), this yields
$X_{n,0,s}=\sum\limits_{\substack{\lambda\in\Par_{n}%
;\\\ell\left(  \lambda\right)  =n-s}}m_{\lambda}$. This proves
(\ref{sol.shareshian-wachs.path.b.8}).

Now, (\ref{sol.shareshian-wachs.path.b.8}) yields $X_{n,0,s}=\sum\limits
_{\substack{\lambda\in\Par_{n};\\\ell\left(
\lambda\right)  =n-s}}\underbrace{m_{\lambda}}_{\in\Lambda}\in\Lambda$, so
that $X_{n,d,s}=X_{n,0,s}\in\Lambda$. This completes the induction step in
Case 2.

Thus, the induction step is complete in both Cases 1 and 2. This finally
completes the induction step. Exercise \ref{exe.shareshian-wachs.path}(b) is
thus solved.

(a) \textit{First solution of Exercise \ref{exe.shareshian-wachs.path}(a):}
Recall the power series $X_{n,d,s}$ defined in Exercise
\ref{exe.shareshian-wachs.path}(b) for any $d\in \NN$ and $s\in
 \NN$. Exercise \ref{exe.shareshian-wachs.path}(b) yields that
$X_{n,k,0}\in\Lambda$. But an $n$-tuple $w$ is Smirnov if and only if its
stagnation set $\operatorname*{Stag}\left(  w\right)  $ is empty, i.e., if and
only if $\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =0$.
Hence, $X_{n,k,0}$ is the sum of the monomials $x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}$ over all Smirnov $n$-tuples $w\in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=k$. This shows that $X_{n,k,0}=X_{n,k}$. Thus, $X_{n,k}=X_{n,k,0}\in\Lambda$,
so that Exercise \ref{exe.shareshian-wachs.path}(a) is solved.

\textit{Second solution of Exercise \ref{exe.shareshian-wachs.path}(a):} Here
is an alternative solution to Exercise \ref{exe.shareshian-wachs.path}(a),
which avoids using part (b). It is an application of \cite[proof of Theorem
4.5]{ShareshianWachs-2014} to our special setting.

We proceed similarly to the proof of Proposition
\ref{Schur-functions-are-symmetric-prop}: It suffices to show that for every
positive integer $p$, the power series $X_{n,k}$ is invariant under swapping
the variables $x_{p}$ and $x_{p+1}$. So let us fix a positive integer $p$.

Let $S_{n,k}$ denote the set of all Smirnov $n$-tuples $w\in\left\{
1,2,3,\ldots\right\}  ^{n}$ satisfying $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =k$. Then, the definition of $X_{n,k}$ rewrites as
follows:%
\begin{equation}
X_{n,k}=\sum_{w=\left(  w_{1},w_{2},...,w_{n}\right)  \in S_{n,k}}x_{w_{1}%
}x_{w_{2}}...x_{w_{n}}.\label{sol.shareshian-wachs.path.a.1}%
\end{equation}


For every $w\in S_{n,k}$, let $ \xx _{w}$ denote the monomial $x_{w_{1}%
}x_{w_{2}}\ldots x_{w_{n}}$, where $w$ is written in the form $w=\left(
w_{1},w_{2},...,w_{n}\right)  $. Then, (\ref{sol.shareshian-wachs.path.a.1})
rewrites as%
\begin{equation}
X_{n,k}=\sum_{w\in S_{n,k}} \xx _{w}%
.\label{sol.shareshian-wachs.path.a.2}%
\end{equation}


We need to show that this power series $X_{n,k}$ is invariant under swapping
the variables $x_{p}$ and $x_{p+1}$. In order to do so, it is clearly enough
to provide an involution $J:S_{n,k}\rightarrow S_{n,k}$ which has the property
that for every $w\in S_{n,k}$, the monomial $ \xx _{J\left(  w\right)  }$
is obtained from the monomial $ \xx _{w}$ by swapping the variables
$x_{p}$ and $x_{p+1}$.

To define the involution $J$, we introduce some notations. If $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n}$ is any
$n$-tuple, then a nonempty interval $I$ of $\left\{  1,2,...,n\right\}  $ will
be called a \emph{$\left(  p,p+1\right)  $-interval} of $w$ if every $i\in
I$ satisfies $w_{i}\in\left\{  p,p+1\right\}  $. A
\emph{$\left(  p,p+1\right)$-run} of $w$ will mean a
$\left(  p,p+1\right)  $-interval of $w$
maximal with respect to inclusion. For instance, if $n=9$, $w=\left(
3,1,2,5,4,2,3,2,4\right)  $ and $p=2$, then the $\left(  p,p+1\right)
$-intervals of $w$ are $\left\{  1\right\}  $, $\left\{  3\right\}  $,
$\left\{  6\right\}  $, $\left\{  7\right\}  $, $\left\{  8\right\}  $,
$\left\{  6,7\right\}  $, $\left\{  7,8\right\}  $ and $\left\{
6,7,8\right\}  $ (since the letters of $w$ belonging to $\left\{
p,p+1\right\}  $ are the $1$-st, $3$-rd, the $6$-th, the $7$-th and the $8$-th
letter), while only $\left\{  1\right\}  $, $\left\{  3\right\}  $ and
$\left\{  6,7,8\right\}  $ are $\left(  p,p+1\right)  $-runs of $w$.

An interval of $\left\{  1,2,...,n\right\}  $ is said to be \emph{even} if
it has even size, and \emph{odd} if it has odd size.

It is easy to see that if $w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ is any Smirnov $n$-tuple, and $\left\{
a+1,a+2,...,b\right\}  $ is any $\left(  p,p+1\right)  $-interval of $w$, then
the $\left(  b-a\right)  $-tuple $\left(  w_{a+1},w_{a+2},...,w_{b}\right)  $
alternates between $p$'s and $\left(  p+1\right)  $'s, that is, has one of the
forms%
\[
\left(  p,p+1,p,p+1,p,...,p+1\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  p+1,p,p+1,p,p+1,...,p\right)
\]
(if $b-a$ is even) or one of the forms%
\[
\left(  p,p+1,p,p+1,p,...,p\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  p+1,p,p+1,p,p+1,...,p+1\right)
\]
(if $b-a$ is odd).

Now, we define a map $J:S_{n,k}\rightarrow S_{n,k}$ as follows: Let $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in S_{n,k}$. We know that $w$ is Smirnov and
satisfies $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =k$. It
is clear\footnote{since $\left(  p,p+1\right)  $-runs are $\left(
p,p+1\right)  $-intervals \textbf{maximal with respect to inclusion}} that the
$\left(  p,p+1\right)  $-runs of $w$ are disjoint and even separated
from each other by at least $1$ (this means that if $\alpha$ and $\beta$ are
elements of two distinct $\left(  p,p+1\right)  $-runs of $w$, then
$\left\vert \alpha-\beta\right\vert >1$). For every $i\in\left\{
1,2,...,n\right\}  $, define a positive integer $w_{i}^{\prime}$ as follows:

\begin{itemize}
\item If $i$ belongs to an odd $\left(  p,p+1\right)  $-run of $w$, then set
$w_{i}^{\prime}=
\begin{cases}
p+1, & \text{if }w_{i}=p;\\
p, & \text{if }w_{i}=p+1
\end{cases} \quad$.

\item Otherwise, set $w_{i}^{\prime}=w_{i}$.
\end{itemize}

Thus, we have defined an $n$-tuple $\left(  w_{1}^{\prime},w_{2}^{\prime
},...,w_{n}^{\prime}\right)  $ of positive integers.\footnote{Informally
speaking, $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime}\right)  $
is simply obtained by changing those $p$'s and $p+1$'s in $w$ whose positions
belong to odd $\left(  p,p+1\right)  $-runs of $w$ into $p+1$'s and $p$'s,
respectively, while leaving all other entries of $w$ intact. For instance, in
our above example of $n=9$, $w=\left(  3,1,2,5,4,2,3,2,4\right)  $ and $p=2$,
we would have $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime
}\right)  =\left(  2,1,3,5,4,3,2,3,4\right)  $. (All letters $2$ and $3$ have
been changed here because all $\left(  p,p+1\right)  $-runs of this $w$ were
odd.) For another example, if $n=5$, $w=\left(  2,3,1,5,2\right)  $ and $p=2$,
then $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime}\right)
=\left(  2,3,1,5,3\right)  $ (the first two letters are unchanged since the
$\left(  p,p+1\right)  $-run $\left\{  1,2\right\}  $ is even).} Denote this
$n$-tuple $\left(  w_{1}^{\prime},w_{2}^{\prime},...,w_{n}^{\prime}\right)  $
by $w^{\prime}$. This new $n$-tuple $w^{\prime}$ is again
Smirnov\footnote{Indeed, every $i\in\left\{  1,2,...,n-1\right\}  $ satisfies
$w_{i}^{\prime}\neq w_{i+1}^{\prime}$, because for every $i\in\left\{
1,2,...,n-1\right\}  $, we must be in one of the following three cases:
\par
\begin{enumerate}
\item both $i$ and $i+1$ belong to an odd $\left(  p,p+1\right)  $-run of $w$;
\par
\item exactly one of $i$ and $i+1$ belongs to an odd $\left(  p,p+1\right)
$-run of $w$;
\par
\item none of $i$ and $i+1$ belongs to an odd $\left(  p,p+1\right)  $-run of
$w$.
\end{enumerate}
\par
In the first of these cases, $w_{i}^{\prime}$ and $w_{i+1}^{\prime}$ are the
numbers $p$ and $p+1$ in some order, and therefore $w_{i}^{\prime}\neq
w_{i+1}^{\prime}$ holds. In the second case, one of $w_{i}^{\prime}$ and
$w_{i+1}^{\prime}$ is $p$ or $p+1$, whereas the other is neither $p$ nor
$p+1$; again, this yields $w_{i}^{\prime}\neq w_{i+1}^{\prime}$. In the third
case, $w_{i}^{\prime}=w_{i}$ and $w_{i+1}^{\prime}=w_{i+1}$, and therefore
\begin{align*}
w_{i}^{\prime} &  =w_{i}\neq w_{i+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left(  w_{1},w_{2},...,w_{n}\right)  =w\text{ is Smirnov}\right)  \\
&  =w_{i+1}^{\prime}%
\end{align*}
holds again. Hence, $w_{i}^{\prime}\neq w_{i+1}^{\prime}$ in every case, qed.}
and satisfies $\left\vert \operatorname{Des}\left(  w^{\prime}\right)
\right\vert =k$\ \ \ \ \footnote{\textit{Proof.} We will show that $\left\vert
\operatorname*{Des}\left(  w\right)  \right\vert =\left\vert
\operatorname{Des}\left(  w^{\prime}\right)  \right\vert $.
\par
Whenever $P=\left\{  a+1,a+2,...,b\right\}  $ is an odd interval of $\left\{
1,2,...,n\right\}  $, let $\underline{P}$ denote the interval $\left\{
a+1,a+2,...,b-1\right\}  $ (this is the interval $P$ without its highest
element). Clearly, $\underline{P}\subset P$ and $\underline{P}\subset\left\{
1,2,...,n-1\right\}  $. Let $Q$ be the set
\[
\left\{  1,2,...,n-1\right\}  \setminus\bigsqcup_{\substack{P\text{ is an
odd}\\\left(  p,p+1\right)  \text{-run of }w}}\underline{P}%
\]
(note that the union is disjoint because the odd $\left(  p,p+1\right)  $-runs
of $w$ are disjoint, and thus their subsets $\underline{P}$ are all the more
disjoint). Then,%
\[
\left\{  1,2,...,n-1\right\}  =\left(  \bigsqcup_{\substack{P\text{ is an
odd}\\\left(  p,p+1\right)  \text{-run of }w}}\underline{P}\right)  \sqcup Q.
\]
Thus, the equality $\operatorname*{Des}\left(  w\right)  =\operatorname*{Des}%
\left(  w\right)  \cap\left\{  1,2,...,n-1\right\}  $ (which is clear, because
$\operatorname*{Des}\left(  w\right)  \subset\left\{  1,2,...,n-1\right\}  $)
rewrites as
\[
\operatorname*{Des}\left(  w\right)  =\operatorname*{Des}\left(  w\right)
\cap\left(  \left(  \bigsqcup_{\substack{P\text{ is an odd}\\\left(
p,p+1\right)  \text{-run of }w}}\underline{P}\right)  \sqcup Q\right)
=\left(  \bigsqcup_{\substack{P\text{ is an odd}\\\left(  p,p+1\right)
\text{-run of }w}}\left(  \operatorname*{Des}\left(  w\right)  \cap
\underline{P}\right)  \right)  \sqcup\left(  \operatorname*{Des}\left(
w\right)  \cap Q\right)  .
\]
Thus,%
\begin{equation}
\left\vert \operatorname*{Des}\left(  w\right)  \right\vert =\left\vert
\left(  \bigsqcup_{\substack{P\text{ is an odd}\\\left(  p,p+1\right)
\text{-run of }w}}\left(  \operatorname*{Des}\left(  w\right)  \cap
\underline{P}\right)  \right)  \sqcup\left(  \operatorname*{Des}\left(
w\right)  \cap Q\right)  \right\vert =\sum\limits_{\substack{P\text{ is an
odd}\\\left(  p,p+1\right)  \text{-run of }w}}\left\vert \operatorname*{Des}%
\left(  w\right)  \cap\underline{P}\right\vert +\left\vert \operatorname*{Des}%
\left(  w\right)  \cap Q\right\vert .\label{sol.shareshian-wachs.path.a.a}%
\end{equation}
The same reasoning, applied to $w^{\prime}$ instead of $w$, yields%
\begin{equation}
\left\vert \operatorname*{Des}\left(  w^{\prime}\right)  \right\vert
=\sum\limits_{\substack{P\text{ is an odd}\\\left(  p,p+1\right)  \text{-run of }%
w}}\left\vert \operatorname*{Des}\left(  w^{\prime}\right)  \cap
\underline{P}\right\vert +\left\vert \operatorname*{Des}\left(  w^{\prime
}\right)  \cap Q\right\vert .\label{sol.shareshian-wachs.path.a.b}%
\end{equation}
Comparing (\ref{sol.shareshian-wachs.path.a.a}) and
(\ref{sol.shareshian-wachs.path.a.b}), it becomes clear that, in order to show
that $\left\vert \operatorname*{Des}\left(  w\right)  \right\vert =\left\vert
\operatorname*{Des}\left(  w^{\prime}\right)  \right\vert $, it is enough to
prove the following assertions:
\par
\begin{statement}
\textit{Assertion 1:} Whenever $P$ is an odd $\left(  p,p+1\right)  $-run of
$w$, we have $\left\vert \operatorname*{Des}\left(  w\right)  \cap
\underline{P}\right\vert =\left\vert \operatorname*{Des}\left(  w^{\prime
}\right)  \cap\underline{P}\right\vert $.
\end{statement}
\par
\begin{statement}
\textit{Assertion 2:} We have $\left\vert \operatorname*{Des}\left(  w\right)
\cap Q\right\vert =\left\vert \operatorname*{Des}\left(  w^{\prime}\right)
\cap Q\right\vert $.
\end{statement}
\par
\textit{Proof of Assertion 1:} Let $P$ be an odd $\left(  p,p+1\right)  $-run
of $w$. Write $P$ in the form $\left\{  a+1,a+2,...,b\right\}  $ for some
$a,b\in\left\{  1,2,...,n\right\}  $; note that $a<b$. Then, $\left(
w_{a+1},w_{a+2},...,w_{b}\right)  $ has one of the forms%
\[
\left(  p,p+1,p,p+1,p,...,p\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  p+1,p,p+1,p,p+1,...,p+1\right)  ,
\]
whereas $\left(  w_{a+1}^{\prime},w_{a+2}^{\prime},...,w_{b}^{\prime}\right)
$ has the respectively other form (by definition of $w_{i}^{\prime}$). It is
now easy to see that both $\left\vert \operatorname*{Des}\left(  w\right)
\cap\underline{P}\right\vert $ and $\left\vert \operatorname*{Des}\left(
w^{\prime}\right)  \cap\underline{P}\right\vert $ equal $\dfrac{b-a-1}{2}$
(more precisely, one of the sets $\operatorname*{Des}\left(  w\right)
\cap\underline{P}$ and $\operatorname*{Des}\left(  w^{\prime}\right)
\cap\underline{P}$ consists of all even elements of $\underline{P}$, whereas
the other consists of all odd elements of $\underline{P}$), and thus
$\left\vert \operatorname*{Des}\left(  w\right)  \cap\underline{P}\right\vert
=\left\vert \operatorname*{Des}\left(  w^{\prime}\right)  \cap\underline{P}%
\right\vert $. This proves Assertion 1.
\par
\textit{Proof of Assertion 2:} We can prove a stronger claim: namely, the sets
$\operatorname*{Des}\left(  w\right)  \cap Q$ and $\operatorname*{Des}\left(
w^{\prime}\right)  \cap Q$ are equal (not just their cardinalities).
\par
Indeed, proving this means proving that for every $q\in Q$, we have
$q\in\operatorname*{Des}\left(  w\right)  $ if and only if $q\in
\operatorname*{Des}\left(  w^{\prime}\right)  $. In order to prove this,
assume the contrary. Thus, there exists a $q\in Q$ such that%
\begin{equation}
\text{one of the statements }q\in\operatorname*{Des}\left(  w\right)  \text{
and }q\in\operatorname*{Des}\left(  w^{\prime}\right)  \text{ is satisfied but
not the other.}\label{sol.shareshian-wachs.path.a.c}%
\end{equation}
Consider this $q$.
\par
We have $q\in Q=\left\{  1,2,...,n-1\right\}  \setminus\bigsqcup
_{\substack{P\text{ is an odd}\\\left(  p,p+1\right)  \text{-run of }%
w}}\underline{P}$, so that $q\notin\bigsqcup_{\substack{P\text{ is an
odd}\\\left(  p,p+1\right)  \text{-run of }w}}\underline{P}$. In other words,
\begin{equation}
\text{for every odd }\left(  p,p+1\right)  \text{-run }P\text{ of }w\text{, we
have }q\notin\underline{P}.\label{sol.shareshian-wachs.path.a.d}%
\end{equation}
\par
But recall that $q\in\operatorname*{Des}\left(  w\right)  $ holds if and only
if $w_{q}>w_{q+1}$, and that $q\in\operatorname*{Des}\left(  w^{\prime
}\right)  $ holds if and only if $w_{q}^{\prime}>w_{q+1}^{\prime}$. Hence,
(\ref{sol.shareshian-wachs.path.a.c}) rewrites as follows:%
\[
\text{one of the statements }w_{q}>w_{q+1}\text{ and }w_{q}^{\prime}%
>w_{q+1}^{\prime}\text{ is satisfied but not the other.}%
\]
We assume WLOG that the statement $w_{q}>w_{q+1}$ is satisfied but
$w_{q}^{\prime}>w_{q+1}^{\prime}$ is not (the other case is proven in the same
way, except that more or less all $w_{i}$ are switched with the corresponding
$w_{i}^{\prime}$). Thus, we have $w_{q}>w_{q+1}$ but $w_{q}^{\prime}\leq
w_{q+1}^{\prime}$. Since $w^{\prime}$ is Smirnov, we have $w_{q}^{\prime}\neq
w_{q+1}^{\prime}$. Combined with $w_{q}^{\prime}\leq w_{q+1}^{\prime}$, this
yields $w_{q}^{\prime}<w_{q+1}^{\prime}$. Thus, $w_{q}^{\prime}\leq
w_{q+1}^{\prime}-1$ (since $w_{q}^{\prime}$ and $w_{q+1}^{\prime}$ are
integers). Also, $w_{q}>w_{q+1}$, so that $w_{q+1}<w_{q}$.
\par
But the construction of $w_{q}^{\prime}$ shows that $w_{q}^{\prime}$ differs
from $w_{q}$ at most by $1$. Thus, $w_{q}^{\prime}\geq w_{q}-1$, so that
$w_{q}-1\leq w_{q}^{\prime}$.
\par
Also, the construction of $w_{q+1}^{\prime}$ shows that $w_{q+1}^{\prime}$
differs from $w_{q+1}$ at most by $1$. Thus, $w_{q+1}^{\prime}\leq w_{q+1}+1$,
so that $w_{q+1}^{\prime}-1\leq w_{q+1}$.
\par
If $q$ does not belong to any odd $\left(  p,p+1\right)  $-run of $w$, then
$w_{q}^{\prime}=w_{q}$ (by the construction of $w_{q}^{\prime}$), which
quickly leads to a contradiction (indeed, it leads to $w_{q}=w_{q}^{\prime
}\leq w_{q+1}^{\prime}-1\leq w_{q+1}<w_{q}$, which is absurd). Hence, $q$ must
belong to an odd $\left(  p,p+1\right)  $-run of $w$.
\par
If $q+1$ does not belong to any odd $\left(  p,p+1\right)  $-run of $w$, then
$w_{q+1}^{\prime}=w_{q+1}$ (by the construction of $w_{q+1}^{\prime}$), which
quickly leads to a contradiction (indeed, it leads to $w_{q}^{\prime}%
\leq\underbrace{w_{q+1}^{\prime}}_{=w_{q+1}}-1=\underbrace{w_{q+1}}_{<w_{q}%
}-1<w_{q}-1\leq w_{q}^{\prime}$, which is absurd). Hence, $q+1$ must belong to
an odd $\left(  p,p+1\right)  $-run of $w$.
\par
So we know that each of $q$ and $q+1$ must belong to an odd $\left(
p,p+1\right)  $-run of $w$. Since $q$ and $q+1$ cannot belong to two different
$\left(  p,p+1\right)  $-runs of $w$ (since $\left(  p,p+1\right)  $-runs of
$w$ are disjoint and even separated from each other by at least $1$),
this shows that $q$ and $q+1$ belong to one and the same odd $\left(
p,p+1\right)  $-run of $w$. Denote this odd $\left(  p,p+1\right)  $-run by
$P$. Then, $q\in P$ and $q+1\in P$; this yields $q\in\underline{P}$, which
contradicts (\ref{sol.shareshian-wachs.path.a.d}). This contradiction
concludes our proof of the fact that the sets $\operatorname*{Des}\left(
w\right)  \cap Q$ and $\operatorname*{Des}\left(  w^{\prime}\right)  \cap Q$
are equal. Thus, Assertion 2 is proven, and this finishes the proof of
$\left\vert \operatorname*{Des}\left(  w\right)  \right\vert =\left\vert
\operatorname*{Des}\left(  w^{\prime}\right)  \right\vert $. Hence,
$\left\vert \operatorname*{Des}\left(  w^{\prime}\right)  \right\vert
=\left\vert \operatorname*{Des}\left(  w\right)  \right\vert =k$, qed.}. In
other words, $w^{\prime}\in S_{n,k}$. Now, set $J\left(  w\right)  =w^{\prime
}$. We have thus defined a map $J:S_{n,k}\rightarrow S_{n,k}$. It is easy to
see that, for every $w\in S_{n,k}$, the $\left(  p,p+1\right)  $-runs of
$J\left(  w\right)  $ are exactly the $\left(  p,p+1\right)  $-runs of $w$,
and applying the map $J$ to $J\left(  w\right)  $ precisely reverts the
changes made by the map $J$ to $w$. In other words, $J\circ
J= \id$, so that the map $J$ is an involution. Finally, it is
clear that for every $w\in S_{n,k}$, we have the following facts:

\begin{enumerate}
\item The number of entries equal to $p+1$ in $J\left(  w\right)  $ equals the
number of entries equal to $p$ in $w$.

\item The number of entries equal to $p$ in $J\left(  w\right)  $ equals the
number of entries equal to $p+1$ in $w$.

\item For every $j\in\left\{  1,2,3,...\right\}  \setminus\left\{
p,p+1\right\}  $, the number of entries equal to $j$ in $J\left(  w\right)  $
equals the number of entries equal to $j$ in $w$.
\end{enumerate}

Indeed, the third of these statements is clear because in the transformation
from $w$ to $J\left(  w\right)  $, the only entries that get changed are
entries equal to $p$ or $p+1$, and entries only get changed to either $p$ or
$p+1$. The first statement is easy to see as well (indeed, every entry equal
to $p$ in $w$ is contributed by a $\left(  p,p+1\right)  $-run\footnote{i.e.,
it has the form $w_{i}$ for $i$ belonging to some $\left( p, p+1 \right)$-run
of $w$}, and the same holds for every entry equal to $p+1$ in $J\left(
w\right)  $, and it is easy to see that \textbf{for each }$\left(
p,p+1\right)  $-run $R$ in $w$, the number of entries equal to $p+1$ in
$J\left(  w\right)  $ contributed by this $\left(  p,p+1\right)  $-run $R$
equals the number of entries equal to $p$ in $w$ contributed by this $\left(
p,p+1\right)  $-run $R$), and the second statement is analogous. The three
statements, combined, show that for every $w\in S_{n,k}$, the monomial
$ \xx _{J\left(  w\right)  }$ is obtained from the monomial
$ \xx _{w}$ by swapping the variables $x_{p}$ and $x_{p+1}$. This
concludes our solution of Exercise \ref{exe.shareshian-wachs.path}(a).

\textit{Remark:} We could have given an alternative solution to
Exercise~\ref{exe.shareshian-wachs.path}(b) that would still rely on
\eqref{sol.shareshian-wachs.path.b.rec}, but proceed by induction on
$n+s$ (rather than $n+d$) and handle the case $s=0$ separately
(rather than the case $d=0$ as we did). In the case $s=0$, the
assertion of Exercise~\ref{exe.shareshian-wachs.path}(b) follows
from Exercise~\ref{exe.shareshian-wachs.path}(a). (But this only works
combined with a solution to Exercise~\ref{exe.shareshian-wachs.path}(a)
that does not rely on Exercise~\ref{exe.shareshian-wachs.path}(b).)

(c) For every $d\in \NN$, $s\in \NN$ and $i\in\left\{
1,2,...,n-1\right\}  $, we define the power series $\mathcal{D}_{i}$,
$\mathcal{S}_{i}$ and $\mathcal{A}_{i}$ as in the solution to
Exercise~\ref{exe.shareshian-wachs.path}(b) above.

Let us first show that if $n$ is a positive integer, then any $d\in
{ \NN }$ and $s\in{ \NN }$ satisfy%
\begin{equation}
\left(  d+1\right)  U_{n,d+1,s}+\left(  s+1\right)  U_{n,d,s+1}+\left(
n-1-d-s\right)  U_{n,d,s}=\left(  d+1\right)  X_{n,d+1,s}.
\label{sol.shareshian-wachs.path.c.rec}%
\end{equation}


\textit{Proof of (\ref{sol.shareshian-wachs.path.c.rec}):} Let $n$ be a
positive integer. Let $d\in{ \NN }$ and $s\in{ \NN }$. Let
$i\in\left\{  1,2,...,n-1\right\}  $ be arbitrary.

Before we make any new definitions, let us recall a statement that we have
shown during our proof of (\ref{sol.shareshian-wachs.path.b.rec}) (back when
we were solving Exercise \ref{exe.shareshian-wachs.path}(b)): The power series
$\mathcal{D}_{i}\in \kk \left[  \left[  {\xx}\right]  \right]  $
is the sum of the monomials $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all
$n$-tuples $w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{n}$ satisfying $\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d$, $\left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s$ and $w_{i}>w_{i+1}$. Written as a formula, this
yields%
\[
\mathcal{D}_{i}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  i\right\}  \right\vert
=d;\ \left\vert \left(  \operatorname*{Stag}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =s;\\w_{i}>w_{i+1}}}x_{w_{1}}%
x_{w_{2}}\cdots x_{w_{n}}.
\]
Applying this to $n-i$ instead of $i$, we obtain%
\begin{align}
\mathcal{D}_{n-i}  &  =\underbrace{\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
n-i\right\}  \right\vert =d;\ \left\vert \left(  \operatorname*{Stag}\left(
w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert =s;\\w_{n-i}%
>w_{n-i+1}}}}_{=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=d;\ \left\vert \left(  \operatorname*{Stag}\left(  w\right)  \right)
\setminus\left\{  n-i\right\}  \right\vert =s;\\w_{n-i+1}<w_{n-i}}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}\left(
w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname*{Stag}\left(  w\right)  \right)  \setminus\left\{
n-i\right\}  \right\vert =s;\\w_{n-i+1}<w_{n-i}}}x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}. \label{sol.shareshian-wachs.path.c.oldrec}%
\end{align}


Now, let us make some more definitions:

\begin{itemize}
\item Define a power series $\mathcal{D}_{i}^{\prime}\in \kk \left[
\left[  {\xx}\right]  \right]  $ as the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d+1$,
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$, $w_{1}%
<w_{n}$ and $w_{i}>w_{i+1}$.

\item Define a power series $\mathcal{S}_{i}^{\prime}\in \kk \left[
\left[  {\xx}\right]  \right]  $ as the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$,
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s+1$,
$w_{1}<w_{n}$ and $w_{i}=w_{i+1}$.

\item Define a power series $\mathcal{A}_{i}^{\prime}\in \kk \left[
\left[  {\xx}\right]  \right]  $ as the sum of the monomials
$x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$
satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d$,
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$, $w_{1}%
<w_{n}$ and $w_{i}<w_{i+1}$.
\end{itemize}

These three power series $\mathcal{D}_{i}^{\prime}$, $\mathcal{S}_{i}^{\prime
}$ and $\mathcal{A}_{i}^{\prime}$ are not necessarily symmetric (but will
nevertheless come useful). Let us notice that the definitions of
$\mathcal{D}_{i}^{\prime}$, $\mathcal{S}_{i}^{\prime}$ and $\mathcal{A}%
_{i}^{\prime}$ can be rewritten as follows:

\begin{itemize}
\item The power series $\mathcal{D}_{i}^{\prime}\in \kk \left[  \left[
{\xx}\right]  \right]  $ is the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$, $w_{1}<w_{n}$ and $w_{i}>w_{i+1}$.

\item The power series $\mathcal{S}_{i}^{\prime}\in \kk \left[  \left[
{\xx}\right]  \right]  $ is the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$, $w_{1}<w_{n}$ and $w_{i}=w_{i+1}$.

\item The power series $\mathcal{A}_{i}^{\prime}\in \kk \left[  \left[
{\xx}\right]  \right]  $ is the sum of the monomials $x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples $w=\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfying
$\left\vert \left(  \operatorname{Des}\left(  w\right)  \right)
\setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$, $w_{1}<w_{n}$ and $w_{i}<w_{i+1}$.
\end{itemize}

These reformulations make it obvious that the sum $\mathcal{D}_{i}^{\prime
}+\mathcal{S}_{i}^{\prime}+\mathcal{A}_{i}^{\prime}$ is precisely the sum of
the monomials $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ over all $n$-tuples
$w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \left(  \operatorname{Des}\left(  w\right)
\right)  \setminus\left\{  i\right\}  \right\vert =d$, $\left\vert \left(
\operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{  i\right\}
\right\vert =s$ and $w_{1}<w_{n}$. In other words,
\begin{equation}
\mathcal{D}_{i}^{\prime}+\mathcal{S}_{i}^{\prime}+\mathcal{A}_{i}^{\prime
}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\label{sol.shareshian-wachs.path.c.rec.pf.1}%
\end{equation}


But the sum on the right hand side of
(\ref{sol.shareshian-wachs.path.c.rec.pf.1}) can be rewritten in terms of more
familiar sums. In fact, the $n$-tuples $w=\left(  w_{1},w_{2},...,w_{n}%
\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}$ are in bijection with the
pairs $\left(  u,v\right)  $ consisting of an $i$-tuple $u=\left(  u_{1}%
,u_{2},...,u_{i}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{i}$ and an
$\left(  n-i\right)  $-tuple $v=\left(  v_{1},v_{2},...,v_{n-i}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n-i}$. This bijection sends an $n$-tuple
$\left(  w_{1},w_{2},...,w_{n}\right)  $ to the pair $\left(  \left(
w_{1},w_{2},...,w_{i}\right)  ,\left(  w_{i+1},w_{i+2},...,w_{n}\right)
\right)  $, and has the property that $\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $\ \ \ \ \footnote{In fact,
$\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
i\right\}  $ is the union of the set $\operatorname*{Des}\left(  u\right)  $
with the set $\operatorname*{Des}\left(  v\right)  $ shifted by $i$ (that is,
the set $\left\{  p+i\ \mid\ p\in\operatorname*{Des}\left(  v\right)
\right\}  $). This is a disjoint union, and thus we find that
$\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $
(since the set $\operatorname*{Des} \left(  v\right)  $ shifted by $i$
has cardinality $\left\vert \operatorname*{Des}\left( v \right)
\right\vert$).
}, $\left\vert \left(  \operatorname*{Stag}\left(  w\right)
\right)  \setminus\left\{  i\right\}  \right\vert =\left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert $ \ \ \ \ \footnote{for
similar reasons}, $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}$, $w_{1}=u_{1}$ and
$w_{n}=v_{n-i}$. Hence,%
\begin{align}
&  \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
} \nonumber \\
&=\underbrace{\sum\limits_{\substack{u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\v=\left(  v_{1},v_{2},...,v_{n-i}%
\right)  \in\left\{  1,2,3,...\right\}  ^{n-i};\\\left\vert
\operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert =d;\\\left\vert
\operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}<v_{n-i}}}}%
_{=\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}}\underbrace{x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}%
}\cdots x_{v_{n-i}}}_{=x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}x_{u_{1}}x_{u_{2}%
}\cdots x_{u_{i}}}\nonumber\\
&  =\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}. \label{sol.shareshian-wachs.path.c.rec.pf.15}%
\end{align}


On the other hand, the pairs $\left(  v,u\right)  $ consisting of an $\left(
n-i\right)  $-tuple $v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n-i}$ and an $i$-tuple $u=\left(  u_{1},u_{2}%
,...,u_{i}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{i}$ are in bijection
with the $n$-tuples $w=\left(  w_{1},w_{2},...,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}$. This bijection sends a pair $\left(  v,u\right)
$ with $v=\left(  v_{1},v_{2},...,v_{n-i}\right)  $ and $u=\left(  u_{1}%
,u_{2},...,u_{i}\right)  $ to the $n$-tuple $\left(  v_{1},v_{2}%
,...,v_{n-i},u_{1},u_{2},...,u_{i}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$, and has the property that $\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $\ \ \ \ \footnote{In fact,
$\left(  \operatorname*{Des}\left(  w\right)  \right)  \setminus\left\{
n-i\right\}  $ is the union of the set $\operatorname*{Des}\left(  v\right)  $
with the set $\operatorname*{Des}\left(  u\right)  $ shifted by $n-i$ (that
is, the set $\left\{  p+\left(  n-i\right)  \ \mid\ p\in\operatorname*{Des}%
\left(  u\right)  \right\}  $). This is a disjoint union; thus, we find
$\left\vert \left(  \operatorname*{Des}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=\left\vert \operatorname*{Des}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Des}\left(  v\right)  \right\vert $
(since the set $\operatorname*{Des} \left( u \right)  $ shifted by $n-i$
has cardinality $\left\vert \operatorname*{Des}\left( u \right)
\right\vert$).
}, $\left\vert \left(  \operatorname*{Stag}%
\left(  w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert
=\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert +\left\vert
\operatorname*{Stag}\left(  v\right)  \right\vert $ \ \ \ \ \footnote{for
similar reasons}, $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}$, $w_{n-i+1}=u_{1}$ and
$w_{n-i}=v_{n-i}$. We can use this bijection to transform the sum on the right
hand side of (\ref{sol.shareshian-wachs.path.c.rec.pf.15}), and obtain
\begin{align}
&  \sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}x_{u_{1}}x_{u_{2}}\cdots x_{u_{i}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{n-i}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname*{Des}\left(
w\right)  \right)  \setminus\left\{  n-i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname*{Stag}\left(  w\right)  \right)  \setminus\left\{
n-i\right\}  \right\vert =s;\\w_{n-i+1}<w_{n-i}}}x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}=\mathcal{D}_{n-i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.oldrec})}\right)  .
\label{sol.shareshian-wachs.path.c.rec.pf.17}%
\end{align}
Now, (\ref{sol.shareshian-wachs.path.c.rec.pf.1}) becomes%
\begin{align}
\mathcal{D}_{i}^{\prime}+\mathcal{S}_{i}^{\prime}+\mathcal{A}_{i}^{\prime}  &
=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \left(  \operatorname{Des}\left(
w\right)  \right)  \setminus\left\{  i\right\}  \right\vert =d;\ \left\vert
\left(  \operatorname{Stag}\left(  w\right)  \right)  \setminus\left\{
i\right\}  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\nonumber\\
&  =\sum\limits_{\substack{v=\left(  v_{1},v_{2},...,v_{n-i}\right)  \in\left\{
1,2,3,...\right\}  ^{n-i};\\u=\left(  u_{1},u_{2},\ldots,u_{i}\right)
\in\left\{  1,2,3,...\right\}  ^{i};\\\left\vert \operatorname*{Des}\left(
u\right)  \right\vert +\left\vert \operatorname*{Des}\left(  v\right)
\right\vert =d;\\\left\vert \operatorname*{Stag}\left(  u\right)  \right\vert
+\left\vert \operatorname*{Stag}\left(  v\right)  \right\vert =s;\\u_{1}%
<v_{n-i}}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{n-i}}x_{u_{1}}x_{u_{2}}\cdots
x_{u_{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.rec.pf.15})}\right) \nonumber\\
&  =\mathcal{D}_{n-i}. \label{sol.shareshian-wachs.path.c.rec.pf.4}%
\end{align}


Now, let us forget that we fixed $i$. We thus have defined $\mathcal{D}%
_{i}^{\prime}$, $\mathcal{S}_{i}^{\prime}$ and $\mathcal{A}_{i}^{\prime}$ and
proven the equality (\ref{sol.shareshian-wachs.path.c.rec.pf.4}) for all
$i\in\left\{  1,2,...,n-1\right\}  $.

Now, let us take a closer look at the sums $\sum_{i=1}^{n-1}\mathcal{D}%
_{i}^{\prime}$, $\sum_{i=1}^{n-1}\mathcal{S}_{i}^{\prime}$ and $\sum
_{i=1}^{n-1}\mathcal{A}_{i}^{\prime}$:

\begin{itemize}
\item The definition of $\mathcal{D}_{i}^{\prime}$ can be rewritten as
follows:%
\[
\mathcal{D}_{i}^{\prime}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d+1;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}<w_{n}%
;\ w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,n-1\right\}  .
\]
Summing up these equations over all $i\in\left\{  1,2,...,n-1\right\}  $, we
obtain%
\begin{align}
\sum_{i=1}^{n-1}\mathcal{D}_{i}^{\prime}  &  =\underbrace{\sum_{i=1}^{n-1}%
\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n};\ w_{i}>w_{i+1}}}}_{=\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d+1;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}<w_{n}}%
}\sum\limits_{\substack{i\in\left\{  1,2,...,n-1\right\}  ;\\w_{i}>w_{i+1}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}\underbrace{\sum\limits_{\substack{i\in\left\{
1,2,...,n-1\right\}  ;\\w_{i}>w_{i+1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
}_{\substack{=\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}
\ \mid\ w_{i}>w_{i+1}\right\}  \right\vert x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\\=\left(  d+1\right)  x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\\text{(by
(\ref{sol.shareshian-wachs.path.b.rec.pf.6a}))}}}\nonumber\\
&  =\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d+1;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}\left(  d+1\right)  x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\nonumber\\
&  =\left(  d+1\right)  \underbrace{\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d+1;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}}_{\substack{=U_{n,d+1,s}\\\text{(by
(\ref{eq.exe.shareshian-wachs.path.c.U}), applied to }d+1\text{ instead of
}d\text{)}}}\nonumber\\
&  =\left(  d+1\right)  U_{n,d+1,s}.
\label{sol.shareshian-wachs.path.c.rec.pf.7a}%
\end{align}


\item Similarly, we can see that%
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{S}_{i}^{\prime}=\left(  s+1\right)  U_{n,d,s+1}.
\label{sol.shareshian-wachs.path.c.rec.pf.7b}%
\end{equation}
(In proving this, we have to replace
(\ref{sol.shareshian-wachs.path.b.rec.pf.6a}) by the following fact: If an
$n$-tuple $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfies $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$ and $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s+1$, then
\[
\left\vert \left\{  i\in\left\{  1,2,...,n-1\right\}  \ \mid\ w_{i}%
=w_{i+1}\right\}  \right\vert =s+1.
\]
This is proven in the same way as (\ref{sol.shareshian-wachs.path.b.rec.pf.6a}).)

\item Similarly, we can see that
\begin{equation}
\sum_{i=1}^{n-1}\mathcal{A}_{i}^{\prime}=\left(  n-1-d-s\right)  U_{n,d,s}.
\label{sol.shareshian-wachs.path.c.rec.pf.7c}%
\end{equation}
(In proving this, we have to replace
(\ref{sol.shareshian-wachs.path.b.rec.pf.6a}) by
(\ref{sol.shareshian-wachs.path.b.rec.pf.6c}).)
\end{itemize}

Now, summing the equality (\ref{sol.shareshian-wachs.path.c.rec.pf.4}) over
all $i\in\left\{  1,2,...,n-1\right\}  $ yields%
\begin{align*}
\sum_{i=1}^{n-1}\left(  \mathcal{D}_{i}^{\prime}+\mathcal{S}_{i}^{\prime
}+\mathcal{A}_{i}^{\prime}\right) 
&=\sum_{i=1}^{n-1}\mathcal{D}_{n-i}
=\sum_{i=1}^{n-1}\mathcal{D}_{i}
\qquad \qquad \left(\text{here, we substituted } i
\text{ for } n-i \text{in the sum}\right) \\
&=\left(
d+1\right)  X_{n,d+1,s}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.b.rec.pf.7a})}\right)  .
\end{align*}
Hence,%
\begin{align*}
 \left(  d+1\right)  X_{n,d+1,s}
&  =\sum_{i=1}^{n-1}\left(
 \mathcal{D}^\prime_{i}+\mathcal{S}^\prime_{i}+\mathcal{A}^\prime_{i}
\right)
 = \underbrace{\sum_{i=1}^{n-1}\mathcal{D}^\prime_{i}}_{\substack{=\left(
d+1\right)  U_{n,d+1,s}\\\text{(by
(\ref{sol.shareshian-wachs.path.c.rec.pf.7a}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{S}^\prime_{i}}_{\substack{=\left(  s+1\right)  U_{n,d,s+1}\\\text{(by
(\ref{sol.shareshian-wachs.path.c.rec.pf.7b}))}}}+\underbrace{\sum_{i=1}%
^{n-1}\mathcal{A}^\prime_{i}}_{\substack{=\left(  n-1-d-s\right)  U_{n,d,s}%
\\\text{(by (\ref{sol.shareshian-wachs.path.c.rec.pf.7c}))}}}\\
&  =\left(  d+1\right)  U_{n,d+1,s}+\left(  s+1\right)  U_{n,d,s+1}+\left(
n-1-d-s\right)  U_{n,d,s}.
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.c.rec}).

In solving Exercise~\ref{exe.shareshian-wachs.path}(c), we WLOG assume that
$\kk ={ \ZZ }$ (because there is a canonical ring homomorphism
${ \ZZ }\left[  \left[  {\mathbf{x}}\right]  \right]  \rightarrow
 \kk \left[  \left[  {\mathbf{x}}\right]  \right]  $ for any commutative
ring $\kk$, and because the power series $U_{n,d,s}$, $V_{n,d,s}$ and
$W_{n,d,s}$ defined over $\kk$ are clearly the image of the
corresponding power series $U_{n,d,s}$, $V_{n,d,s}$ and $W_{n,d,s}$ defined
over $\ZZ$ under this homomorphism, whereas the image of the
ring $\Lambda_\ZZ$ under this homomorphism is a subring of
$\Lambda_\kk$).

Now, we are going to prove $U_{n,d,s}\in\Lambda$ by strong induction over
$n+d$. So (for the induction step) we need to show that $U_{n,d,s}\in\Lambda$,
and we can assume (as the induction hypothesis) that $U_{n^{\prime},d^{\prime
},s^{\prime}}\in\Lambda$ is already known to hold for any nonnegative integers
$n^{\prime}$, $d^{\prime}$ and $s^{\prime}$ satisfying $n^{\prime}+d^{\prime
}<n+d$.

We must be in one of the following two cases:

\textit{Case 1:} We have $d>0$.

\textit{Case 2:} We have $d=0$.

Let us first consider Case 1. In this case, we have $d>0$. Hence,
$d-1\in{ \NN }$. We also WLOG assume that $n$ is positive (otherwise,
$X_{n,d,s}$ is a constant and thus lies in $\Lambda$ for sure). Applying
(\ref{sol.shareshian-wachs.path.c.rec}) to $d-1$ instead of $d$, we obtain%
\[
dU_{n,d,s}+\left(  s+1\right)  U_{n,d-1,s+1}+\left(  n-d-s\right)
U_{n,d-1,s}= d X_{n,d,s}.
\]
Thus,%
\begin{equation}
dU_{n,d,s}= d X_{n,d,s}-\left(  s+1\right)  U_{n,d-1,s+1}%
+\left(  n-d-s\right)  U_{n,d-1,s}.\label{sol.shareshian-wachs.path.c.5}%
\end{equation}
Each of the power series $U_{n,d-1,s+1}$ and $U_{n,d-1,s}$ on the right hand
side of this equality is already known to lie in $\Lambda$ (by the induction
hypothesis), and so is the power series $X_{n,d,s}$ (by Exercise
\ref{exe.shareshian-wachs.path}(b)). Hence,
the right hand side of (\ref{sol.shareshian-wachs.path.c.5}) lies in $\Lambda$
(since $\Lambda$ is a ring), and thus (\ref{sol.shareshian-wachs.path.c.5})
shows that $dU_{n,d,s}\in\Lambda$. Since $d>0$, this yields $U_{n,d,s}%
\in\Lambda$ (because if a power series $Q\in \ZZ \left[  \left[
\xx \right]  \right]  $ satisfies $dQ \in \Lambda$, then $Q$ must belong
to $\Lambda$ itself\footnote{It is here that we are using our assumption that
$\kk ={ \ZZ }$.}). Hence, the induction step is complete in Case 1.

Let us now consider Case 2. In this case, we have $d=0$. Hence, $U_{n,d,s}%
=U_{n,0,s}$.

We now distinguish between two subcases:

\textit{Subcase 2.1:} We have $s=n-1$.

\textit{Subcase 2.2:} We don't have $s=n-1$.

Let us consider Subcase 2.1 first. In this subcase, we have $s=n-1$. We will
show that $U_{n,d,s}=0$ in this subcase.

Indeed, let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}$ be such that $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$, $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$ and $w_{1}<w_{n}$. Then, $\operatorname{Stag}\left(  w\right)
$ is a subset of $\left\{  1,2,...,n-1\right\}  $ whose cardinality is
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s=n-1=\left\vert
\left\{  1,2,...,n-1\right\}  \right\vert $. Obviously, the only such subset
is $\left\{  1,2,...,n-1\right\}  $ itself, and so $\operatorname*{Stag}%
\left(  w\right)  $ must be $\left\{  1,2,...,n-1\right\}  $. Hence, every
$j\in\left\{  1,2,...,n-1\right\}  $ satisfies $j\in\left\{
1,2,...,n-1\right\}  =\operatorname*{Stag}\left(  w\right)  =\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  :w_{i}=w_{i+1}\right\}  $ and thus
$w_{j}=w_{j+1}$. In other words, $w_{1}=w_{2}=...=w_{n}$. This contradicts
$w_{1}<w_{n}$.

Now, forget that we fixed $w$. We thus have obtained a contradiction for every
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d$, $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$ and
$w_{1}<w_{n}$. Therefore, there exists no such $w$. Hence, the sum
$\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ is empty and thus equals
$0$. Consequently, (\ref{eq.exe.shareshian-wachs.path.c.U}) becomes%
\[
U_{n,d,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=0\in\Lambda.
\]
Hence, the induction step is complete in Subcase 2.1.

Let us now consider Subcase 2.2. In this case, we don't have $s=n-1$. In other
words, we have $s\neq n-1$. We will show that $U_{n,d,s}=X_{n,d,s}$ in this subcase.

Indeed, let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n}$ be such that $\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d$ and $\left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s$. We will show that $w_{1}<w_{n}$.

We have $\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d=0$, so
that the set $\operatorname{Des}\left(  w\right)  $ is empty. Now, every
$j\in\left\{  1,2,...,n-1\right\}  $ satisfies $w_{j}\leq w_{j+1}%
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, not every
$j\in\left\{  1,2,...,n-1\right\}  $ satisfies $w_{j}\leq w_{j+1}$. Hence,
there exists some $j\in\left\{  1,2,...,n-1\right\}  $ which does not satisfy
$w_{j}\leq w_{j+1}$. Fix such an $j$. Thus, $j$ does not satisfy $w_{j}\leq
w_{j+1}$. Hence, we have $w_{j}>w_{j+1}$, so that $j\in\left\{  i\in\left\{
1,2,\ldots,n-1\right\}  :w_{i}>w_{i+1}\right\}  =\operatorname*{Des}\left(
w\right)  $ (since $\operatorname*{Des}\left(  w\right)  =\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  :w_{i}>w_{i+1}\right\}  $). Hence, the
set $\operatorname*{Des}\left(  w\right)  $ has at least one element (namely,
$j$), and thus is nonempty. This contradicts the fact that the set
$\operatorname{Des}\left(  w\right)  $ is empty. This contradiction proves
that our assumption was wrong, qed.}. Hence, we have the chain of inequalities
$w_{1}\leq w_{2}\leq...\leq w_{n}$. At least one inequality in this chain must
be strict\footnote{\textit{Proof.} Assume the contrary. Then, none of the
inequalities in the chain $w_{1}\leq w_{2}\leq...\leq w_{n}$ is strict. Thus,
we have $w_{1}=w_{2}=...=w_{n}$. Thus, every $j\in\left\{
1,2,...,n-1\right\}  $ satisfies $w_{j}=w_{j+1}$. In other words, every
$j\in\left\{  1,2,...,n-1\right\}  $ satisfies%
\[
j\in\left\{  i\in\left\{  1,2,\ldots,n-1\right\}  :w_{i}=w_{i+1}\right\}
=\operatorname*{Stag}\left(  w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\operatorname*{Stag}\left(  w\right)  =\left\{  i\in\left\{  1,2,\ldots
,n-1\right\}  :w_{i}=w_{i+1}\right\}  \right)  .
\]
Hence, $\left\{  1,2,...,n-1\right\}  \subset\operatorname*{Stag}\left(
w\right)  $. Combined with the obvious fact $\operatorname*{Stag}\left(
w\right)  \subset\left\{  1,2,...,n-1\right\}  $, this yields
$\operatorname*{Stag}\left(  w\right)  =\left\{  1,2,...,n-1\right\}  $.
Hence, $\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert
=\left\vert \left\{  1,2,...,n-1\right\}  \right\vert =n-1$. This contradicts
$\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =s\neq n-1$.
This contradiction completes the proof.}, and thus we have $w_{1}<w_{n}$.

Now, let us forget that we fixed $w$. We thus have proven that every
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n}$ satisfying $\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d$ and $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$
automatically satisfies $w_{1}<w_{n}$. Hence, the condition $w_{1}<w_{n}$
under the summation sign ``$\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}$'' is redundant (i.e., can be removed
without changing the range of the summation). Thus,%
\[
\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=\sum\limits_{\substack{w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert
=d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}=X_{n,d,s}%
\]
(by (\ref{sol.shareshian-wachs.path.b.def})). Now,
(\ref{eq.exe.shareshian-wachs.path.c.U}) becomes%
\[
U_{n,d,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=X_{n,d,s}%
\in\Lambda
\]
(by Exercise \ref{exe.shareshian-wachs.path}(b)). Hence, the induction step is
complete in Subcase 2.2.

The induction step is thus complete in each of the two Subcases 2.1 and 2.2.
Since these two Subcases cover the whole Case 2, this yields that the
induction step is complete in the whole Case 2.

Thus, the induction step is complete in both Cases 1 and 2. This finally
completes the induction step. This proves that
\begin{equation}
\label{sol.shareshian-wachs.path.c.Udone}
U_{n, d, s} \in \Lambda
\ \ \ \ \ \ \ \ \ \ \text{for all positive integers } n
\text{, all } d \in \NN \text{ and all } s \in \NN .
\end{equation}

In order to complete the solution of Exercise~\ref{exe.shareshian-wachs.path}(c),
we still need to prove that $V_{n, d, s}$ and $W_{n, d, s}$
belong to $\Lambda$ for all positive integers $n$, all
$d \in \NN$ and all $s \in \NN$.

Let $n$ be a positive integer. Let $d \in \NN$. Let $s \in \NN$.

Define a map $\operatorname*{rev}:\left\{  1,2,3,\ldots\right\}
^{n}\rightarrow\left\{  1,2,3,\ldots\right\}  ^{n}$ by
\[
\left(  \operatorname*{rev}\left(  w_{1},w_{2},...,w_{n}\right)  =\left(
w_{n},w_{n-1},...,w_{1}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
w_{1},w_{2},...,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}\right)
.
\]
It is clear that this map $\operatorname*{rev}$ is a bijection. Also, every
$n$-tuple $w\in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfies
\begin{equation}
\left\vert \operatorname*{Des}\left(  \operatorname*{rev}w\right)  \right\vert
=n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert -\left\vert
\operatorname*{Stag}\left(  w\right)  \right\vert
\label{sol.shareshian-wachs.path.c.revDes}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.shareshian-wachs.path.c.revDes}):} Let
$w\in\left\{  1,2,3,\ldots\right\}  ^{n}$. Write $w$ in the form $w=\left(
w_{1},w_{2},...,w_{n}\right)  $. Then, the definition of $\operatorname*{rev}%
w$ yields $\operatorname*{rev}w=\left(  w_{n},w_{n-1},...,w_{1}\right)  $.
Hence, the definition of $\operatorname*{Des}\left(  \operatorname*{rev}%
w\right)  $ yields%
\[
\operatorname*{Des}\left(  \operatorname*{rev}w\right)  =\left\{  i\in\left\{
1,2,\ldots,n-1\right\}  :w_{n+1-i}>w_{n+1-\left(  i+1\right)  }\right\}  .
\]
Thus, for any $i\in\left\{  1,2,...,n-1\right\}  $, we have the following
equivalence of assertions:%
\begin{equation}
\left(  i\in\operatorname*{Des}\left(  \operatorname*{rev}w\right)  \right)
\ \Longleftrightarrow\ \left(  w_{n+1-i}>w_{n+1-\left(  i+1\right)  }\right)
. \label{sol.shareshian-wachs.path.c.revDes.pf.1}%
\end{equation}
\par
Also, $\operatorname*{Des}\left(  w\right)  =\left\{  i\in\left\{
1,2,\ldots,n-1\right\}  :w_{i}>w_{i+1}\right\}  $. Hence, for any
$i\in\left\{  1,2,...,n-1\right\}  $, we have the following equivalence of
assertions:%
\begin{equation}
\left(  i\in\operatorname*{Des}\left(  w\right)  \right)
\ \Longleftrightarrow\ \left(  w_{i}>w_{i+1}\right)  .
\label{sol.shareshian-wachs.path.c.revDes.pf.2}%
\end{equation}
\par
Finally, $\operatorname*{Stag}\left(  w\right)  =\left\{  i\in\left\{
1,2,\ldots,n-1\right\}  :w_{i}=w_{i+1}\right\}  $. Hence, for any
$i\in\left\{  1,2,...,n-1\right\}  $, we have the following equivalence of
assertions:%
\begin{equation}
\left(  i\in\operatorname*{Stag}\left(  w\right)  \right)
\ \Longleftrightarrow\ \left(  w_{i}=w_{i+1}\right)  .
\label{sol.shareshian-wachs.path.c.revDes.pf.3}%
\end{equation}
\par
The sets $\operatorname*{Stag}\left(  w\right)  $ and $\operatorname*{Des}%
\left(  w\right)  $ are disjoint (because these sets $\operatorname*{Stag}%
\left(  w\right)  =\left\{  i\in\left\{  1,2,\ldots,n-1\right\}
:w_{i}=w_{i+1}\right\}  $ and $\operatorname*{Des}\left(  w\right)  =\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  :w_{i}>w_{i+1}\right\}  $ are defined by
two mutually exclusive conditions on their elements). Hence, $\left\vert
\operatorname*{Stag}\left(  w\right)  \cup\operatorname*{Des}\left(  w\right)
\right\vert =\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert
+\left\vert \operatorname*{Des} \left( w \right) \right\vert $.
Also, $\operatorname*{Stag}%
\left(  w\right)  \cup\operatorname*{Des}\left(  w\right)  \subset\left\{
1,2,...,n-1\right\}  $ (since both $\operatorname*{Stag}\left(  w\right)  $
and $\operatorname*{Des}\left(  w\right)  $ are subsets of $\left\{
1,2,...,n-1\right\}  $).
\par
Now, for every $K\subset\left\{  1,2,...,n-1\right\}  $, let $n-K$ denote the
set $\left\{  n-k\ \mid\ k\in K\right\}  $; this set $n-K$ is again a subset
of $\left\{  1,2,...,n-1\right\}  $ and has cardinality $\left\vert
n-K\right\vert =\left\vert K\right\vert $. For any $i\in\left\{
1,2,...,n-1\right\}  $, we have the following equivalence of assertions:%
\begin{align*}
\left(  i\in n-\operatorname*{Des}\left(  \operatorname*{rev}w\right)
\right)  \  &  \Longleftrightarrow\ \left(  i\in\left\{  n-k\ \mid
\ k\in\operatorname*{Des}\left(  \operatorname*{rev}w\right)  \right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-\operatorname*{Des}\left(
\operatorname*{rev}w\right)  =\left\{  n-k\ \mid\ k\in\operatorname*{Des}%
\left(  \operatorname*{rev}w\right)  \right\}  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{i=n-k}_{\text{this is equivalent
to }n-i=k}\text{ for some }k\in\operatorname*{Des}\left(  \operatorname*{rev}%
w\right)  \right) \\
&  \Longleftrightarrow\ \left(  n-i=k\text{ for some }k\in\operatorname*{Des}%
\left(  \operatorname*{rev}w\right)  \right) \\
&  \Longleftrightarrow\ \left(  n-i\in\operatorname*{Des}\left(
\operatorname*{rev}w\right)  \right)  \ \Longleftrightarrow\ \left(
w_{n+1-\left(  n-i\right)  }>w_{n+1-\left(  \left(  n-i\right)  +1\right)
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.revDes.pf.1}), applied to }n-i\text{ instead
of }i\right) \\
&  \Longleftrightarrow\ \left(  w_{i+1}>w_{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n+1-\left(  n-i\right)  =i+1\text{
and }n+1-\left(  \left(  n-i\right)  +1\right)  =i\right) \\
&  \Longleftrightarrow\ \left(  w_{i}<w_{i+1}\right)  \ \Longleftrightarrow
\ \left(  \text{neither }\underbrace{w_{i}=w_{i+1}}%
_{\substack{\Longleftrightarrow\left(  i\in\operatorname*{Stag}\left(
w\right)  \right)  \\\text{(by (\ref{sol.shareshian-wachs.path.c.revDes.pf.3}%
))}}}\text{ nor }\underbrace{w_{i}>w_{i+1}}_{\substack{\Longleftrightarrow
\left(  i\in\operatorname*{Des}\left(  w\right)  \right)  \\\text{(by
(\ref{sol.shareshian-wachs.path.c.revDes.pf.2}))}}}\right) \\
&  \Longleftrightarrow\ \left(  \text{neither }i\in\operatorname*{Stag}\left(
w\right)  \text{ nor }i\in\operatorname*{Des}\left(  w\right)  \right) \\
&  \Longleftrightarrow\ \left(  i\notin\operatorname*{Stag}\left(  w\right)
\cup\operatorname*{Des}\left(  w\right)  \right) \\
&  \Longleftrightarrow\ \left(  i\in\left\{  1,2,...,n-1\right\}
\setminus\left(  \operatorname*{Stag}\left(  w\right)  \cup\operatorname*{Des}%
\left(  w\right)  \right)  \right)  .
\end{align*}
Thus, $n-\operatorname*{Des}\left(  \operatorname*{rev}w\right)  =\left\{
1,2,...,n-1\right\}  \setminus\left(  \operatorname*{Stag}\left(  w\right)
\cup\operatorname*{Des}\left(  w\right)  \right)  $. Hence,%
\begin{align*}
\left\vert n-\operatorname*{Des}\left(  \operatorname*{rev}w\right)
\right\vert  &  =\left\vert \left\{  1,2,...,n-1\right\}  \setminus\left(
\operatorname*{Stag}\left(  w\right)  \cup\operatorname*{Des}\left(  w\right)
\right)  \right\vert \\
&  =\underbrace{\left\vert \left\{  1,2,...,n-1\right\}  \right\vert }%
_{=n-1}-\underbrace{\left\vert \operatorname*{Stag}\left(  w\right)
\cup\operatorname*{Des}\left(  w\right)  \right\vert }_{=\left\vert
\operatorname*{Stag}\left(  w\right)  \right\vert +\left\vert
\operatorname*{Des} \left( w \right)\right\vert }
\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\operatorname*{Stag}\left(  w\right)  \cup\operatorname*{Des}\left(
w\right)  \subset\left\{  1,2,...,n-1\right\}  \right) \\
&  =n-1-\left(  \left\vert \operatorname*{Stag}\left(  w\right)  \right\vert
+\left\vert \operatorname*{Des} \left( w \right) \right\vert \right)
=n-1-\left\vert
\operatorname*{Des}\left(  w\right)  \right\vert -\left\vert
\operatorname*{Stag}\left(  w\right)  \right\vert .
\end{align*}
Compared with%
\[
\left\vert n-\operatorname*{Des}\left(  \operatorname*{rev}w\right)
\right\vert =\left\vert \operatorname*{Des}\left(  \operatorname*{rev}%
w\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert
n-K\right\vert =\left\vert K\right\vert \text{ for every }K\subset\left\{
1,2,...,n-1\right\}  \right)  ,
\]
this yields $\left\vert \operatorname*{Des}\left(  \operatorname*{rev}%
w\right)  \right\vert =n-1-\left\vert \operatorname*{Des}\left(  w\right)
\right\vert -\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert $.
This proves (\ref{sol.shareshian-wachs.path.c.revDes}).}. Furthermore, every
$n$-tuple $w\in\left\{  1,2,3,\ldots\right\}  ^{n}$ satisfies
\begin{equation}
\left\vert \operatorname*{Stag}\left(  \operatorname*{rev}w\right)
\right\vert =\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert
\label{sol.shareshian-wachs.path.c.revStag}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.shareshian-wachs.path.c.revStag}):} Let
$w\in\left\{  1,2,3,\ldots\right\}  ^{n}$. Write $w$ in the form $w=\left(
w_{1},w_{2},...,w_{n}\right)  $. Then, the definition of $\operatorname*{rev}%
w$ yields $\operatorname*{rev}w=\left(  w_{n},w_{n-1},...,w_{1}\right)  $.
Hence, the definition of $\operatorname*{Stag}\left(  \operatorname*{rev}%
w\right)  $ yields%
\[
\operatorname*{Stag}\left(  \operatorname*{rev}w\right)  =\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  :w_{n+1-i}=w_{n+1-\left(  i+1\right)
}\right\}  .
\]
Thus, for any $i\in\left\{  1,2,...,n-1\right\}  $, we have the following
equivalence of assertions:%
\begin{equation}
\left(  i\in\operatorname*{Stag}\left(  \operatorname*{rev}w\right)  \right)
\ \Longleftrightarrow\ \left(  w_{n+1-i}=w_{n+1-\left(  i+1\right)  }\right)
. \label{sol.shareshian-wachs.path.c.revStag.pf.1}%
\end{equation}
\par
But $\operatorname*{Stag}\left(  w\right)  =\left\{  i\in\left\{
1,2,\ldots,n-1\right\}  :w_{i}=w_{i+1}\right\}  $. Hence, for any
$i\in\left\{  1,2,...,n-1\right\}  $, we have the following equivalence of
assertions:%
\begin{equation}
\left(  i\in\operatorname*{Stag}\left(  w\right)  \right)
\ \Longleftrightarrow\ \left(  w_{i}=w_{i+1}\right)  .
\label{sol.shareshian-wachs.path.c.revStag.pf.3}%
\end{equation}
\par
Now, for every $K\subset\left\{  1,2,...,n-1\right\}  $, let $n-K$ denote the
set $\left\{  n-k\ \mid\ k\in K\right\}  $; this set $n-K$ is again a subset
of $\left\{  1,2,...,n-1\right\}  $ and has cardinality $\left\vert
n-K\right\vert =\left\vert K\right\vert $. For any $i\in\left\{
1,2,...,n-1\right\}  $, we have the following equivalence of assertions:%
\begin{align*}
\left(  i\in n-\operatorname*{Stag}\left(  \operatorname*{rev}w\right)
\right)  \  &  \Longleftrightarrow\ \left(  i\in\left\{  n-k\ \mid
\ k\in\operatorname*{Stag}\left(  \operatorname*{rev}w\right)  \right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-\operatorname*{Stag}\left(
\operatorname*{rev}w\right)  =\left\{  n-k\ \mid\ k\in\operatorname*{Stag}%
\left(  \operatorname*{rev}w\right)  \right\}  \right) \\
&  \Longleftrightarrow\ \left(  \underbrace{i=n-k}_{\text{this is equivalent
to }n-i=k}\text{ for some }k\in\operatorname*{Stag}\left(  \operatorname*{rev}%
w\right)  \right) \\
&  \Longleftrightarrow\ \left(  n-i=k\text{ for some }k\in\operatorname*{Stag}%
\left(  \operatorname*{rev}w\right)  \right) \\
&  \Longleftrightarrow\ \left(  n-i\in\operatorname*{Stag}\left(
\operatorname*{rev}w\right)  \right)  \ \Longleftrightarrow\ \left(
w_{n+1-\left(  n-i\right)  }=w_{n+1-\left(  \left(  n-i\right)  +1\right)
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.revStag.pf.1}), applied to }n-i\text{
instead of }i\right) \\
&  \Longleftrightarrow\ \left(  w_{i+1}=w_{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n+1-\left(  n-i\right)  =i+1\text{
and }n+1-\left(  \left(  n-i\right)  +1\right)  =i\right) \\
&  \Longleftrightarrow\ \left(  w_{i}=w_{i+1}\right)  \ \Longleftrightarrow
\ \left(  i\in\operatorname*{Stag}\left(  w\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.revStag.pf.3})}\right)  .
\end{align*}
\par
Thus, $n-\operatorname*{Stag}\left(  \operatorname*{rev}w\right)
=\operatorname*{Stag}\left(  w\right)  $. Hence, $\left\vert
n-\operatorname*{Stag}\left(  \operatorname*{rev}w\right)  \right\vert
=\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert $. Compared with%
\[
\left\vert n-\operatorname*{Stag}\left(  \operatorname*{rev}w\right)
\right\vert =\left\vert \operatorname*{Stag}\left(  \operatorname*{rev}%
w\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert
n-K\right\vert =\left\vert K\right\vert \text{ for every }K\subset\left\{
1,2,...,n-1\right\}  \right)  ,
\]
this yields $\left\vert \operatorname*{Stag}\left(  \operatorname*{rev}%
w\right)  \right\vert =\left\vert \operatorname*{Stag}\left(  w\right)
\right\vert $. This proves (\ref{sol.shareshian-wachs.path.c.revStag}).}.

Furthermore, for every $n$-tuple $w\in\left\{  1,2,3,\ldots\right\}  ^{n}$,
let us define the monomial $\xx_{w}\in \kk \left[  \left[
\xx\right]  \right]  $ by setting $\xx_{w}=x_{w_{1}}x_{w_{2}%
}\cdots x_{w_{n}}$, where $w$ is written as $\left(  w_{1},w_{2}%
,...,w_{n}\right)  $. Then, every $n$-tuple $w\in\left\{  1,2,3,\ldots
\right\}  ^{n}$ satisfies%
\begin{equation}
\xx_{\operatorname*{rev}w}=\xx_{w}.
\label{sol.shareshian-wachs.path.c.revx}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.shareshian-wachs.path.c.revx}):} Let
$w\in\left\{  1,2,3,\ldots\right\}  ^{n}$. Write the $n$-tuple $w$ as $\left(
w_{1},w_{2},...,w_{n}\right)  $. Then, $\operatorname*{rev}\underbrace{w}%
_{=\left(  w_{1},w_{2},...,w_{n}\right)  }=\operatorname*{rev}\left(
w_{1},w_{2},...,w_{n}\right)  =\left(  w_{n},w_{n-1},...,w_{1}\right)  $ (by
the definition of $\operatorname*{rev}$), so that%
\begin{align*}
\xx_{\operatorname*{rev}w}  &  =x_{w_{n}}x_{w_{n-1}}\cdots x_{w_{1}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\xx%
_{\operatorname*{rev}w}\right) \\
&  =x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=\xx_{w}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\xx_{w}=x_{w_{1}}x_{w_{2}%
}\cdots x_{w_{n}}\text{ (by the definition of }\xx_{w}\text{)}\right)
.
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.c.revx}).}

Finally, let us introduce two more pieces of notation. A $n$-tuple $\left(
w_{1},w_{2},...,w_{n}\right)  $ is said to be \emph{U-bounded} if
$w_{1}<w_{n}$. A $n$-tuple $\left(  w_{1},w_{2},...,w_{n}\right)  $ is said to
be \emph{W-bounded} if $w_{1}>w_{n}$. It is easy to see that every $n$-tuple
$w\in\left\{  1,2,3,...\right\}  ^{n}$ satisfies the following equivalence of
assertions:
\begin{equation}
\left(  \text{the }n\text{-tuple }w\text{ is U-bounded}\right)
\ \Longleftrightarrow\ \left(  \text{the }n\text{-tuple }\operatorname*{rev}%
w\text{ is W-bounded}\right)  .\label{sol.shareshian-wachs.path.c.revUV}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.shareshian-wachs.path.c.revUV}):} Let
$w\in\left\{  1,2,3,...\right\}  ^{n}$. Write the $n$-tuple $w$ in the form
$\left(  w_{1},w_{2},...,w_{n}\right)  $. Then, $\operatorname*{rev}%
\underbrace{w}_{=\left(  w_{1},w_{2},...,w_{n}\right)  }=\operatorname*{rev}%
\left(  w_{1},w_{2},...,w_{n}\right)  =\left(  w_{n},w_{n-1},...,w_{1}\right)
$ (by the definition of $\operatorname*{rev}$). Hence, the $n$-tuple
$\operatorname*{rev}w$ is W-bounded if and only if $w_{n}>w_{1}$ (due to the
definition of ``W-bounded''). Thus, we have
the following equivalence of assertions:%
\begin{equation}
\left(  \text{the }n\text{-tuple }\operatorname*{rev}w\text{ is W-bounded}%
\right)  \ \Longleftrightarrow\ \left(  w_{n}>w_{1}\right)
\ \Longleftrightarrow\ \left(  w_{1}<w_{n}\right)
.\label{sol.shareshian-wachs.path.c.revUV.pf.1}%
\end{equation}
\par
But $w = \left( w_1 , w_2 , ... , w_n \right)$. Hence,
the $n$-tuple $w$ is U-bounded if and
only if $w_{1}<w_{n}$ (by the definition of ``U-bounded''). Hence,
we have the following equivalence of assertions:
\begin{align*}
\left(  \text{the }n\text{-tuple }w\text{ is U-bounded}\right)  \   &
\Longleftrightarrow\ \left(  w_{1}<w_{n}\right)  \ \Longleftrightarrow
\ \left(  \text{the }n\text{-tuple }\operatorname*{rev}w\text{ is
W-bounded}\right)  \\
& \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.shareshian-wachs.path.c.revUV.pf.1})}\right)  .
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.c.revUV}).}

Let us also notice that every $n$-tuple $w\in\left\{  1,2,3,...\right\}  ^{n}$
satisfying $\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$
satisfies the following equivalence of assertions:%
\begin{equation}
\left(  n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
-\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =d\right)
\ \Longleftrightarrow\ \left(  \left\vert \operatorname*{Des}\left(  w\right)
\right\vert =n-1-d-s\right)  \label{sol.shareshian-wachs.path.c.revnumbers}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.shareshian-wachs.path.c.revnumbers}):}
Let $w\in\left\{  1,2,3,...\right\}  ^{n}$ be an $n$-tuple satisfying
$\left\vert \operatorname{Stag}\left(  w\right)  \right\vert =s$. Then, we
have the following equivalence of assertions:%
\begin{align*}
\left(  n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
-\underbrace{\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert
}_{=s}=d\right)  \   & \Longleftrightarrow\ \left(  n-1-\left\vert
\operatorname*{Des}\left(  w\right)  \right\vert -s=d\right)
\ \Longleftrightarrow\ \left(  n-1-\left\vert \operatorname*{Des}\left(
w\right)  \right\vert =d+s\right)  \\
& \Longleftrightarrow\ \left(  \left\vert \operatorname*{Des}\left(  w\right)
\right\vert =\underbrace{\left(  n-1\right)  -\left(  d+s\right)  }%
_{=n-1-d-s}\right)  \ \Longleftrightarrow\ \left(  \left\vert
\operatorname*{Des}\left(  w\right)  \right\vert =n-1-d-s\right)  .
\end{align*}
This proves (\ref{sol.shareshian-wachs.path.c.revnumbers}).}.

Now, (\ref{eq.exe.shareshian-wachs.path.c.W}) becomes%
\begin{align*}
W_{n,d,s} &  =\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}>w_{n}}%
}}_{\substack{=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\\text{the }n\text{-tuple }w\text{ is W-bounded}%
}}\\\text{(because for an }n\text{-tuple }w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \text{,}\\\text{the }n\text{-tuple }w\text{ is W-bounded if and
only if }w_{1}>w_{n}\\\text{(by the definition of ``W-bounded''))}}}
\underbrace{x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}}_{\substack{=\xx_{w}\\\text{(because }\xx_{w}=x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}\\\text{(by the definition of }\xx%
_{w}\text{))}}}\\
&  =\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\\text{the }n\text{-tuple }w\text{ is W-bounded}}}}%
_{=\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\\text{the }n\text{-tuple
}w\text{ is W-bounded}}}}\xx_{w}=\sum\limits_{\substack{w\in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\\text{the }n\text{-tuple }w\text{ is W-bounded}}}\xx_{w}\\
&  =\underbrace{\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname{Des}\left(  \operatorname*{rev}w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  \operatorname*{rev}%
w\right)  \right\vert =s;\\\text{the }n\text{-tuple }\operatorname*{rev}%
w\text{ is W-bounded}}}}_{\substack{=\sum\limits_{\substack{w\in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
\operatorname*{rev}w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}%
\left(  \operatorname*{rev}w\right)  \right\vert =s;\\\text{the }n\text{-tuple
}w\text{ is U-bounded}}}\\\text{(because for an }n\text{-tuple }w\text{, the
}n\text{-tuple}\\\operatorname*{rev}w\text{ is W-bounded if and only if
the}\\n\text{-tuple }w\text{ is U-bounded}\\\text{(by
(\ref{sol.shareshian-wachs.path.c.revUV})))}}}\underbrace{\xx%
_{\operatorname*{rev}w}}_{\substack{=\xx_{w}\\\text{(by
(\ref{sol.shareshian-wachs.path.c.revx}))}}}\\
&\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\operatorname*{rev}w\text{ for }w\text{ in
the sum,}\\
\text{since }\operatorname*{rev}:\left\{  1,2,3,\ldots\right\}  ^{n}%
\rightarrow\left\{  1,2,3,\ldots\right\}  ^{n}\text{ is a bijection}%
\end{array}
\right)  \\
&  =\underbrace{\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname{Des}\left(  \operatorname*{rev}w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  \operatorname*{rev}%
w\right)  \right\vert =s;\\\text{the }n\text{-tuple }w\text{ is U-bounded}}%
}}_{\substack{=\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}
^{n};\\n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
-\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\\text{the }n\text{-tuple
}w\text{ is U-bounded}}}\\\text{(because for an }n\text{-tuple }w\text{, we
have}\\\left\vert \operatorname{Des}\left(  \operatorname*{rev}w\right)
\right\vert =n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
-\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert \\\text{(by
(\ref{sol.shareshian-wachs.path.c.revDes}))}\\\text{and }\left\vert
\operatorname{Stag}\left(  \operatorname*{rev}w\right)  \right\vert
=\left\vert \operatorname{Stag}\left(  w\right)  \right\vert \\\text{(by
(\ref{sol.shareshian-wachs.path.c.revStag})))}}}\xx_{w}\\
&  =\underbrace{\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}
^{n};\\n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
-\left\vert \operatorname*{Stag}\left(  w\right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\\text{the }n\text{-tuple
}w\text{ is U-bounded}}}}_{\substack{=\sum\limits_{\substack{w\in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname*{Des}\left(  w\right)
\right\vert =n-1-d-s;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\\text{the }n\text{-tuple }w\text{ is U-bounded}%
}}\\\text{(because for an }n\text{-tuple }w\text{ satisfying }\left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s\text{,}\\\text{we have
}n-1-\left\vert \operatorname*{Des}\left(  w\right)  \right\vert -\left\vert
\operatorname*{Stag}\left(  w\right)  \right\vert =d\text{ if and only
if}\\\text{we have }\left\vert \operatorname*{Des}\left(  w\right)
\right\vert =n-1-d-s\\\text{(by (\ref{sol.shareshian-wachs.path.c.revnumbers}%
)))}}}\xx_{w}=\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}
^{n};\\\left\vert \operatorname*{Des}\left(  w\right)  \right\vert
=n-1-d-s;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s;\\\text{the }n\text{-tuple }w\text{ is U-bounded}}}\xx_{w}.
\end{align*}
Compared with%
\begin{align*}
U_{n,n-1-d-s,s}  & =\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert
\operatorname{Des}\left(  w\right)  \right\vert =n-1-d-s;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}<w_{n}}%
}}_{\substack{=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =n-1-d-s;\ \left\vert \operatorname{Stag}\left(
w\right)  \right\vert =s;\\\text{the }n\text{-tuple }w\text{ is U-bounded}%
}}\\\text{(because for an }n\text{-tuple }w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \text{,}\\\text{the }n\text{-tuple }w\text{ is U-bounded if and
only if }w_{1}<w_{n}\\\text{(by the definition of ``U-bounded''))}}}
\underbrace{x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}}_{\substack{=\xx_{w}\\\text{(because }\xx_{w}=x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}\\\text{(by the definition of }\xx%
_{w}\text{))}}}\\
&\ \ \ \ \ \ \ \ \ \ \left(\text{by \eqref{eq.exe.shareshian-wachs.path.c.U},
applied to } n-1-d-s \text{ instead of } d\right) \\
& =\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =n-1-d-s;\ \left\vert \operatorname{Stag}\left(
w\right)  \right\vert =s;\\\text{the }n\text{-tuple }w\text{ is U-bounded}}%
}}_{=\sum\limits_{\substack{w\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert
\operatorname*{Des}\left(  w\right)  \right\vert =n-1-d-s;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\\text{the }n\text{-tuple
}w\text{ is U-bounded}}}}\xx_{w}=\sum\limits_{\substack{w\in\left\{
1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname*{Des}\left(  w\right)
\right\vert =n-1-d-s;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\\text{the }n\text{-tuple }w\text{ is U-bounded}}%
}\xx_{w},
\end{align*}
this yields $W_{n,d,s}=U_{n,n-1-d-s,s}\in\Lambda$ (by
(\ref{sol.shareshian-wachs.path.c.Udone}), applied to $n-1-d-s$ instead of $d$).

Finally, every $\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n}$ satisfies exactly one of the three statements
$w_{1}<w_{n}$, $w_{1}=w_{n}$ and $w_{1}>w_{n}$. Hence,%
\begin{align*}
& \sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\left\{
1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(  w\right)
\right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)  \right\vert
=s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}\\
& =\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}<w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
}_{\substack{=U_{n,d,s}\\\text{(by (\ref{eq.exe.shareshian-wachs.path.c.U}))}%
}}+\underbrace{\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,\ldots\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s;\\w_{1}=w_{n}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
}_{\substack{=V_{n,d,s}\\\text{(by (\ref{eq.exe.shareshian-wachs.path.c.V}))}%
}}\\
& \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{n}%
;\\\left\vert \operatorname{Des}\left(  w\right)  \right\vert =d;\ \left\vert
\operatorname{Stag}\left(  w\right)  \right\vert =s;\\w_{1}>w_{n}}}x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}}_{\substack{=W_{n,d,s}\\\text{(by
(\ref{eq.exe.shareshian-wachs.path.c.W}))}}}\\
& =U_{n,d,s}+V_{n,d,s}+W_{n,d,s}.
\end{align*}


Now, (\ref{sol.shareshian-wachs.path.b.def}) becomes%
\[
X_{n,d,s}=\sum\limits_{\substack{w=\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\left\{  1,2,3,...\right\}  ^{n};\\\left\vert \operatorname{Des}\left(
w\right)  \right\vert =d;\ \left\vert \operatorname{Stag}\left(  w\right)
\right\vert =s}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=U_{n,d,s}+V_{n,d,s}%
+W_{n,d,s},
\]
so that $V_{n,d,s}=\underbrace{X_{n,d,s}}_{\substack{\in\Lambda\\\text{(by
Exercise \ref{exe.shareshian-wachs.path}(b))}}}-\underbrace{U_{n,d,s}}%
_{\in\Lambda}-\underbrace{W_{n,d,s}}_{\in\Lambda}\in\Lambda-\Lambda
-\Lambda\subset\Lambda$.

We have thus shown that $V_{n,d,s}\in\Lambda$ and $W_{n,d,s}\in\Lambda$.
Combined with $U_{n,d,s}\in\Lambda$ (this follows from
(\ref{sol.shareshian-wachs.path.c.Udone})), this completes the solution of
Exercise~\ref{exe.shareshian-wachs.path}(c).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.pe-ep-determinants}}

\textit{Solution to Exercise \ref{exe.pe-ep-determinants}.} We start
out with two definitions:

\begin{itemize}
\item If $m$ and $q$ are integers satisfying $0\leq q\leq m$, and if
$U=\left(  u_{i,j}\right)  _{i,j=1,2,...,m}\in \kk ^{m\times m}$ is an
$m\times m$-matrix, then $\operatorname{NWsm}_{q}U$ will mean the
matrix $\left(  u_{i,j}\right)  _{i,j=1,2,...,q}\in \kk ^{q\times q}$.
This is the submatrix of $U$ obtained by removing all rows other than the
first $q$ rows and then removing all columns other than the first $q$
columns.\footnote{The notation $\operatorname{NWsm}_{q}U$ stands
short for ``$q$-th northwest submatrix of $U$''.
It is the kind of submatrices whose determinants usually figure in the
Sylvester criterion for the positive definiteness of a matrix.}

\item A square matrix $\left(  u_{i,j}\right)  _{i,j=1,2,...,m}\in
 \kk ^{m\times m}$ is said to be \emph{nearly lower-triangular} if we
have%
\[
\left(  u_{i,j}=0\text{ for every }\left(  i,j\right)  \in\left\{
1,2,...,m\right\}  ^{2}\text{ satisfying }j>i+1\right)  .
\]
(Thus, informally, a square matrix is nearly lower-triangular if and only if
all its entries above the superdiagonal are $0$, where the
\emph{superdiagonal} is the set of all cells which lie just one step north
of a cell on the diagonal.)
\end{itemize}

We will now show a lemma:

\begin{lemma}
\label{lem.sol.exe.pe-ep-determinants}Let $m\in \NN$. Let $U=\left(
u_{i,j}\right)  _{i,j=1,2,...,m}\in \kk ^{m\times m}$ be a nearly
lower-triangular $m\times m$-matrix. Then,%
\[
\det U=\sum_{r=1}^{m}\left(  -1\right)  ^{m-r}u_{m,r}\det\left(
\operatorname{NWsm}_{r-1}U\right)  \cdot\prod_{k=r}^{m-1}u_{k,k+1}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.exe.pe-ep-determinants}.]Fix some $r\in\left\{
1,2,...,m\right\}  $. We define the following four matrices:

\begin{itemize}
\item the $\left(  r-1\right)  \times\left(  r-1\right)  $-matrix $P=\left(
u_{i,j}\right)  _{i,j=1,2,...,r-1}$, which is obtained from the matrix $U$ by
removing all rows other than the first $r-1$ rows and then removing all
columns other than the first $r-1$ columns;

\item the $\left(  r-1\right)  \times\left(  m-r\right)  $-matrix $Q=\left(
u_{i,r+j}\right)  _{i=1,2,...,r-1;\ j=1,2,...,m-r}$, which is obtained from
the matrix $U$ by removing all rows other than the first $r-1$ rows and then
removing all columns other than the last $m-r$ columns;

\item the $\left(  m-r\right)  \times\left(  r-1\right)  $-matrix $R=\left(
u_{r-1+i,j}\right)  _{i=1,2,...,m-r;\ j=1,2,...,r-1}$, which is obtained from
the matrix $U$ by removing all rows other than the $r$-th, the $\left(
r+1\right)  $-st, etc., the $\left(  m-1\right)  $-st row, and then removing
all columns other than the first $r-1$ columns;

\item the $\left(  m-r\right)  \times\left(  m-r\right)  $-matrix $S=\left(
u_{r-1+i,r+j}\right)  _{i,j=1,2,...,m-r}$, which is obtained from the matrix
$U$ by removing all rows other than the $r$-th, the $\left(  r+1\right)  $-st,
etc., the $\left(  m-1\right)  $-st row, and then removing all columns other
than the last $m-r$ columns.
\end{itemize}

Now, the matrix $U$ can be written as a block matrix as follows: $U=\left(
\begin{array}
[c]{ccc}%
P & v & Q\\
R & w & S\\
x & y & z
\end{array}
\right)  $, where $\left(
\begin{array}
[c]{c}%
v\\
w\\
y
\end{array}
\right)  $ is the $r$-th column of $U$, and where $\left(
\begin{array}
[c]{ccc}%
x & y & z
\end{array}
\right)  $ is the $m$-th row of $U$.\ \ \ \ \footnote{Despite being labelled
with lowercase letters, $v$, $w$, $x$, $y$ and $z$ are still blocks, although
each has (at least) one of its dimensions equal to $1$ (and $y$ is a
$1\times1$-block).} Hence,%
\begin{align}
&  \left(  \text{the matrix obtained from }U\text{ by removing the }m\text{-th
row and the }r\text{-th column}\right) \nonumber\\
&  =\left(
\begin{array}
[c]{cc}%
P & Q\\
R & S
\end{array}
\right)  . \label{pf.lem.sol.exe.pe-ep-determinants.1}%
\end{align}


But the matrix $Q=\left(  u_{i,r+j}\right)  _{i=1,2,...,r-1;\ j=1,2,...,m-r}$
is the zero matrix (since $U$ is nearly lower-triangular), and the matrices
$P$ and $S$ are square matrices. Hence, $\left(
\begin{array}
[c]{cc}%
P & Q\\
R & S
\end{array}
\right)  $ is a block lower-triangular matrix with diagonal blocks $P$ and
$S$. Thus, its determinant is%
\begin{equation}
\det\left(
\begin{array}
[c]{cc}%
P & Q\\
R & S
\end{array}
\right)  =\det P\cdot\det S \label{pf.lem.sol.exe.pe-ep-determinants.3}%
\end{equation}
(since the determinant of a block lower-triangular matrix is known to equal
the product of the determinants of its diagonal blocks).

But the matrix $S=\left(  u_{r-1+i,r+j}\right)  _{i,j=1,2,...,m-r}$ is
lower-triangular (since $U$ is nearly lower-triangular). Since it is
well-known that the determinant of a lower-triangular matrix equals the
product of its diagonal entries, we can therefore compute the determinant
$\det S$ of $S$ as follows:%
\begin{equation}
\det S=\prod_{i=1}^{m-r}u_{r-1+i,r+i}=\prod_{k=r}^{m-1}u_{k,k+1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k\text{ for
}r-1+i\text{ in the product}\right)  .
\label{pf.lem.sol.exe.pe-ep-determinants.2}
\end{equation}


Moreover, $P=\operatorname{NWsm}_{r-1}U$ (since comparing the
definitions of $P$ and of $\operatorname{NWsm}_{r-1}U$ shows that
these two matrices are the same). Now, applying the map $\det$ to both sides
of (\ref{pf.lem.sol.exe.pe-ep-determinants.1}), we obtain%
\begin{align}
&  \det\left(  \text{the matrix obtained from }U\text{ by removing the
}m\text{-th row and the }r\text{-th column}\right) \nonumber\\
&  =\det\left(
\begin{array}
[c]{cc}%
P & Q\\
R & S
\end{array}
\right)  =\det\underbrace{P}_{=\operatorname{NWsm}_{r-1}U}%
\cdot\underbrace{\det S}_{\substack{=\prod_{k=r}^{m-1}u_{k,k+1}\\\text{(by
(\ref{pf.lem.sol.exe.pe-ep-determinants.2}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.lem.sol.exe.pe-ep-determinants.3})}\right) \nonumber\\
&  =\det\left(  \operatorname{NWsm}_{r-1}U\right)  \cdot\prod
_{k=r}^{m-1}u_{k,k+1}. \label{pf.lem.sol.exe.pe-ep-determinants.5}%
\end{align}


Now, forget that we fixed $r$. We can compute the determinant $\det U$ by
Laplace expansion along the $m$-th row, thus obtaining%
\begin{align*}
&  \det U\\
&  =\sum_{r=1}^{m}u_{m,r}\cdot\underbrace{\left(  \text{the }\left(
m,r\right)  \text{-th cofactor of the matrix }U\right)  }_{=\left(  -1\right)
^{m+r}\det\left(  \text{the matrix obtained from }U\text{ by removing the
}m\text{-th row and the }r\text{-th column}\right)  }\\
&  =\sum_{r=1}^{m}u_{m,r}\cdot\underbrace{\left(  -1\right)  ^{m+r}}_{=\left(
-1\right)  ^{m-r}}\underbrace{\det\left(  \text{the matrix obtained from
}U\text{ by removing the }m\text{-th row and the }r\text{-th column}\right)
}_{\substack{=\det\left(  \operatorname{NWsm}_{r-1}U\right)
\cdot\prod_{k=r}^{m-1}u_{k,k+1}\\\text{(by
(\ref{pf.lem.sol.exe.pe-ep-determinants.5}))}}}\\
&  =\sum_{r=1}^{m}u_{m,r}\cdot\left(  -1\right)  ^{m-r}\cdot\det\left(
\operatorname{NWsm}_{r-1}U\right)  \cdot\prod_{k=r}^{m-1}%
u_{k,k+1}=\sum_{r=1}^{m}\left(  -1\right)  ^{m-r}u_{m,r}\det\left(
\operatorname{NWsm}_{r-1}U\right)  \cdot\prod_{k=r}^{m-1}u_{k,k+1}.
\end{align*}
This proves Lemma \ref{lem.sol.exe.pe-ep-determinants}.
\end{proof}

Before we solve the actual exercise, we record one further identity that we
will be using twice. Namely, we claim that%
\begin{equation}
me_{m}=\sum_{i=1}^{m}\left(  -1\right)  ^{i-1}e_{m-i}p_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in \NN \text{.}
\label{sol.pe-ep-determinants.identity}%
\end{equation}


\textit{Proof of (\ref{sol.pe-ep-determinants.identity}):} Recall the power
series $H\left(  t\right)  $ defined in (\ref{h-generating-function}), and the
power series $E\left(  t\right)  $ defined in (\ref{e-generating-function}).
From (\ref{E(t)-H(t)-relation}), we know that $E\left(  -t\right)  H\left(
t\right)  =1$. Differentiating both sides of this equation with respect to
$t$, we obtain $\left(  E\left(  -t\right)  H\left(  t\right)  \right)
^{\prime}=1^{\prime}=0$, so that%
\begin{align*}
0  &  =\left(  E\left(  -t\right)  H\left(  t\right)  \right)  ^{\prime
}=\underbrace{\left(  E\left(  -t\right)  \right)  ^{\prime}}_{=-E^{\prime
}\left(  -t\right)  }H\left(  t\right)  +E\left(  -t\right)  H^{\prime}\left(
t\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the Leibniz rule}\right) \\
&  =-E^{\prime}\left(  -t\right)  H\left(  t\right)  +E\left(  -t\right)
H^{\prime}\left(  t\right)  ,
\end{align*}
and thus $E^{\prime}\left(  -t\right)  H\left(  t\right)  =E\left(  -t\right)
H^{\prime}\left(  t\right)  $. Hence, $\dfrac{E^{\prime}\left(  -t\right)
}{E\left(  -t\right)  }=\dfrac{H^{\prime}\left(  t\right)  }{H\left(
t\right)  }$.

But Exercise \ref{exe.E(t)-H(t)-P(t)-relation} yields $\sum_{m\geq0}%
p_{m+1}t^{m}=\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }$.
Compared with $\dfrac{E^{\prime}\left(  -t\right)  }{E\left(  -t\right)
}=\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)  }$, this yields
$\dfrac{E^{\prime}\left(  -t\right)  }{E\left(  -t\right)  }=\sum_{m\geq
0}p_{m+1}t^{m}$. Substituting $-t$ for $t$ in this equality, we obtain
\[
\dfrac{E^{\prime}\left(  t\right)  }{E\left(  t\right)  }=\sum_{m\geq0}%
p_{m+1}\left(  -t\right)  ^{m}=\sum_{m\geq0}p_{m+1}\left(  -1\right)
^{m}t^{m}.
\]
Thus,%
\begin{align*}
E^{\prime}\left(  t\right)   &  =\underbrace{E\left(  t\right)  }%
_{\substack{=\sum_{m\geq0}e_{m}t^{m}\\\text{(by the definition of }E\left(
t\right)  \text{)}}}\cdot\left(  \sum_{m\geq0}p_{m+1}\left(  -1\right)
^{m}t^{m}\right)  =\left(  \sum_{m\geq0}e_{m}t^{m}\right)  \left(  \sum
_{m\geq0}p_{m+1}\left(  -1\right)  ^{m}t^{m}\right) \\
&  =\left(  \sum_{m\geq0}p_{m+1}\left(  -1\right)  ^{m}t^{m}\right)  \left(
\sum_{m\geq0}e_{m}t^{m}\right) \\
&  =\sum_{m\geq0}\left(  \sum_{i=0}^{m}p_{i+1}\left(  -1\right)  ^{i}%
e_{m-i}\right)  t^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
the product of two power series}\right) \\
&  =\sum_{m\geq1}\left(  \sum_{i=0}^{m-1}p_{i+1}\left(  -1\right)
^{i}e_{m-1-i}\right)  t^{m-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we
substituted }m-1\text{ for }m\text{ in the first sum}\right) \\
&  =\sum_{m\geq0}\underbrace{\left(  \sum_{i=0}^{m-1}p_{i+1}\left(  -1\right)
^{i}e_{m-1-i}\right)  }_{\substack{=\sum_{i=1}^{m}p_{i}\left(  -1\right)
^{i-1}e_{m-i}\\\text{(here we substituted }i-1\text{ for }i\text{ in the
sum)}}}t^{m-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have added an }m=0\text{ addend to the first sum;}\\
\text{this did not change the sum since this addend is }0
\end{array}
\right) \\
&  =\sum_{m\geq0}\left(  \sum_{i=1}^{m}p_{i}\left(  -1\right)  ^{i-1}%
e_{m-i}\right)  t^{m-1}=\sum_{m\geq0}\left(  \sum_{i=1}^{m}\left(  -1\right)
^{i-1}e_{m-i}p_{i}\right)  t^{m-1}.
\end{align*}
Compared with
\begin{align*}
E^{\prime}\left(  t\right)   &  =\left(  \sum_{m\geq0}e_{m}t^{m}\right)
^{\prime}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }E\left(  t\right)
=\sum_{m\geq0}e_{m}t^{m}\right) \\
&  =\sum_{m\geq0}e_{m}\cdot mt^{m-1}=\sum_{m\geq0}me_{m}t^{m-1},
\end{align*}
this yields%
\[
\sum_{m\geq0}me_{m}t^{m-1}=\sum_{m\geq0}\left(  \sum_{i=1}^{m}\left(
-1\right)  ^{i-1}e_{m-i}p_{i}\right)  t^{m-1}.
\]
Multiplying both sides of this equality by $t$, we obtain%
\[
\sum_{m\geq0}me_{m}t^{m}=\sum_{m\geq0}\left(  \sum_{i=1}^{m}\left(  -1\right)
^{i-1}e_{m-i}p_{i}\right)  t^{m}.
\]
Comparing coefficients in this equality of power series, we conclude that
every $m\in \NN$ satisfies%
\[
me_{m}=\sum_{i=1}^{m}\left(  -1\right)  ^{i-1}e_{m-i}p_{i}.
\]
This proves (\ref{sol.pe-ep-determinants.identity}).

Now, let us solve the exercise.

(a) We will solve Exercise \ref{exe.pe-ep-determinants}(a) by strong induction
over $n$. Thus, we assume (as the induction hypothesis) that%
\begin{equation}
\det\left(  A_{k}\right)  =k!e_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }%
k\in \NN \text{ satisfying }k<n. \label{sol.pe-ep-determinants.a.indhyp}%
\end{equation}
We now need to prove that $\det\left(  A_{n}\right)  =n!e_{n}$.

The matrix $A_{n}=\left(  a_{i,j}\right)  _{i,j=1,2,\ldots,n}$ is nearly
lower-triangular\footnote{because for every $\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}$ satisfying $j>i+1$, we have%
\[
a_{i,j}=%
\begin{cases}
p_{i-j+1}, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i<j-1\text{ (because }%
j>i+1\text{)}\right)  .
\]
}. Hence, Lemma \ref{lem.sol.exe.pe-ep-determinants} (applied to $m=n$,
$U=A_{n}$ and $u_{i,j}=a_{i,j}$) yields%
\begin{align*}
\det\left(  A_{n}\right)   &  =\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}%
\underbrace{a_{n,r}}_{\substack{=p_{n-r+1}\\\text{(by the definition of
}a_{n,r}\text{,}\\\text{since }n\geq r\text{)}}}\det\underbrace{\left(
\operatorname{NWsm}_{r-1}\left(  A_{n}\right)  \right)
}_{\substack{=A_{r-1}\\\text{(this is easy to see by}\\\text{the definitions
of }A_{n}\text{ and }A_{r-1}\text{)}}}\cdot\prod_{k=r}^{n-1}%
\underbrace{a_{k,k+1}}_{\substack{=k\\\text{(by the definition of }%
a_{k,k+1}\text{,}\\\text{since }k=\left(  k+1\right)  -1\text{)}}}\\
&  =\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}p_{n-r+1}\underbrace{\det\left(
A_{r-1}\right)  }_{\substack{=\left(  r-1\right)  !e_{r-1}\\\text{(by
(\ref{sol.pe-ep-determinants.a.indhyp}), applied to }k=r-1\text{)}}}\cdot
\prod_{k=r}^{n-1}k\\
&  =\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}p_{n-r+1}\left(  r-1\right)
!e_{r-1}\prod_{k=r}^{n-1}k=\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}%
\underbrace{\left(  r-1\right)  !
\left(\prod_{k=r}^{n-1}k\right)}_{=\left(  n-1\right)
!}p_{n-r+1}e_{r-1}\\
&  =\left(  n-1\right)  !\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}%
p_{n-r+1}e_{r-1}=\left(  n-1\right)  !\underbrace{\sum_{i=1}^{n}\left(
-1\right)  ^{i-1}p_{i}e_{n-i}}_{\substack{=\sum_{i=1}^{n}\left(  -1\right)
^{i-1}e_{n-i}p_{i}=ne_{n}\\\text{(because
(\ref{sol.pe-ep-determinants.identity}) (applied to }m=n\text{)}\\\text{yields
}ne_{n}=\sum_{i=1}^{n}\left(  -1\right)  ^{i-1}e_{n-i}p_{i}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n-i+1\text{
for }r\text{ in the sum}\right) \\
&  =\underbrace{\left(  n-1\right)  !n}_{=n!}e_{n}=n!e_{n}.
\end{align*}
This completes the induction step, and so Exercise
\ref{exe.pe-ep-determinants}(a) is solved.

(b) We will solve Exercise \ref{exe.pe-ep-determinants}(b) by strong induction
over $n$. Thus, we assume (as the induction hypothesis) that%
\begin{equation}
\det\left(  B_{k}\right)  =p_{k}\ \ \ \ \ \ \ \ \ \ \text{for all positive
integers }k\text{ satisfying }k<n. \label{sol.pe-ep-determinants.b.indhyp}%
\end{equation}
We now need to prove that $\det\left(  B_{n}\right)  =p_{n}$.

The matrix $B_{n}=\left(  b_{i,j}\right)  _{i,j=1,2,\ldots,n}$ is nearly
lower-triangular\footnote{because for every $\left(  i,j\right)  \in\left\{
1,2,...,n\right\}  ^{2}$ satisfying $j>i+1$, we have%
\begin{align*}
b_{i,j}  &  =%
\begin{cases}
ie_{i}, & \text{if }j=1;\\
e_{i-j+1}, & \text{if }j>1
\end{cases}  \quad
=e_{i-j+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j>1\text{ (because
}j>i+1\geq1\text{)}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i-j+1<0\text{ (since
}j>i+1\text{)}\right)
\end{align*}
}. Hence, Lemma \ref{lem.sol.exe.pe-ep-determinants} (applied to $m=n$,
$U=B_{n}$ and $u_{i,j}=b_{i,j}$) yields%
\begin{align*}
\det\left(  B_{n}\right)   &  =\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}%
b_{n,r}\det\underbrace{\left(  \operatorname{NWsm}_{r-1}\left(
B_{n}\right)  \right)  }_{\substack{=B_{r-1}\\\text{(this is easy to see
by}\\\text{the definitions of }B_{n}\text{ and }B_{r-1}\text{)}}}\cdot
\prod_{k=r}^{n-1}\underbrace{b_{k,k+1}}_{\substack{=e_{k-\left(  k+1\right)
+1}\\\text{(by the definition of }b_{k,k+1}\text{,}\\\text{since }k+1>k\geq
r\geq1\text{)}}}\\
&  =\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}b_{n,r}\det\left(  B_{r-1}\right)
\cdot\prod_{k=r}^{n-1}\underbrace{e_{k-\left(  k+1\right)  +1}}_{=e_{0}=1}\\
&  =\sum_{r=1}^{n}\left(  -1\right)  ^{n-r}b_{n,r}\det\left(  B_{r-1}\right)
\cdot\underbrace{\prod_{k=r}^{n-1}1}_{=1}=\sum_{r=1}^{n}\left(  -1\right)
^{n-r}b_{n,r}\det\left(  B_{r-1}\right) \\
&  =\left(  -1\right)  ^{n-1}\underbrace{b_{n,1}}_{\substack{=ne_{n}%
\\\text{(by the definition}\\\text{of }b_{n,1}\text{)}}}\underbrace{\det
\left(  B_{1-1}\right)  }_{\substack{=1\\\text{(since }B_{1-1}\text{ is
a}\\0\times0\text{-matrix)}}}+\sum_{r=2}^{n}\left(  -1\right)  ^{n-r}%
\underbrace{b_{n,r}}_{\substack{=e_{n-r+1}\\\text{(by the definition of
}b_{n,r}\text{,}\\\text{since }r>1\text{)}}}\underbrace{\det\left(
B_{r-1}\right)  }_{\substack{=p_{r-1}\\\text{(by
(\ref{sol.pe-ep-determinants.b.indhyp}),}\\\text{applied to }k=r-1\text{)}}}\\
&  =\left(  -1\right)  ^{n-1}\underbrace{ne_{n}}_{\substack{=\sum_{i=1}%
^{n}\left(  -1\right)  ^{i-1}e_{n-i}p_{i}\\\text{(by
(\ref{sol.pe-ep-determinants.identity}), applied}\\\text{to }m=n\text{)}%
}}+\underbrace{\sum_{r=2}^{n}\left(  -1\right)  ^{n-r}e_{n-r+1}p_{r-1}%
}_{\substack{=\sum_{i=1}^{n-1}\left(  -1\right)  ^{n-\left(  i+1\right)
}e_{n-i}p_{i}\\\text{(here, we substituted }i+1\text{ for }r\text{ in the
sum)}}}\\
&  =\left(  -1\right)  ^{n-1}\sum_{i=1}^{n}\left(  -1\right)  ^{i-1}%
e_{n-i}p_{i}+\sum_{i=1}^{n-1}\left(  -1\right)  ^{n-\left(  i+1\right)
}e_{n-i}p_{i}\\
&  =\sum_{i=1}^{n}\underbrace{\left(  -1\right)  ^{n-1}\left(  -1\right)
^{i-1}}_{=\left(  -1\right)  ^{n-i}}e_{n-i}p_{i}+\sum_{i=1}^{n-1}%
\underbrace{\left(  -1\right)  ^{n-\left(  i+1\right)  }}_{=-\left(
-1\right)  ^{n-i}}e_{n-i}p_{i}\\
&  =\underbrace{\sum_{i=1}^{n}\left(  -1\right)  ^{n-i}e_{n-i}p_{i}}%
_{=\sum_{i=1}^{n-1}\left(  -1\right)  ^{n-i}e_{n-i}p_{i}+\left(  -1\right)
^{n-n}e_{n-n}p_{n}}-\sum_{i=1}^{n-1}\left(  -1\right)  ^{n-i}e_{n-i}p_{i}\\
&  =\sum_{i=1}^{n-1}\left(  -1\right)  ^{n-i}e_{n-i}p_{i}+\left(  -1\right)
^{n-n}e_{n-n}p_{n}-\sum_{i=1}^{n-1}\left(  -1\right)  ^{n-i}e_{n-i}%
p_{i}=\underbrace{\left(  -1\right)  ^{n-n}}_{=1}\underbrace{e_{n-n}}%
_{=e_0=1}p_{n}=p_{n}.
\end{align*}
This completes the induction step, and so Exercise
\ref{exe.pe-ep-determinants}(b) is solved.

\textit{Remark:} The above solution follows closely the solution of Exercise
9.3 in the first author's ``$\lambda$-rings: Definitions and
basic properties'' (
\url{https://github.com/darijgr/lambda} , version 0.0.21),
which is more or less a restatement of this exercise
(since the elements of $\Lambda_\ZZ$ are in a 1-to-1 correspondence
with unary functorial operations defined on every $\lambda$-ring).

Whenever $\ell\in \NN$ and two partitions $\lambda$ and $\mu$ of length
$\leq\ell$ have the property that the transpose of the matrix $\left(
h_{\lambda_{i}-\mu_{j}-i+j}\right)  _{i,j=1,2,...,\ell}$ is nearly
lower-triangular, we can use Lemma \ref{lem.sol.exe.pe-ep-determinants} to
obtain a recursive formula for the determinant of this matrix, which is
$s_{\lambda/\mu}$ according to (\ref{eq.jacobi-trudi.h}). This does not seem
to be of much use, however.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.hookschur}}

\textit{Solution to Exercise \ref{exe.hookschur}.} For any integers $a$
and $b$, we define an element $\mathbf{s}\left(  a,b\right)  $ of $\Lambda$
by
\[
\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases} \quad .
\]
(We are introducing this $\mathbf{s}\left(  a,b\right)  $ chiefly for reasons
of convenience: it will allow us to unify parts (b) and (c) of the exercise.)

Using the definition of $\mathbf{s}\left(  a,b\right)  $ and straightforward
case analysis, we can see that:%
\begin{equation}
\mathbf{s}\left(  a,b\right)  =s_{\left(  a+1,1^{b}\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in \NN \text{ and }b\in \NN ;
\label{sol.hookschur.sab-pos}%
\end{equation}
\begin{equation}
\mathbf{s}\left(  a,b\right)  =\left(  -1\right)  ^{b}\delta_{a+b,-1}%
\ \ \ \ \ \ \ \ \ \ \text{for every negative integer }a\text{ and every }%
b\in \ZZ ; \label{sol.hookschur.sab-neg}%
\end{equation}%
\begin{equation}
\mathbf{s}\left(  a,0\right)  =h_{a+1}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in \ZZ ; \label{sol.hookschur.b=0}%
\end{equation}%
\begin{equation}
\mathbf{s}\left(  a,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for any }%
a\in \ZZ \text{ and any negative }b\in \ZZ .
\label{sol.hookschur.b<0}%
\end{equation}


Let us now show that%
\begin{equation}
e_{n}h_{m}=\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(  m-1,n\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \ZZ \text{ and every }%
m\in \ZZ . \label{sol.hookschur.general-a}%
\end{equation}


\textit{Proof of (\ref{sol.hookschur.general-a}):} Let $n\in \ZZ$ and
$m\in \ZZ$. It is easy to prove (\ref{sol.hookschur.general-a}) in the
case $n<0$ (since both sides of (\ref{sol.hookschur.general-a}) vanish in this
case) and in the case $n=0$ (here, it follows from (\ref{sol.hookschur.b=0})).
We can therefore WLOG assume that $n>0$. Assume this.

It is easy to prove (\ref{sol.hookschur.general-a}) in the case $m<0$ (in
which case both sides of (\ref{sol.hookschur.general-a}) vanish) and in the
case $m=0$ (in which case both sides of (\ref{sol.hookschur.general-a}) equal
$e_{n}$). Thus, we can WLOG assume that $m>0$. Assume this. Since $m>0$, we
have $h_{m}=s_{\left(  m\right)  }$, so that%
\begin{equation}
e_{n}\underbrace{h_{m}}_{=s_{\left(  m\right)  }}=e_{n}s_{\left(  m\right)
}=s_{\left(  m\right)  }e_{n}=\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\left(
m\right)  \text{ is a }\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.pieri.e}), applied to }%
\lambda=\left(  m\right)  \right)  . \label{sol.hookschur.general-a.pf.1}%
\end{equation}


It is easy to see that there are exactly two partitions $\lambda^{+}$ for
which $\lambda^{+}/\left(  m\right)  $ is a vertical $n$-strip, namely the
partitions $\left(  m,1^{n}\right)  $ and $\left(  m+1,1^{n-1}\right)  $.
Hence, the sum $\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\left(  m\right)
\text{ is a }\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}}$ has exactly
two addends: that for $\lambda^{+}=\left(  m,1^{n}\right)  $ and that for
$\lambda^{+}=\left(  m+1,1^{n-1}\right)  $. Thus, this sum simplifies to
$s_{\left(  m,1^{n}\right)  }+s_{\left(  m+1,1^{n-1}\right)  }$. Hence,
(\ref{sol.hookschur.general-a.pf.1}) rewrites as%
\begin{equation}
e_{n}h_{m}=s_{\left(  m,1^{n}\right)  }+s_{\left(  m+1,1^{n-1}\right)  }.
\label{sol.hookschur.general-a.a}%
\end{equation}
But (\ref{sol.hookschur.sab-pos}) (applied to $a=m-1$ and $b=n$) yields
$\mathbf{s}\left(  m-1,n\right)  =s_{\left(  m,1^{n}\right)  }$. Also,
(\ref{sol.hookschur.sab-pos}) (applied to $a=m$ and $b=n-1$) yields
$\mathbf{s}\left(  m,n-1\right)  =s_{\left(  m+1,1^{n-1}\right)  }$. Thus,
\[
e_{n}h_{m}=\underbrace{s_{\left(  m,1^{n}\right)  }}_{=\mathbf{s}\left(
m-1,n\right)  }+\underbrace{s_{\left(  m+1,1^{n-1}\right)  }}_{=\mathbf{s}%
\left(  m,n-1\right)  }=\mathbf{s}\left(  m-1,n\right)  +\mathbf{s}\left(
m,n-1\right)  =\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(
m-1,n\right)  .
\]
This proves (\ref{sol.hookschur.general-a}).

Let us now prove a statement which encompasses both parts (b) and (c) of
Exercise \ref{exe.hookschur}. Namely, we are going to prove that%
\begin{equation}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}=\mathbf{s}\left(
a,b\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }a\in \ZZ \text{ and }%
b\in \ZZ . \label{sol.hookschur.general-bc}%
\end{equation}


\textit{Proof of (\ref{sol.hookschur.general-bc}):} Let $a\in \ZZ$ and
$b\in \ZZ$. We WLOG assume that $b\in \NN$ (since otherwise, the
left hand side of (\ref{sol.hookschur.general-bc}) is an empty sum, and the
right hand side is $0$ by definition). For every $i\in\left\{
0,1,...,b\right\}  $, we have%
\begin{align}
h_{a+i+1}e_{b-i}  &  =e_{b-i}h_{a+i+1}=\mathbf{s}\left(  \underbrace{a+i+1}%
_{=a+\left(  i+1\right)  },\underbrace{b-i-1}_{=b-\left(  i+1\right)
}\right)  +\mathbf{s}\left(  \underbrace{\left(  a+i+1\right)  -1}%
_{=a+i},b-i\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.hookschur.general-a}),
applied to }n=b-i\text{ and }m=a+i+1\right) \nonumber\\
&  =\mathbf{s}\left(  a+\left(  i+1\right)  ,b-\left(  i+1\right)  \right)
+\mathbf{s}\left(  a+i,b-i\right)  . \label{sol.hookschur.general-bc.pf.1}%
\end{align}
Now,%
\begin{align*}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}\underbrace{h_{a+i+1}e_{b-i}%
}_{\substack{=\mathbf{s}\left(  a+\left(  i+1\right)  ,b-\left(  i+1\right)
\right)  +\mathbf{s}\left(  a+i,b-i\right)  \\\text{(by
(\ref{sol.hookschur.general-bc.pf.1}))}}}  &  =\sum_{i=0}^{b}%
\underbrace{\left(  -1\right)  ^{i}\left(  \mathbf{s}\left(  a+\left(
i+1\right)  ,b-\left(  i+1\right)  \right)  +\mathbf{s}\left(  a+i,b-i\right)
\right)  }_{\substack{=\left(  -1\right)  ^{i}\mathbf{s}\left(  a+\left(
i+1\right)  ,b-\left(  i+1\right)  \right)  +\left(  -1\right)  ^{i}%
\mathbf{s}\left(  a+i,b-i\right)  \\=\left(  -1\right)  ^{i}\mathbf{s}\left(
a+\left(  i+1\right)  ,b-\left(  i+1\right)  \right)  -\left(  -1\right)
^{i-1}\mathbf{s}\left(  a+i,b-i\right)  }}\\
&  =\sum_{i=0}^{b}\left(  \left(  -1\right)  ^{i}\mathbf{s}\left(  a+\left(
i+1\right)  ,b-\left(  i+1\right)  \right)  -\left(  -1\right)  ^{i-1}%
\mathbf{s}\left(  a+i,b-i\right)  \right) \\
&  =\left(  -1\right)  ^{b}\mathbf{s}\left(  a+\left(  b+1\right)
,\underbrace{b-\left(  b+1\right)  }_{=-1}\right)  -\underbrace{\left(
-1\right)  ^{0-1}}_{=-1}\mathbf{s}\left(  \underbrace{a+0}_{=a}%
,\underbrace{b-0}_{=b}\right) \\
& \qquad \qquad \left(\text{by the telescope principle}\right) \\
&  =\left(  -1\right)  ^{b}\underbrace{\mathbf{s}\left(  a,-1\right)
}_{\substack{=0\\\text{(by (\ref{sol.hookschur.b<0}))}}}-\left(  -1\right)
\mathbf{s}\left(  a,b\right)  =-\left(  -1\right)  \mathbf{s}\left(
a,b\right)  =\mathbf{s}\left(  a,b\right)  .
\end{align*}
This proves (\ref{sol.hookschur.general-bc}).

Here is a modified version of (\ref{sol.hookschur.general-bc}) which will be
used in our solution of Exercise \ref{exe.hookschur}(d) further below: For any
$a\in \ZZ$, $b\in \ZZ$ and $c\in \ZZ$ satisfying $c\geq b$,
we have%
\begin{equation}
\sum_{k=0}^{c}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}=\mathbf{s}\left(
a,b\right)  .
\label{sol.hookschur.general-bb'c}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.general-bb'c}):} Let
$a\in \ZZ$, $b\in \ZZ$ and $c\in \ZZ$ be such that $c\geq
b$. We WLOG assume that $b\geq0$ (since otherwise,
both sides of (\ref{sol.hookschur.general-bb'c}) vanish).
Then,
\begin{align*}
\sum_{k=0}^{c}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}  &  =\sum_{k=0}%
^{b}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}%
+\sum_{k=b+1}^{c}\left(  -1\right)  ^{k}h_{a+k+1}\underbrace{e_{b-k}%
}_{\substack{=0\\\text{(since }b-k<0\\\text{(since }k>b\text{))}}}\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}+\underbrace{\sum
_{k=b+1}^{c}\left(  -1\right)  ^{k}h_{a+k+1}0}_{=0}\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}=\sum_{i=0}%
^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}=\mathbf{s}\left(  a,b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.hookschur.general-bc})}\right)
.
\end{align*}
This proves (\ref{sol.hookschur.general-bb'c}).}

Now, the first three parts of Exercise \ref{exe.hookschur} are as good as
solved: The claim of Exercise \ref{exe.hookschur}(a) has already been proven
in (\ref{sol.hookschur.general-a.a}), and the claims of Exercise
\ref{exe.hookschur}(b) and Exercise \ref{exe.hookschur}(c) follow from
(\ref{sol.hookschur.general-bc}).

Before we come to the solution of Exercise \ref{exe.hookschur}(d), we simplify
our life by introducing another definition. Namely, let us define a
$\kk$-module endomorphism $\overline{\id}$ of $\Lambda$
by $\overline{\id}=\id_{\Lambda
}-u\epsilon$. Then, it is easy to see that $\overline{\id}
\left(  s_{\lambda}\right)  =s_{\lambda}$ for every nonempty partition
$\lambda$, whereas $\overline{\id}\left(  c\right)  =0$ for
every $c\in \kk$. Now, straightforward computation shows that%
\begin{equation}
\overline{\id}\left(  \mathbf{s}\left(  a,b\right)  \right)
=s_{\left(  a+1,1^{b}\right)  }\ \ \ \ \ \ \ \ \ \ \text{for all }%
a\in \NN \text{ and }b\in \NN ; \label{sol.hookschur.d.sbar}%
\end{equation}%
\begin{equation}
\overline{\id}\left(  \mathbf{s}\left(  a,b\right)  \right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a,b\right)  \in \ZZ %
^{2}\text{ satisfying }\left(  a,b\right)  \notin \NN ^{2}.
\label{sol.hookschur.d.sbar.NN}%
\end{equation}
But every $x\in\Lambda$ satisfies%
\begin{equation}
\Delta x-1\otimes x-x\otimes1=\left(  \overline{\id}%
\otimes\overline{\id}\right)  \left(  \Delta x\right)
-\epsilon\left(  x\right)  \cdot1\otimes1\ \ \ \ \ \ \ \ \ \ \text{in }%
\Lambda\otimes\Lambda. \label{sol.hookschur.d.Delta}%
\end{equation}
(This is an identity which holds not only in $\Lambda$ but in every
$\kk$-bialgebra, and which is proven by applying the bialgebra axioms.)

(d) Recall that $\Lambda$ is a $\kk$-bialgebra. Hence, $\Delta
:\Lambda\rightarrow\Lambda\otimes\Lambda$ is a $\kk$-algebra homomorphism.

We can rewrite Proposition \ref{symm-comultiplication-formulas-prop}(ii) as
follows: Every $n\in \NN$ satisfies%
\begin{equation}
\Delta e_{n}=\sum_{i=0}^{n}e_{i}\otimes e_{n-i}. \label{sol.hookschur.d.Delen}%
\end{equation}
Thus, every $n\in \NN$ satisfies%
\begin{equation}
\Delta e_{n}=\sum_{i\in \NN }e_{i}\otimes e_{n-i}.
\label{sol.hookschur.d.Delen.inf}%
\end{equation}
\footnote{Indeed, (\ref{sol.hookschur.d.Delen.inf}) follows from
(\ref{sol.hookschur.d.Delen}), because
\[
\sum_{i\in \NN }e_{i}\otimes e_{n-i}
=\sum_{i=0}^{n}e_{i}\otimes e_{n-i}
+\sum_{i=n+1}^{\infty}e_{i}\otimes\underbrace{e_{n-i}}%
_{\substack{=0\\\text{(since }n-i<0\text{)}}}=\sum_{i=0}^{n}e_{i}\otimes
e_{n-i}+\underbrace{\sum_{i=n+1}^{\infty}e_{i}\otimes0}_{=0}
=\sum_{i=0}^{n} e_{i}\otimes e_{n-i}.
\]
} Similarly, every $n\in \NN$ satisfies
\begin{equation}
\Delta h_{n}=\sum_{i\in \NN }h_{i}\otimes h_{n-i}.
\label{sol.hookschur.d.Delhn.inf}
\end{equation}


Let $a\in \NN$ and $b\in \NN$. Exercise \ref{exe.hookschur}(b)
yields $\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}=s_{\left(
a+1,1^{b}\right)  }$, so that
\[
s_{\left(  a+1,1^{b}\right)  }=\sum_{i=0}^{b}\left(  -1\right)  ^{i}%
h_{a+i+1}e_{b-i}=\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}%
\]
(here, we have renamed the summation index $i$ as $k$). Applying the map
$\Delta$ to both sides of this equality, we obtain%
\begin{align}
\Delta s_{\left(  a+1,1^{b}\right)  }  &  =\Delta\left(  \sum_{k=0}^{b}\left(
-1\right)  ^{k}h_{a+k+1}e_{b-k}\right)  \nonumber \\
&=\sum_{k=0}^{b}\left(  -1\right)
^{k}\underbrace{\Delta\left(  h_{a+k+1}\right)  }_{\substack{=\sum
_{i\in \NN }h_{i}\otimes h_{a+k+1-i}\\\text{(by
(\ref{sol.hookschur.d.Delhn.inf}), applied to }n=a+k+1\text{)}}}\cdot
\underbrace{\Delta\left(  e_{b-k}\right)  }_{\substack{=\sum_{i\in \NN %
}e_{i}\otimes e_{b-k-i}\\\text{(by (\ref{sol.hookschur.d.Delen.inf}), applied
to }n=b-k\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a } \kk %
\text{-algebra homomorphism}\right) \nonumber\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}\left(  \sum_{i\in \NN }%
h_{i}\otimes h_{a+k+1-i}\right)  \left(  \sum_{i\in \NN }e_{i}\otimes
e_{b-k-i}\right) \nonumber\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}\left(  \sum_{i\in \NN }%
h_{i}\otimes h_{a+k+1-i}\right)  \left(  \sum_{j\in \NN }e_{j}\otimes
e_{b-k-j}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}i\text{ as }j\text{ in the third sum}\right) \nonumber\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}\sum_{i\in \NN }\sum
_{j\in \NN }\underbrace{\left(  h_{i}\otimes h_{a+k+1-i}\right)  \left(
e_{j}\otimes e_{b-k-j}\right)  }_{=h_{i}e_{j}\otimes h_{a+k+1-i}e_{b-k-j}%
}\nonumber \\
&=\sum_{k=0}^{b}\left(  -1\right)  ^{k}\sum_{i\in \NN }\sum
_{j\in \NN }h_{i}e_{j}\otimes h_{a+k+1-i}e_{b-k-j}\nonumber\\
&  =\underbrace{\sum_{i\in \NN }\sum_{j\in \NN }}_{=\sum_{\left(
i,j\right)  \in \NN ^{2}}}\underbrace{h_{i}e_{j}}_{\substack{=e_{j}%
h_{i}\\=\mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)
\\\text{(by (\ref{sol.hookschur.general-a}), applied to }j\text{ and
}i\\\text{instead of }n\text{ and }m\text{)}}}\otimes\underbrace{\left(
\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a-i+k+1}e_{b-j-k}\right)
}_{\substack{=\mathbf{s}\left(  a-i,b-j\right)  \\\text{(by
(\ref{sol.hookschur.general-bb'c}), applied to }a-i\text{, }b-j\text{ and
}b\\\text{instead of }a\text{, }b\text{ and }c\text{ (since }b\geq
b-j\text{))}}}\nonumber\\
&  =\sum_{\left(  i,j\right)  \in \NN ^{2}}\left(  \mathbf{s}\left(
i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)  \otimes\mathbf{s}%
\left(  a-i,b-j\right)  . \label{sol.hookschur.d.6}%
\end{align}


Now, applying (\ref{sol.hookschur.d.Delta}) to $x=s_{\left(  a+1,1^{b}\right)
}$, we obtain%
\begin{align}
&  \Delta s_{\left(  a+1,1^{b}\right)  }-1\otimes s_{\left(  a+1,1^{b}\right)
}-s_{\left(  a+1,1^{b}\right)  }\otimes1\nonumber\\
&  =\left(  \overline{\id}\otimes\overline{\id}\right)
\underbrace{\left(  \Delta s_{\left(  a+1,1^{b}\right)  }\right)
}_{\substack{=\sum_{\left(  i,j\right)  \in \NN ^{2}}\left(
\mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)
\otimes\mathbf{s}\left(  a-i,b-j\right)  \\\text{(by (\ref{sol.hookschur.d.6}%
))}}}-\underbrace{\epsilon\left(  s_{\left(  a+1,1^{b}\right)  }\right)
}_{\substack{=0\\\text{(since }\left(  a+1,1^{b}\right)  \text{ is
a}\\\text{nonempty partition)}}}1\otimes1\nonumber\\
&  =\left(  \overline{\id}\otimes\overline{\operatorname*{id}%
}\right)  \left(  \sum_{\left(  i,j\right)  \in \NN ^{2}}\left(
\mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)
\otimes\mathbf{s}\left(  a-i,b-j\right)  \right)  -\underbrace{0\cdot
1\otimes1}_{=0}\nonumber\\
&  =\left(  \overline{\id}\otimes\overline{\operatorname*{id}%
}\right)  \left(  \sum_{\left(  i,j\right)  \in \NN ^{2}}\left(
\mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)
\otimes\mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  =\sum_{\left(  i,j\right)  \in \NN ^{2}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
a-i,b-j\right)  \right) \nonumber\\
&  =\sum_{\left(  i,j\right)  \in \NN ^{2}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes
\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
+\sum_{\left(  i,j\right)  \in \NN ^{2}}\overline{\id}
\left(  \mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline{\id}
\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  .
\label{sol.hookschur.d.8}%
\end{align}


We can now observe that
\begin{equation}
\sum_{\left(  i,j\right)  \in \NN ^{2}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes
\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
=\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4};\\c+e=a-1;\\d+f=b}%
}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(  e+1,1^{f}\right)  }.
\label{sol.hookschur.d.reindex.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.reindex.1}):} Let
$\overline{ \NN }=\left\{  -1,0,1,2,...\right\}  =\left\{  -1\right\}
\cup \NN$. We have%
\begin{equation}
\sum\limits_{\substack{\left(  c,d,e,f\right)  \in\overline{ \NN }%
\times \NN \times \ZZ \times \ZZ ;\\c+e=a-1;\\d+f=b}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  e,f\right)
\right)  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a-1;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  .
\label{sol.hookschur.d.reindex.1.pf.step0}%
\end{equation}
(In fact, the sum on the left hand side of
(\ref{sol.hookschur.d.reindex.1.pf.step0}) differs from that on the right hand
side of (\ref{sol.hookschur.d.reindex.1.pf.step0}) only by the presence of
addends of the form $\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  $ for certain quadruples $\left(
c,d,e,f\right)  \in\overline{ \NN }\times \NN \times \ZZ %
\times \ZZ$ which don't belong to $\NN ^{4}$. But all such addends
are $0$ (as can be easily seen using (\ref{sol.hookschur.d.sbar.NN})), and so
the two sums have the same value, and
(\ref{sol.hookschur.d.reindex.1.pf.step0}) is proven.)
\par
The map%
\begin{align*}
 \NN ^{2}  &  \rightarrow\left\{  \left(  c,d,e,f\right)  \in
\overline{ \NN }\times \NN \times \ZZ \times \ZZ %
\ \mid\ c+e=a-1;\ d+f=b\right\}  ,\\
\left(  i,j\right)   &  \mapsto\left(  i-1,j,a-i,b-j\right)
\end{align*}
is a bijection. Hence, we can substitute $\left(  i-1,j,a-i,b-j\right)  $ for
$\left(  c,d,e,f\right)  $ in the sum $\sum\limits_{\substack{\left(  c,d,e,f\right)
\in\overline{ \NN }\times \NN \times \ZZ \times \ZZ %
;\\c+e=a-1;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  $, and thus obtain
\[
\sum\limits_{\substack{\left(  c,d,e,f\right)  \in\overline{ \NN }%
\times \NN \times \ZZ \times \ZZ ;\\c+e=a-1;\\d+f=b}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  e,f\right)
\right)  =\sum_{\left(  i,j\right)  \in \NN ^{2}}
\overline{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  .
\]
Comparing this with (\ref{sol.hookschur.d.reindex.1.pf.step0}), we obtain%
\begin{align*}
&  \sum_{\left(  i,j\right)  \in \NN ^{2}}\overline{\id}
\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes
\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right) \\
&  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a-1;\\d+f=b}}\underbrace{\overline{\id}\left(
\mathbf{s}\left(  c,d\right)  \right)  }_{\substack{=s_{\left(  c+1,1^{d}%
\right)  }\\\text{(by (\ref{sol.hookschur.d.sbar}), applied to}\\
c\text{ and }d\text{ instead of }a\text{ and }b\text{)}}}\otimes
\underbrace{\overline{\id}\left(  \mathbf{s}\left(  e,f\right)  \right)
}_{\substack{=s_{\left(  e+1,1^{f}\right)  }\\\text{(by
(\ref{sol.hookschur.d.sbar}), applied to}\\e\text{ and }f\text{ instead of
}a\text{ and }b\text{)}}}\\
&  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }.
\end{align*}
This proves (\ref{sol.hookschur.d.reindex.1}).} Similarly,%
\begin{equation}
\sum_{\left(  i,j\right)  \in \NN ^{2}} \overline{\id}\left(
\mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  =\sum\limits_{\substack{\left(
c,d,e,f\right)  \in \NN ^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}%
\right)  }\otimes s_{\left(  e+1,1^{f}\right)  }.
\label{sol.hookschur.d.reindex.2}%
\end{equation}
Now, (\ref{sol.hookschur.d.8}) becomes%
\begin{align*}
&  \Delta s_{\left(  a+1,1^{b}\right)  }-1\otimes s_{\left(  a+1,1^{b}\right)
}-s_{\left(  a+1,1^{b}\right)  }\otimes1\\
&  =\underbrace{\sum_{\left(  i,j\right)  \in \NN ^{2}}
\overline{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{\substack{=\sum\limits_{\substack{\left(  c,d,e,f\right)  \in
 \NN ^{4};\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes
s_{\left(  e+1,1^{f}\right)  }\\\text{(by (\ref{sol.hookschur.d.reindex.1}))}%
}}+\underbrace{\sum_{\left(  i,j\right)  \in \NN ^{2}}
\overline{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{\substack{=\sum\limits_{\substack{\left(  c,d,e,f\right)  \in
 \NN ^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}\right)  }\otimes
s_{\left(  e+1,1^{f}\right)  }\\\text{(by (\ref{sol.hookschur.d.reindex.2}))}%
}}\\
&  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }+\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN %
^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }.
\end{align*}
Adding $1\otimes s_{\left(  a+1,1^{b}\right)  }+s_{\left(  a+1,1^{b}\right)
}\otimes1$ to both sides of this equality, we obtain the claim of Exercise
\ref{exe.hookschur}(d).

\textit{Remark:} In our above solution, we have first proved
(\ref{sol.hookschur.general-a}) and then used it to derive
(\ref{sol.hookschur.general-bc}). There is also an alternative way to prove
these two identities, which proceeds the other way round. The main idea is to
obtain (\ref{sol.hookschur.general-bc}) by applying Exercise
\ref{exe.bernstein}(b) to $\lambda=\left(  1^{b}\right)  $ and $m=a+1$. (This
works in the case of $a\in \NN$ and $b\in \NN$ only. The remaining
cases, however, are easy to either check directly or reduce to
(\ref{e-h-relation}).) Once this is done, (\ref{sol.hookschur.general-a}) can
be verified by rewriting both $\mathbf{s}\left(  m,n-1\right)  $ and
$\mathbf{s}\left(  m-1,n\right)  $ using (\ref{sol.hookschur.general-bc}).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.hookschur}}

\textit{Solution to Exercise \ref{exe.hookschur}.} For any integers $a$
and $b$, we define an element $\mathbf{s}\left(  a,b\right)  $ of $\Lambda$
by
\[
\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases} \quad .
\]
(We are introducing this $\mathbf{s}\left(  a,b\right)  $ chiefly for reasons
of convenience: it will allow us to unify parts (b) and (c) of the exercise as
well as other statements.)

We notice that%
\begin{equation}
\mathbf{s}\left(  a,b\right)  =s_{\left(  a+1,1^{b}\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in \NN \text{ and }b\in \NN .
\label{sol.hookschur.sab-pos}%
\end{equation}
\footnote{\textit{Proof.} Let $a\in \NN$ and $b\in \NN$. Then,
$a\geq0$ and $b\geq0$. The definition of $\mathbf{s}\left(  a,b\right)  $
yields $\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=s_{\left(  a+1,1^{b}\right)  }$ (since $a\geq0$ and $b\geq0$). This
proves (\ref{sol.hookschur.sab-pos}).} Also,
\begin{equation}
\mathbf{s}\left(  a,b\right)  =\left(  -1\right)  ^{b}\delta_{a+b,-1}%
\ \ \ \ \ \ \ \ \ \ \text{for every negative integer }a\text{ and every }%
b\in \ZZ . \label{sol.hookschur.sab-neg}%
\end{equation}
\footnote{\textit{Proof.} Let $a$ be a negative integer, and let
$b\in \ZZ$. We need to prove (\ref{sol.hookschur.sab-neg}). If $b\geq0$,
then%
\begin{align*}
\mathbf{s}\left(  a,b\right)   &  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{s}\left(  a,b\right)  \right) \\
&  =\left(  -1\right)  ^{b}\delta_{a+b,-1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }a<0\text{ and }b\geq0\right)  .
\end{align*}
Hence, (\ref{sol.hookschur.sab-neg}) is proven in the case when $b\geq0$.
Thus, for the rest of our proof of (\ref{sol.hookschur.sab-neg}), we can WLOG
assume that we don't have $b\geq0$. Assume this. Hence, we have $b<0$ (since
we don't have $b\geq0$), so that $b\leq-1$. Combined with $a<0$ (since $a$ is
negative), this yields $\underbrace{a}_{<0}+\underbrace{b}_{\leq-1}<0+\left(
-1\right)  =-1$, so that $a+b\neq-1$ and therefore $\delta_{a+b,-1}=0$. Hence,
$\left(  -1\right)  ^{b}\underbrace{\delta_{a+b,-1}}_{=0}=0$. But the
definition of $\mathbf{s}\left(  a,b\right)  $ yields%
\begin{align*}
\mathbf{s}\left(  a,b\right)   &  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }b<0\right) \\
&  =\left(  -1\right)  ^{b}\delta_{a+b,-1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  -1\right)  ^{b}\delta_{a+b,-1}=0\right)  .
\end{align*}
This proves (\ref{sol.hookschur.sab-neg}).} Further,%
\begin{equation}
\mathbf{s}\left(  a,0\right)  =h_{a+1}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in \ZZ . \label{sol.hookschur.b=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.b=0}):} Let $a\in \ZZ$.
We need to prove (\ref{sol.hookschur.b=0}). If $a\in \NN$, then
(\ref{sol.hookschur.b=0}) is true (because if $a\in \NN$, then%
\begin{align*}
\mathbf{s}\left(  a,0\right)   &  =s_{\left(  a+1,1^{0}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.hookschur.sab-pos}), applied to
}b=0\right) \\
&  =s_{\left(  a+1\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
a+1,1^{0}\right)  =\left(  a+1\right)  \right) \\
&  =h_{a+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }s_{\left(  m\right)
}=h_{m}\text{ for every }m\in \NN \right)
\end{align*}
). Hence, for the rest of this proof, we can WLOG assume that we don't have
$a\in \NN$.
\par
We have $a\in \ZZ$, but we don't have $a\in \NN$. Hence, we have
$a\in \ZZ \setminus \NN =\left\{  -1,-2,-3,...\right\}  $. In other
words, $a$ is negative. If $a=-1$, then (\ref{sol.hookschur.b=0}) is true
(because if $a=-1$, then%
\begin{align*}
\mathbf{s}\left(  a,0\right)   &  =\underbrace{\left(  -1\right)  ^{0}}%
_{=1}\delta_{a+0,-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.hookschur.sab-neg}), applied to }b=0\right) \\
&  =\delta_{a+0,-1}=1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a+0=a=-1\right)
\\
&  =h_{0}=h_{a+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0=\underbrace{-1}%
_{=a}+1=a+1\right)
\end{align*}
). Hence, for the rest of this proof, we can WLOG assume that we don't have
$a=-1$. Thus, $a\neq-1$. Also, since $a\in\left\{  -1,-2,-3,...\right\}  $ and
$a\neq-1$, we have $a\in\left\{  -1,-2,-3,...\right\}  \setminus\left\{
-1\right\}  =\left\{  -2,-3,-4,...\right\}  $, and thus $a<-1$, so that
$a+1<0$ and thus $h_{a+1}=0$. Now,%
\begin{align*}
\mathbf{s}\left(  a,0\right)   &  =\underbrace{\left(  -1\right)  ^{0}}%
_{=1}\delta_{a+0,-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.hookschur.sab-neg}), applied to }b=0\right) \\
&  =\delta_{a+0,-1}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a+0=a\neq
-1\right) \\
&  =h_{a+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h_{a+1}=0\right)  ,
\end{align*}
and thus (\ref{sol.hookschur.b=0}) is proven.} Also,%
\begin{equation}
\mathbf{s}\left(  a,b\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for any }%
a\in \ZZ \text{ and any negative }b\in \ZZ .
\label{sol.hookschur.b<0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.b<0}):} Let $a\in \ZZ$.
Let $b\in \ZZ$ be negative. Then, the definition of $\mathbf{s}\left(
a,b\right)  $ yields $\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=0$ (since $b<0$). This proves (\ref{sol.hookschur.b<0}).}

Let us now show that%
\begin{equation}
e_{n}h_{m}=\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(  m-1,n\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in \ZZ \text{ and every }%
m\in \ZZ . \label{sol.hookschur.general-a}%
\end{equation}


\textit{Proof of (\ref{sol.hookschur.general-a}):} Let $n\in \ZZ$ and
$m\in \ZZ$. It is easy to prove (\ref{sol.hookschur.general-a}) in the
case $n<0$ (since both sides of (\ref{sol.hookschur.general-a}) vanish in this
case) and in the case $n=0$ (here, it follows from (\ref{sol.hookschur.b=0})).
We can therefore WLOG assume that $n>0$. Assume this.

It is easy to prove (\ref{sol.hookschur.general-a}) in the case $m<0$ (in
which case both sides of (\ref{sol.hookschur.general-a}) vanish) and in the
case $m=0$ (in which case both sides of (\ref{sol.hookschur.general-a}) equal
$e_{n}$). Thus, we can WLOG assume that $m>0$. Assume this. Since $m>0$, we
have $h_{m}=s_{\left(  m\right)  }$, so that%
\begin{equation}
e_{n}\underbrace{h_{m}}_{=s_{\left(  m\right)  }}=e_{n}s_{\left(  m\right)
}=s_{\left(  m\right)  }e_{n}=\sum\limits_{\substack{\lambda^{+}:\lambda^{+}/\left(
m\right)  \text{ is a }\\\text{vertical }n\text{-strip}}}s_{\lambda^{+}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.pieri.e}), applied to }%
\lambda=\left(  m\right)  \right)  . \label{sol.hookschur.general-a.pf.1}%
\end{equation}


It is easy to see that there are exactly two partitions $\lambda^{+}$ for
which $\lambda^{+}/\left(  m\right)  $ is a vertical $n$-strip, namely the
partitions $\left(  m,1^{n}\right)  $ and $\left(  m+1,1^{n-1}\right)
$\ \ \ \ \footnote{\textit{Proof.} First of all, the partitions $\left(
m,1^{n}\right)  $ and $\left(  m+1,1^{n-1}\right)  $ are clearly well-defined
(since $m>0$ and $n>0$), and are partitions $\lambda^{+}$ for which
$\lambda^{+}/\left(  m\right)  $ is a vertical $n$-strip. It remains to prove
that they are the only such partitions.
\par
Indeed, consider any partition $\lambda^{+}$ for which $\lambda^{+}/\left(
m\right)  $ is a vertical $n$-strip. Thus, the skew shape $\lambda^{+}/\left(
m\right)  $ has exactly $n$ cells and is a vertical strip, i.e., no two cells
of this skew shape lie on the same row. As a consequence, the skew shape
$\lambda^{+}/\left(  m\right)  $ consists of $n$ cells distributed among $n$
rows. If the skew shape $\lambda^{+}/\left(  m\right)  $ has no cells in the
$1$-st row, then these $n$ rows must be the $2$-nd row, the $3$-rd row, etc.,
the $\left(  n+1\right)  $-st rows (because if the skew shape $\lambda
^{+}/\left(  m\right)  $ would miss out on any of these rows -- say, it would
have no cells in the $i$-th row for some $i\in\left\{  2,3,...,n+1\right\}  $
--, then $\lambda_{i}$ would be $0$ for this $i$, and as a consequence the
skew shape $\lambda^{+}/\left(  m\right)  $ would have to miss out on the
$\left(  i+1\right)  $-st row, on the $\left(  i+2\right)  $-nd row, and so on
as well, whence it would not succeed to have $n$ cells); as a consequence,
$\lambda^{+}$ must be $\left(  m,1^{n}\right)  $ in this case. If, on the
other hand, the skew shape $\lambda^{+}/\left(  m\right)  $ does have cells in
the $1$-st row, then these $n$ rows must be the $1$-st row, the $2$-nd row,
etc., the $n$-th row (again, we see by the argument that it cannot miss out on
any of these rows), and thus $\lambda^{+}$ must be $\left(  m+1,1^{n-1}%
\right)  $ in this case. We thus see that $\lambda^{+}$ is $\left(
m+1,1^{n-1}\right)  $ or $\left(  m,1^{n}\right)  $.
\par
Forget now that we fixed $\lambda^{+}$. We thus have shown that the only
partitions $\lambda^{+}$ for which $\lambda^{+}/\left(  m\right)  $ is a
vertical $n$-strip are $\left(  m,1^{n}\right)  $ and $\left(  m+1,1^{n-1}%
\right)  $, qed.}. Hence, the sum $\sum\limits_{\substack{\lambda^{+}:\lambda
^{+}/\left(  m\right)  \text{ is a }\\\text{vertical }n\text{-strip}%
}}s_{\lambda^{+}}$ has exactly two addends: that for $\lambda^{+}=\left(
m,1^{n}\right)  $ and that for $\lambda^{+}=\left(  m+1,1^{n-1}\right)  $.
Thus, this sum simplifies to $s_{\left(  m,1^{n}\right)  }+s_{\left(
m+1,1^{n-1}\right)  }$. Hence, (\ref{sol.hookschur.general-a.pf.1}) rewrites
as%
\begin{equation}
e_{n}h_{m}=s_{\left(  m,1^{n}\right)  }+s_{\left(  m+1,1^{n-1}\right)  }.
\label{sol.hookschur.general-a.a}%
\end{equation}
But (\ref{sol.hookschur.sab-pos}) (applied to $a=m-1$ and $b=n$) yields
$\mathbf{s}\left(  m-1,n\right)  =s_{\left(  m,1^{n}\right)  }$. Also,
(\ref{sol.hookschur.sab-pos}) (applied to $a=m$ and $b=n-1$) yields
$\mathbf{s}\left(  m,n-1\right)  =s_{\left(  m+1,1^{n-1}\right)  }$. Thus,
\[
e_{n}h_{m}=\underbrace{s_{\left(  m,1^{n}\right)  }}_{=\mathbf{s}\left(
m-1,n\right)  }+\underbrace{s_{\left(  m+1,1^{n-1}\right)  }}_{=\mathbf{s}%
\left(  m,n-1\right)  }=\mathbf{s}\left(  m-1,n\right)  +\mathbf{s}\left(
m,n-1\right)  =\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(
m-1,n\right)  .
\]
This proves (\ref{sol.hookschur.general-a}).

Let us now prove a statement which encompasses both parts (b) and (c) of
Exercise \ref{exe.hookschur}. Namely, we are going to prove that%
\begin{equation}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}=\mathbf{s}\left(
a,b\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }a\in \ZZ \text{ and }%
b\in \ZZ . \label{sol.hookschur.general-bc}%
\end{equation}


\textit{Proof of (\ref{sol.hookschur.general-bc}):} Let $a\in \ZZ$ and
$b\in \ZZ$. We WLOG assume that $b\in \NN$ (since otherwise, the
left hand side of (\ref{sol.hookschur.general-bc}) is an empty sum, and the
right hand side is $0$ by definition). For every $i\in\left\{
0,1,...,b-1\right\}  $, the number $b-i$ is a positive integer. Hence, for
every $i\in\left\{  0,1,...,b-1\right\}  $, we have%
\begin{align}
e_{b-i}h_{a+i+1}  &  =\mathbf{s}\left(  a+i+1,\underbrace{b-i-1}_{=b-\left(
i+1\right)  }\right)  +\mathbf{s}\left(  \underbrace{\left(  a+i+1\right)
-1}_{=a+i},b-i\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.hookschur.general-a}),
applied to }n=b-i\text{ and }m=a+i+1\right) \nonumber\\
&  =\mathbf{s}\left(  a+i+1,b-\left(  i+1\right)  \right)  +\mathbf{s}\left(
a+i,b-i\right)  . \label{sol.hookschur.general-bc.pf.1}%
\end{align}
Now,%
\begin{align*}
&  \sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}\\
&  =\sum_{i=0}^{b-1}\left(  -1\right)  ^{i}\underbrace{h_{a+i+1}e_{b-i}%
}_{\substack{=e_{b-i}h_{a+i+1}\\=\mathbf{s}\left(  a+i+1,b-\left(  i+1\right)
\right)  +\mathbf{s}\left(  a+i,b-i\right)  \\\text{(by
(\ref{sol.hookschur.general-bc.pf.1}))}}}+\left(  -1\right)  ^{b}%
h_{a+b+1}\underbrace{e_{b-b}}_{=e_{0}=1}\\
&  =\underbrace{\sum_{i=0}^{b-1}\left(  -1\right)  ^{i}\left(  \mathbf{s}%
\left(  a+i+1,b-\left(  i+1\right)  \right)  +\mathbf{s}\left(
a+i,b-i\right)  \right)  }_{=\sum_{i=0}^{b-1}\left(  -1\right)  ^{i}%
\mathbf{s}\left(  a+i+1,b-\left(  i+1\right)  \right)  +\sum_{i=0}%
^{b-1}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)  }+\left(
-1\right)  ^{b}\underbrace{h_{a+b+1}}_{\substack{=\mathbf{s}\left(
a+b,0\right)  \\\text{(since }\mathbf{s}\left(  a+b,0\right)  =h_{a+b+1}%
\\\text{(by (\ref{sol.hookschur.b=0}), applied to}\\a+b+1\text{ instead of
}a\text{))}}}\\
&  =\sum_{i=0}^{b-1}\underbrace{\left(  -1\right)  ^{i}}_{\substack{=-\left(
-1\right)  ^{i+1}\\\text{(since }\left(  -1\right)  ^{i+1}=-\left(  -1\right)
^{i}\text{)}}}\mathbf{s}\left(  a+i+1,b-\left(  i+1\right)  \right)
+\sum_{i=0}^{b-1}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)
+\left(  -1\right)  ^{b}\mathbf{s}\left(  a+b,\underbrace{0}_{=b-b}\right) \\
&  =\sum_{i=0}^{b-1}\left(  -\left(  -1\right)  ^{i+1}\right)  \mathbf{s}%
\left(  a+i+1,b-\left(  i+1\right)  \right)  +\underbrace{\sum_{i=0}%
^{b-1}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)  +\left(
-1\right)  ^{b}\mathbf{s}\left(  a+b,b-b\right)  }_{\substack{=\sum_{i=0}%
^{b}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)  \\\text{(since
}\sum_{i=0}^{b}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)
\\=\sum_{i=0}^{b-1}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)
+\left(  -1\right)  ^{b}\mathbf{s}\left(  a+b,b-b\right)  \text{)}}}\\
&  =\sum_{i=0}^{b-1}\left(  -\left(  -1\right)  ^{i+1}\right)  \mathbf{s}%
\left(  a+i+1,b-\left(  i+1\right)  \right)  +\sum_{i=0}^{b}\left(  -1\right)
^{i}\mathbf{s}\left(  a+i,b-i\right) \\
&  =-\underbrace{\sum_{i=0}^{b-1}\left(  -1\right)  ^{i+1}\mathbf{s}\left(
a+i+1,b-\left(  i+1\right)  \right)  }_{\substack{=\sum_{i=1}^{b}\left(
-1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)  \\\text{(here, we substituted
}i\text{ for}\\i+1\text{ in the sum)}}}+\underbrace{\sum_{i=0}^{b}\left(
-1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)  }_{=\left(  -1\right)
^{0}\mathbf{s}\left(  a+0,b-0\right)  +\sum_{i=1}^{b}\left(  -1\right)
^{i}\mathbf{s}\left(  a+i,b-i\right)  }\\
&  =-\sum_{i=1}^{b}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right)
+\left(  -1\right)  ^{0}\mathbf{s}\left(  a+0,b-0\right)  +\sum_{i=1}%
^{b}\left(  -1\right)  ^{i}\mathbf{s}\left(  a+i,b-i\right) \\
&  =\underbrace{\left(  -1\right)  ^{0}}_{=1}\mathbf{s}\left(  a+0,b-0\right)
=\mathbf{s}\left(  \underbrace{a+0}_{=a},\underbrace{b-0}_{=b}\right)
=\mathbf{s}\left(  a,b\right)  .
\end{align*}
This proves (\ref{sol.hookschur.general-bc}).

Here is a modified version of (\ref{sol.hookschur.general-bc}) which will be
used in our solution of Exercise \ref{exe.hookschur}(d) further below: For any
$a\in \ZZ$, $b\in \ZZ$ and $c\in \ZZ$ satisfying $c\geq b$,
we have%
\begin{equation}
\sum_{k=0}^{c}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}=\mathbf{s}\left(
a,b\right)  . \label{sol.hookschur.general-bb'c}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.general-bb'c}):} Let
$a\in \ZZ$, $b\in \ZZ$ and $c\in \ZZ$ be such that $c\geq
b$. We need to prove (\ref{sol.hookschur.general-bb'c}).
\par
If $b<0$, then%
\[
\sum_{k=0}^{c}\left(  -1\right)  ^{k}h_{a+k+1}\underbrace{e_{b-k}%
}_{\substack{=0\\\text{(since }b-k<0\\\text{(since }b<0\text{ and }%
k\geq0\text{))}}}=\sum_{k=0}^{c}\left(  -1\right)  ^{k}h_{a+k+1}%
0=0=\mathbf{s}\left(  a,b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.hookschur.b<0})}\right)  .
\]
Hence, if $b<0$, then (\ref{sol.hookschur.general-bb'c}) is proven. Thus, for
the rest of the proof of (\ref{sol.hookschur.general-bb'c}), we can WLOG
assume that we don't have $b<0$.
\par
We have $b\geq0$ (since we don't have $b<0$). Now,%
\begin{align*}
\sum_{k=0}^{c}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}  &  =\sum_{i=0}%
^{c}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed the summation index }k\text{ as }i\right) \\
&  =\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}+\sum_{i=b+1}%
^{c}\left(  -1\right)  ^{i}h_{a+i+1}\underbrace{e_{b-i}}%
_{\substack{=0\\\text{(since }b-i<0\text{ (since }i>b\text{))}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }c\geq b\geq0\right) \\
&  =\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}+\underbrace{\sum
_{i=b+1}^{c}\left(  -1\right)  ^{i}h_{a+i+1}0}_{=0}=\sum_{i=0}^{b}\left(
-1\right)  ^{i}h_{a+i+1}e_{b-i}=\mathbf{s}\left(  a,b\right)
\end{align*}
(by (\ref{sol.hookschur.general-bc})).}

Now, the first three parts of Exercise \ref{exe.hookschur} are as good as
solved: The claim of Exercise \ref{exe.hookschur}(a) has already been proven
in (\ref{sol.hookschur.general-a.a}), and the claims of Exercise
\ref{exe.hookschur}(b) and Exercise \ref{exe.hookschur}(c) follow from
(\ref{sol.hookschur.general-bc}).

Before we come to the solution of Exercise \ref{exe.hookschur}(d), we simplify
our life by introducing another definition. Namely, let us define a
$\kk$-module endomorphism $\overline{\id}$ of $\Lambda$
by $\overline{\id}=\operatorname{id}_{\Lambda
}-u\epsilon$. Then,%
\begin{equation}
\overline{\id}\left(  \mathbf{s}\left(  a,b\right)  \right)
=s_{\left(  a+1,1^{b}\right)  }\ \ \ \ \ \ \ \ \ \ \text{for all }%
a\in \NN \text{ and }b\in \NN . \label{sol.hookschur.d.sbar}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.sbar}):} Let $a\in \NN$
and $b\in \NN$. We need to prove (\ref{sol.hookschur.d.sbar}).
\par
The partition $\left(  a+1,1^{b}\right)  $ is nonempty (since its first entry
is $\underbrace{a}_{\geq0}+1\geq0+1>0$). In other words, $\left(
a+1,1^{b}\right)  \neq\varnothing$.
\par
We know that $\epsilon s_{\lambda}=\delta_{\lambda,\varnothing}$ for every
partition $\lambda$. Applied to $\lambda=\left(  a+1,1^{b}\right)  $, this
yields $\epsilon s_{\left(  a+1,1^{b}\right)  }=\delta_{\left(  a+1,1^{b}%
\right)  ,\varnothing}=0$ (since $\left(  a+1,1^{b}\right)  \neq\varnothing$).
Hence,%
\[
\left(  u\epsilon\right)  \left(  s_{\left(  a+1,1^{b}\right)  }\right)
=u\left(  \underbrace{\epsilon s_{\left(  a+1,1^{b}\right)  }}_{=0}\right)
=u\left(  0\right)  =0.
\]
\par
Now, (\ref{sol.hookschur.sab-pos}) yields\textbf{ }$\mathbf{s}\left(
a,b\right)  =s_{\left(  a+1,1^{b}\right)  }$. Applying the map
$\overline{\id}$ to both sides of this equality, we obtain%
\begin{align*}
\overline{\id}\left(  \mathbf{s}\left(  a,b\right)  \right)
&  =\underbrace{\overline{\id}}_{=
\id_{\Lambda}-u\epsilon}\left(  s_{\left(  a+1,1^{b}\right)  }\right)
=\left(  \operatorname{id}_{\Lambda}-u\epsilon\right)  \left(
s_{\left(  a+1,1^{b}\right)  }\right)  =\underbrace{
\id_{\Lambda}\left(  s_{\left(  a+1,1^{b}\right)  }\right)
}_{=s_{\left(  a+1,1^{b}\right)  }}-\underbrace{\left(  u\epsilon\right)
\left(  s_{\left(  a+1,1^{b}\right)  }\right)  }_{=0}\\
&  =s_{\left(  a+1,1^{b}\right)  }-0=s_{\left(  a+1,1^{b}\right)  }.
\end{align*}
This proves (\ref{sol.hookschur.d.sbar}).} Also,%
\begin{equation}
\overline{\id}\left(  \mathbf{s}\left(  a,b\right)  \right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every negative integer }a\text{ and every
}b\in \ZZ . \label{sol.hookschur.d.sbar.NZ}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.sbar.NZ}):} Let $a$ be a
negative integer. Let $b\in \ZZ$. Notice that $\left(  -1\right)
^{b}\delta_{a+b,-1}\in \kk$, and thus $\epsilon\left(  \left(
-1\right)  ^{b}\delta_{a+b,-1}\right)  =\left(  -1\right)  ^{b}\delta
_{a+b,-1}$ (since every $\lambda\in \kk$ satisfies $\epsilon\left(
\lambda\right)  =\lambda$).
\par
But (\ref{sol.hookschur.sab-neg}) yields $\mathbf{s}\left(  a,b\right)
=\left(  -1\right)  ^{b}\delta_{a+b,-1}$. Hence,%
\begin{align*}
\underbrace{\overline{\id}}_{=
\id_{\Lambda}-u\epsilon}\left(  \underbrace{\mathbf{s}\left(
a,b\right)  }_{=\left(  -1\right)  ^{b}\delta_{a+b,-1}}\right)   &  =\left(
\operatorname{id}_{\Lambda}-u\epsilon\right)  \left(  \left(
-1\right)  ^{b}\delta_{a+b,-1}\right)  =\underbrace{
\id_{\Lambda}\left(  \left(  -1\right)  ^{b}\delta_{a+b,-1}\right)
}_{=\left(  -1\right)  ^{b}\delta_{a+b,-1}}-\underbrace{\left(  u\epsilon
\right)  \left(  \left(  -1\right)  ^{b}\delta_{a+b,-1}\right)  }_{=u\left(
\epsilon\left(  \left(  -1\right)  ^{b}\delta_{a+b,-1}\right)  \right)  }\\
&  =\left(  -1\right)  ^{b}\delta_{a+b,-1}-u\left(  \underbrace{\epsilon
\left(  \left(  -1\right)  ^{b}\delta_{a+b,-1}\right)  }_{=\left(  -1\right)
^{b}\delta_{a+b,-1}}\right)  =\left(  -1\right)  ^{b}\delta_{a+b,-1}%
-\underbrace{u\left(  \left(  -1\right)  ^{b}\delta_{a+b,-1}\right)
}_{\substack{=\left(  -1\right)  ^{b}\delta_{a+b,-1}\cdot1_{\Lambda
}\\\text{(by the definition of }u\left(  \left(  -1\right)  ^{b}%
\delta_{a+b,-1}\right)  \text{)}}}\\
&  =\left(  -1\right)  ^{b}\delta_{a+b,-1}-\underbrace{\left(  -1\right)
^{b}\delta_{a+b,-1}\cdot1_{\Lambda}}_{=\left(  -1\right)  ^{b}\delta_{a+b,-1}%
}=\left(  -1\right)  ^{b}\delta_{a+b,-1}-\left(  -1\right)  ^{b}%
\delta_{a+b,-1}=0;
\end{align*}
this proves (\ref{sol.hookschur.d.sbar.NZ}).} Also,%
\begin{equation}
\overline{\id}\left(  \mathbf{s}\left(  a,b\right)  \right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every negative integer }b\text{ and every
}a\in \ZZ . \label{sol.hookschur.d.sbar.ZN}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.sbar.ZN}):} Let $b$ be a
negative integer. Let $a\in \ZZ$. The definition of $\mathbf{s}\left(
a,b\right)  $ yields%
\[
\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }b<0\right)  ,
\]
so that $\overline{\id}\left(  \underbrace{\mathbf{s}\left(
a,b\right)  }_{=0}\right)  =\overline{\id}\left(  0\right)  =0$
(since $\overline{\id}$ is a $\kk$-linear map). This
proves (\ref{sol.hookschur.d.sbar.ZN}).} But every $x\in\Lambda$ satisfies%
\begin{equation}
\Delta x-1\otimes x-x\otimes1=\left(  \overline{\id}%
\otimes\overline{\id}\right)  \left(  \Delta x\right)
-\epsilon\left(  x\right)  \cdot1\otimes1\ \ \ \ \ \ \ \ \ \ \text{in }%
\Lambda\otimes\Lambda. \label{sol.hookschur.d.Delta}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.Delta}):} Let $x\in\Lambda$.
Let $\operatorname{kan}_{1}:\Lambda\rightarrow\Lambda\otimes
 \kk$ and $\operatorname{kan}_{2}:\Lambda\rightarrow
 \kk \otimes\Lambda$ be the canonical $\kk$-module isomorphisms.
Then, by the axioms of a $\kk$-coalgebra, we have $\left(
\operatorname{id}_{\Lambda}\otimes\epsilon\right)  \circ
\Delta=\operatorname{kan}_{1}$ and $\left(  \epsilon\otimes
\operatorname{id}_{\Lambda}\right)  \circ\Delta
= \operatorname{kan}_{2}$. We have
\[
\underbrace{\overline{\id}}_{=
\id_{\Lambda}-u\epsilon}\left(  x\right)  =\left(  
\id_{\Lambda}-u\epsilon\right)  \left(  x\right)
=\underbrace{\operatorname{id}_{\Lambda}\left(  x\right)  }%
_{=x}-\underbrace{\left(  u\epsilon\right)  \left(  x\right)  }%
_{\substack{=u\left(  \epsilon\left(  x\right)  \right)  =\epsilon\left(
x\right)  1\\\text{(by the definition of }u\left(  \epsilon\left(  x\right)
\right)  \text{)}}}=x-\epsilon\left(  x\right)  1.
\]
Now,%
\begin{align*}
\left(  \underbrace{\operatorname{id}_{\Lambda}}%
_{=\operatorname{id}_{\Lambda}\circ
\id_{\Lambda}}\otimes\underbrace{\left(  u\epsilon\right)  }%
_{=u\circ\epsilon}\right)  \left(  \Delta x\right)   &  =\underbrace{\left(
\left(  \operatorname{id}_{\Lambda}\circ
\id_{\Lambda}\right)  \otimes\left(  u\circ\epsilon\right)  \right)
}_{=\left(  \operatorname{id}_{\Lambda}\otimes u\right)
\circ\left(  \operatorname{id}_{\Lambda}\otimes\epsilon\right)
}\left(  \Delta x\right)  =\left(  \left(  
\id_{\Lambda}\otimes u\right)  \circ\left(  
\id_{\Lambda}\otimes\epsilon\right)  \right)  \left(  \Delta x\right) \\
&  =\left(  \left(  \operatorname{id}_{\Lambda}\otimes u\right)
\circ\underbrace{\left(  \operatorname{id}_{\Lambda}\otimes
\epsilon\right)  \circ\Delta}_{=\operatorname{kan}_{1}}\right)
x=\left(  \left(  \operatorname{id}_{\Lambda}\otimes u\right)
\circ\operatorname{kan}_{1}\right)  x\\
&  =\left(  \operatorname{id}_{\Lambda}\otimes u\right)  \left(
\underbrace{\operatorname{kan}_{1}x}_{\substack{=x\otimes
1\\\text{(by the definition of }\operatorname{kan}_{1}\text{)}%
}}\right)  =\left(  \operatorname{id}_{\Lambda}\otimes u\right)
\left(  x\otimes1\right) \\
&  =\underbrace{\operatorname{id}_{\Lambda}\left(  x\right)  }%
_{=x}\otimes\underbrace{u\left(  1\right)  }_{\substack{=1\\\text{(by the
definition of }u\text{)}}}=x\otimes1
\end{align*}
and%
\begin{align*}
\left(  \underbrace{\left(  u\epsilon\right)  }_{=u\circ\epsilon}%
\otimes\underbrace{\overline{\id}}_{=\overline{\id}
\circ\operatorname{id}_{\Lambda}}\right)
\left(  \Delta x\right)   &  =\underbrace{\left(  \left(  u\circ
\epsilon\right)  \otimes\left(  \overline{\id}\circ
\operatorname{id}_{\Lambda}\right)  \right)  }_{=\left(
u\otimes\overline{\id}\right)  \circ\left(  \epsilon
\otimes\operatorname{id}_{\Lambda}\right)  }\left(  \Delta x\right)
=\left(  \left(  u\otimes\overline{\id}\right)  \circ\left(
\epsilon\otimes\operatorname{id}_{\Lambda}\right)  \right)  \left(
\Delta x\right) \\
&  =\left(  \left(  u\otimes\overline{\id}\right)
\circ\underbrace{\left(  \epsilon\otimes\operatorname{id}_{\Lambda
}\right)  \circ\Delta}_{=\operatorname{kan}_{2}}\right)  x=\left(
\left(  u\otimes\overline{\id}\right)  \circ
\operatorname{kan}_{2}\right)  x\\
&  =\left(  u\otimes\overline{\id}\right)  \left(
\underbrace{\operatorname{kan}_{2}x}_{\substack{=1\otimes
x\\\text{(by the definition of }\operatorname{kan}_{2}\text{)}%
}}\right)  =\left(  u\otimes\overline{\id}\right)  \left(
1\otimes x\right) \\
&  =\underbrace{u\left(  1\right)  }_{\substack{=1\\\text{(by the definition
of }u\text{)}}}\otimes\underbrace{\overline{\id}\left(
x\right)  }_{=x-\epsilon\left(  x\right)  1}=1\otimes\left(  x-\epsilon\left(
x\right)  1\right) \\
&  =1\otimes x-\epsilon\left(  x\right)  \cdot1\otimes1.
\end{align*}
Now,%
\begin{align*}
\left(  \underbrace{\overline{\id}}_{=
\id_{\Lambda}-u\epsilon}\otimes\overline{\id}\right)
\left(  \Delta x\right)   &  =\underbrace{\left(  \left(  
\id_{\Lambda}-u\epsilon\right)  \otimes\overline{\operatorname*{id}%
}\right)  }_{\substack{=\operatorname{id}_{\Lambda}\otimes
\overline{\id}-\left(  u\epsilon\right)  \otimes\overline{\id}\\
\text{(since tensoring of } \kk \text{-linear maps
is } \kk \text{-bilinear)}}}\left(  \Delta x\right)  =\left(
\operatorname{id}_{\Lambda}\otimes\overline{\operatorname*{id}%
}-\left(  u\epsilon\right)  \otimes\overline{\id}\right)
\left(  \Delta x\right) \\
&  =\left(  \id_{\Lambda}\otimes
\underbrace{\overline{\id}}_{=\id_{\Lambda} -u\epsilon}\right)
\left(  \Delta x\right)  -\underbrace{\left(  \left(  u\epsilon
\right)  \otimes\overline{\id}\right)  \left(  \Delta x\right)
}_{=1\otimes x-\epsilon\left(  x\right)  \cdot1\otimes1}\\
&  =\underbrace{\left(  \operatorname{id}_{\Lambda}\otimes\left(
\operatorname{id}_{\Lambda}-u\epsilon\right)  \right)
}_{\substack{=\operatorname{id}_{\Lambda}\otimes
\id_{\Lambda}-\operatorname{id}_{\Lambda}\otimes\left(
u\epsilon\right)  \\\text{(since tensoring of } \kk \text{-linear maps is
} \kk \text{-bilinear)}}}\left(  \Delta x\right)  -\left(  1\otimes
x-\epsilon\left(  x\right)  \cdot1\otimes1\right) \\
&  =\left(  \underbrace{\operatorname{id}_{\Lambda}\otimes
\operatorname{id}_{\Lambda}}_{=\operatorname{id}_{\Lambda
\otimes\Lambda}}-\operatorname{id}_{\Lambda}\otimes\left(
u\epsilon\right)  \right)  \left(  \Delta x\right)  -\left(  1\otimes
x-\epsilon\left(  x\right)  \cdot1\otimes1\right) \\
&  =\underbrace{\left(  \operatorname{id}_{\Lambda\otimes\Lambda
}-\operatorname{id}_{\Lambda}\otimes\left(  u\epsilon\right)
\right)  \left(  \Delta x\right)  }_{=\operatorname{id}_{\Lambda
\otimes\Lambda}\left(  \Delta x\right)  -\left(  
\id_{\Lambda}\otimes\left(  u\epsilon\right)  \right)  \left(  \Delta
x\right)  }-\left(  1\otimes x-\epsilon\left(  x\right)  \cdot1\otimes1\right)
\\
&  =\left(  \underbrace{\operatorname{id}_{\Lambda\otimes\Lambda
}\left(  \Delta x\right)  }_{=\Delta x}-\underbrace{\left(  
\id_{\Lambda}\otimes\left(  u\epsilon\right)  \right)  \left(  \Delta
x\right)  }_{=x\otimes1}\right)  -\left(  1\otimes x-\epsilon\left(  x\right)
\cdot1\otimes1\right) \\
&  =\left(  \Delta x-x\otimes1\right)  -\left(  1\otimes x-\epsilon\left(
x\right)  \cdot1\otimes1\right)  =\Delta x-1\otimes x-x\otimes1+\epsilon
\left(  x\right)  \cdot1\otimes1.
\end{align*}
Subtracting $\epsilon\left(  x\right)  \cdot1\otimes1$ from both sides of this
equality, we obtain%
\[
\left(  \overline{\id}\otimes\overline{\operatorname*{id}%
}\right)  \left(  \Delta x\right)  -\epsilon\left(  x\right)  \cdot
1\otimes1=\Delta x-1\otimes x-x\otimes1.
\]
This proves (\ref{sol.hookschur.d.Delta}).}

(d) Recall that $\Lambda$ is a $\kk$-bialgebra. Hence, $\Delta
:\Lambda\rightarrow\Lambda\otimes\Lambda$ is a $\kk$-algebra
homomorphism (by the axioms of a $\kk$-bialgebra).

Every $n\in \NN$ satisfies%
\begin{equation}
\Delta e_{n}=\sum_{i=0}^{n}e_{i}\otimes e_{n-i}. \label{sol.hookschur.d.Delen}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.Delen}):} Let $n\in
 \NN$.
\par
Let $i\in \NN$ be such that $i\leq n$. Then, $n-i\in \NN$. Hence,
there exists exactly one $j\in \NN$ satisfying $i+j=n$, namely $j=n-i$.
Thus, the sum $\sum\limits_{\substack{j\in \NN ;\\i+j=n}}e_{i}\otimes e_{j}$ has
exactly one addend, namely the one for $j=n-i$. As a consequence,
$\sum\limits_{\substack{j\in \NN ;\\i+j=n}}e_{i}\otimes e_{j}=e_{i}\otimes
e_{n-i}$.
\par
Now, let us forget that we fixed $i$. We thus have shown that%
\begin{equation}
\sum\limits_{\substack{j\in \NN ;\\i+j=n}}e_{i}\otimes e_{j}=e_{i}\otimes
e_{n-i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in \NN \text{ satisfying
}i\leq n. \label{sol.hookschur.d.Delen.pf.1}%
\end{equation}
Now, Proposition \ref{symm-comultiplication-formulas-prop}(ii) yields%
\begin{align*}
\Delta e_{n}  &  =\underbrace{\sum\limits_{\substack{\left(  i,j\right)
\in \NN ^{2};\\i+j=n}}}_{=\sum_{i\in \NN }\sum\limits_{\substack{j\in
 \NN ;\\i+j=n}}}e_{i}\otimes e_{j}=\sum_{i\in \NN }\sum\limits
_{\substack{j\in \NN ;\\i+j=n}}e_{i}\otimes e_{j}
  =\underbrace{\sum\limits_{\substack{i\in \NN ;\\i\leq n}}}_{=\sum_{i=0}^{n}%
}\underbrace{\sum\limits_{\substack{j\in \NN ;\\i+j=n}}e_{i}\otimes e_{j}%
}_{\substack{=e_{i}\otimes e_{n-i}\\\text{(by
(\ref{sol.hookschur.d.Delen.pf.1}))}}}+\sum\limits_{\substack{i\in \NN %
;\\i>n}}\underbrace{\sum\limits_{\substack{j\in \NN ;\\i+j=n}}e_{i}\otimes
e_{j}}_{\substack{=\left(  \text{empty sum}\right)  \\\text{(because there
exists no }j\in \NN \\ \text{ satisfying }i+j=n\\\text{(since }i>n\text{))}%
}}\\
&  =\sum_{i=0}^{n}e_{i}\otimes e_{n-i}+\sum\limits_{\substack{i\in \NN %
;\\i>n}}\underbrace{\left(  \text{empty sum}\right)  }_{=0}=\sum_{i=0}%
^{n}e_{i}\otimes e_{n-i}+\underbrace{\sum\limits_{\substack{i\in \NN ;\\i>n}%
}0}_{=0}=\sum_{i=0}^{n}e_{i}\otimes e_{n-i}.
\end{align*}
This proves (\ref{sol.hookschur.d.Delen}).} Thus, every $n\in \NN$
satisfies%
\begin{equation}
\Delta e_{n}=\sum_{i\in \NN }e_{i}\otimes e_{n-i}.
\label{sol.hookschur.d.Delen.inf}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.Delen.inf}):} Let
$n\in \NN$. Then,%
\begin{align*}
\sum_{i\in \NN }e_{i}\otimes e_{n-i}  &  =\underbrace{\sum\limits
_{\substack{i\in \NN ;\\i\leq n}}}_{=\sum_{i=0}^{n}}e_{i}\otimes
e_{n-i}+\sum\limits_{\substack{i\in \NN ;\\i>n}}e_{i}\otimes\underbrace{e_{n-i}%
}_{\substack{=0\\\text{(since }n-i<0 \\ \text{ (since }i>n\text{))}}}=\sum
_{i=0}^{n}e_{i}\otimes e_{n-i}+\underbrace{\sum\limits_{\substack{i\in \NN %
;\\i>n}}e_{i}\otimes0}_{=0}
  =\sum_{i=0}^{n}e_{i}\otimes e_{n-i} \\
&= \Delta e_{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.hookschur.d.Delen})}\right)  .
\end{align*}
This proves (\ref{sol.hookschur.d.Delen.inf}).}

Also, every $n\in \NN$ satisfies%
\begin{equation}
\Delta h_{n}=\sum_{i=0}^{n}h_{i}\otimes h_{n-i} \label{sol.hookschur.d.Delhn}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.Delhn}):} The proof of
(\ref{sol.hookschur.d.Delhn}) can be obtained from the proof of
(\ref{sol.hookschur.d.Delen}) by replacing every $e_{k}$ (for every
$k\in \NN$) by $h_{k}$, and replacing the reference to Proposition
\ref{symm-comultiplication-formulas-prop}(ii) by a reference to Proposition
\ref{symm-comultiplication-formulas-prop}(iii).} and%
\begin{equation}
\Delta h_{n}=\sum_{i\in \NN }h_{i}\otimes h_{n-i}.
\label{sol.hookschur.d.Delhn.inf}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.Delhn.inf}):} The proof of
(\ref{sol.hookschur.d.Delhn.inf}) can be obtained from the proof of
(\ref{sol.hookschur.d.Delen.inf}) by replacing every $e_{k}$ (for every
$k\in \NN$) by $h_{k}$, and replacing the reference to Proposition
\ref{symm-comultiplication-formulas-prop}(ii) by a reference to Proposition
\ref{symm-comultiplication-formulas-prop}(iii).}

Let $a\in \NN$ and $b\in \NN$. From (\ref{sol.hookschur.sab-pos}),
we obtain $\mathbf{s}\left(  a,b\right)  =s_{\left(  a+1,1^{b}\right)  }$, so
that%
\begin{align*}
s_{\left(  a+1,1^{b}\right)  }  & =\mathbf{s}\left(  a,b\right)  =\sum
_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.hookschur.general-bc})}\right)  \\
& =\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a+k+1}e_{b-k}%
\end{align*}
(here, we have renamed the summation index $i$ as $k$). Applying the map
$\Delta$ to both sides of this equality, we obtain%
\begin{align}
\Delta s_{\left(  a+1,1^{b}\right)  } &  =\Delta\left(  \sum_{k=0}^{b}\left(
-1\right)  ^{k}h_{a+k+1}e_{b-k}\right)  \nonumber \\
&=\sum_{k=0}^{b}\left(  -1\right)
^{k}\underbrace{\Delta\left(  h_{a+k+1}\right)  }_{\substack{=\sum
_{i\in \NN }h_{i}\otimes h_{a+k+1-i}\\\text{(by
(\ref{sol.hookschur.d.Delhn.inf}), applied to }n=a+k+1\text{)}}}\cdot
\underbrace{\Delta\left(  e_{b-k}\right)  }_{\substack{=\sum_{i\in \NN %
}e_{i}\otimes e_{b-k-i}\\\text{(by (\ref{sol.hookschur.d.Delen.inf}), applied
to }n=b-k\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a } \kk %
\text{-algebra homomorphism}\right)  \nonumber\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}\left(  \sum_{i\in \NN }%
h_{i}\otimes h_{a+k+1-i}\right)  \left(  \sum_{i\in \NN }e_{i}\otimes
e_{b-k-i}\right)  \nonumber\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}\left(  \sum_{i\in \NN }%
h_{i}\otimes h_{a+k+1-i}\right)  \left(  \sum_{j\in \NN }e_{j}\otimes
e_{b-k-j}\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}i\text{ as }j\text{ in the third sum}\right)  \nonumber\\
&  =\sum_{k=0}^{b}\left(  -1\right)  ^{k}\sum_{i\in \NN }\sum
_{j\in \NN }\underbrace{\left(  h_{i}\otimes h_{a+k+1-i}\right)  \left(
e_{j}\otimes e_{b-k-j}\right)  }_{=h_{i}e_{j}\otimes h_{a+k+1-i}e_{b-k-j}%
}\nonumber \\
&=\sum_{k=0}^{b}\left(  -1\right)  ^{k}\sum_{i\in \NN }\sum
_{j\in \NN }h_{i}e_{j}\otimes h_{a+k+1-i}e_{b-k-j}\nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\underbrace{\sum_{k=0}^{b}%
h_{i}e_{j}\otimes\left(  -1\right)  ^{k}h_{a+k+1-i}e_{b-k-j}}_{=h_{i}%
e_{j}\otimes\left(  \sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a+k+1-i}%
e_{b-k-j}\right)  }\nonumber \\
&=\sum_{i\in \NN }\sum_{j\in \NN }h_{i}%
e_{j}\otimes\left(  \sum_{k=0}^{b}\left(  -1\right)  ^{k}%
\underbrace{h_{a+k+1-i}}_{=h_{a-i+k+1}}\underbrace{e_{b-k-j}}_{=e_{b-j-k}%
}\right)  \nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }h_{i}e_{j}\otimes\left(
\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a-i+k+1}e_{b-j-k}\right)
.\label{sol.hookschur.d.3}%
\end{align}
But every $i\in \NN$ and $j\in \NN$ satisfy
\begin{equation}
\sum_{k=0}^{b}\left(  -1\right)  ^{k}h_{a-i+k+1}e_{b-j-k}=\mathbf{s}\left(
a-i,b-j\right)  \label{sol.hookschur.d.4}%
\end{equation}
(by (\ref{sol.hookschur.general-bb'c}), applied to $a-i$, $b-j$ and $b$
instead of $a$, $b$ and $c$), since $b\geq b-j$ for every $j\in \NN$.
Hence, (\ref{sol.hookschur.d.3}) becomes%
\begin{align}
\Delta s_{\left(  a+1,1^{b}\right)  } &  =\sum_{i\in \NN }\sum
_{j\in \NN }\underbrace{h_{i}e_{j}}_{\substack{=e_{j}h_{i}\\=\mathbf{s}%
\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \\\text{(by
(\ref{sol.hookschur.general-a}), applied to }j\text{ and }i\\\text{instead of
}n\text{ and }m\text{)}}}\otimes\underbrace{\left(  \sum_{k=0}^{b}\left(
-1\right)  ^{k}h_{a-i+k+1}e_{b-j-k}\right)  }_{\substack{=\mathbf{s}\left(
a-i,b-j\right)  \\\text{(by (\ref{sol.hookschur.d.4}))}}}\nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\left(  \mathbf{s}\left(
i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)  \otimes\mathbf{s}%
\left(  a-i,b-j\right)  .\label{sol.hookschur.d.6}%
\end{align}


We can readily see that%
\begin{equation}
\Delta s_{\left(  a+1,1^{b}\right)  }-1\otimes s_{\left(  a+1,1^{b}\right)
}-s_{\left(  a+1,1^{b}\right)  }\otimes1=\left(  \overline{\operatorname*{id}%
}\otimes\overline{\id}\right)  \left(  \Delta s_{\left(
a+1,1^{b}\right)  }\right)  . \label{sol.hookschur.d.Delta.applied}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.Delta.applied}):} The
partition $\left(  a+1,1^{b}\right)  $ is nonempty (since its first entry is
$\underbrace{a}_{\geq0}+1\geq0+1>0$). In other words, $\left(  a+1,1^{b}%
\right)  \neq\varnothing$.
\par
We know that $\epsilon\left(  s_{\lambda}\right)  =\delta_{\lambda
,\varnothing}$ for every partition $\lambda$. Applied to $\lambda=\left(
a+1,1^{b}\right)  $, this yields $\epsilon\left(  s_{\left(  a+1,1^{b}\right)
}\right)  =\delta_{\left(  a+1,1^{b}\right)  ,\varnothing}=0$ (since $\left(
a+1,1^{b}\right)  \neq\varnothing$). Now, (\ref{sol.hookschur.d.Delta})
(applied to $x=s_{\left(  a+1,1^{b}\right)  }$) yields%
\begin{align*}
\Delta s_{\left(  a+1,1^{b}\right)  }-1\otimes s_{\left(  a+1,1^{b}\right)
}-s_{\left(  a+1,1^{b}\right)  }\otimes1  &  =\left(  \overline
{\id}\otimes\overline{\id}\right)  \left(
\Delta s_{\left(  a+1,1^{b}\right)  }\right)  -\underbrace{\epsilon\left(
s_{\left(  a+1,1^{b}\right)  }\right)  }_{=0}\cdot1\otimes1\\
&  =\left(  \overline{\id}\otimes\overline{\operatorname*{id}%
}\right)  \left(  \Delta s_{\left(  a+1,1^{b}\right)  }\right)  .
\end{align*}
This proves (\ref{sol.hookschur.d.Delta.applied}).} This becomes%
\begin{align}
&  \Delta s_{\left(  a+1,1^{b}\right)  }-1\otimes s_{\left(  a+1,1^{b}\right)
}-s_{\left(  a+1,1^{b}\right)  }\otimes1\nonumber\\
&  =\left(  \overline{\id}\otimes\overline{\operatorname*{id}%
}\right)  \underbrace{\left(  \Delta s_{\left(  a+1,1^{b}\right)  }\right)
}_{\substack{=\sum_{i\in \NN }\sum_{j\in \NN }\left(  \mathbf{s}%
\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)  \otimes
\mathbf{s}\left(  a-i,b-j\right)  \\\text{(by (\ref{sol.hookschur.d.6}))}%
}}\nonumber\\
&  =\left(  \overline{\id}\otimes\overline{\operatorname*{id}%
}\right)  \left(  \sum_{i\in \NN }\sum_{j\in \NN }\left(
\mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(  i-1,j\right)  \right)
\otimes\mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\underbrace{\left(
\overline{\id}\otimes\overline{\id}\right)
\left(  \left(  \mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(
i-1,j\right)  \right)  \otimes\mathbf{s}\left(  a-i,b-j\right)  \right)
}_{=\overline{\id}\left(  \mathbf{s}\left(  i,j-1\right)
+\mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\overline
{\id}\otimes\overline{\id}\text{ is }%
 \kk \text{-linear}\right) \nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\underbrace{\overline
{\id}\left(  \mathbf{s}\left(  i,j-1\right)  +\mathbf{s}\left(
i-1,j\right)  \right)  }_{\substack{=\overline{\id}\left(
\mathbf{s}\left(  i,j-1\right)  \right)  +\overline{\id}\left(
\mathbf{s}\left(  i-1,j\right)  \right)  \\\text{(since the map }%
\overline{\id}\text{ is } \kk \text{-linear)}}%
}\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right) \nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\underbrace{\left(
\overline{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
+\overline{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
a-i,b-j\right)  \right)  }_{=\overline{\id}\left(
\mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  +\overline
{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }\nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\left(  \overline
{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  +\overline{\id}\left(  \mathbf{s}\left(  i-1,j\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
a-i,b-j\right)  \right)  \right) \nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
+\sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
\nonumber\\
&  =\sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
+\sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  .
\label{sol.hookschur.d.8}%
\end{align}


We can now observe that
\begin{equation}
\sum_{i\in \NN }\sum_{j\in \NN }\overline{\id}\left(
\mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  =\sum\limits_{\substack{\left(
c,d,e,f\right)  \in \NN ^{4};\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}%
\right)  }\otimes s_{\left(  e+1,1^{f}\right)  }.
\label{sol.hookschur.d.reindex.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.reindex.1}):} First, we have%
\begin{align}
&  \sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
\nonumber\\
&  =\underbrace{\sum\limits_{\substack{i\in \NN ;\\i=0}}\sum_{j\in \NN %
}\overline{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{\substack{=\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  0-1,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-0,b-j\right)  \right)
\\\text{(since }0\in \NN \text{)}}}+\underbrace{\sum\limits_{\substack{i\in
 \NN ;\\i\neq0}}}_{=\sum_{i\in\left\{  1,2,3,...\right\}  }}\sum
_{j\in \NN }\overline{\id}\left(  \mathbf{s}\left(
i-1,j\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }i\in \NN \text{
satisfies exactly one of the two assertions }i=0\text{ and }i\neq0\right)
\nonumber\\
&  =\sum_{j\in \NN }\underbrace{\overline{\id}\left(
\mathbf{s}\left(  0-1,j\right)  \right)  }_{\substack{=0\\\text{(by
(\ref{sol.hookschur.d.sbar.NZ}), applied to}\\0-1\text{ and }j\text{ instead
of }a\text{ and }b\text{)}}}\otimes\overline{\id}\left(
\mathbf{s}\left(  a-0,b-j\right)  \right)  +\sum_{i\in\left\{
1,2,3,...\right\}  }\sum_{j\in \NN }\overline{\id}\left(
\mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  =\underbrace{\sum_{j\in \NN }0\otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-0,b-j\right)  \right)  }_{=0}+\sum_{i\in\left\{
1,2,3,...\right\}  }\sum_{j\in \NN }\overline{\id}\left(
\mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  =\sum_{i\in\left\{  1,2,3,...\right\}  }\sum_{j\in \NN }%
\overline{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right) \nonumber\\
&  =\sum_{c\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  \underbrace{\left(  c+1\right)  -1}_{=c},j\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
\underbrace{a-\left(  c+1\right)  }_{=\left(  a-1\right)  -c},b-j\right)
\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }c+1\text{ for }i\text{ in the first sum, since the
map}\\
 \NN \rightarrow\left\{  1,2,3,...\right\}  ,\ c\mapsto c+1\text{ is a
bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{c\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)
-c,b-j\right)  \right) \nonumber\\
&  =\sum_{c\in \NN }\sum_{d\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)
-c,b-d\right)  \right) \label{sol.hookschur.d.reindex.1.pf.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}j\text{ as }d\text{ in the second sum}\right)  .\nonumber
\end{align}
\par
Now, fix $c\in \NN$ and $d\in \NN$. There exists exactly one
$e\in \ZZ$ satisfying $c+e=a-1$, namely $e=\left(  a-1\right)  -c$.
Hence, the sum $\sum\limits_{\substack{e\in \ZZ ;\\c+e=a-1}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  $ has exactly one addend, namely the
addend for $e=\left(  a-1\right)  -c$. Thus, this sum simplifies as follows:%
\[
\sum\limits_{\substack{e\in \ZZ ;\\c+e=a-1}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  =\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,f\right)  \right)  .
\]
Compared with%
\begin{align*}
&  \sum\limits_{\substack{e\in \ZZ ;\\c+e=a-1}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right) \\
&  =\underbrace{\sum\limits_{\substack{e\in \ZZ ;\\c+e=a-1;\\e\in \NN }%
}}_{\substack{=\sum\limits_{\substack{e\in \NN ;\\c+e=a-1}}\\\text{(since
} \NN \subset \ZZ \text{)}}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  +\underbrace{\sum\limits_{\substack{e\in \ZZ %
;\\c+e=a-1;\\e\notin \NN }}}_{\substack{=\sum\limits_{\substack{e\in
 \ZZ ;\\c+e=a-1;\\e<0}}\\\text{(because for an }e\in \ZZ \text{, the
condition}\\e\notin \NN \text{ is equivalent to the condition
}e<0\text{)}}}\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  e,f\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }e\in \ZZ \text{
satisfies exactly one of the conditions }e\in \NN \text{ and }%
e\notin \NN \right) \\
&  =\sum\limits_{\substack{e\in \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  +\sum\limits_{\substack{e\in \ZZ %
;\\c+e=a-1;\\e<0}}\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\underbrace{\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  }_{\substack{=0\\\text{(by (\ref{sol.hookschur.d.sbar.NZ}%
), applied}\\\text{to }e\text{ and }f\text{ instead of }a\text{ and }%
b\text{)}}}\\
&  =\sum\limits_{\substack{e\in \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  +\underbrace{\sum\limits_{\substack{e\in
 \ZZ ;\\c+e=a-1;\\e<0}}\sum\limits_{\substack{f\in \NN ;\\d+f=b}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes0}_{=0}\\
&  =\sum\limits_{\substack{e\in \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  ,
\end{align*}
this yields%
\begin{equation}
\sum\limits_{\substack{e\in \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  =\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,f\right)  \right)  . \label{sol.hookschur.d.reindex.1.pf.3}%
\end{equation}
\par
But there exists exactly one $f\in \ZZ$ satisfying $d+f=b$, namely
$f=b-d$. Hence, the sum $\sum\limits_{\substack{f\in \ZZ ;\\d+f=b}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,f\right)  \right)  $ has exactly one addend, namely the addend
for $f=b-d$. Therefore, this sum simplifies as follows:%
\[
\sum\limits_{\substack{f\in \ZZ ;\\d+f=b}}\overline{\id}\left(
\mathbf{s}\left(  c,d\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)  \right)
=\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,b-d\right)  \right)  .
\]
Compared with%
\begin{align*}
&  \sum\limits_{\substack{f\in \ZZ ;\\d+f=b}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)
\right) \\
&  =\underbrace{\sum\limits_{\substack{f\in \ZZ ;\\d+f=b;\\f\in \NN }%
}}_{=\sum\limits_{\substack{f\in \NN ;\\d+f=b}}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)
\right)  +\underbrace{\sum\limits_{\substack{f\in \ZZ ;\\d+f=b;\\f\notin%
 \NN }}}_{\substack{=\sum\limits_{\substack{f\in \ZZ ;\\d+f=b;\\f<0}%
}\\\text{(since for an }f\in \ZZ \text{, the condition}\\f\notin%
 \NN \text{ is equivalent to }f<0\text{)}}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }f\in \ZZ \text{
satisfies exactly one of the conditions }f\in \NN \text{ and }%
f\notin \NN \right) \\
&  =\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)
\right)  +\sum\limits_{\substack{f\in \ZZ ;\\d+f=b;\\f<0}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\underbrace{\overline{\id}\left(  \mathbf{s}\left(
\left(  a-1\right)  -c,f\right)  \right)  }_{\substack{=0\\\text{(by
(\ref{sol.hookschur.d.sbar.ZN}), applied to}\\\left(  a-1\right)  -c\text{ and
}f\text{ instead of }a\text{ and }b\text{)}}}\\
&  =\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)
\right)  +\underbrace{\sum\limits_{\substack{f\in \ZZ ;\\d+f=b;\\f<0}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes0}_{=0}\\
&  =\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)
\right)  ,
\end{align*}
this yields%
\[
\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline{\id}\left(
\mathbf{s}\left(  c,d\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  \left(  a-1\right)  -c,f\right)  \right)
=\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,b-d\right)  \right)  .
\]
Hence, (\ref{sol.hookschur.d.reindex.1.pf.3}) becomes%
\begin{align}
\sum\limits_{\substack{e\in \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)   &  =\sum\limits_{\substack{f\in \NN ;\\d+f=b}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,f\right)  \right) \nonumber\\
&  =\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,b-d\right)  \right)  . \label{sol.hookschur.d.reindex.1.pf.6}%
\end{align}
\par
Now, let us forget that we fixed $c$ and $d$. We thus have proven that every
$c\in \NN$ and $d\in \NN$ satisfy
(\ref{sol.hookschur.d.reindex.1.pf.6}). Now,
(\ref{sol.hookschur.d.reindex.1.pf.1}) becomes%
\begin{align*}
&  \sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right) \\
&  =\sum_{c\in \NN }\sum_{d\in \NN }\underbrace{\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  \left(
a-1\right)  -c,b-d\right)  \right)  }_{\substack{=\sum\limits_{\substack{e\in
 \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in \NN ;\\d+f=b}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  e,f\right)
\right)  \\\text{(by (\ref{sol.hookschur.d.reindex.1.pf.6}))}}}\\
&  =\underbrace{\sum_{c\in \NN }\sum_{d\in \NN }\sum\limits_{\substack{e\in
 \NN ;\\c+e=a-1}}\sum\limits_{\substack{f\in \NN ;\\d+f=b}}}_{=\sum\limits
_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4};\\c+e=a-1;\\d+f=b}%
}}\underbrace{\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  }_{\substack{=s_{\left(  c+1,1^{d}\right)  }\\\text{(by
(\ref{sol.hookschur.d.sbar}), applied to}\\c\text{ and }d\text{ instead of
}a\text{ and }b\text{)}}}\otimes\underbrace{\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  e,f\right)  \right)  }_{\substack{=s_{\left(
e+1,1^{f}\right)  }\\\text{(by (\ref{sol.hookschur.d.sbar}), applied
to}\\e\text{ and }f\text{ instead of }a\text{ and }b\text{)}}}\\
&  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }.
\end{align*}
This proves (\ref{sol.hookschur.d.reindex.1}).} Also,%
\begin{equation}
\sum_{i\in \NN }\sum_{j\in \NN }\overline{\id}\left(
\mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)  =\sum\limits_{\substack{\left(
c,d,e,f\right)  \in \NN ^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}%
\right)  }\otimes s_{\left(  e+1,1^{f}\right)  }.
\label{sol.hookschur.d.reindex.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.d.reindex.2}):} For every
$i\in \NN$, we have%
\begin{align}
&  \sum_{j\in \NN }\overline{\id}\left(  \mathbf{s}\left(
i,j-1\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{j\in \NN ;\\j=0}}\overline
{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{\substack{=\overline{\id}\left(  \mathbf{s}\left(
i,0-1\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  a-i,b-0\right)  \right)  \\\text{(since }0\in \NN %
\text{)}}}+\underbrace{\sum\limits_{\substack{j\in \NN ;\\j\neq0}}}_{=\sum
_{j\in\left\{  1,2,3,...\right\}  }}\overline{\id}\left(
\mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }j\in \NN \text{
satisfies exactly one of the two assertions }j=0\text{ and }j\neq0\right)
\nonumber\\
&  =\underbrace{\overline{\id}\left(  \mathbf{s}\left(
i,0-1\right)  \right)  }_{\substack{=0\\\text{(by
(\ref{sol.hookschur.d.sbar.ZN}), applied to}\\i\text{ and }0-1\text{ instead
of }a\text{ and }b\text{)}}}\otimes\overline{\id}\left(
\mathbf{s}\left(  a-i,b-0\right)  \right)  +\sum_{j\in\left\{
1,2,3,...\right\}  }\overline{\id}\left(  \mathbf{s}\left(
i,j-1\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  a-i,b-j\right)  \right) \nonumber\\
&  =\underbrace{0\otimes\overline{\id}\left(  \mathbf{s}\left(
a-i,b-0\right)  \right)  }_{=0}+\sum_{j\in\left\{  1,2,3,...\right\}
}\overline{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right) \nonumber\\
&  =\sum_{j\in\left\{  1,2,3,...\right\}  }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i,j-1\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right)
\nonumber\\
&  =\sum_{d\in \NN }\overline{\id}\left(  \mathbf{s}%
\left(  i,\underbrace{\left(  d+1\right)  -1}_{=d}\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(
a-i,\underbrace{b-\left(  d+1\right)  }_{=\left(  b-1\right)  -d}\right)
\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }d+1\text{ for }j\text{ in the sum, since the
map}\\
 \NN \rightarrow\left\{  1,2,3,...\right\}  ,\ d\mapsto d+1\text{ is a
bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{d\in \NN }\overline{\id}\left(  \mathbf{s}%
\left(  i,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  a-i,\left(  b-1\right)  -d\right)  \right)  .\nonumber
\end{align}
Hence,%
\begin{align}
&  \sum_{i\in \NN }\underbrace{\sum_{j\in \NN }\overline
{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{=\sum_{d\in \NN }\overline{\id}\left(
\mathbf{s}\left(  i,d\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-i,\left(  b-1\right)  -d\right)  \right)
}\nonumber\\
&  =\sum_{i\in \NN }\sum_{d\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,\left(  b-1\right)
-d\right)  \right) \nonumber\\
&  =\sum_{c\in \NN }\sum_{d\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-c,\left(  b-1\right)
-d\right)  \right) \label{sol.hookschur.d.reindex.2.pf.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}i\text{ as }d\text{ in the first sum}\right)  .\nonumber
\end{align}
\par
Now, fix $c\in \NN$ and $d\in \NN$. There exists exactly one
$e\in \ZZ$ satisfying $c+e=a$, namely $e=a-c$. Hence, the sum
$\sum\limits_{\substack{e\in \ZZ ;\\c+e=a}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  $ has exactly one addend, namely the addend for $e=a-c$.
Thus, this sum simplifies as follows:%
\[
\sum\limits_{\substack{e\in \ZZ ;\\c+e=a}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  =\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-c,f\right)
\right)  .
\]
Compared with%
\begin{align*}
&  \sum\limits_{\substack{e\in \ZZ ;\\c+e=a}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right) \\
&  =\underbrace{\sum\limits_{\substack{e\in \ZZ ;\\c+e=a;\\e\in \NN }%
}}_{\substack{=\sum\limits_{\substack{e\in \NN ;\\c+e=a}}\\\text{(since
} \NN \subset \ZZ \text{)}}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  +\underbrace{\sum\limits_{\substack{e\in \ZZ %
;\\c+e=a;\\e\notin \NN }}}_{\substack{=\sum\limits_{\substack{e\in
 \ZZ ;\\c+e=a;\\e<0}}\\\text{(because for an }e\in \ZZ \text{, the
condition}\\e\notin \NN \text{ is equivalent to the condition
}e<0\text{)}}}\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  e,f\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }e\in \ZZ \text{
satisfies exactly one of the conditions }e\in \NN \text{ and }%
e\notin \NN \right) \\
&  =\sum\limits_{\substack{e\in \NN ;\\c+e=a}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  +\sum\limits_{\substack{e\in \ZZ %
;\\c+e=a;\\e<0}}\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\underbrace{\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  }_{\substack{=0\\\text{(by (\ref{sol.hookschur.d.sbar.NZ}%
), applied}\\\text{to }e\text{ and }f\text{ instead of }a\text{ and }%
b\text{)}}}\\
&  =\sum\limits_{\substack{e\in \NN ;\\c+e=a}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  +\underbrace{\sum\limits_{\substack{e\in
 \ZZ ;\\c+e=a;\\e<0}}\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes0}_{=0}\\
&  =\sum\limits_{\substack{e\in \NN ;\\c+e=a}}\sum\limits_{\substack{f\in
 \NN ;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  e,f\right)  \right)  ,
\end{align*}
this yields%
\begin{equation}
\sum\limits_{\substack{e\in \NN ;\\c+e=a}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)  =\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-c,f\right)
\right)  . \label{sol.hookschur.d.reindex.2.pf.3}%
\end{equation}
\par
But there exists exactly one $f\in \ZZ$ satisfying $d+f=b-1$, namely
$f=\left(  b-1\right)  -d$. Hence, the sum $\sum\limits_{\substack{f\in
 \ZZ ;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  \otimes\overline{\id}\left(
\mathbf{s}\left(  a-c,f\right)  \right)  $ has exactly one addend, namely the
addend for $f=\left(  b-1\right)  -d$. Therefore, this sum simplifies as
follows:%
\[
\sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1}}\overline{\id}\left(
\mathbf{s}\left(  c,d\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-c,f\right)  \right)  =\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-c,\left(
b-1\right)  -d\right)  \right)  .
\]
Compared with%
\begin{align*}
&  \sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-c,f\right)  \right) \\
&  =\underbrace{\sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1;\\f\in \NN }%
}}_{=\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-c,f\right)  \right)
+\underbrace{\sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1;\\f\notin \NN }%
}}_{\substack{=\sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1;\\f<0}}\\\text{(since
for an }f\in \ZZ \text{, the condition}\\f\notin \NN \text{ is
equivalent to }f<0\text{)}}}\overline{\id}\left(
\mathbf{s}\left(  c,d\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-c,f\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }f\in \ZZ \text{
satisfies exactly one of the conditions }f\in \NN \text{ and }%
f\notin \NN \right) \\
&  =\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-c,f\right)  \right)
+\sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1;\\f<0}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\underbrace{\overline
{\id}\left(  \mathbf{s}\left(  a-c,f\right)  \right)
}_{\substack{=0\\\text{(by (\ref{sol.hookschur.d.sbar.ZN}), applied
to}\\a-c\text{ and }f\text{ instead of }a\text{ and }b\text{)}}}\\
&  =\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-c,f\right)  \right)
+\underbrace{\sum\limits_{\substack{f\in \ZZ ;\\d+f=b-1;\\f<0}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes
0}_{=0}\\
&  =\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  c,d\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-c,f\right)  \right)  ,
\end{align*}
this yields%
\[
\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline{\id}\left(
\mathbf{s}\left(  c,d\right)  \right)  \otimes\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  a-c,f\right)  \right)  =\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-c,\left(
b-1\right)  -d\right)  \right)  .
\]
Hence, (\ref{sol.hookschur.d.reindex.2.pf.3}) becomes%
\begin{align}
\sum\limits_{\substack{e\in \NN ;\\c+e=a}}\sum\limits_{\substack{f\in \NN %
;\\d+f=b-1}}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
e,f\right)  \right)   &  =\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}%
}\overline{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-c,f\right)
\right) \nonumber\\
&  =\overline{\id}\left(  \mathbf{s}\left(  c,d\right)
\right)  \otimes\overline{\id}\left(  \mathbf{s}\left(
a-c,\left(  b-1\right)  -d\right)  \right)  .
\label{sol.hookschur.d.reindex.2.pf.6}%
\end{align}
\par
Now, let us forget that we fixed $c$ and $d$. We thus have proven that every
$c\in \NN$ and $d\in \NN$ satisfy
(\ref{sol.hookschur.d.reindex.2.pf.6}). Now,
(\ref{sol.hookschur.d.reindex.2.pf.1}) becomes%
\begin{align*}
&  \sum_{i\in \NN }\sum_{j\in \NN }\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  i-1,j\right)  \right)  \otimes\overline
{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)  \right) \\
&  =\sum_{c\in \NN }\sum_{d\in \NN }\underbrace{\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-c,\left(
b-1\right)  -d\right)  \right)  }_{\substack{=\sum\limits_{\substack{e\in
 \NN ;\\c+e=a}}\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}\overline
{\id}\left(  \mathbf{s}\left(  c,d\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  e,f\right)
\right)  \\\text{(by (\ref{sol.hookschur.d.reindex.2.pf.6}))}}}\\
&  =\underbrace{\sum_{c\in \NN }\sum_{d\in \NN }\sum\limits_{\substack{e\in
 \NN ;\\c+e=a}}\sum\limits_{\substack{f\in \NN ;\\d+f=b-1}}}_{=\sum\limits
_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4};\\c+e=a;\\d+f=b-1}%
}}\underbrace{\overline{\id}\left(  \mathbf{s}\left(
c,d\right)  \right)  }_{\substack{=s_{\left(  c+1,1^{d}\right)  }\\\text{(by
(\ref{sol.hookschur.d.sbar}), applied to}\\c\text{ and }d\text{ instead of
}a\text{ and }b\text{)}}}\otimes\underbrace{\overline{\operatorname*{id}%
}\left(  \mathbf{s}\left(  e,f\right)  \right)  }_{\substack{=s_{\left(
e+1,1^{f}\right)  }\\\text{(by (\ref{sol.hookschur.d.sbar}), applied
to}\\e\text{ and }f\text{ instead of }a\text{ and }b\text{)}}}\\
&  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }.
\end{align*}
This proves (\ref{sol.hookschur.d.reindex.2}).} Now, (\ref{sol.hookschur.d.8})
becomes%
\begin{align*}
&  \Delta s_{\left(  a+1,1^{b}\right)  }-1\otimes s_{\left(  a+1,1^{b}\right)
}-s_{\left(  a+1,1^{b}\right)  }\otimes1\\
&  =\underbrace{\sum_{i\in \NN }\sum_{j\in \NN }\overline
{\id}\left(  \mathbf{s}\left(  i-1,j\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{\substack{=\sum\limits_{\substack{\left(  c,d,e,f\right)  \in
 \NN ^{4};\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes
s_{\left(  e+1,1^{f}\right)  }\\\text{(by (\ref{sol.hookschur.d.reindex.1}))}%
}}+\underbrace{\sum_{i\in \NN }\sum_{j\in \NN }\overline
{\id}\left(  \mathbf{s}\left(  i,j-1\right)  \right)
\otimes\overline{\id}\left(  \mathbf{s}\left(  a-i,b-j\right)
\right)  }_{\substack{=\sum\limits_{\substack{\left(  c,d,e,f\right)  \in
 \NN ^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}\right)  }\otimes
s_{\left(  e+1,1^{f}\right)  }\\\text{(by (\ref{sol.hookschur.d.reindex.2}))}%
}}\\
&  =\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN ^{4}%
;\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }+\sum\limits_{\substack{\left(  c,d,e,f\right)  \in \NN %
^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}\right)  }\otimes s_{\left(
e+1,1^{f}\right)  }.
\end{align*}
Adding $1\otimes s_{\left(  a+1,1^{b}\right)  }+s_{\left(  a+1,1^{b}\right)
}\otimes1$ to both sides of this equality, we obtain%
\begin{align*}
\Delta s_{\left(  a+1,1^{b}\right)  }  &  =1\otimes s_{\left(  a+1,1^{b}%
\right)  }+s_{\left(  a+1,1^{b}\right)  }\otimes1\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  c,d,e,f\right)  \in
 \NN ^{4};\\c+e=a-1;\\d+f=b}}s_{\left(  c+1,1^{d}\right)  }\otimes
s_{\left(  e+1,1^{f}\right)  }+\sum\limits_{\substack{\left(  c,d,e,f\right)
\in \NN ^{4};\\c+e=a;\\d+f=b-1}}s_{\left(  c+1,1^{d}\right)  }\otimes
s_{\left(  e+1,1^{f}\right)  }.
\end{align*}
This solves Exercise \ref{exe.hookschur}(d).

\textit{Remark:} In our above solution, we have first proved
(\ref{sol.hookschur.general-a}) and then used it to derive
(\ref{sol.hookschur.general-bc}). Let me show an alternative argument, which
proceeds the other way round:

\textit{Alternative proof of (\ref{sol.hookschur.general-bc}):} Let
$a\in \ZZ$ and $b\in \ZZ$. We need to prove
(\ref{sol.hookschur.general-bc}).

If $b<0$, then (\ref{sol.hookschur.general-bc}) is
true\footnote{\textit{Proof.} Assume that $b<0$. Then, the definition of
$\mathbf{s}\left(  a,b\right)  $ yields $\mathbf{s}\left(  a,b\right)
=
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=0$ (since $b<0$), so that $0=\mathbf{s}\left(  a,b\right)  $. Now,%
\begin{align*}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}  &  =\left(  \text{empty
sum}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }b<0\right) \\
&  =0=\mathbf{s}\left(  a,b\right)  .
\end{align*}
Hence, (\ref{sol.hookschur.general-bc}) is proven (under the assumption that
$b<0$).}. Hence, for the rest of our proof of (\ref{sol.hookschur.general-bc}%
), we can WLOG assume that we don't have $b<0$. Assume this.

We have $b\geq0$ (since we don't have $b<0$).

We have%
\begin{align}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}  &  =\sum_{i=0}%
^{b}\left(  -1\right)  ^{b-i}\underbrace{h_{a+\left(  b-i\right)  +1}%
}_{=h_{\left(  a+b+1\right)  -i}}\underbrace{e_{b-\left(  b-i\right)  }%
}_{=e_{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted
}b-i\text{ for }i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i=0}^{b}\left(  -1\right)  ^{b-i}h_{\left(  a+b+1\right)  -i}e_{i}.
\label{sol.hookschur.general-bc.altpf.1}%
\end{align}


Now, if $a+b+1\leq0$, then%
\begin{equation}
h_{a+b+1}=\delta_{a+b,-1}. \label{sol.hookschur.general-bc.altpf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.general-bc.altpf.3}):} Assume
that $a+b+1\leq0$. We need to prove (\ref{sol.hookschur.general-bc.altpf.3}).
\par
Assume first that $a+b+1=0$. Then, $a+b+1=0$, so that $a+b=-1$ and thus
$\delta_{a+b,-1}=1$, so that $1=\delta_{a+b,-1}$. But from $a+b+1=0$, we also
obtain $h_{a+b+1}=h_{0}=1=\delta_{a+b,-1}$. Hence,
(\ref{sol.hookschur.general-bc.altpf.3}) is proven under the assumption that
$a+b+1=0$.
\par
Now, let us forget that we assumed that $a+b+1=0$. We thus have proven that
(\ref{sol.hookschur.general-bc.altpf.3}) holds if $a+b+1=0$. Hence, for the
rest of our proof of (\ref{sol.hookschur.general-bc.altpf.3}), we can WLOG
assume that we don't have $a+b+1=0$. Assume this.
\par
We have $a+b+1\neq0$ (since we don't have $a+b+1=0$). Combining this with
$a+b+1\leq0$, we obtain $a+b+1<0$, so that $a+b<-1$ and thus $a+b\neq-1$.
Hence, $\delta_{a+b,-1}=0$, and thus $0=\delta_{a+b,-1}$. But from $a+b+1<0$,
we also obtain $h_{a+b+1}=0=\delta_{a+b,-1}$. This proves
(\ref{sol.hookschur.general-bc.altpf.3}).} Furthermore, it is easy to see that
if $a+b+1\leq0$, then (\ref{sol.hookschur.general-bc}) is
true\footnote{\textit{Proof.} Assume that $a+b+1\leq0$. The equality
(\ref{sol.hookschur.general-bc.altpf.1}) becomes%
\begin{align}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}  &  =\sum_{i=0}%
^{b}\left(  -1\right)  ^{b-i}h_{\left(  a+b+1\right)  -i}e_{i}\nonumber\\
&  =\underbrace{\left(  -1\right)  ^{b-0}}_{=\left(  -1\right)  ^{b}%
}\underbrace{h_{\left(  a+b+1\right)  -0}}_{=h_{a+b+1}}\underbrace{e_{0}}%
_{=1}+\sum_{i=1}^{b}\left(  -1\right)  ^{b-i}\underbrace{h_{\left(
a+b+1\right)  -i}}_{\substack{=0\\\text{(since }\left(  a+b+1\right)
-i<0\\\text{(since }a+b+1\leq0\text{ and }i>0\text{))}}}e_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }b\geq0\right) \nonumber\\
&  =\left(  -1\right)  ^{b}h_{a+b+1}
+\underbrace{\sum_{i=1}^{b}\left(  -1\right) ^{b-i}0e_{i}}_{=0}
=\left(  -1\right)  ^{b}\underbrace{h_{a+b+1}}_{\substack{=\delta
_{a+b,-1}\\\text{(by (\ref{sol.hookschur.general-bc.altpf.3}))}}}=\left(
-1\right)  ^{b}\delta_{a+b,-1}. \label{sol.hookschur.general-bc.altpf.5}%
\end{align}
\par
On the other hand, $a+b+1\leq0$, so that $0\geq a+\underbrace{b}_{\geq0}+1\geq
a+1>a$ and thus $a<0$. The definition of $\mathbf{s}\left(  a,b\right)  $
yields%
\[
\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=\left(  -1\right)  ^{b}\delta_{a+b,-1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }a<0\text{ and }b\geq0\right)
\]
and thus $\left(  -1\right)  ^{b}\delta_{a+b,-1}=\mathbf{s}\left(  a,b\right)
$. Hence, (\ref{sol.hookschur.general-bc.altpf.5}) becomes%
\[
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}=\left(  -1\right)
^{b}\delta_{a+b,-1}=\mathbf{s}\left(  a,b\right)  .
\]
Hence, (\ref{sol.hookschur.general-bc}) is proven (under the assumption that
$a+b+1\leq0$).}. Hence, for the rest of our proof of
(\ref{sol.hookschur.general-bc}), we can WLOG assume that we don't have
$a+b+1\leq0$. Assume this.

We have $a+b+1>0$ (since we don't have $a+b+1\leq0$). Thus, $a+b+1\neq0$, so
that $\delta_{a+b+1,0}=0$. Also, $a+b+1\in \NN$ (since $a+b+1>0$).

Now, we must be in one of the following two cases:

\textit{Case 1:} We have $a<0$.

\textit{Case 2:} We have $a\geq0$.

Let us first consider Case 1. In this case, we have $a<0$. Now, recall that
every $n\in \NN$ satisfies%
\begin{equation}
\epsilon\left(  h_{n}\right)  =\delta_{n,0}.
\label{sol.hookschur.general-bc.altpf.epsihn}%
\end{equation}
Now, it is easy to see that every $n\in \NN$ satisfies%
\begin{equation}
\sum_{i=0}^{n}\left(  -1\right)  ^{i}h_{n-i}e_{i}=\delta_{n,0}.
\label{sol.hookschur.general-bc.altpf.antip}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.general-bc.altpf.antip}):} Let
$n\in \NN$. We know that the antipode $S$ of the Hopf algebra $\Lambda$
is a two-sided $\star$-inverse of the identity map $
\id_{\Lambda}:\Lambda\rightarrow\Lambda$ (by the definition of the
antipode of a Hopf algebra). Thus, $S\star\operatorname{id}_{\Lambda
}=u\epsilon$. Applying both sides of this equality to the element $h_{n}$ of
$\Lambda$, we obtain%
\begin{align*}
\left(  S\star\operatorname{id}_{\Lambda}\right)  \left(
h_{n}\right)   &  =\left(  u\epsilon\right)  \left(  h_{n}\right)  =u\left(
\underbrace{\epsilon\left(  h_{n}\right)  }_{\substack{=\delta_{n,0}%
\\\text{(by (\ref{sol.hookschur.general-bc.altpf.epsihn}))}}}\right)
=u\left(  \delta_{n,0}\right)  =\delta_{n,0}\cdot1\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }u\left(  \delta_{n,0}\right)  \right) \\
&  =\delta_{n,0},
\end{align*}
so that%
\begin{align*}
\delta_{n,0}  &  =\underbrace{\left(  S\star
\id_{\Lambda}\right)  }_{\substack{=m\circ\left(  S\otimes
\operatorname{id}_{\Lambda}\right)  \circ\Delta\\\text{(by the
definition of convolution)}}}\left(  h_{n}\right)  =\left(  m\circ\left(
S\otimes\operatorname{id}_{\Lambda}\right)  \circ\Delta\right)
\left(  h_{n}\right) \\
&  =m\left(  \left(  S\otimes\operatorname{id}_{\Lambda}\right)
\underbrace{\left(  \Delta h_{n}\right)  }_{\substack{=\sum_{i=0}^{n}%
h_{i}\otimes h_{n-i}\\\text{(by (\ref{sol.hookschur.d.Delhn}))}}}\right)
=m\left(  \underbrace{\left(  S\otimes\operatorname{id}_{\Lambda
}\right)  \left(  \sum_{i=0}^{n}h_{i}\otimes h_{n-i}\right)  }%
_{\substack{=\sum_{i=0}^{n}\left(  S\otimes
\id_{\Lambda}\right)  \left(  h_{i}\otimes h_{n-i}\right)
\\\text{(since }S\otimes\operatorname{id}_{\Lambda}\text{ is a
} \kk \text{-linear map)}}}\right) \\
&  =m\left(  \sum_{i=0}^{n}\underbrace{\left(  S\otimes
\id_{\Lambda}\right)  \left(  h_{i}\otimes h_{n-i}\right)  }_{=S\left(
h_{i}\right)  \otimes\operatorname{id}_{\Lambda}\left(
h_{n-i}\right)  }\right)  =m\left(  \sum_{i=0}^{n}\underbrace{S\left(
h_{i}\right)  }_{\substack{=\left(  -1\right)  ^{i}e_{i}\\\text{(by
Proposition \ref{symm-antipode-and-h-basis}(iii),}\\\text{applied to }i\text{
instead of }n\text{)}}}\otimes\underbrace{\operatorname{id}_{\Lambda
}\left(  h_{n-i}\right)  }_{=h_{n-i}}\right) \\
&  =m\left(  \sum_{i=0}^{n}\left(  -1\right)  ^{i}e_{i}\otimes h_{n-i}\right)
=\sum_{i=0}^{n}\left(  -1\right)  ^{i}\underbrace{m\left(  e_{i}\otimes
h_{n-i}\right)  }_{\substack{=e_{i}h_{n-i}\\\text{(since }m\text{ is the
multiplication map)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m\text{ is a
} \kk \text{-linear map}\right) \\
&  =\sum_{i=0}^{n}\left(  -1\right)  ^{i}e_{i}h_{n-i}=\sum_{i=0}^{n}\left(
-1\right)  ^{i}h_{n-i}e_{i}.
\end{align*}
This proves (\ref{sol.hookschur.general-bc.altpf.antip}).}

Now, $a<0$, so that
$a\leq-1$ (since $a$ and $0$ are integers) and thus $a+1\leq0$. Thus,
$a+b+1=b+\underbrace{\left(  a+1\right)  }_{\leq0}\leq b$. Combined with
$a+b+1>0$, this shows that $0<a+b+1\leq b$. Now,
(\ref{sol.hookschur.general-bc.altpf.1}) becomes%
\begin{align*}
&  \sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}\\
& =\sum_{i=0}^{b}\left(  -1\right)  ^{b-i}h_{\left(  a+b+1\right)  -i}
e_{i} \nonumber \\
&  =\sum_{i=0}^{a+b+1}\underbrace{\left(  -1\right)  ^{b-i}}%
_{\substack{=\left(  -1\right)  ^{b+\left(  -i\right)  }\\=\left(  -1\right)
^{b}\left(  -1\right)  ^{-i}}}h_{\left(  a+b+1\right)  -i}e_{i}+\sum
_{i=\left(  a+b+1\right)  +1}^{b}\left(  -1\right)  ^{b-i}%
\underbrace{h_{\left(  a+b+1\right)  -i}}_{\substack{=0\\\text{(since }\left(
a+b+1\right)  -i<0\\\text{(since }i>a+b+1\text{))}}}e_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0<a+b+1\leq b\right) \\
&  =\sum_{i=0}^{a+b+1}\left(  -1\right)  ^{b}\left(  -1\right)  ^{-i}%
h_{\left(  a+b+1\right)  -i}e_{i}+\underbrace{\sum_{i=\left(  a+b+1\right)
+1}^{b}\left(  -1\right)  ^{b-i}0e_{i}}_{=0}\\
&  =\sum_{i=0}^{a+b+1}\left(  -1\right)  ^{b}\underbrace{\left(  -1\right)
^{-i}}_{\substack{=\left(  -1\right)  ^{i}\\\text{(since }-i\equiv
i\operatorname{mod}2\text{)}}}h_{\left(  a+b+1\right)  -i}e_{i}=\sum
_{i=0}^{a+b+1}\left(  -1\right)  ^{b}\left(  -1\right)  ^{i}h_{\left(
a+b+1\right)  -i}e_{i}\\
&  =\left(  -1\right)  ^{b}\underbrace{\sum_{i=0}^{a+b+1}\left(  -1\right)
^{i}h_{\left(  a+b+1\right)  -i}e_{i}}_{\substack{=\delta_{a+b+1,0}\\\text{(by
(\ref{sol.hookschur.general-bc.altpf.antip}), applied to }n=a+b+1\text{)}%
}}=\left(  -1\right)  ^{b}\underbrace{\delta_{a+b+1,0}}_{=0}=0.
\end{align*}
Compared with%
\begin{align*}
\mathbf{s}\left(  a,b\right)   &  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{s}\left(  a,b\right)  \right) \\
&  =\left(  -1\right)  ^{b}\underbrace{\delta_{a+b,-1}}%
_{\substack{=0\\\text{(since }a+b\neq-1\\\text{(since }a+b+1\neq0\text{))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a<0\text{ and }b\geq0\right) \\
&  =0,
\end{align*}
this yields $\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}%
=\mathbf{s}\left(  a,b\right)  $. Thus, (\ref{sol.hookschur.general-bc}) is
proven in Case 1.

Let us now consider Case 2. In this case, we have $a\geq0$. The definition of
$\mathbf{s}\left(  a,b\right)  $ yields%
\begin{equation}
\mathbf{s}\left(  a,b\right)  =
\begin{cases}
s_{\left(  a+1,1^{b}\right)  },& \text{if }a\geq0\text{ and
}b\geq0;\\
\left(  -1\right)  ^{b}\delta_{a+b,-1},& \text{if }a<0\text{
and }b\geq0;\\
0,& \text{if }b<0
\end{cases}  \quad
=s_{\left(  a+1,1^{b}\right)  }
\label{sol.hookschur.general-bc.altpf.9}%
\end{equation}
(since $a\geq0$ and $b\geq0$).

Let us now recall that the basis $\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ is orthonormal with respect to the Hall inner product
$\left(  \cdot,\cdot\right)  $. In other words,%
\begin{equation}
\left(  s_{\mu},s_{\nu}\right)  =\delta_{\mu,\nu}\ \ \ \ \ \ \ \ \ \ \text{for
any two partitions }\mu\text{ and }\nu\text{.}
\label{sol.hookschur.general-bc.altpf.hall.orthonor-s}%
\end{equation}
As a consequence, it is easy to see that%
\begin{equation}
\left(  e_{i},e_{j}\right)  =\delta_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for any
}i\in \NN \text{ and }j\in \NN \text{.}
\label{sol.hookschur.general-bc.altpf.hall.orthonor-e}%
\end{equation}
\footnote{\textit{Proof of
(\ref{sol.hookschur.general-bc.altpf.hall.orthonor-e}):} Let $i\in \NN$
and $j\in \NN$. We need to prove $\left(  e_{i},e_{j}\right)
=\delta_{i,j}$.
\par
We have $s_{\left(  1^{i}\right)  }=e_{i}$ and $s_{\left(  1^{j}\right)
}=e_{j}$. Hence, $\left(  \underbrace{s_{\left(  1^{i}\right)  }}_{=e_{i}%
},\underbrace{s_{\left(  1^{j}\right)  }}_{=e_{j}}\right)  =\left(
e_{i},e_{j}\right)  $ and thus%
\[
\left(  e_{i},e_{j}\right)  =\left(  s_{\left(  1^{i}\right)  },s_{\left(
1^{j}\right)  }\right)  =\delta_{\left(  1^{i}\right)  ,\left(  1^{j}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.hookschur.general-bc.altpf.hall.orthonor-s}), applied to }%
\mu=\left(  1^{i}\right)  \text{ and }\nu=\left(  1^{j}\right)  \right)  .
\]
\par
Let us now assume that $i=j$. Then, $\left(  1^{i}\right)  =\left(
1^{j}\right)  $ and $\left(  e_{i},e_{j}\right)  =\delta_{\left(
1^{i}\right)  ,\left(  1^{j}\right)  }=1$ (since $\left(  1^{i}\right)
=\left(  1^{j}\right)  $). Compared with $\delta_{i,j}=1$ (since $i=j$), this
yields $\left(  e_{i},e_{j}\right)  =\delta_{i,j}$.
\par
Now, let us forget our assumption that $i=j$. We thus have proven that
$\left(  e_{i},e_{j}\right)  =\delta_{i,j}$ when $i=j$. Hence, for the rest of
our proof of $\left(  e_{i},e_{j}\right)  =\delta_{i,j}$, we can WLOG assume
that we don't have $i=j$. Assume this.
\par
We don't have $i=j$. Hence, we have $i\neq j$, so that $\ell\left(  \left(
1^{i}\right)  \right)  =i\neq j=\ell\left(  \left(  1^{j}\right)  \right)  $
and thus $\left(  1^{i}\right)  \neq\left(  1^{j}\right)  $. Now, $\left(
e_{i},e_{j}\right)  =\delta_{\left(  1^{i}\right)  ,\left(  1^{j}\right)  }=0$
(since $\left(  1^{i}\right)  \neq\left(  1^{j}\right)  $). Compared with
$\delta_{i,j}=0$ (since $i\neq j$), this yields $\left(  e_{i},e_{j}\right)
=\delta_{i,j}$. This proves
(\ref{sol.hookschur.general-bc.altpf.hall.orthonor-e}).}

Let $\lambda$ be the partition $\left(  1^{b}\right)  $. Thus, $\lambda
=\left(  1^{b}\right)  =\left(  \underbrace{1,1,...,1}_{b\text{ times}%
},0,0,0,...\right)  $, and therefore $\lambda_{1}$ is either $1$ or $0$ (where
we are using the notations of Exercise \ref{exe.bernstein}). Consequently,
$\lambda_{1}\leq1$, and hence $1\geq\lambda_{1}$. Now, $\underbrace{a}_{\geq
0}+1\geq1\geq\lambda_{1}$. Thus, Exercise \ref{exe.bernstein}(b) (applied to
$m=a+1$) yields%
\[
\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{a+1+i}e_{i}^{\perp}%
s_{\lambda}=s_{\left(  a+1,\lambda_{1},\lambda_{2},\lambda_{3},\ldots\right)
}.
\]
Since
\begin{align*}
s_{\lambda}  &  =s_{\left(  1^{b}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\lambda=\left(  1^{b}\right)  \right) \\
&  =e_{b}%
\end{align*}
and
\[
\left(  a+1,\underbrace{\lambda_{1},\lambda_{2},\lambda_{3},\ldots}%
_{=\lambda=\left(  \underbrace{1,1,...,1}_{b\text{ times}},0,0,0,...\right)
}\right)  =\left(  a+1,\underbrace{1,1,...,1}_{b\text{ times}}%
,0,0,0,...\right)  =\left(  a+1,1^{b}\right)  ,
\]
this rewrites as follows:%
\[
\sum_{i\in{ \NN }}\left(  -1\right)  ^{i}h_{a+1+i}e_{i}^{\perp}%
e_{b}=s_{\left(  a+1,1^{b}\right)  }.
\]
Compared with (\ref{sol.hookschur.general-bc.altpf.9}), this yields%
\begin{equation}
\mathbf{s}\left(  a,b\right)  =\sum_{i\in{ \NN }}\left(  -1\right)
^{i}h_{a+1+i}e_{i}^{\perp}e_{b}=\sum_{j\in{ \NN }}\left(  -1\right)
^{j}h_{a+1+j}e_{j}^{\perp}e_{b} \label{sol.hookschur.general-bc.altpf.11}%
\end{equation}
(here, we renamed the summation index $i$ as $j$).

But every $j\in \NN$ satisfies%
\begin{equation}
e_{j}^{\perp}e_{b}=e_{b-j}. \label{sol.hookschur.general-bc.altpf.14}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hookschur.general-bc.altpf.14}):} Let
$j\in \NN$. We need to prove (\ref{sol.hookschur.general-bc.altpf.14}).
\par
Let us use the Sweedler notation. Then, $\sum_{\left(  e_{b}\right)  }\left(
e_{b}\right)  _{1}\otimes\left(  e_{b}\right)  _{2}=\Delta e_{b}=\sum
_{i\in \NN }e_{i}\otimes e_{b-i}$ (by (\ref{sol.hookschur.d.Delen.inf}),
applied to $b$ instead of $n$). Now, the definition of $e_{j}^{\perp}e_{b}$
yields%
\begin{align*}
e_{j}^{\perp}e_{b}  &  =\sum_{\left(  e_{b}\right)  }\left(  e_{j},\left(
e_{b}\right)  _{1}\right)  \left(  e_{b}\right)  _{2}=\sum_{i\in \NN %
}\underbrace{\left(  e_{j},e_{i}\right)  }_{\substack{=\delta_{j,i}\\\text{(by
(\ref{sol.hookschur.general-bc.altpf.hall.orthonor-e}), applied to }j\text{
and }i\\\text{instead of }i\text{ and }j\text{)}}}e_{b-i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum_{\left(  e_{b}\right)  }\left(
e_{b}\right)  _{1}\otimes\left(  e_{b}\right)  _{2}=\sum_{i\in \NN }%
e_{i}\otimes e_{b-i}\right) \\
&  =\sum_{i\in \NN }\underbrace{\delta_{j,i}}_{=\delta_{i,j}}e_{b-i}%
=\sum_{i\in \NN }\delta_{i,j}e_{b-i}=\sum\limits_{\substack{i\in \NN %
;\\i=j}}\underbrace{\delta_{i,j}}_{\substack{=1\\\text{(since }i=j\text{)}%
}}e_{b-i}+\sum\limits_{\substack{i\in \NN ;\\i\neq j}}\underbrace{\delta_{i,j}%
}_{\substack{=0\\\text{(since }i\neq j\text{)}}}e_{b-i}\\
&  =\sum\limits_{\substack{i\in \NN ;\\i=j}}e_{b-i}+\underbrace{\sum\limits
_{\substack{i\in \NN ;\\i\neq j}}0e_{b-i}}_{=0}=\sum\limits_{\substack{i\in
 \NN ;\\i=j}}e_{b-i}=e_{b-j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
j\in \NN \right)  .
\end{align*}
This proves (\ref{sol.hookschur.general-bc.altpf.14}).} Thus,
(\ref{sol.hookschur.general-bc.altpf.11}) becomes%
\begin{align*}
\mathbf{s}\left(  a,b\right)   &  =\sum_{j\in{ \NN }}\left(  -1\right)
^{j}\underbrace{h_{a+1+j}}_{=h_{a+j+1}}\underbrace{e_{j}^{\perp}e_{b}%
}_{\substack{=e_{b-j}\\\text{(by (\ref{sol.hookschur.general-bc.altpf.14}))}%
}}=\sum_{j\in{ \NN }}\left(  -1\right)  ^{j}h_{a+j+1}e_{b-j}\\
&  =\underbrace{\sum\limits_{\substack{j\in{ \NN };\\j\leq b}}}_{=\sum_{j=0}%
^{b}}\left(  -1\right)  ^{j}h_{a+j+1}e_{b-j}+\sum\limits_{\substack{j\in{ \NN %
};\\j>b}}\left(  -1\right)  ^{j}h_{a+j+1}\underbrace{e_{b-j}}%
_{\substack{=0\\\text{(since }b-j<0\\\text{(since }j>b\text{))}}}\\
&  =\sum_{j=0}^{b}\left(  -1\right)  ^{j}h_{a+j+1}e_{b-j}+\underbrace{\sum\limits
_{\substack{j\in{ \NN };\\j>b}}\left(  -1\right)  ^{j}h_{a+j+1}0}%
_{=0}=\sum_{j=0}^{b}\left(  -1\right)  ^{j}h_{a+j+1}e_{b-j}=\sum_{i=0}%
^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}%
\end{align*}
(here, we have renamed the summation index $j$ as $i$). Thus,
(\ref{sol.hookschur.general-bc}) is proven in Case 2.

We have now proven (\ref{sol.hookschur.general-bc}) in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this shows that
(\ref{sol.hookschur.general-bc}) always holds. This completes the alternative
proof of (\ref{sol.hookschur.general-bc}).

\textit{Alternative proof of (\ref{sol.hookschur.general-a}):} Let
$n\in \ZZ$ and $m\in \ZZ$. We need to prove
(\ref{sol.hookschur.general-a}).

It is easy to see that (\ref{sol.hookschur.general-a}) holds if $n<0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $n<0$. We need to prove that
(\ref{sol.hookschur.general-a}) holds.
\par
Applying (\ref{sol.hookschur.b<0}) to $a=m$ and $b=n-1$, we obtain
$\mathbf{s}\left(  m,n-1\right)  =0$ (since $n-1<n<0$). Applying
(\ref{sol.hookschur.b<0}) to $a=m-1$ and $b=n$, we obtain $\mathbf{s}\left(
m-1,n\right)  =0$. Hence, $\underbrace{\mathbf{s}\left(  m,n-1\right)  }%
_{=0}+\underbrace{\mathbf{s}\left(  m-1,n\right)  }_{=0}=0+0=0$, so that
$0=\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(  m-1,n\right)  $.
\par
But $e_{n}=0$ (since $n<0$) and thus $\underbrace{e_{n}}_{=0}h_{m}%
=0=\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(  m-1,n\right)  $. Thus,
(\ref{sol.hookschur.general-a}) holds (if $n<0$), qed.}. Hence, for the rest
of this proof, we can WLOG assume that we don't have $n<0$. Assume this.

We have $n\geq0$ (since we don't have $n<0$). But
(\ref{sol.hookschur.general-bc}) (applied to $a=m$ and $b=n-1$) yields%
\[
\sum_{i=0}^{n-1}\left(  -1\right)  ^{i}h_{m+i+1}e_{\left(  n-1\right)
-i}=\mathbf{s}\left(  m,n-1\right)  .
\]
Thus,%
\begin{align}
\mathbf{s}\left(  m,n-1\right)   &  =\sum_{i=0}^{n-1}\left(  -1\right)
^{i}h_{m+i+1}e_{\left(  n-1\right)  -i}=\sum_{i=1}^{n}\underbrace{\left(
-1\right)  ^{i-1}}_{=\left(  -1\right)  ^{i}/\left(  -1\right)  =-\left(
-1\right)  ^{i}}\underbrace{h_{m+\left(  i-1\right)  +1}}_{=h_{m+i}%
}\underbrace{e_{\left(  n-1\right)  -\left(  i-1\right)  }}_{=e_{n-i}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }i-1\text{ for
}i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i=1}^{n}\left(  -\left(  -1\right)  ^{i}\right)  h_{m+i}%
e_{n-i}=-\sum_{i=1}^{n}\left(  -1\right)  ^{i}h_{m+i}e_{n-i}.
\label{sol.hookschur.general-a.altpf.1}%
\end{align}
But (\ref{sol.hookschur.general-bc}) (applied to $a=m-1$ and $b=n$) yields%
\[
\sum_{i=0}^{n}\left(  -1\right)  ^{i}h_{\left(  m-1\right)  +i+1}%
e_{n-i}=\mathbf{s}\left(  m-1,n\right)  .
\]
Therefore,%
\begin{align*}
\mathbf{s}\left(  m-1,n\right)   &  =\sum_{i=0}^{n}\left(  -1\right)
^{i}\underbrace{h_{\left(  m-1\right)  +i+1}}_{=h_{m+i}}e_{n-i}=\sum_{i=0}%
^{n}\left(  -1\right)  ^{i}h_{m+i}e_{n-i}\\
&  =\underbrace{\left(  -1\right)  ^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}%
}\underbrace{e_{n-0}}_{=e_{n}}+\sum_{i=1}^{n}\left(  -1\right)  ^{i}%
h_{m+i}e_{n-i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\geq0\right) \\
&  =h_{m}e_{n}+\sum_{i=1}^{n}\left(  -1\right)  ^{i}h_{m+i}e_{n-i}.
\end{align*}
Adding this equality to the equality (\ref{sol.hookschur.general-a.altpf.1}),
we obtain%
\[
\mathbf{s}\left(  m,n-1\right)  +\mathbf{s}\left(  m-1,n\right)  =\left(
-\sum_{i=1}^{n}\left(  -1\right)  ^{i}h_{m+i}e_{n-i}\right)  +\left(
h_{m}e_{n}+\sum_{i=1}^{n}\left(  -1\right)  ^{i}h_{m+i}e_{n-i}\right)
=h_{m}e_{n}
= e_n h_m.
\]
This proves (\ref{sol.hookschur.general-a}) again.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.180}}

\textit{Solution to Exercise \ref{exe.liri.180}.} Consider the partition
$\left(  m^{k}\right)  =\left(  \underbrace{m,m,\ldots,m}_{k\text{ times}%
}\right)  $.

(a) The fact that $\lambda^{\vee}$ and $\mu^{\vee}$ are partitions is easy to
check. It remains to show that $s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}}$.

It is easy to show that if $\mu\subseteq\lambda$ does not hold, then
$\lambda^{\vee}\subseteq\mu^{\vee}$ does not hold either (because if some
positive integer $i$ fails to satisfy $\mu_{i}\leq\lambda_{i}$, then this $i$
belongs to $\left\{  1,2,\ldots,k\right\}  $ and fails to satisfy
$m-\lambda_{i}\leq m-\mu_{i}$ as well). Hence, if $\mu\subseteq\lambda$ does
not hold, then both $s_{\lambda/\mu}$ and $s_{\mu^{\vee}/\lambda^{\vee}}$ are
$0$, and therefore the equality $s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}%
}$ is obvious. Hence, for the rest of the proof of $s_{\lambda/\mu}%
=s_{\mu^{\vee}/\lambda^{\vee}}$, we WLOG assume that $\mu\subseteq\lambda$
does hold. Then, it is easy to see that $\lambda^{\vee}\subseteq\mu^{\vee}$
holds, too.

Now, let $Z$ denote the $180^{\circ}$ rotation around the center of the
Ferrers diagram of $\left(  m^{k}\right)  $. Let $Y\left(  \rho\right)  $
denote the Ferrers diagram of $\rho$ whenever $\rho$ is a partition or a skew
partition. It is straightforward to see that $Z\left(  Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \mu\right)  \right)  =Y\left(
\mu^{\vee}\right)  $ and $Z\left(  Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \lambda\right)  \right)  =Y\left(  \lambda^{\vee}\right)
$. Now,%
\begin{align*}
Y\left(  \mu^{\vee}/\lambda^{\vee}\right)   &  =\underbrace{Y\left(  \mu
^{\vee}\right)  }_{=Z\left(  Y\left(  \left(  m^{k}\right)  \right)  \setminus
Y\left(  \mu\right)  \right)  }\setminus\underbrace{Y\left(  \lambda^{\vee
}\right)  }_{=Z\left(  Y\left(  \left(  m^{k}\right)  \right)  \setminus
Y\left(  \lambda\right)  \right)  }=Z\left(  Y\left(  \left(  m^{k}\right)
\right)  \setminus Y\left(  \mu\right)  \right)  \setminus Z\left(  Y\left(
\left(  m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)  \right) \\
&  =Z\left(  \underbrace{\left(  Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \mu\right)  \right)  \setminus\left(  Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)  \right)
}_{\substack{=Y\left(  \lambda\right)  \setminus Y\left(  \mu\right)
\\\text{(since one can easily see that both }Y\left(  \lambda\right)  \text{
and }Y\left(  \mu\right)  \\\text{are subsets of }Y\left(  \left(
m^{k}\right)  \right)  \text{)}}}\right) \\
&  =Z\left(  \underbrace{Y\left(  \lambda\right)  \setminus Y\left(
\mu\right)  }_{=Y\left(  \lambda/\mu\right)  }\right)  =Z\left(  Y\left(
\lambda/\mu\right)  \right)  .
\end{align*}
Hence, the skew Ferrers diagram $\mu^{\vee}/\lambda^{\vee}$ can be obtained
from the skew Ferrers diagram $\lambda/\mu$ by a $180^{\circ}$ rotation
(namely, by the $180^{\circ}$ rotation $Z$). Thus, Exercise
\ref{exe.schur.rotate180}(b) (applied to $\lambda^{\prime}=\mu^{\vee}$ and
$\mu^{\prime}=\lambda^{\vee}$) yields $s_{\lambda/\mu}=s_{\mu^{\vee}%
/\lambda^{\vee}}$. This completes the solution of Exercise \ref{exe.liri.180}(a).

(b) According to Remark \ref{third-L-R-interpretation-remark}, we have
$s_{\lambda/\mu}=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum ranges
over all partitions $\nu$. In other words, we have%
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in \Par }c_{\mu,\nu}^{\lambda}s_{\nu}.
\label{sol.liri.180.b.1}%
\end{equation}


Exercise \ref{exe.liri.180}(b) yields that $\lambda^{\vee}$ and $\mu^{\vee}$
are partitions, and that $s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}}$.
Applying (\ref{sol.liri.180.b.1}) to $\mu^{\vee}$ and $\lambda^{\vee}$ instead
of $\lambda$ and $\mu$, we obtain%
\[
s_{\mu^{\vee}/\lambda^{\vee}}=\sum_{\nu\in \Par }c_{\lambda
^{\vee},\nu}^{\mu^{\vee}}s_{\nu}.
\]
Now, (\ref{sol.liri.180.b.1}) yields
\[
\sum_{\nu\in \Par }c_{\mu,\nu}^{\lambda}s_{\nu}=s_{\lambda/\mu
}=s_{\mu^{\vee}/\lambda^{\vee}}=\sum_{\nu\in \Par }%
c_{\lambda^{\vee},\nu}^{\mu^{\vee}}s_{\nu}.
\]
Comparing coefficients before $s_{\nu}$ in this equality, we conclude that
$c_{\mu,\nu}^{\lambda}=c_{\lambda^{\vee},\nu}^{\mu^{\vee}}$ for every $\nu
\in \Par $ (since $\left(  s_{\nu}\right)  _{\nu\in
 \Par }$ is a basis of the $\kk$-module $\Lambda$). This
solves Exercise \ref{exe.liri.180}(b).

(c) Exercise \ref{exe.liri.180}(a) (applied to $\nu$ instead of $\mu$) yields
that $\lambda^{\vee}$ and $\nu^{\vee}$ are partitions and that $s_{\lambda
/\nu}=s_{\nu^{\vee}/\lambda^{\vee}}$.

From (\ref{eq.third-L-R-interpretation-remark.commutativity}), we obtain%
\begin{align*}
c_{\mu,\nu}^{\lambda}  &  =c_{\nu,\mu}^{\lambda}=c_{\lambda^{\vee},\mu}%
^{\nu^{\vee}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.liri.180}%
(b) (applied to }\nu\text{ and }\mu\text{ instead of }\mu\text{ and }%
\nu\text{)}\right) \\
&  =c_{\mu,\lambda^{\vee}}^{\nu^{\vee}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.third-L-R-interpretation-remark.commutativity}) (applied to }%
\nu^{\vee}\text{, }\lambda^{\vee}\text{ and }\mu\text{ instead of }%
\lambda\text{, }\mu\text{ and }\nu\text{)}\right)  .
\end{align*}
On the other hand, Exercise \ref{exe.liri.180}(b) yields $c_{\mu,\nu}%
^{\lambda}=c_{\lambda^{\vee},\nu}^{\mu^{\vee}}=c_{\nu,\lambda^{\vee}}%
^{\mu^{\vee}}$ (by (\ref{eq.third-L-R-interpretation-remark.commutativity})
(applied to $\mu^{\vee}$, $\lambda^{\vee}$ and $\nu$ instead of $\lambda$,
$\mu$ and $\nu$)). Thus, $c_{\nu,\lambda^{\vee}}^{\mu^{\vee}}=c_{\lambda^{\vee
},\nu}^{\mu^{\vee}}=c_{\mu,\nu}^{\lambda}=c_{\nu,\mu}^{\lambda}=c_{\lambda
^{\vee},\mu}^{\nu^{\vee}}=c_{\mu,\lambda^{\vee}}^{\nu^{\vee}}$. This solves
Exercise \ref{exe.liri.180}(c).

(d) \textit{First solution to Exercise \ref{exe.liri.180}(d):} Notice that
$\lambda^{\vee}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda
_{1}\right)  $. Thus, $\ell\left(  \lambda^{\vee}\right)  \leq k$.

Let $n=k$. We shall use the notations of Section \ref{bialternant-section}; in
particular, we set $\xx=\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ and
$\rho=\left(  n-1,n-2,\ldots,2,1,0\right)  $. Clearly, $\xx=\left(
x_{1},x_{2},\ldots,x_{n}\right)  =\left(  x_{1},x_{2},\ldots,x_{k}\right)  $
(since $n=k$). Notice that $\ell\left(  \lambda\right)  \leq k=n$. Hence, we
can regard $\lambda$ as an element of $\NN^n$; therefore,
$\lambda+\rho$ is a well-defined element of $\NN^n$, and the
alternant $a_{\lambda+\rho}$ is well-defined. Corollary
\ref{cor.stembridge.slambda} yields that $s_{\lambda}\left(  \xx%
\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$, so that%
\begin{equation}
a_{\rho}\cdot s_{\lambda}\left(  \xx\right)  =a_{\lambda+\rho}.
\label{sol.liri.180.d.s_lambda-as-quotient}%
\end{equation}
Applying this equality to $\lambda^{\vee}$ instead of $\lambda$, we obtain%
\begin{equation}
a_{\rho}\cdot s_{\lambda^{\vee}}\left(  \xx\right)  =a_{\lambda^{\vee
}+\rho} \label{sol.liri.180.d.s_lambda-check-as-quotient}%
\end{equation}
(since $\lambda^{\vee}$ is also a partition satisfying $\ell\left(
\lambda^{\vee}\right)  \leq k=n$).

Substituting the variables $x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}$ for
$x_{1},x_{2},\ldots,x_{n}$ in the equality
(\ref{sol.liri.180.d.s_lambda-as-quotient}), we obtain%
\begin{equation}
a_{\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \cdot
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)
=a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  .
\label{sol.liri.180.d.1}%
\end{equation}


Let $w_{\circ}:\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{
1,2,\ldots,n\right\}  $ be the map which sends every $i\in\left\{
1,2,\ldots,n\right\}  $ to $n+1-i$. Then, $w_{\circ}$ is a permutation of
$\left\{  1,2,\ldots,n\right\}  $, thus an element of $\Symm_n$. For
every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{equation}
\left(  w\circ w_{\circ}\right)  \left(  i\right)  =w\left(  n+1-i\right)  .
\label{sol.liri.180.d.wnach}%
\end{equation}
The map $\Symm_n\rightarrow\Symm_n,\ w\mapsto w\circ
w_{\circ}$ is a bijection (since $\Symm_n$ is a group, and since
$w_{\circ}\in\Symm_n$).

But every $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
\in\NN^n$ satisfies%
\begin{align}
a_{\alpha}  &  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{w\left(  {\mathbf{x}}^{\alpha}\right)  }_{\substack{=\prod
_{i=1}^{n}x_{w\left(  i\right)  }^{\alpha_{i}}\\\text{(by the definition of
}w\left(  \mathbf{x}^{\alpha}\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }a_{\alpha}\right) \nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}x_{w\left(  i\right)  }^{\alpha_{i}}.
\label{sol.liri.180.d.a_alpha-formula}%
\end{align}


But $\lambda=\left(  \lambda_{1},\lambda_{2},\lambda_{3},\ldots\right)  $,
thus $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $
(since $\ell\left(  \lambda\right)  \leq k=n$). Hence,%
\begin{align*}
\underbrace{\lambda}_{=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{n}\right)  }+\underbrace{\rho}_{\substack{=\left(  n-1,n-2,\ldots
,2,1,0\right)  \\=\left(  n-1,n-2,\ldots,n-n\right)  }}  &  =\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  +\left(  n-1,n-2,\ldots
,n-n\right) \\
&  =\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,\ldots,\lambda_{n}+n-n\right)  .
\end{align*}
Thus, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to $\lambda+\rho$ and
$\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,\ldots,\lambda_{n}+n-n\right)  $
instead of $\alpha$ and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  $) yields%
\begin{equation}
a_{\lambda+\rho}=\sum_{w\in\Symm_n}\operatorname{sgn}\left(
w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{\lambda_{i}+n-i}.
\label{sol.liri.180.d.a_lambdaplusrho1}%
\end{equation}


Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$, $x_{n}^{-1}$ for $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ on both sides of this equality, we obtain%
\begin{align*}
&  a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{\left(  x_{w\left(  i\right)  }^{-1}\right)
^{\lambda_{i}+n-i}}_{=x_{w\left(  i\right)  }^{-\left(  \lambda_{i}%
+n-i\right)  }}=\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda
_{i}+n-i\right)  }}_{\substack{=\prod_{i=1}^{n}x_{w\left(  n+1-i\right)
}^{-\left(  \lambda_{n+1-i}+n-\left(  n+1-i\right)  \right)  }\\\text{(here,
we have substituted}\\n+1-i\text{ for }i\text{ in the product)}}}\\
&  =\sum_{w\in\Symm_n}\underbrace{\operatorname{sgn}\left(  w\right)
}_{=\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }%
\operatorname*{sgn}\left(  w\right)  \operatorname*{sgn}\left(  w_{\circ
}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)  }^{-\left(
\lambda_{n+1-i}+n-\left(  n+1-i\right)  \right)  }}_{\substack{=x_{w\left(
n+1-i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)  }\\\text{(since
}n-\left(  n+1-i\right)  =i-1\text{)}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\underbrace{\operatorname*{sgn}\left(  w\right)
\operatorname*{sgn}\left(  w_{\circ}\right)  }_{=\operatorname*{sgn}\left(
w\circ w_{\circ}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }}_{\substack{=x_{\left(  w\circ
w_{\circ}\right)  \left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\\\text{(since }w\left(  n+1-i\right)  =\left(  w\circ w_{\circ}\right)
\left(  i\right)  \\\text{(by (\ref{sol.liri.180.d.wnach})))}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\operatorname*{sgn}\left(  w\circ w_{\circ}\right)  \prod_{i=1}%
^{n}x_{\left(  w\circ w_{\circ}\right)  \left(  i\right)  }^{-\left(
\lambda_{n+1-i}+i-1\right)  }=\sum_{w\in\Symm_n}\dfrac{1}%
{\operatorname*{sgn}\left(  w_{\circ}\right)  }\operatorname*{sgn}\left(
w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda
_{n+1-i}+i-1\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }w\text{ for }w\circ w_{\circ}\text{ in the sum}\\
\text{(since the map }\Symm_n\rightarrow\Symm_n,\ w\mapsto
w\circ w_{\circ}\text{ is a bijection)}%
\end{array}
\right) \\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}%
^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)  }.
\end{align*}
Multiplying both sides of this equality by $\operatorname*{sgn}\left(
w_{\circ}\right)  $, we obtain%
\begin{equation}
\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot a_{\lambda+\rho}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  =\sum_{w\in\Symm_n%
}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }.
\label{sol.liri.180.d.a_lambdaplusrho}%
\end{equation}


But $\lambda^{\vee}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda
_{1}\right)  =\left(  m-\lambda_{n},m-\lambda_{n-1},\ldots,m-\lambda
_{1}\right)  $ (since $k=n$), so that%
\begin{align*}
&  \underbrace{\lambda^{\vee}}_{\substack{=\left(  m-\lambda_{n}%
,m-\lambda_{n-1},\ldots,m-\lambda_{1}\right)  \\=\left(  m-\lambda_{\left(
n+1\right)  -1},m-\lambda_{\left(  n+1\right)  -2},\ldots,m-\lambda_{\left(
n+1\right)  -n}\right)  }}+\underbrace{\rho}_{\substack{=\left(
n-1,n-2,\ldots,2,1,0\right)  \\=\left(  n-1,n-2,\ldots,n-n\right)  }}\\
&  =\left(  m-\lambda_{\left(  n+1\right)  -1},m-\lambda_{\left(  n+1\right)
-2},\ldots,m-\lambda_{\left(  n+1\right)  -n}\right)  +\left(  n-1,n-2,\ldots
,n-n\right) \\
&  =\left(  \left(  m-\lambda_{\left(  n+1\right)  -1}\right)  +\left(
n-1\right)  ,\left(  m-\lambda_{\left(  n+1\right)  -2}\right)  +\left(
n-2\right)  ,\ldots,\left(  m-\lambda_{\left(  n+1\right)  -n}\right)
+\left(  n-n\right)  \right)  .
\end{align*}
Hence, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to $\lambda^{\vee}%
+\rho$ and \newline$\left(  \left(  m-\lambda_{\left(  n+1\right)  -1}\right)
+\left(  n-1\right)  ,\left(  m-\lambda_{\left(  n+1\right)  -2}\right)
+\left(  n-2\right)  ,\ldots,\left(  m-\lambda_{\left(  n+1\right)
-n}\right)  +\left(  n-n\right)  \right)  $ instead of $\alpha$ and
\newline$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $) yields%
\begin{align}
a_{\lambda^{\vee}+\rho}  &  =\sum_{w\in\Symm_n}\operatorname{sgn}%
\left(  w\right)  \prod_{i=1}^{n}\underbrace{x_{w\left(  i\right)  }^{\left(
m-\lambda_{\left(  n+1\right)  -i}\right)  +\left(n-i\right)}}%
_{\substack{_{\substack{=x_{w\left(  i\right)  }^{-\left(  \lambda
_{n+1-i}+i-1\right)  +\left(  n+m-1\right)  }}}\\=x_{w\left(  i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }x_{w\left(  i\right)  }^{n+m-1}%
}}\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\prod_{i=1}^{n}\left(  x_{w\left(  i\right)  }^{-\left(
\lambda_{n+1-i}+i-1\right)  }x_{w\left(  i\right)  }^{n+m-1}\right)
}_{=\left(  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda
_{n+1-i}+i-1\right)  }\right)  \left(  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{n+m-1}\right)  }\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\right)  \underbrace{\left(  \prod_{i=1}^{n}x_{w\left(  i\right)  }%
^{n+m-1}\right)  }_{\substack{=\prod_{i=1}^{n}x_{i}^{n+m-1}\\\text{(here, we
have substituted }i\text{ for }w\left(  i\right)  \text{ in the product}%
\\\text{(since }w\text{ is a permutation of }\left\{  1,2,\ldots,n\right\}
\text{))}}}\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\right)  \left(  \prod_{i=1}^{n}x_{i}^{n+m-1}\right) \nonumber\\
&  =\underbrace{\left(  \prod_{i=1}^{n}x_{i}^{n+m-1}\right)  }_{=\left(
\prod_{i=1}^{n}x_{i}\right)  ^{n+m-1}}\underbrace{\sum_{w\in\Symm_n%
}\operatorname{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }}_{\substack{=\operatorname*{sgn}%
\left(  w_{\circ}\right)  \cdot a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right)  \\\text{(by
(\ref{sol.liri.180.d.a_lambdaplusrho}))}}}\nonumber\\
&  =\left(  \prod_{i=1}^{n}x_{i}\right)  ^{n+m-1}\cdot\operatorname*{sgn}%
\left(  w_{\circ}\right)  \cdot a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right)  .
\label{sol.liri.180.d.a_lambdaplusrho-comes-around}%
\end{align}


Finally, $\rho=\left(  n-1,n-2,\ldots,2,1,0\right)  =\left(  n-1,n-2,\ldots
,n-n\right)  $. Thus, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to
$\rho$ and \newline$\left(  n-1,n-2,\ldots,n-n\right)  $ instead of $\alpha$
and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $) yields%
\begin{equation}
a_{\rho}=\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}. \label{sol.liri.180.d.a_rho}%
\end{equation}
Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$, $x_{n}^{-1}$ for $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ on both sides of this equality, we obtain%
\begin{align*}
&  a_{\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{\left(  x_{w\left(  i\right)  }^{-1}\right)  ^{n-i}%
}_{=x_{w\left(  i\right)  }^{-\left(  n-i\right)  }=x_{w\left(  i\right)
}^{i-n}}=\sum_{w\in\Symm_n}\operatorname{sgn}%
\left(  w\right)  \underbrace{\prod_{i=1}^{n}x_{w\left(  i\right)  }^{i-n}%
}_{\substack{=\prod_{i=1}^{n}x_{w\left(  n+1-i\right)  }^{\left(
n+1-i\right)  -n}\\\text{(here, we have substituted}\\n+1-i\text{ for }i\text{
in the product)}}}\\
&  =\sum_{w\in\Symm_n}\underbrace{\operatorname{sgn}\left(  w\right)
}_{=\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }%
\operatorname*{sgn}\left(  w\right)  \operatorname*{sgn}\left(  w_{\circ
}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)  }^{\left(
n+1-i\right)  -n}}_{=x_{w\left(  n+1-i\right)  }^{-i+1}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\underbrace{\operatorname*{sgn}\left(  w\right)
\operatorname*{sgn}\left(  w_{\circ}\right)  }_{=\operatorname*{sgn}\left(
w\circ w_{\circ}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)
}^{-i+1}}_{\substack{=x_{\left(  w\circ w_{\circ}\right)  \left(  i\right)
}^{-i+1}\\\text{(since }w\left(  n+1-i\right)  =\left(  w\circ w_{\circ
}\right)  \left(  i\right)  \\\text{(by (\ref{sol.liri.180.d.wnach})))}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\operatorname*{sgn}\left(  w\circ w_{\circ}\right)  \prod_{i=1}%
^{n}x_{\left(  w\circ w_{\circ}\right)  \left(  i\right)  }^{-i+1}=\sum
_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)
}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{-i+1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }w\text{ for }w\circ w_{\circ}\text{ in the sum}\\
\text{(since the map }\Symm_n\rightarrow\Symm_n,\ w\mapsto
w\circ w_{\circ}\text{ is a bijection)}%
\end{array}
\right) \\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}%
^{n}\underbrace{x_{w\left(  i\right)  }^{-i+1}}_{\substack{=x_{w\left(
i\right)  }^{\left(  n-i\right)  +\left(  1-n\right)  }\\=x_{w\left(
i\right)  }^{n-i}x_{w\left(  i\right)  }^{1-n}}}=\dfrac{1}{\operatorname*{sgn}%
\left(  w_{\circ}\right)  }\sum_{w\in\Symm_n}\operatorname*{sgn}%
\left(  w\right)  \underbrace{\prod_{i=1}^{n}\left(  x_{w\left(  i\right)
}^{n-i}x_{w\left(  i\right)  }^{1-n}\right)  }_{=\left(  \prod_{i=1}%
^{n}x_{w\left(  i\right)  }^{n-i}\right)  \left(  \prod_{i=1}^{n}x_{w\left(
i\right)  }^{1-n}\right)  }\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}\right)  \underbrace{\left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{1-n}\right)  }_{\substack{=\prod
_{i=1}^{n}x_{i}^{1-n}\\\text{(here, we have substituted }i\text{ for }w\left(
i\right)  \text{ in the product}\\\text{(since }w\text{ is a permutation of
}\left\{  1,2,\ldots,n\right\}  \text{))}}}\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}\right)  \left(  \prod_{i=1}%
^{n}x_{i}^{1-n}\right)  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\underbrace{\left(  \prod_{i=1}^{n}x_{i}^{1-n}\right)  }_{=\left(
\prod_{i=1}^{n}x_{i}\right)  ^{1-n}}\underbrace{\sum_{w\in\Symm_n%
}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{n-i}}_{\substack{=a_{\rho}\\\text{(by (\ref{sol.liri.180.d.a_rho}))}}}\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\left(
\prod_{i=1}^{n}x_{i}\right)  ^{1-n}a_{\rho}.
\end{align*}
Multiplying both sides of this equality by $\operatorname*{sgn}\left(
w_{\circ}\right)  $, we obtain%
\begin{equation}
\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot a_{\rho}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  =\left(  \prod_{i=1}^{n}%
x_{i}\right)  ^{1-n}a_{\rho}. \label{sol.liri.180.d.a_rho-comes-around}%
\end{equation}


Now, (\ref{sol.liri.180.d.s_lambda-check-as-quotient}) becomes%
\begin{align}
a_{\rho}\cdot s_{\lambda^{\vee}}\left(  \xx\right)   &  =a_{\lambda
^{\vee}+\rho} \nonumber\\
&=\left(  \prod_{i=1}^{n}x_{i}\right)  ^{n+m-1}\cdot
\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot\underbrace{a_{\lambda+\rho
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  }_{\substack{=a_{\rho
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \cdot s_{\lambda
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \\\text{(by
(\ref{sol.liri.180.d.1}))}}} \nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.d.a_lambdaplusrho-comes-around})}\right) \nonumber\\
&  =\left(  \prod_{i=1}^{n}x_{i}\right)  ^{n+m-1}\cdot
\underbrace{\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot a_{\rho}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  }_{\substack{=\left(
\prod_{i=1}^{n}x_{i}\right)  ^{1-n}a_{\rho}\\\text{(by
(\ref{sol.liri.180.d.a_rho-comes-around}))}}}\cdot s_{\lambda}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \nonumber\\
&  =\underbrace{\left(  \prod_{i=1}^{n}x_{i}\right)  ^{n+m-1}\cdot\left(
\prod_{i=1}^{n}x_{i}\right)  ^{1-n}}_{=\left(  \prod_{i=1}^{n}x_{i}\right)
^{\left(  n+m-1\right)  +\left(  1-n\right)  }=\left(  \prod_{i=1}^{n}%
x_{i}\right)  ^{m}}a_{\rho}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right) \nonumber\\
&  =\left(  \underbrace{\prod_{i=1}^{n}x_{i}}_{=x_{1}x_{2}\cdots x_{n}%
}\right)  ^{m}a_{\rho}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{n}^{-1}\right)  \nonumber\\
&=\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{m}%
a_{\rho}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}%
^{-1}\right) \nonumber\\
&  =a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  .
\label{sol.liri.180.d.almostthere}%
\end{align}


But\footnote{In the following, a \emph{non-zero-divisor} in a commutative
ring $B$ means an element $b\in B$ such that every element $c\in B$ satisfying
$bc=0$ must satisfy $c=0$.}
\begin{equation}
a_{\rho}\text{ is a non-zero-divisor in the ring }\kk\left[
x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  .
\label{sol.liri.180.d.nonzerodiv}%
\end{equation}
\footnote{\textit{Proof.} We know (from a footnote in Corollary
\ref{cor.stembridge.slambda}) that $a_{\rho}$ is not a zero-divisor in the
ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $. In other words,
$a_{\rho}$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $.
\par
Now, let $b$ be any element of the ring $\kk\left[  x_{1},x_{2}%
,\ldots,x_{n}\right]  $ such that $b$ is a non-zero-divisor in the ring
$\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $. We shall show that $b$
is a non-zero-divisor in the ring $\kk\left[  x_{1},x_{2},\ldots
,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ as well.
\par
Indeed, let $c\in\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ be such that $bc=0$. It is known that
every element of $\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ has the form $\dfrac{p}{x_{1}^{g_{1}%
}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}$ for some $p\in\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $ and some $\left(  g_{1},g_{2},\ldots
,g_{n}\right)  \in\NN^n$. So let us write $c$ in this form, and
consider the corresponding $p$ and $\left(  g_{1},g_{2},\ldots,g_{n}\right)
$. We thus have $c=\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}$.
Now, $bc=0$. This rewrites as $b\cdot\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}%
}\cdots x_{n}^{g_{n}}}=0$ (since $c=\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots
x_{n}^{g_{n}}}$). Multiplying both sides of this equality by $x_{1}^{g_{1}%
}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}$, we obtain $bp=0$. This equality holds in
the ring $\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right]  $, and thus also in the ring $\kk%
\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ (since $\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $ is a subring of $\kk\left[
x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $).
Thus, $p=0$ (since $b$ is a non-zero-divisor in the ring $\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $, and since $p\in\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $). Hence,
\begin{align*}
c  &  =\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}=\dfrac
{0}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }p=0\right) \\
&  =0.
\end{align*}
\par
Now, let us forget that we fixed $c$. We thus have proven that every element
$c\in\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{n}^{-1}\right]  $ satisfying $bc=0$ must satisfy $c=0$. In other
words, $b$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $.
\par
Now, let us forget that we fixed $b$. We thus have proven that if $b$ is any
element of the ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ such
that $b$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $, then $b$ is a non-zero-divisor in the ring
$\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right]  $ as well. Applying this to $b=a_{\rho}$, we conclude that
$a_{\rho}$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ (since
we know that $a_{\rho}$ is a non-zero-divisor in the ring $\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $). This proves
(\ref{sol.liri.180.d.nonzerodiv}).} Hence, we can cancel the factor $a_{\rho}$
from the equality (\ref{sol.liri.180.d.almostthere}). As a result, we obtain
$s_{\lambda^{\vee}}\left(  \xx\right)  =\left(  x_{1}x_{2}\cdots
x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right)  $. Compared with $s_{\lambda^{\vee}}\left(  \xx%
\right)  =s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  $ (since
$\xx=\left(  x_{1},x_{2},\ldots,x_{k}\right)  $), this yields%
\[
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}%
x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right)  =\left(  x_{1}x_{2}\cdots x_{k}\right)
^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\]
(since $n=k$). This solves Exercise \ref{exe.liri.180}(d).

\textit{Second solution to Exercise \ref{exe.liri.180}(d) (sketched):} Let us
give a more combinatorial solution now. In the following, if $\alpha$ is a
partition, then an \emph{$\left(  \alpha,k\right)  $-CST} will mean a
column-strict tableau $T$ of shape $\alpha$ such that all entries of $T$
belong to $\left\{  1,2,\ldots,k\right\}  $.

We have%
\[
\underbrace{s_{\lambda^{\vee}}}_{=s_{\lambda^{\vee}/\varnothing}}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =s_{\lambda^{\vee}/\varnothing}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda^{\vee}/\varnothing\text{;}\\\text{all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,\ldots,k\right\}
}}\mathbf{x}^{\cont \left(  T\right)  }%
\]
(by Exercise \ref{exe.schur.finite}(a), applied to $k$, $\lambda^{\vee}$ and
$\varnothing$ instead of $n$, $\lambda$ and $\mu$). This rewrites as%
\begin{equation}
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\sum_{T\text{ is a
}\left(  \lambda^{\vee},k\right)  \text{-CST}}\mathbf{x}^{\cont
\left(  T\right)  } \label{sol.liri.180.d.alternative1}%
\end{equation}
(because the column-strict tableaux $T$ of shape $\lambda^{\vee}/\varnothing$
such that all entries of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $ are
precisely the $\left(  \lambda^{\vee},k\right)  $-CSTs).

On the other hand,%
\[
\underbrace{s_{\lambda}}_{=s_{\lambda/\varnothing}}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  =s_{\lambda/\varnothing}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\varnothing\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\xx%
^{\cont \left(  T\right)  }%
\]
(by Exercise \ref{exe.schur.finite}(a), applied to $k$ and $\varnothing$
instead of $n$ and $\mu$). Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$,
$x_{k}^{-1}$ for $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ on both sides of this
equality, we obtain%
\begin{equation}
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,k\right\}  }}\left(  \mathbf{x}^{-1}\right)
^{\cont \left(  T\right)  }, \label{sol.liri.180.d.alternative2a}%
\end{equation}
where $\left(  \mathbf{x}^{-1}\right)  ^{\cont \left(  T\right)
}$ is defined as $\prod_{i\geq1}\left(  x_{i}^{-1}\right)  ^{\left\vert
T^{-1}\left(  i\right)  \right\vert }$ for any column-strict tableau $T$. The
equality (\ref{sol.liri.180.d.alternative2a}) rewrites as
\[
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
=\sum_{T\text{ is a }\left(  \lambda,k\right)  \text{-CST}}\left(
\mathbf{x}^{-1}\right)  ^{\cont \left(  T\right)  }%
\]
(since the column-strict tableaux $T$ of shape $\lambda/\varnothing$ such that
all entries of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $ are precisely
the $\left(  \lambda,k\right)  $-CSTs). Hence,%
\begin{align}
\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)   &  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m}\cdot\sum_{T\text{ is a }\left(  \lambda,k\right)
\text{-CST}}\left(  \mathbf{x}^{-1}\right)  ^{\cont \left(
T\right)  }\nonumber\\
&  =\sum_{T\text{ is a }\left(  \lambda,k\right)  \text{-CST}}\left(
x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \mathbf{x}^{-1}\right)
^{\cont \left(  T\right)  }. \label{sol.liri.180.d.alternative2}%
\end{align}


We need to prove that%
\[
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}%
x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  .
\]
Due to (\ref{sol.liri.180.d.alternative1}) and
(\ref{sol.liri.180.d.alternative2}), this rewrites as%
\begin{equation}
\sum_{T\text{ is a }\left(  \lambda^{\vee},k\right)  \text{-CST}}%
\mathbf{x}^{\cont \left(  T\right)  }=\sum_{T\text{ is a }\left(
\lambda,k\right)  \text{-CST}}\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}%
\cdot\left(  \mathbf{x}^{-1}\right)  ^{\cont \left(  T\right)  }.
\label{sol.liri.180.d.alternative4}%
\end{equation}
So it remains to prove (\ref{sol.liri.180.d.alternative4}).

In order to prove (\ref{sol.liri.180.d.alternative4}), it is clearly
sufficient to construct a bijection%
\[
\Omega:\left(  \text{the set of all }\left(  \lambda,k\right)  \text{-CSTs}%
\right)  \rightarrow\left(  \text{the set of all }\left(  \lambda^{\vee
},k\right)  \text{-CSTs}\right)
\]
with the property that every $\left(  \lambda,k\right)  $-CST $T$ satisfies%
\begin{equation}
\mathbf{x}^{\cont \left(  \Omega\left(  T\right)  \right)
}=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \xx%
^{-1}\right)  ^{\cont \left(  T\right)  }.
\label{sol.liri.180.d.alternative.condition-on-Phi}%
\end{equation}


Let us construct such a $\Omega$ now. We begin by doing some elementary combinatorics.

We define a relation $\leq_{\#}$ on the set of all subsets of $\left\{  1,2,
\ldots,k\right\}  $ as follows:

\begin{definition}
Let $k$ be a nonnegative integer. Let $\left[  k\right]  =\left\{  1,2,
\ldots,k\right\}  $. Let $I$ and $J$ be two subsets of $\left[  k\right]  $.
We say that $I\leq_{\#}J$ if the following two properties hold:

-- We have $\left\vert I\right\vert \geq\left\vert J\right\vert $.

-- For every $r\in\left\{  1,2, \ldots,\left\vert J\right\vert \right\}  $,
the $r$-th smallest element of $I$ is $\leq$ to the $r$-th smallest element of
$J$.
\end{definition}

We notice that this relation $\leq_{\#}$ is the less-or-equal relation of a
partial order (as follows easily from the definition); but we will not have
any use for this fact. Instead, we need a symmetry property:

\begin{proposition}
\label{prop.sol.liri.180.d.complement}Let $k$ be a nonnegative integer. Let
$\left[  k\right]  =\left\{  1,2,\ldots,k\right\}  $. Let $I$ and $J$ be two
subsets of $\left[  k\right]  $.

\begin{enumerate}
\item[(a)] We have $I\leq_{\#}J$ if and only if $\left[  k\right]  \setminus
J\leq_{\#}\left[  k\right]  \setminus I$.

\item[(b)] For every $\ell\in\left[  k\right]  $ and $S\subset\left[
k\right]  $, let $\alpha_{S}\left(  \ell\right)  $ denote the number
$\left\vert \left\{  s\in S\ \mid\ s\leq\ell\right\}  \right\vert $. Then,%
\begin{equation}
I\leq_{\#}J\text{ holds if and only if every }\ell\in\left[  k\right]  \text{
satisfies }\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  .
\label{pf.prop.sol.liri.180.d.complement.maj}%
\end{equation}

\end{enumerate}
\end{proposition}

\textit{Proof of Proposition \ref{prop.sol.liri.180.d.complement}.} (b)
\textit{Proof of (\ref{pf.prop.sol.liri.180.d.complement.maj}):}
$\Longrightarrow:$ Assume that $I\leq_{\#}J$. In other words, the following
two properties hold:

\textit{Property $\alpha$:} We have $\left\vert I\right\vert
\geq\left\vert J\right\vert $.

\textit{Property $\beta$:} For every $r\in\left\{  1,2,
\ldots,\left\vert J\right\vert \right\}  $, the $r$-th smallest element of $I$
is $\leq$ to the $r$-th smallest element of $J$.

Now, let $\ell\in\left[  k\right]  $. Then, we need to show that $\alpha
_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  $. Since this is
obvious if $\alpha_{J}\left(  \ell\right)  =0$ (because $\alpha_{I}\left(
\ell\right)  \geq0$), we can WLOG assume that $\alpha_{J}\left(  \ell\right)
\neq0$. Assume this. Thus, $\alpha_{J}\left(  \ell\right)  \geq1$. Also,
$\alpha_{J}\left(  \ell\right)  =\left\vert \underbrace{\left\{  s\in
J\ \mid\ s\leq\ell\right\}  }_{\subset J}\right\vert \leq\left\vert
J\right\vert \leq\left\vert I\right\vert $ (since $\left\vert I\right\vert
\geq\left\vert J\right\vert $). Hence, both the $\alpha_{J}\left(
\ell\right)  $-th smallest element of $J$ and the $\alpha_{J}\left(
\ell\right)  $-th smallest element of $I$ are well-defined.

Since $\alpha_{J}\left(  \ell\right)  =\left\vert \left\{  s\in J\ \mid
\ s\leq\ell\right\}  \right\vert $, we know that the elements of $J$ which are
$\leq\ell$ are precisely the $\alpha_{J}\left(  \ell\right)  $ smallest
elements of $J$. Thus,%
\[
\left(  \text{the }\alpha_{J}\left(  \ell\right)  \text{-th smallest element
of }J\right)  =\left(  \text{the largest element of }J\text{ which is }%
\leq\ell\right)  .
\]
But by Property $\beta$ (applied to $r=\alpha_{J}\left(  \ell\right)  $), we
have%
\begin{align*}
\left(  \text{the }\alpha_{J}\left(  \ell\right)  \text{-th smallest element
of }I\right)   &  \leq\left(  \text{the }\alpha_{J}\left(  \ell\right)
\text{-th smallest element of }J\right) \\
&  =\left(  \text{the largest element of }J\text{ which is }\leq\ell\right)
\leq\ell.
\end{align*}
Hence, there are at least $\alpha_{J}\left(  \ell\right)  $ elements of $I$
which are $\leq\ell$ (namely, the $\alpha_{J}\left(  \ell\right)  $ smallest
ones). In other words, $\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}
\right\vert \geq\alpha_{J}\left(  \ell\right)  $. Now, $\alpha_{I}\left(
\ell\right)  =\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}  \right\vert
\geq\alpha_{J}\left(  \ell\right)  $. We thus have proven the $\Longrightarrow
$ direction of (\ref{pf.prop.sol.liri.180.d.complement.maj}).

$\Longleftarrow:$ Assume that every $\ell\in\left[  k\right]  $ satisfies
$\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  $. We need
to prove that $I\leq_{\#}J$. In other words, we need to prove that the
following two properties hold:

\textit{Property $\alpha$:} We have $\left\vert I\right\vert
\geq\left\vert J\right\vert $.

\textit{Property $\beta$:} For every $r\in\left\{  1,2,
\ldots,\left\vert J\right\vert \right\}  $, the $r$-th smallest element of $I$
is $\leq$ to the $r$-th smallest element of $J$.

First of all, $\left\{  s\in I\ \mid\ s\leq k\right\}  =I$ (since every $s\in
I$ satisfies $s\leq k$), and the definition of $\alpha_{I}\left(  k\right)  $
yields $\alpha_{I}\left(  k\right)  =\left\vert \underbrace{\left\{  s\in
I\ \mid\ s\leq k\right\}  }_{=I}\right\vert =\left\vert I\right\vert $.
Similarly, $\alpha_{J}\left(  k\right)  =\left\vert J\right\vert $. Applying
$\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  $ to
$\ell=k$, we obtain $\alpha_{I}\left(  k\right)  \geq\alpha_{J}\left(
k\right)  $, so that $\left\vert I\right\vert =\alpha_{I}\left(  k\right)
\geq\alpha_{J}\left(  k\right)  =\left\vert J\right\vert $, and thus Property
$\alpha$ is proven.

Now, let $r\in\left\{  1,2, \ldots,\left\vert J\right\vert \right\}  $. The
$r$-th smallest element of $I$ and the $r$-th smallest element of $J$ are then
well-defined (because of $r\leq\left\vert J\right\vert \leq\left\vert
I\right\vert $). Let $\ell$ be the $r$-th smallest element of $J$. Then,
$\left\{  s\in J\ \mid\ s\leq\ell\right\}  $ is the set consisting of the $r$
smallest elements of $J$, so that $\left\vert \left\{  s\in J\ \mid\ s\leq
\ell\right\}  \right\vert =r$. Now, $\alpha_{J}\left(  \ell\right)
=\left\vert \left\{  s\in J\ \mid\ s\leq\ell\right\}  \right\vert =r$.

But $\alpha_{I}\left(  \ell\right)  =\left\vert \left\{  s\in I\ \mid
\ s\leq\ell\right\}  \right\vert $, so that%
\[
\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}  \right\vert =\alpha
_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  =r.
\]
In other words, there exist at least $r$ elements of $I$ which are $\leq\ell$.
Hence, the $r$-th smallest element of $I$ must be $\leq\ell$. Since $\ell$ is
the $r$-th smallest element of $J$, this rewrites as follows: The $r$-th
smallest element of $I$ is $\leq$ to the $r$-th smallest element of $J$. Thus,
Property $\beta$ holds. Now we know that both Properties $\alpha$ and $\beta$
hold. Hence, $I\leq_{\#}J$ holds (which, as we know, is equivalent to the
conjunction of said properties). This proves the $\Longleftarrow$ direction of
(\ref{pf.prop.sol.liri.180.d.complement.maj}). Thus,
(\ref{pf.prop.sol.liri.180.d.complement.maj}) is proven. In other words,
Proposition \ref{prop.sol.liri.180.d.complement}(b) is proven.

(a) For every $\ell\in\left[  k\right]  $ and $S\subset\left[  k\right]  $,
let $\alpha_{S}\left(  \ell\right)  $ denote the number $\left\vert \left\{
s\in S\ \mid\ s\leq\ell\right\}  \right\vert $. Thus, every $\ell\in\left[
k\right]  $ satisfies%
\begin{align*}
\alpha_{I}\left(  \ell\right)  +\alpha_{\left[  k\right]  \setminus I}\left(
\ell\right)   &  =\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}
\right\vert +\left\vert \left\{  s\in\left[  k\right]  \setminus
I\ \mid\ s\leq\ell\right\}  \right\vert \\
&  =\left\vert \left\{  s\in\underbrace{I\cup\left(  \left[  k\right]
\setminus I\right)  }_{=\left[  k\right]  }\ \mid\ s\leq\ell\right\}
\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since }I\text{ and }\left[
k\right]  \setminus I\text{ are disjoint}\right) \\
&  =\left\vert \left\{  s\in\left[  k\right]  \ \mid\ s\leq\ell\right\}
\right\vert =\ell,
\end{align*}
so that $\alpha_{\left[  k\right]  \setminus I}\left(  \ell\right)
=\ell-\alpha_{I}\left(  \ell\right)  $. Similarly, every $\ell\in\left[
k\right]  $ satisfies $\alpha_{\left[  k\right]  \setminus J}\left(
\ell\right)  =\ell-\alpha_{J}\left(  \ell\right)  $.

Applying (\ref{pf.prop.sol.liri.180.d.complement.maj}) to $\left[  k\right]
\setminus J$ and $\left[  k\right]  \setminus I$ in lieu of $I$ and $J$, we
obtain that%
\begin{equation}
\left[  k\right]  \setminus J\leq_{\#}\left[  k\right]  \setminus I\text{
holds if and only if every }\ell\in\left[  k\right]  \text{ satisfies }%
\alpha_{\left[  k\right]  \setminus J}\left(  \ell\right)  \geq\alpha_{\left[
k\right]  \setminus I}\left(  \ell\right)  .
\label{pf.prop.sol.liri.180.d.complement.maj.comp}%
\end{equation}


Now, we have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  \left[  k\right]  \setminus J\leq_{\#}\left[  k\right]  \setminus
I\right) \\
&  \Longleftrightarrow\ \left(  \text{every }\ell\in\left[  k\right]  \text{
satisfies }\underbrace{\alpha_{\left[  k\right]  \setminus J}\left(
\ell\right)  }_{=\ell-\alpha_{J}\left(  \ell\right)  }\geq\underbrace{\alpha
_{\left[  k\right]  \setminus I}\left(  \ell\right)  }_{=\ell-\alpha
_{I}\left(  \ell\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.liri.180.d.complement.maj.comp})}\right) \\
&  \Longleftrightarrow\ \left(  \text{every }\ell\in\left[  k\right]  \text{
satisfies }\ell-\alpha_{J}\left(  \ell\right)  \geq\ell-\alpha_{I}\left(
\ell\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }\ell\in\left[  k\right]  \text{
satisfies }\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)
\right) \\
&  \Longleftrightarrow\ \left(  I\leq_{\#}J\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.liri.180.d.complement.maj})}\right)  .
\end{align*}
This proves Proposition \ref{prop.sol.liri.180.d.complement}(a).

Returning to the solution of Exercise \ref{exe.liri.180}(d), we now define
some more notations.

If $T$ is a column-strict tableau and $i$ is an integer, then the
\emph{$i$-th set column} of $T$ will mean the set of the entries in the
$i$-th column of $T$. Notice that the cardinality of the $i$-th set column of
$T$ is the length of the $i$-th column of the shape of $T$ (since every column
of a column-strict tableau has all its entries distinct), and that the $i$-th
column of $T$ can be uniquely reconstructed from the $i$-th set column of $T$
(because the order of the entries in a column can only be increasing).

For every subset $S$ of $\left[  k\right]  $, we define $\xx_{S}$ to be
the monomial $\prod_{s\in S}x_{s}$ in $\kk\left[  x_{1},x_{2}%
,\ldots,x_{k}\right]  $. If $\left(  S_{1},S_{2},\ldots,S_{m}\right)  $ is an
$m$-tuple of subsets of $\left[  k\right]  $, then we set $\xx_{\left(
S_{1},S_{2},\ldots,S_{m}\right)  }=\prod_{i=1}^{m}\xx_{S_{i}}$.

Let $\lambda^{t}$ denote the conjugate of the partition $\lambda$. Then,
$\left(  \lambda^{t}\right)  _{i}$ is the length of the $i$-th column of the
Ferrers diagram of $\lambda$ for every $i\in\left\{  1,2,\ldots,m\right\}  $.

Let $A\left(  \lambda\right)  $ denote the set of all $m$-tuples $\left(
I_{1},I_{2},\ldots,I_{m}\right)  $ of subsets of $\left[  k\right]  $
satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ and $\left(
\left\vert I_{i}\right\vert =\left(  \lambda^{t}\right)  _{i}\text{ for every
}i\in\left\{  1,2,\ldots,m\right\}  \right)  $. Note that we denote it by
$A\left(  \lambda\right)  $ to stress its dependency on $\lambda$.

Let $\left(  \lambda^{\vee}\right)  ^{t}$ denote the conjugate of the
partition $\lambda^{\vee}$. It is easy to see that%
\begin{equation}
\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}=k-\left(  \lambda
^{t}\right)  _{m+1-i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,\ldots,m\right\}  . \label{sol.liri.180.d.conjugate-of-check}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.d.conjugate-of-check}):} Fix
$i\in\left\{  1,2,\ldots,m\right\}  $. We recall that $\lambda^{\vee}=\left(
m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  $. Hence,
\begin{equation}
\left(  \lambda^{\vee}\right)  _{j}=m-\lambda_{k+1-j}%
\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{  1,2,\ldots,k\right\}  .
\label{sol.liri.180.d.conjugate-of-check.1}%
\end{equation}
Also, $\ell\left(  \lambda^{\vee}\right)  \leq k$ (since $\lambda^{\vee
}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  $), so
that every positive integer $j>k$ satisfies $\left(  \lambda^{\vee}\right)
_{j}=0$. Thus, every positive integer $j$ satisfying $\left(  \lambda^{\vee
}\right)  _{j}\geq i$ must belong to $\left\{  1,2,\ldots,k\right\}  $ (since
otherwise, this $j$ would satisfy $j>k$, and thus $\left(  \lambda^{\vee
}\right)  _{j}=0$, which would contradict $\left(  \lambda^{\vee}\right)
_{j}\geq i>0$). Hence,%
\begin{align*}
\left\{  j\ \mid\ \left(  \lambda^{\vee}\right)  _{j}\geq i\right\}   &
=\left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \underbrace{\left(
\lambda^{\vee}\right)  _{j}}_{\substack{=m-\lambda_{k+1-j}\\\text{(by
(\ref{sol.liri.180.d.conjugate-of-check.1}))}}}\geq i\right\}  =\left\{
j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \underbrace{m-\lambda_{k+1-j}\geq
i}_{\substack{\text{this is equivalent to }\lambda_{k+1-j}\leq m-i\text{, and
thus}\\\text{also equivalent to }\lambda_{k+1-j}<m-i+1\\\text{(since }%
\lambda_{k+1-j}\text{ and }m-i\text{ are integers)}}}\right\} \\
&  =\left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}%
<m-i+1\right\} \\
&  =\left\{  1,2,\ldots,k\right\}  \setminus\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  .
\end{align*}
But applying (\ref{eq.partition.transpose}) to $\lambda^{\vee}$ instead of
$\lambda$, we obtain%
\begin{align}
\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}  &  =\left\vert
\underbrace{\left\{  j\ \mid\ \left(  \lambda^{\vee}\right)  _{j}\geq
i\right\}  }_{=\left\{  1,2,\ldots,k\right\}  \setminus\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  }\right\vert
=\left\vert \left\{  1,2,\ldots,k\right\}  \setminus\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  \right\vert
\nonumber\\
&  =\underbrace{\left\vert \left\{  1,2,\ldots,k\right\}  \right\vert }%
_{=k}-\underbrace{\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}
\ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  \right\vert }%
_{\substack{=\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}
\ \mid\ \lambda_{j}\geq m-i+1\right\}  \right\vert \\\text{(here, we have
substituted }j\text{ for }k+1-j\text{)}}}\nonumber\\
&  =k-\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid
\ \lambda_{j}\geq m-i+1\right\}  \right\vert .
\label{sol.liri.180.d.conjugate-of-check.2}%
\end{align}
\par
But $\ell\left(  \lambda\right)  \leq k$, so that every positive integer $j>k$
satisfies $\lambda_{j}=0$. Thus, every positive integer $j$ satisfying
$\lambda_{j}\geq m-i+1$ must belong to $\left\{  1,2,\ldots,k\right\}  $
(since otherwise, this $j$ would satisfy $j>k$, and thus $\lambda_{j}=0$,
which would contradict $\lambda_{j}\geq m-i+1>0$). Hence,%
\[
\left\{  j\ \mid\ \lambda_{j}\geq m-i+1\right\}  =\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{j}\geq m-i+1\right\}  .
\]
Now, (\ref{eq.partition.transpose}) (applied to $m-i+1$ instead of $i$) yields%
\begin{equation}
\left(  \lambda^{t}\right)  _{m-i+1}=\left\vert \underbrace{\left\{
j\ \mid\ \lambda_{j}\geq m-i+1\right\}  }_{=\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{j}\geq m-i+1\right\}  }\right\vert
=\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \lambda
_{j}\geq m-i+1\right\}  \right\vert .
\label{sol.liri.180.d.conjugate-of-check.3}%
\end{equation}
Thus, (\ref{sol.liri.180.d.conjugate-of-check.2}) becomes%
\[
\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}%
=k-\underbrace{\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}
\ \mid\ \lambda_{j}\geq m-i+1\right\}  \right\vert }_{\substack{=\left(
\lambda^{t}\right)  _{m-i+1}\\\text{(by
(\ref{sol.liri.180.d.conjugate-of-check.3}))}}}=k-\left(  \lambda^{t}\right)
_{m-i+1}=k-\left(  \lambda^{t}\right)  _{m+1-i}.
\]
This proves (\ref{sol.liri.180.d.conjugate-of-check}).}

It is easy to see (from the definition of $\lambda^{\vee}$) that
$\lambda^{\vee}$ is a partition satisfying $\ell\left(  \lambda^{\vee}\right)
\leq k$, and all parts of $\lambda^{\vee}$ are $\leq m$. Hence, we can define
a set $A\left(  \lambda^{\vee}\right)  $ in the same way as we defined the set
$A\left(  \lambda\right)  $ (but with every $\lambda$ replaced by
$\lambda^{\vee}$). Explicitly, $A\left(  \lambda^{\vee}\right)  $ is the set
of all $m$-tuples $\left(  I_{1},I_{2},\ldots,I_{m}\right)  $ of subsets of
$\left[  k\right]  $ satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq
_{\#}I_{m}$ and $\left(  \left\vert I_{i}\right\vert =\left(  \left(
\lambda^{\vee}\right)  ^{t}\right)  _{i}\text{ for every }i\in\left\{
1,2,\ldots,m\right\}  \right)  $.

Define a map
\[
\varphi\left(  \lambda\right)  :\left(  \text{the set of all }\left(
\lambda,k\right)  \text{-CSTs}\right)  \rightarrow A\left(  \lambda\right)
\]
by sending every $\left(  \lambda,k\right)  $-CST $T$ to the $m$-tuple whose
$i$-th entry is the $i$-th set column of $T$. This map is
well-defined\footnote{Indeed, if $T$ is an $\left(  \lambda,k\right)  $-CST,
then the $m$-tuple whose $i$-th entry is the $i$-th set column of $T$ belongs
to $A\left(  \lambda\right)  $ (because if we denote this $m$-tuple by
$\left(  I_{1},I_{2},\ldots,I_{m}\right)  $, then%
\begin{align*}
\left\vert I_{i}\right\vert  &  =\left\vert \left(  \text{the }i\text{-th set
column of }T\right)  \right\vert =\left(  \text{the length of the }i\text{-th
column of }T\right) \\
&  =\left(  \text{the length of the }i\text{-th column of the Ferrers diagram
of }\lambda\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }T\text{ is a }\left(
\lambda,k\right)  \text{-CST, thus a column-strict tableau of shape }%
\lambda\right) \\
&  = \left(\lambda^t\right)_i ,
\end{align*}
and the fact that the entries of $T$ increase weakly along rows (because $T$
is a column-strict tableau) translates precisely into the inequality chain
$I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$).}, injective\footnote{This
is because the $i$-th column of a $\left(  \lambda,k\right)  $-CST $T$ can be
uniquely reconstructed from the $i$-th set column of $T$ (indeed, the entries
of the $i$-th column of $T$ must be strictly increasing down the column, and
therefore the knowledge of the set of these entries is sufficient to recover
the $i$-th column). Here, we are using the fact that $T$ has at most $m$
columns (since every part of $\lambda$ is $\leq m$).} and
surjective\footnote{Indeed, given any $\left(  I_{1},I_{2},\ldots
,I_{m}\right)  \in A\left(  \lambda\right)  $. Then, $\left(  I_{1}%
,I_{2},\ldots,I_{m}\right)  $ is an $m$-tuple of subsets of $\left[  k\right]
$ satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ and $\left(
\left\vert I_{i}\right\vert =\left(  \lambda^t \right) _i \text{ for every
}i\in\left\{  1,2,\ldots,m\right\}  \right)  $. Now, we can fill in each
column of the Ferrers diagram of $\lambda$ with the entries of the
corresponding set $I_{i}$ in increasing order, and then the resulting filling
is a column-strict tableau (indeed, its entries increase weakly along its rows
(due to $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$)), and more
precisely a $\left(  \lambda,k\right)  $-CST. Our $m$-tuple $\left(
I_{1},I_{2},\ldots,I_{m}\right)  $ is the image of this $\left(
\lambda,k\right)  $-CST under $\varphi\left(  \lambda\right)  $; therefore,
$\left(  I_{1},I_{2},\ldots,I_{m}\right)  $ lies in the image of
$\varphi\left(  \lambda\right)  $. Hence, we have shown that every $\left(
I_{1},I_{2},\ldots,I_{m}\right)  \in A\left(  \lambda\right)  $ lies in the
image of $\varphi\left(  \lambda\right)  $. The map $\varphi\left(
\lambda\right)  $ is thus surjective, qed.}. Hence, $\varphi\left(
\lambda\right)  $ is a bijection. This bijection $\varphi\left(
\lambda\right)  $ furthermore satisfies%
\begin{equation}
\xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(  T\right)
}=\mathbf{x}^{\cont \left(  T\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \lambda,k\right)  \text{-CST }T
\label{sol.liri.180.d.phi1}%
\end{equation}
\footnote{Indeed, it is easy to check that both $\xx_{\left(
\varphi\left(  \lambda\right)  \right)  \left(  T\right)  }$ and
$\mathbf{x}^{\cont \left(  T\right)  }$ are equal to the product
$\prod_{c\text{ is a cell of }T}x_{\left(  \text{entry of }T\text{ in
}c\right)  }$.}. Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$,
$x_{k}^{-1}$ for $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ on both sides of this
equality, we obtain%
\begin{equation}
\left(  \xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(
T\right)  }\right)  ^{-1}=\left(  \mathbf{x}^{-1}\right)
^{\cont \left(  T\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  \lambda,k\right)  \text{-CST }T \label{sol.liri.180.d.phi1inv}%
\end{equation}
(in fact, $\xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(
T\right)  }$ is a monomial, so that substituting $x_{1}^{-1}$, $x_{2}^{-1}$,
$\ldots$, $x_{k}^{-1}$ for $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ transforms it
into its inverse $\left(  \xx_{\left(  \varphi\left(  \lambda\right)
\right)  \left(  T\right)  }\right)  ^{-1}$).

Recall that $\lambda^{\vee}$ is a partition satisfying $\ell\left(
\lambda^{\vee}\right)  \leq k$, and all parts of $\lambda^{\vee}$ are $\leq
m$. Hence, we can define a bijection%
\[
\varphi\left(  \lambda^{\vee}\right)  :\left(  \text{the set of all }\left(
\lambda^{\vee},k\right)  \text{-CSTs}\right)  \rightarrow A\left(
\lambda^{\vee}\right)
\]
in the same way as we defined the bijection $\varphi\left(  \lambda\right)
:\left(  \text{the set of all }\left(  \lambda,k\right)  \text{-CSTs}\right)
\rightarrow A\left(  \lambda\right)  $ (but with $\lambda^{\vee}$ instead of
$\lambda$). This bijection satisfies%
\begin{equation}
\xx_{\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  \left(
T\right)  }=\mathbf{x}^{\cont \left(  T\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \lambda^{\vee},k\right)
\text{-CST }T \label{sol.liri.180.d.phi2}%
\end{equation}
(this can be shown in analogy to (\ref{sol.liri.180.d.phi1})).

Finally, we define a map $\psi\left(  \lambda\right)  :A\left(  \lambda
\right)  \rightarrow A\left(  \lambda^{\vee}\right)  $ as follows: For every
$\left(  I_{1},I_{2},\ldots,I_{m}\right)  \in A\left(  \lambda\right)  $, let%
\[
\left(  \psi\left(  \lambda\right)  \right)  \left(  I_{1},I_{2},\ldots
,I_{m}\right)  =\left(  \left[  k\right]  \setminus I_{m},\left[  k\right]
\setminus I_{m-1},\ldots,\left[  k\right]  \setminus I_{1}\right)  .
\]
This map $\psi\left(  \lambda\right)  $ is
well-defined\footnote{\textit{Proof.} Let $\left(  I_{1},I_{2},\ldots
,I_{m}\right)  \in A\left(  \lambda\right)  $. Then, $\left(  I_{1}%
,I_{2},\ldots,I_{m}\right)  $ is an $m$-tuple of subsets of $\left[  k\right]
$ satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ and $\left(
\left\vert I_{i}\right\vert =\left(  \lambda^{t}\right)  _{i}\text{ for every
}i\in\left\{  1,2,\ldots,m\right\}  \right)  $.
\par
In order to prove the well-definedness of $\psi\left(  \lambda\right)  $, we
need to show that $\left(  \left[  k\right]  \setminus I_{m},\left[  k\right]
\setminus I_{m-1},\ldots,\left[  k\right]  \setminus I_{1}\right)  \in
A\left(  \lambda^{\vee}\right)  $. In other words, we need to prove that
$\left(  \left[  k\right]  \setminus I_{m},\left[  k\right]  \setminus
I_{m-1},\ldots,\left[  k\right]  \setminus I_{1}\right)  $ is an $m$-tuple of
subsets of $\left[  k\right]  $ satisfying $\left[  k\right]  \setminus
I_{m}\leq_{\#}\left[  k\right]  \setminus I_{m-1}\leq_{\#}\cdots\leq
_{\#}\left[  k\right]  \setminus I_{1}$ and $\left(  \left\vert \left[
k\right]  \setminus I_{m+1-i}\right\vert =\left(  \left(  \lambda^{\vee
}\right)  ^{t}\right)  _{i}\text{ for every }i\in\left\{  1,2,\ldots
,m\right\}  \right)  $.
\par
First of all, it is clear that $\left(  \left[  k\right]  \setminus
I_{m},\left[  k\right]  \setminus I_{m-1},\ldots,\left[  k\right]  \setminus
I_{1}\right)  $ is an $m$-tuple of subsets of $\left[  k\right]  $.
Furthermore, $\left[  k\right]  \setminus I_{m}\leq_{\#}\left[  k\right]
\setminus I_{m-1}\leq_{\#}\cdots\leq_{\#}\left[  k\right]  \setminus I_{1}$
follows from $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ (according to
Proposition \ref{prop.sol.liri.180.d.complement}(a)). Thus, it remains to
prove that $\left\vert \left[  k\right]  \setminus I_{m+1-i}\right\vert
=\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}$ for every
$i\in\left\{  1,2,\ldots,m\right\}  $.
\par
So fix some $i\in\left\{  1,2,\ldots,m\right\}  $. Then, $\left\vert
I_{m+1-i}\right\vert =\left(  \lambda^{t}\right)  _{m+1-i}$ (this follows from
the $\left\vert I_{i}\right\vert =\left(  \lambda^{t}\right)  _{i}$ formula,
but applied to $m+1-i$ instead of $i$). Since $I_{m+1-i}\subset\left[
k\right]  $, we have
\[
\left\vert \left[  k\right]  \setminus I_{m+1-i}\right\vert
=\underbrace{\left\vert \left[  k\right]  \right\vert }_{=k}%
-\underbrace{\left\vert I_{m+1-i}\right\vert }_{\substack{=\left(  \lambda
^{t}\right)  _{m+1-i}}}=k-\left(  \lambda^{t}\right)  _{m+1-i}=\left(  \left(
\lambda^{\vee}\right)  ^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.d.conjugate-of-check})}\right)  ,
\]
qed.} and bijective\footnote{Indeed, we can define a map $\psi\left(
\lambda^{\vee}\right)  $ in the same way as we have defined $\psi\left(
\lambda\right)  $ (but with $\lambda^{\vee}$ instead of $\lambda$). It is then
easy to see that the maps $\psi\left(  \lambda\right)  $ and $\psi\left(
\lambda^{\vee}\right)  $ are mutually inverse, so that $\psi\left(
\lambda\right)  $ is bijective, qed.}. This bijective map $\psi\left(
\lambda\right)  $ has the property that%
\begin{equation}
\xx_{\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)
}=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \xx%
_{S}\right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \text{for every }S\in A\left(
\lambda\right)  \label{sol.liri.180.d.psi1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.d.psi1}):} Let $S\in A\left(
\lambda\right)  $. Then, $S$ is an $m$-tuple of subsets of $\left[  k\right]
$. Write $S$ as $S=\left(  I_{1},I_{2},\ldots,I_{m}\right)  $. The definition
of $\xx_{S}$ then yields%
\[
\xx_{S}=\prod_{i=1}^{m}\xx_{I_{i}}.
\]
Hence,%
\begin{equation}
\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \xx_{S}\right)
^{-1}=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \prod_{i=1}%
^{m}\xx_{I_{i}}\right)  ^{-1}=\prod_{i=1}^{m}\dfrac{x_{1}x_{2}\cdots
x_{k}}{\xx_{I_{i}}}. \label{sol.liri.180.d.psi1.pf.0}%
\end{equation}
On the other hand, the definition of $\psi\left(  \lambda\right)  $ yields
$\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)  =\left(
\left[  k\right]  \setminus I_{m},\left[  k\right]  \setminus I_{m-1}%
,\ldots,\left[  k\right]  \setminus I_{1}\right)  $, and thus (by the
definition of $\xx_{\left(  \psi\left(  \lambda\right)  \right)
\left(  S\right)  }$) we have%
\begin{equation}
\xx_{\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)
}=\prod_{i=1}^{m}\xx_{\left[  k\right]  \setminus I_{m+1-i}}%
=\prod_{i=1}^{m}\xx_{\left[  k\right]  \setminus I_{i}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }m+1-i\text{ for
}i\text{ in the product}\right)  . \label{sol.liri.180.d.psi1.pf.1}%
\end{equation}
\par
But we need to prove (\ref{sol.liri.180.d.psi1}). In view of
(\ref{sol.liri.180.d.psi1.pf.0}) and (\ref{sol.liri.180.d.psi1.pf.1}), this
boils down to proving that $\prod_{i=1}^{m}\xx_{\left[  k\right]
\setminus I_{i}}=\prod_{i=1}^{m}\dfrac{x_{1}x_{2}\cdots x_{k}}{\xx%
_{I_{i}}}$. But this will immediately follow if we can prove the identity
$\xx_{\left[  k\right]  \setminus I_{i}}=\dfrac{x_{1}x_{2}\cdots x_{k}%
}{\xx_{I_{i}}}$ for every $i\in\left\{  1,2,\ldots,m\right\}  $. But
the latter identity follows from the (obvious) fact that $\xx_{\left[
k\right]  \setminus S}=\dfrac{x_{1}x_{2}\cdots x_{k}}{\xx_{S}}$ for
every $S\subset\left[  k\right]  $. Thus, (\ref{sol.liri.180.d.psi1}) is
proven.}.

Using the three bijective maps%
\begin{align*}
\varphi\left(  \lambda\right)   &  :\left(  \text{the set of all }\left(
\lambda,k\right)  \text{-CSTs}\right)  \rightarrow A\left(  \lambda\right)
,\\
\psi\left(  \lambda\right)   &  :A\left(  \lambda\right)  \rightarrow A\left(
\lambda^{\vee}\right)  ,\\
\varphi\left(  \lambda^{\vee}\right)   &  :\left(  \text{the set of all
}\left(  \lambda^{\vee},k\right)  \text{-CSTs}\right)  \rightarrow A\left(
\lambda^{\vee}\right)  ,
\end{align*}
we can define a map%
\[
\Omega:\left(  \text{the set of all }\left(  \lambda,k\right)  \text{-CSTs}%
\right)  \rightarrow\left(  \text{the set of all }\left(  \lambda^{\vee
},k\right)  \text{-CSTs}\right)
\]
by $\Omega=\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  ^{-1}%
\circ\left(  \psi\left(  \lambda\right)  \right)  \circ\left(  \varphi\left(
\lambda\right)  \right)  $. Clearly, this $\Omega$ is a bijection. If we can
show that every $\left(  \lambda,k\right)  $-CST $T$ satisfies
(\ref{sol.liri.180.d.alternative.condition-on-Phi}), then
(\ref{sol.liri.180.d.alternative4}) will be proven, and thus Exercise
\ref{exe.liri.180}(d) will be solved again. Hence, all that remains to prove
now is that every $\left(  \lambda,k\right)  $-CST $T$ satisfies
(\ref{sol.liri.180.d.alternative.condition-on-Phi}).

Let $T$ be a $\left(  \lambda,k\right)  $-CST. We have $\Omega=\left(
\varphi\left(  \lambda^{\vee}\right)  \right)  ^{-1}\circ\left(  \psi\left(
\lambda\right)  \right)  \circ\left(  \varphi\left(  \lambda\right)  \right)
$, thus $\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  \circ
\Omega=\left(  \psi\left(  \lambda\right)  \right)  \circ\left(
\varphi\left(  \lambda\right)  \right)  $. Applying (\ref{sol.liri.180.d.phi2}%
) to $\Omega\left(  T\right)  $ instead of $T$, we obtain $\xx_{\left(
\varphi\left(  \lambda^{\vee}\right)  \right)  \left(  \Omega\left(  T\right)
\right)  }=\mathbf{x}^{\cont \left(  \Omega\left(  T\right)
\right)  }$, whence%
\begin{align*}
&  \mathbf{x}^{\cont \left(  \Omega\left(  T\right)  \right)
}=\xx_{\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  \left(
\Omega\left(  T\right)  \right)  }=\xx_{\left(  \psi\left(
\lambda\right)  \right)  \left(  \left(  \varphi\left(  \lambda\right)
\right)  \left(  T\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \varphi\left(
\lambda^{\vee}\right)  \right)  \left(  \Omega\left(  T\right)  \right)
=\underbrace{\left(  \left(  \varphi\left(  \lambda^{\vee}\right)  \right)
\circ\Omega\right)  }_{=\left(  \psi\left(  \lambda\right)  \right)
\circ\left(  \varphi\left(  \lambda\right)  \right)  }\left(  T\right)
=\left(  \left(  \psi\left(  \lambda\right)  \right)  \circ\left(
\varphi\left(  \lambda\right)  \right)  \right)  \left(  T\right)  =\left(
\psi\left(  \lambda\right)  \right)  \left(  \left(  \varphi\left(
\lambda\right)  \right)  \left(  T\right)  \right)  \right) \\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\underbrace{\left(
\xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(  T\right)
}\right)  ^{-1}}_{\substack{=\left(  \mathbf{x}^{-1}\right)
^{\cont \left(  T\right)  }\\\text{(by
(\ref{sol.liri.180.d.phi1inv}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.d.psi1}), applied to }S=\left(  \varphi\left(
\lambda\right)  \right)  \left(  T\right)  \right) \\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \xx%
^{-1}\right)  ^{\cont \left(  T\right)  }.
\end{align*}
Thus, (\ref{sol.liri.180.d.alternative.condition-on-Phi}) holds. We thus have
proven that every $\left(  \lambda,k\right)  $-CST $T$ satisfies
(\ref{sol.liri.180.d.alternative.condition-on-Phi}). Exercise
\ref{exe.liri.180}(d) is thus solved.

(e) \textit{First solution to Exercise \ref{exe.liri.180}(e):} Obviously, the
$k$-tuple $\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  $
is a partition. Let us denote this partition by $\lambda^{\left[  r\right]  }%
$. Then, $\lambda^{\left[  r\right]  }=\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  $ and $\ell\left(  \lambda^{\left[
r\right]  }\right)  \leq k$.

Let $n=k$. We shall use the notations of Section \ref{bialternant-section}; in
particular, we set $\xx=\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ and
$\rho=\left(  n-1,n-2,\ldots,2,1,0\right)  $. Clearly, $\xx=\left(
x_{1},x_{2},\ldots,x_{n}\right)  =\left(  x_{1},x_{2},\ldots,x_{k}\right)  $
(since $n=k$). Notice that $\ell\left(  \lambda\right)  \leq k=n$. Hence, we
can regard $\lambda$ as an element of $\NN^n$; therefore,
$\lambda+\rho$ is a well-defined element of $\NN^n$, and the
alternant $a_{\lambda+\rho}$ is well-defined. Corollary
\ref{cor.stembridge.slambda} yields that $s_{\lambda}\left(  \xx%
\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$, so that%
\begin{equation}
a_{\rho}\cdot s_{\lambda}\left(  \xx\right)  =a_{\lambda+\rho}.
\label{sol.liri.180.e.s_lambda-as-quotient}%
\end{equation}


Applying this equality to $\lambda^{\left[  r\right]  }$ instead of $\lambda$,
we obtain%
\begin{equation}
a_{\rho}\cdot s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)
=a_{\lambda^{\left[  r\right]  }+\rho}
\label{sol.liri.180.e.s_lambdar-as-quotient}%
\end{equation}
(since $\lambda^{\left[  r\right]  }$ is also a partition satisfying
$\ell\left(  \lambda^{\left[  r\right]  }\right)  \leq k=n$).

We have (\ref{sol.liri.180.d.a_lambdaplusrho1}), and every $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \in\NN^n$ satisfies
(\ref{sol.liri.180.d.a_alpha-formula}). (This can be proven just as in the
First solution to Exercise \ref{exe.liri.180}(d).)

We have $\lambda^{\left[  r\right]  }=\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  =\left(  r+\lambda_{1},r+\lambda_{2}%
,\ldots,r+\lambda_{n}\right)  $ (since $k=n$), so that%
\begin{align*}
\underbrace{\lambda^{\left[  r\right]  }}_{=\left(  r+\lambda_{1}%
,r+\lambda_{2},\ldots,r+\lambda_{n}\right)  }+\underbrace{\rho}%
_{\substack{=\left(  n-1,n-2,\ldots,2,1,0\right)  \\=\left(  n-1,n-2,\ldots
,n-n\right)  }}  &  =\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda
_{n}\right)  +\left(  n-1,n-2,\ldots,n-n\right) \\
&  =\left(  \left(  r+\lambda_{1}\right)  +\left(  n-1\right)  ,\left(
r+\lambda_{2}\right)  +\left(  n-2\right)  ,\ldots,\left(  r+\lambda
_{n}\right)  +\left(  n-n\right)  \right)  .
\end{align*}
Hence, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to $\lambda^{\left[
r\right]  }+\rho$ and \newline$\left(  \left(  r+\lambda_{1}\right)  +\left(
n-1\right)  ,\left(  r+\lambda_{2}\right)  +\left(  n-2\right)  ,\ldots
,\left(  r+\lambda_{n}\right)  +\left(  n-n\right)  \right)  $ instead of
$\alpha$ and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $)
yields%
\begin{align*}
&  a_{\lambda^{\left[  r\right]  }+\rho}\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{x_{w\left(  i\right)  }^{\left(  r+\lambda_{i}\right)
+\left(  n-i\right)  }}_{=x_{w\left(  i\right)  }^{r+\left(  \lambda
_{i}+n-i\right)  }=x_{w\left(  i\right)  }^{r}x_{w\left(  i\right)  }%
^{\lambda_{i}+n-i}}\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\prod_{i=1}^{n}\left(  x_{w\left(  i\right)  }^{r}x_{w\left(
i\right)  }^{\lambda_{i}+n-i}\right)  }_{=\left(  \prod_{i=1}^{n}x_{w\left(
i\right)  }^{r}\right)  \left(  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{\lambda_{i}+n-i}\right)  }=\sum_{w\in\Symm_n}\operatorname{sgn}%
\left(  w\right)  \underbrace{\left(  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{r}\right)  }_{\substack{=\prod_{i=1}^{n}x_{i}^{r}\\\text{(here, we have
substituted }i\text{ for}\\w\left(  i\right)  \text{ in the product (since
}w\text{ is}\\\text{a permutation of }\left\{  1,2,\ldots,n\right\}
\text{))}}}\left(  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{\lambda_{i}%
+n-i}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{i}^{r}\right)  \left(  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{\lambda_{i}+n-i}\right)  =\underbrace{\left(  \prod_{i=1}^{n}x_{i}%
^{r}\right)  }_{=x_{1}^{r}x_{2}^{r}\cdots x_{n}^{r}=\left(  x_{1}x_{2}\cdots
x_{n}\right)  ^{r}}\cdot\underbrace{\sum_{w\in\Symm_n}%
\operatorname{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{\lambda_{i}+n-i}}_{\substack{=a_{\lambda+\rho}\\\text{(by
(\ref{sol.liri.180.d.a_lambdaplusrho1}))}}}\\
&  =\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot\underbrace{a_{\lambda
+\rho}}_{\substack{=a_{\rho}\cdot s_{\lambda}\left(  \xx\right)
\\\text{(by (\ref{sol.liri.180.e.s_lambda-as-quotient}))}}}=\left(  x_{1}%
x_{2}\cdots x_{n}\right)  ^{r}\cdot a_{\rho}\cdot s_{\lambda}\left(
\xx\right)  =a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)
^{r}\cdot s_{\lambda}\left(  \xx\right)  .
\end{align*}
Hence, $a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot
s_{\lambda}\left(  \xx\right)  =a_{\lambda^{\left[  r\right]  }+\rho
}=a_{\rho}\cdot s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)  $
(by (\ref{sol.liri.180.e.s_lambda-as-quotient})).

But we know (from a footnote in Corollary \ref{cor.stembridge.slambda}) that
$a_{\rho}$ is not a zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $. In other words, $a_{\rho}$ is a
non-zero-divisor in the ring $\kk\left[  x_{1},x_{2},\ldots
,x_{n}\right]  $. Hence, we can cancel the factor $a_{\rho}$ from the equality
$a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot s_{\lambda
}\left(  \xx\right)  =a_{\rho}\cdot s_{\lambda^{\left[  r\right]  }%
}\left(  \xx\right)  $. As a result, we obtain $\left(  x_{1}%
x_{2}\cdots x_{n}\right)  ^{r}\cdot s_{\lambda}\left(  \xx\right)
=s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)  =s_{\lambda
^{\left[  r\right]  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)  $ (since
$\xx=\left(  x_{1},x_{2},\ldots,x_{k}\right)  $), so that%
\[
s_{\lambda^{\left[  r\right]  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)
=\left(  \underbrace{x_{1}x_{2}\cdots x_{n}}_{\substack{=x_{1}x_{2}\cdots
x_{k}\\\text{(since }n=k\text{)}}}\right)  ^{r}\cdot\underbrace{s_{\lambda
}\left(  \xx\right)  }_{\substack{=s_{\lambda}\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  \\\text{(since }\xx=\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  \text{)}}}=\left(  x_{1}x_{2}\cdots x_{k}\right)
^{r}\cdot s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\]
Since $\lambda^{\left[  r\right]  }=s_{\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  }$, this rewrites as
\[
s_{\left(
r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  }\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\]
This solves Exercise \ref{exe.liri.180}(e).

\textit{Second solution to Exercise \ref{exe.liri.180}(e) (sketched):}
Obviously, the $k$-tuple $\left(  r+\lambda_{1},r+\lambda_{2},\ldots
,r+\lambda_{k}\right)  $ is a partition. Let us denote this partition by
$\lambda^{\left[  r\right]  }$. Then, $\lambda^{\left[  r\right]  }=\left(
r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  $ and $\ell\left(
\lambda^{\left[  r\right]  }\right)  \leq k$.

We have $\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)
=\lambda^{\left[  r\right]  }$, thus $s_{\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  }=s_{\lambda^{\left[  r\right]  }%
}=s_{\lambda^{\left[  r\right]  }/\varnothing}$, hence%
\begin{align}
&  \underbrace{s_{\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda
_{k}\right)  }}_{=s_{\lambda^{\left[  r\right]  }/\varnothing}}\left(
x_{1},x_{2},\ldots,x_{k}\right) \nonumber\\
&  =s_{\lambda^{\left[  r\right]  }/\varnothing}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda^{\left[  r\right]  }/\varnothing
\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\mathbf{x}^{\cont \left(  T\right)
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.schur.finite}(a),
applied to }k\text{, }\lambda^{\left[  r\right]  }\text{ and }\varnothing
\text{ instead of }n\text{, }\lambda\text{ and }\mu\right) \nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda^{\left[  r\right]  }\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\xx%
^{\cont \left(  T\right)  } \label{sol.liri.180.e.alternative1}%
\end{align}
(since column-strict tableaux of shape $\lambda^{\left[  r\right]
}/\varnothing$ are the same as column-strict tableaux of shape $\lambda
^{\left[  r\right]  }$).

But an argument analogous to the one we just used to prove
(\ref{sol.liri.180.e.alternative1}) (but with $\lambda$ in place of
$\lambda^{\left[  r\right]  }$) shows that%
\[
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda\text{;}\\\text{all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,\ldots,k\right\}
}}\mathbf{x}^{\cont \left(  T\right)  }.
\]
Multiplied by $\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}$, this equality
becomes%
\begin{align}
\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)   &  =\left(  x_{1}x_{2}\cdots x_{k}\right)
^{r}\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\mathbf{x}^{\cont \left(  T\right)
}\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}%
\cdot\mathbf{x}^{\cont \left(  T\right)  }.
\label{sol.liri.180.e.alternative2}%
\end{align}


We need to prove that%
\[
s_{\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  }\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots x_{k}\right)
^{r}\cdot s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\]
Due to (\ref{sol.liri.180.e.alternative1}) and
(\ref{sol.liri.180.e.alternative2}), this rewrites as%
\begin{equation}
\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda^{\left[  r\right]  }\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\xx%
^{\cont \left(  T\right)  }=\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda\text{;}\\\text{all entries of
}T\text{ belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\left(
x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot\mathbf{x}^{\cont \left(
T\right)  }. \label{sol.liri.180.e.alternative4}%
\end{equation}
So it remains to prove (\ref{sol.liri.180.e.alternative4}).

In order to prove (\ref{sol.liri.180.e.alternative4}), it is clearly
sufficient to construct a bijection%
\begin{align*}
\Omega &  :\left(  \text{the set of all column-strict tableaux }T\text{ of
shape }\lambda^{\left[  r\right]  }\text{ such that all entries of }T\text{
belong to }\left\{  1,2,\ldots,k\right\}  \right) \\
&  \rightarrow\left(  \text{the set of all column-strict tableaux }T\text{ of
shape }\lambda\text{ such that all entries of }T\text{ belong to }\left\{
1,2,\ldots,k\right\}  \right)
\end{align*}
with the property that every column-strict tableau $T$ of shape
$\lambda^{\left[  r\right]  }$ such that all entries of $T$ belong to
$\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot\xx%
^{\cont \left(  \Omega\left(  T\right)  \right)  }=\xx%
^{\cont \left(  T\right)  }.
\label{sol.liri.180.e.alternative.omega-cond}%
\end{equation}


Constructing such an $\Omega$ is easy: The map $\Omega$ just maps every
column-strict tableau $T$ of shape $\lambda^{\left[  r\right]  }$ to the
tableau of shape $\lambda$ obtained by removing the first $r$ entries of each
row of $T$, and moving all other entries to the left by $r$ cells. To prove
that $\Omega$ is bijective, we need to construct a map inverse to $\Omega$;
this latter map sends every column-strict tableau $T$ of shape $\lambda$ to
the tableau of shape $\lambda^{\left[  r\right]  }$ obtained by moving all
entries of $T$ to the right by $r$ cells, and filling the now-vacant first $r$
columns as follows:%
\[%
\begin{array}
[c]{cccc}%
1 & 1 & \cdots & 1\\
2 & 2 & \cdots & 2\\
\vdots & \vdots & \ddots & \vdots\\
k & k & \cdots & k
\end{array}
.
\]
Proving that this map is well-defined and really inverse to $\Omega$ is left
to the reader\footnote{The main ingredient of this proof is the observation
that if $T$ is a column-strict tableau of shape $\lambda^{\left[  r\right]  }$
such that all entries of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $, then
the first $r$ columns of $T$ must look like this:%
\[%
\begin{array}
[c]{cccc}%
1 & 1 & \cdots & 1\\
2 & 2 & \cdots & 2\\
\vdots & \vdots & \ddots & \vdots\\
k & k & \cdots & k
\end{array}
.
\]
(This is because each of the first $r$ columns of $T$ has length $k$, but the
entries in this column must be strictly increasing from top to bottom, and
knowing that these entries belong to $\left\{  1,2,\ldots,k\right\}  $, we see
that this is only possible if the column has the form $%
\begin{array}
[c]{c}%
1\\
2\\
\vdots\\
k
\end{array}
$.)}. Anyway, we now know that $\Omega$ is bijective, and it is easy to check
that (\ref{sol.liri.180.e.alternative.omega-cond}) holds. This completes the
second solution of Exercise \ref{exe.liri.180}(e).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.180}}

\textit{Solution to Exercise \ref{exe.liri.180}.} Consider the partition
$\left(  m^{k}\right)  =\left(  \underbrace{m,m,\ldots,m}_{k\text{ times}%
}\right)  $.

By our assumption, we know that all parts of $\lambda$ are $\leq m$. Thus,%
\begin{equation}
\lambda_{i}\leq m\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.
\label{sol.liri.180.a.partsleq1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.a.partsleq1}):} Let $i$ be a
positive integer. We want to prove that $\lambda_{i}\leq m$.
\par
If $\lambda_{i}=0$, then $\lambda_{i}=0\leq m$ (since $m\in \NN$).
Hence, if $\lambda_{i}=0$, then $\lambda_{i}\leq m$ is proven. Thus, for the
rest of the proof of $\lambda_{i}\leq m$, we can WLOG assume that we don't
have $\lambda_{i}=0$. Assume this.
\par
The number $\lambda_{i}$ is nonzero (since we don't have $\lambda_{i}=0$).
Hence, $\lambda_{i}$ is a part of $\lambda$. Therefore, $\lambda_{i}$ is $\leq
m$ (since all parts of $\lambda$ are $\leq m$). Hence, $\lambda_{i}\leq m$ is
proven. This proves (\ref{sol.liri.180.a.partsleq1}).} Also,
\begin{equation}
\mu_{i}\leq m\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.
\label{sol.liri.180.a.partsleq2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.a.partsleq2}):} The assertion
(\ref{sol.liri.180.a.partsleq2}) follows from the same arguments as
(\ref{sol.liri.180.a.partsleq1}) (but with $\lambda$ replaced by $\mu$).}

(a) We have $\lambda=\left(  \lambda_{1},\lambda_{2},\lambda_{3}%
,\ldots\right)  $, thus $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{k}\right)  $ (since $\ell\left(  \lambda\right)  \leq k$).
Similarly, $\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{k}\right)  $.

Let us first show that $\lambda^{\vee}$ is a partition.

Indeed, for every $i\in\left\{  1,2,\ldots,k\right\}  $, we have
$m-\lambda_{k+1-i}\in \NN$\ \ \ \ \footnote{\textit{Proof.} Let
$i\in\left\{  1,2,\ldots,k\right\}  $. Then, $k+1-i\in\left\{  1,2,\ldots
,k\right\}  $. If $\lambda_{k+1-i}=0$, then $m-\underbrace{\lambda_{k+1-i}%
}_{=0}=m\in \NN$. Hence, for the rest of this proof of $m-\lambda
_{k+1-i}\in \NN$, we can WLOG assume that $\lambda_{k+1-i}\neq0$. Assume
this. Then, $\lambda_{k+1-i}$ is a part of the partition $\lambda$ (since
$\lambda_{k+1-i}\neq0$). Therefore, $\lambda_{k+1-i}\leq m$ (since all parts
of the partition $\lambda$ are $\leq m$), so that $m-\lambda_{k+1-i}\geq0$.
Hence, $m-\lambda_{k+1-i}\in \NN$, qed.}. Thus, $\left(  m-\lambda
_{k},m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  \in\NN^{k}$. Combined
with $m-\lambda_{k}\geq m-\lambda_{k-1}\geq\cdots\geq m-\lambda_{1}%
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k-1\right\}
$. We have $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}$ (since
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  =\lambda$ is a
partition), so that $\lambda_{i}\geq\lambda_{i+1}$. Hence, $m-\lambda_{i}\leq
m-\lambda_{i+1}$.
\par
Now, let us forget that we fixed $i$. We thus have shown that $m-\lambda
_{i}\leq m-\lambda_{i+1}$ for every $i\in\left\{  1,2,\ldots,k-1\right\}  $.
In other words, $m-\lambda_{1}\leq m-\lambda_{2}\leq\cdots\leq m-\lambda_{k}$.
In other words, $m-\lambda_{k}\geq m-\lambda_{k-1}\geq\cdots\geq m-\lambda
_{1}$, qed.}, this yields that $\left(  m-\lambda_{k},m-\lambda_{k-1}%
,\ldots,m-\lambda_{1}\right)  $ is a partition. In other words, $\lambda
^{\vee}$ is a partition (since $\lambda^{\vee}=\left(  m-\lambda_{k}%
,m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  $). Similarly, $\mu^{\vee}$ is a
partition. It now remains to show that $s_{\lambda/\mu}=s_{\mu^{\vee}%
/\lambda^{\vee}}$.

Notice that%
\begin{equation}
\left(  m^{k}\right)  _{i}=m\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,\ldots,k\right\}  . \label{sol.liri.180.a.mki}%
\end{equation}
Also,%
\begin{equation}
\left(  m^{k}\right)  _{i}=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
k+1,k+2,k+3,\ldots\right\}  . \label{sol.liri.180.a.mki2}%
\end{equation}


Whenever $\zeta$ is a partition, let us denote by
$Y\left(  \zeta\right)  $ the Ferrers diagram of $\zeta$. Recall the
definition of $Y\left(  \zeta\right)  $: For every partition $\zeta$, the
Ferrers diagram $Y\left(  \zeta\right)  $ is defined by%
\[
Y\left(  \zeta\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\zeta_{i}\right\}  .
\]

A \emph{skew partition} shall mean a pair $\left(\alpha, \beta\right)$ of
partitions satisfying $\beta \subseteq \alpha$.
We write such a skew partition $\left(\alpha, \beta\right)$
as $\alpha / \beta$.
For every skew partition $\alpha/\beta$, we define the skew Ferrers diagram
$Y\left(  \alpha/\beta\right)  $ of $\alpha / \beta$
by $Y\left(  \alpha/\beta\right)
=Y\left(  \alpha\right)  \setminus Y\left(  \beta\right)  $.

The Ferrers diagram $Y\left(  \left(  m^{k}\right)  \right)  $ of the
partition $\left(  m^{k}\right)  $ satisfies%
\begin{equation}
Y\left(  \left(  m^{k}\right)  \right)  =\left\{  1,2,\ldots,k\right\}
\times\left\{  1,2,\ldots,m\right\}  . \label{sol.liri.180.a.Ymk}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.a.mki}):} By the definition of
$Y\left(  \left(  m^{k}\right)  \right)  $, we have%
\begin{equation}
Y\left(  \left(  m^{k}\right)  \right)  =\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\left(  m^{k}\right)
_{i}\right\}  . \label{sol.liri.180.a.Ymk.pf.1}%
\end{equation}
\par
Now, let $\mathbf{c}$ be any element of $Y\left(  \left(  m^{k}\right)
\right)  $. Then, $\mathbf{c}\in Y\left(  \left(  m^{k}\right)  \right)
=\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}%
\ \mid\ j\leq\left(  m^{k}\right)  _{i}\right\}  $. Hence, $\mathbf{c}$ can be
written in the form $\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\left(  m^{k}\right)
_{i}$. Consider this $\left(  i,j\right)  $. We have $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$. Thus, $i\in\left\{  1,2,3,\ldots
\right\}  $ and $j\in\left\{  1,2,3,\ldots\right\}  $. We have $j\leq\left(
m^{k}\right)  _{i}$, thus $\left(  m^{k}\right)  _{i}\geq j>0$ (since
$j\in\left\{  1,2,3,\ldots\right\}  $). If we had $i\in\left\{
k+1,k+2,k+3,\ldots\right\}  $, then we would have $\left(  m^{k}\right)
_{i}=0$ (by (\ref{sol.liri.180.a.mki2})), which would contradict $\left(
m^{k}\right)  _{i}>0$. Hence, we cannot have $i\in\left\{  k+1,k+2,k+3,\ldots
\right\}  $. Thus, $i\notin\left\{  k+1,k+2,k+3,\ldots\right\}  $.
\par
Combining $i\in\left\{  1,2,3,\ldots\right\}  $ with $i\notin\left\{
k+1,k+2,k+3,\ldots\right\}  $, we obtain $i\in\left\{  1,2,3,\ldots\right\}
\setminus\left\{  k+1,k+2,k+3,\ldots\right\}  =\left\{  1,2,\ldots,k\right\}
$. Thus, $\left(  m^{k}\right)  _{i}=m$ (by (\ref{sol.liri.180.a.mki})). Now,
$j\leq\left(  m^{k}\right)  _{i}=m$, so that $j\in\left\{  1,2,\ldots
,m\right\}  $ (since $j\in\left\{  1,2,3,\ldots\right\}  $). Combined with
$i\in\left\{  1,2,\ldots,k\right\}  $, this yields $\left(  i,j\right)
\in\left\{  1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots,m\right\}  $.
Thus, $\mathbf{c}=\left(  i,j\right)  \in\left\{  1,2,\ldots,k\right\}
\times\left\{  1,2,\ldots,m\right\}  $.
\par
Now, let us forget that we fixed $\mathbf{c}$. We thus have shown that
$\mathbf{c}\in\left\{  1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots
,m\right\}  $ for every $\mathbf{c}\in Y\left(  \left(  m^{k}\right)  \right)
$. In other words, $Y\left(  \left(  m^{k}\right)  \right)  \subset\left\{
1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots,m\right\}  $.
\par
Let now $\mathbf{d}$ be an element of $\left\{  1,2,\ldots,k\right\}
\times\left\{  1,2,\ldots,m\right\}  $. Then, $\mathbf{d}$ has the form
$\left(  a,b\right)  $ for some $a\in\left\{  1,2,\ldots,k\right\}  $ and some
$b\in\left\{  1,2,\ldots,m\right\}  $. Consider these $a$ and $b$.
\par
We have $a\in\left\{  1,2,\ldots,k\right\}  \subset\left\{  1,2,3,\ldots
\right\}  $ and $b\in\left\{  1,2,\ldots,m\right\}  \subset\left\{
1,2,3,\ldots\right\}  $. Hence, $\left(  a,b\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$. Also, $\left(  m^{k}\right)  _{a}=m$ (by
(\ref{sol.liri.180.a.mki}), applied to $i=a$). Now, $b\in\left\{
1,2,\ldots,m\right\}  $, so that $b\leq m=\left(  m^{k}\right)  _{a}$ (since
$\left(  m^{k}\right)  _{a}=m$). Altogether, we now know that $\mathbf{d}%
=\left(  a,b\right)  $ with $\left(  a,b\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}$ satisfying $b\leq\left(  m^{k}\right)  _{a}$. Hence,
$\mathbf{d}$ has the form $\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\left(  m^{k}\right)
_{i}$. Thus,%
\[
\mathbf{d}\in\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}\ \mid\ j\leq\left(  m^{k}\right)  _{i}\right\}  =Y\left(  \left(
m^{k}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.a.Ymk.pf.1})}\right)  .
\]
\par
Now, let us forget that we fixed $\mathbf{d}$. We thus have shown that
$\mathbf{d}\in Y\left(  \left(  m^{k}\right)  \right)  $ for every
$\mathbf{c}\in\left\{  1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots
,m\right\}  $. In other words, $\left\{  1,2,\ldots,k\right\}  \times\left\{
1,2,\ldots,m\right\}  \subset Y\left(  \left(  m^{k}\right)  \right)  $.
Combining this with $Y\left(  \left(  m^{k}\right)  \right)  \subset\left\{
1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots,m\right\}  $, we obtain
$Y\left(  \left(  m^{k}\right)  \right)  =\left\{  1,2,\ldots,k\right\}
\times\left\{  1,2,\ldots,m\right\}  $. This proves (\ref{sol.liri.180.a.mki}%
).} Thus, the Ferrers diagram $Y\left(  \left(  m^{k}\right)  \right)  $ of
the partition $\left(  m^{k}\right)  $ is a rectangle of height $k$ and width
$m$. Let $\mathbf{Z}$ denote the reflection with respect to the center of this
rectangle. Then,%
\begin{equation}
\mathbf{Z}\left(  \left(  i,j\right)  \right)  =\left(  k+1-i,m+1-j\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)  \in \ZZ^{2}.
\label{sol.liri.180.a.Z}%
\end{equation}


Notice that $\mathbf{Z}$ is a reflection with respect to a point. Hence,
$\mathbf{Z}$ is an involution on $\ZZ^{2}$; that is, we have
$\mathbf{Z}\circ\mathbf{Z}=\operatorname*{id}$. Thus, $\mathbf{Z}$ is a
bijection $\ZZ^{2} \rightarrow \ZZ^{2}$. Also, $\mathbf{Z}\left(
Y\left(  \left(  m^{k}\right)  \right)  \right)  =Y\left(  \left(
m^{k}\right)  \right)  $\ \ \ \ \footnote{\textit{Proof.} Let $\mathbf{c}%
\in\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)  \right)  $. Then,
there exists a $\mathbf{d}\in Y\left(  \left(  m^{k}\right)  \right)  $ such
that $\mathbf{c}=\mathbf{Z}\left(  \mathbf{d}\right)  $ (since $\mathbf{c}%
\in\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)  \right)  $).
Consider this $\mathbf{d}$. We have $\mathbf{d}\in Y\left(  \left(
m^{k}\right)  \right)  =\left\{  1,2,\ldots,k\right\}  \times\left\{
1,2,\ldots,m\right\}  $ (by (\ref{sol.liri.180.a.Ymk})). Hence, $\mathbf{d}$
has the form $\mathbf{d}=\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots,m\right\}  $.
Denote this $\left(  i,j\right)  $ as $\left(  a,b\right)  $. Thus,
$\mathbf{d}=\left(  a,b\right)  $ and $\left(  a,b\right)  \in\left\{
1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots,m\right\}  $.
\par
We have $\left(  a,b\right)  \in\left\{  1,2,\ldots,k\right\}  \times\left\{
1,2,\ldots,m\right\}  $. Thus, $a\in\left\{  1,2,\ldots,k\right\}  $ and
$b\in\left\{  1,2,\ldots,m\right\}  $. We have $k+1-a\in\left\{
1,2,\ldots,k\right\}  $ (since $a\in\left\{  1,2,\ldots,k\right\}  $) and
$m+1-b\in\left\{  1,2,\ldots,m\right\}  $ (since $b\in\left\{  1,2,\ldots
,m\right\}  $). Combining these two relations, we obtain
\[
\left(  k+1-a,m+1-b\right)  \in\left\{  1,2,\ldots,k\right\}  \times\left\{
1,2,\ldots,m\right\}  =Y\left(  \left(  m^{k}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.180.a.Ymk})}\right)  .
\]
But (\ref{sol.liri.180.a.Z}) (applied to $i=a$ and $j=b$) yields
$\mathbf{Z}\left(  \left(  a,b\right)  \right)  =\left(  k+1-a,m+1-b\right)
\in Y\left(  \left(  m^{k}\right)  \right)  $, thus $\mathbf{c}=\mathbf{Z}%
\left(  \underbrace{\mathbf{d}}_{=\left(  a,b\right)  }\right)  =\mathbf{Z}%
\left(  \left(  a,b\right)  \right)  \in Y\left(  \left(  m^{k}\right)
\right)  $.
\par
Let us now forget that we fixed $\mathbf{c}$. We thus have proven that
$\mathbf{c}\in Y\left(  \left(  m^{k}\right)  \right)  $ for every
$\mathbf{c}\in\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)
\right)  $. In other words, $\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)
\right)  \right)  \subset Y\left(  \left(  m^{k}\right)  \right)  $. Hence,
$Y\left(  \left(  m^{k}\right)  \right)  \subset\mathbf{Z}^{-1}\left(
Y\left(  \left(  m^{k}\right)  \right)  \right)  $. But $\mathbf{Z}$ is a
bijection, and the map $\mathbf{Z}^{-1}$ satisfies $\mathbf{Z}^{-1}%
=\mathbf{Z}$ (since $\mathbf{Z}\circ\mathbf{Z}=\operatorname*{id}$). Hence,
$Y\left(  \left(  m^{k}\right)  \right)  \subset\underbrace{\mathbf{Z}^{-1}%
}_{=\mathbf{Z}}\left(  Y\left(  \left(  m^{k}\right)  \right)  \right)
=\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)  \right)  $.
Combined with $\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)
\right)  \subset Y\left(  \left(  m^{k}\right)  \right)  $, this yields
$\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)  \right)  =Y\left(
\left(  m^{k}\right)  \right)  $, qed.}.

It is easy to see that $Y\left(  \lambda\right)  \subset Y\left(  \left(
m^{k}\right)  \right)  $\ \ \ \ \footnote{\textit{Proof.} We have $Y\left(
\lambda\right)  =\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $. Now, let $\mathbf{c}$ be
any element of $Y\left(  \lambda\right)  $. Then, $\mathbf{c}\in Y\left(
\lambda\right)  =\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $. Hence, $\mathbf{c}$ can be
written in the form $\left(  i,j\right)  $ for some $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$.
Consider this $\left(  i,j\right)  $. We have $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$. Thus, $i\in\left\{  1,2,3,\ldots\right\}  $ and
$j\in\left\{  1,2,3,\ldots\right\}  $. We have $j\leq\lambda_{i}$, thus
$\lambda_{i}\geq j>0$ (since $j\in\left\{  1,2,3,\ldots\right\}  $). Hence,
$i\leq\ell\left(  \lambda\right)  \leq k$, so that $i\in\left\{
1,2,\ldots,k\right\}  $ (since $i\in\left\{  1,2,3,\ldots\right\}  $ and
$i\leq k$). Also, $\lambda_{i}\leq m$ (according to
(\ref{sol.liri.180.a.partsleq1})), so that $j\leq\lambda_{i}\leq m$ and
therefore $j\in\left\{  1,2,\ldots,m\right\}  $ (since $j\in\left\{
1,2,3,\ldots\right\}  $ and $j\leq m$). Combined with $i\in\left\{
1,2,\ldots,k\right\}  $, this yields $\left(  i,j\right)  \in\left\{
1,2,\ldots,k\right\}  \times\left\{  1,2,\ldots,m\right\}  $. Thus,
$\mathbf{c}=\left(  i,j\right)  \in\left\{  1,2,\ldots,k\right\}
\times\left\{  1,2,\ldots,m\right\}  =Y\left(  \left(  m^{k}\right)  \right)
$ (due to (\ref{sol.liri.180.a.Ymk})).
\par
Now, let us forget that we fixed $\mathbf{c}$. We thus have shown that
$\mathbf{c}\in Y\left(  \left(  m^{k}\right)  \right)  $ for every
$\mathbf{c}\in Y\left(  \lambda\right)  $. In other words, $Y\left(
\lambda\right)  \subset Y\left(  \left(  m^{k}\right)  \right)  $, qed.}. The
same argument (but with every $\lambda$ replaced by $\mu$) yields $Y\left(
\mu\right)  \subset Y\left(  \left(  m^{k}\right)  \right)  $.

Now,%
\begin{equation}
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda\right)  \right)
=Y\left(  \lambda^{\vee}\right)  \label{sol.liri.180.a.u}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.a.u}):} By the definition of
$Y\left(  \left(  m^{k}\right)  /\lambda\right)  $, we have $Y\left(  \left(
m^{k}\right)  /\lambda\right)  =Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \lambda\right)  $.
\par
By the definition of $Y\left(  \lambda^{\vee}\right)  $, we have%
\begin{equation}
Y\left(  \lambda^{\vee}\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\left(  \lambda^{\vee}\right)
_{i}\right\}  . \label{sol.liri.180.a.u0}%
\end{equation}
\par
But $\lambda^{\vee}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda
_{1}\right)  $. Thus,%
\begin{equation}
\left(  \lambda^{\vee}\right)  _{i}=m-\lambda_{k+1-i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k\right\}  .
\label{sol.liri.180.a.u1}%
\end{equation}
\par
Also, $\lambda^{\vee}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots
,m-\lambda_{1}\right)  $ is a $k$-tuple of nonnegative integers. Hence,
$\ell\left(  \lambda^{\vee}\right)  \leq k$. Thus,%
\begin{equation}
\left(  \lambda^{\vee}\right)  _{i}=0\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  k+1,k+2,k+3,\ldots\right\}  . \label{sol.liri.180.a.u2}%
\end{equation}
\par
Now, let $\mathbf{c}$ be an element of $\mathbf{Z}\left(  Y\left(  \left(
m^{k}\right)  /\lambda\right)  \right)  $. Then, $\mathbf{c}=\mathbf{Z}\left(
\mathbf{d}\right)  $ for some $\mathbf{d}\in Y\left(  \left(  m^{k}\right)
/\lambda\right)  $. Consider this $\mathbf{d}$. We have $\mathbf{d}\in
Y\left(  \left(  m^{k}\right)  /\lambda\right)  =Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)  \subset Y\left(
\left(  m^{k}\right)  \right)  =\left\{  1,2,\ldots,k\right\}  \times\left\{
1,2,\ldots,m\right\}  $ (by (\ref{sol.liri.180.a.Ymk})). Hence, we can write
$\mathbf{d}$ in the form $\left(  a,b\right)  $ for some $a\in\left\{
1,2,\ldots,k\right\}  $ and $b\in\left\{  1,2,\ldots,m\right\}  $. Consider
these $a$ and $b$. We have $\mathbf{c}=\mathbf{Z}\left(
\underbrace{\mathbf{d}}_{=\left(  a,b\right)  }\right)  =\mathbf{Z}\left(
\left(  a,b\right)  \right)  =\left(  k+1-a,m+1-b\right)  $ (by
(\ref{sol.liri.180.a.Z}), applied to $i=a$ and $j=b$).
\par
Notice that $\left(  a,b\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
(since $a\in\left\{  1,2,\ldots,k\right\}  \subset\left\{  1,2,3,\ldots
\right\}  $ and $b\in\left\{  1,2,\ldots,m\right\}  \subset\left\{
1,2,3,\ldots\right\}  $).
\par
But we have $\left(  a,b\right)  =\mathbf{d}\in Y\left(  \left(  m^{k}\right)
\right)  \setminus Y\left(  \lambda\right)  $, so that $\left(  a,b\right)
\notin Y\left(  \lambda\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $ (by the
definition of $Y\left(  \lambda\right)  $). In other words, $\left(
a,b\right)  $ is not an element $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$. Hence, we cannot
have $b\leq\lambda_{a}$ (because otherwise, we would have $b\leq\lambda_{a}$,
and therefore $\left(  a,b\right)  $ would be an element $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$, which
would contradict the fact that $\left(  a,b\right)  $ is not an element
$\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying
$j\leq\lambda_{i}$). Thus, we have $b>\lambda_{a}$. Since $b$ and $\lambda
_{a}$ are integers, this yields $b-1\geq\lambda_{a}$. Hence, $\lambda_{a}\leq
b-1$.
\par
Now, $a\in\left\{  1,2,\ldots,k\right\}  $, so that $k+1-a\in\left\{
1,2,\ldots,k\right\}  \subset\left\{  1,2,3,\ldots\right\}  $. Also,
$b\in\left\{  1,2,\ldots,m\right\}  $, and thus $m+1-b\in\left\{
1,2,\ldots,m\right\}  \subset\left\{  1,2,3,\ldots\right\}  $. Now, $\left(
k+1-a,m+1-b\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ (since
$k+1-a\in\left\{  1,2,3,\ldots\right\}  $ and $m+1-b\in\left\{  1,2,3,\ldots
\right\}  $).
\par
Recall that $k+1-a\in\left\{  1,2,\ldots,k\right\}  $. Hence,
(\ref{sol.liri.180.a.u1}) (applied to $i=k+1-a$) yields $\left(  \lambda
^{\vee}\right)  _{k+1-a}=m-\underbrace{\lambda_{k+1-\left(  k+1-a\right)  }%
}_{=\lambda_{a}\leq b-1}=m-\left(  b-1\right)  =m+1-b$. In other words,
$m+1-b\leq\left(  \lambda^{\vee}\right)  _{k+1-a}$. Now, $\left(
k+1-a,m+1-b\right)  $ is an element $\left(  i,j\right)  $ of $\left\{
1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\left(  \lambda^{\vee}\right)
_{i}$. In other words,%
\[
\left(  k+1-a,m+1-b\right)  \in\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\left(  \lambda^{\vee}\right)
_{i}\right\}  =Y\left(  \lambda^{\vee}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.liri.180.a.u0})}\right)  .
\]
Thus, $\mathbf{c}=\left(  k+1-a,m+1-b\right)  \in Y\left(  \lambda^{\vee
}\right)  $.
\par
Let us now forget that we fixed $\mathbf{c}$. We thus have proven that every
$\mathbf{c}\in\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda
\right)  \right)  $ satisfies $\mathbf{c}\in Y\left(  \lambda^{\vee}\right)
$. In other words,%
\begin{equation}
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda\right)  \right)
\subset Y\left(  \lambda^{\vee}\right)  . \label{sol.liri.180.a.upart1}%
\end{equation}
\par
Let now $\mathbf{e}$ be an element of $Y\left(  \lambda^{\vee}\right)  $. Our
next goal is to prove that $\mathbf{e}\in\mathbf{Z}\left(  Y\left(  \left(
m^{k}\right)  /\lambda\right)  \right)  $.
\par
We have $\mathbf{e}\in Y\left(  \lambda^{\vee}\right)  =\left\{  \left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\left(
\lambda^{\vee}\right)  _{i}\right\}  $. Thus, there exists an $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\left(
\lambda^{\vee}\right)  _{i}$ and $\mathbf{e}=\left(  i,j\right)  $. Let
$\left(  a,b\right)  $ denote this $\left(  i,j\right)  $. Thus, $\left(
a,b\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfies $b\leq\left(
\lambda^{\vee}\right)  _{a}$ and $\mathbf{e}=\left(  a,b\right)  $. We have
$\left(  a,b\right)  =\mathbf{e}\in Y\left(  \lambda^{\vee}\right)  \subset
Y\left(  \left(  m^{k}\right)  \right)  =\left\{  1,2,\ldots,k\right\}
\times\left\{  1,2,\ldots,m\right\}  $ (by (\ref{sol.liri.180.a.Ymk})), so
that $a\in\left\{  1,2,\ldots,k\right\}  $ and $b\in\left\{  1,2,\ldots
,m\right\}  $.
\par
We have $k+1-a\in\left\{  1,2,\ldots,k\right\}  $ (since $a\in\left\{
1,2,\ldots,k\right\}  $) and $m+1-b\in\left\{  1,2,\ldots,m\right\}  $ (since
$b\in\left\{  1,2,\ldots,m\right\}  $). Combining these two relations, we
obtain%
\[
\left(  k+1-a,m+1-b\right)  \in\left\{  1,2,\ldots,k\right\}  \times\left\{
1,2,\ldots,m\right\}  =Y\left(  \left(  m^{k}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.180.a.Ymk})}\right)  .
\]
\par
On the other hand, let us assume (for the sake of contradiction) that $\left(
k+1-a,m+1-b\right)  \in Y\left(  \lambda\right)  $. Then, $\left(
k+1-a,m+1-b\right)  \in Y\left(  \lambda\right)  =\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $ (by
the definition of $Y\left(  \lambda\right)  $). In other words, $\left(
k+1-a,m+1-b\right)  $ is an element $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$ satisfying $j\leq\lambda_{i}$. Hence,
$m+1-b\leq\lambda_{k+1-a}$. In other words, $\lambda_{k+1-a}\geq m+1-b$.
\par
But $a\in\left\{  1,2,\ldots,k\right\}  $. Hence, (\ref{sol.liri.180.a.u1})
(applied to $i=a$) yields $\left(  \lambda^{\vee}\right)  _{a}=m-\lambda
_{k+1-a}$. Now, $b\leq\left(  \lambda^{\vee}\right)  _{a}%
=m-\underbrace{\lambda_{k+1-a}}_{\geq m+1-b}\leq m-\left(  m+1-b\right)
=b-1<b$. This is absurd. This contradiction shows that our assumption (that
$\left(  k+1-a,m+1-b\right)  \in Y\left(  \lambda\right)  $) was wrong. Hence,
we cannot have $\left(  k+1-a,m+1-b\right)  \in Y\left(  \lambda\right)  $. In
other words, we have $\left(  k+1-a,m+1-b\right)  \notin Y\left(
\lambda\right)  $.
\par
Combining $\left(  k+1-a,m+1-b\right)  \in Y\left(  \left(  m^{k}\right)
\right)  $ with $\left(  k+1-a,m+1-b\right)  \notin Y\left(  \lambda\right)
$, we obtain
\begin{equation}
\left(  k+1-a,m+1-b\right)  \in Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \lambda\right)  =Y\left(  \left(  m^{k}\right)
/\lambda\right)  \label{sol.liri.180.a.u5}%
\end{equation}
(since $Y\left(  \left(  m^{k}\right)  /\lambda\right)  =Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)  $). Now,
(\ref{sol.liri.180.a.Z}) (applied to $i=k+1-a$ and $j=m+1-b$) yields%
\[
\mathbf{Z}\left(  \left(  k+1-a,m+1-b\right)  \right)  =\left(
\underbrace{k+1-\left(  k+1-a\right)  }_{=a},\underbrace{m+1-\left(
m+1-b\right)  }_{=b}\right)  =\left(  a,b\right)  =\mathbf{e}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{e}=\left(  a,b\right)
\right)  .
\]
Thus, $\mathbf{e}=\mathbf{Z}\left(  \underbrace{\left(  k+1-a,m+1-b\right)
}_{\substack{\in Y\left(  \left(  m^{k}\right)  /\lambda\right)  \\\text{(by
(\ref{sol.liri.180.a.u5}))}}}\right)  \in\mathbf{Z}\left(  Y\left(  \left(
m^{k}\right)  /\lambda\right)  \right)  $.
\par
Now, let us forget that we fixed $\mathbf{e}$. We thus have proven that
$\mathbf{e}\in\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda
\right)  \right)  $ for every $\mathbf{e}\in Y\left(  \lambda^{\vee}\right)
$. In other words,%
\[
Y\left(  \lambda^{\vee}\right)  \subset\mathbf{Z}\left(  Y\left(  \left(
m^{k}\right)  /\lambda\right)  \right)  .
\]
Combining this with (\ref{sol.liri.180.a.upart1}), we obtain $\mathbf{Z}%
\left(  Y\left(  \left(  m^{k}\right)  /\lambda\right)  \right)  =Y\left(
\lambda^{\vee}\right)  $. This proves (\ref{sol.liri.180.a.u}).}. The same
argument (but with every $\lambda$ replaced by $\mu$) yields%
\begin{equation}
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\mu\right)  \right)
=Y\left(  \mu^{\vee}\right)  . \label{sol.liri.180.a.v}%
\end{equation}
Furthermore,%
\begin{equation}
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda^{\vee}\right)
\right)  =Y\left(  \lambda\right)  \label{sol.liri.180.a.w}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.a.w}):} Applying the map
$\mathbf{Z}$ to both sides of the equality (\ref{sol.liri.180.a.u}), we obtain
$\mathbf{Z}\left(  \mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)
/\lambda\right)  \right)  \right)  =\mathbf{Z}\left(  Y\left(  \lambda^{\vee
}\right)  \right)  $, so that%
\[
\mathbf{Z}\left(  Y\left(  \lambda^{\vee}\right)  \right)  =\mathbf{Z}\left(
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda\right)  \right)
\right)  =\underbrace{\left(  \mathbf{Z}\circ\mathbf{Z}\right)  }%
_{=\operatorname*{id}}\left(  Y\left(  \left(  m^{k}\right)  /\lambda\right)
\right)  =\operatorname*{id}\left(  Y\left(  \left(  m^{k}\right)
/\lambda\right)  \right)  =Y\left(  \left(  m^{k}\right)  /\lambda\right)
=Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)
\]
(by the definition of the skew Ferrers diagram $Y\left(  \left(  m^{k}\right)
/\lambda\right)  $ of $\left(  m^{k}\right)  /\lambda$).
\par
The definition of $Y\left(  \left(  m^{k}\right)  /\lambda^{\vee}\right)  $
yields $Y\left(  \left(  m^{k}\right)  /\lambda^{\vee}\right)  =Y\left(
\left(  m^{k}\right)  \right)  \setminus Y\left(  \lambda^{\vee}\right)  $.
Applying the map $\mathbf{Z}$ to both sides of this equality, we obtain%
\begin{align*}
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda^{\vee}\right)
\right)   &  =\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \lambda^{\vee}\right)  \right)  =\underbrace{\mathbf{Z}%
\left(  Y\left(  \left(  m^{k}\right)  \right)  \right)  }_{=Y\left(  \left(
m^{k}\right)  \right)  }\setminus\underbrace{\mathbf{Z}\left(  Y\left(
\lambda^{\vee}\right)  \right)  }_{=Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \lambda\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathbf{Z}\text{ is a
bijection}\right) \\
&  =Y\left(  \left(  m^{k}\right)  \right)  \setminus\left(  Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)  \right)  =Y\left(
\lambda\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }Y\left(
\lambda\right)  \subset Y\left(  \left(  m^{k}\right)  \right)  \right)  .
\end{align*}
This proves (\ref{sol.liri.180.a.w}).}. The same argument (but with every
$\lambda$ replaced by $\mu$) yields%
\begin{equation}
\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\mu^{\vee}\right)  \right)
=Y\left(  \mu\right)  . \label{sol.liri.180.a.x}%
\end{equation}


Now, let us recall that
\begin{equation}
s_{\alpha/\beta}=0\ \ \ \ \ \ \ \ \ \ \text{for any two partitions }%
\alpha\text{ and }\beta\text{ which don't satisfy }\beta\subseteq\alpha.
\label{sol.liri.180.a.0}%
\end{equation}


We need to prove that $s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}}$. If we
don't have $\mu\subseteq\lambda$, then this is easy to
prove\footnote{\textit{Proof.} Assume that we don't have $\mu\subseteq\lambda
$. Then, $s_{\lambda/\mu}=0$ (by (\ref{sol.liri.180.a.0}), applied to
$\alpha=\lambda$ and $\beta=\mu$).
\par
On the other hand, let us assume (for the sake of contradiction) that
$\lambda^{\vee}\subseteq\mu^{\vee}$. Then, $Y\left(  \lambda^{\vee}\right)
\subset Y\left(  \mu^{\vee}\right)  $ \text{(since }$\lambda^{\vee}%
\subseteq\mu^{\vee}$ holds if and only if $Y\left(  \lambda^{\vee}\right)
\subset Y\left(  \mu^{\vee}\right)  $ (by the definition of $\lambda^{\vee
}\subseteq\mu^{\vee}$)).
\par
The definition of the skew Ferrers diagram $Y\left(  \left(  m^{k}\right)
/\lambda^{\vee}\right)  $ of $\left(  m^{k}\right)  /\lambda^{\vee}$ yields
$Y\left(  \left(  m^{k}\right)  /\lambda^{\vee}\right)  =Y\left(  \left(
m^{k}\right)  \right)  \setminus\underbrace{Y\left(  \lambda^{\vee}\right)
}_{\subset Y\left(  \mu^{\vee}\right)  }\supset Y\left(  \left(  m^{k}\right)
\right)  \setminus Y\left(  \mu^{\vee}\right)  $. Hence, $Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \mu^{\vee}\right)  \subset Y\left(
\left(  m^{k}\right)  /\lambda^{\vee}\right)  $.
\par
But the definition of the skew Ferrers diagram $Y\left(  \left(  m^{k}\right)
/\mu^{\vee}\right)  $ of $\left(  m^{k}\right)  /\mu^{\vee}$ yields $Y\left(
\left(  m^{k}\right)  /\mu^{\vee}\right)  =Y\left(  \left(  m^{k}\right)
\right)  \setminus Y\left(  \mu^{\vee}\right)  \subset Y\left(  \left(
m^{k}\right)  /\lambda^{\vee}\right)  $. Now, (\ref{sol.liri.180.a.x}) yields%
\[
Y\left(  \mu\right)  =\mathbf{Z}\left(  \underbrace{Y\left(  \left(
m^{k}\right)  /\mu^{\vee}\right)  }_{\subset Y\left(  \left(  m^{k}\right)
/\lambda^{\vee}\right)  }\right)  \subset\mathbf{Z}\left(  Y\left(  \left(
m^{k}\right)  /\lambda^{\vee}\right)  \right)  =Y\left(  \lambda\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.180.a.w})}\right)  .
\]
In other words, $\mu\subseteq\lambda$ (because $\mu\subseteq\lambda$ holds if
and only if $Y\left(  \mu\right)  \subset Y\left(  \lambda\right)  $ (by the
definition of $\mu\subseteq\lambda$)). This contradicts the fact that we don't
have $\mu\subseteq\lambda$. This contradiction shows that our assumption (that
$\lambda^{\vee}\subseteq\mu^{\vee}$) was wrong. Hence, we don't have
$\lambda^{\vee}\subseteq\mu^{\vee}$. Consequently, $s_{\mu^{\vee}%
/\lambda^{\vee}}=0$ (by (\ref{sol.liri.180.a.0}), applied to $\alpha=\mu
^{\vee}$ and $\beta=\lambda^{\vee}$). Compared with $s_{\lambda/\mu}=0$, this
yields $s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}}$, qed.}. Hence, for the
rest of this proof, we can WLOG assume that we have $\mu\subseteq\lambda$.
Assume this.

We have $\mu\subseteq\lambda$. Hence, the skew partition $\lambda/\mu$ is
well-defined, and we have $Y\left(  \lambda/\mu\right)  =Y\left(
\lambda\right)  \setminus Y\left(  \mu\right)  $ (by the definition of the
skew Ferrers diagram $Y\left(  \lambda/\mu\right)  $ of $\lambda/\mu$).
Furthermore, we have $\lambda^{\vee}\subseteq\mu^{\vee}$%
\ \ \ \ \footnote{\textit{Proof.} We have $\mu\subseteq\lambda$. In other
words, $Y\left(  \mu\right)  \subset Y\left(  \lambda\right)  $ (because
$\mu\subseteq\lambda$ holds if and only if $Y\left(  \mu\right)  \subset
Y\left(  \lambda\right)  $ (by the definition of $\mu\subseteq\lambda$)). In
other words, $Y\left(  \lambda\right)  \supset Y\left(  \mu\right)  $. Now,
$Y\left(  \left(  m^{k}\right)  \right)  \setminus\underbrace{Y\left(
\lambda\right)  }_{\supset Y\left(  \mu\right)  }\subset Y\left(  \left(
m^{k}\right)  \right)  \setminus Y\left(  \mu\right)  $.
\par
The definition of the skew Ferrers diagram $Y\left(  \left(  m^{k}\right)
/\lambda\right)  $ yields $Y\left(  \left(  m^{k}\right)  /\lambda\right)
=Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(  \lambda\right)  $.
The definition of the skew Ferrers diagram $Y\left(  \left(  m^{k}\right)
/\mu\right)  $ yields $Y\left(  \left(  m^{k}\right)  /\mu\right)  =Y\left(
\left(  m^{k}\right)  \right)  \setminus Y\left(  \mu\right)  $. In other
words, $Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(  \mu\right)
=Y\left(  \left(  m^{k}\right)  /\mu\right)  $. Now, (\ref{sol.liri.180.a.u})
yields
\begin{align*}
Y\left(  \lambda^{\vee}\right)   &  =\mathbf{Z}\left(  \underbrace{Y\left(
\left(  m^{k}\right)  /\lambda\right)  }_{=Y\left(  \left(  m^{k}\right)
\right)  \setminus Y\left(  \lambda\right)  }\right)  =\mathbf{Z}\left(
\underbrace{Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(
\lambda\right)  }_{\subset Y\left(  \left(  m^{k}\right)  \right)  \setminus
Y\left(  \mu\right)  }\right)  \subset\mathbf{Z}\left(  \underbrace{Y\left(
\left(  m^{k}\right)  \right)  \setminus Y\left(  \mu\right)  }_{=Y\left(
\left(  m^{k}\right)  /\mu\right)  }\right) \\
&  =\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\mu\right)  \right)
=Y\left(  \mu^{\vee}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.a.v})}\right)  .
\end{align*}
In other words, $\lambda^{\vee}\subseteq\mu^{\vee}$, qed.}. Hence, the skew
partition $\mu^{\vee}/\lambda^{\vee}$ is well-defined, and we have%
\begin{align}
Y\left(  \mu^{\vee}/\lambda^{\vee}\right)   &  =\underbrace{Y\left(  \mu
^{\vee}\right)  }_{\substack{=\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)
/\mu\right)  \right)  \\\text{(by (\ref{sol.liri.180.a.v}))}}}\setminus
\underbrace{Y\left(  \lambda^{\vee}\right)  }_{\substack{=\mathbf{Z}\left(
Y\left(  \left(  m^{k}\right)  /\lambda\right)  \right)  \\\text{(by
(\ref{sol.liri.180.a.u}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the skew Ferrers
diagram }Y\left(  \mu^{\vee}/\lambda^{\vee}\right)  \text{ of }\mu^{\vee
}/\lambda^{\vee}\right) \nonumber\\
&  =\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\mu\right)  \right)
\setminus\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\lambda\right)
\right) \nonumber\\
&  =\mathbf{Z}\left(  Y\left(  \left(  m^{k}\right)  /\mu\right)  \setminus
Y\left(  \left(  m^{k}\right)  /\lambda\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{Z}\text{ is a bijection}%
\right)  . \label{sol.liri.180.a.p1}%
\end{align}
But a very simple set-theoretical fact says that if $\mathfrak{A}$,
$\mathfrak{B}$ and $\mathfrak{C}$ are three sets such that $\mathfrak{B}%
\subset\mathfrak{A}$ and $\mathfrak{C}\subset\mathfrak{A}$, then%
\begin{equation}
\left(  \mathfrak{A}\setminus\mathfrak{B}\right)  \setminus\left(
\mathfrak{A}\setminus\mathfrak{C}\right)  =\mathfrak{C}\setminus\mathfrak{B}.
\label{sol.liri.180.a.setth}%
\end{equation}
Now,%
\begin{align*}
&  \underbrace{Y\left(  \left(  m^{k}\right)  /\mu\right)  }%
_{\substack{=Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(
\mu\right)  \\\text{(by the definition of the skew diagram}\\Y\left(  \left(
m^{k}\right)  /\mu\right)  \text{ of }\left(  m^{k}\right)  /\mu\text{)}%
}}\setminus\underbrace{Y\left(  \left(  m^{k}\right)  /\lambda\right)
}_{\substack{=Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(
\lambda\right)  \\\text{(by the definition of the skew diagram}\\Y\left(
\left(  m^{k}\right)  /\lambda\right)  \text{ of }\left(  m^{k}\right)
/\lambda\text{)}}}\\
&  =\left(  Y\left(  \left(  m^{k}\right)  \right)  \setminus Y\left(
\mu\right)  \right)  \setminus\left(  Y\left(  \left(  m^{k}\right)  \right)
\setminus Y\left(  \lambda\right)  \right)  =Y\left(  \lambda\right)
\setminus Y\left(  \mu\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.180.a.setth}), applied
to }\mathfrak{A}=Y\left(  \left(  m^{k}\right)  \right)  \text{, }%
\mathfrak{B}=Y\left(  \mu\right)  \text{ and }\mathfrak{C}=Y\left(
\lambda\right)  \right) \\
&  =Y\left(  \lambda/\mu\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}Y\left(  \lambda/\mu\right)  =Y\left(  \lambda\right)  \setminus Y\left(
\mu\right)  \right)  .
\end{align*}
Hence, (\ref{sol.liri.180.a.p1}) becomes%
\[
Y\left(  \mu^{\vee}/\lambda^{\vee}\right)  =\mathbf{Z}\left(
\underbrace{Y\left(  \left(  m^{k}\right)  /\mu\right)  \setminus Y\left(
\left(  m^{k}\right)  /\lambda\right)  }_{=Y\left(  \lambda/\mu\right)
}\right)  =\mathbf{Z}\left(  Y\left(  \lambda/\mu\right)  \right)  .
\]
In other words, the set $Y\left(  \mu^{\vee}/\lambda^{\vee}\right)  $ can be
obtained from the set $Y\left(  \lambda/\mu\right)  $ by the map $\mathbf{Z}$.
Hence, the set $Y\left(  \mu^{\vee}/\lambda^{\vee}\right)  $ can be obtained
from the set $Y\left(  \lambda/\mu\right)  $ by a $180^{\circ}$ rotation
(since $\mathbf{Z}$ is a $180^{\circ}$ rotation\footnote{This is because
$\mathbf{Z}$ is a reflection with respect to a point, but a reflection with
respect to a point (in the plane) is precisely the same as a $180^{\circ}$
rotation.}). In other words, the skew Ferrers diagram $\mu^{\vee}%
/\lambda^{\vee}$ can be obtained from the skew Ferrers diagram $\lambda/\mu$
by a $180^{\circ}$ rotation (since the set $Y\left(  \mu^{\vee}/\lambda^{\vee
}\right)  $ is the skew Ferrers diagram $\mu^{\vee}/\lambda^{\vee}$, and since
the set $Y\left(  \lambda/\mu\right)  $ is the skew Ferrers diagram
$\lambda/\mu$). Thus, Exercise \ref{exe.schur.rotate180}(b) (applied to
$\lambda^{\prime}=\mu^{\vee}$ and $\mu^{\prime}=\lambda^{\vee}$) yields
$s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}}$. This completes the solution
of Exercise \ref{exe.liri.180}(a).

(b) According to Remark \ref{third-L-R-interpretation-remark}, we have
$s_{\lambda/\mu}=\sum_{\nu}c_{\mu,\nu}^{\lambda}s_{\nu}$, where the sum ranges
over all partitions $\nu$. In other words, we have%
\begin{equation}
s_{\lambda/\mu}=\sum_{\nu\in \Par }c_{\mu,\nu}^{\lambda}s_{\nu}.
\label{sol.liri.180.b.1}%
\end{equation}


Exercise \ref{exe.liri.180}(b) yields that $\lambda^{\vee}$ and $\mu^{\vee}$
are partitions, and that $s_{\lambda/\mu}=s_{\mu^{\vee}/\lambda^{\vee}}$.
Applying (\ref{sol.liri.180.b.1}) to $\mu^{\vee}$ and $\lambda^{\vee}$ instead
of $\lambda$ and $\mu$, we obtain%
\[
s_{\mu^{\vee}/\lambda^{\vee}}=\sum_{\nu\in \Par }c_{\lambda
^{\vee},\nu}^{\mu^{\vee}}s_{\nu}.
\]
Subtracting this equality from (\ref{sol.liri.180.b.1}), we obtain%
\[
s_{\lambda/\mu}-s_{\mu^{\vee}/\lambda^{\vee}}=\sum_{\nu\in \Par %
}c_{\mu,\nu}^{\lambda}s_{\nu}-\sum_{\nu\in \Par }c_{\lambda
^{\vee},\nu}^{\mu^{\vee}}s_{\nu}=\sum_{\nu\in \Par }\left(
c_{\mu,\nu}^{\lambda}-c_{\lambda^{\vee},\nu}^{\mu^{\vee}}\right)  s_{\nu}.
\]
Thus,%
\begin{equation}
\sum_{\nu\in \Par }\left(  c_{\mu,\nu}^{\lambda}-c_{\lambda
^{\vee},\nu}^{\mu^{\vee}}\right)  s_{\nu}=\underbrace{s_{\lambda/\mu}%
}_{=s_{\mu^{\vee}/\lambda^{\vee}}}-s_{\mu^{\vee}/\lambda^{\vee}}=s_{\mu^{\vee
}/\lambda^{\vee}}-s_{\mu^{\vee}/\lambda^{\vee}}=0. \label{sol.liri.180.b.3}%
\end{equation}


But recall that $\left(  s_{\nu}\right)  _{\nu\in\operatorname{Par}}$ is a
basis of the $\kk$-module $\Lambda$. In particular, the family $\left(
s_{\nu}\right)  _{\nu\in\operatorname{Par}}$ is $\kk$-linearly
independent. Thus, if $\left(  g_{\nu}\right)  _{\nu\in\operatorname{Par}}%
\in\kk^{\operatorname{Par}}$ is a family of elements of $\kk$
such that \newline$\left(  \text{all but finitely many }\nu\in
\operatorname{Par}\text{ satisfy }g_{\nu}=0\right)  $ and $\sum_{\nu
\in\operatorname{Par}}g_{\nu}s_{\nu}=0$, then%
\[
\left(  g_{\nu}\right)  _{\nu\in\operatorname{Par}}=\left(  0\right)  _{\nu
\in\operatorname{Par}}.
\]
Applying this to $g_{\nu}=c_{\mu,\nu}^{\lambda}-c_{\lambda^{\vee},\nu}%
^{\mu^{\vee}}$, we obtain%
\[
\left(  c_{\mu,\nu}^{\lambda}-c_{\lambda^{\vee},\nu}^{\mu^{\vee}}\right)
_{\nu\in\operatorname{Par}}=\left(  0\right)  _{\nu\in\operatorname{Par}}%
\]
(because of (\ref{sol.liri.180.b.3})). In other words, $c_{\mu,\nu}^{\lambda
}-c_{\lambda^{\vee},\nu}^{\mu^{\vee}}=0$ for every $\nu\in \Par %
$. In other words, $c_{\mu,\nu}^{\lambda}=c_{\lambda^{\vee},\nu}^{\mu^{\vee}}$
for every $\nu\in \Par $. In other words, $c_{\mu,\nu}^{\lambda
}=c_{\lambda^{\vee},\nu}^{\mu^{\vee}}$ for any partition $\nu$. This solves
Exercise \ref{exe.liri.180}(b).

(c) Exercise \ref{exe.liri.180}(a) (applied to $\nu$ instead of $\mu$) yields
that $\lambda^{\vee}$ and $\nu^{\vee}$ are partitions and that $s_{\lambda
/\nu}=s_{\nu^{\vee}/\lambda^{\vee}}$.

From (\ref{eq.third-L-R-interpretation-remark.commutativity}), we obtain%
\begin{align*}
c_{\mu,\nu}^{\lambda}  &  =c_{\nu,\mu}^{\lambda}=c_{\lambda^{\vee},\mu}%
^{\nu^{\vee}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.liri.180}%
(b) (applied to }\nu\text{ and }\mu\text{ instead of }\mu\text{ and }%
\nu\text{)}\right) \\
&  =c_{\mu,\lambda^{\vee}}^{\nu^{\vee}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.third-L-R-interpretation-remark.commutativity}) (applied to }%
\nu^{\vee}\text{, }\lambda^{\vee}\text{ and }\mu\text{ instead of }%
\lambda\text{, }\mu\text{ and }\nu\text{)}\right)  .
\end{align*}
On the other hand, Exercise \ref{exe.liri.180}(b) yields $c_{\mu,\nu}%
^{\lambda}=c_{\lambda^{\vee},\nu}^{\mu^{\vee}}=c_{\nu,\lambda^{\vee}}%
^{\mu^{\vee}}$ (by (\ref{eq.third-L-R-interpretation-remark.commutativity})
(applied to $\mu^{\vee}$, $\lambda^{\vee}$ and $\nu$ instead of $\lambda$,
$\mu$ and $\nu$)). Thus, $c_{\nu,\lambda^{\vee}}^{\mu^{\vee}}=c_{\lambda^{\vee
},\nu}^{\mu^{\vee}}=c_{\mu,\nu}^{\lambda}=c_{\nu,\mu}^{\lambda}=c_{\lambda
^{\vee},\mu}^{\nu^{\vee}}=c_{\mu,\lambda^{\vee}}^{\nu^{\vee}}$. In other
words, $c_{\mu,\nu}^{\lambda}=c_{\nu,\mu}^{\lambda}=c_{\lambda^{\vee},\nu
}^{\mu^{\vee}}=c_{\nu,\lambda^{\vee}}^{\mu^{\vee}}=c_{\mu,\lambda^{\vee}}%
^{\nu^{\vee}}=c_{\lambda^{\vee},\mu}^{\nu^{\vee}}$. This solves Exercise
\ref{exe.liri.180}(c).

(d) \textit{First solution to Exercise \ref{exe.liri.180}(d):} Notice that
$\lambda^{\vee}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda
_{1}\right)  $. Thus, $\lambda^{\vee}$ is a $k$-tuple of nonnegative integers;
consequently, $\ell\left(  \lambda^{\vee}\right)  \leq k$.

Let $n=k$. We shall use the notations of Section \ref{bialternant-section}; in
particular, we set $\xx=\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ and
$\rho=\left(  n-1,n-2,\ldots,2,1,0\right)  $. Clearly, $\xx=\left(
x_{1},x_{2},\ldots,x_{n}\right)  =\left(  x_{1},x_{2},\ldots,x_{k}\right)  $
(since $n=k$). Notice that $\ell\left(  \lambda\right)  \leq k=n$. Hence, we
can regard $\lambda$ as an element of $\NN^n$; therefore,
$\lambda+\rho$ is a well-defined element of $\NN^n$, and the
alternant $a_{\lambda+\rho}$ is well-defined. Corollary
\ref{cor.stembridge.slambda} yields that $s_{\lambda}\left(  \xx%
\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$, so that%
\begin{equation}
a_{\rho}\cdot s_{\lambda}\left(  \xx\right)  =a_{\lambda+\rho}.
\label{sol.liri.180.d.s_lambda-as-quotient}%
\end{equation}
Applying this equality to $\lambda^{\vee}$ instead of $\lambda$, we obtain%
\begin{equation}
a_{\rho}\cdot s_{\lambda^{\vee}}\left(  \xx\right)  =a_{\lambda^{\vee
}+\rho} \label{sol.liri.180.d.s_lambda-check-as-quotient}%
\end{equation}
(since $\lambda^{\vee}$ is also a partition satisfying $\ell\left(
\lambda^{\vee}\right)  \leq k=n$).

Substituting the variables $x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}$ for
$x_{1},x_{2},\ldots,x_{n}$ in the equality
(\ref{sol.liri.180.d.s_lambda-as-quotient}), we obtain%
\begin{equation}
a_{\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \cdot
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)
=a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  .
\label{sol.liri.180.d.1}%
\end{equation}


Let $w_{\circ}:\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{
1,2,\ldots,n\right\}  $ be the map which sends every $i\in\left\{
1,2,\ldots,n\right\}  $ to $n+1-i$. Then, $w_{\circ}$ is a permutation of
$\left\{  1,2,\ldots,n\right\}  $, thus an element of $\Symm_n$. For
every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{equation}
\left(  w\circ w_{\circ}\right)  \left(  i\right)  =w\left(
\underbrace{w_{\circ}\left(  i\right)  }_{\substack{=n+1-i\\\text{(by the
definition of }w_{\circ}\left(  i\right)  \text{)}}}\right)  =w\left(
n+1-i\right)  . \label{sol.liri.180.d.wnach}%
\end{equation}
The map $\Symm_n\rightarrow\Symm_n,\ w\mapsto w\circ
w_{\circ}$ is a bijection (since $\Symm_n$ is a group, and since
$w_{\circ}\in\Symm_n$).

But every $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
\in\NN^n$ satisfies%
\begin{align}
a_{\alpha}  &  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{w\left(  {\mathbf{x}}^{\alpha}\right)  }_{\substack{=\xx%
^{\alpha}\left(  x_{w\left(  1\right)  },x_{w\left(  2\right)  }%
,\ldots,x_{w\left(  n\right)  }\right)  \\\text{(by the definition of
}w\left(  {\mathbf{x}}^{\alpha}\right)  \text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }a_{\alpha}\right) \nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\mathbf{x}^{\alpha}}_{\substack{=x_{1}^{\alpha_{1}}x_{2}%
^{\alpha_{2}}\cdots x_{n}^{\alpha_{n}}\\\text{(since }\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  \text{)}}}\left(  x_{w\left(
1\right)  },x_{w\left(  2\right)  },\ldots,x_{w\left(  n\right)  }\right)
\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\left(  x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots x_{n}%
^{\alpha_{n}}\right)  \left(  x_{w\left(  1\right)  },x_{w\left(  2\right)
},\ldots,x_{w\left(  n\right)  }\right)  }_{=x_{w\left(  1\right)  }%
^{\alpha_{1}}x_{w\left(  2\right)  }^{\alpha_{2}}\cdots x_{w\left(  n\right)
}^{\alpha_{n}}=\prod_{i=1}^{n}x_{w\left(  i\right)  }^{\alpha_{i}}}\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}x_{w\left(  i\right)  }^{\alpha_{i}}.
\label{sol.liri.180.d.a_alpha-formula}%
\end{align}


But $\lambda=\left(  \lambda_{1},\lambda_{2},\lambda_{3},\ldots\right)  $,
thus $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $
(since $\ell\left(  \lambda\right)  \leq k=n$). Hence,%
\begin{align*}
\underbrace{\lambda}_{=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{n}\right)  }+\underbrace{\rho}_{\substack{=\left(  n-1,n-2,\ldots
,2,1,0\right)  \\=\left(  n-1,n-2,\ldots,n-n\right)  }}  &  =\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  +\left(  n-1,n-2,\ldots
,n-n\right) \\
&  =\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,\ldots,\lambda_{n}+n-n\right)  .
\end{align*}
Thus, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to $\lambda+\rho$ and
$\left(  \lambda_{1}+n-1,\lambda_{2}+n-2,\ldots,\lambda_{n}+n-n\right)  $
instead of $\alpha$ and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{n}\right)  $) yields%
\begin{equation}
a_{\lambda+\rho}=\sum_{w\in\Symm_n}\operatorname{sgn}\left(
w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{\lambda_{i}+n-i}.
\label{sol.liri.180.d.a_lambdaplusrho1}%
\end{equation}


Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$, $x_{n}^{-1}$ for $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ on both sides of this equality, we obtain%
\begin{align*}
&  a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{\left(  x_{w\left(  i\right)  }^{-1}\right)
^{\lambda_{i}+n-i}}_{=x_{w\left(  i\right)  }^{-\left(  \lambda_{i}%
+n-i\right)  }}=\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda
_{i}+n-i\right)  }}_{\substack{=\prod_{i=1}^{n}x_{w\left(  n+1-i\right)
}^{-\left(  \lambda_{n+1-i}+n-\left(  n+1-i\right)  \right)  }\\\text{(here,
we have substituted}\\n+1-i\text{ for }i\text{ in the product)}}}\\
&  =\sum_{w\in\Symm_n}\underbrace{\operatorname{sgn}\left(  w\right)
}_{=\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }%
\operatorname*{sgn}\left(  w\right)  \operatorname*{sgn}\left(  w_{\circ
}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)  }^{-\left(
\lambda_{n+1-i}+n-\left(  n+1-i\right)  \right)  }}_{\substack{=x_{w\left(
n+1-i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)  }\\\text{(since
}n-\left(  n+1-i\right)  =i-1\text{)}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\underbrace{\operatorname*{sgn}\left(  w\right)
\operatorname*{sgn}\left(  w_{\circ}\right)  }_{=\operatorname*{sgn}\left(
w\circ w_{\circ}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }}_{\substack{=x_{\left(  w\circ
w_{\circ}\right)  \left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\\\text{(since }w\left(  n+1-i\right)  =\left(  w\circ w_{\circ}\right)
\left(  i\right)  \\\text{(by (\ref{sol.liri.180.d.wnach})))}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\operatorname*{sgn}\left(  w\circ w_{\circ}\right)  \prod_{i=1}%
^{n}x_{\left(  w\circ w_{\circ}\right)  \left(  i\right)  }^{-\left(
\lambda_{n+1-i}+i-1\right)  }=\sum_{w\in\Symm_n}\dfrac{1}%
{\operatorname*{sgn}\left(  w_{\circ}\right)  }\operatorname*{sgn}\left(
w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda
_{n+1-i}+i-1\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }w\text{ for }w\circ w_{\circ}\text{ in the sum}\\
\text{(since the map }\Symm_n\rightarrow\Symm_n,\ w\mapsto
w\circ w_{\circ}\text{ is a bijection)}%
\end{array}
\right) \\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}%
^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)  }.
\end{align*}
Multiplying both sides of this equality by $\operatorname*{sgn}\left(
w_{\circ}\right)  $, we obtain%
\begin{equation}
\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot a_{\lambda+\rho}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  =\sum_{w\in\Symm_n%
}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }.
\label{sol.liri.180.d.a_lambdaplusrho}%
\end{equation}


But $\lambda^{\vee}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda
_{1}\right)  =\left(  m-\lambda_{n},m-\lambda_{n-1},\ldots,m-\lambda
_{1}\right)  $ (since $k=n$), so that%
\begin{align*}
&  \underbrace{\lambda^{\vee}}_{\substack{=\left(  m-\lambda_{n}%
,m-\lambda_{n-1},\ldots,m-\lambda_{1}\right)  \\=\left(  m-\lambda_{\left(
n+1\right)  -1},m-\lambda_{\left(  n+1\right)  -2},\ldots,m-\lambda_{\left(
n+1\right)  -n}\right)  }}+\underbrace{\rho}_{\substack{=\left(
n-1,n-2,\ldots,2,1,0\right)  \\=\left(  n-1,n-2,\ldots,n-n\right)  }}\\
&  =\left(  m-\lambda_{\left(  n+1\right)  -1},m-\lambda_{\left(  n+1\right)
-2},\ldots,m-\lambda_{\left(  n+1\right)  -n}\right)  +\left(  n-1,n-2,\ldots
,n-n\right) \\
&  =\left(  \left(  m-\lambda_{\left(  n+1\right)  -1}\right)  +\left(
n-1\right)  ,\left(  m-\lambda_{\left(  n+1\right)  -2}\right)  +\left(
n-2\right)  ,\ldots,\left(  m-\lambda_{\left(  n+1\right)  -n}\right)
+\left(  n-n\right)  \right)  .
\end{align*}
Hence, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to $\lambda^{\vee}%
+\rho$ and \newline$\left(  \left(  m-\lambda_{\left(  n+1\right)  -1}\right)
+\left(  n-1\right)  ,\left(  m-\lambda_{\left(  n+1\right)  -2}\right)
+\left(  n-2\right)  ,\ldots,\left(  m-\lambda_{\left(  n+1\right)
-n}\right)  +\left(  n-n\right)  \right)  $ instead of $\alpha$ and
\newline$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $) yields%
\begin{align}
a_{\lambda^{\vee}+\rho}  &  =\sum_{w\in\Symm_n}\operatorname{sgn}%
\left(  w\right)  \prod_{i=1}^{n}\underbrace{x_{w\left(  i\right)  }^{\left(
m-\lambda_{\left(  n+1\right)  -i}\right)  +\left(n-i\right)}
}_{\substack{=x_{w\left(
i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)  +\left(  n+m-1\right)
}\\\text{(since }\left(  m-\lambda_{\left(  n+1\right)  -i}\right)
+\left(n-i\right)=\left(  m-\lambda_{n+1-i}\right)  +\left(n-i\right)\\
=-\left(  \lambda_{n+1-i}+i-1\right)  +\left(  n+m-1\right)  \text{)}}}
\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}%
+i-1\right)  +\left(  n+m-1\right)  }}_{=x_{w\left(  i\right)  }^{-\left(
\lambda_{n+1-i}+i-1\right)  }x_{w\left(  i\right)  }^{n+m-1}}=\sum
_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \underbrace{\prod
_{i=1}^{n}\left(  x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}%
+i-1\right)  }x_{w\left(  i\right)  }^{n+m-1}\right)  }_{=\left(  \prod
_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\right)  \left(  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{n+m-1}\right)
}\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\right)  \left(  \underbrace{\prod_{i=1}^{n}}_{=\prod_{i\in\left\{
1,2,\ldots,n\right\}  }}x_{w\left(  i\right)  }^{n+m-1}\right) \nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\right)  \underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}
}x_{w\left(  i\right)  }^{n+m-1}\right)  }_{\substack{=\prod_{i\in\left\{
1,2,\ldots,n\right\}  }x_{i}^{n+m-1}\\\text{(here, we have substituted
}i\text{ for }w\left(  i\right)  \text{ in the product}\\\text{(since }w\text{
is a permutation of }\left\{  1,2,\ldots,n\right\}  \text{))}}}\nonumber\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{-\left(  \lambda_{n+1-i}+i-1\right)
}\right)  \left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}%
^{n+m-1}\right) \nonumber\\
&  =\underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
x_{i}^{n+m-1}\right)  }_{=\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}
}x_{i}\right)  ^{n+m-1}}\underbrace{\sum_{w\in\Symm_n}%
\operatorname{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{-\left(  \lambda_{n+1-i}+i-1\right)  }}_{\substack{=\operatorname*{sgn}%
\left(  w_{\circ}\right)  \cdot a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right)  \\\text{(by
(\ref{sol.liri.180.d.a_lambdaplusrho}))}}}\nonumber\\
&  =\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)
^{n+m-1}\cdot\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot a_{\lambda
+\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  .
\label{sol.liri.180.d.a_lambdaplusrho-comes-around}%
\end{align}


Finally, $\rho=\left(  n-1,n-2,\ldots,2,1,0\right)  =\left(  n-1,n-2,\ldots
,n-n\right)  $. Thus, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to
$\rho$ and \newline$\left(  n-1,n-2,\ldots,n-n\right)  $ instead of $\alpha$
and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $) yields%
\begin{equation}
a_{\rho}=\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}. \label{sol.liri.180.d.a_rho}%
\end{equation}
Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$, $x_{n}^{-1}$ for $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ on both sides of this equality, we obtain%
\begin{align*}
&  a_{\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{\left(  x_{w\left(  i\right)  }^{-1}\right)  ^{n-i}%
}_{\substack{=x_{w\left(  i\right)  }^{-\left(  n-i\right)  }=x_{w\left(
i\right)  }^{i-n}\\\text{(since }-\left(  n-i\right)
=i-n  \text{)}}}=\sum_{w\in\Symm_n}\operatorname{sgn}%
\left(  w\right)  \underbrace{\prod_{i=1}^{n}x_{w\left(  i\right)  }^{i-n}%
}_{\substack{=\prod_{i=1}^{n}x_{w\left(  n+1-i\right)  }^{\left(
n+1-i\right)  -n}\\\text{(here, we have substituted}\\n+1-i\text{ for }i\text{
in the product)}}}\\
&  =\sum_{w\in\Symm_n}\underbrace{\operatorname{sgn}\left(  w\right)
}_{=\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }%
\operatorname*{sgn}\left(  w\right)  \operatorname*{sgn}\left(  w_{\circ
}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)  }^{\left(
n+1-i\right)  -n}}_{\substack{=x_{w\left(  n+1-i\right)  }^{-i+1}%
\\\text{(since }\left(  n+1-i\right)  -n=-i+1\text{)}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\underbrace{\operatorname*{sgn}\left(  w\right)
\operatorname*{sgn}\left(  w_{\circ}\right)  }_{=\operatorname*{sgn}\left(
w\circ w_{\circ}\right)  }\prod_{i=1}^{n}\underbrace{x_{w\left(  n+1-i\right)
}^{-i+1}}_{\substack{=x_{\left(  w\circ w_{\circ}\right)  \left(  i\right)
}^{-i+1}\\\text{(since }w\left(  n+1-i\right)  =\left(  w\circ w_{\circ
}\right)  \left(  i\right)  \\\text{(by (\ref{sol.liri.180.d.wnach})))}}}\\
&  =\sum_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ
}\right)  }\operatorname*{sgn}\left(  w\circ w_{\circ}\right)  \prod_{i=1}%
^{n}x_{\left(  w\circ w_{\circ}\right)  \left(  i\right)  }^{-i+1}=\sum
_{w\in\Symm_n}\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)
}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{-i+1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }w\text{ for }w\circ w_{\circ}\text{ in the sum}\\
\text{(since the map }\Symm_n\rightarrow\Symm_n,\ w\mapsto
w\circ w_{\circ}\text{ is a bijection)}%
\end{array}
\right) \\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}%
^{n}\underbrace{x_{w\left(  i\right)  }^{-i+1}}_{\substack{=x_{w\left(
i\right)  }^{\left(  n-i\right)  +\left(  1-n\right)  }\\\text{(since
}-i+1=\left(  n-i\right)  +\left(  1-n\right)  \text{)}}}=\dfrac
{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum_{w\in\Symm_n%
}\operatorname*{sgn}\left(  w\right)  \prod_{i=1}^{n}\underbrace{x_{w\left(
i\right)  }^{\left(  n-i\right)  +\left(  1-n\right)  }}_{=x_{w\left(
i\right)  }^{n-i}x_{w\left(  i\right)  }^{1-n}}\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \underbrace{\prod
_{i=1}^{n}\left(  x_{w\left(  i\right)  }^{n-i}x_{w\left(  i\right)  }%
^{1-n}\right)  }_{=\left(  \prod_{i=1}^{n}x_{w\left(  i\right)  }%
^{n-i}\right)  \left(  \prod_{i=1}^{n}x_{w\left(  i\right)  }^{1-n}\right)
}\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}\right)  \left(  \underbrace{\prod
_{i=1}^{n}}_{=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }}x_{w\left(
i\right)  }^{1-n}\right) \\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}\right)  \underbrace{\left(
\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{w\left(  i\right)  }%
^{1-n}\right)  }_{\substack{=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
x_{i}^{1-n}\\\text{(here, we have substituted }i\text{ for }w\left(  i\right)
\text{ in the product}\\\text{(since }w\text{ is a permutation of }\left\{
1,2,\ldots,n\right\}  \text{))}}}\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\sum
_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}\right)  \left(  \prod
_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}^{1-n}\right)  =\dfrac
{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\underbrace{\left(
\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}^{1-n}\right)  }_{=\left(
\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)  ^{1-n}}%
\underbrace{\sum_{w\in\Symm_n}\operatorname*{sgn}\left(  w\right)
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{n-i}}_{\substack{=a_{\rho}\\\text{(by
(\ref{sol.liri.180.d.a_rho}))}}}\\
&  =\dfrac{1}{\operatorname*{sgn}\left(  w_{\circ}\right)  }\left(
\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)  ^{1-n}a_{\rho}.
\end{align*}
Multiplying both sides of this equality by $\operatorname*{sgn}\left(
w_{\circ}\right)  $, we obtain%
\begin{equation}
\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot a_{\rho}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  =\left(  \prod_{i\in\left\{
1,2,\ldots,n\right\}  }x_{i}\right)  ^{1-n}a_{\rho}.
\label{sol.liri.180.d.a_rho-comes-around}%
\end{equation}


Now, (\ref{sol.liri.180.d.s_lambda-check-as-quotient}) becomes%
\begin{align}
a_{\rho}\cdot s_{\lambda^{\vee}}\left(  \xx\right)
&=a_{\lambda^{\vee}+\rho}
\nonumber\\
&=\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)
^{n+m-1}\cdot\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot
\underbrace{a_{\lambda+\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}%
^{-1}\right)  }_{\substack{=a_{\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right)  \cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{n}^{-1}\right)  \\\text{(by (\ref{sol.liri.180.d.1}))}%
}}\nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.d.a_lambdaplusrho-comes-around})}\right) \nonumber\\
&  =\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)
^{n+m-1}\cdot\underbrace{\operatorname*{sgn}\left(  w_{\circ}\right)  \cdot
a_{\rho}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  }%
_{\substack{=\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)
^{1-n}a_{\rho}\\\text{(by (\ref{sol.liri.180.d.a_rho-comes-around}))}}}\cdot
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \nonumber\\
&  =\underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }%
x_{i}\right)  ^{n+m-1}\cdot\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}
}x_{i}\right)  ^{1-n}}_{\substack{=\left(  \prod_{i\in\left\{  1,2,\ldots
,n\right\}  }x_{i}\right)  ^{\left(  n+m-1\right)  +\left(  1-n\right)
}=\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}\right)
^{m}\\\text{(since }\left(  n+m-1\right)  +\left(  1-n\right)  =m\text{)}%
}}a_{\rho}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}%
^{-1}\right) \nonumber\\
&  =\left(  \underbrace{\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}%
}_{=x_{1}x_{2}\cdots x_{n}}\right)  ^{m}a_{\rho}\cdot s_{\lambda}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \nonumber\\
&=\left(  x_{1}x_{2}\cdots
x_{n}\right)  ^{m}a_{\rho}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right) \nonumber\\
&  =a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  .
\label{sol.liri.180.d.almostthere}%
\end{align}
Hence,%
\begin{align}
&  a_{\rho}\cdot\left(  s_{\lambda^{\vee}}\left(  \xx\right)  -\left(
x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \right) \nonumber\\
&  =\underbrace{a_{\rho}\cdot s_{\lambda^{\vee}}\left(  \xx\right)
}_{\substack{=a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \\\text{(by
(\ref{sol.liri.180.d.almostthere}))}}}-a_{\rho}\cdot\left(  x_{1}x_{2}\cdots
x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right) \nonumber\\
&  =a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right) \nonumber\\
&\ \ \ \ \ \ \ \ \ \   -a_{\rho}\cdot\left(
x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right)  \nonumber\\
&=0. \label{sol.liri.180.d.almostthere2}%
\end{align}


But\footnote{In the following, a \emph{non-zero-divisor} in a commutative
ring $B$ means an element $b\in B$ such that every element $c\in B$ satisfying
$bc=0$ must satisfy $c=0$.}
\begin{equation}
a_{\rho}\text{ is a non-zero-divisor in the ring }\kk\left[
x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  .
\label{sol.liri.180.d.nonzerodiv}%
\end{equation}
\footnote{\textit{Proof.} We know (from a footnote in Corollary
\ref{cor.stembridge.slambda}) that $a_{\rho}$ is not a zero-divisor in the
ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $. In other words,
$a_{\rho}$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $.
\par
Now, let $b$ be any element of the ring $\kk\left[  x_{1},x_{2}%
,\ldots,x_{n}\right]  $ such that $b$ is a non-zero-divisor in the ring
$\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $. We shall show that $b$
is a non-zero-divisor in the ring $\kk\left[  x_{1},x_{2},\ldots
,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ as well.
\par
Indeed, let $c\in\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ be such that $bc=0$. It is known that
every element of $\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ has the form $\dfrac{p}{x_{1}^{g_{1}%
}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}$ for some $p\in\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $ and some $\left(  g_{1},g_{2},\ldots
,g_{n}\right)  \in\NN^n$. Applying this to the element $c$, we
conclude that the element $c$ has the form $\dfrac{p}{x_{1}^{g_{1}}%
x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}$ for some $p\in\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $ and some $\left(  g_{1},g_{2},\ldots
,g_{n}\right)  \in\NN^n$. Consider this $p$ and this $\left(
g_{1},g_{2},\ldots,g_{n}\right)  $. We thus have $c=\dfrac{p}{x_{1}^{g_{1}%
}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}$. Now, $bc=0$. This rewrites as
$b\cdot\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}=0$ (since
$c=\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}$). Multiplying
both sides of this equality by $x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}%
}$, we obtain $bp=0$. This equality holds in the ring $\kk\left[
x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $,
and thus also in the ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]
$ (since $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ is a subring of
$\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right]  $). Thus, $p=0$ (since $b$ is a non-zero-divisor in the
ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $, and since
$p\in\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $). Hence,
\begin{align*}
c  &  =\dfrac{p}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}=\dfrac
{0}{x_{1}^{g_{1}}x_{2}^{g_{2}}\cdots x_{n}^{g_{n}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }p=0\right) \\
&  =0.
\end{align*}
\par
Now, let us forget that we fixed $c$. We thus have proven that every element
$c\in\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{n}^{-1}\right]  $ satisfying $bc=0$ must satisfy $c=0$. In other
words, $b$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $.
\par
Now, let us forget that we fixed $b$. We thus have proven that if $b$ is any
element of the ring $\kk\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ such
that $b$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $, then $b$ is a non-zero-divisor in the ring
$\kk\left[  x_{1},x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right]  $ as well. Applying this to $b=a_{\rho}$, we conclude that
$a_{\rho}$ is a non-zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n},x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right]  $ (since
we know that $a_{\rho}$ is a non-zero-divisor in the ring $\kk\left[
x_{1},x_{2},\ldots,x_{n}\right]  $). This proves
(\ref{sol.liri.180.d.nonzerodiv}).} Hence, we can cancel the factor $a_{\rho}$
from the equality (\ref{sol.liri.180.d.almostthere2}). As a result, we obtain%
\[
s_{\lambda^{\vee}}\left(  \xx\right)  -\left(  x_{1}x_{2}\cdots
x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{n}^{-1}\right)  =0.
\]
In other words, $s_{\lambda^{\vee}}\left(  \xx\right)  =\left(
x_{1}x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{n}^{-1}\right)  $. Compared with $s_{\lambda^{\vee}%
}\left(  \xx\right)  =s_{\lambda^{\vee}}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  $ (since $\xx=\left(  x_{1},x_{2},\ldots
,x_{k}\right)  $), this yields%
\[
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}%
x_{2}\cdots x_{n}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{n}^{-1}\right)  =\left(  x_{1}x_{2}\cdots x_{k}\right)
^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\]
(since $n=k$). This solves Exercise \ref{exe.liri.180}(d).

\textit{Second solution to Exercise \ref{exe.liri.180}(d) (sketched):} Let us
give a more combinatorial solution now. In the following, if $\alpha$ is a
partition, then an \emph{$\left(  \alpha,k\right)  $-CST} will mean a
column-strict tableau $T$ of shape $\alpha$ such that all entries of $T$
belong to $\left\{  1,2,\ldots,k\right\}  $.

We have%
\[
\underbrace{s_{\lambda^{\vee}}}_{=s_{\lambda^{\vee}/\varnothing}}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =s_{\lambda^{\vee}/\varnothing}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}%
\\\text{tableau of shape }\lambda^{\vee}/\varnothing\text{;}\\\text{all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,\ldots,k\right\}
}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }%
\]
(by Exercise \ref{exe.schur.finite}(a), applied to $k$, $\lambda^{\vee}$ and
$\varnothing$ instead of $n$, $\lambda$ and $\mu$), where $\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  ^{\cont \left(  T\right)  }$ is
defined as in Exercise \ref{exe.schur.finite}(a). In other words,%
\begin{align}
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)   &  =\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }\lambda
^{\vee}/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)
^{\cont \left(  T\right)  }=\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda^{\vee}\text{;}\\\text{all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,\ldots,k\right\}
}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since column-strict tableaux of shape }\lambda^{\vee}/\varnothing\text{
are}\\
\text{the same as column-strict tableaux of shape }\lambda^{\vee}%
\end{array}
\right) \nonumber\\
&  =\sum_{T\text{ is a }\left(  \lambda^{\vee},k\right)  \text{-CST}}\left(
x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(  T\right)  }
\label{sol.liri.180.d.alternative1}%
\end{align}
(because the column-strict tableaux $T$ of shape $\lambda^{\vee}$ such that
all entries of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $ are precisely
the $\left(  \lambda^{\vee},k\right)  $-CSTs).

On the other hand,%
\[
\underbrace{s_{\lambda}}_{=s_{\lambda/\varnothing}}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  =s_{\lambda/\varnothing}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of
shape }\lambda/\varnothing\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  ^{\cont \left(  T\right)  }%
\]
(by Exercise \ref{exe.schur.finite}(a), applied to $k$ and $\varnothing$
instead of $n$ and $\mu$), where $\left(  x_{1},x_{2},\ldots,x_{k}\right)
^{\cont \left(  T\right)  }$ is defined as in Exercise
\ref{exe.schur.finite}(a). Substituting $x_{1}^{-1}$, $x_{2}^{-1}$, $\ldots$,
$x_{k}^{-1}$ for $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ on both sides of this
equality, we obtain%
\begin{equation}
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{k}^{-1}\right)  ^{\cont \left(  T\right)  },
\label{sol.liri.180.d.alternative2a}%
\end{equation}
where $\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
^{\cont \left(  T\right)  }$ is defined as $\prod_{i=1}%
^{k}\left(  x_{i}^{-1}\right)  ^{\left\vert T^{-1}\left(  i\right)
\right\vert }$ for any column-strict tableau $T$ having the property that all
entries of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $. The equality
(\ref{sol.liri.180.d.alternative2a}) becomes%
\begin{align*}
s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)   &
=\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{k}^{-1}\right)  ^{\cont \left(  T\right)  }=\sum\limits
_{\substack{T\text{ is a column-strict}\\\text{tableau of shape }%
\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  ^{\cont \left(  T\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since column-strict tableaux of shape }\lambda/\varnothing\text{ are}\\
\text{the same as column-strict tableaux of shape }\lambda
\end{array}
\right) \\
&  =\sum_{T\text{ is a }\left(  \lambda,k\right)  \text{-CST}}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  ^{\cont \left(
T\right)  }%
\end{align*}
(since the column-strict tableaux $T$ of shape $\lambda$ such that all entries
of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $ are precisely the $\left(
\lambda,k\right)  $-CSTs). Multiplying both sides of this equality by $\left(
x_{1}x_{2}\cdots x_{k}\right)  ^{m}$, we obtain%
\begin{align}
&  \left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\lambda}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \nonumber\\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\sum_{T\text{ is a
}\left(  \lambda,k\right)  \text{-CST}}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  ^{\cont \left(  T\right)  }\nonumber\\
&  =\sum_{T\text{ is a }\left(  \lambda,k\right)  \text{-CST}}\left(
x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  ^{\cont \left(  T\right)  }.
\label{sol.liri.180.d.alternative2}%
\end{align}


We need to prove that%
\[
s_{\lambda^{\vee}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}%
x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  .
\]
Due to (\ref{sol.liri.180.d.alternative1}) and
(\ref{sol.liri.180.d.alternative2}), this rewrites as%
\begin{align}
&  \sum_{T\text{ is a }\left(  \lambda^{\vee},k\right)  \text{-CST}}\left(
x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(  T\right)
}\nonumber\\
&  =\sum_{T\text{ is a }\left(  \lambda,k\right)  \text{-CST}}\left(
x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  ^{\cont \left(  T\right)  }.
\label{sol.liri.180.d.alternative4}%
\end{align}
So it remains to prove (\ref{sol.liri.180.d.alternative4}).

In order to prove (\ref{sol.liri.180.d.alternative4}), it is clearly
sufficient to construct a bijection%
\[
\Omega:\left(  \text{the set of all }\left(  \lambda,k\right)  \text{-CSTs}%
\right)  \rightarrow\left(  \text{the set of all }\left(  \lambda^{\vee
},k\right)  \text{-CSTs}\right)
\]
with the property that every $\left(  \lambda,k\right)  $-CST $T$ satisfies%
\begin{equation}
\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
\Omega\left(  T\right)  \right)  }=\left(  x_{1}x_{2}\cdots x_{k}\right)
^{m}\cdot\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
^{\cont \left(  T\right)  }.
\label{sol.liri.180.d.alternative.condition-on-Phi}%
\end{equation}


Let us construct such a $\Omega$ now. We begin by doing some elementary combinatorics.

We define a relation $\leq_{\#}$ on the set of all subsets of $\left\{  1,2,
\ldots,k\right\}  $ as follows:

\begin{definition}
Let $k$ be a nonnegative integer. Let $\left[  k\right]  =\left\{  1,2,
\ldots,k\right\}  $. Let $I$ and $J$ be two subsets of $\left[  k\right]  $.
We say that $I\leq_{\#}J$ if the following two properties hold:

-- We have $\left\vert I\right\vert \geq\left\vert J\right\vert $.

-- For every $r\in\left\{  1,2, \ldots,\left\vert J\right\vert \right\}  $,
the $r$-th smallest element of $I$ is $\leq$ to the $r$-th smallest element of
$J$.
\end{definition}

We notice that this relation $\leq_{\#}$ is the less-or-equal relation of a
partial order (as follows easily from the definition); but we will not have
any use for this fact. Instead, we need a symmetry property:

\begin{proposition}
\label{prop.sol.liri.180.d.complement}Let $k$ be a nonnegative integer. Let
$\left[  k\right]  =\left\{  1,2,\ldots,k\right\}  $. Let $I$ and $J$ be two
subsets of $\left[  k\right]  $.

\begin{enumerate}
\item[(a)] We have $I\leq_{\#}J$ if and only if $\left[  k\right]  \setminus
J\leq_{\#}\left[  k\right]  \setminus I$.

\item[(b)] For every $\ell\in\left[  k\right]  $ and $S\subset\left[
k\right]  $, let $\alpha_{S}\left(  \ell\right)  $ denote the number
$\left\vert \left\{  s\in S\ \mid\ s\leq\ell\right\}  \right\vert $. Then,%
\begin{equation}
I\leq_{\#}J\text{ holds if and only if every }\ell\in\left[  k\right]  \text{
satisfies }\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  .
\label{pf.prop.sol.liri.180.d.complement.maj}%
\end{equation}

\end{enumerate}
\end{proposition}

\textit{Proof of Proposition \ref{prop.sol.liri.180.d.complement}.} (b)
\textit{Proof of (\ref{pf.prop.sol.liri.180.d.complement.maj}):}
$\Longrightarrow:$ Assume that $I\leq_{\#}J$. In other words, the following
two properties hold:

\textit{Property $\alpha$:} We have $\left\vert I\right\vert
\geq\left\vert J\right\vert $.

\textit{Property $\beta$:} For every $r\in\left\{  1,2,
\ldots,\left\vert J\right\vert \right\}  $, the $r$-th smallest element of $I$
is $\leq$ to the $r$-th smallest element of $J$.

Now, let $\ell\in\left[  k\right]  $. Then, we need to show that $\alpha
_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  $. Since this is
obvious if $\alpha_{J}\left(  \ell\right)  =0$ (because $\alpha_{I}\left(
\ell\right)  \geq0$), we can WLOG assume that $\alpha_{J}\left(  \ell\right)
\neq0$. Assume this. Thus, $\alpha_{J}\left(  \ell\right)  \geq1$. Also,
$\alpha_{J}\left(  \ell\right)  =\left\vert \underbrace{\left\{  s\in
J\ \mid\ s\leq\ell\right\}  }_{\subset J}\right\vert \leq\left\vert
J\right\vert \leq\left\vert I\right\vert $ (since $\left\vert I\right\vert
\geq\left\vert J\right\vert $). Hence, both the $\alpha_{J}\left(
\ell\right)  $-th smallest element of $J$ and the $\alpha_{J}\left(
\ell\right)  $-th smallest element of $I$ are well-defined.

Since $\alpha_{J}\left(  \ell\right)  =\left\vert \left\{  s\in J\ \mid
\ s\leq\ell\right\}  \right\vert $, we know that the elements of $J$ which are
$\leq\ell$ are precisely the $\alpha_{J}\left(  \ell\right)  $ smallest
elements of $J$. Thus,%
\[
\left(  \text{the }\alpha_{J}\left(  \ell\right)  \text{-th smallest element
of }J\right)  =\left(  \text{the largest element of }J\text{ which is }%
\leq\ell\right)  .
\]
But by Property $\beta$ (applied to $r=\alpha_{J}\left(  \ell\right)  $), we
have%
\begin{align*}
\left(  \text{the }\alpha_{J}\left(  \ell\right)  \text{-th smallest element
of }I\right)   &  \leq\left(  \text{the }\alpha_{J}\left(  \ell\right)
\text{-th smallest element of }J\right) \\
&  =\left(  \text{the largest element of }J\text{ which is }\leq\ell\right)
\leq\ell.
\end{align*}
Hence, there are at least $\alpha_{J}\left(  \ell\right)  $ elements of $I$
which are $\leq\ell$ (namely, the $\alpha_{J}\left(  \ell\right)  $ smallest
ones). In other words, $\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}
\right\vert \geq\alpha_{J}\left(  \ell\right)  $. Now, $\alpha_{I}\left(
\ell\right)  =\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}  \right\vert
\geq\alpha_{J}\left(  \ell\right)  $. We thus have proven the $\Longrightarrow
$ direction of (\ref{pf.prop.sol.liri.180.d.complement.maj}).

$\Longleftarrow:$ Assume that every $\ell\in\left[  k\right]  $ satisfies
$\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  $. We need
to prove that $I\leq_{\#}J$. In other words, we need to prove that the
following two properties hold:

\textit{Property $\alpha$:} We have $\left\vert I\right\vert
\geq\left\vert J\right\vert $.

\textit{Property $\beta$:} For every $r\in\left\{  1,2,
\ldots,\left\vert J\right\vert \right\}  $, the $r$-th smallest element of $I$
is $\leq$ to the $r$-th smallest element of $J$.

First of all, $\left\{  s\in I\ \mid\ s\leq k\right\}  =I$ (since every $s\in
I$ satisfies $s\leq k$), and the definition of $\alpha_{I}\left(  k\right)  $
yields $\alpha_{I}\left(  k\right)  =\left\vert \underbrace{\left\{  s\in
I\ \mid\ s\leq k\right\}  }_{=I}\right\vert =\left\vert I\right\vert $.
Similarly, $\alpha_{J}\left(  k\right)  =\left\vert J\right\vert $. Applying
$\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  $ to
$\ell=k$, we obtain $\alpha_{I}\left(  k\right)  \geq\alpha_{J}\left(
k\right)  $, so that $\left\vert I\right\vert =\alpha_{I}\left(  k\right)
\geq\alpha_{J}\left(  k\right)  =\left\vert J\right\vert $, and thus Property
$\alpha$ is proven.

Now, let $r\in\left\{  1,2, \ldots,\left\vert J\right\vert \right\}  $. The
$r$-th smallest element of $I$ and the $r$-th smallest element of $J$ are then
well-defined (because of $r\leq\left\vert J\right\vert \leq\left\vert
I\right\vert $). Let $\ell$ be the $r$-th smallest element of $J$. Then,
$\left\{  s\in J\ \mid\ s\leq\ell\right\}  $ is the set consisting of the $r$
smallest elements of $J$, so that $\left\vert \left\{  s\in J\ \mid\ s\leq
\ell\right\}  \right\vert =r$. Now, $\alpha_{J}\left(  \ell\right)
=\left\vert \left\{  s\in J\ \mid\ s\leq\ell\right\}  \right\vert =r$.

But $\alpha_{I}\left(  \ell\right)  =\left\vert \left\{  s\in I\ \mid
\ s\leq\ell\right\}  \right\vert $, so that%
\[
\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}  \right\vert =\alpha
_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)  =r.
\]
In other words, there exist at least $r$ elements of $I$ which are $\leq\ell$.
Hence, the $r$-th smallest element of $I$ must be $\leq\ell$. Since $\ell$ is
the $r$-th smallest element of $J$, this rewrites as follows: The $r$-th
smallest element of $I$ is $\leq$ to the $r$-th smallest element of $J$. Thus,
Property $\beta$ holds. Now we know that both Properties $\alpha$ and $\beta$
hold. Hence, $I\leq_{\#}J$ holds (which, as we know, is equivalent to the
conjunction of said properties). This proves the $\Longleftarrow$ direction of
(\ref{pf.prop.sol.liri.180.d.complement.maj}). Thus,
(\ref{pf.prop.sol.liri.180.d.complement.maj}) is proven. In other words,
Proposition \ref{prop.sol.liri.180.d.complement}(b) is proven.

(a) For every $\ell\in\left[  k\right]  $ and $S\subset\left[  k\right]  $,
let $\alpha_{S}\left(  \ell\right)  $ denote the number $\left\vert \left\{
s\in S\ \mid\ s\leq\ell\right\}  \right\vert $. Thus, every $\ell\in\left[
k\right]  $ satisfies%
\begin{align*}
\alpha_{I}\left(  \ell\right)  +\alpha_{\left[  k\right]  \setminus I}\left(
\ell\right)   &  =\left\vert \left\{  s\in I\ \mid\ s\leq\ell\right\}
\right\vert +\left\vert \left\{  s\in\left[  k\right]  \setminus
I\ \mid\ s\leq\ell\right\}  \right\vert \\
&  =\left\vert \left\{  s\in\underbrace{I\cup\left(  \left[  k\right]
\setminus I\right)  }_{=\left[  k\right]  }\ \mid\ s\leq\ell\right\}
\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since }I\text{ and }\left[
k\right]  \setminus I\text{ are disjoint}\right) \\
&  =\left\vert \left\{  s\in\left[  k\right]  \ \mid\ s\leq\ell\right\}
\right\vert =\ell,
\end{align*}
so that $\alpha_{\left[  k\right]  \setminus I}\left(  \ell\right)
=\ell-\alpha_{I}\left(  \ell\right)  $. Similarly, every $\ell\in\left[
k\right]  $ satisfies $\alpha_{\left[  k\right]  \setminus J}\left(
\ell\right)  =\ell-\alpha_{J}\left(  \ell\right)  $.

Applying (\ref{pf.prop.sol.liri.180.d.complement.maj}) to $\left[  k\right]
\setminus J$ and $\left[  k\right]  \setminus I$ in lieu of $I$ and $J$, we
obtain that%
\begin{equation}
\left[  k\right]  \setminus J\leq_{\#}\left[  k\right]  \setminus I\text{
holds if and only if every }\ell\in\left[  k\right]  \text{ satisfies }%
\alpha_{\left[  k\right]  \setminus J}\left(  \ell\right)  \geq\alpha_{\left[
k\right]  \setminus I}\left(  \ell\right)  .
\label{pf.prop.sol.liri.180.d.complement.maj.comp}%
\end{equation}


Now, we have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  \left[  k\right]  \setminus J\leq_{\#}\left[  k\right]  \setminus
I\right) \\
&  \Longleftrightarrow\ \left(  \text{every }\ell\in\left[  k\right]  \text{
satisfies }\underbrace{\alpha_{\left[  k\right]  \setminus J}\left(
\ell\right)  }_{=\ell-\alpha_{J}\left(  \ell\right)  }\geq\underbrace{\alpha
_{\left[  k\right]  \setminus I}\left(  \ell\right)  }_{=\ell-\alpha
_{I}\left(  \ell\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.liri.180.d.complement.maj.comp})}\right) \\
&  \Longleftrightarrow\ \left(  \text{every }\ell\in\left[  k\right]  \text{
satisfies }\ell-\alpha_{J}\left(  \ell\right)  \geq\ell-\alpha_{I}\left(
\ell\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{every }\ell\in\left[  k\right]  \text{
satisfies }\alpha_{I}\left(  \ell\right)  \geq\alpha_{J}\left(  \ell\right)
\right) \\
&  \Longleftrightarrow\ \left(  I\leq_{\#}J\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.liri.180.d.complement.maj})}\right)  .
\end{align*}
This proves Proposition \ref{prop.sol.liri.180.d.complement}(a).

We now return to the solution of Exercise \ref{exe.liri.180}(d). Let us first
define some more notations.

If $T$ is a column-strict tableau and $i$ is an integer, then the
\emph{$i$-th set column} of $T$ will mean the set of the entries in the
$i$-th column of $T$. Notice that the cardinality of the $i$-th set column of
$T$ is the length of the $i$-th column of the shape of $T$ (since every column
of a column-strict tableau has all its entries distinct), and that the $i$-th
column of $T$ can be uniquely reconstructed from the $i$-th set column of $T$
(because the order of the entries in a column can only be increasing).

For every subset $S$ of $\left[  k\right]  $, we define $\xx_{S}$ to be
the monomial $\prod_{s\in S}x_{s}$ in $\kk\left[  x_{1},x_{2}%
,\ldots,x_{k}\right]  $. If $\left(  S_{1},S_{2},\ldots,S_{m}\right)  $ is an
$m$-tuple of subsets of $\left[  k\right]  $, then we set $\xx_{\left(
S_{1},S_{2},\ldots,S_{m}\right)  }=\prod_{i=1}^{m}\xx_{S_{i}}$.

Let $\lambda^{t}$ denote the conjugate of the partition $\lambda$. Then,
$\left(  \lambda^{t}\right)  _{i}$ is the length of the $i$-th column of the
Ferrers diagram of $\lambda$ for every $i\in\left\{  1,2,\ldots,m\right\}  $.

Let $A\left(  \lambda\right)  $ denote the set of all $m$-tuples $\left(
I_{1},I_{2},\ldots,I_{m}\right)  $ of subsets of $\left[  k\right]  $
satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ and $\left(
\left\vert I_{i}\right\vert =\left(  \lambda^{t}\right)  _{i}\text{ for every
}i\in\left\{  1,2,\ldots,m\right\}  \right)  $. Note that we denote it by
$A\left(  \lambda\right)  $ to stress its dependency on $\lambda$.

Let $\left(  \lambda^{\vee}\right)  ^{t}$ denote the conjugate of the
partition $\lambda^{\vee}$. It is easy to see that%
\begin{equation}
\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}=k-\left(  \lambda
^{t}\right)  _{m+1-i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,\ldots,m\right\}  . \label{sol.liri.180.d.conjugate-of-check}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.d.conjugate-of-check}):} Fix
$i\in\left\{  1,2,\ldots,m\right\}  $. We recall that $\lambda^{\vee}=\left(
m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  $. Hence,
\begin{equation}
\left(  \lambda^{\vee}\right)  _{j}=m-\lambda_{k+1-j}%
\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{  1,2,\ldots,k\right\}  .
\label{sol.liri.180.d.conjugate-of-check.1}%
\end{equation}
Also, $\ell\left(  \lambda^{\vee}\right)  \leq k$ (since $\lambda^{\vee
}=\left(  m-\lambda_{k},m-\lambda_{k-1},\ldots,m-\lambda_{1}\right)  $), so
that every positive integer $j>k$ satisfies $\left(  \lambda^{\vee}\right)
_{j}=0$. Thus, every positive integer $j$ satisfying $\left(  \lambda^{\vee
}\right)  _{j}\geq i$ must satisfy $j\leq k$ (since otherwise, this $j$ would
satisfy $j>k$, and thus $\left(  \lambda^{\vee}\right)  _{j}=0$, which would
contradict $\left(  \lambda^{\vee}\right)  _{j}\geq i>0$). In other words,
every positive integer $j$ satisfying $\left(  \lambda^{\vee}\right)  _{j}\geq
i$ must belong to $\left\{  1,2,\ldots,k\right\}  $. Hence,%
\begin{align*}
\left\{  j\ \mid\ \left(  \lambda^{\vee}\right)  _{j}\geq i\right\}   &
=\left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \underbrace{\left(
\lambda^{\vee}\right)  _{j}}_{\substack{=m-\lambda_{k+1-j}\\\text{(by
(\ref{sol.liri.180.d.conjugate-of-check.1}))}}}\geq i\right\}  =\left\{
j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \underbrace{m-\lambda_{k+1-j}\geq
i}_{\substack{\text{this is equivalent to }\lambda_{k+1-j}\leq m-i\text{, and
thus}\\\text{also equivalent to }\lambda_{k+1-j}<m-i+1\\\text{(since }%
\lambda_{k+1-j}\text{ and }m-i\text{ are integers)}}}\right\} \\
&  =\left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \underbrace{\lambda
_{k+1-j}<m-i+1}_{\text{this is equivalent to }\left(  \text{not }%
\lambda_{k+1-j}\geq m-i+1\right)  }\right\}  =\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \left(  \text{not }\lambda_{k+1-j}\geq
m-i+1\right)  \right\} \\
&  =\left\{  1,2,\ldots,k\right\}  \setminus\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  .
\end{align*}
But applying (\ref{eq.partition.transpose}) to $\lambda^{\vee}$ instead of
$\lambda$, we obtain%
\begin{align}
\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}  &  =\left\vert
\underbrace{\left\{  j\ \mid\ \left(  \lambda^{\vee}\right)  _{j}\geq
i\right\}  }_{=\left\{  1,2,\ldots,k\right\}  \setminus\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  }\right\vert
=\left\vert \left\{  1,2,\ldots,k\right\}  \setminus\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  \right\vert
\nonumber\\
&  =\underbrace{\left\vert \left\{  1,2,\ldots,k\right\}  \right\vert }%
_{=k}-\underbrace{\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}
\ \mid\ \lambda_{k+1-j}\geq m-i+1\right\}  \right\vert }%
_{\substack{=\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}
\ \mid\ \lambda_{j}\geq m-i+1\right\}  \right\vert \\\text{(here, we have
substituted }j\text{ for }k+1-j\text{)}}}\nonumber\\
&  =k-\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid
\ \lambda_{j}\geq m-i+1\right\}  \right\vert .
\label{sol.liri.180.d.conjugate-of-check.2}%
\end{align}
\par
But $\ell\left(  \lambda\right)  \leq k$, so that every positive integer $j>k$
satisfies $\lambda_{j}=0$. Thus, every positive integer $j$ satisfying
$\lambda_{j}\geq m-i+1$ must satisfy $j\leq k$ (since otherwise, this $j$
would satisfy $j>k$, and thus $\lambda_{j}=0$, which would contradict
$\lambda_{j}\geq m-i+1>0$). In other words, every positive integer $j$
satisfying $\lambda_{j}\geq i$ must belong to $\left\{  1,2,\ldots,k\right\}
$. Hence,%
\[
\left\{  j\ \mid\ \lambda_{j}\geq m-i+1\right\}  =\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{j}\geq m-i+1\right\}  .
\]
Now, (\ref{eq.partition.transpose}) (applied to $m-i+1$ instead of $i$) yields%
\begin{equation}
\left(  \lambda^{t}\right)  _{m-i+1}=\left\vert \underbrace{\left\{
j\ \mid\ \lambda_{j}\geq m-i+1\right\}  }_{=\left\{  j\in\left\{
1,2,\ldots,k\right\}  \ \mid\ \lambda_{j}\geq m-i+1\right\}  }\right\vert
=\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}  \ \mid\ \lambda
_{j}\geq m-i+1\right\}  \right\vert .
\label{sol.liri.180.d.conjugate-of-check.3}%
\end{equation}
Thus, (\ref{sol.liri.180.d.conjugate-of-check.2}) becomes%
\[
\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}%
=k-\underbrace{\left\vert \left\{  j\in\left\{  1,2,\ldots,k\right\}
\ \mid\ \lambda_{j}\geq m-i+1\right\}  \right\vert }_{\substack{=\left(
\lambda^{t}\right)  _{m-i+1}\\\text{(by
(\ref{sol.liri.180.d.conjugate-of-check.3}))}}}=k-\left(  \lambda^{t}\right)
_{m-i+1}=k-\left(  \lambda^{t}\right)  _{m+1-i}.
\]
This proves (\ref{sol.liri.180.d.conjugate-of-check}).}

It is easy to see (from the definition of $\lambda^{\vee}$) that
$\lambda^{\vee}$ is a partition satisfying $\ell\left(  \lambda^{\vee}\right)
\leq k$, and all parts of $\lambda^{\vee}$ are $\leq m$. Hence, we can define
a set $A\left(  \lambda^{\vee}\right)  $ in the same way as we defined the set
$A\left(  \lambda\right)  $ (but with every $\lambda$ replaced by
$\lambda^{\vee}$). Explicitly, $A\left(  \lambda^{\vee}\right)  $ is the set
of all $m$-tuples $\left(  I_{1},I_{2},\ldots,I_{m}\right)  $ of subsets of
$\left[  k\right]  $ satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq
_{\#}I_{m}$ and $\left(  \left\vert I_{i}\right\vert =\left(  \left(
\lambda^{\vee}\right)  ^{t}\right)  _{i}\text{ for every }i\in\left\{
1,2,\ldots,m\right\}  \right)  $.

Define a map
\[
\varphi\left(  \lambda\right)  :\left(  \text{the set of all }\left(
\lambda,k\right)  \text{-CSTs}\right)  \rightarrow A\left(  \lambda\right)
\]
by sending every $\left(  \lambda,k\right)  $-CST $T$ to the $m$-tuple whose
$i$-th entry is the $i$-th set column of $T$. This map is
well-defined\footnote{Indeed, if $T$ is an $\left(  \lambda,k\right)  $-CST,
then the $m$-tuple whose $i$-th entry is the $i$-th set column of $T$ belongs
to $A\left(  \lambda\right)  $ (because if we denote this $m$-tuple by
$\left(  I_{1},I_{2},\ldots,I_{m}\right)  $, then%
\begin{align*}
\left\vert I_{i}\right\vert  &  =\left\vert \left(  \text{the }i\text{-th set
column of }T\right)  \right\vert =\left(  \text{the length of the }i\text{-th
column of }T\right) \\
&  =\left(  \text{the length of the }i\text{-th column of the Ferrers diagram
of }\lambda\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }T\text{ is a }\left(
\lambda,k\right)  \text{-CST, thus a column-strict tableau of shape }%
\lambda\right) \\
&  = \left(\lambda^t\right)_i ,
\end{align*}
and the fact that the entries of $T$ increase weakly along rows (because $T$
is a column-strict tableau (since $T$ is a $\left(  \lambda,k\right)  $-CST))
translates precisely into the inequality chain $I_{1}\leq_{\#}I_{2}\leq
_{\#}\cdots\leq_{\#}I_{m}$).}, injective\footnote{This is because the $i$-th
column of a $\left(  \lambda,k\right)  $-CST $T$ can be uniquely reconstructed
from the $i$-th set column of $T$ (indeed, the entries of the $i$-th column of
$T$ must be strictly increasing down the column, and therefore the knowledge
of the set of these entries is sufficient to recover the $i$-th column). Here,
we are using the fact that $T$ has at most $m$ columns (since every part of
$\lambda$ is $\leq m$).} and surjective\footnote{Indeed, given any $\left(
I_{1},I_{2},\ldots,I_{m}\right)  \in A\left(  \lambda\right)  $. Then,
$\left(  I_{1},I_{2},\ldots,I_{m}\right)  $ is an $m$-tuple of subsets of
$\left[  k\right]  $ satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq
_{\#}I_{m}$ and
$\left(  \left\vert I_{i}\right\vert = \left( \lambda^t \right)_i
\text{ for every }i\in\left\{  1,2,\ldots,m\right\}  \right)
$. Now, we can fill in each column of the Ferrers diagram of $\lambda$ with
the entries of the corresponding set $I_{i}$ in increasing order, and then the
resulting filling is a column-strict tableau (indeed, its entries increase
weakly along its rows (due to $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq
_{\#}I_{m}$)), and more precisely a $\left(  \lambda,k\right)  $-CST. Our
$m$-tuple $\left(  I_{1},I_{2},\ldots,I_{m}\right)  $ is the image of this
$\left(  \lambda,k\right)  $-CST under $\varphi\left(  \lambda\right)  $;
therefore, $\left(  I_{1},I_{2},\ldots,I_{m}\right)  $ lies in the image of
$\varphi\left(  \lambda\right)  $. Hence, we have shown that every $\left(
I_{1},I_{2},\ldots,I_{m}\right)  \in A\left(  \lambda\right)  $ lies in the
image of $\varphi\left(  \lambda\right)  $. The map $\varphi\left(
\lambda\right)  $ is thus surjective, qed.}. Hence, $\varphi\left(
\lambda\right)  $ is a bijection. This bijection $\varphi\left(
\lambda\right)  $ furthermore satisfies%
\begin{equation}
\xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(  T\right)
}=\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \lambda,k\right)
\text{-CST }T \label{sol.liri.180.d.phi1}%
\end{equation}
\footnote{Indeed, it is easy to check that both $\xx_{\left(
\varphi\left(  \lambda\right)  \right)  \left(  T\right)  }$ and $\left(
x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(  T\right)  }$
are equal to the product $\prod_{c\text{ is a cell of }T}x_{\left(
\text{entry of }T\text{ in }c\right)  }$.}. Substituting $x_{1}^{-1}$,
$x_{2}^{-1}$, $\ldots$, $x_{k}^{-1}$ for $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$
on both sides of this equality, we obtain%
\begin{equation}
\left(  \xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(
T\right)  }\right)  ^{-1}=\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  ^{\cont \left(  T\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \lambda,k\right)  \text{-CST }T
\label{sol.liri.180.d.phi1inv}%
\end{equation}
(in fact, $\xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(
T\right)  }$ is a monomial, so that substituting $x_{1}^{-1}$, $x_{2}^{-1}$,
$\ldots$, $x_{k}^{-1}$ for $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ transforms it
into its inverse $\left(  \xx_{\left(  \varphi\left(  \lambda\right)
\right)  \left(  T\right)  }\right)  ^{-1}$).

Recall that $\lambda^{\vee}$ is a partition satisfying $\ell\left(
\lambda^{\vee}\right)  \leq k$, and all parts of $\lambda^{\vee}$ are $\leq
m$. Hence, we can define a bijection%
\[
\varphi\left(  \lambda^{\vee}\right)  :\left(  \text{the set of all }\left(
\lambda^{\vee},k\right)  \text{-CSTs}\right)  \rightarrow A\left(
\lambda^{\vee}\right)
\]
in the same way as we defined the bijection $\varphi\left(  \lambda\right)
:\left(  \text{the set of all }\left(  \lambda,k\right)  \text{-CSTs}\right)
\rightarrow A\left(  \lambda\right)  $ (but with $\lambda^{\vee}$ instead of
$\lambda$). This bijection satisfies%
\begin{equation}
\xx_{\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  \left(
T\right)  }=\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont
\left(  T\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \lambda^{\vee
},k\right)  \text{-CST }T \label{sol.liri.180.d.phi2}%
\end{equation}
(this can be shown in analogy to (\ref{sol.liri.180.d.phi1})).

Finally, we define a map $\psi\left(  \lambda\right)  :A\left(  \lambda
\right)  \rightarrow A\left(  \lambda^{\vee}\right)  $ as follows: For every
$\left(  I_{1},I_{2},\ldots,I_{m}\right)  \in A\left(  \lambda\right)  $, let%
\[
\left(  \psi\left(  \lambda\right)  \right)  \left(  I_{1},I_{2},\ldots
,I_{m}\right)  =\left(  \left[  k\right]  \setminus I_{m},\left[  k\right]
\setminus I_{m-1},\ldots,\left[  k\right]  \setminus I_{1}\right)  .
\]
This map $\psi\left(  \lambda\right)  $ is
well-defined\footnote{\textit{Proof.} Let $\left(  I_{1},I_{2},\ldots
,I_{m}\right)  \in A\left(  \lambda\right)  $. Then, $\left(  I_{1}%
,I_{2},\ldots,I_{m}\right)  $ is an $m$-tuple of subsets of $\left[  k\right]
$ satisfying $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ and $\left(
\left\vert I_{i}\right\vert =\left(  \lambda^{t}\right)  _{i}\text{ for every
}i\in\left\{  1,2,\ldots,m\right\}  \right)  $.
\par
In order to prove the well-definedness of $\psi\left(  \lambda\right)  $, we
need to show that $\left(  \left[  k\right]  \setminus I_{m},\left[  k\right]
\setminus I_{m-1},\ldots,\left[  k\right]  \setminus I_{1}\right)  \in
A\left(  \lambda^{\vee}\right)  $. In other words, we need to prove that
$\left(  \left[  k\right]  \setminus I_{m},\left[  k\right]  \setminus
I_{m-1},\ldots,\left[  k\right]  \setminus I_{1}\right)  $ is an $m$-tuple of
subsets of $\left[  k\right]  $ satisfying $\left[  k\right]  \setminus
I_{m}\leq_{\#}\left[  k\right]  \setminus I_{m-1}\leq_{\#}\cdots\leq
_{\#}\left[  k\right]  \setminus I_{1}$ and $\left(  \left\vert \left[
k\right]  \setminus I_{m+1-i}\right\vert =\left(  \left(  \lambda^{\vee
}\right)  ^{t}\right)  _{i}\text{ for every }i\in\left\{  1,2,\ldots
,m\right\}  \right)  $.
\par
First of all, it is clear that $\left(  \left[  k\right]  \setminus
I_{m},\left[  k\right]  \setminus I_{m-1},\ldots,\left[  k\right]  \setminus
I_{1}\right)  $ is an $m$-tuple of subsets of $\left[  k\right]  $.
Furthermore, $\left[  k\right]  \setminus I_{m}\leq_{\#}\left[  k\right]
\setminus I_{m-1}\leq_{\#}\cdots\leq_{\#}\left[  k\right]  \setminus I_{1}$
follows from $I_{1}\leq_{\#}I_{2}\leq_{\#}\cdots\leq_{\#}I_{m}$ (according to
Proposition \ref{prop.sol.liri.180.d.complement}(a)). Thus, it remains to
prove that $\left\vert \left[  k\right]  \setminus I_{m+1-i}\right\vert
=\left(  \left(  \lambda^{\vee}\right)  ^{t}\right)  _{i}$ for every
$i\in\left\{  1,2,\ldots,m\right\}  $.
\par
So fix some $i\in\left\{  1,2,\ldots,m\right\}  $. Then, $\left\vert
I_{m+1-i}\right\vert =\left(  \lambda^{t}\right)  _{m+1-i}$ (this follows from
the $\left\vert I_{i}\right\vert =\left(  \lambda^{t}\right)  _{i}$ formula,
but applied to $m+1-i$ instead of $i$). Since $I_{m+1-i}\subset\left[
k\right]  $, we have
\[
\left\vert \left[  k\right]  \setminus I_{m+1-i}\right\vert
=\underbrace{\left\vert \left[  k\right]  \right\vert }_{=k}%
-\underbrace{\left\vert I_{m+1-i}\right\vert }_{\substack{=\left(  \lambda
^{t}\right)  _{m+1-i}}}=k-\left(  \lambda^{t}\right)  _{m+1-i}=\left(  \left(
\lambda^{\vee}\right)  ^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.d.conjugate-of-check})}\right)  ,
\]
qed.} and bijective\footnote{Indeed, we can define a map $\psi\left(
\lambda^{\vee}\right)  $ in the same way as we have defined $\psi\left(
\lambda\right)  $ (but with $\lambda^{\vee}$ instead of $\lambda$). It is then
easy to see that the maps $\psi\left(  \lambda\right)  $ and $\psi\left(
\lambda^{\vee}\right)  $ are mutually inverse, so that $\psi\left(
\lambda\right)  $ is bijective, qed.}. This bijective map $\psi\left(
\lambda\right)  $ has the property that%
\begin{equation}
\xx_{\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)
}=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \xx%
_{S}\right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \text{for every }S\in A\left(
\lambda\right)  \label{sol.liri.180.d.psi1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180.d.psi1}):} Let $S\in A\left(
\lambda\right)  $. Then, $S$ is an $m$-tuple of subsets of $\left[  k\right]
$. Write $S$ as $S=\left(  I_{1},I_{2},\ldots,I_{m}\right)  $. The definition
of $\xx_{S}$ then yields%
\begin{equation}
\xx_{S}=\prod_{i=1}^{m}\xx_{I_{i}}.
\label{sol.liri.180.d.psi1.pf.0}%
\end{equation}
On the other hand, the definition of $\psi\left(  \lambda\right)  $ yields
$\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)  =\left(
\left[  k\right]  \setminus I_{m},\left[  k\right]  \setminus I_{m-1}%
,\ldots,\left[  k\right]  \setminus I_{1}\right)  $, and thus (by the
definition of $\xx_{\left(  \psi\left(  \lambda\right)  \right)
\left(  S\right)  }$) we have%
\begin{equation}
\xx_{\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)
}=\prod_{i=1}^{m}\xx_{\left[  k\right]  \setminus I_{m+1-i}}%
=\prod_{i=1}^{m}\xx_{\left[  k\right]  \setminus I_{i}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }m+1-i\text{ for
}i\text{ in the product}\right)  . \label{sol.liri.180.d.psi1.pf.1}%
\end{equation}
\par
But every $S\subset\left[  k\right]  $ satisfies%
\begin{align*}
\underbrace{\xx_{S}}_{\substack{=\prod_{s\in S}x_{s}\\\text{(by the
definition of }\xx_{S}\text{)}}}\cdot\underbrace{\xx_{\left[
k\right]  \setminus S}}_{\substack{=\prod_{s\in\left[  k\right]  \setminus
S}x_{s}\\\text{(by the definition of }\xx_{\left[  k\right]  \setminus
S}\text{)}}}  &  =\left(  \prod_{s\in S}x_{s}\right)  \cdot\left(  \prod
_{s\in\left[  k\right]  \setminus S}x_{s}\right) \\
&  =\prod_{s\in\left[  k\right]  }x_{s}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left[  k\right]  \text{ is the union of the two disjoint sets }S\text{ and
}\left[  k\right]  \setminus S\right) \\
&  =x_{1}x_{2}\cdots x_{k}.
\end{align*}
Applying this to $S=I_{i}$, we obtain $\xx_{I_{i}}\cdot\xx%
_{\left[  k\right]  \setminus I_{i}}=x_{1}x_{2}\cdots x_{k}$ for every
$i\in\left\{  1,2,\ldots,m\right\}  $. Hence, $\xx_{\left[  k\right]
\setminus I_{i}}=\dfrac{x_{1}x_{2}\cdots x_{k}}{\xx_{I_{i}}}$ for every
$i\in\left\{  1,2,\ldots,m\right\}  $. Now, (\ref{sol.liri.180.d.psi1.pf.1})
becomes%
\begin{align*}
\xx_{\left(  \psi\left(  \lambda\right)  \right)  \left(  S\right)  }
&  =\prod_{i=1}^{m}\underbrace{\xx_{\left[  k\right]  \setminus I_{i}}%
}_{=\dfrac{x_{1}x_{2}\cdots x_{k}}{\xx_{I_{i}}}}=\prod_{i=1}^{m}%
\dfrac{x_{1}x_{2}\cdots x_{k}}{\xx_{I_{i}}}=\dfrac{\left(  x_{1}%
x_{2}\cdots x_{k}\right)  ^{m}}{\prod_{i=1}^{m}\xx_{I_{i}}}=\left(
x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \underbrace{\prod_{i=1}%
^{m}\xx_{I_{i}}}_{\substack{=\xx_{S}\\\text{(by
(\ref{sol.liri.180.d.psi1.pf.0}))}}}\right)  ^{-1}\\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  \xx%
_{S}\right)  ^{-1}.
\end{align*}
This proves (\ref{sol.liri.180.d.psi1}).}.

Using the three bijective maps%
\begin{align*}
\varphi\left(  \lambda\right)   &  :\left(  \text{the set of all }\left(
\lambda,k\right)  \text{-CSTs}\right)  \rightarrow A\left(  \lambda\right)
,\\
\psi\left(  \lambda\right)   &  :A\left(  \lambda\right)  \rightarrow A\left(
\lambda^{\vee}\right)  ,\\
\varphi\left(  \lambda^{\vee}\right)   &  :\left(  \text{the set of all
}\left(  \lambda^{\vee},k\right)  \text{-CSTs}\right)  \rightarrow A\left(
\lambda^{\vee}\right)  ,
\end{align*}
we can define a map%
\[
\Omega:\left(  \text{the set of all }\left(  \lambda,k\right)  \text{-CSTs}%
\right)  \rightarrow\left(  \text{the set of all }\left(  \lambda^{\vee
},k\right)  \text{-CSTs}\right)
\]
by $\Omega=\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  ^{-1}%
\circ\left(  \psi\left(  \lambda\right)  \right)  \circ\left(  \varphi\left(
\lambda\right)  \right)  $. Clearly, this $\Omega$ is a bijection. If we can
show that every $\left(  \lambda,k\right)  $-CST $T$ satisfies
(\ref{sol.liri.180.d.alternative.condition-on-Phi}), then
(\ref{sol.liri.180.d.alternative4}) will be proven, and thus Exercise
\ref{exe.liri.180}(d) will be solved again. Hence, all that remains to prove
now is that every $\left(  \lambda,k\right)  $-CST $T$ satisfies
(\ref{sol.liri.180.d.alternative.condition-on-Phi}).

Let $T$ be a $\left(  \lambda,k\right)  $-CST. We have $\Omega=\left(
\varphi\left(  \lambda^{\vee}\right)  \right)  ^{-1}\circ\left(  \psi\left(
\lambda\right)  \right)  \circ\left(  \varphi\left(  \lambda\right)  \right)
$, thus $\left(  \varphi\left(  \lambda^{\vee}\right)  \right)  \circ
\Omega=\left(  \psi\left(  \lambda\right)  \right)  \circ\left(
\varphi\left(  \lambda\right)  \right)  $. Applying (\ref{sol.liri.180.d.phi2}%
) to $\Omega\left(  T\right)  $ instead of $T$, we obtain $\xx_{\left(
\varphi\left(  \lambda^{\vee}\right)  \right)  \left(  \Omega\left(  T\right)
\right)  }=\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont
\left(  \Omega\left(  T\right)  \right)  }$, whence%
\begin{align*}
&  \left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
\Omega\left(  T\right)  \right)  }=\xx_{\left(  \varphi\left(
\lambda^{\vee}\right)  \right)  \left(  \Omega\left(  T\right)  \right)
}=\xx_{\left(  \psi\left(  \lambda\right)  \right)  \left(  \left(
\varphi\left(  \lambda\right)  \right)  \left(  T\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \varphi\left(
\lambda^{\vee}\right)  \right)  \left(  \Omega\left(  T\right)  \right)
=\underbrace{\left(  \left(  \varphi\left(  \lambda^{\vee}\right)  \right)
\circ\Omega\right)  }_{=\left(  \psi\left(  \lambda\right)  \right)
\circ\left(  \varphi\left(  \lambda\right)  \right)  }\left(  T\right)
=\left(  \left(  \psi\left(  \lambda\right)  \right)  \circ\left(
\varphi\left(  \lambda\right)  \right)  \right)  \left(  T\right)  =\left(
\psi\left(  \lambda\right)  \right)  \left(  \left(  \varphi\left(
\lambda\right)  \right)  \left(  T\right)  \right)  \right) \\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\underbrace{\left(
\xx_{\left(  \varphi\left(  \lambda\right)  \right)  \left(  T\right)
}\right)  ^{-1}}_{\substack{=\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  ^{\cont \left(  T\right)  }\\\text{(by
(\ref{sol.liri.180.d.phi1inv}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.d.psi1}), applied to }S=\left(  \varphi\left(
\lambda\right)  \right)  \left(  T\right)  \right) \\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot\left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{k}^{-1}\right)  ^{\cont \left(  T\right)
}.
\end{align*}
Thus, (\ref{sol.liri.180.d.alternative.condition-on-Phi}) holds. We thus have
proven that every $\left(  \lambda,k\right)  $-CST $T$ satisfies
(\ref{sol.liri.180.d.alternative.condition-on-Phi}). Exercise
\ref{exe.liri.180}(d) is thus solved.

(e) \textit{First solution to Exercise \ref{exe.liri.180}(e):} The $k$-tuple
$\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  $ is a
partition\footnote{\textit{Proof.} We have $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{k}\right)  $ (since $\ell\left(
\lambda\right)  \leq k$). Thus, $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{k}\right)  =\lambda$ is a partition. Hence, $\lambda_{1}$,
$\lambda_{2}$, $\ldots$, $\lambda_{k}$ are nonnegative integers and satisfy
$\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}$.
\par
Now, $r+\lambda_{1}$, $r+\lambda_{2}$, $\ldots$, $r+\lambda_{k}$ are
nonnegative integers (since $r$ is a nonnegative integer, and since
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{k}$ are nonnegative
integers) and satisfy $r+\lambda_{1}\geq r+\lambda_{2}\geq\cdots\geq
r+\lambda_{k}$ (since $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}$).
In other words, $\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda
_{k}\right)  $ is a partition, qed.}. Let us denote this partition by
$\lambda^{\left[  r\right]  }$. Then, $\lambda^{\left[  r\right]  }=\left(
r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  \in\NN^{k}$;
thus, $\ell\left(  \lambda^{\left[  r\right]  }\right)  \leq k$.

Let $n=k$. We shall use the notations of Section \ref{bialternant-section}; in
particular, we set $\xx=\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ and
$\rho=\left(  n-1,n-2,\ldots,2,1,0\right)  $. Clearly, $\xx=\left(
x_{1},x_{2},\ldots,x_{n}\right)  =\left(  x_{1},x_{2},\ldots,x_{k}\right)  $
(since $n=k$). Notice that $\ell\left(  \lambda\right)  \leq k=n$. Hence, we
can regard $\lambda$ as an element of $\NN^n$; therefore,
$\lambda+\rho$ is a well-defined element of $\NN^n$, and the
alternant $a_{\lambda+\rho}$ is well-defined. Corollary
\ref{cor.stembridge.slambda} yields that $s_{\lambda}\left(  \xx%
\right)  =\dfrac{a_{\lambda+\rho}}{a_{\rho}}$, so that%
\begin{equation}
a_{\rho}\cdot s_{\lambda}\left(  \xx\right)  =a_{\lambda+\rho}.
\label{sol.liri.180.e.s_lambda-as-quotient}%
\end{equation}


Applying this equality to $\lambda^{\left[  r\right]  }$ instead of $\lambda$,
we obtain%
\begin{equation}
a_{\rho}\cdot s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)
=a_{\lambda^{\left[  r\right]  }+\rho}
\label{sol.liri.180.e.s_lambdar-as-quotient}%
\end{equation}
(since $\lambda^{\left[  r\right]  }$ is also a partition satisfying
$\ell\left(  \lambda^{\left[  r\right]  }\right)  \leq k=n$).

Every $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)
\in\NN^n$ satisfies (\ref{sol.liri.180.d.a_alpha-formula}). (This can
be proven just as in the First solution to Exercise \ref{exe.liri.180}(d).)

Also, (\ref{sol.liri.180.d.a_lambdaplusrho1}) holds. (This can be proven just
as in the First solution to Exercise \ref{exe.liri.180}(d).)

We have $\lambda^{\left[  r\right]  }=\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  =\left(  r+\lambda_{1},r+\lambda_{2}%
,\ldots,r+\lambda_{n}\right)  $ (since $k=n$), so that%
\begin{align*}
\underbrace{\lambda^{\left[  r\right]  }}_{=\left(  r+\lambda_{1}%
,r+\lambda_{2},\ldots,r+\lambda_{n}\right)  }+\underbrace{\rho}%
_{\substack{=\left(  n-1,n-2,\ldots,2,1,0\right)  \\=\left(  n-1,n-2,\ldots
,n-n\right)  }}  &  =\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda
_{n}\right)  +\left(  n-1,n-2,\ldots,n-n\right) \\
&  =\left(  \left(  r+\lambda_{1}\right)  +\left(  n-1\right)  ,\left(
r+\lambda_{2}\right)  +\left(  n-2\right)  ,\ldots,\left(  r+\lambda
_{n}\right)  +\left(  n-n\right)  \right)  .
\end{align*}
Hence, (\ref{sol.liri.180.d.a_alpha-formula}) (applied to $\lambda^{\left[
r\right]  }+\rho$ and \newline$\left(  \left(  r+\lambda_{1}\right)  +\left(
n-1\right)  ,\left(  r+\lambda_{2}\right)  +\left(  n-2\right)  ,\ldots
,\left(  r+\lambda_{n}\right)  +\left(  n-n\right)  \right)  $ instead of
$\alpha$ and $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right)  $)
yields%
\begin{align*}
&  a_{\lambda^{\left[  r\right]  }+\rho}\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod
_{i=1}^{n}\underbrace{x_{w\left(  i\right)  }^{\left(  r+\lambda_{i}\right)
+\left(  n-i\right)  }}_{\substack{=x_{w\left(  i\right)  }^{r+\left(
\lambda_{i}+n-i\right)  }\\\text{(since }\left(  r+\lambda_{i}\right)
+\left(  n-i\right)  =r+\left(  \lambda_{i}+n-i\right)  \text{)}}}=\sum
_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \prod_{i=1}%
^{n}\underbrace{x_{w\left(  i\right)  }^{r+\left(  \lambda_{i}+n-i\right)  }%
}_{=x_{w\left(  i\right)  }^{r}x_{w\left(  i\right)  }^{\lambda_{i}+n-i}}\\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\prod_{i=1}^{n}\left(  x_{w\left(  i\right)  }^{r}x_{w\left(
i\right)  }^{\lambda_{i}+n-i}\right)  }_{=\left(  \prod_{i=1}^{n}x_{w\left(
i\right)  }^{r}\right)  \left(  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{\lambda_{i}+n-i}\right)  }=\sum_{w\in\Symm_n}\operatorname{sgn}%
\left(  w\right)  \left(  \underbrace{\prod_{i=1}^{n}}_{=\prod_{i\in\left\{
1,2,\ldots,n\right\}  }}x_{w\left(  i\right)  }^{r}\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{\lambda_{i}+n-i}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)
\underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{w\left(
i\right)  }^{r}\right)  }_{\substack{=\prod_{i\in\left\{  1,2,\ldots
,n\right\}  }x_{i}^{r}\\\text{(here, we have substituted }i\text{ for
}w\left(  i\right)  \text{ in the product}\\\text{(since }w\text{ is a
permutation of }\left\{  1,2,\ldots,n\right\}  \text{))}}}\left(  \prod
_{i=1}^{n}x_{w\left(  i\right)  }^{\lambda_{i}+n-i}\right) \\
&  =\sum_{w\in\Symm_n}\operatorname{sgn}\left(  w\right)  \left(
\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}^{r}\right)  \left(
\prod_{i=1}^{n}x_{w\left(  i\right)  }^{\lambda_{i}+n-i}\right)
=\underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{i}%
^{r}\right)  }_{=x_{1}^{r}x_{2}^{r}\cdots x_{n}^{r}=\left(  x_{1}x_{2}\cdots
x_{n}\right)  ^{r}}\cdot\underbrace{\sum_{w\in\Symm_n}%
\operatorname{sgn}\left(  w\right)  \prod_{i=1}^{n}x_{w\left(  i\right)
}^{\lambda_{i}+n-i}}_{\substack{=a_{\lambda+\rho}\\\text{(by
(\ref{sol.liri.180.d.a_lambdaplusrho1}))}}}\\
&  =\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot\underbrace{a_{\lambda
+\rho}}_{\substack{=a_{\rho}\cdot s_{\lambda}\left(  \xx\right)
\\\text{(by (\ref{sol.liri.180.e.s_lambda-as-quotient}))}}}=\left(  x_{1}%
x_{2}\cdots x_{n}\right)  ^{r}\cdot a_{\rho}\cdot s_{\lambda}\left(
\xx\right)  =a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)
^{r}\cdot s_{\lambda}\left(  \xx\right)  .
\end{align*}
Hence,%
\[
a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot s_{\lambda
}\left(  \xx\right)  =a_{\lambda^{\left[  r\right]  }+\rho}=a_{\rho
}\cdot s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180.e.s_lambda-as-quotient})}\right)  ,
\]
so that
\begin{align}
&  a_{\rho}\cdot\left(  \left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot
s_{\lambda}\left(  \xx\right)  -s_{\lambda^{\left[  r\right]  }}\left(
\xx\right)  \right) \nonumber\\
&  =\underbrace{a_{\rho}\cdot\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot
s_{\lambda}\left(  \xx\right)  }_{=a_{\rho}\cdot s_{\lambda^{\left[
r\right]  }}\left(  \xx\right)  }-a_{\rho}\cdot s_{\lambda^{\left[
r\right]  }}\left(  \xx\right)  =a_{\rho}\cdot s_{\lambda^{\left[
r\right]  }}\left(  \xx\right)  -a_{\rho}\cdot s_{\lambda^{\left[
r\right]  }}\left(  \xx\right)  =0. \label{sol.liri.180.e.almostthere2}%
\end{align}


But we know (from a footnote in Corollary \ref{cor.stembridge.slambda}) that
$a_{\rho}$ is not a zero-divisor in the ring $\kk\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $. In other words, $a_{\rho}$ is a
non-zero-divisor in the ring $\kk\left[  x_{1},x_{2},\ldots
,x_{n}\right]  $. Hence, we can cancel the factor $a_{\rho}$ from the equality
(\ref{sol.liri.180.e.almostthere2}). As a result, we obtain%
\[
\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot s_{\lambda}\left(
\xx\right)  -s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)
=0.
\]
Hence, $\left(  x_{1}x_{2}\cdots x_{n}\right)  ^{r}\cdot s_{\lambda}\left(
\xx\right)  =s_{\lambda^{\left[  r\right]  }}\left(  \xx\right)
=s_{\lambda^{\left[  r\right]  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)  $
(since $\xx=\left(  x_{1},x_{2},\ldots,x_{k}\right)  $), so that%
\[
s_{\lambda^{\left[  r\right]  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)
=\left(  \underbrace{x_{1}x_{2}\cdots x_{n}}_{\substack{=x_{1}x_{2}\cdots
x_{k}\\\text{(since }n=k\text{)}}}\right)  ^{r}\cdot\underbrace{s_{\lambda
}\left(  \xx\right)  }_{\substack{=s_{\lambda}\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  \\\text{(since }\xx=\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  \text{)}}}=\left(  x_{1}x_{2}\cdots x_{k}\right)
^{r}\cdot s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\]
Since $\lambda^{\left[  r\right]  }=s_{\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  }$, this rewrites as
\[
s_{\left(
r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  }\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\]
This solves Exercise \ref{exe.liri.180}(e).

\textit{Second solution to Exercise \ref{exe.liri.180}(e) (sketched):} The
$k$-tuple $\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  $
is a partition\footnote{This can be proven just as in the First solution to
Exercise \ref{exe.liri.180}(e).}. Let us denote this partition by
$\lambda^{\left[  r\right]  }$. Then, $\lambda^{\left[  r\right]  }=\left(
r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  \in\NN^{k}$;
thus, $\ell\left(  \lambda^{\left[  r\right]  }\right)  \leq k$.

We have $\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)
=\lambda^{\left[  r\right]  }$, thus $s_{\left(  r+\lambda_{1},r+\lambda
_{2},\ldots,r+\lambda_{k}\right)  }=s_{\lambda^{\left[  r\right]  }%
}=s_{\lambda^{\left[  r\right]  }/\varnothing}$, hence
\[
\underbrace{s_{\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda
_{k}\right)  }}_{=s_{\lambda^{\left[  r\right]  }/\varnothing}}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =s_{\lambda^{\left[  r\right]  }/\varnothing
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda^{\left[  r\right]
}/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)
^{\cont \left(  T\right)  }%
\]
(by Exercise \ref{exe.schur.finite}(a), applied to $k$, $\lambda^{\left[
r\right]  }$ and $\varnothing$ instead of $n$, $\lambda$ and $\mu$), where
$\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }$ is defined as in Exercise \ref{exe.schur.finite}(a). Thus,%
\begin{align}
s_{\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  }\left(
x_{1},x_{2},\ldots,x_{k}\right)   &  =\sum\limits_{\substack{T\text{ is a
column-strict}\\\text{tableau of shape }\lambda^{\left[  r\right]
}/\varnothing\text{;}\\\text{all entries of }T\text{ belong}\\\text{to
}\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)
^{\cont \left(  T\right)  }\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda^{\left[  r\right]  }\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  ^{\cont \left(  T\right)  }
\label{sol.liri.180.e.alternative1}%
\end{align}
(since column-strict tableaux of shape $\lambda^{\left[  r\right]
}/\varnothing$ are the same as column-strict tableaux of shape $\lambda
^{\left[  r\right]  }$).

But an argument analogous to the one we just used to prove
(\ref{sol.liri.180.e.alternative1}) (but with $\lambda$ in place of
$\lambda^{\left[  r\right]  }$) shows that%
\[
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{T\text{
is a column-strict}\\\text{tableau of shape }\lambda\text{;}\\\text{all
entries of }T\text{ belong}\\\text{to }\left\{  1,2,\ldots,k\right\}
}}\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }.
\]
Multiplied by $\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}$, this equality
becomes%
\begin{align}
\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)   &  =\left(  x_{1}x_{2}\cdots x_{k}\right)
^{r}\cdot\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2},\ldots,x_{k}\right)
^{\cont \left(  T\right)  }\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}%
\cdot\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }. \label{sol.liri.180.e.alternative2}%
\end{align}


We need to prove that%
\[
s_{\left(  r+\lambda_{1},r+\lambda_{2},\ldots,r+\lambda_{k}\right)  }\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots x_{k}\right)
^{r}\cdot s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\]
Due to (\ref{sol.liri.180.e.alternative1}) and
(\ref{sol.liri.180.e.alternative2}), this rewrites as%
\begin{align}
&  \sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda^{\left[  r\right]  }\text{;}\\\text{all entries of }T\text{
belong}\\\text{to }\left\{  1,2,\ldots,k\right\}  }}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  ^{\cont \left(  T\right)  }\nonumber\\
&  =\sum\limits_{\substack{T\text{ is a column-strict}\\\text{tableau of shape
}\lambda\text{;}\\\text{all entries of }T\text{ belong}\\\text{to }\left\{
1,2,\ldots,k\right\}  }}\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}%
\cdot\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont \left(
T\right)  }. \label{sol.liri.180.e.alternative4}%
\end{align}
So it remains to prove (\ref{sol.liri.180.e.alternative4}).

In order to prove (\ref{sol.liri.180.e.alternative4}), it is clearly
sufficient to construct a bijection%
\begin{align*}
\Omega &  :\left(  \text{the set of all column-strict tableaux }T\text{ of
shape }\lambda^{\left[  r\right]  }\text{ such that all entries of }T\text{
belong to }\left\{  1,2,\ldots,k\right\}  \right) \\
&  \rightarrow\left(  \text{the set of all column-strict tableaux }T\text{ of
shape }\lambda\text{ such that all entries of }T\text{ belong to }\left\{
1,2,\ldots,k\right\}  \right)
\end{align*}
with the property that every column-strict tableau $T$ of shape
$\lambda^{\left[  r\right]  }$ such that all entries of $T$ belong to
$\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{equation}
\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{r}\cdot\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  ^{\cont \left(  \Omega\left(  T\right)
\right)  }=\left(  x_{1},x_{2},\ldots,x_{k}\right)  ^{\cont
\left(  T\right)  }. \label{sol.liri.180.e.alternative.omega-cond}%
\end{equation}


Constructing such an $\Omega$ is easy: The map $\Omega$ just maps every
column-strict tableau $T$ of shape $\lambda^{\left[  r\right]  }$ to the
tableau of shape $\lambda$ obtained by removing the first $r$ entries of each
row of $T$, and moving all other entries to the left by $r$ cells. To prove
that $\Omega$ is bijective, we need to construct a map inverse to $\Omega$;
this latter map sends every column-strict tableau $T$ of shape $\lambda$ to
the tableau of shape $\lambda^{\left[  r\right]  }$ obtained by moving all
entries of $T$ to the right by $r$ cells, and filling the now-vacant first $r$
columns as follows:%
\[%
\begin{array}
[c]{cccc}%
1 & 1 & \cdots & 1\\
2 & 2 & \cdots & 2\\
\vdots & \vdots & \ddots & \vdots\\
k & k & \cdots & k
\end{array}
.
\]
Proving that this map is well-defined and really inverse to $\Omega$ is left
to the reader\footnote{The main ingredient of this proof is the observation
that if $T$ is a column-strict tableau of shape $\lambda^{\left[  r\right]  }$
such that all entries of $T$ belong to $\left\{  1,2,\ldots,k\right\}  $, then
the first $r$ columns of $T$ must look like this:%
\[%
\begin{array}
[c]{cccc}%
1 & 1 & \cdots & 1\\
2 & 2 & \cdots & 2\\
\vdots & \vdots & \ddots & \vdots\\
k & k & \cdots & k
\end{array}
.
\]
(This is because each of the first $r$ columns of $T$ has length $k$, but the
entries in this column must be strictly increasing from top to bottom, and
knowing that these entries belong to $\left\{  1,2,\ldots,k\right\}  $, we see
that this is only possible if the column has the form $%
\begin{array}
[c]{c}%
1\\
2\\
\vdots\\
k
\end{array}
$.)}. Anyway, we now know that $\Omega$ is bijective, and it is easy to check
that (\ref{sol.liri.180.e.alternative.omega-cond}) holds. This completes the
second solution of Exercise \ref{exe.liri.180}(e).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.180ab}}

\textit{Solution to Exercise \ref{exe.liri.180ab}.}
(a) We can apply Exercise \ref{exe.liri.180}(a) to $\mu$, $\mu$, $\mu
^{\vee\left\{  m\right\}  }$ and $\mu^{\vee\left\{  m\right\}  }$ instead of
$\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$. As a consequence, we
conclude that $\mu^{\vee\left\{  m\right\}  }$ and
$\mu^{\vee\left\{  m\right\}  }$ are partitions, and that
$s_{\mu/\mu}=s_{\mu^{\vee\left\{  m\right\}  }/\mu^{\vee\left\{  m\right\}  }}$.
Thus, $\mu^{\vee\left\{  m\right\}  }$ is a partition.
The same argument (but with $m$, $\mu$ and $\mu^{\vee\left\{  m\right\}  }$
replaced by $n$, $\nu$ and $\nu^{\vee\left\{  n\right\}  }$) yields that
$\nu^{\vee\left\{  n\right\}  }$ is a partition. This solves Exercise
\ref{exe.liri.180ab}(a).

(b) Assume that not all parts of $\lambda$ are $\leq m+n$. Then, some part of
$\lambda$ is $>m+n$; hence, the greatest part of $\lambda$ is $>m+n$. Thus,
$\lambda_{1}>m+n$ (since $\lambda_{1}$ is the greatest part of $\lambda$). As
a consequence, $\lambda_{1}-\mu_{1}>\nu_{1}$\ \ \ \ \footnote{\textit{Proof.}
All parts of $\mu$ are $\leq m$. Thus, $\mu_{1}\leq m$ (since $\mu_{1}$ is
either a part of $\mu$ and therefore $\leq m$, or is zero and therefore $\leq
m$ as well). Hence, $m\geq\mu_{1}$. Similarly, $n\geq\nu_{1}$. Now,
$\lambda_{1}>\underbrace{m}_{\geq\mu_{1}}+\underbrace{n}_{\geq\nu_{1}}\geq
\mu_{1}+\nu_{1}$, so that $\lambda_{1}-\mu_{1}>\nu_{1}$, qed.}.

Now, we shall show that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{there exists no column-strict tableau }T\text{ of shape }\lambda
/\mu\text{ with }\cont \left(  T\right)  =\nu\\
\text{having the property that each }\cont \left(  T\mid
_{\operatorname{cols}\geq j}\right)  \text{ is a partition}%
\end{array}
\right)  . \label{sol.liri.180ab.b.nonexist}%
\end{equation}


\textit{Proof of (\ref{sol.liri.180ab.b.nonexist}):} Assume the contrary. Thus,
there exists a column-strict tableau $T$ of shape $\lambda/\mu$ with
$\cont \left(  T\right)  =\nu$ having the property that each
$\cont \left(  T\mid_{\operatorname{cols}\geq j}\right)  $ is a
partition. Consider this $T$.

The column-strict tableau $T$ has shape $\lambda/\mu$, and thus its $1$-st row
has $\lambda_{1}-\mu_{1}$ entries. In other words,
\begin{align*}
&  \left(  \text{the number of entries in the }1\text{-st row of }T\right) \\
&  =\lambda_{1}-\mu_{1}>\nu_{1}=\left(  \cont \left(  T\right)
\right)  _{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\nu=\cont
\left(  T\right)  \right) \\
&  =\left\vert T^{-1}\left(  1\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\cont \left(  T\right)  \right) \\
&  =\left(  \text{the number of entries of }T\text{ equal to }1\right)  .
\end{align*}
Hence, the tableau $T$ has more entries in its $1$-st row than it has entries
equal to $1$. Consequently, not every entry in the $1$-st row is equal to $1$.
Thus, there exists an entry in the $1$-st row of $T$ which is $>1$. Let $k$ be
this entry, and let $\mathbf{c}$ be the cell it occupies. Then, $\mathbf{c}$
is a cell in the $1$-st row of $T$, and thus can be written in the form
$\left(  1,j\right)  $ for some positive integer $j$. Consider this $j$. The
cell $\mathbf{c}$ lies in column $j$, and thus is a cell of the skew tableau
$T\mid_{\operatorname{cols}\geq j}$; its entry is $\left(  T\mid
_{\operatorname{cols}\geq j}\right)  \left(  \mathbf{c}\right)  =T\left(
\mathbf{c}\right)  =k$ (since we know that $k$ is the entry of $T$ occupying
cell $\mathbf{c}$). Hence, $\mathbf{c}\in\left(  T\mid_{\operatorname{cols}%
\geq j}\right)  ^{-1}\left(  k\right)  $, so that the set $\left(
T\mid_{\operatorname{cols}\geq j}\right)  ^{-1}\left(  k\right)  $ is
nonempty. Thus, $\left\vert \left(  T\mid_{\operatorname{cols}\geq j}\right)
^{-1}\left(  k\right)  \right\vert >0$.

Recalling the definition of $\cont \left(  T\mid
_{\operatorname{cols}\geq j}\right)  $, we see that $\left(
\cont \left(  T\mid_{\operatorname{cols}\geq j}\right)  \right)
_{k}=\left\vert \left(  T\mid_{\operatorname{cols}\geq j}\right)  ^{-1}\left(
k\right)  \right\vert >0$.

But we know that $\cont \left(  T\mid_{\operatorname{cols}\geq
j}\right)  $ is a partition. Thus,
\[
\left(  \cont \left(
T\mid_{\operatorname{cols}\geq j}\right)  \right)  _{1}\geq\left(
\cont \left(  T\mid_{\operatorname{cols}\geq j}\right)  \right)
_{2}\geq\left(  \cont \left(  T\mid_{\operatorname{cols}\geq
j}\right)  \right)  _{3}\geq\cdots ,
\]
so that $\left(  \cont
\left(  T\mid_{\operatorname{cols}\geq j}\right)  \right)  _{1}\geq\left(
\cont \left(  T\mid_{\operatorname{cols}\geq j}\right)  \right)
_{k}>0$. But the definition of $\cont \left(  T\mid
_{\operatorname{cols}\geq j}\right)  $ yields%
\[
\left(  \cont \left(  T\mid_{\operatorname{cols}\geq j}\right)
\right)  _{1}=\left\vert \left(  T\mid_{\operatorname{cols}\geq j}\right)
^{-1}\left(  1\right)  \right\vert =\left(  \text{the number of entries of
}T\mid_{\operatorname{cols}\geq j}\text{ equal to }1\right)  .
\]
Hence, $\left(  \text{the number of entries of }T\mid_{\operatorname{cols}\geq
j}\text{ equal to }1\right)  =\left(  \cont \left(
T\mid_{\operatorname{cols}\geq j}\right)  \right)  _{1}>0$. Hence, the skew
tableau $T\mid_{\operatorname{cols}\geq j}$ must have at least one entry equal
to $1$.

But each cell of the skew tableau $T\mid_{\operatorname{cols}\geq j}$ lies in
one of the columns $j,j+1,j+2,\ldots$ (by the definition of $T\mid
_{\operatorname{cols}\geq j}$) and in one of the rows $1,2,3,\ldots$
(obviously). Hence, each cell of the skew tableau $T\mid_{\operatorname{cols}%
\geq j}$ lies (weakly) southeast of the cell $\left(  1,j\right)  =\mathbf{c}%
$. As a consequence, each entry of the skew tableau $T\mid
_{\operatorname{cols}\geq j}$ is greater or equal to the entry of
$T\mid_{\operatorname{cols}\geq j}$ in the cell $\mathbf{c}$%
\ \ \ \ \footnote{Indeed, $T$ is a column-strict tableau, and thus the entries
of $T$ increase weakly left-to-right along rows, and increase strictly
top-to-bottom in columns. Consequently, the entries of $T$ increase weakly as
one moves to the southeast. The same holds for the entries of $T\mid
_{\operatorname{cols}\geq j}$ (since $T\mid_{\operatorname{cols}\geq j}$ is a
restriction of $T$), and thus each entry of $T\mid_{\operatorname{cols}\geq
j}$ is greater or equal to the entry of $T\mid_{\operatorname{cols}\geq j}$ in
the cell $\mathbf{c}$ (because each cell of $T\mid_{\operatorname{cols}\geq
j}$ lies (weakly) southeast of the cell $\mathbf{c}$). Qed.}. Since the entry
of $T\mid_{\operatorname{cols}\geq j}$ in the cell $\mathbf{c}$ is $\left(
T\mid_{\operatorname{cols}\geq j}\right)  \left(  \mathbf{c}\right)  =k$, this
yields that each entry of the skew tableau $T\mid_{\operatorname{cols}\geq j}$
is greater or equal to $k$, and thus greater than $1$ (since $k>1$). As a
consequence, no entry of the skew tableau $T\mid_{\operatorname{cols}\geq j}$
can be equal to $1$. This contradicts the fact that the skew tableau
$T\mid_{\operatorname{cols}\geq j}$ must have at least one entry equal to $1$.
This contradiction proves that our assumption was wrong. Hence,
(\ref{sol.liri.180ab.b.nonexist}) is proven.

Now, Corollary \ref{L-R-rule} yields that $c_{\mu,\nu}^{\lambda}$ counts
column-strict tableaux $T$ of shape $\lambda/\mu$ with $\cont
\left(  T\right)  =\nu$ having the property that each $\cont
\left(  T\mid_{\operatorname{cols}\geq j}\right)  $ is a partition. Since
there exists no such $T$ (according to (\ref{sol.liri.180ab.b.nonexist})),
this yields that $c_{\mu,\nu}^{\lambda}=0$. This solves Exercise
\ref{exe.liri.180ab}(b).

\textit{Remark:} An alternative solution of Exercise \ref{exe.liri.180ab}(b)
can be obtained easily from Exercise \ref{exe.liri.simple}(c).

(c) Let us forget that $\lambda$ is fixed.

If $\lambda\in \Par $ is such that $\ell\left(  \lambda\right)
>k$, then%
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =0.
\label{sol.liri.180ab.c.toobig}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.toobig}):} Let $\lambda
\in \Par $ be such that $\ell\left(  \lambda\right)  >k$. Then,
the number of parts of $\lambda$ is $\ell\left(  \lambda\right)  >k$. Hence,
Exercise \ref{exe.schur.finite}(b) (applied to $k$ instead of $n$) yields
$s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =0$, qed.}

We first notice that%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum_{\lambda\in\operatorname{Par}%
;\ \ell\left(  \lambda\right)  \leq k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  . \label{sol.liri.180ab.c.LR1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.LR1}):} We have $s_{\mu
}s_{\nu}=\sum_{\lambda}c_{\mu,\nu}^{\lambda}s_{\lambda}$, where the sum ranges
over all partitions $\lambda$ (according to the definition of the coefficients
$c_{\mu,\nu}^{\lambda}$). In other words, $s_{\mu}s_{\nu}=\sum_{\lambda
\in \Par }c_{\mu,\nu}^{\lambda}s_{\lambda}$. Evaluating both
sides of this equality at $\left(  x_{1},x_{2},\ldots,x_{k}\right)  $, we
obtain%
\begin{align*}
&  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum_{\lambda\in\operatorname{Par}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
+\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  >k}%
c_{\mu,\nu}^{\lambda}\underbrace{s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  }_{\substack{=0\\\text{(by (\ref{sol.liri.180ab.c.toobig}))}%
}}\\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
+\underbrace{\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
>k}c_{\mu,\nu}^{\lambda}0}_{=0}\\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  ,
\end{align*}
qed.} Hence,%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\ \ell\left(  \lambda\right)  \leq k;\\\text{all parts of }\lambda\text{ are
}\leq m+n}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  . \label{sol.liri.180ab.c.LR2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.LR2}):} From
(\ref{sol.liri.180ab.c.LR1}), we obtain%
\begin{align*}
&  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}c_{\mu,\nu}%
^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  +\sum\limits
_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k;\\\text{not all parts of }\lambda\text{ are }\leq m+n}}\underbrace{c_{\mu
,\nu}^{\lambda}}_{\substack{=0\\\text{(by Exercise \ref{exe.liri.180ab}(b))}%
}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}c_{\mu,\nu}%
^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
+\underbrace{\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(
\lambda\right)  \leq k;\\\text{not all parts of }\lambda\text{ are }\leq
m+n}}0s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  }_{=0}\\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}c_{\mu,\nu}%
^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\end{align*}
This proves (\ref{sol.liri.180ab.c.LR2}).}

Define a set $\mathfrak{A}$ by
\begin{equation}
\mathfrak{A}=\left\{  \alpha\in \Par \ \mid\ \ell\left(
\alpha\right)  \leq k\text{; all parts of }\alpha\text{ are }\leq m+n\right\}
. \label{sol.liri.180ab.bc.A}%
\end{equation}
For every partition $\lambda$, let $\lambda^{\vee\left\{  m+n\right\}  }$
denote the $k$-tuple $\left(  m+n-\lambda_{k},m+n-\lambda_{k-1},\ldots
,m+n-\lambda_{1}\right)  $. It is straightforward to see that for every
$\lambda\in\mathfrak{A}$, we have%
\begin{equation}
\lambda^{\vee\left\{  m+n\right\}  }\in\mathfrak{A}.
\label{sol.liri.180ab.bc.is-a-partition}%
\end{equation}
Thus, the map%
\[
\mathfrak{A}\rightarrow\mathfrak{A},\ \ \ \ \ \ \ \ \ \ \lambda\mapsto
\lambda^{\vee\left\{  m+n\right\}  }%
\]
is well-defined. It is easy to see that this map is an involution (i.e., every
$\lambda\in\mathfrak{A}$ satisfies $\left(  \lambda^{\vee\left\{  m+n\right\}
}\right)  ^{\vee\left\{  m+n\right\}  }=\lambda$), thus a bijection.

Now, the summation sign ``$\sum\limits_{\substack{\lambda
\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq k;\\\text{all parts
of }\lambda\text{ are }\leq m+n}}$'' in
(\ref{sol.liri.180ab.c.LR2}) rewrites as ``$\sum_{\lambda
\in\mathfrak{A}}$'' (because of how we defined $\mathfrak{A}$).
Hence, (\ref{sol.liri.180ab.c.LR2}) rewrites as
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\label{sol.liri.180ab.c.LR3}%
\end{equation}
Hence, in the Laurent polynomial ring $\kk\left[  x_{1},x_{2}%
,\ldots,x_{k},x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right]  $, we have%
\begin{equation}
s_{\mu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \cdot s_{\nu
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  =\sum_{\lambda
\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  . \label{sol.liri.180ab.c.LR-3}%
\end{equation}
\footnote{This follows by substituting the variables $x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}$ for $x_{1},x_{2},\ldots,x_{k}$ in the equality
(\ref{sol.liri.180ab.c.LR3}).}

Now, it is straightforward to see that $\mu^{\vee\left\{  m\right\}  }$ and
$\nu^{\vee\left\{  n\right\}  }$ are partitions satisfying $\ell\left(
\mu^{\vee\left\{  m\right\}  }\right)  \leq k$ and $\ell\left(  \nu
^{\vee\left\{  n\right\}  }\right)  \leq k$; also, all parts of the partition
$\mu^{\vee\left\{  m\right\}  }$ are $\leq m$, and all parts of the partition
$\nu^{\vee\left\{  n\right\}  }$ are $\leq n$. Moreover,%
\begin{equation}
\mu=\left(  m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k},m-\left(
\mu^{\vee\left\{  m\right\}  }\right)  _{k-1},\ldots,m-\left(  \mu
^{\vee\left\{  m\right\}  }\right)  _{1}\right)
\label{sol.liri.180ab.c.mu-conj-conj}%
\end{equation}
and%
\begin{equation}
\nu=\left(  n-\left(  \nu^{\vee\left\{  n\right\}  }\right)  _{k},n-\left(
\nu^{\vee\left\{  n\right\}  }\right)  _{k-1},\ldots,n-\left(  \nu
^{\vee\left\{  n\right\}  }\right)  _{1}\right)
\label{sol.liri.180ab.c.nu-conj-conj}%
\end{equation}
(this is still easy to verify). These observations lead us to the conclusion
that we can apply (\ref{sol.liri.180ab.c.LR-3}) to $\mu^{\vee\left\{
m\right\}  }$, $\mu$, $\nu^{\vee\left\{  n\right\}  }$ and $\nu$ instead of
$\mu$, $\mu^{\vee\left\{  m\right\}  }$, $\nu$ and $\nu^{\vee\left\{
n\right\}  }$. As a result, we obtain%
\begin{align}
s_{\mu^{\vee\left\{  m\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots
,x_{k}^{-1}\right)  \cdot s_{\nu^{\vee\left\{  n\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)   &  =\sum_{\lambda\in\mathfrak{A}%
}c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}^{\lambda
}s_{\lambda}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}%
s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  \label{sol.liri.180ab.c.LR-3a}%
\end{align}
(here, we have substituted $\lambda^{\vee\left\{  m+n\right\}  }$ for the
summation index $\lambda$ (since the map $\mathfrak{A}\rightarrow
\mathfrak{A},\ \lambda\mapsto\lambda^{\vee\left\{  m+n\right\}  }$ is a bijection)).

Now, we can apply Exercise \ref{exe.liri.180}(d) to $\mu^{\vee\left\{
m\right\}  }$, $\mu^{\vee\left\{  m\right\}  }$, $\mu$ and $\mu$ instead of
$\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$ (because of
(\ref{sol.liri.180ab.c.mu-conj-conj})). As a result, we obtain%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m}\cdot s_{\mu^{\vee\left\{  m\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\label{sol.liri.180ab.c.mu-reversion}%
\end{equation}


Also, we can apply Exercise \ref{exe.liri.180}(d) to $n$, $\nu^{\vee\left\{
n\right\}  }$, $\nu^{\vee\left\{  n\right\}  }$, $\nu$ and $\nu$ instead of
$m$, $\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$ (because of
(\ref{sol.liri.180ab.c.nu-conj-conj})). As a result, we obtain%
\begin{equation}
s_{\nu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{n}\cdot s_{\nu^{\vee\left\{  n\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\label{sol.liri.180ab.c.nu-reversion}%
\end{equation}


Furthermore, every $\lambda\in\mathfrak{A}$ satisfies%
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m+n}\cdot s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
.\label{sol.liri.180ab.c.lambda-reversion}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.lambda-reversion}):} Let
$\lambda\in\mathfrak{A}$. Then, $\lambda\in\mathfrak{A}=\left\{  \alpha
\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all
parts of }\alpha\text{ are }\leq m+n\right\}  $. In other words, $\lambda$ is
an element of $ \Par $ such that $\ell\left(  \lambda\right)
\leq k$ and such that all parts of $\lambda$ are $\leq m+n$. Now,
(\ref{sol.liri.180ab.bc.is-a-partition}) yields that $\lambda^{\vee\left\{
m+n\right\}  }\in\mathfrak{A}=\left\{  \alpha\in \Par %
\ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all parts of }\alpha\text{
are }\leq m+n\right\}  $. In other words, $\lambda^{\vee\left\{  m+n\right\}
}$ is an element of $ \Par $ such that $\ell\left(  \lambda
^{\vee\left\{  m+n\right\}  }\right)  \leq k$ and such that all parts of
$\lambda^{\vee\left\{  m+n\right\}  }$ are $\leq m+n$.
\par
Recall that $\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)
^{\vee\left\{  m+n\right\}  }=\lambda$ (as we said, this follows easily from
the definitions). Thus,%
\[
\lambda=\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)  ^{\vee\left\{
m+n\right\}  }=\left(  m+n-\left(  \lambda^{\vee\left\{  m+n\right\}
}\right)  _{k},m+n-\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)
_{k-1},\ldots,m+n-\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)
_{1}\right)
\]
(by the definition of $\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)
^{\vee\left\{  m+n\right\}  }$). Hence, we can apply Exercise
\ref{exe.liri.180}(d) to $m+n$, $\lambda^{\vee\left\{  m+n\right\}  }$,
$\lambda^{\vee\left\{  m+n\right\}  }$, $\lambda$ and $\lambda$ instead of
$m$, $\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$. As a result, we
obtain%
\[
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m+n}\cdot s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\]
This proves (\ref{sol.liri.180ab.c.lambda-reversion}).}

Now, (\ref{sol.liri.180ab.c.LR3}) yields%
\begin{align}
&  \sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  \nonumber\\
&  =\underbrace{s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  }%
_{\substack{=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\mu
^{\vee\left\{  m\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  \\\text{(by (\ref{sol.liri.180ab.c.mu-reversion}))}}%
}\cdot\underbrace{s_{\nu}\left(  x_{1},x_{2},\ldots,x_{k}\right)
}_{\substack{=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{n}\cdot s_{\nu
^{\vee\left\{  n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  \\\text{(by (\ref{sol.liri.180ab.c.nu-reversion}))}}}\nonumber\\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\mu^{\vee\left\{
m\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\cdot\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{n}\cdot s_{\nu^{\vee\left\{
n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\nonumber\\
&  =\underbrace{\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\left(  x_{1}%
x_{2}\cdots x_{k}\right)  ^{n}}_{=\left(  x_{1}x_{2}\cdots x_{k}\right)
^{m+n}}\cdot\underbrace{s_{\mu^{\vee\left\{  m\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \cdot s_{\nu^{\vee\left\{
n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
}_{\substack{=\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}
},\nu^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }%
}s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  \\\text{(by (\ref{sol.liri.180ab.c.LR-3a}))}%
}}\nonumber\\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m+n}\cdot\sum_{\lambda
\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}
}}^{\lambda^{\vee\left\{  m+n\right\}  }}s_{\lambda^{\vee\left\{  m+n\right\}
}}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }%
}\underbrace{\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m+n}\cdot s_{\lambda
^{\vee\left\{  m+n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  }_{\substack{=s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \\\text{(by (\ref{sol.liri.180ab.c.lambda-reversion}))}%
}}\nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}s_{\lambda
}\left(  x_{1},x_{2},\ldots,x_{k}\right)
.\label{sol.liri.180ab.c.almostthere}%
\end{align}


But Remark \ref{rmk.many-symmetric-function-bases-prop.finite}(d) (applied to
$N=k$) yields that the set $\left\{  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  \right\}  $, as $\lambda$ runs through all partitions
having length $\leq k$, is a basis of the $\kk$-module $\Lambda\left(
x_{1},x_{2},\ldots,x_{k}\right)  $. In other words, the family $\left(
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  _{\lambda
\in \Par ;\ \ell\left(  \lambda\right)  \leq k}$ is a basis of
the $\kk$-module $\Lambda\left(  x_{1},x_{2},\ldots,x_{k}\right)  $. In
particular, the family $\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \right)  _{\lambda\in \Par ;\ \ell\left(
\lambda\right)  \leq k}$ is $\kk$-linearly independent.

But the set $\mathfrak{A}$ is a subset of $\left\{  \alpha\in
 \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\right\}  $.
Hence, the family $\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
\right)  _{\lambda\in\mathfrak{A}}$ is a subfamily of the family $\left(
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  _{\lambda
\in\left\{  \alpha\in \Par \ \mid\ \ell\left(  \alpha\right)
\leq k\right\}  }=\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
\right)  _{\lambda\in \Par ;\ \ell\left(  \lambda\right)  \leq
k}$. Since the family $\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \right)  _{\lambda\in \Par ;\ \ell\left(
\lambda\right)  \leq k}$ is $\kk$-linearly independent, we thus
conclude that its subfamily $\left(  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  \right)  _{\lambda\in\mathfrak{A}}$ is also $\kk%
$-linearly independent. As a consequence, if two $\kk$-linear
combinations of the family $\left(  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  \right)  _{\lambda\in\mathfrak{A}}$ are equal, then
their respective coefficients must be equal. Thus, from
(\ref{sol.liri.180ab.c.almostthere4}), we conclude that
\begin{equation}
c_{\mu,\nu}^{\lambda}=c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{
n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathfrak{A}\text{.}%
\label{sol.liri.180ab.c.almostthere4}%
\end{equation}


Now, let $\lambda$ be a partition such that $\ell\left(  \lambda\right)  \leq
k$. Assume that all parts of $\lambda$ are $\leq m+n$. Then, $\lambda$ is an
element of $ \Par $ such that $\ell\left(  \lambda\right)  \leq
k$ and such that all parts of $\lambda$ are $\leq m+n$. In other words,
$\lambda\in\mathfrak{A}$ (by the definition of $\mathfrak{A}$). Hence,
(\ref{sol.liri.180ab.c.almostthere4}) yields $c_{\mu,\nu}^{\lambda}%
=c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}%
^{\lambda^{\vee\left\{  m+n\right\}  }}$. This solves Exercise
\ref{exe.liri.180ab}(c).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.180ab}}

\textit{Solution to Exercise \ref{exe.liri.180ab}.} By our assumption, we know
that all parts of $\mu$ are $\leq m$. Thus,%
\begin{equation}
\mu_i \leq m\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.
\label{sol.liri.180ab.allparts1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.allparts1}):} Let $i$ be a
positive integer. We want to prove that $\mu_i \leq m$.
\par
If $\mu_i =0$, then $\mu_i =0\leq m$ (since $m\in \NN$). Hence, if
$\mu_i =0$, then $\mu_i \leq m$ is proven. Thus, for the rest of the proof
of $\mu_i \leq m$, we can WLOG assume that we don't have $\mu_i =0$. Assume
this.
\par
The number $\mu_i$ is nonzero (since we don't have $\mu_i =0$). Hence,
$\mu_i$ is a part of $\mu$. Therefore, $\mu_i $ is $\leq m$ (since all
parts of $\mu$ are $\leq m$). Hence, $\mu_i \leq m$ is proven. This proves
(\ref{sol.liri.180ab.allparts1}).} Also,
\begin{equation}
\nu_{i}\leq n\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }i.
\label{sol.liri.180ab.allparts2}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.allparts2}):} The assertion
(\ref{sol.liri.180ab.allparts2}) follows from the same arguments as
(\ref{sol.liri.180ab.allparts1}) (but with $\mu$ and $m$ replaced by $\nu$ and
$n$).}

(a) We can apply Exercise \ref{exe.liri.180}(a) to $\mu$, $\mu$, $\mu
^{\vee\left\{  m\right\}  }$ and $\mu^{\vee\left\{  m\right\}  }$ instead of
$\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$. As a consequence, we
conclude that $\mu^{\vee\left\{  m\right\}  }$ and
$\mu^{\vee\left\{  m\right\}  }$ are partitions, and that
$s_{\mu/\mu}=s_{\mu^{\vee\left\{  m\right\}  }/\mu^{\vee\left\{  m\right\}  }}$.
Thus, $\mu^{\vee\left\{  m\right\}  }$ is a partition.
The same argument (but with $m$, $\mu$ and $\mu^{\vee\left\{  m\right\}  }$
replaced by $n$, $\nu$ and $\nu^{\vee\left\{  n\right\}  }$) yields that
$\nu^{\vee\left\{  n\right\}  }$ is a partition. This solves Exercise
\ref{exe.liri.180ab}(a).

(b) Assume that not all parts of $\lambda$ are $\leq m+n$. Then, $\lambda
_{1}>m+n$\ \ \ \ \footnote{\textit{Proof.} Not all parts of $\lambda$ are
$\leq m+n$. In other words, there exists a part of $\lambda$ which is $>m+n$.
This part must have the form $\lambda_{i}$ for some positive integer $i$.
Consider this $i$. Thus, $\lambda_{i}$ is a part of $\lambda$ which is $>m+n$.
Hence, $\lambda_{i}>m+n$.
\par
But $\lambda$ is a partition. Hence, $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq\cdots$. In other words, $\lambda_{u}\geq\lambda_{v}$ for any
two positive integers $u$ and $v$ satisfying $u\leq v$. Applying this to $u=1$
and $v=i$, we obtain $\lambda_{1}\geq\lambda_{i}$ (since $1\leq i$ (since $i$
is a positive integer)), so that $\lambda_{1}\geq\lambda_{i}>m+n$, qed.}. As a
consequence, $\lambda_{1}-\mu_{1}>\nu_{1}$\ \ \ \ \footnote{\textit{Proof.}
Applying (\ref{sol.liri.180ab.allparts1}) to $i=1$, we obtain $\mu_{1}\leq m$.
Hence, $m\geq\mu_{1}$.
\par
Applying (\ref{sol.liri.180ab.allparts2}) to $i=1$, we obtain $\nu_{1}\leq n$.
Hence, $n\geq\nu_{1}$.
\par
Now, $\lambda_{1}>\underbrace{m}_{\geq\mu_{1}}+\underbrace{n}_{\geq\nu_{1}%
}\geq\mu_{1}+\nu_{1}$, so that $\lambda_{1}-\mu_{1}>\nu_{1}$, qed.}.

Now, we shall show that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{there exists no column-strict tableau }T\text{ of shape }\lambda
/\mu\text{ with }\cont \left(  T\right)  =\nu\\
\text{having the property that each }\cont \left(  T\mid
_{\operatorname{cols}\geq j}\right)  \text{ is a partition}%
\end{array}
\right)  . \label{sol.liri.180ab.b.nonexist}%
\end{equation}


\textit{Proof of (\ref{sol.liri.180ab.b.nonexist}):} Assume the contrary. Thus,
there exists a column-strict tableau $T$ of shape $\lambda/\mu$ with
$\cont \left(  T\right)  =\nu$ having the property that each
$\cont \left(  T\mid_{\operatorname{cols}\geq j}\right)  $ is a
partition. Consider this $T$.

The column-strict tableau $T$ has shape $\lambda/\mu$, and thus its $1$-st row
has $\lambda_{1}-\mu_{1}$ entries. In other words,%
\[
\left(  \text{the number of entries in the }1\text{-st row of }T\right)
=\lambda_{1}-\mu_{1}.
\]


We know that each $\cont \left(  T\mid_{\operatorname{cols}\geq
j}\right)  $ is a partition. Using this fact, it is easy to see that every
entry in the $1$-st row of $T$ equals $1$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Then, there exists an entry in the $1$-st row of $T$
which does not equal $1$. In other words, there exists a cell $\mathbf{c}$ of
the $1$-st row of $T$ such that the entry of $T$ in $\mathbf{c}$ does not
equal $1$. Consider such a $\mathbf{c}$. We can write $\mathbf{c}$ in the form
$\mathbf{c}=\left(  1,J\right)  $ for some positive integer $J$ (since
$\mathbf{c}$ is a cell of the $1$-st row). Consider this $J$.
\par
Recall that $\mathbf{c}$ is a cell of the $1$-st row of $T$; thus,
$\mathbf{c}$ is a cell of $T$. Let $k$ denote the entry of $T$ in $\mathbf{c}%
$. Then, $k$ does not equal $1$ (since the entry of $T$ in $\mathbf{c}$ does
not equal $1$). In other words, $k\neq1$. But $k$ is a positive integer (since
$k$ is an entry of $T$), so that $k\geq1$. Combined with $k\neq1$, this yields
$k>1$, so that $k\geq2$ (since $k$ is an integer).
\par
The cell $\mathbf{c}$ lies in column $J$ (since $\mathbf{c}=\left(
1,J\right)  $) and is a cell of the tableau $T$ (since it is a cell of the
$1$-st row of $T$). In other words, $\mathbf{c}$ is a cell of column $J$ of
the tableau $T$. Hence, $\mathbf{c}$ is a cell of one of the columns
$J,J+1,J+2,\ldots$ of the tableau $T$. But $T\mid_{\operatorname{cols}\geq J}$
is defined as the restriction of the tableau $T$ to the union of its columns
$J,J+1,J+2,\ldots$. Thus, the cell $\mathbf{c}$ is a cell of $T\mid
_{\operatorname{cols}\geq J}$ (because $\mathbf{c}$ is a cell of one of the
columns $J,J+1,J+2,\ldots$ of the tableau $T$). We have%
\[
\left(  \text{the entry of }T\mid_{\operatorname{cols}\geq J}\text{ in
}\mathbf{c}\right)  =\left(  \text{the entry of }T\text{ in }\mathbf{c}%
\right)  =k
\]
(since the entry of $T$ in $\mathbf{c}$ is $k$). Hence, $k$ is the entry of
$T\mid_{\operatorname{cols}\geq J}$ in $\mathbf{c}$. In particular, $k$ is an
entry of $T\mid_{\operatorname{cols}\geq J}$. Consequently, there exists at
least one entry of $T\mid_{\operatorname{cols}\geq J}$ equal to $k$.
\par
We know that each $\cont \left(  T\mid_{\operatorname{cols}\geq
j}\right)  $ is a partition. Applying this to $j=J$, we conclude that
$\cont \left(  T\mid_{\operatorname{cols}\geq J}\right)  $ is a
partition. Thus,%
\[
\left(  \cont \left(  T\mid_{\operatorname{cols}\geq J}\right)
\right)  _{1}\geq\left(  \cont \left(  T\mid_{\operatorname{cols}%
\geq J}\right)  \right)  _{2}\geq\left(  \cont \left(
T\mid_{\operatorname{cols}\geq J}\right)  \right)  _{3}\geq\cdots.
\]
In other words, $\left(  \cont \left(  T\mid_{\operatorname{cols}%
\geq J}\right)  \right)  _{u}\geq\left(  \cont \left(
T\mid_{\operatorname{cols}\geq J}\right)  \right)  _{v}$ for every two
positive integers $u$ and $v$ satisfying $u\leq v$. Applying this to $u=1$ and
$v=k$, we obtain%
\begin{align*}
\left(  \cont \left(  T\mid_{\operatorname{cols}\geq J}\right)
\right)  _{1}  &  \geq\left(  \cont \left(  T\mid
_{\operatorname{cols}\geq J}\right)  \right)  _{k}=\left\vert \left(
T\mid_{\operatorname{cols}\geq J}\right)  ^{-1}\left(  k\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\cont \left(  T\mid_{\operatorname{cols}\geq J}\right)  \right)
_{k}\right) \\
&  =\left(  \text{the number of entries of }T\mid_{\operatorname{cols}\geq
J}\text{ equal to }k\right)  \geq1
\end{align*}
(since there exists at least one entry of $T\mid_{\operatorname{cols}\geq J}$
equal to $k$). But%
\begin{align*}
\left(  \cont \left(  T\mid_{\operatorname{cols}\geq J}\right)
\right)  _{1}  &  =\left\vert \left(  T\mid_{\operatorname{cols}\geq
J}\right)  ^{-1}\left(  1\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\left(  \cont \left(  T\mid
_{\operatorname{cols}\geq J}\right)  \right)  _{1}\right) \\
&  =\left(  \text{the number of entries of }T\mid_{\operatorname{cols}\geq
J}\text{ equal to }1\right)  ,
\end{align*}
so that%
\[
\left(  \text{the number of entries of }T\mid_{\operatorname{cols}\geq
J}\text{ equal to }1\right)  =\left(  \cont \left(
T\mid_{\operatorname{cols}\geq J}\right)  \right)  _{1}\geq1.
\]
Hence, there exists at least one entry of $T\mid_{\operatorname{cols}\geq J}$
equal to $1$. In other words, there exists at least one cell $\mathbf{d}$ of
$T\mid_{\operatorname{cols}\geq J}$ such that the entry of $T\mid
_{\operatorname{cols}\geq J}$ in $\mathbf{d}$ is equal to $1$. Consider this
$\mathbf{d}$.
\par
The cells of $T\mid_{\operatorname{cols}\geq J}$ are precisely the cells of
the tableau $T$ which lie in its columns $J,J+1,J+2,\ldots$ (since
$T\mid_{\operatorname{cols}\geq J}$ is the restriction of the tableau $T$ to
the union of its columns $J,J+1,J+2,\ldots$). Hence, $\mathbf{d}$ is a cell of
the tableau $T$ which lies in its columns $J,J+1,J+2,\ldots$ (since
$\mathbf{d}$ is a cell of $T\mid_{\operatorname{cols}\geq J}$). We have
\[
\left(  \text{the entry of }T\text{ in }\mathbf{d}\right)  =\left(  \text{the
entry of }T\mid_{\operatorname*{cols}\geq J}\text{ in }\mathbf{d}\right)  =1
\]
(since the entry of $T\mid_{\operatorname{cols}\geq J}$ in $\mathbf{d}$ is
equal to $1$).
\par
Write the cell $\mathbf{d}$ in the form $\mathbf{d}=\left(  u,v\right)  $ for
some positive integers $u$ and $v$. Then, the cell $\mathbf{d}$ lies in column
$v$. In other words,%
\[
v=\left(  \text{the column in which the cell }\mathbf{d}\text{ lies}\right)
.
\]
But we know that $\mathbf{d}$ is a cell of the tableau $T$ which lies in its
columns $J,J+1,J+2,\ldots$. In particular, $\mathbf{d}$ lies in one of the
columns $J,J+1,J+2,\ldots$. Consequently, $\left(  \text{the column in which
the cell }\mathbf{d}\text{ lies}\right)  $ is one of the integers
$J,J+1,J+2,\ldots$. In other words,%
\[
\left(  \text{the column in which the cell }\mathbf{d}\text{ lies}\right)
\in\left\{  J,J+1,J+2,\ldots\right\}  .
\]
Hence, $v=\left(  \text{the column in which the cell }\mathbf{d}\text{
lies}\right)  \in\left\{  J,J+1,J+2,\ldots\right\}  $, so that $v\geq J$.
\par
Let $Y\left(  \lambda/\mu\right)  $ denote the skew Ferrers diagram
$\lambda/\mu$. Then, the cells of $Y\left(  \lambda/\mu\right)  $ are exactly
the cells of $T$ (since $T$ has shape $\lambda/\mu$). Therefore, $\mathbf{c}$
and $\mathbf{d}$ are cells of $Y\left(  \lambda/\mu\right)  $ (since
$\mathbf{c}$ and $\mathbf{d}$ are cells of $T$). In other words, $\mathbf{c}$
and $\mathbf{d}$ are elements of $Y\left(  \lambda/\mu\right)  $. That is, we
have $\mathbf{c}\in Y\left(  \lambda/\mu\right)  $ and $\mathbf{d}\in Y\left(
\lambda/\mu\right)  $.
\par
We recall that the skew Ferrers diagram $\lambda/\mu$ is the set of all
$\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfying
$\mu_{i}<j\leq\lambda_{i}$. In other words, $Y\left(  \lambda/\mu\right)  $ is
the set of all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfying $\mu_{i}<j\leq\lambda_{i}$ (since $Y\left(  \lambda/\mu\right)  $
is the skew Ferrers diagram $\lambda/\mu$). In other words,
\[
Y\left(  \lambda/\mu\right)  =\left\{  \left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}\ \mid\ \mu_{i}<j\leq\lambda_{i}\right\}  .
\]
Now,%
\[
\left(  1,J\right)  =\mathbf{c}\in Y\left(  \lambda/\mu\right)  =\left\{
\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ \mu
_{i}<j\leq\lambda_{i}\right\}  .
\]
In other words, $\left(  1,J\right)  $ is an element of $\left\{
1,2,3,\ldots\right\}  ^{2}$ and satisfies $\mu_{1}<J\leq\lambda_{1}$.
\par
Furthermore,%
\[
\left(  u,v\right)  =\mathbf{d}\in Y\left(  \lambda/\mu\right)  =\left\{
\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ \mu
_{i}<j\leq\lambda_{i}\right\}  .
\]
In other words, $\left(  u,v\right)  $ is an element of $\left\{
1,2,3,\ldots\right\}  ^{2}$ and satisfies $\mu_{u}<v\leq\lambda_{u}$. We have
$\left(  u,v\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$; thus,
$u\in\left\{  1,2,3,\ldots\right\}  $ and $v\in\left\{  1,2,3,\ldots\right\}
$.
\par
Now, $\mu_{1}<J\leq v$ (since $v\geq J$). Also $1\leq u$ (since $u\in\left\{
1,2,3,\ldots\right\}  $), so that $u\geq1$.
\par
Also, $\lambda$ is a partition. Thus, $\lambda_{1}\geq\lambda_{2}\geq
\lambda_{3}\geq\cdots$. Hence, any two positive integers $u^{\prime}$ and
$v^{\prime}$ satisfying $u^{\prime}\leq v^{\prime}$ satisfy $\lambda
_{u^{\prime}}\geq\lambda_{v^{\prime}}$. Applying this to $u^{\prime}=1$ and
$v^{\prime}=u$, we obtain $\lambda_{1}\geq\lambda_{u}$ (since $1\leq u$).
Thus, $\lambda_{u}\leq\lambda_{1}$, so that $v\leq\lambda_{u}\leq\lambda_{1}$.
Now, $\left(  1,v\right)  $ is an element of $\left\{  1,2,3,\ldots\right\}
^{2}$ (since $1\in\left\{  1,2,3,\ldots\right\}  $ and $v\in\left\{
1,2,3,\ldots\right\}  $) and satisfies $\mu_{1}<v\leq\lambda_{1}$. In other
words,%
\[
\left(  1,v\right)  \in\left\{  \left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}\ \mid\ \mu_{i}<j\leq\lambda_{i}\right\}  =Y\left(  \lambda
/\mu\right)  .
\]
In other words, $\left(  1,v\right)  $ is a cell of $Y\left(  \lambda
/\mu\right)  $. In other words, $\left(  1,v\right)  $ is a cell of $T$ (since
the cells of $Y\left(  \lambda/\mu\right)  $ are exactly the cells of $T$).
Recall that $\mathbf{c}$ is also a cell of $T$; in other words, $\left(
1,J\right)  $ is a cell of $T$ (since $\mathbf{c}=\left(  1,J\right)  $).
\par
We know that $T$ is a column-strict tableau. Thus, the entries of $T$ are
weakly increasing left-to-right in rows, and strictly increasing top-to-bottom
in columns.
\par
The entries of $T$ are weakly increasing left-to-right in rows. Consequently,
if $\mathbf{p}$ and $\mathbf{q}$ are two cells of $T$ lying in the same row,
with $\mathbf{q}$ lying weakly to the east of $\mathbf{p}$ (that is,
$\mathbf{q}$ either lies to the east of $\mathbf{p}$ or is identical with
$\mathbf{p}$), then we have%
\[
\left(  \text{the entry of }T\text{ in }\mathbf{p}\right)  \leq\left(
\text{the entry of }T\text{ in }\mathbf{q}\right)  .
\]
Applying this to $\mathbf{p}=\left(  1,J\right)  $ and $\mathbf{q}=\left(
1,v\right)  $, we obtain%
\[
\left(  \text{the entry of }T\text{ in }\left(  1,J\right)  \right)
\leq\left(  \text{the entry of }T\text{ in }\left(  1,v\right)  \right)
\]
(since the cells $\left(  1,J\right)  $ and $\left(  1,v\right)  $ of $T$ lie
in the same row (because $1=1$), and the cell $\left(  1,v\right)  $ lies
weakly to the east of $\left(  1,J\right)  $ (since $v\geq J$)). In other
words,%
\[
\left(  \text{the entry of }T\text{ in }\left(  1,v\right)  \right)  =\left(
\text{the entry of }T\text{ in }\underbrace{\left(  1,J\right)  }%
_{=\mathbf{c}}\right)  =\left(  \text{the entry of }T\text{ in }%
\mathbf{c}\right)  =k>1.
\]
\par
On the other hand, $\mathbf{d}$ is a cell of $T$. In other words, $\left(
u,v\right)  $ is a cell of $T$ (since $\mathbf{d}=\left(  u,v\right)  $).
\par
The entries of $T$ are strictly increasing top-to-bottom in columns.
Consequently, the entries of $T$ are weakly increasing top-to-bottom in
columns. Thus, if $\mathbf{p}$ and $\mathbf{q}$ are two cells of $T$ lying in
the same column, with $\mathbf{q}$ lying weakly to the south of $\mathbf{p}$
(that is, $\mathbf{q}$ either lies to the south of $\mathbf{p}$ or is
identical with $\mathbf{p}$), then we have%
\[
\left(  \text{the entry of }T\text{ in }\mathbf{p}\right)  \leq\left(
\text{the entry of }T\text{ in }\mathbf{q}\right)  .
\]
Applying this to $\mathbf{p}=\left(  1,v\right)  $ and $\mathbf{q}=\left(
u,v\right)  $, we obtain%
\[
\left(  \text{the entry of }T\text{ in }\left(  1,v\right)  \right)
\leq\left(  \text{the entry of }T\text{ in }\left(  u,v\right)  \right)
\]
(since the cells $\left(  1,v\right)  $ and $\left(  u,v\right)  $ of $T$ lie
in the same column (because $v=v$), and the cell $\left(  u,v\right)  $ lies
weakly to the south of $\left(  1,v\right)  $ (since $u\geq1$)). Thus,%
\[
\left(  \text{the entry of }T\text{ in }\left(  1,v\right)  \right)
\leq\left(  \text{the entry of }T\text{ in }\underbrace{\left(  u,v\right)
}_{=\mathbf{d}}\right)  =\left(  \text{the entry of }T\text{ in }%
\mathbf{d}\right)  =1.
\]
This contradicts the inequality $\left(  \text{the entry of }T\text{ in
}\left(  1,v\right)  \right)  >1$ (which was proven above). This contradiction
proves that our assumption was wrong, qed.}. Hence, every entry in the $1$-st
row of $T$ is an entry of $T$ equal to $1$. As a consequence,%
\[
\left(  \text{the number of entries in the }1\text{-st row of }T\right)
\leq\left(  \text{the number of entries of }T\text{ equal to }1\right)  .
\]
Thus,%
\[
\left(  \text{the number of entries of }T\text{ equal to }1\right)
\geq\left(  \text{the number of entries in the }1\text{-st row of }T\right)
=\lambda_{1}-\mu_{1}>\nu_{1}.
\]
But now, recall that $\cont \left(  T\right)  =\nu$, so that
$\nu=\cont \left(  T\right)  $ and therefore%
\begin{align*}
\nu_{1}  &  =\left(  \cont \left(  T\right)  \right)
_{1}=\left\vert T^{-1}\left(  1\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\cont \left(  T\right)  \right)  _{1}\right) \\
&  =\left(  \text{the number of entries of }T\text{ equal to }1\right)
>\nu_{1}.
\end{align*}
This is absurd. This contradiction proves that our assumption was wrong.
Hence, (\ref{sol.liri.180ab.b.nonexist}) is proven.

Now, Corollary \ref{L-R-rule} yields that $c_{\mu,\nu}^{\lambda}$ counts
column-strict tableaux $T$ of shape $\lambda/\mu$ with $\cont
\left(  T\right)  =\nu$ having the property that each $\cont
\left(  T\mid_{\operatorname{cols}\geq j}\right)  $ is a partition. Since
there exists no such $T$ (according to (\ref{sol.liri.180ab.b.nonexist})),
this yields that $c_{\mu,\nu}^{\lambda}=0$. This solves Exercise
\ref{exe.liri.180ab}(b).

\textit{Remark:} An alternative solution of Exercise \ref{exe.liri.180ab}(b)
can be obtained easily from Exercise \ref{exe.liri.simple}(c).

(c) Let us forget that $\lambda$ is fixed.

If $\lambda\in \Par $ is such that $\ell\left(  \lambda\right)
>k$, then%
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =0.
\label{sol.liri.180ab.c.toobig}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.toobig}):} Let $\lambda
\in \Par $ be such that $\ell\left(  \lambda\right)  >k$. We know
that $\ell\left(  \lambda\right)  $ is the number of parts of $\lambda$. Thus,
the number of parts of $\lambda$ is $>k$ (since $\ell\left(  \lambda\right)
>k$). In other words, the partition $\lambda$ has more than $k$ parts. Hence,
Exercise \ref{exe.schur.finite}(b) (applied to $k$ instead of $n$) yields
$s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =0$. This proves
(\ref{sol.liri.180ab.c.toobig}).}

We first notice that%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum_{\lambda\in\operatorname{Par}%
;\ \ell\left(  \lambda\right)  \leq k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  . \label{sol.liri.180ab.c.LR1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.LR1}):} We have $s_{\mu
}s_{\nu}=\sum_{\lambda}c_{\mu,\nu}^{\lambda}s_{\lambda}$, where the sum ranges
over all partitions $\lambda$ (according to the definition of the coefficients
$c_{\mu,\nu}^{\lambda}$). In other words, $s_{\mu}s_{\nu}=\sum_{\lambda
\in \Par }c_{\mu,\nu}^{\lambda}s_{\lambda}$. Thus,
\begin{align*}
\underbrace{\left(  s_{\mu}s_{\nu}\right)  }_{=\sum_{\lambda\in
 \Par }c_{\mu,\nu}^{\lambda}s_{\lambda}}\left(  x_{1}%
,x_{2},\ldots,x_{k}\right)   &  =\left(  \sum_{\lambda\in \Par %
}c_{\mu,\nu}^{\lambda}s_{\lambda}\right)  \left(  x_{1},x_{2},\ldots
,x_{k}\right)  =\sum_{\lambda\in\operatorname{Par}}c_{\mu,\nu}^{\lambda
}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
+\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  >k}%
c_{\mu,\nu}^{\lambda}\underbrace{s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  }_{\substack{=0\\\text{(by (\ref{sol.liri.180ab.c.toobig}))}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }\lambda\in
 \Par \text{ satisfies either }\ell\left(  \lambda\right)  \leq
k\text{ or }\ell\left(  \lambda\right)  >k\text{, but not both}\right) \\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
+\underbrace{\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
>k}c_{\mu,\nu}^{\lambda}0}_{=0}\\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\end{align*}
Compared with%
\[
\left(  s_{\mu}s_{\nu}\right)  \left(  x_{1},x_{2},\ldots,x_{k}\right)
=s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  ,
\]
this yields $s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\sum_{\lambda\in\operatorname{Par}%
;\ \ell\left(  \lambda\right)  \leq k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  $. This proves (\ref{sol.liri.180ab.c.LR1}).}
Hence,%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\ \ell\left(  \lambda\right)  \leq k;\\\text{all parts of }\lambda\text{ are
}\leq m+n}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  . \label{sol.liri.180ab.c.LR2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.LR2}):} From
(\ref{sol.liri.180ab.c.LR1}), we obtain%
\begin{align*}
&  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum_{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}c_{\mu,\nu}%
^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  +\sum\limits
_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq
k;\\\text{not all parts of }\lambda\text{ are }\leq m+n}}\underbrace{c_{\mu
,\nu}^{\lambda}}_{\substack{=0\\\text{(by Exercise \ref{exe.liri.180ab}(b))}%
}}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for every }\lambda\in \Par \text{, we have either
}\left(  \text{all parts of }\lambda\text{ are }\leq m+n\right)  \text{ or}\\
\left(  \text{not all parts of }\lambda\text{ are }\leq m+n\right)  \text{,
but not both}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}c_{\mu,\nu}%
^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)
+\underbrace{\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(
\lambda\right)  \leq k;\\\text{not all parts of }\lambda\text{ are }\leq
m+n}}0s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  }_{=0}\\
&  =\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}c_{\mu,\nu}%
^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\end{align*}
This proves (\ref{sol.liri.180ab.c.LR2}).}

Define a set $\mathfrak{A}$ by
\begin{equation}
\mathfrak{A}=\left\{  \alpha\in \Par \ \mid\ \ell\left(
\alpha\right)  \leq k\text{; all parts of }\alpha\text{ are }\leq m+n\right\}
. \label{sol.liri.180ab.bc.A}%
\end{equation}
For every partition $\lambda$, let $\lambda^{\vee\left\{  m+n\right\}  }$
denote the $k$-tuple $\left(  m+n-\lambda_{k},m+n-\lambda_{k-1},\ldots
,m+n-\lambda_{1}\right)  $. For every $\lambda\in\mathfrak{A}$, we have%
\begin{equation}
\lambda^{\vee\left\{  m+n\right\}  }\in\mathfrak{A}.
\label{sol.liri.180ab.bc.is-a-partition}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.bc.is-a-partition}):} Let
$\lambda\in\mathfrak{A}$. Then, $\lambda\in\mathfrak{A}=\left\{  \alpha
\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all
parts of }\alpha\text{ are }\leq m+n\right\}  $. In other words, $\lambda$ is
an element of $ \Par $ such that $\ell\left(  \lambda\right)
\leq k$ and such that all parts of $\lambda$ are $\leq m+n$. In particular,
$\lambda$ is a partition (since $\lambda\in \Par $). Now, we can
apply Exercise \ref{exe.liri.180}(a) to $m+n$, $\lambda$, $\lambda$,
$\lambda^{\vee\left\{  m+n\right\}  }$ and $\lambda^{\vee\left\{  m+n\right\}
}$ instead of $m$, $\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$. As a
consequence, we conclude that $\lambda^{\vee\left\{  m+n\right\}  }$ and
$\lambda^{\vee\left\{  m+n\right\}  }$ are partitions, and that $s_{\lambda
/\lambda}=s_{\lambda^{\vee\left\{  m+n\right\}  }/\lambda^{\vee\left\{
m+n\right\}  }}$. In particular, $\lambda^{\vee\left\{  m+n\right\}  }$ is a
partition. That is, $\lambda^{\vee\left\{  m+n\right\}  }\in
 \Par $.
\par
The partition $\lambda^{\vee\left\{  m+n\right\}  }$ can be written as a
$k$-tuple (since $\lambda^{\vee\left\{  m+n\right\}  }=\left(  m+n-\lambda
_{k},m+n-\lambda_{k-1},\ldots,m+n-\lambda_{1}\right)  $). Thus, $\ell\left(
\lambda^{\vee\left\{  m+n\right\}  }\right)  \leq k$.
\par
Notice that
\begin{equation}
\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)  _{i}=m+n-\lambda
_{k+1-i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,k\right\}  \label{sol.liri.180ab.bc.is-a-partition.pf.1}%
\end{equation}
(due to $\lambda^{\vee\left\{  m+n\right\}  }=\left(  m+n-\lambda
_{k},m+n-\lambda_{k-1},\ldots,m+n-\lambda_{1}\right)  $).
\par
Now, let $g$ be a part of the partition $\lambda^{\vee\left\{  m+n\right\}  }%
$. We shall show that $g\leq m+n$.
\par
Assume the contrary. Thus, $g>m+n$. We have $g=\left(  \lambda^{\vee\left\{
m+n\right\}  }\right)  _{i}$ for some positive integer $i$ (since $g$ is a
part of the partition $\lambda^{\vee\left\{  m+n\right\}  }$). Consider this
$i$. We have $\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)
_{i}=g>m+n\geq0$. Therefore, $i\leq\ell\left(  \lambda^{\vee\left\{
m+n\right\}  }\right)  \leq k$. Hence, $\left(  \lambda^{\vee\left\{
m+n\right\}  }\right)  _{i}=m+n-\lambda_{k+1-i}$ (by
(\ref{sol.liri.180ab.bc.is-a-partition.pf.1})). Now, $g=\left(  \lambda
^{\vee\left\{  m+n\right\}  }\right)  _{i}=m+n-\underbrace{\lambda_{k+1-i}%
}_{\geq0}\leq m+n$, and this contradicts $g>m+n$. This contradiction proves
that our assumption was wrong. Hence, $g\leq m+n$ is proven.
\par
Now, let us forget that we fixed $g$. We thus have proven that every part $g$
of the partition $\lambda^{\vee\left\{  m+n\right\}  }$ satisfies $g\leq m+n$.
In other words, all parts of the partition $\lambda^{\vee\left\{  m+n\right\}
}$ are $\leq m+n$.
\par
Altogether, we thus know that $\lambda^{\vee\left\{  m+n\right\}  }$ is an
element of $ \Par $ such that $\ell\left(  \lambda^{\vee\left\{
m+n\right\}  }\right)  \leq k$ and such that all parts of the partition
$\lambda^{\vee\left\{  m+n\right\}  }$ are $\leq m+n$. In other words,%
\[
\lambda^{\vee\left\{  m+n\right\}  }\in\left\{  \alpha\in \Par %
\ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all parts of }\alpha\text{
are }\leq m+n\right\}  =\mathfrak{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.180ab.bc.A})}\right)  .
\]
This proves (\ref{sol.liri.180ab.bc.is-a-partition}).} Thus, the map%
\[
\mathfrak{A}\rightarrow\mathfrak{A},\ \ \ \ \ \ \ \ \ \ \lambda\mapsto
\lambda^{\vee\left\{  m+n\right\}  }%
\]
is well-defined. Let us denote this map by $G$. This map $G$ satisfies $G\circ
G=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} We have $G\left(
\lambda\right)  =\lambda^{\vee\left\{  m+n\right\}  }$ for every $\lambda
\in\mathfrak{A}$ (by the definition of $G$).
\par
Let $\lambda\in\mathfrak{A}$. Then, $\lambda\in\mathfrak{A}=\left\{  \alpha
\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all
parts of }\alpha\text{ are }\leq m+n\right\}  $. In other words, $\lambda$ is
an element of $ \Par $ such that $\ell\left(  \lambda\right)
\leq k$ and such that all parts of $\lambda$ are $\leq m+n$.
\par
But $G\left(  \lambda\right)  =\lambda^{\vee\left\{  m+n\right\}  }=\left(
m+n-\lambda_{k},m+n-\lambda_{k-1},\ldots,m+n-\lambda_{1}\right)  $ (by the
definition of $\lambda^{\vee\left\{  m+n\right\}  }$). Thus,
\begin{equation}
\left(  G\left(  \lambda\right)  \right)  _{i}=m+n-\lambda_{k+1-i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k\right\}  .
\label{sol.liri.180ab.A.bij.pf.1}%
\end{equation}
Now, for every $i\in\left\{  1,2,\ldots,k\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the }k\text{-tuple }\left(
m+n-\left(  G\left(  \lambda\right)  \right)  _{k},m+n-\left(  G\left(
\lambda\right)  \right)  _{k-1},\ldots,m+n-\left(  G\left(  \lambda\right)
\right)  _{1}\right)  \right) \\
&  =m+n-\underbrace{\left(  G\left(  \lambda\right)  \right)  _{k+1-i}%
}_{\substack{=m+n-\lambda_{k+1-\left(  k+1-i\right)  }\\\text{(by
(\ref{sol.liri.180ab.A.bij.pf.1}), applied to}\\k+1-i\text{ instead of
}i\text{)}}}=m+n-\left(  m+n-\lambda_{k+1-\left(  k+1-i\right)  }\right)
=\lambda_{k+1-\left(  k+1-i\right)  }=\lambda_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }k+1-\left(  k+1-i\right)  =i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the }k\text{-tuple }\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  \right)  .
\end{align*}
In other words, the two $k$-tuples $\left(  m+n-\left(  G\left(
\lambda\right)  \right)  _{k},m+n-\left(  G\left(  \lambda\right)  \right)
_{k-1},\ldots,m+n-\left(  G\left(  \lambda\right)  \right)  _{1}\right)  $ and
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  $ are equal to
each other in each of their coordinates. Hence, these two $k$-tuples are
identical. That is, we have%
\[
\left(  m+n-\left(  G\left(  \lambda\right)  \right)  _{k},m+n-\left(
G\left(  \lambda\right)  \right)  _{k-1},\ldots,m+n-\left(  G\left(
\lambda\right)  \right)  _{1}\right)  =\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{k}\right)  .
\]
\par
But the definition of $G\left(  G\left(  \lambda\right)  \right)  $ yields%
\begin{align*}
\left(  G\circ G\right)  \left(  \lambda\right)   &  =G\left(  G\left(
\lambda\right)  \right)  =\left(  G\left(  \lambda\right)  \right)
^{\vee\left\{  m+n\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }G\left(  G\left(  \lambda\right)  \right)  \right) \\
&  =\left(  m+n-\left(  G\left(  \lambda\right)  \right)  _{k},m+n-\left(
G\left(  \lambda\right)  \right)  _{k-1},\ldots,m+n-\left(  G\left(
\lambda\right)  \right)  _{1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(  G\left(
\lambda\right)  \right)  ^{\vee\left\{  m+n\right\}  }\right) \\
&  =\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  =\lambda
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{k}\right)  \text{ (since }\ell\left(
\lambda\right)  \leq k\text{)}\right) \\
&  =\operatorname*{id}\left(  \lambda\right)  .
\end{align*}
\par
Let us now forget that we fixed $\lambda$. We thus have shown that $\left(
G\circ G\right)  \left(  \lambda\right)  =\operatorname*{id}\left(
\lambda\right)  $ for every $\lambda\in\mathfrak{A}$. In other words, $G\circ
G=\operatorname*{id}$, qed.}. Hence, the maps $G$ and $G$ are mutually
inverse. Hence, the map $G$ has an inverse, and thus is a bijection.

We notice that every $f\in\kk\left[  \left[  \xx\right]
\right]  $ satisfies%
\begin{equation}
\left(  f\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  \left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  =f\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  \label{sol.liri.180ab.c.subsubs}%
\end{equation}
in the Laurent polynomial ring $\kk\left[  x_{1},x_{2},\ldots
,x_{k},x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right]  $%
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.liri.180ab.c.subsubs}):} Let
$f\in\kk\left[  \left[  \xx\right]  \right]  $. Then, we can
write $f$ in the form $f=\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{
is a weak composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }%
x_{1}^{a_{1}}x_{2}^{a_{2}}x_{3}^{a_{3}}\cdots$ for some family $\left(
\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\right)  _{\left(
a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak composition}}$ of elements
of $\kk$. Consider this family $\left(  \zeta_{\left(  a_{1}%
,a_{2},a_{3},\ldots\right)  }\right)  _{\left(  a_{1},a_{2},a_{3}%
,\ldots\right)  \text{ is a weak composition}}$. Then,%
\begin{align*}
f  &  =\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }%
\underbrace{x_{1}^{a_{1}}x_{2}^{a_{2}}x_{3}^{a_{3}}\cdots}_{=\prod_{i\geq
1}x_{i}^{a_{i}}=\left(  \prod_{i=1}^{k}x_{i}^{a_{i}}\right)  \left(
\prod_{i\geq k+1}x_{i}^{a_{i}}\right)  }\\
&  =\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\left(
\prod_{i=1}^{k}x_{i}^{a_{i}}\right)  \left(  \prod_{i\geq k+1}x_{i}^{a_{i}%
}\right)
\end{align*}
\par
The definition of $f\left(  x_{1},x_{2},\ldots,x_{k}\right)  $ shows that
$f\left(  x_{1},x_{2},\ldots,x_{k}\right)  $ is the result of substituting
$x_{1},x_{2},\ldots,x_{k},0,0,0,\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in $f$.
Hence,%
\begin{align*}
&  f\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\left(  \text{the result of substituting }x_{1},x_{2},\ldots
,x_{k},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{in }\underbrace{f}_{=\sum_{\left(
a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak composition}}\zeta_{\left(
a_{1},a_{2},a_{3},\ldots\right)  }\left(  \prod_{i=1}^{k}x_{i}^{a_{i}}\right)
\left(  \prod_{i\geq k+1}x_{i}^{a_{i}}\right)  }\right) \\
&  =\left(  \text{the result of substituting }x_{1},x_{2},\ldots
,x_{k},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\text{ }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{in }\sum_{\left(  a_{1},a_{2}%
,a_{3},\ldots\right)  \text{ is a weak composition}}\zeta_{\left(  a_{1}%
,a_{2},a_{3},\ldots\right)  }\left(  \prod_{i=1}^{k}x_{i}^{a_{i}}\right)
\left(  \prod_{i\geq k+1}x_{i}^{a_{i}}\right)  \right) \\
&  =\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\left(
\prod_{i=1}^{k}x_{i}^{a_{i}}\right)  \left(  \prod_{i\geq k+1}0^{a_{i}%
}\right)  .
\end{align*}
Thus,%
\begin{align*}
&  \underbrace{\left(  f\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)
}_{=\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\left(
\prod_{i=1}^{k}x_{i}^{a_{i}}\right)  \left(  \prod_{i\geq k+1}0^{a_{i}%
}\right)  }\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \\
&  =\left(  \sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\left(
\prod_{i=1}^{k}x_{i}^{a_{i}}\right)  \left(  \prod_{i\geq k+1}0^{a_{i}%
}\right)  \right)  \left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \\
&  =\left(  \text{the result of substituting }x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\text{ for }x_{1},x_{2},\ldots,x_{k}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{in }\sum_{\left(  a_{1},a_{2}%
,a_{3},\ldots\right)  \text{ is a weak composition}}\zeta_{\left(  a_{1}%
,a_{2},a_{3},\ldots\right)  }\left(  \prod_{i=1}^{k}x_{i}^{a_{i}}\right)
\left(  \prod_{i\geq k+1}0^{a_{i}}\right)  \right) \\
&  =\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\left(
\prod_{i=1}^{k}\left(  x_{i}^{-1}\right)  ^{a_{i}}\right)  \left(
\prod_{i\geq k+1}0^{a_{i}}\right)  .
\end{align*}
Compared with%
\begin{align*}
&  f\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \\
&  =\left(  \text{the result of substituting }x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{in }\underbrace{f}_{=\sum_{\left(
a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak composition}}\zeta_{\left(
a_{1},a_{2},a_{3},\ldots\right)  }\left(  \prod_{i=1}^{k}x_{i}^{a_{i}}\right)
\left(  \prod_{i\geq k+1}x_{i}^{a_{i}}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \right) \\
&  =\left(  \text{the result of substituting }x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1},0,0,0,\ldots\text{ for }x_{1},x_{2},x_{3},\ldots\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{in }\sum_{\left(  a_{1},a_{2}%
,a_{3},\ldots\right)  \text{ is a weak composition}}\zeta_{\left(  a_{1}%
,a_{2},a_{3},\ldots\right)  }\left(  \prod_{i=1}^{k}x_{i}^{a_{i}}\right)
\left(  \prod_{i\geq k+1}x_{i}^{a_{i}}\right)  \right) \\
&  =\sum_{\left(  a_{1},a_{2},a_{3},\ldots\right)  \text{ is a weak
composition}}\zeta_{\left(  a_{1},a_{2},a_{3},\ldots\right)  }\left(
\prod_{i=1}^{k}\left(  x_{i}^{-1}\right)  ^{a_{i}}\right)  \left(
\prod_{i\geq k+1}0^{a_{i}}\right)  ,
\end{align*}
this yields $\left(  f\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)
\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  =f\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  $. Thus,
(\ref{sol.liri.180ab.c.subsubs}) is proven.}.

Now, from (\ref{sol.liri.180ab.c.LR2}), we obtain:%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)  =\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\label{sol.liri.180ab.c.LR3}%
\end{equation}
\footnote{\textit{Proof.} The summation sign $\sum\limits_{\substack{\lambda
\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq k;\\\text{all parts
of }\lambda\text{ are }\leq m+n}}$ in (\ref{sol.liri.180ab.c.LR2}) can be
rewritten as follows:%
\[
\sum\limits_{\substack{\lambda\in\operatorname{Par};\ \ell\left(  \lambda\right)
\leq k;\\\text{all parts of }\lambda\text{ are }\leq m+n}}=\sum_{\lambda
\in\left\{  \alpha\in \Par \ \mid\ \ell\left(  \alpha\right)
\leq k\text{; all parts of }\alpha\text{ are }\leq m+n\right\}  }%
=\sum_{\lambda\in\mathfrak{A}}%
\]
(since $\left\{  \alpha\in \Par \ \mid\ \ell\left(
\alpha\right)  \leq k\text{; all parts of }\alpha\text{ are }\leq m+n\right\}
=\mathfrak{A}$ (by (\ref{sol.liri.180ab.bc.A}))). Hence,
(\ref{sol.liri.180ab.c.LR2}) becomes%
\begin{align*}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu}\left(
x_{1},x_{2},\ldots,x_{k}\right)   &  =\underbrace{\sum\limits_{\substack{\lambda
\in\operatorname{Par};\ \ell\left(  \lambda\right)  \leq k;\\\text{all parts
of }\lambda\text{ are }\leq m+n}}}_{=\sum_{\lambda\in\mathfrak{A}}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right) \\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  .
\end{align*}
This proves (\ref{sol.liri.180ab.c.LR3}).} Hence, in the Laurent polynomial
ring $\kk\left[  x_{1},x_{2},\ldots,x_{k},x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right]  $, we have%
\begin{equation}
s_{\mu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \cdot s_{\nu
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  =\sum_{\lambda
\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  . \label{sol.liri.180ab.c.LR-3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.LR-3}):} Substituting the
variables $x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}$ for $x_{1},x_{2}%
,\ldots,x_{k}$ in the equality (\ref{sol.liri.180ab.c.LR3}), we obtain%
\begin{align*}
&  \left(  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  \left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{k}^{-1}\right) \\
&  =\left(  \sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  \left(  \left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \right)  =\sum_{\lambda
\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}\underbrace{\left(  s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  \right)  \left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  }_{\substack{=s_{\lambda}\left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \\\text{(by
(\ref{sol.liri.180ab.c.subsubs}), applied to }f=s_{\lambda}\text{)}}}\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \left(  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \cdot s_{\nu
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  \left(  x_{1}^{-1}%
,x_{2}^{-1},\ldots,x_{k}^{-1}\right) \\
&  =\underbrace{\left(  s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)
\right)  \left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
}_{\substack{=s_{\mu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\\\text{(by (\ref{sol.liri.180ab.c.subsubs}), applied to }f=s_{\nu}\text{)}%
}}\cdot\underbrace{\left(  s_{\nu}\left(  x_{1},x_{2},\ldots,x_{k}\right)
\right)  \left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
}_{\substack{=s_{\nu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\\\text{(by (\ref{sol.liri.180ab.c.subsubs}), applied to }f=s_{\nu}\text{)}%
}}\\
&  =s_{\mu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \cdot
s_{\nu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  ,
\end{align*}
this yields%
\[
s_{\mu}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \cdot s_{\nu
}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  =\sum_{\lambda
\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  .
\]
This proves (\ref{sol.liri.180ab.c.LR-3}).}

Now, let us notice that $\mu^{\vee\left\{  m\right\}  }$ and $\nu
^{\vee\left\{  n\right\}  }$ are partitions (according to Exercise
\ref{exe.liri.180ab}(a)). We have $\ell\left(  \mu^{\vee\left\{  m\right\}
}\right)  \leq k$\ \ \ \ \footnote{\textit{Proof.} The partition $\mu
^{\vee\left\{  m\right\}  }$ can be written as a $k$-tuple (since $\mu
^{\vee\left\{  m\right\}  }=\left(  m-\mu_{k},m-\mu_{k-1},\ldots,m-\mu
_{1}\right)  $); thus, $\ell\left(  \mu^{\vee\left\{  m\right\}  }\right)
\leq k$, qed.}. The same argument (but applied to $\nu$ and $n$ instead of
$\mu$ and $m$) yields $\ell\left(  \nu^{\vee\left\{  n\right\}  }\right)  \leq
k$. Also, all parts of the partition $\mu^{\vee\left\{  m\right\}  }$ are
$\leq m$\ \ \ \ \footnote{\textit{Proof.} Recall that $\mu^{\vee\left\{
m\right\}  }=\left(  m-\mu_{k},m-\mu_{k-1},\ldots,m-\mu_{1}\right)  $. Hence,%
\begin{equation}
\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{i}=m-\mu_{k+1-i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k\right\}
\text{.} \label{sol.liri.180ab.c.mu-conj.pf.1}%
\end{equation}
\par
Let $i$ be a positive integer. We are going to show that $\left(  \mu
^{\vee\left\{  m\right\}  }\right)  _{i}\leq m$.
\par
Indeed, assume the contrary. Thus, $\left(  \mu^{\vee\left\{  m\right\}
}\right)  _{i}>m$, so that $\left(  \mu^{\vee\left\{  m\right\}  }\right)
_{i}>m\geq0$. Thus, $i\leq\ell\left(  \mu^{\vee\left\{  m\right\}  }\right)
\leq k$, so that $i\in\left\{  1,2,\ldots,k\right\}  $. Hence,
(\ref{sol.liri.180ab.c.mu-conj.pf.1}) shows that $\left(  \mu^{\vee\left\{
m\right\}  }\right)  _{i}=m-\underbrace{\mu_{k+1-i}}_{\geq0}\leq m$; this
contradicts $\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{i}\leq m$. This
contradiction shows that our assumption was wrong. Hence, we have $\left(
\mu^{\vee\left\{  m\right\}  }\right)  _{i}\leq m$.
\par
Now, let us forget that we fixed $i$. We thus have proven that $\left(
\mu^{\vee\left\{  m\right\}  }\right)  _{i}\leq m$ for every positive integer
$i$. Thus, for every positive integer $i$, the $i$-th entry of the partition
$\mu^{\vee\left\{  m\right\}  }$ is $\left(  \mu^{\vee\left\{  m\right\}
}\right)  _{i}\leq m$. In other words, all entries of the partition $\mu
^{\vee\left\{  m\right\}  }$ are $\leq m$. Hence,
all parts of the partition $\mu^{\vee\left\{  m\right\}  }$ are $\leq m$
(since any part of the partition $\mu^{\vee\left\{  m\right\}  }$
is an entry of the partition $\mu^{\vee\left\{  m\right\}  }$). Qed.}.
The same argument (but
applied to $\nu$ and $n$ instead of $\mu$ and $m$) yields that all parts of
the partition $\nu^{\vee\left\{  n\right\}  }$ are $\leq n$. Moreover,%
\begin{equation}
\mu=\left(  m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k},m-\left(
\mu^{\vee\left\{  m\right\}  }\right)  _{k-1},\ldots,m-\left(  \mu
^{\vee\left\{  m\right\}  }\right)  _{1}\right)
\label{sol.liri.180ab.c.mu-conj-conj}%
\end{equation}
\footnote{\textit{Proof.} Recall that $\mu^{\vee\left\{  m\right\}  }=\left(
m-\mu_{k},m-\mu_{k-1},\ldots,m-\mu_{1}\right)  $. Hence,%
\begin{equation}
\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{i}=m-\mu_{k+1-i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k\right\}
\text{.} \label{sol.liri.180ab.c.mu-conj-conj.pf.1}%
\end{equation}
\par
Now, for every $i\in\left\{  1,2,\ldots,k\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th entry of the }k\text{-tuple }\left(
m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k},m-\left(  \mu
^{\vee\left\{  m\right\}  }\right)  _{k-1},\ldots,m-\left(  \mu^{\vee\left\{
m\right\}  }\right)  _{1}\right)  \right) \\
&  =m-\underbrace{\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k+1-i}%
}_{\substack{=m-\mu_{k+1-\left(  k+1-i\right)  }\\\text{(by
(\ref{sol.liri.180ab.c.mu-conj-conj.pf.1}), applied to }k+1-i\text{ instead of
}i\text{)}}}=m-\left(  m-\mu_{k+1-\left(  k+1-i\right)  }\right)
=\mu_{k+1-\left(  k+1-i\right)  }=\mu_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }k+1-\left(  k+1-i\right)  =i\right) \\
&  =\left(  \text{the }i\text{-th entry of the }k\text{-tuple }\left(  \mu
_{1},\mu_{2},\ldots,\mu_{k}\right)  \right)  .
\end{align*}
Thus, the two $k$-tuples $\left(  m-\left(  \mu^{\vee\left\{  m\right\}
}\right)  _{k},m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k-1}%
,\ldots,m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{1}\right)  $ and
$\left(  \mu_{1},\mu_{2},\ldots,\mu_{k}\right)  $ are equal to each other
entry by entry. Thus, these two $k$-tuples are identical. In other words, we
have
\begin{equation}
\left(  m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k},m-\left(
\mu^{\vee\left\{  m\right\}  }\right)  _{k-1},\ldots,m-\left(  \mu
^{\vee\left\{  m\right\}  }\right)  _{1}\right)  =\left(  \mu_{1},\mu
_{2},\ldots,\mu_{k}\right)  . \label{sol.liri.180ab.c.mu-conj-conj.pf.2}%
\end{equation}
\par
But $\ell\left(  \mu\right)  \leq k$, and thus $\mu=\left(  \mu_{1},\mu
_{2},\ldots,\mu_{k}\right)  =\left(  m-\left(  \mu^{\vee\left\{  m\right\}
}\right)  _{k},m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{k-1}%
,\ldots,m-\left(  \mu^{\vee\left\{  m\right\}  }\right)  _{1}\right)  $ (by
(\ref{sol.liri.180ab.c.mu-conj-conj.pf.2})), qed.}. The same argument (but
applied to $\nu$ and $n$ instead of $\mu$ and $m$) yields that%
\begin{equation}
\nu=\left(  n-\left(  \nu^{\vee\left\{  n\right\}  }\right)  _{k},n-\left(
\nu^{\vee\left\{  n\right\}  }\right)  _{k-1},\ldots,n-\left(  \nu
^{\vee\left\{  n\right\}  }\right)  _{1}\right)  .
\label{sol.liri.180ab.c.nu-conj-conj}%
\end{equation}


These observations lead us to the conclusion that we can apply
(\ref{sol.liri.180ab.c.LR-3}) to $\mu^{\vee\left\{  m\right\}  }$, $\mu$,
$\nu^{\vee\left\{  n\right\}  }$ and $\nu$ instead of $\mu$, $\mu
^{\vee\left\{  m\right\}  }$, $\nu$ and $\nu^{\vee\left\{  n\right\}  }$. As a
result, we obtain%
\begin{align}
&  s_{\mu^{\vee\left\{  m\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  \cdot s_{\nu^{\vee\left\{  n\right\}  }}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda}s_{\lambda}\left(  x_{1}^{-1},x_{2}%
^{-1},\ldots,x_{k}^{-1}\right)  =\sum_{\lambda\in\mathfrak{A}}%
\underbrace{c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }%
}^{G\left(  \lambda\right)  }s_{G\left(  \lambda\right)  }\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  }_{\substack{=c_{\mu^{\vee\left\{
m\right\}  },\nu^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{
m+n\right\}  }}s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \\\text{(since }G\left(
\lambda\right)  =\lambda^{\vee\left\{  m+n\right\}  }\text{ (by the definition
of }G\left(  \lambda\right)  \text{))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }G\left(  \lambda\right)  \text{ for the
summation index }\lambda\\
\text{(since the map }G:\mathfrak{A}\rightarrow\mathfrak{A}\text{ is a
bijection)}%
\end{array}
\right) \nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}%
s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  . \label{sol.liri.180ab.c.LR-3a}%
\end{align}


Now, we can apply Exercise \ref{exe.liri.180}(d) to $\mu^{\vee\left\{
m\right\}  }$, $\mu^{\vee\left\{  m\right\}  }$, $\mu$ and $\mu$ instead of
$\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$ (because of
(\ref{sol.liri.180ab.c.mu-conj-conj})). As a result, we obtain%
\begin{equation}
s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m}\cdot s_{\mu^{\vee\left\{  m\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\label{sol.liri.180ab.c.mu-reversion}%
\end{equation}


Also, we can apply Exercise \ref{exe.liri.180}(d) to $n$, $\nu^{\vee\left\{
n\right\}  }$, $\nu^{\vee\left\{  n\right\}  }$, $\nu$ and $\nu$ instead of
$m$, $\lambda$, $\mu$, $\lambda^{\vee}$ and $\mu^{\vee}$ (because of
(\ref{sol.liri.180ab.c.nu-conj-conj})). As a result, we obtain%
\begin{equation}
s_{\nu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{n}\cdot s_{\nu^{\vee\left\{  n\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\label{sol.liri.180ab.c.nu-reversion}%
\end{equation}


Furthermore, every $\lambda\in\mathfrak{A}$ satisfies%
\begin{equation}
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m+n}\cdot s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
.\label{sol.liri.180ab.c.lambda-reversion}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.180ab.c.lambda-reversion}):} Let
$\lambda\in\mathfrak{A}$. Then, $\lambda\in\mathfrak{A}=\left\{  \alpha
\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all
parts of }\alpha\text{ are }\leq m+n\right\}  $. In other words, $\lambda$ is
an element of $ \Par $ such that $\ell\left(  \lambda\right)
\leq k$ and such that all parts of $\lambda$ are $\leq m+n$. In particular,
$\lambda$ is a partition (since $\lambda\in \Par $). Now,
(\ref{sol.liri.180ab.bc.is-a-partition}) yields that $\lambda^{\vee\left\{
m+n\right\}  }\in\mathfrak{A}=\left\{  \alpha\in \Par %
\ \mid\ \ell\left(  \alpha\right)  \leq k\text{; all parts of }\alpha\text{
are }\leq m+n\right\}  $. In other words, $\lambda^{\vee\left\{  m+n\right\}
}$ is an element of $ \Par $ such that $\ell\left(  \lambda
^{\vee\left\{  m+n\right\}  }\right)  \leq k$ and such that all parts of
$\lambda^{\vee\left\{  m+n\right\}  }$ are $\leq m+n$. In particular,
$\lambda^{\vee\left\{  m+n\right\}  }$ is a partition (since $\lambda
^{\vee\left\{  m+n\right\}  }\in \Par $).
\par
The definition of $G\left(  \lambda\right)  $ yields $G\left(  \lambda\right)
=\lambda^{\vee\left\{  m+n\right\}  }$. But we have $G\circ
G=\operatorname*{id}$, thus $\underbrace{\left(  G\circ G\right)
}_{=\operatorname*{id}}\left(  \lambda\right)  =\operatorname*{id}\left(
\lambda\right)  =\lambda$, and thus%
\begin{align*}
\lambda &  =\left(  G\circ G\right)  \left(  \lambda\right)  =G\left(
\underbrace{G\left(  \lambda\right)  }_{=\lambda^{\vee\left\{  m+n\right\}  }%
}\right)  =G\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)  =\left(
\lambda^{\vee\left\{  m+n\right\}  }\right)  ^{\vee\left\{  m+n\right\}
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }G\left(
\lambda^{\vee\left\{  m+n\right\}  }\right)  \right)  \\
&  =\left(  m+n-\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)
_{k},m+n-\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)  _{k-1}%
,\ldots,m+n-\left(  \lambda^{\vee\left\{  m+n\right\}  }\right)  _{1}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\lambda^{\vee\left\{  m+n\right\}  }\right)  ^{\vee\left\{  m+n\right\}
}\right)  .
\end{align*}
Hence, we can apply Exercise \ref{exe.liri.180}(d) to $m+n$, $\lambda
^{\vee\left\{  m+n\right\}  }$, $\lambda^{\vee\left\{  m+n\right\}  }$,
$\lambda$ and $\lambda$ instead of $m$, $\lambda$, $\mu$, $\lambda^{\vee}$ and
$\mu^{\vee}$. As a result, we obtain%
\[
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =\left(  x_{1}x_{2}\cdots
x_{k}\right)  ^{m+n}\cdot s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(
x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  .
\]
This proves (\ref{sol.liri.180ab.c.lambda-reversion}).}

Now, (\ref{sol.liri.180ab.c.LR3}) yields%
\begin{align}
&  \sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right) \nonumber\\
&  =\underbrace{s_{\mu}\left(  x_{1},x_{2},\ldots,x_{k}\right)  }%
_{\substack{=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\mu
^{\vee\left\{  m\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  \\\text{(by (\ref{sol.liri.180ab.c.mu-reversion}))}}%
}\cdot\underbrace{s_{\nu}\left(  x_{1},x_{2},\ldots,x_{k}\right)
}_{\substack{=\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{n}\cdot s_{\nu
^{\vee\left\{  n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  \\\text{(by (\ref{sol.liri.180ab.c.nu-reversion}))}}}\nonumber\\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\cdot s_{\mu^{\vee\left\{
m\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\cdot\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{n}\cdot s_{\nu^{\vee\left\{
n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
\nonumber\\
&  =\underbrace{\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m}\left(  x_{1}%
x_{2}\cdots x_{k}\right)  ^{n}}_{=\left(  x_{1}x_{2}\cdots x_{k}\right)
^{m+n}}\cdot\underbrace{s_{\mu^{\vee\left\{  m\right\}  }}\left(  x_{1}%
^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)  \cdot s_{\nu^{\vee\left\{
n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right)
}_{\substack{=\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}
},\nu^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }%
}s_{\lambda^{\vee\left\{  m+n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1}%
,\ldots,x_{k}^{-1}\right)  \\\text{(by (\ref{sol.liri.180ab.c.LR-3a}))}%
}}\nonumber\\
&  =\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m+n}\cdot\sum_{\lambda
\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}
}}^{\lambda^{\vee\left\{  m+n\right\}  }}s_{\lambda^{\vee\left\{  m+n\right\}
}}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}^{-1}\right) \nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }%
}\underbrace{\left(  x_{1}x_{2}\cdots x_{k}\right)  ^{m+n}\cdot s_{\lambda
^{\vee\left\{  m+n\right\}  }}\left(  x_{1}^{-1},x_{2}^{-1},\ldots,x_{k}%
^{-1}\right)  }_{\substack{=s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \\\text{(by (\ref{sol.liri.180ab.c.lambda-reversion}))}%
}}\nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}s_{\lambda
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  .
\label{sol.liri.180ab.c.almostthere}%
\end{align}
Thus,%
\begin{align}
&  \sum_{\lambda\in\mathfrak{A}}\left(  c_{\mu,\nu}^{\lambda}-c_{\mu
^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}^{\lambda
^{\vee\left\{  m+n\right\}  }}\right)  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right) \nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  -\underbrace{\sum_{\lambda\in\mathfrak{A}%
}c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}%
^{\lambda^{\vee\left\{  m+n\right\}  }}s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  }_{\substack{=\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \\\text{(by
(\ref{sol.liri.180ab.c.almostthere}))}}}\nonumber\\
&  =\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu}^{\lambda}s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  -\sum_{\lambda\in\mathfrak{A}}c_{\mu,\nu
}^{\lambda}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =0.
\label{sol.liri.180ab.almostthere2}%
\end{align}


But Remark \ref{rmk.many-symmetric-function-bases-prop.finite}(d) (applied to
$N=k$) yields that the set $\left\{  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  \right\}  $, as $\lambda$ runs through all partitions
having length $\leq k$, is a basis of the $\kk$-module $\Lambda\left(
x_{1},x_{2},\ldots,x_{k}\right)  $. In other words, the family $\left(
s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  _{\lambda
\in \Par ;\ \ell\left(  \lambda\right)  \leq k}$ is a basis of
the $\kk$-module $\Lambda\left(  x_{1},x_{2},\ldots,x_{k}\right)  $. In
particular, the family $\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \right)  _{\lambda\in \Par ;\ \ell\left(
\lambda\right)  \leq k}$ is $\kk$-linearly independent. In other words,
the family \newline$\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \right)  _{\lambda\in\left\{  \alpha\in \Par %
\ \mid\ \ell\left(  \alpha\right)  \leq k\right\}  }$ is $\kk$-linearly
independent (since the family $\left(  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  \right)  _{\lambda\in \Par ;\ \ell\left(
\lambda\right)  \leq k}$ is precisely the family $\left(  s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  \right)  _{\lambda\in\left\{  \alpha
\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\right\}  }$
(because the elements $\lambda\in \Par $ satisfying $\ell\left(
\lambda\right)  \leq k$ are exactly the elements of the set $\left\{
\alpha\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\right\}
$)).

But the set $\mathfrak{A}$ is a subset of $\left\{  \alpha\in
 \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\right\}  $ (since%
\[
\mathfrak{A}=\left\{  \alpha\in \Par \ \mid\ \underbrace{\ell
\left(  \alpha\right)  \leq k\text{; all parts of }\alpha\text{ are }\leq
m+n}_{\text{this condition implies }\ell\left(  \alpha\right)  \leq
k}\right\}  \subset\left\{  \alpha\in \Par \ \mid\ \ell\left(
\alpha\right)  \leq k\right\}
\]
). Hence, the family $\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \right)  _{\lambda\in\mathfrak{A}}$ is a subfamily of the
family $\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)
_{\lambda\in\left\{  \alpha\in \Par \ \mid\ \ell\left(
\alpha\right)  \leq k\right\}  }$. Since the family $\left(  s_{\lambda
}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)  _{\lambda\in\left\{
\alpha\in \Par \ \mid\ \ell\left(  \alpha\right)  \leq k\right\}
}$ is $\kk$-linearly independent, we thus conclude that its subfamily
$\left(  s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  \right)
_{\lambda\in\mathfrak{A}}$ is also $\kk$-linearly independent. In other
words, if $\left(  g_{\lambda}\right)  _{\lambda\in\mathfrak{A}}\in
\kk^{\mathfrak{A}}$ is a family of elements of $\kk$ such that
$\left(  \text{all but finitely many }\lambda\in\mathfrak{A}\text{ satisfy
}g_{\lambda}=0\right)  $ and $\sum_{\lambda\in\mathfrak{A}}g_{\lambda
}s_{\lambda}\left(  x_{1},x_{2},\ldots,x_{k}\right)  =0$, then%
\begin{equation}
\left(  g_{\lambda}\right)  _{\lambda\in\mathfrak{A}}=\left(  0\right)
_{\lambda\in\mathfrak{A}}. \label{sol.liri.180ab.c.lin-ind}%
\end{equation}


Now, the family $\left(  c_{\mu,\nu}^{\lambda}-c_{\mu^{\vee\left\{  m\right\}
},\nu^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }%
}\right)  _{\lambda\in\mathfrak{A}}\in\kk^{\mathfrak{A}}$ satisfies
$\sum_{\lambda\in\mathfrak{A}}\left(  c_{\mu,\nu}^{\lambda}-c_{\mu
^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}^{\lambda
^{\vee\left\{  m+n\right\}  }}\right)  s_{\lambda}\left(  x_{1},x_{2}%
,\ldots,x_{k}\right)  =0$ (by (\ref{sol.liri.180ab.almostthere2})); in
particular, the sum $\sum_{\lambda\in\mathfrak{A}}\left(  c_{\mu,\nu}%
^{\lambda}-c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }%
}^{\lambda^{\vee\left\{  m+n\right\}  }}\right)  s_{\lambda}\left(
x_{1},x_{2},\ldots,x_{k}\right)  $ is well-defined, and therefore $\left(
\text{all but finitely many }\lambda\in\mathfrak{A}\text{ satisfy }c_{\mu,\nu
}^{\lambda}-c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }%
}^{\lambda^{\vee\left\{  m+n\right\}  }}=0\right)  $. Hence, we can apply
(\ref{sol.liri.180ab.c.lin-ind}) to $g_{\lambda}=c_{\mu,\nu}^{\lambda}%
-c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}%
^{\lambda^{\vee\left\{  m+n\right\}  }}$. As a result, we obtain%
\[
\left(  c_{\mu,\nu}^{\lambda}-c_{\mu^{\vee\left\{  m\right\}  },\nu
^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}\right)
_{\lambda\in\mathfrak{A}}=\left(  0\right)  _{\lambda\in\mathfrak{A}}.
\]
In other words, $c_{\mu,\nu}^{\lambda}-c_{\mu^{\vee\left\{  m\right\}  }%
,\nu^{\vee\left\{  n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}=0$ for
every $\lambda\in\mathfrak{A}$. In other words,%
\begin{equation}
c_{\mu,\nu}^{\lambda}=c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{
n\right\}  }}^{\lambda^{\vee\left\{  m+n\right\}  }}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathfrak{A}\text{.}
\label{sol.liri.180ab.c.almostthere4}%
\end{equation}


Now, let $\lambda$ be a partition such that $\ell\left(  \lambda\right)  \leq
k$. Assume that all parts of $\lambda$ are $\leq m+n$. Then, $\lambda$ is an
element of $ \Par $ (since $\lambda$ is a partition) such that
$\ell\left(  \lambda\right)  \leq k$ and such that all parts of $\lambda$ are
$\leq m+n$. In other words,%
\[
\lambda\in\left\{  \alpha\in \Par \ \mid\ \ell\left(
\alpha\right)  \leq k\text{; all parts of }\alpha\text{ are }\leq m+n\right\}
=\mathfrak{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.180ab.bc.A}%
)}\right)  .
\]
Hence, (\ref{sol.liri.180ab.c.almostthere4}) yields $c_{\mu,\nu}^{\lambda
}=c_{\mu^{\vee\left\{  m\right\}  },\nu^{\vee\left\{  n\right\}  }}%
^{\lambda^{\vee\left\{  m+n\right\}  }}$. This solves Exercise
\ref{exe.liri.180ab}(c).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.simple}}

\textit{Solution to Exercise \ref{exe.liri.simple}.} (a) We notice that
every partition $\lambda$ and every positive integer $i$ satisfy%
\begin{align}
\left(  \lambda^{t}\right)  _{i}  &  =\left\vert \left\{  j\in\left\{
1,2,3,...\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.partition.transpose})}\right)
\nonumber\\
&  =\left\vert \left\{  j\in\left\{  1,2,...,\ell\left(  \lambda\right)
\right\}  \ \mid\ \lambda_{j}\geq i\right\}  \right\vert
\label{sol.liri.simple.eq.partition.transpose}%
\end{align}
(because $\lambda_{j}\geq i$ can happen only when $j\in\left\{  1,2,...,\ell
\left(  \lambda\right)  \right\}  $ (since $i$ is positive)).

Let $\mu$ and $\nu$ be two partitions. For every positive integer $i$, we have%
\begin{align*}
\left(  \left(  \mu\sqcup\nu\right)  ^{t}\right)  _{i}  &  =\left\vert
\left\{  j\in\left\{  1,2,...,\ell\left(  \mu\sqcup\nu\right)  \right\}
\ \mid\ \left(  \mu\sqcup\nu\right)  _{j}\geq i\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.simple.eq.partition.transpose}), applied to }\lambda=\mu
\sqcup\nu\right) \\
&  =\left(  \text{the number of entries of }\mu\sqcup\nu\text{ which are }\geq
i\right) \\
&  =\left(  \text{the number of entries of the list }\left(  \mu_{1},\mu
_{2},\ldots,\mu_{\ell\left(  \mu\right)  },\nu_{1},\nu_{2},\ldots,\nu
_{\ell\left(  \nu\right)  }\right)  \text{ which are }\geq i\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\mu\sqcup\nu\text{ is the result of sorting the list }\left(
\mu_{1},\mu_{2},\ldots,\mu_{\ell\left(  \mu\right)  },\nu_{1},\nu_{2}%
,\ldots,\nu_{\ell\left(  \nu\right)  }\right) \\
\text{in decreasing order, and clearly the procedure of sorting does not
change}\\
\text{the number of entries of the list which are }\geq i
\end{array}
\right) \\
&  =\underbrace{\left(  \text{the number of entries of the list }\left(
\mu_{1},\mu_{2},\ldots,\mu_{\ell\left(  \mu\right)  }\right)  \text{ which are
}\geq i\right)  }_{\substack{=\left\vert \left\{  j\in\left\{  1,2,...,\ell
\left(  \mu\right)  \right\}  \ \mid\ \mu_{j}\geq i\right\}  \right\vert
=\left(  \mu^{t}\right)  _{i}\\\text{(by
(\ref{sol.liri.simple.eq.partition.transpose}), applied to }\lambda
=\mu\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries of the
list }\left(  \nu_{1},\nu_{2},\ldots,\nu_{\ell\left(  \nu\right)  }\right)
\text{ which are }\geq i\right)  }_{\substack{=\left\vert \left\{
j\in\left\{  1,2,...,\ell\left(  \nu\right)  \right\}  \ \mid\ \nu_{j}\geq
i\right\}  \right\vert =\left(  \nu^{t}\right)  _{i}\\\text{(by
(\ref{sol.liri.simple.eq.partition.transpose}), applied to }\lambda
=\nu\text{)}}}\\
&  =\left(  \mu^{t}\right)  _{i}+\left(  \nu^{t}\right)  _{i}=\left(  \mu
^{t}+\nu^{t}\right)  _{i}%
\end{align*}
(since the definition of $\mu^{t}+\nu^{t}$ yields $\left(  \mu^{t}+\nu
^{t}\right)  _{i}=\left(  \mu^{t}\right)  _{i}+\left(  \nu^{t}\right)  _{i}$).
In other words,%
\begin{equation}
\left(  \mu\sqcup\nu\right)  ^{t}=\mu^{t}+\nu^{t}. \label{sol.liri.simple.a.1}%
\end{equation}


Applying this to $\mu^{t}$ and $\nu^{t}$ instead of $\mu$ and $\nu$, we obtain
$\left(  \mu^{t}\sqcup\nu^{t}\right)  ^{t}=\underbrace{\left(  \mu^{t}\right)
^{t}}_{=\mu}+\underbrace{\left(  \nu^{t}\right)  ^{t}}_{=\nu}=\mu+\nu$, so
that $\mu+\nu=\left(  \mu^{t}\sqcup\nu^{t}\right)  ^{t}$ and thus%
\[
\left(  \underbrace{\mu+\nu}_{=\left(  \mu^{t}\sqcup\nu^{t}\right)  ^{t}%
}\right)  ^{t}=\left(  \left(  \mu^{t}\sqcup\nu^{t}\right)  ^{t}\right)
^{t}=\mu^{t}\sqcup\nu^{t}.
\]
This solves Exercise \ref{exe.liri.simple}(a).

(b) Let $\mu$ and $\nu$ be two partitions. We are going to prove that
$c_{\mu,\nu}^{\mu+\nu}=1$.

A \emph{$\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableau} will mean a
column-strict tableau $T$ of shape $\left(  \mu+\nu\right)  /\mu$ with
$ \cont \left(  T\right)  =\nu$ having the property that each
$ \cont \left(  T|_{ \cols \geq j}\right)  $ is a
partition (where we are using the notations of Corollary \ref{L-R-rule}).
Corollary \ref{L-R-rule} shows that $c_{\mu,\nu}^{\mu+\nu}$ is the number of
$\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableaux. Hence, in order to prove that
$c_{\mu,\nu}^{\mu+\nu}=1$, it will be enough to show that there exists one and
only one $\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableau.

First of all, let $T_{0}$ be the filling of the skew shape $\left(  \mu
+\nu\right)  /\mu$ which assigns to every cell in row $i$ the number $i$, for
all $i\in\left\{  1,2,3,...\right\}  $. This $T_{0}$ is clearly a
column-strict tableau, and satisfies $ \cont \left(
T_{0}\right)  =\nu$ (because for every positive integer $i$, the $i$-th row of
the skew shape $\left(  \mu+\nu\right)  /\mu$ has $\underbrace{\left(  \mu
+\nu\right)  _{i}}_{=\mu_{i}+\nu_{i}}-\mu_{i}=\left(  \mu_{i}+\nu_{i}\right)
-\mu_{i}=\nu_{i}$ cells). Moreover, for every $j\in\left\{  1,2,3,...\right\}
$, the weak composition $ \cont \left(  T_{0}%
|_{ \cols \geq j}\right)  $ is a
partition\footnote{\textit{Proof.} In fact, fix $j\in\left\{
1,2,3,...\right\}  $. Then, both $\nu_{i}$ and $\mu_{i}$ decrease with
$i\in\left\{  1,2,3,...\right\}  $ (since $\nu$ and $\mu$ are partitions), and
thus $\min\left\{  \mu_{i},j\right\}  $ also decreases with $i\in\left\{
1,2,3,...\right\}  $.
\par
Now, for every $i\in\left\{  1,2,3,...\right\}  $, the $i$-th entry of
$ \cont \left(  T_{0}|_{ \cols \geq j}\right)  $ is
the number of boxes in columns $j,j+1,j+2,...$ of the $i$-th row of the skew
shape $\left(  \mu+\nu\right)  /\mu$. This number is easily seen to be
$\underbrace{\left(  \mu+\nu\right)  _{i}}_{=\mu_{i}+\nu_{i}}-\underbrace{\max
\left\{  \mu_{i},j\right\}  }_{=\mu_{i}+j-\min\left\{  \mu_{i},j\right\}
}=\mu_{i}+\nu_{i}-\left(  \mu_{i}+j-\min\left\{  \mu_{i},j\right\}  \right)
=\nu_{i}-j+\min\left\{  \mu_{i},j\right\}  $, and therefore decreases with $i$
(because both $\nu_{i}$ and $\min\left\{  \mu_{i},j\right\}  $ decrease with
$i$). Hence, $ \cont \left(  T_{0}|_{ \cols \geq
j}\right)  $ is a partition, qed.}. Therefore, $T_{0}$ is a $\left(  \mu
+\nu,\mu,\nu\right)  $-LR-tableau. It remains to prove that it is the only
$\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableau.

So fix any $\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableau $T$. Thus, $T$ is a
column-strict tableau of shape $\left(  \mu+\nu\right)  /\mu$ with
$ \cont \left(  T\right)  =\nu$ having the property that each
$ \cont \left(  T|_{ \cols \geq j}\right)  $ is a partition.

We shall show that%
\begin{equation}
\text{for every }i\in\left\{  1,2,3,...\right\}  \text{, all entries in the
}i\text{-th row of }T\text{ are }\leq i. \label{sol.liri.simple.b.1}%
\end{equation}


\textit{Proof of (\ref{sol.liri.simple.b.1}):} Assume the contrary. Then,
there exists an $i\in\left\{  1,2,3,...\right\}  $ such that not all entries
in the $i$-th row of $T$ are $\leq i$. Let $p$ be the \textbf{smallest} such
$i$ (this is clearly well-defined). Hence, not all entries in the $p$-th row
of $T$ are $\leq p$. But since $p$ is minimal, we know that
(\ref{sol.liri.simple.b.1}) holds for every $i<p$.

Not all entries in the $p$-th row of $T$ are $\leq p$. In other words,
at least one entry of the $p$-th row of $T$ must be $>p$. Since the entries of
$T$ weakly increase along rows, this yields that the rightmost entry of the
$p$-th row of $T$ is $>p$. Let $q$ be this entry, and let $j$ be the column in
which the rightmost cell of the $p$-th row of $T$ lies. Thus, $q$ is the entry
in cell $\left(  p,j\right)  $ of $T$. Therefore, the entry $q$ appears in
the tableau $T|_{ \cols \geq j}$. Hence, $\cont \left(
T|_{ \cols \geq j}\right)  _{q}\geq 1$. Note that $q > p$ (by the definition
of $q$).

We know that $\cont \left(  T|_{ \cols \geq
j}\right)  $ is a partition, so that
\begin{align*}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{p}  &  \geq \cont \left(  T|_{ \cols \geq
j}\right)  _{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p<q\right) \\
&  \geq1.
\end{align*}
In other words, the entry $p$ appears somewhere in the tableau
$T|_{ \cols \geq j}$. Where can it appear? It cannot appear in any
of the first $p-1$ rows, because all entries in these rows are $<p$ (since
(\ref{sol.liri.simple.b.1}) holds for every $i<p$). Hence, it must appear in
the $p$-th row or further down. In other words, this entry $p$ appears in a
cell $\left(  u,v\right)  $ of $T$ with $u\geq p$ and $v\geq j$. As a
consequence, this entry $p$ is $\geq$ to the entry in cell $\left(
p,j\right)  $ of $T$ (since $T$ is column-strict). Since the entry in cell
$\left(  p,j\right)  $ of $T$ is $q$, this yields $p\geq q$, which contradicts
$q>p$. This contradiction shows that our assumption was wrong, and
(\ref{sol.liri.simple.b.1}) is proven.

We furthermore claim that
\begin{equation}
\text{for every }i\in\left\{  1,2,3,...\right\}  \text{, all entries in the
}i\text{-th row of }T\text{ are }i.
\label{sol.liri.simple.b.2}
\end{equation}


\textit{Proof of (\ref{sol.liri.simple.b.2}):} Assume the contrary. Then,
there exists an $i\in\left\{  1,2,3,...\right\}  $ such that not all entries
in the $i$-th row of $T$ are $i$. Let $p$ be the \textbf{smallest} such $i$
(this is clearly well-defined). Hence, not all entries in the $p$-th row of
$T$ are $p$. But since $p$ is minimal, we know that
(\ref{sol.liri.simple.b.2}) holds for every $i<p$.

All entries in the $p$-th row of $T$ are $\leq p$ (by
(\ref{sol.liri.simple.b.1})), but not all of them are $p$. Hence, the $p$-th
row of $T$ has an entry $<p$. Let $k$ be this entry. Thus, $k < p$.
We can apply
(\ref{sol.liri.simple.b.2}) to $i=k$ (since (\ref{sol.liri.simple.b.2}) holds
for every $i<p$), and conclude that all entries in the $k$-th row of $T$ are
$k$. Thus, in the tableau $T$, the entry $k$ appears $\nu_{k}$ times in row
$k$ (since the length of the $k$-th row of $T$ is $\underbrace{\left(  \mu
+\nu\right)  _{k}}_{=\mu_{k}+\nu_{k}}-\mu_{k}=\left(  \mu_{k}+\nu_{k}\right)
-\mu_{k}=\nu_{k}$) and at least $1$ time in row $p$ (by the definition of
$k$). In total, $k$ must thus appear at least $\nu_{k}+1$ times in $T$, which
contradicts the fact that $ \cont \left(  T\right)  =\nu$. This
contradiction disproves our assumption, and thus (\ref{sol.liri.simple.b.2})
is proven.

Now that (\ref{sol.liri.simple.b.2}) is proven, we immediately conclude that
$T=T_{0}$. Now, forget that we fixed $T$. We thus have shown that every
$\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableau $T$ equals $T_{0}$. Hence, there
exists one and only one $\left(  \mu+\nu,\mu,\nu\right)  $-LR-tableau, namely
$T=T_{0}$ (because we have already seen that $T_{0}$ is a $\left(  \mu+\nu
,\mu,\nu\right)  $-LR-tableau). As we said above, this proves that%
\begin{equation}
c_{\mu,\nu}^{\mu+\nu}=1.
\label{sol.liri.simple.b.3}
\end{equation}


It remains to show that $c_{\mu,\nu}^{\mu\sqcup\nu}=1$. Exercise
\ref{exe.pieri.omega}(c) (applied to $\lambda=\mu\sqcup\nu$) shows that
\begin{align*}
c_{\mu,\nu}^{\mu\sqcup\nu}  &  =c_{\mu^{t},\nu^{t}}^{\left(  \mu\sqcup
\nu\right)  ^{t}}=c_{\mu^{t},\nu^{t}}^{\mu^{t}+\nu^{t}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \mu\sqcup\nu\right)  ^{t}%
=\mu^{t}+\nu^{t}\text{ (by Exercise \ref{exe.liri.simple}(a))}\right) \\
&  =1\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.simple.b.3}), applied
to }\mu^{t}\text{ and }\nu^{t}\text{ instead of }\mu\text{ and }\nu\right)  .
\end{align*}
This solves Exercise~\ref{exe.liri.simple}(b).

(c) Let $k\in \NN$ and $n\in \NN$ satisfy $k\leq n$, and let
$\mu\in \Par_{k}$, $\nu\in \Par_{n-k}$ and
$\lambda\in \Par_{n}$ be such that $c_{\mu,\nu}^{\lambda}\neq0$.
We need to prove that $\mu+\nu\triangleright\lambda\triangleright\mu\sqcup\nu$.

We have $\left\vert \lambda\right\vert =n$, $\left\vert \mu\right\vert =k$ and
$\left\vert \nu\right\vert =n-k$, so that $\left\vert \lambda\right\vert
=n=\underbrace{k}_{=\left\vert \mu\right\vert }+\underbrace{\left(
n-k\right)  }_{=\left\vert \nu\right\vert }=\left\vert \mu\right\vert
+\left\vert \nu\right\vert $. We will first show that
\begin{equation}
\mu+\nu\triangleright\lambda .
\label{sol.liri.simple.c.1}
\end{equation}


A \emph{$\left(  \lambda,\mu,\nu\right)  $-LR-tableau} will mean a
column-strict tableau $T$ of shape $\lambda/\mu$ with $ \cont
\left(  T\right)  =\nu$ having the property that each $ \cont
\left(  T|_{ \cols \geq j}\right)  $ is a partition (where we are
using the notations of Corollary \ref{L-R-rule}). Corollary \ref{L-R-rule}
shows that $c_{\mu,\nu}^{\lambda}$ is the number of $\left(  \lambda,\mu
,\nu\right)  $-LR-tableaux. Since $c_{\mu,\nu}^{\lambda}\neq0$, we thus see
that there exists at least one $\left(  \lambda,\mu,\nu\right)  $-LR-tableau.
Let this $\left(  \lambda,\mu,\nu\right)  $-LR-tableau be $T$.

Just as in the solution of Exercise \ref{exe.liri.simple}(b), we can prove
that (\ref{sol.liri.simple.b.1}) holds. Let $k$ be a positive integer.
Applying (\ref{sol.liri.simple.b.1}) to all $i\in\left\{  1,2,...,k\right\}
$, we see that all entries in the first $k$ rows of $T$ (meaning the $1$-st
row, the $2$-nd row, etc., the $k$-th row) are $\leq k$. Hence,
\begin{align*}
&  \left(  \text{the number of all entries in the first }k\text{ rows of
}T\right) \\
&  \leq\left(  \text{the number of all entries }\leq k\text{ in }T\right)
=\sum_{i=1}^{k}\underbrace{\left(  \text{the number of all entries }i\text{ in
}T\right)  }_{\substack{=\left(   \cont T\right)  _{i}=\nu
_{i}\\\text{(since } \cont T=\nu\text{)}}}=\sum_{i=1}^{k}\nu_{i}.
\end{align*}
Since
\begin{align*}
&  \left(  \text{the number of all entries in the first }k\text{ rows of
}T\right) \\
&  =\sum_{i=1}^{k}\underbrace{\left(  \text{the number of all entries in the
}i\text{-th row of }T\right)  }_{=\lambda_{i}-\mu_{i}}=\sum_{i=1}^{k}\left(
\lambda_{i}-\mu_{i}\right)  =\sum_{i=1}^{k}\lambda_{i}-\sum_{i=1}^{k}\mu_{i},
\end{align*}
this rewrites as $\sum_{i=1}^{k}\lambda_{i}-\sum_{i=1}^{k}\mu_{i}\leq
\sum_{i=1}^{k}\nu_{i}$. Hence,
\[
\sum_{i=1}^{k}\lambda_{i}\leq\sum_{i=1}^{k}\mu_{i}+\sum_{i=1}^{k}\nu_{i}
=\sum_{i=1}^{k}\underbrace{\left(  \mu_{i}+\nu_{i}\right)  }_{=\left(  \mu
+\nu\right)  _{i}}=\sum_{i=1}^{k}\left(  \mu+\nu\right)  _{i},
\]
so that $\sum_{i=1}^{k}\left(  \mu+\nu\right)  _{i}\geq\sum_{i=1}^{k}%
\lambda_{i}$. In other words, $\left(  \mu+\nu\right)  _{1}+\left(  \mu
+\nu\right)  _{2}+\cdots+\left(  \mu+\nu\right)  _{k}\geq\lambda_{1}
+\lambda_{2}+\cdots+\lambda_{k}$.

Now, let us forget that we fixed $k$. We have shown that $\left(  \mu
+\nu\right)  _{1}+\left(  \mu+\nu\right)  _{2}+\cdots+\left(  \mu+\nu\right)
_{k}\geq\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}$ for every positive integer
$k$. Combined with $\left\vert \lambda\right\vert =\left\vert \mu\right\vert
+\left\vert \nu\right\vert =\left\vert \mu+\nu\right\vert $, this yields that
$\mu+\nu\triangleright\lambda$. This proves (\ref{sol.liri.simple.c.1}).

It now remains to prove that $\lambda\triangleright\mu\sqcup\nu$. To do so, we
notice that Exercise \ref{exe.pieri.omega}(c) yields $c_{\mu,\nu}^{\lambda
}=c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$, so that $c_{\mu^{t},\nu^{t}}%
^{\lambda^{t}}=c_{\mu,\nu}^{\lambda}\neq0$. Hence, we can apply
(\ref{sol.liri.simple.c.1}) to $\lambda^{t}$, $\mu^{t}$ and $\nu^{t}$ instead
of $\lambda$, $\mu$ and $\nu$. As a result, we obtain $\mu^{t}+\nu
^{t}\triangleright\lambda^{t}$. Since $\left(  \mu\sqcup\nu\right)  ^{t}
=\mu^{t}+\nu^{t}$ (by Exercise \ref{exe.liri.simple}(a)), this rewrites as
$\left(  \mu\sqcup\nu\right)  ^{t}\triangleright\lambda^{t}$. But Exercise
\ref{exe.partition.dominance} (applied to $\mu\sqcup\nu$ instead of $\mu$)
yields that $\lambda\triangleright\mu\sqcup\nu$ if and only if $\left(
\mu\sqcup\nu\right)  ^{t}\triangleright\lambda^{t}$. Since we already know
that $\left(  \mu\sqcup\nu\right)  ^{t}\triangleright\lambda^{t}$, we can thus
conclude that $\lambda\triangleright\mu\sqcup\nu$. Combined with
(\ref{sol.liri.simple.c.1}), this yields $\mu+\nu\triangleright\lambda
\triangleright\mu\sqcup\nu$. This solves Exercise \ref{exe.liri.simple}(c).

(d) Let $n \in \NN$ and $m \in \NN$ and $\alpha,\beta
\in \Par_{n}$ and $\gamma,\delta\in \Par_{m}$ be
such that $\alpha\triangleright\beta$ and $\gamma\triangleright\delta$. It is
completely straightforward to check that
\begin{equation}
\alpha+\gamma\triangleright\beta+\delta
\label{sol.liri.simple.d.1}
\end{equation}
\footnote{To prove this, just recall how $\alpha+\gamma$ and $\beta+\delta$
are defined, and recall that two partitions $\lambda,\mu\in\Par_{n}$
satisfy $\lambda\triangleright\mu$ if and only if we have
$\left(  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}
+\cdots+\mu_{k}\text{ for every positive integer }k\right)  $.}. It remains to
prove that $\alpha\sqcup\gamma\triangleright\beta\sqcup\delta$.

Exercise \ref{exe.partition.dominance} (applied to $\lambda=\alpha$ and
$\mu=\beta$) yields that $\alpha\triangleright\beta$ if and only if $\beta
^{t}\triangleright\alpha^{t}$. Since $\alpha\triangleright\beta$, we thus have
$\beta^{t}\triangleright\alpha^{t}$. Similarly, $\delta^{t}\triangleright
\gamma^{t}$. Hence, we can apply (\ref{sol.liri.simple.d.1}) to $\beta^{t}$,
$\alpha^{t}$, $\delta^{t}$ and $\gamma^{t}$ instead of $\alpha$, $\beta$,
$\gamma$ and $\delta$. As a result, we obtain $\beta^{t}+\delta^{t}%
\triangleright\alpha^{t}+\gamma^{t}$. Now, (\ref{sol.liri.simple.a.1})
(applied to $\mu=\beta$ and $\nu=\delta$) yields $\left(  \beta\sqcup
\delta\right)  ^{t}=\beta^{t}+\delta^{t}$. Similarly, $\left(  \alpha\sqcup
\gamma\right)  ^{t}=\alpha^{t}+\gamma^{t}$. Thus, $\left(  \beta\sqcup\delta\right)
^{t}=\beta^{t}+\delta^{t}\triangleright\alpha^{t}+\gamma^{t}=\left(
\alpha\sqcup\gamma\right)  ^{t}$.

Finally, Exercise~\ref{exe.partition.dominance} (applied to $n+m$,
$\alpha\sqcup\gamma$ and $\beta\sqcup\delta$ instead of $n$, $\lambda$ and $\mu$)
yields that $\alpha\sqcup\gamma\triangleright\beta\sqcup\delta$ if and only if
$\left(  \beta\sqcup\delta\right)  ^{t}\triangleright\left(  \alpha\sqcup
\gamma\right)  ^{t}$. Since we have $\left(  \beta\sqcup\delta\right)
^{t}\triangleright\left(  \alpha\sqcup\gamma\right)  ^{t}$, we thus obtain
$\alpha\sqcup\gamma\triangleright\beta\sqcup\delta$. This completes the
solution of Exercise \ref{exe.liri.simple}(d).

(e) The Ferrers diagram of the partition $\lambda=\left(  m^{k}\right)  $ is a
rectangle. Let $C$ denote the center of this rectangle. For every partition
$\mu$ satisfying $\mu\subseteq\lambda$, let us define a partition $\mu^{c}$ by
$\mu^{c}=\left(  m-\mu_{k},m-\mu_{k-1},...,m-\mu_{1}\right)  $%
\ \ \ \ \footnote{This is well-defined, since every $i\in\left\{
1,2,...,k\right\}  $ satisfies $\mu_{i}\leq m$ (because $\mu\subseteq
\lambda=\left(  m^{k}\right)  $).}. The Ferrers diagram of this partition
$\mu^{c}$ is obtained from the skew Ferrers diagram of the skew partition
\footnote{A \emph{skew partition} shall mean a pair $\left(\alpha, \beta\right)$ of
partitions satisfying $\beta \subseteq \alpha$.
We write such a skew partition $\left(\alpha, \beta\right)$
as $\alpha / \beta$.
For every skew partition $\alpha/\beta$, we define the skew Ferrers diagram
$Y\left(  \alpha/\beta\right)  $ of $\alpha / \beta$
by $Y\left(  \alpha/\beta\right)
=Y\left(  \alpha\right)  \setminus Y\left(  \beta\right)  $,
where $Y\left(\kappa\right)$ means the Ferrers diagram of a partition
$\kappa$.}
$\lambda/\mu$ by the $180^{\circ}$ rotation around $C$. In other words, the
skew Ferrers diagram of $\mu^{c}/\varnothing$ is obtained from the skew
Ferrers diagram of the skew partition $\lambda/\mu$ by the $180^{\circ}$
rotation around $C$. Hence, Exercise \ref{exe.schur.rotate180}(b) (applied to
$\lambda^{\prime}=\mu^{c}$ and $\mu^{\prime}=\varnothing$) yields that%
\begin{equation}
s_{\lambda/\mu}=s_{\mu^{c}/\varnothing}=s_{\mu^{c}} .
\label{sol.liri.simple.e.1}
\end{equation}


Now, Proposition \ref{symm-comultiplication-formulas-prop}(iv) yields
\[
\Delta s_{\lambda}=\sum_{\mu\subseteq\lambda}s_{\mu}\otimes
\underbrace{s_{\lambda/\mu}}_{\substack{=s_{\mu^{c}}\\\text{(by
(\ref{sol.liri.simple.e.1}))}}}=\sum_{\mu\subseteq\lambda}s_{\mu}\otimes
s_{\mu^{c}}.
\]
Compared with
\begin{align*}
\Delta s_{\lambda}
&= \sum_{\mu,\nu} \underbrace{\hat{c}_{\mu,\nu}^{\lambda}}_{=c_{\mu,\nu}^{\lambda}}
 s_{\mu}\otimes s_{\nu}
 \qquad \left(\text{by \eqref{symm-structure-constants.2}}\right)
\\
&= \sum_{\mu,\nu} c_{\mu,\nu}^{\lambda} s_{\mu}\otimes s_{\nu},
\end{align*}
this yields $\sum_{\mu,\nu}c_{\mu,\nu}^{\lambda}s_{\mu}\otimes s_{\nu}%
=\sum_{\mu\subseteq\lambda}s_{\mu}\otimes s_{\mu^{c}}$. Comparing the
coefficients in front of $s_{\mu}\otimes s_{\nu}$ on both sides of this
equality, we obtain: Any two partitions $\mu$ and $\nu$ satisfy
\[
c_{\mu,\nu}^{\lambda}=
\begin{cases}
1,& \text{if }\mu\subseteq\lambda\text{ and }\nu=\mu^{c};\\
0,& \text{otherwise}%
\end{cases}  \quad
\in\left\{  0,1\right\}  .
\]
This solves Exercise \ref{exe.liri.simple}(e).

(f) We know that $\left(  s_{\mu}\right)  _{\mu\in \Par }$ is a
basis of the $\kk$-module $\Lambda$. Hence, $\left(  s_{\mu}\otimes
s_{\nu}\right)  _{\mu,\nu\in \Par }$ is a basis of the
$\kk$-module $\Lambda\otimes\Lambda$. We will refer to this basis as
the \emph{Schur basis} of $\Lambda\otimes\Lambda$.

The equality \eqref{symm-structure-constants.2} yields
\[
\Delta s_{\lambda}=\sum_{\mu,\nu}\underbrace{\hat{c}_{\mu,\nu}^{\lambda}%
}_{=c_{\mu,\nu}^{\lambda}}s_{\mu}\otimes s_{\nu}=\sum_{\mu,\nu}c_{\mu,\nu
}^{\lambda}s_{\mu}\otimes s_{\nu}.
\]
Thus, for every $\mu,\nu\in \Par $, the
$s_{\mu}\otimes s_{\nu}$-coefficient of $\Delta s_\lambda$
with respect to the Schur basis of $\Lambda\otimes\Lambda$
is $c_{\mu,\nu}^{\lambda}$.

But $\lambda=\left(  a+1,1^{b}\right)  $. Hence, Exercise \ref{exe.hookschur}%
(d) gives a formula for $\Delta s_{\lambda}=\Delta s_{\left(  a+1,1^{b}%
\right)  }$ as a sum of pure tensors of the form $s_{\mu}\otimes s_{\nu}$ with
$\mu,\nu\in \Par $. Every such pure tensor occurs \textbf{at most
once} in this formula (as can be easily verified). In other words, for every
$\mu,\nu\in \Par $, the
$s_{\mu}\otimes s_{\nu}$-coefficient of $\Delta s_\lambda$
with respect to the Schur basis of $\Lambda\otimes\Lambda$ is either $0$ or
$1$. Since the
$s_{\mu}\otimes s_{\nu}$-coefficient of $\Delta s_\lambda$
with respect to the
Schur basis of $\Lambda\otimes\Lambda$ is $c_{\mu,\nu}^{\lambda}$, this
rewrites as follows: For every $\mu,\nu\in \Par $, the scalar
$c_{\mu,\nu}^{\lambda}$ is either $0$ or $1$. This solves Exercise
\ref{exe.liri.simple}(f).

(g) The following solution is inspired by
\cite[proof of Thm. 2.1(iv)]{Stembridge-multfree}.

Before we start with the solution, we recall two formulas. Firstly, any two
partitions $\mu$ and $\nu$ satisfy
\begin{align}
s_{\mu}s_{\nu}  & =\sum_{\lambda\in \Par }c_{\mu,\nu}^{\lambda
}s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(
\text{this is just a restatement of \eqref{symm-structure-constants.1}}
\right)  \nonumber\\
& =\sum_{\tau\in \Par }c_{\mu,\nu}^{\tau}s_{\tau}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\lambda\text{ as }\tau\right)  .
\label{sol.liri.simple.g.ss}
\end{align}
Secondly, any two partitions $\lambda$ and $\mu$ satisfy%
\begin{align}
s_{\lambda/\mu}  & =\sum_{\nu\in \Par }c_{\mu,\nu}^{\lambda
}s_{\nu}\ \ \ \ \ \ \ \ \ \ \left(  \text{this is one of the identities in
Remark \ref{third-L-R-interpretation-remark}}\right)  \nonumber\\
& =\sum_{\tau\in \Par }c_{\mu,\tau}^{\lambda}s_{\tau
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index }%
\nu\text{ as }\tau\right)  .\label{sol.liri.simple.g.s/}%
\end{align}


Now, let $\lambda$ be any partition, and let $\mu$ and $\nu$ be two
rectangular partitions. We need to show that $c_{\mu,\nu}^{\lambda}\in\left\{
0,1\right\}  $.

Assume the contrary. Thus, $c_{\mu,\nu}^{\lambda}\notin\left\{  0,1\right\}
$, so that $c_{\mu,\nu}^{\lambda}\neq0$.

We have $c_{\mu,\nu}^{\lambda}=c_{\nu,\mu}^{\lambda}$. Hence, we can WLOG
assume that $\ell\left(  \mu\right)  \geq\ell\left(  \nu\right)  $ (since
otherwise, we can just switch $\mu$ and $\nu$). Assume this. Assume WLOG that
$\mu\neq\varnothing$ (since otherwise, $\mu=\varnothing$ and thus
$\nu=\varnothing$ (because $\ell\left(  \mu\right)  \geq\ell\left(
\nu\right)  $), which makes the claim $c_{\mu,\nu}^{\lambda}\in\left\{
0,1\right\}  $ a rather obvious fact).

The partition $\mu$ is rectangular, i.e., has the form $\left(  m^{k}\right)
=\left(  \underbrace{m,m,\ldots,m}_{k\text{ times}}\right)  $ for some
$m\in{ \NN }$ and $k\in{ \NN }$. Consider these $m$ and $k$. Both
$m$ and $k$ are positive (since $\left(  m^{k}\right)  =\mu\neq\varnothing$).
Thus, $k=\ell\left(  \mu\right)  $ (since $\mu=\left(  m^{k}\right)  $), so
that $k=\ell\left(  \mu\right)  \geq\ell\left(  \nu\right)  $, thus
$\ell\left(  \nu\right)  \leq k$.

Corollary \ref{L-R-rule} shows that $c_{\mu,\nu}^{\lambda}$ counts
column-strict tableaux $T$ of shape $\lambda/\mu$ with $\cont
\left(  T\right)  =\nu$ having the property that each $\cont
\left(  T|_{\operatorname{cols}\geq j}\right)  $ is a partition (where we are
using the notations of Corollary \ref{L-R-rule}). Since $c_{\mu,\nu}^{\lambda
}\neq0$, we see that there exists at least one such tableau $T$. Consider this
$T$. All entries of the tableau $T$ are $\leq\ell\left(  \nu\right)  $
(because $\cont \left(  T\right)  =\nu$). This quickly yields
that $\mu_{k}\geq\lambda_{k+1}$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, $\mu_{k}<\lambda_{k+1}$. Since $\mu=\left(  m^{k}\right)  $,
we have $\mu_{k}=m$, so that $m=\mu_{k}<\lambda_{k+1}$. Thus,
$m \leq \lambda_{k+1}-1$ (since $m$ and $\lambda_{k+1}$ are integers),
so that $m+1 \leq \lambda_{k+1}$.
Now, for every
$i\in\left\{  1,2,...,k+1\right\}  $, we have
\begin{align*}
\mu_{i}  &  \leq m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu=\left(
m^{k}\right)  \right) \\
&  < m+1 \leq \lambda_{k+1}\leq\lambda_{i}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k+1\geq i\right)  .
\end{align*}
Hence, $\left(  i,m+1\right)  $ is a cell of the skew Ferrers diagram
$\lambda/\mu$ for every $i\in\left\{  1,2,...,k+1\right\}  $. Altogether, the
$\left( m+1 \right)$-th column of the skew Ferrers diagram $\lambda/\mu$
contains at least $k+1$ different cells (namely, the cells $\left(
i,m+1\right)  $ for all $i\in\left\{  1,2,...,k+1\right\}  $). In the tableau
$T$, these $k+1$ different cells must be filled with $k+1$ distinct values
(because the entries of $T$ are strictly decreasing top-to-bottom in columns).
As a consequence, there must be at least $k+1$ distinct values among the
entries of $T$; but this is impossible, because all entries of $T$ are
$\leq\ell\left(  \nu\right)  \leq k$. This contradiction proves that our
assumption was wrong, qed.}. Of course, we also have $\mu\subseteq\lambda$
(since $T$ is a tableau of shape $\lambda/\mu$), and thus $\lambda_{i}\geq
\mu_{i}$ for every $i\in\left\{  1,2,3,...\right\}  $. In particular, every
$i\in\left\{  1,2,...,k\right\}  $ satisfies $\lambda_{i}\geq\mu_{i}=m$ (since
$\mu=\left(  m^{k}\right)  $).

Now, define $F$, $F_{\operatorname*{rows}\leq k}$ and $F_{\operatorname*{rows}%
>k}$ as in Exercise \ref{exe.schur.disconnected}. Define four partitions
$\alpha$, $\beta$, $\gamma$ and $\delta$ by%
\begin{align*}
\alpha &  =\left(  \lambda_{1}-m,\lambda_{2}-m,...,\lambda_{k}-m\right)
,\ \ \ \ \ \ \ \ \ \ \beta=\varnothing,\\
\gamma &  =\left(  \lambda_{k+1},\lambda_{k+2},\lambda_{k+3},...\right)
,\ \ \ \ \ \ \ \ \ \ \delta=\varnothing
\end{align*}
(notice that $\alpha$ is well-defined because every $i\in\left\{
1,2,...,k\right\}  $ satisfies $\lambda_{i}\geq m$). It is now easy to see
that the skew Ferrers diagram $\alpha/\beta$ can be obtained from
$F_{\operatorname{rows}\leq k}$ by parallel translation (namely, the
translation by $m$ steps to the west), and that the skew Ferrers diagram
$\gamma/\delta$ can be obtained from $F_{\operatorname{rows}>k}$ by parallel
translation (namely, the translation by $k$ steps to the north). Hence,
Exercise \ref{exe.schur.disconnected} yields
\begin{align*}
s_{\lambda/\mu} &  =s_{\alpha/\beta}s_{\gamma/\delta}=\underbrace{s_{\alpha
/\varnothing}}_{=s_{\alpha}}\underbrace{s_{\gamma/\varnothing}}_{=s_{\gamma}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\beta=\varnothing\text{ and }%
\delta=\varnothing\right)  \\
&  =s_{\alpha}s_{\gamma}=\sum_{\tau\in \Par }c_{\alpha,\gamma
}^{\tau}s_{\tau}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.simple.g.ss})}\right)  .
\end{align*}
Compared with (\ref{sol.liri.simple.g.s/}), this yields $\sum_{\tau
\in \Par }c_{\mu,\tau}^{\lambda}s_{\tau}=\sum_{\tau
\in \Par }c_{\alpha,\gamma}^{\tau}s_{\tau}$. Since $\left(
s_{\tau}\right)  _{\tau\in \Par }$ is a $\kk$-basis of
$\Lambda$, we can compare coefficients before $s_{\nu}$ in this equality, and
thus obtain $c_{\mu,\nu}^{\lambda}=c_{\alpha,\gamma}^{\nu}$. But $\nu$ is a
rectangular partition, and thus has the form $\left(  \widetilde{m}%
^{\widetilde{k}}\right)  $ for some $\widetilde{m}\in{ \NN }$ and
$\widetilde{k}\in \NN$. Hence, $c_{\alpha,\gamma}^{\nu}\in\left\{
0,1\right\}  $ (by Exercise \ref{exe.liri.simple}(e), applied to
$\widetilde{m}$, $\widetilde{k}$, $\nu$, $\alpha$ and $\gamma$ instead of $m$,
$k$, $\lambda$, $\mu$ and $\nu$). Thus, $c_{\mu,\nu}^{\lambda}=c_{\alpha
,\gamma}^{\nu}\in\left\{  0,1\right\}  $, which contradicts $c_{\mu,\nu
}^{\lambda}\notin\left\{  0,1\right\}  $. This contradiction proves that our
assumption was wrong, and Exercise \ref{exe.liri.simple}(g) is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.yamanouchi}}

\textit{Solution to Exercise \ref{exe.liri.yamanouchi}.} (a) We shall
prove the implications $\mathcal{A}\Longrightarrow\mathcal{B}$ and
$\mathcal{B}\Longrightarrow\mathcal{A}$.

\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{B}%
$:} Assume that Assertion $\mathcal{A}$ holds. That is, there exist a
partition $\lambda$ and a column-strict tableau $T$ of shape $\lambda/\mu$
such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfy (\ref{eq.exe.liri.yamanouchi.A.bij}). Consider this $\lambda$ and this
$T$. Since $T$ is column-strict, the entries of $T$ increase weakly
left-to-right along rows, and increase strictly top-to-bottom along columns.

For every $u\in \NN$ and $j\in\left\{  1,2,3,...\right\}  $, we have%
\begin{align}
&  \left(  \text{the number of all entries }\leq u\text{ in the }j\text{-th
row of }T\right) \nonumber\\
&  =\sum_{i=1}^{u}\underbrace{\left(  \text{the number of all entries }i\text{
in the }j\text{-th row of }T\right)  }_{\substack{=b_{i,j}\\\text{(by
(\ref{eq.exe.liri.yamanouchi.A.bij}))}}}=\sum_{i=1}^{u}b_{i,j}\nonumber\\
&  =b_{1,j}+b_{2,j}+\cdots+b_{u,j}. \label{sol.liri.yamanouchi.a.AB.0}%
\end{align}


Let $\left(  i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}  $.
We are going to prove that
\begin{equation}
\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)  \leq\mu
_{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)  .
\label{sol.liri.yamanouchi.a.AB.1}%
\end{equation}


Indeed, assume the contrary. Then,
\begin{equation}
\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)  >\mu
_{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)  .
\label{sol.liri.yamanouchi.a.AB.non1}%
\end{equation}
Hence,
\begin{align*}
&  \mu_{j+1}+\underbrace{\left(  \text{the number of all entries }\leq
i+1\text{ in the }\left(  j+1\right)  \text{-th row of }T\right)
}_{\substack{=b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\\\text{(by
(\ref{sol.liri.yamanouchi.a.AB.0}), applied to }i+1\text{ and }j+1\text{
instead of }u\text{ and }j\text{)}}}\\
&  =\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right) \\
&  >\mu_{j}+\underbrace{\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)  }%
_{\geq0}>\mu_{j}\geq\mu_{j+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mu\text{ is a partition}\right)  .
\end{align*}
Subtracting $\mu_{j+1}$ from both sides of this inequality, we obtain
\[
\left(  \text{the number of all entries }\leq i+1\text{ in the }\left(
j+1\right)  \text{-th row of }T\right)  >0.
\]
In other words, there exists at least one entry $\leq i+1$ in the $\left(
j+1\right)  $-th row of $T$. Let $\mathbf{c}$ be the rightmost cell of the
$\left(  j+1\right)  $-th row of $T$ which contains such an entry. That is,
$\mathbf{c}$ is the rightmost cell of the $\left(  j+1\right)  $-th row of $T$
which contains an entry $\leq i+1$.

The cell $\mathbf{c}$ lies in the $\left(  j+1\right)  $-th row of $T$. Hence,
we can write the cell $\mathbf{c}$ in the form $\mathbf{c}=\left(
j+1,y\right)  $ for some positive integer $y$. Consider this $y$.

The entries of $T$ increase weakly left-to-right along rows. Thus, the cells
of the $\left(  j+1\right)  $-th row of $T$ which contain entries $\leq i+1$
form a contiguous segment of the $\left(  j+1\right)  $-th row of $T$. This
segment begins in cell $\left(  j+1,\mu_{j+1}+1\right)  $ (since the entries
of the $\left(  j+1\right)  $-th row of $T$ begin in cell $\left(
j+1,\mu_{j+1}+1\right)  $ (because $T$ has shape $\lambda/\mu$)), and ends in
cell $\left(  j+1,y\right)  $ (since $\left(  j+1,y\right)  =\mathbf{c}$ is
the rightmost cell of the $\left(  j+1\right)  $-th row of $T$ which contains
an entry $\leq i+1$). Hence, this segment contains precisely $y-\mu_{j+1}$
cells. In other words, there exist exactly $y-\mu_{j+1}$ cells of the $\left(
j+1\right)  $-th row of $T$ which contain entries $\leq i+1$. In other words,
the number of all cells of the $\left(  j+1\right)  $-th row of $T$ which
contain entries $\leq i+1$ is $y-\mu_{j+1}$. In other words,%
\[
\left(  \text{the number of all entries }\leq i+1\text{ in the }\left(
j+1\right)  \text{-th row of }T\right)  =y-\mu_{j+1}.
\]
Solving this for $y$, we obtain%
\[
y=\mu_{j+1}+\left(  \text{the number of all entries }\leq i+1\text{ in the
}\left(  j+1\right)  \text{-th row of }T\right)  .
\]
Since $\mathbf{c}$ is a cell of the tableau $T$, it is clear that the cell
$\mathbf{c}$ lies inside the Ferrers diagram of $\lambda$, and therefore (due
to $j\in\left\{  1,2,3,...\right\}  $) the cell $\left(  j,y\right)  $ must
also lie inside the Ferrers diagram of $\lambda$ (because the cell $\left(
j,y\right)  $ is the northern neighbor of the cell $\left(  j+1,y\right)
=\mathbf{c}$).

We have%
\begin{align*}
y  &  =\mu_{j+1}+\left(  \text{the number of all entries }\leq i+1\text{ in
the }\left(  j+1\right)  \text{-th row of }T\right) \\
&  =\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)  >\mu_{j}.
\end{align*}
Therefore, the cell $\left(  j,y\right)  $ lies outside the Ferrers diagram of
$\mu$. Since $\left(  j,y\right)  $ lies inside the Ferrers diagram of
$\lambda$ but outside the Ferrers diagram of $\mu$, we see that $\left(
j,y\right)  $ is a cell of $T$.

But recall that the entries of $T$ increase strictly top-to-bottom along
columns. Hence, the entry of $T$ in the cell $\left(  j,y\right)  $ must be
strictly smaller than the entry of $T$ in the cell $\mathbf{c}$ (since the
cell $\left(  j,y\right)  $ is the northern neighbor of the cell $\left(
j+1,y\right)  =\mathbf{c}$). Since the latter entry is $\leq i+1$ (by the
definition of $\mathbf{c}$), this shows that the entry of $T$ in the cell
$\left(  j,y\right)  $ must be strictly smaller than $i+1$. Hence, this entry
must be $\leq i$. As a consequence, all cells in the $j$-th row of $T$ which
lie weakly to the left of the cell $\left(  j,y\right)  $ must also have
entries $\leq i$ (because the entries of $T$ increase weakly left-to-right
along rows). The number of such cells is $y-\mu_{j}$ (because the entries in
the $j$-th row of $T$ begin in cell $\left(  j,\mu_{j}+1\right)  $ (since $T$
has shape $\lambda/\mu$)). Thus, there are at least $y-\mu_{j}$ cells in the
$j$-th row of $T$ which have entries $\leq i$; in other words, the number of
all entries $\leq i$ in the $j$-th row of $T$ is at least $y-\mu_{j}$. In
other words,%
\[
\left(  \text{the number of all entries }\leq i\text{ in the }j\text{-th row
of }T\right)  \geq y-\mu_{j}.
\]
Since
\[
\left(  \text{the number of all entries }\leq i\text{ in the }j\text{-th row
of }T\right)  =b_{1,j}+b_{2,j}+\cdots+b_{i,j}
\]
(by (\ref{sol.liri.yamanouchi.a.AB.0}), applied to $u=i$), this rewrites as
follows:
\[
b_{1,j}+b_{2,j}+\cdots+b_{i,j}\geq y-\mu_{j}.
\]
Now, (\ref{sol.liri.yamanouchi.a.AB.non1}) becomes
\begin{align*}
\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)   &  >\mu
_{j}+\underbrace{\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)  }_{\geq
y-\mu_{j}}\geq\mu_{j}+\left(  y-\mu_{j}\right) \\
&  =y=\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)  .
\end{align*}
This is absurd. This contradiction proves that our assumption was wrong.
Hence, (\ref{sol.liri.yamanouchi.a.AB.1}) holds.

Now, forget that we have fixed $\left(  i,j\right)  $. We thus have proven
that the inequality (\ref{sol.liri.yamanouchi.a.AB.1}) holds for all $\left(
i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}  $. In other
words, Assertion $\mathcal{B}$ holds. We thus have proven the implication
$\mathcal{A}\Longrightarrow\mathcal{B}$.

\textit{Proof of the implication $\mathcal{B}\Longrightarrow\mathcal{A}%
$:} Assume that Assertion $\mathcal{B}$ holds. That is, the inequality
(\ref{eq.exe.liri.yamanouchi.B.bij}) holds for all $\left(  i,j\right)
\in \NN \times\left\{  1,2,3,\ldots\right\}  $.

For every $j\in\left\{  1,2,3,...\right\}  $, the sum $b_{1,j}+b_{2,j}%
+b_{3,j}+\cdots$ has only finitely many nonzero addends (since $b_{i,j}=0$ for
all but finitely many pairs $\left(  i,j\right)  $), and can be computed as
the following limit with respect to the discrete topology:%
\begin{equation}
b_{1,j}+b_{2,j}+b_{3,j}+\cdots=\lim\limits_{i\rightarrow\infty}\left(
b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)  . \label{sol.liri.yamanouchi.a.BA.lim}%
\end{equation}


For every $j\in\left\{  1,2,3,...\right\}  $, we have%
\begin{equation}
\mu_{j}+\left(  b_{1,j}+b_{2,j}+b_{3,j}+\cdots\right)  \geq\mu_{j+1}+\left(
b_{1,j+1}+b_{2,j+1}+b_{3,j+1}+\cdots\right)
\label{sol.liri.yamanouchi.a.BA.parti}%
\end{equation}
\footnote{\textit{Proof.} Let $j\in\left\{  1,2,3,...\right\}  $. Then,
\begin{align*}
&  \mu_{j+1}+\underbrace{\left(  b_{1,j+1}+b_{2,j+1}+b_{3,j+1}+\cdots\right)
}_{\substack{=\lim\limits_{i\rightarrow\infty}\left(  b_{1,j+1}+b_{2,j+1}%
+\cdots+b_{i,j+1}\right)  \\\text{(by (\ref{sol.liri.yamanouchi.a.BA.lim}),
applied to }j+1\text{ instead of }j\text{)}}}\\
&  =\mu_{j+1}+\underbrace{\lim\limits_{i\rightarrow\infty}\left(
b_{1,j+1}+b_{2,j+1}+\cdots+b_{i,j+1}\right)  }_{\substack{=\lim
\limits_{i\rightarrow\infty}\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}%
\right)  \\\text{(here, we substituted }i+1\text{ for }i\text{ in the limit)}%
}}\\
&  =\mu_{j+1}+\lim\limits_{i\rightarrow\infty}\left(  b_{1,j+1}+b_{2,j+1}%
+\cdots+b_{i+1,j+1}\right)  =\lim\limits_{i\rightarrow\infty}\left(
\underbrace{\mu_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)
}_{\substack{\leq\mu_{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
\\\text{(by (\ref{eq.exe.liri.yamanouchi.B.bij}))}}}\right) \\
&  \leq\lim\limits_{i\rightarrow\infty}\left(  \mu_{j}+\left(  b_{1,j}%
+b_{2,j}+\cdots+b_{i,j}\right)  \right)  =\mu_{j}+\underbrace{\lim
\limits_{i\rightarrow\infty}\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
}_{\substack{=b_{1,j}+b_{2,j}+b_{3,j}+\cdots\\\text{(by
(\ref{sol.liri.yamanouchi.a.BA.lim}))}}}=\mu_{j}+\left(  b_{1,j}%
+b_{2,j}+b_{3,j}+\cdots\right)  ,
\end{align*}
and this proves (\ref{sol.liri.yamanouchi.a.BA.parti}).}. Hence, we can define
a partition $\lambda$ by%
\begin{equation}
\left(  \lambda_{j}=\mu_{j}+\left(  b_{1,j}+b_{2,j}+b_{3,j}+\cdots\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{  1,2,3,...\right\}  \right)
\label{sol.liri.yamanouchi.a.BA.gamma-beta}%
\end{equation}
\footnote{Here, we are using the fact that $\mu_{j}+\left(  b_{1,j}%
+b_{2,j}+b_{3,j}+\cdots\right)  =0$ for all sufficiently high positive
integers $j$. The proof of this is easy: We have $\mu_{j}=0$ for all
sufficiently high positive integers $j$ (since $\mu$ is a partition), and we
also have $b_{1,j}+b_{2,j}+b_{3,j}+\cdots=0$ for all sufficiently high
positive integers $j$ (since $b_{i,j}=0$ for all but finitely many pairs
$\left(  i,j\right)  $). Thus, if a positive integer $j$ is sufficiently high,
we have both $\mu_{j}=0$ and $b_{1,j}+b_{2,j}+b_{3,j}+\cdots=0$, and therefore
$\mu_{j}+\left(  b_{1,j}+b_{2,j}+b_{3,j}+\cdots\right)  =0+0=0$, qed.}.
Consider this partition $\lambda$.

We have $\mu\subseteq\lambda$ (since every
$j\in\left\{  1,2,3,...\right\}  $ satisfies $\lambda_{j}=\mu_{j}%
+\underbrace{\left(  b_{1,j}+b_{2,j}+b_{3,j}+\cdots\right)  }_{\geq0}\geq
\mu_{j}$). We also have%
\[
\lambda_{j}-\mu_{j}=b_{1,j}+b_{2,j}+b_{3,j}+\cdots
\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{  1,2,3,...\right\}
\]
(because of (\ref{sol.liri.yamanouchi.a.BA.gamma-beta})).

Now, we construct a filling $T$ of the Ferrers diagram of $\lambda/\mu$ with
positive integers as follows: For every $j\in\left\{  1,2,3,...\right\}  $,
the $j$-th row of this Ferrers diagram of $\lambda/\mu$ has $\lambda_{j}%
-\mu_{j}=b_{1,j}+b_{2,j}+b_{3,j}+\cdots$ cells. We fill in the leftmost
$b_{1,j}$ of these cells with $1$'s, the leftmost $b_{2,j}$ of the remaining
cells with $2$'s, the leftmost $b_{3,j}$ of the still remaining cells with
$3$'s, and so on.
Once this has been done for all positive integers $j$ (of course, for
all sufficiently high $j$, the $j$-th row of the Ferrers diagram of
$\lambda/\mu$ has no cells, and therefore nothing has to be filled), we are
left with a filling of the Ferrers diagram of $\lambda/\mu$ with positive
integers. Denote this filling by $T$. It is clear that the entries of $T$
increase weakly left-to-right in rows (by the construction of $T$). We shall
soon show that the entries of $T$ increase strictly top-to-bottom in columns.

First, however, let us observe that (\ref{eq.exe.liri.yamanouchi.A.bij}) holds
for all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ (by the
construction of $T$). Hence, every $u\in \NN$ and $j\in\left\{
1,2,3,...\right\}  $ satisfy (\ref{sol.liri.yamanouchi.a.AB.0}) (this is
proven just as in our proof of the implication $\mathcal{A}\Longrightarrow
\mathcal{B}$).

Now, we are going to prove that the entries of $T$ increase strictly
top-to-bottom in columns.

Indeed, assume the contrary. Then, there exists at least one column of $T$ in
which the entries don't increase strictly top-to-bottom. Let this be the
$k$-th column. So the entries in the $k$-th column of $T$ don't increase
strictly top-to-bottom. As a consequence, there exist two cells $\mathbf{c}$
and $\mathbf{d}$ in the $k$-th column of $T$ such that $\mathbf{d}$ is the
northern neighbor of $\mathbf{c}$, but the entry of $T$ in cell $\mathbf{d}$
is not smaller than the entry of $T$ in cell $\mathbf{c}$. Consider these
cells $\mathbf{c}$ and $\mathbf{d}$.

Write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x,y\right)  $. Then,
$\mathbf{d}=\left(  x-1,y\right)  $ (since $\mathbf{d}$ is the northern
neighbor of $\mathbf{c}$), so that $x-1\geq1$.

Let $p$ be the entry of $T$ in cell $\mathbf{c}$. Then, the entry of $T$ in
cell $\mathbf{d}$ is not smaller than $p$ (since the entry of $T$ in cell
$\mathbf{d}$ is not smaller than the entry of $T$ in cell $\mathbf{c}$). In
other words, the entry of $T$ in cell $\left(  x-1,y\right)  $ is not smaller
than $p$ (since $\mathbf{d}=\left(  x-1,y\right)  $). In other words,%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\left(  x-1,y\right)  \right)
\geq p. \label{sol.liri.yamanouchi.a.BA.5}%
\end{equation}


But let $i=p-1$. Then, $i+1=p$ is the entry of $T$ in cell $\mathbf{c}=\left(
x,y\right)  $. Thus, the cell $\left(  x,y\right)  $ of $T$ has the entry
$i+1$. Hence, all cells in the $x$-th row of $T$ which lie weakly to the left
of the cell $\left(  x,y\right)  $ must have entries $\leq i+1$ (since the
entries of $T$ increase weakly left-to-right in rows). The number of such cells
is $y-\mu_{x}$ (since the entries in the $x$-th row of $T$ begin in cell
$\left(  x,\mu_{x}+1\right)  $ (since $T$ has shape $\lambda/\mu$)). Thus,
there are at least $y-\mu_{x}$ cells in the $x$-th row of $T$ which have
entries $\leq i+1$; in other words, the number of all entries $\leq i+1$ in
the $x$-th row of $T$ is at least $y-\mu_{x}$. In other words,%
\[
\left(  \text{the number of all entries }\leq i+1\text{ in the }x\text{-th row
of }T\right)  \geq y-\mu_{x}.
\]


Since $\left(  \text{the number of all entries }\leq i+1\text{ in the
}x\text{-th row of }T\right)  =b_{1,x}+b_{2,x}+\cdots+b_{i+1,x}$ (by
(\ref{sol.liri.yamanouchi.a.AB.0}), applied to $u=i+1$ and $j=x$), this
rewrites as follows:%
\[
b_{1,x}+b_{2,x}+\cdots+b_{i+1,x}\geq y-\mu_{x}.
\]
Now, recall that $x-1\geq1$, so that $x-1\in\left\{  1,2,3,...\right\}  $.
Hence, (\ref{eq.exe.liri.yamanouchi.B.bij}) (applied to $j=x-1$) yields%
\[
\mu_{\left(  x-1\right)  +1}+\left(  b_{1,\left(  x-1\right)  +1}+b_{2,\left(
x-1\right)  +1}+\cdots+b_{i+1,\left(  x-1\right)  +1}\right)  \leq\mu
_{x-1}+\left(  b_{1,x-1}+b_{2,x-1}+\cdots+b_{i,x-1}\right)  ,
\]
so that
\begin{align}
\mu_{x-1}+\left(  b_{1,x-1}+b_{2,x-1}+\cdots+b_{i,x-1}\right)   &
\geq\underbrace{\mu_{\left(  x-1\right)  +1}}_{=\mu_{x}}+\underbrace{\left(
b_{1,\left(  x-1\right)  +1}+b_{2,\left(  x-1\right)  +1}+\cdots
+b_{i+1,\left(  x-1\right)  +1}\right)  }_{=b_{1,x}+b_{2,x}+\cdots
+b_{i+1,x}\geq y-\mu_{x}}\nonumber\\
&  \geq \mu_{x}+y-\mu_{x} = y. \label{sol.liri.yamanouchi.a.BA.7}%
\end{align}


But every entry $\leq i$ in the $\left(  x-1\right)  $-st row of $T$ must lie
in a cell strictly left of the cell $\left(  x-1,y\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, there exists an
entry $\leq i$ in the $\left(  x-1\right)  $-st row of $T$ which lies in a
cell not strictly left of the cell $\left(  x-1,y\right)  $. Let $c$ be this
entry. Since $c$ lies in a cell not strictly left of the cell $\left(
x-1,y\right)  $, the entry $c$ must lie in a cell weakly to the right of the
cell $\left(  x-1,y\right)  $, and therefore this entry $c$ must be $\geq$ to
the entry of $T$ in cell $\left(  x-1,y\right)  $ (because the entries of $T$
increase weakly left-to-right in rows). Hence,%
\begin{align*}
c  &  \geq\left(  \text{the entry of }T\text{ in cell }\left(  x-1,y\right)
\right)  \geq p\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.a.BA.5})}\right) \\
&  >i\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i=p-1<p\right)  ,
\end{align*}
which contradicts the fact that $c\leq i$ (by definition of $c$). This
contradiction shows that our assumption was wrong, qed.}. Since the number of
such cells\footnote{Here, ``such cells'' means cells of the
$\left(  x-1\right)  $-st row of $T$ which lie strictly
left of the cell $\left(  x-1,y\right)  $.} is $y-1-\mu_{x-1}$ (because the
entries in the $\left(  x-1\right)  $-st row of $T$ begin in cell $\left(
x-1,\mu_{x-1}+1\right)  $ (since $T$ has shape $\lambda/\mu$)), this yields
that there are at most $y-1-\mu_{x-1}$ entries $\leq i$ in the $\left(
x-1\right)  $-st row of $T$. In other words, the number of all entries $\leq
i$ in the $\left(  x-1\right)  $-st row of $T$ is at most $y-1-\mu_{x-1}$. In
other words,%
\[
\left(  \text{the number of all entries }\leq i\text{ in the }\left(
x-1\right)  \text{-st row of }T\right)  \leq y-1-\mu_{x-1}.
\]
Since $\left(  \text{the number of all entries }\leq i\text{ in the }\left(
x-1\right)  \text{-st row of }T\right)  =b_{1,x-1}+b_{2,x-1}+\cdots+b_{i,x-1}$
(by (\ref{sol.liri.yamanouchi.a.AB.0}), applied to $u=i$ and $j=x-1$), this
rewrites as follows:%
\[
b_{1,x-1}+b_{2,x-1}+\cdots+b_{i,x-1}\leq y-\underbrace{1}_{>0}-\mu_{x-1}%
<y-\mu_{x-1}.
\]
Hence, $\mu_{x-1}+\left(  b_{1,x-1}+b_{2,x-1}+\cdots+b_{i,x-1}\right)  <y$.
This contradicts (\ref{sol.liri.yamanouchi.a.BA.7}). This contradiction shows
that our assumption was wrong. Hence, the entries of $T$ increase strictly
top-to-bottom in columns. Since we also know that the entries of $T$ increase
weakly left-to-right in rows, and that $T$ is a filling of the Ferrers diagram
of $\lambda/\mu$, this yields that $T$ is a column-strict tableau of shape
$\lambda/\mu$. Besides, we already know that all $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfy
(\ref{eq.exe.liri.yamanouchi.A.bij}). Hence, Assertion $\mathcal{A}$ holds
(with the $\lambda$ and $T$ that we have constructed above). We thus have
proven the implication $\mathcal{B}\Longrightarrow\mathcal{A}$.

Now that both implications $\mathcal{A}\Longrightarrow\mathcal{B}$ and
$\mathcal{B}\Longrightarrow\mathcal{A}$ are proven, we conclude that
Assertions $\mathcal{A}$ and $\mathcal{B}$ are equivalent. Exercise
\ref{exe.liri.yamanouchi}(a) is solved.

(b) Since $T$ is a column-strict tableau, we know that the entries of $T$
increase weakly left-to-right in rows and increase strictly top-to-bottom
along columns. Hence, if $\mathbf{c}$ and $\mathbf{d}$ are two cells of $T$
such that the cell $\mathbf{c}$ lies southeast\footnote{A cell $\left(
r,c\right)  $ is said to lie \emph{southeast} of a cell $\left(  r^{\prime
},c^{\prime}\right)  $ if and only if we have $r\geq r^{\prime}$ and $c\geq
c^{\prime}$.} of the cell $\mathbf{d}$, then%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  \geq\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  .
\label{sol.liri.yamanouchi.b.SE}%
\end{equation}
For the same reason, if $\mathbf{c}$ and $\mathbf{d}$ are two cells of $T$
such that the cell $\mathbf{c}$ lies southeast of the cell $\mathbf{d}$ but
not on the same row as $\mathbf{d}$, then%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  >\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  .
\label{sol.liri.yamanouchi.b.SEstrict}%
\end{equation}


We shall first prove the equivalence of Assertions $\mathcal{D}$ and
$\mathcal{G}$:

\textit{Proof of the equivalence $\mathcal{D}\Longleftrightarrow\mathcal{G}%
$:} For every $\left(  i,j\right)  \in\left\{  1,2,3,...\right\}
^{2}$, let $b_{i,j}$ be the number of all entries $j$ in the $i$-th row of
$T$. (This is not a typo; we don't want the number of all entries $i$ in the
$j$-th row of $T$.)

It is clear that $b_{i,j}=0$ for all but finitely many pairs $\left(
i,j\right)  $. Hence, we can apply Exercise \ref{exe.liri.yamanouchi}(a) to
$\varnothing$ instead of $\mu$. We conclude that the following two assertions
$\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ are equivalent\footnote{Note
that we denote by $\nu$ and $S$ the variables that have been called $\lambda$
and $T$ in Exercise \ref{exe.liri.yamanouchi}(a), since in our current
situation the letters $\lambda$ and $T$ already have different meanings.}:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime}$:} There exist a
partition $\nu$ and a column-strict tableau $S$ of shape $\nu/\varnothing$
such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfy
\[
b_{i,j}=\left(  \text{the number of all entries }i\text{ in the }j\text{-th
row of }S\right)  .
\]


\item \textit{Assertion $\mathcal{B}^{\prime}$:} The inequality
\[
\varnothing_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)
\leq\varnothing_{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
\]
holds for all $\left(  i,j\right)  \in{ \NN }\times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}

Now, it is easy to see that Assertion $\mathcal{A}^{\prime}$ is equivalent to
Assertion $\mathcal{G}$\ \ \ \ \footnote{\textit{Proof.} Assertion
$\mathcal{A}^{\prime}$ is equivalent to the following Assertion $\mathcal{A}%
^{\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime}$:} There exists a
column-strict tableau $S$ whose shape is a partition such that all $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfy
\[
b_{i,j}=\left(  \text{the number of all entries }i\text{ in the }j\text{-th
row of }S\right)  .
\]
\end{itemize}
\par
(Indeed, Assertion $\mathcal{A}^{\prime}$ is equivalent to Assertion
$\mathcal{A}^{\prime\prime}$ because a column-strict tableau whose shape is a
partition is the same thing as a column-strict tableau of shape $\nu
/\varnothing$ with $\nu$ being a partition.)
\par
Recall that $b_{i,j}=\left(  \text{the number of all entries }j\text{ in the
}i\text{-th row of }T\right)  $ for all $\left(  i,j\right)  \in\left\{
1,2,3,...\right\}  ^{2}$ (by the definition of $b_{i,j}$). Hence, Assertion
$\mathcal{A}^{\prime\prime}$ is equivalent to the following Assertion
$\mathcal{A}^{\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime\prime}$:} There
exists a column-strict tableau $S$ whose shape is a partition such that all
$\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfy
\[
\left(  \text{the number of all entries }j\text{ in the }i\text{-th row of
}T\right)  =\left(  \text{the number of all entries }i\text{ in the
}j\text{-th row of }S\right)  .
\]
\end{itemize}
\par
Assertion $\mathcal{A}^{\prime\prime\prime}$ is equivalent to the following
Assertion $\mathcal{A}^{\prime\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime\prime\prime}$:}
There exists a column-strict tableau $S$ whose shape is a partition such that
all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfy
\[
\left(  \text{the number of all entries }i\text{ in the }j\text{-th row of
}T\right)  =\left(  \text{the number of all entries }j\text{ in the
}i\text{-th row of }S\right)  .
\]
\end{itemize}
\par
(Indeed, Assertion $\mathcal{A}^{\prime\prime\prime\prime}$ is obtained from
Assertion $\mathcal{A}^{\prime\prime\prime}$ by substituting $\left(
j,i\right)  $ for the index $\left(  i,j\right)  $.)
\par
Assertion $\mathcal{A}^{\prime\prime\prime\prime}$ is obviously equivalent to
Assertion $\mathcal{G}$.
\par
Altogether, we have obtained the chain of equivalences $\mathcal{A}^{\prime
}\Longleftrightarrow\mathcal{A}^{\prime\prime}\Longleftrightarrow
\mathcal{A}^{\prime\prime\prime}\Longleftrightarrow\mathcal{A}^{\prime
\prime\prime\prime}\Longleftrightarrow\mathcal{G}$. Thus, we know that
Assertion $\mathcal{A}^{\prime}$ is equivalent to Assertion $\mathcal{G}$,
qed.}. Also, Assertion $\mathcal{B}^{\prime}$ is equivalent to Assertion
$\mathcal{D}$\ \ \ \ \footnote{\textit{Proof.} Every $\left(  i,j\right)
\in{ \NN }\times\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align}
\underbrace{\varnothing_{j}}_{=0}+\left(  b_{1,j}+b_{2,j}+\cdots
+b_{i,j}\right)   &  =b_{1,j}+b_{2,j}+\cdots+b_{i,j}=\sum_{u=1}^{i}%
\underbrace{b_{u,j}}_{\substack{=\left(  \text{the number of all entries
}j\text{ in the }u\text{-th row of }T\right)  \\\text{(by the definition of
}b_{u,j}\text{)}}}\nonumber\\
&  =\sum_{u=1}^{i}\left(  \text{the number of all entries }j\text{ in the
}u\text{-th row of }T\right) \nonumber\\
&  =\left(  \text{the number of all entries }j\text{ in the first }i\text{
rows of }T\right)  \label{sol.liri.yamanouchi.b.DG.3}%
\end{align}
and%
\begin{equation}
\varnothing_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)
=\left(  \text{the number of all entries }j+1\text{ in the first }i+1\text{
rows of }T\right)  \label{sol.liri.yamanouchi.b.DG.4}%
\end{equation}
(by (\ref{sol.liri.yamanouchi.b.DG.3}), applied to $\left(  i+1,j+1\right)  $
instead of $\left(  i,j\right)  $). Hence, Assertion $\mathcal{B}^{\prime}$ is
equivalent to the following Assertion $\mathcal{B}^{\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{B}^{\prime\prime}$:} The
inequality
\[
\left(  \text{the number of all entries }j+1\text{ in the first }i+1\text{
rows of }T\right)  \leq\left(  \text{the number of all entries }j\text{ in the
first }i\text{ rows of }T\right)
\]
holds for all $\left(  i,j\right)  \in{ \NN }\times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}
\par
This Assertion $\mathcal{B}^{\prime\prime}$, in turn, is equivalent to the
following Assertion $\mathcal{B}^{\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{B}^{\prime\prime\prime}$:} The
inequality
\[
\left(  \text{the number of all entries }i+1\text{ in the first }j\text{ rows
of }T\right)  \leq\left(  \text{the number of all entries }i\text{ in the
first }j-1\text{ rows of }T\right)
\]
holds for all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
\times\left\{  1,2,3,\ldots\right\}  $.
\end{itemize}
\par
(Indeed, Assertion $\mathcal{B}^{\prime\prime\prime}$ is obtained from
Assertion $\mathcal{B}^{\prime\prime}$ by substituting $\left(  j-1,i\right)
$ for the index $\left(  i,j\right)  $.) But Assertion $\mathcal{B}%
^{\prime\prime\prime}$ is clearly equivalent to Assertion $\mathcal{D}$.
\par
We thus have found the chain of equivalences $\mathcal{B}^{\prime
}\Longleftrightarrow\mathcal{B}^{\prime\prime}\Longleftrightarrow
\mathcal{B}^{\prime\prime\prime}\Longleftrightarrow\mathcal{D}$. Thus,
Assertion $\mathcal{B}^{\prime}$ is equivalent to Assertion $\mathcal{D}$,
qed.}. Altogether, we have obtained the chain of equivalences $\mathcal{D}%
\Longleftrightarrow\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{A}^{\prime
}\Longleftrightarrow\mathcal{G}$. Hence, we have proven the equivalence
$\mathcal{D}\Longleftrightarrow\mathcal{G}$. In order to complete the solution
of Exercise \ref{exe.liri.yamanouchi}(b), it thus remains to prove the
equivalence $\mathcal{C}\Longleftrightarrow\mathcal{D}\Longleftrightarrow
\mathcal{E}\Longleftrightarrow\mathcal{F}$.

We will achieve this by splitting each of the assertions $\mathcal{C}$,
$\mathcal{D}$, $\mathcal{E}$ and $\mathcal{F}$ into many sub-assertions.
Namely, for every $i\in\left\{  1,2,3,...\right\}  $, let us define four
assertions $\mathcal{C}_{i}$, $\mathcal{D}_{i}$, $\mathcal{E}_{i}$ and
$\mathcal{F}_{i}$ as follows:

\begin{itemize}
\item \textit{Assertion $\mathcal{C}_{i}$:} For every positive
integer $j$, we have $\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i+1}$.

\item \textit{Assertion $\mathcal{D}_{i}$:} For every positive
integer $j$, the number of entries $i+1$ in the first $j$ rows of $T$ is
$\leq$ to the number of entries $i$ in the first $j-1$ rows of $T$.

\item \textit{Assertion $\mathcal{E}_{i}$:} For every NE-set $S$ of
$T$, we have $\left(   \cont \left(  T|_{S}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{S}\right)  \right)  _{i+1}$.

\item \textit{Assertion $\mathcal{F}_{i}$:} For every prefix $v$ of
the Semitic reading word of $T$, there are at least as many $i$'s among the
letters of $v$ as there are $\left(  i+1\right)  $'s among them.
\end{itemize}

We have the following equivalences:%
\[
\mathcal{C}\Longleftrightarrow\left(  \mathcal{C}_{i}\text{ holds for every
}i\in\left\{  1,2,3,...\right\}  \right)
\]
(since $ \cont \left(  T|_{ \cols \geq j}\right)  $ is
a partition if and only if every $i\in\left\{  1,2,3,...\right\}  $ satisfies
$\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i}\geq\left(   \cont \left(  T|_{ \cols %
\geq j}\right)  \right)  _{i+1}$), and%
\[
\mathcal{D}\Longleftrightarrow\left(  \mathcal{D}_{i}\text{ holds for every
}i\in\left\{  1,2,3,...\right\}  \right)
\]
(obviously), and%
\[
\mathcal{E}\Longleftrightarrow\left(  \mathcal{E}_{i}\text{ holds for every
}i\in\left\{  1,2,3,...\right\}  \right)
\]
(since $ \cont \left(  T|_{S}\right)  $ is a partition if and only
if every $i\in\left\{  1,2,3,...\right\}  $ satisfies $\left(
 \cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1}$), and%
\[
\mathcal{F}\Longleftrightarrow\left(  \mathcal{F}_{i}\text{ holds for every
}i\in\left\{  1,2,3,...\right\}  \right)
\]
(by the definition of a Yamanouchi word). Hence, in order to prove the
equivalence $\mathcal{C}\Longleftrightarrow\mathcal{D}\Longleftrightarrow
\mathcal{E}\Longleftrightarrow\mathcal{F}$, it is enough to show that for
every $i\in\left\{  1,2,3,...\right\}  $, we have an equivalence
$\mathcal{C}_{i}\Longleftrightarrow\mathcal{D}_{i}\Longleftrightarrow
\mathcal{E}_{i}\Longleftrightarrow\mathcal{F}_{i}$. So let us do this now.

Let $i\in\left\{  1,2,3,...\right\}  $. We need to prove the equivalence
$\mathcal{C}_{i}\Longleftrightarrow\mathcal{D}_{i}\Longleftrightarrow
\mathcal{E}_{i}\Longleftrightarrow\mathcal{F}_{i}$. We shall achieve this by
proving the implications $\mathcal{C}_{i}\Longrightarrow\mathcal{E}_{i}$,
$\mathcal{E}_{i}\Longrightarrow\mathcal{F}_{i}$, $\mathcal{F}_{i}%
\Longrightarrow\mathcal{D}_{i}$ and $\mathcal{D}_{i}\Longrightarrow
\mathcal{C}_{i}$.

\textit{Proof of the implication $\mathcal{C}_{i}\Longrightarrow
\mathcal{E}_{i}$:} Assume that Assertion $\mathcal{C}_{i}$ holds. Let
$S$ be an NE-set of $T$. We will prove that $\left(   \cont %
\left(  T|_{S}\right)  \right)  _{i}\geq\left(   \cont \left(
T|_{S}\right)  \right)  _{i+1}$.

Assume the contrary. Thus, $\left(   \cont \left(  T|_{S}\right)
\right)  _{i}<\left(   \cont \left(  T|_{S}\right)  \right)
_{i+1}$. In other words, $\left(   \cont \left(  T|_{S}\right)
\right)  _{i+1}>\left(   \cont \left(  T|_{S}\right)  \right)
_{i}$. Recall that
\[
\left(   \cont \left(  T|_{S}\right)  \right)  _{i}=\left\vert
\left(  T|_{S}\right)  ^{-1}\left(  i\right)  \right\vert =\left(  \text{the
number of entries }i\text{ in }T|_{S}\right)
\]
and%
\[
\left(   \cont \left(  T|_{S}\right)  \right)  _{i+1}=\left\vert
\left(  T|_{S}\right)  ^{-1}\left(  i+1\right)  \right\vert =\left(  \text{the
number of entries }i+1\text{ in }T|_{S}\right)  .
\]
Hence,%
\begin{align*}
\left(  \text{the number of entries }i+1\text{ in }T|_{S}\right)   &  =\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
 \cont \left(  T|_{S}\right)  \right)  _{i}\\
&  =\left(  \text{the number of entries }i\text{ in }T|_{S}\right)  \geq0.
\end{align*}
Hence, there exists at least one entry $i+1$ in $T|_{S}$. In other words,
there exists at least one cell $\mathbf{c}\in S$ such that the entry of $T$ in
$\mathbf{c}$ equals $i+1$. Let $\mathbf{d}$ be the \textbf{leftmost} such cell
$\mathbf{c}$ (or one of the leftmost, if there are several of them\footnote{A
moment's thought reveals that there cannot be several of them, but we don't
actually need to think about this.}). Thus, we have $\mathbf{d}\in S$, and the
entry of $T$ in $\mathbf{d}$ equals $i+1$. Let $j$ be the column in which this
cell $\mathbf{d}$ lies. Assertion $\mathcal{C}_{i}$ then yields $\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}$. It is rather clear that $\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i+1}\geq\left(   \cont \left(  T|_{S}\right)  \right)  _{i+1}%
$\ \ \ \ \footnote{\textit{Proof.} We have defined $\mathbf{d}$ as the
leftmost cell $\mathbf{c}\in S$ such that the entry of $T$ in $\mathbf{c}$
equals $i+1$. Hence, if $\mathbf{c}\in S$ is any cell such that the entry of
$T$ in $\mathbf{c}$ equals $i+1$, then $\mathbf{c}$ lies in the same column as
$\mathbf{d}$ or in some column further right. Since $\mathbf{d}$ lies in
column $j$, this rewrites as follows: If $\mathbf{c}\in S$ is any cell such
that the entry of $T$ in $\mathbf{c}$ equals $i+1$, then $\mathbf{c}$ lies in
column $j$ or in some column further right. In other words, if $\mathbf{c}\in
S$ is any cell such that the entry of $T$ in $\mathbf{c}$ equals $i+1$, then
$\mathbf{c}$ lies in one of the columns $j,j+1,j+2,...$. In other words, if
$\mathbf{c}\in S$ is any cell such that the entry of $T$ in $\mathbf{c}$
equals $i+1$, then $\mathbf{c}$ is a cell of $T|_{ \cols \geq j}$.
Thus, the set%
\[
\left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in }%
\mathbf{c}\text{ equals }i+1\right\}
\]
is a subset of the set%
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols \geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i+1\right\}
.
\]
As a consequence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }%
T|_{ \cols \geq j}\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in }T|_{ \cols %
\geq j}\right)  =\left\vert \left(  T|_{ \cols \geq j}\right)
^{-1}\left(  i+1\right)  \right\vert =\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}%
\end{align*}
(because $\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}=\left\vert \left(  T|_{ \cols \geq
j}\right)  ^{-1}\left(  i+1\right)  \right\vert $ (by the definition of
$ \cont \left(  T|_{ \cols \geq j}\right)  $)). Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in }T|_{S}\right)  =\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1},
\end{align*}
this rewrites as $\left(   \cont \left(  T|_{S}\right)  \right)
_{i+1}\leq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}$. Hence, $\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}\geq\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1}$, qed.}.

We shall now show that $\left(   \cont \left(  T|_{S}\right)
\right)  _{i}\geq\left(   \cont \left(  T|_{ \cols %
\geq j}\right)  \right)  _{i}$. Indeed, let $\mathbf{c}$ be a cell of
$T|_{ \cols \geq j}$ such that the entry of $T$ in $\mathbf{c}$
equals $i$. We shall show that $\mathbf{c}\in S$.

Write the cell $\mathbf{d}$ as $\mathbf{d}=\left(  x,j\right)  $ for some
positive integer $x$ (this is possible, since $\mathbf{d}$ lies in column
$j$). Write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x^{\prime},y^{\prime
}\right)  $ for two positive integers $x^{\prime}$ and $y^{\prime}$. Then,
$y^{\prime}$ is the column in which the cell $\mathbf{c}$ lies. Since this
column is one of the columns $j,j+1,j+2,...$ (because $\mathbf{c}$ is a cell
of $T|_{ \cols \geq j}$), this yields that $y^{\prime}\in\left\{
j,j+1,j+2,...\right\}  $, so that $y^{\prime}\geq j$.

If $x^{\prime}\geq x$, then the cell $\left(  x^{\prime},y^{\prime}\right)  $
lies southeast of the cell $\left(  x,j\right)  $ (since $x^{\prime}\geq x$
and $y^{\prime}\geq j$). Since $\left(  x^{\prime},y^{\prime}\right)
=\mathbf{c}$ and $\left(  x,j\right)  =\mathbf{d}$, this rewrites as follows:
If $x^{\prime}\geq x$, then the cell $\mathbf{c}$ lies southeast of the cell
$\mathbf{d}$. Hence, if $x^{\prime}\geq x$, then
(\ref{sol.liri.yamanouchi.b.SE}) yields
\[
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  \geq\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  =i+1,
\]
which contradicts $\left(  \text{the entry of }T\text{ in cell }%
\mathbf{c}\right)  =i<i+1$. Therefore, we cannot have $x^{\prime}\geq x$. We
thus have $x^{\prime}<x$.

Since $x^{\prime}<x$ and $y^{\prime}\geq j$, the cell $\left(  x^{\prime
},y^{\prime}\right)  $ lies northeast of the cell $\left(  x,j\right)  $. In
other words, the cell $\mathbf{c}$ lies northeast of the cell $\mathbf{d}$
(since $\left(  x^{\prime},y^{\prime}\right)  =\mathbf{c}$ and $\left(
x,j\right)  =\mathbf{d}$). Since $\mathbf{c}$ is a cell of $T$ and since
$\mathbf{d}\in S$, this yields that $\mathbf{c}\in S$ as well (since $S$ is an NE-set).

Now, forget that we fixed $\mathbf{c}$. We thus have shown that if
$\mathbf{c}$ is a cell of $T|_{ \cols \geq j}$ such that the entry
of $T$ in $\mathbf{c}$ equals $i$, then $\mathbf{c}\in S$. Hence, the set%
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols \geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i\right\}
\]
is a subset of the set%
\[
\left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in }%
\mathbf{c}\text{ equals }i\right\}  .
\]
As a consequence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols %
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{S}\right)  =\left(
 \cont \left(  T|_{S}\right)  \right)  _{i}.
\end{align*}
Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols %
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{ \cols %
\geq j}\right)  =\left\vert \left(  T|_{ \cols \geq j}\right)
^{-1}\left(  i\right)  \right\vert =\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}%
\end{align*}
(because the definition of $ \cont \left(  T|_{ \cols %
\geq j}\right)  $ shows that $\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}=\left\vert \left(
T|_{ \cols \geq j}\right)  ^{-1}\left(  i\right)  \right\vert $),
this rewrites as $\left(   \cont \left(  T|_{ \cols %
\geq j}\right)  \right)  _{i}\leq\left(   \cont \left(
T|_{S}\right)  \right)  _{i}$. In other words, $\left(   \cont %
\left(  T|_{S}\right)  \right)  _{i}\geq\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}$. Altogether,%
\[
\left(   \cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}\geq\left(   \cont \left(
T|_{S}\right)  \right)  _{i+1},
\]
which contradicts $\left(   \cont \left(  T|_{S}\right)  \right)
_{i}<\left(   \cont \left(  T|_{S}\right)  \right)  _{i+1}$. This
contradiction shows that our assumption was wrong. Hence, $\left(
 \cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1}$.

Now, forget that we fixed $S$. We thus have proven that for every NE-set $S$
of $T$, we have $\left(   \cont \left(  T|_{S}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{S}\right)  \right)  _{i+1}$.
Thus, Assertion $\mathcal{E}_{i}$ holds. This proves the implication
$\mathcal{C}_{i}\Longrightarrow\mathcal{E}_{i}$.

\textit{Proof of the implication $\mathcal{E}_{i}\Longrightarrow
\mathcal{F}_{i}$:} Assume that Assertion $\mathcal{E}_{i}$ holds. Let
$v$ be a prefix of the Semitic reading word of $T$. We shall prove that there
are at least as many $i$'s among the letters of $v$ as there are $\left(
i+1\right)  $'s among them.

For every $i\in\left\{  1,2,3,...\right\}  $, let $r_{i}$ be the word obtained
by reading the $i$-th row of $T$ from right to left. Then, the Semitic reading
word of $T$ is the concatenation $r_{1}r_{2}r_{3}\cdots$ (according to its
definition). Hence, every prefix of this Semitic reading word must have the
form $r_{1}r_{2}\cdots r_{k}s$ for some $k\in\left\{  0,1,2,...\right\}  $ and
some prefix $s$ of $r_{k+1}$. In particular, $v$ must have this form (since
$v$ is a prefix of the Semitic reading word of $T$). In other words, there
exists some $k\in\left\{  0,1,2,...\right\}  $ and some prefix $s$ of
$r_{k+1}$ such that $v=r_{1}r_{2}\cdots r_{k}s$. Consider this $k$ and this
$s$.

Let $\ell$ be the length of $s$. Since $s$ is a prefix of $r_{k+1}$ and has
length $\ell$, it is evident that the word $s$ consists of the first $\ell$
letters of $r_{k+1}$. These first $\ell$ letters of $r_{k+1}$ are the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$ (since
$r_{k+1}$ is the word obtained by reading the $\left(  k+1\right)  $-st row of
$T$ from right to left). Thus, the word $s$ consists of the rightmost $\ell$
entries of the $\left(  k+1\right)  $-st row of $T$. Hence, the word
$r_{1}r_{2}\cdots r_{k}s$ consists of all entries of the first $k$ rows of $T$
and the rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$.
In other words, the letters of the word $r_{1}r_{2}\cdots r_{k}s$ are
precisely all entries of the first $k$ rows of $T$ and the rightmost $\ell$
entries of the $\left(  k+1\right)  $-st row of $T$.

Let $S$ be the set which consists of all cells of the first $k$ rows of $T$
and the rightmost $\ell$ cells of the $\left(  k+1\right)  $-st row of $T$.
Then, $S$ is an NE-set of $T$, and therefore we have $\left(
 \cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1}$ (by Assertion
$\mathcal{E}_{i}$).

Now, $\left(   \cont \left(  T|_{S}\right)  \right)
_{i}=\left\vert \left(  T|_{S}\right)  ^{-1}\left(  i\right)  \right\vert $
(by the definition of $ \cont \left(  T|_{S}\right)  $), so that
$\left(   \cont \left(  T|_{S}\right)  \right)  _{i}$ is the
number of entries $i$ in $T|_{S}$. In other words, $\left(
 \cont \left(  T|_{S}\right)  \right)  _{i}$ is the number of
$i$'s among the entries of $T|_{S}$. But since the entries of $T|_{S}$ are
precisely the letters of $v$\ \ \ \ \footnote{\textit{Proof.} The set $S$
consists of all cells of the first $k$ rows of $T$ and the rightmost $\ell$
cells of the $\left(  k+1\right)  $-st row of $T$. Hence, the entries of
$T|_{S}$ are precisely all entries of the first $k$ rows of $T$ and the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$.
Comparing this with the fact that the letters of the word $r_{1}r_{2}\cdots
r_{k}s$ are precisely all entries of the first $k$ rows of $T$ and the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$, we
conclude the following: The entries of $T|_{S}$ are precisely the letters of
$r_{1}r_{2}\cdots r_{k}s$. In other words, the entries of $T|_{S}$ are
precisely the letters of $v$ (since $v=r_{1}r_{2}\cdots r_{k}s$), qed.}, this
rewrites as follows: $\left(   \cont \left(  T|_{S}\right)
\right)  _{i}$ is the number of $i$'s among the letters of $v$. So we have%
\[
\left(   \cont \left(  T|_{S}\right)  \right)  _{i}=\left(
\text{the number of }i\text{'s among the letters of }v\right)  .
\]
The same argument, with $i+1$ in place of $i$, shows that%
\[
\left(   \cont \left(  T|_{S}\right)  \right)  _{i+1}=\left(
\text{the number of }\left(  i+1\right)  \text{'s among the letters of
}v\right)  .
\]
Hence,%
\begin{align*}
\left(  \text{the number of }i\text{'s among the letters of }v\right)   &
=\left(   \cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{S}\right)  \right)  _{i+1}\\
&  =\left(  \text{the number of }\left(  i+1\right)  \text{'s among the
letters of }v\right)  .
\end{align*}
In other words, there are at least as many $i$'s among the letters of $v$ as
there are $\left(  i+1\right)  $'s among them.

Now, forget that we fixed $v$. We thus have shown that for every prefix $v$ of
the Semitic reading word of $T$, there are at least as many $i$'s among the
letters of $v$ as there are $\left(  i+1\right)  $'s among them. In other
words, Assertion $\mathcal{F}_{i}$ holds. We thus have shown the implication
$\mathcal{E}_{i}\Longrightarrow\mathcal{F}_{i}$.

\textit{Proof of the implication $\mathcal{F}_{i}\Longrightarrow
\mathcal{D}_{i}$:} Assume that Assertion $\mathcal{F}_{i}$ holds.

For every $i\in\left\{  1,2,3,...\right\}  $, let $r_{i}$ be the word obtained
by reading the $i$-th row of $T$ from right to left. Then, the Semitic reading
word of $T$ is the concatenation $r_{1}r_{2}r_{3}\cdots$ (according to its definition).

Now, let $j$ be a positive integer. The word $r_{j}$ is the word obtained by
reading the $j$-th row of $T$ from right to left. Hence, the letters of this
word $r_{j}$ are in (weakly) decreasing order (since the entries of $T$
increase weakly left-to-right in rows). Thus, the subword of $r_{j}$
consisting of all letters $>i$ in $r_{j}$ is a prefix of $r_{j}$. Denote this
prefix by $s$. By the definition of $s$, all letters of $s$ are $>i$, so that%
\[
\left(  \text{the number of }i\text{'s among the letters of }s\right)  =0.
\]
Also, by the definition of $s$, the word $s$ consists of all letters $>i$ in
$r_{j}$. As a consequence,%
\[
\left(  \text{the number of }\left(  i+1\right)  \text{'s among the letters of
}s\right)  =\left(  \text{the number of }\left(  i+1\right)  \text{'s among
the letters of }r_{j}\right)  .
\]


Since $s$ is a prefix of $r_{j}$, it is clear that the word $r_{1}r_{2}\cdots
r_{j-1}s$ is a prefix of the word $r_{1}r_{2}r_{3}\cdots$. In other words, the
word $r_{1}r_{2}\cdots r_{j-1}s$ is a prefix of the Semitic reading word of
$T$ (since the Semitic reading word of $T$ is the concatenation $r_{1}%
r_{2}r_{3}\cdots$). Hence, there are at least as many $i$'s among the letters
of $r_{1}r_{2}\cdots r_{j-1}s$ as there are $\left(  i+1\right)  $'s among
them (by Assertion $\mathcal{F}_{i}$, applied to $v=r_{1}r_{2}\cdots r_{j-1}%
s$). In other words,%
\begin{align*}
&  \left(  \text{the number of }i\text{'s among the letters of }r_{1}%
r_{2}\cdots r_{j-1}s\right) \\
&  \geq\left(  \text{the number of }\left(  i+1\right)  \text{'s among the
letters of }r_{1}r_{2}\cdots r_{j-1}s\right)  .
\end{align*}
Since
\begin{align*}
&  \left(  \text{the number of }i\text{'s among the letters of }r_{1}%
r_{2}\cdots r_{j-1}s\right) \\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }i\text{'s among the letters
of }r_{k}\right)  +\underbrace{\left(  \text{the number of }i\text{'s among
the letters of }s\right)  }_{=0}\\
&  =\sum_{k=1}^{j-1}\underbrace{\left(  \text{the number of }i\text{'s among
the letters of }r_{k}\right)  }_{\substack{=\left(  \text{the number of
entries }i\text{ in the }k\text{-th row of }T\right)  \\\text{(since }%
r_{k}\text{ is the word obtained by reading the }k\text{-th row of }T\text{
from right to left)}}}\\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of entries }i\text{ in the
}k\text{-th row of }T\right) \\
&  =\left(  \text{the number of entries }i\text{ in the first }j-1\text{ rows
of }T\right)
\end{align*}
and%
\begin{align*}
&  \left(  \text{the number of }\left(  i+1\right)  \text{'s among the letters
of }r_{1}r_{2}\cdots r_{j-1}s\right) \\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{k}\right)  +\underbrace{\left(  \text{the number of
}\left(  i+1\right)  \text{'s among the letters of }s\right)  }_{=\left(
\text{the number of }\left(  i+1\right)  \text{'s among the letters of }%
r_{j}\right)  }\\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{k}\right)  +\left(  \text{the number of }\left(
i+1\right)  \text{'s among the letters of }r_{j}\right) \\
&  =\sum_{k=1}^{j}\underbrace{\left(  \text{the number of }\left(  i+1\right)
\text{'s among the letters of }r_{k}\right)  }_{\substack{=\left(  \text{the
number of entries }i+1\text{ in the }k\text{-th row of }T\right)
\\\text{(since }r_{k}\text{ is the word obtained by reading the }k\text{-th
row of }T\text{ from right to left)}}}\\
&  =\sum_{k=1}^{j}\left(  \text{the number of entries }i+1\text{ in the
}k\text{-th row of }T\right) \\
&  =\left(  \text{the number of entries }i+1\text{ in the first }j\text{ rows
of }T\right)  ,
\end{align*}
this rewrites as follows:%
\begin{align*}
&  \left(  \text{the number of entries }i\text{ in the first }j-1\text{ rows
of }T\right) \\
&  \geq\left(  \text{the number of entries }i+1\text{ in the first }j\text{
rows of }T\right)  .
\end{align*}
In other words, the number of entries $i+1$ in the first $j$ rows of $T$ is
$\leq$ to the number of entries $i$ in the first $j-1$ rows of $T$.

Now, forget that we fixed $j$. We thus have shown that for every positive
integer $j$, the number of entries $i+1$ in the first $j$ rows of $T$ is
$\leq$ to the number of entries $i$ in the first $j-1$ rows of $T$. In other
words, Assertion $\mathcal{D}_{i}$ holds. Thus, we have shown the implication
$\mathcal{F}_{i}\Longrightarrow\mathcal{D}_{i}$.

\textit{Proof of the implication $\mathcal{D}_{i}\Longrightarrow
\mathcal{C}_{i}$:} Assume that Assertion $\mathcal{D}_{i}$ holds. We
want to prove that Assertion $\mathcal{C}_{i}$ holds. In other words, we want
to prove that, for every positive integer $j$, we have $\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}$.

In order to prove this, we assume the contrary. That is, there exists a
positive integer $j$ such that we don't have $\left(   \cont
\left(  T|_{ \cols \geq j}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i+1}$. Consider the \textbf{minimal} such $j$. Then, we don't have $\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}$; but
\begin{equation}
\text{for every }k\in\left\{  1,2,...,j-1\right\}  \text{, we do have }\left(
 \cont \left(  T|_{ \cols \geq k}\right)  \right)
_{i}\geq\left(   \cont \left(  T|_{ \cols \geq
k}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.b.DC.notearlier.general}
\end{equation}
Applying (\ref{sol.liri.yamanouchi.b.DC.notearlier.general}) to $k=j-1$, we
obtain:
\begin{equation}
\text{if }j-1\text{ is positive, then we do have }
\left(  \cont 
\left(  T|_{ \cols \geq j-1}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{ \cols \geq j-1}\right)  \right) _{i+1} .
\label{sol.liri.yamanouchi.b.DC.notearlier}
\end{equation}


We don't have $\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i}\geq\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}$. Thus,
\begin{equation}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i}<\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}. \label{sol.liri.yamanouchi.b.DC.ineq1}%
\end{equation}
In other words, $\left(   \cont \left(  T|_{ \cols %
\geq j}\right)  \right)  _{i+1}>\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}$. Recall that
\begin{equation}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i}=\left(  \text{the number of entries }i\text{ in }%
T|_{ \cols \geq j}\right)  \label{sol.liri.yamanouchi.b.DC.ji}%
\end{equation}
and%
\[
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i+1}=\left(  \text{the number of entries }i+1\text{ in
}T|_{ \cols \geq j}\right)  .
\]
Hence,
\begin{align*}
\left(  \text{the number of entries }i+1\text{ in }T|_{ \cols \geq
j}\right)   &  =\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}>\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}\\
&  =\left(  \text{the number of entries }i\text{ in }T|_{ \cols
\geq j}\right)  \geq0.
\end{align*}
Hence, there exists at least one entry $i+1$ in $T|_{ \cols \geq
j}$. In other words, there exists at least one cell $\mathbf{c}$ of
$T|_{ \cols \geq j}$ such that the entry of
$T|_{ \cols \geq j}$ in $\mathbf{c}$ equals $i+1$. In other words,
there exists at least one cell $\mathbf{c}$ of $T|_{ \cols \geq
j}$ such that the entry of $T$ in $\mathbf{c}$ equals $i+1$ (since the entry
of $T|_{ \cols \geq j}$ in $\mathbf{c}$, when it is defined,
equals the entry of $T$ in $\mathbf{c}$). Let $r$ be the \textbf{bottommost}
row which contains such a cell $\mathbf{c}$, and let $\left(  x,y\right)  $ be
the \textbf{leftmost} such cell $\mathbf{c}$ on this row. Thus, $\left(
x,y\right)  $ is a cell of $T|_{ \cols \geq j}$ such that the
entry of $T$ in $\left(  x,y\right)  $ equals $i+1$. Also, the cell $\left(
x,y\right)  $ lies in the $r$-th row; that is, $r=x$.

The cell $\left(  x,y\right)  $ must lie in one of the columns $j,j+1,j+2,...$
(since it is a cell of $T|_{ \cols \geq j}$), so that we have
$y\geq j$. Also, $\left(  x,y\right)  $ is a cell of $T$ (since it is a cell
of $T|_{ \cols \geq j}$).

Applying Assertion $\mathcal{D}_{i}$ to $x$ instead of $j$, we see that the
number of entries $i+1$ in the first $x$ rows of $T$ is $\leq$ to the number
of entries $i$ in the first $x-1$ rows of $T$. In other words,
\begin{align}
&  \left(  \text{the number of entries }i+1\text{ in the first }x\text{ rows
of }T\right) \nonumber\\
&  \leq\left(  \text{the number of entries }i\text{ in the first }x-1\text{
rows of }T\right)  .
\label{sol.liri.yamanouchi.b.DC.ineq0}
\end{align}


It is easy to see that%
\begin{equation}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i+1}\leq\left(  \text{the number of entries }i+1\text{ in the first
}x\text{ rows of }T\right)
\label{sol.liri.yamanouchi.b.DC.ineq2}
\end{equation}
\footnote{\textit{Proof.} Let $\mathbf{c}$ be a cell of
$T|_{ \cols \geq j}$ such that the entry of $T$ in $\mathbf{c}$
equals $i+1$. Since $r$ is the bottommost row which contains such a cell
$\mathbf{c}$, it is clear that the cell $\mathbf{c}$ must be in the $r$-th row
or in a row further north. In other words, the cell $\mathbf{c}$ must lie in
one of the first $r$ rows of $T$. Since $r=x$, this rewrites as follows: The
cell $\mathbf{c}$ must lie in one of the first $x$ rows of $T$.
\par
Now, let us forget that we fixed $\mathbf{c}$. We thus have proven that if
$\mathbf{c}$ is a cell of $T|_{ \cols \geq j}$ such that the entry
of $T$ in $\mathbf{c}$ equals $i+1$, then the cell $\mathbf{c}$ must lie in
one of the first $x$ rows of $T$. Hence, the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols \geq j}
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i+1\right\}
\]
is a subset of the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i+1;\ \text{the cell }\mathbf{c}\text{ lies in
one of the first }x\text{ rows of }T\right\}  .
\]
Hence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols %
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i+1\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the
entry of }T\text{ in }\mathbf{c}\text{ equals }i+1;\ \text{the cell
}\mathbf{c}\text{ lies in one of the first }x\text{ rows of }T\right\}
\right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in the first }x\text{ rows
of }T\right)  .
\end{align*}
But
\begin{align*}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i+1}  &  =\left(  \text{the number of entries }i+1\text{ in
}T|_{ \cols \geq j}\right) \\
&  =\left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols %
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i+1\right\}  \right\vert \\
&  \leq\left(  \text{the number of entries }i+1\text{ in the first }x\text{
rows of }T\right)  ,
\end{align*}
qed.}.

We shall now show that
\begin{equation}
\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows of
}T\right)  \leq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i} .
\label{sol.liri.yamanouchi.b.DC.ineq3}
\end{equation}


Indeed, let $\mathbf{c}$ be a cell of $T$ such that the entry of $T$ in
$\mathbf{c}$ equals $i$ and such that $\mathbf{c}$ lies in one of the first
$x-1$ rows of $T$. We shall show that $\mathbf{c}$ is a cell of
$T|_{ \cols \geq j}$.

Assume the contrary. Thus, $\mathbf{c}$ is not a cell of
$T|_{ \cols \geq j}$.

Let us write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x^{\prime}%
,y^{\prime}\right)  $ for some integers $x^{\prime}$ and $y^{\prime}$. Then,
the cell $\mathbf{c}$ lies in row $x^{\prime}$, and thus this row $x^{\prime}$
must be one of the first $x-1$ rows of $T$ (since we know that $\mathbf{c}$
lies in one of the first $x-1$ rows of $T$). In other words, $x^{\prime}\leq
x-1$. Hence, $x^{\prime}\leq x-1<x$ and thus $x>x^{\prime}$ and $x\neq
x^{\prime}$ and $x\geq x^{\prime}$.

The cell $\mathbf{c}$ lies in the $y^{\prime}$-th column (since $\mathbf{c}%
=\left(  x^{\prime},y^{\prime}\right)  $). Hence, if $y^{\prime}\geq j$, then
the cell $\mathbf{c}$ lies in one of the columns $j,j+1,j+2,...$ and therefore
is a cell of $T|_{ \cols \geq j}$, which contradicts our
assumption that $\mathbf{c}$ is not a cell of $T|_{ \cols \geq j}%
$. Thus, we cannot have $y^{\prime}\geq j$. We therefore have $y^{\prime}<j$.
That is, $y^{\prime}\leq j-1$ (since $y^{\prime}$ and $j$ are integers), so
that $j-1\geq y^{\prime}$.

Now, the cell $\left(  x,y\right)  $ lies southeast of the cell $\left(
x,j-1\right)  $ (since $x\geq x$ and $y\geq j\geq j-1$), which in turn lies
southeast of the cell $\left(  x^{\prime},y^{\prime}\right)  $ (since $x\geq
x^{\prime}$ and $j-1\geq y^{\prime}$). Since both cells $\left(  x,y\right)  $
and $\left(  x^{\prime},y^{\prime}\right)  $ are cells of $T$ (indeed,
$\left(  x,y\right)  $ is known to be a cell of $T$, and $\left(  x^{\prime
},y^{\prime}\right)  =\mathbf{c}$ also is a cell of $T$), this yields that the
intermediate cell $\left(  x,j-1\right)  $ is also a cell of $T$%
\ \ \ \ \footnote{Here, we are using the following fact: If $\alpha$, $\beta$
and $\gamma$ are three cells such that $\alpha$ lies southeast of the cell
$\beta$, which in turn lies southeast of the cell $\gamma$, and if $\alpha$
and $\gamma$ are cells of $T$, then $\beta$ is also a cell of $T$. This can be
easily derived from the fact that $T$ has shape $\lambda/\mu$.}. Thus, $j-1$
is a positive integer. Moreover, since the cell $\left(  x,y\right)  $ lies
southeast of the cell $\left(  x,j-1\right)  $, we can apply
(\ref{sol.liri.yamanouchi.b.SE}) to $\left(  x,y\right)  $ and $\left(
x,j-1\right)  $ instead of $\mathbf{c}$ and $\mathbf{d}$. We thus obtain
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,y\right)  \right)
\geq\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)
\right)  .
\]
Since $\left(  \text{the entry of }T\text{ in cell }\left(  x,y\right)
\right)  =i+1$, this rewrites as
\begin{equation}
i+1\geq\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)
\right)  .
\label{sol.liri.yamanouchi.b.DC.5}
\end{equation}


But the cell $\left(  x,j-1\right)  $ lies southeast of the cell $\left(
x^{\prime},y^{\prime}\right)  $ and not on the same row as $\left(  x^{\prime
},y^{\prime}\right)  $ (since $x\neq x^{\prime}$). Since $\left(  x^{\prime
},y^{\prime}\right)  =\mathbf{c}$, this rewrites as follows: The cell $\left(
x,j-1\right)  $ lies southeast of the cell $\mathbf{c}$ and not on the same
row as $\mathbf{c}$. Hence, (\ref{sol.liri.yamanouchi.b.SEstrict}) (applied to
$\left(  x,j-1\right)  $ and $\mathbf{c}$ instead of $\mathbf{c}$ and
$\mathbf{d}$) yields
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
>\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  =i,
\]
so that%
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
\geq i+1
\]
(since the entry of $T$ in cell $\left(  x,j-1\right)  $ and the number $i$
are integers). Combined with (\ref{sol.liri.yamanouchi.b.DC.5}), this yields
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
=i+1.
\]
Thus, the number $i+1$ appears in the $\left(  j-1\right)  $-th column of $T$
(since the cell $\left(  x,j-1\right)  $ lies in the $\left(  j-1\right)  $-th
column of $T$). In other words,
\begin{equation}
\left(  \text{the number of entries }i+1\text{ in the }\left(  j-1\right)
\text{-th column of }T\right)  \geq 1 .
\label{sol.liri.yamanouchi.b.DC.7i+1}
\end{equation}


On the other hand, the entries of $T$ increase strictly top-to-bottom along
columns. In particular, in every given column of $T$, the entries are
distinct. Applied to the $\left(  j-1\right)  $-th column, this shows that the
entries of the $\left(  j-1\right)  $-th column of $T$ are distinct.
In particular, for every $k\in\left\{  1,2,3,...\right\}  $, the number $k$
appears at most once in the $\left(  j-1\right)  $-th column of $T$. Applying
this to $k=i$, we conclude that the number $i$ appears at most once in the
$\left(  j-1\right)  $-th column of $T$. That is,
\begin{equation}
\left(  \text{the number of entries }i\text{ in the }\left(  j-1\right)
\text{-th column of }T\right)  \leq1.
\label{sol.liri.yamanouchi.b.DC.7i}
\end{equation}


Now, for every $k\in\left\{  1,2,3,...\right\}  $ and $h\in\left\{
1,2,3,...\right\}  $, we have
\begin{align}
\left(   \cont \left(  T|_{ \cols \geq k}\right)
\right)  _{h}  &  =\left\vert \left(  T|_{ \cols \geq k}\right)
^{-1}\left(  h\right)  \right\vert =\left(  \text{the number of entries
}h\text{ in }T|_{ \cols \geq k}\right) \nonumber\\
&  =\left(  \text{the number of entries }h\text{ in the columns }%
k,k+1,k+2,...\text{ of }T\right)  .
\label{sol.liri.yamanouchi.b.DC.increase-only-slightly.general}
\end{align}
Applying this to $k=j$ and $h=i$, we obtain%
\begin{equation}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i}=\left(  \text{the number of entries }i\text{ in the columns
}j,j+1,j+2,...\text{ of }T\right)  .
\label{sol.liri.yamanouchi.b.DC.increase-only-slightly.1}%
\end{equation}
But applying (\ref{sol.liri.yamanouchi.b.DC.increase-only-slightly.general})
to $k=j-1$ and $h=i$, we obtain%
\begin{align*}
\left(   \cont \left(  T|_{ \cols \geq j-1}\right)
\right)  _{i}  &  =\left(  \text{the number of entries }i\text{ in the columns
}j-1,j,j+1,...\text{ of }T\right) \\
&  =\underbrace{\left(  \text{the number of entries }i\text{ in the }\left(
j-1\right)  \text{-th column of }T\right)  }_{\substack{\leq1\\\text{(by
(\ref{sol.liri.yamanouchi.b.DC.7i}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries
}i\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)  }%
_{\substack{=\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i}\\\text{(by
(\ref{sol.liri.yamanouchi.b.DC.increase-only-slightly.1}))}}}\\
&  \leq1+\underbrace{\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i}}_{\substack{<\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i+1}\\\text{(by (\ref{sol.liri.yamanouchi.b.DC.ineq1}))}}}<1+\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i+1}.
\end{align*}
Since $\left(   \cont \left(  T|_{ \cols \geq
j-1}\right)  \right)  _{i}$ and $1+\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}$ are integers, this
yields
\[
\left(   \cont \left(  T|_{ \cols \geq j-1}\right)
\right)  _{i}\leq\left(  1+\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}\right)  -1=\left(
 \cont \left(  T|_{ \cols \geq j}\right)  \right)
_{i+1},
\]
so that
\begin{align*}
\left(   \cont \left(  T|_{ \cols \geq j}\right)
\right)  _{i+1}  &  \geq\left(   \cont \left(
T|_{ \cols \geq j-1}\right)  \right)  _{i}\geq\left(
 \cont \left(  T|_{ \cols \geq j-1}\right)  \right)
_{i+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.b.DC.notearlier})}\right) \\
&  =\left(  \text{the number of entries }i+1\text{ in the columns
}j-1,j,j+1,...\text{ of }T\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.b.DC.increase-only-slightly.general}), applied to
}k=j-1\text{ and }h=i+1\right) \\
&  =\underbrace{\left(  \text{the number of entries }i+1\text{ in the }\left(
j-1\right)  \text{-th column of }T\right)  }_{\substack{\geq1\\\text{(by
(\ref{sol.liri.yamanouchi.b.DC.7i+1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries
}i+1\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)  }%
_{\substack{=\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}\\\text{(since }\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}=\left(  \text{the
number of entries }i+1\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)
\\\text{(by (\ref{sol.liri.yamanouchi.b.DC.increase-only-slightly.general}),
applied to }k=j\text{ and }h=i+1\text{))}}}\\
&  \geq1+\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i+1}>\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1},
\end{align*}
which is absurd. This contradiction proves that our assumption was wrong, and
thus we have proven that $\mathbf{c}$ is a cell of $T|_{ \cols %
\geq j}$.

Now, let us forget that we fixed $\mathbf{c}$. We thus have proven that if
$\mathbf{c}$ is a cell of $T$ such that the entry of $T$ in $\mathbf{c}$
equals $i$ and such that $\mathbf{c}$ lies in one of the first $x-1$ rows of
$T$, then $\mathbf{c}$ is a cell of $T|_{ \cols \geq j}$. Hence,
the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{ lies in one
of the first }x-1\text{ rows of }T\right\}
\]
is a subset of the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{ \cols \geq j}
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i\right\}  .
\]
Hence,
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry
of }T\text{ in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{
lies in one of the first }x-1\text{ rows of }T\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }%
T|_{ \cols \geq j}\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{ \cols
\geq j}\right)  =\left(   \cont \left(  T|_{ \cols \geq j}\right)
 \right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.b.DC.ji})}\right)  .
\end{align*}
Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry
of }T\text{ in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{
lies in one of the first }x-1\text{ rows of }T\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows
of }T\right)  ,
\end{align*}
this rewrites as
\[
\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows of
}T\right)  \leq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i}.
\]
Thus, (\ref{sol.liri.yamanouchi.b.DC.ineq3}) is proven. Now,
(\ref{sol.liri.yamanouchi.b.DC.ineq3}) becomes
\begin{align*}
&  \left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows
of }T\right) \\
&  \leq\left(   \cont \left(  T|_{ \cols \geq
j}\right)  \right)  _{i}<\left(   \cont \left(
T|_{ \cols \geq j}\right)  \right)  _{i+1}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.yamanouchi.b.DC.ineq1}%
)}\right) \\
&  \leq\left(  \text{the number of entries }i+1\text{ in the first }x\text{
rows of }T\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.b.DC.ineq2})}\right)  ,
\end{align*}
which contradicts (\ref{sol.liri.yamanouchi.b.DC.ineq0}). This contradiction
proves that our assumption was wrong. Hence, Assertion $\mathcal{C}_{i}$ must
hold. We thus have proven the implication $\mathcal{D}_{i}\Longrightarrow
\mathcal{C}_{i}$.

We have now proven the four implications $\mathcal{C}_{i}\Longrightarrow
\mathcal{E}_{i}$, $\mathcal{E}_{i}\Longrightarrow\mathcal{F}_{i}$,
$\mathcal{F}_{i}\Longrightarrow\mathcal{D}_{i}$ and $\mathcal{D}%
_{i}\Longrightarrow\mathcal{C}_{i}$. Combining them, we obtain the equivalence
$\mathcal{C}_{i}\Longleftrightarrow\mathcal{D}_{i}\Longleftrightarrow
\mathcal{E}_{i}\Longleftrightarrow\mathcal{F}_{i}$.

Now, let us forget that we fixed $i$. We thus have proven the equivalence
$\mathcal{C}_{i}\Longleftrightarrow\mathcal{D}_{i}\Longleftrightarrow
\mathcal{E}_{i}\Longleftrightarrow\mathcal{F}_{i}$ for every $i\in\left\{
1,2,3,...\right\}  $. As we know, this shows that we have the equivalence
$\mathcal{C}\Longleftrightarrow\mathcal{D}\Longleftrightarrow\mathcal{E}%
\Longleftrightarrow\mathcal{F}$, and this finishes the solution of Exercise
\ref{exe.liri.yamanouchi}(b).

\begin{commentedout}
[The following remark has been commented out, since it is now
superseded by Exercise \ref{exe.liri.yamanouchi.kappa}.]

\textit{Remark:} Exercise \ref{exe.liri.yamanouchi}(b) can be slightly
generalized. Indeed, the following proposition holds:

\begin{proposition}
\label{prop.sol.liri.yamanouchi.relative}Let $\kappa$, $\lambda$ and $\mu$ be
three partitions, and let $T$ be a column-strict tableau of shape $\lambda
/\mu$. Then, the following five assertions are equivalent:

\begin{itemize}
\item \textit{Assertion $\mathcal{C}^{\left(  \kappa\right)  }$:}
For every positive integer $j$, the weak composition $\kappa
+ \cont \left(  T|_{ \cols \geq j}\right)  $ is a partition.

\item \textit{Assertion $\mathcal{D}^{\left(  \kappa\right)  }$:}
For every positive integers $j$ and $i$, we have%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}


\item \textit{Assertion $\mathcal{E}^{\left(  \kappa\right)  }$:}
For every NE-set $S$ of $T$, the weak composition $\kappa+ \cont %
\left(  T|_{S}\right)  $ is a partition.

\item \textit{Assertion $\mathcal{F}^{\left(  \kappa\right)  }$:}
For every prefix $v$ of the Semitic reading word of $T$, and for every
positive integer $i$, we have
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}


\item \textit{Assertion $\mathcal{G}^{\left(  \kappa\right)  }$:}
There exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ which satisfies the following property: For any positive
integers $i$ and $j$, the number of entries $i$ in the $j$-th row of $T$
equals the number of entries $j$ in the $i$-th row of $S$.
\end{itemize}
\end{proposition}

Exercise \ref{exe.liri.yamanouchi}(b) is the particular case of Proposition
\ref{prop.sol.liri.yamanouchi.relative} when $\kappa=\varnothing$.

The proof of Proposition~\ref{prop.sol.liri.yamanouchi.relative} is a rather
straightforward adaptation of the solution of Exercise
\ref{exe.liri.yamanouchi}(b) we gave above; it is left to the reader. This
proposition can be used to reformulate Theorem \ref{Stembridge-theorem}.

\end{commentedout}
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.yamanouchi.kappa}}

\textit{Solution to Exercise \ref{exe.liri.yamanouchi.kappa}.} (a) The
solution to Exercise \ref{exe.liri.yamanouchi.kappa}(a) is a rather
straightforward adaptation of the solution of Exercise
\ref{exe.liri.yamanouchi}(b) that we gave above. We leave the details to the
reader (who can also look them up in the LaTeX source code of this file, where
they appear in a ``commentedout'' environment starting after this sentence).

\begin{commentedout}
Here are the details:

Since $T$ is a column-strict tableau, we know that the entries of $T$ increase
weakly left-to-right in rows and increase strictly top-to-bottom along
columns. Hence, if $\mathbf{c}$ and $\mathbf{d}$ are two cells of $T$ such
that the cell $\mathbf{c}$ lies southeast\footnote{A cell $\left(  r,c\right)
$ is said to lie \emph{southeast} of a cell $\left(  r^{\prime},c^{\prime
}\right)  $ if and only if we have $r\geq r^{\prime}$ and $c\geq c^{\prime}$.}
of the cell $\mathbf{d}$, then%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  \geq\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  .
\label{sol.liri.yamanouchi.kappa.b.SE}%
\end{equation}
For the same reason, if $\mathbf{c}$ and $\mathbf{d}$ are two cells of $T$
such that the cell $\mathbf{c}$ lies southeast of the cell $\mathbf{d}$ but
not on the same row as $\mathbf{d}$, then%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  >\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  .
\label{sol.liri.yamanouchi.kappa.b.SEstrict}%
\end{equation}


We shall first prove the equivalence of Assertions $\mathcal{D}^{\left(
\kappa\right)  }$ and $\mathcal{G}^{\left(  \kappa\right)  }$:

\textit{Proof of the equivalence $\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{G}^{\left(  \kappa\right)  }$:} For every
$\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}$, let $b_{i,j}$ be
the number of all entries $j$ in the $i$-th row of $T$. (This is not a typo;
we don't want the number of all entries $i$ in the $j$-th row of $T$.)

It is clear that $b_{i,j}=0$ for all but finitely many pairs $\left(
i,j\right)  $. Hence, we can apply Exercise \ref{exe.liri.yamanouchi}(a) to
$\kappa$ instead of $\mu$. We conclude that the following two assertions
$\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ are equivalent\footnote{Note
that we denote by $\zeta$ and $S$ the variables that have been called
$\lambda$ and $T$ in Exercise \ref{exe.liri.yamanouchi}(a), since in our
current situation the letters $\lambda$ and $T$ already have different
meanings.}:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime}$:} There exist a
partition $\zeta$ and a column-strict tableau $S$ of shape $\zeta/\kappa$ such
that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfy
\[
b_{i,j}=\left(  \text{the number of all entries }i\text{ in the }j\text{-th
row of }S\right)  .
\]


\item \textit{Assertion $\mathcal{B}^{\prime}$:} The inequality
\[
\kappa_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)  \leq
\kappa_{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
\]
holds for all $\left(  i,j\right)  \in \NN \times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}

Now, it is easy to see that Assertion $\mathcal{A}^{\prime}$ is equivalent to
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$%
\ \ \ \ \footnote{\textit{Proof.} Recall that $b_{i,j}=\left(  \text{the
number of all entries }j\text{ in the }i\text{-th row of }T\right)  $ for all
$\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ (by the definition
of $b_{i,j}$). Hence, Assertion $\mathcal{A}^{\prime}$ is equivalent to the
following Assertion $\mathcal{A}^{\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime\prime}$:} There
exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}$ satisfy
\[
\left(  \text{the number of all entries }j\text{ in the }i\text{-th row of
}T\right)  =\left(  \text{the number of all entries }i\text{ in the
}j\text{-th row of }S\right)  .
\]
\end{itemize}
\par
Assertion $\mathcal{A}^{\prime\prime\prime}$ is equivalent to the following
Assertion $\mathcal{A}^{\prime\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime\prime\prime}$:}
There exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}$ satisfy
\[
\left(  \text{the number of all entries }i\text{ in the }j\text{-th row of
}T\right)  =\left(  \text{the number of all entries }j\text{ in the
}i\text{-th row of }S\right)  .
\]
\end{itemize}
\par
(Indeed, Assertion $\mathcal{A}^{\prime\prime\prime\prime}$ is obtained from
Assertion $\mathcal{A}^{\prime\prime\prime}$ by substituting $\left(
j,i\right)  $ for the index $\left(  i,j\right)  $.)
\par
Assertion $\mathcal{A}^{\prime\prime\prime\prime}$ is obviously equivalent to
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$.
\par
Altogether, we have obtained the chain of equivalences $\mathcal{A}^{\prime
}\Longleftrightarrow\mathcal{A}^{\prime\prime\prime}\Longleftrightarrow
\mathcal{A}^{\prime\prime\prime\prime}\Longleftrightarrow\mathcal{G}^{\left(
\kappa\right)  }$. Thus, we know that Assertion $\mathcal{A}^{\prime}$ is
equivalent to Assertion $\mathcal{G}^{\left(  \kappa\right)  }$, qed.}. Also,
Assertion $\mathcal{B}^{\prime}$ is equivalent to Assertion $\mathcal{D}%
^{\left(  \kappa\right)  }$\ \ \ \ \footnote{\textit{Proof.} Every $\left(
i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align}
b_{1,j}+b_{2,j}+\cdots+b_{i,j}  &  =\sum_{u=1}^{i}\underbrace{b_{u,j}%
}_{\substack{=\left(  \text{the number of all entries }j\text{ in the
}u\text{-th row of }T\right)  \\\text{(by the definition of }b_{u,j}\text{)}%
}}\nonumber\\
&  =\sum_{u=1}^{i}\left(  \text{the number of all entries }j\text{ in the
}u\text{-th row of }T\right) \nonumber\\
&  =\left(  \text{the number of all entries }j\text{ in the first }i\text{
rows of }T\right)  \label{sol.liri.yamanouchi.kappa.b.DG.3}%
\end{align}
and%
\begin{equation}
b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}=\left(  \text{the number of all entries
}j+1\text{ in the first }i+1\text{ rows of }T\right)
\label{sol.liri.yamanouchi.kappa.b.DG.4}%
\end{equation}
(by (\ref{sol.liri.yamanouchi.kappa.b.DG.3}), applied to $\left(
i+1,j+1\right)  $ instead of $\left(  i,j\right)  $). Hence, Assertion
$\mathcal{B}^{\prime}$ is equivalent to the following Assertion $\mathcal{B}%
^{\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{B}^{\prime\prime}$:} The
inequality
\begin{align*}
&  \kappa_{j+1}+\left(  \text{the number of all entries }j+1\text{ in the
first }i+1\text{ rows of }T\right) \\
&  \leq\kappa_{j}+\left(  \text{the number of all entries }j\text{ in the
first }i\text{ rows of }T\right)
\end{align*}
holds for all $\left(  i,j\right)  \in \NN \times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}
\par
This Assertion $\mathcal{B}^{\prime\prime}$, in turn, is equivalent to the
following Assertion $\mathcal{B}^{\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{B}^{\prime\prime\prime}$:} The
inequality
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of all entries }i+1\text{ in the
first }j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of all entries }i\text{ in the
first }j-1\text{ rows of }T\right)
\end{align*}
holds for all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
\times\left\{  1,2,3,\ldots\right\}  $.
\end{itemize}
\par
(Indeed, Assertion $\mathcal{B}^{\prime\prime\prime}$ is obtained from
Assertion $\mathcal{B}^{\prime\prime}$ by substituting $\left(  j-1,i\right)
$ for the index $\left(  i,j\right)  $.) But Assertion $\mathcal{B}%
^{\prime\prime\prime}$ is clearly equivalent to Assertion $\mathcal{D}%
^{\left(  \kappa\right)  }$.
\par
We thus have found the chain of equivalences $\mathcal{B}^{\prime
}\Longleftrightarrow\mathcal{B}^{\prime\prime}\Longleftrightarrow
\mathcal{B}^{\prime\prime\prime}\Longleftrightarrow\mathcal{D}^{\left(
\kappa\right)  }$. Thus, Assertion $\mathcal{B}^{\prime}$ is equivalent to
Assertion $\mathcal{D}^{\left(  \kappa\right)  }$, qed.}. Altogether, we have
obtained the chain of equivalences $\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{A}%
^{\prime}\Longleftrightarrow\mathcal{G}^{\left(  \kappa\right)  }$. Hence, we
have proven the equivalence $\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{G}^{\left(  \kappa\right)  }$. In order to
complete the solution of Exercise \ref{exe.liri.yamanouchi.kappa}(a), it thus
remains to prove the equivalence $\mathcal{C}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{D}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{E}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{F}^{\left(
\kappa\right)  }$.

We will achieve this by splitting each of the assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}%
^{\left(  \kappa\right)  }$ and $\mathcal{F}^{\left(  \kappa\right)  }$ into
many sub-assertions. Namely, for every $i\in\left\{  1,2,3,...\right\}  $, let
us define four assertions $\mathcal{C}_{i}^{\left(  \kappa\right)  }$,
$\mathcal{D}_{i}^{\left(  \kappa\right)  }$, $\mathcal{E}_{i}^{\left(
\kappa\right)  }$ and $\mathcal{F}_{i}^{\left(  \kappa\right)  }$ as follows:

\begin{itemize}
\item \textit{Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$:}
For every positive integer $j$, we have $\left(  \kappa+\cont
\left(  T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\geq\left(
\kappa+\cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}$.

\item \textit{Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$:}
For every positive integer $j$, we have%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}


\item \textit{Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$:}
For every NE-set $S$ of $T$, we have $\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i}\geq\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i+1}$.

\item \textit{Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$:}
For every prefix $v$ of the Semitic reading word of $T$, we have%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}

\end{itemize}

We have the following equivalences:%
\[
\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{C}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(since $\kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  $ is a partition if and only if every $i\in\left\{
1,2,3,...\right\}  $ satisfies $\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\geq\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}$), and%
\[
\mathcal{D}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{D}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(obviously), and%
\[
\mathcal{E}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{E}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(since $\kappa+\cont \left(  T|_{S}\right)  $ is a partition if
and only if every $i\in\left\{  1,2,3,...\right\}  $ satisfies $\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i+1}$), and%
\[
\mathcal{F}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{F}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(obviously). Hence, in order to prove the equivalence $\mathcal{C}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{E}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{F}^{\left(  \kappa\right)  }$, it is enough to show that for every
$i\in\left\{  1,2,3,...\right\}  $, we have an equivalence $\mathcal{C}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$. So let us do
this now.

Let $i\in\left\{  1,2,3,...\right\}  $. We need to prove the equivalence
$\mathcal{C}_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$.
We shall achieve this by proving the implications $\mathcal{C}_{i}^{\left(
\kappa\right)  }\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$,
$\mathcal{E}_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{F}%
_{i}^{\left(  \kappa\right)  }$, $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$ and $\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{C}_{i}^{\left(
\kappa\right)  }$.

\textit{Proof of the implication $\mathcal{C}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$ holds. Let $S$ be
an NE-set of $T$. We will prove that $\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i}\geq\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i+1}$.

Assume the contrary. Thus, $\left(  \kappa+\cont \left(
T|_{S}\right)  \right)  _{i}<\left(  \kappa+\cont \left(
T|_{S}\right)  \right)  _{i+1}$. In other words, $\left(  \kappa
+\cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i}$.

But $\kappa$ is a partition. Thus, $\kappa_{1}\geq\kappa_{2}\geq\kappa_{3}%
\geq\cdots$. In other words, $\kappa_{u}\geq\kappa_{u+1}$ for every
$u\in\left\{  1,2,3,\ldots\right\}  $. Applying this to $u=i$, we obtain
$\kappa_{i}\geq\kappa_{i+1}$. Now,
\begin{align*}
\kappa_{i+1}+\left(  \cont \left(  T|_{S}\right)  \right)
_{i+1}  &  =\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i+1}\\
&  >\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i}=\underbrace{\kappa_{i}}_{\geq\kappa_{i+1}}+\left(  \cont
\left(  T|_{S}\right)  \right)  _{i}\geq\kappa_{i+1}+\left(
\cont \left(  T|_{S}\right)  \right)  _{i}.
\end{align*}
Subtracting $\kappa_{i+1}$ from both sides of this equality, we obtain
$\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
\cont \left(  T|_{S}\right)  \right)  _{i}$.

Recall that%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i}=\left\vert
\left(  T|_{S}\right)  ^{-1}\left(  i\right)  \right\vert =\left(  \text{the
number of entries }i\text{ in }T|_{S}\right)
\]
and%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}=\left\vert
\left(  T|_{S}\right)  ^{-1}\left(  i+1\right)  \right\vert =\left(  \text{the
number of entries }i+1\text{ in }T|_{S}\right)  .
\]
Hence,%
\begin{align*}
\left(  \text{the number of entries }i+1\text{ in }T|_{S}\right)   &  =\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
\cont \left(  T|_{S}\right)  \right)  _{i}\\
&  =\left(  \text{the number of entries }i\text{ in }T|_{S}\right)  \geq0.
\end{align*}
Hence, there exists at least one entry $i+1$ in $T|_{S}$. In other words,
there exists at least one cell $\mathbf{c}\in S$ such that the entry of $T$ in
$\mathbf{c}$ equals $i+1$. Let $\mathbf{d}$ be the \textbf{leftmost} such cell
$\mathbf{c}$ (or one of the leftmost, if there are several of them\footnote{A
moment's thought reveals that there cannot be several of them, but we don't
actually need to think about this.}). Thus, we have $\mathbf{d}\in S$, and the
entry of $T$ in $\mathbf{d}$ equals $i+1$. Let $j$ be the column in which this
cell $\mathbf{d}$ lies. Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$
then yields $\left(  \kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$. It is rather clear
that $\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\geq\left(  \cont \left(
T|_{S}\right)  \right)  _{i+1}$\ \ \ \ \footnote{\textit{Proof.} We have
defined $\mathbf{d}$ as the leftmost cell $\mathbf{c}\in S$ such that the
entry of $T$ in $\mathbf{c}$ equals $i+1$. Hence, if $\mathbf{c}\in S$ is any
cell such that the entry of $T$ in $\mathbf{c}$ equals $i+1$, then
$\mathbf{c}$ lies in the same column as $\mathbf{d}$ or in some column further
right. Since $\mathbf{d}$ lies in column $j$, this rewrites as follows: If
$\mathbf{c}\in S$ is any cell such that the entry of $T$ in $\mathbf{c}$
equals $i+1$, then $\mathbf{c}$ lies in column $j$ or in some column further
right. In other words, if $\mathbf{c}\in S$ is any cell such that the entry of
$T$ in $\mathbf{c}$ equals $i+1$, then $\mathbf{c}$ lies in one of the columns
$j,j+1,j+2,...$. In other words, if $\mathbf{c}\in S$ is any cell such that
the entry of $T$ in $\mathbf{c}$ equals $i+1$, then $\mathbf{c}$ is a cell of
$T|_{\operatorname{cols}\geq j}$. Thus, the set%
\[
\left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in }%
\mathbf{c}\text{ equals }i+1\right\}
\]
is a subset of the set%
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i+1\right\}
.
\]
As a consequence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }%
T|_{\operatorname{cols}\geq j}\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  =\left\vert \left(  T|_{\operatorname{cols}\geq j}\right)
^{-1}\left(  i+1\right)  \right\vert =\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}%
\end{align*}
(because $\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}=\left\vert \left(  T|_{\operatorname{cols}\geq
j}\right)  ^{-1}\left(  i+1\right)  \right\vert $ (by the definition of
$\cont \left(  T|_{\operatorname{cols}\geq j}\right)  $)). Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in }T|_{S}\right)  =\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1},
\end{align*}
this rewrites as $\left(  \cont \left(  T|_{S}\right)  \right)
_{i+1}\leq\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}$. Hence, $\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\geq\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1}$, qed.}.

We shall now show that $\left(  \cont \left(  T|_{S}\right)
\right)  _{i}\geq\left(  \cont \left(  T|_{ \operatorname{cols}
\geq j}\right)  \right)  _{i}$. Indeed, let $\mathbf{c}$ be a cell of $T|_{
\operatorname{cols} \geq j}$ such that the entry of $T$ in $\mathbf{c}$ equals
$i$. We shall show that $\mathbf{c}\in S$.

Write the cell $\mathbf{d}$ as $\mathbf{d}=\left(  x,j\right)  $ for some
positive integer $x$ (this is possible, since $\mathbf{d}$ lies in column
$j$). Write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x^{\prime},y^{\prime
}\right)  $ for two positive integers $x^{\prime}$ and $y^{\prime}$. Then,
$y^{\prime}$ is the column in which the cell $\mathbf{c}$ lies. Since this
column is one of the columns $j,j+1,j+2,...$ (because $\mathbf{c}$ is a cell
of $T|_{ \operatorname{cols} \geq j}$), this yields that $y^{\prime}%
\in\left\{  j,j+1,j+2,...\right\}  $, so that $y^{\prime}\geq j$.

If $x^{\prime}\geq x$, then the cell $\left(  x^{\prime},y^{\prime}\right)  $
lies southeast of the cell $\left(  x,j\right)  $ (since $x^{\prime}\geq x$
and $y^{\prime}\geq j$). Since $\left(  x^{\prime},y^{\prime}\right)
=\mathbf{c}$ and $\left(  x,j\right)  =\mathbf{d}$, this rewrites as follows:
If $x^{\prime}\geq x$, then the cell $\mathbf{c}$ lies southeast of the cell
$\mathbf{d}$. Hence, if $x^{\prime}\geq x$, then
(\ref{sol.liri.yamanouchi.kappa.b.SE}) yields
\[
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  \geq\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  =i+1,
\]
which contradicts $\left(  \text{the entry of }T\text{ in cell }%
\mathbf{c}\right)  =i<i+1$. Therefore, we cannot have $x^{\prime}\geq x$. We
thus have $x^{\prime}<x$.

Since $x^{\prime}<x$ and $y^{\prime}\geq j$, the cell $\left(  x^{\prime
},y^{\prime}\right)  $ lies northeast of the cell $\left(  x,j\right)  $. In
other words, the cell $\mathbf{c}$ lies northeast of the cell $\mathbf{d}$
(since $\left(  x^{\prime},y^{\prime}\right)  =\mathbf{c}$ and $\left(
x,j\right)  =\mathbf{d}$). Since $\mathbf{c}$ is a cell of $T$ and since
$\mathbf{d}\in S$, this yields that $\mathbf{c}\in S$ as well (since $S$ is an NE-set).

Now, forget that we fixed $\mathbf{c}$. We thus have shown that if
$\mathbf{c}$ is a cell of $T|_{\operatorname{cols}\geq j}$ such that the entry
of $T$ in $\mathbf{c}$ equals $i$, then $\mathbf{c}\in S$. Hence, the set%
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i\right\}
\]
is a subset of the set%
\[
\left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in }%
\mathbf{c}\text{ equals }i\right\}  .
\]
As a consequence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{S}\right)  =\left(
\cont \left(  T|_{S}\right)  \right)  _{i}.
\end{align*}
Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  =\left\vert \left(  T|_{\operatorname{cols}\geq j}\right)
^{-1}\left(  i\right)  \right\vert =\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}%
\end{align*}
(because the definition of $\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  $ shows that $\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}=\left\vert \left(
T|_{\operatorname{cols}\geq j}\right)  ^{-1}\left(  i\right)  \right\vert $),
this rewrites as $\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\leq\left(  \cont \left(
T|_{S}\right)  \right)  _{i}$. In other words, $\left(  \cont
\left(  T|_{S}\right)  \right)  _{i}\geq\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}$. Thus,%
\begin{align*}
\left(  \kappa+\cont \left(  T|_{S}\right)  \right)  _{i}  &
=\kappa_{i}+\underbrace{\left(  \cont \left(  T|_{S}\right)
\right)  _{i}}_{\geq\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}}\geq\kappa_{i}+\left(  \cont
\left(  T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\\
&  =\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\\
&  =\kappa_{i+1}+\underbrace{\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}}_{\geq\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1}}\geq\kappa
_{i+1}+\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}\\
&  =\left(  \kappa+\cont \left(  T|_{S}\right)  \right)  _{i+1},
\end{align*}
which contradicts $\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i}<\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i+1}$. This contradiction shows that our assumption was wrong.
Hence, $\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i+1}$.

Now, forget that we fixed $S$. We thus have proven that for every NE-set $S$
of $T$, we have $\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i}\geq\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i+1}$. Thus, Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$
holds. This proves the implication $\mathcal{C}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$ holds. Let $v$ be a
prefix of the Semitic reading word of $T$. We shall prove that%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}


For every $i\in\left\{  1,2,3,...\right\}  $, let $r_{i}$ be the word obtained
by reading the $i$-th row of $T$ from right to left. Then, the Semitic reading
word of $T$ is the concatenation $r_{1}r_{2}r_{3}\cdots$ (according to its
definition). Hence, every prefix of this Semitic reading word must have the
form $r_{1}r_{2}\cdots r_{k}s$ for some $k\in\left\{  0,1,2,...\right\}  $ and
some prefix $s$ of $r_{k+1}$. In particular, $v$ must have this form (since
$v$ is a prefix of the Semitic reading word of $T$). In other words, there
exists some $k\in\left\{  0,1,2,...\right\}  $ and some prefix $s$ of
$r_{k+1}$ such that $v=r_{1}r_{2}\cdots r_{k}s$. Consider this $k$ and this
$s$.

Let $\ell$ be the length of $s$. Since $s$ is a prefix of $r_{k+1}$ and has
length $\ell$, it is evident that the word $s$ consists of the first $\ell$
letters of $r_{k+1}$. These first $\ell$ letters of $r_{k+1}$ are the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$ (since
$r_{k+1}$ is the word obtained by reading the $\left(  k+1\right)  $-st row of
$T$ from right to left). Thus, the word $s$ consists of the rightmost $\ell$
entries of the $\left(  k+1\right)  $-st row of $T$. Hence, the word
$r_{1}r_{2}\cdots r_{k}s$ consists of all entries of the first $k$ rows of $T$
and the rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$.
In other words, the letters of the word $r_{1}r_{2}\cdots r_{k}s$ are
precisely all entries of the first $k$ rows of $T$ and the rightmost $\ell$
entries of the $\left(  k+1\right)  $-st row of $T$.

Let $S$ be the set which consists of all cells of the first $k$ rows of $T$
and the rightmost $\ell$ cells of the $\left(  k+1\right)  $-st row of $T$.
Then, $S$ is an NE-set of $T$, and therefore we have $\left(  \kappa
+\cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i+1}$ (by
Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$).

Now, $\left(  \cont \left(  T|_{S}\right)  \right)
_{i}=\left\vert \left(  T|_{S}\right)  ^{-1}\left(  i\right)  \right\vert $
(by the definition of $\cont \left(  T|_{S}\right)  $), so that
$\left(  \cont \left(  T|_{S}\right)  \right)  _{i}$ is the
number of entries $i$ in $T|_{S}$. In other words, $\left(
\cont \left(  T|_{S}\right)  \right)  _{i}$ is the number of
$i$'s among the entries of $T|_{S}$. But since the entries of $T|_{S}$ are
precisely the letters of $v$\ \ \ \ \footnote{\textit{Proof.} The set $S$
consists of all cells of the first $k$ rows of $T$ and the rightmost $\ell$
cells of the $\left(  k+1\right)  $-st row of $T$. Hence, the entries of
$T|_{S}$ are precisely all entries of the first $k$ rows of $T$ and the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$.
Comparing this with the fact that the letters of the word $r_{1}r_{2}\cdots
r_{k}s$ are precisely all entries of the first $k$ rows of $T$ and the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$, we
conclude the following: The entries of $T|_{S}$ are precisely the letters of
$r_{1}r_{2}\cdots r_{k}s$. In other words, the entries of $T|_{S}$ are
precisely the letters of $v$ (since $v=r_{1}r_{2}\cdots r_{k}s$), qed.}, this
rewrites as follows: $\left(  \cont \left(  T|_{S}\right)
\right)  _{i}$ is the number of $i$'s among the letters of $v$. So we have%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i}=\left(
\text{the number of }i\text{'s among the letters of }v\right)  .
\]
The same argument, with $i+1$ in place of $i$, shows that%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}=\left(
\text{the number of }\left(  i+1\right)  \text{'s among the letters of
}v\right)  .
\]
Hence,%
\begin{align*}
&  \kappa_{i}+\underbrace{\left(  \text{the number of }i\text{'s among the
letters of }v\right)  }_{=\left(  \cont \left(  T|_{S}\right)
\right)  _{i}}\\
&  =\kappa_{i}+\left(  \cont \left(  T|_{S}\right)  \right)
_{i}=\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i+1}\\
&  =\kappa_{i+1}+\underbrace{\left(  \cont \left(  T|_{S}\right)
\right)  _{i+1}}_{=\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  }\\
&  =\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}


Now, forget that we fixed $v$. We thus have shown that for every prefix $v$ of
the Semitic reading word of $T$, we have%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}
In other words, Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$ holds.
We thus have shown the implication $\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$ holds.

For every $i\in\left\{  1,2,3,...\right\}  $, let $r_{i}$ be the word obtained
by reading the $i$-th row of $T$ from right to left. Then, the Semitic reading
word of $T$ is the concatenation $r_{1}r_{2}r_{3}\cdots$ (according to its definition).

Now, let $j$ be a positive integer. The word $r_{j}$ is the word obtained by
reading the $j$-th row of $T$ from right to left. Hence, the letters of this
word $r_{j}$ are in (weakly) decreasing order (since the entries of $T$
increase weakly left-to-right in rows). Thus, the subword of $r_{j}$
consisting of all letters $>i$ in $r_{j}$ is a prefix of $r_{j}$. Denote this
prefix by $s$. By the definition of $s$, all letters of $s$ are $>i$, so that%
\[
\left(  \text{the number of }i\text{'s among the letters of }s\right)  =0.
\]
Also, by the definition of $s$, the word $s$ consists of all letters $>i$ in
$r_{j}$. As a consequence,%
\[
\left(  \text{the number of }\left(  i+1\right)  \text{'s among the letters of
}s\right)  =\left(  \text{the number of }\left(  i+1\right)  \text{'s among
the letters of }r_{j}\right)  .
\]


Since $s$ is a prefix of $r_{j}$, it is clear that the word $r_{1}r_{2}\cdots
r_{j-1}s$ is a prefix of the word $r_{1}r_{2}r_{3}\cdots$. In other words, the
word $r_{1}r_{2}\cdots r_{j-1}s$ is a prefix of the Semitic reading word of
$T$ (since the Semitic reading word of $T$ is the concatenation $r_{1}%
r_{2}r_{3}\cdots$). Hence, we have%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}r_{1}r_{2}\cdots r_{j-1}s\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{1}r_{2}\cdots r_{j-1}s\right)  .
\end{align*}
(by Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$, applied to
$v=r_{1}r_{2}\cdots r_{j-1}s$). Since
\begin{align*}
&  \left(  \text{the number of }i\text{'s among the letters of }r_{1}%
r_{2}\cdots r_{j-1}s\right) \\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }i\text{'s among the letters
of }r_{k}\right)  +\underbrace{\left(  \text{the number of }i\text{'s among
the letters of }s\right)  }_{=0}\\
&  =\sum_{k=1}^{j-1}\underbrace{\left(  \text{the number of }i\text{'s among
the letters of }r_{k}\right)  }_{\substack{=\left(  \text{the number of
entries }i\text{ in the }k\text{-th row of }T\right)  \\\text{(since }%
r_{k}\text{ is the word obtained by reading the }k\text{-th row of }T\text{
from right to left)}}}\\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of entries }i\text{ in the
}k\text{-th row of }T\right) \\
&  =\left(  \text{the number of entries }i\text{ in the first }j-1\text{ rows
of }T\right)
\end{align*}
and%
\begin{align*}
&  \left(  \text{the number of }\left(  i+1\right)  \text{'s among the letters
of }r_{1}r_{2}\cdots r_{j-1}s\right) \\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{k}\right)  +\underbrace{\left(  \text{the number of
}\left(  i+1\right)  \text{'s among the letters of }s\right)  }_{=\left(
\text{the number of }\left(  i+1\right)  \text{'s among the letters of }%
r_{j}\right)  }\\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{k}\right)  +\left(  \text{the number of }\left(
i+1\right)  \text{'s among the letters of }r_{j}\right) \\
&  =\sum_{k=1}^{j}\underbrace{\left(  \text{the number of }\left(  i+1\right)
\text{'s among the letters of }r_{k}\right)  }_{\substack{=\left(  \text{the
number of entries }i+1\text{ in the }k\text{-th row of }T\right)
\\\text{(since }r_{k}\text{ is the word obtained by reading the }k\text{-th
row of }T\text{ from right to left)}}}\\
&  =\sum_{k=1}^{j}\left(  \text{the number of entries }i+1\text{ in the
}k\text{-th row of }T\right) \\
&  =\left(  \text{the number of entries }i+1\text{ in the first }j\text{ rows
of }T\right)  ,
\end{align*}
this rewrites as follows:%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the
first }j\text{ rows of }T\right)  .
\end{align*}
In other words,%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}


Now, forget that we fixed $j$. We thus have shown that for every positive
integer $j$, we have%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}
In other words, Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$ holds.
Thus, we have shown the implication $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{D}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{C}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$ holds. We want to
prove that Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$ holds. In
other words, we want to prove that, for every positive integer $j$, we have
$\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$.

In order to prove this, we assume the contrary. That is, there exists a
positive integer $j$ such that we don't have $\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}$. Consider the \textbf{minimal} such $j$. Then, we
don't have $\left(  \kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$; but
\begin{equation}
\text{for every }k\in\left\{  1,2,...,j-1\right\}  \text{, we do have }\left(
\kappa+\cont \left(  T|_{\operatorname{cols}\geq k}\right)
\right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq k}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.kappa.b.DC.notearlier.general}%
\end{equation}
Applying (\ref{sol.liri.yamanouchi.kappa.b.DC.notearlier.general}) to $k=j-1$,
we obtain:
\begin{equation}
\text{if }j-1\text{ is positive, then we do have }\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j-1}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.kappa.b.DC.notearlier}%
\end{equation}


We don't have $\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\geq\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}$. Thus,
\begin{equation}
\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}<\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.kappa.b.DC.ineq1}%
\end{equation}
In other words, $\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}>\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}$. Thus, $\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}>\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}$%
\ \ \ \ \footnote{\textit{Proof.} Recall that $\kappa$ is a partition. Thus,
$\kappa_{1}\geq\kappa_{2}\geq\kappa_{3}\geq\cdots$. In other words,
$\kappa_{u}\geq\kappa_{u+1}$ for every $u\in\left\{  1,2,3,\ldots\right\}  $.
Applying this to $u=i$, we obtain $\kappa_{i}\geq\kappa_{i+1}$. Now,
\begin{align*}
\kappa_{i+1}+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}  &  =\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\\
&  >\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}=\underbrace{\kappa_{i}}_{\geq\kappa_{i+1}}+\left(
\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}\geq\kappa_{i+1}+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}.
\end{align*}
Subtracting $\kappa_{i+1}$ from both sides of this equality, we obtain
$\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}>\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}$, qed.}.

Recall that
\begin{equation}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i}=\left(  \text{the number of entries }i\text{ in }%
T|_{\operatorname{cols}\geq j}\right)
\label{sol.liri.yamanouchi.kappa.b.DC.ji}%
\end{equation}
and%
\[
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}=\left(  \text{the number of entries }i+1\text{ in
}T|_{\operatorname{cols}\geq j}\right)  .
\]
Hence,
\begin{align*}
\left(  \text{the number of entries }i+1\text{ in }T|_{\operatorname{cols}\geq
j}\right)   &  =\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}>\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\\
&  =\left(  \text{the number of entries }i\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  \geq0.
\end{align*}
Hence, there exists at least one entry $i+1$ in $T|_{\operatorname{cols}\geq
j}$. In other words, there exists at least one cell $\mathbf{c}$ of
$T|_{\operatorname{cols}\geq j}$ such that the entry of
$T|_{\operatorname{cols}\geq j}$ in $\mathbf{c}$ equals $i+1$. In other words,
there exists at least one cell $\mathbf{c}$ of $T|_{\operatorname{cols}\geq
j}$ such that the entry of $T$ in $\mathbf{c}$ equals $i+1$ (since the entry
of $T|_{\operatorname{cols}\geq j}$ in $\mathbf{c}$, when it is defined,
equals the entry of $T$ in $\mathbf{c}$). Let $r$ be the \textbf{bottommost}
row which contains such a cell $\mathbf{c}$, and let $\left(  x,y\right)  $ be
the \textbf{leftmost} such cell $\mathbf{c}$ on this row. Thus, $\left(
x,y\right)  $ is a cell of $T|_{\operatorname{cols}\geq j}$ such that the
entry of $T$ in $\left(  x,y\right)  $ equals $i+1$. Also, the cell $\left(
x,y\right)  $ lies in the $r$-th row; that is, $r=x$.

The cell $\left(  x,y\right)  $ must lie in one of the columns $j,j+1,j+2,...$
(since it is a cell of $T|_{\operatorname{cols}\geq j}$), so that we have
$y\geq j$. Also, $\left(  x,y\right)  $ is a cell of $T$ (since it is a cell
of $T|_{\operatorname{cols}\geq j}$).

Applying Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$ to $x$ instead
of $j$, we see that
\begin{align}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}x\text{ rows of }T\right) \nonumber\\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}x-1\text{ rows of }T\right)  . \label{sol.liri.yamanouchi.kappa.b.DC.ineq0}%
\end{align}


It is easy to see that%
\begin{equation}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}\leq\left(  \text{the number of entries }i+1\text{ in the first
}x\text{ rows of }T\right)  \label{sol.liri.yamanouchi.kappa.b.DC.ineq2}%
\end{equation}
\footnote{\textit{Proof.} Let $\mathbf{c}$ be a cell of
$T|_{\operatorname{cols}\geq j}$ such that the entry of $T$ in $\mathbf{c}$
equals $i+1$. Since $r$ is the bottommost row which contains such a cell
$\mathbf{c}$, it is clear that the cell $\mathbf{c}$ must be in the $r$-th row
or in a row further north. In other words, the cell $\mathbf{c}$ must lie in
one of the first $r$ rows of $T$. Since $r=x$, this rewrites as follows: The
cell $\mathbf{c}$ must lie in one of the first $x$ rows of $T$.
\par
Now, let us forget that we fixed $\mathbf{c}$. We thus have proven that if
$\mathbf{c}$ is a cell of $T|_{\operatorname{cols}\geq j}$ such that the entry
of $T$ in $\mathbf{c}$ equals $i+1$, then the cell $\mathbf{c}$ must lie in
one of the first $x$ rows of $T$. Hence, the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i+1\right\}
\]
is a subset of the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i+1;\ \text{the cell }\mathbf{c}\text{ lies in
one of the first }x\text{ rows of }T\right\}  .
\]
Hence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i+1\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the
entry of }T\text{ in }\mathbf{c}\text{ equals }i+1;\ \text{the cell
}\mathbf{c}\text{ lies in one of the first }x\text{ rows of }T\right\}
\right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in the first }x\text{ rows
of }T\right)  .
\end{align*}
But
\begin{align*}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}  &  =\left(  \text{the number of entries }i+1\text{ in
}T|_{\operatorname{cols}\geq j}\right) \\
&  =\left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i+1\right\}  \right\vert \\
&  \leq\left(  \text{the number of entries }i+1\text{ in the first }x\text{
rows of }T\right)  ,
\end{align*}
qed.}.

We shall now show that
\begin{equation}
\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows of
}T\right)  \leq\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}. \label{sol.liri.yamanouchi.kappa.b.DC.ineq3}%
\end{equation}


Indeed, let $\mathbf{c}$ be a cell of $T$ such that the entry of $T$ in
$\mathbf{c}$ equals $i$ and such that $\mathbf{c}$ lies in one of the first
$x-1$ rows of $T$. We shall show that $\mathbf{c}$ is a cell of
$T|_{\operatorname{cols}\geq j}$.

Assume the contrary. Thus, $\mathbf{c}$ is not a cell of
$T|_{\operatorname{cols}\geq j}$.

Let us write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x^{\prime}%
,y^{\prime}\right)  $ for some integers $x^{\prime}$ and $y^{\prime}$. Then,
the cell $\mathbf{c}$ lies in row $x^{\prime}$, and thus this row $x^{\prime}$
must be one of the first $x-1$ rows of $T$ (since we know that $\mathbf{c}$
lies in one of the first $x-1$ rows of $T$). In other words, $x^{\prime}\leq
x-1$. Hence, $x^{\prime}\leq x-1<x$ and thus $x>x^{\prime}$ and $x\neq
x^{\prime}$ and $x\geq x^{\prime}$.

The cell $\mathbf{c}$ lies in the $y^{\prime}$-th column (since $\mathbf{c}%
=\left(  x^{\prime},y^{\prime}\right)  $). Hence, if $y^{\prime}\geq j$, then
the cell $\mathbf{c}$ lies in one of the columns $j,j+1,j+2,...$ and therefore
is a cell of $T|_{\operatorname{cols}\geq j}$, which contradicts our
assumption that $\mathbf{c}$ is not a cell of $T|_{\operatorname{cols}\geq j}%
$. Thus, we cannot have $y^{\prime}\geq j$. We therefore have $y^{\prime}<j$.
That is, $y^{\prime}\leq j-1$ (since $y^{\prime}$ and $j$ are integers), so
that $j-1\geq y^{\prime}$.

Now, the cell $\left(  x,y\right)  $ lies southeast of the cell $\left(
x,j-1\right)  $ (since $x\geq x$ and $y\geq j\geq j-1$), which in turn lies
southeast of the cell $\left(  x^{\prime},y^{\prime}\right)  $ (since $x\geq
x^{\prime}$ and $j-1\geq y^{\prime}$). Since both cells $\left(  x,y\right)  $
and $\left(  x^{\prime},y^{\prime}\right)  $ are cells of $T$ (indeed,
$\left(  x,y\right)  $ is known to be a cell of $T$, and $\left(  x^{\prime
},y^{\prime}\right)  =\mathbf{c}$ also is a cell of $T$), this yields that the
intermediate cell $\left(  x,j-1\right)  $ is also a cell of $T$%
\ \ \ \ \footnote{Here, we are using the following fact: If $\alpha$, $\beta$
and $\gamma$ are three cells such that $\alpha$ lies southeast of the cell
$\beta$, which in turn lies southeast of the cell $\gamma$, and if $\alpha$
and $\gamma$ are cells of $T$, then $\beta$ is also a cell of $T$. This can be
easily derived from the fact that $T$ has shape $\lambda/\mu$.}. Thus, $j-1$
is a positive integer. Moreover, since the cell $\left(  x,y\right)  $ lies
southeast of the cell $\left(  x,j-1\right)  $, we can apply
(\ref{sol.liri.yamanouchi.kappa.b.SE}) to $\left(  x,y\right)  $ and $\left(
x,j-1\right)  $ instead of $\mathbf{c}$ and $\mathbf{d}$. We thus obtain
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,y\right)  \right)
\geq\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)
\right)  .
\]
Since $\left(  \text{the entry of }T\text{ in cell }\left(  x,y\right)
\right)  =i+1$, this rewrites as
\begin{equation}
i+1\geq\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)
\right)  . \label{sol.liri.yamanouchi.kappa.b.DC.5}%
\end{equation}


But the cell $\left(  x,j-1\right)  $ lies southeast of the cell $\left(
x^{\prime},y^{\prime}\right)  $ and not on the same row as $\left(  x^{\prime
},y^{\prime}\right)  $ (since $x\neq x^{\prime}$). Since $\left(  x^{\prime
},y^{\prime}\right)  =\mathbf{c}$, this rewrites as follows: The cell $\left(
x,j-1\right)  $ lies southeast of the cell $\mathbf{c}$ and not on the same
row as $\mathbf{c}$. Hence, (\ref{sol.liri.yamanouchi.kappa.b.SEstrict})
(applied to $\left(  x,j-1\right)  $ and $\mathbf{c}$ instead of $\mathbf{c}$
and $\mathbf{d}$) yields
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
>\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  =i,
\]
so that%
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
\geq i+1
\]
(since the entry of $T$ in cell $\left(  x,j-1\right)  $ and the number $i$
are integers). Combined with (\ref{sol.liri.yamanouchi.kappa.b.DC.5}), this
yields
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
=i+1.
\]
Thus, the number $i+1$ appears in the $\left(  j-1\right)  $-th column of $T$
(since the cell $\left(  x,j-1\right)  $ lies in the $\left(  j-1\right)  $-th
column of $T$). In other words,
\begin{equation}
\left(  \text{the number of entries }i+1\text{ in the }\left(  j-1\right)
\text{-th column of }T\right)  \geq1 .
\label{sol.liri.yamanouchi.kappa.b.DC.7i+1}%
\end{equation}


On the other hand, the entries of $T$ increase strictly top-to-bottom along
columns. In particular, in every given column of $T$, the entries are
distinct. Applied to the $\left(  j-1\right)  $-th column, this shows that the
entries of the $\left(  j-1\right)  $-th column of $T$ are distinct.
In particular, for every $k\in\left\{  1,2,3,...\right\}  $, the number $k$
appears at most once in the $\left(  j-1\right)  $-th column of $T$. Applying
this to $k=i$, we conclude that the number $i$ appears at most once in the
$\left(  j-1\right)  $-th column of $T$. That is,
\begin{equation}
\left(  \text{the number of entries }i\text{ in the }\left(  j-1\right)
\text{-th column of }T\right)  \leq1.
\label{sol.liri.yamanouchi.kappa.b.DC.7i}%
\end{equation}


Now, for every $k\in\left\{  1,2,3,...\right\}  $ and $h\in\left\{
1,2,3,...\right\}  $, we have
\begin{align}
\left(  \cont \left(  T|_{\operatorname{cols}\geq k}\right)
\right)  _{h}  &  =\left\vert \left(  T|_{\operatorname{cols}\geq k}\right)
^{-1}\left(  h\right)  \right\vert =\left(  \text{the number of entries
}h\text{ in }T|_{\operatorname{cols}\geq k}\right) \nonumber\\
&  =\left(  \text{the number of entries }h\text{ in the columns }%
k,k+1,k+2,...\text{ of }T\right)  .
\label{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}%
\end{align}
Applying this to $k=j$ and $h=i$, we obtain%
\begin{equation}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i}=\left(  \text{the number of entries }i\text{ in the columns
}j,j+1,j+2,...\text{ of }T\right)  .
\label{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.1}%
\end{equation}
But applying
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}) to
$k=j-1$ and $h=i$, we obtain%
\begin{align*}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j-1}\right)
\right)  _{i}  &  =\left(  \text{the number of entries }i\text{ in the columns
}j-1,j,j+1,...\text{ of }T\right) \\
&  =\underbrace{\left(  \text{the number of entries }i\text{ in the }\left(
j-1\right)  \text{-th column of }T\right)  }_{\substack{\leq1\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.7i}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries
}i\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)  }%
_{\substack{=\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.1}))}}}\\
&  \leq1+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}.
\end{align*}
Thus,%
\begin{align*}
\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}  &  =\kappa_{i}+\underbrace{\left(
\cont \left(  T|_{\operatorname{cols}\geq j-1}\right)  \right)
_{i}}_{\leq1+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}}\leq\kappa_{i}+1+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\\
&  =1+\underbrace{\kappa_{i}+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}}_{\substack{=\left(
\kappa+\cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i}<\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.ineq1}))}}}\\
&  <1+\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}.
\end{align*}
Since $\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}$ and $1+\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$ are integers, this
yields
\[
\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}\leq\left(  1+\left(  \kappa+\cont
\left(  T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\right)
-1=\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1},
\]
so that%
\begin{align*}
&  \left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\\
&  \geq\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j-1}\right)  \right)  _{i+1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.notearlier})}\right) \\
&  =\kappa_{i+1}+\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j-1}\right)  \right)  _{i+1}.
\end{align*}
Since%
\begin{align*}
&  \left(  \cont \left(  T|_{\operatorname{cols}\geq j-1}\right)
\right)  _{i+1}\\
&  =\left(  \text{the number of entries }i+1\text{ in the columns
}j-1,j,j+1,...\text{ of }T\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}), applied
to }k=j-1\text{ and }h=i+1\right) \\
&  =\underbrace{\left(  \text{the number of entries }i+1\text{ in the }\left(
j-1\right)  \text{-th column of }T\right)  }_{\substack{\geq1\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.7i+1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries
}i+1\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)  }%
_{\substack{=\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\\\text{(since }\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}=\left(  \text{the
number of entries }i+1\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)
\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}), applied
to }k=j\text{ and }h=i+1\text{))}}}\\
&  \geq1+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}>\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1},
\end{align*}
this becomes%
\begin{align*}
&  \left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\\
&  \geq\kappa_{i+1}+\underbrace{\left(  \cont \left(
T|_{\operatorname{cols}\geq j-1}\right)  \right)  _{i+1}}_{>\left(
\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}}>\kappa_{i+1}+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}=\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}.
\end{align*}
which is absurd. This contradiction proves that our assumption was wrong, and
thus we have proven that $\mathbf{c}$ is a cell of $T|_{\operatorname{cols}%
\geq j}$.

Now, let us forget that we fixed $\mathbf{c}$. We thus have proven that if
$\mathbf{c}$ is a cell of $T$ such that the entry of $T$ in $\mathbf{c}$
equals $i$ and such that $\mathbf{c}$ lies in one of the first $x-1$ rows of
$T$, then $\mathbf{c}$ is a cell of $T|_{\operatorname{cols}\geq j}$. Hence,
the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{ lies in one
of the first }x-1\text{ rows of }T\right\}
\]
is a subset of the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i\right\}  .
\]
Hence,
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry
of }T\text{ in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{
lies in one of the first }x-1\text{ rows of }T\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }%
T|_{\operatorname{cols}\geq j}\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  =\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.ji})}\right)  .
\end{align*}
Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry
of }T\text{ in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{
lies in one of the first }x-1\text{ rows of }T\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows
of }T\right)  ,
\end{align*}
this rewrites as
\[
\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows of
}T\right)  \leq\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}.
\]
Thus, (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq3}) is proven. Now,%
\begin{align*}
&  \kappa_{i}+\underbrace{\left(  \text{the number of entries }i\text{ in the
first }x-1\text{ rows of }T\right)  }_{\substack{\leq\left(
\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}\\\text{(by (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq3}))}}}\\
&  \leq\kappa_{i}+\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}=\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}<\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.ineq1})}\right) \\
&  =\kappa_{i+1}+\underbrace{\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}}_{\substack{\leq\left(
\text{the number of entries }i+1\text{ in the first }x\text{ rows of
}T\right)  \\\text{(by (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq2}))}}}\\
&  \leq\kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the
first }x\text{ rows of }T\right)  ,
\end{align*}
which contradicts (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq0}). This
contradiction proves that our assumption was wrong. Hence, Assertion
$\mathcal{C}_{i}^{\left(  \kappa\right)  }$ must hold. We thus have proven the
implication $\mathcal{D}_{i}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{C}_{i}^{\left(  \kappa\right)  }$.

We have now proven the four implications $\mathcal{C}_{i}^{\left(
\kappa\right)  }\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$,
$\mathcal{E}_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{F}%
_{i}^{\left(  \kappa\right)  }$, $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$ and $\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{C}_{i}^{\left(
\kappa\right)  }$. Combining them, we obtain the equivalence $\mathcal{C}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$.

Now, let us forget that we fixed $i$. We thus have proven the equivalence
$\mathcal{C}_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$
for every $i\in\left\{  1,2,3,...\right\}  $. As we know, this shows that we
have the equivalence $\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{D}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{E}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{F}^{\left(  \kappa\right)  }$, and
this finishes the solution of Exercise \ref{exe.liri.yamanouchi.kappa}(a).
\end{commentedout}

(b) Exercise \ref{exe.liri.yamanouchi.kappa}(a) yields the equivalence
$\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{G}^{\left(  \kappa\right)  }$. It thus remains to prove the
equivalence $\mathcal{G}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{H}^{\left(  \kappa\right)  }$. In order to do so, we must prove the
implications $\mathcal{G}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{H}^{\left(  \kappa\right)  }$ and $\mathcal{H}^{\left(
\kappa\right)  }\Longrightarrow\mathcal{G}^{\left(  \kappa\right)  }$.

The implication $\mathcal{H}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{G}^{\left(  \kappa\right)  }$ is obvious (since we can just take the
$S$ whose existence is guaranteed by Assertion $\mathcal{H}^{\left(
\kappa\right)  }$, and set $\zeta=\tau$). It thus remains to prove the
implication $\mathcal{G}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{H}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{G}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{H}^{\left(  \kappa\right)  }$:} Assume that
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$ holds. We want to prove that
Assertion $\mathcal{H}^{\left(  \kappa\right)  }$ holds.

We have assumed that Assertion $\mathcal{G}^{\left(  \kappa\right)  }$ holds.
In other words, there exist a partition $\zeta$ and a column-strict tableau
$S$ of shape $\zeta/\kappa$ which satisfies the following property: For any
positive integers $i$ and $j$,
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{the number of entries }i\text{ in the }j\text{-th row of }T\text{
equals}\\
\text{the number of entries }j\text{ in the }i\text{-th row of }S
\end{array}
\right)  . \label{sol.liri.yamanouchi.kappa.b.1}%
\end{equation}
Consider this $\zeta$ and this $S$.

For every $i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\begin{align*}
\left(  \cont T\right)  _{i}  &  =\left\vert T^{-1}\left(
i\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\cont T\right) \\
&  =\left(  \text{the number of entries }i\text{ in }T\right) \\
&  =\sum_{j=1}^{\infty}\underbrace{\left(  \text{the number of entries
}i\text{ in the }j\text{-th row of }T\right)  }_{\substack{=\left(  \text{the
number of entries }j\text{ in the }i\text{-th row of }S\right)  \\\text{(by
\eqref{sol.liri.yamanouchi.kappa.b.1})}}}\\
&  =\sum_{j=1}^{\infty}\left(  \text{the number of entries }j\text{ in the
}i\text{-th row of }S\right) \\
&  =\left(  \text{the number of entries in the }i\text{-th row of }S\right) \\
&  =\left(  \text{the length of the }i\text{-th row of the skew partition
}\zeta/\kappa\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the tableau
}S\text{ has shape }\zeta/\kappa\right) \\
&  =\zeta_{i}-\kappa_{i}.
\end{align*}
Hence, for every $i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\[
\left(  \kappa+\cont T\right)  _{i}=\kappa_{i}%
+\underbrace{\left(  \cont T\right)  _{i}}_{=\zeta_{i}%
-\kappa_{i}}=\kappa_{i}+\left(  \zeta_{i}-\kappa_{i}\right)  =\zeta_{i}.
\]
In other words, $\kappa+\cont T=\zeta$, so that $\zeta
=\kappa+\cont T=\tau$. But $S$ is a column-strict tableau of
shape $\zeta/\kappa$. In other words, $S$ is a column-strict tableau of shape
$\tau/\kappa$ (since $\zeta=\tau$).

We thus have constructed a column-strict tableau $S$ of shape $\tau/\kappa$
such that for any positive integers $i$ and $j$, the property
(\ref{sol.liri.yamanouchi.kappa.b.1}) holds. Therefore, Assertion
$\mathcal{H}^{\left(  \kappa\right)  }$ holds. We thus have proven the
implication $\mathcal{G}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{H}^{\left(  \kappa\right)  }$. As we have said, this finishes the
solution of Exercise \ref{exe.liri.yamanouchi.kappa}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.yamanouchi.kappa}}

\textit{Solution to Exercise~\ref{exe.liri.yamanouchi.kappa}.} (a) The
following solution to Exercise \ref{exe.liri.yamanouchi.kappa}(a) is a rather
straightforward adaptation of the solution of Exercise
\ref{exe.liri.yamanouchi}(b) that we gave above.

Since $T$ is a column-strict tableau, we know that the entries of $T$ increase
weakly left-to-right in rows and increase strictly top-to-bottom along
columns. Hence, if $\mathbf{c}$ and $\mathbf{d}$ are two cells of $T$ such
that the cell $\mathbf{c}$ lies southeast\footnote{A cell $\left(  r,c\right)
$ is said to lie \emph{southeast} of a cell $\left(  r^{\prime},c^{\prime
}\right)  $ if and only if we have $r\geq r^{\prime}$ and $c\geq c^{\prime}$.}
of the cell $\mathbf{d}$, then%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  \geq\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  .
\label{sol.liri.yamanouchi.kappa.b.SE}%
\end{equation}
For the same reason, if $\mathbf{c}$ and $\mathbf{d}$ are two cells of $T$
such that the cell $\mathbf{c}$ lies southeast of the cell $\mathbf{d}$ but
not on the same row as $\mathbf{d}$, then%
\begin{equation}
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  >\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  .
\label{sol.liri.yamanouchi.kappa.b.SEstrict}%
\end{equation}


We shall first prove the equivalence of Assertions $\mathcal{D}^{\left(
\kappa\right)  }$ and $\mathcal{G}^{\left(  \kappa\right)  }$:

\textit{Proof of the equivalence $\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{G}^{\left(  \kappa\right)  }$:} For every
$\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}$, let $b_{i,j}$ be
the number of all entries $j$ in the $i$-th row of $T$. (This is not a typo;
we don't want the number of all entries $i$ in the $j$-th row of $T$.)

It is clear that $b_{i,j}=0$ for all but finitely many pairs $\left(
i,j\right)  $. Hence, we can apply Exercise \ref{exe.liri.yamanouchi}(a) to
$\kappa$ instead of $\mu$. We conclude that the following two assertions
$\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ are equivalent\footnote{Note
that we denote by $\zeta$ and $S$ the variables that have been called
$\lambda$ and $T$ in Exercise \ref{exe.liri.yamanouchi}(a), since in our
current situation the letters $\lambda$ and $T$ already have different
meanings.}:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime}$:} There exist a
partition $\zeta$ and a column-strict tableau $S$ of shape $\zeta/\kappa$ such
that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$ satisfy
\[
b_{i,j}=\left(  \text{the number of all entries }i\text{ in the }j\text{-th
row of }S\right)  .
\]


\item \textit{Assertion $\mathcal{B}^{\prime}$:} The inequality
\[
\kappa_{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}\right)  \leq
\kappa_{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
\]
holds for all $\left(  i,j\right)  \in \NN \times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}

Now, it is easy to see that Assertion $\mathcal{A}^{\prime}$ is equivalent to
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$%
\ \ \ \ \footnote{\textit{Proof.} Recall that $b_{i,j}=\left(  \text{the
number of all entries }j\text{ in the }i\text{-th row of }T\right)  $ for all
$\left(  i,j\right)  \in\left\{  1,2,3,...\right\}  ^{2}$ (by the definition
of $b_{i,j}$). Hence, Assertion $\mathcal{A}^{\prime}$ is equivalent to the
following Assertion $\mathcal{A}^{\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime\prime}$:} There
exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}$ satisfy
\[
\left(  \text{the number of all entries }j\text{ in the }i\text{-th row of
}T\right)  =\left(  \text{the number of all entries }i\text{ in the
}j\text{-th row of }S\right)  .
\]
\end{itemize}
\par
Assertion $\mathcal{A}^{\prime\prime\prime}$ is equivalent to the following
Assertion $\mathcal{A}^{\prime\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime\prime\prime\prime}$:}
There exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{2}$ satisfy
\[
\left(  \text{the number of all entries }i\text{ in the }j\text{-th row of
}T\right)  =\left(  \text{the number of all entries }j\text{ in the
}i\text{-th row of }S\right)  .
\]
\end{itemize}
\par
(Indeed, Assertion $\mathcal{A}^{\prime\prime\prime\prime}$ is obtained from
Assertion $\mathcal{A}^{\prime\prime\prime}$ by substituting $\left(
j,i\right)  $ for the index $\left(  i,j\right)  $.)
\par
Assertion $\mathcal{A}^{\prime\prime\prime\prime}$ is obviously equivalent to
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$.
\par
Altogether, we have obtained the chain of equivalences $\mathcal{A}^{\prime
}\Longleftrightarrow\mathcal{A}^{\prime\prime\prime}\Longleftrightarrow
\mathcal{A}^{\prime\prime\prime\prime}\Longleftrightarrow\mathcal{G}^{\left(
\kappa\right)  }$. Thus, we know that Assertion $\mathcal{A}^{\prime}$ is
equivalent to Assertion $\mathcal{G}^{\left(  \kappa\right)  }$, qed.}. Also,
Assertion $\mathcal{B}^{\prime}$ is equivalent to Assertion $\mathcal{D}%
^{\left(  \kappa\right)  }$\ \ \ \ \footnote{\textit{Proof.} Every $\left(
i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}  $ satisfies%
\begin{align}
b_{1,j}+b_{2,j}+\cdots+b_{i,j}  &  =\sum_{u=1}^{i}\underbrace{b_{u,j}%
}_{\substack{=\left(  \text{the number of all entries }j\text{ in the
}u\text{-th row of }T\right)  \\\text{(by the definition of }b_{u,j}\text{)}%
}}\nonumber\\
&  =\sum_{u=1}^{i}\left(  \text{the number of all entries }j\text{ in the
}u\text{-th row of }T\right) \nonumber\\
&  =\left(  \text{the number of all entries }j\text{ in the first }i\text{
rows of }T\right)  \label{sol.liri.yamanouchi.kappa.b.DG.3}%
\end{align}
and%
\begin{equation}
b_{1,j+1}+b_{2,j+1}+\cdots+b_{i+1,j+1}=\left(  \text{the number of all entries
}j+1\text{ in the first }i+1\text{ rows of }T\right)
\label{sol.liri.yamanouchi.kappa.b.DG.4}%
\end{equation}
(by (\ref{sol.liri.yamanouchi.kappa.b.DG.3}), applied to $\left(
i+1,j+1\right)  $ instead of $\left(  i,j\right)  $). Hence, Assertion
$\mathcal{B}^{\prime}$ is equivalent to the following Assertion $\mathcal{B}%
^{\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{B}^{\prime\prime}$:} The
inequality
\begin{align*}
&  \kappa_{j+1}+\left(  \text{the number of all entries }j+1\text{ in the
first }i+1\text{ rows of }T\right) \\
&  \leq\kappa_{j}+\left(  \text{the number of all entries }j\text{ in the
first }i\text{ rows of }T\right)
\end{align*}
holds for all $\left(  i,j\right)  \in \NN \times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}
\par
This Assertion $\mathcal{B}^{\prime\prime}$, in turn, is equivalent to the
following Assertion $\mathcal{B}^{\prime\prime\prime}$:
\par
\begin{itemize}
\item \textit{Assertion $\mathcal{B}^{\prime\prime\prime}$:} The
inequality
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of all entries }i+1\text{ in the
first }j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of all entries }i\text{ in the
first }j-1\text{ rows of }T\right)
\end{align*}
holds for all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
\times\left\{  1,2,3,\ldots\right\}  $.
\end{itemize}
\par
(Indeed, Assertion $\mathcal{B}^{\prime\prime\prime}$ is obtained from
Assertion $\mathcal{B}^{\prime\prime}$ by substituting $\left(  j-1,i\right)
$ for the index $\left(  i,j\right)  $.) But Assertion $\mathcal{B}%
^{\prime\prime\prime}$ is clearly equivalent to Assertion $\mathcal{D}%
^{\left(  \kappa\right)  }$.
\par
We thus have found the chain of equivalences $\mathcal{B}^{\prime
}\Longleftrightarrow\mathcal{B}^{\prime\prime}\Longleftrightarrow
\mathcal{B}^{\prime\prime\prime}\Longleftrightarrow\mathcal{D}^{\left(
\kappa\right)  }$. Thus, Assertion $\mathcal{B}^{\prime}$ is equivalent to
Assertion $\mathcal{D}^{\left(  \kappa\right)  }$, qed.}. Altogether, we have
obtained the chain of equivalences $\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{A}%
^{\prime}\Longleftrightarrow\mathcal{G}^{\left(  \kappa\right)  }$. Hence, we
have proven the equivalence $\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{G}^{\left(  \kappa\right)  }$. In order to
complete the solution of Exercise \ref{exe.liri.yamanouchi.kappa}(a), it thus
remains to prove the equivalence $\mathcal{C}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{D}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{E}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{F}^{\left(
\kappa\right)  }$.

We will achieve this by splitting each of the assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}%
^{\left(  \kappa\right)  }$ and $\mathcal{F}^{\left(  \kappa\right)  }$ into
many sub-assertions. Namely, for every $i\in\left\{  1,2,3,...\right\}  $, let
us define four assertions $\mathcal{C}_{i}^{\left(  \kappa\right)  }$,
$\mathcal{D}_{i}^{\left(  \kappa\right)  }$, $\mathcal{E}_{i}^{\left(
\kappa\right)  }$ and $\mathcal{F}_{i}^{\left(  \kappa\right)  }$ as follows:

\begin{itemize}
\item \textit{Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$:}
For every positive integer $j$, we have $\left(  \kappa+\cont
\left(  T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\geq\left(
\kappa+\cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}$.

\item \textit{Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$:}
For every positive integer $j$, we have%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}


\item \textit{Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$:}
For every NE-set $S$ of $T$, we have $\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i}\geq\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i+1}$.

\item \textit{Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$:}
For every prefix $v$ of the Semitic reading word of $T$, we have%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}

\end{itemize}

We have the following equivalences:%
\[
\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{C}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(since $\kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  $ is a partition if and only if every $i\in\left\{
1,2,3,...\right\}  $ satisfies $\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\geq\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}$), and%
\[
\mathcal{D}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{D}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(obviously), and%
\[
\mathcal{E}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{E}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(since $\kappa+\cont \left(  T|_{S}\right)  $ is a partition if
and only if every $i\in\left\{  1,2,3,...\right\}  $ satisfies $\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i+1}$), and%
\[
\mathcal{F}^{\left(  \kappa\right)  }\Longleftrightarrow\left(  \mathcal{F}%
_{i}^{\left(  \kappa\right)  }\text{ holds for every }i\in\left\{
1,2,3,...\right\}  \right)
\]
(obviously). Hence, in order to prove the equivalence $\mathcal{C}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{D}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{E}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{F}^{\left(  \kappa\right)  }$, it is enough to show that for every
$i\in\left\{  1,2,3,...\right\}  $, we have an equivalence $\mathcal{C}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$. So let us do
this now.

Let $i\in\left\{  1,2,3,...\right\}  $. We need to prove the equivalence
$\mathcal{C}_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$.
We shall achieve this by proving the implications $\mathcal{C}_{i}^{\left(
\kappa\right)  }\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$,
$\mathcal{E}_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{F}%
_{i}^{\left(  \kappa\right)  }$, $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$ and $\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{C}_{i}^{\left(
\kappa\right)  }$.

\textit{Proof of the implication $\mathcal{C}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$ holds. Let $S$ be
an NE-set of $T$. We will prove that $\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i}\geq\left(  \kappa+\cont
\left(  T|_{S}\right)  \right)  _{i+1}$.

Assume the contrary. Thus, $\left(  \kappa+\cont \left(
T|_{S}\right)  \right)  _{i}<\left(  \kappa+\cont \left(
T|_{S}\right)  \right)  _{i+1}$. In other words, $\left(  \kappa
+\cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i}$.

But $\kappa$ is a partition. Thus, $\kappa_{1}\geq\kappa_{2}\geq\kappa_{3}%
\geq\cdots$. In other words, $\kappa_{u}\geq\kappa_{u+1}$ for every
$u\in\left\{  1,2,3,\ldots\right\}  $. Applying this to $u=i$, we obtain
$\kappa_{i}\geq\kappa_{i+1}$. Now,
\begin{align*}
\kappa_{i+1}+\left(  \cont \left(  T|_{S}\right)  \right)
_{i+1}  &  =\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i+1}\\
&  >\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i}=\underbrace{\kappa_{i}}_{\geq\kappa_{i+1}}+\left(  \cont
\left(  T|_{S}\right)  \right)  _{i}\geq\kappa_{i+1}+\left(
\cont \left(  T|_{S}\right)  \right)  _{i}.
\end{align*}
Subtracting $\kappa_{i+1}$ from both sides of this equality, we obtain
$\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
\cont \left(  T|_{S}\right)  \right)  _{i}$.

Recall that%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i}=\left\vert
\left(  T|_{S}\right)  ^{-1}\left(  i\right)  \right\vert =\left(  \text{the
number of entries }i\text{ in }T|_{S}\right)
\]
and%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}=\left\vert
\left(  T|_{S}\right)  ^{-1}\left(  i+1\right)  \right\vert =\left(  \text{the
number of entries }i+1\text{ in }T|_{S}\right)  .
\]
Hence,%
\begin{align*}
\left(  \text{the number of entries }i+1\text{ in }T|_{S}\right)   &  =\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1}>\left(
\cont \left(  T|_{S}\right)  \right)  _{i}\\
&  =\left(  \text{the number of entries }i\text{ in }T|_{S}\right)  \geq0.
\end{align*}
Hence, there exists at least one entry $i+1$ in $T|_{S}$. In other words,
there exists at least one cell $\mathbf{c}\in S$ such that the entry of $T$ in
$\mathbf{c}$ equals $i+1$. Let $\mathbf{d}$ be the \textbf{leftmost} such cell
$\mathbf{c}$ (or one of the leftmost, if there are several of them\footnote{A
moment's thought reveals that there cannot be several of them, but we don't
actually need to think about this.}). Thus, we have $\mathbf{d}\in S$, and the
entry of $T$ in $\mathbf{d}$ equals $i+1$. Let $j$ be the column in which this
cell $\mathbf{d}$ lies. Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$
then yields $\left(  \kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$. It is rather clear
that $\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\geq\left(  \cont \left(
T|_{S}\right)  \right)  _{i+1}$\ \ \ \ \footnote{\textit{Proof.} We have
defined $\mathbf{d}$ as the leftmost cell $\mathbf{c}\in S$ such that the
entry of $T$ in $\mathbf{c}$ equals $i+1$. Hence, if $\mathbf{c}\in S$ is any
cell such that the entry of $T$ in $\mathbf{c}$ equals $i+1$, then
$\mathbf{c}$ lies in the same column as $\mathbf{d}$ or in some column further
right. Since $\mathbf{d}$ lies in column $j$, this rewrites as follows: If
$\mathbf{c}\in S$ is any cell such that the entry of $T$ in $\mathbf{c}$
equals $i+1$, then $\mathbf{c}$ lies in column $j$ or in some column further
right. In other words, if $\mathbf{c}\in S$ is any cell such that the entry of
$T$ in $\mathbf{c}$ equals $i+1$, then $\mathbf{c}$ lies in one of the columns
$j,j+1,j+2,...$. In other words, if $\mathbf{c}\in S$ is any cell such that
the entry of $T$ in $\mathbf{c}$ equals $i+1$, then $\mathbf{c}$ is a cell of
$T|_{\operatorname{cols}\geq j}$. Thus, the set%
\[
\left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in }%
\mathbf{c}\text{ equals }i+1\right\}
\]
is a subset of the set%
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i+1\right\}
.
\]
As a consequence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }%
T|_{\operatorname{cols}\geq j}\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  =\left\vert \left(  T|_{\operatorname{cols}\geq j}\right)
^{-1}\left(  i+1\right)  \right\vert =\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}%
\end{align*}
(because $\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}=\left\vert \left(  T|_{\operatorname{cols}\geq
j}\right)  ^{-1}\left(  i+1\right)  \right\vert $ (by the definition of
$\cont \left(  T|_{\operatorname{cols}\geq j}\right)  $)). Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i+1\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in }T|_{S}\right)  =\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1},
\end{align*}
this rewrites as $\left(  \cont \left(  T|_{S}\right)  \right)
_{i+1}\leq\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}$. Hence, $\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\geq\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1}$, qed.}.

We shall now show that $\left(  \cont \left(  T|_{S}\right)
\right)  _{i}\geq\left(  \cont \left(  T|_{ \operatorname{cols}
\geq j}\right)  \right)  _{i}$. Indeed, let $\mathbf{c}$ be a cell of $T|_{
\operatorname{cols} \geq j}$ such that the entry of $T$ in $\mathbf{c}$ equals
$i$. We shall show that $\mathbf{c}\in S$.

Write the cell $\mathbf{d}$ as $\mathbf{d}=\left(  x,j\right)  $ for some
positive integer $x$ (this is possible, since $\mathbf{d}$ lies in column
$j$). Write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x^{\prime},y^{\prime
}\right)  $ for two positive integers $x^{\prime}$ and $y^{\prime}$. Then,
$y^{\prime}$ is the column in which the cell $\mathbf{c}$ lies. Since this
column is one of the columns $j,j+1,j+2,...$ (because $\mathbf{c}$ is a cell
of $T|_{ \operatorname{cols} \geq j}$), this yields that $y^{\prime}%
\in\left\{  j,j+1,j+2,...\right\}  $, so that $y^{\prime}\geq j$.

If $x^{\prime}\geq x$, then the cell $\left(  x^{\prime},y^{\prime}\right)  $
lies southeast of the cell $\left(  x,j\right)  $ (since $x^{\prime}\geq x$
and $y^{\prime}\geq j$). Since $\left(  x^{\prime},y^{\prime}\right)
=\mathbf{c}$ and $\left(  x,j\right)  =\mathbf{d}$, this rewrites as follows:
If $x^{\prime}\geq x$, then the cell $\mathbf{c}$ lies southeast of the cell
$\mathbf{d}$. Hence, if $x^{\prime}\geq x$, then
(\ref{sol.liri.yamanouchi.kappa.b.SE}) yields
\[
\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  \geq\left(
\text{the entry of }T\text{ in cell }\mathbf{d}\right)  =i+1,
\]
which contradicts $\left(  \text{the entry of }T\text{ in cell }%
\mathbf{c}\right)  =i<i+1$. Therefore, we cannot have $x^{\prime}\geq x$. We
thus have $x^{\prime}<x$.

Since $x^{\prime}<x$ and $y^{\prime}\geq j$, the cell $\left(  x^{\prime
},y^{\prime}\right)  $ lies northeast of the cell $\left(  x,j\right)  $. In
other words, the cell $\mathbf{c}$ lies northeast of the cell $\mathbf{d}$
(since $\left(  x^{\prime},y^{\prime}\right)  =\mathbf{c}$ and $\left(
x,j\right)  =\mathbf{d}$). Since $\mathbf{c}$ is a cell of $T$ and since
$\mathbf{d}\in S$, this yields that $\mathbf{c}\in S$ as well (since $S$ is an NE-set).

Now, forget that we fixed $\mathbf{c}$. We thus have shown that if
$\mathbf{c}$ is a cell of $T|_{\operatorname{cols}\geq j}$ such that the entry
of $T$ in $\mathbf{c}$ equals $i$, then $\mathbf{c}\in S$. Hence, the set%
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i\right\}
\]
is a subset of the set%
\[
\left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{ in }%
\mathbf{c}\text{ equals }i\right\}  .
\]
As a consequence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\in S\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{S}\right)  =\left(
\cont \left(  T|_{S}\right)  \right)  _{i}.
\end{align*}
Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  =\left\vert \left(  T|_{\operatorname{cols}\geq j}\right)
^{-1}\left(  i\right)  \right\vert =\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}%
\end{align*}
(because the definition of $\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  $ shows that $\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}=\left\vert \left(
T|_{\operatorname{cols}\geq j}\right)  ^{-1}\left(  i\right)  \right\vert $),
this rewrites as $\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\leq\left(  \cont \left(
T|_{S}\right)  \right)  _{i}$. In other words, $\left(  \cont
\left(  T|_{S}\right)  \right)  _{i}\geq\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}$. Thus,%
\begin{align*}
\left(  \kappa+\cont \left(  T|_{S}\right)  \right)  _{i}  &
=\kappa_{i}+\underbrace{\left(  \cont \left(  T|_{S}\right)
\right)  _{i}}_{\geq\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}}\geq\kappa_{i}+\left(  \cont
\left(  T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\\
&  =\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\\
&  =\kappa_{i+1}+\underbrace{\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}}_{\geq\left(
\cont \left(  T|_{S}\right)  \right)  _{i+1}}\geq\kappa
_{i+1}+\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}\\
&  =\left(  \kappa+\cont \left(  T|_{S}\right)  \right)  _{i+1},
\end{align*}
which contradicts $\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i}<\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i+1}$. This contradiction shows that our assumption was wrong.
Hence, $\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i+1}$.

Now, forget that we fixed $S$. We thus have proven that for every NE-set $S$
of $T$, we have $\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i}\geq\left(  \kappa+\cont \left(  T|_{S}\right)
\right)  _{i+1}$. Thus, Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$
holds. This proves the implication $\mathcal{C}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$ holds. Let $v$ be a
prefix of the Semitic reading word of $T$. We shall prove that%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}


For every $i\in\left\{  1,2,3,...\right\}  $, let $r_{i}$ be the word obtained
by reading the $i$-th row of $T$ from right to left. Then, the Semitic reading
word of $T$ is the concatenation $r_{1}r_{2}r_{3}\cdots$ (according to its
definition). Hence, every prefix of this Semitic reading word must have the
form $r_{1}r_{2}\cdots r_{k}s$ for some $k\in\left\{  0,1,2,...\right\}  $ and
some prefix $s$ of $r_{k+1}$. In particular, $v$ must have this form (since
$v$ is a prefix of the Semitic reading word of $T$). In other words, there
exists some $k\in\left\{  0,1,2,...\right\}  $ and some prefix $s$ of
$r_{k+1}$ such that $v=r_{1}r_{2}\cdots r_{k}s$. Consider this $k$ and this
$s$.

Let $\ell$ be the length of $s$. Since $s$ is a prefix of $r_{k+1}$ and has
length $\ell$, it is evident that the word $s$ consists of the first $\ell$
letters of $r_{k+1}$. These first $\ell$ letters of $r_{k+1}$ are the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$ (since
$r_{k+1}$ is the word obtained by reading the $\left(  k+1\right)  $-st row of
$T$ from right to left). Thus, the word $s$ consists of the rightmost $\ell$
entries of the $\left(  k+1\right)  $-st row of $T$. Hence, the word
$r_{1}r_{2}\cdots r_{k}s$ consists of all entries of the first $k$ rows of $T$
and the rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$.
In other words, the letters of the word $r_{1}r_{2}\cdots r_{k}s$ are
precisely all entries of the first $k$ rows of $T$ and the rightmost $\ell$
entries of the $\left(  k+1\right)  $-st row of $T$.

Let $S$ be the set which consists of all cells of the first $k$ rows of $T$
and the rightmost $\ell$ cells of the $\left(  k+1\right)  $-st row of $T$.
Then, $S$ is an NE-set of $T$, and therefore we have $\left(  \kappa
+\cont \left(  T|_{S}\right)  \right)  _{i}\geq\left(
\kappa+\cont \left(  T|_{S}\right)  \right)  _{i+1}$ (by
Assertion $\mathcal{E}_{i}^{\left(  \kappa\right)  }$).

Now, $\left(  \cont \left(  T|_{S}\right)  \right)
_{i}=\left\vert \left(  T|_{S}\right)  ^{-1}\left(  i\right)  \right\vert $
(by the definition of $\cont \left(  T|_{S}\right)  $), so that
$\left(  \cont \left(  T|_{S}\right)  \right)  _{i}$ is the
number of entries $i$ in $T|_{S}$. In other words, $\left(
\cont \left(  T|_{S}\right)  \right)  _{i}$ is the number of
$i$'s among the entries of $T|_{S}$. But since the entries of $T|_{S}$ are
precisely the letters of $v$\ \ \ \ \footnote{\textit{Proof.} The set $S$
consists of all cells of the first $k$ rows of $T$ and the rightmost $\ell$
cells of the $\left(  k+1\right)  $-st row of $T$. Hence, the entries of
$T|_{S}$ are precisely all entries of the first $k$ rows of $T$ and the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$.
Comparing this with the fact that the letters of the word $r_{1}r_{2}\cdots
r_{k}s$ are precisely all entries of the first $k$ rows of $T$ and the
rightmost $\ell$ entries of the $\left(  k+1\right)  $-st row of $T$, we
conclude the following: The entries of $T|_{S}$ are precisely the letters of
$r_{1}r_{2}\cdots r_{k}s$. In other words, the entries of $T|_{S}$ are
precisely the letters of $v$ (since $v=r_{1}r_{2}\cdots r_{k}s$), qed.}, this
rewrites as follows: $\left(  \cont \left(  T|_{S}\right)
\right)  _{i}$ is the number of $i$'s among the letters of $v$. So we have%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i}=\left(
\text{the number of }i\text{'s among the letters of }v\right)  .
\]
The same argument, with $i+1$ in place of $i$, shows that%
\[
\left(  \cont \left(  T|_{S}\right)  \right)  _{i+1}=\left(
\text{the number of }\left(  i+1\right)  \text{'s among the letters of
}v\right)  .
\]
Hence,%
\begin{align*}
&  \kappa_{i}+\underbrace{\left(  \text{the number of }i\text{'s among the
letters of }v\right)  }_{=\left(  \cont \left(  T|_{S}\right)
\right)  _{i}}\\
&  =\kappa_{i}+\left(  \cont \left(  T|_{S}\right)  \right)
_{i}=\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{S}\right)  \right)
_{i+1}\\
&  =\kappa_{i+1}+\underbrace{\left(  \cont \left(  T|_{S}\right)
\right)  _{i+1}}_{=\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  }\\
&  =\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}


Now, forget that we fixed $v$. We thus have shown that for every prefix $v$ of
the Semitic reading word of $T$, we have%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}v\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }v\right)  .
\end{align*}
In other words, Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$ holds.
We thus have shown the implication $\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$ holds.

For every $i\in\left\{  1,2,3,...\right\}  $, let $r_{i}$ be the word obtained
by reading the $i$-th row of $T$ from right to left. Then, the Semitic reading
word of $T$ is the concatenation $r_{1}r_{2}r_{3}\cdots$ (according to its definition).

Now, let $j$ be a positive integer. The word $r_{j}$ is the word obtained by
reading the $j$-th row of $T$ from right to left. Hence, the letters of this
word $r_{j}$ are in (weakly) decreasing order (since the entries of $T$
increase weakly left-to-right in rows). Thus, the subword of $r_{j}$
consisting of all letters $>i$ in $r_{j}$ is a prefix of $r_{j}$. Denote this
prefix by $s$. By the definition of $s$, all letters of $s$ are $>i$, so that%
\[
\left(  \text{the number of }i\text{'s among the letters of }s\right)  =0.
\]
Also, by the definition of $s$, the word $s$ consists of all letters $>i$ in
$r_{j}$. As a consequence,%
\[
\left(  \text{the number of }\left(  i+1\right)  \text{'s among the letters of
}s\right)  =\left(  \text{the number of }\left(  i+1\right)  \text{'s among
the letters of }r_{j}\right)  .
\]


Since $s$ is a prefix of $r_{j}$, it is clear that the word $r_{1}r_{2}\cdots
r_{j-1}s$ is a prefix of the word $r_{1}r_{2}r_{3}\cdots$. In other words, the
word $r_{1}r_{2}\cdots r_{j-1}s$ is a prefix of the Semitic reading word of
$T$ (since the Semitic reading word of $T$ is the concatenation $r_{1}%
r_{2}r_{3}\cdots$). Hence, we have%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of }i\text{'s among the letters of
}r_{1}r_{2}\cdots r_{j-1}s\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{1}r_{2}\cdots r_{j-1}s\right)  .
\end{align*}
(by Assertion $\mathcal{F}_{i}^{\left(  \kappa\right)  }$, applied to
$v=r_{1}r_{2}\cdots r_{j-1}s$). Since
\begin{align*}
&  \left(  \text{the number of }i\text{'s among the letters of }r_{1}%
r_{2}\cdots r_{j-1}s\right) \\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }i\text{'s among the letters
of }r_{k}\right)  +\underbrace{\left(  \text{the number of }i\text{'s among
the letters of }s\right)  }_{=0}\\
&  =\sum_{k=1}^{j-1}\underbrace{\left(  \text{the number of }i\text{'s among
the letters of }r_{k}\right)  }_{\substack{=\left(  \text{the number of
entries }i\text{ in the }k\text{-th row of }T\right)  \\\text{(since }%
r_{k}\text{ is the word obtained by reading the }k\text{-th row of }T\text{
from right to left)}}}\\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of entries }i\text{ in the
}k\text{-th row of }T\right) \\
&  =\left(  \text{the number of entries }i\text{ in the first }j-1\text{ rows
of }T\right)
\end{align*}
and%
\begin{align*}
&  \left(  \text{the number of }\left(  i+1\right)  \text{'s among the letters
of }r_{1}r_{2}\cdots r_{j-1}s\right) \\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{k}\right)  +\underbrace{\left(  \text{the number of
}\left(  i+1\right)  \text{'s among the letters of }s\right)  }_{=\left(
\text{the number of }\left(  i+1\right)  \text{'s among the letters of }%
r_{j}\right)  }\\
&  =\sum_{k=1}^{j-1}\left(  \text{the number of }\left(  i+1\right)  \text{'s
among the letters of }r_{k}\right)  +\left(  \text{the number of }\left(
i+1\right)  \text{'s among the letters of }r_{j}\right) \\
&  =\sum_{k=1}^{j}\underbrace{\left(  \text{the number of }\left(  i+1\right)
\text{'s among the letters of }r_{k}\right)  }_{\substack{=\left(  \text{the
number of entries }i+1\text{ in the }k\text{-th row of }T\right)
\\\text{(since }r_{k}\text{ is the word obtained by reading the }k\text{-th
row of }T\text{ from right to left)}}}\\
&  =\sum_{k=1}^{j}\left(  \text{the number of entries }i+1\text{ in the
}k\text{-th row of }T\right) \\
&  =\left(  \text{the number of entries }i+1\text{ in the first }j\text{ rows
of }T\right)  ,
\end{align*}
this rewrites as follows:%
\begin{align*}
&  \kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right) \\
&  \geq\kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the
first }j\text{ rows of }T\right)  .
\end{align*}
In other words,%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}


Now, forget that we fixed $j$. We thus have shown that for every positive
integer $j$, we have%
\begin{align*}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}j\text{ rows of }T\right) \\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}j-1\text{ rows of }T\right)  .
\end{align*}
In other words, Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$ holds.
Thus, we have shown the implication $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{D}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{C}_{i}^{\left(  \kappa\right)  }$:} Assume
that Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$ holds. We want to
prove that Assertion $\mathcal{C}_{i}^{\left(  \kappa\right)  }$ holds. In
other words, we want to prove that, for every positive integer $j$, we have
$\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$.

In order to prove this, we assume the contrary. That is, there exists a
positive integer $j$ such that we don't have $\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}$. Consider the \textbf{minimal} such $j$. Then, we
don't have $\left(  \kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$; but
\begin{equation}
\text{for every }k\in\left\{  1,2,...,j-1\right\}  \text{, we do have }\left(
\kappa+\cont \left(  T|_{\operatorname{cols}\geq k}\right)
\right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq k}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.kappa.b.DC.notearlier.general}%
\end{equation}
Applying (\ref{sol.liri.yamanouchi.kappa.b.DC.notearlier.general}) to $k=j-1$,
we obtain:
\begin{equation}
\text{if }j-1\text{ is positive, then we do have }\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j-1}\right)  \right)
_{i}\geq\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.kappa.b.DC.notearlier}%
\end{equation}


We don't have $\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\geq\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}$. Thus,
\begin{equation}
\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}<\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}.
\label{sol.liri.yamanouchi.kappa.b.DC.ineq1}%
\end{equation}
In other words, $\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}>\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}$. Thus, $\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}>\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}$%
\ \ \ \ \footnote{\textit{Proof.} Recall that $\kappa$ is a partition. Thus,
$\kappa_{1}\geq\kappa_{2}\geq\kappa_{3}\geq\cdots$. In other words,
$\kappa_{u}\geq\kappa_{u+1}$ for every $u\in\left\{  1,2,3,\ldots\right\}  $.
Applying this to $u=i$, we obtain $\kappa_{i}\geq\kappa_{i+1}$. Now,
\begin{align*}
\kappa_{i+1}+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}  &  =\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\\
&  >\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}=\underbrace{\kappa_{i}}_{\geq\kappa_{i+1}}+\left(
\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}\geq\kappa_{i+1}+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}.
\end{align*}
Subtracting $\kappa_{i+1}$ from both sides of this equality, we obtain
$\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}>\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}$, qed.}.

Recall that
\begin{equation}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i}=\left(  \text{the number of entries }i\text{ in }%
T|_{\operatorname{cols}\geq j}\right)
\label{sol.liri.yamanouchi.kappa.b.DC.ji}%
\end{equation}
and%
\[
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}=\left(  \text{the number of entries }i+1\text{ in
}T|_{\operatorname{cols}\geq j}\right)  .
\]
Hence,
\begin{align*}
\left(  \text{the number of entries }i+1\text{ in }T|_{\operatorname{cols}\geq
j}\right)   &  =\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}>\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\\
&  =\left(  \text{the number of entries }i\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  \geq0.
\end{align*}
Hence, there exists at least one entry $i+1$ in $T|_{\operatorname{cols}\geq
j}$. In other words, there exists at least one cell $\mathbf{c}$ of
$T|_{\operatorname{cols}\geq j}$ such that the entry of
$T|_{\operatorname{cols}\geq j}$ in $\mathbf{c}$ equals $i+1$. In other words,
there exists at least one cell $\mathbf{c}$ of $T|_{\operatorname{cols}\geq
j}$ such that the entry of $T$ in $\mathbf{c}$ equals $i+1$ (since the entry
of $T|_{\operatorname{cols}\geq j}$ in $\mathbf{c}$, when it is defined,
equals the entry of $T$ in $\mathbf{c}$). Let $r$ be the \textbf{bottommost}
row which contains such a cell $\mathbf{c}$, and let $\left(  x,y\right)  $ be
the \textbf{leftmost} such cell $\mathbf{c}$ on this row. Thus, $\left(
x,y\right)  $ is a cell of $T|_{\operatorname{cols}\geq j}$ such that the
entry of $T$ in $\left(  x,y\right)  $ equals $i+1$. Also, the cell $\left(
x,y\right)  $ lies in the $r$-th row; that is, $r=x$.

The cell $\left(  x,y\right)  $ must lie in one of the columns $j,j+1,j+2,...$
(since it is a cell of $T|_{\operatorname{cols}\geq j}$), so that we have
$y\geq j$. Also, $\left(  x,y\right)  $ is a cell of $T$ (since it is a cell
of $T|_{\operatorname{cols}\geq j}$).

Applying Assertion $\mathcal{D}_{i}^{\left(  \kappa\right)  }$ to $x$ instead
of $j$, we see that
\begin{align}
&  \kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the first
}x\text{ rows of }T\right) \nonumber\\
&  \leq\kappa_{i}+\left(  \text{the number of entries }i\text{ in the first
}x-1\text{ rows of }T\right)  . \label{sol.liri.yamanouchi.kappa.b.DC.ineq0}%
\end{align}


It is easy to see that%
\begin{equation}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}\leq\left(  \text{the number of entries }i+1\text{ in the first
}x\text{ rows of }T\right)  \label{sol.liri.yamanouchi.kappa.b.DC.ineq2}%
\end{equation}
\footnote{\textit{Proof.} Let $\mathbf{c}$ be a cell of
$T|_{\operatorname{cols}\geq j}$ such that the entry of $T$ in $\mathbf{c}$
equals $i+1$. Since $r$ is the bottommost row which contains such a cell
$\mathbf{c}$, it is clear that the cell $\mathbf{c}$ must be in the $r$-th row
or in a row further north. In other words, the cell $\mathbf{c}$ must lie in
one of the first $r$ rows of $T$. Since $r=x$, this rewrites as follows: The
cell $\mathbf{c}$ must lie in one of the first $x$ rows of $T$.
\par
Now, let us forget that we fixed $\mathbf{c}$. We thus have proven that if
$\mathbf{c}$ is a cell of $T|_{\operatorname{cols}\geq j}$ such that the entry
of $T$ in $\mathbf{c}$ equals $i+1$, then the cell $\mathbf{c}$ must lie in
one of the first $x$ rows of $T$. Hence, the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i+1\right\}
\]
is a subset of the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i+1;\ \text{the cell }\mathbf{c}\text{ lies in
one of the first }x\text{ rows of }T\right\}  .
\]
Hence,%
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i+1\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the
entry of }T\text{ in }\mathbf{c}\text{ equals }i+1;\ \text{the cell
}\mathbf{c}\text{ lies in one of the first }x\text{ rows of }T\right\}
\right\vert \\
&  =\left(  \text{the number of entries }i+1\text{ in the first }x\text{ rows
of }T\right)  .
\end{align*}
But
\begin{align*}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i+1}  &  =\left(  \text{the number of entries }i+1\text{ in
}T|_{\operatorname{cols}\geq j}\right) \\
&  =\left\vert \left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}%
\geq j}\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals
}i+1\right\}  \right\vert \\
&  \leq\left(  \text{the number of entries }i+1\text{ in the first }x\text{
rows of }T\right)  ,
\end{align*}
qed.}.

We shall now show that
\begin{equation}
\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows of
}T\right)  \leq\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}. \label{sol.liri.yamanouchi.kappa.b.DC.ineq3}%
\end{equation}


Indeed, let $\mathbf{c}$ be a cell of $T$ such that the entry of $T$ in
$\mathbf{c}$ equals $i$ and such that $\mathbf{c}$ lies in one of the first
$x-1$ rows of $T$. We shall show that $\mathbf{c}$ is a cell of
$T|_{\operatorname{cols}\geq j}$.

Assume the contrary. Thus, $\mathbf{c}$ is not a cell of
$T|_{\operatorname{cols}\geq j}$.

Let us write the cell $\mathbf{c}$ as $\mathbf{c}=\left(  x^{\prime}%
,y^{\prime}\right)  $ for some integers $x^{\prime}$ and $y^{\prime}$. Then,
the cell $\mathbf{c}$ lies in row $x^{\prime}$, and thus this row $x^{\prime}$
must be one of the first $x-1$ rows of $T$ (since we know that $\mathbf{c}$
lies in one of the first $x-1$ rows of $T$). In other words, $x^{\prime}\leq
x-1$. Hence, $x^{\prime}\leq x-1<x$ and thus $x>x^{\prime}$ and $x\neq
x^{\prime}$ and $x\geq x^{\prime}$.

The cell $\mathbf{c}$ lies in the $y^{\prime}$-th column (since $\mathbf{c}%
=\left(  x^{\prime},y^{\prime}\right)  $). Hence, if $y^{\prime}\geq j$, then
the cell $\mathbf{c}$ lies in one of the columns $j,j+1,j+2,...$ and therefore
is a cell of $T|_{\operatorname{cols}\geq j}$, which contradicts our
assumption that $\mathbf{c}$ is not a cell of $T|_{\operatorname{cols}\geq j}%
$. Thus, we cannot have $y^{\prime}\geq j$. We therefore have $y^{\prime}<j$.
That is, $y^{\prime}\leq j-1$ (since $y^{\prime}$ and $j$ are integers), so
that $j-1\geq y^{\prime}$.

Now, the cell $\left(  x,y\right)  $ lies southeast of the cell $\left(
x,j-1\right)  $ (since $x\geq x$ and $y\geq j\geq j-1$), which in turn lies
southeast of the cell $\left(  x^{\prime},y^{\prime}\right)  $ (since $x\geq
x^{\prime}$ and $j-1\geq y^{\prime}$). Since both cells $\left(  x,y\right)  $
and $\left(  x^{\prime},y^{\prime}\right)  $ are cells of $T$ (indeed,
$\left(  x,y\right)  $ is known to be a cell of $T$, and $\left(  x^{\prime
},y^{\prime}\right)  =\mathbf{c}$ also is a cell of $T$), this yields that the
intermediate cell $\left(  x,j-1\right)  $ is also a cell of $T$%
\ \ \ \ \footnote{Here, we are using the following fact: If $\alpha$, $\beta$
and $\gamma$ are three cells such that $\alpha$ lies southeast of the cell
$\beta$, which in turn lies southeast of the cell $\gamma$, and if $\alpha$
and $\gamma$ are cells of $T$, then $\beta$ is also a cell of $T$. This can be
easily derived from the fact that $T$ has shape $\lambda/\mu$.}. Thus, $j-1$
is a positive integer. Moreover, since the cell $\left(  x,y\right)  $ lies
southeast of the cell $\left(  x,j-1\right)  $, we can apply
(\ref{sol.liri.yamanouchi.kappa.b.SE}) to $\left(  x,y\right)  $ and $\left(
x,j-1\right)  $ instead of $\mathbf{c}$ and $\mathbf{d}$. We thus obtain
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,y\right)  \right)
\geq\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)
\right)  .
\]
Since $\left(  \text{the entry of }T\text{ in cell }\left(  x,y\right)
\right)  =i+1$, this rewrites as
\begin{equation}
i+1\geq\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)
\right)  . \label{sol.liri.yamanouchi.kappa.b.DC.5}%
\end{equation}


But the cell $\left(  x,j-1\right)  $ lies southeast of the cell $\left(
x^{\prime},y^{\prime}\right)  $ and not on the same row as $\left(  x^{\prime
},y^{\prime}\right)  $ (since $x\neq x^{\prime}$). Since $\left(  x^{\prime
},y^{\prime}\right)  =\mathbf{c}$, this rewrites as follows: The cell $\left(
x,j-1\right)  $ lies southeast of the cell $\mathbf{c}$ and not on the same
row as $\mathbf{c}$. Hence, (\ref{sol.liri.yamanouchi.kappa.b.SEstrict})
(applied to $\left(  x,j-1\right)  $ and $\mathbf{c}$ instead of $\mathbf{c}$
and $\mathbf{d}$) yields
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
>\left(  \text{the entry of }T\text{ in cell }\mathbf{c}\right)  =i,
\]
so that%
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
\geq i+1
\]
(since the entry of $T$ in cell $\left(  x,j-1\right)  $ and the number $i$
are integers). Combined with (\ref{sol.liri.yamanouchi.kappa.b.DC.5}), this
yields
\[
\left(  \text{the entry of }T\text{ in cell }\left(  x,j-1\right)  \right)
=i+1.
\]
Thus, the number $i+1$ appears in the $\left(  j-1\right)  $-th column of $T$
(since the cell $\left(  x,j-1\right)  $ lies in the $\left(  j-1\right)  $-th
column of $T$). In other words,
\begin{equation}
\left(  \text{the number of entries }i+1\text{ in the }\left(  j-1\right)
\text{-th column of }T\right)  \geq1 .
\label{sol.liri.yamanouchi.kappa.b.DC.7i+1}%
\end{equation}


On the other hand, the entries of $T$ increase strictly top-to-bottom along
columns. In particular, in every given column of $T$, the entries are
distinct. Applied to the $\left(  j-1\right)  $-th column, this shows that the
entries of the $\left(  j-1\right)  $-th column of $T$ are distinct.
In particular, for every $k\in\left\{  1,2,3,...\right\}  $, the number $k$
appears at most once in the $\left(  j-1\right)  $-th column of $T$. Applying
this to $k=i$, we conclude that the number $i$ appears at most once in the
$\left(  j-1\right)  $-th column of $T$. That is,
\begin{equation}
\left(  \text{the number of entries }i\text{ in the }\left(  j-1\right)
\text{-th column of }T\right)  \leq1.
\label{sol.liri.yamanouchi.kappa.b.DC.7i}%
\end{equation}


Now, for every $k\in\left\{  1,2,3,...\right\}  $ and $h\in\left\{
1,2,3,...\right\}  $, we have
\begin{align}
\left(  \cont \left(  T|_{\operatorname{cols}\geq k}\right)
\right)  _{h}  &  =\left\vert \left(  T|_{\operatorname{cols}\geq k}\right)
^{-1}\left(  h\right)  \right\vert =\left(  \text{the number of entries
}h\text{ in }T|_{\operatorname{cols}\geq k}\right) \nonumber\\
&  =\left(  \text{the number of entries }h\text{ in the columns }%
k,k+1,k+2,...\text{ of }T\right)  .
\label{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}%
\end{align}
Applying this to $k=j$ and $h=i$, we obtain%
\begin{equation}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i}=\left(  \text{the number of entries }i\text{ in the columns
}j,j+1,j+2,...\text{ of }T\right)  .
\label{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.1}%
\end{equation}
But applying
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}) to
$k=j-1$ and $h=i$, we obtain%
\begin{align*}
\left(  \cont \left(  T|_{\operatorname{cols}\geq j-1}\right)
\right)  _{i}  &  =\left(  \text{the number of entries }i\text{ in the columns
}j-1,j,j+1,...\text{ of }T\right) \\
&  =\underbrace{\left(  \text{the number of entries }i\text{ in the }\left(
j-1\right)  \text{-th column of }T\right)  }_{\substack{\leq1\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.7i}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries
}i\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)  }%
_{\substack{=\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.1}))}}}\\
&  \leq1+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}.
\end{align*}
Thus,%
\begin{align*}
\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}  &  =\kappa_{i}+\underbrace{\left(
\cont \left(  T|_{\operatorname{cols}\geq j-1}\right)  \right)
_{i}}_{\leq1+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}}\leq\kappa_{i}+1+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}\\
&  =1+\underbrace{\kappa_{i}+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}}_{\substack{=\left(
\kappa+\cont \left(  T|_{\operatorname{cols}\geq j}\right)
\right)  _{i}<\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.ineq1}))}}}\\
&  <1+\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}.
\end{align*}
Since $\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}$ and $1+\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}$ are integers, this
yields
\[
\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}\leq\left(  1+\left(  \kappa+\cont
\left(  T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}\right)
-1=\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1},
\]
so that%
\begin{align*}
&  \left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\\
&  \geq\left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j-1}\right)  \right)  _{i}\geq\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j-1}\right)  \right)  _{i+1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.notearlier})}\right) \\
&  =\kappa_{i+1}+\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j-1}\right)  \right)  _{i+1}.
\end{align*}
Since%
\begin{align*}
&  \left(  \cont \left(  T|_{\operatorname{cols}\geq j-1}\right)
\right)  _{i+1}\\
&  =\left(  \text{the number of entries }i+1\text{ in the columns
}j-1,j,j+1,...\text{ of }T\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}), applied
to }k=j-1\text{ and }h=i+1\right) \\
&  =\underbrace{\left(  \text{the number of entries }i+1\text{ in the }\left(
j-1\right)  \text{-th column of }T\right)  }_{\substack{\geq1\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.7i+1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of entries
}i+1\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)  }%
_{\substack{=\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\\\text{(since }\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}=\left(  \text{the
number of entries }i+1\text{ in the columns }j,j+1,j+2,...\text{ of }T\right)
\\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.DC.increase-only-slightly.general}), applied
to }k=j\text{ and }h=i+1\text{))}}}\\
&  \geq1+\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}>\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1},
\end{align*}
this becomes%
\begin{align*}
&  \left(  \kappa+\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i+1}\\
&  \geq\kappa_{i+1}+\underbrace{\left(  \cont \left(
T|_{\operatorname{cols}\geq j-1}\right)  \right)  _{i+1}}_{>\left(
\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}}>\kappa_{i+1}+\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}=\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}.
\end{align*}
which is absurd. This contradiction proves that our assumption was wrong, and
thus we have proven that $\mathbf{c}$ is a cell of $T|_{\operatorname{cols}%
\geq j}$.

Now, let us forget that we fixed $\mathbf{c}$. We thus have proven that if
$\mathbf{c}$ is a cell of $T$ such that the entry of $T$ in $\mathbf{c}$
equals $i$ and such that $\mathbf{c}$ lies in one of the first $x-1$ rows of
$T$, then $\mathbf{c}$ is a cell of $T|_{\operatorname{cols}\geq j}$. Hence,
the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry of }T\text{
in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{ lies in one
of the first }x-1\text{ rows of }T\right\}
\]
is a subset of the set
\[
\left\{  \mathbf{c}\text{ is a cell of }T|_{\operatorname{cols}\geq j}%
\ \mid\ \text{the entry of }T\text{ in }\mathbf{c}\text{ equals }i\right\}  .
\]
Hence,
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry
of }T\text{ in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{
lies in one of the first }x-1\text{ rows of }T\right\}  \right\vert \\
&  \leq\left\vert \left\{  \mathbf{c}\text{ is a cell of }%
T|_{\operatorname{cols}\geq j}\ \mid\ \text{the entry of }T\text{ in
}\mathbf{c}\text{ equals }i\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in }T|_{\operatorname{cols}%
\geq j}\right)  =\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.ji})}\right)  .
\end{align*}
Since
\begin{align*}
&  \left\vert \left\{  \mathbf{c}\text{ is a cell of }T\ \mid\ \text{the entry
of }T\text{ in }\mathbf{c}\text{ equals }i\text{; the cell }\mathbf{c}\text{
lies in one of the first }x-1\text{ rows of }T\right\}  \right\vert \\
&  =\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows
of }T\right)  ,
\end{align*}
this rewrites as
\[
\left(  \text{the number of entries }i\text{ in the first }x-1\text{ rows of
}T\right)  \leq\left(  \cont \left(  T|_{\operatorname{cols}\geq
j}\right)  \right)  _{i}.
\]
Thus, (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq3}) is proven. Now,%
\begin{align*}
&  \kappa_{i}+\underbrace{\left(  \text{the number of entries }i\text{ in the
first }x-1\text{ rows of }T\right)  }_{\substack{\leq\left(
\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i}\\\text{(by (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq3}))}}}\\
&  \leq\kappa_{i}+\left(  \cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \right)  _{i}=\left(  \kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i}<\left(  \kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  \right)
_{i+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.yamanouchi.kappa.b.DC.ineq1})}\right) \\
&  =\kappa_{i+1}+\underbrace{\left(  \cont \left(
T|_{\operatorname{cols}\geq j}\right)  \right)  _{i+1}}_{\substack{\leq\left(
\text{the number of entries }i+1\text{ in the first }x\text{ rows of
}T\right)  \\\text{(by (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq2}))}}}\\
&  \leq\kappa_{i+1}+\left(  \text{the number of entries }i+1\text{ in the
first }x\text{ rows of }T\right)  ,
\end{align*}
which contradicts (\ref{sol.liri.yamanouchi.kappa.b.DC.ineq0}). This
contradiction proves that our assumption was wrong. Hence, Assertion
$\mathcal{C}_{i}^{\left(  \kappa\right)  }$ must hold. We thus have proven the
implication $\mathcal{D}_{i}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{C}_{i}^{\left(  \kappa\right)  }$.

We have now proven the four implications $\mathcal{C}_{i}^{\left(
\kappa\right)  }\Longrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)  }$,
$\mathcal{E}_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{F}%
_{i}^{\left(  \kappa\right)  }$, $\mathcal{F}_{i}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{D}_{i}^{\left(  \kappa\right)  }$ and $\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{C}_{i}^{\left(
\kappa\right)  }$. Combining them, we obtain the equivalence $\mathcal{C}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$.

Now, let us forget that we fixed $i$. We thus have proven the equivalence
$\mathcal{C}_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}%
_{i}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{E}_{i}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{F}_{i}^{\left(  \kappa\right)  }$
for every $i\in\left\{  1,2,3,...\right\}  $. As we know, this shows that we
have the equivalence $\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{D}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{E}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{F}^{\left(  \kappa\right)  }$, and
this finishes the solution of Exercise \ref{exe.liri.yamanouchi.kappa}(a).

(b) Exercise \ref{exe.liri.yamanouchi.kappa}(a) yields the equivalence
$\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{G}^{\left(  \kappa\right)  }$. We shall now prove the equivalence
$\mathcal{G}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{H}^{\left(
\kappa\right)  }$. In order to do so, we will prove the implications
$\mathcal{G}^{\left(  \kappa\right)  }\Longrightarrow\mathcal{H}^{\left(
\kappa\right)  }$ and $\mathcal{H}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{G}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{G}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{H}^{\left(  \kappa\right)  }$:} Assume that
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$ holds. We want to prove that
Assertion $\mathcal{H}^{\left(  \kappa\right)  }$ holds.

We have assumed that Assertion $\mathcal{G}^{\left(  \kappa\right)  }$ holds.
In other words, there exist a partition $\zeta$ and a column-strict tableau
$S$ of shape $\zeta/\kappa$ which satisfies the following property: For any
positive integers $i$ and $j$, the number of entries $i$ in the $j$-th row of
$T$ equals the number of entries $j$ in the $i$-th row of $S$. Consider this
$\zeta$ and this $S$.

We know that, for any positive integers $i$ and $j$, the number of entries $i$
in the $j$-th row of $T$ equals the number of entries $j$ in the $i$-th row of
$S$. In other words, for any positive integers $i$ and $j$, we have%
\begin{align}
&  \left(  \text{the number of entries }i\text{ in the }j\text{-th row of
}T\right) \nonumber\\
&  =\left(  \text{the number of entries }j\text{ in the }i\text{-th row of
}S\right)  . \label{sol.liri.yamanouchi.kappa.b.GK.1}%
\end{align}


Now, for every $i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\begin{align*}
\left(  \cont T\right)  _{i}  &  =\left\vert T^{-1}\left(
i\right)  \right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\cont T\right) \\
&  =\left(  \text{the number of entries }i\text{ in }T\right) \\
&  =\sum_{j=1}^{\infty}\underbrace{\left(  \text{the number of entries
}i\text{ in the }j\text{-th row of }T\right)  }_{\substack{=\left(  \text{the
number of entries }j\text{ in the }i\text{-th row of }S\right)  \\\text{(by
(\ref{sol.liri.yamanouchi.kappa.b.GK.1}))}}}\\
&  =\sum_{j=1}^{\infty}\left(  \text{the number of entries }j\text{ in the
}i\text{-th row of }S\right) \\
&  =\left(  \text{the number of entries in the }i\text{-th row of }S\right)
=\left(  \text{the length of the }i\text{-th row of }S\right) \\
&  =\left(  \text{the length of the }i\text{-th row of the skew partition
}\zeta/\kappa\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the tableau }S\text{ has shape
}\zeta/\kappa\right) \\
&  =\zeta_{i}-\kappa_{i}.
\end{align*}
Hence, for every $i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\[
\left(  \kappa+\cont T\right)  _{i}=\kappa_{i}%
+\underbrace{\left(  \cont T\right)  _{i}}_{=\zeta_{i}%
-\kappa_{i}}=\kappa_{i}+\left(  \zeta_{i}-\kappa_{i}\right)  =\zeta_{i}.
\]
In other words, $\kappa+\cont T=\zeta$, so that $\zeta
=\kappa+\cont T=\tau$. But $S$ is a column-strict tableau of
shape $\zeta/\kappa$. In other words, $S$ is a column-strict tableau of shape
$\tau/\kappa$ (since $\zeta=\tau$).

Now, let us forget that we defined $S$. We thus have constructed a
column-strict tableau $S$ of shape $\tau/\kappa$ such that for any positive
integers $i$ and $j$, the equality (\ref{sol.liri.yamanouchi.kappa.b.GK.1})
holds. Thus, there exists a column-strict tableau $S$ of shape $\tau/\kappa$
such that for any positive integers $i$ and $j$, the equality
(\ref{sol.liri.yamanouchi.kappa.b.GK.1}) holds. In other words, there exists a
column-strict tableau $S$ of shape $\tau/\kappa$ which satisfies the following
property: For any positive integers $i$ and $j$, the number of entries $i$ in
the $j$-th row of $T$ equals the number of entries $j$ in the $i$-th row of
$S$. In other words, Assertion $\mathcal{H}^{\left(  \kappa\right)  }$ holds.
We thus have proven the implication $\mathcal{G}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{H}^{\left(  \kappa\right)  }$.

\textit{Proof of the implication $\mathcal{H}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{G}^{\left(  \kappa\right)  }$:} Assume that
Assertion $\mathcal{H}^{\left(  \kappa\right)  }$ holds. We want to prove that
Assertion $\mathcal{G}^{\left(  \kappa\right)  }$ holds.

We have assumed that Assertion $\mathcal{H}^{\left(  \kappa\right)  }$ holds.
In other words, there exists a column-strict tableau $S$ of shape $\tau
/\kappa$ which satisfies the following property: For any positive integers $i$
and $j$, the number of entries $i$ in the $j$-th row of $T$ equals the number
of entries $j$ in the $i$-th row of $S$. Let us denote this $S$ by $S^{\prime
}$. Thus, $S^{\prime}$ is a column-strict tableau of shape $\tau/\kappa$ which
satisfies the following property: For any positive integers $i$ and $j$, the
number of entries $i$ in the $j$-th row of $T$ equals the number of entries
$j$ in the $i$-th row of $S^{\prime}$.

Now, there exist a partition $\zeta$ and a column-strict tableau $S$ of shape
$\zeta/\kappa$ which satisfies the following property: For any positive
integers $i$ and $j$, the number of entries $i$ in the $j$-th row of $T$
equals the number of entries $j$ in the $i$-th row of $S$. (Namely, we can
take $\zeta=\tau$ and $S=S^{\prime}$.) In other words, Assertion
$\mathcal{G}^{\left(  \kappa\right)  }$ holds. We thus have proven the
implication $\mathcal{H}^{\left(  \kappa\right)  }\Longrightarrow
\mathcal{G}^{\left(  \kappa\right)  }$.

Now, we have proven the implications $\mathcal{G}^{\left(  \kappa\right)
}\Longrightarrow\mathcal{H}^{\left(  \kappa\right)  }$ and $\mathcal{H}%
^{\left(  \kappa\right)  }\Longrightarrow\mathcal{G}^{\left(  \kappa\right)
}$. Combining these two implications, we obtain the equivalence $\mathcal{G}%
^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{H}^{\left(
\kappa\right)  }$. Combining this equivalence with the equivalence
$\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{G}^{\left(  \kappa\right)  }$, we obtain the equivalence
$\mathcal{C}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{D}^{\left(
\kappa\right)  }\Longleftrightarrow\mathcal{E}^{\left(  \kappa\right)
}\Longleftrightarrow\mathcal{F}^{\left(  \kappa\right)  }\Longleftrightarrow
\mathcal{G}^{\left(  \kappa\right)  }\Longleftrightarrow\mathcal{H}^{\left(
\kappa\right)  }$. This solves Exercise \ref{exe.liri.yamanouchi.kappa}(b).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.skew.1}}

\textit{Solution to Exercise \ref{exe.liri.skew.1}.} (a) If $T$ is a
column-strict tableau of shape $\lambda/\mu$, then we have the following chain
of logical equivalences:%
\begin{align}
&  \ \left(  \text{for all }j\in\left\{  1,2,3,\ldots\right\}  \text{, the
weak composition }\kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \text{ is a partition}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{Assertion }\mathcal{C}^{\left(
\kappa\right)  }\text{ holds}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{because this is how we defined Assertion }\mathcal{C}^{\left(
\kappa\right)  }\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the five equivalent assertions
}\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}^{\left(
\kappa\right)  }\text{ hold}\right)  \label{sol.liri.skew.1.a.1}%
\end{align}
(because the five assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ are equivalent (by Exercise \ref{exe.liri.yamanouchi.kappa}(a))).

We can apply (\ref{eq.infinite-L-R-rule.1}) to $\nu=\kappa$. As a result, we
see that%
\[
s_{\kappa}s_{\lambda/\mu}=\sum_{T}s_{\kappa+\cont T},
\]
where $T$ runs through all column-strict tableaux of shape $\lambda/\mu$ with
the property that for each $j=1,2,3,\ldots$, the weak composition $\kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  $ is a
partition. In other words,%
\begin{align}
s_{\kappa}s_{\lambda/\mu} &  =\underbrace{\sum\limits_{\substack{T\text{ is a
column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{for all
}j\in\left\{  1,2,3,\ldots\right\}  \text{, the weak}\\\text{composition
}\kappa+\cont \left(  T|_{\operatorname{cols}\geq j}\right)
\\\text{is a partition}}}}_{\substack{=\sum\limits_{\substack{T\text{ is a
column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{the five
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}}}\\\text{(because of the
equivalence (\ref{sol.liri.skew.1.a.1}))}}}s_{\kappa+\cont
T}\label{sol.liri.skew.1.a.3}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}s_{\kappa+\cont T}.\label{sol.liri.skew.1.a.4}%
\end{align}
In other words,%
\[
s_{\kappa}s_{\lambda/\mu}=\sum_{T}s_{\kappa+\cont T},
\]
where the sum ranges over all column-strict tableaux $T$ of shape $\lambda
/\mu$ satisfying the five equivalent assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}%
^{\left(  \kappa\right)  }$, $\mathcal{F}^{\left(  \kappa\right)  }$ and
$\mathcal{G}^{\left(  \kappa\right)  }$ introduced in
Exercise~\ref{exe.liri.yamanouchi.kappa}(a). This solves Exercise
\ref{exe.liri.skew.1}(a).

(b) If $T$ is a column-strict tableau of shape $\lambda/\mu$ such that the
five equivalent assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ hold, then we have%
\begin{equation}
\left(  s_{\kappa+\cont T},s_{\tau}\right)  _{\Lambda}%
=\delta_{\tau,\kappa+\cont T}\label{sol.liri.skew.1.b.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.skew.1.b.1}):} We know that the
basis $\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ of
$\Lambda$ is orthonormal with respect to the Hall inner product. In other
words, we have%
\begin{equation}
\left(  s_{\alpha},s_{\beta}\right)  _{\Lambda}=\delta_{\alpha,\beta
}\label{sol.liri.skew.1.b.1.pf.1}%
\end{equation}
for any $\alpha\in \Par $ and $\beta\in \Par $.
\par
Now, let $T$ be a column-strict tableau of shape $\lambda/\mu$ such that the
five equivalent assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ hold. Then, Assertion $\mathcal{C}^{\left(  \kappa\right)
}$ holds (since the five equivalent assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}%
^{\left(  \kappa\right)  }$, $\mathcal{F}^{\left(  \kappa\right)  }$ and
$\mathcal{G}^{\left(  \kappa\right)  }$ hold). In other words, for every
positive integer $j$, the weak composition $\kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  $ is a partition. Applying this to
$j=1$, we conclude that the weak composition $\kappa+\cont
\left(  T|_{\operatorname{cols}\geq1}\right)  $ is a partition.
\par
But $T|_{\operatorname{cols}\geq1}$ is the subtableau which is the restriction
of $T$ to the union of its columns $1,2,3,\ldots$. In other words,
$T|_{\operatorname{cols}\geq1}$ is the whole tableau $T$. In other words,
$T|_{\operatorname{cols}\geq1}=T$.
\par
Now, recall that $\kappa+\cont \left(  T|_{\operatorname{cols}%
\geq1}\right)  $ is a partition. In other words, $\kappa+\cont
T$ is a partition (since $T|_{\operatorname{cols}\geq1}=T$). In other words,
$\kappa+\cont T\in \Par $. Also, $\tau
\in \Par $ (since $\tau$ is a partition). Thus,
(\ref{sol.liri.skew.1.b.1.pf.1}) (applied to $\alpha=\kappa
+\cont T$ and $\beta=\tau$) shows that $\left(  s_{\kappa
+\cont T},s_{\tau}\right)  _{\Lambda}
= \delta_{\kappa + \cont T, \tau}
=\delta_{\tau
,\kappa+\cont T}$. This proves (\ref{sol.liri.skew.1.b.1}).}.

But if $T$ is a column-strict tableau of shape $\lambda/\mu$ satisfying
$\tau=\kappa+\cont T$, then we have the following chain of
logical equivalences:%
\begin{align}
&  \ \left(  \text{the five equivalent assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the six equivalent assertions
}\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{, }\mathcal{G}^{\left(
\kappa\right)  }\text{ and }\mathcal{H}^{\left(  \kappa\right)  }\text{
hold}\right)  \label{sol.liri.skew.1.b.2}%
\end{align}
(because the six assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$, $\mathcal{G}^{\left(
\kappa\right)  }$ and $\mathcal{H}^{\left(  \kappa\right)  }$ are equivalent
(by Exercise \ref{exe.liri.yamanouchi.kappa}(b))).

Now, recall that $s_{\mu}^{\perp}\left(  s_{\lambda}\right)  =s_{\lambda/\mu}%
$. Applying this to $\kappa$ and $\tau$ instead of $\mu$ and $\lambda$, we
obtain $s_{\kappa}^{\perp}\left(  s_{\tau}\right)  =s_{\tau/\kappa}$.

Now, Proposition \ref{skewing-properties-prop}(i) (applied to $A=\Lambda$)
shows that every $a\in\Lambda$, $f\in\Lambda$ and $g\in\Lambda$ satisfy
$\left(  g,f^{\perp}\left(  a\right)  \right)  _{\Lambda}=\left(  fg,a\right)
_{\Lambda}$. We can apply this to $f=s_{\kappa}$, $g=s_{\lambda/\mu}$ and
$a=s_{\tau}$. As a result, we obtain $\left(  s_{\lambda/\mu},s_{\kappa
}^{\perp}\left(  s_{\tau}\right)  \right)  _{\Lambda}=\left(  s_{\kappa
}s_{\lambda/\mu},s_{\tau}\right)  _{\Lambda}$. Since $s_{\kappa}^{\perp
}\left(  s_{\tau}\right)  =s_{\tau/\kappa}$, this rewrites as follows:%
\begin{equation}
\left(  s_{\lambda/\mu},s_{\tau/\kappa}\right)  _{\Lambda}=\left(  s_{\kappa
}s_{\lambda/\mu},s_{\tau}\right)  _{\Lambda}.\label{sol.liri.skew.1.b.5}%
\end{equation}
Thus,%
\begin{align*}
\left(  s_{\lambda/\mu},s_{\tau/\kappa}\right)  _{\Lambda} &  =\left(
s_{\kappa}s_{\lambda/\mu},s_{\tau}\right)  _{\Lambda}=\left(  \sum\limits
_{\substack{T\text{ is a column-strict tableau}\\\text{of shape }\lambda
/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}s_{\kappa+\cont T},s_{\tau}\right)  _{\Lambda}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.skew.1.a.4})}\right)  \\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}\underbrace{\left(  s_{\kappa+\cont T},s_{\tau}\right)
_{\Lambda}}_{\substack{=\delta_{\tau,\kappa+\cont T}\\\text{(by
(\ref{sol.liri.skew.1.b.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
Hall inner product is }\kk\text{-bilinear}\right)  \\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}\delta_{\tau,\kappa+\cont T}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold;}\\\tau=\kappa+\cont T}}\underbrace{\delta_{\tau
,\kappa+\cont T}}_{\substack{=1\\\text{(since }\tau
=\kappa+\cont T\text{)}}}+\sum\limits_{\substack{T\text{ is a
column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{the five
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold;}\\\tau\neq\kappa
+\cont T}}\underbrace{\delta_{\tau,\kappa+\cont
T}}_{\substack{=0\\\text{(since }\tau\neq\kappa+\cont T\text{)}%
}}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold;}\\\tau=\kappa+\cont T}}1+\underbrace{\sum\limits
_{\substack{T\text{ is a column-strict tableau}\\\text{of shape }\lambda
/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold;}\\\tau\neq\kappa+\cont T}}0}_{=0}\\
&  =\underbrace{\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of
shape }\lambda/\mu\text{;}\\\text{the five equivalent assertions}%
\\\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}^{\left(
\kappa\right)  }\text{ hold;}\\\tau=\kappa+\cont T}%
}}_{\substack{=\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of
shape }\lambda/\mu\text{;}\\\tau=\kappa+\cont T\text{;}%
\\\text{the five equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)
}\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}}}=\sum\limits_{\substack{T\text{
is a column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\tau
=\kappa+\cont T\text{;}\\\text{the six equivalent assertions}%
\\\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{, }\mathcal{G}^{\left(
\kappa\right)  }\text{ and }\mathcal{H}^{\left(  \kappa\right)  }\text{ hold}%
}}\\\text{(by the equivalence (\ref{sol.liri.skew.1.b.2}))}}}1
\end{align*}%
\begin{align*}
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\tau=\kappa+\cont T\text{;}\\\text{the six
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{,
}\mathcal{G}^{\left(  \kappa\right)  }\text{ and }\mathcal{H}^{\left(
\kappa\right)  }\text{ hold}}}1\\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda/\mu\text{ such that}\right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \tau=\kappa+\cont T\text{ and
such that the six equivalent}\right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{, }\mathcal{G}^{\left(  \kappa\right)  }\text{ and
}\mathcal{H}^{\left(  \kappa\right)  }\text{ hold}\right)  \\
&  =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda/\mu\text{ satisfying}\right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \tau=\kappa+\cont T\text{ and
also satisfying the six equivalent}\right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{, }\mathcal{G}^{\left(  \kappa\right)  }\text{ and
}\mathcal{H}^{\left(  \kappa\right)  }\right)  .
\end{align*}
This solves Exercise \ref{exe.liri.skew.1}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.skew.1}}

\textit{Solution to Exercise \ref{exe.liri.skew.1}.} (a) If $T$ is a
column-strict tableau of shape $\lambda/\mu$, then we have the following chain
of logical equivalences:%
\begin{align}
&  \ \left(  \text{for all }j\in\left\{  1,2,3,\ldots\right\}  \text{, the
weak composition }\kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \text{ is a partition}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{for every positive integer }j\text{, the
weak composition }\kappa+\cont \left(  T|_{\operatorname{cols}%
\geq j}\right)  \text{ is a partition}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{Assertion }\mathcal{C}^{\left(
\kappa\right)  }\text{ holds}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{because this is how we defined Assertion }\mathcal{C}^{\left(
\kappa\right)  }\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the five equivalent assertions
}\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}^{\left(
\kappa\right)  }\text{ hold}\right)  \label{sol.liri.skew.1.a.1}%
\end{align}
(because the five assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ are equivalent (by Exercise \ref{exe.liri.yamanouchi.kappa}(a))).

We can apply (\ref{eq.infinite-L-R-rule.1}) to $\nu=\kappa$. As a result, we
see that%
\[
s_{\kappa}s_{\lambda/\mu}=\sum_{T}s_{\kappa+\cont T},
\]
where $T$ runs through all column-strict tableaux of shape $\lambda/\mu$ with
the property that for each $j=1,2,3,\ldots$, the weak composition $\kappa
+\cont \left(  T|_{\operatorname{cols}\geq j}\right)  $ is a
partition. In other words,%
\begin{align}
s_{\kappa}s_{\lambda/\mu} &  =\underbrace{\sum\limits_{\substack{T\text{ is a
column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{for all
}j\in\left\{  1,2,3,\ldots\right\}  \text{, the weak}\\\text{composition
}\kappa+\cont \left(  T|_{\operatorname{cols}\geq j}\right)
\\\text{is a partition}}}}_{\substack{=\sum\limits_{\substack{T\text{ is a
column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{the five
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}}}\\\text{(because the
statement}\\\left(  \text{for all }j\in\left\{  1,2,3,\ldots\right\}  \text{,
the weak composition }\kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  \text{ is a partition}\right)
\\\text{is equivalent to the statement}\\\left(  \text{the five equivalent
assertions }\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}^{\left(
\kappa\right)  }\text{ hold}\right)  \\\text{(by (\ref{sol.liri.skew.1.a.1}%
)))}}}s_{\kappa+\cont T}\label{sol.liri.skew.1.a.3}\\
&  =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}s_{\kappa+\cont T}.\label{sol.liri.skew.1.a.4}%
\end{align}
In other words,%
\[
s_{\kappa}s_{\lambda/\mu}=\sum_{T}s_{\kappa+\cont T},
\]
where the sum ranges over all column-strict tableaux $T$ of shape $\lambda
/\mu$ satisfying the five equivalent assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}%
^{\left(  \kappa\right)  }$, $\mathcal{F}^{\left(  \kappa\right)  }$ and
$\mathcal{G}^{\left(  \kappa\right)  }$ introduced in
Exercise~\ref{exe.liri.yamanouchi.kappa}(a). This solves Exercise
\ref{exe.liri.skew.1}(a).

(b) If $T$ is a column-strict tableau of shape $\lambda/\mu$ such that the
five equivalent assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ hold, then we have%
\begin{equation}
\left(  s_{\kappa+\cont T},s_{\tau}\right)  _{\Lambda}%
=\delta_{\tau,\kappa+\cont T}\label{sol.liri.skew.1.b.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.skew.1.b.1}):} We know that the
basis $\left(  s_{\lambda}\right)  _{\lambda\in \Par }$ of
$\Lambda$ is orthonormal with respect to the Hall inner product. In other
words, we have%
\begin{equation}
\left(  s_{\alpha},s_{\beta}\right)  _{\Lambda}=\delta_{\alpha,\beta
}\label{sol.liri.skew.1.b.1.pf.1}%
\end{equation}
for any $\alpha\in \Par $ and $\beta\in \Par $.
\par
Now, let $T$ be a column-strict tableau of shape $\lambda/\mu$ such that the
five equivalent assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ hold. Then, Assertion $\mathcal{C}^{\left(  \kappa\right)
}$ holds (since the five equivalent assertions $\mathcal{C}^{\left(
\kappa\right)  }$, $\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}%
^{\left(  \kappa\right)  }$, $\mathcal{F}^{\left(  \kappa\right)  }$ and
$\mathcal{G}^{\left(  \kappa\right)  }$ hold). In other words, for every
positive integer $j$, the weak composition $\kappa+\cont \left(
T|_{\operatorname{cols}\geq j}\right)  $ is a partition. Applying this to
$j=1$, we conclude that the weak composition $\kappa+\cont
\left(  T|_{\operatorname{cols}\geq1}\right)  $ is a partition.
\par
But $T|_{\operatorname{cols}\geq1}$ is the subtableau which is the restriction
of $T$ to the union of its columns $1,2,3,\ldots$ (because this is how
$T|_{\operatorname{cols}\geq1}$ is defined). Thus, $T|_{\operatorname{cols}%
\geq1}$ is the whole tableau $T$ (because the restriction of $T$ to the union
of its columns $1,2,3,\ldots$ is clearly the whole tableau $T$ (since every
cell of $T$ belongs to one of the columns $1,2,3,\ldots$)). In other words,
$T|_{\operatorname{cols}\geq1}=T$.
\par
Now, recall that $\kappa+\cont \left(  T|_{\operatorname{cols}%
\geq1}\right)  $ is a partition. In other words, $\kappa+\cont
T$ is a partition (since $T|_{\operatorname{cols}\geq1}=T$). In other words,
$\kappa+\cont T\in \Par $. Also, $\tau
\in \Par $ (since $\tau$ is a partition). Thus,
(\ref{sol.liri.skew.1.b.1.pf.1}) (applied to $\alpha=\kappa
+\cont T$ and $\beta=\tau$) shows that $\left(  s_{\kappa
+\cont T},s_{\tau}\right)  _{\Lambda}
= \delta_{\kappa + \cont T, \tau}
=\delta_{\tau
,\kappa+\cont T}$. This proves (\ref{sol.liri.skew.1.b.1}).}.

But if $T$ is a column-strict tableau of shape $\lambda/\mu$ satisfying
$\tau=\kappa+\cont T$, then we have the following chain of
logical equivalences:%
\begin{align}
&  \ \left(  \text{the five equivalent assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the six equivalent assertions
}\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{, }\mathcal{G}^{\left(
\kappa\right)  }\text{ and }\mathcal{H}^{\left(  \kappa\right)  }\text{
hold}\right)  \label{sol.liri.skew.1.b.2}%
\end{align}
(because the six assertions $\mathcal{C}^{\left(  \kappa\right)  }$,
$\mathcal{D}^{\left(  \kappa\right)  }$, $\mathcal{E}^{\left(  \kappa\right)
}$, $\mathcal{F}^{\left(  \kappa\right)  }$, $\mathcal{G}^{\left(
\kappa\right)  }$ and $\mathcal{H}^{\left(  \kappa\right)  }$ are equivalent
(by Exercise \ref{exe.liri.yamanouchi.kappa}(b))).

Now, recall that $s_{\mu}^{\perp}\left(  s_{\lambda}\right)  =s_{\lambda/\mu}%
$. Applying this to $\kappa$ and $\tau$ instead of $\mu$ and $\lambda$, we
obtain $s_{\kappa}^{\perp}\left(  s_{\tau}\right)  =s_{\tau/\kappa}$.

Now, Proposition \ref{skewing-properties-prop}(i) (applied to $A=\Lambda$)
shows that every $a\in\Lambda$, $f\in\Lambda$ and $g\in\Lambda$ satisfy
$\left(  g,f^{\perp}\left(  a\right)  \right)  _{\Lambda}=\left(  fg,a\right)
_{\Lambda}$. We can apply this to $f=s_{\kappa}$, $g=s_{\lambda/\mu}$ and
$a=s_{\tau}$. As a result, we obtain $\left(  s_{\lambda/\mu},s_{\kappa
}^{\perp}\left(  s_{\tau}\right)  \right)  _{\Lambda}=\left(  s_{\kappa
}s_{\lambda/\mu},s_{\tau}\right)  _{\Lambda}$. Since $s_{\kappa}^{\perp
}\left(  s_{\tau}\right)  =s_{\tau/\kappa}$, this rewrites as follows:%
\begin{equation}
\left(  s_{\lambda/\mu},s_{\tau/\kappa}\right)  _{\Lambda}=\left(  s_{\kappa
}s_{\lambda/\mu},s_{\tau}\right)  _{\Lambda}.\label{sol.liri.skew.1.b.5}%
\end{equation}
Thus,%
\begin{align*}
\left(  s_{\lambda/\mu},s_{\tau/\kappa}\right)  _{\Lambda}  & =\left(
\underbrace{s_{\kappa}s_{\lambda/\mu}}_{\substack{=\sum\limits_{\substack{T\text{ is
a column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{the five
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}}}s_{\kappa
+\cont T}\\\text{(by (\ref{sol.liri.skew.1.a.4}))}}},s_{\tau
}\right)  _{\Lambda}=\left(  \sum\limits_{\substack{T\text{ is a column-strict
tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{the five equivalent
assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}%
^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)
}\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}%
^{\left(  \kappa\right)  }\text{ hold}}}s_{\kappa+\cont
T},s_{\tau}\right)  _{\Lambda}\\
& =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}\underbrace{\left(  s_{\kappa+\cont T},s_{\tau}\right)
_{\Lambda}}_{\substack{=\delta_{\tau,\kappa+\cont T}\\\text{(by
(\ref{sol.liri.skew.1.b.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
Hall inner product is }\kk\text{-bilinear}\right)  \\
& =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}%
}}\delta_{\tau,\kappa+\cont T}\\
& =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold;}\\\tau=\kappa+\cont T}}\underbrace{\delta_{\tau
,\kappa+\cont T}}_{\substack{=1\\\text{(since }\tau
=\kappa+\cont T\text{)}}}+\sum\limits_{\substack{T\text{ is a
column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\text{the five
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold;}\\\tau\neq\kappa
+\cont T}}\underbrace{\delta_{\tau,\kappa+\cont
T}}_{\substack{=0\\\text{(since }\tau\neq\kappa+\cont T\text{)}%
}}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because if }T\text{ is a column-strict tableau of shape }\lambda
/\mu\text{ such that}\\
\text{the five equivalent assertions }\mathcal{C}^{\left(  \kappa\right)
}\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold,}\\
\text{then we must have either }\tau=\kappa+\cont T\text{ or
}\tau\neq\kappa+\cont T\\
\text{(but not both at the same time)}%
\end{array}
\right)  \\
& =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}%
^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)
}\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold;}\\\tau=\kappa+\cont T}}1+\underbrace{\sum\limits
_{\substack{T\text{ is a column-strict tableau}\\\text{of shape }\lambda
/\mu\text{;}\\\text{the five equivalent assertions}\\\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{ and }\mathcal{G}^{\left(  \kappa\right)  }\text{
hold;}\\\tau\neq\kappa+\cont T}}0}_{=0}\\
& =\underbrace{\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of
shape }\lambda/\mu\text{;}\\\text{the five equivalent assertions}%
\\\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}^{\left(
\kappa\right)  }\text{ hold;}\\\tau=\kappa+\cont T}%
}}_{\substack{=\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of
shape }\lambda/\mu\text{;}\\\tau=\kappa+\cont T\text{;}%
\\\text{the five equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)
}\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{ and
}\mathcal{G}^{\left(  \kappa\right)  }\text{ hold}}}=\sum\limits_{\substack{T\text{
is a column-strict tableau}\\\text{of shape }\lambda/\mu\text{;}\\\tau
=\kappa+\cont T\text{;}\\\text{the six equivalent assertions}%
\\\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{, }\mathcal{G}^{\left(
\kappa\right)  }\text{ and }\mathcal{H}^{\left(  \kappa\right)  }\text{ hold}%
}}\\\text{(because if }T\text{ is a column-strict tableau of shape }%
\lambda/\mu\text{ satisfying }\tau=\kappa+\cont T\text{,}%
\\\text{then the statement}\\\left(  \text{the five equivalent assertions
}\mathcal{C}^{\left(  \kappa\right)  }\text{, }\mathcal{D}^{\left(
\kappa\right)  }\text{, }\mathcal{E}^{\left(  \kappa\right)  }\text{,
}\mathcal{F}^{\left(  \kappa\right)  }\text{ and }\mathcal{G}^{\left(
\kappa\right)  }\text{ hold}\right)  \\\text{is equivalent to the
statement}\\\left(  \text{the six equivalent assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{, }\mathcal{G}^{\left(  \kappa\right)  }\text{ and
}\mathcal{H}^{\left(  \kappa\right)  }\text{ hold}\right)  \\\text{(by
(\ref{sol.liri.skew.1.b.2})))}}}1
\end{align*}
\begin{align*}
& =\sum\limits_{\substack{T\text{ is a column-strict tableau}\\\text{of shape
}\lambda/\mu\text{;}\\\tau=\kappa+\cont T\text{;}\\\text{the six
equivalent assertions}\\\mathcal{C}^{\left(  \kappa\right)  }\text{,
}\mathcal{D}^{\left(  \kappa\right)  }\text{, }\mathcal{E}^{\left(
\kappa\right)  }\text{, }\mathcal{F}^{\left(  \kappa\right)  }\text{,
}\mathcal{G}^{\left(  \kappa\right)  }\text{ and }\mathcal{H}^{\left(
\kappa\right)  }\text{ hold}}}1\\
& =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda/\mu\text{ such that}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \tau=\kappa+\cont T\text{ and such
that the six equivalent}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \text{assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{, }\mathcal{G}^{\left(  \kappa\right)  }\text{ and
}\mathcal{H}^{\left(  \kappa\right)  }\text{ hold}\right)  \cdot1\\
& =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda/\mu\text{ such that}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \tau=\kappa+\cont T\text{ and such
that the six equivalent}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \text{assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{, }\mathcal{G}^{\left(  \kappa\right)  }\text{ and
}\mathcal{H}^{\left(  \kappa\right)  }\text{ hold}\right)  \\
& =\left(  \text{the number of all column-strict tableaux }T\text{ of shape
}\lambda/\mu\text{ satisfying}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \tau=\kappa+\cont T\text{ and also
satisfying the six equivalent}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \text{assertions }\mathcal{C}^{\left(
\kappa\right)  }\text{, }\mathcal{D}^{\left(  \kappa\right)  }\text{,
}\mathcal{E}^{\left(  \kappa\right)  }\text{, }\mathcal{F}^{\left(
\kappa\right)  }\text{, }\mathcal{G}^{\left(  \kappa\right)  }\text{ and
}\mathcal{H}^{\left(  \kappa\right)  }\right)  .
\end{align*}
This solves Exercise \ref{exe.liri.skew.1}(b).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.jordan}}

\textit{Solution to Exercise \ref{exe.liri.jordan}.} (a) We shall first
prove that%
\begin{equation}
\text{if }N\text{ has Jordan type }\lambda\text{, then every }k\in
 \NN \text{ satisfies }\dim\left(  \ker\left(  N^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}. \label{sol.liri.jordan.a.goal}%
\end{equation}


\textit{Proof of (\ref{sol.liri.jordan.a.goal}):} Assume that $N$ has Jordan
type $\lambda$. Let $k\in \NN$.

For each $m \in \NN$, let $J_{m}$ be $m\times m$-matrix $\left(
\begin{array}
[c]{ccccc}%
0 & 1 & 0 & \cdots & 0\\
0 & 0 & 1 & \cdots & 0\\
0 & 0 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & 0
\end{array}
\right) \in \mathbb{K}^{m\times m}$.
This is a Jordan block of size $m$ corresponding to the
eigenvalue $0$ (when $m$ is positive). For future use, we record the
following simple fact:
\begin{equation}
\dim \left( \ker \left( J_m^k \right) \right) = \min \left\{ m, k \right \}
\label{sol.liri.jordan.a.Jm-power}
\end{equation}
for all $m \in \NN$.\ \ \ \ \footnote{\textit{Proof sketch.} Let
$m \in \NN$. 
Let $\left(e_1, e_2, \ldots, e_m\right)$ be the standard basis of
the $\mathbb{K}$-vector space $\mathbb{K}^m$.
Now, it is easy to check that
$J_m^k$ is the $m\times m$-matrix
whose $\left(i, j\right)$-th entry is
$\begin{cases} 1, & \text{if } j = i + k; \\
               0, & \text{if } j \neq i + k
 \end{cases}
$ for all $\left(i, j\right) \in \left\{1, 2, \ldots, m\right\}^2$.
Hence, it is easy to see that $\ker \left( J_m^k \right)$ is the
$\mathbb{K}$-linear span of the basis vectors $e_j$ with $j \leq k$.
These basis vectors are linearly independent and their number is
$\min \left\{ m, k \right \}$; therefore,
\eqref{sol.liri.jordan.a.Jm-power} follows.}.

Since $N$ has Jordan type $\lambda$, the Jordan normal form of $N$ has Jordan
blocks of sizes $\lambda_{1},\lambda_{2},\lambda_{3},\ldots$.
Since the only eigenvalue of $N$ is $0$, this shows that
the Jordan blocks of $N$ are $J_{\lambda_1}, J_{\lambda_2}, J_{\lambda_3},
\ldots$ (or, more precisely, the nonempty matrices among
$J_{\lambda_1}, J_{\lambda_2}, J_{\lambda_3}, \ldots$). 
In other words,
$N$ is similar to the block-diagonal matrix $J_{\lambda}:=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  $. (This block-diagonal matrix
$J_{\lambda}$ is finite, since only finitely many $\lambda_{p}$ are nonzero.)
We can thus WLOG assume that $N$ \textbf{is} this matrix $J_{\lambda}$
(because replacing $N$ by a matrix similar to $N$ changes neither the
dimension $\dim\left(  \ker\left(  N^{k}\right)  \right)  $ nor the Jordan
type of $N$). Assume this. Thus, $N=J_{\lambda}=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  $, so that%
\[
N^{k}=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  ^{k}=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}}^{k} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}}^{k} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}}^{k} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)
\]
and therefore%
\[
\ker\left(  N^{k}\right)  =\ker\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}}^{k} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}}^{k} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}}^{k} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  \cong\bigoplus_{p\geq1}\ker\left(  J_{\lambda_{p}}^{k}\right)  ,
\]
whence%
\begin{equation}
\dim\left(  \ker\left(  N^{k}\right)  \right)  =\dim\left(  \bigoplus_{p\geq
1}\ker\left(  J_{\lambda_{p}}^{k}\right)  \right)  =\sum_{p\geq1}%
\underbrace{\dim\left(  \ker\left(  J_{\lambda_{p}}^{k}\right)  \right)
}_{\substack{=\min\left\{  \lambda_{p},k\right\}  \\\text{(by
\eqref{sol.liri.jordan.a.Jm-power} (applied to }
m = \lambda_p \text{))}}}
=\sum_{p\geq1}\min\left\{  \lambda_{p},k\right\}  .
\label{sol.liri.jordan.a.1}%
\end{equation}


We are now going to prove that the right hand side of this equality is
$\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}$. Indeed, let us use the
\dfn{Iverson bracket notation}: For every assertion $\mathcal{A}$, we let
$\left[  \mathcal{A}\right]  $ denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
(This integer is called the \dfn{truth value} of $\mathcal{A}$.)

Any two nonnegative integers $u$ and $v$ satisfy $\min\left\{  u,v\right\}
=\sum_{i=1}^{v}\left[  i\leq u\right]  $. Thus, $\min\left\{  \lambda
_{p},k\right\}  =\sum_{i=1}^{k}\left[  i\leq\lambda_{p}\right]  $ for every
$p\in\left\{  1,2,3,\ldots\right\}  $. Hence, (\ref{sol.liri.jordan.a.1})
becomes%
\begin{align*}
\dim\left(  \ker\left(  N^{k}\right)  \right)   &  =\sum_{p\geq1}%
\underbrace{\min\left\{  \lambda_{p},k\right\}  }_{=\sum_{i=1}^{k}\left[
i\leq\lambda_{p}\right]  }=\sum_{p\geq1}\sum_{i=1}^{k}\left[  i\leq\lambda
_{p}\right]  =\sum_{i=1}^{k}\underbrace{\sum_{p\geq1}\left[  i\leq\lambda
_{p}\right]  }_{\substack{=\left\vert \left\{  p\geq1\ \mid\ i\leq\lambda
_{p}\right\}  \right\vert \\=\left\vert \left\{  p\geq1\ \mid\ \lambda_{p}\geq
i\right\}  \right\vert \\=\left\vert \left\{  j\geq1\ \mid\ \lambda_{j}\geq
i\right\}  \right\vert =\left(  \lambda^{t}\right)  _{i}\\\text{(by
(\ref{eq.partition.transpose}))}}}\\
&  =\sum_{i=1}^{k}\left(  \lambda^{t}\right)  _{i}=\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}.
\end{align*}
This proves (\ref{sol.liri.jordan.a.goal}).

With (\ref{sol.liri.jordan.a.goal}), we have proven one direction of the
equivalence that Exercise \ref{exe.liri.jordan}(a) requires us to prove. To
prove the other direction, we need to show that
\begin{equation}
\text{if every }k\in \NN \text{ satisfies }\dim\left(  \ker\left(
N^{k}\right)  \right)  =\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}\text{, then }N\text{
has Jordan type }\lambda\text{.} \label{sol.liri.jordan.a.goal2}%
\end{equation}


\textit{Proof of (\ref{sol.liri.jordan.a.goal2}):} Assume that every
$k\in \NN$ satisfies $\dim\left(  \ker\left(  N^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}$. Let $\mu$ be the Jordan type of
$N$. Then, (\ref{sol.liri.jordan.a.goal}) (applied to $\mu$ instead of
$\lambda$) shows that every $k\in \NN$ satisfies $\dim\left(
\ker\left(  N^{k}\right)  \right)  =\left(  \mu^{t}\right)  _{1}+\left(
\mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{k}$. Hence, every
$k\in \NN$ satisfies%
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}=\dim\left(  \ker\left(  N^{k}\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{k}. \label{sol.liri.jordan.a.4}%
\end{equation}
Applying (\ref{sol.liri.jordan.a.4}) to $k-1$ instead of $k$, and subtracting
the result from (\ref{sol.liri.jordan.a.4}), we obtain%
\[
\left(  \mu^{t}\right)  _{k}=\left(  \lambda^{t}\right)  _{k}%
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }k\text{.}%
\]
Hence, $\mu^{t}=\lambda^{t}$, so that $\mu=\lambda$. Thus, $N$ has Jordan type
$\mu=\lambda$. This proves (\ref{sol.liri.jordan.a.goal2}), and thus the
solution of Exercise \ref{exe.liri.jordan}(a) is complete.

Before we come to the solution of Exercise \ref{exe.liri.jordan}(b), let us
show a linear-algebraic lemma:

\begin{lemma}
\label{lem.sol.liri.jordan.b.general}Let $W$ be a finite-dimensional
$\mathbb{K}$-vector space. Let $A$ and $B$ be $\mathbb{K}$-vector subspaces of
$W$. Let $f\in\operatorname*{End}W$ be such that $f\left(  A\right)  \subset
A$ and $f\left(  B\right)  \subset B$. For any $\left(  i,j\right)
\in \NN ^{2}$, let us define a nonnegative integer $w_{i,j}$ by
$w_{i,j}=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  $.

(a) For any $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$, we
have $w_{i,j}+w_{i-1,j-1}\geq w_{i,j-1}+w_{i-1,j}$.

(b) For any $\left(  i,j\right)  \in \NN \times\left\{  1,2,3,\ldots
\right\}  $, we have $w_{i+1,j+1}-w_{i+1,j}\leq w_{i,j}-w_{i,j-1}$.

(c) For any $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
\times \NN$, we have $w_{i+1,j+1}-w_{i,j+1}\leq w_{i,j}-w_{i-1,j}$.

(d) If $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$ are such
that $i>\dim W$, then $w_{i,j}=w_{i-1,j}$.

(e) If $j\in\left\{  1,2,3,\ldots\right\}  $ and $i\in \NN$ are such
that $j>\dim W$, then $w_{i,j}=w_{i,j-1}$.

(f) Assume that $f$ is nilpotent. For every $j\in \NN$, we have
$\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)  =\dim\left(  \left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  -\dim\left(  A\cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  $. (In particular, the sum
$\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)  $ converges with respect
to the discrete topology, i.e., all but finitely many of its terms are $0$.)

(g) Assume that $f$ is nilpotent. For every $i\in \NN$, we have
$\sum_{j=1}^{\infty}\left(  w_{i,j}-w_{i,j-1}\right)  =\dim\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \right)  -\dim\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap B\right)  $. (In particular, the sum
$\sum_{j=1}^{\infty}\left(  w_{i,j}-w_{i,j-1}\right)  $ converges with respect
to the discrete topology, i.e., all but finitely many of its terms are $0$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.liri.jordan.b.general}.]
Notice first that $\left(\underbrace{f^0}_{=\id}\right)^{-1} \left(A\right)
= \id^{-1} \left( A \right) = A$ and
$A = \left(f^0\right)^{-1} \left(A\right) \subset
\left(f^1\right)^{-1} \left(A\right) \subset
\left(f^2\right)^{-1} \left(A\right) \subset \cdots$ (since
$f\left(A\right) \subset A$) and
$B = \left(f^0\right)^{-1} \left(B\right) \subset
\left(f^1\right)^{-1} \left(B\right) \subset
\left(f^2\right)^{-1} \left(B\right) \subset \cdots$ (similarly).

(a) Let $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. We need to prove that
$w_{i,j}+w_{i-1,j-1}\geq w_{i,j-1}+w_{i-1,j}$. In other words, we need to
prove that $w_{i,j}-w_{i,j-1}\geq w_{i-1,j}-w_{i-1,j-1}$. Since
\begin{align}
w_{i,j}-w_{i,j-1}  &  =\dim\left(  \left(  f^{i}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)
-\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definitions of }w_{i,j}\text{ and
}w_{i,j-1}\right) \nonumber\\
&  =\dim\left(  \left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right)  \right) \label{pf.lem.sol.liri.jordan.b.general.a.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  f^{i}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)  \subset\left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right) \nonumber
\end{align}
and%
\[
w_{i-1,j}-w_{i-1,j-1}=\dim\left(  \left(  \left(  f^{i-1}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(
\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)
^{-1}\left(  B\right)  \right)  \right)
\]
(by the same argument as (\ref{pf.lem.sol.liri.jordan.b.general.a.1}), only
with $i-1$ instead of $i$), this is equivalent to showing that%
\begin{align*}
&  \dim\left(  \left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right)  \right) \\
&  \geq\dim\left(  \left(  \left(  f^{i-1}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)
^{-1}\left(  B\right)  \right)  \right)  .
\end{align*}
This will clearly be achieved if we can construct a $\mathbb{K}$-linear
injection%
\begin{align*}
&  \left(  \left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i-1}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  .
\end{align*}
Here is how to construct it: Let $\iota$ denote the canonical inclusion
$\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)
^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  $. Then, $\iota$ restricts to
an inclusion $\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)  $,
and so gives rise to a map%
\begin{align*}
&  \left(  \left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i-1}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  .
\end{align*}
This map is injective because $\iota^{-1}\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  =\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  $, and thus we have found our
$\mathbb{K}$-linear injection. Lemma \ref{lem.sol.liri.jordan.b.general}(a) is proven.

(b) Let $\left(  i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}
$. We need to prove that $w_{i+1,j+1}-w_{i+1,j}\leq w_{i,j}-w_{i,j-1}$. Due to
(\ref{pf.lem.sol.liri.jordan.b.general.a.1}) and due to%
\[
w_{i+1,j+1}-w_{i+1,j}=\dim\left(  \left(  \left(  f^{i+1}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(
\left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)
^{-1}\left(  B\right)  \right)  \right)
\]
(this follows by the same arguments as
(\ref{pf.lem.sol.liri.jordan.b.general.a.1}), only with $i$ and $j$ replaced
by $i+1$ and $j+1$), this is equivalent to showing that%
\begin{align*}
&  \dim\left(  \left(  \left(  f^{i+1}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right)  \right) \\
&  \leq\dim\left(  \left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right)  \right)  .
\end{align*}
This will clearly be achieved if we can construct a $\mathbb{K}$-linear
injection%
\begin{align*}
&  \left(  \left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  .
\end{align*}
Here is how this can be done: The map $f$ restricts to a map $\varphi:\left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j+1}\right)
^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  $, which further restricts to
a map $\left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)  $.
Hence, $\varphi$ gives rise to a map%
\begin{align*}
&  \left(  \left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  ,
\end{align*}
which is injective because%
\begin{align*}
\varphi^{-1}\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  \right)   &  =f^{-1}\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right) \\
&  =\underbrace{f^{-1}\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\right)  }_{=\left(  f^{i+1}\right)  ^{-1}\left(  A\right)  }\cap
\underbrace{f^{-1}\left(  \left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  }_{=\left(  f^{j}\right)  ^{-1}\left(  B\right)  }=\left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  .
\end{align*}
Thus, we have found our $\mathbb{K}$-linear injection. Lemma
\ref{lem.sol.liri.jordan.b.general}(b) is proven.

(c) The proof of Lemma \ref{lem.sol.liri.jordan.b.general}(c) is analogous to
our above proof of Lemma \ref{lem.sol.liri.jordan.b.general}(b) (one merely
has to interchange $A$ with $B$ and $i$ with $j$).

(d) We have%
\[
\left(  f^{0}\right)  ^{-1}\left(  A\right)  \subset\left(  f^{1}\right)
^{-1}\left(  A\right)  \subset\cdots\subset\left(  f^{\dim W+1}\right)
^{-1}\left(  A\right)
\]
and therefore%
\[
\dim\left(  \left(  f^{0}\right)  ^{-1}\left(  A\right)  \right)  \leq
\dim\left(  \left(  f^{1}\right)  ^{-1}\left(  A\right)  \right)  \leq
\cdots\leq\dim\left(  \left(  f^{\dim W+1}\right)  ^{-1}\left(  A\right)
\right)  .
\]
This latter chain of inequalities contains $\dim W+1$ inequality signs, but
only at most $\dim W$ of them can be strict (because each $\dim\left(  \left(
f^{k}\right)  ^{-1}\left(  A\right)  \right)  $ is an integer between $0$ and
$\dim W$ inclusive\footnote{since $\left(  f^{k}\right)  ^{-1}\left(
A\right)  $ is a subspace of the finite-dimensional $\mathbb{K}$-vector space
$W$}, and a sequence of $\dim W+2$ integers between $0$ and $\dim W$ cannot
strictly increase). Thus, at least one of the inequality signs is an equality.
That is, there exists an $I\in\left\{  1,2,\ldots,\dim W+1\right\}  $ such
that $\dim\left(  \left(  f^{I-1}\right)  ^{-1}\left(  A\right)  \right)
=\dim\left(  \left(  f^{I}\right)  ^{-1}\left(  A\right)  \right)  $. Consider
this $I$.

Since $\dim\left(  \left(  f^{I-1}\right)  ^{-1}\left(  A\right)  \right)
=\dim\left(  \left(  f^{I}\right)  ^{-1}\left(  A\right)  \right)  $ and
$\left(  f^{I-1}\right)  ^{-1}\left(  A\right)  \subset\left(  f^{I}\right)
^{-1}\left(  A\right)  $, we must have
\begin{equation}
\left(  f^{I-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{I}\right)
^{-1}\left(  A\right)  . \label{pf.lem.sol.liri.jordan.b.general.c.1}%
\end{equation}


From here, it is easy to see that
\[
\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{i}\right)
^{-1}\left(  A\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,3,\ldots\right\}  \text{ satisfying }i>\dim W.
\]
\footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots\right\}  $ be such
that $i>\dim W$. Thus, $i\geq\dim W+1$. But $I\leq\dim W+1$ (since
$I\in\left\{  1,2,\ldots,\dim W+1\right\}  $) and thus $i\geq\dim W+1\geq I$.
Hence, there exists some $k\in \NN$ such that $i=I+k$. Consider this
$k$. Since $i=I+k$, we have $i-1=I+k-1=\left(  I-1\right)  +k$, so that
$f^{i-1}=f^{\left(  I-1\right)  +k}=f^{I-1}\circ f^{k}$ and thus $\left(
f^{i-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{I-1}\circ f^{k}\right)
^{-1}\left(  A\right)  =\left(  f^{k}\right)  ^{-1}\left(  \left(
f^{I-1}\right)  ^{-1}\left(  A\right)  \right)  $. Similarly, $\left(
f^{i}\right)  ^{-1}\left(  A\right)  =\left(  f^{k}\right)  ^{-1}\left(
\left(  f^{I}\right)  ^{-1}\left(  A\right)  \right)  $. Now,%
\[
\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{k}\right)
^{-1}\left(  \underbrace{\left(  f^{I-1}\right)  ^{-1}\left(  A\right)
}_{\substack{=\left(  f^{I}\right)  ^{-1}\left(  A\right)  \\\text{(by
(\ref{pf.lem.sol.liri.jordan.b.general.c.1}))}}}\right)  =\left(
f^{k}\right)  ^{-1}\left(  \left(  f^{I}\right)  ^{-1}\left(  A\right)
\right)  =\left(  f^{i}\right)  ^{-1}\left(  A\right)  ,
\]
qed.}

Thus, for every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfying $i>\dim W$,
we have%
\begin{align*}
w_{i-1,j}  &  =\dim\left(  \underbrace{\left(  f^{i-1}\right)  ^{-1}\left(
A\right)  }_{=\left(  f^{i}\right)  ^{-1}\left(  A\right)  }\cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }w_{i-1,j}\right) \\
&  =\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  =w_{i,j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }j\in \NN .
\end{align*}
Thus, if $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$ are such
that $i>\dim W$, then $w_{i-1,j}=w_{i-1,j}$. This proves Lemma
\ref{lem.sol.liri.jordan.b.general}(d).

(e) The proof of Lemma \ref{lem.sol.liri.jordan.b.general}(e) is analogous to
our above proof of Lemma \ref{lem.sol.liri.jordan.b.general}(d) (one merely
has to interchange $A$ with $B$ and $i$ with $j$).

(f) Let $j\in \NN$. We have%
\begin{align}
\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)   &  =\sum_{i=1}^{\dim
W}\left(  w_{i,j}-w_{i-1,j}\right)  +\sum_{i=\dim W+1}^{\infty}\left(
\underbrace{w_{i,j}}_{\substack{=w_{i-1,j}\\\text{(by Lemma
\ref{lem.sol.liri.jordan.b.general}(d),}\\\text{since }i>\dim W\text{)}%
}}-w_{i-1,j}\right) \nonumber\\
&  =\sum_{i=1}^{\dim W}\left(  w_{i,j}-w_{i-1,j}\right)  +\sum_{i=\dim
W+1}^{\infty}\underbrace{\left(  w_{i-1,j}-w_{i-1,j}\right)  }_{=0}=\sum
_{i=1}^{\dim W}\left(  w_{i,j}-w_{i-1,j}\right)  +\underbrace{\sum_{i=\dim
W+1}^{\infty}0}_{=0}\nonumber\\
&  =\sum_{i=1}^{\dim W}\left(  w_{i,j}-w_{i-1,j}\right)  =w_{\dim W,j}%
-w_{0,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the telescope principle}\right)
. \label{pf.lem.sol.liri.jordan.b.general.f.1}%
\end{align}
Now, we are going to prove that $w_{\dim W,j}=\dim\left(  \left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  $ and $w_{0,j}=\dim\left(
A\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  $.

Indeed, it is well-known that if $g$ is any nilpotent endomorphism of a
finite-dimensional vector space $V$, then $g^{\dim V}=0$. Applied to $g=f$ and
$V=W$, this yields $f^{\dim W}=0$, and thus $f^{\dim W}\left(  W\right)
=0\left(  W\right)  =0\subset A$, whence $W\subset\left(  f^{\dim W}\right)
^{-1}\left(  A\right)  $, so that $\left(  f^{\dim W}\right)  ^{-1}\left(
A\right)  =W$. Hence,%
\[
\underbrace{\left(  f^{\dim W}\right)  ^{-1}\left(  A\right)  }_{=W}%
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  =W\cap\left(  f^{j}\right)
^{-1}\left(  B\right)  =\left(  f^{j}\right)  ^{-1}\left(  B\right)
\]
(since $\left(  f^{j}\right)  ^{-1}\left(  B\right)  \subset W$). Now, the
definition of $w_{\dim W,j}$ yields%
\[
w_{\dim W,j}=\dim\left(  \underbrace{\left(  f^{\dim W}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  }_{=\left(
f^{j}\right)  ^{-1}\left(  B\right)  }\right)  =\dim\left(  \left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  .
\]
Also, the definition of $w_{0,j}$ yields%
\[
w_{0,j}=\dim\left(  \left(  \underbrace{f^{0}}_{=\id}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)
\right)  =\dim\left(  \underbrace{\id^{-1}\left(
A\right)  }_{=A}\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)
=\dim\left(  A\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  .
\]
Hence, (\ref{pf.lem.sol.liri.jordan.b.general.f.1}) becomes%
\[
\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)  =\underbrace{w_{\dim
W,j}}_{=\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)
}-\underbrace{w_{0,j}}_{=\dim\left(  A\cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right)  }=\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  B\right)
\right)  -\dim\left(  A\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)
\right)  .
\]
This proves Lemma \ref{lem.sol.liri.jordan.b.general}(f).

(g) The proof of Lemma \ref{lem.sol.liri.jordan.b.general}(g) is analogous to
our above proof of Lemma \ref{lem.sol.liri.jordan.b.general}(f) (one merely
has to interchange $A$ with $B$ and $i$ with $j$).
\end{proof}

(b) Assume that $\ZZ$ is a subring of $\kk$.

For any $\left(  i,j\right)  \in \NN ^{2}$, let us define a nonnegative
integer $a_{i,j}$ by%
\[
a_{i,j}=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \cap
\ker\left(  f^{j}\right)  \right)  .
\]
Furthermore, for every $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$, let us define an integer $b_{i,j}$ by%
\[
b_{i,j}=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}.
\]
We first observe that%
\begin{equation}
a_{i,0}=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in \NN .
\label{sol.liri.jordan.b.ai0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.ai0}):} For every
$i\in \NN$, the definition of $a_{i,0}$ yields $a_{i,0}=\dim\left(
\left(  f^{i}\right)  ^{-1}\left(  U\right)  \cap\underbrace{\ker\left(
f^{0}\right)  }_{=\ker\left(  \operatorname*{id}\right)  =0}\right)
=\dim\underbrace{\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap0\right)  }_{=0}=\dim0=0$, qed.} Furthermore,%
\begin{equation}
a_{0,j}=\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}%
+\cdots+\left(  \mu^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for every }%
j\in \NN . \label{sol.liri.jordan.b.a0j}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.a0j}):} Let $j\in
 \NN$. We can represent the endomorphism $f\mid U$ of $U$ as a $k\times
k$-matrix $G\in\mathbb{K}^{k\times k}$ for $k=\dim U$. This $k\times k$-matrix
$G$ is nilpotent (since $f\mid U$ is nilpotent) and has Jordan type $\mu$
(since $f\mid U$ has Jordan type $\mu$). But Exercise \ref{exe.liri.jordan}(a)
(applied to $k$, $G$ and $\mu$ instead of $n$, $N$ and $\lambda$) shows that
$G$ has Jordan type $\mu$ if and only if every $k\in{ \NN }$ satisfies
$\dim\left(  \ker\left(  G^{k}\right)  \right)  =\left(  \mu^{t}\right)
_{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{k}$. Since
we know that $G$ has Jordan type $\mu$, we thus conclude that every
$k\in{ \NN }$ satisfies $\dim\left(  \ker\left(  G^{k}\right)  \right)
=\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}$. Applied to $k=j$, this yields $\dim\left(  \ker\left(
G^{j}\right)  \right)  =\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)
_{2}+\cdots+\left(  \mu^{t}\right)  _{j}$.
\par
Since the matrix $G$ represents the endomorphism $f\mid U$, we have
\[
\dim\left(  \ker\left(  G^{j}\right)  \right)  =\dim\left(  \underbrace{\ker
\left(  \left(  f\mid U\right)  ^{j}\right)  }_{=U\cap\ker\left(
f^{j}\right)  }\right)  =\dim\left(  U\cap\ker\left(  f^{j}\right)  \right)
.
\]
Compared with $\dim\left(  \ker\left(  G^{j}\right)  \right)  =\left(  \mu
^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)
_{j}$, this yields%
\[
\dim\left(  U\cap\ker\left(  f^{j}\right)  \right)  =\left(  \mu^{t}\right)
_{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{j}.
\]
Now, the definition of $a_{0,j}$ yields
\[
a_{0,j}=\dim\left(  \left(  \underbrace{f^{0}}_{=\id}\right)
^{-1}\left(  U\right)  \cap\ker\left(  f^{j}\right)  \right)  =\dim\left(
\underbrace{\id^{-1}\left(  U\right)  }_{=U}\cap
\ker\left(  f^{j}\right)  \right)  =\dim\left(  U\cap\ker\left(  f^{j}\right)
\right)  =\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}%
+\cdots+\left(  \mu^{t}\right)  _{j}.
\]
This proves (\ref{sol.liri.jordan.b.a0j}).} Thus,%
\begin{equation}
a_{0,j}-a_{0,j-1}=\left(  \mu^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for
every }j\in\left\{  1,2,3,\ldots\right\}  . \label{sol.liri.jordan.b.a0j-diff}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.a0j-diff}):} For every
$j\in\left\{  1,2,3,\ldots\right\}  $, we have%
\[
\underbrace{a_{0,j}}_{\substack{=\left(  \mu^{t}\right)  _{1}+\left(  \mu
^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{j}\\\text{(by
(\ref{sol.liri.jordan.b.a0j}))}}}-\underbrace{a_{0,j-1}}_{\substack{=\left(
\mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu
^{t}\right)  _{j-1}\\\text{(by (\ref{sol.liri.jordan.b.a0j}), applied
to}\\j-1\text{ instead of }j\text{)}}}=\left(  \left(  \mu^{t}\right)
_{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{j}\right)
-\left(  \left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}%
+\cdots+\left(  \mu^{t}\right)  _{j-1}\right)  =\left(  \mu^{t}\right)  _{j},
\]
qed.} Moreover, every $\left(  i,j\right)  \in \NN ^{2}$ satisfies%
\[
a_{i,j}=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap\underbrace{\ker\left(  f^{j}\right)  }_{=\left(  f^{j}\right)
^{-1}\left(  0\right)  }\right)  =\dim\left(  \left(  f^{i}\right)
^{-1}\left(  U\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  .
\]
Hence, Lemma \ref{lem.sol.liri.jordan.b.general}(a) (applied to $W=V$, $A=U$,
$B=0$ and $w_{i,j}=a_{i,j}$) shows that for any $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$, we have $a_{i,j}+a_{i-1,j-1}\geq
a_{i,j-1}+a_{i-1,j}$. In other words, for any $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$, we have $a_{i,j}-a_{i,j-1}-a_{i-1,j}%
+a_{i-1,j-1}\geq0$. In other words, for any $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$, we have%
\begin{equation}
b_{i,j}\geq0 \label{sol.liri.jordan.b.bij}%
\end{equation}
(since $b_{i,j}=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}$). Thus, $b_{i,j}$ is
a nonnegative integer for every two positive integers $i$ and $j$. Moreover,%
\begin{equation}
b_{i,j}=0\ \ \ \ \ \ \ \ \ \ \text{for all but finitely many pairs }\left(
i,j\right)  . \label{sol.liri.jordan.b.bij0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.bij0}):} Lemma
\ref{lem.sol.liri.jordan.b.general}(d) (applied to $W=V$, $A=U$, $B=0$ and
$w_{i,j}=a_{i,j}$) shows that if $i\in\left\{  1,2,3,\ldots\right\}  $ and
$j\in \NN$ are such that $i>\dim V$, then%
\begin{equation}
a_{i,j}=a_{i-1,j}. \label{sol.liri.jordan.b.bij0.pf.1}%
\end{equation}
\par
Lemma \ref{lem.sol.liri.jordan.b.general}(e) (applied to $W=V$, $A=U$, $B=0$
and $w_{i,j}=a_{i,j}$) shows that if $j\in\left\{  1,2,3,\ldots\right\}  $ and
$i\in \NN$ satisfy $j>\dim V$, then $a_{i,j}=a_{i,j-1}$.
\par
Now, let $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. If
$i>\dim V$, then%
\[
b_{i,j}=\underbrace{a_{i,j}}_{\substack{=a_{i-1,j}\\\text{(by
(\ref{sol.liri.jordan.b.bij0.pf.1}),}\\\text{since }i>\dim V\text{)}%
}}-\underbrace{a_{i,j-1}}_{\substack{=a_{i-1,j-1}\\\text{(by
(\ref{sol.liri.jordan.b.bij0.pf.1}) (applied to}\\j-1\text{ instead of
}j\text{),}\\\text{since }i> \dim V\text{)}}}-a_{i-1,j}+a_{i-1,j-1}%
=a_{i-1,j}-a_{i-1,j-1}-a_{i-1,j}+a_{i-1,j-1}=0.
\]
Similarly, $b_{i,j}=0$ if $j>\dim V$. Hence, we see that $b_{i,j}=0$ if we
have $i>\dim V$ or $j>\dim V$ (or both). Thus, $b_{i,j}=0$ for all but
finitely many pairs $\left(  i,j\right)  $ (because all but finitely many
pairs $\left(  i,j\right)  $ satisfy $i>\dim V$ or $j>\dim V$). This proves
(\ref{sol.liri.jordan.b.bij0}).} Also,%
\begin{equation}
\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
=a_{i,j}-a_{i,j-1}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in \NN \times\left\{  1,2,3,\ldots\right\}  .
\label{sol.liri.jordan.b.bij-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.bij-sum}):} Let $\left(
i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}  $. Then,%
\begin{align*}
b_{1,j}+b_{2,j}+\cdots+b_{i,j}  &  =\sum_{k=1}^{i}\underbrace{b_{k,j}%
}_{\substack{=a_{k,j}-a_{k,j-1}-a_{k-1,j}+a_{k-1,j-1}\\\text{(by the
definition of }b_{k,j}\text{)}}}=\sum_{k=1}^{i}\underbrace{\left(
a_{k,j}-a_{k,j-1}-a_{k-1,j}+a_{k-1,j-1}\right)  }_{=\left(  a_{k,j}%
-a_{k,j-1}\right)  -\left(  a_{k-1,j}-a_{k-1,j-1}\right)  }\\
&  =\sum_{k=1}^{i}\left(  \left(  a_{k,j}-a_{k,j-1}\right)  -\left(
a_{k-1,j}-a_{k-1,j-1}\right)  \right)  =\left(  a_{i,j}-a_{i,j-1}\right)
-\underbrace{\left(  a_{0,j}-a_{0,j-1}\right)  }_{\substack{=\left(  \mu
^{t}\right)  _{j}\\\text{(by (\ref{sol.liri.jordan.b.a0j-diff}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the telescope principle}\right) \\
&  =\left(  a_{i,j}-a_{i,j-1}\right)  -\left(  \mu^{t}\right)  _{j},
\end{align*}
and thus $\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}+b_{2,j}+\cdots
+b_{i,j}\right)  =a_{i,j}-a_{i,j-1}$, qed.}

Now, Exercise \ref{exe.liri.yamanouchi}(a) (applied to $\mu^{t}$ instead of
$\mu$) yields that the following two assertions are equivalent\footnote{The
partition that we call $\gamma$ in Assertion $\mathcal{A}^{\prime}$ is the
partition that was called $\lambda$ in Assertion $\mathcal{A}$.}:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime}$:} There exist a
partition $\gamma$ and a column-strict tableau $T$ of shape $\gamma/\mu^{t}$
such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfy
\begin{equation}
b_{i,j}=\left(  \text{the number of all entries }i\text{ in the }j\text{-th
row of }T\right)  . \label{sol.liri.jordan.b.A}%
\end{equation}


\item \textit{Assertion $\mathcal{B}^{\prime}$:} The inequality
\begin{equation}
\left(  \mu^{t}\right)  _{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots
+b_{i+1,j+1}\right)  \leq\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}%
+b_{2,j}+\cdots+b_{i,j}\right)  \label{sol.liri.jordan.b.B}%
\end{equation}
holds for all $\left(  i,j\right)  \in{ \NN }\times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}

We shall now prove that Assertion $\mathcal{B}^{\prime}$ holds. Indeed, any
$\left(  i,j\right)  \in{ \NN }\times\left\{  1,2,3,\ldots\right\}  $
satisfies%
\begin{align*}
&  \left(  \mu^{t}\right)  _{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots
+b_{i+1,j+1}\right) \\
&  =a_{i+1,j+1}-a_{i+1,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.bij-sum}), applied to }i+1\text{ and }j+1\text{
instead of }i\text{ and }j\right) \\
&  \leq a_{i,j}-a_{i,j-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.liri.jordan.b.general}(b) (applied to }W=V\text{, }A=U\text{,
}B=0\text{ and }w_{i,j}=a_{i,j}\text{)}\right) \\
&  =\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}%
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.jordan.b.bij-sum}%
)}\right)  .
\end{align*}
Thus, Assertion $\mathcal{B}^{\prime}$ holds. Since Assertions $\mathcal{A}%
^{\prime}$ and $\mathcal{B}^{\prime}$ are equivalent, this yields that
Assertion $\mathcal{A}^{\prime}$ also holds. In other words, there exist a
partition $\gamma$ and a column-strict tableau $T$ of shape $\gamma/\mu^{t}$
such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfy (\ref{sol.liri.jordan.b.A}). Let us consider this $\gamma$ and this
$T$.

We shall soon see that $\gamma=\lambda^{t}$ and $\cont T=\nu
^{t}$. Let us first prepare for this. We have%
\begin{equation}
\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in \NN .
\label{sol.liri.jordan.b.lambda-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.lambda-sum}):} Let
$j\in \NN$. We can represent the endomorphism $f$ of $V$ as an $n\times
n$-matrix $F\in\mathbb{K}^{n\times n}$ for $n=\dim V$. This $n\times n$-matrix
$F$ is nilpotent (since $f$ is nilpotent) and has Jordan type $\lambda$ (since
$f$ has Jordan type $\lambda$). But Exercise \ref{exe.liri.jordan}(a) (applied
to $N=F$) shows that $F$ has Jordan type $\lambda$ if and only if every
$k\in{ \NN }$ satisfies $\dim\left(  \ker\left(  F^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}$. Since we know that $F$ has Jordan
type $\lambda$, we thus conclude that every $k\in{ \NN }$ satisfies
$\dim\left(  \ker\left(  F^{k}\right)  \right)  =\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}%
$. Applied to $k=j$, this yields $\dim\left(  \ker\left(  F^{j}\right)
\right)  =\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\cdots+\left(  \lambda^{t}\right)  _{j}$.
\par
Since the matrix $F$ represents the endomorphism $f$, we have
\[
\dim\left(  \ker\left(  F^{j}\right)  \right)  =\dim\left(  \underbrace{\ker
\left(  f^{j}\right)  }_{=\left(  f^{j}\right)  ^{-1}\left(  0\right)
}\right)  =\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
.
\]
Compared with $\dim\left(  \ker\left(  F^{j}\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j}$, this yields%
\[
\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j},
\]
and thus (\ref{sol.liri.jordan.b.lambda-sum}) is proven.} From here, it is
easy to see that%
\begin{equation}
\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)  -\dim\left(
\left(  f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  =\left(  \lambda
^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{
1,2,3,\ldots\right\}  . \label{sol.liri.jordan.b.lambda-single}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.lambda-single}):} Let
$j\in\left\{  1,2,3,\ldots\right\}  $. Then,%
\begin{align*}
&  \underbrace{\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  }_{\substack{=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{j}\\\text{(by
(\ref{sol.liri.jordan.b.lambda-sum}))}}}-\underbrace{\dim\left(  \left(
f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  }_{\substack{=\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j-1}\\\text{(by (\ref{sol.liri.jordan.b.lambda-sum}),
applied}\\\text{to }j-1\text{ instead of }j\text{)}}}\\
&  =\left(  \left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\cdots+\left(  \lambda^{t}\right)  _{j}\right)  -\left(  \left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j-1}\right)  =\left(  \lambda^{t}\right)  _{j},
\end{align*}
so that (\ref{sol.liri.jordan.b.lambda-single}) is proven.} Now, for every
$j\in\left\{  1,2,3,\ldots\right\}  $, we have%
\[
\left(  \text{the number of all entries in the }j\text{-th row of }T\right)
=\gamma_{j}-\left(  \mu^{t}\right)  _{j}%
\]
(since $T$ has shape $\gamma/\mu^{t}$), so that%
\begin{align*}
\gamma_{j}-\left(  \mu^{t}\right)  _{j}  &  =\left(  \text{the number of all
entries in the }j\text{-th row of }T\right) \\
&  =\sum_{i=1}^{\infty}\underbrace{\left(  \text{the number of all entries
}i\text{ in the }j\text{-th row of }T\right)  }_{\substack{=b_{i,j}\\\text{(by
(\ref{sol.liri.jordan.b.A}))}}}=\sum_{i=1}^{\infty}\underbrace{b_{i,j}%
}_{=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}}\\
&  =\sum_{i=1}^{\infty}\underbrace{\left(  a_{i,j}-a_{i,j-1}-a_{i-1,j}%
+a_{i-1,j-1}\right)  }_{=\left(  a_{i,j}-a_{i-1,j}\right)  -\left(
a_{i,j-1}-a_{i-1,j-1}\right)  }=\sum_{i=1}^{\infty}\left(  \left(
a_{i,j}-a_{i-1,j}\right)  -\left(  a_{i,j-1}-a_{i-1,j-1}\right)  \right) \\
&  =\underbrace{\sum_{i=1}^{\infty}\left(  a_{i,j}-a_{i-1,j}\right)
}_{\substack{=\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  -\dim\left(  U\cap\left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(f), applied
to}\\W=V\text{, }A=U\text{, }B=0\text{ and }w_{i,j}=a_{i,j}\text{)}%
}}-\underbrace{\sum_{i=1}^{\infty}\left(  a_{i,j-1}-a_{i-1,j-1}\right)
}_{\substack{=\dim\left(  \left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  -\dim\left(  U\cap\left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(f), applied
to}\\V\text{, }U\text{, }0\text{, }a_{i,j}\text{ and }j-1\text{
instead of }W\text{, }A\text{, }B\text{, }w_{i,j}\text{ and }j\text{)}}}\\
&  =\left(  \dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
-\dim\left(  U\cap\left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
\right)  -\left(  \dim\left(  \left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  -\dim\left(  U\cap\left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  \right) \\
&  =\underbrace{\left(  \dim\left(  \left(  f^{j}\right)  ^{-1}\left(
0\right)  \right)  -\dim\left(  \left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  \right)  }_{\substack{=\left(  \lambda^{t}\right)  _{j}\\\text{(by
(\ref{sol.liri.jordan.b.lambda-single}))}}}-\left(  \dim\left(  U\cap\left(
f^{j}\right)  ^{-1}\left(  0\right)  \right)  -\dim\left(  U\cap\left(
f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  \right) \\
&  =\left(  \lambda^{t}\right)  _{j}-\left(  \dim\left(  U\cap\left(
f^{j}\right)  ^{-1}\left(  0\right)  \right)  -\dim\left(  U\cap\left(
f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  \right)  .
\end{align*}
Adding this equality to the equality%
\begin{align*}
\left(  \mu^{t}\right)  _{j}  &  =\underbrace{a_{0,j}}_{\substack{=\dim\left(
\left(  f^{0}\right)  ^{-1}\left(  U\right)  \cap\ker\left(  f^{j}\right)
\right)  \\\text{(by the definition of }a_{0,j}\text{)}}%
}-\underbrace{a_{0,j-1}}_{\substack{=\dim\left(  \left(  f^{0}\right)
^{-1}\left(  U\right)  \cap\ker\left(  f^{j-1}\right)  \right)  \\\text{(by
the definition of }a_{0,j-1}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.a0j-diff})}\right) \\
&  =\dim\left(  \left(  \underbrace{f^{0}}_{=\id}\right)
^{-1}\left(  U\right)  \cap\ker\left(  f^{j}\right)  \right)  -\dim\left(
\left(  \underbrace{f^{0}}_{=\id}\right)  ^{-1}\left(
U\right)  \cap\ker\left(  f^{j-1}\right)  \right) \\
&  =\dim\left(  \underbrace{\left(  \operatorname*{id}\right)  ^{-1}\left(
U\right)  }_{=U}\cap\underbrace{\ker\left(  f^{j}\right)  }_{=\left(
f^{j}\right)  ^{-1}\left(  0\right)  }\right)  -\dim\left(
\underbrace{\left(  \operatorname*{id}\right)  ^{-1}\left(  U\right)  }%
_{=U}\cap\underbrace{\ker\left(  f^{j-1}\right)  }_{=\left(  f^{j-1}\right)
^{-1}\left(  0\right)  }\right) \\
&  =\dim\left(  U\cap\left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
-\dim\left(  U\cap\left(  f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  ,
\end{align*}
we obtain $\gamma_{j}=\left(  \lambda^{t}\right)  _{j}$ for every
$j\in\left\{  1,2,3,\ldots\right\}  $. Thus, $\gamma=\lambda^{t}$. Hence, $T$
is a column-strict tableau of shape $\lambda^{t}/\mu^{t}$ (since $T$ is a
column-strict tableau of shape $\gamma/\mu^{t}$).

Next, let us prove that $\cont T=\nu^{t}$. Indeed, we first
notice that%
\begin{equation}
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  =\left(
\dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in \NN . \label{sol.liri.jordan.b.nu-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.nu-sum}):} Let
$i\in \NN$. Recall that the nilpotent endomorphism $\overline{f}$ of the
quotient space $V/U$ (induced by $f\in\operatorname*{End}V$) has Jordan type
$\nu$. We can represent this nilpotent endomorphism $\overline{f}$ of $V/U$ as
an $\ell\times\ell$-matrix $H\in\mathbb{K}^{\ell\times\ell}$ for $\ell
=\dim\left(  V/U\right)  $. This $\ell\times\ell$-matrix $H$ is nilpotent
(since $\overline{f}$ is nilpotent) and has Jordan type $\nu$ (since
$\overline{f}$ has Jordan type $\nu$). But Exercise \ref{exe.liri.jordan}(a)
(applied to $\ell$, $H$ and $\nu$ instead of $n$, $N$ and $\lambda$)
shows that $H$ has Jordan type $\nu$ if and only if every
$k\in{ \NN }$ satisfies $\dim\left(  \ker\left(  H^{k}\right)  \right)
=\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k}$. Since we know that $H$ has Jordan type $\nu$, we thus
conclude that every $k\in{ \NN }$ satisfies $\dim\left(  \ker\left(
H^{k}\right)  \right)  =\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{k}$. Applied to $k=i$, this yields
$\dim\left(  \ker\left(  H^{i}\right)  \right)  =\left(  \nu^{t}\right)
_{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)  _{i}$.
\par
Since the matrix $H$ represents the endomorphism $\overline{f}$, we have
\[
\dim\left(  \ker\left(  \overline{f}^{i}\right)  \right)  =\dim\left(
\ker\left(  H^{i}\right)  \right)  =\left(  \nu^{t}\right)  _{1}+\left(
\nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)  _{i}.
\]
\par
But $\left(  f^{i}\right)  ^{-1}\left(  U\right)  $ is a $\mathbb{K}$-vector
subspace of $V$ containing $U$. Thus, $\left(  f^{i}\right)  ^{-1}\left(
U\right)  /U$ is canonically a $\mathbb{K}$-vector subspace of $V/U$.
Moreover, this subspace $\left(  f^{i}\right)  ^{-1}\left(  U\right)  /U$ is
precisely the kernel $\ker\left(  \overline{f}^{i}\right)  $ (this is
straightforward to check). Hence,
\[
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  /U\right)
=\dim\left(  \ker\left(  \overline{f}^{i}\right)  \right)  =\left(  \nu
^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)
_{i}.
\]
Since
\[
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  /U\right)
=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  -\dim U,
\]
this rewrites as%
\[
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  -\dim
U=\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{i}.
\]
Adding $\dim U$ to both sides of this equality yields
(\ref{sol.liri.jordan.b.nu-sum}).} As a consequence of this, we have%
\begin{equation}
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  -\dim\left(
\left(  f^{i-1}\right)  ^{-1}\left(  U\right)  \right)  =\left(  \nu
^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,3,\ldots\right\}  . \label{sol.liri.jordan.b.nu-single}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.nu-single}):} Let
$i\in\left\{  1,2,3,\ldots\right\}  $. Then,%
\begin{align*}
&  \underbrace{\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\right)  }_{\substack{=\left(  \dim U\right)  +\left(  \nu^{t}\right)
_{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)
_{i}\\\text{(by (\ref{sol.liri.jordan.b.nu-sum}))}}}-\underbrace{\dim\left(
\left(  f^{i-1}\right)  ^{-1}\left(  U\right)  \right)  }_{\substack{=\left(
\dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{i-1}\\\text{(by
(\ref{sol.liri.jordan.b.nu-sum}), applied}\\\text{to }i-1\text{ instead of
}i\text{)}}}\\
&  =\left(  \left(  \dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(
\nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)  _{i}\right)  -\left(
\left(  \dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{i-1}\right)  =\left(  \nu^{t}\right)
_{i}.
\end{align*}
Thus, (\ref{sol.liri.jordan.b.nu-single}) is proven.}

Now, for every $i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\begin{align*}
\left(  \cont T\right)  _{i}  &  =\left\vert T^{-1}\left(
i\right)  \right\vert =\left(  \text{the number of entries }i\text{ in
}T\right) \\
&  =\sum_{j=1}^{\infty}\underbrace{\left(  \text{the number of all entries
}i\text{ in the }j\text{-th row of }T\right)  }_{\substack{=b_{i,j}\\\text{(by
(\ref{sol.liri.jordan.b.A}))}}}=\sum_{j=1}^{\infty}\underbrace{b_{i,j}%
}_{=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}}\\
&  =\sum_{j=1}^{\infty}\underbrace{\left(  a_{i,j}-a_{i,j-1}-a_{i-1,j}%
+a_{i-1,j-1}\right)  }_{=\left(  a_{i,j}-a_{i,j-1}\right)  -\left(
a_{i-1,j}-a_{i-1,j-1}\right)  }=\sum_{j=1}^{\infty}\left(  \left(
a_{i,j}-a_{i,j-1}\right)  -\left(  a_{i-1,j}-a_{i-1,j-1}\right)  \right) \\
&  =\underbrace{\sum_{j=1}^{\infty}\left(  a_{i,j}-a_{i,j-1}\right)
}_{\substack{=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\right)  -\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap0\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(g),
applied to}\\W=V\text{, }A=U\text{, }B=0\text{ and }w_{i,j}=a_{i,j}\text{)}%
}}-\underbrace{\sum_{j=1}^{\infty}\left(  a_{i-1,j}-a_{i-1,j-1}\right)
}_{\substack{=\dim\left(  \left(  f^{i-1}\right)  ^{-1}\left(  U\right)
\right)  -\dim\left(  \left(  f^{i-1}\right)  ^{-1}\left(  U\right)
\cap0\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(f),
applied to}\\V\text{, }U\text{, }0\text{, }a_{i,j}\text{ and
}i-1\text{ instead of }W\text{, }A\text{, }B\text{, }w_{i,j}\text{ and
}i\text{)}}}\\
&  =\left(  \dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)
-\dim\underbrace{\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap0\right)  }_{=0}\right)  -\left(  \dim\left(  \left(  f^{i-1}\right)
^{-1}\left(  U\right)  \right)  -\dim\underbrace{\left(  \left(
f^{i-1}\right)  ^{-1}\left(  U\right)  \cap0\right)  }_{=0}\right) \\
&  =\left(  \dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)
-\underbrace{\dim0}_{=0}\right)  -\left(  \dim\left(  \left(  f^{i-1}\right)
^{-1}\left(  U\right)  \right)  -\underbrace{\dim0}_{=0}\right) \\
&  =\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)
-\dim\left(  \left(  f^{i-1}\right)  ^{-1}\left(  U\right)  \right)  =\left(
\nu^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.nu-single})}\right)  .
\end{align*}
Hence, $\cont T=\nu^{t}$.

We shall next see that for every positive integer $j$, the weak composition
$\cont (T|_{\operatorname{cols}\geq j})$ is a partition. Here,
and in the following, we are using the notations of Exercise
\ref{exe.liri.yamanouchi}.

We first notice that every
$i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$ satisfy%
\begin{equation}
\left(  \text{the number of entries }i\text{ in the first }j\text{ rows of
}T\right)  =a_{i,j}-a_{i-1,j}. \label{sol.liri.jordan.b.entrycount}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.entrycount}):} Let
$i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$. We have%
\begin{align*}
&  \left(  \text{the number of entries }i\text{ in the first }j\text{ rows of
}T\right) \\
&  =\sum_{k=1}^{j}\underbrace{\left(  \text{the number of all entries }i\text{
in the }k\text{-th row of }T\right)  }_{\substack{=b_{i,k}\\\text{(since
(\ref{sol.liri.jordan.b.A}) (applied to }\left(  i,k\right)  \text{
instead of }\left(  i,j\right)  \text{) yields }\\ b_{i,k}=\left(
\text{the number of all entries }i\text{ in the }k\text{-th row of }T\right)
\text{)}}}\\
&  =\sum_{k=1}^{j}\underbrace{b_{i,k}}_{\substack{=a_{i,k}-a_{i,k-1}%
-a_{i-1,k}+a_{i-1,k-1}\\\text{(by the definition of }b_{i,k}\text{)}}%
}=\sum_{k=1}^{j}\underbrace{\left(  a_{i,k}-a_{i,k-1}-a_{i-1,k}+a_{i-1,k-1}%
\right)  }_{=\left(  a_{i,k}-a_{i-1,k}\right)  -\left(  a_{i,k-1}%
-a_{i-1,k-1}\right)  }\\
&  =\sum_{k=1}^{j}\left(  \left(  a_{i,k}-a_{i-1,k}\right)  -\left(
a_{i,k-1}-a_{i-1,k-1}\right)  \right) \\
& =\left(  a_{i,j}-a_{i-1,j}\right)
-\left(  \underbrace{a_{i,0}}_{\substack{=0\\\text{(by
(\ref{sol.liri.jordan.b.ai0}))}}}-\underbrace{a_{i-1,0}}%
_{\substack{=0\\\text{(by (\ref{sol.liri.jordan.b.ai0}), applied}\\\text{to
}i-1\text{ instead of }i\text{)}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the telescope principle}\right) \\
&  =\left(  a_{i,j}-a_{i-1,j}\right)  -\left(  0-0\right)  =a_{i,j}-a_{i-1,j}.
\end{align*}
This proves (\ref{sol.liri.jordan.b.entrycount}).} Now, for every positive
integers $j$ and $i$, we have%
\begin{align*}
&  \left(  \text{the number of entries }i+1\text{ in the first }j\text{ rows
of }T\right) \\
&  =a_{i+1,j}-a_{i,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.entrycount}), applied to }i+1\text{ instead of
}i\right) \\
&  \leq a_{i,j-1}-a_{i-1,j-1}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.sol.liri.jordan.b.general}(c), applied to}\\
V\text{, }U\text{, }0\text{, }a_{i,j}\text{ and }j-1\text{ instead of
}W\text{, }A\text{, }B\text{, }w_{i,j}\text{ and }j
\end{array}
\right) \\
&  =\left(  \text{the number of entries }i\text{ in the first }j-1\text{ rows
of }T\right)
\end{align*}
(since (\ref{sol.liri.jordan.b.entrycount}) (applied to $j-1$ instead of $j$)
yields $\left(  \text{the number of entries }i\text{ in the first }j-1\text{
rows of }T\right)  =a_{i,j-1}-a_{i-1,j-1}$). In other words, for every
positive integers $j$ and $i$, the number of entries $i+1$ in the first $j$
rows of $T$ is $\leq$ to the number of entries $i$ in the first $j-1$ rows of
$T$. In other words, Assertion $\mathcal{D}$ of Exercise
\ref{exe.liri.yamanouchi}(b) (with $\lambda$ and $\mu$ replaced by
$\lambda^{t}$ and $\mu^{t}$) is satisfied. Hence, Assertion $\mathcal{C}$ of
Exercise \ref{exe.liri.yamanouchi}(b) (with $\lambda$ and $\mu$ replaced by
$\lambda^{t}$ and $\mu^{t}$) is satisfied as well (because Exercise
\ref{exe.liri.yamanouchi}(b) yields that these Assertions $\mathcal{C}$ and
$\mathcal{D}$ are equivalent). In other words, for every positive integer $j$,
the weak composition $\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  $ is a partition.

Now, let us forget that we defined $\nu$ and $T$. We thus have found a
column-strict tableau $T$ of shape $\lambda^{t}/\mu^{t}$ with
$\cont T=\nu^{t}$ which has the property that for every positive
integer $j$, the weak composition $\cont \left(
T|_{\operatorname{cols}\geq j}\right)  $ is a partition. But we know that the
number of such tableaux is $c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$ (by Corollary
\ref{L-R-rule}, applied to $\lambda^{t}$, $\mu^{t}$ and $\nu^{t}$ instead of
$\lambda$, $\mu$ and $\nu$). Hence, this number $c_{\mu^{t},\nu^{t}}%
^{\lambda^{t}}$ must be $\neq0$ (because we have found such a tableau $T$). So
we have proven that $c_{\mu^{t},\nu^{t}}^{\lambda^{t}}\neq0$. Now, Exercise
\ref{exe.pieri.omega}(c) yields $c_{\mu,\nu}^{\lambda}=c_{\mu^{t},\nu^{t}%
}^{\lambda^{t}}\neq0$, so that Exercise \ref{exe.liri.jordan}(b) is solved.
%\end{solutions}


\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.liri.jordan}}

\textit{Solution to Exercise \ref{exe.liri.jordan}.} (a) We shall first
prove that%
\begin{equation}
\text{if }N\text{ has Jordan type }\lambda\text{, then every }k\in
 \NN \text{ satisfies }\dim\left(  \ker\left(  N^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}. \label{sol.liri.jordan.a.goal}%
\end{equation}


\textit{Proof of (\ref{sol.liri.jordan.a.goal}):} Assume that $N$ has Jordan
type $\lambda$. Let $k\in \NN$.

For each $m \in \NN$, let $J_{m}$ be $m\times m$-matrix $\left(
\begin{array}
[c]{ccccc}%
0 & 1 & 0 & \cdots & 0\\
0 & 0 & 1 & \cdots & 0\\
0 & 0 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & 0
\end{array}
\right) \in \mathbb{K}^{m\times m}$.
This is a Jordan block of size $m$ corresponding to the
eigenvalue $0$ (when $m$ is positive). For future use, we record the
following simple fact:
\begin{equation}
\dim \left( \ker \left( J_m^k \right) \right) = \min \left\{ m, k \right \}
\label{sol.liri.jordan.a.Jm-power}
\end{equation}
for all $m \in \NN$.\ \ \ \ \footnote{\textit{Proof sketch.} Let
$m \in \NN$. 
Let $\left(e_1, e_2, \ldots, e_m\right)$ be the standard basis of
the $\mathbb{K}$-vector space $\mathbb{K}^m$.
Now, it is easy to check that
$J_m^k$ is the $m\times m$-matrix
whose $\left(i, j\right)$-th entry is
$\begin{cases} 1, & \text{if } j = i + k; \\
               0, & \text{if } j \neq i + k
 \end{cases}
$ for all $\left(i, j\right) \in \left\{1, 2, \ldots, m\right\}^2$.
Hence, it is easy to see that $\ker \left( J_m^k \right)$ is the
$\mathbb{K}$-linear span of the basis vectors $e_j$ with $j \leq k$.
These basis vectors are linearly independent and their number is
$\min \left\{ m, k \right \}$; therefore,
\eqref{sol.liri.jordan.a.Jm-power} follows.}.

Since $N$ has Jordan type $\lambda$, the Jordan normal form of $N$ has Jordan
blocks of sizes $\lambda_{1},\lambda_{2},\lambda_{3},\ldots$.
Since the only eigenvalue of $N$ is $0$, this shows that
the Jordan blocks of $N$ are $J_{\lambda_1}, J_{\lambda_2}, J_{\lambda_3},
\ldots$ (or, more precisely, the nonempty matrices among
$J_{\lambda_1}, J_{\lambda_2}, J_{\lambda_3}, \ldots$). 
In other words,
$N$ is similar to the block-diagonal matrix $J_{\lambda}:=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  $. (This block-diagonal matrix
$J_{\lambda}$ is finite, since only finitely many $\lambda_{p}$ are nonzero.)
We can thus WLOG assume that $N$ \textbf{is} this matrix $J_{\lambda}$
(because replacing $N$ by a matrix similar to $N$ changes neither the
dimension $\dim\left(  \ker\left(  N^{k}\right)  \right)  $ nor the Jordan
type of $N$). Assume this. Thus, $N=J_{\lambda}=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  $, so that%
\[
N^{k}=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  ^{k}=\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}}^{k} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}}^{k} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}}^{k} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)
\]
and therefore%
\[
\ker\left(  N^{k}\right)  =\ker\left(
\begin{array}
[c]{cccc}%
J_{\lambda_{1}}^{k} & 0 & 0 & \cdots\\
0 & J_{\lambda_{2}}^{k} & 0 & \cdots\\
0 & 0 & J_{\lambda_{3}}^{k} & \cdots\\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right)  \cong\bigoplus_{p\geq1}\ker\left(  J_{\lambda_{p}}^{k}\right)  ,
\]
whence%
\begin{equation}
\dim\left(  \ker\left(  N^{k}\right)  \right)  =\dim\left(  \bigoplus_{p\geq
1}\ker\left(  J_{\lambda_{p}}^{k}\right)  \right)  =\sum_{p\geq1}%
\underbrace{\dim\left(  \ker\left(  J_{\lambda_{p}}^{k}\right)  \right)
}_{\substack{=\min\left\{  \lambda_{p},k\right\}  \\\text{(by
\eqref{sol.liri.jordan.a.Jm-power} (applied to }
m = \lambda_p \text{))}}}
=\sum_{p\geq1}\min\left\{  \lambda_{p},k\right\}  .
\label{sol.liri.jordan.a.1}%
\end{equation}


We are now going to prove that the right hand side of this equality is
$\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}$. Indeed, let us use the
\dfn{Iverson bracket notation}: For every assertion $\mathcal{A}$, we let
$\left[  \mathcal{A}\right]  $ denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.
(This integer is called the \dfn{truth value} of $\mathcal{A}$.)

Any two nonnegative integers $u$ and $v$ satisfy $\min\left\{  u,v\right\}
=\sum_{i=1}^{v}\left[  i\leq u\right]  $. Thus, $\min\left\{  \lambda
_{p},k\right\}  =\sum_{i=1}^{k}\left[  i\leq\lambda_{p}\right]  $ for every
$p\in\left\{  1,2,3,\ldots\right\}  $. Hence, (\ref{sol.liri.jordan.a.1})
becomes%
\begin{align*}
\dim\left(  \ker\left(  N^{k}\right)  \right)   &  =\sum_{p\geq1}%
\underbrace{\min\left\{  \lambda_{p},k\right\}  }_{=\sum_{i=1}^{k}\left[
i\leq\lambda_{p}\right]  }=\sum_{p\geq1}\sum_{i=1}^{k}\left[  i\leq\lambda
_{p}\right]  =\sum_{i=1}^{k}\underbrace{\sum_{p\geq1}\left[  i\leq\lambda
_{p}\right]  }_{\substack{=\left\vert \left\{  p\geq1\ \mid\ i\leq\lambda
_{p}\right\}  \right\vert \\=\left\vert \left\{  p\geq1\ \mid\ \lambda_{p}\geq
i\right\}  \right\vert \\=\left\vert \left\{  j\geq1\ \mid\ \lambda_{j}\geq
i\right\}  \right\vert =\left(  \lambda^{t}\right)  _{i}\\\text{(by
(\ref{eq.partition.transpose}))}}}\\
&  =\sum_{i=1}^{k}\left(  \lambda^{t}\right)  _{i}=\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}.
\end{align*}
This proves (\ref{sol.liri.jordan.a.goal}).

With (\ref{sol.liri.jordan.a.goal}), we have proven one direction of the
equivalence that Exercise \ref{exe.liri.jordan}(a) requires us to prove. To
prove the other direction, we need to show that
\begin{equation}
\text{if every }k\in \NN \text{ satisfies }\dim\left(  \ker\left(
N^{k}\right)  \right)  =\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}\text{, then }N\text{
has Jordan type }\lambda\text{.} \label{sol.liri.jordan.a.goal2}%
\end{equation}


\textit{Proof of (\ref{sol.liri.jordan.a.goal2}):} Assume that every
$k\in \NN$ satisfies $\dim\left(  \ker\left(  N^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}$. Let $\mu$ be the Jordan type of
$N$. Then, (\ref{sol.liri.jordan.a.goal}) (applied to $\mu$ instead of
$\lambda$) shows that every $k\in \NN$ satisfies $\dim\left(
\ker\left(  N^{k}\right)  \right)  =\left(  \mu^{t}\right)  _{1}+\left(
\mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{k}$. Hence, every
$k\in \NN$ satisfies%
\begin{equation}
\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}=\dim\left(  \ker\left(  N^{k}\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{k}. \label{sol.liri.jordan.a.4}%
\end{equation}
Applying (\ref{sol.liri.jordan.a.4}) to $k-1$ instead of $k$, and subtracting
the result from (\ref{sol.liri.jordan.a.4}), we obtain%
\[
\left(  \mu^{t}\right)  _{k}=\left(  \lambda^{t}\right)  _{k}%
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }k\text{.}%
\]
Hence, $\mu^{t}=\lambda^{t}$, so that $\mu=\lambda$. Thus, $N$ has Jordan type
$\mu=\lambda$. This proves (\ref{sol.liri.jordan.a.goal2}), and thus the
solution of Exercise \ref{exe.liri.jordan}(a) is complete.

Before we come to the solution of Exercise \ref{exe.liri.jordan}(b), let us
show a linear-algebraic lemma:

\begin{lemma}
\label{lem.sol.liri.jordan.b.general}Let $W$ be a finite-dimensional
$\mathbb{K}$-vector space. Let $A$ and $B$ be $\mathbb{K}$-vector subspaces of
$W$. Let $f\in\operatorname*{End}W$ be such that $f\left(  A\right)  \subset
A$ and $f\left(  B\right)  \subset B$. For any $\left(  i,j\right)
\in \NN ^{2}$, let us define a nonnegative integer $w_{i,j}$ by
$w_{i,j}=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  $.

(a) For any $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$, we
have $w_{i,j}+w_{i-1,j-1}\geq w_{i,j-1}+w_{i-1,j}$.

(b) For any $\left(  i,j\right)  \in \NN \times\left\{  1,2,3,\ldots
\right\}  $, we have $w_{i+1,j+1}-w_{i+1,j}\leq w_{i,j}-w_{i,j-1}$.

(c) For any $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
\times \NN$, we have $w_{i+1,j+1}-w_{i,j+1}\leq w_{i,j}-w_{i-1,j}$.

(d) If $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$ are such
that $i>\dim W$, then $w_{i,j}=w_{i-1,j}$.

(e) If $j\in\left\{  1,2,3,\ldots\right\}  $ and $i\in \NN$ are such
that $j>\dim W$, then $w_{i,j}=w_{i,j-1}$.

(f) Assume that $f$ is nilpotent. For every $j\in \NN$, we have
$\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)  =\dim\left(  \left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  -\dim\left(  A\cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  $. (In particular, the sum
$\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)  $ converges with respect
to the discrete topology, i.e., all but finitely many of its terms are $0$.)

(g) Assume that $f$ is nilpotent. For every $i\in \NN$, we have
$\sum_{j=1}^{\infty}\left(  w_{i,j}-w_{i,j-1}\right)  =\dim\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \right)  -\dim\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap B\right)  $. (In particular, the sum
$\sum_{j=1}^{\infty}\left(  w_{i,j}-w_{i,j-1}\right)  $ converges with respect
to the discrete topology, i.e., all but finitely many of its terms are $0$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.liri.jordan.b.general}.]
Notice first that $\left(\underbrace{f^0}_{=\id}\right)^{-1} \left(A\right)
= \id^{-1} \left( A \right) = A$ and
$A = \left(f^0\right)^{-1} \left(A\right) \subset
\left(f^1\right)^{-1} \left(A\right) \subset
\left(f^2\right)^{-1} \left(A\right) \subset \cdots$ (since
$f\left(A\right) \subset A$) and
$B = \left(f^0\right)^{-1} \left(B\right) \subset
\left(f^1\right)^{-1} \left(B\right) \subset
\left(f^2\right)^{-1} \left(B\right) \subset \cdots$ (similarly).

(a) Let $\left(
i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. We need to prove that
$w_{i,j}+w_{i-1,j-1}\geq w_{i,j-1}+w_{i-1,j}$. In other words, we need to
prove that $w_{i,j}-w_{i,j-1}\geq w_{i-1,j}-w_{i-1,j-1}$. Since
\begin{align}
w_{i,j}-w_{i,j-1}  &  =\dim\left(  \left(  f^{i}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)
-\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definitions of }w_{i,j}\text{ and
}w_{i,j-1}\right) \nonumber\\
&  =\dim\left(  \left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right)  \right) \label{pf.lem.sol.liri.jordan.b.general.a.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  f^{i}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)  \subset\left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right) \nonumber
\end{align}
and%
\[
w_{i-1,j}-w_{i-1,j-1}=\dim\left(  \left(  \left(  f^{i-1}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(
\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)
^{-1}\left(  B\right)  \right)  \right)
\]
(by the same argument as (\ref{pf.lem.sol.liri.jordan.b.general.a.1}), only
with $i-1$ instead of $i$), this is equivalent to showing that%
\begin{align*}
&  \dim\left(  \left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right)  \right) \\
&  \geq\dim\left(  \left(  \left(  f^{i-1}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)
^{-1}\left(  B\right)  \right)  \right)  .
\end{align*}
This will clearly be achieved if we can construct a $\mathbb{K}$-linear
injection%
\begin{align*}
&  \left(  \left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i-1}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  .
\end{align*}
Here is how to construct it: Let $\iota$ denote the canonical inclusion
$\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)
^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  $. Then, $\iota$ restricts to
an inclusion $\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)  $,
and so gives rise to a map%
\begin{align*}
&  \left(  \left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i-1}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  .
\end{align*}
This map is injective because $\iota^{-1}\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  =\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  $, and thus we have found our
$\mathbb{K}$-linear injection. Lemma \ref{lem.sol.liri.jordan.b.general}(a) is proven.

(b) Let $\left(  i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}
$. We need to prove that $w_{i+1,j+1}-w_{i+1,j}\leq w_{i,j}-w_{i,j-1}$. Due to
(\ref{pf.lem.sol.liri.jordan.b.general.a.1}) and due to%
\[
w_{i+1,j+1}-w_{i+1,j}=\dim\left(  \left(  \left(  f^{i+1}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(
\left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)
^{-1}\left(  B\right)  \right)  \right)
\]
(this follows by the same arguments as
(\ref{pf.lem.sol.liri.jordan.b.general.a.1}), only with $i$ and $j$ replaced
by $i+1$ and $j+1$), this is equivalent to showing that%
\begin{align*}
&  \dim\left(  \left(  \left(  f^{i+1}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right)  \right) \\
&  \leq\dim\left(  \left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right)  \right)  .
\end{align*}
This will clearly be achieved if we can construct a $\mathbb{K}$-linear
injection%
\begin{align*}
&  \left(  \left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  .
\end{align*}
Here is how this can be done: The map $f$ restricts to a map $\varphi:\left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j+1}\right)
^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)  ^{-1}\left(  A\right)
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  $, which further restricts to
a map $\left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \rightarrow\left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)  $.
Hence, $\varphi$ gives rise to a map%
\begin{align*}
&  \left(  \left(  f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j+1}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right) \\
&  \rightarrow\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  /\left(  \left(  f^{i}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  ,
\end{align*}
which is injective because%
\begin{align*}
\varphi^{-1}\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j-1}\right)  ^{-1}\left(  B\right)  \right)   &  =f^{-1}\left(  \left(
f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j-1}\right)  ^{-1}\left(
B\right)  \right) \\
&  =\underbrace{f^{-1}\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)
\right)  }_{=\left(  f^{i+1}\right)  ^{-1}\left(  A\right)  }\cap
\underbrace{f^{-1}\left(  \left(  f^{j-1}\right)  ^{-1}\left(  B\right)
\right)  }_{=\left(  f^{j}\right)  ^{-1}\left(  B\right)  }=\left(
f^{i+1}\right)  ^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  .
\end{align*}
Thus, we have found our $\mathbb{K}$-linear injection. Lemma
\ref{lem.sol.liri.jordan.b.general}(b) is proven.

(c) The proof of Lemma \ref{lem.sol.liri.jordan.b.general}(c) is analogous to
our above proof of Lemma \ref{lem.sol.liri.jordan.b.general}(b) (one merely
has to interchange $A$ with $B$ and $i$ with $j$).

(d) We have%
\[
\left(  f^{0}\right)  ^{-1}\left(  A\right)  \subset\left(  f^{1}\right)
^{-1}\left(  A\right)  \subset\cdots\subset\left(  f^{\dim W+1}\right)
^{-1}\left(  A\right)
\]
and therefore%
\[
\dim\left(  \left(  f^{0}\right)  ^{-1}\left(  A\right)  \right)  \leq
\dim\left(  \left(  f^{1}\right)  ^{-1}\left(  A\right)  \right)  \leq
\cdots\leq\dim\left(  \left(  f^{\dim W+1}\right)  ^{-1}\left(  A\right)
\right)  .
\]
This latter chain of inequalities contains $\dim W+1$ inequality signs, but
only at most $\dim W$ of them can be strict (because each $\dim\left(  \left(
f^{k}\right)  ^{-1}\left(  A\right)  \right)  $ is an integer between $0$ and
$\dim W$ inclusive\footnote{since $\left(  f^{k}\right)  ^{-1}\left(
A\right)  $ is a subspace of the finite-dimensional $\mathbb{K}$-vector space
$W$}, and a sequence of $\dim W+2$ integers between $0$ and $\dim W$ cannot
strictly increase). Thus, at least one of the inequality signs is an equality.
That is, there exists an $I\in\left\{  1,2,\ldots,\dim W+1\right\}  $ such
that $\dim\left(  \left(  f^{I-1}\right)  ^{-1}\left(  A\right)  \right)
=\dim\left(  \left(  f^{I}\right)  ^{-1}\left(  A\right)  \right)  $. Consider
this $I$.

Since $\dim\left(  \left(  f^{I-1}\right)  ^{-1}\left(  A\right)  \right)
=\dim\left(  \left(  f^{I}\right)  ^{-1}\left(  A\right)  \right)  $ and
$\left(  f^{I-1}\right)  ^{-1}\left(  A\right)  \subset\left(  f^{I}\right)
^{-1}\left(  A\right)  $, we must have
\begin{equation}
\left(  f^{I-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{I}\right)
^{-1}\left(  A\right)  . \label{pf.lem.sol.liri.jordan.b.general.c.1}%
\end{equation}


From here, it is easy to see that
\[
\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{i}\right)
^{-1}\left(  A\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,3,\ldots\right\}  \text{ satisfying }i>\dim W.
\]
\footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots\right\}  $ be such
that $i>\dim W$. Thus, $i\geq\dim W+1$. But $I\leq\dim W+1$ (since
$I\in\left\{  1,2,\ldots,\dim W+1\right\}  $) and thus $i\geq\dim W+1\geq I$.
Hence, there exists some $k\in \NN$ such that $i=I+k$. Consider this
$k$. Since $i=I+k$, we have $i-1=I+k-1=\left(  I-1\right)  +k$, so that
$f^{i-1}=f^{\left(  I-1\right)  +k}=f^{I-1}\circ f^{k}$ and thus $\left(
f^{i-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{I-1}\circ f^{k}\right)
^{-1}\left(  A\right)  =\left(  f^{k}\right)  ^{-1}\left(  \left(
f^{I-1}\right)  ^{-1}\left(  A\right)  \right)  $. Similarly, $\left(
f^{i}\right)  ^{-1}\left(  A\right)  =\left(  f^{k}\right)  ^{-1}\left(
\left(  f^{I}\right)  ^{-1}\left(  A\right)  \right)  $. Now,%
\[
\left(  f^{i-1}\right)  ^{-1}\left(  A\right)  =\left(  f^{k}\right)
^{-1}\left(  \underbrace{\left(  f^{I-1}\right)  ^{-1}\left(  A\right)
}_{\substack{=\left(  f^{I}\right)  ^{-1}\left(  A\right)  \\\text{(by
(\ref{pf.lem.sol.liri.jordan.b.general.c.1}))}}}\right)  =\left(
f^{k}\right)  ^{-1}\left(  \left(  f^{I}\right)  ^{-1}\left(  A\right)
\right)  =\left(  f^{i}\right)  ^{-1}\left(  A\right)  ,
\]
qed.}

Thus, for every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfying $i>\dim W$,
we have%
\begin{align*}
w_{i-1,j}  &  =\dim\left(  \underbrace{\left(  f^{i-1}\right)  ^{-1}\left(
A\right)  }_{=\left(  f^{i}\right)  ^{-1}\left(  A\right)  }\cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }w_{i-1,j}\right) \\
&  =\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  A\right)  \cap\left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  =w_{i,j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }j\in \NN .
\end{align*}
Thus, if $i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$ are such
that $i>\dim W$, then $w_{i-1,j}=w_{i-1,j}$. This proves Lemma
\ref{lem.sol.liri.jordan.b.general}(d).

(e) The proof of Lemma \ref{lem.sol.liri.jordan.b.general}(e) is analogous to
our above proof of Lemma \ref{lem.sol.liri.jordan.b.general}(d) (one merely
has to interchange $A$ with $B$ and $i$ with $j$).

(f) Let $j\in \NN$. We have%
\begin{align}
\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)   &  =\sum_{i=1}^{\dim
W}\left(  w_{i,j}-w_{i-1,j}\right)  +\sum_{i=\dim W+1}^{\infty}\left(
\underbrace{w_{i,j}}_{\substack{=w_{i-1,j}\\\text{(by Lemma
\ref{lem.sol.liri.jordan.b.general}(d),}\\\text{since }i>\dim W\text{)}%
}}-w_{i-1,j}\right) \nonumber\\
&  =\sum_{i=1}^{\dim W}\left(  w_{i,j}-w_{i-1,j}\right)  +\sum_{i=\dim
W+1}^{\infty}\underbrace{\left(  w_{i-1,j}-w_{i-1,j}\right)  }_{=0}=\sum
_{i=1}^{\dim W}\left(  w_{i,j}-w_{i-1,j}\right)  +\underbrace{\sum_{i=\dim
W+1}^{\infty}0}_{=0}\nonumber\\
&  =\sum_{i=1}^{\dim W}\left(  w_{i,j}-w_{i-1,j}\right)  =w_{\dim W,j}%
-w_{0,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the telescope principle}\right)
. \label{pf.lem.sol.liri.jordan.b.general.f.1}%
\end{align}
Now, we are going to prove that $w_{\dim W,j}=\dim\left(  \left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  $ and $w_{0,j}=\dim\left(
A\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  $.

Indeed, it is well-known that if $g$ is any nilpotent endomorphism of a
finite-dimensional vector space $V$, then $g^{\dim V}=0$. Applied to $g=f$ and
$V=W$, this yields $f^{\dim W}=0$, and thus $f^{\dim W}\left(  W\right)
=0\left(  W\right)  =0\subset A$, whence $W\subset\left(  f^{\dim W}\right)
^{-1}\left(  A\right)  $, so that $\left(  f^{\dim W}\right)  ^{-1}\left(
A\right)  =W$. Hence,%
\[
\underbrace{\left(  f^{\dim W}\right)  ^{-1}\left(  A\right)  }_{=W}%
\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  =W\cap\left(  f^{j}\right)
^{-1}\left(  B\right)  =\left(  f^{j}\right)  ^{-1}\left(  B\right)
\]
(since $\left(  f^{j}\right)  ^{-1}\left(  B\right)  \subset W$). Now, the
definition of $w_{\dim W,j}$ yields%
\[
w_{\dim W,j}=\dim\left(  \underbrace{\left(  f^{\dim W}\right)  ^{-1}\left(
A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  }_{=\left(
f^{j}\right)  ^{-1}\left(  B\right)  }\right)  =\dim\left(  \left(
f^{j}\right)  ^{-1}\left(  B\right)  \right)  .
\]
Also, the definition of $w_{0,j}$ yields%
\[
w_{0,j}=\dim\left(  \left(  \underbrace{f^{0}}_{=\id}\right)
^{-1}\left(  A\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  B\right)
\right)  =\dim\left(  \underbrace{\id^{-1}\left(
A\right)  }_{=A}\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)
=\dim\left(  A\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)  .
\]
Hence, (\ref{pf.lem.sol.liri.jordan.b.general.f.1}) becomes%
\[
\sum_{i=1}^{\infty}\left(  w_{i,j}-w_{i-1,j}\right)  =\underbrace{w_{\dim
W,j}}_{=\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  B\right)  \right)
}-\underbrace{w_{0,j}}_{=\dim\left(  A\cap\left(  f^{j}\right)  ^{-1}\left(
B\right)  \right)  }=\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  B\right)
\right)  -\dim\left(  A\cap\left(  f^{j}\right)  ^{-1}\left(  B\right)
\right)  .
\]
This proves Lemma \ref{lem.sol.liri.jordan.b.general}(f).

(g) The proof of Lemma \ref{lem.sol.liri.jordan.b.general}(g) is analogous to
our above proof of Lemma \ref{lem.sol.liri.jordan.b.general}(f) (one merely
has to interchange $A$ with $B$ and $i$ with $j$).
\end{proof}

(b) Assume that $\ZZ$ is a subring of $\kk$.

For any $\left(  i,j\right)  \in \NN ^{2}$, let us define a nonnegative
integer $a_{i,j}$ by%
\[
a_{i,j}=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \cap
\ker\left(  f^{j}\right)  \right)  .
\]
Furthermore, for every $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}
^{2}$, let us define an integer $b_{i,j}$ by%
\[
b_{i,j}=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}.
\]
We first observe that%
\begin{equation}
a_{i,0}=0\ \ \ \ \ \ \ \ \ \ \text{for every }i\in \NN .
\label{sol.liri.jordan.b.ai0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.ai0}):} For every
$i\in \NN$, the definition of $a_{i,0}$ yields $a_{i,0}=\dim\left(
\left(  f^{i}\right)  ^{-1}\left(  U\right)  \cap\underbrace{\ker\left(
f^{0}\right)  }_{=\ker\left(  \operatorname*{id}\right)  =0}\right)
=\dim\underbrace{\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap0\right)  }_{=0}=\dim0=0$, qed.} Furthermore,%
\begin{equation}
a_{0,j}=\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}%
+\cdots+\left(  \mu^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for every }%
j\in \NN . \label{sol.liri.jordan.b.a0j}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.a0j}):} Let $j\in
 \NN$. We can represent the endomorphism $f\mid U$ of $U$ as a $k\times
k$-matrix $G\in\mathbb{K}^{k\times k}$ for $k=\dim U$. This $k\times k$-matrix
$G$ is nilpotent (since $f\mid U$ is nilpotent) and has Jordan type $\mu$
(since $f\mid U$ has Jordan type $\mu$). But Exercise \ref{exe.liri.jordan}(a)
(applied to $k$, $G$ and $\mu$ instead of $n$, $N$ and $\lambda$) shows that
$G$ has Jordan type $\mu$ if and only if every $k\in{ \NN }$ satisfies
$\dim\left(  \ker\left(  G^{k}\right)  \right)  =\left(  \mu^{t}\right)
_{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{k}$. Since
we know that $G$ has Jordan type $\mu$, we thus conclude that every
$k\in{ \NN }$ satisfies $\dim\left(  \ker\left(  G^{k}\right)  \right)
=\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(
\mu^{t}\right)  _{k}$. Applied to $k=j$, this yields $\dim\left(  \ker\left(
G^{j}\right)  \right)  =\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)
_{2}+\cdots+\left(  \mu^{t}\right)  _{j}$.
\par
Since the matrix $G$ represents the endomorphism $f\mid U$, we have
\[
\dim\left(  \ker\left(  G^{j}\right)  \right)  =\dim\left(  \underbrace{\ker
\left(  \left(  f\mid U\right)  ^{j}\right)  }_{=U\cap\ker\left(
f^{j}\right)  }\right)  =\dim\left(  U\cap\ker\left(  f^{j}\right)  \right)
.
\]
Compared with $\dim\left(  \ker\left(  G^{j}\right)  \right)  =\left(  \mu
^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)
_{j}$, this yields%
\[
\dim\left(  U\cap\ker\left(  f^{j}\right)  \right)  =\left(  \mu^{t}\right)
_{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{j}.
\]
Now, the definition of $a_{0,j}$ yields
\[
a_{0,j}=\dim\left(  \left(  \underbrace{f^{0}}_{=\id}\right)
^{-1}\left(  U\right)  \cap\ker\left(  f^{j}\right)  \right)  =\dim\left(
\underbrace{\id^{-1}\left(  U\right)  }_{=U}\cap
\ker\left(  f^{j}\right)  \right)  =\dim\left(  U\cap\ker\left(  f^{j}\right)
\right)  =\left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}%
+\cdots+\left(  \mu^{t}\right)  _{j}.
\]
This proves (\ref{sol.liri.jordan.b.a0j}).} Thus,%
\begin{equation}
a_{0,j}-a_{0,j-1}=\left(  \mu^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for
every }j\in\left\{  1,2,3,\ldots\right\}  . \label{sol.liri.jordan.b.a0j-diff}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.a0j-diff}):} For every
$j\in\left\{  1,2,3,\ldots\right\}  $, we have%
\[
\underbrace{a_{0,j}}_{\substack{=\left(  \mu^{t}\right)  _{1}+\left(  \mu
^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{j}\\\text{(by
(\ref{sol.liri.jordan.b.a0j}))}}}-\underbrace{a_{0,j-1}}_{\substack{=\left(
\mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu
^{t}\right)  _{j-1}\\\text{(by (\ref{sol.liri.jordan.b.a0j}), applied
to}\\j-1\text{ instead of }j\text{)}}}=\left(  \left(  \mu^{t}\right)
_{1}+\left(  \mu^{t}\right)  _{2}+\cdots+\left(  \mu^{t}\right)  _{j}\right)
-\left(  \left(  \mu^{t}\right)  _{1}+\left(  \mu^{t}\right)  _{2}%
+\cdots+\left(  \mu^{t}\right)  _{j-1}\right)  =\left(  \mu^{t}\right)  _{j},
\]
qed.} Moreover, every $\left(  i,j\right)  \in \NN ^{2}$ satisfies%
\[
a_{i,j}=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap\underbrace{\ker\left(  f^{j}\right)  }_{=\left(  f^{j}\right)
^{-1}\left(  0\right)  }\right)  =\dim\left(  \left(  f^{i}\right)
^{-1}\left(  U\right)  \cap\left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  .
\]
Hence, Lemma \ref{lem.sol.liri.jordan.b.general}(a) (applied to $W=V$, $A=U$,
$B=0$ and $w_{i,j}=a_{i,j}$) shows that for any $\left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}$, we have $a_{i,j}+a_{i-1,j-1}\geq
a_{i,j-1}+a_{i-1,j}$. In other words, for any $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$, we have $a_{i,j}-a_{i,j-1}-a_{i-1,j}%
+a_{i-1,j-1}\geq0$. In other words, for any $\left(  i,j\right)  \in\left\{
1,2,3,\ldots\right\}  ^{2}$, we have%
\begin{equation}
b_{i,j}\geq0 \label{sol.liri.jordan.b.bij}%
\end{equation}
(since $b_{i,j}=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}$). Thus, $b_{i,j}$ is
a nonnegative integer for every two positive integers $i$ and $j$. Moreover,%
\begin{equation}
b_{i,j}=0\ \ \ \ \ \ \ \ \ \ \text{for all but finitely many pairs }\left(
i,j\right)  . \label{sol.liri.jordan.b.bij0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.bij0}):} Lemma
\ref{lem.sol.liri.jordan.b.general}(d) (applied to $W=V$, $A=U$, $B=0$ and
$w_{i,j}=a_{i,j}$) shows that if $i\in\left\{  1,2,3,\ldots\right\}  $ and
$j\in \NN$ are such that $i>\dim V$, then%
\begin{equation}
a_{i,j}=a_{i-1,j}. \label{sol.liri.jordan.b.bij0.pf.1}%
\end{equation}
\par
Lemma \ref{lem.sol.liri.jordan.b.general}(e) (applied to $W=V$, $A=U$, $B=0$
and $w_{i,j}=a_{i,j}$) shows that if $j\in\left\{  1,2,3,\ldots\right\}  $ and
$i\in \NN$ satisfy $j>\dim V$, then $a_{i,j}=a_{i,j-1}$.
\par
Now, let $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$. If
$i>\dim V$, then%
\[
b_{i,j}=\underbrace{a_{i,j}}_{\substack{=a_{i-1,j}\\\text{(by
(\ref{sol.liri.jordan.b.bij0.pf.1}),}\\\text{since }i>\dim V\text{)}%
}}-\underbrace{a_{i,j-1}}_{\substack{=a_{i-1,j-1}\\\text{(by
(\ref{sol.liri.jordan.b.bij0.pf.1}) (applied to}\\j-1\text{ instead of
}j\text{),}\\\text{since }i> \dim V\text{)}}}-a_{i-1,j}+a_{i-1,j-1}%
=a_{i-1,j}-a_{i-1,j-1}-a_{i-1,j}+a_{i-1,j-1}=0.
\]
Similarly, $b_{i,j}=0$ if $j>\dim V$. Hence, we see that $b_{i,j}=0$ if we
have $i>\dim V$ or $j>\dim V$ (or both). Thus, $b_{i,j}=0$ for all but
finitely many pairs $\left(  i,j\right)  $ (because all but finitely many
pairs $\left(  i,j\right)  $ satisfy $i>\dim V$ or $j>\dim V$). This proves
(\ref{sol.liri.jordan.b.bij0}).} Also,%
\begin{equation}
\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}\right)
=a_{i,j}-a_{i,j-1}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in \NN \times\left\{  1,2,3,\ldots\right\}  .
\label{sol.liri.jordan.b.bij-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.bij-sum}):} Let $\left(
i,j\right)  \in \NN \times\left\{  1,2,3,\ldots\right\}  $. Then,%
\begin{align*}
b_{1,j}+b_{2,j}+\cdots+b_{i,j}  &  =\sum_{k=1}^{i}\underbrace{b_{k,j}%
}_{\substack{=a_{k,j}-a_{k,j-1}-a_{k-1,j}+a_{k-1,j-1}\\\text{(by the
definition of }b_{k,j}\text{)}}}=\sum_{k=1}^{i}\underbrace{\left(
a_{k,j}-a_{k,j-1}-a_{k-1,j}+a_{k-1,j-1}\right)  }_{=\left(  a_{k,j}%
-a_{k,j-1}\right)  -\left(  a_{k-1,j}-a_{k-1,j-1}\right)  }\\
&  =\sum_{k=1}^{i}\left(  \left(  a_{k,j}-a_{k,j-1}\right)  -\left(
a_{k-1,j}-a_{k-1,j-1}\right)  \right)  =\left(  a_{i,j}-a_{i,j-1}\right)
-\underbrace{\left(  a_{0,j}-a_{0,j-1}\right)  }_{\substack{=\left(  \mu
^{t}\right)  _{j}\\\text{(by (\ref{sol.liri.jordan.b.a0j-diff}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the telescope principle}\right) \\
&  =\left(  a_{i,j}-a_{i,j-1}\right)  -\left(  \mu^{t}\right)  _{j},
\end{align*}
and thus $\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}+b_{2,j}+\cdots
+b_{i,j}\right)  =a_{i,j}-a_{i,j-1}$, qed.}

Now, Exercise \ref{exe.liri.yamanouchi}(a) (applied to $\mu^{t}$ instead of
$\mu$) yields that the following two assertions are equivalent\footnote{The
partition that we call $\gamma$ in Assertion $\mathcal{A}^{\prime}$ is the
partition that was called $\lambda$ in Assertion $\mathcal{A}$.}:

\begin{itemize}
\item \textit{Assertion $\mathcal{A}^{\prime}$:} There exist a
partition $\gamma$ and a column-strict tableau $T$ of shape $\gamma/\mu^{t}$
such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfy
\begin{equation}
b_{i,j}=\left(  \text{the number of all entries }i\text{ in the }j\text{-th
row of }T\right)  . \label{sol.liri.jordan.b.A}%
\end{equation}


\item \textit{Assertion $\mathcal{B}^{\prime}$:} The inequality
\begin{equation}
\left(  \mu^{t}\right)  _{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots
+b_{i+1,j+1}\right)  \leq\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}%
+b_{2,j}+\cdots+b_{i,j}\right)  \label{sol.liri.jordan.b.B}%
\end{equation}
holds for all $\left(  i,j\right)  \in{ \NN }\times\left\{
1,2,3,\ldots\right\}  $.
\end{itemize}

We shall now prove that Assertion $\mathcal{B}^{\prime}$ holds. Indeed, any
$\left(  i,j\right)  \in{ \NN }\times\left\{  1,2,3,\ldots\right\}  $
satisfies%
\begin{align*}
&  \left(  \mu^{t}\right)  _{j+1}+\left(  b_{1,j+1}+b_{2,j+1}+\cdots
+b_{i+1,j+1}\right) \\
&  =a_{i+1,j+1}-a_{i+1,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.bij-sum}), applied to }i+1\text{ and }j+1\text{
instead of }i\text{ and }j\right) \\
&  \leq a_{i,j}-a_{i,j-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.liri.jordan.b.general}(b) (applied to }W=V\text{, }A=U\text{,
}B=0\text{ and }w_{i,j}=a_{i,j}\text{)}\right) \\
&  =\left(  \mu^{t}\right)  _{j}+\left(  b_{1,j}+b_{2,j}+\cdots+b_{i,j}%
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.liri.jordan.b.bij-sum}%
)}\right)  .
\end{align*}
Thus, Assertion $\mathcal{B}^{\prime}$ holds. Since Assertions $\mathcal{A}%
^{\prime}$ and $\mathcal{B}^{\prime}$ are equivalent, this yields that
Assertion $\mathcal{A}^{\prime}$ also holds. In other words, there exist a
partition $\gamma$ and a column-strict tableau $T$ of shape $\gamma/\mu^{t}$
such that all $\left(  i,j\right)  \in\left\{  1,2,3,\ldots\right\}  ^{2}$
satisfy (\ref{sol.liri.jordan.b.A}). Let us consider this $\gamma$ and this
$T$.

We shall soon see that $\gamma=\lambda^{t}$ and $\cont T=\nu
^{t}$. Let us first prepare for this. We have%
\begin{equation}
\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in \NN .
\label{sol.liri.jordan.b.lambda-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.lambda-sum}):} Let
$j\in \NN$. We can represent the endomorphism $f$ of $V$ as an $n\times
n$-matrix $F\in\mathbb{K}^{n\times n}$ for $n=\dim V$. This $n\times n$-matrix
$F$ is nilpotent (since $f$ is nilpotent) and has Jordan type $\lambda$ (since
$f$ has Jordan type $\lambda$). But Exercise \ref{exe.liri.jordan}(a) (applied
to $N=F$) shows that $F$ has Jordan type $\lambda$ if and only if every
$k\in{ \NN }$ satisfies $\dim\left(  \ker\left(  F^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\cdots+\left(  \lambda^{t}\right)  _{k}$. Since we know that $F$ has Jordan
type $\lambda$, we thus conclude that every $k\in{ \NN }$ satisfies
$\dim\left(  \ker\left(  F^{k}\right)  \right)  =\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{k}%
$. Applied to $k=j$, this yields $\dim\left(  \ker\left(  F^{j}\right)
\right)  =\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\cdots+\left(  \lambda^{t}\right)  _{j}$.
\par
Since the matrix $F$ represents the endomorphism $f$, we have
\[
\dim\left(  \ker\left(  F^{j}\right)  \right)  =\dim\left(  \underbrace{\ker
\left(  f^{j}\right)  }_{=\left(  f^{j}\right)  ^{-1}\left(  0\right)
}\right)  =\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
.
\]
Compared with $\dim\left(  \ker\left(  F^{j}\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j}$, this yields%
\[
\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)  =\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j},
\]
and thus (\ref{sol.liri.jordan.b.lambda-sum}) is proven.} From here, it is
easy to see that%
\begin{equation}
\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)  -\dim\left(
\left(  f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  =\left(  \lambda
^{t}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for every }j\in\left\{
1,2,3,\ldots\right\}  . \label{sol.liri.jordan.b.lambda-single}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.lambda-single}):} Let
$j\in\left\{  1,2,3,\ldots\right\}  $. Then,%
\begin{align*}
&  \underbrace{\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  }_{\substack{=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\cdots+\left(  \lambda^{t}\right)  _{j}\\\text{(by
(\ref{sol.liri.jordan.b.lambda-sum}))}}}-\underbrace{\dim\left(  \left(
f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  }_{\substack{=\left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j-1}\\\text{(by (\ref{sol.liri.jordan.b.lambda-sum}),
applied}\\\text{to }j-1\text{ instead of }j\text{)}}}\\
&  =\left(  \left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\cdots+\left(  \lambda^{t}\right)  _{j}\right)  -\left(  \left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\cdots+\left(
\lambda^{t}\right)  _{j-1}\right)  =\left(  \lambda^{t}\right)  _{j},
\end{align*}
so that (\ref{sol.liri.jordan.b.lambda-single}) is proven.} Now, for every
$j\in\left\{  1,2,3,\ldots\right\}  $, we have%
\[
\left(  \text{the number of all entries in the }j\text{-th row of }T\right)
=\gamma_{j}-\left(  \mu^{t}\right)  _{j}%
\]
(since $T$ has shape $\gamma/\mu^{t}$), so that%
\begin{align*}
\gamma_{j}-\left(  \mu^{t}\right)  _{j}  &  =\left(  \text{the number of all
entries in the }j\text{-th row of }T\right) \\
&  =\sum_{i=1}^{\infty}\underbrace{\left(  \text{the number of all entries
}i\text{ in the }j\text{-th row of }T\right)  }_{\substack{=b_{i,j}\\\text{(by
(\ref{sol.liri.jordan.b.A}))}}}=\sum_{i=1}^{\infty}\underbrace{b_{i,j}%
}_{=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}}\\
&  =\sum_{i=1}^{\infty}\underbrace{\left(  a_{i,j}-a_{i,j-1}-a_{i-1,j}%
+a_{i-1,j-1}\right)  }_{=\left(  a_{i,j}-a_{i-1,j}\right)  -\left(
a_{i,j-1}-a_{i-1,j-1}\right)  }=\sum_{i=1}^{\infty}\left(  \left(
a_{i,j}-a_{i-1,j}\right)  -\left(  a_{i,j-1}-a_{i-1,j-1}\right)  \right) \\
&  =\underbrace{\sum_{i=1}^{\infty}\left(  a_{i,j}-a_{i-1,j}\right)
}_{\substack{=\dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  -\dim\left(  U\cap\left(  f^{j}\right)  ^{-1}\left(  0\right)
\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(f), applied
to}\\W=V\text{, }A=U\text{, }B=0\text{ and }w_{i,j}=a_{i,j}\text{)}%
}}-\underbrace{\sum_{i=1}^{\infty}\left(  a_{i,j-1}-a_{i-1,j-1}\right)
}_{\substack{=\dim\left(  \left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  -\dim\left(  U\cap\left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(f), applied
to}\\V\text{, }U\text{, }0\text{, }a_{i,j}\text{ and }j-1\text{
instead of }W\text{, }A\text{, }B\text{, }w_{i,j}\text{ and }j\text{)}}}\\
&  =\left(  \dim\left(  \left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
-\dim\left(  U\cap\left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
\right)  -\left(  \dim\left(  \left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  -\dim\left(  U\cap\left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  \right) \\
&  =\underbrace{\left(  \dim\left(  \left(  f^{j}\right)  ^{-1}\left(
0\right)  \right)  -\dim\left(  \left(  f^{j-1}\right)  ^{-1}\left(  0\right)
\right)  \right)  }_{\substack{=\left(  \lambda^{t}\right)  _{j}\\\text{(by
(\ref{sol.liri.jordan.b.lambda-single}))}}}-\left(  \dim\left(  U\cap\left(
f^{j}\right)  ^{-1}\left(  0\right)  \right)  -\dim\left(  U\cap\left(
f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  \right) \\
&  =\left(  \lambda^{t}\right)  _{j}-\left(  \dim\left(  U\cap\left(
f^{j}\right)  ^{-1}\left(  0\right)  \right)  -\dim\left(  U\cap\left(
f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  \right)  .
\end{align*}
Adding this equality to the equality%
\begin{align*}
\left(  \mu^{t}\right)  _{j}  &  =\underbrace{a_{0,j}}_{\substack{=\dim\left(
\left(  f^{0}\right)  ^{-1}\left(  U\right)  \cap\ker\left(  f^{j}\right)
\right)  \\\text{(by the definition of }a_{0,j}\text{)}}%
}-\underbrace{a_{0,j-1}}_{\substack{=\dim\left(  \left(  f^{0}\right)
^{-1}\left(  U\right)  \cap\ker\left(  f^{j-1}\right)  \right)  \\\text{(by
the definition of }a_{0,j-1}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.a0j-diff})}\right) \\
&  =\dim\left(  \left(  \underbrace{f^{0}}_{=\id}\right)
^{-1}\left(  U\right)  \cap\ker\left(  f^{j}\right)  \right)  -\dim\left(
\left(  \underbrace{f^{0}}_{=\id}\right)  ^{-1}\left(
U\right)  \cap\ker\left(  f^{j-1}\right)  \right) \\
&  =\dim\left(  \underbrace{\left(  \operatorname*{id}\right)  ^{-1}\left(
U\right)  }_{=U}\cap\underbrace{\ker\left(  f^{j}\right)  }_{=\left(
f^{j}\right)  ^{-1}\left(  0\right)  }\right)  -\dim\left(
\underbrace{\left(  \operatorname*{id}\right)  ^{-1}\left(  U\right)  }%
_{=U}\cap\underbrace{\ker\left(  f^{j-1}\right)  }_{=\left(  f^{j-1}\right)
^{-1}\left(  0\right)  }\right) \\
&  =\dim\left(  U\cap\left(  f^{j}\right)  ^{-1}\left(  0\right)  \right)
-\dim\left(  U\cap\left(  f^{j-1}\right)  ^{-1}\left(  0\right)  \right)  ,
\end{align*}
we obtain $\gamma_{j}=\left(  \lambda^{t}\right)  _{j}$ for every
$j\in\left\{  1,2,3,\ldots\right\}  $. Thus, $\gamma=\lambda^{t}$. Hence, $T$
is a column-strict tableau of shape $\lambda^{t}/\mu^{t}$ (since $T$ is a
column-strict tableau of shape $\gamma/\mu^{t}$).

Next, let us prove that $\cont T=\nu^{t}$. Indeed, we first
notice that%
\begin{equation}
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  =\left(
\dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in \NN . \label{sol.liri.jordan.b.nu-sum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.nu-sum}):} Let
$i\in \NN$. Recall that the nilpotent endomorphism $\overline{f}$ of the
quotient space $V/U$ (induced by $f\in\operatorname*{End}V$) has Jordan type
$\nu$. We can represent this nilpotent endomorphism $\overline{f}$ of $V/U$ as
an $\ell\times\ell$-matrix $H\in\mathbb{K}^{\ell\times\ell}$ for $\ell
=\dim\left(  V/U\right)  $. This $\ell\times\ell$-matrix $H$ is nilpotent
(since $\overline{f}$ is nilpotent) and has Jordan type $\nu$ (since
$\overline{f}$ has Jordan type $\nu$). But Exercise \ref{exe.liri.jordan}(a)
(applied to $\ell$, $H$ and $\nu$ instead of $n$, $N$ and $\lambda$)
shows that $H$ has Jordan type $\nu$ if and only if every
$k\in{ \NN }$ satisfies $\dim\left(  \ker\left(  H^{k}\right)  \right)
=\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{k}$. Since we know that $H$ has Jordan type $\nu$, we thus
conclude that every $k\in{ \NN }$ satisfies $\dim\left(  \ker\left(
H^{k}\right)  \right)  =\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{k}$. Applied to $k=i$, this yields
$\dim\left(  \ker\left(  H^{i}\right)  \right)  =\left(  \nu^{t}\right)
_{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)  _{i}$.
\par
Since the matrix $H$ represents the endomorphism $\overline{f}$, we have
\[
\dim\left(  \ker\left(  \overline{f}^{i}\right)  \right)  =\dim\left(
\ker\left(  H^{i}\right)  \right)  =\left(  \nu^{t}\right)  _{1}+\left(
\nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)  _{i}.
\]
\par
But $\left(  f^{i}\right)  ^{-1}\left(  U\right)  $ is a $\mathbb{K}$-vector
subspace of $V$ containing $U$. Thus, $\left(  f^{i}\right)  ^{-1}\left(
U\right)  /U$ is canonically a $\mathbb{K}$-vector subspace of $V/U$.
Moreover, this subspace $\left(  f^{i}\right)  ^{-1}\left(  U\right)  /U$ is
precisely the kernel $\ker\left(  \overline{f}^{i}\right)  $ (this is
straightforward to check). Hence,
\[
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  /U\right)
=\dim\left(  \ker\left(  \overline{f}^{i}\right)  \right)  =\left(  \nu
^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)
_{i}.
\]
Since
\[
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  /U\right)
=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  -\dim U,
\]
this rewrites as%
\[
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  -\dim
U=\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(
\nu^{t}\right)  _{i}.
\]
Adding $\dim U$ to both sides of this equality yields
(\ref{sol.liri.jordan.b.nu-sum}).} As a consequence of this, we have%
\begin{equation}
\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)  -\dim\left(
\left(  f^{i-1}\right)  ^{-1}\left(  U\right)  \right)  =\left(  \nu
^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,3,\ldots\right\}  . \label{sol.liri.jordan.b.nu-single}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.nu-single}):} Let
$i\in\left\{  1,2,3,\ldots\right\}  $. Then,%
\begin{align*}
&  \underbrace{\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\right)  }_{\substack{=\left(  \dim U\right)  +\left(  \nu^{t}\right)
_{1}+\left(  \nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)
_{i}\\\text{(by (\ref{sol.liri.jordan.b.nu-sum}))}}}-\underbrace{\dim\left(
\left(  f^{i-1}\right)  ^{-1}\left(  U\right)  \right)  }_{\substack{=\left(
\dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{i-1}\\\text{(by
(\ref{sol.liri.jordan.b.nu-sum}), applied}\\\text{to }i-1\text{ instead of
}i\text{)}}}\\
&  =\left(  \left(  \dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(
\nu^{t}\right)  _{2}+\cdots+\left(  \nu^{t}\right)  _{i}\right)  -\left(
\left(  \dim U\right)  +\left(  \nu^{t}\right)  _{1}+\left(  \nu^{t}\right)
_{2}+\cdots+\left(  \nu^{t}\right)  _{i-1}\right)  =\left(  \nu^{t}\right)
_{i}.
\end{align*}
Thus, (\ref{sol.liri.jordan.b.nu-single}) is proven.}

Now, for every $i\in\left\{  1,2,3,\ldots\right\}  $, we have%
\begin{align*}
\left(  \cont T\right)  _{i}  &  =\left\vert T^{-1}\left(
i\right)  \right\vert =\left(  \text{the number of entries }i\text{ in
}T\right) \\
&  =\sum_{j=1}^{\infty}\underbrace{\left(  \text{the number of all entries
}i\text{ in the }j\text{-th row of }T\right)  }_{\substack{=b_{i,j}\\\text{(by
(\ref{sol.liri.jordan.b.A}))}}}=\sum_{j=1}^{\infty}\underbrace{b_{i,j}%
}_{=a_{i,j}-a_{i,j-1}-a_{i-1,j}+a_{i-1,j-1}}\\
&  =\sum_{j=1}^{\infty}\underbrace{\left(  a_{i,j}-a_{i,j-1}-a_{i-1,j}%
+a_{i-1,j-1}\right)  }_{=\left(  a_{i,j}-a_{i,j-1}\right)  -\left(
a_{i-1,j}-a_{i-1,j-1}\right)  }=\sum_{j=1}^{\infty}\left(  \left(
a_{i,j}-a_{i,j-1}\right)  -\left(  a_{i-1,j}-a_{i-1,j-1}\right)  \right) \\
&  =\underbrace{\sum_{j=1}^{\infty}\left(  a_{i,j}-a_{i,j-1}\right)
}_{\substack{=\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\right)  -\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap0\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(g),
applied to}\\W=V\text{, }A=U\text{, }B=0\text{ and }w_{i,j}=a_{i,j}\text{)}%
}}-\underbrace{\sum_{j=1}^{\infty}\left(  a_{i-1,j}-a_{i-1,j-1}\right)
}_{\substack{=\dim\left(  \left(  f^{i-1}\right)  ^{-1}\left(  U\right)
\right)  -\dim\left(  \left(  f^{i-1}\right)  ^{-1}\left(  U\right)
\cap0\right)  \\\text{(by Lemma \ref{lem.sol.liri.jordan.b.general}(f),
applied to}\\V\text{, }U\text{, }0\text{, }a_{i,j}\text{ and
}i-1\text{ instead of }W\text{, }A\text{, }B\text{, }w_{i,j}\text{ and
}i\text{)}}}\\
&  =\left(  \dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)
-\dim\underbrace{\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)
\cap0\right)  }_{=0}\right)  -\left(  \dim\left(  \left(  f^{i-1}\right)
^{-1}\left(  U\right)  \right)  -\dim\underbrace{\left(  \left(
f^{i-1}\right)  ^{-1}\left(  U\right)  \cap0\right)  }_{=0}\right) \\
&  =\left(  \dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)
-\underbrace{\dim0}_{=0}\right)  -\left(  \dim\left(  \left(  f^{i-1}\right)
^{-1}\left(  U\right)  \right)  -\underbrace{\dim0}_{=0}\right) \\
&  =\dim\left(  \left(  f^{i}\right)  ^{-1}\left(  U\right)  \right)
-\dim\left(  \left(  f^{i-1}\right)  ^{-1}\left(  U\right)  \right)  =\left(
\nu^{t}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.nu-single})}\right)  .
\end{align*}
Hence, $\cont T=\nu^{t}$.

We shall next see that for every positive integer $j$, the weak composition
$\cont (T|_{\operatorname{cols}\geq j})$ is a partition. Here,
and in the following, we are using the notations of Exercise
\ref{exe.liri.yamanouchi}.

We first notice that every
$i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$ satisfy%
\begin{equation}
\left(  \text{the number of entries }i\text{ in the first }j\text{ rows of
}T\right)  =a_{i,j}-a_{i-1,j}. \label{sol.liri.jordan.b.entrycount}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.liri.jordan.b.entrycount}):} Let
$i\in\left\{  1,2,3,\ldots\right\}  $ and $j\in \NN$. We have%
\begin{align*}
&  \left(  \text{the number of entries }i\text{ in the first }j\text{ rows of
}T\right) \\
&  =\sum_{k=1}^{j}\underbrace{\left(  \text{the number of all entries }i\text{
in the }k\text{-th row of }T\right)  }_{\substack{=b_{i,k}\\\text{(since
(\ref{sol.liri.jordan.b.A}) (applied to }\left(  i,k\right)  \text{
instead of }\left(  i,j\right)  \text{) yields }\\ b_{i,k}=\left(
\text{the number of all entries }i\text{ in the }k\text{-th row of }T\right)
\text{)}}}\\
&  =\sum_{k=1}^{j}\underbrace{b_{i,k}}_{\substack{=a_{i,k}-a_{i,k-1}%
-a_{i-1,k}+a_{i-1,k-1}\\\text{(by the definition of }b_{i,k}\text{)}}%
}=\sum_{k=1}^{j}\underbrace{\left(  a_{i,k}-a_{i,k-1}-a_{i-1,k}+a_{i-1,k-1}%
\right)  }_{=\left(  a_{i,k}-a_{i-1,k}\right)  -\left(  a_{i,k-1}%
-a_{i-1,k-1}\right)  }\\
&  =\sum_{k=1}^{j}\left(  \left(  a_{i,k}-a_{i-1,k}\right)  -\left(
a_{i,k-1}-a_{i-1,k-1}\right)  \right) \\
& =\left(  a_{i,j}-a_{i-1,j}\right)
-\left(  \underbrace{a_{i,0}}_{\substack{=0\\\text{(by
(\ref{sol.liri.jordan.b.ai0}))}}}-\underbrace{a_{i-1,0}}%
_{\substack{=0\\\text{(by (\ref{sol.liri.jordan.b.ai0}), applied}\\\text{to
}i-1\text{ instead of }i\text{)}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the telescope principle}\right) \\
&  =\left(  a_{i,j}-a_{i-1,j}\right)  -\left(  0-0\right)  =a_{i,j}-a_{i-1,j}.
\end{align*}
This proves (\ref{sol.liri.jordan.b.entrycount}).} Now, for every positive
integers $j$ and $i$, we have%
\begin{align*}
&  \left(  \text{the number of entries }i+1\text{ in the first }j\text{ rows
of }T\right) \\
&  =a_{i+1,j}-a_{i,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.liri.jordan.b.entrycount}), applied to }i+1\text{ instead of
}i\right) \\
&  \leq a_{i,j-1}-a_{i-1,j-1}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.sol.liri.jordan.b.general}(c), applied to}\\
V\text{, }U\text{, }0\text{, }a_{i,j}\text{ and }j-1\text{ instead of
}W\text{, }A\text{, }B\text{, }w_{i,j}\text{ and }j
\end{array}
\right) \\
&  =\left(  \text{the number of entries }i\text{ in the first }j-1\text{ rows
of }T\right)
\end{align*}
(since (\ref{sol.liri.jordan.b.entrycount}) (applied to $j-1$ instead of $j$)
yields $\left(  \text{the number of entries }i\text{ in the first }j-1\text{
rows of }T\right)  =a_{i,j-1}-a_{i-1,j-1}$). In other words, for every
positive integers $j$ and $i$, the number of entries $i+1$ in the first $j$
rows of $T$ is $\leq$ to the number of entries $i$ in the first $j-1$ rows of
$T$. In other words, Assertion $\mathcal{D}$ of Exercise
\ref{exe.liri.yamanouchi}(b) (with $\lambda$ and $\mu$ replaced by
$\lambda^{t}$ and $\mu^{t}$) is satisfied. Hence, Assertion $\mathcal{C}$ of
Exercise \ref{exe.liri.yamanouchi}(b) (with $\lambda$ and $\mu$ replaced by
$\lambda^{t}$ and $\mu^{t}$) is satisfied as well (because Exercise
\ref{exe.liri.yamanouchi}(b) yields that these Assertions $\mathcal{C}$ and
$\mathcal{D}$ are equivalent). In other words, for every positive integer $j$,
the weak composition $\cont \left(  T|_{\operatorname{cols}\geq
j}\right)  $ is a partition.

Now, let us forget that we defined $\nu$ and $T$. We thus have found a
column-strict tableau $T$ of shape $\lambda^{t}/\mu^{t}$ with
$\cont T=\nu^{t}$ which has the property that for every positive
integer $j$, the weak composition $\cont \left(
T|_{\operatorname{cols}\geq j}\right)  $ is a partition. But we know that the
number of such tableaux is $c_{\mu^{t},\nu^{t}}^{\lambda^{t}}$ (by Corollary
\ref{L-R-rule}, applied to $\lambda^{t}$, $\mu^{t}$ and $\nu^{t}$ instead of
$\lambda$, $\mu$ and $\nu$). Hence, this number $c_{\mu^{t},\nu^{t}}%
^{\lambda^{t}}$ must be $\neq0$ (because we have found such a tableau $T$). So
we have proven that $c_{\mu^{t},\nu^{t}}^{\lambda^{t}}\neq0$. Now, Exercise
\ref{exe.pieri.omega}(c) yields $c_{\mu,\nu}^{\lambda}=c_{\mu^{t},\nu^{t}%
}^{\lambda^{t}}\neq0$, so that Exercise \ref{exe.liri.jordan}(b) is solved.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.staircase-skew-1}}

\textit{Solution to Exercise~\ref{exe.Lambda.staircase-skew-1}.} Define
a set $\mathfrak{D}$ by
\begin{equation}
\mathfrak{D}=\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(
g\right)  \right)  ^{\perp}a\right\}  .
\label{sol.Lambda.staircase-skew-1.a.D=}
\end{equation}
We shall show that $\mathfrak{D}$ is a $\kk$-subalgebra of $\Lambda$.


Define a map $\kappa:\Lambda\rightarrow\Lambda$ by%
\[
\left(  \kappa\left(  g\right)  =g^{\perp}a-\left(  \omega\left(  g\right)
\right)  ^{\perp}a\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\Lambda\right)  .
\]
This map $\kappa$ is $\kk$-linear\footnote{\textit{Proof.} For each
$g\in\Lambda$, we have $g^{\perp}a=\sum\left(  g,a_{1}\right)  a_{2}$ (by the
definition of $g^{\perp}$). Hence, the element $g^{\perp}a$ of $\Lambda$
depends $\kk$-linearly on $g$ (because the Hall inner product $\left(
\cdot,\cdot\right)  $ is $\kk$-bilinear). Thus, the element $\left(
\omega\left(  g\right)  \right)  ^{\perp}a$ also depends $\kk$-linearly
on $g$ (since the map $\omega$ is $\kk$-linear). Hence, the difference
$g^{\perp}a-\left(  \omega\left(  g\right)  \right)  ^{\perp}a$ of these two
elements also depends $\kk$-linearly on $g$. In other words,
$\kappa\left(  g\right)  $ depends $\kk$-linearly on $g$ (since
$\kappa\left(  g\right)  =g^{\perp}a-\left(  \omega\left(  g\right)  \right)
^{\perp}a$). In other words, the map $\kappa$ is $\kk$-linear.}. Hence,
its kernel $\ker\kappa$ is a $\kk$-submodule of $\Lambda$. Since%
\begin{align*}
\ker\kappa &  =\left\{  g\in\Lambda\ \mid\ \underbrace{\kappa\left(  g\right)
}_{\substack{=g^{\perp}a-\left(  \omega\left(  g\right)  \right)  ^{\perp
}a\\\text{(by the definition of }\kappa\text{)}}}=0\right\}  =\left\{
g\in\Lambda\ \mid\ \underbrace{g^{\perp}a-\left(  \omega\left(  g\right)
\right)  ^{\perp}a=0}_{\Longleftrightarrow\ \left(  g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a\right)  }\right\} \\
&  =\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\end{align*}
this rewrites as follows: The set $\mathfrak{D}$ is a $\kk$-submodule
of $\Lambda$.

Furthermore, we have the following two observations:

\begin{itemize}
\item We have $1\in\mathfrak{D}$\ \ \ \ \footnote{\textit{Proof.} Recall that
$\omega$ is a $\kk$-algebra endomorphism of $\Lambda$. Hence,
$\omega\left(  1\right)  =1$. Thus, $\left(  \underbrace{\omega\left(
1\right)  }_{=1}\right)  ^{\perp}a=1^{\perp}a$, so that $1^{\perp}a=\left(
\omega\left(  1\right)  \right)  ^{\perp}a$. Hence, $1$ is an element of
$\Lambda$ and satisfies $1^{\perp}a=\left(  \omega\left(  1\right)  \right)
^{\perp}a$. Thus,%
\[
1\in\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\]
qed.}.

\item We have $xy\in\mathfrak{D}$ for each $x\in\mathfrak{D}$ and
$y\in\mathfrak{D}$\ \ \ \ \footnote{\textit{Proof.} Let $x\in\mathfrak{D}$ and
$y\in\mathfrak{D}$.
\par
We have $x\in\mathfrak{D}=\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a\right\}  $. In other words, $x$ is
an element of $\Lambda$ and satisfies $x^{\perp}a=\left(  \omega\left(
x\right)  \right)  ^{\perp}a$.
\par
We have $y\in\mathfrak{D}=\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a\right\}  $. In other words, $y$ is
an element of $\Lambda$ and satisfies $y^{\perp}a=\left(  \omega\left(
y\right)  \right)  ^{\perp}a$.
\par
Proposition~\ref{skewing-properties-prop}(ii) (applied to $f=\omega\left(
y\right)  $ and $g=\omega\left(  x\right)  $) yields%
\begin{equation}
\left(  \omega\left(  y\right)  \omega\left(  x\right)  \right)  ^{\perp
}a=\left(  \omega\left(  x\right)  \right)  ^{\perp}\left(  \left(
\omega\left(  y\right)  \right)  ^{\perp}a\right)  .
\label{sol.Lambda.staircase-skew-1.a.prod.1}%
\end{equation}
But $\omega\left(  xy\right)  =\omega\left(  x\right)  \omega\left(  y\right)
$ (since $\omega$ is an endomorphism of the $\kk$-algebra $\Lambda$).
Hence,%
\begin{equation}
\left(  \underbrace{\omega\left(  xy\right)  }_{\substack{=\omega\left(
x\right)  \omega\left(  y\right)  \\=\omega\left(  y\right)  \omega\left(
x\right)  }}\right)  ^{\perp}a=\left(  \omega\left(  y\right)  \omega\left(
x\right)  \right)  ^{\perp}a=\left(  \omega\left(  x\right)  \right)  ^{\perp
}\left(  \left(  \omega\left(  y\right)  \right)  ^{\perp}a\right)
\label{sol.Lambda.staircase-skew-1.a.prod.2}%
\end{equation}
(by (\ref{sol.Lambda.staircase-skew-1.a.prod.1})).
\par
On the other hand, Proposition~\ref{skewing-properties-prop}(ii) (applied to
$f=x$ and $g=y$) yields%
\[
\left(  xy\right)  ^{\perp}a=y^{\perp}\left(  \underbrace{x^{\perp}%
a}_{=\left(  \omega\left(  x\right)  \right)  ^{\perp}a}\right)  =y^{\perp
}\left(  \left(  \omega\left(  x\right)  \right)  ^{\perp}a\right)  .
\]
Comparing this with%
\[
\left(  \omega\left(  x\right)  y\right)  ^{\perp}a=y^{\perp}\left(  \left(
\omega\left(  x\right)  \right)  ^{\perp}a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{skewing-properties-prop}(ii) (applied to }%
f=\omega\left(  x\right)  \text{ and }g=y\text{)}\right)  ,
\]
we obtain%
\begin{align*}
\left(  xy\right)  ^{\perp}a  &  =\left(  \underbrace{\omega\left(  x\right)
y}_{=y\omega\left(  x\right)  }\right)  ^{\perp}a=\left(  y\omega\left(
x\right)  \right)  ^{\perp}a=\left(  \omega\left(  x\right)  \right)  ^{\perp
}\left(  \underbrace{y^{\perp}a}_{=\left(  \omega\left(  y\right)  \right)
^{\perp}a}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{skewing-properties-prop}(ii) (applied to }f=y\text{ and
}g=\omega\left(  x\right)  \text{)}\right) \\
&  =\left(  \omega\left(  x\right)  \right)  ^{\perp}\left(  \left(
\omega\left(  y\right)  \right)  ^{\perp}a\right)  =\left(  \omega\left(
xy\right)  \right)  ^{\perp}a\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.prod.2})}\right)  .
\end{align*}
Thus, $xy$ is an element of $\Lambda$ and satisfies $\left(  xy\right)
^{\perp}a=\left(  \omega\left(  xy\right)  \right)  ^{\perp}a$. Hence,%
\[
xy\in\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\]
qed.}.
\end{itemize}

Combining these two observations, we conclude that the set $\mathfrak{D}$ is a
$\kk$-subalgebra of $\Lambda$ (since we already know that
$\mathfrak{D}$ is a $\kk$-submodule of $\Lambda$). In view of
(\ref{sol.Lambda.staircase-skew-1.a.D=}), this rewrites as follows: The set
$\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  $ is a $\kk$-subalgebra of $\Lambda$. This
solves Exercise~\ref{exe.Lambda.staircase-skew-1}(a).

(b) We have
\begin{equation}
e_{k}^{\perp}a=h_{k}^{\perp}a\ \ \ \ \ \ \ \ \ \ \text{for each positive
integer }k \label{sol.Lambda.staircase-skew-1.b.ass}%
\end{equation}
(by assumption). Thus,%
\begin{equation}
e_{n}\in\mathfrak{D}\text{ for each }n\in\left\{  1,2,3,\ldots\right\}
\label{sol.Lambda.staircase-skew-1.b.enin}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.staircase-skew-1.b.enin}):} Let
$n\in\left\{  1,2,3,\ldots\right\}  $. Thus, $n$ is a positive integer. Hence,
$\omega\left(  e_{n}\right)  =h_{n}$ (by the definition of $\omega$). Thus,
$\left(  \underbrace{\omega\left(  e_{n}\right)  }_{=h_{n}}\right)  ^{\perp
}a=h_{n}^{\perp}a$. But (\ref{sol.Lambda.staircase-skew-1.b.ass}) (applied to
$k=n$) yields $e_{n}^{\perp}a=h_{n}^{\perp}a$. Comparing this with $\left(
\omega\left(  e_{n}\right)  \right)  ^{\perp}a=h_{n}^{\perp}a$, we find
$e_{n}^{\perp}a=\left(  \omega\left(  e_{n}\right)  \right)  ^{\perp}a$.
Hence, $e_{n}$ is an element of $\Lambda$ and satisfies $e_{n}^{\perp
}a=\left(  \omega\left(  e_{n}\right)  \right)  ^{\perp}a$. Therefore,%
\[
e_{n}\in\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\]
qed.}.

But Proposition~\ref{symm-antipode-and-h-basis} shows that the family $\left(
e_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }$ generates $\Lambda$ as a
$\kk$-algebra. Thus, the smallest $\kk$-subalgebra of $\Lambda$
that contains all elements of the family $\left(  e_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }$ is $\Lambda$ itself. In other words, if
$\mathfrak{B}$ is a $\kk$-subalgebra of $\Lambda$ satisfying%
\[
\left(  e_{n}\in\mathfrak{B}\text{ for each }n\in\left\{  1,2,3,\ldots
\right\}  \right)  ,
\]
then $\mathfrak{B}=\Lambda$. Applying this to $\mathfrak{B}=\mathfrak{D}$, we
conclude that $\mathfrak{D}=\Lambda$ (because $\mathfrak{D}$ is a $\kk%
$-subalgebra of $\Lambda$, and because it satisfies
(\ref{sol.Lambda.staircase-skew-1.b.enin})).

Now, let $f\in\Lambda$. Then, $f\in\Lambda=\mathfrak{D}=\left\{  g\in
\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)  \right)  ^{\perp
}a\right\}  $ (by (\ref{sol.Lambda.staircase-skew-1.a.D=})). In other words,
$f$ is an element of $\Lambda$ and satisfies $f^{\perp}a=\left(  \omega\left(
f\right)  \right)  ^{\perp}a$.

Now, forget that we fixed $f$. We thus have proven that $f^{\perp}a=\left(
\omega\left(  f\right)  \right)  ^{\perp}a$ for each $f\in\Lambda$. Renaming
$f$ as $g$ in this statement, we conclude the following: $g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a$ for each $g\in\Lambda$. This
solves Exercise~\ref{exe.Lambda.staircase-skew-1}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.staircase-skew-1}}

\textit{Solution to Exercise~\ref{exe.Lambda.staircase-skew-1}.}
Define a set $\mathfrak{D}$ by
\begin{equation}
\mathfrak{D}=\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(
g\right)  \right)  ^{\perp}a\right\}  .
\label{sol.Lambda.staircase-skew-1.a.D=}
\end{equation}
We shall show that $\mathfrak{D}$ is a $\kk$-subalgebra of $\Lambda$.


Define a map $\kappa:\Lambda\rightarrow\Lambda$ by%
\[
\left(  \kappa\left(  g\right)  =g^{\perp}a-\left(  \omega\left(  g\right)
\right)  ^{\perp}a\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\Lambda\right)  .
\]
This map $\kappa$ is $\kk$-linear\footnote{\textit{Proof.} Let
$x\in\Lambda$, $y\in\Lambda$, $\lambda\in\kk$ and $\mu\in\kk$.
Then, the definition of $\left(  \lambda x+\mu y\right)  ^{\perp}$ yields
\begin{align}
\left(  \lambda x+\mu y\right)  ^{\perp}a  &  =\sum\underbrace{\left(  \lambda
x+\mu y,a_{1}\right)  }_{\substack{=\lambda\left(  x,a_{1}\right)  +\mu\left(
y,a_{1}\right)  \\\text{(since the Hall inner product}\\\text{is }%
\kk\text{-bilinear)}}}a_{2}=\sum\left(  \lambda\left(  x,a_{1}\right)
+\mu\left(  y,a_{1}\right)  \right)  a_{2}\nonumber\\
&  =\lambda\underbrace{\sum\left(  x,a_{1}\right)  a_{2}}_{\substack{=x^{\perp
}a\\\text{(because }x^{\perp}a=\sum\left(  x,a_{1}\right)  a_{2}\\\text{(by
the definition of }x^{\perp}\text{))}}}+\mu\underbrace{\sum\left(
y,a_{1}\right)  a_{2}}_{\substack{=y^{\perp}a\\\text{(because }y^{\perp}%
a=\sum\left(  y,a_{1}\right)  a_{2}\\\text{(by the definition of }y^{\perp
}\text{))}}}=\lambda x^{\perp}a+\mu y^{\perp}a.
\label{sol.Lambda.staircase-skew-1.a.linearity.1}%
\end{align}
Applying this to $\omega\left(  x\right)  $ and $\omega\left(  y\right)  $
instead of $x$ and $y$, we obtain%
\begin{equation}
\left(  \lambda\omega\left(  x\right)  +\mu\omega\left(  y\right)  \right)
^{\perp}a=\lambda\left(  \omega\left(  x\right)  \right)  ^{\perp}a+\mu\left(
\omega\left(  y\right)  \right)  ^{\perp}a.
\label{sol.Lambda.staircase-skew-1.a.linearity.2}%
\end{equation}
\par
But $\omega\left(  \lambda x+\mu y\right)  =\lambda\omega\left(  x\right)
+\mu\omega\left(  y\right)  $ (since $\omega$ is a $\kk$-algebra
endomorphism of $\Lambda$). Hence,%
\begin{align}
\left(  \underbrace{\omega\left(  \lambda x+\mu y\right)  }_{=\lambda
\omega\left(  x\right)  +\mu\omega\left(  y\right)  }\right)  ^{\perp}a  &
=\left(  \lambda\omega\left(  x\right)  +\mu\omega\left(  y\right)  \right)
^{\perp}a=\lambda\left(  \omega\left(  x\right)  \right)  ^{\perp}a+\mu\left(
\omega\left(  y\right)  \right)  ^{\perp}%
a\label{sol.Lambda.staircase-skew-1.a.linearity.2b}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.linearity.2})}\right)  .\nonumber
\end{align}
\par
Now, the definition of $\kappa$ yields%
\begin{equation}
\kappa\left(  x\right)  =x^{\perp}a-\left(  \omega\left(  x\right)  \right)
^{\perp}a \label{sol.Lambda.staircase-skew-1.a.linearity.4a}%
\end{equation}
and%
\begin{equation}
\kappa\left(  y\right)  =y^{\perp}a-\left(  \omega\left(  y\right)  \right)
^{\perp}a. \label{sol.Lambda.staircase-skew-1.a.linearity.4b}%
\end{equation}
But the definition of $\kappa$ also yields%
\begin{align*}
\kappa\left(  \lambda x+\mu y\right)   &  =\underbrace{\left(  \lambda x+\mu
y\right)  ^{\perp}a}_{\substack{=\lambda x^{\perp}a+\mu y^{\perp}a\\\text{(by
(\ref{sol.Lambda.staircase-skew-1.a.linearity.1}))}}}-\underbrace{\left(
\omega\left(  \lambda x+\mu y\right)  \right)  ^{\perp}a}_{\substack{=\lambda
\left(  \omega\left(  x\right)  \right)  ^{\perp}a+\mu\left(  \omega\left(
y\right)  \right)  ^{\perp}a\\\text{(by
(\ref{sol.Lambda.staircase-skew-1.a.linearity.2b}))}}}=\left(  \lambda
x^{\perp}a+\mu y^{\perp}a\right)  -\left(  \lambda\left(  \omega\left(
x\right)  \right)  ^{\perp}a+\mu\left(  \omega\left(  y\right)  \right)
^{\perp}a\right) \\
&  =\lambda x^{\perp}a+\mu y^{\perp}a-\lambda\left(  \omega\left(  x\right)
\right)  ^{\perp}a-\mu\left(  \omega\left(  y\right)  \right)  ^{\perp
}a=\underbrace{\left(  \lambda x^{\perp}a-\lambda\left(  \omega\left(
x\right)  \right)  ^{\perp}a\right)  }_{=\lambda\left(  x^{\perp}a-\left(
\omega\left(  x\right)  \right)  ^{\perp}a\right)  }+\underbrace{\left(  \mu
y^{\perp}a-\mu\left(  \omega\left(  y\right)  \right)  ^{\perp}a\right)
}_{=\mu\left(  y^{\perp}a-\left(  \omega\left(  y\right)  \right)  ^{\perp
}a\right)  }\\
&  =\lambda\underbrace{\left(  x^{\perp}a-\left(  \omega\left(  x\right)
\right)  ^{\perp}a\right)  }_{\substack{=\kappa\left(  x\right)  \\\text{(by
(\ref{sol.Lambda.staircase-skew-1.a.linearity.4a}))}}}+\mu\underbrace{\left(
y^{\perp}a-\left(  \omega\left(  y\right)  \right)  ^{\perp}a\right)
}_{\substack{=\kappa\left(  y\right)  \\\text{(by
(\ref{sol.Lambda.staircase-skew-1.a.linearity.4b}))}}}=\lambda\kappa\left(
x\right)  +\mu\kappa\left(  y\right)  .
\end{align*}
\par
Now, forget that we fixed $x$, $y$, $\lambda$ and $\mu$. We have thus shown
that $\kappa\left(  \lambda x+\mu y\right)  =\lambda\kappa\left(  x\right)
+\mu\kappa\left(  y\right)  $ for every $x\in\Lambda$, $y\in\Lambda$,
$\lambda\in\kk$ and $\mu\in\kk$. In other words, the map
$\kappa$ is $\kk$-linear. Qed.}. Hence, its kernel $\ker\kappa$ is a
$\kk$-submodule of $\Lambda$. Since%
\begin{align*}
\ker\kappa &  =\left\{  g\in\Lambda\ \mid\ \underbrace{\kappa\left(  g\right)
}_{\substack{=g^{\perp}a-\left(  \omega\left(  g\right)  \right)  ^{\perp
}a\\\text{(by the definition of }\kappa\text{)}}}=0\right\}  =\left\{
g\in\Lambda\ \mid\ \underbrace{g^{\perp}a-\left(  \omega\left(  g\right)
\right)  ^{\perp}a=0}_{\Longleftrightarrow\ \left(  g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a\right)  }\right\} \\
&  =\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\end{align*}
this rewrites as follows: The set $\mathfrak{D}$ is a $\kk$-submodule
of $\Lambda$.

Furthermore, we have the following two observations:

\begin{itemize}
\item We have $1\in\mathfrak{D}$\ \ \ \ \footnote{\textit{Proof.} Recall that
$\omega$ is a $\kk$-algebra endomorphism of $\Lambda$. Hence,
$\omega\left(  1\right)  =1$. Thus, $\left(  \underbrace{\omega\left(
1\right)  }_{=1}\right)  ^{\perp}a=1^{\perp}a$, so that $1^{\perp}a=\left(
\omega\left(  1\right)  \right)  ^{\perp}a$. Hence, $1$ is an element of
$\Lambda$ and satisfies $1^{\perp}a=\left(  \omega\left(  1\right)  \right)
^{\perp}a$. In other words, $1$ is an element $g$ of $\Lambda$ satisfying
$g^{\perp}a=\left(  \omega\left(  g\right)  \right)  ^{\perp}a$. Thus,%
\[
1\in\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\]
qed.}.

\item We have $xy\in\mathfrak{D}$ for each $x\in\mathfrak{D}$ and
$y\in\mathfrak{D}$\ \ \ \ \footnote{\textit{Proof.} Let $x\in\mathfrak{D}$ and
$y\in\mathfrak{D}$.
\par
We have $x\in\mathfrak{D}=\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a\right\}  $. In other words, $x$ is
an element $g$ of $\Lambda$ satisfying $g^{\perp}a=\left(  \omega\left(
g\right)  \right)  ^{\perp}a$. In other words, $x$ is an element of $\Lambda$
and satisfies $x^{\perp}a=\left(  \omega\left(  x\right)  \right)  ^{\perp}a$.
\par
We have $y\in\mathfrak{D}=\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a\right\}  $. In other words, $y$ is
an element $g$ of $\Lambda$ satisfying $g^{\perp}a=\left(  \omega\left(
g\right)  \right)  ^{\perp}a$. In other words, $y$ is an element of $\Lambda$
and satisfies $y^{\perp}a=\left(  \omega\left(  y\right)  \right)  ^{\perp}a$.
\par
Proposition~\ref{skewing-properties-prop}(ii) (applied to $f=\omega\left(
y\right)  $ and $g=\omega\left(  x\right)  $) yields%
\begin{equation}
\left(  \omega\left(  y\right)  \omega\left(  x\right)  \right)  ^{\perp
}a=\left(  \omega\left(  x\right)  \right)  ^{\perp}\left(  \left(
\omega\left(  y\right)  \right)  ^{\perp}a\right)  .
\label{sol.Lambda.staircase-skew-1.a.prod.1}%
\end{equation}
But $\omega\left(  xy\right)  =\omega\left(  x\right)  \omega\left(  y\right)
$ (since $\omega$ is an endomorphism of the $\kk$-algebra $\Lambda$).
Hence,%
\begin{equation}
\left(  \underbrace{\omega\left(  xy\right)  }_{\substack{=\omega\left(
x\right)  \omega\left(  y\right)  \\=\omega\left(  y\right)  \omega\left(
x\right)  }}\right)  ^{\perp}a=\left(  \omega\left(  y\right)  \omega\left(
x\right)  \right)  ^{\perp}a=\left(  \omega\left(  x\right)  \right)  ^{\perp
}\left(  \left(  \omega\left(  y\right)  \right)  ^{\perp}a\right)
\label{sol.Lambda.staircase-skew-1.a.prod.2}%
\end{equation}
(by (\ref{sol.Lambda.staircase-skew-1.a.prod.1})).
\par
On the other hand, Proposition~\ref{skewing-properties-prop}(ii) (applied to
$f=x$ and $g=y$) yields%
\[
\left(  xy\right)  ^{\perp}a=y^{\perp}\left(  \underbrace{x^{\perp}%
a}_{=\left(  \omega\left(  x\right)  \right)  ^{\perp}a}\right)  =y^{\perp
}\left(  \left(  \omega\left(  x\right)  \right)  ^{\perp}a\right)  .
\]
Comparing this with%
\[
\left(  \omega\left(  x\right)  y\right)  ^{\perp}a=y^{\perp}\left(  \left(
\omega\left(  x\right)  \right)  ^{\perp}a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition~\ref{skewing-properties-prop}(ii) (applied to }%
f=\omega\left(  x\right)  \text{ and }g=y\text{)}\right)  ,
\]
we obtain%
\begin{align*}
\left(  xy\right)  ^{\perp}a  &  =\left(  \underbrace{\omega\left(  x\right)
y}_{=y\omega\left(  x\right)  }\right)  ^{\perp}a=\left(  y\omega\left(
x\right)  \right)  ^{\perp}a=\left(  \omega\left(  x\right)  \right)  ^{\perp
}\left(  \underbrace{y^{\perp}a}_{=\left(  \omega\left(  y\right)  \right)
^{\perp}a}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition~\ref{skewing-properties-prop}(ii) (applied to }f=y\text{ and
}g=\omega\left(  x\right)  \text{)}\right) \\
&  =\left(  \omega\left(  x\right)  \right)  ^{\perp}\left(  \left(
\omega\left(  y\right)  \right)  ^{\perp}a\right)  =\left(  \omega\left(
xy\right)  \right)  ^{\perp}a\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.prod.2})}\right)  .
\end{align*}
Thus, $xy$ is an element of $\Lambda$ and satisfies $\left(  xy\right)
^{\perp}a=\left(  \omega\left(  xy\right)  \right)  ^{\perp}a$. In other
words, $xy$ is an element $g$ of $\Lambda$ satisfying $g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a$. Thus,%
\[
xy\in\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\]
qed.}.
\end{itemize}

Combining these two observations, we conclude that the set $\mathfrak{D}$ is a
$\kk$-subalgebra of $\Lambda$ (since we already know that
$\mathfrak{D}$ is a $\kk$-submodule of $\Lambda$). In view of
(\ref{sol.Lambda.staircase-skew-1.a.D=}), this rewrites as follows: The set
$\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  $ is a $\kk$-subalgebra of $\Lambda$. This
solves Exercise~\ref{exe.Lambda.staircase-skew-1}(a).

(b) We have
\begin{equation}
e_{k}^{\perp}a=h_{k}^{\perp}a\ \ \ \ \ \ \ \ \ \ \text{for each positive
integer }k \label{sol.Lambda.staircase-skew-1.b.ass}%
\end{equation}
(by assumption). Thus,%
\begin{equation}
e_{n}\in\mathfrak{D}\text{ for each }n\in\left\{  1,2,3,\ldots\right\}
\label{sol.Lambda.staircase-skew-1.b.enin}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Lambda.staircase-skew-1.b.enin}):} Let
$n\in\left\{  1,2,3,\ldots\right\}  $. Thus, $n$ is a positive integer. Hence,
$\omega\left(  e_{n}\right)  =h_{n}$ (by the definition of $\omega$). Thus,
$\left(  \underbrace{\omega\left(  e_{n}\right)  }_{=h_{n}}\right)  ^{\perp
}a=h_{n}^{\perp}a$. But (\ref{sol.Lambda.staircase-skew-1.b.ass}) (applied to
$k=n$) yields $e_{n}^{\perp}a=h_{n}^{\perp}a$. Comparing this with $\left(
\omega\left(  e_{n}\right)  \right)  ^{\perp}a=h_{n}^{\perp}a$, we find
$e_{n}^{\perp}a=\left(  \omega\left(  e_{n}\right)  \right)  ^{\perp}a$.
Hence, $e_{n}$ is an element of $\Lambda$ and satisfies $e_{n}^{\perp
}a=\left(  \omega\left(  e_{n}\right)  \right)  ^{\perp}a$. In other words,
$e_{n}$ is an element $g$ of $\Lambda$ satisfying $g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a$. Thus,%
\[
e_{n}\in\left\{  g\in\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)
\right)  ^{\perp}a\right\}  =\mathfrak{D}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-1.a.D=})}\right)  ,
\]
qed.}.

But Proposition~\ref{symm-antipode-and-h-basis} shows that the family
$\left\{  e_{n}\right\}  _{n=1,2,\ldots}$ generates $\Lambda_{\kk}$ as
an algebra. In other words, the $\kk$-algebra $\Lambda$ is generated by
the family $\left(  e_{n}\right)  _{n\in\left\{  1,2,3,\ldots\right\}  }$.
Thus, the smallest $\kk$-subalgebra of $\Lambda$ that contains all
elements of the family $\left(  e_{n}\right)  _{n\in\left\{  1,2,3,\ldots
\right\}  }$ is $\Lambda$ itself. In other words, if $\mathfrak{B}$ is a
$\kk$-subalgebra of $\Lambda$ satisfying%
\[
\left(  e_{n}\in\mathfrak{B}\text{ for each }n\in\left\{  1,2,3,\ldots
\right\}  \right)  ,
\]
then $\mathfrak{B}=\Lambda$. Applying this to $\mathfrak{B}=\mathfrak{D}$, we
conclude that $\mathfrak{D}=\Lambda$ (because $\mathfrak{D}$ is a $\kk%
$-subalgebra of $\Lambda$, and because it satisfies
(\ref{sol.Lambda.staircase-skew-1.b.enin})).

Now, let $f\in\Lambda$. Then, $f\in\Lambda=\mathfrak{D}=\left\{  g\in
\Lambda\ \mid\ g^{\perp}a=\left(  \omega\left(  g\right)  \right)  ^{\perp
}a\right\}  $ (by (\ref{sol.Lambda.staircase-skew-1.a.D=})). In other words,
$f$ is an element $g$ of $\Lambda$ satisfying $g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a$. In other words, $f$ is an element
of $\Lambda$ and satisfies $f^{\perp}a=\left(  \omega\left(  f\right)
\right)  ^{\perp}a$.

Now, forget that we fixed $f$. We thus have proven that $f^{\perp}a=\left(
\omega\left(  f\right)  \right)  ^{\perp}a$ for each $f\in\Lambda$. Renaming
$f$ as $g$ in this statement, we conclude the following: $g^{\perp}a=\left(
\omega\left(  g\right)  \right)  ^{\perp}a$ for each $g\in\Lambda$. This
solves Exercise~\ref{exe.Lambda.staircase-skew-1}(b).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.staircase-skew-2}}

\textit{Solution to Exercise~\ref{exe.Lambda.staircase-skew-2}.} Let us begin
by proving a few simple lemmas:

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.strip=strip}Let $n\in \NN$. Let
$\rho$ be the partition $\left(  n-1,n-2,\ldots,1\right)  $. Let $\mu
\in \Par $. Then, $\rho/\mu$ is a horizontal strip if and only if
$\rho/\mu$ is a vertical strip.
\end{lemma}

Lemma \ref{lem.sol.Lambda.staircase-skew-2.strip=strip} becomes visually
obvious if one draws in one's mind the Ferrers diagram of the staircase
partition $\rho$ and attempts to cut off either a horizontal strip or a
vertical strip from it (in either case, the only possibilities are to remove
some of its corners). But let us give a rigorous proof:

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.strip=strip}.]Write the
partition $\mu$ in the form $\mu=\left(  \mu_{1},\mu_{2},\mu_{3}%
,\ldots\right)  $. Write the partition $\rho$ in the form $\rho=\left(
\rho_{1},\rho_{2},\rho_{3},\ldots\right)  $. Then,%
\[
\left(  \rho_{1},\rho_{2},\rho_{3},\ldots\right)  =\rho=\left(  n-1,n-2,\ldots
,1\right)  =\left(  n-1,n-2,\ldots,1,0,0,0,\ldots\right)  .
\]
Hence,%
\begin{equation}
\rho_{i}=%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for each positive integer }i\text{.}
\label{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}%
\end{equation}


Now, we are going to prove the following two claims:

\begin{statement}
\textit{Claim 1:} If $\rho/\mu$ is a horizontal strip, then $\rho/\mu$ is a
vertical strip.
\end{statement}

\begin{statement}
\textit{Claim 2:} If $\rho/\mu$ is a vertical strip, then $\rho/\mu$ is a
horizontal strip.
\end{statement}

\textit{Proof of Claim 1:} Assume that $\rho/\mu$ is a horizontal strip. We
must then prove that $\rho/\mu$ is a vertical strip.

We have $\mu\subseteq\rho$ (since $\rho/\mu$ is a horizontal strip). Exercise
\ref{exe.horistrip.postnikov}(a) (applied to $\lambda=\rho$ and $\lambda
_{i}=\rho_{i}$) yields that $\rho/\mu$ is a horizontal strip if and only if
every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq\rho_{i+1}$.
Thus,
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies }\mu_{i}%
\geq\rho_{i+1}
\label{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c1.pf.ass}%
\end{equation}
(since $\rho/\mu$ is a horizontal strip).

Now, every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\rho_{i}\leq
\mu_{i}+1$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots
\right\}  $. We must show that $\rho_{i}\leq\mu_{i}+1$.
\par
If $i\geq n-1$, then%
\begin{align*}
\rho_{i}  &  =%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=})}\right) \\
&  \leq%
\begin{cases}
1, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-i\leq1\text{ (because }i\geq
n-1\text{)}\right) \\
&  \leq%
\begin{cases}
1, & \text{if }i<n;\\
1, & \text{if }i\geq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq1\right) \\
&  =1=\underbrace{0}_{\leq\mu_{i}}+1\leq\mu_{i}+1.
\end{align*}
Hence, if $i\geq n-1$, then $\rho_{i}\leq\mu_{i}+1$ holds. Thus, for the rest
of this proof of $\rho_{i}\leq\mu_{i}+1$, we WLOG assume that we don't have
$i\geq n-1$.
\par
We have $i<n-1$ (since we don't have $i\geq n-1$). But
(\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) yields $\rho_{i}=%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}  \quad
=n-i$ (since $i<n-1<n$).
\par
But from $i<n-1$, we obtain $i+1<n$. Now,
(\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) (applied to $i+1$
instead of $i$) yields $\rho_{i+1}=%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases}  \quad
=n-\left(  i+1\right)  $ (since $i+1<n$).
\par
But (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c1.pf.ass}) yields
$\mu_{i}\geq\rho_{i+1}=n-\left(  i+1\right)  =n-i-1$. Adding $1$ to both sides
of this inequality, we obtain $\mu_{i}+1\geq n-i=\rho_{i}$ (since $\rho
_{i}=n-i$). Hence, $\rho_{i}\leq\mu_{i}+1$. This completes our proof of
$\rho_{i}\leq\mu_{i}+1$.}. But Exercise \ref{exe.horistrip.postnikov}(b)
(applied to $\lambda=\rho$ and $\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$
is a vertical strip if and only if every $i\in\left\{  1,2,3,\ldots\right\}  $
satisfies $\rho_{i}\leq\mu_{i}+1$. Thus, $\rho/\mu$ is a vertical strip (since
every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\rho_{i}\leq\mu_{i}%
+1$). This proves Claim 1.

\textit{Proof of Claim 2:} Assume that $\rho/\mu$ is a vertical strip. We must
then prove that $\rho/\mu$ is a horizontal strip.

We know that $\rho/\mu$ is a vertical strip. In particular, $\mu\subseteq\rho$.

Exercise \ref{exe.horistrip.postnikov}(b) (applied to $\lambda=\rho$ and
$\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$ is a vertical strip if and only
if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\rho_{i}\leq\mu
_{i}+1$. Thus,
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies }\rho_{i}%
\leq\mu_{i}+1 \label{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c2.pf.ass}%
\end{equation}
(since $\rho/\mu$ is a vertical strip).

Now, every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq
\rho_{i+1}$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots
\right\}  $. We must show that $\mu_{i}\geq\rho_{i+1}$.
\par
Applying (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) to $i+1$
instead of $i$, we obtain $\rho_{i+1}=%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases} \quad
$.
\par
If $i+1\geq n$, then%
\begin{align*}
\rho_{i+1}  &  =%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i+1\geq n\right) \\
&  \leq\mu_{i}.
\end{align*}
In other words, if $i+1\geq n$, then $\mu_{i}\geq\rho_{i+1}$. Thus, for the
rest of this proof of $\mu_{i}\geq\rho_{i+1}$, we WLOG assume that we don't
have $i+1\geq n$.
\par
We have $i+1<n$ (since we don't have $i+1\geq n$). Thus, $\rho_{i+1}=%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases}  \quad
=n-\left(  i+1\right)  $ (since $i+1<n$).
\par
But (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) yields
$\rho_{i}=%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}  \quad
=n-i$ (since $i<i+1<n$).
\par
From (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c2.pf.ass}), we
obtain $\rho_{i}\leq\mu_{i}+1$, so that $\mu_{i}+1\geq\rho_{i}=n-i$.
Subtracting $1$ from both sides of this inequality, we find $\mu_{i}\geq
n-i-1=n-\left(  i+1\right)  =\rho_{i+1}$ (since $\rho_{i+1}=n-\left(
i+1\right)  $). This completes our proof of $\mu_{i}\geq\rho_{i+1}$.}. But
Exercise \ref{exe.horistrip.postnikov}(a) (applied to $\lambda=\rho$ and
$\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$ is a horizontal strip if and
only if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq
\rho_{i+1}$. Thus, $\rho/\mu$ is a horizontal strip (since every $i\in\left\{
1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq\rho_{i+1}$). This proves Claim 2.

Combining Claim 1 with Claim 2, we conclude that $\rho/\mu$ is a horizontal
strip if and only if $\rho/\mu$ is a vertical strip. Lemma
\ref{lem.sol.Lambda.staircase-skew-2.strip=strip} is thus proven.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.ekTrho}Let $n\in \NN$. Let $\rho$
be the partition $\left(  n-1,n-2,\ldots,1\right)  $. Let $k$ be a positive
integer. Then, $e_{k}^{\perp}s_{\rho}=h_{k}^{\perp}s_{\rho}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.ekTrho}.]Fix $\nu
\in \Par $. We have the following logical equivalence:%
\begin{align}
&  \ \left(  \rho/\nu\text{ is a horizontal }k\text{-strip}\right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a horizontal strip and has
size }k\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
a ``horizontal $k$-strip''}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a horizontal strip and
satisfies }\left\vert \rho/\nu\right\vert =k\right)
.\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv1}%
\end{align}
Similarly, we also have the following logical equivalence:%
\begin{align}
&  \ \left(  \rho/\nu\text{ is a vertical }k\text{-strip}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical strip and
satisfies }\left\vert \rho/\nu\right\vert =k\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv2}
\end{align}


But Lemma \ref{lem.sol.Lambda.staircase-skew-2.strip=strip} (applied to
$\mu=\nu$) shows that $\rho/\nu$ is a horizontal strip if and only if
$\rho/\nu$ is a vertical strip. In other words, we have the following logical
equivalence:%
\begin{equation}
\ \left(  \rho/\nu\text{ is a horizontal strip}\right)  \ \Longleftrightarrow
\ \left(  \rho/\nu\text{ is a vertical strip}\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv3}
\end{equation}


Now, we have the following chain of logical equivalences:
\begin{align}
&  \ \left(  \rho/\nu\text{ is a horizontal }k\text{-strip}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\rho/\nu\text{ is a horizontal
strip}}_{\substack{\Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical
strip}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv3}))}}}\text{ and
satisfies }\left\vert \rho/\nu\right\vert =k\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv1})}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical strip and
satisfies }\left\vert \rho/\nu\right\vert =k\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical }k\text{-strip}%
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv2})}\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv5}%
\end{align}


Now, forget that we fixed $\nu$. We thus have proven the equivalence
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv5}) for each $\nu
\in \Par $.

Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}(a) (applied to
$\gamma=\rho$) yields%
\[
h_{k}^{\perp}s_{\rho}=\underbrace{\sum\limits_{\substack{\nu\in \Par %
;\\\rho/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}}_{\substack{=\sum\limits
_{\substack{\nu\in \Par ;\\\rho/\nu\text{ is a}\\\text{vertical
}k\text{-strip}}}\\\text{(by the equivalence
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv5}))}}}s_{\nu}%
=\sum\limits_{\substack{\nu\in \Par ;\\\rho/\nu\text{ is a}%
\\\text{vertical }k\text{-strip}}}s_{\nu}.
\]
Comparing this with%
\[
e_{k}^{\perp}s_{\rho}=\sum\limits_{\substack{\nu\in \Par ;\\\rho
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.Lambda.staircase-skew-2.coPieri}(b) (applied to }\gamma
=\rho\text{)}\right)  ,
\]
we obtain $e_{k}^{\perp}s_{\rho}=h_{k}^{\perp}s_{\rho}$. This proves
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.ekTrho}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.omegas}Let $\lambda\in
 \Par $. Then, $\omega\left(  s_{\lambda}\right)  =s_{\lambda
^{t}}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas}.]We have
$\varnothing\subseteq\lambda$. Hence, the first equation of
(\ref{skew-Schur-antipodes}) (applied to $\mu=\varnothing$) yields
$\omega\left(  s_{\lambda/\varnothing}\right)  =s_{\lambda^{t}/\varnothing
^{t}}=s_{\lambda^{t}/\varnothing}$ (since $\varnothing^{t}=\varnothing$).

But recall that $s_{\lambda}=s_{\lambda/\varnothing}$. The same argument (but
applied to $\lambda^{t}$ instead of $\lambda$) shows that $s_{\lambda^{t}%
}=s_{\lambda^{t}/\varnothing}$. Comparing this with $\omega\left(
\underbrace{s_{\lambda}}_{=s_{\lambda/\varnothing}}\right)  =\omega\left(
s_{\lambda/\varnothing}\right)  =s_{\lambda^{t}/\varnothing}$, we obtain
$\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}$. This proves
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas}.
\end{proof}

We can now solve Exercise~\ref{exe.Lambda.staircase-skew-2} easily:

\textit{Solution to Exercise~\ref{exe.Lambda.staircase-skew-2}.}
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.ekTrho} shows that $e_{k}^{\perp
}s_{\rho}=h_{k}^{\perp}s_{\rho}$ for each positive integer $k$. Hence,
Exercise~\ref{exe.Lambda.staircase-skew-1}(b) (applied to $a=s_{\rho}$) yields
that%
\begin{equation}
g^{\perp}s_{\rho}=\left(  \omega\left(  g\right)  \right)  ^{\perp}s_{\rho
}\ \ \ \ \ \ \ \ \ \ \text{for each }g\in\Lambda.
\label{sol.Lambda.staircase-skew-2.almostthere}%
\end{equation}


Recall that
\begin{equation}
s_{\mu}^{\perp}\left(  s_{\lambda}\right)  =s_{\lambda/\mu}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in \Par \text{ and
}\mu\in \Par . \label{sol.Lambda.staircase-skew-2.perp}%
\end{equation}


Fix $\mu\in \Par $. Then,
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas} (applied to $\mu$ instead
of $\lambda$) shows that $\omega\left(  s_{\mu}\right)  =s_{\mu^{t}}$. Now,
(\ref{sol.Lambda.staircase-skew-2.almostthere}) (applied to $g=s_{\mu}$)
yields%
\[
s_{\mu}^{\perp}s_{\rho}=\left(  \underbrace{\omega\left(  s_{\mu}\right)
}_{=s_{\mu^{t}}}\right)  ^{\perp}s_{\rho}=\left(  s_{\mu^{t}}\right)  ^{\perp
}s_{\rho}=\left(  s_{\mu^{t}}\right)  ^{\perp}\left(  s_{\rho}\right)
=s_{\rho/\mu^{t}}%
\]
(by (\ref{sol.Lambda.staircase-skew-2.perp}) (applied to $\rho$ and $\mu^{t}$
instead of $\lambda$ and $\mu$)). Comparing this with%
\[
s_{\mu}^{\perp}s_{\rho}=s_{\mu}^{\perp}\left(  s_{\rho}\right)  =s_{\rho/\mu
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Lambda.staircase-skew-2.perp})
(applied to }\lambda=\rho\text{)}\right)  ,
\]
this yields $s_{\rho/\mu}=s_{\rho/\mu^{t}}$. This solves
Exercise~\ref{exe.Lambda.staircase-skew-2}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.staircase-skew-2}}

\textit{Solution to Exercise~\ref{exe.Lambda.staircase-skew-2}.} Let us begin
by proving a few simple lemmas:

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.strip=strip}Let $n\in \NN$. Let
$\rho$ be the partition $\left(  n-1,n-2,\ldots,1\right)  $. Let $\mu
\in \Par $. Then, $\rho/\mu$ is a horizontal strip if and only if
$\rho/\mu$ is a vertical strip.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.strip=strip}.]Write the
partition $\mu$ in the form $\mu=\left(  \mu_{1},\mu_{2},\mu_{3}%
,\ldots\right)  $. Write the partition $\rho$ in the form $\rho=\left(
\rho_{1},\rho_{2},\rho_{3},\ldots\right)  $. Then,%
\[
\left(  \rho_{1},\rho_{2},\rho_{3},\ldots\right)  =\rho=\left(  n-1,n-2,\ldots
,1\right)  =\left(  n-1,n-2,\ldots,1,0,0,0,\ldots\right)  .
\]
Hence,%
\begin{equation}
\rho_{i}=%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for each positive integer }i\text{.}
\label{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}%
\end{equation}


Now, we are going to prove the following two claims:

\begin{statement}
\textit{Claim 1:} If $\rho/\mu$ is a horizontal strip, then $\rho/\mu$ is a
vertical strip.
\end{statement}

\begin{statement}
\textit{Claim 2:} If $\rho/\mu$ is a vertical strip, then $\rho/\mu$ is a
horizontal strip.
\end{statement}

\textit{Proof of Claim 1:} Assume that $\rho/\mu$ is a horizontal strip. We
must then prove that $\rho/\mu$ is a vertical strip.

We know that $\rho/\mu$ is a horizontal strip. In particular, $\mu
\subseteq\rho$.

Exercise \ref{exe.horistrip.postnikov}(a) (applied to $\lambda=\rho$ and
$\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$ is a horizontal strip if and
only if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq
\rho_{i+1}$. Thus,
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies }\mu_{i}%
\geq\rho_{i+1}
\label{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c1.pf.ass}%
\end{equation}
(since $\rho/\mu$ is a horizontal strip).

Now, every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\rho_{i}\leq
\mu_{i}+1$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots
\right\}  $. We must show that $\rho_{i}\leq\mu_{i}+1$.
\par
If $i\geq n$, then%
\begin{align*}
\rho_{i}  &  =%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=})}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\geq n\right) \\
&  \leq\mu_{i}\leq\mu_{i}+1.
\end{align*}
Hence, if $i\geq n$, then $\rho_{i}\leq\mu_{i}+1$ holds. Thus, for the rest of
this proof of $\rho_{i}\leq\mu_{i}+1$, we can WLOG assume that we don't have
$i\geq n$. Assume this.
\par
We have $i<n$ (since we don't have $i\geq n$). Thus, $i\leq n-1$ (since $i$
and $n$ are integers). But
(\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) yields $\rho_{i}=%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}  \quad
=n-i$ (since $i<n$).
\par
If $i=n-1$, then $\rho_{i}=n-\underbrace{i}_{=n-1}=n-\left(  n-1\right)
=1=\underbrace{0}_{\leq\mu_{i}}+1\leq\mu_{i}+1$. Hence, if $i=n-1$, then
$\rho_{i}\leq\mu_{i}+1$ holds. Thus, for the rest of this proof of $\rho
_{i}\leq\mu_{i}+1$, we can WLOG assume that we don't have $i=n-1$. Assume
this.
\par
We have $i\neq n-1$ (since we don't have $i=n-1$). Combining this with $i\leq
n-1$, we obtain $i<n-1$. Hence, $i+1<n$. Now,
(\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) (applied to $i+1$
instead of $i$) yields $\rho_{i+1}=%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases}  \quad
=n-\left(  i+1\right)  $ (since $i+1<n$).
\par
But (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c1.pf.ass}) yields
$\mu_{i}\geq\rho_{i+1}=n-\left(  i+1\right)  =n-i-1$. Adding $1$ to both sides
of this inequality, we obtain $\mu_{i}+1\geq n-i=\rho_{i}$ (since $\rho
_{i}=n-i$). Hence, $\rho_{i}\leq\mu_{i}+1$. This completes our proof of
$\rho_{i}\leq\mu_{i}+1$.}. But Exercise \ref{exe.horistrip.postnikov}(b)
(applied to $\lambda=\rho$ and $\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$
is a vertical strip if and only if every $i\in\left\{  1,2,3,\ldots\right\}  $
satisfies $\rho_{i}\leq\mu_{i}+1$. Thus, $\rho/\mu$ is a vertical strip (since
every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\rho_{i}\leq\mu_{i}%
+1$). This proves Claim 1.

\textit{Proof of Claim 2:} Assume that $\rho/\mu$ is a vertical strip. We must
then prove that $\rho/\mu$ is a horizontal strip.

We know that $\rho/\mu$ is a vertical strip. In particular, $\mu\subseteq\rho$.

Exercise \ref{exe.horistrip.postnikov}(b) (applied to $\lambda=\rho$ and
$\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$ is a vertical strip if and only
if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\rho_{i}\leq\mu
_{i}+1$. Thus,
\begin{equation}
\text{every }i\in\left\{  1,2,3,\ldots\right\}  \text{ satisfies }\rho_{i}%
\leq\mu_{i}+1 \label{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c2.pf.ass}%
\end{equation}
(since $\rho/\mu$ is a vertical strip).

Now, every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq
\rho_{i+1}$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,3,\ldots
\right\}  $. We must show that $\mu_{i}\geq\rho_{i+1}$.
\par
Applying (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) to $i+1$
instead of $i$, we obtain $\rho_{i+1}=%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases} \quad
$.
\par
If $i+1\geq n$, then%
\begin{align*}
\rho_{i+1}  &  =%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i+1\geq n\right) \\
&  \leq\mu_{i}.
\end{align*}
In other words, if $i+1\geq n$, then $\mu_{i}\geq\rho_{i+1}$. Thus, for the
rest of this proof of $\mu_{i}\geq\rho_{i+1}$, we can WLOG assume that we
don't have $i+1\geq n$. Assume this.
\par
We have $i+1<n$ (since we don't have $i+1\geq n$). Thus, $\rho_{i+1}=%
\begin{cases}
n-\left(  i+1\right)  , & \text{if }i+1<n;\\
0, & \text{if }i+1\geq n
\end{cases}  \quad
=n-\left(  i+1\right)  $ (since $i+1<n$).
\par
But (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.rhoi=}) yields
$\rho_{i}=%
\begin{cases}
n-i, & \text{if }i<n;\\
0, & \text{if }i\geq n
\end{cases}  \quad
=n-i$ (since $i<i+1<n$).
\par
From (\ref{pf.lem.sol.Lambda.staircase-skew-2.strip=strip.c2.pf.ass}), we
obtain $\rho_{i}\leq\mu_{i}+1$, so that $\mu_{i}+1\geq\rho_{i}=n-i$.
Subtracting $1$ from both sides of this inequality, we find $\mu_{i}\geq
n-i-1=n-\left(  i+1\right)  =\rho_{i+1}$ (since $\rho_{i+1}=n-\left(
i+1\right)  $). This completes our proof of $\mu_{i}\geq\rho_{i+1}$.}. But
Exercise \ref{exe.horistrip.postnikov}(a) (applied to $\lambda=\rho$ and
$\lambda_{i}=\rho_{i}$) yields that $\rho/\mu$ is a horizontal strip if and
only if every $i\in\left\{  1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq
\rho_{i+1}$. Thus, $\rho/\mu$ is a horizontal strip (since every $i\in\left\{
1,2,3,\ldots\right\}  $ satisfies $\mu_{i}\geq\rho_{i+1}$). This proves Claim 2.

Combining Claim 1 with Claim 2, we conclude that $\rho/\mu$ is a horizontal
strip if and only if $\rho/\mu$ is a vertical strip. Lemma
\ref{lem.sol.Lambda.staircase-skew-2.strip=strip} is thus proven.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.ekTrho}Let $n\in \NN$. Let $\rho$
be the partition $\left(  n-1,n-2,\ldots,1\right)  $. Let $k$ be a positive
integer. Then, $e_{k}^{\perp}s_{\rho}=h_{k}^{\perp}s_{\rho}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lambda.staircase-skew-2.ekTrho}.]Fix $\nu
\in \Par $. We have the following logical equivalence:%
\begin{align}
&  \ \left(  \rho/\nu\text{ is a horizontal }k\text{-strip}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a horizontal strip and has
size }k\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
a ``horizontal $k$-strip''}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a horizontal strip and
satisfies }\left\vert \rho/\nu\right\vert =k\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv1}%
\end{align}
We also have the following logical equivalence:%
\begin{align}
&  \ \left(  \rho/\nu\text{ is a vertical }k\text{-strip}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical strip and has
size }k\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
a ``vertical $k$-strip''}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical strip and
satisfies }\left\vert \rho/\nu\right\vert =k\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv2}
\end{align}


But Lemma \ref{lem.sol.Lambda.staircase-skew-2.strip=strip} (applied to
$\mu=\nu$) shows that $\rho/\nu$ is a horizontal strip if and only if
$\rho/\nu$ is a vertical strip. In other words, we have the following logical
equivalence:%
\begin{equation}
\ \left(  \rho/\nu\text{ is a horizontal strip}\right)  \ \Longleftrightarrow
\ \left(  \rho/\nu\text{ is a vertical strip}\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv3}
\end{equation}


Now, we have the following chain of logical equivalences:
\begin{align}
&  \ \left(  \rho/\nu\text{ is a horizontal }k\text{-strip}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\rho/\nu\text{ is a horizontal
strip}}_{\substack{\Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical
strip}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv3}))}}}\text{ and
satisfies }\left\vert \rho/\nu\right\vert =k\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv1})}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical strip and
satisfies }\left\vert \rho/\nu\right\vert =k\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \rho/\nu\text{ is a vertical }k\text{-strip}%
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv2})}\right)  .
\label{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv5}%
\end{align}


Now, forget that we fixed $\nu$. We thus have proven the equivalence
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv5}) for each $\nu
\in \Par $.

Lemma \ref{lem.sol.Lambda.staircase-skew-2.coPieri}(a) (applied to
$\gamma=\rho$) yields%
\[
h_{k}^{\perp}s_{\rho}=\underbrace{\sum\limits_{\substack{\nu\in \Par %
;\\\rho/\nu\text{ is a}\\\text{horizontal }k\text{-strip}}}}_{\substack{=\sum\limits
_{\substack{\nu\in \Par ;\\\rho/\nu\text{ is a}\\\text{vertical
}k\text{-strip}}}\\\text{(because for each }\nu\in \Par \text{,
the}\\\text{condition }\left(  \rho/\nu\text{ is a horizontal }k\text{-strip}%
\right)  \\\text{is equivalent to the condition}\\\left(  \rho/\nu\text{ is a
vertical }k\text{-strip}\right)  \\\text{(by
(\ref{pf.lem.sol.Lambda.staircase-skew-2.ekTrho.equiv5})))}}}s_{\nu}%
=\sum\limits_{\substack{\nu\in \Par ;\\\rho/\nu\text{ is a}%
\\\text{vertical }k\text{-strip}}}s_{\nu}.
\]
Comparing this with%
\[
e_{k}^{\perp}s_{\rho}=\sum\limits_{\substack{\nu\in \Par ;\\\rho
/\nu\text{ is a}\\\text{vertical }k\text{-strip}}}s_{\nu}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.Lambda.staircase-skew-2.coPieri}(b) (applied to }\gamma
=\rho\text{)}\right)  ,
\]
we obtain $e_{k}^{\perp}s_{\rho}=h_{k}^{\perp}s_{\rho}$. This proves
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.ekTrho}.
\end{proof}

\begin{lemma}
\label{lem.sol.Lambda.staircase-skew-2.omegas}Let $\lambda\in
 \Par $. Then, $\omega\left(  s_{\lambda}\right)  =s_{\lambda
^{t}}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas}.]We have
$\varnothing\subseteq\lambda$. Hence, the first equation of
(\ref{skew-Schur-antipodes}) (applied to $\mu=\varnothing$) yields
$\omega\left(  s_{\lambda/\varnothing}\right)  =s_{\lambda^{t}/\varnothing
^{t}}=s_{\lambda^{t}/\varnothing}$ (since $\varnothing^{t}=\varnothing$).

But recall that $s_{\lambda}=s_{\lambda/\varnothing}$. The same argument (but
applied to $\lambda^{t}$ instead of $\lambda$) shows that $s_{\lambda^{t}%
}=s_{\lambda^{t}/\varnothing}$. Comparing this with $\omega\left(
\underbrace{s_{\lambda}}_{=s_{\lambda/\varnothing}}\right)  =\omega\left(
s_{\lambda/\varnothing}\right)  =s_{\lambda^{t}/\varnothing}$, we obtain
$\omega\left(  s_{\lambda}\right)  =s_{\lambda^{t}}$. This proves
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas}.
\end{proof}

We can now solve Exercise~\ref{exe.Lambda.staircase-skew-2} easily:

\textit{Solution to Exercise~\ref{exe.Lambda.staircase-skew-2}.}
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.ekTrho} shows that $e_{k}^{\perp
}s_{\rho}=h_{k}^{\perp}s_{\rho}$ for each positive integer $k$. Hence,
Exercise~\ref{exe.Lambda.staircase-skew-1}(b) (applied to $a=s_{\rho}$) yields
that%
\begin{equation}
g^{\perp}s_{\rho}=\left(  \omega\left(  g\right)  \right)  ^{\perp}s_{\rho
}\ \ \ \ \ \ \ \ \ \ \text{for each }g\in\Lambda.
\label{sol.Lambda.staircase-skew-2.almostthere}%
\end{equation}


Recall that
\begin{equation}
s_{\mu}^{\perp}s_{\lambda}=s_{\lambda/\mu}\ \ \ \ \ \ \ \ \ \ \text{for every
}\lambda\in \Par \text{ and }\mu\in \Par %
.\label{sol.Lambda.staircase-skew-2.perp}%
\end{equation}


Fix $\mu\in \Par $. Then,
Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas} (applied to $\mu$ instead
of $\lambda$) shows that $\omega\left(  s_{\mu}\right)  =s_{\mu^{t}}$. Now,
(\ref{sol.Lambda.staircase-skew-2.almostthere}) (applied to $g=s_{\mu}$)
yields%
\[
s_{\mu}^{\perp}s_{\rho}=\left(  \underbrace{\omega\left(  s_{\mu}\right)
}_{=s_{\mu^{t}}}\right)  ^{\perp}s_{\rho}=\left(  s_{\mu^{t}}\right)  ^{\perp
}s_{\rho}=s_{\rho/\mu^{t}}%
\]
(by (\ref{sol.Lambda.staircase-skew-2.perp}), applied to $\rho$ and $\mu^{t}$
instead of $\lambda$ and $\mu$). Comparing this with%
\[
s_{\mu}^{\perp}s_{\rho}=s_{\rho/\mu}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Lambda.staircase-skew-2.perp}), applied to }\lambda=\rho\right)  ,
\]
this yields $s_{\rho/\mu}=s_{\rho/\mu^{t}}$. This solves
Exercise~\ref{exe.Lambda.staircase-skew-2}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.primitive-split-off-prop.infinite}}

\textit{Solution to Exercise~\ref{exe.primitive-split-off-prop.infinite}.}
Let $A$ be as in
Proposition~\ref{primitives-split-off-prop}, and assume that
$\kk = \QQ$. We have $\Liep \cap I^{2}=0$ (by
Proposition~\ref{primitives-split-off-prop} (b)). Thus,
Lemma~\ref{general-commutativity-lemma} yields that $A$ is commutative.
Since $A$ is self-dual, this yields that $A$ is cocommutative. Hence,
Exercise~\ref{exe.dynkin} (d) shows that the $\kk$-algebra $A$ is
generated by the
$\kk$-submodule $\Liep$. In other words, $A$ is the $\kk$-subalgebra
of $A$ generated by the $\kk$-submodule $\Liep$. Since the
$\kk$-subalgebra of $A$ generated by the $\kk$-submodule $\Liep$ is $\sum_{n\geq
0}\Liep^n$ (this is because generally, if $B$ is a
$\kk$-algebra, and $U$ is a $\kk$-submodule of $B$, then the
$\kk$-subalgebra of $B$ generated by the $\kk$-submodule $U$ is $\sum_{n\geq 0}U^{n}$),
this rewrites as follows: $A$ is $\sum_{n\geq 0}\Liep^{n}$. Hence,
\begin{align*}
A &=\sum_{n\geq 0}\Liep^{n}=\underbrace{\Liep^{0}}_{=\kk
\cdot 1_{A}}+\underbrace{\Liep^{1}}_{=\Liep}+\sum_{n\geq 2}
\underbrace{\Liep^{n}}_{=\Liep\cdot \Liep\cdot
\Liep^{n-2}}=\kk\cdot 1_{A}+\Liep+\sum_{n\geq 2}
\underbrace{\Liep}_{\subset I}\cdot \underbrace{\Liep}
_{\subset I}\cdot \Liep^{n-2} \\
&\subset \kk\cdot 1_{A}+\Liep+\sum_{n\geq 2}I\cdot
\underbrace{I\cdot \Liep^{n-2}}_{\substack{ \subset I \\ \text{(since
}I\text{ is an ideal of }A\text{)}}}\subset \kk\cdot 1_{A}+\Liep
+\underbrace{\sum_{n\geq 2}I\cdot I}_{\subset I\cdot I=I^{2}}\subset
\kk\cdot 1_{A}+\Liep+I^{2}.
\end{align*}
Now, let $a\in I$. Then, $\epsilon \left( a\right) =0$ (by the definition of
$I$). But we know that $a\in I\subset A\subset \kk\cdot 1_{A}+
\Liep+I^{2}$. Hence, there exist some $\lambda \in \kk$ and
some $a^{\prime }\in \Liep+I^{2}$ such that $a=\lambda \cdot
1_{A}+a^{\prime }$. Since $a^{\prime }\in \Liep+I^{2}\subset I$, we
have $\epsilon \left( a^{\prime }\right) =0$. Now, $\epsilon \left( a\right)
=0$, so that
\[
0=\epsilon \left( \underbrace{a}_{=\lambda \cdot 1_{A}+a^{\prime }}\right)
=\epsilon \left( \lambda \cdot 1_{A}+a^{\prime }\right) =\lambda \underbrace{
\epsilon \left( 1_{A}\right) }_{=1}+\underbrace{\epsilon \left( a^{\prime
}\right) }_{=0}=\lambda ,
\]
so that $\lambda =0$ and thus $a=\underbrace{\lambda }_{=0}\cdot
1_{A}+a^{\prime }=a^{\prime }\in \Liep+I^{2}$.
We thus have shown that every $a\in I$ satisfies $a\in \Liep+I^{2}$.
Hence, $I\subset \Liep+I^{2}$. Combined with $\Liep
+I^{2}\subset I$, this yields $I=\Liep+I^{2}$, and thus
$I=\Liep \oplus I^{2}$ (since $\Liep\cap I^{2}=0$). This proves
Proposition~\ref{primitives-split-off-prop}(c).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.primitives}}

\textit{Solution to Exercise \ref{exe.Lambda.primitives}.} Corollary
\ref{dual-bases-h-m-p}(a) yields that $\left(  h_{\lambda}\right)
_{\lambda\in\operatorname{Par}}$ and $\left(  m_{\lambda}\right)  _{\lambda
\in\operatorname{Par}}$ are dual bases with respect to the Hall inner product
on $\Lambda$. In other words,
\[
\left(  h_{\lambda},m_{\mu}\right)  =\delta_{\lambda,\mu}%
\ \ \ \ \ \ \ \ \ \ \text{for any partitions }\lambda\text{ and }\mu.
\]
Thus, every nonempty partition $\lambda$ satisfies $\left(  h_{\lambda
},m_{\varnothing}\right)  =\delta_{\lambda,\varnothing}=0$. Hence, every
nonempty partition $\lambda$ satisfies%
\begin{equation}
\left(  h_{\lambda},\underbrace{1}_{=m_{\varnothing}}\right)  =\left(
h_{\lambda},m_{\varnothing}\right)  =0.
\label{sol.exe.Lambda.primitives.annihilate}%
\end{equation}


Recall the following fundamental fact from linear algebra: If $\kk$ is
a commutative ring, if $A$ is a $\kk$-module, if $\left(  \cdot
,\cdot\right)  :A\times A\rightarrow \kk  $ is a symmetric $\kk 
$-bilinear form on $A$, and if $\left(  u_{\lambda}\right)  _{\lambda\in L}$
and $\left(  v_{\lambda}\right)  _{\lambda\in L}$ are two $\kk  $-bases
of $A$ which are dual to each other with respect to the form $\left(
\cdot,\cdot\right)  $ (where $L$ is some indexing set), then every $a\in A$
satisfies
\begin{equation}
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}.
\label{sol.exe.Lambda.primitives.dual-coords.general}%
\end{equation}
We can apply this fact to $A=\Lambda$, $L= \operatorname{Par} $, $\left(
u_{\lambda}\right)  _{\lambda\in L}=\left(  h_{\lambda}\right)  _{\lambda
\in\operatorname{Par} }$ and $\left(  v_{\lambda}\right)  _{\lambda\in
L}=\left(  m_{\lambda}\right)  _{\lambda\in\operatorname{Par} }$ (since the
bases $\left(  h_{\lambda}\right)  _{\lambda\in\operatorname{Par} }$ and
$\left(  m_{\lambda}\right)  _{\lambda\in\operatorname{Par} }$ of $\Lambda$
are dual to each other with respect to the Hall inner product $\left(
\cdot,\cdot\right)  $). As a result, we obtain that every $a\in\Lambda$
satisfies%
\begin{equation}
a=\sum_{\lambda\in\operatorname{Par} }\left(  h_{\lambda},a\right)
m_{\lambda}. \label{sol.exe.Lambda.primitives.dual-coords}%
\end{equation}


Now, let us solve the exercise. We need to show that for every $a\in\Lambda$,
the element $a$ of $\Lambda$ is primitive if and only if $a$ lies in the
$\kk$-linear span of $p_{1},p_{2},p_{3},...$. The ``if'' direction of
this statement is obvious\footnote{because
Proposition \ref{symm-comultiplication-formulas-prop}(i) shows that each of
$p_{1},p_{2},p_{3},...$ is primitive, and therefore every element of their
$\kk$-linear span is also primitive}. Hence, we only need to prove the
``only if'' statement. In other words, we
need to prove that if $a$ is primitive, then $a$ lies in the
$\kk$-linear span of $p_{1},p_{2},p_{3},...$. So let us assume that $a$ is
primitive. That is, $\Delta\left(  a\right)  =1\otimes a+a\otimes1$.

For every $x\in\Lambda$ and $y\in\Lambda$, we have (using the Sweedler
notation)%
\[
x^{\perp}\left(  a\right)  =\sum_{\left(  a\right)  }\left(  x,a_{1}\right)
a_{2}=\left(  x,1\right)  a+\left(  x,a\right)  1\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sum_{\left(  a\right)  }a_{1}\otimes a_{2}=\Delta\left(
a\right)  =1\otimes a+a\otimes1\right)
\]
and therefore
\begin{align}
\left(  xy,a\right)   &  =\left(  y,\underbrace{x^{\perp}\left(  a\right)
}_{=\left(  x,1\right)  a+\left(  x,a\right)  1}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{skewing-properties-prop}%
(i)}\right)  \nonumber\\
&  =\left(  y,\left(  x,1\right)  a+\left(  x,a\right)  1\right)  =\left(
x,1\right)  \left(  y,a\right)  +\left(  x,a\right)  \left(  y,1\right)
=\left(  x,1\right)  \left(  y,a\right)  +\left(  y,1\right)  \left(
x,a\right)  .\label{sol.exe.Lambda.primitives.xya}%
\end{align}
Using this, we can easily obtain $\left(  1,a\right)  =0$%
\ \ \ \ \footnote{Indeed, applying (\ref{sol.exe.Lambda.primitives.xya}) to
$x=1$ and $y=1$, we obtain $\left(  1,a\right)  =\underbrace{\left(
1,1\right)  }_{=1}\left(  1,a\right)  +\underbrace{\left(  1,1\right)  }%
_{=1}\left(  1,a\right)  =\left(  1,a\right)  +\left(  1,a\right)  =2\left(
1,a\right)  $, so that $\left(  1,a\right)  =0$, qed.}.

Now, using (\ref{sol.exe.Lambda.primitives.xya}), it is easy to see that%
\begin{equation}
\left(  h_{\lambda},a\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every partition
}\lambda\text{ satisfying }\ell\left(  \lambda\right)  \geq
2.\label{sol.exe.Lambda.primitives.0}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.Lambda.primitives.0}):} Let $\lambda$
be a partition satisfying $\ell\left(  \lambda\right)  \geq2$. Write the
partition $\lambda$ in the form $\lambda=\left(  \lambda_{1},\lambda
_{2},...,\lambda_{\ell}\right)  $ with $\ell=\ell\left(  \lambda\right)  $.
Then, $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{\ell}$ are positive
integers. Let $\nu$ be the partition $\left(  \lambda_{2},\lambda
_{3},...,\lambda_{\ell}\right)  $; then, $\nu$ is nonempty (since $\ell
=\ell\left(  \lambda\right)  \geq2$), so that $\left(  h_{\nu},1\right)  =0$
(by (\ref{sol.exe.Lambda.primitives.annihilate}), applied to $\nu$ instead of
$\lambda$). Also, $\left(  h_{\left(  \lambda_{1}\right)  },1\right)  =0$ (by
(\ref{sol.exe.Lambda.primitives.annihilate}), applied to $\left(  \lambda
_{1}\right)  $ instead of $\lambda$).
\par
By the definition of $h_{\lambda}$, we have $h_{\lambda}=h_{\lambda_{1}%
}h_{\lambda_{2}}...h_{\lambda_{\ell}}=\underbrace{h_{\lambda_{1}}%
}_{=h_{\left(  \lambda_{1}\right)  }}\underbrace{\left(  h_{\lambda_{2}%
}h_{\lambda_{3}}...h_{\lambda_{\ell}}\right)  }_{\substack{=h_{\nu
}\\\text{(since }\nu=\left(  \lambda_{2},\lambda_{3},...,\lambda_{\ell
}\right)  \text{)}}}=h_{\left(  \lambda_{1}\right)  }h_{\nu}$. Thus,
\begin{align*}
\left(  \underbrace{h_{\lambda}}_{=h_{\left(  \lambda_{1}\right)  }h_{\nu}%
},a\right)   &  =\left(  h_{\left(\lambda_{1}\right)}h_{\nu},a\right)
=\underbrace{\left(
h_{\left(  \lambda_{1}\right)  },1\right)  }_{=0}\left(  h_{\nu},a\right)
+\underbrace{\left(  h_{\nu},1\right)  }_{=0}\left(  h_{\left(  \lambda
_{1}\right)  },a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.exe.Lambda.primitives.xya}), applied to }x=h_{\left(  \lambda
_{1}\right)  }\text{ and }y=h_{\nu}\right)  \\
&  =0+0=0.
\end{align*}
This proves (\ref{sol.exe.Lambda.primitives.0}).} Now,
(\ref{sol.exe.Lambda.primitives.dual-coords}) becomes%
\begin{align*}
a &  =\sum_{\lambda\in\operatorname{Par}}\left(  h_{\lambda},a\right)
m_{\lambda}=\underbrace{\sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\\\ell\left(  \lambda\right)  =0}}\left(  h_{\lambda},a\right)  m_{\lambda}%
}_{=\left(  h_{\varnothing},a\right)  m_{\varnothing}}+\underbrace{\sum\limits
_{\substack{\lambda\in\operatorname{Par};\\\ell\left(  \lambda\right)
=1}}\left(  h_{\lambda},a\right)  m_{\lambda}}_{=\sum_{n\geq1}\left(
h_{\left(  n\right)  },a\right)  m_{\left(  n\right)  }}+\sum\limits
_{\substack{\lambda\in\operatorname{Par};\\\ell\left(  \lambda\right)  \geq
2}}\underbrace{\left(  h_{\lambda},a\right)  }_{\substack{=0\\\text{(by
(\ref{sol.exe.Lambda.primitives.0}))}}}m_{\lambda}\\
&  =\left(  \underbrace{h_{\varnothing}}_{=1},a\right)  m_{\varnothing}%
+\sum_{n\geq1}\left(  h_{\left(  n\right)  },a\right)  \underbrace{m_{\left(
n\right)  }}_{=p_{n}}+\underbrace{\sum\limits_{\substack{\lambda\in\operatorname{Par}%
;\\\ell\left(  \lambda\right)  \geq2}}0m_{\lambda}}_{=0}=\underbrace{\left(
1,a\right)  }_{=0}m_{\varnothing}+\sum_{n\geq1}\left(  h_{\left(  n\right)
},a\right)  p_{n}\\
&  =\sum_{n\geq1}\left(  h_{\left(  n\right)  },a\right)  p_{n}.
\end{align*}
Thus, $a$ lies in the $\kk$-linear span of $p_{1},p_{2},p_{3},...$.
This completes the solution to Exercise~\ref{exe.Lambda.primitives}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Lambda.primitives}}

\textit{Solution to Exercise \ref{exe.Lambda.primitives}.} Corollary
\ref{dual-bases-h-m-p}(a) yields that $\left(  h_{\lambda}\right)
_{\lambda\in \Par }$ and $\left(  m_{\lambda}\right)  _{\lambda
\in \Par }$ are dual bases with respect to the Hall inner product
on $\Lambda$. In other words,
\begin{equation}
\left(  h_{\lambda},m_{\mu}\right)  =\delta_{\lambda,\mu}%
\ \ \ \ \ \ \ \ \ \ \text{for any partitions }\lambda\text{ and }%
\mu .
\label{sol.exe.Lambda.primitives.dual}
\end{equation}


Recall the following fundamental fact from linear algebra: If $\kk$ is
a commutative ring, if $A$ is a $\kk$-module, if $\left(  \cdot
,\cdot\right)  :A\times A\rightarrow \kk$ is a symmetric $\kk %
$-bilinear form on $A$, and if $\left(  u_{\lambda}\right)  _{\lambda\in L}$
and $\left(  v_{\lambda}\right)  _{\lambda\in L}$ are two $\kk$-bases
of $A$ which are dual to each other with respect to the form $\left(
\cdot,\cdot\right)  $ (where $L$ is some indexing set), then every $a\in A$
satisfies
\begin{equation}
a=\sum_{\lambda\in L}\left(  u_{\lambda},a\right)  v_{\lambda}%
.\label{sol.exe.Lambda.primitives.dual-coords.general}%
\end{equation}
We can apply this fact to $A=\Lambda$, $L= \Par $, $\left(
u_{\lambda}\right)  _{\lambda\in L}=\left(  h_{\lambda}\right)  _{\lambda
\in \Par }$ and $\left(  v_{\lambda}\right)  _{\lambda\in
L}=\left(  m_{\lambda}\right)  _{\lambda\in \Par }$ (since the
bases $\left(  h_{\lambda}\right)  _{\lambda\in \Par }$ and
$\left(  m_{\lambda}\right)  _{\lambda\in \Par }$ of $\Lambda$ are
dual to each other with respect to the Hall inner product $\left(  \cdot
,\cdot\right)  $). As a result, we obtain that every $a\in\Lambda$ satisfies%
\begin{equation}
a=\sum_{\lambda\in \Par }\left(  h_{\lambda},a\right)
m_{\lambda}.\label{sol.exe.Lambda.primitives.dual-coords}%
\end{equation}


Now, let us fix some $a\in\Lambda$. We are going to prove the equivalence%
\begin{equation}
\left(  a\text{ is primitive in }\Lambda\right)  \Longleftrightarrow\left(
a\text{ lies in the } \kk \text{-linear span of }p_{1},p_{2}%
,p_{3},...\right)  .\label{sol.exe.Lambda.primitives.goal}%
\end{equation}


Indeed, let us first prove the implication%
\begin{equation}
\left(  a\text{ is primitive in }\Lambda\right)  \Longrightarrow\left(
a\text{ lies in the } \kk \text{-linear span of }p_{1},p_{2}%
,p_{3},...\right)  .\label{sol.exe.Lambda.primitives.goal1}%
\end{equation}


\textit{Proof of (\ref{sol.exe.Lambda.primitives.goal1}):} Assume that $a$ is
primitive in $\Lambda$. That is, $\Delta\left(  a\right)  =1\otimes
a+a\otimes1$.

It is now easy to see that
\begin{equation}
\left(  xy,a\right)  =\left(  x,1\right)  \left(  y,a\right)  +\left(
y,1\right)  \left(  x,a\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
x\in\Lambda\text{ and }y\in\Lambda.\label{sol.exe.Lambda.primitives.xy}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.Lambda.primitives.xy}):} Let
$x\in\Lambda$ and $y\in\Lambda$. Proposition \ref{skewing-properties-prop}(i)
(applied to $f=x$ and $g=y$) yields $\left(  y,x^{\perp}\left(  a\right)
\right)  =\left(  xy,a\right)  $, so that $\left(  xy,a\right)  =\left(
y,x^{\perp}\left(  a\right)  \right)  $. But by the definition of $x^{\perp
}\left(  a\right)  $, we have $x^{\perp}\left(  a\right)  =\sum_{\left(
a\right)  }\left(  x,a_{1}\right)  a_{2}$, where we are using the Sweedler
notation. Thus,%
\[
x^{\perp}\left(  a\right)  =\sum_{\left(  a\right)  }\left(  x,a_{1}\right)
a_{2}=\left(  x,1\right)  a+\left(  x,a\right)  1\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sum_{\left(  a\right)  }a_{1}\otimes a_{2}=\Delta\left(
a\right)  =1\otimes a+a\otimes1\right)  .
\]
Now,
\begin{align*}
\left(  xy,a\right)    & =\left(  y,\underbrace{x^{\perp}\left(  a\right)
}_{=\left(  x,1\right)  a+\left(  x,a\right)  1}\right)  =\left(  y,\left(
x,1\right)  a+\left(  x,a\right)  1\right)  =\left(  x,1\right)  \left(
y,a\right)  +\left(  x,a\right)  \left(  y,1\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is }%
 \kk \text{-bilinear}\right)  \\
& =\left(  x,1\right)  \left(  y,a\right)  +\left(  y,1\right)  \left(
x,a\right)  .
\end{align*}
This proves (\ref{sol.exe.Lambda.primitives.xy}).} As a consequence, $\left(
1,a\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Applying
(\ref{sol.exe.Lambda.primitives.dual}) to $\lambda=\varnothing$ and
$\mu=\varnothing$, we obtain $\left(  h_{\varnothing},m_{\varnothing}\right)
=\delta_{\varnothing,\varnothing}=1$. Since $h_{\varnothing}=1$ and
$m_{\varnothing}=1$, this rewrites as $\left(  1,1\right)  =1$.
\par
Applying (\ref{sol.exe.Lambda.primitives.xy}) to $x=1$ and $y=1$, we obtain
\[
\left(  1\cdot1,a\right)  =\underbrace{\left(  1,1\right)  }_{=1}\left(
1,a\right)  +\underbrace{\left(  1,1\right)  }_{=1}\left(  1,a\right)
=\left(  1,a\right)  +\left(  1,a\right)  .
\]
Since $1\cdot1=1$, this rewrites as $\left(  1,a\right)  =\left(  1,a\right)
+\left(  1,a\right)  $. Subtracting $\left(  1,a\right)  $ from this equality,
we obtain $0=\left(  1,a\right)  $, qed.}.

Now, let $U=\left\{  \left(  1\right)  ,\left(  2\right)  ,\left(  3\right)
,...\right\}  \subset \Par $. Then, it is easy to see that every
partition $\lambda$ satisfying $\lambda\notin U$ satisfies
\begin{equation}
\left(  h_{\lambda},a\right)  =0.
\label{sol.exe.Lambda.primitives.0}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.Lambda.primitives.0}):} Let $\lambda$
be a partition satisfying $\lambda\notin U$. Write the partition $\lambda$ in
the form $\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)  $
with $\ell=\ell\left(  \lambda\right)  $. Then, $\lambda_{1}$, $\lambda_{2}$,
$...$, $\lambda_{\ell}$ are positive integers. We have $h_{\lambda}
=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{\ell}}$ (by the definition of
$h_{\lambda}$).
\par
We need to prove (\ref{sol.exe.Lambda.primitives.0}). If $\ell=0$, then
(\ref{sol.exe.Lambda.primitives.0}) is true (because if $\ell=0$,
then
\[
\left(  \underbrace{h_{\lambda}}_{\substack{=h_{\lambda_{1}}h_{\lambda_{2}}
...h_{\lambda_{\ell}}\\=\left(  \text{empty product}\right)  \\\text{(since
}\ell=0\text{)}}},a\right)  =\left(  \underbrace{\left(  \text{empty
product}\right)  }_{=1},a\right)  =\left(  1,a\right)  =0
\]
). Hence, for the rest of the proof of (\ref{sol.exe.Lambda.primitives.0}), we
can WLOG assume that we don't have $\ell=0$. Assume this. We have $\ell\geq1$
(since we don't have $\ell=0$). Hence, the positive integer $\lambda_{1}$ and
the partition $\left(  \lambda_{2},\lambda_{3},...,\lambda_{\ell}\right)  $
are well-defined. Denote the partition $\left(  \lambda_{2},\lambda
_{3},...,\lambda_{\ell}\right)  $ by $\nu$.
\par
Since $\lambda_{1}$ is a positive integer, we have $\lambda_{1}\in\left\{
1,2,3,...\right\}  $ and $h_{\lambda_{1}}=h_{\left(  \lambda_{1}\right)  }$
and $\left(  \lambda_{1}\right)  \neq\varnothing$.
\par
Since $\nu=\left(  \lambda_{2},\lambda_{3},...,\lambda_{\ell}\right)  $, we
have $h_{\nu}=h_{\lambda_{2}}h_{\lambda_{3}}...h_{\lambda_{\ell}}$, so that
$h_{\lambda_{2}}h_{\lambda_{3}}...h_{\lambda_{\ell}}=h_{\nu}$. Now,%
\[
h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{\ell}}%
=\underbrace{h_{\lambda_{1}}}_{=h_{\left(  \lambda_{1}\right)  }%
}\underbrace{\left(  h_{\lambda_{2}}h_{\lambda_{3}}...h_{\lambda_{\ell}%
}\right)  }_{=h_{\nu}}=h_{\left(  \lambda_{1}\right)  }h_{\nu}.
\]
\par
If $\ell=1$, then%
\begin{align*}
\lambda & =\left(  \lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)  =\left(
\lambda_{1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell=1\right)
\\
& \subset\left\{  \left(  1\right)  ,\left(  2\right)  ,\left(  3\right)
,...\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda_{1}\in\left\{
1,2,3,...\right\}  \right)  \\
& =U,
\end{align*}
which contradicts $\lambda\notin U$. Hence, we cannot have $\ell=1$. Thus,
$\ell\geq2$ (since $\ell\geq1$), so that $\ell-1\geq1>0$. Hence, the partition
$\left(  \lambda_{2},\lambda_{3},...,\lambda_{\ell}\right)  $ is nonempty
(because this partition has $\ell-1$ parts, and $\ell-1>0$). In other words,
$\left(  \lambda_{2},\lambda_{3},...,\lambda_{\ell}\right)  \neq\varnothing$.
Hence, $\nu=\left(  \lambda_{2},\lambda_{3},...,\lambda_{\ell}\right)
\neq\varnothing$.
\par
Now,
\begin{align*}
\left(  \underbrace{h_{\lambda}}_{=h_{\left(  \lambda_{1}\right)  }h_{\nu}%
},a\right)    & =\left(  h_{\left(\lambda_{1}\right)}h_{\nu},a\right)
=\left(  h_{\left(
\lambda_{1}\right)  },\underbrace{1}_{=m_{\varnothing}}\right)  \left(
h_{\nu},a\right)  +\left(  h_{\nu},\underbrace{1}_{=m_{\varnothing}}\right)
\left(  h_{\left(  \lambda_{1}\right)  },a\right)\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.exe.Lambda.primitives.xy}), applied to }x=h_{\left(
\lambda_{1}\right)  }\text{ and }y=h_{\nu}\right)  \\
& =\underbrace{\left(  h_{\left(  \lambda_{1}\right)  },m_{\varnothing
}\right)  }_{\substack{=\delta_{\left(  \lambda_{1}\right)  ,\varnothing
}\\\text{(by (\ref{sol.exe.Lambda.primitives.dual}), applied to}\\\left(
\lambda_{1}\right)  \text{ and }\varnothing\text{ instead of }\lambda\text{
and }\mu\text{)}}}\left(  h_{\nu},a\right)  +\underbrace{\left(  h_{\nu
},m_{\varnothing}\right)  }_{\substack{=\delta_{\nu,\varnothing}\\\text{(by
(\ref{sol.exe.Lambda.primitives.dual}), applied to}\\\nu\text{ and
}\varnothing\text{ instead of }\lambda\text{ and }\mu\text{)}}}\left(
h_{\left(  \lambda_{1}\right)  },a\right)  \\
& =\underbrace{\delta_{\left(  \lambda_{1}\right)  ,\varnothing}%
}_{\substack{=0\\\text{(since }\left(  \lambda_{1}\right)  \neq\varnothing
\text{)}}}\left(  h_{\nu},a\right)  +\underbrace{\delta_{\nu,\varnothing}%
}_{\substack{=0\\\text{(since }\nu\neq\varnothing\text{)}}}\left(  h_{\left(
\lambda_{1}\right)  },a\right)  =0+0=0.
\end{align*}
This proves (\ref{sol.exe.Lambda.primitives.0}).} On the other hand, every
partition $\lambda$ satisfying $\lambda\in U$ satisfies%
\begin{equation}
m_{\lambda}\in\left\{  p_{1},p_{2},p_{3},...\right\}
\label{sol.exe.Lambda.primitives.U}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.exe.Lambda.primitives.U}):} Let $\lambda$
be a partition satisfying $\lambda\in U$. Then, $\lambda\in U=\left\{  \left(
1\right)  ,\left(  2\right)  ,\left(  3\right)  ,...\right\}  $. Hence, there
exists a positive integer $n$ such that $\lambda=\left(  n\right)  $. Consider
this $n$. Clearly, $\lambda=\left(  n\right)  $, so that $m_{\lambda
}=m_{\left(  n\right)  }=p_{n}\in\left\{  p_{1},p_{2},p_{3},...\right\}  $.
This proves (\ref{sol.exe.Lambda.primitives.U}).}.

Now, (\ref{sol.exe.Lambda.primitives.dual-coords}) becomes
\begin{align*}
a  & =\sum_{\lambda\in \Par }\left(  h_{\lambda},a\right)
m_{\lambda}=\sum\limits_{\substack{\lambda\in \Par ;\\\lambda\in
U}}\left(  h_{\lambda},a\right)  m_{\lambda}+\sum\limits_{\substack{\lambda
\in \Par ;\\\lambda\notin U}}\underbrace{\left(  h_{\lambda
},a\right)  }_{
\substack{=0\\\text{(by (\ref{sol.exe.Lambda.primitives.0}))}}}
m_{\lambda}=\sum\limits_{\substack{\lambda\in \Par ;\\\lambda\in
U}}\left(  h_{\lambda},a\right)  m_{\lambda}+\underbrace{\sum\limits
_{\substack{\lambda\in \Par ;\\\lambda\notin U}}
0m_{\lambda}}_{=0}\\
& =\sum\limits_{\substack{\lambda\in \Par ;\\\lambda\in U}}\left(
h_{\lambda},a\right)  m_{\lambda}.
\end{align*}
Thus, $a$ is a $\kk$-linear combination of the elements $m_{\lambda}$
for partitions $\lambda$ satisfying $\lambda\in U$. Since these elements
$m_{\lambda}$ all are elements of the set $\left\{  p_{1},p_{2},p_{3}%
,...\right\}  $ (by (\ref{sol.exe.Lambda.primitives.U})), this yields that $a$
is a $\kk$-linear combination of elements of the set $\left\{
p_{1},p_{2},p_{3},...\right\}  $. In other words, $a$ lies in the
$\kk$-linear span of $p_{1},p_{2},p_{3},...$. Thus, the
implication (\ref{sol.exe.Lambda.primitives.goal1}) is proven.

Next, we will prove the implication
\begin{equation}
\left(  a\text{ lies in the } \kk \text{-linear span of }p_{1}%
,p_{2},p_{3},...\right)  \Longrightarrow\left(  a\text{ is primitive in
}\Lambda\right)  .
\label{sol.exe.Lambda.primitives.goal2}
\end{equation}


\textit{Proof of (\ref{sol.exe.Lambda.primitives.goal2}):} Assume that $a$
lies in the $\kk$-linear span of $p_{1},p_{2},p_{3},...$. Thus, $a$ is
a $\kk$-linear combination of the elements  $p_{1},p_{2},p_{3},...$. In
other words, there exists a sequence $\left(  a_{1},a_{2},a_{3},...\right)  $
of elements of $\kk$ such that $\left(  \text{all but finitely many
positive integers }n\text{ satisfy }a_{n}=0\right)  $ and $a=\sum
_{n=1}^{\infty}a_{n}p_{n}$. Consider this $\left(  a_{1},a_{2},a_{3}%
,...\right)  $. We have $\sum_{n=1}^{\infty}a_{n}p_{n}=a$ (since $a=\sum
_{n=1}^{\infty}a_{n}p_{n}$).

Applying the map $\Delta$ to both sides of the equality $a=\sum_{n=1}^{\infty
}a_{n}p_{n}$, we obtain
\begin{align*}
\Delta\left(  a\right)    & =\Delta\left(  \sum_{n=1}^{\infty}a_{n}%
p_{n}\right)  =\sum_{n=1}^{\infty}a_{n}\underbrace{\Delta\left(  p_{n}\right)
}_{\substack{=1\otimes p_{n}+p_{n}\otimes1\\\text{(by Proposition
\ref{symm-comultiplication-formulas-prop}(i))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\Delta\text{ is } \kk \text{-linear}\right)  \\
& =\sum_{n=1}^{\infty}a_{n}\left(  1\otimes p_{n}+p_{n}\otimes1\right)
=\underbrace{\sum_{n=1}^{\infty}a_{n}1\otimes p_{n}}_{=1\otimes\left(
\sum_{n=1}^{\infty}a_{n}p_{n}\right)  }+\underbrace{\sum_{n=1}^{\infty}%
a_{n}p_{n}\otimes1}_{=\left(  \sum_{n=1}^{\infty}a_{n}p_{n}\right)  \otimes
1}\\
& =1\otimes\underbrace{\left(  \sum_{n=1}^{\infty}a_{n}p_{n}\right)  }%
_{=a}+\underbrace{\left(  \sum_{n=1}^{\infty}a_{n}p_{n}\right)  }_{=a}%
\otimes1=1\otimes a+a\otimes1.
\end{align*}
In other words, the element $a$ is primitive in $\Lambda$. This proves the
implication (\ref{sol.exe.Lambda.primitives.goal2}).

We have now proven both implications (\ref{sol.exe.Lambda.primitives.goal1})
and (\ref{sol.exe.Lambda.primitives.goal2}). Combining these implications, we
obtain the equivalence (\ref{sol.exe.Lambda.primitives.goal}).

Now, let us forget that we fixed $a$. We thus have proven the equivalence
(\ref{sol.exe.Lambda.primitives.goal}) for every $a\in\Lambda$. This
equivalence shows that the primitive elements of $\Lambda$ are exactly the
elements of $\Lambda$ which lie in the $\kk$-linear span of
$p_{1},p_{2},p_{3},...$. In other words, the primitive elements of $\Lambda$
are precisely the elements of the $\kk$-linear span of $p_{1}%
,p_{2},p_{3},\ldots$. This solves Exercise~\ref{exe.Lambda.primitives}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.induction.characters}}

\textit{Solution to Exercise~\ref{exe.induction.characters}.}

(a) This is straightforward: Let $g_1$ and $g_2$ be two elements of
$G$ belonging to the same conjugacy class. Thus, $g_1$ and $g_2$ are
conjugate. In other words, there exists some $x \in G$ such that
$g_1 = x g_2 x^{-1}$. Consider this $x$.

By the definition of $\Ind_H^G f$, we have
\begin{align*}
\left(\Ind_H^G f\right) \left(g_1\right)
&= \frac{1}{\left|H\right|} \sum\limits_{\substack{k \in G:\\ k g_1 k^{-1} \in H}}
  f \left(k g_1 k^{-1}\right)
= \frac{1}{\left|H\right|} \sum\limits_{\substack{k \in G:\\ k x g_2 x^{-1} k^{-1} \in H}}
  f \left(k x g_2 x^{-1} k^{-1}\right)
\qquad \left(\text{since } g_1 = x g_2 x^{-1}\right) \\
&= \frac{1}{\left|H\right|} \sum\limits_{\substack{k \in G:\\ k x g_2 \left(k x\right)^{-1} \in H}}
  f \left(k x g_2 \left(k x\right)^{-1}\right)
\qquad \left(\text{since } x^{-1} k^{-1} = \left(k x\right)^{-1} \right) \\
&= \frac{1}{\left|H\right|} \sum\limits_{\substack{k \in G:\\ k g_2 k^{-1} \in H}}
  f \left(k g_2 k^{-1}\right)
\end{align*}
(here, we have substituted $k$ for $k x$ in the sum, because the map
$G \to G, \  k \to k x$ is a bijection). Compared with
\[
\left(\Ind_H^G f\right) \left(g_2\right)
= \frac{1}{\left|H\right|} \sum\limits_{\substack{k \in G:\\ k g_2 k^{-1} \in H}}
  f \left(k g_2 k^{-1}\right)
\]
(this follows from the definition of $\Ind_H^G f$), this yields
$\left(\Ind_H^G f\right) \left(g_1\right) = \left(\Ind_H^G f\right) \left(g_2\right)$.

Now, forget that we fixed $g_1$ and $g_2$. We thus have proven that
if $g_1$ and $g_2$ are two elements of $G$ belonging to the same
conjugacy class, then
$\left(\Ind_H^G f\right) \left(g_1\right) = \left(\Ind_H^G f\right) \left(g_2\right)$.
In other words, the map $\Ind_H^G f$ is constant on conjugacy classes.
Hence, $\Ind_H^G f$ is a class function on $G$. In other words,
$\Ind_H^G f \in R_\CC \left(G\right)$. This solves part (a) of the
exercise.

(b) We have $G = \bigsqcup_{j\in J} Hj$. Thus, every $k \in G$ can be
uniquely written in the form $hj$ for some $j \in J$ and $h \in H$. Hence,
\begin{align}
\sum\limits_{\substack{k\in G:\\kgk^{-1}\in H}} f\left(  kgk^{-1}\right)    &
=\underbrace{\sum\limits_{\substack{j\in J;\ h\in H:\\\left(  hj\right)  g\left(
hj\right)  ^{-1}\in H}}}_{\substack{=\sum\limits_{\substack{j\in J;\ h\in
H:\\hjgj^{-1}h^{-1}\in H}}=\sum\limits_{\substack{j\in J;\ h\in H:\\jgj^{-1}\in
H}}\\\text{(because under the condition that }h\in H\text{,}\\\text{the
relation }hjgj^{-1}h^{-1}\in H\text{ is equivalent}\\\text{to the relation
}jgj^{-1}\in H\text{)}}} f\left(  \underbrace{\left(  hj\right)  g\left(
hj\right)  ^{-1}}_{=hjgj^{-1}h^{-1}}\right)  \nonumber\\
& =\underbrace{\sum\limits_{\substack{j\in J;\ h\in H:\\jgj^{-1}\in H}}}
_{=\sum\limits_{\substack{j\in J:\\jgj^{-1}\in H}}\sum_{h\in H}}\underbrace{f
\left(  hjgj^{-1}h^{-1}\right)  }_{\substack{=f\left(  jgj^{-1}\right)
\\\text{(since }f\text{ is a class function on }H\text{,}\\\text{and
}hjgj^{-1}h^{-1}\text{ is }H\text{-conjugate to }jgj^{-1}\text{)}}}
\nonumber\\
& =\sum\limits_{\substack{j\in J:\\jgj^{-1}\in H}}\underbrace{\sum_{h\in H}
f\left(  jgj^{-1}\right)  }_{=\left\vert H\right\vert f\left(
jgj^{-1}\right)  }
= \left\vert H\right\vert \sum\limits_{\substack{j\in J:\\jgj^{-1} \in H}}
f\left(  jgj^{-1}\right)
.\label{sol.induction.characters.1}
\end{align}

Now, the definition of $\Ind^G_H f$ yields
\[
\left(  \Ind^G_H f\right)  \left(  g\right)    
=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{k\in
G:\\kgk^{-1}\in H}} f\left(  kgk^{-1}\right)  }_{\substack{=\left\vert
H\right\vert \sum\limits_{\substack{j\in J:\\jgj^{-1}\in H}} f\left(
jgj^{-1}\right)}}
 =\sum\limits_{\substack{j\in J:\\jgj^{-1}\in H}} f\left(
jgj^{-1}\right).
\]
This solves part (b) of the exercise.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.induction.transitivity}}

\textit{Solution to Exercise~\ref{exe.induction.transitivity}.}
We have $\CC G \otimes_{\CC H} \CC H \cong \CC G$ as
$\left(\CC G, \CC H\right)$-bimodules, and thus also as
$\left(\CC G, \CC I\right)$-bimodules\footnote{since the right
$\CC I$-module structures on $\CC G \otimes_{\CC H} \CC H$ and on
$\CC G$ are obtained from the respective right $\CC H$-module
structures by restriction}.

By the definition of induction, we have
$\Ind^H_I U = \CC H \otimes_{\CC I} U$. But by the definition of
induction, we also have
\begin{align*}
\Ind^G_H \Ind^H_I U
&= \CC G \otimes_{\CC H}
\underbrace{\Ind^H_I U}_{= \CC H \otimes_{\CC I} U}
= \CC G \otimes_{\CC H} \left(\CC H \otimes_{\CC I} U\right) \\
&\cong \underbrace{\left(\CC G \otimes_{\CC H} \CC H\right)}
  _{\substack{\cong \CC G \\
   \text{(as } \left(\CC G, \CC I\right) \text{-bimodules)}}}
\otimes_{\CC I} U
\qquad \qquad \left(\text{by the associativity of the tensor product}\right) \\
&\cong \CC G \otimes_{\CC I} U = \Ind^G_I U
\end{align*}
(since $\Ind^G_I U = \CC G \otimes_{\CC I} U$ (by the definition
of induction)).
This solves Exercise~\ref{exe.induction.transitivity}.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.ind.tensor}}

\textit{Solution to Exercise~\ref{exe.ind.tensor}.} The solution rests
upon the following general fact:

\begin{proposition}
\label{prop.sol.ind.tensor.general}Let $\kk$ be a commutative ring. Let
$A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be six $\kk %
$-algebras. Let $P$ be an $\left(  A,B\right)  $-bimodule\footnote{As usual,
we understand the notion of a bimodule to be defined over $\kk$; that
is, the left $A$-module structure and the right $B$-module structure of an
$\left(  A,B\right)  $-bimodule must restrict to one and the same $\kk %
$-module structure.}. Let $Q$ be a $\left(  B,C\right)  $-bimodule. Let
$P^{\prime}$ be an $\left(  A^{\prime},B^{\prime}\right)  $-bimodule. Let
$Q^{\prime}$ be a $\left(  B^{\prime},C^{\prime}\right)  $-bimodule. Then,%
\[
\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
Q\otimes Q^{\prime}\right)  \cong\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
\]
as $\left(  A\otimes A^{\prime},C\otimes C^{\prime}\right)  $-bimodules. Here,
all $\otimes$ signs without subscript stand for $\otimes_{ \kk }$.
\end{proposition}

This proposition is proven by straightforward (repeated) application of the
universal property of the tensor product. (Of course, the $\left(  A\otimes
A^{\prime},C\otimes C^{\prime}\right)  $-bimodule isomorphism $\left(
P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes
Q^{\prime}\right)  \rightarrow\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  $ sends every $\left(
p\otimes p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q\otimes
q^{\prime}\right)  $ to $\left(  p\otimes_{B}q\right)  \otimes\left(
p^{\prime}\otimes_{B}q^{\prime}\right)  $.) We leave all details to the reader.

Now, let us come to the solution of
Exercise~\ref{exe.ind.tensor}.\footnote{The following solution
involves some handwaving: We are going to use certain isomorphisms
to identify
$\CC \left[ G_1 \times G_2 \right]$ with $\CC G_1 \otimes \CC G_2$
and to identify
$\CC \left[ H_1 \times H_2 \right]$ with $\CC H_1 \otimes \CC H_2$.
See the solution of Exercise~\ref{exe.Indrho.tensor} for an example
of how to avoid this kind of handwaving. (Actually,
Exercise~\ref{exe.Indrho}(b) shows that
Exercise~\ref{exe.Indrho.tensor} is a generalization of
Exercise~\ref{exe.ind.tensor}.)}
We want to prove the isomorphism
\[
\operatorname{Ind}_{H_{1}\times H_{2}}^{G_{1}\times G_{2}}\left(  U_{1}\otimes
U_{2}\right)  \cong\left(  \operatorname{Ind}_{H_{1}}^{G_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{H_{2}}^{G_{2}}U_{2}\right)
\]
of $\CC\left[  G_{1}\times G_{2}\right]  $-modules. Recalling the
definition of $\operatorname*{Ind}$, we rewrite this as%
\begin{equation}
\CC\left[  G_{1}\times G_{2}\right]  \otimes_{\CC \left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \cong\left(
\CC G_{1}\otimes_{\CC H_{1}}U_{1}\right)  \otimes\left(
\CC G_{2}\otimes_{\CC H_{2}}U_{2}\right)  .
\label{sol.ind.tensor.want}%
\end{equation}
But Proposition \ref{prop.sol.ind.tensor.general} (applied to $\kk %
=\CC$, $A=\CC G_{1}$, $B=\CC H_{1}$, $C=\CC$, $A^{\prime}=\CC G_{2}$,
$B^{\prime}=\CC H_{2}$, $C^{\prime}=\CC$,
$P=\CC G_{1}$, $Q=U_{1}$, $P^{\prime}=\CC G_{2}$ and
$Q^{\prime}=U_{2}$) yields%
\[
\left(  \CC G_{1}\otimes \CC G_{2}\right)  \otimes
_{\CC H_{1}\otimes \CC H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  \CC G_{1}\otimes_{\CC H_{1}}U_{1}\right)
\otimes\left(  \CC G_{2}\otimes_{\CC H_{2}}U_{2}\right)
\]
as $\left(  \CC G_{1}\otimes \CC G_{2}, \CC
\otimes \CC \right)  $-bimodules (thus, as left
$\CC G_{1}\otimes \CC G_{2}$-modules). In order to derive
(\ref{sol.ind.tensor.want}) from this isomorphism, we need to realize that:

\begin{itemize}
\item there exist algebra isomorphisms $\CC \left[  G_{1}\times
G_{2}\right]  \rightarrow \CC G_{1}\otimes \CC G_{2}$ and
$ \CC \left[  H_{1}\times H_{2}\right]  \rightarrow \CC %
H_{1}\otimes \CC H_{2}$ which commute with the canonical inclusion maps
$ \CC \left[  H_{1}\times H_{2}\right]  \rightarrow \CC \left[
G_{1}\times G_{2}\right]  $ and $ \CC H_{1}\otimes \CC %
H_{2}\rightarrow \CC G_{1}\otimes \CC G_{2}$;

\item when we identify $ \CC \left[  H_{1}\times H_{2}\right]  $ with
$ \CC H_{1}\otimes \CC H_{2}$ along the isomorphism
$ \CC \left[  H_{1}\times H_{2}\right]  \rightarrow \CC %
H_{1}\otimes \CC H_{2}$, the $ \CC H_{1}\otimes \CC %
H_{2}$-module $U_{1}\otimes U_{2}$ becomes exactly the $\CC \left[
H_{1}\times H_{2}\right]  $-module $U_{1}\otimes U_{2}$;

\item when we identify $ \CC \left[  G_{1}\times G_{2}\right]  $ with
$ \CC G_{1}\otimes \CC G_{2}$ along the isomorphism
$ \CC \left[  G_{1}\times G_{2}\right]  \rightarrow \CC %
G_{1}\otimes \CC G_{2}$, the $ \CC G_{1}\otimes \CC %
G_{2}$-module $\left(   \CC G_{1}\otimes_{\CC H_{1}}%
U_{1}\right)  \otimes\left(   \CC G_{2}\otimes_{\CC H_{2}}%
U_{2}\right)  $ becomes exactly the $\CC \left[  G_{1}\times
G_{2}\right]  $-module $\left(   \CC G_{1}\otimes_{\CC H_{1}%
}U_{1}\right)  \otimes\left(   \CC G_{2}\otimes_{\CC H_{2}%
}U_{2}\right)  $.
\end{itemize}

\noindent
These facts are all trivial to verify (of course, the isomorphism
$ \CC \left[  G_{1}\times G_{2}\right]  \rightarrow \CC %
G_{1}\otimes \CC G_{2}$ is given by sending every $t_{\left(
g_{1},g_{2}\right)  }$ to $t_{g_{1}}\otimes t_{g_{2}}$, and similarly for the
other isomorphism). Thus, (\ref{sol.ind.tensor.want}) holds, and the exercise
is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.ind.tensor}}

\textit{Solution to Exercise~\ref{exe.ind.tensor}.} We shall prove the
following fact:

\begin{proposition}
\label{prop.sol.ind.tensor.general}Let $\kk$ be a commutative ring. Let
$A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be six $\kk %
$-algebras. Let $P$ be an $\left(  A,B\right)  $-bimodule\footnote{As usual,
we understand the notion of a bimodule to be defined over $\kk$; that
is, the left $A$-module structure and the right $B$-module structure of an
$\left(  A,B\right)  $-bimodule must restrict to one and the same $\kk %
$-module structure.}. Let $Q$ be a $\left(  B,C\right)  $-bimodule. Let
$P^{\prime}$ be an $\left(  A^{\prime},B^{\prime}\right)  $-bimodule. Let
$Q^{\prime}$ be a $\left(  B^{\prime},C^{\prime}\right)  $-bimodule. Then,%
\[
\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
Q\otimes Q^{\prime}\right)  \cong\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
\]
as $\left(  A\otimes A^{\prime},C\otimes C^{\prime}\right)  $-bimodules. Here,
all $\otimes$ signs without subscript stand for $\otimes_{ \kk }$.
\end{proposition}

In order to prove this, we shall first show another lemma:

\begin{lemma}
\label{lem.sol.ind.tensor.XYZW}Let $\kk$ be a commutative ring. Let $M$
be a $\kk$-module. Let $U$ and $V$ be two $\kk$-algebras. Let
$X$ be a right $U$-module. Let $Y$ be a left $U$-module. Let $Z$ be a right
$V$-module. Let $W$ be a left $V$-module. Let $f:X\times Y\times Z\times
W\rightarrow M$ be a $\kk$-multilinear map. Assume that%
\begin{equation}
\left(  f\left(  xu,y,z,w\right)  =f\left(  x,uy,z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\text{, }u\in U\text{, }y\in
Y\text{, }z\in Z\text{ and }w\in W\right)
\label{eq.lem.sol.ind.tensor.XYZW.ass1}%
\end{equation}
and%
\begin{equation}
\left(  f\left(  x,y,zv,w\right)  =f\left(  x,y,z,vw\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\text{, }y\in Y\text{, }z\in
Z\text{, }v\in V\text{ and }w\in W\right)  .
\label{eq.lem.sol.ind.tensor.XYZW.ass2}%
\end{equation}


Then, there exists a $\kk$-linear map $F:\left(  X\otimes_{U}Y\right)
\otimes\left(  Z\otimes_{V}W\right)  \rightarrow M$ which satisfies%
\[
\left(  F\left(  \left(  x\otimes_{U}y\right)  \otimes\left(  z\otimes
_{V}w\right)  \right)  =f\left(  x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }\left(  x,y,z,w\right)  \in X\times Y\times Z\times W\right)  .
\]
(Here, just as in Proposition \ref{prop.sol.ind.tensor.general}, all $\otimes$
signs without subscript stand for $\otimes_{ \kk }$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.ind.tensor.XYZW}.]For every $\left(  z,w\right)
\in Z\times W$, let us define a map $f_{z,w}:X\times Y\rightarrow M$ by%
\[
\left(  f_{z,w}\left(  x,y\right)  =f\left(  x,y,z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)  \in X\times Y\right)
.
\]
Fix $\left(  z,w\right)  \in Z\times W$. The map $f_{z,w}$ is $\kk %
$-bilinear\footnote{\textit{Proof.} Let $\lambda_{1}\in \kk$,
$\lambda_{2}\in \kk$, $x_{1}\in X$, $x_{2}\in X$ and $y\in Y$. We are
going to prove that $f_{z,w}\left(  \lambda_{1}x_{1}+\lambda_{2}%
x_{2},y\right)  =\lambda_{1}f_{z,w}\left(  x_{1},y\right)  +\lambda_{2}%
f_{z,w}\left(  x_{2},y\right)  $.
\par
In fact, the definition of $f_{z,w}\left(  x_{1},y\right)  $ yields
$f_{z,w}\left(  x_{1},y\right)  =f\left(  x_{1},y,z,w\right)  $. Hence,
$f\left(  x_{1},y,z,w\right)  =f_{z,w}\left(  x_{1},y\right)  $.
\par
Also, the definition of $f_{z,w}\left(  x_{2},y\right)  $ yields
$f_{z,w}\left(  x_{2},y\right)  =f\left(  x_{2},y,z,w\right)  $. Hence,
$f\left(  x_{2},y,z,w\right)  =f_{z,w}\left(  x_{2},y\right)  $.
\par
The definition of $f_{z,w}\left(  \lambda_{1}x_{1}+\lambda_{2}x_{2},y\right)
$ yields%
\begin{align*}
f_{z,w}\left(  \lambda_{1}x_{1}+\lambda_{2}x_{2},y\right)   &  =f\left(
\lambda_{1}x_{1}+\lambda_{2}x_{2},y,z,w\right)  =\lambda_{1}%
\underbrace{f\left(  x_{1},y,z,w\right)  }_{=f_{z,w}\left(  x_{1},y\right)
}+\lambda_{2}\underbrace{f\left(  x_{2},y,z,w\right)  }_{=f_{z,w}\left(
x_{2},y\right)  }\\
&  =\lambda_{1}f_{z,w}\left(  x_{1},y\right)  +\lambda_{2}f_{z,w}\left(
x_{2},y\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}\in \kk$, $\lambda_{2}%
\in \kk$, $x_{1}\in X$, $x_{2}\in X$ and $y\in Y$. We thus have proven
that $f_{z,w}\left(  \lambda_{1}x_{1}+\lambda_{2}x_{2},y\right)  =\lambda
_{1}f_{z,w}\left(  x_{1},y\right)  +\lambda_{2}f_{z,w}\left(  x_{2},y\right)
$ for every $\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $x_{1}\in
X$, $x_{2}\in X$ and $y\in Y$. In other words, the map $f_{z,w}$ is
$\kk$-linear in its first coordinate. Similarly, it can be shown that
the map $f_{z,w}$ is $\kk$-linear in its second coordinate. Now, the
map $f_{z,w}$ is $\kk$-bilinear (since it is $\kk$-linear in its
first coordinate and $\kk$-linear in its second coordinate), qed.}.
Moreover, the map $f_{z,w}$ is $U$-bilinear with respect to the right
$U$-module structure on $X$ and the left $U$-module structure on
$Y$\ \ \ \ \footnote{\textit{Proof.} Every $x\in X$, $u\in U$ and $y\in Y$
satisfy%
\begin{align*}
f_{z,w}\left(  x,uy\right)   &  =f\left(  xu,y,z,w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f_{z,w}\left(
x,uy\right)  \right) \\
&  =f\left(  x,uy,z,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.lem.sol.ind.tensor.XYZW.ass1})}\right) \\
&  =f_{z,w}\left(  xu,y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}f_{z,w}\left(  xu,y\right)  =f\left(  x,uy,z,w\right)  \text{ (by the
definition of }f_{z,w}\left(  xu,y\right)  \text{)}\right)  .
\end{align*}
Combined with the fact that the map $f_{z,w}$ is $\kk$-bilinear, this
yields that the map $f_{z,w}$ is $U$-bilinear with respect to the right
$U$-module structure on $X$ and the left $U$-module structure on $Y$, qed.}.
Due to the universal property of the tensor product, this yields that there
exists a unique $\kk$-linear map $g_{z,w}:X\otimes_{U}Y\rightarrow M$
such that%
\begin{equation}
\left(  g_{z,w}\left(  x\otimes_{U}y\right)  =f_{z,w}\left(  x,y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)  \in X\times Y\right)  .
\label{pf.lem.sol.ind.tensor.XYZW.gzw}%
\end{equation}
Consider this map $g_{z,w}$. This map $g_{z,w}$ is $\kk$-linear, and
thus belongs to $\Hom_{ \kk }\left(  X\otimes
_{U}Y,M\right)  $.

Now, let us forget that we fixed $\left(  z,w\right)  \in Z\times W$. We thus
have defined an element $g_{z,w}$ of $\Hom_{ \kk %
}\left(  X\otimes_{U}Y,M\right)  $ for every $\left(  z,w\right)  \in Z\times
W$. This element $g_{z,w}$ satisfies%
\begin{align}
g_{z,w}\left(  x\otimes_{U}y\right)   &  =f_{z,w}\left(  x,y\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.sol.ind.tensor.XYZW.gzw}%
)}\right) \nonumber\\
&  =f\left(  x,y,z,w\right)  . \label{pf.lem.sol.ind.tensor.XYZW.gzw.real}%
\end{align}


Now, we define a map $g:Z\times W\rightarrow\Hom_{ \kk }
\left(  X\otimes_{U}Y,M\right)  $ by%
\[
\left(  g\left(  z,w\right)  =g_{z,w}\ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  z,w\right)  \in Z\times W\right)  .
\]
The map $g$ is $\kk$-bilinear\footnote{\textit{Proof.} Let $\lambda
_{1}\in \kk$, $\lambda_{2}\in \kk$, $z_{1}\in Z$, $z_{2}\in Z$ and
$w\in W$ be arbitrary. We shall show that $g\left(  \lambda_{1}z_{1}%
+\lambda_{2}z_{2},w\right)  =\lambda_{1}g\left(  z_{1},w\right)  +\lambda
_{2}g\left(  z_{2},w\right)  $.
\par
Indeed, let $\mathfrak{L}=g\left(  \lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)
$ and $\mathfrak{R}=\lambda_{1}g\left(  z_{1},w\right)  +\lambda_{2}g\left(
z_{2},w\right)  $. Notice that $\mathfrak{L}$ and $\mathfrak{R}$ are elements
of $\Hom_{ \kk }\left(  X\otimes_{U}Y,M\right)  $
(since $g\left(  \lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)  $, $g\left(
z_{1},w\right)  $ and $g\left(  z_{2},w\right)  $ are elements of
$\Hom_{ \kk }\left(  X\otimes_{U}Y,M\right)  $),
therefore $\kk$-linear maps.
\par
Let us first fix $x\in X$ and $y\in Y$. The definition of $g\left(
z_{1},w\right)  $ yields $g\left(  z_{1},w\right)  =g_{z_{1},w}$. Applying
both sides of this equality to $x\otimes_{U}y$, we obtain%
\[
\left(  g\left(  z_{1},w\right)  \right)  \left(  x\otimes_{U}y\right)
=g_{z_{1},w}\left(  x\otimes_{U}y\right)  =f\left(  x,y,z_{1},w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.ind.tensor.XYZW.gzw.real}), applied to }z=z_{1}\right)  ,
\]
and thus $f\left(  x,y,z_{1},w\right)  =\left(  g\left(  z_{1},w\right)
\right)  \left(  x\otimes_{U}y\right)  $. The same argument, with $z_{2}$ used
instead of $z_{1}$, shows that $f\left(  x,y,z_{2},w\right)  =\left(  g\left(
z_{2},w\right)  \right)  \left(  x\otimes_{U}y\right)  $.
\par
We have $\mathfrak{L}=g\left(  \lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)
=g_{\lambda_{1}z_{1}+\lambda_{2}z_{2},w}$ (by the definition of $g\left(
\lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)  $). Applying both sides of this
equality to $x\otimes_{U}y$, we obtain%
\begin{align*}
\mathfrak{L}\left(  x\otimes_{U}y\right)   &  =g_{\lambda_{1}z_{1}+\lambda
_{2}z_{2},w}\left(  x\otimes_{U}y\right) \\
&  =f\left(  x,y,\lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.ind.tensor.XYZW.gzw.real}), applied to }z=\lambda_{1}%
z_{1}+\lambda_{2}z_{2}\right) \\
&  =\lambda_{1}\underbrace{f\left(  x,y,z_{1},w\right)  }_{=\left(  g\left(
z_{1},w\right)  \right)  \left(  x\otimes_{U}y\right)  }+\lambda
_{2}\underbrace{f\left(  x,y,z_{2},w\right)  }_{=\left(  g\left(
z_{2},w\right)  \right)  \left(  x\otimes_{U}y\right)  }
\ \ \ \ \ \ \ \ \ \  \left(\text{since the map } f \text{ is } \kk\text{-multilinear}\right) \\
&  =\lambda_{1}\left(  g\left(  z_{1},w\right)  \right)  \left(  x\otimes
_{U}y\right)  +\lambda_{2}\left(  g\left(  z_{2},w\right)  \right)  \left(
x\otimes_{U}y\right) \\
&  =\underbrace{\left(  \lambda_{1}g\left(  z_{1},w\right)  +\lambda
_{2}g\left(  z_{2},w\right)  \right)  }_{\substack{=\mathfrak{R}\\\text{(since
}\mathfrak{R}=\lambda_{1}g\left(  z_{1},w\right)  +\lambda_{2}g\left(
z_{2},w\right)  \text{)}}}\left(  x\otimes_{U}y\right)  =\mathfrak{R}\left(
x\otimes_{U}y\right)  .
\end{align*}
\par
Now, forget that we fixed $x$ and $y$. We thus have proven that $\mathfrak{L}%
\left(  x\otimes_{U}y\right)  =\mathfrak{R}\left(  x\otimes_{U}y\right)  $ for
all $x\in X$ and $y\in Y$. In other words, the two maps $\mathfrak{L}$ and
$\mathfrak{R}$ are equal to each other on every pure tensor in $X\otimes_{U}%
Y$. Since these two maps $\mathfrak{L}$ and $\mathfrak{R}$ are $\kk %
$-linear, this yields that the two maps $\mathfrak{L}$ and $\mathfrak{R}$ must
be identical (because if two $\kk$-linear maps from a tensor product
are equal to each other on every pure tensor, then these two maps must be
identical). In other words, $\mathfrak{L}=\mathfrak{R}$. Since $\mathfrak{L}%
=g\left(  \lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)  $ and $\mathfrak{R}%
=\lambda_{1}g\left(  z_{1},w\right)  +\lambda_{2}g\left(  z_{2},w\right)  $,
this rewrites as $g\left(  \lambda_{1}z_{1}+\lambda_{2}z_{2},w\right)
=\lambda_{1}g\left(  z_{1},w\right)  +\lambda_{2}g\left(  z_{2},w\right)  $.
\par
Now, let us forget that we fixed $\lambda_{1}$, $\lambda_{2}$, $z_{1}$,
$z_{2}$ and $w$. We thus have proven that $g\left(  \lambda_{1}z_{1}%
+\lambda_{2}z_{2},w\right)  =\lambda_{1}g\left(  z_{1},w\right)  +\lambda
_{2}g\left(  z_{2},w\right)  $ for every $\lambda_{1}\in \kk$,
$\lambda_{2}\in \kk$, $z_{1}\in Z$, $z_{2}\in Z$ and $w\in W$. In other
words, the map $g$ is $\kk$-linear in its first argument. Similarly, it
can be shown that the map $g$ is $\kk$-linear in its second argument.
Now, the map $g$ is $\kk$-bilinear (since it is $\kk$-linear in
its first argument and $\kk$-linear in its second argument), qed.}.
Moreover, the map $g$ is $V$-bilinear with respect to the right $V$-module
structure on $Z$ and the left $V$-module structure on $W$%
\ \ \ \ \footnote{\textit{Proof.} Let $z\in Z$, $v\in V$ and $w\in W$ be
arbitrary. We shall show that $g\left(  zv,w\right)  =g\left(  z,vw\right)  $.
\par
Indeed, let $\mathfrak{L}=g\left(  zv,w\right)  $ and $\mathfrak{R}=g\left(
z,vw\right)  $. Notice that $\mathfrak{L}$ and $\mathfrak{R}$ are elements of
$\Hom_{ \kk }\left(  X\otimes_{U}Y,M\right)  $
(since $g\left(  zv,w\right)  $ and $g\left(  z,vw\right)  $ are elements of
$\Hom_{ \kk }\left(  X\otimes_{U}Y,M\right)  $),
therefore $\kk$-linear maps.
\par
Let us first fix $x\in X$ and $y\in Y$. We have $\mathfrak{L}=g\left(
zv,w\right)  =g_{zv,w}$ (by the definition of $g\left(  zv,w\right)  $).
Applying both sides of this equality to $x\otimes_{U}y$, we obtain%
\begin{align*}
\mathfrak{L}\left(  x\otimes_{U}y\right)   &  =g_{zv,w}\left(  x\otimes
_{U}y\right)  =f\left(  x,y,zv,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.ind.tensor.XYZW.gzw.real}), applied to }zv\text{ instead of
}z\right) \\
&  =f\left(  x,y,z,vw\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.lem.sol.ind.tensor.XYZW.ass2})}\right)  ,
\end{align*}
and thus $f\left(  x,y,z,vw\right)  =\mathfrak{L}\left(  x\otimes_{U}y\right)
$.
\par
We have $\mathfrak{R}=g\left(  z,vw\right)  =g_{z,vw}$ (by the definition of
$g\left(  z,vw\right)  $). Applying both sides of this equality to
$x\otimes_{U}y$, we obtain%
\begin{align*}
\mathfrak{R}\left(  x\otimes_{U}y\right)   &  =g_{z,vw}\left(  x\otimes
_{U}y\right) \\
&  =f\left(  x,y,z,vw\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.ind.tensor.XYZW.gzw.real}), applied to }vw\text{ instead of
}w\right) \\
&  =\mathfrak{L}\left(  x\otimes_{U}y\right)  .
\end{align*}
Hence, $\mathfrak{L}\left(  x\otimes_{U}y\right)  =\mathfrak{R}\left(
x\otimes_{U}y\right)  $.
\par
Now, forget that we fixed $x$ and $y$. We thus have proven that $\mathfrak{L}%
\left(  x\otimes_{U}y\right)  =\mathfrak{R}\left(  x\otimes_{U}y\right)  $ for
all $x\in X$ and $y\in Y$. In other words, the two maps $\mathfrak{L}$ and
$\mathfrak{R}$ are equal to each other on every pure tensor in $X\otimes_{U}%
Y$. Since these two maps $\mathfrak{L}$ and $\mathfrak{R}$ are $\kk %
$-linear, this yields that the two maps $\mathfrak{L}$ and $\mathfrak{R}$ must
be identical (because if two $\kk$-linear maps from a tensor product
are equal to each other on every pure tensor, then these two maps must be
identical). In other words, $\mathfrak{L}=\mathfrak{R}$. Since $\mathfrak{L}%
=g\left(  zv,w\right)  $ and $\mathfrak{R}=g\left(  z,vw\right)  $, this
rewrites as $g\left(  zv,w\right)  =g\left(  z,vw\right)  $.
\par
Now, let us forget that we fixed $z$, $v$ and $w$. We thus have proven that
$g\left(  zv,w\right)  =g\left(  z,vw\right)  $ for every $z\in Z$, $v\in V$
and $w\in W$. Combined with the fact that the map $g$ is $\kk %
$-bilinear, this yields that the map $g$ is $V$-bilinear with respect to the
right $V$-module structure on $Z$ and the left $V$-module structure on $W$,
qed.}. Due to the universal property of the tensor product, this yields that
there exists a unique $\kk$-linear map $h:Z\otimes_{V}W\rightarrow
\Hom_{ \kk }\left(  X\otimes_{U}Y,M\right)  $
such that%
\begin{equation}
\left(  h\left(  z\otimes_{V}w\right)  =g\left(  z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  z,w\right)  \in Z\times W\right)  .
\label{pf.lem.sol.ind.tensor.XYZW.h}%
\end{equation}
Consider this map $h$.

Now, let us define a map $p:\left(  X\otimes_{U}Y\right)  \times\left(
Z\otimes_{V}W\right)  \rightarrow M$ by%
\[
\left(  p\left(  \alpha,\beta\right)  =\left(  h\left(  \beta\right)  \right)
\left(  \alpha\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  \alpha
,\beta\right)  \in\left(  X\otimes_{U}Y\right)  \times\left(  Z\otimes
_{V}W\right)  \right)  .
\]
Then, it is easy to see that the map $p$ is $\kk$-linear in its first
argument\footnote{\textit{Proof.} Let $\lambda_{1}\in \kk$, $\lambda
_{2}\in \kk$, $\alpha_{1}\in X\otimes_{U}Y$, $\alpha_{2}\in X\otimes
_{U}Y$ and $\beta\in Z\otimes_{V}W$. Then, the map $h\left(  \beta\right)  $
is $\kk$-linear (since $h\left(  \beta\right)  \in
\Hom_{ \kk }\left(  X\otimes_{U}Y,M\right)  $).
\par
The definition of $p\left(  \alpha_{1},\beta\right)  $ yields $p\left(
\alpha_{1},\beta\right)  =\left(  h\left(  \beta\right)  \right)  \left(
\alpha_{1}\right)  $, so that $\left(  h\left(  \beta\right)  \right)  \left(
\alpha_{1}\right)  =p\left(  \alpha_{1},\beta\right)  $. The definition of
$p\left(  \alpha_{2},\beta\right)  $ yields $p\left(  \alpha_{2},\beta\right)
=\left(  h\left(  \beta\right)  \right)  \left(  \alpha_{2}\right)  $, so that
$\left(  h\left(  \beta\right)  \right)  \left(  \alpha_{2}\right)  =p\left(
\alpha_{2},\beta\right)  $. But the definition of $p\left(  \lambda_{1}%
\alpha_{1}+\lambda_{2}\alpha_{2},\beta\right)  $ yields%
\begin{align*}
p\left(  \lambda_{1}\alpha_{1}+\lambda_{2}\alpha_{2},\beta\right)   &
=\left(  h\left(  \beta\right)  \right)  \left(  \lambda_{1}\alpha_{1}%
+\lambda_{2}\alpha_{2}\right)  =\lambda_{1}\underbrace{\left(  h\left(
\beta\right)  \right)  \left(  \alpha_{1}\right)  }_{=p\left(  \alpha
_{1},\beta\right)  }+\lambda_{2}\underbrace{\left(  h\left(  \beta\right)
\right)  \left(  \alpha_{2}\right)  }_{=p\left(  \alpha_{2},\beta\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }h\left(  \beta\right)
\text{ is } \kk \text{-linear}\right) \\
&  =\lambda_{1}p\left(  \alpha_{1},\beta\right)  +\lambda_{2}p\left(
\alpha_{2},\beta\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}$, $\lambda_{2}$, $\alpha_{1}$,
$\alpha_{2}$ and $\beta$. We thus have proven that $p\left(  \lambda_{1}%
\alpha_{1}+\lambda_{2}\alpha_{2},\beta\right)  =\lambda_{1}p\left(  \alpha
_{1},\beta\right)  +\lambda_{2}p\left(  \alpha_{2},\beta\right)  $ for every
$\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $\alpha_{1}\in
X\otimes_{U}Y$, $\alpha_{2}\in X\otimes_{U}Y$ and $\beta\in Z\otimes_{V}W$. In
other words, the map $p$ is $\kk$-linear in its first argument, qed.}
and is $\kk$-linear in its second argument\footnote{\textit{Proof.} Let
$\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $\alpha\in
X\otimes_{U}Y$, $\beta_{1}\in Z\otimes_{V}W$ and $\beta_{2}\in Z\otimes_{V}W$.
\par
The definition of $p\left(  \alpha,\beta_{1}\right)  $ yields $p\left(
\alpha,\beta_{1}\right)  =\left(  h\left(  \beta_{1}\right)  \right)  \left(
\alpha\right)  $, so that $\left(  h\left(  \beta_{1}\right)  \right)  \left(
\alpha\right)  =p\left(  \alpha,\beta_{1}\right)  $. The definition of
$p\left(  \alpha,\beta_{2}\right)  $ yields $p\left(  \alpha,\beta_{2}\right)
=\left(  h\left(  \beta_{2}\right)  \right)  \left(  \alpha\right)  $, so that
$\left(  h\left(  \beta_{2}\right)  \right)  \left(  \alpha\right)  =p\left(
\alpha,\beta_{2}\right)  $. But the definition of $p\left(  \alpha,\lambda
_{1}\beta_{1}+\lambda_{2}\beta_{2}\right)  $ yields%
\begin{align*}
p\left(  \alpha,\lambda_{1}\beta_{1}+\lambda_{2}\beta_{2}\right)   &  =\left(
\underbrace{h\left(  \lambda_{1}\beta_{1}+\lambda_{2}\beta_{2}\right)
}_{\substack{=\lambda_{1}h\left(  \beta_{1}\right)  +\lambda_{2}h\left(
\beta_{2}\right)  \\\text{(since the map }h\text{ is } \kk %
\text{-linear)}}}\right)  \left(  \alpha\right)  =\left(  \lambda_{1}h\left(
\beta_{1}\right)  +\lambda_{2}h\left(  \beta_{2}\right)  \right)  \left(
\alpha\right) \\
&  =\lambda_{1}\underbrace{\left(  h\left(  \beta_{1}\right)  \right)  \left(
\alpha\right)  }_{=p\left(  \alpha,\beta_{1}\right)  }+\lambda_{2}%
\underbrace{\left(  h\left(  \beta_{2}\right)  \right)  \left(  \alpha\right)
}_{=p\left(  \alpha,\beta_{2}\right)  }=\lambda_{1}p\left(  \alpha,\beta
_{1}\right)  +\lambda_{2}p\left(  \alpha,\beta_{2}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}$, $\lambda_{2}$, $\alpha$,
$\beta_{1}$ and $\beta_{2}$. We thus have proven that $p\left(  \alpha
,\lambda_{1}\beta_{1}+\lambda_{2}\beta_{2}\right)  =\lambda_{1}p\left(
\alpha,\beta_{1}\right)  +\lambda_{2}p\left(  \alpha,\beta_{2}\right)  $ for
every $\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $\alpha\in
X\otimes_{U}Y$, $\beta_{1}\in Z\otimes_{V}W$ and $\beta_{2}\in Z\otimes_{V}W$.
In other words, the map $p$ is $\kk$-linear in its second argument,
qed.}. So the map $p$ is $\kk$-bilinear\footnote{because it is
$\kk$-linear in its first argument and is $\kk$-linear in its
second argument}. Due to the universal property of the tensor product, this
yields that there exists a unique $\kk$-linear map $P:\left(
X\otimes_{U}Y\right)  \otimes\left(  Z\otimes_{V}W\right)  \rightarrow M$ such
that%
\begin{equation}
\left(  P\left(  \alpha\otimes\beta\right)  =p\left(  \alpha,\beta\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  \alpha,\beta\right)  \in\left(
X\otimes_{U}Y\right)  \times\left(  Z\otimes_{V}W\right)  \right)  .
\label{pf.lem.sol.ind.tensor.XYZW.P}
\end{equation}
Consider this map $P$. For all $\left(  x,y,z,w\right)  \in X\times Y\times
Z\times W$, we have%
\begin{align*}
P\left(  \left(  x\otimes_{U}y\right)  \otimes\left(  z\otimes_{V}w\right)
\right)   &  =p\left(  x\otimes_{U}y,z\otimes_{V}w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.sol.ind.tensor.XYZW.P}),
applied to }\alpha=x\otimes_{U}y\text{ and }\beta=z\otimes_{V}w\right) \\
&  =\underbrace{\left(  h\left(  z\otimes_{V}w\right)  \right)  }%
_{\substack{=g\left(  z,w\right)  \\\text{(by
(\ref{pf.lem.sol.ind.tensor.XYZW.h}))}}}\left(  x\otimes_{U}y\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }p\left(  x\otimes
_{U}y,z\otimes_{V}w\right)  \right) \\
&  =\underbrace{\left(  g\left(  z,w\right)  \right)  }_{\substack{=g_{z,w}%
\\\text{(by the definition of }g\left(  z,w\right)  \text{)}}}\left(
x\otimes_{U}y\right)  =g_{z,w}\left(  x\otimes_{U}y\right)  =f\left(
x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.ind.tensor.XYZW.gzw.real})}\right)  .
\end{align*}
Thus, there exists a $\kk$-linear map $F:\left(  X\otimes_{U}Y\right)
\otimes\left(  Z\otimes_{V}W\right)  \rightarrow M$ which satisfies%
\[
\left(  F\left(  \left(  x\otimes_{U}y\right)  \otimes\left(  z\otimes
_{V}w\right)  \right)  =f\left(  x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }\left(  x,y,z,w\right)  \in X\times Y\times Z\times W\right)
\]
(namely, $F=P$). This proves Lemma \ref{lem.sol.ind.tensor.XYZW}.
\end{proof}

We state a simple corollary of Lemma \ref{lem.sol.ind.tensor.XYZW}:

\begin{lemma}
\label{lem.sol.ind.tensor.simple}Let $\kk$ be a commutative ring. Let
$M$ be a $\kk$-module. Let $X$, $Y$, $Z$ and $W$ be four $\kk %
$-modules. Let $g:X\times Y\times Z\times W\rightarrow M$ be a $\kk %
$-multilinear map. Then, there exists a $\kk$-linear map $\mathfrak{G}%
:\left(  X\otimes Y\right)  \otimes\left(  Z\otimes W\right)  \rightarrow M$
which satisfies%
\[
\left(  \mathfrak{G}\left(  \left(  x\otimes y\right)  \otimes\left(  z\otimes
w\right)  \right)  =g\left(  x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  x,y,z,w\right)  \in X\times Y\times Z\times W\right)  .
\]
(Here, just as in Proposition \ref{prop.sol.ind.tensor.general}, all $\otimes$
signs without subscript stand for $\otimes_{ \kk }$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.ind.tensor.simple}.]We know that $X$ and $Z$ are
$\kk$-modules, and thus right $\kk$-modules. Also, $Y$ and $W$
are $\kk$-modules, thus left $\kk$-modules. Now, the map $g$ is
$\kk$-multilinear; hence, the map $g$ is $\kk$-linear in each of
its four arguments. In other words, the map $g$ is $\kk$-linear in its
first argument, is $\kk$-linear in its second argument, is $\kk %
$-linear in its third argument, and is $\kk$-linear in its fourth argument.

Now,%
\[
\left(  g\left(  xu,y,z,w\right)  =g\left(  x,uy,z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\text{, }u\in \kk \text{,
}y\in Y\text{, }z\in Z\text{ and }w\in W\right)
\]
\footnote{\textit{Proof.} Let $x\in X$, $u\in \kk$, $y\in Y$, $z\in Z$
and $w\in W$. Then, $g\left(  \underbrace{xu}_{=ux},y,z,w\right)  =g\left(
ux,y,z,w\right)  =ug\left(  x,y,z,w\right)  $ (since the map $g$ is
$\kk$-linear in its first argument) and $g\left(  x,uy,z,w\right)
=ug\left(  x,y,z,w\right)  $ (since the map $g$ is $\kk$-linear in its
second argument). Hence, $g\left(  xu,y,z,w\right)  =ug\left(  x,y,z,w\right)
=g\left(  x,uy,z,w\right)  $, qed.} and
\[
\left(  g\left(  x,y,zv,w\right)  =g\left(  x,y,z,vw\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\text{, }y\in Y\text{, }z\in
Z\text{, }v\in \kk \text{ and }w\in W\right)  .
\]
\footnote{\textit{Proof.} Let $x\in X$, $y\in Y$, $z\in Z$, $v\in \kk$
and $w\in W$. Then, $g\left(  x,y,\underbrace{zv}_{=vz},w\right)  =g\left(
x,y,vz,w\right)  =vg\left(  x,y,z,w\right)  $ (since the map $g$ is
$\kk$-linear in its third argument) and $g\left(  x,y,z,vw\right)
=vg\left(  x,y,z,w\right)  $ (since the map $g$ is $\kk$-linear in its
fourth argument). Hence, $g\left(  x,y,zv,w\right)  =vg\left(  x,y,z,w\right)
=g\left(  x,y,z,vw\right)  $, qed.}. Thus, Lemma \ref{lem.sol.ind.tensor.XYZW}
(applied to $U= \kk$, $V= \kk$ and $f=g$) yields that there exists
a $\kk$-linear map $F:\left(  X\otimes_{ \kk }Y\right)
\otimes\left(  Z\otimes_{ \kk }W\right)  \rightarrow M$ which satisfies%
\[
\left(  F\left(  \left(  x\otimes_{ \kk }y\right)  \otimes\left(
z\otimes_{ \kk }w\right)  \right)  =g\left(  x,y,z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y,z,w\right)  \in X\times Y\times
Z\times W\right)  .
\]
Since the $\otimes_{ \kk }$ sign means the same as the $\otimes$ sign
without subscript, we can rewrite this as follows: There exists a $\kk %
$-linear map $F:\left(  X\otimes Y\right)  \otimes\left(  Z\otimes W\right)
\rightarrow M$ which satisfies%
\[
\left(  F\left(  \left(  x\otimes y\right)  \otimes\left(  z\otimes w\right)
\right)  =g\left(  x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(
x,y,z,w\right)  \in X\times Y\times Z\times W\right)  .
\]
Renaming the index $F$ as $\mathfrak{G}$ in this result, we obtain the
following: There exists a $\kk$-linear map $\mathfrak{G}:\left(
X\otimes Y\right)  \otimes\left(  Z\otimes W\right)  \rightarrow M$ which
satisfies%
\[
\left(  \mathfrak{G}\left(  \left(  x\otimes y\right)  \otimes\left(  z\otimes
w\right)  \right)  =g\left(  x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  x,y,z,w\right)  \in X\times Y\times Z\times W\right)  .
\]
This proves Lemma \ref{lem.sol.ind.tensor.simple}.
\end{proof}

Here is another lemma that we will use:

\begin{lemma}
\label{lem.sol.ind.tensor.kU}Let $\kk$ be a commutative ring. Let $M$
be a $\kk$-module. Let $U$ be a $\kk$-algebra. Let $X$ be a
right $U$-module. Let $Y$ be a left $U$-module. Let $\mathfrak{G}:X\otimes
Y\rightarrow M$ be a $\kk$-linear map. Assume that%
\begin{equation}
\mathfrak{G}\left(  xu\otimes y\right)  =\mathfrak{G}\left(  x\otimes
uy\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }x\in X\text{, }u\in U\text{ and
}y\in Y. \label{eq.lem.sol.ind.tensor.kU.ass1}%
\end{equation}
Then, there exists a $\kk$-linear map $G:X\otimes_{U}Y\rightarrow M$
such that%
\[
\left(  G\left(  x\otimes_{U}y\right)  =\mathfrak{G}\left(  x\otimes y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)  \in X\times Y\right)
.
\]
(Here, just as in Proposition \ref{prop.sol.ind.tensor.general}, all $\otimes$
signs without subscript stand for $\otimes_{ \kk }$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.ind.tensor.kU}.]Let us define a map $g:X\times
Y\rightarrow M$ by%
\[
\left(  g\left(  x,y\right)  =\mathfrak{G}\left(  x\otimes y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)  \in X\times Y\right)
.
\]
This map $g$ is $\kk$-bilinear\footnote{\textit{Proof.} Let
$\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $x_{1}\in X$,
$x_{2}\in X$ and $y\in Y$ be arbitrary. Then, the definition of $g\left(
x_{1},y\right)  $ yields $g\left(  x_{1},y\right)  =\mathfrak{G}\left(
x_{1}\otimes y\right)  $. Thus, $\mathfrak{G}\left(  x_{1}\otimes y\right)
=g\left(  x_{1},y\right)  $. Also, the definition of $g\left(  x_{2},y\right)
$ yields $g\left(  x_{2},y\right)  =\mathfrak{G}\left(  x_{2}\otimes y\right)
$. Thus, $\mathfrak{G}\left(  x_{2}\otimes y\right)  =g\left(  x_{2},y\right)
$. But the definition of $g\left(  \lambda_{1}x_{1}+\lambda_{2}x_{2},y\right)
$ yields%
\begin{align*}
g\left(  \lambda_{1}x_{1}+\lambda_{2}x_{2},y\right)   &  =\mathfrak{G}\left(
\underbrace{\left(  \lambda_{1}x_{1}+\lambda_{2}x_{2}\right)  \otimes
y}_{=\lambda_{1}x_{1}\otimes y+\lambda_{2}x_{2}\otimes y}\right)
=\mathfrak{G}\left(  \lambda_{1}x_{1}\otimes y+\lambda_{2}x_{2}\otimes
y\right) \\
&  =\lambda_{1}\underbrace{\mathfrak{G}\left(  x_{1}\otimes y\right)
}_{=g\left(  x_{1},y\right)  }+\lambda_{2}\underbrace{\mathfrak{G}\left(
x_{2}\otimes y\right)  }_{=g\left(  x_{2},y\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathfrak{g}\text{ is
} \kk \text{-linear}\right) \\
&  =\lambda_{1}g\left(  x_{1},y\right)  +\lambda_{2}g\left(  x_{2},y\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}$, $\lambda_{2}$, $x_{1}$,
$x_{2}$ and $y$. We thus have proven that $g\left(  \lambda_{1}x_{1}%
+\lambda_{2}x_{2},y\right)  =\lambda_{1}g\left(  x_{1},y\right)  +\lambda
_{2}g\left(  x_{2},y\right)  $ for any $\lambda_{1}\in \kk$,
$\lambda_{2}\in \kk$, $x_{1}\in X$, $x_{2}\in X$ and $y\in Y$. In other
words, the map $g$ is $\kk$-linear in its first argument. Similarly,
one can see that the map $g$ is $\kk$-linear in its second argument.
\par
Now, the map $g$ is $\kk$-bilinear (since it is $\kk$-linear in
its first argument and $\kk$-linear in its second argument), qed.}.
Moreover, the map $g$ is $U$-bilinear with respect to the right $U$-module
structure on $X$ and the left $U$-module structure on $Y$%
\ \ \ \ \footnote{\textit{Proof.} Every $x\in X$, $u\in U$ and $y\in Y$
satisfy%
\begin{align*}
g\left(  xu,y\right)   &  =\mathfrak{G}\left(  xu\otimes y\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }g\left(  xu,y\right)
\right) \\
&  =\mathfrak{G}\left(  x\otimes uy\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.lem.sol.ind.tensor.kU.ass1})}\right) \\
&  =g\left(  x,uy\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }g\left(
x,uy\right)  =\mathfrak{G}\left(  x\otimes uy\right)  \text{ (by the
definition of }g\left(  x,uy\right)  \text{)}\right)  .
\end{align*}
Since the map $g$ is $\kk$-bilinear, this yields that the map $g$ is
$U$-bilinear with respect to the right $U$-module structure on $X$ and the
left $U$-module structure on $Y$, qed.}. Due to the universal property of the
tensor product, this yields that there exists a unique $\kk$-linear map
$h:X\otimes_{U}Y\rightarrow M$ such that%
\begin{equation}
\left(  h\left(  x\otimes_{U}y\right)  =g\left(  x,y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)  \in X\times Y\right)  .
\label{pf.lem.sol.ind.tensor.kU.h}%
\end{equation}
Consider this map $h$. We have%
\[
h\left(  x\otimes_{U}y\right)  =g\left(  x,y\right)  =\mathfrak{G}\left(
x\otimes y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}g\left(  x,y\right)  \right)
\]
for all $\left(  x,y\right)  \in X\times Y$. Hence, there exists a
$\kk$-linear map $G:X\otimes_{U}Y\rightarrow M$ such that%
\[
\left(  G\left(  x\otimes_{U}y\right)  =\mathfrak{G}\left(  x\otimes y\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)  \in X\times Y\right)
\]
(namely, $G=h$). This proves Lemma \ref{lem.sol.ind.tensor.kU}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.sol.ind.tensor.general}.]Let us define a map
$f:P\times Q\times P^{\prime}\times Q^{\prime}\rightarrow\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  $ by%
\[
\left(  f\left(  p,q,p^{\prime},q^{\prime}\right)  =\left(  p\otimes
p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q\otimes q^{\prime
}\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  p,q,p^{\prime},q^{\prime
}\right)  \in P\times Q\times P^{\prime}\times Q^{\prime}\right)  .
\]
This map $f$ is $\kk$-multilinear\footnote{\textit{Proof.} Let
$\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $p_{1}\in P$,
$p_{2}\in P$, $q\in Q$, $p^{\prime}\in P^{\prime}$ and $q^{\prime}\in
Q^{\prime}$.
\par
By the definition of $f\left(  p_{1},q,p^{\prime},q^{\prime}\right)  $, we
have $f\left(  p_{1},q,p^{\prime},q^{\prime}\right)  =\left(  p_{1}\otimes
p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q\otimes q^{\prime
}\right)  $. Thus, $\left(  p_{1}\otimes p^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  q\otimes q^{\prime}\right)  =f\left(  p_{1},q,p^{\prime
},q^{\prime}\right)  $.
\par
By the definition of $f\left(  p_{2},q,p^{\prime},q^{\prime}\right)  $, we
have $f\left(  p_{2},q,p^{\prime},q^{\prime}\right)  =\left(  p_{2}\otimes
p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q\otimes q^{\prime
}\right)  $. Thus, $\left(  p_{2}\otimes p^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  q\otimes q^{\prime}\right)  =f\left(  p_{2},q,p^{\prime
},q^{\prime}\right)  $.
\par
By the definition of $f\left(  \lambda_{1}p_{1}+\lambda_{2}p_{2},q,p^{\prime
},q^{\prime}\right)  $, we have%
\begin{align*}
&  f\left(  \lambda_{1}p_{1}+\lambda_{2}p_{2},q,p^{\prime},q^{\prime}\right)
\\
&  =\underbrace{\left(  \left(  \lambda_{1}p_{1}+\lambda_{2}p_{2}\right)
\otimes p^{\prime}\right)  }_{=\lambda_{1}p_{1}\otimes p^{\prime}+\lambda
_{2}p_{2}\otimes p^{\prime}}\otimes_{B\otimes B^{\prime}}\left(  q\otimes
q^{\prime}\right) \\
&  =\left(  \lambda_{1}p_{1}\otimes p^{\prime}+\lambda_{2}p_{2}\otimes
p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q\otimes q^{\prime
}\right) \\
&  =\lambda_{1}\underbrace{\left(  p_{1}\otimes p^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  q\otimes q^{\prime}\right)  }_{=f\left(
p_{1},q,p^{\prime},q^{\prime}\right)  }+\lambda_{2}\underbrace{\left(
p_{2}\otimes p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q\otimes
q^{\prime}\right)  }_{=f\left(  p_{2},q,p^{\prime},q^{\prime}\right)  }\\
&  =\lambda_{1}f\left(  p_{1},q,p^{\prime},q^{\prime}\right)  +\lambda
_{2}f\left(  p_{2},q,p^{\prime},q^{\prime}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}$, $\lambda_{2}$, $p_{1}$,
$p_{2}$, $q$, $p^{\prime}$ and $q^{\prime}$. We thus have proven that
$f\left(  \lambda_{1}p_{1}+\lambda_{2}p_{2},q,p^{\prime},q^{\prime}\right)
=\lambda_{1}f\left(  p_{1},q,p^{\prime},q^{\prime}\right)  +\lambda
_{2}f\left(  p_{2},q,p^{\prime},q^{\prime}\right)  $ for every $\lambda_{1}%
\in \kk$, $\lambda_{2}\in \kk$, $p_{1}\in P$, $p_{2}\in P$, $q\in
Q$, $p^{\prime}\in P^{\prime}$ and $q^{\prime}\in Q^{\prime}$. In other words,
the map $f$ is $\kk$-linear in its first argument. Similarly, the map
$f$ is $\kk$-linear in its second argument. Similarly, the map $f$ is
$\kk$-linear in its fourth argument. Also, the map $f$ is $\kk %
$-linear in its third argument (this is proven in the same way as we have
showed that the map $f$ is $\kk$-linear in its first argument).
\par
We now know that the map $f$ is $\kk$-linear in its first argument, is
$\kk$-linear in its second argument, is $\kk$-linear in its
third argument, and $\kk$-linear in its fourth argument. In other
words, the map $f$ is $\kk$-linear in each of its four arguments. In
other words, the map $f$ is $\kk$-multilinear, qed.}. Moreover, we have%
\[
\left(  f\left(  xu,y,z,w\right)  =f\left(  x,uy,z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in P\text{, }u\in B\text{, }y\in
Q\text{, }z\in P^{\prime}\text{ and }w\in Q^{\prime}\right)
\]
\footnote{\textit{Proof.} Let $x\in P$, $u\in B$, $y\in Q$, $z\in P^{\prime}$
and $w\in Q^{\prime}$. Then, the definition of $f\left(  xu,y,z,w\right)  $
yields
\begin{align*}
f\left(  xu,y,z,w\right)   &  =\left(  xu\otimes\underbrace{z}_{=z1_{B^{\prime
}}}\right)  \otimes_{B\otimes B^{\prime}}\left(  y\otimes w\right)
=\underbrace{\left(  xu\otimes z1_{B^{\prime}}\right)  }_{=\left(  x\otimes
z\right)  \left(  u\otimes1_{B^{\prime}}\right)  }\otimes_{B\otimes B^{\prime
}}\left(  y\otimes w\right) \\
&  =\left(  x\otimes z\right)  \left(  u\otimes1_{B^{\prime}}\right)
\otimes_{B\otimes B^{\prime}}\left(  y\otimes w\right)  =\left(  x\otimes
z\right)  \otimes_{B\otimes B^{\prime}}\underbrace{\left(  u\otimes
1_{B^{\prime}}\right)  \left(  y\otimes w\right)  }_{=uy\otimes1_{B^{\prime}%
}w}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }u\otimes1_{B^{\prime}}\text{ can
be moved past the }\otimes_{B\otimes B^{\prime}}\text{ sign (because }%
u\otimes1_{B^{\prime}}\in B\otimes B^{\prime}\text{)}\right) \\
&  =\left(  x\otimes z\right)  \otimes_{B\otimes B^{\prime}}\left(
uy\otimes\underbrace{1_{B^{\prime}}w}_{=w}\right)  =\left(  x\otimes z\right)
\otimes_{B\otimes B^{\prime}}\left(  uy\otimes w\right)  .
\end{align*}
Compared with%
\[
f\left(  x,uy,z,w\right)  =\left(  x\otimes z\right)  \otimes_{B\otimes
B^{\prime}}\left(  uy\otimes w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }f\left(  x,uy,z,w\right)  \right)  ,
\]
this yields $f\left(  xu,y,z,w\right)  =f\left(  x,uy,z,w\right)  $, qed.} and%
\[
\left(  f\left(  x,y,zv,w\right)  =f\left(  x,y,z,vw\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in P\text{, }y\in Q\text{, }z\in
P^{\prime}\text{, }v\in B^{\prime}\text{ and }w\in Q^{\prime}\right)  .
\]
\footnote{\textit{Proof.} Let $x\in P$, $y\in Q$, $z\in P^{\prime}$, $v\in
B^{\prime}$ and $w\in Q^{\prime}$. Then, the definition of $f\left(
x,y,zv,w\right)  $ yields
\begin{align*}
f\left(  x,y,zv,w\right)   &  =\left(  \underbrace{x}_{=x1_{B}}\otimes
zv\right)  \otimes_{B\otimes B^{\prime}}\left(  y\otimes w\right)
=\underbrace{\left(  x1_{B}\otimes zv\right)  }_{=\left(  x\otimes z\right)
\left(  1_{B}\otimes v\right)  }\otimes_{B\otimes B^{\prime}}\left(  y\otimes
w\right) \\
&  =\left(  x\otimes z\right)  \left(  1_{B}\otimes v\right)  \otimes
_{B\otimes B^{\prime}}\left(  y\otimes w\right)  =\left(  x\otimes z\right)
\otimes_{B\otimes B^{\prime}}\underbrace{\left(  1_{B}\otimes v\right)
\left(  y\otimes w\right)  }_{=1_{B}y\otimes vw}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }1_{B}\otimes v\text{ can be moved
past the }\otimes_{B\otimes B^{\prime}}\text{ sign (because }1_{B}\otimes v\in
B\otimes B^{\prime}\text{)}\right) \\
&  =\left(  x\otimes z\right)  \otimes_{B\otimes B^{\prime}}\left(
\underbrace{1_{B}y}_{=y}\otimes vw\right)  =\left(  x\otimes z\right)
\otimes_{B\otimes B^{\prime}}\left(  y\otimes vw\right)  .
\end{align*}
Compared with%
\[
f\left(  x,y,z,vw\right)  =\left(  x\otimes z\right)  \otimes_{B\otimes
B^{\prime}}\left(  y\otimes vw\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }f\left(  x,y,z,vw\right)  \right)  ,
\]
this yields $f\left(  x,y,zv,w\right)  =f\left(  x,y,z,vw\right)  $, qed.}
Thus, Lemma \ref{lem.sol.ind.tensor.XYZW} (applied to $M=\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  $, $U=B$, $V=B^{\prime}$, $X=P$, $Y=Q$, $Z=P^{\prime}$ and
$W=Q^{\prime}$) yields that there exists a $\kk$-linear map $F:\left(
P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime
}\right)  \rightarrow\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  $ which satisfies%
\begin{equation}
\left(  F\left(  \left(  x\otimes_{B}y\right)  \otimes\left(  z\otimes
_{B^{\prime}}w\right)  \right)  =f\left(  x,y,z,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y,z,w\right)  \in P\times Q\times
P^{\prime}\times Q^{\prime}\right)  . \label{pf.prop.sol.ind.tensor.general.F}%
\end{equation}
Consider this $F$. Every $\left(  p,q,p^{\prime},q^{\prime}\right)  \in
P\times Q\times P^{\prime}\times Q^{\prime}$ satisfies%
\begin{align}
F\left(  \left(  p\otimes_{B}q\right)  \otimes\left(  p^{\prime}%
\otimes_{B^{\prime}}q^{\prime}\right)  \right)   &  =f\left(  p,q,p^{\prime
},q^{\prime}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.ind.tensor.general.F}), applied to }\left(  x,y,z,w\right)
=\left(  p,q,p^{\prime},q^{\prime}\right)  \right) \nonumber\\
&  =\left(  p\otimes p^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q\otimes q^{\prime}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }f\right)  . \label{pf.prop.sol.ind.tensor.general.F.real}%
\end{align}


We have%
\begin{equation}
F\left(  \alpha\left(  \varphi\otimes\zeta\right)  \right)  =\alpha\cdot
F\left(  \varphi\otimes\zeta\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}\alpha\in A\otimes A^{\prime}\text{, }\varphi\in P\otimes_{B}Q\text{ and
}\psi\in P^{\prime}\otimes_{B^{\prime}}Q^{\prime}.
\label{pf.prop.sol.ind.tensor.general.F.leftmod.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.ind.tensor.general.F.leftmod.1}%
):} Let $\alpha\in A\otimes A^{\prime}$, $\varphi\in P\otimes_{B}Q$ and
$\psi\in P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$.
\par
Since $\alpha$ is a tensor in $A\otimes A^{\prime}$, we can write $\alpha$ in
the form $\alpha=\sum_{n=1}^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}$ for some
$N\in \NN$, some elements $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$ of
$\kk$, some elements $a_{1}$, $a_{2}$, $...$, $a_{N}$ of $A$, and some
elements $a_{1}^{\prime}$, $a_{2}^{\prime}$, $...$, $a_{N}^{\prime}$ of
$A^{\prime}$. Consider this $N$, these $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$,
these $a_{1}$, $a_{2}$, $...$, $a_{N}$, and these $a_{1}^{\prime}$,
$a_{2}^{\prime}$, $...$, $a_{N}^{\prime}$.
\par
Since $\varphi$ is a tensor in $P\otimes_{B}Q$, we can write $\varphi$ in the
form $\varphi=\sum_{m=1}^{M}\mu_{m}p_{m}\otimes_{B}q_{m}$ for some
$M\in \NN$, some elements $\mu_{1}$, $\mu_{2}$, $...$, $\mu_{M}$ of
$\kk$, some elements $p_{1}$, $p_{2}$, $...$, $p_{M}$ of $P$, and some
elements $q_{1}$, $q_{2}$, $...$, $q_{M}$ of $Q$.
\par
Since $\zeta$ is a tensor in $P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$, we
can write $\zeta$ in the form $\zeta=\sum_{k=1}^{K}\kappa_{k}p_{k}^{\prime
}\otimes_{B^{\prime}}q_{k}^{\prime}$ for some $K\in \NN$, some elements
$\kappa_{1}$, $\kappa_{2}$, $...$, $\kappa_{K}$ of $\kk$, some elements
$p_{1}^{\prime}$, $p_{2}^{\prime}$, $...$, $p_{K}^{\prime}$ of $P^{\prime}$,
and some elements $q_{1}^{\prime}$, $q_{2}^{\prime}$, $...$, $q_{K}^{\prime}$
of $Q^{\prime}$.
\par
Now,%
\begin{align}
\underbrace{\varphi}_{=\sum_{m=1}^{M}\mu_{m}p_{m}\otimes_{B}q_{m}}%
\otimes\underbrace{\zeta}_{=\sum_{k=1}^{K}\kappa_{k}p_{k}^{\prime}%
\otimes_{B^{\prime}}q_{k}^{\prime}}  &  =\left(  \sum_{m=1}^{M}\mu_{m}%
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  \sum_{k=1}^{K}\kappa_{k}%
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right) \nonumber\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}%
q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime
}\right)  \label{pf.prop.sol.ind.tensor.general.F.leftmod.1.pf.1}%
\end{align}
(since the tensor product is $\kk$-bilinear). Hence,
\begin{align*}
&  \underbrace{\alpha}_{=\sum_{n=1}^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}%
}\underbrace{\left(  \varphi\otimes\zeta\right)  }_{=\sum_{m=1}^{M}\mu_{m}%
\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}q_{m}\right)  \otimes\left(
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  }\\
&  =\left(  \sum_{n=1}^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}\right)  \left(
\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}%
q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime
}\right)  \right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{\left(
\sum_{n=1}^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}\right)  \left(  \left(
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime
}}q_{k}^{\prime}\right)  \right)  }_{\substack{=\sum_{n=1}^{N}\nu_{n}\left(
a_{n}\otimes a_{n}^{\prime}\right)  \left(  \left(  p_{m}\otimes_{B}%
q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime
}\right)  \right)  \\\text{(since the left action of }A\otimes A^{\prime
}\text{ on }\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  \text{ is } \kk \text{-bilinear)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the left action of }\sum_{n=1}%
^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}\in A\otimes A^{\prime}\text{ on
}\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}%
}Q^{\prime}\right)  \text{ is } \kk \text{-linear}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{\left(  a_{n}\otimes a_{n}^{\prime}\right)  \left(  \left(
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime
}}q_{k}^{\prime}\right)  \right)  }_{\substack{=a_{n}\left(  p_{m}\otimes
_{B}q_{m}\right)  \otimes a_{n}^{\prime}\left(  p_{k}^{\prime}\otimes
_{B^{\prime}}q_{k}^{\prime}\right)  \\\text{(by the definition of the left
}A\otimes A^{\prime}\text{-module}\\\text{structure on }\left(  P\otimes
_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
\text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{a_{n}\left(  p_{m}\otimes_{B}q_{m}\right)  }_{\substack{=a_{n}%
p_{m}\otimes_{B}q_{m}\\\text{(by the definition of the left}\\A\text{-module
structure on }P\otimes_{B}Q\text{)}}}\otimes\underbrace{a_{n}^{\prime}\left(
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  }_{\substack{=a_{n}%
^{\prime}p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\\\text{(by the
definition of the left}\\A^{\prime}\text{-module structure on }P^{\prime
}\otimes_{B^{\prime}}Q^{\prime}\text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
a_{n}p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  a_{n}^{\prime}p_{k}^{\prime
}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  .
\end{align*}
Applying the map $F$ to both sides of this equality, we obtain%
\begin{align}
&  F\left(  \alpha\left(  \varphi\otimes\zeta\right)  \right) \nonumber\\
&  =F\left(  \sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}%
\nu_{n}\left(  a_{n}p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  a_{n}%
^{\prime}p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  \right)
\nonumber\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{F\left(  \left(  a_{n}p_{m}\otimes_{B}q_{m}\right)
\otimes\left(  a_{n}^{\prime}p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime
}\right)  \right)  }_{\substack{=\left(  a_{n}p_{m}\otimes a_{n}^{\prime}%
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.real}), applied to}\\\left(
p,q,p^{\prime},q^{\prime}\right)  =\left(  a_{n}p_{m},q_{m},a_{n}^{\prime
}p_{k}^{\prime},q_{k}^{\prime}\right)  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is }%
 \kk \text{-linear}\right) \nonumber\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
a_{n}p_{m}\otimes a_{n}^{\prime}p_{k}^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime}\right)  .
\label{pf.prop.sol.ind.tensor.general.F.leftmod.1.pf.4}%
\end{align}
\par
On the other hand, applying the map $F$ to both sides of the equality
(\ref{pf.prop.sol.ind.tensor.general.F.leftmod.1.pf.1}), we obtain%
\begin{align*}
F\left(  \varphi\otimes\zeta\right)   &  =F\left(  \sum_{m=1}^{M}\mu_{m}%
\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}q_{m}\right)  \otimes\left(
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  \right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{F\left(  \left(
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime
}}q_{k}^{\prime}\right)  \right)  }_{\substack{=\left(  p_{m}\otimes
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.real}), applied to}\\\left(
p,q,p^{\prime},q^{\prime}\right)  =\left(  p_{m},q_{m},p_{k}^{\prime}%
,q_{k}^{\prime}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is }%
 \kk \text{-linear}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \alpha\cdot\underbrace{F\left(  \varphi\otimes\zeta\right)  }_{=\sum
_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes p_{k}^{\prime
}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime
}\right)  }\\
&  =\alpha\cdot\left(  \sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}\otimes q_{k}^{\prime}\right)  \right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{\alpha}%
_{=\sum_{n=1}^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}}\cdot\left(  \left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}\otimes q_{k}^{\prime}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the left action of }\alpha\in
A\otimes A^{\prime}\text{ on }\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  \text{ is
} \kk \text{-linear}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{\left(
\sum_{n=1}^{N}\nu_{n}a_{n}\otimes a_{n}^{\prime}\right)  \cdot\left(  \left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}\otimes q_{k}^{\prime}\right)  \right)  }_{\substack{=\sum_{n=1}^{N}%
\nu_{n}\left(  a_{n}\otimes a_{n}^{\prime}\right)  \cdot\left(  \left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}\otimes q_{k}^{\prime}\right)  \right)  \\\text{(since the left action of
}A\otimes A^{\prime}\text{ on }\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  \text{ is
} \kk \text{-bilinear)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{\left(  a_{n}\otimes a_{n}^{\prime}\right)  \cdot\left(
\left(  p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}%
}\left(  q_{m}\otimes q_{k}^{\prime}\right)  \right)  }_{\substack{=\left(
\left(  a_{n}\otimes a_{n}^{\prime}\right)  \cdot\left(  p_{m}\otimes
p_{k}^{\prime}\right)  \right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}\otimes q_{k}^{\prime}\right)  \\\text{(by the definition of the left
}A\otimes A^{\prime}\text{-module structure}\\\text{on }\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  \text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
\underbrace{\left(  a_{n}\otimes a_{n}^{\prime}\right)  \cdot\left(
p_{m}\otimes p_{k}^{\prime}\right)  }_{\substack{=a_{n}p_{m}\otimes
a_{n}^{\prime}p_{k}^{\prime}\\\text{(by the definition of the left}\\A\otimes
A^{\prime}\text{-module structure on }P\otimes P^{\prime}\text{)}}}\right)
\otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
a_{n}p_{m}\otimes a_{n}^{\prime}p_{k}^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime}\right)  .
\end{align*}
Compared with (\ref{pf.prop.sol.ind.tensor.general.F.leftmod.1.pf.4}), this
yields $F\left(  \alpha\left(  \varphi\otimes\zeta\right)  \right)
=\alpha\cdot F\left(  \varphi\otimes\zeta\right)  $. This proves
(\ref{pf.prop.sol.ind.tensor.general.F.leftmod.1}).} It is now easy to see
that the map $F$ is a left $A\otimes A^{\prime}$-module
homomorphism\footnote{\textit{Proof.} Let $\alpha\in A\otimes A^{\prime}$ and
$\rho\in\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes
_{B^{\prime}}Q^{\prime}\right)  $.
\par
The element $\rho\in\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime
}\otimes_{B^{\prime}}Q^{\prime}\right)  $ is a tensor. Hence, we can write
$\rho$ in the form $\rho=\sum_{n=1}^{N}\lambda_{n}\varphi_{n}\otimes\zeta_{n}$
for some $N\in \NN$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{N}$ of $\kk$, some elements $\varphi_{1}$, $\varphi_{2}$,
$...$, $\varphi_{N}$ of $P\otimes_{B}Q$, and some elements $\zeta_{1}$,
$\zeta_{2}$, $...$, $\zeta_{N}$ of $P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$.
Consider this $N$, these $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{N}$,
these $\varphi_{1}$, $\varphi_{2}$, $...$, $\varphi_{N}$, and these $\zeta
_{1}$, $\zeta_{2}$, $...$, $\zeta_{N}$.
\par
We have%
\begin{align*}
F\left(  \alpha\underbrace{\rho}_{=\sum_{n=1}^{N}\lambda_{n}\varphi_{n}%
\otimes\zeta_{n}}\right)   &  =F\left(  \underbrace{\alpha\left(  \sum
_{n=1}^{N}\lambda_{n}\varphi_{n}\otimes\zeta_{n}\right)  }_{\substack{=\sum
_{n=1}^{N}\lambda_{n}\alpha\left(  \varphi_{n}\otimes\zeta_{n}\right)
\\\text{(since the left action of }A\otimes A^{\prime}\text{ on}\\\left(
P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime
}\right)  \text{ is } \kk \text{-bilinear)}}}\right)  =F\left(
\sum_{n=1}^{N}\lambda_{n}\alpha\left(  \varphi_{n}\otimes\zeta_{n}\right)
\right) \\
&  =\sum_{n=1}^{N}\lambda_{n}\underbrace{F\left(  \alpha\left(  \varphi
_{n}\otimes\zeta_{n}\right)  \right)  }_{\substack{=\alpha\cdot F\left(
\varphi_{n}\otimes\zeta_{n}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.leftmod.1}), applied to }%
\varphi=\varphi_{n}\text{ and }\zeta=\zeta_{n}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is } \kk %
\text{-linear}\right) \\
&  =\sum_{n=1}^{N}\lambda_{n}\alpha\cdot F\left(  \varphi_{n}\otimes\zeta
_{n}\right)  .
\end{align*}
Compared with%
\begin{align*}
\alpha\cdot F\left(  \underbrace{\rho}_{=\sum_{n=1}^{N}\lambda_{n}\varphi
_{n}\otimes\zeta_{n}}\right)   &  =\alpha\cdot\underbrace{F\left(  \sum
_{n=1}^{N}\lambda_{n}\varphi_{n}\otimes\zeta_{n}\right)  }_{\substack{=\sum
_{n=1}^{N}\lambda_{n}F\left(  \varphi_{n}\otimes\zeta_{n}\right)
\\\text{(since the map }F\text{ is } \kk \text{-linear)}}}=\alpha
\cdot\sum_{n=1}^{N}\lambda_{n}F\left(  \varphi_{n}\otimes\zeta_{n}\right)
=\sum_{n=1}^{N}\lambda_{n}\alpha\cdot F\left(  \varphi_{n}\otimes\zeta
_{n}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the left action of }A\otimes
A^{\prime}\text{ on }\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  \text{ is } \kk %
\text{-bilinear}\right)  ,
\end{align*}
this yields $F\left(  \alpha\rho\right)  =\alpha\cdot F\left(  \rho\right)  $.
\par
Now, let us forget that we fixed $\alpha$ and $\rho$. We thus have proven that
$F\left(  \alpha\rho\right)  =\alpha\cdot F\left(  \rho\right)  $ for every
$\alpha\in A\otimes A^{\prime}$ and $\rho\in\left(  P\otimes_{B}Q\right)
\otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  $. In other
words, the map $F$ is a left $A\otimes A^{\prime}$-module homomorphism, qed.}.

Also, we have%
\begin{equation}
F\left(  \left(  \varphi\otimes\zeta\right)  \gamma\right)  =F\left(
\varphi\otimes\zeta\right)  \cdot\gamma\ \ \ \ \ \ \ \ \ \ \text{for every
}\varphi\in P\otimes_{B}Q\text{, }\psi\in P^{\prime}\otimes_{B^{\prime}%
}Q^{\prime}\text{ and }\gamma\in C\otimes C^{\prime}.
\label{pf.prop.sol.ind.tensor.general.F.rightmod.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.ind.tensor.general.F.rightmod.1}%
):} Let $\varphi\in P\otimes_{B}Q$, $\psi\in P^{\prime}\otimes_{B^{\prime}%
}Q^{\prime}$ and $\gamma\in C\otimes C^{\prime}$.
\par
Since $\gamma$ is a tensor in $C\otimes C^{\prime}$, we can write $\gamma$ in
the form $\gamma=\sum_{n=1}^{N}\nu_{n}c_{n}\otimes c_{n}^{\prime}$ for some
$N\in \NN$, some elements $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$ of
$\kk$, some elements $c_{1}$, $c_{2}$, $...$, $c_{N}$ of $C$, and some
elements $c_{1}^{\prime}$, $c_{2}^{\prime}$, $...$, $c_{N}^{\prime}$ of
$C^{\prime}$. Consider this $N$, these $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$,
these $c_{1}$, $c_{2}$, $...$, $c_{N}$, and these $c_{1}^{\prime}$,
$c_{2}^{\prime}$, $...$, $c_{N}^{\prime}$.
\par
Since $\varphi$ is a tensor in $P\otimes_{B}Q$, we can write $\varphi$ in the
form $\varphi=\sum_{m=1}^{M}\mu_{m}p_{m}\otimes_{B}q_{m}$ for some
$M\in \NN$, some elements $\mu_{1}$, $\mu_{2}$, $...$, $\mu_{M}$ of
$\kk$, some elements $p_{1}$, $p_{2}$, $...$, $p_{M}$ of $P$, and some
elements $q_{1}$, $q_{2}$, $...$, $q_{M}$ of $Q$.
\par
Since $\zeta$ is a tensor in $P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$, we
can write $\zeta$ in the form $\zeta=\sum_{k=1}^{K}\kappa_{k}p_{k}^{\prime
}\otimes_{B^{\prime}}q_{k}^{\prime}$ for some $K\in \NN$, some elements
$\kappa_{1}$, $\kappa_{2}$, $...$, $\kappa_{K}$ of $\kk$, some elements
$p_{1}^{\prime}$, $p_{2}^{\prime}$, $...$, $p_{K}^{\prime}$ of $P^{\prime}$,
and some elements $q_{1}^{\prime}$, $q_{2}^{\prime}$, $...$, $q_{K}^{\prime}$
of $Q^{\prime}$.
\par
Now,%
\begin{align}
\underbrace{\varphi}_{=\sum_{m=1}^{M}\mu_{m}p_{m}\otimes_{B}q_{m}}%
\otimes\underbrace{\zeta}_{=\sum_{k=1}^{K}\kappa_{k}p_{k}^{\prime}%
\otimes_{B^{\prime}}q_{k}^{\prime}}  &  =\left(  \sum_{m=1}^{M}\mu_{m}%
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  \sum_{k=1}^{K}\kappa_{k}%
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right) \nonumber\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}%
q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime
}\right)  \label{pf.prop.sol.ind.tensor.general.F.rightmod.1.pf.1}%
\end{align}
(since the tensor product is $\kk$-bilinear). Hence,
\begin{align*}
&  \underbrace{\left(  \varphi\otimes\zeta\right)  }_{=\sum_{m=1}^{M}\mu
_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}q_{m}\right)
\otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)
}\underbrace{\gamma}_{=\sum_{n=1}^{N}\nu_{n}c_{n}\otimes c_{n}^{\prime}}\\
&  =\left(  \sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}%
\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}%
}q_{k}^{\prime}\right)  \right)  \left(  \sum_{n=1}^{N}\nu_{n}c_{n}\otimes
c_{n}^{\prime}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{\left(  \left(
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime
}}q_{k}^{\prime}\right)  \right)  \left(  \sum_{n=1}^{N}\nu_{n}c_{n}\otimes
c_{n}^{\prime}\right)  }_{\substack{=\sum_{n=1}^{N}\nu_{n}\left(  \left(
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime
}}q_{k}^{\prime}\right)  \right)  \left(  c_{n}\otimes c_{n}^{\prime}\right)
\\\text{(since the right action of }C\otimes C^{\prime}\text{ on }\left(
P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime
}\right)  \text{ is } \kk \text{-bilinear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the right action of }\sum_{n=1}%
^{N}\nu_{n}c_{n}\otimes c_{n}^{\prime}\in C\otimes C^{\prime}\text{ on
}\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}%
}Q^{\prime}\right)  \text{ is } \kk \text{-linear}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{\left(  \left(  p_{m}\otimes_{B}q_{m}\right)  \otimes\left(
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  \right)  \left(
c_{n}\otimes c_{n}^{\prime}\right)  }_{\substack{=\left(  p_{m}\otimes
_{B}q_{m}\right)  c_{n}\otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}%
q_{k}^{\prime}\right)  c_{n}^{\prime}\\\text{(by the definition of the right
}C\otimes C^{\prime}\text{-module}\\\text{structure on }\left(  P\otimes
_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
\text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{\left(  p_{m}\otimes_{B}q_{m}\right)  c_{n}}_{\substack{=p_{m}%
\otimes_{B}q_{m}c_{n}\\\text{(by the definition of the right}\\C\text{-module
structure on }P\otimes_{B}Q\text{)}}}\otimes\underbrace{\left(  p_{k}^{\prime
}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  c_{n}^{\prime}}_{\substack{=p_{k}%
^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}c_{n}^{\prime}\\\text{(by the
definition of the right}\\C^{\prime}\text{-module structure on }P^{\prime
}\otimes_{B^{\prime}}Q^{\prime}\text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
p_{m}\otimes_{B}q_{m}c_{n}\right)  \otimes\left(  p_{k}^{\prime}%
\otimes_{B^{\prime}}q_{k}^{\prime}c_{n}^{\prime}\right)  .
\end{align*}
Applying the map $F$ to both sides of this equality, we obtain%
\begin{align}
&  F\left(  \left(  \varphi\otimes\zeta\right)  \gamma\right) \nonumber\\
&  =F\left(  \sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}%
\nu_{n}\left(  p_{m}\otimes_{B}q_{m}c_{n}\right)  \otimes\left(  p_{k}%
^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}c_{n}^{\prime}\right)  \right)
\nonumber\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{F\left(  \left(  p_{m}\otimes_{B}q_{m}c_{n}\right)
\otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}c_{n}^{\prime
}\right)  \right)  }_{\substack{=\left(  p_{m}\otimes p_{k}^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  q_{m}c_{n}\otimes q_{k}^{\prime}%
c_{n}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.real}), applied to}\\\left(
p,q,p^{\prime},q^{\prime}\right)  =\left(  p_{m},q_{m}c_{n},p_{k}^{\prime
},q_{k}^{\prime}c_{n}^{\prime}\right)  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is }%
 \kk \text{-linear}\right) \nonumber\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}c_{n}\otimes q_{k}^{\prime}c_{n}^{\prime}\right)  .
\label{pf.prop.sol.ind.tensor.general.F.rightmod.1.pf.4}%
\end{align}
\par
On the other hand, applying the map $F$ to both sides of the equality
(\ref{pf.prop.sol.ind.tensor.general.F.rightmod.1.pf.1}), we obtain%
\begin{align*}
F\left(  \varphi\otimes\zeta\right)   &  =F\left(  \sum_{m=1}^{M}\mu_{m}%
\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes_{B}q_{m}\right)  \otimes\left(
p_{k}^{\prime}\otimes_{B^{\prime}}q_{k}^{\prime}\right)  \right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{F\left(  \left(
p_{m}\otimes_{B}q_{m}\right)  \otimes\left(  p_{k}^{\prime}\otimes_{B^{\prime
}}q_{k}^{\prime}\right)  \right)  }_{\substack{=\left(  p_{m}\otimes
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.real}), applied to}\\\left(
p,q,p^{\prime},q^{\prime}\right)  =\left(  p_{m},q_{m},p_{k}^{\prime}%
,q_{k}^{\prime}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is }%
 \kk \text{-linear}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  .
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{F\left(  \varphi\otimes\zeta\right)  }_{=\sum_{m=1}^{M}\mu
_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes p_{k}^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime}\right)
}\cdot\gamma\\
&  =\left(  \sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  p_{m}\otimes
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  \right)  \cdot\gamma\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(  \left(  p_{m}\otimes
p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{k}^{\prime}\right)  \right)  \cdot\underbrace{\gamma}_{=\sum_{n=1}^{N}%
\nu_{n}c_{n}\otimes c_{n}^{\prime}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the right action of }\gamma\in
C\otimes C^{\prime}\text{ on }\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  \text{ is
} \kk \text{-linear}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{\left(  \left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}\otimes q_{k}^{\prime}\right)  \right)  \cdot\left(  \sum_{n=1}^{N}%
\nu_{n}c_{n}\otimes c_{n}^{\prime}\right)  }_{\substack{=\sum_{n=1}^{N}\nu
_{n}\left(  \left(  p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime}\right)  \right)  \cdot\left(
c_{n}\otimes c_{n}^{\prime}\right)  \\\text{(since the right action of
}C\otimes C^{\prime}\text{ on }\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  \text{ is
} \kk \text{-bilinear)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu
_{n}\underbrace{\left(  \left(  p_{m}\otimes p_{k}^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes q_{k}^{\prime}\right)
\right)  \cdot\left(  c_{n}\otimes c_{n}^{\prime}\right)  }%
_{\substack{=\left(  p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  \left(  q_{m}\otimes q_{k}^{\prime}\right)  \cdot\left(
c_{n}\otimes c_{n}^{\prime}\right)  \right)  \\\text{(by the definition of the
left }C\otimes C^{\prime}\text{-module structure}\\\text{on }\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  \text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
\underbrace{\left(  q_{m}\otimes q_{k}^{\prime}\right)  \cdot\left(
c_{n}\otimes c_{n}^{\prime}\right)  }_{\substack{=q_{m}c_{n}\otimes
q_{k}^{\prime}c_{n}^{\prime}\\\text{(by the definition of the right}\\C\otimes
C^{\prime}\text{-module structure on }P\otimes P^{\prime}\text{)}}}\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\sum_{n=1}^{N}\nu_{n}\left(
p_{m}\otimes p_{k}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{m}c_{n}\otimes q_{k}^{\prime}c_{n}^{\prime}\right)  .
\end{align*}
Compared with (\ref{pf.prop.sol.ind.tensor.general.F.rightmod.1.pf.4}), this
yields $F\left(  \left(  \varphi\otimes\zeta\right)  \gamma\right)  =F\left(
\varphi\otimes\zeta\right)  \cdot\gamma$. This proves
(\ref{pf.prop.sol.ind.tensor.general.F.rightmod.1}).} It is now easy to see
that the map $F$ is a right $C\otimes C^{\prime}$-module
homomorphism\footnote{\textit{Proof.} Let $\rho\in\left(  P\otimes
_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
$ and $\gamma\in C\otimes C^{\prime}$.
\par
The element $\rho\in\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime
}\otimes_{B^{\prime}}Q^{\prime}\right)  $ is a tensor. Hence, we can write
$\rho$ in the form $\rho=\sum_{n=1}^{N}\lambda_{n}\varphi_{n}\otimes\zeta_{n}$
for some $N\in \NN$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{N}$ of $\kk$, some elements $\varphi_{1}$, $\varphi_{2}$,
$...$, $\varphi_{N}$ of $P\otimes_{B}Q$, and some elements $\zeta_{1}$,
$\zeta_{2}$, $...$, $\zeta_{N}$ of $P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$.
Consider this $N$, these $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{N}$,
these $\varphi_{1}$, $\varphi_{2}$, $...$, $\varphi_{N}$, and these $\zeta
_{1}$, $\zeta_{2}$, $...$, $\zeta_{N}$.
\par
We have%
\begin{align*}
F\left(  \underbrace{\rho}_{=\sum_{n=1}^{N}\lambda_{n}\varphi_{n}\otimes
\zeta_{n}}\gamma\right)   &  =F\left(  \underbrace{\left(  \sum_{n=1}%
^{N}\lambda_{n}\varphi_{n}\otimes\zeta_{n}\right)  \gamma}_{\substack{=\sum
_{n=1}^{N}\lambda_{n}\left(  \varphi_{n}\otimes\zeta_{n}\right)
\gamma\\\text{(since the right action of }C\otimes C^{\prime}\text{
on}\\\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes
_{B^{\prime}}Q^{\prime}\right)  \text{ is } \kk \text{-bilinear)}%
}}\right)  =F\left(  \sum_{n=1}^{N}\lambda_{n}\left(  \varphi_{n}\otimes
\zeta_{n}\right)  \gamma\right) \\
&  =\sum_{n=1}^{N}\lambda_{n}\underbrace{F\left(  \left(  \varphi_{n}%
\otimes\zeta_{n}\right)  \gamma\right)  }_{\substack{=F\left(  \varphi
_{n}\otimes\zeta_{n}\right)  \cdot\gamma\\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.rightmod.1}), applied to }%
\varphi=\varphi_{n}\text{ and }\zeta=\zeta_{n}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is } \kk %
\text{-linear}\right) \\
&  =\sum_{n=1}^{N}\lambda_{n}F\left(  \varphi_{n}\otimes\zeta_{n}\right)
\cdot\gamma.
\end{align*}
Compared with%
\begin{align*}
F\left(  \underbrace{\rho}_{=\sum_{n=1}^{N}\lambda_{n}\varphi_{n}\otimes
\zeta_{n}}\right)  \cdot\gamma &  =\underbrace{F\left(  \sum_{n=1}^{N}%
\lambda_{n}\varphi_{n}\otimes\zeta_{n}\right)  }_{\substack{=\sum_{n=1}%
^{N}\lambda_{n}F\left(  \varphi_{n}\otimes\zeta_{n}\right)  \\\text{(since the
map }F\text{ is } \kk \text{-linear)}}}\cdot\gamma=\left(  \sum_{n=1}%
^{N}\lambda_{n}F\left(  \varphi_{n}\otimes\zeta_{n}\right)  \right)
\cdot\gamma=\sum_{n=1}^{N}\lambda_{n}F\left(  \varphi_{n}\otimes\zeta
_{n}\right)  \cdot\gamma\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the right action of }C\otimes
C^{\prime}\text{ on }\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes
B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  \text{ is } \kk %
\text{-bilinear}\right)  ,
\end{align*}
this yields $F\left(  \rho\gamma\right)  =F\left(  \rho\right)  \cdot\gamma$.
\par
Now, let us forget that we fixed $\gamma$ and $\rho$. We thus have proven that
$F\left(  \rho\gamma\right)  =F\left(  \rho\right)  \cdot\gamma$ for every
$\rho\in\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes
_{B^{\prime}}Q^{\prime}\right)  $ and $\gamma\in C\otimes C^{\prime}$. In
other words, the map $F$ is a right $C\otimes C^{\prime}$-module homomorphism,
qed.}.

Now, we know that the map $F$ is an $\left(  A\otimes A^{\prime},C\otimes
C^{\prime}\right)  $-bimodule homomorphism (since $F$ is a left $A\otimes
A^{\prime}$-module homomorphism and is a right $C\otimes C^{\prime}$-module homomorphism).

Now, let us define a map $g:P\times P^{\prime}\times Q\times Q^{\prime
}\rightarrow\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  $ by%
\[
\left(  g\left(  p,p^{\prime},q,q^{\prime}\right)  =\left(  p\otimes
_{B}q\right)  \otimes\left(  p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  p,p^{\prime},q,q^{\prime}\right)
\in P\times P^{\prime}\times Q\times Q^{\prime}\right)  .
\]
This map $g$ is $\kk$-multilinear\footnote{\textit{Proof.} Let
$\lambda_{1}\in \kk$, $\lambda_{2}\in \kk$, $p_{1}\in P$,
$p_{2}\in P$, $p^{\prime}\in P^{\prime}$, $q\in Q$ and $q^{\prime}\in
Q^{\prime}$.
\par
By the definition of $g\left(  p_{1},p^{\prime},q,q^{\prime}\right)  $, we
have $g\left(  p_{1},p^{\prime},q,q^{\prime}\right)  =\left(  p_{1}\otimes
_{B}q\right)  \otimes\left(  p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right)
$. Thus, $\left(  p_{1}\otimes_{B}q\right)  \otimes\left(  p^{\prime}%
\otimes_{B^{\prime}}q^{\prime}\right)  =g\left(  p_{1},p^{\prime},q,q^{\prime
}\right)  $.
\par
By the definition of $g\left(  p_{2},p^{\prime},q,q^{\prime}\right)  $, we
have $g\left(  p_{2},p^{\prime},q,q^{\prime}\right)  =\left(  p_{2}\otimes
_{B}q\right)  \otimes\left(  p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right)
$. Thus, $\left(  p_{2}\otimes_{B}q\right)  \otimes\left(  p^{\prime}%
\otimes_{B^{\prime}}q^{\prime}\right)  =g\left(  p_{2},p^{\prime},q,q^{\prime
}\right)  $.
\par
By the definition of $g\left(  \lambda_{1}p_{1}+\lambda_{2}p_{2},p^{\prime
},q,q^{\prime}\right)  $, we have%
\begin{align*}
&  g\left(  \lambda_{1}p_{1}+\lambda_{2}p_{2},p^{\prime},q,q^{\prime}\right)
\\
&  =\underbrace{\left(  \left(  \lambda_{1}p_{1}+\lambda_{2}p_{2}\right)
\otimes_{B}q\right)  }_{=\lambda_{1}p_{1}\otimes_{B}q+\lambda_{2}p_{2}%
\otimes_{B}q}\otimes\left(  p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right) \\
&  =\left(  \lambda_{1}p_{1}\otimes_{B}q+\lambda_{2}p_{2}\otimes_{B}q\right)
\otimes\left(  p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right) \\
&  =\lambda_{1}\underbrace{\left(  p_{1}\otimes_{B}q\right)  \otimes\left(
p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right)  }_{=g\left(  p_{1},p^{\prime
},q,q^{\prime}\right)  }+\lambda_{2}\underbrace{\left(  p_{2}\otimes
_{B}q\right)  \otimes\left(  p^{\prime}\otimes_{B^{\prime}}q^{\prime}\right)
}_{=g\left(  p_{2},p^{\prime},q,q^{\prime}\right)  }\\
&  =\lambda_{1}g\left(  p_{1},p^{\prime},q,q^{\prime}\right)  +\lambda
_{2}g\left(  p_{2},p^{\prime},q,q^{\prime}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}$, $\lambda_{2}$, $p_{1}$,
$p_{2}$, $p^{\prime}$, $q$ and $q^{\prime}$. We thus have proven that
$g\left(  \lambda_{1}p_{1}+\lambda_{2}p_{2},p^{\prime},q,q^{\prime}\right)
=\lambda_{1}g\left(  p_{1},p^{\prime},q,q^{\prime}\right)  +\lambda
_{2}g\left(  p_{2},p^{\prime},q,q^{\prime}\right)  $ for every $\lambda_{1}%
\in \kk$, $\lambda_{2}\in \kk$, $p_{1}\in P$, $p_{2}\in P$,
$p^{\prime}\in P^{\prime}$, $q\in Q$ and $q^{\prime}\in Q^{\prime}$. In other
words, the map $g$ is $\kk$-linear in its first argument. Similarly,
the map $g$ is $\kk$-linear in its second argument. Similarly, the map
$g$ is $\kk$-linear in its fourth argument. Also, the map $g$ is
$\kk$-linear in its third argument (this is proven in the same way as
we have showed that the map $g$ is $\kk$-linear in its first argument).
\par
We now know that the map $g$ is $\kk$-linear in its first argument, is
$\kk$-linear in its second argument, is $\kk$-linear in its
third argument, and $\kk$-linear in its fourth argument. In other
words, the map $g$ is $\kk$-linear in each of its four arguments. In
other words, the map $g$ is $\kk$-multilinear, qed.}. Hence, Lemma
\ref{lem.sol.ind.tensor.simple} (applied to $X=P$, $Y=P^{\prime}$, $Z=Q$,
$W=Q^{\prime}$ and $M=\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime
}\otimes_{B^{\prime}}Q^{\prime}\right)  $) yields that there exists a
$\kk$-linear map $\mathfrak{G}:\left(  P\otimes P^{\prime}\right)
\otimes\left(  Q\otimes Q^{\prime}\right)  \rightarrow\left(  P\otimes
_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
$ which satisfies%
\begin{equation}
\left(  \mathfrak{G}\left(  \left(  x\otimes y\right)  \otimes\left(  z\otimes
w\right)  \right)  =g\left(  x,y,z,w\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  x,y,z,w\right)  \in P\times P^{\prime}\times Q\times Q^{\prime
}\right)  . \label{pf.prop.sol.ind.tensor.general.Gfrak}%
\end{equation}
Consider this $\mathfrak{G}$. Every $\left(  p,p^{\prime},q,q^{\prime}\right)
\in P\times P^{\prime}\times Q\times Q^{\prime}$ satisfies%
\begin{align}
\mathfrak{G}\left(  \left(  p\otimes p^{\prime}\right)  \otimes\left(
q\otimes q^{\prime}\right)  \right)   &  =g\left(  p,p^{\prime},q,q^{\prime
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.ind.tensor.general.Gfrak}), applied to }\left(
x,y,z,w\right)  =\left(  p,p^{\prime},q,q^{\prime}\right)  \right) \nonumber\\
&  =\left(  p\otimes_{B}q\right)  \otimes\left(  p^{\prime}\otimes_{B^{\prime
}}q^{\prime}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}g\right)  . \label{pf.prop.sol.ind.tensor.general.Gfrak.real}%
\end{align}


Now, we can see that%
\[
\mathfrak{G}\left(  xu\otimes y\right)  =\mathfrak{G}\left(  x\otimes
uy\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }x\in P\otimes P^{\prime}\text{,
}u\in B\otimes B^{\prime}\text{ and }y\in Q\otimes Q^{\prime}%
\]
\footnote{\textit{Proof.} Let $x\in P\otimes P^{\prime}$, $u\in B\otimes
B^{\prime}$ and $y\in Q\otimes Q^{\prime}$.
\par
The element $x\in P\otimes P^{\prime}$ is a tensor. Hence, there exist an
$N\in \NN$, some elements $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$ of
$\kk$, some elements $p_{1}$, $p_{2}$, $...$, $p_{N}$ of $P$, and some
elements $p_{1}^{\prime}$, $p_{2}^{\prime}$, $...$, $p_{N}^{\prime}$ of
$P^{\prime}$ such that $x=\sum_{n=1}^{N}\nu_{n}p_{n}\otimes p_{n}^{\prime}$.
Consider this $N$, these $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$, these $p_{1}%
$, $p_{2}$, $...$, $p_{N}$, and these $p_{1}^{\prime}$, $p_{2}^{\prime}$,
$...$, $p_{N}^{\prime}$.
\par
The element $y\in Q\otimes Q^{\prime}$ is a tensor. Hence, there exist an
$M\in \NN$, some elements $\mu_{1}$, $\mu_{2}$, $...$, $\mu_{M}$ of
$\kk$, some elements $q_{1}$, $q_{2}$, $...$, $q_{M}$ of $Q$, and some
elements $q_{1}^{\prime}$, $q_{2}^{\prime}$, $...$, $q_{M}^{\prime}$ of
$Q^{\prime}$ such that $y=\sum_{m=1}^{M}\mu_{m}q_{m}\otimes q_{m}^{\prime}$.
Consider this $M$, these $\mu_{1}$, $\mu_{2}$, $...$, $\mu_{M}$, these $q_{1}%
$, $q_{2}$, $...$, $q_{M}$, and these $q_{1}^{\prime}$, $q_{2}^{\prime}$,
$...$, $q_{M}^{\prime}$.
\par
The element $u\in B\otimes B^{\prime}$ is a tensor. Hence, there exist an
$K\in \NN$, some elements $\kappa_{1}$, $\kappa_{2}$, $...$, $\kappa
_{K}$ of $\kk$, some elements $b_{1}$, $b_{2}$, $...$, $b_{K}$ of $B$,
and some elements $b_{1}^{\prime}$, $b_{2}^{\prime}$, $...$, $b_{K}^{\prime}$
of $B^{\prime}$ such that $u=\sum_{k=1}^{K}\kappa_{k}b_{k}\otimes
b_{k}^{\prime}$. Consider this $K$, these $\kappa_{1}$, $\kappa_{2}$, $...$,
$\kappa_{K}$, these $b_{1}$, $b_{2}$, $...$, $b_{K}$, and these $b_{1}%
^{\prime}$, $b_{2}^{\prime}$, $...$, $b_{K}^{\prime}$.
\par
We have%
\begin{align*}
\underbrace{x}_{=\sum_{n=1}^{N}\nu_{n}p_{n}\otimes p_{n}^{\prime}}u  &
=\left(  \sum_{n=1}^{N}\nu_{n}p_{n}\otimes p_{n}^{\prime}\right)  u=\sum
_{n=1}^{N}\nu_{n}\left(  p_{n}\otimes p_{n}^{\prime}\right)  \underbrace{u}%
_{=\sum_{k=1}^{K}\kappa_{k}b_{k}\otimes b_{k}^{\prime}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the right action of }u\in B\otimes
B^{\prime}\text{ on }P\otimes P^{\prime}\text{ is } \kk \text{-linear}%
\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\underbrace{\left(  p_{n}\otimes p_{n}^{\prime
}\right)  \left(  \sum_{k=1}^{K}\kappa_{k}b_{k}\otimes b_{k}^{\prime}\right)
}_{\substack{=\sum_{k=1}^{K}\kappa_{k}\left(  p_{n}\otimes p_{n}^{\prime
}\right)  \left(  b_{k}\otimes b_{k}^{\prime}\right)  \\\text{(since the right
action of }B\otimes B^{\prime}\text{ on}\\P\otimes P^{\prime}\text{ is
} \kk \text{-bilinear)}}}\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\underbrace{\left(
p_{n}\otimes p_{n}^{\prime}\right)  \left(  b_{k}\otimes b_{k}^{\prime
}\right)  }_{\substack{=p_{n}b_{k}\otimes p_{n}^{\prime}b_{k}^{\prime
}\\\text{(by the definition of the right}\\B\otimes B^{\prime}\text{-module
structure on }P\otimes P^{\prime}\text{)}}}=\sum_{n=1}^{N}\nu_{n}\sum
_{k=1}^{K}\kappa_{k}p_{n}b_{k}\otimes p_{n}^{\prime}b_{k}^{\prime}.
\end{align*}
Hence,%
\begin{align*}
\underbrace{xu}_{=\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}p_{n}%
b_{k}\otimes p_{n}^{\prime}b_{k}^{\prime}}\otimes\underbrace{y}_{=\sum
_{m=1}^{M}\mu_{m}q_{m}\otimes q_{m}^{\prime}}  &  =\left(  \sum_{n=1}^{N}%
\nu_{n}\sum_{k=1}^{K}\kappa_{k}p_{n}b_{k}\otimes p_{n}^{\prime}b_{k}^{\prime
}\right)  \otimes\left(  \sum_{m=1}^{M}\mu_{m}q_{m}\otimes q_{m}^{\prime
}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}b_{k}\otimes p_{n}^{\prime}b_{k}^{\prime}\right)  \otimes\left(
q_{m}\otimes q_{m}^{\prime}\right)
\end{align*}
(since tensoring is $\kk$-bilinear). Applying the map $\mathfrak{G}$ to
both sides of this equality, we obtain%
\begin{align}
\mathfrak{G}\left(  xu\otimes y\right)   &  =\mathfrak{G}\left(  \sum
_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}b_{k}\otimes p_{n}^{\prime}b_{k}^{\prime}\right)  \otimes\left(
q_{m}\otimes q_{m}^{\prime}\right)  \right) \nonumber\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu
_{m}\underbrace{\mathfrak{G}\left(  \left(  p_{n}b_{k}\otimes p_{n}^{\prime
}b_{k}^{\prime}\right)  \otimes\left(  q_{m}\otimes q_{m}^{\prime}\right)
\right)  }_{\substack{=\left(  p_{n}b_{k}\otimes_{B}q_{m}\right)
\otimes\left(  p_{n}^{\prime}b_{k}^{\prime}\otimes_{B^{\prime}}q_{m}^{\prime
}\right)  \\\text{(by (\ref{pf.prop.sol.ind.tensor.general.Gfrak.real}),
applied to }\left(  p,p^{\prime},q,q^{\prime}\right)  =\left(  p_{n}%
b_{k},p_{n}^{\prime}b_{k}^{\prime},q_{m},q_{m}^{\prime}\right)  \text{)}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathfrak{G}\text{ is
} \kk \text{-linear}\right) \nonumber\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu
_{m}\underbrace{\left(  p_{n}b_{k}\otimes_{B}q_{m}\right)  }_{\substack{=p_{n}%
\otimes_{B}b_{k}q_{m}\\\text{(since the element }b_{k}\text{ can
be}\\\text{moved past the }\otimes_{B}\text{ sign}\\\text{(because }b_{k}\in
B\text{))}}}\otimes\underbrace{\left(  p_{n}^{\prime}b_{k}^{\prime}%
\otimes_{B^{\prime}}q_{m}^{\prime}\right)  }_{\substack{=p_{n}^{\prime}%
\otimes_{B^{\prime}}b_{k}^{\prime}q_{m}^{\prime}\\\text{(since the element
}b_{k}^{\prime}\text{ can be}\\\text{moved past the }\otimes_{B^{\prime}%
}\text{ sign}\\\text{(because }b_{k}^{\prime}\in B^{\prime}\text{))}%
}}\nonumber\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}\otimes_{B}b_{k}q_{m}\right)  \otimes\left(  p_{n}^{\prime}%
\otimes_{B^{\prime}}b_{k}^{\prime}q_{m}^{\prime}\right)  .
\label{pf.prop.sol.ind.tensor.general.Gfrak.bil.1}%
\end{align}
\par
On the other hand,%
\begin{align*}
u\underbrace{y}_{=\sum_{m=1}^{M}\mu_{m}q_{m}\otimes q_{m}^{\prime}}  &
=u\left(  \sum_{m=1}^{M}\mu_{m}q_{m}\otimes q_{m}^{\prime}\right)  =\sum
_{m=1}^{M}\mu_{m}\underbrace{u}_{=\sum_{k=1}^{K}\kappa_{k}b_{k}\otimes
b_{k}^{\prime}}\left(  q_{m}\otimes q_{m}^{\prime}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the left action of }u\in B\otimes
B^{\prime}\text{ on }Q\otimes Q^{\prime}\text{ is } \kk \text{-linear}%
\right) \\
&  =\sum_{m=1}^{M}\mu_{m}\underbrace{\left(  \sum_{k=1}^{K}\kappa_{k}%
b_{k}\otimes b_{k}^{\prime}\right)  \left(  q_{m}\otimes q_{m}^{\prime
}\right)  }_{\substack{=\sum_{k=1}^{K}\kappa_{k}\left(  b_{k}\otimes
b_{k}^{\prime}\right)  \left(  q_{m}\otimes q_{m}^{\prime}\right)
\\\text{(since the left action of }B\otimes B^{\prime}\text{ on}\\Q\otimes
Q^{\prime}\text{ is } \kk \text{-bilinear)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\underbrace{\left(
b_{k}\otimes b_{k}^{\prime}\right)  \left(  q_{m}\otimes q_{m}^{\prime
}\right)  }_{\substack{=b_{k}q_{m}\otimes b_{k}^{\prime}q_{m}^{\prime
}\\\text{(by the definition of the left}\\B\otimes B^{\prime}\text{-module
structure on }Q\otimes Q^{\prime}\text{)}}}\\
&  =\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}b_{k}q_{m}\otimes
b_{k}^{\prime}q_{m}^{\prime}.
\end{align*}
Hence,%
\begin{align*}
\underbrace{x}_{=\sum_{n=1}^{N}\nu_{n}p_{n}\otimes p_{n}^{\prime}}%
\otimes\underbrace{uy}_{=\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}%
b_{k}q_{m}\otimes b_{k}^{\prime}q_{m}^{\prime}}  &  =\left(  \sum_{n=1}^{N}%
\nu_{n}p_{n}\otimes p_{n}^{\prime}\right)  \otimes\left(  \sum_{m=1}^{M}%
\mu_{m}\sum_{k=1}^{K}\kappa_{k}b_{k}q_{m}\otimes b_{k}^{\prime}q_{m}^{\prime
}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\sum_{k=1}^{K}\kappa_{k}\left(
p_{n}\otimes p_{n}^{\prime}\right)  \otimes\left(  b_{k}q_{m}\otimes
b_{k}^{\prime}q_{m}^{\prime}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since tensoring is } \kk %
\text{-bilinear}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}\otimes p_{n}^{\prime}\right)  \otimes\left(  b_{k}q_{m}\otimes
b_{k}^{\prime}q_{m}^{\prime}\right)  .
\end{align*}
Applying the map $\mathfrak{G}$ to both sides of this equality, we obtain%
\begin{align*}
\mathfrak{G}\left(  x\otimes uy\right)   &  =\mathfrak{G}\left(  \sum
_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}\otimes p_{n}^{\prime}\right)  \otimes\left(  b_{k}q_{m}\otimes
b_{k}^{\prime}q_{m}^{\prime}\right)  \right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu
_{m}\underbrace{\mathfrak{G}\left(  \left(  p_{n}\otimes p_{n}^{\prime
}\right)  \otimes\left(  b_{k}q_{m}\otimes b_{k}^{\prime}q_{m}^{\prime
}\right)  \right)  }_{\substack{=\left(  p_{n}\otimes_{B}b_{k}q_{m}\right)
\otimes\left(  p_{n}^{\prime}\otimes_{B^{\prime}}b_{k}^{\prime}q_{m}^{\prime
}\right)  \\\text{(by (\ref{pf.prop.sol.ind.tensor.general.Gfrak.real}),
applied to }\left(  p,p^{\prime},q,q^{\prime}\right)  =\left(  p_{n}%
,p_{n}^{\prime},b_{k}q_{m},b_{k}^{\prime}q_{m}^{\prime}\right)  \text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\mathfrak{G}\text{ is
} \kk \text{-linear}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{k=1}^{K}\kappa_{k}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}\otimes_{B}b_{k}q_{m}\right)  \otimes\left(  p_{n}^{\prime}%
\otimes_{B^{\prime}}b_{k}^{\prime}q_{m}^{\prime}\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.sol.ind.tensor.general.Gfrak.bil.1}), we
obtain $\mathfrak{G}\left(  xu\otimes y\right)  =\mathfrak{G}\left(  x\otimes
uy\right)  $, qed.} Thus, Lemma \ref{lem.sol.ind.tensor.kU} (applied to
$M=\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}%
}Q^{\prime}\right)  $, $U=B\otimes B^{\prime}$, $X=P\otimes P^{\prime}$ and
$Y=Q\otimes Q^{\prime}$) yields that there exists a $\kk$-linear map
$G:\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
Q\otimes Q^{\prime}\right)  \rightarrow\left(  P\otimes_{B}Q\right)
\otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  $ such that%
\begin{equation}
\left(  G\left(  x\otimes_{B\otimes B^{\prime}}y\right)  =\mathfrak{G}\left(
x\otimes y\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  x,y\right)
\in\left(  P\otimes P^{\prime}\right)  \times\left(  Q\otimes Q^{\prime
}\right)  \right)  . \label{pf.prop.sol.ind.tensor.general.G}%
\end{equation}
Consider this map $G$. Every $\left(  p,p^{\prime},q,q^{\prime}\right)  \in
P\times P^{\prime}\times Q\times Q^{\prime}$ satisfies%
\begin{align}
G\left(  \left(  p\otimes p^{\prime}\right)  \otimes_{B\otimes B^{\prime}%
}\left(  q\otimes q^{\prime}\right)  \right)   &  =\mathfrak{G}\left(  \left(
p\otimes p^{\prime}\right)  \otimes\left(  q\otimes q^{\prime}\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.ind.tensor.general.G}), applied to }\left(  x,y\right)
=\left(  p\otimes p^{\prime},q\otimes q^{\prime}\right)  \right) \nonumber\\
&  =\left(  p\otimes_{B}q\right)  \otimes\left(  p^{\prime}\otimes_{B^{\prime
}}q^{\prime}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.ind.tensor.general.Gfrak.real})}\right)  .
\label{pf.prop.sol.ind.tensor.general.G.real}%
\end{align}


We shall now prove that the maps $F$ and $G$ are mutually inverse.

Notice that the maps $F\circ G$ and $G\circ F$ are $\kk$-linear (since
$F$ and $G$ are $\kk$-linear). We have
$F\circ G=\id_{\left(  P\otimes P^{\prime}\right) 
 \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right) 
 }$\ \ \ \ \footnote{\textit{Proof.} Let
$x\in P\otimes P^{\prime}$ and $y\in Q\otimes Q^{\prime}$. We are going to
prove that $\left(  F\circ G\right)  \left(  x\otimes_{B\otimes B^{\prime}%
}y\right)  =\id_{\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  }\left(
x\otimes_{B\otimes B^{\prime}}y\right)  $.
\par
The element $x\in P\otimes P^{\prime}$ is a tensor. Hence, there exist an
$N\in \NN$, some elements $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$ of
$\kk$, some elements $p_{1}$, $p_{2}$, $...$, $p_{N}$ of $P$, and some
elements $p_{1}^{\prime}$, $p_{2}^{\prime}$, $...$, $p_{N}^{\prime}$ of
$P^{\prime}$ such that $x=\sum_{n=1}^{N}\nu_{n}p_{n}\otimes p_{n}^{\prime}$.
Consider this $N$, these $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$, these $p_{1}%
$, $p_{2}$, $...$, $p_{N}$, and these $p_{1}^{\prime}$, $p_{2}^{\prime}$,
$...$, $p_{N}^{\prime}$.
\par
The element $y\in Q\otimes Q^{\prime}$ is a tensor. Hence, there exist an
$M\in \NN$, some elements $\mu_{1}$, $\mu_{2}$, $...$, $\mu_{M}$ of
$\kk$, some elements $q_{1}$, $q_{2}$, $...$, $q_{M}$ of $Q$, and some
elements $q_{1}^{\prime}$, $q_{2}^{\prime}$, $...$, $q_{M}^{\prime}$ of
$Q^{\prime}$ such that $y=\sum_{m=1}^{M}\mu_{m}q_{m}\otimes q_{m}^{\prime}$.
Consider this $M$, these $\mu_{1}$, $\mu_{2}$, $...$, $\mu_{M}$, these $q_{1}%
$, $q_{2}$, $...$, $q_{M}$, and these $q_{1}^{\prime}$, $q_{2}^{\prime}$,
$...$, $q_{M}^{\prime}$.
\par
We have%
\begin{align}
\underbrace{x}_{=\sum_{n=1}^{N}\nu_{n}p_{n}\otimes p_{n}^{\prime}}%
\otimes_{B\otimes B^{\prime}}\underbrace{y}_{=\sum_{m=1}^{M}\mu_{m}%
q_{m}\otimes q_{m}^{\prime}}  &  =\left(  \sum_{n=1}^{N}\nu_{n}p_{n}\otimes
p_{n}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  \sum_{m=1}^{M}%
\mu_{m}q_{m}\otimes q_{m}^{\prime}\right) \nonumber\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\left(  p_{n}\otimes
p_{n}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{m}^{\prime}\right)  \label{pf.prop.sol.ind.tensor.general.FG.1}%
\end{align}
(since tensoring is $\kk$-bilinear). Applying the map $F\circ G$ to
both sides of this equality, we obtain%
\begin{align*}
\left(  F\circ G\right)  \left(  x\otimes_{B\otimes B^{\prime}}y\right)   &
=\left(  F\circ G\right)  \left(  \sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu
_{m}\left(  p_{n}\otimes p_{n}^{\prime}\right)  \otimes_{B\otimes B^{\prime}%
}\left(  q_{m}\otimes q_{m}^{\prime}\right)  \right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\underbrace{\left(  F\circ
G\right)  \left(  \left(  p_{n}\otimes p_{n}^{\prime}\right)  \otimes
_{B\otimes B^{\prime}}\left(  q_{m}\otimes q_{m}^{\prime}\right)  \right)
}_{=F\left(  G\left(  \left(  p_{n}\otimes p_{n}^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes q_{m}^{\prime}\right)
\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\circ
G\text{ is } \kk \text{-linear}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}F\left(  \underbrace{G\left(
\left(  p_{n}\otimes p_{n}^{\prime}\right)  \otimes_{B\otimes B^{\prime}%
}\left(  q_{m}\otimes q_{m}^{\prime}\right)  \right)  }_{\substack{=\left(
p_{n}\otimes_{B}q_{m}\right)  \otimes\left(  p_{n}^{\prime}\otimes_{B^{\prime
}}q_{m}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.G.real}), applied to}\\\left(
p,p^{\prime},q,q^{\prime}\right)  =\left(  p_{n},p_{n}^{\prime},q_{m}%
,q_{m}^{\prime}\right)  \text{)}}}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\underbrace{F\left(  \left(
p_{n}\otimes_{B}q_{m}\right)  \otimes\left(  p_{n}^{\prime}\otimes_{B^{\prime
}}q_{m}^{\prime}\right)  \right)  }_{\substack{=\left(  p_{n}\otimes
p_{n}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{m}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.real}), applied to}\\\left(
p,q,p^{\prime},q^{\prime}\right)  =\left(  p_{n},q_{m},p_{n}^{\prime}%
,q_{m}^{\prime}\right)  \text{)}}}\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\left(  p_{n}\otimes
p_{n}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  q_{m}\otimes
q_{m}^{\prime}\right)  =x\otimes_{B\otimes B^{\prime}}%
y\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.ind.tensor.general.FG.1})}\right) \\
&  =\id_{\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  }\left(
x\otimes_{B\otimes B^{\prime}}y\right)  .
\end{align*}
\par
Let us now forget that we fixed $x\in P\otimes P^{\prime}$ and $y\in Q\otimes
Q^{\prime}$. We thus have proven that $\left(  F\circ G\right)  \left(
x\otimes_{B\otimes B^{\prime}}y\right)  =\id_{\left(
P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes
Q^{\prime}\right)  }\left(  x\otimes_{B\otimes B^{\prime}}y\right)  $ for
every $x\in P\otimes P^{\prime}$ and $y\in Q\otimes Q^{\prime}$. In other
words, the maps $F\circ G$ and $\id_{\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  }$ are equal to each other on every pure tensor in $\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  $. Since these maps $F\circ G$ and
$\id_{\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}%
}\left(  Q\otimes Q^{\prime}\right)  }$ are $\kk$-linear, this yields
that these maps $F\circ G$ and $\id_{\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  }$ must be identical (because if two $\kk$-linear maps from a
tensor product are equal to each other on every pure tensor, then these maps
must be identical). In other words, $F\circ G
=\id_{\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}%
}\left(  Q\otimes Q^{\prime}\right)  }$, qed.} and $G\circ
F=\id_{\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  }$%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in P\otimes_{B}Q$ and $y\in
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$. We are going to prove that $\left(
G\circ F\right)  \left(  x\otimes y\right)  =
\id_{\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  }\left(  x\otimes y\right)  $.
\par
The element $x\in P\otimes_{B}Q$ is a tensor. Hence, there exist an
$N\in \NN$, some elements $\nu_{1}$, $\nu_{2}$, $...$, $\nu_{N}$ of
$\kk$, some elements $p_{1}$, $p_{2}$, $...$, $p_{N}$ of $P$, and some
elements $q_{1}$, $q_{2}$, $...$, $q_{N}$ of $Q$ such that $x=\sum_{n=1}%
^{N}\nu_{n}p_{n}\otimes_{B}q_{n}$. Consider this $N$, these $\nu_{1}$,
$\nu_{2}$, $...$, $\nu_{N}$, these $p_{1}$, $p_{2}$, $...$, $p_{N}$, and these
$q_{1}$, $q_{2}$, $...$, $q_{N}$.
\par
The element $y\in P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$ is a tensor.
Hence, there exist an $M\in \NN$, some elements $\mu_{1}$, $\mu_{2}$,
$...$, $\mu_{M}$ of $\kk$, some elements $p_{1}^{\prime}$,
$p_{2}^{\prime}$, $...$, $p_{M}^{\prime}$ of $P^{\prime}$, and some elements
$q_{1}^{\prime}$, $q_{2}^{\prime}$, $...$, $q_{M}^{\prime}$ of $Q^{\prime}$
such that $y=\sum_{m=1}^{M}\mu_{m}p_{m}^{\prime}\otimes_{B^{\prime}}%
q_{m}^{\prime}$. Consider this $M$, these $\mu_{1}$, $\mu_{2}$, $...$,
$\mu_{M}$, these $p_{1}^{\prime}$, $p_{2}^{\prime}$, $...$, $p_{M}^{\prime}$,
and these $q_{1}^{\prime}$, $q_{2}^{\prime}$, $...$, $q_{M}^{\prime}$.
\par
We have%
\begin{align}
\underbrace{x}_{=\sum_{n=1}^{N}\nu_{n}p_{n}\otimes_{B}q_{n}}\otimes
\underbrace{y}_{=\sum_{m=1}^{M}\mu_{m}p_{m}^{\prime}\otimes_{B^{\prime}}%
q_{m}^{\prime}}  &  =\left(  \sum_{n=1}^{N}\nu_{n}p_{n}\otimes_{B}%
q_{n}\right)  \otimes\left(  \sum_{m=1}^{M}\mu_{m}p_{m}^{\prime}%
\otimes_{B^{\prime}}q_{m}^{\prime}\right) \nonumber\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\left(  p_{n}\otimes_{B}%
q_{n}\right)  \otimes\left(  p_{m}^{\prime}\otimes_{B^{\prime}}q_{m}^{\prime
}\right)  \label{pf.prop.sol.ind.tensor.general.GF.1}%
\end{align}
(since tensoring is $\kk$-bilinear). Applying the map $G\circ F$ to
both sides of this equality, we obtain%
\begin{align*}
\left(  G\circ F\right)  \left(  x\otimes y\right)   &  =\left(  G\circ
F\right)  \left(  \sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\left(
p_{n}\otimes_{B}q_{n}\right)  \otimes\left(  p_{m}^{\prime}\otimes_{B^{\prime
}}q_{m}^{\prime}\right)  \right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\underbrace{\left(  G\circ
F\right)  \left(  \left(  p_{n}\otimes_{B}q_{n}\right)  \otimes\left(
p_{m}^{\prime}\otimes_{B^{\prime}}q_{m}^{\prime}\right)  \right)  }_{=G\left(
F\left(  \left(  p_{n}\otimes_{B}q_{n}\right)  \otimes\left(  p_{m}^{\prime
}\otimes_{B^{\prime}}q_{m}^{\prime}\right)  \right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }G\circ F\text{ is }%
 \kk \text{-linear}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}G\left(  \underbrace{F\left(
\left(  p_{n}\otimes_{B}q_{n}\right)  \otimes\left(  p_{m}^{\prime}%
\otimes_{B^{\prime}}q_{m}^{\prime}\right)  \right)  }_{\substack{=\left(
p_{n}\otimes p_{m}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{n}\otimes q_{m}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.F.real}), applied to}\\\left(
p,q,p^{\prime},q^{\prime}\right)  =\left(  p_{n},q_{n},p_{m}^{\prime}%
,q_{m}^{\prime}\right)  \text{)}}}\right) \\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\underbrace{G\left(  \left(
p_{n}\otimes p_{m}^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
q_{n}\otimes q_{m}^{\prime}\right)  \right)  }_{\substack{=\left(
p_{n}\otimes_{B}q_{n}\right)  \otimes\left(  p_{m}^{\prime}\otimes_{B^{\prime
}}q_{m}^{\prime}\right)  \\\text{(by
(\ref{pf.prop.sol.ind.tensor.general.G.real}), applied to}\\\left(
p,p^{\prime},q,q^{\prime}\right)  =\left(  p_{n},p_{m}^{\prime},q_{n}%
,q_{m}^{\prime}\right)  \text{)}}}\\
&  =\sum_{n=1}^{N}\nu_{n}\sum_{m=1}^{M}\mu_{m}\left(  p_{n}\otimes_{B}%
q_{n}\right)  \otimes\left(  p_{m}^{\prime}\otimes_{B^{\prime}}q_{m}^{\prime
}\right)  =x\otimes y\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.ind.tensor.general.GF.1})}\right) \\
&  =\id_{\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  }\left(  x\otimes y\right)  .
\end{align*}
\par
Let us now forget that we fixed $x$ and $y$. We thus have proven that $\left(
G\circ F\right)  \left(  x\otimes y\right)  =
\id_{\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  }\left(  x\otimes y\right)  $ for every
$x\in P\otimes_{B}Q$ and $y\in P^{\prime}\otimes_{B^{\prime}}Q^{\prime}$. In
other words, the maps $G\circ F$ and $\id_{\left(
P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime
}\right)  }$ are equal to each other on every pure tensor in $\left(
P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime
}\right)  $. Since these maps $G\circ F$ and
$\id_{\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  }$ are $\kk$-linear, this yields
that these maps $G\circ F$ and $\id_{\left(
P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime
}\right)  }$ must be identical (because if two $\kk$-linear maps from a
tensor product are equal to each other on every pure tensor, then these maps
must be identical). In other words,
$G\circ F=\id_{\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  }$, qed.}.

Now, the maps $F$ and $G$ are mutually inverse (since $F\circ
G=\id_{\left(  P\otimes P^{\prime}\right)
\otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  }$ and
$G\circ F=\id_{\left(  P\otimes_{B}Q\right)
\otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  }$). Hence,
the map $F$ is invertible. But we know that $F$ is an $\left(  A\otimes
A^{\prime},C\otimes C^{\prime}\right)  $-bimodule homomorphism. Thus, $F$ is
an invertible $\left(  A\otimes A^{\prime},C\otimes C^{\prime}\right)
$-bimodule homomorphism. In other words, $F$ is an $\left(  A\otimes
A^{\prime},C\otimes C^{\prime}\right)  $-bimodule isomorphism. Since $F$ is a
map from $\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  $ to $\left(  P\otimes P^{\prime
}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime}\right)  $,
this yields that there exists an $\left(  A\otimes A^{\prime},C\otimes
C^{\prime}\right)  $-bimodule isomorphism from $\left(  P\otimes_{B}Q\right)
\otimes\left(  P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)  $ to $\left(
P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes
Q^{\prime}\right)  $ (namely, $F$). In other words, $\left(  P\otimes
P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(  Q\otimes Q^{\prime
}\right)  \cong\left(  P\otimes_{B}Q\right)  \otimes\left(  P^{\prime}%
\otimes_{B^{\prime}}Q^{\prime}\right)  $ as $\left(  A\otimes A^{\prime
},C\otimes C^{\prime}\right)  $-bimodules. This proves Proposition
\ref{prop.sol.ind.tensor.general}.
\end{proof}

Now, let us come to the solution of
Exercise~\ref{exe.ind.tensor}.\footnote{The following solution
involves some handwaving: We are going to use certain isomorphisms
to identify
$\CC \left[ G_1 \times G_2 \right]$ with $\CC G_1 \otimes \CC G_2$
and to identify
$\CC \left[ H_1 \times H_2 \right]$ with $\CC H_1 \otimes \CC H_2$.
See the solution of Exercise~\ref{exe.Indrho.tensor} for an example
of how to avoid this kind of handwaving. (Actually,
Exercise~\ref{exe.Indrho}(b) shows that
Exercise~\ref{exe.Indrho.tensor} is a generalization of
Exercise~\ref{exe.ind.tensor}.)}
By the definition of $\Ind_{H_{1}}^{G_{1}}U_{1}$, we have
$\Ind_{H_{1}}^{G_{1}}U_{1}= \CC G_{1}\otimes_{ \CC %
H_{1}}U_{1}$. By the definition of $\Ind_{H_{2}}^{G_{2}}U_{2}$,
we have $\Ind_{H_{2}}^{G_{2}}U_{2}= \CC G_{2}%
\otimes_{ \CC H_{2}}U_{2}$.

We identify the $ \CC $-algebra $ \CC \left[  G_{1}\times
G_{2}\right]  $ with the $ \CC $-algebra $ \CC G_{1}\otimes
 \CC G_{2}$ (by identifying the basis element $t_{\left(  g_{1}%
,g_{2}\right)  }$ of the former $ \CC $-algebra with the basis element
$t_{g_{1}}\otimes t_{g_{2}}$ of the latter $ \CC $-algebra for every
$\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}$). Similarly, we identify
the $ \CC $-algebra $ \CC \left[  H_{1}\times H_{2}\right]  $ with
the $ \CC $-algebra $ \CC H_{1}\otimes \CC H_{2}$. These two
identifications play nicely with each other (this means that the canonical
injection $ \CC \left[  H_{1}\times H_{2}\right]  \rightarrow
 \CC \left[  G_{1}\times G_{2}\right]  $ becomes identified with the
tensor product of the canonical injections $ \CC H_{1}\rightarrow
 \CC G_{1}$ and $ \CC H_{2}\rightarrow \CC G_{2}$). In
particular, the $\left(   \CC \left[  G_{1}\times G_{2}\right]
, \CC \left[  H_{1}\times H_{2}\right]  \right)  $-bimodule
$ \CC \left[  G_{1}\times G_{2}\right]  $ becomes identified with the
$\left(   \CC G_{1}\otimes \CC G_{2}, \CC H_{1}\otimes
 \CC H_{2}\right)  $-bimodule $ \CC G_{1}\otimes \CC G_{2}$.
Moreover, whenever $K_{1}$ is a $ \CC G_{1}$-module and $K_{2}$ is a
$ \CC G_{2}$-module, the $ \CC \left[  G_{1}\times G_{2}\right]
$-module $K_{1}\otimes K_{2}$ becomes identified with the $ \CC %
G_{1}\otimes \CC G_{2}$-module $K_{1}\otimes K_{2}$. The same statement
holds with $G_{1}$ and $G_{2}$ replaced by $H_{1}$ and $H_{2}$.

We are identifying $ \CC \left[  G_{1}\times G_{2}\right]  $ with
$ \CC G_{1}\otimes \CC G_{2}$ and identifying $ \CC \left[
H_{1}\times H_{2}\right]  $ with $ \CC H_{1}\otimes \CC H_{2}$.
Thus,
\begin{align*}
& \left(  \text{the left } \CC \left[  G_{1}\times G_{2}\right]
\text{-module } \CC \left[  G_{1}\times G_{2}\right]  \otimes
_{ \CC \left[  H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes
U_{2}\right)  \right)  \\
& =\left(  \text{the left } \CC G_{1}\otimes \CC G_{2}\text{-module
}\left(   \CC G_{1}\otimes \CC G_{2}\right)  \otimes_{ \CC %
H_{1}\otimes \CC H_{2}}\left(  U_{1}\otimes U_{2}\right)  \right)
\end{align*}
(since our identifications are such that the $\left(   \CC \left[
G_{1}\times G_{2}\right]  , \CC \left[  H_{1}\times H_{2}\right]
\right)  $-bimodule $ \CC \left[  G_{1}\times G_{2}\right]  $ becomes
identified with the $\left(   \CC G_{1}\otimes \CC G_{2}%
, \CC H_{1}\otimes \CC H_{2}\right)  $-bimodule $ \CC %
G_{1}\otimes \CC G_{2}$, and the left $ \CC \left[  G_{1}\times
G_{2}\right]  $-module $U_{1}\otimes U_{2}$ becomes identified with the left
$ \CC G_{1}\otimes \CC G_{2}$-module $U_{1}\otimes U_{2}$).

But Proposition \ref{prop.sol.ind.tensor.general} (applied to $\kk %
= \CC $, $A= \CC G_{1}$, $B= \CC H_{1}$, $C= \CC $,
$A^{\prime}= \CC G_{2}$, $B^{\prime}= \CC H_{2}$, $C^{\prime
}= \CC $, $P= \CC G_{1}$, $Q=U_{1}$, $P^{\prime}= \CC G_{2}$
and $Q^{\prime}=U_{2}$) yields%
\[
\left(   \CC G_{1}\otimes \CC G_{2}\right)  \otimes_{ \CC %
H_{1}\otimes \CC H_{2}}\left(  U_{1}\otimes U_{2}\right)  \cong\left(
 \CC G_{1}\otimes_{ \CC H_{1}}U_{1}\right)  \otimes\left(
 \CC G_{2}\otimes_{ \CC H_{2}}U_{2}\right)
\]
as $\left(   \CC G_{1}\otimes \CC G_{2}, \CC \otimes
 \CC \right)  $-bimodules, hence also as left $ \CC G_{1}%
\otimes \CC G_{2}$-modules. In other words,%
\begin{align*}
& \left(  \text{the left } \CC G_{1}\otimes \CC G_{2}\text{-module
}\left(   \CC G_{1}\otimes \CC G_{2}\right)  \otimes_{ \CC %
H_{1}\otimes \CC H_{2}}\left(  U_{1}\otimes U_{2}\right)  \right)  \\
& \cong\left(  \text{the left } \CC G_{1}\otimes \CC G_{2}%
\text{-module }\left(   \CC G_{1}\otimes_{ \CC H_{1}}U_{1}\right)
\otimes\left(   \CC G_{2}\otimes_{ \CC H_{2}}U_{2}\right)  \right)
.
\end{align*}


Now, by the definition of $\Ind_{H_{1}\times H_{2}}^{G_{1}\times
G_{2}}\left(  U_{1}\otimes U_{2}\right)  $, we have
$\Ind_{H_{1}\times H_{2}}^{G_{1}\times G_{2}}
\left(  U_{1}\otimes U_{2}\right)
= \CC \left[  G_{1}\times G_{2}\right]  \otimes_{ \CC \left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  $. Thus,%
\begin{align*}
& \left(  \text{the left } \CC \left[  G_{1}\times G_{2}\right]
\text{-module }\Ind_{H_{1}\times H_{2}}^{G_{1}\times G_{2}%
}\left(  U_{1}\otimes U_{2}\right)  \right)  \\
& =\left(  \text{the left } \CC \left[  G_{1}\times G_{2}\right]
\text{-module } \CC \left[  G_{1}\times G_{2}\right]  \otimes
_{ \CC \left[  H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes
U_{2}\right)  \right)  \\
& =\left(  \text{the left } \CC G_{1}\otimes \CC G_{2}\text{-module
}\left(   \CC G_{1}\otimes \CC G_{2}\right)  \otimes_{ \CC %
H_{1}\otimes \CC H_{2}}\left(  U_{1}\otimes U_{2}\right)  \right)  \\
& \cong\left(  \text{the left } \CC G_{1}\otimes \CC G_{2}%
\text{-module }\underbrace{\left(   \CC G_{1}\otimes_{ \CC H_{1}%
}U_{1}\right)  }_{\substack{=\Ind_{H_{1}}^{G_{1}}%
U_{1}\\\text{(since }\Ind_{H_{1}}^{G_{1}}U_{1}= \CC %
G_{1}\otimes_{ \CC H_{1}}U_{1}\text{)}}}\otimes\underbrace{\left(
 \CC G_{2}\otimes_{ \CC H_{2}}U_{2}\right)  }%
_{\substack{=\Ind_{H_{2}}^{G_{2}}U_{2}\\\text{(since
}\Ind_{H_{2}}^{G_{2}}U_{2}= \CC G_{2}\otimes_{ \CC %
H_{2}}U_{2}\text{)}}}\right)  \\
& \cong\left(  \text{the left } \CC G_{1}\otimes \CC G_{2}%
\text{-module }\left(  \Ind_{H_{1}}^{G_{1}}%
U_{1}\right)  \otimes\left(  \Ind_{H_{2}}^{G_{2}}U_{2}\right)
\right)  \\
& =\left(  \text{the left } \CC \left[  G_{1}\times G_{2}\right]
\text{-module }\left(  \Ind_{H_{1}}^{G_{1}}%
U_{1}\right)  \otimes\left(  \Ind_{H_{2}}^{G_{2}}U_{2}\right)
\right)
\end{align*}
(because the left $ \CC G_{1}\otimes \CC G_{2}$-module $\left(
\Ind_{H_{1}}^{G_{1}}U_{1}\right)  \otimes\left(
\Ind_{H_{2}}^{G_{2}}U_{2}\right)  $ is identified with the left
$ \CC \left[  G_{1}\times G_{2}\right]  $-module $\left(
\Ind_{H_{1}}^{G_{1}}U_{1}\right)  \otimes\left(
\Ind_{H_{2}}^{G_{2}}U_{2}\right)  $). In other words,%
\[
\Ind_{H_{1}\times H_{2}}^{G_{1}\times G_{2}}\left(  U_{1}\otimes
U_{2}\right)  \cong\left(  \Ind_{H_{1}}^{G_{1}}U_{1}\right)
\otimes\left(  \Ind_{H_{2}}^{G_{2}}U_{2}\right)
\]
as $ \CC \left[  G_{1}\times G_{2}\right]  $-modules. This solves
Exercise~\ref{exe.ind.tensor}.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.induction.ind-as-hom}}

\textit{Solution to Exercise~\ref{exe.induction.ind-as-hom}.} In
the following, we will write $g$ for the element $t_g$ of $\CC G$
whenever $g$ is an element of $G$. This is a relatively common abuse
of notation, and it is harmless because the map $G \to \CC G, \ 
g \mapsto t_g$ is an injective homomorphism of multiplicative monoids
(so $t_{gh} = t_g t_h$ and $t_1 = 1$, which means that we won't run
into ambiguities denoting $t_g$ by $g$)
and because every $\CC G$-module $M$, every $m \in M$ and every $g \in G$
satisfy $gm = t_g m$.

Recall that $\Ind^G_H U$ is defined as the $\CC G$-module
$\CC G \otimes_{\CC H} U$, where $\CC G$ is regarded as a
$\left(\CC G, \CC H\right)$-bimodule.

The $\CC$-vector space $\CC G$ is endowed with both a
$\left(\CC G, \CC H\right)$-bimodule structure (this is the structure
used in the definition of $\Ind^G_H U$) and a
$\left(\CC H, \CC G\right)$-bimodule structure (this is the structure
used in the statement of Exercise~\ref{exe.induction.ind-as-hom}).
Thus, $\CC G$ has a left $\CC G$-module structure, a right
$\CC H$-module structure, a left $\CC H$-module structure, and a right
$\CC G$-module structure. All four of these structures are simply
given by multiplication inside $\CC G$ (since $\CC H$ is a
$\CC$-subalgebra of $\CC G$). Therefore, notations like $xy$ with
$x$ and $y$ being two elements of $\CC G$ or $\CC H$ will never be
ambiguous: While they might be interpreted in different
ways, all of the possible interpretations will produce identical results.

We recall that $\Hom_{\CC H}\left(\CC G, U\right)$ is the left
$\CC G$-module consisting of all left $\CC H$-module homomorphisms
from $\CC G$ to $U$. This uses only the
$\left(\CC H, \CC G\right)$-bimodule structure on $\CC G$ that was
used in the statement of Exercise~\ref{exe.induction.ind-as-hom}
(but not the $\left(\CC G, \CC H\right)$-bimodule structure that
was used in the definition of $\Ind^G_H U$).

Let $J$
be a system of distinct representatives for the right $H$-cosets in
$G$. Then, $G=\bigsqcup_{j\in J}Hj$.

We now define a map $\alpha:\Hom_{\CC H}\left(
{\CC}G,U\right)  \rightarrow\CC G\otimes_{\CC H}U$ by
setting
\[
\alpha\left(  f\right)  =\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  \qquad\text{ for all }f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  .
\]
This $\alpha$ is a map $\Hom_{\CC H}\left( \CC G,U\right) 
 \rightarrow\Ind_{H}^{G}U$ (since
$\Ind_{H}^{G}U={\CC}G\otimes_{\CC H}U$). It is easy to see that this
map $\alpha$ does not depend on the choice of $J$. (In fact, if
$j_1$ and $j_2$ are two elements of $G$ lying in the same right
$H$-coset,
and if $f \in \Hom_{\CC H} \left(\CC G, U\right)$, then it is easy
to see that
$j_1^{-1} \otimes_{\CC H} f\left(j_1\right)
= j_2^{-1} \otimes_{\CC H} f\left(j_2\right)$.) In other words, if
$J'$ is any system of distinct representatives for the right
$H$-cosets in $G$ (which may and may not be equal to $J$), then
\begin{equation}
\label{sol.induction.ind-as-hom.othersystem}
\alpha\left(  f\right)  =\sum_{j\in J'}j^{-1}\otimes_{\CC H}f\left(
j\right)  \qquad\text{ for all }f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  .
\end{equation}
Notice that if $g \in G$ is arbitrary, then $Jg = \left\{ jg \mid
j \in J\right\}$ is also a system of distinct representatives for the
right $H$-cosets in $G$.

We will show that $\alpha$ is a $\CC G$-module isomorphism.

First, let us prove that $\alpha$ is a left $\CC G$-module homomorphism. In
fact, any $f\in\Hom_{\CC H}\left( \CC G,U\right)  $ and $g\in G$
satisfy
\begin{align*}
\alpha\left(  gf\right)   &  =\sum_{j\in J}
\underbrace{j^{-1}}_{=g \left(jg\right)^{-1}}
\otimes_{\CC H}\underbrace{\left(  gf\right)  \left(  j\right)  }_{\substack{=f\left(
jg\right)  \\\text{(by the definition of }gf\text{)}}}
= \sum_{j\in J} g \left(jg\right)^{-1} \otimes_{\CC H} f\left(jg\right) \\
&  =\sum_{j\in Jg} g j^{-1} \otimes_{\CC H} f\left(j\right)
\qquad \qquad \left(\text{here, we substituted } j \text{ for } jg
\text{ in the sum}\right) \\
&= g \cdot \underbrace{\left(\sum_{j\in Jg} j^{-1} \otimes_{\CC H} f\left(j\right)\right)}
  _{\substack{=\alpha\left(f\right) \\
    \text{(by \eqref{sol.induction.ind-as-hom.othersystem}, applied to }J'=Jg\text{)}}}
= g\cdot\alpha\left(  f\right)  .
\end{align*}
The map $\alpha$ is thus a homomorphism of left $G$-sets. Since $\alpha$ is
furthermore $\CC$-linear, this yields that $\alpha$ is a
left $\CC G$-module homomorphism.

We now are going to construct an inverse for $\alpha$. This will be more cumbersome.

For every $g\in G$ and every $p\in\CC G$, we denote by $\epsilon
_{g}\left(  p\right)  $ the $g$-coordinate of $p$ with respect to the basis
$G$ of the $\CC$-vector space $\CC G$. By the definition of
``coordinate'', we have
\begin{equation}
q=\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\ \ \ \ \ \ \ \ \ \ \text{for every }q\in\CC 
G.
\label{sol.induction.ind-as-hom.epsilong.coordinates}
\end{equation}
For every $g\in G$, we have defined a map $\epsilon_{g}:\CC 
G\rightarrow\CC $ (because we have defined an element $\epsilon
_{g}\left(  p\right)  $ for every $p\in\CC G$). This map $\epsilon_{g}$
is $\CC$-linear. Here are some simple properties of this map:

\begin{itemize}
\item
For every $g\in G$ and $h\in G$, we have
\begin{equation}
\epsilon_{g}\left(  h\right)  =\delta_{g,h}.
\label{sol.induction.ind-as-hom.epsilong.g}
\end{equation}
\footnote{\textit{Proof.} Let $g \in G$ and $h \in G$. Then,
$\epsilon_{g}\left(  h\right)$ is defined as the $g$-coordinate
of $h$ with respect to the basis $G$ of the $\CC$-vector space
$\CC G$. This $g$-coordinate is precisely $1$ (since $h$ is an
element of this basis $G$). Thus,
$\epsilon_{g}\left(  h\right) = 1$, qed.}

\item We have
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in\CC G\text{ and }q\in
\CC G.\label{sol.induction.ind-as-hom.trace}
\end{equation}
\footnote{\textit{Proof.} Let $p \in \CC G$ and $q \in \CC G$.
We need to prove the equality
\eqref{sol.induction.ind-as-hom.trace}. This equality is
$\CC$-linear in each of $p$ and $q$, and thus we can WLOG
assume that both $p$ and $q$ belong to the basis $G$ of the
$\CC$-vector space $\CC G$. Assume this. Then, $pq$ and $qp$
belong to $G$ as well. Hence,
\eqref{sol.induction.ind-as-hom.epsilong.g} yields
$\epsilon_1\left(pq\right) = \delta_{1, pq}$ and
$\epsilon_1\left(qp\right) = \delta_{1, qp}$. But $p$ and
$q$ are elements of the group $G$. Hence, we have
$1 = pq$ if and only if $1 = qp$ (because both of these
statements are equivalent to $q = p^{-1}$). Therefore,
$\delta_{1, pq} = \delta_{1, qp}$, so that
$\epsilon_1\left(pq\right) = \delta_{1, pq} = \delta_{1, qp}
= \epsilon_1\left(qp\right)$. This proves
\eqref{sol.induction.ind-as-hom.trace}.}

\item Moreover,
\begin{equation}
\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(  q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\text{ and }q\in\CC 
G.
\label{sol.induction.ind-as-hom.epsi1g}
\end{equation}
\footnote{\textit{Proof.} Let $g \in G$ and $q \in \CC G$.
We need to prove the equality
\eqref{sol.induction.ind-as-hom.epsi1g}. This equality is
$\CC$-linear in $q$, and thus we can WLOG
assume that $q$ belongs to the basis $G$ of the
$\CC$-vector space $\CC G$. Assume this. Then, $g^{-1} q
\in G$ as well. Hence,
\eqref{sol.induction.ind-as-hom.epsilong.g} yields
$\epsilon_1\left(g^{-1} q\right) = \delta_{1, g^{-1} q}$ and
$\epsilon_g\left(q\right) = \delta_{g, q}$. But $g$ and
$q$ are elements of the group $G$. Hence, we have
$1 = g^{-1} q$ if and only if $g = q$. Therefore,
$\delta_{1, g^{-1} q} = \delta_{g, q}$, so that
$\epsilon_1\left(g^{-1} q\right) = \delta_{1, g^{-1} q}
= \delta_{g, q} = \epsilon_g\left(q\right)$. This proves
\eqref{sol.induction.ind-as-hom.epsi1g}.}
\end{itemize}

Now, fix $q\in\CC G$ and $u\in U$. We let $f_{q,u}$ be the map
$\CC G\rightarrow U$ defined by
\begin{equation}
f_{q,u}\left(  p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpq\right)
h^{-1}u\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\CC G.
\label{sol.induction.ind-as-hom.fqu}
\end{equation}
It is obvious that this map $f_{q,u}$ is $\CC$-linear. We will show
that $f_{q,u}$ is a left $\CC H$-module homomorphism.

The map $f_{q,u}$ is a homomorphism of left $H$-sets\footnote{\textit{Proof.} In
fact, for every $h^{\prime}\in H$ and every $p\in\CC G$, we have
\begin{align*}
f_{q,u}\left(  h^{\prime}p\right)   &  =\sum\limits_{h\in H}\epsilon
_{1}\left(  hh^{\prime}pq\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }f_{q,u}\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  h\underbrace{\left(  h^{\prime
}\right)  ^{-1}h^{\prime}}_{=1}pq\right)  \underbrace{\left(  h\left(
h^{\prime}\right)  ^{-1}\right)  ^{-1}}_{=\left(  \left(  h^{\prime}\right)
^{-1}\right)  ^{-1}h^{-1}=h^{\prime}h^{-1}}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we have substituted }h\left(  h^{\prime}\right)  ^{-1}\text{ for
}h\text{ in the sum,}\\
\text{because the map }H\rightarrow H,\ h\mapsto h\left(  h^{\prime}\right)
^{-1}\text{ is a bijection}\\
\text{(since }H\text{ is a group and since }h^{\prime}\in H\text{)}
\end{array}
\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpq\right)  h^{\prime}
h^{-1}u=h^{\prime}\cdot\underbrace{\sum\limits_{h\in H}\epsilon_{1}\left(
hpq\right)  h^{-1}u}_{\substack{=f_{q,u}\left(  p\right)  \\\text{(by
\eqref{sol.induction.ind-as-hom.fqu})}}}=h^{\prime}\cdot f_{q,u}\left(
p\right)  .
\end{align*}
In other words, $f_{q,u}$ is a homomorphism of left $H$-sets, qed.}. Since
$f_{q,u}$ is furthermore $\CC$-linear, this yields that $f_{q,u}$ is a
left $\CC H$-module homomorphism. Hence,
$f_{q,u}\in\Hom_{\CC H}\left(  {\CC}G,U\right)  $.

Now, forget that we fixed $q$ and $u$. We thus have defined a map
$f_{q,u} \in \Hom_{\CC H}\left( \CC G,U\right)$ for
every $q \in \CC G$ and $u\in U$. It is easy to see that this map
$f_{q,u}$ depends $\CC$-linearly on each of $q$ and $u$. Now, define a
map $\widetilde{\beta} : \CC G\times U\rightarrow
\Hom_{\CC H}\left( \CC G,U\right)  $ by
\[
\widetilde{\beta}\left(  q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ 
\text{for every }\left(  q,u\right)  \in\CC G\times U.
\]
Then, $\widetilde{\beta}$ is a $\CC$-bilinear map (because
$\widetilde{\beta}\left(  q,u\right) = f_{q,u}$ depends $\CC$-linearly
on each of $q$ and $u$). We are now going to prove that the map
$\widetilde{\beta}$ is $\CC H$-bilinear with respect to the right
$\CC H$-module structure on $\CC G$ and the left $\CC H$-module
structure on $U$.

In fact, every $h^{\prime}\in H$, $q \in \CC G$ and $u \in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta}\left(
qh^{\prime},u\right)  .
\label{sol.induction.ind-as-hom.Rg.9}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.induction.ind-as-hom.Rg.9}:} Let
$h^{\prime}\in H$, $q \in \CC G$ and $u \in U$.
Then, the map $H\rightarrow H,\ h\mapsto h^{\prime}h$ is a
bijection (since $H$ is a group). Now, let $p\in\CC G$. The definition
of $\widetilde{\beta}$ yields $\widetilde{\beta}\left(  q,h^{\prime}u\right)
=f_{q,h^{\prime}u}$. Hence,
\begin{align}
\left(  \underbrace{\widetilde{\beta}\left(  q,h^{\prime}u\right)
}_{=f_{q,h^{\prime}u}}\right)  \left(  p\right)   &  =f_{q,h^{\prime}u}\left(
p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpq\right)  h^{-1}
h^{\prime}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}f_{q,h^{\prime}u}\right) \nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  h^{\prime}hpq\right)
\underbrace{\left(  h^{\prime}h\right)  ^{-1}}_{=h^{-1}\left(  h^{\prime
}\right)  ^{-1}}h^{\prime}u\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }h^{\prime}h\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{\prime}h\text{ is a
bijection}
\end{array}
\right) \nonumber\\
&  =\sum\limits_{h\in H}\underbrace{\epsilon_{1}\left(  h^{\prime}hpq\right)
}_{\substack{=\epsilon_{1}\left(  hpqh^{\prime}\right)  \\\text{(by
\eqref{sol.induction.ind-as-hom.trace}, applied to}\\h^{\prime}\text{ and
}hpq\text{ instead of }p\text{ and }q\text{)}}}h^{-1}\underbrace{\left(
h^{\prime}\right)  ^{-1}h^{\prime}}_{=1}u\nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpqh^{\prime}\right)  h^{-1}u.
\label{sol.induction.ind-as-hom.Rg.9.pf.1}
\end{align}
But the definition of $\widetilde{\beta}$ also yields $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =f_{qh^{\prime},u}$. Hence,
\begin{align*}
\left(  \widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(
p\right)   &  =f_{qh^{\prime},u}\left(  p\right)  =\sum\limits_{h\in
H}\epsilon_{1}\left(  hpqh^{\prime}\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }f_{qh^{\prime},u}\right) \\
&  =\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)  \right)  \left(
p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.induction.ind-as-hom.Rg.9.pf.1}}\right)  .
\end{align*}
\par
Now, forget that we fixed $p$. We have thus proven that $\left(
\widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(  p\right)
=\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)  \right)  \left(
p\right)  $ for every $p\in\CC G$. In other words, $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =\widetilde{\beta}\left(  q,h^{\prime}u\right)
$, so that $\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta
}\left(  qh^{\prime},u\right)  $. This proves
\eqref{sol.induction.ind-as-hom.Rg.9}.} As a consequence of this,
every $r\in\CC H$, $q \in \CC G$ and $u \in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,ru\right)  =\widetilde{\beta}\left(  qr,u\right)  
\label{sol.induction.ind-as-hom.Rg.10}
\end{equation}
(by $\CC$-bilinearity of $\widetilde{\beta}$). In other words, the map
$\widetilde{\beta}$ is $\CC H$-bilinear with respect to the right
$\CC H$-module structure on $\CC G$ and the left $\CC H$-module
structure on $U$. Hence, by the universal property of the tensor
product, we conclude that there exists a unique $\CC$-linear map
$\beta:\CC G\otimes_{\CC H}U
\rightarrow \Hom_{\CC H}\left( \CC G,U \right)$ such that every
$\left( q,u\right)  \in\CC G\times U$ satisfies
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  .
\label{sol.induction.ind-as-hom.beta}
\end{equation}
Consider this map $\beta$. Clearly, every $q\in\CC G$ and $u\in U$
satisfy
\[
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{\beta}\right)  .
\]
Hence, every $q\in\CC G$, $u\in U$ and $p\in\CC G$ satisfy
\begin{equation}
\left(  \underbrace{\beta\left(  q\otimes_{\CC H}u\right)  }_{=f_{q,u}}
\right)  \left(  p\right)  =f_{q,u}\left(  p\right)  =\sum\limits_{h\in
H}\epsilon_{1}\left(  hpq\right)  h^{-1}u
\label{sol.induction.ind-as-hom.beta.explicit}
\end{equation}
(by \eqref{sol.induction.ind-as-hom.fqu}).

We shall now show that the maps $\alpha$ and $\beta$ are mutually inverse. To
do so, we will show that $\alpha\circ\beta=\id$ and $\beta
\circ\alpha=\id$.

Let us first show that $\alpha\circ\beta=\id$. In fact, every
$q\in\CC G$ and $u\in U$ satisfy
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)   &
=\alpha\left(  \beta\left(  q\otimes_{\CC H}u\right)  \right)
=\sum_{j\in J}j^{-1}\otimes_{\CC H}\underbrace{\left(  \beta\left(
q\otimes_{\CC H}u\right)  \right)  \left(  j\right)  }_{\substack{=\sum
\limits_{h\in H}\epsilon_{1}\left(  hjq\right)  h^{-1}u\\\text{(by
\eqref{sol.induction.ind-as-hom.beta.explicit}, applied to }p=j\text{)}}}\\
& \qquad \qquad \left(\text{by the definition of } \alpha\right) \\
&  =\sum_{j\in J}j^{-1}\otimes_{\CC H}\left(  \sum\limits_{h\in
H}\epsilon_{1}\left(  hjq\right)  h^{-1}u\right)  =\sum_{j\in J}
\sum\limits_{h\in H}\epsilon_{1}\left(  hjq\right)  \underbrace{j^{-1}
\otimes_{\CC H}h^{-1}u}_{\substack{=j^{-1}h^{-1}\otimes_{\CC H}
u\\\text{(since }h^{-1}\in H\text{ (since }h\in H\text{))}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  hjq\right)
\underbrace{j^{-1}h^{-1}}_{=\left(  hj\right)  ^{-1}}\otimes_{\CC H}
u=\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  hjq\right)  \left(
hj\right)  ^{-1}\otimes_{\CC H}u\\
&  =\underbrace{\sum_{j\in J}\sum\limits_{g\in Hj}}_{\substack{=\sum
\limits_{g\in G}\\\text{(since }\bigsqcup_{j\in J}Hj=G\text{)}}}\epsilon
_{1}\left(  gq\right)  g^{-1}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }g\text{ for }hj\text{ in the second sum, since the
map}\\
H\rightarrow Hj,\ h\mapsto hj\text{ is a bijection (because }G\text{ is a
group)}
\end{array}
\right) \\
&  =\sum\limits_{g\in G}\epsilon_{1}\left(  gq\right)  g^{-1}\otimes
_{\CC H}u=\sum\limits_{g\in G}\underbrace{\epsilon_{1}\left(
g^{-1}q\right)  }_{\substack{=\epsilon_{g}\left(  q\right)  \\\text{(by
\eqref{sol.induction.ind-as-hom.epsi1g})}}}\underbrace{\left(  g^{-1}\right)
^{-1}}_{=g}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}
\end{array}
\right) \\
&  =\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)  g\otimes_{\CC H}
u=\underbrace{\left(  \sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\right)  }_{\substack{=q\\\text{(by
\eqref{sol.induction.ind-as-hom.epsilong.coordinates})}}}
\otimes_{\CC H}u\\
&  =q\otimes_{\CC H}u=\id \left(  q\otimes_{\CC H}u
\right)  .
\end{align*}
Thus, the two maps $\alpha\circ\beta$ and $\id$ are equal
on every pure tensor. Since these two maps are $\CC$-linear, this yields
that $\alpha\circ\beta=\id$.

Next, we are going to show that $\beta\circ\alpha=\id$.

Let $f\in\Hom_{\CC H}\left(  {\CC}G,U\right)  $.
Let $p\in\CC G$. The map $f$ is left $\CC H$-linear (since
$f\in\Hom_{\CC H}\left(  {\CC}G,U\right)  $),
hence $\CC$-linear. We have $\alpha\left(  f\right)  =\sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  $ (by the definition of
$\alpha$). Applying the map $\beta$ to this equality, we obtain
\[
\beta\left(  \alpha\left(  f\right)  \right)  =\beta\left(  \sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)  =\sum_{j\in J}
\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)
\]
(since $\beta$ is a $\CC$-linear map). Thus,
\begin{align*}
& \underbrace{\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
}_{=\sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)
\right)  }\left(  p\right)  \\
& =\left(  \sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  =\sum_{j\in J}
\underbrace{\left(  \beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  }_{\substack{=\sum\limits_{h\in
H}\epsilon_{1}\left(  hpj^{-1}\right)  h^{-1}f\left(  j\right)  \\\text{(by
\eqref{sol.induction.ind-as-hom.beta.explicit}, applied to}\\q=j^{-1}\text{
and }u=f\left(  j\right)  \text{)}}}\\
& =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  hpj^{-1}\right)
h^{-1}f\left(  j\right)  =\sum_{j\in J}\sum\limits_{h\in H}
\underbrace{\epsilon_{1}\left(  h^{-1}pj^{-1}\right)  }_{\substack{=\epsilon
_{1}\left(  j^{-1}h^{-1}p\right)  \\\text{(by
\eqref{sol.induction.ind-as-hom.trace}, applied to}\\h^{-1}p\text{ and }
j^{-1}\text{ instead of }p\text{ and }q\text{)}}}\underbrace{\left(
h^{-1}\right)  ^{-1}}_{=h}f\left(  j\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }h^{-1}\text{ for }h\text{ in the second sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{-1}\text{ is a bijection}\\
\text{(because }H\text{ is a group)}
\end{array}
\right)  \\
& =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{j^{-1}
h^{-1}}_{=\left(  hj\right)  ^{-1}}p\right)  \underbrace{hf\left(  j\right)
}_{\substack{=f\left(  hj\right)  \\\text{(since }f\text{ is left }\CC 
H\text{-linear and since }h\in H\subset\CC H\text{)}}}\\
& =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \left(  hj\right)
^{-1}p\right)  f\left(  hj\right)  =\underbrace{\sum_{j\in J}\sum\limits_{g\in
Hj}}_{\substack{=\sum\limits_{g\in G}\\\text{(since }\bigsqcup_{j\in
J}Hj=G\text{)}}}\epsilon_{1}\left(  g^{-1}p\right)  f\left(  g\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }g\text{ for }hj\text{ in the second sum, since the
map}\\
H\rightarrow Hj,\ h\mapsto hj\text{ is a bijection (because }G\text{ is a
group)}
\end{array}
\right)  \\
& =\sum\limits_{g\in G}\underbrace{\epsilon_{1}\left(  g^{-1}p\right)
}_{\substack{=\epsilon_{g}\left(  p\right)  \\\text{(by
\eqref{sol.induction.ind-as-hom.epsi1g}, applied to }q=p\text{)}}}f\left(
g\right)  =\sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  f\left(
g\right)  \\
& =f\left(  \underbrace{\sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  g}
_{\substack{=p \\ \text{(by \eqref{sol.induction.ind-as-hom.epsilong.coordinates})}}}
\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is }\CC \text{-linear}\right) \\
& = f\left(p\right)  .
\end{align*}

Now, forget that we fixed $p$. We thus have proven that $\left(  \beta\left(
\alpha\left(  f\right)  \right)  \right)  \left(  p\right)  =f\left(
p\right)  $ for every $p\in\CC G$. In other words, $\beta\left(
\alpha\left(  f\right)  \right)  =f$. Hence, $\left(  \beta\circ\alpha\right)
\left(  f\right)  =\beta\left(  \alpha\left(  f\right)  \right)
=f=\id \left(  f\right)$.

Since we have shown this for every $f$, we can thus conclude that
$\beta\circ\alpha=\id$.
Combined with $\alpha\circ\beta=\id$, this yields that the maps
$\alpha$ and $\beta$ are mutually inverse. Hence, the map $\alpha$ is invertible,
and thus a left $\CC G$-module isomorphism (as we already know that $\alpha$ is
a left $\CC G$-module homomorphism). Hence,
$\Hom_{\CC H}\left( \CC G,U\right)
\cong\CC G\otimes_{\CC H}U=\Ind_{H}^{G}U$
as left $\CC G$-modules. This solves Exercise~\ref{exe.induction.ind-as-hom}.

\textit{Remark:} In our solution, we explicitly constructed a
$\CC G$-module isomorphism $\alpha:\Hom_{\CC H}\left(
\CC G,U\right)  \rightarrow\Ind_{H}^{G}U$.
This isomorphism is functorial with respect to $U$. It is also independent on
the choice of $J$ (this is not immediately clear from its definition, but it
can be shown very easily, by observing that the tensor $j^{-1}\otimes
_{\CC H}f\left( j \right)$ for $j\in G$ depends only on the coset
$Hj$ and not on $j$ itself). One might ask whether this isomorphism is
functorial in $G$ and $H$; but to make sense of this question, one has to
define the category with respect to which this functoriality is to be
understood. I don't know a good answer.

It is also worth noting that in our solution, $\CC$ could be replaced
by any commutative ring. We used neither that $\CC$ is a field, nor
that $\CC$ has characteristic $0$.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.induction.ind-as-hom}}

\textit{Solution to Exercise~\ref{exe.induction.ind-as-hom}.} In
the following, we will write $g$ for the element $t_g$ of $\CC G$
whenever $g$ is an element of $G$. This is a relatively common abuse
of notation, and it is harmless because the map $G \to \CC G, \ 
g \mapsto t_g$ is an injective homomorphism of multiplicative monoids
(so $t_{gh} = t_g t_h$ and $t_1 = 1$, which means that we won't run
into ambiguities denoting $t_g$ by $g$) and because every
$\CC G$-module $M$, every $m \in M$ and every $g \in G$
satisfy $gm = t_g m$.

Recall that $\Ind^G_H U$ is defined as the $\CC G$-module
$\CC G \otimes_{\CC H} U$, where $\CC G$ is regarded as a
$\left(\CC G, \CC H\right)$-bimodule.

The $\CC$-vector space $\CC G$ is endowed with both a
$\left(\CC G, \CC H\right)$-bimodule structure (this is the structure
used in the definition of $\Ind^G_H U$) and a
$\left(\CC H, \CC G\right)$-bimodule structure (this is the structure
used in the statement of Exercise~\ref{exe.induction.ind-as-hom}).
Thus, $\CC G$ has a left $\CC G$-module structure, a right
$\CC H$-module structure, a left $\CC H$-module structure, and a right
$\CC G$-module structure. All four of these structures are simply
given by multiplication inside $\CC G$ (since $\CC H$ is a
$\CC$-subalgebra of $\CC G$). Therefore, notations like $xy$ with
$x$ and $y$ being two elements of $\CC G$ or $\CC H$ will never be
ambiguous: While they might be interpreted in different
ways\footnote{For example, if $x \in \CC G$ and $y \in \CC H$, then
we can read the expression $xy$ in three different ways:
\begin{itemize}
\item as the product of $x$ and $y$ inside $\CC G$;
\item as the result of acting by $y$ on $x$ from the right using the
right $\CC H$-module structure on $\CC G$;
\item as the result of acting by $y$ on $x$ from the right using the
right $\CC G$-module structure on $\CC G$;
\item as the result of acting by $x$ on $y$ from the left using the
left $\CC G$-module structure on $\CC G$.
\end{itemize}}, all of the possible
interpretations will produce identical results. Also, the
right $\CC H$-module structure on $\CC G$ and the left $\CC H$-module
structure on $\CC G$ satisfy the equality
\[
\left(xy\right) z = x \left(yz\right)
\]
whenever each of $x$, $y$ and $z$ is an element of either $\CC G$
or $\CC H$. In fact, this equality follows from the associative law
in $\CC G$ (since the right $\CC H$-module structure on $\CC G$ and
the left $\CC H$-module structure on $\CC G$ are both given by
multiplication inside $\CC G$). This equality will be used tacitly.

We recall that $\Hom_{\CC H}\left(\CC G, U\right)$ is the left
$\CC G$-module consisting of all left $\CC H$-module homomorphisms
from $\CC G$ to $U$. This uses only the
$\left(\CC H, \CC G\right)$-bimodule structure on $\CC G$ that was
used in the statement of Exercise~\ref{exe.induction.ind-as-hom}
(but not the $\left(\CC G, \CC H\right)$-bimodule structure that
was used in the definition of $\Ind^G_H U$).

Let $J$
be a system of distinct representatives for the right $H$-cosets in
$G$. Then, $G=\bigsqcup_{j\in J}Hj$.

For every $g\in G$, define a map $\mathfrak{R}_{g}:J\rightarrow J$ as follows:
Let $i\in J$. Then, $ig\in G=\bigsqcup_{j\in J}Hj$. Thus, there exists a
unique $j\in J$ such that $ig\in Hj$. Define $\mathfrak{R}_{g}\left(
i\right)  $ to be this $j$. Hence, we have defined $\mathfrak{R}_{g}\left(
i\right)  $ for every $i\in J$. Thus, we have defined a map
$\mathfrak{R}_{g}:J\rightarrow J$.

For every $g\in G$ and $i\in J$, we have
\begin{equation}
ig\in H\cdot\mathfrak{R}_{g}\left(  i\right)
\label{sol.induction.ind-as-hom.Rg.def}
\end{equation}
(because $\mathfrak{R}_{g}\left(  i\right)  $ is defined as the $j\in J$
satisfying $ig\in Hj$).

It is easy to see that for every $g\in G$, the map
$\mathfrak{R}_{g}:J\rightarrow J$ is a
bijection.\footnote{\textit{Proof.} Fix $g\in G$.
\par
The sets $Hj$ for all $j\in J$ are disjoint (since $\bigsqcup_{j\in
J}Hj$ is well-defined). In other words, if $i$ and $i^{\prime}$ are two
elements of $J$ such that $Hi$ and $Hi^{\prime}$ are not disjoint, then
\begin{equation}
i=i^{\prime}. \label{sol.induction.ind-as-hom-Lg.1}
\end{equation}
\par
Let us now prove that $\mathfrak{R}_{g}$ is injective.
\par
Let $i$ and $i^{\prime}$ be two elements of $J$. Assume that
$\mathfrak{R}_{g}\left(  i\right) 
 =\mathfrak{R}_{g}\left(  i^{\prime}\right)  $. We will
show that $i=i^{\prime}$.
\par
From (\ref{sol.induction.ind-as-hom.Rg.def}), we have $ig\in H\cdot
\mathfrak{R}_{g}\left(  i\right)  $. Hence, there exists some $h\in H$ such
that $ig=h\cdot\mathfrak{R}_{g}\left(  i\right)  $. Consider this $h$. We have
\[
H\underbrace{ig}_{=h\cdot\mathfrak{R}_{g}\left(  i\right)  }
=\underbrace{Hh}_{\substack{=H\\\text{(since }h\in H\text{)}}}
\cdot\mathfrak{R}_{g}\left( i\right) 
=H\cdot\mathfrak{R}_{g}\left(  i\right)  .
\]
The same argument (but for $i^{\prime}$ instead of $i$) yields $Hi^{\prime
}g=H\cdot\mathfrak{R}_{g}\left(  i^{\prime}\right)  $. Hence, $Hig=H\cdot
\underbrace{\mathfrak{R}_{g}\left(  i\right)  }_{=\mathfrak{R}_{g}\left(
i^{\prime}\right)  }=H\cdot\mathfrak{R}_{g}\left(  i^{\prime}\right)
=Hi^{\prime}g$. Thus, $\underbrace{Hig}_{=Hi^{\prime}g}g^{-1}=Hi^{\prime
}gg^{-1}=Hi^{\prime}$, so that $Hi^{\prime}=Higg^{-1}=Hi$. Thus, the sets $Hi$
and $Hi^{\prime}$ are not disjoint (because $\left(  Hi\right)  \cap
\underbrace{\left(  Hi^{\prime}\right)  }_{=Hi}=\left(  Hi\right)  \cap\left(
Hi\right)  =Hi\neq\varnothing$). Therefore, $i=i^{\prime}$ (by
(\ref{sol.induction.ind-as-hom-Lg.1})).
\par
Now, forget that we fixed $i$ and $i^{\prime}$. We thus have shown that if $i$
and $i^{\prime}$ are two elements of $J$ such that $\mathfrak{R}_{g}\left(
i\right)  =\mathfrak{R}_{g}\left(  i^{\prime}\right)  $, then $i=i^{\prime}$.
In other words, the map $\mathfrak{R}_{g}$ is injective. But the set $J$ is
finite (since it is a subset of the finite set $G$). Hence, $\mathfrak{R}_{g}$
is a map from a finite set (namely, $J$) to itself. Since we know that
$\mathfrak{R}_{g}$ is injective, this yields that $\mathfrak{R}_{g}$ is
surjective (because any injective map from a finite set to itself must be
surjective). Hence, $\mathfrak{R}_{g}$ is bijective (since $\mathfrak{R}_{g}$
is injective and surjective), that is, a bijection, qed.} We have
\begin{equation}
j^{-1}\otimes_{\CC H}f\left(  jg\right) 
=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}
 \otimes_{\CC H}f\left( \mathfrak{R}_{g}\left(  j\right)  \right) 
 \ \ \ \ \ \ \ \ \ \ \text{in } \CC G\otimes_{\CC H}U
\label{sol.induction.ind-as-hom.Rg.property}
\end{equation}
for every $g\in G$, every $j\in J$ and every
$f\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $.\ \ \ \ 
\footnote{\textit{Proof of
(\ref{sol.induction.ind-as-hom.Rg.property}):} Let $g\in G$, $j\in J$ and
$f\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $.
Applying (\ref{sol.induction.ind-as-hom.Rg.def}) to $i=j$, we obtain $jg\in
H\cdot\mathfrak{R}_{g}\left(  j\right)  $. In other words, there exists an
$h\in H$ such that $jg=h\cdot\mathfrak{R}_{g}\left(  j\right)  $. Consider
this $h$. We have
\[
j^{-1}\underbrace{jg}_{=h\cdot\mathfrak{R}_{g}\left(  j\right)  }\cdot\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}=j^{-1}h\cdot
\underbrace{\mathfrak{R}_{g}\left(  j\right) 
 \cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}}_{=1}
=j^{-1}h,
\]
hence
\[
j^{-1}h=j^{-1}jg\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}.
\]
But
\[
f\left(  \underbrace{jg}_{=h\cdot\mathfrak{R}_{g}\left(  j\right)  }\right)
=f\left(  h\cdot\mathfrak{R}_{g}\left(  j\right)  \right)  =h\cdot f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)
\]
(because $f$ is left $\CC H$-linear (since
$f \in \Hom_{\CC H}\left(\CC G, U\right)$) and because $h\in H\subset\CC H$).
Hence,
\begin{align*}
j^{-1}\otimes_{\CC H}\underbrace{f\left(  jg\right)  }_{=h\cdot f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  } 
&  =j^{-1}\otimes_{\CC H} h
\cdot f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
=\underbrace{j^{-1}h}_{=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)
\right)  ^{-1}}\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)
\right) \\
& \ \ \ \ \ \ \ \ \ \ \left(\text{here, we have moved } h \text{ past
the } \otimes_{\CC H} \text{ sign, since } h \in H \subset \CC H\right) \\
&  =g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}
\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  .
\end{align*}
This proves (\ref{sol.induction.ind-as-hom.Rg.property}).}

We now define a map $\alpha:\Hom_{\CC H}\left(
{\CC }G,U\right)  \rightarrow\CC G\otimes_{\CC H}U$ by
setting
\[
\alpha\left(  f\right)  =\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  \qquad\text{ for all }f\in\Hom_{\CC H}\left(
{\CC }G,U\right)  .
\]
This $\alpha$ is a map $\Hom_{\CC H}\left( \CC G,U\right) 
 \rightarrow\Ind_{H}^{G}U$ (since
$\Ind_{H}^{G}U={\CC }G\otimes_{\CC H}U$).

We will show that $\alpha$ is a $\CC G$-module isomorphism.

First, let us prove that $\alpha$ is a left $\CC G$-module homomorphism. In
fact, any $f\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $ and $g\in G$
satisfy
\begin{align*}
\alpha\left(  gf\right)   &  =\sum_{j\in J}j^{-1}\otimes_{\CC H}
\underbrace{\left(  gf\right)  \left(  j\right)  }_{\substack{=f\left(
jg\right)  \\\text{(by the definition of }gf\text{)}}}
\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of }
\alpha\left(gf\right)\right) \\
& =\sum_{j\in
J}\underbrace{j^{-1}\otimes_{\CC H}f\left(  jg\right)  }
_{\substack{=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
\\\text{(by (\ref{sol.induction.ind-as-hom.Rg.property}))}}}
=\sum_{j\in J}g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
=\sum_{j\in J}g\cdot j^{-1}\otimes_{\CC H}f\left(  j\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we have substituted }j\text{ for }\mathfrak{R}_{g}\left(
j\right)  \text{ in the sum, since}\\
\text{the map }\mathfrak{R}_{g}:J\rightarrow J\text{ is a bijection}
\end{array}
\right) \\
&  =g\cdot\underbrace{\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  }_{=\alpha\left(  f\right)  }=g\cdot\alpha\left(  f\right)  .
\end{align*}
The map $\alpha$ is thus a homomorphism of left $G$-sets. Since $\alpha$ is
furthermore $\CC$-linear, this yields that $\alpha$ is a left
$\CC G$-module homomorphism.

We now are going to construct an inverse for $\alpha$. This will be more cumbersome.

For every $g\in G$ and every $p\in\CC G$, we denote by $\epsilon
_{g}\left(  p\right)  $ the $g$-coordinate of $p$ with respect to the basis
$G$ of the $\CC$-vector space $\CC G$. By the definition of
``coordinate'', we have
\begin{equation}
q=\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\ \ \ \ \ \ \ \ \ \ \text{for every }q\in\CC G.
\label{sol.induction.ind-as-hom.epsilong.coordinates}
\end{equation}
For every $g\in G$, we have defined a map
$\epsilon_{g}:\CC G\rightarrow\CC$ (because we have defined an element
$\epsilon_{g}\left(  p\right)  $ for every $p\in\CC G$). This map
$\epsilon_{g}$ is $\CC$-linear.

For every $g\in G$ and $h\in G$, we know that $\epsilon_{g}\left(  h\right)  $
is the $g$-coordinate of $h$ with respect to the basis $G$ of the
$\CC$-vector space $\CC G$ (because this is how $\epsilon_{g}\left(
h\right)  $ was defined). Since this $g$-coordinate of $h$ is clearly
$\delta_{g,h}$, this rewrites as follows: For every $g\in G$ and $h\in G$, we
know that $\epsilon_{g}\left(  h\right)  $ is $\delta_{g,h}$. In other words,
for every $g\in G$ and $h\in G$, we have
\begin{equation}
\epsilon_{g}\left(  h\right)  =\delta_{g,h}.
\label{sol.induction.ind-as-hom.epsilong.g}
\end{equation}


We notice that
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in\CC G\text{ and }q\in
\CC G.\label{sol.induction.ind-as-hom.trace}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.induction.ind-as-hom.trace}):} Let
$p\in\CC G$ and $q\in\CC G$.
\par
Since $p\in\CC G$, we can write $p$ in the form $p=\sum\limits_{g\in
G}p_{g}g$ for some family $\left(  p_{g}\right)  _{g\in G}$ of elements of
$\CC$. Consider this family $\left(  p_{g}\right)  _{g\in G}$.
\par
Since $q\in\CC G$, we can write $q$ in the form $q=\sum\limits_{g\in
G}q_{g}g$ for some family $\left(  q_{g}\right)  _{g\in G}$ of elements of
$\CC$. Consider this family $\left(  q_{g}\right)  _{g\in G}$.
\par
Multiplying the equalities $p=\sum\limits_{g\in G}p_{g}g$ and $q=\sum
\limits_{g\in G}q_{g}g=\sum\limits_{h\in G}q_{h}h$ (here, we renamed the
summation index $g$ as $h$), we obtain
\[
pq=\left(  \sum\limits_{g\in G}p_{g}g\right)  \left(  \sum\limits_{h\in
G}q_{h}h\right)  =\underbrace{\sum\limits_{g\in G}\sum\limits_{h\in G}}
_{=\sum\limits_{\left(  g,h\right)  \in G\times G}}p_{g}q_{h}gh=\sum
\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}gh.
\]
Applying the map $\epsilon_{1}$ to this equality, we obtain
\begin{align}
\epsilon_{1}\left(  pq\right)   &  =\epsilon_{1}\left(  \sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}gh\right)  =\sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}\underbrace{\epsilon_{1}\left(  gh\right)
}_{\substack{=\delta_{1,gh}\\\text{(by
(\ref{sol.induction.ind-as-hom.epsilong.g}), applied to } \\
1 \text{ and } gh \text{ instead of } \\
g \text{ and } h \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
}\CC \text{-linear}\right)  \nonumber\\
&  =\sum\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}\delta
_{1,gh}=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}
p_{g}q_{h}\underbrace{\delta_{1,gh}}_{\substack{=1\\\text{(since
}1=gh\\\text{(since }gh=1\text{))}}}+\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\gh\neq1}}p_{g}q_{h}\underbrace{\delta_{1,gh}}
_{\substack{=0\\\text{(since }1\neq gh\\\text{(since }gh\neq1\text{))}}}
\nonumber\\
&  =\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}
p_{g}q_{h}1+\underbrace{\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\gh\neq1}}p_{g}q_{h}0}_{=0}=\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\gh=1}}p_{g}q_{h}1\nonumber\\
&  =\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}
p_{g}q_{h}.\label{sol.induction.ind-as-hom.Rg.6}
\end{align}
\par
However, for any $\left(  g,h\right)  \in G\times G$, we have the following
chain of equivalences:
\[
\left(  gh=1\right)  \ \Longleftrightarrow\ \left(  g=h^{-1}\right)
\ \Longleftrightarrow\ \left(  hg=1\right)  .
\]
Hence, we can replace the summation sign ``$\sum
\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}$''
by ``$\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg=1}}$''. We thus obtain
$\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}p_{g}
q_{h}=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}}
p_{g}q_{h}$. Thus, (\ref{sol.induction.ind-as-hom.Rg.6}) becomes
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\gh=1}}p_{g}q_{h}=\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\hg=1}}p_{g}q_{h}.\label{sol.induction.ind-as-hom.Rg.7}
\end{equation}
\par
Now, multiplying the equalities $q=\sum\limits_{g\in G}q_{g}g$ and
$p=\sum\limits_{g\in G}p_{g}g=\sum\limits_{h\in G}p_{h}h$ (here, we renamed
the summation index $g$ as $h$), we obtain
\begin{align*}
qp &  =\left(  \sum\limits_{g\in G}q_{g}g\right)  \left(  \sum\limits_{h\in
G}p_{h}h\right)  =\underbrace{\sum\limits_{g\in G}\sum\limits_{h\in G}}
_{=\sum\limits_{h\in G}\sum\limits_{g\in G}=\sum\limits_{\left(  h,g\right)
\in G\times G}}\underbrace{q_{g}p_{h}}_{=p_{h}g_{q}}gh\\
&  =\sum\limits_{\left(  h,g\right)  \in G\times G}p_{h}q_{g}gh=\sum
\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}hg
\end{align*}
(here, we renamed the summation index $\left(  h,g\right)  $ as $\left(
g,h\right)  $). Applying the map $\epsilon_{1}$ to this equality, we obtain
\begin{align*}
\epsilon_{1}\left(  qp\right)   &  =\epsilon_{1}\left(  \sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}hg\right)  =\sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}\underbrace{\epsilon_{1}\left(  hg\right)
}_{\substack{=\delta_{1,hg}\\\text{(by
(\ref{sol.induction.ind-as-hom.epsilong.g}), applied to } \\
1 \text{ and } hg \text{ instead of } \\
g \text{ and } h \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
}\CC \text{-linear}\right)  \\
&  =\sum\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}\delta
_{1,hg}=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}}
p_{g}q_{h}\underbrace{\delta_{1,hg}}_{\substack{=1\\\text{(since
}1=hg\\\text{(since }hg=1\text{))}}}+\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg\neq1}}p_{g}q_{h}
\underbrace{\delta_{1,hg}}_{\substack{=0\\\text{(since }1\neq hg\\
\text{(since }hg\neq1\text{))}}}\\
&  =\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}}
p_{g}q_{h}1+\underbrace{\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\hg\neq1}}p_{g}q_{h}0}_{=0}=\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg=1}}p_{g}q_{h}1
=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}}
p_{g}q_{h}=\epsilon_{1}\left(  pq\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.induction.ind-as-hom.Rg.7})}\right)  .
\end{align*}
Thus, $\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)  $, so
that (\ref{sol.induction.ind-as-hom.trace}) is proven.} Moreover,
\begin{equation}
\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(  q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\text{ and }q\in\CC G.
\label{sol.induction.ind-as-hom.epsi1g}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.induction.ind-as-hom.epsi1g}):} Let $g\in
G$ and $q\in\CC G$.
\par
Since $q\in\CC G$, we can write $q$ in the form $q=\sum\limits_{h\in
G}q_{h}h$ for some family $\left(  q_{h}\right)  _{h\in G}$ of elements of
$\CC$. Consider this family $\left(  q_{h}\right)  _{h\in G}$.
\par
We have
\begin{align}
\epsilon_{1}\left(  g^{-1}\underbrace{q}_{=\sum\limits_{h\in G}q_{h}h}\right)
&  =\epsilon_{1}\left(  g^{-1}\sum\limits_{h\in G}q_{h}h\right)  =\epsilon
_{1}\left(  \sum\limits_{h\in G}q_{h}g^{-1}h\right)  =\sum\limits_{h\in
G}q_{h}\underbrace{\epsilon_{1}\left(  g^{-1}h\right)  }_{\substack{=\delta
_{1,g^{-1}h}\\\text{(by (\ref{sol.induction.ind-as-hom.epsilong.g}), applied
to}\\1\text{ and }g^{-1}h\text{ instead of }g\text{ and }h\text{)}}}
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
}\CC \text{-linear}\right)  \nonumber\\
&  =\sum\limits_{h\in G}q_{h}\delta_{1,g^{-1}h}=\sum\limits_{\substack{h\in
G;\\1=g^{-1}h}}q_{h}\underbrace{\delta_{1,g^{-1}h}}
_{\substack{=1\\\text{(since }1=g^{-1}h\text{)}}}+\sum\limits_{\substack{h\in
G;\\1\neq g^{-1}h}}q_{h}\underbrace{\delta_{1,g^{-1}h}}
_{\substack{=0\\\text{(since }1\neq g^{-1}h\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{h\in G;\\1=g^{-1}h}}q_{h}1+\underbrace{\sum
\limits_{\substack{h\in G;\\1\neq g^{-1}h}}q_{h}0}_{=0}=\sum
\limits_{\substack{h\in G;\\1=g^{-1}h}}q_{h}1=\underbrace{\sum
\limits_{\substack{h\in G;\\1=g^{-1}h}}}_{\substack{=\sum
\limits_{\substack{h\in G;\\g=h}}\\\text{(because for every}\\h\in G\text{,
the assertions}\\\left(  1=g^{-1}h\right)  \text{ and }\left(  g=h\right)
\\\text{are equivalent)}}}q_{h}=\sum\limits_{\substack{h\in G;\\g=h}}
q_{h}.
\label{sol.induction.ind-as-hom.epsi1g.pf.1}
\end{align}
On the other hand, applying the map $\epsilon_{g}$ to the identity
$q=\sum\limits_{h\in G}q_{h}h$, we obtain
\begin{align*}
\epsilon_{g}\left(  q\right)   &  =\epsilon_{g}\left(  \sum\limits_{h\in
G}q_{h}h\right)  =\sum\limits_{h\in G}q_{h}\underbrace{\epsilon_{g}\left(
h\right)  }_{\substack{=\delta_{g,h}\\\text{(by
(\ref{sol.induction.ind-as-hom.epsilong.g}))}}}
= \sum\limits_{h\in G} q_{h}\delta_{g,h}\\
&  =\sum\limits_{\substack{h\in G;\\g=h}}q_{h}\underbrace{\delta_{g,h}}
_{\substack{=1\\\text{(since }g=h\text{)}}}+\sum\limits_{\substack{h\in
G;\\g\neq h}}q_{h}\underbrace{\delta_{g,h}}_{\substack{=0\\\text{(since }g\neq
h\text{)}}}=\sum\limits_{\substack{h\in G;\\g=h}}q_{h}1+\underbrace{\sum
\limits_{\substack{h\in G;\\g\neq h}}q_{h}0}_{=0}\\
&  =\sum\limits_{\substack{h\in G;\\g=h}}q_{h}1=\sum\limits_{\substack{h\in
G;\\g=h}}q_{h}=\epsilon_{1}\left(  g^{-1}q\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.induction.ind-as-hom.epsi1g.pf.1})}\right)  .
\end{align*}
In other words, $\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(
q\right)  $. This proves (\ref{sol.induction.ind-as-hom.epsi1g}).}

Now, fix $q\in\CC G$ and $u\in U$. We let $f_{q,u}$ be the map
$\CC G\rightarrow U$ defined by
\begin{equation}
f_{q,u}\left(  p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpq\right)
h^{-1}u\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\CC G.
\label{sol.induction.ind-as-hom.fqu}
\end{equation}
It is obvious that this map $f_{q,u}$ is $\CC$-linear. We will show
that $f_{q,u}$ is a left $\CC H$-module homomorphism.

The map $f_{q,u}$ is a homomorphism of left $H$-sets\footnote{\textit{Proof.} In
fact, for every $h^{\prime}\in H$ and every $p\in\CC G$, we have
\begin{align*}
f_{q,u}\left(  h^{\prime}p\right)   &  =\sum\limits_{h\in H}\epsilon
_{1}\left(  hh^{\prime}pq\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }f_{q,u}\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  h\underbrace{\left(  h^{\prime
}\right)  ^{-1}h^{\prime}}_{=1}pq\right)  \underbrace{\left(  h\left(
h^{\prime}\right)  ^{-1}\right)  ^{-1}}_{=\left(  \left(  h^{\prime}\right)
^{-1}\right)  ^{-1}h^{-1}=h^{\prime}h^{-1}}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we have substituted }h\left(  h^{\prime}\right)  ^{-1}\text{ for
}h\text{ in the sum,}\\
\text{because the map }H\rightarrow H,\ h\mapsto h\left(  h^{\prime}\right)
^{-1}\text{ is a bijection}\\
\text{(since }H\text{ is a group and since }h^{\prime}\in H\text{)}
\end{array}
\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpq\right)  h^{\prime}
h^{-1}u=h^{\prime}\cdot\underbrace{\sum\limits_{h\in H}\epsilon_{1}\left(
hpq\right)  h^{-1}u}_{\substack{=f_{q,u}\left(  p\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.fqu}))}}}=h^{\prime}\cdot f_{q,u}\left(
p\right)  .
\end{align*}
In other words, $f_{q,u}$ is a homomorphism of left $H$-sets, qed.}. Since
$f_{q,u}$ is furthermore $\CC$-linear, this yields that $f_{q,u}$ is a
left $\CC H$-module homomorphism. Hence,
$f_{q,u} \in \Hom_{\CC H}\left( \CC G,U\right)$.

Now, forget that we fixed $q$ and $u$. We thus have defined a map
$f_{q,u} \in \Hom_{\CC H}\left( \CC G,U\right)$ for
every $q \in \CC G$ and $u\in U$. It is easy to see that this map
$f_{q,u}$ depends $\CC$-linearly on each of $q$ and $u$. Now, define a
map $\widetilde{\beta} : \CC G\times U\rightarrow
\Hom_{\CC H}\left( \CC G,U\right)  $ by
\[
\widetilde{\beta}\left(  q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ 
\text{for every }\left(  q,u\right)  \in\CC G\times U.
\]
Then, $\widetilde{\beta}$ is a $\CC$-bilinear map (because
$\widetilde{\beta}\left(  q,u\right) = f_{q,u}$ depends $\CC$-linearly
on each of $q$ and $u$). We are now going to prove that the map
$\widetilde{\beta}$ is $\CC H$-bilinear with respect to the right
$\CC H$-module structure on $\CC G$ and the left $\CC H$-module
structure on $U$.

In fact, every $h^{\prime}\in H$, $q \in \CC G$ and $u \in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta}\left(
qh^{\prime},u\right)  .
\label{sol.induction.ind-as-hom.Rg.9}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.induction.ind-as-hom.Rg.9}):} Let
$h^{\prime}\in H$, $q \in \CC G$ and $u \in U$. Then, the map
$H\rightarrow H,\ h\mapsto h^{\prime}h$ is a
bijection (since $H$ is a group). Now, let $p\in\CC G$. The definition
of $\widetilde{\beta}$ yields $\widetilde{\beta}\left(  q,h^{\prime}u\right)
=f_{q,h^{\prime}u}$. Hence,
\begin{align}
\left(  \underbrace{\widetilde{\beta}\left(  q,h^{\prime}u\right)
}_{=f_{q,h^{\prime}u}}\right)  \left(  p\right)   &  =f_{q,h^{\prime}u}\left(
p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpq\right)  h^{-1}
h^{\prime}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}f_{q,h^{\prime}u}\right) \nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  h^{\prime}hpq\right)
\underbrace{\left(  h^{\prime}h\right)  ^{-1}}_{=h^{-1}\left(  h^{\prime
}\right)  ^{-1}}h^{\prime}u\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }h^{\prime}h\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{\prime}h\text{ is a
bijection}
\end{array}
\right) \nonumber\\
&  =\sum\limits_{h\in H}\underbrace{\epsilon_{1}\left(  h^{\prime}hpq\right)
}_{\substack{=\epsilon_{1}\left(  hpqh^{\prime}\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.trace}), applied to}\\h^{\prime}\text{ and
}hpq\text{ instead of }p\text{ and }q\text{)}}}h^{-1}\underbrace{\left(
h^{\prime}\right)  ^{-1}h^{\prime}}_{=1}u\nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  hpqh^{\prime}\right)  h^{-1}u.
\label{sol.induction.ind-as-hom.Rg.9.pf.1}
\end{align}
But the definition of $\widetilde{\beta}$ also yields $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =f_{qh^{\prime},u}$. Hence,
\begin{align*}
\left(  \widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(
p\right)   &  =f_{qh^{\prime},u}\left(  p\right)  =\sum\limits_{h\in
H}\epsilon_{1}\left(  hpqh^{\prime}\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }f_{qh^{\prime},u}\right) \\
&  =\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)  \right)  \left(
p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.induction.ind-as-hom.Rg.9.pf.1})}\right)  .
\end{align*}
\par
Now, forget that we fixed $p$. We have thus proven that $\left(
\widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(  p\right)
=\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)  \right)  \left(
p\right)  $ for every $p\in\CC G$. In other words, $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =\widetilde{\beta}\left(  q,h^{\prime}u\right)
$, so that $\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta
}\left(  qh^{\prime},u\right)  $. This proves
(\ref{sol.induction.ind-as-hom.Rg.9}).} As a consequence of this, we can see
that every $r\in\CC H$, $q \in \CC G$ and $u \in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,ru\right)  =\widetilde{\beta}\left(  qr,u\right)  .
\label{sol.induction.ind-as-hom.Rg.10}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.induction.ind-as-hom.Rg.10}):} Let
$r\in\CC H$, $q \in \CC G$ and $u \in U$. Then, we can write $r$ in the form $r=\sum\limits_{h\in
H}r_{h}h$ for some family $\left(  r_{h}\right)  _{h\in H}$ of elements of
$\CC$. Consider this family $\left(  r_{h}\right)  _{h\in H}$. We have
\begin{align*}
\widetilde{\beta}\left(  q,\underbrace{r}_{=\sum\limits_{h\in H}r_{h}h}
u\right)   &  =\widetilde{\beta}\left(  q,\sum\limits_{h\in H}
r_{h}hu\right)  =\sum\limits_{h\in H}r_{h}\underbrace{\widetilde{\beta}\left(
q,hu\right)  }_{\substack{=\widetilde{\beta}\left(  qh,u\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.Rg.9}), applied to }h^{\prime}=h\text{)}}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\widetilde{\beta}\text{ is
}\CC \text{-bilinear}\right) \\
&  =\sum\limits_{h\in H}r_{h}\widetilde{\beta}\left(  qh,u\right)
=\widetilde{\beta}\left(  q\underbrace{\sum\limits_{h\in H}r_{h}h}
_{=r},u\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\widetilde{\beta}\text{ is
}\CC \text{-bilinear}\right) \\
&= \widetilde{\beta}\left(  qr,u\right)  .
\end{align*}
This proves (\ref{sol.induction.ind-as-hom.Rg.10}).} In other words, the map
$\widetilde{\beta}$ is $\CC H$-bilinear with respect to the right
$\CC H$-module structure on $\CC G$ and the left $\CC H$-module structure
on $U$ (since we already know that $\widetilde{\beta}$ is $\CC$-bilinear).
Hence, by the universal property of the tensor
product, we conclude that there exists a unique $\CC$-linear map
$\beta : \CC G \otimes_{\CC H} U
\rightarrow \Hom_{\CC H} \left( \CC G, U \right)$ such that every
$\left( q,u\right)  \in\CC G\times U$ satisfies
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  . \label{sol.induction.ind-as-hom.beta}
\end{equation}
Consider this map $\beta$. Clearly, every $q\in\CC G$ and $u\in U$
satisfy
\[
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{\beta}\right)  .
\]
Hence, every $q\in\CC G$, $u\in U$ and $p\in\CC G$ satisfy
\begin{equation}
\left(  \underbrace{\beta\left(  q\otimes_{\CC H}u\right)  }_{=f_{q,u}}
\right)  \left(  p\right)  =f_{q,u}\left(  p\right)  =\sum\limits_{h\in
H}\epsilon_{1}\left(  hpq\right)  h^{-1}u
\label{sol.induction.ind-as-hom.beta.explicit}
\end{equation}
(by (\ref{sol.induction.ind-as-hom.fqu})).

We shall now show that the maps $\alpha$ and $\beta$ are mutually inverse. To
do so, we will show that $\alpha\circ\beta=\id$ and $\beta
\circ\alpha=\id$.

Let us first show that $\alpha\circ\beta=\id$. In fact, let
$q\in\CC G$ and $u\in U$ be arbitrary. Then,
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)   &
=\alpha\left(  \beta\left(  q\otimes_{\CC H}u\right)  \right)
=\sum_{j\in J}j^{-1}\otimes_{\CC H}\underbrace{\left(  \beta\left(
q\otimes_{\CC H}u\right)  \right)  \left(  j\right)  }_{\substack{=\sum
\limits_{h\in H}\epsilon_{1}\left(  hjq\right)  h^{-1}u\\\text{(by
(\ref{sol.induction.ind-as-hom.beta.explicit}), applied to }p=j\text{)}}}\\
& \ \ \ \ \  \ \ \ \ \ \left(\text{by the definition of } \alpha\right) \\
&  =\sum_{j\in J}j^{-1}\otimes_{\CC H}\left(  \sum\limits_{h\in
H}\epsilon_{1}\left(  hjq\right)  h^{-1}u\right)  =\sum_{j\in J}
\sum\limits_{h\in H}\epsilon_{1}\left(  hjq\right)  \underbrace{j^{-1}
\otimes_{\CC H}h^{-1}u}_{\substack{=j^{-1}h^{-1}\otimes_{\CC H}
u\\\text{(since }h^{-1}\in \CC H\text{ (since }h\in H \\ \text{ and thus }
h^{-1} \in H \subset \CC H \text{))}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  hjq\right)
\underbrace{j^{-1}h^{-1}}_{=\left(  hj\right)  ^{-1}}\otimes_{\CC H}
u=\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  hjq\right)  \left(
hj\right)  ^{-1}\otimes_{\CC H}u\\
&  =\underbrace{\sum_{j\in J}\sum\limits_{g\in Hj}}_{\substack{=\sum
\limits_{g\in G}\\\text{(since }\bigsqcup_{j\in J}Hj=G\text{)}}}\epsilon
_{1}\left(  gq\right)  g^{-1}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }g\text{ for }hj\text{ in the second sum, since the
map}\\
H\rightarrow Hj,\ h\mapsto hj\text{ is a bijection (because }G\text{ is a
group)}
\end{array}
\right) \\
&  =\sum\limits_{g\in G}\epsilon_{1}\left(  gq\right)  g^{-1}\otimes
_{\CC H}u=\sum\limits_{g\in G}\underbrace{\epsilon_{1}\left(
g^{-1}q\right)  }_{\substack{=\epsilon_{g}\left(  q\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.epsi1g}))}}}\underbrace{\left(  g^{-1}\right)
^{-1}}_{=g}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}
\end{array}
\right) \\
&  =\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)  g\otimes_{\CC H}
u=\underbrace{\left(  \sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\right)  }_{\substack{=q\\\text{(by
(\ref{sol.induction.ind-as-hom.epsilong.coordinates}))}}}\otimes_{\CC H}
u\\
&  =q\otimes_{\CC H}u=\id \left(  q\otimes_{\CC H}u\right)  .
\end{align*}


Now, let us forget that we fixed $q$ and $u$. We thus have shown that $\left(
\alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)
=\id \left(  q\otimes_{\CC H}u\right)  $ for all
$q\in\CC G$ and $u\in U$. In other words, the two maps $\alpha\circ
\beta:\CC G\otimes_{\CC H}U\rightarrow\CC G\otimes
_{\CC H}U$ and $\id :\CC G\otimes_{\CC H}U\rightarrow\CC G\otimes_{\CC H}U$
are equal to each other on
each pure tensor. Since these two maps are $\CC$-linear, this yields
that these two maps $\alpha\circ\beta:\CC G\otimes_{\CC H}U
\rightarrow\CC G\otimes_{\CC H}U$ and
$\id :\CC G\otimes_{\CC H}U\rightarrow\CC G\otimes_{\CC H}U$ must be
identical\footnote{This is because a well-known fact from linear
algebra states the following: If two $\CC$-linear maps from a tensor
product are equal to each other on each pure tensor, then these two maps must
be identical.}. In other words, $\alpha\circ\beta=\id$.

Next, we are going to show that $\beta\circ\alpha=\id$.

Let $f\in\Hom_{\CC H}\left( \CC G, U\right)  $.
Let $p\in\CC G$. The map $f$ is left $\CC H$-linear (since
$f\in\Hom_{\CC H}\left( \CC G, U\right)  $),
hence $\CC$-linear. We have $\alpha\left(  f\right)  =\sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  $ (by the definition of
$\alpha$). Applying the map $\beta$ to this equality, we obtain
\[
\beta\left(  \alpha\left(  f\right)  \right)  =\beta\left(  \sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)  =\sum_{j\in J}
\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)
\]
(since $\beta$ is a $\CC$-linear map). Thus,
\begin{align*}
& \underbrace{\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
}_{=\sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)
\right)  }\left(  p\right)  \\
& =\left(  \sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  =\sum_{j\in J}
\underbrace{\left(  \beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  }_{\substack{=\sum\limits_{h\in
H}\epsilon_{1}\left(  hpj^{-1}\right)  h^{-1}f\left(  j\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.beta.explicit}), applied to}\\q=j^{-1}\text{
and }u=f\left(  j\right)  \text{)}}}\\
& =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  hpj^{-1}\right)
h^{-1}f\left(  j\right)  =\sum_{j\in J}\sum\limits_{h\in H}
\underbrace{\epsilon_{1}\left(  h^{-1}pj^{-1}\right)  }_{\substack{=\epsilon
_{1}\left(  j^{-1}h^{-1}p\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.trace}), applied to}\\h^{-1}p\text{ and }
j^{-1}\text{ instead of }p\text{ and }q\text{)}}}\underbrace{\left(
h^{-1}\right)  ^{-1}}_{=h}f\left(  j\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }h^{-1}\text{ for }h\text{ in the second sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{-1}\text{ is a bijection}\\
\text{(because }H\text{ is a group)}
\end{array}
\right)  \\
& =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{j^{-1}
h^{-1}}_{=\left(  hj\right)  ^{-1}}p\right)  \underbrace{hf\left(  j\right)
}_{\substack{=f\left(  hj\right)  \\\text{(since }f\text{ is left }
\CC H\text{-linear and since }h\in H\subset\CC H\text{)}}}\\
& =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \left(  hj\right)
^{-1}p\right)  f\left(  hj\right)  =\underbrace{\sum_{j\in J}\sum\limits_{g\in
Hj}}_{\substack{=\sum\limits_{g\in G}\\\text{(since }\bigsqcup_{j\in
J}Hj=G\text{)}}}\epsilon_{1}\left(  g^{-1}p\right)  f\left(  g\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we substituted }g\text{ for }hj\text{ in the second sum, since the
map}\\
H\rightarrow Hj,\ h\mapsto hj\text{ is a bijection (because }G\text{ is a
group)}
\end{array}
\right)  \\
& =\sum\limits_{g\in G}\underbrace{\epsilon_{1}\left(  g^{-1}p\right)
}_{\substack{=\epsilon_{g}\left(  p\right)  \\\text{(by
(\ref{sol.induction.ind-as-hom.epsi1g}), applied to }q=p\text{)}}}f\left(
g\right)  =\sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  f\left(
g\right)  \\
& =f\left(  \sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  g\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is }\CC \text{-linear}
\right)  .
\end{align*}
Compared with
\[
f\left(  \underbrace{p}_{\substack{=\sum\limits_{g\in G}\epsilon_{g}\left(
p\right)  g\\\text{(by (\ref{sol.induction.ind-as-hom.epsilong.coordinates}),
applied to }q=p\text{)}}}\right)  =f\left(  \sum\limits_{g\in G}\epsilon
_{g}\left(  p\right)  g\right)  ,
\]
this yields $\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
\left(  p\right)  =f\left(  p\right)  $.

Now, forget that we fixed $p$. We thus have proven that $\left(  \beta\left(
\alpha\left(  f\right)  \right)  \right)  \left(  p\right)  =f\left(
p\right)  $ for every $p\in\CC G$. In other words, $\beta\left(
\alpha\left(  f\right)  \right)  =f$. Hence, $\left(  \beta\circ\alpha\right)
\left(  f\right)  =\beta\left(  \alpha\left(  f\right)  \right)
=f=\id \left(  f\right)  $.

Now, forget that we fixed $f$. We thus have shown that $\left(  \beta
\circ\alpha\right)  \left(  f\right)  =\id \left(  f\right)  $
for every $f\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $. In other words,
$\beta\circ\alpha=\id$.
Combined with $\alpha\circ\beta=\id$, this yields that the maps
$\alpha$ and $\beta$ are mutually inverse. Hence, the map $\alpha$ is invertible.

Now, we know that the map $\alpha$ is an invertible left $\CC G$-module
homomorphism. Hence, $\alpha$ is a left $\CC G$-module isomorphism.
Therefore, there exists a left $\CC G$-module isomorphism
$\Hom_{\CC H}\left( \CC G, U \right)
\rightarrow\CC G\otimes_{\CC H}U$ (namely, $\alpha$). Therefore,
$\Hom_{\CC H}\left( \CC G, U \right)
\cong\CC G\otimes_{\CC H}U=\Ind_{H}^{G}U$
as left $\CC G$-modules. This solves Exercise~\ref{exe.induction.ind-as-hom}.

\textit{Remark:} In our solution, we explicitly constructed a
$\CC G$-module isomorphism $\alpha:\Hom_{\CC H}\left(
\CC G,U\right)  \rightarrow\Ind_{H}^{G}U$.
This isomorphism is functorial with respect to $U$. It is also independent on
the choice of $J$ (this is not immediately clear from its definition, but it
can be shown very easily, by observing that the tensor $j^{-1}\otimes
_{\CC H}f\left( j \right)$ for $j\in G$ depends only on the coset
$Hj$ and not on $j$ itself). One might ask whether this isomorphism is
functorial in $G$ and $H$; but to make sense of this question, one has to
define the category with respect to which this functoriality is to be
understood. I don't know a good answer.

It is also worth noting that in our solution, $\CC$ could be replaced
by any commutative ring. We used neither that $\CC$ is a field, nor
that $\CC$ has characteristic $0$.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.induction.wrong-way-frob}}

\textit{Solution to Exercise~\ref{exe.induction.wrong-way-frob}.}
Consider the $\CC G$-module $\Hom_{\CC H}\left(\CC G, V\right)$ defined
as in Exercise~\ref{exe.induction.ind-as-hom}. Then,
Exercise~\ref{exe.induction.ind-as-hom} (applied to $V$ instead of
$U$) yields that this module $\Hom_{\CC H}\left(\CC G, V\right)$ is
isomorphic to $\Ind^G_H V$. That is, $\Ind^G_H V \cong
\Hom_{\CC H}\left(\CC G, V\right)$.
Hence, \newline
$\Hom_{\CC G}\left(U, \Ind^G_H V\right)
\cong \Hom_{\CC G}\left(U, \Hom_{\CC H}\left(\CC G, V\right)\right)$.
But \eqref{Hom-tensor-adj} (applied to $R = \CC G$, $S = \CC H$,
$A = \CC G$, $B = U$ and $C = V$) yields
$\Hom_{\CC H}\left(\CC G \otimes_{\CC G} U, V\right)
\cong \Hom_{\CC G}\left(U, \Hom_{\CC H}\left(\CC G, V\right)\right)$.
Altogether, we thus have
\[
\Hom_{\CC G}\left(U, \Ind^G_H V\right)
\cong \Hom_{\CC G}\left(U, \Hom_{\CC H}\left(\CC G, V\right)\right)
\cong \Hom_{\CC H}\left(\CC G \otimes_{\CC G} U, V\right).
\]
It remains to see what the $\CC H$-module $\CC G \otimes_{\CC G} U$ is.
By the basic properties of tensor products, the $\CC H$-module
$\CC G \otimes_{\CC G} U$ is isomorphic to the $\CC H$-module
obtained by restricting the action of $\CC G$ on the $\CC G$-module
$U$ to the subalgebra $\CC H$. Since this latter $\CC H$-module is
precisely $\Res^G_H U$ (in fact, this is how $\Res^G_H U$ was
defined!), this rewrites as follows: The $\CC H$-module
$\CC G \otimes_{\CC G} U$ is isomorphic to $\Res^G_H U$. In other
words, $\CC G \otimes_{\CC G} U \cong \Res^G_H U$. Now,
\[
\Hom_{\CC G}\left(U, \Ind^G_H V\right)
\cong \Hom_{\CC H}\left(\underbrace{\CC G \otimes_{\CC G} U}_{\cong \Res^G_H U}, V\right)
\cong \Hom_{\CC H}\left(\Res^G_H U, V\right).
\]
This solves the exercise.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.HomxHom}}

\textit{Solution to Exercise~\ref{exe.HomxHom}.} In the following, a
``$\Hom$'' symbol without a
subscript means ``$\Hom_{\CC}$'' (rather than ``$\Hom_{\CC G}$''
or whatever other meaning this symbol
could possibly have in the context).

If $G$ is a group and if $M$ and $N$ are two ${\CC}G$-modules, then
$\Hom\left(  M,N\right)  $ becomes a ${\CC}G$-module,
with $G$ acting as follows: If $g\in G$ and $f\in\Hom \left(
M,N\right)  $, then $t_{g}f$ is the ${\CC}$-linear map $M\rightarrow N$
sending every $m\in M$ to $t_{g}f\left(  t_{g^{-1}}m\right)  $. This
${\CC}G$-module structure is precisely the one we know from
Remark~\ref{antipode-as-dual-action}.

Now, it is well-known (and straightforward to verify) that every two
${\CC}G$-modules $M$ and $N$ satisfy
\begin{equation}
\Hom_{\CC G}\left(  M,N\right)  =\left(
\Hom \left(  M,N\right)  \right)  ^{G}
\label{sol.HomxHom.lemma}
\end{equation}
(where we regard $\Hom_{\CC G}\left(  M,N\right)  $ as a
${\CC}$-vector subspace of $\Hom \left(  M,N\right)  $
because every ${\CC}G$-linear map $M\rightarrow N$ is a
$\CC$-linear map $M\rightarrow N$).

Let us now come to the solution of the exercise.

(a) Let $\psi$ denote the $\CC$-linear map
\[
\Hom \left(  V_{1},W_{1}\right)  \otimes\Hom \left(
V_{2},W_{2}\right)  \rightarrow\Hom \left(  V_{1}\otimes
V_{2},W_{1}\otimes W_{2}\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms. This map $\psi$ is completely independent of $G_{1}$ and
$G_{2}$ (it is defined whenever $V_{1}$, $V_{2}$, $W_{1}$ and $W_{2}$ are four
$\CC$-vector spaces) and is a vector space isomorphism (this is a basic
fact from linear algebra, relying only on the finite-dimensionality of
$V_{1}$, $V_{2}$, $W_{1}$ and $W_{2}$). But we can regard $\Hom \left(
V_{1},W_{1}\right)  $ as a $\CC G_{1}$-module,
$\Hom \left(  V_{2},W_{2}\right)  $ as a $\CC G_{2}$-module and
$\Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)  $ as
a $\CC \left[  G_{1}\times G_{2}\right]  $-module. Then, the map $\psi$
is a homomorphism of $\CC \left[  G_{1}\times G_{2}\right]
$-modules\footnote{This is easy to verify by checking that $t_{\left(
h_{1},h_{2}\right)  }\left(  \psi\left(  f\otimes g\right)  \right)
=\psi\left(  t_{h_{1}}f\otimes t_{h_{2}}g\right)  $ for all $\left(
h_{1},h_{2}\right)  \in G_{1}\times G_{2}$, $f\in\Hom \left(
V_{1},W_{1}\right)  $ and $g\in\Hom \left(  V_{2},W_{2}\right)
$.}. Hence, this map $\psi$ must be an isomorphism of $\CC \left[
G_{1}\times G_{2}\right]  $-modules (being a vector space isomorphism). As a
consequence, it sends the $G_{1}\times G_{2}$-fixed space of its domain to the
$G_{1}\times G_{2}$-fixed space of its target:
\[
\psi\left(  \left(  \Hom \left(  V_{1},W_{1}\right)
\otimes\Hom \left(  V_{2},W_{2}\right)  \right)  ^{G_{1}\times
G_{2}}\right)  =\left(  \Hom \left(  V_{1}\otimes V_{2}
,W_{1}\otimes W_{2}\right)  \right)  ^{G_{1}\times G_{2}}.
\]
Since
\begin{align*}
&  \left(  \Hom \left(  V_{1},W_{1}\right)  \otimes
\Hom \left(  V_{2},W_{2}\right)  \right)  ^{G_{1}\times G_{2}}\\
&  =\underbrace{\left(  \Hom \left(  V_{1},W_{1}\right)  \right)
^{G_{1}}}_{\substack{=\Hom_{\CC G_1}\left(
V_{1},W_{1}\right)  \\\text{(by (\ref{sol.HomxHom.lemma}),}\\\text{applied to
}G=G_{1}\text{, }M=V_{1}\text{ and }N=W_{1}\text{)}}}\otimes
\underbrace{\left(  \Hom \left(  V_{2},W_{2}\right)  \right)
^{G_{2}}}_{\substack{=\Hom_{\CC G_2}\left(
V_{2},W_{2}\right)  \\\text{(by (\ref{sol.HomxHom.lemma}),}\\\text{applied to
}G=G_{2}\text{, }M=V_{2}\text{ and }N=W_{2}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by (\ref{invariants-vs-tensor-product}), applied to }K_{1}=G_{1}\text{,
}K_{2}=G_{2}\text{,}\\
U_{1}=\Hom \left(  V_{1},W_{1}\right)  \text{ and }
U_{2}=\Hom \left(  V_{2},W_{2}\right)
\end{array}
\right)  \\
&  =\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  ,
\end{align*}
this rewrites as
\[
\psi\left(  \Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)
\right)  =\left(  \Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  \right)  ^{G_{1}\times G_{2}}.
\]
Hence, the isomorphism $\psi$ restricts to an isomorphism from
$\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  $ to
\newline
$\left(  \Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  \right)  ^{G_{1}\times G_{2}}$. This restriction is precisely
the $\CC$-linear map
\[
\Hom_{\CC G_{1}}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)
\rightarrow\Hom_{{\CC}\left[  G_{1}\times G_{2}\right]
}\left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms (i.e., the map alleged to be an isomorphism in the statement of
the exercise). So we know now that this map is an isomorphism. This solves
Exercise~\ref{exe.HomxHom}(a).

(b) Let $G_{1}$ and $G_{2}$ be two groups. Let $V_{i}$ and $W_{i}$ be
finite-dimensional ${\CC}G_{i}$-modules for every $i\in\left\{
1,2\right\}$. Exercise~\ref{exe.HomxHom}(a) provides a vector space
isomorphism from $\Hom_{\CC G_{1}}\left(  V_{1}
,W_{1}\right)  \otimes\Hom_{\CC G_{2}}\left(
V_{2},W_{2}\right)  $ to $\Hom_{{\CC}\left[
G_{1}\times G_{2}\right]  }\left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  $. As a consequence,
\begin{align}
&  \dim_{\CC}\left(  \Hom_{\CC G_{1}}\left(
V_{1},W_{1}\right)  \otimes\Hom_{\CC G_{2}}\left(
V_{2},W_{2}\right)  \right)  \nonumber\\
&  =\dim_{\CC}\left(  \Hom_{{\CC}\left[
G_{1}\times G_{2}\right]  }\left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  \right)  .\label{sol.HomxHom.b.1}
\end{align}


Applying (\ref{character-scalprod}) to $G_{1}$, $V_{1}$ and $W_{1}$ instead of
$G$, $V_{1}$ and $V_{2}$, we obtain
\begin{equation}
\left(  \chi_{V_{1}},\chi_{W_{1}}\right)  _{G_{1}}=\dim_{{\CC} }\left(
\Hom_{{\CC} G_{1}}\left(  V_{1},W_{1}\right)  \right)  .
\label{sol.HomxHom.b.2}
\end{equation}
Applying (\ref{character-scalprod}) to $G_{2}$, $V_{2}$ and $W_{2}$ instead of
$G$, $V_{1}$ and $V_{2}$, we obtain
\begin{equation}
\left(  \chi_{V_{2}},\chi_{W_{2}}\right)  _{G_{2}}=\dim_{{\CC} }\left(
\Hom_{{\CC} G_{2}}\left(  V_{2},W_{2}\right)  \right)  .
\label{sol.HomxHom.b.3}
\end{equation}


Applying (\ref{character-scalprod}) to $G_{1}\times G_{2}$, $V_{1}\otimes
V_{2}$ and $W_{1}\otimes W_{2}$ instead of $G$, $V_1$ and $V_2$, we obtain
\begin{align*}
\left(  \chi_{V_{1}\otimes V_{2}},\chi_{W_{1}\otimes W_{2}}\right)
_{G_{1}\times G_{2}} 
&  =\dim_{\CC} \left(  \Hom_{\CC \left[  G_{1}\times G_{2}\right]  }
\left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)  \right) \\
&  =\dim_{\CC} \left(  \Hom_{\CC G_1} \left(  V_{1},W_{1}\right) 
 \otimes \Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  \right)
 \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.HomxHom.b.1})}\right) \\
&  =\underbrace{\dim_{{\CC} }\left(  \Hom  _{{{\CC}
}G_{1}}\left(  V_{1},W_{1}\right)  \right)  }
_{\substack{=\left(  \chi_{V_{1}},\chi_{W_{1}}\right) _{G_{1}}\\
\text{(by (\ref{sol.HomxHom.b.2}))}}}
\cdot\underbrace{\dim_{{\CC} }\left(  \Hom 
_{{{\CC} }G_{2}}\left(  V_{2},W_{2}\right)  \right)  }
_{\substack{=\left(  \chi_{V_{2}},\chi_{W_{2}}\right)  _{G_{2}}\\\text{(by
(\ref{sol.HomxHom.b.3}))}}}\\
&  =\left(  \chi_{V_{1}},\chi_{W_{1}}\right)  _{G_{1}}
\left(  \chi_{V_{2}},\chi_{W_{2}}\right)  _{G_{2}}.
\end{align*}
This proves (\ref{character-tensorprod}). Thus, Exercise~\ref{exe.HomxHom}(b)
is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.HomxHom}}

\textit{Solution to Exercise~\ref{exe.HomxHom}.} Before we start
solving the exercise, let us show do some general representation theory.

In the following, a ``$\Hom$'' symbol without a subscript means
``$\Hom_\CC$'' (rather than ``$\Hom_{\CC G}$'' or whatever
other meaning this symbol could possibly have in the context).

Let $G$ be a group. In the following, we will write $g$ for the element
$t_{g}$ of ${\CC }G$ whenever $g$ is an element of $G$. This is a
relatively common abuse of notation, and it is harmless because the map
$G\rightarrow{\CC }G,\ g\mapsto t_{g}$ is an injective homomorphism of
multiplicative monoids (so $t_{gh}=t_{g}t_{h}$ and $t_{1}=1$, which means that
we won't run into ambiguities denoting $t_{g}$ by $g$) and because every
$\CC G$-module $M$, every $m\in M$ and every $g\in G$ satisfy
$gm=t_{g}m$.

If $M$ and $N$ are two $\CC G$-modules, then $\Hom \left(
M,N\right)  $ becomes a $\CC G$-module, with $G$ acting as follows: If
$g\in G$ and $f\in\Hom \left(  M,N\right)  $, then $gf$ is the
$\CC$-linear map $M\rightarrow N$ sending every $m\in M$ to $gf\left(
g^{-1}m\right)  $. This $\CC G$-module structure is precisely the one we
know from Remark~\ref{antipode-as-dual-action}.

Now, it is easy to see that every two $\CC G$-modules $M$ and $N$
satisfy
\begin{equation}
\Hom_{\CC G}\left(  M,N\right)  =\left(
\Hom \left(  M,N\right)  \right)  ^{G} \label{sol.HomxHom.lemma}
\end{equation}
(where we regard $\Hom_{\CC G}\left(
M,N\right)  $ as a $\CC$-vector subspace of $\Hom \left(
M,N\right)  $ because every $\CC G$-linear map $M\rightarrow N$ is a
$\CC$-linear map $M\rightarrow N$).\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.HomxHom.lemma}):} Let $M$ and $N$ be two $\CC G$-modules.
Every $g\in G$, every $f\in\Hom \left(  M,N\right)  $ and every
$m\in M$ satisfy
\begin{equation}
\left(  gf\right)  \left(  m\right)  =gf\left(  g^{-1}m\right)
\label{sol.HomxHom.lemma.pf.intro}
\end{equation}
(by the definition of the $\CC G$-module structure on
$\Hom \left(  M,N\right)  $). The definition of $\left(
\Hom \left(  M,N\right)  \right)  ^{G}$ yields
\begin{equation}
\left(  \Hom \left(  M,N\right)  \right)  ^{G}=\left\{
\varphi\in\Hom \left(  M,N\right)  \ \mid\ g\varphi
=\varphi\text{ for every }g\in G\right\}  .
\label{sol.HomxHom.lemma.pf.1}
\end{equation}
\par
Let $f\in\Hom_{\CC G}\left(  M,N\right)  $.
Then, $f$ is a $\CC G$-module homomorphism $M\rightarrow N$. Thus, $f$
is a $\CC$-linear homomorphism of $G$-sets $M\rightarrow N$.
\par
Since $f$ is $\CC$-linear, we have $f\in\Hom \left(
M,N\right)  $. Now, let $g\in G$. Every $m\in M$ satisfies
\begin{align*}
\left(  gf\right)  \left(  m\right)   &  =g\underbrace{f\left(  g^{-1}
m\right)  }_{\substack{=g^{-1}f\left(  m\right)  \\\text{(since }f\text{ is a
homomorphism}\\\text{of }G\text{-sets)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.HomxHom.lemma.pf.intro})}\right) \\
&  =\underbrace{gg^{-1}}_{=1_{G}}f\left(  m\right)  =f\left(  m\right)  .
\end{align*}
In other words, $gf=f$. Now, forget that we fixed $g$. We thus have proven
that $gf=f$ for every $g\in G$. So $f$ is an element of $\Hom 
\left(  M,N\right)  $ satisfying $\left(  gf=f\text{ for every }g\in G\right)
$. In other words,
\[
f\in\left\{  \varphi\in\Hom \left(  M,N\right)  \ \mid
\ g\varphi=\varphi\text{ for every }g\in G\right\}  =\left(
\Hom \left(  M,N\right)  \right)  ^{G}
\]
(by (\ref{sol.HomxHom.lemma.pf.1})).
\par
Now, forget that we fixed $f$. We thus have shown that every $f\in
\Hom_{\CC G}\left(  M,N\right)  $ satisfies
$f\in\left(  \Hom \left(  M,N\right)  \right)  ^{G}$. In other
words,
\begin{equation}
\Hom_{\CC G}\left(  M,N\right)  \subset\left(
\Hom \left(  M,N\right)  \right)  ^{G}.
\label{sol.HomxHom.lemma.pf.2}
\end{equation}
\par
Now, let $f\in\left(  \Hom \left(  M,N\right)  \right)  ^{G}$ be
arbitrary. Then,
\[
f\in\left(  \Hom \left(  M,N\right)  \right)  ^{G}=\left\{
\varphi\in\Hom \left(  M,N\right)  \ \mid\ g\varphi
=\varphi\text{ for every }g\in G\right\}
\]
(by (\ref{sol.HomxHom.lemma.pf.1})). In other words, $f$ is an element of
$\Hom \left(  M,N\right)  $ satisfying $\left(  gf=f\text{ for
every }g\in G\right)  $.
\par
Now, let $m\in M$ and $g\in G$. We have $gf=f$ (because we assumed that $gf=f$
for every $g\in G$). Hence, $\underbrace{\left(  gf\right)  }_{=f}\left(
gm\right)  =f\left(  gm\right)  $, therefore
\begin{align*}
f\left(  gm\right)   &  =\left(  gf\right)  \left(  gm\right)  =gf\left(
\underbrace{g^{-1}g}_{=1_{G}}m\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.HomxHom.lemma.pf.intro}), applied to }gm\text{ instead of }m\right)
\\
&  =gf\left(  m\right)  .
\end{align*}
\par
Now, forget that we fixed $m$ and $g$. We thus have proven that $f\left(
gm\right)  =gf\left(  m\right)  $ for every $m\in M$ and $g\in G$. In other
words, $f$ is a homomorphism of $G$-sets $M\rightarrow N$. We also know that
$f$ is $\CC$-linear (since $f$ is an element of $\Hom 
\left(  M,N\right)  $). Hence, $f$ is a $\CC$-linear homomorphism of
$G$-sets $M\rightarrow N$. In other words, $f$ is a $\CC G$-module
homomorphism $M\rightarrow N$. In other words, $f\in\Hom_{\CC G}\left(  M,N\right)  $.
\par
Now, forget that we fixed $f$. We thus have shown that every $f\in\left(
\Hom \left(  M,N\right)  \right)  ^{G}$ satisfies $f\in
\Hom_{\CC G}\left(  M,N\right)  $. In other
words,
\[
\left(  \Hom \left(  M,N\right)  \right)  ^{G}\subset
\Hom_{\CC G}\left(  M,N\right)  .
\]
Combined with (\ref{sol.HomxHom.lemma.pf.2}), this yields
$\Hom_{\CC G}\left(  M,N\right)  =\left(  \Hom \left(
M,N\right)  \right)  ^{G}$. This proves (\ref{sol.HomxHom.lemma}).}

Now, forget that we fixed $G$. We thus have proven that
(\ref{sol.HomxHom.lemma}) holds for every group $G$ and any two
$\CC G$-modules $M$ and $N$.

Let us now recall a fact from linear algebra: If $M_{1}$, $M_{2}$, $N_{1}$ and
$N_{2}$ are four $\CC$-vector spaces, then there is a well-defined
$\CC$-linear map
\[
\psi_{M_{1},M_{2},N_{1},N_{2}}:\Hom \left(  M_{1},N_{1}\right)
\otimes\Hom \left(  M_{2},N_{2}\right)  \rightarrow
\Hom \left(  M_{1}\otimes M_{2},N_{1}\otimes N_{2}\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms. If these four vector spaces $M_{1}$, $M_{2}$, $N_{1}$ and
$N_{2}$ are finite-dimensional, then this map $\psi_{M_{1},M_{2},N_{1},N_{2}}$
is a vector space isomorphism.

Let us now come to the solution of the exercise.

(a) Let $\Psi$ be the $\CC$-linear map
\[
\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)
\rightarrow\Hom_{{\CC }\left[  G_{1}\times G_{2}\right]
}\left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms. We want to prove that $\Psi$ is a vector space isomorphism.

First, let us recall that we have defined a $\CC$-linear map
$\psi_{M_{1},M_{2},N_{1},N_{2}}$ for any four $\CC$-vector spaces
$M_{1}$, $M_{2}$, $N_{1}$ and $N_{2}$. If these four vector spaces $M_{1}$,
$M_{2}$, $N_{1}$ and $N_{2}$ are finite-dimensional, then this map
$\psi_{M_{1},M_{2},N_{1},N_{2}}$ is a vector space isomorphism. Applying this
to $M_{1}=V_{1}$, $M_{2}=V_{2}$, $N_{1}=W_{1}$ and $N_{2}=W_{2}$, this yields
the following: The map $\psi_{V_{1},V_{2},W_{1},W_{2}}$ is a vector space
isomorphism. In particular, the map $\psi_{V_{1},V_{2},W_{1},W_{2}}$ is
injective. Notice that $\psi_{V_{1},V_{2},W_{1},W_{2}}$ is a map
\[
\Hom \left(  V_{1},W_{1}\right)  \otimes\Hom \left(
V_{2},W_{2}\right)  \rightarrow\Hom \left(  V_{1}\otimes
V_{2},W_{1}\otimes W_{2}\right)  .
\]


Whenever $G$ is a group and $M$ and $N$ are two $\CC G$-modules, we
regard $\Hom_{\CC G}\left(  M,N\right)  $ as a
$\CC$-vector subspace of $\Hom \left(  M,N\right)  $
(because every $\CC G$-linear map $M\rightarrow N$ is a
$\CC$-linear map $M\rightarrow N$). Thus,
$\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)  $ is a
$\CC$-vector subspace of
$\Hom \left(  V_{1},W_{1}\right)  $, and
$\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  $ is a $\CC$-vector
subspace of $\Hom \left(  V_{2},W_{2}\right)  $. Consequently,
the tensor product
$\Hom_{\CC G_1}\left(  V_{1},W_{1}\right) 
 \otimes\Hom_{\CC G_2}\left( V_{2},W_{2}\right)  $
canonically embeds into $\Hom \left(
V_{1},W_{1}\right)  \otimes\Hom \left(  V_{2},W_{2}\right)  $
(because we are working over $\CC$, which is a field). We will regard
this embedding as an inclusion.

We have
\begin{equation}
\Psi\left(  p\right)  =\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  p\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\Hom_{\CC G_1}
\left(  V_{1},W_{1}\right)  \otimes\Hom_{\CC G_2}
\left(  V_{2},W_{2}\right)  .
\label{sol.HomxHom.restr}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.HomxHom.restr}):} Let $p\in
\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  $.
Then, $p$ is a tensor; hence, we can write $p$ in the form
$p=\sum_{i=1}^{N}\lambda_{i}f_{i}\otimes g_{i}$ for some
$N\in\NN$, some scalars
$\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{N}$ in $\CC$, some
elements $f_{1}$, $f_{2}$, $...$, $f_{N}$ in $\Hom_{\CC G_1}
\left(  V_{1},W_{1}\right)  $, and some elements $g_{1}$, $g_{2}$,
$...$, $g_{N}$ in $\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  $.
Consider this $N$, these $\lambda_{1}$, $\lambda_{2}$,
$...$, $\lambda_{N}$, these $f_{1}$, $f_{2}$, $...$, $f_{N}$, and these
$g_{1}$, $g_{2}$, $...$, $g_{N}$. Applying the map $\Psi$ to both sides of the
identity $p=\sum_{i=1}^{N}\lambda_{i}f_{i}\otimes g_{i}$, we obtain
\begin{align}
\Psi\left(  p\right)   &  =\Psi\left(  \sum_{i=1}^{N}\lambda_{i}f_{i}\otimes
g_{i}\right)  =\sum_{i=1}^{N}\lambda_{i}\underbrace{\Psi\left(  f_{i}\otimes
g_{i}\right)  }_{\substack{=f_{i}\otimes g_{i}\\\text{(by the definition of
}\Psi\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\Psi\text{ is
}\CC \text{-linear}\right) \nonumber\\
&  =\sum_{i=1}^{N}\lambda_{i}f_{i}\otimes g_{i}.
\label{sol.HomxHom.restr.pf.1}
\end{align}
\par
On the other hand, applying the map $\psi_{V_{1},V_{2},W_{1},W_{2}}$ to both
sides of the identity $p=\sum_{i=1}^{N}\lambda_{i}f_{i}\otimes g_{i}$, we
obtain
\begin{align*}
\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  p\right)   &  =\psi_{V_{1},V_{2}
,W_{1},W_{2}}\left(  \sum_{i=1}^{N}\lambda_{i}f_{i}\otimes g_{i}\right) \\
&  =\sum_{i=1}^{N}\lambda_{i}\underbrace{\psi_{V_{1},V_{2},W_{1},W_{2}}\left(
f_{i}\otimes g_{i}\right)  }_{\substack{=f_{i}\otimes g_{i}\\\text{(by the
definition of }\psi_{V_{1},V_{2},W_{1},W_{2}}\text{)}}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\psi_{V_{1},V_{2}
,W_{1},W_{2}}\text{ is }\CC \text{-linear}\right) \\
&  =\sum_{i=1}^{N}\lambda_{i}f_{i}\otimes g_{i}.
\end{align*}
Compared with (\ref{sol.HomxHom.restr.pf.1}), this yields $\Psi\left(
p\right)  =\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  p\right)  $. This proves
(\ref{sol.HomxHom.restr}).} Thus, the map $\Psi$ is obtained from the map
$\psi_{V_{1},V_{2},W_{1},W_{2}}$ by restricting its domain to
$\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  $ and
restricting its codomain to $\Hom_{{\CC }\left[
G_{1}\times G_{2}\right]  }\left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  $. Since restriction of a map preserves injectivity, this yields
that the map $\Psi$ is injective (since the map $\psi_{V_{1},V_{2},W_{1}
,W_{2}}$ is injective).

Now, let us notice that $\Hom \left(  V_{1},W_{1}\right)  $ is a
$\CC G_{1}$-module (since $V_{1}$ and $W_{1}$ are $\CC G_{1}$-modules),
and $\Hom \left(  V_{2},W_{2}\right)  $ is a
$\CC G_{2}$-module (since $V_{2}$ and $W_{2}$ are $\CC G_{2}$-modules).
Thus, the tensor product
$\Hom \left(  V_{1},W_{1}\right)  \otimes\Hom \left(  V_{2},W_{2}\right) $
becomes
a $\CC \left[  G_{1}\times G_{2}\right]  $-module (being a tensor
product of a $\CC G_{1}$-module with a $\CC G_{2}$-module). It is
easy to see that the map $\psi_{V_{1},V_{2},W_{1},W_{2}}$ is a
$\CC \left[  G_{1}\times G_{2}\right]  $-module
homomorphism\footnote{\textit{Proof.} Let $p\in\Hom \left(
V_{1},W_{1}\right)  \otimes\Hom \left(  V_{2},W_{2}\right)  $
and $g\in G_{1}\times G_{2}$.
\par
Write $g\in G_{1}\times G_{2}$ in the form $g=\left(  g_{1},g_{2}\right)  $.
\par
The element $p$ is a tensor; hence, we can write $p$ in the form $p=\sum
_{i=1}^{N}\lambda_{i}u_{i}\otimes v_{i}$ for some $N\in\NN$, some
scalars $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{N}$ in $\CC$,
some elements $u_{1}$, $u_{2}$, $...$, $u_{N}$ in $\Hom \left(
V_{1},W_{1}\right)  $, and some elements $v_{1}$, $v_{2}$, $...$, $v_{N}$ in
$\Hom \left(  V_{2},W_{2}\right)  $. Consider this $N$, these
$\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{N}$, these $u_{1}$, $u_{2}$,
$...$, $u_{N}$, and these $v_{1}$, $v_{2}$, $...$, $v_{N}$. Applying the map
$\psi_{V_{1},V_{2},W_{1},W_{2}}$ to both sides of the identity $p=\sum
_{i=1}^{N}\lambda_{i}u_{i}\otimes v_{i}$, we obtain
\begin{align*}
\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  p\right)   &  =\psi_{V_{1},V_{2}
,W_{1},W_{2}}\left(  \sum_{i=1}^{N}\lambda_{i}u_{i}\otimes v_{i}\right)  \\
&  =\sum_{i=1}^{N}\lambda_{i}\underbrace{\psi_{V_{1},V_{2},W_{1},W_{2}}\left(
u_{i}\otimes v_{i}\right)  }_{\substack{=u_{i}\otimes v_{i}\\\text{(by the
definition of }\psi_{V_{1},V_{2},W_{1},W_{2}}\text{)}}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\psi_{V_{1},V_{2}
,W_{1},W_{2}}\text{ is }\CC \text{-linear}\right)  \\
&  =\sum_{i=1}^{N}\lambda_{i}u_{i}\otimes v_{i}.
\end{align*}
Hence,
\begin{align}
\underbrace{g}_{=\left(  g_{1},g_{2}\right)  }\underbrace{\psi_{V_{1}
,V_{2},W_{1},W_{2}}\left(  p\right)  }_{=\sum_{i=1}^{N}\lambda_{i}u_{i}\otimes
v_{i}} &  =\left(  g_{1},g_{2}\right)  \left(  \sum_{i=1}^{N}\lambda_{i}
u_{i}\otimes v_{i}\right)  =\sum_{i=1}^{N}\lambda_{i}\underbrace{\left(
g_{1},g_{2}\right)  \left(  u_{i}\otimes v_{i}\right)  }_{\substack{=g_{1}
u_{i}\otimes g_{2}v_{i}\\\text{(by the definition of the}\\\CC \left[
G_{1}\times G_{2}\right]  \text{-module structure}\\\text{on }
\Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the action of }\left(  g_{1}
,g_{2}\right)  \text{ on }\Hom \left(  V_{1}\otimes V_{2}
,W_{1}\otimes W_{2}\right)  \text{ is }\CC \text{-linear}\right)
\nonumber\\
&  =\sum_{i=1}^{N}\lambda_{i}g_{1}u_{i}\otimes g_{2}v_{i}
.
\label{sol.HomxHom.modmap.1}
\end{align}
\par
On the other hand,
\begin{align*}
\underbrace{g}_{=\left(  g_{1},g_{2}\right)  }
\underbrace{p}_{=\sum_{i=1}^{N}\lambda_{i}u_{i}\otimes v_{i}} 
&  =\left(  g_{1},g_{2}\right)  \left(
\sum_{i=1}^{N}\lambda_{i}u_{i}\otimes v_{i}\right)  =\sum_{i=1}^{N}\lambda
_{i}\underbrace{\left(  g_{1},g_{2}\right)  \left(  u_{i}\otimes v_{i}\right)
}_{\substack{=g_{1}u_{i}\otimes g_{2}v_{i}\\\text{(by the definition of
the}\\\CC \left[  G_{1}\times G_{2}\right]  \text{-module structure}
\\\text{on }\Hom \left(  V_{1},W_{1}\right)  \otimes
\Hom \left(  V_{2},W_{2}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the action of }\left(  g_{1}
,g_{2}\right)  \text{ on }\Hom \left(  V_{1},W_{1}\right)
\otimes\Hom \left(  V_{2},W_{2}\right)  \text{ is }
\CC \text{-linear}\right)  \\
&  =\sum_{i=1}^{N}\lambda_{i}g_{1}u_{i}\otimes g_{2}v_{i}.
\end{align*}
Applying the map $\psi_{V_{1},V_{2},W_{1},W_{2}}$ to both sides of this
equality, we obtain
\begin{align*}
\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  gp\right)   &  =\psi_{V_{1},V_{2}
,W_{1},W_{2}}\left(  \sum_{i=1}^{N}\lambda_{i}g_{1}u_{i}\otimes g_{2}
v_{i}\right)  \\
&  =\sum_{i=1}^{N}\lambda_{i}\underbrace{\psi_{V_{1},V_{2},W_{1},W_{2}}\left(
g_{1}u_{i}\otimes g_{2}v_{i}\right)  }_{\substack{=g_{1}u_{i}\otimes
g_{2}v_{i}\\\text{(by the definition of}\\\text{the map }\psi_{V_{1}
,V_{2},W_{1},W_{2}}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\psi_{V_{1},V_{2},W_{1},W_{2}}\text{ is }\CC \text{-linear}\right)  \\
&  =\sum_{i=1}^{N}\lambda_{i}g_{1}u_{i}\otimes g_{2}v_{i}=g\psi_{V_{1}
,V_{2},W_{1},W_{2}}\left(  p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.HomxHom.modmap.1})}\right)  .
\end{align*}
\par
Now, forget that we fixed $p$ and $g$. We thus have shown that
$\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  gp\right)  
=g\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  p\right)  $ for every
$p\in\Hom \left(  V_{1},W_{1}\right) 
 \otimes\Hom \left(  V_{2},W_{2}\right)  $ and
$g\in G_{1}\times G_{2}$. In other words, $\psi_{V_{1},V_{2},W_{1},W_{2}}$ is
a homomorphism of $G_{1}\times G_{2}$-sets. Thus, $\psi_{V_{1},V_{2}
,W_{1},W_{2}}$ is a $\CC$-linear homomorphism of
$G_{1}\times G_{2}$-sets. In other words,
$\psi_{V_{1},V_{2},W_{1},W_{2}}$ is a
$\CC \left[  G_{1}\times G_{2}\right]  $-module homomorphism, qed.}. As a
consequence, $\psi_{V_{1},V_{2},W_{1},W_{2}}$ is a $\CC \left[
G_{1}\times G_{2}\right]  $-module isomorphism (since $\psi_{V_{1},V_{2}
,W_{1},W_{2}}$ is a $\CC \left[  G_{1}\times G_{2}\right]  $-module
homomorphism and a vector space isomorphism).

Applying (\ref{sol.HomxHom.lemma}) to $G=G_{1}$, $M=V_{1}$ and $N=W_{1}$, we
obtain
\begin{equation}
\Hom_{\CC G_{1}}\left(  V_{1},W_{1}\right)
=\left(  \Hom \left(  V_{1},W_{1}\right)  \right)  ^{G_{1}}.
\label{sol.HomxHom.5}
\end{equation}


Applying (\ref{sol.HomxHom.lemma}) to $G=G_{2}$, $M=V_{2}$ and $N=W_{2}$, we
obtain
\begin{equation}
\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)
=\left(  \Hom \left(  V_{2},W_{2}\right)  \right)  ^{G_{2}}
.\label{sol.HomxHom.6}
\end{equation}


Let us now recall the following trivial fact: If $G$ is a group and $M$ and
$N$ are two $\CC G$-modules, and if $\phi:M\rightarrow N$ is a
$\CC G$-module isomorphism, then $\phi\left(  M^{G}\right)  =N^{G}$.
Applying this to $G=G_{1}\times G_{2}$, $M=\Hom \left(
V_{1},W_{1}\right)  \otimes\Hom \left(  V_{2},W_{2}\right)  $,
$N=\Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)  $
and $\phi=\psi_{V_{1},V_{2},W_{1},W_{2}}$, we obtain
\[
\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  \left(  \Hom \left(
V_{1},W_{1}\right)  \otimes\Hom \left(  V_{2},W_{2}\right)
\right)  ^{G_{1}\times G_{2}}\right)  =\left(  \Hom \left(
V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)  \right)  ^{G_{1}\times G_{2}}.
\]
Since
\begin{align*}
& \left(  \Hom \left(  V_{1},W_{1}\right)  \otimes
\Hom \left(  V_{2},W_{2}\right)  \right)  ^{G_{1}\times G_{2}} \\
& =\underbrace{\left(  \Hom \left(  V_{1},W_{1}\right)  \right)
^{G_{1}}}_{\substack{=\Hom_{\CC G_{1}}\left(
V_{1},W_{1}\right)  \\\text{(by (\ref{sol.HomxHom.5}))}}}\otimes
\underbrace{\left(  \Hom \left(  V_{2},W_{2}\right)  \right)
^{G_{2}}}_{\substack{=\Hom_{\CC G_{2}}\left(
V_{2},W_{2}\right)  \\\text{(by (\ref{sol.HomxHom.6}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by (\ref{invariants-vs-tensor-product}), applied to }K_{1}=G_{1}\text{,
}K_{2}=G_{2}\text{,}\\
U_{1}=\Hom \left(  V_{1},W_{1}\right)  \text{ and }
U_{2}=\Hom \left(  V_{2},W_{2}\right)
\end{array}
\right)  \\
& =\Hom_{\CC G_{1}}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)  ,
\end{align*}
this rewrites as
\[
\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  \Hom_{\CC G_{1}}
\left(  V_{1},W_{1}\right)  \otimes
\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)  \right)  =\left(
\Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\right)  ^{G_{1}\times G_{2}}.
\]
Hence,
\begin{align*}
& \left(  \Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  \right)  ^{G_{1}\times G_{2}}\\
& =\psi_{V_{1},V_{2},W_{1},W_{2}}
\left(  \Hom_{\CC G_{1}}\left(  V_{1},W_{1}\right)  \otimes
\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)
\right)  \\
& =\left\{  \underbrace{\psi_{V_{1},V_{2},W_{1},W_{2}}\left(  p\right)
}_{\substack{=\Psi\left(  p\right)  \\
\text{(by (\ref{sol.HomxHom.restr}))}}}\ \mid\ 
p\in\Hom_{\CC G_{1}}\left( V_{1},W_{1}\right)  \otimes
\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)  \right\}  \\
& =\left\{  \Psi\left(  p\right)  \ \mid\ 
p\in\Hom_{\CC G_{1}}\left(  V_{1},W_{1}\right)  \otimes
\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)
\right\}  \\
& =\Psi\left(  \Hom_{\CC G_{1}}\left(
V_{1},W_{1}\right)  \otimes\Hom_{\CC G_{2}}
\left(  V_{2},W_{2}\right)  \right)  .
\end{align*}
But applying (\ref{sol.HomxHom.lemma}) to $G=G_{1}\times G_{2}$,
$M=V_{1}\otimes V_{2}$ and $N=W_{1}\otimes W_{2}$, we obtain
\begin{align*}
\Hom_{\CC \left[  G_{1}\times G_{2}\right]
}\left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)    & =\left(
\Hom \left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\right)  ^{G_{1}\times G_{2}}\\
& =\Psi\left(  \Hom_{\CC G_{1}}\left(
V_{1},W_{1}\right)  \otimes\Hom_{\CC G_{2}}
\left(  V_{2},W_{2}\right)  \right)  .
\end{align*}
In other words, the map $\Psi$ is surjective. Combined with the fact that
$\Psi$ is injective, this yields that $\Psi$ is bijective. Thus, we know that
$\Psi$ is a bijective $\CC$-linear map. Therefore, $\Psi$ is a vector
space isomorphism. In other words, the $\CC$-linear map
\[
\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)
\rightarrow\Hom_{{\CC }\left[  G_{1}\times G_{2}\right]
}\left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms is a vector space isomorphism (because $\Psi$ is precisely this
$\CC$-linear map). This solves Exercise \ref{exe.HomxHom}(a).

(b) Let $G_{1}$ and $G_{2}$ be two groups. Let $V_{i}$ and $W_{i}$ be
finite-dimensional ${\CC }G_{i}$-modules for every $i\in\left\{
1,2\right\}  $.  Exercise \ref{exe.HomxHom}(a) yields that the
$\CC$-linear map
\[
\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)
\rightarrow\Hom_{{\CC }\left[  G_{1}\times G_{2}\right]
}\left(  V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)
\]
sending each tensor $f\otimes g$ to the tensor product $f\otimes g$ of
homomorphisms is a vector space isomorphism. Hence, the $\CC$-vector
spaces $\Hom_{\CC G_1}\left(  V_{1},W_{1}\right)
\otimes\Hom_{\CC G_2}\left(  V_{2},W_{2}\right)  $ and
$\Hom_{{\CC }\left[  G_{1}\times G_{2}\right]  }\left(
V_{1}\otimes V_{2},W_{1}\otimes W_{2}\right)  $ are isomorphic. Therefore,
\begin{align}
& \dim_{\CC }\left(  \Hom_{\CC G_1}\left(
V_{1},W_{1}\right)  \otimes\Hom_{\CC G_2}\left(
V_{2},W_{2}\right)  \right)  \nonumber\\
& =\dim_{\CC }\left(  \Hom_{{\CC }\left[
G_{1}\times G_{2}\right]  }\left(  V_{1}\otimes V_{2},W_{1}\otimes
W_{2}\right)  \right)  .\label{sol.HomxHom.b.1}
\end{align}


Applying (\ref{character-scalprod}) to $G_{1}$, $V_{1}$ and $W_{1}$ instead of
$G$, $V_{1}$ and $V_{2}$, we obtain
\begin{equation}
\left(  \chi_{V_{1}},\chi_{W_{1}}\right)  _{G_{1}}=\dim_{\CC }\left(
\Hom_{\CC G_{1}}\left(  V_{1},W_{1}\right)
\right)  .
\label{sol.HomxHom.b.2}
\end{equation}
Applying (\ref{character-scalprod}) to $G_{2}$, $V_{2}$ and $W_{2}$ instead of
$G$, $V_{1}$ and $V_{2}$, we obtain
\begin{equation}
\left(  \chi_{V_{2}},\chi_{W_{2}}\right)  _{G_{2}}=\dim_{\CC }\left(
\Hom_{\CC G_{2}}\left(  V_{2},W_{2}\right)
\right)  .
\label{sol.HomxHom.b.3}
\end{equation}


Applying (\ref{character-scalprod}) to $G_{1}\times G_{2}$, $V_{1}\otimes
V_{2}$ and $W_{1}\otimes W_{2}$ instead of $G$, $V_1$ and $V_2$, we obtain
\begin{align*}
\left(  \chi_{V_{1}\otimes V_{2}},\chi_{W_{1}\otimes W_{2}}\right)
_{G_{1}\times G_{2}}  
& =\dim_{\CC }\left( 
 \Hom_{\CC \left[  G_{1}\times G_{2}\right]  }\left(  V_{1}\otimes
V_{2},W_{1}\otimes W_{2}\right)  \right)  \\
& =\dim_{\CC }\left(  \Hom_{\CC G_1}\left(
V_{1},W_{1}\right)  \otimes\Hom_{\CC G_2}\left(
V_{2},W_{2}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.HomxHom.b.1})}\right)  \\
& =\underbrace{\dim_{\CC }\left(  \Hom_{\CC G_1}
\left(  V_{1},W_{1}\right)  \right)  }_{\substack{=\left(  \chi_{V_{1}}
,\chi_{W_{1}}\right)  _{G_{1}}\\\text{(by (\ref{sol.HomxHom.b.2}))}}}
\cdot\underbrace{\dim_{\CC }\left(  \Hom_{\CC G_2}
\left(  V_{2},W_{2}\right)  \right)  }_{\substack{=\left(  \chi_{V_{2}}
,\chi_{W_{2}}\right)  _{G_{2}}\\\text{(by (\ref{sol.HomxHom.b.3}))}}}\\
& =\left(  \chi_{V_{1}},\chi_{W_{1}}\right)  _{G_{1}}\left(  \chi_{V_{2}}
,\chi_{W_{2}}\right)  _{G_{2}}.
\end{align*}
This proves (\ref{character-tensorprod}). Thus, Exercise \ref{exe.HomxHom}(b)
is solved.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.ind-fixed-points}}

\textit{Solution to Exercise~\ref{exe.ind-fixed-points}.} If $A$ and
$B$ are two algebras, $P$ is a $\left(  B,A\right)  $-bimodule and $Q$ is a
left $B$-module, then $\Hom_{B}\left(  P,Q\right)  $ is a left
$A$-module. Consequently, $\Hom_{\CC H}\left(
CG,U\right)  $ is a left $\CC G$-module, and
$\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(
\CC \left[  G/K\right]  ,U^{H\cap K}\right)  $ is a left $\CC 
\left[  G/K\right]  $-module. Exercise~\ref{exe.induction.ind-as-hom} yields
that the $\CC G$-module $\Hom_{\CC H}\left(
{\CC}G,U\right)  $ is isomorphic to $\Ind_{H}^{G}U$. In
other words,
\begin{equation}
\Hom_{\CC H}\left(  {\CC}G,U\right)
\cong\Ind_{H}^{G}U\ \ \ \ \ \ \ \ \ \ \text{as }
\CC G\text{-modules.} \label{sol.ind-fixed-points.1}
\end{equation}


Also, Exercise~\ref{exe.induction.ind-as-hom} (applied to $G/K$, $H/\left(
H\cap K\right)  $ and $U^{H\cap K}$ instead of $G$, $H$ and $U$) yields that
the $\CC \left[  G/K\right]  $-module
$\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(
\CC \left[  G/K\right]  ,U^{H\cap K}\right)  $ is isomorphic to
$\Ind_{H/\left(  H\cap K\right)  }^{G/K}\left(
U^{H\cap K}\right)  $. In other words,
\begin{equation}
\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
\cong\Ind_{H/\left(  H\cap K\right)  }^{G/K}\left(
U^{H\cap K}\right)  \ \ \ \ \ \ \ \ \ \ \text{as }\CC \left[
G/K\right]  \text{-modules.} \label{sol.ind-fixed-points.2}
\end{equation}


But we also have
\begin{equation}
\left(  \Hom_{\CC H}\left(  {\CC}G,U\right)
\right)  ^{K}\cong \Hom_{\CC \left[  H/\left(
H\cap K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap
K}\right)  \ \ \ \ \ \ \ \ \ \ \text{as }\CC \left[  G/K\right]
\text{-modules.} \label{sol.ind-fixed-points.3}
\end{equation}

Here is just a brief sketch of the proof of \eqref{sol.ind-fixed-points.3}:
Let $\pi$ be the canonical projection $G\rightarrow G/K$, and let
$\CC \left[  \pi\right]  $ be the $\CC$-linear map
$\CC G\rightarrow{\CC}\left[  G/K\right]  $ obtained by
$\CC$-linearly extending $\pi$. Clearly, $\CC \left[  \pi\right]  $ is a
surjective $\CC G$-linear $\CC$-algebra homomorphism, and we have
\begin{align}
\ker\left(  \CC \left[  \pi\right]  \right)    & =\left\langle
g-g^{\prime}\ \mid\ g\in G,\ g^{\prime}\in G,\ gK=g^{\prime}K\right\rangle
\label{sol.ind-fixed-points.ker.1}\\
& =\left\langle g-gk\ \mid\ g\in G,\ k\in K\right\rangle
\label{sol.ind-fixed-points.ker.2}\\
& =\left\langle g-g^{\prime}\ \mid\ g\in G,\ g^{\prime}\in G,\ Kg=Kg^{\prime
}\right\rangle \label{sol.ind-fixed-points.ker.3}\\
& =\left\langle g-kg\ \mid\ g\in G,\ k\in K\right\rangle
\label{sol.ind-fixed-points.ker.4}
\end{align}
(where the $\left\langle \cdot\right\rangle $ brackets stand for
``$\CC$-span'').

Let $f$ be an element of $\left(  \Hom_{\CC H}\left(
\CC G, U\right)  \right)  ^{K}$. Then, $f\in\left(
\Hom_{\CC H}\left(  \CC G,U\right)  \right)
^{K}\subset\Hom_{\CC H}\left(  \CC G,U\right)  $ is $\CC H$-linear,
and it can easily be shown that
$\ker\left(  \CC \left[  \pi\right]  \right)  \subset\ker f$
\ \ \ \ \footnote{Indeed, $f\in\left(  \Hom_{\CC H}
\left( \CC G, U\right)  \right) ^K$. Hence, every $k\in K$
satisfies $kf=f$. Thus, every $g\in G$ and $k\in K$ satisfy $\left(
kf\right)  \left(  g\right)  =f\left(  g\right)  $. But since $\left(
kf\right)  \left(  g\right)  =f\left(  gk\right)  $ (by the definition of the
action of $G$ on $\Hom_{\CC H}\left( \CC G, U\right)  $), this becomes
$f\left(  gk\right)  =f\left(  g\right)  $, so
that $f\left(  g-gk\right)  =0$. The map $f$ therefore annihilates $g-gk$ for
all $g\in G$ and $k\in K$. Due to \eqref{sol.ind-fixed-points.ker.2}, this
yields $f\left(  \ker\left(  \CC \left[  \pi\right]  \right)  \right)
=0$, so that $\ker\left(  \CC \left[  \pi\right]  \right)  \subset\ker
f$, qed.}. Hence, the map $f$ factors through the surjective map
$\CC \left[  \pi\right]  $. The resulting map ${\CC}\left[
G/K\right]  \rightarrow U$ is a ${\CC}H$-module homomorphism (since
$\CC \left[  \pi\right]  $ and $f$ were both $\CC H$-linear), and
it is easy to see that its image (i.e., the image of $f$) is contained in
$U^{H\cap K}$\ \ \ \ \footnote{\textit{Proof.} We want to show that the image
of $f$ is contained in $U^{H\cap K}$.
\par
For this, it clearly suffices to prove that $f\left(  g\right)  \in U^{H\cap
K}$ for every $g\in G$. So fix $g\in G$. Let $k\in H\cap K$. Then, $k\cdot
f\left(  g\right)  =f\left(  kg\right)  $ (since $f$ is $\CC H$-linear
and $k\in H$) and $f\left(  g\right)  =f\left(  kg\right)  $ (since $k\in K$,
so that \eqref{sol.ind-fixed-points.ker.4} yields $g-kg\in\ker\left(
\CC \left[  \pi\right]  \right)  \subset\ker f$ and thus $f\left(
g-kg\right)  =0$). Hence, $k\cdot f\left(  g\right)  =f\left(  kg\right)
=f\left(  g\right)  $. Since this has been proven for all $k\in H\cap K$, we
thus have $f\left(  g\right)  \in U^{H\cap K}$, qed.}. Hence, this map factors
through the canonical inclusion $U^{H\cap K}\rightarrow U$, leaving behind a
map ${\CC}\left[  G/K\right]  \rightarrow U^{H\cap K}$ which we denote
by $\Phi\left(  f\right)  $. This resulting map $\Phi\left(  f\right)  $ turns
out to be ${\CC}\left[  H/\left(  H\cap K\right)  \right]
$-linear\footnote{This is straightforward to see (everything in sight is
$\CC H$-linear).}, thus belongs to
$\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(  \CC \left[
G/K\right]  ,U^{H\cap K}\right)  $. Since this holds for every $f$, we thus
obtain a ${\CC}$-linear map
\begin{align*}
\left(  \Hom_{\CC H}\left(  {\CC}G,U\right)
\right)  ^{K}  & \rightarrow\Hom_{{\CC}\left[  H/\left(
H\cap K\right)  \right]  }\left(  {\CC}\left[  G/K\right]  ,U^{H\cap
K}\right)  ,\\
f  & \mapsto\Phi\left(  f\right)  .
\end{align*}
This map is invertible\footnote{In fact, defining the inverse is very easy
(just send every map $g\in\Hom_{{\CC}\left[  H/\left(
H\cap K\right)  \right]  }\left(  {\CC}\left[  G/K\right]  ,U^{H\cap
K}\right)  $ to the composition $\CC G\overset{\CC \left[
\pi\right]  }{\longrightarrow}{\CC}\left[  G/K\right]
\overset{g}{\longrightarrow}U^{H\cap K}
\overset{\text{inclusion}}{\longrightarrow}U$). Checking that these maps
are mutually inverse is also
straightforward.} and
$\CC \left[ G/K\right]$-linear\footnote{Indeed, it is easier to check
that its inverse is
$\CC \left[ G/K\right]$-linear (this can be proven by straightforward
computations).}, therefore an isomorphism of ${\CC}\left[  G/K\right]
$-modules. This proves \eqref{sol.ind-fixed-points.3}. All steps that were
left to the reader are straightforward.

Now,
\begin{align*}
\left(  \underbrace{\Ind_{H}^{G}U}_{\substack{\cong
\Hom_{\CC H}\left(  {\CC}G,U\right)  \text{ as
}\CC G\text{-modules}\\\text{(by \eqref{sol.ind-fixed-points.1})}
}}\right)  ^{K}  & \cong\left(  \Hom_{\CC H}\left(
{\CC}G,U\right)  \right)  ^{K}\\
& \cong\Hom_{\CC \left[  H/\left(  H\cap
K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.ind-fixed-points.3}}\right)
\\
& \cong\Ind_{H/\left(  H\cap K\right)  }^{G/K}\left(  U^{H\cap K}\right)  
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.ind-fixed-points.2}}\right)
\end{align*}
as $\CC \left[ G/K \right]$-modules. This solves
Exercise~\ref{exe.ind-fixed-points}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.ind-fixed-points}}

\textit{Solution to Exercise~\ref{exe.ind-fixed-points}.} If $A$ and
$B$ are two algebras, $P$ is a $\left(  B,A\right)  $-bimodule and $Q$ is a
left $B$-module, then $\Hom_{B}\left(  P,Q\right)  $ is a left
$A$-module. Consequently, $\Hom_{\CC H}\left(
CG,U\right)  $ is a left $\CC G$-module, and
$\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(
\CC \left[  G/K\right]  ,U^{H\cap K}\right)  $ is a left
$\CC \left[  G/K\right]$-module. Exercise~\ref{exe.induction.ind-as-hom}
yields that the $\CC G$-module $\Hom_{\CC H}\left(
{\CC }G,U\right)  $ is isomorphic to $\Ind_{H}^{G}U$. In
other words,
\begin{equation}
\Hom_{\CC H}\left(  {\CC }G,U\right)
\cong\Ind_{H}^{G}U\ \ \ \ \ \ \ \ \ \ \text{as }\CC G\text{-modules.}
\label{sol.ind-fixed-points.1}
\end{equation}


Also, Exercise~\ref{exe.induction.ind-as-hom} (applied to $G/K$, $H/\left(
H\cap K\right)  $ and $U^{H\cap K}$ instead of $G$, $H$ and $U$) yields that
the $\CC \left[  G/K\right]  $-module
$\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(
\CC \left[  G/K\right]  ,U^{H\cap K}\right)  $ is isomorphic to
$\Ind_{H/\left(  H\cap K\right)  }^{G/K}\left(
U^{H\cap K}\right)  $. In other words,
\begin{equation}
\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
\cong\Ind_{H/\left(  H\cap K\right)  }^{G/K}\left(
U^{H\cap K}\right)  \ \ \ \ \ \ \ \ \ \ \text{as }\CC \left[
G/K\right]  \text{-modules.}
\label{sol.ind-fixed-points.2}
\end{equation}


We shall now show that
\begin{equation}
\left(  \Hom_{\CC H}\left(  {\CC }G,U\right)
\right)  ^{K}\cong \Hom_{\CC \left[  H/\left(
H\cap K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap
K}\right)  \ \ \ \ \ \ \ \ \ \ \text{as }\CC \left[  G/K\right]
\text{-modules.} \label{sol.ind-fixed-points.3}
\end{equation}


\textit{Proof of (\ref{sol.ind-fixed-points.3}):} Let us first introduce some
notation. For any set $S$ and any field $\kk$, we let $\kk S$
denote the $\kk$-vector space with basis $\left\{  t_{s}\right\}
_{s\in S}$. We sometimes write $\kk\left[  S\right]  $ instead of
$\kk S$. We will also write $s$ for the basis element $t_{s}$ whenever
$s\in S$ (this is an abuse of notation, but it is harmless since the map
$S\rightarrow \kk S$ sending each $s$ to $t_{s}$ is injective\footnote{And
even if this map were not injective (e.g., if we would allow $\kk$ to
be an arbitrary ring, which includes the possibility of it being
the zero ring), then this would be harmless, since in the arguments
that follow, we will never deduce the equality of two elements of
$S$ from that of their corresponding basis vectors in $\kk S$
(but only the other way round).}). When $S$
is a group $G$, the $\kk$-vector space $\kk S$ is identical with
the underlying vector space of the $\kk$-algebra $\kk G$ defined
in Example~\ref{group-algebra-example}.

If $\kk$ is a field, if $S$ and $T$ are two sets, and if $\varphi
:S\rightarrow T$ is a map, then we let $\kk \left[  \varphi\right]  $
denote the $\kk$-linear map $\kk S \rightarrow \kk T$ which
sends the basis element $s$ of $\kk S$ to $\varphi\left(  s\right)
\in \kk T$ for every $s\in S$. This map $\kk \left[  \varphi
\right]  $ is clearly well-defined. It satisfies
\begin{equation}
\left(  \kk \left[  \varphi\right]  \right)  \left(  s\right)
=\varphi\left(  s\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }s\in S\text{.}
\label{sol.ind-fixed-points.s}
\end{equation}
Moreover,
\begin{equation}
\text{this map }\kk \left[  \varphi\right]  \text{ is surjective if
}\varphi\text{ is surjective.} \label{sol.ind-fixed-points.surj}
\end{equation}
\footnote{\textit{Proof.} Let $\kk$ be a field. Let $S$ and $T$ be two
sets. Let $\varphi:S\rightarrow T$ be a map. Assume that $\varphi$ is
surjective. We must prove that the map $\kk \left[  \varphi\right]  $ is
surjective.
\par
Let $v\in\kk T$ be arbitrary. Then, $v$ is a $\kk$-linear
combination of elements of $T$ (since $T$ is a basis of the
$\kk$-vector space $\kk T$). In other words, there exist some $n\in
\NN$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$
of $\kk$ and some elements $p_{1}$, $p_{2}$, $...$, $p_{n}$ of $T$ such
that $v=\sum_{i=1}^{n}\lambda_{i}p_{i}$. Consider this $n$, these $\lambda
_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$, and these $p_{1}$, $p_{2}$, $...$,
$p_{n}$.
\par
For every $i\in\left\{  1,2,...,n\right\}  $, there exists an $s\in S$ such
that $p_{i}=\varphi\left(  s\right)  $ (since $\varphi$ is surjective). Choose
such an $s$ (for every given $i\in\left\{  1,2,...,n\right\}  $) and denote it
by $s_{i}$. Hence, for every $i\in\left\{  1,2,...,n\right\}  $, the element
$s_{i}$ is an element of $S$ such that $p_{i}=\varphi\left(  s_{i}\right)  $.
\par
But every $i\in\left\{  1,2,...,n\right\}  $ satisfies
\begin{align*}
\left(  \kk \left[  \varphi\right]  \right)  \left(  s_{i}\right)   &
=\varphi\left(  s_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.ind-fixed-points.s}), applied to }s=s_{i}\right) \\
&  =p_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{i}=\varphi\left(
s_{i}\right)  \right)  .
\end{align*}
\par
Now, $\sum_{i=1}^{n}\lambda_{i}s_{i}$ is an element of $\kk S$ and
satisfies
\[
\left(  \kk \left[  \varphi\right]  \right)  \left(  \sum_{i=1}^{n}
\lambda_{i}s_{i}\right)  =\sum_{i=1}^{n}\lambda_{i}\underbrace{\left(
\kk \left[  \varphi\right]  \right)  \left(  s_{i}\right)  }_{=p_{i}}
=\sum_{i=1}^{n}\lambda_{i}p_{i}=v.
\]
Hence, $v=\left(  \kk \left[  \varphi\right]  \right)  \left(
\underbrace{\sum_{i=1}^{n}\lambda_{i}s_{i}}_{\in\kk S}\right)
\in\left(  \kk \left[  \varphi\right]  \right) \left( \kk S\right)$.
\par
Now, forget that we fixed $v$. We thus have shown that every $v\in\kk T$
satisfies $v\in\left(  \kk \left[  \varphi\right]  \right)  \left(
\kk S\right)  $. In other words, $\kk T\subset\left(
\kk \left[  \varphi\right]  \right)  \left(  \kk S\right)  $. In
other words, the map $\kk \left[  \varphi\right]  $ is surjective. This
proves (\ref{sol.ind-fixed-points.surj}).} Furthermore, if $S$ and $T$ are two
groups, and if $\varphi:S\rightarrow T$ is a group homomorphism, then
\begin{equation}
\text{the map }\kk \left[  \varphi\right]  :\kk S\rightarrow
\kk T\text{ is a }\kk \text{-algebra homomorphism.}
\label{sol.ind-fixed-points.alg}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.alg}):} Let $S$ and $T$
be two groups. Let $\varphi:S\rightarrow T$ be a group homomorphism.
\par
Whenever $P$ is a group, we denote by $1_{P}$ the neutral element of the group
$P$.
\par
We have
\begin{align*}
\left(  \kk \left[  \varphi\right]  \right)  \left(  1_{S}\right)   &
=\varphi\left(  1_{S}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.ind-fixed-points.s}), applied to }s=1_{S}\right) \\
&  =1_{T}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is a group
homomorphism}\right)  .
\end{align*}
In other words, the map $\kk \left[  \varphi\right]  $ sends $1_{S}$ to
$1_{T}$. Since $1_{S}$ is the unity of the $\kk$-algebra $\kk S$,
whereas $1_{T}$ is the unity of the $\kk$-algebra $\kk T$, this
rewrites as follows: The map $\kk \left[  \varphi\right]  $ sends the
unity of the $\kk$-algebra $\kk S$ to the unity of the
$\kk$-algebra $\kk T$.
\par
Now, let $v$ and $w$ be any two elements of $\kk S$.
\par
We have $v\in\kk S$. Thus, we can write $v$ in the form $v=\sum_{s\in
S}\lambda_{s}s$ for some family $\left(  \lambda_{s}\right)  _{s\in S}$ of
elements of $\kk$ (since $S$ is a $\kk$-vector space basis of
$\kk S$). Consider this family $\left(  \lambda_{s}\right)  _{s\in S}$.
\par
We have $w\in\kk S$. Thus, we can write $w$ in the form $w=\sum_{s\in
S}\mu_{s}s$ for some family $\left(  \mu_{s}\right)  _{s\in S}$ of elements of
$\kk$ (since $S$ is a $\kk$-vector space basis of $\kk S$). Consider
this family $\left(  \mu_{s}\right)  _{s\in S}$.
\par
We have $w=\sum_{s\in S}\mu_{s}s=\sum_{t\in S}\mu_{t}t$ (here, we renamed the
index $s$ as $t$ in the sum).
\par
Now,
\[
\underbrace{v}_{=\sum_{s\in S}\lambda_{s}s}\underbrace{w}_{=\sum_{t\in S}
\mu_{t}t}=\left(  \sum_{s\in S}\lambda_{s}s\right)  \left(  \sum_{t\in S}
\mu_{t}t\right)  =\sum_{s\in S}\sum_{t\in S}\lambda_{s}\mu_{t}st.
\]
Applying the map $\kk \left[  \varphi\right]  $ to both sides of this
equation, we obtain
\begin{align*}
\left(  \kk \left[  \varphi\right]  \right)  \left(  vw\right)   &
=\left(  \kk \left[  \varphi\right]  \right)  \left(  \sum_{s\in S}
\sum_{t\in S}\lambda_{s}\mu_{t}st\right)  =\sum_{s\in S}\underbrace{\left(
\kk \left[  \varphi\right]  \right)  \left(  \sum_{t\in S}\lambda_{s}
\mu_{t}st\right)  }_{\substack{=\sum_{t\in S}\lambda_{s}\mu_{t}\left(
\kk \left[  \varphi\right]  \right)  \left(  st\right)  \\\text{(since
the map }\kk \left[  \varphi\right]  \text{ is }\kk \text{-linear)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\kk \left[
\varphi\right]  \text{ is }\kk \text{-linear}\right) \\
&  =\sum_{s\in S}\sum_{t\in S}\lambda_{s}\mu_{t}\underbrace{\left(
\kk \left[  \varphi\right]  \right)  \left(  st\right)  }
_{\substack{=\varphi\left(  st\right)  \\\text{(by
(\ref{sol.ind-fixed-points.s}), applied to }st\\
\text{instead of }s\text{)}}}
=\sum_{s\in S}\sum_{t\in S}\lambda_{s}\mu_{t}\underbrace{\varphi\left(
st\right)  }_{\substack{=\varphi\left(  s\right)  \cdot\varphi\left(
t\right)  \\\text{(since }\varphi\text{ is a group}\\\text{homomorphism)}}}\\
&  =\sum_{s\in S}\sum_{t\in S}\lambda_{s}\mu_{t}\varphi\left(  s\right)
\cdot\varphi\left(  t\right)  =\left(  \sum_{s\in S}\lambda_{s}\varphi\left(
s\right)  \right)  \cdot\left(  \sum_{t\in S}\mu_{t}\varphi\left(  t\right)
\right)  .
\end{align*}
Compared with
\begin{align*}
&  \left(  \kk \left[  \varphi\right]  \right)  \left(  \underbrace{v}
_{=\sum_{s\in S}\lambda_{s}s}\right)  \cdot\left(  \kk \left[
\varphi\right]  \right)  \left(  \underbrace{w}_{=\sum_{t\in S}\mu_{t}
t}\right) \\
&  =\underbrace{\left(  \kk \left[  \varphi\right]  \right)  \left(
\sum_{s\in S}\lambda_{s}s\right)  }_{\substack{=\sum_{s\in S}\lambda
_{s}\left(  \kk \left[  \varphi\right]  \right)  \left(  s\right)
\\\text{(since the map }\kk \left[  \varphi\right]  \text{ is
}\kk \text{-linear)}}}\cdot\underbrace{\left(  \kk \left[
\varphi\right]  \right)  \left(  \sum_{t\in S}\mu_{t}t\right)  }
_{\substack{=\sum_{t\in S}\mu_{t}\left(  \kk \left[  \varphi\right]
\right)  \left(  t\right)  \\\text{(since the map }\kk \left[
\varphi\right]  \text{ is }\kk \text{-linear)}}}\\
&  =\left(  \sum_{s\in S}\lambda_{s}\underbrace{\left(  \kk \left[
\varphi\right]  \right)  \left(  s\right)  }_{\substack{=\varphi\left(
s\right)  \\\text{(by (\ref{sol.ind-fixed-points.s}))}}}\right)  \cdot\left(
\sum_{t\in S}\mu_{t}\underbrace{\left(  \kk \left[  \varphi\right]
\right)  \left(  t\right)  }_{\substack{=\varphi\left(  t\right)  \\\text{(by
(\ref{sol.ind-fixed-points.s}), applied to }t\\
\text{instead of }s\text{)}}}\right) \\
&  =\left(  \sum_{s\in S}\lambda_{s}\varphi\left(  s\right)  \right)
\cdot\left(  \sum_{t\in S}\mu_{t}\varphi\left(  t\right)  \right)  ,
\end{align*}
this yields $\left(  \kk \left[  \varphi\right]  \right)  \left(
vw\right)  =\left(  \kk \left[  \varphi\right]  \right)  \left(
v\right)  \cdot\left(  \kk \left[  \varphi\right]  \right)  \left(
w\right)  $.
\par
Now, forget that we fixed $v$ and $w$. We thus have shown that every
$v\in\kk S$ and $w\in\kk S$ satisfy $\left(  \kk \left[
\varphi\right]  \right)  \left(  vw\right)  =\left(  \kk \left[
\varphi\right]  \right)  \left(  v\right)  \cdot\left(  \kk \left[
\varphi\right]  \right)  \left(  w\right)  $. This (combined with the fact
that the map $\kk \left[  \varphi\right]  $ sends the unity of the
$\kk$-algebra $\kk S$ to the unity of the $\kk$-algebra
$\kk T$) yields that $\kk \left[ \varphi \right]$ is a
$\kk$-algebra homomorphism (since $\kk \left[ \varphi \right]  $
is $\kk$-linear). This proves (\ref{sol.ind-fixed-points.alg}).}

Let $\iota:U^{H\cap K}\rightarrow U$ be the canonical inclusion map. Notice
that $\iota$ is a $\CC$-linear map. Let $\pi_{G}:G\rightarrow G/K$ be
the canonical projection map. Let $\pi_{H}:H \to H/\left(  H\cap K\right)  $ be the
canonical projection map. The maps $\pi_{G}$ and $\pi_{H}$ are surjective
group homomorphisms (since each of them is a projection map from a group onto
its quotient). As we know, the map $\pi_{G}:G\rightarrow G/K$ induces a
$\CC$-linear map $\CC \left[  \pi_{G}\right]  :\CC G\rightarrow
\CC \left[  G/K\right]  $. This map $\CC \left[
\pi_{G}\right]  $ is surjective (by (\ref{sol.ind-fixed-points.surj}), applied
to $\kk =\CC $, $S=G$, $T=G/K$ and $\varphi=\pi_{G}$), since the
map $\pi_{G}$ is surjective. Also, we know that the map $\CC \left[
\pi_{G}\right]  :\CC G\rightarrow\CC \left[  G/K\right]  $ is a
$\CC$-algebra homomorphism\footnote{This follows from
(\ref{sol.ind-fixed-points.alg}) (applied to $\kk =\CC $, $S=G$,
$T=G/K$ and $\varphi=\pi_{G}$), since $\pi_{G}:G\rightarrow G/K$ is a group
homomorphism.}.

Let us state a further general observation. Namely, if $P$ is any
$\CC$-vector space, and if $\rho:\CC G\rightarrow P$ is a $\CC$-linear
map satisfying
\begin{equation}
\left(  \rho\left(  gk\right)  =\rho\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }k\in K\text{ and }g\in G\right)  ,
\label{sol.ind-fixed-points.factor1cond}
\end{equation}
then
\begin{equation}
\text{there exists a }\CC \text{-linear map }\widetilde{\rho}
:\CC \left[  G/K\right]  \rightarrow P\text{ satisfying }\rho
=\widetilde{\rho}\circ\left(  \CC \left[  \pi_{G}\right]  \right)  .
\label{sol.ind-fixed-points.factor1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.factor1}):} Let $P$ be
any $\CC$-vector space. Let $\rho:\CC G\rightarrow P$ be any
$\CC$-linear map satisfying (\ref{sol.ind-fixed-points.factor1cond}).
\par
Let $\gamma\in G/K$. We are going to define a certain element $z_{\gamma}$ of
$P$ depending only on $\gamma$.
\par
Indeed, we have $\gamma\in G/K=\pi_{G}\left(  G\right)  $ (since the map
$\pi_{G}$ is surjective). In other words, there exists a $\gamma^{\prime}\in
G$ such that $\gamma=\pi_{G}\left(  \gamma^{\prime}\right)  $. Consider this
$\gamma^{\prime}$.
\par
We want to define the element $z_{\gamma}$ of $P$ by $z_{\gamma}=\rho\left(
\gamma^{\prime}\right)  $. In order to make sure that this definition is
well-defined, we need to check that the element $\rho\left(  \gamma^{\prime
}\right)  $ of $P$ does not depend on the choice of $\gamma^{\prime}$. Let us
prove this now.
\par
We have $\gamma\in\pi_{G}\left(  G\right)  $. Hence, there exists a $\delta\in
G$ such that $\gamma=\pi_{G}\left(  \delta\right)  $. Consider this $\delta$.
Since $\pi_{G}$ is a group homomorphism, we have
\[
\pi_{G}\left(  \delta^{-1}\gamma^{\prime}\right)  =\left(  \underbrace{\pi
_{G}\left(  \delta\right)  }_{=\gamma}\right)  ^{-1}\cdot\left(
\underbrace{\pi_{G}\left(  \gamma^{\prime}\right)  }_{=\gamma}\right)
=\gamma^{-1}\cdot\gamma=1,
\]
so that $\delta^{-1}\gamma^{\prime}\in\ker\left(  \pi_{G}\right)  =K$ (since
$\pi_{G}$ is the canonical projection $G\rightarrow G/K$). Hence,
(\ref{sol.ind-fixed-points.factor1cond}) (applied to $g=\delta$ and $k=\delta
^{-1}\gamma^{\prime}$) yields $\rho\left(  \delta\delta^{-1}\gamma^{\prime
}\right)  =\rho\left(  \delta\right)  $. Since
$\underbrace{\delta\delta^{-1}}_{=1}\gamma^{\prime}=\gamma^{\prime}$,
this rewrites as
\begin{equation}
\rho\left(  \gamma^{\prime}\right)  =\rho\left(  \delta\right)  .
\label{sol.ind-fixed-points.factor1.pf.1}
\end{equation}
Hence, $\rho\left(  \gamma^{\prime}\right)  $ does not depend on the choice of
$\gamma^{\prime}$ (since $\rho\left(  \delta\right)  $ does not depend on the
choice of $\gamma^{\prime}$ (because even though the choice of $\delta$
involved some freedom, it was not influenced by the choice of $\gamma^{\prime
}$)).
\par
Thus, we have shown that the element $\rho\left(  \gamma^{\prime}\right)  $ of
$P$ does not depend on the choice of $\gamma^{\prime}$. Hence, we can define
an element $z_{\gamma}$ of $P$ by $z_{\gamma}=\rho\left(  \gamma^{\prime
}\right)  $, and this element $z_{\gamma}$ does not depend on the choice of
$\gamma^{\prime}$.
\par
Moreover, this $z_{\gamma}$ satisfies
\begin{equation}
z_{\gamma}=\rho\left(  \delta\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}\delta\in G\text{ such that }\gamma=\pi_{G}\left(  \delta\right)  .
\label{sol.ind-fixed-points.factor1.pf.2}
\end{equation}
\par
(\textit{Proof of (\ref{sol.ind-fixed-points.factor1.pf.2}):} Let $\delta\in
G$ be such that $\gamma=\pi_{G}\left(  \delta\right)  $. Since $\gamma\in
\pi_{G}\left(  G\right)  $, we know that there exists a $\gamma^{\prime}\in G$
such that $\gamma=\pi_{G}\left(  \gamma^{\prime}\right)  $. Consider this
$\gamma^{\prime}$. The definition of $z_{\gamma}$ yields $z_{\gamma}
=\rho\left(  \gamma^{\prime}\right)  =\rho\left(  \delta\right)  $ (by
(\ref{sol.ind-fixed-points.factor1.pf.1})). Hence,
(\ref{sol.ind-fixed-points.factor1.pf.2}) is proven.)
\par
Now, forget that we fixed $\gamma$. Thus, for every $\gamma\in G/K$, we have
defined an element $z_{\gamma}$ of $P$ (depending only on $\gamma$), and we
have shown that this element $z_{\gamma}$ satisfies
(\ref{sol.ind-fixed-points.factor1.pf.2}).
\par
Now, we can define a $\CC$-linear map $\xi:\CC \left[
G/K\right]  \rightarrow P$ by setting
\begin{equation}
\left(  \xi\left(  \gamma\right)  =z_{\gamma}\ \ \ \ \ \ \ \ \ \ \text{for
every }\gamma\in G/K\right)  . \label{sol.ind-fixed-points.factor1.pf.3}
\end{equation}
(This is well-defined because $G/K$ is a basis of the $\CC$-vector
space $\CC \left[  G/K\right]  $.) Consider this $\CC$-linear map
$\xi$.
\par
Every $\delta\in G$ satisfies
\begin{align*}
\left(  \xi\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \right)
\left(  \delta\right)   &  =\xi\left(  \underbrace{\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  \delta\right)  }_{\substack{=\pi_{G}\left(
\delta\right)  \\\text{(by (\ref{sol.ind-fixed-points.s}), applied
to}\\\kk =\CC \text{, }S=G\text{, }T=G/K\text{, }\varphi=\pi
_{G}\text{ and }s=\delta\text{)}}}\right)  =\xi\left(  \pi_{G}\left(
\delta\right)  \right) \\
&  =z_{\pi_{G}\left(  \delta\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by (\ref{sol.ind-fixed-points.factor1.pf.3}) (applied to }\gamma=\pi
_{G}\left(  \delta\right)  \text{)}\\
\text{(since }\pi_{G}\left(  \delta\right)  \in G/K\text{)}
\end{array}
\right) \\
&  =\rho\left(  \delta\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by (\ref{sol.ind-fixed-points.factor1.pf.2}) (applied to }\gamma=\pi
_{G}\left(  \delta\right)  \text{)}\\
\text{(since }\pi_{G}\left(  \delta\right)  =\pi_{G}\left(  \delta\right)
\text{)}
\end{array}
\right)  .
\end{align*}
In other words, the two maps $\xi\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  $ and $\rho$ are equal to each other on every element of
$G$. Since $G$ is a basis of the $\CC$-vector space $\CC G$, this
yields that the two maps $\xi\circ\left(  \CC \left[  \pi_{G}\right]
\right)  $ and $\rho$ are equal to each other on every element of a certain
basis of the $\CC$-vector space $\CC G$. Since these two maps
$\xi\circ\left(  \CC \left[  \pi_{G}\right]  \right)  $ and $\rho$ are
$\CC$-linear (because $\xi$, $\CC \left[  \pi_{G}\right]  $ and
$\rho$ are $\CC$-linear), this yields that these two maps $\xi
\circ\left(  \CC \left[  \pi_{G}\right]  \right)  $ and $\rho$ must be
identical (because if two $\CC$-linear maps $V\rightarrow W$ between
two $\CC$-vector spaces $V$ and $W$ are equal to each other on every
element of a certain basis of $V$, then these two maps must be identical). In
other words, $\xi\circ\left(  \CC \left[  \pi_{G}\right]  \right)
=\rho$. Hence, $\rho=\xi\circ\left(  \CC \left[  \pi_{G}\right]
\right)  $. Hence, there exists a $\CC$-linear map $\widetilde{\rho
}:\CC \left[  G/K\right]  \rightarrow P$ satisfying $\rho
=\widetilde{\rho}\circ\CC \left[  \pi_{G}\right]  $ (namely,
$\widetilde{\rho}=\xi$). This proves (\ref{sol.ind-fixed-points.factor1}).}

Since we are identifying $H/\left(  H\cap K\right)  $ with a subgroup of
$G/K$, we have $\pi_{H}=\pi_{G}\mid_{H}$
(up to the fact that the targets of these maps are different).
Hence, $\pi_{H}\left(  H\right)
=\left(  \pi_{G}\mid_{H}\right)  \left(  H\right)  =\pi_{G}\left(  H\right)
$, so that $\pi_{G}\left(  H\right)  =\pi_{H}\left(  H\right)  =H/\left(
H\cap K\right)  $ (since the map $\pi_{H}$ is surjective).

Also, every $h\in H$ satisfies
\begin{equation}
\underbrace{\pi_{H}}_{=\pi_{G}\mid_{H}}\left(  h\right)  =\left(  \pi_{G}
\mid_{H}\right)  \left(  h\right)  =\pi_{G}\left(  h\right)  .
\label{sol.ind-fixed-points.piGh}
\end{equation}


Furthermore, every $g\in G$ and every $\alpha\in\left( 
\Hom_{\CC H}\left(  \CC G, U\right)  \right)  ^{K}$ satisfy
\begin{equation}
\pi_{G}\left(  g\right)  \cdot\alpha=g\alpha
\label{sol.ind-fixed-points.fix-act.G}
\end{equation}
(by the definition of the $\CC \left[  G/K\right]  $-module structure on
$\left(  \Hom_{\CC H}\left(  {\CC }G,U\right)
\right)  ^{K}$). Moreover, every $h\in H$ and every $\beta\in U^{H\cap K}$
satisfy
\begin{equation}
\pi_{H}\left(  h\right)  \cdot\beta=h\beta
\label{sol.ind-fixed-points.fix-act.H}
\end{equation}
(by the definition of the $\CC \left[  H/\left(  H\cap K\right)
\right]  $-module structure on $U^{H\cap K}$).

We have
\begin{equation}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  wk\right)  =\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }w\in\CC G\text{ and }k\in K.
\label{sol.ind-fixed-points.Kignore}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.Kignore}):} Let
$w\in\CC G$ and $k\in K$. We know that $\pi_{G}$ is the canonical
projection $G\rightarrow G/K$; hence, $\ker\left(  \pi_{G}\right)  =K$. Now,
$k\in K=\ker\left(  \pi_{G}\right)  $, so that $\pi_{G}\left(  k\right)  =1$.
But $\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  k\right)
=\pi_{G}\left(  k\right)  $ (by (\ref{sol.ind-fixed-points.s}), applied to
$\kk =\CC $, $S=G$, $T=G/K$, $\varphi=\pi_{G}$ and $s=k$). Hence,
$\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  k\right)  =\pi
_{G}\left(  k\right)  =1$.
\par
We know that the map $\CC \left[  \pi_{G}\right]  :\CC G
\rightarrow\CC \left[  G/K\right]  $ is a $\CC$-algebra
homomorphism. Hence,
\[
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  wk\right)  =\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)  \cdot
\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
k\right)  }_{=1}=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  .
\]
This proves (\ref{sol.ind-fixed-points.Kignore}).}

Every $f\in\Hom_{\CC \left[  H/\left(  H\cap
K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
$ satisfies
\begin{equation}
\iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \in\left(
\Hom_{\CC H}\left(  {\CC }G,U\right)  \right)
^{K}. \label{sol.ind-fixed-points.wd}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.wd}):} Let
$f\in\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)  $. Then,
$f$ is a $\CC \left[  H/\left(  H\cap K\right)  \right]  $-module
homomorphism $\CC \left[  G/K\right]  \rightarrow U^{H\cap K}$. Hence,
$f$ is a $\CC$-linear homomorphism of $H/\left(  H\cap K\right)  $-sets
$\CC \left[  G/K\right]  \rightarrow U^{H\cap K}$.
\par
We know that $f$ is a homomorphism of $H/\left(  H\cap K\right)  $-sets. In
other words,
\begin{equation}
f\left(  jv\right)  =jf\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}v\in\CC \left[  G/K\right]  \text{ and }j\in H/\left(  H\cap K\right)
. \label{sol.ind-fixed-points.wd.1}
\end{equation}
\par
The composition $\iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]
\right)  $ of the $\CC$-linear maps $\CC \left[  \pi_{G}\right]
:\CC G\rightarrow\CC \left[  G/K\right]  $, $f:\CC \left[
G/K\right]  \rightarrow U^{H\cap K}$ and $\iota:U^{H\cap K}\rightarrow U$ is
clearly a $\CC$-linear map $\CC G\rightarrow U$. We will now show
that this composition is a $\CC H$-linear map.
\par
Let $F$ be the map $\iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]
\right)  $. Then, $F$ is a $\CC$-linear map.
\par
In fact, let $h\in H$. Further, let $g\in G$ be arbitrary. Then,
\begin{align*}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  hg\right)   &
=\pi_{G}\left(  hg\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by (\ref{sol.ind-fixed-points.s}), applied to}\\
\kk =\CC \text{, }S=G\text{, }T=G/K\text{, }\varphi=\pi_{G}\text{
and }s=hg
\end{array}
\right) \\
&  =\pi_{G}\left(  h\right)  \cdot\pi_{G}\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{G}\text{ is a group
homomorphism}\right)
\end{align*}
and $\pi_{G}\left(  \underbrace{h}_{\in H}\right)  \in\pi_{G}\left(  H\right)
=H/\left(  H\cap K\right)  $. Now,
\begin{align*}
\underbrace{F}_{=\iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]
\right)  }\left(  hg\right)   &  =\left(  \iota\circ f\circ\left(
\CC \left[  \pi_{G}\right]  \right)  \right)  \left(  hg\right)
=\left(  \iota\circ f\right)  \left(  \underbrace{\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  hg\right)  }_{=\pi_{G}\left(  h\right)
\cdot\pi_{G}\left(  g\right)  }\right)  =\iota\left(  f\left(  \pi_{G}\left(
h\right)  \cdot\pi_{G}\left(  g\right)  \right)  \right) \\
&  =f\left(  \pi_{G}\left(  h\right)  \cdot\pi_{G}\left(  g\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\iota\text{ is just an inclusion
map}\right) \\
&  =\underbrace{\pi_{G}\left(  h\right)  }_{\substack{=\pi_{H}\left(
h\right)  \\\text{(by (\ref{sol.ind-fixed-points.piGh}))}}}\cdot f\left(
\pi_{G}\left(  g\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by (\ref{sol.ind-fixed-points.wd.1}) (applied to }v=\pi_{G}\left(
g\right)  \text{ and }j=\pi_{G}\left(  h\right)  \text{)}\\
\text{(since }\pi_{G}\left(  h\right)  \in H/\left(  H\cap K\right)  \text{)}
\end{array}
\right) \\
&  =\pi_{H}\left(  h\right)  \cdot f\left(  \pi_{G}\left(  g\right)  \right)
\\
&  =h\cdot f\left(  \pi_{G}\left(  g\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.ind-fixed-points.fix-act.H}),
applied to }\beta=f\left(  \pi_{G}\left(  g\right)  \right)  \right)  .
\end{align*}
Since
\begin{align}
\underbrace{F}_{=\iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]
\right)  }\left(  g\right)   &  =\left(  \iota\circ f\circ\left(
\CC \left[  \pi_{G}\right]  \right)  \right)  \left(  g\right)  =\left(
\iota\circ f\right)  \left(  \underbrace{\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  g\right)  }_{\substack{=\pi_{G}\left(  g\right)
\\\text{(by (\ref{sol.ind-fixed-points.s}), applied to}\\\kk 
=\CC \text{, }S=G\text{, }T=G/K\text{, }\varphi=\pi_{G}\text{ and
}s=g\text{)}}}\right)  =\left(  \iota\circ f\right)  \left(  \pi_{G}\left(
g\right)  \right) \nonumber\\
&  =\iota\left(  f\left(  \pi_{G}\left(  g\right)  \right)  \right)  =f\left(
\pi_{G}\left(  g\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\iota\text{ is just an inclusion map}\right)  ,
\label{sol.ind-fixed-points.wd.3}
\end{align}
this rewrites as
\[
F\left(  hg\right)  =h\cdot\underbrace{f\left(  \pi_{G}\left(  g\right)
\right)  }_{\substack{=\left(  \iota\circ f\circ\left(  \CC \left[
\pi_{G}\right]  \right)  \right)  \left(  g\right)  \\\text{(by
(\ref{sol.ind-fixed-points.wd.3}))}}}=h\cdot F\left(  g\right)  .
\]
\par
Now, forget that we fixed $g$. We thus have shown that
\begin{equation}
F\left(  hg\right)  =h\cdot F\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }g\in G. \label{sol.ind-fixed-points.wd.4}
\end{equation}
\par
Now, let $w\in\CC G$. Then, we can write $w$ in the form $w=\sum_{g\in
G}\lambda_{g}g$ for some family $\left(  \lambda_{g}\right)  _{g\in G}$ of
elements of $\CC$ (because $G$ is a basis of $\CC G$). Consider
this family $\left(  \lambda_{g}\right)  _{g\in G}$. Now,
\begin{align*}
F\left(  h\underbrace{w}_{=\sum_{g\in G}\lambda_{g}g}\right)   &  =F\left(
h\sum_{g\in G}\lambda_{g}g\right)  =F\left(  \sum_{g\in G}\lambda
_{g}hg\right)  =\sum_{g\in G}\lambda_{g}\underbrace{F\left(  hg\right)
}_{\substack{=h\cdot F\left(  g\right)  \\\text{(by
(\ref{sol.ind-fixed-points.wd.4}))}}}
  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }F\text{ is }
\CC \text{-linear}\right) \\
&  =\sum_{g\in G}\lambda_{g}h\cdot F\left(  g\right)  =h\underbrace{\sum_{g\in
G}\lambda_{g}F\left(  g\right)  }_{\substack{=F\left(  \sum_{g\in G}
\lambda_{g}g\right)  \\\text{(since the map }F\text{ is }
\CC \text{-linear)}}}
= hF\left(  \underbrace{\sum_{g\in G}\lambda_{g}g}_{=w}\right)
= hF\left(  w\right)  .
\end{align*}
\par
Now, forget that we fixed $w$ and $h$. We thus have shown that every $h\in H$
and $w\in\CC G$ satisfy $F\left(  hw\right)  =hF\left(  w\right)  $. In
other words, $F$ is a homomorphism of $H$-sets. Since $F$ is $\CC$-linear,
this shows that $F$ is a $\CC H$-module homomorphism. Thus,
$F \in \Hom_{\CC H} \left( \CC G, U \right)$.
\par
Now, let $k\in K$ be arbitrary. We shall show that $kF=F$.
\par
Indeed, every $w\in\CC G$ satisfies
\begin{align*}
\left(  kF\right)  \left(  w\right)   &  =\underbrace{F}_{=\iota\circ
f\circ\left(  \CC \left[  \pi_{G}\right]  \right)  }\left(  wk\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the }
\CC G\text{-module structure on }\Hom_{\CC H}\left(
\CC G,U\right)  \right) \\
&  =\left(  \iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]
\right)  \right)  \left(  wk\right)  =\left(  \iota\circ f\right)  \left(
\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
wk\right)  }_{\substack{=\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  w\right)  \\\text{(by (\ref{sol.ind-fixed-points.Kignore}))}}}\right)
=\left(  \iota\circ f\right)  \left(  \left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  w\right)  \right) \\
&  =\underbrace{\left(  \iota\circ f\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  \right)  }_{=F}\left(  w\right)  =F\left(  w\right)  .
\end{align*}
In other words, $kF=F$.
\par
Now, forget that we fixed $k$. We thus have proven that $kF=F$ for every $k\in
K$. Thus, $F$ is an element of $\Hom_{\CC H}\left(
{\CC }G,U\right)  $ and satisfies $\left(  kF=F\text{ for every }k\in
K\right)  $. In other words,
\[
F \in \left\{ \psi \in \Hom_{\CC H} \left( \CC G, U \right)
 \ \mid\ k\psi = \psi \text{ for every } k \in K \right\}
= \left( \Hom_{\CC H} \left( \CC G, U \right) \right) ^K
\]
(since $\left( \Hom_{\CC H}\left( \CC G,U\right) \right) ^{K}
= \left\{ \psi \in \Hom_{\CC H} \left( \CC G, U \right)
 \ \mid\ k\psi = \psi \text{ for every } k \in K \right\}$
(by the definition of
$\left(  \Hom_{\CC H}\left(  {\CC }G,U\right)  \right)  ^{K}$)).
This proves
(\ref{sol.ind-fixed-points.wd}).} Hence, we can define a map $\Phi
:\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
\rightarrow\left( \Hom_{\CC H}\left( \CC G,U\right) \right) ^{K}$ by
\[
\left(  \Phi\left(  f\right)  =\iota\circ f\circ\left(  \CC \left[
\pi_{G}\right]  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every }f\in
\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)  \right)
.
\]
This map $\Phi$ is $\CC$-linear (since $\iota\circ f\circ\left(
\CC \left[  \pi_{G}\right]  \right)  $ depends $\CC$-linearly on
$f$ (because composition of $\CC$-linear map is $\CC$-bilinear
in these maps)). Furthermore, the map $\Phi$ is a $\CC \left[
G/K\right]  $-module homomorphism\footnote{\textit{Proof.} Let $\gamma\in G/K$
and $f\in\Hom_{\CC \left[  H/\left(  H\cap
K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
$. Let $w\in\CC G$. Then,
\[
\Phi\left(  \gamma f\right)  =\iota\circ\left(  \gamma f\right)  \circ\left(
\CC \left[  \pi_{G}\right]  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\Phi\right)  .
\]
Applying both sides of this equality to $w$, we obtain
\begin{align}
\left(  \Phi\left(  \gamma f\right)  \right)  \left(  w\right)   &  =\left(
\iota\circ\left(  \gamma f\right)  \circ\left(  \CC \left[  \pi
_{G}\right]  \right)  \right)  \left(  w\right)  =\iota\left(  \left(  \gamma
f\right)  \left(  \left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  \right)  \right) \nonumber\\
&  =\left(  \gamma f\right)  \left(  \left(  \CC \left[  \pi_{G}\right]
\right)  \left(  w\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\iota\text{ is just an inclusion map}\right) \nonumber\\
&  =f\left(  \left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  \cdot\gamma\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by the definition of the left }\CC \left[  G/K\right]
\text{-module}\\
\text{structure on }\Hom_{\CC \left[  H/\left(
H\cap K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap
K}\right)
\end{array}
\right)  .
\label{sol.ind-fixed-points.hom.pf.2}
\end{align}
\par
But the map $\pi_{G}$ is surjective. Hence, $G/K=\pi_{G}\left(  G\right)  $.
Thus, $\gamma\in G/K=\pi_{G}\left(  G\right)  $. Therefore, there exists some
$\gamma^{\prime}\in G$ such that $\gamma=\pi_{G}\left(  \gamma^{\prime
}\right)  $. Consider this $\gamma^{\prime}$. From
(\ref{sol.ind-fixed-points.s}) (applied to $\kk =\CC $, $S=G$,
$T=G/K$, $\varphi=\pi_{G}$ and $s=\gamma^{\prime}$), we obtain $\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  \gamma^{\prime}\right)
=\pi_{G}\left(  \gamma^{\prime}\right)  =\gamma$. But
(\ref{sol.ind-fixed-points.alg}) (applied to $\kk =\CC $, $S=G$,
$T=G/K$ and $\varphi=\pi_{G}$) yields that $\CC \left[  \pi_{G}\right]
$ is a $\CC$-algebra homomorphism. Hence,
\begin{equation}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  w\gamma^{\prime
}\right)  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  \cdot\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  \gamma^{\prime}\right)  }_{=\gamma}=\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  w\right)  \cdot\gamma.
\label{sol.ind-fixed-points.hom.pf.4}
\end{equation}
\par
Now,
\[
\underbrace{\gamma}_{=\pi_{G}\left(  \gamma^{\prime}\right)  }\cdot\left(
\Phi\left(  f\right)  \right)  =\left(  \pi_{G}\left(  \gamma^{\prime}\right)
\right)  \cdot\left(  \Phi\left(  f\right)  \right)  =\gamma^{\prime}
\cdot\left(  \Phi\left(  f\right)  \right)
\]
(by (\ref{sol.ind-fixed-points.fix-act.G}), applied to $g=\gamma^{\prime}$ and
$\alpha=\Phi\left(  f\right)  $). Applying both sides of this equality to $w$,
we obtain
\begin{align*}
\left(  \gamma\cdot\left(  \Phi\left(  f\right)  \right)  \right)  \left(
w\right)   &  =\left(  \gamma^{\prime}\cdot\left(  \Phi\left(  f\right)
\right)  \right)  \left(  w\right) \\
&  =\underbrace{\left(  \Phi\left(  f\right)  \right)  }_{\substack{=\iota
\circ f\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \\\text{(by the
definition of }\Phi\left(  f\right)  \text{)}}}\left(  w\gamma^{\prime
}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by the definition of the left }\CC G\text{-module}\\
\text{structure on }\Hom_{\CC H}\left( \CC G,U\right)
\end{array}
\right) \\
&  =\left(  \iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]
\right)  \right)  \left(  w\gamma^{\prime}\right)  =\iota\left(  \left(
f\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \right)  \left(
w\gamma^{\prime}\right)  \right) \\
&  =\left(  f\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \right)
\left(  w\gamma^{\prime}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\iota\text{ is just an inclusion map}\right) \\
&  =f\left(  \underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  w\gamma^{\prime}\right)  }_{\substack{=\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  w\right)  \cdot\gamma\\\text{(by
(\ref{sol.ind-fixed-points.hom.pf.4}))}}}\right)  =f\left(  \left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)  \cdot
\gamma\right)  .
\end{align*}
Compared with (\ref{sol.ind-fixed-points.hom.pf.2}), this yields $\left(
\Phi\left(  \gamma f\right)  \right)  \left(  w\right)  =\left(  \gamma
\cdot\left(  \Phi\left(  f\right)  \right)  \right)  \left(  w\right)  $.
\par
Now, forget that we fixed $w$. We thus have proven that $\left(  \Phi\left(
\gamma f\right)  \right)  \left(  w\right)  =\left(  \gamma\cdot\left(
\Phi\left(  f\right)  \right)  \right)  \left(  w\right)  $ for every
$w\in\CC G$. In other words, $\Phi\left(  \gamma f\right)  =\gamma
\cdot\left(  \Phi\left(  f\right)  \right)  $.
\par
Now, forget that we fixed $\gamma$ and $f$. We have thus shown that
$\Phi\left(  \gamma f\right)  =\gamma\cdot\left(  \Phi\left(  f\right)
\right)  $ for every $\gamma\in G/K$ and $f\in
\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(
\CC \left[  G/K\right]  ,U^{H\cap K}\right)  $. In other words, $\Phi$
is a homomorphism of $G/K$-sets. Since $\Phi$ is $\CC$-linear, this
yields that $\Phi$ is a $\CC \left[  G/K\right]  $-module homomorphism,
qed.}.

Now, it is easy to see that the map $\Phi$ is
injective\footnote{\textit{Proof.} We know that $\Phi$ is $\CC$-linear.
Hence, $\ker\Phi$ is well-defined.
\par
Let $f\in\ker\Phi$. Then, $\Phi\left(  f\right)  =0$ (since $f\in\ker\Phi$).
Since $\Phi\left(  f\right)  =\iota\circ f\circ\left(  \CC \left[
\pi_{G}\right]  \right)  $ (by the definition of $\Phi$), this rewrites as
$\iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]  \right)  =0$.
\par
Now, let $v\in\CC \left[  G/K\right]  $ be arbitrary. We know that the
map $\CC \left[  \pi_{G}\right]  $ is surjective. Hence,
$\CC \left[  G/K\right]  =\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  \CC G\right)  $. Thus, $v\in\CC \left[  G/K\right]
=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  \CC G\right) $.
Hence, there exists some $w\in\CC G$ such that $v=\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)  $. Consider this
$w$. We have
\begin{align*}
\left(  \underbrace{\Phi\left(  f\right)  }_{=\iota\circ f\circ\left(
\CC \left[  \pi_{G}\right]  \right)  }\right)  \left(  w\right)   &
=\left(  \iota\circ f\circ\left(  \CC \left[  \pi_{G}\right]  \right)
\right)  \left(  w\right)  =\iota\left(  f\left(  \underbrace{\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)  }
_{\substack{=v\\\text{(since }v=\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  w\right)  \text{)}}}\right)  \right) \\
&  =\iota\left(  f\left(  v\right)  \right)  =f\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\iota\text{ is just an inclusion
map}\right)  .
\end{align*}
Thus, $f\left(  v\right)  =\underbrace{\left(  \Phi\left(  f\right)  \right)
}_{=0}\left(  v\right)  =0\left(  v\right)  =0$.
\par
Now, forget that we fixed $v$. We thus have shown that $f\left(  v\right)  =0$
for every $v\in\CC \left[  G/K\right]  $. In other words, $f=0$.
\par
Now, forget that we fixed $f$. We have thus proven that $f=0$ for every
$f\in\ker\Phi$. Hence, $\ker\Phi=0$. In other words, the map $\Phi$ is
injective (since $\Phi$ is $\CC$-linear), qed.}. We shall now show that
the map $\Phi$ is surjective.

In fact, let $F\in\left(  \Hom_{\CC H}\left(
{\CC }G,U\right)  \right)  ^{K}$ be arbitrary. We are going to prove
that $F\in\Phi\left(  \Hom_{\CC \left[
H/\left(  H\cap K\right)  \right]  }\left(  \CC \left[  G/K\right]
,U^{H\cap K}\right)  \right)  $.

We have
\[
F\in\left(  \Hom_{\CC H}\left(  {\CC }G,U\right)
\right)  ^{K}=\left\{  \varphi\in\Hom_{\CC H}\left(
{\CC }G,U\right)  \ \mid\ k\varphi=\varphi\text{ for all }k\in
K\right\}
\]
(by the definition of $\left(  \Hom_{\CC H}\left(
{\CC }G,U\right)  \right)  ^{K}$). In other words, $F$ is an element of
$\Hom_{\CC H}\left(  {\CC }G,U\right)  $ and
satisfies
\begin{equation}
\left(  kF=F\text{ for all }k\in K\right)  .
\label{sol.ind-fixed-points.surj.0}
\end{equation}
Now, we can easily see that
\begin{equation}
F\left(  gk\right)  =F\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}k\in K\text{ and }g\in G\text{.} \label{sol.ind-fixed-points.surj.1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.surj.1}):} Let $k\in K$
and $g\in K$. By the definition of the action of $\CC G$ on
$\Hom_{\CC H}\left(  {\CC }G,U\right)  $, we have
$\left(  kF\right)  \left(  g\right)  =F\left(  gk\right)  $. Hence, $F\left(
gk\right)  =\underbrace{\left(  kF\right)  }_{\substack{=F\\\text{(by
(\ref{sol.ind-fixed-points.surj.0}))}}}\left(  g\right)  =F\left(  g\right)
$. This proves (\ref{sol.ind-fixed-points.surj.1}).} Therefore,
(\ref{sol.ind-fixed-points.factor1}) (applied to $P=U$ and $\rho=F$) yields
that
\[
\text{there exists a }\CC \text{-linear map }\widetilde{\rho}
:\CC \left[  G/K\right]  \rightarrow U\text{ satisfying }
F=\widetilde{\rho}\circ\left(  \CC \left[  \pi_{G}\right]  \right)  .
\]
Denote this map $\widetilde{\rho}$ by $\widetilde{F}$. Hence, $\widetilde{F}$
is a $\CC$-linear map $\CC \left[  G/K\right]  \rightarrow U$
satisfying $F=\widetilde{F}\circ\left(  \CC \left[  \pi_{G}\right]
\right)  $.

It is easy to see that
\begin{equation}
\widetilde{F}\left(  v\right)  \in U^{H\cap K}\ \ \ \ \ \ \ \ \ \ \text{for
every }v\in\CC \left[  G/K\right]  .\label{sol.ind-fixed-points.surj.2}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.surj.2}):} Let
$v\in\CC \left[  G/K\right]  $. Then, $v\in\CC \left[  G/K\right]
=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  \CC G\right) $
(since the map $\CC \left[  \pi_{G}\right]  $ is
surjective). Hence, there exists a $w\in\CC G$ such that $v=\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)  $. Consider this
$w$. Recall that $F=\widetilde{F}\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  $. Applying both sides of this equality to $w$, we
obtain
\[
F\left(  w\right)  =\left(  \widetilde{F}\circ\left(  \CC \left[
\pi_{G}\right]  \right)  \right)  \left(  w\right)  =\widetilde{F}\left(
\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  }_{\substack{=v\\\text{(since }v=\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  w\right)  \text{)}}}\right)  =\widetilde{F}
\left(  v\right)  .
\]
Hence, $\widetilde{F}\left(  v\right)  =F\left(  w\right)  $.
\par
Let $k\in H\cap K$. We have $F\in\Hom_{\CC H}\left(
{\CC }G,U\right)  $; therefore, the map $F$ is a $\CC H$-module
homomorphism. In other words, $F$ is a $\CC$-linear homomorphism of
$H$-sets. Now, we have $k\in H\cap K\subset H$. Since $F$ is a homomorphism of
$H$-sets, this yields $F\left(  kw\right)  =k\cdot F\left(  w\right)  $. Since
$F\left(  w\right)  =\widetilde{F}\left(  v\right)  $, this rewrites as
$F\left(  kw\right)  =k\cdot\widetilde{F}\left(  v\right)  $.
\par
But $k\in H\cap K\subset K=\ker\left(  \pi_{G}\right)  $ (since $\pi_{G}$ is
the canonical projection $G\rightarrow G/K$), so that $\pi_{G}\left(
k\right)  =1$. Now, (\ref{sol.ind-fixed-points.s}) (applied to $\kk 
=\CC $, $S=G$, $T=G/K$, $\varphi=\pi_{G}$ and $s=k$) yields $\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  k\right)  =\pi_{G}\left(
k\right)  =1$. But we know that $\CC \left[  \pi_{G}\right]
:\CC G\rightarrow\CC \left[  G/K\right]  $ is a $\CC$-algebra
homomorphism. Thus,
\[
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  kw\right)
=\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
k\right)  }_{=1}\cdot\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  w\right)  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  =v
\]
(since $v=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  w\right)
$).
\par
Recall again that $F=\widetilde{F}\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  $. Applying both sides of this equality to $kw$, we
obtain
\[
F\left(  kw\right)  =\left(  \widetilde{F}\circ\left(  \CC \left[
\pi_{G}\right]  \right)  \right)  \left(  kw\right)  =\widetilde{F}\left(
\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
kw\right)  }_{=v}\right)  =\widetilde{F}\left(  v\right)  .
\]
Thus, $\widetilde{F}\left(  v\right)  =F\left(  kw\right)  =k\cdot
\widetilde{F}\left(  v\right)  $. In other words, $k\cdot\widetilde{F}\left(
v\right)  =\widetilde{F}\left(  v\right)  $.
\par
Now, forget that we fixed $k$. We thus have proven that $k\cdot\widetilde{F}
\left(  v\right)  =\widetilde{F}\left(  v\right)  $ for every $k\in H\cap K$.
Hence, $\widetilde{F}\left(  v\right)  $ is an element of $U$ and satisfies
$\left(  k\cdot\widetilde{F}\left(  v\right)  =\widetilde{F}\left(  v\right)
\text{ for every }k\in H\cap K\right)  $. In other words,
\[
\widetilde{F}\left(  v\right)  \in\left\{  z\in U\ \mid\ k\cdot z=z\text{ for
every }k\in H\cap K\right\}  =U^{H\cap K}
\]
(since $U^{H\cap K}=\left\{  z\in U\ \mid\ k\cdot z=z\text{ for every }k\in
H\cap K\right\}  $ (by the definition of $U^{H\cap K}$)). This proves
(\ref{sol.ind-fixed-points.surj.2}).} Therefore, we can define a map
$F^{\prime}:\CC \left[  G/K\right]  \rightarrow U^{H\cap K}$ by
\begin{equation}
\left(  F^{\prime}\left(  v\right)  =\widetilde{F}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\CC \left[  G/K\right]
\right)  .
\label{sol.ind-fixed-points.surj.3}
\end{equation}
Consider this map $F^{\prime}$. This map $F^{\prime}$ is obtained from the map
$\widetilde{F}$ by restricting the codomain to $U^{H\cap K}$ (because of
(\ref{sol.ind-fixed-points.surj.3})). Hence, this map $F^{\prime}$ is
$\CC$-linear (because the map $\widetilde{F}$ is $\CC$-linear).
It is easy to see that
\begin{equation}
\iota\circ F^{\prime}\circ\left(  \CC \left[  \pi_{G}\right]  \right)
=F.
\label{sol.ind-fixed-points.surj.4}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.surj.4}):} Every
$v\in\CC G$ satisfies
\begin{align*}
\left(  \iota\circ F^{\prime}\circ\left(  \CC \left[  \pi_{G}\right]
\right)  \right)  \left(  v\right)   &  =\iota\left(  \left(  F^{\prime}
\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \right)  \left(
v\right)  \right)  =\left(  F^{\prime}\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  \right)  \left(  v\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\iota\text{ is just an inclusion
map}\right)  \\
&  =F^{\prime}\left(  \left(  \CC \left[  \pi_{G}\right]  \right)
\left(  v\right)  \right)  =\widetilde{F}\left(  \left(  \CC \left[
\pi_{G}\right]  \right)  \left(  v\right)  \right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.ind-fixed-points.surj.3}),
applied to }\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
v\right)  \text{ instead of }v\right)  \\
&  =\underbrace{\left(  \widetilde{F}\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  \right)  }_{\substack{=F\\\text{(since }F=\widetilde{F}
\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \text{)}}}\left(
v\right)  =F\left(  v\right)  .
\end{align*}
In other words, $\iota\circ F^{\prime}\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  =F$. This proves (\ref{sol.ind-fixed-points.surj.4}).}
Finally, we have
\begin{equation}
F^{\prime}\in\Hom_{\CC \left[  H/\left(  H\cap
K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
.
\label{sol.ind-fixed-points.surj.5}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-fixed-points.surj.5}):} Let
$v\in\CC \left[  G/K\right]  $ and $\gamma\in H/\left(  H\cap K\right)
$. We shall show that $F^{\prime}\left(  \gamma v\right)  =\gamma\cdot
F^{\prime}\left(  v\right)  $.
\par
We have $\gamma\in H/\left(  H\cap K\right)  =\pi_{H}\left(  H\right)  $
(since the map $\pi_{H}$ is surjective). Hence, there exists some $\delta\in
H$ such that $\gamma=\pi_{H}\left(  \delta\right)  $. Consider this $\delta$.
Applying (\ref{sol.ind-fixed-points.piGh}) to $h=\delta$, we obtain $\pi
_{H}\left(  \delta\right)  =\pi_{G}\left(  \delta\right)  $, so that
$\gamma=\pi_{H}\left(  \delta\right)  =\pi_{G}\left(  \delta\right)  $. Hence,
$\pi_{G}\left(  \delta\right)  =\gamma$.
\par
But $v\in\CC \left[  G/K\right]  =\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  \CC G\right)  $ (since the map
$\CC \left[  \pi_{G}\right]  $ is surjective). Hence, there exists a
$w\in\CC G$ such that $v=\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  w\right)  $. Consider this $w$.
\par
We know that $\CC \left[  \pi_{G}\right]  :\CC G\rightarrow
\CC \left[  G/K\right]  $ is a $\CC$-algebra homomorphism. Hence,
\begin{align}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  \delta w\right)
&  =\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
\delta\right)  }_{\substack{=\pi_{G}\left(  \delta\right)  \\\text{(by
(\ref{sol.ind-fixed-points.s}), applied to}\\\kk =\CC \text{,
}S=G\text{, }T=G/K\text{, }\varphi=\pi_{G}\text{ and }
s=\delta\text{)}}}
\cdot\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
w\right)  }_{\substack{=v\\\text{(since }v=\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  w\right)  \text{)}}}\nonumber\\
&  =\underbrace{\pi_{G}\left(  \delta\right)  }_{=\gamma}\cdot v=\gamma
v.
\label{sol.ind-fixed-points.surj.5.pf.1}
\end{align}
But (\ref{sol.ind-fixed-points.surj.4}) yields $F=\iota\circ F^{\prime}
\circ\left(  \CC \left[  \pi_{G}\right]  \right)  $. Applying both sides
of this equality to the element $\delta w$, we obtain
\begin{align}
F\left(  \delta w\right)   &  =\left(  \iota\circ F^{\prime}\circ\left(
\CC \left[  \pi_{G}\right]  \right)  \right)  \left(  \delta w\right)
=\iota\left(  \left(  F^{\prime}\circ\left(  \CC \left[  \pi_{G}\right]
\right)  \right)  \left(  \delta w\right)  \right)  \nonumber\\
&  =\left(  F^{\prime}\circ\left(  \CC \left[  \pi_{G}\right]  \right)
\right)  \left(  \delta w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\iota\text{ is just an inclusion map}\right)  \nonumber\\
&  =F^{\prime}\left(  \underbrace{\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  \delta w\right)  }_{\substack{=\gamma v\\\text{(by
(\ref{sol.ind-fixed-points.surj.5.pf.1}))}}}\right)  =F^{\prime}\left(  \gamma
v\right)  .\label{sol.ind-fixed-points.surj.5.pf.2}
\end{align}
\par
But $F\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $.
Hence, $F$ is a $\CC H$-module homomorphism $\CC G\rightarrow U$.
In other words, $F$ is a $\CC$-linear homomorphism of $H$-sets
$\CC G\rightarrow U$. Since $F$ is a homomorphism of $H$-sets, we have
$F\left(  \delta w\right)  =\delta\cdot F\left(  w\right)  $ (because
$\delta\in H$). Since
\begin{align*}
\underbrace{F}_{=\iota\circ F^{\prime}\circ\left(  \CC \left[  \pi
_{G}\right]  \right)  }\left(  w\right)    & =\left(  \iota\circ F^{\prime
}\circ\left(  \CC \left[  \pi_{G}\right]  \right)  \right)  \left(
w\right)  =\iota\left(  \left(  F^{\prime}\circ\left(  \CC \left[
\pi_{G}\right]  \right)  \right)  \left(  w\right)  \right)  \\
& =\left(  F^{\prime}\circ\left(  \CC \left[  \pi_{G}\right]  \right)
\right)  \left(  w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\iota\text{ is just an inclusion map}\right)  \\
& =F^{\prime}\left(  \underbrace{\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  w\right)  }_{\substack{=v\\\text{(because }v=\left(
\CC \left[  \pi_{G}\right]  \right)  \left(  w\right)  \text{)}}}
\right)  =F^{\prime}\left(  v\right)  ,
\end{align*}
this rewrites as $F\left(  \delta w\right)  =\delta\cdot F^{\prime}\left(
v\right)  $. Compared with (\ref{sol.ind-fixed-points.surj.5.pf.2}), this
yields
\[
F^{\prime}\left(  \gamma v\right)  =\delta\cdot F^{\prime}\left(  v\right)  .
\]
Compared with
\[
\underbrace{\gamma}_{=\pi_{H}\left(  \delta\right)  }\cdot F^{\prime}\left(
v\right)  =\pi_{H}\left(  \delta\right)  \cdot F^{\prime}\left(  v\right)
=\delta\cdot F^{\prime}\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.ind-fixed-points.fix-act.H}), applied to }h=\delta\text{ and }
\beta=F^{\prime}\left(  v\right)  \right)  ,
\]
this yields $F^{\prime}\left(  \gamma v\right)  =\gamma\cdot F^{\prime}\left(
v\right)  $.
\par
Now, forget that we fixed $v$ and $\gamma$. We thus have proven that every
$v\in\CC \left[  G/K\right]  $ and $\gamma\in H/\left(  H\cap K\right)
$ satisfy $F^{\prime}\left(  \gamma v\right)  =\gamma\cdot F^{\prime}\left(
v\right)  $. In other words, $F^{\prime}$ is a homomorphism of $H/\left(
H\cap K\right)  $-sets. Since $F^{\prime}$ is also $\CC$-linear, this
yields that $F^{\prime}$ is a homomorphism of $\CC \left[  H/\left(
H\cap K\right)  \right]  $-modules. In other words, $F^{\prime}\in
\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)  $. This
proves (\ref{sol.ind-fixed-points.surj.5}).} Hence, $\Phi\left(  F^{\prime
}\right)  $ is well-defined, and we have
\begin{align*}
\Phi\left(  F^{\prime}\right)    & =\iota\circ F^{\prime}\circ\left(
\CC \left[  \pi_{G}\right]  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\Phi\right)  \\
& =F\ \ \ \ \ \ \ \ \ \ 
\left(  \text{by (\ref{sol.ind-fixed-points.surj.4})}\right)  ,
\end{align*}
so that $F=\Phi\left(  \underbrace{F^{\prime}}_{\in
\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }\left(
\CC \left[  G/K\right]  ,U^{H\cap K}\right)  }\right)  \in\Phi\left(
\Hom_{\CC \left[  H/\left(  H\cap K\right)
\right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)  \right)
$.

Now, forget that we fixed $F$. We thus have proven that every $F\in\left(
\Hom_{\CC H}\left(  {\CC }G,U\right)  \right)
^{K}$ satisfies $F\in\Phi\left(  
\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }
\left(  \CC \left[ G/K\right]  ,U^{H\cap K}\right)  \right)  $.
In other words,
\[
\left(  \Hom_{\CC H}\left(  {\CC }G,U\right)
\right)  ^{K}\subset\Phi\left( 
 \Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }
\left(  \CC \left[ G/K\right]  ,U^{H\cap K}\right)  \right)  .
\]
In other words, the map $\Phi$ is surjective. Combined with the fact that
$\Phi$ is injective, this yields that $\Phi$ is bijective. Combining this with
the fact that the map $\Phi$ is a $\CC \left[  G/K\right]  $-module
homomorphism, we conclude that the map $\Phi$ is a $\CC \left[
G/K\right]  $-module isomorphism. We have thus found a $\CC \left[
G/K\right]  $-module isomorphism
$\Hom_{\CC \left[  H/\left(  H\cap K\right)  \right]  }
\left(  \CC \left[ G/K\right]  ,U^{H\cap K}\right)  \rightarrow\left( 
\Hom_{\CC H}\left(  \CC G,U\right)  \right)  ^{K}$ (namely, the
map $\Phi$). Thus,
\[
\left(  \Hom_{\CC H}\left(  {\CC }G,U\right)
\right)  ^{K}\cong\Hom_{\CC \left[  H/\left(
H\cap K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap
K}\right)  \ \ \ \ \ \ \ \ \ \ \text{as }\CC \left[  G/K\right]
\text{-modules.}
\]
This proves (\ref{sol.ind-fixed-points.3}).

Now,
\begin{align*}
\left(  \underbrace{\Ind_{H}^{G}U}_{\substack{\cong
\Hom_{\CC H}\left(  {\CC }G,U\right)  \text{ as
}\CC G\text{-modules}\\\text{(by (\ref{sol.ind-fixed-points.1}))}}}
\right)  ^{K}  & \cong\left(  \Hom_{\CC H}\left(
{\CC }G,U\right)  \right)  ^{K}\\
& \cong\Hom_{\CC \left[  H/\left(  H\cap
K\right)  \right]  }\left(  \CC \left[  G/K\right]  ,U^{H\cap K}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.ind-fixed-points.3})}\right)
\\
& \cong\Ind_{H/\left(  H\cap K\right)  }^{G/K}
\left(  U^{H\cap K}\right)  \ \ \ \ \ \ \ \ \ \ 
\left(  \text{by (\ref{sol.ind-fixed-points.2})}\right)
\end{align*}
as ${\CC }\left[  G/K\right]  $-modules. This solves
Exercise~\ref{exe.ind-fixed-points}.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.ind-infl}}

\textit{Solution to Exercise \ref{exe.ind-infl}.} In the following, we
will write $g$ for the element $t_{g}$ of $\CC G$ whenever $g$ is an
element of $G$. This is a relatively common abuse of notation, and it is
harmless because the map $G\rightarrow \CC G,\ g\mapsto t_{g}$ is an
injective homomorphism of multiplicative monoids (so $t_{gh}=t_{g}t_{h}$ and
$t_{1}=1$, which means that we won't run into ambiguities denoting $t_{g}$ by
$g$) and because every $ \CC G$-module $M$, every $m\in M$ and every
$g\in G$ satisfy $gm=t_{g}m$. We will do the same abuse of notation for
elements of $H$ and of $K$.

Inflation does not change the underlying $\CC$-vector space of a
representation. Thus, $\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}%
_{H/K}^{G/K}V=\operatorname{Ind}_{H/K}^{G/K}V$ as $\CC$-vector spaces.
For the same reason, $\operatorname{Infl}_{H/K}^{H}V=V$ as $\CC$-vector spaces.

Let $\pi_{G}$ be the canonical projection map $G\rightarrow G/K$. This gives
rise to a surjective $\CC$-algebra homomorphism $\CC \left[
\pi_{G}\right]  : \CC G\rightarrow \CC \left[  G/K\right]  $ (which
sends every $g\in G$ to $\pi_{G}\left(  g\right)  $). We now define a
$\CC$-linear map%
\begin{align*}
\beta: \CC G\otimes_{\CC H}\operatorname{Infl}_{H/K}^{H}V  &
\rightarrow \CC \left[  G/K\right]  \otimes_{\CC \left[
H/K\right]  }V,\\
s\otimes_{\CC H}v  &  \mapsto\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  s\right)  \otimes_{ \CC \left[  H/K\right]  }v.
\end{align*}
This is easily seen to be well-defined (using the universal property of the
tensor product and the observation that every $t\in \CC H$ satisfies
$\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  t\right)
\in \CC \left[  H/K\right]  $).

We also want to define a map $\alpha$ in the opposite direction, but this will
require some more work. First, for every $g\in G$, we define a
$\CC$-linear map
\begin{align*}
\mathbf{a}_{g}:V  &  \rightarrow \CC G\otimes_{\CC H}%
\operatorname{Infl}_{H/K}^{H}V,\\
v  &  \mapsto g\otimes_{\CC H}v.
\end{align*}
It is easily seen that if $g_{1}$ and $g_{2}$ are two elements of $G$
satisfying $\pi_{G}\left(  g_{1}\right)  =\pi_{G}\left(  g_{2}\right)  $, then
$\mathbf{a}_{g_{1}}=\mathbf{a}_{g_{2}}$\ \ \ \ \footnote{Indeed, let $g_{1}$
and $g_{2}$ be two elements of $G$ satisfying $\pi_{G}\left(  g_{1}\right)
=\pi_{G}\left(  g_{2}\right)  $. Then, $g_{1}\in g_{2}K$, so that there exists
some $k\in K$ such that $g_{1}=g_{2}k$. Consider this $k$. Then, every $v\in
V$ satisfies
\begin{align*}
\mathbf{a}_{g_{1}}\left(  v\right)   &  =\underbrace{g_{1}}_{=g_{2}k}%
\otimes_{\CC H}v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{a}_{g_{1}}\right) \\
&  =g_{2}k\otimes_{\CC H}v=g_{2}\otimes_{\CC H}\underbrace{kv}%
_{\substack{=v\\\text{(since }k\in K\text{ acts trivially}\\\text{on
}\operatorname{Infl}_{H/K}^{H}V\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }k\text{ lies in }K\subset H\text{, and thus can be moved past the
}\otimes_{\CC H}\text{ sign}\right) \\
&  =g_{2}\otimes_{\CC H}v=\mathbf{a}_{g_{2}}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{a}_{g_{2}%
}\right)  .
\end{align*}
Thus, $\mathbf{a}_{g_{1}}=\mathbf{a}_{g_{2}}$, qed.}. In other words, the map
$\mathbf{a}_{g}$ depends only on $\pi_{G}\left(  g\right)  $ rather than on
$g$ itself. Thus, we can define a $\CC$-linear map
$\widetilde{\mathbf{a}}_{p}:V\rightarrow \CC G\otimes_{\CC H}
\operatorname{Infl}_{H/K}^{H}V$ for every $p\in G/K$ by choosing any $g\in
G$ satisfying $\pi_{G}\left(  g\right)  =p$, and then setting
$\widetilde{\mathbf{a}}_{p}=\mathbf{a}_{g}$; the resulting map
$\widetilde{\mathbf{a}}_{p}$ does not depend on the choice of $g$.

Hence, we have defined a $\CC$-linear map $\widetilde{\mathbf{a}}%
_{p}:V\rightarrow \CC G\otimes_{\CC H}\operatorname{Infl}_{H/K}%
^{H}V$ for every $p\in G/K$. In other words, we have defined an element
$\widetilde{\mathbf{a}}_{p}$ of $\Hom_{\CC}
\left(  V, \CC G\otimes_{\CC H}\operatorname{Infl}_{H/K}%
^{H}V\right)  $ for every $p\in G/K$. Hence, we can define a
$\CC$-linear map
\begin{align*}
\mathbf{A}: \CC \left[  G/K\right]   &  \rightarrow
\Hom_{\CC} \left(  V, \CC G\otimes_{\CC H}%
\operatorname{Infl}_{H/K}^{H}V\right)  ,\\
p &  \mapsto\widetilde{\mathbf{a}}_{p}\ \ \ \ \ \ \ \ \ \ \text{for every
}p\in G/K
\end{align*}
(because in order to define a $\CC$-linear map from the
$\CC$-vector space $\CC \left[  G/K\right]  $, it is enough to assign its
values on the basis $G/K$). Using this map $\mathbf{A}$, we can now define a
$\CC$-linear map
\begin{align*}
\alpha: \CC \left[  G/K\right]  \otimes_{ \CC \left[  H/K\right]  }V
&  \rightarrow \CC G\otimes_{\CC H}\operatorname{Infl}_{H/K}^{H}V,\\
s\otimes_{\CC \left[ H/K \right]}v &
\mapsto\left(  \mathbf{A}\left(  s\right)  \right)
\left(  v\right)  .
\end{align*}
It is again easy to check that this is well-defined\footnote{\textit{Proof.}
In order to prove this well-definedness, we have to check that $\left(
\mathbf{A}\left(  s\right)  \right)  \left(  v\right)  $ depends
$\CC \left[ H/K \right]$-bilinearly on $\left(  s,v\right)  $.
It is very easy to see that $\left(
\mathbf{A}\left(  s\right)  \right)  \left(  v\right)  $ depends
$\CC$-bilinearly on $\left(  s,v\right)  $;
therefore, it only remains to prove
that $\left(  \mathbf{A}\left(  st\right)  \right)  \left(  v\right)  =\left(
\mathbf{A}\left(  s\right)  \right)  \left(  tv\right)  $ for all
$s\in \CC \left[  G/K\right]  $, $t\in \CC \left[  H/K\right]  $ and
$v\in V$. So let $s\in \CC \left[  G/K\right]  $, $t\in \CC \left[
H/K\right]  $ and $v\in V$ be arbitrary. We want to prove that $\left(
\mathbf{A}\left(  st\right)  \right)  \left(  v\right)  =\left(
\mathbf{A}\left(  s\right)  \right)  \left(  tv\right)  $. Since both sides of
this equality are $\CC$-linear in $s$, we can WLOG assume that $s$
belongs to the basis $G/K$ of $\CC \left[  G/K\right]  $. Assume this,
and pick $g\in G$ such that $s=\pi_{G}\left(  g\right)  $. (This $g$ exists
since $\pi_{G}$ is surjective.)
\par
Similarly, we can WLOG assume that $t$ belongs to the basis $H/K$ of
$\CC \left[  H/K\right]  $. Assume this and pick $h\in H$ such that
$t=\pi_{G}\left(  h\right)  $. (This $t$ exists since $H/K=\pi_{G}\left(
H\right)  $.) Since $s=\pi_{G}\left(  g\right)  $ and $t=\pi_{G}\left(
h\right)  $, we have $st=\pi_{G}\left(  g\right)  \cdot\pi_{G}\left(
h\right)  =\pi_{G}\left(  gh\right)  $ (as $\pi_{G}$ is a group homomorphism).
Notice that $\underbrace{t}_{=\pi_{G}\left(  h\right)  }v=\pi_{G}\left(
h\right)  v=hv$ (because the action of $H$ on $\operatorname{Infl}_{H/K}^{H}V$
factors through $\pi_{G}$).
\par
Now, the definition of $\mathbf{A}\left(  st\right)  $ yields $\mathbf{A}%
\left(  st\right)  =\widetilde{\mathbf{a}}_{st}=\mathbf{a}_{gh}$ (by the
definition of $\widetilde{\mathbf{a}}_{st}$, since $st=\pi_{G}\left(
gh\right)  $). Similarly, $\mathbf{A}\left(  s\right)  =\mathbf{a}_{g}$. Now,
comparing $\underbrace{\left(  \mathbf{A}\left(  st\right)  \right)
}_{=\mathbf{a}_{gh}}\left(  v\right)  =\mathbf{a}_{gh}\left(  v\right)
=gh\otimes_{\CC H}v=g\otimes_{\CC H}hv$ (since $h$ belongs to
$\CC H$ and thus can be moved past the $\otimes_{\CC H}$ sign)
with $\underbrace{\left(  \mathbf{A}\left(  s\right)  \right)  }%
_{=\mathbf{a}_{g}}\left(  \underbrace{tv}_{=hv}\right)  =\mathbf{a}_{g}\left(
hv\right)  =g\otimes_{\CC H}hv$, we obtain $\left(  \mathbf{A}\left(
st\right)  \right)  \left(  v\right)  =\left(  \mathbf{A}\left(  s\right)
\right)  \left(  tv\right)  $, which is precisely what we needed to prove.}.
It is straightforward to show that $\alpha\circ\beta= \id$
(indeed, this only needs to be proven on tensors of the form $g\otimes
_{\CC H}v$ for $g\in G$ and $v\in\operatorname{Infl}_{H/K}^{H}V$,
because the $\CC$-vector space $\CC G\otimes_{\CC H}
\operatorname{Infl}_{H/K}^{H}V$ is spanned by such tensors; but on such
tensors it is very easy to check) and $\beta\circ\alpha= \id$
(using a similar argument). Thus, the maps $\alpha$ and $\beta$ are mutually
inverse, and therefore $\beta$ is invertible. But we can regard $\beta$ as a
map from $\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ to
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$ (since
$\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V= \CC G
\otimes_{\CC H}\operatorname{Infl}_{H/K}^{H}V$ and $\operatorname{Infl}%
_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V=\operatorname{Ind}_{H/K}%
^{G/K}V= \CC \left[  G/K\right]  \otimes_{ \CC \left[  H/K\right]
}V$ as $\CC$-vector spaces), and it is easy to verify that $\beta$
becomes a $\CC G$-module homomorphism when regarded this way. Thus,
$\beta$ is an invertible $\CC G$-module homomorphism from
$\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ to
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$, hence a
$\CC G$-module isomorphism. Thus, such an isomorphism exists, i.e., we
have $\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}%
V\cong\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ as
$\CC G$-modules. This solves Exercise \ref{exe.ind-infl}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.ind-infl}}

\textit{Solution to Exercise \ref{exe.ind-infl}.}

\underline{\textit{Step 1: Introducing general notation.}}

Let us first introduce some notation.

For any set $S$ and any field $\kk$, we let $\kk S$ denote the
$\kk$-vector space with basis $\left\{  t_{s}\right\}  _{s\in S}$. We
sometimes write $\kk \left[  S\right]  $ instead of $\kk S$. We
will also write $s$ for the basis element $t_{s}$ whenever $s\in S$ (this is
an abuse of notation, but it is harmless since the map $S\rightarrow
 \kk S$ sending each $s$ to $t_{s}$ is injective\footnote{And even if
this map were not injective (e.g., if we would allow $\kk$ to be an
arbitrary ring, which includes the possibility of it being the zero ring),
then this would be harmless, since in the arguments that follow, we will never
deduce the equality of two elements of $S$ from that of their corresponding
basis vectors in $\kk S$ (but only the other way round).}). When $S$ is
a group $G$, the $\kk$-vector space $\kk S$ is identical with the
underlying vector space of the $\kk$-algebra $\kk G$ defined in
Example~\ref{group-algebra-example}.

If $\kk$ is a field, if $S$ is a set, if $U$ is a $\kk$-vector
space, and if $\rho:S\rightarrow U$ is any map, then
\begin{equation}
\text{there exists exactly one } \kk \text{-linear map }\widetilde{\rho
}: \kk S\rightarrow U\text{ satisfying }\left(  \widetilde{\rho}\left(
s\right)  =\rho\left(  s\right)  \text{ for every }s\in S\right)  .
\label{sol.ind-infl.uniprop-kS}%
\end{equation}
(In fact, this is the universal property of the free $\kk$-vector space
$\kk S$.)

If $\kk$ is a field, if $S$ and $T$ are two sets, and if $\varphi
:S\rightarrow T$ is a map, then we let $\kk \left[  \varphi\right]  $
denote the $\kk$-linear map $\kk S\rightarrow \kk T$ which
sends the basis element $s$ of $\kk S$ to $\varphi\left(  s\right)
\in \kk T$ for every $s\in S$. This map $\kk \left[  \varphi
\right]  $ is clearly well-defined. It satisfies
\begin{equation}
\left(   \kk \left[  \varphi\right]  \right)  \left(  s\right)
=\varphi\left(  s\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }s\in S\text{.}
\label{sol.ind-infl.s}%
\end{equation}
Moreover,
\begin{equation}
\text{this map } \kk \left[  \varphi\right]  \text{ is surjective if
}\varphi\text{ is surjective.} \label{sol.ind-infl.surj}%
\end{equation}
\footnote{For the proof of this fact, see the proof of
(\ref{sol.ind-fixed-points.surj}) (in the solution of Exercise
\ref{exe.ind-fixed-points}).} Furthermore, if $S$ and $T$ are two groups, and
if $\varphi:S\rightarrow T$ is a group homomorphism, then
\begin{equation}
\text{the map } \kk \left[  \varphi\right]  : \kk S\rightarrow
 \kk T\text{ is a } \kk \text{-algebra homomorphism.}
\label{sol.ind-infl.alg}%
\end{equation}
\footnote{For the proof of this fact, see the proof of
(\ref{sol.ind-fixed-points.alg}) (in the solution of Exercise
\ref{exe.ind-fixed-points}).}

\underline{\textit{Step 2: The maps }$\pi_{G}$\textit{ and }$\pi_{H}$.}

Now, let us come to the solution of the exercise.
We have $K\subset H$ (since $K<H$) and $H\subset G$
(since $H<G$).

Let $\pi_{G}:G\rightarrow G/K$ be the canonical projection map. Let $\pi
_{H}:H\rightarrow H/K$ be the canonical projection map. The maps $\pi_{G}$ and
$\pi_{H}$ are surjective group homomorphisms (since each of them is a
projection map from a group onto its quotient). As we know, the map $\pi
_{G}:G\rightarrow G/K$ induces a $\CC$-linear map $\CC
\left[  \pi_{G}\right]  : \CC G\rightarrow \CC \left[
G/K\right]  $. This map $\CC \left[  \pi_{G}\right]  $ is surjective
(by (\ref{sol.ind-infl.surj}), applied to $\kk = \CC$, $S=G$,
$T=G/K$ and $\varphi=\pi_{G}$), since the map $\pi_{G}$ is surjective. Also,
we know that the map $\CC \left[  \pi_{G}\right]  : \CC G
\rightarrow \CC \left[  G/K\right]  $ is a $\CC$-algebra
homomorphism\footnote{This follows from (\ref{sol.ind-infl.alg}) (applied to
$\kk = \CC$, $S=G$, $T=G/K$ and $\varphi=\pi_{G}$), since
$\pi_{G}:G\rightarrow G/K$ is a group homomorphism.}.

Since we are identifying $H/K$ with a subgroup of $G/K$, we have $\pi_{H}%
=\pi_{G}\mid_{H}$ (up to the fact that the targets of these maps are
different). Hence, $\pi_{H}\left(  H\right)  =\left(  \pi_{G}\mid_{H}\right)
\left(  H\right)  =\pi_{G}\left(  H\right)  $, so that $\pi_{G}\left(
H\right)  =\pi_{H}\left(  H\right)  =H/K$ (since the map $\pi_{H}$ is
surjective). Also, $\pi_{G}\left(  G\right)  =G/K$ (since $\pi_{G}$ is surjective).

The map $\pi_{G}$ is the canonical projection map $G\rightarrow G/K$, and thus
its kernel is $\ker\left(  \pi_{G}\right)  =K$.

The map $\pi_{H}$ is the canonical projection map $H\rightarrow H/K$, and thus
its kernel is $\ker\left(  \pi_{H}\right)  =K$.

Since $H$ is a subgroup of $G$, the $\CC$-algebra $\CC H$ is a
subalgebra of $\CC G$. Since $H/K$ is a subgroup of $G/K$, the
$\CC$-algebra $\CC \left[  H/K\right]  $ is a subalgebra of
$\CC \left[  G/K\right]  $. We have%
\begin{equation}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  y\right)  =\left(
\CC \left[  \pi_{H}\right]  \right)  \left(  y\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }y\in \CC H.
\label{sol.ind-infl.CGCH}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.CGCH}):} Let $y\in \CC H$.
Then, $y$ is a $\CC$-linear combination of elements of $H$ (since $H$
is a basis of the $\CC$-vector space $\CC H$). In other words,
there exists a family $\left(  y_{h}\right)  _{h\in H}\in \CC^{H}$ of
elements of $\CC$ such that $y=\sum_{h\in H}y_{h}h$. Consider this
family $\left(  y_{h}\right)  _{h\in H}$.
\par
Applying the map $\CC \left[  \pi_{H}\right]  $ to both sides of the
equality $y=\sum_{h\in H}y_{h}h$, we obtain
\begin{align}
\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  y\right)   &
=\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  \sum_{h\in H}%
y_{h}h\right)  =\sum_{h\in H}y_{h}\underbrace{\left(  \CC \left[
\pi_{H}\right]  \right)  \left(  h\right)  }_{\substack{=\pi_{H}\left(
h\right)  \\\text{(by (\ref{sol.ind-infl.s}), applied to}\\ \CC \text{,
}H\text{, }H/K\text{, }\pi_{H}\text{ and }h\text{ instead of}\\ \kk %
\text{, }S\text{, }T\text{, }\rho\text{ and }s\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map } \CC \left[  \pi
_{H}\right]  \text{ is } \CC \text{-linear}\right) \nonumber\\
&  =\sum_{h\in H}y_{h}\underbrace{\pi_{H}}_{=\pi_{G}\mid_{H}}\left(  h\right)
=\sum_{h\in H}y_{h}\underbrace{\left(  \pi_{G}\mid_{H}\right)  \left(
h\right)  }_{=\pi_{G}\left(  h\right)  }=\sum_{h\in H}y_{h}\pi_{G}\left(
h\right)  . \label{sol.ind-infl.CGCH.pf.1}%
\end{align}
On the other hand, applying the map $\CC \left[  \pi_{G}\right]  $ to
both sides of the equality $y=\sum_{h\in H}y_{h}h$, we obtain
\begin{align*}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  y\right)   &
=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  \sum_{h\in H}%
y_{h}h\right)  =\sum_{h\in H}y_{h}\underbrace{\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  h\right)  }_{\substack{=\pi_{G}\left(
h\right)  \\\text{(by (\ref{sol.ind-infl.s}), applied to}\\ \CC \text{,
}G\text{, }G/K\text{, }\pi_{G}\text{ and }h\text{ instead of}\\ \kk %
\text{, }S\text{, }T\text{, }\rho\text{ and }s\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map } \CC \left[  \pi
_{G}\right]  \text{ is } \CC \text{-linear}\right) \\
&  =\sum_{h\in H}y_{h}\pi_{G}\left(  h\right)  .
\end{align*}
Compared with (\ref{sol.ind-infl.CGCH.pf.1}), this yields $\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  y\right)  =\left(
 \CC \left[  \pi_{H}\right]  \right)  \left(  y\right)  $. This proves
(\ref{sol.ind-infl.CGCH}).}

\underline{\textit{Step 3: Writing }
$\Inf_{G/K}^{G}\Ind_{H/K}^{G/K}V$ \textit{and}
$\Ind_{H}^{G}\Inf_{H/K}^{H}V$ \textit{as tensor
products.}}

Inflation does not change the underlying $\CC$-vector space of a
representation. Thus, $\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}%
_{H/K}^{G/K}V=\operatorname{Ind}_{H/K}^{G/K}V$ as $\CC$-vector spaces.
For the same reason, $\operatorname{Infl}_{H/K}^{H}V=V$ as $\CC$-vector spaces.

By the definition of the inflation $\operatorname{Infl}_{G/K}^{G}%
\operatorname{Ind}_{H/K}^{G/K}V$, we have%
\begin{equation}
g\cdot w=\pi_{G}\left(  g\right)  \cdot w\ \ \ \ \ \ \ \ \ \ \text{for every
}g\in G\text{ and }w\in\operatorname{Ind}_{H/K}^{G/K}V
\label{sol.ind-infl.infl1}%
\end{equation}
(where the $w$ on the left hand side is regarded as an element of
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$, whereas the
$w$ on the right hand side is regarded as an element of $\operatorname{Ind}%
_{H/K}^{G/K}V$). It is now easy to see that%
\begin{equation}
x\cdot w=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  x\right)
\cdot w\ \ \ \ \ \ \ \ \ \ \text{for every }x\in \CC G\text{ and }%
w\in\operatorname{Ind}_{H/K}^{G/K}V \label{sol.ind-infl.infl1a}%
\end{equation}
(where the $w$ on the left hand side is regarded as an element of
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$, whereas the
$w$ on the right hand side is regarded as an element of $\operatorname{Ind}%
_{H/K}^{G/K}V$).\ \ \ \ \footnote{\textit{Proof of (\ref{sol.ind-infl.infl1a}%
):} Let $x\in \CC G$ and $w\in\operatorname{Ind}_{H/K}^{G/K}V$.
\par
We have $x\in \CC G$. Thus, $x$ is a $\CC$-linear combination of
the elements of $G$ (since $G$ is a basis of the $\CC$-vector space
$\CC G$). In other words, we can write $x$ in the form $x=\sum_{g\in
G}x_{g}g$ for some family $\left(  x_{g}\right)  _{g\in G}\in \CC^{G}$
of elements of $\CC$. Consider this family $\left(  x_{g}\right)
_{g\in G}$.
\par
We have $x=\sum_{g\in G}x_{g}g$. Applying the map $\CC \left[  \pi
_{G}\right]  $ to both sides of this equality, we obtain
\begin{align*}
\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  x\right)   &
=\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  \sum_{g\in G}%
x_{g}g\right)  =\sum_{g\in G}x_{g}\underbrace{\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  g\right)  }_{\substack{=\pi_{G}\left(
g\right)  \\\text{(by (\ref{sol.ind-infl.s}), applied to}\\ \CC \text{,
}G\text{, }G/K\text{, }\pi_{G}\text{ and }g\text{ instead of}\\ \kk %
\text{, }S\text{, }T\text{, }\rho\text{ and }s\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map } \CC \left[  \pi
_{G}\right]  \text{ is } \CC \text{-linear}\right) \\
&  =\sum_{g\in G}x_{g}\pi_{G}\left(  g\right)  .
\end{align*}
Hence,%
\begin{align*}
\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
x\right)  }_{=\sum_{g\in G}x_{g}\pi_{G}\left(  g\right)  }\cdot w  &
=\sum_{g\in G}x_{g}\pi_{G}\left(  g\right)  \cdot w=\sum_{g\in G}%
x_{g}\underbrace{\left(  \pi_{G}\left(  g\right)  \cdot w\right)
}_{\substack{=g\cdot w\\\text{(since (\ref{sol.ind-infl.infl1}) yields}%
\\g\cdot w=\pi_{G}\left(  g\right)  \cdot w\text{)}}}\\
&  =\sum_{g\in G}x_{g}g\cdot w=\underbrace{\left(  \sum_{g\in G}x_{g}g\right)
}_{\substack{=x\\\text{(since }x=\sum_{g\in G}x_{g}g\text{)}}}\cdot w=x\cdot
w.
\end{align*}
This proves (\ref{sol.ind-infl.infl1a}).}

By the definition of the inflation $\operatorname{Infl}_{H/K}^{H}V$, we have%
\begin{equation}
h\cdot v=\pi_{H}\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \text{for every
}h\in H\text{ and }v\in V \label{sol.ind-infl.infl2}%
\end{equation}
(where the $v$ on the left hand side is regarded as an element of
$\operatorname{Infl}_{H/K}^{H}V$, whereas the $v$ on the right hand side is
regarded as an element of $V$). It is now easy to see that%
\begin{equation}
x\cdot v=\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  x\right)
\cdot v\ \ \ \ \ \ \ \ \ \ \text{for every }x\in \CC H\text{ and }v\in V
\label{sol.ind-infl.infl2a}%
\end{equation}
(where the $v$ on the left hand side is regarded as an element of
$\operatorname{Infl}_{H/K}^{H}V$, whereas the $v$ on the right hand side is
regarded as an element of $V$).\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.ind-infl.infl2a}):} Let $x\in \CC H$ and $v\in V$.
\par
We have $x\in \CC H$. Thus, $x$ is a $\CC$-linear combination of
the elements of $H$ (since $H$ is a basis of the $\CC$-vector space
$\CC H$). In other words, we can write $x$ in the form $x=\sum_{h\in
H}x_{h}h$ for some family $\left(  x_{h}\right)  _{h\in H}\in \CC^{H}$
of elements of $\CC$. Consider this family $\left(  x_{h}\right)
_{h\in H}$.
\par
We have $x=\sum_{h\in H}x_{h}h$. Applying the map $\CC \left[  \pi
_{H}\right]  $ to both sides of this equality, we obtain
\begin{align*}
\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  x\right)   &
=\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  \sum_{h\in H}%
x_{h}h\right)  =\sum_{h\in H}x_{h}\underbrace{\left(  \CC \left[
\pi_{H}\right]  \right)  \left(  h\right)  }_{\substack{=\pi_{H}\left(
h\right)  \\\text{(by (\ref{sol.ind-infl.s}), applied to}\\ \CC \text{,
}H\text{, }H/K\text{, }\pi_{H}\text{ and }h\text{ instead of}\\ \kk %
\text{, }S\text{, }T\text{, }\rho\text{ and }s\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map } \CC \left[  \pi
_{H}\right]  \text{ is } \CC \text{-linear}\right) \\
&  =\sum_{h\in H}x_{h}\pi_{H}\left(  h\right)  .
\end{align*}
Hence,%
\begin{align*}
\underbrace{\left(  \CC \left[  \pi_{H}\right]  \right)  \left(
x\right)  }_{=\sum_{h\in H}x_{h}\pi_{H}\left(  h\right)  }\cdot v  &
=\sum_{h\in H}x_{h}\pi_{H}\left(  h\right)  \cdot v=\sum_{h\in H}%
x_{h}\underbrace{\left(  \pi_{H}\left(  h\right)  \cdot v\right)
}_{\substack{=h\cdot v\\\text{(since (\ref{sol.ind-infl.infl2}) yields}%
\\h\cdot v=\pi_{H}\left(  h\right)  \cdot v\text{)}}}\\
&  =\sum_{h\in H}x_{h}h\cdot v=\underbrace{\left(  \sum_{h\in H}x_{h}h\right)
}_{\substack{=x\\\text{(since }x=\sum_{h\in H}x_{h}h\text{)}}}\cdot v=x\cdot
v.
\end{align*}
This proves (\ref{sol.ind-infl.infl2a}).}

We have $\Ind_{H}^{G}\Inf_{H/K}^{H}V
= \CC G\otimes_{\CC H}\Inf_{H/K}^{H}V$
(by the definition of $\Ind_{H}^{G}\Inf_{H/K}^{H}V$) and
$\operatorname{Ind}_{H/K}^{G/K}V= \CC \left[  G/K\right]  \otimes
_{ \CC \left[  H/K\right]  }V$ (by the definition of $\operatorname{Ind}%
_{H/K}^{G/K}V$). Hence, $\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}%
_{H/K}^{G/K}V=\operatorname{Ind}_{H/K}^{G/K}V= \CC \left[  G/K\right]
\otimes_{ \CC \left[  H/K\right]  }V$ as $\CC$-vector spaces.

In the following, we shall construct two maps $\alpha:\operatorname{Infl}%
_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V\rightarrow\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V$ and $\beta:\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V\rightarrow\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V$ and show that they are inverse and one of
them is a $\CC G$-module homomorphism (which will, of course, imply that
so is the other).

\underline{\textit{Step 4: Defining a map }$\alpha$ \textit{from
}$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$ \textit{to}
$\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$\textit{.}}

For every $g\in G$, let us define a map $\mathbf{a}_{g}:V\rightarrow
\Ind_{H}^{G}\Inf_{H/K}^{H}V$
as follows: For any element $v$ of $V$, the element $g\otimes_{\CC H}v$
of $\Ind_{H}^{G}\Inf_{H/K}^{H}V$ is well-defined (since
$\underbrace{g}_{\in \CC G}\otimes_{\CC H}
\underbrace{v}_{\in V=\Inf_{H/K}^{H}V}\in \CC G\otimes_{\CC H}
\Inf_{H/K}^{H}V=\Ind_{H}^{G}\Inf_{H/K}^{H}V$). Thus, we can define a map
$\mathbf{a}_{g}:V\rightarrow \Ind_{H}^{G}\Inf_{H/K}^{H}V$
by setting
\[
\left(  \mathbf{a}_{g}\left(  v\right)  =g\otimes_{\CC H}%
v\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V\right)  .
\]
Consider this map $\mathbf{a}_{g}$. This map $\mathbf{a}_{g}$ is $\CC %
$-linear (since the element $\mathbf{a}_{g}\left(  v\right)  =g\otimes
_{\CC H}v$ depends $\CC$-linearly on $v$). We thus have defined a
$\CC$-linear map $\mathbf{a}_{g}:V\rightarrow\Ind_{H}^{G}
\Inf_{H/K}^{H}V$ for every $g\in G$.

It is easy to see that%
\begin{equation}
\mathbf{a}_{g}=\mathbf{a}_{g^{\prime}}\ \ \ \ \ \ \ \ \ \ \text{for any two
elements }g\text{ and }g^{\prime}\text{ of }G\text{ satisfying }\pi_{G}\left(
g\right)  =\pi_{G}\left(  g^{\prime}\right)  . \label{sol.ind-infl.a.toGK}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.a.toGK}):} Let $g$ and
$g^{\prime}$ be two elements of $G$ satisfying $\pi_{G}\left(  g\right)
=\pi_{G}\left(  g^{\prime}\right)  $. The definition of $\pi_{G}$ yields
$\pi_{G}\left(  g\right)  =gK$ (as a left coset in $G/K$) and $\pi_{G}\left(
g^{\prime}\right)  =g^{\prime}K$ (as a left coset in $G/K$). Since $\pi
_{G}\left(  g\right)  =gK$, we have $gK=\pi_{G}\left(  g\right)  =\pi
_{G}\left(  g^{\prime}\right)  =g^{\prime}K$ and thus $g=g\underbrace{1}_{\in
K}\in gK=g^{\prime}K$. In other words, there exists a $k\in K$ satisfying
$g=g^{\prime}k$. Consider this $k$.
\par
Let $v\in V$. The definition of $\mathbf{a}_{g}$ yields $\mathbf{a}_{g}\left(
v\right)  =\underbrace{g}_{=g^{\prime}k}\otimes_{\CC H}v=g^{\prime
}k\otimes_{\CC H}v$. But $k\in K\subset H\subset \CC H$, and thus
we can move $k$ past the tensor sign $\otimes_{\CC H}$ in $g^{\prime
}k\otimes_{\CC H}v$. Thus, $g^{\prime}k\otimes_{\CC H}v=g^{\prime
}\otimes_{\CC H}kv$.
\par
But we have $k\in K=\ker\left(  \pi_{H}\right)  $ and thus $\pi_{H}\left(
k\right)  =1$. But $k\in H$. Thus, (\ref{sol.ind-infl.infl2}) (applied to
$h=k$) yields $k\cdot v=\underbrace{\pi_{H}\left(  k\right)  }_{=1}\cdot v=v$.
Altogether, we now have%
\[
\mathbf{a}_{g}\left(  v\right)  =g^{\prime}k\otimes_{\CC H}v=g^{\prime
}\otimes_{\CC H}\underbrace{kv}_{=k\cdot v=v}=g^{\prime}\otimes
_{\CC H}v=\mathbf{a}_{g^{\prime}}\left(  v\right)
\]
(since $\mathbf{a}_{g^{\prime}}\left(  v\right)  =g^{\prime}\otimes
_{\CC H}v$ (by the definition of $\mathbf{a}_{g^{\prime}}$)).
\par
Now, forget that we fixed $v$. We thus have proven that $\mathbf{a}_{g}\left(
v\right)  =\mathbf{a}_{g^{\prime}}\left(  v\right)  $ for every $v\in V$. In
other words, $\mathbf{a}_{g}=\mathbf{a}_{g^{\prime}}$. This proves
(\ref{sol.ind-infl.a.toGK}).}

Now, for every $s\in G/K$, let us define a $\CC$-linear map
$\mathbf{a}_{s}^{\prime}:V\rightarrow\Ind_{H}%
^{G}\Inf_{H/K}^{H}V$ as follows: Since $s\in
G/K=\pi_{G}\left(  G\right)  $ (because $\pi_{G}$ is surjective), there must
exist some $g\in G$ such that $s=\pi_{G}\left(  g\right)  $. Consider this
$g$. Now, set $\mathbf{a}_{s}^{\prime}$ to be the $\CC$-linear map
$\mathbf{a}_{g}$. This is clearly well-defined, because the map $\mathbf{a}%
_{g}$ does not depend on the choice of $g$.\ \ \ \ \footnote{\textit{Proof.}
Let $g_{1}$ and $g_{2}$ be two elements $g$ of $G$ such that $s=\pi_{G}\left(
g\right)  $. Thus, $g_{1}$ and $g_{2}$ are elements of $G$ such that
$s=\pi_{G}\left(  g_{1}\right)  $ and $s=\pi_{G}\left(  g_{2}\right)  $.
Hence, $\pi_{G}\left(  g_{1}\right)  =s=\pi_{G}\left(  g_{2}\right)  $.
Therefore, (\ref{sol.ind-infl.a.toGK}) (applied to $g=g_{1}$ and $g^{\prime
}=g_{2}$) yields $\mathbf{a}_{g_{1}}=\mathbf{a}_{g_{2}}$.
\par
Now, forget that we fixed $g_{1}$ and $g_{2}$. We thus have shown that
$\mathbf{a}_{g_{1}}=\mathbf{a}_{g_{2}}$ whenever $g_{1}$ and $g_{2}$ are two
elements $g$ of $G$ such that $s=\pi_{G}\left(  g\right)  $. Hence, the map
$\mathbf{a}_{g}$ does not depend on the choice of $g$ when $g$ is chosen to be
an element of $G$ such that $s=\pi_{G}\left(  g\right)  $. Qed.}

So we have defined a $\CC$-linear map $\mathbf{a}_{s}^{\prime
}:V\rightarrow\Ind_{H}^{G}\Inf_{H/K}^{H}V$ for every
$s\in G/K$. Thus, $\mathbf{a}_{s}^{\prime}%
\in\Hom_{\CC} \left(  V,\Ind_{H}^{G}\Inf_{H/K}^{H}V\right)  $
for every
$s\in G/K$ (since $\mathbf{a}_{s}^{\prime}$ is a $\CC$-linear map
$V\rightarrow\Ind_{H}^{G}\Inf_{H/K}^{H}V$ for every $s\in G/K$). Thus,
we can define a map
$\mathbf{a}:G/K\rightarrow\Hom_{\CC}\left(
V,\Ind_{H}^{G}\Inf_{H/K}^{H}V\right)  $ by setting
\[
\left(  \mathbf{a}\left(  s\right)  =\mathbf{a}_{s}^{\prime}
\ \ \ \ \ \ \ \ \ \ \text{for every }s\in G/K\right)  .
\]
Consider this map $\mathbf{a}$. Applying (\ref{sol.ind-infl.uniprop-kS}) to
$\kk = \CC$, $S=G/K$, $U=\Hom_{\CC}
\left(  V,\Ind_{H}^{G}\Inf_{H/K}^{H}V\right)  $ and $\rho=\mathbf{a}$,
we conclude that there
exists exactly one $\CC$-linear map $\widetilde{\mathbf{a}}
: \CC \left[  G/K\right]  \rightarrow
\Hom_{\CC}\left(  V,\Ind_{H}%
^{G}\Inf_{H/K}^{H}V\right)  $ satisfying $\left(
\widetilde{\mathbf{a}}\left(  s\right)  =\mathbf{a}\left(  s\right)  \text{
for every }s\in G/K\right)  $. Consider this $\CC$-linear map
$\widetilde{\mathbf{a}}$.

For every $s\in G/K$, we have
\begin{equation}
\widetilde{\mathbf{a}}\left(  s\right)  =\mathbf{a}\left(  s\right)
=\mathbf{a}_{s}^{\prime}. \label{sol.ind-infl.a.tilde}%
\end{equation}
For every $x\in G$, we have%
\begin{equation}
\widetilde{\mathbf{a}}\left(  \pi_{G}\left(  x\right)  \right)  =\mathbf{a}%
_{x}. \label{sol.ind-infl.a.tilde-x}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.a.tilde-x}):} Let $x\in G$. Let
$s=\pi_{G}\left(  x\right)  $. Then, $s\in G/K$. The definition of
$\mathbf{a}_{s}^{\prime}$ shows that if $g$ is any element of $G$ such that
$s=\pi_{G}\left(  g\right)  $, then $\mathbf{a}_{s}^{\prime}=\mathbf{a}_{g}$.
Applying this to $g=x$, we obtain $\mathbf{a}_{s}^{\prime}=\mathbf{a}_{x}$
(since $s=\pi_{G}\left(  x\right)  $). Now, (\ref{sol.ind-infl.a.tilde})
yields $\widetilde{\mathbf{a}}\left(  s\right)  =\mathbf{a}_{s}^{\prime
}=\mathbf{a}_{x}$. Since $s=\pi_{G}\left(  x\right)  $, this rewrites as
$\widetilde{\mathbf{a}}\left(  \pi_{G}\left(  x\right)  \right)
=\mathbf{a}_{x}$. This proves (\ref{sol.ind-infl.a.tilde-x}).} For every $x\in
G$ and $v\in V$, we have%
\begin{equation}
\left(  \underbrace{\widetilde{\mathbf{a}}\left(  \pi_{G}\left(  x\right)
\right)  }_{\substack{=\mathbf{a}_{x}\\\text{(by (\ref{sol.ind-infl.a.tilde-x}%
))}}}\right)  \left(  v\right)  =\mathbf{a}_{x}\left(  v\right)
=x\otimes_{\CC H}v \label{sol.ind-infl.a.tilde-x.(v)}%
\end{equation}
(by the definition of $\mathbf{a}_{x}\left(  v\right)  $).

It is well-known that if $A$ is a $\CC$-algebra, if $P$ is a left
$A$-module and if $Q$ is a $\CC$-vector space, then the
$\CC$-vector space $\Hom_{\CC}\left(  P,Q\right)  $
canonically becomes a right $A$-module\footnote{Indeed, this right $A$-module
structure is defined as follows: If
$f\in\Hom_{\CC}\left(  P,Q\right)  $ and $a\in A$, then $fa$ is the
$\CC$-linear map $P\rightarrow Q$ which sends every $p\in P$ to
$f\left(  ap\right)  $.}. Applying this to $A= \CC \left[  H/K\right]  $,
$P=V$ and $Q=\Ind_{H}^{G}\Inf_{H/K}^{H}V$, we conclude that the
$\CC$-vector space
$\Hom_{\CC}\left(  V,\Ind_{H}^{G}\Inf_{H/K}^{H}V\right)  $
canonically becomes a right $\CC \left[  H/K\right]  $-module. Also,
$\CC \left[  G/K\right]  $ is a right $\CC \left[  H/K\right]
$-module (since $H/K$ is a subgroup of $G/K$). The map $\widetilde{\mathbf{a}%
}$ is a right $\CC \left[  H/K\right]  $-module
homomorphism\footnote{\textit{Proof.} Let $t\in H/K$. We are first going to
prove that
\begin{equation}
\widetilde{\mathbf{a}}\left(  st\right)  =\left(  \widetilde{\mathbf{a}%
}\left(  s\right)  \right)  \cdot t\ \ \ \ \ \ \ \ \ \ \text{for every }s\in
G/K. \label{sol.ind-inf.a.Hequ}%
\end{equation}
\par
\textit{Proof of (\ref{sol.ind-inf.a.Hequ}):} Let $s\in G/K$. Since $s\in
G/K=\pi_{G}\left(  G\right)  $ (because $\pi_{G}\left(  G\right)  =G/K$),
there exists a $g\in G$ satisfying $s=\pi_{G}\left(  g\right)  $. Consider
this $g$.
\par
Since $t\in H/K=\pi_{G}\left(  H\right)  $ (because $\pi_{G}\left(  H\right)
=H/K$), there exists an $h\in H$ satisfying $t=\pi_{G}\left(  h\right)  $.
Consider this $h$. Since $h\in H$, we have $\pi_{G}\left(  h\right)
=\underbrace{\left(  \pi_{G}\mid_{H}\right)  }_{=\pi_{H}}\left(  h\right)
=\pi_{H}\left(  h\right)  $. Thus, $t=\pi_{G}\left(  h\right)  =\pi_{H}\left(
h\right)  $.
\par
Let $v\in V$. We have $h\cdot v=\pi_{H}\left(  h\right)  \cdot v$ (by
(\ref{sol.ind-infl.infl2})). Thus, $\pi_{H}\left(  h\right)  \cdot v=h\cdot
v=hv$. Hence, $tv=\underbrace{t}_{=\pi_{H}\left(  h\right)  }\cdot v=\pi
_{H}\left(  h\right)  \cdot v=hv$.
\par
Multiplying the equalities $s=\pi_{G}\left(  g\right)  $ and $t=\pi_{G}\left(
h\right)  $, we obtain $st=\pi_{G}\left(  g\right)  \cdot\pi_{G}\left(
h\right)  =\pi_{G}\left(  gh\right)  $ (since $\pi_{G}$ is a group
homomorphism). Applying the map $\widetilde{\mathbf{a}}$ to both sides of this
equality, we obtain $\widetilde{\mathbf{a}}\left(  st\right)
=\widetilde{\mathbf{a}}\left(  \pi_{G}\left(  gh\right)  \right)
=\mathbf{a}_{gh}$ (by (\ref{sol.ind-infl.a.tilde-x}), applied to $x=gh$).
Applying both sides of this equality to $v$, we obtain%
\begin{align*}
\left(  \widetilde{\mathbf{a}}\left(  st\right)  \right)  \left(  v\right)
&  =\mathbf{a}_{gh}\left(  v\right)  =gh\otimes_{\CC H}%
v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{a}%
_{gh}\left(  v\right)  \right) \\
&  =g\otimes_{\CC H}hv\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h\text{
can be moved past the }\otimes_{\CC H}\text{ sign (because }h\in
H\subset \CC H\text{)}\right)  .
\end{align*}
\par
On the other hand, $\widetilde{\mathbf{a}}\left(  \underbrace{s}_{=\pi
_{G}\left(  g\right)  }\right)  =\widetilde{\mathbf{a}}\left(  \pi_{G}\left(
g\right)  \right)  =\mathbf{a}_{g}$ (by (\ref{sol.ind-infl.a.tilde-x}),
applied to $x=g$). Hence,%
\begin{align*}
\left(  \underbrace{\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)
}_{=\mathbf{a}_{g}}\cdot t\right)  \left(  v\right)   &  =\left(
\mathbf{a}_{g}t\right)  \left(  v\right)  =\mathbf{a}_{g}\left(
\underbrace{tv}_{=hv}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{by the definition of }\mathbf{a}_{g}t\text{ (that is, by the definition
of}\\
\text{the right } \CC \left[  H/K\right]  \text{-module structure on
}\Hom_\CC \left(  V,\Ind_{H}^{G}\Inf_{H/K}^{H}V\right)  \text{)}
\end{array}
\right) \\
&  =\mathbf{a}_{g}\left(  hv\right)  =g\otimes_{\CC H}
hv\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }
\mathbf{a}_{g}\left(  hv\right)  \right)  .
\end{align*}
Comparing this with $\left(  \widetilde{\mathbf{a}}\left(  st\right)  \right)
\left(  v\right)  =g\otimes_{\CC H}hv$, we obtain $\left(
\widetilde{\mathbf{a}}\left(  st\right)  \right)  \left(  v\right)  =\left(
\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)  \cdot t\right)
\left(  v\right)  $.
\par
Now, forget that we fixed $v\in V$. We thus have proven that $\left(
\widetilde{\mathbf{a}}\left(  st\right)  \right)  \left(  v\right)  =\left(
\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)  \cdot t\right)
\left(  v\right)  $ for every $v\in V$. In other words, $\widetilde{\mathbf{a}%
}\left(  st\right)  =\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)
\cdot t$. This proves (\ref{sol.ind-inf.a.Hequ}).
\par
Now, let $x\in \CC \left[  G/K\right]  $ be arbitrary. Then, $x$ is a
$\CC$-linear combination of the elements of $G/K$ (since $G/K$ is a
basis of the $\CC$-vector space $\CC \left[  G/K\right]  $). In
other words, $x$ can be written in the form $x=\sum_{s\in G/K}x_{s}s$ for a
family $\left(  x_{s}\right)  _{s\in G/K}\in \CC^{G/K}$ of complex
numbers. Consider this family $\left(  x_{s}\right)  _{s\in G/K}$.
\par
Multiplying the equality $x=\sum_{s\in G/K}x_{s}s$ with $t$ from the right, we
obtain $xt=\left(  \sum_{s\in G/K}x_{s}s\right)  t=\sum_{s\in G/K}x_{s}st$.
Applying the map $\widetilde{\mathbf{a}}$ to both sides of this equality, we
obtain%
\begin{align*}
\widetilde{\mathbf{a}}\left(  xt\right)   &  =\widetilde{\mathbf{a}}\left(
\sum_{s\in G/K}x_{s}st\right)  =\sum_{s\in G/K}x_{s}%
\underbrace{\widetilde{\mathbf{a}}\left(  st\right)  }_{\substack{=\left(
\widetilde{\mathbf{a}}\left(  s\right)  \right)  \cdot t\\\text{(by
(\ref{sol.ind-inf.a.Hequ}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\widetilde{\mathbf{a}}\text{ is } \CC \text{-linear}\right) \\
&  =\sum_{s\in G/K}x_{s}\left(  \widetilde{\mathbf{a}}\left(  s\right)
\right)  \cdot t=\underbrace{\left(  \sum_{s\in G/K}x_{s}\widetilde{\mathbf{a}%
}\left(  s\right)  \right)  }_{\substack{=\widetilde{\mathbf{a}}\left(
\sum_{s\in G/K}x_{s}s\right)  \\\text{(since the map }\widetilde{\mathbf{a}%
}\text{ is } \CC \text{-linear)}}}t=\widetilde{\mathbf{a}}\left(
\underbrace{\sum_{s\in G/K}x_{s}s}_{=x}\right)  \cdot t=\widetilde{\mathbf{a}%
}\left(  x\right)  \cdot t.
\end{align*}
\par
Now, let us forget that we fixed $x$ and $t$. We thus have proven that
$\widetilde{\mathbf{a}}\left(  xt\right)  =\widetilde{\mathbf{a}}\left(
x\right)  \cdot t$ for every $t\in H/K$ and every $x\in \CC \left[
G/K\right]  $. In other words, the map $\widetilde{\mathbf{a}}$ is a
homomorphism of right $H/K$-sets. Combining this with the fact that this map
$\widetilde{\mathbf{a}}$ is $\CC$-linear, we conclude that
$\widetilde{\mathbf{a}}$ is a right $\CC \left[  H/K\right]  $-module
homomorphism, qed.}. Thus,%
\begin{equation}
\widetilde{\mathbf{a}}\left(  sy\right)  =\widetilde{\mathbf{a}}\left(
s\right)  \cdot y\ \ \ \ \ \ \ \ \ \ \text{for every }s\in \CC \left[
G/K\right]  \text{ and }y\in \CC \left[  H/K\right]  .
\label{sol.ind-infl.a.rimo}%
\end{equation}


Now, let us define a map $\mathbf{A}: \CC \left[  G/K\right]  \times
V\rightarrow\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ by%
\begin{equation}
\left(  \mathbf{A}\left(  s,v\right)  =\left(  \widetilde{\mathbf{a}}\left(
s\right)  \right)  \left(  v\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  s,v\right)  \in \CC \left[  G/K\right]  \times V\right)  .
\label{sol.ind-infl.a.A}%
\end{equation}
Then, the map $\mathbf{A}$ is $\CC$-linear in its first
argument\footnote{\textit{Proof.} Let $\lambda_{1}\in \CC$, $\lambda
_{2}\in \CC$, $s_{1}\in \CC \left[  G/K\right]  $ and $s_{2}%
\in \CC \left[  G/K\right]  $ be arbitrary. Then, $\widetilde{\mathbf{a}%
}\left(  \lambda_{1}s_{1}+\lambda_{2}s_{2}\right)  =\lambda_{1}%
\widetilde{\mathbf{a}}\left(  s_{1}\right)  +\lambda_{2}\widetilde{\mathbf{a}%
}\left(  s_{2}\right)  $ (since the map $\widetilde{\mathbf{a}}$ is
$\CC$-linear).
\par
Now, let $v\in V$. Then, $\mathbf{A}\left(  s_{1},v\right)  =\left(
\widetilde{\mathbf{a}}\left(  s_{1}\right)  \right)  \left(  v\right)  $ (by
the definition of $\mathbf{A}$) and $\mathbf{A}\left(  s_{2},v\right)
=\left(  \widetilde{\mathbf{a}}\left(  s_{2}\right)  \right)  \left(
v\right)  $ (by the definition of $\mathbf{A}$). But by the definition of
$\mathbf{A}$, we have%
\begin{align*}
\mathbf{A}\left(  \lambda_{1}s_{1}+\lambda_{2}s_{2},v\right)   &  =\left(
\underbrace{\widetilde{\mathbf{a}}\left(  \lambda_{1}s_{1}+\lambda_{2}%
s_{2}\right)  }_{=\lambda_{1}\widetilde{\mathbf{a}}\left(  s_{1}\right)
+\lambda_{2}\widetilde{\mathbf{a}}\left(  s_{2}\right)  }\right)  \left(
v\right)  =\left(  \lambda_{1}\widetilde{\mathbf{a}}\left(  s_{1}\right)
+\lambda_{2}\widetilde{\mathbf{a}}\left(  s_{2}\right)  \right)  \left(
v\right) \\
&  =\lambda_{1}\left(  \widetilde{\mathbf{a}}\left(  s_{1}\right)  \right)
\left(  v\right)  +\lambda_{2}\left(  \widetilde{\mathbf{a}}\left(
s_{2}\right)  \right)  \left(  v\right)  .
\end{align*}
Compared with%
\[
\lambda_{1}\underbrace{\mathbf{A}\left(  s_{1},v\right)  }_{=\left(
\widetilde{\mathbf{a}}\left(  s_{1}\right)  \right)  \left(  v\right)
}+\lambda_{2}\underbrace{\mathbf{A}\left(  s_{2},v\right)  }_{=\left(
\widetilde{\mathbf{a}}\left(  s_{2}\right)  \right)  \left(  v\right)
}=\lambda_{1}\left(  \widetilde{\mathbf{a}}\left(  s_{1}\right)  \right)
\left(  v\right)  +\lambda_{2}\left(  \widetilde{\mathbf{a}}\left(
s_{2}\right)  \right)  \left(  v\right)  ,
\]
this yields $\mathbf{A}\left(  \lambda_{1}s_{1}+\lambda_{2}s_{2},v\right)
=\lambda_{1}\mathbf{A}\left(  s_{1},v\right)  +\lambda_{2}\mathbf{A}\left(
s_{2},v\right)  $.
\par
Now, let us forget that we fixed $\lambda_{1}\in \CC$, $\lambda_{2}%
\in \CC$, $s_{1}\in \CC \left[  G/K\right]  $, $s_{2}\in
 \CC \left[  G/K\right]  $ and $v\in V$. We thus have proven that
$\mathbf{A}\left(  \lambda_{1}s_{1}+\lambda_{2}s_{2},v\right)  =\lambda
_{1}\mathbf{A}\left(  s_{1},v\right)  +\lambda_{2}\mathbf{A}\left(
s_{2},v\right)  $ for every $\lambda_{1}\in \CC$, $\lambda_{2}%
\in \CC$, $s_{1}\in \CC \left[  G/K\right]  $, $s_{2}\in
 \CC \left[  G/K\right]  $ and $v\in V$. In other words, the map
$\mathbf{A}$ is $\CC$-linear in its first argument, qed.} and is
$\CC$-linear in its second argument\footnote{\textit{Proof.} Let
$s\in \CC \left[  G/K\right]  $, $\lambda_{1}\in \CC$, $\lambda
_{2}\in \CC$, $v_{1}\in V$ and $v_{2}\in V$ be arbitrary. Then,
$\mathbf{A}\left(  s,v_{1}\right)  =\left(  \widetilde{\mathbf{a}}\left(
s\right)  \right)  \left(  v_{1}\right)  $ (by the definition of $\mathbf{A}%
$), so that $\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(
v_{1}\right)  =\mathbf{A}\left(  s,v_{1}\right)  $. Also, $\mathbf{A}\left(
s,v_{2}\right)  =\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)
\left(  v_{2}\right)  $ (by the definition of $\mathbf{A}$), so that $\left(
\widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(  v_{2}\right)
=\mathbf{A}\left(  s,v_{2}\right)  $. But by the definition of $\mathbf{A}$,
we have
\begin{align*}
\mathbf{A}\left(  s,\lambda_{1}v_{1}+\lambda_{2}v_{2}\right)   &  =\left(
\widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(  \lambda_{1}%
v_{1}+\lambda_{2}v_{2}\right)  =\lambda_{1}\underbrace{\left(
\widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(  v_{1}\right)
}_{=\mathbf{A}\left(  s,v_{1}\right)  }+\lambda_{2}\underbrace{\left(
\widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(  v_{2}\right)
}_{=\mathbf{A}\left(  s,v_{2}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\widetilde{\mathbf{a}%
}\left(  s\right)  \text{ is } \CC \text{-linear (because }%
\widetilde{\mathbf{a}}\left(  s\right)  \in
\Hom_{\CC}\left(  V,\Ind_{H}%
^{G}\Inf_{H/K}^{H}V\right)  \text{)}\right) \\
&  =\lambda_{1}\mathbf{A}\left(  s,v_{1}\right)  +\lambda_{2}\mathbf{A}\left(
s,v_{2}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $s\in \CC \left[  G/K\right]  $,
$\lambda_{1}\in \CC$, $\lambda_{2}\in \CC$, $v_{1}\in V$ and
$v_{2}\in V$. We thus have proven that $\mathbf{A}\left(  s,\lambda_{1}%
v_{1}+\lambda_{2}v_{2}\right)  =\lambda_{1}\mathbf{A}\left(  s,v_{1}\right)
+\lambda_{2}\mathbf{A}\left(  s,v_{2}\right)  $ for every $s\in \CC %
\left[  G/K\right]  $, $\lambda_{1}\in \CC$, $\lambda_{2}\in \CC$,
$v_{1}\in V$ and $v_{2}\in V$. In other words, the map $\mathbf{A}$ is
$\CC$-linear in its second argument, qed.}. Thus, the map $\mathbf{A}$
is $\CC$-bilinear (since it is $\CC$-linear in its first
argument and is $\CC$-linear in its second argument). Furthermore, the
map $\mathbf{A}$ is $\CC \left[  H/K\right]  $-bilinear with respect to
the right $\CC \left[  H/K\right]  $-module structure on $\CC %
\left[  G/K\right]  $ and the left $\CC \left[  H/K\right]  $-module
structure on $V$\ \ \ \ \footnote{\textit{Proof.} Every $y\in \CC \left[
H/K\right]  $, $s\in \CC \left[  G/K\right]  $ and $v\in V$ satisfy%
\begin{align*}
\mathbf{A}\left(  sy,v\right)   &  =\left(  \underbrace{\widetilde{\mathbf{a}%
}\left(  sy\right)  }_{\substack{=\widetilde{\mathbf{a}}\left(  s\right)
\cdot y\\\text{(by (\ref{sol.ind-infl.a.rimo}))}}}\right)  \left(  v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{A}\left(
sy,v\right)  \right) \\
&  =\left(  \widetilde{\mathbf{a}}\left(  s\right)  \cdot y\right)  \left(
v\right)  =\left(  \widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(
yv\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of }\widetilde{\mathbf{a}}\left(  s\right)  \cdot
y\text{ (that is, by the definition of}\\
\text{the right } \CC \left[  H/K\right]  \text{-module structure on
}\Hom_{\CC}\left(  V,\Ind_{H}^{G}\Inf_{H/K}^{H}V\right)  \text{)}
\end{array}
\right) \\
&  =\mathbf{A}\left(  s,yv\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{A}\left(  s,yv\right)  =\left(  \widetilde{\mathbf{a}}\left(
s\right)  \right)  \left(  yv\right)  \text{ (by the definition of }%
\mathbf{A}\left(  s,yv\right)  \text{)}\right)  .
\end{align*}
In other words, the map $\mathbf{A}$ is $\CC \left[  H/K\right]
$-bilinear with respect to the right $\CC \left[  H/K\right]  $-module
structure on $\CC \left[  G/K\right]  $ and the left $\CC \left[
H/K\right]  $-module structure on $V$ (because we already know that
$\mathbf{A}$ is $\CC$-bilinear), qed.}. Hence, by the universal
property of the tensor product, we conclude that there exists a unique
$\CC$-linear map $\alpha: \CC \left[  G/K\right]  \otimes
_{ \CC \left[  H/K\right]  }V\rightarrow\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V$ such that every $\left(  s,v\right)
\in \CC \left[  G/K\right]  \times V$ satisfies%
\begin{equation}
\alpha\left(  s\otimes_{ \CC \left[  H/K\right]  }v\right)
=\mathbf{A}\left(  s,v\right)  . \label{sol.ind-infl.alpha}%
\end{equation}
Consider this map $\alpha$. This map $\alpha$ is a $\CC$-linear map
from $\CC \left[  G/K\right]  \otimes_{ \CC \left[  H/K\right]  }V$
to $\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$. In other words,
the map $\alpha$ is a $\CC$-linear map from $\operatorname{Infl}%
_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$ to $\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V$ (since $\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V= \CC \left[  G/K\right]
\otimes_{ \CC \left[  H/K\right]  }V$ as $\CC$-vector spaces).

\underline{\textit{Step 5: Defining a map }$\beta$ \textit{from }%
$\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ \textit{to}
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$\textit{.}}

We shall now construct another map.

For every $\left(  s,v\right)  \in \CC G\times\operatorname{Infl}%
_{H/K}^{H}V$, we have%
\[
\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
s\right)  }_{\in \CC \left[  G/K\right]  }\otimes_{ \CC \left[
H/K\right]  }\underbrace{v}_{\in\operatorname{Infl}_{H/K}^{H}V=V}\in
 \CC \left[  G/K\right]  \otimes_{ \CC \left[  H/K\right]
}V=\operatorname{Ind}_{H/K}^{G/K}V=\operatorname{Infl}_{G/K}^{G}%
\operatorname{Ind}_{H/K}^{G/K}V.
\]
Hence, we can define a map $\mathbf{B}: \CC G\times\operatorname{Infl}%
_{H/K}^{H}V\rightarrow\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}%
_{H/K}^{G/K}V$ by%
\[
\left(  \mathbf{B}\left(  s,v\right)  =\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  s\right)  \otimes_{ \CC \left[  H/K\right]
}v\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  s,v\right)  \in \CC G
\times\operatorname{Infl}_{H/K}^{H}V\right)  .
\]
Consider this map $\mathbf{B}$. The map $\mathbf{B}$ is $\CC$-linear in
its first argument\footnote{\textit{Proof.} Let $\lambda_{1}\in \CC$,
$\lambda_{2}\in \CC$, $s_{1}\in \CC G$, $s_{2}\in \CC G$ and
$v\in\operatorname{Infl}_{H/K}^{H}V$ be arbitrary. We have $\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  \lambda_{1}s_{1}%
+\lambda_{2}s_{2}\right)  =\lambda_{1}\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  s_{1}\right)  +\lambda_{2}\left(  \CC 
\left[  \pi_{G}\right]  \right)  \left(  s_{2}\right)  $ (since the map
$\CC \left[  \pi_{G}\right]  $ is $\CC$-linear). The definition
of $\mathbf{B}\left(  \lambda_{1}s_{1}+\lambda_{2}s_{2},v\right)  $ yields%
\begin{align*}
\mathbf{B}\left(  \lambda_{1}s_{1}+\lambda_{2}s_{2},v\right)   &
=\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
\lambda_{1}s_{1}+\lambda_{2}s_{2}\right)  }_{=\lambda_{1}\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s_{1}\right)  +\lambda
_{2}\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  s_{2}\right)
}\otimes_{ \CC \left[  H/K\right]  }v=\left(  \lambda_{1}\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s_{1}\right)  +\lambda
_{2}\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  s_{2}\right)
\right)  \otimes_{ \CC \left[  H/K\right]  }v\\
&  =\lambda_{1}\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  s_{1}\right)  \otimes_{ \CC \left[  H/K\right]  }v}%
_{\substack{=\mathbf{B}\left(  s_{1},v\right)  \\\text{(because }%
\mathbf{B}\left(  s_{1},v\right)  \\=\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  s_{1}\right)  \otimes_{ \CC \left[  H/K\right]
}v\\\text{(by the definition of }\mathbf{B}\left(  s_{1},v\right)  \text{))}%
}}+\lambda_{2}\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  s_{2}\right)  \otimes_{ \CC \left[  H/K\right]  }v}%
_{\substack{=\mathbf{B}\left(  s_{2},v\right)  \\\text{(because }%
\mathbf{B}\left(  s_{2},v\right)  \\=\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  s_{2}\right)  \otimes_{ \CC \left[  H/K\right]
}v\\\text{(by the definition of }\mathbf{B}\left(  s_{2},v\right)  \text{))}%
}}\\
&  =\lambda_{1}\mathbf{B}\left(  s_{1},v\right)  +\lambda_{2}\mathbf{B}\left(
s_{2},v\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\lambda_{1}\in \CC$, $\lambda_{2}%
\in \CC$, $s_{1}\in \CC G$, $s_{2}\in \CC G$ and
$v\in\operatorname{Infl}_{H/K}^{H}V$. We
thus have proven that $\mathbf{B}\left(  \lambda_{1}s_{1}+\lambda_{2}%
s_{2},v\right)  =\lambda_{1}\mathbf{B}\left(  s_{1},v\right)  +\lambda
_{2}\mathbf{B}\left(  s_{2},v\right)  $ for every $\lambda_{1}\in \CC$,
$\lambda_{2}\in \CC$, $s_{1}\in \CC G$, $s_{2}\in \CC G$ and
$v\in\operatorname{Infl}_{H/K}^{H}V$. In other words, the map $\mathbf{B}$ is
$\CC$-linear in its first argument, qed.} and is $\CC$-linear in
its second argument\footnote{\textit{Proof.} Let $s\in \CC G$,
$\lambda_{1}\in \CC$, $\lambda_{2}\in \CC$, $v_{1}\in
\operatorname{Infl}_{H/K}^{H}V$ and $v_{2}\in\operatorname{Infl}_{H/K}^{H}V$
be arbitrary. Then, $\mathbf{B}\left(  s,v_{1}\right)  =\left(  \CC
\left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes_{ \CC \left[
H/K\right]  }v_{1}$ (by the definition of $\mathbf{B}$), so that $\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes
_{ \CC \left[  H/K\right]  }v_{1}=\mathbf{B}\left(  s,v_{1}\right)  $.
Also, $\mathbf{B}\left(  s,v_{2}\right)  =\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  s\right)  \otimes_{ \CC \left[  H/K\right]
}v_{2}$ (by the definition of $\mathbf{B}$), so that $\left(  \CC
\left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes_{ \CC \left[
H/K\right]  }v_{2}=\mathbf{B}\left(  s,v_{2}\right)  $. But by the definition
of $\mathbf{B}$, we have
\begin{align*}
\mathbf{B}\left(  s,\lambda_{1}v_{1}+\lambda_{2}v_{2}\right)   &  =\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes
_{ \CC \left[  H/K\right]  }\left(  \lambda_{1}v_{1}+\lambda_{2}%
v_{2}\right) \\
&  =\lambda_{1}\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  s\right)  \otimes_{ \CC \left[  H/K\right]  }v_{1}}%
_{=\mathbf{B}\left(  s,v_{1}\right)  }+\lambda_{2}\underbrace{\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes
_{ \CC \left[  H/K\right]  }v_{2}}_{=\mathbf{B}\left(  s,v_{2}\right)
}=\lambda_{1}\mathbf{B}\left(  s,v_{1}\right)  +\lambda_{2}\mathbf{B}\left(
s,v_{2}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $s\in \CC G$,
$\lambda_{1}\in \CC$, $\lambda_{2}\in \CC$, $v_{1}\in
\operatorname{Infl}_{H/K}^{H}V$ and $v_{2}\in\operatorname{Infl}_{H/K}^{H}V$.
We thus have proven that $\mathbf{B}\left(  s,\lambda_{1}v_{1}+\lambda
_{2}v_{2}\right)  =\lambda_{1}\mathbf{B}\left(  s,v_{1}\right)  +\lambda
_{2}\mathbf{B}\left(  s,v_{2}\right)  $ for every $s\in \CC G$,
$\lambda_{1}\in \CC$, $\lambda_{2}\in \CC$, $v_{1}\in
\operatorname{Infl}_{H/K}^{H}V$ and $v_{2}\in\operatorname{Infl}_{H/K}^{H}V$.
In other words, the map $\mathbf{B}$ is $\CC$-linear in its second
argument, qed.}. Thus, the map $\mathbf{B}$ is $\CC$-bilinear (since it
is $\CC$-linear in its first argument and is $\CC$-linear in its
second argument). Furthermore, the map $\mathbf{B}$ is $\CC H$-bilinear
with respect to the right $\CC H$-module structure on $\CC G$ and
the left $\CC H$-module structure on $\operatorname{Infl}_{H/K}^{H}%
V$\ \ \ \ \footnote{\textit{Proof.} Every $y\in \CC H$, $s\in \CC G$
and $v\in V$ satisfy%
\begin{align*}
\mathbf{B}\left(  sy,v\right)   &  =\underbrace{\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  sy\right)  }_{\substack{=\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)  \cdot\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  y\right)  \\\text{(since
} \CC \left[  \pi_{G}\right]  \text{ is a}\\ \CC \text{-algebra
homomorphism)}}}\otimes_{ \CC \left[  H/K\right]  }%
v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{B}\left(
sy,v\right)  \right) \\
&  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)
\cdot\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
y\right)  }_{\substack{=\left(  \CC \left[  \pi_{H}\right]  \right)
\left(  y\right)  \\\text{(by (\ref{sol.ind-infl.CGCH}))}}}\otimes
_{ \CC \left[  H/K\right]  }v\\
&  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)
\cdot\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  y\right)
\otimes_{ \CC \left[  H/K\right]  }v=\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  s\right)  \otimes_{ \CC \left[  H/K\right]
}\underbrace{\left(  \CC \left[  \pi_{H}\right]  \right)  \left(
y\right)  \cdot v}_{\substack{=y\cdot v\\\text{(because
(\ref{sol.ind-infl.infl2a}) (applied to }x=y\text{)}\\\text{yields }y\cdot
v=\left(  \CC \left[  \pi_{H}\right]  \right)  \left(  y\right)  \cdot
v\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because we can move }\left(  \CC \left[  \pi_{H}\right]  \right)
\left(  y\right)  \text{ past the }\otimes_{ \CC \left[  H/K\right]
}\text{ sign}\\
\text{(since }\left(  \CC \left[  \pi_{H}\right]  \right)  \left(
y\right)  \in \CC \left[  H/K\right]  \text{)}%
\end{array}
\right) \\
&  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)
\otimes_{ \CC \left[  H/K\right]  }\underbrace{y\cdot v}_{=yv}=\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes
_{ \CC \left[  H/K\right]  }yv\\
&  =\mathbf{B}\left(  s,yv\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{B}\left(  s,yv\right)  =\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  s\right)  \otimes_{ \CC \left[  H/K\right]  }yv\text{
(by the definition of }\mathbf{B}\left(  s,yv\right)  \text{)}\right)  .
\end{align*}
In other words, the map $\mathbf{B}$ is $\CC \left[  H/K\right]
$-bilinear with respect to the right $\CC \left[  H/K\right]  $-module
structure on $\CC \left[  G/K\right]  $ and the left $\CC \left[
H/K\right]  $-module structure on $V$ (since we already know that $\mathbf{B}$
is $\CC$-bilinear), qed.}. Hence, by the universal property of the
tensor product, we conclude that there exists a unique $\CC$-linear map
$\beta: \CC G\otimes_{\CC H}\operatorname{Infl}_{H/K}%
^{H}V\rightarrow\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$
such that every $\left(  s,v\right)  \in \CC G\times\operatorname{Infl}%
_{H/K}^{H}V$ satisfies%
\begin{equation}
\beta\left(  s\otimes_{\CC H}v\right)  =\mathbf{B}\left(  s,v\right)  .
\label{sol.ind-infl.beta}%
\end{equation}
Consider this map $\beta$. This map $\beta$ is a $\CC$-linear map from
$\CC G\otimes_{\CC H}\operatorname{Infl}_{H/K}^{H}V$ to
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$. In other
words, the map $\beta$ is a $\CC$-linear map from $\operatorname{Ind}%
_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ to $\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V$ (since $\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V= \CC G\otimes_{\CC H}
\operatorname{Infl}_{H/K}^{H}V$ as $\CC$-vector spaces).

\underline{\textit{Step 6: Showing that }$\beta$ \textit{is a }%
$\CC G$\textit{-module homomorphism.}}

It is easy to see that the map $\beta:\operatorname{Ind}_{H}^{G}%
\operatorname{Infl}_{H/K}^{H}V\rightarrow\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V$ is a $\CC G$-module
homomorphism.\footnote{\textit{Proof.} Let $g\in G$. We shall first show that
\begin{equation}
\beta\left(  g\left(  s\otimes_{\CC H}v\right)  \right)  =g\beta\left(
s\otimes_{\CC H}v\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }%
s\in \CC G\text{ and }v\in\operatorname{Infl}_{H/K}^{H}V.
\label{sol.ind-infl.beta.CGhom.1}%
\end{equation}
\par
\textit{Proof of (\ref{sol.ind-infl.beta}):} Let $s\in \CC G$ and
$v\in\operatorname{Infl}_{H/K}^{H}V$. By the definition of the
$\CC G$-module structure on $\CC G\otimes_{\CC H}\operatorname{Infl}%
_{H/K}^{H}V$, we have $g\left(  s\otimes_{\CC H}v\right)  =gs\otimes
_{\CC H}v$. Applying the map $\beta$ to both sides of this equality, we
obtain%
\begin{align}
\beta\left(  g\left(  s\otimes_{\CC H}v\right)  \right)   &
=\beta\left(  gs\otimes_{\CC H}v\right)  =\mathbf{B}\left(  gs,v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.ind-infl.beta}), applied to
}gs\text{ instead of }s\right) \nonumber\\
&  =\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)  \left(
gs\right)  }_{\substack{=\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  g\right)  \cdot\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  s\right)  \\\text{(since } \CC \left[  \pi_{G}\right]  \text{ is
a } \CC \text{-algebra}\\\text{homomorphism)}}}\otimes_{ \CC \left[
H/K\right]  }v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{B}\left(  gs,v\right)  \right) \nonumber\\
&  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  g\right)
\cdot\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)
\otimes_{ \CC \left[  H/K\right]  }v.
\label{sol.ind-infl.beta.CGhom.1.pf.1}%
\end{align}
\par
But (\ref{sol.ind-infl.beta}) yields $\beta\left(  s\otimes_{\CC H}
v\right)  =\mathbf{B}\left(  s,v\right)  =\left(  \CC \left[  \pi
_{G}\right]  \right)  \left(  s\right)  \otimes_{ \CC \left[  H/K\right]
}v$ (by the definition of $\mathbf{B}\left(  s,v\right)  $). Meanwhile, we
have $g\in G\subset \CC G$ and $\beta\left(  s\otimes_{ \CC H}v
\right)  \in\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}%
^{G/K}V=\operatorname{Ind}_{H/K}^{G/K}V$. Hence, (\ref{sol.ind-infl.infl1a})
(applied to $g$ and $\beta\left(  s\otimes_{\CC H}v\right)  $ instead of
$x$ and $w$) yields%
\[
g\cdot\beta\left(  s\otimes_{\CC H}v\right)  =\left(  \CC \left[
\pi_{G}\right]  \right)  \left(  g\right)  \cdot\beta\left(  s\otimes
_{\CC H}v\right)
\]
(where the $\beta\left(  s\otimes_{\CC H}v\right)  $ on the left hand
side is regarded as an element of $\operatorname{Infl}_{G/K}^{G}%
\operatorname{Ind}_{H/K}^{G/K}V$, whereas the $\beta\left(  s\otimes
_{\CC H}v\right)  $ on the right hand side is regarded as an element of
$\operatorname{Ind}_{H/K}^{G/K}V$). Thus,%
\begin{align*}
g\beta\left(  s\otimes_{\CC H}v\right)   &  =g\cdot\beta\left(
s\otimes_{\CC H}v\right)  =\left(  \CC \left[  \pi_{G}\right]
\right)  \left(  g\right)  \cdot\underbrace{\beta\left(  s\otimes
_{\CC H}v\right)  }_{=\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  s\right)  \otimes_{ \CC \left[  H/K\right]  }v}\\
&  =\left(  \CC \left[  \pi_{G}\right]  \right)  \left(  g\right)
\cdot\left(  \left(  \CC \left[  \pi_{G}\right]  \right)  \left(
s\right)  \otimes_{ \CC \left[  H/K\right]  }v\right)  =\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  g\right)  \cdot\left(
 \CC \left[  \pi_{G}\right]  \right)  \left(  s\right)  \otimes
_{ \CC \left[  H/K\right]  }v
\end{align*}
(by the definition of the left $\CC \left[  G/K\right]  $-module
structure on $\CC \left[  G/K\right]  \otimes_{ \CC \left[
H/K\right]  }V$). Compared with (\ref{sol.ind-infl.beta.CGhom.1.pf.1}), this
yields $\beta\left(  g\left(  s\otimes_{\CC H}v\right)  \right)
=g\beta\left(  s\otimes_{\CC H}v\right)  $. This proves
(\ref{sol.ind-infl.beta.CGhom.1}).
\par
Now, let $p\in\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ be
arbitrary. We will show that $\beta\left(  gp\right)  =g\beta\left(  p\right)
$.
\par
Indeed, $p\in\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}%
^{H}V= \CC G\otimes_{\CC H}\Inf_{H/K}^{H}V$.
Hence, $p$ is a tensor in $\CC G\otimes_{\CC H}\Inf_{H/K}^{H}V$.
Thus, $p$ can be written in the
form $p=\sum_{i=1}^{N}\lambda_{i}s_{i}\otimes_{\CC H}v_{i}$ for some
$N\in \NN$, some elements $\lambda_{1}$, $\lambda_{2}$, $...$,
$\lambda_{N}$ of $\CC$, some elements $s_{1}$, $s_{2}$, $...$, $s_{N}$
of $\CC G$, and some elements $v_{1}$, $v_{2}$, $...$, $v_{N}$ of
$\Inf_{H/K}^{H}V$. Consider this $N$, these
$\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{N}$, these $s_{1}$, $s_{2}$,
$...$, $s_{N}$, and these $v_{1}$, $v_{2}$, $...$, $v_{N}$.
\par
Applying the map $\beta$ to both sides of the equality $p=\sum_{i=1}%
^{N}\lambda_{i}s_{i}\otimes_{\CC H}v_{i}$, we obtain%
\begin{equation}
\beta\left(  p\right)  =\beta\left(  \sum_{i=1}^{N}\lambda_{i}s_{i}%
\otimes_{\CC H}v_{i}\right)  =\sum_{i=1}^{N}\lambda_{i}\beta\left(
s_{i}\otimes_{\CC H}v_{i}\right)  \label{sol.ind-infl.beta.CGhom.2}%
\end{equation}
(since the map $\beta$ is $\CC$-linear).
\par
We have%
\begin{align*}
g\underbrace{p}_{=\sum_{i=1}^{N}\lambda_{i}s_{i}\otimes_{\CC H}v_{i}}
&  =g\left(  \sum_{i=1}^{N}\lambda_{i}s_{i}\otimes_{\CC H}v_{i}\right)
=\sum_{i=1}^{N}\underbrace{g\left(  \lambda_{i}s_{i}\right)  }_{=\lambda
_{i}gs_{i}}\otimes_{\CC H}v_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\CC G\text{-module structure on } \CC G\otimes_{\CC H}
\Inf_{H/K}^{H}V\right) \\
&  =\sum_{i=1}^{N}\lambda_{i}gs_{i}\otimes_{\CC H}v_{i}.
\end{align*}
Applying the map $\beta$ to both sides of this equality, we obtain%
\begin{align*}
\beta\left(  gp\right)   &  =\beta\left(  \sum_{i=1}^{N}\lambda_{i}%
gs_{i}\otimes_{\CC H}v_{i}\right)  =\sum_{i=1}^{N}\lambda_{i}%
\underbrace{\beta\left(  gs_{i}\otimes_{\CC H}v_{i}\right)
}_{\substack{=g\beta\left(  s_{i}\otimes_{\CC H}v_{i}\right)
\\\text{(by (\ref{sol.ind-infl.beta.CGhom.1}), applied}\\\text{to }s_{i}\text{
and }v_{i}\text{ instead of }s\text{ and }v\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\beta\text{ is }%
\CC \text{-linear}\right) \\
&  =\sum_{i=1}^{N}\lambda_{i}g\beta\left(  s_{i}\otimes_{\CC H}%
v_{i}\right)  =g\underbrace{\sum_{i=1}^{N}\lambda_{i}\beta\left(  s_{i}%
\otimes_{\CC H}v_{i}\right)  }_{\substack{=\beta\left(  p\right)
\\\text{(by (\ref{sol.ind-infl.beta.CGhom.2}))}}}=g\beta\left(  p\right)  .
\end{align*}
\par
Now, forget that we fixed $p$ and $g$. We thus have proven that $\beta\left(
gp\right)  =g\beta\left(  p\right)  $ for every $g\in G$ and $p\in
\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$. In other words,
$\beta$ is a homomorphism of left $G$-sets. Combined with the fact that
$\beta$ is a $\CC$-linear map, this yields that $\beta$ is a
$\CC G$-module homomorphism, qed.}

\underline{\textit{Step 7: Showing that }$\alpha\circ\beta= \id %
$\textit{.}}

We next will show that $\alpha\circ\beta
=\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V}$. Before
we do this, we will prove some auxiliary results.

First, we notice that
\begin{equation}
\alpha\left(  \beta\left(  g\otimes_{\CC H}v\right)  \right)
=g\otimes_{\CC H}v\ \ \ \ \ \ \ \ \ \ \text{for all }g\in G\text{ and
}v\in\operatorname{Infl}_{H/K}^{H}V.\label{sol.ind-infl.alphabeta.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.alphabeta.1}):} Let $g\in G$ and
$v\in\operatorname{Infl}_{H/K}^{H}V$. Since $g\in G\subset \CC G$ and
$v\in\operatorname{Infl}_{H/K}^{H}V$, we have $\left(  g,v\right)
\in \CC G\times\operatorname{Infl}_{H/K}^{H}V$. Thus,
(\ref{sol.ind-infl.beta}) (applied to $s=g$) shows that%
\begin{align*}
\beta\left(  g\otimes_{\CC H}v\right)    & =\mathbf{B}\left(
g,v\right)  =\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  g\right)  }_{\substack{=\pi_{G}\left(  g\right)  \\\text{(by
(\ref{sol.ind-infl.s}), applied to}\\ \CC \text{, }G\text{, }G/K\text{,
}\pi_{G}\text{ and }g\text{ instead of}\\ \kk \text{, }S\text{, }T\text{,
}\rho\text{ and }s\text{)}}}\otimes_{ \CC \left[  H/K\right]
}v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{B}\left(
g,v\right)  \right)  \\
& =\pi_{G}\left(  g\right)  \otimes_{ \CC \left[  H/K\right]  }v.
\end{align*}
Applying the map $\alpha$ to both sides of this equality, we obtain
$\alpha\left(  \beta\left(  g\otimes_{\CC H}v\right)  \right)
=\alpha\left(  \pi_{G}\left(  g\right)  \otimes_{ \CC \left[  H/K\right]
}v\right)  $.
\par
But since $\pi_{G}\left(  g\right)  \in G/K\subset \CC \left[
G/K\right]  $ and $v\in V$, we have $\left(  \pi_{G}\left(  g\right)
,v\right)  \in \CC \left[  G/K\right]  \times V$. Hence,
(\ref{sol.ind-infl.alpha}) (applied to $\pi_{G}\left(  g\right)  $ instead of
$s$) yields $\alpha\left(  \pi_{G}\left(  g\right)  \otimes_{ \CC \left[
H/K\right]  }v\right)  =\mathbf{A}\left(  \pi_{G}\left(  g\right)  ,v\right)
$. Altogether,%
\begin{align*}
\alpha\left(  \beta\left(  g\otimes_{\CC H}v\right)  \right)    &
=\alpha\left(  \pi_{G}\left(  g\right)  \otimes_{ \CC \left[  H/K\right]
}v\right)  =\mathbf{A}\left(  \pi_{G}\left(  g\right)  ,v\right)  =\left(
\underbrace{\widetilde{\mathbf{a}}\left(  \pi_{G}\left(  g\right)  \right)
}_{\substack{=\mathbf{a}_{g}\\\text{(by (\ref{sol.ind-infl.a.tilde-x}%
),}\\\text{applied to }x=g\text{)}}}\right)  \left(  v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.ind-infl.a.A}), applied to
}s=\pi_{G}\left(  g\right)  \right)  \\
& =\mathbf{a}_{g}\left(  v\right)  =g\otimes_{\CC H}%
v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{a}_{g}\left(
v\right)  \right)  .
\end{align*}
This proves (\ref{sol.ind-infl.alphabeta.1}).} Next, we can see that
\begin{equation}
\left(  \alpha\circ\beta\right)  \left(  s\otimes_{\CC H}v\right)
=\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}%
_{H/K}^{H}V}\left(  s\otimes_{\CC H}v\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }s\in \CC G\text{ and }v\in
\operatorname{Infl}_{H/K}^{H}V.\label{sol.ind-infl.alphabeta.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.alphabeta.2}):} Let
$s\in \CC G$ and $v\in\operatorname{Infl}_{H/K}^{H}V$. We have
$s\in \CC G$. Thus, $s$ is a $\CC$-linear combination of elements
of $G$ (since $G$ is a basis of the $\CC$-vector space $\CC G$).
In other words, there exists a family $\left(  s_{g}\right)  _{g\in G}%
\in \CC^{G}$ of elements of $\CC$ satisfying $s=\sum_{g\in
G}s_{g}g$. Consider this family $\left(  s_{g}\right)  _{g\in G}$. We have%
\[
\underbrace{s}_{=\sum_{g\in G}s_{g}g}\otimes_{\CC H}v=\left(  \sum_{g\in
G}s_{g}g\right)  \otimes_{\CC H}v=\sum_{g\in G}s_{g}g\otimes
_{\CC H}v.
\]
Applying the map $\beta$ to both sides of this equality, we obtain%
\[
\beta\left(  s\otimes_{\CC H}v\right)  =\beta\left(  \sum_{g\in G}%
s_{g}g\otimes_{\CC H}v\right)  =\sum_{g\in G}s_{g}\beta\left(
g\otimes_{\CC H}v\right)
\]
(since the map $\beta$ is $\CC$-linear). Applying the map $\alpha$ to
both sides of this equality, we obtain%
\begin{align*}
\alpha\left(  \beta\left(  s\otimes_{\CC H}v\right)  \right)    &
=\alpha\left(  \sum_{g\in G}s_{g}\beta\left(  g\otimes_{\CC H}v\right)
\right)  =\sum_{g\in G}s_{g}\underbrace{\alpha\left(  \beta\left(
g\otimes_{\CC H}v\right)  \right)  }_{\substack{=g\otimes_{\CC H}v
\\\text{(by (\ref{sol.ind-infl.alphabeta.1}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\alpha\text{ is }%
\CC \text{-linear}\right)  \\
& =\sum_{g\in G}s_{g}g\otimes_{\CC H}v=s\otimes_{\CC H}v
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }s\otimes_{\CC H}v=\sum_{g\in
G}s_{g}g\otimes_{\CC H}v\right)  \\
& =\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}%
_{H/K}^{H}V}\left(  s\otimes_{\CC H}v\right)  .
\end{align*}
Now,%
\[
\left(  \alpha\circ\beta\right)  \left(  s\otimes_{\CC H}v\right)
=\alpha\left(  \beta\left(  s\otimes_{\CC H}v\right)  \right)
=\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}%
_{H/K}^{H}V}\left(  s\otimes_{\CC H}v\right)  .
\]
This proves (\ref{sol.ind-infl.alphabeta.2}).} In other words, the two maps
$\alpha\circ\beta:\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}%
^{H}V\rightarrow\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ and
$\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}%
_{H/K}^{H}V}:\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}%
^{H}V\rightarrow\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ are
equal to each other on each pure tensor. Since these two maps are $\CC %
$-linear\footnote{In fact, the map $\alpha\circ\beta$ is $\CC$-linear
(since $\alpha$ and $\beta$ are $\CC$-linear), and the map
$\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}%
_{H/K}^{H}V}$ is $\CC$-linear.}, this yields that these two maps
$\alpha\circ\beta:\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}%
^{H}V\rightarrow\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ and
$\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}%
_{H/K}^{H}V}:\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}%
^{H}V\rightarrow\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ must
be identical\footnote{This is because a well-known fact from linear algebra
states the following: If two $\CC$-linear maps from a tensor product
are equal to each other on each pure tensor, then these two maps must be
identical.}. In other words, $\alpha\circ\beta
=\id_{\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V}$.

\underline{\textit{Step 8: Showing that }$\beta\circ\alpha= \id %
$\textit{.}}

Next, we will prove that $\beta\circ\alpha
=\id_{\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V}$.
Before we do this, we will prove some auxiliary results.

First, we notice that
\begin{equation}
\beta\left(  \alpha\left(  s\otimes_{ \CC \left[  H/K\right]  }v\right)
\right)  =s\otimes_{ \CC \left[  H/K\right]  }%
v\ \ \ \ \ \ \ \ \ \ \text{for all }s\in G/K\text{ and }v\in
V.\label{sol.ind-infl.betaalpha.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.betaalpha.1}):} Let $s\in G/K$
and $v\in V$. Since $s\in G/K\subset \CC \left[  G/K\right]  $ and $v\in
V$, we have $\left(  s,v\right)  \in \CC \left[  G/K\right]  \times V$.
Thus, (\ref{sol.ind-infl.alpha}) yields $\alpha\left(  s\otimes_{ \CC %
\left[  H/K\right]  }v\right)  =\mathbf{A}\left(  s,v\right)  =\left(
\widetilde{\mathbf{a}}\left(  s\right)  \right)  \left(  v\right)  $ (by
(\ref{sol.ind-infl.a.A})).
\par
But $s\in G/K=\pi_{G}\left(  G\right)  $ (since $\pi_{G}\left(  G\right)
=G/K$). Hence, there exists a $g\in G$ satisfying $s=\pi_{G}\left(  g\right)
$. Consider this $g$. Since $g\in G\subset \CC G$ and $v\in
V=\operatorname{Infl}_{H/K}^{H}V$, we have $\left(  g,v\right)  \in
\CC G\times\operatorname{Infl}_{H/K}^{H}V$. Thus,
(\ref{sol.ind-infl.beta}) (applied to $g$ instead of) shows that%
\begin{align*}
\beta\left(  g\otimes_{\CC H}v\right)    & =\mathbf{B}\left(
g,v\right)  =\underbrace{\left(  \CC \left[  \pi_{G}\right]  \right)
\left(  g\right)  }_{\substack{=\pi_{G}\left(  g\right)  \\\text{(by
(\ref{sol.ind-infl.s}), applied to}\\ \CC \text{, }G\text{, }G/K\text{,
}\pi_{G}\text{ and }g\text{ instead of}\\ \kk \text{, }S\text{, }T\text{,
}\rho\text{ and }s\text{)}}}\otimes_{ \CC \left[  H/K\right]
}v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{B}\left(
g,v\right)  \right)  \\
& =\underbrace{\pi_{G}\left(  g\right)  }_{=s}\otimes_{ \CC \left[
H/K\right]  }v=s\otimes_{ \CC \left[  H/K\right]  }v.
\end{align*}
\par
But $\widetilde{\mathbf{a}}\left(  \underbrace{s}_{=\pi_{G}\left(  g\right)
}\right)  =\widetilde{\mathbf{a}}\left(  \pi_{G}\left(  g\right)  \right)
=\mathbf{a}_{g}$ (by (\ref{sol.ind-infl.a.tilde-x}), applied to $x=g$). Now,%
\[
\alpha\left(  s\otimes_{ \CC \left[  H/K\right]  }v\right)  =\left(
\underbrace{\widetilde{\mathbf{a}}\left(  s\right)  }_{=\mathbf{a}_{g}%
}\right)  \left(  v\right)  =\mathbf{a}_{g}\left(  v\right)  =g\otimes
_{\CC H}v
\]
(by the definition of $\mathbf{a}_{g}\left(  v\right)  $). Applying the map
$\beta$ to both sides of this equality, we obtain $\beta\left(  \alpha\left(
s\otimes_{ \CC \left[  H/K\right]  }v\right)  \right)  =\beta\left(
g\otimes_{\CC H}v\right)  =s\otimes_{ \CC \left[  H/K\right]  }v$.
This proves (\ref{sol.ind-infl.betaalpha.1}).} Next, we can see that
\begin{equation}
\left(  \beta\circ\alpha\right)  \left(  t\otimes_{ \CC \left[
H/K\right]  }v\right)  =\id_{\operatorname{Infl}_{G/K}^{G}
\operatorname{Ind}_{H/K}^{G/K}V}\left(  t\otimes_{ \CC \left[
H/K\right]  }v\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }t\in \CC
\left[  G/K\right]  \text{ and }v\in V.\label{sol.ind-infl.betaalpha.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.ind-infl.betaalpha.2}):} Let
$t\in \CC \left[  G/K\right]  $ and $v\in V$. We have $t\in
\CC \left[  G/K\right]  $. Thus, $t$ is a $\CC$-linear
combination of elements of $G/K$ (since $G/K$ is a basis of the
$\CC$-vector space $\CC \left[  G/K\right]  $). In other words, there exists
a family $\left(  t_{s}\right)  _{s\in G/K}\in \CC^{G/K}$ of elements of
$\CC$ satisfying $t=\sum_{s\in G/K}t_{s}s$. Consider this family
$\left(  t_{s}\right)  _{s\in G/K}$. We have%
\[
\underbrace{t}_{=\sum_{s\in G/K}t_{s}s}\otimes_{ \CC \left[  H/K\right]
}v=\left(  \sum_{s\in G/K}t_{s}s\right)  \otimes_{ \CC \left[
H/K\right]  }v=\sum_{s\in G/K}t_{s}s\otimes_{ \CC \left[  H/K\right]  }v.
\]
Applying the map $\beta$ to both sides of this equality, we obtain%
\[
\alpha\left(  t\otimes_{ \CC \left[  H/K\right]  }v\right)
=\alpha\left(  \sum_{s\in G/K}t_{s}s\otimes_{ \CC \left[  H/K\right]
}v\right)  =\sum_{s\in G/K}t_{s}\alpha\left(  s\otimes_{ \CC \left[
H/K\right]  }v\right)
\]
(since the map $\alpha$ is $\CC$-linear). Applying the map $\beta$ to
both sides of this equality, we obtain%
\begin{align*}
\beta\left(  \alpha\left(  t\otimes_{ \CC \left[  H/K\right]  }v\right)
\right)    & =\beta\left(  \sum_{s\in G/K}t_{s}\alpha\left(  s\otimes
_{ \CC \left[  H/K\right]  }v\right)  \right)  =\sum_{s\in G/K}%
t_{s}\underbrace{\beta\left(  \alpha\left(  s\otimes_{ \CC \left[
H/K\right]  }v\right)  \right)  }_{\substack{=s\otimes_{ \CC \left[
H/K\right]  }v\\\text{(by (\ref{sol.ind-infl.betaalpha.1}))}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\beta\text{ is }%
\CC \text{-linear}\right)  \\
& =\sum_{s\in G/K}t_{s}s\otimes_{ \CC \left[  H/K\right]  }%
v=t\otimes_{ \CC \left[  H/K\right]  }v\ \ \ \ \ \ \ \ \ \ \left(
\text{since }t\otimes_{ \CC \left[  H/K\right]  }v=\sum_{s\in G/K}%
t_{s}s\otimes_{ \CC \left[  H/K\right]  }v\right)  \\
& =\id_{\operatorname{Infl}_{G/K}^{G}%
\operatorname{Ind}_{H/K}^{G/K}V}\left(  t\otimes_{ \CC \left[
H/K\right]  }v\right)  .
\end{align*}
Now,%
\[
\left(  \beta\circ\alpha\right)  \left(  t\otimes_{ \CC \left[
H/K\right]  }v\right)  =\beta\left(  \alpha\left(  t\otimes_{ \CC \left[
H/K\right]  }v\right)  \right)
=\id_{\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}%
V}\left(  t\otimes_{ \CC \left[  H/K\right]  }v\right)  .
\]
This proves (\ref{sol.ind-infl.betaalpha.2}).} In other words, the two maps
$\beta\circ\alpha:\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}%
^{G/K}V\rightarrow\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}%
^{G/K}V$ and $\id_{\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V}:\operatorname{Infl}_{G/K}^{G}%
\operatorname{Ind}_{H/K}^{G/K}V\rightarrow\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V$ are equal to each other on each pure
tensor. Since these two maps are $\CC$-linear\footnote{In fact, the map
$\beta\circ\alpha$ is $\CC$-linear (since $\beta$ and $\alpha$ are
$\CC$-linear), and the map
$\id_{\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V}$ is
$\CC$-linear.}, this yields that these two maps $\beta\circ
\alpha:\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}%
V\rightarrow\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$ and
$\id_{\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}%
_{H/K}^{G/K}V}:\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}%
^{G/K}V\rightarrow\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}%
^{G/K}V$ must be identical\footnote{This is because a well-known fact from
linear algebra states the following: If two $\CC$-linear maps from a
tensor product are equal to each other on each pure tensor, then these two
maps must be identical.}. In other words, $\beta\circ\alpha
=\id_{\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V}$.

\underline{\textit{Step 9: Finishing the proof.}}

Now, the maps $\alpha$ and $\beta$ are mutually inverse (since $\alpha
\circ\beta=\id_{\operatorname{Ind}_{H}^{G}%
\operatorname{Infl}_{H/K}^{H}V}$ and $\beta\circ\alpha
=\id_{\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V}$). As
a consequence, the map $\beta$ is invertible. Since $\beta$ is a
$\CC G$-module homomorphism from
$\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$ to
$\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K} V$,
we thus have shown that $\beta$ is an invertible $\CC G$-module
homomorphism from $\operatorname{Ind}_{H}^{G}\operatorname{Infl}_{H/K}^{H}V$
to $\operatorname{Infl}_{G/K}^{G}\operatorname{Ind}_{H/K}^{G/K}V$. Hence,
$\beta$ is a $\CC G$-module isomorphism from $\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V$ to $\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V$ (since every invertible
$\CC G$-module homomorphism is a $\CC G$-module isomorphism). Thus, there
exists a $\CC G$-module isomorphism from $\operatorname{Ind}_{H}%
^{G}\operatorname{Infl}_{H/K}^{H}V$ to $\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V$. Hence, $\operatorname{Infl}_{G/K}%
^{G}\operatorname{Ind}_{H/K}^{G/K}V\cong\operatorname{Ind}_{H}^{G}%
\operatorname{Infl}_{H/K}^{H}V$ as $\CC G$-modules. This solves
Exercise \ref{exe.ind-infl}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.fixed-vs-coinvariants}}

\textit{Solution to Exercise~\ref{exe.fixed-vs-coinvariants}.}
(a) It is clearly enough to show that $g\left(v-kv\right) \in
I_{V,K}$ for all $g \in G$, $k \in K$ and $v \in V$. But if
$g \in G$, $k \in K$ and $v \in V$, then $gk$ has the form
$gk = k'g$ for some $k' \in K$ (since $K \triangleleft G$), and
thus we have $g\left(v-kv\right) = gv - \underbrace{gk}_{=k'g}v
= gv - k'gv \in I_{V,K}$ (by the definition of $I_{V,K}$, since
$k' \in K$ and $gv \in V$). Hence, part (a) of the exercise is
solved.

(b) For every $v \in V$, let $\overline{v}$ denote the projection of
$v$ onto the quotient space $V / I_{V,K} = V_K$.

We can define a map $\Phi : V^K \to V_K$
by sending every
$v \in V^K$ to $\overline{v} \in V_K$. This map $\Phi$ is easily seen
to be a $\CC G$-module homomorphism
$\Inf^G_{G/K} \left( V^K \right) \to V_K$. If we can show that $\Phi$
is also bijective, then it will follow that $\Phi$ is a $\CC G$-module
isomorphism
$\Inf^G_{G/K} \left( V^K \right) \to V_K$, whence part (b) of the
exercise will be solved. Hence, all that remains to be done is proving
that $\Phi$ is bijective.

Let us construct an inverse map to $\Phi$. First, let us notice that every
$v\in V$ satisfies
\[
\frac{1}{\left\vert K\right\vert }\sum_{j\in K}jv\in V^{K}.
\]
\footnote{\textit{Proof.} Let $v\in V$. Let $j\in K$. Then, the map
$K\rightarrow K,\ s\mapsto js$ is a bijection (since $j\in K$ and since $K$ is
a group). Now,
\[
j\cdot\left(  \frac{1}{\left\vert K\right\vert }\sum_{s\in K}sv\right)
=\frac{1}{\left\vert K\right\vert }\sum_{s\in K}jsv=\frac{1}{\left\vert
K\right\vert }\sum_{s\in K}sv
\]
(here, we have substituted $s$ for $js$ in the sum, because the map
$K\rightarrow K,\ s\mapsto js$ is a bijection). Now, forget that we fixed $j$.
We thus have shown that $j\cdot\left(  \frac{1}{\left\vert K\right\vert }
\sum_{s\in K}sv\right)  =\frac{1}{\left\vert K\right\vert }\sum_{s\in K}sv$
for every $j\in K$. Hence, $\frac{1}{\left\vert K\right\vert }\sum_{s\in
K}sv\in V^{K}$, qed.} Thus, we can define a map $\psi:V\rightarrow V^{K}$ by
sending every $v\in V$ to $\frac{1}{\left\vert K\right\vert }\sum_{j\in K}jv$.
This map $\psi$ is $\CC$-linear and vanishes on $I_{V,K}$
\ \ \ \ \footnote{\textit{Proof.} We want to show that $\psi$ vanishes on
$I_{V,K}$. In order to do so, we only need to check that $\psi\left(
v-kv\right)  =0$ for all $k\in K$ and $v\in V$ (since $I_{V,K}$ is spanned by
all $v-kv$ with $k\in K$ and $v\in V$). But this follows from the fact that
all $k\in K$ and $v\in V$ satisfy
\begin{align*}
\psi\left(  v-kv\right)    & =\frac{1}{\left\vert K\right\vert }\sum_{j\in
K}\underbrace{j\left(  v-kv\right)  }_{=jv-jkv}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\psi\right)  \\
& =\frac{1}{\left\vert K\right\vert }\sum_{j\in K}\left(  jv-jkv\right)
=\frac{1}{\left\vert K\right\vert }\sum_{j\in K}jv-\frac{1}{\left\vert
K\right\vert }\sum_{j\in K}jkv=\frac{1}{\left\vert K\right\vert }\sum_{j\in
K}jv-\frac{1}{\left\vert K\right\vert }\sum_{j\in K}jv\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}
\text{here, we have substituted }j\text{ for }jk\text{ in the second sum, since the map }K\rightarrow K,\ j\mapsto jk \\
\text{ is a bijection (because }k\in K\text{ and because }K\text{ is a group)}
\end{array}
\right)  \\
& =0.
\end{align*}
}. Hence, the map $\psi$ factors through the quotient $V/I_{V,K}=V_{K}$. Let
us denote the resulting $\CC$-linear map $V_{K}\rightarrow V^{K}$ by
$\Psi$. We are now going to show that the maps $\Phi$ and $\Psi$ are mutually inverse.

We have $\Phi\circ\Psi=\id$\ \ \ \ \footnote{\textit{Proof.}
Let $w\in V_{K}$. Then, there exists some $v\in V$ such that $w=\overline{v}$.
Consider this $v$.
\par
Since $\Psi$ was defined as a quotient of the map $\psi$, we have $\Psi\left(
\overline{v}\right)  =\psi\left(  v\right)  $. Now,
\begin{align*}
\left(  \Phi\circ\Psi\right)  \left(  \underbrace{w}_{=\overline{v}}\right)
& =\left(  \Phi\circ\Psi\right)  \left(  \overline{v}\right)  =\Phi\left(
\underbrace{\Psi\left(  \overline{v}\right)  }_{=\psi\left(  v\right)
}\right)  =\Phi\left(  \underbrace{\psi\left(  v\right)  }_{\substack{=\frac
{1}{\left\vert K\right\vert }\sum_{j\in K}jv\\\text{(by the definition of
}\psi\text{)}}}\right)  =\Phi\left(  \frac{1}{\left\vert K\right\vert }
\sum_{j\in K}jv\right)  \\
& =\overline{\frac{1}{\left\vert K\right\vert }\sum_{j\in K}jv}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\right)  \\
& =\frac{1}{\left\vert K\right\vert }\sum_{j\in K}
 \underbrace{\overline{jv}}_
  {\substack{=\overline{v}\\\text{(since }v-jv\in I_{V,K}\\\text{(because }
  j\in K\text{))}}}
= \frac{1}{\left\vert K\right\vert }\underbrace{\sum_{j\in
K}\overline{v}}_{=\left\vert K\right\vert \cdot\overline{v}}=\dfrac
{1}{\left\vert K\right\vert }\left\vert K\right\vert \cdot\overline
{v}=\overline{v}=w.
\end{align*}
\par
Thus we have shown that $\left(  \Phi\circ\Psi\right)  \left(  w\right)  =w$
for every $w\in V_{K}$. In other words, $\Phi\circ\Psi=\id$,
qed.} and $\Psi\circ\Phi=\id$\ \ \ \ \footnote{\textit{Proof.}
For every $v\in V^{K}$, we have
\begin{align*}
\left(  \Psi\circ\Phi\right)  \left(  v\right)    & =\Psi\left(
\underbrace{\Phi\left(  v\right)  }_{\substack{=\overline{v}\\\text{(by the
definition of }\Phi\text{)}}}\right)  =\Psi\left(  \overline{v}\right)
=\psi\left(  v\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Psi\text{ was defined as the
quotient of the map }\psi\right)  \\
& =\frac{1}{\left\vert K\right\vert }\sum_{j\in K}\underbrace{jv}
_{\substack{=v\\\text{(since }v\in V^{K}\text{ and }j\in K\text{)}}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\psi\right)  \\
& =\frac{1}{\left\vert K\right\vert }\underbrace{\sum_{j\in K}v}_{=\left\vert
K\right\vert \cdot v}=\dfrac{1}{\left\vert K\right\vert }\left\vert
K\right\vert \cdot v=v.
\end{align*}
Thus, $\Psi\circ\Phi=\id$, qed.}. Hence, the maps $\Phi$ and
$\Psi$ are mutually inverse. It follows that $\Phi$ is bijective, and so the
solution is complete.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Indrho}}

\textit{Solution to Exercise \ref{exe.Indrho}.} In the following, we will use
the following convention: Whenever $K$ is a group, and $k$ is an element of
$K$, we shall write $k$ for the element $t_{k}$ of ${\CC }K$. This is a
relatively common abuse of notation, and it is harmless because the map
$K\rightarrow{\CC }K,\ k\mapsto t_{k}$ is an injective homomorphism of
multiplicative monoids (so $t_{gh}=t_{g}t_{h}$ and $t_{1}=1$, which means that
we won't run into ambiguities denoting $t_{k}$ by $k$) and because every
${\CC }K$-module $M$, every $m\in M$ and every $k\in K$ satisfy
$km=t_{k}m$.

Let us first notice a trivial fact: If $K$ is a finite group, and if
$f:K\rightarrow\CC $ is any function, then we have the following
equivalence:%
\begin{align}
&  \left(  f\in R_{\CC }\left(  K\right)  \right) \nonumber\\
\Longleftrightarrow\  &  \left(  f\text{ is a class function on }K\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }R_{\CC }\left(  K\right)  \text{ is defined to be}\\
\text{the set of all class functions on }K
\end{array}
\right) \nonumber\\
\Longleftrightarrow\  &  \left(  f\text{ is constant on }K\text{-conjugacy
classes}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since a class function on }K\text{ is defined to mean a function}\\
K\rightarrow\CC \text{ which is constant on }K\text{-conjugacy classes}%
\end{array}
\right) \nonumber\\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }K\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime
}\right)  \right)  . \label{sol.Indrho.RCK}%
\end{align}


(a) Let $f\in R_{\CC }\left(  H\right)  $.

We can apply (\ref{sol.Indrho.RCK}) to $K=H$. As a consequence, we obtain the
following equivalence:%
\begin{align*}
&  \left(  f\in R_{\CC }\left(  H\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }H\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime
}\right)  \right)  .
\end{align*}
Hence,
\begin{equation}
\text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of }H\text{
satisfy }f\left(  k\right)  =f\left(  k^{\prime}\right)
\label{sol.Indrho.a.1}%
\end{equation}
(because we know that $f\in R_{\CC }\left(  H\right)  $).

Now, let $g$ and $g^{\prime}$ be two conjugate elements of $G$. Then, there
exists a $p\in G$ such that $g^{\prime}=pgp^{-1}$ (since $g$ and $g^{\prime}$
are conjugate). Consider this $p$.

Applying the map $\Ind_{\rho}f$ to both sides of the
equality $g^{\prime}=pgp^{-1}$, we obtain%
\begin{align*}
\left(  \Ind_{\rho}f\right)  \left(  g^{\prime
}\right)   &  =\left(  \Ind_{\rho}f\right)  \left(
pgp^{-1}\right)  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits
_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(  h\right)
k^{-1}=pgp^{-1}}}}_{=\sum_{h\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
h\right)  k^{-1}=pgp^{-1}}}}f\left(  h\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Ind}%
\nolimits_{\rho}f\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\underbrace{\sum\limits
_{\substack{k\in G;\\k\rho\left(  h\right)  k^{-1}=pgp^{-1}}}f\left(
h\right)  }_{\substack{=\sum\limits_{\substack{k\in G;\\pk\rho\left(  h\right)
\left(  pk\right)  ^{-1}=pgp^{-1}}}f\left(  h\right)  \\\text{(here, we have
substituted }pk\text{ for }k\text{ in the sum,}\\\text{since the map
}G\rightarrow G,\ k\mapsto pk\text{ is a bijection)}}}\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\underbrace{\sum\limits
_{\substack{k\in G;\\pk\rho\left(  h\right)  \left(  pk\right)  ^{-1}%
=pgp^{-1}}}}_{\substack{=\sum\limits_{\substack{k\in G;\\pk\rho\left(  h\right)
k^{-1}p^{-1}=pgp^{-1}}}\\\text{(since }\left(  pk\right)  ^{-1}=k^{-1}%
p^{-1}\text{)}}}f\left(  h\right)  =\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\underbrace{\sum\limits_{\substack{k\in G;\\pk\rho\left(  h\right)
k^{-1}p^{-1}=pgp^{-1}}}}_{\substack{=\sum\limits_{\substack{k\in G;\\pk\rho\left(
h\right)  k^{-1}=pg}}\\\text{(since }pk\rho\left(  h\right)  k^{-1}%
p^{-1}=pgp^{-1}\text{ is}\\\text{equivalent to }pk\rho\left(  h\right)
k^{-1}=pg\text{)}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\underbrace{\sum\limits
_{\substack{k\in G;\\pk\rho\left(  h\right)  k^{-1}=pg}}}_{\substack{=\sum\limits
_{\substack{k\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\\\text{(since }%
pk\rho\left(  h\right)  k^{-1}=pg\text{ is}\\\text{equivalent to }k\rho\left(
h\right)  k^{-1}=g\text{)}}}f\left(  h\right)  =\dfrac{1}{\left\vert
H\right\vert }\underbrace{\sum_{h\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
h\right)  k^{-1}=g}}}_{=\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\k\rho\left(  h\right)  k^{-1}=g}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}f\left(  h\right)  =\left(
\Ind_{\rho}f\right)  \left(  g\right)
\end{align*}
(since the definition of $\Ind_{\rho}f$ yields
$\left(  \Ind_{\rho}f\right)  \left(  g\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}f\left(  h\right)  $).

Let us now forget that we fixed $g$ and $g^{\prime}$. We thus have shown that
any two conjugate elements $g$ and $g^{\prime}$ of $G$ satisfy $\left(
\Ind_{\rho}f\right)  \left(  g\right)  =\left(
\Ind_{\rho}f\right)  \left(  g^{\prime}\right)  $.
Renaming $g$ and $g^{\prime}$ as $k$ and $k^{\prime}$ in this statement, we
obtain the following: Any two conjugate elements $k$ and $k^{\prime}$ of $G$
satisfy $\left(  \Ind_{\rho}f\right)  \left(
k\right)  =\left(  \Ind_{\rho}f\right)  \left(
k^{\prime}\right)  $.

But (\ref{sol.Indrho.RCK}) (applied to $G$ and $\operatorname*{Ind}%
\nolimits_{\rho}f$ instead of $K$ and $f$) yields the following equivalence:%
\begin{align*}
&  \left(  \Ind_{\rho}f\in R_{\CC }\left(
G\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }G\text{ satisfy }\left(  \operatorname*{Ind}%
\nolimits_{\rho}f\right)  \left(  k\right)  =\left(  \operatorname*{Ind}%
\nolimits_{\rho}f\right)  \left(  k^{\prime}\right)  \right)  .
\end{align*}
Thus, $\Ind_{\rho}f\in R_{\CC }\left(  G\right)
$ (because we know that any two conjugate elements $k$ and $k^{\prime}$ of $G$
satisfy $\left(  \Ind_{\rho}f\right)  \left(
k\right)  =\left(  \Ind_{\rho}f\right)  \left(
k^{\prime}\right)  $). This solves Exercise \ref{exe.Indrho}(a).

(b) Let us first introduce an elementary (but apocryphal) notion from linear
algebra: the notion of \emph{finite dual generating systems}.

\begin{definition}
Let $\mathbb{K}$ be a commutative ring. Let $V$ be a $\mathbb{K}$-module. A
\dfn{finite dual generating system} for $V$ means a triple $\left(
I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $, where

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}f_{i}\left(
v\right)  a_{i}$.
\end{definition}

In the following, we shall only use finite dual generating systems in the case
when $\mathbb{K}$ is a field; nevertheless, they are more useful in the
general case.

The first question one might ask about finite dual generating systems for $V$
is when they exist. The answer is very simple when $\mathbb{K}$ is a field:

\begin{proposition}
\label{prop.sol.Indrho.b.dualgens.findim}Let $\mathbb{K}$ be a field. Let $V$
be a $\mathbb{K}$-vector space. Then, a finite dual generating system for $V$
exists if and only if the vector space $V$ is finite-dimensional.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.b.dualgens.findim}.]Proposition
\ref{prop.sol.Indrho.b.dualgens.findim} is an ``if and only if''
statement. Hence, in order to prove Proposition
\ref{prop.sol.Indrho.b.dualgens.findim}, it is sufficient to verify the
following two claims:

\begin{statement}
\textit{Claim 1:} If a finite dual generating system for $V$ exists, then the
vector space $V$ is finite-dimensional.
\end{statement}

\begin{statement}
\textit{Claim 2:} If the vector space $V$ is finite-dimensional, then a finite
dual generating system for $V$ exists.
\end{statement}

Let us now prove these two claims.

\textit{Proof of Claim 1.} Assume that a finite dual generating system for $V$
exists. Let $\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)
_{i\in I}\right)  $ be such a finite dual generating system for $V$.

We know that $I$ is a finite set, that $\left(  a_{i}\right)  _{i\in I}$ is a
family of elements of $V$, and that every $v\in V$ satisfies $v=\sum
\limits_{i\in I}f_{i}\left(  v\right)  a_{i}$. (Indeed, this is part of what
it means for $\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)
_{i\in I}\right)  $ to be a finite dual generating system for $V$.)

Now, every $v\in V$ satisfies $v=\sum\limits_{i\in I}\underbrace{f_{i}\left(
v\right)  }_{\in\mathbb{K}}a_{i}\in\sum\limits_{i\in I}\mathbb{K}a_{i}$. Thus,
$V\subset\sum\limits_{i\in I}\mathbb{K}a_{i}$. Combined with the (obvious)
inclusion $\sum\limits_{i\in I}\mathbb{K}a_{i}\subset V$, this yields
$V=\sum\limits_{i\in I}\mathbb{K}a_{i}$. But the vector space $\sum
\limits_{i\in I}\mathbb{K}a_{i}$ is finite-dimensional (since $I$ is a finite
set). In other words, the vector space $V$ is finite-dimensional (since
$V=\sum\limits_{i\in I}\mathbb{K}a_{i}$). This proves Claim 1.

\textit{Proof of Claim 2.} Assume that $V$ is a finite-dimensional vector
space. Then, $V$ has a finite basis. Let $\left(  e_{i}\right)  _{i\in I}$ be
such a basis. Thus, $I$ is a finite set, and $\left(  e_{i}\right)  _{i\in I}$
is a basis of the $\mathbb{K}$-vector space $V$. Let $\left(  e_{i}^{\ast
}\right)  _{i\in I}$ be the basis of $V^{\ast}$ dual to the basis $\left(
e_{i}\right)  _{i\in I}$ of $V$. (This is well-defined, since $V$ is finite-dimensional.)

We know that $\left(  e_{i}^{\ast}\right)  _{i\in I}$ is the basis of
$V^{\ast}$ dual to the basis $\left(  e_{i}\right)  _{i\in I}$ of $V$. Thus,%
\begin{equation}
e_{i}^{\ast}\left(  \sum\limits_{j\in I}\lambda_{j}e_{j}\right)  =\lambda
_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }i\in I\text{ and }\left(  \lambda
_{j}\right)  _{j\in I}\in\mathbb{K}^{I}.
\label{pf.prop.sol.Indrho.b.dualgens.findim.4}%
\end{equation}
(Indeed, this is one of the ways to define a dual basis.)

Now, every $v\in V$ satisfies $v=\sum\limits_{i\in I}e_{i}^{\ast}\left(
v\right)  e_{i}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$. Then, we can
write $v$ in the form $v=\sum\limits_{i\in I}\lambda_{i}e_{i}$ for some family
$\left(  \lambda_{i}\right)  _{i\in I}\in\mathbb{K}^{I}$ (since $\left(
e_{i}\right)  _{i\in I}$ is a basis of $V$). Consider this family $\left(
\lambda_{i}\right)  _{i\in I}$. Now, $v=\sum\limits_{i\in I}\lambda_{i}%
e_{i}=\sum\limits_{j\in I}\lambda_{j}e_{j}$ (here, we have renamed the
summation index $i$ as $j$ in the sum). For each $i\in I$, we now have%
\[
e_{i}^{\ast}\left(  \underbrace{v}_{=\sum\limits_{j\in I}\lambda_{j}e_{j}%
}\right)  =e_{i}^{\ast}\left(  \sum\limits_{j\in I}\lambda_{j}e_{j}\right)
=\lambda_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Indrho.b.dualgens.findim.4})}\right)  .
\]
Thus, $\sum\limits_{i\in I}\underbrace{e_{i}^{\ast}\left(  v\right)
}_{=\lambda_{i}}e_{i}=\sum\limits_{i\in I}\lambda_{i}e_{i}$. Compared with
$v=\sum\limits_{i\in I}\lambda_{i}e_{i}$, this yields $v=\sum\limits_{i\in
I}e_{i}^{\ast}\left(  v\right)  e_{i}$, qed.}.

So we know that $\left(  I,\left(  e_{i}\right)  _{i\in I},\left(  e_{i}%
^{\ast}\right)  _{i\in I}\right)  $ is a triple such that

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  e_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  e_{i}^{\ast}\right)  _{i\in I}$ is a family of elements of
$V^{\ast}$ (where $V^{\ast}$ means $\Hom_{\mathbb{K}%
}\left(  V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}e_{i}^{\ast}\left(
v\right)  e_{i}$. In other words, $\left(  I,\left(  e_{i}\right)  _{i\in
I},\left(  e_{i}^{\ast}\right)  _{i\in I}\right)  $ is a finite dual
generating system for $V$ (by the definition of a ``finite
dual generating system''). Thus, a finite dual generating
system for $V$ exists (namely, $\left(  I,\left(  e_{i}\right)  _{i\in
I},\left(  e_{i}^{\ast}\right)  _{i\in I}\right)  $). This proves Claim 2.

Now, both Claim 1 and Claim 2 are proven. Hence, the proof of Proposition
\ref{prop.sol.Indrho.b.dualgens.findim} is complete.
\end{proof}

[\textit{Remark:} The notion of finite dual generating system for $V$ is more
versatile than the notion of a finite basis of $V$. One difference between
these notions is that all bases of $V$ have the same size, while the set $I$
in a finite dual generating system $\left(  I,\left(  a_{i}\right)  _{i\in
I},\left(  f_{i}\right)  _{i\in I}\right)  $ of $V$ can have any (finite) size
$\geq\dim V$. Another difference manifests itself in the general setting when
$\mathbb{K}$ is a commutative ring, not necessarily a field. In this
generality, a finite basis of $V$ exists if and only if $V$ is a finite free
$\mathbb{K}$-module (this is the definition of a finite free $\mathbb{K}%
$-module), whereas a finite dual generating system for $V$ exists if and only
if $V$ is a finitely generated projective $\mathbb{K}$-module. Projective
$\mathbb{K}$-modules are a more frequent occurrence in commutative algebra
than free $\mathbb{K}$-modules, and in the absence of a finite basis, a finite
dual generating system is the thing that comes closest to allowing
``computing in a basis''.]

One significant application of finite dual generating systems is computing
traces of endomorphisms:

\begin{proposition}
\label{prop.sol.Indrho.b.dualgens.Tr}Let $\mathbb{K}$ be a field. Let $V$ be a
finite-dimensional $\mathbb{K}$-vector space. Let $\left(  I,\left(
a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ be a finite
dual generating system for $V$. Let $T:V\rightarrow V$ be a $\mathbb{K}%
$-linear map. Then,%
\[
\operatorname*{trace}T=\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  .
\]

\end{proposition}

Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} can be easily proven directly,
but let us take a slight detour and derive it from the following more general fact:

\begin{proposition}
\label{prop.sol.Indrho.b.dualgens.Tr.ring}Let $\mathbb{K}$ be a commutative
ring. Let $V$ be a $\mathbb{K}$-module. Let $\left(  I,\left(  a_{i}\right)
_{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ be a finite dual
generating system for $V$. Let $\left(  J,\left(  b_{j}\right)  _{j\in
J},\left(  g_{j}\right)  _{j\in J}\right)  $ be a further finite dual
generating system for $V$. Let $T:V\rightarrow V$ be a $\mathbb{K}$-linear
map. Then,%
\[
\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  =\sum_{j\in J}g_{j}\left(
Tb_{j}\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.b.dualgens.Tr.ring}.]We know that
$\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in
I}\right)  $ is a finite dual generating system for $V$. In other words,
$\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in
I}\right)  $ is a triple such that

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies%
\begin{equation}
v=\sum\limits_{i\in I}f_{i}\left(  v\right)  a_{i} .
\label{pf.prop.sol.Indrho.b.dualgens.Tr.ring.1}
\end{equation}


We know that $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)
_{j\in J}\right)  $ is a finite dual generating system for $V$. In other
words, $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  $ is a triple such that

\begin{itemize}
\item $J$ is a finite set;

\item $\left(  b_{j}\right)  _{j\in J}$ is a family of elements of $V$;

\item $\left(  g_{j}\right)  _{j\in J}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies%
\begin{equation}
v=\sum\limits_{j\in J}g_{j}\left(  v\right)  b_{j}.
\label{pf.prop.sol.Indrho.b.dualgens.Tr.ring.2}%
\end{equation}


Now,%
\begin{align*}
\sum_{i\in I}f_{i}\left(  \underbrace{Ta_{i}}_{\substack{=\sum\limits_{j\in
J}g_{j}\left(  Ta_{i}\right)  b_{j}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.ring.2}), applied}\\\text{to }%
v=Ta_{i}\text{)}}}\right)
&  =\sum_{i\in I}\underbrace{f_{i}\left(  \sum\limits_{j\in J}g_{j}\left(
Ta_{i}\right)  b_{j}\right)  }_{\substack{=\sum\limits_{j\in J}g_{j}\left(
Ta_{i}\right)  f_{i}\left(  b_{j}\right)  \\\text{(since the map }f_{i}\text{
is }\mathbb{K}\text{-linear)}}}=\underbrace{\sum_{i\in I}\sum\limits_{j\in J}%
}_{=\sum_{j\in J}\sum_{i\in I}}\underbrace{g_{j}\left(  Ta_{i}\right)
f_{i}\left(  b_{j}\right)  }_{\substack{=f_{i}\left(  b_{j}\right)
g_{j}\left(  Ta_{i}\right)  }}\\
&  =\sum_{j\in J}\sum_{i\in I}f_{i}\left(  b_{j}\right)  g_{j}\left(
Ta_{i}\right)  .
\end{align*}
Compared with%
\begin{align*}
 \sum_{j\in J}g_{j}\left(  T\underbrace{b_{j}}_{\substack{=\sum\limits_{i\in
I}f_{i}\left(  b_{j}\right)  a_{i}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.ring.1}), applied}\\\text{to }%
v=b_{j}\text{)}}}\right)
&  =\sum_{j\in J}g_{j}\left(  \underbrace{T\sum\limits_{i\in I}f_{i}\left(
b_{j}\right)  a_{i}}_{\substack{=\sum_{i\in I}f_{i}\left(  b_{j}\right)
Ta_{i}\\\text{(since the map }T\text{ is }\mathbb{K}\text{-linear)}}}\right)
=\sum_{j\in J}\underbrace{g_{j}\left(  \sum_{i\in I}f_{i}\left(  b_{j}\right)
Ta_{i}\right)  }_{\substack{=\sum\limits_{i\in I}f_{i}\left(  b_{j}\right)
g_{j}\left(  Ta_{i}\right)  \\\text{(since the map }g_{j}\text{ is }%
\mathbb{K}\text{-linear)}}} \\
& = \sum_{j\in J}\sum\limits_{i\in I}f_{i}\left(
b_{j}\right)  g_{j}\left(  Ta_{i}\right)  ,
\end{align*}
this yields $\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  =\sum_{j\in J}%
g_{j}\left(  Tb_{j}\right)  $. This proves Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr.ring}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.b.dualgens.Tr}.]The vector space
$V$ has a finite basis (since it is finite-dimensional). Let $\left(
e_{1},e_{2},\ldots,e_{n}\right)  $ be such a basis. Let $\left(
m_{i,j}\right)  _{1\leq i,j\leq n}\in\mathbb{K}^{n\times n}$ be the matrix
which represents the map $T:V\rightarrow V$ with respect to this basis
$\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ of $V$. Then,%
\begin{equation}
Te_{j}=\sum_{i=1}^{n}m_{i,j}e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
j\in\left\{  1,2,\ldots,n\right\}  \label{pf.prop.sol.Indrho.b.dualgens.Tr.1}%
\end{equation}
(due to the definition of ``the matrix which represents the
map $T:V\rightarrow V$ with respect to this basis $\left(  e_{1},e_{2}%
,\ldots,e_{n}\right)  $ of $V$'').

Let $\left(  e_{1}^{\ast},e_{2}^{\ast},\ldots,e_{n}^{\ast}\right)  $ be the
basis of $V^{\ast}$ dual to the basis $\left(  e_{1},e_{2},\ldots
,e_{n}\right)  $ of $V$. (This is well-defined, since $V$ is
finite-dimensional.) Thus,%
\begin{equation}
e_{k}^{\ast}\left(  \sum_{i=1}^{n}\lambda_{i}e_{i}\right)  =\lambda
_{k}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in\left\{  1,2,\ldots,n\right\}
\text{ and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)
\in\mathbb{K}^{n}. \label{pf.prop.sol.Indrho.b.dualgens.Tr.sumform}%
\end{equation}
(This follows immediately from the definition of a ``dual
basis''.) Now, every $k\in\left\{  1,2,\ldots,n\right\}  $ and
$j\in\left\{  1,2,\ldots,n\right\}  $ satisfy%
\begin{equation}
e_{k}^{\ast}\left(  Te_{j}\right)  =m_{k,j}.
\label{pf.prop.sol.Indrho.b.dualgens.Tr.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.Indrho.b.dualgens.Tr.2}):} Fix
$k\in\left\{  1,2,\ldots,n\right\}  $ and $j\in\left\{  1,2,\ldots,n\right\}
$. Applying the map $e_{k}^{\ast}$ to both sides of
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.1}), we obtain%
\[
e_{k}^{\ast}\left(  Te_{j}\right)  =e_{k}^{\ast}\left(  \sum_{i=1}^{n}%
m_{i,j}e_{i}\right)  =m_{k,j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.sumform}), applied to }\lambda
_{i}=m_{i,j}\right)  .
\]
This proves (\ref{pf.prop.sol.Indrho.b.dualgens.Tr.2}).} Now, every $v\in V$
satisfies $v=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }e_{i}^{\ast
}\left(  v\right)  e_{i}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$. Then,
we can write $v$ in the form $v=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}$ for
some $n$-tuple $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)
\in\mathbb{K}^{n}$ (since $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ is a
basis of $V$). Consider this $n$-tuple $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{n}\right)  $. Now, $v=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}$.
Hence, for each $k\in\left\{  1,2,\ldots,n\right\}  $, we have%
\[
e_{k}^{\ast}\left(  \underbrace{v}_{=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}%
}\right)  =e_{k}^{\ast}\left(  \sum\limits_{i=1}^{n}\lambda_{i}e_{i}\right)
=\lambda_{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.sumform})}\right)  .
\]
Renaming $k$ as $i$ in this statement, we obtain the following: For each
$i\in\left\{  1,2,\ldots,n\right\}  $, we have $e_{i}^{\ast}\left(  v\right)
=\lambda_{i}$. Thus, $\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }}_{=\sum_{i=1}^{n}}\underbrace{e_{i}^{\ast}\left(  v\right)
}_{=\lambda_{i}}e_{i}=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}$. Compared with
$v=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}$, this yields $v=\sum\limits_{i\in
\left\{  1,2,\ldots,n\right\}  }e_{i}^{\ast}\left(  v\right)  e_{i}$, qed.}.

So we know that $\left(  \left\{  1,2,\ldots,n\right\}  ,\left(  e_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  },\left(  e_{i}^{\ast}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }\right)  $ is a triple such that

\begin{itemize}
\item $\left\{  1,2,\ldots,n\right\}  $ is a finite set;

\item $\left(  e_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a
family of elements of $V$;

\item $\left(  e_{i}^{\ast}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is
a family of elements of $V^{\ast}$ (where $V^{\ast}$ means
$\Hom_{\mathbb{K}}\left(  V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }e_{i}^{\ast}\left(  v\right)  e_{i}$. In other words,
\newline$\left(  \left\{  1,2,\ldots,n\right\}  ,\left(  e_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  },\left(  e_{i}^{\ast}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }\right)  $ is a finite dual generating
system for $V$. Therefore, Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr.ring} (applied to $\left(  J,\left(
b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  =\left(
\left\{  1,2,\ldots,n\right\}  ,\left(  e_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  },\left(  e_{i}^{\ast}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }\right)  $) yields%
\begin{align}
\sum_{i\in I}f_{i}\left(  Ta_{i}\right)   &  =\underbrace{\sum_{j\in\left\{
1,2,\ldots,n\right\}  }}_{=\sum_{j=1}^{n}}\underbrace{e_{j}^{\ast}\left(
Te_{j}\right)  }_{\substack{=m_{j,j}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.2}), applied}\\\text{to }k=j\text{)}%
}}=\sum_{j=1}^{n}m_{j,j}\nonumber\\
&  =\sum_{i=1}^{n}m_{i,i}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
the summation index }j\text{ as }i\right)  .
\label{pf.prop.sol.Indrho.b.dualgens.Tr.5}%
\end{align}


But recall that if $\mathfrak{G}$ is any endomorphism of the $\mathbb{K}%
$-vector space $V$, then the trace of $\mathfrak{G}$ equals the trace of any
matrix which represents $\mathfrak{G}$ with respect to a basis of
$V$\ \ \ \ \footnote{Indeed, this is how the trace of $\mathfrak{G}$ is
defined.}. Applying this to $\mathfrak{G}=T$, we conclude that the trace of
$T$ equals the trace of any matrix which represents $T$ with respect to a
basis of $V$. In particular, the trace of $T$ equals the trace of the matrix
$\left(  m_{i,j}\right)  _{1\leq i,j\leq n}$ (since $\left(  m_{i,j}\right)
_{1\leq i,j\leq n}$ is a matrix which represents $T$ with respect to the basis
$\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ of $V$). In other words,
$\operatorname*{trace}T=\operatorname*{trace}\left(  \left(  m_{i,j}\right)
_{1\leq i,j\leq n}\right)  =\sum_{i=1}^{n}m_{i,i}$ (by the definition of
$\operatorname*{trace}\left(  \left(  m_{i,j}\right)  _{1\leq i,j\leq
n}\right)  $). Compared with (\ref{pf.prop.sol.Indrho.b.dualgens.Tr.5}), this
yields $\operatorname*{trace}T=\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  $.
This proves Proposition \ref{prop.sol.Indrho.b.dualgens.Tr}.
\end{proof}

[\textit{Remark:} Proposition \ref{prop.sol.Indrho.b.dualgens.Tr.ring} can be
used to define the trace of an endomorphism of a finitely generated projective
$\mathbb{K}$-module when $\mathbb{K}$ is a commutative ring. Indeed, if
$\mathbb{K}$ is a commutative ring and if $V$ is a finitely generated
projective $\mathbb{K}$-module, and if $T:V\rightarrow V$ is a $\mathbb{K}%
$-linear map, then the trace $\operatorname*{trace}\left(  T\right)  $ of $T$
can be defined as $\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  $, where $\left(
I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ is
a finite dual generating system for $V$. This notion of trace is
well-defined\footnote{because Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr.ring} shows that $\sum_{i\in I}f_{i}\left(
Ta_{i}\right)  $ does not depend on the choice of $\left(  I,\left(
a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $} and
generalizes the classical notion from linear algebra (which is defined only
for finitely generated \textbf{free} $\mathbb{K}$-modules)\footnote{because
Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} (or, more precisely, its
straightforward generalization to free $\mathbb{K}$-modules over commutative
rings) shows that these two notions give the same result when $V$ is a free
$\mathbb{K}$-module}. But we will not concern ourselves with these
generalizations, since our exercise deals only with representations of groups
over a field.]

After all these preparations, we finally come to the actual solution of
Exercise \ref{exe.Indrho}(b). Let $U$ be any finite-dimensional $\CC %
H$-module. We want to prove $\chi_{\Ind_{\rho}%
U}=\Ind_{\rho}\chi_{U}$.

For every $g\in G$, we define a $\CC$-linear map $g^{\ast}%
:\CC G\rightarrow\CC $ by
\begin{equation}
\left(  g^{\ast}\left(  k\right)  =\delta_{g,k}\ \ \ \ \ \ \ \ \ \ \text{for
all }k\in G\right)  . \label{sol.Indrho.b.g*.def}%
\end{equation}
(This is well-defined, since $\left(  k\right)  _{k\in G}$ is a basis of the
$\CC$-vector space $\CC G$.) Then,%
\begin{equation}
\sum_{g\in G}g^{\ast}\left(  \gamma\right)  \cdot g=\gamma
\ \ \ \ \ \ \ \ \ \ \text{for every }\gamma\in\CC G.
\label{sol.Indrho.b.g*.completeness}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.b.g*.completeness}):} Let
$\gamma\in\CC G$. We need to prove the equality
(\ref{sol.Indrho.b.g*.completeness}). We notice that this equality is
$\CC$-linear in $\gamma$. Hence, we can WLOG assume that $\gamma\in G$
(since $G$ is a basis of the $\CC$-vector space $\CC G$). Assume
this. Now,%
\begin{align*}
\sum_{g\in G}g^{\ast}\left(  \gamma\right)  \cdot g  &  =\sum_{g\in
G}\underbrace{g^{\ast}\left(  \gamma\right)  }_{\substack{=\delta_{g,\gamma
}\\\text{(by (\ref{sol.Indrho.b.g*.def}), applied}\\\text{to }k=\gamma\text{
(since }\gamma\in G\text{))}}}\cdot g=\sum_{g\in G}\delta_{g,\gamma}\cdot g\\
&  =\sum\limits_{\substack{g\in G;\\g=\gamma}}\underbrace{\delta_{g,\gamma}%
}_{\substack{=1\\\text{(since }g=\gamma\text{)}}}\cdot g+\sum\limits_{\substack{g\in
G;\\g\neq\gamma}}\underbrace{\delta_{g,\gamma}}_{\substack{=0\\\text{(since
}g\neq\gamma\text{)}}}\cdot g=\sum\limits_{\substack{g\in G;\\g=\gamma}}1\cdot
g+\underbrace{\sum\limits_{\substack{g\in G;\\g\neq\gamma}}0\cdot g}_{=0}\\
&  =\sum\limits_{\substack{g\in G;\\g=\gamma}}\underbrace{1\cdot g}_{=g}%
=\sum\limits_{\substack{g\in G;\\g=\gamma}}g=\gamma\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\gamma\in G\right)  .
\end{align*}
This proves (\ref{sol.Indrho.b.g*.completeness}).} Also,%
\begin{equation}
\left(  gr\right)  ^{\ast}\left(  \gamma r\right)  =g^{\ast}\left(
\gamma\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in G\text{, }r\in G\text{
and }\gamma\in\CC G \label{sol.Indrho.b.g*.r-mul}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.b.g*.r-mul}):} Let $g\in G$, $r\in
G$ and $\gamma\in\CC G$. We need to prove the equality
(\ref{sol.Indrho.b.g*.r-mul}). Since this equality is $\CC$-linear in
$\gamma$ (because the maps $\left(  gr\right)  ^{\ast}$ and $g^{\ast}$ are
$\CC$-linear), we can WLOG assume that $\gamma\in G$ (since $G$ is a
basis of the $\CC$-vector space $\CC G$). Assume this.
\par
We have $\gamma\in G$, and thus we can apply (\ref{sol.Indrho.b.g*.def}) to
$k=\gamma$. We thus obtain $g^{\ast}\left(  \gamma\right)  =\delta_{g,\gamma}%
$. Also, $\underbrace{\gamma}_{\in G}\underbrace{r}_{\in G}\in GG\subset G$.
Hence, (\ref{sol.Indrho.b.g*.def}) (applied to $gr$ and $\gamma r$ instead of
$g$ and $k$) yields $\left(  gr\right)  ^{\ast}\left(  \gamma r\right)
=\delta_{gr,\gamma r}$.
\par
Now, we have $g=\gamma$ if and only if $gr=\gamma r$ (because $G$ is a group).
Now,%
\begin{align*}
g^{\ast}\left(  \gamma\right)   &  =\delta_{g,\gamma}=
\begin{cases}
1,& \text{if }g=\gamma;\\
0,& \text{if }g\neq\gamma
\end{cases}  \quad
=
\begin{cases}
1,& \text{if }gr=\gamma r;\\
0,& \text{if }gr\neq\gamma r
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g=\gamma\text{ if and only
if }gr=\gamma r\right) \\
&  =\delta_{gr,\gamma r}=\left(  gr\right)  ^{\ast}\left(  \gamma r\right)  .
\end{align*}
This proves (\ref{sol.Indrho.b.g*.r-mul}).}.

Proposition \ref{prop.sol.Indrho.b.dualgens.findim} (applied to $\mathbb{K}%
=\CC $ and $V=U$) yields that a finite dual generating system for $U$
exists if and only if the vector space $U$ is finite-dimensional. Thus, a
finite dual generating system for $U$ exists (since the vector space $U$ is
finite-dimensional). Let us fix such a finite dual generating system for $U$,
and denote it by $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(
g_{j}\right)  _{j\in J}\right)  $.

Hence, $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  $ is a finite dual generating system for $U$. In other words,
$\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  $ is a triple such that

\begin{itemize}
\item $J$ is a finite set;

\item $\left(  b_{j}\right)  _{j\in J}$ is a family of elements of $U$;

\item $\left(  g_{j}\right)  _{j\in J}$ is a family of elements of $U^{\ast}$
(where $U^{\ast}$ means $\Hom_{\CC }\left(
U,\CC \right)  $)
\end{itemize}

\noindent
such that every $v\in U$ satisfies%
\begin{equation}
v=\sum\limits_{j\in J}g_{j}\left(  v\right)  b_{j}.
\label{sol.Indrho.b.gj.completeness}%
\end{equation}


We now endow $\CC G$ with the $\left(  \CC G,\CC H\right)
$-bimodule structure that was used to define $\operatorname*{Ind}%
\nolimits_{\rho}U$. We recall that $\Ind_{\rho
}U=\CC G\otimes_{\CC H}U$ (by the definition of
$\Ind_{\rho}U$).

Let us fix $g\in G$. We define a map $\widetilde{F}_{g}:\CC G\times
U\rightarrow U$ by setting%
\[
\left(  \widetilde{F}_{g}\left(  \gamma,u\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  hu\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
\gamma,u\right)  \in\CC G\times U\right)  .
\]
The map $\widetilde{F}_{g}$ is $\CC H$-bilinear with respect to the
right $\CC H$-module structure on $\CC G$ and the left
$\CC H$-module structure on $U$\ \ \ \ \footnote{\textit{Proof.} We
notice that $\widetilde{F}_{g}\left(  \gamma,u\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  hu$ depends $\CC$-linearly on each of
$\gamma$ and $u$ (for obvious reasons). In other words, the map $\widetilde{F}%
_{g}$ is $\CC$-bilinear.
\par
Now, let us fix $\gamma\in\CC G$, $u\in U$ and $\kappa\in\CC H$.
We are going to prove the equality $\widetilde{F}_{g}\left(  \gamma
\kappa,u\right)  =\widetilde{F}_{g}\left(  \gamma,\kappa u\right)  $.
\par
Since this equality is $\CC$-linear in $\kappa$ (because the map
$\widetilde{F}_{g}$ is $\CC$-bilinear), we can WLOG assume that
$\kappa\in H$ (since $H$ is a basis of the $\CC$-vector space
$\CC H$). Assume this.
\par
We have $\kappa\in H$. Thus, the map $H\rightarrow H,\ h\mapsto h\kappa$ is a
bijection (since $H$ is a group). Hence, we can substitute $h\kappa$ for $h$
in the sum $\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\rho\left(  \kappa\right)  \right)  hu$. As a result, we
obtain $\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)  ^{\ast
}\left(  \gamma\rho\left(  \kappa\right)  \right)  hu=\sum\limits_{h\in
H}\left(  g\rho\left(  h\kappa\right)  \right)  ^{\ast}\left(  \gamma
\rho\left(  \kappa\right)  \right)  \left(  h\kappa\right)  u$.
\par
Now, the definition of $\widetilde{F}_{g}\left(  \gamma\kappa,u\right)  $
yields%
\begin{align*}
\widetilde{F}_{g}\left(  \gamma\kappa,u\right)   &  =\dfrac{1}{\left\vert
H\right\vert }
\underbrace{\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\rho\left(  \kappa\right)  \right)  hu}_{
=\sum\limits_{h\in
H}\left(  g\rho\left(  h\kappa\right)  \right)  ^{\ast}\left(  \gamma
\rho\left(  \kappa\right)  \right)  \left(  h\kappa\right)  u
}
=\dfrac
{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\left(  g\underbrace{\rho
\left(  h\kappa\right)  }_{\substack{=\rho\left(  h\right)  \rho\left(
\kappa\right)  \\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}%
}}\right)  ^{\ast}\left(  \gamma\rho\left(  \kappa\right)  \right)  \left(
h\kappa\right)  u\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\underbrace{\left(
g\rho\left(  h\right)  \rho\left(  \kappa\right)  \right)  ^{\ast}\left(
\gamma\rho\left(  \kappa\right)  \right)  }_{\substack{=\left(  g\rho\left(
h\right)  \right)  ^{\ast}\left(  \gamma\right)  \\\text{(by
(\ref{sol.Indrho.b.g*.r-mul}), applied to}\\g\rho\left(  h\right)  \text{ and
}\rho\left(  \kappa\right)  \text{ instead of }g\text{ and }r\text{)}}}h\kappa
u=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(
h\right)  \right)  ^{\ast}\left(  \gamma\right)  h\kappa u.
\end{align*}
Compared with%
\[
\widetilde{F}_{g}\left(  \gamma,\kappa u\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  h\kappa u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\widetilde{F}_{g}\left(  \gamma,\kappa u\right)  \right)
,
\]
this yields $\widetilde{F}_{g}\left(  \gamma\kappa,u\right)  =\widetilde{F}%
_{g}\left(  \gamma,\kappa u\right)  $.
\par
Now let us forget that we fixed $\gamma$, $u$ and $\kappa$. Thus, we have
shown that $\widetilde{F}_{g}\left(  \gamma\kappa,u\right)  =\widetilde{F}%
_{g}\left(  \gamma,\kappa u\right)  $ for all $\gamma\in\CC G$, $u\in U$
and $\kappa\in\CC H$. This yields that the map $\widetilde{F}_{g}$ is
$\CC H$-bilinear with respect to the right $\CC H$-module
structure on $\CC G$ and the left $\CC H$-module structure on $U$
(since we already know that this map $\widetilde{F}_{g}$ is $\CC %
$-bilinear). Qed.}. According to the universal property of the tensor product,
this yields that there exists a unique $\CC$-linear map $F_{g}%
:\CC G\otimes_{\CC H}U\rightarrow U$ such that%
\[
\left(  F_{g}\left(  \gamma\otimes_{\CC H}u\right)  =\widetilde{F}%
_{g}\left(  \gamma,u\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(
\gamma,u\right)  \in\CC G\times U\right)  .
\]
Consider this map $F_{g}$.

So we know that any $\left(  \gamma,u\right)  \in\CC G\times U$
satisfies%
\begin{equation}
F_{g}\left(  \gamma\otimes_{\CC H}u\right)  =\widetilde{F}_{g}\left(
\gamma,u\right)  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in
H}\left(  g\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)  hu.
\label{sol.Indrho.b.Fg.def}%
\end{equation}


Let us now forget that we fixed $g$. We thus have constructed a $\CC %
$-linear map $F_{g}:\CC G\otimes_{\CC H}U\rightarrow U$ for each
$g\in G$. We have shown that this map satisfies (\ref{sol.Indrho.b.Fg.def})
for any $\left(  \gamma,u\right)  \in\CC G\times U$.

The set $G\times J$ is finite (since the sets $G$ and $J$ are finite). Now, we
define a family $\left(  a_{i}\right)  _{i\in G\times J}$ of elements of
$\CC G\otimes_{\CC H}U$ by%
\[
\left(  a_{\left(  k,j\right)  }=k\otimes_{\CC H}b_{j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  k,j\right)  \in G\times J\right)
.
\]
Furthermore, we define a family $\left(  f_{i}\right)  _{i\in G\times J}$ of
elements of $\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ by%
\[
\left(  f_{\left(  k,j\right)  }=g_{j}\circ F_{k}\ \ \ \ \ \ \ \ \ \ \text{for
all }\left(  k,j\right)  \in G\times J\right)  .
\]
(This is well-defined because, for any $\left(  k,j\right)  \in G\times J$,
the composition $g_{j}\circ F_{k}$ of the $\CC$-linear maps
$F_{k}:\CC G\otimes_{\CC H}U\rightarrow U$ and $g_{j}%
:U\rightarrow\CC $ is a $\CC$-linear map $\CC %
G\otimes_{\CC H}U\rightarrow\CC $.)

Our goal is now to prove that $\left(  G\times J,\left(  a_{i}\right)  _{i\in
G\times J},\left(  f_{i}\right)  _{i\in G\times J}\right)  $ is a finite dual
generating system for $\CC G\otimes_{\CC H}U$.

Indeed, every $v\in\CC G\otimes_{\CC H}U$ satisfies $v=\sum
\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$%
\ \ \ \ \footnote{\textit{Proof.} Let $v\in\CC G\otimes_{\CC H}U$.
We need to prove the equality $v=\sum\limits_{i\in G\times J}f_{i}\left(
v\right)  a_{i}$. Since this equality is $\CC$-linear in $v$, we can
WLOG assume that $v$ is a pure tensor (since the pure tensors in
$\CC G\otimes_{\CC H}U$ span the whole $\CC$-vector space
$\CC G\otimes_{\CC H}U$). Assume this. Thus, $v$ is a pure tensor.
In other words, $v=\gamma\otimes_{\CC H}u$ for some $\gamma\in
\CC G$ and $u\in U$. Consider these $\gamma$ and $u$.
\par
We notice that for any $w\in G$, the map $G\rightarrow G,\ k\mapsto kw$ is a
bijection (since $G$ is a group). Hence, for any $w\in G$, we have%
\begin{align}
\sum_{k\in G}\left(  kw\right)  ^{\ast}\left(  \gamma\right)  kw  &
=\sum_{k\in G}k^{\ast}\left(  \gamma\right)  k\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }k\text{ for }kw\text{ in the sum,}\\
\text{since the map }G\rightarrow G,\ k\mapsto kw\text{ is a bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{g\in G}g^{\ast}\left(  \gamma\right)  g\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed the summation index }k\text{ as }g\right) \nonumber\\
&  =\sum_{g\in G}g^{\ast}\left(  \gamma\right)  \cdot g=\gamma
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.b.g*.completeness}%
)}\right)  . \label{sol.Indrho.b.dual.pf.1}%
\end{align}
Now, every $h\in H$ satisfies%
\begin{align}
\sum_{k\in G}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
\gamma\right)  \underbrace{kh}_{\substack{=k\cdot\left(  \CC \left[
\rho\right]  \right)  \left(  h\right)  \\\text{(by the definition of
the}\\\text{right }\CC H\text{-module}\\\text{structure on }%
\CC G\text{)}}}  &  =\sum_{k\in G}\left(  k\rho\left(  h\right)
\right)  ^{\ast}\left(  \gamma\right)  k\cdot\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  }_{\substack{=\rho
\left(  h\right)  \\\text{(since }h\in H\text{)}}}=\sum_{k\in G}\left(
k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)  k\rho\left(
h\right) \nonumber\\
&  =\gamma\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.b.dual.pf.1}),
applied to }w=\rho\left(  h\right)  \right)  . \label{sol.Indrho.b.dual.pf.2}%
\end{align}
\par
Now,%
\begin{align*}
\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}  &  =\underbrace{\sum
\limits_{\left(  k,j\right)  \in G\times J}}_{=\sum_{k\in G}\sum_{j\in J}%
}\underbrace{f_{\left(  k,j\right)  }}_{=g_{j}\circ F_{k}}\left(  v\right)
\underbrace{a_{\left(  k,j\right)  }}_{=k\otimes_{\CC H}b_{j}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }\left(  k,j\right)
\text{ for the summation index }i\right) \\
&  =\sum_{k\in G}\underbrace{\sum_{j\in J}\left(  g_{j}\circ F_{k}\right)
\left(  v\right)  k\otimes_{\CC H}b_{j}}_{\substack{=k\otimes
_{\CC H}\left(  \sum_{j\in J}\left(  g_{j}\circ F_{k}\right)  \left(
v\right)  b_{j}\right)  \\\text{(since the tensor product is }\CC %
\text{-bilinear)}}}=\sum_{k\in G}k\otimes_{\CC H}\left(  \sum_{j\in
J}\underbrace{\left(  g_{j}\circ F_{k}\right)  \left(  v\right)  }%
_{=g_{j}\left(  F_{k}\left(  v\right)  \right)  }b_{j}\right) \\
&  =\sum_{k\in G}k\otimes_{\CC H}\underbrace{\left(  \sum_{j\in J}%
g_{j}\left(  F_{k}\left(  v\right)  \right)  b_{j}\right)  }_{\substack{=F_{k}%
\left(  v\right)  \\\text{(because }F_{k}\left(  v\right)  =\sum_{j\in J}%
g_{j}\left(  F_{k}\left(  v\right)  \right)  b_{j}\\\text{(by
(\ref{sol.Indrho.b.gj.completeness}), applied to }F_{k}\left(  v\right)
\text{ instead of }v\text{))}}}=\sum_{k\in G}k\otimes_{\CC H}%
F_{k}\left(  \underbrace{v}_{=\gamma\otimes_{\CC H}u}\right) \\
&  =\sum_{k\in G}k\otimes_{\CC H}\underbrace{F_{k}\left(  \gamma
\otimes_{\CC H}u\right)  }_{\substack{=\dfrac{1}{\left\vert H\right\vert
}\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
\gamma\right)  hu\\\text{(by (\ref{sol.Indrho.b.Fg.def}), applied to
}g=k\text{)}}}=\sum_{k\in G}k\otimes_{\CC H}\left(  \dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  hu\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }
\underbrace{\sum_{k\in G} \sum_{h\in H}}_{= \sum_{h\in H} \sum_{k\in G}}
\left(
k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)
 k\otimes_{\CC H}hu\ \ \ \ \ \ \ \ \ \ \left(  \text{since the tensor
product is }\CC \text{-bilinear}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\sum_{k\in
G}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)
\underbrace{k\otimes_{\CC H}hu}_{\substack{=kh\otimes_{\CC %
H}u\\\text{(since }h\text{ can be moved past}\\\text{the }\otimes
_{\CC H}\text{ sign (since }h\in H\subset\CC H\text{))}}%
}=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\sum_{k\in G}\left(
k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)  kh\otimes
_{\CC H}u\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\underbrace{\left(
\sum_{k\in G}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
\gamma\right)  kh\right)  }_{\substack{=\gamma\\\text{(by
(\ref{sol.Indrho.b.dual.pf.2}))}}}\otimes_{\CC H}u=\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\underbrace{\gamma\otimes_{\CC H}%
u}_{=v}=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{h\in H}%
v}_{=\left\vert H\right\vert \cdot v}=\dfrac{1}{\left\vert H\right\vert
}\left\vert H\right\vert \cdot v=v.
\end{align*}
Thus, $v=\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$ is proven,
qed.}. Thus, we know that $\left(  G\times J,\left(  a_{i}\right)  _{i\in
G\times J},\left(  f_{i}\right)  _{i\in G\times J}\right)  $ is a triple such that

\begin{itemize}
\item $G\times J$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in G\times J}$ is a family of elements of
$\CC G\otimes_{\CC H}U$;

\item $\left(  f_{i}\right)  _{i\in G\times J}$ is a family of elements of
$\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ (where $\left(
\CC G\otimes_{\CC H}U\right)  ^{\ast}$ means $\Hom_{\CC}
\left(  \CC G\otimes_{\CC H}U,\CC \right)  $)
\end{itemize}

\noindent
such that every $v\in\CC G\otimes_{\CC H}U$ satisfies
$v=\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$. In other words,
$\left(  G\times J,\left(  a_{i}\right)  _{i\in G\times J},\left(
f_{i}\right)  _{i\in G\times J}\right)  $ is a finite dual generating system
for $\CC G\otimes_{\CC H}U$ (because this is precisely how a
``finite dual generating system'' was
defined). In other words, $\left(  G\times J,\left(  a_{i}\right)  _{i\in
G\times J},\left(  f_{i}\right)  _{i\in G\times J}\right)  $ is a finite dual
generating system for $\Ind_{\rho}U$ (since
$\Ind_{\rho}U=\CC G\otimes_{\CC H}U$).

Let us now notice that%
\begin{equation}
\sum_{j\in J}g_{j}\left(  hb_{j}\right)  =\chi_{U}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H \label{sol.Indrho.b.chiU}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.b.chiU}):} Let $h\in H$. The
definition of $\chi_{U}\left(  h\right)  $ yields $\chi_{U}\left(  h\right)
=\operatorname*{trace}\left(  h:U\rightarrow U\right)  $. But Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr} (applied to $\CC$, $U$, $\left(
J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  $
and $\left(  h:U\rightarrow U\right)  $ instead of $\CC$, $U$, $\left(
I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $
and $T$) yields
\[
\operatorname*{trace}\left(  h:U\rightarrow U\right)  =\sum_{j\in J}%
g_{j}\left(  \underbrace{\left(  h:U\rightarrow U\right)  b_{j}}_{=hb_{j}%
}\right)  =\sum_{j\in J}g_{j}\left(  hb_{j}\right)  .
\]
Thus, $\chi_{U}\left(  h\right)  =\operatorname*{trace}\left(  h:U\rightarrow
U\right)  =\sum_{j\in J}g_{j}\left(  hb_{j}\right)  $. This proves
(\ref{sol.Indrho.b.chiU}).}.

Now, let us fix $g\in G$. We want to compute $\chi_{\operatorname*{Ind}%
\nolimits_{\rho}U}\left(  g\right)  $. The definition of $\chi
_{\Ind_{\rho}U}\left(  g\right)  $ yields
$\chi_{\Ind_{\rho}U}\left(  g\right)
=\operatorname*{trace}\left(  g:\Ind_{\rho
}U\rightarrow\Ind_{\rho}U\right)  $. But Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr} (applied to $\mathbb{K}=\CC $,
$V=\Ind_{\rho}U$, $I=G\times J$ and $T=\left(
g:\Ind_{\rho}U\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U\right)  $) yields%
\begin{align}
&  \operatorname*{trace}\left(  g:\Ind_{\rho
}U\rightarrow\Ind_{\rho}U\right) \nonumber\\
&  =\sum_{i\in G\times J}f_{i}\left(  \underbrace{\left(
g:\Ind_{\rho}U\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U\right)  a_{i}}_{=ga_{i}}\right)  =\sum_{i\in G\times J}%
f_{i}\left(  ga_{i}\right) \nonumber\\
&  =\sum_{\left(  k,j\right)  \in G\times J}\underbrace{f_{\left(  k,j\right)
}}_{=g_{j}\circ F_{k}}\left(  g\underbrace{a_{\left(  k,j\right)  }%
}_{=k\otimes_{\CC H}b_{j}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{here, we substituted }\left(  k,j\right)  \text{ for the summation index
}i\right) \nonumber\\
&  =\sum_{\left(  k,j\right)  \in G\times J}\left(  g_{j}\circ F_{k}\right)
\underbrace{\left(  g\left(  k\otimes_{\CC H}b_{j}\right)  \right)
}_{=gk\otimes_{\CC H}b_{j}}=\sum_{\left(  k,j\right)  \in G\times
J}\underbrace{\left(  g_{j}\circ F_{k}\right)  \left(  gk\otimes_{\CC %
H}b_{j}\right)  }_{=g_{j}\left(  F_{k}\left(  gk\otimes_{\CC H}%
b_{j}\right)  \right)  }\nonumber\\
&  =\sum_{\left(  k,j\right)  \in G\times J}g_{j}\left(  \underbrace{F_{k}%
\left(  gk\otimes_{\CC H}b_{j}\right)  }_{\substack{=\dfrac
{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\left(  k\rho\left(
h\right)  \right)  ^{\ast}\left(  gk\right)  hb_{j}\\\text{(by
(\ref{sol.Indrho.b.Fg.def}), applied to }k\text{, }gk\text{ and }%
b_{j}\\\text{instead of }g\text{, }\gamma\text{ and }u\text{)}}}\right)
\nonumber\\
&  =\underbrace{\sum_{\left(  k,j\right)  \in G\times J}}_{=\sum_{k\in G}%
\sum_{j\in J}}\underbrace{g_{j}\left(  \dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
gk\right)  hb_{j}\right)  }_{\substack{=\dfrac{1}{\left\vert H\right\vert
}\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
gk\right)  g_{j}\left(  hb_{j}\right)  \\\text{(since the map }g_{j}\text{ is
}\CC \text{-linear)}}}=\sum_{k\in G}\sum_{j\in J}\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)
^{\ast}\left(  gk\right)  g_{j}\left(  hb_{j}\right) \nonumber\\
&  =\sum_{k\in G}\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in
H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(  gk\right)  \sum_{j\in
J}g_{j}\left(  hb_{j}\right)  . \label{sol.Indrho.b.computing-the-trace.1}%
\end{align}
But every $k\in G$ satisfies%
\begin{align}
&  \sum\limits_{h\in H}\underbrace{\left(  k\rho\left(  h\right)  \right)
^{\ast}\left(  gk\right)  }_{\substack{=\delta_{k\rho\left(  h\right)
,gk}\\\text{(by (\ref{sol.Indrho.b.g*.def}), applied to }k\rho\left(
h\right)  \\\text{and }gk\text{ instead of }g\text{ and }k\text{)}%
}}\underbrace{\sum_{j\in J}g_{j}\left(  hb_{j}\right)  }_{\substack{=\chi
_{U}\left(  h\right)  \\\text{(by (\ref{sol.Indrho.b.chiU}))}}}\nonumber\\
&  =\sum\limits_{h\in H}\delta_{k\rho\left(  h\right)  ,gk}\chi_{U}\left(
h\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  =gk}%
}}_{\substack{=\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}%
=g}}\\\text{(because for every }h\in H\text{,}\\\text{the statement }\left(
k\rho\left(  h\right)  =gk\right)  \\\text{is equivalent}\\\text{to }\left(
k\rho\left(  h\right)  k^{-1}=g\right)  \text{)}}}\underbrace{\delta
_{k\rho\left(  h\right)  ,gk}}_{\substack{=1\\\text{(since }k\rho\left(
h\right)  =gk\text{)}}}\chi_{U}\left(  h\right)  +\sum\limits_{\substack{h\in
H;\\k\rho\left(  h\right)  \neq gk}}\underbrace{\delta_{k\rho\left(  h\right)
,gk}}_{\substack{=0\\\text{(since }k\rho\left(  h\right)  \neq gk\text{)}%
}}\chi_{U}\left(  h\right) \nonumber\\
&  =\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}=g}}1\chi_{U}\left(
h\right)  +\underbrace{\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)
\neq gk}}0\chi_{U}\left(  h\right)  }_{=0}=\sum\limits_{\substack{h\in H;\\k\rho
\left(  h\right)  k^{-1}=g}}1\chi_{U}\left(  h\right) \nonumber\\
&  =\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(
h\right)  . \label{sol.Indrho.b.computing-the-trace.lem2}%
\end{align}
Now, recall that%
\begin{align*}
\chi_{\Ind_{\rho}U}\left(  g\right)   &
=\operatorname*{trace}\left(  g:\Ind_{\rho
}U\rightarrow\Ind_{\rho}U\right) \\
&  =\sum_{k\in G}\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum
\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
gk\right)  \sum_{j\in J}g_{j}\left(  hb_{j}\right)  }_{\substack{=\sum\limits
_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(
h\right)  \\\text{(by (\ref{sol.Indrho.b.computing-the-trace.lem2}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.b.computing-the-trace.1})}\right) \\
&  =\sum_{k\in G}\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{h\in
H;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(  h\right)  =\dfrac
{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{h\in H;\\k\rho\left(
h\right)  k^{-1}=g}}\sum_{k\in G}}_{=\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}}\chi_{U}\left(  h\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(  h\right)  .
\end{align*}
Compared with
\[
\left(  \Ind_{\rho}\chi_{U}\right)  \left(  g\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\Ind_{\rho}\chi_{U}\right)  \left(  g\right)  \right)
,
\]
this yields $\chi_{\Ind_{\rho}U}\left(  g\right)
=\left(  \Ind_{\rho}\chi_{U}\right)  \left(  g\right)
$.

Now, let us forget that we fixed $g$. We thus have proven that $\chi
_{\Ind_{\rho}U}\left(  g\right)  =\left(
\Ind_{\rho}\chi_{U}\right)  \left(  g\right)  $ for
every $g\in G$. In other words, $\chi_{\Ind_{\rho}%
U}=\Ind_{\rho}\chi_{U}$. This solves Exercise
\ref{exe.Indrho}(b).

(c) Assume that $H$ is a subgroup of $G$, and that $\rho:H\rightarrow G$ is
the inclusion map. We need to show that $\Ind_{\rho
}f=\Ind_{H}^{G}f$ for every $f\in R_{\CC %
}\left(  H\right)  $.

We notice that the map $G\rightarrow G,\ k\mapsto k^{-1}$ is a bijection
(since $G$ is a group).

Let $f\in R_{\CC }\left(  H\right)  $. Let $g\in G$. Then, the
definition of $\left(  \Ind_{\rho}f\right)  \left(
g\right)  $ yields%
\begin{align*}
\left(  \Ind_{\rho}f\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}}%
_{\substack{=\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\khk^{-1}%
=g}}\\\text{(since }\rho\left(  h\right)  =h\text{ for every }h\in
H\\\text{(since }\rho:H\rightarrow G\text{ is the inclusion map))}}}f\left(
h\right)  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits
_{\substack{\left(  h,k\right)  \in H\times G;\\khk^{-1}=g}}}_{=\sum_{k\in
G}\sum\limits_{\substack{h\in H;\\khk^{-1}=g}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{k\in G}\underbrace{\sum\limits
_{\substack{h\in H;\\khk^{-1}=g}}}_{\substack{=\sum\limits_{\substack{h\in
H;\\h=k^{-1}gk}}\\\text{(because for every }h\in H\text{,}\\\text{the
statement }\left(  khk^{-1}=g\right)  \text{ is}\\\text{equivalent to }\left(
h=k^{-1}gk\right)  \text{)}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum_{k\in G}%
\sum\limits_{\substack{h\in H;\\h=k^{-1}gk}}f\left(  h\right)  }_{\substack{=\sum\limits
_{\substack{k\in G;\\k^{-1}gk\in H}}\sum\limits_{\substack{h\in H;\\h=k^{-1}%
gk}}f\left(  h\right)  +\sum\limits_{\substack{k\in G;\\k^{-1}gk\notin H}%
}\sum\limits_{\substack{h\in H;\\h=k^{-1}gk}}f\left(  h\right)  \\\text{(since every
}k\in G\text{ satisfies either }k^{-1}gk\in H\text{ or }k^{-1}gk\notin H\text{
(but never both))}}}\\
&  =\dfrac{1}{\left\vert H\right\vert }\left(  \sum\limits_{\substack{k\in
G;\\k^{-1}gk\in H}}\underbrace{\sum\limits_{\substack{h\in H;\\h=k^{-1}gk}}f\left(
h\right)  }_{\substack{=f\left(  k^{-1}gk\right)  \\\text{(since }k^{-1}gk\in
H\text{)}}}+\sum\limits_{\substack{k\in G;\\k^{-1}gk\notin H}}\underbrace{\sum\limits
_{\substack{h\in H;\\h=k^{-1}gk}}f\left(  h\right)  }_{\substack{=\left(
\text{empty sum}\right)  \\\text{(since there is no }h\in H\text{ satisfying
}h=k^{-1}gk\\\text{(because }k^{-1}gk\notin H\text{))}}}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\left(  \sum\limits_{\substack{k\in
G;\\k^{-1}gk\in H}}f\left(  k^{-1}gk\right)  +\sum\limits_{\substack{k\in
G;\\k^{-1}gk\notin H}}\underbrace{\left(  \text{empty sum}\right)  }%
_{=0}\right)  =\dfrac{1}{\left\vert H\right\vert }\left(  \sum\limits_{\substack{k\in
G;\\k^{-1}gk\in H}}f\left(  k^{-1}gk\right)  +\underbrace{\sum\limits_{\substack{k\in
G;\\k^{-1}gk\notin H}}0}_{=0}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G;\\k^{-1}gk\in
H}}f\left(  k^{-1}gk\right)  =\dfrac{1}{\left\vert H\right\vert }%
\underbrace{\sum\limits_{\substack{k\in G;\\\left(  k^{-1}\right)  ^{-1}gk^{-1}\in
H}}}_{\substack{=\sum\limits_{\substack{k\in G;\\kgk^{-1}\in H}}\\\text{(since
}\left(  k^{-1}\right)  ^{-1}=k\text{ for every }k\in G\text{)}}}f\left(
\underbrace{\left(  k^{-1}\right)  ^{-1}}_{=k}gk^{-1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }k^{-1}\text{ for }k\text{ in the sum, since}\\
\text{the map }G\rightarrow G,\ k\mapsto k^{-1}\text{ is a bijection}%
\end{array}
\right) \\
&  =\frac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G;\\kgk^{-1}\in
H}}f\left(  kgk^{-1}\right)  =\left(  \operatorname{Ind}_{H}^{G}f\right)
\left(  g\right)
\end{align*}
(since (\ref{eq.exe.induction.characters.eq}) yields $\left(
\operatorname{Ind}_{H}^{G}f\right)  \left(  g\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{\substack{k\in G;\\kgk^{-1}\in H}}f\left(  kgk^{-1}%
\right)  $).

Let us now forget that we fixed $g$. We thus have proven that $\left(
\Ind_{\rho}f\right)  \left(  g\right)  =\left(
\operatorname{Ind}_{H}^{G}f\right)  \left(  g\right)  $ for every $g\in G$. In
other words, $\Ind_{\rho}f=\operatorname*{Ind}%
\nolimits_{H}^{G}f$. This solves Exercise \ref{exe.Indrho}(c).

(d) Assume that $H$ is a subgroup of $G$, and that $\rho:H\rightarrow G$ is
the inclusion map. We need to show that $\Ind_{\rho
}U=\Ind_{H}^{G}U$ for every $\CC H$-module $U$.

We notice that $\rho:H\rightarrow G$ is the inclusion map. Hence,
$\CC \left[  \rho\right]  :\CC H\rightarrow\CC G$ is also
the inclusion map (since we identify $\CC H$ with a $\CC %
$-subalgebra of $\CC G$ along this map $\CC \left[  \rho\right]
$). Thus,%
\begin{equation}
\left(  \CC \left[  \rho\right]  \right)  \eta=\eta
\ \ \ \ \ \ \ \ \ \ \text{for every }\eta\in\CC H.
\label{sol.Indrho.d.incmap}%
\end{equation}


Let $U$ be a $\CC H$-module. Both $\operatorname{Ind}_{H}^{G}U$ and
$\Ind_{\rho}U$ are defined as ${\CC }%
G\otimes_{\CC H}U$ for some $\left(  \CC G,\CC H\right)
$-bimodule structure on $\CC G$; however, their definitions differ at
how this $\left(  \CC G,\CC H\right)  $-bimodule structure is
defined. We are now going to prove that these two $\left(  \CC %
G,\CC H\right)  $-bimodule structures on $\CC G$ are identical.

The definition of $\Ind_{\rho}U$ shows that we have%
\begin{equation}
\Ind_{\rho}U=\CC G\otimes_{\CC H}U,
\label{sol.Indrho.d.1}%
\end{equation}
where $\CC G$ is regarded as a $\left(  \CC G,\CC H\right)
$-bimodule according to the following rule: The left $\CC G$-module
structure on $\CC G$ is plain multiplication inside $\CC G$; the
right $\CC H$-module structure on $\CC G$ is induced by the
$\CC$-algebra homomorphism $\CC \left[  \rho\right]
:\CC H\rightarrow\CC G$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$
for all $\gamma\in\CC G$ and $\eta\in\CC H$). We denote this
$\left(  \CC G,\CC H\right)  $-bimodule structure on $\CC G$
as the \emph{first structure}.

On the other hand, the definition of $\Ind_{H}^{G}U$
shows that we have%
\begin{equation}
\operatorname{Ind}_{H}^{G}U={\CC }G\otimes_{\CC H}U,
\label{sol.Indrho.d.2}%
\end{equation}
where $\CC G$ is regarded as a $\left(  \CC G,\CC H\right)
$-bimodule in the usual way (i.e., the left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$, and the right
$\CC H$-module structure on $\CC G$ is also plain multiplication
inside $\CC G$ because $\CC H\subset\CC G$). We denote this
$\left(  \CC G,\CC H\right)  $-bimodule structure on $\CC G$
as the \emph{second structure}.

The right hand sides of the equalities (\ref{sol.Indrho.d.1}) and
(\ref{sol.Indrho.d.2}) appear identical, but so far we do not know if they
actually mean the same thing, because the meanings of ``$\CC G$'' possibly
differ. Namely, we have two $\left(
\CC G,\CC H\right)  $-bimodule structures on the $\CC %
$-vector space $\CC G$: the first structure (used in
(\ref{sol.Indrho.d.1})) and the second structure (used in
(\ref{sol.Indrho.d.2})). These two structures clearly have the same left
$\CC G$-module structure. But they also have the same right
$\CC H$-module structure, because every $\gamma\in\CC G$ and
$\eta\in\CC H$ satisfy%
\begin{align*}
&  \left(  \text{the result of the right action of }\eta\text{ on }%
\gamma\text{ according to the first structure}\right) \\
&  =\gamma\cdot\underbrace{\left(  \CC \left[  \rho\right]  \right)
\eta}_{\substack{=\eta\\\text{(by (\ref{sol.Indrho.d.incmap}))}}}=\gamma
\cdot\eta\\
&  =\left(  \text{the result of the right action of }\eta\text{ on }%
\gamma\text{ according to the second structure}\right)  .
\end{align*}
Hence, the first structure and the second structure are identical. Thus, the
right hand sides of the equalities (\ref{sol.Indrho.d.1}) and
(\ref{sol.Indrho.d.2}) really mean the same thing. Thus, comparing the
equalities (\ref{sol.Indrho.d.1}) and (\ref{sol.Indrho.d.2}), we obtain
$\Ind_{\rho}U=\Ind_{H}^{G}U$
as left $\CC G$-modules. This solves Exercise \ref{exe.Indrho}(d).

(e) Assume that $G=H/K$ for some normal subgroup $K$ of $H$. Let
$\rho:H\rightarrow G$ be the projection map. We want to prove that
$\Ind_{\rho}f=f^{K}$ for every $f\in R_{\CC %
}\left(  H\right)  $.

Let $f\in R_{\CC }\left(  H\right)  $. Applying (\ref{sol.Indrho.RCK})
to $H$ instead of $K$, we obtain the following equivalence:%
\begin{align*}
&  \left(  f\in R_{\CC }\left(  H\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }H\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime
}\right)  \right)  .
\end{align*}
Hence,
\begin{equation}
\left(  \text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of
}H\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime}\right)  \right)
\label{sol.Indrho.e.conj}%
\end{equation}
(since we know that $f\in R_{\CC }\left(  H\right)  $). Thus,%
\begin{equation}
f\left(  y^{-1}zy\right)  =f\left(  z\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }z\in H\text{ and }y\in H \label{sol.Indrho.e.conj2}%
\end{equation}
(because if $z\in H$ and $y\in H$, then $y^{-1}zy$ and $z$ are two conjugate
elements of $H$).

Let $g\in G$. Then, $g\in G=H/K$. Hence, there exists an $x\in H$ such that
$g=xK$. Consider this $x$. The map $\rho:H\rightarrow G$ is the projection map
from $H$ to $G=H/K$, and thus sends every $h\in H$ to the coset $hK\in G$. In
other words, $\rho\left(  h\right)  =hK$ for every $h\in H$. Applied to $h=x$,
this yields $\rho\left(  x\right)  =xK=g$. Of course, $\ker%
\rho=K$ (since $\rho$ is the projection map from $H$ to $H/K$).

We have $\left\vert G\right\vert =\left\vert H/K\right\vert =\left[
H:K\right]  =\left\vert H\right\vert /\left\vert K\right\vert $.

We make another simple observation: If $h\in H$ and $y\in H$, then we have the
following logical equivalence:%
\begin{equation}
\left(  \rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  \right)  \ \Longleftrightarrow\ \left(  h\in
y^{-1}xKy\right)  . \label{sol.Indrho.e.equiv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.e.equiv1}):} Let $h\in H$ and
$y\in H$. Recall that the map $\rho:H\rightarrow G$ is the projection map from
$H$ to $G=H/K$. Thus, $\rho\left(  w\right)  =wK$ for every $w\in H$.
\par
We have the following chain of equivalences:%
\begin{align*}
&  \ \left(  \rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  \right) \\
&  \Longleftrightarrow\ \left(  \rho\left(  y\right)  \rho\left(  h\right)
=g\rho\left(  y\right)  \right)  \ \Longleftrightarrow\ \left(
\underbrace{\rho\left(  y\right)  \rho\left(  h\right)  \left(  \rho\left(
y\right)  \right)  ^{-1}}_{\substack{=\rho\left(  yhy^{-1}\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}=\underbrace{g}%
_{=\rho\left(  x\right)  }\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\rho\left(  yhy^{-1}\right)
}_{\substack{=yhy^{-1}K\\\text{(since }\rho\left(  w\right)  =wK\text{ for
every }w\in H\text{)}}}=\underbrace{\rho\left(  x\right)  }%
_{\substack{=xK\\\text{(since }\rho\left(  w\right)  =wK\text{ for every }w\in
H\text{)}}}\right)  \ \Longleftrightarrow\ \left(  yhy^{-1}K=xK\right) \\
&  \Longleftrightarrow\ \left(  yhy^{-1}\in xK\right)  \ \Longleftrightarrow
\ \left(  yh\in xKy\right)  \ \Longleftrightarrow\ \left(  h\in y^{-1}%
xKy\right)  .
\end{align*}
This proves (\ref{sol.Indrho.e.equiv1}).}

The definition of $f^{K}$ yields $f^{K}\left(  xK\right)  =\dfrac
{1}{\left\vert K\right\vert }\sum\limits_{k\in K}f\left(  xk\right)  $. Hence,%
\begin{equation}
f^{K}\left(  \underbrace{g}_{=xK}\right)  =f^{K}\left(  xK\right)  =\dfrac
{1}{\left\vert K\right\vert }\sum\limits_{k\in K}f\left(  xk\right)  .
\label{sol.Indrho.e.fK}%
\end{equation}
But the definition of $\left(  \Ind_{\rho}f\right)
\left(  g\right)  $ yields%
\begin{align}
\left(  \Ind_{\rho}f\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}}%
_{\substack{=\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\\rho\left(
h\right)  =k^{-1}gk}}\\\text{(because for any }\left(  h,k\right)  \in H\times
G\text{,}\\\text{the statement }\left(  k\rho\left(  h\right)  k^{-1}%
=g\right)  \\\text{is equivalent to }\left(  \rho\left(  h\right)
=k^{-1}gk\right)  \text{)}}}f\left(  h\right)  =\dfrac{1}{\left\vert
H\right\vert }\underbrace{\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\\rho\left(  h\right)  =k^{-1}gk}}}_{=\sum\limits_{k\in G}\sum
\limits_{\substack{h\in H;\\\rho\left(  h\right)  =k^{-1}gk}}}f\left(
h\right) \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{k\in G}\sum
\limits_{\substack{h\in H;\\\rho\left(  h\right)  =k^{-1}gk}}f\left(
h\right)  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{p\in G}%
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}}f\left(
h\right)  \label{sol.Indrho.e.4}%
\end{align}
(here, we renamed the index $k$ as $p$ in the first sum). But every $p\in G$
satisfies $\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}%
gp}}f\left(  h\right)  =\sum\limits_{k\in K}f\left(  xk\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $p\in G$. Then, $p\in G=H/K$. Hence,
there exists a $y\in H$ such that $p=yK$. Consider this $y$.
\par
Recall that $\rho\left(  h\right)  =hK$ for every $h\in H$. Applying this to
$h=y$, we obtain $\rho\left(  y\right)  =yK=p$.
\par
We notice that the map $K\rightarrow y^{-1}xKy,\ k\mapsto y^{-1}xky$ is a
bijection (since $H$ is a group).
\par
Now, for every $h\in H$, we have the following equivalence:%
\begin{align}
\left(  \rho\left(  h\right)  =p^{-1}gp\right)  \  &  \Longleftrightarrow
\ \left(  \rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}p=\rho\left(  y\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  h\in y^{-1}xKy\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{according to (\ref{sol.Indrho.e.equiv1}%
)}\right)  . \label{sol.Indrho.e.equiv2}%
\end{align}
\par
Now,%
\[
\underbrace{\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}%
}}_{\substack{=\sum\limits_{\substack{h\in H;\\h\in y^{-1}xKy}%
}\\\text{(because for every }h\in H\text{, the statement}\\\left(  \rho\left(
h\right)  =p^{-1}gp\right)  \text{ is equivalent to }\left(  h\in
y^{-1}xKy\right)  \\\text{(because of (\ref{sol.Indrho.e.equiv2})))}}}f\left(
h\right)  =\underbrace{\sum\limits_{\substack{h\in H;\\h\in y^{-1}xKy}%
}}_{\substack{=\sum\limits_{h\in y^{-1}xKy}\\\text{(since }y^{-1}xKy\subset
H\text{)}}}f\left(  h\right)  =\sum\limits_{h\in y^{-1}xKy}f\left(  h\right)
=\sum\limits_{k\in K}f\left(  y^{-1}xky\right)
\]
(here, we substituted $y^{-1}xky$ for $h$ in the sum, since the map
$K\rightarrow y^{-1}xKy,\ k\mapsto y^{-1}xky$ is a bijection). Thus,%
\[
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}}f\left(
h\right)  =\sum\limits_{k\in K}\underbrace{f\left(  y^{-1}xky\right)
}_{\substack{=f\left(  xk\right)  \\\text{(by (\ref{sol.Indrho.e.conj2}),
applied to }z=xk\text{)}}}=\sum\limits_{k\in K}f\left(  xk\right)  ,
\]
qed.}. Hence, (\ref{sol.Indrho.e.4}) becomes%
\begin{align*}
\left(  \Ind_{\rho}f\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{p\in G}\underbrace{\sum
\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}}f\left(
h\right)  }_{=\sum\limits_{k\in K}f\left(  xk\right)  }=\dfrac{1}{\left\vert
H\right\vert }\underbrace{\sum\limits_{p\in G}\sum\limits_{k\in K}f\left(
xk\right)  }_{=\left\vert G\right\vert \cdot\sum\limits_{k\in K}f\left(
xk\right)  }=\dfrac{1}{\left\vert H\right\vert }\underbrace{\left\vert
G\right\vert }_{=\left\vert H\right\vert /\left\vert K\right\vert }\cdot
\sum\limits_{k\in K}f\left(  xk\right) \\
&  =\underbrace{\dfrac{1}{\left\vert H\right\vert }\cdot\left\vert
H\right\vert /\left\vert K\right\vert }_{=\dfrac{1}{\left\vert K\right\vert }%
}\cdot\sum\limits_{k\in K}f\left(  xk\right)  =\dfrac{1}{\left\vert
K\right\vert }\sum\limits_{k\in K}f\left(  xk\right)  =f^{K}\left(  g\right)
\end{align*}
(by (\ref{sol.Indrho.e.fK})).

Let us now forget that we fixed $g$. We thus have shown that $\left(
\Ind_{\rho}f\right)  \left(  g\right)  =f^{K}\left(
g\right)  $ for every $g\in G$. In other words, $\operatorname*{Ind}%
\nolimits_{\rho}f=f^{K}$. This solves Exercise \ref{exe.Indrho}(e).

(f) Assume that $G=H/K$ for some normal subgroup $K$ of $H$. Let
$\rho:H\rightarrow G$ be the projection map. We want to prove that
$\Ind_{\rho}U\cong U^{K}$ for every $\CC %
H$-module $U$.

The map $\rho$ is the projection map from $H$ to $G=H/K$. Thus, the map $\rho$
is surjective and has kernel $\ker\rho=K$. Furthermore,%
\begin{equation}
\left\vert \rho^{-1}\left(  g\right)  \right\vert =\left\vert K\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G.
\label{sol.Indrho.f.sol1.preimage}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.preimage}):} Let $g\in G$.
The map $\rho$ is surjective. Hence, $\rho\left(  H\right)  =G$. Thus, $g\in
G=\rho\left(  H\right)  $. Therefore, there exists some $x\in H$ such that
$g=\rho\left(  x\right)  $. Let us fix such an $x$.
\par
We know that $H$ is a group. Hence, the map $K\rightarrow xK,\ k\mapsto xk$ is
a bijection. Thus, the sets $K$ and $xK$ are in bijection. Therefore,
$\left\vert xK\right\vert =\left\vert K\right\vert $.
\par
However, for every $y\in H$, we have the following logical equivalence:
\begin{align*}
\left(  y\in\rho^{-1}\left(  g\right)  \right)  \  &  \Longleftrightarrow
\ \left(  \rho\left(  y\right)  =\underbrace{g}_{=\rho\left(  x\right)
}\right)  \ \Longleftrightarrow\ \left(  \rho\left(  y\right)  =\rho\left(
x\right)  \right)  \ \Longleftrightarrow\ \left(  \underbrace{\rho\left(
y\right)  \cdot\left(  \rho\left(  x\right)  \right)  ^{-1}}_{\substack{=\rho
\left(  yx^{-1}\right)  \\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}=1\right) \\
&  \Longleftrightarrow\ \left(  \rho\left(  yx^{-1}\right)  =1\right)
\ \Longleftrightarrow\ \left(  yx^{-1}\in\underbrace{\ker\rho
}_{=K}\right)  \ \Longleftrightarrow\ \left(  yx^{-1}\in K\right)
\ \Longleftrightarrow\ \left(  y\in xK\right)  .
\end{align*}
Hence, $\rho^{-1}\left(  g\right)  =xK$, so that $\left\vert \rho^{-1}\left(
g\right)  \right\vert =\left\vert xK\right\vert =\left\vert K\right\vert $.
This proves (\ref{sol.Indrho.f.sol1.preimage}).}

Let $U$ be a $\CC H$-module. Recall that $U^{K}$ is a $\CC \left[
H/K\right]  $-module, thus a $\CC G$-module (since $H/K=G$).

Recall that $\Ind_{\rho}U$ is defined as the
$\CC G$-module $\CC G\otimes_{\CC H}U$, where $\CC G$
is regarded as a $\left(  \CC G,\CC H\right)  $-bimodule according
to the following rule: The left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$; the right
$\CC H$-module structure on $\CC G$ is induced by the $\CC %
$-algebra homomorphism $\CC \left[  \rho\right]  :\CC %
H\rightarrow\CC G$ (thus, it is explicitly given by $\gamma\eta
=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$ for all
$\gamma\in\CC G$ and $\eta\in\CC H$). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC G,\CC H\right)
$-bimodule structure.

We define a map $\alpha:U^{K}\rightarrow\CC G\otimes_{\CC H}U$ by
setting%
\[
\alpha\left(  u\right)  =1\otimes_{\CC H}u\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in U^{K}.
\]
This $\alpha$ is a map $U^{K}\rightarrow\Ind_{\rho}U$
(since $\Ind_{\rho}U=\CC G\otimes_{\CC %
H}U$) and is $\CC$-linear (since $\alpha\left(  u\right)
=1\otimes_{\CC H}u$ depends $\CC$-linearly on $u$).

We will show that $\alpha$ is a $\CC G$-module isomorphism.

The $\CC G$-module structure on $U^{K}$ has the property that%
\begin{equation}
\rho\left(  h\right)  \cdot v=hv\ \ \ \ \ \ \ \ \ \ \text{for any }h\in
H\text{ and }v\in U^{K}. \label{sol.Indrho.f.sol1.UK-modstr}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.UK-modstr}):} The map
$\rho$ is the projection map from $H$ to $H/K$. Thus, the map $\rho$ sends
every $h\in H$ to the coset $hK\in H/K$. In other words,%
\begin{equation}
\rho\left(  h\right)  =hK\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H,
\label{sol.Indrho.f.sol1.hK}%
\end{equation}
where $hK$ means the coset $hK\in H/K$. But by the definition of the
$\CC \left[  H/K\right]  $-module structure on $U^{K}$, we have%
\[
\left(  hK\right)  \cdot v=hv\ \ \ \ \ \ \ \ \ \ \text{for any }h\in H\text{
and }v\in U^{K},
\]
where $hK$ means the coset $hK\in H/K$. Thus, any $h\in H$ and $v\in U^{K}$
satisfy%
\[
\underbrace{\rho\left(  h\right)  }_{\substack{=hK\\\text{(by
(\ref{sol.Indrho.f.sol1.hK}))}}}\cdot v=\left(  hK\right)  \cdot v=hv,
\]
where $hK$ means the coset $hK\in H/K$. This proves
(\ref{sol.Indrho.f.sol1.UK-modstr}).}

Now, $\alpha$ is a $\CC G$-module homomorphism\footnote{\textit{Proof.}
Let $u\in U^{K}$ and $g\in G$. We have $g\in G=\rho\left(  H\right)  $ (since
the map $\rho:H\rightarrow G$ is surjective). Thus, $g=\rho\left(  y\right)  $
for some $y\in H$. Let us consider this $y$.
\par
Applying (\ref{sol.Indrho.f.sol1.UK-modstr}) to $h=y$ and $v=u$, we obtain
$\rho\left(  y\right)  \cdot u=yu$. Thus, $\underbrace{g}_{=\rho\left(
y\right)  }u=\rho\left(  y\right)  \cdot u=yu$.
\par
Now, the definition of $\alpha\left(  gu\right)  $ yields
\[
\alpha\left(  gu\right)  =1\otimes_{\CC H}\underbrace{gu}_{=yu}%
=1\otimes_{\CC H}yu=1y\otimes_{\CC H}u
\]
(here, we moved the $y$ past the tensor sign; this is allowed because $y\in
H\subset\CC H$).
\par
By the definition of the left $\CC G$-module structure on $\CC G$,
we have $g\cdot1=g1=g$.
\par
By the definition of the right $\CC H$-module structure on $\CC %
G$, we have $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying this to $\gamma=1$ and $\eta=y$, we obtain $1y=1\cdot\left(
\CC \left[  \rho\right]  \right)  y$. But $y\in H$ and thus $\left(
\CC \left[  \rho\right]  \right)  y=\rho\left(  y\right)  =g$.
Therefore, $1y=1\cdot\underbrace{\left(  \CC \left[  \rho\right]
\right)  y}_{=g}=1\cdot g=g=g\cdot1$.
\par
Now, $\alpha\left(  gu\right)  =\underbrace{1y}_{=g\cdot1}\otimes
_{\CC H}u=g\cdot1\otimes_{\CC H}u$. Compared with $g\cdot
\underbrace{\alpha\left(  u\right)  }_{=1\otimes_{\CC H}u}=g\cdot\left(
1\otimes_{\CC H}u\right)  =g\cdot1\otimes_{\CC H}u$, this yields
$\alpha\left(  gu\right)  =g\cdot\alpha\left(  u\right)  $.
\par
Now, let us forget that we fixed $u$ and $g$. We thus have shown that
$\alpha\left(  gu\right)  =g\cdot\alpha\left(  u\right)  $ for all $u\in
U^{K}$ and $g\in G$. Thus, the map $\alpha$ is a homomorphism of $G$-sets.
Since the map $\alpha$ is also $\CC$-linear, this
yields that $\alpha$ is a $\CC G$-module homomorphism. Qed.}. We will
eventually construct an inverse to $\alpha$; but first we need to prepare.

For every $g\in G$ and every $p\in{\CC }G$, we denote by $\epsilon
_{g}\left(  p\right)  $ the $g$-coordinate of $p$ with respect to the basis
$G$ of the $\CC$-vector space ${\CC }G$. By the definition of
``coordinate'', we have
\begin{equation}
q=\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\ \ \ \ \ \ \ \ \ \ \text{for every }q\in{\CC }G.
\label{sol.Indrho.f.sol1.epsilong.coordinates}%
\end{equation}
For every $g\in G$, we have defined a map $\epsilon_{g}:{\CC %
}G\rightarrow{\CC }$ (because we have defined an element $\epsilon
_{g}\left(  p\right)  $ for every $p\in{\CC }G$). This map $\epsilon
_{g}$ is $\CC$-linear. We notice some basic properties of these maps:

\begin{itemize}
\item For every $g\in G$ and $h\in G$, we have
\begin{equation}
\epsilon_{g}\left(  h\right)  =\delta_{g,h}.
\label{sol.Indrho.f.sol1.epsilong.g}%
\end{equation}
\footnote{This equality has been proven in our solution of Exercise
\ref{exe.induction.ind-as-hom}. (Namely, it appeared there as
(\ref{sol.induction.ind-as-hom.epsilong.g}).)}

\item We have
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in{\CC }G\text{ and }%
q\in{\CC }G. \label{sol.Indrho.f.sol1.trace}%
\end{equation}
\footnote{This equality has been proven in our solution of Exercise
\ref{exe.induction.ind-as-hom}. (Namely, it appeared there as
(\ref{sol.induction.ind-as-hom.trace}).)}

\item We have
\begin{equation}
\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(  q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\text{ and }q\in{\CC }G.
\label{sol.Indrho.f.sol1.epsi1g}%
\end{equation}
\footnote{This equality has been proven in our solution of Exercise
\ref{exe.induction.ind-as-hom}. (Namely, it appeared there as
(\ref{sol.induction.ind-as-hom.epsi1g}).)}
\end{itemize}

Now, for every $\left(  q,u\right)  \in{\CC }G\times U$, we have
$\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u\in U^{K}$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  q,u\right)
\in{\CC }G\times U$. Let $x=\sum\limits_{h\in H}\epsilon_{1}\left(
\rho\left(  h\right)  q\right)  h^{-1}u$. We shall show that $x\in U^{K}$.
\par
Let $k\in K$. Then, the map $H\rightarrow H,\ h\mapsto hk$ is a bijection
(since $H$ is a group). Now, multiplying both sides of the equality
$x=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u$ with $k$ from the left, we obtain%
\begin{align*}
kx  &  =k\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
q\right)  h^{-1}u=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  q\right)  kh^{-1}u\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{\rho\left(  hk\right)
}_{\substack{=\rho\left(  h\right)  \rho\left(  k\right)  \\\text{(since }%
\rho\text{ is a group}\\\text{homomorphism)}}}q\right)  k\underbrace{\left(
hk\right)  ^{-1}}_{=k^{-1}h^{-1}}u\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }hk\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto hk\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
\underbrace{\rho\left(  k\right)  }_{\substack{=1\\\text{(since }k\in
K=\ker\rho\text{)}}}q\right)  \underbrace{kk^{-1}}_{=1}%
h^{-1}u=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
q\right)  h^{-1}u=x.
\end{align*}
\par
Let us now forget that we fixed $k$. We thus have shown that $kx=x$ for every
$k\in K$. In other words, $x$ is an element $y\in U$ satisfying $ky=y$ for
every $k\in K$. Hence,%
\[
x\in\left\{  y\in U\ \mid\ ky=y\text{ for every }k\in K\right\}  =U^{K}.
\]
Thus, $\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u=x\in U^{K}$, qed.}. Hence, we can define a map $\widetilde{\beta
}:{\CC }G\times U\rightarrow U^{K}$ by setting
\[
\widetilde{\beta}\left(  q,u\right)  =\sum_{h\in H}\epsilon_{1}\left(
\rho\left(  h\right)  q\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  q,u\right)  \in{\CC }G\times U.
\]
Consider this map $\widetilde{\beta}$. Then, $\widetilde{\beta}$ is a
$\CC$-bilinear map (because $\widetilde{\beta}\left(  q,u\right)
=\sum_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)  h^{-1}u$
depends $\CC$-linearly on each of $q$ and $u$). We are now going to
prove that the map $\widetilde{\beta}$ is ${\CC }H$-bilinear with
respect to the right ${\CC }H$-module structure on ${\CC }G$ and
the left ${\CC }H$-module structure on $U$.

In fact, every $h^{\prime}\in H$, $q\in{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta}\left(
qh^{\prime},u\right)  . \label{sol.Indrho.f.sol1.Rg.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.Rg.9}):} Let $h^{\prime}\in
H$, $q\in{\CC }G$ and $u\in U$. Then, the map $H\rightarrow H,\ h\mapsto
h^{\prime}h$ is a bijection (since $H$ is a group). The definition of
$\widetilde{\beta}$ yields
\begin{align}
\widetilde{\beta}\left(  q,h^{\prime}u\right)   &  =\sum_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  q\right)  h^{-1}h^{\prime}u=\sum_{h\in
H}\epsilon_{1}\left(  \underbrace{\rho\left(  h^{\prime}h\right)
}_{\substack{=\rho\left(  h^{\prime}\right)  \rho\left(  h\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}q\right)
\underbrace{\left(  h^{\prime}h\right)  ^{-1}}_{=h^{-1}\left(  h^{\prime
}\right)  ^{-1}}h^{\prime}u\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }h^{\prime}h\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{\prime}h\text{ is a
bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{h\in H}\underbrace{\epsilon_{1}\left(  \rho\left(  h^{\prime
}\right)  \rho\left(  h\right)  q\right)  }_{\substack{=\epsilon_{1}\left(
\rho\left(  h\right)  q\rho\left(  h^{\prime}\right)  \right)  \\\text{(by
(\ref{sol.Indrho.f.sol1.trace}), applied to}\\\rho\left(  h^{\prime}\right)
\text{ and }\rho\left(  h\right)  q\text{ instead of }p\text{ and }q\text{)}%
}}h^{-1}\underbrace{\left(  h^{\prime}\right)  ^{-1}h^{\prime}}_{=1}%
u\nonumber\\
&  =\sum_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\rho\left(
h^{\prime}\right)  \right)  h^{-1}u. \label{sol.Indrho.f.sol1.Rg.9.pf.1}%
\end{align}
\par
But the definition of the right $\CC H$-module structure on
$\CC G$ yields $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$. Applying this to $\gamma=q$ and $\eta=h^{\prime}$, we obtain
$qh^{\prime}=q\cdot\left(  \CC \left[  \rho\right]  \right)  \left(
h^{\prime}\right)  $. But $h^{\prime}\in H$ and thus $\left(  \CC %
\left[  \rho\right]  \right)  \left(  h^{\prime}\right)  =\rho\left(
h^{\prime}\right)  $. Hence, $qh^{\prime}=q\cdot\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h^{\prime}\right)  }%
_{=\rho\left(  h^{\prime}\right)  }=q\rho\left(  h^{\prime}\right)  $.
\par
But the definition of $\widetilde{\beta}$ also yields%
\begin{align*}
\widetilde{\beta}\left(  qh^{\prime},u\right)   &  =\sum_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  \underbrace{qh^{\prime}}_{=q\rho\left(
h^{\prime}\right)  }\right)  h^{-1}u=\sum_{h\in H}\epsilon_{1}\left(
\rho\left(  h\right)  q\rho\left(  h^{\prime}\right)  \right)  h^{-1}u\\
&  =\widetilde{\beta}\left(  q,h^{\prime}u\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Indrho.f.sol1.Rg.9.pf.1})}\right)  .
\end{align*}
\par
In other words, $\widetilde{\beta}\left(  q,h^{\prime}u\right)
=\widetilde{\beta}\left(  qh^{\prime},u\right)  $. This proves
(\ref{sol.Indrho.f.sol1.Rg.9}).} As a consequence of this, we can see that
every $r\in{\CC }H$, $q\in{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,ru\right)  =\widetilde{\beta}\left(  qr,u\right)  .
\label{sol.Indrho.f.sol1.Rg.10}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.Rg.10}):} Let
$r\in{\CC }H$, $q\in{\CC }G$ and $u\in U$. We need to prove the
equality (\ref{sol.Indrho.f.sol1.Rg.10}). But this equality is $\CC %
$-linear in $r$. Hence, we can WLOG assume that $r$ belongs to the basis $H$
of the $\CC$-vector space $\CC H$. Assume this. Then,
(\ref{sol.Indrho.f.sol1.Rg.10}) follows from (\ref{sol.Indrho.f.sol1.Rg.9})
(applied to $h^{\prime}=r$).} In other words, the map $\widetilde{\beta}$ is
${\CC }H$-bilinear with respect to the right ${\CC }H$-module
structure on ${\CC }G$ and the left ${\CC }H$-module structure on
$U$ (since we already know that $\widetilde{\beta}$ is ${\CC }%
$-bilinear). Hence, by the universal property of the tensor product, we
conclude that there exists a unique $\CC$-linear map $\beta
:{\CC }G\otimes_{\CC H}U\rightarrow U^{K}$ such that every
$\left(  q,u\right)  \in{\CC }G\times U$ satisfies
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  . \label{sol.Indrho.f.sol1.beta}%
\end{equation}
Consider this map $\beta$. Clearly, every $q\in{\CC }G$ and $u\in U$
satisfy
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  =\sum_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\widetilde{\beta}\right)  . \label{sol.Indrho.f.sol1.beta.explicit}%
\end{equation}


We shall now show that the maps $\alpha$ and $\dfrac{1}{\left\vert
K\right\vert }\beta$ are mutually inverse. To do so, we will show that
$\alpha\circ\beta=\left\vert K\right\vert \operatorname{id}$ and $\beta
\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

Let us first notice that%
\begin{equation}
1\otimes_{\CC H}h^{-1}u=\left(  \rho\left(  h\right)  \right)
^{-1}\otimes_{\CC H}u\ \ \ \ \ \ \ \ \ \ \text{for any }h\in H\text{ and
}u\in U. \label{sol.Indrho.f.sol1.rhohu}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.rhohu}):} Let $h\in H$ and
$u\in U$. We have $h\in H$. Since $H$ is a group, this yields $h^{-1}\in
H\subset\CC H$. Hence, we can move the $h^{-1}$ past the tensor sign in
$1\otimes_{\CC H}h^{-1}u$. We thus obtain $1\otimes_{\CC H}%
h^{-1}u=1h^{-1}\otimes_{\CC H}u$.
\par
But the definition of the right $\CC H$-module structure on
$\CC G$ yields $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$. Applying this to $\gamma=1$ and $\eta=h^{-1}$, we obtain
$1h^{-1}=1\cdot\left(  \CC \left[  \rho\right]  \right)  \left(
h^{-1}\right)  $. But $h^{-1}\in H$ and thus $\left(  \CC \left[
\rho\right]  \right)  \left(  h^{-1}\right)  =\rho\left(  h^{-1}\right)
=\left(  \rho\left(  h\right)  \right)  ^{-1}$ (since $\rho$ is a group
homomorphism). Hence, $1h^{-1}=1\cdot\underbrace{\left(  \CC \left[
\rho\right]  \right)  \left(  h^{-1}\right)  }_{=\left(  \rho\left(  h\right)
\right)  ^{-1}}=1\cdot\left(  \rho\left(  h\right)  \right)  ^{-1}=\left(
\rho\left(  h\right)  \right)  ^{-1}$. Now, $1\otimes_{\CC H}%
h^{-1}u=\underbrace{1h^{-1}}_{=\left(  \rho\left(  h\right)  \right)  ^{-1}%
}\otimes_{\CC H}u=\left(  \rho\left(  h\right)  \right)  ^{-1}%
\otimes_{\CC H}u$. This proves (\ref{sol.Indrho.f.sol1.rhohu}).}

Now, we are going to show that $\alpha\circ\beta=\left\vert K\right\vert
\operatorname{id}$. In fact, let $q\in{\CC }G$ and $u\in U$ be
arbitrary. Then,
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)   &
=\alpha\left(  \beta\left(  q\otimes_{\CC H}u\right)  \right)
=1\otimes_{\CC H}\underbrace{\beta\left(  q\otimes_{{\CC }%
H}u\right)  }_{\substack{=\sum_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  q\right)  h^{-1}u\\\text{(by (\ref{sol.Indrho.f.sol1.beta.explicit}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\right) \\
&  =1\otimes_{\CC H}\left(  \sum_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  q\right)  h^{-1}u\right)  =\underbrace{\sum_{h\in H}}_{=\sum_{g\in
G}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}}\epsilon_{1}\left(
\rho\left(  h\right)  q\right)  \underbrace{1\otimes_{\CC H}h^{-1}%
u}_{\substack{=\left(  \rho\left(  h\right)  \right)  ^{-1}\otimes
_{\CC H}u\\\text{(by (\ref{sol.Indrho.f.sol1.rhohu}))}}}\\
&  =\sum_{g\in G}\underbrace{\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}}_{\substack{=\sum_{h\in\rho^{-1}\left(  g\right)  }}}\epsilon_{1}\left(
\underbrace{\rho\left(  h\right)  }_{=g}q\right)  \left(  \underbrace{\rho
\left(  h\right)  }_{=g}\right)  ^{-1}\otimes_{\CC H}u=\sum_{g\in
G}\underbrace{\sum_{h\in\rho^{-1}\left(  g\right)  }\epsilon_{1}\left(
gq\right)  g^{-1}\otimes_{\CC H}u}_{=\left\vert \rho^{-1}\left(
g\right)  \right\vert \epsilon_{1}\left(  gq\right)  g^{-1}\otimes
_{\CC H}u}\\
&  =\sum_{g\in G}\underbrace{\left\vert \rho^{-1}\left(  g\right)  \right\vert
}_{\substack{=\left\vert K\right\vert \\\text{(by
(\ref{sol.Indrho.f.sol1.preimage}))}}}\epsilon_{1}\left(  gq\right)
g^{-1}\otimes_{\CC H}u=\sum_{g\in G}\left\vert K\right\vert \epsilon
_{1}\left(  gq\right)  g^{-1}\otimes_{\CC H}u\\
&  =\left\vert K\right\vert \left(  \sum_{g\in G}\epsilon_{1}\left(
gq\right)  g^{-1}\right)  \otimes_{\CC H}u=\left\vert K\right\vert
\left(  \sum_{g\in G}\epsilon_{1}\left(  g^{-1}q\right)  \underbrace{\left(
g^{-1}\right)  ^{-1}}_{=g}\right)  \otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \left(  \sum_{g\in G}\underbrace{\epsilon
_{1}\left(  g^{-1}q\right)  }_{\substack{=\epsilon_{g}\left(  q\right)
\\\text{(by (\ref{sol.Indrho.f.sol1.epsi1g}))}}}g\right)  \otimes
_{\CC H}u=\left\vert K\right\vert \underbrace{\left(  \sum_{g\in
G}\epsilon_{g}\left(  q\right)  g\right)  }_{\substack{=q\\\text{(by
(\ref{sol.Indrho.f.sol1.epsilong.coordinates}))}}}\otimes_{\CC H}u\\
&  =\left\vert K\right\vert \underbrace{q\otimes_{\CC H}u}%
_{=\operatorname*{id}\left(  q\otimes_{\CC H}u\right)  }=\left\vert
K\right\vert \operatorname*{id}\left(  q\otimes_{\CC H}u\right)  .
\end{align*}


Now, let us forget that we fixed $q$ and $u$. We thus have shown that $\left(
\alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)  =\left\vert
K\right\vert \operatorname{id}\left(  q\otimes_{\CC H}u\right)  $ for
all $q\in{\CC }G$ and $u\in U$. In other words, the two maps
$\alpha\circ\beta:{\CC }G\otimes_{\CC H}U\rightarrow{\CC %
}G\otimes_{\CC H}U$ and $\left\vert K\right\vert \operatorname{id}%
:{\CC }G\otimes_{\CC H}U\rightarrow{\CC }G\otimes
_{\CC H}U$ are equal to each other on each pure tensor. Since these
two maps are $\CC$-linear, this yields that these two maps
$\alpha\circ\beta:{\CC }G\otimes_{\CC H}U\rightarrow{\CC %
}G\otimes_{\CC H}U$ and $\left\vert K\right\vert \operatorname{id}%
:{\CC }G\otimes_{\CC H}U\rightarrow{\CC }G\otimes
_{\CC H}U$ must be identical. In other words, $\alpha\circ
\beta=\left\vert K\right\vert \operatorname{id}$.

Next, we are going to show that $\beta\circ\alpha=\left\vert K\right\vert
\operatorname{id}$.

We first notice that%
\begin{equation}
h^{-1}u=\left(  \rho\left(  h\right)  \right)  ^{-1}%
u\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H\text{ and }u\in U^{K}.
\label{sol.Indrho.f.sol1.h-1u}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.h-1u}):} Let $h\in H$ and
$u\in U^{K}$. Applying (\ref{sol.Indrho.f.sol1.UK-modstr}) to $h^{-1}$ and $u$
instead of $h$ and $v$, we obtain $\rho\left(  h^{-1}\right)  \cdot u=h^{-1}%
u$. But $\rho$ is a group homomorphism, and thus we have $\left(  \rho\left(
h\right)  \right)  ^{-1}=\rho\left(  h^{-1}\right)  $. Hence,
$\underbrace{\left(  \rho\left(  h\right)  \right)  ^{-1}}_{=\rho\left(
h^{-1}\right)  }u=\rho\left(  h^{-1}\right)  u=\rho\left(  h^{-1}\right)
\cdot u=h^{-1}u$. This proves (\ref{sol.Indrho.f.sol1.h-1u}).}

Let $u\in U^{K}$. The definition of $\alpha\left(  u\right)  $ yields
$\alpha\left(  u\right)  =1\otimes_{\CC H}u$. Applying the map $\beta$
to this equality, we obtain
\begin{align*}
\beta\left(  \alpha\left(  u\right)  \right)   &  =\beta\left(  1\otimes
_{\CC H}u\right)  =\underbrace{\sum_{h\in H}}_{=\sum_{g\in G}%
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}}\epsilon_{1}\left(
\rho\left(  h\right)  1\right)  \underbrace{h^{-1}u}_{\substack{=\left(
\rho\left(  h\right)  \right)  ^{-1}u\\\text{(by (\ref{sol.Indrho.f.sol1.h-1u}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.f.sol1.beta.explicit}), applied to }q=1\right) \\
&  =\sum_{g\in G}\underbrace{\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}}_{=\sum_{h\in\rho^{-1}\left(  g\right)  }}\epsilon_{1}\left(
\underbrace{\rho\left(  h\right)  }_{=g}1\right)  \left(  \underbrace{\rho
\left(  h\right)  }_{=g}\right)  ^{-1}u=\sum_{g\in G}\underbrace{\sum
_{h\in\rho^{-1}\left(  g\right)  }\epsilon_{1}\left(  g1\right)  g^{-1}%
u}_{=\left\vert \rho^{-1}\left(  g\right)  \right\vert \epsilon_{1}\left(
g1\right)  g^{-1}u}\\
&  =\sum_{g\in G}\underbrace{\left\vert \rho^{-1}\left(  g\right)  \right\vert
}_{\substack{=\left\vert K\right\vert \\\text{(by
(\ref{sol.Indrho.f.sol1.preimage}))}}}\epsilon_{1}\left(  g1\right)
g^{-1}u=\sum_{g\in G}\left\vert K\right\vert \epsilon_{1}\left(  g1\right)
g^{-1}u\\
&  =\left\vert K\right\vert \sum_{g\in G}\epsilon_{1}\left(  g1\right)
g^{-1}u=\left\vert K\right\vert \sum_{g\in G}\epsilon_{1}\left(
g^{-1}1\right)  \underbrace{\left(  g^{-1}\right)  ^{-1}}_{=g}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \sum_{g\in G}\underbrace{\epsilon_{1}\left(
g^{-1}1\right)  }_{\substack{=\epsilon_{g}\left(  1\right)  \\\text{(by
(\ref{sol.Indrho.f.sol1.epsi1g}), applied to }q=1\text{)}}}gu=\left\vert
K\right\vert \sum_{g\in G}\epsilon_{g}\left(  1\right)  gu=\left\vert
K\right\vert \underbrace{\left(  \sum_{g\in G}\epsilon_{g}\left(  1\right)
g\right)  }_{\substack{=1\\\text{(because }1=\sum_{g\in G}\epsilon_{g}\left(
1\right)  g\\\text{(by (\ref{sol.Indrho.f.sol1.epsilong.coordinates}), applied
to }q=1\text{))}}}u\\
&  =\left\vert K\right\vert \underbrace{u}_{=\operatorname*{id}\left(
u\right)  }=\left\vert K\right\vert \operatorname*{id}\left(  u\right)  .
\end{align*}
In other words,
$\left(\beta \circ \alpha\right)\left(u\right)
= \left\vert K\right\vert \operatorname*{id}\left(  u\right)$
(since $\left(\beta \circ \alpha\right)\left(u\right)
= \beta\left(\alpha\left(u\right)\right)$).

Now, forget that we fixed $u$. We thus have shown that $\left(  \beta
\circ\alpha\right)  \left(  u\right)  =\left\vert K\right\vert
\operatorname{id}\left(  u\right)  $ for every $u\in U^{K}$. In other words,
$\beta\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

The equalities $\alpha\circ\left(  \dfrac{1}{\left\vert K\right\vert }%
\beta\right)  =\dfrac{1}{\left\vert K\right\vert }\underbrace{\alpha\circ
\beta}_{=\left\vert K\right\vert \operatorname*{id}}=\dfrac{1}{\left\vert
K\right\vert }\cdot\left\vert K\right\vert \operatorname*{id}%
=\operatorname*{id}$ and $\left(  \dfrac{1}{\left\vert K\right\vert }%
\beta\right)  \circ\alpha=\dfrac{1}{\left\vert K\right\vert }\underbrace{\beta
\circ\alpha}_{=\left\vert K\right\vert \operatorname*{id}}=\dfrac
{1}{\left\vert K\right\vert }\cdot\left\vert K\right\vert \operatorname*{id}%
=\operatorname*{id}$ show that the maps $\alpha$ and $\dfrac{1}{\left\vert
K\right\vert }\beta$ are mutually inverse. Hence, the map $\alpha$ is invertible.

Now, we know that the map $\alpha$ is an invertible ${\CC }G$-module
homomorphism. Hence, $\alpha$ is a ${\CC }G$-module isomorphism.
Therefore, there exists a ${\CC }G$-module isomorphism $U^{K}%
\rightarrow{\CC }G\otimes_{\CC H}U$ (namely, $\alpha$).
Therefore, $U^{K}\cong{\CC }G\otimes_{\CC H}U=\operatorname{Ind}%
_{\rho}U$ as ${\CC }G$-modules. This solves Exercise~\ref{exe.Indrho}(f).

[\textit{Remark:} There is another solution of Exercise \ref{exe.Indrho}(f),
which uses the result of Exercise \ref{exe.fixed-vs-coinvariants}(b). Yet
another solution of Exercise \ref{exe.Indrho}(f) relies on
Exercise~\ref{exe.Indrho}(i), and will be given after the solution of the latter.]

(g) Let $\alpha\in R_{\CC}\left(  H\right)  $ and $\beta\in
R_{\CC }\left(  G\right)  $. We notice that%
\begin{equation}
\text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of }G\text{
satisfy }\beta\left(  k\right)  =\beta\left(  k^{\prime}\right)
\label{sol.Indrho.g.bil.beta}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.g.bil.beta}):} We can apply
(\ref{sol.Indrho.RCK}) to $K=G$ and $f=\beta$. As a consequence, we obtain the
following equivalence:%
\begin{align*}
&  \left(  \beta\in R_{\CC }\left(  G\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }G\text{ satisfy }\beta\left(  k\right)  =\beta\left(
k^{\prime}\right)  \right)  .
\end{align*}
Hence, any two conjugate elements $k$ and $k^{\prime}$ of $G$ satisfy
$\beta\left(  k\right)  =\beta\left(  k^{\prime}\right)  $ (because we know
that $\beta\in R_{\CC }\left(  G\right)  $).}.

Let us first prove (\ref{eq.exe.Indrho.g.bil}). The definition of
$\left\langle \Ind_{\rho}\alpha,\beta\right\rangle
_{G}$ yields%
\begin{align}
\left\langle \Ind_{\rho}\alpha,\beta\right\rangle
_{G}  &  =\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}\underbrace{\left(
\Ind_{\rho}\alpha\right)  \left(  g\right)
}_{\substack{=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \\\text{(by the definition of }\Ind_{\rho
}\alpha\text{)}}}\beta\left(  g^{-1}\right)  =\dfrac{1}{\left\vert
G\right\vert }\sum_{g\in G}\left(  \dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \right)  \beta\left(  g^{-1}\right)
\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{g\in G}\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\k\rho\left(  h\right)  k^{-1}=g}}}_{=\sum_{\left(  h,k\right)  \in H\times
G}\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}}\alpha\left(
h\right)  \beta\left(  \underbrace{g^{-1}}_{\substack{=\left(  k\rho\left(
h\right)  k^{-1}\right)  ^{-1}\\\text{(since }g=k\rho\left(  h\right)
k^{-1}\\\text{(since }k\rho\left(  h\right)  k^{-1}=g\text{))}}}\right)
\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{\left(  h,k\right)  \in H\times G}\sum\limits_{\substack{g\in G;\\k\rho\left(
h\right)  k^{-1}=g}}\alpha\left(  h\right)  \beta\left(  \left(  k\rho\left(
h\right)  k^{-1}\right)  ^{-1}\right)  . \label{sol.Indrho.g.bil.1}%
\end{align}
However, for every $\left(  h,k\right)  \in H\times G$, we have
\begin{equation}
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)
^{-1}\right)  =\alpha\left(  h\right)  \left(  \Res_{\rho}\beta\right)
\left(  h^{-1}\right)
\label{sol.Indrho.g.bil.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.g.bil.2}):} Fix a $\left(
h,k\right)  \in H\times G$. Thus, $h\in H$ and $k\in G$. Now, there exists
only one $g\in G$ such that $k\rho\left(  h\right)  k^{-1}=g$ (namely,
$g=k\rho\left(  h\right)  k^{-1}$). Hence, the sum $\sum\limits_{\substack{g\in
G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(  h\right)  \beta\left(
\left(  k\rho\left(  h\right)  k^{-1}\right)  ^{-1}\right)  $ has only one
addend. Hence, this sum simplifies as follows:%
\begin{align}
&  \sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)
^{-1}\right) \nonumber\\
&  =\alpha\left(  h\right)  \beta\left(  \underbrace{\left(  k\rho\left(
h\right)  k^{-1}\right)  ^{-1}}_{=\left(  k^{-1}\right)  ^{-1}\left(
\rho\left(  h\right)  \right)  ^{-1}k^{-1}}\right)  =\alpha\left(  h\right)
\beta\left(  \underbrace{\left(  k^{-1}\right)  ^{-1}}_{=k}\underbrace{\left(
\rho\left(  h\right)  \right)  ^{-1}}_{\substack{=\rho\left(  h^{-1}\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}k^{-1}\right)
=\alpha\left(  h\right)  \beta\left(  k\rho\left(  h^{-1}\right)
k^{-1}\right)  . \label{sol.Indrho.g.bil.2.pf.1}%
\end{align}
\par
But $\rho\left(  h^{-1}\right)  $ and $k\rho\left(  h^{-1}\right)  k^{-1}$ are
two conjugate elements of $G$. Hence, (\ref{sol.Indrho.g.bil.beta}) (applied
to $\rho\left(  h^{-1}\right)  $ and $k\rho\left(  h^{-1}\right)  k^{-1}$
instead of $k$ and $k^{\prime}$) yields $\beta\left(  \rho\left(
h^{-1}\right)  \right)  =\beta\left(  k\rho\left(  h^{-1}\right)
k^{-1}\right)  $. Thus,%
\[
\beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)  ^{-1}\right)
=\beta\left(  \rho\left(  h^{-1}\right)  \right)  =\underbrace{\left(
\beta\circ\rho\right)  }_{\substack{=\Res_{\rho}%
\beta\\\text{(since }\Res_{\rho}\beta\text{ is defined
as }\beta\circ\rho\text{)}}}\left(  h^{-1}\right)  =\left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)  .
\]
Hence, (\ref{sol.Indrho.g.bil.2.pf.1}) becomes%
\[
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)
^{-1}\right)  =\alpha\left(  h\right)  \underbrace{\beta\left(  k\rho\left(
h^{-1}\right)  k^{-1}\right)  }_{=\left(  \Res_{\rho
}\beta\right)  \left(  h^{-1}\right)  }=\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)  .
\]
This proves (\ref{sol.Indrho.g.bil.2}).}. Thus, (\ref{sol.Indrho.g.bil.1})
becomes%
\begin{align*}
\left\langle \Ind_{\rho}\alpha,\beta\right\rangle
_{G}  &  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{\left(  h,k\right)  \in H\times G}}_{=\sum_{h\in H}%
\sum_{k\in G}}\underbrace{\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \beta\left(  \left(  k\rho\left(  h\right)
k^{-1}\right)  ^{-1}\right)  }_{=\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)  }\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\underbrace{\sum_{k\in G}\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)
}_{=\left\vert G\right\vert \cdot\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)
}=\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\left\vert G\right\vert \cdot\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\left(  \Res_{\rho}\beta\right)  \left(
h^{-1}\right)  .
\end{align*}
Compared with%
\begin{align*}
\left\langle \alpha,\Res_{\rho}\beta\right\rangle
_{H}  &  =\dfrac{1}{\left\vert H\right\vert }\sum_{g\in H}\alpha\left(
g\right)  \left(  \Res_{\rho}\beta\right)  \left(
g^{-1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\left\langle \alpha,\Res_{\rho}\beta\right\rangle
_{H}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\left(  \Res_{\rho}\beta\right)  \left(  h^{-1}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}g\text{ as }h\right)  ,
\end{align*}
this yields $\left\langle \Ind_{\rho}\alpha
,\beta\right\rangle _{G}=\left\langle \alpha,\Res_{\rho}\beta
\right\rangle _{H}$. Thus, (\ref{eq.exe.Indrho.g.bil}) is proven.

Let us now prove (\ref{eq.exe.Indrho.g.herm}). (This proof will be very much
similar to the proof of (\ref{eq.exe.Indrho.g.bil}) above, but in a few places
even easier.) The definition of $\left(  \Ind_{\rho
}\alpha,\beta\right)  _{G}$ yields%
\begin{align}
\left(  \Ind_{\rho}\alpha,\beta\right)  _{G}  &
=\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}\underbrace{\left(
\Ind_{\rho}\alpha\right)  \left(  g\right)
}_{\substack{=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \\\text{(by the definition of }\Ind_{\rho
}\alpha\text{)}}}\overline{\beta\left(  g\right)  }=\dfrac{1}{\left\vert
G\right\vert }\sum_{g\in G}\left(  \dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \right)  \overline{\beta\left(  g\right)
}\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{g\in G}\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\k\rho\left(  h\right)  k^{-1}=g}}}_{=\sum_{\left(  h,k\right)  \in H\times
G}\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}}\alpha\left(
h\right)  \overline{\beta\left(  \underbrace{g}_{\substack{=k\rho\left(
h\right)  k^{-1}\\\text{(since }k\rho\left(  h\right)  k^{-1}=g\text{)}%
}}\right)  }\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{\left(  h,k\right)  \in H\times G}\sum\limits_{\substack{g\in G;\\k\rho\left(
h\right)  k^{-1}=g}}\alpha\left(  h\right)  \overline{\beta\left(
k\rho\left(  h\right)  k^{-1}\right)  }. \label{sol.Indrho.g.herm.1}%
\end{align}
However, for every $\left(  h,k\right)  \in H\times G$, we have
\begin{equation}
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \overline{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)
}=\alpha\left(  h\right)  \overline{\left(  \Res_{\rho
}\beta\right)  \left(  h\right)  } \label{sol.Indrho.g.herm.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.g.herm.2}):} Fix a $\left(
h,k\right)  \in H\times G$. Thus, $h\in H$ and $k\in G$. Now, there exists
only one $g\in G$ such that $k\rho\left(  h\right)  k^{-1}=g$ (namely,
$g=k\rho\left(  h\right)  k^{-1}$). Hence, the sum $\sum\limits_{\substack{g\in
G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(  h\right)  \overline
{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)  }$ has only one addend.
Hence, this sum simplifies as follows:%
\begin{equation}
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \overline{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)
}=\alpha\left(  h\right)  \overline{\beta\left(  k\rho\left(  h\right)
k^{-1}\right)  }. \label{sol.Indrho.g.herm.2.pf.1}%
\end{equation}
\par
But $\rho\left(  h\right)  $ and $k\rho\left(  h\right)  k^{-1}$ are two
conjugate elements of $G$. Hence, (\ref{sol.Indrho.g.bil.beta}) (applied to
$\rho\left(  h\right)  $ and $k\rho\left(  h\right)  k^{-1}$ instead of $k$
and $k^{\prime}$) yields $\beta\left(  \rho\left(  h\right)  \right)
=\beta\left(  k\rho\left(  h\right)  k^{-1}\right)  $. Thus,%
\[
\beta\left(  k\rho\left(  h\right)  k^{-1}\right)  =\beta\left(  \rho\left(
h\right)  \right)  =\underbrace{\left(  \beta\circ\rho\right)  }%
_{\substack{=\Res_{\rho}\beta\\\text{(since
}\Res_{\rho}\beta\text{ is defined as }\beta\circ
\rho\text{)}}}\left(  h\right)  =\left(  \Res_{\rho
}\beta\right)  \left(  h\right)  .
\]
Hence, (\ref{sol.Indrho.g.herm.2.pf.1}) becomes%
\[
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \overline{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)
}=\alpha\left(  h\right)  \underbrace{\overline{\beta\left(  k\rho\left(
h\right)  k^{-1}\right)  }}_{\substack{=\overline{\left(  \Res_{\rho}
\beta\right)  \left(  h\right)  }\\\text{(since }\beta\left(
k\rho\left(  h\right)  k^{-1}\right)  =\left(  \Res_{\rho}\beta\right)
\left(  h\right)  \text{)}}}=\alpha\left(
h\right)  \overline{\left(  \Res_{\rho}\beta\right)
\left(  h\right)  }.
\]
This proves (\ref{sol.Indrho.g.herm.2}).}. Thus, (\ref{sol.Indrho.g.herm.1})
becomes%
\begin{align*}
\left(  \Ind_{\rho}\alpha,\beta\right)  _{G}  &
=\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{\left(  h,k\right)  \in H\times G}}_{=\sum_{h\in H}%
\sum_{k\in G}}\underbrace{\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \overline{\beta\left(  k\rho\left(
h\right)  k^{-1}\right)  }}_{=\alpha\left(  h\right)  \overline{\left(
\Res_{\rho}\beta\right)  \left(  h\right)  }}\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\underbrace{\sum_{k\in G}\alpha\left(  h\right)  \overline
{\left(  \Res_{\rho}\beta\right)  \left(  h\right)  }%
}_{=\left\vert G\right\vert \cdot\alpha\left(  h\right)  \overline{\left(
\Res_{\rho}\beta\right)  \left(  h\right)  }}%
=\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\left\vert G\right\vert \cdot\alpha\left(  h\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
h\right)  }\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
h\right)  }.
\end{align*}
Compared with%
\begin{align*}
\left(  \alpha,\Res_{\rho}\beta\right)  _{H}  &
=\dfrac{1}{\left\vert H\right\vert }\sum_{g\in H}\alpha\left(  g\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
g\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\alpha,\Res_{\rho}\beta\right)  _{H}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
h\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}g\text{ as }h\right)  ,
\end{align*}
this yields $\left(  \Ind_{\rho}\alpha,\beta\right)
_{G}=\left(  \alpha,\Res_{\rho}\beta\right)  _{H}$.
Thus, (\ref{eq.exe.Indrho.g.herm}) is proven. The solution of Exercise
\ref{exe.Indrho}(g) is thus complete.

(h) Let $U$ be a $\CC H$-module, and let $V$ be a $\CC G$-module.

Recall that $\Ind_{\rho}U$ is defined as the
$\CC G$-module $\CC G\otimes_{\CC H}U$, where $\CC G$
is regarded as a $\left(  \CC G,\CC H\right)  $-bimodule according
to the following rule: The left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$; the right
$\CC H$-module structure on $\CC G$ is induced by the $\CC %
$-algebra homomorphism $\CC \left[  \rho\right]  :\CC %
H\rightarrow\CC G$ (thus, it is explicitly given by $\gamma\eta
=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$ for all
$\gamma\in\CC G$ and $\eta\in\CC H$). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC G,\CC H\right)
$-bimodule structure.

Now, (\ref{Hom-tensor-adj}) (applied to $R=\CC H$, $S=\CC G$,
$A=\CC G$, $B=U$ and $C=V$) yields
\begin{equation}
\Hom_{\CC G}\left(  \CC G\otimes
_{\CC H}U,V\right)  \cong\Hom_{\CC %
H}\left(  U,\Hom_{\CC G}\left(  \CC %
G,V\right)  \right)  . \label{sol.Indrho.h.adj}%
\end{equation}


We shall now prove that $\Hom_{\CC G}\left(
\CC G,V\right)  \cong\Res_{\rho}V$ as left
$\CC H$-modules.

Indeed, a fundamental fact in abstract algebra says the following: If
$\mathfrak{A}$ is a $\CC$-algebra, and if $\mathfrak{M}$ is a left
$\mathfrak{A}$-module, then there exists a $\CC$-vector space
isomorphism $\Xi:\Hom_{\mathfrak{A}}\left(
\mathfrak{A},\mathfrak{M}\right)  \rightarrow\mathfrak{M}$ which satisfies
$\left(  \Xi\left(  f\right)  =f\left(  1\right)  \text{ for every }%
f\in\Hom_{\mathfrak{A}}\left(  \mathfrak{A}%
,\mathfrak{M}\right)  \right)  $. Applying this fact to $\mathfrak{A}%
=\CC G$ and $\mathfrak{M}=V$, we conclude that there exists a
$\CC$-vector space isomorphism
$\Xi:\Hom_{\CC G}\left(  \CC G,V\right)  \rightarrow V$ which
satisfies \newline
$\left(  \Xi\left(  f\right)  =f\left(  1\right)  \text{ for every
}f\in\Hom_{\CC G}\left(  \CC G,V\right)
\right)  $. Consider this $\Xi$.

The map $\Xi$ is a homomorphism of $H$-sets from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
\ \ \ \ \footnote{\textit{Proof.} The map $\Xi$ is clearly a
map from $\Hom_{\CC G}\left(  \CC G,V\right)  $ to $V$,
therefore a map from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
(since $\Res_{\rho}V=V$ as sets).
\par
Recall that the $\CC H$-module structure on $\Res_{\rho}V$ is given by
\begin{equation}
h\cdot v=\rho\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \text{for every
}h\in H\text{ and }v\in V. \label{sol.Indrho.h.Res}%
\end{equation}
\par
Now, let $f\in\Hom_{\CC G}\left(  \CC %
G,V\right)  $ and $h\in H$. We are going to prove that $\Xi\left(  h\cdot
f\right)  =h\cdot\Xi\left(  f\right)  $, where $h\cdot\Xi\left(  f\right)  $
is computed in the $\CC H$-module $\Res_{\rho}%
V$.
\par
Indeed, the definition of the left $\CC H$-module structure on
$\Hom_{\CC G}\left(  \CC G,V\right)  $
yields $\left(  \eta\cdot\alpha\right)  \left(  p\right)  =\alpha\left(
p\eta\right)  $ for every $\eta\in\CC H$,
$\alpha\in\Hom_{\CC G}\left(  \CC G,V\right)  $ and $p\in\CC G$.
Applying this to $\eta=h$, $\alpha=f$ and $p=1$, we obtain $\left(  h\cdot
f\right)  \left(  1\right)  =f\left(  1h\right)  $ (since $h\in H\subset
\CC H$). But the definition of $\Xi$ yields $\Xi\left(  h\cdot f\right)
=\left(  h\cdot f\right)  \left(  1\right)  =f\left(  1h\right)  $.
\par
Recall that the right $\CC H$-module structure on $\CC G$ is given
by the equality $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying this equality to $\gamma=1$ and $\eta=h$, we obtain $1h=1\cdot\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  $ (since $h\in
H\subset\CC H$). But $h\in H$ and thus $\left(  \CC \left[
\rho\right]  \right)  \left(  h\right)  =\rho\left(  h\right)  $. Thus,
$1h=1\cdot\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
=\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
=\rho\left(  h\right)  =\rho\left(  h\right)  \cdot1$. Now, $\Xi\left(  h\cdot
f\right)  =f\left(  \underbrace{1h}_{=\rho\left(  h\right)  \cdot1}\right)
=f\left(  \rho\left(  h\right)  \cdot1\right)  =\rho\left(  h\right)  \cdot
f\left(  1\right)  $ (since $f$ is a $\CC G$-module homomorphism
(because $f\in\Hom_{\CC G}\left(  \CC %
G,V\right)  $) and since $\rho\left(  h\right)  \in G\subset\CC G$).
\par
On the other hand, (\ref{sol.Indrho.h.Res}) (applied to $v=f\left(  1\right)
$) yields $h\cdot f\left(  1\right)  =\rho\left(  h\right)  \cdot f\left(
1\right)  $. Compared with $\Xi\left(  h\cdot f\right)  =\rho\left(  h\right)
\cdot f\left(  1\right)  $, this yields $\Xi\left(  h\cdot f\right)  =h\cdot
f\left(  1\right)  $.
\par
But the definition of $\Xi$ yields $\Xi\left(  f\right)  =f\left(  1\right)
$. Thus, $f\left(  1\right)  =\Xi\left(  f\right)  $, so that $\Xi\left(
h\cdot f\right)  =h\cdot\underbrace{f\left(  1\right)  }_{=\Xi\left(
f\right)  }=h\cdot\Xi\left(  f\right)  $.
\par
Now, let us forget that we fixed $f$ and $h$. We thus have shown that
$\Xi\left(  h\cdot f\right)  =h\cdot\Xi\left(  f\right)  $ for every
$f\in\Hom_{\CC G}\left(  \CC G,V\right)  $
and $h\in H$. In other words, $\Xi$ is a homomorphism of $H$-sets from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to
$\Res_{\rho}V$, qed.}. Therefore, $\Xi$ is a
$\CC H$-module homomorphism from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
(since $\Xi$ is a $\CC$-linear map). Consequently,
$\Xi$ is a $\CC H$-module isomorphism from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
(since $\Xi$ is a $\CC$-vector space isomorphism).
Therefore, $\Hom_{\CC G}\left(  \CC %
G,V\right)  \cong\Res_{\rho}V$ as $\CC %
H$-modules. Now,%
\begin{align*}
\Hom_{\CC G}\left(
\underbrace{\Ind_{\rho}U}_{=\CC G\otimes
_{\CC H}U},V\right)   &  =\Hom_{\CC %
G}\left(  \CC G\otimes_{\CC H}U,V\right) \\
&  \cong\Hom_{\CC H}\left(
U,\underbrace{\Hom_{\CC G}\left(  \CC %
G,V\right)  }_{\cong\Res_{\rho}V\text{ as }%
\CC H\text{-modules}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.h.adj})}\right) \\
&  \cong\Hom_{\CC H}\left(
U,\Res_{\rho}V\right)  .
\end{align*}
This solves Exercise \ref{exe.Indrho}(h).

(i) The following solution will mostly be an imitation of the solution of
Exercise \ref{exe.induction.ind-as-hom}.

Let $U$ be any $\CC H$-module. Recall that $\operatorname*{Ind}%
\nolimits_{\rho}U$ is defined as the $\CC G$-module $\CC %
G\otimes_{\CC H}U$, where $\CC G$ is regarded as a $\left(
\CC G,\CC H\right)  $-bimodule according to the following rule:
The left $\CC G$-module structure on $\CC G$ is plain
multiplication inside $\CC G$; the right $\CC H$-module structure
on $\CC G$ is induced by the $\CC$-algebra homomorphism
$\CC \left[  \rho\right]  :\CC H\rightarrow\CC G$ (thus, it
is explicitly given by $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$). From now on, we regard $\CC G$ as endowed with this
$\left(  \CC G,\CC H\right)  $-bimodule structure (besides the
$\left(  \CC H,\CC G\right)  $-bimodule structure that was
introduced in the statement of Exercise \ref{exe.Indrho}(i)).

The $\CC$-vector space $\CC G$ is thus endowed with a left
$\CC H$-module structure (which is part of the $\left(  \CC %
G,\CC H\right)  $-bimodule structure) and with a right $\CC %
H$-module structure (which is part of the $\left(  \CC H,\CC %
G\right)  $-bimodule structure). These two structures, combined, form a
$\left(  \CC H,\CC H\right)  $-bimodule structure\footnote{This is
easy to check (we leave the details of this verification to the reader).}.
This allows us to write expressions like $xyz$ with $x\in\CC H$,
$y\in\CC G$ and $z\in\CC H$, without having to disambiguate
whether they mean $\left(  xy\right)  z$ or $x\left(  yz\right)  $.

We recall that $\Hom_{\CC H}\left(
\CC G,U\right)  $ is the left $\CC G$-module consisting of all
left $\CC H$-module homomorphisms from $\CC G$ to $U$. This uses
only the $\left(  \CC H,\CC G\right)  $-bimodule structure on
$\CC G$ that was introduced in the statement of Exercise
\ref{exe.Indrho}(i) (but not the $\left(  \CC G,\CC H\right)
$-bimodule structure on $\CC G$ that was introduced in the definition of
$\Ind_{\rho}U$).

Recall that the right $\CC H$-module structure on $\CC G$ is given
by%
\begin{equation}
\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)
\eta\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\CC G\text{ and }\eta
\in\CC H. \label{sol.Indrho.i.rightCH}%
\end{equation}
For every $\gamma\in\CC G$ and $\eta\in H$, we have%
\begin{align}
\gamma\eta &  =\gamma\cdot\underbrace{\left(  \CC \left[  \rho\right]
\right)  \eta}_{\substack{=\rho\left(  \eta\right)  \\\text{(since }\eta\in
H\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.i.rightCH})
(since }\eta\in H\subset\CC H\text{)}\right) \nonumber\\
&  =\gamma\cdot\rho\left(  \eta\right)  . \label{sol.Indrho.i.rightCH2}%
\end{align}
On the other hand, the left $\CC H$-module structure on $\CC G$ is
induced by the $\CC$-algebra homomorphism $\CC \left[
\rho\right]  :\CC H\rightarrow\CC G$. In other words, this
structure is given by
\begin{equation}
\eta\gamma=\left(  \CC \left[  \rho\right]  \right)  \eta\cdot
\gamma\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\CC G\text{ and }%
\eta\in\CC H. \label{sol.Indrho.i.leftCH}%
\end{equation}
For every $\gamma\in\CC G$ and $\eta\in H$, we have%
\begin{align}
\eta\gamma &  =\underbrace{\left(  \CC \left[  \rho\right]  \right)
\eta}_{\substack{=\rho\left(  \eta\right)  \\\text{(since }\eta\in H\text{)}%
}}\cdot\gamma\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.i.leftCH})
(since }\eta\in H\subset\CC H\text{)}\right) \nonumber\\
&  =\rho\left(  \eta\right)  \cdot\gamma. \label{sol.Indrho.i.leftCH2}%
\end{align}


The map $\rho$ is a group homomorphism. Hence, $\rho\left(  H\right)  $ is a
subgroup of $G$. Let us denote this subgroup by $\overline{H}$. Thus,
$\overline{H}=\rho\left(  H\right)  $.

Let $J$ be a system of distinct representatives for the right $\overline{H}%
$-cosets in $G$. Then, $G=\bigsqcup_{j\in J}\overline{H}j$.

For every $g\in G$, define a map $\mathfrak{R}_{g}:J\rightarrow J$ as follows:
Let $i\in J$. Then, $ig\in G=\bigsqcup_{j\in J}\overline{H}j$. Thus, there
exists a unique $j\in J$ such that $ig\in\overline{H}j$. Define $\mathfrak{R}%
_{g}\left(  i\right)  $ to be this $j$. Hence, we have defined $\mathfrak{R}%
_{g}\left(  i\right)  $ for every $i\in J$. Thus, we have defined a map
$\mathfrak{R}_{g}:J\rightarrow J$.

For every $g\in G$ and $i\in J$, we have
\begin{equation}
ig\in\overline{H}\cdot\mathfrak{R}_{g}\left(  i\right)
\label{sol.Indrho.i.Rg.def}%
\end{equation}
(because $\mathfrak{R}_{g}\left(  i\right)  $ is defined as the $j\in J$
satisfying $ig\in\overline{H}j$).

It is easy to see that for every $g\in G$, the map $\mathfrak{R}%
_{g}:J\rightarrow J$ is a bijection.\footnote{\textit{Proof.} Fix $g\in G$.
\par
The sets $\overline{H}j$ for all $j\in J$ are disjoint (since
$\bigsqcup_{j\in J}\overline{H}j$ is well-defined). In other words, if $i$ and
$i^{\prime}$ are two elements of $J$ such that $\overline{H}i$ and
$\overline{H}i^{\prime}$ are not disjoint, then
\begin{equation}
i=i^{\prime}. \label{sol.Indrho.i.Rg.ind-as-hom-Lg.1}%
\end{equation}
\par
Let us now prove that $\mathfrak{R}_{g}$ is injective.
\par
Let $i$ and $i^{\prime}$ be two elements of $J$. Assume that $\mathfrak{R}%
_{g}\left(  i\right)  =\mathfrak{R}_{g}\left(  i^{\prime}\right)  $. We will
show that $i=i^{\prime}$.
\par
From (\ref{sol.Indrho.i.Rg.def}), we have $ig\in\overline{H}\cdot
\mathfrak{R}_{g}\left(  i\right)  $. Hence, there exists some $h\in
\overline{H}$ such that $ig=h\cdot\mathfrak{R}_{g}\left(  i\right)  $.
Consider this $h$. We have
\[
\overline{H}\underbrace{ig}_{=h\cdot\mathfrak{R}_{g}\left(  i\right)
}=\underbrace{\overline{H}h}_{\substack{=\overline{H}\\\text{(since }%
h\in\overline{H}\text{ and since}\\\overline{H}\text{ is a group)}}%
}\cdot\mathfrak{R}_{g}\left(  i\right)  =\overline{H}\cdot\mathfrak{R}%
_{g}\left(  i\right)  .
\]
The same argument (but for $i^{\prime}$ instead of $i$) yields $\overline
{H}i^{\prime}g=\overline{H}\cdot\mathfrak{R}_{g}\left(  i^{\prime}\right)  $.
Hence, $\overline{H}ig=\overline{H}\cdot\underbrace{\mathfrak{R}_{g}\left(
i\right)  }_{=\mathfrak{R}_{g}\left(  i^{\prime}\right)  }=\overline{H}%
\cdot\mathfrak{R}_{g}\left(  i^{\prime}\right)  =\overline{H}i^{\prime}g$.
Thus, $\underbrace{\overline{H}ig}_{=\overline{H}i^{\prime}g}g^{-1}%
=\overline{H}i^{\prime}gg^{-1}=\overline{H}i^{\prime}$, so that $\overline
{H}i^{\prime}=\overline{H}igg^{-1}=\overline{H}i$. Thus, the sets
$\overline{H}i$ and $\overline{H}i^{\prime}$ are not disjoint (because
$\left(  \overline{H}i\right)  \cap\underbrace{\left(  \overline{H}i^{\prime
}\right)  }_{= \overline{H}i}
=\left(  \overline{H}i\right)  \cap\left(  \overline
{H}i\right)  =\overline{H}i\neq\varnothing$). Therefore, $i=i^{\prime}$ (by
(\ref{sol.Indrho.i.Rg.ind-as-hom-Lg.1})).
\par
Now, forget that we fixed $i$ and $i^{\prime}$. We thus have shown that if $i$
and $i^{\prime}$ are two elements of $J$ such that $\mathfrak{R}_{g}\left(
i\right)  =\mathfrak{R}_{g}\left(  i^{\prime}\right)  $, then $i=i^{\prime}$.
In other words, the map $\mathfrak{R}_{g}$ is injective. But the set $J$ is
finite (since it is a subset of the finite set $G$). Hence, $\mathfrak{R}_{g}$
is a map from a finite set (namely, $J$) to itself. Since we know that
$\mathfrak{R}_{g}$ is injective, this yields that $\mathfrak{R}_{g}$ is
surjective (because any injective map from a finite set to itself must be
surjective). Hence, $\mathfrak{R}_{g}$ is bijective (since $\mathfrak{R}_{g}$
is injective and surjective), that is, a bijection, qed.} We have
\begin{equation}
j^{-1}\otimes_{\CC H}f\left(  jg\right)  =g\cdot\left(  \mathfrak{R}%
_{g}\left(  j\right)  \right)  ^{-1}\otimes_{\CC H}f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{in
}{\CC }G\otimes_{\CC H}U \label{sol.Indrho.i.Rg.property}%
\end{equation}
for every $g\in G$, every $j\in J$ and every $f\in\Hom %
_{\CC H}\left(  {\CC}G,U\right)  $%
.\ \ \ \ \footnote{\textit{Proof of (\ref{sol.Indrho.i.Rg.property}):} Let
$g\in G$, $j\in J$ and $f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  $. Applying (\ref{sol.Indrho.i.Rg.def}) to $i=j$, we
obtain $jg\in\overline{H}\cdot\mathfrak{R}_{g}\left(  j\right)  $. In other
words, there exists an $h\in\overline{H}$ such that $jg=h\cdot\mathfrak{R}%
_{g}\left(  j\right)  $. Consider this $h$. We have
\[
j^{-1}\underbrace{jg}_{=h\cdot\mathfrak{R}_{g}\left(  j\right)  }\cdot\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}=j^{-1}h\cdot
\underbrace{\mathfrak{R}_{g}\left(  j\right)  \cdot\left(  \mathfrak{R}%
_{g}\left(  j\right)  \right)  ^{-1}}_{=1}=j^{-1}h,
\]
hence
\[
j^{-1}h=j^{-1}jg\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}.
\]
\par
But $h\in\overline{H}=\rho\left(  H\right)  $. Hence, there exists a
$h^{\prime}\in H$ such that $h=\rho\left(  h^{\prime}\right)  $. Consider this
$h^{\prime}$.
\par
Applying (\ref{sol.Indrho.i.leftCH2}) to $\eta=h^{\prime}$ and $\gamma
=\mathfrak{R}_{g}\left(  j\right)  $, we obtain $h^{\prime}\cdot
\mathfrak{R}_{g}\left(  j\right)  =\underbrace{\rho\left(  h^{\prime}\right)
}_{=h}\cdot\mathfrak{R}_{g}\left(  j\right)  =h\cdot\mathfrak{R}_{g}\left(
j\right)  =jg$ (since $jg=h\cdot\mathfrak{R}_{g}\left(  j\right)  $).
\par
But the map $f$ is left $\CC H$-linear (since $f\in\Hom %
_{\CC H}\left(  {\CC}G,U\right)  $). Thus, $f\left(
h^{\prime}\cdot\mathfrak{R}_{g}\left(  j\right)  \right)  =h^{\prime}\cdot
f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  $ (since $h^{\prime}\in
H\subset\CC H$). Since $h^{\prime}\cdot\mathfrak{R}_{g}\left(  j\right)
=jg$, this rewrites as $f\left(  jg\right)  =h^{\prime}\cdot f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  $. Hence,
\begin{equation}
j^{-1}\otimes_{\CC H}\underbrace{f\left(  jg\right)  }_{=h^{\prime
}\cdot f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  }=j^{-1}%
\otimes_{\CC H}h^{\prime}\cdot f\left(  \mathfrak{R}_{g}\left(
j\right)  \right)  =j^{-1}h^{\prime}\otimes_{\CC H}f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)
\label{sol.Indrho.i.Rg.property.pf.3}%
\end{equation}
(here, we have moved $h^{\prime}$ past the $\otimes_{\CC H}$ sign,
since $h^{\prime}\in H\subset{\CC }H$).
\par
On the other hand, applying (\ref{sol.Indrho.i.rightCH2}) to $\gamma=j^{-1}$
and $\eta=h^{\prime}$, we obtain $j^{-1}h^{\prime}=j^{-1}\cdot\underbrace{\rho
\left(  h^{\prime}\right)  }_{=h}=j^{-1}h=g\cdot\left(  \mathfrak{R}%
_{g}\left(  j\right)  \right)  ^{-1}$. Hence,
(\ref{sol.Indrho.i.Rg.property.pf.3}) becomes%
\[
j^{-1}\otimes_{\CC H}f\left(  jg\right)  =\underbrace{j^{-1}h^{\prime
}}_{=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}}%
\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}\otimes
_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  .
\]
This proves (\ref{sol.Indrho.i.Rg.property}).}

We now define a map $\alpha:\Hom_{\CC H}\left(
{\CC}G,U\right)  \rightarrow{\CC }G\otimes_{\CC H}U$ by
setting
\[
\alpha\left(  f\right)  =\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  \qquad\text{ for all }f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  .
\]
This $\alpha$ is a map $\Hom_{\CC H}\left(  {\CC %
}G,U\right)  \rightarrow\operatorname{Ind}_{\rho}U$ (since $\operatorname{Ind}%
_{\rho}U={\CC}G\otimes_{\CC H}U$).

We will show that $\alpha$ is a ${\CC }G$-module isomorphism.

First, let us prove that $\alpha$ is a left $\CC  G$-module
homomorphism. In fact, any $f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  $ and $g\in G$ satisfy
\begin{align*}
\alpha\left(  gf\right)   &  =\sum_{j\in J}j^{-1}\otimes_{{\CC }%
H}\underbrace{\left(  gf\right)  \left(  j\right)  }_{\substack{=f\left(
jg\right)  \\\text{(by the definition of }gf\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\left(
gf\right)  \right) \\
&  =\sum_{j\in J}\underbrace{j^{-1}\otimes_{\CC H}f\left(  jg\right)
}_{\substack{=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)
\right)  \\\text{(by (\ref{sol.Indrho.i.Rg.property}))}}}=\sum_{j\in J}%
g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}\otimes
_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
=\sum_{j\in J}g\cdot j^{-1}\otimes_{\CC H}f\left(  j\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }j\text{ for }\mathfrak{R}_{g}\left(
j\right)  \text{ in the sum, since}\\
\text{the map }\mathfrak{R}_{g}:J\rightarrow J\text{ is a bijection}%
\end{array}
\right) \\
&  =g\cdot\underbrace{\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  }_{=\alpha\left(  f\right)  }=g\cdot\alpha\left(  f\right)  .
\end{align*}
The map $\alpha$ is thus a homomorphism of left $G$-sets. Since $\alpha$ is
furthermore $\CC$-linear, this yields that $\alpha$ is a left
${\CC }G$-module homomorphism.

We now are going to construct an inverse for $\alpha$. This will be more cumbersome.

For every $g\in G$ and every $p\in{\CC }G$, we denote by $\epsilon
_{g}\left(  p\right)  $ the $g$-coordinate of $p$ with respect to the basis
$G$ of the $\CC$-vector space ${\CC }G$. By the definition of
``coordinate'', we have
\begin{equation}
q=\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\ \ \ \ \ \ \ \ \ \ \text{for every }q\in{\CC }G.
\label{sol.Indrho.i.epsilong.coordinates}%
\end{equation}
For every $g\in G$, we have defined a map $\epsilon_{g}:{\CC %
}G\rightarrow{\CC }$ (because we have defined an element $\epsilon
_{g}\left(  p\right)  $ for every $p\in{\CC }G$). This map $\epsilon
_{g}$ is $\CC$-linear. We record some properties of these maps:

\begin{itemize}
\item For every $g\in G$ and $h\in G$, we have
\begin{equation}
\epsilon_{g}\left(  h\right)  =\delta_{g,h}. \label{sol.Indrho.i.epsilong.g}%
\end{equation}
\footnote{This equality has been proven in our solution of Exercise
\ref{exe.induction.ind-as-hom}. (Namely, it appeared there as
(\ref{sol.induction.ind-as-hom.epsilong.g}).)}

\item We have%
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in{\CC }G\text{ and }%
q\in{\CC }G. \label{sol.Indrho.i.trace}%
\end{equation}
\footnote{This equality has been proven in our solution of Exercise
\ref{exe.induction.ind-as-hom}. (Namely, it appeared there as
(\ref{sol.induction.ind-as-hom.trace}).)}

\item We have%
\begin{equation}
\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(  q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\text{ and }q\in{\CC }G.
\label{sol.Indrho.i.epsi1g}%
\end{equation}
\footnote{This equality has been proven in our solution of Exercise
\ref{exe.induction.ind-as-hom}. (Namely, it appeared there as
(\ref{sol.induction.ind-as-hom.epsi1g}).)}
\end{itemize}

Now, fix $q\in{\CC }G$ and $u\in U$. We let $f_{q,u}$ be the map
${\CC }G\rightarrow U$ defined by
\begin{equation}
f_{q,u}\left(  p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  pq\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \text{for every }%
p\in{\CC }G. \label{sol.Indrho.i.fqu}%
\end{equation}
It is obvious that this map $f_{q,u}$ is $\CC$-linear. We will show
that $f_{q,u}$ is a left ${\CC }H$-module homomorphism.

The map $f_{q,u}$ is a homomorphism of left $H$-sets\footnote{\textit{Proof.}
In fact, for every $h^{\prime}\in H$ and every $p\in{\CC }G$, we have
\begin{align*}
f_{q,u}\left(  h^{\prime}p\right)   &  =\sum\limits_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  \underbrace{\left(  h^{\prime}p\right)
}_{\substack{=\rho\left(  h^{\prime}\right)  \cdot p\\\text{(by
(\ref{sol.Indrho.i.leftCH2}), applied}\\\text{to }\gamma=p\text{ and }%
\eta=h^{\prime}\text{)}}}q\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }f_{q,u}\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  \rho\left(
h^{\prime}\right)  \cdot pq\right)  h^{-1}u=\sum\limits_{h\in H}\epsilon
_{1}\left(  \underbrace{\rho\left(  h\left(  h^{\prime}\right)  ^{-1}\right)
}_{\substack{=\rho\left(  h\right)  \left(  \rho\left(  h^{\prime}\right)
\right)  ^{-1}\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}%
}}\rho\left(  h^{\prime}\right)  \cdot pq\right)  \underbrace{\left(  h\left(
h^{\prime}\right)  ^{-1}\right)  ^{-1}}_{=\left(  \left(  h^{\prime}\right)
^{-1}\right)  ^{-1}h^{-1}=h^{\prime}h^{-1}}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }h\left(  h^{\prime}\right)  ^{-1}\text{ for
}h\text{ in the sum,}\\
\text{because the map }H\rightarrow H,\ h\mapsto h\left(  h^{\prime}\right)
^{-1}\text{ is a bijection}\\
\text{(since }H\text{ is a group and since }h^{\prime}\in H\text{)}%
\end{array}
\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
\underbrace{\left(  \rho\left(  h^{\prime}\right)  \right)  ^{-1}\rho\left(
h^{\prime}\right)  }_{=1}\cdot pq\right)  h^{\prime}h^{-1}u\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  pq\right)
h^{\prime}h^{-1}u=h^{\prime}\cdot\underbrace{\sum\limits_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  pq\right)  h^{-1}u}_{\substack{=f_{q,u}%
\left(  p\right)  \\\text{(by (\ref{sol.Indrho.i.fqu}))}}}=h^{\prime}\cdot
f_{q,u}\left(  p\right)  .
\end{align*}
In other words, $f_{q,u}$ is a homomorphism of left $H$-sets, qed.}. Since
$f_{q,u}$ is furthermore $\CC$-linear, this yields that $f_{q,u}$ is
a left $\CC H$-module homomorphism. Hence, $f_{q,u}\in\Hom %
_{\CC H}\left(  {\CC }G,U\right)  $.

Now, forget that we fixed $q$ and $u$. We thus have defined a map $f_{q,u}%
\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $ for
every $q\in{\CC }G$ and $u\in U$. It is easy to see that this map
$f_{q,u}$ depends $\CC$-linearly on each of $q$ and $u$. Now, define
a map $\widetilde{\beta}:{\CC }G\times U\rightarrow\Hom %
_{\CC H}\left(  {\CC }G,U\right)  $ by
\[
\widetilde{\beta}\left(  q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ \text{for
every }\left(  q,u\right)  \in{\CC }G\times U.
\]
Then, $\widetilde{\beta}$ is a $\CC$-bilinear map (because
$\widetilde{\beta}\left(  q,u\right)  =f_{q,u}$ depends ${\CC }%
$-linearly on each of $q$ and $u$). We are now going to prove that the map
$\widetilde{\beta}$ is ${\CC }H$-bilinear with respect to the right
${\CC }H$-module structure on ${\CC }G$ and the left ${\CC %
}H$-module structure on $U$.

In fact, every $h^{\prime}\in H$, $q\in{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta}\left(
qh^{\prime},u\right)  . \label{sol.Indrho.i.Rg.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.Rg.9}):} Let $h^{\prime}\in H$,
$q\in{\CC }G$ and $u\in U$. Then, the map $H\rightarrow H,\ h\mapsto
h^{\prime}h$ is a bijection (since $H$ is a group). Now, let $p\in{\CC %
}G$. The definition of $\widetilde{\beta}$ yields $\widetilde{\beta}\left(
q,h^{\prime}u\right)  =f_{q,h^{\prime}u}$. Hence,
\begin{align}
\left(  \underbrace{\widetilde{\beta}\left(  q,h^{\prime}u\right)
}_{=f_{q,h^{\prime}u}}\right)  \left(  p\right)   &  =f_{q,h^{\prime}u}\left(
p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
pq\right)  h^{-1}h^{\prime}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }f_{q,h^{\prime}u}\right) \nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{\rho\left(  h^{\prime
}h\right)  }_{\substack{=\rho\left(  h^{\prime}\right)  \rho\left(  h\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}pq\right)
\underbrace{\left(  h^{\prime}h\right)  ^{-1}}_{=h^{-1}\left(  h^{\prime
}\right)  ^{-1}}h^{\prime}u\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }h^{\prime}h\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{\prime}h\text{ is a
bijection}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{h\in H}\underbrace{\epsilon_{1}\left(  \rho\left(  h^{\prime
}\right)  \rho\left(  h\right)  pq\right)  }_{\substack{=\epsilon_{1}\left(
\rho\left(  h\right)  pq\rho\left(  h^{\prime}\right)  \right)  \\\text{(by
(\ref{sol.Indrho.i.trace}), applied to}\\\rho\left(  h^{\prime}\right)  \text{
and }\rho\left(  h\right)  pq\text{ instead of }p\text{ and }q\text{)}}%
}h^{-1}\underbrace{\left(  h^{\prime}\right)  ^{-1}h^{\prime}}_{=1}%
u\nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
pq\rho\left(  h^{\prime}\right)  \right)  h^{-1}u.
\label{sol.Indrho.i.Rg.9.pf.1}%
\end{align}
But the definition of $\widetilde{\beta}$ also yields $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =f_{qh^{\prime},u}$. Hence,
\begin{align*}
\left(  \widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(
p\right)   &  =f_{qh^{\prime},u}\left(  p\right)  =\sum\limits_{h\in
H}\epsilon_{1}\left(  \rho\left(  h\right)  p\underbrace{\left(  qh^{\prime
}\right)  }_{\substack{=q\cdot\rho\left(  h^{\prime}\right)  \\\text{(by
(\ref{sol.Indrho.i.rightCH2}), applied to}\\\gamma=q\text{ and }\eta
=h^{\prime}\text{)}}}\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }f_{qh^{\prime},u}\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  pq\cdot
\rho\left(  h^{\prime}\right)  \right)  h^{-1}u=\sum\limits_{h\in H}%
\epsilon_{1}\left(  \rho\left(  h\right)  pq\rho\left(  h^{\prime}\right)
\right)  h^{-1}u=\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)
\right)  \left(  p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.i.Rg.9.pf.1})}\right)  .
\end{align*}
\par
Now, forget that we fixed $p$. We have thus proven that $\left(
\widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(  p\right)
=\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)  \right)  \left(
p\right)  $ for every $p\in{\CC }G$. In other words, $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =\widetilde{\beta}\left(  q,h^{\prime}u\right)
$, so that $\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta
}\left(  qh^{\prime},u\right)  $. This proves (\ref{sol.Indrho.i.Rg.9}).} As a
consequence of this, we can see that every $r\in{\CC }H$, $q\in
{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,ru\right)  =\widetilde{\beta}\left(  qr,u\right)  .
\label{sol.Indrho.i.Rg.10}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.Rg.10}):} Let $r\in\CC H$,
$q\in\CC G$ and $u\in U$. We need to prove the equality
(\ref{sol.Indrho.i.Rg.10}). But this equality is $\CC$-linear in $r$.
Hence, we can WLOG assume that $r$ belongs to the basis $H$ of the
$\CC$-vector space $\CC H$. Assume this. Then,
(\ref{sol.Indrho.i.Rg.10}) follows from (\ref{sol.Indrho.i.Rg.9}) (applied to
$h^{\prime}=r$).} In other words, the map $\widetilde{\beta}$ is ${\CC %
}H$-bilinear with respect to the right ${\CC }H$-module structure on
${\CC }G$ and the left ${\CC }H$-module structure on $U$ (since we
already know that $\widetilde{\beta}$ is $\CC$-bilinear). Hence, by
the universal property of the tensor product, we conclude that there exists a
unique $\CC$-linear map $\beta:{\CC }G\otimes_{{\CC }%
H}U\rightarrow\Hom_{\CC H}\left(  {\CC %
}G,U\right)  $ such that every $\left(  q,u\right)  \in{\CC }G\times U$
satisfies
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  . \label{sol.Indrho.i.beta}%
\end{equation}
Consider this map $\beta$. Clearly, every $q\in{\CC }G$ and $u\in U$
satisfy
\[
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{\beta}\right)  .
\]
Hence, every $q\in{\CC }G$, $u\in U$ and $p\in{\CC }G$ satisfy
\begin{equation}
\left(  \underbrace{\beta\left(  q\otimes_{\CC H}u\right)  }%
_{=f_{q,u}}\right)  \left(  p\right)  =f_{q,u}\left(  p\right)  =\sum
\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  pq\right)  h^{-1}u
\label{sol.Indrho.i.beta.explicit}%
\end{equation}
(by (\ref{sol.Indrho.i.fqu})).

Let $K=\ker\rho$. Thus, $K$ is the kernel of a group
homomorphism out of $H$ (since $\rho$ is a group homomorphism out of $H$), and
therefore a normal subgroup of $H$.

We notice that%
\begin{equation}
\left\vert \rho^{-1}\left(  g\right)  \right\vert =\left\vert K\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\overline{H}.
\label{sol.Indrho.i.preimage}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.preimage}):} Let $g\in
\overline{H}$. Then, $g\in\overline{H}=\rho\left(  H\right)  $. Hence,
$g=\rho\left(  x\right)  $ for some $x\in H$. Let us fix such a $x$.
\par
We know that $H$ is a group. Hence, the map $K\rightarrow xK,\ k\mapsto xk$ is
a bijection. Thus, the sets $K$ and $xK$ are in bijection. Therefore,
$\left\vert xK\right\vert =\left\vert K\right\vert $.
\par
However, for every $y\in H$, we have the following logical equivalence:
\begin{align*}
\left(  y\in\rho^{-1}\left(  g\right)  \right)  \  &  \Longleftrightarrow
\ \left(  \rho\left(  y\right)  =\underbrace{g}_{=\rho\left(  x\right)
}\right)  \ \Longleftrightarrow\ \left(  \rho\left(  y\right)  =\rho\left(
x\right)  \right)  \ \Longleftrightarrow\ \left(  \underbrace{\rho\left(
y\right)  \cdot\left(  \rho\left(  x\right)  \right)  ^{-1}}_{\substack{=\rho
\left(  yx^{-1}\right)  \\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}=1\right) \\
&  \Longleftrightarrow\ \left(  \rho\left(  yx^{-1}\right)  =1\right)
\ \Longleftrightarrow\ \left(  yx^{-1}\in\underbrace{\ker\rho
}_{=K}\right)  \ \Longleftrightarrow\ \left(  yx^{-1}\in K\right)
\ \Longleftrightarrow\ \left(  y\in xK\right)  .
\end{align*}
Hence, $\rho^{-1}\left(  g\right)  =xK$, so that $\left\vert \rho^{-1}\left(
g\right)  \right\vert =\left\vert xK\right\vert =\left\vert K\right\vert $.
This proves (\ref{sol.Indrho.i.preimage}).}

We shall now show that the maps $\alpha$ and $\dfrac{1}{\left\vert
K\right\vert }\beta$ are mutually inverse. To do so, we will show that
$\alpha\circ\beta=\left\vert K\right\vert \operatorname{id}$ and $\beta
\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

Let us first show that $\alpha\circ\beta=\operatorname{id}$. In fact, let
$q\in{\CC }G$ and $u\in U$ be arbitrary. Then,
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)   &
=\alpha\left(  \beta\left(  q\otimes_{\CC H}u\right)  \right)
=\sum_{j\in J}j^{-1}\otimes_{\CC H}\underbrace{\left(  \beta\left(
q\otimes_{\CC H}u\right)  \right)  \left(  j\right)  }%
_{\substack{=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
jq\right)  h^{-1}u\\\text{(by (\ref{sol.Indrho.i.beta.explicit}), applied to
}p=j\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\right) \\
&  =\sum_{j\in J}j^{-1}\otimes_{\CC H}\left(  \sum\limits_{h\in
H}\epsilon_{1}\left(  \rho\left(  h\right)  jq\right)  h^{-1}u\right)
=\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
jq\right)  \underbrace{j^{-1}\otimes_{\CC H}h^{-1}u}%
_{\substack{=j^{-1}h^{-1}\otimes_{\CC H}u\\\text{(since }h^{-1}%
\in\CC H\text{ (since }h\in H\\\text{and thus }h^{-1}\in H\subset
\CC H\text{))}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
jq\right)  \underbrace{j^{-1}h^{-1}}_{\substack{=j^{-1}\cdot\rho\left(
h^{-1}\right)  \\\text{(by (\ref{sol.Indrho.i.rightCH2}), applied to }%
\gamma=j^{-1}\\\text{and }\eta=h^{-1}\text{ (since }h^{-1}\in H\text{ (since
}h\in H\text{)))}}}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\underbrace{\sum\limits_{h\in H}}_{\substack{=\sum
_{g\in\overline{H}}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}\\\text{(since every }h\in H\text{ satisfies}\\\rho\left(  h\right)
\in\rho\left(  H\right)  =\overline{H}\text{)}}}\epsilon_{1}\left(
\rho\left(  h\right)  jq\right)  j^{-1}\cdot\underbrace{\rho\left(
h^{-1}\right)  }_{\substack{=\left(  \rho\left(  h\right)  \right)
^{-1}\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}%
}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}}_{=\sum_{h\in\rho^{-1}\left(  g\right)  }%
}\epsilon_{1}\left(  \underbrace{\rho\left(  h\right)  }_{=g}jq\right)
j^{-1}\cdot\left(  \underbrace{\rho\left(  h\right)  }_{=g}\right)
^{-1}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum_{h\in\rho^{-1}\left(
g\right)  }\epsilon_{1}\left(  gjq\right)  j^{-1}\cdot g^{-1}\otimes
_{\CC H}u}_{=\left\vert \rho^{-1}\left(  g\right)  \right\vert
\cdot\epsilon_{1}\left(  gjq\right)  j^{-1}\cdot g^{-1}\otimes_{{\CC }%
H}u}\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\left\vert \rho
^{-1}\left(  g\right)  \right\vert }_{\substack{=\left\vert K\right\vert
\\\text{(by (\ref{sol.Indrho.i.preimage}))}}}\cdot\epsilon_{1}\left(
gjq\right)  \underbrace{j^{-1}\cdot g^{-1}}_{=\left(  gj\right)  ^{-1}}%
\otimes_{\CC H}u
\end{align*}%
\begin{align*}
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\left\vert K\right\vert \cdot
\epsilon_{1}\left(  gjq\right)  \left(  gj\right)  ^{-1}\otimes_{{\CC %
}H}u=\left\vert K\right\vert \cdot\sum_{j\in J}\sum_{g\in\overline{H}}%
\epsilon_{1}\left(  gjq\right)  \left(  gj\right)  ^{-1}\otimes_{{\CC %
}H}u\\
&  =\left\vert K\right\vert \cdot\underbrace{\sum_{j\in J}\sum_{g\in
\overline{H}j}}_{\substack{=\sum\limits_{g\in G}\\\text{(since }%
\bigsqcup_{j\in J}\overline{H}j=G\text{)}}}\epsilon_{1}\left(  gq\right)
g^{-1}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g\text{ for }gj\text{ in the second sum, since the
map}\\
\overline{H}\rightarrow\overline{H}j,\ g\mapsto gj\text{ is a bijection
(because }G\text{ is a group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \cdot\sum\limits_{g\in G}\epsilon_{1}\left(
gq\right)  g^{-1}\otimes_{\CC H}u=\left\vert K\right\vert \cdot
\sum\limits_{g\in G}\underbrace{\epsilon_{1}\left(  g^{-1}q\right)
}_{\substack{=\epsilon_{g}\left(  q\right)  \\\text{(by
(\ref{sol.Indrho.i.epsi1g}))}}}\underbrace{\left(  g^{-1}\right)  ^{-1}}%
_{=g}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \cdot\sum\limits_{g\in G}\epsilon_{g}\left(
q\right)  g\otimes_{\CC H}u=\left\vert K\right\vert \cdot
\underbrace{\left(  \sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\right)  }_{\substack{=q\\\text{(by (\ref{sol.Indrho.i.epsilong.coordinates}%
))}}}\otimes_{\CC H}u\\
&  =\left\vert K\right\vert \cdot\underbrace{q\otimes_{\CC H}%
u}_{=\operatorname{id}\left(  q\otimes_{\CC H}u\right)  }=\left\vert
K\right\vert \cdot\operatorname{id}\left(  q\otimes_{\CC H}u\right)
=\left(  \left\vert K\right\vert \operatorname{id}\right)  \left(
q\otimes_{\CC H}u\right)  .
\end{align*}


Now, let us forget that we fixed $q$ and $u$. We thus have shown that $\left(
\alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)  =\left(
\left\vert K\right\vert \operatorname{id}\right)  \left(  q\otimes
_{\CC H}u\right)  $ for all $q\in{\CC }G$ and $u\in U$. In other
words, the two maps $\alpha\circ\beta:{\CC }G\otimes_{{\CC }%
H}U\rightarrow{\CC }G\otimes_{\CC H}U$ and $\left\vert
K\right\vert \operatorname{id}:{\CC }G\otimes_{{\CC }%
H}U\rightarrow{\CC }G\otimes_{\CC H}U$ are equal to each other
on each pure tensor. Since these two maps are $\CC$-linear, this
yields that these two maps $\alpha\circ\beta:{\CC }G\otimes
_{\CC H}U\rightarrow{\CC }G\otimes_{\CC H}U$ and
$\left\vert K\right\vert \operatorname{id}:{\CC }G\otimes_{{\CC %
}H}U\rightarrow{\CC }G\otimes_{\CC H}U$ must be identical. In
other words, $\alpha\circ\beta=\left\vert K\right\vert \operatorname{id}$.

Next, we are going to show that $\beta\circ\alpha=\left\vert K\right\vert
\operatorname{id}$.

Let $f\in\Hom_{\CC H}\left(  {\CC}G,U\right)
$. Let $p\in{\CC }G$. The map $f$ is left ${\CC }H$-linear (since
$f\in\Hom_{\CC H}\left(  {\CC}G,U\right)  $),
hence $\CC$-linear. We have $\alpha\left(  f\right)  =\sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  $ (by the definition of
$\alpha$). Applying the map $\beta$ to this equality, we obtain
\[
\beta\left(  \alpha\left(  f\right)  \right)  =\beta\left(  \sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)  =\sum_{j\in J}%
\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)
\]
(since $\beta$ is a $\CC$-linear map). Thus,
\begin{align*}
&  \underbrace{\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
}_{=\sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)
\right)  }\left(  p\right) \\
&  =\left(  \sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  =\sum_{j\in J}%
\underbrace{\left(  \beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  }_{\substack{=\sum\limits_{h\in
H}\epsilon_{1}\left(  \rho\left(  h\right)  pj^{-1}\right)  h^{-1}f\left(
j\right)  \\\text{(by (\ref{sol.Indrho.i.beta.explicit}), applied
to}\\q=j^{-1}\text{ and }u=f\left(  j\right)  \text{)}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
pj^{-1}\right)  h^{-1}f\left(  j\right)  =\sum_{j\in J}\sum\limits_{h\in
H}\underbrace{\epsilon_{1}\left(  \rho\left(  h^{-1}\right)  pj^{-1}\right)
}_{\substack{=\epsilon_{1}\left(  j^{-1}\rho\left(  h^{-1}\right)  p\right)
\\\text{(by (\ref{sol.Indrho.i.trace}), applied to}\\\rho\left(
h^{-1}\right)  p\text{ and }j^{-1}\text{ instead of }p\text{ and }q\text{)}%
}}\underbrace{\left(  h^{-1}\right)  ^{-1}}_{=h}f\left(  j\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }h^{-1}\text{ for }h\text{ in the second sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{-1}\text{ is a bijection}\\
\text{(because }H\text{ is a group)}%
\end{array}
\right) \\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  j^{-1}%
\underbrace{\rho\left(  h^{-1}\right)  }_{\substack{=\left(  \rho\left(
h\right)  \right)  ^{-1}\\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}p\right)  \underbrace{hf\left(  j\right)
}_{\substack{=f\left(  hj\right)  \\\text{(since }f\text{ is left }%
{\CC }H\text{-linear and since }h\in H\subset{\CC }H\text{)}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{j^{-1}%
\left(  \rho\left(  h\right)  \right)  ^{-1}}_{=\left(  \rho\left(  h\right)
j\right)  ^{-1}}p\right)  f\left(  \underbrace{hj}_{\substack{=\rho\left(
h\right)  \cdot j\\\text{(by (\ref{sol.Indrho.i.leftCH2}), applied to}%
\\\eta=h\text{ and }\gamma=j\text{)}}}\right) \\
&  =\sum_{j\in J}\underbrace{\sum\limits_{h\in H}}_{\substack{=\sum
_{g\in\overline{H}}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}\\\text{(since every }h\in H\text{ satisfies}\\\rho\left(  h\right)
\in\rho\left(  H\right)  =\overline{H}\text{)}}}\epsilon_{1}\left(  \left(
\rho\left(  h\right)  j\right)  ^{-1}p\right)  f\left(  \rho\left(  h\right)
j\right) \\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}}_{=\sum_{h\in\rho^{-1}\left(  g\right)  }%
}\epsilon_{1}\left(  \left(  \underbrace{\rho\left(  h\right)  }_{=g}j\right)
^{-1}p\right)  f\left(  \underbrace{\rho\left(  h\right)  }_{=g}j\right)
=\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum_{h\in\rho^{-1}\left(
g\right)  }\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(
gj\right)  }_{=\left\vert \rho^{-1}\left(  g\right)  \right\vert \cdot
\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(  gj\right)  }\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\left\vert \rho
^{-1}\left(  g\right)  \right\vert }_{\substack{=\left\vert K\right\vert
\\\text{(by (\ref{sol.Indrho.i.preimage}))}}}\cdot\epsilon_{1}\left(  \left(
gj\right)  ^{-1}p\right)  f\left(  gj\right)  =\sum_{j\in J}\sum
_{g\in\overline{H}}\left\vert K\right\vert \cdot\epsilon_{1}\left(  \left(
gj\right)  ^{-1}p\right)  f\left(  gj\right)
\end{align*}%
\begin{align*}
&  =\left\vert K\right\vert \cdot\sum_{j\in J}\sum_{g\in\overline{H}}%
\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(  gj\right)
=\left\vert K\right\vert \cdot\underbrace{\sum_{j\in J}\sum_{g\in\overline
{H}j}}_{\substack{=\sum\limits_{g\in G}\\\text{(since }\bigsqcup_{j\in
J}\overline{H}j=G\text{)}}}\epsilon_{1}\left(  g^{-1}p\right)  f\left(
g\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g\text{ for }gj\text{ in the second sum, since the
map}\\
\overline{H}\rightarrow\overline{H}j,\ g\mapsto gj\text{ is a bijection
(because }G\text{ is a group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \cdot\sum\limits_{g\in G}\underbrace{\epsilon
_{1}\left(  g^{-1}p\right)  }_{\substack{=\epsilon_{g}\left(  p\right)
\\\text{(by (\ref{sol.Indrho.i.epsi1g}), applied to }q=p\text{)}}}f\left(
g\right)  =\left\vert K\right\vert \cdot\underbrace{\sum\limits_{g\in
G}\epsilon_{g}\left(  p\right)  f\left(  g\right)  }_{\substack{=f\left(
\sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  g\right)  \\\text{(since
}f\text{ is }\CC \text{-linear)}}}=\left\vert K\right\vert \cdot
f\left(  \sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  g\right)  .
\end{align*}
Compared with
\[
\left(  \left\vert K\right\vert f\right)  \left(  p\right)  =\left\vert
K\right\vert \cdot f\left(  \underbrace{p}_{\substack{=\sum\limits_{g\in
G}\epsilon_{g}\left(  p\right)  g\\\text{(by
(\ref{sol.Indrho.i.epsilong.coordinates}), applied to }q=p\text{)}}}\right)
=\left\vert K\right\vert \cdot f\left(  \sum\limits_{g\in G}\epsilon
_{g}\left(  p\right)  g\right)  ,
\]
this yields $\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
\left(  p\right)  =\left(  \left\vert K\right\vert f\right)  \left(  p\right)
$.

Now, forget that we fixed $p$. We thus have proven that $\left(  \beta\left(
\alpha\left(  f\right)  \right)  \right)  \left(  p\right)  =\left(
\left\vert K\right\vert f\right)  \left(  p\right)  $ for every $p\in
{\CC }G$. In other words, $\beta\left(  \alpha\left(  f\right)  \right)
=\left\vert K\right\vert f$. Hence, $\left(  \beta\circ\alpha\right)  \left(
f\right)  =\beta\left(  \alpha\left(  f\right)  \right)  =\left\vert
K\right\vert f=\left(  \left\vert K\right\vert \operatorname{id}\right)
\left(  f\right)  $.

Now, forget that we fixed $f$. We thus have shown that $\left(  \beta
\circ\alpha\right)  \left(  f\right)  =\left(  \left\vert K\right\vert
\operatorname{id}\right)  \left(  f\right)  $ for every $f\in
\Hom_{\CC H}\left(  {\CC}G,U\right)  $. In
other words, $\beta\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

We now have $\beta\circ\alpha=\left\vert K\right\vert \operatorname{id}$ and
thus $\left(  \dfrac{1}{\left\vert K\right\vert }\beta\right)  \circ
\alpha=\dfrac{1}{\left\vert K\right\vert }\underbrace{\left(  \beta\circ
\alpha\right)  }_{=\left\vert K\right\vert \operatorname{id}}=\dfrac
{1}{\left\vert K\right\vert }\left\vert K\right\vert \operatorname{id}%
=\operatorname*{id}$.

On the other hand, recall that $\alpha\circ\beta=\left\vert K\right\vert
\operatorname{id}$. Hence, $\alpha\circ\left(  \dfrac{1}{\left\vert
K\right\vert }\beta\right)  =\dfrac{1}{\left\vert K\right\vert }%
\underbrace{\left(  \alpha\circ\beta\right)  }_{=\left\vert K\right\vert
\operatorname{id}}=\dfrac{1}{\left\vert K\right\vert }\left\vert K\right\vert
\operatorname{id}=\operatorname*{id}$.

Combining the equalities $\left(  \dfrac{1}{\left\vert K\right\vert }%
\beta\right)  \circ\alpha=\operatorname*{id}$ and $\alpha\circ\left(
\dfrac{1}{\left\vert K\right\vert }\beta\right)  =\operatorname*{id}$, we
conclude that the maps $\alpha$ and $\dfrac{1}{\left\vert K\right\vert }\beta$
are mutually inverse. Hence, the map $\alpha$ is invertible.

Now, we know that the map $\alpha$ is an invertible left $\CC G$-module
homomorphism. Hence, $\alpha$ is a left ${\CC }G$-module isomorphism.
Therefore, there exists a left ${\CC }G$-module isomorphism
$\Hom_{\CC H}\left(  {\CC }G,U\right)
\rightarrow{\CC }G\otimes_{\CC H}U$ (namely, $\alpha$).
Therefore, $\Hom_{\CC H}\left(  {\CC }G,U\right)
\cong{\CC }G\otimes_{\CC H}U=\operatorname{Ind}_{\rho}U$ as left
${\CC }G$-modules. This solves Exercise~\ref{exe.Indrho}(i).

[\textit{Remark:} In our above solution of Exercise~\ref{exe.Indrho}(i), we
explicitly constructed a ${\CC }G$-module isomorphism $\alpha
:\Hom_{\CC H}\left(  {\CC}G,U\right)
\rightarrow\operatorname{Ind}_{\rho}U$. This isomorphism is functorial with
respect to $U$. It is also independent on the choice of $J$ (this is not
immediately clear from its definition, but it can be shown very easily, by
observing that the tensor $j^{-1}\otimes_{\CC H}f\left(  j\right)  $
for $j\in G$ depends only on the coset $\overline{H}j$ and not on $j$ itself).]

[\textit{Remark:} Exercise \ref{exe.Indrho}(i) allows us to give yet another
solution to Exercise \ref{exe.Indrho}(f):

\textit{Second solution to Exercise \ref{exe.Indrho}(f):} Assume that $G=H/K$
for some normal subgroup $K$ of $H$. Let $\rho:H\rightarrow G$ be the
projection map. We want to prove that $\Ind_{\rho
}U\cong U^{K}$ for every $\CC H$-module $U$.

We know that $\rho$ is the projection map from $H$ to $H/K$. Hence, $\rho$ is
a surjective group homomorphism and has kernel $\ker\rho=K$.

Let $U$ be a $\CC H$-module. Recall that $U^{K}$ is a $\CC \left[
H/K\right]  $-module, thus a $\CC G$-module (since $H/K=G$). This
$\CC G$-module structure has the property that%
\begin{equation}
\rho\left(  h\right)  \cdot v=hv\ \ \ \ \ \ \ \ \ \ \text{for any }h\in
H\text{ and }v\in U^{K}. \label{sol.Indrho.f.sol3.UK-modstr}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol3.UK-modstr}): }The proof of
(\ref{sol.Indrho.f.sol3.UK-modstr}) is identical with the proof of
(\ref{sol.Indrho.f.sol1.UK-modstr}) in the First solution to Exercise
\ref{exe.Indrho}(f).}

Let us make $\CC G$ into a $\left(  \CC H,\CC G\right)
$-bimodule as in Exercise \ref{exe.Indrho}(i). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC H,\CC G\right)
$-bimodule structure. Then, the $\CC G$-module
$\Hom_{\CC H}\left(  \CC G,U\right)  $ (defined as in Exercise
\ref{exe.induction.ind-as-hom} using the $\left(  \CC H,\CC %
G\right)  $-bimodule structure on $\CC G$) is isomorphic to
$\Ind_{\rho}U$ (because of Exercise \ref{exe.Indrho}(i)).
In other words,
$\Hom_{\CC H}\left( \CC G,U\right)  \cong\Ind_{\rho}U$ as left
$\CC G$-modules.

We recall that $\Hom_{\CC H}\left(
\CC G,U\right)  $ is the left $\CC G$-module consisting of all
left $\CC H$-module homomorphisms from $\CC G$ to $U$. This uses
only the $\left(  \CC H,\CC G\right)  $-bimodule structure on
$\CC G$ that was introduced in the statement of Exercise
\ref{exe.Indrho}(i) (but not the $\left(  \CC G,\CC H\right)
$-bimodule structure on $\CC G$ that was introduced in the definition of
$\Ind_{\rho}U$).

We have $f\left(  1\right)  \in U^{K}$ for every
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $. Let $k\in K$. Thus,
$k\in K=\ker\rho$, so that $\rho\left(  k\right)  =1_{G}$.
\par
But $k\in K\subset H$. Thus, (\ref{sol.Indrho.i.leftCH2}) (applied to
$\gamma=1_{\CC G}$ and $\eta=k$) yields $k1_{\CC G}%
=\underbrace{\rho\left(  k\right)  }_{=1_{G}}1_{\CC G}=1_{G}%
1_{\CC G}=1_{G}$.
\par
But $f\in\Hom_{\CC H}\left(  \CC %
G,U\right)  $. Hence, the map $f$ is left $\CC H$-linear. Thus,
$f\left(  k1_{\CC G}\right)  =kf\left(  1_{\CC G}\right)  $ (since
$k\in H\subset\CC H$). Since $k1_{\CC G}=1_{G}$, this rewrites as
$f\left(  1_{G}\right)  =kf\left(  \underbrace{1_{\CC G}}_{=1}\right)
=kf\left(  1\right)  $, so that $kf\left(  1\right)  =f\left(
\underbrace{1_{G}}_{=1}\right)  =f\left(  1\right)  $.
\par
Now, let us forget that we fixed $k$. We thus have shown that $kf\left(
1\right)  =f\left(  1\right)  $ for every $k\in K$. In other words, $f\left(
1\right)  $ is an element $y$ of $U$ which satisfies $ky=y$ for every $k\in
K$. In other words,%
\[
f\left(  1\right)  \in\left\{  y\in U\ \mid\ ky=y\text{ for every }k\in
K\right\}  =U^{K},
\]
qed.}. Thus, we can define a map
$\phi:\Hom_{\CC H}\left(  \CC G,U\right)  \rightarrow U^{K}$ by
\[
\left(  \phi\left(  f\right)  =f\left(  1\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }f\in\Hom_{\CC H}
\left(  \CC G,U\right)  \right)  .
\]
Consider this $\phi$. This map $\phi$ is clearly $\CC$-linear (since
$f\left(  1\right)  $ depends $\CC$-linearly on $f$). Also, the map
$\phi$ is surjective\footnote{\textit{Proof.} Let $u\in U^{K}$. We are going
to construct a map $f\in\Hom_{\CC H}\left(
\CC G,U\right)  $ which satisfies $\phi\left(  f\right)  =u$.
\par
Indeed, let us define a map $f:\CC G\rightarrow U$ by%
\begin{equation}
\left(  f\left(  \gamma\right)  =\gamma u\ \ \ \ \ \ \ \ \ \ \text{for every
}\gamma\in\CC G\right)  . \label{sol.Indrho.f.sol3.surj.pf.def-f}%
\end{equation}
(This is well-defined, since $U^{K}$ is a $\CC G$-module.) This map $f$
is $\CC$-linear (since $\gamma u$ depends $\CC$-linearly on
$\gamma$). We shall now check that the map $f$ is a homomorphism of left
$H$-sets.
\par
Indeed, let $h\in H$ and $\gamma\in\CC G$. We have $f\left(
h\gamma\right)  =\left(  h\gamma\right)  u$ (by the definition of $f$).
\par
On the other hand, (\ref{sol.Indrho.i.leftCH2}) (applied to $\eta=h$) yields
$h\gamma=\rho\left(  h\right)  \cdot\gamma$.
\par
But $u\in U^{K}$ and therefore $\gamma u\in U^{K}$ (since $\gamma\in
\CC G$ and since $U^{K}$ is a $\CC G$-module). Thus,
(\ref{sol.Indrho.f.sol3.UK-modstr}) (applied to $\gamma u$ instead of $v$)
yields $\rho\left(  h\right)  \cdot\left(  \gamma u\right)  =h\left(  \gamma
u\right)  $. Now,%
\[
f\left(  h\gamma\right)  =\underbrace{\left(  h\gamma\right)  }_{=\rho\left(
h\right)  \cdot\gamma}u=\left(  \rho\left(  h\right)  \cdot\gamma\right)
u=\rho\left(  h\right)  \cdot\left(  \gamma u\right)  =h\underbrace{\left(
\gamma u\right)  }_{\substack{=f\left(  \gamma\right)  \\\text{(by
(\ref{sol.Indrho.f.sol3.surj.pf.def-f}))}}}=hf\left(  \gamma\right)  .
\]
\par
Now, let us forget that we fixed $h$ and $\gamma$. We thus have proven that
$f\left(  h\gamma\right)  =hf\left(  \gamma\right)  $ for all $h\in H$ and
$\gamma\in\CC G$. In other words, $f$ is a homomorphism of left
$H$-sets. Hence, $f$ is a left $\CC H$-module homomorphism (since $f$ is
$\CC$-linear). In other words,
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $. Now, the definition of
$\phi$ yields%
\begin{align*}
\phi\left(  f\right)   &  =f\left(  1\right)  =1u\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Indrho.f.sol3.surj.pf.def-f}), applied to }\gamma=1\right)
\\
&  =u.
\end{align*}
Thus, $u=\phi\left(  \underbrace{f}_{\in\Hom_{\CC H}
\left(  \CC G,U\right)  }\right)  \in\phi\left(
\Hom_{\CC H}\left(  \CC G,U\right)
\right)  $.
\par
Let us now forget that we fixed $u$. We thus have shown that $u\in\phi\left(
\Hom_{\CC H}\left(  \CC G,U\right)
\right)  $ for every $u\in U^{K}$. In other words, $U^{K}\subset\phi\left(
\Hom_{\CC H}\left(  \CC G,U\right)
\right)  $. This proves that the map $\phi$ is surjective. Qed.} and
injective\footnote{\textit{Proof.} Let $f\in\ker\phi$. Then,
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $
and $\phi\left(  f\right)  =0$ (since $f\in\ker\phi$). The
definition of $\phi$ yields $\phi\left(  f\right)  =f\left(  \underbrace{1}%
_{=1_{\CC G}}\right)  =f\left(  1_{\CC G}\right)  $, so that
$f\left(  1_{\CC G}\right)  =\phi\left(  f\right)  =0$.
\par
Now, let $g\in G$ be arbitrary. Then, there exists some $h\in H$ such that
$g=\rho\left(  h\right)  $ (since $\rho$ is surjective). Consider this $h$.
Applying (\ref{sol.Indrho.i.leftCH2}) to $\gamma=1_{\CC G}$ and $\eta
=h$, we obtain $h1_{\CC G}=\rho\left(  h\right)  \cdot1_{\CC %
G}=\rho\left(  h\right)  =g$. But the map $f$ is left $\CC H$-linear
(since $f\in\Hom_{\CC H}\left(  \CC %
G,U\right)  $). Thus, $f\left(  h1_{\CC G}\right)  =hf\left(
1_{\CC G}\right)  $ (since $h\in H\subset\CC H$), so that
$f\left(  h1_{\CC G}\right)  =h\underbrace{f\left(  1_{\CC %
G}\right)  }_{=0}=0$. Compared with $f\left(  \underbrace{h1_{\CC G}%
}_{=g}\right)  =f\left(  g\right)  $, this yields $f\left(  g\right)  =0$.
\par
Now, let us forget that we fixed $g$. Thus, we have shown that $f\left(
g\right)  =0$ for every $g\in G$. In other words, the map $f$ sends every
element of the basis $G$ of the $\CC$-vector space $\CC G$ to
$0$. Hence, $f=0$ (since the map $f$ is $\CC$-linear).
\par
Now, let us forget that we fixed $f$. We thus have shown that $f=0$ for every
$f\in\ker\phi$. In other words, $\ker\phi=0$.
Since $\phi$ is a $\CC$-linear map, this shows that $\phi$ is
injective. Qed.}. Hence, the map $\phi$ is bijective. Thus, $\phi$ is a
$\CC$-vector space isomorphism (since $\phi$ is $\CC$-linear).

Moreover, $\phi$ is a homomorphism of left $G$-sets\footnote{\textit{Proof.}
Let $g\in G$ and $f\in\Hom_{\CC H}\left(
\CC G,U\right)  $. The definition of the left $\CC G$-module
structure on $\Hom_{\CC H}\left(  \CC %
G,U\right)  $ yields%
\[
\left(  gf\right)  \left(  1\right)  =f\left(  \underbrace{1g}_{=g}\right)
=f\left(  g\right)  .
\]
Now, the definition of $\phi$ yields $\phi\left(  gf\right)  =\left(
gf\right)  \left(  1\right)  =f\left(  g\right)  $.
\par
But there exists some $h\in H$ such that $g=\rho\left(  h\right)  $ (since
$\rho$ is surjective). Consider this $h$. The map $f$ is left $\CC %
H$-linear (since $f\in\Hom_{\CC H}\left(
\CC G,U\right)  $). Thus, we have $f\left(  h1_{\CC G}\right)
=hf\left(  1_{\CC G}\right)  $. But (\ref{sol.Indrho.i.leftCH2})
(applied to $\gamma=1_{\CC G}$ and $\eta=h$) yields $h1_{\CC %
G}=\underbrace{\rho\left(  h\right)  }_{=g}\cdot1_{\CC G}=g\cdot
1_{\CC G}=g$, so that $g=h1_{\CC G}$. Hence, $f\left(
\underbrace{g}_{=h1_{\CC G}}\right)  =f\left(  h1_{\CC G}\right)
=hf\left(  1_{\CC G}\right)  $.
\par
The definition of $\phi$ yields $\phi\left(  f\right)  =f\left(
\underbrace{1}_{=1_{\CC G}}\right)  =f\left(  1_{\CC G}\right)  $,
so that $f\left(  1_{\CC G}\right)  =\phi\left(  f\right)  $. Now,
$f\left(  g\right)  =h\underbrace{f\left(  1_{\CC G}\right)  }%
_{=\phi\left(  f\right)  }=h\phi\left(  f\right)  $. Hence,%
\[
\phi\left(  gf\right)  =f\left(  g\right)  =h\phi\left(  f\right)  .
\]
\par
The equality (\ref{sol.Indrho.f.sol3.UK-modstr}) (applied to $\phi\left(
f\right)  $ instead of $v$) yields $\rho\left(  h\right)  \cdot\phi\left(
f\right)  =h\phi\left(  f\right)  $. Thus, $h\phi\left(  f\right)
=\underbrace{\rho\left(  h\right)  }_{=g}\cdot\phi\left(  f\right)
=g\cdot\phi\left(  f\right)  =g\phi\left(  f\right)  $. Hence, $\phi\left(
gf\right)  =h\phi\left(  f\right)  =g\phi\left(  f\right)  $.
\par
Now, let us forget that we fixed $g$ and $f$. We thus have proven that
$\phi\left(  gf\right)  =g\phi\left(  f\right)  $ for all $g\in G$ and
$f\in\Hom_{\CC H}\left(  \CC G,U\right)
$. In other words, the map $\phi$ is a homomorphism of left $G$-sets. Qed.},
and therefore a left $\CC G$-module homomorphism (since $f$ is
$\CC$-linear). Thus, $\phi$ is a left $\CC G$-module isomorphism
(since $\phi$ is a $\CC$-vector space isomorphism). We thus have found
a left $\CC G$-module isomorphism from
$\Hom_{\CC H}\left(  \CC G,U\right)  $ to $U^{K}$ (namely,
$\phi$). Hence, $U^{K}\cong\Hom_{\CC H}\left(
\CC G,U\right)  $ as left $\CC G$-modules. Hence, $U^{K}%
\cong\Hom_{\CC H}\left(  \CC G,U\right)
\cong\Ind_{\rho}U$ as left $\CC G$-modules. This
solves Exercise \ref{exe.Indrho}(f) again.]

(j) Let $U$ be a $\CC G$-module, and let $V$ be a $\CC H$-module.

Let us make $\CC G$ into a $\left(  \CC H,\CC G\right)
$-bimodule as in Exercise \ref{exe.Indrho}(i). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC H,\CC G\right)
$-bimodule structure. Then, the $\CC G$-module
$\Hom_{\CC H}\left(  \CC G,V\right)  $ (defined as in Exercise
\ref{exe.induction.ind-as-hom} using the $\left(  \CC H,\CC %
G\right)  $-bimodule structure on $\CC G$) is isomorphic to
$\Ind_{\rho}V$ (because of Exercise \ref{exe.Indrho}%
(i), applied to $V$ instead of $U$). In other words,
\begin{equation}
\Hom_{\CC H}\left(  \CC G,V\right)
\cong\Ind_{\rho}V\ \ \ \ \ \ \ \ \ \ \text{as left
}\CC G\text{-modules.} \label{sol.Indrho.j.i}%
\end{equation}


Now, (\ref{Hom-tensor-adj}) (applied to $R=\CC G$, $S=\CC H$,
$A=\CC G$, $B=U$ and $C=V$) yields
\begin{equation}
\Hom_{\CC H}\left(  \CC G\otimes
_{\CC G}U,V\right)  \cong\Hom_{\CC %
G}\left(  U,\Hom_{\CC H}\left(  \CC %
G,V\right)  \right)  . \label{sol.Indrho.j.adj}%
\end{equation}


We shall now prove that $\CC G\otimes_{\CC G}U\cong%
\Res_{\rho}U$ as left $\CC H$-modules.

Indeed, a fundamental fact in abstract algebra says the following: If
$\mathfrak{A}$ is a $\CC$-algebra, and if $\mathfrak{M}$ is a left
$\mathfrak{A}$-module, then there exists a $\CC$-vector space
isomorphism $\Xi:\mathfrak{A}\otimes_{\mathfrak{A}}\mathfrak{M}\rightarrow
\mathfrak{M}$ which satisfies $\left(  \Xi\left(  a\otimes_{\mathfrak{A}%
}m\right)  =am\text{ for every }a\in\mathfrak{A}\text{ and }m\in
\mathfrak{M}\right)  $. Applying this fact to $\mathfrak{A}=\CC G$ and
$\mathfrak{M}=U$, we conclude that there exists a $\CC$-vector space
isomorphism $\Xi:\CC G\otimes_{\CC G}U\rightarrow U$ which
satisfies \newline
$\left(  \Xi\left(  a\otimes_{\CC G}m\right)  =am\text{ for
every }a\in\CC G\text{ and }m\in U\right)  $. Consider this $\Xi$.

The map $\Xi$ is a $\CC$-vector space isomorphism. Hence, its inverse
$\Xi^{-1}$ exists and also is a $\CC$-vector space isomorphism, and
therefore a $\CC$-linear map. It furthermore satisfies%
\begin{equation}
\Xi^{-1}\left(  u\right)  =1\otimes_{\CC G}%
u\ \ \ \ \ \ \ \ \ \ \text{for every }u\in U. \label{sol.Indrho.j.Xi-1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.j.Xi-1}):} Let $u\in U$. Recall
that $\Xi\left(  a\otimes_{\CC G}m\right)  =am$ for every $a\in
\CC G$ and $m\in U$. Applying this to $a=1$ and $m=u$, we obtain
$\Xi\left(  1\otimes_{\CC G}u\right)  =1u=u$. Hence, $1\otimes
_{\CC G}u=\Xi^{-1}\left(  u\right)  $ (since $\Xi$ is invertible). This
proves (\ref{sol.Indrho.j.Xi-1}).}

The map $\Xi^{-1}$ is a homomorphism of $H$-sets from $\Res_{\rho}U$
to $\CC G\otimes_{\CC G}U$%
\ \ \ \ \footnote{\textit{Proof.} The map $\Xi^{-1}$ is a map from $U$ to
$\CC G\otimes_{\CC G}U$, therefore a map from $\Res_{\rho}U$ to
$\CC G\otimes_{\CC G}U$ (since
$\Res_{\rho}U=U$ as sets).
\par
Recall that the $\CC H$-module structure on $\Res_{\rho}U$ is given by
\begin{equation}
h\cdot v=\rho\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \text{for every
}h\in H\text{ and }v\in U. \label{sol.Indrho.j.Res}%
\end{equation}
\par
Now, let $v\in\Res_{\rho}U$ and $h\in H$. We are going
to prove that $\Xi^{-1}\left(  h\cdot v\right)  =h\cdot\Xi^{-1}\left(
v\right)  $, where $h\cdot\Xi^{-1}\left(  v\right)  $ is computed in the
$\CC H$-module $\CC G\otimes_{\CC G}U$.
\par
Indeed, recall that the left $\CC H$-module structure on $\CC G$
is induced by the
$\CC$-algebra homomorphism $\CC \left[  \rho\right]
:\CC H\rightarrow\CC G$. Thus, it is explicitly given by%
\[
\eta\gamma=\left(  \CC \left[  \rho\right]  \right)  \eta\cdot
\gamma\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\CC G\text{ and }%
\eta\in\CC H.
\]
Applying this to $\eta=h$ and $\gamma=1_{\CC G}$, we obtain
$h1_{\CC G}=\left(  \CC \left[  \rho\right]  \right)  \left(
h\right)  \cdot1_{\CC G}$ (since $h\in H\subset\CC H$). But $h\in
H$ and thus $\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
=\rho\left(  h\right)  $. Hence, $h1_{\CC G}=\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  }_{=\rho\left(
h\right)  }\cdot1_{\CC G}=\rho\left(  h\right)  \cdot1_{\CC %
G}=\rho\left(  h\right)  =1_{\CC G}\cdot\rho\left(  h\right)  $. Now,
$v\in\Res_{\rho}U=U$. Hence, (\ref{sol.Indrho.j.Xi-1})
(applied to $u=v$) yields $\Xi^{-1}\left(  v\right)  =1\otimes_{\CC G}%
v$. Thus,%
\begin{align*}
h\cdot\underbrace{\Xi^{-1}\left(  v\right)  }_{=1\otimes_{\CC G}v}  &
=h\cdot\left(  1\otimes_{\CC G}v\right)  =h\underbrace{1}%
_{=1_{\CC G}}\otimes_{\CC G}v=\underbrace{h1_{\CC G}%
}_{=1_{\CC G}\cdot\rho\left(  h\right)  }\otimes_{\CC %
G}v=1_{\CC G}\cdot\rho\left(  h\right)  \otimes_{\CC G}v\\
&  =\underbrace{1_{\CC G}}_{=1}\otimes_{\CC G}\underbrace{\rho
\left(  h\right)  \cdot v}_{\substack{=h\cdot v\\\text{(by
(\ref{sol.Indrho.j.Res}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have moved }\rho\left(  h\right)  \text{ past the tensor
sign}\\
\text{(this is allowed since }\rho\left(  h\right)  \in G\subset
\CC G\text{)}%
\end{array}
\right) \\
&  =1\otimes_{\CC G}h\cdot v.
\end{align*}
Compared with $\Xi^{-1}\left(  h\cdot v\right)  =1\otimes_{\CC G}h\cdot
v$ (by (\ref{sol.Indrho.j.Xi-1}), applied to $u=h\cdot v$), this yields
$\Xi^{-1}\left(  h\cdot v\right)  =h\cdot\Xi^{-1}\left(  v\right)  $.
\par
Now, let us forget that we fixed $v$ and $h$. We thus have shown that
$\Xi^{-1}\left(  h\cdot v\right)  =h\cdot\Xi^{-1}\left(  v\right)  $ for every
$v\in\Res_{\rho}U$ and $h\in H$. In other words,
$\Xi^{-1}$ is a homomorphism of $H$-sets from $\Res_{\rho}U$ to
$\CC G\otimes_{\CC G}U$, qed.}. Therefore,
$\Xi^{-1}$ is a $\CC H$-module homomorphism from $\Res_{\rho}U$ to
$\CC G\otimes_{\CC G}U$ (since $\Xi^{-1}$ is
a $\CC$-linear map). Consequently, $\Xi^{-1}$ is a $\CC H$-module
isomorphism from $\Res_{\rho}U$ to $\CC %
G\otimes_{\CC G}U$ (since $\Xi^{-1}$ is a $\CC$-vector space
isomorphism). Therefore, $\CC G\otimes_{\CC G}U\cong%
\Res_{\rho}U$ as $\CC H$-modules.
Now, \eqref{Hom-tensor-adj} (applied to $\CC G$, $\CC H$,
$\CC G$, $U$ and $V$ instead of $R$, $S$, $A$, $B$ and $C$)
yields
\[
\Hom_{\CC H}\left( \CC G\otimes_{\CC G}U ,V\right)
\cong
\Hom_{\CC G}\left(  U,\Hom_{\CC H}\left(  \CC G,V\right)  \right) .
\]
Thus,
\begin{align*}
\Hom_{\CC G}\left(  U,\Hom_{\CC H}\left(  \CC G,V\right)  \right)
& \cong\Hom_{\CC H}
\left(  \underbrace{\CC G\otimes_{\CC G}U}_{\cong\Res_{\rho}U
\text{ as }\CC H\text{-modules}},V\right) \\
&  \cong\Hom_{\CC H}\left(  \Res_{\rho}U,V\right)  .
\end{align*}
Thus,%
\[
\Hom_{\CC H}\left(  \Res_{\rho}U,V\right)  \cong\Hom_{\CC G}
\left(  U,\underbrace{\Hom_{\CC H}\left(
\CC G,V\right)  }_{\substack{\cong\Ind_{\rho
}V\text{ as left }\CC G\text{-modules}\\\text{(by (\ref{sol.Indrho.j.i}%
))}}}\right)  \cong\Hom_{\CC G}\left(
U,\Ind_{\rho}V\right)  .
\]
This solves Exercise \ref{exe.Indrho}(j).

(k) \textit{Alternative proof of (\ref{induced-character-formula}):} Let $G$
be a finite group, and let $H$ be a subgroup of $G$. Let $U$ be a
finite-dimensional $\CC H$-module. We need to prove the identity
(\ref{induced-character-formula}).

Let $\rho$ denote the inclusion map $H\rightarrow G$. Clearly, $\rho$ is a
group homomorphism. But Exercise \ref{exe.Indrho}(d) yields
$\Ind_{\rho}U=\Ind_{H}^{G}U$.
Hence, $\chi_{\Ind_{\rho}U}=\chi_{\operatorname*{Ind}%
\nolimits_{H}^{G}U}$, so that $\chi_{\Ind_{H}^{G}%
U}=\chi_{\Ind_{\rho}U}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}$ (by Exercise \ref{exe.Indrho}(b)). But Exercise
\ref{exe.Indrho}(c) (applied to $f=\chi_{U}$) yields $\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}=\Ind_{H}^{G}\chi_{U}$. Thus,
$\chi_{\Ind_{H}^{G}U}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}=\Ind_{H}^{G}\chi_{U}$. Thus,
every $g\in G$ satisfies%
\[
\underbrace{\chi_{\Ind_{H}^{G}U}}%
_{=\Ind_{H}^{G}\chi_{U}}\left(  g\right)  =\left(
\Ind_{H}^{G}\chi_{U}\right)  \left(  g\right)
=\frac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G:\\kgk^{-1}\in
H}}\chi_{U}\left(  kgk^{-1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\Ind_{H}^{G}\chi_{U}\right)  .
\]
This proves (\ref{induced-character-formula}). Thus, Exercise \ref{exe.Indrho}%
(k) is solved.

(l) \textit{Alternative proof of (\ref{deflation-character-formula}):} Let $G$
be a finite group, and let $K$ be a normal subgroup of $G$. Let $V$ be a
finite-dimensional $\CC G$-module. We need to prove the identity
(\ref{deflation-character-formula}).

Let $\rho$ denote the projection map $G\rightarrow G/K$. Exercise
\ref{exe.Indrho}(f) (applied to $G/K$, $G$ and $V$ instead of $G$, $H$ and
$U$) yields $\Ind_{\rho}V\cong V^{K}$ as
$\CC \left[  G/K\right]  $-modules. Hence, $\chi_{\operatorname*{Ind}%
\nolimits_{\rho}V}=\chi_{V^{K}}$ (since isomorphic $\CC \left[
G/K\right]  $-modules have equal characters). Thus, $\chi_{V^{K}}%
=\chi_{\Ind_{\rho}V}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{V}$ (by Exercise \ref{exe.Indrho}(b), applied to $G/K$,
$G$ and $V$ instead of $G$, $H$ and $U$). But Exercise \ref{exe.Indrho}(e)
(applied to $G/K$, $G$, $V$ and $\chi_{V}$ instead of $G$, $H$, $U$ and $f$)
yields $\Ind_{\rho}\chi_{V}=\left(  \chi_{V}\right)
^{K}$. Hence, $\chi_{V^{K}}=\Ind_{\rho}\chi
_{V}=\left(  \chi_{V}\right)  ^{K}$. Thus, every $g\in G$ satisfies%
\[
\underbrace{\chi_{V^{K}}}_{=\left(  \chi_{V}\right)  ^{K}}\left(  gK\right)
=\left(  \chi_{V}\right)  ^{K}\left(  gK\right)  =\frac{1}{|K|}\sum_{k\in
K}\chi_{V}\left(  gk\right)
\]
(by the definition of $\left(  \chi_{V}\right)  ^{K}$). This proves
(\ref{deflation-character-formula}). Thus, Exercise \ref{exe.Indrho}(l) is solved.

[\textit{Remark:} Most parts of Exercise \ref{exe.Indrho} work in a far
greater generality than they are stated in. Let us briefly survey the
straightforward generalizations. We begin with the generalizations of the
notions of induction, restriction, inflation and fixed point constructions
defined in Chapter \ref{character-theory-review-section}:

\begin{itemize}
\item Recall that if $G$ is a finite group and $H$ is a subgroup of $G$, then
we have defined a $\CC G$-module $\Ind_{H}^{G}U$
for every $\CC H$-module $U$, and we have also defined a $\CC %
H$-module $\Res_{H}^{G}V$ for every $\CC %
G$-module $V$. Both of these definitions are also valid when $\CC$ is
replaced by any commutative ring. Basic properties of induction and
restriction (such as the equality (\ref{Frobenius-reciprocity}), the statement
of Exercise \ref{exe.induction.transitivity} and the statement of Exercise
\ref{exe.ind.tensor}) are still true in this generality (and the proofs that
we gave can be transferred to this generality with almost no changes).
However, more advanced properties might fail in this generality, and some can
not even be stated over a general commutative ring $\CC$%
\ \ \ \ \footnote{For instance, any properties of characters of modules can
only be stated when these characters are well-defined. Since characters are
defined as traces of certain endomorphisms of $\CC$-vector spaces, they
are not automatically well-defined when $\CC$ is replaced by a
commutative ring. They are, however, well-defined if our modules are
projective modules over this base ring; see the Remark after the proof of
Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} for details.}.

\item Recall that if $G$ is a finite group and $K$ is a normal subgroup of
$G$, then we have defined a $\CC G$-module $\operatorname*{Infl}%
\nolimits_{G/K}^{G}U$ for every $\CC \left[  G/K\right]  $-module $U$,
and we have also defined a $\CC \left[  G/K\right]  $-module $V^{K}$ for
every $\CC G$-module $V$. Both of these definitions are also valid when
$\CC$ is replaced by any commutative ring. Some basic properties of
inflation and fixed points (such as the isomorphisms
(\ref{restriction-inflation-commutation}) and (\ref{inflation-reciprocity}))
are still correct in this generality (again, they can be proven in the same
way as above), but some others are not (e.g., Exercise
\ref{exe.fixed-vs-coinvariants}(b) is generally false\footnote{However,
Exercise \ref{exe.fixed-vs-coinvariants}(b) holds if we additionally assume
that $\left\vert K\right\vert $ is invertible in the ring that replaces
$\CC$. Again, this can be proven by repeating our solution of Exercise
\ref{exe.fixed-vs-coinvariants}(b).}).

\item If $G$ is a finite group, then we defined $R_{\CC }\left(
G\right)  $ as the $\CC$-vector space of all class functions
$G\rightarrow\CC $. This definition still applies when $\CC$ is
replaced by any commutative ring. (Of course, in this case,
``$\CC$-vector space'' will have to be replaced by
``$\CC$-module''.)

\item Recall that if $G$ is a finite group and $H$ is a subgroup of $G$, then
we have defined a class function $\Ind_{H}^{G}f\in
R_{\CC }\left(  G\right)  $ for every $f\in R_{\CC }\left(
H\right)  $, and we have also defined a class function
$\Res_{H}^{G}f\in R_{\CC }\left(  H\right)  $ for every $f\in
R_{\CC }\left(  G\right)  $. The definition of $\Res_{H}^{G}f$
still works when $\CC$ is replaced by any
commutative ring. Our definition of $\Ind_{H}^{G}f$
cannot be reasonably interpreted in this generality (due to the denominator
$\left\vert H\right\vert $ in (\ref{eq.exe.induction.characters.eq}));
however, Exercise \ref{exe.induction.characters}(b) can be used as an
alternative definition of $\Ind_{H}^{G}f$ when
$\CC$ is replaced by any commutative ring. (Of course, the necessity of
using Exercise \ref{exe.induction.characters}(b) as a definition of
$\Ind_{H}^{G}f$ entails that most of our proofs
concerning $\Ind_{H}^{G}f$ no longer are valid in this
generality, because they use the equality
(\ref{eq.exe.induction.characters.eq}) as the definition of
$\Ind_{H}^{G}f$. Even the fact that
$\Ind_{H}^{G}f$ is well-defined needs to be proven
anew, since it is not immediately obvious that the sum $\sum\limits_{\substack{j\in
J:\\jgj^{-1}\in H}}f\left(  jgj^{-1}\right)  $ in Exercise
\ref{exe.induction.characters}(b) is independent on $J$.)
\end{itemize}

Now, let us move on to generalizing Remark \ref{rmk.Resrho} and Exercise
\ref{exe.Indrho}. Let $G$ and $H$ be two finite groups, and let $\rho
:H\rightarrow G$ be a group homomorphism.

\begin{itemize}
\item Remark \ref{rmk.Resrho} still holds if $\CC$ is replaced by any
commutative ring.

\item In Exercise \ref{exe.Indrho}, we defined a map $\operatorname*{Ind}%
\nolimits_{\rho}f:G\rightarrow\CC $ for every $f\in R_{\CC %
}\left(  H\right)  $. This definition still works when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. There is an alternative definition which works in even greater
generality: Namely, as long as $\left\vert \ker\rho\right\vert
$ is invertible\footnote{This is a weaker condition than $\left\vert
H\right\vert $ being invertible.}, we can define $\operatorname*{Ind}%
\nolimits_{\rho}f:G\rightarrow\CC $ by the equality
\begin{equation}
\left(  \Ind_{\rho}f\right)  \left(  g\right)
=\dfrac{1}{\left\vert \ker\rho\right\vert }\sum\limits
_{\substack{\left(  h,k\right)  \in H\times J;\\k\rho\left(  h\right)
k^{-1}=g}}f\left(  h\right)  , \label{sol.Indrho.generalize.Indf}%
\end{equation}
where $J$ is a system of left coset representatives for $G/\rho\left(
H\right)  $ (so that $G=\bigsqcup_{j\in J}j\rho\left(  H\right)  $). We leave
it to the reader to show that this definition is still well-defined (i.e.,
that $\sum\limits_{\substack{\left(  h,k\right)  \in H\times J;\\k\rho\left(
h\right)  k^{-1}=g}}f\left(  h\right)  $ is independent of the choice of $J$),
and that it defines the same function $\Ind_{\rho}f$
as the definition made in Exercise \ref{exe.Indrho} when $\left\vert
H\right\vert $ is invertible in the base ring. To my knowledge, there exists
no reasonable definition of $\Ind_{\rho}f$ that
completely avoids making any requirements on the base ring.

\item In Exercise \ref{exe.Indrho}, we defined a $\CC G$-module
$\Ind_{\rho}U$ for every $\CC H$-module $U$.
This definition is still valid when $\CC$ is replaced by a commutative
ring. (Notice the slightly surprising fact that if $\CC$ is replaced by
a commutative ring, then the $\rho$-induction of a $\CC H$-module is
always well-defined, whereas the $\rho$-induction of a class function might
not be.)

\item Exercise \ref{exe.Indrho}(a) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. And the solution that we gave still applies in this generality.
More generally, if we define $\Ind_{\rho}f$ by
(\ref{sol.Indrho.generalize.Indf}), then Exercise \ref{exe.Indrho}(a) is still
valid when $\CC$ is replaced by any commutative ring in which
$\left\vert \ker\rho\right\vert $ is invertible. However, this
can no longer be proven by blindly copying our solution.

\item Exercise \ref{exe.Indrho}(b) is still valid when $\CC$ is
replaced by any field in which $\left\vert H\right\vert $ is invertible.
Again, our solution is still valid in this generality.\footnote{We can also
replace $\CC$ by a commutative ring $A$ rather than by a field; but
then we need to replace ``finite-dimensional $\CC H$-module $U$'' by
``$AH$-module $U$ which is
a finitely generated projective $A$-module'' in the statement
of the exercise. The ``finitely generated projective
$A$-module'' condition is needed in order to make sure that
$\chi_{U}$ and $\chi_{\Ind_{\rho}U}$ are well-defined.
(See the Remark after the proof of Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr} for details.)} More generally, Exercise
\ref{exe.Indrho}(b) is still valid when $\CC$ is replaced by any field
in which $\left\vert \ker\rho\right\vert $ is invertible (as
long as $\Ind_{\rho}f$ is defined through
(\ref{sol.Indrho.generalize.Indf})); but our solution is not sufficient to
prove this\footnote{However, a certain modification of this solution does the
trick. Namely, we replace the definition of $\widetilde{F}_{g}$ by%
\[
\left(  \widetilde{F}_{g}\left(  \gamma,u\right)  =\dfrac{1}{\left\vert
\ker\rho\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(
h\right)  \right)  ^{\ast}\left(  \gamma\right)
hu\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  \gamma,u\right)  \in
\CC G\times U\right)  .
\]
Then, we should pick a system $J^{\prime}$ of left coset representatives for
$G/\rho\left(  H\right)  $ (so that $G=\bigsqcup_{j\in J^{\prime}}j\rho\left(
H\right)  $). (Notice that we do not call it $J$ because the letter $J$
already has a different meaning in this solution.) Then, we define a family
$\left(  a_{i}\right)  _{i\in J^{\prime}\times J}$ of elements of
$\CC G\otimes_{\CC H}U$ by%
\[
\left(  a_{\left(  k,j\right)  }=k\otimes_{\CC H}b_{j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  k,j\right)  \in J^{\prime}\times
J\right)  ,
\]
and a family $\left(  f_{i}\right)  _{i\in J^{\prime}\times J}$ of elements of
$\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ by%
\[
\left(  f_{\left(  k,j\right)  }=g_{j}\circ F_{k}\ \ \ \ \ \ \ \ \ \ \text{for
all }\left(  k,j\right)  \in J^{\prime}\times J\right)  .
\]
(Notice that these families $\left(  a_{i}\right)  _{i\in J^{\prime}\times J}$
and $\left(  f_{i}\right)  _{i\in J^{\prime}\times J}$ are subfamilies of the
families $\left(  a_{i}\right)  _{i\in G\times J}$ and $\left(  f_{i}\right)
_{i\in G\times J}$ from the original solution to Exercise \ref{exe.Indrho}%
(b).) It can then be shown that $\left(  J^{\prime}\times J,\left(
a_{i}\right)  _{i\in J^{\prime}\times J},\left(  f_{i}\right)  _{i\in
J^{\prime}\times J}\right)  $ is a finite dual generating system for
$\Ind_{\rho}U$ (though the proof is somewhat different
from the way this was shown in the solution to Exercise \ref{exe.Indrho}(b)).
This allows us to use Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} to
derive $\chi_{\Ind_{\rho}U}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}$. All details are left to the reader.}. Even more
generally, Exercise \ref{exe.Indrho}(b) holds whenever $\CC$ is
replaced by any \textbf{commutative ring} $A$ in which $\left\vert
\ker\rho\right\vert $ is invertible, as long as $U$ is assumed
to be a finitely generated projective $A$-module. However, in order to make
sense of this statement, one needs to know that $\operatorname*{Ind}%
\nolimits_{\rho}U$ is a finitely generated projective $A$-module as well, and
one needs to know how the trace of an endomorphism of a finitely generated
projective $A$-module is defined. We essentially did most of this in our
solution above\footnote{The only missing link is the fact that an $A$-module
is finitely generated and projective if and only if it has a finite dual
generating system; this fact is easy to prove.}.

\item Exercise \ref{exe.Indrho}(c) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. Again, our solution is still valid in this generality. Again, if
we define $\Ind_{\rho}f$ by
(\ref{sol.Indrho.generalize.Indf}) and define $\operatorname*{Ind}%
\nolimits_{H}^{G}f$ by Exercise \ref{exe.induction.characters}(b), then
Exercise \ref{exe.Indrho}(c) even holds when $\CC$ is replaced by any
arbitrary commutative ring. (The definition of $\operatorname*{Ind}%
\nolimits_{\rho}f$ requires $\left\vert \ker\rho\right\vert $
to be invertible, but this is automatically satisfied since $\rho$ is injective.)

\item Exercise \ref{exe.Indrho}(d) is still valid when $\CC$ is
replaced by any commutative ring whatsoever. Again, our solution is still
valid in this generality.

\item Exercise \ref{exe.Indrho}(e) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. Again, our solution is still valid in this generality. Again, it
is possible to prove Exercise \ref{exe.Indrho}(e) also when $\CC$ is
replaced by any commutative ring in which $\left\vert \ker%
\rho\right\vert $ is invertible (as long as $\operatorname*{Ind}%
\nolimits_{\rho}f$ is defined using (\ref{sol.Indrho.generalize.Indf})). (This
is actually very easy to check -- arguably even easier than our above solution
of Exercise \ref{exe.Indrho}(e).)

\item Exercise \ref{exe.Indrho}(f) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert K\right\vert $ is
invertible. Both solutions of Exercise \ref{exe.Indrho}(f) given above still
remain valid in this generality.

\item The second claim of Exercise \ref{exe.Indrho}(g) (that is, the equality
(\ref{eq.exe.Indrho.g.bil})) is still valid when $\CC$ is replaced by
any commutative ring in which $\left\vert G\right\vert $ and $\left\vert
H\right\vert $ are invertible. The first claim, a priori, makes no sense when
$\CC$ is replaced by an arbitrary commutative ring, because the
definition of the Hermitian forms $\left(  \cdot,\cdot\right)  _{G}$ and
$\left(  \cdot,\cdot\right)  _{H}$ involves complex conjugation (which is only
defined on $\CC$). However, it turns out that this complex conjugation
can be replaced by \textbf{any map from the base ring to itself} and Exercise
\ref{exe.Indrho}(g) remains valid\footnote{More precisely: Let $A$ be any
commutative ring in which $\left\vert G\right\vert $ and $\left\vert
H\right\vert $ are invertible. Fix an arbitrary map $\operatorname*{conj}%
:A\rightarrow A$ (not necessarily linear). For every $a\in A$, let
$\overline{a}$ denote the image $\operatorname*{conj}\left(  a\right)  $ of
$a$ under this map. For any $f_{1}\in R_{A}\left(  G\right)  $ and $f_{2}\in
R_{A}\left(  G\right)  $, define $\left(  f_{1},f_{2}\right)  _{G}\in A$ by
$\left(  f_{1},f_{2}\right)  _{G}=\dfrac{1}{\left\vert G\right\vert }%
\sum_{g\in G}f_{1}\left(  g\right)  \overline{f_{2}\left(  g\right)  }$ (using
the notation $\overline{a}$ that we just introduced). Similarly, define
$\left(  f_{1},f_{2}\right)  _{H}\in A$ for any $f_{1}\in R_{A}\left(
H\right)  $ and $f_{2}\in R_{A}\left(  H\right)  $. Then, Exercise
\ref{exe.Indrho}(g) is still valid when $\CC$ is replaced by $A$. (And
our solution to this exercise still applies.)}.

\item Exercise \ref{exe.Indrho}(h) is still valid when $\CC$ is
replaced by any commutative ring whatsoever. Again, our solution is still
valid in this generality.

\item Exercise \ref{exe.Indrho}(i) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert \ker%
\rho\right\vert $ is invertible. Again, our solution is still valid in this
generality (because if we set $K=\ker\rho$ as we did in our
solution, then $\left\vert \underbrace{K}_{=\ker\rho
}\right\vert =\left\vert \ker\rho\right\vert $ is invertible).

\item Exercise \ref{exe.Indrho}(j) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert \ker%
\rho\right\vert $ is invertible.

\item Exercise \ref{exe.Indrho}(k) still gives a proof of the formula
(\ref{induced-character-formula}) when $\CC$ is replaced by any field
in which $\left\vert H\right\vert $ is invertible.

\item Exercise \ref{exe.Indrho}(l) still gives a proof of the formula
(\ref{deflation-character-formula}) when $\CC$ is replaced by any field
in which $\left\vert G\right\vert $ and $\left\vert K\right\vert $ are invertible.
\end{itemize}

We actually tailored our above solutions to apply in a reasonably high
generality (although we did not state Exercise \ref{exe.Indrho} in this
generality). Had we contented ourselves with solving no more and no less than
what Exercise \ref{exe.Indrho} demanded of us, we could have obtained much
shorter solutions of some parts of it:

\begin{itemize}
\item Exercise \ref{exe.Indrho}(b) can be obtained by combining parts (g) and
(h) of Exercise \ref{exe.Indrho} with the following two
observations\footnote{We only sketch this argument; the details are left to
the reader.}:

\begin{itemize}
\item Every $\CC G$-module $V$ satisfies
$\chi_{\Res_{\rho}V}=\Res_{\rho}\chi_{V}$. (This is obvious.)

\item If $\alpha$ and $\beta$ are two elements of $R_{\CC }\left(
G\right)  $ such that every finite-dimensional $\CC G$-module $V$
satisfies $\left(  \alpha,\chi_{V}\right)  _{G}=\left(  \beta,\chi_{V}\right)
_{G}$, then $\alpha=\beta$. (This follows from the fact that $R\left(
G\right)  $ spans the $\CC$-vector space $R_{\CC }\left(
G\right)  $ and from the fact that the Hermitian form $\left(  \cdot
,\cdot\right)  _{G}$ is nondegenerate.)
\end{itemize}

However, this quick argument only works when the base ring is $\CC$. It
cannot be easily generalized to the case when $\CC$ is replaced by any
field in which $\left\vert H\right\vert $ is invertible.

\item Part (e) (at least in the case when $U$ is finite-dimensional) and part
(f) of Exercise \ref{exe.Indrho} can be derived from each other using the
correspondence between irreducible representations and irreducible characters.
Again, this argument is quick but hard to generalize.
\end{itemize}

]
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Indrho}}

\textit{Solution to Exercise \ref{exe.Indrho}.} In the following, we will use
the following convention: Whenever $K$ is a group, and $k$ is an element of
$K$, we shall write $k$ for the element $t_{k}$ of ${\CC }K$. This is a
relatively common abuse of notation, and it is harmless because the map
$K\rightarrow{\CC }K,\ k\mapsto t_{k}$ is an injective homomorphism of
multiplicative monoids (so $t_{gh}=t_{g}t_{h}$ and $t_{1}=1$, which means that
we won't run into ambiguities denoting $t_{k}$ by $k$) and because every
${\CC }K$-module $M$, every $m\in M$ and every $k\in K$ satisfy
$km=t_{k}m$.

Let us first notice a trivial fact: If $K$ is a finite group, and if
$f:K\rightarrow\CC $ is any function, then we have the following
equivalence:%
\begin{align}
&  \left(  f\in R_{\CC }\left(  K\right)  \right) \nonumber\\
\Longleftrightarrow\  &  \left(  f\text{ is a class function on }K\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }R_{\CC }\left(  K\right)  \text{ is the set of all class
functions on }K\\
\text{(by the definition of }R_{\CC }\left(  K\right)  \text{)}%
\end{array}
\right) \nonumber\\
\Longleftrightarrow\  &  \left(  f\text{ is constant on }K\text{-conjugacy
classes}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since a class function on }K\text{ is defined to mean a function}\\
K\rightarrow\CC \text{ which is constant on }K\text{-conjugacy classes}%
\end{array}
\right) \nonumber\\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }K\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime
}\right)  \right)  . \label{sol.Indrho.RCK}%
\end{align}


(a) Let $f\in R_{\CC }\left(  H\right)  $.

We can apply (\ref{sol.Indrho.RCK}) to $K=H$. As a consequence, we obtain the
following equivalence:%
\begin{align*}
&  \left(  f\in R_{\CC }\left(  H\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }H\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime
}\right)  \right)  .
\end{align*}
Hence,
\begin{equation}
\text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of }H\text{
satisfy }f\left(  k\right)  =f\left(  k^{\prime}\right)
\label{sol.Indrho.a.1}%
\end{equation}
(because we know that $f\in R_{\CC }\left(  H\right)  $).

Now, let $g$ and $g^{\prime}$ be two conjugate elements of $G$. Then, there
exists a $p\in G$ such that $g^{\prime}=pgp^{-1}$ (since $g$ and $g^{\prime}$
are conjugate). Consider this $p$.

Applying the map $\Ind_{\rho}f$ to both sides of the
equality $g^{\prime}=pgp^{-1}$, we obtain%
\begin{align*}
\left(  \Ind_{\rho}f\right)  \left(  g^{\prime
}\right)   &  =\left(  \Ind_{\rho}f\right)  \left(
pgp^{-1}\right)  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits
_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(  h\right)
k^{-1}=pgp^{-1}}}}_{=\sum_{h\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
h\right)  k^{-1}=pgp^{-1}}}}f\left(  h\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Ind}%
\nolimits_{\rho}f\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\underbrace{\sum\limits
_{\substack{k\in G;\\k\rho\left(  h\right)  k^{-1}=pgp^{-1}}}f\left(
h\right)  }_{\substack{=\sum\limits_{\substack{k\in G;\\pk\rho\left(  h\right)
\left(  pk\right)  ^{-1}=pgp^{-1}}}f\left(  h\right)  \\\text{(here, we have
substituted }pk\text{ for }k\text{ in the sum,}\\\text{since the map
}G\rightarrow G,\ k\mapsto pk\text{ is a bijection)}}}\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\underbrace{\sum\limits
_{\substack{k\in G;\\pk\rho\left(  h\right)  \left(  pk\right)  ^{-1}%
=pgp^{-1}}}}_{\substack{=\sum\limits_{\substack{k\in G;\\pk\rho\left(  h\right)
k^{-1}p^{-1}=pgp^{-1}}}\\\text{(since }\left(  pk\right)  ^{-1}=k^{-1}%
p^{-1}\text{)}}}f\left(  h\right)  =\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\underbrace{\sum\limits_{\substack{k\in G;\\pk\rho\left(  h\right)
k^{-1}p^{-1}=pgp^{-1}}}}_{\substack{=\sum\limits_{\substack{k\in G;\\pk\rho\left(
h\right)  k^{-1}=pg}}\\\text{(since }pk\rho\left(  h\right)  k^{-1}%
p^{-1}=pgp^{-1}\text{ is}\\\text{equivalent to }pk\rho\left(  h\right)
k^{-1}=pg\text{)}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\underbrace{\sum\limits
_{\substack{k\in G;\\pk\rho\left(  h\right)  k^{-1}=pg}}}_{\substack{=\sum\limits
_{\substack{k\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\\\text{(since }%
pk\rho\left(  h\right)  k^{-1}=pg\text{ is}\\\text{equivalent to }k\rho\left(
h\right)  k^{-1}=g\text{)}}}f\left(  h\right)  =\dfrac{1}{\left\vert
H\right\vert }\underbrace{\sum_{h\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
h\right)  k^{-1}=g}}}_{=\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\k\rho\left(  h\right)  k^{-1}=g}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}f\left(  h\right)  =\left(
\Ind_{\rho}f\right)  \left(  g\right)
\end{align*}
(since the definition of $\Ind_{\rho}f$ yields
$\left(  \Ind_{\rho}f\right)  \left(  g\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}f\left(  h\right)  $).

Let us now forget that we fixed $g$ and $g^{\prime}$. We thus have shown that
any two conjugate elements $g$ and $g^{\prime}$ of $G$ satisfy $\left(
\Ind_{\rho}f\right)  \left(  g\right)  =\left(
\Ind_{\rho}f\right)  \left(  g^{\prime}\right)  $.
Renaming $g$ and $g^{\prime}$ as $k$ and $k^{\prime}$ in this statement, we
obtain the following: Any two conjugate elements $k$ and $k^{\prime}$ of $G$
satisfy $\left(  \Ind_{\rho}f\right)  \left(
k\right)  =\left(  \Ind_{\rho}f\right)  \left(
k^{\prime}\right)  $.

But (\ref{sol.Indrho.RCK}) (applied to $G$ and $\operatorname*{Ind}%
\nolimits_{\rho}f$ instead of $K$ and $f$) yields the following equivalence:%
\begin{align*}
&  \left(  \Ind_{\rho}f\in R_{\CC }\left(
G\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }G\text{ satisfy }\left(  \operatorname*{Ind}%
\nolimits_{\rho}f\right)  \left(  k\right)  =\left(  \operatorname*{Ind}%
\nolimits_{\rho}f\right)  \left(  k^{\prime}\right)  \right)  .
\end{align*}
Thus, $\Ind_{\rho}f\in R_{\CC }\left(  G\right)
$ (because we know that any two conjugate elements $k$ and $k^{\prime}$ of $G$
satisfy $\left(  \Ind_{\rho}f\right)  \left(
k\right)  =\left(  \Ind_{\rho}f\right)  \left(
k^{\prime}\right)  $). This solves Exercise \ref{exe.Indrho}(a).

(b) Let us first introduce an elementary notion from linear algebra: the
notion of \emph{finite dual generating systems}.

\begin{definition}
Let $\mathbb{K}$ be a commutative ring. Let $V$ be a $\mathbb{K}$-module. A
\dfn{finite dual generating system} for $V$ means a triple $\left(
I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $, where

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}f_{i}\left(
v\right)  a_{i}$.
\end{definition}

In the following, we shall only use finite dual generating systems in the case
when $\mathbb{K}$ is a field; nevertheless, they are more useful in the
general case.

The first question one might ask about finite dual generating systems for $V$
is when they exist. The answer is very simple when $\mathbb{K}$ is a field:

\begin{proposition}
\label{prop.sol.Indrho.b.dualgens.findim}Let $\mathbb{K}$ be a field. Let $V$
be a $\mathbb{K}$-vector space. Then, a finite dual generating system for $V$
exists if and only if the vector space $V$ is finite-dimensional.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.b.dualgens.findim}.]Proposition
\ref{prop.sol.Indrho.b.dualgens.findim} is an ``if and only if''
statement. Hence, in order to prove Proposition
\ref{prop.sol.Indrho.b.dualgens.findim}, it is sufficient to verify the
following two claims:

\begin{statement}
\textit{Claim 1:} If a finite dual generating system for $V$ exists, then the
vector space $V$ is finite-dimensional.
\end{statement}

\begin{statement}
\textit{Claim 2:} If the vector space $V$ is finite-dimensional, then a finite
dual generating system for $V$ exists.
\end{statement}

Let us now prove these two claims.

\textit{Proof of Claim 1.} Assume that a finite dual generating system for $V$
exists. Let $\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)
_{i\in I}\right)  $ be such a finite dual generating system for $V$.

We know that $\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)
_{i\in I}\right)  $ is a finite dual generating system for $V$. In other
words, $\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in
I}\right)  $ is a triple such that

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}f_{i}\left(
v\right)  a_{i}$\ \ \ \ \footnote{because $\left(  I,\left(  a_{i}\right)
_{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ is a finite dual
generating system for $V$ if and only if $\left(  I,\left(  a_{i}\right)
_{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ is a triple such that
\par
\begin{itemize}
\item $I$ is a finite set;
\par
\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;
\par
\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}
\par
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}f_{i}\left(
v\right)  a_{i}$ (according to the definition of a ``finite
dual generating system for $V$'')}.

Now, every $v\in V$ satisfies $v=\sum\limits_{i\in I}\underbrace{f_{i}\left(
v\right)  }_{\in\mathbb{K}}a_{i}\in\sum\limits_{i\in I}\mathbb{K}a_{i}$. Thus,
$V\subset\sum\limits_{i\in I}\mathbb{K}a_{i}$. Combined with the (obvious)
inclusion $\sum\limits_{i\in I}\mathbb{K}a_{i}\subset V$, this yields
$V=\sum\limits_{i\in I}\mathbb{K}a_{i}$. But the vector space $\sum
\limits_{i\in I}\mathbb{K}a_{i}$ is finite-dimensional (since $I$ is a finite
set). In other words, the vector space $V$ is finite-dimensional (since
$V=\sum\limits_{i\in I}\mathbb{K}a_{i}$). This proves Claim 1.

\textit{Proof of Claim 2.} Assume that $V$ is a finite-dimensional vector
space. Then, $V$ has a finite basis. Let $\left(  e_{i}\right)  _{i\in I}$ be
such a basis. Thus, $I$ is a finite set, and $\left(  e_{i}\right)  _{i\in I}$
is a basis of the $\mathbb{K}$-vector space $V$. Let $\left(  e_{i}^{\ast
}\right)  _{i\in I}$ be the basis of $V^{\ast}$ dual to the basis $\left(
e_{i}\right)  _{i\in I}$ of $V$. (This is well-defined, since $V$ is finite-dimensional.)

We know that $\left(  e_{i}^{\ast}\right)  _{i\in I}$ is the basis of
$V^{\ast}$ dual to the basis $\left(  e_{i}\right)  _{i\in I}$ of $V$. Thus,%
\begin{equation}
e_{i}^{\ast}\left(  e_{j}\right)  =\delta_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for
all }i\in I\text{ and }j\in I. \label{pf.prop.sol.Indrho.b.dualgens.findim.1}%
\end{equation}


Now, every $v\in V$ satisfies $v=\sum\limits_{i\in I}e_{i}^{\ast}\left(
v\right)  e_{i}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$. Then, we can
write $v$ in the form $v=\sum\limits_{i\in I}\lambda_{i}e_{i}$ for some family
$\left(  \lambda_{i}\right)  _{i\in I}\in\mathbb{K}^{I}$ (since $\left(
e_{i}\right)  _{i\in I}$ is a basis of $V$). Consider this family $\left(
\lambda_{i}\right)  _{i\in I}$. Now, $v=\sum\limits_{i\in I}\lambda_{i}%
e_{i}=\sum\limits_{j\in I}\lambda_{j}e_{j}$ (here, we have renamed the
summation index $i$ as $j$ in the sum). For each $i\in I$, we now have%
\begin{align*}
e_{i}^{\ast}\left(  \underbrace{v}_{=\sum\limits_{j\in I}\lambda_{j}e_{j}%
}\right)   &  =e_{i}^{\ast}\left(  \sum\limits_{j\in I}\lambda_{j}%
e_{j}\right)  =\sum\limits_{j\in I}\lambda_{j}\underbrace{e_{i}^{\ast}\left(
e_{j}\right)  }_{\substack{=\delta_{i,j}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.findim.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }e_{i}^{\ast}\text{ is }\mathbb{K}\text{-linear}\right) \\
&  =\sum\limits_{j\in I}\lambda_{j}\delta_{i,j}=\sum\limits_{\substack{j\in
I;\\j=i}}\lambda_{j}\underbrace{\delta_{i,j}}_{\substack{=1\\\text{(since
}i=j\\\text{(since }j=i\text{))}}}+\sum\limits_{\substack{j\in I;\\j\neq
i}}\lambda_{j}\underbrace{\delta_{i,j}}_{\substack{=0\\\text{(since }i\neq
j\\\text{(since }j\neq i\text{))}}}=\sum\limits_{\substack{j\in I;\\j=i}%
}\lambda_{j}1+\underbrace{\sum\limits_{\substack{j\in I;\\j\neq i}}\lambda
_{j}0}_{=0}\\
&  =\sum\limits_{\substack{j\in I;\\j=i}}\lambda_{j}1=\lambda_{i}1=\lambda
_{i}.
\end{align*}
Thus, $\sum\limits_{i\in I}\underbrace{e_{i}^{\ast}\left(  v\right)
}_{=\lambda_{i}}e_{i}=\sum\limits_{i\in I}\lambda_{i}e_{i}$. Compared with
$v=\sum\limits_{i\in I}\lambda_{i}e_{i}$, this yields $v=\sum\limits_{i\in
I}e_{i}^{\ast}\left(  v\right)  e_{i}$, qed.}.

So we know that $\left(  I,\left(  e_{i}\right)  _{i\in I},\left(  e_{i}%
^{\ast}\right)  _{i\in I}\right)  $ is a triple such that

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  e_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  e_{i}^{\ast}\right)  _{i\in I}$ is a family of elements of
$V^{\ast}$ (where $V^{\ast}$ means $\Hom_{\mathbb{K}%
}\left(  V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}e_{i}^{\ast}\left(
v\right)  e_{i}$. In other words, $\left(  I,\left(  e_{i}\right)  _{i\in
I},\left(  e_{i}^{\ast}\right)  _{i\in I}\right)  $ is a finite dual
generating system for $V$\ \ \ \ \footnote{because $\left(  I,\left(
e_{i}\right)  _{i\in I},\left(  e_{i}^{\ast}\right)  _{i\in I}\right)  $ is a
finite dual generating system for $V$ if and only if $\left(  I,\left(
e_{i}\right)  _{i\in I},\left(  e_{i}^{\ast}\right)  _{i\in I}\right)  $ is a
triple such that
\par
\begin{itemize}
\item $I$ is a finite set;
\par
\item $\left(  e_{i}\right)  _{i\in I}$ is a family of elements of $V$;
\par
\item $\left(  e_{i}^{\ast}\right)  _{i\in I}$ is a family of elements of
$V^{\ast}$ (where $V^{\ast}$ means $\Hom_{\mathbb{K}%
}\left(  V,\mathbb{K}\right)  $)
\end{itemize}
\par
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}e_{i}^{\ast}\left(
v\right)  e_{i}$ (according to the definition of a ``finite
dual generating system for $V$'')}. Thus, a finite dual
generating system for $V$ exists (namely, $\left(  I,\left(  e_{i}\right)
_{i\in I},\left(  e_{i}^{\ast}\right)  _{i\in I}\right)  $). This proves Claim 2.

Now, both Claim 1 and Claim 2 are proven. Hence, the proof of Proposition
\ref{prop.sol.Indrho.b.dualgens.findim} is complete.
\end{proof}

[\textit{Remark:} The notion of finite dual generating system for $V$ is more
versatile than the notion of a finite basis of $V$. One difference between
these notions is that all bases of $V$ have the same size, while the set $I$
in a finite dual generating system $\left(  I,\left(  a_{i}\right)  _{i\in
I},\left(  f_{i}\right)  _{i\in I}\right)  $ of $V$ can have any (finite) size
$\geq\dim V$. Another difference manifests itself in the general setting when
$\mathbb{K}$ is a commutative ring, not necessarily a field. In this
generality, a finite basis of $V$ exists if and only if $V$ is a finite free
$\mathbb{K}$-module (this is the definition of a finite free $\mathbb{K}%
$-module), whereas a finite dual generating system for $V$ exists if and only
if $V$ is a finitely generated projective $\mathbb{K}$-module. Projective
$\mathbb{K}$-modules are a more frequent occurrence in commutative algebra
than free $\mathbb{K}$-modules, and in the absence of a finite basis, a finite
dual generating system is the thing that comes closest to allowing
``computing in a basis''.]

One significant application of finite dual generating systems is computing
traces of endomorphisms:

\begin{proposition}
\label{prop.sol.Indrho.b.dualgens.Tr}Let $\mathbb{K}$ be a field. Let $V$ be a
finite-dimensional $\mathbb{K}$-vector space. Let $\left(  I,\left(
a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ be a finite
dual generating system for $V$. Let $T:V\rightarrow V$ be a $\mathbb{K}%
$-linear map. Then,%
\[
\operatorname*{trace}T=\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  .
\]

\end{proposition}

Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} can be easily proven directly,
but let us take a slight detour and derive it from the following more general fact:

\begin{proposition}
\label{prop.sol.Indrho.b.dualgens.Tr.ring}Let $\mathbb{K}$ be a commutative
ring. Let $V$ be a $\mathbb{K}$-module. Let $\left(  I,\left(  a_{i}\right)
_{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ be a finite dual
generating system for $V$. Let $\left(  J,\left(  b_{j}\right)  _{j\in
J},\left(  g_{j}\right)  _{j\in J}\right)  $ be a further finite dual
generating system for $V$. Let $T:V\rightarrow V$ be a $\mathbb{K}$-linear
map. Then,%
\[
\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  =\sum_{j\in J}g_{j}\left(
Tb_{j}\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.b.dualgens.Tr.ring}.]We know that
$\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in
I}\right)  $ is a finite dual generating system for $V$. In other words,
$\left(  I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in
I}\right)  $ is a triple such that

\begin{itemize}
\item $I$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;

\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}f_{i}\left(
v\right)  a_{i}$\ \ \ \ \footnote{because $\left(  I,\left(  a_{i}\right)
_{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ is a finite dual
generating system for $V$ if and only if $\left(  I,\left(  a_{i}\right)
_{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ is a triple such that
\par
\begin{itemize}
\item $I$ is a finite set;
\par
\item $\left(  a_{i}\right)  _{i\in I}$ is a family of elements of $V$;
\par
\item $\left(  f_{i}\right)  _{i\in I}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}
\par
such that every $v\in V$ satisfies $v=\sum\limits_{i\in I}f_{i}\left(
v\right)  a_{i}$ (according to the definition of a ``finite
dual generating system for $V$'')}.

We know that $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)
_{j\in J}\right)  $ is a finite dual generating system for $V$. In other
words, $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  $ is a triple such that

\begin{itemize}
\item $J$ is a finite set;

\item $\left(  b_{j}\right)  _{j\in J}$ is a family of elements of $V$;

\item $\left(  g_{j}\right)  _{j\in J}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{j\in J}g_{j}\left(
v\right)  b_{j}$\ \ \ \ \footnote{because $\left(  J,\left(  b_{j}\right)
_{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  $ is a finite dual
generating system for $V$ if and only if $\left(  J,\left(  b_{j}\right)
_{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  $ is a triple such that
\par
\begin{itemize}
\item $J$ is a finite set;
\par
\item $\left(  b_{j}\right)  _{j\in J}$ is a family of elements of $V$;
\par
\item $\left(  g_{j}\right)  _{j\in J}$ is a family of elements of $V^{\ast}$
(where $V^{\ast}$ means $\Hom_{\mathbb{K}}\left(
V,\mathbb{K}\right)  $)
\end{itemize}
\par
such that every $v\in V$ satisfies $v=\sum\limits_{j\in J}g_{j}\left(
v\right)  b_{j}$ (according to the definition of a ``finite
dual generating system for $V$'')}.

So we know that every $v\in V$ satisfies%
\begin{equation}
v=\sum\limits_{i\in I}f_{i}\left(  v\right)  a_{i}
\label{pf.prop.sol.Indrho.b.dualgens.Tr.ring.1}%
\end{equation}
and%
\begin{equation}
v=\sum\limits_{j\in J}g_{j}\left(  v\right)  b_{j}.
\label{pf.prop.sol.Indrho.b.dualgens.Tr.ring.2}%
\end{equation}


Now,%
\begin{align*}
&  \sum_{i\in I}f_{i}\left(  \underbrace{Ta_{i}}_{\substack{=\sum\limits_{j\in
J}g_{j}\left(  Ta_{i}\right)  b_{j}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.ring.2}), applied}\\\text{to }%
v=Ta_{i}\text{)}}}\right) \\
&  =\sum_{i\in I}\underbrace{f_{i}\left(  \sum\limits_{j\in J}g_{j}\left(
Ta_{i}\right)  b_{j}\right)  }_{\substack{=\sum\limits_{j\in J}g_{j}\left(
Ta_{i}\right)  f_{i}\left(  b_{j}\right)  \\\text{(since the map }f_{i}\text{
is }\mathbb{K}\text{-linear)}}}=\underbrace{\sum_{i\in I}\sum\limits_{j\in J}%
}_{=\sum_{j\in J}\sum_{i\in I}}\underbrace{g_{j}\left(  Ta_{i}\right)
f_{i}\left(  b_{j}\right)  }_{\substack{=f_{i}\left(  b_{j}\right)
g_{j}\left(  Ta_{i}\right)  }}\\
&  =\sum_{j\in J}\sum_{i\in I}f_{i}\left(  b_{j}\right)  g_{j}\left(
Ta_{i}\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \sum_{j\in J}g_{j}\left(  T\underbrace{b_{j}}_{\substack{=\sum\limits_{i\in
I}f_{i}\left(  b_{j}\right)  a_{i}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.ring.1}), applied}\\\text{to }%
v=b_{j}\text{)}}}\right) \\
&  =\sum_{j\in J}g_{j}\left(  \underbrace{T\sum\limits_{i\in I}f_{i}\left(
b_{j}\right)  a_{i}}_{\substack{=\sum_{i\in I}f_{i}\left(  b_{j}\right)
Ta_{i}\\\text{(since the map }T\text{ is }\mathbb{K}\text{-linear)}}}\right)
=\sum_{j\in J}\underbrace{g_{j}\left(  \sum_{i\in I}f_{i}\left(  b_{j}\right)
Ta_{i}\right)  }_{\substack{=\sum\limits_{i\in I}f_{i}\left(  b_{j}\right)
g_{j}\left(  Ta_{i}\right)  \\\text{(since the map }g_{j}\text{ is }%
\mathbb{K}\text{-linear)}}}=\sum_{j\in J}\sum\limits_{i\in I}f_{i}\left(
b_{j}\right)  g_{j}\left(  Ta_{i}\right)  ,
\end{align*}
this yields $\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  =\sum_{j\in J}%
g_{j}\left(  Tb_{j}\right)  $. This proves Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr.ring}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.b.dualgens.Tr}.]The vector space
$V$ has a finite basis (since it is finite-dimensional). Let $\left(
e_{1},e_{2},\ldots,e_{n}\right)  $ be such a basis. Let $\left(
m_{i,j}\right)  _{1\leq i,j\leq n}\in\mathbb{K}^{n\times n}$ be the matrix
which represents the map $T:V\rightarrow V$ with respect to this basis
$\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ of $V$. Then,%
\begin{equation}
Te_{j}=\sum_{i=1}^{n}m_{i,j}e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
j\in\left\{  1,2,\ldots,n\right\}  \label{pf.prop.sol.Indrho.b.dualgens.Tr.1}%
\end{equation}
(due to the definition of ``the matrix which represents the
map $T:V\rightarrow V$ with respect to this basis $\left(  e_{1},e_{2}%
,\ldots,e_{n}\right)  $ of $V$'').

Let $\left(  e_{1}^{\ast},e_{2}^{\ast},\ldots,e_{n}^{\ast}\right)  $ be the
basis of $V^{\ast}$ dual to the basis $\left(  e_{1},e_{2},\ldots
,e_{n}\right)  $ of $V$. (This is well-defined, since $V$ is
finite-dimensional.) Thus,%
\begin{equation}
e_{i}^{\ast}\left(  e_{j}\right)  =\delta_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for
all }i\in\left\{  1,2,\ldots,n\right\}  \text{ and }j\in\left\{
1,2,\ldots,n\right\}  . \label{pf.prop.sol.Indrho.b.dualgens.Tr.delta}%
\end{equation}
(This follows immediately from the definition of a ``dual
basis''.) Now, every $k\in\left\{  1,2,\ldots,n\right\}  $ and
$j\in\left\{  1,2,\ldots,n\right\}  $ satisfy%
\begin{equation}
e_{k}^{\ast}\left(  Te_{j}\right)  =m_{k,j}.
\label{pf.prop.sol.Indrho.b.dualgens.Tr.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.Indrho.b.dualgens.Tr.2}):} Fix
$k\in\left\{  1,2,\ldots,n\right\}  $ and $j\in\left\{  1,2,\ldots,n\right\}
$. Applying the map $e_{k}^{\ast}$ to both sides of
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.1}), we obtain%
\begin{align*}
e_{k}^{\ast}\left(  Te_{j}\right)   &  =e_{k}^{\ast}\left(  \sum_{i=1}%
^{n}m_{i,j}e_{i}\right)  =\underbrace{\sum_{i=1}^{n}}_{=\sum_{i\in\left\{
1,2,\ldots,n\right\}  }}m_{i,j}\underbrace{e_{k}^{\ast}\left(  e_{i}\right)
}_{\substack{=\delta_{k,i}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.delta}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }e_{k}^{\ast}\text{ is }\mathbb{K}\text{-linear}\right) \\
&  =\sum_{i\in\left\{  1,2,\ldots,n\right\}  }m_{i,j}\delta_{k,i}%
=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}}m_{i,j}%
\underbrace{\delta_{k,i}}_{\substack{=1\\\text{(since }k=i\\\text{(since
}i=k\text{))}}}+\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq
k}}m_{i,j}\underbrace{\delta_{k,i}}_{\substack{=0\\\text{(since }k\neq
i\\\text{(since }i\neq k\text{))}}}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}}m_{i,j}%
1+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq
k}}m_{i,j}0}_{=0}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i=k}}m_{i,j}1=m_{k,j}1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\in\left\{
1,2,\ldots,n\right\}  \right) \\
&  =m_{k,j}.
\end{align*}
This proves (\ref{pf.prop.sol.Indrho.b.dualgens.Tr.2}).} Now, every $v\in V$
satisfies $v=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }e_{i}^{\ast
}\left(  v\right)  e_{i}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in V$. Then,
we can write $v$ in the form $v=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}$ for
some $n$-tuple $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)
\in\mathbb{K}^{n}$ (since $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ is a
basis of $V$). Consider this $n$-tuple $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{n}\right)  $. Now, $v=\sum\limits_{i=1}^{n}\lambda_{i}%
e_{i}=\sum\limits_{j=1}^{n}\lambda_{j}e_{j}$ (here, we have renamed the
summation index $i$ as $j$ in the sum). For each $i\in\left\{  1,2,\ldots
,n\right\}  $, we now have%
\begin{align*}
e_{i}^{\ast}\left(  \underbrace{v}_{=\sum\limits_{j=1}^{n}\lambda_{j}e_{j}%
}\right)   &  =e_{i}^{\ast}\left(  \sum\limits_{j=1}^{n}\lambda_{j}%
e_{j}\right)  =\underbrace{\sum\limits_{j=1}^{n}}_{=\sum_{j\in\left\{
1,2,\ldots,n\right\}  }}\lambda_{j}\underbrace{e_{i}^{\ast}\left(
e_{j}\right)  }_{\substack{=\delta_{i,j}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.delta}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }e_{i}^{\ast}\text{ is }\mathbb{K}\text{-linear}\right) \\
&  =\sum_{j\in\left\{  1,2,\ldots,n\right\}  }\lambda_{j}\delta_{i,j}%
=\sum\limits_{\substack{j\in\left\{  1,2,\ldots,n\right\}  ;\\j=i}}\lambda
_{j}\underbrace{\delta_{i,j}}_{\substack{=1\\\text{(since }i=j\\\text{(since
}j=i\text{))}}}+\sum\limits_{\substack{j\in\left\{  1,2,\ldots,n\right\}
;\\j\neq i}}\lambda_{j}\underbrace{\delta_{i,j}}_{\substack{=0\\\text{(since
}i\neq j\\\text{(since }j\neq i\text{))}}}=\sum\limits_{\substack{j\in
I;\\j=i}}\lambda_{j}1+\underbrace{\sum\limits_{\substack{j\in I;\\j\neq
i}}\lambda_{j}0}_{=0}\\
&  =\sum\limits_{\substack{j\in\left\{  1,2,\ldots,n\right\}  ;\\j=i}%
}\lambda_{j}1=\lambda_{i}1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
i\in\left\{  1,2,\ldots,n\right\}  \right) \\
&  =\lambda_{i}.
\end{align*}
Thus, $\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }}%
_{=\sum_{i=1}^{n}}\underbrace{e_{i}^{\ast}\left(  v\right)  }_{=\lambda_{i}%
}e_{i}=\sum\limits_{i=1}^{n}\lambda_{i}e_{i}$. Compared with $v=\sum
\limits_{i=1}^{n}\lambda_{i}e_{i}$, this yields $v=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }e_{i}^{\ast}\left(  v\right)  e_{i}$, qed.}.

So we know that $\left(  \left\{  1,2,\ldots,n\right\}  ,\left(  e_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  },\left(  e_{i}^{\ast}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }\right)  $ is a triple such that

\begin{itemize}
\item $\left\{  1,2,\ldots,n\right\}  $ is a finite set;

\item $\left(  e_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a
family of elements of $V$;

\item $\left(  e_{i}^{\ast}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is
a family of elements of $V^{\ast}$ (where $V^{\ast}$ means
$\Hom_{\mathbb{K}}\left(  V,\mathbb{K}\right)  $)
\end{itemize}

\noindent
such that every $v\in V$ satisfies $v=\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }e_{i}^{\ast}\left(  v\right)  e_{i}$. In other words,
\newline$\left(  \left\{  1,2,\ldots,n\right\}  ,\left(  e_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  },\left(  e_{i}^{\ast}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }\right)  $ is a finite dual generating
system for $V$\ \ \ \ \footnote{because $\left(  \left\{  1,2,\ldots
,n\right\}  ,\left(  e_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
},\left(  e_{i}^{\ast}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }\right)
$ is a finite dual generating system for $V$ if and only if $\left(  \left\{
1,2,\ldots,n\right\}  ,\left(  e_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  },\left(  e_{i}^{\ast}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }\right)  $ is a triple such that
\par
\begin{itemize}
\item $\left\{  1,2,\ldots,n\right\}  $ is a finite set;
\par
\item $\left(  e_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a
family of elements of $V$;
\par
\item $\left(  e_{i}^{\ast}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is
a family of elements of $V^{\ast}$ (where $V^{\ast}$ means
$\Hom_{\mathbb{K}}\left(  V,\mathbb{K}\right)  $)
\end{itemize}
\par
such that every $v\in V$ satisfies $v=\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }e_{i}^{\ast}\left(  v\right)  e_{i}$ (according to the definition
of a ``finite dual generating system for $V$'')}.
Therefore, Proposition \ref{prop.sol.Indrho.b.dualgens.Tr.ring} (applied
to $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  =\left(  \left\{  1,2,\ldots,n\right\}  ,\left(  e_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  },\left(  e_{i}^{\ast}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }\right)  $) yields%
\begin{align}
\sum_{i\in I}f_{i}\left(  Ta_{i}\right)   &  =\underbrace{\sum_{j\in\left\{
1,2,\ldots,n\right\}  }}_{=\sum_{j=1}^{n}}\underbrace{e_{j}^{\ast}\left(
Te_{j}\right)  }_{\substack{=m_{j,j}\\\text{(by
(\ref{pf.prop.sol.Indrho.b.dualgens.Tr.2}), applied}\\\text{to }k=j\text{)}%
}}=\sum_{j=1}^{n}m_{j,j}\nonumber\\
&  =\sum_{i=1}^{n}m_{i,i}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
the summation index }j\text{ as }i\right)  .
\label{pf.prop.sol.Indrho.b.dualgens.Tr.5}%
\end{align}


But recall that if $\mathfrak{G}$ is any endomorphism of the $\mathbb{K}%
$-vector space $V$, then the trace of $\mathfrak{G}$ equals the trace of any
matrix which represents $\mathfrak{G}$ with respect to a basis of
$V$\ \ \ \ \footnote{Indeed, this is how the trace of $\mathfrak{G}$ is
defined.}. Applying this to $\mathfrak{G}=T$, we conclude that the trace of
$T$ equals the trace of any matrix which represents $T$ with respect to a
basis of $V$. In other words, the trace of $T$ equals the trace of $M$
whenever $M$ is a matrix which represents $T$ with respect to a basis of $V$.
We can apply this to $M=\left(  m_{i,j}\right)  _{1\leq i,j\leq n}$ (since
$\left(  m_{i,j}\right)  _{1\leq i,j\leq n}$ is a matrix which represents $T$
with respect to a basis of $V$\ \ \ \ \footnote{namely, with respect to the
basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $}). As a consequence, we
conclude that the trace of $T$ equals the trace of $\left(  m_{i,j}\right)
_{1\leq i,j\leq n}$. In other words, $\operatorname*{trace}%
T=\operatorname*{trace}\left(  \left(  m_{i,j}\right)  _{1\leq i,j\leq
n}\right)  =\sum_{i=1}^{n}m_{i,i}$ (by the definition of
$\operatorname*{trace}\left(  \left(  m_{i,j}\right)  _{1\leq i,j\leq
n}\right)  $). Compared with (\ref{pf.prop.sol.Indrho.b.dualgens.Tr.5}), this
yields $\operatorname*{trace}T=\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  $.
This proves Proposition \ref{prop.sol.Indrho.b.dualgens.Tr}.
\end{proof}

[\textit{Remark:} Proposition \ref{prop.sol.Indrho.b.dualgens.Tr.ring} can be
used to define the trace of an endomorphism of a finitely generated projective
$\mathbb{K}$-module when $\mathbb{K}$ is a commutative ring. Indeed, if
$\mathbb{K}$ is a commutative ring and if $V$ is a finitely generated
projective $\mathbb{K}$-module, and if $T:V\rightarrow V$ is a $\mathbb{K}%
$-linear map, then the trace $\operatorname*{trace}\left(  T\right)  $ of $T$
can be defined as $\sum_{i\in I}f_{i}\left(  Ta_{i}\right)  $, where $\left(
I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $ is
a finite dual generating system for $V$. This notion of trace is
well-defined\footnote{because Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr.ring} shows that $\sum_{i\in I}f_{i}\left(
Ta_{i}\right)  $ does not depend on the choice of $\left(  I,\left(
a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $} and
generalizes the classical notion from linear algebra (which is defined only
for finitely generated \textbf{free} $\mathbb{K}$-modules)\footnote{because
Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} (or, more precisely, its
straightforward generalization to free $\mathbb{K}$-modules over commutative
rings) shows that these two notions give the same result when $V$ is a free
$\mathbb{K}$-module}. But we will not concern ourselves with these
generalizations, since our exercise deals only with representations of groups
over a field.]

After all these preparations, we finally come to the actual solution of
Exercise \ref{exe.Indrho}(b). Let $U$ be any finite-dimensional $\CC %
H$-module. We want to prove $\chi_{\Ind_{\rho}%
U}=\Ind_{\rho}\chi_{U}$.

For every $g\in G$, we define a $\CC$-linear map $g^{\ast}%
:\CC G\rightarrow\CC $ by
\begin{equation}
\left(  g^{\ast}\left(  k\right)  =\delta_{g,k}\ \ \ \ \ \ \ \ \ \ \text{for
all }k\in G\right)  . \label{sol.Indrho.b.g*.def}%
\end{equation}
(This is well-defined, since $\left(  k\right)  _{k\in G}$ is a basis of the
$\CC$-vector space $\CC G$.) Then,%
\begin{equation}
\sum_{g\in G}g^{\ast}\left(  \gamma\right)  \cdot g=\gamma
\ \ \ \ \ \ \ \ \ \ \text{for every }\gamma\in\CC G.
\label{sol.Indrho.b.g*.completeness}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.b.g*.completeness}):} Let
$\gamma\in\CC G$. We need to prove the equality
(\ref{sol.Indrho.b.g*.completeness}). We notice that this equality is
$\CC$-linear in $\gamma$. Hence, we can WLOG assume that $\gamma\in G$
(since $G$ is a basis of the $\CC$-vector space $\CC G$). Assume
this. Now,%
\begin{align*}
\sum_{g\in G}g^{\ast}\left(  \gamma\right)  \cdot g  &  =\sum_{g\in
G}\underbrace{g^{\ast}\left(  \gamma\right)  }_{\substack{=\delta_{g,\gamma
}\\\text{(by (\ref{sol.Indrho.b.g*.def}), applied}\\\text{to }k=\gamma\text{
(since }\gamma\in G\text{))}}}\cdot g=\sum_{g\in G}\delta_{g,\gamma}\cdot g\\
&  =\sum\limits_{\substack{g\in G;\\g=\gamma}}\underbrace{\delta_{g,\gamma}%
}_{\substack{=1\\\text{(since }g=\gamma\text{)}}}\cdot g+\sum\limits_{\substack{g\in
G;\\g\neq\gamma}}\underbrace{\delta_{g,\gamma}}_{\substack{=0\\\text{(since
}g\neq\gamma\text{)}}}\cdot g=\sum\limits_{\substack{g\in G;\\g=\gamma}}1\cdot
g+\underbrace{\sum\limits_{\substack{g\in G;\\g\neq\gamma}}0\cdot g}_{=0}\\
&  =\sum\limits_{\substack{g\in G;\\g=\gamma}}\underbrace{1\cdot g}_{=g}%
=\sum\limits_{\substack{g\in G;\\g=\gamma}}g=\gamma\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\gamma\in G\right)  .
\end{align*}
This proves (\ref{sol.Indrho.b.g*.completeness}).} Also,%
\begin{equation}
\left(  gr\right)  ^{\ast}\left(  \gamma r\right)  =g^{\ast}\left(
\gamma\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in G\text{, }r\in G\text{
and }\gamma\in\CC G \label{sol.Indrho.b.g*.r-mul}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.b.g*.r-mul}):} Let $g\in G$, $r\in
G$ and $\gamma\in\CC G$. We need to prove the equality
(\ref{sol.Indrho.b.g*.r-mul}). Since this equality is $\CC$-linear in
$\gamma$ (because the maps $\left(  gr\right)  ^{\ast}$ and $g^{\ast}$ are
$\CC$-linear), we can WLOG assume that $\gamma\in G$ (since $G$ is a
basis of the $\CC$-vector space $\CC G$). Assume this.
\par
We have $\gamma\in G$, and thus we can apply (\ref{sol.Indrho.b.g*.def}) to
$k=\gamma$. We thus obtain $g^{\ast}\left(  \gamma\right)  =\delta_{g,\gamma}%
$.
\par
We have $\gamma\in G$ and $r\in G$. Hence, $\underbrace{\gamma}_{\in
G}\underbrace{r}_{\in G}\in GG\subset G$. Hence, (\ref{sol.Indrho.b.g*.def})
(applied to $gr$ and $\gamma r$ instead of $g$ and $k$) yields $\left(
gr\right)  ^{\ast}\left(  \gamma r\right)  =\delta_{gr,\gamma r}$.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $gr=\gamma r$.
\par
\textit{Case 2:} We don't have $gr=\gamma r$.
\par
Let us first consider Case 1. In this case, we have $gr=\gamma r$. Thus,
$\delta_{gr,\gamma r}=1$. But the element $r$ of $G$ is invertible (since $G$
is a group). Hence, we can cancel $r$ from the equality $gr=\gamma r$, and
thus we obtain $g=\gamma$, so that $\delta_{g,\gamma}=1$. Hence, $\left(
gr\right)  ^{\ast}\left(  \gamma r\right)  =\delta_{gr,\gamma r}%
=1=\delta_{g,\gamma}=g^{\ast}\left(  \gamma\right)  $. Thus, $\left(
gr\right)  ^{\ast}\left(  \gamma r\right)  =g^{\ast}\left(  \gamma\right)  $
is proven in Case 1.
\par
Let us now consider Case 2. In this case, we don't have $gr=\gamma r$. Hence,
we have $gr\neq\gamma r$ and thus $\delta_{gr,\gamma r}=0$. But if we had
$g=\gamma$, then we would have $\underbrace{g}_{=\gamma}r=\gamma r$, which
would contradict the fact that we don't have $gr=\gamma r$. Hence, we don't
have $g=\gamma$. Thus, we have $g\neq\gamma$. Hence, $\delta_{g,\gamma}=0$.
Consequently, $\left(  gr\right)  ^{\ast}\left(  \gamma r\right)
=\delta_{gr,\gamma r}=0=\delta_{g,\gamma}=g^{\ast}\left(  \gamma\right)  $.
Thus, $\left(  gr\right)  ^{\ast}\left(  \gamma r\right)  =g^{\ast}\left(
\gamma\right)  $ is proven in Case 2.
\par
We now have proven $\left(  gr\right)  ^{\ast}\left(  \gamma r\right)
=g^{\ast}\left(  \gamma\right)  $ in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, we thus conclude that $\left(
gr\right)  ^{\ast}\left(  \gamma r\right)  =g^{\ast}\left(  \gamma\right)  $
always holds. This proves (\ref{sol.Indrho.b.g*.r-mul}).}.

Proposition \ref{prop.sol.Indrho.b.dualgens.findim} (applied to $\mathbb{K}%
=\CC $ and $V=U$) yields that a finite dual generating system for $U$
exists if and only if the vector space $U$ is finite-dimensional. Thus, a
finite dual generating system for $U$ exists (since the vector space $U$ is
finite-dimensional). Let us fix such a finite dual generating system for $U$,
and denote it by $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(
g_{j}\right)  _{j\in J}\right)  $.

Hence, $\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  $ is a finite dual generating system for $U$. In other words,
$\left(  J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in
J}\right)  $ is a triple such that

\begin{itemize}
\item $J$ is a finite set;

\item $\left(  b_{j}\right)  _{j\in J}$ is a family of elements of $U$;

\item $\left(  g_{j}\right)  _{j\in J}$ is a family of elements of $U^{\ast}$
(where $U^{\ast}$ means $\Hom_{\CC }\left(
U,\CC \right)  $)
\end{itemize}

\noindent
such that every $v\in U$ satisfies $v=\sum\limits_{j\in J}g_{j}\left(
v\right)  b_{j}$\ \ \ \ \footnote{because $\left(  J,\left(  b_{j}\right)
_{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  $ is a finite dual
generating system for $U$ if and only if $\left(  J,\left(  b_{j}\right)
_{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  $ is a triple such that
\par
\begin{itemize}
\item $J$ is a finite set;
\par
\item $\left(  b_{j}\right)  _{j\in J}$ is a family of elements of $U$;
\par
\item $\left(  g_{j}\right)  _{j\in J}$ is a family of elements of $U^{\ast}$
(where $U^{\ast}$ means $\Hom_{\CC }\left(
U,\CC \right)  $)
\end{itemize}
\par
such that every $v\in U$ satisfies $v=\sum\limits_{j\in J}g_{j}\left(
v\right)  b_{j}$ (according to the definition of a ``finite
dual generating system for $U$'')}.

We record the fact that every $v\in U$ satisfies%
\begin{equation}
v=\sum\limits_{j\in J}g_{j}\left(  v\right)  b_{j}.
\label{sol.Indrho.b.gj.completeness}%
\end{equation}


We now endow $\CC G$ with the $\left(  \CC G,\CC H\right)
$-bimodule structure that was used to define $\operatorname*{Ind}%
\nolimits_{\rho}U$. We recall that $\Ind_{\rho
}U=\CC G\otimes_{\CC H}U$ (by the definition of
$\Ind_{\rho}U$).

Let us fix $g\in G$. We define a map $\widetilde{F}_{g}:\CC G\times
U\rightarrow U$ by setting%
\[
\left(  \widetilde{F}_{g}\left(  \gamma,u\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  hu\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
\gamma,u\right)  \in\CC G\times U\right)  .
\]
The map $\widetilde{F}_{g}$ is $\CC H$-bilinear with respect to the
right $\CC H$-module structure on $\CC G$ and the left
$\CC H$-module structure on $U$\ \ \ \ \footnote{\textit{Proof.} We
notice that $\widetilde{F}_{g}\left(  \gamma,u\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  hu$ depends $\CC$-linearly on each of
$\gamma$ and $u$ (for obvious reasons). In other words, the map $\widetilde{F}%
_{g}$ is $\CC$-bilinear.
\par
Now, let us fix $\gamma\in\CC G$, $u\in U$ and $\kappa\in\CC H$.
We are going to prove the equality $\widetilde{F}_{g}\left(  \gamma
\kappa,u\right)  =\widetilde{F}_{g}\left(  \gamma,\kappa u\right)  $.
\par
Since this equality is $\CC$-linear in $\kappa$ (because the map
$\widetilde{F}_{g}$ is $\CC$-bilinear), we can WLOG assume that
$\kappa\in H$ (since $H$ is a basis of the $\CC$-vector space
$\CC H$). Assume this.
\par
We have $\kappa\in H$. Thus, the map $H\rightarrow H,\ h\mapsto h\kappa$ is a
bijection (since $H$ is a group). Hence, we can substitute $h\kappa$ for $h$
in the sum $\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\rho\left(  \kappa\right)  \right)  hu$. As a result, we
obtain $\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)  ^{\ast
}\left(  \gamma\rho\left(  \kappa\right)  \right)  hu=\sum\limits_{h\in
H}\left(  g\rho\left(  h\kappa\right)  \right)  ^{\ast}\left(  \gamma
\rho\left(  \kappa\right)  \right)  \left(  h\kappa\right)  u$.
\par
Now, the definition of $\widetilde{F}_{g}\left(  \gamma\kappa,u\right)  $
yields%
\begin{align*}
\widetilde{F}_{g}\left(  \gamma\kappa,u\right)   &  =\dfrac{1}{\left\vert
H\right\vert }
\underbrace{\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\rho\left(  \kappa\right)  \right)  hu}_{
=\sum\limits_{h\in
H}\left(  g\rho\left(  h\kappa\right)  \right)  ^{\ast}\left(  \gamma
\rho\left(  \kappa\right)  \right)  \left(  h\kappa\right)  u
}
=\dfrac
{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\left(  g\underbrace{\rho
\left(  h\kappa\right)  }_{\substack{=\rho\left(  h\right)  \rho\left(
\kappa\right)  \\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}%
}}\right)  ^{\ast}\left(  \gamma\rho\left(  \kappa\right)  \right)  \left(
h\kappa\right)  u\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\underbrace{\left(
g\rho\left(  h\right)  \rho\left(  \kappa\right)  \right)  ^{\ast}\left(
\gamma\rho\left(  \kappa\right)  \right)  }_{\substack{=\left(  g\rho\left(
h\right)  \right)  ^{\ast}\left(  \gamma\right)  \\\text{(by
(\ref{sol.Indrho.b.g*.r-mul}), applied to}\\g\rho\left(  h\right)  \text{ and
}\rho\left(  \kappa\right)  \text{ instead of }g\text{ and }r\text{)}}}h\kappa
u=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(
h\right)  \right)  ^{\ast}\left(  \gamma\right)  h\kappa u.
\end{align*}
Compared with%
\[
\widetilde{F}_{g}\left(  \gamma,\kappa u\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  h\kappa u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\widetilde{F}_{g}\left(  \gamma,\kappa u\right)  \right)
,
\]
this yields $\widetilde{F}_{g}\left(  \gamma\kappa,u\right)  =\widetilde{F}%
_{g}\left(  \gamma,\kappa u\right)  $.
\par
Now let us forget that we fixed $\gamma$, $u$ and $\kappa$. Thus, we have
shown that $\widetilde{F}_{g}\left(  \gamma\kappa,u\right)  =\widetilde{F}%
_{g}\left(  \gamma,\kappa u\right)  $ for all $\gamma\in\CC G$, $u\in U$
and $\kappa\in\CC H$. This yields that the map $\widetilde{F}_{g}$ is
$\CC H$-bilinear with respect to the right $\CC H$-module
structure on $\CC G$ and the left $\CC H$-module structure on $U$
(since we already know that this map $\widetilde{F}_{g}$ is $\CC %
$-bilinear). Qed.}. According to the universal property of the tensor product,
this yields that there exists a unique $\CC$-linear map $F_{g}%
:\CC G\otimes_{\CC H}U\rightarrow U$ such that%
\[
\left(  F_{g}\left(  \gamma\otimes_{\CC H}u\right)  =\widetilde{F}%
_{g}\left(  \gamma,u\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(
\gamma,u\right)  \in\CC G\times U\right)  .
\]
Consider this map $F_{g}$.

So we know that any $\left(  \gamma,u\right)  \in\CC G\times U$
satisfies%
\begin{equation}
F_{g}\left(  \gamma\otimes_{\CC H}u\right)  =\widetilde{F}_{g}\left(
\gamma,u\right)  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in
H}\left(  g\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)  hu.
\label{sol.Indrho.b.Fg.def}%
\end{equation}


Let us now forget that we fixed $g$. We thus have constructed a $\CC %
$-linear map $F_{g}:\CC G\otimes_{\CC H}U\rightarrow U$ for each
$g\in G$. We have shown that this map satisfies (\ref{sol.Indrho.b.Fg.def})
for any $\left(  \gamma,u\right)  \in\CC G\times U$.

The set $G\times J$ is finite (since the sets $G$ and $J$ are finite). Now, we
define a family $\left(  a_{i}\right)  _{i\in G\times J}$ of elements of
$\CC G\otimes_{\CC H}U$ by%
\[
\left(  a_{\left(  k,j\right)  }=k\otimes_{\CC H}b_{j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  k,j\right)  \in G\times J\right)
.
\]
Furthermore, we define a family $\left(  f_{i}\right)  _{i\in G\times J}$ of
elements of $\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ by%
\[
\left(  f_{\left(  k,j\right)  }=g_{j}\circ F_{k}\ \ \ \ \ \ \ \ \ \ \text{for
all }\left(  k,j\right)  \in G\times J\right)  .
\]
(This is well-defined because, for any $\left(  k,j\right)  \in G\times J$,
the composition $g_{j}\circ F_{k}$ of the $\CC$-linear maps
$F_{k}:\CC G\otimes_{\CC H}U\rightarrow U$ and $g_{j}%
:U\rightarrow\CC $ is a $\CC$-linear map $\CC %
G\otimes_{\CC H}U\rightarrow\CC $.)

Our goal is now to prove that $\left(  G\times J,\left(  a_{i}\right)  _{i\in
G\times J},\left(  f_{i}\right)  _{i\in G\times J}\right)  $ is a finite dual
generating system for $\CC G\otimes_{\CC H}U$.

Indeed, every $v\in\CC G\otimes_{\CC H}U$ satisfies $v=\sum
\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$%
\ \ \ \ \footnote{\textit{Proof.} Let $v\in\CC G\otimes_{\CC H}U$.
We need to prove the equality $v=\sum\limits_{i\in G\times J}f_{i}\left(
v\right)  a_{i}$. Since this equality is $\CC$-linear in $v$, we can
WLOG assume that $v$ is a pure tensor (since the pure tensors in
$\CC G\otimes_{\CC H}U$ span the whole $\CC$-vector space
$\CC G\otimes_{\CC H}U$). Assume this. Thus, $v$ is a pure tensor.
In other words, $v=\gamma\otimes_{\CC H}u$ for some $\gamma\in
\CC G$ and $u\in U$. Consider these $\gamma$ and $u$.
\par
We notice that for any $w\in G$, the map $G\rightarrow G,\ k\mapsto kw$ is a
bijection (since $G$ is a group). Hence, for any $w\in G$, we have%
\begin{align}
\sum_{k\in G}\left(  kw\right)  ^{\ast}\left(  \gamma\right)  kw  &
=\sum_{k\in G}k^{\ast}\left(  \gamma\right)  k\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }k\text{ for }kw\text{ in the sum,}\\
\text{since the map }G\rightarrow G,\ k\mapsto kw\text{ is a bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{g\in G}g^{\ast}\left(  \gamma\right)  g\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we renamed the summation index }k\text{ as }g\right) \nonumber\\
&  =\sum_{g\in G}g^{\ast}\left(  \gamma\right)  \cdot g=\gamma
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.b.g*.completeness}%
)}\right)  . \label{sol.Indrho.b.dual.pf.1}%
\end{align}
Now, every $h\in H$ satisfies%
\begin{align}
\sum_{k\in G}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
\gamma\right)  \underbrace{kh}_{\substack{=k\cdot\left(  \CC \left[
\rho\right]  \right)  \left(  h\right)  \\\text{(by the definition of
the}\\\text{right }\CC H\text{-module}\\\text{structure on }%
\CC G\text{)}}}  &  =\sum_{k\in G}\left(  k\rho\left(  h\right)
\right)  ^{\ast}\left(  \gamma\right)  k\cdot\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  }_{\substack{=\rho
\left(  h\right)  \\\text{(since }h\in H\text{)}}}=\sum_{k\in G}\left(
k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)  k\rho\left(
h\right) \nonumber\\
&  =\gamma\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.b.dual.pf.1}),
applied to }w=\rho\left(  h\right)  \right)  . \label{sol.Indrho.b.dual.pf.2}%
\end{align}
\par
Now,%
\begin{align*}
\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}  &  =\underbrace{\sum
\limits_{\left(  k,j\right)  \in G\times J}}_{=\sum_{k\in G}\sum_{j\in J}%
}\underbrace{f_{\left(  k,j\right)  }}_{=g_{j}\circ F_{k}}\left(  v\right)
\underbrace{a_{\left(  k,j\right)  }}_{=k\otimes_{\CC H}b_{j}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }\left(  k,j\right)
\text{ for the summation index }i\right) \\
&  =\sum_{k\in G}\underbrace{\sum_{j\in J}\left(  g_{j}\circ F_{k}\right)
\left(  v\right)  k\otimes_{\CC H}b_{j}}_{\substack{=k\otimes
_{\CC H}\left(  \sum_{j\in J}\left(  g_{j}\circ F_{k}\right)  \left(
v\right)  b_{j}\right)  \\\text{(since the tensor product is }\CC %
\text{-bilinear)}}}=\sum_{k\in G}k\otimes_{\CC H}\left(  \sum_{j\in
J}\underbrace{\left(  g_{j}\circ F_{k}\right)  \left(  v\right)  }%
_{=g_{j}\left(  F_{k}\left(  v\right)  \right)  }b_{j}\right) \\
&  =\sum_{k\in G}k\otimes_{\CC H}\underbrace{\left(  \sum_{j\in J}%
g_{j}\left(  F_{k}\left(  v\right)  \right)  b_{j}\right)  }_{\substack{=F_{k}%
\left(  v\right)  \\\text{(because }F_{k}\left(  v\right)  =\sum_{j\in J}%
g_{j}\left(  F_{k}\left(  v\right)  \right)  b_{j}\\\text{(by
(\ref{sol.Indrho.b.gj.completeness}), applied to }F_{k}\left(  v\right)
\text{ instead of }v\text{))}}}=\sum_{k\in G}k\otimes_{\CC H}%
F_{k}\left(  \underbrace{v}_{=\gamma\otimes_{\CC H}u}\right) \\
&  =\sum_{k\in G}k\otimes_{\CC H}\underbrace{F_{k}\left(  \gamma
\otimes_{\CC H}u\right)  }_{\substack{=\dfrac{1}{\left\vert H\right\vert
}\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
\gamma\right)  hu\\\text{(by (\ref{sol.Indrho.b.Fg.def}), applied to
}g=k\text{)}}}=\sum_{k\in G}k\otimes_{\CC H}\left(  \dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)
^{\ast}\left(  \gamma\right)  hu\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }
\underbrace{\sum_{k\in G} \sum_{h\in H}}_{= \sum_{h\in H} \sum_{k\in G}}
\left(
k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)
 k\otimes_{\CC H}hu\ \ \ \ \ \ \ \ \ \ \left(  \text{since the tensor
product is }\CC \text{-bilinear}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\sum_{k\in
G}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)
\underbrace{k\otimes_{\CC H}hu}_{\substack{=kh\otimes_{\CC %
H}u\\\text{(since }h\text{ can be moved past}\\\text{the }\otimes
_{\CC H}\text{ sign (since }h\in H\subset\CC H\text{))}}%
}=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\sum_{k\in G}\left(
k\rho\left(  h\right)  \right)  ^{\ast}\left(  \gamma\right)  kh\otimes
_{\CC H}u\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\underbrace{\left(
\sum_{k\in G}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
\gamma\right)  kh\right)  }_{\substack{=\gamma\\\text{(by
(\ref{sol.Indrho.b.dual.pf.2}))}}}\otimes_{\CC H}u=\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\underbrace{\gamma\otimes_{\CC H}%
u}_{=v}=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{h\in H}%
v}_{=\left\vert H\right\vert \cdot v}=\dfrac{1}{\left\vert H\right\vert
}\left\vert H\right\vert \cdot v=v.
\end{align*}
Thus, $v=\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$ is proven,
qed.}. Thus, we know that $\left(  G\times J,\left(  a_{i}\right)  _{i\in
G\times J},\left(  f_{i}\right)  _{i\in G\times J}\right)  $ is a triple such that

\begin{itemize}
\item $G\times J$ is a finite set;

\item $\left(  a_{i}\right)  _{i\in G\times J}$ is a family of elements of
$\CC G\otimes_{\CC H}U$;

\item $\left(  f_{i}\right)  _{i\in G\times J}$ is a family of elements of
$\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ (where $\left(
\CC G\otimes_{\CC H}U\right)  ^{\ast}$ means
$\Hom_{\CC}\left(  \CC G\otimes_{\CC H}U,\CC \right)  $)
\end{itemize}

\noindent
such that every $v\in\CC G\otimes_{\CC H}U$ satisfies
$v=\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$. In other words,
$\left(  G\times J,\left(  a_{i}\right)  _{i\in G\times J},\left(
f_{i}\right)  _{i\in G\times J}\right)  $ is a finite dual generating system
for $\CC G\otimes_{\CC H}U$ \ \ \ \ \footnote{because $\left(
G\times J,\left(  a_{i}\right)  _{i\in G\times J},\left(  f_{i}\right)  _{i\in
G\times J}\right)  $ is a finite dual generating system for $\CC %
G\otimes_{\CC H}U$ if and only if $\left(  G\times J,\left(
a_{i}\right)  _{i\in G\times J},\left(  f_{i}\right)  _{i\in G\times
J}\right)  $ is a triple such that
\par
\begin{itemize}
\item $G\times J$ is a finite set;
\par
\item $\left(  a_{i}\right)  _{i\in G\times J}$ is a family of elements of
$\CC G\otimes_{\CC H}U$;
\par
\item $\left(  f_{i}\right)  _{i\in G\times J}$ is a family of elements of
$\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ (where $\left(
\CC G\otimes_{\CC H}U\right)  ^{\ast}$ means
$\Hom_{\CC} \left(  \CC G\otimes_{\CC H}U,\CC \right)  $)
\end{itemize}
\par
such that every $v\in\CC G\otimes_{\CC H}U$ satisfies
$v=\sum\limits_{i\in G\times J}f_{i}\left(  v\right)  a_{i}$ (according to the
definition of a ``finite dual generating system for
$\CC G\otimes_{\CC H}U$'')}. In other words,
$\left(  G\times J,\left(  a_{i}\right)  _{i\in G\times J},\left(
f_{i}\right)  _{i\in G\times J}\right)  $ is a finite dual generating system
for $\Ind_{\rho}U$ (since $\operatorname*{Ind}%
\nolimits_{\rho}U=\CC G\otimes_{\CC H}U$).

Let us now notice that%
\begin{equation}
\sum_{j\in J}g_{j}\left(  hb_{j}\right)  =\chi_{U}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H \label{sol.Indrho.b.chiU}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.b.chiU}):} Let $h\in H$. The
definition of $\chi_{U}\left(  h\right)  $ yields $\chi_{U}\left(  h\right)
=\operatorname*{trace}\left(  h:U\rightarrow U\right)  $. But Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr} (applied to $\CC$, $U$, $\left(
J,\left(  b_{j}\right)  _{j\in J},\left(  g_{j}\right)  _{j\in J}\right)  $
and $\left(  h:U\rightarrow U\right)  $ instead of $\CC$, $U$, $\left(
I,\left(  a_{i}\right)  _{i\in I},\left(  f_{i}\right)  _{i\in I}\right)  $
and $T$) yields
\[
\operatorname*{trace}\left(  h:U\rightarrow U\right)  =\sum_{j\in J}%
g_{j}\left(  \underbrace{\left(  h:U\rightarrow U\right)  b_{j}}_{=hb_{j}%
}\right)  =\sum_{j\in J}g_{j}\left(  hb_{j}\right)  .
\]
Thus, $\chi_{U}\left(  h\right)  =\operatorname*{trace}\left(  h:U\rightarrow
U\right)  =\sum_{j\in J}g_{j}\left(  hb_{j}\right)  $. This proves
(\ref{sol.Indrho.b.chiU}).}.

Now, let us fix $g\in G$. We want to compute $\chi_{\operatorname*{Ind}%
\nolimits_{\rho}U}\left(  g\right)  $. The definition of $\chi
_{\Ind_{\rho}U}\left(  g\right)  $ yields
$\chi_{\Ind_{\rho}U}\left(  g\right)
=\operatorname*{trace}\left(  g:\Ind_{\rho
}U\rightarrow\Ind_{\rho}U\right)  $. But Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr} (applied to $\mathbb{K}=\CC $,
$V=\Ind_{\rho}U$, $I=G\times J$ and $T=\left(
g:\Ind_{\rho}U\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U\right)  $) yields%
\begin{align}
&  \operatorname*{trace}\left(  g:\Ind_{\rho
}U\rightarrow\Ind_{\rho}U\right) \nonumber\\
&  =\sum_{i\in G\times J}f_{i}\left(  \underbrace{\left(
g:\Ind_{\rho}U\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U\right)  a_{i}}_{=ga_{i}}\right)  =\sum_{i\in G\times J}%
f_{i}\left(  ga_{i}\right) \nonumber\\
&  =\sum_{\left(  k,j\right)  \in G\times J}\underbrace{f_{\left(  k,j\right)
}}_{=g_{j}\circ F_{k}}\left(  g\underbrace{a_{\left(  k,j\right)  }%
}_{=k\otimes_{\CC H}b_{j}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{here, we substituted }\left(  k,j\right)  \text{ for the summation index
}i\right) \nonumber\\
&  =\sum_{\left(  k,j\right)  \in G\times J}\left(  g_{j}\circ F_{k}\right)
\underbrace{\left(  g\left(  k\otimes_{\CC H}b_{j}\right)  \right)
}_{=gk\otimes_{\CC H}b_{j}}=\sum_{\left(  k,j\right)  \in G\times
J}\underbrace{\left(  g_{j}\circ F_{k}\right)  \left(  gk\otimes_{\CC %
H}b_{j}\right)  }_{=g_{j}\left(  F_{k}\left(  gk\otimes_{\CC H}%
b_{j}\right)  \right)  }\nonumber\\
&  =\sum_{\left(  k,j\right)  \in G\times J}g_{j}\left(  \underbrace{F_{k}%
\left(  gk\otimes_{\CC H}b_{j}\right)  }_{\substack{=\dfrac
{1}{\left\vert H\right\vert }\sum\limits_{h\in H}\left(  k\rho\left(
h\right)  \right)  ^{\ast}\left(  gk\right)  hb_{j}\\\text{(by
(\ref{sol.Indrho.b.Fg.def}), applied to }k\text{, }gk\text{ and }%
b_{j}\\\text{instead of }g\text{, }\gamma\text{ and }u\text{)}}}\right)
\nonumber\\
&  =\underbrace{\sum_{\left(  k,j\right)  \in G\times J}}_{=\sum_{k\in G}%
\sum_{j\in J}}\underbrace{g_{j}\left(  \dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
gk\right)  hb_{j}\right)  }_{\substack{=\dfrac{1}{\left\vert H\right\vert
}\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
gk\right)  g_{j}\left(  hb_{j}\right)  \\\text{(since the map }g_{j}\text{ is
}\CC \text{-linear)}}}=\sum_{k\in G}\sum_{j\in J}\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)
^{\ast}\left(  gk\right)  g_{j}\left(  hb_{j}\right) \nonumber\\
&  =\sum_{k\in G}\dfrac{1}{\left\vert H\right\vert }\sum\limits_{h\in
H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(  gk\right)  \sum_{j\in
J}g_{j}\left(  hb_{j}\right)  . \label{sol.Indrho.b.computing-the-trace.1}%
\end{align}
But every $k\in G$ satisfies%
\begin{align}
&  \sum\limits_{h\in H}\underbrace{\left(  k\rho\left(  h\right)  \right)
^{\ast}\left(  gk\right)  }_{\substack{=\delta_{k\rho\left(  h\right)
,gk}\\\text{(by (\ref{sol.Indrho.b.g*.def}), applied to }k\rho\left(
h\right)  \\\text{and }gk\text{ instead of }g\text{ and }k\text{)}%
}}\underbrace{\sum_{j\in J}g_{j}\left(  hb_{j}\right)  }_{\substack{=\chi
_{U}\left(  h\right)  \\\text{(by (\ref{sol.Indrho.b.chiU}))}}}\nonumber\\
&  =\sum\limits_{h\in H}\delta_{k\rho\left(  h\right)  ,gk}\chi_{U}\left(
h\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  =gk}%
}}_{\substack{=\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}%
=g}}\\\text{(because for every }h\in H\text{,}\\\text{the statement }\left(
k\rho\left(  h\right)  =gk\right)  \\\text{is equivalent}\\\text{to }\left(
k\rho\left(  h\right)  k^{-1}=g\right)  \text{)}}}\underbrace{\delta
_{k\rho\left(  h\right)  ,gk}}_{\substack{=1\\\text{(since }k\rho\left(
h\right)  =gk\text{)}}}\chi_{U}\left(  h\right)  +\sum\limits_{\substack{h\in
H;\\k\rho\left(  h\right)  \neq gk}}\underbrace{\delta_{k\rho\left(  h\right)
,gk}}_{\substack{=0\\\text{(since }k\rho\left(  h\right)  \neq gk\text{)}%
}}\chi_{U}\left(  h\right) \nonumber\\
&  =\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}=g}}1\chi_{U}\left(
h\right)  +\underbrace{\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)
\neq gk}}0\chi_{U}\left(  h\right)  }_{=0}=\sum\limits_{\substack{h\in H;\\k\rho
\left(  h\right)  k^{-1}=g}}1\chi_{U}\left(  h\right) \nonumber\\
&  =\sum\limits_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(
h\right)  . \label{sol.Indrho.b.computing-the-trace.lem2}%
\end{align}
Now, recall that%
\begin{align*}
\chi_{\Ind_{\rho}U}\left(  g\right)   &
=\operatorname*{trace}\left(  g:\Ind_{\rho
}U\rightarrow\Ind_{\rho}U\right) \\
&  =\sum_{k\in G}\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum
\limits_{h\in H}\left(  k\rho\left(  h\right)  \right)  ^{\ast}\left(
gk\right)  \sum_{j\in J}g_{j}\left(  hb_{j}\right)  }_{\substack{=\sum\limits
_{\substack{h\in H;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(
h\right)  \\\text{(by (\ref{sol.Indrho.b.computing-the-trace.lem2}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.b.computing-the-trace.1})}\right) \\
&  =\sum_{k\in G}\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{h\in
H;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(  h\right)  =\dfrac
{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{h\in H;\\k\rho\left(
h\right)  k^{-1}=g}}\sum_{k\in G}}_{=\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}}\chi_{U}\left(  h\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(  h\right)  .
\end{align*}
Compared with
\[
\left(  \Ind_{\rho}\chi_{U}\right)  \left(  g\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in
H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\chi_{U}\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\Ind_{\rho}\chi_{U}\right)  \left(  g\right)  \right)
,
\]
this yields $\chi_{\Ind_{\rho}U}\left(  g\right)
=\left(  \Ind_{\rho}\chi_{U}\right)  \left(  g\right)
$.

Now, let us forget that we fixed $g$. We thus have proven that $\chi
_{\Ind_{\rho}U}\left(  g\right)  =\left(
\Ind_{\rho}\chi_{U}\right)  \left(  g\right)  $ for
every $g\in G$. In other words, $\chi_{\Ind_{\rho}%
U}=\Ind_{\rho}\chi_{U}$. This solves Exercise
\ref{exe.Indrho}(b).

(c) Assume that $H$ is a subgroup of $G$, and that $\rho:H\rightarrow G$ is
the inclusion map. We need to show that $\Ind_{\rho
}f=\Ind_{H}^{G}f$ for every $f\in R_{\CC %
}\left(  H\right)  $.

We notice that the map $G\rightarrow G,\ k\mapsto k^{-1}$ is a bijection
(since $G$ is a group).

Let $f\in R_{\CC }\left(  H\right)  $. Let $g\in G$. Then, the
definition of $\left(  \Ind_{\rho}f\right)  \left(
g\right)  $ yields%
\begin{align*}
\left(  \Ind_{\rho}f\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}}%
_{\substack{=\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\khk^{-1}%
=g}}\\\text{(since }\rho\left(  h\right)  =h\text{ for every }h\in
H\\\text{(since }\rho:H\rightarrow G\text{ is the inclusion map))}}}f\left(
h\right)  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits
_{\substack{\left(  h,k\right)  \in H\times G;\\khk^{-1}=g}}}_{=\sum_{k\in
G}\sum\limits_{\substack{h\in H;\\khk^{-1}=g}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{k\in G}\underbrace{\sum\limits
_{\substack{h\in H;\\khk^{-1}=g}}}_{\substack{=\sum\limits_{\substack{h\in
H;\\h=k^{-1}gk}}\\\text{(because for every }h\in H\text{,}\\\text{the
statement }\left(  khk^{-1}=g\right)  \text{ is}\\\text{equivalent to }\left(
h=k^{-1}gk\right)  \text{)}}}f\left(  h\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum_{k\in G}%
\sum\limits_{\substack{h\in H;\\h=k^{-1}gk}}f\left(  h\right)  }_{\substack{=\sum\limits
_{\substack{k\in G;\\k^{-1}gk\in H}}\sum\limits_{\substack{h\in H;\\h=k^{-1}%
gk}}f\left(  h\right)  +\sum\limits_{\substack{k\in G;\\k^{-1}gk\notin H}%
}\sum\limits_{\substack{h\in H;\\h=k^{-1}gk}}f\left(  h\right)  \\\text{(since every
}k\in G\text{ satisfies either }k^{-1}gk\in H\text{ or }k^{-1}gk\notin H\text{
(but never both))}}}\\
&  =\dfrac{1}{\left\vert H\right\vert }\left(  \sum\limits_{\substack{k\in
G;\\k^{-1}gk\in H}}\underbrace{\sum\limits_{\substack{h\in H;\\h=k^{-1}gk}}f\left(
h\right)  }_{\substack{=f\left(  k^{-1}gk\right)  \\\text{(since }k^{-1}gk\in
H\text{)}}}+\sum\limits_{\substack{k\in G;\\k^{-1}gk\notin H}}\underbrace{\sum\limits
_{\substack{h\in H;\\h=k^{-1}gk}}f\left(  h\right)  }_{\substack{=\left(
\text{empty sum}\right)  \\\text{(since there is no }h\in H\text{ satisfying
}h=k^{-1}gk\\\text{(because }k^{-1}gk\notin H\text{))}}}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\left(  \sum\limits_{\substack{k\in
G;\\k^{-1}gk\in H}}f\left(  k^{-1}gk\right)  +\sum\limits_{\substack{k\in
G;\\k^{-1}gk\notin H}}\underbrace{\left(  \text{empty sum}\right)  }%
_{=0}\right)  =\dfrac{1}{\left\vert H\right\vert }\left(  \sum\limits_{\substack{k\in
G;\\k^{-1}gk\in H}}f\left(  k^{-1}gk\right)  +\underbrace{\sum\limits_{\substack{k\in
G;\\k^{-1}gk\notin H}}0}_{=0}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G;\\k^{-1}gk\in
H}}f\left(  k^{-1}gk\right)  =\dfrac{1}{\left\vert H\right\vert }%
\underbrace{\sum\limits_{\substack{k\in G;\\\left(  k^{-1}\right)  ^{-1}gk^{-1}\in
H}}}_{\substack{=\sum\limits_{\substack{k\in G;\\kgk^{-1}\in H}}\\\text{(since
}\left(  k^{-1}\right)  ^{-1}=k\text{ for every }k\in G\text{)}}}f\left(
\underbrace{\left(  k^{-1}\right)  ^{-1}}_{=k}gk^{-1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }k^{-1}\text{ for }k\text{ in the sum, since}\\
\text{the map }G\rightarrow G,\ k\mapsto k^{-1}\text{ is a bijection}%
\end{array}
\right) \\
&  =\frac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G;\\kgk^{-1}\in
H}}f\left(  kgk^{-1}\right)  =\left(  \operatorname{Ind}_{H}^{G}f\right)
\left(  g\right)
\end{align*}
(since (\ref{eq.exe.induction.characters.eq}) yields $\left(
\operatorname{Ind}_{H}^{G}f\right)  \left(  g\right)  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{\substack{k\in G;\\kgk^{-1}\in H}}f\left(  kgk^{-1}%
\right)  $).

Let us now forget that we fixed $g$. We thus have proven that $\left(
\Ind_{\rho}f\right)  \left(  g\right)  =\left(
\operatorname{Ind}_{H}^{G}f\right)  \left(  g\right)  $ for every $g\in G$. In
other words, $\Ind_{\rho}f=\operatorname*{Ind}%
\nolimits_{H}^{G}f$. This solves Exercise \ref{exe.Indrho}(c).

(d) Assume that $H$ is a subgroup of $G$, and that $\rho:H\rightarrow G$ is
the inclusion map. We need to show that $\Ind_{\rho
}U=\Ind_{H}^{G}U$ for every $\CC H$-module $U$.

We notice that $\rho:H\rightarrow G$ is the inclusion map. Hence,
$\CC \left[  \rho\right]  :\CC H\rightarrow\CC G$ is also
the inclusion map (since we identify $\CC H$ with a $\CC %
$-subalgebra of $\CC G$ along this map $\CC \left[  \rho\right]
$). Thus,%
\begin{equation}
\left(  \CC \left[  \rho\right]  \right)  \eta=\eta
\ \ \ \ \ \ \ \ \ \ \text{for every }\eta\in\CC H.
\label{sol.Indrho.d.incmap}%
\end{equation}


Let $U$ be a $\CC H$-module. Both $\operatorname{Ind}_{H}^{G}U$ and
$\Ind_{\rho}U$ are defined as ${\CC }%
G\otimes_{\CC H}U$ for some $\left(  \CC G,\CC H\right)
$-bimodule structure on $\CC G$; however, their definitions differ at
how this $\left(  \CC G,\CC H\right)  $-bimodule structure is
defined. We are now going to prove that these two $\left(  \CC %
G,\CC H\right)  $-bimodule structures on $\CC G$ are identical.

The definition of $\Ind_{\rho}U$ shows that we have%
\begin{equation}
\Ind_{\rho}U=\CC G\otimes_{\CC H}U,
\label{sol.Indrho.d.1}%
\end{equation}
where $\CC G$ is regarded as a $\left(  \CC G,\CC H\right)
$-bimodule according to the following rule: The left $\CC G$-module
structure on $\CC G$ is plain multiplication inside $\CC G$; the
right $\CC H$-module structure on $\CC G$ is induced by the
$\CC$-algebra homomorphism $\CC \left[  \rho\right]
:\CC H\rightarrow\CC G$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$
for all $\gamma\in\CC G$ and $\eta\in\CC H$). Let us denote the
$\left(  \CC G,\CC H\right)  $-bimodule $\CC G$ defined in
the previous sentence by $A$. Thus, the $\left(  \CC G,\CC %
H\right)  $-bimodule $A$ has underlying $\CC$-vector space
$\CC G$, and its $\left(  \CC G,\CC H\right)  $-bimodule
structure is given as follows: The left $\CC G$-module structure is
plain multiplication inside $\CC G$; the right $\CC H$-module
structure is induced by the $\CC$-algebra homomorphism $\CC %
\left[  \rho\right]  :\CC H\rightarrow\CC G$ (thus, it is
explicitly given by $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in A$ and $\eta\in\CC H$).

The $\left(  \CC G,\CC H\right)  $-bimodule $A$ is precisely the
$\left(  \CC G,\CC H\right)  $-bimodule denoted by $\CC G$
in (\ref{sol.Indrho.d.1}). Hence, (\ref{sol.Indrho.d.1}) can be rewritten as
follows:%
\begin{equation}
\Ind_{\rho}U=A\otimes_{\CC H}U.
\label{sol.Indrho.d.1a}%
\end{equation}
This is an equality of left $\CC G$-modules.

On the other hand, the definition of $\Ind_{H}^{G}U$
shows that we have%
\begin{equation}
\operatorname{Ind}_{H}^{G}U={\CC }G\otimes_{\CC H}U,
\label{sol.Indrho.d.2}%
\end{equation}
where $\CC G$ is regarded as a $\left(  \CC G,\CC H\right)
$-bimodule in the usual way (i.e., the left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$, and the right
$\CC H$-module structure on $\CC G$ is also plain multiplication
inside $\CC G$ because $\CC H\subset\CC G$). Let us denote
the $\left(  \CC G,\CC H\right)  $-bimodule $\CC G$ defined
in the previous sentence by $B$. Thus, $B$ is the $\CC$-vector space
$\CC G$, regarded as a $\left(  \CC G,\CC H\right)
$-bimodule in the usual way (i.e., the left $\CC G$-module structure is
plain multiplication inside $\CC G$, and the right $\CC H$-module
structure is also plain multiplication inside $\CC G$ because
$\CC H\subset\CC G$).

The $\left(  \CC G,\CC H\right)  $-bimodule $B$ is precisely the
$\left(  \CC G,\CC H\right)  $-bimodule denoted by $\CC G$
in (\ref{sol.Indrho.d.2}). Hence, (\ref{sol.Indrho.d.2}) can be rewritten as
follows:%
\begin{equation}
\operatorname{Ind}_{H}^{G}U=B\otimes_{\CC H}U. \label{sol.Indrho.d.2a}%
\end{equation}
This is an equality of left $\CC G$-modules.

We shall now show that $A=B$ as $\left(  \CC G,\CC H\right)
$-bimodules. Indeed, $A$ and $B$ have the same underlying $\CC$-vector
space (namely, $\CC G$) and the same left $\CC G$-module structure
(namely, the one given by plain multiplication inside $\CC G$).

For every $\gamma\in\CC G$ and $\eta\in\CC H$, we have%
\begin{align}
&  \left(  \text{the image of }\gamma\text{ under the action of }\eta\text{
according to the right }\CC H\text{-module structure on }B\right)
\nonumber\\
&  =\gamma\cdot\eta\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the right }\CC H\text{-module structure on }B\text{ is}\\
\text{plain multiplication inside }\CC G
\end{array}
\right)  . \label{sol.Indrho.d.5}%
\end{align}
However, for every $\gamma\in\CC G$ and $\eta\in\CC H$, we have
\begin{align*}
&  \left(  \text{the image of }\gamma\text{ under the action of }\eta\text{
according to the right }\CC H\text{-module structure on }A\right) \\
&  =\gamma\cdot\underbrace{\left(  \CC \left[  \rho\right]  \right)
\eta}_{\substack{=\eta\\\text{(by (\ref{sol.Indrho.d.incmap}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the right }\CC H\text{-module structure on }A\text{ is
induced by}\\
\text{the }\CC \text{-algebra homomorphism }\CC \left[
\rho\right]  :\CC H\rightarrow\CC G
\end{array}
\right) \\
&  =\gamma\cdot\eta\\
&  =\left(  \text{the image of }\gamma\text{ under the action of }\eta\text{
according to the right }\CC H\text{-module structure on }B\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.d.5})}\right)  .
\end{align*}
Thus, the right $\CC H$-module structure on $A$ is identical to the
right $\CC H$-module structure on $B$ (since these two right
$\CC H$-module structures also share the same addition (because $A$ and
$B$ have the same underlying $\CC$-vector space)). In other words, $A$
and $B$ have the same right $\CC H$-module structure.

Altogether, we know that $A$ and $B$ have the same underlying $\CC %
$-vector space, the same left $\CC G$-module structure, and the same
right $\CC H$-module structure. In other words, $A=B$ as $\left(
\CC G,\CC H\right)  $-bimodules.

Now, (\ref{sol.Indrho.d.1}) yields $\Ind_{\rho
}U=\underbrace{A}_{=B}\otimes_{\CC H}U=B\otimes_{\CC %
H}U=\Ind_{H}^{G}U$ (by (\ref{sol.Indrho.d.2})) as left
$\CC G$-modules. This solves Exercise \ref{exe.Indrho}(d).

(e) Assume that $G=H/K$ for some normal subgroup $K$ of $H$. Let
$\rho:H\rightarrow G$ be the projection map. We want to prove that
$\Ind_{\rho}f=f^{K}$ for every $f\in R_{\CC %
}\left(  H\right)  $.

Let $f\in R_{\CC }\left(  H\right)  $. Applying (\ref{sol.Indrho.RCK})
to $H$ instead of $K$, we obtain the following equivalence:%
\begin{align*}
&  \left(  f\in R_{\CC }\left(  H\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }H\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime
}\right)  \right)  .
\end{align*}
Hence,
\begin{equation}
\left(  \text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of
}H\text{ satisfy }f\left(  k\right)  =f\left(  k^{\prime}\right)  \right)
\label{sol.Indrho.e.conj}%
\end{equation}
(since we know that $f\in R_{\CC }\left(  H\right)  $). In other words,%
\begin{equation}
f\left(  y^{-1}zy\right)  =f\left(  z\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }z\in H\text{ and }y\in H \label{sol.Indrho.e.conj2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.e.conj2}):} Let $z\in H$ and $y\in
H$. Then, the elements $z$ and $y^{-1}zy$ of $H$ are conjugate (by the
definition of ``conjugate elements''). Hence,
$f\left(  z\right)  =f\left(  y^{-1}zy\right)  $ (by (\ref{sol.Indrho.e.conj}%
), applied to $k=z$ and $k^{\prime}=y^{-1}zy$). This proves
(\ref{sol.Indrho.e.conj2}).}.

Let $g\in G$. Then, $g\in G=H/K$. Hence, there exists an $x\in H$ such that
$g=xK$. Consider this $x$. The map $\rho:H\rightarrow G$ is the projection map
from $H$ to $G=H/K$, and thus sends every $h\in H$ to the coset $hK\in G$. In
other words, $\rho\left(  h\right)  =hK$ for every $h\in H$. Applied to $h=x$,
this yields $\rho\left(  x\right)  =xK=g$. Of course, $\ker%
\rho=K$ (since $\rho$ is the projection map from $H$ to $H/K$).

We have $\left\vert G\right\vert =\left\vert H/K\right\vert =\left[
H:K\right]  =\left\vert H\right\vert /\left\vert K\right\vert $.

We make another simple observation: If $h\in H$ and $y\in H$, then we have the
following logical equivalence:%
\begin{equation}
\left(  \rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  \right)  \ \Longleftrightarrow\ \left(  h\in
y^{-1}xKy\right)  . \label{sol.Indrho.e.equiv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.e.equiv1}):} Let $h\in H$ and
$y\in H$. We need to prove the equivalence (\ref{sol.Indrho.e.equiv1}).
\par
Let us first assume that $\rho\left(  h\right)  =\left(  \rho\left(  y\right)
\right)  ^{-1}g\rho\left(  y\right)  $. Then, $\left(  \underbrace{\left(
\rho\left(  y\right)  \right)  ^{-1}g\rho\left(  y\right)  }_{=\rho\left(
h\right)  }\right)  ^{-1}\cdot\rho\left(  h\right)  =\left(  \rho\left(
h\right)  \right)  ^{-1}\cdot\rho\left(  h\right)  =1$, so that
\begin{align*}
1  &  =\left(  \left(  \rho\left(  y\right)  \right)  ^{-1}\underbrace{g}%
_{=\rho\left(  x\right)  }\rho\left(  y\right)  \right)  ^{-1}\cdot\rho\left(
h\right)  =\left(  \left(  \rho\left(  y\right)  \right)  ^{-1}\rho\left(
x\right)  \rho\left(  y\right)  \right)  ^{-1}\cdot\rho\left(  h\right) \\
&  =\left(  \rho\left(  y\right)  \right)  ^{-1}\left(  \rho\left(  x\right)
\right)  ^{-1}\rho\left(  y\right)  \rho\left(  h\right) \\
&  =\rho\left(  y^{-1}x^{-1}yh\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\rho\text{ is a group homomorphism}\right)  .
\end{align*}
Hence, $\rho\left(  y^{-1}x^{-1}yh\right)  =1$, so that $y^{-1}x^{-1}%
yh\in\ker\rho=K$. Thus, $x^{-1}yh\in yK$. But $K$ is a normal
subgroup of $H$, and thus we have $yK=Ky$. Hence, $x^{-1}yh\in yK=Ky$, so that
$yh\in xKy$, and thus $h\in y^{-1}xKy$.
\par
Now, let us forget that we assumed that $\rho\left(  h\right)  =\left(
\rho\left(  y\right)  \right)  ^{-1}g\rho\left(  y\right)  $. We thus have
shown that
\begin{equation}
\text{if }\rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  \text{, then }h\in y^{-1}xKy.
\label{sol.Indrho.e.equiv1.pf.1}%
\end{equation}
\par
Let us now assume that $h\in y^{-1}xKy$. Thus, $yh\in xKy$, so that
$x^{-1}yh\in Ky=yK$ (since $K$ is a normal subgroup of $H$). Hence,
$y^{-1}x^{-1}yh\in K=\ker\rho$, so that $\rho\left(
y^{-1}x^{-1}yh\right)  =1$. Thus,
\begin{align*}
1  &  =\rho\left(  y^{-1}x^{-1}yh\right)  =\left(  \rho\left(  y\right)
\right)  ^{-1}\cdot\left(  \underbrace{\rho\left(  x\right)  }_{=g}\right)
^{-1}\cdot\rho\left(  y\right)  \cdot\rho\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho\text{ is a group homomorphism}%
\right) \\
&  =\left(  \rho\left(  y\right)  \right)  ^{-1}\cdot g^{-1}\cdot\rho\left(
y\right)  \cdot\rho\left(  h\right)  .
\end{align*}
Multiplying both sides of this identity with $\left(  \rho\left(  y\right)
\right)  ^{-1}g\rho\left(  y\right)  $ from the left, we obtain%
\begin{align*}
\left(  \rho\left(  y\right)  \right)  ^{-1}g\rho\left(  y\right)   &
=\left(  \rho\left(  y\right)  \right)  ^{-1}g\underbrace{\rho\left(
y\right)  \cdot\left(  \rho\left(  y\right)  \right)  ^{-1}}_{=1}\cdot
g^{-1}\cdot\rho\left(  y\right)  \cdot\rho\left(  h\right) \\
&  =\left(  \rho\left(  y\right)  \right)  ^{-1}\underbrace{g\cdot g^{-1}%
}_{=1}\cdot\rho\left(  y\right)  \cdot\rho\left(  h\right)
=\underbrace{\left(  \rho\left(  y\right)  \right)  ^{-1}\cdot\rho\left(
y\right)  }_{=1}\cdot\rho\left(  h\right)  =\rho\left(  h\right)  .
\end{align*}
In other words, $\rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  $. Now, let us forget that we assumed that $h\in
y^{-1}xKy$. We thus have shown that
\[
\text{if }h\in y^{-1}xKy\text{, then }\rho\left(  h\right)  =\left(
\rho\left(  y\right)  \right)  ^{-1}g\rho\left(  y\right)  .
\]
Combining this logical implication with (\ref{sol.Indrho.e.equiv1.pf.1}), we
conclude that $\rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  $ holds if and only if $h\in y^{-1}xKy$. In other
words, the equivalence (\ref{sol.Indrho.e.equiv1}) is proven.}

The definition of $f^{K}$ yields $f^{K}\left(  xK\right)  =\dfrac
{1}{\left\vert K\right\vert }\sum\limits_{k\in K}f\left(  xk\right)  $. Hence,%
\begin{equation}
f^{K}\left(  \underbrace{g}_{=xK}\right)  =f^{K}\left(  xK\right)  =\dfrac
{1}{\left\vert K\right\vert }\sum\limits_{k\in K}f\left(  xk\right)  .
\label{sol.Indrho.e.fK}%
\end{equation}
But the definition of $\left(  \Ind_{\rho}f\right)
\left(  g\right)  $ yields%
\begin{align}
\left(  \Ind_{\rho}f\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}}%
_{\substack{=\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\\rho\left(
h\right)  =k^{-1}gk}}\\\text{(because for any }\left(  h,k\right)  \in H\times
G\text{,}\\\text{the statement }\left(  k\rho\left(  h\right)  k^{-1}%
=g\right)  \\\text{is equivalent to }\left(  \rho\left(  h\right)
=k^{-1}gk\right)  \text{)}}}f\left(  h\right)  =\dfrac{1}{\left\vert
H\right\vert }\underbrace{\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\\rho\left(  h\right)  =k^{-1}gk}}}_{=\sum\limits_{k\in G}\sum
\limits_{\substack{h\in H;\\\rho\left(  h\right)  =k^{-1}gk}}}f\left(
h\right) \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{k\in G}\sum
\limits_{\substack{h\in H;\\\rho\left(  h\right)  =k^{-1}gk}}f\left(
h\right)  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{p\in G}%
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}}f\left(
h\right)  \label{sol.Indrho.e.4}%
\end{align}
(here, we renamed the index $k$ as $p$ in the first sum). But every $p\in G$
satisfies $\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}%
gp}}f\left(  h\right)  =\sum\limits_{k\in K}f\left(  xk\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $p\in G$. Then, $p\in G=H/K$. Hence,
there exists a $y\in H$ such that $p=yK$. Consider this $y$.
\par
Recall that $\rho\left(  h\right)  =hK$ for every $h\in H$. Applying this to
$h=y$, we obtain $\rho\left(  y\right)  =yK=p$.
\par
We notice that the map $K\rightarrow y^{-1}xKy,\ k\mapsto y^{-1}xky$ is a
bijection (since $H$ is a group).
\par
Now, for every $h\in H$, we have the following equivalence:%
\begin{align}
\left(  \rho\left(  h\right)  =p^{-1}gp\right)  \  &  \Longleftrightarrow
\ \left(  \rho\left(  h\right)  =\left(  \rho\left(  y\right)  \right)
^{-1}g\rho\left(  y\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}p=\rho\left(  y\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  h\in y^{-1}xKy\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{according to (\ref{sol.Indrho.e.equiv1}%
)}\right)  . \label{sol.Indrho.e.equiv2}%
\end{align}
\par
Now,%
\[
\underbrace{\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}%
}}_{\substack{=\sum\limits_{\substack{h\in H;\\h\in y^{-1}xKy}%
}\\\text{(because for every }h\in H\text{, the statement}\\\left(  \rho\left(
h\right)  =p^{-1}gp\right)  \text{ is equivalent to }\left(  h\in
y^{-1}xKy\right)  \\\text{(because of (\ref{sol.Indrho.e.equiv2})))}}}f\left(
h\right)  =\underbrace{\sum\limits_{\substack{h\in H;\\h\in y^{-1}xKy}%
}}_{\substack{=\sum\limits_{h\in y^{-1}xKy}\\\text{(since }y^{-1}xKy\subset
H\text{)}}}f\left(  h\right)  =\sum\limits_{h\in y^{-1}xKy}f\left(  h\right)
=\sum\limits_{k\in K}f\left(  y^{-1}xky\right)
\]
(here, we substituted $y^{-1}xky$ for $h$ in the sum, since the map
$K\rightarrow y^{-1}xKy,\ k\mapsto y^{-1}xky$ is a bijection). Thus,%
\[
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}}f\left(
h\right)  =\sum\limits_{k\in K}\underbrace{f\left(  y^{-1}xky\right)
}_{\substack{=f\left(  xk\right)  \\\text{(by (\ref{sol.Indrho.e.conj2}),
applied to }z=xk\text{)}}}=\sum\limits_{k\in K}f\left(  xk\right)  ,
\]
qed.}. Hence, (\ref{sol.Indrho.e.4}) becomes%
\begin{align*}
\left(  \Ind_{\rho}f\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{p\in G}\underbrace{\sum
\limits_{\substack{h\in H;\\\rho\left(  h\right)  =p^{-1}gp}}f\left(
h\right)  }_{=\sum\limits_{k\in K}f\left(  xk\right)  }=\dfrac{1}{\left\vert
H\right\vert }\underbrace{\sum\limits_{p\in G}\sum\limits_{k\in K}f\left(
xk\right)  }_{=\left\vert G\right\vert \cdot\sum\limits_{k\in K}f\left(
xk\right)  }=\dfrac{1}{\left\vert H\right\vert }\underbrace{\left\vert
G\right\vert }_{=\left\vert H\right\vert /\left\vert K\right\vert }\cdot
\sum\limits_{k\in K}f\left(  xk\right) \\
&  =\underbrace{\dfrac{1}{\left\vert H\right\vert }\cdot\left\vert
H\right\vert /\left\vert K\right\vert }_{=\dfrac{1}{\left\vert K\right\vert }%
}\cdot\sum\limits_{k\in K}f\left(  xk\right)  =\dfrac{1}{\left\vert
K\right\vert }\sum\limits_{k\in K}f\left(  xk\right)  =f^{K}\left(  g\right)
\end{align*}
(by (\ref{sol.Indrho.e.fK})).

Let us now forget that we fixed $g$. We thus have shown that $\left(
\Ind_{\rho}f\right)  \left(  g\right)  =f^{K}\left(
g\right)  $ for every $g\in G$. In other words, $\operatorname*{Ind}%
\nolimits_{\rho}f=f^{K}$. This solves Exercise \ref{exe.Indrho}(e).

(f) \textit{First solution to Exercise \ref{exe.Indrho}(f).} Assume that
$G=H/K$ for some normal subgroup $K$ of $H$. Let $\rho:H\rightarrow G$ be the
projection map. We want to prove that $\Ind_{\rho
}U\cong U^{K}$ for every $\CC H$-module $U$.

The map $\rho$ is the projection map from $H$ to $G=H/K$. Hence, the map
$\rho$ is surjective and has kernel $\ker\rho=K$. Next, we
notice that%
\begin{equation}
\left\vert \rho^{-1}\left(  g\right)  \right\vert =\left\vert K\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G.
\label{sol.Indrho.f.sol1.preimage}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.preimage}):} Let $g\in G$.
The map $\rho$ is surjective (since $\rho$ is the projection map from $H$ to
$H/K$). Hence, $\rho\left(  H\right)  =G$. Thus, $g\in G=\rho\left(  H\right)
$. Therefore, there exists some $x\in H$ such that $g=\rho\left(  x\right)  $.
Let us fix such an $x$.
\par
We know that $H$ is a group. Hence, the map $K\rightarrow xK,\ k\mapsto xk$ is
a bijection. Thus, the sets $K$ and $xK$ are in bijection. Therefore,
$\left\vert xK\right\vert =\left\vert K\right\vert $.
\par
However, for every $y\in H$, we have the following logical equivalence:
\begin{align*}
\left(  y\in\rho^{-1}\left(  g\right)  \right)  \  &  \Longleftrightarrow
\ \left(  \rho\left(  y\right)  =\underbrace{g}_{=\rho\left(  x\right)
}\right)  \ \Longleftrightarrow\ \left(  \rho\left(  y\right)  =\rho\left(
x\right)  \right)  \ \Longleftrightarrow\ \left(  \underbrace{\rho\left(
y\right)  \cdot\left(  \rho\left(  x\right)  \right)  ^{-1}}_{\substack{=\rho
\left(  yx^{-1}\right)  \\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}=1\right) \\
&  \Longleftrightarrow\ \left(  \rho\left(  yx^{-1}\right)  =1\right)
\ \Longleftrightarrow\ \left(  yx^{-1}\in\underbrace{\ker\rho
}_{=K}\right)  \ \Longleftrightarrow\ \left(  yx^{-1}\in K\right)
\ \Longleftrightarrow\ \left(  y\in xK\right)  .
\end{align*}
Hence, $\rho^{-1}\left(  g\right)  =xK$, so that $\left\vert \rho^{-1}\left(
g\right)  \right\vert =\left\vert xK\right\vert =\left\vert K\right\vert $.
This proves (\ref{sol.Indrho.f.sol1.preimage}).}

Let $U$ be a $\CC H$-module. Recall that $U^{K}$ is a $\CC \left[
H/K\right]  $-module, thus a $\CC G$-module (since $H/K=G$).

Recall that $\Ind_{\rho}U$ is defined as the
$\CC G$-module $\CC G\otimes_{\CC H}U$, where $\CC G$
is regarded as a $\left(  \CC G,\CC H\right)  $-bimodule according
to the following rule: The left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$; the right
$\CC H$-module structure on $\CC G$ is induced by the $\CC %
$-algebra homomorphism $\CC \left[  \rho\right]  :\CC %
H\rightarrow\CC G$ (thus, it is explicitly given by $\gamma\eta
=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$ for all
$\gamma\in\CC G$ and $\eta\in\CC H$). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC G,\CC H\right)
$-bimodule structure.

We define a map $\alpha:U^{K}\rightarrow\CC G\otimes_{\CC H}U$ by
setting%
\[
\alpha\left(  u\right)  =1\otimes_{\CC H}u\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in U^{K}.
\]
This $\alpha$ is a map $U^{K}\rightarrow\Ind_{\rho}U$
(since $\Ind_{\rho}U=\CC G\otimes_{\CC %
H}U$) and is $\CC$-linear (since $\alpha\left(  u\right)
=1\otimes_{\CC H}u$ depends $\CC$-linearly on $u$).

We will show that $\alpha$ is a $\CC G$-module isomorphism.

The $\CC G$-module structure on $U^{K}$ has the property that%
\begin{equation}
\rho\left(  h\right)  \cdot v=hv\ \ \ \ \ \ \ \ \ \ \text{for any }h\in
H\text{ and }v\in U^{K}. \label{sol.Indrho.f.sol1.UK-modstr}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.UK-modstr}):} The map
$\rho$ is the projection map from $H$ to $H/K$. Thus, the map $\rho$ sends
every $h\in H$ to the coset $hK\in H/K$. In other words,%
\begin{equation}
\rho\left(  h\right)  =hK\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H,
\label{sol.Indrho.f.sol1.hK}%
\end{equation}
where $hK$ means the coset $hK\in H/K$. But by the definition of the
$\CC \left[  H/K\right]  $-module structure on $U^{K}$, we have%
\[
\left(  hK\right)  \cdot v=hv\ \ \ \ \ \ \ \ \ \ \text{for any }h\in H\text{
and }v\in U^{K},
\]
where $hK$ means the coset $hK\in H/K$. Thus, any $h\in H$ and $v\in U^{K}$
satisfy%
\[
\underbrace{\rho\left(  h\right)  }_{\substack{=hK\\\text{(by
(\ref{sol.Indrho.f.sol1.hK}))}}}\cdot v=\left(  hK\right)  \cdot v=hv,
\]
where $hK$ means the coset $hK\in H/K$. This proves
(\ref{sol.Indrho.f.sol1.UK-modstr}).}

Now, $\alpha$ is a $\CC G$-module homomorphism\footnote{\textit{Proof.}
Let $u\in U^{K}$ and $g\in G$. We have $g\in G=\rho\left(  H\right)  $ (since
the map $\rho:H\rightarrow G$ is surjective). Thus, $g=\rho\left(  y\right)  $
for some $y\in H$. Let us consider this $y$.
\par
Applying (\ref{sol.Indrho.f.sol1.UK-modstr}) to $h=y$ and $v=u$, we obtain
$\rho\left(  y\right)  \cdot u=yu$. Thus, $\underbrace{g}_{=\rho\left(
y\right)  }u=\rho\left(  y\right)  \cdot u=yu$.
\par
Now, the definition of $\alpha\left(  gu\right)  $ yields
\[
\alpha\left(  gu\right)  =1\otimes_{\CC H}\underbrace{gu}_{=yu}%
=1\otimes_{\CC H}yu=1y\otimes_{\CC H}u
\]
(here, we moved the $y$ past the tensor sign; this is allowed because $y\in
H\subset\CC H$).
\par
By the definition of the left $\CC G$-module structure on $\CC G$,
we have $g\cdot1=g1=g$.
\par
By the definition of the right $\CC H$-module structure on $\CC %
G$, we have $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying this to $\gamma=1$ and $\eta=y$, we obtain $1y=1\cdot\left(
\CC \left[  \rho\right]  \right)  y$. But $y\in H$ and thus $\left(
\CC \left[  \rho\right]  \right)  y=\rho\left(  y\right)  =g$.
Therefore, $1y=1\cdot\underbrace{\left(  \CC \left[  \rho\right]
\right)  y}_{=g}=1\cdot g=g=g\cdot1$.
\par
Now, $\alpha\left(  gu\right)  =\underbrace{1y}_{=g\cdot1}\otimes
_{\CC H}u=g\cdot1\otimes_{\CC H}u$. Compared with $g\cdot
\underbrace{\alpha\left(  u\right)  }_{=1\otimes_{\CC H}u}=g\cdot\left(
1\otimes_{\CC H}u\right)  =g\cdot1\otimes_{\CC H}u$, this yields
$\alpha\left(  gu\right)  =g\cdot\alpha\left(  u\right)  $.
\par
Now, let us forget that we fixed $u$ and $g$. We thus have shown that
$\alpha\left(  gu\right)  =g\cdot\alpha\left(  u\right)  $ for all $u\in
U^{K}$ and $g\in G$. Thus, the map $\alpha$ is a homomorphism of $G$-sets.
Since the map $\alpha$ is also $\CC$-linear, this
yields that $\alpha$ is a $\CC G$-module homomorphism. Qed.}. We will
eventually construct an inverse to $\alpha$; but first we need to prepare.

For every $g\in G$ and every $p\in{\CC }G$, we denote by $\epsilon
_{g}\left(  p\right)  $ the $g$-coordinate of $p$ with respect to the basis
$G$ of the $\CC$-vector space ${\CC }G$. By the definition of
``coordinate'', we have
\begin{equation}
q=\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\ \ \ \ \ \ \ \ \ \ \text{for every }q\in{\CC }G.
\label{sol.Indrho.f.sol1.epsilong.coordinates}%
\end{equation}
For every $g\in G$, we have defined a map $\epsilon_{g}:{\CC %
}G\rightarrow{\CC }$ (because we have defined an element $\epsilon
_{g}\left(  p\right)  $ for every $p\in{\CC }G$). This map $\epsilon
_{g}$ is $\CC$-linear.

For every $g\in G$ and $h\in G$, we know that $\epsilon_{g}\left(  h\right)  $
is the $g$-coordinate of $h$ with respect to the basis $G$ of the
$\CC$-vector space ${\CC }G$ (because this is how $\epsilon
_{g}\left(  h\right)  $ was defined). Since this $g$-coordinate of $h$ is
clearly $\delta_{g,h}$ (because $h$ is an element of the basis $G$), this
rewrites as follows: For every $g\in G$ and $h\in G$, we know that
$\epsilon_{g}\left(  h\right)  $ is $\delta_{g,h}$. In other words, for every
$g\in G$ and $h\in G$, we have
\begin{equation}
\epsilon_{g}\left(  h\right)  =\delta_{g,h}.
\label{sol.Indrho.f.sol1.epsilong.g}%
\end{equation}


We notice that
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in{\CC }G\text{ and }%
q\in{\CC }G. \label{sol.Indrho.f.sol1.trace}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.trace}):} Let
$p\in{\CC }G$ and $q\in{\CC }G$.
\par
Since $p\in{\CC }G$, we can write $p$ in the form $p=\sum\limits_{g\in
G}p_{g}g$ for some family $\left(  p_{g}\right)  _{g\in G}$ of elements of
$\CC$. Consider this family $\left(  p_{g}\right)  _{g\in G}$.
\par
Since $q\in{\CC }G$, we can write $q$ in the form $q=\sum\limits_{g\in
G}q_{g}g$ for some family $\left(  q_{g}\right)  _{g\in G}$ of elements of
$\CC$. Consider this family $\left(  q_{g}\right)  _{g\in G}$.
\par
Multiplying the equalities $p=\sum\limits_{g\in G}p_{g}g$ and $q=\sum
\limits_{g\in G}q_{g}g=\sum\limits_{h\in G}q_{h}h$ (here, we renamed the
summation index $g$ as $h$), we obtain
\[
pq=\left(  \sum\limits_{g\in G}p_{g}g\right)  \left(  \sum\limits_{h\in
G}q_{h}h\right)  =\underbrace{\sum\limits_{g\in G}\sum\limits_{h\in G}}%
_{=\sum\limits_{\left(  g,h\right)  \in G\times G}}p_{g}q_{h}gh=\sum
\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}gh.
\]
Applying the map $\epsilon_{1}$ to this equality, we obtain
\begin{align}
\epsilon_{1}\left(  pq\right)   &  =\epsilon_{1}\left(  \sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}gh\right)  =\sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}\underbrace{\epsilon_{1}\left(  gh\right)
}_{\substack{=\delta_{1,gh}\\\text{(by (\ref{sol.Indrho.f.sol1.epsilong.g}),
applied to }\\1\text{ and }gh\text{ instead of }\\g\text{ and }h\text{)}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
}{\CC }\text{-linear}\right) \nonumber\\
&  =\sum\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}\delta
_{1,gh}=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}%
}p_{g}q_{h}\underbrace{\delta_{1,gh}}_{\substack{=1\\\text{(since
}1=gh\\\text{(since }gh=1\text{))}}}+\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\gh\neq1}}p_{g}q_{h}\underbrace{\delta_{1,gh}%
}_{\substack{=0\\\text{(since }1\neq gh\\\text{(since }gh\neq1\text{))}%
}}\nonumber\\
&  =\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}%
}p_{g}q_{h}1+\underbrace{\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\gh\neq1}}p_{g}q_{h}0}_{=0}=\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\gh=1}}p_{g}q_{h}1\nonumber\\
&  =\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}%
}p_{g}q_{h}. \label{sol.Indrho.f.sol1.Rg.6}%
\end{align}
\par
However, for any $\left(  g,h\right)  \in G\times G$, we have the following
chain of equivalences:
\[
\left(  gh=1\right)  \ \Longleftrightarrow\ \left(  g=h^{-1}\right)
\ \Longleftrightarrow\ \left(  hg=1\right)  .
\]
Hence, we can replace the summation sign ``$\sum
\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}$''
by ``$\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg=1}}$''. We thus obtain
$\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\gh=1}}p_{g}%
q_{h}=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}%
}p_{g}q_{h}$. Thus, (\ref{sol.Indrho.f.sol1.Rg.6}) becomes
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\gh=1}}p_{g}q_{h}=\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\hg=1}}p_{g}q_{h}. \label{sol.Indrho.f.sol1.Rg.7}%
\end{equation}
\par
Now, multiplying the equalities $q=\sum\limits_{g\in G}q_{g}g$ and
$p=\sum\limits_{g\in G}p_{g}g=\sum\limits_{h\in G}p_{h}h$ (here, we renamed
the summation index $g$ as $h$), we obtain
\begin{align*}
qp  &  =\left(  \sum\limits_{g\in G}q_{g}g\right)  \left(  \sum\limits_{h\in
G}p_{h}h\right)  =\underbrace{\sum\limits_{g\in G}\sum\limits_{h\in G}}%
_{=\sum\limits_{h\in G}\sum\limits_{g\in G}=\sum\limits_{\left(  h,g\right)
\in G\times G}}\underbrace{q_{g}p_{h}}_{=p_{h}g_{q}}gh\\
&  =\sum\limits_{\left(  h,g\right)  \in G\times G}p_{h}q_{g}gh=\sum
\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}hg
\end{align*}
(here, we renamed the summation index $\left(  h,g\right)  $ as $\left(
g,h\right)  $). Applying the map $\epsilon_{1}$ to this equality, we obtain
\begin{align*}
\epsilon_{1}\left(  qp\right)   &  =\epsilon_{1}\left(  \sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}hg\right)  =\sum\limits_{\left(
g,h\right)  \in G\times G}p_{g}q_{h}\underbrace{\epsilon_{1}\left(  hg\right)
}_{\substack{=\delta_{1,hg}\\\text{(by (\ref{sol.Indrho.f.sol1.epsilong.g}),
applied to }\\1\text{ and }hg\text{ instead of }\\g\text{ and }h\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
}{\CC }\text{-linear}\right) \\
&  =\sum\limits_{\left(  g,h\right)  \in G\times G}p_{g}q_{h}\delta
_{1,hg}=\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}%
}p_{g}q_{h}\underbrace{\delta_{1,hg}}_{\substack{=1\\\text{(since
}1=hg\\\text{(since }hg=1\text{))}}}+\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg\neq1}}p_{g}q_{h}\underbrace{\delta_{1,hg}%
}_{\substack{=0\\\text{(since }1\neq hg\\\text{(since }hg\neq1\text{))}}}\\
&  =\sum\limits_{\substack{\left(  g,h\right)  \in G\times G;\\hg=1}%
}p_{g}q_{h}1+\underbrace{\sum\limits_{\substack{\left(  g,h\right)  \in
G\times G;\\hg\neq1}}p_{g}q_{h}0}_{=0}=\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg=1}}p_{g}q_{h}1=\sum\limits_{\substack{\left(
g,h\right)  \in G\times G;\\hg=1}}p_{g}q_{h}=\epsilon_{1}\left(  pq\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.f.sol1.Rg.7})}\right)  .
\end{align*}
Thus, $\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)  $, so
that (\ref{sol.Indrho.f.sol1.trace}) is proven.} Moreover,
\begin{equation}
\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(  q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\text{ and }q\in{\CC }G.
\label{sol.Indrho.f.sol1.epsi1g}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.epsi1g}):} Let $g\in G$ and
$q\in{\CC }G$.
\par
Since $q\in{\CC }G$, we can write $q$ in the form $q=\sum\limits_{h\in
G}q_{h}h$ for some family $\left(  q_{h}\right)  _{h\in G}$ of elements of
$\CC$. Consider this family $\left(  q_{h}\right)  _{h\in G}$.
\par
We have
\begin{align}
\epsilon_{1}\left(  g^{-1}\underbrace{q}_{=\sum\limits_{h\in G}q_{h}h}\right)
&  =\epsilon_{1}\left(  g^{-1}\sum\limits_{h\in G}q_{h}h\right)  =\epsilon
_{1}\left(  \sum\limits_{h\in G}q_{h}g^{-1}h\right)  =\sum\limits_{h\in
G}q_{h}\underbrace{\epsilon_{1}\left(  g^{-1}h\right)  }_{\substack{=\delta
_{1,g^{-1}h}\\\text{(by (\ref{sol.Indrho.f.sol1.epsilong.g}), applied
to}\\1\text{ and }g^{-1}h\text{ instead of }g\text{ and }h\text{)}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\epsilon_{1}\text{ is
}{\CC }\text{-linear}\right) \nonumber\\
&  =\sum\limits_{h\in G}q_{h}\delta_{1,g^{-1}h}=\sum\limits_{\substack{h\in
G;\\1=g^{-1}h}}q_{h}\underbrace{\delta_{1,g^{-1}h}}%
_{\substack{=1\\\text{(since }1=g^{-1}h\text{)}}}+\sum\limits_{\substack{h\in
G;\\1\neq g^{-1}h}}q_{h}\underbrace{\delta_{1,g^{-1}h}}%
_{\substack{=0\\\text{(since }1\neq g^{-1}h\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{h\in G;\\1=g^{-1}h}}q_{h}1+\underbrace{\sum
\limits_{\substack{h\in G;\\1\neq g^{-1}h}}q_{h}0}_{=0}=\sum
\limits_{\substack{h\in G;\\1=g^{-1}h}}q_{h}1=\underbrace{\sum
\limits_{\substack{h\in G;\\1=g^{-1}h}}}_{\substack{=\sum
\limits_{\substack{h\in G;\\g=h}}\\\text{(because for every}\\h\in G\text{,
the assertions}\\\left(  1=g^{-1}h\right)  \text{ and }\left(  g=h\right)
\\\text{are equivalent)}}}q_{h}=\sum\limits_{\substack{h\in G;\\g=h}}q_{h}.
\label{sol.Indrho.f.sol1.epsi1g.pf.1}%
\end{align}
On the other hand, applying the map $\epsilon_{g}$ to the identity
$q=\sum\limits_{h\in G}q_{h}h$, we obtain
\begin{align*}
\epsilon_{g}\left(  q\right)   &  =\epsilon_{g}\left(  \sum\limits_{h\in
G}q_{h}h\right)  =\sum\limits_{h\in G}q_{h}\underbrace{\epsilon_{g}\left(
h\right)  }_{\substack{=\delta_{g,h}\\\text{(by
(\ref{sol.Indrho.f.sol1.epsilong.g}))}}}=\sum\limits_{h\in G}q_{h}\delta
_{g,h}\\
&  =\sum\limits_{\substack{h\in G;\\g=h}}q_{h}\underbrace{\delta_{g,h}%
}_{\substack{=1\\\text{(since }g=h\text{)}}}+\sum\limits_{\substack{h\in
G;\\g\neq h}}q_{h}\underbrace{\delta_{g,h}}_{\substack{=0\\\text{(since }g\neq
h\text{)}}}=\sum\limits_{\substack{h\in G;\\g=h}}q_{h}1+\underbrace{\sum
\limits_{\substack{h\in G;\\g\neq h}}q_{h}0}_{=0}\\
&  =\sum\limits_{\substack{h\in G;\\g=h}}q_{h}1=\sum\limits_{\substack{h\in
G;\\g=h}}q_{h}=\epsilon_{1}\left(  g^{-1}q\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Indrho.f.sol1.epsi1g.pf.1})}\right)  .
\end{align*}
In other words, $\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(
q\right)  $. This proves (\ref{sol.Indrho.f.sol1.epsi1g}).}

Now, for every $\left(  q,u\right)  \in{\CC }G\times U$, we have
$\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u\in U^{K}$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  q,u\right)
\in{\CC }G\times U$. Let $x=\sum\limits_{h\in H}\epsilon_{1}\left(
\rho\left(  h\right)  q\right)  h^{-1}u$. We shall show that $x\in U^{K}$.
\par
We recall that $\rho$ is the projection from $H$ to $G=H/K$. Hence,
$\ker\rho=K$.
\par
Let $k\in K$. Then, the map $H\rightarrow H,\ h\mapsto hk$ is a bijection
(since $H$ is a group). Now, multiplying both sides of the equality
$x=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u$ with $k$ from the left, we obtain%
\begin{align*}
kx  &  =k\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
q\right)  h^{-1}u=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  q\right)  kh^{-1}u\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{\rho\left(  hk\right)
}_{\substack{=\rho\left(  h\right)  \rho\left(  k\right)  \\\text{(since }%
\rho\text{ is a group}\\\text{homomorphism)}}}q\right)  k\underbrace{\left(
hk\right)  ^{-1}}_{=k^{-1}h^{-1}}u\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }hk\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto hk\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
\underbrace{\rho\left(  k\right)  }_{\substack{=1\\\text{(since }k\in
K=\ker\rho\text{)}}}q\right)  \underbrace{kk^{-1}}_{=1}%
h^{-1}u=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
q\right)  h^{-1}u=x.
\end{align*}
\par
Let us now forget that we fixed $k$. We thus have shown that $kx=x$ for every
$k\in K$. In other words, $x$ is an element $y\in U$ satisfying $ky=y$ for
every $k\in K$. Hence,%
\[
x\in\left\{  y\in U\ \mid\ ky=y\text{ for every }k\in K\right\}  =U^{K}.
\]
Thus, $\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u=x\in U^{K}$, qed.}. Hence, we can define a map $\widetilde{\beta
}:{\CC }G\times U\rightarrow U^{K}$ by setting
\[
\widetilde{\beta}\left(  q,u\right)  =\sum_{h\in H}\epsilon_{1}\left(
\rho\left(  h\right)  q\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  q,u\right)  \in{\CC }G\times U.
\]
Consider this map $\widetilde{\beta}$. Then, $\widetilde{\beta}$ is a
$\CC$-bilinear map (because $\widetilde{\beta}\left(  q,u\right)
=\sum_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)  h^{-1}u$
depends $\CC$-linearly on each of $q$ and $u$). We are now going to
prove that the map $\widetilde{\beta}$ is ${\CC }H$-bilinear with
respect to the right ${\CC }H$-module structure on ${\CC }G$ and
the left ${\CC }H$-module structure on $U$.

In fact, every $h^{\prime}\in H$, $q\in{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta}\left(
qh^{\prime},u\right)  . \label{sol.Indrho.f.sol1.Rg.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.Rg.9}):} Let $h^{\prime}\in
H$, $q\in{\CC }G$ and $u\in U$. Then, the map $H\rightarrow H,\ h\mapsto
h^{\prime}h$ is a bijection (since $H$ is a group). The definition of
$\widetilde{\beta}$ yields
\begin{align}
\widetilde{\beta}\left(  q,h^{\prime}u\right)   &  =\sum_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  q\right)  h^{-1}h^{\prime}u=\sum_{h\in
H}\epsilon_{1}\left(  \underbrace{\rho\left(  h^{\prime}h\right)
}_{\substack{=\rho\left(  h^{\prime}\right)  \rho\left(  h\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}q\right)
\underbrace{\left(  h^{\prime}h\right)  ^{-1}}_{=h^{-1}\left(  h^{\prime
}\right)  ^{-1}}h^{\prime}u\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }h^{\prime}h\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{\prime}h\text{ is a
bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{h\in H}\underbrace{\epsilon_{1}\left(  \rho\left(  h^{\prime
}\right)  \rho\left(  h\right)  q\right)  }_{\substack{=\epsilon_{1}\left(
\rho\left(  h\right)  q\rho\left(  h^{\prime}\right)  \right)  \\\text{(by
(\ref{sol.Indrho.f.sol1.trace}), applied to}\\\rho\left(  h^{\prime}\right)
\text{ and }\rho\left(  h\right)  q\text{ instead of }p\text{ and }q\text{)}%
}}h^{-1}\underbrace{\left(  h^{\prime}\right)  ^{-1}h^{\prime}}_{=1}%
u\nonumber\\
&  =\sum_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\rho\left(
h^{\prime}\right)  \right)  h^{-1}u. \label{sol.Indrho.f.sol1.Rg.9.pf.1}%
\end{align}
\par
But the definition of the right $\CC H$-module structure on
$\CC G$ yields $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$. Applying this to $\gamma=q$ and $\eta=h^{\prime}$, we obtain
$qh^{\prime}=q\cdot\left(  \CC \left[  \rho\right]  \right)  \left(
h^{\prime}\right)  $. But $h^{\prime}\in H$ and thus $\left(  \CC %
\left[  \rho\right]  \right)  \left(  h^{\prime}\right)  =\rho\left(
h^{\prime}\right)  $. Hence, $qh^{\prime}=q\cdot\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h^{\prime}\right)  }%
_{=\rho\left(  h^{\prime}\right)  }=q\rho\left(  h^{\prime}\right)  $.
\par
But the definition of $\widetilde{\beta}$ also yields%
\begin{align*}
\widetilde{\beta}\left(  qh^{\prime},u\right)   &  =\sum_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  \underbrace{qh^{\prime}}_{=q\rho\left(
h^{\prime}\right)  }\right)  h^{-1}u=\sum_{h\in H}\epsilon_{1}\left(
\rho\left(  h\right)  q\rho\left(  h^{\prime}\right)  \right)  h^{-1}u\\
&  =\widetilde{\beta}\left(  q,h^{\prime}u\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Indrho.f.sol1.Rg.9.pf.1})}\right)  .
\end{align*}
\par
In other words, $\widetilde{\beta}\left(  q,h^{\prime}u\right)
=\widetilde{\beta}\left(  qh^{\prime},u\right)  $. This proves
(\ref{sol.Indrho.f.sol1.Rg.9}).} As a consequence of this, we can see that
every $r\in{\CC }H$, $q\in{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,ru\right)  =\widetilde{\beta}\left(  qr,u\right)  .
\label{sol.Indrho.f.sol1.Rg.10}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.Rg.10}):} Let
$r\in{\CC }H$, $q\in{\CC }G$ and $u\in U$. Then, we can write $r$
in the form $r=\sum\limits_{h\in H}r_{h}h$ for some family $\left(
r_{h}\right)  _{h\in H}$ of elements of $\CC$. Consider this family
$\left(  r_{h}\right)  _{h\in H}$. We have
\begin{align*}
\widetilde{\beta}\left(  q,\underbrace{r}_{=\sum\limits_{h\in H}r_{h}%
h}u\right)   &  =\widetilde{\beta}\left(  q,\sum\limits_{h\in H}%
r_{h}hu\right)  =\sum\limits_{h\in H}r_{h}\underbrace{\widetilde{\beta}\left(
q,hu\right)  }_{\substack{=\widetilde{\beta}\left(  qh,u\right)  \\\text{(by
(\ref{sol.Indrho.f.sol1.Rg.9}), applied to }h^{\prime}=h\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\widetilde{\beta}\text{ is
}{\CC }\text{-bilinear}\right) \\
&  =\sum\limits_{h\in H}r_{h}\widetilde{\beta}\left(  qh,u\right)
=\widetilde{\beta}\left(  q\underbrace{\sum\limits_{h\in H}r_{h}h}%
_{=r},u\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\widetilde{\beta}\text{ is }\CC \text{-bilinear}\right) \\
&  =\widetilde{\beta}\left(  qr,u\right)  .
\end{align*}
This proves (\ref{sol.Indrho.f.sol1.Rg.10}).} In other words, the map
$\widetilde{\beta}$ is ${\CC }H$-bilinear with respect to the right
${\CC }H$-module structure on ${\CC }G$ and the left ${\CC %
}H$-module structure on $U$ (since we already know that $\widetilde{\beta}$ is
$\CC$-bilinear). Hence, by the universal property of the tensor
product, we conclude that there exists a unique $\CC$-linear map
$\beta:{\CC }G\otimes_{\CC H}U\rightarrow U^{K}$ such that every
$\left(  q,u\right)  \in{\CC }G\times U$ satisfies
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  . \label{sol.Indrho.f.sol1.beta}%
\end{equation}
Consider this map $\beta$. Clearly, every $q\in{\CC }G$ and $u\in U$
satisfy
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  =\sum_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  q\right)
h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\widetilde{\beta}\right)  . \label{sol.Indrho.f.sol1.beta.explicit}%
\end{equation}


We shall now show that the maps $\alpha$ and $\dfrac{1}{\left\vert
K\right\vert }\beta$ are mutually inverse. To do so, we will show that
$\alpha\circ\beta=\left\vert K\right\vert \operatorname{id}$ and $\beta
\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

Let us first notice that%
\begin{equation}
1\otimes_{\CC H}h^{-1}u=\left(  \rho\left(  h\right)  \right)
^{-1}\otimes_{\CC H}u\ \ \ \ \ \ \ \ \ \ \text{for any }h\in H\text{ and
}u\in U. \label{sol.Indrho.f.sol1.rhohu}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.rhohu}):} Let $h\in H$ and
$u\in U$. We have $h\in H$. Since $H$ is a group, this yields $h^{-1}\in
H\subset\CC H$. Hence, we can move the $h^{-1}$ past the tensor sign in
$1\otimes_{\CC H}h^{-1}u$. We thus obtain $1\otimes_{\CC H}%
h^{-1}u=1h^{-1}\otimes_{\CC H}u$.
\par
But the definition of the right $\CC H$-module structure on
$\CC G$ yields $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$. Applying this to $\gamma=1$ and $\eta=h^{-1}$, we obtain
$1h^{-1}=1\cdot\left(  \CC \left[  \rho\right]  \right)  \left(
h^{-1}\right)  $. But $h^{-1}\in H$ and thus $\left(  \CC \left[
\rho\right]  \right)  \left(  h^{-1}\right)  =\rho\left(  h^{-1}\right)
=\left(  \rho\left(  h\right)  \right)  ^{-1}$ (since $\rho$ is a group
homomorphism). Hence, $1h^{-1}=1\cdot\underbrace{\left(  \CC \left[
\rho\right]  \right)  \left(  h^{-1}\right)  }_{=\left(  \rho\left(  h\right)
\right)  ^{-1}}=1\cdot\left(  \rho\left(  h\right)  \right)  ^{-1}=\left(
\rho\left(  h\right)  \right)  ^{-1}$. Now, $1\otimes_{\CC H}%
h^{-1}u=\underbrace{1h^{-1}}_{=\left(  \rho\left(  h\right)  \right)  ^{-1}%
}\otimes_{\CC H}u=\left(  \rho\left(  h\right)  \right)  ^{-1}%
\otimes_{\CC H}u$. This proves (\ref{sol.Indrho.f.sol1.rhohu}).}

Now, we are going to show that $\alpha\circ\beta=\left\vert K\right\vert
\operatorname{id}$. In fact, let $q\in{\CC }G$ and $u\in U$ be
arbitrary. Then,
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)   &
=\alpha\left(  \beta\left(  q\otimes_{\CC H}u\right)  \right)
=1\otimes_{\CC H}\underbrace{\beta\left(  q\otimes_{{\CC }%
H}u\right)  }_{\substack{=\sum_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  q\right)  h^{-1}u\\\text{(by (\ref{sol.Indrho.f.sol1.beta.explicit}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\right) \\
&  =1\otimes_{\CC H}\left(  \sum_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  q\right)  h^{-1}u\right)  =\underbrace{\sum_{h\in H}}_{=\sum_{g\in
G}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}}\epsilon_{1}\left(
\rho\left(  h\right)  q\right)  \underbrace{1\otimes_{\CC H}h^{-1}%
u}_{\substack{=\left(  \rho\left(  h\right)  \right)  ^{-1}\otimes
_{\CC H}u\\\text{(by (\ref{sol.Indrho.f.sol1.rhohu}))}}}\\
&  =\sum_{g\in G}\underbrace{\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}}_{\substack{=\sum\limits_{\substack{h\in H;\\h\in\rho^{-1}\left(  g\right)
}}\\\text{(because for any }h\in H\text{, the}\\\text{statement }\left(
\rho\left(  h\right)  =g\right)  \text{ is}\\\text{equivalent to }\left(
h\in\rho^{-1}\left(  g\right)  \right)  \text{)}}}\epsilon_{1}\left(
\underbrace{\rho\left(  h\right)  }_{=g}q\right)  \left(  \underbrace{\rho
\left(  h\right)  }_{=g}\right)  ^{-1}\otimes_{\CC H}u\\
&  =\sum_{g\in G}\underbrace{\sum\limits_{\substack{h\in H;\\h\in\rho^{-1}\left(
g\right)  }}}_{\substack{=\sum_{h\in\rho^{-1}\left(  g\right)  }\\\text{(since
}\rho^{-1}\left(  g\right)  \subset H\text{)}}}\epsilon_{1}\left(  gq\right)
g^{-1}\otimes_{\CC H}u=\sum_{g\in G}\underbrace{\sum_{h\in\rho
^{-1}\left(  g\right)  }\epsilon_{1}\left(  gq\right)  g^{-1}\otimes
_{\CC H}u}_{=\left\vert \rho^{-1}\left(  g\right)  \right\vert
\epsilon_{1}\left(  gq\right)  g^{-1}\otimes_{\CC H}u}\\
&  =\sum_{g\in G}\underbrace{\left\vert \rho^{-1}\left(  g\right)  \right\vert
}_{\substack{=\left\vert K\right\vert \\\text{(by
(\ref{sol.Indrho.f.sol1.preimage}))}}}\epsilon_{1}\left(  gq\right)
g^{-1}\otimes_{\CC H}u=\sum_{g\in G}\left\vert K\right\vert \epsilon
_{1}\left(  gq\right)  g^{-1}\otimes_{\CC H}u\\
&  =\left\vert K\right\vert \left(  \sum_{g\in G}\epsilon_{1}\left(
gq\right)  g^{-1}\right)  \otimes_{\CC H}u=\left\vert K\right\vert
\left(  \sum_{g\in G}\epsilon_{1}\left(  g^{-1}q\right)  \underbrace{\left(
g^{-1}\right)  ^{-1}}_{=g}\right)  \otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \left(  \sum_{g\in G}\underbrace{\epsilon
_{1}\left(  g^{-1}q\right)  }_{\substack{=\epsilon_{g}\left(  q\right)
\\\text{(by (\ref{sol.Indrho.f.sol1.epsi1g}))}}}g\right)  \otimes
_{\CC H}u=\left\vert K\right\vert \underbrace{\left(  \sum_{g\in
G}\epsilon_{g}\left(  q\right)  g\right)  }_{\substack{=q\\\text{(by
(\ref{sol.Indrho.f.sol1.epsilong.coordinates}))}}}\otimes_{\CC H}u\\
&  =\left\vert K\right\vert \underbrace{q\otimes_{\CC H}u}%
_{=\operatorname*{id}\left(  q\otimes_{\CC H}u\right)  }=\left\vert
K\right\vert \operatorname*{id}\left(  q\otimes_{\CC H}u\right)  .
\end{align*}


Now, let us forget that we fixed $q$ and $u$. We thus have shown that $\left(
\alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)  =\left\vert
K\right\vert \operatorname{id}\left(  q\otimes_{\CC H}u\right)  $ for
all $q\in{\CC }G$ and $u\in U$. In other words, the two maps
$\alpha\circ\beta:{\CC }G\otimes_{\CC H}U\rightarrow{\CC %
}G\otimes_{\CC H}U$ and $\left\vert K\right\vert \operatorname{id}%
:{\CC }G\otimes_{\CC H}U\rightarrow{\CC }G\otimes
_{\CC H}U$ are equal to each other on each pure tensor. Since these
two maps are $\CC$-linear, this yields that these two maps
$\alpha\circ\beta:{\CC }G\otimes_{\CC H}U\rightarrow{\CC %
}G\otimes_{\CC H}U$ and $\left\vert K\right\vert \operatorname{id}%
:{\CC }G\otimes_{\CC H}U\rightarrow{\CC }G\otimes
_{\CC H}U$ must be identical\footnote{This is because a well-known
fact from linear algebra states the following: If two $\CC$-linear
maps from a tensor product are equal to each other on each pure tensor, then
these two maps must be identical.}. In other words, $\alpha\circ
\beta=\left\vert K\right\vert \operatorname{id}$.

Next, we are going to show that $\beta\circ\alpha=\left\vert K\right\vert
\operatorname{id}$.

We first notice that%
\begin{equation}
h^{-1}u=\left(  \rho\left(  h\right)  \right)  ^{-1}%
u\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H\text{ and }u\in U^{K}.
\label{sol.Indrho.f.sol1.h-1u}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol1.h-1u}):} Let $h\in H$ and
$u\in U^{K}$. Applying (\ref{sol.Indrho.f.sol1.UK-modstr}) to $h^{-1}$ and $u$
instead of $h$ and $v$, we obtain $\rho\left(  h^{-1}\right)  \cdot u=h^{-1}%
u$. But $\rho$ is a group homomorphism, and thus we have $\left(  \rho\left(
h\right)  \right)  ^{-1}=\rho\left(  h^{-1}\right)  $. Hence,
$\underbrace{\left(  \rho\left(  h\right)  \right)  ^{-1}}_{=\rho\left(
h^{-1}\right)  }u=\rho\left(  h^{-1}\right)  u=\rho\left(  h^{-1}\right)
\cdot u=h^{-1}u$. This proves (\ref{sol.Indrho.f.sol1.h-1u}).}

Let $u\in U^{K}$. The definition of $\alpha\left(  u\right)  $ yields
$\alpha\left(  u\right)  =1\otimes_{\CC H}u$. Applying the map $\beta$
to this equality, we obtain
\begin{align*}
\beta\left(  \alpha\left(  u\right)  \right)   &  =\beta\left(  1\otimes
_{\CC H}u\right)  =\underbrace{\sum_{h\in H}}_{=\sum_{g\in G}%
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}}\epsilon_{1}\left(
\rho\left(  h\right)  1\right)  \underbrace{h^{-1}u}_{\substack{=\left(
\rho\left(  h\right)  \right)  ^{-1}u\\\text{(by (\ref{sol.Indrho.f.sol1.h-1u}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.f.sol1.beta.explicit}), applied to }q=1\right) \\
&  =\sum_{g\in G}\underbrace{\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}}_{\substack{=\sum\limits_{\substack{h\in H;\\h\in\rho^{-1}\left(  g\right)
}}\\\text{(because for any }h\in H\text{, the}\\\text{statement }\left(
\rho\left(  h\right)  =g\right)  \text{ is}\\\text{equivalent to }\left(
h\in\rho^{-1}\left(  g\right)  \right)  \text{)}}}\epsilon_{1}\left(
\underbrace{\rho\left(  h\right)  }_{=g}1\right)  \left(  \underbrace{\rho
\left(  h\right)  }_{=g}\right)  ^{-1}u=\sum_{g\in G}\underbrace{\sum\limits
_{\substack{h\in H;\\h\in\rho^{-1}\left(  g\right)  }}}_{\substack{=\sum
_{h\in\rho^{-1}\left(  g\right)  }\\\text{(since }\rho^{-1}\left(  g\right)
\subset H\text{)}}}\epsilon_{1}\left(  g1\right)  g^{-1}u\\
&  =\sum_{g\in G}\underbrace{\sum_{h\in\rho^{-1}\left(  g\right)  }%
\epsilon_{1}\left(  g1\right)  g^{-1}u}_{=\left\vert \rho^{-1}\left(
g\right)  \right\vert \epsilon_{1}\left(  g1\right)  g^{-1}u}=\sum_{g\in
G}\underbrace{\left\vert \rho^{-1}\left(  g\right)  \right\vert }%
_{\substack{=\left\vert K\right\vert \\\text{(by
(\ref{sol.Indrho.f.sol1.preimage}))}}}\epsilon_{1}\left(  g1\right)
g^{-1}u=\sum_{g\in G}\left\vert K\right\vert \epsilon_{1}\left(  g1\right)
g^{-1}u\\
&  =\left\vert K\right\vert \sum_{g\in G}\epsilon_{1}\left(  g1\right)
g^{-1}u=\left\vert K\right\vert \sum_{g\in G}\epsilon_{1}\left(
g^{-1}1\right)  \underbrace{\left(  g^{-1}\right)  ^{-1}}_{=g}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \sum_{g\in G}\underbrace{\epsilon_{1}\left(
g^{-1}1\right)  }_{\substack{=\epsilon_{g}\left(  1\right)  \\\text{(by
(\ref{sol.Indrho.f.sol1.epsi1g}), applied to }q=1\text{)}}}gu=\left\vert
K\right\vert \sum_{g\in G}\epsilon_{g}\left(  1\right)  gu=\left\vert
K\right\vert \underbrace{\left(  \sum_{g\in G}\epsilon_{g}\left(  1\right)
g\right)  }_{\substack{=1\\\text{(because }1=\sum_{g\in G}\epsilon_{g}\left(
1\right)  g\\\text{(by (\ref{sol.Indrho.f.sol1.epsilong.coordinates}), applied
to }q=1\text{))}}}u\\
&  =\left\vert K\right\vert \underbrace{u}_{=\operatorname*{id}\left(
u\right)  }=\left\vert K\right\vert \operatorname*{id}\left(  u\right)  .
\end{align*}
In other words,
$\left(\beta \circ \alpha\right)\left(u\right)
= \left\vert K\right\vert \operatorname*{id}\left(  u\right)$
(since $\left(\beta \circ \alpha\right)\left(u\right)
= \beta\left(\alpha\left(u\right)\right)$).

Now, forget that we fixed $u$. We thus have shown that $\left(  \beta
\circ\alpha\right)  \left(  u\right)  =\left\vert K\right\vert
\operatorname{id}\left(  u\right)  $ for every $u\in U^{K}$. In other words,
$\beta\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

Composition of $\CC$-linear maps is $\CC$-bilinear. Thus,
$\left(  \dfrac{1}{\left\vert K\right\vert }\beta\right)  \circ\alpha
=\dfrac{1}{\left\vert K\right\vert }\underbrace{\beta\circ\alpha}_{=\left\vert
K\right\vert \operatorname*{id}}=\dfrac{1}{\left\vert K\right\vert }%
\cdot\left\vert K\right\vert \operatorname*{id}=\operatorname*{id}$. For the
same reason, $\alpha\circ\left(  \dfrac{1}{\left\vert K\right\vert }%
\beta\right)  =\dfrac{1}{\left\vert K\right\vert }\underbrace{\alpha\circ
\beta}_{=\left\vert K\right\vert \operatorname*{id}}=\dfrac{1}{\left\vert
K\right\vert }\cdot\left\vert K\right\vert \operatorname*{id}%
=\operatorname*{id}$.

The equalities $\alpha\circ\left(  \dfrac{1}{\left\vert K\right\vert }%
\beta\right)  =\operatorname*{id}$ and $\left(  \dfrac{1}{\left\vert
K\right\vert }\beta\right)  \circ\alpha=\operatorname*{id}$ yield that the
maps $\alpha$ and $\dfrac{1}{\left\vert K\right\vert }\beta$ are mutually
inverse. Hence, the map $\alpha$ is invertible.

Now, we know that the map $\alpha$ is an invertible ${\CC }G$-module
homomorphism. Hence, $\alpha$ is a ${\CC }G$-module isomorphism.
Therefore, there exists a ${\CC }G$-module isomorphism $U^{K}%
\rightarrow{\CC }G\otimes_{\CC H}U$ (namely, $\alpha$).
Therefore, $U^{K}\cong{\CC }G\otimes_{\CC H}U=\operatorname{Ind}%
_{\rho}U$ as ${\CC }G$-modules. This solves Exercise~\ref{exe.Indrho}(f).

\textit{Second solution to Exercise \ref{exe.Indrho}(f) (sketched).} Assume
that $G=H/K$ for some normal subgroup $K$ of $H$. Let $\rho:H\rightarrow G$ be
the projection map. We want to prove that $\Ind_{\rho
}U\cong U^{K}$ for every $\CC H$-module $U$.

We know that $\rho$ is the projection from $H$ to $G=H/K$. Hence,
$\ker\rho=K$.

Let $U$ be any $\CC H$-module. Recall that $U^{K}$ is a $\CC %
\left[  H/K\right]  $-module, thus a $\CC G$-module (since $H/K=G$).

Let $I_{U,K}$ denote the $\CC$-vector subspace of $U$ spanned by
elements of the form $v-kv$ for $k\in K$ and $v\in U$. Exercise
\ref{exe.fixed-vs-coinvariants}(a) (applied to $H$, $U$ and $I_{U,K}$ instead
of $G$, $V$ and $I_{V,K}$) yields that $I_{U,K}$ is a $\CC H$-submodule
of $U$. Let $U_{K}$ denote the quotient $\CC H$-module $U/I_{U,K}$.

Notice that%
\begin{equation}
v-kv\in I_{U,K}\ \ \ \ \ \ \ \ \ \ \text{for all }k\in K\text{ and }v\in U.
\label{sol.Indrho.f.sol2.IUK1}%
\end{equation}
(This is because $I_{U,K}$ is the $\CC$-vector subspace of $U$ spanned
by elements of the form $v-kv$ for $k\in K$ and $v\in U$.)

For every $v\in U$, let $\overline{v}$ denote the projection of $v$ onto the
quotient space $U/I_{U,K}=U_{K}$.

We can define a map $\Phi:U^{K}\rightarrow U_{K}$ by sending every $v\in
U^{K}$ to $\overline{v}\in U_{K}$. Then, the map $\Phi$ is bijective. (In
fact, the proof of this fact proceeds precisely like the proof of the
bijectivity of $\Phi$ in the solution of Exercise
\ref{exe.fixed-vs-coinvariants}(b), with the only difference that $G$, $V$,
$I_{V,K}$ and $V_{K}$ have to be replaced by $H$, $U$, $I_{U,K}$ and $U_{K}$.)

The $\CC$-algebra homomorphism $\CC \left[  \rho\right]  $ is
surjective\footnote{\textit{Proof.} Let $g$ be any element of $G$. Then, there
exists an element $h\in H$ such that $g=\rho\left(  h\right)  $ (since $\rho$
is surjective (because $\rho$ is a projection map)). Consider this $h$. Since
$h\in H$, we have $\left(  \CC \left[  \rho\right]  \right)  \left(
h\right)  =\rho\left(  h\right)  =g$, so that $g=\left(  \CC \left[
\rho\right]  \right)  \left(  h\right)  \in\left(  \CC \left[
\rho\right]  \right)  \left(  \CC H\right)  $.
\par
Let us now forget that we fixed $g$. We thus have shown that $g\in\left(
\CC \left[  \rho\right]  \right)  \left(  \CC H\right)  $ for
every $g\in G$. In other words, $G$ (viewed as a subset of $\CC G$) is
contained in $\left(  \CC \left[  \rho\right]  \right)  \left(
\CC H\right)  $. Hence, $\left(  \CC \left[  \rho\right]  \right)
\left(  \CC H\right)  $ is a $\CC$-vector subspace of
$\CC G$ which contains $G$ as a subset.
\par
But the set $G$ spans the $\CC$-vector space $\CC G$. Hence,
every $\CC$-vector subspace of $\CC G$ which contains $G$ as a
subset must be equal to $\CC G$. Applying this to the $\CC %
$-vector subspace $\left(  \CC \left[  \rho\right]  \right)  \left(
\CC H\right)  $, we conclude that $\left(  \CC \left[
\rho\right]  \right)  \left(  \CC H\right)  $ must be equal to
$\CC G$ (since $\left(  \CC \left[  \rho\right]  \right)  \left(
\CC H\right)  $ contains $G$ as a subset). In other words, $\left(
\CC \left[  \rho\right]  \right)  \left(  \CC H\right)
=\CC G$. In other words, the map $\CC \left[  \rho\right]  $ is
surjective, qed.}.

Let $J=\ker\left(  \CC \left[  \rho\right]  \right)  $.
Then, $J$ is a two-sided ideal of the $\CC$-algebra $\CC H$
(since it is the kernel of the $\CC$-algebra homomorphism
$\CC \left[  \rho\right]  $). Let $\iota:J\rightarrow\CC H$ be the
canonical inclusion map. Then, $\iota\left(  J\right)  =J$ (since $\iota$ is
the inclusion map). Now, the sequence
\[
\xymatrix@C=5pc{
J \ar[r]^{\iota} & \CC H \ar[r]^{\CC\left[\rho\right]} & \CC G \ar[r] & 0
}
\]
is exact (since $\iota\left(  J\right)  =J=\ker\left(
\CC \left[  \rho\right]  \right)  $, and since $\CC \left[
\rho\right]  $ is surjective). Hence, the sequence
\[
\xymatrix@C=5pc{
J \otimes_{\CC H} U \ar[r]^{\iota\otimes_{\CC H} \id_U} & \CC H \otimes_{\CC
H} U \ar[r]^{\CC\left[\rho\right] \otimes_{\CC H} \id_U} & \CC G \otimes_{\CC
H} U \ar[r] & 0
}
\]
is exact (since the functor $-\otimes_{\CC H}U$ is right-exact). Thus,
we have
\begin{equation}
\ker\left(  \CC \left[  \rho\right]  \otimes
_{\CC H}\id_{U}\right)  =\left(  \iota
\otimes_{\CC H}\id_{U}\right)  \left(
J\otimes_{\CC H}U\right)  \label{sol.Indrho.f.sol2.exactness}%
\end{equation}
(since this sequence is exact at $\CC H\otimes_{\CC H}U$) and the
map $\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}$ is surjective (since this sequence is exact at $\CC %
G\otimes_{\CC H}U$).

Next, we notice that $J$ is the $\CC$-vector subspace of $\CC H$
spanned by elements of the form $h-kh$ for $k\in K$ and $h\in H$%
\ \ \ \ \footnote{\textit{Proof.} Let $Z$ be the $\CC$-vector subspace
of $\CC H$ spanned by elements of the form $h-kh$ for $k\in K$ and $h\in
H$. We shall show that $J=Z$.
\par
For all $k\in K$ and $h\in H$, we have%
\begin{align*}
\left(  \CC \left[  \rho\right]  \right)  \left(  h-kh\right)   &
=\underbrace{\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
}_{\substack{=\rho\left(  h\right)  \\\text{(since }h\in H\text{)}%
}}-\underbrace{\left(  \CC \left[  \rho\right]  \right)  \left(
kh\right)  }_{\substack{=\rho\left(  kh\right)  \\\text{(since }kh\in
H\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\CC \left[
\rho\right]  \text{ is }\CC \text{-linear}\right) \\
&  =\rho\left(  h\right)  -\underbrace{\rho\left(  kh\right)  }%
_{\substack{=\rho\left(  k\right)  \rho\left(  h\right)  \\\text{(since }%
\rho\text{ is a group}\\\text{homomorphism)}}}=\rho\left(  h\right)
-\underbrace{\rho\left(  k\right)  }_{\substack{=1\\\text{(since }k\in
K=\ker\rho\text{)}}}\rho\left(  h\right)  =\rho\left(
h\right)  -\rho\left(  h\right)  =0.
\end{align*}
In other words, for all $k\in K$ and $h\in H$, we have $h-kh\in
\ker\left(  \CC \left[  \rho\right]  \right)  =J$. In
other words,
\begin{equation}
\left\{  h-kh\ \mid\ k\in K;\ h\in H\right\}  \subset J.
\label{sol.Indrho.f.sol2.spanJ.pf.inJ}%
\end{equation}
\par
Now, $Z$ is the $\CC$-vector subspace of $\CC H$ spanned by
elements of the form $h-kh$ for $k\in K$ and $h\in H$. In other words, $Z$ is
the $\CC$-vector subspace of $\CC H$ spanned by the set $\left\{
h-kh\ \mid\ k\in K;\ h\in H\right\}  $. In other words, $Z$ is the smallest
$\CC$-vector subspace of $\CC H$ which contains $\left\{
h-kh\ \mid\ k\in K;\ h\in H\right\}  $ as a subset. In other words, if $S$ is
any $\CC$-vector subspace of $\CC H$ which contains $\left\{
h-kh\ \mid\ k\in K;\ h\in H\right\}  $ as a subset, then%
\begin{equation}
Z\subset S. \label{sol.Indrho.f.sol2.spanJ.pf.ZinS}%
\end{equation}
\par
But the $\CC$-vector subspace $J$ of $\CC H$ contains $\left\{
h-kh\ \mid\ k\in K;\ h\in H\right\}  $ as a subset (according to
(\ref{sol.Indrho.f.sol2.spanJ.pf.inJ})). Thus, $Z\subset J$ (by
(\ref{sol.Indrho.f.sol2.spanJ.pf.ZinS}), applied to $S=J$).
\par
The $\CC$-vector space $Z$ is spanned by elements of the form $h-kh$
for $k\in K$ and $h\in H$ (due to the definition of $Z$). Hence,%
\begin{equation}
h-kh\in Z\ \ \ \ \ \ \ \ \ \ \text{for all }k\in K\text{ and }h\in H.
\label{sol.Indrho.f.sol2.spanJ.pf.h-kh}%
\end{equation}
\par
Now, let $j\in J$. Then, $j\in J\subset\CC H$. Hence, we can write $j$
in the form $j=\sum_{h\in H}j_{h}h$ for some family $\left(  j_{h}\right)
_{h\in H}$ of elements of $\CC$. Consider this family $\left(
j_{h}\right)  _{h\in H}$.
\par
The map $\rho$ is surjective (since $\rho$ is a projection map). Hence, for
every $g\in G$, there exists an element $h\in H$ such that $g=\rho\left(
h\right)  $. Fix such a $h$ (for every $g\in G$) and denote it by $r_{g}$.
Thus, for every $g\in G$, the element $r_{g}$ of $H$ satisfies%
\begin{equation}
g=\rho\left(  r_{g}\right)  . \label{sol.Indrho.f.sol2.spanJ.pf.rg}%
\end{equation}
\par
But $\sum_{h\in H}j_{h}h=j\in J=\ker\left(  \CC \left[
\rho\right]  \right)  $, so that $\left(  \CC \left[  \rho\right]
\right)  \left(  \sum_{h\in H}j_{h}h\right)  =0$. Comparing this with%
\begin{align*}
\left(  \CC \left[  \rho\right]  \right)  \left(  \sum_{h\in H}%
j_{h}h\right)   &  =\underbrace{\sum_{h\in H}}_{=\sum_{g\in G}\sum\limits
_{\substack{h\in H;\\\rho\left(  h\right)  =g}}}j_{h}\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  }_{\substack{=\rho
\left(  h\right)  \\\text{(since }h\in H\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\CC \left[  \rho\right]  \text{ is }\CC %
\text{-linear}\right) \\
&  =\sum_{g\in G}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}%
}j_{h}\underbrace{\rho\left(  h\right)  }_{\substack{=g\\\text{(since }%
\rho\left(  h\right)  =g\text{)}}}=\sum_{g\in G}\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}j_{h}g=\sum_{g\in G}\left(  \sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}j_{h}\right)  g,
\end{align*}
we obtain $\sum_{g\in G}\left(  \sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}j_{h}\right)  g=0$.
\par
But $G$ is a basis of the $\CC$-vector space $\CC G$, and thus a
$\CC$-linearly independent subset of $\CC G$. In other words, if
$\left(  \lambda_{g}\right)  _{g\in G}\in\CC ^{G}$ is any family of
elements of $\CC$ satisfying $\sum_{g\in G}\lambda_{g}g=0$, then
$\left(  \lambda_{g}\right)  _{g\in G}=\left(  0\right)  _{g\in G}$. Applying
this to the family $\left(  \lambda_{g}\right)  _{g\in G}=\left(
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}j_{h}\right)  _{g\in G}$,
we obtain $\left(  \sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}%
}j_{h}\right)  _{g\in G}=\left(  0\right)  _{g\in G}$ (since $\sum_{g\in
G}\left(  \sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}j_{h}\right)
g=0$). In other words,%
\begin{equation}
\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}}j_{h}%
=0\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G.
\label{sol.Indrho.f.sol2.spanJ.pf.sum0}%
\end{equation}
\par
On the other hand, fix $h\in H$. Applying (\ref{sol.Indrho.f.sol2.spanJ.pf.rg}%
) to $g=\rho\left(  h\right)  $, we obtain $\rho\left(  h\right)  =\rho\left(
r_{\rho\left(  h\right)  }\right)  $. Since $\rho$ is a group homomorphism, we
have $\rho\left(  r_{\rho\left(  h\right)  }h^{-1}\right)  =\underbrace{\rho
\left(  r_{\rho\left(  h\right)  }\right)  }_{=\rho\left(  h\right)  }\left(
\rho\left(  h\right)  \right)  ^{-1}=\rho\left(  h\right)  \left(  \rho\left(
h\right)  \right)  ^{-1}=1$, so that $r_{\rho\left(  h\right)  }h^{-1}%
\in\ker\rho=K$ and thus $r_{\rho\left(  h\right)  }\in Kh$. In
other words, there exists a $k\in K$ such that $r_{\rho\left(  h\right)  }%
=kh$. Consider this $k$. Then, $h-\underbrace{r_{\rho\left(  h\right)  }%
}_{=kh}=h-kh\in Z$ (by (\ref{sol.Indrho.f.sol2.spanJ.pf.h-kh})).
\par
Let us now forget that we fixed $h$. We thus have shown that%
\begin{equation}
h-r_{\rho\left(  h\right)  }\in Z\ \ \ \ \ \ \ \ \ \ \text{for every }h\in H.
\label{sol.Indrho.f.sol2.spanJ.pf.modZ}%
\end{equation}
Now, subtracting $\sum_{h\in H}j_{h}\left(  h-r_{\rho\left(  h\right)
}\right)  $ from both sides of the equality $j=\sum_{h\in H}j_{h}h$, we
obtain
\begin{align*}
j-\sum_{h\in H}j_{h}\left(  h-r_{\rho\left(  h\right)  }\right)   &
=\sum_{h\in H}j_{h}h-\sum_{h\in H}j_{h}\left(  h-r_{\rho\left(  h\right)
}\right)  =\underbrace{\sum_{h\in H}}_{=\sum_{g\in G}\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}}j_{h}\underbrace{\left(  h-\left(
h-r_{\rho\left(  h\right)  }\right)  \right)  }_{=r_{\rho\left(  h\right)  }%
}\\
&  =\sum_{g\in G}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)  =g}%
}j_{h}\underbrace{r_{\rho\left(  h\right)  }}_{\substack{=r_{g}\\\text{(since
}\rho\left(  h\right)  =g\text{)}}}=\sum_{g\in G}\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}j_{h}r_{g}\\
&  =\sum_{g\in G}\underbrace{\left(  \sum\limits_{\substack{h\in H;\\\rho\left(
h\right)  =g}}j_{h}\right)  }_{\substack{=0\\\text{(by
(\ref{sol.Indrho.f.sol2.spanJ.pf.sum0}))}}}r_{g}=\sum_{g\in G}0r_{g}=0.
\end{align*}
Hence,%
\[
j=\sum_{h\in H}j_{h}\underbrace{\left(  h-r_{\rho\left(  h\right)  }\right)
}_{\substack{\in Z\\\text{(by (\ref{sol.Indrho.f.sol2.spanJ.pf.modZ}))}}%
}\in\sum_{h\in H}j_{h}Z\subset Z
\]
(since $Z$ is a $\CC$-vector space).
\par
Let us now forget that we fixed $j$. We thus have shown that $j\in Z$ for
every $j\in J$. In other words, $J\subset Z$. Combined with $Z\subset J$, this
yields $J=Z$.
\par
Recall that $Z$ is the $\CC$-vector subspace of $\CC H$ spanned
by elements of the form $h-kh$ for $k\in K$ and $h\in H$. In other words, $J$
is the $\CC$-vector subspace of $\CC H$ spanned by elements of
the form $h-kh$ for $k\in K$ and $h\in H$ (since $J=Z$). Qed.}. In other
words, $J$ is the $\CC$-vector subspace of $\CC H$ spanned by
elements of the form $h-kh$ for $\left(  k,h\right)  \in K\times H$. In other
words,%
\begin{equation}
\text{the family }\left(  h-kh\right)  _{\left(  k,h\right)  \in K\times
H}\text{ spans the }\CC \text{-vector space }J.
\label{sol.Indrho.f.sol2.Jspan}%
\end{equation}


Recall that $\Ind_{\rho}U$ is defined as the
$\CC G$-module $\CC G\otimes_{\CC H}U$, where $\CC G$
is regarded as a $\left(  \CC G,\CC H\right)  $-bimodule according
to the following rule: The left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$; the right
$\CC H$-module structure on $\CC G$ is induced by the $\CC %
$-algebra homomorphism $\CC \left[  \rho\right]  :\CC %
H\rightarrow\CC G$ (thus, it is explicitly given by $\gamma\eta
=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$ for all
$\gamma\in\CC G$ and $\eta\in\CC H$). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC G,\CC H\right)
$-bimodule structure.

We define a map $\alpha:U^{K}\rightarrow\CC G\otimes_{\CC H}U$ by
setting%
\begin{equation}
\alpha\left(  u\right)  =1\otimes_{\CC H}u\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in U^{K}. \label{sol.Indrho.f.sol2.alpha}%
\end{equation}
This $\alpha$ is a map $U^{K}\rightarrow\Ind_{\rho}U$
(since $\Ind_{\rho}U=\CC G\otimes_{\CC %
H}U$). This map $\alpha$ is a $\CC G$-module homomorphism\footnote{This
can be proven in the same way as in the First solution to Exercise
\ref{exe.Indrho}(f).}.

Define a $\CC$-linear map $\zeta:U\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U$ by%
\[
\left(  \zeta\left(  u\right)  =1\otimes_{\CC H}%
u\ \ \ \ \ \ \ \ \ \ \text{for all }u\in U\right)  .
\]
(This is well-defined, since every $u\in U$ satisfies $1\otimes_{\CC %
H}u\in\CC G\otimes_{\CC H}U=\Ind_{\rho}%
U$.) Clearly, the map $\zeta$ is $\CC$-linear (since $\zeta\left(
u\right)  =1\otimes_{\CC H}u$ depends $\CC$-linearly on $u$).
Moreover, $\zeta\mid_{U^{K}}=\alpha$\ \ \ \ \footnote{\textit{Proof.} Every
$u\in U^{K}$ satisfies $\left(  \zeta\mid_{U^{K}}\right)  \left(  u\right)
=\zeta\left(  u\right)  =1\otimes_{\CC H}u=\alpha\left(  u\right)  $
(because of (\ref{sol.Indrho.f.sol2.alpha})). Hence, $\zeta\mid_{U^{K}}%
=\alpha$, qed.}. Hence, the map $\zeta\mid_{U^{K}}:U^{K}\rightarrow
\Ind_{\rho}U$ is a $\CC G$-module homomorphism
(since the map $\alpha:U^{K}\rightarrow\Ind_{\rho}U$
is a $\CC G$-module homomorphism). We are going to show that this map
$\zeta\mid_{U^{K}}$ is actually an isomorphism.

For every $\CC$-algebra $A$ and every $A$-module $M$, there is a
canonical $A$-module isomorphism $A\otimes_{A}M\rightarrow M$ which sends
every $a\otimes_{A}m\in A\otimes_{A}M$ to $am\in M$. Applying this to
$A=\CC H$ and $M=U$, we conclude that there is a canonical $\CC %
H$-module isomorphism $\CC H\otimes_{\CC H}U\rightarrow U$ which
sends every $a\otimes_{\CC H}m\in\CC H\otimes_{\CC H}U$ to
$am\in U$. Let us denote this $\CC H$-module isomorphism by $\psi$.
Thus, $\psi:\CC H\otimes_{\CC H}U\rightarrow U$ is a $\CC %
H$-module isomorphism, and satisfies%
\begin{equation}
\psi\left(  a\otimes_{\CC H}m\right)  =am\ \ \ \ \ \ \ \ \ \ \text{for
all }a\in\CC H\text{ and }m\in U. \label{sol.Indrho.f.sol2.psi}%
\end{equation}
Since $\psi$ is an isomorphism, we have
\begin{equation}
\ker\zeta=\psi\left(  \ker\left(  \zeta\circ
\psi\right)  \right)  \label{sol.Indrho.f.sol2.KerKer}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol2.KerKer}):} We have%
\[
\psi^{-1}\left(  \underbrace{\ker\zeta}_{=\zeta^{-1}\left(
0\right)  }\right)  =\psi^{-1}\left(  \zeta^{-1}\left(  0\right)  \right)
=\left(  \zeta\circ\psi\right)  ^{-1}\left(  0\right)  =\ker%
\left(  \zeta\circ\psi\right)  ,
\]
so that $\psi\left(  \underbrace{\psi^{-1}\left(  \ker%
\zeta\right)  }_{=\ker\left(  \zeta\circ\psi\right)  }\right)
=\psi\left(  \ker\left(  \zeta\circ\psi\right)  \right)  $.
\par
But $\psi$ is an $\CC H$-module isomorphism, and thus a bijection.
Hence, $\psi\left(  \psi^{-1}\left(  \ker\zeta\right)  \right)
=\ker\zeta$. Compared with $\psi\left(  \psi^{-1}\left(
\ker\zeta\right)  \right)  =\psi\left(  \ker%
\left(  \zeta\circ\psi\right)  \right)  $, this yields $\ker%
\zeta=\psi\left(  \ker\left(  \zeta\circ\psi\right)  \right)
$. This proves (\ref{sol.Indrho.f.sol2.KerKer}).}. But%
\begin{equation}
\zeta\circ\psi=\CC \left[  \rho\right]  \otimes_{\CC %
H}\id_{U} \label{sol.Indrho.f.sol2.zetapsi}%
\end{equation}
as maps from $\CC H\otimes_{\CC H}U$ to $\operatorname*{Ind}%
\nolimits_{\rho}U$\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.Indrho.f.sol2.zetapsi}):} The map $\psi$ is a $\CC H$-module
isomorphism, and thus a bijection. Hence, $\psi^{-1}$ is a well-defined map
$U\rightarrow\CC H\otimes_{\CC H}U$. Therefore, $\left(
\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \circ\psi^{-1}$ is a map $U\rightarrow\CC %
G\otimes_{\CC H}U$, therefore a map $U\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U$ (since $\CC G\otimes_{\CC H}%
U=\Ind_{\rho}U$). Hence, both $\zeta$ and $\left(
\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \circ\psi^{-1}$ are maps $U\rightarrow
\Ind_{\rho}U$.
\par
We shall now show that $\zeta=\left(  \CC \left[  \rho\right]
\otimes_{\CC H}\id_{U}\right)  \circ\psi^{-1}$.
Indeed, let $u\in U$. The definition of $\psi$ yields $\psi\left(
1\otimes_{\CC H}u\right)  =1u=u$. Hence, $1\otimes_{\CC H}%
u=\psi^{-1}\left(  u\right)  $ (since $\psi$ is invertible (because $\psi$ is
an isomorphism)). Now,%
\[
\left(  \CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  1\otimes_{\CC H}u\right)
=\underbrace{\left(  \CC \left[  \rho\right]  \right)  \left(  1\right)
}_{\substack{=1\\\text{(since }\CC \left[  \rho\right]  \text{ is
a}\\\CC \text{-algebra homomorphism)}}}\otimes_{\CC H}%
u=1\otimes_{\CC H}u=\zeta\left(  u\right)
\]
(since $\zeta\left(  u\right)  =1\otimes_{\CC H}u$ (by the definition of
$\zeta$)). Hence,%
\begin{align*}
\zeta\left(  u\right)   &  =\left(  \CC \left[  \rho\right]
\otimes_{\CC H}\id_{U}\right)  \left(
\underbrace{1\otimes_{\CC H}u}_{=\psi^{-1}\left(  u\right)  }\right)
=\left(  \CC \left[  \rho\right]  \otimes_{\CC H}%
\id_{U}\right)  \left(  \psi^{-1}\left(  u\right)
\right) \\
&  =\left(  \left(  \CC \left[  \rho\right]  \otimes_{\CC %
H}\id_{U}\right)  \circ\psi^{-1}\right)  \left(
u\right)  .
\end{align*}
\par
Let us now forget that we fixed $u$. We thus have shown that $\zeta\left(
u\right)  =\left(  \left(  \CC \left[  \rho\right]  \otimes
_{\CC H}\id_{U}\right)  \circ\psi^{-1}\right)
\left(  u\right)  $ for every $u\in U$. In other words, $\zeta=\left(
\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \circ\psi^{-1}$ (since both $\zeta$ and $\left(
\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \circ\psi^{-1}$ are maps $U\rightarrow
\Ind_{\rho}U$). Hence, $\underbrace{\zeta}_{=\left(
\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \circ\psi^{-1}}\circ\psi=\left(  \CC \left[
\rho\right]  \otimes_{\CC H}\id_{U}\right)
\circ\underbrace{\psi^{-1}\circ\psi}_{=\operatorname*{id}}=\CC \left[
\rho\right]  \otimes_{\CC H}\id_{U}$. This
proves (\ref{sol.Indrho.f.sol2.zetapsi}).}. Hence,
(\ref{sol.Indrho.f.sol2.KerKer}) becomes%
\begin{align}
\ker\zeta &  =\psi\left(  \ker\left(
\underbrace{\zeta\circ\psi}_{=\CC \left[  \rho\right]  \otimes
_{\CC H}\id_{U}}\right)  \right)  =\psi\left(
\underbrace{\ker\left(  \CC \left[  \rho\right]
\otimes_{\CC H}\id_{U}\right)  }%
_{\substack{=\left(  \iota\otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  J\otimes_{\CC H}U\right)  \\\text{(by
(\ref{sol.Indrho.f.sol2.exactness}))}}}\right) \nonumber\\
&  =\psi\left(  \left(  \iota\otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  J\otimes_{\CC H}U\right)  \right)  .
\label{sol.Indrho.f.sol2.Ker4}%
\end{align}


Now, using (\ref{sol.Indrho.f.sol2.Jspan}), it is not hard to see that%
\begin{equation}
\psi\left(  \left(  \iota\otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  J\otimes_{\CC H}U\right)  \right)  \subset
I_{U,K}. \label{sol.Indrho.f.sol2.Ker6}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol2.Ker6}):} Clearly, the
family $\left(  u\right)  _{u\in U}$ spans the $\CC$-vector space $U$.
On the other hand, the family $\left(  h-kh\right)  _{\left(  k,h\right)  \in
K\times H}$ spans the $\CC$-vector space $J$ (according to
(\ref{sol.Indrho.f.sol2.Jspan})).
\par
But a very simple fact from linear algebra says the following: If
$\mathfrak{A}$ is a $\CC$-algebra, if $\mathfrak{M}$ is a right
$A$-module, if $\mathfrak{N}$ is a left $A$-module, if $\left(  \mathfrak{m}%
_{p}\right)  _{p\in P}$ is a family of elements of $\mathfrak{M}$ which spans
the $\CC$-vector space $\mathfrak{M}$, and if $\left(  \mathfrak{n}%
_{q}\right)  _{q\in Q}$ is a family of elements of $\mathfrak{N}$ which spans
the $\CC$-vector space $\mathfrak{N}$, then the family $\left(
\mathfrak{m}_{p}\otimes_{\mathfrak{A}}\mathfrak{n}_{q}\right)  _{\left(
p,q\right)  \in P\times Q}$ spans the $\CC$-vector space $\mathfrak{M}%
\otimes_{\mathfrak{A}}\mathfrak{N}$. We can apply this fact to $\mathfrak{A}%
=\CC H$, $\mathfrak{M}=J$, $\mathfrak{N}=U$, $P=K\times H$, $p=\left(
k,h\right)  $, $\mathfrak{m}_{\left(  k,h\right)  }=h-kh$, $Q=U$, $q=u$, and
$\mathfrak{n}_{u}=u$ (since the family $\left(  h-kh\right)  _{\left(
k,h\right)  \in K\times H}$ spans the $\CC$-vector space $J$, and since
the family $\left(  u\right)  _{u\in U}$ spans the $\CC$-vector space
$U$). As a result, we conclude that the family $\left(  \left(  h-kh\right)
\otimes_{\CC H}u\right)  _{\left(  \left(  k,h\right)  ,u\right)
\in\left(  K\times H\right)  \times U}$ spans the $\CC$-vector space
$J\otimes_{\CC H}U$.
\par
We shall now prove that $\psi\left(  \left(  \iota\otimes_{\CC %
H}\id_{U}\right)  \left(  x\right)  \right)  \in
I_{U,K}$ for every $x\in J\otimes_{\CC H}U$. Indeed, let $x\in
J\otimes_{\CC H}U$. We need to prove the relation $\psi\left(  \left(
\iota\otimes_{\CC H}\id_{U}\right)  \left(
x\right)  \right)  \in I_{U,K}$. This relation is $\CC$-linear in $x$
(since $I_{U,K}$ is a $\CC$-vector space). Hence, we can WLOG assume
that $x$ belongs to the family $\left(  \left(  h-kh\right)  \otimes
_{\CC H}u\right)  _{\left(  \left(  k,h\right)  ,u\right)  \in\left(
K\times H\right)  \times U}$ (since this family $\left(  \left(  h-kh\right)
\otimes_{\CC H}u\right)  _{\left(  \left(  k,h\right)  ,u\right)
\in\left(  K\times H\right)  \times U}$ spans the $\CC$-vector space
$J\otimes_{\CC H}U$). Assume this. Thus, $x$ belongs to the family
$\left(  \left(  h-kh\right)  \otimes_{\CC H}u\right)  _{\left(  \left(
k,h\right)  ,u\right)  \in\left(  K\times H\right)  \times U}$. In other
words, there exists an $\left(  \left(  k,h\right)  ,u\right)  \in\left(
K\times H\right)  \times U$ such that $x=\left(  h-kh\right)  \otimes
_{\CC H}u$. Consider this $\left(  \left(  k,h\right)  ,u\right)  $.
Now, applying the map $\iota\otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}$ to both sides of the equality $x=\left(  h-kh\right)
\otimes_{\CC H}u$, we obtain%
\begin{align*}
\left(  \iota\otimes_{\CC H}\id_{U}\right)
\left(  x\right)   &  =\left(  \iota\otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  \left(  h-kh\right)  \otimes_{\CC %
H}u\right) \\
&  =\underbrace{\iota\left(  h-kh\right)  }_{\substack{=h-kh\\\text{(since
}\iota\text{ is the inclusion map)}}}\otimes_{\CC H}%
\underbrace{\id_{U}\left(  u\right)  }_{=u}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }\iota
\otimes_{\CC H}\id_{U}\right) \\
&  =\left(  h-kh\right)  \otimes_{\CC H}u.
\end{align*}
Applying the map $\psi$ to both sides of this identity, we obtain%
\begin{align*}
\psi\left(  \left(  \iota\otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  x\right)  \right)   &  =\psi\left(  \left(
h-kh\right)  \otimes_{\CC H}u\right)  =\left(  h-kh\right)  u\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }\psi\right)
\\
&  =hu-khu\in I_{U,K}%
\end{align*}
(by (\ref{sol.Indrho.f.sol2.IUK1}), applied to $v=hu$). Thus, $\psi\left(
\left(  \iota\otimes_{\CC H}\id_{U}\right)
\left(  x\right)  \right)  \in I_{U,K}$ is proven.
\par
Now, let us forget that we fixed $x$. We thus have shown that $\psi\left(
\left(  \iota\otimes_{\CC H}\id_{U}\right)
\left(  x\right)  \right)  \in I_{U,K}$ for every $x\in J\otimes_{\CC %
H}U$. In other words, $\psi\left(  \left(  \iota\otimes_{\CC %
H}\id_{U}\right)  \left(  J\otimes_{\CC %
H}U\right)  \right)  \subset I_{U,K}$. This proves
(\ref{sol.Indrho.f.sol2.Ker6}).} Hence, (\ref{sol.Indrho.f.sol2.Ker4}) becomes%
\[
\ker\zeta=\psi\left(  \left(  \iota\otimes_{\CC %
H}\id_{U}\right)  \left(  J\otimes_{\CC %
H}U\right)  \right)  \subset I_{U,K}.
\]
Combined with $I_{U,K}\subset\ker\zeta$%
\ \ \ \ \footnote{\textit{Proof.} Recall that $I_{U,K}$ is the $\CC %
$-vector subspace of $U$ spanned by elements of the form $v-kv$ for $k\in K$
and $v\in U$. In other words, $I_{U,K}$ is the $\CC$-vector subspace of
$U$ spanned by elements of the form $v-kv$ for $\left(  k,v\right)  \in
K\times U$. In other words, the family $\left(  v-kv\right)  _{\left(
k,v\right)  \in K\times U}$ spans the $\CC$-vector space $I_{U,K}$.
\par
We shall now show that every $x\in I_{U,K}$ satisfies $x\in\ker%
\zeta$.
\par
Indeed, let $x\in I_{U,K}$. We need to prove the relation $x\in
\ker\zeta$. This relation is $\CC$-linear in $x$ (since
$\ker\zeta$ is a $\CC$-vector space). Hence, we can WLOG
assume that $x$ belongs to the family $\left(  v-kv\right)  _{\left(
k,v\right)  \in K\times U}$ (since this family $\left(  v-kv\right)  _{\left(
k,v\right)  \in K\times U}$ spans the $\CC$-vector space $I_{U,K}$).
Assume this.
\par
There exists a $\left(  k,v\right)  \in K\times U$ such that $x=v-kv$ (since
$x$ belongs to the family $\left(  v-kv\right)  _{\left(  k,v\right)  \in
K\times U}$). Consider this $\left(  k,v\right)  $.
\par
Recall that the right $\CC H$-module structure on $\CC G$ is given
by the equality $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying the latter equality to $\gamma=1_{\CC G}$ and $\eta=k$, we
obtain $1_{\CC G}k=1_{\CC G}\cdot\left(  \CC \left[
\rho\right]  \right)  k$. But $k\in K$ and thus $\left(  \CC \left[
\rho\right]  \right)  k=\rho\left(  k\right)  =1_{G}$ (since $k\in
K=\ker\rho$). Hence, $1_{\CC G}k=1_{\CC G}%
\cdot\underbrace{\left(  \CC \left[  \rho\right]  \right)  k}_{=1_{G}%
}=1_{\CC G}\cdot1_{G}=1_{G}=1_{\CC G}$.
\par
But $k\in K\subset H\subset\CC H$. Thus, the factor $k$ can be moved
past the tensor sign in the expression $1_{\CC G}\otimes_{\CC %
H}kv\in\CC G\otimes_{\CC H}U$. We thus obtain $1_{\CC %
G}\otimes_{\CC H}kv=\underbrace{1_{\CC G}k}_{=1_{\CC G}%
}\otimes_{\CC H}v=1_{\CC G}\otimes_{\CC H}v$.
\par
Applying the map $\zeta$ to both sides of the equality $x=v-kv$, we obtain%
\begin{align*}
\zeta\left(  x\right)   &  =\zeta\left(  v-kv\right)  =\underbrace{1}%
_{=1_{\CC G}}\otimes_{\CC H}\left(  v-kv\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\zeta\right) \\
&  =1_{\CC G}\otimes_{\CC H}\left(  v-kv\right)  =1_{\CC %
G}\otimes_{\CC H}v-\underbrace{1_{\CC G}\otimes_{\CC H}%
kv}_{=1_{\CC G}\otimes_{\CC H}v}=1_{\CC G}\otimes
_{\CC H}v-1_{\CC G}\otimes_{\CC H}v=0.
\end{align*}
Hence, $x\in\ker\zeta$.
\par
Now, let us forget that we fixed $x$. We thus have shown that $x\in
\ker\zeta$ for every $x\in I_{U,K}$. In other words,
$I_{U,K}\subset\ker\zeta$, qed.}, this yields
\begin{equation}
\ker\zeta=I_{U,K}. \label{sol.Indrho.f.sol2.Ker9}%
\end{equation}


Also,%
\begin{equation}
\zeta\left(  U\right)  =\Ind_{\rho}U
\label{sol.Indrho.f.sol2.Im1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol2.Im1}):} We know that the
map $\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}$ is surjective. Thus, $\CC G\otimes_{\CC H}U=\left(
\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}\right)  \left(  \CC H\otimes_{\CC H}%
\id_{U}\right)  $ (since the codomain of this map
$\CC \left[  \rho\right]  \otimes_{\CC H}\operatorname*{id}%
\nolimits_{U}$ is $\CC G\otimes_{\CC H}U$). Hence,%
\begin{align*}
\Ind_{\rho}U  &  =\CC G\otimes_{\CC %
H}U=\left(  \underbrace{\CC \left[  \rho\right]  \otimes_{\CC %
H}\id_{U}}_{\substack{=\zeta\circ\psi\\\text{(by
(\ref{sol.Indrho.f.sol2.zetapsi}))}}}\right)  \left(  \CC %
H\otimes_{\CC H}\id_{U}\right) \\
&  =\left(  \zeta\circ\psi\right)  \left(  \CC H\otimes_{\CC %
H}\id_{U}\right)  =\zeta\left(  \underbrace{\psi
\left(  \CC H\otimes_{\CC H}\id_{U}%
\right)  }_{\subset U}\right)  \subset\zeta\left(  U\right)  .
\end{align*}
Combined with the obvious inclusion $\zeta\left(  U\right)  \subset
\Ind_{\rho}U$, this yields $\zeta\left(  U\right)
=\Ind_{\rho}U$. This proves
(\ref{sol.Indrho.f.sol2.Im1}).}.

Let us now introduce a notation: If $\mathfrak{M}$ is any $\CC$-vector
space, if $\mathfrak{J}$ is a $\CC$-vector subspace of $\mathfrak{M}$,
and if $\mathfrak{m}$ is an element of $\mathfrak{M}$, then we let $\left[
\mathfrak{m}\right]  _{\mathfrak{M},\mathfrak{J}}$ denote the projection of
$\mathfrak{m}\in\mathfrak{M}$ onto the quotient space $\mathfrak{M}%
/\mathfrak{J}$.

Let us now recall a basic fact from abstract algebra: If $\mathfrak{M}$ and
$\mathfrak{N}$ are two $\CC$-vector spaces, and if $\mathfrak{f}%
:\mathfrak{M}\rightarrow\mathfrak{N}$ is a $\CC$-linear map, then:

\begin{enumerate}
\item there exists a unique $\CC$-linear map $\overline{\mathfrak{f}%
}:\mathfrak{M}/\ker\mathfrak{f}\rightarrow\mathfrak{N}$ which
satisfies%
\[
\left(  \overline{\mathfrak{f}}\left(  \left[  \mathfrak{m}\right]
_{\mathfrak{M},\ker\mathfrak{f}}\right)  =\mathfrak{f}\left(
\mathfrak{m}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\mathfrak{m}%
\in\mathfrak{M}\right)  ;
\]


\item this map $\overline{\mathfrak{f}}$ is injective;

\item this map $\overline{\mathfrak{f}}$ satisfies $\overline{\mathfrak{f}%
}\left(  \mathfrak{M}/\ker\mathfrak{f}\right)  =\mathfrak{f}%
\left(  \mathfrak{M}\right)  $.
\end{enumerate}

We now apply this fact to $\mathfrak{M}=U$, $\mathfrak{N}=\operatorname*{Ind}%
\nolimits_{\rho}U$ and $\mathfrak{f}=\zeta$. As a result, we conclude that:

\begin{enumerate}
\item there exists a unique $\CC$-linear map $\overline{\mathfrak{f}%
}:U/\ker\zeta\rightarrow\Ind_{\rho}U$
which satisfies%
\[
\left(  \overline{\mathfrak{f}}\left(  \left[  \mathfrak{m}\right]
_{U,\ker\zeta}\right)  =\zeta\left(  \mathfrak{m}%
\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\mathfrak{m}\in U\right)  ;
\]


\item this map $\overline{\mathfrak{f}}$ is injective;

\item this map $\overline{\mathfrak{f}}$ satisfies $\overline{\mathfrak{f}%
}\left(  U/\ker\zeta\right)  =\zeta\left(  U\right)  $.
\end{enumerate}

Because of (\ref{sol.Indrho.f.sol2.Ker9}), this rewrites as follows:

\begin{enumerate}
\item There exists a unique $\CC$-linear map $\overline{\mathfrak{f}%
}:U/I_{U,K}\rightarrow\Ind_{\rho}U$ which satisfies%
\[
\left(  \overline{\mathfrak{f}}\left(  \left[  \mathfrak{m}\right]
_{U,I_{U,K}}\right)  =\zeta\left(  \mathfrak{m}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\mathfrak{m}\in U\right)  ;
\]


\item this map $\overline{\mathfrak{f}}$ is injective;

\item this map $\overline{\mathfrak{f}}$ satisfies $\overline{\mathfrak{f}%
}\left(  U/I_{U,K}\right)  =\zeta\left(  U\right)  $.
\end{enumerate}

Let us denote this map $\overline{\mathfrak{f}}$ by $\overline{\zeta}$. Thus,
$\overline{\zeta}$ is a $\CC$-linear map $U/I_{U,K}\rightarrow
\Ind_{\rho}U$ which satisfies%
\begin{equation}
\left(  \overline{\zeta}\left(  \left[  \mathfrak{m}\right]  _{U,I_{U,K}%
}\right)  =\zeta\left(  \mathfrak{m}\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }\mathfrak{m}\in U\right)  ; \label{sol.Indrho.f.sol2.zetabar}%
\end{equation}
and this map $\overline{\zeta}$ is injective and satisfies $\overline{\zeta
}\left(  U/I_{U,K}\right)  =\zeta\left(  U\right)  $. Hence, the map
$\overline{\zeta}$ is bijective\footnote{\textit{Proof.} We have
$\overline{\zeta}\left(  U/I_{U,K}\right)  =\zeta\left(  U\right)
=\Ind_{\rho}U$ (due to (\ref{sol.Indrho.f.sol2.Im1})).
Hence, $\overline{\zeta}$ is surjective. Since $\overline{\zeta}$ is also
injective, we can thus conclude that $\overline{\zeta}$ is bijective, qed.}.

On the other hand, $\overline{\zeta}$ is a map $U/I_{U,K}\rightarrow
\Ind_{\rho}U$, thus a map $U_{K}\rightarrow
\Ind_{\rho}U$ (since $U/I_{U,K}=U_{K}$). Hence, the
composition $\overline{\zeta}\circ\Phi:U^{K}\rightarrow\operatorname*{Ind}%
\nolimits_{\rho}U$ is well-defined. It is easy to see that $\overline{\zeta
}\circ\Phi=\zeta\mid_{U^{K}}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in
U^{K}$. The definition of $\Phi$ yields $\Phi\left(  v\right)  =\overline{v}$.
\par
But let us recall that $\overline{v}$ is the projection of $v$ onto the
quotient space $U/I_{U,K}$ (by the definition of $\overline{v}$). In other
words, $\overline{v}$ is $\left[  v\right]  _{U,I_{U,K}}$ (because $\left[
v\right]  _{U,I_{U,K}}$ is defined as the projection of $v$ onto the quotient
space $U/I_{U,K}$). Thus, $\overline{v}=\left[  v\right]  _{U,I_{U,K}}$, so
that $\Phi\left(  v\right)  =\overline{v}=\left[  v\right]  _{U,I_{U,K}}$.
Applying the map $\overline{\zeta}$ to both sides of this equality, we obtain
$\overline{\zeta}\left(  \Phi\left(  v\right)  \right)  =\overline{\zeta
}\left(  \left[  v\right]  _{U,I_{U,K}}\right)  =\zeta\left(  v\right)  $ (by
(\ref{sol.Indrho.f.sol2.zetabar}), applied to $\mathfrak{m}=v$). Hence,
$\left(  \overline{\zeta}\circ\Phi\right)  \left(  v\right)  =\overline{\zeta
}\left(  \Phi\left(  v\right)  \right)  =\zeta\left(  v\right)  =\left(
\zeta\mid_{U^{K}}\right)  \left(  v\right)  $ (since $v\in U^{K}$).
\par
Now, let us forget that we fixed $v$. We thus have shown that $\left(
\overline{\zeta}\circ\Phi\right)  \left(  v\right)  =\left(  \zeta\mid_{U^{K}%
}\right)  \left(  v\right)  $ for every $v\in U^{K}$. In other words,
$\overline{\zeta}\circ\Phi=\zeta\mid_{U^{K}}$, qed.}. Thus, $\zeta\mid_{U^{K}%
}$ is the composition of the maps $\overline{\zeta}$ and $\Phi$. Hence,
$\zeta\mid_{U^{K}}$ is the composition of two bijective maps (since both maps
$\overline{\zeta}$ and $\Phi$ are bijective), and therefore itself bijective.
Hence, $\zeta\mid_{U^{K}}$ is a $\CC$-vector space isomorphism (since
$\zeta\mid_{U^{K}}$ is $\CC$-linear) and therefore a $\CC %
G$-module isomorphism (since we already know that $\zeta\mid_{U^{K}}%
:U^{K}\rightarrow\Ind_{\rho}U$ is a $\CC %
G$-module homomorphism). Hence, we have found a $\CC G$-module
isomorphism $U^{K}\rightarrow\Ind_{\rho}U$ (namely,
$\zeta\mid_{U^{K}}$). Thus, $U^{K}\cong\operatorname{Ind}_{\rho}U$ as
${\CC }G$-modules. This solves Exercise~\ref{exe.Indrho}(f) again.

[\textit{Remark:} After the solution to Exercise~\ref{exe.Indrho}(i), we shall
give yet another (a third) solution for Exercise~\ref{exe.Indrho}(f).]

(g) Let $\alpha\in R_{\CC}\left(  H\right)  $ and $\beta\in
R_{\CC }\left(  G\right)  $. We notice that%
\begin{equation}
\text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of }G\text{
satisfy }\beta\left(  k\right)  =\beta\left(  k^{\prime}\right)
\label{sol.Indrho.g.bil.beta}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.g.bil.beta}):} We can apply
(\ref{sol.Indrho.RCK}) to $K=G$ and $f=\beta$. As a consequence, we obtain the
following equivalence:%
\begin{align*}
&  \left(  \beta\in R_{\CC }\left(  G\right)  \right) \\
\Longleftrightarrow\  &  \left(  \text{any two conjugate elements }k\text{ and
}k^{\prime}\text{ of }G\text{ satisfy }\beta\left(  k\right)  =\beta\left(
k^{\prime}\right)  \right)  .
\end{align*}
Hence,
\[
\text{any two conjugate elements }k\text{ and }k^{\prime}\text{ of }G\text{
satisfy }\beta\left(  k\right)  =\beta\left(  k^{\prime}\right)
\]
(because we know that $\beta\in R_{\CC }\left(  G\right)  $).}.

Let us first prove (\ref{eq.exe.Indrho.g.bil}). The definition of
$\left\langle \Ind_{\rho}\alpha,\beta\right\rangle
_{G}$ yields%
\begin{align}
\left\langle \Ind_{\rho}\alpha,\beta\right\rangle
_{G}  &  =\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}\underbrace{\left(
\Ind_{\rho}\alpha\right)  \left(  g\right)
}_{\substack{=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \\\text{(by the definition of }\Ind_{\rho
}\alpha\text{)}}}\beta\left(  g^{-1}\right)  =\dfrac{1}{\left\vert
G\right\vert }\sum_{g\in G}\left(  \dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \right)  \beta\left(  g^{-1}\right)
\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{g\in G}\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\k\rho\left(  h\right)  k^{-1}=g}}}_{=\sum_{\left(  h,k\right)  \in H\times
G}\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}}\alpha\left(
h\right)  \beta\left(  \underbrace{g^{-1}}_{\substack{=\left(  k\rho\left(
h\right)  k^{-1}\right)  ^{-1}\\\text{(since }g=k\rho\left(  h\right)
k^{-1}\\\text{(since }k\rho\left(  h\right)  k^{-1}=g\text{))}}}\right)
\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{\left(  h,k\right)  \in H\times G}\sum\limits_{\substack{g\in G;\\k\rho\left(
h\right)  k^{-1}=g}}\alpha\left(  h\right)  \beta\left(  \left(  k\rho\left(
h\right)  k^{-1}\right)  ^{-1}\right)  . \label{sol.Indrho.g.bil.1}%
\end{align}
However, for every $\left(  h,k\right)  \in H\times G$, we have
\begin{equation}
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)
^{-1}\right)  =\alpha\left(  h\right)  \left(  \Res_{\rho}\beta\right)
\left(  h^{-1}\right)
\label{sol.Indrho.g.bil.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.g.bil.2}):} Fix a $\left(
h,k\right)  \in H\times G$. Thus, $h\in H$ and $k\in G$. Now, there exists
only one $g\in G$ such that $k\rho\left(  h\right)  k^{-1}=g$ (namely,
$g=k\rho\left(  h\right)  k^{-1}$). Hence, the sum $\sum\limits_{\substack{g\in
G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(  h\right)  \beta\left(
\left(  k\rho\left(  h\right)  k^{-1}\right)  ^{-1}\right)  $ has only one
addend (namely, the addend for $g=k\rho\left(  h\right)  k^{-1}$). Hence, this
sum simplifies as follows:%
\begin{align}
&  \sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)
^{-1}\right) \nonumber\\
&  =\alpha\left(  h\right)  \beta\left(  \underbrace{\left(  k\rho\left(
h\right)  k^{-1}\right)  ^{-1}}_{=\left(  k^{-1}\right)  ^{-1}\left(
\rho\left(  h\right)  \right)  ^{-1}k^{-1}}\right)  =\alpha\left(  h\right)
\beta\left(  \underbrace{\left(  k^{-1}\right)  ^{-1}}_{=k}\underbrace{\left(
\rho\left(  h\right)  \right)  ^{-1}}_{\substack{=\rho\left(  h^{-1}\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}k^{-1}\right)
=\alpha\left(  h\right)  \beta\left(  k\rho\left(  h^{-1}\right)
k^{-1}\right)  . \label{sol.Indrho.g.bil.2.pf.1}%
\end{align}
\par
But $\rho\left(  h^{-1}\right)  $ and $k\rho\left(  h^{-1}\right)  k^{-1}$ are
two conjugate elements of $G$. Hence, (\ref{sol.Indrho.g.bil.beta}) (applied
to $\rho\left(  h^{-1}\right)  $ and $k\rho\left(  h^{-1}\right)  k^{-1}$
instead of $k$ and $k^{\prime}$) yields $\beta\left(  \rho\left(
h^{-1}\right)  \right)  =\beta\left(  k\rho\left(  h^{-1}\right)
k^{-1}\right)  $. Thus,%
\[
\beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)  ^{-1}\right)
=\beta\left(  \rho\left(  h^{-1}\right)  \right)  =\underbrace{\left(
\beta\circ\rho\right)  }_{\substack{=\Res_{\rho}%
\beta\\\text{(since }\Res_{\rho}\beta=\beta\circ
\rho\\\text{(by the definition of }\Res_{\rho}%
\beta\text{))}}}\left(  h^{-1}\right)  =\left(  \Res_{\rho}\beta\right)
\left(  h^{-1}\right)  .
\]
Hence, (\ref{sol.Indrho.g.bil.2.pf.1}) becomes%
\[
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \beta\left(  \left(  k\rho\left(  h\right)  k^{-1}\right)
^{-1}\right)  =\alpha\left(  h\right)  \underbrace{\beta\left(  k\rho\left(
h^{-1}\right)  k^{-1}\right)  }_{=\left(  \Res_{\rho
}\beta\right)  \left(  h^{-1}\right)  }=\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)  .
\]
This proves (\ref{sol.Indrho.g.bil.2}).}. Thus, (\ref{sol.Indrho.g.bil.1})
becomes%
\begin{align*}
\left\langle \Ind_{\rho}\alpha,\beta\right\rangle
_{G}  &  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{\left(  h,k\right)  \in H\times G}}_{=\sum_{h\in H}%
\sum_{k\in G}}\underbrace{\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \beta\left(  \left(  k\rho\left(  h\right)
k^{-1}\right)  ^{-1}\right)  }_{=\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)  }\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\underbrace{\sum_{k\in G}\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)
}_{=\left\vert G\right\vert \cdot\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right)
}=\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\left\vert G\right\vert \cdot\alpha\left(  h\right)  \left(
\Res_{\rho}\beta\right)  \left(  h^{-1}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\left(  \Res_{\rho}\beta\right)  \left(
h^{-1}\right)  .
\end{align*}
Compared with%
\begin{align*}
\left\langle \alpha,\Res_{\rho}\beta\right\rangle
_{H}  &  =\dfrac{1}{\left\vert H\right\vert }\sum_{g\in H}\alpha\left(
g\right)  \left(  \Res_{\rho}\beta\right)  \left(
g^{-1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\left\langle \alpha,\Res_{\rho}\beta\right\rangle
_{H}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\left(  \Res_{\rho}\beta\right)  \left(  h^{-1}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}g\text{ as }h\right)  ,
\end{align*}
this yields $\left\langle \Ind_{\rho}\alpha
,\beta\right\rangle _{G}=\left\langle \alpha,\Res_{\rho}\beta
\right\rangle _{H}$. Thus, (\ref{eq.exe.Indrho.g.bil}) is proven.

Let us now prove (\ref{eq.exe.Indrho.g.herm}). (This proof will be very much
similar to the proof of (\ref{eq.exe.Indrho.g.bil}) above, but in a few places
even easier.) The definition of $\left(  \Ind_{\rho
}\alpha,\beta\right)  _{G}$ yields%
\begin{align}
\left(  \Ind_{\rho}\alpha,\beta\right)  _{G}  &
=\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}\underbrace{\left(
\Ind_{\rho}\alpha\right)  \left(  g\right)
}_{\substack{=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \\\text{(by the definition of }\Ind_{\rho
}\alpha\text{)}}}\overline{\beta\left(  g\right)  }=\dfrac{1}{\left\vert
G\right\vert }\sum_{g\in G}\left(  \dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \right)  \overline{\beta\left(  g\right)
}\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{g\in G}\sum\limits_{\substack{\left(  h,k\right)  \in H\times
G;\\k\rho\left(  h\right)  k^{-1}=g}}}_{=\sum_{\left(  h,k\right)  \in H\times
G}\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}}\alpha\left(
h\right)  \overline{\beta\left(  \underbrace{g}_{\substack{=k\rho\left(
h\right)  k^{-1}\\\text{(since }k\rho\left(  h\right)  k^{-1}=g\text{)}%
}}\right)  }\nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{\left(  h,k\right)  \in H\times G}\sum\limits_{\substack{g\in G;\\k\rho\left(
h\right)  k^{-1}=g}}\alpha\left(  h\right)  \overline{\beta\left(
k\rho\left(  h\right)  k^{-1}\right)  }. \label{sol.Indrho.g.herm.1}%
\end{align}
However, for every $\left(  h,k\right)  \in H\times G$, we have
\begin{equation}
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \overline{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)
}=\alpha\left(  h\right)  \overline{\left(  \Res_{\rho
}\beta\right)  \left(  h\right)  } \label{sol.Indrho.g.herm.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.g.herm.2}):} Fix a $\left(
h,k\right)  \in H\times G$. Thus, $h\in H$ and $k\in G$. Now, there exists
only one $g\in G$ such that $k\rho\left(  h\right)  k^{-1}=g$ (namely,
$g=k\rho\left(  h\right)  k^{-1}$). Hence, the sum $\sum\limits_{\substack{g\in
G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(  h\right)  \overline
{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)  }$ has only one addend
(namely, the addend for $g=k\rho\left(  h\right)  k^{-1}$). Hence, this sum
simplifies as follows:%
\begin{equation}
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \overline{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)
}=\alpha\left(  h\right)  \overline{\beta\left(  k\rho\left(  h\right)
k^{-1}\right)  }. \label{sol.Indrho.g.herm.2.pf.1}%
\end{equation}
\par
But $\rho\left(  h\right)  $ and $k\rho\left(  h\right)  k^{-1}$ are two
conjugate elements of $G$. Hence, (\ref{sol.Indrho.g.bil.beta}) (applied to
$\rho\left(  h\right)  $ and $k\rho\left(  h\right)  k^{-1}$ instead of $k$
and $k^{\prime}$) yields $\beta\left(  \rho\left(  h\right)  \right)
=\beta\left(  k\rho\left(  h\right)  k^{-1}\right)  $. Thus,%
\[
\beta\left(  k\rho\left(  h\right)  k^{-1}\right)  =\beta\left(  \rho\left(
h\right)  \right)  =\underbrace{\left(  \beta\circ\rho\right)  }%
_{\substack{=\Res_{\rho}\beta\\\text{(since
}\Res_{\rho}\beta=\beta\circ\rho\\\text{(by the
definition of }\Res_{\rho}\beta\text{))}}}\left(
h\right)  =\left(  \Res_{\rho}\beta\right)  \left(
h\right)  .
\]
Hence, (\ref{sol.Indrho.g.herm.2.pf.1}) becomes%
\[
\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)  k^{-1}=g}}\alpha\left(
h\right)  \overline{\beta\left(  k\rho\left(  h\right)  k^{-1}\right)
}=\alpha\left(  h\right)  \underbrace{\overline{\beta\left(  k\rho\left(
h\right)  k^{-1}\right)  }}_{\substack{=\overline{\left(  \Res_{\rho}
\beta\right)  \left(  h\right)  }\\\text{(since }\beta\left(
k\rho\left(  h\right)  k^{-1}\right)  =\left(  \Res_{\rho}\beta\right)
\left(  h\right)  \text{)}}}=\alpha\left(
h\right)  \overline{\left(  \Res_{\rho}\beta\right)
\left(  h\right)  }.
\]
This proves (\ref{sol.Indrho.g.herm.2}).}. Thus, (\ref{sol.Indrho.g.herm.1})
becomes%
\begin{align*}
\left(  \Ind_{\rho}\alpha,\beta\right)  _{G}  &
=\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{\left(  h,k\right)  \in H\times G}}_{=\sum_{h\in H}%
\sum_{k\in G}}\underbrace{\sum\limits_{\substack{g\in G;\\k\rho\left(  h\right)
k^{-1}=g}}\alpha\left(  h\right)  \overline{\beta\left(  k\rho\left(
h\right)  k^{-1}\right)  }}_{=\alpha\left(  h\right)  \overline{\left(
\Res_{\rho}\beta\right)  \left(  h\right)  }}\\
&  =\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\underbrace{\sum_{k\in G}\alpha\left(  h\right)  \overline
{\left(  \Res_{\rho}\beta\right)  \left(  h\right)  }%
}_{=\left\vert G\right\vert \cdot\alpha\left(  h\right)  \overline{\left(
\Res_{\rho}\beta\right)  \left(  h\right)  }}%
=\dfrac{1}{\left\vert G\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{h\in H}\left\vert G\right\vert \cdot\alpha\left(  h\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
h\right)  }\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
h\right)  }.
\end{align*}
Compared with%
\begin{align*}
\left(  \alpha,\Res_{\rho}\beta\right)  _{H}  &
=\dfrac{1}{\left\vert H\right\vert }\sum_{g\in H}\alpha\left(  g\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
g\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(
\alpha,\Res_{\rho}\beta\right)  _{H}\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{h\in H}\alpha\left(  h\right)
\overline{\left(  \Res_{\rho}\beta\right)  \left(
h\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}g\text{ as }h\right)  ,
\end{align*}
this yields $\left(  \Ind_{\rho}\alpha,\beta\right)
_{G}=\left(  \alpha,\Res_{\rho}\beta\right)  _{H}$.
Thus, (\ref{eq.exe.Indrho.g.herm}) is proven. The solution of Exercise
\ref{exe.Indrho}(g) is thus complete.

(h) Let $U$ be a $\CC H$-module, and let $V$ be a $\CC G$-module.

Recall that $\Ind_{\rho}U$ is defined as the
$\CC G$-module $\CC G\otimes_{\CC H}U$, where $\CC G$
is regarded as a $\left(  \CC G,\CC H\right)  $-bimodule according
to the following rule: The left $\CC G$-module structure on
$\CC G$ is plain multiplication inside $\CC G$; the right
$\CC H$-module structure on $\CC G$ is induced by the $\CC %
$-algebra homomorphism $\CC \left[  \rho\right]  :\CC %
H\rightarrow\CC G$ (thus, it is explicitly given by $\gamma\eta
=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)  \eta$ for all
$\gamma\in\CC G$ and $\eta\in\CC H$). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC G,\CC H\right)
$-bimodule structure.

Now, (\ref{Hom-tensor-adj}) (applied to $R=\CC H$, $S=\CC G$,
$A=\CC G$, $B=U$ and $C=V$) yields
\begin{equation}
\Hom_{\CC G}\left(  \CC G\otimes
_{\CC H}U,V\right)  \cong\Hom_{\CC %
H}\left(  U,\Hom_{\CC G}\left(  \CC %
G,V\right)  \right)  . \label{sol.Indrho.h.adj}%
\end{equation}


We shall now prove that $\Hom_{\CC G}\left(
\CC G,V\right)  \cong\Res_{\rho}V$ as left
$\CC H$-modules.

Indeed, a fundamental fact in abstract algebra says the following: If
$\mathfrak{A}$ is a $\CC$-algebra, and if $\mathfrak{M}$ is a left
$\mathfrak{A}$-module, then there exists a $\CC$-vector space
isomorphism $\Xi:\Hom_{\mathfrak{A}}\left(
\mathfrak{A},\mathfrak{M}\right)  \rightarrow\mathfrak{M}$ which satisfies
$\left(  \Xi\left(  f\right)  =f\left(  1\right)  \text{ for every }%
f\in\Hom_{\mathfrak{A}}\left(  \mathfrak{A}%
,\mathfrak{M}\right)  \right)  $. Applying this fact to $\mathfrak{A}%
=\CC G$ and $\mathfrak{M}=V$, we conclude that there exists a
$\CC$-vector space isomorphism
$\Xi:\Hom_{\CC G}\left(  \CC G,V\right)  \rightarrow V$ which
satisfies \newline
$\left(  \Xi\left(  f\right)  =f\left(  1\right)  \text{ for every
}f\in\Hom_{\CC G}\left(  \CC G,V\right)
\right)  $. Consider this $\Xi$.

The map $\Xi$ is a homomorphism of $H$-sets from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
\ \ \ \ \footnote{\textit{Proof.} The map $\Xi$ is clearly a
map from $\Hom_{\CC G}\left(  \CC G,V\right)  $ to $V$,
therefore a map from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
(since $\Res_{\rho}V=V$ as sets).
\par
Recall that the $\CC H$-module structure on $\Res_{\rho}V$ is
given by
\begin{equation}
h\cdot v=\rho\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \text{for every
}h\in H\text{ and }v\in V. \label{sol.Indrho.h.Res}%
\end{equation}
\par
Now, let $f\in\Hom_{\CC G}\left(  \CC %
G,V\right)  $ and $h\in H$. We are going to prove that $\Xi\left(  h\cdot
f\right)  =h\cdot\Xi\left(  f\right)  $, where $h\cdot\Xi\left(  f\right)  $
is computed in the $\CC H$-module $\Res_{\rho}%
V$.
\par
Indeed, the definition of the left $\CC H$-module structure on
$\Hom_{\CC G}\left(  \CC G,V\right)  $
yields $\left(  \eta\cdot\alpha\right)  \left(  p\right)  =\alpha\left(
p\eta\right)  $ for every $\eta\in\CC H$,
$\alpha\in\Hom_{\CC G}\left(  \CC G,V\right)  $ and $p\in\CC G$.
Applying this to $\eta=h$, $\alpha=f$ and $p=1$, we obtain $\left(  h\cdot
f\right)  \left(  1\right)  =f\left(  1h\right)  $ (since $h\in H\subset
\CC H$). But the definition of $\Xi$ yields $\Xi\left(  h\cdot f\right)
=\left(  h\cdot f\right)  \left(  1\right)  =f\left(  1h\right)  $.
\par
Recall that the right $\CC H$-module structure on $\CC G$ is given
by the equality $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying this equality to $\gamma=1$ and $\eta=h$, we obtain $1h=1\cdot\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  $ (since $h\in
H\subset\CC H$). But $h\in H$ and thus $\left(  \CC \left[
\rho\right]  \right)  \left(  h\right)  =\rho\left(  h\right)  $. Thus,
$1h=1\cdot\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
=\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
=\rho\left(  h\right)  =\rho\left(  h\right)  \cdot1$. Now, $\Xi\left(  h\cdot
f\right)  =f\left(  \underbrace{1h}_{=\rho\left(  h\right)  \cdot1}\right)
=f\left(  \rho\left(  h\right)  \cdot1\right)  =\rho\left(  h\right)  \cdot
f\left(  1\right)  $ (since $f$ is a $\CC G$-module homomorphism
(because $f\in\Hom_{\CC G}\left(  \CC %
G,V\right)  $) and since $\rho\left(  h\right)  \in G\subset\CC G$).
\par
On the other hand, (\ref{sol.Indrho.h.Res}) (applied to $v=f\left(  1\right)
$) yields $h\cdot f\left(  1\right)  =\rho\left(  h\right)  \cdot f\left(
1\right)  $. Compared with $\Xi\left(  h\cdot f\right)  =\rho\left(  h\right)
\cdot f\left(  1\right)  $, this yields $\Xi\left(  h\cdot f\right)  =h\cdot
f\left(  1\right)  $.
\par
But the definition of $\Xi$ yields $\Xi\left(  f\right)  =f\left(  1\right)
$. Thus, $f\left(  1\right)  =\Xi\left(  f\right)  $, so that $\Xi\left(
h\cdot f\right)  =h\cdot\underbrace{f\left(  1\right)  }_{=\Xi\left(
f\right)  }=h\cdot\Xi\left(  f\right)  $.
\par
Now, let us forget that we fixed $f$ and $h$. We thus have shown that
$\Xi\left(  h\cdot f\right)  =h\cdot\Xi\left(  f\right)  $ for every
$f\in\Hom_{\CC G}\left(  \CC G,V\right)  $
and $h\in H$. In other words, $\Xi$ is a homomorphism of $H$-sets from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to
$\Res_{\rho}V$, qed.}. Therefore, $\Xi$ is a
$\CC H$-module homomorphism from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
(since $\Xi$ is a $\CC$-linear map). Consequently,
$\Xi$ is a $\CC H$-module isomorphism from
$\Hom_{\CC G}\left(  \CC G,V\right)  $ to $\Res_{\rho}V$
(since $\Xi$ is a $\CC$-vector space isomorphism).
Therefore,
$\Hom_{\CC G}\left(  \CC G,V\right)  \cong\Res_{\rho}V$ as
$\CC H$-modules. Now,
\begin{align*}
\Hom_{\CC G}\left(
\underbrace{\Ind_{\rho}U}_{=\CC G\otimes
_{\CC H}U},V\right)   &  =\Hom_{\CC %
G}\left(  \CC G\otimes_{\CC H}U,V\right) \\
&  \cong\Hom_{\CC H}\left(
U,\underbrace{\Hom_{\CC G}\left(  \CC %
G,V\right)  }_{\cong\Res_{\rho}V\text{ as }%
\CC H\text{-modules}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.h.adj})}\right) \\
&  \cong\Hom_{\CC H}\left(
U,\Res_{\rho}V\right)  .
\end{align*}
This solves Exercise \ref{exe.Indrho}(h).

(i) The following solution will mostly be an imitation of the solution of
Exercise \ref{exe.induction.ind-as-hom}.

Let $U$ be any $\CC H$-module. Recall that $\operatorname*{Ind}%
\nolimits_{\rho}U$ is defined as the $\CC G$-module $\CC %
G\otimes_{\CC H}U$, where $\CC G$ is regarded as a $\left(
\CC G,\CC H\right)  $-bimodule according to the following rule:
The left $\CC G$-module structure on $\CC G$ is plain
multiplication inside $\CC G$; the right $\CC H$-module structure
on $\CC G$ is induced by the $\CC$-algebra homomorphism
$\CC \left[  \rho\right]  :\CC H\rightarrow\CC G$ (thus, it
is explicitly given by $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$). From now on, we regard $\CC G$ as endowed with this
$\left(  \CC G,\CC H\right)  $-bimodule structure (besides the
$\left(  \CC H,\CC G\right)  $-bimodule structure that was
introduced in the statement of Exercise \ref{exe.Indrho}(i)).

The $\CC$-vector space $\CC G$ is thus endowed with a left
$\CC H$-module structure (which is part of the $\left(  \CC %
G,\CC H\right)  $-bimodule structure) and with a right $\CC %
H$-module structure (which is part of the $\left(  \CC H,\CC %
G\right)  $-bimodule structure). These two structures, combined, form a
$\left(  \CC H,\CC H\right)  $-bimodule
structure\footnote{\textit{Proof.} Let $x\in\CC H$, $y\in\CC G$
and $z\in\CC H$. We are going to show that $\left(  xy\right)
z=x\left(  yz\right)  $.
\par
Recall that the right $\CC H$-module structure on $\CC G$ is given
by the equality $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying this equality to $\gamma=xy$ and $\eta=z$, we obtain $\left(
xy\right)  z=xy\cdot\left(  \CC \left[  \rho\right]  \right)  \left(
z\right)  $.
\par
Recall that the left $\CC H$-module structure on $\CC G$ is
induced by the $\CC$-algebra homomorphism $\CC \left[
\rho\right]  :\CC H\rightarrow\CC G$. In other words, this
structure is given by the equality $\eta\gamma=\left(  \CC \left[
\rho\right]  \right)  \eta\cdot\gamma$ for all $\gamma\in\CC G$ and
$\eta\in\CC H$. Applying this equality to $\gamma=y$ and $\eta=x$, we
obtain $xy=\left(  \CC \left[  \rho\right]  \right)  \left(  x\right)
\cdot y$.
\par
Now,
\[
\left(  xy\right)  z=\underbrace{xy}_{=\left(  \CC \left[  \rho\right]
\right)  \left(  x\right)  \cdot y}\cdot\left(  \CC \left[  \rho\right]
\right)  \left(  z\right)  =\left(  \CC \left[  \rho\right]  \right)
\left(  x\right)  \cdot y\cdot\left(  \CC \left[  \rho\right]  \right)
\left(  z\right)
\]
\par
Recall again the equality $\gamma\eta=\gamma\cdot\left(  \CC \left[
\rho\right]  \right)  \eta$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$. Applying this equality to $\gamma=y$ and $\eta=z$, we obtain
$yz=y\cdot\left(  \CC \left[  \rho\right]  \right)  \left(  z\right)  $.
\par
Recall again the equality $\eta\gamma=\left(  \CC \left[  \rho\right]
\right)  \eta\cdot\gamma$ for all $\gamma\in\CC G$ and $\eta
\in\CC H$. Applying this equality to $\gamma=yz$ and $\eta=x$, we obtain%
\[
x\left(  yz\right)  =\left(  \CC \left[  \rho\right]  \right)  \left(
x\right)  \cdot\underbrace{yz}_{=y\cdot\left(  \CC \left[  \rho\right]
\right)  \left(  z\right)  }=\left(  \CC \left[  \rho\right]  \right)
\left(  x\right)  \cdot y\cdot\left(  \CC \left[  \rho\right]  \right)
\left(  z\right)  .
\]
Comparing with $\left(  xy\right)  z=\left(  \CC \left[  \rho\right]
\right)  \left(  x\right)  \cdot y\cdot\left(  \CC \left[  \rho\right]
\right)  \left(  z\right)  $, we obtain $\left(  xy\right)  z=x\left(
yz\right)  $.
\par
Now, let us forget that we fixed $x$, $y$ and $z$. We thus have proven that
$\left(  xy\right)  z=x\left(  yz\right)  $ for every $x\in\CC H$,
$y\in\CC G$ and $z\in\CC H$. Hence, the left $\CC H$-module
structure on $\CC G$ and the right $\CC H$-module structure on
$\CC G$, combined, form a $\left(  \CC H,\CC H\right)
$-bimodule structure (since these two structures both have the same underlying
vector space $\CC$). Qed.}. This allows us to write expressions like
$xyz$ with $x\in\CC H$, $y\in\CC G$ and $z\in\CC H$, without
having to disambiguate whether they mean $\left(  xy\right)  z$ or $x\left(
yz\right)  $.

We recall that $\Hom_{\CC H}\left(
\CC G,U\right)  $ is the left $\CC G$-module consisting of all
left $\CC H$-module homomorphisms from $\CC G$ to $U$. This uses
only the $\left(  \CC H,\CC G\right)  $-bimodule structure on
$\CC G$ that was introduced in the statement of Exercise
\ref{exe.Indrho}(i) (but not the $\left(  \CC G,\CC H\right)
$-bimodule structure on $\CC G$ that was introduced in the definition of
$\Ind_{\rho}U$).

Recall that the right $\CC H$-module structure on $\CC G$ is given
by%
\begin{equation}
\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]  \right)
\eta\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\CC G\text{ and }\eta
\in\CC H. \label{sol.Indrho.i.rightCH}%
\end{equation}
For every $\gamma\in\CC G$ and $\eta\in H$, we have%
\begin{align}
\gamma\eta &  =\gamma\cdot\underbrace{\left(  \CC \left[  \rho\right]
\right)  \eta}_{\substack{=\rho\left(  \eta\right)  \\\text{(since }\eta\in
H\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.i.rightCH})
(since }\eta\in H\subset\CC H\text{)}\right) \nonumber\\
&  =\gamma\cdot\rho\left(  \eta\right)  . \label{sol.Indrho.i.rightCH2}%
\end{align}
On the other hand, the left $\CC H$-module structure on $\CC G$ is
induced by the $\CC$-algebra homomorphism $\CC \left[
\rho\right]  :\CC H\rightarrow\CC G$. In other words, this
structure is given by
\begin{equation}
\eta\gamma=\left(  \CC \left[  \rho\right]  \right)  \eta\cdot
\gamma\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\CC G\text{ and }%
\eta\in\CC H. \label{sol.Indrho.i.leftCH}%
\end{equation}
For every $\gamma\in\CC G$ and $\eta\in H$, we have%
\begin{align}
\eta\gamma &  =\underbrace{\left(  \CC \left[  \rho\right]  \right)
\eta}_{\substack{=\rho\left(  \eta\right)  \\\text{(since }\eta\in H\text{)}%
}}\cdot\gamma\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.i.leftCH})
(since }\eta\in H\subset\CC H\text{)}\right) \nonumber\\
&  =\rho\left(  \eta\right)  \cdot\gamma. \label{sol.Indrho.i.leftCH2}%
\end{align}


The map $\rho$ is a group homomorphism. Hence, $\rho\left(  H\right)  $ is a
subgroup of $G$. Let us denote this subgroup by $\overline{H}$. Thus,
$\overline{H}=\rho\left(  H\right)  $.

Let $J$ be a system of distinct representatives for the right $\overline{H}%
$-cosets in $G$. Then, $G=\bigsqcup_{j\in J}\overline{H}j$.

For every $g\in G$, define a map $\mathfrak{R}_{g}:J\rightarrow J$ as follows:
Let $i\in J$. Then, $ig\in G=\bigsqcup_{j\in J}\overline{H}j$. Thus, there
exists a unique $j\in J$ such that $ig\in\overline{H}j$. Define $\mathfrak{R}%
_{g}\left(  i\right)  $ to be this $j$. Hence, we have defined $\mathfrak{R}%
_{g}\left(  i\right)  $ for every $i\in J$. Thus, we have defined a map
$\mathfrak{R}_{g}:J\rightarrow J$.

For every $g\in G$ and $i\in J$, we have
\begin{equation}
ig\in\overline{H}\cdot\mathfrak{R}_{g}\left(  i\right)
\label{sol.Indrho.i.Rg.def}%
\end{equation}
(because $\mathfrak{R}_{g}\left(  i\right)  $ is defined as the $j\in J$
satisfying $ig\in\overline{H}j$).

It is easy to see that for every $g\in G$, the map $\mathfrak{R}%
_{g}:J\rightarrow J$ is a bijection.\footnote{\textit{Proof.} Fix $g\in G$.
\par
The sets $\overline{H}j$ for all $j\in J$ are disjoint (since
$\bigsqcup_{j\in J}\overline{H}j$ is well-defined). In other words, if $i$ and
$i^{\prime}$ are two elements of $J$ such that $\overline{H}i$ and
$\overline{H}i^{\prime}$ are not disjoint, then
\begin{equation}
i=i^{\prime}. \label{sol.Indrho.i.Rg.ind-as-hom-Lg.1}%
\end{equation}
\par
Let us now prove that $\mathfrak{R}_{g}$ is injective.
\par
Let $i$ and $i^{\prime}$ be two elements of $J$. Assume that $\mathfrak{R}%
_{g}\left(  i\right)  =\mathfrak{R}_{g}\left(  i^{\prime}\right)  $. We will
show that $i=i^{\prime}$.
\par
From (\ref{sol.Indrho.i.Rg.def}), we have $ig\in\overline{H}\cdot
\mathfrak{R}_{g}\left(  i\right)  $. Hence, there exists some $h\in
\overline{H}$ such that $ig=h\cdot\mathfrak{R}_{g}\left(  i\right)  $.
Consider this $h$. We have
\[
\overline{H}\underbrace{ig}_{=h\cdot\mathfrak{R}_{g}\left(  i\right)
}=\underbrace{\overline{H}h}_{\substack{=\overline{H}\\\text{(since }%
h\in\overline{H}\text{ and since}\\\overline{H}\text{ is a group)}}%
}\cdot\mathfrak{R}_{g}\left(  i\right)  =\overline{H}\cdot\mathfrak{R}%
_{g}\left(  i\right)  .
\]
The same argument (but for $i^{\prime}$ instead of $i$) yields $\overline
{H}i^{\prime}g=\overline{H}\cdot\mathfrak{R}_{g}\left(  i^{\prime}\right)  $.
Hence, $\overline{H}ig=\overline{H}\cdot\underbrace{\mathfrak{R}_{g}\left(
i\right)  }_{=\mathfrak{R}_{g}\left(  i^{\prime}\right)  }=\overline{H}%
\cdot\mathfrak{R}_{g}\left(  i^{\prime}\right)  =\overline{H}i^{\prime}g$.
Thus, $\underbrace{\overline{H}ig}_{=\overline{H}i^{\prime}g}g^{-1}%
=\overline{H}i^{\prime}gg^{-1}=\overline{H}i^{\prime}$, so that $\overline
{H}i^{\prime}=\overline{H}igg^{-1}=\overline{H}i$. Thus, the sets
$\overline{H}i$ and $\overline{H}i^{\prime}$ are not disjoint (because
$\left(  \overline{H}i\right)  \cap\underbrace{\left(  \overline{H}i^{\prime
}\right)  }_{= \overline{H}i}
=\left(  \overline{H}i\right)  \cap\left(  \overline
{H}i\right)  =\overline{H}i\neq\varnothing$). Therefore, $i=i^{\prime}$ (by
(\ref{sol.Indrho.i.Rg.ind-as-hom-Lg.1})).
\par
Now, forget that we fixed $i$ and $i^{\prime}$. We thus have shown that if $i$
and $i^{\prime}$ are two elements of $J$ such that $\mathfrak{R}_{g}\left(
i\right)  =\mathfrak{R}_{g}\left(  i^{\prime}\right)  $, then $i=i^{\prime}$.
In other words, the map $\mathfrak{R}_{g}$ is injective. But the set $J$ is
finite (since it is a subset of the finite set $G$). Hence, $\mathfrak{R}_{g}$
is a map from a finite set (namely, $J$) to itself. Since we know that
$\mathfrak{R}_{g}$ is injective, this yields that $\mathfrak{R}_{g}$ is
surjective (because any injective map from a finite set to itself must be
surjective). Hence, $\mathfrak{R}_{g}$ is bijective (since $\mathfrak{R}_{g}$
is injective and surjective), that is, a bijection, qed.} We have
\begin{equation}
j^{-1}\otimes_{\CC H}f\left(  jg\right)  =g\cdot\left(  \mathfrak{R}%
_{g}\left(  j\right)  \right)  ^{-1}\otimes_{\CC H}f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{in
}{\CC }G\otimes_{\CC H}U \label{sol.Indrho.i.Rg.property}%
\end{equation}
for every $g\in G$, every $j\in J$ and every $f\in\Hom %
_{\CC H}\left(  {\CC}G,U\right)  $%
.\ \ \ \ \footnote{\textit{Proof of (\ref{sol.Indrho.i.Rg.property}):} Let
$g\in G$, $j\in J$ and $f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  $. Applying (\ref{sol.Indrho.i.Rg.def}) to $i=j$, we
obtain $jg\in\overline{H}\cdot\mathfrak{R}_{g}\left(  j\right)  $. In other
words, there exists an $h\in\overline{H}$ such that $jg=h\cdot\mathfrak{R}%
_{g}\left(  j\right)  $. Consider this $h$. We have
\[
j^{-1}\underbrace{jg}_{=h\cdot\mathfrak{R}_{g}\left(  j\right)  }\cdot\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}=j^{-1}h\cdot
\underbrace{\mathfrak{R}_{g}\left(  j\right)  \cdot\left(  \mathfrak{R}%
_{g}\left(  j\right)  \right)  ^{-1}}_{=1}=j^{-1}h,
\]
hence
\[
j^{-1}h=j^{-1}jg\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}.
\]
\par
But $h\in\overline{H}=\rho\left(  H\right)  $. Hence, there exists a
$h^{\prime}\in H$ such that $h=\rho\left(  h^{\prime}\right)  $. Consider this
$h^{\prime}$. We have $h^{\prime}\in H$ and thus $\left(  \CC \left[
\rho\right]  \right)  \left(  h^{\prime}\right)  =\rho\left(  h^{\prime
}\right)  =h$.
\par
Now, recall that the left $\CC H$-module structure on $\CC G$ is
induced by the $\CC$-algebra homomorphism $\CC \left[
\rho\right]  :\CC H\rightarrow\CC G$. In other words, this
structure is given by the equality $\eta\gamma=\left(  \CC \left[
\rho\right]  \right)  \eta\cdot\gamma$ for all $\gamma\in\CC G$ and
$\eta\in\CC H$. Applying this equality to $\eta=h^{\prime}$ and
$\gamma=\mathfrak{R}_{g}\left(  j\right)  $, we obtain $h^{\prime}%
\cdot\mathfrak{R}_{g}\left(  j\right)  =\underbrace{\left(  \CC \left[
\rho\right]  \right)  \left(  h^{\prime}\right)  }_{=h}\cdot\mathfrak{R}%
_{g}\left(  j\right)  =h\cdot\mathfrak{R}_{g}\left(  j\right)  =jg$ (since
$jg=h\cdot\mathfrak{R}_{g}\left(  j\right)  $).
\par
But the map $f$ is left $\CC H$-linear (since $f\in\Hom %
_{\CC H}\left(  {\CC}G,U\right)  $). Thus, $f\left(
h^{\prime}\cdot\mathfrak{R}_{g}\left(  j\right)  \right)  =h^{\prime}\cdot
f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  $ (since $h^{\prime}\in
H\subset\CC H$). Since $h^{\prime}\cdot\mathfrak{R}_{g}\left(  j\right)
=jg$, this rewrites as $f\left(  jg\right)  =h^{\prime}\cdot f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)  $. Hence,
\begin{equation}
j^{-1}\otimes_{\CC H}\underbrace{f\left(  jg\right)  }_{=h^{\prime
}\cdot f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  }=j^{-1}%
\otimes_{\CC H}h^{\prime}\cdot f\left(  \mathfrak{R}_{g}\left(
j\right)  \right)  =j^{-1}h^{\prime}\otimes_{\CC H}f\left(
\mathfrak{R}_{g}\left(  j\right)  \right)
\label{sol.Indrho.i.Rg.property.pf.3}%
\end{equation}
(here, we have moved $h^{\prime}$ past the $\otimes_{\CC H}$ sign,
since $h^{\prime}\in H\subset{\CC }H$).
\par
On the other hand, we recall that the right $\CC H$-module structure on
$\CC G$ is given by the equality $\gamma\eta=\gamma\cdot\left(
\CC \left[  \rho\right]  \right)  \eta$ for all $\gamma\in\CC G$
and $\eta\in\CC H$. Applying this equality to $\gamma=j^{-1}$ and
$\eta=h^{\prime}$, we obtain $j^{-1}h^{\prime}=j^{-1}\cdot\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h^{\prime}\right)  }%
_{=h}=j^{-1}h=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}$.
Hence, (\ref{sol.Indrho.i.Rg.property.pf.3}) becomes%
\[
j^{-1}\otimes_{\CC H}f\left(  jg\right)  =\underbrace{j^{-1}h^{\prime
}}_{=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}}%
\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}\otimes
_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  .
\]
This proves (\ref{sol.Indrho.i.Rg.property}).}

We now define a map $\alpha:\Hom_{\CC H}\left(
{\CC}G,U\right)  \rightarrow{\CC }G\otimes_{\CC H}U$ by
setting
\[
\alpha\left(  f\right)  =\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  \qquad\text{ for all }f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  .
\]
This $\alpha$ is a map $\Hom_{\CC H}\left(  {\CC %
}G,U\right)  \rightarrow\operatorname{Ind}_{\rho}U$ (since $\operatorname{Ind}%
_{\rho}U={\CC}G\otimes_{\CC H}U$).

We will show that $\alpha$ is a ${\CC }G$-module isomorphism.

First, let us prove that $\alpha$ is a left $\CC  G$-module
homomorphism. In fact, any $f\in\Hom_{\CC H}\left(
{\CC}G,U\right)  $ and $g\in G$ satisfy
\begin{align*}
\alpha\left(  gf\right)   &  =\sum_{j\in J}j^{-1}\otimes_{{\CC }%
H}\underbrace{\left(  gf\right)  \left(  j\right)  }_{\substack{=f\left(
jg\right)  \\\text{(by the definition of }gf\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\left(
gf\right)  \right) \\
&  =\sum_{j\in J}\underbrace{j^{-1}\otimes_{\CC H}f\left(  jg\right)
}_{\substack{=g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
^{-1}\otimes_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)
\right)  \\\text{(by (\ref{sol.Indrho.i.Rg.property}))}}}=\sum_{j\in J}%
g\cdot\left(  \mathfrak{R}_{g}\left(  j\right)  \right)  ^{-1}\otimes
_{\CC H}f\left(  \mathfrak{R}_{g}\left(  j\right)  \right)
=\sum_{j\in J}g\cdot j^{-1}\otimes_{\CC H}f\left(  j\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }j\text{ for }\mathfrak{R}_{g}\left(
j\right)  \text{ in the sum, since}\\
\text{the map }\mathfrak{R}_{g}:J\rightarrow J\text{ is a bijection}%
\end{array}
\right) \\
&  =g\cdot\underbrace{\sum_{j\in J}j^{-1}\otimes_{\CC H}f\left(
j\right)  }_{=\alpha\left(  f\right)  }=g\cdot\alpha\left(  f\right)  .
\end{align*}
The map $\alpha$ is thus a homomorphism of left $G$-sets. Since $\alpha$ is
furthermore $\CC$-linear, this yields that $\alpha$ is a left
${\CC }G$-module homomorphism.

We now are going to construct an inverse for $\alpha$. This will be more cumbersome.

For every $g\in G$ and every $p\in{\CC }G$, we denote by $\epsilon
_{g}\left(  p\right)  $ the $g$-coordinate of $p$ with respect to the basis
$G$ of the $\CC$-vector space ${\CC }G$. By the definition of
``coordinate'', we have
\begin{equation}
q=\sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\ \ \ \ \ \ \ \ \ \ \text{for every }q\in{\CC }G.
\label{sol.Indrho.i.epsilong.coordinates}%
\end{equation}
For every $g\in G$, we have defined a map $\epsilon_{g}:{\CC %
}G\rightarrow{\CC }$ (because we have defined an element $\epsilon
_{g}\left(  p\right)  $ for every $p\in{\CC }G$). This map $\epsilon
_{g}$ is $\CC$-linear.

For every $g\in G$ and $h\in G$, we know that $\epsilon_{g}\left(  h\right)  $
is the $g$-coordinate of $h$ with respect to the basis $G$ of the
$\CC$-vector space ${\CC }G$ (because this is how $\epsilon
_{g}\left(  h\right)  $ was defined). Since this $g$-coordinate of $h$ is
clearly $\delta_{g,h}$, this rewrites as follows: For every $g\in G$ and $h\in
G$, we know that $\epsilon_{g}\left(  h\right)  $ is $\delta_{g,h}$. In other
words, for every $g\in G$ and $h\in G$, we have
\begin{equation}
\epsilon_{g}\left(  h\right)  =\delta_{g,h}. \label{sol.Indrho.i.epsilong.g}%
\end{equation}


We notice that
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in{\CC }G\text{ and }%
q\in{\CC }G. \label{sol.Indrho.i.trace}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.trace}):} The equality
(\ref{sol.Indrho.i.trace}) is identical to
(\ref{sol.induction.ind-as-hom.trace}). Its proof can thus be found in the
Solution to Exercise \ref{exe.induction.ind-as-hom}.}

\begin{commentedout}
Here is a useless corollary: We have
\begin{equation}
\epsilon_{1}\left(  pq\right)  =\epsilon_{1}\left(  qp\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in{\CC }H\text{ and }%
q\in{\CC }G \label{sol.Indrho.i.traceHG}%
\end{equation}
(where the left $\CC H$-module structure on $\CC G$ is used in the
expression $pq$, and where the right $\CC H$-module structure on
$\CC G$ is used in the expression $qp$)\ \ \ \ \footnote{\textit{Proof
of (\ref{sol.Indrho.i.traceHG}):} Let $p\in{\CC }H$ and $q\in
{\CC }G$.
\par
Recall that the right $\CC H$-module structure on $\CC G$ is given
by the equality $\gamma\eta=\gamma\cdot\left(  \CC \left[  \rho\right]
\right)  \eta$ for all $\gamma\in\CC G$ and $\eta\in\CC H$.
Applying this equality to $\gamma=q$ and $\eta=p$, we obtain $qp=q\cdot\left(
\CC \left[  \rho\right]  \right)  p$. Thus, $q\cdot\left(
\CC \left[  \rho\right]  \right)  p=qp$.
\par
Recall that the left $\CC H$-module structure on $\CC G$ is
induced by the $\CC$-algebra homomorphism $\CC \left[
\rho\right]  :\CC H\rightarrow\CC G$. In other words, this
structure is given by the equality $\eta\gamma=\left(  \CC \left[
\rho\right]  \right)  \eta\cdot\gamma$ for all $\gamma\in\CC G$ and
$\eta\in\CC H$. Applying this equality to $\gamma=q$ and $\eta=p$, we
obtain $pq=\left(  \CC \left[  \rho\right]  \right)  p\cdot q$.
\par
Now,%
\begin{align*}
\epsilon_{1}\left(  \underbrace{pq}_{=\left(  \CC \left[  \rho\right]
\right)  p\cdot q}\right)   &  =\epsilon_{1}\left(  \left(  \CC \left[
\rho\right]  \right)  p\cdot q\right)  =\epsilon_{1}\left(  \underbrace{q\cdot
\left(  \CC \left[  \rho\right]  \right)  p}_{=qp}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.i.trace}), applied to
}\left(  \CC \left[  \rho\right]  \right)  p\text{ instead of }p\right)
\\
&  =\epsilon_{1}\left(  qp\right)  .
\end{align*}
This proves (\ref{sol.Indrho.i.traceHG}).}.
\end{commentedout}

Moreover,
\begin{equation}
\epsilon_{1}\left(  g^{-1}q\right)  =\epsilon_{g}\left(  q\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\text{ and }q\in{\CC }G.
\label{sol.Indrho.i.epsi1g}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.epsi1g}):} The equality
(\ref{sol.Indrho.i.epsi1g}) is identical to
(\ref{sol.induction.ind-as-hom.epsi1g}). Its proof can thus be found in the
Solution to Exercise \ref{exe.induction.ind-as-hom}.}

Now, fix $q\in{\CC }G$ and $u\in U$. We let $f_{q,u}$ be the map
${\CC }G\rightarrow U$ defined by
\begin{equation}
f_{q,u}\left(  p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(
h\right)  pq\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \text{for every }%
p\in{\CC }G. \label{sol.Indrho.i.fqu}%
\end{equation}
It is obvious that this map $f_{q,u}$ is $\CC$-linear. We will show
that $f_{q,u}$ is a left ${\CC }H$-module homomorphism.

The map $f_{q,u}$ is a homomorphism of left $H$-sets\footnote{\textit{Proof.}
In fact, for every $h^{\prime}\in H$ and every $p\in{\CC }G$, we have
\begin{align*}
f_{q,u}\left(  h^{\prime}p\right)   &  =\sum\limits_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  \underbrace{\left(  h^{\prime}p\right)
}_{\substack{=\rho\left(  h^{\prime}\right)  \cdot p\\\text{(by
(\ref{sol.Indrho.i.leftCH2}), applied}\\\text{to }\gamma=p\text{ and }%
\eta=h^{\prime}\text{)}}}q\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }f_{q,u}\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  \rho\left(
h^{\prime}\right)  \cdot pq\right)  h^{-1}u=\sum\limits_{h\in H}\epsilon
_{1}\left(  \underbrace{\rho\left(  h\left(  h^{\prime}\right)  ^{-1}\right)
}_{\substack{=\rho\left(  h\right)  \left(  \rho\left(  h^{\prime}\right)
\right)  ^{-1}\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}%
}}\rho\left(  h^{\prime}\right)  \cdot pq\right)  \underbrace{\left(  h\left(
h^{\prime}\right)  ^{-1}\right)  ^{-1}}_{=\left(  \left(  h^{\prime}\right)
^{-1}\right)  ^{-1}h^{-1}=h^{\prime}h^{-1}}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }h\left(  h^{\prime}\right)  ^{-1}\text{ for
}h\text{ in the sum,}\\
\text{because the map }H\rightarrow H,\ h\mapsto h\left(  h^{\prime}\right)
^{-1}\text{ is a bijection}\\
\text{(since }H\text{ is a group and since }h^{\prime}\in H\text{)}%
\end{array}
\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
\underbrace{\left(  \rho\left(  h^{\prime}\right)  \right)  ^{-1}\rho\left(
h^{\prime}\right)  }_{=1}\cdot pq\right)  h^{\prime}h^{-1}u\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  pq\right)
h^{\prime}h^{-1}u=h^{\prime}\cdot\underbrace{\sum\limits_{h\in H}\epsilon
_{1}\left(  \rho\left(  h\right)  pq\right)  h^{-1}u}_{\substack{=f_{q,u}%
\left(  p\right)  \\\text{(by (\ref{sol.Indrho.i.fqu}))}}}=h^{\prime}\cdot
f_{q,u}\left(  p\right)  .
\end{align*}
In other words, $f_{q,u}$ is a homomorphism of left $H$-sets, qed.}. Since
$f_{q,u}$ is furthermore $\CC$-linear, this yields that $f_{q,u}$ is
a left $\CC H$-module homomorphism. Hence, $f_{q,u}\in\Hom %
_{\CC H}\left(  {\CC }G,U\right)  $.

Now, forget that we fixed $q$ and $u$. We thus have defined a map $f_{q,u}%
\in\Hom_{\CC H}\left(  {\CC }G,U\right)  $ for
every $q\in{\CC }G$ and $u\in U$. It is easy to see that this map
$f_{q,u}$ depends $\CC$-linearly on each of $q$ and $u$. Now, define
a map $\widetilde{\beta}:{\CC }G\times U\rightarrow\Hom %
_{\CC H}\left(  {\CC }G,U\right)  $ by
\[
\widetilde{\beta}\left(  q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ \text{for
every }\left(  q,u\right)  \in{\CC }G\times U.
\]
Then, $\widetilde{\beta}$ is a $\CC$-bilinear map (because
$\widetilde{\beta}\left(  q,u\right)  =f_{q,u}$ depends ${\CC }%
$-linearly on each of $q$ and $u$). We are now going to prove that the map
$\widetilde{\beta}$ is ${\CC }H$-bilinear with respect to the right
${\CC }H$-module structure on ${\CC }G$ and the left ${\CC %
}H$-module structure on $U$.

In fact, every $h^{\prime}\in H$, $q\in{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta}\left(
qh^{\prime},u\right)  . \label{sol.Indrho.i.Rg.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.Rg.9}):} Let $h^{\prime}\in H$,
$q\in{\CC }G$ and $u\in U$. Then, the map $H\rightarrow H,\ h\mapsto
h^{\prime}h$ is a bijection (since $H$ is a group). Now, let $p\in{\CC %
}G$. The definition of $\widetilde{\beta}$ yields $\widetilde{\beta}\left(
q,h^{\prime}u\right)  =f_{q,h^{\prime}u}$. Hence,
\begin{align}
\left(  \underbrace{\widetilde{\beta}\left(  q,h^{\prime}u\right)
}_{=f_{q,h^{\prime}u}}\right)  \left(  p\right)   &  =f_{q,h^{\prime}u}\left(
p\right)  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
pq\right)  h^{-1}h^{\prime}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }f_{q,h^{\prime}u}\right) \nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{\rho\left(  h^{\prime
}h\right)  }_{\substack{=\rho\left(  h^{\prime}\right)  \rho\left(  h\right)
\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}}pq\right)
\underbrace{\left(  h^{\prime}h\right)  ^{-1}}_{=h^{-1}\left(  h^{\prime
}\right)  ^{-1}}h^{\prime}u\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }h^{\prime}h\text{ for }h\text{ in the sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{\prime}h\text{ is a
bijection}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{h\in H}\underbrace{\epsilon_{1}\left(  \rho\left(  h^{\prime
}\right)  \rho\left(  h\right)  pq\right)  }_{\substack{=\epsilon_{1}\left(
\rho\left(  h\right)  pq\rho\left(  h^{\prime}\right)  \right)  \\\text{(by
(\ref{sol.Indrho.i.trace}), applied to}\\\rho\left(  h^{\prime}\right)  \text{
and }\rho\left(  h\right)  pq\text{ instead of }p\text{ and }q\text{)}}%
}h^{-1}\underbrace{\left(  h^{\prime}\right)  ^{-1}h^{\prime}}_{=1}%
u\nonumber\\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
pq\rho\left(  h^{\prime}\right)  \right)  h^{-1}u.
\label{sol.Indrho.i.Rg.9.pf.1}%
\end{align}
But the definition of $\widetilde{\beta}$ also yields $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =f_{qh^{\prime},u}$. Hence,
\begin{align*}
\left(  \widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(
p\right)   &  =f_{qh^{\prime},u}\left(  p\right)  =\sum\limits_{h\in
H}\epsilon_{1}\left(  \rho\left(  h\right)  p\underbrace{\left(  qh^{\prime
}\right)  }_{\substack{=q\cdot\rho\left(  h^{\prime}\right)  \\\text{(by
(\ref{sol.Indrho.i.rightCH2}), applied to}\\\gamma=q\text{ and }\eta
=h^{\prime}\text{)}}}\right)  h^{-1}u\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }f_{qh^{\prime},u}\right) \\
&  =\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  pq\cdot
\rho\left(  h^{\prime}\right)  \right)  h^{-1}u=\sum\limits_{h\in H}%
\epsilon_{1}\left(  \rho\left(  h\right)  pq\rho\left(  h^{\prime}\right)
\right)  h^{-1}u=\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)
\right)  \left(  p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.i.Rg.9.pf.1})}\right)  .
\end{align*}
\par
Now, forget that we fixed $p$. We have thus proven that $\left(
\widetilde{\beta}\left(  qh^{\prime},u\right)  \right)  \left(  p\right)
=\left(  \widetilde{\beta}\left(  q,h^{\prime}u\right)  \right)  \left(
p\right)  $ for every $p\in{\CC }G$. In other words, $\widetilde{\beta
}\left(  qh^{\prime},u\right)  =\widetilde{\beta}\left(  q,h^{\prime}u\right)
$, so that $\widetilde{\beta}\left(  q,h^{\prime}u\right)  =\widetilde{\beta
}\left(  qh^{\prime},u\right)  $. This proves (\ref{sol.Indrho.i.Rg.9}).} As a
consequence of this, we can see that every $r\in{\CC }H$, $q\in
{\CC }G$ and $u\in U$ satisfy
\begin{equation}
\widetilde{\beta}\left(  q,ru\right)  =\widetilde{\beta}\left(  qr,u\right)  .
\label{sol.Indrho.i.Rg.10}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.Rg.10}):} Let $r\in{\CC %
}H$, $q\in{\CC }G$ and $u\in U$. Then, we can write $r$ in the form
$r=\sum\limits_{h\in H}r_{h}h$ for some family $\left(  r_{h}\right)  _{h\in
H}$ of elements of $\CC$. Consider this family $\left(  r_{h}\right)
_{h\in H}$. We have
\begin{align*}
\widetilde{\beta}\left(  q,\underbrace{r}_{=\sum\limits_{h\in H}r_{h}%
h}u\right)   &  =\widetilde{\beta}\left(  q,\sum\limits_{h\in H}%
r_{h}hu\right)  =\sum\limits_{h\in H}r_{h}\underbrace{\widetilde{\beta}\left(
q,hu\right)  }_{\substack{=\widetilde{\beta}\left(  qh,u\right)  \\\text{(by
(\ref{sol.Indrho.i.Rg.9}), applied to }h^{\prime}=h\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\widetilde{\beta}\text{ is
}{\CC }\text{-bilinear}\right) \\
&  =\sum\limits_{h\in H}r_{h}\widetilde{\beta}\left(  qh,u\right)
=\widetilde{\beta}\left(  q\underbrace{\sum\limits_{h\in H}r_{h}h}%
_{=r},u\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }%
\widetilde{\beta}\text{ is }{\CC }\text{-bilinear}\right) \\
&  =\widetilde{\beta}\left(  qr,u\right)  .
\end{align*}
This proves (\ref{sol.Indrho.i.Rg.10}).} In other words, the map
$\widetilde{\beta}$ is ${\CC }H$-bilinear with respect to the right
${\CC }H$-module structure on ${\CC }G$ and the left ${\CC %
}H$-module structure on $U$ (since we already know that $\widetilde{\beta}$ is
$\CC$-bilinear). Hence, by the universal property of the tensor
product, we conclude that there exists a unique $\CC$-linear map
$\beta:{\CC }G\otimes_{\CC H}U\rightarrow\Hom %
_{\CC H}\left(  {\CC }G,U\right)  $ such that every $\left(
q,u\right)  \in{\CC }G\times U$ satisfies
\begin{equation}
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  . \label{sol.Indrho.i.beta}%
\end{equation}
Consider this map $\beta$. Clearly, every $q\in{\CC }G$ and $u\in U$
satisfy
\[
\beta\left(  q\otimes_{\CC H}u\right)  =\widetilde{\beta}\left(
q,u\right)  =f_{q,u}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{\beta}\right)  .
\]
Hence, every $q\in{\CC }G$, $u\in U$ and $p\in{\CC }G$ satisfy
\begin{equation}
\left(  \underbrace{\beta\left(  q\otimes_{\CC H}u\right)  }%
_{=f_{q,u}}\right)  \left(  p\right)  =f_{q,u}\left(  p\right)  =\sum
\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)  pq\right)  h^{-1}u
\label{sol.Indrho.i.beta.explicit}%
\end{equation}
(by (\ref{sol.Indrho.i.fqu})).

Let $K=\ker\rho$. Thus, $K$ is the kernel of a group
homomorphism out of $H$ (since $\rho$ is a group homomorphism out of $H$), and
therefore a normal subgroup of $H$.

We notice that%
\begin{equation}
\left\vert \rho^{-1}\left(  g\right)  \right\vert =\left\vert K\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\overline{H}.
\label{sol.Indrho.i.preimage}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.i.preimage}):} Let $g\in
\overline{H}$. Then, $g\in\overline{H}=\rho\left(  H\right)  $. Hence,
$g=\rho\left(  x\right)  $ for some $x\in H$. Let us fix such a $x$.
\par
We know that $H$ is a group. Hence, the map $K\rightarrow xK,\ k\mapsto xk$ is
a bijection. Thus, the sets $K$ and $xK$ are in bijection. Therefore,
$\left\vert xK\right\vert =\left\vert K\right\vert $.
\par
However, for every $y\in H$, we have the following logical equivalence:
\begin{align*}
\left(  y\in\rho^{-1}\left(  g\right)  \right)  \  &  \Longleftrightarrow
\ \left(  \rho\left(  y\right)  =\underbrace{g}_{=\rho\left(  x\right)
}\right)  \ \Longleftrightarrow\ \left(  \rho\left(  y\right)  =\rho\left(
x\right)  \right)  \ \Longleftrightarrow\ \left(  \underbrace{\rho\left(
y\right)  \cdot\left(  \rho\left(  x\right)  \right)  ^{-1}}_{\substack{=\rho
\left(  yx^{-1}\right)  \\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}=1\right) \\
&  \Longleftrightarrow\ \left(  \rho\left(  yx^{-1}\right)  =1\right)
\ \Longleftrightarrow\ \left(  yx^{-1}\in\underbrace{\ker\rho
}_{=K}\right)  \ \Longleftrightarrow\ \left(  yx^{-1}\in K\right)
\ \Longleftrightarrow\ \left(  y\in xK\right)  .
\end{align*}
Hence, $\rho^{-1}\left(  g\right)  =xK$, so that $\left\vert \rho^{-1}\left(
g\right)  \right\vert =\left\vert xK\right\vert =\left\vert K\right\vert $.
This proves (\ref{sol.Indrho.i.preimage}).}

We shall now show that the maps $\alpha$ and $\dfrac{1}{\left\vert
K\right\vert }\beta$ are mutually inverse. To do so, we will show that
$\alpha\circ\beta=\left\vert K\right\vert \operatorname{id}$ and $\beta
\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

Let us first show that $\alpha\circ\beta=\operatorname{id}$. In fact, let
$q\in{\CC }G$ and $u\in U$ be arbitrary. Then,
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)   &
=\alpha\left(  \beta\left(  q\otimes_{\CC H}u\right)  \right)
=\sum_{j\in J}j^{-1}\otimes_{\CC H}\underbrace{\left(  \beta\left(
q\otimes_{\CC H}u\right)  \right)  \left(  j\right)  }%
_{\substack{=\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
jq\right)  h^{-1}u\\\text{(by (\ref{sol.Indrho.i.beta.explicit}), applied to
}p=j\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\right) \\
&  =\sum_{j\in J}j^{-1}\otimes_{\CC H}\left(  \sum\limits_{h\in
H}\epsilon_{1}\left(  \rho\left(  h\right)  jq\right)  h^{-1}u\right)
=\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
jq\right)  \underbrace{j^{-1}\otimes_{\CC H}h^{-1}u}%
_{\substack{=j^{-1}h^{-1}\otimes_{\CC H}u\\\text{(since }h^{-1}%
\in\CC H\text{ (since }h\in H\\\text{and thus }h^{-1}\in H\subset
\CC H\text{))}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
jq\right)  \underbrace{j^{-1}h^{-1}}_{\substack{=j^{-1}\cdot\rho\left(
h^{-1}\right)  \\\text{(by (\ref{sol.Indrho.i.rightCH2}), applied to }%
\gamma=j^{-1}\\\text{and }\eta=h^{-1}\text{ (since }h^{-1}\in H\text{ (since
}h\in H\text{)))}}}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\underbrace{\sum\limits_{h\in H}}_{\substack{=\sum
_{g\in\overline{H}}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}\\\text{(since every }h\in H\text{ satisfies}\\\rho\left(  h\right)
\in\rho\left(  H\right)  =\overline{H}\text{)}}}\epsilon_{1}\left(
\rho\left(  h\right)  jq\right)  j^{-1}\cdot\underbrace{\rho\left(
h^{-1}\right)  }_{\substack{=\left(  \rho\left(  h\right)  \right)
^{-1}\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}%
}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}}_{\substack{=\sum\limits_{\substack{h\in H;\\h\in
\rho^{-1}\left(  g\right)  }}\\\text{(because for every }h\in H\text{,}%
\\\text{the statement }\left(  \rho\left(  h\right)  =g\right)  \\\text{is
equivalent to}\\\text{the statement }\left(  h\in\rho^{-1}\left(  g\right)
\right)  \text{)}}}\epsilon_{1}\left(  \underbrace{\rho\left(  h\right)
}_{=g}jq\right)  j^{-1}\cdot\left(  \underbrace{\rho\left(  h\right)  }%
_{=g}\right)  ^{-1}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum\limits_{\substack{h\in
H;\\h\in\rho^{-1}\left(  g\right)  }}}_{\substack{=\sum_{h\in\rho^{-1}\left(
g\right)  }\\\text{(since }\rho^{-1}\left(  g\right)  \subset H\text{)}%
}}\epsilon_{1}\left(  gjq\right)  \underbrace{j^{-1}\cdot g^{-1}}_{=\left(
gj\right)  ^{-1}}\otimes_{\CC H}u\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum_{h\in\rho^{-1}\left(
g\right)  }\epsilon_{1}\left(  gjq\right)  \left(  gj\right)  ^{-1}%
\otimes_{\CC H}u}_{=\left\vert \rho^{-1}\left(  g\right)  \right\vert
\cdot\epsilon_{1}\left(  gjq\right)  \left(  gj\right)  ^{-1}\otimes
_{\CC H}u}\\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\left\vert \rho
^{-1}\left(  g\right)  \right\vert }_{\substack{=\left\vert K\right\vert
\\\text{(by (\ref{sol.Indrho.i.preimage}))}}}\cdot\epsilon_{1}\left(
gjq\right)  \left(  gj\right)  ^{-1}\otimes_{\CC H}u
\end{align*}%
\begin{align*}
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\left\vert K\right\vert \cdot
\epsilon_{1}\left(  gjq\right)  \left(  gj\right)  ^{-1}\otimes_{{\CC %
}H}u=\left\vert K\right\vert \cdot\sum_{j\in J}\sum_{g\in\overline{H}}%
\epsilon_{1}\left(  gjq\right)  \left(  gj\right)  ^{-1}\otimes_{{\CC %
}H}u\\
&  =\left\vert K\right\vert \cdot\underbrace{\sum_{j\in J}\sum_{g\in
\overline{H}j}}_{\substack{=\sum\limits_{g\in G}\\\text{(since }%
\bigsqcup_{j\in J}\overline{H}j=G\text{)}}}\epsilon_{1}\left(  gq\right)
g^{-1}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g\text{ for }gj\text{ in the second sum, since the
map}\\
\overline{H}\rightarrow\overline{H}j,\ g\mapsto gj\text{ is a bijection
(because }G\text{ is a group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \cdot\sum\limits_{g\in G}\epsilon_{1}\left(
gq\right)  g^{-1}\otimes_{\CC H}u=\left\vert K\right\vert \cdot
\sum\limits_{g\in G}\underbrace{\epsilon_{1}\left(  g^{-1}q\right)
}_{\substack{=\epsilon_{g}\left(  q\right)  \\\text{(by
(\ref{sol.Indrho.i.epsi1g}))}}}\underbrace{\left(  g^{-1}\right)  ^{-1}}%
_{=g}\otimes_{\CC H}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g^{-1}\text{ for }g\text{ in the sum, since the
map}\\
G\rightarrow G,\ g\mapsto g^{-1}\text{ is a bijection (since }G\text{ is a
group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \cdot\sum\limits_{g\in G}\epsilon_{g}\left(
q\right)  g\otimes_{\CC H}u=\left\vert K\right\vert \cdot
\underbrace{\left(  \sum\limits_{g\in G}\epsilon_{g}\left(  q\right)
g\right)  }_{\substack{=q\\\text{(by (\ref{sol.Indrho.i.epsilong.coordinates}%
))}}}\otimes_{\CC H}u\\
&  =\left\vert K\right\vert \cdot\underbrace{q\otimes_{\CC H}%
u}_{=\operatorname{id}\left(  q\otimes_{\CC H}u\right)  }=\left\vert
K\right\vert \cdot\operatorname{id}\left(  q\otimes_{\CC H}u\right)
=\left(  \left\vert K\right\vert \operatorname{id}\right)  \left(
q\otimes_{\CC H}u\right)  .
\end{align*}


Now, let us forget that we fixed $q$ and $u$. We thus have shown that $\left(
\alpha\circ\beta\right)  \left(  q\otimes_{\CC H}u\right)  =\left(
\left\vert K\right\vert \operatorname{id}\right)  \left(  q\otimes
_{\CC H}u\right)  $ for all $q\in{\CC }G$ and $u\in U$. In other
words, the two maps $\alpha\circ\beta:{\CC }G\otimes_{{\CC }%
H}U\rightarrow{\CC }G\otimes_{\CC H}U$ and $\left\vert
K\right\vert \operatorname{id}:{\CC }G\otimes_{{\CC }%
H}U\rightarrow{\CC }G\otimes_{\CC H}U$ are equal to each other
on each pure tensor. Since these two maps are $\CC$-linear, this
yields that these two maps $\alpha\circ\beta:{\CC }G\otimes
_{\CC H}U\rightarrow{\CC }G\otimes_{\CC H}U$ and
$\left\vert K\right\vert \operatorname{id}:{\CC }G\otimes_{{\CC %
}H}U\rightarrow{\CC }G\otimes_{\CC H}U$ must be
identical\footnote{This is because a well-known fact from linear algebra
states the following: If two $\CC$-linear maps from a tensor product
are equal to each other on each pure tensor, then these two maps must be
identical.}. In other words, $\alpha\circ\beta=\left\vert K\right\vert
\operatorname{id}$.

Next, we are going to show that $\beta\circ\alpha=\left\vert K\right\vert
\operatorname{id}$.

Let $f\in\Hom_{\CC H}\left(  {\CC}G,U\right)
$. Let $p\in{\CC }G$. The map $f$ is left ${\CC }H$-linear (since
$f\in\Hom_{\CC H}\left(  {\CC}G,U\right)  $),
hence $\CC$-linear. We have $\alpha\left(  f\right)  =\sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  $ (by the definition of
$\alpha$). Applying the map $\beta$ to this equality, we obtain
\[
\beta\left(  \alpha\left(  f\right)  \right)  =\beta\left(  \sum_{j\in
J}j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)  =\sum_{j\in J}%
\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)  \right)
\]
(since $\beta$ is a $\CC$-linear map). Thus,
\begin{align*}
&  \underbrace{\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
}_{=\sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(  j\right)
\right)  }\left(  p\right) \\
&  =\left(  \sum_{j\in J}\beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  =\sum_{j\in J}%
\underbrace{\left(  \beta\left(  j^{-1}\otimes_{\CC H}f\left(
j\right)  \right)  \right)  \left(  p\right)  }_{\substack{=\sum\limits_{h\in
H}\epsilon_{1}\left(  \rho\left(  h\right)  pj^{-1}\right)  h^{-1}f\left(
j\right)  \\\text{(by (\ref{sol.Indrho.i.beta.explicit}), applied
to}\\q=j^{-1}\text{ and }u=f\left(  j\right)  \text{)}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \rho\left(  h\right)
pj^{-1}\right)  h^{-1}f\left(  j\right)  =\sum_{j\in J}\sum\limits_{h\in
H}\underbrace{\epsilon_{1}\left(  \rho\left(  h^{-1}\right)  pj^{-1}\right)
}_{\substack{=\epsilon_{1}\left(  j^{-1}\rho\left(  h^{-1}\right)  p\right)
\\\text{(by (\ref{sol.Indrho.i.trace}), applied to}\\\rho\left(
h^{-1}\right)  p\text{ and }j^{-1}\text{ instead of }p\text{ and }q\text{)}%
}}\underbrace{\left(  h^{-1}\right)  ^{-1}}_{=h}f\left(  j\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }h^{-1}\text{ for }h\text{ in the second sum,}\\
\text{since the map }H\rightarrow H,\ h\mapsto h^{-1}\text{ is a bijection}\\
\text{(because }H\text{ is a group)}%
\end{array}
\right) \\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  j^{-1}%
\underbrace{\rho\left(  h^{-1}\right)  }_{\substack{=\left(  \rho\left(
h\right)  \right)  ^{-1}\\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}p\right)  \underbrace{hf\left(  j\right)
}_{\substack{=f\left(  hj\right)  \\\text{(since }f\text{ is left }%
{\CC }H\text{-linear and since }h\in H\subset{\CC }H\text{)}}}\\
&  =\sum_{j\in J}\sum\limits_{h\in H}\epsilon_{1}\left(  \underbrace{j^{-1}%
\left(  \rho\left(  h\right)  \right)  ^{-1}}_{=\left(  \rho\left(  h\right)
j\right)  ^{-1}}p\right)  f\left(  \underbrace{hj}_{\substack{=\rho\left(
h\right)  \cdot j\\\text{(by (\ref{sol.Indrho.i.leftCH2}), applied to}%
\\\eta=h\text{ and }\gamma=j\text{)}}}\right)
\end{align*}%
\begin{align*}
&  =\sum_{j\in J}\underbrace{\sum\limits_{h\in H}}_{\substack{=\sum
_{g\in\overline{H}}\sum\limits_{\substack{h\in H;\\\rho\left(  h\right)
=g}}\\\text{(since every }h\in H\text{ satisfies}\\\rho\left(  h\right)
\in\rho\left(  H\right)  =\overline{H}\text{)}}}\epsilon_{1}\left(  \left(
\rho\left(  h\right)  j\right)  ^{-1}p\right)  f\left(  \rho\left(  h\right)
j\right) \\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum\limits_{\substack{h\in
H;\\\rho\left(  h\right)  =g}}}_{\substack{=\sum\limits_{\substack{h\in H;\\h\in
\rho^{-1}\left(  g\right)  }}\\\text{(because for every }h\in H\text{,}%
\\\text{the statement }\left(  \rho\left(  h\right)  =g\right)  \\\text{is
equivalent to}\\\text{the statement }\left(  h\in\rho^{-1}\left(  g\right)
\right)  \text{)}}}\epsilon_{1}\left(  \left(  \underbrace{\rho\left(
h\right)  }_{=g}j\right)  ^{-1}p\right)  f\left(  \underbrace{\rho\left(
h\right)  }_{=g}j\right) \\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum\limits_{\substack{h\in
H;\\h\in\rho^{-1}\left(  g\right)  }}}_{\substack{=\sum_{h\in\rho^{-1}\left(
g\right)  }\\\text{(since }\rho^{-1}\left(  g\right)  \subset H\text{)}%
}}\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(  gj\right) \\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\sum_{h\in\rho^{-1}\left(
g\right)  }\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(
gj\right)  }_{=\left\vert \rho^{-1}\left(  g\right)  \right\vert \cdot
\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(  gj\right)
}=\sum_{j\in J}\sum_{g\in\overline{H}}\underbrace{\left\vert \rho^{-1}\left(
g\right)  \right\vert }_{\substack{=\left\vert K\right\vert \\\text{(by
(\ref{sol.Indrho.i.preimage}))}}}\cdot\epsilon_{1}\left(  \left(  gj\right)
^{-1}p\right)  f\left(  gj\right) \\
&  =\sum_{j\in J}\sum_{g\in\overline{H}}\left\vert K\right\vert \cdot
\epsilon_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(  gj\right)
=\left\vert K\right\vert \cdot\sum_{j\in J}\sum_{g\in\overline{H}}\epsilon
_{1}\left(  \left(  gj\right)  ^{-1}p\right)  f\left(  gj\right) \\
&  =\left\vert K\right\vert \cdot\underbrace{\sum_{j\in J}\sum_{g\in
\overline{H}j}}_{\substack{=\sum\limits_{g\in G}\\\text{(since }%
\bigsqcup_{j\in J}\overline{H}j=G\text{)}}}\epsilon_{1}\left(  g^{-1}p\right)
f\left(  g\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }g\text{ for }gj\text{ in the second sum, since the
map}\\
\overline{H}\rightarrow\overline{H}j,\ g\mapsto gj\text{ is a bijection
(because }G\text{ is a group)}%
\end{array}
\right) \\
&  =\left\vert K\right\vert \cdot\sum\limits_{g\in G}\underbrace{\epsilon
_{1}\left(  g^{-1}p\right)  }_{\substack{=\epsilon_{g}\left(  p\right)
\\\text{(by (\ref{sol.Indrho.i.epsi1g}), applied to }q=p\text{)}}}f\left(
g\right)  =\left\vert K\right\vert \cdot\underbrace{\sum\limits_{g\in
G}\epsilon_{g}\left(  p\right)  f\left(  g\right)  }_{\substack{=f\left(
\sum\limits_{g\in G}\epsilon_{g}\left(  p\right)  g\right)  \\\text{(since
}f\text{ is }\CC \text{-linear)}}}\\
&  =\left\vert K\right\vert \cdot f\left(  \sum\limits_{g\in G}\epsilon
_{g}\left(  p\right)  g\right)  .
\end{align*}
Compared with
\[
\left(  \left\vert K\right\vert f\right)  \left(  p\right)  =\left\vert
K\right\vert \cdot f\left(  \underbrace{p}_{\substack{=\sum\limits_{g\in
G}\epsilon_{g}\left(  p\right)  g\\\text{(by
(\ref{sol.Indrho.i.epsilong.coordinates}), applied to }q=p\text{)}}}\right)
=\left\vert K\right\vert \cdot f\left(  \sum\limits_{g\in G}\epsilon
_{g}\left(  p\right)  g\right)  ,
\]
this yields $\left(  \beta\left(  \alpha\left(  f\right)  \right)  \right)
\left(  p\right)  =\left(  \left\vert K\right\vert f\right)  \left(  p\right)
$.

Now, forget that we fixed $p$. We thus have proven that $\left(  \beta\left(
\alpha\left(  f\right)  \right)  \right)  \left(  p\right)  =\left(
\left\vert K\right\vert f\right)  \left(  p\right)  $ for every $p\in
{\CC }G$. In other words, $\beta\left(  \alpha\left(  f\right)  \right)
=\left\vert K\right\vert f$. Hence, $\left(  \beta\circ\alpha\right)  \left(
f\right)  =\beta\left(  \alpha\left(  f\right)  \right)  =\left\vert
K\right\vert f=\left(  \left\vert K\right\vert \operatorname{id}\right)
\left(  f\right)  $.

Now, forget that we fixed $f$. We thus have shown that $\left(  \beta
\circ\alpha\right)  \left(  f\right)  =\left(  \left\vert K\right\vert
\operatorname{id}\right)  \left(  f\right)  $ for every $f\in
\Hom_{\CC H}\left(  {\CC}G,U\right)  $. In
other words, $\beta\circ\alpha=\left\vert K\right\vert \operatorname{id}$.

We now have $\beta\circ\alpha=\left\vert K\right\vert \operatorname{id}$.
Since composition of $\CC$-linear maps is a $\CC$-bilinear
operation, we have $\left(  \dfrac{1}{\left\vert K\right\vert }\beta\right)
\circ\alpha=\dfrac{1}{\left\vert K\right\vert }\underbrace{\left(  \beta
\circ\alpha\right)  }_{=\left\vert K\right\vert \operatorname{id}}=\dfrac
{1}{\left\vert K\right\vert }\left\vert K\right\vert \operatorname{id}%
=\operatorname*{id}$.

On the other hand, recall that $\alpha\circ\beta=\left\vert K\right\vert
\operatorname{id}$. Since composition of $\CC$-linear maps is a
$\CC$-bilinear operation, we have $\alpha\circ\left(  \dfrac
{1}{\left\vert K\right\vert }\beta\right)  =\dfrac{1}{\left\vert K\right\vert
}\underbrace{\left(  \alpha\circ\beta\right)  }_{=\left\vert K\right\vert
\operatorname{id}}=\dfrac{1}{\left\vert K\right\vert }\left\vert K\right\vert
\operatorname{id}=\operatorname*{id}$.

Combining the equalities $\left(  \dfrac{1}{\left\vert K\right\vert }%
\beta\right)  \circ\alpha=\operatorname*{id}$ and $\alpha\circ\left(
\dfrac{1}{\left\vert K\right\vert }\beta\right)  =\operatorname*{id}$, we
conclude that the maps $\alpha$ and $\dfrac{1}{\left\vert K\right\vert }\beta$
are mutually inverse. Hence, the map $\alpha$ is invertible.

Now, we know that the map $\alpha$ is an invertible left $\CC G$-module
homomorphism. Hence, $\alpha$ is a left ${\CC }G$-module isomorphism.
Therefore, there exists a left ${\CC }G$-module isomorphism
$\Hom_{\CC H}\left(  {\CC }G,U\right)
\rightarrow{\CC }G\otimes_{\CC H}U$ (namely, $\alpha$).
Therefore, $\Hom_{\CC H}\left(  {\CC }G,U\right)
\cong{\CC }G\otimes_{\CC H}U=\operatorname{Ind}_{\rho}U$ as left
${\CC }G$-modules. This solves Exercise~\ref{exe.Indrho}(i).

[\textit{Remark:} In our above solution of Exercise~\ref{exe.Indrho}(i), we
explicitly constructed a ${\CC }G$-module isomorphism $\alpha
:\Hom_{\CC H}\left(  {\CC}G,U\right)
\rightarrow\operatorname{Ind}_{\rho}U$. This isomorphism is functorial with
respect to $U$. It is also independent on the choice of $J$ (this is not
immediately clear from its definition, but it can be shown very easily, by
observing that the tensor $j^{-1}\otimes_{\CC H}f\left(  j\right)  $
for $j\in G$ depends only on the coset $\overline{H}j$ and not on $j$ itself).]

[\textit{Remark:} Exercise \ref{exe.Indrho}(i) allows us to give yet another
solution to Exercise \ref{exe.Indrho}(f):

\textit{Third solution to Exercise \ref{exe.Indrho}(f):} Assume that $G=H/K$
for some normal subgroup $K$ of $H$. Let $\rho:H\rightarrow G$ be the
projection map. We want to prove that $\Ind_{\rho
}U\cong U^{K}$ for every $\CC H$-module $U$.

We know that $\rho$ is the projection map from $H$ to $H/K$. Hence, $\rho$ is
a surjective group homomorphism and has kernel $\ker\rho=K$.

Let $U$ be a $\CC H$-module. Recall that $U^{K}$ is a $\CC \left[
H/K\right]  $-module, thus a $\CC G$-module (since $H/K=G$). This
$\CC G$-module structure has the property that%
\begin{equation}
\rho\left(  h\right)  \cdot v=hv\ \ \ \ \ \ \ \ \ \ \text{for any }h\in
H\text{ and }v\in U^{K}. \label{sol.Indrho.f.sol3.UK-modstr}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.f.sol3.UK-modstr}): }The proof of
(\ref{sol.Indrho.f.sol3.UK-modstr}) is identical with the proof of
(\ref{sol.Indrho.f.sol1.UK-modstr}) in the First solution to Exercise
\ref{exe.Indrho}(f).}

Let us make $\CC G$ into a $\left(  \CC H,\CC G\right)
$-bimodule as in Exercise \ref{exe.Indrho}(i). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC H,\CC G\right)
$-bimodule structure. Then, the $\CC G$-module
$\Hom_{\CC H}\left(  \CC G,U\right)  $ (defined as in Exercise
\ref{exe.induction.ind-as-hom} using the $\left(  \CC H,\CC %
G\right)  $-bimodule structure on $\CC G$) is isomorphic to
$\Ind_{\rho}U$ (because of Exercise \ref{exe.Indrho}%
(i)). In other words, $\Hom_{\CC H}\left(
\CC G,U\right)  \cong\Ind_{\rho}U$ as left
$\CC G$-modules.

We recall that $\Hom_{\CC H}\left(
\CC G,U\right)  $ is the left $\CC G$-module consisting of all
left $\CC H$-module homomorphisms from $\CC G$ to $U$. This uses
only the $\left(  \CC H,\CC G\right)  $-bimodule structure on
$\CC G$ that was introduced in the statement of Exercise
\ref{exe.Indrho}(i) (but not the $\left(  \CC G,\CC H\right)
$-bimodule structure on $\CC G$ that was introduced in the definition of
$\Ind_{\rho}U$).

We have $f\left(  1\right)  \in U^{K}$ for every
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $. Let $k\in K$. Thus,
$k\in K=\ker\rho$, so that $\rho\left(  k\right)  =1_{G}$.
\par
But $k\in K\subset H$. Thus, (\ref{sol.Indrho.i.leftCH2}) (applied to
$\gamma=1_{\CC G}$ and $\eta=k$) yields $k1_{\CC G}%
=\underbrace{\rho\left(  k\right)  }_{=1_{G}}1_{\CC G}=1_{G}%
1_{\CC G}=1_{G}$.
\par
But $f\in\Hom_{\CC H}\left(  \CC %
G,U\right)  $. Hence, the map $f$ is left $\CC H$-linear. Thus,
$f\left(  k1_{\CC G}\right)  =kf\left(  1_{\CC G}\right)  $ (since
$k\in H\subset\CC H$). Since $k1_{\CC G}=1_{G}$, this rewrites as
$f\left(  1_{G}\right)  =kf\left(  \underbrace{1_{\CC G}}_{=1}\right)
=kf\left(  1\right)  $, so that $kf\left(  1\right)  =f\left(
\underbrace{1_{G}}_{=1}\right)  =f\left(  1\right)  $.
\par
Now, let us forget that we fixed $k$. We thus have shown that $kf\left(
1\right)  =f\left(  1\right)  $ for every $k\in K$. In other words, $f\left(
1\right)  $ is an element $y$ of $U$ which satisfies $ky=y$ for every $k\in
K$. In other words,%
\[
f\left(  1\right)  \in\left\{  y\in U\ \mid\ ky=y\text{ for every }k\in
K\right\}  =U^{K},
\]
qed.}. Thus, we can define a map
$\phi:\Hom_{\CC H}\left(  \CC G,U\right)  \rightarrow U^{K}$ by
\[
\left(  \phi\left(  f\right)  =f\left(  1\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }f\in\Hom_{\CC H}
\left(  \CC G,U\right)  \right)  .
\]
Consider this $\phi$. This map $\phi$ is clearly $\CC$-linear (since
$f\left(  1\right)  $ depends $\CC$-linearly on $f$). Also, the map
$\phi$ is surjective\footnote{\textit{Proof.} Let $u\in U^{K}$. We are going
to construct a map $f\in\Hom_{\CC H}\left(
\CC G,U\right)  $ which satisfies $\phi\left(  f\right)  =u$.
\par
Indeed, let us define a map $f:\CC G\rightarrow U$ by%
\begin{equation}
\left(  f\left(  \gamma\right)  =\gamma u\ \ \ \ \ \ \ \ \ \ \text{for every
}\gamma\in\CC G\right)  . \label{sol.Indrho.f.sol3.surj.pf.def-f}%
\end{equation}
(This is well-defined, since $U^{K}$ is a $\CC G$-module.) This map $f$
is $\CC$-linear (since $\gamma u$ depends $\CC$-linearly on
$\gamma$). We shall now check that the map $f$ is a homomorphism of left
$H$-sets.
\par
Indeed, let $h\in H$ and $\gamma\in\CC G$. We have $f\left(
h\gamma\right)  =\left(  h\gamma\right)  u$ (by the definition of $f$).
\par
On the other hand, (\ref{sol.Indrho.i.leftCH2}) (applied to $\eta=h$) yields
$h\gamma=\rho\left(  h\right)  \cdot\gamma$.
\par
But $u\in U^{K}$ and therefore $\gamma u\in U^{K}$ (since $\gamma\in
\CC G$ and since $U^{K}$ is a $\CC G$-module). Thus,
(\ref{sol.Indrho.f.sol3.UK-modstr}) (applied to $\gamma u$ instead of $v$)
yields $\rho\left(  h\right)  \cdot\left(  \gamma u\right)  =h\left(  \gamma
u\right)  $. Now,%
\[
f\left(  h\gamma\right)  =\underbrace{\left(  h\gamma\right)  }_{=\rho\left(
h\right)  \cdot\gamma}u=\left(  \rho\left(  h\right)  \cdot\gamma\right)
u=\rho\left(  h\right)  \cdot\left(  \gamma u\right)  =h\underbrace{\left(
\gamma u\right)  }_{\substack{=f\left(  \gamma\right)  \\\text{(by
(\ref{sol.Indrho.f.sol3.surj.pf.def-f}))}}}=hf\left(  \gamma\right)  .
\]
\par
Now, let us forget that we fixed $h$ and $\gamma$. We thus have proven that
$f\left(  h\gamma\right)  =hf\left(  \gamma\right)  $ for all $h\in H$ and
$\gamma\in\CC G$. In other words, $f$ is a homomorphism of left
$H$-sets. Hence, $f$ is a left $\CC H$-module homomorphism (since $f$ is
$\CC$-linear). In other words,
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $. Now, the definition of
$\phi$ yields%
\begin{align*}
\phi\left(  f\right)   &  =f\left(  1\right)  =1u\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Indrho.f.sol3.surj.pf.def-f}), applied to }\gamma=1\right)
\\
&  =u.
\end{align*}
Thus, $u=\phi\left(  \underbrace{f}_{\in\Hom_{\CC H}
\left(  \CC G,U\right)  }\right)  \in\phi\left(
\Hom_{\CC H}\left(  \CC G,U\right)
\right)  $.
\par
Let us now forget that we fixed $u$. We thus have shown that $u\in\phi\left(
\Hom_{\CC H}\left(  \CC G,U\right)
\right)  $ for every $u\in U^{K}$. In other words, $U^{K}\subset\phi\left(
\Hom_{\CC H}\left(  \CC G,U\right)
\right)  $. This proves that the map $\phi$ is surjective. Qed.} and
injective\footnote{\textit{Proof.} Let $f\in\ker\phi$. Then,
$f\in\Hom_{\CC H}\left(  \CC G,U\right)  $
and $\phi\left(  f\right)  =0$ (since $f\in\ker\phi$). The
definition of $\phi$ yields $\phi\left(  f\right)  =f\left(  \underbrace{1}%
_{=1_{\CC G}}\right)  =f\left(  1_{\CC G}\right)  $, so that
$f\left(  1_{\CC G}\right)  =\phi\left(  f\right)  =0$.
\par
Now, let $g\in G$ be arbitrary. Then, there exists some $h\in H$ such that
$g=\rho\left(  h\right)  $ (since $\rho$ is surjective). Consider this $h$.
Applying (\ref{sol.Indrho.i.leftCH2}) to $\gamma=1_{\CC G}$ and $\eta
=h$, we obtain $h1_{\CC G}=\rho\left(  h\right)  \cdot1_{\CC %
G}=\rho\left(  h\right)  =g$. But the map $f$ is left $\CC H$-linear
(since $f\in\Hom_{\CC H}\left(  \CC %
G,U\right)  $). Thus, $f\left(  h1_{\CC G}\right)  =hf\left(
1_{\CC G}\right)  $ (since $h\in H\subset\CC H$), so that
$f\left(  h1_{\CC G}\right)  =h\underbrace{f\left(  1_{\CC %
G}\right)  }_{=0}=0$. Compared with $f\left(  \underbrace{h1_{\CC G}%
}_{=g}\right)  =f\left(  g\right)  $, this yields $f\left(  g\right)
=0=0\left(  g\right)  $ (where the $0$ on the right hand side means the zero
map $0:\CC G\rightarrow U$).
\par
Now, let us forget that we fixed $g$. Thus, we have shown that $f\left(
g\right)  =0\left(  g\right)  $ for every $g\in G$. In other words, the two
$\CC$-linear maps $f:\CC G\rightarrow U$ and $0:\CC %
G\rightarrow U$ are equal to each other on every element of the set $G$. Since
$G$ is a basis of the $\CC$-vector space $\CC G$, this shows that
the maps $f:\CC G\rightarrow U$ and $0:\CC G\rightarrow U$ must be
identical (because if two $\CC$-linear maps having the same domain and
the same codomain are equal to each other on every element of some basis of
their domain, then these two maps must be identical). In other words, $f=0$.
\par
Now, let us forget that we fixed $f$. We thus have shown that $f=0$ for every
$f\in\ker\phi$. In other words, $\ker\phi=0$.
Since $\phi$ is a $\CC$-linear map, this shows that $\phi$ is
injective. Qed.}. Hence, the map $\phi$ is bijective. Thus, $\phi$ is a
$\CC$-vector space isomorphism (since $\phi$ is $\CC$-linear).

Moreover, $\phi$ is a homomorphism of left $G$-sets\footnote{\textit{Proof.}
Let $g\in G$ and $f\in\Hom_{\CC H}\left(
\CC G,U\right)  $. The definition of the left $\CC G$-module
structure on $\Hom_{\CC H}\left(  \CC %
G,U\right)  $ yields%
\[
\left(  gf\right)  \left(  1\right)  =f\left(  \underbrace{1g}_{=g}\right)
=f\left(  g\right)  .
\]
Now, the definition of $\phi$ yields $\phi\left(  gf\right)  =\left(
gf\right)  \left(  1\right)  =f\left(  g\right)  $.
\par
But there exists some $h\in H$ such that $g=\rho\left(  h\right)  $ (since
$\rho$ is surjective). Consider this $h$. The map $f$ is left $\CC %
H$-linear (since $f\in\Hom_{\CC H}\left(
\CC G,U\right)  $). Thus, we have $f\left(  h1_{\CC G}\right)
=hf\left(  1_{\CC G}\right)  $. But (\ref{sol.Indrho.i.leftCH2})
(applied to $\gamma=1_{\CC G}$ and $\eta=h$) yields $h1_{\CC %
G}=\underbrace{\rho\left(  h\right)  }_{=g}\cdot1_{\CC G}=g\cdot
1_{\CC G}=g$, so that $g=h1_{\CC G}$. Hence, $f\left(
\underbrace{g}_{=h1_{\CC G}}\right)  =f\left(  h1_{\CC G}\right)
=hf\left(  1_{\CC G}\right)  $.
\par
The definition of $\phi$ yields $\phi\left(  f\right)  =f\left(
\underbrace{1}_{=1_{\CC G}}\right)  =f\left(  1_{\CC G}\right)  $,
so that $f\left(  1_{\CC G}\right)  =\phi\left(  f\right)  $. Now,
$f\left(  g\right)  =h\underbrace{f\left(  1_{\CC G}\right)  }%
_{=\phi\left(  f\right)  }=h\phi\left(  f\right)  $. Hence,%
\[
\phi\left(  gf\right)  =f\left(  g\right)  =h\phi\left(  f\right)  .
\]
\par
The equality (\ref{sol.Indrho.f.sol3.UK-modstr}) (applied to $\phi\left(
f\right)  $ instead of $v$) yields $\rho\left(  h\right)  \cdot\phi\left(
f\right)  =h\phi\left(  f\right)  $. Thus, $h\phi\left(  f\right)
=\underbrace{\rho\left(  h\right)  }_{=g}\cdot\phi\left(  f\right)
=g\cdot\phi\left(  f\right)  =g\phi\left(  f\right)  $. Hence, $\phi\left(
gf\right)  =h\phi\left(  f\right)  =g\phi\left(  f\right)  $.
\par
Now, let us forget that we fixed $g$ and $f$. We thus have proven that
$\phi\left(  gf\right)  =g\phi\left(  f\right)  $ for all $g\in G$ and
$f\in\Hom_{\CC H}\left(  \CC G,U\right)
$. In other words, the map $\phi$ is a homomorphism of left $G$-sets. Qed.},
and therefore a left $\CC G$-module homomorphism (since $f$ is
$\CC$-linear). Thus, $\phi$ is a left $\CC G$-module isomorphism
(since $\phi$ is a $\CC$-vector space isomorphism). We thus have found
a left $\CC G$-module isomorphism from
$\Hom_{\CC H}\left(  \CC G,U\right)  $ to $U^{K}$ (namely,
$\phi$). Hence, $U^{K}\cong\Hom_{\CC H}\left(
\CC G,U\right)  $ as left $\CC G$-modules. Hence, $U^{K}%
\cong\Hom_{\CC H}\left(  \CC G,U\right)
\cong\Ind_{\rho}U$ as left $\CC G$-modules. This
solves Exercise \ref{exe.Indrho}(f) again.]

(j) Let $U$ be a $\CC G$-module, and let $V$ be a $\CC H$-module.

Let us make $\CC G$ into a $\left(  \CC H,\CC G\right)
$-bimodule as in Exercise \ref{exe.Indrho}(i). From now on, we regard
$\CC G$ as endowed with this $\left(  \CC H,\CC G\right)
$-bimodule structure. Then, the $\CC G$-module
$\Hom_{\CC H}\left(  \CC G,V\right)  $ (defined as in Exercise
\ref{exe.induction.ind-as-hom} using the $\left(  \CC H,\CC %
G\right)  $-bimodule structure on $\CC G$) is isomorphic to
$\Ind_{\rho}V$ (because of Exercise \ref{exe.Indrho}%
(i), applied to $V$ instead of $U$). In other words,
\begin{equation}
\Hom_{\CC H}\left(  \CC G,V\right)
\cong\Ind_{\rho}V\ \ \ \ \ \ \ \ \ \ \text{as left
}\CC G\text{-modules.} \label{sol.Indrho.j.i}%
\end{equation}


Now, (\ref{Hom-tensor-adj}) (applied to $R=\CC G$, $S=\CC H$,
$A=\CC G$, $B=U$ and $C=V$) yields
\begin{equation}
\Hom_{\CC H}\left(  \CC G\otimes
_{\CC G}U,V\right)  \cong\Hom_{\CC %
G}\left(  U,\Hom_{\CC H}\left(  \CC %
G,V\right)  \right)  . \label{sol.Indrho.j.adj}%
\end{equation}


We shall now prove that $\CC G\otimes_{\CC G}U\cong%
\Res_{\rho}U$ as left $\CC H$-modules.

Indeed, a fundamental fact in abstract algebra says the following: If
$\mathfrak{A}$ is a $\CC$-algebra, and if $\mathfrak{M}$ is a left
$\mathfrak{A}$-module, then there exists a $\CC$-vector space
isomorphism $\Xi:\mathfrak{A}\otimes_{\mathfrak{A}}\mathfrak{M}\rightarrow
\mathfrak{M}$ which satisfies
\newline
$\left(  \Xi\left(  a\otimes_{\mathfrak{A}%
}m\right)  =am\text{ for every }a\in\mathfrak{A}\text{ and }m\in
\mathfrak{M}\right)  $. Applying this fact to $\mathfrak{A}=\CC G$ and
$\mathfrak{M}=U$, we conclude that there exists a $\CC$-vector space
isomorphism $\Xi:\CC G\otimes_{\CC G}U\rightarrow U$ which
satisfies
\newline
$\left(  \Xi\left(  a\otimes_{\CC G}m\right)  =am\text{ for
every }a\in\CC G\text{ and }m\in U\right)  $. Consider this $\Xi$.

The map $\Xi$ is a $\CC$-vector space isomorphism. Hence, its inverse
$\Xi^{-1}$ exists and also is a $\CC$-vector space isomorphism, and
therefore a $\CC$-linear map. It furthermore satisfies%
\begin{equation}
\Xi^{-1}\left(  u\right)  =1\otimes_{\CC G}%
u\ \ \ \ \ \ \ \ \ \ \text{for every }u\in U. \label{sol.Indrho.j.Xi-1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.j.Xi-1}):} Let $u\in U$. Recall
that $\Xi\left(  a\otimes_{\CC G}m\right)  =am$ for every $a\in
\CC G$ and $m\in U$. Applying this to $a=1$ and $m=u$, we obtain
$\Xi\left(  1\otimes_{\CC G}u\right)  =1u=u$. Hence, $1\otimes
_{\CC G}u=\Xi^{-1}\left(  u\right)  $ (since $\Xi$ is invertible). This
proves (\ref{sol.Indrho.j.Xi-1}).}

The map $\Xi^{-1}$ is a homomorphism of $H$-sets from $\Res_{\rho}U$
to $\CC G\otimes_{\CC G}U$%
\ \ \ \ \footnote{\textit{Proof.} The map $\Xi^{-1}$ is a map from $U$ to
$\CC G\otimes_{\CC G}U$, therefore a map from $\Res_{\rho}U$ to
$\CC G\otimes_{\CC G}U$ (since $\Res_{\rho}U=U$ as sets).
\par
Recall that the $\CC H$-module structure on $\Res_{\rho}U$ is given by
\begin{equation}
h\cdot v=\rho\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \text{for every
}h\in H\text{ and }v\in U. \label{sol.Indrho.j.Res}%
\end{equation}
\par
Now, let $v\in\Res_{\rho}U$ and $h\in H$. We are going
to prove that $\Xi^{-1}\left(  h\cdot v\right)  =h\cdot\Xi^{-1}\left(
v\right)  $, where $h\cdot\Xi^{-1}\left(  v\right)  $ is computed in the
$\CC H$-module $\CC G\otimes_{\CC G}U$.
\par
Indeed, recall that the left $\CC H$-module structure on $\CC G$
is induced by the
$\CC$-algebra homomorphism $\CC \left[  \rho\right]
:\CC H\rightarrow\CC G$. Thus, it is explicitly given by%
\[
\eta\gamma=\left(  \CC \left[  \rho\right]  \right)  \eta\cdot
\gamma\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\CC G\text{ and }%
\eta\in\CC H.
\]
Applying this to $\eta=h$ and $\gamma=1_{\CC G}$, we obtain
$h1_{\CC G}=\left(  \CC \left[  \rho\right]  \right)  \left(
h\right)  \cdot1_{\CC G}$ (since $h\in H\subset\CC H$). But $h\in
H$ and thus $\left(  \CC \left[  \rho\right]  \right)  \left(  h\right)
=\rho\left(  h\right)  $. Hence, $h1_{\CC G}=\underbrace{\left(
\CC \left[  \rho\right]  \right)  \left(  h\right)  }_{=\rho\left(
h\right)  }\cdot1_{\CC G}=\rho\left(  h\right)  \cdot1_{\CC %
G}=\rho\left(  h\right)  =1_{\CC G}\cdot\rho\left(  h\right)  $. Now,
$v\in\Res_{\rho}U=U$. Hence, (\ref{sol.Indrho.j.Xi-1})
(applied to $u=v$) yields $\Xi^{-1}\left(  v\right)  =1\otimes_{\CC G}%
v$. Thus,%
\begin{align*}
h\cdot\underbrace{\Xi^{-1}\left(  v\right)  }_{=1\otimes_{\CC G}v}  &
=h\cdot\left(  1\otimes_{\CC G}v\right)  =h\underbrace{1}%
_{=1_{\CC G}}\otimes_{\CC G}v=\underbrace{h1_{\CC G}%
}_{=1_{\CC G}\cdot\rho\left(  h\right)  }\otimes_{\CC %
G}v=1_{\CC G}\cdot\rho\left(  h\right)  \otimes_{\CC G}v\\
&  =\underbrace{1_{\CC G}}_{=1}\otimes_{\CC G}\underbrace{\rho
\left(  h\right)  \cdot v}_{\substack{=h\cdot v\\\text{(by
(\ref{sol.Indrho.j.Res}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have moved }\rho\left(  h\right)  \text{ past the tensor
sign}\\
\text{(this is allowed since }\rho\left(  h\right)  \in G\subset
\CC G\text{)}%
\end{array}
\right) \\
&  =1\otimes_{\CC G}h\cdot v.
\end{align*}
Compared with $\Xi^{-1}\left(  h\cdot v\right)  =1\otimes_{\CC G}h\cdot
v$ (by (\ref{sol.Indrho.j.Xi-1}), applied to $u=h\cdot v$), this yields
$\Xi^{-1}\left(  h\cdot v\right)  =h\cdot\Xi^{-1}\left(  v\right)  $.
\par
Now, let us forget that we fixed $v$ and $h$. We thus have shown that
$\Xi^{-1}\left(  h\cdot v\right)  =h\cdot\Xi^{-1}\left(  v\right)  $ for every
$v\in\Res_{\rho}U$ and $h\in H$. In other words,
$\Xi^{-1}$ is a homomorphism of $H$-sets from $\Res_{\rho}U$ to
$\CC G\otimes_{\CC G}U$, qed.}. Therefore,
$\Xi^{-1}$ is a $\CC H$-module homomorphism from $\Res_{\rho}U$ to
$\CC G\otimes_{\CC G}U$ (since $\Xi^{-1}$ is
a $\CC$-linear map). Consequently, $\Xi^{-1}$ is a $\CC H$-module
isomorphism from $\Res_{\rho}U$ to $\CC G\otimes_{\CC G}U$
(since $\Xi^{-1}$ is a $\CC$-vector space
isomorphism). Therefore, $\CC G\otimes_{\CC G}U\cong%
\Res_{\rho}U$ as $\CC H$-modules.
Now, \eqref{Hom-tensor-adj} (applied to $\CC G$, $\CC H$,
$\CC G$, $U$ and $V$ instead of $R$, $S$, $A$, $B$ and $C$)
yields
\[
\Hom_{\CC H}\left( \CC G\otimes_{\CC G}U ,V\right)
\cong
\Hom_{\CC G}\left(  U,\Hom_{\CC H}\left(  \CC G,V\right)  \right) .
\]
Thus,
\begin{align*}
\Hom_{\CC G}\left(  U,\Hom_{\CC H}\left(  \CC G,V\right)  \right)
& \cong\Hom_{\CC H}\left(  \underbrace{\CC %
G\otimes_{\CC G}U}_{\cong\Res_{\rho}U\text{ as
}\CC H\text{-modules}},V\right) \\
&  \cong\Hom_{\CC H}\left(  \Res_{\rho}U,V\right)  .
\end{align*}
Thus,%
\[
\Hom_{\CC H}\left(  \Res_{\rho}U,V\right)  \cong\Hom_{\CC G}
\left(  U,\underbrace{\Hom_{\CC H}\left(
\CC G,V\right)  }_{\substack{\cong\Ind_{\rho
}V\text{ as left }\CC G\text{-modules}\\\text{(by (\ref{sol.Indrho.j.i}%
))}}}\right)  \cong\Hom_{\CC G}\left(
U,\Ind_{\rho}V\right)  .
\]
This solves Exercise \ref{exe.Indrho}(j).

(k) \textit{Alternative proof of (\ref{induced-character-formula}):} Let $G$
be a finite group, and let $H$ be a subgroup of $G$. Let $U$ be a
finite-dimensional $\CC H$-module. We need to prove the identity
(\ref{induced-character-formula}).

Let $\rho$ denote the inclusion map $H\rightarrow G$. Clearly, $\rho$ is a
group homomorphism. But Exercise \ref{exe.Indrho}(d) yields
$\Ind_{\rho}U=\Ind_{H}^{G}U$.
Hence, $\chi_{\Ind_{\rho}U}=\chi_{\operatorname*{Ind}%
\nolimits_{H}^{G}U}$, so that $\chi_{\Ind_{H}^{G}%
U}=\chi_{\Ind_{\rho}U}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}$ (by Exercise \ref{exe.Indrho}(b)). But Exercise
\ref{exe.Indrho}(c) (applied to $f=\chi_{U}$) yields $\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}=\Ind_{H}^{G}\chi_{U}$. Thus,
$\chi_{\Ind_{H}^{G}U}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}=\Ind_{H}^{G}\chi_{U}$. Thus,
every $g\in G$ satisfies%
\[
\underbrace{\chi_{\Ind_{H}^{G}U}}%
_{=\Ind_{H}^{G}\chi_{U}}\left(  g\right)  =\left(
\Ind_{H}^{G}\chi_{U}\right)  \left(  g\right)
=\frac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G:\\kgk^{-1}\in
H}}\chi_{U}\left(  kgk^{-1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\Ind_{H}^{G}\chi_{U}\right)  .
\]
This proves (\ref{induced-character-formula}). Thus, Exercise \ref{exe.Indrho}%
(k) is solved.

(l) \textit{Alternative proof of (\ref{deflation-character-formula}):} Let $G$
be a finite group, and let $K$ be a normal subgroup of $G$. Let $V$ be a
finite-dimensional $\CC G$-module. We need to prove the identity
(\ref{deflation-character-formula}).

Let $\rho$ denote the projection map $G\rightarrow G/K$. Exercise
\ref{exe.Indrho}(f) (applied to $G/K$, $G$ and $V$ instead of $G$, $H$ and
$U$) yields $\Ind_{\rho}V\cong V^{K}$ as
$\CC \left[  G/K\right]  $-modules. Hence, $\chi_{\operatorname*{Ind}%
\nolimits_{\rho}V}=\chi_{V^{K}}$ (since isomorphic $\CC \left[
G/K\right]  $-modules have equal characters). Thus, $\chi_{V^{K}}%
=\chi_{\Ind_{\rho}V}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{V}$ (by Exercise \ref{exe.Indrho}(b), applied to $G/K$,
$G$ and $V$ instead of $G$, $H$ and $U$). But Exercise \ref{exe.Indrho}(e)
(applied to $G/K$, $G$, $V$ and $\chi_{V}$ instead of $G$, $H$, $U$ and $f$)
yields $\Ind_{\rho}\chi_{V}=\left(  \chi_{V}\right)
^{K}$. Hence, $\chi_{V^{K}}=\Ind_{\rho}\chi
_{V}=\left(  \chi_{V}\right)  ^{K}$. Thus, every $g\in G$ satisfies%
\[
\underbrace{\chi_{V^{K}}}_{=\left(  \chi_{V}\right)  ^{K}}\left(  gK\right)
=\left(  \chi_{V}\right)  ^{K}\left(  gK\right)  =\frac{1}{|K|}\sum_{k\in
K}\chi_{V}\left(  gk\right)
\]
(by the definition of $\left(  \chi_{V}\right)  ^{K}$). This proves
(\ref{deflation-character-formula}). Thus, Exercise \ref{exe.Indrho}(l) is solved.

[\textit{Remark:} Most parts of Exercise \ref{exe.Indrho} work in a far
greater generality than they are stated in. Let us briefly survey the
straightforward generalizations. We begin with the generalizations of the
notions of induction, restriction, inflation and fixed point constructions
defined in Chapter \ref{character-theory-review-section}:

\begin{itemize}
\item Recall that if $G$ is a finite group and $H$ is a subgroup of $G$, then
we have defined a $\CC G$-module $\Ind_{H}^{G}U$
for every $\CC H$-module $U$, and we have also defined a $\CC %
H$-module $\Res_{H}^{G}V$ for every $\CC %
G$-module $V$. Both of these definitions are also valid when $\CC$ is
replaced by any commutative ring. Basic properties of induction and
restriction (such as the equality (\ref{Frobenius-reciprocity}), the statement
of Exercise \ref{exe.induction.transitivity} and the statement of Exercise
\ref{exe.ind.tensor}) are still true in this generality (and the proofs that
we gave can be transferred to this generality with almost no changes).
However, more advanced properties might fail in this generality, and some can
not even be stated over a general commutative ring $\CC$%
\ \ \ \ \footnote{For instance, any properties of characters of modules can
only be stated when these characters are well-defined. Since characters are
defined as traces of certain endomorphisms of $\CC$-vector spaces, they
are not automatically well-defined when $\CC$ is replaced by a
commutative ring. They are, however, well-defined if our modules are
projective modules over this base ring; see the Remark after the proof of
Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} for details.}.

\item Recall that if $G$ is a finite group and $K$ is a normal subgroup of
$G$, then we have defined a $\CC G$-module $\operatorname*{Infl}%
\nolimits_{G/K}^{G}U$ for every $\CC \left[  G/K\right]  $-module $U$,
and we have also defined a $\CC \left[  G/K\right]  $-module $V^{K}$ for
every $\CC G$-module $V$. Both of these definitions are also valid when
$\CC$ is replaced by any commutative ring. Some basic properties of
inflation and fixed points (such as the isomorphisms
(\ref{restriction-inflation-commutation}) and (\ref{inflation-reciprocity}))
are still correct in this generality (again, they can be proven in the same
way as above), but some others are not (e.g., Exercise
\ref{exe.fixed-vs-coinvariants}(b) is generally false\footnote{However,
Exercise \ref{exe.fixed-vs-coinvariants}(b) holds if we additionally assume
that $\left\vert K\right\vert $ is invertible in the ring that replaces
$\CC$. Again, this can be proven by repeating our solution of Exercise
\ref{exe.fixed-vs-coinvariants}(b).}).

\item If $G$ is a finite group, then we defined $R_{\CC }\left(
G\right)  $ as the $\CC$-vector space of all class functions
$G\rightarrow\CC $. This definition still applies when $\CC$ is
replaced by any commutative ring. (Of course, in this case,
``$\CC$-vector space'' will have to be replaced by
``$\CC$-module''.)

\item Recall that if $G$ is a finite group and $H$ is a subgroup of $G$, then
we have defined a class function $\Ind_{H}^{G}f\in
R_{\CC }\left(  G\right)  $ for every $f\in R_{\CC }\left(
H\right)  $, and we have also defined a class function
$\Res_{H}^{G}f\in R_{\CC }\left(  H\right)  $ for every $f\in
R_{\CC }\left(  G\right)  $. The definition of $\Res_{H}^{G}f$
still works when $\CC$ is replaced by any
commutative ring. Our definition of $\Ind_{H}^{G}f$
cannot be reasonably interpreted in this generality (due to the denominator
$\left\vert H\right\vert $ in (\ref{eq.exe.induction.characters.eq}));
however, Exercise \ref{exe.induction.characters}(b) can be used as an
alternative definition of $\Ind_{H}^{G}f$ when
$\CC$ is replaced by any commutative ring. (Of course, the necessity of
using Exercise \ref{exe.induction.characters}(b) as a definition of
$\Ind_{H}^{G}f$ entails that most of our proofs
concerning $\Ind_{H}^{G}f$ no longer are valid in this
generality, because they use the equality
(\ref{eq.exe.induction.characters.eq}) as the definition of
$\Ind_{H}^{G}f$. Even the fact that
$\Ind_{H}^{G}f$ is well-defined needs to be proven
anew, since it is not immediately obvious that the sum $\sum\limits_{\substack{j\in
J:\\jgj^{-1}\in H}}f\left(  jgj^{-1}\right)  $ in Exercise
\ref{exe.induction.characters}(b) is independent on $J$.)
\end{itemize}

Now, let us move on to generalizing Remark \ref{rmk.Resrho} and Exercise
\ref{exe.Indrho}. Let $G$ and $H$ be two finite groups, and let $\rho
:H\rightarrow G$ be a group homomorphism.

\begin{itemize}
\item Remark \ref{rmk.Resrho} still holds if $\CC$ is replaced by any
commutative ring.

\item In Exercise \ref{exe.Indrho}, we defined a map $\operatorname*{Ind}%
\nolimits_{\rho}f:G\rightarrow\CC $ for every $f\in R_{\CC %
}\left(  H\right)  $. This definition still works when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. There is an alternative definition which works in even greater
generality: Namely, as long as $\left\vert \ker\rho\right\vert
$ is invertible\footnote{This is a weaker condition than $\left\vert
H\right\vert $ being invertible.}, we can define $\operatorname*{Ind}%
\nolimits_{\rho}f:G\rightarrow\CC $ by the equality
\begin{equation}
\left(  \Ind_{\rho}f\right)  \left(  g\right)
=\dfrac{1}{\left\vert \ker\rho\right\vert }\sum\limits
_{\substack{\left(  h,k\right)  \in H\times J;\\k\rho\left(  h\right)
k^{-1}=g}}f\left(  h\right)  , \label{sol.Indrho.generalize.Indf}%
\end{equation}
where $J$ is a system of left coset representatives for $G/\rho\left(
H\right)  $ (so that $G=\bigsqcup_{j\in J}j\rho\left(  H\right)  $). We leave
it to the reader to show that this definition is still well-defined (i.e.,
that $\sum\limits_{\substack{\left(  h,k\right)  \in H\times J;\\k\rho\left(
h\right)  k^{-1}=g}}f\left(  h\right)  $ is independent of the choice of $J$),
and that it defines the same function $\Ind_{\rho}f$
as the definition made in Exercise \ref{exe.Indrho} when $\left\vert
H\right\vert $ is invertible in the base ring. To my knowledge, there exists
no reasonable definition of $\Ind_{\rho}f$ that
completely avoids making any requirements on the base ring.

\item In Exercise \ref{exe.Indrho}, we defined a $\CC G$-module
$\Ind_{\rho}U$ for every $\CC H$-module $U$.
This definition is still valid when $\CC$ is replaced by a commutative
ring. (Notice the slightly surprising fact that if $\CC$ is replaced by
a commutative ring, then the $\rho$-induction of a $\CC H$-module is
always well-defined, whereas the $\rho$-induction of a class function might
not be.)

\item Exercise \ref{exe.Indrho}(a) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. And the solution that we gave still applies in this generality.
More generally, if we define $\Ind_{\rho}f$ by
(\ref{sol.Indrho.generalize.Indf}), then Exercise \ref{exe.Indrho}(a) is still
valid when $\CC$ is replaced by any commutative ring in which
$\left\vert \ker\rho\right\vert $ is invertible. However, this
can no longer be proven by blindly copying our solution.

\item Exercise \ref{exe.Indrho}(b) is still valid when $\CC$ is
replaced by any field in which $\left\vert H\right\vert $ is invertible.
Again, our solution is still valid in this generality.\footnote{We can also
replace $\CC$ by a commutative ring $A$ rather than by a field; but
then we need to replace ``finite-dimensional $\CC H$-module $U$''
by ``$AH$-module $U$ which is
a finitely generated projective $A$-module'' in the statement
of the exercise. The ``finitely generated projective
$A$-module'' condition is needed in order to make sure that
$\chi_{U}$ and $\chi_{\Ind_{\rho}U}$ are well-defined.
(See the Remark after the proof of Proposition
\ref{prop.sol.Indrho.b.dualgens.Tr} for details.)} More generally, Exercise
\ref{exe.Indrho}(b) is still valid when $\CC$ is replaced by any field
in which $\left\vert \ker\rho\right\vert $ is invertible (as
long as $\Ind_{\rho}f$ is defined through
(\ref{sol.Indrho.generalize.Indf})); but our solution is not sufficient to
prove this\footnote{However, a certain modification of this solution does the
trick. Namely, we replace the definition of $\widetilde{F}_{g}$ by%
\[
\left(  \widetilde{F}_{g}\left(  \gamma,u\right)  =\dfrac{1}{\left\vert
\ker\rho\right\vert }\sum\limits_{h\in H}\left(  g\rho\left(
h\right)  \right)  ^{\ast}\left(  \gamma\right)
hu\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  \gamma,u\right)  \in
\CC G\times U\right)  .
\]
Then, we should pick a system $J^{\prime}$ of left coset representatives for
$G/\rho\left(  H\right)  $ (so that $G=\bigsqcup_{j\in J^{\prime}}j\rho\left(
H\right)  $). (Notice that we do not call it $J$ because the letter $J$
already has a different meaning in this solution.) Then, we define a family
$\left(  a_{i}\right)  _{i\in J^{\prime}\times J}$ of elements of
$\CC G\otimes_{\CC H}U$ by%
\[
\left(  a_{\left(  k,j\right)  }=k\otimes_{\CC H}b_{j}%
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  k,j\right)  \in J^{\prime}\times
J\right)  ,
\]
and a family $\left(  f_{i}\right)  _{i\in J^{\prime}\times J}$ of elements of
$\left(  \CC G\otimes_{\CC H}U\right)  ^{\ast}$ by%
\[
\left(  f_{\left(  k,j\right)  }=g_{j}\circ F_{k}\ \ \ \ \ \ \ \ \ \ \text{for
all }\left(  k,j\right)  \in J^{\prime}\times J\right)  .
\]
(Notice that these families $\left(  a_{i}\right)  _{i\in J^{\prime}\times J}$
and $\left(  f_{i}\right)  _{i\in J^{\prime}\times J}$ are subfamilies of the
families $\left(  a_{i}\right)  _{i\in G\times J}$ and $\left(  f_{i}\right)
_{i\in G\times J}$ from the original solution to Exercise \ref{exe.Indrho}%
(b).) It can then be shown that $\left(  J^{\prime}\times J,\left(
a_{i}\right)  _{i\in J^{\prime}\times J},\left(  f_{i}\right)  _{i\in
J^{\prime}\times J}\right)  $ is a finite dual generating system for
$\Ind_{\rho}U$ (though the proof is somewhat different
from the way this was shown in the solution to Exercise \ref{exe.Indrho}(b)).
This allows us to use Proposition \ref{prop.sol.Indrho.b.dualgens.Tr} to
derive $\chi_{\Ind_{\rho}U}=\operatorname*{Ind}%
\nolimits_{\rho}\chi_{U}$. All details are left to the reader.}. Even more
generally, Exercise \ref{exe.Indrho}(b) holds whenever $\CC$ is
replaced by any \textbf{commutative ring} $A$ in which $\left\vert
\ker\rho\right\vert $ is invertible, as long as $U$ is assumed
to be a finitely generated projective $A$-module. However, in order to make
sense of this statement, one needs to know that $\operatorname*{Ind}%
\nolimits_{\rho}U$ is a finitely generated projective $A$-module as well, and
one needs to know how the trace of an endomorphism of a finitely generated
projective $A$-module is defined. We essentially did most of this in our
solution above\footnote{The only missing link is the fact that an $A$-module
is finitely generated and projective if and only if it has a finite dual
generating system; this fact is easy to prove.}.

\item Exercise \ref{exe.Indrho}(c) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. Again, our solution is still valid in this generality. Again, if
we define $\Ind_{\rho}f$ by
(\ref{sol.Indrho.generalize.Indf}) and define $\operatorname*{Ind}%
\nolimits_{H}^{G}f$ by Exercise \ref{exe.induction.characters}(b), then
Exercise \ref{exe.Indrho}(c) even holds when $\CC$ is replaced by any
arbitrary commutative ring. (The definition of $\operatorname*{Ind}%
\nolimits_{\rho}f$ requires $\left\vert \ker\rho\right\vert $
to be invertible, but this is automatically satisfied since $\rho$ is injective.)

\item Exercise \ref{exe.Indrho}(d) is still valid when $\CC$ is
replaced by any commutative ring whatsoever. Again, our solution is still
valid in this generality.

\item Exercise \ref{exe.Indrho}(e) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert H\right\vert $ is
invertible. Again, our solution is still valid in this generality. Again, it
is possible to prove Exercise \ref{exe.Indrho}(e) also when $\CC$ is
replaced by any commutative ring in which $\left\vert \ker%
\rho\right\vert $ is invertible (as long as $\operatorname*{Ind}%
\nolimits_{\rho}f$ is defined using (\ref{sol.Indrho.generalize.Indf})). (This
is actually very easy to check -- arguably even easier than our above solution
of Exercise \ref{exe.Indrho}(e).)

\item Exercise \ref{exe.Indrho}(f) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert K\right\vert $ is
invertible. All three solutions of Exercise \ref{exe.Indrho}(f) that we have
given above still remain valid in this generality.

\item The second claim of Exercise \ref{exe.Indrho}(g) (that is, the equality
(\ref{eq.exe.Indrho.g.bil})) is still valid when $\CC$ is replaced by
any commutative ring in which $\left\vert G\right\vert $ and $\left\vert
H\right\vert $ are invertible. The first claim, a priori, makes no sense when
$\CC$ is replaced by an arbitrary commutative ring, because the
definition of the Hermitian forms $\left(  \cdot,\cdot\right)  _{G}$ and
$\left(  \cdot,\cdot\right)  _{H}$ involves complex conjugation (which is only
defined on $\CC$). However, it turns out that this complex conjugation
can be replaced by \textbf{any map from the base ring to itself} and Exercise
\ref{exe.Indrho}(g) remains valid\footnote{More precisely: Let $A$ be any
commutative ring in which $\left\vert G\right\vert $ and $\left\vert
H\right\vert $ are invertible. Fix an arbitrary map $\operatorname*{conj}%
:A\rightarrow A$ (not necessarily linear). For every $a\in A$, let
$\overline{a}$ denote the image $\operatorname*{conj}\left(  a\right)  $ of
$a$ under this map. For any $f_{1}\in R_{A}\left(  G\right)  $ and $f_{2}\in
R_{A}\left(  G\right)  $, define $\left(  f_{1},f_{2}\right)  _{G}\in A$ by
$\left(  f_{1},f_{2}\right)  _{G}=\dfrac{1}{\left\vert G\right\vert }%
\sum_{g\in G}f_{1}\left(  g\right)  \overline{f_{2}\left(  g\right)  }$ (using
the notation $\overline{a}$ that we just introduced). Similarly, define
$\left(  f_{1},f_{2}\right)  _{H}\in A$ for any $f_{1}\in R_{A}\left(
H\right)  $ and $f_{2}\in R_{A}\left(  H\right)  $. Then, Exercise
\ref{exe.Indrho}(g) is still valid when $\CC$ is replaced by $A$. (And
our solution to this exercise still applies.)}.

\item Exercise \ref{exe.Indrho}(h) is still valid when $\CC$ is
replaced by any commutative ring whatsoever. Again, our solution is still
valid in this generality.

\item Exercise \ref{exe.Indrho}(i) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert \ker%
\rho\right\vert $ is invertible. Again, our solution is still valid in this
generality (because if we set $K=\ker\rho$ as we did in our
solution, then $\left\vert \underbrace{K}_{=\ker\rho
}\right\vert =\left\vert \ker\rho\right\vert $ is invertible).

\item Exercise \ref{exe.Indrho}(j) is still valid when $\CC$ is
replaced by any commutative ring in which $\left\vert \ker%
\rho\right\vert $ is invertible.

\item Exercise \ref{exe.Indrho}(k) still gives a proof of the formula
(\ref{induced-character-formula}) when $\CC$ is replaced by any field
in which $\left\vert H\right\vert $ is invertible.

\item Exercise \ref{exe.Indrho}(l) still gives a proof of the formula
(\ref{deflation-character-formula}) when $\CC$ is replaced by any field
in which $\left\vert G\right\vert $ and $\left\vert K\right\vert $ are invertible.
\end{itemize}

We actually tailored our above solutions to apply in a reasonably high
generality (although we did not state Exercise \ref{exe.Indrho} in this
generality). Had we contented ourselves with solving no more and no less than
what Exercise \ref{exe.Indrho} demanded of us, we could have obtained much
shorter solutions of some parts of it:

\begin{itemize}
\item Exercise \ref{exe.Indrho}(b) can be obtained by combining parts (g) and
(h) of Exercise \ref{exe.Indrho} with the following two
observations\footnote{We only sketch this argument; the details are left to
the reader.}:

\begin{itemize}
\item Every $\CC G$-module $V$ satisfies
$\chi_{\Res_{\rho}V}=\Res_{\rho}\chi_{V}$. (This is obvious.)

\item If $\alpha$ and $\beta$ are two elements of $R_{\CC }\left(
G\right)  $ such that every finite-dimensional $\CC G$-module $V$
satisfies $\left(  \alpha,\chi_{V}\right)  _{G}=\left(  \beta,\chi_{V}\right)
_{G}$, then $\alpha=\beta$. (This follows from the fact that $R\left(
G\right)  $ spans the $\CC$-vector space $R_{\CC }\left(
G\right)  $ and from the fact that the Hermitian form $\left(  \cdot
,\cdot\right)  _{G}$ is nondegenerate.)
\end{itemize}

However, this quick argument only works when the base ring is $\CC$. It
cannot be easily generalized to the case when $\CC$ is replaced by any
field in which $\left\vert H\right\vert $ is invertible.

\item Part (e) (at least in the case when $U$ is finite-dimensional) and part
(f) of Exercise \ref{exe.Indrho} can be derived from each other using the
correspondence between irreducible representations and irreducible characters.
Again, this argument is quick but hard to generalize.
\end{itemize}

]
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Indrho.tensor}}

\textit{Solution to Exercise \ref{exe.Indrho.tensor}.} We shall use the
following fact:

\begin{proposition}
\label{prop.sol.Indrho.tensor.general}Let $\kk$ be a commutative ring.
Let $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be six
$\kk$-algebras. Let $P$ be an $\left(  A,B\right)  $%
-bimodule\footnote{As usual, we understand the notion of a bimodule to be
defined over $\kk$; that is, the left $A$-module structure and the
right $B$-module structure of an $\left(  A,B\right)  $-bimodule must restrict
to one and the same $\kk$-module structure.}. Let $Q$ be a $\left(
B,C\right)  $-bimodule. Let $P^{\prime}$ be an $\left(  A^{\prime},B^{\prime
}\right)  $-bimodule. Let $Q^{\prime}$ be a $\left(  B^{\prime},C^{\prime
}\right)  $-bimodule. Then,%
\[
\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
Q\otimes Q^{\prime}\right)  \cong\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
\]
as $\left(  A\otimes A^{\prime},C\otimes C^{\prime}\right)  $-bimodules. Here,
all $\otimes$ signs without subscript stand for $\otimes_{\kk}$.
\end{proposition}

Proposition \ref{prop.sol.Indrho.tensor.general} is identical with the
Proposition \ref{prop.sol.ind.tensor.general} that appeared in our solution to
Exercise \ref{exe.ind.tensor}. It is proven by straightforward (repeated) use
of the universal property of the tensor product.

We have a canonical $\CC$-algebra isomorphism%
\[
\mathfrak{H}:\mathbb{C}\left[  H_{1}\times H_{2}\right]  \rightarrow
\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}%
\]
which satisfies%
\[
\mathfrak{H}\left(  t_{\left(  h_{1},h_{2}\right)  }\right)  =t_{h_{1}}\otimes
t_{h_{2}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  h_{1},h_{2}\right)  \in
H_{1}\times H_{2}.
\]
We also have a canonical $\CC$-algebra isomorphism%
\[
\mathfrak{G}:\mathbb{C}\left[  G_{1}\times G_{2}\right]  \rightarrow
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}%
\]
which satisfies%
\[
\mathfrak{G}\left(  t_{\left(  g_{1},g_{2}\right)  }\right)  =t_{g_{1}}\otimes
t_{g_{2}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}.
\]


Recall how the left $\mathbb{C}\left[  H_{1}\times H_{2}\right]  $-module
$U_{1}\otimes U_{2}$ is defined: Its left $\mathbb{C}\left[  H_{1}\times
H_{2}\right]  $-module structure is given by%
\[
\left(  t_{\left(  h_{1},h_{2}\right)  }\left(  u_{1}\otimes u_{2}\right)
=t_{h_{1}}u_{1}\otimes t_{h_{2}}u_{2}\ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  h_{1},h_{2}\right)  \in H_{1}\times H_{2}\text{ and }\left(
u_{1},u_{2}\right)  \in U_{1}\times U_{2}\right)  .
\]


We shall now recall the definitions of $\operatorname*{Ind}\nolimits_{\rho
_{1}}U_{1}$, $\operatorname*{Ind}\nolimits_{\rho_{2}}U_{2}$ and
$\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(  U_{1}\otimes U_{2}\right)
$:

\begin{itemize}
\item We defined $\operatorname{Ind}_{\rho_{1}}\left(  U_{1}\right)  $ as the
${\mathbb{C}}G_{1}$-module ${\mathbb{C}}G_{1}\otimes_{\CC H_{1}}U_{1}%
$, where ${\mathbb{C}}G_{1}$ is regarded as a $\left(  {\mathbb{C}}%
G_{1},{\mathbb{C}}H_{1}\right)  $-bimodule according to the following rule:
The left ${\mathbb{C}}G_{1}$-module structure on ${\mathbb{C}}G_{1}$ is plain
multiplication inside ${\mathbb{C}}G_{1}$; the right ${\mathbb{C}}H_{1}%
$-module structure on ${\mathbb{C}}G_{1}$ is induced by the ${\mathbb{C}}%
$-algebra homomorphism ${\mathbb{C}}\left[  \rho_{1}\right]  :{\mathbb{C}%
}H_{1}\rightarrow{\mathbb{C}}G_{1}$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho_{1}\right]  \right)
\eta$ for all $\gamma\in{\mathbb{C}}G_{1}$ and $\eta\in{\mathbb{C}}H_{1}$).

\item We defined $\operatorname{Ind}_{\rho_{2}}\left(  U_{2}\right)  $ as the
${\mathbb{C}}G_{2}$-module ${\mathbb{C}}G_{2}\otimes_{\CC H_{2}}U_{2}%
$, where ${\mathbb{C}}G_{2}$ is regarded as a $\left(  {\mathbb{C}}%
G_{2},{\mathbb{C}}H_{2}\right)  $-bimodule in a similar fashion.

\item We defined $\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(
U_{1}\otimes U_{2}\right)  $ as the ${\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  $-module ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]
\otimes_{\CC \left[  H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes
U_{2}\right)  $, where ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $ is
regarded as a $\left(  {\mathbb{C}}\left[  G_{1}\times G_{2}\right]
,{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  \right)  $-bimodule in a
similar fashion.
\end{itemize}

Proposition \ref{prop.sol.ind.tensor.general} (applied to $\kk%
={\mathbb{C}}$, $A={\mathbb{C}}G_{1}$, $B={\mathbb{C}}H_{1}$, $C={\mathbb{C}}%
$, $A^{\prime}={\mathbb{C}}G_{2}$, $B^{\prime}={\mathbb{C}}H_{2}$, $C^{\prime
}={\mathbb{C}}$, $P={\mathbb{C}}G_{1}$, $Q=U_{1}$, $P^{\prime}={\mathbb{C}%
}G_{2}$ and $Q^{\prime}=U_{2}$) yields%
\[
\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  {\mathbb{C}}G_{1}\otimes_{\CC H_{1}}U_{1}\right)
\otimes\left(  {\mathbb{C}}G_{2}\otimes_{\CC H_{2}}U_{2}\right)
\]
as $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2},{\mathbb{C}}%
\otimes{\mathbb{C}}\right)  $-bimodules, hence also as left ${\mathbb{C}}%
G_{1}\otimes{\mathbb{C}}G_{2}$-modules. Thus,%
\begin{align}
&  \left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\nonumber\\
&  \cong\underbrace{\left(  {\mathbb{C}}G_{1}\otimes_{\CC H_{1}}%
U_{1}\right)  }_{=\operatorname{Ind}_{\rho_{1}}U_{1}}\otimes
\underbrace{\left(  {\mathbb{C}}G_{2}\otimes_{\CC H_{2}}U_{2}\right)
}_{=\operatorname{Ind}_{\rho_{2}}U_{2}}\nonumber\\
&  =\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)
\label{sol.Indrho.tensor.finish-part-1}%
\end{align}
as left ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$-modules.

At this point, there is a quick way to finish the solution using handwaving:
We use the $\CC$-algebra isomorphism $\mathfrak{H}:\mathbb{C}\left[
H_{1}\times H_{2}\right]  \rightarrow\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$ to
identify the $\CC$-algebra $\mathbb{C}\left[  H_{1}\times H_{2}\right]
$ with the $\CC$-algebra $\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$, and
we use the $\CC$-algebra isomorphism $\mathfrak{G}:\mathbb{C}\left[
G_{1}\times G_{2}\right]  \rightarrow\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ to
identify the $\CC$-algebra $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$ with the $\CC$-algebra $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$. It is
``easy to see'' that these two identifications ``play nicely with each other
and with the module structures on $U_{1}\otimes U_{2}$'' (this is the part
where we wave our hands). Now, (\ref{sol.Indrho.tensor.finish-part-1}) yields
that $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)  $ as left ${\mathbb{C}}G_{1}%
\otimes{\mathbb{C}}G_{2}$-modules, and therefore also as left $\mathbb{C}%
\left[  G_{1}\times G_{2}\right]  $-modules (since $\mathbb{C}\left[
G_{1}\times G_{2}\right]  ={\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$). Now,%
\begin{align*}
\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(  U_{1}\otimes U_{2}\right)
&  ={\mathbb{C}}\left[  G_{1}\times G_{2}\right]  \otimes_{\CC \left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right) \\
&  =\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }{\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  ={\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\text{ and
}{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  ={\mathbb{C}}H_{1}%
\otimes{\mathbb{C}}H_{2}\right) \\
&  \cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)
\end{align*}
as left ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $-modules. This solves
Exercise \ref{exe.Indrho.tensor} if you believe the handwaving I have done above.

The handwaving we have done is slightly questionable, since we have made two
identifications which turned two isomorphisms into identities. In reality,
they are merely isomorphisms, not identities, and it is not immediately clear
that regarding them as identities will not lead to contradictions. (Indeed,
this is the meaning of our vague claim that the two identifications
``play nicely with each other''.)

Let us now show a way to formalize the above questionable argument. We shall
focus on explaining how to do this proof in a clean fashion (in particular, we
shall avoid identifying any things that are not already identical; instead, we
will work with the isomorphisms $\mathfrak{G}$ and $\mathfrak{H}$ explicitly);
we will leave straightforward computations and arguments to the reader.

We first state a general fact:

\begin{proposition}
\label{prop.sol.Indrho.tensor.MAN}Let $\kk$ be a commutative ring. In
the following, all $\otimes$ signs without subscript stand for $\otimes
_{\kk}$.

Let $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be six
$\kk$-algebras. Let $M$ be an $\left(  A,B\right)  $%
-bimodule\footnote{As usual, we understand the notion of a bimodule to be
defined over $\kk$; that is, the left $A$-module structure and the
right $B$-module structure of an $\left(  A,B\right)  $-bimodule must restrict
to one and the same $\kk$-module structure.}. Let $N$ be a $\left(
B,C\right)  $-bimodule. Let $M^{\prime}$ be an $\left(  A^{\prime},B^{\prime
}\right)  $-bimodule. Let $N^{\prime}$ be a $\left(  B^{\prime},C^{\prime
}\right)  $-bimodule. Let $\beta:B\rightarrow B^{\prime}$, $\mu:M\rightarrow
M^{\prime}$ and $\nu:N\rightarrow N^{\prime}$ be three $\kk$-module
homomorphisms. Assume that we have%
\begin{equation}
\left(  \mu\left(  mb\right)  =\mu\left(  m\right)  \beta\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B\text{ and }m\in M\right)
\label{eq.prop.sol.Indrho.tensor.MAN.ass1}%
\end{equation}
and%
\begin{equation}
\left(  \nu\left(  bn\right)  =\beta\left(  b\right)  \nu\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B\text{ and }n\in N\right)  .
\label{eq.prop.sol.Indrho.tensor.MAN.ass2}%
\end{equation}
Then:

\begin{itemize}
\item[(a)] There exists a unique $\kk$-module homomorphism%
\[
\Omega:M\otimes_{B}N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}%
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{B}n\right)  =\mu\left(  m\right)
\otimes_{B^{\prime}}\nu\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  m,n\right)  \in M\times N\right)  .
\]


We shall denote this homomorphism $\Omega$ by $\Omega_{\beta,\mu,\nu}$.

\item[(b)] If the maps $\beta$, $\mu$ and $\nu$ are invertible,
then $\Omega_{\beta,\mu,\nu}$ is a $\kk$-module isomorphism.

\item[(c)] Let $\alpha:A\rightarrow A^{\prime}$ be a $\kk$-module
homomorphism. Assume that%
\begin{equation}
\mu\left(  am\right)  =\alpha\left(  a\right)  \mu\left(  m\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in A\text{ and }m\in
M.\label{eq.prop.sol.Indrho.tensor.MAN.ass3}%
\end{equation}
Assume also that the $\kk$-module $M^{\prime}$ is endowed with a left
$A$-module structure. Assume that this left $A$-module structure on
$M^{\prime}$ and the right $B^{\prime}$-module structure on $M^{\prime}$
together form an $\left(  A,B^{\prime}\right)  $-bimodule structure on
$M^{\prime}$. Thus, $M^{\prime}\otimes_{B^{\prime}}N^{\prime}$ becomes a left
$A$-module. Assume furthermore that%
\begin{equation}
am=\alpha\left(  a\right)  m\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{
and }m\in M^{\prime}\label{eq.prop.sol.Indrho.tensor.MAN.ass4}%
\end{equation}
\footnote{Here the $\alpha\left(  a\right)  m$ on the right hand side is
defined using the action of $A^{\prime}$ on the left $A^{\prime}$-module
$M^{\prime}$, whereas the $am$ on the left hand side is defined using the
action of $A$ on the left $A$-module $M^{\prime}$.}. Then, $\Omega_{\beta
,\mu,\nu}$ is a left $A$-module homomorphism.
\end{itemize}
\end{proposition}

The proof of Proposition \ref{prop.sol.Indrho.tensor.MAN} is
straightforward\footnote{For part (b), the inverse of $\Omega_{\beta,\mu,\nu}$
is $\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}$, of course.} and is left to the
reader. We could also add a part (d) to Proposition
\ref{prop.sol.Indrho.tensor.MAN}, which would give a criterion for
$\Omega_{\beta,\mu,\nu}$ to be a right $C$-module homomorphism given an
appropriate right $C$-module structure on $N^{\prime}$.

Let us now return to solving Exercise \ref{exe.Indrho.tensor}. We have four bimodules:

\begin{itemize}
\item The $\CC$-vector space $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ is
a $\left(  \mathbb{C}G_{1}\otimes\mathbb{C}G_{2},\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}\right)  $-bimodule (because it is the tensor product of
the $\left(  \mathbb{C}G_{1},\mathbb{C}H_{1}\right)  $-bimodule $\mathbb{C}%
G_{1}$ with the $\left(  \mathbb{C}G_{2},\mathbb{C}H_{2}\right)  $-bimodule
$\mathbb{C}G_{2}$). As a left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-module,
it is thus the tensor product of the left $\mathbb{C}G_{1}$-module
$\mathbb{C}G_{1}$ with the left $\mathbb{C}G_{2}$-module $\mathbb{C}G_{2}$.
Hence, its left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-module structure is
given by plain multiplication inside $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}%
$\ \ \ \ \footnote{That is, the action of any $a\in\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$ on any element $m$ of the left $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$-module $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ equals
the product of $a$ with $m$ in the $\CC$-algebra $\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$.}. (This is straightforward to check.)

\item The $\CC$-vector space $U_{1}\otimes U_{2}$ is a left
$\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$-module (because it is the tensor
product of the left $\mathbb{C}H_{1}$-module $U_{1}$ with the left
$\mathbb{C}H_{2}$-module $U_{2}$), and thus a $\left(  \mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2},\mathbb{C}\right)  $-bimodule.

\item The $\CC$-vector space ${\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  $ is a $\left(  {\mathbb{C}}\left[  G_{1}\times G_{2}\right]
,{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  \right)  $-bimodule (as we
already know).

\item The $\CC$-vector space $U_{1}\otimes U_{2}$ is a left
${\mathbb{C}}\left[  H_{1}\times H_{2}\right]  $-module (since it is the
tensor product of the left $\mathbb{C}H_{1}$-module $U_{1}$ with the left
$\mathbb{C}H_{2}$-module $U_{2}$), and thus a $\left(  {\mathbb{C}}\left[
H_{1}\times H_{2}\right]  ,\mathbb{C}\right)  $-bimodule.
\end{itemize}

We have%
\begin{equation}
\left(  \mathfrak{G}\left(  mb\right)  =\mathfrak{G}\left(  m\right)
\mathfrak{H}\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }%
b\in\mathbb{C}\left[  H_{1}\times H_{2}\right]  \text{ and }m\in
\mathbb{C}\left[  G_{1}\times G_{2}\right]  \right)
\label{sol.Indrho.tensor.axiom1}%
\end{equation}
and%
\begin{equation}
\left(  \operatorname*{id}\left(  bn\right)  =\mathfrak{H}\left(  b\right)
\operatorname*{id}\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }%
b\in\mathbb{C}\left[  H_{1}\times H_{2}\right]  \text{ and }n\in U_{1}\otimes
U_{2}\right)  .\label{sol.Indrho.tensor.axiom2}%
\end{equation}
(In fact, both of these equalities are easily checked on basis elements and
pure tensors.) Hence, we can apply Proposition
\ref{prop.sol.Indrho.tensor.MAN}(a) to $\kk=\mathbb{C}$, $A=\mathbb{C}%
\left[  G_{1}\times G_{2}\right]  $, $B=\mathbb{C}\left[  H_{1}\times
H_{2}\right]  $, $C=\mathbb{C}$, $A^{\prime}=\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$, $B^{\prime}=\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$,
$C^{\prime}=\mathbb{C}$, $M=\mathbb{C}\left[  G_{1}\times G_{2}\right]  $,
$N=U_{1}\otimes U_{2}$, $M^{\prime}=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$,
$N^{\prime}=U_{1}\otimes U_{2}$, $\beta=\mathfrak{H}$, $\mu=\mathfrak{G}$ and
$\nu=\operatorname*{id}$. As a consequence, we conclude that there exists a
unique $\CC$-module homomorphism%
\[
\Omega:\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \rightarrow
\left(  \mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\right)  \otimes_{\mathbb{C}%
H_{1}\otimes\mathbb{C}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{\mathbb{C}\left[  H_{1}\times H_{2}\right]
}n\right)
= \mathfrak{G} \left(  m\right)  \otimes_{\mathbb{C}H_{1}\otimes
\mathbb{C}H_{2}} \id \left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  m,n\right)  \in\mathbb{C}\left[  G_{1}\times G_{2}\right]
\times\left(  U_{1}\otimes U_{2}\right)  \right)  .
\]
According to Proposition \ref{prop.sol.Indrho.tensor.MAN}(a), this
homomorphism $\Omega$ is denoted by $\Omega_{\mathfrak{H},\mathfrak{G}%
,\operatorname*{id}}$.

The maps $\mathfrak{H}$, $\mathfrak{G}$ and $\operatorname*{id}$ are
$\CC$-algebra isomorphisms, and therefore invertible. Hence,
Proposition \ref{prop.sol.Indrho.tensor.MAN}(b) (applied to $\kk%
=\mathbb{C}$, $A=\mathbb{C}\left[  G_{1}\times G_{2}\right]  $, $B=\mathbb{C}%
\left[  H_{1}\times H_{2}\right]  $, $C=\mathbb{C}$, $A^{\prime}%
=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $B^{\prime}=\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}$, $C^{\prime}=\mathbb{C}$, $M=\mathbb{C}\left[
G_{1}\times G_{2}\right]  $, $N=U_{1}\otimes U_{2}$, $M^{\prime}%
=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $N^{\prime}=U_{1}\otimes U_{2}$,
$\beta=\mathfrak{H}$, $\mu=\mathfrak{G}$ and $\nu=\operatorname*{id}$) yields
that $\Omega_{\mathfrak{H},\mathfrak{G},\operatorname*{id}}$ is a $\mathbb{C}%
$-module isomorphism.

Next, we notice that the $\CC$-vector space $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$ is a left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module (since it is the tensor product of the left $\mathbb{C}G_{1}$-module
$\mathbb{C}G_{1}$ with the left $\mathbb{C}G_{2}$-module $\mathbb{C}G_{2}$).
This left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module structure is
defined by the rule%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }\left(  u_{1}\otimes u_{2}\right)  =t_{g_{1}%
}u_{1}\otimes t_{g_{2}}u_{2}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
g_{1},g_{2}\right)  \in G_{1}\times G_{2}\text{ and }\left(  u_{1}%
,u_{2}\right)  \in\mathbb{C}G_{1}\times\mathbb{C}G_{2}%
.\label{sol.Indrho.tensor.left-mod}%
\end{equation}
This left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module structure on
$\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ and the right $\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}$-module structure on $\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$ are connected by the equality%
\begin{equation}
\left(  \left(  gm\right)  h=g\left(  mh\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }g\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  \text{, }m\in
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\text{ and }h\in\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}\right)  \label{sol.Indrho.tensor.bimod}%
\end{equation}
(which is easily checked). Hence, these two structures together form an
$\left(  \mathbb{C}\left[  G_{1}\times G_{2}\right]  ,\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}\right)  $-bimodule structure on $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$ (because both of these structures are $\mathbb{C}%
$-bilinear). Thus, the tensor product $\left(  \mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}\right)  \otimes_{\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}}\left(
U_{1}\otimes U_{2}\right)  $ becomes a left $\mathbb{C}\left[  G_{1}\times
G_{2}\right]  $-module.

Next, it is easy to check that%
\begin{equation}
\mathfrak{G}\left(  am\right)  =\mathfrak{G}\left(  a\right)  \mathfrak{G}%
\left(  m\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathbb{C}\left[
G_{1}\times G_{2}\right]  \text{ and }m\in\mathbb{C}\left[  G_{1}\times
G_{2}\right]  .\label{sol.Indrho.tensor.axiom3}%
\end{equation}
It is also easy to see that%
\begin{equation}
am=\mathfrak{G}\left(  a\right)  m\ \ \ \ \ \ \ \ \ \ \text{for every }%
a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  \text{ and }m\in
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}.\label{sol.Indrho.tensor.axiom4}%
\end{equation}


Thus, we can apply Proposition \ref{prop.sol.Indrho.tensor.MAN}(c) to
$\kk=\mathbb{C}$, $A=\mathbb{C}\left[  G_{1}\times G_{2}\right]  $,
$B=\mathbb{C}\left[  H_{1}\times H_{2}\right]  $, $C=\mathbb{C}$, $A^{\prime
}=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $B^{\prime}=\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}$, $C^{\prime}=\mathbb{C}$, $M=\mathbb{C}\left[
G_{1}\times G_{2}\right]  $, $N=U_{1}\otimes U_{2}$, $M^{\prime}%
=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $N^{\prime}=U_{1}\otimes U_{2}$,
$\beta=\mathfrak{H}$, $\mu=\mathfrak{G}$, $\nu=\operatorname*{id}$ and
$\alpha=\mathfrak{G}$. As a result, we obtain that $\Omega_{\mathfrak{H}%
,\mathfrak{G},\operatorname*{id}}$ is a left $\mathbb{C}\left[  G_{1}\times
G_{2}\right]  $-module homomorphism. Hence, $\Omega_{\mathfrak{H}%
,\mathfrak{G},\operatorname*{id}}$ is a left $\mathbb{C}\left[  G_{1}\times
G_{2}\right]  $-module isomorphism (since $\Omega_{\mathfrak{H},\mathfrak{G}%
,\operatorname*{id}}$ is invertible). Thus,%
\begin{equation}
\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \cong\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\label{sol.Indrho.tensor.finish-part-2}%
\end{equation}
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules.

Now, let us recall the isomorphism (\ref{sol.Indrho.tensor.finish-part-1}). It
is an isomorphism of left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-modules, and
thus cannot be immediately combined with
(\ref{sol.Indrho.tensor.finish-part-2}). However, it is easy to see that the
corresponding isomorphism of left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-modules holds as well: Namely, we have%
\begin{equation}
\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)
\label{sol.Indrho.tensor.finish-part-1a}%
\end{equation}
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. Before we
prove this, let us make three auxiliary observations which connect the left
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module structures on the
modules appearing in (\ref{sol.Indrho.tensor.finish-part-1a}) with the left
${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$-module structures on the same modules:

\begin{itemize}
\item We have%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }m=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
m\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}\text{ and }m\in{\mathbb{C}}G_{1}\otimes{\mathbb{C}}%
G_{2}\label{sol.Indrho.tensor.CG1CG2CG1G2-1}%
\end{equation}
(where the expression $t_{\left(  g_{1},g_{2}\right)  }m$ on the left hand
side is defined using the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$, whereas the
expression $\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  m$ on the right hand
side is defined using the left ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}%
$-module structure on ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$). This is
easy to prove.

\item We have%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }n=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
n\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}\text{ and }n\in\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}%
}G_{2}\right)  \otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(
U_{1}\otimes U_{2}\right)  \label{sol.Indrho.tensor.CG1CG2CG1G2-2}%
\end{equation}
(where the expression $t_{\left(  g_{1},g_{2}\right)  }n$ on the left hand
side is defined using the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}%
G_{2}\right)  \otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(
U_{1}\otimes U_{2}\right)  $, whereas the expression $\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  n$ on the right hand side is defined using the left
${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$-module structure on $\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $). This is
easy to prove using (\ref{sol.Indrho.tensor.CG1CG2CG1G2-1}).

\item Finally, we have%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }m=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
m\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}\text{ and }m\in\left(  \operatorname{Ind}_{\rho_{1}}%
U_{1}\right)  \otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)
\label{sol.Indrho.tensor.CG1CG2CG1G2-3}%
\end{equation}
(where the expression $t_{\left(  g_{1},g_{2}\right)  }m$ on the left hand
side is defined using the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $, whereas the
expression $\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  m$ on the right hand
side is defined using the left ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}%
$-module structure on $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $). Again, this is
easy to check.
\end{itemize}

Now, it is easy to see that (\ref{sol.Indrho.tensor.finish-part-1a})
holds\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.finish-part-1a}):}
From (\ref{sol.Indrho.tensor.finish-part-1}), we conclude that there exists an
isomorphism%
\[
\mathbf{T}:\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)
\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes
U_{2}\right)  \rightarrow\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)
\]
of left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-modules. We shall now show
that this map $\mathbf{T}$ is an isomorphism of left $\mathbb{C}\left[
G_{1}\times G_{2}\right]  $-modules as well. In order to do so, it is
sufficient to show that the map $\mathbf{T}$ is a homomorphism of left
$\mathbb{C}\left[ G_{1}\times G_{2}\right]  $-modules
(because we already know that $\mathbf{T}$ is
invertible). In other words, it is sufficient to show
that%
\begin{equation}
\mathbf{T}\left(  an\right)  =a\mathbf{T}\left(  n\right)
\label{sol.Indrho.tensor.finish-part-1a.pf.1}
\end{equation}
for every $a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $ and $n\in\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $
(since we already know that the map $\mathbf{T}$ is $\CC$-linear).
\par
\textit{Proof of (\ref{sol.Indrho.tensor.finish-part-1a.pf.1}):} Let
$a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $ and $n\in\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $. We need to
prove the equality $\mathbf{T}\left(  an\right)  =a\mathbf{T}\left(  n\right)
$. This equality is $\CC$-linear in $a$. Hence, we can WLOG assume that
$a$ belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)  }\right)
_{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\mathbb{C}%
$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. Assume this.
Thus, $a=t_{\left(  g_{1},g_{2}\right)  }$ for some $\left(  g_{1}%
,g_{2}\right)  \in G_{1}\times G_{2}$. Consider this $\left(  g_{1}%
,g_{2}\right)  $. We have $\underbrace{a}_{=t_{\left(  g_{1},g_{2}\right)  }%
}n=t_{\left(  g_{1},g_{2}\right)  }n=\left(  t_{g_{1}}\otimes t_{g_{2}%
}\right)  n$ (by (\ref{sol.Indrho.tensor.CG1CG2CG1G2-2})). Applying the map
$\mathbf{T}$ to both sides of this equality, we obtain%
\[
\mathbf{T}\left(  an\right)  =\mathbf{T}\left(  \left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  n\right)  =\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
\mathbf{T}\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{T}\text{ is a homomorphism of left }\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}\text{-modules}\right)  .
\]
Compared with%
\[
\underbrace{a}_{=t_{\left(  g_{1},g_{2}\right)  }}\mathbf{T}\left(  n\right)
=t_{\left(  g_{1},g_{2}\right)  }\mathbf{T}\left(  n\right)  =\left(
t_{g_{1}}\otimes t_{g_{2}}\right)  \mathbf{T}\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.tensor.CG1CG2CG1G2-3}),
applied to }m=\mathbf{T}\left(  n\right)  \right)  ,
\]
this yields $\mathbf{T}\left(  an\right)  =a\mathbf{T}\left(  n\right)  $.
This proves (\ref{sol.Indrho.tensor.finish-part-1a.pf.1}).
\par
Now, (\ref{sol.Indrho.tensor.finish-part-1a.pf.1}) shows that $\mathbf{T}$ is
a homomorphism
of left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules (since
$\mathbf{T}$ is $\CC$-linear), and therefore an isomorphism of left
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules (since $\mathbf{T}$ is
invertible). Thus, $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)
\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes
U_{2}\right)  \cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $ as left
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. This proves
(\ref{sol.Indrho.tensor.finish-part-1a}).}.

Now, (\ref{sol.Indrho.tensor.finish-part-2}) becomes%
\begin{align*}
&  \mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right) \\
&  \cong\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)
\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes
U_{2}\right)  \cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.tensor.finish-part-1a}%
)}\right)
\end{align*}
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. Therefore,%
\[
\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(  U_{1}\otimes U_{2}\right)
=\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \cong\left(
\operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(  \operatorname{Ind}%
_{\rho_{2}}U_{2}\right)
\]
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. This finishes
the solution of Exercise \ref{exe.Indrho.tensor}.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Indrho.tensor}}

\textit{Solution to Exercise \ref{exe.Indrho.tensor}.} We shall use the
following fact:

\begin{proposition}
\label{prop.sol.Indrho.tensor.general}Let $\kk$ be a commutative ring.
Let $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be six
$\kk$-algebras. Let $P$ be an $\left(  A,B\right)  $%
-bimodule\footnote{As usual, we understand the notion of a bimodule to be
defined over $\kk$; that is, the left $A$-module structure and the
right $B$-module structure of an $\left(  A,B\right)  $-bimodule must restrict
to one and the same $\kk$-module structure.}. Let $Q$ be a $\left(
B,C\right)  $-bimodule. Let $P^{\prime}$ be an $\left(  A^{\prime},B^{\prime
}\right)  $-bimodule. Let $Q^{\prime}$ be a $\left(  B^{\prime},C^{\prime
}\right)  $-bimodule. Then,%
\[
\left(  P\otimes P^{\prime}\right)  \otimes_{B\otimes B^{\prime}}\left(
Q\otimes Q^{\prime}\right)  \cong\left(  P\otimes_{B}Q\right)  \otimes\left(
P^{\prime}\otimes_{B^{\prime}}Q^{\prime}\right)
\]
as $\left(  A\otimes A^{\prime},C\otimes C^{\prime}\right)  $-bimodules. Here,
all $\otimes$ signs without subscript stand for $\otimes_{\kk}$.
\end{proposition}

Proposition \ref{prop.sol.Indrho.tensor.general} is identical with the
Proposition \ref{prop.sol.ind.tensor.general} that appeared in our solution to
Exercise \ref{exe.ind.tensor}. Hence, we refer to the latter solution for a
proof of Proposition \ref{prop.sol.Indrho.tensor.general}.

We have a canonical $\CC$-algebra isomorphism%
\[
\mathfrak{H}:\mathbb{C}\left[  H_{1}\times H_{2}\right]  \rightarrow
\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}%
\]
which satisfies%
\[
\left(  \mathfrak{H}\left(  t_{\left(  h_{1},h_{2}\right)  }\right)
=t_{h_{1}}\otimes t_{h_{2}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
h_{1},h_{2}\right)  \in H_{1}\times H_{2}\right)  .
\]
We also have a canonical $\CC$-algebra isomorphism%
\[
\mathfrak{G}:\mathbb{C}\left[  G_{1}\times G_{2}\right]  \rightarrow
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}%
\]
which satisfies%
\[
\left(  \mathfrak{G}\left(  t_{\left(  g_{1},g_{2}\right)  }\right)
=t_{g_{1}}\otimes t_{g_{2}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
g_{1},g_{2}\right)  \in G_{1}\times G_{2}\right)  .
\]


Recall how the left $\mathbb{C}\left[  H_{1}\times H_{2}\right]  $-module
$U_{1}\otimes U_{2}$ is defined: Its left $\mathbb{C}\left[  H_{1}\times
H_{2}\right]  $-module structure is given by%
\[
\left(  t_{\left(  h_{1},h_{2}\right)  }\left(  u_{1}\otimes u_{2}\right)
=t_{h_{1}}u_{1}\otimes t_{h_{2}}u_{2}\ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  h_{1},h_{2}\right)  \in H_{1}\times H_{2}\text{ and }\left(
u_{1},u_{2}\right)  \in U_{1}\times U_{2}\right)  .
\]


We shall now recall the definitions of $\operatorname*{Ind}\nolimits_{\rho
_{1}}U_{1}$, $\operatorname*{Ind}\nolimits_{\rho_{2}}U_{2}$ and
$\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(  U_{1}\otimes U_{2}\right)
$:

\begin{itemize}
\item We defined $\operatorname{Ind}_{\rho_{1}}\left(  U_{1}\right)  $ as the
${\mathbb{C}}G_{1}$-module ${\mathbb{C}}G_{1}\otimes_{\CC H_{1}}U_{1}%
$, where ${\mathbb{C}}G_{1}$ is regarded as a $\left(  {\mathbb{C}}%
G_{1},{\mathbb{C}}H_{1}\right)  $-bimodule according to the following rule:
The left ${\mathbb{C}}G_{1}$-module structure on ${\mathbb{C}}G_{1}$ is plain
multiplication inside ${\mathbb{C}}G_{1}$; the right ${\mathbb{C}}H_{1}%
$-module structure on ${\mathbb{C}}G_{1}$ is induced by the ${\mathbb{C}}%
$-algebra homomorphism ${\mathbb{C}}\left[  \rho_{1}\right]  :{\mathbb{C}%
}H_{1}\rightarrow{\mathbb{C}}G_{1}$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho_{1}\right]  \right)
\eta$ for all $\gamma\in{\mathbb{C}}G_{1}$ and $\eta\in{\mathbb{C}}H_{1}$).

\item We defined $\operatorname{Ind}_{\rho_{2}}\left(  U_{2}\right)  $ as the
${\mathbb{C}}G_{2}$-module ${\mathbb{C}}G_{2}\otimes_{\CC H_{2}}U_{2}%
$, where ${\mathbb{C}}G_{2}$ is regarded as a $\left(  {\mathbb{C}}%
G_{2},{\mathbb{C}}H_{2}\right)  $-bimodule according to the following rule:
The left ${\mathbb{C}}G_{2}$-module structure on ${\mathbb{C}}G_{2}$ is plain
multiplication inside ${\mathbb{C}}G_{2}$; the right ${\mathbb{C}}H_{2}%
$-module structure on ${\mathbb{C}}G_{2}$ is induced by the ${\mathbb{C}}%
$-algebra homomorphism ${\mathbb{C}}\left[  \rho_{2}\right]  :{\mathbb{C}%
}H_{2}\rightarrow{\mathbb{C}}G_{2}$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho_{2}\right]  \right)
\eta$ for all $\gamma\in{\mathbb{C}}G_{2}$ and $\eta\in{\mathbb{C}}H_{2}$).

\item We defined $\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(
U_{1}\otimes U_{2}\right)  $ as the ${\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  $-module ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]
\otimes_{\CC \left[  H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes
U_{2}\right)  $, where ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $ is
regarded as a $\left(  {\mathbb{C}}\left[  G_{1}\times G_{2}\right]
,{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  \right)  $-bimodule according
to the following rule: The left ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]
$-module structure on ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $ is
plain multiplication inside ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $;
the right ${\mathbb{C}}\left[  H_{1}\times H_{2}\right]  $-module structure on
${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $ is induced by the
${\mathbb{C}}$-algebra homomorphism ${\mathbb{C}}\left[  \rho_{1}\times
\rho_{2}\right]  :{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  \rightarrow
{\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $ (thus, it is explicitly given
by $\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho_{1}\times\rho
_{2}\right]  \right)  \eta$ for all $\gamma\in{\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  $ and $\eta\in{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  $).
\end{itemize}

Proposition \ref{prop.sol.ind.tensor.general} (applied to $\kk%
={\mathbb{C}}$, $A={\mathbb{C}}G_{1}$, $B={\mathbb{C}}H_{1}$, $C={\mathbb{C}}%
$, $A^{\prime}={\mathbb{C}}G_{2}$, $B^{\prime}={\mathbb{C}}H_{2}$, $C^{\prime
}={\mathbb{C}}$, $P={\mathbb{C}}G_{1}$, $Q=U_{1}$, $P^{\prime}={\mathbb{C}%
}G_{2}$ and $Q^{\prime}=U_{2}$) yields%
\[
\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  {\mathbb{C}}G_{1}\otimes_{\CC H_{1}}U_{1}\right)
\otimes\left(  {\mathbb{C}}G_{2}\otimes_{\CC H_{2}}U_{2}\right)
\]
as $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2},{\mathbb{C}}%
\otimes{\mathbb{C}}\right)  $-bimodules, hence also as left ${\mathbb{C}}%
G_{1}\otimes{\mathbb{C}}G_{2}$-modules. Thus,%
\begin{align}
&  \left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\nonumber\\
&  \cong\underbrace{\left(  {\mathbb{C}}G_{1}\otimes_{\CC H_{1}}%
U_{1}\right)  }_{\substack{=\operatorname{Ind}_{\rho_{1}}U_{1}\\\text{(since
}\operatorname{Ind}_{\rho_{1}}U_{1}={\mathbb{C}}G_{1}\otimes_{\CC
H_{1}}U_{1}\text{)}}}\otimes\underbrace{\left(  {\mathbb{C}}G_{2}%
\otimes_{\CC H_{2}}U_{2}\right)  }_{\substack{=\operatorname{Ind}%
_{\rho_{2}}U_{2}\\\text{(since }\operatorname{Ind}_{\rho_{2}}U_{2}%
={\mathbb{C}}G_{2}\otimes_{\CC H_{2}}U_{2}\text{)}}}\nonumber\\
&  =\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)
\label{sol.Indrho.tensor.finish-part-1}%
\end{align}
as left ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$-modules.

At this point, there is a quick way to finish the solution using handwaving:
We use the $\CC$-algebra isomorphism $\mathfrak{H}:\mathbb{C}\left[
H_{1}\times H_{2}\right]  \rightarrow\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$ to
identify the $\CC$-algebra $\mathbb{C}\left[  H_{1}\times H_{2}\right]
$ with the $\CC$-algebra $\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$, and
we use the $\CC$-algebra isomorphism $\mathfrak{G}:\mathbb{C}\left[
G_{1}\times G_{2}\right]  \rightarrow\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ to
identify the $\CC$-algebra $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$ with the $\CC$-algebra $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$. It is
``easy to see'' that these two identifications ``play nicely with each other
and with the module structures on $U_{1}\otimes U_{2}$'' (this is the part
where we wave our hands). Now, (\ref{sol.Indrho.tensor.finish-part-1}) yields
that $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)  $ as left ${\mathbb{C}}G_{1}%
\otimes{\mathbb{C}}G_{2}$-modules, and therefore also as left $\mathbb{C}%
\left[  G_{1}\times G_{2}\right]  $-modules (since $\mathbb{C}\left[
G_{1}\times G_{2}\right]  ={\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$). Now,%
\begin{align*}
\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(  U_{1}\otimes U_{2}\right)
&  ={\mathbb{C}}\left[  G_{1}\times G_{2}\right]  \otimes_{\CC \left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right) \\
&  =\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }{\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  ={\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\text{ and
}{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  ={\mathbb{C}}H_{1}%
\otimes{\mathbb{C}}H_{2}\right) \\
&  \cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)
\end{align*}
as left ${\mathbb{C}}\left[  G_{1}\times G_{2}\right]  $-modules. This solves
Exercise \ref{exe.Indrho.tensor} if you believe the handwaving I have done above.

The handwaving we have done is slightly questionable, since we have made two
identifications which turned two isomorphisms into identities. In reality,
they are merely isomorphisms, not identities, and it is not immediately clear
that regarding them as identities will not lead to contradictions. (Indeed,
this is the meaning of our vague claim that the two identifications
``play nicely with each other''.)

Let us now show a way to formalize the above questionable argument. We shall
avoid identifying any things that are not already identical; instead, we will
work with the isomorphisms $\mathfrak{G}$ and $\mathfrak{H}$ explicitly.

We first state a general fact:

\begin{proposition}
\label{prop.sol.Indrho.tensor.MAN}Let $\kk$ be a commutative ring. In
the following, all $\otimes$ signs without subscript stand for $\otimes
_{\kk}$.

Let $A$, $B$, $C$, $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$ be six
$\kk$-algebras. Let $M$ be an $\left(  A,B\right)  $%
-bimodule\footnote{As usual, we understand the notion of a bimodule to be
defined over $\kk$; that is, the left $A$-module structure and the
right $B$-module structure of an $\left(  A,B\right)  $-bimodule must restrict
to one and the same $\kk$-module structure.}. Let $N$ be a $\left(
B,C\right)  $-bimodule. Let $M^{\prime}$ be an $\left(  A^{\prime},B^{\prime
}\right)  $-bimodule. Let $N^{\prime}$ be a $\left(  B^{\prime},C^{\prime
}\right)  $-bimodule. Let $\beta:B\rightarrow B^{\prime}$, $\mu:M\rightarrow
M^{\prime}$ and $\nu:N\rightarrow N^{\prime}$ be three $\kk$-module
homomorphisms. Assume that we have%
\begin{equation}
\left(  \mu\left(  mb\right)  =\mu\left(  m\right)  \beta\left(  b\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B\text{ and }m\in M\right)
\label{eq.prop.sol.Indrho.tensor.MAN.ass1}%
\end{equation}
and%
\begin{equation}
\left(  \nu\left(  bn\right)  =\beta\left(  b\right)  \nu\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }b\in B\text{ and }n\in N\right)  .
\label{eq.prop.sol.Indrho.tensor.MAN.ass2}%
\end{equation}
Then:

\begin{itemize}
\item[(a)] There exists a unique $\kk$-module homomorphism%
\[
\Omega:M\otimes_{B}N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}%
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{B}n\right)  =\mu\left(  m\right)
\otimes_{B^{\prime}}\nu\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  m,n\right)  \in M\times N\right)  .
\]


We shall denote this homomorphism $\Omega$ by $\Omega_{\beta,\mu,\nu}$.

\item[(b)] If the maps $\beta$, $\mu$ and $\nu$ are invertible,
then $\Omega_{\beta,\mu,\nu}$ is a $\kk$-module isomorphism.

\item[(c)] Let $\alpha:A\rightarrow A^{\prime}$ be a $\kk$-module
homomorphism. Assume that%
\begin{equation}
\mu\left(  am\right)  =\alpha\left(  a\right)  \mu\left(  m\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in A\text{ and }m\in M.
\label{eq.prop.sol.Indrho.tensor.MAN.ass3}%
\end{equation}
Assume also that the $\kk$-module $M^{\prime}$ is endowed with a left
$A$-module structure. Assume that this left $A$-module structure on
$M^{\prime}$ and the right $B^{\prime}$-module structure on $M^{\prime}$
together form an $\left(  A,B^{\prime}\right)  $-bimodule structure on
$M^{\prime}$. Thus, $M^{\prime}\otimes_{B^{\prime}}N^{\prime}$ becomes a left
$A$-module. Assume furthermore that%
\begin{equation}
am=\alpha\left(  a\right)  m\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{
and }m\in M^{\prime} \label{eq.prop.sol.Indrho.tensor.MAN.ass4}%
\end{equation}
\footnote{Here the $\alpha\left(  a\right)  m$ on the right hand side is
defined using the action of $A^{\prime}$ on the left $A^{\prime}$-module
$M^{\prime}$, whereas the $am$ on the left hand side is defined using the
action of $A$ on the left $A$-module $M^{\prime}$.}. Then, $\Omega_{\beta
,\mu,\nu}$ is a left $A$-module homomorphism.
\end{itemize}
\end{proposition}

We could also add a part (d) to Proposition \ref{prop.sol.Indrho.tensor.MAN},
which would give a criterion for $\Omega_{\beta,\mu,\nu}$ to be a right
$C$-module homomorphism given an appropriate right $C$-module structure on
$N^{\prime}$. We leave both stating and proving this result\footnote{which is
completely analogous to Proposition \ref{prop.sol.Indrho.tensor.MAN}(c) and
can also be combined with Proposition \ref{prop.sol.Indrho.tensor.MAN}(c) to
obtain a criterion for when $\Omega_{\beta,\mu,\nu}$ is an $\left(
A,C\right)  $-bimodule homomorphism} to the reader, since we will not use it.

\begin{proof}
[Proof of Proposition \ref{prop.sol.Indrho.tensor.MAN}.](a) We define a map%
\[
\omega:M\times N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}%
\]
by%
\[
\left(  \omega\left(  m,n\right)  =\mu\left(  m\right)  \otimes_{B^{\prime}%
}\nu\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\left(  m,n\right)
\in M\times N\right)  .
\]
This map $\omega$ is $\kk$-bilinear (because $\omega\left(  m,n\right)
=\mu\left(  m\right)  \otimes_{B^{\prime}}\nu\left(  n\right)  $ depends
$\kk$-linearly on each of $m$ and $n$). Moreover, this map $\omega$ is
$B$-bilinear with respect to the right $B$-module structure on $M$ and the
left $B$-module structure on $N$\ \ \ \ \footnote{\textit{Proof.} Let $m\in
M$, $n\in N$ and $b\in B$. Then, the definition of $\omega\left(  mb,n\right)
$ yields%
\begin{align*}
\omega\left(  mb,n\right)   &  =\underbrace{\mu\left(  mb\right)
}_{\substack{=\mu\left(  m\right)  \beta\left(  b\right)  \\\text{(by
(\ref{eq.prop.sol.Indrho.tensor.MAN.ass1}))}}}\otimes_{B^{\prime}}\nu\left(
n\right)  =\mu\left(  m\right)  \beta\left(  b\right)  \otimes_{B^{\prime}}%
\nu\left(  n\right)  =\mu\left(  m\right)  \otimes_{B^{\prime}}%
\underbrace{\beta\left(  b\right)  \nu\left(  n\right)  }_{\substack{=\nu
\left(  bn\right)  \\\text{(by (\ref{eq.prop.sol.Indrho.tensor.MAN.ass2}))}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have moved the }\beta\left(
b\right)  \text{ past the tensor sign }\left(  \text{because }\beta\left(
b\right)  \in B^{\prime}\right)  \right) \\
&  =\mu\left(  m\right)  \otimes_{B}\nu\left(  bn\right)  =\omega\left(
m,bn\right)
\end{align*}
(because $\omega\left(  m,bn\right)  =\mu\left(  m\right)  \otimes_{B}%
\nu\left(  bn\right)  $ (by the definition of $\omega\left(  m,bn\right)  $)).
\par
Let us now forget that we fixed $m$, $n$ and $b$. We thus have shown that
$\omega\left(  mb,n\right)  =\omega\left(  m,bn\right)  $ for all $m\in M$,
$n\in N$ and $b\in B$. Therefore, the map $\omega$ is $B$-bilinear with
respect to the right $B$-module structure on $M$ and the left $B$-module
structure on $N$ (since we already know that the map $\omega$ is $\kk%
$-bilinear). Qed.}. Hence, by the universal property of the tensor product, we
conclude that there exists a unique $\kk$-module homomorphism%
\[
\Omega:M\otimes_{B}N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}%
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{B}n\right)  =\omega\left(  m,n\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  m,n\right)  \in M\times N\right)
.
\]
In other words, there exists a unique $\kk$-module homomorphism%
\[
\Omega:M\otimes_{B}N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}%
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{B}n\right)  =\mu\left(  m\right)
\otimes_{B^{\prime}}\nu\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  m,n\right)  \in M\times N\right)
\]
(because $\omega\left(  m,n\right)  =\mu\left(  m\right)  \otimes_{B^{\prime}%
}\nu\left(  n\right)  $ for all $\left(  m,n\right)  \in M\times N$). This
proves Proposition \ref{prop.sol.Indrho.tensor.MAN}(a).

(b) Assume that $\beta$, $\mu$ and $\nu$ are invertible. These maps $\beta$,
$\mu$ and $\gamma$ must therefore be $\kk$-module isomorphisms (since
they are $\kk$-module homomorphisms). Hence, their inverses $\beta
^{-1}$, $\mu^{-1}$ and $\nu^{-1}$ are well-defined $\kk$-module
isomorphisms as well.

We have%
\[
\left(  \mu^{-1}\left(  mb\right)  =\mu^{-1}\left(  m\right)  \beta
^{-1}\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }b\in B^{\prime
}\text{ and }m\in M^{\prime}\right)
\]
\footnote{\textit{Proof.} Let $b\in B^{\prime}$ and $m\in M^{\prime}$. Then,
(\ref{eq.prop.sol.Indrho.tensor.MAN.ass1}) (applied to $\beta^{-1}\left(
b\right)  $ and $\mu^{-1}\left(  m\right)  $ instead of $b$ and $m$) shows
that
\[
\mu\left(  \mu^{-1}\left(  m\right)  \beta^{-1}\left(  b\right)  \right)
=\underbrace{\mu\left(  \mu^{-1}\left(  m\right)  \right)  }_{=m}%
\underbrace{\beta\left(  \beta^{-1}\left(  b\right)  \right)  }_{=b}=mb.
\]
Applying the map $\mu^{-1}$ to both sides of this equality, we find $\mu
^{-1}\left(  \mu\left(  \mu^{-1}\left(  m\right)  \beta^{-1}\left(  b\right)
\right)  \right)  =\mu^{-1}\left(  mb\right)  $. Hence, $\mu^{-1}\left(
mb\right)  =\mu^{-1}\left(  \mu\left(  \mu^{-1}\left(  m\right)  \beta
^{-1}\left(  b\right)  \right)  \right)  =\mu^{-1}\left(  m\right)  \beta
^{-1}\left(  b\right)  $. Qed.} and%
\[
\left(  \nu^{-1}\left(  bn\right)  =\beta^{-1}\left(  b\right)  \nu
^{-1}\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }b\in B^{\prime
}\text{ and }n\in N^{\prime}\right)
\]
\footnote{\textit{Proof.} Let $b\in B^{\prime}$ and $n\in N^{\prime}$. Then,
(\ref{eq.prop.sol.Indrho.tensor.MAN.ass2}) (applied to $\beta^{-1}\left(
b\right)  $ and $\nu^{-1}\left(  n\right)  $ instead of $b$ and $n$) shows
that
\[
\nu\left(  \beta^{-1}\left(  b\right)  \nu^{-1}\left(  n\right)  \right)
=\underbrace{\beta\left(  \beta^{-1}\left(  b\right)  \right)  }%
_{=b}\underbrace{\nu\left(  \nu^{-1}\left(  n\right)  \right)  }_{=n}=bn.
\]
Applying the map $\nu^{-1}$ to both sides of this equality, we find $\nu
^{-1}\left(  \nu\left(  \beta^{-1}\left(  b\right)  \nu^{-1}\left(  n\right)
\right)  \right)  =\nu^{-1}\left(  bn\right)  $. Hence, $\nu^{-1}\left(
bn\right)  =\nu^{-1}\left(  \nu\left(  \beta^{-1}\left(  b\right)  \nu
^{-1}\left(  n\right)  \right)  \right)  =\beta^{-1}\left(  b\right)  \nu
^{-1}\left(  n\right)  $. Qed.}.

Recall that $\Omega_{\beta,\mu,\nu}$ is defined as the unique $\kk%
$-module homomorphism%
\[
\Omega:M\otimes_{B}N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}%
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{B}n\right)  =\mu\left(  m\right)
\otimes_{B^{\prime}}\nu\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  m,n\right)  \in M\times N\right)  .
\]
Hence, $\Omega_{\beta,\mu,\nu}$ is a $\kk$-module homomorphism
$M\otimes_{B}N\rightarrow M^{\prime}\otimes_{B^{\prime}}N^{\prime}$ and
satisfies%
\begin{equation}
\left(  \Omega_{\beta,\mu,\nu}\left(  m\otimes_{B}n\right)  =\mu\left(
m\right)  \otimes_{B^{\prime}}\nu\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  m,n\right)  \in M\times N\right)  .
\label{pf.prop.sol.Indrho.tensor.MAN.omega1}%
\end{equation}
We can apply the preceding statement to $A^{\prime}$, $B^{\prime}$,
$C^{\prime}$, $A$, $B$, $C$, $M^{\prime}$, $N^{\prime}$, $M$, $N$, $\beta
^{-1}$, $\mu^{-1}$ and $\nu^{-1}$ instead of $A$, $B$, $C$, $A^{\prime}$,
$B^{\prime}$, $C^{\prime}$, $M$, $N$, $M^{\prime}$, $N^{\prime}$, $\beta$,
$\mu$ and $\nu$. As a result, we obtain that $\Omega_{\beta^{-1},\mu^{-1}%
,\nu^{-1}}$ is a $\kk$-module homomorphism $M^{\prime}\otimes
_{B^{\prime}}N^{\prime}\rightarrow M\otimes_{B}N$ and satisfies%
\begin{equation}
\left(  \Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}\left(  m\otimes_{B^{\prime}%
}n\right)  =\mu^{-1}\left(  m\right)  \otimes_{B}\nu^{-1}\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  m,n\right)  \in M^{\prime}\times
N^{\prime}\right)  . \label{pf.prop.sol.Indrho.tensor.MAN.omega2}%
\end{equation}


Both $\Omega_{\beta,\mu,\nu}$ and $\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}$ are
$\kk$-module homomorphisms, thus $\kk$-linear maps. Let us now
prove that the maps $\Omega_{\beta,\mu,\nu}$ and $\Omega_{\beta^{-1},\mu
^{-1},\nu^{-1}}$ are mutually inverse. Indeed, we have $\Omega_{\beta,\mu,\nu
}\circ\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}=\operatorname*{id}$%
\ \ \ \ \footnote{\textit{Proof.} Every $\left(  m,n\right)  \in M^{\prime
}\times N^{\prime}$ satisfies%
\begin{align*}
\left(  \Omega_{\beta,\mu,\nu}\circ\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}%
}\right)  \left(  m\otimes_{B^{\prime}}n\right)   &  =\Omega_{\beta,\mu,\nu
}\left(  \underbrace{\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}\left(
m\otimes_{B^{\prime}}n\right)  }_{\substack{=\mu^{-1}\left(  m\right)
\otimes_{B}\nu^{-1}\left(  n\right)  \\\text{(by
(\ref{pf.prop.sol.Indrho.tensor.MAN.omega2}))}}}\right) \\
&  =\Omega_{\beta,\mu,\nu}\left(  \mu^{-1}\left(  m\right)  \otimes_{B}%
\nu^{-1}\left(  n\right)  \right)  =\underbrace{\mu\left(  \mu^{-1}\left(
m\right)  \right)  }_{=m}\otimes_{B^{\prime}}\underbrace{\nu\left(  \nu
^{-1}\left(  n\right)  \right)  }_{=n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Indrho.tensor.MAN.omega1}), applied to }\left(  \mu^{\prime
}\left(  m\right)  ,\nu^{\prime}\left(  n\right)  \right)  \text{ instead of
}\left(  m,n\right)  \right) \\
&  =m\otimes_{B}n=\operatorname*{id}\left(  m\otimes_{B}n\right)  .
\end{align*}
In other words, the two maps $\Omega_{\beta,\mu,\nu}\circ\Omega_{\beta
^{-1},\mu^{-1},\nu^{-1}}$ and $\operatorname*{id}$ (from $M^{\prime}%
\otimes_{B^{\prime}}N^{\prime}$ to $M^{\prime}\otimes_{B^{\prime}}N^{\prime}$)
are equal to each other on every pure tensor. But these two maps are both
$\kk$-linear (since $\Omega_{\beta,\mu,\nu}$ and $\Omega_{\beta
^{-1},\mu^{-1},\nu^{-1}}$ are $\kk$-linear). Recall that if two
$\kk$-linear maps from a tensor product to a given codomain are equal
to each other on every pure tensor, then these two maps must be identical.
Applying this fact to the two $\kk$-linear maps $\Omega_{\beta,\mu,\nu
}\circ\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}$ and $\operatorname*{id}$, we
conclude that the two maps $\Omega_{\beta,\mu,\nu}\circ\Omega_{\beta^{-1}%
,\mu^{-1},\nu^{-1}}$ and $\operatorname*{id}$ must be identical (because they
are equal to each other on every pure tensor). In other words, $\Omega
_{\beta,\mu,\nu}\circ\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}=\operatorname*{id}%
$, qed.} and $\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}\circ\Omega_{\beta,\mu,\nu
}=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} We have already shown
that $\Omega_{\beta,\mu,\nu}\circ\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}%
}=\operatorname*{id}$. We can also apply this statement to $A^{\prime}$,
$B^{\prime}$, $C^{\prime}$, $A$, $B$, $C$, $M^{\prime}$, $N^{\prime}$, $M$,
$N$, $\beta^{-1}$, $\mu^{-1}$ and $\nu^{-1}$ instead of $A$, $B$, $C$,
$A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $M$, $N$, $M^{\prime}$, $N^{\prime}%
$, $\beta$, $\mu$ and $\nu$. As a result, we obtain $\Omega_{\beta^{-1}%
,\mu^{-1},\nu^{-1}}\circ\Omega_{\left(  \beta^{-1}\right)  ^{-1},\left(
\mu^{-1}\right)  ^{-1},\left(  \nu^{-1}\right)  ^{-1}}=\operatorname*{id}$.
This rewrites as $\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}\circ\Omega_{\beta
,\mu,\nu}=\operatorname*{id}$ (since $\left(  \beta^{-1}\right)  ^{-1}=\beta$,
$\left(  \mu^{-1}\right)  ^{-1}=\mu$ and $\left(  \nu^{-1}\right)  ^{-1}=\nu
$). Qed.}. These two equalities yield that the maps $\Omega_{\beta,\mu,\nu}$
and $\Omega_{\beta^{-1},\mu^{-1},\nu^{-1}}$ are mutually inverse. Hence, the
map $\Omega_{\beta,\mu,\nu}$ is invertible. Thus, this map $\Omega_{\beta
,\mu,\nu}$ is a $\kk$-module isomorphism (since we already know that
$\Omega_{\beta,\mu,\nu}$ is a $\kk$-module homomorphism). Proposition
\ref{prop.sol.Indrho.tensor.MAN}(b) is proven.

(c) First, we notice that (\ref{pf.prop.sol.Indrho.tensor.MAN.omega1}) holds.
(Indeed, this can be proven in the same way as in our proof of Proposition
\ref{prop.sol.Indrho.tensor.MAN}(b) above.)

Now, every $a\in A$ and $h\in M\otimes_{B}N$ satisfy%
\begin{equation}
\Omega_{\beta,\mu,\nu}\left(  ah\right)  =a\cdot\Omega_{\beta,\mu,\nu}\left(
h\right)  \label{pf.prop.sol.Indrho.tensor.MAN.c.goal}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.Indrho.tensor.MAN.c.goal}):} Let
$a\in A$ and $h\in M\otimes_{B}N$. We need to prove the equality
(\ref{pf.prop.sol.Indrho.tensor.MAN.c.goal}). This equality is $\kk%
$-linear in $h$. Hence, we can WLOG assume that $h$ is a pure tensor (because
the $\kk$-module $M\otimes_{B}N$ is spanned by the pure tensors).
Assume this.
\par
There exist $m\in M$ and $n\in N$ such that $h=m\otimes_{B}n$ (since $h$ is a
pure tensor). Consider these $m$ and $n$. The definition of the left
$A$-module structure on $M\otimes_{B}N$ yields $a\left(  m\otimes_{B}n\right)
=am\otimes_{B}n$, so that $a\underbrace{h}_{=m\otimes_{B}n}=a\left(
m\otimes_{B}n\right)  =am\otimes_{B}n$. Applying the map $\Omega_{\beta
,\mu,\nu}$ to both sides of this equality, we obtain%
\begin{align*}
\Omega_{\beta,\mu,\nu}\left(  ah\right)   &  =\Omega_{\beta,\mu,\nu}\left(
am\otimes_{B}n\right)  =\underbrace{\mu\left(  am\right)  }_{\substack{=\alpha
\left(  a\right)  \mu\left(  m\right)  \\\text{(by
(\ref{eq.prop.sol.Indrho.tensor.MAN.ass3}))}}}\otimes_{B^{\prime}}\nu\left(
n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Indrho.tensor.MAN.omega1}), applied to }am\text{ instead of
}m\right) \\
&  =\alpha\left(  a\right)  \mu\left(  m\right)  \otimes_{B^{\prime}}%
\nu\left(  n\right)  .
\end{align*}
Compared with%
\begin{align*}
a\cdot\Omega_{\beta,\mu,\nu}\left(  \underbrace{h}_{=m\otimes_{B}n}\right)
&  =a\cdot\underbrace{\Omega_{\beta,\mu,\nu}\left(  m\otimes_{B}n\right)
}_{\substack{=\mu\left(  m\right)  \otimes_{B^{\prime}}\nu\left(  n\right)
\\\text{(by (\ref{pf.prop.sol.Indrho.tensor.MAN.omega1}))}}}=a\cdot\left(
\mu\left(  m\right)  \otimes_{B^{\prime}}\nu\left(  n\right)  \right) \\
&  =\alpha\left(  a\right)  \cdot\left(  \mu\left(  m\right)  \otimes
_{B^{\prime}}\nu\left(  n\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.prop.sol.Indrho.tensor.MAN.ass4}), applied to }g=\mu\left(
m\right)  \otimes_{B^{\prime}}\nu\left(  n\right)  \right) \\
&  =\alpha\left(  a\right)  \mu\left(  m\right)  \otimes_{B^{\prime}}%
\nu\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
the left }A^{\prime}\text{-module structure on }M^{\prime}\otimes_{B}%
N^{\prime}\right)  ,
\end{align*}
this yields $\Omega_{\beta,\mu,\nu}\left(  ah\right)  =a\cdot\Omega_{\beta
,\mu,\nu}\left(  h\right)  $. This proves
(\ref{pf.prop.sol.Indrho.tensor.MAN.c.goal}).}. Hence, $\Omega_{\beta,\mu,\nu
}$ is a left $A$-module homomorphism (since we already know that
$\Omega_{\beta,\mu,\nu}$ is $\kk$-linear). This proves Proposition
\ref{prop.sol.Indrho.tensor.MAN}(c).
\end{proof}

Let us now return to solving Exercise \ref{exe.Indrho.tensor}. The family
$\left(  t_{\left(  h_{1},h_{2}\right)  }\right)  _{\left(  h_{1}%
,h_{2}\right)  \in H_{1}\times H_{2}}$ is a basis of the $\CC$-vector
space $\mathbb{C}\left[  H_{1}\times H_{2}\right]  $%
\ \ \ \ \footnote{\textit{Proof.} The family $\left(  t_{h}\right)  _{h\in
H_{1}\times H_{2}}$ is a basis of the $\CC$-vector space $\mathbb{C}%
\left[  H_{1}\times H_{2}\right]  $. Renaming the index $h$ as $\left(
h_{1},h_{2}\right)  $ in this statement, we can rewrite it as follows: The
family $\left(  t_{\left(  h_{1},h_{2}\right)  }\right)  _{\left(  h_{1}%
,h_{2}\right)  \in H_{1}\times H_{2}}$ is a basis of the $\CC$-vector
space $\mathbb{C}\left[  H_{1}\times H_{2}\right]  $. Qed.}, whereas the
family $\left(  t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1}%
,g_{2}\right)  \in G_{1}\times G_{2}}$ is a basis of the $\CC$-vector
space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $%
\ \ \ \ \footnote{\textit{Proof.} The family $\left(  t_{g}\right)  _{g\in
G_{1}\times G_{2}}$ is a basis of the $\CC$-vector space $\mathbb{C}%
\left[  G_{1}\times G_{2}\right]  $. Renaming the index $g$ as $\left(
g_{1},g_{2}\right)  $ in this statement, we can rewrite it as follows: The
family $\left(  t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1}%
,g_{2}\right)  \in G_{1}\times G_{2}}$ is a basis of the $\CC$-vector
space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. Qed.}.

We now discuss four bimodules:

\begin{itemize}
\item The $\CC$-vector space $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ is
a $\left(  \mathbb{C}G_{1}\otimes\mathbb{C}G_{2},\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}\right)  $-bimodule (because it is the tensor product of
the $\left(  \mathbb{C}G_{1},\mathbb{C}H_{1}\right)  $-bimodule $\mathbb{C}%
G_{1}$ with the $\left(  \mathbb{C}G_{2},\mathbb{C}H_{2}\right)  $-bimodule
$\mathbb{C}G_{2}$). In particular, it is thus a left $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$-module. Its left $\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$-module structure satisfies%
\begin{equation}
\left(  am=a\cdot m\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}\text{ and }m\in\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}\right)  \label{sol.Indrho.tensor.am=a*m}%
\end{equation}
(where the $am$ on the left hand side means the action of $a\in\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$ on the element $m$ of the left $\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$-module $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$,
whereas the $a\cdot m$ on the right hand side means the product of
$a\in\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ with $m\in\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$ in the $\CC$-algebra $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$)\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.Indrho.tensor.am=a*m}):} Let $a\in\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$ and $m\in\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$. For this
proof, we shall always use the dot $\cdot$ to denote multiplication inside a
$\CC$-algebra (such as $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$). In
particular, the expression $a\cdot m$ will be understood as the product of $a$
and $m$ in the $\CC$-algebra $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$.
In contrast, when we write ``$bn$'' for some
element $b$ of a $\CC$-algebra $B$ and some element $n$ of a left
$B$-module $N$, then we mean the action of $b\in B$ on the element $n$ of the
left $B$-module $N$. In particular, the expression $am$ will be understood as
the action of $a\in\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ on the element $m$
of the left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-module $\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$.
\par
We need to prove the equality $am=a\cdot m$. This is equality is $\mathbb{C}%
$-linear in each of $a$ and $m$. Hence, we can WLOG assume that both $a$ and
$m$ are pure tensors (since the $\CC$-vector space $\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$ is spanned by pure tensors). Assume this.
\par
We know that $a$ is a pure tensor. In other words, $a=a_{1}\otimes a_{2}$ for
some $a_{1}\in\mathbb{C}G_{1}$ and $a_{2}\in\mathbb{C}G_{2}$. Consider these
$a_{1}$ and $a_{2}$.
\par
We know that $m$ is a pure tensor. In other words, $m=m_{1}\otimes m_{2}$ for
some $m_{1}\in\mathbb{C}G_{1}$ and $m_{2}\in\mathbb{C}G_{2}$. Consider these
$m_{1}$ and $m_{2}$.
\par
We have%
\begin{align*}
\underbrace{a}_{=a_{1}\otimes a_{2}}\underbrace{m}_{=m_{1}\otimes m_{2}}  &
=\left(  a_{1}\otimes a_{2}\right)  \left(  m_{1}\otimes m_{2}\right)
=\underbrace{a_{1}m_{1}}_{\substack{=a_{1}\cdot m_{1}\\\text{(since the left
}\mathbb{C}G_{1}\text{-module}\\\text{structure on }\mathbb{C}G_{1}\text{ is
plain}\\\text{multiplication inside }\mathbb{C}G_{1}\text{)}}}\otimes
\underbrace{a_{2}m_{2}}_{\substack{=a_{2}\cdot m_{2}\\\text{(since the left
}\mathbb{C}G_{2}\text{-module}\\\text{structure on }\mathbb{C}G_{2}\text{ is
plain}\\\text{multiplication inside }\mathbb{C}G_{2}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\text{-module structure on }%
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\right) \\
&  =\left(  a_{1}\cdot m_{1}\right)  \otimes\left(  a_{2}\cdot m_{2}\right)  .
\end{align*}
Compared with%
\[
\underbrace{a}_{=a_{1}\otimes a_{2}}\cdot\underbrace{m}_{=m_{1}\otimes m_{2}%
}=\left(  a_{1}\otimes a_{2}\right)  \cdot\left(  m_{1}\otimes m_{2}\right)
=\left(  a_{1}\cdot m_{1}\right)  \otimes\left(  a_{2}\cdot m_{2}\right)  ,
\]
this yields $am=a\cdot m$. This proves (\ref{sol.Indrho.tensor.am=a*m}).}. In
other words, this left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-module
structure on $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ is plain multiplication
inside $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$. Thus, we do not need to
distinguish the action of an element $a\in\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$ on the element $m$ of the left $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$-module $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ from
the product of $a\in\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ with $m\in
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ in the $\CC$-algebra
$\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$.

\item The $\CC$-vector space $U_{1}\otimes U_{2}$ is a left
$\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$-module (because it is the tensor
product of the left $\mathbb{C}H_{1}$-module $U_{1}$ with the left
$\mathbb{C}H_{2}$-module $U_{2}$), and thus a $\left(  \mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2},\mathbb{C}\right)  $-bimodule.

\item The $\CC$-vector space ${\mathbb{C}}\left[  G_{1}\times
G_{2}\right]  $ is a $\left(  {\mathbb{C}}\left[  G_{1}\times G_{2}\right]
,{\mathbb{C}}\left[  H_{1}\times H_{2}\right]  \right)  $-bimodule (as we
already know).

\item The $\CC$-vector space $U_{1}\otimes U_{2}$ is a left
${\mathbb{C}}\left[  H_{1}\times H_{2}\right]  $-module (since it is the
tensor product of the left $\mathbb{C}H_{1}$-module $U_{1}$ with the left
$\mathbb{C}H_{2}$-module $U_{2}$), and thus a $\left(  {\mathbb{C}}\left[
H_{1}\times H_{2}\right]  ,\mathbb{C}\right)  $-bimodule.
\end{itemize}

We have%
\begin{equation}
\left(  \mathfrak{G}\left(  mb\right)  =\mathfrak{G}\left(  m\right)
\mathfrak{H}\left(  b\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }%
b\in\mathbb{C}\left[  H_{1}\times H_{2}\right]  \text{ and }m\in
\mathbb{C}\left[  G_{1}\times G_{2}\right]  \right)
\label{sol.Indrho.tensor.axiom1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.axiom1}):} Let
$b\in\mathbb{C}\left[  H_{1}\times H_{2}\right]  $ and $m\in\mathbb{C}\left[
G_{1}\times G_{2}\right]  $.
\par
We need to prove the equality $\mathfrak{G}\left(  mb\right)  =\mathfrak{G}%
\left(  m\right)  \mathfrak{H}\left(  b\right)  $. This equality is
$\CC$-linear in $m$ (because the map $\mathfrak{G}$ is $\mathbb{C}%
$-linear). Hence, we can WLOG assume that $m$ belongs to the basis $\left(
t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}}$ of the $\CC$-vector space $\mathbb{C}\left[
G_{1}\times G_{2}\right]  $. Assume this. Thus, $m$ belongs to the basis
$\left(  t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1}%
,g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\CC$-vector space
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. In other words, there exists
some $\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}$ such that
$m=t_{\left(  g_{1},g_{2}\right)  }$. Consider this $\left(  g_{1}%
,g_{2}\right)  $. Applying the map $\mathfrak{G}$ to both sides of the
equality $m=t_{\left(  g_{1},g_{2}\right)  }$, we obtain%
\[
\mathfrak{G}\left(  m\right)  =\mathfrak{G}\left(  t_{\left(  g_{1}%
,g_{2}\right)  }\right)  =t_{g_{1}}\otimes t_{g_{2}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\mathfrak{G}\right)  .
\]
\par
We need to prove the equality $\mathfrak{G}\left(  mb\right)  =\mathfrak{G}%
\left(  m\right)  \mathfrak{H}\left(  b\right)  $. This equality is
$\CC$-linear in $b$ (because the maps $\mathfrak{G}$ and $\mathfrak{H}$
are $\CC$-linear). Hence, we can WLOG assume that $b$ belongs to the
basis $\left(  t_{\left(  h_{1},h_{2}\right)  }\right)  _{\left(  h_{1}%
,h_{2}\right)  \in H_{1}\times H_{2}}$ of the $\CC$-vector space
$\mathbb{C}\left[  H_{1}\times H_{2}\right]  $. Assume this. Thus, $b$ belongs
to the basis $\left(  t_{\left(  h_{1},h_{2}\right)  }\right)  _{\left(
h_{1},h_{2}\right)  \in H_{1}\times H_{2}}$ of the $\CC$-vector space
$\mathbb{C}\left[  H_{1}\times H_{2}\right]  $. In other words, there exists
some $\left(  h_{1},h_{2}\right)  \in H_{1}\times H_{2}$ such that
$b=t_{\left(  h_{1},h_{2}\right)  }$. Consider this $\left(  h_{1}%
,h_{2}\right)  $. Applying the map $\mathfrak{H}$ to both sides of the
equality $b=t_{\left(  h_{1},h_{2}\right)  }$, we obtain%
\[
\mathfrak{H}\left(  b\right)  =\mathfrak{H}\left(  t_{\left(  h_{1}%
,h_{2}\right)  }\right)  =t_{h_{1}}\otimes t_{h_{2}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\mathfrak{H}\right)  .
\]
\par
The definition of the map $\rho_{1}\times\rho_{2}$ yields $\left(  \rho
_{1}\times\rho_{2}\right)  \left(  h_{1},h_{2}\right)  =\left(  \rho
_{1}\left(  h_{1}\right)  ,\rho_{2}\left(  h_{2}\right)  \right)  $. The
definition of the map ${\mathbb{C}}\left[  \rho_{1}\times\rho_{2}\right]  $
yields $\left(  {\mathbb{C}}\left[  \rho_{1}\times\rho_{2}\right]  \right)
\left(  t_{\left(  h_{1},h_{2}\right)  }\right)  =t_{\left(  \rho_{1}%
\times\rho_{2}\right)  \left(  h_{1},h_{2}\right)  }=t_{\left(  \rho
_{1}\left(  h_{1}\right)  ,\rho_{2}\left(  h_{2}\right)  \right)  }$ (since
$\left(  \rho_{1}\times\rho_{2}\right)  \left(  h_{1},h_{2}\right)  =\left(
\rho_{1}\left(  h_{1}\right)  ,\rho_{2}\left(  h_{2}\right)  \right)  $). Now,%
\begin{align*}
\underbrace{m}_{=t_{\left(  g_{1},g_{2}\right)  }}\underbrace{b}_{=t_{\left(
h_{1},h_{2}\right)  }}  &  =t_{\left(  g_{1},g_{2}\right)  }t_{\left(
h_{1},h_{2}\right)  }=t_{\left(  g_{1},g_{2}\right)  }\cdot\underbrace{\left(
{\mathbb{C}}\left[  \rho_{1}\times\rho_{2}\right]  \right)  \left(  t_{\left(
h_{1},h_{2}\right)  }\right)  }_{=t_{\left(  \rho_{1}\left(  h_{1}\right)
,\rho_{2}\left(  h_{2}\right)  \right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the right }{\mathbb{C}}\left[  H_{1}\times H_{2}\right]
\text{-module structure on }{\mathbb{C}}\left[  G_{1}\times G_{2}\right]
\text{ is}\\
\text{induced by the }\mathbb{C}\text{-algebra homomorphism }{\mathbb{C}%
}\left[  \rho_{1}\times\rho_{2}\right]  :{\mathbb{C}}\left[  H_{1}\times
H_{2}\right]  \rightarrow{\mathbb{C}}\left[  G_{1}\times G_{2}\right]
\end{array}
\right) \\
&  =t_{\left(  g_{1},g_{2}\right)  }\cdot t_{\left(  \rho_{1}\left(
h_{1}\right)  ,\rho_{2}\left(  h_{2}\right)  \right)  }=t_{\left(  g_{1},
g_2\right) \left( \rho_{1}\left(  h_{1}\right)  , \rho_{2}\left(  h_{2}
\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the multiplication
on }{\mathbb{C}}\left[  G_{1}\times G_{2}\right]  \right)  \\
&= t_{\left(  g_{1}%
\rho_{1}\left(  h_{1}\right)  ,g_{2}\rho_{2}\left(  h_{2}\right)  \right)  } .
\end{align*}
Applying the map $\mathfrak{G}$ to both sides of this equality, we obtain%
\[
\mathfrak{G}\left(  mb\right)  =\mathfrak{G}\left(  t_{\left(  g_{1}\rho
_{1}\left(  h_{1}\right)  ,g_{2}\rho_{2}\left(  h_{2}\right)  \right)
}\right)  =t_{g_{1}\rho_{1}\left(  h_{1}\right)  }\otimes t_{g_{2}\rho
_{2}\left(  h_{2}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of the map }\mathfrak{G}\right)  .
\]
Compared with%
\begin{align*}
\underbrace{\mathfrak{G}\left(  m\right)  }_{=t_{g_{1}}\otimes t_{g_{2}}%
}\underbrace{\mathfrak{H}\left(  b\right)  }_{=t_{h_{1}}\otimes t_{h_{2}}}  &
=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \left(  t_{h_{1}}\otimes t_{h_{2}%
}\right)  =\underbrace{t_{g_{1}}t_{h_{1}}}_{\substack{=t_{g_{1}}\cdot\left(
\mathbb{C}\left[  \rho_{1}\right]  \right)  \left(  t_{h_{1}}\right)
\\\text{(since the right }{\mathbb{C}}H_{1}\text{-module}\\\text{structure on
}{\mathbb{C}}G_{1}\text{ is induced}\\\text{by the }\mathbb{C}\text{-algebra
homomorphism}\\\mathbb{C}\left[  \rho_{1}\right]  :\mathbb{C}H_{1}%
\rightarrow\mathbb{C}G_{1}\text{)}}}\otimes\underbrace{t_{g_{2}}t_{h_{2}}%
}_{\substack{=t_{g_{2}}\cdot\left(  \mathbb{C}\left[  \rho_{2}\right]
\right)  \left(  t_{h_{2}}\right)  \\\text{(since the right }{\mathbb{C}}%
H_{2}\text{-module}\\\text{structure on }{\mathbb{C}}G_{2}\text{ is
induced}\\\text{by the }\mathbb{C}\text{-algebra homomorphism}\\\mathbb{C}%
\left[  \rho_{2}\right]  :\mathbb{C}H_{2}\rightarrow\mathbb{C}G_{2}\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the right }\mathbb{C}H_{1}\otimes\mathbb{C}%
H_{2}\text{-module}\\
\text{structure on }\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}%
\end{array}
\right) \\
&  =\left(  t_{g_{1}}\cdot\underbrace{\left(  \mathbb{C}\left[  \rho
_{1}\right]  \right)  \left(  t_{h_{1}}\right)  }_{\substack{=t_{\rho
_{1}\left(  h_{1}\right)  }\\\text{(by the definition}\\\text{of the map
}\mathbb{C}\left[  \rho_{1}\right]  \text{)}}}\right)  \otimes\left(
t_{g_{2}}\cdot\underbrace{\left(  \mathbb{C}\left[  \rho_{2}\right]  \right)
\left(  t_{h_{2}}\right)  }_{\substack{=t_{\rho_{2}\left(  h_{2}\right)
}\\\text{(by the definition}\\\text{of the map }\mathbb{C}\left[  \rho
_{2}\right]  \text{)}}}\right)  =\underbrace{\left(  t_{g_{1}}\cdot
t_{\rho_{1}\left(  h_{1}\right)  }\right)  }_{\substack{=t_{g_{1}\rho
_{1}\left(  h_{1}\right)  }\\\text{(by the definition}\\\text{of the
multiplication}\\\text{on }\mathbb{C}G_{1}\text{)}}}\otimes\underbrace{\left(
t_{g_{2}}\cdot t_{\rho_{2}\left(  h_{2}\right)  }\right)  }%
_{\substack{=t_{g_{2}\rho_{2}\left(  h_{2}\right)  }\\\text{(by the
definition}\\\text{of the multiplication}\\\text{on }\mathbb{C}G_{2}\text{)}%
}}\\
&  =t_{g_{1}\rho_{1}\left(  h_{1}\right)  }\otimes t_{g_{2}\rho_{2}\left(
h_{2}\right)  },
\end{align*}
this yields $\mathfrak{G}\left(  mb\right)  =\mathfrak{G}\left(  m\right)
\mathfrak{H}\left(  b\right)  $. This proves (\ref{sol.Indrho.tensor.axiom1}%
).} and%
\begin{equation}
\left(  \operatorname*{id}\left(  bn\right)  =\mathfrak{H}\left(  b\right)
\operatorname*{id}\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }%
b\in\mathbb{C}\left[  H_{1}\times H_{2}\right]  \text{ and }n\in U_{1}\otimes
U_{2}\right)  \label{sol.Indrho.tensor.axiom2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.axiom2}):} Let
$\mathbb{C}\left[  H_{1}\times H_{2}\right]  $ and $n\in U_{1}\otimes U_{2}$.
\par
We need to prove the equality $\operatorname*{id}\left(  bn\right)
=\mathfrak{H}\left(  b\right)  \operatorname*{id}\left(  n\right)  $. This
equality is $\CC$-linear in $n$ (because the map $\operatorname*{id}$
is $\CC$-linear). Hence, we can WLOG assume that $n$ is a pure tensor
(since every element of $U_{1}\otimes U_{2}$ is a $\CC$-linear
combination of pure tensors). Assume this. Thus, $n$ is a pure tensor. In
other words, there exist $u_{1}\in U_{1}$ and $u_{2}\in U_{2}$ such that
$n=u_{1}\otimes u_{2}$. Consider these $u_{1}$ and $u_{2}$.
\par
We need to prove the equality $\operatorname*{id}\left(  bn\right)
=\mathfrak{H}\left(  b\right)  \operatorname*{id}\left(  n\right)  $. This
equality is $\CC$-linear in $b$ (because the maps $\operatorname*{id}$
and $\mathfrak{H}$ are $\CC$-linear). Hence, we can WLOG assume that
$b$ belongs to the basis $\left(  t_{\left(  h_{1},h_{2}\right)  }\right)
_{\left(  h_{1},h_{2}\right)  \in H_{1}\times H_{2}}$ of the $\mathbb{C}%
$-vector space $\mathbb{C}\left[  H_{1}\times H_{2}\right]  $. Assume this.
Thus, $b$ belongs to the basis $\left(  t_{\left(  h_{1},h_{2}\right)
}\right)  _{\left(  h_{1},h_{2}\right)  \in H_{1}\times H_{2}}$ of the
$\CC$-vector space $\mathbb{C}\left[  H_{1}\times H_{2}\right]  $. In
other words, there exists some $\left(  h_{1},h_{2}\right)  \in H_{1}\times
H_{2}$ such that $b=t_{\left(  h_{1},h_{2}\right)  }$. Consider this $\left(
h_{1},h_{2}\right)  $. Applying the map $\mathfrak{H}$ to both sides of the
equality $b=t_{\left(  h_{1},h_{2}\right)  }$, we obtain%
\[
\mathfrak{H}\left(  b\right)  =\mathfrak{H}\left(  t_{\left(  h_{1}%
,h_{2}\right)  }\right)  =t_{h_{1}}\otimes t_{h_{2}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\mathfrak{H}\right)  .
\]
\par
Now,
\begin{align*}
\underbrace{\mathfrak{H}\left(  b\right)  }_{=t_{h_{1}}\otimes t_{h_{2}}%
}\underbrace{\operatorname*{id}\left(  n\right)  }_{=n=u_{1}\otimes u_{2}}  &
=\left(  t_{h_{1}}\otimes t_{h_{2}}\right)  \left(  u_{1}\otimes u_{2}\right)
=t_{h_{1}}u_{1}\otimes t_{h_{2}}u_{2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}\text{-module structure on }U_{1}\otimes
U_{2}\right)  .
\end{align*}
Compared with%
\begin{align*}
\operatorname*{id}\left(  bn\right)   &  =\underbrace{b}_{=t_{\left(
h_{1},h_{2}\right)  }}\underbrace{n}_{=u_{1}\otimes u_{2}}=t_{\left(
h_{1},h_{2}\right)  }\left(  u_{1}\otimes u_{2}\right)  =t_{h_{1}}u_{1}\otimes
t_{h_{2}}u_{2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}\left[  H_{1}\otimes H_{2}\right]  \text{-module structure on }%
U_{1}\otimes U_{2}\right)  ,
\end{align*}
this yields $\operatorname*{id}\left(  bn\right)  =\mathfrak{H}\left(
b\right)  \operatorname*{id}\left(  n\right)  $. This proves
(\ref{sol.Indrho.tensor.axiom2}).}. Hence, we can apply Proposition
\ref{prop.sol.Indrho.tensor.MAN}(a) to $\kk=\mathbb{C}$, $A=\mathbb{C}%
\left[  G_{1}\times G_{2}\right]  $, $B=\mathbb{C}\left[  H_{1}\times
H_{2}\right]  $, $C=\mathbb{C}$, $A^{\prime}=\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$, $B^{\prime}=\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$,
$C^{\prime}=\mathbb{C}$, $M=\mathbb{C}\left[  G_{1}\times G_{2}\right]  $,
$N=U_{1}\otimes U_{2}$, $M^{\prime}=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$,
$N^{\prime}=U_{1}\otimes U_{2}$, $\beta=\mathfrak{H}$, $\mu=\mathfrak{G}$ and
$\nu=\operatorname*{id}$. As a consequence, we conclude that there exists a
unique $\CC$-module homomorphism%
\[
\Omega:\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \rightarrow
\left(  \mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\right)  \otimes_{\mathbb{C}%
H_{1}\otimes\mathbb{C}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\]
which satisfies
\[
\left(  \Omega\left(  m\otimes_{\mathbb{C}\left[  H_{1}\times H_{2}\right]
}n\right)
= \mathfrak{G}\left(  m\right)  \otimes_{\mathbb{C}H_{1}\otimes
\mathbb{C}H_{2}}\id\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for all
}\left(  m,n\right)  \in\mathbb{C}\left[  G_{1}\times G_{2}\right]
\times\left(  U_{1}\otimes U_{2}\right)  \right)  .
\]
According to Proposition \ref{prop.sol.Indrho.tensor.MAN}(a), this
homomorphism $\Omega$ is denoted by $\Omega_{\mathfrak{H},\mathfrak{G}%
,\operatorname*{id}}$. Thus, $\Omega_{\mathfrak{H},\mathfrak{G}%
,\operatorname*{id}}$ is a $\CC$-module homomorphism
\[
\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \rightarrow
\left(  \mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\right)  \otimes_{\mathbb{C}%
H_{1}\otimes\mathbb{C}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\]
which satisfies
\[
\left(  \Omega_{\mathfrak{H},\mathfrak{G},\operatorname*{id}}\left(
m\otimes_{\mathbb{C}\left[  H_{1}\times H_{2}\right]  }n\right)
= \mathfrak{G} \left(
m\right)  \otimes_{\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}}\id\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }\left(  m,n\right)  \in\mathbb{C}\left[
G_{1}\times G_{2}\right]  \times\left(  U_{1}\otimes U_{2}\right)  \right)  .
\]


The maps $\mathfrak{H}$, $\mathfrak{G}$ and $\operatorname*{id}$ are
$\CC$-algebra isomorphisms, and therefore invertible. Hence,
Proposition \ref{prop.sol.Indrho.tensor.MAN}(b) (applied to $\kk%
=\mathbb{C}$, $A=\mathbb{C}\left[  G_{1}\times G_{2}\right]  $, $B=\mathbb{C}%
\left[  H_{1}\times H_{2}\right]  $, $C=\mathbb{C}$, $A^{\prime}%
=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $B^{\prime}=\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}$, $C^{\prime}=\mathbb{C}$, $M=\mathbb{C}\left[
G_{1}\times G_{2}\right]  $, $N=U_{1}\otimes U_{2}$, $M^{\prime}%
=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $N^{\prime}=U_{1}\otimes U_{2}$,
$\beta=\mathfrak{H}$, $\mu=\mathfrak{G}$ and $\nu=\operatorname*{id}$) yields
that $\Omega_{\mathfrak{H},\mathfrak{G},\operatorname*{id}}$ is a $\mathbb{C}%
$-module isomorphism. Thus, the map $\Omega_{\mathfrak{H},\mathfrak{G}%
,\operatorname*{id}}$ is invertible.

Next, we notice that the $\CC$-vector space $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$ is a left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module (since it is the tensor product of the left $\mathbb{C}G_{1}$-module
$\mathbb{C}G_{1}$ with the left $\mathbb{C}G_{2}$-module $\mathbb{C}G_{2}$).
This left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module structure is
defined by the rule%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }\left(  u_{1}\otimes u_{2}\right)  =t_{g_{1}%
}u_{1}\otimes t_{g_{2}}u_{2}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
g_{1},g_{2}\right)  \in G_{1}\times G_{2}\text{ and }\left(  u_{1}%
,u_{2}\right)  \in\mathbb{C}G_{1}\times\mathbb{C}G_{2}.
\label{sol.Indrho.tensor.left-mod}%
\end{equation}
This left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module structure on
$\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ and the right $\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}$-module structure on $\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$ are connected by the equality%
\begin{equation}
\left(  \left(  gm\right)  h=g\left(  mh\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }g\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  \text{, }m\in
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\text{ and }h\in\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}\right)  \label{sol.Indrho.tensor.bimod}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.bimod}):} Let
$g\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $, $m\in\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$ and $h\in\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$.
\par
We need to prove the equality $\left(  gm\right)  h=g\left(  mh\right)  $.
This equality is $\CC$-linear in $g$. Hence, we can WLOG assume that
$g$ belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)  }\right)
_{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\mathbb{C}%
$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. Assume this.
Thus, $g$ belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)
}\right)  _{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the
$\CC$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. In
other words, there exists a $\left(  g_{1},g_{2}\right)  \in G_{1}\times
G_{2}$ such that $g=t_{\left(  g_{1},g_{2}\right)  }$. Consider this $\left(
g_{1},g_{2}\right)  $.
\par
We need to prove the equality $\left(  gm\right)  h=g\left(  mh\right)  $.
This equality is $\CC$-linear in each of $m$ and $h$. Hence, we can
WLOG assume that $m$ and $h$ are pure tensors (because each of the
$\CC$-vector spaces $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ and
$\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$ is spanned by pure tensors). Assume
this. We know that $m$ is a pure tensor. In other words, $m=u_{1}\otimes
u_{2}$ for some $u_{1}\in\mathbb{C}G_{1}$ and some $u_{2}\in\mathbb{C}G_{2}$.
Consider these $u_{1}$ and $u_{2}$. Also, we know that $h$ is a pure tensor.
In other words, $h=p_{1}\otimes p_{2}$ for some $p_{1}\in\mathbb{C}H_{1}$ and
some $p_{2}\in\mathbb{C}H_{2}$. Consider these $p_{1}$ and $p_{2}$.
\par
Now,%
\[
\underbrace{g}_{=t_{\left(  g_{1},g_{2}\right)  }}\underbrace{m}%
_{=u_{1}\otimes u_{2}}=t_{\left(  g_{1},g_{2}\right)  }\left(  u_{1}\otimes
u_{2}\right)  =t_{g_{1}}u_{1}\otimes t_{g_{2}}u_{2}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.Indrho.tensor.left-mod})}\right)  ,
\]
so that%
\[
\underbrace{\left(  gm\right)  }_{=t_{g_{1}}u_{1}\otimes t_{g_{2}}u_{2}%
}\underbrace{h}_{=p_{1}\otimes p_{2}}=\left(  t_{g_{1}}u_{1}\otimes t_{g_{2}%
}u_{2}\right)  \left(  p_{1}\otimes p_{2}\right)  =t_{g_{1}}u_{1}p_{1}\otimes
t_{g_{2}}u_{2}p_{2}%
\]
(by the definition of the right $\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$-module
structure on $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$). On the other hand,%
\[
\underbrace{m}_{=u_{1}\otimes u_{2}}\underbrace{h}_{=p_{1}\otimes p_{2}%
}=\left(  u_{1}\otimes u_{2}\right)  \left(  p_{1}\otimes p_{2}\right)
=u_{1}p_{1}\otimes u_{2}p_{2}%
\]
(by the definition of the right $\mathbb{C}H_{1}\otimes\mathbb{C}H_{2}$-module
structure on $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$), so that%
\[
\underbrace{g}_{=t_{\left(  g_{1},g_{2}\right)  }}\underbrace{\left(
mh\right)  }_{=u_{1}p_{1}\otimes u_{2}p_{2}}=t_{\left(  g_{1},g_{2}\right)
}\left(  u_{1}p_{1}\otimes u_{2}p_{2}\right)  =t_{g_{1}}u_{1}p_{1}\otimes
t_{g_{2}}u_{2}p_{2}%
\]
(by (\ref{sol.Indrho.tensor.left-mod}), applied to $u_{1}p_{1}$ and
$u_{2}p_{2}$ instead of $u_{1}$ and $u_{2}$). Compared with $\left(
gm\right)  h=t_{g_{1}}u_{1}p_{1}\otimes t_{g_{2}}u_{2}p_{2}$, this yields
$\left(  gm\right)  h=g\left(  mh\right)  $. This proves
(\ref{sol.Indrho.tensor.bimod}).}. Hence, these two structures together form
an $\left(  \mathbb{C}\left[  G_{1}\times G_{2}\right]  ,\mathbb{C}%
H_{1}\otimes\mathbb{C}H_{2}\right)  $-bimodule structure on $\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$ (because both of these structures are
$\CC$-bilinear). Thus, the tensor product $\left(  \mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}\right)  \otimes_{\mathbb{C}H_{1}\otimes
\mathbb{C}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $ becomes a left
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module.

Next, we recall that $\mathfrak{G}:\mathbb{C}\left[  G_{1}\times G_{2}\right]
\rightarrow\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ is a $\CC$-algebra
homomorphism, thus a $\CC$-module homomorphism. Furthermore,%
\begin{equation}
\mathfrak{G}\left(  am\right)  =\mathfrak{G}\left(  a\right)  \mathfrak{G}%
\left(  m\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathbb{C}\left[
G_{1}\times G_{2}\right]  \text{ and }m\in\mathbb{C}\left[  G_{1}\times
G_{2}\right]  \label{sol.Indrho.tensor.axiom3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.axiom3}):} Let
$a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $ and $m\in\mathbb{C}\left[
G_{1}\times G_{2}\right]  $.
\par
We need to prove the equality $\mathfrak{G}\left(  am\right)  =\mathfrak{G}%
\left(  a\right)  \mathfrak{G}\left(  m\right)  $. This equality is
$\CC$-linear in $m$ (because the map $\mathfrak{G}$ is $\mathbb{C}%
$-linear). Hence, we can WLOG assume that $m$ belongs to the basis $\left(
t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}}$ of the $\CC$-vector space $\mathbb{C}\left[
G_{1}\times G_{2}\right]  $. Assume this. Thus, $m$ belongs to the basis
$\left(  t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1}%
,g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\CC$-vector space
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. In other words, there exists
some $\left(  f_{1},f_{2}\right)  \in G_{1}\times G_{2}$ such that
$m=t_{\left(  f_{1},f_{2}\right)  }$. Consider this $\left(  f_{1}%
,f_{2}\right)  $. Applying the map $\mathfrak{G}$ to both sides of the
equality $m=t_{\left(  f_{1},f_{2}\right)  }$, we obtain%
\[
\mathfrak{G}\left(  m\right)  =\mathfrak{G}\left(  t_{\left(  f_{1}%
,f_{2}\right)  }\right)  =t_{f_{1}}\otimes t_{f_{2}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\mathfrak{G}\right)  .
\]
\par
We need to prove the equality $\mathfrak{G}\left(  am\right)  =\mathfrak{G}%
\left(  a\right)  \mathfrak{G}\left(  m\right)  $. This equality is
$\CC$-linear in $a$ (because the map $\mathfrak{G}$ is $\mathbb{C}%
$-linear). Hence, we can WLOG assume that $a$ belongs to the basis $\left(
t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}}$ of the $\CC$-vector space $\mathbb{C}\left[
G_{1}\times G_{2}\right]  $. Assume this. Thus, $a$ belongs to the basis
$\left(  t_{\left(  g_{1},g_{2}\right)  }\right)  _{\left(  g_{1}%
,g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\CC$-vector space
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. In other words, there exists
some $\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}$ such that
$a=t_{\left(  g_{1},g_{2}\right)  }$. Consider this $\left(  g_{1}%
,g_{2}\right)  $. Applying the map $\mathfrak{G}$ to both sides of the
equality $a=t_{\left(  g_{1},g_{2}\right)  }$, we obtain%
\[
\mathfrak{G}\left(  a\right)  =\mathfrak{G}\left(  t_{\left(  g_{1}%
,g_{2}\right)  }\right)  =t_{g_{1}}\otimes t_{g_{2}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\mathfrak{G}\right)  .
\]
\par
Now,%
\begin{align*}
\underbrace{a}_{=t_{\left(  g_{1},g_{2}\right)  }}\underbrace{m}_{=t_{\left(
f_{1},f_{2}\right)  }}  &  =t_{\left(  g_{1},g_{2}\right)  }t_{\left(
f_{1},f_{2}\right)  }=t_{\left(  g_{1},g_{2}\right)  }\cdot t_{\left(
f_{1},f_{2}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the left }{\mathbb{C}}\left[  G_{1}\times G_{2}\right]
\text{-module structure on }{\mathbb{C}}\left[  G_{1}\times G_{2}\right]
\text{ is}\\
\text{plain multiplication inside }{\mathbb{C}}\left[  G_{1}\times
G_{2}\right]
\end{array}
\right) \\
&  =t_{\left(  g_{1},g_{2}\right)  \left( f_1, f_2\right)  }
\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the multiplication on }{\mathbb{C}}\left[
G_{1}\times G_{2}\right]  \right)  \\
&  =t_{\left(  g_{1}f_{1},g_{2}f_{2}\right)  } .
\end{align*}
Applying the map $\mathfrak{G}$ to both sides of this equality, we obtain%
\[
\mathfrak{G}\left(  am\right)  =\mathfrak{G}\left(  t_{\left(  g_{1}%
f_{1},g_{2}f_{2}\right)  }\right)  =t_{g_{1}f_{1}}\otimes t_{g_{2}f_{2}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the map }%
\mathfrak{G}\right)  .
\]
Compared with%
\begin{align*}
\underbrace{\mathfrak{G}\left(  a\right)  }_{=t_{g_{1}}\otimes t_{g_{2}}%
}\underbrace{\mathfrak{G}\left(  m\right)  }_{=t_{f_{1}}\otimes t_{f_{2}}}  &
=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \left(  t_{f_{1}}\otimes t_{f_{2}%
}\right)  =\underbrace{t_{g_{1}}t_{f_{1}}}_{\substack{=t_{g_{1}}\cdot
t_{f_{1}}\\\text{(since the left }\mathbb{C}G_{1}\text{-module}%
\\\text{structure on }\mathbb{C}G_{1}\text{ is plain}\\\text{multiplication
inside }\mathbb{C}G_{1}\text{)}}}\otimes\underbrace{t_{g_{2}}t_{f_{2}}%
}_{\substack{=t_{g_{2}}\cdot t_{f_{2}}\\\text{(since the left }\mathbb{C}%
G_{2}\text{-module}\\\text{structure on }\mathbb{C}G_{2}\text{ is
plain}\\\text{multiplication inside }\mathbb{C}G_{2}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the left }\mathbb{C}G_{1}\otimes\mathbb{C}%
G_{2}\text{-module}\\
\text{structure on }\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}%
\end{array}
\right) \\
&  =\left(  t_{g_{1}}\cdot t_{f_{1}}\right)  \otimes\left(  t_{g_{2}}\cdot
t_{f_{2}}\right)  =\underbrace{\left(  t_{g_{1}}\cdot t_{f_{1}}\right)
}_{\substack{=t_{g_{1}f_{1}}\\\text{(by the definition}\\\text{of the
multiplication}\\\text{on }\mathbb{C}G_{1}\text{)}}}\otimes\underbrace{\left(
t_{g_{2}}\cdot t_{f_{2}}\right)  }_{\substack{=t_{g_{2}f_{2}}\\\text{(by the
definition}\\\text{of the multiplication}\\\text{on }\mathbb{C}G_{2}\text{)}%
}}=t_{g_{1}f_{1}}\otimes t_{g_{2}f_{2}},
\end{align*}
this yields $\mathfrak{G}\left(  am\right)  =\mathfrak{G}\left(  a\right)
\mathfrak{G}\left(  m\right)  $. This proves (\ref{sol.Indrho.tensor.axiom3}%
).}. Moreover,%
\begin{equation}
am=\mathfrak{G}\left(  a\right)  m\ \ \ \ \ \ \ \ \ \ \text{for every }%
a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  \text{ and }m\in
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2} \label{sol.Indrho.tensor.axiom4}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.axiom4}):} Let
$a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $ and $m\in\mathbb{C}%
G_{1}\otimes\mathbb{C}G_{2}$.
\par
We need to prove the equality $am=\mathfrak{G}\left(  a\right)  m$. This
equality is $\CC$-linear in $m$. Hence, we can WLOG assume that $m$ is
a pure tensor (since the $\CC$-vector space $\mathbb{C}G_{1}%
\otimes\mathbb{C}G_{2}$ is spanned by the pure tensors). Assume this. We thus
know that $m$ is a pure tensor. In other words, there exist $f_{1}%
\in\mathbb{C}G_{1}$ and $f_{2}\in\mathbb{C}G_{2}$ such that $m=f_{1}\otimes
f_{2}$. Consider these $f_{1}$ and $f_{2}$.
\par
We need to prove the equality $am=\mathfrak{G}\left(  a\right)  m$. This
equality is $\CC$-linear in $a$. Hence, we can WLOG assume that $a$
belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)  }\right)
_{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\mathbb{C}%
$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. Assume this.
Thus, $a$ belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)
}\right)  _{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the
$\CC$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. In
other words, there exists a $\left(  g_{1},g_{2}\right)  \in G_{1}\times
G_{2}$ such that $a=t_{\left(  g_{1},g_{2}\right)  }$. Consider this $\left(
g_{1},g_{2}\right)  $.
\par
The definition of the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$ yields
$t_{\left(  g_{1},g_{2}\right)  }\left(  f_{1}\otimes f_{2}\right)  =t_{g_{1}%
}f_{1}\otimes t_{g_{2}}f_{2}$. But applying the map $\mathfrak{G}$ to the
equality $a=t_{\left(  g_{1},g_{2}\right)  }$, we obtain $\mathfrak{G}\left(
a\right)  =\mathfrak{G}\left(  t_{\left(  g_{1},g_{2}\right)  }\right)
=t_{g_{1}}\otimes t_{g_{2}}$ (by the definition of the map $\mathfrak{G}$).
Hence, $\underbrace{\mathfrak{G}\left(  a\right)  }_{=t_{g_{1}}\otimes
t_{g_{2}}}\underbrace{m}_{=f_{1}\otimes f_{2}}=\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  \left(  f_{1}\otimes f_{2}\right)  =t_{g_{1}}f_{1}\otimes
t_{g_{2}}f_{2}$ (by the definition of the left $\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}$-module structure on $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$).
Compared with $\underbrace{a}_{=t_{\left(  g_{1},g_{2}\right)  }%
}\underbrace{m}_{=f_{1}\otimes f_{2}}=t_{\left(  g_{1},g_{2}\right)  }\left(
f_{1}\otimes f_{2}\right)  =t_{g_{1}}f_{1}\otimes t_{g_{2}}f_{2}$, this yields
$am=\mathfrak{G}\left(  a\right)  m$. This proves
(\ref{sol.Indrho.tensor.axiom4}).}.

Now, we can apply Proposition \ref{prop.sol.Indrho.tensor.MAN}(c) to
$\kk=\mathbb{C}$, $A=\mathbb{C}\left[  G_{1}\times G_{2}\right]  $,
$B=\mathbb{C}\left[  H_{1}\times H_{2}\right]  $, $C=\mathbb{C}$, $A^{\prime
}=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $B^{\prime}=\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}$, $C^{\prime}=\mathbb{C}$, $M=\mathbb{C}\left[
G_{1}\times G_{2}\right]  $, $N=U_{1}\otimes U_{2}$, $M^{\prime}%
=\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$, $N^{\prime}=U_{1}\otimes U_{2}$,
$\beta=\mathfrak{H}$, $\mu=\mathfrak{G}$, $\nu=\operatorname*{id}$ and
$\alpha=\mathfrak{G}$. As a result, we obtain that $\Omega_{\mathfrak{H}%
,\mathfrak{G},\operatorname*{id}}$ is a left $\mathbb{C}\left[  G_{1}\times
G_{2}\right]  $-module homomorphism. Hence, $\Omega_{\mathfrak{H}%
,\mathfrak{G},\operatorname*{id}}$ is a left $\mathbb{C}\left[  G_{1}\times
G_{2}\right]  $-module isomorphism (since $\Omega_{\mathfrak{H},\mathfrak{G}%
,\operatorname*{id}}$ is invertible). Thus, we have found a left
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module isomorphism from
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  $ to $\left(
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\right)  \otimes_{\mathbb{C}H_{1}%
\otimes\mathbb{C}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $ (namely,
$\Omega_{\mathfrak{H},\mathfrak{G},\operatorname*{id}}$). Thus,%
\begin{equation}
\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \cong\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\label{sol.Indrho.tensor.finish-part-2}%
\end{equation}
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules.

Now, let us recall the isomorphism (\ref{sol.Indrho.tensor.finish-part-1}). It
is an isomorphism of left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-modules, and
thus cannot be immediately combined with
(\ref{sol.Indrho.tensor.finish-part-2}). However, it is easy to see that the
corresponding isomorphism of left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-modules holds as well: Namely, we have%
\begin{equation}
\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)
\label{sol.Indrho.tensor.finish-part-1a}%
\end{equation}
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. Before we
prove this, let us make some auxiliary observations which connect the left
$\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-module structures on the
modules appearing in (\ref{sol.Indrho.tensor.finish-part-1a}) with the left
${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$-module structures on the same modules.

First, let us notice that%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }m=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
m\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}\text{ and }m\in{\mathbb{C}}G_{1}\otimes{\mathbb{C}}%
G_{2}\label{sol.Indrho.tensor.CG1CG2CG1G2-1}%
\end{equation}
(where the expression $t_{\left(  g_{1},g_{2}\right)  }m$ on the left hand
side is defined using the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$, whereas the
expression $\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  m$ on the right hand
side is defined using the left ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}%
$-module structure on ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}%
$)\ \ \ \ \footnote{\textit{Proof of (\ref{sol.Indrho.tensor.CG1CG2CG1G2-1}):}
Let $\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}$ and $m\in{\mathbb{C}%
}G_{1}\otimes{\mathbb{C}}G_{2}$.
\par
We need to prove the equality $t_{\left(  g_{1},g_{2}\right)  }m=\left(
t_{g_{1}}\otimes t_{g_{2}}\right)  m$. This equality is $\CC$-linear in
$m$. Hence, we can WLOG assume that $m$ is a pure tensor (since the
$\CC$-vector space ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$ is
spanned by the pure tensors). Assume this. Thus, $m$ is a pure tensor. In
other words, $m=p_{1}\otimes p_{2}$ for some $p_{1}\in\mathbb{C}G_{1}$ and
$p_{2}\in\mathbb{C}G_{2}$. Consider these $p_{1}$ and $p_{2}$.
\par
We have%
\[
t_{\left(  g_{1},g_{2}\right)  }\underbrace{m}_{=p_{1}\otimes p_{2}%
}=t_{\left(  g_{1},g_{2}\right)  }\left(  p_{1}\otimes p_{2}\right)
=t_{g_{1}}p_{1}\otimes t_{g_{2}}p_{2}%
\]
(by the definition of the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$). Comparing
this with%
\begin{align*}
\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \underbrace{m}_{=p_{1}\otimes
p_{2}}  & =\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \left(  p_{1}\otimes
p_{2}\right)  =t_{g_{1}}p_{1}\otimes t_{g_{2}}p_{2}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\text{-module structure on }%
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\right)  ,
\end{align*}
we obtain $t_{\left(  g_{1},g_{2}\right)  }m=\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  m$. This proves (\ref{sol.Indrho.tensor.CG1CG2CG1G2-1}).}.
Next, we observe that%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }n=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
n\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}\text{ and }n\in\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}%
}G_{2}\right)  \otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(
U_{1}\otimes U_{2}\right)  \label{sol.Indrho.tensor.CG1CG2CG1G2-2}%
\end{equation}
(where the expression $t_{\left(  g_{1},g_{2}\right)  }n$ on the left hand
side is defined using the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}%
G_{2}\right)  \otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(
U_{1}\otimes U_{2}\right)  $, whereas the expression $\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  n$ on the right hand side is defined using the left
${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}$-module structure on $\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
$)\ \ \ \ \footnote{\textit{Proof of (\ref{sol.Indrho.tensor.CG1CG2CG1G2-1}):}
Let $\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}$ and $n\in\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $.
\par
We need to prove the equality $t_{\left(  g_{1},g_{2}\right)  }n=\left(
t_{g_{1}}\otimes t_{g_{2}}\right)  n$. This equality is $\CC$-linear in
$n$. Hence, we can WLOG assume that $n$ is a pure tensor (since the
$\CC$-vector space $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}%
G_{2}\right)  \otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(
U_{1}\otimes U_{2}\right)  $ is spanned by the pure tensors). Assume this.
Thus, $n$ is a pure tensor. In other words, $n=m\otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}u$ for some $m\in{\mathbb{C}}G_{1}%
\otimes{\mathbb{C}}G_{2}$ and $u\in U_{1}\otimes U_{2}$. Consider these
$m$ and $u$.
\par
We have%
\begin{align*}
t_{\left(  g_{1},g_{2}\right)  }\underbrace{n}_{=m\otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}u} &  =t_{\left(  g_{1},g_{2}\right)  }\left(
m\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}u\right)
=\underbrace{t_{\left(  g_{1},g_{2}\right)  }m}_{\substack{=\left(  t_{g_{1}%
}\otimes t_{g_{2}}\right)  m\\\text{(by (\ref{sol.Indrho.tensor.CG1CG2CG1G2-1}%
))}}}\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the left }\mathbb{C}\left[  G_{1}\times
G_{2}\right]  \text{-module structure on}\\
\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\end{array}
\right)  \\
&  =\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  m\otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}u.
\end{align*}
Comparing this with%
\begin{align*}
\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \underbrace{n}_{=m\otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}u} &  =\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  \left(  m\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}%
}u\right)  =\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  m\otimes_{\CC
H_{1}\otimes \CC H_{2}}u\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the left }\mathbb{C}G_{1}\otimes\mathbb{C}%
G_{2}\text{-module structure on}\\
\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\end{array}
\right)  ,
\end{align*}
we obtain $t_{\left(  g_{1},g_{2}\right)  }n=\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  n$. This proves (\ref{sol.Indrho.tensor.CG1CG2CG1G2-2}).}.
Furthermore, we can see that%
\begin{equation}
t_{\left(  g_{1},g_{2}\right)  }m=\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
m\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  g_{1},g_{2}\right)  \in
G_{1}\times G_{2}\text{ and }m\in\left(  \operatorname{Ind}_{\rho_{1}}%
U_{1}\right)  \otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)
\label{sol.Indrho.tensor.CG1CG2CG1G2-3}%
\end{equation}
(where the expression $t_{\left(  g_{1},g_{2}\right)  }m$ on the left hand
side is defined using the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $, whereas the
expression $\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  m$ on the right hand
side is defined using the left ${\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}%
$-module structure on $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $%
)\ \ \ \ \footnote{\textit{Proof of (\ref{sol.Indrho.tensor.CG1CG2CG1G2-3}):}
Let $\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}$ and $m\in\left(
\operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(  \operatorname{Ind}%
_{\rho_{2}}U_{2}\right)  $.
\par
We need to prove the equality $t_{\left(  g_{1},g_{2}\right)  }m=\left(
t_{g_{1}}\otimes t_{g_{2}}\right)  m$. This equality is $\CC$-linear in
$m$. Hence, we can WLOG assume that $m$ is a pure tensor (since the
$\CC$-vector space $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $ is spanned by the
pure tensors). Assume this. Thus, $m$ is a pure tensor. In other words,
$m=p_{1}\otimes p_{2}$ for some $p_{1}\in\operatorname{Ind}_{\rho_{1}}U_{1}$
and $p_{2}\in\operatorname{Ind}_{\rho_{2}}U_{2}$. Consider these $p_{1}$ and
$p_{2}$.
\par
We have%
\[
t_{\left(  g_{1},g_{2}\right)  }\underbrace{m}_{=p_{1}\otimes p_{2}%
}=t_{\left(  g_{1},g_{2}\right)  }\left(  p_{1}\otimes p_{2}\right)
=t_{g_{1}}p_{1}\otimes t_{g_{2}}p_{2}%
\]
(by the definition of the left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-module structure on $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)  $). Comparing this
with%
\begin{align*}
\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \underbrace{m}_{=p_{1}\otimes
p_{2}} &  =\left(  t_{g_{1}}\otimes t_{g_{2}}\right)  \left(  p_{1}\otimes
p_{2}\right)  =t_{g_{1}}p_{1}\otimes t_{g_{2}}p_{2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}\text{-module structure on }\left(
\operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(  \operatorname{Ind}%
_{\rho_{2}}U_{2}\right)  \right)  ,
\end{align*}
we obtain $t_{\left(  g_{1},g_{2}\right)  }m=\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  m$. This proves (\ref{sol.Indrho.tensor.CG1CG2CG1G2-3}).}.

Now, it is easy to see that (\ref{sol.Indrho.tensor.finish-part-1a})
holds\footnote{\textit{Proof of (\ref{sol.Indrho.tensor.finish-part-1a}):}
From (\ref{sol.Indrho.tensor.finish-part-1}), we conclude that there exists an
isomorphism%
\[
\mathbf{T}:\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)
\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes
U_{2}\right)  \rightarrow\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)
\]
of left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-modules. We shall now show
that this map $\mathbf{T}$ is an isomorphism of left $\mathbb{C}\left[
G_{1}\times G_{2}\right]  $-modules as well.
\par
First, we notice that the map $\mathbf{T}$ is invertible (because it is an
isomorphism of left $\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-modules) and
$\CC$-linear (for the same reason) and a homomorphism of left
$\mathbb{C}G_{1}\otimes\mathbb{C}G_{2}$-modules (for the same reason). We
shall now prove that%
\begin{equation}
\mathbf{T}\left(  an\right)  =a\mathbf{T}\left(  n\right)
\label{sol.Indrho.tensor.finish-part-1a.pf.1}%
\end{equation}
for every $a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $ and $n\in\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $.
\par
\textit{Proof of (\ref{sol.Indrho.tensor.finish-part-1a.pf.1}):} Let
$a\in\mathbb{C}\left[  G_{1}\times G_{2}\right]  $ and $n\in\left(
{\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes_{\CC %
H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)  $. We need to
prove the equality $\mathbf{T}\left(  an\right)  =a\mathbf{T}\left(  n\right)
$. This equality is $\CC$-linear in $a$. Hence, we can WLOG assume that
$a$ belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)  }\right)
_{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the $\mathbb{C}%
$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. Assume this.
Thus, $a$ belongs to the basis $\left(  t_{\left(  g_{1},g_{2}\right)
}\right)  _{\left(  g_{1},g_{2}\right)  \in G_{1}\times G_{2}}$ of the
$\CC$-vector space $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $. In
other words, there exists some $\left(  g_{1},g_{2}\right)  \in G_{1}\times
G_{2}$ such that $a=t_{\left(  g_{1},g_{2}\right)  }$. Consider this $\left(
g_{1},g_{2}\right)  $. We have $\underbrace{a}_{=t_{\left(  g_{1}%
,g_{2}\right)  }}n=t_{\left(  g_{1},g_{2}\right)  }n=\left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  n$ (by (\ref{sol.Indrho.tensor.CG1CG2CG1G2-2})). Applying
the map $\mathbf{T}$ to both sides of this equality, we obtain%
\[
\mathbf{T}\left(  an\right)  =\mathbf{T}\left(  \left(  t_{g_{1}}\otimes
t_{g_{2}}\right)  n\right)  =\left(  t_{g_{1}}\otimes t_{g_{2}}\right)
\mathbf{T}\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{T}\text{ is a homomorphism of left }\mathbb{C}G_{1}\otimes
\mathbb{C}G_{2}\text{-modules}\right)  .
\]
Compared with%
\[
\underbrace{a}_{=t_{\left(  g_{1},g_{2}\right)  }}\mathbf{T}\left(  n\right)
=t_{\left(  g_{1},g_{2}\right)  }\mathbf{T}\left(  n\right)  =\left(
t_{g_{1}}\otimes t_{g_{2}}\right)  \mathbf{T}\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.tensor.CG1CG2CG1G2-3}),
applied to }m=\mathbf{T}\left(  n\right)  \right)  ,
\]
this yields $\mathbf{T}\left(  an\right)  =a\mathbf{T}\left(  n\right)  $.
This proves (\ref{sol.Indrho.tensor.finish-part-1a.pf.1}).
\par
Now, the map $\mathbf{T}$ is $\CC$-linear, and therefore
is a homomorphism of left $\mathbb{C}\left[  G_{1}\times
G_{2}\right]  $-modules
(because (\ref{sol.Indrho.tensor.finish-part-1a.pf.1}) holds).
Hence, $\mathbf{T}$ is
an isomorphism of left $\mathbb{C}\left[  G_{1}\times G_{2}\right]
$-modules (since $\mathbf{T}$ is invertible). Thus, there exists an
isomorphism of left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules
from $\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
$ to $\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)  $ (namely, $\mathbf{T}$). Hence,
$\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)  \otimes
_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes U_{2}\right)
\cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(
\operatorname{Ind}_{\rho_{2}}U_{2}\right)  $ as left $\mathbb{C}\left[
G_{1}\times G_{2}\right]  $-modules. This proves
(\ref{sol.Indrho.tensor.finish-part-1a}).}.

Now, (\ref{sol.Indrho.tensor.finish-part-2}) becomes%
\begin{align*}
&  \mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right) \\
&  \cong\left(  {\mathbb{C}}G_{1}\otimes{\mathbb{C}}G_{2}\right)
\otimes_{\CC H_{1}\otimes{\mathbb{C}}H_{2}}\left(  U_{1}\otimes
U_{2}\right)  \cong\left(  \operatorname{Ind}_{\rho_{1}}U_{1}\right)
\otimes\left(  \operatorname{Ind}_{\rho_{2}}U_{2}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.tensor.finish-part-1a}%
)}\right)
\end{align*}
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. Therefore,%
\[
\operatorname{Ind}_{\rho_{1}\times\rho_{2}}\left(  U_{1}\otimes U_{2}\right)
=\mathbb{C}\left[  G_{1}\times G_{2}\right]  \otimes_{\mathbb{C}\left[
H_{1}\times H_{2}\right]  }\left(  U_{1}\otimes U_{2}\right)  \cong\left(
\operatorname{Ind}_{\rho_{1}}U_{1}\right)  \otimes\left(  \operatorname{Ind}%
_{\rho_{2}}U_{2}\right)
\]
as left $\mathbb{C}\left[  G_{1}\times G_{2}\right]  $-modules. This finishes
the solution of Exercise \ref{exe.Indrho.tensor}.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Indrho.transitivity}}

\textit{Solution to Exercise~\ref{exe.Indrho.transitivity}.} In the following,
we will use the following convention: Whenever $K$ is a group, and $k$ is an
element of $K$, we shall write $k$ for the element $t_{k}$ of $\CC K$.
This is a relatively common abuse of notation, and it is harmless because
the map $K\rightarrow \CC K,\ k\mapsto t_{k}$ is an injective
homomorphism of multiplicative monoids (so $t_{gh}=t_{g}t_{h}$ and $t_{1}=1$,
which means that we won't run into ambiguities denoting $t_{k}$ by $k$) and
because every $\CC K$-module $M$, every $m\in M$ and every $k\in K$
satisfy $km=t_{k}m$.

We solve the four parts of Exercise \ref{exe.Indrho.transitivity} in the
following order: first, part (c); then, part (d); then, part (a); finally,
part (b).

(c) The definition of $\Res_{\rho}V$ yields
$\Res_{\rho}V=V$ as $\CC$-vector spaces.
Similarly, $\operatorname{Res}_{\tau}\Res_{\rho
}V=\Res_{\rho}V$ as $\CC$-vector spaces, and
$\operatorname{Res}_{\rho\circ\tau}V=V$ as $\CC$-vector spaces. Thus,
$\operatorname{Res}_{\tau}\Res_{\rho}%
V=\Res_{\rho}V=V=\operatorname{Res}_{\rho\circ\tau}V$
as $\CC$-vector spaces. But our goal is to show that
$\operatorname{Res}_{\tau}\Res_{\rho}%
V=\operatorname{Res}_{\rho\circ\tau}V$ as $\mathbb{C}I$-modules. Thus, it
suffices to prove that the left $\mathbb{C}I$-module structures on
$\operatorname{Res}_{\tau}\Res_{\rho}V$ and
$\operatorname{Res}_{\rho\circ\tau}V$ are identical. In other words, it
suffices to show that every $i\in\mathbb{C}I$ and $v\in V$ satisfy the
equality%
\begin{align}
&  \left(  \text{the action of }i\in\mathbb{C}I\text{ on the element }v\text{
of the left }\mathbb{C}I\text{-module }\operatorname{Res}_{\tau}%
\Res_{\rho}V\right) \nonumber\\
&  =\left(  \text{the action of }i\in\mathbb{C}I\text{ on the element }v\text{
of the left }\mathbb{C}I\text{-module }\operatorname{Res}_{\rho\circ\tau
}V\right)  . \label{sol.Indrho.transitivity.c.goal}%
\end{align}
(Of course, both sides of the equality (\ref{sol.Indrho.transitivity.c.goal})
could be rewritten as $iv$, but this notation is ambiguous, because $i$
simultaneously belongs to two $\mathbb{C}I$-modules $\operatorname{Res}_{\tau
}\Res_{\rho}V$ and $\operatorname{Res}_{\rho\circ\tau
}V$ which are not yet known to be identical.)

\textit{Proof of (\ref{sol.Indrho.transitivity.c.goal}):} Let $i\in
\mathbb{C}I$ and $v\in V$. We need to prove the equality
(\ref{sol.Indrho.transitivity.c.goal}). Since this equality is $\mathbb{C}%
$-linear in $i$, we can WLOG assume that $i$ belongs to the basis $I$ of the
$\CC$-vector space $\mathbb{C}I$. Assume this. Now, the definition of
the left $\mathbb{C}I$-module $\operatorname{Res}_{\tau}\Res_{\rho}V$ yields%
\begin{align*}
&  \left(  \text{the action of }i\in\mathbb{C}I\text{ on the element }v\text{
of the left }\mathbb{C}I\text{-module }\operatorname{Res}_{\tau}%
\Res_{\rho}V\right) \\
&  =\left(  \text{the action of }\tau\left(  i\right)  \in\mathbb{C}H\text{ on
the element }v\text{ of the left }\mathbb{C}H\text{-module }%
\Res_{\rho}V\right) \\
&  =\left(  \text{the action of }\underbrace{\rho\left(  \tau\left(  i\right)
\right)  }_{=\left(  \rho\circ\tau\right)  \left(  i\right)  }\in
\mathbb{C}G\text{ on the element }v\text{ of the left }\mathbb{C}%
G\text{-module }V\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}H\text{-module }\Res_{\rho}V\right) \\
&  =\left(  \text{the action of }\left(  \rho\circ\tau\right)  \left(
i\right)  \in\mathbb{C}G\text{ on the element }v\text{ of the left }%
\mathbb{C}G\text{-module }V\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \left(  \text{the action of }i\in\mathbb{C}I\text{ on the element }v\text{
of the left }\mathbb{C}I\text{-module }\operatorname{Res}_{\rho\circ\tau
}V\right) \\
&  =\left(  \text{the action of }\left(  \rho\circ\tau\right)  \left(
i\right)  \in\mathbb{C}G\text{ on the element }v\text{ of the left }%
\mathbb{C}G\text{-module }V\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left }%
\mathbb{C}I\text{-module }\operatorname{Res}_{\rho\circ\tau}V\right)  ,
\end{align*}
this yields%
\begin{align*}
&  \left(  \text{the action of }i\in\mathbb{C}I\text{ on the element }v\text{
of the left }\mathbb{C}I\text{-module }\operatorname{Res}_{\tau}%
\Res_{\rho}V\right) \\
&  =\left(  \text{the action of }i\in\mathbb{C}I\text{ on the element }v\text{
of the left }\mathbb{C}I\text{-module }\operatorname{Res}_{\rho\circ\tau
}V\right)  .
\end{align*}
This proves (\ref{sol.Indrho.transitivity.c.goal}).

Now, we know that $\operatorname{Res}_{\tau}\Res_{\rho
}V=\operatorname{Res}_{\rho\circ\tau}V$ as $\CC$-vector spaces. Thus,
(\ref{sol.Indrho.transitivity.c.goal}) shows that $\operatorname{Res}_{\tau
}\Res_{\rho}V=\operatorname{Res}_{\rho\circ\tau}V$ as
left $\mathbb{C}I$-modules as well. Exercise \ref{exe.Indrho.transitivity}(c)
is thus solved.

(d) Let $f\in R_{\mathbb{C}}\left(  G\right)  $. The definition of
$\operatorname{Res}_{\rho\circ\tau}f$ yields $\operatorname{Res}_{\rho
\circ\tau}f=f\circ\left(  \rho\circ\tau\right)  $. But the definition of
$\Res_{\rho}f$ yields $\Res_{\rho}f=f\circ\rho$. The definition of $\operatorname{Res}_{\tau
}\Res_{\rho}f$ yields%
\[
\operatorname{Res}_{\tau}\Res_{\rho}%
f=\underbrace{\left(  \Res_{\rho}f\right)  }%
_{=f\circ\rho}\circ\tau=\left(  f\circ\rho\right)  \circ\tau=f\circ\left(
\rho\circ\tau\right)  =\operatorname{Res}_{\rho\circ\tau}f.
\]
This solves Exercise \ref{exe.Indrho.transitivity}(d).

(a) Let $U$ be any $\mathbb{C}I$-module.

Recall that $\operatorname{Ind}_{\tau}U$ is defined as the ${\mathbb{C}}%
H$-module ${\mathbb{C}}H\otimes_{\CC I}U$, where ${\mathbb{C}}H$ is
regarded as a $\left(  {\mathbb{C}}H,{\mathbb{C}}I\right)  $-bimodule
according to the following rule: The left ${\mathbb{C}}H$-module structure on
${\mathbb{C}}H$ is plain multiplication inside ${\mathbb{C}}H$; the right
${\mathbb{C}}I$-module structure on ${\mathbb{C}}H$ is induced by the
${\mathbb{C}}$-algebra homomorphism ${\mathbb{C}}\left[  \tau\right]
:{\mathbb{C}}I\rightarrow{\mathbb{C}}H$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \tau\right]  \right)  \eta$
for all $\gamma\in\mathbb{C}H$ and $\eta\in\mathbb{C}I$).

Furthermore, $\operatorname{Ind}_{\rho}\left(  \Ind_{\tau}U\right)  $ is defined as the ${\mathbb{C}}G$-module
${\mathbb{C}}G\otimes_{\CC H}\left(  \Ind_{\tau}U\right)  $, where ${\mathbb{C}}G$ is regarded as a $\left(
{\mathbb{C}}G,{\mathbb{C}}H\right)  $-bimodule according to the following
rule: The left ${\mathbb{C}}G$-module structure on ${\mathbb{C}}G$ is plain
multiplication inside ${\mathbb{C}}G$; the right ${\mathbb{C}}H$-module
structure on ${\mathbb{C}}G$ is induced by the ${\mathbb{C}}$-algebra
homomorphism ${\mathbb{C}}\left[  \rho\right]  :{\mathbb{C}}H\rightarrow
{\mathbb{C}}G$ (thus, it is explicitly given by $\gamma\eta=\gamma\cdot\left(
{\mathbb{C}}\left[  \rho\right]  \right)  \eta$ for all $\gamma\in{\mathbb{C}%
}G$ and $\eta\in{\mathbb{C}}H$).

Finally, $\operatorname{Ind}_{\rho\circ\tau}U$ is defined as the ${\mathbb{C}%
}G$-module ${\mathbb{C}}G\otimes_{\CC I}U$, where ${\mathbb{C}}G$ is
regarded as a $\left(  {\mathbb{C}}G,{\mathbb{C}}I\right)  $-bimodule
according to the following rule: The left ${\mathbb{C}}G$-module structure on
${\mathbb{C}}G$ is plain multiplication inside ${\mathbb{C}}G$; the right
${\mathbb{C}}I$-module structure on ${\mathbb{C}}G$ is induced by the
${\mathbb{C}}$-algebra homomorphism ${\mathbb{C}}\left[  \rho\circ\tau\right]
:{\mathbb{C}}I\rightarrow{\mathbb{C}}G$ (thus, it is explicitly given by
$\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho\circ\tau\right]
\right)  \eta$ for all $\gamma\in{\mathbb{C}}G$ and $\eta\in{\mathbb{C}}I$).

Thus, we have introduced a $\left(  {\mathbb{C}}H,{\mathbb{C}}I\right)
$-bimodule structure on $\mathbb{C}H$, a $\left(  {\mathbb{C}}G,{\mathbb{C}%
}H\right)  $-bimodule structure on $\mathbb{C}G$ and a $\left(  {\mathbb{C}%
}G,{\mathbb{C}}I\right)  $-bimodule structure on $\mathbb{C}G$. The left
$\mathbb{C}G$-module structure that underlies the $\left(  {\mathbb{C}%
}G,{\mathbb{C}}H\right)  $-bimodule structure on $\mathbb{C}G$ is identical
with the left $\mathbb{C}G$-module structure that underlies the $\left(
{\mathbb{C}}G,{\mathbb{C}}I\right)  $-bimodule structure on $\mathbb{C}G$
(because both of these left $\mathbb{C}G$-module structures are defined to be
plain multiplication inside $\mathbb{C}G$). Therefore, we will not run into
ambiguities if we write expressions such as $ab$ for $a\in\mathbb{C}G$ and
$b\in\mathbb{C}G$.

We shall now show that%
\begin{equation}
\mathbb{C}G\otimes_{\CC H}\mathbb{C}H\cong\mathbb{C}%
G\ \ \ \ \ \ \ \ \ \ \text{as }\left(  \mathbb{C}G,\mathbb{C}I\right)
\text{-bimodules.} \label{sol.Indrho.transitivity.a.1}%
\end{equation}
(Here, on the left hand side, $\mathbb{C}G$ is regarded as a $\left(
{\mathbb{C}}G,{\mathbb{C}}H\right)  $-bimodule and $\mathbb{C}H$ is regarded
as $\left(  {\mathbb{C}}H,{\mathbb{C}}I\right)  $-bimodule, whereas on the
right hand side, $\mathbb{C}G$ is regarded as a $\left(  {\mathbb{C}%
}G,{\mathbb{C}}I\right)  $-bimodule.)

\textit{Proof of (\ref{sol.Indrho.transitivity.a.1}):} There is clearly a
unique $\CC$-vector space isomorphism $\Phi:\mathbb{C}G\rightarrow
\mathbb{C}G\otimes_{\CC H}\mathbb{C}H$ which satisfies%
\begin{equation}
\left(  \Phi\left(  m\right)  =m\otimes_{\CC H}1_{\mathbb{C}%
H}\ \ \ \ \ \ \ \ \ \ \text{for all }m\in\mathbb{C}G\right)  .
\label{sol.Indrho.transitivity.a.1.pf.1}%
\end{equation}
\footnote{Indeed, this is a particular case of the following fundamental fact
from linear algebra: If $A$ is a $\CC$-algebra, and if $M$ is a right
$A$-module, then there is a unique $\CC$-vector space isomorphism
$\Phi:M\rightarrow M\otimes_{A}A$ which satisfies%
\[
\left(  \Phi\left(  m\right)  =m\otimes_{A}1_{A}\ \ \ \ \ \ \ \ \ \ \text{for
all }m\in M\right)  .
\]
} Consider this $\Phi$. It is straightforward to see that $\Phi$ is a
homomorphism of left $G$-sets and a homomorphism of right $I$-sets. Hence,
$\Phi$ is a homomorphism of $\left(  \mathbb{C}G,\mathbb{C}I\right)
$-bimodules (since $\Phi$ is $\CC$-linear), thus an isomorphism of
$\left(  \mathbb{C}G,\mathbb{C}I\right)  $-bimodules (since $\Phi$ is
invertible). Therefore, there exists an isomorphism of $\left(  \mathbb{C}%
G,\mathbb{C}I\right)  $-bimodules from $\mathbb{C}G$ to $\mathbb{C}%
G\otimes_{\CC H}\mathbb{C}H$ (namely, $\Phi$). In other words,
$\mathbb{C}G\otimes_{\CC H}\mathbb{C}H\cong\mathbb{C}G$ as $\left(
\mathbb{C}G,\mathbb{C}I\right)  $-bimodules. This proves
(\ref{sol.Indrho.transitivity.a.1}).

Now,%
\begin{align*}
\operatorname{Ind}_{\rho}\left(  \Ind_{\tau}U\right)
&  ={\mathbb{C}}G\otimes_{\CC H}\underbrace{\left(
\Ind_{\tau}U\right)  }_{={\mathbb{C}}H\otimes
_{\CC I}U}={\mathbb{C}}G\otimes_{\CC H}\left(  {\mathbb{C}%
}H\otimes_{\CC I}U\right) \\
&  \cong\underbrace{\left(  \mathbb{C}G\otimes_{\CC H}\mathbb{C}%
H\right)  }_{\substack{\cong\mathbb{C}G\text{ as }\left(  \mathbb{C}%
G,\mathbb{C}I\right)  \text{-bimodules}\\\text{(by
(\ref{sol.Indrho.transitivity.a.1}))}}}\otimes_{\mathbb{C}I}%
U\ \ \ \ \ \ \ \ \ \ \left(  \text{by the associativity of the tensor
product}\right) \\
&  \cong\mathbb{C}G\otimes_{\mathbb{C}I}U=\operatorname{Ind}_{\rho\circ\tau
}U\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname{Ind}_{\rho\circ\tau
}U={\mathbb{C}}G\otimes_{\CC I}U\text{ as left }\mathbb{C}%
G\text{-modules}\right)
\end{align*}
as left $\mathbb{C}G$-modules. This solves Exercise
\ref{exe.Indrho.transitivity}(a).

(b) \textit{First solution to Exercise \ref{exe.Indrho.transitivity}(b).} Let
$f\in R_{\mathbb{C}}\left(  I\right)  $. Every $r\in H$ satisfies%
\begin{align}
\left(  \operatorname{Ind}_{\tau}f\right)  \left(  r\right)   &  =\dfrac
{1}{\left\vert I\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in I\times
H;\\k\tau\left(  h\right)  k^{-1}=r}}f\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Ind_{\tau}f\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\underbrace{\sum\limits_{\substack{\left(
i,v\right)  \in I\times H;\\v\tau\left(  i\right)  v^{-1}=r}}}_{=\sum_{i\in
I}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}}f\left(
i\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation
index }\left(  h,k\right)  \text{ as }\left(  i,v\right)  \right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\sum_{i\in I}\sum\limits_{\substack{v\in
H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  .
\label{sol.Indrho.transitivity.b.1}%
\end{align}


Now, let $g\in G$. Then, the definition of $\Ind_{\rho
}\left(  \Ind_{\tau}f\right)  $ yields%
\begin{align}
\left(  \Ind_{\rho}\left(  \Ind_{\tau}f\right)  \right)  \left(  g\right)   &  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(
h\right)  k^{-1}=g}}\left(  \Ind_{\tau}f\right)
\left(  h\right)  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits
_{\substack{\left(  r,k\right)  \in H\times G;\\k\rho\left(  r\right)
k^{-1}=g}}}_{=\sum_{k\in G}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g}}}\underbrace{\left(  \Ind_{\tau}f\right)
\left(  r\right)  }_{\substack{=\dfrac{1}{\left\vert I\right\vert }\sum_{i\in
I}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)
\\\text{(by (\ref{sol.Indrho.transitivity.b.1}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  h,k\right)  \text{ as }\left(  r,k\right)  \right) \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{k\in G}\sum\limits_{\substack{r\in
H;\\k\rho\left(  r\right)  k^{-1}=g}}\dfrac{1}{\left\vert I\right\vert }%
\sum_{i\in I}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{k\in G}\underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g}}\sum_{i\in I}}_{=\sum_{i\in I}\sum\limits_{\substack{r\in H;\\k\rho\left(
r\right)  k^{-1}=g}}}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{k\in G}\sum_{i\in I}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g}}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(
i\right)  . \label{sol.Indrho.transitivity.b.3}%
\end{align}
But every $k\in G$ and $i\in I$ satisfy%
\begin{equation}
\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}}\sum\limits_{\substack{v\in
H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  =\sum\limits_{\substack{v\in
H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)
\label{sol.Indrho.transitivity.b.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.b.5}):} Let $k\in G$
and $i\in I$. We must prove (\ref{sol.Indrho.transitivity.b.5}). We have%
\begin{align}
&  \underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}%
}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}}_{=\sum_{v\in
H}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}}f\left(  i\right) \nonumber\\
&  =\sum_{v\in H}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  =\sum\limits
_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}%
}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  +\sum\limits_{\substack{v\in H;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}\neq g}}\sum\limits_{\substack{r\in
H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right)  \label{sol.Indrho.transitivity.b.5.pf.0}%
\end{align}
(because every $v\in H$ satisfies either $k\rho\left(  v\right)  \cdot
\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)
\right)  ^{-1}=g$ or $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g$,
but never both).
\par
Now, we shall show that%
\begin{equation}
\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  =0
\label{sol.Indrho.transitivity.b.5.pf.1}%
\end{equation}
for every $v\in H$ which satisfies $k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}\neq g$.
\par
\textit{Proof of (\ref{sol.Indrho.transitivity.b.5.pf.1}):} Let $v\in H$ be
such that $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)
\right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g$. If some
$r\in H$ satisfies $k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(
i\right)  v^{-1}=r$, then this $r$ must satisfy%
\begin{align*}
k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\underbrace{\left(  k\rho\left(  v\right)  \right)  ^{-1}}_{=\left(
\rho\left(  v\right)  \right)  ^{-1}k^{-1}}  &  =k\underbrace{\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
\rho\left(  v\right)  \right)  ^{-1}}_{\substack{=\rho\left(  v\tau\left(
i\right)  v^{-1}\right)  \\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}k^{-1}\\
&  =k\rho\left(  \underbrace{v\tau\left(  i\right)  v^{-1}}_{=r}\right)
k^{-1}=k\rho\left(  r\right)  k^{-1}=g,
\end{align*}
which contradicts $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g$.
Hence, we have obtained a contradiction for every $r\in H$ which satisfies
$k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$. Thus,
there exists no $r\in H$ which satisfies $k\rho\left(  r\right)  k^{-1}=g$ and
$v\tau\left(  i\right)  v^{-1}=r$. Therefore, the sum $\sum\limits_{\substack{r\in
H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right)  $ is an empty sum, and thus its value is $0$. This
proves (\ref{sol.Indrho.transitivity.b.5.pf.1}).
\par
On the other hand, let us show that
\begin{equation}
\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  =f\left(  i\right)
\label{sol.Indrho.transitivity.b.5.pf.2}%
\end{equation}
for every $v\in H$ which satisfies $k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g$.
\par
\textit{Proof of (\ref{sol.Indrho.transitivity.b.5.pf.2}):} Let $v\in H$ be
such that $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)
\right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g$. If some $r\in
H$ satisfies $k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)
v^{-1}=r$, then this $r$ must equal $v\tau\left(  i\right)  v^{-1}$ (because
$v\tau\left(  i\right)  v^{-1}=r$). Hence, there exists \textbf{at most one}
$r\in H$ which satisfies $k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(
i\right)  v^{-1}=r$.
\par
On the other hand, the element $v\tau\left(  i\right)  v^{-1}$ of $H$
satisfies%
\[
k\underbrace{\rho\left(  v\tau\left(  i\right)  v^{-1}\right)  }%
_{\substack{=\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)
\right)  \cdot\rho\left(  v\right)  ^{-1}\\\text{(since }\rho\text{ is a
group}\\\text{homomorphism)}}}k^{-1}=k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\underbrace{\rho\left(  v\right)
^{-1}k^{-1}}_{=\left(  k\rho\left(  v\right)  \right)  ^{-1}}=k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g
\]
and $v\tau\left(  i\right)  v^{-1}=v\tau\left(  i\right)  v^{-1}$. In other
words, $v\tau\left(  i\right)  v^{-1}$ is an element $r\in H$ which satisfies
$k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$.
Hence, there exists \textbf{at least one} $r\in H$ which satisfies
$k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$
(namely, $v\tau\left(  i\right)  v^{-1}$). Since we also have shown that there
exists \textbf{at most one} such $r\in H$, we can thus conclude that there
exists \textbf{exactly one} $r\in H$ which satisfies $k\rho\left(  r\right)
k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$. In other words, the sum
$\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  $ has precisely one addend. Hence, this
sum rewrites as follows: $\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  =f\left(
i\right)  $. This proves (\ref{sol.Indrho.transitivity.b.5.pf.2}).
\par
Now, (\ref{sol.Indrho.transitivity.b.5.pf.0}) becomes%
\begin{align*}
&  \sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}}\sum\limits
_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right) \\
&  =\sum\limits_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g}}\underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  }%
_{\substack{=f\left(  i\right)  \\\text{(by
(\ref{sol.Indrho.transitivity.b.5.pf.2}))}}}+\sum\limits_{\substack{v\in
H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g}}\underbrace{\sum\limits
_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(  i\right)
v^{-1}=r}}f\left(  i\right)  }_{\substack{=0\\\text{(by
(\ref{sol.Indrho.transitivity.b.5.pf.1}))}}}\\
&  =\sum\limits_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g}}f\left(  i\right)  +\underbrace{\sum\limits_{\substack{v\in H;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}\neq g}}0}_{=0}=\sum\limits_{\substack{v\in
H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)  .
\end{align*}
This proves (\ref{sol.Indrho.transitivity.b.5}).}.

Thus, (\ref{sol.Indrho.transitivity.b.3}) becomes%
\begin{align}
&  \left(  \Ind_{\rho}\left(  \Ind_{\tau}f\right)  \right)  \left(  g\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{k\in G}\sum_{i\in I}}_{=\sum_{i\in I}\sum_{k\in G}%
}\underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}%
}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)
}_{\substack{=\sum\limits_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g}}f\left(  i\right)  \\\text{(by (\ref{sol.Indrho.transitivity.b.5}))}%
}} \nonumber \\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\underbrace{\sum_{k\in G}\sum\limits_{\substack{v\in H;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g}}}_{=\sum_{v\in H}\sum\limits_{\substack{k\in
G;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}}}f\left(  i\right)
\nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\sum_{v\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(  v\right)
\cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(
v\right)  \right)  ^{-1}=g}}f\left(  i\right)  .
\label{sol.Indrho.transitivity.b.7}%
\end{align}


However, for every $i\in I$ and $v\in H$, we have
\begin{equation}
\sum\limits_{\substack{k\in G;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}%
=g}}f\left(  i\right)  =\sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(
i\right)  \right)  k^{-1}=g}}f\left(  i\right)
\label{sol.Indrho.transitivity.b.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.b.9}):} Let $i\in I$
and $v\in H$. We have $\rho\left(  v\right)  \in G$. Therefore, the map
$G\rightarrow G,\ k\mapsto k\rho\left(  v\right)  $ is a bijection (since $G$
is a group). Therefore, we can substitute $k\rho\left(  v\right)  $ for $k$ in
the sum $\sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)  \right)
k^{-1}=g}}f\left(  i\right)  $. We thus obtain%
\[
\sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)  \right)  k^{-1}%
=g}}f\left(  i\right)  =\sum\limits_{\substack{k\in G;\\k\rho\left(  v\right)
\rho\left(  \tau\left(  i\right)  \right)  \left(  k\rho\left(  v\right)
\right)  ^{-1}=g}}f\left(  i\right)  =\sum\limits_{\substack{k\in G;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)  .
\]
This proves (\ref{sol.Indrho.transitivity.b.9}).}.

Thus, (\ref{sol.Indrho.transitivity.b.7}) becomes%
\begin{align*}
&  \left(  \Ind_{\rho}\left(  \Ind_{\tau}f\right)  \right)  \left(  g\right) \\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\sum_{v\in H}\underbrace{\sum\limits_{\substack{k\in G;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)  }_{\substack{=\sum\limits
_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)  \right)  k^{-1}%
=g}}f\left(  i\right)  \\\text{(by (\ref{sol.Indrho.transitivity.b.9}))}%
}}=\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\underbrace{\sum_{v\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
\tau\left(  i\right)  \right)  k^{-1}=g}}f\left(  i\right)  }_{=\left\vert
H\right\vert \sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)
\right)  k^{-1}=g}}f\left(  i\right)  }\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\left\vert H\right\vert \sum\limits_{\substack{k\in G;\\k\rho\left(
\tau\left(  i\right)  \right)  k^{-1}=g}}f\left(  i\right)  =\dfrac
{1}{\left\vert I\right\vert }\underbrace{\dfrac{1}{\left\vert H\right\vert
}\left\vert H\right\vert }_{=1}\underbrace{\sum_{i\in I}\sum\limits_{\substack{k\in
G;\\k\rho\left(  \tau\left(  i\right)  \right)  k^{-1}=g}}}_{\substack{=\sum\limits
_{\substack{\left(  i,k\right)  \in I\times G;\\k\rho\left(  \tau\left(
i\right)  \right)  k^{-1}=g}}=\sum\limits_{\substack{\left(  i,k\right)  \in I\times
G;\\k\left(  \rho\circ\tau\right)  \left(  i\right)  k^{-1}=g}}\\\text{(since
}\rho\left(  \tau\left(  i\right)  \right)  =\left(  \rho\circ\tau\right)
\left(  i\right)  \text{)}}}f\left(  i\right) \\
&  =\dfrac{1}{\left\vert I\right\vert }\sum\limits_{\substack{\left(  i,k\right)  \in
I\times G;\\k\left(  \rho\circ\tau\right)  \left(  i\right)  k^{-1}%
=g}}f\left(  i\right)  =\dfrac{1}{\left\vert I\right\vert }\sum\limits
_{\substack{\left(  h,k\right)  \in I\times G;\\k\left(  \rho\circ\tau\right)
\left(  h\right)  k^{-1}=g}}f\left(  h\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  i,k\right)  \text{ as }\left(  h,k\right)  \right)  .
\end{align*}
Compared with%
\[
\left(  \Ind_{\rho\circ\tau}f\right)  \left(
g\right)  =\dfrac{1}{\left\vert I\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in I\times G;\\k\left(  \rho\circ\tau\right)  \left(  h\right)
k^{-1}=g}}f\left(  h\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\left(  \Ind_{\rho\circ\tau}f\right)
\left(  g\right)  \right)  ,
\]
this yields $\left(  \Ind_{\rho}\left(
\Ind_{\tau}f\right)  \right)  \left(  g\right)
=\left(  \Ind_{\rho\circ\tau}f\right)  \left(
g\right)  $.

Let us now forget that we fixed $g$. We thus have shown that $\left(
\Ind_{\rho}\left(  \Ind_{\tau
}f\right)  \right)  \left(  g\right)  =\left(  \Ind_{\rho\circ\tau}f\right)  \left(  g\right)  $ for every $g\in G$. In
other words, $\operatorname{Ind}_{\rho}\Ind_{\tau
}f=\operatorname{Ind}_{\rho\circ\tau}f$. This solves Exercise
\ref{exe.Indrho.transitivity}(b).

\textit{Second solution to Exercise \ref{exe.Indrho.transitivity}(b).} We
shall now give an alternative solution of Exercise
\ref{exe.Indrho.transitivity}(b) which relies on Exercise
\ref{exe.Indrho.transitivity}(d) and a certain fact about class functions:

\begin{lemma}
\label{lem.sol.Indrho.transitivity.d.sol2.nondeg}Let $G$ be a finite group.
Let $u\in R_{\mathbb{C}}\left(  G\right)  $. Assume that every $v\in
R_{\mathbb{C}}\left(  G\right)  $ satisfies $\left\langle u,v\right\rangle
_{G}=0$. Then, $u=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Indrho.transitivity.d.sol2.nondeg}.]Let us use
the Iverson bracket notation; that is, for any statement $\mathcal{A}$, we
define $\left[  \mathcal{A}\right]  $ to be the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.

Fix an element $h\in G$. We define a map $\alpha_{h}:G\rightarrow\mathbb{C}$
by%
\[
\left(  \alpha_{h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\right)  .
\]
(Notice that this map $\alpha_{h}$ is identical with the map $\alpha_{G,h}$
defined in Exercise \ref{exe.frob-char.plambda}, but we will not use this.)
Then, $\alpha_{h}\in R_{\mathbb{C}}\left(  G\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $p$ and $q$ be two conjugate elements of
$G$. Then, there exists some $r\in G$ such that $p=rqr^{-1}$ (since $p$ and
$q$ are conjugate). Consider this $r$. The definition of $\alpha_{h}$ yields%
\[
\alpha_{h}\left(  p\right)  =\sum_{k\in G}\left[  khk^{-1}=\underbrace{p}%
_{=rqr^{-1}}\right]  =\sum_{k\in G}\left[  \underbrace{khk^{-1}=rqr^{-1}%
}_{\substack{\text{this is equivalent to}\\\left(  r^{-1}khk^{-1}r=q\right)
}}\right]  =\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]  .
\]
\par
Now, recall that $r\in G$. Hence, the map $G\rightarrow G,\ k\mapsto rk$ is a
bijection (since $G$ is a group). Hence, we can substitute $rk$ for $k$ in the
sum $\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]  $. We thus obtain%
\begin{align*}
\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]   &  =\sum_{k\in G}\left[
r^{-1}rkh\underbrace{\left(  rk\right)  ^{-1}}_{=k^{-1}r^{-1}}r=q\right]
=\sum_{k\in G}\left[  \underbrace{r^{-1}r}_{=1}khk^{-1}\underbrace{r^{-1}%
r}_{=1}=q\right]  \\
&  =\sum_{k\in G}\left[  khk^{-1}=q\right]  =\alpha_{h}\left(  q\right)
\end{align*}
(since $\alpha_{h}\left(  q\right)  =\sum_{k\in G}\left[  khk^{-1}=q\right]  $
(by the definition of $\alpha_{h}$)). Hence, $\alpha_{h}\left(  p\right)
=\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]  =\alpha_{h}\left(  q\right)
$.
\par
Let us now forget that we fixed $p$ and $q$. We thus have shown that
$\alpha_{h}\left(  p\right)  =\alpha_{h}\left(  q\right)  $ whenever $p$ and
$q$ are two conjugate elements of $G$. In other words, the function
$\alpha_{h}$ is constant on $G$-conjugacy classes. In other words, $\alpha
_{h}$ is a class function on $G$ (because the class functions on $G$ are
defined to be the functions $G\rightarrow\mathbb{C}$ which are constant on
$G$-conjugacy classes). In other words, $\alpha_{h}\in R_{\mathbb{C}}\left(
G\right)  $ (since $R_{\mathbb{C}}\left(  G\right)  $ is the set of all class
functions on $G$). Qed.}. Hence, $\left\langle u,\alpha_{h}\right\rangle _{G}$
is well-defined. But recall that every $v\in R_{\mathbb{C}}\left(  G\right)  $
satisfies $\left\langle u,v\right\rangle _{G}=0$. Applying this to
$v=\alpha_{h}$, we obtain $\left\langle u,\alpha_{h}\right\rangle _{G}=0$.
Thus,%
\begin{align}
0 &  =\left\langle u,\alpha_{h}\right\rangle _{G}=\dfrac{1}{\left\vert
G\right\vert }\sum_{g\in G}u\left(  g\right)  \underbrace{\alpha_{h}\left(
g^{-1}\right)  }_{\substack{=\sum_{k\in G}\left[  khk^{-1}=g^{-1}\right]
\\\text{(by the definition of }\alpha_{h}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the bilinear form }\left\langle \cdot
,\cdot\right\rangle _{G}\right)  \nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}u\left(  g\right)
\sum_{k\in G}\left[  khk^{-1}=g^{-1}\right]  =\dfrac{1}{\left\vert
G\right\vert }\sum_{k\in G}\sum_{g\in G}u\left(  g\right)  \left[
\underbrace{khk^{-1}=g^{-1}}_{\text{this is equivalent to }\left(  g=\left(
khk^{-1}\right)  ^{-1}\right)  }\right]  \nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\sum_{k\in G}\sum_{g\in G}u\left(
g\right)  \left[  g=\left(  khk^{-1}\right)  ^{-1}\right]
.\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.0}%
\end{align}
But every $k\in G$ satisfies%
\begin{equation}
\sum_{g\in G}u\left(  g\right)  \left[  g=\left(  khk^{-1}\right)
^{-1}\right]  =u\left(  h^{-1}\right)
\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}):} Let us first recall
that $u$ belongs to the set $R_{\mathbb{C}}\left(  G\right)  $. In other
words, $u$ is a class function on $G$ (since $R_{\mathbb{C}}\left(  G\right)
$ is the set of all class functions on $G$). In other words, the function $u$
is constant on $G$-conjugacy classes (because the class functions on $G$ are
defined to be the functions $G\rightarrow\mathbb{C}$ which are constant on
$G$-conjugacy classes). In other words,%
\begin{equation}
u\left(  p\right)  =u\left(  q\right)
\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1.pf.1}%
\end{equation}
whenever $p$ and $q$ are two conjugate elements of $G$.
\par
Now, let $k\in G$. Then, the elements $kh^{-1}k^{-1}$ and $h^{-1}$ of $G$ are
conjugate. Hence, $u\left(  kh^{-1}k^{-1}\right)  =u\left(  h^{-1}\right)  $
(by (\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1.pf.1}), applied to
$p=kh^{-1}k^{-1}$ and $q=h^{-1}$).
\par
All addends of the sum $\sum_{g\in G}u\left(  g\right)  \left[  g=\left(
khk^{-1}\right)  ^{-1}\right]  $ are zero except for the addend for $g=\left(
khk^{-1}\right)  ^{-1}$ (because the factor $\left[  g=\left(  khk^{-1}%
\right)  ^{-1}\right]  $ is zero unless $g=\left(  khk^{-1}\right)  ^{-1}$).
Hence, this sum simplifies as follows:%
\[
\sum_{g\in G}u\left(  g\right)  \left[  g=\left(  khk^{-1}\right)
^{-1}\right]  =u\left(  \underbrace{\left(  khk^{-1}\right)  ^{-1}}%
_{=kh^{-1}k^{-1}}\right)  \underbrace{\left[  \left(  khk^{-1}\right)
^{-1}=\left(  khk^{-1}\right)  ^{-1}\right]  }_{=1}=u\left(  kh^{-1}%
k^{-1}\right)  =u\left(  h^{-1}\right)  .
\]
This proves (\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}).}. Hence,
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.0}) becomes%
\[
0=\dfrac{1}{\left\vert G\right\vert }\sum_{k\in G}\underbrace{\sum_{g\in
G}u\left(  g\right)  \left[  g=\left(  khk^{-1}\right)  ^{-1}\right]
}_{\substack{=u\left(  h^{-1}\right)  \\\text{(by
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}))}}}=\dfrac
{1}{\left\vert G\right\vert }\underbrace{\sum_{k\in G}u\left(  h^{-1}\right)
}_{=\left\vert G\right\vert u\left(  h^{-1}\right)  }=\dfrac{1}{\left\vert
G\right\vert }\left\vert G\right\vert u\left(  h^{-1}\right)  =u\left(
h^{-1}\right)  .
\]
Thus, $u\left(  h^{-1}\right)  =0$.

Let us now forget that we fixed $h$. We thus have shown that
\begin{equation}
u\left(  h^{-1}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }h\in G\text{.}
\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.5}%
\end{equation}


Let us now fix $h\in G$. Then,
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.5}) (applied to $h^{-1}$
instead of $h$) yields $u\left(  \left(  h^{-1}\right)  ^{-1}\right)  =0$. In
other words, $u\left(  h\right)  =0$ (since $\left(  h^{-1}\right)  ^{-1}=h$).

Let us now forget that we fixed $h$. We thus have shown that $u\left(
h\right)  =0$ for every $h\in G$. In other words, $u=0$. This proves Lemma
\ref{lem.sol.Indrho.transitivity.d.sol2.nondeg}.
\end{proof}

Now, let us return to solving Exercise \ref{exe.Indrho.transitivity}(b). Let
$f\in R_{\mathbb{C}}\left(  I\right)  $. Let $v\in R_{\mathbb{C}}\left(
G\right)  $. Then,%
\begin{align*}
\left\langle \Ind_{\rho}\Ind_{\tau}f,v\right\rangle _{G}  &  =\left\langle \Ind_{\tau}f,\Res_{\rho}v\right\rangle
_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.Indrho.g.bil}), applied
to }\alpha=\Ind_{\tau}f\text{ and }\beta=v\right) \\
&  =\left\langle f,\underbrace{\Res_{\tau
}\Res_{\rho}v}_{\substack{=\Res_{\rho\circ\tau}v\\\text{(by Exercise \ref{exe.Indrho.transitivity}%
(d),}\\\text{applied to }v\text{ instead of }f\text{)}}}\right\rangle _{I}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.Indrho.g.bil}), applied
to }H\text{, }I\text{, }\tau\text{, }f\text{ and }\Res_{\rho}v\text{ instead of }G\text{, }H\text{, }\rho\text{, }%
\alpha\text{ and }\beta\right) \\
&  =\left\langle f,\Res_{\rho\circ\tau}v\right\rangle
_{I}.
\end{align*}
Compared with%
\[
\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=\left\langle f,\Res_{\rho\circ\tau}v\right\rangle
_{I}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.Indrho.g.bil}), applied
to }I\text{, }\rho\circ\tau\text{, }f\text{ and }v\text{ instead of }H\text{,
}\rho\text{, }\alpha\text{ and }\beta\right)  ,
\]
this yields $\left\langle \Ind_{\rho}%
\Ind_{\tau}f,v\right\rangle _{G}=\left\langle
\Ind_{\rho\circ\tau}f,v\right\rangle _{G}$. Now, the
form $\left\langle \cdot,\cdot\right\rangle _{G}$ is $\CC$-bilinear,
and therefore we have%
\[
\left\langle \Ind_{\rho}\Ind_{\tau}f-\Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=\underbrace{\left\langle \Ind_{\rho
}\Ind_{\tau}f,v\right\rangle _{G}}_{=\left\langle
\Ind_{\rho\circ\tau}f,v\right\rangle _{G}%
}-\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}-\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=0.
\]


Now, let us forget that we fixed $v$. We thus have shown that every $v\in
R_{\mathbb{C}}\left(  G\right)  $ satisfies $\left\langle \Ind_{\rho}\Ind_{\tau}f-\Ind_{\rho\circ\tau}f,v\right\rangle _{G}=0$. Hence, Lemma
\ref{lem.sol.Indrho.transitivity.d.sol2.nondeg} (applied to
$u=\Ind_{\rho}\Ind_{\tau
}f-\Ind_{\rho\circ\tau}f$) yields $\Ind_{\rho}\Ind_{\tau}f-\Ind_{\rho\circ\tau}f=0$. Thus, $\Ind_{\rho
}\Ind_{\tau}f=\Ind_{\rho
\circ\tau}f$. This solves Exercise \ref{exe.Indrho.transitivity}(b) again.

[\textit{Remark:} Recall that Exercise \ref{exe.Indrho} (specifically, its
parts (c), (d), (e) and (f)) shows that $\rho$-induction (of modules and of
class functions) generalizes both the usual notion of induction and the fixed
point construction. More precisely, $\rho$-induction becomes usual induction
(at least up to isomorphism) when $\rho$ is injective, and becomes fixed point
construction when $\rho$ is surjective. Exercise \ref{exe.Indrho.transitivity}
(specifically, its parts (a) and (b)), on the other hand, shows that in the
general case, $\rho$-induction can be reduced to a composition of usual
induction and fixed point construction, because every group homomorphism
$\rho:H\rightarrow G$ can be factored as a composition $\alpha\circ\beta$ of a
surjective group homomorphism $\beta:H\rightarrow\overline{H}$ with an
injective group homomorphism $\alpha:\overline{H}\rightarrow G$. This allows
some alternative proofs of some parts of Exercise \ref{exe.Indrho}, which the
interested reader can find.]
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Indrho.transitivity}}

\textit{Solution to Exercise~\ref{exe.Indrho.transitivity}.} In the following,
we will use the following convention: Whenever $K$ is a group, and $k$ is an
element of $K$, we shall write $k$ for the element $t_{k}$ of $\CC K$.
This is a relatively common abuse of notation, and it is harmless because
the map $K\rightarrow \CC K,\ k\mapsto t_{k}$ is an injective
homomorphism of multiplicative monoids (so $t_{gh}=t_{g}t_{h}$ and $t_{1}=1$,
which means that we won't run into ambiguities denoting $t_{k}$ by $k$) and
because every $\CC K$-module $M$, every $m\in M$ and every $k\in K$
satisfy $km=t_{k}m$.

We solve the four parts of Exercise \ref{exe.Indrho.transitivity} in the
following order: first, part (c); then, part (d); then, part (a); finally,
part (b).

(c) Let us first introduce a notation. Namely, if $A$ is a $\mathbb{C}%
$-algebra, and if $M$ is a left $A$-module, and if $a\in A$ and $m\in M$, then
we let $\operatorname*{act}\nolimits_{M}\left(  a,m\right)  $ denote the
action of $a\in A$ on $m\in M$. This is what is usually denoted by $am$.
However, we will encounter situations in which the notation $am$ can be
ambiguous (namely, when $m$ is an element of a $\CC$-vector space which
is endowed with two $A$-module structures, and we do not know a priori that
these two structures are identical) and therefore a more explicit notation
such as $\operatorname*{act}\nolimits_{M}\left(  a,m\right)  $ is needed to
disambiguate its meaning.

Let us summarize: If $A$ is a $\CC$-algebra, and if $M$ is a left
$A$-module, and if $a\in A$ and $m\in M$, then%
\begin{equation}
\operatorname*{act}\nolimits_{M}\left(  a,m\right)  =am
\label{sol.Indrho.transitivity.c.act}%
\end{equation}
(where the $am$ on the right hand side means the action of $a\in A$ on $m\in
M$)\ \ \ \ \footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.c.act}):}
Let $a\in A$ and $m\in M$. Both $\operatorname*{act}\nolimits_{M}\left(
a,m\right)  $ and $am$ are defined to be the action of $a\in A$ on $m\in M$.
Hence, $\operatorname*{act}\nolimits_{M}\left(  a,m\right)  =am$. This proves
(\ref{sol.Indrho.transitivity.c.act}).}.

Recall that $G$ and $H$ are two finite groups, and $\rho:H\rightarrow G$ is a
group homomorphism. For every $\mathbb{C}G$-module $V$, the $\mathbb{C}%
H$-module structure on $\Res_{\rho}V$ is given by the
equality%
\begin{equation}
\operatorname*{act}\nolimits_{\Res_{\rho}V}\left(
h,v\right)  =\operatorname*{act}\nolimits_{V}\left(  \rho\left(  h\right)
,v\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }h\in H\text{ and }v\in V
\label{sol.Indrho.transitivity.c.defact}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.c.defact}):} Recall
the definition of $\Res_{\rho}V$. It says that the
$\mathbb{C}H$-module structure on $\Res_{\rho}V$ is
given by the equality%
\begin{equation}
h\cdot v=\rho\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \text{for every
}h\in H\text{ and }v\in V. \label{sol.Indrho.transitivity.c.defact.pf.1}%
\end{equation}
\par
Now, let $h\in H$ and $v\in V$. Then,%
\begin{align*}
&  \operatorname*{act}\nolimits_{\Res_{\rho}V}\left(
h,v\right) \\
&  =h\cdot v\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.transitivity.c.act}), applied to }A=\mathbb{C}H\text{,
}M=V\text{, }a=h\text{ and }m=v\right) \\
&  =\rho\left(  h\right)  \cdot v\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.transitivity.c.defact.pf.1})}\right) \\
&  =\operatorname*{act}\nolimits_{V}\left(  \rho\left(  h\right)  ,v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{act}\nolimits_{V}%
\left(  \rho\left(  h\right)  ,v\right)  =\rho\left(  h\right)  \cdot v\text{
(by (\ref{sol.Indrho.transitivity.c.act}), applied to }A=\mathbb{C}G\text{,
}M=V\text{, }a=\rho\left(  h\right)  \text{ and }m=v\text{)}\right)  .
\end{align*}
This proves (\ref{sol.Indrho.transitivity.c.defact}).}.

Furthermore, $H$ and $I$ are two finite groups, and $\tau:I\rightarrow H$ is a
group homomorphism. For every $\mathbb{C}H$-module $U$, the $\mathbb{C}%
I$-module structure on $\Res_{\tau}U$ is given by the
equality%
\begin{equation}
\operatorname*{act}\nolimits_{\Res_{\tau}U}\left(
h,v\right)  =\operatorname*{act}\nolimits_{U}\left(  \tau\left(  h\right)
,v\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }h\in I\text{ and }v\in V
\label{sol.Indrho.transitivity.c.defact2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.c.defact2}):} The
equality (\ref{sol.Indrho.transitivity.c.defact2}) is obtained by applying the
equality (\ref{sol.Indrho.transitivity.c.defact}) to $H$, $I$, $\tau$ and $U$
instead of $G$, $H$, $\rho$ and $V$.}.

Finally, $G$ and $I$ are two finite groups, and $\rho\circ\tau:I\rightarrow G$
is a group homomorphism (since it is the composition of the two group
homomorphisms $\rho$ and $\tau$). For every $\mathbb{C}G$-module $V$, the
$\mathbb{C}I$-module structure on $\Res_{\rho\circ
\tau}V$ is given by the equality%
\begin{equation}
\operatorname*{act}\nolimits_{\Res_{\rho\circ\tau}%
V}\left(  h,v\right)  =\operatorname*{act}\nolimits_{V}\left(  \left(
\rho\circ\tau\right)  \left(  h\right)  ,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in I\text{ and }v\in V
\label{sol.Indrho.transitivity.c.defact3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.c.defact3}):} The
equality (\ref{sol.Indrho.transitivity.c.defact3}) is obtained by applying the
equality (\ref{sol.Indrho.transitivity.c.defact}) to $I$ and $\rho\circ\tau$
and $U$ instead of $H$ and $\rho$.}.

Now, let $V$ be a $\mathbb{C}G$-module. As $\CC$-vector spaces, we have%
\begin{align*}
\Res_{\tau}\Res_{\rho}V  &
=\Res_{\rho}V\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\Res_{\tau}\Res_{\rho}V\right) \\
&  =V\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }V\right) \\
&  =\Res_{\rho\circ\tau}V\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\Res_{\rho\circ\tau}V=V\text{ as
}\mathbb{C}\text{-vector spaces (by the definition of }\Res_{\rho\circ\tau}V\text{)}\right)  .
\end{align*}
Let us now show that the left $\mathbb{C}I$-module structure on
$\Res_{\tau}\Res_{\rho}V$ is identical
with the left $\mathbb{C}I$-module structure on $\Res_{\rho
\circ\tau}V$. Clearly, both of these structures have the same underlying
$\CC$-vector space (because $\Res_{\tau}%
\Res_{\rho}V=\Res_{\rho
\circ\tau}V$ as $\CC$-vector spaces), and thus share the same addition.

Now, every $h\in\mathbb{C}I$ and $v\in\Res_{\tau}%
\Res_{\rho}V$ satisfy%
\begin{equation}
\operatorname*{act}\nolimits_{\Res_{\tau}\Res_{\rho}V}\left(  h,v\right)  =\operatorname*{act}%
\nolimits_{\Res_{\rho\circ\tau}V}\left(  h,v\right)
\label{sol.Indrho.transitivity.c.acts-equal}%
\end{equation}
\footnote{\textit{Proof.} Let $h\in\mathbb{C}I$ and $v\in\operatorname{Res}%
_{\tau}\Res_{\rho}V$. We need to prove the equality
(\ref{sol.Indrho.transitivity.c.acts-equal}). This equality is clearly
$\CC$-linear in $h$. Hence, we can WLOG assume that $h$ belongs to the
basis $I$ of the $\CC$-vector space $\mathbb{C}I$. Assume this. Now,
(\ref{sol.Indrho.transitivity.c.defact2}) (applied to $U=\Res_{\rho}V$) yields%
\begin{align*}
\operatorname*{act}\nolimits_{\Res_{\tau}\Res_{\rho}V}\left(  h,v\right)   &  =\operatorname*{act}%
\nolimits_{\Res_{\rho}V}\left(  \tau\left(  h\right)
,v\right)  =\operatorname*{act}\nolimits_{V}\left(  \underbrace{\rho\left(
\tau\left(  h\right)  \right)  }_{=\left(  \rho\circ\tau\right)  \left(
h\right)  },v\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Indrho.transitivity.c.defact}), applied to }\tau\left(  h\right)
\text{ instead of }h\right) \\
&  =\operatorname*{act}\nolimits_{V}\left(  \left(  \rho\circ\tau\right)
\left(  h\right)  ,v\right)  =\operatorname*{act}%
\nolimits_{\Res_{\rho\circ\tau}V}\left(  h,v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.Indrho.transitivity.c.defact3}%
)}\right)  .
\end{align*}
This proves (\ref{sol.Indrho.transitivity.c.acts-equal}).}. Therefore, the
left $\mathbb{C}I$-module structure on $\Res_{\tau
}\Res_{\rho}V$ is identical with the left
$\mathbb{C}I$-module structure on $\Res_{\rho\circ\tau}V$
(because these two structures share the same addition). Thus,
$\Res_{\tau}\Res_{\rho}%
V=\Res_{\rho\circ\tau}V$ as left $\mathbb{C}I$-modules. This
solves Exercise \ref{exe.Indrho.transitivity}(c).

(d) Let $f\in R_{\mathbb{C}}\left(  G\right)  $. The definition of
$\Res_{\rho\circ\tau}f$ yields $\Res_{\rho
\circ\tau}f=f\circ\left(  \rho\circ\tau\right)  $. But the definition of
$\Res_{\rho}f$ yields $\Res_{\rho}f=f\circ\rho$. The definition of $\Res_{\tau
}\Res_{\rho}f$ yields%
\[
\Res_{\tau}\Res_{\rho}%
f=\underbrace{\left(  \Res_{\rho}f\right)  }%
_{=f\circ\rho}\circ\tau=\left(  f\circ\rho\right)  \circ\tau=f\circ\left(
\rho\circ\tau\right)  =\Res_{\rho\circ\tau}f.
\]
This solves Exercise \ref{exe.Indrho.transitivity}(d).

(a) We notice that%
\begin{equation}
\left(  \mathbb{C}\left[  \rho\right]  \right)  \circ\left(  \mathbb{C}\left[
\tau\right]  \right)  =\mathbb{C}\left[  \rho\circ\tau\right]  .
\label{sol.Indrho.transitivity.a.composition}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.a.composition}):} Let
$i\in I$. Then, $\left(  \mathbb{C}\left[  \tau\right]  \right)  \left(
i\right)  =\tau\left(  i\right)  $ (since $i\in I$). Applying the map
$\mathbb{C}\left[  \rho\right]  $ to both sides of this equality, we obtain
\begin{align*}
\left(  \mathbb{C}\left[  \rho\right]  \right)  \left(  \left(  \mathbb{C}%
\left[  \tau\right]  \right)  \left(  i\right)  \right)   &  =\left(
\mathbb{C}\left[  \rho\right]  \right)  \left(  \tau\left(  i\right)  \right)
=\rho\left(  \tau\left(  i\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau\left(  i\right)  \in H\right) \\
&  =\left(  \rho\circ\tau\right)  \left(  i\right)  .
\end{align*}
Compared with%
\[
\left(  \mathbb{C}\left[  \rho\circ\tau\right]  \right)  \left(  i\right)
=\left(  \rho\circ\tau\right)  \left(  i\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }i\in I\right)  ,
\]
this yields $\left(  \mathbb{C}\left[  \rho\right]  \right)  \left(  \left(
\mathbb{C}\left[  \tau\right]  \right)  \left(  i\right)  \right)  =\left(
\mathbb{C}\left[  \rho\circ\tau\right]  \right)  \left(  i\right)  $. Thus,
$\left(  \left(  \mathbb{C}\left[  \rho\right]  \right)  \circ\left(
\mathbb{C}\left[  \tau\right]  \right)  \right)  \left(  i\right)  =\left(
\mathbb{C}\left[  \rho\right]  \right)  \left(  \left(  \mathbb{C}\left[
\tau\right]  \right)  \left(  i\right)  \right)  =\left(  \mathbb{C}\left[
\rho\circ\tau\right]  \right)  \left(  i\right)  $.
\par
Let us now forget that we fixed $i$. We thus have shown that $\left(  \left(
\mathbb{C}\left[  \rho\right]  \right)  \circ\left(  \mathbb{C}\left[
\tau\right]  \right)  \right)  \left(  i\right)  =\left(  \mathbb{C}\left[
\rho\circ\tau\right]  \right)  \left(  i\right)  $ for every $i\in I$. In
other words, the two maps $\left(  \mathbb{C}\left[  \rho\right]  \right)
\circ\left(  \mathbb{C}\left[  \tau\right]  \right)  $ and $\mathbb{C}\left[
\rho\circ\tau\right]  $ are equal to each other on every element of $I$. Since
$I$ is a basis of the $\CC$-vector space $\mathbb{C}I$, and since these
two maps $\left(  \mathbb{C}\left[  \rho\right]  \right)  \circ\left(
\mathbb{C}\left[  \tau\right]  \right)  $ and $\mathbb{C}\left[  \rho\circ
\tau\right]  $ are $\CC$-linear, this shows that the two maps $\left(
\mathbb{C}\left[  \rho\right]  \right)  \circ\left(  \mathbb{C}\left[
\tau\right]  \right)  $ and $\mathbb{C}\left[  \rho\circ\tau\right]  $ are
identical (because if two $\CC$-linear maps from a given domain to a
given codomain are equal to each other on every element of a given basis of
their domain, then these two maps are identical). In other words, $\left(
\mathbb{C}\left[  \rho\right]  \right)  \circ\left(  \mathbb{C}\left[
\tau\right]  \right)  =\mathbb{C}\left[  \rho\circ\tau\right]  $. This proves
(\ref{sol.Indrho.transitivity.a.composition}).}

Let $U$ be any $\mathbb{C}I$-module.

Recall that $\Ind_{\tau}U$ is defined as the ${\mathbb{C}}%
H$-module ${\mathbb{C}}H\otimes_{\CC I}U$, where ${\mathbb{C}}H$ is
regarded as a $\left(  {\mathbb{C}}H,{\mathbb{C}}I\right)  $-bimodule
according to the following rule: The left ${\mathbb{C}}H$-module structure on
${\mathbb{C}}H$ is plain multiplication inside ${\mathbb{C}}H$; the right
${\mathbb{C}}I$-module structure on ${\mathbb{C}}H$ is induced by the
${\mathbb{C}}$-algebra homomorphism ${\mathbb{C}}\left[  \tau\right]
:{\mathbb{C}}I\rightarrow{\mathbb{C}}H$ (thus, it is explicitly given by
\begin{equation}
\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \tau\right]  \right)
\eta\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in\mathbb{C}H\text{ and }\eta
\in\mathbb{C}I \label{sol.Indrho.transitivity.a.structHI}%
\end{equation}
).

Furthermore, $\Ind_{\rho}\left(  \Ind_{\tau}U\right)  $ is defined as the ${\mathbb{C}}G$-module
${\mathbb{C}}G\otimes_{\CC H}\left(  \Ind_{\tau}U\right)  $, where ${\mathbb{C}}G$ is regarded as a $\left(
{\mathbb{C}}G,{\mathbb{C}}H\right)  $-bimodule according to the following
rule: The left ${\mathbb{C}}G$-module structure on ${\mathbb{C}}G$ is plain
multiplication inside ${\mathbb{C}}G$; the right ${\mathbb{C}}H$-module
structure on ${\mathbb{C}}G$ is induced by the ${\mathbb{C}}$-algebra
homomorphism ${\mathbb{C}}\left[  \rho\right]  :{\mathbb{C}}H\rightarrow
{\mathbb{C}}G$ (thus, it is explicitly given by%
\begin{equation}
\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho\right]  \right)
\eta\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in{\mathbb{C}}G\text{ and }%
\eta\in{\mathbb{C}}H \label{sol.Indrho.transitivity.a.structGH}%
\end{equation}
).

Finally, $\Ind_{\rho\circ\tau}U$ is defined as the ${\mathbb{C}%
}G$-module ${\mathbb{C}}G\otimes_{\CC I}U$, where ${\mathbb{C}}G$ is
regarded as a $\left(  {\mathbb{C}}G,{\mathbb{C}}I\right)  $-bimodule
according to the following rule: The left ${\mathbb{C}}G$-module structure on
${\mathbb{C}}G$ is plain multiplication inside ${\mathbb{C}}G$; the right
${\mathbb{C}}I$-module structure on ${\mathbb{C}}G$ is induced by the
${\mathbb{C}}$-algebra homomorphism ${\mathbb{C}}\left[  \rho\circ\tau\right]
:{\mathbb{C}}I\rightarrow{\mathbb{C}}G$ (thus, it is explicitly given by%
\begin{equation}
\gamma\eta=\gamma\cdot\left(  {\mathbb{C}}\left[  \rho\circ\tau\right]
\right)  \eta\ \ \ \ \ \ \ \ \ \ \text{for all }\gamma\in{\mathbb{C}}G\text{
and }\eta\in{\mathbb{C}}I \label{sol.Indrho.transitivity.a.structGI}%
\end{equation}
).

Thus, we have introduced a $\left(  {\mathbb{C}}H,{\mathbb{C}}I\right)
$-bimodule structure on $\mathbb{C}H$, a $\left(  {\mathbb{C}}G,{\mathbb{C}%
}H\right)  $-bimodule structure on $\mathbb{C}G$ and a $\left(  {\mathbb{C}%
}G,{\mathbb{C}}I\right)  $-bimodule structure on $\mathbb{C}G$. The left
$\mathbb{C}G$-module structure that underlies the $\left(  {\mathbb{C}%
}G,{\mathbb{C}}H\right)  $-bimodule structure on $\mathbb{C}G$ is identical
with the left $\mathbb{C}G$-module structure that underlies the $\left(
{\mathbb{C}}G,{\mathbb{C}}I\right)  $-bimodule structure on $\mathbb{C}G$
(because both of these left $\mathbb{C}G$-module structures are defined to be
plain multiplication inside $\mathbb{C}G$). Therefore, we will not run into
ambiguities if we write expressions such as $ab$ for $a\in\mathbb{C}G$ and
$b\in\mathbb{C}G$.

We shall now show that%
\begin{equation}
\mathbb{C}G\otimes_{\CC H}\mathbb{C}H\cong\mathbb{C}%
G\ \ \ \ \ \ \ \ \ \ \text{as }\left(  \mathbb{C}G,\mathbb{C}I\right)
\text{-bimodules.} \label{sol.Indrho.transitivity.a.1}%
\end{equation}
(Here, on the left hand side, $\mathbb{C}G$ is regarded as a $\left(
{\mathbb{C}}G,{\mathbb{C}}H\right)  $-bimodule and $\mathbb{C}H$ is regarded
as $\left(  {\mathbb{C}}H,{\mathbb{C}}I\right)  $-bimodule, whereas on the
right hand side, $\mathbb{C}G$ is regarded as a $\left(  {\mathbb{C}%
}G,{\mathbb{C}}I\right)  $-bimodule.)

\textit{Proof of (\ref{sol.Indrho.transitivity.a.1}):} A fundamental fact from
linear algebra says the following: If $A$ is a $\CC$-algebra, and if
$M$ is a right $A$-module, then there is a unique $\CC$-vector space
isomorphism $\Phi:M\rightarrow M\otimes_{A}A$ which satisfies%
\[
\left(  \Phi\left(  m\right)  =m\otimes_{A}1_{A}\ \ \ \ \ \ \ \ \ \ \text{for
all }m\in M\right)  .
\]
Applying this fact to $A=\mathbb{C}H$ and $M=\mathbb{C}G$, we conclude that
there is a unique $\CC$-vector space isomorphism $\Phi:\mathbb{C}%
G\rightarrow\mathbb{C}G\otimes_{\CC H}\mathbb{C}H$ which satisfies%
\begin{equation}
\left(  \Phi\left(  m\right)  =m\otimes_{\CC H}1_{\mathbb{C}%
H}\ \ \ \ \ \ \ \ \ \ \text{for all }m\in\mathbb{C}G\right)  .
\label{sol.Indrho.transitivity.a.1.pf.1}%
\end{equation}
Consider this $\Phi$. This map $\Phi$ is $\CC$-linear (because it is a
$\CC$-vector space isomorphism) and invertible (for the same reason).

We have%
\begin{equation}
\Phi\left(  my\right)  =\Phi\left(  m\right)  y\ \ \ \ \ \ \ \ \ \ \text{for
all }m\in\mathbb{C}G\text{ and }y\in\mathbb{C}I
\label{sol.Indrho.transitivity.a.1.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.a.1.pf.1}):} Let
$m\in\mathbb{C}G$ and $y\in\mathbb{C}I$. We have to prove the equality
$\Phi\left(  my\right)  =\Phi\left(  m\right)  y$. The definition of
$\Phi\left(  m\right)  $ yields $\Phi\left(  m\right)  =m\otimes_{\mathbb{C}%
H}1_{\CC H}$. Hence,%
\begin{align*}
\underbrace{\Phi\left(  m\right)  }_{=m\otimes_{\CC H}1_{\CC H}}y
&  =\left(  m\otimes_{\CC H}1_{\CC H}\right)  y=m\otimes
_{\CC H}\underbrace{1_{\CC H}y}_{\substack{=1_{\CC H}\left(
\mathbb{C}\left[  \tau\right]  \right)  \left(  y\right)  \\\text{(by
(\ref{sol.Indrho.transitivity.a.structHI}), applied to}\\\gamma=1_{\mathbb{C}%
H}\text{ and }\eta=y\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the right
}\mathbb{C}I\text{-module structure on }\mathbb{C}G\otimes_{\mathbb{C}%
H}\mathbb{C}H\right) \\
&  =m\otimes_{\CC H}\underbrace{1_{\CC H}\left(  \mathbb{C}\left[
\tau\right]  \right)  \left(  y\right)  }_{=\left(  \mathbb{C}\left[
\tau\right]  \right)  \left(  y\right)  =\left(  \mathbb{C}\left[
\tau\right]  \right)  \left(  y\right)  1_{\CC H}}=m\otimes
_{\CC H}\left(  \mathbb{C}\left[  \tau\right]  \right)  \left(
y\right)  1_{\CC H}=\underbrace{m\left(  \mathbb{C}\left[  \tau\right]
\right)  \left(  y\right)  }_{\substack{=m\left(  \mathbb{C}\left[
\rho\right]  \right)  \left(  \left(  \mathbb{C}\left[  \tau\right]  \right)
\left(  y\right)  \right)  \\\text{(by
(\ref{sol.Indrho.transitivity.a.structGH}), applied to}\\\gamma=m\text{ and
}\eta=\left(  \mathbb{C}\left[  \tau\right]  \right)  \left(  y\right)
\text{)}}}\otimes_{\CC H}1_{\CC H}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have moved }\left(
\mathbb{C}\left[  \tau\right]  \right)  \left(  y\right)  \text{ past the
tensor sign, since }\left(  \mathbb{C}\left[  \tau\right]  \right)  \left(
y\right)  \in\mathbb{C}H\right) \\
&  =m\underbrace{\left(  \mathbb{C}\left[  \rho\right]  \right)  \left(
\left(  \mathbb{C}\left[  \tau\right]  \right)  \left(  y\right)  \right)
}_{=\left(  \left(  \mathbb{C}\left[  \rho\right]  \right)  \circ\left(
\mathbb{C}\left[  \tau\right]  \right)  \right)  \left(  y\right)  }%
\otimes_{\CC H}1_{\CC H}=m\underbrace{\left(  \left(
\mathbb{C}\left[  \rho\right]  \right)  \circ\left(  \mathbb{C}\left[
\tau\right]  \right)  \right)  }_{\substack{=\mathbb{C}\left[  \rho\circ
\tau\right]  \\\text{(by (\ref{sol.Indrho.transitivity.a.composition}))}%
}}\left(  y\right)  \otimes_{\CC H}1_{\CC H}\\
&  =m\left(  \mathbb{C}\left[  \rho\circ\tau\right]  \right)  \left(
y\right)  \otimes_{\CC H}1_{\CC H}.
\end{align*}
Compared with%
\begin{align*}
\Phi\left(  my\right)   &  =\underbrace{my}_{\substack{=m\left(
\mathbb{C}\left[  \rho\circ\tau\right]  \right)  \left(  y\right)  \\\text{(by
(\ref{sol.Indrho.transitivity.a.structGI}), applied to}\\\gamma=m\text{ and
}\eta=y\text{)}}}\otimes_{\CC H}1_{\CC H}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\left(  my\right)
\right) \\
&  =m\left(  \mathbb{C}\left[  \rho\circ\tau\right]  \right)  \left(
y\right)  \otimes_{\CC H}1_{\CC H},
\end{align*}
this yields $\Phi\left(  m\right)  y=\Phi\left(  my\right)  $. This proves
(\ref{sol.Indrho.transitivity.a.1.pf.1}).}. Hence, $\Phi$ is a right
$\mathbb{C}I$-module homomorphism (since $\Phi$ is $\CC$-linear).
Furthermore, we have%
\begin{equation}
\Phi\left(  xm\right)  =x\Phi\left(  m\right)  \ \ \ \ \ \ \ \ \ \ \text{for
all }m\in\mathbb{C}G\text{ and }x\in\mathbb{C}G
\label{sol.Indrho.transitivity.a.1.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.a.1.pf.3}):} Let
$m\in\mathbb{C}G$ and $x\in\mathbb{C}G$. The definition of $\Phi\left(
xm\right)  $ yields%
\begin{align*}
\Phi\left(  xm\right)   &  =xm\otimes_{\CC H}1_{\mathbb{C}%
H}=x\underbrace{\left(  m\otimes_{\CC H}1_{\CC H}\right)
}_{\substack{=\Phi\left(  m\right)  \\\text{(by
(\ref{sol.Indrho.transitivity.a.1.pf.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }x\left(  m\otimes_{\CC H}1_{\CC H}\right)
=xm\otimes_{\CC H}1_{\CC H}\text{ (by the definition of}\\
\text{the left }\mathbb{C}G\text{-module structure on }\mathbb{C}%
G\otimes_{\CC H}\mathbb{C}H\text{)}%
\end{array}
\right) \\
&  =x\Phi\left(  m\right)  .
\end{align*}
This proves (\ref{sol.Indrho.transitivity.a.1.pf.3}).}. Hence, $\Phi$ is a
left $\mathbb{C}G$-module homomorphism (since $\Phi$ is a $\mathbb{C}%
$-linear). Now, $\Phi$ is a homomorphism of $\left(  \mathbb{C}G,\mathbb{C}%
I\right)  $-bimodules (since $\Phi$ is a left $\mathbb{C}G$-module
homomorphism and a right $\mathbb{C}I$-module homomorphism), thus an
isomorphism of $\left(  \mathbb{C}G,\mathbb{C}I\right)  $-bimodules (since
$\Phi$ is invertible). Therefore, there exists an isomorphism of $\left(
\mathbb{C}G,\mathbb{C}I\right)  $-bimodules from $\mathbb{C}G$ to
$\mathbb{C}G\otimes_{\CC H}\mathbb{C}H$ (namely, $\Phi$). Therefore,
$\mathbb{C}G\otimes_{\CC H}\mathbb{C}H\cong\mathbb{C}G$ as $\left(
\mathbb{C}G,\mathbb{C}I\right)  $-bimodules. This proves
(\ref{sol.Indrho.transitivity.a.1}).

Now, recall that we have $\Ind_{\tau}U={\mathbb{C}}%
H\otimes_{\CC I}U$ as left $\mathbb{C}H$-modules. Recall also that
$\Ind_{\rho}\left(  \Ind_{\tau}U\right)
={\mathbb{C}}G\otimes_{\CC H}\left(  \Ind_{\tau}U\right)  $ as left $\mathbb{C}G$-modules. Hence,%
\begin{align*}
\Ind_{\rho}\left(  \Ind_{\tau}U\right)
&  ={\mathbb{C}}G\otimes_{\CC H}\underbrace{\left(
\Ind_{\tau}U\right)  }_{={\mathbb{C}}H\otimes
_{\CC I}U}={\mathbb{C}}G\otimes_{\CC H}\left(  {\mathbb{C}%
}H\otimes_{\CC I}U\right) \\
&  \cong\underbrace{\left(  \mathbb{C}G\otimes_{\CC H}\mathbb{C}%
H\right)  }_{\substack{\cong\mathbb{C}G\text{ as }\left(  \mathbb{C}%
G,\mathbb{C}I\right)  \text{-bimodules}\\\text{(by
(\ref{sol.Indrho.transitivity.a.1}))}}}\otimes_{\mathbb{C}I}%
U\ \ \ \ \ \ \ \ \ \ \left(  \text{by the associativity of the tensor
product}\right) \\
&  \cong\mathbb{C}G\otimes_{\mathbb{C}I}U=\Ind_{\rho\circ\tau
}U\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Ind_{\rho\circ\tau
}U={\mathbb{C}}G\otimes_{\CC I}U\text{ as left }\mathbb{C}%
G\text{-modules}\right)
\end{align*}
as left $\mathbb{C}G$-modules. This solves Exercise
\ref{exe.Indrho.transitivity}(a).

(b) \textit{First solution to Exercise \ref{exe.Indrho.transitivity}(b).} Let
$f\in R_{\mathbb{C}}\left(  I\right)  $. Every $r\in H$ satisfies%
\begin{align}
\left(  \Ind_{\tau}f\right)  \left(  r\right)   &  =\dfrac
{1}{\left\vert I\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in I\times
H;\\k\tau\left(  h\right)  k^{-1}=r}}f\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Ind_{\tau}f\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\underbrace{\sum\limits_{\substack{\left(
i,v\right)  \in I\times H;\\v\tau\left(  i\right)  v^{-1}=r}}}_{=\sum_{i\in
I}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}}f\left(
i\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation
index }\left(  h,k\right)  \text{ as }\left(  i,v\right)  \right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\sum_{i\in I}\sum\limits_{\substack{v\in
H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  .
\label{sol.Indrho.transitivity.b.1}%
\end{align}


Now, let $g\in G$. Then, the definition of $\Ind_{\rho
}\left(  \Ind_{\tau}f\right)  $ yields%
\begin{align}
\left(  \Ind_{\rho}\left(  \Ind_{\tau}f\right)  \right)  \left(  g\right)   &  =\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{\substack{\left(  h,k\right)  \in H\times G;\\k\rho\left(
h\right)  k^{-1}=g}}\left(  \Ind_{\tau}f\right)
\left(  h\right)  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits
_{\substack{\left(  r,k\right)  \in H\times G;\\k\rho\left(  r\right)
k^{-1}=g}}}_{\substack{=\sum_{r\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
r\right)  k^{-1}=g}}\\=\sum_{k\in G}\sum\limits_{\substack{r\in H;\\k\rho\left(
r\right)  k^{-1}=g}}}}\underbrace{\left(  \Ind_{\tau
}f\right)  \left(  r\right)  }_{\substack{=\dfrac{1}{\left\vert I\right\vert
}\sum_{i\in I}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right)  \\\text{(by (\ref{sol.Indrho.transitivity.b.1}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  h,k\right)  \text{ as }\left(  r,k\right)  \right) \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{k\in G}\sum\limits_{\substack{r\in
H;\\k\rho\left(  r\right)  k^{-1}=g}}\dfrac{1}{\left\vert I\right\vert }%
\sum_{i\in I}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{k\in G}\underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g}}\sum_{i\in I}}_{=\sum_{i\in I}\sum\limits_{\substack{r\in H;\\k\rho\left(
r\right)  k^{-1}=g}}}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{k\in G}\sum_{i\in I}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g}}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(
i\right)  . \label{sol.Indrho.transitivity.b.3}%
\end{align}
But every $k\in G$ and $i\in I$ satisfy%
\begin{equation}
\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}}\sum\limits_{\substack{v\in
H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  =\sum\limits_{\substack{v\in
H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)
\label{sol.Indrho.transitivity.b.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.b.5}):} Let $k\in G$
and $i\in I$. We must prove (\ref{sol.Indrho.transitivity.b.5}). We have%
\begin{align}
&  \underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}%
}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}}_{=\sum_{v\in
H}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}}f\left(  i\right) \nonumber\\
&  =\sum_{v\in H}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  =\sum\limits
_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}%
}\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  +\sum\limits_{\substack{v\in H;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}\neq g}}\sum\limits_{\substack{r\in
H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right)  \label{sol.Indrho.transitivity.b.5.pf.0}%
\end{align}
(because every $v\in H$ satisfies either $k\rho\left(  v\right)  \cdot
\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)
\right)  ^{-1}=g$ or $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g$,
but never both).
\par
Now, we shall show that%
\begin{equation}
\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  =0
\label{sol.Indrho.transitivity.b.5.pf.1}%
\end{equation}
for every $v\in H$ which satisfies $k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}\neq g$.
\par
\textit{Proof of (\ref{sol.Indrho.transitivity.b.5.pf.1}):} Let $v\in H$ be
such that $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)
\right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g$. If some
$r\in H$ satisfies $k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(
i\right)  v^{-1}=r$, then this $r$ must satisfy%
\begin{align*}
k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\underbrace{\left(  k\rho\left(  v\right)  \right)  ^{-1}}_{=\left(
\rho\left(  v\right)  \right)  ^{-1}k^{-1}}  &  =k\underbrace{\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
\rho\left(  v\right)  \right)  ^{-1}}_{\substack{=\rho\left(  v\tau\left(
i\right)  v^{-1}\right)  \\\text{(since }\rho\text{ is a group}%
\\\text{homomorphism)}}}k^{-1}\\
&  =k\rho\left(  \underbrace{v\tau\left(  i\right)  v^{-1}}_{=r}\right)
k^{-1}=k\rho\left(  r\right)  k^{-1}=g,
\end{align*}
which contradicts $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g$.
Hence, we have obtained a contradiction for every $r\in H$ which satisfies
$k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$. Thus,
there exists no $r\in H$ which satisfies $k\rho\left(  r\right)  k^{-1}=g$ and
$v\tau\left(  i\right)  v^{-1}=r$. Therefore, the sum $\sum\limits_{\substack{r\in
H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}%
=r}}f\left(  i\right)  $ is an empty sum, and thus its value is $0$. This
proves (\ref{sol.Indrho.transitivity.b.5.pf.1}).
\par
On the other hand, let us show that
\begin{equation}
\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  =f\left(  i\right)
\label{sol.Indrho.transitivity.b.5.pf.2}%
\end{equation}
for every $v\in H$ which satisfies $k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g$.
\par
\textit{Proof of (\ref{sol.Indrho.transitivity.b.5.pf.2}):} Let $v\in H$ be
such that $k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)
\right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g$. If some $r\in
H$ satisfies $k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)
v^{-1}=r$, then this $r$ must equal $v\tau\left(  i\right)  v^{-1}$ (because
$v\tau\left(  i\right)  v^{-1}=r$). Hence, there exists \textbf{at most one}
$r\in H$ which satisfies $k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(
i\right)  v^{-1}=r$.
\par
On the other hand, the element $v\tau\left(  i\right)  v^{-1}$ of $H$
satisfies%
\begin{align*}
k\underbrace{\rho\left(  v\tau\left(  i\right)  v^{-1}\right)  }%
_{\substack{=\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)
\right)  \cdot\rho\left(  v\right)  ^{-1}\\\text{(since }\rho\text{ is a
group}\\\text{homomorphism)}}}k^{-1}  &  =k\rho\left(  v\right)  \cdot
\rho\left(  \tau\left(  i\right)  \right)  \cdot\underbrace{\rho\left(
v\right)  ^{-1}k^{-1}}_{=\left(  k\rho\left(  v\right)  \right)  ^{-1}}\\
&  =k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g
\end{align*}
and $v\tau\left(  i\right)  v^{-1}=v\tau\left(  i\right)  v^{-1}$. In other
words, $v\tau\left(  i\right)  v^{-1}$ is an element $r\in H$ which satisfies
$k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$.
Hence, there exists \textbf{at least one} $r\in H$ which satisfies
$k\rho\left(  r\right)  k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$
(namely, $v\tau\left(  i\right)  v^{-1}$). Since we also have shown that there
exists \textbf{at most one} such $r\in H$, we can thus conclude that there
exists \textbf{exactly one} $r\in H$ which satisfies $k\rho\left(  r\right)
k^{-1}=g$ and $v\tau\left(  i\right)  v^{-1}=r$. In other words, the sum
$\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(
i\right)  v^{-1}=r}}f\left(  i\right)  $ has precisely one addend. Hence, this
sum rewrites as follows: $\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  =f\left(
i\right)  $. This proves (\ref{sol.Indrho.transitivity.b.5.pf.2}).
\par
Now, (\ref{sol.Indrho.transitivity.b.5.pf.0}) becomes%
\begin{align*}
&  \sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}}\sum\limits
_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right) \\
&  =\sum\limits_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g}}\underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)
k^{-1}=g;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)  }%
_{\substack{=f\left(  i\right)  \\\text{(by
(\ref{sol.Indrho.transitivity.b.5.pf.2}))}}}+\sum\limits_{\substack{v\in
H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}\neq g}}\underbrace{\sum\limits
_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g;\\v\tau\left(  i\right)
v^{-1}=r}}f\left(  i\right)  }_{\substack{=0\\\text{(by
(\ref{sol.Indrho.transitivity.b.5.pf.1}))}}}\\
&  =\sum\limits_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g}}f\left(  i\right)  +\underbrace{\sum\limits_{\substack{v\in H;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}\neq g}}0}_{=0}=\sum\limits_{\substack{v\in
H;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)  .
\end{align*}
This proves (\ref{sol.Indrho.transitivity.b.5}).}.

Thus, (\ref{sol.Indrho.transitivity.b.3}) becomes%
\begin{align}
&  \left(  \Ind_{\rho}\left(  \Ind_{\tau}f\right)  \right)  \left(  g\right) \nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert
}\underbrace{\sum_{k\in G}\sum_{i\in I}}_{=\sum_{i\in I}\sum_{k\in G}%
}\underbrace{\sum\limits_{\substack{r\in H;\\k\rho\left(  r\right)  k^{-1}=g}%
}\sum\limits_{\substack{v\in H;\\v\tau\left(  i\right)  v^{-1}=r}}f\left(  i\right)
}_{\substack{=\sum\limits_{\substack{v\in H;\\k\rho\left(  v\right)  \cdot\rho\left(
\tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)
^{-1}=g}}f\left(  i\right)  \\\text{(by (\ref{sol.Indrho.transitivity.b.5}))}%
}} \nonumber \\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\underbrace{\sum_{k\in G}\sum\limits_{\substack{v\in H;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g}}}_{=\sum_{v\in H}\sum\limits_{\substack{k\in
G;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)
\cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}=g}}}f\left(  i\right)
\nonumber\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\sum_{v\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(  v\right)
\cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(  k\rho\left(
v\right)  \right)  ^{-1}=g}}f\left(  i\right)  .
\label{sol.Indrho.transitivity.b.7}%
\end{align}


However, for every $i\in I$ and $v\in H$, we have
\begin{equation}
\sum\limits_{\substack{k\in G;\\k\rho\left(  v\right)  \cdot\rho\left(  \tau\left(
i\right)  \right)  \cdot\left(  k\rho\left(  v\right)  \right)  ^{-1}%
=g}}f\left(  i\right)  =\sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(
i\right)  \right)  k^{-1}=g}}f\left(  i\right)
\label{sol.Indrho.transitivity.b.9}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.Indrho.transitivity.b.9}):} Let $i\in I$
and $v\in H$. We have $\rho\left(  v\right)  \in G$. Therefore, the map
$G\rightarrow G,\ k\mapsto k\rho\left(  v\right)  $ is a bijection (since $G$
is a group). Therefore, we can substitute $k\rho\left(  v\right)  $ for $k$ in
the sum $\sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)  \right)
k^{-1}=g}}f\left(  i\right)  $. We thus obtain%
\[
\sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)  \right)  k^{-1}%
=g}}f\left(  i\right)  =\sum\limits_{\substack{k\in G;\\k\rho\left(  v\right)
\rho\left(  \tau\left(  i\right)  \right)  \left(  k\rho\left(  v\right)
\right)  ^{-1}=g}}f\left(  i\right)  =\sum\limits_{\substack{k\in G;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)  .
\]
This proves (\ref{sol.Indrho.transitivity.b.9}).}.

Thus, (\ref{sol.Indrho.transitivity.b.7}) becomes%
\begin{align*}
&  \left(  \Ind_{\rho}\left(  \Ind_{\tau}f\right)  \right)  \left(  g\right) \\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\sum_{v\in H}\underbrace{\sum\limits_{\substack{k\in G;\\k\rho\left(
v\right)  \cdot\rho\left(  \tau\left(  i\right)  \right)  \cdot\left(
k\rho\left(  v\right)  \right)  ^{-1}=g}}f\left(  i\right)  }_{\substack{=\sum\limits
_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)  \right)  k^{-1}%
=g}}f\left(  i\right)  \\\text{(by (\ref{sol.Indrho.transitivity.b.9}))}%
}}=\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\underbrace{\sum_{v\in H}\sum\limits_{\substack{k\in G;\\k\rho\left(
\tau\left(  i\right)  \right)  k^{-1}=g}}f\left(  i\right)  }_{=\left\vert
H\right\vert \sum\limits_{\substack{k\in G;\\k\rho\left(  \tau\left(  i\right)
\right)  k^{-1}=g}}f\left(  i\right)  }\\
&  =\dfrac{1}{\left\vert I\right\vert }\dfrac{1}{\left\vert H\right\vert }%
\sum_{i\in I}\left\vert H\right\vert \sum\limits_{\substack{k\in G;\\k\rho\left(
\tau\left(  i\right)  \right)  k^{-1}=g}}f\left(  i\right)  =\dfrac
{1}{\left\vert I\right\vert }\underbrace{\dfrac{1}{\left\vert H\right\vert
}\left\vert H\right\vert }_{=1}\underbrace{\sum_{i\in I}\sum\limits_{\substack{k\in
G;\\k\rho\left(  \tau\left(  i\right)  \right)  k^{-1}=g}}}_{\substack{=\sum\limits
_{\substack{\left(  i,k\right)  \in I\times G;\\k\rho\left(  \tau\left(
i\right)  \right)  k^{-1}=g}}=\sum\limits_{\substack{\left(  i,k\right)  \in I\times
G;\\k\left(  \rho\circ\tau\right)  \left(  i\right)  k^{-1}=g}}\\\text{(since
}\rho\left(  \tau\left(  i\right)  \right)  =\left(  \rho\circ\tau\right)
\left(  i\right)  \text{)}}}f\left(  i\right) \\
&  =\dfrac{1}{\left\vert I\right\vert }\sum\limits_{\substack{\left(  i,k\right)  \in
I\times G;\\k\left(  \rho\circ\tau\right)  \left(  i\right)  k^{-1}%
=g}}f\left(  i\right)  =\dfrac{1}{\left\vert I\right\vert }\sum\limits
_{\substack{\left(  h,k\right)  \in I\times G;\\k\left(  \rho\circ\tau\right)
\left(  h\right)  k^{-1}=g}}f\left(  h\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  i,k\right)  \text{ as }\left(  h,k\right)  \right)  .
\end{align*}
Compared with%
\[
\left(  \Ind_{\rho\circ\tau}f\right)  \left(
g\right)  =\dfrac{1}{\left\vert I\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in I\times G;\\k\left(  \rho\circ\tau\right)  \left(  h\right)
k^{-1}=g}}f\left(  h\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\left(  \Ind_{\rho\circ\tau}f\right)
\left(  g\right)  \right)  ,
\]
this yields $\left(  \Ind_{\rho}\left(
\Ind_{\tau}f\right)  \right)  \left(  g\right)
=\left(  \Ind_{\rho\circ\tau}f\right)  \left(
g\right)  $.

Let us now forget that we fixed $g$. We thus have shown that $\left(
\Ind_{\rho}\left(  \Ind_{\tau
}f\right)  \right)  \left(  g\right)  =\left(  \Ind_{\rho\circ\tau}f\right)  \left(  g\right)  $ for every $g\in G$. In
other words, $\Ind_{\rho}\Ind_{\tau
}f=\Ind_{\rho\circ\tau}f$. This solves Exercise
\ref{exe.Indrho.transitivity}(b).

\textit{Second solution to Exercise \ref{exe.Indrho.transitivity}(b).} We
shall now give an alternative solution of Exercise
\ref{exe.Indrho.transitivity}(b) which relies on Exercise
\ref{exe.Indrho.transitivity}(d) and a certain fact about class functions:

\begin{lemma}
\label{lem.sol.Indrho.transitivity.d.sol2.nondeg}Let $G$ be a finite group.
Let $u\in R_{\mathbb{C}}\left(  G\right)  $. Assume that every $v\in
R_{\mathbb{C}}\left(  G\right)  $ satisfies $\left\langle u,v\right\rangle
_{G}=0$. Then, $u=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Indrho.transitivity.d.sol2.nondeg}.]Let us use
the Iverson bracket notation; that is, for any statement $\mathcal{A}$, we
define $\left[  \mathcal{A}\right]  $ to be the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.

Fix an element $h\in G$. We define a map $\alpha_{h}:G\rightarrow\mathbb{C}$
by%
\[
\left(  \alpha_{h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\right)  .
\]
(Notice that this map $\alpha_{h}$ is identical with the map $\alpha_{G,h}$
defined in Exercise \ref{exe.frob-char.plambda}, but we will not use this.)
Then, $\alpha_{h}\in R_{\mathbb{C}}\left(  G\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $p$ and $q$ be two conjugate elements of
$G$. Then, there exists some $r\in G$ such that $p=rqr^{-1}$ (since $p$ and
$q$ are conjugate). Consider this $r$. The definition of $\alpha_{h}$ yields%
\begin{align*}
\alpha_{h}\left(  p\right)   &  =\sum_{k\in G}\left[  khk^{-1}=\underbrace{p}%
_{=rqr^{-1}}\right]  =\sum_{k\in G}\left[  \underbrace{khk^{-1}=rqr^{-1}%
}_{\substack{\text{this is equivalent to}\\\left(  r^{-1}khk^{-1}%
=qr^{-1}\right)  }}\right]  =\sum_{k\in G}\left[  \underbrace{r^{-1}%
khk^{-1}=qr^{-1}}_{\substack{\text{this is equivalent to}\\\left(
r^{-1}khk^{-1}r=q\right)  }}\right]  \\
&  =\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]  .
\end{align*}
\par
Now, recall that $r\in G$. Hence, the map $G\rightarrow G,\ k\mapsto rk$ is a
bijection (since $G$ is a group). Hence, we can substitute $rk$ for $k$ in the
sum $\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]  $. We thus obtain%
\begin{align*}
\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]   &  =\sum_{k\in G}\left[
r^{-1}rkh\underbrace{\left(  rk\right)  ^{-1}}_{=k^{-1}r^{-1}}r=q\right]
=\sum_{k\in G}\left[  \underbrace{r^{-1}r}_{=1}khk^{-1}\underbrace{r^{-1}%
r}_{=1}=q\right]  \\
&  =\sum_{k\in G}\left[  khk^{-1}=q\right]  =\alpha_{h}\left(  q\right)
\end{align*}
(since $\alpha_{h}\left(  q\right)  =\sum_{k\in G}\left[  khk^{-1}=q\right]  $
(by the definition of $\alpha_{h}$)). Hence, $\alpha_{h}\left(  p\right)
=\sum_{k\in G}\left[  r^{-1}khk^{-1}r=q\right]  =\alpha_{h}\left(  q\right)
$.
\par
Let us now forget that we fixed $p$ and $q$. We thus have shown that
$\alpha_{h}\left(  p\right)  =\alpha_{h}\left(  q\right)  $ whenever $p$ and
$q$ are two conjugate elements of $G$. In other words, the function
$\alpha_{h}$ is constant on $G$-conjugacy classes. In other words, $\alpha
_{h}$ is a class function on $G$ (because $\alpha_{h}$ is a class function on
$G$ if and only if $\alpha_{h}$ is constant on $G$-conjugacy classes (by the
definition of a ``class function'')). In other
words, $\alpha_{h}$ belongs to the set of all class functions on $G$. In other
words, $\alpha_{h}$ belongs to $R_{\mathbb{C}}\left(  G\right)  $ (since
$R_{\mathbb{C}}\left(  G\right)  $ is the set of all class functions on $G$).
Qed.}. Hence, $\left\langle u,\alpha_{h}\right\rangle _{G}$ is well-defined.
But recall that every $v\in R_{\mathbb{C}}\left(  G\right)  $ satisfies
$\left\langle u,v\right\rangle _{G}=0$. Applying this to $v=\alpha_{h}$, we
obtain $\left\langle u,\alpha_{h}\right\rangle _{G}=0$. Thus,%
\begin{align}
0 &  =\left\langle u,\alpha_{h}\right\rangle _{G}=\dfrac{1}{\left\vert
G\right\vert }\sum_{g\in G}u\left(  g\right)  \underbrace{\alpha_{h}\left(
g^{-1}\right)  }_{\substack{=\sum_{k\in G}\left[  khk^{-1}=g^{-1}\right]
\\\text{(by the definition of }\alpha_{h}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the bilinear form }\left\langle \cdot
,\cdot\right\rangle _{G}\right)  \nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\sum_{g\in G}u\left(  g\right)
\sum_{k\in G}\left[  khk^{-1}=g^{-1}\right]  =\dfrac{1}{\left\vert
G\right\vert }\sum_{k\in G}\sum_{g\in G}u\left(  g\right)  \left[
\underbrace{khk^{-1}=g^{-1}}_{\text{this is equivalent to }\left(
khk^{-1}g=1\right)  }\right]  \nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\sum_{k\in G}\sum_{g\in G}u\left(
g\right)  \left[  \underbrace{khk^{-1}g=1}_{\text{this is equivalent to
}\left(  g=\left(  khk^{-1}\right)  ^{-1}\right)  }\right]  \nonumber\\
&  =\dfrac{1}{\left\vert G\right\vert }\sum_{k\in G}\sum_{g\in G}u\left(
g\right)  \left[  g=\left(  khk^{-1}\right)  ^{-1}\right]
.\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.0}%
\end{align}
But every $k\in G$ satisfies%
\begin{equation}
\sum_{g\in G}u\left(  g\right)  \left[  g=\left(  khk^{-1}\right)
^{-1}\right]  =u\left(  h^{-1}\right)
\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}):} Let us first recall
that $u$ belongs to the set $R_{\mathbb{C}}\left(  G\right)  $. In other
words, $u$ belongs to the set of all class functions on $G$ (since the set
$R_{\mathbb{C}}\left(  G\right)  $ is the set of all class functions on $G$).
In other words, $u$ is a class function on $G$. In other words, the function
$u$ is constant on $G$-conjugacy classes (because $u$ is a class function on
$G$ if and only if $u$ is constant on $G$-conjugacy classes (by the definition
of a ``class function'')). In other words,%
\begin{equation}
u\left(  p\right)  =u\left(  q\right)
\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1.pf.1}%
\end{equation}
whenever $p$ and $q$ are two conjugate elements of $G$.
\par
Now, let $k\in G$. Then, the elements $kh^{-1}k^{-1}$ and $h^{-1}$ of $G$ are
conjugate. Hence, $u\left(  kh^{-1}k^{-1}\right)  =u\left(  h^{-1}\right)  $
(by (\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1.pf.1}), applied to
$p=kh^{-1}k^{-1}$ and $q=h^{-1}$).
\par
Every $g\in G$ satisfies either $g=\left(  khk^{-1}\right)  ^{-1}$ or
$g\neq\left(  khk^{-1}\right)  ^{-1}$ (but not both). Thus,%
\begin{align*}
&  \sum_{g\in G}u\left(  g\right)  \left[  g=\left(  khk^{-1}\right)
^{-1}\right]  \\
&  =\sum\limits_{\substack{g\in G;\\g=\left(  khk^{-1}\right)  ^{-1}}}u\left(
g\right)  \underbrace{\left[  g=\left(  khk^{-1}\right)  ^{-1}\right]
}_{\substack{=1\\\text{(since }g=\left(  khk^{-1}\right)  ^{-1}\text{ is
true)}}}+\sum\limits_{\substack{g\in G;\\g\neq\left(  khk^{-1}\right)  ^{-1}%
}}u\left(  g\right)  \underbrace{\left[  g=\left(  khk^{-1}\right)
^{-1}\right]  }_{\substack{=0\\\text{(since }g=\left(  khk^{-1}\right)
^{-1}\text{ is false}\\\text{(since we have }g\neq\left(  khk^{-1}\right)
^{-1}\text{))}}}\\
&  =\sum\limits_{\substack{g\in G;\\g=\left(  khk^{-1}\right)  ^{-1}}}u\left(
g\right)  +\underbrace{\sum\limits_{\substack{g\in G;\\g\neq\left(  khk^{-1}\right)
^{-1}}}u\left(  g\right)  0}_{=0}=\sum\limits_{\substack{g\in G;\\g=\left(
khk^{-1}\right)  ^{-1}}}u\left(  g\right)  =u\left(  \underbrace{\left(
khk^{-1}\right)  ^{-1}}_{=\left(  k^{-1}\right)  ^{-1}h^{-1}k^{-1}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  khk^{-1}\right)  ^{-1}\in
G\right)  \\
&  =u\left(  \underbrace{\left(  k^{-1}\right)  ^{-1}}_{=k}h^{-1}%
k^{-1}\right)  =u\left(  kh^{-1}k^{-1}\right)  =u\left(  h^{-1}\right)  .
\end{align*}
This proves (\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}).}. Hence,
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.0}) becomes%
\[
0=\dfrac{1}{\left\vert G\right\vert }\sum_{k\in G}\underbrace{\sum_{g\in
G}u\left(  g\right)  \left[  g=\left(  khk^{-1}\right)  ^{-1}\right]
}_{\substack{=u\left(  h^{-1}\right)  \\\text{(by
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.1}))}}}=\dfrac
{1}{\left\vert G\right\vert }\underbrace{\sum_{k\in G}u\left(  h^{-1}\right)
}_{=\left\vert G\right\vert u\left(  h^{-1}\right)  }=\dfrac{1}{\left\vert
G\right\vert }\left\vert G\right\vert u\left(  h^{-1}\right)  =u\left(
h^{-1}\right)  .
\]
Thus, $u\left(  h^{-1}\right)  =0$.

Let us now forget that we fixed $h$. We thus have shown that
\begin{equation}
u\left(  h^{-1}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }h\in G\text{.}
\label{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.5}%
\end{equation}


Let us now fix $h\in G$. Then,
(\ref{pf.lem.sol.Indrho.transitivity.d.sol2.nondeg.5}) (applied to $h^{-1}$
instead of $h$) yields $u\left(  \left(  h^{-1}\right)  ^{-1}\right)  =0$. In
other words, $u\left(  h\right)  =0$ (since $\left(  h^{-1}\right)  ^{-1}=h$).

Let us now forget that we fixed $h$. We thus have shown that $u\left(
h\right)  =0$ for every $h\in G$. In other words, $u=0$. This proves Lemma
\ref{lem.sol.Indrho.transitivity.d.sol2.nondeg}.
\end{proof}

Now, let us return to solving Exercise \ref{exe.Indrho.transitivity}(b). Let
$f\in R_{\mathbb{C}}\left(  I\right)  $. Let $v\in R_{\mathbb{C}}\left(
G\right)  $. Then,%
\begin{align*}
\left\langle \Ind_{\rho}\Ind_{\tau}f,v\right\rangle _{G}  &  =\left\langle \Ind_{\tau}f,\Res_{\rho}v\right\rangle
_{H}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.Indrho.g.bil}), applied
to }\alpha=\Ind_{\tau}f\text{ and }\beta=v\right) \\
&  =\left\langle f,\underbrace{\Res_{\tau
}\Res_{\rho}v}_{\substack{=\Res_{\rho\circ\tau}v\\\text{(by Exercise \ref{exe.Indrho.transitivity}%
(d),}\\\text{applied to }v\text{ instead of }f\text{)}}}\right\rangle _{I}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.Indrho.g.bil}), applied
to }H\text{, }I\text{, }\tau\text{, }f\text{ and }\Res_{\rho}v\text{ instead of }G\text{, }H\text{, }\rho\text{, }%
\alpha\text{ and }\beta\right) \\
&  =\left\langle f,\Res_{\rho\circ\tau}v\right\rangle
_{I}.
\end{align*}
Compared with%
\[
\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=\left\langle f,\Res_{\rho\circ\tau}v\right\rangle
_{I}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.exe.Indrho.g.bil}), applied
to }I\text{, }\rho\circ\tau\text{, }f\text{ and }v\text{ instead of }H\text{,
}\rho\text{, }\alpha\text{ and }\beta\right)  ,
\]
this yields $\left\langle \Ind_{\rho}%
\Ind_{\tau}f,v\right\rangle _{G}=\left\langle
\Ind_{\rho\circ\tau}f,v\right\rangle _{G}$. Now, the
form $\left\langle \cdot,\cdot\right\rangle _{G}$ is $\CC$-bilinear,
and therefore we have%
\[
\left\langle \Ind_{\rho}\Ind_{\tau}f-\Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=\underbrace{\left\langle \Ind_{\rho
}\Ind_{\tau}f,v\right\rangle _{G}}_{=\left\langle
\Ind_{\rho\circ\tau}f,v\right\rangle _{G}%
}-\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}-\left\langle \Ind_{\rho\circ\tau}f,v\right\rangle
_{G}=0.
\]


Now, let us forget that we fixed $v$. We thus have shown that every $v\in
R_{\mathbb{C}}\left(  G\right)  $ satisfies $\left\langle \Ind_{\rho}\Ind_{\tau}f-\Ind_{\rho\circ\tau}f,v\right\rangle _{G}=0$. Hence, Lemma
\ref{lem.sol.Indrho.transitivity.d.sol2.nondeg} (applied to
$u=\Ind_{\rho}\Ind_{\tau
}f-\Ind_{\rho\circ\tau}f$) yields $\Ind_{\rho}\Ind_{\tau}f-\Ind_{\rho\circ\tau}f=0$. Thus, $\Ind_{\rho
}\Ind_{\tau}f=\Ind_{\rho
\circ\tau}f$. This solves Exercise \ref{exe.Indrho.transitivity}(b) again.

[\textit{Remark:} Recall that Exercise \ref{exe.Indrho} (specifically, its
parts (c), (d), (e) and (f)) shows that $\rho$-induction (of modules and of
class functions) generalizes both the usual notion of induction and the fixed
point construction. More precisely, $\rho$-induction becomes usual induction
(at least up to isomorphism) when $\rho$ is injective, and becomes fixed point
construction when $\rho$ is surjective. Exercise \ref{exe.Indrho.transitivity}
(specifically, its parts (a) and (b)), on the other hand, shows that in the
general case, $\rho$-induction can be reduced to a composition of usual
induction and fixed point construction, because every group homomorphism
$\rho:H\rightarrow G$ can be factored as a composition $\alpha\circ\beta$ of a
surjective group homomorphism $\beta:H\rightarrow\overline{H}$ with an
injective group homomorphism $\alpha:\overline{H}\rightarrow G$. This allows
some alternative proofs of some parts of Exercise \ref{exe.Indrho}, which the
interested reader can find.]
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Ind-ass}}

\textit{Solution to Exercise~\ref{exe.Ind-ass}.} We need to prove that
\begin{equation}
\operatorname{Ind}_{\Symm_{i+j}\times\Symm_k}^{\Symm_{i+j+k}}
\left(  \operatorname{Ind}_{\Symm_i\times\Symm_j%
}^{\Symm_{i+j}}\left(  U\otimes V\right)  \otimes W\right)
\cong\operatorname{Ind}_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right)
\label{sol.Ind-ass.1}%
\end{equation}
and
\begin{equation}
\operatorname{Ind}_{\Symm_i\times\Symm_{j+k}}^{\Symm_{i+j+k}}
\left(  U\otimes\operatorname{Ind}_{\Symm_j\times
\Symm_k}^{\Symm_{j+k}}\left(  V\otimes W\right)  \right)
\cong\operatorname{Ind}_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right)
\label{sol.Ind-ass.2}%
\end{equation}
as ${\mathbb{C}}\left[  \Symm_i\times\Symm_j\times
\Symm_k\right]  $-modules. We will only prove \eqref{sol.Ind-ass.1},
since \eqref{sol.Ind-ass.2} is analogous.

It is easy to see that every finite group $G$ and every $\mathbb{C}G$-module
$P$ satisfy $\Ind_{G}^{G}P\cong P$. Applied to
$G=\Symm_k$ and $P=W$, this yields
$\Ind_{\Symm_k}^{\Symm_k}W\cong W$.

Now, Exercise~\ref{exe.induction.transitivity} (applied to
$\Symm_{i+j+k}$, $\Symm_{i+j}\times\Symm_k$, $\Symm_i%
\times\Symm_j\times\Symm_k$ and $U\otimes V\otimes W$
instead of $G$, $H$, $I$ and $U$) yields%
\[
\operatorname{Ind}_{\Symm_{i+j}\times\Symm_k}^{\Symm_{i+j+k}}
\left(  \operatorname{Ind}_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j}\times\Symm_k}\left(
U\otimes V\otimes W\right)  \right)  \cong
\operatorname{Ind}_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right)  .
\]
Hence,%
\begin{align*}
\operatorname{Ind}_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right)
&  \cong\operatorname{Ind}_{\Symm_{i+j}\times\Symm_k%
}^{\Symm_{i+j+k}}\left(  \underbrace{\operatorname{Ind}_{\Symm_i
\times\Symm_j\times\Symm_k}^{\Symm_{i+j}%
\times\Symm_k}\left(  U\otimes V\otimes W\right)  }_{\substack{\cong%
\left(  \operatorname{Ind}_{\Symm_i\times\Symm_j%
}^{\Symm_{i+j}}\left(  U\otimes V\right)  \right)  \otimes\left(
\Ind_{\Symm_k}^{\Symm_k}W\right)
\\\text{(by (\ref{induction-vs-tensor-product}), applied to }G_{1}%
=\Symm_{i+j}\text{, }G_{2}=\Symm_k\text{,}\\H_{1}%
=\Symm_i\times\Symm_j\text{, }H_{2}=\Symm_k\text{,
}U_{1}=U\otimes V\text{ and }U_{2}=W\text{)}}}\right) \\
&  \cong\operatorname{Ind}_{\Symm_{i+j}\times\Symm_k%
}^{\Symm_{i+j+k}}\left(  \left(  \operatorname{Ind}_{\Symm_i
\times\Symm_j}^{\Symm_{i+j}}\left(  U\otimes V\right)
\right)  \otimes\underbrace{\left(
\Ind_{\Symm_k}^{\Symm_k}W\right)  }_{\cong W}\right) \\
&  \cong\operatorname{Ind}_{\Symm_{i+j}\times\Symm_k%
}^{\Symm_{i+j+k}}\left(  \operatorname{Ind}_{\Symm_i%
\times\Symm_j}^{\Symm_{i+j}}\left(  U\otimes V\right)  \otimes
W\right)  .
\end{align*}
This proves \eqref{sol.Ind-ass.1}. Thus, the solution of Exercise
\ref{exe.Ind-ass} is complete.

\begin{comment}
\textit{Second solution to Exercise~\ref{exe.Ind-ass}.}
[\textbf{Note:} The following solution is \emph{not} any clearer
than the First solution above. It was written before
Exercise~\ref{exe.ind.tensor} (on which the First solution rests),
and is preserved only for archival reasons.]

In the following, we will write $g$ for the element $t_g$ of $\CC G$
whenever $g$ is an element of some group $G$. This is a relatively common abuse
of notation, and it is harmless because the map $G \to \CC G, \ 
g \mapsto t_g$ is an injective homomorphism of multiplicative monoids
(so $t_{gh} = t_g t_h$ and $t_1 = 1$, which means that we won't run
into ambiguities denoting $t_g$ by $g$)
and because every $\CC G$-module $M$, every $m \in M$ and every $g \in G$
satisfy $gm = t_g m$.

We need to prove that
\begin{equation}
\label{sol.Ind-ass.a1}
\Ind^{\Symm_{i+j+k}}_{\Symm_{i+j} \times \Symm_k}\left(\Ind^{\Symm_{i+j}}_{\Symm_i \times \Symm_j}\left(U\otimes V\right)\otimes W\right)
\cong \Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_j \times \Symm_k}\left(U\otimes V\otimes W\right)
\end{equation}
and
\begin{equation}
\label{sol.Ind-ass.a2}
\Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_{j+k}}\left(U \otimes \Ind^{\Symm_{j+k}}_{\Symm_j \times \Symm_k}\left(V\otimes W\right)\right)
\cong \Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_j \times \Symm_k}\left(U\otimes V\otimes W\right)
\end{equation}
as $\CC \left[\Symm_i \times \Symm_j \times \Symm_k\right]$-modules. We
will only prove \eqref{sol.Ind-ass.a1}, since \eqref{sol.Ind-ass.a2} is
analogous.

Here is the proof: It is easy to check that the $\CC$-linear maps
\begin{align*}
\left(  \Ind_{\Symm_{i}\times\Symm_{j}}
^{\Symm_{i+j}}\left(  U\otimes V\right)  \right)  \otimes W  &
\rightarrow\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}\left(
U\otimes V\otimes W\right)  ,\\
\left(  \sigma\otimes_{\CC\left[ 
\Symm_{i}\times\Symm_{j}\right]  }
\left(  u\otimes v\right)  \right)  \otimes w  &  \mapsto\left(
\sigma,e\right)  \otimes_{\CC\left[  \Symm_{i}\times
\Symm_{j}\times\Symm_{k}\right]  }\left(  u\otimes v\otimes
w\right)
\end{align*}
(where $e$ is the neutral element of $\Symm_{k}$) and
\begin{align*}
\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}\left(
U\otimes V\otimes W\right)   &  \rightarrow\left(  
\Ind_{\Symm_{i}\times\Symm_{j}}^{\Symm_{i+j}}\left(
U\otimes V\right)  \right)  \otimes W,\\
\left(  \sigma,\tau\right)  \otimes_{\CC\left[  \Symm_{i}
\times\Symm_{j}\times\Symm_{k}\right]  }\left(  u\otimes
v\otimes w\right)   &  \mapsto\left(  \sigma\otimes_{\CC\left[
\Symm_{i}\times\Symm_{j}\right]  }\left(  u\otimes v\right)
\right)  \otimes\tau w \ \ \ \ \ \ \ \ \ \ \ \left(  \text{for }\left(
\sigma,\tau\right)  \in\Symm_{i+j}\times\Symm_{k}\right)
\end{align*}
are well-defined mutually inverse $\CC\left[  \Symm_{i+j}
\times\Symm_{k}\right]  $-module isomorphisms. Thus,
\[
\left(  \Ind_{\Symm_{i}\times\Symm_{j}}
^{\Symm_{i+j}}\left(  U\otimes V\right)  \right)  \otimes
W\cong\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}\left(
U\otimes V\otimes W\right)
\]
as $\CC\left[  \Symm_{i+j}\times\Symm_{k}\right]
$-modules. Thus,
\begin{align*}
&  \Ind_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(  \underbrace{\left(  
\Ind_{\Symm_{i}\times\Symm_{j}}^{\Symm_{i+j}}\left(
U\otimes V\right)  \right)  \otimes W}_{\cong
\Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}
^{\Symm_{i+j}\times\Symm_{k}}\left(  U\otimes V\otimes
W\right)  }\right) \\
&  \cong\Ind_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(  \Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}
^{\Symm_{i+j}\times\Symm_{k}}
\left(  U\otimes V\otimes W\right)  \right)
\cong\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes
W\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the transitivity of induction}\right)
\end{align*}
as $\CC\left[  \Symm_{i+j+k}\right]  $-modules. This proves
\eqref{sol.Ind-ass.a1}.
\end{comment}
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Ind-ass}}

\textit{Solution to Exercise~\ref{exe.Ind-ass}.}

\textit{First solution to Exercise~\ref{exe.Ind-ass}.}
We need to prove that
\begin{equation}
\operatorname{Ind}_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(  \operatorname{Ind}_{\Symm_{i}\times\Symm_{j}%
}^{\Symm_{i+j}}\left(  U\otimes V\right)  \otimes W\right)
\cong\operatorname{Ind}_{\Symm_{i}\times\Symm_{j}\times
\Symm_{k}}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right)
\label{sol.Ind-ass.1}
\end{equation}
and
\begin{equation}
\operatorname{Ind}_{\Symm_{i}\times\Symm_{j+k}}^{\Symm_{i+j+k}}
\left(  U\otimes\operatorname{Ind}_{\Symm_{j}\times
\Symm_{k}}^{\Symm_{j+k}}\left(  V\otimes W\right)  \right)
\cong\operatorname{Ind}_{\Symm_{i}\times\Symm_{j}\times
\Symm_{k}}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right)
\label{sol.Ind-ass.2}%
\end{equation}
as ${\mathbb{C}}\left[  \Symm_{i}\times\Symm_{j}\times
\Symm_{k}\right]  $-modules. We will only prove \eqref{sol.Ind-ass.1},
since \eqref{sol.Ind-ass.2} is analogous.

It is easy to see that every finite group $G$ and every $\mathbb{C}G$-module
$P$ satisfy $\Ind_{G}^{G}P\cong P$. Applied to
$G=\Symm_{k}$ and $P=W$, this yields
$\Ind_{\Symm_{k}}^{\Symm_{k}}W\cong W$.

Now, Exercise~\ref{exe.induction.transitivity} (applied to
$\Symm_{i+j+k}$, $\Symm_{i+j}\times\Symm_{k}$,
$\Symm_{i}\times\Symm_{j}\times\Symm_{k}$ and $U\otimes V\otimes W$
instead of $G$, $H$, $I$ and $U$) yields
\[
\Ind_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(
 \Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}
 \left( U\otimes V\otimes W\right)
\right)
\cong\Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(  U\otimes V\otimes W\right)  .
\]
Hence,
\begin{align*}
\Ind_{\Symm_{i}\times\Symm_{j}\times \Symm_{k}}^{\Symm_{i+j+k}}
\left(  U\otimes V\otimes W\right)
& \cong \Ind_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left( \underbrace{
 \Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}
  \left(  U\otimes V\otimes W\right)
 }_{\substack{\cong \left(
  \Ind_{\Symm_{i}\times\Symm_{j}}^{\Symm_{i+j}}\left(  U\otimes V\right) 
 \right)  \otimes\left( \Ind_{\Symm_{k}}^{\Symm_{k}}W\right)
\\\text{(by (\ref{induction-vs-tensor-product}), applied to }G_{1}
=\Symm_{i+j}\text{, }G_{2}=\Symm_{k}\text{,}\\H_{1}
=\Symm_{i}\times\Symm_{j}\text{, }H_{2}=\Symm_{k}\text{,
}U_{1}=U\otimes V\text{ and }U_{2}=W\text{)}}}\right) \\
&  \cong\operatorname{Ind}_{\Symm_{i+j}\times\Symm_{k}%
}^{\Symm_{i+j+k}}\left(  \left(
 \operatorname{Ind}_{\Symm_{i}\times\Symm_{j}}^{\Symm_{i+j}}
 \left(  U\otimes V\right)
\right)  \otimes\underbrace{\left(
 \Ind_{\Symm_{k}}^{\Symm_{k}}W\right)  }_{\cong W}\right) \\
&  \cong\operatorname{Ind}_{\Symm_{i+j}\times\Symm_{k}%
}^{\Symm_{i+j+k}}\left(  \operatorname{Ind}_{\Symm_{i}%
\times\Symm_{j}}^{\Symm_{i+j}}\left(  U\otimes V\right)  \otimes
W\right)  .
\end{align*}
This proves \eqref{sol.Ind-ass.1}. Thus, the solution of Exercise
\ref{exe.Ind-ass} is complete.

\textit{Second solution to Exercise~\ref{exe.Ind-ass}.}
[\textbf{Note:} The following solution is \emph{not} any clearer
than the First solution above. It was written before
Exercise~\ref{exe.ind.tensor} (on which the First solution rests),
and is preserved only for archival reasons.]

In the following, we will write $g$ for the element $t_g$ of $\CC G$
whenever $g$ is an element of some group $G$. This is a relatively common abuse
of notation, and it is harmless because the map $G \to \CC G, \ 
g \mapsto t_g$ is an injective homomorphism of multiplicative monoids
(so $t_{gh} = t_g t_h$ and $t_1 = 1$, which means that we won't run
into ambiguities denoting $t_g$ by $g$)
and because every $\CC G$-module $M$, every $m \in M$ and every $g \in G$
satisfy $gm = t_g m$.

We need to prove that
\begin{equation}
\label{sol.Ind-ass.a1}
\Ind^{\Symm_{i+j+k}}_{\Symm_{i+j} \times \Symm_k}\left(\Ind^{\Symm_{i+j}}_{\Symm_i \times \Symm_j}\left(U\otimes V\right)\otimes W\right)
\cong \Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_j \times \Symm_k}\left(U\otimes V\otimes W\right)
\end{equation}
and
\begin{equation}
\label{sol.Ind-ass.a2}
\Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_{j+k}}\left(U \otimes \Ind^{\Symm_{j+k}}_{\Symm_j \times \Symm_k}\left(V\otimes W\right)\right)
\cong \Ind^{\Symm_{i+j+k}}_{\Symm_i \times \Symm_j \times \Symm_k}\left(U\otimes V\otimes W\right)
\end{equation}
as $\CC \left[\Symm_i \times \Symm_j \times \Symm_k\right]$-modules. We
will only prove \eqref{sol.Ind-ass.a1}, since \eqref{sol.Ind-ass.a2} is
analogous.

Here is the proof: It is easy to check that the $\CC$-linear maps
\begin{align*}
\left(  \Ind_{\Symm_{i}\times\Symm_{j}}
^{\Symm_{i+j}}\left(  U\otimes V\right)  \right)  \otimes W  &
\rightarrow\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}\left(
U\otimes V\otimes W\right)  ,\\
\left(  \sigma\otimes_{\CC\left[ 
\Symm_{i}\times\Symm_{j}\right]  }
\left(  u\otimes v\right)  \right)  \otimes w  &  \mapsto\left(
\sigma,e\right)  \otimes_{\CC\left[  \Symm_{i}\times
\Symm_{j}\times\Symm_{k}\right]  }\left(  u\otimes v\otimes
w\right)
\end{align*}
(where $e$ is the neutral element of $\Symm_{k}$) and
\begin{align*}
\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}\left(
U\otimes V\otimes W\right)   &  \rightarrow\left(  
\Ind_{\Symm_{i}\times\Symm_{j}}^{\Symm_{i+j}}\left(
U\otimes V\right)  \right)  \otimes W,\\
\left(  \sigma,\tau\right)  \otimes_{\CC\left[  \Symm_{i}
\times\Symm_{j}\times\Symm_{k}\right]  }\left(  u\otimes
v\otimes w\right)   &  \mapsto\left(  \sigma\otimes_{\CC\left[
\Symm_{i}\times\Symm_{j}\right]  }\left(  u\otimes v\right)
\right)  \otimes\tau w \ \ \ \ \ \ \ \ \ \ \ \left(  \text{for }\left(
\sigma,\tau\right)  \in\Symm_{i+j}\times\Symm_{k}\right)
\end{align*}
are well-defined mutually inverse $\CC\left[  \Symm_{i+j}
\times\Symm_{k}\right]  $-module isomorphisms. Thus,
\[
\left(  \Ind_{\Symm_{i}\times\Symm_{j}}
^{\Symm_{i+j}}\left(  U\otimes V\right)  \right)  \otimes
W\cong\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j}\times\Symm_{k}}\left(
U\otimes V\otimes W\right)
\]
as $\CC\left[  \Symm_{i+j}\times\Symm_{k}\right]
$-modules. Thus,
\begin{align*}
&  \Ind_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(  \underbrace{\left(  
\Ind_{\Symm_{i}\times\Symm_{j}}^{\Symm_{i+j}}\left(
U\otimes V\right)  \right)  \otimes W}_{\cong
\Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}
^{\Symm_{i+j}\times\Symm_{k}}\left(  U\otimes V\otimes
W\right)  }\right) \\
&  \cong\Ind_{\Symm_{i+j}\times\Symm_{k}}^{\Symm_{i+j+k}}
\left(  \Ind_{\Symm_{i}\times\Symm_{j}\times\Symm_{k}}
^{\Symm_{i+j}\times\Symm_{k}}
\left(  U\otimes V\otimes W\right)  \right)
\cong\Ind_{\Symm_{i}\times\Symm_{j}
\times\Symm_{k}}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes
W\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the transitivity of induction}\right)
\end{align*}
as $\CC\left[  \Symm_{i+j+k}\right]  $-modules. This proves
\eqref{sol.Ind-ass.a1}.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.bruhat.variations}}

\textit{Solution to Exercise~\ref{exe.bruhat.variations}.}
(a) Let $n \in \NN$.
Let $B$ denote the subgroup of $GL_n(\FF)$ consisting of all
upper-triangular matrices. Then,
$GL_n(\FF) = \bigsqcup_{w \in \Symm_n} BwB$. (Indeed, this can be
proven in the same way as we have shown the equality
$GL_n = \bigsqcup_{w \in \Symm_n} BwB$ in our above proof of
Proposition~\ref{double-coset-prop}(c), with the only difference that
we are now working over $\FF$ instead of $\FF_q$.)

Let $w_0$ denote the permutation in $\Symm_n$ which sends every
$i \in \left\{ 1, 2, \ldots, n \right\}$ to $n+1-i$; then,
$w_0^2 = \id$. (The permutation $w_0$ is written as
$\left(n, n-1, \ldots, 1\right)$ in one-line notation.)

Let $A \in GL_n(\FF)$.
We have $w_0 A \in GL_n(\FF) = \bigsqcup_{w \in \Symm_n} BwB$. Thus,
there exists some $w \in \Symm_n$ such that $w_0 A \in BwB$. Consider
this $w$. There exist two upper-triangular invertible matrices $L'$
and $U$ such that $w_0 A = L' w U$ (since $w_0 A \in BwB$). Consider
these $L'$ and $U$. The matrix $w_0 L' w_0$ is lower-triangular
(since $L'$ is upper-triangular, and since $w_0$ is what it is). Set
$L = w_0 L' w_0$. Then,
$\underbrace{L}_{=w_0 L' w_0} w_0 w U
= w_0 L' \underbrace{w_0 w_0}_{= w_0^2 = \id} w U
= w_0 \underbrace{L' w U}_{= w_0 A}
= \underbrace{w_0 w_0}_{= w_0^2 = \id} A = A$. In other words,
$L P U = A$ with $P = w_0 w$. Since $P$ is clearly a permutation
matrix (being the product of the permutation matrices $w_0$ and $w$),
we thus have shown that $A = LPU$ for a
lower-triangular matrix $L \in GL_n(\FF)$, an upper-triangular
matrix $U \in GL_n(\FF)$ and a permutation matrix
$P \in \Symm_n \subset GL_n(\FF)$. This solves
Exercise~\ref{exe.bruhat.variations}(a).

(b) In the proof that follows, we shall essentially mimic the arguments
used to prove $GL_n = \bigsqcup_{w \in \Symm_n} BwB$ in our proof of
Proposition~\ref{double-coset-prop}(c) (but we will add some details
in a few places).

We recall that $B_n$ is a subgroup of $GL_n(\FF)$.
Hence, every element of $B_n$ is invertible in $B_n$.
Likewise, every element of $B_m$ is invertible in $B_m$.

Let us first show that the disjoint union
$\bigsqcup_{f \in F_{n, m}} B_n f B_m$ is well-defined. This means
proving that the sets $B_n f B_m$ for $f \in F_{n, m}$ are disjoint.

For every $A \in \FF^{n \times m}$, every $i \in \left\{1, 2, \ldots,
n+1\right\}$ and every $j \in \left\{0, 1, \ldots, m\right\}$, let
$r_{i,j}\left(A\right)$ denote the rank of the matrix obtained by
restricting $A$ to the rows $i, i+1, \ldots, n$ and columns
$1, 2, \ldots, j$. This rank $r_{i,j}\left(A\right)$ does not change
when we replace $A$ by $XA$ for some $X \in B_n$ (because replacing $A$
by $XA$ for some $X \in B_n$ means that we add to each row of $A$ a
linear combination of the rows further below; but this row
transformation does not change $r_{i,j}\left(A\right)$); neither does
it change when we replace $A$ by $AY$ for some $Y \in B_m$ (for a
similar reason). Hence, for every $f \in F_{n, m}$, $X \in B_n$,
$Y \in B_m$, $i \in \left\{1, 2, \ldots, n+1\right\}$ and
$j \in \left\{0, 1, \ldots, m\right\}$, we have
\begin{equation}
\label{sol.bruhat.variations.b.rinv}
r_{i,j}\left(f\right) = r_{i,j}\left(Xf\right)
= r_{i,j}\left(XfY\right) .
\end{equation}

Now, we claim that if $f \in F_{n, m}$ and $A \in B_n f B_m$, then we
can reconstruct $f$ from $A$. Indeed, let $f \in F_{n, m}$ and
$A \in B_n f B_m$. Then, there exist $X \in B_n$ and
$Y \in B_m$ such that $A = XfY$ (since $A \in B_n f B_m$). Consider
these $X$ and $Y$. Then, \eqref{sol.bruhat.variations.b.rinv} yields
$r_{i,j}\left(f\right) = r_{i,j}\left(\underbrace{XfY}_{=A}\right)
= r_{i,j}\left(A\right)$ for every
$i \in \left\{1, 2, \ldots, n+1\right\}$ and
$j \in \left\{0, 1, \ldots, m\right\}$. Thus, from $A$ we can
reconstruct the ranks $r_{i,j}\left(f\right)$ for all
$i \in \left\{1, 2, \ldots, n+1\right\}$ and
$j \in \left\{0, 1, \ldots, m\right\}$.

Notice that $f \in F_{n, m}$. Thus, $f$ is a matrix
in $\left\{0, 1\right\}^{n \times m}$ such that
each row of $f$ contains at most one $1$ and each column of $f$
contains at most one $1$.

Now, fix $i \in \left\{1, 2, \ldots, n\right\}$ and
$j \in \left\{1, 2, \ldots, m\right\}$. Let $f_{i, j}$ be the entry
of $f$ in row $i$ and column $j$.
We will now show that
\begin{equation}
\label{sol.bruhat.variations.b.req}
f_{i, j} = r_{i,j}\left(f\right) - r_{i,j-1}\left(f\right)
- r_{i+1,j}\left(f\right) + r_{i+1,j-1}\left(f\right) .
\end{equation}

\textit{Proof of \eqref{sol.bruhat.variations.b.req}.} First of
all, if the first $j$ entries of the $i$-th row of $f$ are all $0$,
then \eqref{sol.bruhat.variations.b.req} holds for obvious reasons
(in fact, in this case, it is clear that $f_{i, j} = 0$,
$r_{i,j}\left(f\right) = r_{i+1,j}\left(f\right)$ and
$r_{i,j-1}\left(f\right) = r_{i+1,j-1}\left(f\right)$). We thus
WLOG assume that the first $j$ entries of the $i$-th row of $f$ are
not all $0$. Then, there must be a $1$ among these entries. It must
lie in a different column than the $1$'s appearing in all other
rows of $f$ (because each column of $f$ contains at most one $1$).
Therefore, the first row of the matrix obtained by
restricting $f$ to the rows $i, i+1, \ldots, n$ and columns
$1, 2, \ldots, j$ is linearly independent from its other rows.
Thus,
\[
r_{i,j}\left(f\right) = r_{i+1,j}\left(f\right) + 1 .
\]
If $f_{i,j} = 0$, then the same argument yields
$r_{i,j-1}\left(f\right) = r_{i+1,j-1}\left(f\right) + 1$
(because if $f_{i,j} = 0$, then the $1$ among the
first $j$ entries of the $i$-th row of $f$ must not be the last of
these entries, and so it is one of the first $j-1$ entries of the
$i$-th row of $f$). On the other hand, if $f_{i,j} = 1$, then
we have $r_{i,j-1}\left(f\right) = r_{i+1,j-1}\left(f\right)$
(because if $f_{i,j} = 1$, then the first $j-1$ entries of the
$i$-th row of $f$ must be all $0$ (since the $j$-th entry of this
row is a $1$, but each row of $f$ contains at most one $1$)). We
can subsume both of these statements in one equation, which holds
both if $f_{i,j} = 0$ and if $f_{i,j} = 1$: namely, we have
\[
r_{i,j-1}\left(f\right)
=
\begin{cases}
r_{i+1,j-1}\left(f\right) + 1 & \mbox{if } f_{i,j} = 0 ; \\
r_{i+1,j-1}\left(f\right) & \mbox{if } f_{i,j} = 1
\end{cases}  \quad
= r_{i+1,j-1}\left(f\right) + \left(1-f_{i,j}\right) .
\]
Subtracting this equality from
$r_{i,j}\left(f\right) = r_{i+1,j}\left(f\right) + 1$,
we obtain
\[
r_{i,j}\left(f\right) - r_{i,j-1}\left(f\right)
= \left(r_{i+1,j}\left(f\right) + 1\right)
- \left(r_{i+1,j-1}\left(f\right) + \left(1-f_{i,j}\right)\right)
= r_{i+1,j}\left(f\right)
- r_{i+1,j-1}\left(f\right) + f_{i,j} .
\]
This is easily seen to be equivalent to
\eqref{sol.bruhat.variations.b.req}. Thus,
\eqref{sol.bruhat.variations.b.req} is proven.

\begin{comment}
If the last $n-i+1$ entries of the $j$-th column of $f$ are all $0$,
then \eqref{sol.bruhat.variations.b.req} holds for obvious reasons
(in fact, in this case, it is clear that $f_{i, j} = 0$,
$r_{i,j}\left(f\right) = r_{i,j-1}\left(f\right)$ and
$r_{i+1,j}\left(f\right) = r_{i+1,j-1}\left(f\right)$). We thus
WLOG assume that the last $n-i+1$ entries of the $j$-th column of $f$
are not all $0$. Then, there must be a $1$ among these entries. It
must lie in a different row than the $1$'s appearing in all other
columns of $f$ (because each row of $f$ contains at most one $1$).
Therefore, the last column of the matrix obtained by
restricting $f$ to the rows $i, i+1, \ldots, n$ and columns
$1, 2, \ldots, j$ is linearly independent from its other columns.
Thus, $r_{i,j}\left(f\right) = r_{i,j-1}\left(f\right) + 1$.
In other words, $r_{i,j-1}\left(f\right) + 1 = r_{i,j}\left(f\right)$.
Hence,
\[
r_{i,j}\left(f\right) - 1
= r_{i,j-1}\left(f\right)
= r_{i+1,j-1}\left(f\right) + \left(1-f_{i,j}\right) .
\]
\end{comment}

Now, we can reconstruct
the ranks $r_{i,j}\left(f\right)$, $r_{i,j-1}\left(f\right)$,
$r_{i+1,j}\left(f\right)$ and $r_{i+1,j-1}\left(f\right)$ from $A$
(since from $A$ we can
reconstruct the ranks $r_{i,j}\left(f\right)$ for all
$i \in \left\{1, 2, \ldots, n+1\right\}$ and
$j \in \left\{0, 1, \ldots, m\right\}$). Hence, we can reconstruct
$f_{i,j}$ from $A$ (due to \eqref{sol.bruhat.variations.b.req}).

Let us now forget that we fixed $i$ and $j$. We thus have seen that
we can reconstruct $f_{i,j}$ from $A$ for every
$i \in \left\{1, 2, \ldots, n\right\}$ and
$j \in \left\{1, 2, \ldots, m\right\}$. In other words, we can
reconstruct $f$ from $A$.

Thus we have proven that if $f \in F_{n, m}$ and $A \in B_n f B_m$,
then we can reconstruct $f$ from $A$. In other words,
the sets $B_n f B_m$ for $f \in F_{n, m}$ are disjoint. Thus,
the disjoint union $\bigsqcup_{f \in F_{n, m}} B_n f B_m$ is well-defined.

It remains to prove that this disjoint union
$\bigsqcup_{f \in F_{n, m}} B_n f B_m$ is $\FF^{n \times m}$. In
order to do so, it is clearly enough to show that every
$g \in \FF^{n \times m}$ belongs to
$\bigsqcup_{f \in F_{n, m}} B_n f B_m$.

Let $g \in \FF^{n \times m}$. We need to show that
$g \in \bigsqcup_{f \in F_{n, m}} B_n f B_m$. In other words, we need
to find some $f \in F_{n, m}$ such that $g \in B_n f B_m$. In order
to do so, we shall find a matrix $f \in F_{n, m}$
which lies in $B_n g B_m$. Once this is done, it will follow that
$g \in B_n f B_m$ (since every element of $B_n$ is invertible in
$B_n$, and every element of $B_m$ is invertible in $B_m$),
and we will be done.

We refer to $g B_m$ and $B_n g$ as cosets,
despite the fact that they are not subsets of a group.

The freedom to alter $g$ within the coset $gB_m$ allows one to scale columns and
add scalar multiples of earlier columns to later columns.  We claim that
using such column operations, one can
always find a representative $g'$ for coset $gB_m$ in which
\begin{enumerate}
\item[$\bullet$]
the bottommost nonzero entry of each nonzero column is $1$ (call this entry a \emph{pivot}),
\item[$\bullet$]
the entries to right of each pivot within its row are all $0$, and
\item[$\bullet$]
there is at most one pivot in each row and at most one pivot in each column (so that
their positions are the positions of the $1$'s in
some matrix $f \in F_{n, m}$).
\end{enumerate}
In fact, we will see below that $B_ngB_m=B_nfB_m$ in this case.
The algorithm which produces $g'$ from $g$ is simple:  starting with the
leftmost nonzero column, find its bottommost nonzero entry, and scale the column
to make this entry a $1$, creating the pivot in this column.  Now use this
pivot to clear out all entries in its row to its right, using
column operations that subtract multiples of this column from later columns.
Having done this, move on to the next nonzero column to the right, and repeat, scaling to create a pivot, and using it to eliminate entries to its right.\footnote{To see that this works, we need to check three facts:
\begin{itemize}
\item[\textbf{(a)}] We will find a nonzero entry in every nonzero column during our
algorithm.
\item[\textbf{(b)}] Our column operations preserve
the zeroes lying to the right of already existing pivots.
\item[\textbf{(c)}] Every row contains at most one pivot at the end
of the algorithm.
\end{itemize}
But fact \textbf{(a)} is a tautology.
Fact \textbf{(b)} holds because all our operations either
scale columns (which clearly preserves zero entries) or subtract a
multiple of the column $c$ containing the current pivot from a later
column $d$ (which will preserve every zero lying to the right of an
already existing pivot, because any already existing pivot must lie
in a column $b < c$ and therefore both
columns $c$ and $d$ have zeroes in its row). Fact \textbf{(c)} follows
from noticing that the entries to the right of a pivot in its row
are $0$.}

Having found this $g'$ in $gB_m$, a similar algorithm using left multiplication
by $B_n$ shows that $f$ lies in
$B_ng' \subset B_ng'B_m=B_ngB_m$.  This time no scalings are required
to create the pivot entries:  starting with the bottommost nonzero row, one
uses its pivot to eliminate all the entries above it in the same column
%(shown by stars $*$ above)
by adding multiples of this row to higher rows. Then do the same using the pivot
in the next-to-bottom nonzero row, etc.\footnote{One thing that requires
verification is the fact that these row operations preserve the
following three properties of $g'$:
\begin{enumerate}
\item[$\bullet$]
the bottommost nonzero entry of each nonzero column is $1$ (call this entry a \emph{pivot}),
\item[$\bullet$]
the entries to right of each pivot within its row are all $0$, and
\item[$\bullet$]
there is at most one pivot in each row and at most one pivot in each column (so that
their positions are the positions of the $1$'s in
some matrix $f \in F_{n, m}$).
\end{enumerate}
Let us show this, and also show that the positions of the pivots are
preserved. Indeed, it is clear that the positions of the pivots are
preserved
(because zero columns stay zero, nonzero columns stay nonzero, and the
bottommost entries of nonzero columns do not move); therefore it is
enough to prove the following fact:
\begin{enumerate}
\item[\textbf{(d)}]
Consider an \emph{elimination step}, by which we mean a step in
which a pivot in some position $a$ is used to eliminate all the entries
above it in the same column. Assume
that, before the elimination step, the entries to the right of $a$ within
its row were all $0$. Let $b$ be
the position of another pivot that existed before this elimination step.
Assume that, before the elimination step, the pivot at $b$ equalled $1$,
and the entries to the right of $b$ within its row were all $0$.
Then, after the elimination step, $b$ is still a position of a pivot and
this pivot still equals $1$, and the entries to the right of $b$
within its row are still $0$.
\end{enumerate}
So let us prove fact \textbf{(d)}. We prove it by contradiction: Assume
that it is not the case. Then, the elimination step must have changed at
least one of the entries weakly to the right of $b$ within its row
(because we already know that
$b$ is still a position of a pivot after the elimination step).
In particular, the elimination step must have changed at
least one entry in the row of $b$. Thus, $b$ must lie in a row strictly
above $a$ (since otherwise, the elimination step would have not changed any
entry in the row of $b$). Let $c$ be the position in the same row as
$b$ and in the same column as $a$. Then, the entry at $c$ before the
elimination step must not have been $0$ (since otherwise, the elimination step
would not have changed any entry in the row of $b$). As a consequence,
$c$ must not lie to the right of $b$ within its row (because before
the elimination step, the entries to the right of $b$ within its row were
all $0$). Thus, $c$ lies weakly to the left of $b$ within its row. In
other words, the column containing $b$ lies weakly right of the column
containing $c$. In other words,
the column containing $b$ lies weakly right of the column
containing $a$ (since the column containing $c$ is the column containing
$a$). Hence, the column containing $b$ must lie \textbf{strictly} right
of the column containing $a$ (since
otherwise, $a$ and $b$ would lie in the same column, which is absurd
because $a$ and $b$ are two distinct pivots). But let us recall that,
before the elimination step, the entries to the right of $a$ within
its row were all $0$. Let us call these entries the
\emph{silent entries}. Now, what did the elimination step do to the entries
weakly to the right of $b$ within its row? It changed them by adding
multiples of the corresponding entries of the row containing $a$. But
all these corresponding entries were silent entries (because the column
containing $b$ lies strictly right of the column containing $a$) and
thus were $0$. Hence, the elimination step did not change the entries weakly
to the right of $b$ within its row. This contradicts the fact that
the elimination step must have changed at
least one of the entries weakly to the right of $b$ within its row.
This contradiction finishes the proof of fact \textbf{(d)}.}  The result
is the matrix $f$.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.towers.indres}}

\textit{Solution to Exercise~\ref{exe.towers.indres}.} (a) Let us begin
this solution by stating some trivialities. It is easy to see that every
finite group $G$ and every $ \CC G$-module $P$ satisfy
\begin{equation}
\Ind_{G}^{G}P\cong P \label{sol.towers.indres.a.triv}%
\end{equation}
and%
\begin{equation}
\Inf_{G}^{G}P\cong P.
\label{sol.towers.indres.a.triv2}%
\end{equation}


Furthermore, if $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be two groups, if
$\mathcal{K}_{1}\triangleleft\mathcal{G}_{1}$ and $\mathcal{K}_{2}%
\triangleleft\mathcal{G}_{2}$ are normal subgroups, if $\mathcal{U}_{1}$ is a
$ \CC \left[  \mathcal{G}_{1}/\mathcal{K}_{1}\right]  $-module, and if
$\mathcal{U}_{2}$ is a $ \CC \left[  \mathcal{G}_{2}/\mathcal{K}%
_{2}\right]  $-module, then
\begin{equation}
\Inf_{\left(  \mathcal{G}_{1}/\mathcal{K}_{1}\right)
\times\left(  \mathcal{G}_{2}/\mathcal{K}_{2}\right)  }^{\mathcal{G}_{1}%
\times\mathcal{G}_{2}}\left(  \mathcal{U}_{1}\otimes\mathcal{U}_{2}\right)
\cong\left(  \Inf_{\mathcal{G}_{1}/\mathcal{K}_{1}%
}^{\mathcal{G}_{1}}\mathcal{U}_{1}\right)  \otimes\left( 
\Inf_{\mathcal{G}_{2}/\mathcal{K}_{2}}^{\mathcal{G}_{2}}
\mathcal{U}_{2}\right)
\label{sol.towers.indres.a.infl}
\end{equation}
as ${ \CC }\left[  \mathcal{G}_{1}\times\mathcal{G}_{2}\right]
$-modules. (This is an analogue of (\ref{induction-vs-tensor-product}), but is
trivial to prove.) Also, if $\mathcal{G}$, $\mathcal{H}$ and $\mathcal{I}$ are
three groups such that $\mathcal{I}<\mathcal{H}<\mathcal{G}$, and if
$\mathcal{U}$ is a $ \CC \mathcal{I}$-module, then%
\begin{equation}
\Inf_{\mathcal{H}}^{\mathcal{G}}
\Inf_{\mathcal{I}}^{\mathcal{H}}\mathcal{U}
=\Inf_{\mathcal{I}}^{\mathcal{G}}\mathcal{U}.
\label{sol.towers.indres.a.infl.trans}
\end{equation}
(This is an analogue of Exercise \ref{exe.induction.transitivity}, and is
again trivial\footnote{Note that the equality sign in
(\ref{sol.towers.indres.a.infl.trans}) is a honest equality, not just a
canonical isomorphism.}.)

Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be an
almost-composition of an $n\in \NN$ satisfying $\ell\geq1$. Let $V_{i}$
be a ${ \CC }G_{i}$-module for every $i\in\left\{  1,2,\ldots
,\ell\right\}  $. We need to prove the two isomorphisms
\begin{align}
&  \ind_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}%
,\alpha_{\ell}}^{n}\left(  \ind_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell-1}\right)  }^{\alpha_{1}+\alpha_{2}+\cdots
+\alpha_{\ell-1}}\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell
-1}\right)  \otimes V_{\ell}\right) \nonumber\\
&  \cong\ind_{\alpha}^{n}\left(  V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{\ell}\right)  \label{sol.towers.indres.a.goal1}%
\end{align}
and%
\begin{align}
&  \ind_{\alpha_{1},\alpha_{2}+\alpha_{3}+\cdots+\alpha_{\ell}%
}^{n}\left(  V_{1}\otimes\ind_{\left(  \alpha_{2},\alpha
_{3},\ldots,\alpha_{\ell}\right)  }^{\alpha_{2}+\alpha_{3}+\cdots+\alpha
_{\ell}}\left(  V_{2}\otimes V_{3}\otimes\cdots\otimes V_{\ell}\right)
\right) \nonumber\\
&  \cong\ind_{\alpha}^{n}\left(  V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{\ell}\right)  . \label{sol.towers.indres.a.goal2}%
\end{align}
We will only prove (\ref{sol.towers.indres.a.goal1}), since the proof of
(\ref{sol.towers.indres.a.goal2}) is analogous.

Let $m$ be the nonnegative integer $\alpha_{1}+\alpha_{2}+\cdots+\alpha
_{\ell-1}$. Then, $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  $ is an almost-composition of $m$, and we have $\alpha_{1}%
+\alpha_{2}+\cdots+\alpha_{\ell-1}=m$.

Let $W$ denote the $ \CC \left[  G_{1}\times G_{2}\times\cdots\times
G_{\ell-1}\right]  $-module $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell
-1}$. Then, $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell-1}=W$ and
$V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell}=\underbrace{\left(
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell-1}\right)  }_{=W}\otimes
V_{\ell}=W\otimes V_{\ell}$ and $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}=m$. Hence, the relation (\ref{sol.towers.indres.a.goal1}) (which we want
to prove) rewrites as%
\begin{equation}
\ind_{m,\alpha_{\ell}}^{n}\left(  \ind_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }^{m}W\otimes V_{\ell
}\right)  \cong\ind_{\alpha}^{n}\left(  W\otimes V_{\ell
}\right)  . \label{sol.towers.indres.a.goal1a}%
\end{equation}
It thus remains to prove (\ref{sol.towers.indres.a.goal1a}).

We distinguish between two cases:

\textit{Case 1:} We have $G_{\ast}=\Symm_{\ast}$ or $G_{\ast
}=\Symm_{\ast}\left[  \Gamma\right]  $.

\textit{Case 2:} We have $G_{\ast}=GL_{\ast}$.

Let us consider Case 1 first. In this case, we have
$G_{\ast}=\Symm_{\ast}$ or
$G_{\ast}=\Symm_{\ast}\left[  \Gamma\right]  $. Thus,
$\ind_{\beta}^{N}=\Ind_{G_{\beta}}^{G_{N}}$ for
every $N\in \NN$ and every almost-composition $\beta$ of $N$ (by the
definition of $\ind_{\beta}^{N}$), and
$\ind_{i,j}^{N}=\Ind_{G_{i}\times G_{j}}^{G_{N}}$
for every $N\in \NN$ and every $i,j\in \NN$ satisfying $i+j=N$ (by
the definition of $\ind_{i,j}^{N}$).

Since $\ind_{\beta}^{N}=\Ind_{G_{\beta}}^{G_{N}}$
for every $N\in \NN$ and every almost-composition $\beta$ of $N$, we
have $\ind_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }^{m}=\Ind_{G_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right)  }}^{G_{m}}$. Thus,
\begin{align}
&  \ind_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }^{m}W\otimes V_{\ell}\nonumber\\
&  =\underbrace{\Ind_{G_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right)  }}^{G_{m}}W}_{\substack{
=\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots
\times G_{\alpha_{\ell-1}}}^{G_{m}}W\\
\text{(since }G_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{\ell-1}\right)  }=G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}\text{)}}}\otimes\underbrace{V_{\ell}}_{\substack{\cong%
\Ind_{G_{\alpha_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell
}\\\text{(because (\ref{sol.towers.indres.a.triv}) (applied}\\\text{to
}G=G_{\alpha_{\ell}}\text{ and }P=V_{\ell}\text{) yields }
\Ind_{G_{\alpha_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell}\cong V_{\ell
}\text{)}}}\nonumber\\
&  \cong\left(  \Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}%
\times\cdots\times G_{\alpha_{\ell-1}}}^{G_{m}}W\right)  \otimes\left(
\Ind_{G_{\alpha_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell
}\right)  \cong\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}%
\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Exercise \ref{exe.ind.tensor} (applied to }G_{m}\text{, }%
G_{\alpha_{\ell}}\text{, }G_{\alpha_{1}}\times G_{\alpha_{2}}\times
\cdots\times G_{\alpha_{\ell-1}}\text{, }G_{\alpha_{\ell}}\text{,}\\
W\text{ and }V_{\ell}\text{ instead of }G_{1}\text{, }G_{2}\text{, }%
H_{1}\text{, }H_{2}\text{, }U_{1}\text{ and }U_{2}\text{) yields}\\
\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{G_{\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}\left(
W\otimes V_{\ell}\right) \\
\cong\left(  \Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}%
\times\cdots\times G_{\alpha_{\ell-1}}}^{G_{\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}}} W  \right)  \otimes\left(
 \Ind_{G_{\alpha_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell}\right)
\end{array}
\right) \nonumber\\
&  =\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell}}}^{G_{m}\times G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell
}\right)  . \label{sol.towers.indres.a.1}%
\end{align}
But from $m = \alpha_1 + \alpha_2 + \cdots + \alpha_{\ell-1}$,
we obtain
$m + \alpha_\ell
= \left(\alpha_1 + \alpha_2 + \cdots + \alpha_{\ell-1}\right) + \alpha_\ell
= \alpha_1 + \alpha_2 + \cdots + \alpha_\ell = n$
(since $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $
is an almost-composition of $n$).
Thus, since $\ind_{i,j}^{N}=\Ind_{G_{i}\times G_{j}}^{G_{N}}$ for
every $N\in \NN$ and every
$i,j\in \NN$ satisfying $i+j=N$, we have
$\ind_{m,\alpha_{\ell}}^{n}=\Ind_{G_{m}\times G_{\alpha_{\ell}}%
}^{G_{n}}$. Thus,
\begin{align*}
&  \ind_{m,\alpha_{\ell}}^{n}\left(  \ind_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }^{m}W\otimes V_{\ell
}\right) \\
&  =\Ind_{G_{m}\times G_{\alpha_{\ell}}}^{G_{n}}\left(
\underbrace{\ind_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }^{m}W\otimes V_{\ell}}_{\substack{\cong%
\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell}}}^{G_{m}\times G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell
}\right)  \\\text{(by (\ref{sol.towers.indres.a.1}))}}}\right)  \cong%
\Ind_{G_{m}\times G_{\alpha_{\ell}}}^{G_{n}}
\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots
\times G_{\alpha_{\ell}}%
}^{G_{m}\times G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right) \\
&  \cong\Ind_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times
\cdots\times G_{\alpha_{\ell}}}^{G_{n}}\left(  W\otimes V_{\ell}\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Exercise \ref{exe.induction.transitivity}, applied to }G=G_{n}\text{,
}H=G_{m}\times G_{\alpha_{\ell}}\text{,}\\
I=G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell}%
}\text{ and } U = W \otimes V_\ell
\end{array}
\right) \\
&  =\ind_{\alpha}^{n}\left(  W\otimes V_{\ell}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Ind_{G_{\alpha_{1}%
}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell}}}^{G_{n}%
}=\ind_{\alpha}^{n}\text{ (by the definition of }%
\ind_{\alpha}^{n}\text{)}\right)  .
\end{align*}
This proves (\ref{sol.towers.indres.a.goal1}). Hence,
(\ref{sol.towers.indres.a.goal1}) is proven in Case 1.

Let us now consider Case 2. In this case, we have $G_{\ast}=GL_{\ast}$. Hence,
$\ind_{\beta}^{N}=\Ind_{P_{\beta}}^{G_{N}%
}\Inf_{G_{\beta}}^{P_{\beta}}$ for every $N\in \NN$ and
every almost-composition $\beta$ of $N$ (by the definition of
$\ind_{\beta}^{N}$), and $\ind_{i,j}^{N}
=\Ind_{P_{i,j}}^{G_{N}}\Inf_{G_{i}\times G_{j}}^{P_{i,j}}$
for every $N\in \NN$ and every
$i,j\in \NN$ satisfying $i+j=N$ (by the definition of
$\ind_{i,j}^{N}$).

Since $\ind_{\beta}^{N}=\Ind_{P_{\beta}}^{G_{N}%
}\Inf_{G_{\beta}}^{P_{\beta}}$ for every $N\in \NN$ and
every almost-composition $\beta$ of $N$, we have
\[
\ind_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }^{m}=\Ind_{P_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right)  }}^{G_{m}}
\Inf_{G_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }%
}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }%
}=\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }}^{G_{m}}\Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}}^{P_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }}%
\]
(since $G_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)
}=G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}$).
Therefore,
\begin{align}
&  \ind_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }^{m}W\otimes V_{\ell}\nonumber\\
&  =\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{\ell-1}\right)  }}^{G_{m}}\Inf_{G_{\alpha_{1}%
}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}}^{P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }}W\otimes
\underbrace{V_{\ell}}_{\substack{\cong\Ind_{G_{\alpha
_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell}\\\text{(because
(\ref{sol.towers.indres.a.triv}) (applied}\\\text{to }G=G_{\alpha_{\ell}%
}\text{ and }P=V_{\ell}\text{) yields }\Ind_{G_{\alpha
_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell}\cong V_{\ell}\text{)}}}\nonumber\\
&  \cong\left(  \Ind_{P_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right)  }}^{G_{m}}
\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }}W\right)  \otimes\left(  \Ind_{G_{\alpha
_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell}\right) \nonumber\\
&  \cong\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\left(  \Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}}^{P_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }}W\otimes\underbrace{V_{\ell}%
}_{\substack{\cong\Inf_{G_{\alpha_{\ell}}}%
^{G_{\alpha_{\ell}}}V_{\ell}\\\text{(because (\ref{sol.towers.indres.a.triv})
(applied}\\\text{to }G=G_{\alpha_{\ell}}\text{ and }P=V_{\ell}\text{) yields
}\Inf_{G_{\alpha_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell
}\cong V_{\ell}\text{)}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Exercise \ref{exe.ind.tensor} (applied to }G_{m}\text{, }%
G_{\alpha_{\ell}}\text{, }P_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{\ell
-1}\right)  }\text{, }G_{\alpha_{\ell}}\text{,}\\
\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times
\cdots\times G_{\alpha_{\ell-1}}}^{P_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right)  }}W\text{ and }V_{\ell}\text{ instead of
}G_{1}\text{, }G_{2}\text{, }H_{1}\text{, }H_{2}\text{, }U_{1}\text{ and
}U_{2}\text{) yields}\\
\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\left(  \Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}}^{P_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }}W\otimes
V_{\ell} \right) \\
= \left(  \Ind_{P_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right)  }}^{G_{m}}
\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }}W\right)  \otimes\left(  \Ind_{G_{\alpha
_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell}\right)
\end{array}
\right) \nonumber\\
&  \cong\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\underbrace{\left(  \Inf_{G_{\alpha
_{1}}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}}^{P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }}W\otimes
\Inf_{G_{\alpha_{\ell}}}^{G_{\alpha_{\ell}}}V_{\ell
}\right)  }_{\substack{\cong\Inf_{G_{\alpha_{1}%
}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\times
G_{\alpha_{\ell}}}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }\times G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right)
\\\text{(by (\ref{sol.towers.indres.a.infl}), applied to}\\\mathcal{G}%
_{1}=P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\text{,
}\mathcal{G}_{2}=G_{\alpha_{\ell}}\text{, }\mathcal{K}_{1}=K_{\left(
\alpha_{1},\alpha_{2},...,\alpha_{\ell-1}\right)  }\text{,}\\\mathcal{K}%
_{2}=G_{\alpha_{\ell}}\text{, }\mathcal{U}_{1}=W\text{ and }\mathcal{U}%
_{2}=V_{\ell}\\\text{(because }K_{\left(  \alpha_{1},\alpha_{2},...,\alpha
_{\ell-1}\right)  }\text{ is a normal subgroup}\\\text{of }P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\text{ and the quotient
is}\\P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)
}/K_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{\ell-1}\right)  }=G_{\alpha
_{1}}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\text{))}%
}}\nonumber\\
&  \cong\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}%
}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times
G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right)  .
\label{sol.towers.indres.a.1hard}%
\end{align}
But from $m = \alpha_1 + \alpha_2 + \cdots + \alpha_{\ell-1}$,
we obtain
$m + \alpha_\ell
= \left(\alpha_1 + \alpha_2 + \cdots + \alpha_{\ell-1}\right) + \alpha_\ell
= \alpha_1 + \alpha_2 + \cdots + \alpha_\ell = n$
(since $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $
is an almost-composition of $n$).
Thus, since $\ind_{i,j}^{N}=\Ind_{P_{i,j}}^{G_{N}}\Inf_{G_{i}\times G_{j}%
}^{P_{i,j}}$ for every $N\in \NN$ and every $i,j\in \NN$
satisfying $i+j=N$, we have $\ind_{m,\alpha_{\ell}}^{n}
=\Ind_{P_{m,\alpha_{\ell}}}^{G_{n}}%
\Inf_{G_{m}\times G_{\alpha_{\ell}}}^{P_{m,\alpha
_{\ell}}}$, so that%
\begin{align}
&  \ind_{m,\alpha_{\ell}}^{n}\left(  \ind_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }^{m}W\otimes V_{\ell
}\right) \nonumber\\
&  =\Ind_{P_{m,\alpha_{\ell}}}^{G_{n}}%
\Inf_{G_{m}\times G_{\alpha_{\ell}}}^{P_{m,\alpha
_{\ell}}}\left(  \underbrace{\ind_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell-1}\right)  }^{m}W\otimes V_{\ell}}_{\substack{\cong%
\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times G_{\alpha_{\ell}}%
}\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}%
\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}%
}}\left(  W\otimes V_{\ell}\right)  \\\text{(by
(\ref{sol.towers.indres.a.1hard}))}}}\right) \nonumber\\
&  \cong\Ind_{P_{m,\alpha_{\ell}}}^{G_{n}%
}\Inf_{G_{m}\times G_{\alpha_{\ell}}}^{P_{m,\alpha
_{\ell}}}\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}%
}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times
G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right)
\label{sol.towers.indres.a.3hard}%
\end{align}


Now, we are going to prove that%
\begin{equation}
\Inf_{G_{m}\times G_{\alpha_{\ell}}}^{P_{m,\alpha
_{\ell}}}\Ind_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}Z\cong\Ind_{P_{\alpha}}%
^{P_{m,\alpha_{\ell}}}\Inf_{P_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}%
}^{P_{\alpha}}Z \label{sol.towers.indres.a.4hard}%
\end{equation}
for every $ \CC \left[  P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}\right]  $-module $Z$. In
order to do so, we try to apply Exercise \ref{exe.ind-infl} to $G=P_{m,\alpha
_{\ell}}$, $H=P_{\alpha}$, $K=K_{m,\alpha_{\ell}}$ and $V=Z$. This yields
(\ref{sol.towers.indres.a.4hard}) if we can prove the following two statements:

\begin{enumerate}
\item We have $K_{m,\alpha_{\ell}}<P_{\alpha}<P_{m,\alpha_{\ell}}$ and
$K_{m,\alpha_{\ell}}\triangleleft P_{m,\alpha_{\ell}}$.

\item The quotient $P_{m,\alpha_{\ell}}/K_{m,\alpha_{\ell}}$ is canonically
identified with $G_{m}\times G_{\alpha_{\ell}}$ in such a way that its
subgroup $P_{\alpha}/K_{m,\alpha_{\ell}}$ is canonically identified with
$P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times
G_{\alpha_{\ell}}$.
\end{enumerate}

The first of these two statements is clear (using $m=\alpha_{1}+\alpha
_{2}+\cdots+\alpha_{\ell-1}$ and $\alpha=\left(  \alpha_{1},\alpha
_{2},...,\alpha_{\ell}\right)  $). It remains to prove the second statement.
We know how $P_{m,\alpha_{\ell}}/K_{m,\alpha_{\ell}}$ is identified with
$G_{m}\times G_{\alpha_{\ell}}$ already. The thing that we need to prove is
that the subgroup $P_{\alpha}/K_{m,\alpha_{\ell}}$ of $P_{m,\alpha_{\ell}%
}/K_{m,\alpha_{\ell}}$ is canonically identified with $P_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}$. In
other words, we need to prove that the projection of the subgroup $P_{\alpha}$
of $P_{m,\alpha_{\ell}}$ onto $G_{m}\times G_{\alpha_{\ell}}$ is precisely
$P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times
G_{\alpha_{\ell}}$. But this is obvious. Hence, the second statement is
proven, and so we are able to apply Exercise \ref{exe.ind-infl} and therefore
obtain (\ref{sol.towers.indres.a.4hard}).

Now, (\ref{sol.towers.indres.a.3hard}) becomes%
\begin{align*}
&  \ind_{m,\alpha_{\ell}}^{n}\left(  \ind_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }^{m}W\otimes V_{\ell
}\right) \\
&  \cong\Ind_{P_{m,\alpha_{\ell}}}^{G_{n}%
}\underbrace{\Inf_{G_{m}\times G_{\alpha_{\ell}}%
}^{P_{m,\alpha_{\ell}}}\Ind_{P_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{G_{m}\times
G_{\alpha_{\ell}}}\Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}%
}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times
G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right)  }_{\substack{\cong%
\Ind_{P_{\alpha}}^{P_{m,\alpha_{\ell}}}%
\Inf_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{P_{\alpha}}%
\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times
\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}%
}}\left(  W\otimes V_{\ell}\right)  \\\text{(by
(\ref{sol.towers.indres.a.4hard}), applied to }Z
= \Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}\left(  W\otimes
V_{\ell}\right)  \text{)}}}\\
&  \cong\Ind_{P_{m,\alpha_{\ell}}}^{G_{n}%
}\Ind_{P_{\alpha}}^{P_{m,\alpha_{\ell}}}%
\Inf_{P_{\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{P_{\alpha}}%
\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times
\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}%
}}\left(  W\otimes V_{\ell}\right) \\
&  \cong\Ind_{P_{\alpha}}^{G_{n}}%
\underbrace{\Inf_{P_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}}^{P_{\alpha}%
}\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}%
\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}%
}}\left(  W\otimes V_{\ell}\right)  }_{\substack{=
\Inf_{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\alpha}}\left(  W\otimes
V_{\ell}\right)  \\\text{(by (\ref{sol.towers.indres.a.infl.trans}), applied
to }\mathcal{G}=P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
}\text{,}\\\mathcal{H}=P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }\times G_{\alpha_{\ell}}\text{, }\mathcal{I}=G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}%
}\text{ and }\mathcal{U}=W\otimes V_{\ell}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Exercise \ref{exe.induction.transitivity}, applied to }G=G_{n}\text{,
}H=P_{m,\alpha_{\ell}}\text{,}\\
I=P_{\alpha}\text{ and }U=\Inf_{P_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times G_{\alpha_{\ell}}%
}^{P_{\alpha}}\Inf_{G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}%
}^{P_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }\times
G_{\alpha_{\ell}}}\left(  W\otimes V_{\ell}\right)
\end{array}
\right) \\
&  =\Ind_{P_{\alpha}}^{G_{n}}\Inf_{G_{\alpha_{1}}\times
 G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}}^{P_{\alpha}}\left(  W\otimes
V_{\ell}\right) \\
&  =\Ind_{P_{\alpha}}^{G_{n}}\Inf_{G_{\alpha}}^{P_{\alpha}}
\left(  W\otimes V_{\ell}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }G_{\alpha_{1}}\times G_{\alpha_{2}%
}\times\cdots\times G_{\alpha_{\ell-1}}\times G_{\alpha_{\ell}}=G_{\alpha_{1}%
}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell}}=G_{\alpha}\right)
\\
&  =\ind_{\alpha}^{n}\left(  W\otimes V_{\ell}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Ind_{P_{\alpha}}^{G_{n}}
\Inf_{G_{\alpha}}^{P_{\alpha}}
=\ind_{\alpha}^{n}\text{ (by the definition of
}\ind_{\alpha}^{n}\text{)}\right)  .
\end{align*}
This proves (\ref{sol.towers.indres.a.goal1}). Hence,
(\ref{sol.towers.indres.a.goal1}) is proven in Case 2.

Now, (\ref{sol.towers.indres.a.goal1}) is proven in both Cases 1 and 2. Hence,
(\ref{sol.towers.indres.a.goal1}) always holds. This completes the solution to
Exercise \ref{exe.towers.indres}(a).

(b) \textit{Alternative solution to Exercise \ref{exe.Ind-ass}:} Let $G_{\ast
}$ be the tower $\Symm_{\ast}$ of groups. We can then apply Exercise
\ref{exe.towers.indres}(a) to $\ell=3$, $n=i+j+k$, $\alpha=\left(
i,j,k\right)  $, $V_{1}=U$, $V_{2}=V$ and $V_{3}=W$. As the result, we obtain%
\begin{align*}
\ind_{i+j,k}^{i+j+k}\left(  \ind_{\left(
i,j\right)  }^{i+j}\left(  U\otimes V\right)  \otimes W\right)   &
\cong\ind_{\left(  i,j,k\right)  }^{i+j+k}\left(  U\otimes
V\otimes W\right) \\
&  \cong\ind_{i,j+k}^{i+j+k}\left(  U
\otimes\ind_{\left(  j,k\right)  }^{j+k}
\left(  V\otimes W\right)  \right)  .
\end{align*}
Since $\ind_{i+j,k}^{i+j+k}
=\Ind_{\Symm_{i+j}\times\Symm_k}^{\Symm_{i+j+k}}$,
$\ind_{\left(  i,j\right)  }^{i+j}=\ind_{i,j}^{i+j}%
=\Ind_{\Symm_i\times\Symm_j}^{\Symm_{i+j}}$,
$\ind_{\left(  i,j,k\right)  }^{i+j+k}%
=\Ind_{G_{i,j,k}}^{\Symm_{i+j+k}}%
=\Ind_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j+k}}$
(because $G_{i,j,k}=\Symm_i\times\Symm_j\times\Symm_k$),
$\ind_{i,j+k}^{i+j+k}=\Ind_{\Symm_i\times\Symm_{j+k}%
}^{\Symm_{i+j+k}}$ and $\ind_{\left(  j,k\right)  }%
^{j+k}=\ind_{j,k}^{j+k}
=\Ind_{\Symm_j\times\Symm_k}^{\Symm_{j+k}}$,
this rewrites as follows:
\begin{align*}
\Ind_{\Symm_{i+j}\times\Symm_k}^{\Symm_{i+j+k}}
\left(  \Ind_{\Symm_i\times\Symm_j%
}^{\Symm_{i+j}}\left(  U\otimes V\right)  \otimes W\right)   &
\cong\Ind_{\Symm_i\times\Symm_j\times
\Symm_k}^{\Symm_{i+j+k}}\left(  U\otimes V\otimes W\right) \\
&  \cong\Ind_{\Symm_i\times\Symm_{j+k}%
}^{\Symm_{i+j+k}}\left(  U\otimes
\Ind_{\Symm_j\times\Symm_k}^{\Symm_{j+k}}
\left(  V\otimes W\right)
\right)  .
\end{align*}
Thus, Exercise \ref{exe.Ind-ass} is solved again. Hence, we have solved
Exercise \ref{exe.towers.indres}(b).

(c) Let $\Sigma=\bigsqcup_{n\geq0}\Irr \left(  G_{n}\right)  $.
We have to prove that the map $m$ is associative. In other words, we have to
prove that $m\left(  m\left(  \alpha\otimes\beta\right)  \otimes\gamma\right)
=m\left(  \alpha\otimes m\left(  \beta\otimes\gamma\right)  \right)  $ for any
three elements $\alpha$, $\beta$ and $\gamma$ of $A$. In order to do so, it is
clearly enough to show that $m\left(  m\left(  \alpha\otimes\beta\right)
\otimes\gamma\right)  =m\left(  \alpha\otimes m\left(  \beta\otimes
\gamma\right)  \right)  $ for any three elements $\alpha$, $\beta$ and
$\gamma$ of $\Sigma$ (because $\Sigma$ is a $\ZZ$-module basis of $A$,
and the equality $m\left(  m\left(  \alpha\otimes\beta\right)  \otimes
\gamma\right)  =m\left(  \alpha\otimes m\left(  \beta\otimes\gamma\right)
\right)  $ is $\ZZ$-linear in each of $\alpha$, $\beta$ and $\gamma$).
So let $\alpha$, $\beta$ and $\gamma$ be three elements of $\Sigma$. Then,
there exists $i\in \NN$, $j\in \NN$ and $k\in \NN$
satisfying $\alpha\in\Irr \left(  G_{i}\right)  $, $\beta
\in\Irr \left(  G_{j}\right)  $ and $\gamma\in
\Irr \left(  G_{k}\right)  $ (since $\alpha$, $\beta$ and
$\gamma$ belong to $\Sigma=\bigsqcup_{n\geq0}\Irr \left(
G_{n}\right)  $). Consider these $i$, $j$ and $k$.

There exists an irreducible $ \CC G_{i}$-module $U$ satisfying
$\alpha=\chi_{U}$ (since $\alpha\in\Irr \left(  G_{i}\right)
$). Similarly, there exists an irreducible $ \CC G_{j}$-module $V$
satisfying $\beta=\chi_{V}$, and an irreducible $ \CC G_{k}$-module $W$
satisfying $\gamma=\chi_{W}$. Consider these $U$, $V$ and $W$.

We can apply Exercise \ref{exe.towers.indres}(a) to $\ell=3$, $n=i+j+k$,
$\alpha=\left(  i,j,k\right)  $, $V_{1}=U$, $V_{2}=V$ and $V_{3}=W$. As the
result, we obtain%
\begin{align*}
\ind_{i+j,k}^{i+j+k}\left(  \ind_{\left(
i,j\right)  }^{i+j}\left(  U\otimes V\right)  \otimes W\right)   &
\cong\ind_{\left(  i,j,k\right)  }^{i+j+k}\left(  U\otimes
V\otimes W\right) \\
&  \cong\ind_{i,j+k}^{i+j+k}\left(  U\otimes
\ind_{\left(  j,k\right)  }^{j+k}\left(  V\otimes W\right)  \right)  .
\end{align*}
Thus,%
\[
\ind_{i+j,k}^{i+j+k}\left(  \ind_{\left(
i,j\right)  }^{i+j}\left(  U\otimes V\right)  \otimes W\right)  \cong%
\ind_{i,j+k}^{i+j+k}\left(  U\otimes\ind_{\left(
j,k\right)  }^{j+k}\left(  V\otimes W\right)  \right)  .
\]
Since isomorphic representations have equal characters, this yields%
\begin{equation}
\chi_{\ind_{i+j,k}^{i+j+k}\left(  \ind_{\left(
i,j\right)  }^{i+j}\left(  U\otimes V\right)  \otimes W\right)  }%
=\chi_{\ind_{i,j+k}^{i+j+k}\left(  U\otimes
\ind_{\left(  j,k\right)  }^{j+k}\left(  V\otimes W\right)
 \right)  }.
\label{sol.towers.indres.c.equal-chars}%
\end{equation}


Since $\alpha=\chi_{U}$ and $\beta=\chi_{V}$, we have%
\begin{align*}
m\left(  \alpha\otimes\beta\right)   &  =m\left(  \chi_{U}\otimes\chi
_{V}\right)  =\ind_{i,j}^{i+j}\left(  \chi_{U}\otimes\chi
_{V}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }U\text{ is a }%
 \CC G_{i}\text{-module and }V\text{ is a } \CC G_{j}\text{-module}%
\right) \\
&  =\chi_{\ind_{i,j}^{i+j}\left(  U\otimes V\right)  }%
=\chi_{\ind_{\left(  i,j\right)  }^{i+j}\left(  U\otimes
V\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\ind_{i,j}^{i+j}=\ind_{\left(  i,j\right)  }^{i+j}\right)  .
\end{align*}
Thus,%
\begin{align}
&  m\left(  \underbrace{m\left(  \alpha\otimes\beta\right)  }_{=\chi
_{\ind_{\left(  i,j\right)  }^{i+j}\left(  U\otimes V\right)  }%
}\otimes\underbrace{\gamma}_{=\chi_{W}}\right) \nonumber\\
&  =m\left(  \chi_{\ind_{\left(  i,j\right)  }^{i+j}\left(
U\otimes V\right)  }\otimes\chi_{W}\right)  =\ind_{i+j,k}%
^{i+j+k}\left(  \chi_{\ind_{\left(  i,j\right)  }^{i+j}\left(
U\otimes V\right)  }\otimes\chi_{W}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ind_{\left(
i,j\right)  }^{i+j}\left(  U\otimes V\right)  \text{ is a } \CC \left[
G_{i+j}\right]  \text{-module and }W\text{ is a } \CC G_{k}%
\text{-module}\right) \nonumber\\
&  =\chi_{\ind_{i+j,k}^{i+j+k}\left(
\ind_{\left(  i,j\right)  }^{i+j}\left(  U\otimes V\right)
 \otimes W\right)  }.
\label{sol.towers.indres.c.1}%
\end{align}
Similarly,%
\begin{equation}
m\left(  \alpha\otimes m\left(  \beta\otimes\gamma\right)  \right)
=\chi_{\ind_{i,j+k}^{i+j+k}\left(  U\otimes
\ind_{\left(  j,k\right)  }^{j+k}
\left(  V\otimes W\right)  \right)  }.
\label{sol.towers.indres.c.2}
\end{equation}
Now, (\ref{sol.towers.indres.c.1}) becomes%
\begin{align*}
m\left(  m\left(  \alpha\otimes\beta\right)  \otimes\gamma\right)   &
=\chi_{\ind_{i+j,k}^{i+j+k}\left(  \ind_{\left(
i,j\right)  }^{i+j}\left(  U\otimes V\right)  \otimes W\right)  }%
=\chi_{\ind_{i,j+k}^{i+j+k}\left(  U\otimes
\ind_{\left(  j,k\right)  }^{j+k}
\left(  V\otimes W\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.towers.indres.c.equal-chars}%
)}\right) \\
&  =m\left(  \alpha\otimes m\left(  \beta\otimes\gamma\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.towers.indres.c.2})}\right)  .
\end{align*}
This is what we wanted to prove. Thus, Exercise \ref{exe.towers.indres}(c) is solved.

(d) Let $\Sigma=\bigsqcup_{n\geq0}\Irr \left(  G_{n}\right)  $.

We will solve Exercise \ref{exe.towers.indres}(d) by induction over $\ell$.
The induction base (the case $\ell=0$) is trivial, so we come to the induction
step. Fix a positive integer $\ell$. We assume that Exercise
\ref{exe.towers.indres}(d) is already solved for $\ell-1$ instead of $\ell$.
We now need to solve Exercise \ref{exe.towers.indres}(d) for our $\ell$.

Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be an
almost-composition of an $n\in{ \NN }$. Let $\chi_{i}\in R\left(
G_{\alpha_{i}}\right)  $ for every $i\in\left\{  1,2,\ldots,\ell\right\}  $.
We will show that%
\[
\chi_{1}\chi_{2}\cdots\chi_{\ell}=\ind_{\alpha}^{n}\left(
\chi_{1}\otimes\chi_{2}\otimes\cdots\otimes\chi_{\ell}\right)  .
\]
Since this equality is $\ZZ$-linear in each of $\chi_{1}$, $\chi_{2}$,
$...$, $\chi_{\ell}$, we can WLOG assume that $\chi_{1}$, $\chi_{2}$, $...$,
$\chi_{\ell}$ all lie in $\Sigma$ (since $\Sigma$ is a basis of $A$). Assume
this. Then, $\chi_{i}\in\Sigma\cap R\left(  G_{\alpha_{i}}\right)
=\Irr \left(  G_{\alpha_{i}}\right)  $ for every $i\in\left\{
1,2,...,\ell\right\}  $. Hence, for every $i\in\left\{  1,2,...,\ell\right\}
$, there exists an irreducible $ \CC G_{\alpha_{i}}$-module $V_{i}$ such
that $\chi_{i}=\chi_{V_{i}}$. Consider this $V_{i}$.

Let $m=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}$. By the induction
hypothesis, we can apply Exercise \ref{exe.towers.indres}(d) to the
almost-composition $\left(  \alpha_{1},\alpha_{2},...,\alpha_{\ell-1}\right)
$ of $m$. As a result, we obtain
\begin{align*}
\chi_{1}\chi_{2}\cdots\chi_{\ell-1}  &  =\ind_{\left(
\alpha_{1},\alpha_{2},...,\alpha_{\ell-1}\right)  }^{m}\left(
\underbrace{\chi_{1}}_{=\chi_{V_{1}}}\otimes\underbrace{\chi_{2}}%
_{=\chi_{V_{2}}}\otimes\cdots\otimes\underbrace{\chi_{\ell-1}}_{=\chi
_{V_{\ell-1}}}\right) \\
&  =\ind_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{\ell
-1}\right)  }^{m}\underbrace{\left(  \chi_{V_{1}}\otimes\chi_{V_{2}}%
\otimes\cdots\otimes\chi_{V_{\ell-1}}\right)  }_{=\chi_{V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{\ell-1}}}=\ind_{\left(  \alpha
_{1},\alpha_{2},...,\alpha_{\ell-1}\right)  }^{m}\left(  \chi_{V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{\ell-1}}\right) \\
&  =\chi_{\ind_{\left(  \alpha_{1},\alpha_{2},...,\alpha
_{\ell-1}\right)  }^{m}\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell-1}\right)  }.
\end{align*}


Now,
\begin{align*}
\chi_{1}\chi_{2}\cdots\chi_{\ell}  &  =\underbrace{\left(  \chi_{1}\chi
_{2}\cdots\chi_{\ell-1}\right)  }_{=\chi_{\ind_{\left(
\alpha_{1},\alpha_{2},...,\alpha_{\ell-1}\right)  }^{m}\left(  V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{\ell-1}\right)  }}\underbrace{\chi_{\ell}}%
_{=\chi_{V_{\ell}}}\\
&  =\chi_{\ind_{\left(  \alpha_{1},\alpha_{2},...,\alpha
_{\ell-1}\right)  }^{m}\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell-1}\right)  }\chi_{V_{\ell}}=\ind_{m,\alpha
_{\ell}}^{n}\left(  \chi_{\ind_{\left(  \alpha_{1},\alpha
_{2},...,\alpha_{\ell-1}\right)  }^{m}\left(  V_{1}\otimes V_{2}\otimes
\cdots\otimes V_{\ell-1}\right)  }\otimes\chi_{V_{\ell}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\ind_{\left(  \alpha_{1},\alpha_{2},...,\alpha
_{\ell-1}\right)  }^{m}\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell-1}\right)  \text{ is a } \CC G_{m}\text{-module, and}\\
V_{\ell}\text{ is a } \CC G_{\alpha_{\ell}}\text{-module}%
\end{array}
\right) \\
&  =\chi_{\ind_{m,\alpha_{\ell}}^{n}\left(
\ind_{\left(  \alpha_{1},\alpha_{2},...,\alpha_{\ell-1}\right)
}^{m}\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell-1}\right)  \otimes
V_{\ell}\right)  }\\
&  =\chi_{\ind_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1},\alpha_{\ell}}^{n}\left(  \ind_{\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell-1}\right)  }^{\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}}\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell-1}\right)  \otimes V_{\ell}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }m=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}\right) \\
&  =\chi_{\ind_{\alpha}^{n}\left(  V_{1}\otimes V_{2}%
\otimes\cdots\otimes V_{\ell}\right)  }%
\end{align*}
(since Exercise \ref{exe.towers.indres}(a) yields \newline
$\ind_{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1},\alpha_{\ell}}^{n}
\left( \ind_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right)  }^{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}}\left(
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell-1}\right)  \otimes V_{\ell
}\right)  \cong\ind_{\alpha}^{n}\left(  V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{\ell}\right)  $). Compared with%
\[
\ind_{\alpha}^{n}\left(  \underbrace{\chi_{1}}_{=\chi_{V_{1}}%
}\otimes\underbrace{\chi_{2}}_{=\chi_{V_{2}}}\otimes\cdots\otimes
\underbrace{\chi_{\ell}}_{=\chi_{V_{\ell}}}\right)
= \ind_{\alpha}^{n}\left(  \chi_{V_{1}}\otimes\chi_{V_{2}}
\otimes\cdots\otimes
\chi_{V_\ell}\right)  =\chi_{\ind_{\alpha}^{n}\left(  V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{\ell}\right)  },
\]
this yields $\chi_{1}\chi_{2}\cdots\chi_{\ell}=\ind_{\alpha}%
^{n}\left(  \chi_{1}\otimes\chi_{2}\otimes\cdots\otimes\chi_{\ell}\right)  $.
Thus, Exercise \ref{exe.towers.indres}(d) is solved for our $\ell$. This
completes the induction step, and so Exercise \ref{exe.towers.indres}(d) is solved.

(e) Let $n\in{ \NN }$, $\ell\in{ \NN }$ and $\chi\in R\left(
G_{n}\right)  $. We need to prove that $\Delta^{\left(  \ell-1\right)  }%
\chi=\sum\res_{\alpha}^{n}\chi$. Since this equality is
$\ZZ$-linear in $\chi$, we can WLOG assume that $\chi\in
\Irr \left(  G_{n}\right)  $ (since $\Irr \left(
G_{n}\right)  $ is a $\ZZ$-module basis of $R\left(  G_{n}\right)  $).
Assume this. Then, $\chi=\chi_{P}$ for some irreducible $ \CC G_{n}%
$-module $P$. Consider this $P$.

Similarly to how we showed (\ref{ind-res-adjointness}), we can prove that%
\begin{equation}
\Hom_{ \CC G_{n}}\left(  \ind_{\alpha}^{n}U,V\right)
 \cong\Hom_{ \CC G_{\alpha}}\left(  U,\res_{\alpha
}^{n}V\right)
\label{sol.towers.indres.e.recip}
\end{equation}
for every almost-composition $\alpha$ of $n$, every
$\CC G_{\alpha}$-module $U$ and every $\CC G_{n}$-module $V$. Thus,
\begin{equation}
\left(  \ind_{\alpha}^{n}\varphi,\psi\right)
_{R\left(  G_{n}\right)  }=\left(  \varphi,\res_{\alpha}^{n}\psi
\right)  _{R\left(  G_{\alpha}\right) }
\label{sol.towers.indres.e.recip.char}
\end{equation}
for every almost-composition $\alpha$ of $n$, every $\varphi\in R\left(
G_{\alpha}\right)  $ and every $\psi\in R\left(  G_{n}\right)  $%
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.towers.indres.e.recip.char}):}
Let $\alpha$ be an almost-composition of $n$. Let $\varphi\in R\left(
G_{\alpha}\right)  $ and $\psi\in R\left(  G_{n}\right)  $. We need to prove
the equality (\ref{sol.towers.indres.e.recip.char}). Since this equality is
$\ZZ$-linear in each of $\varphi$ and $\psi$, we can WLOG assume that
$\varphi\in\Irr \left(  G_{\alpha}\right)  $ and $\psi
\in\Irr \left(  G_{n}\right)  $ (since $\Irr %
\left(  G_{\alpha}\right)  $ and $\Irr \left(  G_{n}\right)  $
are $\ZZ$-module bases of $R\left(  G_{\alpha}\right)  $ and $R\left(
G_{n}\right)  $, respectively). Assume this. Thus, there exist an irreducible
$ \CC G_{\alpha}$-module $U$ and an irreducible $ \CC G_{n}$-module
$V$ such that $\varphi=\chi_{U}$ and $\psi=\chi_{V}$. Consider these $U$ and
$V$. We have
\begin{align*}
\left(  \ind_{\alpha}^{n}\underbrace{\varphi}%
_{=\chi_{U}},\underbrace{\psi}_{=\chi_{V}}\right)  _{R\left(  G_{n}\right)  }
& =\left(  \underbrace{\ind_{\alpha}^{n}\chi_{U}%
}_{=\chi_{\ind_{\alpha}^{n}U}},\chi_{V}\right)
_{R\left(  G_{n}\right)  }=\left(  \chi_{\ind_{\alpha
}^{n}U},\chi_{V}\right)  _{R\left(  G_{n}\right)  }\\
& =\dim_{ \CC }\Hom_{ \CC G_{n}}\left(
\ind_{\alpha}^{n}U,V\right)  =\dim_{ \CC %
}\Hom_{ \CC G_{\alpha}}\left(
U,\res_{\alpha}^{n}V\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.towers.indres.e.recip}%
)}\right)  .
\end{align*}
Compared with%
\[
\left(  \underbrace{\varphi}_{=\chi_{U}},\res_{\alpha
}^{n}\underbrace{\psi}_{=\chi_{V}}\right)  _{R\left(  G_{\alpha}\right)
}=\left(  \chi_{U},\underbrace{\res_{\alpha}^{n}%
\chi_{V}}_{=\chi_{\res_{\alpha}^{n}V}}\right)
_{R\left(  G_{\alpha}\right)  }
=\left(  \chi_{U},\chi_{\res_{\alpha}^{n}V}\right)
 _{R\left(  G_{\alpha}\right)  }
=\dim_{ \CC }\Hom_{ \CC G_{\alpha}}\left(
U,\res_{\alpha}^{n}V\right)  ,
\]
this yields $\left(  \ind_{\alpha}^{n}\varphi
,\psi\right)  _{R\left(  G_{n}\right)  }=\left(
 \varphi,\res_{\alpha}^{n}\psi\right)  _{R\left(  G_{\alpha}\right)  }$.
This proves (\ref{sol.towers.indres.e.recip.char}).}.

The bilinear form $\left(  \cdot,\cdot\right)  _{A}$ on $A$ induces a bilinear
form $\left(  \cdot,\cdot\right)  _{A^{\otimes\ell}}$ on $A^{\otimes\ell}$. It
is easy to see that the maps $m^{\left(  \ell-1\right)  }$ and $\Delta
^{\left(  \ell-1\right)  }$ are adjoint with respect to the forms $\left(
\cdot,\cdot\right)  _{A}$ and $\left(  \cdot,\cdot\right)  _{A^{\otimes\ell}}%
$.\ \ \ \ \footnote{Indeed, this follows from the inductive definitions of
$m^{\left(  \ell-1\right)  }$ and $\Delta^{\left(  \ell-1\right)  }$, if one
recalls that the maps $m$ and $\Delta$ are adjoint with respect to the forms
$\left(  \cdot,\cdot\right)  _{A}$ and $\left(  \cdot,\cdot\right)  _{A\otimes
A}$.} Hence, any $\varphi\in A$ and $\rho\in A^{\otimes\ell}$ satisfy%
\begin{equation}
\left(  \Delta^{\left(  \ell-1\right)  }\varphi,\rho\right)  _{A^{\otimes\ell
}}=\left(  \varphi,m^{\left(  \ell-1\right)  }\rho\right)  _{A}%
.\label{sol.towers.indres.e.adj}%
\end{equation}


Since $A=\bigoplus_{n\geq0}R\left(  G_{n}\right)  $, we have%
\begin{align}
A^{\otimes\ell} &  =\left(  \bigoplus_{n\geq0}R\left(  G_{n}\right)  \right)
^{\otimes\ell}=\bigoplus_{n_{1},n_{2},...,n_{\ell}\geq0}\underbrace{R\left(
G_{n_{1}}\right)  \otimes R\left(  G_{n_{2}}\right)  \otimes\cdots\otimes
R\left(  G_{n_{\ell}}\right)  }_{=R\left(  G_{n_{1}}\times G_{n_{2}}%
\times\cdots\times G_{n_{\ell}}\right)  }\nonumber\\
&  =\bigoplus_{n_{1},n_{2},...,n_{\ell}\geq0}R\left(  G_{n_{1}}\times
G_{n_{2}}\times\cdots\times G_{n_{\ell}}\right)  \nonumber\\
&  =\bigoplus_{\substack{\alpha=\left(  \alpha_{1},\alpha_{2},...,\alpha
_{\ell}\right)  \text{ is an}\\\text{almost-composition of length }\ell
}}R\left(  \underbrace{G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell}}}_{=G_{\alpha}}\right) \nonumber\\
&=  \bigoplus_{\substack{\alpha\text{ is
an almost-composition}\\\text{of length }\ell}}R\left(  G_{\alpha}\right)
.\label{sol.towers.indres.e.decomp}%
\end{align}
This direct sum decomposition of $A^{\otimes\ell}$ is orthogonal with respect
to the bilinear form $\left(  \cdot,\cdot\right)  _{A^{\otimes\ell}}$; that
is, if $\alpha$ and $\beta$ are two distinct almost-compositions of length
$\ell$, then
\begin{equation}
\left(  R\left(  G_{\alpha}\right)  ,R\left(  G_{\beta}\right)  \right)
_{A^{\otimes\ell}}=0.\label{sol.towers.indres.e.orthog}%
\end{equation}
\footnote{This follows from the definition of $\left(  \cdot,\cdot\right)
_{A^{\otimes\ell}}$ and the fact that the direct sum decomposition
$A=\bigoplus_{n\geq0}R\left(  G_{n}\right)  $ is orthogonal with respect to
the bilinear form $\left(  \cdot,\cdot\right)  _{A}$.} Moreover, for every
almost-composition $\alpha$ of length $\ell$, we have%
\begin{equation}
\left(  \varphi,\psi\right)  _{A^{\otimes\ell}}=\left(  \varphi,\psi\right)
_{R\left(  G_{\alpha}\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }\varphi\in
R\left(  G_{\alpha}\right)  \text{ and }\psi\in R\left(  G_{\alpha}\right)
.\label{sol.towers.indres.e.form=form}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.towers.indres.e.form=form}):} Let
$\alpha$ be an almost-composition of length $\ell$. Let $\varphi\in R\left(
G_{\alpha}\right)  $ and $\psi\in R\left(  G_{\alpha}\right)  $. We need to
prove the equality (\ref{sol.towers.indres.e.form=form}). Since this equality
is $\ZZ$-linear in each of $\varphi$ and $\psi$, we can WLOG assume
that $\varphi\in\Irr \left(  G_{\alpha}\right)  $ and $\psi
\in\Irr \left(  G_{\alpha}\right)  $ (since $\Irr %
\left(  G_{\alpha}\right)  $ is a $\ZZ$-module basis of $R\left(
G_{\alpha}\right)  $). Assume this. Thus, there exist an irreducible
$ \CC G_{\alpha}$-module $V$ and an irreducible $ \CC G_{\alpha}%
$-module $W$ such that $\varphi=\chi_{V}$ and $\psi=\chi_{W}$. Consider these
$V$ and $W$.
\par
Write the almost-composition $\alpha$ in the form $\left(  \alpha_{1}%
,\alpha_{2},...,\alpha_{\ell}\right)  $. Since $V$ is an irreducible
representation of $G_{\alpha}=G_{\alpha_{1}}\times G_{\alpha_{2}}\times
\cdots\times G_{\alpha_{\ell}}$, we can write $V$ in the form $V=V_{1}\otimes
V_{2}\otimes\cdots\otimes V_{\ell}$, where each $V_{i}$ is an irreducible
representation of $G_{\alpha_{i}}$. Similarly, we can write $W$ in the form
$W=W_{1}\otimes W_{2}\otimes\cdots\otimes W_{\ell}$, where each $W_{i}$ is an
irreducible representation of $G_{\alpha_{i}}$. Consider these $V_{i}$ and
$W_{i}$.
\par
Now, there exists a $ \CC $-vector space isomorphism%
\begin{align*}
& \Hom_{{ \CC }G_{\alpha_{1}}}\left(  V_{1},W_{1}\right)
\otimes\Hom_{{ \CC }G_{\alpha_{2}}}\left(  V_{2}%
,W_{2}\right)  \otimes\cdots\otimes\Hom_{ \CC %
G_{\alpha_{\ell}}}\left(  V_{\ell},W_{\ell}\right)  \\
& \rightarrow\Hom_{{ \CC }\left[  G_{\alpha_{1}}\times
G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell}}\right]  }\left(
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell},W_{1}\otimes W_{2}%
\otimes\cdots\otimes W_{\ell}\right)  .
\end{align*}
(In fact, when $\ell=2$, the existence of such an isomorphism follows from
Exercise \ref{exe.HomxHom}(a); otherwise it follows by induction over $\ell$
using Exercise \ref{exe.HomxHom}(a).) The existence of this isomorphism yields%
\begin{align*}
& \dim_{ \CC }\left(  \Hom_{{ \CC }\left[
G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell}%
}\right]  }\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell}%
,W_{1}\otimes W_{2}\otimes\cdots\otimes W_{\ell}\right)  \right)  \\
& =\dim_{ \CC }\left(  \Hom_{{ \CC }G_{\alpha_{1}}%
}\left(  V_{1},W_{1}\right)  \otimes\Hom_{{ \CC }%
G_{\alpha_{2}}}\left(  V_{2},W_{2}\right)  \otimes\cdots\otimes
\Hom_{ \CC G_{\alpha_{\ell}}}\left(  V_{\ell
},W_{\ell}\right)  \right)  \\
& =\prod_{i=1}^{\ell}\underbrace{\dim_{ \CC }
\Hom_{{ \CC }G_{\alpha_{i}}}\left(  V_{i},W_{i}\right)  }_{=\left(
\chi_{V_{i}},\chi_{W_{i}}\right)  _{R\left(  G_{\alpha_{i}}\right)  }=\left(
\chi_{V_{i}},\chi_{W_{i}}\right)  _{A}}=\prod_{i=1}^{\ell}\left(  \chi_{V_{i}%
},\chi_{W_{i}}\right)  _{A}\\
& =\left(  \underbrace{\chi_{V_{1}}\otimes\chi_{V_{2}}\otimes\cdots\otimes
\chi_{V_{\ell}}}_{\substack{=\chi_{V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell}}=\chi_{V}\\\text{(since }V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell}=V\text{)}}},\underbrace{\chi_{W_{1}}\otimes\chi_{W_{2}}\otimes
\cdots\otimes\chi_{W_{\ell}}}_{\substack{=\chi_{W_{1}\otimes W_{2}%
\otimes\cdots\otimes W_{\ell}}=\chi_{W}\\\text{(since }W_{1}\otimes
W_{2}\otimes\cdots\otimes W_{\ell}=W\text{)}}}\right)  _{A^{\otimes\ell}}\\
& =\left(  \underbrace{\chi_{V}}_{=\varphi},\underbrace{\chi_{W}}_{=\psi
}\right)  _{A^{\otimes\ell}}=\left(  \varphi,\psi\right)  _{A^{\otimes\ell}}.
\end{align*}
Thus,%
\begin{align*}
& \left(  \varphi,\psi\right)  _{A^{\otimes\ell}}\\
& =\dim_{ \CC }\left(  \Hom_{{ \CC }\left[
G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times G_{\alpha_{\ell}%
}\right]  }\left(  V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell}%
,W_{1}\otimes W_{2}\otimes\cdots\otimes W_{\ell}\right)  \right)  \\
& =\dim_{ \CC }\left(  \Hom_{{ \CC }G_{\alpha}}\left(
V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell},W_{1}\otimes W_{2}%
\otimes\cdots\otimes W_{\ell}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }G_{\alpha_{1}}\times G_{\alpha_{2}}\times\cdots\times
G_{\alpha_{\ell}}=G_{\alpha}\right)  \\
& =\left(  \underbrace{\chi_{V_{1}\otimes V_{2}\otimes\cdots\otimes V_{\ell}}%
}_{\substack{=\chi_{V}\\\text{(since }V_{1}\otimes V_{2}\otimes\cdots\otimes
V_{\ell}=V\text{)}}},\underbrace{\chi_{W_{1}\otimes W_{2}\otimes\cdots\otimes
W_{\ell}}}_{\substack{=\chi_{W}\\\text{(since }W_{1}\otimes W_{2}\otimes
\cdots\otimes W_{\ell}=W\text{)}}}\right)  _{R\left(  G_{\alpha}\right)
}=\left(  \underbrace{\chi_{V}}_{=\varphi},\underbrace{\chi_{W}}_{=\psi
}\right)  _{R\left(  G_{\alpha}\right)  }=\left(  \varphi,\psi\right)
_{R\left(  G_{\alpha}\right)  }.
\end{align*}
This proves (\ref{sol.towers.indres.e.form=form}).}

We need to show that $\Delta^{\left(  \ell-1\right)  }\chi=\sum
\res_{\alpha}^{n}\chi$ (where the sum ranges over all
almost-compositions $\alpha$ of $n$ having length $\ell$). In order to do so,
it is clearly enough to prove that%
\begin{equation}
\left(  \Delta^{\left(  \ell-1\right)  }\chi,\rho\right)  _{A^{\otimes\ell}%
}=\left(  \sum\res_{\alpha}^{n}\chi,\rho\right)  _{A^{\otimes
\ell}}\ \ \ \ \ \ \ \ \ \ \text{for every }\rho\in A^{\otimes\ell
}\label{sol.towers.indres.e.goal}%
\end{equation}
(since the bilinear form $\left(  \cdot,\cdot\right)  _{A^{\otimes\ell}}$ is
nondegenerate). So, let $\rho\in A^{\otimes\ell}$. It remains to prove
(\ref{sol.towers.indres.e.goal}).

The equality (\ref{sol.towers.indres.e.goal}) is $\ZZ$-linear in $\rho
$. Since $\bigsqcup_{\substack{\alpha\text{ is an almost-composition}%
\\\text{of length }\ell}}\Irr \left(  G_{\alpha}\right)  $ is a
$\ZZ$-module basis of $A^{\otimes\ell}$ (this follows from
(\ref{sol.towers.indres.e.decomp}) and the fact that each $R\left(  G_{\alpha
}\right)  $ has $\ZZ$-module basis $\Irr \left(
G_{\alpha}\right)  $), we can therefore WLOG assume that $\rho\in
\bigsqcup_{\substack{\alpha\text{ is an almost-composition}\\\text{of length
}\ell}}\Irr \left(  G_{\alpha}\right)  $. Assume this. Then,
there exists an almost-composition $\beta$ of length $\ell$ such that $\rho
\in\Irr \left(  G_{\beta}\right)  $. Consider this $\beta$, and
notice that $\rho\in\Irr \left(  G_{\beta}\right)  \subset
R\left(  G_{\beta}\right)  $.

For every almost-composition $\alpha$ of $n$ having length $\ell$ satisfying
$\alpha\neq\beta$, we have
\begin{equation}
\left(  \res_{\alpha}^{n}\chi,\rho\right)  _{A^{\otimes\ell}%
}=0
\label{sol.towers.indres.e.3}
\end{equation}
\footnote{since
\[
\left(  \underbrace{\res_{\alpha}^{n}\chi}_{\in R\left(
G_{\alpha}\right)  },\underbrace{\rho}_{\in\Irr \left(
G_{\beta}\right)  \subset R\left(  G_{\beta}\right)  }\right)  _{A^{\otimes
\ell}}\in\left(  R\left(  G_{\alpha}\right)  ,R\left(  G_{\beta}\right)
\right)  _{A^{\otimes\ell}}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.towers.indres.e.orthog})}\right)
\]
}. Now,
\begin{align}
\left(  \sum\res_{\alpha}^{n}\chi,\rho\right)  _{A^{\otimes\ell
}}  & =\sum\left(  \res_{\alpha}^{n}\chi,\rho\right)
_{A^{\otimes\ell}}=\left(  \res_{\beta}^{n}\chi,\rho\right)
_{A^{\otimes\ell}}+\sum_{\alpha\neq\beta}\underbrace{\left(
\res_{\alpha}^{n}\chi,\rho\right)  _{A^{\otimes\ell}}%
}_{\substack{=0\\\text{(by (\ref{sol.towers.indres.e.3}))}}}
=\left(  \res_{\beta}^{n}\chi,\rho\right)  _{A^{\otimes\ell}}
\nonumber\\
&= \left(  \res_{\beta}^{n}\chi,\rho\right)  _{R\left(
G_{\beta}\right)  }
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.towers.indres.e.form=form}),
applied to }\varphi=\res_{\beta}^{n}\chi\text{, }%
\psi=\rho\text{ and }\alpha=\beta\right)  \nonumber\\
& =\left(  \rho,\res_{\beta}^{n}\chi\right)
_{R\left(  G_{\beta}\right)  }\nonumber\\
& =\left(  \ind_{\beta}^{n}\rho,\chi\right)
_{R\left(  G_{n}\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since (\ref{sol.towers.indres.e.recip.char}) (applied }\varphi
=\rho\text{, }\psi=\chi\text{ and }\alpha=\beta\text{)}\\
\text{yields }\left(  \ind_{\beta}^{n}\rho
,\chi\right)  _{R\left(  G_{n}\right)  }=\left(  \rho,
\res_{\beta}^{n}\chi\right)  _{R\left(  G_{\beta}\right)  }%
\end{array}
\right)  .\label{sol.towers.indres.e.5}%
\end{align}
But let us write the almost-composition $\beta$ as $\left(  \beta_{1}%
,\beta_{2},...,\beta_{\ell}\right)  $. Then, $\rho\in\Irr %
\left(  G_{\beta}\right)  $ is an irreducible character of $G_{\beta}%
=G_{\beta_{1}}\times G_{\beta_{2}}\times\cdots\times G_{\beta_{\ell}}$, and
thus has the form $\rho=\rho_{1}\otimes\rho_{2}\otimes\cdots\otimes\rho_{\ell
}$, where each $\rho_{i}$ is an irreducible character of $G_{\beta_{i}}$.
Consider these $\rho_{i}$. We have
\begin{align*}
m^{\left(  \ell-1\right)  }\underbrace{\rho}_{=\rho_{1}\otimes\rho_{2}%
\otimes\cdots\otimes\rho_{\ell}}  & =m^{\left(  \ell-1\right)  }\left(
\rho_{1}\otimes\rho_{2}\otimes\cdots\otimes\rho_{\ell}\right)  =\rho_{1}%
\rho_{2}\cdots\rho_{\ell}=\ind_{\beta}^{n}\left(
\underbrace{\rho_{1}\otimes\rho_{2}\otimes\cdots\otimes\rho_{\ell}}_{=\rho
}\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.towers.indres}(d),
applied to }\beta\text{ and }\rho_{i}\text{ instead of }\alpha\text{ and }%
\chi_{i}\right)  \\
& =\ind_{\beta}^{n}\rho.
\end{align*}


Now,
\begin{align*}
\left(  \Delta^{\left(  \ell-1\right)  }\chi,\rho\right)  _{A^{\otimes\ell}}
& =\left(  \chi,\underbrace{m^{\left(  \ell-1\right)  }\rho}%
_{=\ind_{\beta}^{n}\rho}\right)  _{A}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.towers.indres.e.adj}), applied
to }\varphi=\chi\right)  \\
& =\left(  \chi,\ind_{\beta}^{n}\rho\right)
_{A}=\left(  \chi,\ind_{\beta}^{n}\rho\right)
_{R\left(  G_{n}\right)  }=\left(  \ind_{\beta}%
^{n}\rho,\chi\right)  _{R\left(  G_{n}\right)  }\\
& =\left(  \sum\res_{\alpha}^{n}\chi,\rho\right)  _{A^{\otimes
\ell}}\ \ \ \ \ \ \ \ \ \ \left( 
 \text{by (\ref{sol.towers.indres.e.5})}\right)  .
\end{align*}
Thus, (\ref{sol.towers.indres.e.goal}) is proven. This completes our solution
of Exercise \ref{exe.towers.indres}(e).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.frob-char.plambda}}

\textit{Solution to Exercise \ref{exe.frob-char.plambda}.} Define the Iverson
bracket notation as in Exercise \ref{exe.frob-char.plambda}(a).

(a) Let $G$ be a finite group. For every subset $P$ of $G$, the definition of
$\underline{1}_{P}$ yields
\begin{equation}
\underline{1}_{P}\left(  g\right)  =\left[  g\in P\right]
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G.
\label{sol.frob-char.plambda.a.iverson}%
\end{equation}
Also, for every subset $P$ of $G$, we have%
\begin{equation}
\left\vert P\right\vert =\sum_{k\in G}\left[  k\in P\right]  .
\label{sol.frob-char.plambda.a.iversonsum}%
\end{equation}


Let $h\in G$ and $g\in G$. Then, (\ref{sol.frob-char.plambda.a.iversonsum})
(applied to $P=Z_{G}\left(  h\right)  $) yields%
\[
\left\vert Z_{G}\left(  h\right)  \right\vert =\sum_{k\in G}\left[
\underbrace{k\in Z_{G}\left(  h\right)  }_{\substack{\text{this is equivalent
to }khk^{-1}=h\\\text{(by the definition of }Z_{G}\left(  h\right)  \text{)}%
}}\right]  =\sum_{k\in G}\left[  khk^{-1}=h\right]  .
\]


Applying both sides of the identity $\alpha_{G,h}=\left\vert Z_{G}\left(
h\right)  \right\vert \underline{1}_{\operatorname{Conj}_{G}\left(  h\right)
}$ to $g$, we obtain%
\begin{equation}
\alpha_{G,h}\left(  g\right)  =\underbrace{\left\vert Z_{G}\left(  h\right)
\right\vert }_{=\sum_{k\in G}\left[  khk^{-1}=h\right]  }%
\underbrace{\underline{1}_{\operatorname{Conj}_{G}\left(  h\right)  }\left(
g\right)  }_{\substack{=\left[  g\in\operatorname{Conj}_{G}\left(  h\right)
\right]  \\\text{(by (\ref{sol.frob-char.plambda.a.iverson}), applied}%
\\\text{to }P=\operatorname{Conj}_{G}\left(  h\right)  \text{)}}}=\left(
\sum_{k\in G}\left[  khk^{-1}=h\right]  \right)  \left[  g\in
\operatorname{Conj}_{G}\left(  h\right)  \right]  .
\label{sol.frob-char.plambda.a.1}%
\end{equation}


We need to prove that $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[
khk^{-1}=g\right]  $. We must be in one of the following two cases:

\textit{Case 1:} We have $g\in\operatorname{Conj}_{G}\left(  h\right)  $.

\textit{Case 2:} We don't have $g\in\operatorname{Conj}_{G}\left(  h\right)  $.

Let us first consider Case 1. In this case, we have $g\in\operatorname{Conj}%
_{G}\left(  h\right)  $. Thus, there exists a $p\in G$ satisfying $g=php^{-1}%
$. Consider this $p$. Since $g=php^{-1}$, we have $h=p^{-1}gp$. Hence,
(\ref{sol.frob-char.plambda.a.1}) becomes%
\begin{align*}
\alpha_{G,h}\left(  g\right)   &  =\left(  \sum_{k\in G}\left[  khk^{-1}%
=h\right]  \right)  \underbrace{\left[  g\in\operatorname{Conj}_{G}\left(
h\right)  \right]  }_{\substack{=1\\\text{(since }g\in\operatorname{Conj}%
_{G}\left(  h\right)  \text{)}}}=\sum_{k\in G}\left[  k\underbrace{h}%
_{=p^{-1}gp}k^{-1}=h\right] \\
&  =\sum_{k\in G}\left[  \underbrace{kp^{-1}}_{=\left(  pk^{-1}\right)  ^{-1}%
}gpk^{-1}=h\right]  =\sum_{k\in G}\left[  \left(  pk^{-1}\right)
^{-1}g\left(  kp^{-1}\right)  ^{-1}=h\right] \\
&  =\sum_{k\in G}\left[  \underbrace{k^{-1}gk=h}_{\substack{\text{this is
equivalent to}\\khk^{-1}=g}}\right]  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }k\text{ for }pk^{-1}\text{ in the sum}\\
\text{(since the map }G\rightarrow G,\ k\mapsto pk^{-1}\text{ is a bijection)}%
\end{array}
\right) \\
&  =\sum_{k\in G}\left[  khk^{-1}=g\right]  .
\end{align*}
Thus, $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]
$ is proven in Case 1.

Let us now consider Case 2. In this case, we don't have $g\in
\operatorname{Conj}_{G}\left(  h\right)  $. In other words, $g$ is not in the
conjugacy class of $h$. In other words, $g$ is not conjugate to $h$. In other
words, there exists no $k\in G$ satisfying $khk^{-1}=g$. In other words, for
every $k\in G$, we do not have $khk^{-1}=g$. Hence, for every $k\in G$, we
have $\left[  khk^{-1}=g\right]  =0$. Thus, $\sum_{k\in G}\underbrace{\left[
khk^{-1}=g\right]  }_{=0}=\sum_{k\in G}0=0$. Comparing this with%
\[
\alpha_{G,h}\left(  g\right)  =\left(  \sum_{k\in G}\left[  khk^{-1}=h\right]
\right)  \underbrace{\left[  g\in\operatorname{Conj}_{G}\left(  h\right)
\right]  }_{\substack{=0\\\text{(since we don't have }g\in\operatorname{Conj}%
_{G}\left(  h\right)  \text{)}}}=0,
\]
we obtain $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}%
=g\right]  $. Thus, $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[
khk^{-1}=g\right]  $ is proven in Case 2.

Thus, $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]
$ is proven in both Cases 1 and 2. Hence, Exercise \ref{exe.frob-char.plambda}%
(a) is solved.

(b) Let $H$ be a subgroup of a finite group $G$. Let $h\in H$. Let $g\in G$.
Then, the definition of $\operatorname{Ind}_{H}^{G}\alpha_{H,h}$ given in
Exercise \ref{exe.induction.characters} yields%
\begin{align}
\left(  \operatorname{Ind}_{H}^{G}\alpha_{H,h}\right)  \left(  g\right)   &
=\frac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G:\\kgk^{-1}\in
H}}\alpha_{H,h}\left(  kgk^{-1}\right)  =\frac{1}{\left\vert H\right\vert
}\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\underbrace{\alpha_{H,h}\left(
pgp^{-1}\right)  }_{\substack{=\sum_{k\in H}\left[  khk^{-1}=pgp^{-1}\right]
\\\text{(by Exercise \ref{exe.frob-char.plambda}(a), applied}\\\text{to
}H\text{ and }pgp^{-1}\text{ instead of }G\text{ and }g\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}k\text{ as }p\right) \nonumber\\
&  =\frac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{p\in
G:\\pgp^{-1}\in H}}\sum_{k\in H}}_{=\sum_{k\in H}\sum\limits_{\substack{p\in
G:\\pgp^{-1}\in H}}}\left[  khk^{-1}=pgp^{-1}\right]  =\frac{1}{\left\vert
H\right\vert }\sum_{k\in H}\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\left[
khk^{-1}=pgp^{-1}\right]  . \label{sol.frob-char.plambda.b.1}%
\end{align}
But every $k\in H$ satisfies%
\begin{equation}
\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]
=\sum_{p\in G}\left[  khk^{-1}=pgp^{-1}\right]
\label{sol.frob-char.plambda.b.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.b.2}):} Let $k\in H$.
Then,%
\begin{align*}
\sum_{p\in G}\left[  khk^{-1}=pgp^{-1}\right]   &  =\sum\limits_{\substack{p\in
G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]  +\sum\limits_{\substack{p\in
G:\\pgp^{-1}\notin H}}\underbrace{\left[  khk^{-1}=pgp^{-1}\right]
}_{\substack{=0\\\text{(since we don't have }khk^{-1}=pgp^{-1}\\\text{(since
}khk^{-1}\in H\text{ (since }k\in H\text{ and }h\in H\text{) and }%
pgp^{-1}\notin H\text{))}}}\\
&  =\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]
+\underbrace{\sum\limits_{\substack{p\in G:\\pgp^{-1}\notin H}}0}_{=0}=\sum\limits
_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]  .
\end{align*}
This proves (\ref{sol.frob-char.plambda.b.2}).}. Now,
(\ref{sol.frob-char.plambda.b.1}) becomes%
\begin{align*}
\left(  \operatorname{Ind}_{H}^{G}\alpha_{H,h}\right)  \left(  g\right)   &
=\frac{1}{\left\vert H\right\vert }\sum_{k\in H}\underbrace{\sum\limits
_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]
}_{\substack{=\sum_{p\in G}\left[  khk^{-1}=pgp^{-1}\right]  \\\text{(by
(\ref{sol.frob-char.plambda.b.2}))}}}=\frac{1}{\left\vert H\right\vert }%
\sum_{k\in H}\sum_{p\in G}\left[  \underbrace{khk^{-1}=pgp^{-1}}%
_{\substack{\text{this is equivalent to}\\p^{-1}khk^{-1}p=g}}\right] \\
&  =\frac{1}{\left\vert H\right\vert }\sum_{k\in H}\sum_{p\in G}\left[
p^{-1}kh\underbrace{k^{-1}p}_{=\left(  p^{-1}k\right)  ^{-1}}=g\right]
=\frac{1}{\left\vert H\right\vert }\sum_{k\in H}\sum_{p\in G}\left[
p^{-1}kh\left(  p^{-1}k\right)  ^{-1}=g\right] \\
&  =\frac{1}{\left\vert H\right\vert }\underbrace{\sum_{k\in H}\sum_{p\in
G}\left[  php^{-1}=g\right]  }_{=\left\vert H\right\vert \sum_{p\in G}\left[
php^{-1}=g\right]  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }p\text{ for }p^{-1}k\text{ in the inner sum}\\
\text{(since the map }G\rightarrow G,\ p\mapsto p^{-1}k\text{ is a bijection)}%
\end{array}
\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\left\vert H\right\vert \sum_{p\in
G}\left[  php^{-1}=g\right]  =\sum_{p\in G}\left[  php^{-1}=g\right]
=\sum_{k\in G}\left[  khk^{-1}=g\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k\text{ for
}p\text{ in the sum}\right) \\
&  =\alpha_{G,h}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.frob-char.plambda}(a)}\right)  .
\end{align*}


Let us now forget that we fixed $g$. We thus have proven that $\left(
\operatorname{Ind}_{H}^{G}\alpha_{H,h}\right)  \left(  g\right)  =\alpha
_{G,h}\left(  g\right)  $ for every $g\in G$. Hence, $\operatorname{Ind}%
_{H}^{G}\alpha_{H,h}=\alpha_{G,h}$. This solves Exercise
\ref{exe.frob-char.plambda}(b).

(c) Let $G_{1}$ and $G_{2}$ be two finite groups. Let $h_{1}\in G_{1}$ and
$h_{2}\in G_{2}$. Let $\mathbf{I}$ denote the canonical isomorphism
$R_\CC \left(  G_{1}\right)  \otimes R_{\mathbb{C}}\left(
G_{2}\right)  \rightarrow R_{\mathbb{C}}\left(  G_{1}\times G_{2}\right)  $.
We need to prove that%
\begin{equation}
\mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}}\right)
=\alpha_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }.
\label{sol.frob-char.plambda.c.goal}%
\end{equation}


Let $g\in G_{1}\times G_{2}$. Let us write $g$ in the form $g=\left(
g_{1},g_{2}\right)  $. Then, Exercise \ref{exe.frob-char.plambda}(a) (applied
to $G_{1}$, $h_{1}$ and $g_{1}$ instead of $G$, $h$ and $g$) yields%
\[
\alpha_{G_{1},h_{1}}\left(  g_{1}\right)  =\sum_{k\in G_{1}}\left[
kh_{1}k^{-1}=g_{1}\right]  =\sum_{k_{1}\in G_{1}}\left[  k_{1}h_{1}k_{1}%
^{-1}=g_{1}\right]
\]
(here, we have renamed the summation index $k$ as $k_{1}$). Similarly,%
\[
\alpha_{G_{2},h_{2}}\left(  g_{2}\right)  =\sum_{k_{2}\in G_{2}}\left[
k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  .
\]
Now,%
\begin{align*}
&  \left(  \mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}%
}\right)  \right)  \left(  \underbrace{g}_{=\left(  g_{1},g_{2}\right)
}\right) \\
&  =\left(  \mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}%
}\right)  \right)  \left(  \left(  g_{1},g_{2}\right)  \right)
=\underbrace{\alpha_{G_{1},h_{1}}\left(  g_{1}\right)  }_{=\sum_{k_{1}\in
G_{1}}\left[  k_{1}h_{1}k_{1}^{-1}=g_{1}\right]  }\underbrace{\alpha
_{G_{2},h_{2}}\left(  g_{2}\right)  }_{=\sum_{k_{2}\in G_{2}}\left[
k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathbf{I}\right) \\
&  =\left(  \sum_{k_{1}\in G_{1}}\left[  k_{1}h_{1}k_{1}^{-1}=g_{1}\right]
\right)  \left(  \sum_{k_{2}\in G_{2}}\left[  k_{2}h_{2}k_{2}^{-2}%
=g_{2}\right]  \right)  =\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times
G_{2}}\underbrace{\left[  k_{1}h_{1}k_{1}^{-1}=g_{1}\right]  \left[
k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  }_{=\left[  k_{1}h_{1}k_{1}^{-1}%
=g_{1}\text{ and }k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  }\\
&  =\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times G_{2}}\left[
\underbrace{k_{1}h_{1}k_{1}^{-1}=g_{1}\text{ and }k_{2}h_{2}k_{2}^{-2}=g_{2}%
}_{\substack{\text{this is equivalent to}\\\left(  k_{1}h_{1}k_{1}^{-1}%
,k_{2}h_{2}k_{2}^{-1}\right)  =\left(  g_{1},g_{2}\right)  }}\right]
=\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times G_{2}}\left[
\underbrace{\left(  k_{1}h_{1}k_{1}^{-1},k_{2}h_{2}k_{2}^{-1}\right)
}_{=\left(  k_{1},k_{2}\right)  \left(  h_{1},h_{2}\right)  \left(
k_{1},k_{2}\right)  ^{-1}}=\underbrace{\left(  g_{1},g_{2}\right)  }%
_{=g}\right] \\
&  =\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times G_{2}}\left[  \left(
k_{1},k_{2}\right)  \left(  h_{1},h_{2}\right)  \left(  k_{1},k_{2}\right)
^{-1}=g\right]  =\sum_{k\in G_{1}\times G_{2}}\left[  k\left(  h_{1}%
,h_{2}\right)  k^{-1}=g\right]
\end{align*}
(here, we have renamed the summation index $\left(  k_{1},k_{2}\right)  $ as
$k$). Compared with%
\begin{align*}
&  \alpha_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }\left(  g\right)
=\sum_{k\in G_{1}\times G_{2}}\left[  k\left(  h_{1},h_{2}\right)
k^{-1}=g\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.frob-char.plambda}%
(a), applied to }G_{1}\times G_{2}\text{ and }\left(  h_{1},h_{2}\right)
\text{ instead of }G\text{ and }h\right)  ,
\end{align*}
this yields $\left(  \mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes
\alpha_{G_{2},h_{2}}\right)  \right)  \left(  g\right)  =\alpha_{G_{1}\times
G_{2},\left(  h_{1},h_{2}\right)  }\left(  g\right)  $.

Now, let us forget that we fixed $g$. We thus have proven that $\left(
\mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}}\right)
\right)  \left(  g\right)  =\alpha_{G_{1}\times G_{2},\left(  h_{1}%
,h_{2}\right)  }\left(  g\right)  $ for every $g\in G_{1}\times G_{2}$. In
other words, $\mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2}%
,h_{2}}\right)  =\alpha_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }$.
This proves (\ref{sol.frob-char.plambda.c.goal}). Exercise
\ref{exe.frob-char.plambda}(c) is thus solved.

(d) For every partition $\lambda$, let us define $\widetilde{z}_{\lambda}$ as
the size of the centralizer of a permutation in $\Symm_{\left\vert
\lambda\right\vert }$ having cycle type $\lambda$. Note that this does not
depend on the choice of said permutation, since all permutations in
$\Symm_{\left\vert \lambda\right\vert }$ having cycle type $\lambda$
are mutually conjugate (and thus their centralizers are of the same size). It
is well-known that $\widetilde{z}_{\lambda}=z_{\lambda}$ (see
Remark~\ref{rmk.zlambda.centralizer}), but we shall avoid using this fact, as
we can obtain an alternative proof of it from the following argument.

We shall use the same notations as in the proof of Theorem
\ref{symmetric-group-Frobenius-map-theorem}. In particular, $A_{\mathbb{C}%
}=\bigoplus_{n\geq0}R_\CC \left(  \Symm_n\right)  $; this
$\CC$-vector space $A_{\mathbb{C}}$ becomes a $\CC$-algebra as
explained in the proof of Theorem \ref{symmetric-group-Frobenius-map-theorem}.
Its multiplication is given by $\ind_{i,j}^{i+j}$;
more precisely, every $n\in \NN$ and $m\in \NN$, and every
$\beta\in R_\CC \left(  \Symm_n\right)  $ and $\gamma\in
R_\CC \left(  \Symm_m\right)  $ satisfy%
\begin{equation}
\beta\gamma=\underbrace{\ind_{n,m}^{n+m}%
}_{=\operatorname*{Ind}\nolimits_{\Symm_n\times\Symm_m%
}^{\Symm_{n+m}}}\left(  \beta\otimes\gamma\right)  =\operatorname*{Ind}%
\nolimits_{\Symm_n\times\Symm_m}^{\Symm_{n+m}}\left(
\beta\otimes\gamma\right)  . \label{sol.frob-char.plambda.d.prod}%
\end{equation}


We define a $\CC$-linear map $\Phi:\Lambda_{\mathbb{C}}\rightarrow
A_{\mathbb{C}}$ by setting%
\[
\Phi\left(  p_{\lambda}\right)  =\widetilde{z}_{\lambda}\underline{1}%
_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in \Par .
\]
(This is well-defined, since $\left(  p_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\CC$-vector space
$\Lambda_{\mathbb{C}}$.) We notice that if $\lambda$ is a partition of a
nonnegative integer $n$, and if $g\in\Symm_n$ is a permutation having
cycle type $\lambda$, then%
\begin{equation}
\Phi\left(  p_{\lambda}\right)  =\alpha_{\Symm_n,g}.
\label{sol.frob-char.plambda.d.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.d.1}):} Let $\lambda$
be a partition of a nonnegative integer $n$. Let $g\in\Symm_n$ be a
permutation having cycle type $\lambda$. Then, $g$ is a permutation in
$\Symm_{\left\vert \lambda\right\vert }$ having cycle type $\lambda$.
Thus, $\widetilde{z}_{\lambda}$ is the size of the centralizer of $g$ (by the
definition of $\widetilde{z}_{\lambda}$). In other words, $\widetilde{z}%
_{\lambda}=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert $.
\par
But any two permutations in $\Symm_n$ having the same cycle type are
mutually conjugate. Hence, if $h$ is any permutation in $\Symm_n$
having cycle type $\lambda$, then $h$ and $g$ are conjugate (since $h$ and $g$
are permutations in $\Symm_n$ having the same cycle type).
Conversely, if $h$ is a permutation in $\Symm_n$ such that $h$ and
$g$ are conjugate, then $h$ has cycle type $\lambda$ (because $h$ and $g$ are
conjugate permutations and thus have the same cycle type, but the cycle type
of $g$ is $\lambda$). Combining these two statements, we conclude that if $h$
is a permutation in $\Symm_n$, then $h$ has cycle type $\lambda$ if
and only if $h$ and $g$ are conjugate. Hence,%
\begin{align*}
\left\{  h\in\Symm_n\ \mid\ \underbrace{h\text{ has cycle type
}\lambda}_{\substack{\text{this is equivalent to}\\\left(  h\text{ and
}g\text{ are conjugate}\right)  }}\right\}   &  =\left\{  h\in\Symm_n
\ \mid\ \underbrace{h\text{ and }g\text{ are conjugate}}_{\text{this is
equivalent to }h\in\operatorname*{Conj}\nolimits_{\Symm_n}\left(
g\right)  }\right\}  =\left\{  h\in\Symm_n\ \mid\ h\in
\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  \right\} \\
&  =\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  .
\end{align*}
\par
On the other hand, $\underline{1}_{\lambda}$ is defined as the characteristic
function for the $\Symm_n$-conjugacy class of permutations of cycle
type $\lambda$. In other words, $\underline{1}_{\lambda}$ is the indicator
function of the subset $\left\{  h\in\Symm_n\ \mid\ h\text{ has cycle
type }\lambda\right\}  $ of $\Symm_n$. In other words, $\underline{1}%
_{\lambda}$ is the indicator function of the subset $\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  $ (since $\left\{
h\in\Symm_n\ \mid\ h\text{ has cycle type }\lambda\right\}
=\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  $). In
other words, $\underline{1}_{\lambda}=\underline{1}_{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }$.
\par
Now, $\Phi\left(  p_{\lambda}\right)  =\underbrace{\widetilde{z}_{\lambda}%
}_{=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert
}\underbrace{\underline{1}_{\lambda}}_{=\underline{1}_{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }}
=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert
\underline{1}_{\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  }$.
Compared with $\alpha_{\Symm_n,g}=\left\vert Z_{\Symm_n}\left(  g\right)
\right\vert \underline{1}_{\operatorname*{Conj}\nolimits_{\Symm_n%
}\left(  g\right)  }$ (by the definition of $\alpha_{\Symm_n,g}$),
this yields $\Phi\left(  p_{\lambda}\right)  =\alpha_{\Symm_n,g}$.
This proves (\ref{sol.frob-char.plambda.d.1}).} Also, if $\lambda$ is a
partition of a nonnegative integer $n$, then%
\begin{equation}
\left\vert \left\{  h\in\Symm_n\ \mid\ h\text{ has cycle type
}\lambda\right\}  \right\vert =n!/\widetilde{z}_{\lambda}.
\label{sol.frob-char.plambda.d.1b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.d.1b}):} Let $\lambda$
be a partition of a nonnegative integer $n$. Fix a permutation $g\in
\Symm_n$ having cycle type $\lambda$. (Such a $g$ clearly exists.) In
the proof of (\ref{sol.frob-char.plambda.d.1}), we have seen that
$\widetilde{z}_{\lambda}=\left\vert Z_{\Symm_n}\left(  g\right)
\right\vert $ and that $\left\{  h\in\Symm_n\ \mid\ h\text{ has cycle
type }\lambda\right\}  =\operatorname*{Conj}\nolimits_{\Symm_n%
}\left(  g\right)  $.
\par
Now, it is well-known that for every finite group $G$ and every element $f\in
G$, we have $\left\vert G/Z_{G}\left(  f\right)  \right\vert =\left\vert
\operatorname*{Conj}\nolimits_{G}\left(  f\right)  \right\vert $ (in fact,
there is a canonical bijection from the $G$-set $G/Z_{G}\left(  f\right)  $ to
$\operatorname*{Conj}\nolimits_{G}\left(  f\right)  $, which sends the
equivalence class $\left[  \gamma\right]  \in G/Z_{G}\left(  f\right)  $ of
every $\gamma\in G$ to $\gamma f\gamma^{-1}\in\operatorname*{Conj}%
\nolimits_{G}\left(  f\right)  $). Applying this to $G=\Symm_n$ and
$f=g$, we obtain $\left\vert \Symm_n/Z_{\Symm_n}\left(
g\right)  \right\vert =
\left\vert \operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)
\right\vert $. Thus, $\left\vert \operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  \right\vert =\left\vert
\Symm_n/Z_{\Symm_n}\left(  g\right)  \right\vert
=\underbrace{\left\vert \Symm_n\right\vert }_{=n!}%
/\underbrace{\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert
}_{=\widetilde{z}_{\lambda}}=n!/\widetilde{z}_{\lambda}$, so that%
\[
n!/\widetilde{z}_{\lambda}=\left\vert \underbrace{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }_{=\left\{  h\in\Symm_n
\ \mid\ h\text{ has cycle type }\lambda\right\}  }\right\vert =\left\vert
\left\{  h\in\Symm_n\ \mid\ h\text{ has cycle type }\lambda\right\}
\right\vert .
\]
This proves (\ref{sol.frob-char.plambda.d.1b}).}

We shall now prove that $\Phi$ is a $\CC$-algebra homomorphism. Since
$\Phi\left(  1\right)  =1$ is true\footnote{This is because
\begin{align*}
\Phi\left(  \underbrace{1}_{=p_{\varnothing}}\right)   &  =\Phi\left(
p_{\varnothing}\right)  =\underbrace{\widetilde{z}_{\varnothing}}%
_{=1}\underbrace{\underline{1}_{\varnothing}}_{=1}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\Phi\left(  p_{\varnothing}\right)  \right) \\
&  =1.
\end{align*}
}, we only need to verify that $\Phi\left(  uv\right)  =\Phi\left(  u\right)
\Phi\left(  v\right)  $ for any $u\in\Lambda_{\mathbb{C}}$ and $v\in
\Lambda_{\mathbb{C}}$. Let us prove this now. Fix $u\in\Lambda_{\mathbb{C}}$
and $v\in\Lambda_{\mathbb{C}}$. Since the equality $\Phi\left(  uv\right)
=\Phi\left(  u\right)  \Phi\left(  v\right)  $ is $\CC$-linear in each
of $u$ and $v$, we can WLOG assume that $u$ and $v$ are elements of the basis
$\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ of the
$\CC$-vector space $\Lambda_{\mathbb{C}}$. Assume this, and set
$u=p_{\mu}$ and $v=p_{\nu}$ for two partitions $\mu$ and $\nu$. Let
$n=\left\vert \mu\right\vert $ and $m=\left\vert \nu\right\vert $. Let $g$ be
a permutation in $\Symm_n$ having cycle type $\mu$. (Such a $g$
clearly exists.) Let $h$ be a permutation in $\Symm_m$ having cycle
type $\nu$. (Such an $h$ clearly exists.) Applying
(\ref{sol.frob-char.plambda.d.1}) to $\mu$ instead of $\lambda$, we obtain
$\Phi\left(  p_{\mu}\right)  =\alpha_{\Symm_n,g}$. Thus,
\begin{equation}
\Phi\left(  \underbrace{u}_{=p_{\mu}}\right)  =\Phi\left(  p_{\mu}\right)
=\alpha_{\Symm_n,g}. \label{sol.frob-char.plambda.d.Phiu}%
\end{equation}
Applying (\ref{sol.frob-char.plambda.d.1}) to $\nu$, $m$ and $h$ instead of
$\lambda$, $n$ and $g$, we obtain $\Phi\left(  p_{\nu}\right)  =\alpha
_{\Symm_m,h}$. Thus,%
\begin{equation}
\Phi\left(  \underbrace{v}_{=p_{\nu}}\right)  =\Phi\left(  p_{\nu}\right)
=\alpha_{\Symm_m,h}. \label{sol.frob-char.plambda.d.Phiv}%
\end{equation}


The canonical isomorphism $R_\CC \left(  \Symm_n\right)
\otimes R_{\mathbb{C}}\left(  \Symm_m\right)  \rightarrow
R_{\mathbb{C}}\left(  \Symm_n\times\Symm_m\right)  $ sends
$\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}$ to
$\alpha_{\Symm_n\times\Symm_m,\left(  g,h\right)  }$
(according to Exercise \ref{exe.frob-char.plambda}(c), applied to
$G_{1}=\Symm_n$, $G_{2}=\Symm_m$, $h_{1}=g$ and $h_{2}=h$).
Let us identify $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  $ with $R_{\mathbb{C}}\left(
\Symm_n\times\Symm_m\right)  $ along this isomorphism. Then,
the statement we just made rewrites as follows:%
\begin{equation}
\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}=\alpha
_{\Symm_n\times\Symm_m,\left(  g,h\right)  }.
\label{sol.frob-char.plambda.d.2}%
\end{equation}


Now, multiplying the equalities (\ref{sol.frob-char.plambda.d.Phiu}) and
(\ref{sol.frob-char.plambda.d.Phiv}), we obtain%
\begin{align}
\Phi\left(  u\right)  \Phi\left(  v\right)   &  =\alpha_{\Symm_n%
,g}\alpha_{\Symm_m,h}\nonumber\\
&  =\operatorname*{Ind}\nolimits_{\Symm_n\times\Symm_m%
}^{\Symm_{n+m}}\underbrace{\left(  \alpha_{\Symm_n,g}%
\otimes\alpha_{\Symm_m,h}\right)  }_{\substack{=\alpha_{\Symm_n
\times\Symm_m,\left(  g,h\right)  }\\\text{(by
(\ref{sol.frob-char.plambda.d.2}))}}}\nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.frob-char.plambda.d.prod}), applied to }\beta=\alpha_{\Symm_n,
g}\text{ and }\gamma=\alpha_{\Symm_m,h}\right) \nonumber\\
&  =\operatorname*{Ind}\nolimits_{\Symm_n\times\Symm_m%
}^{\Symm_{n+m}}\alpha_{\Symm_n\times\Symm_m,\left(
g,h\right)  }=\alpha_{\Symm_{n+m},\left(  g,h\right)  }
\label{sol.frob-char.plambda.d.5}%
\end{align}
(by Exercise \ref{exe.frob-char.plambda}(b), applied to $\Symm_{n+m}$,
$\Symm_n\times\Symm_m$ and $\left(  g,h\right)  $ instead of
$G$, $H$ and $h$).

Let $\lambda$ be the partition whose parts are $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{\ell\left(  \mu\right)  }$, $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu
_{\ell\left(  \nu\right)  }$. Then, $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell\left(  \lambda\right)  }\right)  $ is a permutation of
the list $\left(  \mu_{1},\mu_{2},\ldots,\mu_{\ell\left(  \mu\right)  }%
,\nu_{1},\nu_{2},\ldots,\nu_{\ell\left(  \nu\right)  }\right)  $, and the
partition $\lambda$ has size
\[
\left\vert \lambda\right\vert =\underbrace{\mu_{1}+\mu_{2}+\cdots+\mu
_{\ell\left(  \mu\right)  }}_{=\left\vert \mu\right\vert =n}+\underbrace{\nu
_{1}+\nu_{2}+\cdots+\nu_{\ell\left(  \nu\right)  }}_{=\left\vert
\nu\right\vert =m}=n+m.
\]
We also have $p_{\lambda}=p_{\mu}p_{\nu}$\ \ \ \ \footnote{\textit{Proof.} By
the definition of $p_{\mu}$, we have $p_{\mu}=p_{\mu_{1}}p_{\mu_{2}}\cdots
p_{\mu_{\ell\left(  \mu\right)  }}$. Similarly, $p_{\nu}=p_{\nu_{1}}p_{\nu
_{2}}\cdots p_{\nu_{\ell\left(  \nu\right)  }}$. Multiplying these two
equalities, we obtain $p_{\mu}p_{\nu}=\left(  p_{\mu_{1}}p_{\mu_{2}}\cdots
p_{\mu_{\ell\left(  \mu\right)  }}\right)  \left(  p_{\nu_{1}}p_{\nu_{2}%
}\cdots p_{\nu_{\ell\left(  \nu\right)  }}\right)  $.
\par
But the definition of $p_{\lambda}$ yields $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(  \lambda\right)  }}$. The
product $p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(
\lambda\right)  }}$ has the same factors as the product $p_{\mu_{1}}p_{\mu
_{2}}\cdots p_{\mu_{\ell\left(  \mu\right)  }}p_{\nu_{1}}p_{\nu_{2}}\cdots
p_{\nu_{\ell\left(  \nu\right)  }}$ but possibly in a different order (since
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)
}\right)  $ is a permutation of the list $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{\ell\left(  \mu\right)  },\nu_{1},\nu_{2},\ldots,\nu_{\ell\left(
\nu\right)  }\right)  $). Hence, $p_{\lambda_{1}}p_{\lambda_{2}}\cdots
p_{\lambda_{\ell\left(  \lambda\right)  }}=p_{\mu_{1}}p_{\mu_{2}}\cdots
p_{\mu_{\ell\left(  \mu\right)  }}p_{\nu_{1}}p_{\nu_{2}}\cdots p_{\nu
_{\ell\left(  \nu\right)  }}$, so that%
\begin{align*}
p_{\lambda}  &  =p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(
\lambda\right)  }}=p_{\mu_{1}}p_{\mu_{2}}\cdots p_{\mu_{\ell\left(
\mu\right)  }}p_{\nu_{1}}p_{\nu_{2}}\cdots p_{\nu_{\ell\left(  \nu\right)  }}
=\left(  p_{\mu_{1}}p_{\mu_{2}}\cdots p_{\mu_{\ell\left(  \mu\right)  }%
}\right)  \left(  p_{\nu_{1}}p_{\nu_{2}}\cdots p_{\nu_{\ell\left(  \nu\right)
}}\right)  =p_{\mu}p_{\nu},
\end{align*}
qed.}. Since $p_{\mu}=u$ and $p_{\nu}=v$, this rewrites as $p_{\lambda}=uv$.

Recall that the permutation $g$ has cycle type $\mu$. In other words, the
cycles of $g$ have lengths $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{\ell\left(
\mu\right)  }$. Similarly, the cycles of $h$ have lengths $\nu_{1}$, $\nu_{2}%
$, $\ldots$, $\nu_{\ell\left(  \nu\right)  }$. Hence, the cycles of the
permutation $\left(  g,h\right)  \in\Symm_{n+m}$ (which, as we recall,
sends every $i\in\left\{  1,2,\ldots,n+m\right\}  $ to $\begin{cases}
g\left(  i\right)  ,& \text{if }i\leq n;\\
n+h\left(  i-n\right)  ,& \text{if }i>n
\end{cases}$) have lengths $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{\ell\left(
\mu\right)  }$, $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{\ell\left(  \nu\right)
}$ (in fact, on the subset $\left\{  1,2,\ldots,n\right\}  $ of $\left\{
1,2,\ldots,n+m\right\}  $, the permutation $\left(  g,h\right)  $ acts as $g$
and thus has cycles of lengths $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu
_{\ell\left(  \mu\right)  }$, whereas on the complementary subset $\left\{
n+1,n+2,\ldots,n+m\right\}  $ of $\left\{  1,2,\ldots,n+m\right\}  $, the
permutation $\left(  g,h\right)  $ acts as (a shifted version of) $h$ and thus
has cycles of lengths $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{\ell\left(
\nu\right)  }$). In other words, the cycles of the permutation $\left(
g,h\right)  \in\Symm_{n+m}$ have lengths $\lambda_{1}$, $\lambda_{2}$,
$\ldots$, $\lambda_{\ell\left(  \lambda\right)  }$ (since $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right)  $ is a
permutation of the list $\left(  \mu_{1},\mu_{2},\ldots,\mu_{\ell\left(
\mu\right)  },\nu_{1},\nu_{2},\ldots,\nu_{\ell\left(  \nu\right)  }\right)
$). In other words, the permutation $\left(  g,h\right)  \in\Symm_{n+m}$
has cycle type $\lambda$. Thus, (\ref{sol.frob-char.plambda.d.1})
(applied to $n+m$ and $\left(  g,h\right)  $ instead of $n$ and $g$) yields
$\Phi\left(  p_{\lambda}\right)  =\alpha_{\Symm_{n+m},\left(
g,h\right)  }$. Compared with (\ref{sol.frob-char.plambda.d.5}), this yields
$\Phi\left(  u\right)  \Phi\left(  v\right)  =\Phi\left(  p_{\lambda}\right)
$. Since $p_{\lambda}=uv$, this rewrites as $\Phi\left(  u\right)  \Phi\left(
v\right)  =\Phi\left(  uv\right)  $. Thus, $\Phi\left(  uv\right)
=\Phi\left(  u\right)  \Phi\left(  v\right)  $ is proven. As explained, this
completes the proof of the fact that $\Phi$ is a $\CC$-algebra homomorphism.

But we know that $\ch : A \rightarrow \Lambda$ is a
$\ZZ$-Hopf algebra isomorphism, and thus the extension of
$\ch$ to a $\CC$-linear map $A_{\CC}\rightarrow\Lambda_{\CC}$ is a
$\CC$-Hopf algebra isomorphism. We shall denote this extension by
$\ch_{\CC}$.

We now shall show that $\ch_{\CC} \circ
\Phi=\id_{\Lambda_{\mathbb{C}}}$. Indeed, let $n$ be a
positive integer. Then, $\widetilde{z}_{\left(  n\right)  }=n$%
\ \ \ \ \footnote{\textit{Proof.} Let $g$ denote the $n$-cycle $\left(
1,2,\ldots,n\right)  $ in $\Symm_n$. Then, $g$ is a permutation in
$\Symm_n$ having cycle type $\left(  n\right)  $. But $\widetilde{z}%
_{\left(  n\right)  }$ is defined as the size of the centralizer of a
permutation in $\Symm_{\left\vert \left(  n\right)  \right\vert }$
having cycle type $\left(  n\right)  $. Hence, $\widetilde{z}_{\left(
n\right)  }$ is the size of the centralizer of $g$ in $\Symm_n$
(since $g$ is a permutation in $\Symm_n=\Symm_{\left\vert
\left(  n\right)  \right\vert }$ having cycle type $\left(  n\right)  $). In
other words, $\widetilde{z}_{\left(  n\right)  }=\left\vert Z_{\Symm_n}
\left(  g\right)  \right\vert $.
\par
It is clear that the subgroup $\left\langle g\right\rangle $ of $\Symm_n$
generated by $g$ satisfies $\left\langle g\right\rangle \subset
Z_{\Symm_n}\left(  g\right)  $ (since every power of $g$ centralizes
$g$). We shall now show that $\left\langle g\right\rangle =
Z_{\Symm_{n}}\left(  g\right)  $.
\par
Indeed, let $z\in Z_{\Symm_n}\left(  g\right)  $. Then, $z$ must
centralize $g$. That is, we have $zgz^{-1}=g$. Hence, $zg=gz$, so that the
elements $g$ and $z$ of $\Symm_n$ commute. Hence, these elements $z$
and $g$ generate a commutative subgroup of $\Symm_n$. Denote this
subgroup by $T$.
\par
Now, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
\begin{equation}
g^{i-1}\left(  1\right)  =i \label{sol.frob-char.plambda.d.n-cycle.i}%
\end{equation}
(since $g$ is the $n$-cycle $\left(  1,2,\ldots,n\right)  $). Now, let
$j=z\left(  1\right)  $. Then, $g^{j-1}\left(  1\right)  =j$ (by
(\ref{sol.frob-char.plambda.d.n-cycle.i}), applied to $i=j$). On the other
hand, let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, $g^{i-1}$ commutes with
$z$ (since $g$ commutes with $z$); in other words, $g^{i-1}z=zg^{i-1}$. Now,%
\begin{align*}
z\left(  \underbrace{i}_{\substack{=g^{i-1}\left(  1\right)  \\\text{(by
(\ref{sol.frob-char.plambda.d.n-cycle.i}))}}}\right)   &  =z\left(
g^{i-1}\left(  1\right)  \right)  =\underbrace{\left(  zg^{i-1}\right)
}_{=g^{i-1}z}\left(  1\right)  =\left(  g^{i-1}z\right)  \left(  1\right)
=g^{i-1}\left(  \underbrace{z\left(  1\right)  }_{=j=g^{j-1}\left(  1\right)
}\right) \\
&  =g^{i-1}\left(  g^{j-1}\left(  1\right)  \right)  =\underbrace{\left(
g^{i-1}g^{j-1}\right)  }_{=g^{\left(  i-1\right)  +\left(  j-1\right)
}=g^{j-1}g^{i-1}}\left(  1\right)  =\left(  g^{j-1}g^{i-1}\right)  \left(
1\right)  =g^{j-1}\left(  \underbrace{g^{i-1}\left(  1\right)  }%
_{\substack{=i\\\text{(by (\ref{sol.frob-char.plambda.d.n-cycle.i}))}%
}}\right)  =g^{j-1}\left(  i\right)  .
\end{align*}
\par
Let us now forget that we fixed $i$. We thus have shown that $z\left(
i\right)  =g^{j-1}\left(  i\right)  $ for every $i\in\left\{  1,2,\ldots
,n\right\}  $. Hence, $z=g^{j-1}\in\left\langle g\right\rangle $.
\par
Now, let us forget that we fixed $z$. We thus have proven that $z\in
\left\langle g\right\rangle $ for every $z\in Z_{\Symm_n}\left(
g\right)  $. Hence, $Z_{\Symm_n}\left(  g\right)  \subset\left\langle
g\right\rangle $. Combined with $\left\langle g\right\rangle \subset
Z_{\Symm_n}\left(  g\right)  $, this yields $\left\langle
g\right\rangle =Z_{\Symm_n}\left(  g\right)  $. Hence, $\left\vert
\left\langle g\right\rangle \right\vert =\left\vert Z_{\Symm_n%
}\left(  g\right)  \right\vert $. Compared with $\widetilde{z}_{\left(
n\right)  }=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert $,
this yields $\widetilde{z}_{\left(  n\right)  }=\left\vert \left\langle
g\right\rangle \right\vert =\left(  \text{the order of }g\text{ in
}\Symm_n\right)  =n$ (since $g$ is an $n$-cycle), qed.}. Now, the
definition of $\Phi\left(  p_{\left(  n\right)  }\right)  $ yields
$\Phi\left(  p_{\left(  n\right)  }\right)  =\underbrace{\widetilde{z}%
_{\left(  n\right)  }}_{=n}\underline{1}_{\left(  n\right)  }=n\underline{1}%
_{\left(  n\right)  }$, so that%
\begin{align*}
\left(  \ch_{\CC} \circ\Phi\right)  \left(
p_{n}\right)   &  =\ch_{\CC} \left(  \Phi\left(
\underbrace{p_{n}}_{=p_{\left(  n\right)  }}\right)  \right)
=\ch_{\CC} \left(  \underbrace{\Phi\left(
p_{\left(  n\right)  }\right)  }_{=n\underline{1}_{\left(  n\right)  }%
}\right)  =\ch_{\CC} \left(  n\underline{1}%
_{\left(  n\right)  }\right)  =n\underbrace{\ch_{\CC}
\left(  \underline{1}_{\left(  n\right)  }\right)
}_{\substack{=\dfrac{p_{n}}{n}\\\text{(by a part of
Theorem~\ref{symmetric-group-Frobenius-map-theorem}}\\\text{that is already
proven)}}}\\
&  =n\cdot\dfrac{p_{n}}{n}=p_{n}=\id_{\Lambda
_{\mathbb{C}}}\left(  p_{n}\right)  .
\end{align*}


Now, let us forget that we fixed $n$. We thus have shown that $\left(
\ch_{\CC} \circ\Phi\right)  \left(
p_{n}\right)  =\id_{\Lambda_{\mathbb{C}}}\left(
p_{n}\right)  $ for every positive integer $n$. Thus, the $\CC$-algebra
homomorphisms $\ch_{\CC} \circ\Phi$ and
$\id_{\Lambda_{\mathbb{C}}}$ are equal to each other
on $p_{n}$ for every positive integer $n$. Since $\left(  p_{n}\right)
_{n\geq1}$ is a generating set of the $\CC$-algebra $\Lambda
_{\mathbb{C}}$, this yields that these homomorphisms are equal to each other
on a generating set of the $\CC$-algebra $\Lambda_{\mathbb{C}}$. Hence,
these homomorphisms must be identical. That is, we have
$\ch_{\CC} \circ\Phi = \id_{\Lambda_{\mathbb{C}}}$.

Since $\ch_{\CC} $ is an isomorphism, this
yields that $\Phi$ is the inverse of $\ch_{\CC}$.
That is, $\Phi=\left(  \ch_{\CC} \right) ^{-1}$.

Corollary \ref{dual-bases-h-m-p}(b) (applied to $\kk=\mathbb{C}$)
yields that $\{p_{\lambda}\}$ and $\left\{  z_{\lambda}^{-1}p_{\lambda
}\right\}  $ are dual bases of $\Lambda_{\mathbb{C}}$ with respect to the Hall
inner product on $\Lambda$. Thus,%
\begin{equation}
\left(  p_{\lambda},z_{\mu}^{-1}p_{\mu}\right)  _{\Lambda_{\mathbb{C}}}%
=\delta_{\lambda,\mu}\ \ \ \ \ \ \ \ \ \ \text{for any partitions }%
\lambda\text{ and }\mu\text{.} \label{sol.frob-char.plambda.d.dual-bases-1}%
\end{equation}
Hence, every partition $\lambda$ satisfies%
\begin{equation}
\left(  p_{\lambda},\underbrace{p_{\lambda}}_{=z_{\lambda}z_{\lambda}%
^{-1}p_{\lambda}}\right)  _{\Lambda_{\mathbb{C}}}=\left(  p_{\lambda
},z_{\lambda}z_{\lambda}^{-1}p_{\lambda}\right)  _{\Lambda_{\mathbb{C}}%
}=z_{\lambda}\underbrace{\left(  p_{\lambda},z_{\lambda}^{-1}p_{\lambda
}\right)  _{\Lambda_{\mathbb{C}}}}_{\substack{=\delta_{\lambda,\lambda
}\\\text{(by (\ref{sol.frob-char.plambda.d.dual-bases-1}), applied}\\\text{to
}\mu=\lambda\text{)}}}=z_{\lambda}\underbrace{\delta_{\lambda,\lambda}}%
_{=1}=z_{\lambda}. \label{sol.frob-char.plambda.d.dual-bases-2}%
\end{equation}


Our goal is to prove that $\ch_{\CC} \left(
\underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$ for every
partition $\lambda$.

Let $\lambda$ be a partition. Let $n$ be the size of $\lambda$. Then,
$\lambda\in\Par_{n}\subset \Par $ and
$\underline{1}_{\lambda}\in R_{\mathbb{C}}\left(  \Symm_n\right)  $.
We know that $\widetilde{z}_{\lambda}$ is a positive integer (since
$\widetilde{z}_{\lambda}$ is defined as the size of a centralizer, and
centralizers are subgroups). Hence, we can divide by $\widetilde{z}_{\lambda}%
$, and we have%
\[
\Phi\left(  \widetilde{z}_{\lambda}^{-1}p_{\lambda}\right)  =\widetilde{z}%
_{\lambda}^{-1}\underbrace{\Phi\left(  p_{\lambda}\right)  }%
_{\substack{=\widetilde{z}_{\lambda}\underline{1}_{\lambda}\\\text{(by the
definition of }\Phi\left(  p_{\lambda}\right)  \text{)}}}=\widetilde{z}%
_{\lambda}^{-1}\widetilde{z}_{\lambda}\underline{1}_{\lambda}=\underline{1}%
_{\lambda}.
\]
Since $\Phi=\left(  \ch_{\CC} \right)  ^{-1}$,
this rewrites as $\left(  \ch_{\CC} \right)
^{-1}\left(  \widetilde{z}_{\lambda}^{-1}p_{\lambda}\right)  =\underline{1}%
_{\lambda}$. Hence,
\begin{equation}
\ch_{\CC} \left(  \underline{1}_{\lambda
}\right)  =\widetilde{z}_{\lambda}^{-1}p_{\lambda}.
\label{sol.frob-char.plambda.d.halfthere}%
\end{equation}


Recall that we want to prove that $\ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$.
If we can show that $\widetilde{z}_{\lambda}=z_{\lambda}$, then
(\ref{sol.frob-char.plambda.d.halfthere}) becomes $\ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  =\left(
\underbrace{\widetilde{z}_{\lambda}}_{=z_{\lambda}}\right)  ^{-1}p_{\lambda
}=z_{\lambda}^{-1}p_{\lambda}=\dfrac{p_{\lambda}}{z_{\lambda}}$, and thus
$\ch_{\CC} \left(  \underline{1}_{\lambda
}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$ will be proven. Hence, all that
remains to be done is proving $\widetilde{z}_{\lambda}=z_{\lambda}$.

But $\ch$ is a PSH-isomorphism, thus an isometry. Hence,
$\ch_{\CC}$ (being the extension of
$\ch$ to a $\CC$-linear map) must also be an isometry,
i.e., we must have
\[
\left(  \ch_{\CC} \beta, \ch_{\CC} \gamma
\right)  _{\Lambda_{\mathbb{C}}}=\left(
\beta,\gamma\right)  _{A_{\mathbb{C}}}\ \ \ \ \ \ \ \ \ \ \text{for all }%
\beta\in A_{\mathbb{C}}\text{ and }\gamma\in A_{\mathbb{C}}.
\]
Applying this to $\beta=\underline{1}_{\lambda}$ and $\gamma=\underline{1}%
_{\lambda}$, we obtain%
\begin{align*}
&  \left(  \ch_{\CC} \left(  \underline{1}%
_{\lambda}\right)  ,\ch_{\CC} \left(
\underline{1}_{\lambda}\right)  \right)  _{\Lambda_{\mathbb{C}}}\\
&  =\left(  \underline{1}_{\lambda},\underline{1}_{\lambda}\right)
_{A_{\mathbb{C}}}=\left\langle \underline{1}_{\lambda},\underline{1}_{\lambda
}\right\rangle _{\Symm_n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\underline{1}_{\lambda}\in R_{\mathbb{C}}\left(  \Symm_n
\right)  \text{, and since the bilinear form}\\
\left(  \cdot,\cdot\right)  _{A_{\mathbb{C}}}\text{ on }A_{\mathbb{C}}\text{
extends the bilinear form }\left\langle \cdot,\cdot\right\rangle
_{\Symm_n}\text{ on }R_{\mathbb{C}}\left(  \Symm_n\right)
\end{array}
\right) \\
&  =\underbrace{\dfrac{1}{\left\vert \Symm_n\right\vert }}%
_{=\dfrac{1}{n!}}\sum_{g\in\Symm_n}\underbrace{\underline{1}%
_{\lambda}\left(  g\right)  }_{\substack{=\left[  g\text{ has cycle type
}\lambda\right]  \\\text{(by the definition of }\underline{1}_{\lambda
}\text{)}}}\underbrace{\underline{1}_{\lambda}\left(  g^{-1}\right)
}_{\substack{=\left[  g^{-1}\text{ has cycle type }\lambda\right]  \\\text{(by
the definition of }\underline{1}_{\lambda}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the bilinear form
}\left\langle \cdot,\cdot\right\rangle _{\Symm_n}\right) \\
&  =\dfrac{1}{n!}\sum_{g\in\Symm_n}\left[  g\text{ has cycle type
}\lambda\right]  \left[  \underbrace{g^{-1}\text{ has cycle type }\lambda
}_{\substack{\text{this is equivalent to}\\\left(  g\text{ has cycle type
}\lambda\right)  \\\text{(since the cycle type of }g^{-1}\\\text{equals the
cycle type of }g\text{)}}}\right]  =\dfrac{1}{n!}\sum_{g\in\Symm_n%
}\underbrace{\left[  g\text{ has cycle type }\lambda\right]  \left[  g\text{
has cycle type }\lambda\right]  }_{\substack{=\left[  \left(  g\text{ has
cycle type }\lambda\right)  \text{ and }\left(  g\text{ has cycle type
}\lambda\right)  \right]  \\=\left[  g\text{ has cycle type }\lambda\right]
}}\\
&  =\dfrac{1}{n!}\underbrace{\sum_{g\in\Symm_n}\left[  g\text{ has
cycle type }\lambda\right]  }_{\substack{=\left\vert \left\{  h\in
\Symm_n\ \mid\ h\text{ has cycle type }\lambda\right\}  \right\vert
=n!/\widetilde{z}_{\lambda}\\\text{(by (\ref{sol.frob-char.plambda.d.1b}))}%
}}=\dfrac{1}{n!}\cdot\dfrac{n!}{\widetilde{z}_{\lambda}}=\dfrac{1}%
{\widetilde{z}_{\lambda}}.
\end{align*}
Hence,%
\[
\dfrac{1}{\widetilde{z}_{\lambda}}=\left(  \underbrace{\ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  }%
_{\substack{=\widetilde{z}_{\lambda}^{-1}p_{\lambda}\\\text{(by
(\ref{sol.frob-char.plambda.d.halfthere}))}}},\underbrace{\ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  }%
_{\substack{=\widetilde{z}_{\lambda}^{-1}p_{\lambda}\\\text{(by
(\ref{sol.frob-char.plambda.d.halfthere}))}}}\right)  _{\Lambda_{\mathbb{C}}%
}=\left(  \widetilde{z}_{\lambda}^{-1}p_{\lambda},\widetilde{z}_{\lambda}%
^{-1}p_{\lambda}\right)  _{\Lambda_{\mathbb{C}}}=\left(  \widetilde{z}%
_{\lambda}^{-1}\right)  ^{2}\left(  p_{\lambda},p_{\lambda}\right)
_{\Lambda_{\mathbb{C}}}.
\]
Multiplying this equality with $\widetilde{z}_{\lambda}^{2}$, we obtain%
\[
\widetilde{z}_{\lambda}=\left(  p_{\lambda},p_{\lambda}\right)  _{\Lambda
_{\mathbb{C}}}=z_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.frob-char.plambda.d.dual-bases-2})}\right)  .
\]
Thus, $\widetilde{z}_{\lambda}=z_{\lambda}$ is proven. As we have explained,
this concludes the proof of $\ch_{\CC} \left(
\underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$, and thus
Exercise \ref{exe.frob-char.plambda}(d) is solved.

(e) Let $\lambda$ be a partition. Let us work with the notations of the
solution of Exercise \ref{exe.frob-char.plambda}(d) above. In the latter
solution, we have shown that $\widetilde{z}_{\lambda}=z_{\lambda}$. Thus,%
\begin{align*}
z_{\lambda}  &  =\widetilde{z}_{\lambda}=\left(  \text{the size of the
centralizer of a permutation in }\Symm_{\left\vert \lambda\right\vert
}\text{ having cycle type }\lambda\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{z}%
_{\lambda}\right) \\
&  =\left(  \text{the size of the centralizer in }\Symm_n\text{ of a
permutation having cycle type }\lambda\text{, where }n=\left\vert
\lambda\right\vert \right)  .
\end{align*}
This proves Remark \ref{rmk.zlambda.centralizer}. Thus, Exercise
\ref{exe.frob-char.plambda}(e) is solved.

(f) Let $G$ and $H$ be two finite groups. Let $\rho:H\rightarrow G$ be a group
homomorphism. Let $y\in H$. We shall show that $\operatorname*{Ind}%
\nolimits_{\rho}\alpha_{H,y}=\alpha_{G,\rho\left(  y\right)  }$.

Let $u\in G$. Then, the definition of $\Ind_{\rho
}\alpha_{H,y}$ yields%
\begin{align}
\left(  \Ind_{\rho}\alpha_{H,y}\right)  \left(
u\right)   &  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=u}}\alpha
_{H,y}\left(  h\right)  =\dfrac{1}{\left\vert H\right\vert }\sum\limits
_{\substack{\left(  h,x\right)  \in H\times G;\\x\rho\left(  h\right)
x^{-1}=u}}\underbrace{\alpha_{H,y}\left(  h\right)  }_{\substack{=\sum_{k\in
H}\left[  kyk^{-1}=h\right]  \\\text{(by Exercise \ref{exe.frob-char.plambda}%
(a), applied to}\\H\text{, }y\text{ and }h\text{ instead of }G\text{, }h\text{
and }g\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  h,k\right)  \text{ as }\left(  h,x\right)  \right) \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{\left(
h,x\right)  \in H\times G;\\x\rho\left(  h\right)  x^{-1}=u}}}_{=\sum_{x\in
G}\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}}\sum_{k\in
H}\left[  kyk^{-1}=h\right] \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{x\in G}\underbrace{\sum\limits
_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\sum_{k\in H}}%
_{=\sum_{k\in H}\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}%
}}\left[  kyk^{-1}=h\right] \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{x\in G}\sum_{k\in H}%
\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]  . \label{sol.frob-char.plambda.1}%
\end{align}
But every $x\in G$ and $k\in H$ satisfy%
\begin{equation}
\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]  =\left[  x\rho\left(  k\right)  \rho\left(  y\right)
\left(  x\rho\left(  k\right)  \right)  ^{-1}=u\right]
\label{sol.frob-char.plambda.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.2}):} Let $x\in G$ and
$k\in H$. We have%
\begin{align*}
&  \sum_{h\in H}\left[  x\rho\left(  h\right)  x^{-1}=u\right]  \left[
kyk^{-1}=h\right] \\
&  =\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}%
}\underbrace{\left[  x\rho\left(  h\right)  x^{-1}=u\right]  }%
_{\substack{=1\\\text{(since we have }x\rho\left(  h\right)  x^{-1}=u\text{)}%
}}\left[  kyk^{-1}=h\right]  +\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)
x^{-1}\neq u}}\underbrace{\left[  x\rho\left(  h\right)  x^{-1}=u\right]
}_{\substack{=0\\\text{(since we don't have }x\rho\left(  h\right)
x^{-1}=u\\\text{(since }x\rho\left(  h\right)  x^{-1}\neq u\text{))}}}\left[
kyk^{-1}=h\right] \\
&  =\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]  +\underbrace{\sum\limits_{\substack{h\in H;\\x\rho\left(
h\right)  x^{-1}\neq u}}0\left[  kyk^{-1}=h\right]  }_{=0}=\sum\limits
_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[  kyk^{-1}%
=h\right]  ,
\end{align*}
so that%
\begin{align*}
\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]   &  =\sum_{h\in H}\left[  x\rho\left(  h\right)
x^{-1}=u\right]  \left[  kyk^{-1}=h\right] \\
&  =\sum\limits_{\substack{h\in H;\\h=kyk^{-1}}}\left[  x\rho\left(  h\right)
x^{-1}=u\right]  \underbrace{\left[  kyk^{-1}=h\right]  }%
_{\substack{=1\\\text{(since we have }kyk^{-1}=h\\\text{(since }%
h=kyk^{-1}\text{))}}}+\sum\limits_{\substack{h\in H;\\h\neq kyk^{-1}}}\left[
x\rho\left(  h\right)  x^{-1}=u\right]  \underbrace{\left[  kyk^{-1}=h\right]
}_{\substack{=0\\\text{(since we don't have }kyk^{-1}=h\\\text{(since
}kyk^{-1}\neq h\text{ (since }h\neq kyk^{-1}\text{)))}}}\\
&  =\sum\limits_{\substack{h\in H;\\h=kyk^{-1}}}\left[  x\rho\left(  h\right)
x^{-1}=u\right]  +\underbrace{\sum\limits_{\substack{h\in H;\\h\neq kyk^{-1}}}\left[
x\rho\left(  h\right)  x^{-1}=u\right]  0}_{=0}=\sum\limits_{\substack{h\in
H;\\h=kyk^{-1}}}\left[  x\rho\left(  h\right)  x^{-1}=u\right] \\
&  =\left[  x\underbrace{\rho\left(  kyk^{-1}\right)  }_{\substack{=\rho
\left(  k\right)  \rho\left(  y\right)  \left(  \rho\left(  k\right)  \right)
^{-1}\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}%
}x^{-1}=u\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }kyk^{-1}\in H\text{
(since }k\in H\text{ and }y\in H\text{)}\right) \\
&  =\left[  x\rho\left(  k\right)  \rho\left(  y\right)  \underbrace{\left(
\rho\left(  k\right)  \right)  ^{-1}x^{-1}}_{=\left(  x\rho\left(  k\right)
\right)  ^{-1}}=u\right]  =\left[  x\rho\left(  k\right)  \rho\left(
y\right)  \left(  x\rho\left(  k\right)  \right)  ^{-1}=u\right]  .
\end{align*}
This proves (\ref{sol.frob-char.plambda.2}).}. Hence,
(\ref{sol.frob-char.plambda.1}) becomes%
\begin{align*}
\left(  \Ind_{\rho}\alpha_{H,y}\right)  \left(
u\right)   &  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum_{x\in
G}\sum_{k\in H}}_{=\sum_{k\in H}\sum_{x\in G}}\underbrace{\sum\limits_{\substack{h\in
H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[  kyk^{-1}=h\right]  }%
_{\substack{=\left[  x\rho\left(  k\right)  \rho\left(  y\right)  \left(
x\rho\left(  k\right)  \right)  ^{-1}=u\right]  \\\text{(by
(\ref{sol.frob-char.plambda.2}))}}}\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{k\in H}\underbrace{\sum_{x\in
G}\left[  x\rho\left(  k\right)  \rho\left(  y\right)  \left(  x\rho\left(
k\right)  \right)  ^{-1}=u\right]  }_{\substack{=\sum_{x\in G}\left[
x\rho\left(  y\right)  x^{-1}=u\right]  \\\text{(here, we have substituted
}x\text{ for }x\rho\left(  k\right)  \text{ in the sum}\\\text{(because the
map }G\rightarrow G,\ x\mapsto x\rho\left(  k\right)  \text{ is a
bijection}\\\text{(since }G\text{ is a group, and since }\rho\left(  k\right)
\in G\text{)))}}}=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum_{k\in
H}\sum_{x\in G}\left[  x\rho\left(  y\right)  x^{-1}=u\right]  }_{=\left\vert
H\right\vert \cdot\sum_{x\in G}\left[  x\rho\left(  y\right)  x^{-1}=u\right]
}\\
&  =\dfrac{1}{\left\vert H\right\vert }\left\vert H\right\vert \cdot\sum_{x\in
G}\left[  x\rho\left(  y\right)  x^{-1}=u\right]  =\sum_{x\in G}\left[
x\rho\left(  y\right)  x^{-1}=u\right] \\
&  =\sum_{k\in G}\left[  k\rho\left(  y\right)  k^{-1}=u\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}x\text{ as }k\right)  .
\end{align*}
Compared with%
\[
\alpha_{G,\rho\left(  y\right)  }\left(  u\right)  =\sum_{k\in G}\left[
k\rho\left(  y\right)  k^{-1}=u\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.frob-char.plambda}(a), applied to }\rho\left(  y\right)
\text{ and }u\text{ instead of }h\text{ and }g\right)  ,
\]
this yields $\left(  \Ind_{\rho}\alpha_{H,y}\right)
\left(  u\right)  =\alpha_{G,\rho\left(  y\right)  }\left(  u\right)  $.

Let us now forget that we fixed $u$. We thus have shown that $\left(
\Ind_{\rho}\alpha_{H,y}\right)  \left(  u\right)
=\alpha_{G,\rho\left(  y\right)  }\left(  u\right)  $ for every $u\in G$. In
other words, $\Ind_{\rho}\alpha_{H,y}=\alpha
_{G,\rho\left(  y\right)  }$.

Let us now forget that we fixed $y$. We thus have shown that
$\Ind_{\rho}\alpha_{H,y}=\alpha_{G,\rho\left(
y\right)  }$ for every $y\in H$. Renaming $y$ as $h$ in this statement, we
obtain that $\Ind_{\rho}\alpha_{H,h}=\alpha
_{G,\rho\left(  h\right)  }$ for every $h\in H$. This solves Exercise
\ref{exe.frob-char.plambda}(f).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.frob-char.plambda}}

\textit{Solution to Exercise \ref{exe.frob-char.plambda}.} Define the Iverson
bracket notation as in Exercise \ref{exe.frob-char.plambda}(a).

(a) Let $G$ be a finite group. For every subset $P$ of $G$, the definition of
$\underline{1}_{P}$ yields
\begin{equation}
\underline{1}_{P}\left(  g\right)  =\left[  g\in P\right]
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G.
\label{sol.frob-char.plambda.a.iverson}%
\end{equation}
Also, for every subset $P$ of $G$, we have%
\begin{equation}
\left\vert P\right\vert =\sum_{k\in G}\left[  k\in P\right]  .
\label{sol.frob-char.plambda.a.iversonsum}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.a.iversonsum}):} We
have
\begin{align*}
\sum_{k\in G}\left[  k\in P\right]   &  =\sum\limits_{\substack{k\in G;\\k\in
P}}\underbrace{\left[  k\in P\right]  }_{\substack{=1\\\text{(since }k\in
P\text{ is true)}}}+\sum\limits_{\substack{k\in G;\\k\notin P}}\underbrace{\left[
k\in P\right]  }_{\substack{=0\\\text{(since }k\in P\text{ is false}%
\\\text{(since }k\notin P\text{))}}}=\underbrace{\sum\limits_{\substack{k\in G;\\k\in
P}}}_{\substack{=\sum_{k\in P}\\\text{(since every }k\in P\\\text{satisfies
}k\in G\text{)}}}1+\underbrace{\sum\limits_{\substack{k\in G;\\k\notin P}}0}_{=0}\\
&  =\sum_{k\in P}1+0=\sum_{k\in P}1=\left\vert P\right\vert \cdot1=\left\vert
P\right\vert .
\end{align*}
Thus, (\ref{sol.frob-char.plambda.a.iversonsum}) is proven.}

Let $h\in G$ and $g\in G$. Then, (\ref{sol.frob-char.plambda.a.iversonsum})
(applied to $P=Z_{G}\left(  h\right)  $) yields%
\[
\left\vert Z_{G}\left(  h\right)  \right\vert =\sum_{k\in G}\left[
\underbrace{k\in Z_{G}\left(  h\right)  }_{\substack{\text{this is equivalent
to }khk^{-1}=h\\\text{(by the definition of }Z_{G}\left(  h\right)  \text{)}%
}}\right]  =\sum_{k\in G}\left[  khk^{-1}=h\right]  .
\]


Applying both sides of the identity $\alpha_{G,h}=\left\vert Z_{G}\left(
h\right)  \right\vert \underline{1}_{\operatorname{Conj}_{G}\left(  h\right)
}$ to $g$, we obtain $\alpha_{G,h}\left(  g\right)  =\left(  \left\vert
Z_{G}\left(  h\right)  \right\vert \underline{1}_{\operatorname{Conj}%
_{G}\left(  h\right)  }\right)  \left(  g\right)  =\left\vert Z_{G}\left(
h\right)  \right\vert \underline{1}_{\operatorname{Conj}_{G}\left(  h\right)
}\left(  g\right)  $, so that%
\begin{equation}
\alpha_{G,h}\left(  g\right)  =\underbrace{\left\vert Z_{G}\left(  h\right)
\right\vert }_{=\sum_{k\in G}\left[  khk^{-1}=h\right]  }%
\underbrace{\underline{1}_{\operatorname{Conj}_{G}\left(  h\right)  }\left(
g\right)  }_{\substack{=\left[  g\in\operatorname{Conj}_{G}\left(  h\right)
\right]  \\\text{(by (\ref{sol.frob-char.plambda.a.iverson}), applied}%
\\\text{to }P=\operatorname{Conj}_{G}\left(  h\right)  \text{)}}}=\left(
\sum_{k\in G}\left[  khk^{-1}=h\right]  \right)  \left[  g\in
\operatorname{Conj}_{G}\left(  h\right)  \right]  .
\label{sol.frob-char.plambda.a.1}%
\end{equation}


We need to prove that $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[
khk^{-1}=g\right]  $. We must be in one of the following two cases:

\textit{Case 1:} We have $g\in\operatorname{Conj}_{G}\left(  h\right)  $.

\textit{Case 2:} We don't have $g\in\operatorname{Conj}_{G}\left(  h\right)  $.

Let us first consider Case 1. In this case, we have $g\in\operatorname{Conj}%
_{G}\left(  h\right)  $. Thus, there exists a $p\in G$ satisfying $g=php^{-1}%
$. Consider this $p$. Since $g=php^{-1}$, we have $h=p^{-1}gp$. Hence,
(\ref{sol.frob-char.plambda.a.1}) becomes%
\begin{align*}
\alpha_{G,h}\left(  g\right)   &  =\left(  \sum_{k\in G}\left[  khk^{-1}%
=h\right]  \right)  \underbrace{\left[  g\in\operatorname{Conj}_{G}\left(
h\right)  \right]  }_{\substack{=1\\\text{(since }g\in\operatorname{Conj}%
_{G}\left(  h\right)  \text{)}}}=\sum_{k\in G}\left[  k\underbrace{h}%
_{=p^{-1}gp}k^{-1}=h\right] \\
&  =\sum_{k\in G}\left[  \underbrace{kp^{-1}}_{=\left(  pk^{-1}\right)  ^{-1}%
}gpk^{-1}=h\right]  =\sum_{k\in G}\left[  \left(  pk^{-1}\right)
^{-1}g\left(  kp^{-1}\right)  ^{-1}=h\right] \\
&  =\sum_{k\in G}\left[  \underbrace{k^{-1}gk=h}_{\substack{\text{this is
equivalent to}\\khk^{-1}=g}}\right]  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }k\text{ for }pk^{-1}\text{ in the sum}\\
\text{(since the map }G\rightarrow G,\ k\mapsto pk^{-1}\text{ is a bijection)}%
\end{array}
\right) \\
&  =\sum_{k\in G}\left[  khk^{-1}=g\right]  .
\end{align*}
Thus, $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]
$ is proven in Case 1.

Let us now consider Case 2. In this case, we don't have $g\in
\operatorname{Conj}_{G}\left(  h\right)  $. In other words, $g$ is not in the
conjugacy class of $h$. In other words, $g$ is not conjugate to $h$. In other
words, there exists no $k\in G$ satisfying $khk^{-1}=g$. In other words, for
every $k\in G$, we do not have $khk^{-1}=g$. Hence, for every $k\in G$, we
have $\left[  khk^{-1}=g\right]  =0$. Thus, $\sum_{k\in G}\underbrace{\left[
khk^{-1}=g\right]  }_{=0}=\sum_{k\in G}0=0$. Comparing this with%
\[
\alpha_{G,h}\left(  g\right)  =\left(  \sum_{k\in G}\left[  khk^{-1}=h\right]
\right)  \underbrace{\left[  g\in\operatorname{Conj}_{G}\left(  h\right)
\right]  }_{\substack{=0\\\text{(since we don't have }g\in\operatorname{Conj}%
_{G}\left(  h\right)  \text{)}}}=0,
\]
we obtain $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}%
=g\right]  $. Thus, $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[
khk^{-1}=g\right]  $ is proven in Case 2.

Thus, $\alpha_{G,h}\left(  g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]
$ is proven in both Cases 1 and 2. Consequently, $\alpha_{G,h}\left(
g\right)  =\sum_{k\in G}\left[  khk^{-1}=g\right]  $ always holds, and
Exercise \ref{exe.frob-char.plambda}(a) is solved.

(b) Let $H$ be a subgroup of a finite group $G$. Let $h\in H$. Let $g\in G$.
Then, the definition of $\operatorname{Ind}_{H}^{G}\alpha_{H,h}$ given in
Exercise \ref{exe.induction.characters} yields%
\begin{align}
\left(  \operatorname{Ind}_{H}^{G}\alpha_{H,h}\right)  \left(  g\right)   &
=\frac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k\in G:\\kgk^{-1}\in
H}}\alpha_{H,h}\left(  kgk^{-1}\right)  =\frac{1}{\left\vert H\right\vert
}\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\underbrace{\alpha_{H,h}\left(
pgp^{-1}\right)  }_{\substack{=\sum_{k\in H}\left[  khk^{-1}=pgp^{-1}\right]
\\\text{(by Exercise \ref{exe.frob-char.plambda}(a), applied}\\\text{to
}H\text{ and }pgp^{-1}\text{ instead of }G\text{ and }g\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}k\text{ as }p\right) \nonumber\\
&  =\frac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{p\in
G:\\pgp^{-1}\in H}}\sum_{k\in H}}_{=\sum_{k\in H}\sum\limits_{\substack{p\in
G:\\pgp^{-1}\in H}}}\left[  khk^{-1}=pgp^{-1}\right]  =\frac{1}{\left\vert
H\right\vert }\sum_{k\in H}\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\left[
khk^{-1}=pgp^{-1}\right]  . \label{sol.frob-char.plambda.b.1}%
\end{align}
But every $k\in H$ satisfies%
\begin{equation}
\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]
=\sum_{p\in G}\left[  khk^{-1}=pgp^{-1}\right]
\label{sol.frob-char.plambda.b.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.b.2}):} Let $k\in H$.
Then,%
\begin{align*}
\sum_{p\in G}\left[  khk^{-1}=pgp^{-1}\right]   &  =\sum\limits_{\substack{p\in
G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]  +\sum\limits_{\substack{p\in
G:\\pgp^{-1}\notin H}}\underbrace{\left[  khk^{-1}=pgp^{-1}\right]
}_{\substack{=0\\\text{(since we don't have }khk^{-1}=pgp^{-1}\\\text{(since
}khk^{-1}\in H\text{ (since }k\in H\text{ and }h\in H\text{) and }%
pgp^{-1}\notin H\text{))}}}\\
&  =\sum\limits_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]
+\underbrace{\sum\limits_{\substack{p\in G:\\pgp^{-1}\notin H}}0}_{=0}=\sum\limits
_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]  .
\end{align*}
This proves (\ref{sol.frob-char.plambda.b.2}).}. Now,
(\ref{sol.frob-char.plambda.b.1}) becomes%
\begin{align*}
\left(  \operatorname{Ind}_{H}^{G}\alpha_{H,h}\right)  \left(  g\right)   &
=\frac{1}{\left\vert H\right\vert }\sum_{k\in H}\underbrace{\sum\limits
_{\substack{p\in G:\\pgp^{-1}\in H}}\left[  khk^{-1}=pgp^{-1}\right]
}_{\substack{=\sum_{p\in G}\left[  khk^{-1}=pgp^{-1}\right]  \\\text{(by
(\ref{sol.frob-char.plambda.b.2}))}}}=\frac{1}{\left\vert H\right\vert }%
\sum_{k\in H}\sum_{p\in G}\left[  \underbrace{khk^{-1}=pgp^{-1}}%
_{\substack{\text{this is equivalent to}\\p^{-1}khk^{-1}p=g}}\right] \\
&  =\frac{1}{\left\vert H\right\vert }\sum_{k\in H}\sum_{p\in G}\left[
p^{-1}kh\underbrace{k^{-1}p}_{=\left(  p^{-1}k\right)  ^{-1}}=g\right]
=\frac{1}{\left\vert H\right\vert }\sum_{k\in H}\sum_{p\in G}\left[
p^{-1}kh\left(  p^{-1}k\right)  ^{-1}=g\right] \\
&  =\frac{1}{\left\vert H\right\vert }\underbrace{\sum_{k\in H}\sum_{p\in
G}\left[  php^{-1}=g\right]  }_{=\left\vert H\right\vert \sum_{p\in G}\left[
php^{-1}=g\right]  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }p\text{ for }p^{-1}k\text{ in the inner sum}\\
\text{(since the map }G\rightarrow G,\ p\mapsto p^{-1}k\text{ is a bijection)}%
\end{array}
\right) \\
&  =\dfrac{1}{\left\vert H\right\vert }\left\vert H\right\vert \sum_{p\in
G}\left[  php^{-1}=g\right]  =\sum_{p\in G}\left[  php^{-1}=g\right]
=\sum_{k\in G}\left[  khk^{-1}=g\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k\text{ for
}p\text{ in the sum}\right) \\
&  =\alpha_{G,h}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.frob-char.plambda}(a)}\right)  .
\end{align*}


Let us now forget that we fixed $g$. We thus have proven that $\left(
\operatorname{Ind}_{H}^{G}\alpha_{H,h}\right)  \left(  g\right)  =\alpha
_{G,h}\left(  g\right)  $ for every $g\in G$. Hence, $\operatorname{Ind}%
_{H}^{G}\alpha_{H,h}=\alpha_{G,h}$. This solves Exercise
\ref{exe.frob-char.plambda}(b).

(c) Let $G_{1}$ and $G_{2}$ be two finite groups. Let $h_{1}\in G_{1}$ and
$h_{2}\in G_{2}$. Let $\mathbf{I}$ denote the canonical isomorphism
$R_\CC \left(  G_{1}\right)  \otimes R_{\mathbb{C}}\left(
G_{2}\right)  \rightarrow R_{\mathbb{C}}\left(  G_{1}\times G_{2}\right)  $.
We need to prove that%
\begin{equation}
\mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}}\right)
=\alpha_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }.
\label{sol.frob-char.plambda.c.goal}%
\end{equation}


Let $g\in G_{1}\times G_{2}$. Let us write $g$ in the form $g=\left(
g_{1},g_{2}\right)  $. Then, Exercise \ref{exe.frob-char.plambda}(a) (applied
to $G_{1}$, $h_{1}$ and $g_{1}$ instead of $G$, $h$ and $g$) yields%
\[
\alpha_{G_{1},h_{1}}\left(  g_{1}\right)  =\sum_{k\in G_{1}}\left[
kh_{1}k^{-1}=g_{1}\right]  =\sum_{k_{1}\in G_{1}}\left[  k_{1}h_{1}k_{1}%
^{-1}=g_{1}\right]
\]
(here, we have renamed the summation index $k$ as $k_{1}$). Similarly,%
\[
\alpha_{G_{2},h_{2}}\left(  g_{2}\right)  =\sum_{k_{2}\in G_{2}}\left[
k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  .
\]
Now,%
\begin{align*}
&  \left(  \mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}%
}\right)  \right)  \left(  \underbrace{g}_{=\left(  g_{1},g_{2}\right)
}\right) \\
&  =\left(  \mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}%
}\right)  \right)  \left(  \left(  g_{1},g_{2}\right)  \right)
=\underbrace{\alpha_{G_{1},h_{1}}\left(  g_{1}\right)  }_{=\sum_{k_{1}\in
G_{1}}\left[  k_{1}h_{1}k_{1}^{-1}=g_{1}\right]  }\underbrace{\alpha
_{G_{2},h_{2}}\left(  g_{2}\right)  }_{=\sum_{k_{2}\in G_{2}}\left[
k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathbf{I}\right) \\
&  =\left(  \sum_{k_{1}\in G_{1}}\left[  k_{1}h_{1}k_{1}^{-1}=g_{1}\right]
\right)  \left(  \sum_{k_{2}\in G_{2}}\left[  k_{2}h_{2}k_{2}^{-2}%
=g_{2}\right]  \right)  =\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times
G_{2}}\underbrace{\left[  k_{1}h_{1}k_{1}^{-1}=g_{1}\right]  \left[
k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  }_{=\left[  k_{1}h_{1}k_{1}^{-1}%
=g_{1}\text{ and }k_{2}h_{2}k_{2}^{-2}=g_{2}\right]  }\\
&  =\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times G_{2}}\left[
\underbrace{k_{1}h_{1}k_{1}^{-1}=g_{1}\text{ and }k_{2}h_{2}k_{2}^{-2}=g_{2}%
}_{\substack{\text{this is equivalent to}\\\left(  k_{1}h_{1}k_{1}^{-1}%
,k_{2}h_{2}k_{2}^{-1}\right)  =\left(  g_{1},g_{2}\right)  }}\right]
=\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times G_{2}}\left[
\underbrace{\left(  k_{1}h_{1}k_{1}^{-1},k_{2}h_{2}k_{2}^{-1}\right)
}_{=\left(  k_{1},k_{2}\right)  \left(  h_{1},h_{2}\right)  \left(
k_{1},k_{2}\right)  ^{-1}}=\underbrace{\left(  g_{1},g_{2}\right)  }%
_{=g}\right] \\
&  =\sum_{\left(  k_{1},k_{2}\right)  \in G_{1}\times G_{2}}\left[  \left(
k_{1},k_{2}\right)  \left(  h_{1},h_{2}\right)  \left(  k_{1},k_{2}\right)
^{-1}=g\right]  =\sum_{k\in G_{1}\times G_{2}}\left[  k\left(  h_{1}%
,h_{2}\right)  k^{-1}=g\right]
\end{align*}
(here, we have renamed the summation index $\left(  k_{1},k_{2}\right)  $ as
$k$). Compared with%
\begin{align*}
&  \alpha_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }\left(  g\right)
=\sum_{k\in G_{1}\times G_{2}}\left[  k\left(  h_{1},h_{2}\right)
k^{-1}=g\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.frob-char.plambda}%
(a), applied to }G_{1}\times G_{2}\text{ and }\left(  h_{1},h_{2}\right)
\text{ instead of }G\text{ and }h\right)  ,
\end{align*}
this yields $\left(  \mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes
\alpha_{G_{2},h_{2}}\right)  \right)  \left(  g\right)  =\alpha_{G_{1}\times
G_{2},\left(  h_{1},h_{2}\right)  }\left(  g\right)  $.

Now, let us forget that we fixed $g$. We thus have proven that $\left(
\mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2},h_{2}}\right)
\right)  \left(  g\right)  =\alpha_{G_{1}\times G_{2},\left(  h_{1}%
,h_{2}\right)  }\left(  g\right)  $ for every $g\in G_{1}\times G_{2}$. In
other words, $\mathbf{I}\left(  \alpha_{G_{1},h_{1}}\otimes\alpha_{G_{2}%
,h_{2}}\right)  =\alpha_{G_{1}\times G_{2},\left(  h_{1},h_{2}\right)  }$.
This proves (\ref{sol.frob-char.plambda.c.goal}). Exercise
\ref{exe.frob-char.plambda}(c) is thus solved.

(d) For every partition $\lambda$, let us define $\widetilde{z}_{\lambda}$ as
the size of the centralizer of a permutation in $\Symm_{\left\vert
\lambda\right\vert }$ having cycle type $\lambda$. Note that this does not
depend on the choice of said permutation, since all permutations in
$\Symm_{\left\vert \lambda\right\vert }$ having cycle type $\lambda$
are mutually conjugate (and thus their centralizers are of the same size). It
is well-known that $\widetilde{z}_{\lambda}=z_{\lambda}$ (see
Remark~\ref{rmk.zlambda.centralizer}), but we shall avoid using this fact, as
we can obtain an alternative proof of it from the following argument.

We shall use the same notations as in the proof of Theorem
\ref{symmetric-group-Frobenius-map-theorem}. In particular, $A_{\mathbb{C}%
}=\bigoplus_{n\geq0}R_\CC \left(  \Symm_n\right)  $; this
$\CC$-vector space $A_{\mathbb{C}}$ becomes a $\CC$-algebra as
explained in the proof of Theorem \ref{symmetric-group-Frobenius-map-theorem}.
Its multiplication is given by $\ind_{i,j}^{i+j}$;
more precisely, every $n\in \NN$ and $m\in \NN$, and every
$\beta\in R_\CC \left(  \Symm_n\right)  $ and $\gamma\in
R_\CC \left(  \Symm_m\right)  $ satisfy%
\begin{equation}
\beta\gamma=\underbrace{\ind_{n,m}^{n+m}%
}_{=\operatorname*{Ind}\nolimits_{\Symm_n\times\Symm_m%
}^{\Symm_{n+m}}}\left(  \beta\otimes\gamma\right)  =\operatorname*{Ind}%
\nolimits_{\Symm_n\times\Symm_m}^{\Symm_{n+m}}\left(
\beta\otimes\gamma\right)  . \label{sol.frob-char.plambda.d.prod}%
\end{equation}


We define a $\CC$-linear map $\Phi:\Lambda_{\mathbb{C}}\rightarrow
A_{\mathbb{C}}$ by setting%
\[
\Phi\left(  p_{\lambda}\right)  =\widetilde{z}_{\lambda}\underline{1}%
_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in \Par .
\]
(This is well-defined, since $\left(  p_{\lambda}\right)  _{\lambda
\in \Par }$ is a basis of the $\CC$-vector space
$\Lambda_{\mathbb{C}}$.) We notice that if $\lambda$ is a partition of a
nonnegative integer $n$, and if $g\in\Symm_n$ is a permutation having
cycle type $\lambda$, then%
\begin{equation}
\Phi\left(  p_{\lambda}\right)  =\alpha_{\Symm_n,g}.
\label{sol.frob-char.plambda.d.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.d.1}):} Let $\lambda$
be a partition of a nonnegative integer $n$. Let $g\in\Symm_n$ be a
permutation having cycle type $\lambda$. Then, $g$ is a permutation in
$\Symm_{\left\vert \lambda\right\vert }$ having cycle type $\lambda$.
Thus, $\widetilde{z}_{\lambda}$ is the size of the centralizer of $g$ (by the
definition of $\widetilde{z}_{\lambda}$). In other words, $\widetilde{z}%
_{\lambda}=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert $.
\par
But any two permutations in $\Symm_n$ having the same cycle type are
mutually conjugate. Hence, if $h$ is any permutation in $\Symm_n$
having cycle type $\lambda$, then $h$ and $g$ are conjugate (since $h$ and $g$
are permutations in $\Symm_n$ having the same cycle type).
Conversely, if $h$ is a permutation in $\Symm_n$ such that $h$ and
$g$ are conjugate, then $h$ has cycle type $\lambda$ (because $h$ and $g$ are
conjugate permutations and thus have the same cycle type, but the cycle type
of $g$ is $\lambda$). Combining these two statements, we conclude that if $h$
is a permutation in $\Symm_n$, then $h$ has cycle type $\lambda$ if
and only if $h$ and $g$ are conjugate. Hence,%
\begin{align*}
\left\{  h\in\Symm_n\ \mid\ \underbrace{h\text{ has cycle type
}\lambda}_{\substack{\text{this is equivalent to}\\\left(  h\text{ and
}g\text{ are conjugate}\right)  }}\right\}   &  =\left\{  h\in\Symm_n
\ \mid\ \underbrace{h\text{ and }g\text{ are conjugate}}_{\text{this is
equivalent to }h\in\operatorname*{Conj}\nolimits_{\Symm_n}\left(
g\right)  }\right\}  =\left\{  h\in\Symm_n\ \mid\ h\in
\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  \right\} \\
&  =\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  .
\end{align*}
\par
On the other hand, $\underline{1}_{\lambda}$ is defined as the characteristic
function for the $\Symm_n$-conjugacy class of permutations of cycle
type $\lambda$. In other words, $\underline{1}_{\lambda}$ is the indicator
function of the subset $\left\{  h\in\Symm_n\ \mid\ h\text{ has cycle
type }\lambda\right\}  $ of $\Symm_n$. In other words, $\underline{1}%
_{\lambda}$ is the indicator function of the subset $\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  $ (since $\left\{
h\in\Symm_n\ \mid\ h\text{ has cycle type }\lambda\right\}
=\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  $). In
other words, $\underline{1}_{\lambda}=\underline{1}_{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }$ (since $\underline{1}%
_{\operatorname*{Conj}\nolimits_{\Symm_n}\left(  g\right)  }$ is
clearly the indicator function of the subset $\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  $).
\par
Now, $\Phi\left(  p_{\lambda}\right)  =\underbrace{\widetilde{z}_{\lambda}%
}_{=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert
}\underbrace{\underline{1}_{\lambda}}_{=\underline{1}_{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }}=\left\vert Z_{\Symm_n}
\left(  g\right)  \right\vert \underline{1}_{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }$. Compared with $\alpha
_{\Symm_n,g}=\left\vert Z_{\Symm_n}\left(  g\right)
\right\vert \underline{1}_{\operatorname*{Conj}\nolimits_{\Symm_n%
}\left(  g\right)  }$ (by the definition of $\alpha_{\Symm_n,g}$),
this yields $\Phi\left(  p_{\lambda}\right)  =\alpha_{\Symm_n,g}$.
This proves (\ref{sol.frob-char.plambda.d.1}).} Also, if $\lambda$ is a
partition of a nonnegative integer $n$, then%
\begin{equation}
\left\vert \left\{  h\in\Symm_n\ \mid\ h\text{ has cycle type
}\lambda\right\}  \right\vert =n!/\widetilde{z}_{\lambda}.
\label{sol.frob-char.plambda.d.1b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.d.1b}):} Let $\lambda$
be a partition of a nonnegative integer $n$. Fix a permutation $g\in
\Symm_n$ having cycle type $\lambda$. (Such a $g$ clearly exists.) In
the proof of (\ref{sol.frob-char.plambda.d.1}), we have seen that
$\widetilde{z}_{\lambda}=\left\vert Z_{\Symm_n}\left(  g\right)
\right\vert $ and that $\left\{  h\in\Symm_n\ \mid\ h\text{ has cycle
type }\lambda\right\}  =\operatorname*{Conj}\nolimits_{\Symm_n%
}\left(  g\right)  $.
\par
Now, it is well-known that for every finite group $G$ and every element $f\in
G$, we have $\left\vert G/Z_{G}\left(  f\right)  \right\vert =\left\vert
\operatorname*{Conj}\nolimits_{G}\left(  f\right)  \right\vert $ (in fact,
there is a canonical bijection from the $G$-set $G/Z_{G}\left(  f\right)  $ to
$\operatorname*{Conj}\nolimits_{G}\left(  f\right)  $, which sends the
equivalence class $\left[  \gamma\right]  \in G/Z_{G}\left(  f\right)  $ of
every $\gamma\in G$ to $\gamma f\gamma^{-1}\in\operatorname*{Conj}%
\nolimits_{G}\left(  f\right)  $). Applying this to $G=\Symm_n$ and
$f=g$, we obtain $\left\vert \Symm_n/Z_{\Symm_n}\left(
g\right)  \right\vert =\left\vert \operatorname*{Conj}\nolimits_{\Symm_n}
\left(  g\right)  \right\vert $. Thus, $\left\vert \operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  \right\vert =\left\vert
\Symm_n/Z_{\Symm_n}\left(  g\right)  \right\vert
=\underbrace{\left\vert \Symm_n\right\vert }_{=n!}%
/\underbrace{\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert
}_{=\widetilde{z}_{\lambda}}=n!/\widetilde{z}_{\lambda}$, so that%
\[
n!/\widetilde{z}_{\lambda}=\left\vert \underbrace{\operatorname*{Conj}%
\nolimits_{\Symm_n}\left(  g\right)  }_{=\left\{  h\in\Symm_n
\ \mid\ h\text{ has cycle type }\lambda\right\}  }\right\vert =\left\vert
\left\{  h\in\Symm_n\ \mid\ h\text{ has cycle type }\lambda\right\}
\right\vert .
\]
This proves (\ref{sol.frob-char.plambda.d.1b}).}

We shall now prove that $\Phi$ is a $\CC$-algebra homomorphism. Since
$\Phi\left(  1\right)  =1$ is true\footnote{This is because
\begin{align*}
\Phi\left(  \underbrace{1}_{=p_{\varnothing}}\right)   &  =\Phi\left(
p_{\varnothing}\right)  =\underbrace{\widetilde{z}_{\varnothing}}%
_{=1}\underbrace{\underline{1}_{\varnothing}}_{=1}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\Phi\left(  p_{\varnothing}\right)  \right) \\
&  =1.
\end{align*}
}, we only need to verify that $\Phi\left(  uv\right)  =\Phi\left(  u\right)
\Phi\left(  v\right)  $ for any $u\in\Lambda_{\mathbb{C}}$ and $v\in
\Lambda_{\mathbb{C}}$. Let us prove this now. Fix $u\in\Lambda_{\mathbb{C}}$
and $v\in\Lambda_{\mathbb{C}}$. Since the equality $\Phi\left(  uv\right)
=\Phi\left(  u\right)  \Phi\left(  v\right)  $ is $\CC$-linear in each
of $u$ and $v$, we can WLOG assume that $u$ and $v$ are elements of the basis
$\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ of the
$\CC$-vector space $\Lambda_{\mathbb{C}}$. Assume this, and set
$u=p_{\mu}$ and $v=p_{\nu}$ for two partitions $\mu$ and $\nu$. Let
$n=\left\vert \mu\right\vert $ and $m=\left\vert \nu\right\vert $. Let $g$ be
a permutation in $\Symm_n$ having cycle type $\mu$. (Such a $g$
clearly exists.) Let $h$ be a permutation in $\Symm_m$ having cycle
type $\nu$. (Such an $h$ clearly exists.) Applying
(\ref{sol.frob-char.plambda.d.1}) to $\mu$ instead of $\lambda$, we obtain
$\Phi\left(  p_{\mu}\right)  =\alpha_{\Symm_n,g}$. Thus,
\begin{equation}
\Phi\left(  \underbrace{u}_{=p_{\mu}}\right)  =\Phi\left(  p_{\mu}\right)
=\alpha_{\Symm_n,g}. \label{sol.frob-char.plambda.d.Phiu}%
\end{equation}
Applying (\ref{sol.frob-char.plambda.d.1}) to $\nu$, $m$ and $h$ instead of
$\lambda$, $n$ and $g$, we obtain $\Phi\left(  p_{\nu}\right)  =\alpha
_{\Symm_m,h}$. Thus,%
\begin{equation}
\Phi\left(  \underbrace{v}_{=p_{\nu}}\right)  =\Phi\left(  p_{\nu}\right)
=\alpha_{\Symm_m,h}. \label{sol.frob-char.plambda.d.Phiv}%
\end{equation}


The canonical isomorphism $R_\CC \left(  \Symm_n\right)
\otimes R_{\mathbb{C}}\left(  \Symm_m\right)  \rightarrow
R_{\mathbb{C}}\left(  \Symm_n\times\Symm_m\right)  $ sends
$\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}$ to
$\alpha_{\Symm_n\times\Symm_m,\left(  g,h\right)  }$
(according to Exercise \ref{exe.frob-char.plambda}(c), applied to
$G_{1}=\Symm_n$, $G_{2}=\Symm_m$, $h_{1}=g$ and $h_{2}=h$).
Let us identify $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  $ with $R_{\mathbb{C}}\left(
\Symm_n\times\Symm_m\right)  $ along this isomorphism. Then,
the statement we just made rewrites as follows:%
\begin{equation}
\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}=\alpha
_{\Symm_n\times\Symm_m,\left(  g,h\right)  }.
\label{sol.frob-char.plambda.d.2}%
\end{equation}


Now, multiplying the equalities (\ref{sol.frob-char.plambda.d.Phiu}) and
(\ref{sol.frob-char.plambda.d.Phiv}), we obtain%
\begin{align}
\Phi\left(  u\right)  \Phi\left(  v\right)   &  =\alpha_{\Symm_n%
,g}\alpha_{\Symm_m,h}\nonumber\\
&  =\operatorname*{Ind}\nolimits_{\Symm_n\times\Symm_m%
}^{\Symm_{n+m}}\underbrace{\left(  \alpha_{\Symm_n,g}%
\otimes\alpha_{\Symm_m,h}\right)  }_{\substack{=\alpha_{\Symm_n
\times\Symm_m,\left(  g,h\right)  }\\\text{(by
(\ref{sol.frob-char.plambda.d.2}))}}}\nonumber\\
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.frob-char.plambda.d.prod}), applied to }\beta=\alpha_{\Symm_n,
g}\text{ and }\gamma=\alpha_{\Symm_m,h}\right) \nonumber\\
&  =\operatorname*{Ind}\nolimits_{\Symm_n\times\Symm_m%
}^{\Symm_{n+m}}\alpha_{\Symm_n\times\Symm_m,\left(
g,h\right)  }=\alpha_{\Symm_{n+m},\left(  g,h\right)  }
\label{sol.frob-char.plambda.d.5}%
\end{align}
(by Exercise \ref{exe.frob-char.plambda}(b), applied to $\Symm_{n+m}$,
$\Symm_n\times\Symm_m$ and $\left(  g,h\right)  $ instead of
$G$, $H$ and $h$).

Let $\lambda$ be the partition whose parts are $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{\ell\left(  \mu\right)  }$, $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu
_{\ell\left(  \nu\right)  }$. Then, $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell\left(  \lambda\right)  }\right)  $ is a permutation of
the list $\left(  \mu_{1},\mu_{2},\ldots,\mu_{\ell\left(  \mu\right)  }%
,\nu_{1},\nu_{2},\ldots,\nu_{\ell\left(  \nu\right)  }\right)  $, and the
partition $\lambda$ has size
\[
\left\vert \lambda\right\vert =\underbrace{\mu_{1}+\mu_{2}+\cdots+\mu
_{\ell\left(  \mu\right)  }}_{=\left\vert \mu\right\vert =n}+\underbrace{\nu
_{1}+\nu_{2}+\cdots+\nu_{\ell\left(  \nu\right)  }}_{=\left\vert
\nu\right\vert =m}=n+m.
\]
We also have $p_{\lambda}=p_{\mu}p_{\nu}$\ \ \ \ \footnote{\textit{Proof.} By
the definition of $p_{\mu}$, we have $p_{\mu}=p_{\mu_{1}}p_{\mu_{2}}\cdots
p_{\mu_{\ell\left(  \mu\right)  }}$. Similarly, $p_{\nu}=p_{\nu_{1}}p_{\nu
_{2}}\cdots p_{\nu_{\ell\left(  \nu\right)  }}$. Multiplying these two
equalities, we obtain $p_{\mu}p_{\nu}=\left(  p_{\mu_{1}}p_{\mu_{2}}\cdots
p_{\mu_{\ell\left(  \mu\right)  }}\right)  \left(  p_{\nu_{1}}p_{\nu_{2}%
}\cdots p_{\nu_{\ell\left(  \nu\right)  }}\right)  $.
\par
But the definition of $p_{\lambda}$ yields $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(  \lambda\right)  }}$. The
product $p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(
\lambda\right)  }}$ has the same factors as the product $p_{\mu_{1}}p_{\mu
_{2}}\cdots p_{\mu_{\ell\left(  \mu\right)  }}p_{\nu_{1}}p_{\nu_{2}}\cdots
p_{\nu_{\ell\left(  \nu\right)  }}$ but possibly in a different order (since
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)
}\right)  $ is a permutation of the list $\left(  \mu_{1},\mu_{2},\ldots
,\mu_{\ell\left(  \mu\right)  },\nu_{1},\nu_{2},\ldots,\nu_{\ell\left(
\nu\right)  }\right)  $). Hence, $p_{\lambda_{1}}p_{\lambda_{2}}\cdots
p_{\lambda_{\ell\left(  \lambda\right)  }}=p_{\mu_{1}}p_{\mu_{2}}\cdots
p_{\mu_{\ell\left(  \mu\right)  }}p_{\nu_{1}}p_{\nu_{2}}\cdots p_{\nu
_{\ell\left(  \nu\right)  }}$, so that%
\begin{align*}
p_{\lambda}  &  =p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(
\lambda\right)  }}=p_{\mu_{1}}p_{\mu_{2}}\cdots p_{\mu_{\ell\left(
\mu\right)  }}p_{\nu_{1}}p_{\nu_{2}}\cdots p_{\nu_{\ell\left(  \nu\right)  }}
=\left(  p_{\mu_{1}}p_{\mu_{2}}\cdots p_{\mu_{\ell\left(  \mu\right)  }%
}\right)  \left(  p_{\nu_{1}}p_{\nu_{2}}\cdots p_{\nu_{\ell\left(  \nu\right)
}}\right)  =p_{\mu}p_{\nu},
\end{align*}
qed.}. Since $p_{\mu}=u$ and $p_{\nu}=v$, this rewrites as $p_{\lambda}=uv$.

Recall that the permutation $g$ has cycle type $\mu$. In other words, the
cycles of $g$ have lengths $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{\ell\left(
\mu\right)  }$. Similarly, the cycles of $h$ have lengths $\nu_{1}$, $\nu_{2}%
$, $\ldots$, $\nu_{\ell\left(  \nu\right)  }$. Hence, the cycles of the
permutation $\left(  g,h\right)  \in\Symm_{n+m}$ (which, as we recall,
sends every $i\in\left\{  1,2,\ldots,n+m\right\}  $ to $\begin{cases}
g\left(  i\right)  ,& \text{if }i\leq n;\\
n+h\left(  i-n\right)  ,& \text{if }i>n
\end{cases}$) have lengths $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{\ell\left(
\mu\right)  }$, $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{\ell\left(  \nu\right)
}$ (in fact, on the subset $\left\{  1,2,\ldots,n\right\}  $ of $\left\{
1,2,\ldots,n+m\right\}  $, the permutation $\left(  g,h\right)  $ acts as $g$
and thus has cycles of lengths $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu
_{\ell\left(  \mu\right)  }$, whereas on the complementary subset $\left\{
n+1,n+2,\ldots,n+m\right\}  $ of $\left\{  1,2,\ldots,n+m\right\}  $, the
permutation $\left(  g,h\right)  $ acts as (a shifted version of) $h$ and thus
has cycles of lengths $\nu_{1}$, $\nu_{2}$, $\ldots$, $\nu_{\ell\left(
\nu\right)  }$). In other words, the cycles of the permutation $\left(
g,h\right)  \in\Symm_{n+m}$ have lengths $\lambda_{1}$, $\lambda_{2}$,
$\ldots$, $\lambda_{\ell\left(  \lambda\right)  }$ (since $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right)  $ is a
permutation of the list $\left(  \mu_{1},\mu_{2},\ldots,\mu_{\ell\left(
\mu\right)  },\nu_{1},\nu_{2},\ldots,\nu_{\ell\left(  \nu\right)  }\right)
$). In other words, the permutation $\left(  g,h\right)  \in\Symm_{n+m}$
has cycle type $\lambda$. Thus, (\ref{sol.frob-char.plambda.d.1})
(applied to $n+m$ and $\left(  g,h\right)  $ instead of $n$ and $g$) yields
$\Phi\left(  p_{\lambda}\right)  =\alpha_{\Symm_{n+m},\left(
g,h\right)  }$. Compared with (\ref{sol.frob-char.plambda.d.5}), this yields
$\Phi\left(  u\right)  \Phi\left(  v\right)  =\Phi\left(  p_{\lambda}\right)
$. Since $p_{\lambda}=uv$, this rewrites as $\Phi\left(  u\right)  \Phi\left(
v\right)  =\Phi\left(  uv\right)  $. Thus, $\Phi\left(  uv\right)
=\Phi\left(  u\right)  \Phi\left(  v\right)  $ is proven. As explained, this
completes the proof of the fact that $\Phi$ is a $\CC$-algebra homomorphism.

But we know that $\ch : A \rightarrow \Lambda$ is a
$\ZZ$-Hopf algebra isomorphism, and thus the extension of
$\ch$ to a $\CC$-linear map $A_{\CC}\rightarrow\Lambda_{\CC}$ is a
$\CC$-Hopf algebra isomorphism. We shall denote this extension by
$\ch_{\CC}$.

We now shall show that $\ch_{\CC} \circ
\Phi=\id_{\Lambda_{\mathbb{C}}}$. Indeed, let $n$ be a
positive integer. Then, $\widetilde{z}_{\left(  n\right)  }=n$%
\ \ \ \ \footnote{\textit{Proof.} Let $g$ denote the $n$-cycle $\left(
1,2,\ldots,n\right)  $ in $\Symm_n$. Then, $g$ is a permutation in
$\Symm_n$ having cycle type $\left(  n\right)  $. But we know that
$\widetilde{z}_{\left(  n\right)  }$ is the size of the centralizer of a
permutation in $\Symm_{\left\vert \left(  n\right)  \right\vert }$
having cycle type $\left(  n\right)  $ (by the definition of $\widetilde{z}%
_{\left(  n\right)  }$). In other words, $\widetilde{z}_{\left(  n\right)  }$
is the size of the centralizer of a permutation in $\Symm_n$ having
cycle type $\left(  n\right)  $ (since $\left\vert \left(  n\right)
\right\vert =n$). Thus, $\widetilde{z}_{\left(  n\right)  }$ is the size of
the centralizer of $g$ in $\Symm_n$ (since $g$ is a permutation in
$\Symm_n$ having cycle type $\left(  n\right)  $). In other words,
$\widetilde{z}_{\left(  n\right)  }=\left\vert Z_{\Symm_n}\left(
g\right)  \right\vert $.
\par
It is clear that the subgroup $\left\langle g\right\rangle $ of $\Symm_n$
generated by $g$ satisfies $\left\langle g\right\rangle \subset
Z_{\Symm_n}\left(  g\right)  $ (since every power of $g$ centralizes
$g$). We shall now show that $\left\langle g\right\rangle =
Z_{\Symm_n}\left(  g\right)  $.
\par
Indeed, let $z\in Z_{\Symm_n}\left(  g\right)  $. Then, $z$ must
centralize $g$. That is, we have $zgz^{-1}=g$. Hence, $zg=gz$, so that the
elements $g$ and $z$ of $\Symm_n$ commute. Hence, these elements $z$
and $g$ generate a commutative subgroup of $\Symm_n$. Denote this
subgroup by $T$.
\par
Now, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
\begin{equation}
g^{i-1}\left(  1\right)  =i \label{sol.frob-char.plambda.d.n-cycle.i}%
\end{equation}
(since $g$ is the $n$-cycle $\left(  1,2,\ldots,n\right)  $). Now, let
$j=z\left(  1\right)  $. Then, $g^{j-1}\left(  1\right)  =j$ (by
(\ref{sol.frob-char.plambda.d.n-cycle.i}), applied to $i=j$). On the other
hand, let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, the elements $g^{i-1}$
and $z$ belong to the subgroup $T$ of $\Symm_n$ (since $g$ and $z$
belong to $T$) and thus must commute (since $T$ is commutative); that is, we
have $g^{i-1}z=zg^{i-1}$. Now,%
\begin{align*}
z\left(  \underbrace{i}_{\substack{=g^{i-1}\left(  1\right)  \\\text{(by
(\ref{sol.frob-char.plambda.d.n-cycle.i}))}}}\right)   &  =z\left(
g^{i-1}\left(  1\right)  \right)  =\underbrace{\left(  zg^{i-1}\right)
}_{=g^{i-1}z}\left(  1\right)  =\left(  g^{i-1}z\right)  \left(  1\right)
=g^{i-1}\left(  \underbrace{z\left(  1\right)  }_{=j=g^{j-1}\left(  1\right)
}\right) \\
&  =g^{i-1}\left(  g^{j-1}\left(  1\right)  \right)  =\underbrace{\left(
g^{i-1}g^{j-1}\right)  }_{=g^{\left(  i-1\right)  +\left(  j-1\right)
}=g^{j-1}g^{i-1}}\left(  1\right)  =\left(  g^{j-1}g^{i-1}\right)  \left(
1\right)  =g^{j-1}\left(  \underbrace{g^{i-1}\left(  1\right)  }%
_{\substack{=i\\\text{(by (\ref{sol.frob-char.plambda.d.n-cycle.i}))}%
}}\right)  =g^{j-1}\left(  i\right)  .
\end{align*}
\par
Let us now forget that we fixed $i$. We thus have shown that $z\left(
i\right)  =g^{j-1}\left(  i\right)  $ for every $i\in\left\{  1,2,\ldots
,n\right\}  $. Hence, $z=g^{j-1}\in\left\langle g\right\rangle $.
\par
Now, let us forget that we fixed $z$. We thus have proven that $z\in
\left\langle g\right\rangle $ for every $z\in Z_{\Symm_n}\left(
g\right)  $. Hence, $Z_{\Symm_n}\left(  g\right)  \subset\left\langle
g\right\rangle $. Combined with $\left\langle g\right\rangle \subset
Z_{\Symm_n}\left(  g\right)  $, this yields $\left\langle
g\right\rangle =Z_{\Symm_n}\left(  g\right)  $. Hence, $\left\vert
\left\langle g\right\rangle \right\vert =\left\vert Z_{\Symm_n%
}\left(  g\right)  \right\vert $. Compared with $\widetilde{z}_{\left(
n\right)  }=\left\vert Z_{\Symm_n}\left(  g\right)  \right\vert $,
this yields $\widetilde{z}_{\left(  n\right)  }=\left\vert \left\langle
g\right\rangle \right\vert =\left(  \text{the order of }g\text{ in
}\Symm_n\right)  =n$ (since $g$ is an $n$-cycle), qed.}. Now, the
definition of $\Phi\left(  p_{\left(  n\right)  }\right)  $ yields
$\Phi\left(  p_{\left(  n\right)  }\right)  =\underbrace{\widetilde{z}%
_{\left(  n\right)  }}_{=n}\underline{1}_{\left(  n\right)  }=n\underline{1}%
_{\left(  n\right)  }$, so that%
\begin{align*}
\left(  \ch_{\CC} \circ\Phi\right)  \left(
p_{n}\right)   &  =\ch_{\CC} \left(  \Phi\left(
\underbrace{p_{n}}_{=p_{\left(  n\right)  }}\right)  \right)
=\ch_{\CC} \left(  \underbrace{\Phi\left(
p_{\left(  n\right)  }\right)  }_{=n\underline{1}_{\left(  n\right)  }%
}\right)  =\ch_{\CC} \left(  n\underline{1}%
_{\left(  n\right)  }\right)  =n\underbrace{\ch_{\CC}
\left(  \underline{1}_{\left(  n\right)  }\right)
}_{\substack{=\dfrac{p_{n}}{n}\\\text{(by a part of
Theorem~\ref{symmetric-group-Frobenius-map-theorem}}\\\text{that is already
proven)}}}\\
&  =n\cdot\dfrac{p_{n}}{n}=p_{n}=\id_{\Lambda
_{\mathbb{C}}}\left(  p_{n}\right)  .
\end{align*}


Now, let us forget that we fixed $n$. We thus have shown that $\left(
\ch_{\CC} \circ\Phi\right)  \left(
p_{n}\right)  =\id_{\Lambda_{\mathbb{C}}}\left(
p_{n}\right)  $ for every positive integer $n$. Thus, the $\CC$-algebra
homomorphisms $\ch_{\CC} \circ\Phi$ and
$\id_{\Lambda_{\mathbb{C}}}$ are equal to each other
on $p_{n}$ for every positive integer $n$. Since $\left(  p_{n}\right)
_{n\geq1}$ is a generating set of the $\CC$-algebra $\Lambda
_{\mathbb{C}}$, this yields that these homomorphisms are equal to each other
on a generating set of the $\CC$-algebra $\Lambda_{\mathbb{C}}$. Hence,
these homomorphisms must be identical. That is, we have
$\ch_{\CC} \circ\Phi = \id_{\Lambda_{\mathbb{C}}}$.

Since $\ch_{\CC}$ is an isomorphism, this
yields that $\Phi$ is the inverse of $\ch_{\CC}$.
That is, $\Phi=\left(  \ch_{\CC} \right) ^{-1}$.

Corollary \ref{dual-bases-h-m-p}(b) (applied to $\kk=\mathbb{C}$)
yields that $\{p_{\lambda}\}$ and $\left\{  z_{\lambda}^{-1}p_{\lambda
}\right\}  $ are dual bases of $\Lambda_{\mathbb{C}}$ with respect to the Hall
inner product on $\Lambda$. Thus,%
\begin{equation}
\left(  p_{\lambda},z_{\mu}^{-1}p_{\mu}\right)  _{\Lambda_{\mathbb{C}}}%
=\delta_{\lambda,\mu}\ \ \ \ \ \ \ \ \ \ \text{for any partitions }%
\lambda\text{ and }\mu\text{.} \label{sol.frob-char.plambda.d.dual-bases-1}%
\end{equation}
Hence, every partition $\lambda$ satisfies%
\begin{equation}
\left(  p_{\lambda},\underbrace{p_{\lambda}}_{=z_{\lambda}z_{\lambda}%
^{-1}p_{\lambda}}\right)  _{\Lambda_{\mathbb{C}}}=\left(  p_{\lambda
},z_{\lambda}z_{\lambda}^{-1}p_{\lambda}\right)  _{\Lambda_{\mathbb{C}}%
}=z_{\lambda}\underbrace{\left(  p_{\lambda},z_{\lambda}^{-1}p_{\lambda
}\right)  _{\Lambda_{\mathbb{C}}}}_{\substack{=\delta_{\lambda,\lambda
}\\\text{(by (\ref{sol.frob-char.plambda.d.dual-bases-1}), applied}\\\text{to
}\mu=\lambda\text{)}}}=z_{\lambda}\underbrace{\delta_{\lambda,\lambda}}%
_{=1}=z_{\lambda}. \label{sol.frob-char.plambda.d.dual-bases-2}%
\end{equation}


Our goal is to prove that $\ch_{\CC} \left(
\underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$ for every
partition $\lambda$.

Let $\lambda$ be a partition. Let $n$ be the size of $\lambda$. Then,
$\lambda\in\Par_{n}\subset \Par $ and
$\underline{1}_{\lambda}\in R_{\mathbb{C}}\left(  \Symm_n\right)  $.
We know that $\widetilde{z}_{\lambda}$ is a positive integer (since
$\widetilde{z}_{\lambda}$ is defined as the size of a centralizer, and
centralizers are subgroups). Hence, we can divide by $\widetilde{z}_{\lambda}%
$, and we have%
\begin{align*}
\Phi\left(  \widetilde{z}_{\lambda}^{-1}p_{\lambda}\right)   &  =\widetilde{z}%
_{\lambda}^{-1}\underbrace{\Phi\left(  p_{\lambda}\right)  }%
_{\substack{=\widetilde{z}_{\lambda}\underline{1}_{\lambda}\\\text{(by the
definition of }\Phi\left(  p_{\lambda}\right)  \text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\text{ is }\mathbb{C}%
\text{-linear}\right) \\
&  =\widetilde{z}_{\lambda}^{-1}\widetilde{z}_{\lambda}\underline{1}_{\lambda
}=\underline{1}_{\lambda}.
\end{align*}
Since $\Phi=\left(  \ch_{\CC} \right)  ^{-1}$,
this rewrites as $\left(  \ch_{\CC} \right)
^{-1}\left(  \widetilde{z}_{\lambda}^{-1}p_{\lambda}\right)  =\underline{1}%
_{\lambda}$. Hence,
\begin{equation}
\ch_{\CC} \left(  \underline{1}_{\lambda
}\right)  =\widetilde{z}_{\lambda}^{-1}p_{\lambda}.
\label{sol.frob-char.plambda.d.halfthere}%
\end{equation}


Recall that we want to prove that $\ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$.
If we can show that $\widetilde{z}_{\lambda}=z_{\lambda}$, then
(\ref{sol.frob-char.plambda.d.halfthere}) becomes $\ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  =\left(
\underbrace{\widetilde{z}_{\lambda}}_{=z_{\lambda}}\right)  ^{-1}p_{\lambda
}=z_{\lambda}^{-1}p_{\lambda}=\dfrac{p_{\lambda}}{z_{\lambda}}$, and thus
$\ch_{\CC} \left(  \underline{1}_{\lambda
}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$ will be proven. Hence, all that
remains to be done is proving $\widetilde{z}_{\lambda}=z_{\lambda}$.

But $\ch$ is a PSH-isomorphism, thus an isometry. Hence,
$\ch_{\CC} $ (being the extension of
$\ch$ to a $\CC$-linear map) must also be an isometry,
i.e., we must have
\[
\left(  \ch_{\CC} \beta, \ch_{\CC} \gamma
\right)  _{\Lambda_{\mathbb{C}}}=\left(
\beta,\gamma\right)  _{A_{\mathbb{C}}}\ \ \ \ \ \ \ \ \ \ \text{for all }%
\beta\in A_{\mathbb{C}}\text{ and }\gamma\in A_{\mathbb{C}}.
\]
Applying this to $\beta=\underline{1}_{\lambda}$ and $\gamma=\underline{1}%
_{\lambda}$, we obtain%
\begin{align*}
&  \left(  \ch_{\CC} \left(  \underline{1}%
_{\lambda}\right)  ,\ch_{\CC} \left(
\underline{1}_{\lambda}\right)  \right)  _{\Lambda_{\mathbb{C}}}\\
&  =\left(  \underline{1}_{\lambda},\underline{1}_{\lambda}\right)
_{A_{\mathbb{C}}}=\left\langle \underline{1}_{\lambda},\underline{1}_{\lambda
}\right\rangle _{\Symm_n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\underline{1}_{\lambda}\in R_{\mathbb{C}}\left(  \Symm_n
\right)  \text{, and since the bilinear form}\\
\left(  \cdot,\cdot\right)  _{A_{\mathbb{C}}}\text{ on }A_{\mathbb{C}}\text{
extends the bilinear form }\left\langle \cdot,\cdot\right\rangle
_{\Symm_n}\text{ on }R_{\mathbb{C}}\left(  \Symm_n\right)
\end{array}
\right) \\
&  =\underbrace{\dfrac{1}{\left\vert \Symm_n\right\vert }}%
_{=\dfrac{1}{n!}}\sum_{g\in\Symm_n}\underbrace{\underline{1}%
_{\lambda}\left(  g\right)  }_{\substack{=\left[  g\text{ has cycle type
}\lambda\right]  \\\text{(by the definition of }\underline{1}_{\lambda
}\text{)}}}\underbrace{\underline{1}_{\lambda}\left(  g^{-1}\right)
}_{\substack{=\left[  g^{-1}\text{ has cycle type }\lambda\right]  \\\text{(by
the definition of }\underline{1}_{\lambda}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the bilinear form
}\left\langle \cdot,\cdot\right\rangle _{\Symm_n}\right) \\
&  =\dfrac{1}{n!}\sum_{g\in\Symm_n}\left[  g\text{ has cycle type
}\lambda\right]  \left[  \underbrace{g^{-1}\text{ has cycle type }\lambda
}_{\substack{\text{this is equivalent to}\\\left(  g\text{ has cycle type
}\lambda\right)  \\\text{(since the cycle type of }g^{-1}\\\text{equals the
cycle type of }g\text{)}}}\right]  =\dfrac{1}{n!}\sum_{g\in\Symm_n%
}\underbrace{\left[  g\text{ has cycle type }\lambda\right]  \left[  g\text{
has cycle type }\lambda\right]  }_{=\left[  \left(  g\text{ has cycle type
}\lambda\right)  \text{ and }\left(  g\text{ has cycle type }\lambda\right)
\right]  }\\
&  =\dfrac{1}{n!}\sum_{g\in\Symm_n}\left[  \underbrace{\left(
g\text{ has cycle type }\lambda\right)  \text{ and }\left(  g\text{ has cycle
type }\lambda\right)  }_{\substack{\text{this is equivalent to}\\\left(
g\text{ has cycle type }\lambda\right)  }}\right]  =\dfrac{1}{n!}\sum
_{g\in\Symm_n}\left[  g\text{ has cycle type }\lambda\right]  ,
\end{align*}
so that%
\begin{align*}
&  n!\cdot\left(  \ch_{\CC} \left(
\underline{1}_{\lambda}\right)  , \ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  \right)  _{\Lambda_{\mathbb{C}}}\\
&  =\sum_{g\in\Symm_n}\left[  g\text{ has cycle type }\lambda\right]
\\
&  =\sum\limits_{\substack{g\in\Symm_n;\\g\text{ has cycle type }\lambda
}}\underbrace{\left[  g\text{ has cycle type }\lambda\right]  }%
_{\substack{=1\\\text{(since }g\text{ has cycle type }\lambda\text{)}}%
}+\sum\limits_{\substack{g\in\Symm_n;\\g\text{ does not have cycle type
}\lambda}}\underbrace{\left[  g\text{ has cycle type }\lambda\right]
}_{\substack{=0\\\text{(since }g\text{ does not have cycle type }%
\lambda\text{)}}}\\
&  =\sum\limits_{\substack{g\in\Symm_n;\\g\text{ has cycle type }\lambda
}}1+\underbrace{\sum\limits_{\substack{g\in\Symm_n;\\g\text{ does not have
cycle type }\lambda}}0}_{=0}=\sum\limits_{\substack{g\in\Symm_n;\\g\text{
has cycle type }\lambda}}1=\sum\limits_{\substack{h\in\Symm_n;\\h\text{ has
cycle type }\lambda}}1\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}g\text{ as }h\right) \\
&  =\left\vert \left\{  h\in\Symm_n\ \mid\ h\text{ has cycle type
}\lambda\right\}  \right\vert =n!/\widetilde{z}_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.frob-char.plambda.d.1b}%
)}\right)  .
\end{align*}
Dividing both sides of this equality by $n!$, we obtain $\left(
\ch_{\CC} \left(  \underline{1}_{\lambda
}\right)  ,\ch_{\CC} \left(  \underline{1}%
_{\lambda}\right)  \right)  _{\Lambda_{\mathbb{C}}}=1/\widetilde{z}_{\lambda}%
$. Hence,%
\[
1/\widetilde{z}_{\lambda}=\left(  \underbrace{ \ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  }%
_{\substack{=\widetilde{z}_{\lambda}^{-1}p_{\lambda}\\\text{(by
(\ref{sol.frob-char.plambda.d.halfthere}))}}},\underbrace{ \ch_{\CC}
\left(  \underline{1}_{\lambda}\right)  }%
_{\substack{=\widetilde{z}_{\lambda}^{-1}p_{\lambda}\\\text{(by
(\ref{sol.frob-char.plambda.d.halfthere}))}}}\right)  _{\Lambda_{\mathbb{C}}%
}=\left(  \widetilde{z}_{\lambda}^{-1}p_{\lambda},\widetilde{z}_{\lambda}%
^{-1}p_{\lambda}\right)  _{\Lambda_{\mathbb{C}}}=\left(  \widetilde{z}%
_{\lambda}^{-1}\right)  ^{2}\left(  p_{\lambda},p_{\lambda}\right)
_{\Lambda_{\mathbb{C}}}.
\]
Multiplying this equality with $\widetilde{z}_{\lambda}^{2}$, we obtain%
\[
\widetilde{z}_{\lambda}=\left(  p_{\lambda},p_{\lambda}\right)  _{\Lambda
_{\mathbb{C}}}=z_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.frob-char.plambda.d.dual-bases-2})}\right)  .
\]
Thus, $\widetilde{z}_{\lambda}=z_{\lambda}$ is proven. As we have explained,
this concludes the proof of $\ch_{\CC} \left(
\underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}}{z_{\lambda}}$, and thus
Exercise \ref{exe.frob-char.plambda}(d) is solved.

(e) Let $\lambda$ be a partition. Let us work with the notations of the
solution of Exercise \ref{exe.frob-char.plambda}(d) above. In the latter
solution, we have shown that $\widetilde{z}_{\lambda}=z_{\lambda}$. Thus,%
\begin{align*}
z_{\lambda}  &  =\widetilde{z}_{\lambda}=\left(  \text{the size of the
centralizer of a permutation in }\Symm_{\left\vert \lambda\right\vert
}\text{ having cycle type }\lambda\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\widetilde{z}%
_{\lambda}\right) \\
&  =\left(  \text{the size of the centralizer in }\Symm_n\text{ of a
permutation having cycle type }\lambda\text{, where }n=\left\vert
\lambda\right\vert \right)  .
\end{align*}
In other words, $z_{\lambda}$ is the size of the centralizer in $\Symm_n$
of a permutation having cycle type $\lambda$, where $n=\left\vert
\lambda\right\vert $. This proves Remark \ref{rmk.zlambda.centralizer}. Thus,
Exercise \ref{exe.frob-char.plambda}(e) is solved.

(f) Let $G$ and $H$ be two finite groups. Let $\rho:H\rightarrow G$ be a group
homomorphism. Let $y\in H$. We shall show that $\operatorname*{Ind}%
\nolimits_{\rho}\alpha_{H,y}=\alpha_{G,\rho\left(  y\right)  }$.

Let $u\in G$. Then, the definition of $\Ind_{\rho
}\alpha_{H,y}$ yields%
\begin{align}
\left(  \Ind_{\rho}\alpha_{H,y}\right)  \left(
u\right)   &  =\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{\left(
h,k\right)  \in H\times G;\\k\rho\left(  h\right)  k^{-1}=u}}\alpha
_{H,y}\left(  h\right)  =\dfrac{1}{\left\vert H\right\vert }\sum\limits
_{\substack{\left(  h,x\right)  \in H\times G;\\x\rho\left(  h\right)
x^{-1}=u}}\underbrace{\alpha_{H,y}\left(  h\right)  }_{\substack{=\sum_{k\in
H}\left[  kyk^{-1}=h\right]  \\\text{(by Exercise \ref{exe.frob-char.plambda}%
(a), applied to}\\H\text{, }y\text{ and }h\text{ instead of }G\text{, }h\text{
and }g\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  h,k\right)  \text{ as }\left(  h,x\right)  \right) \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum\limits_{\substack{\left(
h,x\right)  \in H\times G;\\x\rho\left(  h\right)  x^{-1}=u}}}_{=\sum_{x\in
G}\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}}\sum_{k\in
H}\left[  kyk^{-1}=h\right] \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{x\in G}\underbrace{\sum\limits
_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\sum_{k\in H}}%
_{=\sum_{k\in H}\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}%
}}\left[  kyk^{-1}=h\right] \nonumber\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{x\in G}\sum_{k\in H}%
\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]  . \label{sol.frob-char.plambda.1}%
\end{align}
But every $x\in G$ and $k\in H$ satisfy%
\begin{equation}
\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]  =\left[  x\rho\left(  k\right)  \rho\left(  y\right)
\left(  x\rho\left(  k\right)  \right)  ^{-1}=u\right]
\label{sol.frob-char.plambda.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.frob-char.plambda.2}):} Let $x\in G$ and
$k\in H$. We have%
\begin{align*}
&  \sum_{h\in H}\left[  x\rho\left(  h\right)  x^{-1}=u\right]  \left[
kyk^{-1}=h\right] \\
&  =\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}%
}\underbrace{\left[  x\rho\left(  h\right)  x^{-1}=u\right]  }%
_{\substack{=1\\\text{(since we have }x\rho\left(  h\right)  x^{-1}=u\text{)}%
}}\left[  kyk^{-1}=h\right]  +\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)
x^{-1}\neq u}}\underbrace{\left[  x\rho\left(  h\right)  x^{-1}=u\right]
}_{\substack{=0\\\text{(since we don't have }x\rho\left(  h\right)
x^{-1}=u\\\text{(since }x\rho\left(  h\right)  x^{-1}\neq u\text{))}}}\left[
kyk^{-1}=h\right] \\
&  =\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]  +\underbrace{\sum\limits_{\substack{h\in H;\\x\rho\left(
h\right)  x^{-1}\neq u}}0\left[  kyk^{-1}=h\right]  }_{=0}=\sum\limits
_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[  kyk^{-1}%
=h\right]  ,
\end{align*}
so that%
\begin{align*}
\sum\limits_{\substack{h\in H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[
kyk^{-1}=h\right]   &  =\sum_{h\in H}\left[  x\rho\left(  h\right)
x^{-1}=u\right]  \left[  kyk^{-1}=h\right] \\
&  =\sum\limits_{\substack{h\in H;\\h=kyk^{-1}}}\left[  x\rho\left(  h\right)
x^{-1}=u\right]  \underbrace{\left[  kyk^{-1}=h\right]  }%
_{\substack{=1\\\text{(since we have }kyk^{-1}=h\\\text{(since }%
h=kyk^{-1}\text{))}}}+\sum\limits_{\substack{h\in H;\\h\neq kyk^{-1}}}\left[
x\rho\left(  h\right)  x^{-1}=u\right]  \underbrace{\left[  kyk^{-1}=h\right]
}_{\substack{=0\\\text{(since we don't have }kyk^{-1}=h\\\text{(since
}kyk^{-1}\neq h\text{ (since }h\neq kyk^{-1}\text{)))}}}\\
&  =\sum\limits_{\substack{h\in H;\\h=kyk^{-1}}}\left[  x\rho\left(  h\right)
x^{-1}=u\right]  +\underbrace{\sum\limits_{\substack{h\in H;\\h\neq kyk^{-1}}}\left[
x\rho\left(  h\right)  x^{-1}=u\right]  0}_{=0}=\sum\limits_{\substack{h\in
H;\\h=kyk^{-1}}}\left[  x\rho\left(  h\right)  x^{-1}=u\right] \\
&  =\left[  x\underbrace{\rho\left(  kyk^{-1}\right)  }_{\substack{=\rho
\left(  k\right)  \rho\left(  y\right)  \left(  \rho\left(  k\right)  \right)
^{-1}\\\text{(since }\rho\text{ is a group}\\\text{homomorphism)}}%
}x^{-1}=u\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }kyk^{-1}\in H\text{
(since }k\in H\text{ and }y\in H\text{)}\right) \\
&  =\left[  x\rho\left(  k\right)  \rho\left(  y\right)  \underbrace{\left(
\rho\left(  k\right)  \right)  ^{-1}x^{-1}}_{=\left(  x\rho\left(  k\right)
\right)  ^{-1}}=u\right]  =\left[  x\rho\left(  k\right)  \rho\left(
y\right)  \left(  x\rho\left(  k\right)  \right)  ^{-1}=u\right]  .
\end{align*}
This proves (\ref{sol.frob-char.plambda.2}).}. Hence,
(\ref{sol.frob-char.plambda.1}) becomes%
\begin{align*}
\left(  \Ind_{\rho}\alpha_{H,y}\right)  \left(
u\right)   &  =\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum_{x\in
G}\sum_{k\in H}}_{=\sum_{k\in H}\sum_{x\in G}}\underbrace{\sum\limits_{\substack{h\in
H;\\x\rho\left(  h\right)  x^{-1}=u}}\left[  kyk^{-1}=h\right]  }%
_{\substack{=\left[  x\rho\left(  k\right)  \rho\left(  y\right)  \left(
x\rho\left(  k\right)  \right)  ^{-1}=u\right]  \\\text{(by
(\ref{sol.frob-char.plambda.2}))}}}\\
&  =\dfrac{1}{\left\vert H\right\vert }\sum_{k\in H}\underbrace{\sum_{x\in
G}\left[  x\rho\left(  k\right)  \rho\left(  y\right)  \left(  x\rho\left(
k\right)  \right)  ^{-1}=u\right]  }_{\substack{=\sum_{x\in G}\left[
x\rho\left(  y\right)  x^{-1}=u\right]  \\\text{(here, we have substituted
}x\text{ for }x\rho\left(  k\right)  \text{ in the sum}\\\text{(because the
map }G\rightarrow G,\ x\mapsto x\rho\left(  k\right)  \text{ is a
bijection}\\\text{(since }G\text{ is a group, and since }\rho\left(  k\right)
\in G\text{)))}}}=\dfrac{1}{\left\vert H\right\vert }\underbrace{\sum_{k\in
H}\sum_{x\in G}\left[  x\rho\left(  y\right)  x^{-1}=u\right]  }_{=\left\vert
H\right\vert \cdot\sum_{x\in G}\left[  x\rho\left(  y\right)  x^{-1}=u\right]
}\\
&  =\dfrac{1}{\left\vert H\right\vert }\left\vert H\right\vert \cdot\sum_{x\in
G}\left[  x\rho\left(  y\right)  x^{-1}=u\right]  =\sum_{x\in G}\left[
x\rho\left(  y\right)  x^{-1}=u\right] \\
&  =\sum_{k\in G}\left[  k\rho\left(  y\right)  k^{-1}=u\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}x\text{ as }k\right)  .
\end{align*}
Compared with%
\[
\alpha_{G,\rho\left(  y\right)  }\left(  u\right)  =\sum_{k\in G}\left[
k\rho\left(  y\right)  k^{-1}=u\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.frob-char.plambda}(a), applied to }\rho\left(  y\right)
\text{ and }u\text{ instead of }h\text{ and }g\right)  ,
\]
this yields $\left(  \Ind_{\rho}\alpha_{H,y}\right)
\left(  u\right)  =\alpha_{G,\rho\left(  y\right)  }\left(  u\right)  $.

Let us now forget that we fixed $u$. We thus have shown that $\left(
\Ind_{\rho}\alpha_{H,y}\right)  \left(  u\right)
=\alpha_{G,\rho\left(  y\right)  }\left(  u\right)  $ for every $u\in G$. In
other words, $\Ind_{\rho}\alpha_{H,y}=\alpha
_{G,\rho\left(  y\right)  }$.

Let us now forget that we fixed $y$. We thus have shown that
$\Ind_{\rho}\alpha_{H,y}=\alpha_{G,\rho\left(
y\right)  }$ for every $y\in H$. Renaming $y$ as $h$ in this statement, we
obtain that $\Ind_{\rho}\alpha_{H,h}=\alpha
_{G,\rho\left(  h\right)  }$ for every $h\in H$. This solves Exercise
\ref{exe.frob-char.plambda}(f).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}}

\textit{Solution to Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}.}

\underline{\textit{Step 1: Study of inner tensor products.}}

The well-definedness of the inner tensor product is clear (since the inclusion
map $G\rightarrow G\times G,\ g\mapsto\left(  g,g\right)  $ is a group
homomorphism). We notice that if $G$ is a group and $U_{1}$ and $U_{2}$ are
two ${\CC}G$-modules, then the character $\chi_{U_{1}\otimes U_{2}}$ of
the inner tensor product $U_{1}\otimes U_{2}$ of $U_{1}$ and $U_{2}$ is given
by
\begin{equation}
\chi_{U_{1}\otimes U_{2}}\left(  g\right)  =\chi_{U_{1}}\left(  g\right)
\chi_{U_{2}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in G.
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}
\end{equation}
\footnote{This is because every $g\in G$ satisfies
\begin{align*}
\chi_{U_{1}\otimes U_{2}}\left(  g\right)   &  =\trace 
\underbrace{\left(  g:U_{1}\otimes U_{2}\rightarrow U_{1}\otimes U_{2}\right)
}_{\substack{=\left(  g,g\right) : U_1 \otimes U_2 \to U_1 \otimes U_2
\\\text{(by the definition of the inner tensor product)}}}\\
&  =\trace \underbrace{\left(  \left(  g,g\right)  :
U_1 \otimes U_2 \to U_1 \otimes U_2\right)}_{=\left(  g:U_{1}\rightarrow
U_{1}\right)  \otimes\left(  g:U_{2}\rightarrow U_{2}\right)  }
 =\trace \left(  \left(  g:U_{1}\rightarrow U_{1}\right)
\otimes\left(  g:U_{2}\rightarrow U_{2}\right)  \right) \\
&  =\underbrace{\trace \left(  g:U_{1}\rightarrow U_{1}\right)
}_{=\chi_{U_{1}}\left(  g\right)  }\cdot\underbrace{\trace 
\left(  g:U_{2}\rightarrow U_{2}\right)  }_{=\chi_{U_{2}}\left(  g\right)
}=\chi_{U_{1}}\left(  g\right)  \cdot\chi_{U_{2}}\left(  g\right)  .
\end{align*}
}

\underline{\textit{Step 2: The involutions }$\widetilde{\omega}_n$.}

Now let $n\geq 0$. For every $f\in R_{\CC}\left( \Symm_n\right)$,
it is easy to see that the map $\Symm_n \rightarrow \CC$
which sends every $g\in\Symm_n$ to
$\sgn \left(  g\right)  f\left(  g\right)  $ is a class function
(because both $\sgn \left(  g\right)  $ and $f\left(  g\right)  $
are uniquely determined by the conjugacy class of $g$). This class function
$\Symm_{n}\rightarrow{\CC}$ is denoted by $\sgn 
_{\Symm_{n}}\ast f$ and belongs to $R_{\CC}\left(
\Symm_{n}\right)  $ (being a class function). We can thus define a map
$\widetilde{\omega}_{n}:R_{\CC}\left(  \Symm_{n}\right)
\rightarrow R_{\CC}\left(  \Symm_{n}\right)  $ as follows:
\[
\widetilde{\omega}_{n}\left(  f\right)  = \sgn_{\Symm_{n}}\ast f
\ \ \ \ \ \ \ \ \ \ \text{for all }f\in R_{\CC}\left(
\Symm_{n}\right)  .
\]
Consider this map $\widetilde{\omega}_{n}$. It is ${\CC}$-linear
(obviously) and an involution\footnote{This is because multiplying a scalar by
$\sgn\left(  g\right)  $ twice (for fixed $g\in\Symm_{n}$) does
nothing (since $\left(  \sgn\left(  g\right)
\right)  ^{2}=1$).}. Hence, $\widetilde{\omega}_{n}$ is precisely the
involution on class functions $f:\Symm_{n}\rightarrow{\CC}$
sending $f\mapsto\sgn _{\Symm_{n}}\ast f$.

Now, let $V$ be any finite-dimensional $\CC \Symm_{n}$-module. Then,
every $g\in\Symm_{n}$ satisfies
\[
\underbrace{\left(  \widetilde{\omega}_{n}\left(  \chi_{V}\right)  \right)
}_{\substack{=\sgn _{\Symm_{n}}\ast\chi_{V}\\\text{(by the
definition of }\widetilde{\omega}_{n}\text{)}}}\left(  g\right)  =\left(
\sgn _{\Symm_{n}}\ast\chi_{V}\right)  \left(  g\right)
=\sgn \left(  g\right)  \chi_{V}\left(  g\right)
\]
and
\begin{align*}
\chi_{\sgn _{\Symm_{n}}\otimes V}\left(  g\right)   &
=\underbrace{\chi_{\sgn _{\Symm_{n}}}\left(  g\right)
}_{\substack{=\sgn \left(  g\right)  }}\chi_{V}\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}}\right) \\
&  =\sgn \left(  g\right)  \chi_{V}\left(  g\right)  .
\end{align*}
Hence, every $g\in\Symm_n$ satisfies $\left(  \widetilde{\omega}
_{n}\left(  \chi_{V}\right)  \right)  \left(  g\right)  =\sgn 
\left(  g\right)  \chi_{V}\left(  g\right)  =\chi_{\sgn 
_{\Symm_{n}}\otimes V}\left(  g\right)  $. In other words,
$\widetilde{\omega}_{n}\left(  \chi_{V}\right)  =\chi_{\sgn 
_{\Symm_{n}}\otimes V}$.

Forget that we fixed $V$. We thus have proven that $\widetilde{\omega}
_{n}\left(  \chi_{V}\right)  =\chi_{\sgn_{\Symm_{n}}\otimes V}$ for
every finite-dimensional $\CC \Symm_{n}$-module $V$. In particular, for
every irreducible $\CC \Symm_{n}$-module $V$, we have
$\widetilde{\omega}_{n}\left(  \chi
_{V}\right)  =\chi_{\sgn _{\Symm_{n}}\otimes V}\in
R\left(  \Symm_{n}\right)  $. Since the $\chi_{V}$ span $R\left(
\Symm_{n}\right)  $ as a $\ZZ$-module as $V$ ranges through (a
set of representatives of the isomorphism classes of) the irreducible
${\CC}\Symm_{n}$-modules $V$, this entails that the involution
$\widetilde{\omega}_{n}$ preserves the ${\ZZ}$-lattice $R\left(
\Symm_{n}\right)  $. Since $\widetilde{\omega}_{n}$ is the involution
on class functions $f:\Symm_{n}\rightarrow{\CC}$ sending
$f\mapsto\sgn _{\Symm_{n}}\ast f$, this rewrites as
follows: The involution on class functions $f:\Symm_{n}\rightarrow
{\CC}$ sending $f\mapsto\sgn _{\Symm_{n}}\ast f$
preserves the ${\ZZ}$-lattice $R\left(  \Symm_{n}\right)  $.
This proves one claim of Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b).

Recall that the involution $\widetilde{\omega}_{n}$ preserves the
${\ZZ}$-lattice $R\left(  \Symm_{n}\right)  $. Thus,
$\widetilde{\omega}_{n}$ restricts to a ${\ZZ}$-linear involution
$R\left(  \Symm_{n}\right)  \rightarrow R\left(  \Symm_{n}\right)  $. Denote
this involution by $\widetilde{\omega}_{n}^{\prime}$.
It clearly satisfies
\begin{align}
\widetilde{\omega}_{n}^{\prime}\left(  f\right)   &  =\widetilde{\omega}_{n}
\left(  f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{\omega}_{n}^{\prime}\right)
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wn'.1}\\
&  =\sgn _{\Symm_{n}}\ast f\ \ \ \ \ \ \ \ \ \ \text{for
all }f\in R\left(  \Symm_{n}\right)  .
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wn'.2}
\end{align}


\underline{\textit{Step 3: The involution }$\widetilde{\omega}_{\ZZ}$.}

Now, forget that we fixed $n$. We thus have constructed a ${\ZZ}
$-linear involution $\widetilde{\omega}_{n}^{\prime}:R\left(  \Symm 
_{n}\right)  \rightarrow R\left(  \Symm  _{n}\right)  $ for every
$n\geq0$. The direct sum of these involutions over all $n\geq0$ is a graded
${\ZZ} $-linear involution $\bigoplus_{n\geq0}\widetilde{\omega}
_{n}^{\prime}:\bigoplus_{n\geq0}R\left(  \Symm  _{n}\right)
\rightarrow\bigoplus_{n\geq0}R\left(  \Symm  _{n}\right)  $. Denote this
involution $\bigoplus_{n\geq0}\widetilde{\omega}_{n}^{\prime}$ by
$\widetilde{\omega}_{\ZZ}$. Then, $\widetilde{\omega}_{\ZZ}$
is a graded $\ZZ$-linear involution $A\left(  \Symm 
\right)  \rightarrow A\left(  \Symm  \right)  $ (since $\bigoplus
_{n\geq0}R\left(  \Symm  _{n}\right)  =A\left(  \Symm  \right)
$), and is precisely the involution on $A\left(  \Symm  \right)  $
defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b).

We have $\widetilde{\omega}_{\ZZ}=\bigoplus_{n\geq0}
\widetilde{\omega}_{n}^{\prime}$. Hence, for every $n\in{\NN} $ and
$f\in R_{{\CC} } \left(  \Symm  _{n}\right)  $, we have
\begin{align}
\widetilde{\omega}_{\ZZ}\left(  f\right)   &  =\widetilde{\omega}
_{n}^{\prime}\left(  f\right)  =\widetilde{\omega}_{n}\left(  f\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{\omega}_{n}^{\prime}\text{
is defined as a restriction of }\widetilde{\omega}_{n}\right)
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.1}\\
&  =\sgn _{\Symm  _{n}}\ast f.
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.2}
\end{align}


For every $n\in{\NN} $ and every finite-dimensional ${\CC}
\Symm_{n}$-module $V$, we have
\begin{align}
\widetilde{\omega}_{\ZZ}\left(  \chi_{V}\right)   &
=\widetilde{\omega}_{n}\left(  \chi_{V}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.1},
applied to }f=\chi_{V}\right) \nonumber\\
&  =\chi_{\sgn  _{\Symm  _{n}}\otimes V}.
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZxV}
\end{align}
In other words, the involution $\widetilde{\omega}_{\ZZ}$ sends
$\chi_{V}$ to $\chi_{\sgn  _{\Symm  _{n}}\otimes V}$ for
every $n\in{\NN} $ and every finite-dimensional ${\CC}
\Symm  _{n}$-module $V$. In other words, the involution on $A\left(
\Symm  \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b) sends $\chi_{V}$ to
$\chi_{\sgn  _{\Symm  _{n}}\otimes V}$ for every
$n\in{\NN} $ and every finite-dimensional ${\CC} \Symm 
_{n}$-module $V$ (since $\widetilde{\omega}_{\ZZ}$ is precisely the
involution on $A\left(  \Symm  \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)). This proves one of
the claims of
Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}.

\underline{\textit{Step 4: Properties of }
$\widetilde{\omega}_{\ZZ}$.}

We are now going to prove that $\ch  \circ\widetilde{\omega
}_{\ZZ}=\omega\circ\ch  $ as maps $A\left(
\Symm  \right)  \rightarrow\Lambda$.

Indeed, let $\lambda$ be a partition. Let $n=\left\vert \lambda\right\vert $.
Then, $\lambda\in\Par_{n}$.
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(a) yields
$\ch  \left(  \Ind _{\Symm  _{\lambda}}
^{\Symm  _{n}}\triv_{\Symm  _{\lambda}}\right)  =h_{\lambda}$ and
$\ch  \left(  \Ind _{\Symm  _{\lambda}}
^{\Symm  _{n}}\sgn _{\Symm  _{\lambda}}\right)
=e_{\lambda}$. Since $\ch  \left(  \Ind 
_{\Symm  _{\lambda}}^{\Symm  _{n}} \triv_{\Symm  _{\lambda}}
\right)  =h_{\lambda}$, we have $\ch^{-1}\left(  h_{\lambda
}\right)  =\Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\triv_{\Symm_{\lambda}}$. Since $\ch  \left(
\Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\sgn _{\Symm_{\lambda}}\right)  = e_{\lambda}$, we have
$\ch^{-1}\left(  e_{\lambda}\right)  =\Ind 
_{\Symm  _{\lambda}}^{\Symm  _{n}} \sgn 
_{\Symm  _{\lambda}}$.

On the other hand, $\omega\left(  h_{\lambda}\right)  =e_{\lambda}$
\ \ \ \ \footnote{This follows from the fact that $\omega$ is an algebra
homomorphism and that $\omega\left(  h_{m}\right)  =e_{m}$ for every
$m\in{\NN}$.}.

But it is easy to see that $\widetilde{\omega}_{\ZZ}\left(
\Ind _{\Symm_{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)
 =\Ind _{\Symm_{\lambda}}^{\Symm_{n}}\sgn _{\Symm_{\lambda}}$.
\ \ \ \ \footnote{\textit{Proof.} Let $g\in\Symm_n$. We have
$\widetilde{\omega}_{\ZZ}\left( 
 \Ind _{\Symm _{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)
=\sgn _{\Symm_{n}}\ast\left(  \Ind 
_{\Symm_{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)
$ (by \eqref{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.1},
applied to
$f=\Ind _{\Symm_{\lambda}}^{\Symm_{n}} \triv_{\Symm_{\lambda}}$).
Hence,
\begin{align*}
\left(  \widetilde{\omega}_{\ZZ}\left(  \Ind 
_{\Symm_{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)
\right)  \left(  g\right)   &  =\left(  \sgn _{\Symm_{n}}
\ast\left(  \Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\triv_{\Symm_{\lambda}}\right)  \right)  \left(  g\right)
  =\sgn \left(  g\right)  \cdot\left(  \Ind 
_{\Symm_{\lambda}}^{\Symm_n}\triv_{\Symm_{\lambda}}\right)
\left(  g\right)
\end{align*}
(by the definition of $\sgn _{\Symm_{n}}\ast\left(
\Ind _{\Symm_{\lambda}}^{\Symm_n}\triv_{\Symm _{\lambda}}\right) $).
Since (by the definition of
$\Ind _{\Symm_{\lambda}}^{\Symm_n}\triv_{\Symm _{\lambda}}$)
we have
\begin{align*}
\left(  \Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\triv_{\Symm_{\lambda}}\right)  \left(  g\right)   &  =\dfrac
{1}{\left\vert \Symm_{\lambda}\right\vert }\sum\limits_{\substack{k\in
\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}
\underbrace{\triv_{\Symm_{\lambda}}\left(  kgk^{-1}\right)  }_{=1}
  =\dfrac{1}{\left\vert \Symm_{\lambda}\right\vert }\sum\limits
_{\substack{k\in\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}1,
\end{align*}
this rewrites as
\[
\left(  \widetilde{\omega}_{\ZZ}\left(  \Ind 
_{\Symm_{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)
\right)  \left(  g\right)  =\sgn \left(  g\right)  \cdot
\underbrace{\left(  \Ind _{\Symm_{\lambda}}^{\Symm 
_{n}}\triv_{\Symm_{\lambda}}\right)  \left(  g\right)  }_{=\dfrac
{1}{\left\vert \Symm_{\lambda}\right\vert }\sum\limits_{\substack{k\in
\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}1}=\sgn 
\left(  g\right)  \cdot\dfrac{1}{\left\vert \Symm_{\lambda}\right\vert
}\sum\limits_{\substack{k\in\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}1.
\]
Compared with
\begin{align*}
\left(  \Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\sgn _{\Symm_{\lambda}}\right)  \left(  g\right)   &
=\dfrac{1}{\left\vert \Symm_{\lambda}\right\vert }\sum\limits_{\substack{k\in
\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}
\underbrace{\sgn _{\Symm_{\lambda}}\left(  kgk^{-1}
\right)  }_{=\sgn \left(  kgk^{-1}\right)  }
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Ind 
_{\Symm_{\lambda}}^{\Symm_{n}}\sgn _{\Symm 
_{\lambda}}\right) \\
&  =\dfrac{1}{\left\vert \Symm_{\lambda}\right\vert }\sum\limits
_{\substack{k\in\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}
\underbrace{\sgn \left(  kgk^{-1}\right)  }
_{\substack{=\sgn \left(  k\right)  \cdot\sgn 
\left(  g\right)  \cdot\left(  \sgn \left(  k\right)  \right)
^{-1}\\=\sgn \left(  g\right)  =\sgn \left(
g\right)  \cdot1}}=\dfrac{1}{\left\vert \Symm_{\lambda}\right\vert
}\sum\limits_{\substack{k\in\Symm_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}
\sgn \left(  g\right)  \cdot 1
 =\sgn \left(  g\right)  \cdot\dfrac{1}{\left\vert
\Symm_{\lambda}\right\vert }\sum\limits_{\substack{k\in\Symm 
_{n}:\\kgk^{-1}\in\Symm_{\lambda}}}1,
\end{align*}
this yields $\left(  \widetilde{\omega}_{\ZZ}\left(
\Ind _{\Symm_{\lambda}}^{\Symm_{n}}\triv_{\Symm 
_{\lambda}}\right)  \right)  \left(  g\right)  =\left(  \Ind 
_{\Symm_{\lambda}}^{\Symm_{n}}\sgn _{\Symm 
_{\lambda}}\right)  \left(  g\right)  $.
\par
Now, forget that we fixed $g$. We thus have shown that $\left(
\widetilde{\omega}_{\ZZ}\left(  \Ind _{\Symm 
_{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)  \right)
\left(  g\right)  =\left(  \Ind _{\Symm_{\lambda}}
^{\Symm_{n}}\sgn _{\Symm_{\lambda}}\right)
\left(  g\right)  $ for every $g\in\Symm_{n}$. In other words,
$\widetilde{\omega}_{\ZZ}\left(  \Ind _{\Symm 
_{\lambda}}^{\Symm_{n}}\triv_{\Symm_{\lambda}}\right)
=\Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\sgn _{\Symm_{\lambda}}$, qed.}

Now, comparing
\[
\left(  \widetilde{\omega}_{\ZZ}\circ\ch^{-1}\right)
\left(  h_{\lambda}\right)  =\widetilde{\omega}_{\ZZ}\left(
\underbrace{\ch^{-1}\left(  h_{\lambda}\right)  }
_{=\Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\triv_{\Symm_{\lambda}}}\right)  =\widetilde{\omega}_{\ZZ}\left(
\Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\triv_{\Symm_{\lambda}}\right)  =\Ind_{\Symm_{\lambda}}
^{\Symm_{n}}\sgn _{\Symm_{\lambda}}
\]
with
\[
\left(  \ch^{-1}\circ\omega\right)  \left(  h_{\lambda}\right)
=\ch^{-1}\left(  \underbrace{\omega\left(  h_{\lambda}\right)
}_{=e_{\lambda}}\right)  =\ch^{-1}\left(  e_{\lambda}\right)
=\Ind _{\Symm_{\lambda}}^{\Symm_{n}}
\sgn _{\Symm_{\lambda}},
\]
we obtain $\left(  \widetilde{\omega}_{\ZZ}\circ\ch^{-1}\right)
 \left(  h_{\lambda}\right)  =\left(  \ch^{-1}
\circ\omega\right)  \left(  h_{\lambda}\right)  $.

So we have shown that $\left(  \widetilde{\omega}_{\ZZ}
\circ\ch^{-1}\right)  \left(  h_{\lambda}\right)  =\left(
\ch^{-1}\circ\omega\right)  \left(  h_{\lambda}\right)  $ for
every partition $\lambda$. Thus, $\widetilde{\omega}_{\ZZ}
\circ\ch^{-1}=\ch^{-1}\circ\omega$ (since the
$h_{\lambda}$ form a $\ZZ$-module basis of $\Lambda$). Hence,
$\ch^{-1}\circ\omega=\widetilde{\omega}_{\ZZ}
\circ\ch^{-1}$ so that $\ch^{-1}\circ\omega
\circ\ch =\widetilde{\omega}_{\ZZ}$ and $\omega
\circ\ch =\ch \circ\widetilde{\omega}_{\ZZ}$.

We thus have $\ch  \circ\widetilde{\omega}_{\ZZ}
=\omega\circ\ch  $ as maps $A\left(  \Symm  \right)
\rightarrow\Lambda$. In other words, the map $\widetilde{\omega}_{\ZZ}$
corresponds under $\ch$ to the involution $\omega$ on
$\Lambda$. In other words, the involution on $A\left(  \Symm  \right)  $
defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b) corresponds
under $\ch$ to the involution $\omega$ on $\Lambda$ (since
$\widetilde{\omega}_{\ZZ}$ is precisely the involution on $A\left(
\Symm  \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)). This completes the
proof of Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b).

\underline{\textit{Step 5: The structure-preserving properties of
}$\widetilde{\omega}_{\ZZ}$.}

Now, it only remains to show that the involution on $A\left(  \Symm\right) $
defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)
is a nontrivial PSH-automorphism of $A\left(  \Symm  \right)  $.

Recall that $\ch$ is a PSH-isomorphism. Hence, its inverse
$\ch^{-1}$ also is a PSH-isomorphism.

Proposition~\ref{prop.Lambda.omega-and-S}(f) shows that $\omega:\Lambda
\rightarrow\Lambda$ is a Hopf automorphism. Moreover, the map $\omega$ is
clearly graded. Let $\Sigma$ denote the PSH-basis $\left\{  s_{\lambda}
\ \mid\ \lambda\in\Par \right\}  $ of $\Lambda$. Then, $\omega$
restricts to a bijection $\Sigma\rightarrow\Sigma$ \ \ \ \ \footnote{This
follows from Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas}.}.
As a consequence,
$\omega\left(  \Sigma\right)  \subset\Sigma$, so that $\omega\left(
{\NN}\Sigma\right)  ={\NN}\underbrace{\omega\left(
\Sigma\right)  }_{\subset\Sigma}\subset{\NN}\Sigma$. Hence, $\omega$ is
a PSH-morphism $\Lambda\rightarrow\Lambda$ (since $\omega$ is a graded Hopf
algebra morphism), and thus a PSH-isomorphism $\Lambda\rightarrow\Lambda$
(since $\omega$ is an isomorphism and restricts to a bijection $\Sigma
\rightarrow\Sigma$).

Now, $\ch^{-1}\circ\omega\circ\ch $ is the
composition of three PSH-isomorphisms (since $\ch^{-1}$,
$\omega$ and $\ch$ are PSH-isomorphisms), therefore a
PSH-isomorphism itself.
In other words, $\widetilde{\omega}_{\ZZ}$ is a
PSH-isomorphism (because $\widetilde{\omega}_{\ZZ}=\ch 
^{-1}\circ\omega\circ\ch $). Thus, $\widetilde{\omega
}_{\ZZ}$ is a PSH-automorphism.

Since $\omega\left(  h_{2}\right)  =e_{2}\neq h_{2}$, we have $\omega
\neq\id$ and therefore $\widetilde{\omega}_{\ZZ}
\neq\id$ (since $\widetilde{\omega}_{\ZZ}
=\ch^{-1}\circ\omega\circ\ch $). In other words,
$\widetilde{\omega}_{\ZZ}$ is nontrivial.

So we know that $\widetilde{\omega}_{\ZZ}$ is a nontrivial
PSH-automorphism of $A\left(  \Symm  \right)  $. In other words, the
involution on $A\left(  \Symm  \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b) is a nontrivial
PSH-automorphism of $A\left(  \Symm  \right)  $ (since
$\widetilde{\omega}_{\ZZ}$ is precisely the involution on $A\left(
\Symm  \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)). This completes the
solution of Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}.

\textit{Remark.} There are some alternative ways to solve parts of this exercise.

For example, in order to prove that $\widetilde{\omega}_{\ZZ}
\circ\ch^{-1}=\ch^{-1} \circ\omega$, we showed
that $\left(  \widetilde{\omega}_{\ZZ} \circ\ch 
^{-1}\right)  \left(  h_{\lambda}\right)  =\left(  \ch^{-1}
\circ\omega\right)  \left(  h_{\lambda}\right)  $ for every partition
$\lambda$. Instead of doing this, it is possible to prove that
$\widetilde{\omega}_{\ZZ}\circ\ch^{-1}
=\ch^{-1}\circ\omega$ by showing that $\left(  \widetilde{\omega
}_{\ZZ}\circ\ch^{-1}\right)  \left(  p_{\lambda
}\right)  =\left(  \ch^{-1}\circ\omega\right)  \left(
p_{\lambda}\right)  $ for every partition $\lambda$. This, however, requires a
little technicality (the $p_{\lambda}$ do not span $\Lambda$ as a
${\ZZ}$-module, only as a $\QQ$-module).

We showed that $\widetilde{\omega}_{\ZZ}\circ\ch^{-1}
=\ch^{-1}\circ\omega$ and used this to conclude that
$\widetilde{\omega}_{\ZZ}$ is a PSH-automorphism of $A\left(
\Symm  \right)  $. An alternative argument proceeds the other way round:
The map $\widetilde{\omega}_{\ZZ}$ is ${\ZZ} $-linear and
graded and is easily seen to be self-adjoint with respect to the inner product
on $A(\Symm )$. It also is a coalgebra morphism, as
$\Res_{\Symm_{i}\times\Symm_{j}}^{\Symm_{n}} \omega_{n}\left( f\right)
=\left(  \omega_{i}\otimes\omega_{j}\right)  \left(
\Res_{\Symm_{i}\times\Symm_{j}}^{\Symm_{n}}f\right)  $
for all $n=i+j$ and all $f\in R(\Symm  _{n})$. Hence,
this map $\widetilde{\omega}_{\ZZ}$ also is an algebra morphism
(since it is self-adjoint, and $A(\Symm  )$ is self-dual), hence a
bialgebra morphism and thus a Hopf morphism (by
Corollary~\ref{cor.bialg-mor-is-Hopf}). It also sends
irreducible characters to irreducible characters (by
\eqref{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZxV}), and thus
restricts to a bijection $\left\{  \chi^{\lambda}\right\}  \rightarrow\left\{
\chi^{\lambda}\right\}  $. Hence, $\widetilde{\omega}_{\ZZ}$ is a
PSH-automorphism of $A\left(  \Symm  \right)  $. Every $n\geq0$
satisfies
\begin{align*}
\left(  \widetilde{\omega}_{\ZZ}\circ\ch^{-1}\right)
\left(  h_{n}\right)   &  =\widetilde{\omega}_{\ZZ}\left(
\underbrace{\ch^{-1}\left( h_{n}\right) }_{=\triv_{\Symm _n}}\right)
 =\widetilde{\omega}_{\ZZ} \left(  \triv_{\Symm _{n}}\right) 
 =\omega_{n}\left(  \triv_{\Symm _{n}}\right)
=\sgn _{\Symm  _{n}}
=\ch^{-1}\left(  e_n \right)  \\
&= \ch^{-1}\left( \omega\left(h_n\right) \right)
\ \ \ \ \ \ \ \ \ \ \left(\text{since
Proposition~\ref{prop.Lambda.omega-and-S}(b) yields
$e_n = \omega\left(h_n\right)$}\right)
 \\
&=\left(  \ch^{-1}\circ\omega\right)
\left(  h_{n}\right)  .
\end{align*}
Since $\widetilde{\omega}_{\ZZ}\circ\ch^{-1}$ and
$\ch^{-1}\circ\omega$ are algebra morphisms whereas the $h_{n}$
generate $\Lambda$, this yields $\widetilde{\omega}_{\ZZ}
\circ\ch^{-1}=\ch^{-1}\circ\omega$.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}}

\textit{Solution to Exercise~\ref{internal-tensor-product-with-sgn-is-omega-exercise}.}

\underline{\textit{Step 1: Study of inner tensor products.}}

The well-definedness of the inner tensor product is clear (since the inclusion
map $G\rightarrow G\times G,\ g\mapsto\left(  g,g\right)  $ is a group
homomorphism). We notice that if $G$ is a group and $U_{1}$ and $U_{2}$ are
two $\CC G$-modules, then the character $\chi_{U_{1}\otimes U_{2}}$ of
the inner tensor product $U_{1}\otimes U_{2}$ of $U_{1}$ and $U_{2}$ is given
by
\begin{equation}
\chi_{U_{1}\otimes U_{2}}\left(  g\right)  =\chi_{U_{1}}\left(  g\right)
\chi_{U_{2}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in G.
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}
\end{equation}
\footnote{\textit{Proof of
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}):} Let
$g\in G$. Then, by the definition of $\chi_{U_{1}}$, we have $\chi_{U_{1}
}\left(  g\right)  =\trace \left(  g:U_{1}\rightarrow U_{1}\right)
$. Similarly, $\chi_{U_{2}}\left(  g\right)  =\trace \left(
g:U_{2}\rightarrow U_{2}\right)  $. Also, by the definition of $\chi
_{U_{1}\otimes U_{2}}$, we have $\chi_{U_{1}\otimes U_{2}}\left(  g\right)
=\trace \left(  g:U_{1}\otimes U_{2}\rightarrow U_{1}\otimes
U_{2}\right)  $.
\par
But recall how the $\CC G$-module $U_{1}\otimes U_{2}$ was defined: It
was defined as the restriction of the outer tensor product $U_{1}\otimes
U_{2}$ using the inclusion map $G\rightarrow G\times G,\ g\mapsto\left(
g,g\right)  $. Hence, for every $u_{1}\in U_{1}$ and $u_{2}\in U_{2}$, we have
\[
g\left(  u_{1}\otimes u_{2}\right)  =\left(  g,g\right)  \left(  u_{1}\otimes
u_{2}\right)  =gu_{1}\otimes gu_{2}
\]
(by the definition of the outer tensor product). In other words, the action of
$g$ on $U_{1}\otimes U_{2}$ is the tensor product of the action of $g$ on
$U_{1}$ with the action of $g$ on $U_{2}$. In other words,
\[
\left(  g:U_{1}\otimes U_{2}\rightarrow U_{1}\otimes U_{2}\right)  =\left(
g:U_{1}\rightarrow U_{1}\right)  \otimes\left(  g:U_{2}\rightarrow
U_{2}\right)  .
\]
Now,
\begin{align*}
\chi_{U_{1}\otimes U_{2}}\left(  g\right)   &  =\trace 
\underbrace{\left(  g:U_{1}\otimes U_{2}\rightarrow U_{1}\otimes U_{2}\right)
}_{=\left(  g:U_{1}\rightarrow U_{1}\right)  \otimes\left(  g:U_{2}\rightarrow
U_{2}\right)  }\\
&  =\trace \left(  \left(  g:U_{1}\rightarrow U_{1}\right)
\otimes\left(  g:U_{2}\rightarrow U_{2}\right)  \right) \\
&  =\underbrace{\trace \left(  g:U_{1}\rightarrow U_{1}\right)
}_{=\chi_{U_{1}}\left(  g\right)  }\cdot\underbrace{\trace \left(
g:U_{2}\rightarrow U_{2}\right)  }_{=\chi_{U_{2}}\left(  g\right)  }
=\chi_{U_{1}}\left(  g\right)  \cdot\chi_{U_{2}}\left(  g\right)  .
\end{align*}
This proves
\eqref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}.}

\underline{\textit{Step 2: The involutions }$\widetilde{\omega}_{n}$.}

Now let $n\geq0$. For every
$f\in R_{\CC }\left(  \Symm_{n}\right)  $, it is easy to see
that the map $\Symm _{n}\rightarrow\CC $ which sends
every $g\in\Symm _{n}$ to
$\sgn \left(  g\right)  f\left(  g\right)  $ is a class function
(because both $\sgn \left(  g\right)  $ and $f\left(  g\right)
$ are uniquely determined by the conjugacy class of $g$). This class function
$\Symm _{n}\rightarrow\CC $ is denoted by
$\sgn_{\Symm _{n}}\ast f$. Hence,
\[
\left(  \sgn_{\Symm _{n}}\ast f\right)  \left(
g\right)  =\sgn \left(  g\right)  f\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }g\in\Symm _{n}\text{ for all }f\in
R_{\CC }\left(  \Symm _{n}\right)  .
\]
Since $\sgn_{\Symm _{n}}\ast f$ is a class
function, we have $\sgn_{\Symm _{n}}\ast f\in
R_{\CC }\left(  \Symm _{n}\right)  $. We can thus define a map
$\widetilde{\omega}_{n}:R_{\CC }\left(  \Symm _{n}\right)
\rightarrow R_{\CC }\left(  \Symm _{n}\right)  $ as follows:
\[
\widetilde{\omega}_{n}\left(  f\right)  =
\sgn_{\Symm _{n}}\ast f\ \ \ \ \ \ \ \ \ \ \text{for all }f\in
R_{\CC }\left(  \Symm _{n}\right)  .
\]
Consider this map $\widetilde{\omega}_{n}$. It is $\CC$-linear
(obviously) and an involution\footnote{\textit{Proof.} Let $f\in
R_{\CC }\left(  \Symm _{n}\right)  $. Then,
\begin{align*}
\left(  \left(  \widetilde{\omega}_{n}\circ\widetilde{\omega}_{n}\right)
\left(  f\right)  \right)  \left(  g\right)   &  =\underbrace{\left(
\widetilde{\omega}_{n}\left(  \widetilde{\omega}_{n}\left(  f\right)  \right)
\right)  }_{\substack{=\sgn_{\Symm _{n}}
\ast\left(  \widetilde{\omega}_{n}\left(  f\right)  \right)  \\\text{(by the
definition of }\widetilde{\omega}_{n}\text{)}}}\left(  g\right)  =\left(
\sgn_{\Symm _{n}}\ast\left(  \widetilde{\omega
}_{n}\left(  f\right)  \right)  \right)  \left(  g\right) \\
&  =\sgn \left(  g\right)  \underbrace{\left(  \widetilde{\omega
}_{n}\left(  f\right)  \right)  }
_{\substack{=\sgn_{\Symm _{n}}\ast f\\\text{(by the definition of }
\widetilde{\omega}_{n}\text{)}}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\sgn_{\Symm _{n}}
\ast\left(  \widetilde{\omega}_{n}\left(  f\right)  \right)  \right) \\
&  =\sgn \left(  g\right)  \underbrace{\left(
\sgn_{\Symm _{n}}\ast f\right)  \left(
g\right)  }_{=\sgn \left(  g\right)  f\left(  g\right)
}=\underbrace{\sgn \left(  g\right)  \sgn \left(
g\right)  }_{\substack{=\left(  \sgn \left(  g\right)  \right)
^{2}=1\\\text{(since }\sgn \left(  g\right)  \in\left\{
1,-1\right\}  \text{)}}}f\left(  g\right)  =f\left(  g\right)
\end{align*}
for every $g\in\Symm _{n}$. In other words, $\left(  \widetilde{\omega
}_{n}\circ\widetilde{\omega}_{n}\right)  \left(  f\right)  =f$. Since this
holds for every $f\in R_{\CC }\left(  \Symm _{n}\right)  $, we can
thus conclude that $\widetilde{\omega}_{n}\circ\widetilde{\omega}_{n}
= \id_{R_{\CC }\left(  \Symm _{n}\right)  }$. Hence,
$\widetilde{\omega}_{n}$ is an involution, qed.}. Hence, $\widetilde{\omega
}_{n}$ is precisely the involution on class functions $f:\Symm _{n}
\rightarrow\CC $ sending $f\mapsto\sgn_{\Symm _{n}}\ast f$.

Now, let $V$ be any finite-dimensional $\CC \Symm _{n}$-module.
Then, every $g\in\Symm _{n}$ satisfies
\[
\underbrace{\left(  \widetilde{\omega}_{n}\left(  \chi_{V}\right)  \right)
}_{\substack{=\sgn_{\Symm _{n}}\ast\chi
_{V}\\\text{(by the definition of }\widetilde{\omega}_{n}\text{)}}}\left(
g\right)  =\left(  \sgn_{\Symm _{n}}\ast\chi
_{V}\right)  \left(  g\right)  =\sgn \left(  g\right)  \chi
_{V}\left(  g\right)
\]
and
\begin{align*}
\chi_{\sgn _{\Symm _{n}}\otimes V}\left(  g\right)   &
=\underbrace{\chi_{\sgn_{\Symm _{n}}}\left(
g\right)  }_{\substack{=\sgn \left(  g\right)  }}\chi_{V}\left(
g\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char})}\right) \\
&  =\sgn \left(  g\right)  \chi_{V}\left(  g\right)  .
\end{align*}
Hence, every $g\in\Symm _{n}$ satisfies $\left(  \widetilde{\omega}
_{n}\left(  \chi_{V}\right)  \right)  \left(  g\right)  =\sgn 
\left(  g\right)  \chi_{V}\left(  g\right)  =\chi_{\sgn 
_{\Symm _{n}}\otimes V}\left(  g\right)  $. In other words,
$\widetilde{\omega}_{n}\left(  \chi_{V}\right)  =\chi_{\sgn 
_{\Symm _{n}}\otimes V}$. Forget that we fixed $V$. We thus have proven
that $\widetilde{\omega}_{n}\left(  \chi_{V}\right)  =\chi_{\sgn 
_{\Symm _{n}}\otimes V}$ for every finite-dimensional
$\CC \Symm_n$-module $V$. Hence, for every finite-dimensional
${\CC }\Symm _{n}$-module $V$, we have $\widetilde{\omega}_{n}
\left(  \chi_{V}\right)  =\chi_{\sgn _{\Symm_n}\otimes V}\in 
R\left(  \Symm _{n}\right)  $. Since the $\chi_{V}$ span
$R\left(  \Symm _{n}\right)  $ as a $\ZZ$-module, we have
\[
R\left(  \Symm _{n}\right)  =\sum\limits_{\substack{V\text{ is a
finite-dimensional}\\\CC \Symm _{n}\text{-module}}}\ZZ 
\chi_{V},
\]
so that
\begin{align*}
\widetilde{\omega}_{n}\left(  R\left(  \Symm _{n}\right)  \right)   &
=\widetilde{\omega}_{n}\left(  \sum\limits_{\substack{V\text{ is a
finite-dimensional}\\\CC \Symm _{n}\text{-module}}}\ZZ 
\chi_{V}\right)  =\sum\limits_{\substack{V\text{ is a finite-dimensional}
\\\CC \Symm _{n}\text{-module}}}\ZZ 
\underbrace{\widetilde{\omega}_{n}\left(  \chi_{V}\right)  }_{\in R\left(
\Symm _{n}\right)  }\\
&  \subset\sum\limits_{\substack{V\text{ is a finite-dimensional}\\\CC 
\Symm _{n}\text{-module}}}\ZZ R\left(  \Symm _{n}\right)
\subset R\left(  \Symm _{n}\right)
\end{align*}
(since $R\left(  \Symm _{n}\right)  $ is a $\ZZ$-module). Hence,
the involution $\widetilde{\omega}_{n}$ preserves the $\ZZ$-lattice
$R\left(  \Symm _{n}\right)  $. Since $\widetilde{\omega}_{n}$ is the
involution on class functions $f:\Symm _{n}\rightarrow\CC $
sending $f\mapsto\sgn_{\Symm _{n}}\ast f$, this
rewrites as follows: The involution on class functions
$f:\Symm_{n}\rightarrow\CC$ sending $f\mapsto\sgn_{\Symm_n}\ast f$
preserves the $\ZZ$-lattice
$R\left(  \Symm _{n}\right)  $. This proves one claim of
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b). Recall that the
involution $\widetilde{\omega}_{n}$ preserves the $\ZZ$-lattice
$R\left(  \Symm _{n}\right)  $. Thus, $\widetilde{\omega}_{n}$ restricts
to a $\ZZ$-linear involution $R\left(  \Symm _{n}\right)
\rightarrow R\left(  \Symm _{n}\right)  $. Denote this involution by
$\widetilde{\omega}_{n}^{\prime}$. It clearly satisfies
\begin{align}
\widetilde{\omega}_{n}^{\prime}\left(  f\right)   &  =\widetilde{\omega}
_{n}\left(  f\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\widetilde{\omega}_{n}^{\prime}\right)
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wn'.1}\\
&  =\sgn_{\Symm _{n}}\ast
f\ \ \ \ \ \ \ \ \ \ \text{for all }f\in R\left( \Symm_n \right) .
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wn'.2}
\end{align}


\underline{\textit{Step 3: The involution }$\widetilde{\omega}_{\ZZ }$.}

Now, forget that we fixed $n$. We thus have constructed a $\ZZ$-linear
involution $\widetilde{\omega}_{n}^{\prime}:R\left(  \Symm _{n}\right)
\rightarrow R\left(  \Symm _{n}\right)  $ for every $n\geq0$. The direct
sum of these involutions over all $n\geq0$ is a graded $\ZZ$-linear
involution $\bigoplus_{n\geq0}\widetilde{\omega}_{n}^{\prime}:\bigoplus
_{n\geq0}R\left(  \Symm _{n}\right)  \rightarrow\bigoplus_{n\geq
0}R\left(  \Symm _{n}\right)  $. Denote this involution $\bigoplus
_{n\geq0}\widetilde{\omega}_{n}^{\prime}$ by $\widetilde{\omega}_{\ZZ}$.
Then, $\widetilde{\omega}_{\ZZ }$ is a graded $\ZZ$-linear
involution $A\left(  \Symm \right)  \rightarrow A\left(  \Symm 
\right)  $ (since $\bigoplus_{n\geq0}R\left(  \Symm _{n}\right)
=A\left(  \Symm \right)  $), and is precisely the involution on
$A\left(  \Symm \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b).

We have $\widetilde{\omega}_{\ZZ }=\bigoplus_{n\geq0}\widetilde{\omega
}_{n}^{\prime}$. Hence, for every $n\in\NN$ and $f\in R_{\CC }
\left(  \Symm _{n}\right)  $, we have
\begin{align}
\widetilde{\omega}_{\ZZ }\left(  f\right)    & =\widetilde{\omega}
_{n}^{\prime}\left(  f\right)  =\widetilde{\omega}_{n}\left(  f\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{\omega}_{n}^{\prime}\text{
is defined as a restriction of }\widetilde{\omega}_{n}\right)
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.1}\\
& =\sgn_{\Symm _{n}}\ast
f.\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.2}
\end{align}


For every $n\in\NN$ and every finite-dimensional
$\CC \Symm_n$-module $V$, we have
\begin{align}
\widetilde{\omega}_{\ZZ }\left(  \chi_{V}\right)   &  =\widetilde{\omega
}_{n}\left(  \chi_{V}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.1}), applied
to }f=\chi_{V}\right) \nonumber\\
&  =\chi_{\sgn _{\Symm _{n}}\otimes V}.
\label{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZxV}
\end{align}
In other words, the involution $\widetilde{\omega}_{\ZZ }$ sends
$\chi_{V}$ to $\chi_{\sgn _{\Symm _{n}}\otimes V}$ for
every $n\in\NN$ and every finite-dimensional $\CC \Symm 
_{n}$-module $V$. In other words, the involution on $A\left(  \Symm 
\right)  $ defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)
sends $\chi_{V}$ to $\chi_{\sgn _{\Symm _{n}}\otimes V}$
for every $n\in\NN$ and every finite-dimensional $\CC 
\Symm _{n}$-module $V$ (since $\widetilde{\omega}_{\ZZ }$ is
precisely the involution on $A\left(  \Symm \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)). This proves one of
the claims of Exercise
\ref{internal-tensor-product-with-sgn-is-omega-exercise}.

\underline{\textit{Step 4: Properties of }$\widetilde{\omega}_{\ZZ }$.}

We are now going to prove that $\ch \circ\widetilde{\omega
}_{\ZZ }=\omega\circ\ch $ as maps $A\left(  \Symm 
\right)  \rightarrow\Lambda$.

Indeed, let $\lambda$ be a partition. Let $n=\left\vert \lambda\right\vert$.
Then, $\lambda\in \Par_{n}$.
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(a) yields
$\ch \left(  \Ind _{\Symm 
_{\lambda}}^{\Symm _{n}} \triv_{\Symm _{\lambda}}\right)  =h_{\lambda}$
and $\ch \left(  \Ind _{\Symm 
_{\lambda}}^{\Symm _{n}}\sgn_{\Symm 
_{\lambda}}\right)  =e_{\lambda}$. Since $\ch \left(
\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  =h_{\lambda}$, we have
$\ch^{-1}\left(  h_{\lambda}\right)  =\Ind_{\Symm_{\lambda}}^{\Symm_n}
\triv_{\Symm_{\lambda}}$. Since $\ch \left( \Ind_{\Symm_{\lambda}}^{\Symm_n}
\sgn_{\Symm_{\lambda}}\right) = e_{\lambda}$, we have
$\ch^{-1}\left(  e_{\lambda}\right)
=\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}}$.

On the other hand, $\omega\left(  h_{\lambda}\right)  =e_{\lambda}$
\ \ \ \ \footnote{\textit{Proof.} Write the partition $\lambda$ in the form
$\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)  $. Then,
$h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}...h_{\lambda_{\ell}}$ (by the
definition of $h_{\lambda}$), so that
\begin{align*}
\omega\left(  h_{\lambda}\right)    & =\omega\left(  h_{\lambda_{1}}
h_{\lambda_{2}}...h_{\lambda_{\ell}}\right)  =\omega\left(  h_{\lambda_{1}}
\right)  \omega\left(  h_{\lambda_{2}}\right)  ...\omega\left(
h_{\lambda_{\ell}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\omega\text{ is an algebra homomorphism}\right)  \\
& =e_{\lambda_{1}}e_{\lambda_{2}}...e_{\lambda_{\ell}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\left(  h_{m}\right)
=e_{m}\text{ for every }m\in\NN \right)  .
\end{align*}
Compared with
$e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}...e_{\lambda_{\ell}}$ (by
the definition of $e_{\lambda}$, since $\lambda=\left(  \lambda
_{1},\lambda_{2},...,\lambda_{\ell}\right)  $), this yields $\omega\left(
h_{\lambda}\right)  =e_{\lambda}$, qed.}.

But it is easy to see that $\widetilde{\omega}_{\ZZ }\left(
\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  
=\Ind_{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}}$.\ \ \ \ \footnote{\textit{Proof.} Let
$g\in\Symm _{n}$. We have $\widetilde{\omega}_{\ZZ }\left(
\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  
=\sgn_{\Symm_n}\ast\left(  \Ind_{\Symm_{\lambda}}^{\Symm_n}
\triv_{\Symm_{\lambda}}\right)  $ (by
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZ.1}), applied
to $f=\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}$). Hence,
\begin{align*}
\left(  \underbrace{\widetilde{\omega}_{\ZZ }\left(  
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}\triv_{\Symm _{\lambda
}}\right)  }_{=\sgn_{\Symm _{n}}\ast\left(
\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  }\right)  \left(  g\right)    & =\left(
\sgn_{\Symm _{n}}\ast\left(  
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}\triv_{\Symm _{\lambda
}}\right)  \right)  \left(  g\right)  \\
& =\sgn \left(  g\right)  \cdot\left(  
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}\triv_{\Symm _{\lambda
}}\right)  \left(  g\right)
\end{align*}
(by the definition of $\sgn_{\Symm _{n}}
\ast\left(  \Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  $). Since
\begin{align*}
\left(  \Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  \left(  g\right)    & =\dfrac
{1}{\left\vert \Symm _{\lambda}\right\vert }\sum\limits_{\substack{k\in
\Symm _{n}:\\kgk^{-1}\in\Symm _{\lambda}}}
\underbrace{\triv_{\Symm _{\lambda}}\left(  kgk^{-1}\right)  }
_{\substack{=1\\\text{(by the definition of }\triv_{\Symm _{\lambda}}
\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  \\
& =\dfrac{1}{\left\vert \Symm _{\lambda}\right\vert }\sum\limits
_{\substack{k\in\Symm _{n}:\\kgk^{-1}\in\Symm _{\lambda}}}1,
\end{align*}
this rewrites as
\[
\left(  \widetilde{\omega}_{\ZZ }\left(  
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}\triv_{\Symm _{\lambda
}}\right)  \right)  \left(  g\right)  =\sgn \left(  g\right)
\cdot\underbrace{\left(  \Ind _{\Symm _{\lambda}}
^{\Symm _{n}}\triv_{\Symm _{\lambda}}\right)  \left(  g\right)
}_{=\dfrac{1}{\left\vert \Symm _{\lambda}\right\vert }\sum\limits
_{\substack{k\in\Symm _{n}:\\kgk^{-1}\in\Symm _{\lambda}}}
1}=\sgn \left(  g\right)  \cdot\dfrac{1}{\left\vert
\Symm _{\lambda}\right\vert }\sum\limits_{\substack{k\in\Symm_{n}:
\\kgk^{-1}\in\Symm _{\lambda}}}1.
\]
Compared with
\begin{align*}
\left(  \Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}}\right)  \left(
g\right)    & =\dfrac{1}{\left\vert \Symm _{\lambda}\right\vert }
\sum\limits_{\substack{k\in\Symm _{n}:\\kgk^{-1}\in\Symm _{\lambda}}}
\underbrace{\sgn_{\Symm _{\lambda}}\left(
kgk^{-1}\right)  }_{\substack{=\sgn \left(  kgk^{-1}\right)
\\\text{(by the definition of }\sgn_{\Symm 
_{\lambda}}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}}\right)  \\
& =\dfrac{1}{\left\vert \Symm _{\lambda}\right\vert }\sum\limits
_{\substack{k\in\Symm _{n}:\\kgk^{-1}\in\Symm _{\lambda}}}
\underbrace{\sgn \left(  kgk^{-1}\right)  }
_{\substack{=\sgn \left(  k\right)  \cdot\sgn 
\left(  g\right)  \cdot\left(  \sgn \left(  k\right)  \right)
^{-1}\\=\sgn \left(  g\right)  =\sgn \left(
g\right)  \cdot1}}=\dfrac{1}{\left\vert \Symm _{\lambda}\right\vert
}\sum\limits_{\substack{k\in\Symm _{n}:\\kgk^{-1}\in\Symm _{\lambda}}}
\sgn \left(  g\right)  \cdot1\\
& =\sgn \left(  g\right)  \cdot\dfrac{1}{\left\vert
\Symm _{\lambda}\right\vert }\sum\limits_{\substack{k\in\Symm 
_{n}:\\kgk^{-1}\in\Symm _{\lambda}}}1,
\end{align*}
this yields $\left(  \widetilde{\omega}_{\ZZ }\left(
\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  \right)  \left(  g\right)  =\left(
\Ind _{\Symm _{\lambda}}^{\Symm _{n}
}\sgn_{\Symm _{\lambda}}\right)  \left(
g\right)  $.
\par
Now, forget that we fixed $g$. We thus have shown that $\left(
\widetilde{\omega}_{\ZZ }\left(  
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}\triv_{\Symm _{\lambda
}}\right)  \right)  \left(  g\right)  =\left(  
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}}\right)  \left(  g\right)  $ for every
$g\in\Symm _{n}$. In other words, $\widetilde{\omega}_{\ZZ}
\left(  \Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}\right)  =
\Ind_{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}}$, qed.} Now, comparing
\[
\left(  \widetilde{\omega}_{\ZZ }\circ\ch^{-1}
\right)  \left(  h_{\lambda}\right)  =\widetilde{\omega}_{\ZZ }\left(
\underbrace{\ch^{-1}\left(  h_{\lambda}\right)
}_{=\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\triv_{\Symm _{\lambda}}}\right)  =\widetilde{\omega}_{\ZZ }\left(
\Ind_{\Symm _{\lambda}}^{\Symm _{n}} \triv_{\Symm_{\lambda}}\right)
= \Ind_{\Symm _{\lambda}}^{\Symm _{n}}\sgn_{\Symm_{\lambda}}
\]
with
\[
\left(  \ch^{-1}\circ\omega\right)  \left(
h_{\lambda}\right)  =\ch^{-1}\left(
\underbrace{\omega\left(  h_{\lambda}\right)  }_{=e_{\lambda}}\right)
=\ch^{-1}\left(  e_{\lambda}\right)
=\Ind _{\Symm _{\lambda}}^{\Symm _{n}}
\sgn_{\Symm _{\lambda}},
\]
we obtain $\left(  \widetilde{\omega}_{\ZZ }\circ\ch^{-1}\right)
  \left(  h_{\lambda}\right)  =\left(  \ch^{-1}\circ\omega\right)  \left(  h_{\lambda}\right)  $.

Now, forget that we fixed $\lambda$. We thus have proven that $\left(
\widetilde{\omega}_{\ZZ }\circ\ch^{-1}\right)
\left(  h_{\lambda}\right)  =\left(  \ch^{-1}
\circ\omega\right)  \left(  h_{\lambda}\right)  $ for every partition
$\lambda$. Thus, the two maps $\widetilde{\omega}_{\ZZ }\circ
\ch^{-1}$ and $\ch^{-1}\circ\omega$ are
equal to each other on the $\ZZ$-basis $\left(
h_{\lambda}\right)  _{\lambda\in\Par }$ of $\Lambda$. Since
these two maps $\widetilde{\omega}_{\ZZ }\circ
\ch^{-1}$ and $\ch^{-1}\circ\omega$ are
$\ZZ$-linear, this yields that the two maps $\widetilde{\omega
}_{\ZZ }\circ\ch^{-1}$ and $\ch^{-1}\circ\omega$ are
identical (because if two $\ZZ$-linear
maps between two $\ZZ$-modules $A$ and $B$ are equal to each other on a
$\ZZ$-basis of $A$, then these two maps must be identical). In other
words, $\widetilde{\omega}_{\ZZ }\circ\ch^{-1}
=\ch^{-1}\circ\omega$. Hence, $\ch 
\circ\underbrace{\widetilde{\omega}_{\ZZ }\circ\ch^{-1}}_{
=\ch^{-1}\circ\omega}\circ
\ch =\underbrace{\ch \circ\ch^{-1}}_{=\id }
\circ\omega\circ\ch =\omega\circ\ch $. Thus, $\omega\circ\ch
=\ch \circ\widetilde{\omega}_{\ZZ }\circ
\underbrace{\ch^{-1}\circ\ch}
_{=\id }=\ch \circ\widetilde{\omega}_{\ZZ }$.

We thus have $\ch \circ\widetilde{\omega}_{\ZZ }
=\omega\circ\ch $ as maps $A\left(  \Symm \right)
\rightarrow\Lambda$. In other words, the map $\widetilde{\omega}_{\ZZ }$
corresponds under $\ch$ to the involution $\omega$ on $\Lambda
$. In other words, the involution on $A\left(  \Symm \right)  $ defined
in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b) corresponds under
$\ch$ to the involution $\omega$ on $\Lambda$ (since
$\widetilde{\omega}_{\ZZ }$ is precisely the involution on $A\left(
\Symm \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)). This completes the
proof of Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b).

We have $\ch \circ\widetilde{\omega}_{\ZZ }=\omega
\circ\ch $, so that $\ch^{-1}
\circ\underbrace{\ch \circ\widetilde{\omega}_{\ZZ }}
_{=\omega\circ\ch }=\ch^{-1}\circ
\omega\circ\ch $. Thus, $\ch^{-1}
\circ\omega\circ\ch =\underbrace{\ch^{-1}\circ\ch 
}_{=\id }\circ
\widetilde{\omega}_{\ZZ }=\widetilde{\omega}_{\ZZ }$.

\underline{\textit{Step 5: The structure-preserving properties of
}$\widetilde{\omega}_{\ZZ}$.}

Now, it only remains to show that the involution on $A\left( \Symm\right)  $
defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)
is a nontrivial PSH-automorphism of $A\left(  \Symm \right)  $.

Recall that $\ch$ is a PSH-isomorphism. Hence, its inverse
$\ch^{-1}$ also is a PSH-isomorphism.

Proposition~\ref{prop.Lambda.omega-and-S}(f) shows that $\omega:\Lambda
\rightarrow\Lambda$ is a Hopf automorphism. Moreover, the map $\omega$ is
clearly graded. Let $\Sigma$ denote the PSH-basis $\left\{  s_{\lambda}
\ \mid\ \lambda\in\Par \right\}  $ of $\Lambda$. Then, $\omega$
restricts to a bijection $\Sigma\rightarrow\Sigma$
\ \ \ \ \footnote{\textit{Proof.}
We have $\Sigma=\left\{  s_{\lambda}\ \mid\ \lambda
\in\Par \right\}  $, so that
\begin{align*}
\omega\left(  \Sigma\right)    & =\omega\left(  \left\{  s_{\lambda}
\ \mid\ \lambda\in\Par \right\}  \right)  =\left\{
\underbrace{\omega\left(  s_{\lambda}\right)  }_{\substack{
=s_{\lambda^t} \\
\text{(by Lemma~\ref{lem.sol.Lambda.staircase-skew-2.omegas})}}}
\ \mid\ \lambda\in\Par \right\}  =\left\{  s_{\lambda^{t}}
\ \mid\ \lambda\in\Par \right\}  \\
& \subset\left\{  s_{\lambda}\ \mid\ \lambda\in\Par \right\}
=\Sigma.
\end{align*}
Hence, the map $\omega$ restricts to a map $\Sigma\rightarrow\Sigma$. This map
$\Sigma\rightarrow\Sigma$ is an involution (because $\omega^{2}
=\id$), and thus is bijective. Thus, the map $\omega$ restricts
to a bijection $\Sigma\rightarrow\Sigma$, qed.}. As a consequence,
$\omega\left(  \Sigma\right)  \subset\Sigma$, so that $\omega\left(
\NN \Sigma\right)  =\NN \underbrace{\omega\left(  \Sigma\right)
}_{\subset\Sigma}\subset\NN \Sigma$. Hence, $\omega$ is a PSH-morphism
$\Lambda\rightarrow\Lambda$ (since $\omega$ is a graded Hopf algebra
morphism), and thus a PSH-isomorphism $\Lambda\rightarrow\Lambda$ (since
$\omega$ is an isomorphism and restricts to a bijection $\Sigma\rightarrow
\Sigma$).

Now, $\ch^{-1}\circ\omega\circ\ch$ is
the composition of three PSH-isomorphisms (since $\ch^{-1}$, $\omega$ and
$\ch$ are PSH-isomorphisms),
therefore a PSH-isomorphism (since any composition of PSH-isomorphisms is a
PSH-isomorphism). In other words, $\widetilde{\omega}_{\ZZ }$ is a
PSH-isomorphism (because $\widetilde{\omega}_{\ZZ }
=\ch^{-1}\circ\omega\circ\ch $). Thus, $\widetilde{\omega
}_{\ZZ }$ is a PSH-automorphism (since it is a PSH-isomorphism and an endomorphism).

Assume (for the sake of contradiction) that $\widetilde{\omega}_{\ZZ}
=\id$. Since $\widetilde{\omega}_{\ZZ } =\ch^{-1}\circ\omega\circ\ch $,
this rewrites as $\ch^{-1}\circ\omega\circ \ch =\id$. Hence,
$\omega\circ\ch =\ch $, so that $\omega=\ch \circ
\ch^{-1}=\id$ and thus $\omega\left(
h_{2}\right)  =\id \left(  h_{2}\right)  =h_{2}$, which
contradicts $\omega\left(  h_{2}\right)  =e_{2}\neq h_{2}$. This contradiction
shows that our assumption (that $\widetilde{\omega}_{\ZZ }
=\id$) was wrong. Hence, $\widetilde{\omega}_{\ZZ }
\neq\id$. In other words, $\widetilde{\omega}_{\ZZ }$ is nontrivial.

So we know that $\widetilde{\omega}_{\ZZ }$ is a nontrivial
PSH-automorphism of $A\left(  \Symm \right)  $. In other words, the
involution on $A\left(  \Symm \right)  $ defined in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b) is a nontrivial
PSH-automorphism of $A\left(  \Symm \right)  $ (since $\widetilde{\omega
}_{\ZZ }$ is precisely the involution on $A\left(  \Symm \right)
$ defined in Theorem~\ref{symmetric-group-Frobenius-map-theorem}(b)). This
completes the solution of Exercise
\ref{internal-tensor-product-with-sgn-is-omega-exercise}.

\textit{Remark.} There are some alternative ways to solve parts of this exercise.

For example, in order to prove that $\widetilde{\omega}_{\ZZ}
\circ\ch^{-1}=\ch^{-1} \circ\omega$, we showed that
$\left(  \widetilde{\omega}_{\ZZ}
\circ\ch^{-1}\right)  \left(  h_{\lambda}\right)
=\left(  \ch^{-1}\circ\omega\right)  \left(
h_{\lambda}\right)  $ for every partition $\lambda$. Instead of doing this, it
is possible to prove that  $\widetilde{\omega}_{\ZZ }\circ
\ch^{-1}=\ch^{-1}\circ\omega$
by showing that $\left(  \widetilde{\omega}_{\ZZ }\circ
\ch^{-1}\right)  \left(  p_{\lambda}\right)  =\left(
\ch^{-1}\circ\omega\right)  \left(  p_{\lambda
}\right)  $ for every partition $\lambda$. This, however, requires a little
technicality (the $p_{\lambda}$ do not span $\Lambda$ as a
$\ZZ$-module, only as a $\QQ$-module).

We showed that $\widetilde{\omega}_{\ZZ }\circ\ch^{-1}
=\ch^{-1}\circ\omega$ and used this to
conclude that $\widetilde{\omega}_{\ZZ }$ is a PSH-automorphism of
$A\left(  \Symm \right)  $. An alternative argument proceeds the other
way round: The map $\widetilde{\omega}_{\ZZ }$ is $\ZZ$-linear
and graded and is easily seen to be self-adjoint with respect to the inner
product on $A(\Symm )$. It also is a coalgebra morphism, as
$\Res _{\Symm _{i}\times\Symm _{j}}^{\Symm _{n}}
\omega_{n}\left(  f\right)  =\left(  \omega_{i}\otimes\omega_{j}\right)
\left(  \Res _{\Symm _{i}\times\Symm _{j}}
^{\Symm _{n}}f\right)  $ for all $n=i+j$ and all $f\in R(\Symm 
_{n})$. Hence, this map $\widetilde{\omega}_{\ZZ }$ also is an algebra
morphism (since it is self-adjoint, and $A(\Symm )$ is self-dual), hence
a bialgebra morphism and thus a Hopf morphism (by
Corollary~\ref{cor.bialg-mor-is-Hopf}). It also sends
irreducible characters to irreducible characters (by
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.wZxV})), and thus
restricts to a bijection $\left\{  \chi^{\lambda}\right\}  \rightarrow\left\{
\chi^{\lambda}\right\}  $. Hence, $\widetilde{\omega}_{\ZZ }$ is a
PSH-automorphism of $A\left(  \Symm \right)  $. Every $n\geq0$
satisfies
\begin{align*}
\left(  \widetilde{\omega}_{\ZZ}\circ\ch^{-1}\right)
\left(  h_{n}\right)   &  =\widetilde{\omega}_{\ZZ}\left(
\underbrace{\ch^{-1}\left( h_{n}\right) }_{=\triv_{\Symm _n}}\right)
 =\widetilde{\omega}_{\ZZ} \left(  \triv_{\Symm _{n}}\right) 
 =\omega_{n}\left(  \triv_{\Symm _{n}}\right)
=\sgn _{\Symm  _{n}}
=\ch^{-1}\left(  e_n \right)  \\
&= \ch^{-1}\left( \omega\left(h_n\right) \right)
\ \ \ \ \ \ \ \ \ \ \left(\text{since
Proposition~\ref{prop.Lambda.omega-and-S}(b) yields
$e_n = \omega\left(h_n\right)$}\right)
 \\
&=\left(  \ch^{-1}\circ\omega\right)
\left(  h_{n}\right)  .
\end{align*}
Since $\widetilde{\omega}_{\ZZ }\circ\ch^{-1}$ and
$\ch^{-1}\circ\omega$ are algebra morphisms whereas the $h_{n}$
generate $\Lambda$, this yields $\widetilde{\omega}_{\ZZ }
\circ\ch^{-1}=\ch^{-1}\circ\omega$.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.frob-char.ptype}}

\textit{Solution to Exercise \ref{exe.frob-char.ptype}.} It is known that two
permutations in $\Symm_n$ have the same cycle type if and only if
they are conjugate. In other words, if $g$ and $h$ are two permutations in
$\Symm_n$, then we have the following logical equivalence:%
\begin{equation}
\left(  g\text{ and }h\text{ have the same cycle type}\right)
\ \Longleftrightarrow\ \left(  g\text{ and }h\text{ are conjugate}\right)  .
\label{sol.frob-char.ptype.a.ctc}%
\end{equation}


(a) Let $f\in R_{\CC}\left(   \Symm_{n}\right)  $.

\begin{verlong}
We must prove the equality%
\[
\operatorname*{ch}\left(  f\right)  =\dfrac{1}{n!}\sum_{\sigma\in
 \Symm_{n}}f\left(  \sigma\right)  p_{\operatorname*{type}\sigma}.
\]

\end{verlong}

\begin{vershort}
If $\sigma\in \Symm_{n}$, then the cycle type of $\sigma$ is a partition
of $n$. In other words, we have $\operatorname*{type}\sigma\in
\Par_n$ for each $\sigma\in \Symm_{n}$ (since
$\operatorname*{type}\sigma$ denotes the cycle type of $\sigma$, while
$\Par_n$ denotes the set of all partitions of $n$).
\end{vershort}

\begin{verlong}
If $\sigma\in \Symm_{n}$, then the cycle type of $\sigma$ is a partition
of $n$. In other words, if $\sigma\in \Symm_{n}$, then
$\operatorname*{type}\sigma$ is a partition of $n$ (since
$\operatorname*{type}\sigma$ is defined to be the cycle type of $\sigma$). In
other words, if $\sigma\in \Symm_{n}$, then $\operatorname*{type}%
\sigma\in\Par_n$ (since $\operatorname*{Par}%
\nolimits_{n}$ is the set of all partitions of $n$). In other words, we have
$\operatorname*{type}\sigma\in\Par_n$ for each
$\sigma\in \Symm_{n}$.
\end{verlong}

For each partition $\lambda$, we define a positive integer $z_{\lambda}$ as in
Proposition \ref{prop-Two-other-Cauchy-expansions}.

Recall the following fact from finite group theory:

\begin{statement}
\textit{Claim 1:} Let $G$ be any finite group. Let $g\in G$ be any element.
Let $C_{g}$ denote the conjugacy class of $g$. Let $Z_{g}$ denote the
centralizer of $g$. Then,%
\begin{equation}
\left\vert C_{g}\right\vert =\dfrac{\left\vert G\right\vert }{\left\vert
Z_{g}\right\vert }. \label{sol.frob-char.ptype.a.cc}%
\end{equation}

\end{statement}

\begin{verlong}
[\textit{Proof of Claim 1:} The group $G$ acts on itself from the left by
conjugation -- i.e., by the formula $x\rightharpoonup y=xyx^{-1}$, where we
are using the symbol $\rightharpoonup$ for the action of $G$ on $G$ (that is,
$x\rightharpoonup y$ denotes the image of $y\in G$ under the action of $x\in
G$). This is called the \textit{conjugation action}. Applying the
orbit-stabilizer theorem to this action, we obtain%
\begin{equation}
\left\vert \text{the orbit of }g\right\vert =\dfrac{\left\vert G\right\vert
}{\left\vert \text{the stabilizer of }g\right\vert }.
\label{sol.frob-char.ptype.a.cc.pf.1}%
\end{equation}
But the orbit of $g$ under the conjugation action is the set
\[
\left\{  \underbrace{x\rightharpoonup g}_{\substack{=xgx^{-1}\\\text{(by the
definition of the conjugation action)}}}\ \mid\ x\in G\right\}  =\left\{
xgx^{-1}\ \mid\ x\in G\right\}  =\left(  \text{the conjugacy class of
}g\right)  =C_{g}%
\]
(since $C_{g}$ was defined to be the conjugacy class of $g$). In other words,%
\begin{equation}
\left(  \text{the orbit of }g\right)  =C_{g}.
\label{sol.frob-char.ptype.a.cc.pf.2}%
\end{equation}
Also, the stabilizer of $g$ under the conjugation action is the set%
\[
\left\{  x\in G\ \mid\ \underbrace{x\rightharpoonup g}_{\substack{=xgx^{-1}%
\\\text{(by the definition of the conjugation action)}}}=g\right\}  =\left\{
x\in G\ \mid\ xgx^{-1}=g\right\}  =\left(  \text{the centralizer of }g\right)
=Z_{g}%
\]
(since $Z_{g}$ was defined to be the centralizer of $g$). In other words,%
\begin{equation}
\left(  \text{the stabilizer of }g\right)  =Z_{g}.
\label{sol.frob-char.ptype.a.cc.pf.3}%
\end{equation}


In view of \eqref{sol.frob-char.ptype.a.cc.pf.2} and
\eqref{sol.frob-char.ptype.a.cc.pf.3}, we can rewrite
\eqref{sol.frob-char.ptype.a.cc.pf.1} as $\left\vert C_{g}\right\vert
=\dfrac{\left\vert G\right\vert }{\left\vert Z_{g}\right\vert }$. This proves
Claim 1.]
\end{verlong}

Applying Claim 1 to the symmetric group $\Symm_n$, we quickly arrive
at the following:

\begin{statement}
\textit{Claim 2:} For each $\lambda\in\Par_n$, we
have%
\[
\left(  \text{the number of all }\sigma\in \Symm_{n}\text{ satisfying
}\operatorname*{type}\sigma=\lambda\right)  =\dfrac{n!}{z_{\lambda}}.
\]

\end{statement}

[\textit{Proof of Claim 2:} Let $\lambda\in\Par_n$.
Thus, $\lambda$ is a partition of $n$. Hence, there exists a permutation in
$\Symm_n$ that has cycle type $\lambda$.\ \ \ \ \footnote{Indeed, we
can obtain such a permutation as follows: Let $\ell=\ell\left(  \lambda
\right)  $, so that $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}=\left\vert
\lambda\right\vert =n$. Partition the $n$-element set $\left\{  1,2,\ldots
,n\right\}  $ into $\ell$ disjoint subsets $K_{1},K_{2},\ldots,K_{\ell}$ of
sizes $\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}$, respectively. For each
$i\in\left\{  1,2,\ldots,\ell\right\}  $, pick an arbitrary permutation in
$\Symm_n$ that fixes each element of $\left\{  1,2,\ldots,n\right\}
\setminus K_{i}$ while consisting of a single $\lambda_{i}$-cycle on the set
$K_{i}$. (Thus, $c_{i}$ cycles through the $\lambda_{i}$ elements of $K_{i}$
in some order while leaving all remaining elements of $\left\{  1,2,\ldots
,n\right\}  $ unchanged.) Then, the composition $c_{1}c_{2}\cdots c_{\ell}$ of
these cycles is a permutation in $\Symm_n$ that has cycle type
$\lambda$.} Choose such a permutation, and denote it by $g$.

Let $C_{g}$ denote the conjugacy class of $g$ in $\Symm_n$. Let
$Z_{g}$ denote the centralizer of $g$ in $\Symm_n$.

\begin{vershort}
We have $\left\vert \lambda\right\vert =n$ (since $\lambda$ is a partition of
$n$). Hence, from Remark \ref{rmk.zlambda.centralizer}, we know that
$z_{\lambda}$ is the size of the $\Symm_n$-centralizer subgroup for a
permutation having cycle type $\lambda$. Thus, $z_{\lambda}$ is the size of
the $\Symm_n$-centralizer subgroup of $g$ (since $g \in \Symm_n$
is a permutation having cycle type $\lambda$). In other words,
$z_{\lambda}=\left\vert Z_{g}\right\vert $ (since $Z_{g}$ is the
$\Symm_n$-centralizer subgroup of $g$).
\end{vershort}

\begin{verlong}
We have $\left\vert \lambda\right\vert =n$ (since $\lambda$ is a partition of
$n$). Hence, from Remark \ref{rmk.zlambda.centralizer}, we know that
$z_{\lambda}$ is the size of the $\Symm_n$-centralizer subgroup for a
permutation having cycle type $\lambda$. In other words, if $\sigma
\in \Symm_{n}$ is a permutation having cycle type $\lambda$, then
$z_{\lambda}$ is the size of the centralizer of $\sigma$. Applying this to
$\sigma=g$, we conclude that $z_{\lambda}$ is the size of the centralizer of
$g$ (since $g\in \Symm_{n}$ is a permutation having cycle type $\lambda
$). In other words, $z_{\lambda}$ is the size of $Z_{g}$ (since $Z_{g}$ is the
centralizer of $g$). In other words, $z_{\lambda}=\left\vert Z_{g}\right\vert
$. In other words, $\left\vert Z_{g}\right\vert =z_{\lambda}$.
\end{verlong}

\begin{vershort}
On the other hand, the definition of $C_{g}$ yields%
\begin{align*}
C_{g}  &  =\left(  \text{the conjugacy class of }g\right) \\
&  =\left\{  h\in \Symm_{n}\ \mid\ \underbrace{h\text{ is conjugate to
}g}_{\substack{\Longleftrightarrow\ \left(  g\text{ and }h\text{ are
conjugate}\right)  \\\Longleftrightarrow\ \left(  g\text{ and }h\text{ have
the same cycle type}\right)  \\\text{(by the equivalence
\eqref{sol.frob-char.ptype.a.ctc})}}}\right\}  =\left\{  h \in \Symm_n
\ \mid\ \underbrace{g\text{ and }h\text{ have the same cycle type}%
}_{\substack{\Longleftrightarrow\ \left(  h\text{ has cycle type }%
\lambda\right)  \\\text{(since }g\text{ has cycle type }\lambda\text{)}%
}}\right\} \\
&  =\left\{  h\in \Symm_{n}\ \mid\ \underbrace{h\text{ has cycle type
}\lambda}_{\Longleftrightarrow\ \left(  \operatorname*{type}h=\lambda\right)
}\right\}  =\left\{  h\in \Symm_{n}\ \mid\ \operatorname*{type}%
h=\lambda\right\}  =\left\{  \sigma\in \Symm_{n}\ \mid
\ \operatorname*{type}\sigma=\lambda\right\}
\end{align*}
(here, we have renamed the index $h$ as $\sigma$). Thus,%
\[
\left\vert C_{g}\right\vert =\left\vert \left\{  \sigma \in \Symm_n
\ \mid\ \operatorname*{type}\sigma=\lambda\right\}  \right\vert =\left(
\text{the number of all }\sigma\in \Symm_{n}\text{ satisfying
}\operatorname*{type}\sigma=\lambda\right)  .
\]

\end{vershort}

\begin{verlong}
On the other hand, the definition of $C_{g}$ yields%
\begin{align*}
C_{g}  &  =\left(  \text{the conjugacy class of }g\right) \\
&  =\left\{  h\in \Symm_{n}\ \mid\ \underbrace{h\text{ is conjugate to
}g}_{\substack{\Longleftrightarrow\ \left(  g\text{ and }h\text{ are
conjugate}\right)  \\\Longleftrightarrow\ \left(  g\text{ and }h\text{ have
the same cycle type}\right)  \\\text{(by the equivalence
\eqref{sol.frob-char.ptype.a.ctc})}}}\right\} \\
&  =\left\{  h\in \Symm_{n}\ \mid\ \underbrace{g\text{ and }h\text{ have
the same cycle type}}_{\Longleftrightarrow\ \left(  \left(  \text{the cycle
type of }g\right)  =\left(  \text{the cycle type of }h\right)  \right)
}\right\} \\
&  =\left\{  h\in \Symm_{n}\ \mid\ \underbrace{\left(  \text{the cycle
type of }g\right)  }_{\substack{=\lambda\\\text{(since }g\text{ has cycle type
}\lambda\text{)}}}=\underbrace{\left(  \text{the cycle type of }h\right)
}_{\substack{=\operatorname*{type}h\\\text{(since }\operatorname*{type}h\text{
is the cycle type of }h\\\text{(by the definition of }\operatorname*{type}%
h\text{))}}}\right\} \\
&  =\left\{  h\in \Symm_{n}\ \mid\ \underbrace{\lambda
=\operatorname*{type}h}_{\Longleftrightarrow\ \left(  \operatorname*{type}%
h=\lambda\right)  }\right\}  =\left\{  h\in \Symm_{n}\ \mid
\ \operatorname*{type}h=\lambda\right\}  =\left\{  \sigma \in \Symm_n
\ \mid\ \operatorname*{type}\sigma=\lambda\right\}
\end{align*}
(here, we have renamed the index $h$ as $\sigma$). Thus,%
\[
\left\vert C_{g}\right\vert =\left\vert \left\{  \sigma \in \Symm_n
\ \mid\ \operatorname*{type}\sigma=\lambda\right\}  \right\vert =\left(
\text{the number of all }\sigma\in \Symm_{n}\text{ satisfying
}\operatorname*{type}\sigma=\lambda\right)  .
\]

\end{verlong}

\noindent Comparing this with%
\begin{align*}
\left\vert C_{g}\right\vert  &  =\dfrac{\left\vert  \Symm_{n}\right\vert
}{\left\vert Z_{g}\right\vert }\ \ \ \ \ \ \ \ \ \ \left(  \text{by Claim 1,
applied to }G= \Symm_{n}\right) \\
&  =\dfrac{n!}{z_{\lambda}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert
 \Symm_{n}\right\vert =n!\text{ and }\left\vert Z_{g}\right\vert
=z_{\lambda}\right)  ,
\end{align*}
we obtain%
\[
\left(  \text{the number of all }\sigma\in \Symm_{n}\text{ satisfying
}\operatorname*{type}\sigma=\lambda\right)  =\dfrac{n!}{z_{\lambda}}.
\]
This proves Claim 2.]

For each partition $\lambda$ of $n$, there exists a permutation in
$\Symm_n$ that has cycle type $\lambda$. Choose such a permutation,
and denote it by $g_{\lambda}$. Every $h\in \Symm_{n}$ satisfies%
\begin{equation}
f\left(  h\right)  =f\left(  g_{\operatorname*{type}h}\right)  .
\label{sol.frob-char.ptype.a.lc.pf.0}%
\end{equation}


\begin{vershort}
[\textit{Proof of \eqref{sol.frob-char.ptype.a.lc.pf.0}:} Let $h\in
 \Symm_{n}$. Hence, $\operatorname*{type}h\in\operatorname*{Par}%
\nolimits_{n}$ (since $\operatorname*{type}\sigma\in\operatorname*{Par}%
\nolimits_{n}$ for each $\sigma\in \Symm_{n}$). Therefore, the
permutation $g_{\operatorname*{type}h}\in \Symm_{n}$ is well-defined.

The definition of $\operatorname*{type}h$ shows that the permutation $h$ has
cycle type $\operatorname*{type}h$. On the other hand, the permutation
$g_{\operatorname*{type}h}$ also has cycle type $\operatorname*{type}h$ (by
the definition of $g_{\operatorname*{type}h}$). Hence, the two permutations
$g_{\operatorname*{type}h}$ and $h$ in $\Symm_n$ have the same cycle
type (namely, $\operatorname*{type}h$). But (\ref{sol.frob-char.ptype.a.ctc})
(applied to $g_{\operatorname*{type}h}$ instead of $g$) shows that we have the
equivalence%
\[
\left(  g_{\operatorname*{type}h}\text{ and }h\text{ have the same cycle
type}\right)  \ \Longleftrightarrow\ \left(  g_{\operatorname*{type}h}\text{
and }h\text{ are conjugate}\right)  .
\]
Hence, $g_{\operatorname*{type}h}$ and $h$ are conjugate (since
$g_{\operatorname*{type}h}$ and $h$ have the same cycle type).

But we have $f\in R_{\CC}\left(   \Symm_{n}\right)  $. In other
words, $f$ is a class function of $\Symm_n$ (since $R_{\CC}
\left(   \Symm_{n}\right)  $ is the space of all class functions
$\Symm_{n} \rightarrow \CC$ of $\Symm_{n}$). In other
words, $f$ is a map $\Symm_{n} \rightarrow \CC$ that is constant
on $\Symm_n$-conjugacy classes (because this is what it means to be a
class function of $\Symm_n$). Thus, in particular, $f$ is constant on
$\Symm_n$-conjugacy classes. In other words, if $x$ and $y$ are two
conjugate elements of $\Symm_n$, then $f\left(  x\right)  =f\left(
y\right)  $. Applying this to $x=g_{\operatorname*{type}h}$ and $y=h$, we
obtain $f\left(  g_{\operatorname*{type}h}\right)  =f\left(  h\right)  $
(since the elements $g_{\operatorname*{type}h}$ and $h$ of $\Symm_n$
are conjugate). This proves \eqref{sol.frob-char.ptype.a.lc.pf.0}.]
\end{vershort}

\begin{verlong}
[\textit{Proof of \eqref{sol.frob-char.ptype.a.lc.pf.0}:} Let $h\in
 \Symm_{n}$. Recall that $\operatorname*{type}\sigma\in
\Par_n$ for each $\sigma\in \Symm_{n}$.
Applying this to $\sigma=h$, we obtain $\operatorname*{type}h\in
\Par_n$. Therefore, the permutation
$g_{\operatorname*{type}h}\in \Symm_{n}$ is well-defined.

But $\operatorname*{type}h$ is the cycle type of $h$ (by the definition of
$\operatorname*{type}h$). In other words, the permutation $h$ has cycle type
$\operatorname*{type}h$. On the other hand, the permutation
$g_{\operatorname*{type}h}$ also has cycle type $\operatorname*{type}h$
(because $g_{\operatorname*{type}h}$ was defined to be a permutation in
$\Symm_n$ that has cycle type $\operatorname*{type}h$). Hence, the
two permutations $g_{\operatorname*{type}h}$ and $h$ in $\Symm_n$
have the same cycle type (namely, $\operatorname*{type}h$). But
(\ref{sol.frob-char.ptype.a.ctc}) (applied to $g_{\operatorname*{type}h}$
instead of $g$) shows that we have the equivalence%
\[
\left(  g_{\operatorname*{type}h}\text{ and }h\text{ have the same cycle
type}\right)  \ \Longleftrightarrow\ \left(  g_{\operatorname*{type}h}\text{
and }h\text{ are conjugate}\right)  .
\]
Hence, $g_{\operatorname*{type}h}$ and $h$ are conjugate (since
$g_{\operatorname*{type}h}$ and $h$ have the same cycle type). In other words,
$g_{\operatorname*{type}h}$ and $h$ belong to the same $ \Symm_{n}%
$-conjugacy class. Let $D$ be this $\Symm_n$-conjugacy class. Thus,
$g_{\operatorname*{type}h}$ and $h$ both belong to $D$. In other words,
$g_{\operatorname*{type}h}$ and $h$ are elements of $D$.

But we have $f\in R_{\CC}\left(   \Symm_{n}\right)  $. In other
words, $f$ is a class function of $\Symm_n$ (since $R_{\CC}
\left(   \Symm_{n}\right)  $ is the space of all class functions
$\Symm_{n} \rightarrow \CC$ of $\Symm_n$). In other
words, $f$ is a map $\Symm_{n} \rightarrow \CC$ that is constant
on $\Symm_n$-conjugacy classes (because this is what it means to be a
class function of $\Symm_n$). Thus, in particular, $f$ is constant on
$\Symm_n$-conjugacy classes. In other words, if $C$ is any
$\Symm_n$-conjugacy class, then $f$ is constant on $C$. Applying this
to $C=D$, we conclude that $f$ is constant on $D$ (since $D$ is a
$\Symm_n$-conjugacy class). Therefore, $f\left(  x\right)  =f\left(
y\right)  $ whenever $x$ and $y$ are elements of $D$. Applying this to
$x=g_{\operatorname*{type}h}$ and $y=h$, we obtain $f\left(
g_{\operatorname*{type}h}\right)  =f\left(  h\right)  $ (since
$g_{\operatorname*{type}h}$ and $h$ are elements of $D$). This proves \eqref{sol.frob-char.ptype.a.lc.pf.0}.]
\end{verlong}

Recall that $\operatorname*{type}\sigma\in\Par_n$
for each $\sigma\in \Symm_{n}$. Thus, we can split the sum $\sum
_{\sigma\in \Symm_{n}}f\left(  \sigma\right)  p_{\operatorname*{type}%
\sigma}$ according to the value of $\operatorname*{type}\sigma$. We thus
obtain%
\begin{align*}
\underbrace{\sum_{\sigma\in \Symm_{n}}}_{\substack{=\sum\limits_{\lambda
\in\Par_n }\ \ \sum\limits_{\substack{\sigma
\in \Symm_{n};\\\operatorname*{type}\sigma=\lambda}}\\\text{(since
}\operatorname*{type}\sigma\in\Par_n \\\text{for each
}\sigma\in \Symm_{n}\text{)}}}f\left(  \sigma\right)
p_{\operatorname*{type}\sigma}  &  =\sum\limits_{\lambda\in\operatorname*{Par}%
\nolimits_{n}}\ \ \sum\limits_{\substack{\sigma\in \Symm_{n}%
;\\\operatorname*{type}\sigma=\lambda}}\ \ \underbrace{f\left(  \sigma\right)
}_{\substack{=f\left(  g_{\operatorname*{type}\sigma}\right)  \\\text{(by
\eqref{sol.frob-char.ptype.a.lc.pf.0}, applied to }h=\sigma\text{)}%
}}p_{\operatorname*{type}\sigma}\\
&  =\sum\limits_{\lambda\in\Par_n }\ \ \sum
\limits_{\substack{\sigma\in \Symm_{n};\\\operatorname*{type}%
\sigma=\lambda}}\ \ \underbrace{f\left(  g_{\operatorname*{type}\sigma
}\right)  }_{\substack{=f\left(  g_{\lambda}\right)  \\\text{(since
}\operatorname*{type}\sigma=\lambda\text{)}}%
}\ \ \underbrace{p_{\operatorname*{type}\sigma}}_{\substack{=p_{\lambda
}\\\text{(since }\operatorname*{type}\sigma=\lambda\text{)}}}\\
&  =\sum\limits_{\lambda\in\Par_n }%
\ \ \underbrace{\sum\limits_{\substack{\sigma\in \Symm_{n}%
;\\\operatorname*{type}\sigma=\lambda}}f\left(  g_{\lambda}\right)
p_{\lambda}}_{=\left(  \text{the number of all }\sigma \in \Symm_n
\text{ satisfying }\operatorname*{type}\sigma=\lambda\right)  \cdot
f\left(  g_{\lambda}\right)  p_{\lambda}}\\
&  =\sum_{\lambda\in\Par_n }\underbrace{\left(
\text{the number of all }\sigma\in \Symm_{n}\text{ satisfying
}\operatorname*{type}\sigma=\lambda\right)  }_{\substack{=\dfrac
{n!}{z_{\lambda}}\\\text{(by Claim 2)}}}\cdot f\left(  g_{\lambda}\right)
p_{\lambda}\\
&  =\sum_{\lambda\in\Par_n }\dfrac{n!}{z_{\lambda}%
}\cdot f\left(  g_{\lambda}\right)  p_{\lambda}.
\end{align*}
Multiplying both sides of this equality by $\dfrac{1}{n!}$, we obtain%
\begin{align}
\dfrac{1}{n!}\sum_{\sigma\in \Symm_{n}}f\left(  \sigma\right)
p_{\operatorname*{type}\sigma}  &  =\dfrac{1}{n!}\cdot\sum_{\lambda
\in\Par_n }\dfrac{n!}{z_{\lambda}}\cdot f\left(
g_{\lambda}\right)  p_{\lambda}=\sum_{\lambda\in\operatorname*{Par}%
\nolimits_{n}}\underbrace{\dfrac{1}{n!}\cdot\dfrac{n!}{z_{\lambda}}\cdot
f\left(  g_{\lambda}\right)  p_{\lambda}}_{=f\left(  g_{\lambda}\right)
\cdot\dfrac{p_{\lambda}}{z_{\lambda}}}\nonumber\\
&  =\sum_{\lambda\in\Par_n }f\left(  g_{\lambda
}\right)  \cdot\dfrac{p_{\lambda}}{z_{\lambda}}.
\label{sol.frob-char.ptype.a.rhs}%
\end{align}


We shall now show that%
\begin{equation}
f=\sum_{\lambda\in\Par_n }f\left(  g_{\lambda
}\right)  \underline{1}_{\lambda}. \label{sol.frob-char.ptype.a.lc}%
\end{equation}


[\textit{Proof of \eqref{sol.frob-char.ptype.a.lc}:} Let $h \in \Symm_n$.
Then, $\operatorname*{type}h$ is the cycle type of $h$ (by the
definition of $\operatorname*{type}h$). In other words, $h$ has cycle type
$\operatorname*{type}h$.

Recall that $\operatorname*{type}\sigma\in\Par_n$
for each $\sigma\in \Symm_{n}$. Applying this to $\sigma=h$, we obtain
$\operatorname*{type}h\in\Par_n$.

For each partition $\lambda\in\Par_n$, we have%
\begin{equation}
\underline{1}_{\lambda}\left(  h\right)  =%
\begin{cases}
1, & \text{if }h\text{ has cycle type }\lambda;\\
0, & \text{otherwise}%
\end{cases}
\label{sol.frob-char.ptype.a.lc.pf.1}%
\end{equation}
(since $\underline{1}_{\lambda}$ was defined as the characteristic function
for the $\Symm_n$-conjugacy class of permutations of cycle type
$\lambda$). Applying this to $\lambda=\operatorname*{type}h$, we obtain
\begin{equation}
\underline{1}_{\operatorname*{type}h}\left(  h\right)  =%
\begin{cases}
1, & \text{if }h\text{ has cycle type }\operatorname*{type}h;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ =1 \label{sol.frob-char.ptype.a.lc.pf.2}%
\end{equation}
(since $h$ has cycle type $\operatorname*{type}h$). On the other hand, if
$\lambda\in\Par_n$ is a partition such that
$\lambda\neq\operatorname*{type}h$, then $h$ does not have cycle type
$\lambda$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $h$ has
cycle type $\lambda$. In other words, the cycle type of $h$ is $\lambda$. In
other words, $\operatorname*{type}h$ is $\lambda$ (since $\operatorname*{type}%
h$ is the cycle type of $h$). In other words, $\operatorname*{type}h=\lambda$.
But this contradicts $\lambda\neq\operatorname*{type}h$. This contradiction
shows that our assumption was false. Qed.}, and therefore we have%
\begin{align}
\underline{1}_{\lambda}\left(  h\right)   &  =%
\begin{cases}
1, & \text{if }h\text{ has cycle type }\lambda;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by \eqref{sol.frob-char.ptype.a.lc.pf.1}}%
\right) \nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h\text{ does not have cycle type
}\lambda\right)  . \label{sol.frob-char.ptype.a.lc.pf.3}%
\end{align}


Now,%
\begin{align*}
\left(  \sum_{\lambda\in\Par_n }f\left(  g_{\lambda
}\right)  \underline{1}_{\lambda}\right)  \left(  h\right)   &  =\sum
_{\lambda\in\Par_n }f\left(  g_{\lambda}\right)
\underline{1}_{\lambda}\left(  h\right) \\
&  =f\left(  g_{\operatorname*{type}h}\right)  \underbrace{\underline{1}%
_{\operatorname*{type}h}\left(  h\right)  }_{\substack{=1\\\text{(by
\eqref{sol.frob-char.ptype.a.lc.pf.2})}}}+\sum_{\substack{\lambda
\in\Par_n ;\\\lambda\neq\operatorname*{type}%
h}}f\left(  g_{\lambda}\right)  \underbrace{\underline{1}_{\lambda}\left(
h\right)  }_{\substack{=0\\\text{(by \eqref{sol.frob-char.ptype.a.lc.pf.3})}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }\lambda=\operatorname*{type}%
h\text{ from the sum,}\\
\text{since }\operatorname*{type}h\in\Par_n %
\end{array}
\right) \\
&  =f\left(  g_{\operatorname*{type}h}\right)  +\underbrace{\sum
_{\substack{\lambda\in\Par_n ;\\\lambda
\neq\operatorname*{type}h}}f\left(  g_{\lambda}\right)  0}_{=0}=f\left(
g_{\operatorname*{type}h}\right)  =f\left(  h\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.frob-char.ptype.a.lc.pf.0}%
)}\right)  .
\end{align*}


Forget that we fixed $h$. We thus have shown that $\left(  \sum_{\lambda
\in\Par_n }f\left(  g_{\lambda}\right)
\underline{1}_{\lambda}\right)  \left(  h\right)  =f\left(  h\right)  $ for
each $h\in \Symm_{n}$. In other words, $\sum_{\lambda\in
\Par_n }f\left(  g_{\lambda}\right)  \underline{1}%
_{\lambda}=f$ (since both $\sum_{\lambda\in\Par_n %
}f\left(  g_{\lambda}\right)  \underline{1}_{\lambda}$ and $f$ are maps from
$\Symm_{n}$ to $\CC$). This proves \eqref{sol.frob-char.ptype.a.lc}.]

One of the claims of Theorem \ref{symmetric-group-Frobenius-map-theorem}(a) is
the formula%
\begin{equation}
\operatorname*{ch}\left(  \underline{1}_{\lambda}\right)  =\dfrac{p_{\lambda}%
}{z_{\lambda}}\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in
\Par_n . \label{sol.frob-char.ptype.a.ch1}%
\end{equation}


Now, applying the map $\operatorname*{ch}:A_{\CC}\rightarrow
\Lambda_{\CC}$ to both sides of the equality
\eqref{sol.frob-char.ptype.a.lc}, we find%
\begin{align*}
\operatorname*{ch}\left(  f\right)   &  =\operatorname*{ch}\left(
\sum_{\lambda\in\Par_n }f\left(  g_{\lambda}\right)
\underline{1}_{\lambda}\right)  =\sum_{\lambda\in\operatorname*{Par}%
\nolimits_{n}}f\left(  g_{\lambda}\right)  \underbrace{\operatorname*{ch}%
\left(  \underline{1}_{\lambda}\right)  }_{\substack{=\dfrac{p_{\lambda}%
}{z_{\lambda}}\\\text{(by \eqref{sol.frob-char.ptype.a.ch1})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\operatorname*{ch}%
:A_{\CC}\rightarrow\Lambda_{\CC}\text{ is } \CC\text{-linear}\right) \\
&  =\sum_{\lambda\in\Par_n }f\left(  g_{\lambda
}\right)  \cdot\dfrac{p_{\lambda}}{z_{\lambda}}=\dfrac{1}{n!}\sum_{\sigma
\in \Symm_{n}}f\left(  \sigma\right)  p_{\operatorname*{type}\sigma}%
\end{align*}
(by \eqref{sol.frob-char.ptype.a.rhs}). This solves Exercise
\ref{exe.frob-char.ptype}(a).

(b) It is easy to see that every $k\in \Symm_{n}$ and $\sigma
\in \Symm_{n}$ satisfy%
\begin{equation}
p_{\operatorname*{type}\sigma}=p_{\operatorname*{type}\left(  k\sigma
k^{-1}\right)  }. \label{sol.frob-char.ptype.b.0}%
\end{equation}


\begin{vershort}
[\textit{Proof of \eqref{sol.frob-char.ptype.b.0}:} Let $k\in \Symm_{n}$
and $\sigma\in \Symm_{n}$. Then, \eqref{sol.frob-char.ptype.a.ctc}
(applied to $g=\sigma$ and $h=k\sigma k^{-1}$) yields the equivalence%
\[
\left(  \sigma\text{ and }k\sigma k^{-1}\text{ have the same cycle
type}\right)  \ \Longleftrightarrow\ \left(  \sigma\text{ and }k\sigma
k^{-1}\text{ are conjugate}\right)  .
\]
Hence, $\sigma$ and $k\sigma k^{-1}$ have the same cycle type (since $\sigma$
and $k\sigma k^{-1}$ are conjugate). In other words, $\operatorname*{type}%
\sigma=\operatorname*{type}\left(  k\sigma k^{-1}\right)  $. Hence,
$p_{\operatorname*{type}\sigma}=p_{\operatorname*{type}\left(  k\sigma
k^{-1}\right)  }$. This proves \eqref{sol.frob-char.ptype.b.0}.]
\end{vershort}

\begin{verlong}
[\textit{Proof of \eqref{sol.frob-char.ptype.b.0}:} Let $k\in \Symm_{n}$
and $\sigma\in \Symm_{n}$. Then, the permutations $\sigma\in
 \Symm_{n}$ and $k\sigma k^{-1}\in \Symm_{n}$ are conjugate
(obviously). But \eqref{sol.frob-char.ptype.a.ctc} (applied to $g=\sigma$ and
$h=k\sigma k^{-1}$) yields the equivalence%
\[
\left(  \sigma\text{ and }k\sigma k^{-1}\text{ have the same cycle
type}\right)  \ \Longleftrightarrow\ \left(  \sigma\text{ and }k\sigma
k^{-1}\text{ are conjugate}\right)  .
\]
Hence, $\sigma$ and $k\sigma k^{-1}$ have the same cycle type (since $\sigma$
and $k\sigma k^{-1}$ are conjugate). In other words, $\left(  \text{the cycle
type of }\sigma\right)  =\left(  \text{the cycle type of }k\sigma
k^{-1}\right)  $. But the definition of $\operatorname*{type}\sigma$ yields
$\operatorname*{type}\sigma=\left(  \text{the cycle type of }\sigma\right)  $.
Furthermore, the definition of $\operatorname*{type}\left(  k\sigma
k^{-1}\right)  $ yields $\operatorname*{type}\left(  k\sigma k^{-1}\right)
=\left(  \text{the cycle type of }k\sigma k^{-1}\right)  $. Thus,%
\[
\operatorname*{type}\sigma=\left(  \text{the cycle type of }\sigma\right)
=\left(  \text{the cycle type of }k\sigma k^{-1}\right)  =\operatorname*{type}%
\left(  k\sigma k^{-1}\right)  .
\]
Hence, $p_{\operatorname*{type}\sigma}=p_{\operatorname*{type}\left(  k\sigma
k^{-1}\right)  }$. This proves \eqref{sol.frob-char.ptype.b.0}.]
\end{verlong}

Let $f\in R_{\CC}\left(  H\right)  $. Applying
\eqref{eq.exe.induction.characters.eq} to $G= \Symm_{n}$, we obtain%
\begin{equation}
\left(  \operatorname{Ind}_{H}^{ \Symm_{n}}f\right)  \left(  g\right)
=\dfrac{1}{\left\vert H\right\vert }\sum\limits_{\substack{k \in \Symm_n
:\\kgk^{-1}\in H}}f\left(  kgk^{-1}\right)
\label{sol.frob-char.ptype.b.1}%
\end{equation}
for all $g\in \Symm_{n}$. Furthermore, Exercise
\ref{exe.induction.characters}(a) (applied to $G= \Symm_{n}$) shows that
$\operatorname{Ind}_{H}^{ \Symm_{n}}f$ is a class function on
$\Symm_n$, hence belongs to $R_{\CC}\left(  \Symm_n \right)  $.
Hence, Exercise \ref{exe.frob-char.ptype}(a) (applied to
$\operatorname{Ind}_{H}^{ \Symm_{n}}f$ instead of $f$) yields%
\begin{align*}
\operatorname*{ch}\left(  \operatorname*{Ind}\nolimits_{H}^{ \Symm_{n}%
}f\right)   &  =\dfrac{1}{n!}\sum_{\sigma\in \Symm_{n}}%
\underbrace{\left(  \operatorname{Ind}_{H}^{ \Symm_{n}}f\right)  \left(
\sigma\right)  }_{\substack{=\dfrac{1}{\left\vert H\right\vert }%
\sum\limits_{\substack{k\in \Symm_{n}:\\k\sigma k^{-1}\in H}}f\left(
k\sigma k^{-1}\right)  \\\text{(by \eqref{sol.frob-char.ptype.b.1}, applied to
}g=\sigma\text{)}}}p_{\operatorname*{type}\sigma}\\
&  =\dfrac{1}{n!}\sum_{\sigma\in \Symm_{n}}\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{\substack{k\in \Symm_{n}:\\k\sigma k^{-1}\in
H}}f\left(  k\sigma k^{-1}\right)  \underbrace{p_{\operatorname*{type}\sigma}%
}_{\substack{=p_{\operatorname*{type}\left(  k\sigma k^{-1}\right)
}\\\text{(by \eqref{sol.frob-char.ptype.b.0})}}}\\
&  =\dfrac{1}{n!}\sum_{\sigma\in \Symm_{n}}\dfrac{1}{\left\vert
H\right\vert }\sum\limits_{\substack{k\in \Symm_{n}:\\k\sigma k^{-1}\in
H}}f\left(  k\sigma k^{-1}\right)  p_{\operatorname*{type}\left(  k\sigma
k^{-1}\right)  }\\
&  =\dfrac{1}{n!}\underbrace{\sum_{\sigma\in \Symm_{n}}\sum
\limits_{\substack{k\in \Symm_{n}:\\k\sigma k^{-1}\in H}}}%
_{=\sum\limits_{k\in \Symm_{n}}\sum\limits_{\substack{\sigma
\in \Symm_{n}:\\k\sigma k^{-1}\in H}}}\dfrac{1}{\left\vert H\right\vert
}f\left(  k\sigma k^{-1}\right)  p_{\operatorname*{type}\left(  k\sigma
k^{-1}\right)  }\\
&  =\dfrac{1}{n!}\sum\limits_{k\in \Symm_{n}}\underbrace{\sum
\limits_{\substack{\sigma\in \Symm_{n}:\\k\sigma k^{-1}\in H}}\dfrac
{1}{\left\vert H\right\vert }f\left(  k\sigma k^{-1}\right)
p_{\operatorname*{type}\left(  k\sigma k^{-1}\right)  }}_{\substack{=\sum
\limits_{\substack{h\in \Symm_{n}:\\h\in H}}\dfrac{1}{\left\vert
H\right\vert }f\left(  h\right)  p_{\operatorname*{type}h}\\\text{(here, we
have substituted }h\text{ for }k\sigma k^{-1}\text{ in the sum,}\\\text{since
the map } \Symm_{n}\rightarrow \Symm_{n},\ \sigma\mapsto k\sigma
k^{-1}\text{ is a bijection)}}}\\
&  =\dfrac{1}{n!}\sum\limits_{k\in \Symm_{n}}\underbrace{\sum
\limits_{\substack{h\in \Symm_{n}:\\h\in H}}}_{\substack{=\sum_{h\in
H}\\\text{(since }H\subset \Symm_{n}\text{)}}}\dfrac{1}{\left\vert
H\right\vert }f\left(  h\right)  p_{\operatorname*{type}h}=\dfrac{1}%
{n!}\underbrace{\sum\limits_{k\in \Symm_{n}}\sum_{h\in H}\dfrac
{1}{\left\vert H\right\vert }f\left(  h\right)  p_{\operatorname*{type}h}%
}_{=\left\vert  \Symm_{n}\right\vert \cdot\sum_{h\in H}\dfrac
{1}{\left\vert H\right\vert }f\left(  h\right)  p_{\operatorname*{type}h}}\\
&  =\dfrac{1}{n!}\cdot\underbrace{\left\vert  \Symm_{n}\right\vert
}_{=n!}\cdot\sum_{h\in H}\dfrac{1}{\left\vert H\right\vert }f\left(  h\right)
p_{\operatorname*{type}h}=\underbrace{\dfrac{1}{n!}\cdot n!}_{=1}\cdot
\sum_{h\in H}\dfrac{1}{\left\vert H\right\vert }f\left(  h\right)
p_{\operatorname*{type}h}\\
&  =\sum_{h\in H}\dfrac{1}{\left\vert H\right\vert }f\left(  h\right)
p_{\operatorname*{type}h}=\dfrac{1}{\left\vert H\right\vert }\sum_{h\in
H}f\left(  h\right)  p_{\operatorname*{type}h}.
\end{align*}
This solves Exercise \ref{exe.frob-char.ptype}(b).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.symmetric-group-reps-rational}}

\textit{Solution to Exercise~\ref{exe.symmetric-group-reps-rational}.}
We need an auxiliary observation first. If $G$ is a
finite group, then a class function $\chi\in R_\CC\left(  G\right)  $ of
$G$ will be called \emph{integral} if every $g\in G$ satisfies
$\chi\left(  g\right) \in \ZZ$. Then, if $G$ and $H$ are two groups and if
$\phi$ and $\psi$ are two integral class functions of $G$ and $H$,
respectively, then
\begin{equation}
\phi\otimes\psi\text{ is an integral class function of }G\times
H.\label{sol.symmetric-group-reps-rational.1}
\end{equation}
(This is a consequence of the fact that $\left(  \phi\otimes\psi\right)
\left(  g,h\right)  =\phi\left(  g\right)  \psi\left(  h\right)  $ for all
$\left(  g,h\right)  \in G\times H$.) Moreover, if $H$ is a subgroup of a
group $G$ and if $\chi$ is an integral class function of $H$, then
\begin{equation}
\Ind_{H}^{G}\chi\text{ is an integral class function
of }G.\label{sol.symmetric-group-reps-rational.2}
\end{equation}
(\textit{Proof of \eqref{sol.symmetric-group-reps-rational.2}:}
Let $J$ be a system of coset representatives for $H\backslash G$, so that
$G=\bigsqcup_{j\in J}Hj$. Then, Exercise~\ref{exe.induction.characters}(a)
(applied to $f = \chi$) shows that $\Ind_{H}^{G}\chi$ is a class
function on $G$.
Also, Exercise~\ref{exe.induction.characters}(b)
(applied to $f = \chi$) yields
\[
\left(  \Ind_{H}^{G}\chi\right)  \left(  g\right)    
 =\sum\limits_{\substack{j\in J:\\jgj^{-1}\in H}}\underbrace{\chi\left(
jgj^{-1}\right)  }_{\substack{\in \ZZ \\\text{(since }\chi\text{ is an
integral}\\\text{class function)}}}\in \ZZ
\qquad \text{ for all } g \in G.
\]
In other words, $\Ind^G_H\chi$ is an integral class function on $G$.
This proves \eqref{sol.symmetric-group-reps-rational.2}.)

Now let $A=A\left(  \Symm\right)  $. For every $a\in A$, every $n\geq 0$
and every $g\in\Symm_n$, we define $a\left(  g\right)  $ to mean the
value at $g$ of the $n$-th homogeneous component of $a$. Let $\widetilde{A}$
denote the subset of $A$ formed by all $a\in A$ such that:
\begin{equation}
\left(  \text{for every }n\geq0\text{ and every }g\in\Symm_{n}\text{,
we have }a\left(  g\right)  \in \ZZ \right)
.\label{sol.symmetric-group-reps-rational.4}
\end{equation}
It is clear that $\widetilde{A}$ is a graded $\ZZ$-submodule of $A$ and
contains $1=\triv_{\Symm_0}$.

We are now going to show that $\widetilde{A}$ is closed under multiplication.
In order to do so, it is clearly enough to prove that if $a\in\widetilde{A}$
is homogeneous of degree $n$ and $b\in\widetilde{A}$ is homogeneous of degree
$m$, then $ab\in\widetilde{A}$. But this is now easy: Since $a\in
\widetilde{A}$, we know that $a$ is an integral class function on
$\Symm_{n}$. Similarly, $b$ is an integral class function on
$\Symm_{m}$. Hence, \eqref{sol.symmetric-group-reps-rational.1} shows
that $a\otimes b$ is an integral class function on
$\Symm_{n}\times\Symm_{m}$. Thus,
$\Ind_{\Symm_{n}\times\Symm_{m}}^{\Symm_{n+m}}\left(  a\otimes b\right)  $
is an integral class function on $\Symm_{n+m}$ (by
\eqref{sol.symmetric-group-reps-rational.2}). In other words, $ab$ is an
integral class function on $\Symm_{n+m}$ (since
$ab = \ind_{n,m}^{n+m} \left(  a\otimes b\right)
 = \Ind_{\Symm_n \times \Symm_m}^{\Symm_{n+m}}\left(a\otimes b\right)  $).
In other words, $ab \in \widetilde{A}$. This proves that
$\widetilde{A}$ is closed under multiplication. Hence, $\widetilde{A}$ is a
subring of $A$ (since $\widetilde{A}$ is a $\ZZ$-submodule of $A$ and
contains $1=\triv_{\Symm_0}$). Thus, $\ch\left(  \widetilde{A}\right)  $
(where $\ch$ is defined as in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}) is a subring of
$\Lambda$. For
every $n\geq 1$, we have $\triv_{\Symm_n}\in\widetilde{A}$ (by the
definitions), so that $\ch\left(  \triv_{\Symm_n}\right)
\in\ch\left(  \widetilde{A}\right)  $. Since
$\ch\left(  \triv_{\Symm_n}\right)  = h_n$, this rewrites
as $h_n \in\ch\left(  \widetilde{A}\right)  $. Thus, the
subring $\ch\left(  \widetilde{A}\right)  $ of $\Lambda$
contains $h_n$ for all $n\geq 1$. Hence, $\ch\left(
\widetilde{A}\right)  =\Lambda$ (because the $h_n$ generate $\Lambda$ as a
ring). Therefore, $\widetilde{A}=A$ (since $\ch$ is an
isomorphism). If we recall how $\widetilde{A}$ was defined, we thus see that
every $a\in A$ satisfies \eqref{sol.symmetric-group-reps-rational.4}.

Now, fix $n\geq0$ and $g\in\Symm_{n}$ and a finite-dimensional
$\CC \Symm_n$-module $V$. Then,
$\chi_{V}\in R\left( \Symm_n\right)  \subset A$. Thus,
\eqref{sol.symmetric-group-reps-rational.4} (which holds for every $a\in A$,
as we now know) yields that $\chi_{V}\left(  g\right)  \in\ZZ$, which
solves part (a) of the problem.

Note that a different way to solve part (a) would be by showing that $\Lambda$
is contained in the $\ZZ$-submodule of $\Lambda_\QQ$ spanned
by the $\dfrac{p_{\lambda}}{z_{\lambda}}$ for all partitions $\lambda$. This
yields (by taking preimages under $\ch$) that $A$ is contained
in the $\ZZ$-submodule of $R_{\CC}\left(  \Symm\right)  $
generated by the $\triv_{\lambda}$. This way, we wouldn't have to show that
$\widetilde{A}$ is closed under multiplication; instead, we would obtain
$\widetilde{A}=A$ by noticing that all $\triv_{\lambda}$ are integral class functions.

(b) This can be done using the Noether-Deuring theorem (in fact, it is
easy to show that there are two $\CC \Symm_n$-modules $U$ and $U'$
defined over $\QQ$ satisfying $U' \oplus V \cong U$, and then the
Noether-Deuring theorem allows to ``pull back'' $V$ to a
$\QQ \Symm_n$-module as well, showing that $V$ is also defined
over $\QQ$). We are omitting this argument because it is somewhat
technical and not very enlightening. (The ``right'' approach, in
my opinion, is to construct
the required $\QQ \Symm_n$-module $W$ explicitly; this is done, e.g., in
\cite[\S 7, between Proposition 1 and Lemma 3]{Fulton} or in
\cite[Corollaire 2.2.26]{Krob}. Of course, this
does not have much to do with what we are doing in our notes.)
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.repSn.inner-tensor.intprod}}

\textit{Solution to Exercise~\ref{exe.repSn.inner-tensor.intprod}.} (a)
Let $G$ be any group. If $U_{1}$ and $U_{2}$ are two
$\CC G$-modules, then the character $\chi_{U_1 \boxtimes U_2}$ of the
inner tensor product $U_1 \boxtimes U_2$ of $U_1$ and $U_2$ is given by
\begin{equation}
\chi_{U_{1}\boxtimes U_{2}}\left(  g\right)  =\chi_{U_{1}}\left(  g\right)
\chi_{U_{2}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in G.
\label{sol.repSn.inner-tensor.intprod.a.charprod}
\end{equation}
(This is just a restatement of
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}) using our
notation $U_{1}\boxtimes U_{2}$ for what, in
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}), was
called $U_{1}\otimes U_{2}$.)

Define a map $\ast:R_\CC \left(  G\right)  \times R_{\CC %
}\left(  G\right)  \rightarrow R_\CC \left(  G\right)  $, which will
be written in infix notation (that is, we will write $a\ast b$ instead of
$\ast\left(  a,b\right)  $), by setting%
\[
\left(  a\ast b\right)  \left(  g\right)  =a\left(  g\right)  b\left(
g\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }a\in R_\CC \left(
G\right)  \text{, }b\in R_\CC \left(  G\right)  \text{ and }g\in G.
\]
(This notation $a\ast b$ generalizes the notation $\operatorname{sgn}%
_{\Symm_{n}}\ast f$ used in Theorem
\ref{symmetric-group-Frobenius-map-theorem}.) The map $\ast$ is clearly
$\CC$-bilinear.

Notice that%
\begin{equation}
\chi_{U_{1}}\ast\chi_{U_{2}}=\chi_{U_{1}\boxtimes U_{2}}%
\ \ \ \ \ \ \ \ \ \ \text{for any two }\mathbb{C}G\text{-modules }U_{1}\text{
and }U_{2} \label{sol.repSn.inner-tensor.intprod.a.charprod2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.inner-tensor.intprod.a.charprod2}%
):} Let $U_{1}$ and $U_{2}$ be two $\mathbb{C}G$-modules. Then, every $g\in G$
satisfies%
\begin{align*}
\left(  \chi_{U_{1}}\ast\chi_{U_{2}}\right)  \left(  g\right)   &
=\chi_{U_{1}}\left(  g\right)  \chi_{U_{2}}\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\ast\right) \\
&  =\chi_{U_{1}\boxtimes U_{2}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.repSn.inner-tensor.intprod.a.charprod})}\right)  .
\end{align*}
In other words, $\chi_{U_{1}}\ast\chi_{U_{2}}=\chi_{U_{1}\boxtimes U_{2}}$.
This proves (\ref{sol.repSn.inner-tensor.intprod.a.charprod2}).}

Let $R_{ \QQ }\left(  G\right)  $ denote the set of class functions
$G\rightarrow \QQ $. This is clearly a subset of $R_{\mathbb{C}}\left(
G\right)  $. In general, it is not true that $R\left(  G\right)  \subset
R_{ \QQ }\left(  G\right)  $, but we will see that this holds for
$G=\Symm_{n}$ for any $n\in \NN$.

It is clear that $a\ast b\in R_{ \QQ }\left(  G\right)  $ for any $a\in
R_{ \QQ }\left(  G\right)  $ and $b\in R_{ \QQ }\left(  G\right)  $.

Now, forget that we fixed $G$. We have thus introduced a map $\ast
:R_\CC \left(  G\right)  \times R_\CC \left(  G\right)
\rightarrow R_\CC \left(  G\right)  $ for every group $G$, and we
have proved some properties of this map.

Let now $n\in \NN$. Exercise~\ref{exe.symmetric-group-reps-rational}(a)
yields that $\chi_{V} \left( g \right) \in \ZZ \subset \QQ$ for
every $g \in \Symm_n$ and every finite-dimensional
$\CC \Symm_n$-module $V$. Hence, $\chi_{V}$ is a map from $\Symm_n$ to
$\QQ$ for every finite-dimensional $\CC \Symm_n$-module $V$.
In other words, $\chi_{V} \in R_{\QQ} \left( \Symm_n \right)$
for every finite-dimensional
$\CC \Symm_n$-module $V$. Thus, $R\left(  \Symm_{n}\right)
\subset R_{\QQ}\left(  \Symm_{n}\right)  $ (since $R\left(
\Symm_{n}\right)  $ is the $\ZZ$-module generated by the
$\chi_{V}$ for $V$ ranging over the irreducible $\mathbb{C}\Symm_{n}$-modules).

Every element of $R_{ \QQ }\left(  \Symm_{n}\right)  $ is a class
function $\Symm_{n}\rightarrow \QQ $, and thus a $ \QQ %
$-linear combination of the $\triv_{\lambda}$ for $\lambda
\in\Par_{n}$ (because these functions $\triv%
_{\lambda}$ are the indicator functions for the conjugacy classes of
$\Symm_{n}$). Thus, the $\QQ$-module $R_{ \QQ }\left(
\Symm_{n}\right)  $ is spanned by the $\triv_{\lambda}$ for
$\lambda\in\Par_{n}$.

As in Theorem~\ref{symmetric-group-Frobenius-map-theorem}, we
extend the PSH-isomorphism $\ch : A \to \Lambda$ to a $\CC$-Hopf
algebra isomorphism $A_\CC \to \Lambda_\CC$; we shall denote the
latter isomorphism by $\ch$ as well.
Now, let us notice that
\begin{equation}
\ch \left(  R_{ \QQ }\left(  \Symm_{n}\right)
\right)  \subset\Lambda_{ \QQ } .
\label{sol.repSn.inner-tensor.intprod.a.inLQ}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.inner-tensor.intprod.a.inLQ}):} It
is clearly enough to show that
$\ch \left(  \triv _{\lambda}\right)  \in\Lambda_{\QQ}$ for
every $\lambda\in \Par_{n}$ (because the $\QQ$-module
$R_{ \QQ }\left(  \Symm_{n}\right)  $ is spanned by the
$\triv_{\lambda}$ for $\lambda\in\Par_{n}$).
But this follows from the fact that $\ch \left(  \triv%
_{\lambda}\right)  =\frac{p_{\lambda}}{z_{\lambda}}$ for every $\lambda
\in\Par_{n}$ (this is part of
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(a)). Thus,
(\ref{sol.repSn.inner-tensor.intprod.a.inLQ}) is proven.} Furthermore,
\begin{equation}
\ch \left(  a\ast b\right)  =\left(  \ch a\right)
\ast\left(  \ch b\right)  \ \ \ \ \ \ \ \ \ \ \text{for any
}a\in R_{ \QQ }\left(  \Symm_{n}\right)  \text{ and }b\in
R_{ \QQ }\left(  \Symm_{n}\right)
.\label{sol.repSn.inner-tensor.intprod.a.ch*}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.inner-tensor.intprod.a.ch*}):} Let
$a\in R_{ \QQ }\left(  \Symm_{n}\right)  $ and $b\in
R_{ \QQ }\left(  \Symm_{n}\right)  $. The equality
(\ref{sol.repSn.inner-tensor.intprod.a.ch*}) is clearly $\QQ$-linear in
each of $a$ and $b$. Hence, in proving this equality, we can WLOG assume that
$a$ and $b$ are two of the functions $\triv_{\lambda}$ for $\lambda
\in\Par_{n}$ (because the $\QQ$-module
$R_{ \QQ }\left(  \Symm_{n}\right)  $ is spanned by the
$\triv_{\lambda}$ for $\lambda\in\Par_{n}$).
In other words, we can WLOG assume that $a=\triv_{\lambda}$ and
$b=\triv_{\mu}$ for some $\lambda\in\Par_{n}$
and $\mu\in\Par_{n}$. Assume this, and consider these
$\lambda$ and $\mu$.
\par
Recall that $\triv_{\lambda}$ is the indicator function for the set of
all permutations in $\Symm_{n}$ having cycle type $\lambda$, and that
$\triv_{\mu}$ is the indicator function for the set of all
permutations in $\Symm_{n}$ having cycle type $\mu$. Hence, every
$g\in\Symm_{n}$ satisfies
\[
\triv_{\lambda}\left(  g\right)  \cdot\triv_{\mu}\left(
g\right)  =\delta_{\lambda,\mu}\cdot\triv_{\lambda}\left(  g\right)
\]
(because both sides of this equality vanish if $g$ does not have cycle type
$\lambda$, and also vanish if $\lambda\neq\mu$, but in the remaining case are
both equal to $1$). Thus, every $g\in\Symm_{n}$ satisfies
\begin{align*}
\left(  \triv_{\lambda}\ast\triv_{\mu}\right)  \left(
g\right)   &  =\triv_{\lambda}\left(  g\right)  \cdot\triv%
_{\mu}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\triv_{\lambda}\ast\triv_{\mu}\right)  \\
&  =\delta_{\lambda,\mu}\cdot\triv_{\lambda}\left(  g\right)  =\left(
\delta_{\lambda,\mu}\triv_{\lambda}\right)  \left(  g\right)  .
\end{align*}
Hence,
$\triv_{\lambda}\ast\triv_{\mu}=\delta_{\lambda,\mu}\triv_{\lambda}$.
Applying the map $\ch$ to this equality, we obtain
\[
\ch \left(  \triv_{\lambda}\ast\triv_{\mu
}\right)  =\ch \left(  \delta_{\lambda,\mu}\triv%
_{\lambda}\right)  =\delta_{\lambda,\mu}\underbrace{\ch \left(
\triv_{\lambda}\right)  }_{\substack{=\dfrac{p_{\lambda}}{z_{\lambda}%
}\\\text{(by Theorem \ref{symmetric-group-Frobenius-map-theorem}(a))}}%
}=\delta_{\lambda,\mu}\dfrac{p_{\lambda}}{z_{\lambda}}=\delta_{\lambda,\mu
}z_{\lambda}^{-1}p_{\lambda}.
\]
Compared with
\begin{align*}
&  \underbrace{\left(  \ch \left(  \triv_{\lambda
}\right)  \right)  }_{\substack{=\dfrac{p_{\lambda}}{z_{\lambda}}\\\text{(by
Theorem \ref{symmetric-group-Frobenius-map-theorem}(a))}}}\ast
\underbrace{\left(  \ch \left(  \triv_{\mu}\right)
\right)  }_{\substack{=\dfrac{p_{\mu}}{z_{\mu}}\\\text{(by Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a))}}}\\
&  =\dfrac{p_{\lambda}}{z_{\lambda}}\ast\dfrac{p_{\mu}}{z_{\mu}}=z_{\lambda
}^{-1}z_{\mu}^{-1}\underbrace{p_{\lambda}\ast p_{\mu}}_{\substack{=\delta
_{\lambda,\mu}z_{\lambda}p_{\lambda}\\\text{(by the definition of }%
\ast\text{)}}}=z_{\lambda}^{-1}z_{\mu}^{-1}\delta_{\lambda,\mu}z_{\lambda
}p_{\lambda}=\underbrace{\delta_{\lambda,\mu}z_{\mu}^{-1}}_{\substack{=\delta
_{\lambda,\mu}z_{\lambda}^{-1}\\\text{(because the two sides of this}%
\\\text{equality are equal if }\lambda=\mu\text{, and}\\\text{vanish
otherwise)}}}p_{\lambda}=\delta_{\lambda,\mu}z_{\lambda}^{-1}p_{\lambda},
\end{align*}
this yields $\ch \left(  \triv_{\lambda}\ast
\triv_{\mu}\right)  =\left(  \ch \left(  \triv _{\lambda}\right) 
 \right)  \ast\left(  \ch \left( \triv_{\mu}\right)  \right)$.
This rewrites as $\ch \left(  a\ast b\right) 
 =\left(  \ch a\right)  \ast\left( \ch b\right)  $
(since $a=\triv_{\lambda}$ and $b=\triv_{\mu}$). Thus,
(\ref{sol.repSn.inner-tensor.intprod.a.ch*}) is proven.}

Now, let $U_{1}$ and $U_{2}$ be two $\mathbb{C}\Symm_{n}$-modules.
Then, $\chi_{U_{1}}\in R\left(  \Symm_{n}\right)  \subset
R_{\QQ}\left(  \Symm_{n}\right)  $ and similarly $\chi_{U_{2}%
}\in R_{\QQ}\left(  \Symm_{n}\right)  $, so that
(\ref{sol.repSn.inner-tensor.intprod.a.ch*}) (applied to $a=\chi_{U_{1}}$ and
$b=\chi_{U_{2}}$) yields $\ch \left(  \chi_{U_{1}}\ast
\chi_{U_{2}}\right)  = \ch \left(  \chi_{U_{1}}\right)
\ast \ch \left(  \chi_{U_{2}}\right)  $. Due to
(\ref{sol.repSn.inner-tensor.intprod.a.charprod2}), this rewrites as
$ \ch \left(  \chi_{U_{1}\boxtimes U_{2}}\right)
= \ch \left(  \chi_{U_{1}}\right)  \ast \ch \left(
\chi_{U_{2}}\right)  $. This solves
Exercise~\ref{exe.repSn.inner-tensor.intprod}(a).

(c) Let us first recall that $p_{\lambda}\ast p_{\mu}=\delta_{\lambda,\mu
}z_{\lambda}p_{\lambda}$ for any two partitions $\lambda$ and $\mu$. Thus,
$p_{\lambda}\ast p_{\mu}=0$ whenever $\lambda\neq\mu$. This yields, in
particular, that $p_{\lambda}\ast p_{\mu}=0$ whenever $\left\vert
\lambda\right\vert \neq\left\vert \mu\right\vert $. In other words, for any
two distinct integers $n$ and $m$, we have $p_{\lambda}\ast p_{\mu}=0$ for
every $\lambda\in\Par_{n}$ and every $\mu
\in\Par_{m}$. This yields that, for any two distinct
integers $n$ and $m$, we have $a\ast b=0$ for every $a\in\Lambda_{n}$ and
$b\in\Lambda_{m}$ (because $a$ is a $\QQ$-linear combination of the
$p_{\lambda}$ with $\lambda\in\Par_{n}$, whereas $b$
is a $\QQ$-linear combination of the $p_{\mu}$ with $\mu\in
\Par_{m}$). In particular, for any two distinct
integers $n$ and $m$, we have
\[
s_{\mu}\ast s_{\nu}=0\ \ \ \ \ \ \ \ \ \ \text{for any } \nu \in
\Par_n \text{ and } \mu \in \Par_m
\]
(because $s_{\mu}\in\Lambda_{m}$ and $s_{\nu}\in\Lambda_{n}$). In other
words,
\begin{equation}
s_{\mu}\ast s_{\nu}=0\ \ \ \ \ \ \ \ \ \ \text{for any partitions }\mu\text{
and }\nu\text{ satisfying }\left\vert \mu\right\vert \neq\left\vert
\nu\right\vert .
\label{sol.repSn.inner-tensor.intprod.b.gr0}
\end{equation}


Now, let $\mu$ and $\nu$ be two partitions. We need to prove that $s_{\mu}\ast
s_{\nu}\in\sum_{\lambda\in \Par }{ \NN }s_{\lambda}$. This is
obvious (because of (\ref{sol.repSn.inner-tensor.intprod.b.gr0})) in the case
when $\left\vert \mu\right\vert \neq\left\vert \nu\right\vert $, so we can
WLOG assume that $\left\vert \mu\right\vert =\left\vert \nu\right\vert $.
Assume this, and let $n=\left\vert \mu\right\vert =\left\vert \nu\right\vert
$. Consider the two irreducible characters $\chi^{\mu}$ and $\chi^{\nu}$ of
$\CC \Symm_{n}$ defined as in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(a). Then,
$\chi^{\mu}=\chi_{U_{1}}$ and $\chi^{\nu}=\chi_{U_{2}}$ for two
$\CC \Symm_{n}$-modules
$U_{1}$ and $U_{2}$. Consider these $U_{1}$ and $U_{2}$. We have
$\ch \left(  \underbrace{\chi_{U_{1}}}_{=\chi^{\mu}}\right)
=\ch \left(  \chi^{\mu}\right)  =s_{\mu}$ (by Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a)) and similarly
$\ch \left(  \chi_{U_{2}}\right)  =s_{\nu}$. But $U_{1}\boxtimes
U_{2}$ (being a $\mathbb{C}\Symm_{n}$-module) must be a direct sum of
finitely many irreducible $\mathbb{C}\Symm_{n}$-modules, and thus the
character $\chi_{U_{1}\boxtimes U_{2}}$ is the sum of finitely many
irreducible characters of $\mathbb{C}\Symm_{n}$. Since the irreducible
characters of $\mathbb{C}\Symm_{n}$ are the $\chi^{\lambda}$ for
$\lambda\in\Par_{n}$, this shows that $\chi
_{U_{1}\boxtimes U_{2}}$ is the sum of finitely many $\chi^{\lambda}$. In
other words, $\chi_{U_{1}\boxtimes U_{2}} \in \sum_{\lambda\in\Par}
{\NN}\chi^{\lambda}$. Applying the map $\ch$ to both
sides of this relation, we obtain
\begin{align*}
\ch \left(  \chi_{U_{1}\boxtimes U_{2}}\right)    &
\in\ch \left(  \sum_{\lambda\in \Par } \NN \chi^{\lambda}\right)
 =\sum_{\lambda\in \Par } \NN
\underbrace{\ch \left(  \chi^{\lambda}\right)  }
_{\substack{=s_{\lambda}\\\text{(by Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\ch \text{ is } \ZZ \text{-linear}\right)  \\
& =\sum_{\lambda\in \Par }{ \NN }s_{\lambda}.
\end{align*}
Since
\begin{align*}
\ch \left(  \chi_{U_{1}\boxtimes U_{2}}\right)    &
=\underbrace{ \ch \left(  \chi_{U_{1}}\right)  }_{=s_{\mu}}
\ast
\underbrace{ \ch \left(  \chi_{U_{2}}\right)  }_{=s_{\nu}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.repSn.inner-tensor.intprod}(a)}\right)  \\
& =s_{\mu}\ast s_{\nu},
\end{align*}
this rewrites as $s_{\mu}\ast s_{\nu}\in\sum_{\lambda\in \Par %
}{ \NN }s_{\lambda}$. This solves Exercise
\ref{exe.repSn.inner-tensor.intprod}(c).

(b) \textit{Alternative solution of Exercise~\ref{exe.Lambda.maps-on-pn}(h).}
We need to prove that $f\ast g\in\Lambda$ for any $f\in\Lambda$ and
$g\in\Lambda$. Since the binary operation $\ast$ is $\ZZ$-bilinear, it
is clearly enough to prove that $s_{\mu}\ast s_{\nu}\in\Lambda$ for any
partitions $\mu$ and $\nu$ (since $\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ is a $\ZZ$-basis of $\Lambda$). But this
follows from the fact that%
\begin{align*}
s_{\mu}\ast s_{\nu}  & \in\sum_{\lambda\in \Par }{ \NN %
}s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.repSn.inner-tensor.intprod}(c)}\right)  \\
& \subset\Lambda
\end{align*}
for any partitions $\mu$ and $\nu$. Thus,
Exercise~\ref{exe.Lambda.maps-on-pn}~(h) is solved again.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.repSn.inner-tensor.intprod}}

\textit{Solution to Exercise~\ref{exe.repSn.inner-tensor.intprod}.} (a)
Let $G$ be any group. If $U_{1}$ and $U_{2}$ are two
$\CC G$-modules, then the character $\chi_{U_1 \boxtimes U_2}$ of the
inner tensor product $U_1 \boxtimes U_2$ of $U_1$ and $U_2$ is given by
\begin{equation}
\chi_{U_{1}\boxtimes U_{2}}\left(  g\right)  =\chi_{U_{1}}\left(  g\right)
\chi_{U_{2}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }g\in G.
\label{sol.repSn.inner-tensor.intprod.a.charprod}
\end{equation}
(This is just a restatement of
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}) using our
notation $U_{1}\boxtimes U_{2}$ for what, in
(\ref{sol.internal-tensor-product-with-sgn-is-omega-exercise.char}), was
called $U_{1}\otimes U_{2}$.)

Define a map $\ast:R_\CC \left(  G\right)  \times R_{\CC %
}\left(  G\right)  \rightarrow R_\CC \left(  G\right)  $, which will
be written in infix notation (that is, we will write $a\ast b$ instead of
$\ast\left(  a,b\right)  $), by setting%
\[
\left(  a\ast b\right)  \left(  g\right)  =a\left(  g\right)  b\left(
g\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }a\in R_\CC \left(
G\right)  \text{, }b\in R_\CC \left(  G\right)  \text{ and }g\in G.
\]
(This notation $a\ast b$ generalizes the notation $\operatorname{sgn}%
_{\Symm_{n}}\ast f$ used in Theorem
\ref{symmetric-group-Frobenius-map-theorem}.) The map $\ast$ is clearly
$\CC$-bilinear.

Notice that%
\begin{equation}
\chi_{U_{1}}\ast\chi_{U_{2}}=\chi_{U_{1}\boxtimes U_{2}}%
\ \ \ \ \ \ \ \ \ \ \text{for any two }\mathbb{C}G\text{-modules }U_{1}\text{
and }U_{2} \label{sol.repSn.inner-tensor.intprod.a.charprod2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.inner-tensor.intprod.a.charprod2}%
):} Let $U_{1}$ and $U_{2}$ be two $\mathbb{C}G$-modules. Then, every $g\in G$
satisfies%
\begin{align*}
\left(  \chi_{U_{1}}\ast\chi_{U_{2}}\right)  \left(  g\right)   &
=\chi_{U_{1}}\left(  g\right)  \chi_{U_{2}}\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\ast\right) \\
&  =\chi_{U_{1}\boxtimes U_{2}}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.repSn.inner-tensor.intprod.a.charprod})}\right)  .
\end{align*}
In other words, $\chi_{U_{1}}\ast\chi_{U_{2}}=\chi_{U_{1}\boxtimes U_{2}}$.
This proves (\ref{sol.repSn.inner-tensor.intprod.a.charprod2}).}

Let $R_{ \QQ }\left(  G\right)  $ denote the set of class functions
$G\rightarrow \QQ $. This is clearly a subset of $R_{\mathbb{C}}\left(
G\right)  $. In general, it is not true that $R\left(  G\right)  \subset
R_{ \QQ }\left(  G\right)  $, but we will see that this holds for
$G=\Symm_{n}$ for any $n\in \NN$.

It is clear that $a\ast b\in R_{ \QQ }\left(  G\right)  $ for any $a\in
R_{ \QQ }\left(  G\right)  $ and $b\in R_{ \QQ }\left(  G\right)  $.

Now, forget that we fixed $G$. We have thus introduced a map $\ast
:R_\CC \left(  G\right)  \times R_\CC \left(  G\right)
\rightarrow R_\CC \left(  G\right)  $ for every group $G$, and we
have proved some properties of this map.

Let now $n\in \NN$. Exercise~\ref{exe.symmetric-group-reps-rational}(a)
yields that $\chi_{V} \left( g \right) \in \ZZ \subset \QQ$ for
every $g \in \Symm_n$ and every finite-dimensional
$\CC \Symm_n$-module $V$. Hence, $\chi_{V}$ is a map from $\Symm_n$ to
$\QQ$ for every finite-dimensional $\CC \Symm_n$-module $V$.
In other words, $\chi_{V} \in R_{\QQ} \left( \Symm_n \right)$
for every finite-dimensional
$\CC \Symm_n$-module $V$. Thus, $R\left(  \Symm_{n}\right)
\subset R_{\QQ}\left(  \Symm_{n}\right)  $ (since $R\left(
\Symm_{n}\right)  $ is the $\ZZ$-module generated by the
$\chi_{V}$ for $V$ ranging over the irreducible $\mathbb{C}\Symm_{n}$-modules).

Every element of $R_{ \QQ }\left(  \Symm_{n}\right)  $ is a class
function $\Symm_{n}\rightarrow \QQ $, and thus a $ \QQ %
$-linear combination of the $\triv_{\lambda}$ for $\lambda
\in\Par_{n}$ (because these functions $\triv%
_{\lambda}$ are the indicator functions for the conjugacy classes of
$\Symm_{n}$). Thus, the $\QQ$-module $R_{ \QQ }\left(
\Symm_{n}\right)  $ is spanned by the $\triv_{\lambda}$ for
$\lambda\in\Par_{n}$.

As in Theorem~\ref{symmetric-group-Frobenius-map-theorem}, we
extend the PSH-isomorphism $\ch : A \to \Lambda$ to a $\CC$-Hopf
algebra isomorphism $A_\CC \to \Lambda_\CC$; we shall denote the
latter isomorphism by $\ch$ as well.

Now, let us notice that
\begin{equation}
\ch \left(  R_{ \QQ }\left(  \Symm_{n}\right)
\right)  \subset\Lambda_{ \QQ } .
\label{sol.repSn.inner-tensor.intprod.a.inLQ}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.inner-tensor.intprod.a.inLQ}):} It
is clearly enough to show that
$\ch \left(  \triv _{\lambda}\right)  \in\Lambda_{\QQ}$ for
every $\lambda\in \Par_{n}$ (because the $\QQ$-module
$R_{ \QQ }\left(  \Symm_{n}\right)  $ is spanned by the
$\triv_{\lambda}$ for $\lambda\in\Par_{n}$).
But this follows from the fact that $\ch \left(  \triv%
_{\lambda}\right)  =\frac{p_{\lambda}}{z_{\lambda}}$ for every $\lambda
\in\Par_{n}$ (this is part of
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(a)). Thus,
(\ref{sol.repSn.inner-tensor.intprod.a.inLQ}) is proven.} Furthermore,
\begin{equation}
\ch \left(  a\ast b\right)  =\left(  \ch a\right)
\ast\left(  \ch b\right)  \ \ \ \ \ \ \ \ \ \ \text{for any
}a\in R_{ \QQ }\left(  \Symm_{n}\right)  \text{ and }b\in
R_{ \QQ }\left(  \Symm_{n}\right)
.\label{sol.repSn.inner-tensor.intprod.a.ch*}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.inner-tensor.intprod.a.ch*}):} Let
$a\in R_{ \QQ }\left(  \Symm_{n}\right)  $ and $b\in
R_{ \QQ }\left(  \Symm_{n}\right)  $. The equality
(\ref{sol.repSn.inner-tensor.intprod.a.ch*}) is clearly $\QQ$-linear in
each of $a$ and $b$. Hence, in proving this equality, we can WLOG assume that
$a$ and $b$ are two of the functions $\triv_{\lambda}$ for $\lambda
\in\Par_{n}$ (because the $\QQ$-module
$R_{ \QQ }\left(  \Symm_{n}\right)  $ is spanned by the
$\triv_{\lambda}$ for $\lambda\in\Par_{n}$).
In other words, we can WLOG assume that $a=\triv_{\lambda}$ and
$b=\triv_{\mu}$ for some $\lambda\in\Par_{n}$
and $\mu\in\Par_{n}$. Assume this, and consider these
$\lambda$ and $\mu$.
\par
Recall that $\triv_{\lambda}$ is the indicator function for the set of
all permutations in $\Symm_{n}$ having cycle type $\lambda$, and that
$\triv_{\mu}$ is the indicator function for the set of all
permutations in $\Symm_{n}$ having cycle type $\mu$. Hence, every
$g\in\Symm_{n}$ satisfies
\[
\triv_{\lambda}\left(  g\right)  \cdot\triv_{\mu}\left(
g\right)  =\delta_{\lambda,\mu}\cdot\triv_{\lambda}\left(  g\right)
\]
(because both sides of this equality vanish if $g$ does not have cycle type
$\lambda$, and also vanish if $\lambda\neq\mu$, but in the remaining case are
both equal to $1$). Thus, every $g\in\Symm_{n}$ satisfies
\begin{align*}
\left(  \triv_{\lambda}\ast\triv_{\mu}\right)  \left(
g\right)   &  =\triv_{\lambda}\left(  g\right)  \cdot\triv%
_{\mu}\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\triv_{\lambda}\ast\triv_{\mu}\right)  \\
&  =\delta_{\lambda,\mu}\cdot\triv_{\lambda}\left(  g\right)  =\left(
\delta_{\lambda,\mu}\triv_{\lambda}\right)  \left(  g\right)  .
\end{align*}
Hence,
$\triv_{\lambda}\ast\triv_{\mu}=\delta_{\lambda,\mu}\triv_{\lambda}$.
Applying the map $\ch$ to this equality, we obtain
\[
\ch \left(  \triv_{\lambda}\ast\triv_{\mu
}\right)  =\ch \left(  \delta_{\lambda,\mu}\triv%
_{\lambda}\right)  =\delta_{\lambda,\mu}\underbrace{\ch \left(
\triv_{\lambda}\right)  }_{\substack{=\dfrac{p_{\lambda}}{z_{\lambda}%
}\\\text{(by Theorem \ref{symmetric-group-Frobenius-map-theorem}(a))}}%
}=\delta_{\lambda,\mu}\dfrac{p_{\lambda}}{z_{\lambda}}=\delta_{\lambda,\mu
}z_{\lambda}^{-1}p_{\lambda}.
\]
Compared with
\begin{align*}
&  \underbrace{\left(  \ch \left(  \triv_{\lambda
}\right)  \right)  }_{\substack{=\dfrac{p_{\lambda}}{z_{\lambda}}\\\text{(by
Theorem \ref{symmetric-group-Frobenius-map-theorem}(a))}}}\ast
\underbrace{\left(  \ch \left(  \triv_{\mu}\right)
\right)  }_{\substack{=\dfrac{p_{\mu}}{z_{\mu}}\\\text{(by Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a))}}}\\
&  =\dfrac{p_{\lambda}}{z_{\lambda}}\ast\dfrac{p_{\mu}}{z_{\mu}}=z_{\lambda
}^{-1}z_{\mu}^{-1}\underbrace{p_{\lambda}\ast p_{\mu}}_{\substack{=\delta
_{\lambda,\mu}z_{\lambda}p_{\lambda}\\\text{(by the definition of }%
\ast\text{)}}}=z_{\lambda}^{-1}z_{\mu}^{-1}\delta_{\lambda,\mu}z_{\lambda
}p_{\lambda}=\underbrace{\delta_{\lambda,\mu}z_{\mu}^{-1}}_{\substack{=\delta
_{\lambda,\mu}z_{\lambda}^{-1}\\\text{(because the two sides of this}%
\\\text{equality are equal if }\lambda=\mu\text{, and}\\\text{vanish
otherwise)}}}p_{\lambda}=\delta_{\lambda,\mu}z_{\lambda}^{-1}p_{\lambda},
\end{align*}
this yields $\ch \left(  \triv_{\lambda}\ast
\triv_{\mu}\right)  =\left(  \ch \left(  \triv _{\lambda}\right) 
 \right)  \ast\left(  \ch \left( \triv_{\mu}\right)  \right)$.
This rewrites as $\ch \left(  a\ast b\right) 
 =\left(  \ch a\right)  \ast\left( \ch b\right)  $
(since $a=\triv_{\lambda}$ and $b=\triv_{\mu}$). Thus,
(\ref{sol.repSn.inner-tensor.intprod.a.ch*}) is proven.}

Now, let $U_{1}$ and $U_{2}$ be two $\mathbb{C}\Symm_{n}$-modules.
Then, $\chi_{U_{1}}\in R\left(  \Symm_{n}\right)  \subset
R_{\QQ}\left(  \Symm_{n}\right)  $ and similarly $\chi_{U_{2}%
}\in R_{\QQ}\left(  \Symm_{n}\right)  $, so that
(\ref{sol.repSn.inner-tensor.intprod.a.ch*}) (applied to $a=\chi_{U_{1}}$ and
$b=\chi_{U_{2}}$) yields $\ch \left(  \chi_{U_{1}}\ast
\chi_{U_{2}}\right)  = \ch \left(  \chi_{U_{1}}\right)
\ast \ch \left(  \chi_{U_{2}}\right)  $. Due to
(\ref{sol.repSn.inner-tensor.intprod.a.charprod2}), this rewrites as
$ \ch \left(  \chi_{U_{1}\boxtimes U_{2}}\right)
= \ch \left(  \chi_{U_{1}}\right)  \ast \ch \left(
\chi_{U_{2}}\right)  $. This solves
Exercise~\ref{exe.repSn.inner-tensor.intprod}(a).

(c) Let us first recall that $p_{\lambda}\ast p_{\mu}=\delta_{\lambda,\mu
}z_{\lambda}p_{\lambda}$ for any two partitions $\lambda$ and $\mu$. Thus,
$p_{\lambda}\ast p_{\mu}=0$ whenever $\lambda\neq\mu$. This yields, in
particular, that $p_{\lambda}\ast p_{\mu}=0$ whenever $\left\vert
\lambda\right\vert \neq\left\vert \mu\right\vert $. In other words, for any
two distinct integers $n$ and $m$, we have $p_{\lambda}\ast p_{\mu}=0$ for
every $\lambda\in\Par_{n}$ and every $\mu
\in\Par_{m}$. This yields that, for any two distinct
integers $n$ and $m$, we have $a\ast b=0$ for every $a\in\Lambda_{n}$ and
$b\in\Lambda_{m}$ (because $a$ is a $\QQ$-linear combination of the
$p_{\lambda}$ with $\lambda\in\Par_{n}$, whereas $b$
is a $\QQ$-linear combination of the $p_{\mu}$ with $\mu\in
\Par_{m}$). In particular, for any two distinct
integers $n$ and $m$, we have
\[
s_{\mu}\ast s_{\nu}=0\ \ \ \ \ \ \ \ \ \ \text{for any } \nu \in
\Par_n \text{ and } \mu \in \Par_m
\]
(because $s_{\mu}\in\Lambda_{m}$ and $s_{\nu}\in\Lambda_{n}$). In other
words,
\begin{equation}
s_{\mu}\ast s_{\nu}=0\ \ \ \ \ \ \ \ \ \ \text{for any partitions }\mu\text{
and }\nu\text{ satisfying }\left\vert \mu\right\vert \neq\left\vert
\nu\right\vert .
\label{sol.repSn.inner-tensor.intprod.b.gr0}
\end{equation}


Now, let $\mu$ and $\nu$ be two partitions. We need to prove that $s_{\mu}\ast
s_{\nu}\in\sum_{\lambda\in \Par }{ \NN }s_{\lambda}$. This is
obvious (because of (\ref{sol.repSn.inner-tensor.intprod.b.gr0})) in the case
when $\left\vert \mu\right\vert \neq\left\vert \nu\right\vert $, so we can
WLOG assume that $\left\vert \mu\right\vert =\left\vert \nu\right\vert $.
Assume this, and let $n=\left\vert \mu\right\vert =\left\vert \nu\right\vert
$. Consider the two irreducible characters $\chi^{\mu}$ and $\chi^{\nu}$ of
$\CC \Symm_{n}$ defined as in
Theorem~\ref{symmetric-group-Frobenius-map-theorem}(a). Then,
$\chi^{\mu}=\chi_{U_{1}}$ and $\chi^{\nu}=\chi_{U_{2}}$ for two
$\CC \Symm_{n}$-modules
$U_{1}$ and $U_{2}$. Consider these $U_{1}$ and $U_{2}$. We have
$\ch \left(  \underbrace{\chi_{U_{1}}}_{=\chi^{\mu}}\right)
=\ch \left(  \chi^{\mu}\right)  =s_{\mu}$ (by Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a)) and similarly
$\ch \left(  \chi_{U_{2}}\right)  =s_{\nu}$. But $U_{1}\boxtimes
U_{2}$ (being a $\mathbb{C}\Symm_{n}$-module) must be a direct sum of
finitely many irreducible $\mathbb{C}\Symm_{n}$-modules, and thus the
character $\chi_{U_{1}\boxtimes U_{2}}$ is the sum of finitely many
irreducible characters of $\mathbb{C}\Symm_{n}$. Since the irreducible
characters of $\mathbb{C}\Symm_{n}$ are the $\chi^{\lambda}$ for
$\lambda\in\Par_{n}$, this shows that $\chi
_{U_{1}\boxtimes U_{2}}$ is the sum of finitely many $\chi^{\lambda}$. In
other words, $\chi_{U_{1}\boxtimes U_{2}} \in \sum_{\lambda\in\Par}
{\NN}\chi^{\lambda}$. Applying the map $\ch$ to both
sides of this relation, we obtain
\begin{align*}
\ch \left(  \chi_{U_{1}\boxtimes U_{2}}\right)    &
\in\ch \left(  \sum_{\lambda\in \Par } \NN \chi^{\lambda}\right)
 =\sum_{\lambda\in \Par } \NN
\underbrace{\ch \left(  \chi^{\lambda}\right)  }
_{\substack{=s_{\lambda}\\\text{(by Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\ch \text{ is } \ZZ \text{-linear}\right)  \\
& =\sum_{\lambda\in \Par }{ \NN }s_{\lambda}.
\end{align*}
Since
\begin{align*}
\ch \left(  \chi_{U_{1}\boxtimes U_{2}}\right)    &
=\underbrace{ \ch \left(  \chi_{U_{1}}\right)  }_{=s_{\mu}}
\ast
\underbrace{ \ch \left(  \chi_{U_{2}}\right)  }_{=s_{\nu}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.repSn.inner-tensor.intprod}(a)}\right)  \\
& =s_{\mu}\ast s_{\nu},
\end{align*}
this rewrites as $s_{\mu}\ast s_{\nu}\in\sum_{\lambda\in \Par %
}{ \NN }s_{\lambda}$. This solves Exercise
\ref{exe.repSn.inner-tensor.intprod}(c).

(b) \textit{Alternative solution of Exercise~\ref{exe.Lambda.maps-on-pn}(h).}
We need to prove that $f\ast g\in\Lambda$ for any $f\in\Lambda$ and
$g\in\Lambda$. Since the binary operation $\ast$ is $\ZZ$-bilinear, it
is clearly enough to prove that $s_{\mu}\ast s_{\nu}\in\Lambda$ for any
partitions $\mu$ and $\nu$ (since $\left(  s_{\lambda}\right)  _{\lambda
\in \Par }$ is a $\ZZ$-basis of $\Lambda$). But this
follows from the fact that%
\begin{align*}
s_{\mu}\ast s_{\nu}  & \in\sum_{\lambda\in \Par }{ \NN %
}s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.repSn.inner-tensor.intprod}(c)}\right)  \\
& \subset\Lambda
\end{align*}
for any partitions $\mu$ and $\nu$. Thus,
Exercise~\ref{exe.Lambda.maps-on-pn}~(h) is solved again.
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.repSn.arithmetic-prod}}

\textit{Solution to Exercise \ref{exe.repSn.arithmetic-prod}.} Let us
introduce a fundamental construction. If $X$ is any set, then we let
$\Symm_X$ denote the symmetric group on $X$ (that is, the group of
all permutations of $X$). For any two sets $X$ and $Y$ and any permutations
$\sigma\in\Symm_X$ and $\tau\in\Symm_Y$, we define a
permutation $\sigma\times\tau$ of $X\times Y$ by setting%
\[
\left(  \sigma\times\tau\right)  \left(  \left(  x,y\right)  \right)  =\left(
\sigma\left(  x\right)  ,\tau\left(  y\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\text{ and }y\in Y.
\]
Thus, for any two sets $X$ and $Y$, we can define a map%
\begin{align*}
\operatorname*{cross}:\Symm_X\times\Symm_Y  &
\rightarrow\Symm_{X\times Y},\\
\left(  \sigma,\tau\right)   &  \mapsto\sigma\times\tau.
\end{align*}
We notice a few properties of this map (whose simple proof we leave to the reader):

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.cross}Let $X$ and $Y$ be two sets.

\begin{enumerate}
\item[(a)] The map $\operatorname*{cross}:\Symm_X\times
\Symm_Y\rightarrow\Symm_{X\times Y}$ is a group homomorphism.

\item[(b)] If the sets $X$ and $Y$ are nonempty, then the map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ is injective.
\end{enumerate}
\end{lemma}

The following deeper fact provides the first hint of a combinatorial
interpretation of $\boxdot$:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.cross2p}Let $X$ and $Y$ be two finite
sets. Let $\sigma\in\Symm_X$ and $\tau\in\Symm_Y$ be two
permutations. Let $\lambda$, $\mu$ and $\kappa$ be the cycle types of the
permutations $\sigma\in\Symm_X$, $\tau\in\Symm_Y$ and
$\sigma\times\tau\in\Symm_{X\times Y}$. Then,%
\[
p_{\lambda}\boxdot p_{\mu}=p_{\kappa}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2p}.]The partition
$\lambda$ is the cycle type of the permutation $\sigma$. In other words,
$\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }$ are
the lengths of the cycles of the permutation $\sigma$. Let $C_{1}%
,C_{2},\ldots,C_{\ell\left(  \lambda\right)  }$ denote these cycles, labelled
in such a way that each $C_{i}$ has length $\lambda_{i}$. Clearly, $\left(
C_{1},C_{2},\ldots,C_{\ell\left(  \lambda\right)  }\right)  $ is a set
partition of the set $X$; thus, $X=\bigsqcup_{i=1}^{\ell\left(  \lambda
\right)  }C_{i}$.

The partition $\mu$ is the cycle type of the permutation $\tau$. In other
words, $\mu_{1},\mu_{2},\ldots,\mu_{\ell\left(  \mu\right)  }$ are the lengths
of the cycles of the permutation $\tau$. Let $D_{1},D_{2},\ldots,D_{\ell\left(
\mu\right)  }$ denote these cycles, labelled in such a way that each $D_{j}$
has length $\mu_{j}$. Clearly, $\left(  D_{1},D_{2},\ldots,D_{\ell\left(
\mu\right)  }\right)  $ is a set partition of the set $Y$; thus,
$Y=\bigsqcup_{j=1}^{\ell\left(  \mu\right)  }D_{j}$.

Since $X=\bigsqcup_{i=1}^{\ell\left(  \lambda\right)  }C_{i}$ and
$Y=\bigsqcup_{j=1}^{\ell\left(  \mu\right)  }D_{j}$, we have
\[
X\times Y=\left(  \bigsqcup_{i=1}^{\ell\left(  \lambda\right)  }C_{i}\right)
\times\left(  \bigsqcup_{j=1}^{\ell\left(  \mu\right)  }D_{j}\right)
=\bigsqcup_{i=1}^{\ell\left(  \lambda\right)  }\bigsqcup_{j=1}^{\ell\left(
\mu\right)  }\left(  C_{i}\times D_{j}\right)  .
\]
In particular, the sets $C_{i}\times D_{j}$ for $\left(  i,j\right)  $ ranging
over all $\left(  i,j\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  $ are disjoint.

We now make the following claim:

\begin{statement}
\textit{Claim A:} Let $i\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  $ and $j\in\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  $.
Then, the subset $C_{i}\times D_{j}$ of $X\times Y$ is the union of
$\gcd\left(  \lambda_{i},\mu_{j}\right)  $ disjoint cycles of the permutation
$\sigma\times\tau$, and each of these cycles has length $\operatorname{lcm}%
\left(  \lambda_{i},\mu_{j}\right)  $.
\end{statement}

\textit{Proof of Claim A:} We know that $C_{i}$ is a cycle of $\sigma$; thus,
$\sigma\left(  C_{i}\right)  \subset C_{i}$. Similarly, $\tau\left(
D_{j}\right)  \subset D_{j}$. Now, $\left(  \sigma\times\tau\right)  \left(
C_{i}\times D_{j}\right)  =\underbrace{\sigma\left(  C_{i}\right)  }_{\subset
C_{i}}\times\underbrace{\tau\left(  D_{j}\right)  }_{\subset D_{j}}\subset
C_{i}\times D_{j}$. Hence, the permutation $\sigma\times\tau$ restricts to a
permutation of the subset $C_{i}\times D_{j}$ of $X\times Y$.

Let us first show that, for every $h\in C_{i}\times D_{j}$, the cycle of
$\sigma\times\tau$ containing $h$ is a subset of $C_{i}\times D_{j}$ and has
length $\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  $.

Indeed, fix $h\in C_{i}\times D_{j}$. Then, all of the elements $h,\left(
\sigma\times\tau\right)  \left(  h\right)  ,\left(  \sigma\times\tau\right)
^{2}\left(  h\right)  ,\left(  \sigma\times\tau\right)  ^{3}\left(  h\right)
,\ldots$ belong to $C_{i}\times D_{j}$\ \ \ \ \footnote{since the permutation
$\sigma\times\tau$ restricts to a permutation of the subset $C_{i}\times
D_{j}$ of $X\times Y$}. Hence, the cycle of $\sigma\times\tau$ containing $h$
is a subset of $C_{i}\times D_{j}$ (because this cycle consists of these very
elements $h,\left(  \sigma\times\tau\right)  \left(  h\right)  ,\left(
\sigma\times\tau\right)  ^{2}\left(  h\right)  ,\left(  \sigma\times
\tau\right)  ^{3}\left(  h\right)  ,\ldots$). The length of this cycle is the
smallest positive integer $N$ such that $\left(  \sigma\times\tau\right)
^{N}\left(  h\right)  =h$. We shall now prove that this smallest positive
integer is $\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  $.

Indeed, let us write $h$ in the form $h=\left(  c,d\right)  $ for some $c\in
C_{i}$ and $d\in D_{j}$. Then, the element $c$ belongs to a cycle of $\sigma$
which has length $\lambda_{i}$ (namely, $C_{i}$). Hence, the sequence
$c,\sigma\left(  c\right)  ,\sigma^{2}\left(  c\right)  ,\sigma^{3}\left(
c\right)  ,\ldots$ repeats every $\lambda_i$
elements (and not more frequently). Thus,
for any $N\in \NN$, we have the following equivalence of statements:
\[
\left(  \sigma^{N}\left(  c\right)  =c\right)  \ \Longleftrightarrow\ \left(
\lambda_{i}\mid N\right)  .
\]
Similarly, for any $N\in \NN$, we have the following equivalence of
statements:
\[
\left(  \tau^{N}\left(  d\right)  =d\right)  \ \Longleftrightarrow\ \left(
\mu_{j}\mid N\right)  .
\]


Now, for any $N\in \NN$, we have the following equivalence of
statements:%
\begin{align*}
\left(  \left(  \sigma\times\tau\right)  ^{N}\left(  h\right)  =h\right)  \
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \sigma\times\tau\right)
^{N}\left(  \left(  c,d\right)  \right)  }_{=\left(  \sigma^{N}\left(
c\right)  ,\tau^{N}\left(  d\right)  \right)  }=\left(  c,d\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h=\left(  c,d\right)  \right) \\
&  \Longleftrightarrow\ \left(  \left(  \sigma^{N}\left(  c\right)  ,\tau
^{N}\left(  d\right)  \right)  =\left(  c,d\right)  \right)
\ \Longleftrightarrow\ \left(  \underbrace{\left(  \sigma^{N}\left(  c\right)
=c\right)  }_{\Longleftrightarrow\left(  \lambda_{i}\mid N\right)  }\text{ and
}\underbrace{\left(  \tau^{N}\left(  d\right)  =d\right)  }%
_{\Longleftrightarrow\left(  \mu_{j}\mid N\right)  }\right) \\
&  \Longleftrightarrow\ \left(  \lambda_{i}\mid N\text{ and }\mu_{j}\mid
N\right)  \ \Longleftrightarrow\ \left(  \operatorname{lcm}\left(  \lambda
_{i},\mu_{j}\right)  \mid N\right)  .
\end{align*}
Thus, the smallest positive integer $N$ such that $\left(  \sigma\times
\tau\right)  ^{N}\left(  h\right)  =h$ is $\operatorname{lcm}\left(
\lambda_{i},\mu_{j}\right)  $. In other words, the length of the cycle of
$\sigma\times\tau$ containing $h$ is $\operatorname{lcm}\left(  \lambda
_{i},\mu_{j}\right)  $ (since the length of the cycle of $\sigma\times\tau$
containing $h$ is the smallest positive integer $N$ such that $\left(
\sigma\times\tau\right)  ^{N}\left(  h\right)  =h$).

Now, let us forget that we fixed $h$. We thus have proven that, for every
$h\in C_{i}\times D_{j}$,%
\begin{equation}
\text{the cycle of }\sigma\times\tau\text{ containing }h\text{ is a subset of
}C_{i}\times D_{j}\text{ and has length }\operatorname{lcm}\left(  \lambda
_{i},\mu_{j}\right)  . \label{pf.lem.sol.repSn.arithmetic-prod.cross2p.subset}%
\end{equation}
These cycles (for $h$ ranging over all $C_{i}\times D_{j}$) clearly cover the
set $C_{i}\times D_{j}$ (because each $h\in C_{i}\times D_{j}$ is contained in
its corresponding cycle). Thus, $C_{i}\times D_{j}$ is the union of several
cycles of the permutation $\sigma\times\tau$, and each of these cycles has
length $\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  $. Since any two
cycles of $\sigma\times\tau$ are either disjoint or identical, we can get rid
of redundant cycles in this union, and thus obtain the following conclusion:
$C_{i}\times D_{j}$ is the union of several disjoint cycles of the permutation
$\sigma\times\tau$, and each of these cycles has length $\operatorname{lcm}%
\left(  \lambda_{i},\mu_{j}\right)  $. In order to complete the proof of Claim
A, it thus remains only to show that the number of these cycles is
$\gcd\left(  \lambda_{i},\mu_{j}\right)  $. But this is easy: These cycles are
all disjoint, and cover a set of size $\lambda_{i}\mu_{j}$ (in fact, they
cover the set $C_{i}\times D_{j}$, which has size $\left\vert C_{i}\times
D_{j}\right\vert =\underbrace{\left\vert C_{i}\right\vert }_{=\lambda_{i}%
}\cdot\underbrace{\left\vert D_{j}\right\vert }_{=\mu_{j}}=\lambda_{i}\mu_{j}%
$); since they have length $\operatorname{lcm}\left(  \lambda_{i},\mu
_{j}\right)  $ each, their number must be $\dfrac{\lambda_{i}\mu_{j}%
}{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  }=\gcd\left(
\lambda_{i},\mu_{j}\right)  $. Thus, the proof of Claim A is complete.

We shall now continue our proof of Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross2p}.

The partition $\kappa$ is the cycle type of the permutation $\sigma\times\tau
$. In other words, $\kappa_{1},\kappa_{2},\ldots,\kappa_{\ell\left(
\kappa\right)  }$ are the lengths of the cycles of the permutation
$\sigma\times\tau$. Hence,
\[
\prod_{u=1}^{\ell\left(  \kappa\right)  }p_{\kappa_{u}}=\prod_{E\text{ is a
cycle of }\sigma\times\tau}p_{\left\vert E\right\vert },
\]
where $\left\vert E\right\vert $ denotes the length of any cycle $E$. But
every cycle $E$ of $\sigma\times\tau$ must satisfy $E\subset C_{i}\times
D_{j}$ (where we regard $E$ as a set) for precisely one $\left(  i,j\right)
\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{
1,2,\ldots,\ell\left(  \mu\right)  \right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $E$ be a cycle of $\sigma\times\tau$.
Then, $E$ is nonempty, so that there exists an element $h$ of $E$. Fix such an
$h$. We have $h\in E\subset X\times Y=\bigsqcup_{i=1}^{\ell\left(
\lambda\right)  }\bigsqcup_{j=1}^{\ell\left(  \mu\right)  }\left(  C_{i}\times
D_{j}\right)  $. Hence, there exists some $\left(  i,j\right)  \in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots
,\ell\left(  \mu\right)  \right\}  $ such that $h\in C_{i}\times D_{j}$.
Consider this $\left(  i,j\right)  $. Then, $E$ is the cycle of $\sigma
\times\tau$ containing $h$. Hence,
(\ref{pf.lem.sol.repSn.arithmetic-prod.cross2p.subset}) shows that $E$ is a
subset of $C_{i}\times D_{j}$ and has length $\operatorname{lcm}\left(
\lambda_{i},\mu_{j}\right)  $. So we know that $E\subset C_{i}\times D_{j}$.
\par
So we have found a pair $\left(  i,j\right)  \in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(
\mu\right)  \right\}  $ such that $E\subset C_{i}\times D_{j}$. There clearly
cannot be two distinct such pairs $\left(  i,j\right)  $ (since the sets
$C_{i}\times D_{j}$ for $\left(  i,j\right)  $ ranging over all $\left(
i,j\right)  \in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  $ are disjoint),
and so this pair $\left(  i,j\right)  $ is unique.}. Hence,%
\[
\prod_{E\text{ is a cycle of }\sigma\times\tau}p_{\left\vert E\right\vert
}=\prod_{i=1}^{\ell\left(  \lambda\right)  }\prod_{j=1}^{\ell\left(
\mu\right)  }\prod\limits_{\substack{E\text{ is a cycle of }\sigma\times
\tau;\\E\subset C_{i}\times D_{j}}}p_{\left\vert E\right\vert }.
\]


But every $\left(  i,j\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  $ satisfies
\[
\prod\limits_{\substack{E\text{ is a cycle of }\sigma\times\tau;\\E\subset
C_{i}\times D_{j}}}p_{\left\vert E\right\vert }=p_{\operatorname{lcm}\left(
\lambda_{i},\mu_{j}\right)  }^{\gcd\left(  \lambda_{i},\mu_{j}\right)  }%
\]
\footnote{\textit{Proof.} Let $\left(  i,j\right)  \in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(
\mu\right)  \right\}  $. Claim A yields that the subset $C_{i}\times D_{j}$ of
$X\times Y$ is the union of $\gcd\left(  \lambda_{i},\mu_{j}\right)  $
disjoint cycles of the permutation $\sigma\times\tau$, and each of these
cycles has length $\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  $.
Obviously, these $\gcd\left(  \lambda_{i},\mu_{j}\right)  $ disjoint cycles
are exactly all the cycles $E$ of $\sigma\times\tau$ which satisfy $E\subset
C_{i}\times D_{j}$. Thus, there are precisely $\gcd\left(  \lambda_{i},\mu
_{j}\right)  $ cycles $E$ of $\sigma\times\tau$ which satisfy $E\subset
C_{i}\times D_{j}$, and each of these cycles $E$ has length
$\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  $. Hence,%
\[
\prod\limits_{\substack{E\text{ is a cycle of }\sigma\times\tau;\\E\subset
C_{i}\times D_{j}}}\underbrace{p_{\left\vert E\right\vert }}%
_{\substack{=p_{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)
}\\\text{(since }E\text{ has length }\operatorname{lcm}\left(  \lambda_{i}%
,\mu_{j}\right)  \text{,}\\\text{that is, we have }\left\vert E\right\vert
=\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  \text{)}}%
}=\prod\limits_{\substack{E\text{ is a cycle of }\sigma\times\tau;\\E\subset
C_{i}\times D_{j}}}p_{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)
}=p_{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  }^{\gcd\left(
\lambda_{i},\mu_{j}\right)  }%
\]
(since there are precisely $\gcd\left(  \lambda_{i},\mu_{j}\right)  $ cycles
$E$ of $\sigma\times\tau$ which satisfy $E\subset C_{i}\times D_{j}$), qed.}.
Now, the definition of $p_{\kappa}$ yields%
\begin{align*}
p_{\kappa}  &  =p_{\kappa_{1}}p_{\kappa_{2}}\cdots p_{\kappa_{\ell\left(
\kappa\right)  }}=\prod_{u=1}^{\ell\left(  \kappa\right)  }p_{\kappa_{u}%
}=\prod_{E\text{ is a cycle of }\sigma\times\tau}p_{\left\vert E\right\vert
}=\prod_{i=1}^{\ell\left(  \lambda\right)  }\prod_{j=1}^{\ell\left(
\mu\right)  }\underbrace{\prod\limits_{\substack{E\text{ is a cycle of }\sigma
\times\tau;\\E\subset C_{i}\times D_{j}}}p_{\left\vert E\right\vert }%
}_{=p_{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  }^{\gcd\left(
\lambda_{i},\mu_{j}\right)  }}\\
&  =\prod_{i=1}^{\ell\left(  \lambda\right)  }\prod_{j=1}^{\ell\left(
\mu\right)  }p_{\operatorname{lcm}\left(  \lambda_{i},\mu_{j}\right)  }%
^{\gcd\left(  \lambda_{i},\mu_{j}\right)  }=p_{\lambda}\boxdot p_{\mu
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{\lambda}\boxdot p_{\mu}\text{ is
defined to be }\prod_{i=1}^{\ell\left(  \lambda\right)  }\prod_{j=1}%
^{\ell\left(  \mu\right)  }p_{\operatorname{lcm}\left(  \lambda_{i},\mu
_{j}\right)  }^{\gcd\left(  \lambda_{i},\mu_{j}\right)  }\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2p}.
\end{proof}

(a) We already know that the operation $\boxdot$ is $\QQ$-bilinear.
Thus, we only need to prove that the binary operation $\boxdot$ is commutative
and associative and has unity $p_{1}$.

We shall only prove the associativity of $\boxdot$ (since the other two
properties can be proven similarly). In other words, we shall prove that
$\left(  u\boxdot v\right)  \boxdot w=u\boxdot\left(  v\boxdot w\right)  $ for
any three elements $u$, $v$ and $w$ of $\Lambda_{\QQ}$.

So let $u$, $v$ and $w$ be three elements of $\Lambda_{\QQ}$. We need
to prove the equality $\left(  u\boxdot v\right)  \boxdot w=u\boxdot\left(
v\boxdot w\right)  $. Since this equality is $\QQ$-linear in each of
$u$, $v$ and $w$ (this is because the operation $\boxdot$ is
$\QQ$-bilinear), we can WLOG assume that $u$, $v$ and $w$ belong to the
basis $\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ of the
$\QQ$-vector space $\Lambda_{\QQ}$. Assume this. Then, there
exist three partitions $\lambda$, $\mu$ and $\nu$ such that $u=p_{\lambda}$,
$v=p_{\mu}$ and $w=p_{\nu}$. Consider these $\lambda$, $\mu$ and $\nu$.

There exist a finite set $A$ and a permutation $\alpha\in\Symm_A$
such that $\lambda$ is the cycle type of $\alpha$. Consider these $A$ and
$\alpha$.

There exist a finite set $B$ and a permutation $\beta\in\Symm_B$ such
that $\mu$ is the cycle type of $\beta$. Consider these $B$ and $\beta$.

There exist a finite set $C$ and a permutation $\gamma\in\Symm_C$
such that $\nu$ is the cycle type of $\gamma$. Consider these $C$ and $\gamma$.

For every permutation $\pi$ of any finite set $X$, we let
$\operatorname*{type}\pi$ denote the cycle type of $\pi$. Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross2p} (applied to $X=A$, $Y=B$,
$\sigma=\alpha$, $\tau=\beta$ and $\kappa=\operatorname*{type}\left(
\alpha\times\beta\right)  $) yields $p_{\lambda}\boxdot p_{\mu}%
=p_{\operatorname*{type}\left(  \alpha\times\beta\right)  }$. Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross2p} (applied to $A\times B$, $C$,
$\alpha\times\beta$, $\gamma$, $\operatorname*{type}\left(  \alpha\times
\beta\right)  $, $\nu$ and $\operatorname*{type}\left(  \left(  \alpha
\times\beta\right)  \times\gamma\right)  $ instead of $X$, $Y$, $\sigma$,
$\tau$, $\lambda$, $\mu$ and $\kappa$) yields $p_{\operatorname*{type}\left(
\alpha\times\beta\right)  }\boxdot p_{\nu}=p_{\operatorname*{type}\left(
\left(  \alpha\times\beta\right)  \times\gamma\right)  }$. Thus,%
\[
\left(  \underbrace{u}_{=p_{\lambda}}\boxdot\underbrace{v}_{=p_{\mu}}\right)
\boxdot\underbrace{w}_{=p_{\nu}}=\underbrace{\left(  p_{\lambda}\boxdot
p_{\mu}\right)  }_{=p_{\operatorname*{type}\left(  \alpha\times\beta\right)
}}\boxdot p_{\nu}=p_{\operatorname*{type}\left(  \alpha\times\beta\right)
}\boxdot p_{\nu}=p_{\operatorname*{type}\left(  \left(  \alpha\times
\beta\right)  \times\gamma\right)  }.
\]
A similar argument shows that $u\boxdot\left(  v\boxdot w\right)
=p_{\operatorname*{type}\left(  \alpha\times\left(  \beta\times\gamma\right)
\right)  }$.

Let us now say that if $U$ and $V$ are two finite sets, and if $\sigma
\in\Symm_U$ and $\tau\in\Symm_V$ are two permutations, then
the permutations $\sigma$ and $\tau$ are \emph{isomorphic} if and only if
there exists a bijection $\varphi:U\rightarrow V$ such that $\varphi
\circ\sigma=\tau\circ\varphi$. The intuition behind this meaning of
``isomorphism'' is that two permutations are
isomorphic if one of them becomes the other after a relabelling of its ground
set. It is clear that two isomorphic permutations of finite sets must have the
same cycle type.

But the permutations $\left(  \alpha\times\beta\right)  \times\gamma
\in\Symm_{\left(  A\times B\right)  \times C}$ and $\alpha\times\left(
\beta\times\gamma\right)  \in\Symm_{A\times\left(  B\times C\right)  }$
are isomorphic (as witnessed by the bijection $\varphi:\left(  A\times
B\right)  \times C\rightarrow A\times\left(  B\times C\right)  $ sending every
$\left(  \left(  a,b\right)  ,c\right)  \in\left(  A\times B\right)  \times C$
to $\left(  a,\left(  b,c\right)  \right)  $). Hence, $\operatorname*{type}%
\left(  \left(  \alpha\times\beta\right)  \times\gamma\right)
=\operatorname*{type}\left(  \alpha\times\left(  \beta\times\gamma\right)
\right)  $ (since two isomorphic permutations of finite sets must have the
same cycle type). Thus,
\begin{align*}
\left(  u\boxdot v\right)  \boxdot w  &  =p_{\operatorname*{type}\left(
\left(  \alpha\times\beta\right)  \times\gamma\right)  }%
=p_{\operatorname*{type}\left(  \alpha\times\left(  \beta\times\gamma\right)
\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{type}\left(
\left(  \alpha\times\beta\right)  \times\gamma\right)  =\operatorname*{type}%
\left(  \alpha\times\left(  \beta\times\gamma\right)  \right)  \right) \\
&  =u\boxdot\left(  v\boxdot w\right)  .
\end{align*}
This finishes the proof of the equality $\left(  u\boxdot v\right)  \boxdot
w=u\boxdot\left(  v\boxdot w\right)  $, and thus Exercise
\ref{exe.repSn.arithmetic-prod}(a) is solved.

(b) Let $f\in\Lambda_{\QQ}$. We need to prove the equality $1\boxdot
f=\epsilon_{1}\left(  f\right)  1$. Since this equality is $\QQ$-linear
in $f$ (because the operation $\boxdot$ is $\QQ$-bilinear and the map
$\epsilon_{1}$ is $\QQ$-linear), we can WLOG assume that $f$ belongs to
the basis $\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ of
the $\QQ$-vector space $\Lambda_{\QQ}$. Assume this. Then, there
exists a partition $\lambda$ such that $f=p_{\lambda}$. Consider this
$\lambda$. The definition of $p_{\varnothing}\boxdot p_{\lambda}$ yields
\begin{align*}
p_{\varnothing}\boxdot p_{\lambda}  &  =\prod_{i=1}^{\ell\left(
\varnothing\right)  }\prod_{j=1}^{\ell\left(  \lambda\right)  }%
p_{\operatorname{lcm}\left(  \varnothing_{i},\lambda_{j}\right)  }%
^{\gcd\left(  \varnothing_{i},\lambda_{j}\right)  }=\prod_{i=1}^{0}\prod
_{j=1}^{\ell\left(  \lambda\right)  }p_{\operatorname{lcm}\left(
\varnothing_{i},\lambda_{j}\right)  }^{\gcd\left(  \varnothing_{i},\lambda
_{j}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(
\varnothing\right)  =0\right) \\
&  =\left(  \text{empty product}\right)  =1.
\end{align*}
Hence, $\underbrace{1}_{=p_{\varnothing}}\boxdot\underbrace{f}_{=p_{\lambda}%
}=p_{\varnothing}\boxdot p_{\lambda}=1$.

On the other hand, $\epsilon_{1}\left(  p_{\lambda}\right)  =1$%
\ \ \ \ \footnote{\textit{Proof.} We have $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right)  $. Thus,
the definition of $p_{\lambda}$ yields $p_{\lambda}=p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(  \lambda\right)  }}=\prod
_{i=1}^{\ell\left(  \lambda\right)  }p_{\lambda_{i}}$. Applying the map
$\epsilon_{1}$ to both sides of this equality, we conclude%
\begin{align*}
\epsilon_{1}\left(  p_{\lambda}\right)   &  =\epsilon_{1}\left(  \prod
_{i=1}^{\ell\left(  \lambda\right)  }p_{\lambda_{i}}\right)  =\prod
_{i=1}^{\ell\left(  \lambda\right)  }\underbrace{\epsilon_{1}\left(
p_{\lambda_{i}}\right)  }_{\substack{=1\\\text{(by the definition}\\\text{of
}\epsilon_{1}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\epsilon
_{1}\text{ is a } \QQ\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{\ell\left(  \lambda\right)  }1=1,
\end{align*}
qed.}. Thus, $\epsilon_{1}\left(  \underbrace{f}_{=p_{\lambda}}\right)
=\epsilon_{1}\left(  p_{\lambda}\right)  =1$. Hence, $1\boxdot f=1=\epsilon
_{1}\left(  f\right) = \epsilon_1\left(f\right) 1$.
This solves Exercise \ref{exe.repSn.arithmetic-prod}(b).

(c) For every finite group $G$ and every $h\in G$, we define a class function
$\alpha_{G,h}\in R_{\mathbb{C}}\left(  G\right)  $ as in Exercise
\ref{exe.frob-char.plambda}.

We next recall the notation $U^{\tau}$ defined in Definition
\ref{general-restriction}. We state a few properties of this notation:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.general-restriction1}Let $K$ and $H$ be
two finite groups. Let $\tau:K\rightarrow H$ be a group homomorphism.

\begin{enumerate}
\item[(a)] For every $f\in R_{\mathbb{C}}\left(  H\right)  $, the map
$f\circ\tau:K\rightarrow\mathbb{C}$ belongs to $R_{\mathbb{C}}\left(
K\right)  $. We can thus define a $\CC$-linear map $\tau^{\ast
}:R_{\mathbb{C}}\left(  H\right)  \rightarrow R_{\mathbb{C}}\left(  K\right)
$ by%
\[
\left(  \tau^{\ast}\left(  f\right)  =f\circ\tau\ \ \ \ \ \ \ \ \ \ \text{for
every }f\in R_{\mathbb{C}}\left(  H\right)  \right)  .
\]


\item[(b)] For every (finite-dimensional) $\mathbb{C}H$-module $U$, we have
$\chi_{U^{\tau}}=\tau^{\ast}\left(  \chi_{U}\right)  $.

\item[(c)] We have $\tau^{\ast}\left(  R\left(  H\right)  \right)  \subset
R\left(  K\right)  $.

\item[(d)] Assume that $\tau:K\rightarrow H$ is a group isomorphism. Then,
$\tau^{\ast}\left(  \alpha_{H,\tau\left(  g\right)  }\right)  =\alpha_{K,g}$
for every $g\in K$.
\end{enumerate}
\end{lemma}

The proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1} is
straightforward and left to the reader. (We will not use its part (c).)

Finally, here come two more simple facts whose proofs we leave to the reader:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.core}Let $G$ and $H$ be two groups. Let
$\Omega:G\rightarrow H$ be an injective group homomorphism. Define a map
$\overline{\Omega}:G\rightarrow\Omega\left(  G\right)  $ by%
\[
\left(  \overline{\Omega}\left(  g\right)  =\Omega\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\right)  .
\]
Then, $\overline{\Omega}$ is a well-defined group isomorphism.
\end{lemma}

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.Sf}Let $U$ and $V$ be two finite sets.
Let $\varphi:U\rightarrow V$ be a bijection. Define a map $\varphi^{\ast
}:\Symm_U\rightarrow\Symm_V$ by
\[
\left(  \varphi^{\ast}\left(  \pi\right)  =\varphi\circ\pi\circ\varphi
^{-1}\ \ \ \ \ \ \ \ \ \ \text{for every }\pi\in\Symm_U\right)  .
\]


\begin{enumerate}
\item[(a)] This map $\varphi^{\ast}$ is well-defined and a group isomorphism.

\item[(b)] Let $\pi\in\Symm_U$. Then, the cycle type of
$\varphi^{\ast}\left(  \pi\right)  $ equals the cycle type of $\pi$.
\end{enumerate}
\end{lemma}

Recall that $\boxdot$ is a $\QQ$-bilinear map
$\Lambda_{\QQ}\times\Lambda_{\QQ}\rightarrow\Lambda_{\QQ}$. Let us extend
$\boxdot$ to a $\CC$-bilinear map $\Lambda_{\mathbb{C}}\times
\Lambda_{\mathbb{C}}\rightarrow\Lambda_{\mathbb{C}}$; we will still denote
this extended map by $\boxdot$.

Recall that $\ch : A \rightarrow \Lambda$ is a $\ZZ$-Hopf
algebra isomorphism. Hence, the extension of $\ch$ to a
$\CC$-linear map $A_{\CC} \rightarrow \Lambda_{\CC}$ is a
$\CC$-Hopf algebra isomorphism. We shall denote this extension by
$\ch_{\CC}$.

We are going to repeatedly use the following fact (which is easy to obtain
from the solution of Exercise \ref{exe.frob-char.plambda}): If $\lambda$ is a
partition of a nonnegative integer $n$, and if $g\in\Symm_n$ is a
permutation having cycle type $\lambda$, then%
\begin{equation}
\ch_{\CC} \left(  \alpha_{\Symm_n%
,g}\right)  =p_{\lambda}. \label{sol.repSn.arithmetic-prod.import}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.import}):} Let
$\lambda$ be a partition of a nonnegative integer $n$. Let $g\in
\Symm_n$ be a permutation having cycle type $\lambda$. Define a map
$\Phi:\Lambda_{\mathbb{C}}\rightarrow A_{\mathbb{C}}$ as in the solution of
Exercise \ref{exe.frob-char.plambda}(d). Then,
(\ref{sol.frob-char.plambda.d.1}) yields $\Phi\left(  p_{\lambda}\right)
=\alpha_{\Symm_n,g}$, so that $\alpha_{\Symm_n,g}%
=\Phi\left(  p_{\lambda}\right)  $. But
$\ch_{\CC} \circ\Phi = \id_{\Lambda_{\CC}}$
(this was shown in the solution of Exercise
\ref{exe.frob-char.plambda}(d)). Now, $\ch_{\CC}
\left(  \underbrace{\alpha_{\Symm_n,g}}_{=\Phi\left(  p_{\lambda
}\right)  }\right)  =\ch_{\CC} \left(
\Phi\left(  p_{\lambda}\right)  \right)  =\underbrace{\left(
\ch_{\CC} \circ\Phi\right)  }%
_{=\id_{\Lambda_{\mathbb{C}}}}\left(  p_{\lambda
}\right)  =\id_{\Lambda_{\mathbb{C}}}\left(
p_{\lambda}\right)  =p_{\lambda}$. This proves
(\ref{sol.repSn.arithmetic-prod.import}).}

Theorem \ref{symmetric-group-Frobenius-map-theorem}(a) yields
$\ch \left(  \chi^{\lambda}\right)  =s_{\lambda}$ for every
partition $\lambda$ (where $\chi^{\lambda}$ is defined as in Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a)). Since $\ch_{\CC}$
is the extension of $\ch$ to a
$\CC$-linear map $A_{\mathbb{C}}\rightarrow\Lambda_{\mathbb{C}}$, we
have%
\begin{equation}
\ch_{\CC} \left(  \chi^{\lambda}\right)
=\ch \left(  \chi^{\lambda}\right)  =s_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda.
\label{sol.repSn.arithmetic-prod.chC}%
\end{equation}


Let us now come back to solving Exercise \ref{exe.repSn.arithmetic-prod}(c).
Let $\mu$ and $\nu$ be two partitions. We need to show that $s_{\mu}\boxdot
s_{\nu}\in\sum_{\lambda\in\operatorname{Par}}{\NN}s_{\lambda}$.

Let $m=\left\vert \mu\right\vert $ and $n=\left\vert \nu\right\vert $. Then,
$\mu\in\Par_{m}$ and $\nu\in \Par_{n}$.
If $\min\left\{  m,n\right\}  =0$, then $s_{\mu}\boxdot
s_{\nu}\in\sum_{\lambda\in\operatorname{Par}}{\NN}s_{\lambda}$ is
easily seen to hold\footnote{\textit{Proof.} Assume that $\min\left\{
m,n\right\}  =0$.
\par
Recall that $\Lambda_{\QQ}$, equipped with the binary operation
$\boxdot$, becomes a commutative $\QQ$-algebra with unity $p_{1}$
(according to Exercise \ref{exe.repSn.arithmetic-prod}(a)). Thus, the
operation $\boxdot$ is commutative. Hence, $s_{\mu}\boxdot s_{\nu}=s_{\nu
}\boxdot s_{\mu}$.
\par
We can WLOG assume that $m\leq n$ (since otherwise, we can just interchange
$\mu$ and $m$ with $\nu$ and $n$ (because $s_{\mu}\boxdot s_{\nu}=s_{\nu
}\boxdot s_{\mu}$)). Assume this. Then, $\min\left\{  m,n\right\}  =m$, so
that $m=\min\left\{  m,n\right\}  =0$. Hence,
$\mu\in \Par_{m}=\Par_{0}$ (since $m=0$), so that
$\mu=\varnothing$ and thus $s_{\mu}=s_{\varnothing}=1$.
\par
Let us use the notations of Exercise \ref{exe.repSn.arithmetic-prod}(b). We
have $\underbrace{s_{\mu}}_{=1}\boxdot s_{\nu}=1\boxdot s_{\nu}=\epsilon
_{1}\left(  s_{\nu}\right)  1$ (by Exercise \ref{exe.repSn.arithmetic-prod}%
(b), applied to $f=s_{\nu}$). But $\epsilon_{1}\left(  s_{\nu}\right)
=s_{\nu}\left(  1\right)  $ (by Exercise \ref{exe.Lambda.maps-on-pn}(i),
applied to $f=s_{\nu}$) and $s_{\nu}\left(  1\right)  \in \NN$ (since
$s_{\nu}$ is a sum of monomials). Hence, $s_{\mu}\boxdot s_{\nu}%
=\underbrace{\epsilon_{1}\left(  s_{\nu}\right)  }_{\in\NN %
}\underbrace{1}_{=s_{\varnothing}}\in\NN s_{\varnothing}\subset
\sum_{\lambda\in\operatorname{Par}}{\NN}s_{\lambda}$, qed.}. Hence, for
the rest of this proof, we can WLOG assume that we don't have $\min\left\{
m,n\right\}  =0$. Assume this. Thus, $\min\left\{  m,n\right\}  \geq1$. Hence,
$m\geq1$ and $n\geq1$.

We notice that%
\begin{equation}
\ch_{\CC} \left(  \chi_{P}\right)  \in
\sum_{\lambda\in\operatorname{Par}} \NN s_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \text{for every finite-dimensional }\mathbb{C}%
\Symm_{nm}\text{-module }P\text{.}
\label{sol.repSn.arithmetic-prod.chiP}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.chiP}):} Let $P$ be
a finite-dimensional $\mathbb{C}\Symm_{nm}$-module. Then, $P$ must be a
direct sum of finitely many simple $\mathbb{C}\Symm_{nm}$-modules. In
other words, there exist some simple $\mathbb{C}\Symm_{nm}$-modules
$V_{1},V_{2},\ldots,V_{j}$ such that $P\cong V_{1}\oplus V_{2}\oplus
\cdots\oplus V_{j}$ as $\mathbb{C}\Symm_{nm}$-modules. Consider these
$V_{1},V_{2},\ldots,V_{j}$. We shall now show that $\chi_{V_{i}}\in
\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}$ for every
$i\in\left\{  1,2,\ldots,j\right\}  $.
\par
Indeed, let $i\in\left\{  1,2,\ldots,j\right\}  $. Then, $V_{i}$ is a simple
$\mathbb{C}\Symm_{nm}$-module. Hence, $\chi_{V_{i}}$ is an irreducible
character of $\mathbb{C}\Symm_{nm}$. But (a part of) Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a) (applied to $nm$ instead of
$n$) says that all irreducible characters of $\Symm_{nm}$ have the form
$\chi^{\lambda}$ with $\lambda\in\Par_{nm}$. Thus,
$\chi_{V_{i}}$ has the form $\chi_{V_{i}}=\chi^{\lambda}$ for some $\lambda
\in\Par_{nm}$. In other words, $\chi_{V_{i}}%
\in\left\{  \chi^{\lambda}\ \mid\ \lambda
\in\underbrace{ \Par_{nm}}_{\subset \Par }\right\}  \subset\left\{
\chi^{\lambda}\ \mid\ \lambda\in \Par \right\}  \subset
\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}$.
\par
Now, let us forget that we fixed $i$. We thus have shown that $\chi_{V_{i}}%
\in\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}$ for every
$i\in\left\{  1,2,\ldots,j\right\}  $. Thus, $\sum_{i=1}^{j}\underbrace{\chi
_{V_{i}}}_{\in\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}%
}\in\sum_{i=1}^{j}\sum_{\lambda\in\operatorname{Par}} \NN %
\chi^{\lambda}\subset\sum_{\lambda\in\operatorname{Par}} \NN %
\chi^{\lambda}$ (since $\sum_{\lambda\in\operatorname{Par}} \NN %
\chi^{\lambda}$ is closed under addition).
\par
But isomorphic $\mathbb{C}\Symm_{nm}$-modules have equal characters.
Hence, since $P\cong V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}$, we have%
\[
\chi_{P}=\chi_{V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}}=\chi_{V_{1}}%
+\chi_{V_{2}}+\cdots+\chi_{V_{j}}=\sum_{i=1}^{j}\chi_{V_{i}}\in\sum
_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}.
\]
Applying the map $\ch_{\CC} $ to both sides of
this relation, we obtain%
\begin{align*}
\ch_{\CC} \left(  \chi_{P}\right)   &
\in\ch_{\CC} \left(  \sum_{\lambda
\in\operatorname{Par}} \NN \chi^{\lambda}\right)  \subset
\sum_{\lambda\in\operatorname{Par}} \NN %
\underbrace{\ch_{\CC} \left(  \chi^{\lambda
}\right)  }_{\substack{=s_{\lambda}\\\text{(by
(\ref{sol.repSn.arithmetic-prod.chC}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\ch_{\CC} \text{ is
} \ZZ\text{-linear}\right) \\
&  =\sum_{\lambda\in\operatorname{Par}} \NN s_{\lambda}.
\end{align*}
This proves (\ref{sol.repSn.arithmetic-prod.chiP}).}

We have $\Symm_n=\Symm_{\left\{  1,2,\ldots,n\right\}  }$
(by the definition of $\Symm_n$) and
$\Symm_m=\Symm_{\left\{  1,2,\ldots,m\right\}  }$
(by the definition of $\Symm_m$)
and $\Symm_{nm}=\Symm_{\left\{  1,2,\ldots,nm\right\}  }$ (by
the definition of$\mathfrak{\ S}_{nm}$). Recall that we have defined a map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ for any two sets $X$ and $Y$. Now, set $X=\left\{
1,2,\ldots,n\right\}  $ and $Y=\left\{  1,2,\ldots,m\right\}  $, and consider
the map $\operatorname*{cross}:\Symm_X\times\Symm_Y
\rightarrow\Symm_{X\times Y}$. In the following, when we speak of
$\operatorname*{cross}$, we will always mean this map.

We have $\left\vert \underbrace{X}_{=\left\{  1,2,\ldots,n\right\}
}\right\vert =\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =n$ and
$\left\vert \underbrace{Y}_{=\left\{  1,2,\ldots,m\right\}  }\right\vert
=\left\vert \left\{  1,2,\ldots,m\right\}  \right\vert =m$. Also, since
$X=\left\{  1,2,\ldots,n\right\}  $, we have
$\Symm_X=\Symm_{\left\{  1,2,\ldots,n\right\}  }=\Symm_n$. Similarly,
$\Symm_Y=\Symm_m$.

The set $X$ is nonempty (since $\left\vert X\right\vert =n\geq1$), and the set
$Y$ is nonempty (similarly). Hence, Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross}(b) yields that the map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ is injective. Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross}(a) yields that the map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ is a group homomorphism. Thus, the map
$\operatorname*{cross}$ is a group homomorphism from $\Symm_X%
\times\Symm_Y$ to $\Symm_{X\times Y}$. Since $\Symm_X=\Symm_n$
and $\Symm_Y=\Symm_m$, this rewrites
as follows: The map $\operatorname*{cross}$ is a group homomorphism from
$\Symm_n\times\Symm_m$ to $\Symm_{X\times Y}$.

The set $X\times Y$ has cardinality $\left\vert X\times Y\right\vert
=\underbrace{\left\vert X\right\vert }_{=n}\cdot\underbrace{\left\vert
Y\right\vert }_{=m}=nm$, and thus is in bijection with the set $\left\{
1,2,\ldots,nm\right\}  $. In other words, there exists a bijection
$\varphi:X\times Y\rightarrow\left\{  1,2,\ldots,nm\right\}  $. Fix such a
bijection $\varphi$. Define a map $\varphi^{\ast}:\Symm_{X\times
Y}\rightarrow\Symm_{\left\{  1,2,\ldots,nm\right\}  }$ as in Lemma
\ref{lem.sol.repSn.arithmetic-prod.Sf} (applied to $U=X\times Y$ and
$V=\left\{  1,2,\ldots,nm\right\}  $). Then, Lemma
\ref{lem.sol.repSn.arithmetic-prod.Sf}(a) (applied to $U=X\times Y$ and
$V=\left\{  1,2,\ldots,nm\right\}  $) yields that this map $\varphi^{\ast}$ is
well-defined and a group isomorphism. In particular, $\varphi^{\ast}$ is an
injective group homomorphism. Also, $\varphi^{\ast}$ is a group homomorphism
from $\Symm_{X\times Y}$ to $\Symm_{\left\{  1,2,\ldots
,nm\right\}  }$, therefore a group homomorphism from $\Symm_{X\times
Y}$ to $\Symm_{nm}$ (since $\Symm_{nm}=\Symm_{\left\{
1,2,\ldots,nm\right\}  }$).

Hence, $\varphi^{\ast}\circ\operatorname*{cross}$ is a group homomorphism from
$\Symm_n\times\Symm_m$ to $\Symm_{nm}$ (since
$\varphi^{\ast}$ is a group homomorphism from $\Symm_{X\times Y}$ to
$\Symm_{nm}$, and since $\operatorname*{cross}$ is a group homomorphism
from $\Symm_n\times\Symm_m$ to $\Symm_{X\times Y}$).
Also, $\varphi^{\ast}\circ\operatorname*{cross}$ is injective (since
$\varphi^{\ast}$ and $\operatorname*{cross}$ are injective). Let
$\Omega=\varphi^{\ast}\circ\operatorname*{cross}$. Then, $\Omega$ is an
injective group homomorphism from $\Symm_n\times\Symm_m$ to
$\Symm_{nm}$ (since $\varphi^{\ast}\circ\operatorname*{cross}$ is an
injective group homomorphism from $\Symm_n\times\Symm_m$ to
$\Symm_{nm}$). We can define a map $\overline{\Omega}:\Symm_n
\times\Symm_m\rightarrow\Omega\left(  \Symm_n%
\times\Symm_m\right)  $ by%
\[
\left(  \overline{\Omega}\left(  g\right)  =\Omega\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\Symm_n\times\Symm_m \right)
\]
(since $\Omega\left(  g\right)  \in\Omega\left(  \Symm_n%
\times\Symm_m\right)  $ for every $g\in\Symm_n%
\times\Symm_m$). Consider this $\overline{\Omega}$. Lemma
\ref{lem.sol.repSn.arithmetic-prod.core} (applied to $\Symm_n%
\times\Symm_m$ and $\Symm_{nm}$ instead of $G$ and $H$) yields
that $\overline{\Omega}$ is a well-defined group isomorphism. Hence, the
inverse $\overline{\Omega}^{-1}$ of $\overline{\Omega}$ is well-defined and
also a group isomorphism. Let $\tau$ denote this inverse $\overline{\Omega
}^{-1}$. Thus,
\[
\tau=\overline{\Omega}^{-1}\text{ is a group isomorphism }\Omega\left(
\Symm_n\times\Symm_m\right)  \rightarrow\Symm_n\times\Symm_m.
\]
Hence, a $\CC$-linear map $\tau^{\ast}:R_{\mathbb{C}}\left(
\Symm_n\times\Symm_m\right)  \rightarrow R_{\mathbb{C}%
}\left(  \Omega\left(  \Symm_n\times\Symm_m\right)  \right)
$ is defined (according to Lemma
\ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(a), applied to
$\Omega\left(  \Symm_n\times\Symm_m\right)  $ and
$\Symm_n\times\Symm_m$ instead of $K$ and $H$).

The following commutative diagram illustrates the group homomorphisms we have
just introduced:%
\[\xymatrix{
\Symm_n \times\Symm_m \arinjrev[r]^{\operatorname{cross}} \ar@
/^3pc/[rr]^{\Omega} \ar@/_1pc/[dr]^{\overline{\Omega}}_{\cong} & \Symm
_{X \times Y} \ar[r]^{\varphi^{\ast}}_{\cong} & \Symm_{nm} \\
& \Omega\left(\Symm_n \times\Symm_m\right) \ar@/^4pc/[ul]^{\tau}_{\cong}
\arinjrev[ru]_{\text{inclusion}}
}
\]
(where the cycle formed by the $\overline{\Omega}$ and $\tau$ arrows is not a
mistake: these two maps are mutually inverse!).

In the following, for every $k\in \NN$, we let $\left(  \Lambda
_{\mathbb{C}}\right)  _{k}$ denote the $k$-th homogeneous component of the
graded $\CC$-algebra $\Lambda_{\mathbb{C}}$. Note that $\left(
p_{\lambda}\right)  _{\lambda\in\Par_{k}}$ is a basis
of this $\CC$-vector space $\left(  \Lambda_{\mathbb{C}}\right)  _{k}$.

Let us now claim that every $a\in\left(  \Lambda_{\mathbb{C}}\right)  _{n}$
and $b\in\left(  \Lambda_{\mathbb{C}}\right)  _{m}$ satisfy%
\begin{equation}
a\boxdot b=\ch_{\CC} \left(  \operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \left(  \ch_{\CC}
\right)  ^{-1}\left(  a\right)  \otimes\left(
\ch_{\CC} \right)  ^{-1}\left(  b\right)
\right)  \right)  \right)
\label{sol.repSn.arithmetic-prod.chch}
\end{equation}
(where we identify $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  $ with $R_\CC \left(
\Symm_n\times\Symm_m\right)  $ along the canonical
isomorphism $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  \rightarrow R_{\mathbb{C}%
}\left(  \Symm_n\times\Symm_m\right)  $%
)\ \ \ \ \footnote{Notice that the term $\tau^{\ast}\left(  \left(
\ch_{\CC} \right)  ^{-1}\left(  a\right)
\otimes\left(  \ch_{\CC} \right)  ^{-1}\left(
b\right)  \right)  $ on the right hand side of
(\ref{sol.repSn.arithmetic-prod.chch}) is well-defined. (This is because
\begin{align*}
\left(  \ch_{\CC} \right)  ^{-1}\left(
\underbrace{a}_{\in\left(  \Lambda_{\mathbb{C}}\right)  _{n}}\right)
\otimes\left(  \ch_{\CC} \right)  ^{-1}\left(
\underbrace{b}_{\in\left(  \Lambda_{\mathbb{C}}\right)  _{m}}\right)   &
\in\underbrace{\left(  \ch_{\CC} \right)
^{-1}\left(  \left(  \Lambda_{\mathbb{C}}\right)  _{n}\right)  }%
_{=R_{\mathbb{C}}\left(  \Symm_n\right)  }\otimes\underbrace{\left(
\ch_{\CC} \right)  ^{-1}\left(  \left(
\Lambda_{\mathbb{C}}\right)  _{m}\right)  }_{=R_{\mathbb{C}}\left(
\Symm_m\right)  }\\
&  =R_\CC \left(  \Symm_n\right)  \otimes R_{\mathbb{C}%
}\left(  \Symm_m\right)
=R_{\mathbb{C}}\left(  \Symm_n\times\Symm_m\right)  .
\end{align*}
)}.

\textit{Proof of (\ref{sol.repSn.arithmetic-prod.chch}):} Let $a\in\left(
\Lambda_{\mathbb{C}}\right)  _{n}$ and $b\in\left(  \Lambda_{\CC}
\right)  _{m}$. We need to prove the equality
(\ref{sol.repSn.arithmetic-prod.chch}). Since this equality is $\CC%
$-linear in each of $a$ and $b$ (because the operations $\boxdot$ and
$\otimes$ are $\CC$-bilinear, and the maps $\ch_{\CC}$
$\left( \ch_{\CC} \right)  ^{-1}$, $\Ind_{\Omega\left(  \Symm_n
\times\Symm_m\right)  }^{\Symm_{nm}}$ and $\tau^{\ast}$
are $\CC$-linear), we can WLOG assume that $a$ is an element of the
basis $\left(  p_{\lambda}\right)  _{\lambda\in \Par_{n}}$
of the $\CC$-vector space $\left(  \Lambda_{\mathbb{C}%
}\right)  _{n}$, and that $b$ is an element of the basis $\left(  p_{\lambda
}\right)  _{\lambda\in\Par_{m}}$ of the $\mathbb{C}%
$-vector space $\left(  \Lambda_{\mathbb{C}}\right)  _{m}$. Assume this. Then,
we can write $a$ and $b$ as $a=p_{\gamma}$ and $b=p_{\eta}$ for some
$\gamma\in\Par_{n}$ and some $\eta\in
\Par_{m}$. Consider these $\gamma$ and $\eta$.

Choose some permutation $g\in\Symm_n$ which has cycle type $\gamma$.
(Such a $g$ clearly exists.) Then, $\operatorname*{ch}\nolimits_{\mathbb{C}%
}\left(  \alpha_{\Symm_n,g}\right)  =p_{\gamma}$ (according to
(\ref{sol.repSn.arithmetic-prod.import})). Hence, $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  p_{\gamma}\right)  =\alpha
_{\Symm_n,g}$, so that $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  \underbrace{a}_{=p_{\gamma}%
}\right)  =\left(  \ch_{\CC} \right)
^{-1}\left(  p_{\gamma}\right)  =\alpha_{\Symm_n,g}$.

Choose some permutation $h\in\Symm_m$ which has cycle type $\eta$.
(Such an $h$ clearly exists.) Then, $\operatorname*{ch}\nolimits_{\mathbb{C}%
}\left(  \alpha_{\Symm_m,h}\right)  =p_{\eta}$ (according to
(\ref{sol.repSn.arithmetic-prod.import}), applied to $m$, $h$ and $\eta$
instead of $n$, $g$ and $\lambda$). Hence, $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  p_{\eta}\right)  =\alpha
_{\Symm_m,h}$, so that $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  \underbrace{b}_{=p_{\eta}}\right)
=\left(  \ch_{\CC} \right)  ^{-1}\left(
p_{\eta}\right)  =\alpha_{\Symm_m,h}$. Thus,%
\begin{equation}
\underbrace{\left(  \ch_{\CC} \right)
^{-1}\left(  a\right)  }_{=\alpha_{\Symm_n,g}}\otimes\underbrace{\left(
\ch_{\CC} \right)  ^{-1}\left(  b\right)
}_{=\alpha_{\Symm_m,h}}=\alpha_{\Symm_n,g}\otimes
\alpha_{\Symm_m,h}. \label{sol.repSn.arithmetic-prod.chch.pf.1}%
\end{equation}


Exercise \ref{exe.frob-char.plambda}(c) (applied to $G_{1}=\Symm_n$,
$G_{2}=\Symm_m$, $h_{1}=g$ and $h_{2}=h$) yields that the canonical
isomorphism $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  \rightarrow R_{\mathbb{C}%
}\left(  \Symm_n\times\Symm_m\right)  $ sends
$\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}$ to
$\alpha_{\Symm_n\times\Symm_m,\left(  g,h\right)  }$. Since
we are regarding this isomorphism as an identity (because we have identified
$R_\CC \left(  \Symm_n\right)  \otimes R_{\mathbb{C}%
}\left(  \Symm_m\right)  $ with $R_\CC \left(
\Symm_n\times\Symm_m\right)  $ along this isomorphism), this
yields $\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}%
=\alpha_{\Symm_n\times\Symm_m,\left(  g,h\right)  }$. Thus,
(\ref{sol.repSn.arithmetic-prod.chch.pf.1}) becomes%
\[
\left(  \ch_{\CC} \right)  ^{-1}\left(
a\right)  \otimes\left(  \ch_{\CC} \right)
^{-1}\left(  b\right)  =\alpha_{\Symm_n,g}\otimes\alpha
_{\Symm_m,h}=\alpha_{\Symm_n\times\Symm_m,\left(
g,h\right)  }.
\]
Applying the map $\tau^{\ast}$ to both sides of this equality, we obtain%
\begin{equation}
\tau^{\ast}\left(  \left(  \ch_{\CC} \right)
^{-1}\left(  a\right)  \otimes\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  b\right)  \right)  =\tau^{\ast}\left(  \alpha
_{\Symm_n\times\Symm_m,\left(  g,h\right)  }\right)  .
\label{sol.repSn.arithmetic-prod.chch.pf.2}%
\end{equation}


But $\overline{\Omega}\left(  \left(  g,h\right)  \right)  =\Omega\left(
\left(  g,h\right)  \right)  $ (by the definition of $\overline{\Omega}\left(
\left(  g,h\right)  \right)  $), and so $\left(  g,h\right)
=\underbrace{\overline{\Omega}^{-1}}_{=\tau}\left(  \Omega\left(  \left(
g,h\right)  \right)  \right)  =\tau\left(  \Omega\left(  \left(  g,h\right)
\right)  \right)  $. Thus,
\[
\tau^{\ast}\left(  \alpha_{\Symm_n\times\Symm_m,\left(
g,h\right)  }\right)  =\tau^{\ast}\left(  \alpha_{\Symm_n%
\times\Symm_m,\tau\left(  \Omega\left(  \left(  g,h\right)  \right)
\right)  }\right)  =\alpha_{\Omega\left(  \Symm_n\times
\Symm_m\right)  ,\Omega\left(  \left(  g,h\right)  \right)  }%
\]
(by Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(d), applied
to $\Omega\left(  \Symm_n\times\Symm_m\right)  $,
$\Symm_n\times\Symm_m$ and $\Omega\left(  \left(
g,h\right)  \right)  $ instead of $K$, $H$ and $g$). Thus,
(\ref{sol.repSn.arithmetic-prod.chch.pf.2}) becomes%
\[
\tau^{\ast}\left(  \left(  \ch_{\CC} \right)
^{-1}\left(  a\right)  \otimes\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  b\right)  \right)  =\tau^{\ast}\left(  \alpha
_{\Symm_n\times\Symm_m,\left(  g,h\right)  }\right)
=\alpha_{\Omega\left(  \Symm_n\times\Symm_m\right)
,\Omega\left(  \left(  g,h\right)  \right)  }.
\]
Applying the map $\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n
\times\Symm_m\right)  }^{\Symm_{nm}}$ to both sides of
this equality, we obtain%
\begin{align}
\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n\times
\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\left(  \ch_{\CC} \right)  ^{-1}\left(
a\right)  \otimes\left(  \ch_{\CC} \right)
^{-1}\left(  b\right)  \right)  \right)   &  =\operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \alpha_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  ,\Omega\left(  \left(  g,h\right)  \right)
}\right) \nonumber\\
&  =\alpha_{\Symm_{nm},\Omega\left(  \left(  g,h\right)  \right)  }
\label{sol.repSn.arithmetic-prod.chch.pf.6}%
\end{align}
(by Exercise \ref{exe.frob-char.plambda}(b), applied to $\Symm_{nm}$,
$\Omega\left(  \Symm_n\times\Symm_m\right)  $ and
$\Omega\left(  \left(  g,h\right)  \right)  $ instead of $G$, $H$ and $h$).

Now, $\Omega\left(  \left(  g,h\right)  \right)  \in\Symm_{nm}$, so
that $\Omega\left(  \left(  g,h\right)  \right)  $ is a permutation of
$\left\{  1,2,\ldots,nm\right\}  $. Let $\kappa$ be the cycle type of this
permutation $\Omega\left(  \left(  g,h\right)  \right)  $. Then,
(\ref{sol.repSn.arithmetic-prod.import}) (applied to $nm$, $\kappa$ and
$\Omega\left(  \left(  g,h\right)  \right)  $ instead of $n$, $\lambda$ and
$g$) yields $\ch_{\CC} \left(  \alpha
_{\Symm_{nm},\Omega\left(  \left(  g,h\right)  \right)  }\right)
=p_{\kappa}$. But applying the map $\ch_{\CC} $
to both sides of the identity (\ref{sol.repSn.arithmetic-prod.chch.pf.6}), we
obtain%
\begin{align}
&  \ch_{\CC} \left(  \operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  a\right)  \otimes\left(
\ch_{\CC} \right)  ^{-1}\left(  b\right)
\right)  \right)  \right) \nonumber\\
&  =\ch_{\CC} \left(
\alpha_{\Symm_{nm},\Omega\left(  \left(  g,h\right)  \right)  }\right)
=p_{\kappa}.
\label{sol.repSn.arithmetic-prod.chch.pf.8}%
\end{align}


Let us now show that $a\boxdot b=p_{\kappa}$.

In fact, $\underbrace{\Omega}_{=\varphi^{\ast}\circ\operatorname*{cross}%
}\left(  \left(  g,h\right)  \right)  =\left(  \varphi^{\ast}\circ
\operatorname*{cross}\right)  \left(  \left(  g,h\right)  \right)
=\varphi^{\ast}\left(  \underbrace{\operatorname*{cross}\left(  \left(
g,h\right)  \right)  }_{\substack{=g\times h\\\text{(by the definition}%
\\\text{of }\operatorname*{cross}\text{)}}}\right)  =\varphi^{\ast}\left(
g\times h\right)  $. But Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}(b)
(applied to $U=X\times Y$, $V=\left\{  1,2,\ldots,nm\right\}  $ and
$\pi=g\times h$) yields that the cycle type of $\varphi^{\ast}\left(  g\times
h\right)  $ equals the cycle type of $g\times h$. In other words, the cycle
type of $\Omega\left(  \left(  g,h\right)  \right)  $ equals the cycle type of
$g\times h$ (since $\Omega\left(  \left(  g,h\right)  \right)  =\varphi^{\ast
}\left(  g\times h\right)  $). In other words, $\kappa$ equals the cycle type
of $g\times h$ (since $\kappa$ is the cycle type of $\Omega\left(  \left(
g,h\right)  \right)  $).

So we know that $g\in\Symm_n=\Symm_X$ and $h\in
\Symm_m=\Symm_Y$ are permutations, and that $\gamma$, $\eta$
and $\kappa$ are the cycle types of the permutations $g$, $h$ and $g\times h$.
Thus, Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2p} (applied to $\gamma$
and $\eta$ instead of $\lambda$ and $\mu$) yields $p_{\gamma}\boxdot p_{\eta
}=p_{\kappa}$. Now,%
\[
\underbrace{a}_{=p_{\gamma}}\boxdot\underbrace{b}_{=p_{\eta}}=p_{\gamma
}\boxdot p_{\eta}=p_{\kappa}=\ch_{\CC} \left(
\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n\times
\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\left(  \ch_{\CC} \right)  ^{-1}\left(
a\right)  \otimes\left(  \ch_{\CC} \right)
^{-1}\left(  b\right)  \right)  \right)  \right)
\]
(by (\ref{sol.repSn.arithmetic-prod.chch.pf.8})). This proves
(\ref{sol.repSn.arithmetic-prod.chch}).

Now that (\ref{sol.repSn.arithmetic-prod.chch}) is proven, it is easy to
complete the solution of Exercise \ref{exe.repSn.arithmetic-prod}(c). Recall
that $\mu$ and $\nu$ are two partitions such that $\mu\in \Par_{m}$,
$\nu\in\Par_{n}$, $m\geq1$ and
$n\geq1$. We need to show that $s_{\mu}\boxdot s_{\nu}\in\sum_{\lambda
\in\operatorname{Par}}{\NN}s_{\lambda}$.

Applying (\ref{sol.repSn.arithmetic-prod.chC}) to $\lambda=\mu$, we obtain
$\ch_{\CC} \left(  \chi^{\mu}\right)  =s_{\mu}$,
so that $\left(  \ch_{\CC} \right)  ^{-1}\left(
s_{\mu}\right)  =\chi^{\mu}$. The same argument (but with $\mu$ replaced by
$\nu$) shows that $\left(  \ch_{\CC} \right)
^{-1}\left(  s_{\nu}\right)  =\chi^{\nu}$.

We know that $\chi^{\mu}$ is an irreducible complex character of
$\mathbb{C}\Symm_m$. In other words, there exists a simple
$\mathbb{C}\Symm_m$-module $M$ such that $\chi^{\mu}=\chi_{M}$.
Consider this $M$. Then, $\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  s_{\mu}\right)  =\chi^{\mu}=\chi_{M}$.

We know that $\chi^{\nu}$ is an irreducible complex character of
$\mathbb{C}\Symm_n$. In other words, there exists a simple
$\mathbb{C}\Symm_n$-module $N$ such that $\chi^{\nu}=\chi_{N}$.
Consider this $N$. Then, $\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  s_{\nu}\right)  =\chi^{\nu}=\chi_{N}$.

Recall that we are identifying $R_\CC \left(  \Symm_n\right)
\otimes R_{\mathbb{C}}\left(  \Symm_m\right)  $ with
$R_\CC \left(  \Symm_n\times\Symm_m\right)  $
along the canonical isomorphism $R_\CC \left(  \Symm_n\right)
\otimes R_{\mathbb{C}}\left(  \Symm_m\right)
\rightarrow R_{\mathbb{C}}\left(  \Symm_n\times\Symm_m\right)  $.
Thus, for any finite-dimensional $\mathbb{C}\Symm_n%
$-module $U$ and any finite-dimensional $\mathbb{C}\Symm_m$-module
$V$, we have $\chi_{U\otimes V}=\chi_{U}\otimes\chi_{V}$ (since this
isomorphism sends $\chi_{U}\otimes\chi_{V}$ to $\chi_{U\otimes V}$). Applying
this to $U=N$ and $V=M$, we obtain $\chi_{N\otimes M}=\chi_{N}\otimes\chi_{M}%
$. Thus, $\chi_{N}\otimes\chi_{M}=\chi_{N\otimes M}$.

Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(b) (applied to
$\Omega\left(  \Symm_n\times\Symm_m\right)  $,
$\Symm_n\times\Symm_m$ and $N\otimes M$ instead of $K$, $H$
and $U$) yields $\chi_{\left(  N\otimes M\right)  ^{\tau}}=\tau^{\ast}\left(
\chi_{N\otimes M}\right)  $. Thus, $\tau^{\ast}\left(  \chi_{N\otimes
M}\right)  =\chi_{\left(  N\otimes M\right)  ^{\tau}}$. Hence,%
\begin{equation}
\tau^{\ast}\left(  \underbrace{\chi_{N}\otimes\chi_{M}}_{=\chi_{N\otimes M}%
}\right)  =\tau^{\ast}\left(  \chi_{N\otimes M}\right)  =\chi_{\left(
N\otimes M\right)  ^{\tau}}. \label{sol.repSn.arithmetic-prod.characalc4}%
\end{equation}


But (\ref{eq.induction.mods-and-class-functs}) (applied to $\Symm_{nm}%
$, $\Omega\left(  \Symm_n\times\Symm_m\right)  $ and
$\left(  N\otimes M\right)  ^{\tau}$ instead of $G$, $H$ and $U$) yields
$\chi_{\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \left(  N\otimes
M\right)  ^{\tau}\right)  }=\operatorname*{Ind}\nolimits_{\Omega\left(
\Symm_n\times\Symm_m\right)  }^{\Symm_{nm}}\left(
\chi_{\left(  N\otimes M\right)  ^{\tau}}\right)  $. Let $P$ denote the
finite-dimensional $\mathbb{C}\Symm_{nm}$-module $\operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \left(  N\otimes M\right)  ^{\tau}\right)  $.
Then, $P=\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \left(  N\otimes
M\right)  ^{\tau}\right)  $, so that%
\[
\chi_{P}=\chi_{\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \left(
N\otimes M\right)  ^{\tau}\right)  }=\operatorname*{Ind}\nolimits_{\Omega
\left(  \Symm_n\times\Symm_m\right)  }^{\Symm_{nm}%
}\left(  \underbrace{\chi_{\left(  N\otimes M\right)  ^{\tau}}}%
_{\substack{=\tau^{\ast}\left(  \chi_{N}\otimes\chi_{M}\right)  \\\text{(by
(\ref{sol.repSn.arithmetic-prod.characalc4}))}}}\right)  =\operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \chi_{N}\otimes\chi
_{M}\right)  \right)  .
\]
Hence,%
\begin{equation}
\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n\times
\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\chi_{N}\otimes\chi_{M}\right)  \right)  =\chi_{P}.
\label{sol.repSn.arithmetic-prod.characalc7}%
\end{equation}


We have $s_{\mu}\in\left(  \Lambda_{\mathbb{C}}\right)  _{m}$ (since $\mu
\in\Par_{m}$) and $s_{\nu}\in\left(  \Lambda
_{\mathbb{C}}\right)  _{n}$ (since $\nu\in\Par_{n}$).
Thus, we can apply (\ref{sol.repSn.arithmetic-prod.chch}) to $a=s_{\nu}$ and
$b=s_{\mu}$.

Exercise \ref{exe.repSn.arithmetic-prod}(a) yields that $\Lambda_{\QQ}%
$, equipped with the binary operation $\boxdot$, becomes a commutative
$\QQ$-algebra with unity $p_{1}$. Thus, the operation $\boxdot$ is
commutative. Hence,%
\begin{align*}
s_{\mu}\boxdot s_{\nu}  &  =s_{\nu}\boxdot s_{\mu}=\operatorname*{ch}%
\nolimits_{\mathbb{C}}\left(  \operatorname*{Ind}\nolimits_{\Omega\left(
\Symm_n\times\Symm_m\right)  }^{\Symm_{nm}}\left(
\tau^{\ast}\left(  \underbrace{\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  s_{\nu}\right)  }_{=\chi_{N}}\otimes\underbrace{\left(
\ch_{\CC} \right)  ^{-1}\left( s_\mu \right)
}_{=\chi_{M}}\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.repSn.arithmetic-prod.chch}%
), applied to }a=s_{\nu}\text{ and }b=s_{\mu}\right) \\
&  =\ch_{\CC} \left(
\underbrace{\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\chi_{N}\otimes\chi_{M}\right)  \right)  }_{\substack{=\chi_{P}\\\text{(by
(\ref{sol.repSn.arithmetic-prod.characalc7}))}}}\right)  =\operatorname*{ch}%
\nolimits_{\mathbb{C}}\left(  \chi_{P}\right)  \in\sum_{\lambda\in
\operatorname{Par}} \NN s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.repSn.arithmetic-prod.chiP})}\right)  .
\end{align*}
This solves Exercise \ref{exe.repSn.arithmetic-prod}(c).

[\textit{Remark:} The above solution can be simplified using the results
of Exercise~\ref{exe.Indrho}. We leave the details of this simplification
to the reader, and only mention some highlights. Let us use the notations
of Exercise~\ref{exe.Indrho}. Then, the awkward equality
\eqref{sol.repSn.arithmetic-prod.chch} can be replaced by the simpler
equality
\begin{equation}
a \boxdot b
= \ch_\CC \left( \Ind_\Omega
                 \left( \left(\ch_\CC\right)^{-1} \left(a\right)
                        \otimes
                        \left(\ch_\CC\right)^{-1} \left(b\right)
                 \right)
          \right) .
\end{equation}
This makes the definition of the maps $\overline{\Omega}$ and $\tau$
unnecessary. The WLOG assumption that we don't have $\min\left\{
m,n\right\}  =0$ becomes unnecessary as well; the injectivity of the
map $\operatorname{cross}$ can no longer be guaranteed without this
assumption, but we do not need this map to be injective anymore.
We no longer need
Lemma~\ref{lem.sol.repSn.arithmetic-prod.cross}(b),
Lemma~\ref{lem.sol.repSn.arithmetic-prod.general-restriction1} and
Lemma~\ref{lem.sol.repSn.arithmetic-prod.core}.
However, we need to use Exercise~\ref{exe.Indrho}(b) instead of
\eqref{eq.induction.mods-and-class-functs}, and we have to use
Exercise~\ref{exe.frob-char.plambda}(f) instead of
Exercise~\ref{exe.frob-char.plambda}(b).]

(d) We need to prove that $f\boxdot g\in\Lambda$ for any $f\in\Lambda$ and
$g\in\Lambda$. Since the binary operation $\boxdot$ is $\ZZ$-bilinear
(because it is $\QQ$-bilinear), it is clearly enough to prove that
$s_{\mu}\boxdot s_{\nu}\in\Lambda$ for any partitions $\mu$ and $\nu$ (since
$\left(  s_{\lambda}\right)  _{\lambda\in\operatorname{Par}}$ is a
$\ZZ$-basis of $\Lambda$). But this follows from the fact that%
\begin{align*}
s_{\mu}\boxdot s_{\nu}  &  \in\sum_{\lambda\in\operatorname{Par}}
\NN s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.repSn.arithmetic-prod}(c)}\right) \\
&  \subset\Lambda
\end{align*}
for any partitions $\mu$ and $\nu$. Thus,
Exercise~\ref{exe.repSn.arithmetic-prod}(d) is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.repSn.arithmetic-prod}}

\textit{Solution to Exercise \ref{exe.repSn.arithmetic-prod}.} (a) We first
introduce some notation pertaining to multisets:

\begin{itemize}
\item Given a list $\left(  b_{1},b_{2},\ldots,b_{\ell}\right)  $ of objects,
we let $\left\{  b_{1},b_{2},\ldots,b_{\ell}\right\}
_{\operatorname*{multiset}}$ denote the multiset which consists of the objects
$b_{1}$, $b_{2}$, $\ldots$, $b_{\ell}$ (containing each of them as often as it
appears in the list $\left(  b_{1},b_{2},\ldots,b_{\ell}\right)  $).

\item Given a finite multiset $M$ of positive integers, we define the
\emph{decreasing list}\index{decreasing list of a multiset}
of $M$ to be the list of all elements of $M$ in
weakly decreasing order (such that each element appears in the list as often
as it appears in $M$). For instance, the decreasing list of the multiset
$\left\{  4,1,1,3,1,6,2,3\right\}  _{\operatorname*{multiset}}$ is $\left(
6,4,3,3,2,1,1,1\right)  $. The decreasing list of a finite multiset of
positive integers is always a partition.

\item We let $\mathfrak{M}$ denote the set of all finite multisets of positive integers.

\item We define a map $\operatorname*{muls}: \Par \rightarrow
\mathfrak{M}$ by setting
\[
\operatorname*{muls}\lambda=\left\{  \lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell\left(  \lambda\right)  }\right\}  _{\operatorname*{multiset}%
}\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in \Par .
\]
(In other words, $\operatorname*{muls}\lambda$ is the multiset whose entries
are the (positive) parts of $\lambda$, each appearing as often as it appears
in $\lambda$.) It is clear that the map $\operatorname*{muls}$ is a bijection,
and its inverse is given by%
\[
\operatorname*{muls}\nolimits^{-1}M=\left(  \text{the decreasing list of
}M\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }M\in\mathfrak{M}.
\]


\item Recall that $x \QQ\left[  x\right]  $ is the $\QQ$-vector
space of all polynomials with rational coefficients having constant term $0$.
Such polynomials can be written uniquely in the form $\sum_{i\geq1}\alpha
_{i}x^{i}$ for a sequence $\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  $ of rational numbers such that all but finitely many $i\geq1$
satisfy $\alpha_{i}=0$.

\item Let $x\NN \left[  x\right]  $ be the set of all polynomials in
$x \QQ\left[  x\right]  $ whose coefficients are nonnegative integers.
Such polynomials can be written uniquely in the form $\sum_{i\geq1}\alpha
_{i}x^{i}$ for a sequence $\left(  \alpha_{1},\alpha_{2},\alpha_{3}%
,\ldots\right)  $ of nonnegative integers such that all but finitely many
$i\geq1$ satisfy $\alpha_{i}=0$. Define a map $\operatorname*{pol}%
:\mathfrak{M}\rightarrow x\NN \left[  x\right]  $ by%
\[
\operatorname*{pol}M=\sum_{i\geq1}\left(  \text{the multiplicity of }i\text{
in }M\right)  x^{i}\ \ \ \ \ \ \ \ \ \ \text{for every }M\in\mathfrak{M}.
\]
(This is well-defined because every $M\in\mathfrak{M}$ is a finite multiset of
positive integers.) This map $\operatorname*{pol}$ is a bijection\footnote{Its
inverse can be computed as follows: Given a polynomial $\alpha=\sum_{i\geq
1}\alpha_{i}x^{i}\in x\NN \left[  x\right]  $, we have%
\[
\operatorname*{pol}\nolimits^{-1}\alpha=\left(  \text{the multiset of positive
integers which contains every positive integer }i\text{ exactly }\alpha
_{i}\text{ times}\right)  .
\]
}.

The map $\operatorname*{pol}\circ\operatorname*{muls}: \Par %
\rightarrow x\NN \left[  x\right]  $ is a bijection (since it is the
composition of the bijections $\operatorname*{pol}$ and $\operatorname*{muls}$).

We notice that every partition $\lambda$ satisfies%
\begin{equation}
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  =\sum_{i\geq
1}\left\vert \left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \ \mid\ \lambda_{j}=i\right\}  \right\vert x^{i}
\label{sol.repSn.arithmetic-prod.polmuls1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.polmuls1}):} Let
$\lambda$ be a partition. Then, $\operatorname*{muls}\lambda=\left\{
\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right\}
_{\operatorname*{multiset}}$ (by the definition of $\operatorname*{muls}%
\lambda$). Thus, for every positive integer $i$, we have%
\begin{align}
&  \left(  \text{the multiplicity of }i\text{ in }%
\underbrace{\operatorname*{muls}\lambda}_{=\left\{  \lambda_{1},\lambda
_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right\}
_{\operatorname*{multiset}}}\right) \nonumber\\
&  =\left(  \text{the multiplicity of }i\text{ in }\left\{  \lambda
_{1},\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right\}
_{\operatorname*{multiset}}\right) \nonumber\\
&  =\left(  \text{the number of all }j\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \text{ such that }\lambda_{j}=i\right) \nonumber\\
&  =\left\vert \left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \ \mid\ \lambda_{j}=i\right\}  \right\vert .
\label{sol.repSn.arithmetic-prod.polmuls1.pf.1}%
\end{align}
But the definition of $\operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  $ yields
\[
\operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  =\sum_{i\geq1}\underbrace{\left(  \text{the multiplicity of
}i\text{ in }\operatorname*{muls}\lambda\right)  }_{\substack{=\left\vert
\left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\ \mid\ \lambda_{j}=i\right\}  \right\vert \\\text{(by
(\ref{sol.repSn.arithmetic-prod.polmuls1.pf.1}))}}}x^{i}=\sum_{i\geq
1}\left\vert \left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \ \mid\ \lambda_{j}=i\right\}  \right\vert x^{i} .
\]
This proves
(\ref{sol.repSn.arithmetic-prod.polmuls1}).}. Also, every partition $\lambda$
satisfies%
\begin{equation}
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  =\sum
_{j=1}^{\ell\left(  \lambda\right)  }x^{\lambda_{j}}
\label{sol.repSn.arithmetic-prod.polmuls2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.polmuls2}):} Let
$\lambda$ be a partition. Then,%
\begin{align*}
\sum_{j=1}^{\ell\left(  \lambda\right)  }x^{\lambda_{j}}  &  =\sum
_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  }%
x^{\lambda_{j}}=\sum_{i\geq1}\sum\limits_{\substack{j\in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{j}=i}%
}\underbrace{x^{\lambda_{j}}}_{\substack{=x^{i}\\\text{(since }\lambda
_{j}=i\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda_{j}%
\geq1\text{ for every }j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \right) \\
&  =\sum_{i\geq1}\underbrace{\sum\limits_{\substack{j\in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{j}=i}}x^{i}}_{=\left\vert
\left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\ \mid\ \lambda_{j}=i\right\}  \right\vert x^{i}}=\sum_{i\geq1}\left\vert
\left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\ \mid\ \lambda_{j}=i\right\}  \right\vert x^{i}=\operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)
\end{align*}
(by (\ref{sol.repSn.arithmetic-prod.polmuls1})). This proves
(\ref{sol.repSn.arithmetic-prod.polmuls2}).}.

\item We define a binary operation $\squarebin$ on the set $x \QQ\left[
x\right]  $ (written infix) as follows: For any two polynomials $\alpha
=\sum_{i\geq1}\alpha_{i}x^{i}\in x \QQ\left[  x\right]  $ and
$\beta=\sum_{i\geq1}\beta_{i}x^{i}\in x \QQ\left[  x\right]  $ (where
$\alpha_{i}$ and $\beta_{i}$ are rational numbers), we set%
\[
\alpha\squarebin\beta=\sum_{i\geq1}\sum_{j\geq1}\alpha_{i}\beta_{j}\gcd\left(
i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)  }.
\]
This operation is clearly well-defined (that is, $\alpha\squarebin\beta$ indeed
lies in $x \QQ\left[  x\right]  $) and $\QQ$-bilinear (because
$\alpha_{i}$ and $\beta_{j}$ enter linearly into the sum $\sum_{i\geq1}%
\sum_{j\geq1}\alpha_{i}\beta_{j}\gcd\left(  i,j\right)  x^{\operatorname{lcm}%
\left(  i,j\right)  }$). It is furthermore clear that $\alpha\squarebin\beta\in
x\NN \left[  x\right]  $ whenever $\alpha$ and $\beta$ belong to
$x\NN \left[  x\right]  $. Moreover, we have%
\begin{equation}
x^{u}\squarebin x^{v}=\gcd\left(  u,v\right)  x^{\operatorname{lcm}\left(
u,v\right)  }\ \ \ \ \ \ \ \ \ \ \text{for any positive integers }u\text{ and
}v. \label{sol.repSn.arithmetic-prod.boxdot-on-xes}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}):}
Let $u$ and $v$ be positive integers. We have
\[
\sum_{i\geq1}\delta_{i,u}x^{i}=\sum\limits_{\substack{i\geq1;\\i=u}%
}\underbrace{\delta_{i,u}}_{\substack{=1\\\text{(since }i=u\text{)}}%
}x^{i}+\sum\limits_{\substack{i\geq1;\\i\neq u}}\underbrace{\delta_{i,u}%
}_{\substack{=0\\\text{(since }i\neq u\text{)}}}x^{i}=\sum\limits_{\substack{i\geq
1;\\i=u}}1x^{i}+\underbrace{\sum\limits_{\substack{i\geq1;\\i\neq u}}0x^{i}}%
_{=0}=\sum\limits_{\substack{i\geq1;\\i=u}}1x^{i}=1x^{u}=x^{u}.
\]
The same argument (but with $u$ replaced by $v$) yields $\sum_{i\geq1}%
\delta_{i,v}x^{i}=x^{v}$. Now, the definition of the operation $\squarebin$
yields%
\begin{align*}
&  \left(  \sum_{i\geq1}\delta_{i,u}x^{i}\right)  \squarebin\left(  \sum_{i\geq
1}\delta_{i,v}x^{i}\right)  =\sum_{i\geq1}\sum_{j\geq1}\delta_{i,u}%
\delta_{j,v}\gcd\left(  i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)
}\\
&  =\sum\limits_{\substack{i\geq1;\\i=u}}\sum_{j\geq1}\underbrace{\delta_{i,u}%
}_{\substack{=1\\\text{(since }i=u\text{)}}}\delta_{j,v}\gcd\left(
i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)  }+\sum\limits_{\substack{i\geq
1;\\i\neq u}}\sum_{j\geq1}\underbrace{\delta_{i,u}}%
_{\substack{=0\\\text{(since }i\neq u\text{)}}}\delta_{j,v}\gcd\left(
i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)  }\\
&  =\sum\limits_{\substack{i\geq1;\\i=u}}\sum_{j\geq1}1\delta_{j,v}\gcd\left(
i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)  }+\underbrace{\sum\limits
_{\substack{i\geq1;\\i\neq u}}\sum_{j\geq1}0\delta_{j,v}\gcd\left(
i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)  }}_{=0}\\
&  =\sum\limits_{\substack{i\geq1;\\i=u}}\sum_{j\geq1}1\delta_{j,v}\gcd\left(
i,j\right)  x^{\operatorname{lcm}\left(  i,j\right)  }=\sum_{j\geq1}%
1\delta_{j,v}\gcd\left(  u,j\right)  x^{\operatorname{lcm}\left(  u,j\right)
}\\
&  =\sum_{j\geq1}\delta_{j,v}\gcd\left(  u,j\right)  x^{\operatorname{lcm}%
\left(  u,j\right)  }=\sum\limits_{\substack{j\geq1;\\j=v}}\underbrace{\delta_{j,v}%
}_{\substack{=1\\\text{(since }j=v\text{)}}}\gcd\left(  u,j\right)
x^{\operatorname{lcm}\left(  u,j\right)  }+\sum\limits_{\substack{j\geq1;\\j\neq
v}}\underbrace{\delta_{j,v}}_{\substack{=0\\\text{(since }j\neq v\text{)}%
}}\gcd\left(  u,j\right)  x^{\operatorname{lcm}\left(  u,j\right)  }\\
&  =\sum\limits_{\substack{j\geq1;\\j=v}}1\gcd\left(  u,j\right)
x^{\operatorname{lcm}\left(  u,j\right)  }+\underbrace{\sum\limits_{\substack{j\geq
1;\\j\neq v}}0\gcd\left(  u,j\right)  x^{\operatorname{lcm}\left(  u,j\right)
}}_{=0}=\sum\limits_{\substack{j\geq1;\\j=v}}1\gcd\left(  u,j\right)
x^{\operatorname{lcm}\left(  u,j\right)  }\\
&  =1\gcd\left(  u,v\right)  x^{\operatorname{lcm}\left(  u,v\right)  }%
=\gcd\left(  u,v\right)  x^{\operatorname{lcm}\left(  u,v\right)  }.
\end{align*}
Thus,%
\[
\gcd\left(  u,v\right)  x^{\operatorname{lcm}\left(  u,v\right)
}=\underbrace{\left(  \sum_{i\geq1}\delta_{i,u}x^{i}\right)  }_{=x^{u}}%
\squarebin\underbrace{\left(  \sum_{i\geq1}\delta_{i,v}x^{i}\right)  }_{=x^{v}%
}=x^{u}\squarebin x^{v}.
\]
This proves (\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}).} Furthermore, the
operation $\squarebin$ on $x \QQ\left[  x\right]  $ is
associative\footnote{\textit{Proof.} We need to prove that the operation
$\squarebin$ on $x \QQ\left[  x\right]  $ is associative. In other words,
we need to prove that $\left(  \alpha\squarebin\beta\right)  \squarebin\gamma
=\alpha\squarebin\left(  \beta\squarebin\gamma\right)  $ for any three elements
$\alpha$, $\beta$ and $\gamma$ of $x \QQ\left[  x\right]  $.
\par
So let $\alpha$, $\beta$ and $\gamma$ be three elements of $x \QQ\left[
x\right]  $. We need to prove the equality $\left(  \alpha\squarebin\beta\right)
\squarebin\gamma=\alpha\squarebin\left(  \beta\squarebin\gamma\right)  $. Since this
equality is $\QQ$-linear in each of $\alpha$, $\beta$ and $\gamma$
(this is because the operation $\squarebin$ is $\QQ$-bilinear), we can
WLOG assume that $\alpha$, $\beta$ and $\gamma$ belong to the basis $\left(
x^{i}\right)  _{i\geq1}$ of the $\QQ$-vector space $x \QQ\left[
x\right]  $. Assume this. Then, $\alpha$, $\beta$ and $\gamma$ belong to the
basis $\left(  x^{i}\right)  _{i\geq1}$ of the $\QQ$-vector space
$x \QQ\left[  x\right]  $. In other words, there exist positive integers
$i$, $j$ and $k$ satisfying $\alpha=x^{i}$, $\beta=x^{j}$ and $\gamma=x^{k}$.
Consider these $i$, $j$ and $k$.
\par
Now,%
\begin{align*}
\left(  \underbrace{\alpha}_{=x^{i}}\squarebin\underbrace{\beta}_{=x^{j}}\right)
\squarebin\underbrace{\gamma}_{=x^{k}}  &  =\underbrace{\left(  x^{i}\squarebin
x^{j}\right)  }_{\substack{=\gcd\left(  i,j\right)  x^{\operatorname{lcm}%
\left(  i,j\right)  }\\\text{(by
(\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}), applied to}\\u=i\text{ and
}v=j\text{)}}}\squarebin x^{k}=\left(  \gcd\left(  i,j\right)
x^{\operatorname{lcm}\left(  i,j\right)  }\right)  \squarebin x^{k}\\
&  =\gcd\left(  i,j\right)  \underbrace{\left(  x^{\operatorname{lcm}\left(
i,j\right)  }\squarebin x^{k}\right)  }_{\substack{=\gcd\left(
\operatorname{lcm}\left(  i,j\right)  ,k\right)  x^{\operatorname{lcm}\left(
\operatorname{lcm}\left(  i,j\right)  ,k\right)  }\\\text{(by
(\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}), applied to}%
\\u=\operatorname{lcm}\left(  i,j\right)  \text{ and }v=k\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the operation }\squarebin\text{ is
} \QQ\text{-bilinear}\right) \\
&  =\gcd\left(  i,j\right)  \gcd\left(  \operatorname{lcm}\left(  i,j\right)
,k\right)  x^{\operatorname{lcm}\left(  \operatorname{lcm}\left(  i,j\right)
,k\right)  }.
\end{align*}
But any two positive integers $p$ and $q$ satisfy $\gcd\left(  p,q\right)
=\dfrac{pq}{\operatorname{lcm}\left(  p,q\right)  }$. Applying this to
$p=\operatorname{lcm}\left(  i,j\right)  $ and $q=k$, we obtain $\gcd\left(
\operatorname{lcm}\left(  i,j\right)  ,k\right)  =\dfrac{\operatorname{lcm}%
\left(  i,j\right)  \cdot k}{\operatorname{lcm}\left(  \operatorname{lcm}%
\left(  i,j\right)  ,k\right)  }$. On the other hand, applying the equality
$\gcd\left(  p,q\right)  =\dfrac{pq}{\operatorname{lcm}\left(  p,q\right)  }$
to $p=i$ and $q=j$, we obtain $\gcd\left(  i,j\right)  =\dfrac{ij}%
{\operatorname{lcm}\left(  i,j\right)  }$. Now,%
\begin{align*}
\left(  \alpha\squarebin\beta\right)  \squarebin\gamma &  =\underbrace{\gcd\left(
i,j\right)  }_{=\dfrac{ij}{\operatorname{lcm}\left(  i,j\right)  }%
}\underbrace{\gcd\left(  \operatorname{lcm}\left(  i,j\right)  ,k\right)
}_{=\dfrac{\operatorname{lcm}\left(  i,j\right)  \cdot k}{\operatorname{lcm}%
\left(  \operatorname{lcm}\left(  i,j\right)  ,k\right)  }}%
x^{\operatorname{lcm}\left(  \operatorname{lcm}\left(  i,j\right)  ,k\right)
}\\
&  =\dfrac{ij}{\operatorname{lcm}\left(  i,j\right)  }\cdot\dfrac
{\operatorname{lcm}\left(  i,j\right)  \cdot k}{\operatorname{lcm}\left(
\operatorname{lcm}\left(  i,j\right)  ,k\right)  }\cdot x^{\operatorname{lcm}%
\left(  \operatorname{lcm}\left(  i,j\right)  ,k\right)  }=\dfrac
{ijk}{\operatorname{lcm}\left(  \operatorname{lcm}\left(  i,j\right)
,k\right)  }\cdot x^{\operatorname{lcm}\left(  \operatorname{lcm}\left(
i,j\right)  ,k\right)  }=\dfrac{ijk}{\operatorname{lcm}\left\{  i,j,k\right\}
}x^{\operatorname{lcm}\left\{  i,j,k\right\}  }%
\end{align*}
(since $\operatorname{lcm}\left(  \operatorname{lcm}\left(  i,j\right)
,k\right)  =\operatorname{lcm}\left\{  i,j,k\right\}  $). A similar argument
can be used to prove that $\alpha\squarebin\left(  \beta\squarebin\gamma\right)
=\dfrac{ijk}{\operatorname{lcm}\left\{  i,j,k\right\}  }x^{\operatorname{lcm}%
\left\{  i,j,k\right\}  }$. Compared with $\left(  \alpha\squarebin\beta\right)
\squarebin\gamma=\dfrac{ijk}{\operatorname{lcm}\left\{  i,j,k\right\}
}x^{\operatorname{lcm}\left\{  i,j,k\right\}  }$, we obtain $\left(
\alpha\squarebin\beta\right)  \squarebin\gamma=\alpha\squarebin\left(  \beta
\squarebin\gamma\right)  $. Thus, $\left(  \alpha\squarebin\beta\right)
\squarebin\gamma=\alpha\squarebin\left(  \beta\squarebin\gamma\right)  $ is proven, and
so we have shown that the operation $\squarebin$ on $x \QQ\left[  x\right]
$ is associative. Qed.} and commutative\footnote{The proof of this is similar
to the proof of the associativity of $\squarebin$, but easier (it boils down to
the identities $\gcd\left(  i,j\right)  =\gcd\left(  j,i\right)  $ and
$\operatorname{lcm}\left(  i,j\right)  =\operatorname{lcm}\left(  j,i\right)
$).}. Hence, $\left(  x \QQ\left[  x\right]  ,\squarebin\right)  $ is a
commutative semigroup. This semigroup has neutral element $x$%
\ \ \ \ \footnote{\textit{Proof.} We need to prove that $x$ is a neutral
element of the semigroup $\left(  x \QQ\left[  x\right]  ,\squarebin
\right)  $. In other words, we need to prove that every $\alpha\in
x \QQ\left[  x\right]  $ satisfies $\alpha\squarebin x=\alpha$ and
$x\squarebin\alpha=\alpha$.
\par
So, let $\alpha\in x \QQ\left[  x\right]  $ be arbitrary. We must prove
the equalities $\alpha\squarebin x=\alpha$ and $x\squarebin\alpha=\alpha$. Since
these two equalities are $\QQ$-linear in $\alpha$, we can WLOG assume
that $\alpha$ belongs to the basis $\left(  x^{i}\right)  _{i\geq1}$ of the
$\QQ$-vector space $x \QQ\left[  x\right]  $. Assume this. Then,
$\alpha$ belongs to the basis $\left(  x^{i}\right)  _{i\geq1}$ of the
$\QQ$-vector space $x \QQ\left[  x\right]  $. In other words,
there exists a positive integer $i$ satisfying $\alpha=x^{i}$. Consider this
$i$. Now,%
\begin{align*}
\underbrace{\alpha}_{=x^{i}}\squarebin\underbrace{x}_{=x^{1}}  &  =x^{i}\squarebin
x^{1}=\underbrace{\gcd\left(  i,1\right)  }_{=1}x^{\operatorname{lcm}\left(
i,1\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}), applied to }u=i\text{ and
}v=1\right) \\
&  =x^{\operatorname{lcm}\left(  i,1\right)  }=x^{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname{lcm}\left(  i,1\right)
=i\right) \\
&  =\alpha,
\end{align*}
and similarly $x\squarebin\alpha=\alpha$. This completes our proof of the fact
that $x$ is a neutral element of the semigroup $\left(  x \QQ\left[
x\right]  ,\squarebin\right)  $, qed.}, and thus is a commutative monoid.

\item For every $\alpha\in x\NN \left[  x\right]  $, we define an
element $p_{\alpha}$ of $\Lambda$ as follows: Writing $\alpha$ in the form
$\alpha=\sum_{i\geq1}\alpha_{i}x^{i}$ (with $\alpha_{i}$ being nonnegative
integers), we set $p_{\alpha}=\prod_{i\geq1}p_{i}^{\alpha_{i}}$. Then,%
\begin{equation}
p_{\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  }%
=p_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda
\label{sol.repSn.arithmetic-prod.ppolmuls}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.ppolmuls}):} Let
$\lambda$ be a partition. Then, the definition of $p_{\lambda}$ yields
\begin{align}
p_{\lambda}  &  =p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(
\lambda\right)  }}=\prod_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }p_{\lambda_{j}} \nonumber\\
&=\prod_{i\geq1}\prod\limits_{\substack{j\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{j}=i}%
}\underbrace{p_{\lambda_{j}}}_{\substack{=p_{i}\\\text{(since }\lambda
_{j}=i\text{)}}} \nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda_{j}%
\geq1\text{ for every }j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \right) \nonumber\\
&  =\prod_{i\geq1}\underbrace{\prod\limits_{\substack{j\in\left\{  1,2,\ldots
,\ell\left(  \lambda\right)  \right\}  ;\\\lambda_{j}=i}}p_{i}}_{=p_{i}%
^{\left\vert \left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \ \mid\ \lambda_{j}=i\right\}  \right\vert }} \nonumber\\
&=\prod_{i\geq1}%
p_{i}^{\left\vert \left\{  j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \ \mid\ \lambda_{j}=i\right\}  \right\vert }.
\label{sol.repSn.arithmetic-prod.ppolmuls.pf.2}%
\end{align}
\par
But (\ref{sol.repSn.arithmetic-prod.polmuls1}) yields $\operatorname*{pol}%
\left(  \operatorname*{muls}\lambda\right)  =\sum_{i\geq1}\left\vert \left\{
j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \ \mid
\ \lambda_{j}=i\right\}  \right\vert x^{i}$. Hence, the definition of
$p_{\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  }$ yields
$p_{\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  }%
=\prod_{i\geq1}p_{i}^{\left\vert \left\{  j\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \ \mid\ \lambda_{j}=i\right\}  \right\vert }$.
Compared with (\ref{sol.repSn.arithmetic-prod.ppolmuls.pf.2}), this yields
$p_{\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)
}=p_{\lambda}$. This proves (\ref{sol.repSn.arithmetic-prod.ppolmuls}).}.
Also, the map $\operatorname*{muls}$ is a bijection $ \Par %
\rightarrow\mathfrak{M}$; therefore, the family $\left(
p_{\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  }\right)
_{\lambda\in \Par }$ is a relabelling of the family $\left(
p_{\operatorname*{pol}M}\right)  _{M\in\mathfrak{M}}$. Due to
(\ref{sol.repSn.arithmetic-prod.ppolmuls}), this rewrites as follows: The
family $\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ is a
relabelling of the family $\left(  p_{\operatorname*{pol}M}\right)
_{M\in\mathfrak{M}}$. Therefore, the family $\left(  p_{\operatorname*{pol}%
M}\right)  _{M\in\mathfrak{M}}$ is a basis of the $\QQ$-vector space
$\Lambda_{\QQ}$ (since the family $\left(  p_{\lambda}\right)
_{\lambda\in \Par }$ is a basis of the $\QQ$-vector space
$\Lambda_{\QQ}$).

On the other hand, the family $\left(  p_{\operatorname*{pol}M}\right)
_{M\in\mathfrak{M}}$ is a relabelling of the family $\left(  p_{\alpha
}\right)  _{\alpha\in x\NN \left[  x\right]  }$ (since the map
$\operatorname*{pol}$ is a bijection $\mathfrak{M}\rightarrow x\NN %
\left[  x\right]  $). Thus, the family $\left(  p_{\alpha}\right)  _{\alpha\in
x\NN \left[  x\right]  }$ is a basis of the $\QQ$-vector space
$\Lambda_{\QQ}$ (since the family $\left(  p_{\operatorname*{pol}%
M}\right)  _{M\in\mathfrak{M}}$ is a basis of the $\QQ$-vector space
$\Lambda_{\QQ}$).
\end{itemize}

We now notice that%
\begin{equation}
p_{\alpha}\boxdot p_{\beta}=p_{\alpha\squarebin\beta}%
\ \ \ \ \ \ \ \ \ \ \text{for any }\alpha\in x\NN \left[  x\right]
\text{ and }\beta\in x\NN \left[  x\right]  .
\label{sol.repSn.arithmetic-prod.ariprod2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.ariprod2}):} Let
$\alpha\in x\NN \left[  x\right]  $ and $\beta\in x\NN \left[
x\right]  $.
\par
The map $\operatorname*{pol}\circ\operatorname*{muls}: \Par %
\rightarrow x\NN \left[  x\right]  $ is a bijection, and thus
surjective. Hence, there exists a $\lambda\in \Par $ such that
$\alpha=\left(  \operatorname*{pol}\circ\operatorname*{muls}\right)  \left(
\lambda\right)  $. Consider this $\lambda$.
\par
Also, there exists a $\mu\in \Par $ such that $\beta=\left(
\operatorname*{pol}\circ\operatorname*{muls}\right)  \left(  \mu\right)  $.
Consider this $\mu$.
\par
We have
\begin{align*}
\alpha &  =\left(  \operatorname*{pol}\circ\operatorname*{muls}\right)
\left(  \lambda\right)  =\operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  =\sum_{j=1}^{\ell\left(  \lambda\right)  }x^{\lambda_{j}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.repSn.arithmetic-prod.polmuls2})}\right) \\
&  =\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
}x^{\lambda_{j}}%
\end{align*}
and
\begin{align*}
\beta &  =\left(  \operatorname*{pol}\circ\operatorname*{muls}\right)  \left(
\mu\right)  =\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)
=\sum_{j=1}^{\ell\left(  \mu\right)  }x^{\mu_{j}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.repSn.arithmetic-prod.polmuls2}), applied to }\mu\text{
instead of }\lambda\right) \\
&  =\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  }%
x^{\mu_{j}}=\sum_{k\in\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}
}x^{\mu_{k}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation
index }j\text{ as }k\right)  .
\end{align*}
Thus,%
\begin{align*}
&  \underbrace{\alpha}_{=\sum_{j\in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  }x^{\lambda_{j}}}\squarebin\underbrace{\beta}%
_{=\sum_{k\in\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  }x^{\mu
_{k}}}\\
&  =\left(  \sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }x^{\lambda_{j}}\right)  \squarebin\left(  \sum_{k\in\left\{
1,2,\ldots,\ell\left(  \mu\right)  \right\}  }x^{\mu_{k}}\right) \\
&  =\underbrace{\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }\sum_{k\in\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  }%
}_{=\sum_{\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  }}\underbrace{x^{\lambda_{j}}\squarebin x^{\mu_{k}}}_{\substack{=\gcd
\left(  \lambda_{j},\mu_{k}\right)  x^{\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  }\\\text{(by
(\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}), applied to}\\u=\lambda
_{j}\text{ and }v=\mu_{k}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
operation }\squarebin\text{ is } \QQ\text{-bilinear}\right) \\
&  =\sum_{\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  }\gcd\left(  \lambda_{j},\mu_{k}\right)  x^{\operatorname{lcm}%
\left(  \lambda_{j},\mu_{k}\right)  }=\sum_{i\geq1}\sum\limits_{\substack{\left(
j,k\right)  \in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}
;\\\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  =i}}\gcd\left(
\lambda_{j},\mu_{k}\right)  \underbrace{x^{\operatorname{lcm}\left(
\lambda_{j},\mu_{k}\right)  }}_{\substack{=x^{i}\\\text{(since }%
\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  =i\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  \geq1\text{
for every }\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\} \\
\text{(since }\lambda_{j}\geq1\text{ and }\mu_{k}\geq1\text{ for every
}\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  \text{)}%
\end{array}
\right) \\
&  =\sum_{i\geq1}\sum\limits_{\substack{\left(  j,k\right)  \in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots
,\ell\left(  \mu\right)  \right\}  ;\\\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  =i}}\gcd\left(  \lambda_{j},\mu_{k}\right)  x^{i}%
=\sum_{i\geq1}\left(  \sum\limits_{\substack{\left(  j,k\right)  \in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots
,\ell\left(  \mu\right)  \right\}  ;\\\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  =i}}\gcd\left(  \lambda_{j},\mu_{k}\right)  \right)
x^{i}.
\end{align*}
Hence, the definition of $p_{\alpha\squarebin\beta}$ yields%
\begin{align}
p_{\alpha\squarebin\beta}  &  =\prod_{i\geq1}\underbrace{p_{i}^{\sum\limits
_{\substack{\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  ;\\\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  =i}%
}\gcd\left(  \lambda_{j},\mu_{k}\right)  }}_{=\prod\limits_{\substack{\left(
j,k\right)  \in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
\times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}
;\\\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  =i}}p_{i}%
^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }}\nonumber\\
&  =\prod_{i\geq1}\prod\limits_{\substack{\left(  j,k\right)  \in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots
,\ell\left(  \mu\right)  \right\}  ;\\\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  =i}}p_{i}^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }.
\label{sol.repSn.arithmetic-prod.ariprod2.pf.2}%
\end{align}
\par
On the other hand, $\alpha=\operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  $ and thus $p_{\alpha}=p_{\operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  }=p_{\lambda}$ (by
(\ref{sol.repSn.arithmetic-prod.ppolmuls})). Also, $\beta=\operatorname*{pol}%
\left(  \operatorname*{muls}\mu\right)  $ and thus $p_{\beta}%
=p_{\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  }=p_{\mu}$ (by
(\ref{sol.repSn.arithmetic-prod.ppolmuls}), applied to $\mu$ instead of
$\lambda$). Hence,%
\begin{align*}
\underbrace{p_{\alpha}}_{=p_{\lambda}}\boxdot\underbrace{p_{\beta}}_{=p_{\mu
}}  &  =p_{\lambda}\boxdot p_{\mu}=\prod_{i=1}^{\ell\left(  \lambda\right)
}\prod_{j=1}^{\ell\left(  \mu\right)  }p_{\operatorname{lcm}\left(
\lambda_{i},\mu_{j}\right)  }^{\gcd\left(  \lambda_{i},\mu_{j}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the operation }%
\boxdot\right) \\
&  =\underbrace{\prod_{j=1}^{\ell\left(  \lambda\right)  }}_{=\prod
_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  }%
}\underbrace{\prod_{k=1}^{\ell\left(  \mu\right)  }}_{=\prod_{k\in\left\{
1,2,\ldots,\ell\left(  \mu\right)  \right\}  }}p_{\operatorname{lcm}\left(
\lambda_{j},\mu_{k}\right)  }^{\gcd\left(  \lambda_{j},\mu_{k}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation indices
}i\text{ and }j\text{ as }j\text{ and }k\right) \\
&  =\underbrace{\prod_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }\prod_{k\in\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  }%
}_{=\prod_{\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\}  }}p_{\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  }%
^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }=\prod_{\left(  j,k\right)
\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{
1,2,\ldots,\ell\left(  \mu\right)  \right\}  }p_{\operatorname{lcm}\left(
\lambda_{j},\mu_{k}\right)  }^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }\\
&  =\prod_{i\geq1}\prod\limits_{\substack{\left(  j,k\right)  \in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots
,\ell\left(  \mu\right)  \right\}  ;\\\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  =i}}\underbrace{p_{\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  }^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }%
}_{\substack{=p_{i}^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }\\\text{(since
}\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  =i\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname{lcm}\left(  \lambda_{j},\mu_{k}\right)  \geq1\text{
for every }\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(
\lambda\right)  \right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)
\right\} \\
\text{(since }\lambda_{j}\geq1\text{ and }\mu_{k}\geq1\text{ for every
}\left(  j,k\right)  \in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  \times\left\{  1,2,\ldots,\ell\left(  \mu\right)  \right\}  \text{)}%
\end{array}
\right) \\
&  =\prod_{i\geq1}\prod\limits_{\substack{\left(  j,k\right)  \in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  \times\left\{  1,2,\ldots
,\ell\left(  \mu\right)  \right\}  ;\\\operatorname{lcm}\left(  \lambda
_{j},\mu_{k}\right)  =i}}p_{i}^{\gcd\left(  \lambda_{j},\mu_{k}\right)  }.
\end{align*}
Compared with (\ref{sol.repSn.arithmetic-prod.ariprod2.pf.2}), this yields
$p_{\alpha}\boxdot p_{\beta}=p_{\alpha\squarebin\beta}$. This proves
(\ref{sol.repSn.arithmetic-prod.ariprod2}).}

Now, it is easy to see that the operation $\boxdot$ on the $\QQ$-vector
space $\Lambda_{\QQ}$ is associative\footnote{\textit{Proof.} We need
to prove that the operation $\boxdot$ on $\Lambda_{\QQ}$ is
associative. In other words, we need to prove that $\left(  u\boxdot v\right)
\boxdot w=u\boxdot\left(  v\boxdot w\right)  $ for any three elements $u$, $v$
and $w$ of $\Lambda_{\QQ}$.
\par
So let $u$, $v$ and $w$ be three elements of $\Lambda_{\QQ}$. We need
to prove the equality $\left(  u\boxdot v\right)  \boxdot w=u\boxdot\left(
v\boxdot w\right)  $. Since this equality is $\QQ$-linear in each of
$u$, $v$ and $w$ (this is because the operation $\boxdot$ is
$\QQ$-bilinear), we can WLOG assume that $u$, $v$ and $w$ belong to the
basis $\left(  p_{\alpha}\right)  _{\alpha\in x\NN \left[  x\right]  }$
of the $\QQ$-vector space $\Lambda_{\QQ}$. Assume this. Then, $u$, $v$
and $w$ belong to the basis $\left(  p_{\alpha}\right)  _{\alpha\in
x\NN \left[  x\right]  }$ of the $\QQ$-vector space
$\Lambda_{\QQ}$. In other words, there exist polynomials $\alpha\in
x\NN \left[  x\right]  $, $\beta\in x\NN \left[  x\right]  $ and
$\gamma\in x\NN \left[  x\right]  $ satisfying $u=p_{\alpha}$,
$v=p_{\beta}$ and $w=p_{\gamma}$. Consider these $\alpha$, $\beta$ and
$\gamma$.
\par
We have $\alpha\squarebin\beta\in x\NN \left[  x\right]  $ (since
$\alpha\in x\NN \left[  x\right]  $ and $\beta\in x\NN \left[
x\right]  $) and $\beta\squarebin\gamma\in x\NN \left[  x\right]  $ (for
similar reasons). We have%
\begin{align*}
\left(  \underbrace{u}_{=p_{\alpha}}\boxdot\underbrace{v}_{=p_{\beta}}\right)
\boxdot\underbrace{w}_{=p_{\gamma}}  &  =\underbrace{\left(  p_{\alpha}\boxdot
p_{\beta}\right)  }_{\substack{=p_{\alpha\squarebin\beta}\\\text{(by
(\ref{sol.repSn.arithmetic-prod.ariprod2}))}}}\boxdot p_{\gamma}%
=p_{\alpha\squarebin\beta}\boxdot p_{\gamma}\\
&  =p_{\left(  \alpha\squarebin\beta\right)  \squarebin\gamma}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.repSn.arithmetic-prod.ariprod2}%
), applied to }\alpha\squarebin\beta\text{ and }\gamma\text{ instead of }%
\alpha\text{ and }\beta\right)  .
\end{align*}
Since $\left(  \alpha\squarebin\beta\right)  \squarebin\gamma=\alpha\squarebin\left(
\beta\squarebin\gamma\right)  $ (this is because the operation $\squarebin$ on
$x \QQ\left[  x\right]  $ is associative), this rewrites as $\left(
u\boxdot v\right)  \boxdot w=p_{\alpha\squarebin\left(  \beta\squarebin
\gamma\right)  }$. Compared with%
\begin{align*}
\underbrace{u}_{=p_{\alpha}}\boxdot\left(  \underbrace{v}_{=p_{\beta}}%
\boxdot\underbrace{w}_{=p_{\gamma}}\right)   &  =p_{\alpha}\boxdot
\underbrace{\left(  p_{\beta}\boxdot p_{\gamma}\right)  }_{\substack{=p_{\beta
\squarebin\gamma}\\\text{(by (\ref{sol.repSn.arithmetic-prod.ariprod2}), applied
to }\beta\text{ and }\gamma\\\text{instead of }\alpha\text{ and }\beta
\text{)}}}=p_{\alpha}\boxdot p_{\beta\squarebin\gamma}\\
&  =p_{\alpha\squarebin\left(  \beta\squarebin\gamma\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.repSn.arithmetic-prod.ariprod2}%
), applied to }\beta\squarebin\gamma\text{ instead of }\beta\right)  ,
\end{align*}
this yields $\left(  u\boxdot v\right)  \boxdot w=u\boxdot\left(  v\boxdot
w\right)  $. Thus, $\left(  u\boxdot v\right)  \boxdot w=u\boxdot\left(
v\boxdot w\right)  $ is proven, and so we have shown that the operation
$\boxdot$ on $\Lambda_{\QQ}$ is associative. Qed.} and
commutative\footnote{The proof of this is similar to the proof of the
associativity of $\boxdot$, but easier (it boils down to the commutativity of
$\squarebin$).}, and has neutral element $p_{1}$\ \ \ \ \footnote{\textit{Proof.}
We need to prove that $p_{1}$ is a neutral element for the operation $\boxdot$
on $\Lambda_{\QQ}$. In other words, we need to prove that every
$u\in\Lambda_{\QQ}$ satisfies $u\boxdot p_{1}=u$ and $p_{1}\boxdot
u=u$.
\par
So, let $u\in\Lambda_{\QQ}$ be arbitrary. We must prove the equalities
$u\boxdot p_{1}=u$ and $p_{1}\boxdot u=u$. Since these two equalities are
$\QQ$-linear in $u$, we can WLOG assume that $u$ belongs to the
basis $\left(  p_{\alpha}\right)  _{\alpha\in x\NN \left[  x\right]  }$
of the $\QQ$-vector space $\Lambda_{\QQ}$. Assume this. Then,
$u$ belongs to the basis $\left(  p_{\alpha}\right)  _{\alpha\in
x\NN \left[  x\right]  }$ of the $\QQ$-vector space
$\Lambda_{\QQ}$. In other words, there exists a polynomial $\alpha\in
x\NN \left[  x\right]  $ satisfying $u=p_{\alpha}$. Consider this
$\alpha$.
\par
Recall that $x$ is a neutral element of the semigroup $\left(  x \QQ%
\left[  x\right]  ,\squarebin\right)  $. Hence, $\alpha\squarebin x=\alpha$ and
$x\squarebin\alpha=\alpha$.
\par
Now, (\ref{sol.repSn.arithmetic-prod.polmuls2}) (applied to $\lambda=\left(
1\right)  $) yields%
\begin{align*}
\operatorname*{pol}\left(  \operatorname*{muls}\left(  1\right)  \right)   &
=\sum_{j=1}^{\ell\left(  \left(  1\right)  \right)  }x^{\left(  1\right)
_{j}}=\sum_{j=1}^{1}x^{\left(  1\right)  _{j}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\ell\left(  \left(  1\right)  \right)  =1\right) \\
&  =x^{\left(  1\right)  _{1}}=x^{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left(  1\right)  _{1}=1\right) \\
&  =x.
\end{align*}
Hence, $p_{\operatorname*{pol}\left(  \operatorname*{muls}\left(  1\right)
\right)  }=p_{x}$, so that%
\begin{align*}
p_{x}  &  =p_{\operatorname*{pol}\left(  \operatorname*{muls}\left(  1\right)
\right)  }=p_{\left(  1\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.repSn.arithmetic-prod.ppolmuls}), applied to }\lambda=\left(
1\right)  \right) \\
&  =p_{1},
\end{align*}
hence $p_{1}=p_{x}$.
\par
But now%
\begin{align*}
\underbrace{u}_{=p_{\alpha}}\boxdot\underbrace{p_{1}}_{=p_{x}}  &  =p_{\alpha
}\boxdot p_{x}=p_{\alpha\squarebin x}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.repSn.arithmetic-prod.ariprod2}), applied to }\beta=x\right) \\
&  =p_{\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\alpha\squarebin
x=\alpha\right) \\
&  =u.
\end{align*}
Similarly, $p_{1}\boxdot u=u$. This completes our proof of the fact that
$p_{1}$ is a neutral element for the operation $\boxdot$ on $\Lambda
_{\QQ}$, qed.}. Since this operation $\boxdot$ is also $\QQ$-bilinear,
this yields that $\Lambda_{\QQ}$, equipped with the binary
operation $\boxdot$, becomes a commutative $\QQ$-algebra with unity
$p_{1}$. This solves Exercise \ref{exe.repSn.arithmetic-prod}(a).

(b) Let $f\in\Lambda_{\QQ}$. We need to prove the equality $1\boxdot
f=\epsilon_{1}\left(  f\right)  1$. Since this equality is $\QQ$-linear
in $f$ (because the operation $\boxdot$ is $\QQ$-bilinear and the map
$\epsilon_{1}$ is $\QQ$-linear), we can WLOG assume that $f$ belongs to
the basis $\left(  p_{\lambda}\right)  _{\lambda\in \Par }$ of
the $\QQ$-vector space $\Lambda_{\QQ}$. Assume this.

So $f$ belongs to the basis $\left(  p_{\lambda}\right)  _{\lambda
\in \Par }$ of the $\QQ$-vector space $\Lambda
_{\QQ}$. In other words, there exists a partition $\lambda$ such that
$f=p_{\lambda}$. Consider this $\lambda$. The definition of $p_{\varnothing
}\boxdot p_{\lambda}$ yields
\begin{align*}
p_{\varnothing}\boxdot p_{\lambda}  &  =\prod_{i=1}^{\ell\left(
\varnothing\right)  }\prod_{j=1}^{\ell\left(  \lambda\right)  }%
p_{\operatorname{lcm}\left(  \varnothing_{i},\lambda_{j}\right)  }%
^{\gcd\left(  \varnothing_{i},\lambda_{j}\right)  }=\prod_{i=1}^{0}\prod
_{j=1}^{\ell\left(  \lambda\right)  }p_{\operatorname{lcm}\left(
\varnothing_{i},\lambda_{j}\right)  }^{\gcd\left(  \varnothing_{i},\lambda
_{j}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(
\varnothing\right)  =0\right) \\
&  =\left(  \text{empty product}\right)  =1.
\end{align*}
Hence, $\underbrace{1}_{=p_{\varnothing}}\boxdot\underbrace{f}_{=p_{\lambda}%
}=p_{\varnothing}\boxdot p_{\lambda}=1$.

On the other hand, $\epsilon_{1}\left(  p_{\lambda}\right)  =1$%
\ \ \ \ \footnote{\textit{Proof.} We have $\lambda=\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{\ell\left(  \lambda\right)  }\right)  $. Thus,
the definition of $p_{\lambda}$ yields $p_{\lambda}=p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell\left(  \lambda\right)  }}=\prod
_{i=1}^{\ell\left(  \lambda\right)  }p_{\lambda_{i}}$. Applying the map
$\epsilon_{1}$ to both sides of this equality, we conclude%
\begin{align*}
\epsilon_{1}\left(  p_{\lambda}\right)   &  =\epsilon_{1}\left(  \prod
_{i=1}^{\ell\left(  \lambda\right)  }p_{\lambda_{i}}\right)  =\prod
_{i=1}^{\ell\left(  \lambda\right)  }\underbrace{\epsilon_{1}\left(
p_{\lambda_{i}}\right)  }_{\substack{=1\\\text{(by the definition}\\\text{of
}\epsilon_{1}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\epsilon
_{1}\text{ is a } \QQ\text{-algebra homomorphism}\right) \\
&  =\prod_{i=1}^{\ell\left(  \lambda\right)  }1=1,
\end{align*}
qed.}. Thus, $\epsilon_{1}\left(  \underbrace{f}_{=p_{\lambda}}\right)
=\epsilon_{1}\left(  p_{\lambda}\right)  =1$. Hence, $1\boxdot f=1=\epsilon
_{1}\left(  f\right) = \epsilon_1\left(f\right) 1$.
This solves Exercise \ref{exe.repSn.arithmetic-prod}(b).

(c) For every finite group $G$ and every $h\in G$, we define a class function
$\alpha_{G,h}\in R_{\mathbb{C}}\left(  G\right)  $ as in Exercise
\ref{exe.frob-char.plambda}.

Let us introduce a further construction. If $X$ is any set, then we let
$\Symm_X$ denote the symmetric group on $X$ (that is, the group of
all permutations of $X$). For any two sets $X$ and $Y$ and any permutations
$\sigma\in\Symm_X$ and $\tau\in\Symm_Y$, we define a
permutation $\sigma\times\tau$ of $X\times Y$ by setting%
\[
\left(  \sigma\times\tau\right)  \left(  \left(  x,y\right)  \right)  =\left(
\sigma\left(  x\right)  ,\tau\left(  y\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\text{ and }y\in Y.
\]
Thus, for any two sets $X$ and $Y$, we can define a map%
\begin{align*}
\operatorname*{cross}:\Symm_X\times\Symm_Y  &
\rightarrow\Symm_{X\times Y},\\
\left(  \sigma,\tau\right)   &  \mapsto\sigma\times\tau.
\end{align*}
We notice a few properties of this map:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.cross}Let $X$ and $Y$ be two sets.

\begin{enumerate}
\item[(a)] The map $\operatorname*{cross}:\Symm_X\times
\Symm_Y\rightarrow\Symm_{X\times Y}$ is a group homomorphism.

\item[(b)] If the sets $X$ and $Y$ are nonempty, then the map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ is injective.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.cross}.](a) For the sake of
completeness, here is the straightforward proof:

Let $g\in\Symm_X\times\Symm_Y$ and $h\in\Symm_X\times\Symm_Y$.
Write $g$ in the form $g=\left(  \sigma
,\tau\right)  $ for some $\sigma\in\Symm_X$ and $\tau\in
\Symm_Y$. Write $h$ in the form $h=\left(  \sigma^{\prime}%
,\tau^{\prime}\right)  $ for some $\sigma^{\prime}\in\Symm_X$ and
$\tau^{\prime}\in\Symm_Y$. Then,%
\begin{align*}
\operatorname*{cross}\left(  \underbrace{g}_{=\left(  \sigma,\tau\right)
}\underbrace{h}_{=\left(  \sigma^{\prime},\tau^{\prime}\right)  }\right)   &
=\operatorname*{cross}\left(  \underbrace{\left(  \sigma,\tau\right)  \left(
\sigma^{\prime},\tau^{\prime}\right)  }_{=\left(  \sigma\sigma^{\prime}%
,\tau\tau^{\prime}\right)  }\right)  =\operatorname*{cross}\left(  \left(
\sigma\sigma^{\prime},\tau\tau^{\prime}\right)  \right) \\
&  =\left(  \sigma\sigma^{\prime}\right)  \times\left(  \tau\tau^{\prime
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname*{cross}\right) \\
&  =\underbrace{\left(  \sigma\times\tau\right)  }%
_{\substack{=\operatorname*{cross}\left(  \left(  \sigma,\tau\right)  \right)
\\\text{(since }\operatorname*{cross}\left(  \left(  \sigma,\tau\right)
\right)  =\sigma\times\tau\\\text{(by the definition of }\operatorname*{cross}%
\text{))}}}\underbrace{\left(  \sigma^{\prime}\times\tau^{\prime}\right)
}_{\substack{=\operatorname*{cross}\left(  \left(  \sigma^{\prime}%
,\tau^{\prime}\right)  \right)  \\\text{(since }\operatorname*{cross}\left(
\left(  \sigma^{\prime},\tau^{\prime}\right)  \right)  =\sigma^{\prime}%
\times\tau^{\prime}\\\text{(by the definition of }\operatorname*{cross}%
\text{))}}}=\operatorname*{cross}\left(  \underbrace{\left(  \sigma
,\tau\right)  }_{=g}\right)  \operatorname*{cross}\left(  \underbrace{\left(
\sigma^{\prime},\tau^{\prime}\right)  }_{=h}\right) \\
&  =\operatorname*{cross}g\cdot\operatorname*{cross}h.
\end{align*}


Now, let us forget that we fixed $g$ and $h$. We thus have shown that
$\operatorname*{cross}\left(  gh\right)  =\operatorname*{cross}g\cdot
\operatorname*{cross}h$ for all $g\in\Symm_X\times\Symm_Y$
and $h\in\Symm_X\times\Symm_Y$. In other words, the map
$\operatorname*{cross}$ preserves products. Consequently,
$\operatorname*{cross}$ is a group homomorphism (because any map between
groups which preserves products is a group homomorphism). Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross}(a) is proven.

(b) Assume that the sets $X$ and $Y$ are nonempty. Let $g$ and $h$ be two
elements of $\Symm_X\times\Symm_Y$ such that
$\operatorname*{cross}g=\operatorname*{cross}h$.

Write $g$ in the form $g=\left(  \sigma,\tau\right)  $ for some $\sigma
\in\Symm_X$ and $\tau\in\Symm_Y$. Write $h$ in the form
$h=\left(  \sigma^{\prime},\tau^{\prime}\right)  $ for some $\sigma^{\prime
}\in\Symm_X$ and $\tau^{\prime}\in\Symm_Y$.

We shall now prove that $\sigma=\sigma^{\prime}$.

Indeed, fix some $y_{0}\in Y$. (Such a $y_{0}$ exists, since $Y$ is nonempty.)
Let $x\in X$ be arbitrary. We have $\operatorname*{cross}\underbrace{g}%
_{=\left(  \sigma,\tau\right)  }=\operatorname*{cross}\left(  \left(
\sigma,\tau\right)  \right)  =\sigma\times\tau$ (by the definition of
$\operatorname*{cross}\left(  \left(  \sigma,\tau\right)  \right)  $), so that%
\[
\underbrace{\left(  \operatorname*{cross}g\right)  }_{=\sigma\times\tau
}\left(  \left(  x,y_{0}\right)  \right)  =\left(  \sigma\times\tau\right)
\left(  \left(  x,y_{0}\right)  \right)  =\left(  \sigma\left(  x\right)
,\tau\left(  y_{0}\right)  \right)  .
\]
Thus, $\left(  \sigma\left(  x\right)  ,\tau\left(  y_{0}\right)  \right)
=\underbrace{\left(  \operatorname*{cross}g\right)  }_{=\operatorname*{cross}%
h}\left(  \left(  x,y_{0}\right)  \right)  =\left(  \operatorname*{cross}%
h\right)  \left(  \left(  x,y_{0}\right)  \right)  $.

On the other hand, $\operatorname*{cross}\underbrace{h}_{=\left(
\sigma^{\prime},\tau^{\prime}\right)  }=\operatorname*{cross}\left(  \left(
\sigma^{\prime},\tau^{\prime}\right)  \right)  =\sigma^{\prime}\times
\tau^{\prime}$ (by the definition of $\operatorname*{cross}\left(  \left(
\sigma^{\prime},\tau^{\prime}\right)  \right)  $), so that%
\[
\underbrace{\left(  \operatorname*{cross}h\right)  }_{=\sigma^{\prime}%
\times\tau^{\prime}}\left(  \left(  x,y_{0}\right)  \right)  =\left(
\sigma^{\prime}\times\tau^{\prime}\right)  \left(  \left(  x,y_{0}\right)
\right)  =\left(  \sigma^{\prime}\left(  x\right)  ,\tau^{\prime}\left(
y_{0}\right)  \right)  .
\]
Thus, $\left(  \sigma\left(  x\right)  ,\tau\left(  y_{0}\right)  \right)
=\left(  \operatorname*{cross}h\right)  \left(  \left(  x,y_{0}\right)
\right)  =\left(  \sigma^{\prime}\left(  x\right)  ,\tau^{\prime}\left(
y_{0}\right)  \right)  $. In other words, $\sigma\left(  x\right)
=\sigma^{\prime}\left(  x\right)  $ and $\tau\left(  y_{0}\right)
=\tau^{\prime}\left(  y_{0}\right)  $.

Let us now forget that we fixed $x$. We thus have proven that $\sigma\left(
x\right)  =\sigma^{\prime}\left(  x\right)  $ for every $x\in X$. In other
words, $\sigma=\sigma^{\prime}$. Similarly, $\tau=\tau^{\prime}$. Now,
$g=\left(  \underbrace{\sigma}_{=\sigma^{\prime}},\underbrace{\tau}%
_{=\tau^{\prime}}\right)  =\left(  \sigma^{\prime},\tau^{\prime}\right)  =h$
(since $h=\left(  \sigma^{\prime},\tau^{\prime}\right)  $).

Let us now forget that we fixed $g$ and $h$. We thus have proven that if $g$
and $h$ are two elements of $\Symm_X\times\Symm_Y$ such that
$\operatorname*{cross}g=\operatorname*{cross}h$, then $g=h$. In other words,
the map $\operatorname*{cross}$ is injective. This proves Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross}(b).
\end{proof}

Now, we shall introduce some more notation and prove a few more simple facts.

If $X$ is a finite set and $x\in X$, and if $\sigma\in\Symm_X$, then
we let $O_{\sigma}\left(  x\right)  $ denote the set of all elements of the
cycle of $\sigma$ which contains $x$. This $O_{\sigma}\left(  x\right)  $ is a
subset of the set $X$; it contains $x$ and is thus nonempty. Thus,%
\begin{equation}
\left\vert O_{\sigma}\left(  x\right)  \right\vert \geq1.
\label{sol.repSn.arithmetic-prod.Ogeq1}%
\end{equation}
We shall now show the following simple fact:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.O1}
Let $X$ be a finite set. Let $x\in X$.
Let $\sigma\in\Symm_X$. Let $n\in \NN$. Then, we have the
following equivalence of assertions:\footnote{Here, the
assertion
``$\left\vert O_{\sigma}\left(  x\right)  \right\vert \mid n$''
should be read as
``the size of $O_{\sigma}\left(  x\right)$ divides $n$''.}
\[
\left(  \left\vert O_{\sigma}\left(  x\right)  \right\vert \mid n\right)
\ \Longleftrightarrow\ \left(  \sigma^{n}\left(  x\right)  =x\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.O1}.]Let $\left(
x_{1},x_{2},\ldots,x_{k}\right)  $ denote the cycle of $\sigma$ which contains
$x$. Then, $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ are $k$ distinct
elements of $X$, and satisfy
\begin{equation}
\left(  \sigma\left(  x_{i}\right)  =x_{i+1}\text{ for every }i\in\left\{
1,2,\ldots,k-1\right\}  \right)
\label{pf.lem.sol.repSn.arithmetic-prod.O1.x1}%
\end{equation}
and%
\begin{equation}
\sigma\left(  x_{k}\right)  =x_{1}.
\label{pf.lem.sol.repSn.arithmetic-prod.O1.x2}%
\end{equation}


We have $O_{\sigma}\left(  x\right)  =\left\{  x_{1},x_{2},\ldots
,x_{k}\right\}  $\ \ \ \ \footnote{\textit{Proof.} Recall that $O_{\sigma
}\left(  x\right)  $ is defined as the set of all elements of the cycle of
$\sigma$ which contains $x$. In other words,%
\begin{align*}
O_{\sigma}\left(  x\right)   &  =\left(  \text{the set of all elements of
}\underbrace{\text{the cycle of }\sigma\text{ which contains }x}_{=\left(
x_{1},x_{2},\ldots,x_{k}\right)  }\right) \\
&  =\left(  \text{the set of all elements of }\left(  x_{1},x_{2},\ldots
,x_{k}\right)  \right)  =\left\{  x_{1},x_{2},\ldots,x_{k}\right\}  ,
\end{align*}
qed.} and $\left\vert O_{\sigma}\left(  x\right)  \right\vert =k$%
\ \ \ \ \footnote{\textit{Proof.} We have $\left\vert \underbrace{O_{\sigma
}\left(  x\right)  }_{=\left\{  x_{1},x_{2},\ldots,x_{k}\right\}  }\right\vert
=\left\vert \left\{  x_{1},x_{2},\ldots,x_{k}\right\}  \right\vert =k$ (since
the $k$ elements $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ are distinct),
qed.}. Also, there exists a $j\in\left\{  1,2,\ldots,k\right\}  $ such that
$x=x_{j}$.\ \ \ \ \footnote{\textit{Proof.} We know that the set $O_{\sigma
}\left(  x\right)  $ contains $x$. Thus, $x\in O_{\sigma}\left(  x\right)
=\left\{  x_{1},x_{2},\ldots,x_{k}\right\}  $. Hence, there exists a
$j\in\left\{  1,2,\ldots,k\right\}  $ such that $x=x_{j}$. Qed.} Consider this
$j$.

For every integer $N$, we let $\overline{N}$ denote the remainder of $N$
modulo $k$. Then, for every integer $N$, we have%
\begin{equation}
\overline{N}\equiv N\operatorname{mod}k
\label{pf.lem.sol.repSn.arithmetic-prod.O1.res1}%
\end{equation}
and%
\begin{equation}
\overline{N}\in\left\{  0,1,\ldots,k-1\right\}
\label{pf.lem.sol.repSn.arithmetic-prod.O1.res2}%
\end{equation}
(since $\overline{N}$ is the remainder of $N$ modulo $k$). Also, if $N$ is an
integer, then $\overline{N}$ is the \textbf{only} integer $M\in\left\{
0,1,\ldots,k-1\right\}  $ satisfying $M\equiv N\operatorname{mod}k$ (since
$\overline{N}$ is the remainder of $N$ modulo $k$). In other words, if
$M\in\left\{  0,1,\ldots,k-1\right\}  $ is an integer satisfying $M\equiv
N\operatorname{mod}k$, then%
\begin{equation}
M=\overline{N}. \label{pf.lem.sol.repSn.arithmetic-prod.O1.res3}%
\end{equation}


Now,%
\begin{equation}
\sigma^{r}\left(  x\right)  =x_{\overline{j+r-1}+1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }r\in\NN .
\label{pf.lem.sol.repSn.arithmetic-prod.O1.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}):} We
shall prove (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) by induction over
$r$:
\par
\textit{Induction base:} We have $j\in\left\{  1,2,\ldots,k\right\}  $, so
that $j-1\in\left\{  0,1,\ldots,k-1\right\}  $. Since $j-1\equiv
j-1\operatorname{mod}k$, this shows that $j-1=\overline{j-1}$ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res3}), applied to $N=j-1$ and
$M=j-1$). Now, $\overline{j+0-1}=\overline{j-1}=j-1$ (since $j-1=\overline
{j-1}$) and thus $\overline{j+0-1}+1=j$. Hence, $x_{\overline{j+0-1}+1}%
=x_{j}=x$ (since $x=x_{j}$). Now, $\underbrace{\sigma^{0}}%
_{=\operatorname*{id}}\left(  x\right)  =\operatorname*{id}\left(  x\right)
=x=x_{\overline{j+0-1}+1}$ (since $x_{\overline{j+0-1}+1}=x$). In other words,
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) is proven for $r=0$. This
completes the induction base.
\par
\textit{Induction step:} Let $R\in \NN$. Assume that
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) is proven for $r=R$. We now need
to show that (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) holds for $r=R+1$.
\par
We know that (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) is proven for
$r=R$. In other words, $\sigma^{R}\left(  x\right)  =x_{\overline{j+R-1}+1}$.
We are going to prove that $\sigma^{R+1}\left(  x\right)  =x_{\overline
{j+R}+1}$.
\par
Let us distinguish between two cases:
\par
\textit{Case 1:} We have $\overline{j+R-1}+1=k$.
\par
\textit{Case 2:} We have $\overline{j+R-1}+1\neq k$.
\par
Let us consider Case 1 first. In this case, we have $\overline{j+R-1}+1=k$.
Thus, $x_{\overline{j+R-1}+1}=x_{k}$. But $\underbrace{\sigma^{R+1}}%
_{=\sigma\circ\sigma^{R}}\left(  x\right)  =\left(  \sigma\circ\sigma
^{R}\right)  \left(  x\right)  =\sigma\left(  \underbrace{\sigma^{R}\left(
x\right)  }_{=x_{\overline{j+R-1}+1}=x_{k}}\right)  =\sigma\left(
x_{k}\right)  =x_{1}$ (by (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.x2})).
\par
But $\overline{j+R-1}+1=k$, so that $\overline{j+R-1}=k-1$ and thus
$k-1=\overline{j+R-1}\equiv j+R-1\operatorname{mod}k$ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res1}), applied to $N=j+R-1$).
Adding $1$ to both sides of this congruence, we obtain $k\equiv
j+R\operatorname{mod}k$. Thus, $0\equiv k\equiv j+R\operatorname{mod}k$. Since
$0\in\left\{  0,1,\ldots,k-1\right\}  $, this leads to $0=\overline{j+R}$ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res3}), applied to $N=j+R$ and
$M=0$). Hence, $\overline{j+R}=0$, so that $\overline{j+R}+1=0+1=1$ and thus
$x_{\overline{j+R}+1}=x_{1}$. Compared with $\sigma^{R+1}\left(  x\right)
=x_{1}$, this yields $\sigma^{R+1}\left(  x\right)  =x_{\overline{j+R}+1}$.
Thus, $\sigma^{R+1}\left(  x\right)  =x_{\overline{j+R}+1}$ is proven in Case
1.
\par
Let us now consider Case 2. In this case, we have $\overline{j+R-1}+1\neq k$.
Thus, $\overline{j+R-1}\neq k-1$. But $\overline{j+R-1}\in\left\{
0,1,\ldots,k-1\right\}  $ (by (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res2}%
), applied to $N=j+R-1$). Thus,
\begin{align*}
\overline{j+R-1}  &  \in\left\{  0,1,\ldots,k-1\right\}  \setminus\left\{
k-1\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\overline{j+R-1}%
\in\left\{  0,1,\ldots,k-1\right\}  \text{ and }\overline{j+R-1}\neq
k-1\right) \\
&  =\left\{  0,1,\ldots,k-2\right\}
\end{align*}
and therefore $\overline{j+R-1}+1\in\left\{  1,2,\ldots,k-1\right\}
\subset\left\{  0,1,\ldots,k-1\right\}  $. Also, $\overline{j+R-1}\equiv
j+R-1\operatorname{mod}k$ (by (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res1}%
), applied to $N=j+R-1$). Hence, $\underbrace{\overline{j+R-1}}_{\equiv
j+R-1\operatorname{mod}k}+1\equiv\left(  j+R-1\right)
+1=j+R\operatorname{mod}k$. Thus, $\overline{j+R-1}+1=\overline{j+R}$ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res3}), applied to $N=j+R$ and
$M=\overline{j+R-1}+1$).
\par
But $\underbrace{\sigma^{R+1}}_{=\sigma\circ\sigma^{R}}\left(  x\right)
=\left(  \sigma\circ\sigma^{R}\right)  \left(  x\right)  =\sigma\left(
\underbrace{\sigma^{R}\left(  x\right)  }_{=x_{\overline{j+R-1}+1}}\right)
=\sigma\left(  x_{\overline{j+R-1}+1}\right)  =x_{\left(  \overline
{j+R-1}+1\right)  +1}$ (by (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.x1}),
applied to $i=\overline{j+R-1}+1$). Since $\overline{j+R-1}+1=\overline{j+R}$,
this rewrites as $\sigma^{R+1}\left(  x\right)  =x_{\overline{j+R}+1}$. Thus,
$\sigma^{R+1}\left(  x\right)  =x_{\overline{j+R}+1}$ is proven in Case 2.
\par
Hence, $\sigma^{R+1}\left(  x\right)  =x_{\overline{j+R}+1}$ is proven in each
of the two Cases 1 and 2. Since these two Cases cover all possibilities, this
shows that $\sigma^{R+1}\left(  x\right)  =x_{\overline{j+R}+1}$ always holds.
In other words, $\sigma^{R+1}\left(  x\right)  =x_{\overline{j+\left(
R+1\right)  -1}+1}$ (since $j+R=j+\left(  R+1\right)  -1$). In other words,
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) holds for $r=R+1$. This
completes the induction proof. The induction proof of
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) is thus finished.}

Let us now assume that $k\mid n$. Then, $n\equiv0\operatorname{mod}k$. Also,
$j\in\left\{  1,2,\ldots,k\right\}  $, hence $j-1\in\left\{  0,1,\ldots
,k-1\right\}  $. Now, $j+\underbrace{n}_{\equiv0\operatorname{mod}k}-1\equiv
j-1\operatorname{mod}k$, so that $j-1\equiv j+n-1\operatorname{mod}k$. Hence,
$j-1=\overline{j+n-1}$ (by (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res3}),
applied to $M=j-1$ and $N=j+n-k$), so that $\overline{j+n-1}=j-1$ and thus
$\overline{j+n-1}+1=j$. Now, (\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1})
(applied to $r=n$) yields $\sigma^{n}\left(  x\right)  =x_{\overline{j+n-1}%
+1}=x_{j}$ (since $\overline{j+n-1}+1=j$), so that $\sigma^{n}\left(
x\right)  =x_{j}=x$ (since $x=x_{j}$).

Let us now forget that we have assumed that $k\mid n$. We thus have shown that%
\begin{equation}
\left(  \text{if }k\mid n\text{, then }\sigma^{n}\left(  x\right)  =x\right)
. \label{pf.lem.sol.repSn.arithmetic-prod.O1.impl1}%
\end{equation}


Let us now assume that $\sigma^{n}\left(  x\right)  =x$. But
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.1}) (applied to $r=n$) yields
$\sigma^{n}\left(  x\right)  =x_{\overline{j+n-1}+1}$. Hence,
$x_{\overline{j+n-1}+1}=\sigma^{n}\left(  x\right)  =x=x_{j}$.

But $\overline{j+n-1}\in\left\{  0,1,\ldots,k-1\right\}  $ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res2}), applied to $N=j+n-1$), so
that $\overline{j+n-1}+1\in\left\{  1,2,\ldots,k\right\}  $. Now, recall that
the elements $x_{1}$, $x_{2}$, $\ldots$, $x_{k}$ are distinct. In
other words, if two elements $u$ and $v$ of $\left\{  1,2,\ldots,k\right\}  $
satisfy $x_{u}=x_{v}$, then $u=v$. Applying this to $u=\overline{j+n-1}+1$ and
$v=j$, we obtain $\overline{j+n-1}+1=j$. Hence, $\overline{j+n-1}+1=j-1$, so
that $j-1=\overline{j+n-1}\equiv j+n-1\operatorname{mod}k$ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.res1}), applied to $N=j+n-1$).
Subtracting $j-1$ from both sides of this congruence, we obtain $0\equiv
n\operatorname{mod}k$, so that $n\equiv0\operatorname{mod}k$. In other words,
$k\mid n$.

Let us now forget that we have assumed that
$\sigma^{n}\left(  x\right)  =x$. We thus have shown that%
\[
\left(  \text{if }\sigma^{n}\left(  x\right)  =x\text{, then }k\mid n\right)
.
\]
Combining this implication with
(\ref{pf.lem.sol.repSn.arithmetic-prod.O1.impl1}), we obtain the equivalence%
\[
\left(  k\mid n\right)  \ \Longleftrightarrow\ \left(  \sigma^{n}\left(
x\right)  =x\right)  .
\]
This rewrites as
\[
\left(  \left\vert O_{\sigma}\left(  x\right)  \right\vert \mid n\right)
\ \Longleftrightarrow\ \left(  \sigma^{n}\left(  x\right)  =x\right)
\]
(since $\left\vert O_{\sigma}\left(  x\right)  \right\vert =k$).
\end{proof}

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.O2}
Let $X$ and $Y$ be two finite sets.
Let $x\in X$ and $y\in Y$. Let $\sigma\in\Symm_X$ and $\tau
\in\Symm_Y$. Then,%
\[
\left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)  \right)
\right\vert =\operatorname{lcm}\left(  \left\vert O_{\sigma}\left(  x\right)
\right\vert ,\left\vert O_{\tau}\left(  y\right)  \right\vert \right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.O2}.]We have $\left\vert
O_{\sigma}\left(  x\right)  \right\vert \geq1$ (by
(\ref{sol.repSn.arithmetic-prod.Ogeq1})) and $\left\vert O_{\tau}\left(
y\right)  \right\vert \geq1$ (similarly). Thus, the positive
integer $\operatorname{lcm}\left(
\left\vert O_{\sigma}\left(  x\right)  \right\vert ,\left\vert O_{\tau}\left(
y\right)  \right\vert \right)  $ is well-defined.

For every $n\in \NN$, we have%
\begin{equation}
\left(  \sigma\times\tau\right)  ^{n}\left(  \left(  x,y\right)  \right)
=\left(  \sigma^{n}\left(  x\right)  ,\tau^{n}\left(  y\right)  \right)  .
\label{pf.lem.sol.repSn.arithmetic-prod.O2.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.repSn.arithmetic-prod.O2.1}):} Let
$n\in \NN$. The map $\operatorname*{cross}:\Symm_X%
\times\Symm_Y\rightarrow\Symm_{X\times Y}$ is a group
homomorphism (according to Lemma \ref{lem.sol.repSn.arithmetic-prod.cross}%
(a)). Hence,
\[
\left(  \operatorname*{cross}\left(  \left(  \sigma,\tau\right)  \right)
\right)  ^{n}=\operatorname*{cross}\left(  \underbrace{\left(  \sigma
,\tau\right)  ^{n}}_{=\left(  \sigma^{n},\tau^{n}\right)  }\right)
=\operatorname*{cross}\left(  \left(  \sigma^{n},\tau^{n}\right)  \right)
=\sigma^{n}\times\tau^{n}%
\]
(by the definition of $\operatorname*{cross}$). Since $\operatorname*{cross}%
\left(  \left(  \sigma,\tau\right)  \right)  =\sigma\times\tau$ (by the
definition of $\operatorname*{cross}$), this rewrites as $\left(  \sigma
\times\tau\right)  ^{n}=\sigma^{n}\times\tau^{n}$. Thus,%
\[
\underbrace{\left(  \sigma\times\tau\right)  ^{n}}_{=\sigma^{n}\times\tau^{n}%
}\left(  \left(  x,y\right)  \right)  =\left(  \sigma^{n}\times\tau
^{n}\right)  \left(  \left(  x,y\right)  \right)  =\left(  \sigma^{n}\left(
x\right)  ,\tau^{n}\left(  y\right)  \right)  .
\]
This proves (\ref{pf.lem.sol.repSn.arithmetic-prod.O2.1}).}

Let $N=\left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)  \right)
\right\vert $ and $L=\operatorname{lcm}\left(  \left\vert O_{\sigma}\left(
x\right)  \right\vert ,\left\vert O_{\tau}\left(  y\right)  \right\vert
\right)  $.

Clearly, $N=\left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)
\right)  \right\vert \geq1$ (by (\ref{sol.repSn.arithmetic-prod.Ogeq1}),
applied to $X\times Y$, $\left(  x,y\right)  $ and $\sigma\times\tau$ instead
of $X$, $x$ and $\sigma$). From Lemma \ref{lem.sol.repSn.arithmetic-prod.O1}
(applied to $X\times Y$, $\left(  x,y\right)  $, $\sigma\times\tau$ and $N$
instead of $X$, $x$, $\sigma$ and $n$), we obtain the following equivalence of
assertions:%
\[
\left(  \left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)  \right)
\right\vert \mid N\right)  \ \Longleftrightarrow\ \left(  \left(  \sigma
\times\tau\right)  ^{N}\left(  \left(  x,y\right)  \right)  =\left(
x,y\right)  \right)  .
\]
Since $\left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)  \right)
\right\vert \mid N$ is satisfied (because $N=\left\vert O_{\sigma\times\tau
}\left(  \left(  x,y\right)  \right)  \right\vert $), this shows that we have
$\left(  \sigma\times\tau\right)  ^{N}\left(  \left(  x,y\right)  \right)
=\left(  x,y\right)  $, hence $\left(  x,y\right)  =\left(  \sigma\times
\tau\right)  ^{N}\left(  \left(  x,y\right)  \right)  =\left(  \sigma
^{N}\left(  x\right)  ,\tau^{N}\left(  y\right)  \right)  $ (by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O2.1}), applied to $n=N$), so that
$\left(  \sigma^{N}\left(  x\right)  ,\tau^{N}\left(  y\right)  \right)
=\left(  x,y\right)  $. In other words, we have $\sigma^{N}\left(  x\right)
=x$ and $\tau^{N}\left(  y\right)  =y$.

But Lemma \ref{lem.sol.repSn.arithmetic-prod.O1} (applied to $n=N$) yields the
equivalence of assertions%
\[
\left(  \left\vert O_{\sigma}\left(  x\right)  \right\vert \mid N\right)
\ \Longleftrightarrow\ \left(  \sigma^{N}\left(  x\right)  =x\right)  .
\]
Since $\sigma^{N}\left(  x\right)  =x$ is satisfied, we can thus conclude that
$\left\vert O_{\sigma}\left(  x\right)  \right\vert \mid N$. The same argument
(but with $X$, $x$ and $\sigma$ replaced by $Y$, $y$ and $\tau$) yields
$\left\vert O_{\tau}\left(  y\right)  \right\vert \mid N$.

The integer $N$ is divisible by both integers $\left\vert O_{\sigma}\left(
x\right)  \right\vert $ and $\left\vert O_{\tau}\left(  y\right)  \right\vert
$ (since $\left\vert O_{\sigma}\left(  x\right)  \right\vert \mid N$ and
$\left\vert O_{\tau}\left(  y\right)  \right\vert \mid N$), and thus must be
divisible by their lowest common multiple. That is, we have
$\operatorname{lcm}\left(  \left\vert O_{\sigma}\left(  x\right)  \right\vert
,\left\vert O_{\tau}\left(  y\right)  \right\vert \right)  \mid N$. Hence,
$L=\operatorname{lcm}\left(  \left\vert O_{\sigma}\left(  x\right)
\right\vert ,\left\vert O_{\tau}\left(  y\right)  \right\vert \right)  \mid N$.

On the other hand, $\left\vert O_{\sigma}\left(  x\right)  \right\vert
\mid\operatorname{lcm}\left(  \left\vert O_{\sigma}\left(  x\right)
\right\vert ,\left\vert O_{\tau}\left(  y\right)  \right\vert \right)  =L$.
But Lemma \ref{lem.sol.repSn.arithmetic-prod.O1} (applied to $n=L$) yields the
equivalence of assertions%
\[
\left(  \left\vert O_{\sigma}\left(  x\right)  \right\vert \mid L\right)
\ \Longleftrightarrow\ \left(  \sigma^{L}\left(  x\right)  =x\right)  .
\]
Since $\left\vert O_{\sigma}\left(  x\right)  \right\vert \mid L$ is
satisfied, we thus must have $\sigma^{L}\left(  x\right)  =x$. The same
argument (but with $X$, $x$ and $\sigma$ replaced by $Y$, $y$ and $\tau$)
yields $\tau^{L}\left(  y\right)  =y$. Now,
(\ref{pf.lem.sol.repSn.arithmetic-prod.O2.1}) (applied to $n=L$) yields
$\left(  \sigma\times\tau\right)  ^{L}\left(  \left(  x,y\right)  \right)
=\left(  \underbrace{\sigma^{L}\left(  x\right)  }_{=x},\underbrace{\tau
^{L}\left(  y\right)  }_{=y}\right)  =\left(  x,y\right)  $.

From Lemma \ref{lem.sol.repSn.arithmetic-prod.O1} (applied to $X\times Y$,
$\left(  x,y\right)  $, $\sigma\times\tau$ and $L$ instead of $X$, $x$,
$\sigma$ and $n$), we obtain the following equivalence of assertions:%
\[
\left(  \left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)  \right)
\right\vert \mid L\right)  \ \Longleftrightarrow\ \left(  \left(  \sigma
\times\tau\right)  ^{L}\left(  \left(  x,y\right)  \right)  =\left(
x,y\right)  \right)  .
\]
Since $\left(  \sigma\times\tau\right)  ^{L}\left(  \left(  x,y\right)
\right)  =\left(  x,y\right)  $ is satisfied, we can thus conclude that
$\left\vert O_{\sigma\times\tau}\left(  \left(  x,y\right)  \right)
\right\vert \mid L$. Now, $N=\left\vert O_{\sigma\times\tau}\left(  \left(
x,y\right)  \right)  \right\vert \mid L$.

So we have shown that the positive integers $N$ and $L$ satisfy both $N\mid L$
and $L\mid N$. This yields that these positive integers must be equal, i.e.,
we have $N=L$. Since $N=\left\vert O_{\sigma\times\tau}\left(  \left(
x,y\right)  \right)  \right\vert $ and $L=\operatorname{lcm}\left(  \left\vert
O_{\sigma}\left(  x\right)  \right\vert ,\left\vert O_{\tau}\left(  y\right)
\right\vert \right)  $, this rewrites as $\left\vert O_{\sigma\times\tau
}\left(  \left(  x,y\right)  \right)  \right\vert =\operatorname{lcm}\left(
\left\vert O_{\sigma}\left(  x\right)  \right\vert ,\left\vert O_{\tau}\left(
y\right)  \right\vert \right)  $. This proves Lemma
\ref{lem.sol.repSn.arithmetic-prod.O2}.
\end{proof}

In the following, we shall use the notations introduced in the solution of
Exercise \ref{exe.repSn.arithmetic-prod}(a).

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.O3}Let $X$ be a finite set. Let
$\sigma\in\Symm_X$. Let $\lambda$ be the cycle type of the
permutation $\sigma$. Then,%
\[
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  =\sum
_{\mathbf{x}\in X}\dfrac{1}{\left\vert O_{\sigma}\left(  \xx\right)
\right\vert }x^{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert
}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.O3}.]The permutation
$\sigma$ has cycle type $\lambda$. In other words, the cycles of the
permutation $\sigma$ have lengths $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{\ell\left(  \lambda\right)  }$. Let $c_{1}$, $c_{2}$, $\ldots$,
$c_{\ell\left(  \lambda\right)  }$ be these cycles, labelled in such a way
that for every $j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
$, the cycle $c_{j}$ has length $\lambda_{j}$. For every $j\in\left\{
1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $, let $C_{j}$ denote the set
of all elements of the cycle $c_{j}$. Then, the sets $C_{1}$, $C_{2}$,
$\ldots$, $C_{\ell\left(  \lambda\right)  }$ are the cycles of the permutation
$\sigma$ as sets (since $c_{1}$, $c_{2}$, $\ldots$, $c_{\ell\left(
\lambda\right)  }$ are the cycles of the permutation $\sigma$). Hence, these
sets $C_{1}$, $C_{2}$, $\ldots$, $C_{\ell\left(  \lambda\right)  }$ are
disjoint and their union is $X$. In other words, $X=C_{1}\sqcup
C_{2}\sqcup\cdots\sqcup C_{\ell\left(  \lambda\right)  }$.

For every $j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $, we
have%
\begin{equation}
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert =\left\vert
C_{j}\right\vert =\lambda_{j}\ \ \ \ \ \ \ \ \ \ \text{for every }%
\mathbf{x}\in C_{j}. \label{pf.lem.sol.repSn.arithmetic-prod.O3.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.repSn.arithmetic-prod.O3.1}):} Let
$j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}  $. Let
$\mathbf{x}\in C_{j}$. Then, $\xx$ belongs to $C_{j}$. In other words,
$\xx$ belongs to the set of all elements of the cycle $c_{j}$ (since
$C_{j}$ is the set of all elements of the cycle $c_{j}$ (by the definition of
$C_{j}$)). In other words, $\xx$ is an element of the cycle $c_{j}$.
Thus, the cycle $c_{j}$ contains $\xx$. In other words, the cycle
$c_{j}$ is the cycle of $\sigma$ which contains $\xx$.
\par
But recall that $O_{\sigma}\left(  \xx\right)  $ is the set of all
elements of the cycle of $\sigma$ which contains $\xx$ (by the
definition of $O_{\sigma}\left(  \xx\right)  $). In other words,
$O_{\sigma}\left(  \xx\right)  $ is the set of all elements of the
cycle $c_{j}$ (since the cycle $c_{j}$ is the cycle of $\sigma$ which contains
$\xx$). In other words, $O_{\sigma}\left(  \xx\right)  $ is
$C_{j}$ (since $C_{j}$ is the set of all elements of the cycle $c_{j}$ (by the
definition of $C_{j}$)). Thus, $O_{\sigma}\left(  \xx\right)  =C_{j}$,
so that
\begin{align*}
\left\vert \underbrace{O_{\sigma}\left(  \xx\right)  }_{=C_{j}%
}\right\vert  &  =\left\vert C_{j}\right\vert =\left\vert \left(  \text{the
set of all elements of the cycle }c_{j}\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }C_{j}\text{ is the set of all
elements of the cycle }c_{j}\right) \\
&  =\left(  \text{the number of all elements of the cycle }c_{j}\right) \\
&  =\left(  \text{the length of the cycle }c_{j}\right)  =\lambda
_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the cycle }c_{j}\text{ has length
}\lambda_{j}\right)  .
\end{align*}
This proves (\ref{pf.lem.sol.repSn.arithmetic-prod.O3.1}).}

But $X=C_{1}\sqcup C_{2}\sqcup\cdots\sqcup C_{\ell\left(  \lambda\right)  }$.
Thus, the summation sign ``$\sum_{\mathbf{x}\in X}$''
in the sum $\sum_{\mathbf{x}\in X}\dfrac{1}{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert }x^{\left\vert O_{\sigma
}\left(  \xx\right)  \right\vert }$ can be replaced by
``$\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }\sum_{\mathbf{x}\in C_{j}}$''. Hence, we obtain%
\begin{align*}
\sum_{\mathbf{x}\in X}\dfrac{1}{\left\vert O_{\sigma}\left(  \xx%
\right)  \right\vert }x^{\left\vert O_{\sigma}\left(  \xx\right)
\right\vert }  &  =\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }\sum_{\mathbf{x}\in C_{j}}\underbrace{\dfrac{1}{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert }x^{\left\vert O_{\sigma
}\left(  \xx\right)  \right\vert }}_{\substack{=\dfrac{1}{\lambda_{j}%
}x^{\lambda_{j}}\\\text{(since }\left\vert O_{\sigma}\left(  \xx%
\right)  \right\vert =\lambda_{j}\\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.O3.1}))}}}\\
&  =\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
}\underbrace{\sum_{\mathbf{x}\in C_{j}}\dfrac{1}{\lambda_{j}}x^{\lambda_{j}}%
}_{=\left\vert C_{j}\right\vert \cdot\dfrac{1}{\lambda_{j}}x^{\lambda_{j}}%
}=\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)  \right\}
}\underbrace{\left\vert C_{j}\right\vert }_{\substack{=\lambda_{j}\\\text{(by
the second equality}\\\text{sign in
(\ref{pf.lem.sol.repSn.arithmetic-prod.O3.1}))}}}\cdot\dfrac{1}{\lambda_{j}%
}x^{\lambda_{j}}\\
&  =\underbrace{\sum_{j\in\left\{  1,2,\ldots,\ell\left(  \lambda\right)
\right\}  }}_{=\sum_{j=1}^{\ell\left(  \lambda\right)  }}\underbrace{\lambda
_{j}\cdot\dfrac{1}{\lambda_{j}}}_{=1}x^{\lambda_{j}}=\sum_{j=1}^{\ell\left(
\lambda\right)  }x^{\lambda_{j}}=\operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)
\end{align*}
(by (\ref{sol.repSn.arithmetic-prod.polmuls2})). This proves Lemma
\ref{lem.sol.repSn.arithmetic-prod.O3}.
\end{proof}

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.cross2}Let $X$ and $Y$ be two finite
sets. Let $\sigma\in\Symm_X$ and $\tau\in\Symm_Y$ be two
permutations. Let $\lambda$, $\mu$ and $\kappa$ be the cycle types of the
permutations $\sigma\in\Symm_X$, $\tau\in\Symm_Y$ and
$\sigma\times\tau\in\Symm_{X\times Y}$. Then,%
\[
\operatorname*{pol}\left(  \operatorname*{muls}\kappa\right)  =\left(
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  \right)
\squarebin\left(  \operatorname*{pol}\left(  \operatorname*{muls}\mu\right)
\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2}.]Lemma
\ref{lem.sol.repSn.arithmetic-prod.O3} yields%
\begin{equation}
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  =\sum
_{\mathbf{x}\in X}\dfrac{1}{\left\vert O_{\sigma}\left(  \xx\right)
\right\vert }x^{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert }.
\label{pf.lem.sol.repSn.arithmetic-prod.cross2.pmlambda}%
\end{equation}
Also, Lemma \ref{lem.sol.repSn.arithmetic-prod.O3} (applied to $Y$, $\mu$ and
$\tau$ instead of $X$, $\lambda$ and $\sigma$) yields%
\begin{equation}
\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  =\sum
_{\mathbf{x}\in Y}\dfrac{1}{\left\vert O_{\tau}\left(  \xx\right)
\right\vert }x^{\left\vert O_{\tau}\left(  \xx\right)  \right\vert
}=\sum_{\mathbf{y}\in Y}\dfrac{1}{\left\vert O_{\tau}\left(  \mathbf{y}%
\right)  \right\vert }x^{\left\vert O_{\tau}\left(  \mathbf{y}\right)
\right\vert } \label{pf.lem.sol.repSn.arithmetic-prod.cross2.pmmu}%
\end{equation}
(here, we renamed the summation index $\xx$ as $\mathbf{y}$). Hence,%
\begin{align}
&  \underbrace{\left(  \operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  \right)  }_{\substack{=\sum_{\mathbf{x}\in X}\dfrac
{1}{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert }x^{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert }\\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.cross2.pmlambda}))}}}\squarebin
\underbrace{\left(  \operatorname*{pol}\left(  \operatorname*{muls}\mu\right)
\right)  }_{\substack{=\sum_{\mathbf{y}\in Y}\dfrac{1}{\left\vert O_{\tau
}\left(  \mathbf{y}\right)  \right\vert }x^{\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert }\\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.cross2.pmmu}))}}}\nonumber\\
&  =\left(  \sum_{\mathbf{x}\in X}\dfrac{1}{\left\vert O_{\sigma}\left(
\xx\right)  \right\vert }x^{\left\vert O_{\sigma}\left(  \xx%
\right)  \right\vert }\right)  \squarebin\left(  \sum_{\mathbf{y}\in Y}\dfrac
{1}{\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert }x^{\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }\right) \nonumber\\
&  =\sum_{\mathbf{x}\in X}\dfrac{1}{\left\vert O_{\sigma}\left(
\xx\right)  \right\vert }\sum_{\mathbf{y}\in Y}\dfrac{1}{\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }\underbrace{x^{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert }\squarebin x^{\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }}_{\substack{=\gcd\left(
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  x^{\operatorname{lcm}%
\left(  \left\vert O_{\sigma}\left(  \xx\right)  \right\vert
,\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)
}\\\text{(by (\ref{sol.repSn.arithmetic-prod.boxdot-on-xes}), applied
to}\\u=\left\vert O_{\sigma}\left(  \xx\right)  \right\vert \text{ and
}v=\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert \text{)}%
}}\nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{since the operation }\squarebin\text{ is
} \QQ\text{-bilinear}\right) \nonumber\\
&  =\sum_{\mathbf{x}\in X}\dfrac{1}{\left\vert O_{\sigma}\left(
\xx\right)  \right\vert }\sum_{\mathbf{y}\in Y}\dfrac{1}{\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }\gcd\left(  \left\vert
O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)  x^{\operatorname{lcm}\left(
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  }\nonumber\\
&  =\underbrace{\sum_{\mathbf{x}\in X}\sum_{\mathbf{y}\in Y}}_{
=\sum\limits_{\left(
\xx,\mathbf{y}\right)  \in X\times Y}}\underbrace{\dfrac{1}{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert }\dfrac{1}{\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }\gcd\left(  \left\vert
O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)  }_{=\dfrac{\gcd\left(  \left\vert
O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)  }{\left\vert O_{\sigma}\left(
\xx\right)  \right\vert \cdot\left\vert O_{\tau}\left(  \mathbf{y}%
\right)  \right\vert }}x^{\operatorname{lcm}\left(  \left\vert O_{\sigma
}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)  }\nonumber\\
&  =\sum_{\left(  \xx,\mathbf{y}\right)  \in X\times Y}\dfrac
{\gcd\left(  \left\vert O_{\sigma}\left(  \xx\right)  \right\vert
,\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)
}{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert \cdot\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }x^{\operatorname{lcm}\left(
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  }.
\label{pf.lem.sol.repSn.arithmetic-prod.cross2.oneside}%
\end{align}
On the other hand, it is well-known that any two positive integers $a$ and $b$
satisfy%
\begin{equation}
\operatorname{lcm}\left(  a,b\right)  =\dfrac{a\cdot b}{\gcd\left(
a,b\right)  }. \label{pf.lem.sol.repSn.arithmetic-prod.cross2.lcm}%
\end{equation}


But any $\mathbf{x}\in X$ and $\mathbf{y}\in Y$ satisfy%
\begin{align}
\left\vert O_{\sigma\times\tau}\left(  \left(  \xx,\mathbf{y}\right)
\right)  \right\vert  &  =\operatorname{lcm}\left(  \left\vert O_{\sigma
}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)
\label{pf.lem.sol.repSn.arithmetic-prod.cross2.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.repSn.arithmetic-prod.O2}, applied to }\mathbf{x}\text{ and
}\mathbf{y}\text{ instead of }x\text{ and }y\right) \nonumber\\
&  =\dfrac{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert
\cdot\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert }{\gcd\left(
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  }%
\label{pf.lem.sol.repSn.arithmetic-prod.cross2.3}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.repSn.arithmetic-prod.cross2.lcm}), applied to }a=\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert \text{ and }b=\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  .\nonumber
\end{align}
Taking inverses of both sides of this equality, we conclude that any
$\mathbf{x}\in X$ and $\mathbf{y}\in Y$ satisfy%
\begin{equation}
\dfrac{1}{\left\vert O_{\sigma\times\tau}\left(  \left(  \xx%
,\mathbf{y}\right)  \right)  \right\vert }=\dfrac{1}{\left(  \dfrac{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert \cdot\left\vert O_{\tau
}\left(  \mathbf{y}\right)  \right\vert }{\gcd\left(  \left\vert O_{\sigma
}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)  }\right)  }=\dfrac{\gcd\left(
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  }{\left\vert
O_{\sigma}\left(  \xx\right)  \right\vert \cdot\left\vert O_{\tau
}\left(  \mathbf{y}\right)  \right\vert }.
\label{pf.lem.sol.repSn.arithmetic-prod.cross2.4}%
\end{equation}


Now, Lemma \ref{lem.sol.repSn.arithmetic-prod.O3} (applied to $X\times Y$,
$\kappa$ and $\sigma\times\tau$ instead of $X$, $\lambda$ and $\sigma$) yields%
\begin{align*}
\operatorname*{pol}\left(  \operatorname*{muls}\kappa\right)   &
=\sum_{\mathbf{x}\in X\times Y}\dfrac{1}{\left\vert O_{\sigma\times\tau
}\left(  \xx\right)  \right\vert }x^{\left\vert O_{\sigma\times\tau
}\left(  \xx\right)  \right\vert }\\
&  =\sum_{\left(  \xx,\mathbf{y}\right)  \in X\times Y}%
\underbrace{\dfrac{1}{\left\vert O_{\sigma\times\tau}\left(  \left(
\xx,\mathbf{y}\right)  \right)  \right\vert }}_{\substack{=\dfrac
{\gcd\left(  \left\vert O_{\sigma}\left(  \xx\right)  \right\vert
,\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)
}{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert \cdot\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }\\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.cross2.4})))}}%
}\underbrace{x^{\left\vert O_{\sigma\times\tau}\left(  \left(  \xx%
,\mathbf{y}\right)  \right)  \right\vert }}_{\substack{=x^{\operatorname{lcm}%
\left(  \left\vert O_{\sigma}\left(  \xx\right)  \right\vert
,\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)
}\\\text{(since }\left\vert O_{\sigma\times\tau}\left(  \left(  \xx%
,\mathbf{y}\right)  \right)  \right\vert =\operatorname{lcm}\left(  \left\vert
O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert O_{\tau}\left(
\mathbf{y}\right)  \right\vert \right)  \\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.cross2.1})))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\mathbf{x}\text{ as }\left(  \xx,\mathbf{y}\right)  \right) \\
&  =\sum_{\left(  \xx,\mathbf{y}\right)  \in X\times Y}\dfrac
{\gcd\left(  \left\vert O_{\sigma}\left(  \xx\right)  \right\vert
,\left\vert O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)
}{\left\vert O_{\sigma}\left(  \xx\right)  \right\vert \cdot\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert }x^{\operatorname{lcm}\left(
\left\vert O_{\sigma}\left(  \xx\right)  \right\vert ,\left\vert
O_{\tau}\left(  \mathbf{y}\right)  \right\vert \right)  }=\left(
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  \right)
\squarebin\left(  \operatorname*{pol}\left(  \operatorname*{muls}\mu\right)
\right)
\end{align*}
(by (\ref{pf.lem.sol.repSn.arithmetic-prod.cross2.oneside})). This proves
Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2}.
\end{proof}

As a consequence of Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2}, we can conclude:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.cross2p}Let $X$ and $Y$ be two finite
sets. Let $\sigma\in\Symm_X$ and $\tau\in\Symm_Y$ be two
permutations. Let $\lambda$, $\mu$ and $\kappa$ be the cycle types of the
permutations $\sigma\in\Symm_X$, $\tau\in\Symm_Y$ and
$\sigma\times\tau\in\Symm_{X\times Y}$. Then,%
\[
p_{\kappa}=p_{\lambda}\boxdot p_{\mu}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2p}.]Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross2} yields $\operatorname*{pol}\left(
\operatorname*{muls}\kappa\right)  =\left(  \operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  \right)  \squarebin\left(
\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  \right)  $.

We have $p_{\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)
}=p_{\lambda}$ (by (\ref{sol.repSn.arithmetic-prod.ppolmuls})) and
$p_{\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  }=p_{\mu}$ (by
(\ref{sol.repSn.arithmetic-prod.ppolmuls}), applied to $\mu$ instead of
$\lambda$) and $p_{\operatorname*{pol}\left(  \operatorname*{muls}%
\kappa\right)  }=p_{\kappa}$ (by (\ref{sol.repSn.arithmetic-prod.ppolmuls}),
applied to $\kappa$ instead of $\lambda$). Now, from $p_{\operatorname*{pol}%
\left(  \operatorname*{muls}\kappa\right)  }=p_{\kappa}$, we obtain
$p_{\kappa}=p_{\operatorname*{pol}\left(  \operatorname*{muls}\kappa\right)
}=p_{\left(  \operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)
\right)  \squarebin\left(  \operatorname*{pol}\left(  \operatorname*{muls}%
\mu\right)  \right)  }$ (since $\operatorname*{pol}\left(
\operatorname*{muls}\kappa\right)  =\left(  \operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  \right)  \squarebin\left(
\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  \right)  $).

On the other hand, $p_{\lambda}=p_{\operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  }$ (since $p_{\operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  }=p_{\lambda}$) and $p_{\mu}%
=p_{\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  }$ (since
$p_{\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  }=p_{\mu}$), so
that%
\[
\underbrace{p_{\lambda}}_{=p_{\operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  }}\boxdot\underbrace{p_{\mu}}_{=p_{\operatorname*{pol}\left(
\operatorname*{muls}\mu\right)  }}=p_{\operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  }\boxdot p_{\operatorname*{pol}\left(
\operatorname*{muls}\mu\right)  }=p_{\left(  \operatorname*{pol}\left(
\operatorname*{muls}\lambda\right)  \right)  \squarebin\left(
\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  \right)  }%
\]
(by (\ref{sol.repSn.arithmetic-prod.ariprod2}), applied to $\alpha
=\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)  $ and
$\beta=\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  $). Compared
with $p_{\kappa}=p_{\left(  \operatorname*{pol}\left(  \operatorname*{muls}%
\lambda\right)  \right)  \squarebin\left(  \operatorname*{pol}\left(
\operatorname*{muls}\mu\right)  \right)  }$, this yields $p_{\kappa
}=p_{\lambda}\boxdot p_{\mu}$. Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross2p} is thus proven.
\end{proof}

We next recall the notation $U^{\tau}$ defined in Definition
\ref{general-restriction}. We state a few properties of this notation:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.general-restriction1}Let $K$ and $H$ be
two finite groups. Let $\tau:K\rightarrow H$ be a group homomorphism.

\begin{enumerate}
\item[(a)] For every $f\in R_{\mathbb{C}}\left(  H\right)  $, the map
$f\circ\tau:K\rightarrow\mathbb{C}$ belongs to $R_{\mathbb{C}}\left(
K\right)  $. We can thus define a $\CC$-linear map $\tau^{\ast
}:R_{\mathbb{C}}\left(  H\right)  \rightarrow R_{\mathbb{C}}\left(  K\right)
$ by%
\[
\left(  \tau^{\ast}\left(  f\right)  =f\circ\tau\ \ \ \ \ \ \ \ \ \ \text{for
every }f\in R_{\mathbb{C}}\left(  H\right)  \right)  .
\]


\item[(b)] For every (finite-dimensional) $\mathbb{C}H$-module $U$, we have
$\chi_{U^{\tau}}=\tau^{\ast}\left(  \chi_{U}\right)  $.

\item[(c)] We have $\tau^{\ast}\left(  R\left(  H\right)  \right)  \subset
R\left(  K\right)  $.

\item[(d)] Assume that $\tau:K\rightarrow H$ is a group isomorphism. Then,
$\tau^{\ast}\left(  \alpha_{H,\tau\left(  g\right)  }\right)  =\alpha_{K,g}$
for every $g\in K$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}.](a)
Let $f\in R_{\mathbb{C}}\left(  H\right)  $. Then, $f$ is a class function on
$H$. In other words, the function $f$ is constant on conjugacy classes of $H$.

Now, let $k_{1}$ and $k_{2}$ be two elements of $K$ belonging to the same
conjugacy class. Thus, $k_{1}$ and $k_{2}$ are conjugate. In other words,
there exists some $z\in K$ such that $k_{1}=zk_{2}z^{-1}$. Consider this $z$.
Applying the map $\tau$ to both sides of the identity $k_{1}=zk_{2}z^{-1}$, we
obtain $\tau\left(  k_{1}\right)  =\tau\left(  zk_{2}z^{-1}\right)
=\tau\left(  z\right)  \tau\left(  k_{2}\right)  \tau\left(  z\right)  ^{-1}$
(since $\tau$ is a group homomorphism). Thus, there exists some $y\in H$ such
that $\tau\left(  k_{1}\right)  =y\tau\left(  k_{2}\right)  y^{-1}$ (namely,
$y=\tau\left(  z\right)  $). In other words, the elements $\tau\left(
k_{1}\right)  $ and $\tau\left(  k_{2}\right)  $ of $H$ are conjugate. In
other words, the elements $\tau\left(  k_{1}\right)  $ and $\tau\left(
k_{2}\right)  $ belong to the same conjugacy class of $H$. Hence, $f\left(
\tau\left(  k_{1}\right)  \right)  =f\left(  \tau\left(  k_{2}\right)
\right)  $ (since the function $f$ is constant on conjugacy classes of $H$).
Thus, $\left(  f\circ\tau\right)  \left(  k_{1}\right)  =f\left(  \tau\left(
k_{1}\right)  \right)  =f\left(  \tau\left(  k_{2}\right)  \right)  =\left(
f\circ\tau\right)  \left(  k_{2}\right)  $.

Now, let us forget that we fixed $k_{1}$ and $k_{2}$. We have thus shown that
$\left(  f\circ\tau\right)  \left(  k_{1}\right)  =\left(  f\circ\tau\right)
\left(  k_{2}\right)  $ whenever $k_{1}$ and $k_{2}$ are two elements of $K$
belonging to the same conjugacy class. In other words, the function
$f\circ\tau$ is constant on conjugacy classes of $K$. In other words,
$f\circ\tau$ is a class function on $K$. In other words, $f\circ\tau\in
R_{\mathbb{C}}\left(  K\right)  $.

Let us now forget that we fixed $f$. We thus have proven that for every $f\in
R_{\mathbb{C}}\left(  H\right)  $, the map $f\circ\tau:K\rightarrow\mathbb{C}$
belongs to $R_{\mathbb{C}}\left(  K\right)  $. We can thus define a map
$\tau^{\ast}:R_{\mathbb{C}}\left(  H\right)  \rightarrow R_{\mathbb{C}}\left(
K\right)  $ by%
\[
\left(  \tau^{\ast}\left(  f\right)  =f\circ\tau\ \ \ \ \ \ \ \ \ \ \text{for
every }f\in R_{\mathbb{C}}\left(  H\right)  \right)  .
\]
This map $\tau^{\ast}$ is easily seen to be $\CC$-linear. Lemma
\ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(a) is thus proven.

(b) Let $U$ be a finite-dimensional $\mathbb{C}H$-module. Let $g\in K$. Then,
the definition of $\chi_{U^{\tau}}\left(  g\right)  $ yields $\chi_{U^{\tau}%
}\left(  g\right)  =\operatorname*{Tr}\left(  g:U^{\tau}\rightarrow U^{\tau
}\right)  $. But the definition of $U^{\tau}$ yields that $k\cdot
u=\tau\left(  k\right)  \cdot u$ for all $k\in K$ and $u\in U$. Thus, for
every $k\in K$, we have $\left(  k:U^{\tau}\rightarrow U^{\tau}\right)
=\left(  \tau\left(  k\right)  :U\rightarrow U\right)  $ as linear
maps\footnote{\textit{Proof.} Let $k\in K$. Then, every $u\in U$ satisfies
$k\cdot u=\tau\left(  k\right)  \cdot u$ (as we know). Now, every $u\in U$
satisfies $\left(  k:U^{\tau}\rightarrow U^{\tau}\right)  \left(  u\right)
=k\cdot u=\tau\left(  k\right)  \cdot u=\left(  \tau\left(  k\right)
:U\rightarrow U\right)  $. Hence, $\left(  k:U^{\tau}\rightarrow U^{\tau
}\right)  =\left(  \tau\left(  k\right)  :U\rightarrow U\right)  $, qed.}.
Applying this to $k=g$, we obtain $\left(  g:U^{\tau}\rightarrow U^{\tau
}\right)  =\left(  \tau\left(  g\right)  :U\rightarrow U\right)  $. Hence,
$\chi_{U^{\tau}}\left(  g\right)  =\operatorname*{Tr}\underbrace{\left(
g:U^{\tau}\rightarrow U^{\tau}\right)  }_{=\left(  \tau\left(  g\right)
:U\rightarrow U\right)  }=\operatorname*{Tr}\left(  \tau\left(  g\right)
:U\rightarrow U\right)  $.

But the definition of $\chi_{U}\left(  \tau\left(  g\right)  \right)  $ yields
$\chi_{U}\left(  \tau\left(  g\right)  \right)  =\operatorname*{Tr}\left(
\tau\left(  g\right)  :U\rightarrow U\right)  $. Comparing this
with the equality $\chi_{U^{\tau
}}\left(  g\right)  =\operatorname*{Tr}\left(  \tau\left(  g\right)
:U\rightarrow U\right)  $, we obtain $\chi_{U^{\tau}}\left(  g\right)
=\chi_{U}\left(  \tau\left(  g\right)  \right)  $.

The definition of $\tau^{\ast}$ yields $\tau^{\ast}\left(  \chi_{U}\right)
=\chi_{U}\circ\tau$, so that $\underbrace{\left(  \tau^{\ast}\left(  \chi
_{U}\right)  \right)  }_{=\chi_{U}\circ\tau}\left(  g\right)  =\left(
\chi_{U}\circ\tau\right)  \left(  g\right)  =\chi_{U}\left(  \tau\left(
g\right)  \right)  $. Compared with $\chi_{U^{\tau}}\left(  g\right)
=\chi_{U}\left(  \tau\left(  g\right)  \right)  $, this yields $\chi_{U^{\tau
}}\left(  g\right)  =\left(  \tau^{\ast}\left(  \chi_{U}\right)  \right)
\left(  g\right)  $.

Let us now forget that we fixed $g$. We thus have proven that $\chi_{U^{\tau}%
}\left(  g\right)  =\left(  \tau^{\ast}\left(  \chi_{U}\right)  \right)
\left(  g\right)  $ for every $g\in K$. In other words, $\chi_{U^{\tau}}%
=\tau^{\ast}\left(  \chi_{U}\right)  $. This proves Lemma
\ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(b).

(c) By the definition of $R\left(  H\right)  $, we have $R\left(  H\right)
= \ZZ \operatorname*{Irr}\left(  H\right)  $. By the definition of
$R\left(  K\right)  $, we have $R\left(  K\right)  = \ZZ%
\operatorname*{Irr}\left(  K\right)  $, so that $\ZZ \operatorname*{Irr}%
\left(  K\right)  =R\left(  K\right)  $.

The map $\tau^{\ast}$ is $\CC$-linear, thus $\ZZ$-linear.

Let now $f\in\operatorname*{Irr}\left(  H\right)  $. Then, $f\in
\operatorname*{Irr}\left(  H\right)  =\left\{  \chi_{V}\ :\ V\text{ is a
simple }\mathbb{C}H\text{-module}\right\}  $ (by the definition of
$\operatorname*{Irr}\left(  H\right)  $). In other words, there exists a
simple $\mathbb{C}H$-module $V$ such that $f=\chi_{V}$. Denote this $V$ by
$U$. Thus, $U$ is a simple $\mathbb{C}H$-module such that $f=\chi_{U}$. Now,
$\tau^{\ast}\underbrace{f}_{=\chi_{U}}=\tau^{\ast}\left(  \chi_{U}\right)
=\chi_{U^{\tau}}$ (by Lemma
\ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(b)).

We have
\begin{equation}
\operatorname*{Irr}\left(  K\right)  =\left\{  \chi
_{V}\ :\ V\text{ is a simple }\mathbb{C}K\text{-module}\right\}
\label{pf.lem.sol.repSn.arithmetic-prod.general-restriction1.c.0}
\end{equation}
(by the definition of $\operatorname*{Irr}\left(  K\right)$).

But $U^{\tau}$ is a finite-dimensional $\mathbb{C}K$-module, and thus a direct
sum of finitely many simple $\mathbb{C}K$-modules. In other words, there exist
some simple $\mathbb{C}K$-modules $V_{1},V_{2},\ldots,V_{j}$ such that
$U^{\tau}\cong V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}$ as $\mathbb{C}%
K$-modules. Consider these $V_{1},V_{2},\ldots,V_{j}$. For every $i\in\left\{
1,2,\ldots,j\right\}  $, we know that
\begin{align}
\chi_{V_i}  &  \in\left\{  \chi_{V}\ :\ V\text{ is a simple }\mathbb{C}%
K\text{-module}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }V_{j}\text{
is a simple }\mathbb{C}K\text{-module}\right) \nonumber\\
&  =\operatorname*{Irr}\left(  K\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{pf.lem.sol.repSn.arithmetic-prod.general-restriction1.c.0}}\right)
\nonumber\\
&  \subset \ZZ \operatorname*{Irr}\left(  K\right)  =R\left(  K\right)  .
\label{pf.lem.sol.repSn.arithmetic-prod.general-restriction1.c.1}%
\end{align}
But isomorphic $\mathbb{C}K$-modules have equal characters. Hence, since
$U^{\tau}\cong V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}$, we have%
\[
\chi_{U^{\tau}}=\chi_{V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}}=\chi_{V_{1}%
}+\chi_{V_{2}}+\cdots+\chi_{V_{j}}=\sum_{i=1}^{j}\underbrace{\chi_{V_{i}}%
}_{\substack{\in R\left(  K\right)  \\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.general-restriction1.c.1}))}}}\in
\sum_{i=1}^{j}R\left(  K\right)  \subset R\left(  K\right)
\]
(since $R\left(  K\right)  $ is a $\ZZ$-module). Now, $\tau^{\ast
}f=\chi_{U^{\tau}}\in R\left(  K\right)  $.

Let us now forget that we fixed $f$. We thus have proven that $\tau^{\ast}f\in
R\left(  K\right)  $ for every $f\in\operatorname*{Irr}\left(  H\right)  $. In
other words, $\tau^{\ast}\left(  \operatorname*{Irr}\left(  H\right)  \right)
\subset R\left(  K\right)  $. Now,%
\begin{align*}
\tau^{\ast}\left(  \underbrace{R\left(  H\right)  }_{= \ZZ%
\operatorname*{Irr}\left(  H\right)  }\right)   &  =\tau^{\ast}\left(
\ZZ \operatorname*{Irr}\left(  H\right)  \right)  \subset \ZZ%
\underbrace{\tau^{\ast}\left(  \operatorname*{Irr}\left(  H\right)  \right)
}_{\subset R\left(  K\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\tau^{\ast}\text{ is } \ZZ\text{-linear}\right) \\
&  \subset \ZZ R\left(  K\right)  \subset R\left(  K\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R\left(  K\right)  \text{ is a
} \ZZ\text{-module}\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(c).

(d) The map $\tau:K\rightarrow H$ is a group isomorphism, and thus a
bijection. In particular, $\tau$ is injective.

Let $g\in K$. We shall use the notations of Exercise
\ref{exe.frob-char.plambda}.

Let $s\in K$. Exercise \ref{exe.frob-char.plambda}(a) (applied to $K$, $g$ and
$s$ instead of $G$, $h$ and $g$) yields%
\begin{align*}
\alpha_{K,g}\left(  s\right)   &  =\sum_{k\in K}\left[  \underbrace{kgk^{-1}%
=s}_{\substack{\text{this is equivalent to }\left(  \tau\left(  kgk^{-1}%
\right)  =\tau\left(  s\right)  \right)  \\\text{(since }\tau\text{ is
injective)}}}\right]  =\sum_{k\in K}\left[  \underbrace{\tau\left(
kgk^{-1}\right)  }_{\substack{=\tau\left(  k\right)  \tau\left(  g\right)
\tau\left(  k\right)  ^{-1}\\\text{(since }\tau\text{ is a group}%
\\\text{homomorphism)}}}=\tau\left(  s\right)  \right] \\
&  =\sum_{k\in K}\left[  \tau\left(  k\right)  \tau\left(  g\right)
\tau\left(  k\right)  ^{-1}=\tau\left(  s\right)  \right] \\
&  =\sum_{k\in H}\left[  k\tau\left(  g\right)  k^{-1}=\tau\left(  s\right)
\right]  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }k\text{ for }\tau\left(  k\right)  \text{ in
the sum,}\\
\text{since the map }\tau:K\rightarrow H\text{ is a bijection}%
\end{array}
\right)  .
\end{align*}
Compared with%
\begin{align*}
\left(  \alpha_{H,\tau\left(  g\right)  }\circ\tau\right)  \left(  s\right)
&  =\alpha_{H,\tau\left(  g\right)  }\left(  \tau\left(  s\right)  \right)
=\sum_{k\in H}\left[  k\tau\left(  g\right)  k^{-1}=\tau\left(  s\right)
\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.frob-char.plambda}%
(a), applied to }H\text{, }\tau\left(  g\right)  \text{ and }\tau\left(
s\right)  \text{ instead of }G\text{, }h\text{ and }g\right)  ,
\end{align*}
this yields $\alpha_{K,g}\left(  s\right)  =\left(  \alpha_{H,\tau\left(
g\right)  }\circ\tau\right)  \left(  s\right)  $.

Let us now forget that we fixed $s$. We thus have shown that $\alpha
_{K,g}\left(  s\right)  =\left(  \alpha_{H,\tau\left(  g\right)  }\circ
\tau\right)  \left(  s\right)  $ for every $s\in K$. In other words,
$\alpha_{K,g}=\alpha_{H,\tau\left(  g\right)  }\circ\tau$. But the definition
of $\tau^{\ast}$ yields $\tau^{\ast}\left(  \alpha_{H,\tau\left(  g\right)
}\right)  =\alpha_{H,\tau\left(  g\right)  }\circ\tau$. Compared with
$\alpha_{K,g}=\alpha_{H,\tau\left(  g\right)  }\circ\tau$, this yields
$\tau^{\ast}\left(  \alpha_{H,\tau\left(  g\right)  }\right)  =\alpha_{K,g}$.
This proves Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(d).
\end{proof}

Finally, here come two very simple facts:

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.core}Let $G$ and $H$ be two groups. Let
$\Omega:G\rightarrow H$ be an injective group homomorphism. Define a map
$\overline{\Omega}:G\rightarrow\Omega\left(  G\right)  $ by%
\[
\left(  \overline{\Omega}\left(  g\right)  =\Omega\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in G\right)  .
\]
Then, $\overline{\Omega}$ is a well-defined group isomorphism.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.core}.]Lemma
\ref{lem.sol.repSn.arithmetic-prod.core} is obvious.
\end{proof}

\begin{lemma}
\label{lem.sol.repSn.arithmetic-prod.Sf}Let $U$ and $V$ be two finite sets.
Let $\varphi:U\rightarrow V$ be a bijection. Define a map $\varphi^{\ast
}:\Symm_U\rightarrow\Symm_V$ by%
\[
\left(  \varphi^{\ast}\left(  \pi\right)  =\varphi\circ\pi\circ\varphi
^{-1}\ \ \ \ \ \ \ \ \ \ \text{for every }\pi\in\Symm_U\right)  .
\]


\begin{enumerate}
\item[(a)] This map $\varphi^{\ast}$ is well-defined and a group isomorphism.

\item[(b)] Let $\pi\in\Symm_U$. Then, the cycle type of
$\varphi^{\ast}\left(  \pi\right)  $ equals the cycle type of $\pi$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}.](a) For every $\pi
\in\Symm_U$, we have $\varphi\circ\pi\circ\varphi^{-1}\in
\Symm_V$\ \ \ \ \footnote{\textit{Proof.} Let $\pi\in\Symm_U$.
Then, $\pi$ is a permutation of the set $U$. In other words, $\pi$ is a
bijection $U\rightarrow U$. Also, $\varphi$ is a bijection $U\rightarrow V$,
and thus $\varphi^{-1}$ is a bijection $V\rightarrow U$. Hence, $\varphi
\circ\pi\circ\varphi^{-1}$ is a bijection $V\rightarrow V$ (since $\varphi$ is
a bijection $U\rightarrow V$, since $\pi$ is a bijection $U\rightarrow U$, and
since $\varphi^{-1}$ is a bijection $V\rightarrow U$). In other words,
$\varphi\circ\pi\circ\varphi^{-1}$ is a permutation of the set $V$. In other
words, $\varphi\circ\pi\circ\varphi^{-1}\in\Symm_V$, qed}. Thus, the
map $\varphi^{\ast}$ is well-defined.

Now, any two elements $\pi_{1}$ and $\pi_{2}$ of $\Symm_U$ satisfy
$\varphi^{\ast}\left(  \pi_{1}\pi_{2}\right)  =\varphi^{\ast}\left(  \pi
_{1}\right)  \varphi^{\ast}\left(  \pi_{2}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $\pi_{1}$ and $\pi_{2}$ be two elements
of $\Symm_U$. Then, the definition of $\varphi^{\ast}\left(  \pi
_{1}\pi_{2}\right)  $ yields $\varphi^{\ast}\left(  \pi_{1}\pi_{2}\right)
=\varphi\circ\underbrace{\left(  \pi_{1}\pi_{2}\right)  }_{=\pi_{1}\circ
\pi_{2}}\circ\varphi^{-1}=\varphi\circ\pi_{1}\circ\pi_{2}\circ\varphi^{-1}$,
so that $\varphi\circ\pi_{1}\circ\pi_{2}\circ\varphi^{-1}=\varphi^{\ast
}\left(  \pi_{1}\pi_{2}\right)  $. But we have $\varphi^{\ast}\left(  \pi
_{1}\right)  =\varphi\circ\pi_{1}\circ\varphi^{-1}$ (by the definition of
$\varphi^{\ast}\left(  \pi_{1}\right)  $) and $\varphi^{\ast}\left(  \pi
_{2}\right)  =\varphi\circ\pi_{2}\circ\varphi^{-1}$ (by the definition of
$\varphi^{\ast}\left(  \pi_{2}\right)  $). Multiplying the latter two
equalities, we obtain%
\begin{align*}
\varphi^{\ast}\left(  \pi_{1}\right)  \varphi^{\ast}\left(  \pi_{2}\right)
&  =\left(  \varphi\circ\pi_{1}\circ\varphi^{-1}\right)  \left(  \varphi
\circ\pi_{2}\circ\varphi^{-1}\right)  =\left(  \varphi\circ\pi_{1}\circ
\varphi^{-1}\right)  \circ\left(  \varphi\circ\pi_{2}\circ\varphi^{-1}\right)
\\
&  =\varphi\circ\pi_{1}\circ\underbrace{\varphi^{-1}\circ\varphi
}_{=\operatorname*{id}}\circ\pi_{2}\circ\varphi^{-1}=\varphi\circ\pi_{1}%
\circ\pi_{2}\circ\varphi^{-1}=\varphi^{\ast}\left(  \pi_{1}\pi_{2}\right)  ,
\end{align*}
so that $\varphi^{\ast}\left(  \pi_{1}\pi_{2}\right)  =\varphi^{\ast}\left(
\pi_{1}\right)  \varphi^{\ast}\left(  \pi_{2}\right)  $, qed.}. In other
words, the map $\varphi^{\ast}$ preserves products. Hence, the map
$\varphi^{\ast}$ is a group homomorphism (since any map between two groups
which preserves products must be a group homomorphism).

We have $\ker\left(  \varphi^{\ast}\right)  =\left\{
\id_{U}\right\}  $\ \ \ \ \footnote{\textit{Proof.}
Let $\pi\in\ker\left(  \varphi^{\ast}\right)  $. Then, the
definition of $\varphi^{\ast}\left(  \pi\right)  $ yields $\varphi^{\ast
}\left(  \pi\right)  =\varphi\circ\pi\circ\varphi^{-1}$. But since $\pi
\in\ker\left(  \varphi^{\ast}\right)  $, we have $\varphi
^{\ast}\left(  \pi\right)  =1_{\Symm_V}=
\id_{V}$, thus $\id_{V}=\varphi^{\ast}\left(
\pi\right)  =\varphi\circ\pi\circ\varphi^{-1}$. Multiplying both sides of this
equality by $\varphi$ from the right, we obtain $\varphi=\varphi\circ\pi
\circ\underbrace{\varphi^{-1}\circ\varphi}_{=\operatorname*{id}}=\varphi
\circ\pi$. Multiplying both sides of this equality by $\varphi^{-1}$ from the
left, we obtain $\varphi^{-1}\circ\varphi=\underbrace{\varphi^{-1}\circ
\varphi}_{=\operatorname*{id}}\circ\pi=\pi$, so that $\pi=\varphi^{-1}%
\circ\varphi=\id_{U}$. Thus, $\pi\in\left\{
\id_{U}\right\}  $.
\par
Let us now forget that we fixed $\pi$. We thus have proven that $\pi
\in\left\{  \id_{U}\right\}  $ for every $\pi
\in\ker\left(  \varphi^{\ast}\right)  $. In other words,
$\ker\left(  \varphi^{\ast}\right)  \subset\left\{
\id_{U}\right\}  $. Combining this with the inclusion
$\left\{  \id_{U}\right\}  \subset\ker%
\left(  \varphi^{\ast}\right)  $ (which is clear since $\ker%
\left(  \varphi^{\ast}\right)  $ is a group), this yields $\ker%
\left(  \varphi^{\ast}\right)  =\left\{  \id_{U}%
\right\}  $, qed.}. Thus, $\varphi^{\ast}$ is injective (since $\varphi^{\ast
}$ is a group homomorphism). Also, $\varphi^{\ast}$ is
surjective.\footnote{\textit{Proof.} Let $\gamma\in\Symm_V$. Then,
$\gamma$ is a permutation of the set $V$. In other words, $\gamma$ is a
bijection $V\rightarrow V$. Also, $\varphi$ is a bijection $U\rightarrow V$,
and thus $\varphi^{-1}$ is a bijection $V\rightarrow U$. Hence, $\varphi
^{-1}\circ\gamma\circ\varphi$ is a bijection $U\rightarrow U$ (since
$\varphi^{-1}$ is a bijection $V\rightarrow U$, since $\gamma$ is a bijection
$V\rightarrow V$, and since $\varphi$ is a bijection $U\rightarrow V$). In
other words, $\varphi^{-1}\circ\gamma\circ\varphi$ is a permutation of the set
$U$. In other words, $\varphi^{-1}\circ\gamma\circ\varphi\in\Symm_U$.
Now, the definition of $\varphi^{\ast}\left(  \varphi^{-1}\circ\gamma
\circ\varphi\right)  $ yields%
\[
\varphi^{\ast}\left(  \varphi^{-1}\circ\gamma\circ\varphi\right)
=\varphi\circ\left(  \varphi^{-1}\circ\gamma\circ\varphi\right)  \circ
\varphi^{-1}=\underbrace{\varphi\circ\varphi^{-1}}_{=\operatorname*{id}}%
\circ\gamma\circ\underbrace{\varphi\circ\varphi^{-1}}_{=\operatorname*{id}%
}=\gamma,
\]
so that $\gamma=\varphi^{\ast}\left(  \underbrace{\varphi^{-1}\circ\gamma
\circ\varphi}_{\in\Symm_U}\right)  \in\varphi^{\ast}\left(
\Symm_U\right)  $.
\par
Let us now forget that we fixed $\gamma$. We thus have proven that $\gamma
\in\varphi^{\ast}\left(  \Symm_U\right)  $ for every $\gamma
\in\Symm_V$. In other words, $\Symm_V\subset\varphi^{\ast
}\left(  \Symm_U\right)  $. In other words, the map $\varphi^{\ast}$
is surjective, qed.} The map $\varphi^{\ast}$ is bijective (since it is
surjective and injective) and a group homomorphism; thus, $\varphi^{\ast}$ is
a group isomorphism. This proves Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}(a).

(b) Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}(b) is very easy to prove with
a bit of handwaving (the idea of the lemma is that the cycle type of a
permutation is invariant under relabelling the objects on which this
permutation acts -- and stated in these terms, this is obvious). But let us
give a formal proof which is somewhat less direct.

Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}(a) yields that $\varphi^{\ast}$
is a group isomorphism. Hence, $\varphi^{\ast}$ is a group homomorphism.

Let $\lambda$ be the cycle type of $\pi$. Let $\mu$ be the cycle type of
$\varphi^{\ast}\left(  \pi\right)  $. We are going to prove that $\lambda=\mu$.

Set $\xi=\varphi^{\ast}\left(  \pi\right)  \in\Symm_V$. Then,
$\xi=\varphi^{\ast}\left(  \pi\right)  $, so that $\mu$ is the cycle type of
$\xi$ (since $\mu$ is the cycle type of $\varphi^{\ast}\left(  \pi\right)  $).

Notice that $\varphi$ is a permutation, thus bijective, hence injective.

Let $u\in U$. Set $v=\varphi\left(  u\right)  \in V$. The set $O_{\pi}\left(
u\right)  $ is the set of all elements of the cycle of $\pi$ which contains
$u$ (according to the definition of $O_{\pi}\left(  u\right)  $). Hence, the
set $O_{\pi}\left(  u\right)  $ is nonempty (since it contains $u$); thus,
$\left\vert O_{\pi}\left(  u\right)  \right\vert \geq1$. Now, Lemma
\ref{lem.sol.repSn.arithmetic-prod.O1} (applied to $U$, $u$, $\pi$ and
$\left\vert O_{\pi}\left(  u\right)  \right\vert $ instead of $X$, $x$,
$\sigma$ and $n$) yields the equivalence of assertions $\left(  \left\vert
O_{\pi}\left(  u\right)  \right\vert \mid\left\vert O_{\pi}\left(  u\right)
\right\vert \right)  \ \Longleftrightarrow\ \left(  \pi^{\left\vert O_{\pi
}\left(  u\right)  \right\vert }\left(  u\right)  =u\right)  $. Thus, we have
$\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }\left(  u\right)  =u$
(since we clearly have $\left\vert O_{\pi}\left(  u\right)  \right\vert
\mid\left\vert O_{\pi}\left(  u\right)  \right\vert $). But $\varphi^{\ast}$
is a group homomorphism, and thus we have $\varphi^{\ast}\left(
\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }\right)  =\left(
\underbrace{\varphi^{\ast}\left(  \pi\right)  }_{=\xi}\right)  ^{\left\vert
O_{\pi}\left(  u\right)  \right\vert }=\xi^{\left\vert O_{\pi}\left(
u\right)  \right\vert }$, so that $\xi^{\left\vert O_{\pi}\left(  u\right)
\right\vert }=\varphi^{\ast}\left(  \pi^{\left\vert O_{\pi}\left(  u\right)
\right\vert }\right)  =\varphi\circ\pi^{\left\vert O_{\pi}\left(  u\right)
\right\vert }\circ\varphi^{-1}$ (by the definition of $\varphi^{\ast}\left(
\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }\right)  $). Hence,
\begin{align*}
\underbrace{\xi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }}%
_{=\varphi\circ\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }%
\circ\varphi^{-1}}\left(  \underbrace{v}_{=\varphi\left(  u\right)  }\right)
&  =\left(  \varphi\circ\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert
}\circ\varphi^{-1}\right)  \left(  \varphi\left(  u\right)  \right)  =\left(
\varphi\circ\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }%
\circ\underbrace{\varphi^{-1}\circ\varphi}_{=\operatorname*{id}}\right)
\left(  u\right) \\
&  =\left(  \varphi\circ\pi^{\left\vert O_{\pi}\left(  u\right)  \right\vert
}\right)  \left(  u\right)  =\varphi\left(  \underbrace{\pi^{\left\vert
O_{\pi}\left(  u\right)  \right\vert }\left(  u\right)  }_{=u}\right)
=\varphi\left(  u\right)  =v.
\end{align*}


But Lemma \ref{lem.sol.repSn.arithmetic-prod.O1} (applied to $V$, $v$, $\xi$
and $\left\vert O_{\pi}\left(  u\right)  \right\vert $ instead of $X$, $x$,
$\sigma$ and $n$) yields the equivalence of assertions $\left(  \left\vert
O_{\xi}\left(  v\right)  \right\vert \mid\left\vert O_{\pi}\left(  u\right)
\right\vert \right)  \ \Longleftrightarrow\ \left(  \xi^{\left\vert O_{\pi
}\left(  u\right)  \right\vert }\left(  v\right)  =v\right)  $. Hence, we
have
\begin{equation}
\left\vert O_{\xi}\left(  v\right)  \right\vert \mid\left\vert O_{\pi}\left(
u\right)  \right\vert \label{pf.lem.sol.repSn.arithmetic-prod.Sf.1}%
\end{equation}
(since we have $\xi^{\left\vert O_{\pi}\left(  u\right)  \right\vert }\left(
v\right)  =v$).

On the other hand, $O_{\xi}\left(  v\right)  $ is the set of all elements of
the cycle of $\xi$ which contains $v$ (according to the definition of $O_{\xi
}\left(  v\right)  $). Hence, the set $O_{\xi}\left(  v\right)  $ is nonempty
(since it contains $v$); thus, $\left\vert O_{\xi}\left(  v\right)
\right\vert \geq1$. Now, Lemma \ref{lem.sol.repSn.arithmetic-prod.O1} (applied
to $V$, $v$, $\xi$ and $\left\vert O_{\xi}\left(  v\right)  \right\vert $
instead of $X$, $x$, $\sigma$ and $n$) yields the equivalence of assertions
$\left(  \left\vert O_{\xi}\left(  v\right)  \right\vert \mid\left\vert
O_{\xi}\left(  v\right)  \right\vert \right)  \ \Longleftrightarrow\ \left(
\xi^{\left\vert O_{\xi}\left(  v\right)  \right\vert }\left(  v\right)
=v\right)  $. Thus, we have $\xi^{\left\vert O_{\xi}\left(  v\right)
\right\vert }\left(  v\right)  =v$ (since we clearly have $\left\vert O_{\xi
}\left(  v\right)  \right\vert \mid\left\vert O_{\xi}\left(  v\right)
\right\vert $). But $\varphi^{\ast}$ is a group homomorphism, and thus we have
$\varphi^{\ast}\left(  \pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert
}\right)  =\left(  \underbrace{\varphi^{\ast}\left(  \pi\right)  }_{=\xi
}\right)  ^{\left\vert O_{\xi}\left(  v\right)  \right\vert }=\xi^{\left\vert
O_{\xi}\left(  v\right)  \right\vert }$, so that $\xi^{\left\vert O_{\xi
}\left(  v\right)  \right\vert }=\varphi^{\ast}\left(  \pi^{\left\vert O_{\xi
}\left(  v\right)  \right\vert }\right)  =\varphi\circ\pi^{\left\vert O_{\xi
}\left(  v\right)  \right\vert }\circ\varphi^{-1}$ (by the definition of
$\varphi^{\ast}\left(  \pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert
}\right)  $). Hence,
\begin{align*}
\underbrace{\xi^{\left\vert O_{\xi}\left(  v\right)  \right\vert }}%
_{=\varphi\circ\pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert }%
\circ\varphi^{-1}}\left(  \underbrace{v}_{=\varphi\left(  u\right)  }\right)
&  =\left(  \varphi\circ\pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert
}\circ\varphi^{-1}\right)  \left(  \varphi\left(  u\right)  \right)  =\left(
\varphi\circ\pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert }%
\circ\underbrace{\varphi^{-1}\circ\varphi}_{=\operatorname*{id}}\right)
\left(  u\right) \\
&  =\left(  \varphi\circ\pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert
}\right)  \left(  u\right)  =\varphi\left(  \pi^{\left\vert O_{\xi}\left(
v\right)  \right\vert }\left(  u\right)  \right)  .
\end{align*}
Thus, $\varphi\left(  \pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert
}\left(  u\right)  \right)  =\xi^{\left\vert O_{\xi}\left(  v\right)
\right\vert }\left(  v\right)  =v=\varphi\left(  u\right)  $. Therefore,
$\pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert }\left(  u\right)  =u$
(since $\varphi$ is injective). But Lemma
\ref{lem.sol.repSn.arithmetic-prod.O1} (applied to $U$, $u$, $\pi$ and
$\left\vert O_{\xi}\left(  v\right)  \right\vert $ instead of $X$, $x$,
$\sigma$ and $n$) yields the equivalence of assertions $\left(  \left\vert
O_{\pi}\left(  u\right)  \right\vert \mid\left\vert O_{\xi}\left(  v\right)
\right\vert \right)  \ \Longleftrightarrow\ \left(  \pi^{\left\vert O_{\xi
}\left(  v\right)  \right\vert }\left(  u\right)  =u\right)  $. Hence, we
have
\[
\left\vert O_{\pi}\left(  u\right)  \right\vert \mid\left\vert O_{\xi}\left(
v\right)  \right\vert
\]
(since we have $\pi^{\left\vert O_{\xi}\left(  v\right)  \right\vert }\left(
u\right)  =u$). Combining this divisibility with
(\ref{pf.lem.sol.repSn.arithmetic-prod.Sf.1}), we obtain $\left\vert O_{\pi
}\left(  u\right)  \right\vert =\left\vert O_{\xi}\left(  v\right)
\right\vert $ (since $\left\vert O_{\pi}\left(  u\right)  \right\vert $ and
$\left\vert O_{\xi}\left(  v\right)  \right\vert $ are positive integers).
Since $v=\varphi\left(  u\right)  $, this rewrites as $\left\vert O_{\pi
}\left(  u\right)  \right\vert =\left\vert O_{\xi}\left(  \varphi\left(
u\right)  \right)  \right\vert $.

Now, let us forget that we fixed $u\in U$. We thus have proven that%
\begin{equation}
\left\vert O_{\pi}\left(  u\right)  \right\vert =\left\vert O_{\xi}\left(
\varphi\left(  u\right)  \right)  \right\vert \ \ \ \ \ \ \ \ \ \ \text{for
every }u\in U. \label{pf.lem.sol.repSn.arithmetic-prod.Sf.Oeq}%
\end{equation}


Now, Lemma \ref{lem.sol.repSn.arithmetic-prod.O3} (applied to $X=U$ and
$\sigma=\pi$) yields%
\begin{align*}
\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)   &
=\sum_{\mathbf{x}\in U}\dfrac{1}{\left\vert O_{\pi}\left(  \xx\right)
\right\vert }x^{\left\vert O_{\pi}\left(  \xx\right)  \right\vert
}=\sum_{u\in U}\underbrace{\dfrac{1}{\left\vert O_{\pi}\left(  u\right)
\right\vert }}_{\substack{=\dfrac{1}{\left\vert O_{\xi}\left(  \varphi\left(
u\right)  \right)  \right\vert }\\\text{(since }\left\vert O_{\pi}\left(
u\right)  \right\vert =\left\vert O_{\xi}\left(  \varphi\left(  u\right)
\right)  \right\vert \\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.Sf.Oeq})))}}}\underbrace{x^{\left\vert
O_{\pi}\left(  u\right)  \right\vert }}_{\substack{=x^{\left\vert O_{\xi
}\left(  \varphi\left(  u\right)  \right)  \right\vert }\\\text{(since
}\left\vert O_{\pi}\left(  u\right)  \right\vert =\left\vert O_{\xi}\left(
\varphi\left(  u\right)  \right)  \right\vert \\\text{(by
(\ref{pf.lem.sol.repSn.arithmetic-prod.Sf.Oeq})))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\mathbf{x}\text{ as }u\right) \\
&  =\sum_{u\in U}\dfrac{1}{\left\vert O_{\xi}\left(  \varphi\left(  u\right)
\right)  \right\vert }x^{\left\vert O_{\xi}\left(  \varphi\left(  u\right)
\right)  \right\vert }=\sum_{\mathbf{x}\in V}\dfrac{1}{\left\vert O_{\xi
}\left(  \xx\right)  \right\vert }x^{\left\vert O_{\xi}\left(
\xx\right)  \right\vert }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\mathbf{x}\text{ for }\varphi\left(
u\right)  \text{ in the sum}\\
\text{(since the map }\varphi:U\rightarrow V\text{ is bijective)}%
\end{array}
\right)  .
\end{align*}
Compared with%
\begin{align*}
\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)   &  =\sum
_{\mathbf{x}\in V}\dfrac{1}{\left\vert O_{\xi}\left(  \xx\right)
\right\vert }x^{\left\vert O_{\xi}\left(  \xx\right)  \right\vert }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.repSn.arithmetic-prod.O3}, applied to }V\text{, }\xi\text{ and
}\mu\text{ instead of }X\text{, }\sigma\text{ and }\lambda\right)  ,
\end{align*}
this yields $\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)
=\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  $. Thus, $\left(
\operatorname*{pol}\circ\operatorname*{muls}\right)  \left(  \lambda\right)
=\operatorname*{pol}\left(  \operatorname*{muls}\lambda\right)
=\operatorname*{pol}\left(  \operatorname*{muls}\mu\right)  =\left(
\operatorname*{pol}\circ\operatorname*{muls}\right)  \left(  \mu\right)  $.

But the map $\operatorname*{pol}\circ\operatorname*{muls}: \Par %
\rightarrow x\NN \left[  x\right]  $ is a bijection, and thus injective.
Hence, $\lambda=\mu$ (since $\left(  \operatorname*{pol}\circ
\operatorname*{muls}\right)  \left(  \lambda\right)  =\left(
\operatorname*{pol}\circ\operatorname*{muls}\right)  \left(  \mu\right)  $).

But $\lambda$ is the cycle type of $\pi$. Hence, $\left(  \text{the cycle type
of }\pi\right)  =\lambda$. But $\mu$ is the cycle type of $\varphi^{\ast
}\left(  \pi\right)  $. Hence, $\mu=\left(  \text{the cycle type of }%
\varphi^{\ast}\left(  \pi\right)  \right)  $. Now, we have $\left(  \text{the
cycle type of }\pi\right)  =\lambda=\mu=\left(  \text{the cycle type of
}\varphi^{\ast}\left(  \pi\right)  \right)  $. In other words, the cycle type
of $\varphi^{\ast}\left(  \pi\right)  $ equals the cycle type of $\pi$. This
proves Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}(b).
\end{proof}

Let us now define a $\CC$-bilinear map $\widetilde{\boxdot}%
:\Lambda_\CC \times\Lambda_\CC \rightarrow\Lambda
_\CC $, which will be written in infix notation (that is, we will
write $a\widetilde{\boxdot}b$ instead of $\widetilde{\boxdot}\left(
a,b\right)  $), by setting
\begin{equation}
p_{\lambda}\widetilde{\boxdot}p_{\mu}=p_{\lambda}\boxdot p_{\mu}\qquad
\qquad\text{for any partitions }\lambda\text{ and }\mu.
\label{sol.repSn.arithmetic-prod.bot2}%
\end{equation}
\footnote{This is well-defined, since $\left(  p_{\lambda}\right)
_{\lambda\in\operatorname{Par}}$ is a ${\mathbb{C}}$-module basis of
$\Lambda_\CC $.} Then, the operation $\widetilde{\boxdot}$ is
$\CC$-bilinear, and thus $\QQ$-bilinear. We have
\begin{equation}
a\widetilde{\boxdot}b=a\boxdot b\ \ \ \ \ \ \ \ \ \ \text{for all }a\in
\Lambda_{\QQ}\text{ and }b\in\Lambda_{\QQ}
\label{sol.repSn.arithmetic-prod.botbot2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.botbot2}):} Let
$a\in\Lambda_{\QQ}$ and $b\in\Lambda_{\QQ}$. We need to prove
that the equality $a\widetilde{\boxdot}b=a\boxdot b$ holds. Since this
equality is $\QQ$-linear in each of $a$ and $b$ (since both operations
$\widetilde{\boxdot}$ and $\boxdot$ are $\QQ$-bilinear), we can WLOG
assume that $a$ and $b$ are elements of the basis $\left(  p_{\lambda}\right)
_{\lambda\in \Par }$ of the $\QQ$-vector space
$\Lambda_{\QQ}$. Assume this. Then, $a=p_{\lambda}$ and $b=p_{\mu}$ for
some $\lambda\in \Par $ and $\mu\in \Par $. Consider
these $\lambda$ and $\mu$. We have $\underbrace{a}_{=p_{\lambda}%
}\widetilde{\boxdot}\underbrace{b}_{=p_{\mu}}=p_{\lambda}\widetilde{\boxdot
}p_{\mu}=\underbrace{p_{\lambda}}_{=a}\boxdot\underbrace{p_{\mu}}%
_{=b}=a\boxdot b$. Thus, the equality $a\widetilde{\boxdot}b=a\boxdot b$ is
proven, qed.}.

Recall that $\operatorname*{ch}:A\rightarrow\Lambda$ is a $\ZZ$-Hopf
algebra isomorphism. Hence, the extension of $\operatorname*{ch}$ to a
$\CC$-linear map $A_{\mathbb{C}}\rightarrow\Lambda_{\mathbb{C}}$ is a
$\CC$-Hopf algebra isomorphism. We shall denote this extension by
$\ch_{\CC} $.

We are going to repeatedly use the following fact (which is easy to obtain
from the solution of Exercise \ref{exe.frob-char.plambda}): If $\lambda$ is a
partition of a nonnegative integer $n$, and if $g\in\Symm_n$ is a
permutation having cycle type $\lambda$, then%
\begin{equation}
\ch_{\CC} \left(  \alpha_{\Symm_n%
,g}\right)  =p_{\lambda}. \label{sol.repSn.arithmetic-prod.import}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.import}):} Let
$\lambda$ be a partition of a nonnegative integer $n$. Let $g\in
\Symm_n$ be a permutation having cycle type $\lambda$. Define a map
$\Phi:\Lambda_{\mathbb{C}}\rightarrow A_{\mathbb{C}}$ as in the solution of
Exercise \ref{exe.frob-char.plambda}(d). Then,
(\ref{sol.frob-char.plambda.d.1}) yields $\Phi\left(  p_{\lambda}\right)
=\alpha_{\Symm_n,g}$, so that $\alpha_{\Symm_n,g}%
=\Phi\left(  p_{\lambda}\right)  $. But $\operatorname*{ch}%
\nolimits_{\mathbb{C}}\circ\Phi=\id_{\Lambda
_{\mathbb{C}}}$ (this was shown in the solution of Exercise
\ref{exe.frob-char.plambda}(d)). Now, $\operatorname*{ch}\nolimits_{\mathbb{C}%
}\left(  \underbrace{\alpha_{\Symm_n,g}}_{=\Phi\left(  p_{\lambda
}\right)  }\right)  =\ch_{\CC} \left(
\Phi\left(  p_{\lambda}\right)  \right)  =\underbrace{\left(
\ch_{\CC} \circ\Phi\right)  }%
_{=\id_{\Lambda_{\mathbb{C}}}}\left(  p_{\lambda
}\right)  =\id_{\Lambda_{\mathbb{C}}}\left(
p_{\lambda}\right)  =p_{\lambda}$. This proves
(\ref{sol.repSn.arithmetic-prod.import}).}

Theorem \ref{symmetric-group-Frobenius-map-theorem}(a) yields
$\operatorname*{ch}\left(  \chi^{\lambda}\right)  =s_{\lambda}$ for every
partition $\lambda$ (where $\chi^{\lambda}$ is defined as in Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a)). Thus,
\begin{equation}
\ch_{\CC} \left(  \chi^{\lambda}\right)
=s_{\lambda}\ \ \ \ \ \ \ \ \ \ \text{for every partition }\lambda
\label{sol.repSn.arithmetic-prod.chC}%
\end{equation}
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.chC}):} Let
$\lambda$ be a partition. The map $\operatorname*{ch}$ is the restriction of
the map $\ch_{\CC} $ to $A$ (since
$\ch_{\CC} $ is the $\CC$-linear
extension of $\operatorname*{ch}$ to $A_{\mathbb{C}}$). Thus,
$\operatorname*{ch}\left(  \chi^{\lambda}\right)  =\operatorname*{ch}%
\nolimits_{\mathbb{C}}\left(  \chi^{\lambda}\right)  $, so that
$\ch_{\CC} \left(  \chi^{\lambda}\right)
=\operatorname*{ch}\left(  \chi^{\lambda}\right)  =s_{\lambda}$. This proves
(\ref{sol.repSn.arithmetic-prod.chC}).}.

Let us now come back to solving Exercise \ref{exe.repSn.arithmetic-prod}(c).
Let $\mu$ and $\nu$ be two partitions. We need to show that $s_{\mu}\boxdot
s_{\nu}\in\sum_{\lambda\in\operatorname{Par}}{\NN}s_{\lambda}$.

Let $m=\left\vert \mu\right\vert $ and $n=\left\vert \nu\right\vert $. Then,
$\mu\in\Par_{m}$ and $\nu\in \Par_{n}$.
If $\min\left\{  m,n\right\}  =0$, then $s_{\mu}\boxdot
s_{\nu}\in\sum_{\lambda\in\operatorname{Par}}{\NN}s_{\lambda}$ is
easily seen to hold\footnote{\textit{Proof.} Assume that $\min\left\{
m,n\right\}  =0$.
\par
Recall that $\Lambda_{\QQ}$, equipped with the binary operation
$\boxdot$, becomes a commutative $\QQ$-algebra with unity $p_{1}$
(according to Exercise \ref{exe.repSn.arithmetic-prod}(a)). Thus, the
operation $\boxdot$ is commutative. Hence, $s_{\mu}\boxdot s_{\nu}=s_{\nu
}\boxdot s_{\mu}$.
\par
We can WLOG assume that $m\leq n$ (since otherwise, we can just interchange
$\mu$ and $m$ with $\nu$ and $n$ (because $s_{\mu}\boxdot s_{\nu}=s_{\nu
}\boxdot s_{\mu}$)). Assume this. Then, $\min\left\{  m,n\right\}  =m$, so
that $m=\min\left\{  m,n\right\}  =0$. Hence,
$\mu\in \Par_{m}=\Par_{0}$ (since $m=0$), so that $\mu
\in\Par_{0}=\left\{  \varnothing\right\}  $. Hence,
$\mu=\varnothing$ and thus $s_{\mu}=s_{\varnothing}=1$.
\par
Let us use the notations of Exercise \ref{exe.repSn.arithmetic-prod}(b). We
have $\underbrace{s_{\mu}}_{=1}\boxdot s_{\nu}=1\boxdot s_{\nu}=\epsilon
_{1}\left(  s_{\nu}\right)  1$ (by Exercise \ref{exe.repSn.arithmetic-prod}%
(b), applied to $f=s_{\nu}$). But $\epsilon_{1}\left(  s_{\nu}\right)
=s_{\nu}\left(  1\right)  $ (by Exercise \ref{exe.Lambda.maps-on-pn}(i),
applied to $f=s_{\nu}$). But $\nu\in\Par_{n}$ and thus
$s_{\nu}\in\left(  \Lambda_{\QQ}\right)  _{n}$. Hence, Exercise
\ref{exe.Lambda.hall.gr}(b) (applied to $f=s_{\nu}$) yields $\left(
h_{n},s_{\nu}\right)  =s_{\nu}\left(  1\right)  $. But since $\left(
s_{\lambda}\right)  _{\lambda\in \Par }$ is an orthonormal basis
of $\Lambda_{\QQ}$ with respect to the Hall inner product, we have%
\begin{equation}
\left(  s_{\alpha},s_{\beta}\right)  =\delta_{\alpha,\beta}%
\ \ \ \ \ \ \ \ \ \ \text{for any }\alpha\in \Par \text{ and
}\beta\in \Par . \label{sol.repSn.arithmetic-prod.s-orthonor}%
\end{equation}
\par
Comparing $\epsilon_{1}\left(  s_{\nu}\right)  =s_{\nu}\left(  1\right)  $
with $\left(  h_{n},s_{\nu}\right)  =s_{\nu}\left(  1\right)  $, we obtain
$\epsilon_{1}\left(  s_{\nu}\right)  =\left(  \underbrace{h_{n}}_{=s_{\left(
n\right)  }},s_{\nu}\right)  =\left(  s_{\left(  n\right)  },s_{\nu}\right)
=\delta_{\left(  n\right)  ,\nu}$ (by
(\ref{sol.repSn.arithmetic-prod.s-orthonor}), applied to $\alpha=\left(
n\right)  $ and $\beta=\nu$). Hence, $s_{\mu}\boxdot s_{\nu}%
=\underbrace{\epsilon_{1}\left(  s_{\nu}\right)  }_{=\delta_{\left(  n\right)
,\nu}\in\NN}\underbrace{1}_{=s_{\varnothing}}\in\NN %
s_{\varnothing}
\subset\sum_{\lambda\in\operatorname{Par}} \NN s_{\lambda}$
(since $\NN s_{\varnothing}$ is an addend of the sum
$\sum_{\lambda\in\operatorname{Par}}{\NN}s_{\lambda}$), qed.}. Hence,
for the rest of this proof, we can WLOG assume that we don't have
$\min\left\{  m,n\right\}  =0$. Assume this. We have $\min\left\{
m,n\right\}  \neq0$ (since we don't have $\min\left\{  m,n\right\}  =0$).
Thus, $\min\left\{  m,n\right\}  \geq1$ (since $\min\left\{  m,n\right\}
\in \NN$). Thus, $m\geq\min\left\{  m,n\right\}  \geq1$ and $n\geq
\min\left\{  m,n\right\}  \geq1$.

We notice that%
\begin{equation}
\ch_{\CC} \left(  \chi_{P}\right)  \in
\sum_{\lambda\in\operatorname{Par}} \NN s_{\lambda}%
\ \ \ \ \ \ \ \ \ \ \text{for every finite-dimensional }\mathbb{C}%
\Symm_{nm}\text{-module }P\text{.}
\label{sol.repSn.arithmetic-prod.chiP}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.repSn.arithmetic-prod.chiP}):} Let $P$ be
a finite-dimensional $\mathbb{C}\Symm_{nm}$-module. Then, $P$ must be a
direct sum of finitely many simple $\mathbb{C}\Symm_{nm}$-modules
(since every finite-dimensional $\mathbb{C}\Symm_{nm}$-module is a
direct sum of finitely many simple $\mathbb{C}\Symm_{nm}$-modules). In
other words, there exist some simple $\mathbb{C}\Symm_{nm}$-modules
$V_{1},V_{2},\ldots,V_{j}$ such that $P\cong V_{1}\oplus V_{2}\oplus
\cdots\oplus V_{j}$ as $\mathbb{C}\Symm_{nm}$-modules. Consider these
$V_{1},V_{2},\ldots,V_{j}$. We shall now show that $\chi_{V_{i}}\in
\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}$ for every
$i\in\left\{  1,2,\ldots,j\right\}  $.
\par
Indeed, let $i\in\left\{  1,2,\ldots,j\right\}  $. Then, $V_{i}$ is a simple
$\mathbb{C}\Symm_{nm}$-module. Hence, $\chi_{V_{i}}$ is an irreducible
character of $\mathbb{C}\Symm_{nm}$. But (a part of) Theorem
\ref{symmetric-group-Frobenius-map-theorem}(a) (applied to $nm$ instead of
$n$) says that all irreducible characters of $\Symm_{nm}$ have the form
$\chi^{\lambda}$ with $\lambda\in\Par_{nm}$. Thus,
$\chi_{V_{i}}$ has the form $\chi_{V_{i}}=\chi^{\lambda}$ for some $\lambda
\in\Par_{nm}$. Let $\rho$ be this $\lambda$. Thus,
$\rho\in\Par_{nm}$ and $\chi_{V_{i}}=\chi^{\rho}$.
Now, $\rho\in\Par_{nm}\subset \Par $ and
$\chi_{V_{i}}=\chi^{\rho}\in\NN \chi^{\rho}\subset\sum_{\lambda
\in\operatorname{Par}} \NN \chi^{\lambda}$ (since $\NN %
\chi^{\rho}$ is an addend of the sum $\sum_{\lambda\in\operatorname{Par}%
} \NN \chi^{\lambda}$ (since $\rho\in \Par $)).
\par
Now, let us forget that we fixed $i$. We thus have shown that $\chi_{V_{i}}%
\in\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}$ for every
$i\in\left\{  1,2,\ldots,j\right\}  $. Thus, $\sum_{i=1}^{j}\underbrace{\chi
_{V_{i}}}_{\in\sum_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}%
}\in\sum_{i=1}^{j}\sum_{\lambda\in\operatorname{Par}} \NN %
\chi^{\lambda}\subset\sum_{\lambda\in\operatorname{Par}} \NN %
\chi^{\lambda}$ (since $\sum_{\lambda\in\operatorname{Par}} \NN %
\chi^{\lambda}$ is closed under addition).
\par
But isomorphic $\mathbb{C}\Symm_{nm}$-modules have equal characters.
Hence, since $P\cong V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}$, we have%
\[
\chi_{P}=\chi_{V_{1}\oplus V_{2}\oplus\cdots\oplus V_{j}}=\chi_{V_{1}}%
+\chi_{V_{2}}+\cdots+\chi_{V_{j}}=\sum_{i=1}^{j}\chi_{V_{i}}\in\sum
_{\lambda\in\operatorname{Par}} \NN \chi^{\lambda}.
\]
Applying the map $\ch_{\CC} $ to both sides of
this relation, we obtain%
\begin{align*}
\ch_{\CC} \left(  \chi_{P}\right)   &
\in\ch_{\CC} \left(  \sum_{\lambda
\in\operatorname{Par}} \NN \chi^{\lambda}\right)  \subset
\sum_{\lambda\in\operatorname{Par}} \NN %
\underbrace{\ch_{\CC} \left(  \chi^{\lambda
}\right)  }_{\substack{=s_{\lambda}\\\text{(by
(\ref{sol.repSn.arithmetic-prod.chC}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\ch_{\CC} \text{ is
} \ZZ\text{-linear}\right) \\
&  =\sum_{\lambda\in\operatorname{Par}} \NN s_{\lambda}.
\end{align*}
This proves (\ref{sol.repSn.arithmetic-prod.chiP}).}

We have $\Symm_n=\Symm_{\left\{  1,2,\ldots,n\right\}  }$
(by the definition of $\Symm_n$) and
$\Symm_m=\Symm_{\left\{  1,2,\ldots,m\right\}  }$
(by the definition of $\Symm_m$)
and $\Symm_{nm}=\Symm_{\left\{  1,2,\ldots,nm\right\}  }$ (by
the definition of$\mathfrak{\ S}_{nm}$). Recall that we have defined a map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ for any two sets $X$ and $Y$. Now, set $X=\left\{
1,2,\ldots,n\right\}  $ and $Y=\left\{  1,2,\ldots,m\right\}  $, and consider
the map $\operatorname*{cross}:\Symm_X\times\Symm_Y
\rightarrow\Symm_{X\times Y}$. In the following, when we speak of
$\operatorname*{cross}$, we will always mean this map.

We have $\left\vert \underbrace{X}_{=\left\{  1,2,\ldots,n\right\}
}\right\vert =\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =n$ and
$\left\vert \underbrace{Y}_{=\left\{  1,2,\ldots,m\right\}  }\right\vert
=\left\vert \left\{  1,2,\ldots,m\right\}  \right\vert =m$. Also, since
$X=\left\{  1,2,\ldots,n\right\}  $, we have
$\Symm_X=\Symm_{\left\{  1,2,\ldots,n\right\}  }=\Symm_n$ (since
$\Symm_n=\Symm_{\left\{  1,2,\ldots,n\right\}  }$). Similarly,
$\Symm_Y=\Symm_m$.

The set $X$ is nonempty (since $\left\vert X\right\vert =n\geq1$), and the set
$Y$ is nonempty (similarly). Hence, Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross}(b) yields that the map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ is injective. Lemma
\ref{lem.sol.repSn.arithmetic-prod.cross}(a) yields that the map
$\operatorname*{cross}:\Symm_X\times\Symm_Y\rightarrow
\Symm_{X\times Y}$ is a group homomorphism. Thus, the map
$\operatorname*{cross}$ is a group homomorphism from $\Symm_X%
\times\Symm_Y$ to $\Symm_{X\times Y}$. Since $\Symm_X=\Symm_n$ and
$\Symm_Y=\Symm_m$, this rewrites
as follows: The map $\operatorname*{cross}$ is a group homomorphism from
$\Symm_n\times\Symm_m$ to $\Symm_{X\times Y}$.

The set $X\times Y$ has cardinality $\left\vert X\times Y\right\vert
=\underbrace{\left\vert X\right\vert }_{=n}\cdot\underbrace{\left\vert
Y\right\vert }_{=m}=nm$, and thus is in bijection with the set $\left\{
1,2,\ldots,nm\right\}  $. In other words, there exists a bijection
$\varphi:X\times Y\rightarrow\left\{  1,2,\ldots,nm\right\}  $. Fix such a
bijection $\varphi$. Define a map $\varphi^{\ast}:\Symm_{X\times
Y}\rightarrow\Symm_{\left\{  1,2,\ldots,nm\right\}  }$ as in Lemma
\ref{lem.sol.repSn.arithmetic-prod.Sf} (applied to $U=X\times Y$ and
$V=\left\{  1,2,\ldots,nm\right\}  $). Then, Lemma
\ref{lem.sol.repSn.arithmetic-prod.Sf}(a) (applied to $U=X\times Y$ and
$V=\left\{  1,2,\ldots,nm\right\}  $) yields that this map $\varphi^{\ast}$ is
well-defined and a group isomorphism. In particular, $\varphi^{\ast}$ is an
injective group homomorphism. Also, $\varphi^{\ast}$ is a group homomorphism
from $\Symm_{X\times Y}$ to $\Symm_{\left\{  1,2,\ldots
,nm\right\}  }$, therefore a group homomorphism from $\Symm_{X\times
Y}$ to $\Symm_{nm}$ (since $\Symm_{nm}=\Symm_{\left\{
1,2,\ldots,nm\right\}  }$).

Hence, $\varphi^{\ast}\circ\operatorname*{cross}$ is a group homomorphism from
$\Symm_n\times\Symm_m$ to $\Symm_{nm}$ (since
$\varphi^{\ast}$ is a group homomorphism from $\Symm_{X\times Y}$ to
$\Symm_{nm}$, and since $\operatorname*{cross}$ is a group homomorphism
from $\Symm_n\times\Symm_m$ to $\Symm_{X\times Y}$).
Also, $\varphi^{\ast}\circ\operatorname*{cross}$ is injective (since
$\varphi^{\ast}$ and $\operatorname*{cross}$ are injective). Let
$\Omega=\varphi^{\ast}\circ\operatorname*{cross}$. Then, $\Omega$ is an
injective group homomorphism from $\Symm_n\times\Symm_m$ to
$\Symm_{nm}$ (since $\varphi^{\ast}\circ\operatorname*{cross}$ is an
injective group homomorphism from $\Symm_n\times\Symm_m$ to
$\Symm_{nm}$). We can define a map $\overline{\Omega}:\Symm_n
\times\Symm_m\rightarrow\Omega\left(  \Symm_n%
\times\Symm_m\right)  $ by%
\[
\left(  \overline{\Omega}\left(  g\right)  =\Omega\left(  g\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\Symm_n\times\Symm_m \right)
\]
(since $\Omega\left(  g\right)  \in\Omega\left(  \Symm_n%
\times\Symm_m\right)  $ for every $g\in\Symm_n%
\times\Symm_m$). Consider this $\overline{\Omega}$. Lemma
\ref{lem.sol.repSn.arithmetic-prod.core} (applied to $\Symm_n%
\times\Symm_m$ and $\Symm_{nm}$ instead of $G$ and $H$) yields
that $\overline{\Omega}$ is a well-defined group isomorphism. Hence, the
inverse $\overline{\Omega}^{-1}$ of $\overline{\Omega}$ is well-defined and
also a group isomorphism. Let $\tau$ denote this inverse $\overline{\Omega
}^{-1}$. Thus,
\[
\tau=\overline{\Omega}^{-1}\text{ is a group isomorphism }\Omega\left(
\Symm_n\times\Symm_m\right)  \rightarrow\Symm_n\times\Symm_m.
\]
Hence, a $\CC$-linear map $\tau^{\ast}:R_{\mathbb{C}}\left(
\Symm_n\times\Symm_m\right)  \rightarrow R_{\mathbb{C}%
}\left(  \Omega\left(  \Symm_n\times\Symm_m\right)  \right)
$ is defined (according to Lemma
\ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(a), applied to
$\Omega\left(  \Symm_n\times\Symm_m\right)  $ and
$\Symm_n\times\Symm_m$ instead of $K$ and $H$).

The following commutative diagram illustrates the group homomorphisms we have
just introduced:
\[\xymatrix{
\Symm_n \times\Symm_m \arinjrev[r]^{\operatorname{cross}} \ar@
/^3pc/[rr]^{\Omega} \ar@/_1pc/[dr]^{\overline{\Omega}}_{\cong} & \Symm
_{X \times Y} \ar[r]^{\varphi^{\ast}}_{\cong} & \Symm_{nm} \\
& \Omega\left(\Symm_n \times\Symm_m\right) \ar@/^4pc/[ul]^{\tau}_{\cong}
\arinjrev[ru]_{\text{inclusion}}
}
\]
(where the cycle formed by the $\overline{\Omega}$ and $\tau$ arrows is not a
mistake: these two maps are mutually inverse!).

In the following, for every $k\in \NN$, we let $\left(  \Lambda
_{\mathbb{C}}\right)  _{k}$ denote the $k$-th homogeneous component of the
graded $\CC$-algebra $\Lambda_{\mathbb{C}}$. Note that $\left(
p_{\lambda}\right)  _{\lambda\in\Par_{k}}$ is a basis
of this $\CC$-vector space $\left(  \Lambda_{\mathbb{C}}\right)  _{k}$.
Thus, $\left(  p_{\lambda}\right)  _{\lambda\in \Par_{m}}$
is a basis of the $\CC$-vector space $\left(
\Lambda_{\mathbb{C}}\right)  _{m}$, and $\left(  p_{\lambda}\right)
_{\lambda\in\Par_{n}}$ is a basis of the $\mathbb{C}%
$-vector space $\left(  \Lambda_{\mathbb{C}}\right)  _{n}$.

Let us now claim that every $a\in\left(  \Lambda_{\mathbb{C}}\right)  _{n}$
and $b\in\left(  \Lambda_{\mathbb{C}}\right)  _{m}$ satisfy%
\begin{equation}
a\widetilde{\boxdot}b=\ch_{\CC} \left(
\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n\times
\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\left(  \ch_{\CC} \right)  ^{-1}\left(
a\right)  \otimes\left(  \ch_{\CC} \right)
^{-1}\left(  b\right)  \right)  \right)  \right)
\label{sol.repSn.arithmetic-prod.chch}
\end{equation}
(where we identify $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  $ with $R_\CC \left(
\Symm_n\times\Symm_m\right)  $ along the canonical
isomorphism $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  \rightarrow R_{\mathbb{C}%
}\left(  \Symm_n\times\Symm_m\right)  $%
)\ \ \ \ \footnote{Notice that the term $\tau^{\ast}\left(  \left(
\ch_{\CC} \right)  ^{-1}\left(  a\right)
\otimes\left(  \ch_{\CC} \right)  ^{-1}\left(
b\right)  \right)  $ on the right hand side of
(\ref{sol.repSn.arithmetic-prod.chch}) is well-defined. (This is because
\begin{align*}
\left(  \ch_{\CC} \right)  ^{-1}\left(
\underbrace{a}_{\in\left(  \Lambda_{\mathbb{C}}\right)  _{n}}\right)
\otimes\left(  \ch_{\CC} \right)  ^{-1}\left(
\underbrace{b}_{\in\left(  \Lambda_{\mathbb{C}}\right)  _{m}}\right)   &
\in\underbrace{\left(  \ch_{\CC} \right)
^{-1}\left(  \left(  \Lambda_{\mathbb{C}}\right)  _{n}\right)  }%
_{=R_{\mathbb{C}}\left(  \Symm_n\right)  }\otimes\underbrace{\left(
\ch_{\CC} \right)  ^{-1}\left(  \left(
\Lambda_{\mathbb{C}}\right)  _{m}\right)  }_{=R_{\mathbb{C}}\left(
\Symm_m\right)  }\\
&  =R_\CC \left(  \Symm_n\right)  \otimes R_{\mathbb{C}%
}\left(  \Symm_m\right)  =R_{\mathbb{C}}\left(  \Symm_n
\times\Symm_m\right)  .
\end{align*}
)}.

\textit{Proof of (\ref{sol.repSn.arithmetic-prod.chch}):} Let $a\in\left(
\Lambda_{\mathbb{C}}\right)  _{n}$ and $b\in\left(  \Lambda_{\mathbb{C}%
}\right)  _{m}$. We need to prove the equality
(\ref{sol.repSn.arithmetic-prod.chch}). Since this equality is $\mathbb{C}%
$-linear in each of $a$ and $b$ (because the operations $\widetilde{\boxdot}$
and $\otimes$ are $\CC$-bilinear, and the maps $\operatorname*{ch}%
\nolimits_{\mathbb{C}}$, $\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}$, $\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n
\times\Symm_m\right)  }^{\Symm_{nm}}$ and $\tau^{\ast}$
are $\CC$-linear), we can WLOG assume that $a$ is an element of the
basis $\left(  p_{\lambda}\right)  _{\lambda\in \Par_{n}}$
of the $\CC$-vector space $\left(  \Lambda_{\mathbb{C}%
}\right)  _{n}$, and that $b$ is an element of the basis $\left(  p_{\lambda
}\right)  _{\lambda\in\Par_{m}}$ of the $\mathbb{C}%
$-vector space $\left(  \Lambda_{\mathbb{C}}\right)  _{m}$. Assume this. Then,
we can write $a$ and $b$ as $a=p_{\gamma}$ and $b=p_{\eta}$ for some
$\gamma\in\Par_{n}$ and some $\eta\in
\Par_{m}$. Consider these $\gamma$ and $\eta$.

Choose some permutation $g\in\Symm_n$ which has cycle type $\gamma$.
(Such a $g$ clearly exists, since $\gamma\in\Par_{n}%
$.) Then, $\ch_{\CC} \left(  \alpha
_{\Symm_n,g}\right)  =p_{\gamma}$ (according to
(\ref{sol.repSn.arithmetic-prod.import})). Hence, $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  p_{\gamma}\right)  =\alpha
_{\Symm_n,g}$, so that $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  \underbrace{a}_{=p_{\gamma}%
}\right)  =\left(  \ch_{\CC} \right)
^{-1}\left(  p_{\gamma}\right)  =\alpha_{\Symm_n,g}$.

Choose some permutation $h\in\Symm_m$ which has cycle type $\eta$.
(Such an $h$ clearly exists, since $\eta\in\Par_{m}$.)
Then, $\ch_{\CC} \left(  \alpha_{\Symm_m,
h}\right)  =p_{\eta}$ (according to
(\ref{sol.repSn.arithmetic-prod.import}), applied to $m$, $h$ and $\eta$
instead of $n$, $g$ and $\lambda$). Hence, $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  p_{\eta}\right)  =\alpha
_{\Symm_m,h}$, so that $\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  \underbrace{b}_{=p_{\eta}}\right)
=\left(  \ch_{\CC} \right)  ^{-1}\left(
p_{\eta}\right)  =\alpha_{\Symm_m,h}$. Thus,%
\begin{equation}
\underbrace{\left(  \ch_{\CC} \right)
^{-1}\left(  a\right)  }_{=\alpha_{\Symm_n,g}}\otimes\underbrace{\left(
\ch_{\CC} \right)  ^{-1}\left(  b\right)
}_{=\alpha_{\Symm_m,h}}=\alpha_{\Symm_n,g}\otimes
\alpha_{\Symm_m,h}. \label{sol.repSn.arithmetic-prod.chch.pf.1}%
\end{equation}


Exercise \ref{exe.frob-char.plambda}(c) (applied to $G_{1}=\Symm_n$,
$G_{2}=\Symm_m$, $h_{1}=g$ and $h_{2}=h$) yields that the canonical
isomorphism $R_\CC \left(  \Symm_n\right)  \otimes
R_{\mathbb{C}}\left(  \Symm_m\right)  \rightarrow R_{\mathbb{C}%
}\left(  \Symm_n\times\Symm_m\right)  $ sends
$\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}$ to
$\alpha_{\Symm_n\times\Symm_m,\left(  g,h\right)  }$. Since
we are regarding this isomorphism as an identity (because we have identified
$R_\CC \left(  \Symm_n\right)  \otimes R_{\mathbb{C}%
}\left(  \Symm_m\right)  $ with $R_\CC \left(
\Symm_n\times\Symm_m\right)  $ along this isomorphism), this
yields $\alpha_{\Symm_n,g}\otimes\alpha_{\Symm_m,h}%
=\alpha_{\Symm_n\times\Symm_m,\left(  g,h\right)  }$. Thus,
(\ref{sol.repSn.arithmetic-prod.chch.pf.1}) becomes%
\[
\left(  \ch_{\CC} \right)  ^{-1}\left(
a\right)  \otimes\left(  \ch_{\CC} \right)
^{-1}\left(  b\right)  =\alpha_{\Symm_n,g}\otimes\alpha
_{\Symm_m,h}=\alpha_{\Symm_n\times\Symm_m,\left(
g,h\right)  }.
\]
Applying the map $\tau^{\ast}$ to both sides of this equality, we obtain%
\begin{equation}
\tau^{\ast}\left(  \left(  \ch_{\CC} \right)
^{-1}\left(  a\right)  \otimes\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  b\right)  \right)  =\tau^{\ast}\left(  \alpha
_{\Symm_n\times\Symm_m,\left(  g,h\right)  }\right)  .
\label{sol.repSn.arithmetic-prod.chch.pf.2}%
\end{equation}


But $\overline{\Omega}\left(  \left(  g,h\right)  \right)  =\Omega\left(
\left(  g,h\right)  \right)  $ (by the definition of $\overline{\Omega}\left(
\left(  g,h\right)  \right)  $), and so $\left(  g,h\right)
=\underbrace{\overline{\Omega}^{-1}}_{=\tau}\left(  \Omega\left(  \left(
g,h\right)  \right)  \right)  =\tau\left(  \Omega\left(  \left(  g,h\right)
\right)  \right)  $. Thus,
\[
\tau^{\ast}\left(  \alpha_{\Symm_n\times\Symm_m,\left(
g,h\right)  }\right)  =\tau^{\ast}\left(  \alpha_{\Symm_n%
\times\Symm_m,\tau\left(  \Omega\left(  \left(  g,h\right)  \right)
\right)  }\right)  =\alpha_{\Omega\left(  \Symm_n\times
\Symm_m\right)  ,\Omega\left(  \left(  g,h\right)  \right)  }%
\]
(by Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(d), applied
to $\Omega\left(  \Symm_n\times\Symm_m\right)  $,
$\Symm_n\times\Symm_m$ and $\Omega\left(  \left(
g,h\right)  \right)  $ instead of $K$, $H$ and $g$). Thus,
(\ref{sol.repSn.arithmetic-prod.chch.pf.2}) becomes%
\[
\tau^{\ast}\left(  \left(  \ch_{\CC} \right)
^{-1}\left(  a\right)  \otimes\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  b\right)  \right)  =\tau^{\ast}\left(  \alpha
_{\Symm_n\times\Symm_m,\left(  g,h\right)  }\right)
=\alpha_{\Omega\left(  \Symm_n\times\Symm_m\right)
,\Omega\left(  \left(  g,h\right)  \right)  }.
\]
Applying the map $\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n
\times\Symm_m\right)  }^{\Symm_{nm}}$ to both sides of
this equality, we obtain%
\begin{align}
\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n\times
\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\left(  \ch_{\CC} \right)  ^{-1}\left(
a\right)  \otimes\left(  \ch_{\CC} \right)
^{-1}\left(  b\right)  \right)  \right)   &  =\operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \alpha_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  ,\Omega\left(  \left(  g,h\right)  \right)
}\right) \nonumber\\
&  =\alpha_{\Symm_{nm},\Omega\left(  \left(  g,h\right)  \right)  }
\label{sol.repSn.arithmetic-prod.chch.pf.6}%
\end{align}
(by Exercise \ref{exe.frob-char.plambda}(b), applied to $\Symm_{nm}$,
$\Omega\left(  \Symm_n\times\Symm_m\right)  $ and
$\Omega\left(  \left(  g,h\right)  \right)  $ instead of $G$, $H$ and $h$).

Now, $\Omega\left(  \left(  g,h\right)  \right)  \in\Symm_{nm}$, so
that $\Omega\left(  \left(  g,h\right)  \right)  $ is a permutation of
$\left\{  1,2,\ldots,nm\right\}  $. Let $\kappa$ be the cycle type of this
permutation $\Omega\left(  \left(  g,h\right)  \right)  $. Then,
(\ref{sol.repSn.arithmetic-prod.import}) (applied to $nm$, $\kappa$ and
$\Omega\left(  \left(  g,h\right)  \right)  $ instead of $n$, $\lambda$ and
$g$) yields $\ch_{\CC} \left(  \alpha
_{\Symm_{nm},\Omega\left(  \left(  g,h\right)  \right)  }\right)
=p_{\kappa}$. But applying the map $\ch_{\CC} $
to both sides of the identity (\ref{sol.repSn.arithmetic-prod.chch.pf.6}), we
obtain%
\begin{align}
&  \ch_{\CC} \left(  \operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  a\right)  \otimes\left(
\ch_{\CC} \right)  ^{-1}\left(  b\right)
\right)  \right)  \right) \nonumber\\
&  =\ch_{\CC} \left(  \alpha_{
\Symm_{nm},\Omega\left(  \left(  g,h\right)  \right)  }\right)
=p_{\kappa}.
\label{sol.repSn.arithmetic-prod.chch.pf.8}%
\end{align}


Let us now show that $a\widetilde{\boxdot}b=p_{\kappa}$.

In fact, $\underbrace{\Omega}_{=\varphi^{\ast}\circ\operatorname*{cross}%
}\left(  \left(  g,h\right)  \right)  =\left(  \varphi^{\ast}\circ
\operatorname*{cross}\right)  \left(  \left(  g,h\right)  \right)
=\varphi^{\ast}\left(  \underbrace{\operatorname*{cross}\left(  \left(
g,h\right)  \right)  }_{\substack{=g\times h\\\text{(by the definition}%
\\\text{of }\operatorname*{cross}\text{)}}}\right)  =\varphi^{\ast}\left(
g\times h\right)  $. But Lemma \ref{lem.sol.repSn.arithmetic-prod.Sf}(b)
(applied to $U=X\times Y$, $V=\left\{  1,2,\ldots,nm\right\}  $ and
$\pi=g\times h$) yields that the cycle type of $\varphi^{\ast}\left(  g\times
h\right)  $ equals the cycle type of $g\times h$. In other words, the cycle
type of $\Omega\left(  \left(  g,h\right)  \right)  $ equals the cycle type of
$g\times h$ (since $\Omega\left(  \left(  g,h\right)  \right)  =\varphi^{\ast
}\left(  g\times h\right)  $). In other words, $\kappa$ equals the cycle type
of $g\times h$ (since $\kappa$ is the cycle type of $\Omega\left(  \left(
g,h\right)  \right)  $).

So we know that $g\in\Symm_n=\Symm_X$ and $h\in
\Symm_m=\Symm_Y$ are permutations, and that $\gamma$, $\eta$
and $\kappa$ are the cycle types of the permutations $g$, $h$ and $g\times h$.
Thus, Lemma \ref{lem.sol.repSn.arithmetic-prod.cross2p} (applied to $\gamma$
and $\eta$ instead of $\lambda$ and $\mu$) yields $p_{\kappa}=p_{\gamma
}\boxdot p_{\eta}$. Now,%
\begin{align*}
\underbrace{a}_{=p_{\gamma}}\widetilde{\boxdot}\underbrace{b}_{=p_{\eta}}  &
=p_{\gamma}\widetilde{\boxdot}p_{\eta}=p_{\gamma}\boxdot p_{\eta
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.repSn.arithmetic-prod.bot2}),
applied to }\gamma\text{ and }\eta\text{ instead of }\lambda\text{ and }%
\mu\right) \\
&  =p_{\kappa}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{\kappa}=p_{\gamma
}\boxdot p_{\eta}\right) \\
&  =\ch_{\CC} \left(  \operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left(  a\right)  \otimes\left(
\ch_{\CC} \right)  ^{-1}\left(  b\right)
\right)  \right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.repSn.arithmetic-prod.chch.pf.8})}\right)  .
\end{align*}
This proves (\ref{sol.repSn.arithmetic-prod.chch}).

Now that (\ref{sol.repSn.arithmetic-prod.chch}) is proven, it is easy to
complete the solution of Exercise \ref{exe.repSn.arithmetic-prod}(c). Recall
that $\mu$ and $\nu$ are two partitions such that $\mu\in \Par_{m}$,
$\nu\in\Par_{n}$, $m\geq1$ and
$n\geq1$. We need to show that $s_{\mu}\boxdot s_{\nu}\in\sum_{\lambda
\in\operatorname{Par}}{\NN}s_{\lambda}$.

Applying (\ref{sol.repSn.arithmetic-prod.chC}) to $\lambda=\mu$, we obtain
$\ch_{\CC} \left(  \chi^{\mu}\right)  =s_{\mu}$,
so that $\left(  \ch_{\CC} \right)  ^{-1}\left(
s_{\mu}\right)  =\chi^{\mu}$. The same argument (but with $\mu$ replaced by
$\nu$) shows that $\left(  \ch_{\CC} \right)
^{-1}\left(  s_{\nu}\right)  =\chi^{\nu}$.

We know that $\chi^{\mu}$ is an irreducible complex character of
$\mathbb{C}\Symm_m$. In other words, there exists a simple
$\mathbb{C}\Symm_m$-module $M$ such that $\chi^{\mu}=\chi_{M}$.
Consider this $M$. Then, $\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  s_{\mu}\right)  =\chi^{\mu}=\chi_{M}$.

We know that $\chi^{\nu}$ is an irreducible complex character of
$\mathbb{C}\Symm_n$. In other words, there exists a simple
$\mathbb{C}\Symm_n$-module $N$ such that $\chi^{\nu}=\chi_{N}$.
Consider this $N$. Then, $\left(  \operatorname*{ch}\nolimits_{\mathbb{C}%
}\right)  ^{-1}\left(  s_{\nu}\right)  =\chi^{\nu}=\chi_{N}$.

Recall that we are identifying $R_\CC \left(  \Symm_n\right)
\otimes R_{\mathbb{C}}\left(  \Symm_m\right)  $ with
$R_\CC \left(  \Symm_n\times\Symm_m\right)  $
along the canonical isomorphism $R_\CC \left(  \Symm_n\right)
\otimes R_\CC \left(  \Symm_m\right)
\rightarrow R_{\mathbb{C}}\left(  \Symm_n\times\Symm_m\right)  $.
Thus, for any finite-dimensional $\mathbb{C}\Symm_n%
$-module $U$ and any finite-dimensional $\mathbb{C}\Symm_m$-module
$V$, we have $\chi_{U\otimes V}=\chi_{U}\otimes\chi_{V}$ (since this
isomorphism sends $\chi_{U}\otimes\chi_{V}$ to $\chi_{U\otimes V}$). Applying
this to $U=N$ and $V=M$, we obtain $\chi_{N\otimes M}=\chi_{N}\otimes\chi_{M}%
$. Thus, $\chi_{N}\otimes\chi_{M}=\chi_{N\otimes M}$.

Lemma \ref{lem.sol.repSn.arithmetic-prod.general-restriction1}(b) (applied to
$\Omega\left(  \Symm_n\times\Symm_m\right)  $,
$\Symm_n\times\Symm_m$ and $N\otimes M$ instead of $K$, $H$
and $U$) yields $\chi_{\left(  N\otimes M\right)  ^{\tau}}=\tau^{\ast}\left(
\chi_{N\otimes M}\right)  $. Thus, $\tau^{\ast}\left(  \chi_{N\otimes
M}\right)  =\chi_{\left(  N\otimes M\right)  ^{\tau}}$. Hence,%
\begin{equation}
\tau^{\ast}\left(  \underbrace{\chi_{N}\otimes\chi_{M}}_{=\chi_{N\otimes M}%
}\right)  =\tau^{\ast}\left(  \chi_{N\otimes M}\right)  =\chi_{\left(
N\otimes M\right)  ^{\tau}}. \label{sol.repSn.arithmetic-prod.characalc4}%
\end{equation}


But (\ref{eq.induction.mods-and-class-functs}) (applied to $\Symm_{nm}%
$, $\Omega\left(  \Symm_n\times\Symm_m\right)  $ and
$\left(  N\otimes M\right)  ^{\tau}$ instead of $G$, $H$ and $U$) yields
$\chi_{\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \left(  N\otimes
M\right)  ^{\tau}\right)  }=\operatorname*{Ind}\nolimits_{\Omega\left(
\Symm_n\times\Symm_m\right)  }^{\Symm_{nm}}\left(
\chi_{\left(  N\otimes M\right)  ^{\tau}}\right)  $. Let $P$ denote the
finite-dimensional $\mathbb{C}\Symm_{nm}$-module $\operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \left(  N\otimes M\right)  ^{\tau}\right)  $.
Then, $P=\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \left(  N\otimes
M\right)  ^{\tau}\right)  $, so that%
\[
\chi_{P}=\chi_{\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \left(
N\otimes M\right)  ^{\tau}\right)  }=\operatorname*{Ind}\nolimits_{\Omega
\left(  \Symm_n\times\Symm_m\right)  }^{\Symm_{nm}%
}\left(  \underbrace{\chi_{\left(  N\otimes M\right)  ^{\tau}}}%
_{\substack{=\tau^{\ast}\left(  \chi_{N}\otimes\chi_{M}\right)  \\\text{(by
(\ref{sol.repSn.arithmetic-prod.characalc4}))}}}\right)  =\operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \chi_{N}\otimes\chi
_{M}\right)  \right)  .
\]
Hence,%
\begin{equation}
\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n\times
\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\chi_{N}\otimes\chi_{M}\right)  \right)  =\chi_{P}.
\label{sol.repSn.arithmetic-prod.characalc7}%
\end{equation}


We have $s_{\mu}\in\left(  \Lambda_{\mathbb{C}}\right)  _{m}$ (since $\mu
\in\Par_{m}$) and $s_{\nu}\in\left(  \Lambda
_{\mathbb{C}}\right)  _{n}$ (since $\nu\in\Par_{n}$).
Thus, we can apply (\ref{sol.repSn.arithmetic-prod.chch}) to $a=s_{\nu}$ and
$b=s_{\mu}$.

Exercise \ref{exe.repSn.arithmetic-prod}(a) yields that $\Lambda_{\QQ}%
$, equipped with the binary operation $\boxdot$, becomes a commutative
$\QQ$-algebra with unity $p_{1}$. Thus, the operation $\boxdot$ is
commutative. Hence,%
\begin{align*}
s_{\mu}\boxdot s_{\nu}  &  =s_{\nu}\boxdot s_{\mu}=s_{\nu}\widetilde{\boxdot
}s_{\mu}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }s_{\nu}\widetilde{\boxdot
}s_{\mu}=s_{\nu}\boxdot s_{\mu}\text{ (by
(\ref{sol.repSn.arithmetic-prod.botbot2}), applied to }a=s_{\nu}\text{ and
}b=s_{\mu}\text{)}\right) \\
&  =\ch_{\CC} \left(  \operatorname*{Ind}%
\nolimits_{\Omega\left(  \Symm_n\times\Symm_m\right)
}^{\Symm_{nm}}\left(  \tau^{\ast}\left(  \underbrace{\left(
\ch_{\CC} \right)  ^{-1}\left(  s_{\nu}\right)
}_{=\chi_{N}}\otimes\underbrace{\left(  \operatorname*{ch}%
\nolimits_{\mathbb{C}}\right)  ^{-1}\left( s_\mu \right)  }_{=\chi_{M}}\right)
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.repSn.arithmetic-prod.chch}%
), applied to }a=s_{\nu}\text{ and }b=s_{\mu}\right) \\
&  =\ch_{\CC} \left(
\underbrace{\operatorname*{Ind}\nolimits_{\Omega\left(  \Symm_n%
\times\Symm_m\right)  }^{\Symm_{nm}}\left(  \tau^{\ast}\left(
\chi_{N}\otimes\chi_{M}\right)  \right)  }_{\substack{=\chi_{P}\\\text{(by
(\ref{sol.repSn.arithmetic-prod.characalc7}))}}}\right)  =\operatorname*{ch}%
\nolimits_{\mathbb{C}}\left(  \chi_{P}\right)  \in\sum_{\lambda\in
\operatorname{Par}} \NN s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.repSn.arithmetic-prod.chiP})}\right)  .
\end{align*}
This solves Exercise \ref{exe.repSn.arithmetic-prod}(c).

[\textit{Remark:} The above solution can be simplified using the results
of Exercise~\ref{exe.Indrho}. We leave the details of this simplification
to the reader, and only mention some highlights. Let us use the notations
of Exercise~\ref{exe.Indrho}. Then, the awkward equality
\eqref{sol.repSn.arithmetic-prod.chch} can be replaced by the simpler
equality
\begin{equation}
a \widetilde{\boxdot} b
= \ch_\CC \left( \Ind_\Omega
                 \left( \left(\ch_\CC\right)^{-1} \left(a\right)
                        \otimes
                        \left(\ch_\CC\right)^{-1} \left(b\right)
                 \right)
          \right) .
\end{equation}
This makes the definition of the maps $\overline{\Omega}$ and $\tau$
unnecessary. The WLOG assumption that we don't have $\min\left\{
m,n\right\}  =0$ becomes unnecessary as well; the injectivity of the
map $\operatorname{cross}$ can no longer be guaranteed without this
assumption, but we do not need this map to be injective anymore.
We no longer need
Lemma~\ref{lem.sol.repSn.arithmetic-prod.cross}(b),
Lemma~\ref{lem.sol.repSn.arithmetic-prod.general-restriction1} and
Lemma~\ref{lem.sol.repSn.arithmetic-prod.core}.
However, we need to use Exercise~\ref{exe.Indrho}(b) instead of
\eqref{eq.induction.mods-and-class-functs}, and we have to use
Exercise~\ref{exe.frob-char.plambda}(f) instead of
Exercise~\ref{exe.frob-char.plambda}(b).]

(d) We need to prove that $f\boxdot g\in\Lambda$ for any $f\in\Lambda$ and
$g\in\Lambda$. Since the binary operation $\boxdot$ is $\ZZ$-bilinear
(because it is $\QQ$-bilinear), it is clearly enough to prove that
$s_{\mu}\boxdot s_{\nu}\in\Lambda$ for any partitions $\mu$ and $\nu$ (since
$\left(  s_{\lambda}\right)  _{\lambda\in\operatorname{Par}}$ is a
$\ZZ$-basis of $\Lambda$). But this follows from the fact that%
\begin{align*}
s_{\mu}\boxdot s_{\nu}  &  \in\sum_{\lambda\in\operatorname{Par}}
\NN s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise
\ref{exe.repSn.arithmetic-prod}(c)}\right) \\
&  \subset\Lambda
\end{align*}
for any partitions $\mu$ and $\nu$. Thus,
Exercise~\ref{exe.repSn.arithmetic-prod}(d) is solved.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.polynomials-and-necklaces}}

\textit{Solution to Exercise \ref{exe.polynomials-and-necklaces}.} (a) We
shall prove the following fact:

\begin{lemma}
\label{lem.sol.polynomials-and-necklaces.a}Let $q$ be a prime power. For every
positive integer $n$, we have%
\[
\left(  \text{the number of all irreducible monic degree-}n\text{ polynomials
in } \FF_{q}\left[  x\right]  \right)  =\dfrac{1}{n}\sum_{d\mid n}%
\mu\left(  \dfrac{n}{d}\right)  q^{d}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.polynomials-and-necklaces.a}.]We first recall
that every positive integer $n$ satisfies%
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  =\delta_{n,1}.
\label{pf.lem.sol.polynomials-and-necklaces.a.mob}%
\end{equation}
(This is precisely the equality (\ref{sol.exe.witt.ghost-equiv.mu*1}), which
was proven in the solution of Exercise \ref{exe.witt.ghost-equiv}.)

For every positive integer $n$, define a nonnegative integer
$\operatorname*{irr} n$ by
\begin{equation}
\operatorname*{irr}n=\left(  \text{the number of all irreducible monic
degree-}n\text{ polynomials in } \FF_{q}\left[  x\right]  \right)  .
\label{pf.lem.sol.polynomials-and-necklaces.a.0}%
\end{equation}


Let $\mathfrak{P}$ denote the set of all irreducible monic polynomials in
$ \FF_{q}\left[  x\right]  $. Then, every $p\in\mathfrak{P}$ is an
irreducible polynomial and thus satisfies $\deg p\geq1$. Also, the irreducible
monic polynomials in $ \FF_{q}\left[  x\right]  $ are exactly the
elements of $\mathfrak{P}$ (since $\mathfrak{P}$ is the set of all irreducible
monic polynomials in $ \FF_{q}\left[  x\right]  $). Hence, for every
positive integer $n$, we have%
\begin{align*}
\operatorname*{irr}n  &  =\left(  \text{the number of all irreducible monic
degree-}n\text{ polynomials in } \FF_{q}\left[  x\right]  \right)
\\
&  =\left(  \text{the number of all irreducible monic polynomials }p\text{ in
} \FF_{q}\left[  x\right]  \text{ such that }\deg p=n\right) \\
&  =\left(  \text{the number of all }p\in\mathfrak{P}\text{ such that }\deg
p=n\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the irreducible monic polynomials
in } \FF_{q}\left[  x\right]  \text{ are exactly the elements of
}\mathfrak{P}\right)  .
\end{align*}
In other words, for every positive integer $n$, we have
\begin{align}
\left(  \text{the number of all }p\in\mathfrak{P}\text{ such that }\deg
p=n\right)
= \operatorname*{irr}n .
\label{pf.lem.sol.polynomials-and-necklaces.a.1}
\end{align}


Let $\mathfrak{N}$ be the set of all families $\left(  k_{p}\right)
_{p\in\mathfrak{P}}\in \NN ^{\mathfrak{P}}$ of nonnegative integers
(indexed by the polynomials belonging to $\mathfrak{P}$) such that all but
finitely many $p\in\mathfrak{P}$ satisfy $k_{p}=0$. Every monic polynomial
$P\in \FF_{q}\left[  x\right]  $ has a unique factorization into
irreducible monic polynomials. In other words, every monic polynomial
$P\in \FF_{q}\left[  x\right]  $ can be written in the form
$P=\prod_{p\in\mathfrak{P}}p^{k_{p}}$ for a unique family $\left(
k_{p}\right)  _{p\in\mathfrak{P}}\in\mathfrak{N}$. Thus, the map%
\begin{align*}
\mathfrak{N}  &  \rightarrow\left\{  P\in \FF_{q}\left[  x\right]
\ \mid\ P\text{ is monic}\right\}  ,\\
\left(  k_{p}\right)  _{p\in\mathfrak{P}}  &  \mapsto\prod_{p\in\mathfrak{P}%
}p^{k_{p}}%
\end{align*}
is bijective. Thus, in the ring $ \QQ \left[  \left[  t\right]  \right]  $
of formal power series, we have
\begin{equation}
\sum\limits_{\substack{P\in \FF_{q}\left[  x\right]  ;\\P\text{ is monic}%
}}t^{\deg P}=\sum_{\left(  k_{p}\right)  _{p\in\mathfrak{P}}\in\mathfrak{N}%
}t^{\deg\left(  \prod_{p\in\mathfrak{P}}p^{k_{p}}\right)  } .
\label{pf.lem.sol.polynomials-and-necklaces.a.2}
\end{equation}
But every $\left(  k_{p}\right)  _{p\in\mathfrak{P}}%
\in\mathfrak{N}$ satisfies%
\begin{align*}
t^{\deg\left(  \prod_{p\in\mathfrak{P}}p^{k_{p}}\right)  }  &  =t^{\sum
_{p\in\mathfrak{P}}k_{p}\deg p}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\deg\left(  \prod_{p\in\mathfrak{P}}p^{k_{p}}\right)  =\sum_{p\in\mathfrak{P}%
}k_{p}\deg p\right) \\
&  =\prod_{p\in\mathfrak{P}}t^{k_{p}\deg p}=\prod_{p\in\mathfrak{P}}\left(
t^{\deg p}\right)  ^{k_{p}}.
\end{align*}
Thus, (\ref{pf.lem.sol.polynomials-and-necklaces.a.2}) becomes%
\begin{align*}
\sum\limits_{\substack{P\in \FF_{q}\left[  x\right]  ;\\P\text{ is monic}%
}}t^{\deg P}  &  =\sum_{\left(  k_{p}\right)  _{p\in\mathfrak{P}}%
\in\mathfrak{N}}\underbrace{t^{\deg\left(  \prod_{p\in\mathfrak{P}}p^{k_{p}%
}\right)  }}_{=\prod_{p\in\mathfrak{P}}\left(  t^{\deg p}\right)  ^{k_{p}}%
}=\sum_{\left(  k_{p}\right)  _{p\in\mathfrak{P}}\in\mathfrak{N}}\prod
_{p\in\mathfrak{P}}\left(  t^{\deg p}\right)  ^{k_{p}}\\
&  =\prod_{p\in\mathfrak{P}}\sum_{k\in \NN }\left(  t^{\deg p}\right)
^{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \\
&  =\prod_{n\geq1}\prod\limits_{\substack{p\in\mathfrak{P};\\\deg p=n}}\sum
_{k\in \NN }\left(  \underbrace{t^{\deg p}}_{\substack{=t^{n}%
\\\text{(since }\deg p=n\text{)}}}\right)  ^{k}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\deg p\geq1\text{ for every }p\in\mathfrak{P}\right) \\
&  =\prod_{n\geq1}\prod\limits_{\substack{p\in\mathfrak{P};\\\deg p=n}%
}\underbrace{\sum_{k\in \NN }\left(  t^{n}\right)  ^{k}}_{=\dfrac
{1}{1-t^{n}}}=\prod_{n\geq1}\underbrace{\prod\limits_{\substack{p\in\mathfrak{P}%
;\\\deg p=n}}\dfrac{1}{1-t^{n}}}_{\substack{=\left(  \dfrac{1}{1-t^{n}%
}\right)  ^{\left(  \text{the number of all }p\in\mathfrak{P}\text{ such that
}\deg p=n\right)  }\\=\left(  \dfrac{1}{1-t^{n}}\right)  ^{\operatorname*{irr}%
n}\\\text{(by (\ref{pf.lem.sol.polynomials-and-necklaces.a.1}))}}}\\
&  =\prod_{n\geq1}\left(  \dfrac{1}{1-t^{n}}\right)  ^{\operatorname*{irr}n}.
\end{align*}
Compared with%
\begin{align*}
\sum\limits_{\substack{P\in \FF_{q}\left[  x\right]  ;\\P\text{ is monic}%
}}t^{\deg P}  &  =\sum_{n\in \NN }\sum\limits_{\substack{P\in \FF_{q}
\left[  x\right]  ;\\P\text{ is monic;}\\\deg P=n}}\underbrace{t^{\deg P}%
}_{\substack{=t^{n}\\\text{(since }\deg P=n\text{)}}}=\sum_{n\in \NN %
}\underbrace{\sum\limits_{\substack{P\in \FF_{q}\left[  x\right]  ;\\P\text{ is
monic;}\\\deg P=n}}t^{n}}_{=\left(  \text{the number of all monic }%
P\in \FF_{q}\left[  x\right]  \text{ such that }\deg P=n\right)  \cdot
t^{n}}\\
&  =\sum_{n\in \NN }\underbrace{\left(  \text{the number of all monic
}P\in \FF_{q}\left[  x\right]  \text{ such that }\deg P=n\right)
}_{\substack{=q^{n}\\\text{(because specifying a monic }P\in \FF_{q}
\left[  x\right]  \text{ such that }\deg P=n\\\text{is equivalent to
specifying its coefficients before }x^{0}\text{, }x^{1}\text{, }...\text{,
}x^{n-1}\text{,}\\\text{and each of these coefficients can be chosen freely
from }q\\\text{possible values)}}}\cdot t^{n}\\
&  =\sum_{n\in \NN }q^{n}t^{n}=\sum_{n\in \NN }\left(  qt\right)
^{n}=\dfrac{1}{1-qt},
\end{align*}
this yields%
\[
\dfrac{1}{1-qt}=\prod_{n\geq1}\left(  \dfrac{1}{1-t^{n}}\right)
^{\operatorname*{irr}n}.
\]
Taking the logarithm of both sides of this identity, we obtain%
\begin{align*}
\log\dfrac{1}{1-qt}  &  =\log\left(  \prod_{n\geq1}\left(  \dfrac{1}{1-t^{n}%
}\right)  ^{\operatorname*{irr}n}\right)  =\sum_{n\geq1}\left(
\operatorname*{irr}n\right)  \cdot\underbrace{\log\left(  \dfrac{1}{1-t^{n}%
}\right)  }_{\substack{=-\log\left(  1-t^{n}\right)  =\sum_{u\geq1}\dfrac
{1}{u}\left(  t^{n}\right)  ^{u}\\\text{(by the Mercator series for the
logarithm)}}}\\
&  =\sum_{n\geq1}\left(  \operatorname*{irr}n\right)  \cdot\sum_{u\geq1}%
\dfrac{1}{u}\left(  t^{n}\right)  ^{u}=\sum_{n\geq1}\sum_{u\geq1}\left(
\operatorname*{irr}n\right)  \dfrac{1}{u}\underbrace{\left(  t^{n}\right)
^{u}}_{=t^{nu}}=\sum_{n\geq1}\sum_{u\geq1}\left(  \operatorname*{irr}n\right)
\dfrac{1}{u}t^{nu}\\
&  =\underbrace{\sum_{n\geq1}\sum\limits_{\substack{v\geq1;\\n\mid v}}}_{=\sum
_{v\geq1}\sum_{n\mid v}}\left(  \operatorname*{irr}n\right)
\underbrace{\dfrac{1}{v/n}}_{=\dfrac{n}{v}}t^{v}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we substituted }v/n\text{ for }u\text{ in the second sum}\right)
\\
&  =\sum_{v\geq1}\sum_{n\mid v}\left(  \operatorname*{irr}n\right)  \dfrac
{n}{v}t^{v}=\sum_{n\geq1}\sum_{d\mid n}\left(  \operatorname*{irr}d\right)
\dfrac{d}{n}t^{n}%
\end{align*}
(here, we renamed the summation indices $v$ and $n$ as $n$ and $d$). Since%
\begin{align*}
\log\dfrac{1}{1-qt}  &  =-\log\left(  1-qt\right)  =\sum_{n\geq1}\dfrac{1}%
{n}\left(  qt\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the Mercator
series for the logarithm}\right) \\
&  =\sum_{n\geq1}\dfrac{1}{n}q^{n}t^{n},
\end{align*}
this rewrites as%
\[
\sum_{n\geq1}\dfrac{1}{n}q^{n}t^{n}=\sum_{n\geq1}\sum_{d\mid n}\left(
\operatorname*{irr}d\right)  \dfrac{d}{n}t^{n}.
\]
Comparing coefficients, we conclude that every positive integer $n$ satisfies%
\[
\dfrac{1}{n}q^{n}=\sum_{d\mid n}\left(  \operatorname*{irr}d\right)  \dfrac
{d}{n}.
\]
Multiplying this with $n$, we obtain%
\begin{equation}
q^{n}=\sum_{d\mid n}\left(  \operatorname*{irr}d\right)  d.
\label{pf.lem.sol.polynomials-and-necklaces.a.5}%
\end{equation}


Now, every positive integer $n$ satisfies%
\begin{align*}
\sum\limits_{d\mid n}\mu\left(  d\right)  q^{n/d}  &  =\sum\limits_{e\mid
n}\mu\left(  e\right)  \underbrace{q^{n/e}}_{\substack{=\sum_{d\mid
n/e}\left(  \operatorname*{irr}d\right)  d\\\text{(by
(\ref{pf.lem.sol.polynomials-and-necklaces.a.5}), applied}\\\text{to
}n/e\text{ instead of }n\text{)}}}=\sum\limits_{e\mid n}\mu\left(  e\right)
\sum_{d\mid n/e}\left(  \operatorname*{irr}d\right)  d\\
&  =\underbrace{\sum\limits_{e\mid n}\sum_{d\mid n/e}}_{=\sum_{d\mid n}%
\sum_{e\mid n/d}}\mu\left(  e\right)  \left(  \operatorname*{irr}d\right)
d=\sum_{d\mid n}\underbrace{\sum_{e\mid n/d}\mu\left(  e\right)
}_{\substack{=\delta_{n/d,1}\\\text{(by
(\ref{pf.lem.sol.polynomials-and-necklaces.a.mob}), applied}\\\text{to
}n/d\text{ instead of }n\text{)}}}\left(  \operatorname*{irr}d\right)  d\\
&  =\sum_{d\mid n}\underbrace{\delta_{n/d,1}}_{=\delta_{n,d}}\left(
\operatorname*{irr}d\right)  d=\sum_{d\mid n}\delta_{n,d}\left(
\operatorname*{irr}d\right)  d=\left(  \operatorname*{irr}n\right)  n.
\end{align*}
Dividing this by $n$, we obtain $\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  q^{n/d}=\operatorname*{irr}n$. Now,
(\ref{pf.lem.sol.polynomials-and-necklaces.a.0}) yields%
\[
\left(  \text{the number of all irreducible monic degree-}n\text{ polynomials
in } \FF_{q}\left[  x\right]  \right)  =\operatorname*{irr}n=\dfrac
{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  q^{n/d}.
\]
This proves Lemma \ref{lem.sol.polynomials-and-necklaces.a}.
\end{proof}

Now, let us solve Exercise \ref{exe.polynomials-and-necklaces}(a). Let
$n\geq2$ be an integer. We know that $\left\vert {\mathcal{F}}_{n}\right\vert
$ is the number of irreducible monic degree-$n$ polynomials $f(x)\neq x$ in
$\FF_{q}\left[  x\right]  $ with nonzero constant term. Since the
irreducible monic degree-$n$ polynomials $f(x)\neq x$ in
$\FF_{q}\left[  x\right]  $ with nonzero constant term are precisely the
irreducible monic degree-$n$ polynomials in $ \FF_{q}\left[  x\right]
$\ \ \ \ \footnote{\textit{Proof.} Let $f$ be an irreducible polynomial monic
degree-$n$ polynomial in $ \FF_{q}\left[  x\right]  $. We will now show
that $f\left(  x\right)  \neq x$ and that $f$ has nonzero constant term.
\par
Since $f$ is a degree-$n$ polynomial, we have $\deg f=n\geq2>1=\deg x$, thus
$\deg f\neq\deg x$ and therefore $f\neq x$. In other words, $f\left(
x\right)  \neq x$.
\par
Now, let us assume (for the sake of contradiction) that the constant term of
$f$ is zero. Then, the polynomial $x$ divides $f$ in $\FF_{q}\left[
x \right]$. But since $f$ is irreducible, the polynomial $f$ is a scalar
multiple of every non-constant polynomial which divides $f$. In particular,
$f$ is a scalar multiple of $x$ (since $x$ is a non-constant polynomial which
divides $f$). Consequently, $\deg f=\deg x$, which contradicts $\deg f\neq\deg
x$. This contradiction proves that our assumption (that the constant term of
$f$ is zero) was wrong. Hence, the constant term of $f$ is nonzero. In other
words, $f$ has nonzero constant term.
\par
Now, let us forget that we fixed $f$. We thus have proven that every
irreducible monic degree-$n$ polynomial $f$ in $ \FF_{q}\left[
x\right]  $ satisfies $f\left(  x\right)  \neq x$ and has nonzero constant
term. Thus, all irreducible monic degree-$n$ polynomials in $\FF_{q}
\left[  x\right]  $ are irreducible monic degree-$n$ polynomials $f(x)\neq
x$ in $\FF_{q}\left[  x\right]  $ with nonzero constant term.
Combining this statement with the (obvious) converse statement (which states
that all irreducible monic degree-$n$ polynomials $f(x)\neq x$ in
$ \FF_q\left[  x\right]  $ with nonzero constant term are
irreducible monic degree-$n$ polynomials in $ \FF_{q}\left[  x\right]
$), we conclude that the irreducible monic degree-$n$ polynomials $f(x)\neq x$
in $ \FF_q\left[  x\right]  $ with nonzero constant term are
precisely the irreducible monic degree-$n$ polynomials in $\FF_q
\left[  x\right]  $. Qed.}, this statement rewrites as follows:
$\left\vert {\mathcal{F}}_{n}\right\vert $ is the number of irreducible monic
degree-$n$ polynomials in $ \FF_q\left[  x\right]  $. In other
words,%
\begin{align*}
\left\vert {\mathcal{F}}_{n}\right\vert  &  =\left(  \text{the number of all
irreducible monic degree-}n\text{ polynomials in } \FF_{q}\left[
x\right]  \right) \\
&  =\dfrac{1}{n}\sum_{d\mid n}\mu\left(  \dfrac{n}{d}\right)  q^{d}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.polynomials-and-necklaces.a}}\right)  .
\end{align*}
This solves Exercise \ref{exe.polynomials-and-necklaces}(a).

(b) We shall use the results of Exercise \ref{exe.words.necklaces} (which
provides a more systematic introduction to necklaces).

Let $\mathfrak{A}$ be the set $ \FF_{q}$. Consider the notion of an
$n$-necklace defined in Exercise \ref{exe.words.necklaces}. The
``$n$-necklaces'' (as defined in Exercise
\ref{exe.words.necklaces}) are precisely the ``necklaces with
$n$ beads of $q$ colors'' (as defined in Exercise
\ref{exe.polynomials-and-necklaces}(b))\footnote{\textit{Proof.} Consider the
group $C$, its generator $c$ and the action of $C$ on $\mathfrak{A}^{n}$ which
are defined in Exercise \ref{exe.words.necklaces}. Then, the
``$n$-necklaces'' (as defined in Exercise
\ref{exe.words.necklaces}) are the orbits of the $C$-action on $\mathfrak{A}%
^{n}$. In other words, the ``$n$-necklaces'' (as defined in
Exercise \ref{exe.words.necklaces}) are the equivalence
classes of $n$-tuples $\left(  a_{1},a_{2},...,a_{n}\right)  \in
\mathfrak{A}^{n}$ with respect to cyclic rotation (because $C$ acts on
$\mathfrak{A}^{n}$ by cyclic rotation). But the same can be said about the
``necklaces with $n$ beads of $q$ colors'' (as defined in
Exercise \ref{exe.polynomials-and-necklaces}(b)). Thus, the
``$n$-necklaces'' (as defined in Exercise
\ref{exe.words.necklaces}) are precisely the ``necklaces with
$n$ beads of $q$ colors'' (as defined in Exercise
\ref{exe.polynomials-and-necklaces}(b)), qed.}. Moreover, it is easy to see
that the ``aperiodic $n$-necklaces'' (as
defined in Exercise \ref{exe.words.necklaces}) are precisely the
``primitive necklaces with $n$ beads of $q$
colors'' (as defined in Exercise
\ref{exe.polynomials-and-necklaces}(b))\footnote{This follows from Exercise
\ref{exe.words.necklaces}(b) (because for an $n$-tuple $\left(  w_{1}%
,w_{2},...,w_{n}\right)  \in\mathfrak{A}^{n}$, the statement that no
nontrivial rotation (in $\ZZ / n \ZZ$) fixes $w$ is equivalent to
the statement that every $k\in\left\{  1,2,\ldots,n-1\right\}  $ satisfies
$\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  \neq
w$).}. Hence,%
\begin{align*}
&  \left(  \text{the number of all aperiodic }n\text{-necklaces}\right)  \\
&  =\left(  \text{the number of all primitive necklaces with }n\text{ beads of
}q\text{ colors}\right)  ,
\end{align*}
so that%
\begin{align*}
&  \left(  \text{the number of all primitive necklaces with }n\text{ beads of
}q\text{ colors}\right)  \\
&  =\left(  \text{the number of all aperiodic }n\text{-necklaces}\right)
=\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)  \underbrace{\left\vert
\mathfrak{A}\right\vert ^{n/d}}_{\substack{=q^{n/d}\\\text{(since }\left\vert
\mathfrak{A}\right\vert =q\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.words.necklaces}(f)}\right)  \\
&  =\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)  q^{n/d}.
\end{align*}
This solves Exercise \ref{exe.polynomials-and-necklaces}(b).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.hall-alg.dominance}}

\textit{Solution to Exercise~\ref{exe.hall-alg.dominance}.} Let us
first notice that any $k \in \NN$ and any $k$ partitions
$\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}$ satisfy
\begin{equation}
\triv_{J_{\lambda^{(1)}}}\triv_{J_{\lambda^{(2)}}}
\cdots\triv_{J_{\lambda^{(k)}}}
= \sum_{\lambda\in\Par}
g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}}^{\lambda}
\left( q\right) \triv_{J_{\lambda}}
= \sum_{\mu\in\Par}
g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}}^{\mu}
\left(  q\right) \triv_{J_{\mu}}
\label{sol.hall-alg.dominance.1}
\end{equation}
(here, we renamed the summation index $\lambda$ as $\mu$). Any two partitions
$\mu$ and $\nu$ satisfy
\begin{equation}
\triv_{J_{\mu}}\triv_{J_{\nu}}
= \sum_{\lambda\in \Par} g_{\mu,\nu}^{\lambda} \left(q\right)
\triv_{J_{\lambda}}
= \sum_{\tau\in \Par} g_{\mu,\nu}^{\tau} \left(q\right)
\triv_{J_{\tau}}
\label{sol.hall-alg.dominance.2}
\end{equation}
(here, we renamed the summation index $\lambda$ as $\tau$).

(a) We shall prove the statement of Exercise \ref{exe.hall-alg.dominance}(a)
by induction over $k$. The base cases ($k=0$ and $k=1$) are left to the
reader. We will now handle the induction step. So let us solve Exercise
\ref{exe.hall-alg.dominance}(a) for some positive integer $k>1$, assuming (as
the induction hypothesis) that Exercise \ref{exe.hall-alg.dominance}(a) is
already solved for $k-1$ instead of $k$.

From (\ref{sol.hall-alg.dominance.1}), we obtain
\begin{align*}
&  \sum_{\mu\in \Par }g_{\lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k)}}^{\mu}\left(  q\right)  \triv_{J_{\mu}}\\
&  =\triv_{J_{\lambda^{(1)}}}\triv_{J_{\lambda^{(2)}}}%
\cdots\triv_{J_{\lambda^{(k)}}}
=\underbrace{\triv_{J_{\lambda^{(1)}}}
\triv_{J_{\lambda^{(2)}}}\cdots\triv_{J_{\lambda^{(k-1)}}}
}_{\substack{=\sum_{\mu\in\Par}
g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)
\triv_{J_{\mu}}\\\text{(by (\ref{sol.hall-alg.dominance.1}), applied
to }k-1\text{ instead of }k\text{)}}}\cdot\triv_{J_{\lambda^{(k)}}}\\
&  =\left(  \sum_{\mu\in \Par }g_{\lambda^{(1)},\lambda
^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)  \triv_{J_{\mu}%
}\right)  \cdot\triv_{J_{\lambda^{(k)}}}=\sum_{\mu\in
 \Par }g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}}%
^{\mu}\left(  q\right)  \underbrace{\triv_{J_{\mu}}
\triv_{J_{\lambda^{(k)}}}}_{\substack{=\sum_{\tau\in \Par }%
g_{\mu,\lambda^{(k)}}^{\tau}\left(  q\right)  \triv_{J_{\tau}}\\
\text{(by (\ref{sol.hall-alg.dominance.2}), applied to }\nu=\lambda
^{(k)}\text{)}}}\\
&  =\sum_{\mu\in \Par }g_{\lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)  \sum_{\tau\in
 \Par }g_{\mu,\lambda^{(k)}}^{\tau}\left(  q\right)
\triv_{J_{\tau}}=\sum_{\tau\in \Par }\sum_{\mu
\in \Par }g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}%
}^{\mu}\left(  q\right)  g_{\mu,\lambda^{(k)}}^{\tau}\left(  q\right)
\triv_{J_{\tau}}.
\end{align*}
Comparing coefficients in front of $\triv_{J_{\lambda}}$ on both sides
of this equality, we obtain%
\begin{equation}
g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}}^{\lambda}\left(
q\right)  =\sum_{\mu\in \Par }g_{\lambda^{(1)},\lambda
^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)  g_{\mu,\lambda^{(k)}%
}^{\lambda}\left(  q\right)  .\label{sol.hall-alg.dominance.a.rec}%
\end{equation}


Our $\mathbb{F}_{q}$-vector space $V$ is $n$-dimensional. Hence, we can WLOG
assume that $V=\mathbb{F}_{q}^{n}$ (because nothing changes if we map $V$
isomorphically to $\mathbb{F}_{q}^{n}$ and change $g$ accordingly). Assume this.

In the following, an \emph{ender} will mean a $g$-stable $\mathbb{F}_{q}%
$-vector subspace $W\subset{\mathbb{F}}_{q}^{n}$ for which the induced map
$\bar{g}$ on the quotient space ${\mathbb{F}}_{q}^{n}/W$ has Jordan type
$\lambda^{(k)}$. Notice that if $0=V_{0}\subset V_{1}\subset V_{2}%
\subset\cdots\subset V_{k}=V$ is a $\left(  \lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k)}\right)  $-compatible $g$-flag, then $V_{k-1}$ is an
ender (because the definition of a $\left(  \lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k)}\right)  $-compatible $g$-flag shows that the
endomorphism of $\underbrace{\mathbb{F}_{q}^{n}}_{=V=V_{k}}/V_{k-1}%
=V_{k}/V_{k-1}$ induced by $g$ has Jordan type $\lambda^{(k)}$). (This is why
we have chosen the name ``ender'' -- an ender
is the last proper subspace in a $\left(  \lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k)}\right)  $-compatible $g$-flag.)

Whenever $h$ is a unipotent endomorphism of a finite-dimensional vector space,
we let $\operatorname*{type}h$ denote the Jordan type of $h$.

Proposition \ref{prop.hall-alg.product} (applied to $\nu=\lambda^{(k)}$)
yields that for every $\mu\in \Par $, the number $g_{\mu
,\lambda^{(k)}}^{\lambda}\left(  q\right)  $ counts the $g$-stable
${\mathbb{F}}_{q}$-subspaces $W\subset{\mathbb{F}}_{q}^{n}$ for which the
restriction $g|W$ acts with Jordan type $\mu$, and the induced map $\bar{g}$
on the quotient space ${\mathbb{F}}_{q}^{n}/W$ has Jordan type $\lambda^{(k)}%
$\ \ \ \ \footnote{Note that the variable that I am calling $W$ here has been
denoted by $V$ in Proposition \ref{prop.hall-alg.product}.}. In other words,
for every $\mu\in \Par $, the number $g_{\mu,\lambda^{(k)}%
}^{\lambda}\left(  q\right)  $ counts the enders $W$ for which the restriction
$g|W$ acts with Jordan type $\mu$. In other words, for every $\mu
\in \Par $, the number $g_{\mu,\lambda^{(k)}}^{\lambda}\left(
q\right)  $ counts the enders $W$ for which $\operatorname*{type}\left(
g|W\right)  =\mu$. In other words, for every $\mu\in \Par $, we
have%
\[
g_{\mu,\lambda^{(k)}}^{\lambda}\left(  q\right)  =\sum\limits_{\substack{W\text{ is
an ender;}\\\operatorname*{type}\left(  g|W\right)  =\mu}}1.
\]
Hence, (\ref{sol.hall-alg.dominance.a.rec}) becomes%
\begin{align}
g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}}^{\lambda}\left(
q\right)   &  =\sum_{\mu\in \Par }g_{\lambda^{(1)},\lambda
^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)  \underbrace{g_{\mu
,\lambda^{(k)}}^{\lambda}\left(  q\right)  }_{=\sum\limits_{\substack{W\text{ is an
ender;}\\\operatorname*{type}\left(  g|W\right)  =\mu}}1}=\sum_{\mu
\in \Par }g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}%
}^{\mu}\left(  q\right)  \sum\limits_{\substack{W\text{ is an ender;}%
\\\operatorname*{type}\left(  g|W\right)  =\mu}}1\nonumber\\
&  =\sum_{\mu\in \Par }\sum\limits_{\substack{W\text{ is an
ender;}\\\operatorname*{type}\left(  g|W\right)  =\mu}}g_{\lambda
^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)
=\sum_{W\text{ is an ender}}g_{\lambda^{(1)},\lambda^{(2)},\ldots
,\lambda^{(k-1)}}^{\operatorname*{type}\left(  g|W\right)  }\left(  q\right)
.\label{sol.hall-alg.dominance.a.3}%
\end{align}


Now, let us fix an ender $W$. By the definition of an ender, this $W$ is a
$g$-stable ${\mathbb{F}}_{q}$-subspace of ${\mathbb{F}}_{q}^{n}$ for which the
induced map $\bar{g}$ on the quotient space ${\mathbb{F}}_{q}^{n}/W$ has
Jordan type $\lambda^{(k)}$. Let $m=\dim W$.

By the induction hypothesis, we can apply Exercise
\ref{exe.hall-alg.dominance}(a) to $\operatorname*{type}\left(  g|W\right)  $,
$m$, $W$, $g|W$, $k-1$ and $\left(  \lambda^{(1)},\lambda^{(2)},\ldots
,\lambda^{(k-1)}\right)  $ instead of $\lambda$, $n$, $V$, $g$, $k$ and
$\left(  \lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}\right)  $. As a
result, we conclude that
\begin{equation}
g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}}^{\operatorname*{type}%
\left(  g|W\right)  }\left(  q\right)  \text{ is the number of }\left(
\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}\right)  \text{-compatible
}g|W\text{-flags.}\label{sol.hall-alg.dominance.a.4}%
\end{equation}


Now, forget that we fixed $W$. Combining (\ref{sol.hall-alg.dominance.a.3})
with (\ref{sol.hall-alg.dominance.a.4}), we see that $g_{\lambda^{(1)}%
,\lambda^{(2)},\ldots,\lambda^{(k)}}^{\lambda}\left(  q\right)  $ is the
number of all pairs $\left(  W,\mathbf{F}\right)  $, where $W$ is an ender and
$\mathbf{F}$ is a $\left(  \lambda^{(1)},\lambda^{(2)},\ldots,\lambda
^{(k-1)}\right)  $-compatible $g|W$-flag. But (in order to complete the
induction step) we have to prove that $g_{\lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k)}}^{\lambda}\left(  q\right)  $ is the number of $\left(
\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}\right)  $-compatible
$g$-flags. So an obvious way to prove this would be to find a bijection
between $\left(  \lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}\right)
$-compatible $g$-flags and pairs $\left(  W,\mathbf{F}\right)  $, where $W$ is
an ender and $\mathbf{F}$ is a $\left(  \lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k-1)}\right)  $-compatible $g|W$-flag. This bijection is
very easy to define: It sends a $\left(  \lambda^{(1)},\lambda^{(2)}%
,\ldots,\lambda^{(k)}\right)  $-compatible $g$-flag $0=V_{0}\subset
V_{1}\subset V_{2}\subset\cdots\subset V_{k}=V$ to the pair $\left(
V_{k-1},\mathbf{F}\right)  $, where $\mathbf{F}$ is the $\left(  \lambda
^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}\right)  $-compatible $g|V_{k-1}%
$-flag $0=V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset V_{k-1}=V_{k-1}%
$. Showing that this bijection is well-defined and bijective is completely
straightforward\footnote{The inverse map takes a pair $\left(  W,\mathbf{F}%
\right)  $, where $W$ is an ender and $\mathbf{F}$ is a $\left(  \lambda
^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}\right)  $-compatible $g|W$-flag
$0=V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset V_{k-1}=W$, and maps it
to the $\left(  \lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}\right)
$-compatible $g$-flag $0=V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset
V_{k-1}\subset V=V$.}; thus, the induction step is complete, and the solution
of Exercise \ref{exe.hall-alg.dominance}(a) is finished.

(b) We shall prove the statement of Exercise \ref{exe.hall-alg.dominance}(b)
by induction over $k$. The base cases ($k=0$ and $k=1$) are left to the
reader. We will now handle the induction step. So let us solve Exercise
\ref{exe.hall-alg.dominance}(b) for some positive integer $k>1$, assuming (as
the induction hypothesis) that Exercise \ref{exe.hall-alg.dominance}(b) is
already solved for $k-1$ instead of $k$.

We need to prove that $g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}%
}^{\lambda}\left(  q\right)  =0$ unless $\left\vert \lambda^{(1)}\right\vert
+\left\vert \lambda^{(2)}\right\vert +\cdots+\left\vert \lambda^{(k)}%
\right\vert =\left\vert \lambda\right\vert $ and $\lambda^{(1)}+\lambda
^{(2)}+\cdots+\lambda^{(k)}\triangleright\lambda$. In other words, we need to
prove that if $g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}}^{\lambda
}\left(  q\right)  \neq0$, then
\begin{equation}
\left\vert \lambda^{(1)}\right\vert +\left\vert \lambda^{(2)}\right\vert
+\cdots+\left\vert \lambda^{(k)}\right\vert =\left\vert \lambda\right\vert
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \lambda^{(1)}+\lambda
^{(2)}+\cdots+\lambda^{(k)}\triangleright\lambda
.\label{sol.hall-alg.dominance.b.goal}%
\end{equation}
So let us assume that $g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}%
}^{\lambda}\left(  q\right)  \neq0$. Due to
(\ref{sol.hall-alg.dominance.a.rec}), this rewrites as
\[
\sum_{\mu\in \Par }g_{\lambda^{(1)},\lambda^{(2)},\ldots
,\lambda^{(k-1)}}^{\mu}\left(  q\right)  g_{\mu,\lambda^{(k)}}^{\lambda
}\left(  q\right)  \neq0.
\]
Hence, there exists a $\mu\in \Par $ satisfying $g_{\lambda
^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)  \neq0$ and
$g_{\mu,\lambda^{(k)}}^{\lambda}\left(  q\right)  \neq0$. Consider this $\mu$.

By the induction hypothesis, we can apply Exercise
\ref{exe.hall-alg.dominance}(b) to $\mu$, $k-1$ and $\left(  \lambda
^{(1)},\lambda^{(2)},...,\lambda^{(k-1)}\right)  $ instead of $\lambda$, $k$
and $\left(  \lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k)}\right)  $. Thus,
we conclude that $g_{\lambda^{(1)},\lambda^{(2)},\ldots,\lambda^{(k-1)}}^{\mu
}\left(  q\right)  =0$ unless $\left\vert \lambda^{(1)}\right\vert +\left\vert
\lambda^{(2)}\right\vert +\cdots+\left\vert \lambda^{(k-1)}\right\vert
=\left\vert \mu\right\vert $ and $\lambda^{(1)}+\lambda^{(2)}+\cdots
+\lambda^{(k-1)}\triangleright\mu$. Since we have $g_{\lambda^{(1)}%
,\lambda^{(2)},\ldots,\lambda^{(k-1)}}^{\mu}\left(  q\right)  \neq0$, we
therefore must have $\left\vert \lambda^{(1)}\right\vert +\left\vert
\lambda^{(2)}\right\vert +\cdots+\left\vert \lambda^{(k-1)}\right\vert
=\left\vert \mu\right\vert $ and $\lambda^{(1)}+\lambda^{(2)}+\cdots
+\lambda^{(k-1)}\triangleright\mu$.

But let $n=\left\vert \lambda\right\vert $. Fix a unipotent endomorphism $g$
of $\mathbb{F}_{q}^{n}$ having Jordan type $\lambda$ (such a $g$ clearly
exists). The number $g_{\mu,\lambda^{(k)}}^{\lambda}\left(  q\right)  $ counts
the $g$-stable ${\mathbb{F}}_{q}$-subspaces $W\subset{\mathbb{F}}_{q}^{n}$ for
which the restriction $g|W$ acts with Jordan type $\mu$, and the induced map
$\bar{g}$ on the quotient space ${\mathbb{F}}_{q}^{n}/W$ has Jordan type
$\lambda^{(k)}$ (by Proposition \ref{prop.hall-alg.product}, applied to
$\nu=\lambda^{(k)}$). Since $g_{\mu,\lambda^{(k)}}^{\lambda}\left(  q\right)
\neq0$, this yields that there exists such a subspace $W$. If we define a
nilpotent endomorphism $f$ of $\mathbb{F}_{q}^{n}$ by
$f=g-\id_{\mathbb{F}_{q}^{n}}$ (this is indeed nilpotent because $g$ is
unipotent), then this $W$ is also an $f$-stable $\mathbb{F}_{q}$-subspace of
$\mathbb{F}_{q}^{n}$ (because any $g$-stable subspace of $\mathbb{F}_{q}^{n}$
is $f$-stable) for which the restriction $f|W$ acts with Jordan type $\mu$
(since $f|W=g|W-\id_{W}$), and the induced map
$\overline{f}$ on the quotient space ${\mathbb{F}}_{q}^{n}/W$ has Jordan type
$\lambda^{(k)}$ (since $\overline{f}=\overline{g}-\id_{\FF_q^n/W}$).
Therefore, Exercise \ref{exe.liri.jordan}(b) (applied to
$\mathbb{K}=\mathbb{F}_{q}$, $V=\mathbb{F}_{q}^{n}$, $U=W$ and $\nu
=\lambda^{(k)}$) yields $c_{\mu,\lambda^{(k)}}^{\lambda}\neq0$. Consequently,
$\left\vert \lambda\right\vert =\left\vert \mu\right\vert +\left\vert
\lambda^{(k)}\right\vert $. Exercise \ref{exe.liri.simple}(c) (applied to
$\left\vert \lambda\right\vert $, $\left\vert \mu\right\vert $ and
$\lambda^{(k)}$ instead of $n$, $k$ and $\nu$) thus yields $\mu+\lambda
^{(k)}\triangleright\lambda\triangleright\mu\sqcup\lambda^{(k)}$.

Exercise \ref{exe.liri.simple}(d) (applied to $\left\vert \mu\right\vert $,
$\left\vert \lambda^{(k)}\right\vert $, $\lambda^{(1)}+\lambda^{(2)}%
+\cdots+\lambda^{(k-1)}$, $\mu$, $\lambda^{(k)}$ and $\lambda^{(k)}$ instead
of $n$, $m$, $\alpha$, $\beta$, $\gamma$ and $\delta$) yields%
\[
\left(  \lambda^{(1)}+\lambda^{(2)}+\cdots+\lambda^{(k-1)}\right)
+\lambda^{(k)}\triangleright\mu+\lambda^{(k)}
\]
(since $\lambda^{(1)}+\lambda^{(2)}+\cdots+\lambda^{(k-1)}\triangleright\mu$
and $\lambda^{(k)}\triangleright\lambda^{(k)}$).

Now,
\[
\left\vert \lambda^{(1)}\right\vert +\left\vert \lambda^{(2)}\right\vert
+\cdots+\left\vert \lambda^{(k)}\right\vert =\underbrace{\left\vert
\lambda^{(1)}\right\vert +\left\vert \lambda^{(2)}\right\vert +\cdots
+\left\vert \lambda^{(k-1)}\right\vert }_{=\left\vert \mu\right\vert
}+\left\vert \lambda^{(k)}\right\vert =\left\vert \mu\right\vert +\left\vert
\lambda^{(k)}\right\vert =\left\vert \lambda\right\vert
\]
and
\[
\lambda^{(1)}+\lambda^{(2)}+\cdots+\lambda^{(k)}=\left(  \lambda^{(1)}
+\lambda^{(2)}+\cdots+\lambda^{(k-1)}\right)  +\lambda^{(k)}\triangleright
\mu+\lambda^{(k)}\triangleright\lambda.
\]
Thus, we have proven (\ref{sol.hall-alg.dominance.b.goal}), and so the
induction step is complete. We thus have solved
Exercise~\ref{exe.hall-alg.dominance}(b).

(c) Let $n=\left\vert \lambda\right\vert $. Fix a unipotent endomorphism $g$
of $\mathbb{F}_{q}^{n}$ having Jordan type $\lambda$ (such a $g$ clearly exists).

Exercise \ref{exe.hall-alg.dominance}(a) (applied to $V=\mathbb{F}_{q}^{n}$,
$k=\ell$ and $\lambda^{(i)}=\left(  1^{\left(  \lambda
^{t}\right)  _{i}}\right)  $) shows that $g_{\left(  1^{\left(  \lambda
^{t}\right)  _{1}}\right)  ,\left(  1^{\left(  \lambda^{t}\right)  _{2}%
}\right)  ,\ldots,\left(  1^{\left(  \lambda^{t}\right)  _{\ell}}\right)
}^{\lambda}\left(  q\right)  $ is the number of $\left(  \left(  1^{\left(
\lambda^{t}\right)  _{1}}\right)  ,\left(  1^{\left(  \lambda^{t}\right)
_{2}}\right)  ,\ldots,\left(  1^{\left(  \lambda^{t}\right)  _{\ell}}\right)
\right)  $-compatible $g$-flags. Hence, in order to prove \newline
$g_{\left(
1^{\left(  \lambda^{t}\right)  _{1}}\right)  ,\left(  1^{\left(  \lambda
^{t}\right)  _{2}}\right)  ,\ldots,\left(  1^{\left(  \lambda^{t}\right)
_{\ell}}\right)  }^{\lambda}\left(  q\right)  \neq0$ (and thus, to solve
Exercise \ref{exe.hall-alg.dominance}(c)), it will be enough to prove that
there exists at least one $\left(  \left(  1^{\left(  \lambda^{t}\right)
_{1}}\right)  ,\left(  1^{\left(  \lambda^{t}\right)  _{2}}\right)
,\ldots,\left(  1^{\left(  \lambda^{t}\right)  _{\ell}}\right)  \right)
$-compatible $g$-flag.

Let $f=g-\id_{\mathbb{F}_{q}^{n}}$. Then, $f$ is a
nilpotent endomorphism of $\mathbb{F}_{q}^{n}$ having Jordan type $\lambda$
(since $g$ is a unipotent endomorphism of $\mathbb{F}_{q}^{n}$ having Jordan
type $\lambda$). Also, $g=f+\id_{\mathbb{F}_{q}^{n}}$
(since $f=g-\id_{\mathbb{F}_{q}^{n}}$).

Let us define a sequence $V_{0}\subset V_{1}\subset V_{2}\subset\cdots\subset
V_{\ell}$ of $\mathbb{F}_{q}$-vector subspaces of $\mathbb{F}_{q}^{n}$ by
setting%
\[
V_{i}=\ker\left(  f^{i}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  0,1,...,\ell\right\}  .
\]
Then, $V_{0}=\ker\left(  \underbrace{f^{0}}_{=\id}\right)
=\ker\operatorname*{id}=0$. Also, it follows readily from Exercise
\ref{exe.liri.jordan}(a) that%
\begin{equation}
\dim\left(  \ker\left(  f^{k}\right)  \right)  =\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\ldots+\left(  \lambda^{t}\right)
_{k}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\NN .
\label{sol.hall-alg.dominance.c.1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hall-alg.dominance.c.1}):} Let
$k\in \NN$. Let $N\in\mathbb{F}_{q}^{n\times n}$ be the matrix
representing the endomorphism $f$ of $\mathbb{F}_{q}^{n}$. Then, $N$ is
nilpotent (since $f$ is nilpotent) and has Jordan type $\lambda$ (since $f$
has Jordan type $\lambda$), and thus satisfies $\dim\left(  \ker\left(
N^{k}\right)  \right)  =\left(  \lambda^{t}\right)  _{1}+\left(  \lambda
^{t}\right)  _{2}+\ldots+\left(  \lambda^{t}\right)  _{k}$ (by Exercise
\ref{exe.liri.jordan}(a)). But since $N$ is a matrix representing the map $f$,
we have $\dim\left(  \ker\left(  N^{k}\right)  \right)  =\dim\left(
\ker\left(  f^{k}\right)  \right)  $, so that $\dim\left(  \ker\left(
f^{k}\right)  \right)  =\dim\left(  \ker\left(  N^{k}\right)  \right)
=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}%
+\ldots+\left(  \lambda^{t}\right)  _{k}$. This proves
(\ref{sol.hall-alg.dominance.c.1}).} Applying this to $k=\ell$, we obtain%
\begin{align*}
\dim\left(  \ker\left(  f^{\ell}\right)  \right)    & =\left(  \lambda
^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\ldots+\left(  \lambda
^{t}\right)  _{\ell}=\left\vert \lambda^{t}\right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda^{t}=\left(  \left(
\lambda^{t}\right)  _{1},\left(  \lambda^{t}\right)  _{2},\ldots,\left(
\lambda^{t}\right)  _{\ell}\right)  \right)  \\
& =\left\vert \lambda\right\vert =n=\dim\left(  \mathbb{F}_{q}^{n}\right)
\end{align*}
which yields $\ker\left(  f^{\ell}\right)  =\mathbb{F}_{q}^{n}$ (since
$\ker\left(  f^{\ell}\right)  \subset\mathbb{F}_{q}^{n}$). Thus, $V_{\ell
}=\ker\left(  f^{\ell}\right)  =\mathbb{F}_{q}^{n}$. Since $V_{0}=0$ and
$V_{\ell}=\mathbb{F}_{q}^{n}$, our sequence $V_{0}\subset V_{1}\subset
V_{2}\subset\cdots\subset V_{\ell}$ thus can be written as $0=V_{0}\subset
V_{1}\subset V_{2}\subset\cdots\subset V_{\ell}=\mathbb{F}_{q}^{n}$.

For every $i\in\left\{  0,1,...,\ell\right\}  $, the $\mathbb{F}_{q}$-vector
subspace $V_{i}$ of $\mathbb{F}_{q}^{n}$ is $g$-invariant (since $V_{i}%
=\ker\left(  f^{i}\right)  $ is $f$-invariant and thus $g$-invariant). Also,
for every $i\in\left\{  1,2,...,\ell\right\}  $, we have $\underbrace{V_{i}%
}_{=\ker\left(  f^{i}\right)  }/\underbrace{V_{i-1}}_{=\ker\left(
f^{i-1}\right)  }=\ker\left(  f^{i}\right)  /\ker\left(  f^{i-1}\right)  $;
thus, the endomorphism of $V_{i}/V_{i-1}$ induced by $f$ is the zero map, and
therefore the endomorphism of $V_{i}/V_{i-1}$ induced by $g$ is the identity
map (since $g=f+\id_{\mathbb{F}_{q}^{n}}$). Hence, for
every $i\in\left\{  1,2,...,\ell\right\}  $, this latter endomorphism has
Jordan type $\left(  1^{\dim\left(  V_{i}/V_{i-1}\right)  }\right)  $. But
since every $i\in\left\{  1,2,...,\ell\right\}  $ satisfies%
\begin{align*}
\dim\left(  \underbrace{V_{i}/V_{i-1}}_{=\ker\left(  f^{i}\right)
/\ker\left(  f^{i-1}\right)  }\right)    & =\dim\left(  \ker\left(
f^{i}\right)  /\ker\left(  f^{i-1}\right)  \right)  =\underbrace{\dim\left(
\ker\left(  f^{i}\right)  \right)  }_{\substack{=\left(  \lambda^{t}\right)
_{1}+\left(  \lambda^{t}\right)  _{2}+\ldots+\left(  \lambda^{t}\right)
_{i}\\\text{(by (\ref{sol.hall-alg.dominance.c.1}),}\\\text{applied to
}k=i\text{)}}}-\underbrace{\dim\left(  \ker\left(  f^{i-1}\right)  \right)
}_{\substack{=\left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\ldots+\left(  \lambda^{t}\right)  _{i-1}\\\text{(by
(\ref{sol.hall-alg.dominance.c.1}),}\\\text{applied to }k=i-1\text{)}}}\\
& =\left(  \left(  \lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)
_{2}+\ldots+\left(  \lambda^{t}\right)  _{i}\right)  -\left(  \left(
\lambda^{t}\right)  _{1}+\left(  \lambda^{t}\right)  _{2}+\ldots+\left(
\lambda^{t}\right)  _{i-1}\right)  =\left(  \lambda^{t}\right)  _{i},
\end{align*}
this rewrites as follows: For every $i\in\left\{  1,2,...,\ell\right\}  $, the
endomorphism of $V_{i}/V_{i-1}$ induced by $g$ has Jordan type $\left(
1^{\left(  \lambda^{t}\right)  _{i}}\right)  $.

Altogether, we now know that $0=V_{0}\subset V_{1}\subset V_{2}\subset
\cdots\subset V_{\ell}=\mathbb{F}_{q}^{n}$ is a sequence of $g$-invariant
${\mathbb{F}}_{q}$-vector subspaces $V_{i}$ of $\mathbb{F}_{q}^{n}$ such that
for every $i\in\left\{  1,2,\ldots,\ell\right\}  $, the endomorphism of
$V_{i}/V_{i-1}$ induced by $g$ has Jordan type $\left(  1^{\left(  \lambda
^{t}\right)  _{i}}\right)  $. In other words, $0=V_{0}\subset V_{1}\subset
V_{2}\subset\cdots\subset V_{\ell}=\mathbb{F}_{q}^{n}$ is a $\left(  \left(
1^{\left(  \lambda^{t}\right)  _{1}}\right)  ,\left(  1^{\left(  \lambda
^{t}\right)  _{2}}\right)  ,\ldots,\left(  1^{\left(  \lambda^{t}\right)
_{\ell}}\right)  \right)  $-compatible $g$-flag (according the definition of
the latter notion). Hence, there exists at least one \newline
$\left(  \left(
1^{\left(  \lambda^{t}\right)  _{1}}\right)  ,\left(  1^{\left(  \lambda
^{t}\right)  _{2}}\right)  ,\ldots,\left(  1^{\left(  \lambda^{t}\right)
_{\ell}}\right)  \right)  $-compatible $g$-flag. Exercise
\ref{exe.hall-alg.dominance}(c) is solved.

(d) Write the partition $\lambda$ as $\lambda=\left(  \lambda_{1},\lambda
_{2},...,\lambda_{\ell}\right)  $ with $\ell=\ell\left(  \lambda\right)  $.
Then, $\left(  \lambda^{t}\right)  ^{t}=\lambda=\left(  \lambda_{1}
,\lambda_{2},...,\lambda_{\ell}\right)  $. Hence, Exercise
\ref{exe.hall-alg.dominance}(c) (applied to $\lambda^{t}$ instead of $\lambda
$) yields
\begin{equation}
g_{\left(  1^{\lambda_{1}}\right)  ,\left(  1^{\lambda_{2}}\right)
,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\lambda^{t}}\left(  q\right)
\neq 0.
\label{sol.hall-alg.dominance.d.0}
\end{equation}


But since $\lambda=\left(  \lambda_{1},\lambda_{2},...,\lambda_{\ell}\right)
$, we have $e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots e_{\lambda_{\ell
}}$ and thus
\begin{align}
\varphi\left(  e_{\lambda}\right)    & =\varphi\left(  e_{\lambda_{1}%
}e_{\lambda_{2}}\cdots e_{\lambda_{\ell}}\right)  \nonumber\\
&=\varphi\left(
e_{\lambda_{1}}\right)  \varphi\left(  e_{\lambda_{2}}\right)  \cdots
\varphi\left(  e_{\lambda_{\ell}}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\varphi\text{ is a } \CC \text{-algebra homomorphism}%
\right)  \nonumber\\
& =\prod_{i=1}^{\ell}\underbrace{\varphi\left(  e_{\lambda_{i}}\right)
}_{\substack{=q^{\dbinom{\lambda_{i}}{2}}\triv_{J_{(1^{\lambda_{i}})}%
}\\\text{(since Theorem \ref{Sym-to-Hall-isomorphism-thm} yields}%
\\\varphi\left(  e_{p}\right)  =q^{\dbinom{p}{2}}\triv_{J_{(1^{p})}%
}\text{ for every }p\in \NN \text{)}}}=\prod_{i=1}^{\ell}\left(
q^{\dbinom{\lambda_{i}}{2}}\triv_{J_{(1^{\lambda_{i}})}}\right)
\nonumber\\
& =\left(  \prod_{i=1}^{\ell}q^{\dbinom{\lambda_{i}}{2}}\right)
\underbrace{\prod_{i=1}^{\ell}\triv_{J_{(1^{\lambda_{i}})}}%
}_{\substack{=\triv_{J_{(1^{\lambda_{1}})}}
\triv_{J_{(1^{\lambda_{2}})}}\cdots\triv_{J_{(1^{\lambda_{\ell}})}}\\
=\sum_{\mu\in\Par}g_{\left(  1^{\lambda_{1}}\right)  ,\left(
1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\mu
}\left(  q\right)  \triv_{J_{\mu}}\\\text{(by
(\ref{sol.hall-alg.dominance.1}), applied to }k=\ell\text{ and }\lambda
^{(i)}=\left(  1^{\lambda_{i}}\right)  \text{)}}}\nonumber\\
& =\left(  \prod_{i=1}^{\ell}q^{\dbinom{\lambda_{i}}{2}}\right)  \sum_{\mu
\in \Par }g_{\left(  1^{\lambda_{1}}\right)  ,\left(
1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\mu
}\left(  q\right)  \triv_{J_{\mu}}.
\label{sol.hall-alg.dominance.d.1}
\end{align}


Now, we notice that
\begin{equation}
\lambda^{t}=\left(  1^{\lambda_{1}}\right)  +\left(  1^{\lambda_{2}}\right)
+\cdots+\left(  1^{\lambda_{\ell}}\right)
\label{sol.hall-alg.dominance.d.2}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.hall-alg.dominance.d.2}):} In the
following, we use the so-called \dfn{Iverson bracket notation}: For every
assertion $\mathcal{A}$, we let $\left[  \mathcal{A}\right]  $ denote the
integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$. (This integer is called the \dfn{truth value} of
$\mathcal{A}$.)
\par
For every $p\in \NN$ and $i\in\left\{  1,2,3,...\right\}  $, we have%
\begin{equation}
\left(  1^{p}\right)  _{i}=\left[  p\geq i\right] .
\label{sol.hall-alg.dominance.d.2.pf.1}
\end{equation}
Now, every $i\in\left\{  1,2,3,...\right\}  $ satisfies%
\begin{align*}
\left(  \left(  1^{\lambda_{1}}\right)  +\left(  1^{\lambda_{2}}\right)
+\cdots+\left(  1^{\lambda_{\ell}}\right)  \right)  _{i}  & =\left(
1^{\lambda_{1}}\right)  _{i}+\left(  1^{\lambda_{2}}\right)  _{i}%
+\cdots+\left(  1^{\lambda_{\ell}}\right)  _{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\mu+\nu\text{ for two partitions }\mu\text{ and
}\nu\right)  \\
& =\sum_{k=1}^{\ell}\underbrace{\left(  1^{\lambda_{k}}\right)  _{i}
}_{\substack{=\left[  \lambda_{k}\geq i\right]  \\\text{(by
(\ref{sol.hall-alg.dominance.d.2.pf.1}))}}}=\sum_{k=1}^{\ell}\left[
\lambda_{k}\geq i\right]  \\
& =\left\vert \left\{  j\in\left\{  1,2,...,\ell\right\}  \ \mid\ \lambda
_{j}\geq i\right\}  \right\vert =\left(  \lambda^{t}\right)  _{i}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.partition.transpose})}\right)  .
\end{align*}
Hence, $\left(  1^{\lambda_{1}}\right)  +\left(  1^{\lambda_{2}}\right)
+\cdots+\left(  1^{\lambda_{\ell}}\right)  =\lambda^{t}$, qed.}. Now, for
every partition $\mu$, we have $g_{\left(  1^{\lambda_{1}}\right)  ,\left(
1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\mu
}\left(  q\right)  =0$ unless $\mu\in\Par_{n}$ and
$\lambda^{t}\triangleright\mu$\ \ \ \ \footnote{\textit{Proof.} Let $\mu$ be a
partition. Exercise \ref{exe.hall-alg.dominance}(b) (applied to $\ell$,
$\left(  1^{\lambda_{i}}\right)  $ and $\mu$ instead of $k$, $\lambda^{(i)}$
and $\lambda$) shows that we have $g_{\left(  1^{\lambda_{1}}\right)  ,\left(
1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\mu
}\left(  q\right)  =0$ unless $\left\vert \left(  1^{\lambda_{1}}\right)
\right\vert +\left\vert \left(  1^{\lambda_{2}}\right)  \right\vert
+\cdots+\left\vert \left(  1^{\lambda_{\ell}}\right)  \right\vert =\left\vert
\mu\right\vert $ and $\left(  1^{\lambda_{1}}\right)  +\left(  1^{\lambda_{2}%
}\right)  +\cdots+\left(  1^{\lambda_{\ell}}\right)  \triangleright\mu$. Since
$\left\vert \left(  1^{\lambda_{1}}\right)  \right\vert +\left\vert \left(
1^{\lambda_{2}}\right)  \right\vert +\cdots+\left\vert \left(  1^{\lambda
_{\ell}}\right)  \right\vert =\left\vert \underbrace{\left(  1^{\lambda_{1}%
}\right)  +\left(  1^{\lambda_{2}}\right)  +\cdots+\left(  1^{\lambda_{\ell}%
}\right)  }_{\substack{=\lambda^{t}\\\text{(by
(\ref{sol.hall-alg.dominance.d.2}))}}}\right\vert =\left\vert \lambda
^{t}\right\vert =\left\vert \lambda\right\vert =n$ and $\left(  1^{\lambda
_{1}}\right)  +\left(  1^{\lambda_{2}}\right)  +\cdots+\left(  1^{\lambda
_{\ell}}\right)  =\lambda^{t}$ (by (\ref{sol.hall-alg.dominance.d.2})), this
rewrites as follows: We have $g_{\left(  1^{\lambda_{1}}\right)  ,\left(
1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\mu
}\left(  q\right)  =0$ unless $n=\left\vert \mu\right\vert $ and $\lambda
^{t}\triangleright\mu$. In other words, we have $g_{\left(  1^{\lambda_{1}%
}\right)  ,\left(  1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}%
}\right)  }^{\mu}\left(  q\right)  =0$ unless $\mu \in \Par_n$ and
$\lambda^{t}\triangleright\mu$, qed.}. Hence, we can
replace the summation sign
``$\sum_{\mu \in \Par}$'' on the right hand side of
(\ref{sol.hall-alg.dominance.d.1}) by a more restricted summation
``$\sum_{\mu\in \Par_{n};\ \lambda^{t}%
\triangleright\mu}$'' without changing the value of the sum
(since all addends that we lose are $0$). Thus,
(\ref{sol.hall-alg.dominance.d.1}) rewrites as
\begin{align*}
\varphi\left(  e_{\lambda}\right)    & =\left(  \prod_{i=1}^{\ell}%
q^{\dbinom{\lambda_{i}}{2}}\right)  \sum_{\mu\in \Par %
_{n};\ \lambda^{t}\triangleright\mu}g_{\left(  1^{\lambda_{1}}\right)
,\left(  1^{\lambda_{2}}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)
}^{\mu}\left(  q\right)  \triv_{J_{\mu}}\\
& =\sum_{\mu\in \Par_{n};\ \lambda^{t}\triangleright\mu}\left(
\prod_{i=1}^{\ell}q^{\dbinom{\lambda_{i}}{2}}\right)  g_{\left(
1^{\lambda_{1}}\right)  ,\left(  1^{\lambda_{2}}\right)  ,\ldots,\left(
1^{\lambda_{\ell}}\right)  }^{\mu}\left(  q\right)  \triv_{J_{\mu}}.
\end{align*}
Setting $\alpha_{\lambda,\mu}=\left(  \prod_{i=1}^{\ell}q^{\dbinom{\lambda
_{i}}{2}}\right)  g_{\left(  1^{\lambda_{1}}\right)  ,\left(  1^{\lambda_{2}%
}\right)  ,\ldots,\left(  1^{\lambda_{\ell}}\right)  }^{\mu}\left(  q\right)
$, we can rewrite this as $\varphi\left(  e_{\lambda}\right)  =\sum_{\mu
\in \Par_{n};\ \lambda^{t}\triangleright\mu}\alpha_{\lambda,\mu
}\triv_{J_{\mu}}$. Thus, we will be done solving Exercise
\ref{exe.hall-alg.dominance}(d) as soon as we can prove the inequality
\[
\left(  \prod_{i=1}^{\ell}q^{\dbinom{\lambda_{i}}{2}}\right)  g_{\left(
1^{\lambda_{1}}\right)  ,\left(  1^{\lambda_{2}}\right)  ,\ldots,\left(
1^{\lambda_{\ell}}\right)  }^{\lambda^{t}}\left(  q\right)  \neq0.
\]
But the latter inequality follows from $q\neq0$ and
(\ref{sol.hall-alg.dominance.d.0}). Thus, Exercise
\ref{exe.hall-alg.dominance}(d) is solved.

(e) The map $\varphi : \Lambda_\CC \rightarrow \mathcal{H}$ is graded,
and thus, in order to prove that $\varphi$ is injective, it is enough to show
that the restriction $\varphi\mid_{\left( \Lambda_\CC \right) _n}$
of $\varphi$ to $\left( \Lambda_\CC \right) _n$ is injective for
every $n \in \NN$. So let us fix $n \in \NN$.

We know that $\left(  e_{\lambda}\right)  _{\lambda\in\Par_n}$ is a
basis of the $ \CC $-vector space $\left(
\Lambda_{ \CC }\right)  _{n}$.\ \ \ \ \footnote{This has been proven in
the proof of Proposition \ref{many-symmetric-function-bases-prop}.
(Alternatively, this can be easily concluded from Proposition
\ref{many-symmetric-function-bases-prop}.)} Hence,
$\left( e_{\lambda^{t}} \right) _{\lambda \in \Par_n}$ also is a basis of
the $ \CC $-vector space $\left( \Lambda_\CC \right) _n$.
Every $\lambda \in \Par_n$ satisfies
\[
\varphi\left(  e_{\lambda^{t}}\right)  =\sum_{\mu\in \Par_n;
\ \left(  \lambda^{t}\right)  ^{t}\triangleright\mu}\alpha_{\lambda
^{t},\mu}\triv_{J_{\mu}}
\]
for some coefficients $\alpha_{\lambda^{t},\mu}\in{ \CC }$ satisfying
$\alpha_{\lambda^{t},\left(  \lambda^{t}\right)  ^{t}}\neq0$ (according to
Exercise \ref{exe.hall-alg.dominance}(d)). In other words, every $\lambda
\in\Par_{n}$ satisfies
\begin{equation}
\varphi\left(  e_{\lambda^{t}}\right)  =\sum_{\mu\in \Par_n;
\ \lambda\triangleright\mu}\alpha_{\lambda^{t},\mu}
\triv_{J_{\mu}}
\label{sol.hall-alg.dominance.e.1}
\end{equation}
for some coefficients $\alpha_{\lambda^{t},\mu} \in \CC$ satisfying
$\alpha_{\lambda^{t},\lambda}\neq0$ (since $\left(  \lambda^{t}\right)
^{t}=\lambda$).

Now, regard the set $\Par_n$ as a poset with the smaller-or-equal
relation $\triangleright$.

The $ \CC $-vector space basis
$\left( e_{\lambda^{t}} \right) _{\lambda \in \Par_n}$
of $\left( \Lambda_\CC \right) _n$
and the $ \CC $-vector space basis
$\left( \triv_{J_{\lambda}} \right) _{\lambda \in \Par_n}$
of $\mathcal{H}_n$ are both indexed by the poset $\Par_n$, and the
$\Par_n \times \Par_n$-matrix that represents the map
$\varphi\mid_{\left( \Lambda_\CC\right) _n}$ with respect to these
bases\footnote{i.e., the $\Par_n \times \Par_n$-matrix whose
$\left(\mu, \lambda\right)$-th entry is the
$\triv_{J_{\mu}}$-coordinate of $\varphi\left(e_{\lambda^t}\right)$}
is triangular\footnote{See Definition~\ref{def.STmat.tria}
for the notation we are using here.}
(by \eqref{sol.hall-alg.dominance.e.1}). Furthermore, the
diagonal entries of this triangular matrix are nonzero
(due to $\alpha_{\lambda^{t},\lambda}\neq0$), and therefore
invertible (in $\CC$). Hence, this matrix is
invertibly triangular, and thus invertible\footnote{by
Proposition~\ref{prop.STmat.triangular}(d)}. Therefore,
the map $\varphi\mid_{\left( \Lambda_{\CC}\right)  _{n}}$ (which
is represented by this matrix) is invertible (as a linear map
$\left( \Lambda_{\CC}\right)  _{n}\rightarrow\mathcal{H}_{n}$)
and thus injective. This completes the solution to
Exercise~\ref{exe.hall-alg.dominance}(e).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.P-partition.stanley.altproof}}

\textit{Solution to Exercise~\ref{exe.P-partition.stanley.altproof}.}

\begin{proof}
[Alternative proof of Theorem~\ref{Stanley's-P-partition-theorem}.]Let $P$ be
a labelled poset.

First of all, let $f$ be any map $P\rightarrow\left\{  1,2,3,\ldots\right\}
$. We define a binary relation $\prec_{f}$ on the set $P$ by letting
$i\prec_{f}j$ hold if and only if
\[
\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  .
\]
It is straightforward to see that this binary relation $\prec_{f}$ is the
smaller relation of a total order on $P$. Let us define $\mathbf{w}\left(
f\right)  $ to be the set $P$ endowed with this total order. Thus,
$\mathbf{w}\left(  f\right)  =P$ as sets, but the smaller relation
$<_{\mathbf{w}\left(  f\right)  }$ of the totally ordered set $\mathbf{w}%
\left(  f\right)  $ is the relation $\prec_{f}$.

Let us now forget that we fixed $f$. Thus, for every map $f:P\rightarrow
\left\{  1,2,3,\ldots\right\}  $, we have constructed a binary relation
$\prec_{f}$ on the set $P$ and a totally ordered poset $\mathbf{w}\left(
f\right)  $. It is easy to see that, for every $f\in\mathcal{A}\left(
P\right)  $, we have%
\begin{equation}
\mathbf{w}\left(  f\right)  \in\mathcal{L}\left(  P\right)
\label{sol.P-partition.stanley.altproof.wflin}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.wflin}):}
Let $f\in\mathcal{A}\left(  P\right)  $. We need to show that $\mathbf{w}%
\left(  f\right)  \in\mathcal{L}\left(  P\right)  $. In other words, we need
to show that $\mathbf{w}\left(  f\right)  $ is a linear extension of $P$. In
order to show this, it clearly suffices to prove that every two elements
$i$ and $j$ of $P$ satisfying $i<_{P}j$ satisfy
$i<_{\mathbf{w}\left(  f\right)  }j$
(because we already know that $\mathbf{w}\left(  f\right)  $ is a totally
ordered set). So, let us fix two elements $i$ and $j$ of $P$ satisfying
$i<_{P}j$. We need to prove that $i<_{\mathbf{w}\left(  f\right)  }j$.
\par
We have $i<_{P}j$ and thus $i\neq j$. Hence, either $i<_{\ZZ}j$ or
$i>_{\ZZ}j$. In other words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $i<_{\ZZ}j$.
\par
\textit{Case 2:} We have $i>_{\ZZ}j$.
\par
Let us first consider Case 1. In this case, $i<_{\ZZ}j$. But $f$ is a
$P$-partition (since $f\in\mathcal{A}\left(  P\right)  $), and thus, by the
definition of a $P$-partition, we conclude that $f\left(  i\right)  \leq
f\left(  j\right)  $ (since $i<_{P}j$ and $i<_{\ZZ}j$). In other words,
either $f\left(  i\right)  <f\left(  j\right)  $ or $f\left(  i\right)
=f\left(  j\right)  $. Therefore, either $f\left(  i\right)  <f\left(
j\right)  $ or $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  $ (because we have $i<_{\ZZ}j$ by
assumption). In other words, $i\prec_{f}j$. This rewrites as $i<_{\mathbf{w}%
\left(  f\right)  }j$ (since the relation $<_{\mathbf{w}\left(  f\right)  }$
is the relation $\prec_{f}$). Thus, $i<_{\mathbf{w}\left(  f\right)  }j$ is
proven in Case 1.
\par
Let us now consider Case 2. In this case, $i>_{\ZZ}j$. But $f$ is a
$P$-partition (since $f\in\mathcal{A}\left(  P\right)  $), and thus, by the
definition of a $P$-partition, we conclude that $f\left(  i\right)  <f\left(
j\right)  $ (since $i<_{P}j$ and $i>_{\ZZ}j$). Therefore, either
$f\left(  i\right)  <f\left(  j\right)  $ or $\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  $. In other words,
$i\prec_{f}j$. This rewrites as $i<_{\mathbf{w}\left(  f\right)  }j$ (since
the relation $<_{\mathbf{w}\left(  f\right)  }$ is the relation $\prec_{f}$).
Thus, $i<_{\mathbf{w}\left(  f\right)  }j$ is proven in Case 2.
\par
We have now proven $i<_{\mathbf{w}\left(  f\right)  }j$ in both Cases 1 and 2;
thus, $i<_{\mathbf{w}\left(  f\right)  }j$ always holds. This completes the
proof of (\ref{sol.P-partition.stanley.altproof.wflin}).}.

Now, fix $w\in\mathcal{L}\left(  P\right)  $. Thus, $w$ is a linear extension
of $P$. That is, $w$ is a totally ordered set with ground set $P$, and extends
the poset $P$.

Let us first show that
\begin{equation}
\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  \subset\mathcal{A}\left(  w\right)
.\label{sol.P-partition.stanley.altproof.4}%
\end{equation}


[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4}):} Let
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $. Hence, $f$ is an element of $\mathcal{A}\left(
P\right)  $ and satisfies $\mathbf{w}\left(  f\right)  =w$. Therefore,
$w=\mathbf{w}\left(  f\right)  =P$ as sets.

Our next goal is to prove $f\in\mathcal{A}\left(  w\right)  $. In other words,
we want to prove that $f$ is a $w$-partition.

Indeed, we claim that%
\begin{equation}
\left(  \text{if } i \in w \text{ and } j \in w \text{ satisfy } i<_{w}j
\text{ and }i<_{\ZZ}j\text{, then }f\left(  i\right)
\leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.4.pf.1a}
\end{equation}
and
\begin{equation}
\left(  \text{if } i \in w \text{ and } j \in w \text{ satisfy } i<_{w}j
\text{ and }i>_{\ZZ}j\text{, then }f\left(  i\right)
< f\left(  j\right)  \right)  .
\label{sol.P-partition.stanley.altproof.4.pf.1b}
\end{equation}


Let us prove (\ref{sol.P-partition.stanley.altproof.4.pf.1a}) first. So let
$i$ and $j$ be two elements of $w$ satisfying $i<_{w}j$ and $i<_{\ZZ}j$.
We have $i<_{w}j$, thus $i<_{\mathbf{w}\left(  f\right)  }j$ (since
$w=\mathbf{w}\left(  f\right)  $), and thus $i\prec_{f}j$ (since the relation
$<_{\mathbf{w}\left(  f\right)  }$ is the relation $\prec_{f}$). By the
definition of $\prec_{f}$, this means that we have $\left(  f\left(  i\right)
<f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i<_{\ZZ}j\right)  \right)  $. From this, we immediately
obtain $f\left(  i\right)  \leq f\left(  j\right)  $. Thus,
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}) is proven.

The proof of (\ref{sol.P-partition.stanley.altproof.4.pf.1b}) is similar to
the proof that we just gave for
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}), but with a minor twist: In
order to derive $f\left(  i\right)  <f\left(  j\right)  $ from $\left(
f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $, we need
to recall the assumption $i>_{\ZZ}j$ (which rules out the possibility
$\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i<_{\ZZ}j\right) $).

Thus, both (\ref{sol.P-partition.stanley.altproof.4.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.4.pf.1b}) are proven.
In other words, $f$ is a
$w$-partition. In yet other words, $f\in\mathcal{A}\left(  w\right)  $.

Let us now forget that we fixed $f$. We thus have proven that every
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $ satisfies $f\in\mathcal{A}\left(  w\right)  $. In
other words, (\ref{sol.P-partition.stanley.altproof.4}) is proven.]

Let us next show that
\begin{equation}
\mathcal{A}\left(  w\right)  \subset\left\{  g\in\mathcal{A}\left(
P\right)  \ \mid\ \mathbf{w}\left(  g\right)  =w\right\}
.\label{sol.P-partition.stanley.altproof.5}%
\end{equation}


[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5}):} Let
$f\in\mathcal{A}\left(  w\right)  $. Thus, $f$ is a $w$-partition. We shall
next prove that \newline
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid
\ \mathbf{w}\left(  g\right)  =w\right\}  $.

Indeed, let us first make a general and trivial observation: If $Q$ and $R$
are two labelled posets such that $Q=R$ as sets, and if every two elements $i$
and $j$ of $Q$ satisfying $i<_{Q}j$ satisfy $i<_{R}j$ (that is, the poset $R$
is an extension of the poset $Q$), then every $R$-partition is a
$Q$-partition.\footnote{This is clear, because the requirements for an
$R$-partition are at least as strong as the requirements for a $Q$-partition.}
Applying this to $Q=P$ and $R=w$, we conclude that every
$w$-partition is a $P$-partition. Thus, $f$ is a $P$-partition (since $f$ is a
$w$-partition). In other words, $f\in\mathcal{A}\left(  P\right)  $.

Next, we want to check that $\mathbf{w}\left(  f\right)  =w$.

It is easy to check that every two elements $i$ and $j$ of $w$ satisfying
$i<_{w}j$ satisfy
\begin{equation}
i<_{\mathbf{w}\left(  f\right)  }%
j\label{sol.P-partition.stanley.altproof.5.pf.1c}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.1c}):}
Let $i$ and $j$ be two elements of $w$ satisfying $i<_{w}j$. We must prove
that $i<_{\mathbf{w}\left(  f\right)  }j$.
\par
We have $i<_{w}j$ and thus $i\neq j$. Hence, either $i<_{\ZZ}j$ or
$i>_{\ZZ}j$. In other words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $i<_{\ZZ}j$.
\par
\textit{Case 2:} We have $i>_{\ZZ}j$.
\par
Let us consider Case 1 first. In this case, we have $i<_{\ZZ}j$. Since
$f$ is a $w$-partition, we have $f\left(  i\right)  \leq f\left(  j\right)  $
(because $i<_{w}j$ and $i<_{\ZZ}j$). In other words, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }f\left(  i\right)  =f\left(
j\right)  \right)  $. Hence, $\left(  f\left(  i\right)  <f\left(  j\right)
\text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (because we have assumed that
$i<_{\ZZ}j$). Therefore, $i\prec_{f}j$ (because of the definition of
``$i\prec_{f}j$''). In other words,
$i<_{\mathbf{w}\left(  f\right)  }j$ (since the relation $<_{\mathbf{w}\left(
f\right)  }$ is the relation $\prec_{f}$). Thus, $i<_{\mathbf{w}\left(
f\right)  }j$ is proven in Case 1.
\par
We can similarly prove $i<_{\mathbf{w}\left(  f\right)  }j$ in Case 2 (but now
we obtain $f\left(  i\right)  <f\left(  j\right)  $ instead of $f\left(
i\right)  \leq f\left(  j\right)  $).
\par
We have now shown that $i<_{\mathbf{w}\left(  f\right)  }j$ in both Cases 1
and 2. Thus, (\ref{sol.P-partition.stanley.altproof.5.pf.1c}) is proven.}.
Now, let us again state a triviality: If $Q$ and $R$ are two totally
ordered sets such that $Q=R$ as sets, and if every two elements $i$ and $j$ of
$Q$ satisfying $i<_{Q}j$ satisfy $i<_{R}j$, then $Q=R$ as totally ordered
sets.\footnote{In other words: If a totally ordered set $R$ is an extension
of a totally ordered set $Q$, then $Q = R$.}
Applying this to $Q=w$ and $R=\mathbf{w}\left(  f\right)  $, we conclude
that $w=\mathbf{w}\left(  f\right)  $ as totally ordered sets (since every two
elements $i$ and $j$ of $w$ satisfying $i<_{w}j$ satisfy $i<_{\mathbf{w}%
\left(  f\right)  }j$). In other words, $\mathbf{w}\left(  f\right)  =w$.

Now, we know that $f\in\mathcal{A}\left(  P\right)  $ and $\mathbf{w}\left(
f\right)  =w$. In other words, $f\in\left\{  g\in\mathcal{A}\left(  P\right)
\ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  $.

Let us now forget that we fixed $f$. We thus have shown that $f\in\left\{
g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  $ for every $f\in\mathcal{A}\left(  w\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.5}).]

Combining (\ref{sol.P-partition.stanley.altproof.4}) with
(\ref{sol.P-partition.stanley.altproof.5}), we obtain
\begin{equation}
\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  =\mathcal{A}\left(  w\right)
.\label{sol.P-partition.stanley.altproof.6}%
\end{equation}


Let us now forget that we fixed $w$. We thus have proven
(\ref{sol.P-partition.stanley.altproof.6}) for every $w\in\mathcal{L}\left(
P\right)  $.

Now, the definition of $F_{P}\left(  \xx\right)  $ yields
\begin{align*}
F_{P}\left(  \xx\right)   &  =\sum_{f\in\mathcal{A}\left(  P\right)
}{\mathbf{x}}_{f}=\sum_{w\in\mathcal{L}\left(  P\right)  }\underbrace{\sum\limits
_{\substack{f\in\mathcal{A}\left(  P\right)  ;\\\mathbf{w}\left(  f\right)
=w}}}_{\substack{=\sum_{f\in\left\{  g\in\mathcal{A}\left(  P\right)
\ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  }\\=\sum_{f\in\mathcal{A}%
\left(  w\right)  }\\\text{(by (\ref{sol.P-partition.stanley.altproof.6}))}%
}}{\mathbf{x}}_{f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{w}\left(  f\right)
\in\mathcal{L}\left(  P\right)  \text{ for every }f\in\mathcal{A}\left(
P\right)  \text{ (by (\ref{sol.P-partition.stanley.altproof.wflin}))}\right)
\\
&  =\sum_{w\in\mathcal{L}\left(  P\right)  }\underbrace{\sum_{f\in
\mathcal{A}\left(  w\right)  }\xx_{f}}_{\substack{=F_{w}\left(
\xx\right)  \\\text{(since }F_{w}\left(  \xx\right)  =\sum
_{f\in\mathcal{A}\left(  w\right)  }\xx_{f}\\\text{(by the definition
of }F_{w}\left(  \xx\right)  \text{))}}}=\sum_{w\in\mathcal{L}\left(
P\right)  }F_{w}\left(  \xx\right)  .
\end{align*}
This completes our proof of Theorem \ref{Stanley's-P-partition-theorem}.
\end{proof}
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.P-partition.stanley.altproof}}

In the following, we shall write a poset as a pair $\left(  X,\prec\right)  $,
where $X$ is the underlying set, and where $\prec$ is the smaller relation of
the poset. All binary relations will be written in infix notation (i.e., if
$R$ is a binary relation on a set $X$, and if $a$ and $b$ are two elements of
$X$, then we shall write $aRb$ for $\left(  a,b\right)  \in R$).

We begin with a general lemma about posets:

\begin{lemma}
\label{lem.sol.P-partition.stanley.altproof.wf}Let $\left(  P,\vartriangleleft
\right)  $ and $\left(  Q,<\right)  $ be two posets. Let $f$ be any map
$P\rightarrow Q$ (not necessarily a poset homomorphism).
We define a binary relation $\prec_{f}$ on the set $P$ by
letting $i\prec_{f}j$ hold if and only if
\[
\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i\vartriangleleft j\right)  \right)
.
\]
Then:

\begin{enumerate}
\item[(a)] Let $\vartriangleright$ be the greater relation of the poset
$\left(  P,\vartriangleleft\right)  $. (Thus, two elements $i$ and $j$ of $P$
satisfy $i\vartriangleright j$ if and only if $j\vartriangleleft i$.)

Let $\leq$ be the smaller-or-equal relation of the poset $\left(  Q,<\right)
$. (Thus, two elements $u$ and $v$ of $Q$ satisfy $u\leq v$ if and only if
$\left(  u<v\text{ or }u=v\right)  $.)

If $i$ and $j$ are two elements of $P$, then
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{, then }f\left(  i\right)  \leq f\left(
j\right)  \right)  \label{sol.P-partition.stanley.altproof.lem.1}%
\end{equation}
and%
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{ and }i\vartriangleleft j\text{, then
}f\left(  i\right)  \leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.lem.1a}%
\end{equation}
and%
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{ and }i\vartriangleright j\text{, then
}f\left(  i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.lem.1b}%
\end{equation}
and%
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{ and }f\left(  i\right)  =f\left(
j\right)  \text{, then }i\vartriangleleft j\right)  .
\label{sol.P-partition.stanley.altproof.lem.1c}%
\end{equation}


\item[(b)] The binary relation $\prec_{f}$ is the smaller relation of a
partial order on $P$ (that is, this relation $\prec_{f}$ is irreflexive,
transitive and asymmetric).

\item[(c)] Let $\mathbf{w}\left(  f\right)  $ denote the poset $\left(
P,\prec_{f}\right)  $. (This is well-defined, since Lemma
\ref{lem.sol.P-partition.stanley.altproof.wf}(b) shows that the relation
$\prec_{f}$ is the smaller relation of a partial order on $P$.) Assume that
the poset $\left(  P,\vartriangleleft\right)  $ is totally ordered (i.e.,
every pair of two distinct elements $u$ and $v$ of $P$ satisfies either
$u\vartriangleleft v$ or $v\vartriangleleft u$) and that the poset $\left(
Q,<\right)  $ is totally ordered. Then, the poset $\mathbf{w}\left(  f\right)
$ is totally ordered.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.P-partition.stanley.altproof.wf}.](a) Let $i$ and
$j$ be two elements of $P$. We shall now prove
(\ref{sol.P-partition.stanley.altproof.lem.1}),
(\ref{sol.P-partition.stanley.altproof.lem.1a}),
(\ref{sol.P-partition.stanley.altproof.lem.1b}) and
(\ref{sol.P-partition.stanley.altproof.lem.1c}).

[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.lem.1}):} Assume that
$i\prec_{f}j$. Then, our goal is to show that $f\left(  i\right)  \leq
f\left(  j\right)  $.

We have $i\prec_{f}j$ if and only if $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  \right)  $ (by the definition of
``$i\prec_{f}j$''). Thus, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i\vartriangleleft j\right)  \right)  $ (since we have
$i\prec_{f}j$). Thus, we are in one of the following two cases:

\textit{Case 1:} We have $f\left(  i\right)  <f\left(  j\right)  $.

\textit{Case 2:} We have $\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i\vartriangleleft j\right)  $.

Let us first consider Case 1. In this case, we have $f\left(  i\right)
<f\left(  j\right)  $. Thus, $f\left(  i\right)  \leq f\left(  j\right)  $.
Hence, $f\left(  i\right)  \leq f\left(  j\right)  $ is proven in Case 1.

Let us now consider Case 2. In this case, we have $\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i\vartriangleleft j\right)  $. Hence,
$f\left(  i\right)  =f\left(  j\right)  $, so that $f\left(  i\right)  \leq
f\left(  j\right)  $. Thus, $f\left(  i\right)  \leq f\left(  j\right)  $ is
proven in Case 2.

We have now proven that $f\left(  i\right)  \leq f\left(  j\right)  $ holds in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
we thus conclude that $f\left(  i\right)  \leq f\left(  j\right)  $ always
holds. This completes the proof of
(\ref{sol.P-partition.stanley.altproof.lem.1}).]

[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.lem.1a}):} Assume
that $i\prec_{f}j$ and $i\vartriangleleft j$. Then,
(\ref{sol.P-partition.stanley.altproof.lem.1}) shows that $f\left(  i\right)
\leq f\left(  j\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.lem.1a}).]

[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.lem.1b}):} Assume
that $i\prec_{f}j$ and $i\vartriangleright j$. Then, our goal is to show that
$f\left(  i\right)  <f\left(  j\right)  $.

We have $i\prec_{f}j$ if and only if $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  \right)  $ (by the definition of
``$i\prec_{f}j$''). Thus, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i\vartriangleleft j\right)  \right)  $ (since we have
$i\prec_{f}j$). Thus, we are in one of the following two cases:

\textit{Case 1:} We have $f\left(  i\right)  <f\left(  j\right)  $.

\textit{Case 2:} We have $\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i\vartriangleleft j\right)  $.

Let us first consider Case 1. In this case, we have $f\left(  i\right)
<f\left(  j\right)  $. Hence, $f\left(  i\right)  <f\left(  j\right)  $ is
proven in Case 1.

Let us now consider Case 2. In this case, we have $\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i\vartriangleleft j\right)  $. Hence,
$i\vartriangleleft j$. But we have assumed that $i\vartriangleright j$; in
other words, $j\vartriangleleft i$. This contradicts $i\vartriangleleft j$
(since $\left(  P,\vartriangleleft\right)  $ is a poset). Hence, $f\left(
i\right)  <f\left(  j\right)  $ (since anything follows from a contradiction).
Thus, $f\left(  i\right)  <f\left(  j\right)  $ is proven in Case 2.

We have now proven that $f\left(  i\right)  <f\left(  j\right)  $ holds in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
we thus conclude that $f\left(  i\right)  <f\left(  j\right)  $ always holds.
This completes the proof of (\ref{sol.P-partition.stanley.altproof.lem.1b}).]

[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.lem.1c}):} Assume
that $i\prec_{f}j$ and $f\left(  i\right)  =f\left(  j\right)  $. We must
prove that $i\vartriangleleft j$.

Assume the contrary. Thus, we don't have $i\vartriangleleft j$. Hence, we
don't have $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  $.

But we have $i\prec_{f}j$ if and only if $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  \right)  $ (by the definition of
``$i\prec_{f}j$''). Thus, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i\vartriangleleft j\right)  \right)  $ (since we have
$i\prec_{f}j$). Therefore, we must have $f\left(  i\right)  <f\left(
j\right)  $ (since we don't have $\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i\vartriangleleft j\right)  $). But this contradicts
$f\left(  i\right)  =f\left(  j\right)  $ (since $\left(  Q,<\right)  $ is a
poset). This contradiction shows that our assumption was false. Thus,
(\ref{sol.P-partition.stanley.altproof.lem.1c}) is proven.]

Thus, the four statements (\ref{sol.P-partition.stanley.altproof.lem.1}),
(\ref{sol.P-partition.stanley.altproof.lem.1a}),
(\ref{sol.P-partition.stanley.altproof.lem.1b}) and
(\ref{sol.P-partition.stanley.altproof.lem.1c}) are proven. This proves Lemma
\ref{lem.sol.P-partition.stanley.altproof.wf}(a).

(b) First, we claim that
\begin{equation}
\text{the binary relation }\prec_{f}\text{ is transitive.}%
\label{pf.lem.P-partition.stanley.altproof.lem.trans}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.P-partition.stanley.altproof.lem.trans}):} Let
$i$, $j$ and $k$ be three elements of $P$ such that $i\prec_{f}j$ and
$j\prec_{f}k$. We shall show that $i\prec_{f}k$.

If $f\left(i\right) < f\left(k\right)$, then this
holds\footnote{\textit{Proof.}
Assume that $f\left(i\right) < f\left(k\right)$.
We must prove that $i \prec_f k$.

From $f\left(  i\right)  <f\left(  k\right)  $, we obtain
$\left(  f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  k\right)  \text{ and }i\vartriangleleft k\right)  \right)
$. Hence, $i\prec_{f}k$ (since we have $i\prec_{f}k$ if and only if $\left(
f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(  i\right)
=f\left(  k\right)  \text{ and }i\vartriangleleft k\right)  \right)  $ (by the
definition of ``$i\prec_{f}k$'')).
Thus, we have proven that $i \prec_f k$ under the assumption that
$f\left(i\right) < f\left(k\right)$. Qed.}.
Hence, for the rest of this proof of
$i\prec_{f}k$, we can WLOG assume that we don't have $f\left(  i\right)
<f\left(  k\right)  $. Assume this.

We have $i\prec_{f}j$ and thus $f\left(  i\right)  \leq f\left(  j\right)  $
(by (\ref{sol.P-partition.stanley.altproof.lem.1})). Also, $j\prec_{f}k$ and
thus $f\left(  j\right)  \leq f\left(  k\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.lem.1}), applied to $j$ and $k$ instead
of $i$ and $j$).

If we had $f\left(  i\right)  \neq f\left(  j\right)  $, then we would have%
\begin{align*}
f\left(  i\right)   &  <f\left(  j\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\left(  i\right)  \leq f\left(  j\right)  \text{ and }f\left(
i\right)  \neq f\left(  j\right)  \right)  \\
&  \leq f\left(  k\right)
\end{align*}
and therefore $f\left(i\right) < f\left(k\right)$ (since $\left(Q, <\right)$ is
a poset),
which would contradict the fact that we don't have $f\left(  i\right)
<f\left(  k\right)  $. Hence, we cannot have $f\left(  i\right)  \neq f\left(
j\right)  $. Thus, we have $f\left(  i\right)  =f\left(  j\right)  $. Hence,
(\ref{sol.P-partition.stanley.altproof.lem.1c}) yields $i\vartriangleleft j$
(since $i\prec_{f}j$).

If we had $f\left(  j\right)  \neq f\left(  k\right)  $, then we would have%
\[
f\left(  i\right)  \leq f\left(  j\right)  <f\left(  k\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\left(  j\right)  \leq f\left(
k\right)  \text{ and }f\left(  j\right)  \neq f\left(  k\right)  \right)
\]
and therefore $f\left(i\right) < f\left(k\right)$ (since $\left(Q, <\right)$ is
a poset),
which would contradict the fact that we don't have $f\left(  i\right)
<f\left(  k\right)  $. Hence, we cannot have $f\left(  j\right)  \neq f\left(
k\right)  $. Thus, we have $f\left(  j\right)  =f\left(  k\right)  $. Hence,
(\ref{sol.P-partition.stanley.altproof.lem.1c}) (applied to $j$ and $k$
instead of $i$ and $j$) yields $j\vartriangleleft k$ (since $j\prec_{f}k$).

The relation $\vartriangleleft$ is transitive (since $\left(
P,\vartriangleleft\right)  $ is a poset). Thus, from $i\vartriangleleft j$ and
$j\vartriangleleft k$, we obtain $i\vartriangleleft k$. Also, $f\left(
i\right)  =f\left(  j\right)  =f\left(  k\right)  $. Hence, $\left(  f\left(
i\right)  =f\left(  k\right)  \text{ and }i\vartriangleleft k\right)  $. Thus,
$\left(  f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  k\right)  \text{ and }i\vartriangleleft k\right)  \right)
$. Hence, $i\prec_{f}k$ (since we have $i\prec_{f}k$ if and only if $\left(
f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(  i\right)
=f\left(  k\right)  \text{ and }i\vartriangleleft k\right)  \right)  $ (by the
definition of ``$i\prec_{f}k$'')). Our proof of
$i\prec_{f}k$ is thus complete.

Now, let us forget that we fixed $i$, $j$ and $k$. We thus have shown that if
$i$, $j$ and $k$ are three elements of $P$ such that $i\prec_{f}j$ and
$j\prec_{f}k$, then $i\prec_{f}k$. In other words, the binary relation
$\prec_{f}$ is transitive. This proves
(\ref{pf.lem.P-partition.stanley.altproof.lem.trans}).]

Now, the binary relation $\prec_{f}$ is irreflexive\footnote{\textit{Proof.}
Let $i\in P$ be such that $i\prec_{f}i$. We shall derive a contradiction.
\par
We have $i\prec_{f}i$ if and only if $\left(  f\left(  i\right)  <f\left(
i\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  i\right)  \text{ and
}i\vartriangleleft i\right)  \right)  $ (by the definition of
``$i\prec_{f}i$''). Thus, $\left(  f\left(
i\right)  <f\left(  i\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
i\right)  \text{ and }i\vartriangleleft i\right)  \right)  $ (since we have
$i\prec_{f}i$). Since $f\left(  i\right)  <f\left(  i\right)  $ is impossible
(because $\left(  Q,<\right)  $ is a poset), this shows that we must have
$\left(  f\left(  i\right)  =f\left(  i\right)  \text{ and }i\vartriangleleft
i\right)  $. Thus, $i\vartriangleleft i$. But this is absurd (since $\left(
P,\vartriangleleft\right)  $ is a poset).
\par
Let us forget that we fixed $i$. We thus have derived a contradiction for
every $i\in P$ satisfying $i\prec_{f}i$. Thus, there exists no $i\in P$
satisfying $i\prec_{f}i$. In other words, the binary relation $\prec_{f}$ is
irreflexive. Qed.}, transitive\footnote{by
(\ref{pf.lem.P-partition.stanley.altproof.lem.trans})} and
asymmetric\footnote{\textit{Proof.} It is known that every irreflexive and
transitive relation is asymmetric. Applying this to the relation $\prec_{f}$,
we conclude that the relation $\prec_{f}$ is asymmetric (since $\prec_{f}$ is
irreflexive and transitive). Qed.}. In other words, the relation $\prec_{f}$
is the smaller relation of a partial order on $P$ (because the smaller
relations of partial orders are characterized by being irreflexive, transitive
and asymmetric). This proves Lemma
\ref{lem.sol.P-partition.stanley.altproof.wf}(b).

(c) Let $i$ and $j$ be two distinct elements of $\mathbf{w}\left(  f\right)
$. Thus, $i$ and $j$ are two distinct elements of $P$ (since $\mathbf{w}%
\left(  f\right)  =P$ as sets). We shall show that we have
\begin{equation}
\text{either }i\prec_{f}j\text{ or }j\prec_{f}i.
\label{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal}):}
Let $\leq$ denote the smaller-or-equal relation of the poset $\left(
Q,<\right)  $.

The poset $\left(  Q,<\right)  $ is totally ordered. Thus, we have either
$f\left(  i\right)  \leq f\left(  j\right)  $ or $f\left(  j\right)  \leq
f\left(  i\right)  $. Hence, we can WLOG assume that $f\left(  i\right)  \leq
f\left(  j\right)  $ (since otherwise, we can simply interchange $i$ with
$j$). Assume this.

Let us first assume that $f\left(  i\right)  <f\left(  j\right)  $. Thus,
$\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i\vartriangleleft j\right)  \right)
$. But recall that $i\prec_{f}j$ holds if and only if $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i\vartriangleleft j\right)  \right)  $ (because of the
definition of ``$i\prec_{f}j$''). Thus,
$i\prec_{f}j$ holds (since $\left(  f\left(  i\right)  <f\left(  j\right)
\text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  \right)  $ holds). Thus, we have $\left(
\text{either }i\prec_{f}j\text{ or }j\prec_{f}i\right)  $.
In other words, \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal} holds.

Now, let us forget that we assumed that $f\left(  i\right)  <f\left(
j\right)  $. We therefore have shown that if $f\left(  i\right)  <f\left(
j\right)  $, then \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal}
holds.
Therefore, for the rest of this proof of
\eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal}, we can WLOG assume
that we don't have $f\left(  i\right)  <f\left(  j\right)  $. Assume this.

We don't have $f\left(  i\right)  <f\left(  j\right)  $. In other words, we
have $f\left(  j\right)  \leq f\left(  i\right)  $ (since the poset $\left(
Q,<\right)  $ is totally ordered). Combined with $f\left(  i\right)  \leq
f\left(  j\right)  $, this yields $f\left(  i\right)  =f\left(  j\right)  $
(since $\left(  Q,<\right)  $ is a poset).

But the elements $i$ and $j$ are distinct. Hence, either $i\vartriangleleft j$
or $i\vartriangleright j$ (since the poset $\left(  P,\vartriangleleft\right)
$ is totally ordered). In other words, we are in one of the following two Cases:

\textit{Case 1:} We have $i\vartriangleleft j$.

\textit{Case 2:} We have $i\vartriangleright j$.

Let us consider Case 1 first. In this case, we have $i\vartriangleleft j$.
Thus, $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  $. Hence, $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  \right)  $. Therefore, $i\prec_{f}j$ (since
$i\prec_{f}j$ holds if and only if \newline
$\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i\vartriangleleft j\right)  \right)  $ (because of the definition of
``$i\prec_{f}j$'')). Thus, we have
\newline
$\left( \text{either }i\prec_{f}j\text{ or }j\prec_{f}i\right)  $.
In other words, \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal} holds.
We therefore have
shown that \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal} holds
in Case 1.

Let us now consider Case 2. In this case, we have $i\vartriangleright j$. In
other words, $j\vartriangleleft i$. Combined with $f\left(  j\right)
=f\left(  i\right)  $ (since $f\left(  i\right)  =f\left(  j\right)  $), this
yields $\left(  f\left(  j\right)  =f\left(  i\right)  \text{ and
}j\vartriangleleft i\right)  $. Hence, $\left(  f\left(  j\right)  <f\left(
i\right)  \text{ or }\left(  f\left(  j\right)  =f\left(  i\right)  \text{ and
}j\vartriangleleft i\right)  \right)  $. Therefore, $j\prec_{f}i$ (since
$j\prec_{f}i$ holds if and only if $\left(  f\left(  j\right)  <f\left(
i\right)  \text{ or }\left(  f\left(  j\right)  =f\left(  i\right)  \text{ and
}j\vartriangleleft i\right)  \right)  $ (because of the definition of
``$j\prec_{f}i$'')). Thus, we have $\left(
\text{either }i\prec_{f}j\text{ or }j\prec_{f}i\right)  $.
In other words, \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal} holds.
We therefore have
shown that \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal} holds
in Case 2.

We have now shown that \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal}
holds in both Cases 1 and 2. Since these two Cases cover all
possibilities, this yields that
\eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal} always holds.
This proves \eqref{pf.lem.sol.P-partition.stanley.altproof.wf.c.goal}.]

Now, forget that we fixed $i$ and $j$. We thus have shown that if $i$ and $j$
are two distinct elements of $\mathbf{w}\left(  f\right)  $, then either
$i\prec_{f}j$ or $j\prec_{f}i$. In other words, the poset $\mathbf{w}\left(
f\right)  $ is totally ordered (since $\prec_{f}$ is the smaller relation of
the poset $\mathbf{w}\left(  f\right)  $). This proves Lemma
\ref{lem.sol.P-partition.stanley.altproof.wf}(c).
\end{proof}

Let us state another simple property of posets:

\begin{lemma}
\label{lem.sol.P-partition.stanley.altproof.UV}
Let $U$ and $V$ be two posets such that $U = V$ as sets.
Assume that $U$ is totally ordered
(i.e., any two distinct elements $i$ and $j$ of $U$ satisfy either
$i <_U j$ or $j <_U i$).
Assume that every two
elements $i$ and $j$ of $U$ satisfying $i<_U j$ satisfy
\begin{equation}
i <_V j . \label{eq.lem.sol.P-partition.stanley.altproof.UV.1}
\end{equation}
Then, $U = V$ (as posets).
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem.sol.P-partition.stanley.altproof.UV}.]
For any two elements $i$ and $j$ of $U$, we have
\begin{equation}
\left(  i<_V j\text{ if and only if } i<_U j\right) .
\label{sol.P-partition.stanley.altproof.5.pf.3}
\end{equation}
[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.3}):}
Let $i$ and $j$ be two elements of $U$. We need to prove
(\ref{sol.P-partition.stanley.altproof.5.pf.3}).

If $i=j$, then $\left(  i<_V j\text{ if and only
if }i<_U j\right)  $ holds (because if $i=j$, then neither $i<_V j$ nor $i<_U j$
is true). Hence, if $i=j$, then
(\ref{sol.P-partition.stanley.altproof.5.pf.3}) is proven. Thus, for the rest
of this proof of (\ref{sol.P-partition.stanley.altproof.5.pf.3}), we can WLOG
assume that we don't have $i=j$. Assume this.

Clearly, $i$ and $j$ are elements of $U$, and therefore elements of $V$ (since
$U=V$ as sets). These elements $i$ and $j$ are distinct (since we don't have
$i=j$). Hence, we have either $i<_U j$ or $i>_U j$ (because $U$ is totally
ordered). In other words, we must be in one of the following two Cases:

\textit{Case 1:} We have $i<_U j$.

\textit{Case 2:} We have $i>_U j$.

Let us consider Case 1 first. In this case, we have $i<_U j$. Thus,
$i<_V j$ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1c})). Hence, both $i<_V j$ and
$i <_U j$ hold. Thus, $\left(  i<_V j\text{ if and only if } i<_U j\right)$
holds. We have therefore
proven (\ref{sol.P-partition.stanley.altproof.5.pf.3}) in Case 1.

Let us now consider Case 2. In this case, we have $i>_U j$. Hence, $i<_U j$
is false. But from $i>_Uj$, we also obtain $j<_U i$. Thus, $j<_V i$
(by (\ref{sol.P-partition.stanley.altproof.5.pf.1c}),
applied to $j$ and $i$ instead of $i$ and $j$). Therefore, $i>_V j$.
Hence, $i<_V j$ is false.
Thus, we know that both $i<_V j$ and $i<_U j$ are
false. Thus, $\left(  i<_V j\text{ if and only if } i<_U j\right)$
holds. We have therefore proven
(\ref{sol.P-partition.stanley.altproof.5.pf.3}) in Case 2.

We have now proven (\ref{sol.P-partition.stanley.altproof.5.pf.3}) in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, this
shows that (\ref{sol.P-partition.stanley.altproof.5.pf.3}) always holds.
Qed.].

Recall that $U = V$ as sets.
Hence, \eqref{sol.P-partition.stanley.altproof.5.pf.3}
shows that the relations $<_V$ and $<_U$ are identical. In other words, the
relations $<_U$ and $<_V$ are identical. Hence, $U = V$ as posets (since $U = V$
as sets). This proves Lemma~\ref{lem.sol.P-partition.stanley.altproof.UV}.
\end{proof}

\begin{proof}
[Alternative proof of Theorem~\ref{Stanley's-P-partition-theorem}.]Let $P$ be
a labelled poset.

Let $f$ be any map $P\rightarrow\left\{  1,2,3,\ldots\right\}  $. We define a
binary relation $\prec_{f}$ on the set $P$ by letting $i\prec_{f}j$ hold if
and only if
\[
\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  .
\]


The poset $\left(  P,<_{\ZZ}\right)  $ is totally ordered (since
$<_{\ZZ}$ is a total order on $\ZZ$), and the poset $\left(
\left\{  1,2,3,\ldots\right\}  ,<\right)  $ is totally ordered as well. Also,
$>_{\ZZ}$ is the greater relation of the poset
$\left(  P,<_{\ZZ}\right)  $.
Furthermore, $\leq$ is the smaller-or-equal relation of the poset
$\left(\left\{  1,2,3,\ldots\right\}  ,<\right)  $. Hence, Lemma
\ref{lem.sol.P-partition.stanley.altproof.wf}(a) (applied to $\left(
P,<_{\ZZ}\right)  $, $\left(  \left\{  1,2,3,\ldots\right\}  ,<\right)
$, $>_{\ZZ}$ and $\leq$ instead of $\left(  P,\vartriangleleft\right)
$, $\left(  Q,<\right)  $, $\vartriangleright$ and $\leq$) yields the
following: If $i$ and $j$ are two elements of $P$, then
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{, then }f\left(  i\right)  \leq f\left(
j\right)  \right)  \label{sol.P-partition.stanley.altproof.1}%
\end{equation}
and
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{ and }i<_{\ZZ}j\text{, then
}f\left(  i\right)  \leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.1a}%
\end{equation}
and
\begin{equation}
\left(  \text{if }i\prec_{f}j\text{ and }i>_{\ZZ}j\text{, then
}f\left(  i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.1b}%
\end{equation}
and%
\[
\left(  \text{if }i\prec_{f}j\text{ and }f\left(  i\right)  =f\left(
j\right)  \text{, then }i<_{\ZZ}j\right)  .
\]
Furthermore, Lemma \ref{lem.sol.P-partition.stanley.altproof.wf}(b) (applied
to $\left(  P,<_{\ZZ}\right)  $ and $\left(  \left\{  1,2,3,\ldots
\right\}  ,<\right)  $ instead of $\left(  P,\vartriangleleft\right)  $ and
$\left(  Q,<\right)  $) yields that the binary relation $\prec_{f}$ is the
smaller relation of a partial order on $P$. Let us define $\mathbf{w}\left(
f\right)  $ to be the set $P$ endowed with this partial order. Thus,
$\mathbf{w}\left(  f\right)  =P$ as sets, but the smaller relation
$<_{\mathbf{w}\left(  f\right)  }$ of the poset $\mathbf{w}\left(  f\right)  $
is the relation $\prec_{f}$. In other words, $\mathbf{w}\left(  f\right)  $ is
the poset $\left(  P,\prec_{f}\right)  $. Hence, Lemma
\ref{lem.sol.P-partition.stanley.altproof.wf}(c) (applied to $\left(
P,<_{\ZZ}\right)  $ and $\left(  \left\{  1,2,3,\ldots\right\}
,<\right)  $ instead of $\left(  P,\vartriangleleft\right)  $ and $\left(
Q,<\right)  $) shows that the poset $\mathbf{w}\left(  f\right)  $ is
totally ordered.

Let us now forget that we fixed $f$. Thus, for every map $f:P\rightarrow
\left\{  1,2,3,\ldots\right\}  $, we have constructed a binary relation
$\prec_{f}$ on the set $P$ and a totally ordered poset $\mathbf{w}\left(
f\right)  $ that satisfies $\mathbf{w}\left(f\right) = P$ as sets and
whose smaller relation
$<_{\mathbf{w}\left(  f\right)  }$
is the relation $\prec_{f}$.

We now notice that if $f\in\mathcal{A}\left(  P\right)  $, then
\begin{equation}
\mathbf{w}\left(  f\right)  \in\mathcal{L}\left(  P\right)  .
\label{sol.P-partition.stanley.altproof.wflin}%
\end{equation}


[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.wflin}):} Let
$f\in\mathcal{A}\left(  P\right)  $. We need to show that $\mathbf{w}\left(
f\right)  \in\mathcal{L}\left(  P\right)  $.

We have $f\in\mathcal{A}\left(  P\right)  $. In other words, $f$ is a
$P$-partition
(since $\mathcal{A}\left(  P\right)  $ is the set of all $P$-partitions).

But recall that $f$ is a $P$-partition if and only if $f$ is a map
$P \to \left\{1,2,3,\ldots\right\}$ with the properties
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{sol.P-partition.stanley.altproof.2.pf.1a}%
\end{equation}
and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.2.pf.1b}%
\end{equation}
(due to the definition of a ``$P$-partition'').
Thus, $f$ is a map $P \to \left\{1,2,3,\ldots\right\}$
with the properties (\ref{sol.P-partition.stanley.altproof.2.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.2.pf.1b}) (since $f$ is a $P$-partition).

Let $i$ and $j$ be two elements of $P$. Assume that $i<_{P}j$. We are now
going to show that $i<_{\mathbf{w}\left(  f\right)  }j$.

We have $i<_{P}j$ and thus $i\neq j$. Hence, either $i<_{\ZZ}j$ or
$i>_{\ZZ}j$. In other words, we are in one of the following two Cases:

\textit{Case 1:} We have $i<_{\ZZ}j$.

\textit{Case 2:} We have $i>_{\ZZ}j$.

Let us consider Case 1 first. In this case, we have $i<_{\ZZ}j$.
Therefore, $f\left(  i\right)  =f\left(  j\right)  $ implies $\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  $ (in a
logical sense -- i.e., if $f\left(  i\right)  =f\left(  j\right)  $ is true,
then so is $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  $). On the other hand, $i<_{P}j$ and $i<_{\ZZ}j$;
thus, $f\left(  i\right)  \leq f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.2.pf.1a})). In other words, $\left(
f\left(  i\right)  <f\left(  j\right)  \text{ or }f\left(  i\right)  =f\left(
j\right)  \right)  $. Therefore, $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (because $f\left(  i\right)  =f\left(
j\right)  $ implies $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  $). Hence, $i\prec_{f}j$ (since $i\prec_{f}j$ holds
if and only if $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or
}\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j
\right)  \right)  $ (because of the definition of ``$i\prec_{f}j$'')).
In other words, $i<_{\mathbf{w}\left(
f\right)  }j$ (since the relation $<_{\mathbf{w}\left(  f\right)  }$ is the
relation $\prec_{f}$). Thus, we have shown that $i<_{\mathbf{w}\left(
f\right)  }j$ in Case 1.

Let us now consider Case 2. In this case, we have $i>_{\ZZ}j$. Thus,
$i<_{P}j$ and $i>_{\ZZ}j$. Hence, $f\left(  i\right)  <f\left(
j\right)  $ (by (\ref{sol.P-partition.stanley.altproof.2.pf.1b})). Hence,
$\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $.
Hence, $i\prec_{f}j$ (since $i\prec_{f}j$ holds if and only if $\left(
f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $ (because
of the definition of ``$i\prec_{f}j$'')). In
other words, $i<_{\mathbf{w}\left(  f\right)  }j$ (since the relation
$<_{\mathbf{w}\left(  f\right)  }$ is the relation $\prec_{f}$). Thus, we have
shown that $i<_{\mathbf{w}\left(  f\right)  }j$ in Case 2.

Now, we have shown $i<_{\mathbf{w}\left(  f\right)  }j$ in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, this shows that
$i<_{\mathbf{w}\left(  f\right)  }j$ always holds.

Let us now forget that we fixed $i$ and $j$. We thus have shown that if $i$
and $j$ are two elements of $P$ satisfying $i<_{P}j$, then $i<_{\mathbf{w}%
\left(  f\right)  }j$. In other words, every two elements $i$ and $j$ of $P$
satisfying $i<_{P}j$ satisfy $i<_{\mathbf{w}\left(  f\right)  }j$. But this
yields that $\mathbf{w}\left(  f\right)  $ is a linear extension of $P$ (by
the definition of a ``linear extension''),
since we already know that $\mathbf{w}\left(  f\right)  $ is totally ordered.

So we have shown that $\mathbf{w}\left(  f\right)  $ is a linear extension of
$P$. In other words, $\mathbf{w}\left(  f\right)  \in\mathcal{L}\left(
P\right)  $ (since $\mathcal{L}\left(  P\right)  $ is the set of all linear
extensions of $P$). This proves \eqref{sol.P-partition.stanley.altproof.wflin}.]

Now, fix $w\in\mathcal{L}\left(  P\right)  $. Then, $w$ is a linear extension
of $P$ (since $\mathcal{L}\left(  P\right)  $ is the set of all linear
extensions of $P$). In other words, $w$ is a totally ordered poset satisfying
$w = P$ as sets, and if $i$ and $j$
are two elements of $P$ satisfying $i<_{P}j$, then
\begin{equation}
i<_{w}j. \label{sol.P-partition.stanley.altproof.linex}%
\end{equation}
We notice that $\mathcal{A}\left(  w\right)  $ is the set of all
$w$-partitions (due to the definition of $\mathcal{A}\left(  w\right)  $).

Let us first show that
\begin{equation}
\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  \subset\mathcal{A}\left(  w\right)  .
\label{sol.P-partition.stanley.altproof.4}%
\end{equation}


[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4}):} Let
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $. Then, $f$ is an element $g$ of $\mathcal{A}\left(
P\right)  $ satisfying $\mathbf{w}\left(  g\right)  =w$. In other words, $f$
is an element of $\mathcal{A}\left(  P\right)  $ and satisfies $\mathbf{w}%
\left(  f\right)  =w$. Hence, the relation $<_{w}$ is the relation $\prec_{f}%
$\ \ \ \ \footnote{\textit{Proof.} We know that the relation $<_{\mathbf{w}%
\left(  f\right)  }$ is the relation $\prec_{f}$. In other words, the relation
$<_{w}$ is the relation $\prec_{f}$ (since $\mathbf{w}\left(  f\right)  =w$).
Qed.}. Also, $w=\mathbf{w}\left(  f\right)  =P$ as sets.

The map $f$ has the properties
\begin{equation}
\left(  \text{if }i\in w\text{ and }j\in w\text{ satisfy }i<_w j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{sol.P-partition.stanley.altproof.4.pf.1a}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4.pf.1a}):}
Let $i\in w$ and $j\in w$ be such that $i<_w j$ and $i<_{\ZZ}j$. We
need to prove that $f\left(  i\right)  \leq f\left(  j\right)  $. Notice that
$i$ and $j$ are elements of $w$, and thus elements of $P$ (since $w=P$ as
sets). Also, we have $i<_{w}j$. In other words, $i\prec_{f}j$ (since the
relation $<_{w}$ is the relation $\prec_{f}$). Thus, using
(\ref{sol.P-partition.stanley.altproof.1a}), we obtain $f\left(  i\right)
\leq f\left(  j\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}).} and
\begin{equation}
\left(  \text{if }i\in w\text{ and }j\in w\text{ satisfy }i<_w j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.4.pf.1b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4.pf.1b}):}
Let $i\in w$ and $j\in w$ be such that $i<_w j$ and $i>_{\ZZ}j$. We
need to prove that $f\left(  i\right)  <f\left(  j\right)  $. Notice that $i$
and $j$ are elements of $w$, and thus elements of $P$ (since $w=P$ as sets).
Also, we have $i<_{w}j$. In other words, $i\prec_{f}j$ (since the relation
$<_{w}$ is the relation $\prec_{f}$). Thus, using
(\ref{sol.P-partition.stanley.altproof.1b}), we obtain $f\left(  i\right)
<f\left(  j\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.4.pf.1b}).}. Thus, $f\in\mathcal{A}%
\left(  w\right)  $\ \ \ \ \footnote{\textit{Proof.} The map $f$ is a map
$P \to \left\{1,2,3,\ldots\right\}$, therefore is a map
$w \to \left\{1,2,3,\ldots\right\}$ (since $w = P$ as sets).
Now, recall that $f$ is a
$w$-partition if and only if $f$ is a map
$w \to \left\{1,2,3,\ldots\right\}$
with the properties
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.4.pf.1b}) (due to the definition of a
``$w$-partition''). Thus, $f$ is a
$w$-partition (since $f$ is a map $w \to \left\{1,2,3,\ldots\right\}$ with
the properties
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.4.pf.1b})). In other words,
$f\in\mathcal{A}\left(  w\right)  $ (since $\mathcal{A}\left(  w\right)  $ is
the set of all $w$-partitions), qed.}.

Let us now forget that we fixed $f$. We thus have proven that every
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $ satisfies $f\in\mathcal{A}\left(  w\right)  $. In
other words, $\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}%
\left(  g\right)  =w\right\}  \subset\mathcal{A}\left(  w\right)  $. This
proves (\ref{sol.P-partition.stanley.altproof.4}).]

Let us next show that
\begin{equation}
\mathcal{A}\left(  w\right)  \subset\left\{  g\in\mathcal{A}\left(  P\right)
\ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  .
\label{sol.P-partition.stanley.altproof.5}%
\end{equation}


[\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5}):} Let
$f\in\mathcal{A}\left(  w\right)  $. Then, $f$ is an element of $\mathcal{A}%
\left(  w\right)  $. In other words, $f$ is a $w$-partition (since
$\mathcal{A}\left(  w\right)  $ is the set of all $w$-partitions). In other
words, $f$ is a map $w \to \left\{1,2,3,\ldots\right\}$ with the properties
that
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{w}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{sol.P-partition.stanley.altproof.5.pf.1a}%
\end{equation}
and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{w}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.5.pf.1b}%
\end{equation}
(because $f$ is a $w$-partition if and only if it is a map
$w \to \left\{1,2,3,\ldots\right\}$
with the properties (\ref{sol.P-partition.stanley.altproof.5.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.5.pf.1b}) (due to the definition of a
``$w$-partition'')).

In particular, $f$ is a map $w \to \left\{1,2,3,\ldots\right\}$, thus is a
map $P \to \left\{1,2,3,\ldots\right\}$ (since $w = P$ as sets).
Also,
every two
elements $i$ and $j$ of $w$ satisfying $i<_{w}j$ satisfy
\begin{equation}
i<_{\mathbf{w}\left(  f\right)  }j
\label{sol.P-partition.stanley.altproof.5.pf.1c}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.1c}):}
Let $i$ and $j$ be two elements of $w$ satisfying $i<_{w}j$. We must prove
that $i<_{\mathbf{w}\left(  f\right)  }j$.
\par
Clearly, $i$ and $j$ are elements of $w$, and therefore elements of $P$ (since
$w=P$ as sets). We have $i<_{w}j$ and thus $i\neq j$. Hence, either
$i<_{\ZZ}j$ or $i>_{\ZZ}j$. In other words, we are in one of the
following two Cases:
\par
\textit{Case 1:} We have $i<_{\ZZ}j$.
\par
\textit{Case 2:} We have $i>_{\ZZ}j$.
\par
Let us consider Case 1 first. In this case, we have $i<_{\ZZ}j$. Thus,
$f\left(  i\right)  =f\left(  j\right)  $ implies $\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  $ (in a logical sense
-- i.e., if $f\left(  i\right)  =f\left(  j\right)  $ is true, then so is
$\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j
\right)  $). But $i<_{w}j$ and $i<_{\ZZ}j$, and thus $f\left(
i\right)  \leq f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1a})). In other words, $\left(
f\left(  i\right)  <f\left(  j\right)  \text{ or }f\left(  i\right)  =f\left(
j\right)  \right)  $. Hence, $\left(  f\left(  i\right)  <f\left(  j\right)
\text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (since $f\left(  i\right)  =f\left(
j\right)  $ implies $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  $). Therefore, $i\prec_{f}j$ (since $i\prec_{f}j$
holds if and only if $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or
}\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j
\right)  \right)  $ (because of the definition of ``$i\prec_{f}j$'')).
In other words, $i<_{\mathbf{w}\left(
f\right)  }j$ (since the relation $<_{\mathbf{w}\left(  f\right)  }$ is the
relation $\prec_{f}$). Thus, $i<_{\mathbf{w}\left(  f\right)  }j$ is proven in
Case 1.
\par
Let us now consider Case 2. In this case, we have $i>_{\ZZ}j$. Hence,
$i<_{w}j$ and $i>_{\ZZ}j$. Thus, $f\left(  i\right)  <f\left(
j\right)  $ (according to (\ref{sol.P-partition.stanley.altproof.5.pf.1b})).
Hence, $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(
f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)
\right)  $. Therefore, $i\prec_{f}j$ (since $i\prec_{f}j$ holds if and only if
$\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $
(because of the definition of ``$i\prec_{f}j$'')). In other words,
$i<_{\mathbf{w}\left(  f\right)  }j$ (since the relation
$<_{\mathbf{w}\left(  f\right)  }$ is the relation $\prec_{f}$). Thus,
$i<_{\mathbf{w}\left(  f\right)  }j$ is proven in Case 2.
\par
We have now shown that $i<_{\mathbf{w}\left(  f\right)  }j$ in both Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
$i<_{\mathbf{w}\left(  f\right)  }j$ always holds. Thus,
(\ref{sol.P-partition.stanley.altproof.5.pf.1c}) is proven.}.
Hence, Lemma~\ref{lem.sol.P-partition.stanley.altproof.UV} (applied to
$U = w$ and $V = \mathbf{w}\left(f\right)$) shows that
$w = \mathbf{w}\left(f\right)$ (as posets)
(since $w = \mathbf{w}\left(f\right)$ as sets).
In other words, $\mathbf{w}\left(  f\right)  =w$.

We shall now show that that $f\in\mathcal{A}\left(  P\right)  $.
Indeed,%
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{sol.P-partition.stanley.altproof.5.pf.2a}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.2a}):}
Let $i\in P$ and $j\in P$ be such that $i<_{P}j$ and $i<_{\ZZ}j$. We
need to prove that $f\left(  i\right)  \leq f\left(  j\right)  $. Notice that
$i$ and $j$ are elements of $P$, and thus elements of $w$ (since $w=P$ as
sets). Also, recall that $i<_{P}j$. Thus, $i<_{w}j$ (by
(\ref{sol.P-partition.stanley.altproof.linex})), and therefore $f\left(
i\right)  \leq f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1a})). This proves
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}).} and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.5.pf.2b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.2b}):}
Let $i\in P$ and $j\in P$ be such that $i<_{P}j$ and $i>_{\ZZ}j$. We
need to prove that $f\left(  i\right)  <f\left(  j\right)  $. Notice that $i$
and $j$ are elements of $P$, and thus elements of $w$ (since $w=P$ as sets).
Also, recall that $i<_{P}j$. Thus, $i<_{w}j$ (by
(\ref{sol.P-partition.stanley.altproof.linex})), and therefore $f\left(
i\right)  <f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1b})). This proves
(\ref{sol.P-partition.stanley.altproof.5.pf.2b}).}. Thus, $f\in\mathcal{A}%
\left(  P\right)  $\ \ \ \ \footnote{\textit{Proof.} Recall that $f$ is a
$P$-partition if and only if $f$ is a map $P \to \left\{1,2,3,\ldots\right\}$
with the properties
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}) and
(\ref{sol.P-partition.stanley.altproof.5.pf.2b}) (due to the definition of a
``$P$-partition''). Thus, $f$ is a
$P$-partition (since $f$ is a map $P \to \left\{1,2,3,\ldots\right\}$
with the properties
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}) and
(\ref{sol.P-partition.stanley.altproof.5.pf.2b})). In other words,
$f\in\mathcal{A}\left(  P\right)  $ (since $\mathcal{A}\left(  P\right)  $ is
the set of all $P$-partitions), qed.}.

Now, we know that $f\in\mathcal{A}\left(  P\right)  $ and $\mathbf{w}\left(
f\right)  =w$. In other words, $f$ is an element $g$ of $\mathcal{A}\left(
P\right)  $ satisfying $\mathbf{w}\left(  g\right)  =w$. In other words,
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $.

Let us now forget that we fixed $f$. We thus have shown that $f\in\left\{
g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  $ for every $f\in\mathcal{A}\left(  w\right)  $. In other words,
\[
\mathcal{A}\left(  w\right)  \subset\left\{  g\in\mathcal{A}\left(  P\right)
\ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  .
\]
This proves (\ref{sol.P-partition.stanley.altproof.5}).]

Combining (\ref{sol.P-partition.stanley.altproof.4}) with
(\ref{sol.P-partition.stanley.altproof.5}), we obtain
\begin{equation}
\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  =\mathcal{A}\left(  w\right)  .
\label{sol.P-partition.stanley.altproof.6}%
\end{equation}


Let us now forget that we fixed $w$. We thus have proven
(\ref{sol.P-partition.stanley.altproof.6}) for every $w\in\mathcal{L}\left(
P\right)  $.

Now, the definition of $F_{P}\left(  \xx\right)  $ yields
\begin{align*}
F_{P}\left(  \xx\right)   &  =\sum_{f\in\mathcal{A}\left(  P\right)
}{\mathbf{x}}_{f}=\sum_{w\in\mathcal{L}\left(  P\right)  }\underbrace{\sum\limits
_{\substack{f\in\mathcal{A}\left(  P\right)  ;\\\mathbf{w}\left(  f\right)
=w}}}_{\substack{=\sum_{f\in\left\{  g\in\mathcal{A}\left(  P\right)
\ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  }\\=\sum_{f\in\mathcal{A}%
\left(  w\right)  }\\\text{(by (\ref{sol.P-partition.stanley.altproof.6}))}%
}}{\mathbf{x}}_{f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{w}\left(  f\right)
\in\mathcal{L}\left(  P\right)  \text{ for every }f\in\mathcal{A}\left(
P\right)  \text{ (by (\ref{sol.P-partition.stanley.altproof.wflin}))}\right)
\\
&  =\sum_{w\in\mathcal{L}\left(  P\right)  }\underbrace{\sum_{f\in
\mathcal{A}\left(  w\right)  }\xx_{f}}_{\substack{=F_{w}\left(
\xx\right)  \\\text{(since }F_{w}\left(  \xx\right)  =\sum
_{f\in\mathcal{A}\left(  w\right)  }\xx_{f}\\\text{(by the definition
of }F_{w}\left(  \xx\right)  \text{))}}}=\sum_{w\in\mathcal{L}\left(
P\right)  }F_{w}\left(  \xx\right)  .
\end{align*}
This completes our proof of Theorem \ref{Stanley's-P-partition-theorem}.
\end{proof}

\begin{comment}
What follows is the old version of the above solution (written back
when we used the awkward ``implies'' notation in the definition of
$P$-partitions).

\textit{Solution to Exercise~\ref{exe.P-partition.stanley.altproof}.}

\begin{proof}
[Alternative proof of Theorem~\ref{Stanley's-P-partition-theorem}.]
Let $P$ be a labelled poset.

First of all, let $f$ be any map $P\rightarrow\left\{  1,2,3,\ldots\right\}
$. We define a binary relation $\prec_{f}$ on the set $P$ by letting
$i\prec_{f}j$ hold if and only if
\[
\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  .
\]


We will shortly see that this binary relation $\prec_{f}$ is the smaller
relation of a total order on $P$. First of all, let us notice a simple fact:
If $i$ and $j$ are two elements of $P$, then
\begin{equation}
\left(  i\prec_{f}j\text{ implies }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{sol.P-partition.stanley.altproof.1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.1}):} Let
$i$ and $j$ be two elements of $P$. We need to prove
(\ref{sol.P-partition.stanley.altproof.1}). Assume that $i\prec_{f}j$. Then,
our goal is to show that $f\left(  i\right)  \leq f\left(  j\right)  $.
\par
Indeed, we assume the contrary. Then, $f\left(  i\right)  >f\left(  j\right)
$.
\par
We have $i\prec_{f}j$ if and only if $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (by the definition of ``$i\prec_{f}j$'').
Thus, $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (since we have $i\prec_{f}j$). Since we
do not have $f\left(  i\right)  <f\left(  j\right)  $ (because this would
contradict $f\left(  i\right)  >f\left(  j\right)  $), we therefore must have
$\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i<_{\ZZ}j\right)  $. Hence,
$f\left(  i\right)  =f\left(  j\right)  $. But this
contradicts $f\left(  i\right)  >f\left(  j\right)  $. This contradiction
shows that our assumption was wrong. Hence, $f\left(  i\right)  \leq f\left(
j\right)  $ is proven. This completes the proof of
(\ref{sol.P-partition.stanley.altproof.1}).} and
\begin{equation}
\left(  i\prec_{f}j\text{ and }i<_{\ZZ}j\text{ implies }f\left(
i\right)  \leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.1a}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.1a}):} Let
$i$ and $j$ be two elements of $P$. We need to prove
(\ref{sol.P-partition.stanley.altproof.1a}). Assume that $i\prec_{f}j$ and
$i<_{\ZZ}j$. Then, (\ref{sol.P-partition.stanley.altproof.1}) shows
that $f\left(  i\right)  \leq f\left(  j\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.1a}).} and
\begin{equation}
\left(  i\prec_{f}j\text{ and }i>_{\ZZ}j\text{ implies }f\left(
i\right)  <f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.1b}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.1b}):} Let
$i$ and $j$ be two elements of $P$. We need to prove
(\ref{sol.P-partition.stanley.altproof.1b}). Assume that $i\prec_{f}j$ and
$i>_{\ZZ}j$. Then, our goal is to show that $f\left(  i\right)
<f\left(  j\right)  $.
\par
Indeed, we assume the contrary. Then, $f\left(  i\right)  \geq f\left(
j\right)  $.
\par
We have $i\prec_{f}j$ if and only if $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (by the definition of ``$i\prec_{f}j$''). Thus,
$\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (since we have $i\prec_{f}j$). Since we
do not have $f\left(  i\right)  <f\left(  j\right)  $ (because this would
contradict $f\left(  i\right)  \geq f\left(  j\right)  $), we therefore must
have $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }
i<_{\ZZ}j\right)  $. Hence, $i<_{\ZZ}j$. But this contradicts
$i>_{\ZZ}j$. This contradiction shows that our assumption was wrong.
Hence, $f\left(  i\right)  <f\left(  j\right)  $ is proven. This completes the
proof of (\ref{sol.P-partition.stanley.altproof.1b}).}. Now, the binary
relation $\prec_{f}$ is irreflexive\footnote{\textit{Proof.} Let $i\in P$ be
such that $i\prec_{f}i$. We shall derive a contradiction.
\par
We have $i\prec_{f}i$ if and only if $\left(  f\left(  i\right)  <f\left(
i\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  i\right)  \text{ and
}i<_{\ZZ}i\right)  \right)  $ (by the definition of ``$i\prec_{f}i$''). Thus,
$\left(  f\left(  i\right)  <f\left(
i\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  i\right)  \text{ and
}i<_{\ZZ}i\right)  \right)  $ (since we have $i\prec_{f}i$). Since
$f\left(  i\right)  <f\left(  i\right)  $ is impossible, this shows that we
must have $\left(  f\left(  i\right)  =f\left(  i\right)  \text{ and
}i<_{\ZZ}i\right)  $. Thus, $i<_{\ZZ}i$. But this is absurd.
\par
Let us forget that we fixed $i$. We thus have derived a contradiction for
every $i\in P$ satisfying $i\prec_{f}i$. Thus, there exists no $i\in P$
satisfying $i\prec_{f}i$. In other words, the binary relation $\prec_{f}$ is
irreflexive. Qed.}, transitive\footnote{\textit{Proof.} Let $i$, $j$ and $k$
be three elements of $P$ such that $i\prec_{f}j$ and $j\prec_{f}k$. We shall
show that $i\prec_{f}k$.
\par
Let us first assume that $f\left(  i\right)  <f\left(  k\right)  $. Then,
$\left(  f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  k\right)  \text{ and }i<_{\ZZ}k\right)  \right)  $.
Hence, $i\prec_{f}k$ (since we have $i\prec_{f}k$ if and only if $\left(
f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(  i\right)
=f\left(  k\right)  \text{ and }i<_{\ZZ}k\right)  \right)  $ (by the
definition of ``$i\prec_{f}k$'')).
\par
Let us now forget that we assumed that $f\left(  i\right)  <f\left(  k\right)
$. We thus have proven $i\prec_{f}k$ under the assumption that $f\left(
i\right)  <f\left(  k\right)  $. Hence, for the rest of this proof of
$i\prec_{f}k$, we can WLOG assume that we don't have $f\left(  i\right)
<f\left(  k\right)  $. Assume this.
\par
We don't have $f\left(  i\right)  <f\left(  k\right)  $. Thus, we have
$f\left(  i\right)  \geq f\left(  k\right)  $, so that $f\left(  k\right)
\leq f\left(  i\right)  $.
\par
We have $i\prec_{f}j$ and thus $f\left(  i\right)  \leq f\left(  j\right)  $
(by (\ref{sol.P-partition.stanley.altproof.1})). Also, $j\prec_{f}k$ and thus
$f\left(  j\right)  \leq f\left(  k\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.1}), applied to $j$ and $k$ instead of
$i$ and $j$). Combining $f\left(  i\right)  \leq f\left(  j\right)  $ with
$f\left(  j\right)  \leq f\left(  k\right)  \leq f\left(  i\right)  $, we
obtain $f\left(  i\right)  =f\left(  j\right)  $. Combining $f\left(
j\right)  \leq f\left(  k\right)  $ with $f\left(  k\right)  \leq f\left(
i\right)  =f\left(  j\right)  $, we obtain $f\left(  j\right)  =f\left(
k\right)  $.
\par
We have $i\prec_{f}j$ if and only if $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (by the definition of ``$i\prec_{f}j$''). Thus,
$\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (since we have $i\prec_{f}j$). Since
$f\left(  i\right)  <f\left(  j\right)  $ is impossible (because it would
contradict $f\left(  i\right)  =f\left(  j\right)  $), we thus must have
$\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i<_{\ZZ}j\right)  $. Thus, $i<_{\ZZ}j$.
\par
We have $j\prec_{f}k$ if and only if $\left(  f\left(  j\right)  <f\left(
k\right)  \text{ or }\left(  f\left(  j\right)  =f\left(  k\right)  \text{ and
}j<_{\ZZ}k\right)  \right)  $ (by the definition of ``$j\prec_{f}k$'').
Thus, $\left(  f\left(  j\right)  <f\left(
k\right)  \text{ or }\left(  f\left(  j\right)  =f\left(  k\right)  \text{ and
}j<_{\ZZ}k\right)  \right)  $ (since we have $j\prec_{f}k$). Since
$f\left(  j\right)  <f\left(  k\right)  $ is impossible (because it would
contradict $f\left(  j\right)  =f\left(  k\right)  $), we thus must have
$\left(  f\left(  j\right)  =f\left(  k\right)  \text{ and }j<_{\ZZ}
k\right)  $. Thus, $j<_{\ZZ}k$.
\par
Now, we have $f\left(  i\right)  =f\left(  j\right)  =f\left(  k\right)  $ and
$i<_{\ZZ}j<_{\ZZ}k$. Hence, $\left(  f\left(  i\right)
=f\left(  k\right)  \text{ and }i<_{\ZZ}k\right)  $. Thus, $\left(
f\left(  i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(  i\right)
=f\left(  k\right)  \text{ and }i<_{\ZZ}k\right)  \right)  $. Hence,
$i\prec_{f}k$ (since we have $i\prec_{f}k$ if and only if $\left(  f\left(
i\right)  <f\left(  k\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
k\right)  \text{ and }i<_{\ZZ}k\right)  \right)  $ (by the definition
of ``$i\prec_{f}k$'')). Our proof of $i\prec_{f}k$ is thus complete.
\par
Now, let us forget that we fixed $i$, $j$ and $k$. We thus have shown that if
$i$, $j$ and $k$ are three elements of $P$ such that $i\prec_{f}j$ and
$j\prec_{f}k$, then $i\prec_{f}k$. In other words, the binary relation
$\prec_{f}$ is transitive. Qed.} and asymmetric\footnote{\textit{Proof.} It is
known that every irreflexive and transitive relation is asymmetric. Applying
this to the relation $\prec_{f}$, we conclude that the relation $\prec_{f}$ is
asymmetric (since $\prec_{f}$ is irreflexive and transitive). Qed.}. In other
words, the relation $\prec_{f}$ is the smaller relation of a partial order on
$P$ (because the smaller relations of partial orders are characterized by
being irreflexive, transitive and asymmetric). Let us define $\mathbf{w}
\left(  f\right)  $ to be the set $P$ endowed with this partial order. Thus,
$\mathbf{w}\left(  f\right)  =P$ as sets, but the smaller relation
$<_{\mathbf{w}\left(  f\right)  }$ of the poset $\mathbf{w}\left(  f\right)  $
is the relation $\prec_{f}$. It is now easy to see that the poset
$\mathbf{w}\left(  f\right)  $ is totally ordered\footnote{\textit{Proof.} Let
$i$ and $j$ be two distinct elements of $\mathbf{w}\left(  f\right)  $. We
shall show that we have $\left(  \text{either }i<_{\mathbf{w}\left(  f\right)
}j\text{ or }j<_{\mathbf{w}\left(  f\right)  }i\right)  $.
\par
Indeed, we can WLOG assume that $f\left(  i\right)  \leq f\left(  j\right)  $
(since otherwise, we can simply interchange $i$ with $j$). Assume this.
\par
Let us first assume that $f\left(  i\right)  <f\left(  j\right)  $. Then,
$\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $.
But recall that $i\prec_{f}j$ holds if and only if $\left(  f\left(  i\right)
<f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i<_{\ZZ}j\right)  \right)  $ (because of the definition of
``$i\prec_{f}j$''). Thus, $i\prec_{f}j$ holds
(since $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(
f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)
\right)  $ holds). In other words, $i<_{\mathbf{w}\left(  f\right)  }j$ holds
(since the relation $<_{\mathbf{w}\left(  f\right)  }$ is the relation
$\prec_{f}$). Thus, we have $\left(  \text{either }i<_{\mathbf{w}\left(
f\right)  }j\text{ or }j<_{\mathbf{w}\left(  f\right)  }i\right)  $.
\par
Now, let us forget that we assumed that $f\left(  i\right)  <f\left(
j\right)  $. We therefore have shown that if $f\left(  i\right)  <f\left(
j\right)  $, then we have $\left(  \text{either }i<_{\mathbf{w}\left(
f\right)  }j\text{ or }j<_{\mathbf{w}\left(  f\right)  }i\right)  $.
Therefore, for the rest of this proof of $\left(  \text{either }
i<_{\mathbf{w}\left(  f\right)  }j\text{ or }j<_{\mathbf{w}\left(  f\right)
}i\right)  $, we can WLOG assume that we don't have $f\left(  i\right)
<f\left(  j\right)  $. Assume this.
\par
We don't have $f\left(  i\right)  <f\left(  j\right)  $. In other words, we
have $f\left(  i\right)  \geq f\left(  j\right)  $. Combined with $f\left(
i\right)  \leq f\left(  j\right)  $, this yields $f\left(  i\right)  =f\left(
j\right)  $.
\par
But the elements $i$ and $j$ are distinct. Hence, either $i<_{\ZZ}j$ or
$i>_{\ZZ}j$. In other words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $i<_{\ZZ}j$.
\par
\textit{Case 2:} We have $i>_{\ZZ}j$.
\par
Let us consider Case 1 first. In this case, we have $i<_{\ZZ}j$. Thus,
$\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}
j\right)  $. Hence, $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or
}\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}
j\right)  \right)  $. Therefore, $i\prec_{f}j$ (since $i\prec_{f}j$ holds if
and only if $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(
f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)
\right)  $ (because of the definition of ``$i\prec_{f}j$'')).
In other words, $i<_{\mathbf{w}\left(  f\right)  }j$
(since the relation $<_{\mathbf{w}\left(  f\right)  }$ is the relation
$\prec_{f}$). Thus, we have $\left(  \text{either }i<_{\mathbf{w}\left(
f\right)  }j\text{ or }j<_{\mathbf{w}\left(  f\right)  }i\right)  $. We
therefore have shown that $\left(  \text{either }i<_{\mathbf{w}\left(
f\right)  }j\text{ or }j<_{\mathbf{w}\left(  f\right)  }i\right)  $ in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>_{\ZZ}j$. Thus,
$j<_{\ZZ}i$. Combined with $f\left(  j\right)  =f\left(  i\right)  $
(since $f\left(  i\right)  =f\left(  j\right)  $), this yields $\left(
f\left(  j\right)  =f\left(  i\right)  \text{ and }j<_{\ZZ}i\right)  $.
Hence, $\left(  f\left(  j\right)  <f\left(  i\right)  \text{ or }\left(
f\left(  j\right)  =f\left(  i\right)  \text{ and }j<_{\ZZ}i\right)
\right)  $. Therefore, $j\prec_{f}i$ (since $j\prec_{f}i$ holds if and only if
$\left(  f\left(  j\right)  <f\left(  i\right)  \text{ or }\left(  f\left(
j\right)  =f\left(  i\right)  \text{ and }j<_{\ZZ}i\right)  \right)  $
(because of the definition of ``$j\prec_{f}i$'')). In other words,
$j<_{\mathbf{w}\left(  f\right)  }i$ (since the relation
$<_{\mathbf{w}\left(  f\right)  }$ is the relation $\prec_{f}$). Thus, we have
$\left(  \text{either }i<_{\mathbf{w}\left(  f\right)  }j\text{ or
}j<_{\mathbf{w}\left(  f\right)  }i\right)  $. We therefore have shown that
$\left(  \text{either }i<_{\mathbf{w}\left(  f\right)  }j\text{ or
}j<_{\mathbf{w}\left(  f\right)  }i\right)  $ in Case 2.
\par
We have now shown that $\left(  \text{either }i<_{\mathbf{w}\left(  f\right)
}j\text{ or }j<_{\mathbf{w}\left(  f\right)  }i\right)  $ in both Cases 1 and
2. Since these two Cases cover all possibilities, this yields that $\left(
\text{either }i<_{\mathbf{w}\left(  f\right)  }j\text{ or }j<_{\mathbf{w}
\left(  f\right)  }i\right)  $ always holds.
\par
Now, let us forget that we fixed $i$ and $j$. We thus have shown that if $i$
and $j$ are two distinct elements of $\mathbf{w}\left(  f\right)  $, then
either $i<_{\mathbf{w}\left(  f\right)  }j$ or $j<_{\mathbf{w}\left(
f\right)  }i$. In other words, the poset $\mathbf{w}\left(  f\right)  $ is
totally ordered. Qed.}.

Let us now forget that we fixed $f$. Thus, for every map $f:P\rightarrow
\left\{  1,2,3,\ldots\right\}  $, we have constructed a binary relation
$\prec_{f}$ on the set $P$ and a totally ordered poset $\mathbf{w}\left(
f\right)  $. We now notice that if $f\in\mathcal{A}\left(  P\right)  $, then
\begin{equation}
\mathbf{w}\left(  f\right)  \in\mathcal{L}\left(  P\right)
\label{sol.P-partition.stanley.altproof.wflin}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.wflin}):}
Let $f\in\mathcal{A}\left(  P\right)  $. We need to show that $\mathbf{w}
\left(  f\right)  \in\mathcal{L}\left(  P\right)  $.
\par
We have $f\in\mathcal{A}\left(  P\right)  $. In other words, $f$ is a
$P$-partition (since $\mathcal{A}\left(  P\right)  $ is the set of all
$P$-partitions).
\par
But recall that $f$ is a $P$-partition if and only if every two elements $i$
and $j$ of $P$ satisfy
\begin{equation}
\left(  i<_{P}j\text{ and }i<_{\ZZ}j\text{ implies }f\left(  i\right)
\leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.2.pf.1a}
\end{equation}
and
\begin{equation}
\left(  i<_{P}j\text{ and }i>_{\ZZ}j\text{ implies }f\left(  i\right)
<f\left(  j\right)  \right)  \label{sol.P-partition.stanley.altproof.2.pf.1b}
\end{equation}
(due to the definition of a ``$P$-partition'').
Thus, every two elements $i$ and $j$ of $P$ satisfy
(\ref{sol.P-partition.stanley.altproof.2.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.2.pf.1b}) (since $f$ is a
$P$-partition).
\par
Let $i$ and $j$ be two elements of $P$. Assume that $i<_{P}j$. We are now
going to show that $i<_{\mathbf{w}\left(  f\right)  }j$.
\par
We have $i<_{P}j$ and thus $i\neq j$. Hence, either $i<_{\ZZ}j$ or
$i>_{\ZZ}j$. In other words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $i<_{\ZZ}j$.
\par
\textit{Case 2:} We have $i>_{\ZZ}j$.
\par
Let us consider Case 1 first. In this case, we have $i<_{\ZZ}j$.
Therefore, $f\left(  i\right)  =f\left(  j\right)  $ implies $\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  $ (in a
logical sense -- i.e., if $f\left(  i\right)  =f\left(  j\right)  $ is true,
then so is $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  $). On the other hand, $i<_{\ZZ}j$, and
thus $f\left(  i\right)  \leq f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.2.pf.1a})). In other words, $\left(
f\left(  i\right)  <f\left(  j\right)  \text{ or }f\left(  i\right)  =f\left(
j\right)  \right)  $. Therefore, $\left(  f\left(  i\right)  <f\left(
j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  \right)  $ (because $f\left(  i\right)  =f\left(
j\right)  $ implies $\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and
}i<_{\ZZ}j\right)  $). Hence, $i\prec_{f}j$ (since $i\prec_{f}j$ holds
if and only if $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or
}\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}
j\right)  \right)  $ (because of the definition of ``$i\prec_{f}j$'')).
In other words, $i<_{\mathbf{w}\left(
f\right)  }j$ (since the relation $<_{\mathbf{w}\left(  f\right)  }$ is the
relation $\prec_{f}$). Thus, we have shown that $i<_{\mathbf{w}\left(
f\right)  }j$ in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>_{\ZZ}j$. Thus,
$f\left(  i\right)  <f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.2.pf.1b})). Hence, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $. Hence, $i\prec
_{f}j$ (since $i\prec_{f}j$ holds if and only if $\left(  f\left(  i\right)
<f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i<_{\ZZ}j\right)  \right)  $ (because of the definition of
``$i\prec_{f}j$'')). In other words,
$i<_{\mathbf{w}\left(  f\right)  }j$ (since the relation $<_{\mathbf{w}\left(
f\right)  }$ is the relation $\prec_{f}$). Thus, we have shown that
$i<_{\mathbf{w}\left(  f\right)  }j$ in Case 2.
\par
Now, we have shown $i<_{\mathbf{w}\left(  f\right)  }j$ in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, this shows that
$i<_{\mathbf{w}\left(  f\right)  }j$ always holds.
\par
Let us now forget that we fixed $i$ and $j$. We thus have shown that if $i$
and $j$ are two elements of $P$ satisfying $i<_{P}j$, then $i<_{\mathbf{w}
\left(  f\right)  }j$. In other words, $i<_{P}j$ (for two elements $i$ and $j$
of $P$) implies $i<_{\mathbf{w}\left(  f\right)  }j$. But this yields that
$\mathbf{w}\left(  f\right)  $ is a linear extension of $P$ (by the definition
of a ``linear extension''), since we already
know that $\mathbf{w}\left(  f\right)  $ is totally ordered.
\par
So we have shown that $\mathbf{w}\left(  f\right)  $ is a linear extension of
$P$. In other words, $\mathbf{w}\left(  f\right)  \in\mathcal{L}\left(
P\right)  $ (since $\mathcal{L}\left(  P\right)  $ is the set of all linear
extensions of $P$). This proves
(\ref{sol.P-partition.stanley.altproof.wflin}).}.

Now, fix $w\in\mathcal{L}\left(  P\right)  $. Then, $w$ is a linear extension
of $P$ (since $\mathcal{L}\left(  P\right)  $ is the set of all linear
extensions of $P$). In other words, $w$ is totally ordered, and if $i$ and $j$
are two elements of $P$ satisfying $i<_{P}j$, then
\begin{equation}
i<_{w}j. \label{sol.P-partition.stanley.altproof.linex}
\end{equation}


Let us first show that
\begin{equation}
\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  \subset\mathcal{A}\left(  w\right)  .
\label{sol.P-partition.stanley.altproof.4}
\end{equation}


\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4}):} Let
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $. Then, $f$ is an element $g$ of $\mathcal{A}\left(
P\right)  $ satisfying $\mathbf{w}\left(  g\right)  =w$. In other words, $f$
is an element of $\mathcal{A}\left(  P\right)  $ and satisfies $\mathbf{w}
\left(  f\right)  =w$. Hence, the relation $<_{w}$ is the relation
$\prec_{f}$\ \ \ \ \footnote{\textit{Proof.} We know that the
relation $<_{\mathbf{w} \left(  f\right)  }$ is the relation
$\prec_{f}$. In other words, the relation
$<_{w}$ is the relation $\prec_{f}$ (since $\mathbf{w}\left(  f\right)  =w$).
Qed.}. Also, $w=\mathbf{w}\left(  f\right)  =P$ as sets.

If $i$ and $j$ are two elements of $w$, then
\begin{equation}
\left(  i<_{w}j\text{ and }i<_{\ZZ}j\text{ implies }f\left(  i\right)
\leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.4.pf.1a}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4.pf.1a}):}
Let $i$ and $j$ be two elements of $w$. We need to prove
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}). Assume that $i<_{w}j$ and
$i<_{\ZZ}j$. Notice that $i$ and $j$ are elements of $w$, and thus
elements of $P$ (since $w=P$ as sets). Also, we have $i<_{w}j$. In other
words, $i\prec_{f}j$ (since the relation $<_{w}$ is the relation $\prec_{f}$).
Thus, using (\ref{sol.P-partition.stanley.altproof.1a}), we obtain $f\left(
i\right)  \leq f\left(  j\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}).} and
\begin{equation}
\left(  i<_{w}j\text{ and }i>_{\ZZ}j\text{ implies }f\left(  i\right)
<f\left(  j\right)  \right)  \label{sol.P-partition.stanley.altproof.4.pf.1b}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.4.pf.1b}):}
Let $i$ and $j$ be two elements of $w$. We need to prove
(\ref{sol.P-partition.stanley.altproof.4.pf.1b}). Assume that $i<_{w}j$ and
$i>_{\ZZ}j$. Notice that $i$ and $j$ are elements of $w$, and thus
elements of $P$ (since $w=P$ as sets). Also, we have $i<_{w}j$. In other
words, $i\prec_{f}j$ (since the relation $<_{w}$ is the relation $\prec_{f}$).
Thus, using (\ref{sol.P-partition.stanley.altproof.1b}), we obtain $f\left(
i\right)  <f\left(  j\right)  $. This proves
(\ref{sol.P-partition.stanley.altproof.4.pf.1b}).}. Thus, $f\in\mathcal{A}
\left(  w\right)  $\ \ \ \ \footnote{\textit{Proof.} Recall that $f$ is a
$w$-partition if and only if every two elements $i$ and $j$ of $w$ satisfy
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.4.pf.1b}) (due to the definition of a
``$w$-partition''). Thus, $f$ is a
$w$-partition (since every two elements $i$ and $j$ of $w$ satisfy
(\ref{sol.P-partition.stanley.altproof.4.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.4.pf.1b})). In other words,
$f\in\mathcal{A}\left(  w\right)  $ (since $\mathcal{A}\left(  w\right)  $ is
the set of all $w$-partitions (by the definition of $\mathcal{A}\left(
w\right)  $)), qed.}.

Let us now forget that we fixed $f$. We thus have proven that every
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $ satisfies $f\in\mathcal{A}\left(  w\right)  $. In
other words, $\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}
\left(  g\right)  =w\right\}  \subset\mathcal{A}\left(  w\right)  $. This
proves (\ref{sol.P-partition.stanley.altproof.4}).

Let us next show that
\begin{equation}
\mathcal{A}\left(  w\right)  \subset\left\{  g\in\mathcal{A}\left(
P\right)  \ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  .
\label{sol.P-partition.stanley.altproof.5}
\end{equation}


\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5}):} Let
$f\in\mathcal{A}\left(  w\right)  $. Then, $f$ is an element of $\mathcal{A}
\left(  w\right)  $. In other words, $f$ is a $w$-partition (since
$\mathcal{A}\left(  w\right)  $ is the set of all $w$-partitions (due to the
definition of $\mathcal{A}\left(  w\right)  $)). In other words, every two
elements $i$ and $j$ of $w$ satisfy
\begin{equation}
\left(  i<_{w}j\text{ and }i<_{\ZZ}j\text{ implies }f\left(  i\right)
\leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.5.pf.1a}
\end{equation}
and
\begin{equation}
\left(  i<_{w}j\text{ and }i>_{\ZZ}j\text{ implies }f\left(  i\right)
<f\left(  j\right)  \right)  \label{sol.P-partition.stanley.altproof.5.pf.1b}
\end{equation}
(because $f$ is a $w$-partition if and only if every two elements $i$ and $j$
of $w$ satisfy (\ref{sol.P-partition.stanley.altproof.5.pf.1a}) and
(\ref{sol.P-partition.stanley.altproof.5.pf.1b}) (due to the definition of a
``$w$-partition'')). Consequently, every two
elements $i$ and $j$ of $w$ satisfying $i<_{w}j$ satisfy
\begin{equation}
i<_{\mathbf{w}\left(  f\right)  }j
\label{sol.P-partition.stanley.altproof.5.pf.1c}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.1c}):}
Let $i$ and $j$ be two elements of $w$ satisfying $i<_{w}j$. We must prove
that $i<_{\mathbf{w}\left(  f\right)  }j$.
\par
Clearly, $i$ and $j$ are elements of $w$, and therefore elements of $P$ (since
$w=P$ as sets). We have $i<_{w}j$ and thus $i\neq j$. Hence, either
$i<_{\ZZ}j$ or $i>_{\ZZ}j$. In other words, we are in one of the
following two Cases:
\par
\textit{Case 1:} We have $i<_{\ZZ}j$.
\par
\textit{Case 2:} We have $i>_{\ZZ}j$.
\par
Let us consider Case 1 first. In this case, we have $i<_{\ZZ}j$. Thus,
$f\left(  i\right)  =f\left(  j\right)  $ implies $\left(  f\left(  i\right)
=f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  $ (in a logical sense
-- i.e., if $f\left(  i\right)  =f\left(  j\right)  $ is true, then so is
$\left(  f\left(  i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}
j\right)  $). But $i<_{\ZZ}j$, and thus $f\left(  i\right)
\leq f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1a})).
In other words, $\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or
}f\left(  i\right)  =f\left(  j\right)  \right)  $. Hence, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $ (since $f\left(
i\right)  =f\left(  j\right)  $ implies $\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i<_{\ZZ}j\right)  $). Therefore, $i\prec_{f}j$
(since $i\prec_{f}j$ holds if and only if $\left(  f\left(  i\right)
<f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(  j\right)
\text{ and }i<_{\ZZ}j\right)  \right)  $ (because of the definition of
``$i\prec_{f}j$'')). In other words,
$i<_{\mathbf{w}\left(  f\right)  }j$ (since the relation $<_{\mathbf{w}\left(
f\right)  }$ is the relation $\prec_{f}$). Thus, $i<_{\mathbf{w}\left(
f\right)  }j$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>_{\ZZ}j$. Thus,
$f\left(  i\right)  <f\left(  j\right)  $ (according to
(\ref{sol.P-partition.stanley.altproof.5.pf.1b})). Hence, $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $. Therefore,
$i\prec_{f}j$ (since $i\prec_{f}j$ holds if and only if $\left(  f\left(
i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(  i\right)  =f\left(
j\right)  \text{ and }i<_{\ZZ}j\right)  \right)  $ (because of the
definition of ``$i\prec_{f}j$'')). In other
words, $i<_{\mathbf{w}\left(  f\right)  }j$ (since the relation $<_{\mathbf{w}
\left(  f\right)  }$ is the relation $\prec_{f}$). Thus, $i<_{\mathbf{w}
\left(  f\right)  }j$ is proven in Case 2.
\par
We have now shown that $i<_{\mathbf{w}\left(  f\right)  }j$ in both Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
$i<_{\mathbf{w}\left(  f\right)  }j$ always holds. Thus,
(\ref{sol.P-partition.stanley.altproof.5.pf.1c}) is proven.}.

We shall now show that that $f\in\mathcal{A}\left(  P\right)  $ and that
$\mathbf{w}\left(  f\right)  =w$. Indeed, if $i$ and $j$ are two elements of
$P$, then
\begin{equation}
\left(  i<_{P}j\text{ and }i<_{\ZZ}j\text{ implies }f\left(  i\right)
\leq f\left(  j\right)  \right)
\label{sol.P-partition.stanley.altproof.5.pf.2a}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.2a}):}
Let $i$ and $j$ be two elements of $P$. We need to prove
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}). Assume that $i<_{P}j$ and
$i<_{\ZZ}j$. Notice that $i$ and $j$ are elements of $P$, and thus
elements of $w$ (since $w=P$ as sets). Also, recall that $i<_{P}j$. Thus,
$i<_{w}j$ (by (\ref{sol.P-partition.stanley.altproof.linex})), and therefore
$f\left(  i\right)  \leq f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1a})). This proves
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}).} and
\begin{equation}
\left(  i<_{P}j\text{ and }i>_{\ZZ}j\text{ implies }f\left(  i\right)
<f\left(  j\right)  \right)  \label{sol.P-partition.stanley.altproof.5.pf.2b}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.2b}):}
Let $i$ and $j$ be two elements of $P$. We need to prove
(\ref{sol.P-partition.stanley.altproof.5.pf.2b}). Assume that $i<_{P}j$ and
$i>_{\ZZ}j$. Notice that $i$ and $j$ are elements of $P$, and thus
elements of $w$ (since $w=P$ as sets). Also, recall that $i<_{P}j$. Thus,
$i<_{w}j$ (by (\ref{sol.P-partition.stanley.altproof.linex})), and therefore
$f\left(  i\right)  <f\left(  j\right)  $ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1b})). This proves
(\ref{sol.P-partition.stanley.altproof.5.pf.2b}).}. Thus, $f\in\mathcal{A}
\left(  P\right)  $\ \ \ \ \footnote{\textit{Proof.} Recall that $f$ is a
$P$-partition if and only if every two elements $i$ and $j$ of $P$ satisfy
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}) and
(\ref{sol.P-partition.stanley.altproof.5.pf.2b}) (due to the definition of a
``$P$-partition''). Thus, $f$ is a
$P$-partition (since every two elements $i$ and $j$ of $P$ satisfy
(\ref{sol.P-partition.stanley.altproof.5.pf.2a}) and
(\ref{sol.P-partition.stanley.altproof.5.pf.2b})). In other words,
$f\in\mathcal{A}\left(  P\right)  $ (since $\mathcal{A}\left(  P\right)  $ is
the set of all $P$-partitions), qed.}. Furthermore, for any two elements $i$
and $j$ of $P$, we have
\begin{equation}
\left(  i<_{\mathbf{w}\left(  f\right)  }j\text{ if and only if }
i<_{w}j\right)  \label{sol.P-partition.stanley.altproof.5.pf.3}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.P-partition.stanley.altproof.5.pf.3}):}
Let $i$ and $j$ be two elements of $P$. We need to prove
(\ref{sol.P-partition.stanley.altproof.5.pf.3}).
\par
If $i=j$, then $\left(  i<_{\mathbf{w}\left(  f\right)  }j\text{ if and only
if }i<_{w}j\right)  $ holds (because if $i=j$, then neither $i<_{\mathbf{w}
\left(  f\right)  }j$ nor $i<_{w}j$ is true). Hence, if $i=j$, then
(\ref{sol.P-partition.stanley.altproof.5.pf.3}) is proven. Thus, for the rest
of this proof of (\ref{sol.P-partition.stanley.altproof.5.pf.3}), we can WLOG
assume that we don't have $i=j$. Assume this.
\par
Clearly, $i$ and $j$ are elements of $P$, and therefore elements of $w$ (since
$w=P$ as sets). These elements $i$ and $j$ are distinct (since we don't have
$i=j$). Hence, we have either $i<_{w}j$ or $i>_{w}j$ (because $w$ is totally
ordered). In other words, we must be in one of the following two Cases:
\par
\textit{Case 1:} We have $i<_{w}j$.
\par
\textit{Case 2:} We have $i>_{w}j$.
\par
Let us consider Case 1 first. In this case, we have $i<_{w}j$. Thus,
$i<_{\mathbf{w}\left(  f\right)  }j$ (by
(\ref{sol.P-partition.stanley.altproof.5.pf.1c})). Hence, both $i<_{\mathbf{w}
\left(  f\right)  }j$ and $i<_{w}j$ hold. Thus, $\left(  i<_{\mathbf{w}\left(
f\right)  }j\text{ if and only if }i<_{w}j\right)  $ holds. We have therefore
proven (\ref{sol.P-partition.stanley.altproof.5.pf.3}) in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>_{w}j$. Hence, $i<_{w}j$
is false. But from $i>_{w}j$, we also obtain $j<_{w}i$. Thus, $j<_{\mathbf{w}
\left(  f\right)  }i$ (by (\ref{sol.P-partition.stanley.altproof.5.pf.1c}),
applied to $j$ and $i$ instead of $i$ and $j$). Therefore, $i>_{\mathbf{w}
\left(  f\right)  }j$. Hence, $i<_{\mathbf{w}\left(  f\right)  }j$ is false.
Thus, we know that both $i<_{\mathbf{w}\left(  f\right)  }j$ and $i<_{w}j$ are
false. Thus, $\left(  i<_{\mathbf{w}\left(  f\right)  }j\text{ if and only if
}i<_{w}j\right)  $ holds. We have therefore proven
(\ref{sol.P-partition.stanley.altproof.5.pf.3}) in Case 2.
\par
We have now proven (\ref{sol.P-partition.stanley.altproof.5.pf.3}) in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, this
shows that (\ref{sol.P-partition.stanley.altproof.5.pf.3}) always holds.
Qed.}. In other words, the relations $<_{\mathbf{w}\left(  f\right)  }$ and
$<_{w}$ are identical. Thus, $\mathbf{w}\left(  f\right)  =w$.

Now, we know that $f\in\mathcal{A}\left(  P\right)  $ and $\mathbf{w}\left(
f\right)  =w$. In other words, $f$ is an element $g$ of $\mathcal{A}\left(
P\right)  $ satisfying $\mathbf{w}\left(  g\right)  =w$. In other words,
$f\in\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(
g\right)  =w\right\}  $.

Let us now forget that we fixed $f$. We thus have shown that $f\in\left\{
g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  $ for every $f\in\mathcal{A}\left(  w\right)  $. In other words,
\[
\mathcal{A}\left(  w\right)  \subset\left\{  g\in\mathcal{A}\left(
P\right)  \ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  .
\]
This proves (\ref{sol.P-partition.stanley.altproof.5}).

Combining (\ref{sol.P-partition.stanley.altproof.4}) with
(\ref{sol.P-partition.stanley.altproof.5}), we obtain
\begin{equation}
\left\{  g\in\mathcal{A}\left(  P\right)  \ \mid\ \mathbf{w}\left(  g\right)
=w\right\}  =\mathcal{A}\left(  w\right)  .
\label{sol.P-partition.stanley.altproof.6}
\end{equation}


Let us now forget that we fixed $w$. We thus have proven
(\ref{sol.P-partition.stanley.altproof.6}) for every $w\in\mathcal{L}\left(
P\right)  $.

Now, the definition of $F_{P}\left(  \xx\right)  $ yields
\begin{align*}
F_{P}\left(  \xx\right)   &  =\sum_{f\in\mathcal{A}\left(  P\right)
}{\mathbf{x}}_{f}=\sum_{w\in\mathcal{L}\left(  P\right)  }\underbrace{\sum\limits
_{\substack{f\in\mathcal{A}\left(  P\right)  ;\\\mathbf{w}\left(  f\right)
=w}}}_{\substack{=\sum_{f\in\left\{  g\in\mathcal{A}\left(  P\right)
\ \mid\ \mathbf{w}\left(  g\right)  =w\right\}  }\\=\sum_{f\in\mathcal{A}
\left(  w\right)  }\\
\text{(by (\ref{sol.P-partition.stanley.altproof.6}))}}}
{\mathbf{x}}_{f}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{w}\left(  f\right)
\in\mathcal{L}\left(  P\right)  \text{ for every }f\in\mathcal{A}\left(
P\right)  \text{ (by (\ref{sol.P-partition.stanley.altproof.wflin}))}\right)
\\
&  =\sum_{w\in\mathcal{L}\left(  P\right)  }\underbrace{\sum_{f\in
\mathcal{A}\left(  w\right)  }\xx_{f}}_{\substack{=F_{w}\left(
\xx\right)  \\\text{(since }F_{w}\left(  \xx\right)  =\sum
_{f\in\mathcal{A}\left(  w\right)  }\xx_{f}\\\text{(by the definition
of }F_{w}\left(  \xx\right)  \text{))}}}=\sum_{w\in\mathcal{L}\left(
P\right)  }F_{w}\left(  \xx\right)  .
\end{align*}
This completes our proof of Theorem \ref{Stanley's-P-partition-theorem}.
\end{proof}
\end{comment}
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lalpha-std.proofs}}

\textit{Solution to Exercise \ref{exe.Lalpha-std.proofs}.} Let us first state
a basic fact about totally ordered sets:

\begin{lemma}
\label{lem.sol.Lalpha-std.proofs.sort}Let $T$ be a totally ordered set, and
let $<_{T}$ be the smaller relation of $T$. Let $\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  $ be a finite list of \textbf{distinct} elements of $T$.
Then, there is a \textbf{unique} permutation $\sigma\in\Symm_n$ such
that $a_{\sigma\left(  1\right)  }<_{T}a_{\sigma\left(  2\right)  }<_{T}%
\cdots<_{T}a_{\sigma\left(  n\right)  }$.
\end{lemma}

Lemma \ref{lem.sol.Lalpha-std.proofs.sort} is well-known (it essentially says
that a finite list of distinct elements of a totally ordered set can be sorted
into increasing order by a unique permutation). We shall use it to prove
Proposition \ref{prop.words.std-exists} later.

Next, we state an elementary property of permutations:

\begin{proposition}
\label{prop.sol.Lalpha-std.proofs.Inv=1}Let $n\in \NN$. Let $\varphi$
and $\psi$ be two elements of $\Symm_n$. Assume that for every two
elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }\psi\left(  a\right)  <\psi\left(  b\right)  \right)  .
\label{eq.prop.sol.Lalpha-std.proofs.Inv=1.ass}%
\end{equation}
Then, $\varphi=\psi$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1}.]If $a$ and $b$
are two elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, then
we have the logical equivalence%
\begin{equation}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)
\ \Longleftrightarrow\ \left(  \psi\left(  a\right)  <\psi\left(  b\right)
\right)  \label{pf.prop.sol.Lalpha-std.proofs.Inv=1.-1}%
\end{equation}
(by (\ref{eq.prop.sol.Lalpha-std.proofs.Inv=1.ass})). We next will show that
this equivalence holds even if we don't require $a<b$:

\begin{statement}
\textit{Observation 1:} Let $p\in\left\{  1,2,\ldots,n\right\}  $ and
$q\in\left\{  1,2,\ldots,n\right\}  $. Then, we have the logical equivalence
\[
\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)  \right)
\ \Longleftrightarrow\ \left(  \psi\left(  p\right)  <\psi\left(  q\right)
\right)  .
\]

\end{statement}

[\textit{Proof of Observation 1:} Let us first prove the logical implication%
\begin{equation}
\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)  \right)
\ \Longrightarrow\ \left(  \psi\left(  p\right)  <\psi\left(  q\right)
\right)  . \label{pf.prop.sol.Lalpha-std.proofs.Inv=1.a}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.a}):} Assume that
$\varphi\left(  p\right)  <\varphi\left(  q\right)  $. We want to show that
$\psi\left(  p\right)  <\psi\left(  q\right)  $.

If $p<q$, then (\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.-1}) (applied to
$a=p$ and $b=q$) yields the equivalence $\left(  \varphi\left(  p\right)
<\varphi\left(  q\right)  \right)  \ \Longleftrightarrow\ \left(  \psi\left(
p\right)  <\psi\left(  q\right)  \right)  $, and thus $\psi\left(  p\right)
<\psi\left(  q\right)  $ follows (since $\varphi\left(  p\right)
<\varphi\left(  q\right)  $). Hence, for the rest of the proof of $\psi\left(
p\right)  <\psi\left(  q\right)  $, we WLOG assume that we don't have $p<q$.
Hence, we have $p\geq q$.

From $\varphi\left(  p\right)  <\varphi\left(  q\right)  $, we also obtain
$\varphi\left(  p\right)  \neq\varphi\left(  q\right)  $, so that $p\neq q$.
Combining this with $p\geq q$, we obtain $p>q$. Hence, $q<p$. Thus,
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.-1}) (applied to $a=q$ and $b=p$)
yields the logical equivalence $\left(  \varphi\left(  q\right)
<\varphi\left(  p\right)  \right)  \ \Longleftrightarrow\ \left(  \psi\left(
q\right)  <\psi\left(  p\right)  \right)  $. Since we don't have
$\varphi\left(  q\right)  <\varphi\left(  p\right)  $ (because $\varphi\left(
p\right)  <\varphi\left(  q\right)  $), we thus conclude that we don't have
$\psi\left(  q\right)  <\psi\left(  p\right)  $. Therefore, we have
$\psi\left(  p\right)  \leq\psi\left(  q\right)  $. But the map $\psi$ is
injective (since $\psi\in\Symm_n$); therefore, from $p\neq q$, we
obtain $\psi\left(  p\right)  \neq\psi\left(  q\right)  $. Combining this with
$\psi\left(  p\right)  \leq\psi\left(  q\right)  $, we obtain $\psi\left(
p\right)  <\psi\left(  q\right)  $. This completes the proof of
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.a}).]

So we have proven (\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.a}). The same
argument (but with the roles of $\varphi$ and $\psi$ interchanged) yields the
logical implication%
\[
\left(  \psi\left(  p\right)  <\psi\left(  q\right)  \right)
\ \Longrightarrow\ \left(  \varphi\left(  p\right)  <\varphi\left(  q\right)
\right)  .
\]
Combining this with (\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.a}), we obtain
the equivalence $\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)
\right)  \ \Longleftrightarrow\ \left(  \psi\left(  p\right)  <\psi\left(
q\right)  \right)  $. This proves Observation 1.]

Now, let $q\in\left\{  1,2,\ldots,n\right\}  $. Then, the map $\varphi$ is a
bijection $\left\{  1,2,\ldots,n\right\}  \rightarrow\left\{  1,2,\ldots
,n\right\}  $ (since $\varphi\in\Symm_n$). Thus, we can substitute
$i$ for $\varphi\left(  p\right)  $ in the sum $\sum\limits_{\substack{p\in\left\{
1,2,\ldots,n\right\}  ;\\\varphi\left(  p\right)  <\varphi\left(  q\right)
}}1$. We thus obtain
\begin{equation}
\sum\limits_{\substack{p\in\left\{  1,2,\ldots,n\right\}  ;\\\varphi\left(  p\right)
<\varphi\left(  q\right)  }}1=\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,n\right\}  ;\\i<\varphi\left(  q\right)  }}1=\sum_{i\in\left\{
1,2,\ldots,\varphi\left(  q\right)  -1\right\}  }1=\left(  \varphi\left(
q\right)  -1\right)  \cdot1=\varphi\left(  q\right)  -1.
\label{pf.prop.sol.Lalpha-std.proofs.Inv=1.d}%
\end{equation}
The same argument (applied to $\psi$ instead of $\varphi$) yields%
\[
\sum\limits_{\substack{p\in\left\{  1,2,\ldots,n\right\}  ;\\\psi\left(  p\right)
<\psi\left(  q\right)  }}1=\psi\left(  q\right)  -1.
\]
However, Observation 1 shows that the sums $\sum\limits_{\substack{p\in\left\{
1,2,\ldots,n\right\}  ;\\\varphi\left(  p\right)  <\varphi\left(  q\right)
}}1$ and $\sum\limits_{\substack{p\in\left\{  1,2,\ldots,n\right\}  ;\\\psi\left(
p\right)  <\psi\left(  q\right)  }}1$ range over the same values of $p$. Thus,%
\[
\sum\limits_{\substack{p\in\left\{  1,2,\ldots,n\right\}  ;\\\varphi\left(  p\right)
<\varphi\left(  q\right)  }}1=\sum\limits_{\substack{p\in\left\{  1,2,\ldots
,n\right\}  ;\\\psi\left(  p\right)  <\psi\left(  q\right)  }}1=\psi\left(
q\right)  -1.
\]
Comparing this with (\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.d}), we obtain
$\varphi\left(  q\right)  -1=\psi\left(  q\right)  -1$. Hence, $\varphi\left(
q\right)  =\psi\left(  q\right)  $.

Now, forget that we fixed $q$. We thus have shown that $\varphi\left(
q\right)  =\psi\left(  q\right)  $ for each $q\in\left\{  1,2,\ldots
,n\right\}  $. In other words, $\varphi=\psi$. This proves Proposition
\ref{prop.sol.Lalpha-std.proofs.Inv=1}.
\end{proof}

For later use (in a different solution further below), let us derive another
proposition from Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1}. It relies
on the following notation:

\begin{definition}
\label{def.sol.Lalpha-std.proofs.Inv} Let $n \in{\NN}$. Let $\varphi
\in\Symm_n$. Define the
\emph{inversion set}\index{inversion set of a permutation}
\dfn{$\operatorname*{Inv} \varphi$}
of $\varphi$ to be the set $\left\{  \left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j;\ \varphi\left(  i\right)
>\varphi\left(  j\right)  \right\}  $.
\end{definition}

(Note that this notation is not completely standard. Some authors, instead,
define $\operatorname*{Inv}\varphi$ to be \newline
$\left\{  \left(  \varphi\left(
i\right)  ,\varphi\left(  j\right)  \right)  \ \mid\ \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2};\ i<j;\ \varphi\left(  i\right)
>\varphi\left(  j\right)  \right\}  $. This is a different set, although of
the same size.)

\begin{proposition}
\label{prop.sol.Lalpha-std.proofs.Inv=}Let $n\in \NN$. Let $\varphi$ and
$\psi$ be two elements of $\Symm_n$ satisfying $\operatorname*{Inv}%
\varphi=\operatorname*{Inv}\psi$. Then, $\varphi=\psi$.
\end{proposition}

(Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=} can be restated as follows:
If $n\in \NN$, then a permutation in $\Symm_n$ is uniquely
determined by its inversion set.)

Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=} is a known fact in
elementary combinatorics; let us quickly derive it from Proposition
\ref{prop.sol.Lalpha-std.proofs.Inv=1}:

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=}.]Let $a\in\left\{
1,2,\ldots,n\right\}  $ and $b\in\left\{  1,2,\ldots,n\right\}  $ be such that
$a<b$. We shall prove that $\varphi\left(  a\right)  <\varphi\left(  b\right)
$ if and only if $\psi\left(  a\right)  <\psi\left(  b\right)  $.

The pair $\left(  a,b\right)  $ is an element of $\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $a<b$. Thus, $\left(  a,b\right)  $ belongs to
$\operatorname*{Inv}\varphi$ if and only if it satisfies $\varphi\left(
a\right)  >\varphi\left(  b\right)  $ (by the definition of
$\operatorname*{Inv}\varphi$). In other words, we have the following logical
equivalence:%
\begin{equation}
\left(  \left(  a,b\right)  \in\operatorname*{Inv}\varphi\right)
\ \Longleftrightarrow\ \left(  \varphi\left(  a\right)  >\varphi\left(
b\right)  \right)  . \label{pf.prop.sol.Lalpha-std.proofs.Inv=.1}%
\end{equation}
But the map $\varphi$ is injective (since $\varphi\in\Symm_n$). Thus,
from $a\neq b$ (which follows from $a<b$), we obtain $\varphi\left(  a\right)
\neq\varphi\left(  b\right)  $. Hence, $\varphi\left(  a\right)
<\varphi\left(  b\right)  $ holds if and only if $\varphi\left(  a\right)
\leq\varphi\left(  b\right)  $. Hence, we have the following chain of
equivalences:%
\begin{align}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)  \  &
\Longleftrightarrow\ \left(  \varphi\left(  a\right)  \leq\varphi\left(
b\right)  \right)  \ \Longleftrightarrow\ \left(  \text{not }%
\underbrace{\left(  \varphi\left(  a\right)  >\varphi\left(  b\right)
\right)  }_{\substack{\Longleftrightarrow\ \left(  a,b\right)  \in
\operatorname*{Inv}\varphi\\\text{(by
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=.1}))}}}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\operatorname*{Inv}\varphi\right)  \right)  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.3}%
\end{align}
The same argument (applied to $\psi$ instead of $\varphi$) yields the
equivalence%
\begin{equation}
\left(  \psi\left(  a\right)  <\psi\left(  b\right)  \right)
\ \Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\operatorname*{Inv}\psi\right)  \right)  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.2}%
\end{equation}


Now, due to \eqref{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.3}, we have the
following chain of equivalences:%
\begin{align*}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)  \  &
\Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\operatorname*{Inv}\varphi\right)  \right)  \ \Longleftrightarrow\ \left(
\text{not }\left(  \left(  a,b\right)  \in\operatorname*{Inv}\psi\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{Inv}%
\varphi=\operatorname*{Inv}\psi\right) \\
\  &  \Longleftrightarrow\ \left(  \psi\left(  a\right)  <\psi\left(
b\right)  \right)
\end{align*}
(by \eqref{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.2}). In other words, we
have $\varphi\left(  a\right)  <\varphi\left(  b\right)  $ if and only if
$\psi\left(  a\right)  <\psi\left(  b\right)  $.

Now, forget that we fixed $a$ and $b$. We thus have shown that for every two
elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\[
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }\psi\left(  a\right)  <\psi\left(  b\right)  \right)  .
\]
Thus, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1} yields $\varphi=\psi
$. This proves Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=}.
\end{proof}

We are now ready to prove Proposition \ref{prop.words.std-exists}:

\begin{proof}
[Proof of Proposition \ref{prop.words.std-exists}.]Define a binary relation
$\prec$ on the set $\left\{  1,2,\ldots,n\right\}  $ by letting $i\prec j$
hold if and only if
\[
\left(  w_{i}<w_{j}\text{ or }\left(  w_{i}=w_{j}\text{ and }i<_{\ZZ%
}j\right)  \right)  .
\]
It is easy to see that this relation $\prec$ is the smaller relation of a
total order. Let $T$ denote the set $\left\{  1,2,\ldots,n\right\}  $ endowed
with this total order. Thus, $T=\left\{  1,2,\ldots,n\right\}  $ as sets, but
the smaller relation $<_{T}$ of the poset $T$ is the relation $\prec$.

Clearly, $\left(  1,2,\ldots,n\right)  $ is a finite list of distinct elements
of this totally ordered set $T$. Hence, Lemma
\ref{lem.sol.Lalpha-std.proofs.sort} (applied to $a_{i}=i$) yields that there
is a \textbf{unique} permutation $\sigma\in\Symm_n$ such that
$\sigma\left(  1\right)  <_{T}\sigma\left(  2\right)  <_{T}\cdots<_{T}%
\sigma\left(  n\right)  $. Consider this $\sigma$, and denote it by $\gamma$.
Thus, $\gamma$ is a permutation in $\Symm_n$ and satisfies
$\gamma\left(  1\right)  <_{T}\gamma\left(  2\right)  <_{T}\cdots<_{T}%
\gamma\left(  n\right)  $. Hence, $\gamma^{-1}\in\Symm_n$ as well.

We have $\gamma\left(  1\right)  <_{T}\gamma\left(  2\right)  <_{T}\cdots
<_{T}\gamma\left(  n\right)  $. In other words, if $i$ and $j$ are two
elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $i<j$, then%
\begin{equation}
\gamma\left(  i\right)  <_{T}\gamma\left(  j\right)
.\label{pf.prop.words.std-exists.exi.1}%
\end{equation}
Thus, for every two elements $a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$, we have
\begin{equation}
\left(  \gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(  b\right)  \text{ if
and only if }w_{a}\leq w_{b}\right)  .\label{pf.prop.words.std-exists.exi.2}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.words.std-exists.exi.2}):} Let $a$ and $b$ be
two elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$. Set
$i=\gamma^{-1}\left(  a\right)  $ and $j=\gamma^{-1}\left(  b\right)  $. Thus,
$i$ and $j$ are two elements of $\left\{  1,2,\ldots,n\right\}  $ (since
$\gamma^{-1}\in\Symm_n$). Also, from $i=\gamma^{-1}\left(  a\right)
$, we obtain $\gamma\left(  i\right)  =a$. From $j=\gamma^{-1}\left(
b\right)  $, we obtain $\gamma\left(  j\right)  =b$.

We are in one of the following two cases:

\textit{Case 1:} We have $i<j$.

\textit{Case 2:} We have $i\geq j$.

Let us first consider Case 1. In this case, we have $i<j$. This rewrites as
$\gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(  b\right)  $ (since
$i=\gamma^{-1}\left(  a\right)  $ and $j=\gamma^{-1}\left(  b\right)  $). On
the other hand, from $i<j$, we obtain $\gamma\left(  i\right)  <_{T}%
\gamma\left(  j\right)  $ (by (\ref{pf.prop.words.std-exists.exi.1})). This
rewrites as $a<_{T}b$ (since $\gamma\left(  i\right)  =a$ and $\gamma\left(
j\right)  =b$). This rewrites as $a\prec b$ (since the relation $<_{T}$ is the
relation $\prec$). In other words, $\left(  f\left(  a\right)  <f\left(
b\right)  \text{ or }\left(  f\left(  a\right)  =f\left(  b\right)  \text{ and
}a<_{\ZZ}b\right)  \right)  $ (by the definition of the relation
$\prec$). Thus, $f\left(  a\right)  \leq f\left(  b\right)  $. But the
definition of $f$ yields $f\left(  a\right)  =w_{a}$ and $f\left(  b\right)
=w_{b}$. Thus, $w_{a}=f\left(  a\right)  \leq f\left(  b\right)  =w_{b}$. Now,
we have shown that both statements $\left(  \gamma^{-1}\left(  a\right)
<\gamma^{-1}\left(  b\right)  \right)  $ and $\left(  w_{a}\leq w_{b}\right)
$ are true. Hence, we have $\left(  \gamma^{-1}\left(  a\right)  <\gamma
^{-1}\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $. This
proves (\ref{pf.prop.words.std-exists.exi.2}) in Case 1.

Let us now consider Case 2. In this case, we have $i\geq j$. This rewrites as
$\gamma^{-1}\left(  a\right)  \geq\gamma^{-1}\left(  b\right)  $ (since
$i=\gamma^{-1}\left(  a\right)  $ and $j=\gamma^{-1}\left(  b\right)  $).
Hence, the statement $\left(  \gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(
b\right)  \right)  $ is false.

We have $a<b$. Thus, we cannot have $b<a$. In other words, we cannot have
$b<_{\ZZ}a$. Thus, we cannot have $\left(  f\left(  b\right)  =f\left(
a\right)  \text{ and }b<_{\ZZ}a\right)  $.

If we had $i=j$, then we would have $a=\gamma\left(  \underbrace{i}%
_{=j}\right)  =\gamma\left(  j\right)  =b$, which would contradict $a<b$.
Hence, we cannot have $i=j$. Thus, we have $i\neq j$. Combining this with
$i\geq j$, we obtain $i>j$. Therefore, $j<i$. Therefore,
(\ref{pf.prop.words.std-exists.exi.1}) (applied to $j$ and $i$ instead of $i$
and $j$) yields $\gamma\left(  j\right)  <_{T}\gamma\left(  i\right)  $. This
rewrites as $b<_{T}a$ (since $\gamma\left(  i\right)  =a$ and $\gamma\left(
j\right)  =b$). This rewrites as $b\prec a$ (since the relation $<_{T}$ is the
relation $\prec$). In other words, $\left(  f\left(  b\right)  <f\left(
a\right)  \text{ or }\left(  f\left(  b\right)  =f\left(  a\right)  \text{ and
}b<_{\ZZ}a\right)  \right)  $ (by the definition of the relation
$\prec$). Hence, we must have $f\left(  b\right)  <f\left(  a\right)  $ (since
we cannot have $\left(  f\left(  b\right)  =f\left(  a\right)  \text{ and
}b<_{\ZZ}a\right)  $). But the definition of $f$ yields $f\left(
a\right)  =w_{a}$ and $f\left(  b\right)  =w_{b}$. Thus, $w_{b}=f\left(
b\right)  <f\left(  a\right)  =w_{a}$. Hence, the statement $\left(  w_{a}\leq
w_{b}\right)  $ is false. Now, we have shown that both statements $\left(
\gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(  b\right)  \right)  $ and
$\left(  w_{a}\leq w_{b}\right)  $ are false. Hence, we have $\left(
\gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(  b\right)  \text{ if and only
if }w_{a}\leq w_{b}\right)  $. This proves
(\ref{pf.prop.words.std-exists.exi.2}) in Case 2.

We have now proven (\ref{pf.prop.words.std-exists.exi.2}) in each of the two
Cases 1 and 2. Hence, (\ref{pf.prop.words.std-exists.exi.2}) is always proven.]

So we know that $\gamma^{-1}$ is a permutation in $\Symm_n$, and that
for every two elements $a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$, we have $\left(  \gamma^{-1}\left(  a\right)  <\gamma
^{-1}\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $ (by
(\ref{pf.prop.words.std-exists.exi.2})). Thus, there exists \textbf{at least
one} permutation $\sigma\in\Symm_n$ such that for every two elements
$a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
$\left(  \sigma\left(  a\right)  <\sigma\left(  b\right)  \text{ if and only
if }w_{a}\leq w_{b}\right)  $ (namely, $\sigma=\gamma^{-1}$).

It remains to prove that there exists \textbf{at most one} such permutation
$\sigma$. In other words, it remains to prove that any two such permutations
$\sigma$ are equal. In other words, it remains to prove the following claim:

\begin{statement}
\textit{Claim 1:} Let $\varphi$ and $\psi$ be two permutations $\sigma
\in\Symm_n$ such that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(
a\right)  <\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq
w_{b}\right)  $. Then, $\varphi=\psi$.
\end{statement}

[\textit{Proof of Claim 1:} We know that $\varphi$ is a permutation $\sigma
\in\Symm_n$ such that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(
a\right)  <\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq
w_{b}\right)  $. In other words, $\varphi$ is a permutation in $\Symm_n$,
and has the property that for every two elements $a$ and $b$ of
$\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }w_{a}\leq w_{b}\right)  . \label{pf.prop.words.std-exists.uni.1}%
\end{equation}


Now, let $a$ and $b$ be two elements of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$. Then, from (\ref{pf.prop.words.std-exists.uni.1}), we obtain
the logical equivalence $\left(  \varphi\left(  a\right)  <\varphi\left(
b\right)  \right)  \Longleftrightarrow\left(  w_{a}\leq w_{b}\right)  $. The
same argument (applied to $\psi$ instead of $\varphi$) yields the logical
equivalence $\left(  \psi\left(  a\right)  <\psi\left(  b\right)  \right)
\Longleftrightarrow\left(  w_{a}\leq w_{b}\right)  $. Hence, we have the
following chain of logical equivalences:%
\[
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)
\ \Longleftrightarrow\ \left(  w_{a}\leq w_{b}\right)  \ \Longleftrightarrow
\ \left(  \psi\left(  a\right)  <\psi\left(  b\right)  \right)
\]
(because of the equivalence $\left(  \psi\left(  a\right)  <\psi\left(
b\right)  \right)  \Longleftrightarrow\left(  w_{a}\leq w_{b}\right)  $). In
other words, we have
\[
\left(  \varphi\left(  a\right)  <\varphi\left(
b\right)  \text{ if and only if }\psi\left(  a\right)  <\psi\left(  b\right)
\right) .
\]

Now, forget that we fixed $a$ and $b$. We thus have proven that for every two
elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\[
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }\psi\left(  a\right)  <\psi\left(  b\right)  \right)  .
\]
Thus, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1} yields that
$\varphi=\psi$. This proves Claim 1.]

Hence, Proposition \ref{prop.words.std-exists} is proven.
\end{proof}

Next, we shall show a lemma that will be crucial in our proof of Lemma
\ref{lem.Lalpha-std.std-L}:

\begin{lemma}
\label{lem.sol.Lalpha-std.proofs.P-pa}Let $n\in \NN$. Let $\tau
\in\Symm_n$. Let $P$ be the labelled poset whose underlying set is
$\left\{  1,2,\ldots,n\right\}  $ and which (as a poset) is the total order
$\left(  \tau\left(  1\right)  <\tau\left(  2\right)  <\cdots<\tau\left(
n\right)  \right)  $ (that is, the order $<_{P}$ is given by $\tau\left(
1\right)  <_{P}\tau\left(  2\right)  <_{P}\cdots<_{P}\tau\left(  n\right)  $).

Let $\mathfrak{A}$ denote the totally ordered set $\left\{  1<2<3<\cdots
\right\}  $ of positive integers. Let $f:P\rightarrow\mathfrak{A}$ be any map.
Then, we have the following logical equivalence:%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\tau^{-1}\right)
\]
(where we treat $\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots
,f\left(  n\right)  \right)  $ as a word in $\mathfrak{A}^{n}$).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lalpha-std.proofs.P-pa}.]We have $P=\left\{
1,2,\ldots,n\right\}  $ as sets. Also, the definition of the order $<_{P}$
yields%
\[
\tau\left(  1\right)  <_{P}\tau\left(  2\right)  <_{P}\cdots<_{P}\tau\left(
n\right)  .
\]


Hence, for any two elements $i$ and $j$ of $\left\{  1,2,\ldots,n\right\}  $,
we have%
\begin{equation}
\left(  \tau\left(  i\right)  <_{P}\tau\left(  j\right)  \text{ if and only if
}i<j\right)  \label{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}%
\end{equation}
(where the ``$<$'' sign in ``$i<j$'' refers to the usual smaller relation
$<_{\ZZ}$ of the totally ordered set $\ZZ$).

The map $f$ is a map from $P$ to $\mathfrak{A}$. In other words, the map $f$
is a map from $P$ to $\left\{  1,2,3,\ldots\right\}  $ (since $\mathfrak{A}%
=\left\{  1,2,3,\ldots\right\}  $).

Also, $\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(
n\right)  \right)  $ is a word in $\mathfrak{A}^{n}$. Denote this word by $w$.
Thus,
\[
w=\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)
\right)  \in\mathfrak{A}^{n}.
\]


For each $i\in\left\{  1,2,\ldots,n\right\}  $, the $i$-th letter of the word
$w$ has been denoted by $w_{i}$. Thus, $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $, so that $\left(  w_{1},w_{2},\ldots,w_{n}\right)  =w=\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  $.
In other words,%
\begin{equation}
w_{i}=f\left(  i\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{
1,2,\ldots,n\right\}  . \label{pf.lem.sol.Lalpha-std.proofs.P-pa.wi=}%
\end{equation}
The definition of the standardization $\operatorname*{std}w$ shows that
$\operatorname*{std}w$ is the unique permutation $\sigma\in\Symm_n$
such that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(  a\right)
<\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $. In
particular, $\operatorname*{std}w$ is such a permutation. In other words,
$\operatorname*{std}w$ is a permutation in $\Symm_n$ and has the
property that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
w_{a}\leq w_{b}\right)  . \label{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw}%
\end{equation}


Hence, for every two elements $a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$, we have
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }f\left(
a\right)  \leq f\left(  b\right)  \right)
.\label{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}%
\end{equation}
(Indeed, this just a restatement of
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw}), since
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.wi=}) yields $w_{a}=f\left(  a\right)
$ and $w_{b}=f\left(  b\right)  $.)

We shall next prove the following two claims:

\begin{statement}
\textit{Claim 1:} If $f\in\mathcal{A}\left(  P\right)  $, then
$\operatorname*{std}w=\tau^{-1}$.
\end{statement}

\begin{statement}
\textit{Claim 2:} If $\operatorname*{std}w=\tau^{-1}$, then $f\in
\mathcal{A}\left(  P\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} Assume that $f\in\mathcal{A}\left(  P\right)  $.
Thus, $f$ is a $P$-partition (since $\mathcal{A}\left(  P\right)  $ is the set
of all $P$-partitions). In other words, $f$ is a map $P\rightarrow\left\{
1,2,3,\ldots\right\}  $ with the properties
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.1}%
\end{equation}
and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.2}%
\end{equation}
(because this is how a $P$-partition is defined).

For every two elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}%
):} Let $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{  1,2,\ldots
,n\right\}  $ be such that $a<b$. We must prove
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}).
\par
We have $a\in\left\{  1,2,\ldots,n\right\}  =P$ and $b\in\left\{
1,2,\ldots,n\right\}  =P$.
\par
Note that $a<b$. In other words, $a<_{\ZZ}b$. In other words,
$b>_{\ZZ}a$.
\par
Let $i=\tau^{-1}\left(  a\right)  $ and $j=\tau^{-1}\left(  b\right)  $. Thus,
$i$ and $j$ belong to $\left\{  1,2,\ldots,n\right\}  $. From $i=\tau
^{-1}\left(  a\right)  $, we obtain $a=\tau\left(  i\right)  $. From
$j=\tau^{-1}\left(  b\right)  $, we obtain $b=\tau\left(  j\right)  $. If we
had $i=j$, then we would have $a=\tau\left(  \underbrace{i}_{=j}\right)
=\tau\left(  j\right)  =b$, which would contradict $a<b$. Thus, $i\neq j$.
Hence, we are in one of the following two cases:
\par
\textit{Case 1:} We have $i<j$.
\par
\textit{Case 2:} We have $i>j$.
\par
Let us first consider Case 1. In this case, we have $i<j$. Hence,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) shows that $\tau\left(
i\right)  <_{P}\tau\left(  j\right)  $. In view of $\tau\left(  i\right)  =a$
and $\tau\left(  j\right)  =b$, this rewrites as $a<_{P}b$. Also,
$a<_{\ZZ}b$. Hence, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.1})
(applied to $a$ and $b$ instead of $i$ and $j$) yields $f\left(  a\right)
\leq f\left(  b\right)  $. Because of
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}), this shows that $\left(
\operatorname*{std}w\right)  \left(  a\right)  <\left(  \operatorname*{std}%
w\right)  \left(  b\right)  $. Also, $\tau^{-1}\left(  a\right)
=i<j=\tau^{-1}\left(  b\right)  $. Thus, both statements $\left(  \left(
\operatorname*{std}w\right)  \left(  a\right)  <\left(  \operatorname*{std}%
w\right)  \left(  b\right)  \right)  $ and $\left(  \tau^{-1}\left(  a\right)
<\tau^{-1}\left(  b\right)  \right)  $ are true. Hence, we have $\left(
\left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $. Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>j$. Hence, $j<i$. But
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) (applied to $j$ and $i$
instead of $i$ and $j$) shows that $\tau\left(  j\right)  <_{P}\tau\left(
i\right)  $ if and only if $j<i$. Hence, we have $\tau\left(  j\right)
<_{P}\tau\left(  i\right)  $ (since we have $j<i$). In view of $\tau\left(
i\right)  =a$ and $\tau\left(  j\right)  =b$, this rewrites as $b<_{P}a$.
Also, $b>_{\ZZ}a$. Hence,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.2}) (applied to $b$ and $a$
instead of $i$ and $j$) yields $f\left(  b\right)  <f\left(  a\right)  $. In
other words, we don't have $f\left(  a\right)  \leq f\left(  b\right)  $.
Because of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}), this shows that we
don't have $\left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  $. Also, $\tau^{-1}\left(
a\right)  =i>j=\tau^{-1}\left(  b\right)  $. Hence, we don't have $\tau
^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  $. Thus, both statements
$\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \right)  $ and $\left(
\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $ are false.
Hence, we have $\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)
<\left(  \operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if
}\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $. Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) is proven in Case 2.
\par
We have now proven (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) in
both Cases 1 and 2. Thus, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main})
always holds.}. Hence, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1}
(applied to $\varphi=\operatorname*{std}w$ and $\psi=\tau^{-1}$) yields
$\operatorname*{std}w=\tau^{-1}$. Thus, Claim 1 is proven.]

[\textit{Proof of Claim 2:} Assume that $\operatorname*{std}w=\tau^{-1}$. The
map $f:P\rightarrow\left\{  1,2,3,\ldots\right\}  $ has the following
properties:%
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}):}
Let $i\in P$ and $j\in P$ be such that $i<_{P}j$ and $i<_{\ZZ}j$. We
must prove that $f\left(  i\right)  \leq f\left(  j\right)  $.
\par
Let $a=\tau^{-1}\left(  i\right)  $ and $b=\tau^{-1}\left(  j\right)  $. Thus,
$a=\tau^{-1}\left(  i\right)  \in\left\{  1,2,\ldots,n\right\}  =P$ and
$b=\tau^{-1}\left(  j\right)  \in\left\{  1,2,\ldots,n\right\}  =P$. From
$a=\tau^{-1}\left(  i\right)  $, we obtain $\tau\left(  a\right)  =i$. From
$b=\tau^{-1}\left(  j\right)  $, we obtain $\tau\left(  b\right)  =j$.
\par
We have $i<_{P}j$. In view of $i=\tau\left(  a\right)  $ and $j=\tau\left(
b\right)  $, this rewrites as $\tau\left(  a\right)  <_{P}\tau\left(
b\right)  $.
\par
But (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) (applied to $a$ and $b$
instead of $i$ and $j$) shows that $\left(  \tau\left(  a\right)  <_{P}%
\tau\left(  b\right)  \text{ if and only if }a<b\right)  $. Hence, we have
$a<b$ (since we have $\tau\left(  a\right)  <_{P}\tau\left(  b\right)  $).
Because of $\operatorname*{std}w=\tau^{-1}$, we now have $\underbrace{\left(
\operatorname*{std}w\right)  }_{=\tau^{-1}}\left(  i\right)  =\tau^{-1}\left(
i\right)  =a<b=\underbrace{\tau^{-1}}_{=\operatorname*{std}w}\left(  j\right)
=\left(  \operatorname*{std}w\right)  \left(  j\right)  $.
\par
But $i<_{\ZZ}j$. In other words, $i<j$. Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}) (applied to $i$ and $j$
instead of $a$ and $b$) shows that $\left(  \left(  \operatorname*{std}%
w\right)  \left(  i\right)  <\left(  \operatorname*{std}w\right)  \left(
j\right)  \text{ if and only if }f\left(  i\right)  \leq f\left(  j\right)
\right)  $. Thus, $f\left(  i\right)  \leq f\left(  j\right)  $ (since
$\left(  \operatorname*{std}w\right)  \left(  i\right)  <\left(
\operatorname*{std}w\right)  \left(  j\right)  $). This proves
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}).} and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}):}
Let $i\in P$ and $j\in P$ be such that $i<_{P}j$ and $i>_{\ZZ}j$. We
must prove that $f\left(  i\right)  <f\left(  j\right)  $.
\par
Assume the contrary. Thus, $f\left(  i\right)  \geq f\left(  j\right)  $. In
other words, $f\left(  j\right)  \leq f\left(  i\right)  $.
\par
Let $a=\tau^{-1}\left(  i\right)  $ and $b=\tau^{-1}\left(  j\right)  $. Thus,
$a=\tau^{-1}\left(  i\right)  \in\left\{  1,2,\ldots,n\right\}  =P$ and
$b=\tau^{-1}\left(  j\right)  \in\left\{  1,2,\ldots,n\right\}  =P$. From
$a=\tau^{-1}\left(  i\right)  $, we obtain $\tau\left(  a\right)  =i$. From
$b=\tau^{-1}\left(  j\right)  $, we obtain $\tau\left(  b\right)  =j$.
\par
We have $i<_{P}j$. In view of $i=\tau\left(  a\right)  $ and $j=\tau\left(
b\right)  $, this rewrites as $\tau\left(  a\right)  <_{P}\tau\left(
b\right)  $.
\par
But (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) (applied to $a$ and $b$
instead of $i$ and $j$) shows that $\left(  \tau\left(  a\right)  <_{P}%
\tau\left(  b\right)  \text{ if and only if }a<b\right)  $. Hence, we have
$a<b$ (since we have $\tau\left(  a\right)  <_{P}\tau\left(  b\right)  $).
\par
We have $i>_{\ZZ}j$. In other words, $i>j$. In other words, $j<i$.
Hence, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}) (applied to $j$ and $i$
instead of $a$ and $b$) shows that $\left(  \left(  \operatorname*{std}%
w\right)  \left(  j\right)  <\left(  \operatorname*{std}w\right)  \left(
i\right)  \text{ if and only if }f\left(  j\right)  \leq f\left(  i\right)
\right)  $. Thus, $\left(  \operatorname*{std}w\right)  \left(  j\right)
<\left(  \operatorname*{std}w\right)  \left(  i\right)  $ (since $f\left(
j\right)  \leq f\left(  i\right)  $). In view of $\operatorname*{std}%
w=\tau^{-1}$, this rewrites as $\tau^{-1}\left(  j\right)  <\tau^{-1}\left(
i\right)  $. Thus, $b=\tau^{-1}\left(  j\right)  <\tau^{-1}\left(  i\right)
=a$. This contradicts $a<b$. This contradiction shows that our assumption was
wrong. Hence, $f\left(  i\right)  <f\left(  j\right)  $. This proves
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}).}. Thus, $f$ is a
$P$-partition (because this is how a $P$-partition is defined). In other
words, $f\in\mathcal{A}\left(  P\right)  $ (since $\mathcal{A}\left(
P\right)  $ is the set of all $P$-partitions). This proves Claim 2.]

Combining Claim 1 with Claim 2, we obtain the logical equivalence $\left(
f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow\ \left(
\operatorname*{std}w=\tau^{-1}\right)  $. In view of $w=\left(  f\left(
1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  $, this
rewrites as%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\tau^{-1}\right)  .
\]
This proves Lemma \ref{lem.sol.Lalpha-std.proofs.P-pa}.
\end{proof}

Next, we will use a trivial consequence of Proposition
\ref{fundamental-P-partition-prop}:

\begin{lemma}
\label{lem.sol.Lalpha-std.proofs.4}Let $n\in \NN$. Let $\sigma
\in\Symm_n$. Let $P$ be the labelled poset whose underlying set is
$\left\{  1,2,\ldots,n\right\}  $ and which (as a poset) is the total order
$\left(  \sigma\left(  1\right)  <\sigma\left(  2\right)  <\cdots
<\sigma\left(  n\right)  \right)  $ (that is, the order $<_{P}$ is given by
$\sigma\left(  1\right)  <_{P}\sigma\left(  2\right)  <_{P}\cdots<_{P}%
\sigma\left(  n\right)  $). Then, $F_{P}\left(  \xx\right)
=L_{\gamma\left(  \sigma\right)  }$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lalpha-std.proofs.4}.]In Proposition
\ref{fundamental-P-partition-prop}, we have defined $\operatorname*{Des}w$ for
any labelled poset $w$ that is a total order. Applying this definition to
$w=P$, we obtain%
\[
\operatorname*{Des}P=\left\{  i\in\left\{  1,2,\ldots,n-1\right\}
\ \mid\ \sigma\left(  i\right)  >_{\ZZ}\sigma\left(  i+1\right)
\right\}
\]
(since $P$ is the total order $\left(  \sigma\left(  1\right)  <\sigma\left(
2\right)  <\cdots<\sigma\left(  n\right)  \right)  $). Comparing this with%
\begin{align*}
\operatorname*{Des}\sigma &  =\left\{  i\in\left\{  1,2,\ldots,n-1\right\}
\ \mid\ \sigma\left(  i\right)  >\sigma\left(  i+1\right)  \right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Des}%
\sigma\right) \\
&  =\left\{  i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ \sigma\left(
i\right)  >_{\ZZ}\sigma\left(  i+1\right)  \right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the greater relation }>\text{ of
} \ZZ \text{ is the relation }>_{\ZZ}\right)  ,
\end{align*}
we obtain $\operatorname*{Des}\sigma=\operatorname*{Des}P$.

Recall that $\gamma\left(  \sigma\right)  $ is the unique composition $\alpha$
of $n$ satisfying $D(\alpha)=\operatorname{Des}\sigma$ (by the definition of
$\gamma\left(  \sigma\right)  $). In other words, $\gamma\left(
\sigma\right)  $ is the unique composition $\alpha\in\operatorname{Comp}_{n}$
having partial sums $D(\alpha)=\operatorname{Des}\sigma$. In other words,
$\gamma\left(  \sigma\right)  $ is the unique composition $\alpha
\in\operatorname{Comp}_{n}$ having partial sums $D(\alpha)=\operatorname{Des}%
P$ (since $\operatorname{Des}\sigma=\operatorname{Des}P$).

Hence, Proposition \ref{fundamental-P-partition-prop} (applied to $P$,
$\sigma\left(  i\right)  $ and $\gamma\left(  \sigma\right)  $ instead of $w$,
$w_{i}$ and $\alpha$) yields that the generating function $F_{P}\left(
\xx\right)  $ equals the fundamental quasisymmetric function
$L_{\gamma\left(  \sigma\right)  }$. Thus, $F_{P}\left(  \xx\right)
=L_{\gamma\left(  \sigma\right)  }$. This proves Lemma
\ref{lem.sol.Lalpha-std.proofs.4}.
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.Lalpha-std.std-L}.]We have $\mathfrak{A}=\left\{
1<2<3<\cdots\right\}  $; thus, $\mathfrak{A}=\left\{  1,2,3,\ldots\right\}  $
as sets.

Let $P$ be the labelled poset whose underlying set is $\left\{  1,2,\ldots
,n\right\}  $ and which (as a poset) is the total order $\left(  \sigma\left(
1\right)  <\sigma\left(  2\right)  <\cdots<\sigma\left(  n\right)  \right)  $
(that is, the order $<_{P}$ is given by $\sigma\left(  1\right)  <_{P}%
\sigma\left(  2\right)  <_{P}\cdots<_{P}\sigma\left(  n\right)  $). Lemma
\ref{lem.sol.Lalpha-std.proofs.P-pa} (applied to $\tau=\sigma$) yields that if
$f:P\rightarrow\mathfrak{A}$ is any map, then we have the following logical
equivalence:%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}\right)
\]
(where we treat $\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots
,f\left(  n\right)  \right)  $ as a word in $\mathfrak{A}^{n}$). Hence, we
have the following equality of summation signs:%
\begin{equation}
\sum\limits_{\substack{f:P\rightarrow\mathfrak{A};\\f\in\mathcal{A}\left(  P\right)
}}=\sum\limits_{\substack{f:P\rightarrow\mathfrak{A};\\\operatorname*{std}\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)
=\sigma^{-1}}}.\label{pf.lem.lyndon.grfun.std-L.2}%
\end{equation}


But every $P$-partition is a function $P\rightarrow\left\{  1,2,3,\ldots
\right\}  $. In other words, every $P$-partition is a function $P\rightarrow
\mathfrak{A}$ (since $\left\{  1,2,3,\ldots\right\}  =\mathfrak{A}$). In other
words, every $f\in\mathcal{A}\left(  P\right)  $ is a function $P\rightarrow
\mathfrak{A}$ (since $\mathcal{A}\left(  P\right)  $ is the set of all
$P$-partitions). Hence, we have the following equality of summation sums:%
\[
\sum_{f\in\mathcal{A}\left(  P\right)  }=\sum\limits_{\substack{f:P\rightarrow
\mathfrak{A};\\f\in\mathcal{A}\left(  P\right)  }}=\sum\limits
_{\substack{f:P\rightarrow\mathfrak{A};\\\operatorname*{std}\left(  f\left(
1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma
^{-1}}}
\]
(by (\ref{pf.lem.lyndon.grfun.std-L.2})). But Lemma
\ref{lem.sol.Lalpha-std.proofs.4} yields $F_{P}\left(  \xx\right)
=L_{\gamma\left(  \sigma\right)  }$. Hence,%
\begin{align}
L_{\gamma\left(  \sigma\right)  } &  =F_{P}\left(  \xx\right)
=\underbrace{\sum_{f\in{\mathcal{A}}(P)}}_{=\sum\limits_{\substack{f:P\rightarrow
\mathfrak{A};\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(
2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}}%
\underbrace{\xx_{f}}_{\substack{=\prod_{i\in P}x_{f\left(  i\right)
}\\\text{(by the definition of }\xx_{f}\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }F_{P}\left(
\xx\right)  \right)  \nonumber\\
&  =\underbrace{\sum\limits_{\substack{f:P\rightarrow\mathfrak{A}%
;\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}}_{\substack{=\sum\limits
_{\substack{f:\left\{  1,2,\ldots,n\right\}  \rightarrow\mathfrak{A}%
;\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}\\\text{(since }P=\left\{
1,2,\ldots,n\right\}  \text{)}}}\underbrace{\prod_{i\in P}x_{f\left(
i\right)  }}_{\substack{=x_{f\left(  1\right)  }x_{f\left(  2\right)  }\cdots
x_{f\left(  n\right)  }\\\text{(since }P=\left\{  1,2,\ldots,n\right\}
\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{f:\left\{  1,2,\ldots,n\right\}  \rightarrow
\mathfrak{A};\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(
2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}x_{f\left(
1\right)  }x_{f\left(  2\right)  }\cdots x_{f\left(  n\right)  }%
.\label{pf.lem.lyndon.grfun.std-L.4}%
\end{align}


But the map%
\begin{align*}
\left\{  \text{functions }\left\{  1,2,\ldots,n\right\}  \rightarrow
\mathfrak{A}\right\}   &  \rightarrow\mathfrak{A}^{n},\\
f &  \mapsto\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(
n\right)  \right)
\end{align*}
is a bijection (indeed, this is just the standard bijection between the
functions $\left\{  1,2,\ldots,n\right\}  \rightarrow\mathfrak{A}$ and the
$n$-tuples of elements of $\mathfrak{A}$). Hence, we can substitute $\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ for $\left(  f\left(  1\right)  ,f\left(
2\right)  ,\ldots,f\left(  n\right)  \right)  $ in the sum on the right hand
side of (\ref{pf.lem.lyndon.grfun.std-L.4}). We thus obtain%
\[
\sum\limits_{\substack{f:\left\{  1,2,\ldots,n\right\}  \rightarrow\mathfrak{A}%
;\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}x_{f\left(  1\right)
}x_{f\left(  2\right)  }\cdots x_{f\left(  n\right)  }=\sum\limits_{\substack{\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n};\\\operatorname*{std}%
\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\sigma^{-1}}}x_{w_{1}}x_{w_{2}%
}\cdots x_{w_{n}}.
\]
Hence, (\ref{pf.lem.lyndon.grfun.std-L.4}) becomes%
\[
L_{\gamma\left(  \sigma\right)  }=\sum\limits_{\substack{f:\left\{  1,2,\ldots
,n\right\}  \rightarrow\mathfrak{A};\\\operatorname*{std}\left(  f\left(
1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma
^{-1}}}x_{f\left(  1\right)  }x_{f\left(  2\right)  }\cdots x_{f\left(
n\right)  }=\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\mathfrak{A}^{n};\\\operatorname*{std}\left(  w_{1},w_{2},\ldots
,w_{n}\right)  =\sigma^{-1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\]
Comparing this with%
\begin{align*}
\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}%
}}\xx_{w} &  =\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\mathfrak{A}^{n};\\\operatorname*{std}\left(  w_{1},w_{2},\ldots
,w_{n}\right)  =\sigma^{-1}}}\underbrace{\xx_{\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  }}_{\substack{=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\\\text{(by the definition of }\xx_{\left(  w_{1},w_{2},\ldots
,w_{n}\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}w\text{ as }\left(  w_{1},w_{2},\ldots,w_{n}\right)  \right)  \\
&  =\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in
\mathfrak{A}^{n};\\\operatorname*{std}\left(  w_{1},w_{2},\ldots,w_{n}\right)
=\sigma^{-1}}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}},
\end{align*}
we obtain $L_{\gamma\left(  \sigma\right)  }=\sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}$. This
proves Lemma \ref{lem.Lalpha-std.std-L}.
\end{proof}

Proposition \ref{prop.words.std-exists} and Lemma \ref{lem.Lalpha-std.std-L}
have now been proven. Thus, Exercise \ref{exe.Lalpha-std.proofs} is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.Lalpha-std.proofs}}

\textit{Solution to Exercise \ref{exe.Lalpha-std.proofs}.} Let us first state
a basic fact about totally ordered sets:

\begin{lemma}
\label{lem.sol.Lalpha-std.proofs.sort}Let $T$ be a totally ordered set, and
let $<_{T}$ be the smaller relation of $T$. Let $\left(  a_{1},a_{2}%
,\ldots,a_{n}\right)  $ be a finite list of \textbf{distinct} elements of $T$.
Then, there is a \textbf{unique} permutation $\sigma\in\Symm_n$ such
that $a_{\sigma\left(  1\right)  }<_{T}a_{\sigma\left(  2\right)  }<_{T}%
\cdots<_{T}a_{\sigma\left(  n\right)  }$.
\end{lemma}

Lemma \ref{lem.sol.Lalpha-std.proofs.sort} is well-known (it essentially says
that a finite list of distinct elements of a totally ordered set can be sorted
into increasing order by a unique permutation). We shall use it to prove
Proposition \ref{prop.words.std-exists} later.

Next, we state an elementary property of permutations:

\begin{proposition}
\label{prop.sol.Lalpha-std.proofs.Inv=1}Let $n\in \NN$. Let $\varphi$
and $\psi$ be two elements of $\Symm_n$. Assume that for every two
elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }\psi\left(  a\right)  <\psi\left(  b\right)  \right)  .
\label{eq.prop.sol.Lalpha-std.proofs.Inv=1.ass}%
\end{equation}
Then, $\varphi=\psi$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1}.]The map
$\varphi$ is a permutation of $\left\{  1,2,\ldots,n\right\}  $ (since
$\varphi\in\Symm_n$), and thus is bijective. Hence, $\varphi$ is
injective. Now, we claim the following:

\begin{statement}
\textit{Observation 1:} Let $p\in\left\{  1,2,\ldots,n\right\}  $ and
$q\in\left\{  1,2,\ldots,n\right\}  $. Then, we have the logical equivalence
\[
\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)  \right)
\ \Longleftrightarrow\ \left(  \psi\left(  p\right)  <\psi\left(  q\right)
\right)  .
\]

\end{statement}

[\textit{Proof of Observation 1:} We are in one of the following three cases:

\textit{Case 1:} We have $p<q$.

\textit{Case 2:} We have $p=q$.

\textit{Case 3:} We have $p>q$.

Let us first consider Case 1. In this case, we have $p<q$. Hence,
(\ref{eq.prop.sol.Lalpha-std.proofs.Inv=1.ass}) (applied to $a=p$ and $b=q$)
yields that we have $\left(  \varphi\left(  p\right)  <\varphi\left(
q\right)  \text{ if and only if }\psi\left(  p\right)  <\psi\left(  q\right)
\right)  $. In other words, we have the logical equivalence $\left(
\varphi\left(  p\right)  <\varphi\left(  q\right)  \right)
\ \Longleftrightarrow\ \left(  \psi\left(  p\right)  <\psi\left(  q\right)
\right)  $. Thus, Observation 1 is proven in Case 1.

Let us next consider Case 2. In this case, we have $p=q$. Hence,
$\varphi\left(  p\right)  =\varphi\left(  q\right)  $. Thus, the statement
$\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)  \right)  $ is
false. The same argument (applied to $\psi$ instead of $\varphi$) shows that
the statement $\left(  \psi\left(  p\right)  <\psi\left(  q\right)  \right)  $
is false. Hence, the statements $\left(  \varphi\left(  p\right)
<\varphi\left(  q\right)  \right)  $ and $\left(  \psi\left(  p\right)
<\psi\left(  q\right)  \right)  $ are equivalent (since they are both false).
In other words, we have the logical equivalence $\left(  \varphi\left(
p\right)  <\varphi\left(  q\right)  \right)  \ \Longleftrightarrow\ \left(
\psi\left(  p\right)  <\psi\left(  q\right)  \right)  $. Thus, Observation 1
is proven in Case 2.

Let us finally consider Case 3. In this case, we have $p>q$. Hence, $q<p$.
Thus, (\ref{eq.prop.sol.Lalpha-std.proofs.Inv=1.ass}) (applied to $a=q$ and
$b=p$) yields that we have $\left(  \varphi\left(  q\right)  <\varphi\left(
p\right)  \text{ if and only if }\psi\left(  q\right)  <\psi\left(  p\right)
\right)  $. In other words, we have the logical equivalence
\begin{equation}
\left(  \varphi\left(  q\right)  <\varphi\left(  p\right)  \right)
\ \Longleftrightarrow\ \left(  \psi\left(  q\right)  <\psi\left(  p\right)
\right)  . \label{pf.prop.sol.Lalpha-std.proofs.Inv=1.0}%
\end{equation}


From $p>q$, we obtain $p\neq q$, and thus $\varphi\left(  p\right)
\neq\varphi\left(  q\right)  $ (since $\varphi$ is injective). Now, we have
the following chain of logical equivalences:%
\[
\left(  \varphi\left(  p\right)  \leq\varphi\left(  q\right)  \right)
\ \Longleftrightarrow\ \left(  \varphi\left(  p\right)  <\varphi\left(
q\right)  \text{ or }\underbrace{\varphi\left(  p\right)  =\varphi\left(
q\right)  }_{\substack{\text{This is false}\\\text{(since }\varphi\left(
p\right)  \neq\varphi\left(  q\right)  \text{)}}}\right)
\ \Longleftrightarrow\ \left(  \varphi\left(  p\right)  <\varphi\left(
q\right)  \right)  .
\]
Therefore, we have the following chain of equivalences:%
\begin{align}
\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)  \right)  \  &
\Longleftrightarrow\ \left(  \varphi\left(  p\right)  \leq\varphi\left(
q\right)  \right)  \ \Longleftrightarrow\ \left(  \varphi\left(  q\right)
\geq\varphi\left(  p\right)  \right) \nonumber\\
\  &  \Longleftrightarrow\ \left(  \text{not }\left(  \varphi\left(  q\right)
<\varphi\left(  p\right)  \right)  \right)  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=1.1}%
\end{align}
The same argument (applied to $\psi$ instead of $\varphi$) yields the
equivalence%
\begin{equation}
\left(  \psi\left(  p\right)  <\psi\left(  q\right)  \right)
\ \Longleftrightarrow\ \left(  \text{not }\left(  \psi\left(  q\right)
<\psi\left(  p\right)  \right)  \right)  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=1.2}%
\end{equation}
Now, we have the following chain of equivalences:%
\begin{align*}
\left(  \varphi\left(  p\right)  <\varphi\left(  q\right)  \right)  \  &
\Longleftrightarrow\ \left(  \text{not }\underbrace{\left(  \varphi\left(
q\right)  <\varphi\left(  p\right)  \right)  }_{\substack{\Longleftrightarrow
\ \left(  \psi\left(  q\right)  <\psi\left(  p\right)  \right)  \\\text{(by
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.0}))}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.1})}\right) \\
&  \Longleftrightarrow\ \left(  \text{not }\left(  \psi\left(  q\right)
<\psi\left(  p\right)  \right)  \right)  \ \Longleftrightarrow\ \left(
\psi\left(  p\right)  <\psi\left(  q\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.2})}\right)  .
\end{align*}
Thus, Observation 1 is proven in Case 3.

We have now proven Observation 1 in each of the three Cases 1, 2 and 3. Since
these three Cases cover all possibilities, we thus conclude that Observation 1
always holds.]

In the following, we shall use the so-called \dfn{Iverson bracket
notation}: For every assertion $\mathcal{A}$, we let $\left[  \mathcal{A}%
\right]  $ denote the integer $\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}
\end{cases} \quad$.

We next observe the following:

\begin{statement}
\textit{Observation 2:} Let $q\in\left\{  1,2,\ldots,n\right\}  $. Then,%
\[
\varphi\left(  q\right)  -1=\sum_{p=1}^{n}\left[  \varphi\left(  p\right)
<\varphi\left(  q\right)  \right]  .
\]

\end{statement}

[\textit{Proof of Observation 2:} We have $\varphi\left(  q\right)
\in\left\{  1,2,\ldots,n\right\}  $; thus, $1\leq\varphi\left(  q\right)  \leq
n$. But the map $\varphi$ is a bijection $\left\{  1,2,\ldots,n\right\}
\rightarrow\left\{  1,2,\ldots,n\right\}  $ (since $\varphi$ is a
permutation). Thus, we can substitute $i$ for $\varphi\left(  p\right)  $ in
the sum $\sum_{p\in\left\{  1,2,\ldots,n\right\}  }\left[  \varphi\left(
p\right)  <\varphi\left(  q\right)  \right]  $. We thus obtain
\begin{align*}
\sum_{p\in\left\{  1,2,\ldots,n\right\}  }\left[  \varphi\left(  p\right)
<\varphi\left(  q\right)  \right]   &  =\underbrace{\sum_{i\in\left\{
1,2,\ldots,n\right\}  }}_{=\sum_{i=1}^{n}}\left[  i<\varphi\left(  q\right)
\right]  =\sum_{i=1}^{n}\left[  i<\varphi\left(  q\right)  \right] \\
&  =\sum_{i=1}^{\varphi\left(  q\right)  -1}\underbrace{\left[  i<\varphi
\left(  q\right)  \right]  }_{\substack{=1\\\text{(since }i<\varphi\left(
q\right)  \\\text{(since }i\leq\varphi\left(  q\right)  -1<\varphi\left(
q\right)  \text{))}}}+\sum_{i=\varphi\left(  q\right)  }^{n}%
\underbrace{\left[  i<\varphi\left(  q\right)  \right]  }%
_{\substack{=0\\\text{(since we don't have }i<\varphi\left(  q\right)
\\\text{(since }i\geq\varphi\left(  q\right)  \text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split the sum at }i=\varphi\left(  q\right) \\
\text{(since }1\leq\varphi\left(  q\right)  \leq n\text{)}%
\end{array}
\right) \\
&  =\sum_{i=1}^{\varphi\left(  q\right)  -1}1+\underbrace{\sum_{i=\varphi
\left(  q\right)  }^{n}0}_{=0}=\sum_{i=1}^{\varphi\left(  q\right)
-1}1=\left(  \varphi\left(  q\right)  -1\right)  \cdot1=\varphi\left(
q\right)  -1.
\end{align*}
Hence,%
\[
\varphi\left(  q\right)  -1=\underbrace{\sum_{p\in\left\{  1,2,\ldots
,n\right\}  }}_{=\sum_{p=1}^{n}}\left[  \varphi\left(  p\right)
<\varphi\left(  q\right)  \right]  =\sum_{p=1}^{n}\left[  \varphi\left(
p\right)  <\varphi\left(  q\right)  \right]  .
\]
This proves Observation 2.]

Now, let $q\in\left\{  1,2,\ldots,n\right\}  $. Then, Observation 2 yields%
\[
\varphi\left(  q\right)  -1=\sum_{p=1}^{n}\left[  \varphi\left(  p\right)
<\varphi\left(  q\right)  \right]  .
\]
The same argument (applied to $\psi$ instead of $\varphi$) yields%
\begin{equation}
\psi\left(  q\right)  -1=\sum_{p=1}^{n}\left[  \psi\left(  p\right)
<\psi\left(  q\right)  \right]  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=1.6}%
\end{equation}
Hence,%
\[
\varphi\left(  q\right)  -1=\sum_{p=1}^{n}\left[  \underbrace{\varphi\left(
p\right)  <\varphi\left(  q\right)  }_{\substack{\Longleftrightarrow\ \left(
\psi\left(  p\right)  <\psi\left(  q\right)  \right)  \\\text{(by Observation
1)}}}\right]  =\sum_{p=1}^{n}\left[  \psi\left(  p\right)  <\psi\left(
q\right)  \right]  =\psi\left(  q\right)  -1
\]
(by (\ref{pf.prop.sol.Lalpha-std.proofs.Inv=1.6})). Adding $1$ on both sides
of this equality, we find $\varphi\left(  q\right)  =\psi\left(  q\right)  $.

Now, forget that we fixed $q$. We thus have shown that $\varphi\left(
q\right)  =\psi\left(  q\right)  $ for each $q\in\left\{  1,2,\ldots
,n\right\}  $. In other words, $\varphi=\psi$. This proves Proposition
\ref{prop.sol.Lalpha-std.proofs.Inv=1}.
\end{proof}

Note that Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1} is a restatement
of a known fact from elementary combinatorics. In order to formulate this
known fact (which will be used in a different solution further below), we
introduce a simple notation:

\begin{definition}
\label{def.sol.Lalpha-std.proofs.Inv} Let $n \in{\NN}$. Let $\varphi
\in\Symm_n$. Define the
\emph{inversion set}\index{inversion set of a permutation}
\dfn{$\operatorname*{Inv} \varphi$}
of $\varphi$ to be the set $\left\{  \left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j;\ \varphi\left(  i\right)
>\varphi\left(  j\right)  \right\}  $.
\end{definition}

(Note that this notation is not completely standard; some authors use the word
``inversion set'' for a different
set\footnote{Namely, they use it for the set $\left\{  \left(  \varphi\left(
i\right)  ,\varphi\left(  j\right)  \right)  \ \mid\ \left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2};\ i<j;\ \varphi\left(  i\right)
>\varphi\left(  j\right)  \right\}  $. This is a different set, although of
the same size as our $\operatorname*{Inv}\varphi$.}.)

Then, a well-known fact states the following:

\begin{proposition}
\label{prop.sol.Lalpha-std.proofs.Inv=}Let $n\in \NN$. Let $\varphi$ and
$\psi$ be two elements of $\Symm_n$ satisfying $\operatorname*{Inv}%
\varphi=\operatorname*{Inv}\psi$. Then, $\varphi=\psi$.
\end{proposition}

(Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=} can be restated as follows:
If $n\in \NN$, then a permutation in $\Symm_n$ is uniquely
determined by its inversion set.)

As we have said, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=} is just a
restatement of Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1}. Here is how
to derive the former from the latter:

\begin{proof}
[Proof of Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=}.]Let $a\in\left\{
1,2,\ldots,n\right\}  $ and $b\in\left\{  1,2,\ldots,n\right\}  $ be such that
$a<b$. We shall prove that $\varphi\left(  a\right)  <\varphi\left(  b\right)
$ if and only if $\psi\left(  a\right)  <\psi\left(  b\right)  $.

Indeed, $\left(  a,b\right)  $ is an element of $\left\{  1,2,\ldots
,n\right\}  ^{2}$ (since $a\in\left\{  1,2,\ldots,n\right\}  $ and
$b\in\left\{  1,2,\ldots,n\right\}  $). We have the following chain of
equivalences:%
\begin{align}
&  \ \left(  \left(  a,b\right)  \in\underbrace{\operatorname*{Inv}\varphi
}_{\substack{=\left\{  \left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}\ \mid\ i<j;\ \varphi\left(  i\right)  >\varphi\left(  j\right)  \right\}
\\\text{(by the definition of }\operatorname*{Inv}\varphi\text{)}}}\right)
\nonumber\\
&  \Longleftrightarrow\ \left(  \left(  a,b\right)  \in\left\{  \left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\ \mid\ i<j;\ \varphi\left(
i\right)  >\varphi\left(  j\right)  \right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  a,b\right)  \text{ is an }\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}\text{ satisfying }i<j\text{
and }\varphi\left(  i\right)  >\varphi\left(  j\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  a,b\right)  \text{ is an element of
}\left\{  1,2,\ldots,n\right\}  ^{2}\text{ and satisfies }a<b\text{ and
}\varphi\left(  a\right)  >\varphi\left(  b\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \varphi\left(  a\right)  >\varphi\left(
b\right)  \right)  \label{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.1}%
\end{align}
(because we already know that $\left(  a,b\right)  $ is an element of
$\left\{  1,2,\ldots,n\right\}  ^{2}$ and satisfies $a<b$).

But $\varphi$ is a permutation of $\left\{  1,2,\ldots,n\right\}  $ (since
$\varphi\in\Symm_n$), thus a bijective map. Hence, $\varphi$ is
injective. But from $a<b$, we obtain $a\neq b$, so that $\varphi\left(
a\right)  \neq\varphi\left(  b\right)  $ (since $\varphi$ is injective). We
have the following chain of equivalences:%
\[
\left(  \varphi\left(  a\right)  \leq\varphi\left(  b\right)  \right)
\ \Longleftrightarrow\ \left(  \varphi\left(  a\right)  <\varphi\left(
b\right)  \text{ or }\underbrace{\varphi\left(  a\right)  =\varphi\left(
b\right)  }_{\substack{\text{This is false}\\\text{(since }\varphi\left(
a\right)  \neq\varphi\left(  b\right)  \text{)}}}\right)
\ \Longleftrightarrow\ \left(  \varphi\left(  a\right)  <\varphi\left(
b\right)  \right)  .
\]
Hence, we have the following chain of equivalences:%
\begin{align}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)  \  &
\Longleftrightarrow\ \left(  \varphi\left(  a\right)  \leq\varphi\left(
b\right)  \right)  \ \Longleftrightarrow\ \left(  \text{not }%
\underbrace{\left(  \varphi\left(  a\right)  >\varphi\left(  b\right)
\right)  }_{\substack{\Longleftrightarrow\ \left(  a,b\right)  \in
\operatorname*{Inv}\varphi\\\text{(by
(\ref{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.1}))}}}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\operatorname*{Inv}\varphi\right)  \right)  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.3}%
\end{align}
The same argument (applied to $\psi$ instead of $\varphi$) yields the
equivalence%
\begin{equation}
\left(  \psi\left(  a\right)  <\psi\left(  b\right)  \right)
\ \Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\operatorname*{Inv}\psi\right)  \right)  .
\label{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.2}%
\end{equation}


Now, we have the following chain of equivalences:%
\begin{align*}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)  \  &
\Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\underbrace{\operatorname*{Inv}\varphi}_{=\operatorname*{Inv}\psi}\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.3}}\right) \\
\  &  \Longleftrightarrow\ \left(  \text{not }\left(  \left(  a,b\right)
\in\operatorname*{Inv}\psi\right)  \right)  \ \Longleftrightarrow\ \left(
\psi\left(  a\right)  <\psi\left(  b\right)  \right)
\end{align*}
(by \eqref{pf.prop.sol.Lalpha-std.proofs.Inv=.o1.pf.2}). In other words, we
have $\varphi\left(  a\right)  <\varphi\left(  b\right)  $ if and only if
$\psi\left(  a\right)  <\psi\left(  b\right)  $.

Now, forget that we fixed $a$ and $b$. We thus have shown that for every two
elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\[
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }\psi\left(  a\right)  <\psi\left(  b\right)  \right)  .
\]
Thus, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1} yields $\varphi=\psi
$. This proves Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=}.
\end{proof}

In the following, we shall write a poset as a pair $\left(  X,\prec\right)  $,
where $X$ is the underlying set, and where $\prec$ is the smaller relation of
the poset. All binary relations will be written in infix notation (i.e., if
$R$ is a binary relation on a set $X$, and if $a$ and $b$ are two elements of
$X$, then we shall write $aRb$ for $\left(  a,b\right)  \in R$).

We are now ready to prove Proposition \ref{prop.words.std-exists}:

\begin{proof}
[Proof of Proposition \ref{prop.words.std-exists}.]Define a map $f:\left\{
1,2,\ldots,n\right\}  \rightarrow\mathfrak{A}$ by%
\[
\left(  f\left(  i\right)  =w_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\]
The poset $\left(  \left\{  1,2,\ldots,n\right\}  ,<_{\ZZ}\right)  $ is
totally ordered (since $<_{\ZZ}$ is the smaller relation of a total
order on $\ZZ$). The poset $\left(  \mathfrak{A},<\right)  $ is totally
ordered as well. We define a binary relation $\prec_{f}$ on the set $\left\{
1,2,\ldots,n\right\}  $ by letting $i\prec_{f}j$ hold if and only if
\[
\left(  f\left(  i\right)  <f\left(  j\right)  \text{ or }\left(  f\left(
i\right)  =f\left(  j\right)  \text{ and }i<_{\ZZ}j\right)  \right)
.
\]
Thus, Lemma \ref{lem.sol.P-partition.stanley.altproof.wf}(b) (applied to
$\left(  \left\{  1,2,\ldots,n\right\}  ,<_{\ZZ}\right)  $ and $\left(
\mathfrak{A},<\right)  $ instead of $\left(  P,\vartriangleleft\right)  $ and
$\left(  Q,<\right)  $) yields that the binary relation $\prec_{f}$ is the
smaller relation of a partial order on $\left\{  1,2,\ldots,n\right\}  $ (that
is, this relation $\prec_{f}$ is irreflexive, transitive and asymmetric). Let
$T$ denote the set $\left\{  1,2,\ldots,n\right\}  $ endowed with this partial
order. Thus, $T=\left\{  1,2,\ldots,n\right\}  $ as sets, but the smaller
relation $<_{T}$ of the poset $T$ is the relation $\prec_{f}$. In other words,
$T$ is the poset $\left(  \left\{  1,2,\ldots,n\right\}  ,\prec_{f}\right)  $.
Hence, Lemma \ref{lem.sol.P-partition.stanley.altproof.wf}(c) (applied to
$\left(  \left\{  1,2,\ldots,n\right\}  ,<_{\ZZ}\right)  $, $\left(
\mathfrak{A},<\right)  $ and $T$ instead of $\left(  P,\vartriangleleft
\right)  $, $\left(  Q,<\right)  $ and $\mathbf{w}\left(  f\right)  $) yields
that the poset $T$ is totally ordered.

We have $T=\left\{  1,2,\ldots,n\right\}  $. Thus, $1,2,\ldots,n$ are $n$
elements of $T$. Clearly, $\left(  1,2,\ldots,n\right)  $ is a finite list of
\textbf{distinct} elements of this totally ordered set $T$ (since the elements
$1,2,\ldots,n$ of $T$ are distinct). Hence, Lemma
\ref{lem.sol.Lalpha-std.proofs.sort} (applied to $a_{i}=i$) yields that there
is a \textbf{unique} permutation $\sigma\in\Symm_n$ such that
$\sigma\left(  1\right)  <_{T}\sigma\left(  2\right)  <_{T}\cdots<_{T}%
\sigma\left(  n\right)  $. Consider this $\sigma$, and denote it by $\gamma$.
Thus, $\gamma$ is a permutation in $\Symm_n$ and satisfies
$\gamma\left(  1\right)  <_{T}\gamma\left(  2\right)  <_{T}\cdots<_{T}%
\gamma\left(  n\right)  $.

We know that $\gamma$ is a permutation in $\Symm_n$; thus, the
inverse map $\gamma^{-1}$ of $\gamma$ is well-defined and is a permutation in
$\Symm_n$ as well.

We have $\gamma\left(  1\right)  <_{T}\gamma\left(  2\right)  <_{T}\cdots
<_{T}\gamma\left(  n\right)  $. In other words, if $i$ and $j$ are two
elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $i<j$, then%
\begin{equation}
\gamma\left(  i\right)  <_{T}\gamma\left(  j\right)  .
\label{pf.prop.words.std-exists.exi.1}%
\end{equation}


Now, we claim that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(  b\right)  \text{ if
and only if }w_{a}\leq w_{b}\right)  . \label{pf.prop.words.std-exists.exi.2}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.words.std-exists.exi.2}):} Let $a$ and $b$ be
two elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$. Set
$i=\gamma^{-1}\left(  a\right)  $ and $j=\gamma^{-1}\left(  b\right)  $. Thus,
$i$ and $j$ are two elements of $\left\{  1,2,\ldots,n\right\}  $ (since
$\gamma^{-1}\in\Symm_n$). Also, from $i=\gamma^{-1}\left(  a\right)
$, we obtain $\gamma\left(  i\right)  =a$. From $j=\gamma^{-1}\left(
b\right)  $, we obtain $\gamma\left(  j\right)  =b$.

We are in one of the following two cases:

\textit{Case 1:} We have $i<j$.

\textit{Case 2:} We have $i\geq j$.

Let us first consider Case 1. In this case, we have $i<j$. This rewrites as
$\gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(  b\right)  $ (since
$i=\gamma^{-1}\left(  a\right)  $ and $j=\gamma^{-1}\left(  b\right)  $). On
the other hand, from $i<j$, we obtain $\gamma\left(  i\right)  <_{T}%
\gamma\left(  j\right)  $ (by (\ref{pf.prop.words.std-exists.exi.1})). This
rewrites as $a<_{T}b$ (since $\gamma\left(  i\right)  =a$ and $\gamma\left(
j\right)  =b$). This rewrites as $a\prec_{f}b$ (since the relation $<_{T}$ is
the relation $\prec_{f}$). In other words, $\left(  f\left(  a\right)
<f\left(  b\right)  \text{ or }\left(  f\left(  a\right)  =f\left(  b\right)
\text{ and }a<_{\ZZ}b\right)  \right)  $\ \ \ \ \footnote{because
$a\prec_{f}b$ holds if and only if $\left(  f\left(  a\right)  <f\left(
b\right)  \text{ or }\left(  f\left(  a\right)  =f\left(  b\right)  \text{ and
}a<_{\ZZ}b\right)  \right)  $ (by the definition of the relation
$\prec_{f}$)}. Hence, $\left(  f\left(  a\right)  <f\left(  b\right)  \text{
or }f\left(  a\right)  =f\left(  b\right)  \right)  $ (because if we have
$\left(  f\left(  a\right)  =f\left(  b\right)  \text{ and }a<_{\ZZ}b
\right)  $, then we also have $f\left(  a\right)  =f\left(  b\right)  $). In
other words, $f\left(  a\right)  \leq f\left(  b\right)  $. But the definition
of $f$ yields $f\left(  a\right)  =w_{a}$ and $f\left(  b\right)  =w_{b}$.
Thus, $w_{a}=f\left(  a\right)  \leq f\left(  b\right)  =w_{b}$. Now, we have
shown that both statements $\left(  \gamma^{-1}\left(  a\right)  <\gamma
^{-1}\left(  b\right)  \right)  $ and $\left(  w_{a}\leq w_{b}\right)  $ are
true. Hence, we have $\left(  \gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(
b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $. This proves
(\ref{pf.prop.words.std-exists.exi.2}) in Case 1.

Let us now consider Case 2. In this case, we have $i\geq j$. This rewrites as
$\gamma^{-1}\left(  a\right)  \geq\gamma^{-1}\left(  b\right)  $ (since
$i=\gamma^{-1}\left(  a\right)  $ and $j=\gamma^{-1}\left(  b\right)  $).
Hence, we don't have $\gamma^{-1}\left(  a\right)  <\gamma^{-1}\left(
b\right)  $. In other words, the statement $\left(  \gamma^{-1}\left(
a\right)  <\gamma^{-1}\left(  b\right)  \right)  $ is false.

We have $a<b$. Thus, we cannot have $b<a$. In other words, we cannot have
$b<_{\ZZ}a$ (since the relation $<_{\ZZ}$ is just the standard
smaller relation $<$). Thus, we cannot have $\left(  f\left(  b\right)
=f\left(  a\right)  \text{ and }b<_{\ZZ}a\right)  $.

Notice that $a=\gamma\left(  i\right)  $ (since $\gamma\left(  i\right)  =a$).
If we had $i=j$, then we would have $a=\gamma\left(  \underbrace{i}%
_{=j}\right)  =\gamma\left(  j\right)  =b$, which would contradict $a<b$.
Hence, we cannot have $i=j$. Thus, we have $i\neq j$. Combining this with
$i\geq j$, we obtain $i>j$. Therefore, $j<i$. Therefore,
(\ref{pf.prop.words.std-exists.exi.1}) (applied to $j$ and $i$ instead of $i$
and $j$) yields $\gamma\left(  j\right)  <_{T}\gamma\left(  i\right)  $. This
rewrites as $b<_{T}a$ (since $\gamma\left(  i\right)  =a$ and $\gamma\left(
j\right)  =b$). This rewrites as $b\prec_{f}a$ (since the relation $<_{T}$ is
the relation $\prec_{f}$). In other words, $\left(  f\left(  b\right)
<f\left(  a\right)  \text{ or }\left(  f\left(  b\right)  =f\left(  a\right)
\text{ and }b<_{\ZZ}a\right)  \right)  $\ \ \ \ \footnote{because
$b\prec_{f}a$ holds if and only if $\left(  f\left(  b\right)  <f\left(
a\right)  \text{ or }\left(  f\left(  b\right)  =f\left(  a\right)  \text{ and
}b<_{\ZZ}a\right)  \right)  $ (by the definition of the relation
$\prec_{f}$)}. Hence, we must have $f\left(  b\right)  <f\left(  a\right)  $
(since we cannot have $\left(  f\left(  b\right)  =f\left(  a\right)  \text{
and }b<_{\ZZ}a\right)  $). But the definition of $f$ yields $f\left(
a\right)  =w_{a}$ and $f\left(  b\right)  =w_{b}$. Thus, $w_{b}=f\left(
b\right)  <f\left(  a\right)  =w_{a}$. Hence, we cannot have $w_{a}\leq w_{b}%
$. In other words, the statement $\left(  w_{a}\leq w_{b}\right)  $ is false.
Now, we have shown that both statements $\left(  \gamma^{-1}\left(  a\right)
<\gamma^{-1}\left(  b\right)  \right)  $ and $\left(  w_{a}\leq w_{b}\right)
$ are false. Hence, we have $\left(  \gamma^{-1}\left(  a\right)  <\gamma
^{-1}\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $. This
proves (\ref{pf.prop.words.std-exists.exi.2}) in Case 2.

We have now proven (\ref{pf.prop.words.std-exists.exi.2}) in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, we thus conclude
that (\ref{pf.prop.words.std-exists.exi.2}) always holds. Thus,
(\ref{pf.prop.words.std-exists.exi.2}) is proven.]

So we know that $\gamma^{-1}$ is a permutation in $\Symm_n$, and that
for every two elements $a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$, we have $\left(  \gamma^{-1}\left(  a\right)  <\gamma
^{-1}\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $ (by
(\ref{pf.prop.words.std-exists.exi.2})). Thus, there exists \textbf{at least
one} permutation $\sigma\in\Symm_n$ such that for every two elements
$a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
$\left(  \sigma\left(  a\right)  <\sigma\left(  b\right)  \text{ if and only
if }w_{a}\leq w_{b}\right)  $ (namely, $\sigma=\gamma^{-1}$).

Next, we shall show that there exists \textbf{at most one} such permutation.
Indeed, we shall prove the following claim:

\begin{statement}
\textit{Claim 1:} Let $\varphi$ and $\psi$ be two permutations $\sigma
\in\Symm_n$ such that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(
a\right)  <\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq
w_{b}\right)  $. Then, $\varphi=\psi$.
\end{statement}

[\textit{Proof of Claim 1:} We know that $\varphi$ is a permutation $\sigma
\in\Symm_n$ such that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(
a\right)  <\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq
w_{b}\right)  $. In other words, $\varphi$ is a permutation in $\Symm_n$,
and has the property that for every two elements $a$ and $b$ of
$\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }w_{a}\leq w_{b}\right)  . \label{pf.prop.words.std-exists.uni.1}%
\end{equation}


Now, let $a$ and $b$ be two elements of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$. Then, from (\ref{pf.prop.words.std-exists.uni.1}), we
conclude that $\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)
\text{ if and only if }w_{a}\leq w_{b}\right)  $. In other words, we have the
logical equivalence $\left(  \varphi\left(  a\right)  <\varphi\left(
b\right)  \right)  \Longleftrightarrow\left(  w_{a}\leq w_{b}\right)  $. The
same argument (applied to $\psi$ instead of $\varphi$) yields the logical
equivalence $\left(  \psi\left(  a\right)  <\psi\left(  b\right)  \right)
\Longleftrightarrow\left(  w_{a}\leq w_{b}\right)  $. Hence, we have the
following chain of logical equivalences:%
\[
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \right)
\ \Longleftrightarrow\ \left(  w_{a}\leq w_{b}\right)  \ \Longleftrightarrow
\ \left(  \psi\left(  a\right)  <\psi\left(  b\right)  \right)
\]
(because of the equivalence $\left(  \psi\left(  a\right)  <\psi\left(
b\right)  \right)  \Longleftrightarrow\left(  w_{a}\leq w_{b}\right)  $). In
other words, we have
\[
\left(  \varphi\left(  a\right)  <\varphi\left(
b\right)  \text{ if and only if }\psi\left(  a\right)  <\psi\left(  b\right)
\right) .
\]

Now, forget that we fixed $a$ and $b$. We thus have proven that for every two
elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\[
\left(  \varphi\left(  a\right)  <\varphi\left(  b\right)  \text{ if and only
if }\psi\left(  a\right)  <\psi\left(  b\right)  \right)  .
\]
Thus, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1} yields that
$\varphi=\psi$. This proves Claim 1.]

There exists \textbf{at most one} permutation $\sigma\in\Symm_n$ such
that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$, we have $\left(  \sigma\left(  a\right)  <\sigma\left(
b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $ (because Claim 1
says that any two such permutations $\sigma$ must be equal). We thus have
proven the following two facts:

\begin{itemize}
\item There exists \textbf{at least one} permutation $\sigma\in\Symm_n$
such that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(  a\right)
<\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $.

\item There exists \textbf{at most one} permutation $\sigma\in\Symm_n$
such that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(  a\right)
<\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $.
\end{itemize}

Combining these two facts, we conclude that there exists a \textbf{unique}
permutation $\sigma\in\Symm_n$ such that for every two elements $a$
and $b$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
$\left(  \sigma\left(  a\right)  <\sigma\left(  b\right)  \text{ if and only
if }w_{a}\leq w_{b}\right)  $. This proves Proposition
\ref{prop.words.std-exists}.
\end{proof}

Next, we shall show a lemma that will be crucial in our proof of Lemma
\ref{lem.Lalpha-std.std-L}:

\begin{lemma}
\label{lem.sol.Lalpha-std.proofs.P-pa}Let $n\in \NN$. Let $\tau
\in\Symm_n$. Let $P$ be the labelled poset whose underlying set is
$\left\{  1,2,\ldots,n\right\}  $ and which (as a poset) is the total order
$\left(  \tau\left(  1\right)  <\tau\left(  2\right)  <\cdots<\tau\left(
n\right)  \right)  $ (that is, the order $<_{P}$ is given by $\tau\left(
1\right)  <_{P}\tau\left(  2\right)  <_{P}\cdots<_{P}\tau\left(  n\right)  $).

Let $\mathfrak{A}$ denote the totally ordered set $\left\{  1<2<3<\cdots
\right\}  $ of positive integers. Let $f:P\rightarrow\mathfrak{A}$ be any map.
Then, we have the following logical equivalence:%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\tau^{-1}\right)
\]
(where we treat $\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots
,f\left(  n\right)  \right)  $ as a word in $\mathfrak{A}^{n}$).
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lalpha-std.proofs.P-pa}.]We have $P=\left\{
1,2,\ldots,n\right\}  $ as sets. Also, the definition of the order $<_{P}$
yields%
\[
\tau\left(  1\right)  <_{P}\tau\left(  2\right)  <_{P}\cdots<_{P}\tau\left(
n\right)  .
\]
In other words, if $i$ and $j$ are two elements of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $i<j$, then%
\begin{equation}
\tau\left(  i\right)  <_{P}\tau\left(  j\right)  .
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij0}%
\end{equation}


Hence, for any two elements $i$ and $j$ of $\left\{  1,2,\ldots,n\right\}  $,
we have%
\begin{equation}
\left(  \tau\left(  i\right)  <_{P}\tau\left(  j\right)  \text{ if and only if
}i<j\right)  \label{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}%
\end{equation}
(where the ``$<$'' sign in ``$i<j$'' refers to the usual smaller relation
of the totally ordered set $\ZZ$).

[\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}):} Let $i$
and $j$ be two elements of $\left\{  1,2,\ldots,n\right\}  $. We must prove
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}).

We are in one of the following three cases:

\textit{Case 1:} We have $i<j$.

\textit{Case 2:} We have $i=j$.

\textit{Case 3:} We have $i>j$.

Let us first consider Case 1. In this case, we have $i<j$. Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij0}) yields $\tau\left(  i\right)
<_{P}\tau\left(  j\right)  $. Thus, both statements $\left(  i<j\right)  $ and
$\left(  \tau\left(  i\right)  <_{P}\tau\left(  j\right)  \right)  $ are true.
Hence, $\left(  \tau\left(  i\right)  <_{P}\tau\left(  j\right)  \text{ if and
only if }i<j\right)  $. Thus, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij})
is proven in Case 1.

Let us next consider Case 2. In this case, we have $i=j$. Hence, we don't have
$i<j$. Also, we don't have $\tau\left(  i\right)  <_{P}\tau\left(  j\right)  $
(since we have $\tau\left(  \underbrace{i}_{=j}\right)  =\tau\left(  j\right)
$). Thus, both statements $\left(  i<j\right)  $ and $\left(  \tau\left(
i\right)  <_{P}\tau\left(  j\right)  \right)  $ are false. Hence, $\left(
\tau\left(  i\right)  <_{P}\tau\left(  j\right)  \text{ if and only if
}i<j\right)  $. Thus, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) is
proven in Case 2.

Let us now consider Case 3. In this case, we have $i>j$. Thus, $j<i$. Hence,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij0}) (applied to $j$ and $i$
instead of $i$ and $j$) yields $\tau\left(  j\right)  <_{P}\tau\left(
i\right)  $. Thus, we don't have $\tau\left(  i\right)  <_{P}\tau\left(
j\right)  $. Also, we don't have $i<j$ (since we have $i>j$). Thus, both
statements $\left(  i<j\right)  $ and $\left(  \tau\left(  i\right)  <_{P}%
\tau\left(  j\right)  \right)  $ are false. Hence, $\left(  \tau\left(
i\right)  <_{P}\tau\left(  j\right)  \text{ if and only if }i<j\right)  $.
Thus, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) is proven in Case 3.

We have now proven (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) in each of
the three Cases 1, 2 and 3. Since these three Cases cover all possibilities,
this yields that (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) always holds.
Thus, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) is proven.]

The map $f$ is a map from $P$ to $\mathfrak{A}$. In other words, the map $f$
is a map from $P$ to $\left\{  1,2,3,\ldots\right\}  $ (since $\mathfrak{A}%
=\left\{  1,2,3,\ldots\right\}  $).

Also, $\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(
n\right)  \right)  $ is a word in $\mathfrak{A}^{n}$ (since $f\left(
i\right)  \in\mathfrak{A}$ for each $i\in\left\{  1,2,\ldots,n\right\}  $).
Denote this word by $w$. Thus,
\[
w=\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)
\right)  \in\mathfrak{A}^{n}.
\]


For each $i\in\left\{  1,2,\ldots,n\right\}  $, the $i$-th letter of the word
$w$ has been denoted by $w_{i}$. Thus, $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $, so that $\left(  w_{1},w_{2},\ldots,w_{n}\right)  =w=\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  $.
In other words,%
\begin{equation}
w_{i}=f\left(  i\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{
1,2,\ldots,n\right\}  . \label{pf.lem.sol.Lalpha-std.proofs.P-pa.wi=}%
\end{equation}
The definition of the standardization $\operatorname*{std}w$ shows that
$\operatorname*{std}w$ is the unique permutation $\sigma\in\Symm_n$
such that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(  a\right)
<\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq w_{b}\right)  $. In
particular, $\operatorname*{std}w$ is such a permutation. In other words,
$\operatorname*{std}w$ is a permutation in $\Symm_n$ and has the
property that for every two elements $a$ and $b$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
w_{a}\leq w_{b}\right)  . \label{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw}%
\end{equation}


Hence, for every two elements $a$ and $b$ of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$, we have
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }f\left(
a\right)  \leq f\left(  b\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}):}
Let $a$ and $b$ be two elements of $\left\{  1,2,\ldots,n\right\}  $
satisfying $a<b$. Then, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.wi=}) (applied
to $i=a$) yields $w_{a}=f\left(  a\right)  $. Also,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.wi=}) (applied to $i=b$) yields
$w_{b}=f\left(  b\right)  $. But (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw}%
) shows that%
\[
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
w_{a}\leq w_{b}\right)  .
\]
In view of $w_{a}=f\left(  a\right)  $ and $w_{b}=f\left(  b\right)  $, this
rewrites as
\[
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }f\left(
a\right)  \leq f\left(  b\right)  \right)  .
\]
This proves (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}).}.

We shall prove the following two claims:

\begin{statement}
\textit{Claim 1:} If $f\in\mathcal{A}\left(  P\right)  $, then
$\operatorname*{std}w=\tau^{-1}$.
\end{statement}

\begin{statement}
\textit{Claim 2:} If $\operatorname*{std}w=\tau^{-1}$, then $f\in
\mathcal{A}\left(  P\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} Assume that $f\in\mathcal{A}\left(  P\right)  $.
Thus, $f$ is a $P$-partition (since $\mathcal{A}\left(  P\right)  $ is the set
of all $P$-partitions).

But recall that $f$ is a $P$-partition if and only if $f$ is a map
$P\rightarrow\left\{  1,2,3,\ldots\right\}  $ with the properties
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.1}%
\end{equation}
and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.2}%
\end{equation}
(due to the definition of a ``$P$-partition'').
Thus, $f$ is a map $P\rightarrow\left\{  1,2,3,\ldots\right\}  $ with the
properties (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.1}) and
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.2}) (since $f$ is a $P$-partition).

For every two elements $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}%
):} Let $a\in\left\{  1,2,\ldots,n\right\}  $ and $b\in\left\{  1,2,\ldots
,n\right\}  $ be such that $a<b$. We must prove
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}).
\par
We have $a\in\left\{  1,2,\ldots,n\right\}  =P$ and $b\in\left\{
1,2,\ldots,n\right\}  =P$.
\par
Note that $a<b$. In other words, $a<_{\ZZ}b$ (since the relation $<$ on
$\ZZ$ is precisely the relation $<_{\ZZ}$). In other words,
$b>_{\ZZ}a$.
\par
Let $i=\tau^{-1}\left(  a\right)  $ and $j=\tau^{-1}\left(  b\right)  $. We
have $i=\tau^{-1}\left(  a\right)  \in\left\{  1,2,\ldots,n\right\}  $ and
$j=\tau^{-1}\left(  b\right)  \in\left\{  1,2,\ldots,n\right\}  $. From
$i=\tau^{-1}\left(  a\right)  $, we obtain $a=\tau\left(  i\right)  $. From
$j=\tau^{-1}\left(  b\right)  $, we obtain $b=\tau\left(  j\right)  $. If we
had $i=j$, then we would have $a=\tau\left(  \underbrace{i}_{=j}\right)
=\tau\left(  j\right)  =b$, which would contradict $a<b$. Hence, we cannot
have $i=j$. Thus, we have $i\neq j$. Hence, we are in one of the following two
cases:
\par
\textit{Case 1:} We have $i<j$.
\par
\textit{Case 2:} We have $i>j$.
\par
Let us first consider Case 1. In this case, we have $i<j$. But
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) shows that $\tau\left(
i\right)  <_{P}\tau\left(  j\right)  $ if and only if $i<j$. Hence, we have
$\tau\left(  i\right)  <_{P}\tau\left(  j\right)  $ (since we have $i<j$). In
view of $\tau\left(  i\right)  =a$ and $\tau\left(  j\right)  =b$, this
rewrites as $a<_{P}b$. Also, $a<_{\ZZ}b$. Hence,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.1}) (applied to $a$ and $b$
instead of $i$ and $j$) yields $f\left(  a\right)  \leq f\left(  b\right)  $.
Because of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}), this shows that
$\left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  $. Also, $\tau^{-1}\left(
a\right)  =i<j=\tau^{-1}\left(  b\right)  $. Thus, both statements $\left(
\left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \right)  $ and $\left(
\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $ are true.
Hence, we have $\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)
<\left(  \operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if
}\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $. Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>j$. Hence, $j<i$. But
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) (applied to $j$ and $i$
instead of $i$ and $j$) shows that $\tau\left(  j\right)  <_{P}\tau\left(
i\right)  $ if and only if $j<i$. Hence, we have $\tau\left(  j\right)
<_{P}\tau\left(  i\right)  $ (since we have $j<i$). In view of $\tau\left(
i\right)  =a$ and $\tau\left(  j\right)  =b$, this rewrites as $b<_{P}a$.
Also, $b>_{\ZZ}a$. Hence,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.2}) (applied to $b$ and $a$
instead of $i$ and $j$) yields $f\left(  b\right)  <f\left(  a\right)  $. In
other words, $f\left(  a\right)  >f\left(  b\right)  $. In other words, we
don't have $f\left(  a\right)  \leq f\left(  b\right)  $. Because of
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}), this shows that we don't have
$\left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  $. Also, $\tau^{-1}\left(
a\right)  =i>j=\tau^{-1}\left(  b\right)  $. Hence, we don't have $\tau
^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  $. Thus, both statements
$\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \right)  $ and $\left(
\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $ are false.
Hence, we have $\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)
<\left(  \operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if
}\tau^{-1}\left(  a\right)  <\tau^{-1}\left(  b\right)  \right)  $. Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) is proven in Case 2.
\par
We have now proven (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) in
both Cases 1 and 2. Since these two Cases cover all possibilities, this shows
that (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c1.pf.main}) always holds. Qed.}.
Hence, Proposition \ref{prop.sol.Lalpha-std.proofs.Inv=1} (applied to
$\varphi=\operatorname*{std}w$ and $\psi=\tau^{-1}$) yields
$\operatorname*{std}w=\tau^{-1}$. Thus, Claim 1 is proven.]

[\textit{Proof of Claim 2:} Assume that $\operatorname*{std}w=\tau^{-1}$. The
map $f:P\rightarrow\left\{  1,2,3,\ldots\right\}  $ has the following
properties:%
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i<_{\ZZ}j\text{, then }f\left(  i\right)  \leq f\left(  j\right)
\right)  \label{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}):}
Let $i\in P$ and $j\in P$ be such that $i<_{P}j$ and $i<_{\ZZ}j$. We
must prove that $f\left(  i\right)  \leq f\left(  j\right)  $.
\par
Let $a=\tau^{-1}\left(  i\right)  $ and $b=\tau^{-1}\left(  j\right)  $. Thus,
$a=\tau^{-1}\left(  i\right)  \in\left\{  1,2,\ldots,n\right\}  =P$ and
$b=\tau^{-1}\left(  j\right)  \in\left\{  1,2,\ldots,n\right\}  =P$. From
$a=\tau^{-1}\left(  i\right)  $, we obtain $\tau\left(  a\right)  =i$. From
$b=\tau^{-1}\left(  j\right)  $, we obtain $\tau\left(  b\right)  =j$.
\par
We have $i<_{P}j$. In view of $i=\tau\left(  a\right)  $ and $j=\tau\left(
b\right)  $, this rewrites as $\tau\left(  a\right)  <_{P}\tau\left(
b\right)  $.
\par
But (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) (applied to $a$ and $b$
instead of $i$ and $j$) shows that $\left(  \tau\left(  a\right)  <_{P}%
\tau\left(  b\right)  \text{ if and only if }a<b\right)  $. Hence, we have
$a<b$ (since we have $\tau\left(  a\right)  <_{P}\tau\left(  b\right)  $).
Because of $\operatorname*{std}w=\tau^{-1}$, we now have $\underbrace{\left(
\operatorname*{std}w\right)  }_{=\tau^{-1}}\left(  i\right)  =\tau^{-1}\left(
i\right)  =a<b=\underbrace{\tau^{-1}}_{=\operatorname*{std}w}\left(  j\right)
=\left(  \operatorname*{std}w\right)  \left(  j\right)  $.
\par
But $i<_{\ZZ}j$. In other words, $i<j$ (since the relation
$<_{\ZZ}$ is precisely the smaller relation $<$ on $\ZZ$). Thus,
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2}) (applied to $i$ and $j$
instead of $a$ and $b$) shows that $\left(  \left(  \operatorname*{std}%
w\right)  \left(  i\right)  <\left(  \operatorname*{std}w\right)  \left(
j\right)  \text{ if and only if }f\left(  i\right)  \leq f\left(  j\right)
\right)  $. Thus, $f\left(  i\right)  \leq f\left(  j\right)  $ (since
$\left(  \operatorname*{std}w\right)  \left(  i\right)  <\left(
\operatorname*{std}w\right)  \left(  j\right)  $). This proves
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}).} and
\begin{equation}
\left(  \text{if }i\in P\text{ and }j\in P\text{ satisfy }i<_{P}j\text{ and
}i>_{\ZZ}j\text{, then }f\left(  i\right)  <f\left(  j\right)  \right)
\label{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}):}
Let $i\in P$ and $j\in P$ be such that $i<_{P}j$ and $i>_{\ZZ}j$. We
must prove that $f\left(  i\right)  <f\left(  j\right)  $.
\par
Assume the contrary. Thus, $f\left(  i\right)  \geq f\left(  j\right)  $. In
other words, $f\left(  j\right)  \leq f\left(  i\right)  $.
\par
Let $a=\tau^{-1}\left(  i\right)  $ and $b=\tau^{-1}\left(  j\right)  $. Thus,
$a=\tau^{-1}\left(  i\right)  \in\left\{  1,2,\ldots,n\right\}  =P$ and
$b=\tau^{-1}\left(  j\right)  \in\left\{  1,2,\ldots,n\right\}  =P$. From
$a=\tau^{-1}\left(  i\right)  $, we obtain $\tau\left(  a\right)  =i$. From
$b=\tau^{-1}\left(  j\right)  $, we obtain $\tau\left(  b\right)  =j$.
\par
We have $i<_{P}j$. In view of $i=\tau\left(  a\right)  $ and $j=\tau\left(
b\right)  $, this rewrites as $\tau\left(  a\right)  <_{P}\tau\left(
b\right)  $.
\par
But (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.tauij}) (applied to $a$ and $b$
instead of $i$ and $j$) shows that $\left(  \tau\left(  a\right)  <_{P}%
\tau\left(  b\right)  \text{ if and only if }a<b\right)  $. Hence, we have
$a<b$ (since we have $\tau\left(  a\right)  <_{P}\tau\left(  b\right)  $).
\par
We have $i>_{\ZZ}j$. In other words, $i>j$ (since the relation
$>_{\ZZ}$ is precisely the greater relation $>$ of $\ZZ$). In
other words, $j<i$. Hence, (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.stdw2})
(applied to $j$ and $i$ instead of $a$ and $b$) shows that $\left(  \left(
\operatorname*{std}w\right)  \left(  j\right)  <\left(  \operatorname*{std}%
w\right)  \left(  i\right)  \text{ if and only if }f\left(  j\right)  \leq
f\left(  i\right)  \right)  $. Thus, $\left(  \operatorname*{std}w\right)
\left(  j\right)  <\left(  \operatorname*{std}w\right)  \left(  i\right)  $
(since $f\left(  j\right)  \leq f\left(  i\right)  $). In view of
$\operatorname*{std}w=\tau^{-1}$, this rewrites as $\tau^{-1}\left(  j\right)
<\tau^{-1}\left(  i\right)  $. Thus, $b=\tau^{-1}\left(  j\right)  <\tau
^{-1}\left(  i\right)  =a$. This contradicts $a<b$. This contradiction shows
that our assumption was wrong. Hence, $f\left(  i\right)  <f\left(  j\right)
$. This proves (\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}).}.

But recall that $f$ is a $P$-partition if and only if $f$ is a map
$P\rightarrow\left\{  1,2,3,\ldots\right\}  $ with the properties
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}) and
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2}) (due to the definition of a
``$P$-partition''). Thus, $f$ is a
$P$-partition (since we know that $f$ is a map $P\rightarrow\left\{
1,2,3,\ldots\right\}  $ with the properties
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.1}) and
(\ref{pf.lem.sol.Lalpha-std.proofs.P-pa.c2.pf.2})). In other words,
$f\in\mathcal{A}\left(  P\right)  $ (since $\mathcal{A}\left(  P\right)  $ is
the set of all $P$-partitions). This proves Claim 2.]

Combining Claim 1 with Claim 2, we conclude that $f\in\mathcal{A}\left(
P\right)  $ holds if and only if $\operatorname*{std}w=\tau^{-1}$. In other
words, we have the logical equivalence $\left(  f\in\mathcal{A}\left(
P\right)  \right)  \ \Longleftrightarrow\ \left(  \operatorname*{std}%
w=\tau^{-1}\right)  $. Thus, we have the following chain of logical
equivalences:%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\underbrace{w}_{=\left(  f\left(  1\right)
,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  }=\tau^{-1}\right)
\ \Longleftrightarrow\ \left(  \operatorname*{std}\left(  f\left(  1\right)
,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  =\tau^{-1}\right)  .
\]
This proves Lemma \ref{lem.sol.Lalpha-std.proofs.P-pa}.
\end{proof}

Next, we state a trivial consequence of Proposition
\ref{fundamental-P-partition-prop}:

\begin{lemma}
\label{lem.sol.Lalpha-std.proofs.4}Let $n\in \NN$. Let $\sigma
\in\Symm_n$. Let $P$ be the labelled poset whose underlying set is
$\left\{  1,2,\ldots,n\right\}  $ and which (as a poset) is the total order
$\left(  \sigma\left(  1\right)  <\sigma\left(  2\right)  <\cdots
<\sigma\left(  n\right)  \right)  $ (that is, the order $<_{P}$ is given by
$\sigma\left(  1\right)  <_{P}\sigma\left(  2\right)  <_{P}\cdots<_{P}%
\sigma\left(  n\right)  $). Then, $F_{P}\left(  \xx\right)
=L_{\gamma\left(  \sigma\right)  }$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.Lalpha-std.proofs.4}.]In Proposition
\ref{fundamental-P-partition-prop}, we have defined $\operatorname*{Des}w$ for
any labelled poset $w$ that is a total order. Applying this definition to
$w=P$, we obtain%
\[
\operatorname*{Des}P=\left\{  i\in\left\{  1,2,\ldots,n-1\right\}
\ \mid\ \sigma\left(  i\right)  >_{\ZZ}\sigma\left(  i+1\right)
\right\}
\]
(since $P$ is the total order $\left(  \sigma\left(  1\right)  <\sigma\left(
2\right)  <\cdots<\sigma\left(  n\right)  \right)  $). Comparing this with%
\begin{align*}
\operatorname*{Des}\sigma &  =\left\{  i\in\left\{  1,2,\ldots,n-1\right\}
\ \mid\ \sigma\left(  i\right)  >\sigma\left(  i+1\right)  \right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Des}%
\sigma\right) \\
&  =\left\{  i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ \sigma\left(
i\right)  >_{\ZZ}\sigma\left(  i+1\right)  \right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the greater relation }>\text{ of
} \ZZ \text{ is the relation }>_{\ZZ}\right)  ,
\end{align*}
we obtain $\operatorname*{Des}\sigma=\operatorname*{Des}P$.

Recall that $\gamma\left(  \sigma\right)  $ is the unique composition $\alpha$
of $n$ satisfying $D(\alpha)=\operatorname{Des}\sigma$ (by the definition of
$\gamma\left(  \sigma\right)  $). In other words, $\gamma\left(
\sigma\right)  $ is the unique composition $\alpha\in\operatorname{Comp}_{n}$
having partial sums $D(\alpha)=\operatorname{Des}\sigma$. In other words,
$\gamma\left(  \sigma\right)  $ is the unique composition $\alpha
\in\operatorname{Comp}_{n}$ having partial sums $D(\alpha)=\operatorname{Des}%
P$ (since $\operatorname{Des}\sigma=\operatorname{Des}P$).

Hence, Proposition \ref{fundamental-P-partition-prop} (applied to $P$,
$\sigma\left(  i\right)  $ and $\gamma\left(  \sigma\right)  $ instead of $w$,
$w_{i}$ and $\alpha$) yields that the generating function $F_{P}\left(
\xx\right)  $ equals the fundamental quasisymmetric function
$L_{\gamma\left(  \sigma\right)  }$ (and, in particular, $F_{P}\left(
\xx\right)  $ depends only upon the descent set $\operatorname*{Des}%
P$). Thus, $F_{P}\left(  \xx\right)  =L_{\gamma\left(  \sigma\right)
}$. This proves Lemma \ref{lem.sol.Lalpha-std.proofs.4}.
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.Lalpha-std.std-L}.]We have $\mathfrak{A}=\left\{
1<2<3<\cdots\right\}  $; thus, $\mathfrak{A}=\left\{  1,2,3,\ldots\right\}  $
as sets.

Let $P$ be the labelled poset whose underlying set is $\left\{  1,2,\ldots
,n\right\}  $ and which (as a poset) is the total order $\left(  \sigma\left(
1\right)  <\sigma\left(  2\right)  <\cdots<\sigma\left(  n\right)  \right)  $
(that is, the order $<_{P}$ is given by $\sigma\left(  1\right)  <_{P}%
\sigma\left(  2\right)  <_{P}\cdots<_{P}\sigma\left(  n\right)  $). Let
$f:P\rightarrow\mathfrak{A}$ be any map. Then, Lemma
\ref{lem.sol.Lalpha-std.proofs.P-pa} (applied to $\tau=\sigma$) yields that we
have the following logical equivalence:%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}\right)
\]
(where we treat $\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots
,f\left(  n\right)  \right)  $ as a word in $\mathfrak{A}^{n}$).

Now, forget that we fixed $f$. We thus have proven that for every map
$f:P\rightarrow\mathfrak{A}$, we have the logical equivalence%
\[
\left(  f\in\mathcal{A}\left(  P\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}\right)  .
\]
Hence, we have the following equality of summation signs:%
\begin{equation}
\sum\limits_{\substack{f:P\rightarrow\mathfrak{A};\\f\in\mathcal{A}\left(  P\right)
}}=\sum\limits_{\substack{f:P\rightarrow\mathfrak{A};\\\operatorname*{std}\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)
=\sigma^{-1}}}. \label{pf.lem.lyndon.grfun.std-L.2}%
\end{equation}


But every $P$-partition is a function $P\rightarrow\left\{  1,2,3,\ldots
\right\}  $ (by the definition of a $P$-partition). In other words, every
$P$-partition is a function $P\rightarrow\mathfrak{A}$ (since $\left\{
1,2,3,\ldots\right\}  =\mathfrak{A}$).

On the other hand, every $f\in\mathcal{A}\left(  P\right)  $ is a
$P$-partition (since $\mathcal{A}\left(  P\right)  $ is the set of all
$P$-partitions) and therefore is a function $P\rightarrow\mathfrak{A}$ (since
every $P$-partition is a function $P\rightarrow\mathfrak{A}$). Hence, we have
the following equality of summation sums:%
\[
\sum_{f\in\mathcal{A}\left(  P\right)  }=\sum\limits_{\substack{f\text{ is a function
}P\rightarrow\mathfrak{A};\\f\in\mathcal{A}\left(  P\right)  }}=\sum\limits
_{\substack{f:P\rightarrow\mathfrak{A};\\f\in\mathcal{A}\left(  P\right)
}}=\sum\limits_{\substack{f:P\rightarrow\mathfrak{A};\\\operatorname*{std}\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)
=\sigma^{-1}}}
\]
(by (\ref{pf.lem.lyndon.grfun.std-L.2})).

Lemma \ref{lem.sol.Lalpha-std.proofs.4} yields $F_{P}\left(  \xx%
\right)  =L_{\gamma\left(  \sigma\right)  }$. Hence,%
\begin{align}
L_{\gamma\left(  \sigma\right)  } &  =F_{P}\left(  \xx\right)
=\underbrace{\sum_{f\in{\mathcal{A}}(P)}}_{=\sum\limits_{\substack{f:P\rightarrow
\mathfrak{A};\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(
2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}}%
\underbrace{\xx_{f}}_{\substack{=\prod_{i\in P}x_{f\left(  i\right)
}\\\text{(by the definition of }\xx_{f}\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }F_{P}\left(
\xx\right)  \right)  \nonumber\\
&  =\underbrace{\sum\limits_{\substack{f:P\rightarrow\mathfrak{A}%
;\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}}_{\substack{=\sum\limits
_{\substack{f:\left\{  1,2,\ldots,n\right\}  \rightarrow\mathfrak{A}%
;\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}\\\text{(since }P=\left\{
1,2,\ldots,n\right\}  \text{)}}}\underbrace{\prod_{i\in P}}_{\substack{=\prod
_{i\in\left\{  1,2,\ldots,n\right\}  }\\\text{(since }P=\left\{
1,2,\ldots,n\right\}  \text{)}}}x_{f\left(  i\right)  }\nonumber\\
&  =\underbrace{\sum\limits_{\substack{f:\left\{  1,2,\ldots,n\right\}
\rightarrow\mathfrak{A};\\\operatorname*{std}\left(  f\left(  1\right)
,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}}%
_{=\sum\limits_{\substack{f\in\left\{  \text{functions }\left\{  1,2,\ldots
,n\right\}  \rightarrow\mathfrak{A}\right\}  ;\\\operatorname*{std}\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)
=\sigma^{-1}}}}\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{f\left(
i\right)  }\nonumber\\
&  =\sum\limits_{\substack{f\in\left\{  \text{functions }\left\{  1,2,\ldots
,n\right\}  \rightarrow\mathfrak{A}\right\}  ;\\\operatorname*{std}\left(
f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)
=\sigma^{-1}}}\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{f\left(  i\right)
}.\label{pf.lem.lyndon.grfun.std-L.4}%
\end{align}


But if $Z$ is any set, then the map%
\begin{align*}
\left\{  \text{functions }\left\{  1,2,\ldots,n\right\}  \rightarrow
Z\right\}   &  \rightarrow Z^{n},\\
f &  \mapsto\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(
n\right)  \right)
\end{align*}
is a bijection\footnote{Indeed, this map is the standard bijection between
functions $\left\{  1,2,\ldots,n\right\}  \rightarrow Z$ and $n$-tuples of
elements of $Z$.}. Applying this to $Z=\mathfrak{A}$, we conclude that the map%
\begin{align*}
\left\{  \text{functions }\left\{  1,2,\ldots,n\right\}  \rightarrow
\mathfrak{A}\right\}   &  \rightarrow\mathfrak{A}^{n},\\
f &  \mapsto\left(  f\left(  1\right)  ,f\left(  2\right)  ,\ldots,f\left(
n\right)  \right)
\end{align*}
is a bijection. Hence, we can substitute $\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $ for $\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  $ in the sum on the right hand side of
(\ref{pf.lem.lyndon.grfun.std-L.4}). We thus obtain%
\[
\sum\limits_{\substack{f\in\left\{  \text{functions }\left\{  1,2,\ldots,n\right\}
\rightarrow\mathfrak{A}\right\}  ;\\\operatorname*{std}\left(  f\left(
1\right)  ,f\left(  2\right)  ,\ldots,f\left(  n\right)  \right)  =\sigma
^{-1}}}\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{f\left(  i\right)  }%
=\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}%
^{n};\\\operatorname*{std}\left(  w_{1},w_{2},\ldots,w_{n}\right)
=\sigma^{-1}}}\prod_{i\in\left\{  1,2,\ldots,n\right\}  }x_{w_{i}}.
\]
Hence, (\ref{pf.lem.lyndon.grfun.std-L.4}) becomes%
\[
L_{\gamma\left(  \sigma\right)  }=\sum\limits_{\substack{f\in\left\{  \text{functions
}\left\{  1,2,\ldots,n\right\}  \rightarrow\mathfrak{A}\right\}
;\\\operatorname*{std}\left(  f\left(  1\right)  ,f\left(  2\right)
,\ldots,f\left(  n\right)  \right)  =\sigma^{-1}}}\prod_{i\in\left\{
1,2,\ldots,n\right\}  }x_{f\left(  i\right)  }=\sum\limits_{\substack{\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n};\\\operatorname*{std}%
\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\sigma^{-1}}}\prod_{i\in\left\{
1,2,\ldots,n\right\}  }x_{w_{i}}.
\]
Comparing this with%
\begin{align*}
\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}%
}}\xx_{w} &  =\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)
\in\mathfrak{A}^{n};\\\operatorname*{std}\left(  w_{1},w_{2},\ldots
,w_{n}\right)  =\sigma^{-1}}}\underbrace{\xx_{\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  }}_{\substack{=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}%
}\\\text{(by the definition of }\xx_{\left(  w_{1},w_{2},\ldots
,w_{n}\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}w\text{ as }\left(  w_{1},w_{2},\ldots,w_{n}\right)  \right)  \\
&  =\sum\limits_{\substack{\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in
\mathfrak{A}^{n};\\\operatorname*{std}\left(  w_{1},w_{2},\ldots,w_{n}\right)
=\sigma^{-1}}}\underbrace{x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}}_{=\prod
_{i\in\left\{  1,2,\ldots,n\right\}  }x_{w_{i}}}=\sum\limits_{\substack{\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n};\\\operatorname*{std}%
\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\sigma^{-1}}}\prod_{i\in\left\{
1,2,\ldots,n\right\}  }x_{w_{i}},
\end{align*}
we obtain $L_{\gamma\left(  \sigma\right)  }=\sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}$. This
proves Lemma \ref{lem.Lalpha-std.std-L}.
\end{proof}

Proposition \ref{prop.words.std-exists} and Lemma \ref{lem.Lalpha-std.std-L}
have now been proven. Thus, Exercise \ref{exe.Lalpha-std.proofs} is solved.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.xi-lifts-p}}

\textit{Solution to Exercise~\ref{exe.xi-lifts-p}.}
Consider the power series $\widetilde{H}\left(t\right)$ and $\xi\left(t\right)$
defined in \eqref{relation-definining-pi's}. From
\eqref{relation-definining-pi's}, we know that
\begin{equation}
\label{sol.xi-lifts-p.1}
\sum_{n \geq 1} \xi_n t^n =
\log \widetilde{H}\left(t\right) = \log \left(\sum_{n \geq 0} H_n t^n \right).
\end{equation}

(a) The fact that $\xi_n$ is primitive was proven for $\kk = \QQ$ in
Remark~\ref{second-antipode-proof}, and can be proven in the same way
for general $\kk$. It remains to show that $\xi_n$ is homogeneous of
degree $n$ for each $n \geq 1$. This can be done as follows:

Let us say that a power series
$f\in A\left[ \left[ t\right] \right]$ over a graded $\kk$-algebra
$A$ is \emph{equigraded}\index{equigraded power series}
if, for every $n \in \NN$, the
coefficient of $f$ before $t^n$ is homogeneous of degree $n$. Then,
the set of all equigraded power series in
$A\left[\left[ t\right]\right]$ is a
$\kk$-subalgebra of $A\left[\left[ t\right]\right]$ which is closed
under the usual topology on $A\left[\left[ t\right]\right]$. In
particular, if $g$ is an equigraded power series in
$A\left[\left[ t\right]\right]$ having constant term $1$, then
$\log g$ is equigraded. Applied to $A = \Nsym$ and
$g = \sum_{n \geq 0} H_n t^n$, this yields that the power series
$\log \left(\sum_{n \geq 0} H_n t^n\right)$ is equigraded. Due to
\eqref{sol.xi-lifts-p.1}, this rewrites as follows: The power series
$\sum_{n \geq 1} \xi_n t^n$ is equigraded. In other words,
$\xi_n$ is homogeneous of degree $n$ for each $n \geq 1$. This
completes the solution of Exercise~\ref{exe.xi-lifts-p}(a).

For an alternative proof of Exercise~\ref{exe.xi-lifts-p}(a), one can
simply notice that it follows immediately from
Exercise~\ref{exe.xi-lifts-p}(c).

(b) The ring homomorphism $\pi:\Nsym\rightarrow\Lambda$
induces a ring homomorphism $\Nsym\left[  \left[  t\right]
\right]  \rightarrow\Lambda\left[  \left[  t\right]  \right]  $ which
is continuous with respect to the usual topology on power series.
Applying this
latter homomorphism to the equality \eqref{sol.xi-lifts-p.1},
we obtain
\[
\sum_{n \geq 1} \pi\left(\xi_n\right) t^n
= \log \left(\sum_{n \geq 0} \pi\left(H_n\right) t^n\right)
= \log \left(\sum_{n \geq 0} h_n t^n\right)
= \sum_{m=1}^\infty \frac{1}{m} p_m t^m
\]
(where in the last step, we have used the equality
\eqref{H(t)-logarithm}).
Comparing coefficients, we obtain
$\pi\left(\xi_n\right) = \frac{1}{n} p_n$ for all $n \geq 1$. Multiplying
this by $n$, we obtain $\pi\left(n\xi_n\right) = p_n$ for all $n \geq 1$.
This solves part (b) of the exercise.

\textit{Remark:} Another way to solve Exercise~\ref{exe.xi-lifts-p}(b)
proceeds as follows: We know that $\pi\left(n\xi_n\right)$ is a
primitive homogeneous element of $\Lambda$ of degree $n$ (by
Exercise~\ref{exe.xi-lifts-p}(a) and since $\pi$ is a graded
homomorphism of Hopf algebras). But
Exercise~\ref{exe.Lambda.primitives} shows that all such elements
are scalar multiples of $p_n$. Thus, $\pi\left(n\xi_n\right)$ is a
scalar multiple of $p_n$. Finding the scalar is easy (e.g., it can
be obtained by specializing at $\left(1\right)$).

(c) From \eqref{sol.xi-lifts-p.1}, we have
\begin{align*}
\sum_{n\geq1}\xi_{n}t^{n} 
& =\log\left(  \underbrace{\sum_{n\geq0}H_{n}t^{n}}
_{=1+\sum_{n\geq1}H_{n}t^{n}}\right)  =\log\left(  1+\sum_{n\geq1}H_{n}
t^{n}\right)  \\
& =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{\left(
\sum_{n\geq1}H_{n}t^{n}\right)  ^{i}}_{=\sum_{n_{1},n_{2},...,n_{i}\geq
1}\left(  H_{n_{1}}t^{n_{1}}\right)  \left(  H_{n_{2}}t^{n_{2}}\right)
\cdots\left(  H_{n_{i}}t^{n_{i}}\right)  }\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by the Mercator series for the
logarithm}\right)  \\
& =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\underbrace{\sum
_{n_{1},n_{2},...,n_{i}\geq 1}}_{=\sum\limits_{\substack{\left(  n_{1},n_{2}
,...,n_{i}\right)  \text{ is a composition}\\\text{of length }i}}}
\underbrace{\left( H_{n_{1}}t^{n_{1}}\right)
 \left( H_{n_{2}}t^{n_{2}}\right)
 \cdots\left( H_{n_{i}}t^{n_{i}}\right)  }
_{=H_{n_{1}}H_{n_{2}}\cdots H_{n_{i}}t^{n_{1}+n_{2}+...+n_{i}}}\\
& =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\sum\limits_{\substack{\left(
n_{1},n_{2},...,n_{i}\right)  \text{ is a composition}\\
\text{of length }i}}
\underbrace{H_{n_{1}}H_{n_{2}}\cdots H_{n_{i}}}_{=H_{\left(  n_{1},
n_{2},...,n_{i}\right)  }}\underbrace{t^{n_{1}+n_{2}+...+n_{i}}}
_{=t^{\left\vert \left(  n_{1},n_{2},...,n_{i}\right)  \right\vert }}\\
& =\sum_{i\geq1}\dfrac{\left(  -1\right)  ^{i-1}}{i}\sum\limits_{\substack{\left(
n_{1},n_{2},...,n_{i}\right)  \text{ is a composition}\\
\text{of length }i}}H_{\left(  n_{1},n_{2},...,n_{i}\right)  }
t^{\left\vert \left(  n_{1},n_{2},...,n_{i}\right)  \right\vert } \\
&=\sum_{i\geq1}\dfrac{\left(  -1\right)
^{i-1}}{i}\sum\limits_{\substack{\alpha\text{ is a composition}\\
\text{of length }i}}H_{\alpha}t^{\left\vert \alpha\right\vert }\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  n_{1},n_{2},...,n_{i}\right)  \text{ as }\alpha
\text{ in the inner sum}\right)  \\
& =\sum_{i\geq1}\sum\limits_{\substack{\alpha\text{ is a composition}\\\text{of
length }i}}\underbrace{\dfrac{\left(  -1\right)  ^{i-1}}{i}}_{\substack{=\dfrac
{\left(  -1\right)  ^{\ell\left(  \alpha\right)  -1}}{\ell\left(
\alpha\right)  }\\\text{(since }i=\ell\left(  \alpha\right)  \text{)}}}
H_{\alpha}t^{\left\vert \alpha\right\vert }=\underbrace{\sum_{i\geq1}
\sum\limits_{\substack{\alpha\text{ is a composition}\\\text{of length }i}}}
_{\substack{=\sum\limits_{\substack{\alpha\text{ is a nonempty}\\
\text{composition}}}\\ =\sum_{n\geq1}\sum_{\alpha\in\Comp_{n}}}}
\dfrac{\left(  -1\right)  ^{\ell\left(  \alpha\right)  -1}}{\ell\left(
\alpha\right)  }H_{\alpha}t^{\left\vert \alpha\right\vert }\\
& =\sum_{n\geq1}\sum_{\alpha\in\Comp_{n}}
\dfrac{\left(  -1\right)  ^{\ell\left(  \alpha\right)  -1}}{\ell\left(
\alpha\right)  }H_{\alpha}\underbrace{t^{\left\vert \alpha\right\vert }}
_{\substack{=t^{n}\\\text{(since }\alpha\in\Comp_{n}\text{)}}}
=\sum_{n\geq1}\sum_{\alpha\in\Comp_{n}}
\dfrac{\left(  -1\right)  ^{\ell\left(  \alpha\right)  -1}}
{\ell\left(  \alpha\right)  }H_{\alpha}t^{n}\\
& =\sum_{n\geq1}\left(  \sum_{\alpha\in\Comp_{n}}
\dfrac{\left(  -1\right)  ^{\ell\left(  \alpha\right)  -1}}{\ell\left(
\alpha\right)  }H_{\alpha}\right)  t^{n}.
\end{align*}
Comparing coefficients in this identity, we conclude that every $n\geq1$
satisfies
\[
\xi_{n}=\sum_{\alpha\in\Comp_{n}}\dfrac{\left(
-1\right)  ^{\ell\left(  \alpha\right)  -1}}{\ell\left(  \alpha\right)
}H_{\alpha}=\sum_{\alpha\in\Comp_{n}}\left(  -1\right)
^{\ell\left(  \alpha\right)  -1}\frac{1}{\ell\left(  \alpha\right)  }
H_{\alpha}.
\]
This solves part (c) of the exercise.

(d) Applying the exponential to $\sum_{n\geq1}\xi_{n}t^{n}=\log H\left(
t\right)  $, we obtain $\exp\left(  \sum_{n\geq1}\xi_{n}t^{n}\right)
=H\left(  t\right)  $. Thus,
\begin{align*}
H\left(  t\right)    & =\exp\left(  \sum_{n\geq1}\xi_{n}t^{n}\right)
=\sum_{i\geq0}\dfrac{1}{i!}\underbrace{\left(  \sum_{n\geq1}\xi_{n}
t^{n}\right)  ^{i}}_{=\sum_{n_{1},n_{2},...,n_{i}\geq1}
\left(  \xi_{n_{1}}t^{n_{1}}\right)  \left(  \xi_{n_{2}}t^{n_{2}}\right) 
 \cdots\left( \xi_{n_{i}}t^{n_{i}}\right)  }\\
& =\sum_{i\geq0}\dfrac{1}{i!}\underbrace{\sum_{n_{1},n_{2},...,n_{i}\geq1}}
_{=\sum\limits_{\substack{\left(  n_{1},n_{2},...,n_{i}\right)  \text{ is a
composition}\\\text{of length }i}}}\underbrace{\left(  \xi_{n_{1}}t^{n_{1}}
\right)  \left(  \xi_{n_{2}}t^{n_{2}}\right)  \cdots\left(  \xi_{n_{i}}
t^{n_{i}}\right)  }_{=\xi_{n_{1}}\xi_{n_{2}}\cdots\xi_{n_{i}}t^{n_{1}
+n_{2}+...+n_{i}}}\\
& =\sum_{i\geq0}\dfrac{1}{i!}\sum\limits_{\substack{\left(  n_{1},n_{2}
,...,n_{i}\right)  \text{ is a composition}\\\text{of length }i}}
\underbrace{\xi_{n_{1}}\xi_{n_{2}}\cdots\xi_{n_{i}}}_{\substack{=\xi_{\left(
n_{1},n_{2},...,n_{i}\right)  }\\\text{(because we defined }\xi_{\left(
n_{1},n_{2},...,n_{i}\right)  }\\\text{in such a way that this holds)}}}
\underbrace{t^{n_{1}+n_{2}+...+n_{i}}}_{=t^{\left\vert \left(  n_{1}
,n_{2},...,n_{i}\right)  \right\vert }}\\
& =\sum_{i\geq0}\dfrac{1}{i!}\sum\limits_{\substack{\left(  n_{1},n_{2}
,...,n_{i}\right)  \text{ is a composition}\\\text{of length }i}}\xi_{\left(
n_{1},n_{2},...,n_{i}\right)  }t^{\left\vert \left(  n_{1},n_{2}
,...,n_{i}\right)  \right\vert }=\sum_{i\geq0}\dfrac{1}{i!}\sum\limits
_{\substack{\alpha\text{ is a composition}\\\text{of length }i}}\xi_{\alpha
}t^{\left\vert \alpha\right\vert }\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  n_{1},n_{2},...,n_{i}\right)  \text{ as }\alpha
\text{ in the inner sum}\right)  \\
& =\sum_{i\geq0}\sum\limits_{\substack{\alpha\text{ is a composition}\\\text{of
length }i}}\underbrace{\dfrac{1}{i!}}_{\substack{=\dfrac{1}{\ell\left(
\alpha\right)  !}\\\text{(since }i=\ell\left(  \alpha\right)  \text{)}}}
\xi_{\alpha}t^{\left\vert \alpha\right\vert }=\underbrace{\sum_{i\geq0}
\sum\limits_{\substack{\alpha\text{ is a composition}\\\text{of length }i}}}
_{\substack{=\sum_{\alpha\text{ is a composition}}\\=\sum_{n\geq0}
\sum_{\alpha\in\Comp_{n}}}}\dfrac{1}{\ell\left(
\alpha\right)  !}\xi_{\alpha}t^{\left\vert \alpha\right\vert }\\
& =\sum_{n\geq0}\sum_{\alpha\in\Comp_{n}}\dfrac
{1}{\ell\left(  \alpha\right)  !}\xi_{\alpha}\underbrace{t^{\left\vert
\alpha\right\vert }}_{\substack{=t^{n}\\\text{(since }\alpha\in
\Comp_{n}\text{)}}}=\sum_{n\geq0}\sum_{\alpha
\in\Comp_{n}}\dfrac{1}{\ell\left(  \alpha\right)
!}\xi_{\alpha}t^{n}=\sum_{n\geq0}\left(  \sum_{\alpha\in\Comp_{n}}
\dfrac{1}{\ell\left(  \alpha\right)  !}\xi_{\alpha}\right)
t^{n}.
\end{align*}
Comparing coefficients in this identity, we conclude that every $n\geq0$
satisfies
\[
H_{n}=\sum_{\alpha\in\Comp_{n}}\frac{1}{\ell\left(
\alpha\right)  !}\xi_{\alpha}
\]
(since the coefficient of $t^{n}$ in $H\left(  t\right)  $ is $H_{n}$). This
proves \eqref{eq.exe.xi-lifts-p.d.1}.

Notice that, for every $n\geq 1$, the element $\xi_n$ of
$\Nsym$ is homogeneous of degree $n$ (by
Exercise~\ref{exe.xi-lifts-p}(a)). Hence, for every composition $\alpha$, the
element $\xi_{\alpha}$ of $\Nsym$ is homogeneous of degree
$\left\vert \alpha\right\vert $. In particular, for every $n\in\NN$, it
is clear that $\left( \xi_{\alpha}\right)_{\alpha\in\Comp_n}$ is a
family of elements of $\Nsym_n$. We now need to
prove that this family is a $\kk$-basis of $\Nsym_n$
for every $n\in \NN$.

Let $\mathfrak{A}$ be the $\kk$-subalgebra of $\Nsym$
generated by the elements $\xi_{1}$, $\xi_{2}$, $\xi_{3}$, $...$. Then,
$\mathfrak{A}$ contains $\xi_{\alpha}$ for every composition $\alpha$ (by the
definition of $\xi_{\alpha}$). Therefore, $\mathfrak{A}$ contains $H_{n}$ for
every $n\geq1$ (by \eqref{eq.exe.xi-lifts-p.c.1}). Consequently,
$\mathfrak{A}=\Nsym$ (because $\Nsym$ is generated
as a $\kk$-algebra by $H_{1}$, $H_{2}$, $H_{3}$, $...$). In other
words, the $\kk$-algebra $\Nsym$ is generated by the
elements $\xi_{1}$, $\xi_{2}$, $\xi_{3}$, $...$ (since we defined
$\mathfrak{A}$ as the $\kk$-subalgebra of $\Nsym$
generated by the elements $\xi_1$, $\xi_2$, $\xi_3$, $...$). In other
words, the $\kk$-module $\Nsym$ is spanned by all
possible products of the elements $\xi_1$, $\xi_2$, $\xi_3$, $...$. In
other words, the $\kk$-module $\Nsym$ is spanned by the
elements $\xi_{\alpha}$ with $\alpha\in\Comp $ (because the
elements $\xi_{\alpha}$ with $\alpha\in\Comp $ are precisely all
possible products of the elements $\xi_1$, $\xi_2$, $\xi_3$, $...$). In
yet other words, the family
$\left( \xi_{\alpha}\right) _{\alpha \in\Comp}$ spans the $\kk$-module
$\Nsym$.

Now, fix $n \in \NN$. Every element of $\Nsym_n$ can be
written as a $\kk$-linear combination of the elements $\xi_{\alpha}$
with $\alpha \in \Comp$ (since the family
$\left( \xi_{\alpha} \right)_{\alpha\in\Comp}$ spans the $\kk$-module
$\Nsym$). In this $\kk$-linear combination, we can remove all
terms $\xi_{\alpha}$ with $\alpha \notin \Comp_n$ without
changing the result
(by gradedness, because $\xi_{\alpha}$ is homogeneous of
degree $\left\vert \alpha\right\vert$), and so we conclude that every
element of $\Nsym_n$ can be written as a $\kk$-linear
combination of the elements $\xi_{\alpha}$ with $\alpha \in \Comp_n$.
In other words, the family $\left(
\xi_{\alpha}\right) _{\alpha \in \Comp_n}$ spans the
$\kk$-module $\Nsym_n$.

Now, we can apply Exercise~\ref{exe.linalg.sur-bij}(b) to $A=\Nsym_n$,
$I=\Comp_n$, $\left(  \gamma_i\right) _{i \in I}=\left(
H_{\alpha}\right) _{\alpha \in \Comp_n}$ and
$\left( \beta_i\right) _{i\in I}=\left( \xi_{\alpha}\right) _{\alpha
\in \Comp_n}$ (since we know that $\left( H_{\alpha}\right)
_{\alpha \in \Comp_n}$ is a $\kk$-basis of
$\Nsym_n$, whereas $\left( \xi_{\alpha}\right) _{\alpha
\in \Comp_n}$ spans the $\kk$-module $\Nsym_n$). We conclude that
$\left( \xi_{\alpha}\right) _{\alpha \in \Comp_n}$ is a $\kk$-basis
of $\Nsym_n$. Thus, Exercise~\ref{exe.xi-lifts-p}(d) is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.eulerian-idp}}

\textit{Solution to Exercise~\ref{exe.eulerian-idp}.} We follow the hint.

Let $f$ be the endomorphism $\id_{A}-u\epsilon$ of $A$. Then, $f=\sum_{n\geq
1}\pi_{n}$ (because the definition of the $\pi_{n}$ yields that
$\id_A
=\sum_{n\geq 0}\pi_n
=\underbrace{\pi_{0}}_{=u\epsilon} + \sum_{n\geq 1}\pi_n
=u\epsilon+\sum_{n\geq 1}\pi_n$
and thus $\id_{A}-u\epsilon=\sum_{n\geq 1}\pi_{n}$).
Notice that $f=\id_{A}-u\epsilon$, so that $\id_{A}=f+u\epsilon$.
Now, $\mathfrak{e}=\log^{\star}\left(  \underbrace{\id_{A}}_{=f+u\epsilon}
\right)  =\log^{\star}\left(  f+u\epsilon\right)  =\sum_{n\geq1}\left(
-1\right)  ^{n-1}\frac{1}{n}f^{\star n}$.

We let $\End_{\operatorname{gr}}A$ be the
$\kk$-submodule of $\End A$ consisting of all
graded $\kk$-linear maps $A\rightarrow A$. Then, it is easy to see that
this $\kk$-submodule $\End_{\operatorname{gr}}A$ is closed
under convolution (i.e., if $g_{1}$ and $g_{2}$ are two graded
$\kk$-linear maps $A\rightarrow A$, then $g_{1}\star g_{2}$ is also a graded
$\kk$-linear map $A\rightarrow A$) and contains the unity $u\epsilon$
of the algebra $\left(  \End A,\star\right)  $ (since
$u\epsilon:A\rightarrow A$ is a graded $\kk$-linear map). Hence,
$\left(  \End_{\operatorname{gr}}A,\star\right)  $ is
a $\kk$-subalgebra of $\left(  \End A,\star\right)  $.
Moreover, $\left(  \End_{\operatorname{gr}} A,\star\right)  $ contains
$\pi_{n}$ for every $n\geq 1$. Hence, the
$\kk$-algebra homomorphism $\mathfrak{W}:\Nsym
\rightarrow\left(  \operatorname{End}A,\star\right)  $ maps the generators
$H_{n}$ of $\Nsym$ to elements of
$\left(  \End_{\operatorname{gr}}A,\star\right)  $. Thus, the image of
$\mathfrak{W}$ is contained in
$\left( \End_{\operatorname{gr}}A,\star\right)  $. In other words,
\begin{equation}
\mathfrak{W}\left(  x\right)  \in
\End_{\operatorname{gr}}A\ \ \ \ \ \ \ \ \ \ \text{for every }
x\in\Nsym . \label{sol.eulerian-idp.Wxgr}
\end{equation}
It is immediate to check that the $\kk$-subalgebra $\left(
\End_{\operatorname{gr}}A,\star\right)  $ of $\left(
\End A,\star\right)  $ is closed under the topology of pointwise
convergence. Hence, it is closed under taking logarithms. In other words,
$\log^{\star}g\in\End_{\operatorname{gr}}A$ for every
$g\in\End_{\operatorname{gr}}A$ for which $\log^\star g$ makes
sense. Applied to $g = \id_A$,
this yields that $\log^{\star}\left( \id_A\right)  \in
\End_{\operatorname{gr}}A$ (since $\id_A\in
\End_{\operatorname{gr}}A$). Since $\log^{\star}\left( \id_A\right)
=\mathfrak{e}$, this yields that $\mathfrak{e}\in
\End_{\operatorname{gr}}A$. In other words, $\mathfrak{e}$ is a graded
$\kk$-linear map. The definition of $\mathfrak{e}_{n}$ is legitimate
because the gradedness of $\mathfrak{e}$ yields $\pi_{n}\circ\mathfrak{e}
=\mathfrak{e}\circ\pi_n$. This solves Exercise~\ref{exe.eulerian-idp}(a).

We notice for future use the fact that $\mathfrak{e}_0 = 0$
\ \ \ \ \footnote{\textit{Proof.} The definition of $\mathfrak{e}_0$ yields
$\mathfrak{e}_0 = \pi_0\circ\mathfrak{e}=\mathfrak{e}\circ\pi_0$. Thus,
$\mathfrak{e}_0\left( A\right) = \left( \mathfrak{e}\circ\pi_0\right)
\left(  A\right)  =\mathfrak{e}\left(  \underbrace{\pi_0 \left( A\right)
}_{=\kk \cdot 1_{A}}\right)  =\mathfrak{e}\left(  \kk\cdot
1_{A}\right)  =\kk\cdot\mathfrak{e}\left(  1_{A}\right)  $. But every
$n\geq 1$ satisfies $f^{\star n}\left(  1_{A}\right)  =0$ (since $\Delta
^{\left(  n-1\right)  }\left(  1_{A}\right)  =\underbrace{1_{A}\otimes
1_{A}\otimes...\otimes1_{A}}_{n\text{ times}}$ and $f\left(  1_{A}\right)
=0$). Since $\mathfrak{e}=\sum_{n\geq 1}\left(  -1\right)  ^{n-1}
\frac{1}{n} f^{\star n}$, we have
\[
\mathfrak{e}\left(  1_{A}\right)  =\left(  \sum_{n\geq 1}
\left(  -1\right)^{n-1}\frac{1}{n}f^{\star n}\right)  \left( 1_{A}\right)
=\sum_{n\geq 1}\left(  -1\right)  ^{n-1}\frac{1}{n}
 \underbrace{f^{\star n}\left(1_{A}\right)  }_{=0}
=\sum_{n\geq 1}\left(  -1\right)  ^{n-1}\frac{1}{n}0 = 0.
\]
Now, $\mathfrak{e}_{0}\left(  A\right)  =\kk\cdot
\underbrace{\mathfrak{e}\left(  1_{A}\right)  }_{=0}=0$, so that
$\mathfrak{e}_{0}=0$, qed.}.

We also record the fact that
\begin{equation}
\label{sol.eulerian-idp.projs}
\Delta_{A}\circ\pi_n = \left(  \sum_{k=0}^{n}
\pi_{k} \otimes \pi_{n-k} \right) \circ \Delta_{A} \ \ \ \ \ \ \ \ \ \ 
\text{for all } n \in \NN .
\end{equation}
(This follows by checking that both sides of \eqref{sol.eulerian-idp.projs}
are equal to $\Delta_A$ on the $n$-th homogeneous component $A_n$, while
vanishing on all other components\footnote{And this is because $\Delta_A$
is graded.}.)

(b) From \eqref{sol.xi-lifts-p.1}, we have
\begin{align*}
\sum_{n\geq 1}\xi_{n}t^{n}  &  =\log\left(  \sum_{n\geq0}H_{n}t^{n}\right)
=\log\left(  1+\sum_{n\geq 1}H_{n}t^{n}\right) \\
&  =\sum_{i\geq1}\left(  -1\right)  ^{i-1}\dfrac{1}{i}\underbrace{\left(
\sum_{n\geq1}H_{n}t^{n}\right)  ^{i}}_{=\sum_{n_{1},n_{2},...,n_{i}\geq
1}H_{n_{1}}H_{n_{2}}...H_{n_{i}}t^{n_{1}+n_{2}+...+n_{i}}}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the Mercator series for the
logarithm}\right) \\
&  =\sum_{i\geq1}\left(  -1\right)  ^{i-1}\dfrac{1}{i}
 \sum_{n_1,n_2,...,n_i \geq 1} H_{n_1}H_{n_2}...H_{n_i}
   t^{n_1+n_2+...+n_i} \\
&  =\sum_{n\geq1}\sum_{i\geq1}\left(  -1\right)  ^{i-1}\dfrac{1}{i}
 \sum\limits_{\substack{n_1,n_2,...,n_i \geq 1; \\ n_1+n_2+...+n_i=n}}
   H_{n_{1}}H_{n_{2}}...H_{n_{i}}t^n
\end{align*}
in the power series ring $\Nsym \left[  \left[  t\right]
\right]  $. Comparing coefficients in this equality, we obtain
\begin{equation}
\xi_{n}=\sum_{i\geq1}\left(  -1\right)  ^{i-1}\dfrac{1}{i}\sum
 _{\substack{n_1,n_2,...,n_i \geq 1; \\ n_1+n_2+...+n_i=n}}
   H_{n_1}H_{n_2}...H_{n_i}
\label{sol.eulerian-idp.b.1}
\end{equation}
for every $n\geq1$. Hence, every $n\geq 1$ satisfies
\begin{align}
\mathfrak{W}\left( \xi_n\right)
&  =\mathfrak{W}\left(  \sum_{i\geq 1}\left(  -1\right) ^{i-1}
   \dfrac{1}{i}\sum\limits_{\substack{n_{1},n_{2},...,n_{i} \geq 1;\\ 
   n_1+n_2+...+n_i=n}}
   H_{n_{1}}H_{n_{2}}...H_{n_{i}}\right) \nonumber\\
&  =\sum_{i\geq1}\left(  -1\right)  ^{i-1}\dfrac{1}{i}
 \sum\limits_{\substack{n_1,n_2,...,n_i \geq 1; \\ n_1+n_2+...+n_i=n}}
   \mathfrak{W}\left( H_{n_1}\right)
   \star \mathfrak{W}\left( H_{n_2}\right)
   \star ...
   \star \mathfrak{W}\left(  H_{n_i}\right) \nonumber \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{W}
\text{ is a }\kk\text{-algebra homomorphism}\right) \nonumber\\
&  =\sum_{i\geq1}\left(  -1\right)  ^{i-1}\dfrac{1}{i}
 \sum\limits_{\substack{n_1,n_2,...,n_i \geq 1; \\ n_1+n_2+...+n_i=n}}
   \pi_{n_1}\star\pi_{n_2}\star ... \star\pi_{n_i}
\label{sol.eulerian-idp.b.2}
\end{align}
(since $\mathfrak{W}$ maps every $H_m$ to $\pi_m$).

On the other hand, $f=\sum_{n\geq1}\pi_{n}$. Thus,
\[
f^{\star i}=\left(  \sum_{n\geq1}\pi_{n}\right)  ^{\star i}
=\sum_{n_1,n_2,...,n_i \geq 1}
  \pi_{n_1} \star \pi_{n_2} \star ... \star \pi_{n_i}
\]
for every $i \in \NN$. Now,
\begin{align}
\mathfrak{e} 
&  =\sum_{n\geq1}\left(  -1\right)  ^{n-1}\frac{1}{n}f^{\star n}
= \sum_{i\geq1}\left(  -1\right)  ^{i-1}\frac{1}{i}
\underbrace{f^{\star i}}
_{=\sum_{n_{1},n_{2},...,n_{i}\geq1}\pi_{n_{1}}\star\pi_{n_{2}}\star
...\star\pi_{n_{i}}}\nonumber\\
&  =\sum_{i\geq1}\left(  -1\right)  ^{i-1}\frac{1}{i}
\sum_{n_{1},n_{2},...,n_{i}\geq1}
\pi_{n_{1}}\star\pi_{n_{2}}\star...\star\pi_{n_{i}}.
\label{sol.eulerian-idp.b.4}
\end{align}
Now,
\begin{align*}
\mathfrak{e}_{n}  &  =\pi_{n}\circ\mathfrak{e}\\
&  =\pi_{n}\circ\left(  \sum_{i\geq1}\left(  -1\right)  ^{i-1}\frac{1}{i}
\sum_{n_{1},n_{2},...,n_{i}\geq1}\pi_{n_{1}}\star\pi_{n_{2}}\star...\star
\pi_{n_{i}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.eulerian-idp.b.4}}\right) \\
&  =\sum_{i\geq1}\left(  -1\right)  ^{i-1}\frac{1}{i}
\sum_{n_{1},n_{2},...,n_{i}\geq1}\pi_{n}\circ
\left(  \pi_{n_{1}}\star\pi_{n_{2}} \star...\star\pi_{n_{i}}\right) \\
&  =\underbrace{\sum_{i\geq1}\left(  -1\right)  ^{i-1}\frac{1}{i}
\sum\limits_{\substack{n_{1},n_{2},...,n_{i}\geq1;\\ n_{1}+n_{2}+...+n_{i}=n}}\pi
_{n}\circ\left(  \pi_{n_{1}}\star\pi_{n_{2}}\star...\star\pi_{n_{i}}\right)
}_{\substack{=\mathfrak{W}\left(  \xi_{n}\right)  \\\text{(by
\eqref{sol.eulerian-idp.b.2})}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i\geq1}\left(  -1\right)  ^{i-1}\frac{1}{i}
\sum\limits_{\substack{n_{1},n_{2},...,n_{i}\geq1;\\ n_{1}+n_{2}+...+n_{i}\neq
n}}\underbrace{\pi_{n}\circ\left(  \pi_{n_{1}}\star\pi_{n_{2}}\star...\star
\pi_{n_{i}}\right)  }_{\substack{=0\\\text{(because the image of}\\
\pi_{n_{1}}\star\pi_{n_{2}}\star...\star\pi_{n_{i}}
=m^{\left(  i-1\right) }\circ\left(
\pi_{n_{1}}\otimes\pi_{n_{2}}\otimes...\otimes\pi_{n_{i}}\right)  \circ
\Delta^{\left(  i-1\right)  }\\\text{is contained in}\\m^{\left(  i-1\right)
}\left(  \left(  \pi_{n_1}\otimes\pi_{n_2}\otimes...\otimes\pi_{n_i}
\right)  \left(  A^{\otimes i}\right)  \right) 
=A_{n_{1}}A_{n_{2}}...A_{n_{i}} \subset A_{n_{1}+n_{2}+...+n_{i}}\text{,}
\\\text{which is a different
homogeneous component of }A\text{ than }A_{n}\text{)}}}\\
&  =\mathfrak{W}\left(  \xi_{n}\right)  +\underbrace{\sum_{i\geq1}\left(
-1\right)  ^{i-1}\frac{1}{i}\sum\limits_{\substack{n_{1},n_{2},...,n_{i}\geq
1;\\ n_{1}+n_{2}+...+n_{i}\neq n}}0}_{=0}=\mathfrak{W}\left(  \xi_{n}\right)  .
\end{align*}
This solves Exercise~\ref{exe.eulerian-idp}(b).

(c) Exercise~\ref{exe.cocommutative.coalghom}(a) (applied to $C=A$) yields that
the comultiplication $\Delta_{A}:A\rightarrow A\otimes A$ is a
$\kk$-coalgebra homomorphism (since $A$ is cocommutative).
%Exercise
%\ref{exe.cocommutative.coalghom}(a) (applied to $C=\Nsym$) yields
%that the comultiplication $\Delta_{\Nsym}:\Nsym 
%\rightarrow\Nsym \otimes\Nsym$ is a
%$\kk$-coalgebra homomorphism (since $\Nsym$ is cocommutative).
Also, the comultiplications $\Delta_{A}$ and $\Delta_{\Nsym }$
are $\kk$-algebra homomorphisms (as is the comultiplication of any
$\kk$-bialgebra).

Let $R$ be the $\kk$-linear map $\left(  \End A,\star\right)  
\otimes\left(  \End A,\star\right)
\rightarrow\left(  \End \left(  A\otimes A\right)
,\star\right)  $ which sends every tensor $f\otimes g\in\left(
\End A,\star\right)  \otimes\left(  \End A,\star\right)  $ to the
map $f\otimes g:A\otimes A\rightarrow A\otimes A$.
This map $R$ is a $\kk$-algebra homomorphism\footnote{In fact, this
is a particular case of Exercise~\ref{exe.convolution.tensor}(b).}.

Let $\Omega_{1}$ be the composition
\[
\Nsym \overset{\mathfrak{W}}{\longrightarrow}\left(
\End A,\star\right)  \overset{\operatorname{post}\left(
\Delta_{A}\right)  }{\longrightarrow}\left(  \Hom\left(
A,A\otimes A\right)  ,\star\right)  ,
\]
where $\operatorname{post}\left(  \Delta_{A}\right)  $ denotes the
$\kk$-linear map sending every $\gamma\in\left(  \End A,\star\right) $
to $\Delta_{A}\circ\gamma\in\left(  \Hom
\left(  A,A\otimes A\right)  ,\star\right)  $. Since the map $\mathfrak{W}$ is
a $\kk$-algebra homomorphism, and since the map $\operatorname{post}
\left(  \Delta_{A}\right)  $ is a $\kk$-algebra
homomorphism\footnote{This follows from
Proposition~\ref{prop.convolution.functor}
(applied to $A$, $A$, $A$, $A \otimes A$, $\id_A$ and
$\Delta_A$ instead of $C$, $C'$, $A$, $A'$, $\gamma$ and
$\alpha$)
(because we know
that $\Delta_{A}$ is a $\kk$-algebra homomorphism).}, their composition
$\Omega_{1}$ also is a $\kk$-algebra homomorphism.

Let $\Omega_{2}$ be the composition
\[
\Nsym \overset{\Delta_{\Nsym }}{\longrightarrow
}\Nsym \otimes\Nsym \overset{\mathfrak{W}
\otimes\mathfrak{W}}{\longrightarrow}\left(  \End A,\star\right) 
 \otimes\left(  \End A,\star\right)
\overset{R}{\longrightarrow}\left(  \End \left(  A\otimes
A\right)  ,\star\right)  \overset{\operatorname{pre}\left(  \Delta
_{A}\right)  }{\longrightarrow}\left(  \Hom \left(  A,A\otimes
A\right)  ,\star\right)  ,
\]
where $\operatorname{pre}\left(  \Delta_{A}\right)  $ denotes the
$\kk$-linear map sending every $\gamma\in\left(  \End 
\left(  A\otimes A\right)  ,\star\right)  $ to $\gamma\circ\Delta_{A}
\in\left(  \Hom \left(  A,A\otimes A\right)  ,\star\right)  $.
Since the maps $\Delta_{\Nsym }$, $\mathfrak{W}\otimes
\mathfrak{W}$ and $R$ are $\kk$-algebra homomorphisms, and since the
map $\operatorname{pre}\left(  \Delta_{A}\right)  $ is a $\kk$-algebra
homomorphism\footnote{This follows from
Proposition~\ref{prop.convolution.functor}
(applied to $A$, $A \otimes A$, $A \otimes A$, $A \otimes A$, $\Delta_A$ and
$\id_{A \otimes A}$ instead of $C$, $C'$, $A$, $A'$, $\gamma$ and
$\alpha$)
(because we know
that $\Delta_{A}$ is a $\kk$-coalgebra homomorphism).}, their
composition $\Omega_{2}$ also is a $\kk$-algebra homomorphism.

In order to solve Exercise~\ref{exe.eulerian-idp}(c), we need to show that
every $w\in\Nsym$ satisfies $\Delta\circ\left(  \mathfrak{W}
\left(  w\right)  \right)  =\left(  \sum_{(w)}\mathfrak{W}\left(
w_{1}\right)  \otimes\mathfrak{W}\left(  w_{2}\right)  \right)  \circ\Delta$.
Since
\begin{align*}
\Delta\circ\left(  \mathfrak{W}\left(  w\right)  \right)   =\Delta_{A}
\circ\left(  \mathfrak{W}\left(  w\right)  \right)  =\left(
\operatorname{post}\left(  \Delta_{A}\right)  \right)  \left(  \mathfrak{W}
\left(  w\right)  \right) 
 =\underbrace{\left(  \operatorname{post}\left(  \Delta_{A}\right)
\circ\mathfrak{W}\right)  }_{=\Omega_{1}}\left(  w\right)  =\Omega_{1}\left(
w\right)
\end{align*}
and
\begin{align*}
\underbrace{\left(  \sum_{(w)}\mathfrak{W}\left(  w_{1}\right)  \otimes
\mathfrak{W}\left(  w_{2}\right)  \right)  }_{=R\left(  \left(  \mathfrak{W}
\otimes\mathfrak{W}\right)  \left(  \Delta_{\Nsym }\left(
w\right)  \right)  \right)  }\circ\underbrace{\Delta}_{=\Delta_{A}}  &
=\left(  R\left(  \left(  \mathfrak{W}\otimes\mathfrak{W}\right)  \left(
\Delta_{\Nsym }\left(  w\right)  \right)  \right)  \right)
\circ\Delta_{A}\\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \right)  \left(
R\left(  \left(  \mathfrak{W}\otimes\mathfrak{W}\right)  \left(
\Delta_{\Nsym }\left(  w\right)  \right)  \right)  \right) \\
&  =\underbrace{\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \circ
R\circ\left(  \mathfrak{W}\otimes\mathfrak{W}\right)  \circ\Delta
_{\Nsym }\right)  }_{=\Omega_{2}}\left(  w\right)  =\Omega
_{2}\left(  w\right)  ,
\end{align*}
this is equivalent to showing that every $w\in\Nsym$ satisfies
$\Omega_{1}\left(  w\right)  =\Omega_{2}\left(  w\right)  $. In other words,
we need to prove that $\Omega_{1}=\Omega_{2}$. Since $\Omega_{1}$ and
$\Omega_{2}$ are $\kk$-algebra homomorphisms, it will be enough to
verify this on the generators $H_{1},H_{2},H_{3},...$ of the
$\kk$-algebra $\Nsym$. But on said generators, this is easily seen
to hold, because every $n\geq1$ satisfies
\begin{align*}
\Omega_{1}\left(  H_{n}\right)   &  =\left(  \operatorname{post}\left(
\Delta_{A}\right)  \circ\mathfrak{W}\right)  \left(  H_{n}\right)  =\left(
\operatorname{post}\left(  \Delta_{A}\right)  \right)  \left(
\underbrace{\mathfrak{W}\left(  H_{n}\right)  }_{=\pi_{n}}\right) \\
&  =\left(  \operatorname{post}\left(  \Delta_{A}\right)  \right)  \left(
\pi_{n}\right)  =\Delta_{A}\circ\pi_{n}=\left(  \sum_{k=0}^{n}\pi_{k}
\otimes\pi_{n-k}\right)  \circ\Delta_{A}
  \ \ \ \ \ \ \ \ \ \ \left(  \text{by }\eqref{sol.eulerian-idp.projs}\right)
\end{align*}
and
\begin{align*}
\Omega_{2}\left(  H_{n}\right)   &  =\left(  \operatorname{pre}\left(
\Delta_{A}\right)  \circ R\circ\left(  \mathfrak{W}\otimes\mathfrak{W}\right)
\circ\Delta_{\Nsym }\right)  \left(  H_{n}\right) \\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \circ R\circ\left(
\mathfrak{W}\otimes\mathfrak{W}\right)  \right)  \left(  \underbrace{\Delta
_{\Nsym }\left(  H_{n}\right)  }_{=\sum_{k=0}^{n}H_{k}\otimes
H_{n-k}}\right) \\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \circ R\circ\left(
\mathfrak{W}\otimes\mathfrak{W}\right)  \right)  \left(  \sum_{k=0}^{n}
H_{k}\otimes H_{n-k}\right) \\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \circ R\right)
\left(  \left(  \mathfrak{W}\otimes\mathfrak{W}\right)  \left( 
 \sum_{k=0}^{n}H_{k}\otimes H_{n-k}\right)  \right) \\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \circ R\right)
\left(  \sum_{k=0}^{n}\underbrace{\mathfrak{W}\left(  H_{k}\right)  }
_{=\pi_{k}}\otimes\underbrace{\mathfrak{W}\left(  H_{n-k}\right)  }
_{=\pi_{n-k}}\right) \\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \circ R\right)
\left(  \sum_{k=0}^{n}\pi_{k}\otimes\pi_{n-k}\right) \\
&  =\left(  \operatorname{pre}\left(  \Delta_{A}\right)  \right)  \left(
R\left(  \sum_{k=0}^{n}\pi_{k}\otimes\pi_{n-k}\right)  \right)
=\underbrace{R\left(  \sum_{k=0}^{n}\pi_{k}\otimes\pi_{n-k}\right)  }
_{=\sum_{k=0}^{n}\pi_{k}\otimes\pi_{n-k}}\circ\Delta_{A}\\
&  =\left(  \sum_{k=0}^{n}\pi_{k}\otimes\pi_{n-k}\right)  \circ\Delta_{A}.
\end{align*}
Exercise~\ref{exe.eulerian-idp}(c) is proven.

(d) Let $n\geq 0$. We need to prove that $\mathfrak{e}_n\left( A\right)
\subset\Liep$. If $n=0$, then this is clear because
$\mathfrak{e}_{0}=0$.
Hence, we assume WLOG that we don't have $n=0$. Thus, $n\geq 1$.
Hence, Exercise~\ref{exe.xi-lifts-p}(a) yields that $\xi_n$ is primitive, so
that $\Delta\left( \xi_n\right) = \xi_n \otimes 1 + 1 \otimes \xi_n$.

Applying Exercise~\ref{exe.eulerian-idp}(c) to $w=\xi_{n}$, we obtain
\begin{align*}
\Delta\circ\left(  \mathfrak{W}\left(  \xi_{n}\right)  \right)   &  =\left(
\sum_{(\xi_n)}\mathfrak{W}\left(  \left(  \xi_{n}\right)  _{1}\right)
\otimes\mathfrak{W}\left(  \left(  \xi_{n}\right)  _{2}\right)  \right)
\circ\Delta\\
&  =\left(  \mathfrak{W}\left(  \xi_{n}\right)  \otimes\mathfrak{W}\left(
1\right)  +\mathfrak{W}\left(  1\right)  \otimes\mathfrak{W}\left(  \xi
_{n}\right)  \right)  \circ\Delta\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\sum_{(\xi_n)}\left( \xi_n\right)_{1}
\otimes\left(  \xi_{n}\right) _{2}
=\Delta\left(  \xi_{n}\right) = \xi_n \otimes 1 + 1 \otimes \xi_n
\right) .
\end{align*}
Since $\mathfrak{W}\left(  1\right)  =u\epsilon$ and $\mathfrak{W}\left(
\xi_{n}\right)  =\mathfrak{e}_{n}$, this rewrites as
\[
\Delta\circ\mathfrak{e}_{n}=\left(  \mathfrak{e}_{n}\otimes\left(
u\epsilon\right)  +\left(  u\epsilon\right)  \otimes\mathfrak{e}_{n}\right)
\circ\Delta.
\]


Now, let $x\in A$. Then,
\begin{align*}
\Delta\left(  \mathfrak{e}_{n}\left(  x\right)  \right)   &
=\underbrace{\left(  \Delta\circ\mathfrak{e}_{n}\right)  }_{=\left(
\mathfrak{e}_{n}\otimes\left(  u\epsilon\right)  +\left(  u\epsilon\right)
\otimes\mathfrak{e}_{n}\right)  \circ\Delta}\left(  x\right)  =\left(  \left(
\mathfrak{e}_{n}\otimes\left(  u\epsilon\right)  +\left(  u\epsilon\right)
\otimes\mathfrak{e}_{n}\right)  \circ\Delta\right)  \left(  x\right) \\
&  =\left(  \mathfrak{e}_{n}\otimes\left(  u\epsilon\right)  \right)  \left(
\Delta\left(  x\right)  \right)  +\left(  \left(  u\epsilon\right)
\otimes\mathfrak{e}_{n}\right)  \left(  \Delta\left(  x\right)  \right) \\
&  =\sum_{\left(  x\right)  }\mathfrak{e}_{n}\left(  x_{1}\right)
\otimes\underbrace{\left(  u\epsilon\right)  \left(  x_{2}\right)  }
_{=1_{A}\epsilon\left(  x_{2}\right)  }+\sum_{\left(  x\right)  }
\underbrace{\left(  u\epsilon\right)  \left(  x_{1}\right)  }_{=1_{A}
\epsilon\left(  x_{1}\right)  }\otimes\mathfrak{e}_{n}\left(  x_{2}\right) \\
&  =\sum_{\left(  x\right)  }\mathfrak{e}_{n}\left(  x_{1}\right)
\otimes 1_{A}\epsilon\left(  x_{2}\right)  +\sum_{\left(  x\right)  }
1_{A}\epsilon\left(  x_{1}\right)  \otimes\mathfrak{e}_{n}\left(  x_{2}\right)
\\
&  =\mathfrak{e}_{n}\left(  \underbrace{\sum_{\left(  x\right)  }x_{1}
\epsilon\left(  x_{2}\right)  }_{=x}\right)  \otimes 1_{A}+1_{A}\otimes
\mathfrak{e}_{n}\left(  \underbrace{\sum_{\left(  x\right)  }\epsilon\left(
x_{1}\right)  x_{2}}_{=x}\right) \\
&  =\mathfrak{e}_{n}\left(  x\right)  \otimes1_{A}
+1_{A}\otimes\mathfrak{e}_{n}\left(  x\right)  .
\end{align*}
Hence, $\mathfrak{e}_{n}\left(  x\right)  $ is a primitive element of $A$;
thus, $\mathfrak{e}_{n}\left(  x\right)  \in\Liep$. Forget now that we
fixed $x$. We thus have seen that every $x\in A$ satisfies $\mathfrak{e}
_{n}\left(  x\right)  \in\Liep$. In other words, $\mathfrak{e}
_{n}\left(  A\right)  \subset\Liep$. This solves
Exercise~\ref{exe.eulerian-idp}(d).

(e) Every $n \geq 0$ satisfies $\mathfrak{e}_n\left(A\right) \subset
\Liep$ (by Exercise~\ref{exe.eulerian-idp}(d)).

We have $\id_A = \sum_{n \geq 0}\pi_n$ (by the definition of $\pi_n$),
and
\[
\mathfrak{e}=\underbrace{\id_{A}}_{=\sum_{n\geq0}\pi_{n}}\circ\mathfrak{e=}
\sum_{n\geq0}\underbrace{\pi_{n}\circ\mathfrak{e}}_{\substack{=\mathfrak{e}
_{n}\\\text{(since the definition of }\mathfrak{e}_{n}\\\text{yields }
\mathfrak{e}_{n}=\pi_{n}\circ\mathfrak{e}\text{)}}}=\sum_{n\geq0}
\mathfrak{e}_{n}.
\]
Hence, $\mathfrak{e}\left(  A\right)  =\left(  \sum_{n\geq0}\mathfrak{e}
_{n}\right)  \left(  A\right)  \subset\sum_{n\geq0}\underbrace{\mathfrak{e}
_{n}\left(  A\right)  }_{\subset\Liep}\subset\sum_{n\geq0}
\Liep\subset\Liep$. This proves part (e) of the exercise.

(f) Let $x\in\Liep$. Then, $\Delta\left(  x\right)  =x\otimes
1_{A}+1_{A}\otimes x$.

Let $\mathfrak{g}\left(  A,A\right)  $ denote the $\kk$-submodule
of $\End A$ which consists of all $g\in\End A$
satisfying $g\left(  1_{A}\right)  =0$. Then, it is easy to see that
$\mathfrak{g}\left(  A,A\right)  $ is an ideal of the algebra $\left(
\End A,\star\right)  $ (indeed, any $g_{1}\in\End A$ and
$g_{2}\in\mathfrak{g}\left(  A,A\right)  $ satisfy $g_{1}\star g_{2}
\in\mathfrak{g}\left(  A,A\right)  $ and $g_{2}\star g_{1}\in\mathfrak{g}
\left(  A,A\right)  $). We have $f\in\mathfrak{g}\left(  A,A\right)  $. Since
$\mathfrak{g}\left(  A,A\right)  $ is an ideal of $\left(
\End A,\star\right)  $, this
yields that every $n\geq2$ satisfies $f^{\star n}\in\left(  \mathfrak{g}
\left(  A,A\right)  \right)  ^{\star2}$ (where $I^{\star2}$ denotes the square
of the ideal $I$ for any ideal $I$ of the $\kk$-algebra $\left(
\End A,\star\right)  $). But it is easy to see that
\begin{equation}
g\left(  x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\left(
\mathfrak{g}\left(  A,A\right)  \right)  ^{\star 2}.
\label{sol.eulerian-idp.f.1}
\end{equation}
\footnote{\textit{Proof of \eqref{sol.eulerian-idp.f.1}:} It is clearly enough
to check that $\left(  g_{1}\star g_{2}\right)  \left(  x\right)  =0$ for
every $g_{1}\in\mathfrak{g}\left(  A,A\right)  $ and $g_{2}\in\mathfrak{g}
\left(  A,A\right)  $. But this is easy: If $g_{1}\in\mathfrak{g}\left(
A,A\right)  $ and $g_{2}\in\mathfrak{g}\left(  A,A\right)  $, then
$g_{1}\left(  1_{A}\right)  =0$ and $g_{2}\left(  1_{A}\right)  =0$, so that
\begin{align*}
\left(  g_{1}\star g_{2}\right)  \left(  x\right)   &  =g_{1}\left(  x\right)
\underbrace{g_{2}\left(  1_{A}\right)  }_{=0}+\underbrace{g_{1}\left(
1_{A}\right)  }_{=0}g_{2}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\Delta\left(  x\right)  =x\otimes1_{A}+1_{A}\otimes x\right)  \\
&  =0+0=0.
\end{align*}
This proves \eqref{sol.eulerian-idp.f.1}.} In particular, this yields that
\begin{equation}
f^{\star n}\left(  x\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }
n\geq2\label{sol.eulerian-idp.f.2}
\end{equation}
(since $f^{\star n}\in\left(  \mathfrak{g}\left(  A,A\right)  \right)
^{\star2}$ for every $n\geq2$). Now, $\mathfrak{e}=\sum_{n\geq 1}\left(
-1\right)  ^{n-1}\frac{1}{n}f^{\star n}$, so that
\begin{align*}
\mathfrak{e}\left(  x\right)   &  =\left(  \sum_{n\geq1}\left(  -1\right)
^{n-1}\frac{1}{n}f^{\star n}\right)  \left(  x\right)  =\sum_{n\geq1}\left(
-1\right)  ^{n-1}\frac{1}{n}f^{\star n}\left(  x\right)  \\
&  =\underbrace{\left(  -1\right)  ^{1-1}\dfrac{1}{1}}_{=1}
\underbrace{f^{\star1}}_{=f=\id_A-u\epsilon}\left(  x\right)  +\sum_{n\geq
2}\left(  -1\right)  ^{n-1}\frac{1}{n}\underbrace{f^{\star n}\left(  x\right)
}_{\substack{=0\\\text{(by \eqref{sol.eulerian-idp.f.2})}}}\\
&  =\left( \id_{A}-u\epsilon\right)  \left(  x\right)  +\underbrace{\sum
_{n\geq2}\left(  -1\right)  ^{n-1}\frac{1}{n}0}_{=0}
=\left(  \id_A -u\epsilon\right)  \left(  x\right)
=x-u\left(  \underbrace{\epsilon\left(  x\right)  }
_{\substack{=0\\\text{(by Proposition \ref{antipode-on-primitives})}}}\right)
=x-\underbrace{u\left(  0\right)  }_{=0}=x.
\end{align*}


Now, forget that we fixed $x$. We thus have shown that $\mathfrak{e}\left(
x\right)  =x$ for every $x\in\Liep$. In other words, the map
$\mathfrak{e}$ fixes any element of $\Liep$. This solves
Exercise~\ref{exe.eulerian-idp}(f).

Combining the results of parts (e) and (f) of Exercise~\ref{exe.eulerian-idp},
we conclude that $\mathfrak{e}$ is a projection from $A$ to the $\kk$-submodule
$\Liep$. This completes the solution of the exercise.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.eulerian-idp.2}}

\textit{Solution to Exercise~\ref{exe.eulerian-idp.2}.} Let us first
show the following two lemmas, which have nothing to do with Hopf algebras:

\begin{lemma}
\label{lem.sol.eulerian-idp.2.poly1bis}
Let $V$ be any torsionfree abelian group
(written additively). Let $N\in \NN$. For every $k\in\left\{
0,1,...,N\right\}  $, let $w_{k}$ be an element of $V$. Assume that
\begin{equation}
\sum_{k=0}^{N}w_{k}n^{k}=0\ \ \ \ \ \ \ \ \ \ \text{for all }n\in \NN .
\end{equation}
Then, $w_{k}=0$ for every $k\in\left\{  0,1,...,N\right\}  $.
\end{lemma}

\begin{lemma}
\label{lem.sol.eulerian-idp.2.poly2}
Let $V$ be any torsionfree abelian group (written additively).
Let $N\in \NN$.
For every $\left(  k,\ell\right)
\in\left\{  0,1,...,N\right\}  ^{2}$, let $v_{k,\ell}$ be an
element of $V$.
Assume that
\begin{equation}
\sum_{k=0}^{N}\sum_{\ell=0}^{N}v_{k,\ell}n^{k}m^{\ell}
=0\ \ \ \ \ \ \ \ \ \ \text{for all }n \in \NN \text{ and }
m \in \NN .
\label{eq.lem.sol.eulerian-idp.2.poly2.ass}
\end{equation}
Then, $v_{k,\ell}=0$ for every $\left(  k,\ell\right)  \in\left\{
0,1,...,N\right\}  ^{2}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.eulerian-idp.2.poly1bis}.]
Lemma~\ref{lem.sol.eulerian-idp.2.poly1bis} has already appeared
above (namely, as Lemma~\ref{lem.sol.eulerian-idp.2.poly1}),
and has already been proven
(in the solution to Exercise~\ref{exe.leray.exp-alg}).
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.sol.eulerian-idp.2.poly2}.]Fix $m\in \NN$.
Every $n\in \NN$ satisfies%
\[
\sum_{k=0}^{N}\left(  \sum_{\ell=0}^{N}v_{k,\ell}m^{\ell}\right)  n^{k}%
=\sum_{k=0}^{N}\sum_{\ell=0}^{N}v_{k,\ell}n^{k}m^{\ell}=0.
\]
Thus, Lemma \ref{lem.sol.eulerian-idp.2.poly1bis} (applied to $w_{k}=\sum
_{\ell=0}^{N}v_{k,\ell}m^{\ell}$) yields that $\sum_{\ell=0}^{N}v_{k,\ell
}m^{\ell}=0$ for every $k\in\left\{  0,1,...,N\right\}  $.

Now, forget that we fixed $m$. We thus have proven that%
\begin{equation}
\sum_{\ell=0}^{N}v_{k,\ell}m^{\ell}=0\ \ \ \ \ \ \ \ \ \ \text{for every }%
m\in \NN \text{ and }k\in\left\{  0,1,...,N\right\}  .
\label{pf.lem.sol.eulerian-idp.2.poly2.1}%
\end{equation}


Now, fix $g\in\left\{  0,1,...,N\right\}  $. For every $n\in \NN$, we
have
\begin{align*}
\sum_{k=0}^{N}v_{g,k}n^{k}  &  =\sum_{\ell=0}^{N}v_{g,\ell}n^{\ell
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}k\text{ as }\ell\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.eulerian-idp.2.poly2.1}), applied to }k=g\text{ and
}m=n\right)  .
\end{align*}
Hence, Lemma \ref{lem.sol.eulerian-idp.2.poly1bis} (applied to $w_{k}=v_{g,k}$)
yields that $v_{g,k}=0$ for every $k\in\left\{  0,1,...,N\right\}  $.

Now, forget that we fixed $g$. We thus have shown that $v_{g,k}=0$ for every
$g\in\left\{  0,1,...,N\right\}  $ and $k\in\left\{  0,1,...,N\right\}  $.
Renaming the indices $g$ and $k$ as $k$ and $\ell$
in this statement, we obtain the following:
We have $v_{k,\ell}=0$ for every $k\in\left\{  0,1,...,N\right\}  $ and
$\ell\in\left\{  0,1,...,N\right\}  $. In other words, $v_{k,\ell}=0$ for
every $\left(  k,\ell\right)  \in\left\{  0,1,...,N\right\}  ^{2}$. This
proves Lemma \ref{lem.sol.eulerian-idp.2.poly2}.
\end{proof}

Now, let us come to the solution of Exercise \ref{exe.eulerian-idp.2}.

Define $\operatorname{End}_{\operatorname{gr}}A$ as in the solution of
Exercise~\ref{exe.eulerian-idp}. We can prove (just as in the solution of
Exercise~\ref{exe.eulerian-idp}) that $\left(  \operatorname{End}%
_{\operatorname{gr}}A,\star\right)  $ is a $\kk$-subalgebra of $\left(
\operatorname{End}A,\star\right)  $. Since $ \id_{A}%
\in\operatorname{End}_{\operatorname{gr}}A$, we thus have
$\id_{A}^{\star\ell} \in \End_{\operatorname{gr}}A$ for every $\ell
\in \NN$. In other words,
\begin{equation}
 \id_{A}^{\star\ell}\text{ is a graded } \kk \text{-linear
map for every }\ell\in \NN  .
\label{sol.eulerian-idp.2.gr-id}
\end{equation}


We have $\mathfrak{e}\left(  1_{A}\right)  =0$ (as was proven in a footnote in
the solution of Exercise~\ref{exe.eulerian-idp}). Hence, $\mathfrak{e}\left(
\underbrace{A_{0}}_{= \kk \cdot1_{A}}\right)  =\mathfrak{e}\left(
 \kk \cdot1_{A}\right)  = \kk \cdot\underbrace{\mathfrak{e}\left(
1_{A}\right)  }_{=0}=0$.

Notice that $A$ is a $\kk$-module, hence a $\QQ$-module
(since $\QQ$ is a subring of $\kk$), thus a torsionfree
abelian group.

(a) For every $\kk$-linear map $f:A\rightarrow A$ which annihilates
$A_{0}$, we can define an endomorphism $\exp^{\star}f$ of $A$ by setting
$\exp^{\star}f=\sum_{\ell\geq0}\dfrac{1}{\ell!}f^{\star\ell}$. (This follows
from the same argument as the well-definedness of $\log^{\star}\left(
f+u\epsilon\right)  $.\ \ \ \ \footnote{This definition of
$\exp^\star f$ is actually a particular case of
Definition~\ref{def.leray.pwnilp}(d).
This can be shown as follows:
If $f : A \to A$ is a $\kk$-linear map which annihilates
$A_0$, then
Proposition~\ref{prop.convolution-series}(h) (applied to $C=A$)
yields $f \in \mathfrak{n}\left(A, A\right)$.
Therefore, Definition~\ref{def.leray.pwnilp}(d) defines a
map $\exp^\star f \in \mathfrak{n}\left(A, A\right)$.
This map is identical to the map
$\exp^\star f := \sum_{\ell\geq0}\dfrac{1}{\ell!}f^{\star\ell}$
we have just defined, because the map
$\exp^\star f$ defined using
Definition~\ref{def.leray.pwnilp}(d) satisfies
\begin{align*}
\exp^\star f
&= \sum_{n \geq 0} \dfrac{1}{n!} f^{\star n}
\qquad \left(\text{since } \exp = \sum_{n \geq 0} \dfrac{1}{n!} T^n \right) \\
&= \sum_{\ell\geq0}\dfrac{1}{\ell!} f^{\star\ell} .
\end{align*}
}) The usual rules for exponentials and logarithms apply:
\begin{itemize}
\item We have
$\exp^{\star} \left( \log^{\star}\left(  f+u\epsilon\right) \right)
 =f+u\epsilon$ for every $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(b) (applied to $C = A$
and $g = f + u \epsilon$), after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$).}
\item We have $\log^{\star} \left( \exp^{\star}f \right) =f$
for every $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(a) (applied to $C = A$),
after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$).}
\item We have
$\exp^{\star} \left( f + g \right)
= \left(\exp^{\star} f\right) \star \left(\exp^{\star} g\right)$ for any two
$\kk$-linear maps $f:A\rightarrow A$ and $g:A\rightarrow A$ which
annihilate $A_{0}$ and satisfy $f\star g=g\star f$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(c) (applied to $C = A$),
after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$) and $g \in \mathfrak{n}\left(A, A\right)$
(for similar reasons).}
\item We have $\exp^{\star}\left(
nf\right)  =\left(  \exp^{\star}f\right)  ^{\star n}$ for every $n\in
 \NN$ and any $\kk$-linear map $f:A\rightarrow A$ which
annihilates $A_{0}$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(e) (applied to $C = A$),
after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$).}
\end{itemize}

The map $\mathfrak{e}$ annihilates $A_{0}$ (since $\mathfrak{e}\left(
A_{0}\right)  =0$), and thus an endomorphism $\exp^{\star}\mathfrak{e}$ of $A$
is well-defined. We have
$\mathfrak{e} = \log^{\star} \left( \id_A \right)$,
so that $\exp^{\star} \mathfrak{e} =
\exp^\star \left( \log^{\star} \left( \id_A \right) \right)
= \id_{A}$
(since
$\exp^{\star} \left( \log^{\star}\left(  f+u\epsilon\right) \right) =f+u\epsilon$
for every $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$).

Let us recall that any $f$ annihilating $A_{0}$ has the property that for each
$n$ one has that $A_{n}$ is annihilated by $f^{\star m}$ for every $m>n$ (we
saw this in the proof of Proposition \ref{Takeuchi-prop}). Applying this to
$f=\mathfrak{e}$ (which, as we know, annihilates $A_{0}$), and renaming $m$
and $n$ as $n$ and $N$, we obtain%
\begin{equation}
\mathfrak{e}^{\star n}\left(  A_{N}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for
every }N\in{ \NN }\text{ and every }n\in{ \NN }\text{ satisfying
}n>N.
\label{sol.eulerian-idp.2.finiteness}
\end{equation}


Recall that $\exp^{\star}\left(  nf\right)  =\left(  \exp^{\star}f\right)
^{\star n}$ for every $n\in \NN$ and any $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$. Applying this to $f=\mathfrak{e}%
$, we see that every $n\in \NN$ satisfies
\begin{equation}
\exp^{\star}\left(  n\mathfrak{e}\right)  =\left(  \underbrace{\exp^{\star
}\mathfrak{e}}_{= \id_{A}}\right)  ^{\star n}
= \id_{A}^{\star n}.
\label{sol.eulerian-idp.2.a.expid}
\end{equation}


Now, let us fix $M\in \NN$. Let also $N$ be any integer satisfying
$N\geq M$. It is easy to see that
\begin{equation}
 \id_{A}^{\star m}\left(  v\right)  =\sum_{\ell=0}^{N}\dfrac
{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in A_{M}\text{ and every }%
m \in \NN .
\label{sol.eulerian-idp.2.a.id}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.eulerian-idp.2.a.id}):} Let $v\in A_{M}$
and $m\in{ \NN } $. Every integer $\ell$ satisfying $\ell>M$ satisfies
$\mathfrak{e}^{\star\ell}\left(  \underbrace{v}_{\in A_{M}}\right)
\in\mathfrak{e}^{\star\ell}\left(  A_{M}\right)  =0$ (by
(\ref{sol.eulerian-idp.2.finiteness}), applied to $\ell$ and $M$ instead of
$n$ and $N$). In other words,
\begin{equation}
\text{every integer }\ell\text{ satisfying }\ell>M\text{ satisfies
}\mathfrak{e}^{\star\ell}\left(  v\right)  =0.
\label{sol.eulerian-idp.2.a.id.pf.4}
\end{equation}
\par
From (\ref{sol.eulerian-idp.2.a.expid}) (applied to $m$ instead of $n$), we
obtain $\exp^{\star} \left(  m \mathfrak{e} \right)  =
\id_{A}^{\star m}$, so that
\[
 \id_{A}^{\star m}=\exp^{\star}\left(  m\mathfrak{e}\right)
=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}\left(  m\mathfrak{e}\right)
^{\star\ell}=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}%
^{\star\ell}
\]
and thus%
\begin{align*}
 \id_{A}^{\star m}\left(  v\right)   &  =\sum_{\ell=0}^{\infty
}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)  =\sum
_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)
+\sum_{\ell=N+1}^{\infty}\dfrac{1}{\ell!}m^{\ell}\underbrace{\mathfrak{e}%
^{\star\ell}\left(  v\right)  }_{\substack{=0\\\text{(by
(\ref{sol.eulerian-idp.2.a.id.pf.4}),}\\\text{since }\ell>N\geq M\text{)}}}\\
&  =\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  +\underbrace{\sum_{\ell=N+1}^{\infty}\dfrac{1}{\ell!}m^{\ell}0}%
_{=0}=\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  .
\end{align*}
This proves (\ref{sol.eulerian-idp.2.a.id}).}

Now, let $v\in A_{M}$, $n\in{ \NN }$ and $m\in{ \NN }$ be arbitrary.
Since $ \id_{A}^{\star m}$ is a graded map (by
(\ref{sol.eulerian-idp.2.gr-id}), applied to $\ell=m$), we have
$ \id_{A}^{\star m}\left(  v\right)  \in A_{M}$ (since $v\in
A_{M}$). Hence, (\ref{sol.eulerian-idp.2.a.id}) (applied to $n$ and
$ \id_{A}^{\star m}\left(  v\right)  $ instead of $m$ and $v$)
yields%
\begin{align}
 \id_{A}^{\star n}\left(   \id_{A}^{\star m}\left(
v\right)  \right)   &  =\sum_{\ell=0}^{N}\dfrac{1}{\ell!}n^{\ell}%
\mathfrak{e}^{\star\ell}\left(   \id_{A}^{\star m}\left(
v\right)  \right)  =\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\mathfrak{e}^{\star
k}\left(  \underbrace{ \id_{A}^{\star m}\left(  v\right)
}_{\substack{=\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star
\ell}\left(  v\right)  \\\text{(by (\ref{sol.eulerian-idp.2.a.id}))}}}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\ell\text{ as }k\right) \nonumber\\
&  =\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\underbrace{\mathfrak{e}^{\star k}\left(
\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  \right)  }_{\substack{=\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell
}\mathfrak{e}^{\star k}\left(  \mathfrak{e}^{\star\ell}\left(  v\right)
\right)  \\\text{(since }\mathfrak{e}^{\star k}\text{ is a } \kk %
\text{-linear map)}}}=\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\sum_{\ell=0}^{N}%
\dfrac{1}{\ell!}m^{\ell}\underbrace{\mathfrak{e}^{\star k}\left(
\mathfrak{e}^{\star\ell}\left(  v\right)  \right)  }_{=\left(  \mathfrak{e}%
^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }\nonumber\\
&  =\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell
}\left(  \mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(
v\right)  =\sum_{k=0}^{N}\sum_{\ell=0}^{N}\dfrac{\left(  \mathfrak{e}^{\star
k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell!}%
n^{k}m^{\ell}. \label{sol.eulerian-idp.2.a.5}%
\end{align}
But%
\begin{align}
 \id_{A}^{\star n}\left(   \id_{A}^{\star m}\left(
v\right)  \right)   &  =\underbrace{\left(   \id_{A}^{\star
n}\circ \id_{A}^{\star m}\right)  }_{\substack{
 =\id_{A}^{\star\left(  nm\right)  }\\
\text{(by the dual of Exercise \ref{exe.alghoms.comm}(f),}\\
\text{applied to }k=m\text{ and }\ell=n\text{)}}}\left(  v\right)
= \id_{A}^{\star\left(  nm\right)  }\left(  v\right) \nonumber\\
&  =\sum_{\ell=0}^{N}\dfrac{1}{\ell!}\left(  nm\right)  ^{\ell}\mathfrak{e}%
^{\star\ell}\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.eulerian-idp.2.a.id}), applied to }nm\text{ instead of }m\right)  ,
\label{sol.eulerian-idp.2.a.7}
\end{align}
Now,
\begin{align*}
\sum_{k=0}^{N}\sum_{\ell=0}^{N}\left(  \dfrac{\left(  \mathfrak{e}^{\star
k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell!}%
-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }{k!}\right)
n^{k}m^{\ell}  &  =\underbrace{\sum_{k=0}^{N}\sum_{\ell=0}^{N}\dfrac{\left(
\mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)
}{k!\ell!}n^{k}m^{\ell}}_{\substack{= \id_{A}^{\star n}\left(
 \id_{A}^{\star m}\left(  v\right)  \right)  \\\text{(by
(\ref{sol.eulerian-idp.2.a.5}))}}}-\underbrace{\sum_{k=0}^{N}\sum_{\ell=0}%
^{N}\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }{k!}%
n^{k}m^{\ell}}_{\substack{=\sum_{\ell=0}^{N}\dfrac{\mathfrak{e}^{\star\ell
}\left(  v\right)  }{\ell!}n^{\ell}m^{\ell}\\=\sum_{\ell=0}^{N}\dfrac{1}%
{\ell!}\left(  nm\right)  ^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)
\\= \id_{A}^{\star n}\left(   \id_{A}^{\star
m}\left(  v\right)  \right)  \\\text{(by (\ref{sol.eulerian-idp.2.a.7}))}}}\\
&  = \id_{A}^{\star n}\left(   \id_{A}^{\star
m}\left(  v\right)  \right)  - \id_{A}^{\star n}\left(
 \id_{A}^{\star m}\left(  v\right)  \right)  =0.
\end{align*}


Now, let us forget that we fixed $n$ and $m$. We thus have proven that%
\[
\sum_{k=0}^{N}\sum_{\ell=0}^{N}\left(  \dfrac{\left(  \mathfrak{e}^{\star
k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell!}%
-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }{k!}\right)
n^{k}m^{\ell}=0
\]
for all $n\in{ \NN } $ and $m\in{ \NN } $. Hence, we can apply Lemma
\ref{lem.sol.eulerian-idp.2.poly2} to $V=A$ and $v_{k,\ell}=\dfrac{\left(
\mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)
}{k!\ell!}-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }%
{k!}$ (because $A$ is a torsionfree abelian group). As a result, we obtain
$\dfrac{\left(  \mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)
\left(  v\right)  }{k!\ell!}-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  }{k!}=0$ for every $\left(  k,\ell\right)  \in\left\{
0,1,...,N\right\}  ^{2}$. In other words,%
\begin{equation}
\dfrac{\left(  \mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)
\left(  v\right)  }{k!\ell!}=\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  }{k!}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
k,\ell\right)  \in\left\{  0,1,...,N\right\}  ^{2}.
\label{sol.eulerian-idp.2.a.mostlydone}%
\end{equation}


Now, forget that we fixed $v$, $M$ and $N$. We thus have proven that every
$M\in{ \NN } $, every integer $N$ satisfying $N\geq M$, and every $v\in
A_{M}$ satisfy (\ref{sol.eulerian-idp.2.a.mostlydone}).

Now, let us fix two elements $n\in{ \NN } $ and $m\in{ \NN } $. In
order to finish the solution of Exercise \ref{exe.eulerian-idp.2}(a), it
remains to show that $\mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star
m}=n!\delta_{n,m}\mathfrak{e}^{\star n}$. In order to show this, it is clearly
enough to prove that $\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star
m}\right)  \left(  v\right)  =n!\delta_{n,m}\mathfrak{e}^{\star n}\left(
v\right)  $ for every $v\in A$. So let us fix some $v\in A$, and let us show
that $\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}\right)
\left(  v\right)  =n!\delta_{n,m}\mathfrak{e}^{\star n}\left(  v\right)  $.

Since both sides of the identity $\left(  \mathfrak{e}^{\star n}%
\circ\mathfrak{e}^{\star m}\right)  \left(  v\right)  =n!\delta_{n,m}%
\mathfrak{e}^{\star n}\left(  v\right)  $ are $\kk$-linear in $v$, we
can WLOG assume that $v$ is a homogeneous element of $A$ (because every
element of $A$ is a $\kk$-linear combination of homogeneous elements).
Assume this. Thus, $v\in A_{M}$ for some $M\in{ \NN }$. Consider this
$M$. Choose some $N\in{ \NN }$ satisfying $N\geq M$, $N\geq n$ and $N\geq
m$. (Such an $N$ clearly exists.) Then, $N\geq M$ and $\left(  n,m\right)
\in\left\{  0,1,...,N\right\}  ^{2}$, and therefore we can apply
(\ref{sol.eulerian-idp.2.a.mostlydone}) to $k=n$ and $\ell=m$. As a result, we
obtain $\dfrac{\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star
m}\right)  \left(  v\right)  }{n!m!}=\dfrac{\delta_{n,m}\mathfrak{e}^{\star
n}\left(  v\right)  }{n!}$. Multiplying this identity with $n!m!$, we obtain
\[
\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}\right)  \left(
v\right)  =\underbrace{m!\delta_{n,m}}_{\substack{=n!\delta_{n,m}%
\\\text{(indeed, this is clear if }n=m\text{,}\\\text{and otherwise follows
from }\delta_{n,m}=0\text{)}}}\mathfrak{e}^{\star n}\left(  v\right)
=n!\delta_{n,m}\mathfrak{e}^{\star n}\left(  v\right)  .
\]
So we have proven $\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star
m}\right)  \left(  v\right)  =n!\delta_{n,m}\mathfrak{e}^{\star n}\left(
v\right)  $. As we have seen, this completes the solution of Exercise
\ref{exe.eulerian-idp.2}(a).

(b) Let $n\in \NN$ and $m\in \NN$. From
(\ref{sol.eulerian-idp.2.a.expid}) (applied to $m$ instead of $n$), we obtain
$\exp^{\star}\left(  m\mathfrak{e}\right)  = \id_{A}^{\star m}$,
so that
\[
 \id_{A}^{\star m}=\exp^{\star}\left(  m\mathfrak{e}\right)
=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}\left(  m\mathfrak{e}\right)
^{\star\ell}=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}%
^{\star\ell} .
\]
Thus,
\begin{align*}
\mathfrak{e}^{\star n}\circ\underbrace{ \id_{A}^{\star m}}%
_{=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}} &
=\mathfrak{e}^{\star n}\circ\left(  \sum_{\ell=0}^{\infty}\dfrac{1}{\ell
!}m^{\ell}\mathfrak{e}^{\star\ell}\right)  =\underbrace{\sum_{\ell=0}^{\infty
}}_{=\sum_{\ell\in{ \NN }}}\dfrac{1}{\ell!}m^{\ell}%
\underbrace{\mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star\ell}}%
_{\substack{=n!\delta_{n,\ell}\mathfrak{e}^{\star n}\\\text{(by Exercise
\ref{exe.eulerian-idp.2}(a),}\\\text{applied to }\ell\text{ instead of
}m\text{)}}}\\
&  =\sum_{\ell\in{ \NN }}\dfrac{1}{\ell!}m^{\ell}n!\delta_{n,\ell
}\mathfrak{e}^{\star n}=\dfrac{1}{n!}m^{n}n!\mathfrak{e}^{\star n}%
=m^{n}\mathfrak{e}^{\star n}%
\end{align*}
and similarly $ \id_{A}^{\star m}\circ\mathfrak{e}^{\star
n}=m^{n}\mathfrak{e}^{\star n}$. This solves Exercise~\ref{exe.eulerian-idp.2}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.eulerian-idp.2}}

\textit{Solution to Exercise~\ref{exe.eulerian-idp.2}.} Let us first
show the following two lemmas, which have nothing to do with Hopf algebras:

\begin{lemma}
\label{lem.sol.eulerian-idp.2.poly1bis}
Let $V$ be any torsionfree abelian group
(written additively). Let $N\in \NN$. For every $k\in\left\{
0,1,...,N\right\}  $, let $w_{k}$ be an element of $V$. Assume that
\begin{equation}
\sum_{k=0}^{N}w_{k}n^{k}=0\ \ \ \ \ \ \ \ \ \ \text{for all }n\in \NN .
\end{equation}
Then, $w_{k}=0$ for every $k\in\left\{  0,1,...,N\right\}  $.
\end{lemma}

\begin{lemma}
\label{lem.sol.eulerian-idp.2.poly2}
Let $V$ be any torsionfree abelian group (written additively).
Let $N\in \NN$.
For every $\left(  k,\ell\right)
\in\left\{  0,1,...,N\right\}  ^{2}$, let $v_{k,\ell}$ be an
element of $V$.
Assume that
\begin{equation}
\sum_{k=0}^{N}\sum_{\ell=0}^{N}v_{k,\ell}n^{k}m^{\ell}
=0\ \ \ \ \ \ \ \ \ \ \text{for all }n \in \NN \text{ and }
m \in \NN .
\label{eq.lem.sol.eulerian-idp.2.poly2.ass}
\end{equation}
Then, $v_{k,\ell}=0$ for every $\left(  k,\ell\right)  \in\left\{
0,1,...,N\right\}  ^{2}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.eulerian-idp.2.poly1bis}.]
Lemma~\ref{lem.sol.eulerian-idp.2.poly1bis} has already appeared
above (namely, as Lemma~\ref{lem.sol.eulerian-idp.2.poly1}),
and has already been proven
(in the solution to Exercise~\ref{exe.leray.exp-alg}).
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.sol.eulerian-idp.2.poly2}.]Fix $m\in \NN$.
Every $n\in \NN$ satisfies%
\[
\sum_{k=0}^{N}\left(  \sum_{\ell=0}^{N}v_{k,\ell}m^{\ell}\right)  n^{k}%
=\sum_{k=0}^{N}\sum_{\ell=0}^{N}v_{k,\ell}n^{k}m^{\ell}=0.
\]
Thus, Lemma \ref{lem.sol.eulerian-idp.2.poly1bis} (applied to $w_{k}=\sum
_{\ell=0}^{N}v_{k,\ell}m^{\ell}$) yields that $\sum_{\ell=0}^{N}v_{k,\ell
}m^{\ell}=0$ for every $k\in\left\{  0,1,...,N\right\}  $.

Now, forget that we fixed $m$. We thus have proven that%
\begin{equation}
\sum_{\ell=0}^{N}v_{k,\ell}m^{\ell}=0\ \ \ \ \ \ \ \ \ \ \text{for every }%
m\in \NN \text{ and }k\in\left\{  0,1,...,N\right\}  .
\label{pf.lem.sol.eulerian-idp.2.poly2.1}%
\end{equation}


Now, fix $g\in\left\{  0,1,...,N\right\}  $. For every $n\in \NN$, we
have
\begin{align*}
\sum_{k=0}^{N}v_{g,k}n^{k}  &  =\sum_{\ell=0}^{N}v_{g,\ell}n^{\ell
}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}k\text{ as }\ell\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.eulerian-idp.2.poly2.1}), applied to }k=g\text{ and
}m=n\right)  .
\end{align*}
Hence, Lemma \ref{lem.sol.eulerian-idp.2.poly1bis} (applied to $w_{k}=v_{g,k}$)
yields that $v_{g,k}=0$ for every $k\in\left\{  0,1,...,N\right\}  $.

Now, forget that we fixed $g$. We thus have shown that $v_{g,k}=0$ for every
$g\in\left\{  0,1,...,N\right\}  $ and $k\in\left\{  0,1,...,N\right\}  $.
Renaming the indices $g$ and $k$ as $k$ and $\ell$
in this statement, we obtain the following:
We have $v_{k,\ell}=0$ for every $k\in\left\{  0,1,...,N\right\}  $ and
$\ell\in\left\{  0,1,...,N\right\}  $. In other words, $v_{k,\ell}=0$ for
every $\left(  k,\ell\right)  \in\left\{  0,1,...,N\right\}  ^{2}$. This
proves Lemma \ref{lem.sol.eulerian-idp.2.poly2}.
\end{proof}

Now, let us come to the solution of Exercise \ref{exe.eulerian-idp.2}.

Define $\operatorname{End}_{\operatorname{gr}}A$ as in the solution of
Exercise~\ref{exe.eulerian-idp}. We can prove (just as in the solution of
Exercise~\ref{exe.eulerian-idp}) that $\left(  \operatorname{End}%
_{\operatorname{gr}}A,\star\right)  $ is a $\kk$-subalgebra of $\left(
\operatorname{End}A,\star\right)  $. Since $\id_{A}%
\in\operatorname{End}_{\operatorname{gr}}A$, we thus have
$\id_{A}^{\star\ell}\in\operatorname{End}_{\operatorname{gr}}A$ for every
$\ell\in \NN$. In other words,%
\begin{equation}
\id_{A}^{\star\ell}\text{ is a graded } \kk %
\text{-linear map for every }\ell\in \NN .
\label{sol.eulerian-idp.2.gr-id}
\end{equation}


We have $\mathfrak{e}\left(  1_{A}\right)  =0$ (as was proven in a footnote in
the solution of Exercise~\ref{exe.eulerian-idp}). Hence, $\mathfrak{e}\left(
\underbrace{A_{0}}_{= \kk \cdot1_{A}}\right)  =\mathfrak{e}\left(
 \kk \cdot1_{A}\right)  = \kk \cdot\underbrace{\mathfrak{e}\left(
1_{A}\right)  }_{=0}=0$.

Notice that $A$ is a $\kk$-module, hence a $\QQ$-module
(since $\QQ$ is a subring of $\kk$), thus a torsionfree
abelian group.

(a) For every $\kk$-linear map $f:A\rightarrow A$ which annihilates
$A_{0}$, we can define an endomorphism $\exp^{\star}f$ of $A$ by setting
$\exp^{\star}f=\sum_{\ell\geq0}\dfrac{1}{\ell!}f^{\star\ell}$. (This follows
from the same argument as the well-definedness of $\log^{\star}\left(
f+u\epsilon\right)  $.\ \ \ \ \footnote{This definition of
$\exp^\star f$ is actually a particular case of
Definition~\ref{def.leray.pwnilp}(d).
This can be shown as follows:
If $f : A \to A$ is a $\kk$-linear map which annihilates
$A_0$, then
Proposition~\ref{prop.convolution-series}(h) (applied to $C=A$)
yields $f \in \mathfrak{n}\left(A, A\right)$.
Therefore, Definition~\ref{def.leray.pwnilp}(d) defines a
map $\exp^\star f \in \mathfrak{n}\left(A, A\right)$.
This map is identical to the map
$\exp^\star f := \sum_{\ell\geq0}\dfrac{1}{\ell!}f^{\star\ell}$
we have just defined, because the map
$\exp^\star f$ defined using
Definition~\ref{def.leray.pwnilp}(d) satisfies
\begin{align*}
\exp^\star f
&= \sum_{n \geq 0} \dfrac{1}{n!} f^{\star n}
\qquad \left(\text{since } \exp = \sum_{n \geq 0} \dfrac{1}{n!} T^n \right) \\
&= \sum_{\ell\geq0}\dfrac{1}{\ell!} f^{\star\ell} .
\end{align*}
}) The usual rules for exponentials and logarithms apply:
\begin{itemize}
\item We have
$\exp^{\star} \left( \log^{\star}\left(  f+u\epsilon\right) \right)
 =f+u\epsilon$ for every $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(b) (applied to $C = A$
and $g = f + u \epsilon$), after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$).}
\item We have $\log^{\star} \left( \exp^{\star}f \right) =f$
for every $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(a) (applied to $C = A$),
after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$).}
\item We have
$\exp^{\star} \left( f + g \right)
= \left(\exp^{\star} f\right) \star \left(\exp^{\star} g\right)$ for any two
$\kk$-linear maps $f:A\rightarrow A$ and $g:A\rightarrow A$ which
annihilate $A_{0}$ and satisfy $f\star g=g\star f$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(c) (applied to $C = A$),
after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$) and $g \in \mathfrak{n}\left(A, A\right)$
(for similar reasons).}
\item We have $\exp^{\star}\left(
nf\right)  =\left(  \exp^{\star}f\right)  ^{\star n}$ for every $n\in
 \NN$ and any $\kk$-linear map $f:A\rightarrow A$ which
annihilates $A_{0}$.
\ \ \ \ \footnote{Indeed, this follows from
Proposition~\ref{prop.exp-log}(e) (applied to $C = A$),
after first observing
that $f \in \mathfrak{n}\left(A, A\right)$
(by Proposition \ref{prop.convolution-series}(h), applied to
$C=A$).}
\end{itemize}

The map $\mathfrak{e}$ annihilates $A_{0}$ (since $\mathfrak{e}\left(
A_{0}\right)  =0$), and thus an endomorphism $\exp^{\star}\mathfrak{e}$ of $A$
is well-defined. We have
$\mathfrak{e} = \log^{\star} \left( \id_A \right)$,
so that $\exp^{\star} \mathfrak{e} =
\exp^\star \left( \log^{\star} \left( \id_A \right) \right)
= \id_{A}$
(since
$\exp^{\star} \left( \log^{\star}\left(  f+u\epsilon\right) \right) =f+u\epsilon$
for every $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$).

Let us notice first that%
\begin{equation}
\mathfrak{e}^{\star n}\left(  A_{N}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for
every }N\in \NN \text{ and every }n\in \NN \text{ satisfying }n>N.
\label{sol.eulerian-idp.2.finiteness}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.eulerian-idp.2.finiteness}):} We shall
prove (\ref{sol.eulerian-idp.2.finiteness}) by induction over $n$. The
induction base (the case $n=0$) is vacuously true (since no $N\in \NN$
satisfies $0>N$). For the induction step, we fix some $\nu\in \NN$, and
assume that (\ref{sol.eulerian-idp.2.finiteness}) holds for $n=\nu$. We now
must prove that (\ref{sol.eulerian-idp.2.finiteness}) holds for $n=\nu+1$.
\par
We have assumed that (\ref{sol.eulerian-idp.2.finiteness}) holds for $n=\nu$.
In other words,%
\begin{equation}
\mathfrak{e}^{\star\nu}\left(  A_{N}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for
every }N\in \NN \text{ satisfying }\nu>N.
\label{sol.eulerian-idp.2.finiteness.pf.1}%
\end{equation}
\par
Let $N\in \NN$ be such that $\nu+1>N$. Then, $\nu>N-1$, so that every
$k\in\left\{  1,2,...,N\right\}  $ satisfies $\nu>N-k$. Hence,%
\begin{equation}
\text{every }k\in\left\{  1,2,...,N\right\}  \text{ satisfies }\mathfrak{e}%
^{\star\nu}\left(  A_{N-k}\right)  =0
\label{sol.eulerian-idp.2.finiteness.pf.2}%
\end{equation}
(by (\ref{sol.eulerian-idp.2.finiteness.pf.1}), applied to $N-k$ instead of
$N$). Also, $\mathfrak{e}^{\star\left(  \nu+1\right)  }=\mathfrak{e}%
\star\mathfrak{e}^{\star\nu}=m_{A}\circ\left(  \mathfrak{e}\otimes
\mathfrak{e}^{\star \nu}\right)  \circ\Delta_{A}$ (by the definition of
convolution) and $\Delta_{A}\left(  A_{N}\right)  \subset\sum_{k=0}^{N}%
A_{k}\otimes A_{N-k}$ (since the coalgebra $A$ is graded). Hence,%
\begin{align*}
\underbrace{\mathfrak{e}^{\star\left(  \nu+1\right)  }}_{=m_{A}\circ\left(
\mathfrak{e}\otimes\mathfrak{e}^{\star \nu}\right)  \circ\Delta_{A}}\left(
A_{N}\right)   &  =\left(  m_{A}\circ\left(  \mathfrak{e}\otimes
\mathfrak{e}^{\star \nu}\right)  \circ\Delta_{A}\right)  \left(  A_{N}\right)
=m_{A}\left(  \left(  \mathfrak{e}\otimes\mathfrak{e}^{\star \nu}\right)
\left(  \underbrace{\Delta_{A}\left(  A_{N}\right)  }_{\subset\sum_{k=0}%
^{N}A_{k}\otimes A_{N-k}}\right)  \right) \\
&  \subset m_{A}\left(  \underbrace{\left(  \mathfrak{e}\otimes\mathfrak{e}%
^{\star \nu}\right)  \left(  \sum_{k=0}^{N}A_{k}\otimes A_{N-k}\right)  }%
_{=\sum_{k=0}^{N}\mathfrak{e}\left(  A_{k}\right)  \otimes\mathfrak{e}%
^{\star\nu}\left(  A_{N-k}\right)  }\right) \\
&  =m_{A}\left(  \sum_{k=0}^{N}\mathfrak{e}\left(  A_{k}\right)
\otimes\mathfrak{e}^{\star\nu}\left(  A_{N-k}\right)  \right)  =\sum_{k=0}%
^{N}\mathfrak{e}\left(  A_{k}\right)  \mathfrak{e}^{\star\nu}\left(
A_{N-k}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }m_{A}\text{ is the multiplication
map}\right) \\
&  =\underbrace{\mathfrak{e}\left(  A_{0}\right)  }_{=0}\mathfrak{e}^{\star
\nu}\left(  A_{N-0}\right)  +\sum_{k=1}^{N}\mathfrak{e}\left(  A_{k}\right)
\underbrace{\mathfrak{e}^{\star\nu}\left(  A_{N-k}\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.eulerian-idp.2.finiteness.pf.2}))}}}\\
&  =0\mathfrak{e}^{\star\nu}\left(  A_{N-0}\right)  +\sum_{k=1}^{N}%
\mathfrak{e}\left(  A_{k}\right)  0=0.
\end{align*}
Hence, $\mathfrak{e}^{\star\left(  \nu+1\right)  }\left(  A_{N}\right)  =0$.
\par
Now, forget that we fixed $N$. We thus have shown that $\mathfrak{e}%
^{\star\left(  \nu+1\right)  }\left(  A_{N}\right)  =0$ for every
$N\in \NN$ satisfying $\nu+1>N$. In other words,
(\ref{sol.eulerian-idp.2.finiteness}) holds for $n=\nu+1$. This completes the
induction step and thus the inductive proof of
(\ref{sol.eulerian-idp.2.finiteness}).}

Recall that $\exp^{\star}\left(  nf\right)  =\left(  \exp^{\star}f\right)
^{\star n}$ for every $n\in \NN$ and any $\kk$-linear map
$f:A\rightarrow A$ which annihilates $A_{0}$. Applying this to $f=\mathfrak{e}%
$, we see that every $n\in \NN$ satisfies
\begin{equation}
\exp^{\star}\left(  n\mathfrak{e}\right)  =\left(  \underbrace{\exp^{\star
}\mathfrak{e}}_{=\id_{A}}\right)  ^{\star
n}=\id_{A}^{\star n}.
\label{sol.eulerian-idp.2.a.expid}
\end{equation}


Now, let us fix $M\in \NN$. Let also $N$ be any integer satisfying
$N\geq M$. It is easy to see that
\begin{equation}
\id_{A}^{\star m}\left(  v\right)  =\sum_{\ell=0}%
^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in A_{M}\text{ and every }%
m\in \NN .
\label{sol.eulerian-idp.2.a.id}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.eulerian-idp.2.a.id}):} Let $v\in A_{M}$
and $m\in \NN$. Every integer $\ell$ satisfying $\ell>M$ satisfies
$\mathfrak{e}^{\star\ell}\left(  \underbrace{v}_{\in A_{M}}\right)
\in\mathfrak{e}^{\star\ell}\left(  A_{M}\right)  =0$ (by
(\ref{sol.eulerian-idp.2.finiteness}), applied to $\ell$ and $M$ instead of
$n$ and $N$). In other words,
\begin{equation}
\text{every integer }\ell\text{ satisfying }\ell>M\text{ satisfies
}\mathfrak{e}^{\star\ell}\left(  v\right)  =0.
\label{sol.eulerian-idp.2.a.id.pf.4}
\end{equation}
\par
From (\ref{sol.eulerian-idp.2.a.expid}) (applied to $m$ instead of $n$), we
obtain $\exp^{\star} \left( m \mathfrak{e} \right) = \id_A^{\star m}$,
so that
\[
\id_{A}^{\star m}=\exp^{\star}\left(  m\mathfrak{e}%
\right)  =\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}\left(  m\mathfrak{e}\right)
^{\star\ell}=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}%
^{\star\ell}
\]
and thus%
\begin{align*}
\id_{A}^{\star m}\left(  v\right)   &  =\sum_{\ell
=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)
=\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  +\sum_{\ell=N+1}^{\infty}\dfrac{1}{\ell!}m^{\ell}%
\underbrace{\mathfrak{e}^{\star\ell}\left(  v\right)  }%
_{\substack{=0\\\text{(by (\ref{sol.eulerian-idp.2.a.id.pf.4}),}\\\text{since
}\ell>N\geq M\text{)}}}\\
&  =\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  +\underbrace{\sum_{\ell=N+1}^{\infty}\dfrac{1}{\ell!}m^{\ell}0}%
_{=0}=\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  .
\end{align*}
This proves (\ref{sol.eulerian-idp.2.a.id}).}

Now, let $v\in A_{M}$, $n\in \NN$ and $m\in \NN$ be arbitrary.
Since $\id_{A}^{\star m}$ is a graded map (by
(\ref{sol.eulerian-idp.2.gr-id}), applied to $\ell=m$), we have
$\id_{A}^{\star m}\left(  v\right)  \in A_{M}$ (since
$v\in A_{M}$). Hence, (\ref{sol.eulerian-idp.2.a.id}) (applied to $n$ and
$\id_{A}^{\star m}\left(  v\right)  $ instead of $m$
and $v$) yields%
\begin{align}
\id_{A}^{\star n}\left(  \id_{A}^{\star m}\left(  v\right)  \right)
&  =\sum_{\ell=0}^{N}%
\dfrac{1}{\ell!}n^{\ell}\mathfrak{e}^{\star\ell}\left(  \id_{A}^{\star m}
\left(  v\right)  \right)  =\sum_{k=0}^{N}\dfrac{1}%
{k!}n^{k}\mathfrak{e}^{\star k}\left(  \underbrace{\id_{A}^{\star m}
\left(  v\right)  }_{\substack{=\sum_{\ell=0}^{N}%
\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)  \\\text{(by
(\ref{sol.eulerian-idp.2.a.id}))}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\ell\text{ as }k\right) \nonumber\\
&  =\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\underbrace{\mathfrak{e}^{\star k}\left(
\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}\left(
v\right)  \right)  }_{\substack{=\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell
}\mathfrak{e}^{\star k}\left(  \mathfrak{e}^{\star\ell}\left(  v\right)
\right)  \\\text{(since }\mathfrak{e}^{\star k}\text{ is a } \kk %
\text{-linear map)}}}=\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\sum_{\ell=0}^{N}%
\dfrac{1}{\ell!}m^{\ell}\underbrace{\mathfrak{e}^{\star k}\left(
\mathfrak{e}^{\star\ell}\left(  v\right)  \right)  }_{=\left(  \mathfrak{e}%
^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }\nonumber\\
&  =\sum_{k=0}^{N}\dfrac{1}{k!}n^{k}\sum_{\ell=0}^{N}\dfrac{1}{\ell!}m^{\ell
}\left(  \mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(
v\right)  =\sum_{k=0}^{N}\sum_{\ell=0}^{N}\dfrac{\left(  \mathfrak{e}^{\star
k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell!}%
n^{k}m^{\ell}. \label{sol.eulerian-idp.2.a.5}%
\end{align}
But we know (from Exercise~\ref{exe.alghoms.comm}(h)) that the dual of
Exercise~\ref{exe.alghoms.comm}(f) is true. This dual states that if
$H$ is a cocommutative $\kk$-bialgebra, and $k$ and $\ell$ are two
nonnegative integers, then $\id_H^{\star \ell} \circ \id_H^{\star k}
= \id_H^{\star \left(k\ell\right)}$. Applying this to $H = A$,
$k = m$ and $\ell = n$, we obtain
$\id_{A}^{\star n}\circ\id_{A}^{\star m}
= \id_{A}^{\star\left(  nm\right)}$. Now,
\begin{align}
\id_{A}^{\star n}\left(  \id_{A}^{\star m}\left(  v\right)  \right)
&  =\underbrace{\left(
\id_{A}^{\star n}\circ\id_{A}^{\star m}\right) 
 }_{=\id_{A}^{\star\left(  nm\right)}} \left(  v\right)
=\id_{A}^{\star\left(  nm\right)
}\left(  v\right) \nonumber\\
&  =\sum_{\ell=0}^{N}\dfrac{1}{\ell!}\left(  nm\right)  ^{\ell}\mathfrak{e}%
^{\star\ell}\left(  v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.eulerian-idp.2.a.id}), applied to }nm\text{ instead of }m\right)  ,
\label{sol.eulerian-idp.2.a.7}%
\end{align}
and%
\begin{align}
&  \sum_{k=0}^{N}\underbrace{\sum_{\ell=0}^{N}}_{=\sum_{\ell\in\left\{
0,1,...,N\right\}  }}\underbrace{\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  }{k!}}_{=\dfrac{1}{k!}\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  }n^{k}m^{\ell}\nonumber\\
&  =\sum_{k=0}^{N}\underbrace{\sum_{\ell\in\left\{  0,1,...,N\right\}  }%
\dfrac{1}{k!}\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)
n^{k}m^{\ell}}_{=\sum\limits_{\substack{\ell\in\left\{  0,1,...,N\right\}  ;\\\ell
=k}}\dfrac{1}{k!}\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)
n^{k}m^{\ell}+\sum\limits_{\substack{\ell\in\left\{  0,1,...,N\right\}  ;\\\ell\neq
k}}\dfrac{1}{k!}\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)
n^{k}m^{\ell}}\nonumber\\
&  =\sum_{k=0}^{N}\left(  \underbrace{\sum\limits_{\substack{\ell\in\left\{
0,1,...,N\right\}  ;\\\ell=k}}\dfrac{1}{k!}\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  n^{k}m^{\ell}}_{=\dfrac{1}{k!}\delta_{k,k}\mathfrak{e}%
^{\star k}\left(  v\right)  n^{k}m^{k}}+\sum\limits_{\substack{\ell\in\left\{
0,1,...,N\right\}  ;\\\ell\neq k}}\dfrac{1}{k!}\underbrace{\delta_{k,\ell}%
}_{\substack{=0\\\text{(since }\ell\neq k\text{)}}}\mathfrak{e}^{\star
k}\left(  v\right)  n^{k}m^{\ell}\right) \nonumber\\
&  =\sum_{k=0}^{N}\left(  \dfrac{1}{k!}\delta_{k,k}\mathfrak{e}^{\star
k}\left(  v\right)  n^{k}m^{k}+\underbrace{\sum\limits_{\substack{\ell\in\left\{
0,1,...,N\right\}  ;\\\ell\neq k}}\dfrac{1}{k!}0\mathfrak{e}^{\star k}\left(
v\right)  n^{k}m^{\ell}}_{=0}\right)  =\sum_{k=0}^{N}\dfrac{1}{k!}%
\underbrace{\delta_{k,k}}_{=1}\mathfrak{e}^{\star k}\left(  v\right)
\underbrace{n^{k}m^{k}}_{=\left(  nm\right)  ^{k}}\nonumber\\
&  =\sum_{k=0}^{N}\dfrac{1}{k!}\mathfrak{e}^{\star k}\left(  v\right)  \left(
nm\right)  ^{k}=\sum_{k=0}^{N}\dfrac{1}{k!}\left(  nm\right)  ^{k}%
\mathfrak{e}^{\star k}\left(  v\right)  =\sum_{\ell=0}^{N}\dfrac{1}{\ell
!}\left(  nm\right)  ^{\ell}\mathfrak{e}^{\star\ell}\left(  v\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}k\text{ as }\ell\right) \nonumber\\
&  =\id_{A}^{\star n}\left(  \id_{A}^{\star m}\left(  v\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.eulerian-idp.2.a.7})}\right)  .
\label{sol.eulerian-idp.2.a.9}
\end{align}
Now,%
\begin{align*}
&  \sum_{k=0}^{N}\sum_{\ell=0}^{N}\left(  \dfrac{\left(  \mathfrak{e}^{\star
k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell!}%
-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }{k!}\right)
n^{k}m^{\ell}\\
&  =\underbrace{\sum_{k=0}^{N}\sum_{\ell=0}^{N}\dfrac{\left(  \mathfrak{e}%
^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell
!}n^{k}m^{\ell}}_{\substack{=\id_{A}^{\star n}\left(
\id_{A}^{\star m}\left(  v\right)  \right)
\\\text{(by (\ref{sol.eulerian-idp.2.a.5}))}}}-\underbrace{\sum_{k=0}^{N}%
\sum_{\ell=0}^{N}\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)
}{k!}n^{k}m^{\ell}}_{\substack{=\id_{A}^{\star
n}\left(  \id_{A}^{\star m}\left(  v\right)  \right)
\\\text{(by (\ref{sol.eulerian-idp.2.a.9}))}}}\\
&  =\id_{A}^{\star n}\left(  \id_{A}^{\star m}\left(  v\right)  \right)
-\id_{A}^{\star n}\left(  \id_{A}^{\star m}\left(  v\right)  \right)
=0.
\end{align*}


Now, let us forget that we fixed $n$ and $m$. We thus have proven that%
\[
\sum_{k=0}^{N}\sum_{\ell=0}^{N}\left(  \dfrac{\left(  \mathfrak{e}^{\star
k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)  }{k!\ell!}%
-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }{k!}\right)
n^{k}m^{\ell}=0
\]
for all $n\in \NN$ and $m\in \NN$. Hence, we can apply Lemma
\ref{lem.sol.eulerian-idp.2.poly2} to $V=A$ and $v_{k,\ell}=\dfrac{\left(
\mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)  \left(  v\right)
}{k!\ell!}-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star k}\left(  v\right)  }%
{k!}$ (because $A$ is a torsionfree abelian group). As a result, we obtain
$\dfrac{\left(  \mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)
\left(  v\right)  }{k!\ell!}-\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  }{k!}=0$ for every $\left(  k,\ell\right)  \in\left\{
0,1,...,N\right\}  ^{2}$. In other words,%
\begin{equation}
\dfrac{\left(  \mathfrak{e}^{\star k}\circ\mathfrak{e}^{\star\ell}\right)
\left(  v\right)  }{k!\ell!}=\dfrac{\delta_{k,\ell}\mathfrak{e}^{\star
k}\left(  v\right)  }{k!}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(
k,\ell\right)  \in\left\{  0,1,...,N\right\}  ^{2}.
\label{sol.eulerian-idp.2.a.mostlydone}%
\end{equation}


Now, forget that we fixed $v$, $M$ and $N$. We thus have proven that every
$M\in \NN$, every integer $N$ satisfying $N\geq M$, and every $v\in
A_{M}$ satisfy (\ref{sol.eulerian-idp.2.a.mostlydone}).

Now, let us fix two elements $n\in \NN$ and $m\in \NN$. In order
to finish the solution of Exercise \ref{exe.eulerian-idp.2}(a), it remains to
show that $\mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}=n!\delta
_{n,m}\mathfrak{e}^{\star n}$. In order to show this, it is clearly enough to
prove that $\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}\right)
\left(  v\right)  =n!\delta_{n,m}\mathfrak{e}^{\star n}\left(  v\right)  $ for
every $v\in A$. So let us fix some $v\in A$, and let us show that $\left(
\mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}\right)  \left(  v\right)
=n!\delta_{n,m}\mathfrak{e}^{\star n}\left(  v\right)  $.

Since both sides of the identity $\left(  \mathfrak{e}^{\star n}%
\circ\mathfrak{e}^{\star m}\right)  \left(  v\right)  =n!\delta_{n,m}%
\mathfrak{e}^{\star n}\left(  v\right)  $ are $\kk$-linear in $v$, we
can WLOG assume that $v$ is a homogeneous element of $A$ (because every
element of $A$ is a $\kk$-linear combination of homogeneous elements).
Assume this. Thus, $v\in A_{M}$ for some $M\in \NN$. Consider this $M$.
Choose some $N\in \NN$ satisfying $N\geq M$, $N\geq n$ and $N\geq m$.
(Such an $N$ clearly exists.) Then, $N\geq M$ and $\left(  n,m\right)
\in\left\{  0,1,...,N\right\}  ^{2}$, and therefore we can apply
(\ref{sol.eulerian-idp.2.a.mostlydone}) to $k=n$ and $\ell=m$. As a result, we
obtain $\dfrac{\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star
m}\right)  \left(  v\right)  }{n!m!}=\dfrac{\delta_{n,m}\mathfrak{e}^{\star
n}\left(  v\right)  }{n!}$. Multiplying this identity with $n!m!$, we obtain
\[
\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}\right)  \left(
v\right)  =m!\delta_{n,m}\mathfrak{e}^{\star n}\left(  v\right)  .
\]


But we have $m!\delta_{n,m}=n!\delta_{n,m}$. (Indeed, this is obvious when
$n=m$, whereas in the case $n\neq m$, it follows from $\delta_{n,m}=0$.)
Hence,
\[
\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star m}\right)  \left(
v\right)  =\underbrace{m!\delta_{n,m}}_{=n!\delta_{n,m}}\mathfrak{e}^{\star
n}\left(  v\right)  =n!\delta_{n,m}\mathfrak{e}^{\star n}\left(  v\right)  .
\]
So we have proven $\left(  \mathfrak{e}^{\star n}\circ\mathfrak{e}^{\star
m}\right)  \left(  v\right)  =n!\delta_{n,m}\mathfrak{e}^{\star n}\left(
v\right)  $. As we have seen, this completes the solution of Exercise
\ref{exe.eulerian-idp.2}(a).

(b) Let $n\in \NN$ and $m\in \NN$. From
(\ref{sol.eulerian-idp.2.a.expid}) (applied to $m$ instead of $n$), we obtain
$\exp^{\star}\left(  m\mathfrak{e}\right)  = \id_{A}^{\star m}$,
so that
\[
\id_{A}^{\star m}=\exp^{\star}\left(  m\mathfrak{e}%
\right)  =\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}\left(  m\mathfrak{e}\right)
^{\star\ell}=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}%
^{\star\ell} .
\]
Thus,
\begin{align*}
\mathfrak{e}^{\star n}\circ\underbrace{\id_{A}^{\star
m}}_{=\sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}}
&  =\mathfrak{e}^{\star n}\circ\left(  \sum_{\ell=0}^{\infty}\dfrac{1}{\ell
!}m^{\ell}\mathfrak{e}^{\star\ell}\right)  \\
&  =\underbrace{\sum_{\ell=0}^{\infty}}_{=\sum_{\ell\in \NN }}\dfrac
{1}{\ell!}m^{\ell}\underbrace{\mathfrak{e}^{\star n}\circ\mathfrak{e}%
^{\star\ell}}_{\substack{=n!\delta_{n,\ell}\mathfrak{e}^{\star n}\\\text{(by
Exercise \ref{exe.eulerian-idp.2}(a),}\\\text{applied to }\ell\text{ instead
of }m\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
} \kk \text{-linear maps is } \kk \text{-bilinear}\right)  \\
&  =\sum_{\ell\in \NN }\dfrac{1}{\ell!}m^{\ell}n!\delta_{n,\ell
}\mathfrak{e}^{\star n}=\underbrace{\sum\limits_{\substack{\ell\in \NN %
;\\\ell=n}}\dfrac{1}{\ell!}m^{\ell}n!\delta_{n,\ell}\mathfrak{e}^{\star n}%
}_{\substack{=\dfrac{1}{n!}m^{n}n!\delta_{n,n}\mathfrak{e}^{\star
n}\\\text{(since }n\in \NN \text{)}}}+\sum\limits_{\substack{\ell\in
 \NN ;\\\ell\neq n}}\dfrac{1}{\ell!}m^{\ell}n!\underbrace{\delta_{n,\ell
}}_{\substack{=0\\\text{(since }\ell\neq n\text{)}}}\mathfrak{e}^{\star n}\\
&  =\dfrac{1}{n!}m^{n}n!\delta_{n,n}\mathfrak{e}^{\star n}+\underbrace{\sum\limits
_{\substack{\ell\in \NN ;\\\ell\neq n}}\dfrac{1}{\ell!}m^{\ell
}n!0\mathfrak{e}^{\star n}}_{=0}=\dfrac{1}{n!}m^{n}n!\underbrace{\delta_{n,n}%
}_{=1}\mathfrak{e}^{\star n}=\dfrac{1}{n!}m^{n}n!\mathfrak{e}^{\star n}%
=m^{n}\mathfrak{e}^{\star n}%
\end{align*}
and%
\begin{align*}
\underbrace{\id_{A}^{\star m}}_{=\sum_{\ell=0}%
^{\infty}\dfrac{1}{\ell!}m^{\ell}\mathfrak{e}^{\star\ell}}\circ\mathfrak{e}%
^{\star n} &  =\left(  \sum_{\ell=0}^{\infty}\dfrac{1}{\ell!}m^{\ell
}\mathfrak{e}^{\star\ell}\right)  \circ\mathfrak{e}^{\star n}\\
&  =\underbrace{\sum_{\ell=0}^{\infty}}_{=\sum_{\ell\in \NN }}\dfrac
{1}{\ell!}m^{\ell}\underbrace{\mathfrak{e}^{\star\ell}\circ\mathfrak{e}^{\star
n}}_{\substack{=\ell!\delta_{\ell,n}\mathfrak{e}^{\star\ell}\\\text{(by
Exercise \ref{exe.eulerian-idp.2}(a),}\\\text{applied to }\ell\text{ instead
of }m\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
} \kk \text{-linear maps is } \kk \text{-bilinear}\right)  \\
&  =\sum_{\ell\in \NN }\dfrac{1}{\ell!}m^{\ell}\ell!\delta_{\ell
,n}\mathfrak{e}^{\star\ell}=\underbrace{\sum\limits_{\substack{\ell\in \NN %
;\\\ell=n}}\dfrac{1}{\ell!}m^{\ell}\ell!\delta_{\ell,n}\mathfrak{e}^{\star
\ell}}_{\substack{=\dfrac{1}{n!}m^{n}n!\delta_{n,n}\mathfrak{e}^{\star
n}\\\text{(since }n\in \NN \text{)}}}+\sum\limits_{\substack{\ell\in
 \NN ;\\\ell\neq n}}\dfrac{1}{\ell!}m^{\ell}\ell!\underbrace{\delta
_{\ell,n}}_{\substack{=0\\\text{(since }\ell\neq n\text{)}}}\mathfrak{e}%
^{\star\ell}\\
&  =\dfrac{1}{n!}m^{n}n!\delta_{n,n}\mathfrak{e}^{\star n}+\underbrace{\sum\limits
_{\substack{\ell\in \NN ;\\\ell\neq n}}\dfrac{1}{\ell!}m^{\ell}%
\ell!0\mathfrak{e}^{\star\ell}}_{=0}=\dfrac{1}{n!}m^{n}n!\underbrace{\delta
_{n,n}}_{=1}\mathfrak{e}^{\star n}=\dfrac{1}{n!}m^{n}n!\mathfrak{e}^{\star
n}=m^{n}\mathfrak{e}^{\star n}.
\end{align*}
This solves Exercise~\ref{exe.eulerian-idp.2}(b).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.Psi}}

\textit{Solution to Exercise~\ref{exe.Psi}.}
(a) This is proven by induction over
$i$. The induction step relies on the observation
that $R_{\left(  1^{i},n-i\right)}+R_{\left(  1^{i-1},n-i+1\right)  }
=R_{\left(  1^{i}\right)  }H_{n-i}$
(where $R_{\left(1^{-1},n+1\right)}$ is to be understood to mean $0$
in the $i = 0$ case). Let us prove this observation. We assume
WLOG that $i > 0$ (the proof in the $i = 0$ case is analogous
but simpler).
The equality
\eqref{H-as-sum-of-ribbons} yields $H_{n-i}=R_{\left(  n-i\right)  }$
(since only $\left( n-i\right)  $ coarsens $\left(  n-i\right)  $), so that
\begin{align*}
R_{\left(  1^{i}\right)  }H_{n-i}  &  =R_{\left(  1^{i}\right)  }R_{\left(
n-i\right)  }=R_{\left(  1^{i}\right)  \cdot\left(  n-i\right)  }+R_{\left(
1^{i}\right)  \odot\left(  n-i\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{Nsym-product-in-ribbons}}\right) \\
&  =R_{\left(  1^{i},n-i\right)  }+R_{\left(  1^{i-1},n-i+1\right)  },
\end{align*}
qed.

(b) Part (a) yields
\begin{align*}
\left(  -1\right)  ^{i}R_{\left(  1^{i},n-i\right)  }  &  =\left(  -1\right)
^{i}\sum_{j=0}^{i}\left(  -1\right)  ^{i-j}R_{\left(  1^{j}\right)  }
H_{n-j}=\sum_{j=0}^{i}\underbrace{\left(  -1\right)  ^{j}R_{\left(
1^{j}\right)  }}_{\substack{=S\left(  R_{\left(  j\right)  }\right)
\\\text{(by \eqref{Nsym-antipode-in-ribbons}, since }\omega\left(  \left(  j\right)  \right)  =\left(
1^{j}\right)  \text{)}}}H_{n-j}\\
&  =\sum_{j=0}^{i}S\left(  \underbrace{R_{\left(  j\right)  }}
_{\substack{=H_{j}\\\text{(by \eqref{H-as-sum-of-ribbons})}}}\right)  H_{n-j}=\sum_{j=0}
^{i}S\left(  H_{j}\right)  H_{n-j}.
\end{align*}
This proves (b).

(c) We have
\begin{align*}
\Psi_{n}  &  = \sum_{i=0}^{n-1}\underbrace{\left(  -1\right)  ^{i}R_{\left(
1^{i},n-i\right)  }}_{\substack{=\sum_{j=0}^{i}S\left(  H_{j}\right)
H_{n-j}\\\text{(by part (b))}}}
= \sum_{i=0}^{n-1}\sum_{j=0}^{i}S\left(  H_{j}\right)  H_{n-j}
= \sum_{j=0}^{n}\sum_{i=j}^{n-1}S\left(  H_{j}\right)  H_{n-j} \\
&= \sum_{j=0}^{n}\left(  n-j\right)  S\left( H_{j}\right)  H_{n-j}
 = \sum_{j=0}^{n} S\left( H_{j}\right) \underbrace{\left(  n-j\right)  H_{n-j}}_{= \deg\left(H_{n-j}\right) H_{n-j} = E\left(H_{n-j}\right)} \\
&= \sum_{j=0}^{n} S\left( H_{j}\right) E\left(H_{n-j}\right)
 = \sum_{\left(  H_{n}\right)  } S\left(  \left(  H_{n}\right)  _{1}\right)
 E \left( \left( H_{n}\right)  _{2} \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{using Sweedler notation, since }
\Delta\left(  H_{n}\right)  =\sum_{j=0}^{n}H_{j}\otimes H_{n-j}\right) \\
&  =\left(  S\star E\right)  \left(  H_{n}\right)  ,
\end{align*}
and thus $\Psi_{n}$ is primitive (by Exercise~\ref{exe.dynkin} (a)).
(That said, there are other ways to prove the primitivity of $\Psi_{n}$.)

(d) Let $n\in\NN$. Then,
\begin{align*}
\sum_{k=0}^{n-1}H_{k}\underbrace{\Psi_{n-k}}_{\substack{=\left(  S\star
E\right)  \left(  H_{n-k}\right)  \\\text{(by part (c))}}}  &
=\sum_{k=0}^{n-1}H_{k}\left(  S\star E\right)  \left(  H_{n-k}\right)
=\sum_{k=0}^{n}H_{k}\left(  S\star E\right)  \left(  H_{n-k}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{we added a }k=n\text{ term, which does
not matter since it vanishes}\right) \\
&  =\sum_{\left(  H_{n}\right)  }\left(  H_{n}\right)  _{1}\left(  S\star
E\right)  \left(  \left(  H_{n}\right)  _{2}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{using Sweedler notation, since }
\Delta\left(  H_{n}\right)  =\sum_{k=0}^{n}H_{k}\otimes H_{n-k}\right) \\
&  =\left(  \underbrace{\id \star S}_{=u\epsilon}\star E\right)
\left(  H_{n}\right)  =\underbrace{\left(  u\epsilon\star E\right)  }
_{=E}\left(  H_{n}\right)  =E\left(  H_{n}\right)  =nH_{n},
\end{align*}
qed.

(e) Comparing coefficients reduces this to part (d).

(f) The ring homomorphism $\pi:\Nsym\rightarrow\Lambda$
induces a ring homomorphism $\Nsym\left[  \left[  t\right]
\right]  \rightarrow\Lambda\left[  \left[  t\right]  \right]  $. Applying this
latter homomorphism to the equality $\dfrac{d}{dt}\widetilde{H}\left(
t\right)  =\widetilde{H}\left(  t\right)  \cdot\psi\left(  t\right)  $ of part
(e), we obtain $\dfrac{d}{dt}H\left(  t\right)  =H\left(  t\right)
\cdot\overline{\psi}\left(  t\right)  $, where $H\left(  t\right)  $ is
defined as in \eqref{h-generating-function}, whereas
$\overline{\psi}\left(  t\right)  \in
\Lambda\left[  \left[  t\right]  \right]  $ is defined by
\[
\overline{\psi}\left(  t\right)  =\sum_{n\geq1}\pi\left(  \Psi_n\right)
t^{n-1}.
\]
Hence,
\begin{align*}
\overline{\psi}\left(  t\right)   &  =\dfrac{\dfrac{d}{dt}H\left(  t\right)
}{H\left(  t\right)  }=\dfrac{H^{\prime}\left(  t\right)  }{H\left(  t\right)
}=\sum_{m\geq0}p_{m+1}t^{m}\ \ \ \ \ \ \ \ \ \ 
\left(  \text{by \eqref{E(t)-H(t)-P(t)-relation}}\right)
\\
&  =\sum_{n\geq1}p_{n}t^{n-1}.
\end{align*}
Comparing coefficients in this equality yields $\pi\left(
\Psi_{n}\right)  =p_{n}$ for every positive integer $n$. This solves part
(f).

\textit{Remark:} Another way to solve Exercise~\ref{exe.Psi}(f)
proceeds as follows: We know that $\Psi_n$ is a primitive
homogeneous element of $\Nsym$ of degree $n$ (by
Exercise~\ref{exe.Psi}(c)). Thus, $\pi\left(\Psi_n\right)$ is a
primitive homogeneous element of $\Lambda$ of degree $n$ (since
$\pi$ is a graded homomorphism of Hopf algebras). But
Exercise~\ref{exe.Lambda.primitives} shows that all such elements
are scalar multiples of $p_n$. Thus, $\pi\left(\Psi_n\right)$ is a
scalar multiple of $p_n$. Finding the scalar is easy (e.g., it can
be obtained by specializing at $\left(1\right)$).

(g) Let $n$ be a positive integer. We know that
$\Psi_{n} = \sum\limits_{i=0}^{n-1}\left(  -1\right)  ^{i}R_{\left(  1^{i},n-i\right)}$.
Applying $\pi$ to this equation, we obtain
$\pi\left(\Psi_{n}\right) = \sum\limits_{i=0}^{n-1}\left(  -1\right)  ^{i} \pi\left(R_{\left(  1^{i},n-i\right)}\right)$.
Since $\pi\left(\Psi_{n}\right) = p_n$ (by part (f)), this rewrites as
$p_n = \sum\limits_{i=0}^{n-1}\left(  -1\right)  ^{i} \pi\left(R_{\left(  1^{i},n-i\right)}\right)$.
But we need to show that
$p_n = \sum\limits_{i=0}^{n-1} \left(-1\right)^i s_{\left(n-i, 1^i\right)}$.
Hence, it is enough to prove that every
$i \in \left\{0, 1, \ldots, n-1\right\}$
satisfies $\pi\left(R_{\left(  1^{i},n-i\right)}\right) = s_{\left(n-i, 1^i\right)}$.

So let $i \in \left\{0, 1, \ldots, n-1\right\}$ be arbitrary.
Theorem~\ref{Nsym-structure-on-ribbons-theorem}(b) (applied to
$\alpha = \left(  1^{i},n-i\right)$) shows that
$\pi\left(R_{\left(  1^{i},n-i\right)}\right) = s_\alpha$, where
$\alpha$ is the ribbon diagram of the composition
$\left(  1^{i},n-i\right)$. But since the ribbon diagram of
the composition $\left(  1^{i},n-i\right)$ is the Ferrers diagram
for the partition $\left(n-i, 1^i\right)$ (because its row lengths
are $\underbrace{1, 1, \ldots, 1}_{i\text{ times}}, n-i$ going
from bottom to top, with an overlap of $1$ between every two
adjacent rows), we have $s_\alpha = s_{\left(n-i, 1^i\right)}$.
Hence,
$\pi\left(R_{\left(  1^{i},n-i\right)}\right) = s_\alpha
 = s_{\left(n-i, 1^i\right)}$.
As we have seen, this completes the solution of part (g).

(h) We will prove \eqref{eq.exe.Psi.h.0} by strong induction over $n$. So let
$N$ be an arbitrary positive integer, and let us assume that
\eqref{eq.exe.Psi.h.0} has been proven for every $n<N$. We now need to prove
\eqref{eq.exe.Psi.h.0} for $n=N$.

We notice that the family $\left(  H_{\alpha}\right)  _{\alpha\in
\Nsym}$ is multiplicative, in the sense that any two
compositions $\beta$ and $\gamma$ satisfy $H_{\beta\cdot\gamma}=H_{\beta}\cdot
H_{\gamma}$ (where, as we recall, $\beta\cdot\gamma$ denotes the concatenation
of the compositions $\beta$ and $\gamma$). This follows from the definition of
the $H_{\alpha}$.

We have $N>0$. Hence, every $\alpha\in\Comp_N$ can
be written uniquely in the form $\alpha=\left(  q\right)  \cdot\beta$ for some
$q\in\left\{  1,2,...,N\right\}  $ and some
$\beta\in\Comp_{N-q}$ (indeed, the $q$ is just the first entry of the
composition
$\alpha$, and $\beta$ is the composition obtained by erasing this first
entry). Hence,
\begin{align}
& \sum_{\alpha\in\Comp_{N}}\left(  -1\right)  ^{\ell\left(
\alpha\right)  -1}\operatorname{lp}\left(  \alpha\right)  H_{\alpha
}\nonumber\\
& =\underbrace{\sum\limits_{\substack{q\in\left\{  1,2,...,N\right\}  ;\\\beta
\in\Comp_{N-q}}}}_{=\sum_{q=1}^{N}\sum_{\beta
\in\Comp_{N-q}}}\underbrace{\left(  -1\right)
^{\ell\left(  \left(  q\right)  \cdot\beta\right)  -1}}_{\substack{=\left(
-1\right)  ^{\ell\left(  \beta\right)  }\\\text{(since it is easy to see
that}\\\ell\left(  \left(  q\right)  \cdot\beta\right)  -1=\ell\left(
\beta\right)  \text{)}}}\operatorname{lp}\left(  \left(  q\right)  \cdot
\beta\right)  \underbrace{H_{\left(  q\right)  \cdot\beta}}
_{\substack{=H_{\left(  q\right)  }\cdot H_{\beta}\\\text{(since the
family}\\\left(  H_{\alpha}\right)  _{\alpha\in\Nsym}\text{
is}\\\text{multiplicative)}}}\nonumber\\
& =\sum_{q=1}^{N}\sum_{\beta\in\Comp_{N-q}}\left(
-1\right)  ^{\ell\left(  \beta\right)  }\operatorname{lp}\left(  \left(
q\right)  \cdot\beta\right)  \cdot H_{\left(  q\right)  }\cdot H_{\beta
}\nonumber\\
& =\sum_{q=1}^{N-1}\sum_{\beta\in\Comp_{N-q}}\left(
-1\right)  ^{\ell\left(  \beta\right)  }\underbrace{\operatorname{lp}\left(
\left(  q\right)  \cdot\beta\right)  }_{\substack{=\operatorname{lp}\left(
\beta\right)  \\\text{(since }\beta\text{ is}\\\text{nonempty)}}}\cdot
H_{\left(  q\right)  }\cdot H_{\beta}+\underbrace{\sum_{\beta\in
\Comp_{N-N}}\left(  -1\right)  ^{\ell\left(
\beta\right)  }\operatorname{lp}\left(  \left(  N\right)  \cdot\beta\right)
\cdot H_{\left(  N\right)  }\cdot H_{\beta}}_{\substack{=\left(  -1\right)
^{\ell\left(  \varnothing\right)  }\operatorname{lp}\left(  \left(  N\right)
\cdot\varnothing\right)  \cdot H_{\left(  N\right)  }\cdot H_{\varnothing
}\\\text{(since the only element of }\Comp_{N-N}
\\\text{is the empty composition }\varnothing\text{)}}}\nonumber\\
& =\sum_{q=1}^{N-1}\underbrace{\sum_{\beta\in\Comp_{N-q}}
\left(  -1\right)  ^{\ell\left(  \beta\right)  }
\operatorname{lp}\left(  \beta\right)  \cdot H_{\left(  q\right)  }\cdot
H_{\beta}}_{=H_{\left(  q\right)  }\cdot\left(  \sum_{\beta\in
\Comp_{N-q}}\left(  -1\right)  ^{\ell\left(
\beta\right)  }\operatorname{lp}\left(  \beta\right)  H_{\beta}\right)
}+\underbrace{\left(  -1\right)  ^{\ell\left(  \varnothing\right)  }}
_{=1}\underbrace{\operatorname{lp}\left(  \left(  N\right)  \cdot
\varnothing\right)  }_{=\operatorname{lp}\left(\left(N\right)\right) = N}
\cdot\underbrace{H_{\left(  N\right)  }}_{=H_{N}}
\cdot\underbrace{H_{\varnothing}}_{=1}\nonumber\\
& =\sum_{q=1}^{N-1}H_{\left(  q\right)  }\cdot\left(  \sum_{\beta
\in\Comp_{N-q}}\left(  -1\right)  ^{\ell\left(
\beta\right)} \operatorname{lp}\left( \beta\right) H_{\beta}\right)
+ NH_N .
\label{sol.Psi.h.0.1}
\end{align}
However, for every $q\in\left\{  1,2,...,N-1\right\}  $, we can apply
\eqref{eq.exe.Psi.h.0} to $n=N-q$ (because $N-q<N$, and because
\eqref{eq.exe.Psi.h.0} has been proven for every $n<N$). Thus, for every
$q\in\left\{  1,2,...,N-1\right\}  $, we obtain
\[
\Psi_{N-q}=\sum_{\alpha\in\Comp_{N-q}}
\underbrace{\left(  -1\right)  ^{\ell\left(  \alpha\right)  -1}}_{=-\left(
-1\right)  ^{\ell\left(  \alpha\right)  }}\operatorname{lp}\left(
\alpha\right)  H_{\alpha}=-\sum_{\alpha\in\Comp_{N-q}}
\left(  -1\right)  ^{\ell\left(  \alpha\right)  }\operatorname{lp}\left(
\alpha\right)  H_{\alpha}=-\sum_{\beta\in\Comp_{N-q}}
\left(  -1\right)  ^{\ell\left(  \beta\right)  }\operatorname{lp}\left(
\beta\right)  H_{\beta}
\]
(here, we renamed the summation index $\alpha$ as $\beta$), so that
\begin{equation}
-\Psi_{N-q}=\sum_{\beta\in\Comp_{N-q}}\left(
-1\right)  ^{\ell\left(  \beta\right)  }\operatorname{lp}\left(
\beta\right)  H_{\beta}.\label{sol.Psi.h.0.3}
\end{equation}
Thus, \eqref{sol.Psi.h.0.1} becomes
\begin{align}
\sum_{\alpha\in\Comp_{N}}\left(  -1\right)  ^{\ell\left(
\alpha\right)  -1}\operatorname{lp}\left(  \alpha\right)  H_{\alpha}  &
=\sum_{q=1}^{N-1}\underbrace{H_{\left(  q\right)  }}_{=H_{q}}\cdot
\underbrace{\left(  \sum_{\beta\in\Comp_{N-q}}\left(
-1\right)  ^{\ell\left(  \beta\right)  }\operatorname{lp}\left(
\beta\right)  H_{\beta}\right)  }_{\substack{=-\Psi_{N-q}\\\text{(this follows
from \eqref{sol.Psi.h.0.3})}}}+NH_{N}\nonumber\\
& =\sum_{q=1}^{N-1}H_{q}\cdot\left(  -\Psi_{N-q}\right)  +NH_{N}=-\sum
_{q=1}^{N-1}H_{q}\Psi_{N-q}+NH_{N}.\label{sol.Psi.h.0.4}
\end{align}
However, Exercise~\ref{exe.Psi}(d) (applied to $n=N$) yields
$\sum_{k=0}^{N-1}H_{k}\Psi_{N-k}=NH_{N}$, so that
\[
NH_{N}
= \sum_{k=0}^{N-1}H_{k}\Psi_{N-k}
= \sum_{q=0}^{N-1}H_{q}\Psi_{N-q}
= \underbrace{H_0}_{=1} \underbrace{\Psi_{N-0}}_{=\Psi_N}
 + \sum_{q=1}^{N-1}H_{q}\Psi_{N-q}
= \Psi_N + \sum_{q=1}^{N-1}H_{q}\Psi_{N-q}.
\]
Now, \eqref{sol.Psi.h.0.4} becomes
\begin{align*}
& \sum_{\alpha\in\Comp_{N}}\left(  -1\right)  ^{\ell\left(
\alpha\right)  -1}\operatorname{lp}\left(  \alpha\right)  H_{\alpha}\\
& =-\sum_{q=1}^{N-1}H_{q}\Psi_{N-q}+\underbrace{NH_{N}}
_{=\Psi_N + \sum_{q=1}^{N-1}H_{q}\Psi_{N-q}}
=-\sum_{q=1}^{N-1}H_{q}\Psi_{N-q} + \Psi_N + \sum_{q=1}^{N-1}H_{q}\Psi_{N-q}
=\Psi_{N}.
\end{align*}


In other words, $\Psi_{N}=\sum_{\alpha\in\Comp_{N}}\left(
-1\right)  ^{\ell\left(  \alpha\right)  -1}\operatorname{lp}\left(
\alpha\right)  H_{\alpha}$. Thus, \eqref{eq.exe.Psi.h.0} is proven for $n=N$.
This completes the induction step, and therefore the proof of
\eqref{eq.exe.Psi.h.0} is complete. That is, part (h) of the exercise is solved.

(i) First of all, for every positive integer $n$, the element $\Psi_{n}$ of
$\Nsym$ is homogeneous of degree $n$ (this follows from the
definition of $\Psi_{n}$ or, alternatively, from Exercise~\ref{exe.Psi}(h)).
Hence, for every composition $\alpha$, the element $\Psi_{\alpha}$ of
$\Nsym$ is homogeneous of degree $\left\vert \alpha\right\vert
$.

We notice that the family $\left(  \Psi_{\alpha}\right)  _{\alpha
\in\Nsym }$ is multiplicative, in the sense that any two
compositions $\beta$ and $\gamma$ satisfy $\Psi_{\beta\cdot\gamma}=\Psi
_{\beta}\cdot\Psi_{\gamma}$ (where, as we recall, $\beta\cdot\gamma$ denotes
the concatenation of the compositions $\beta$ and $\gamma$). This follows from
the definition of the $\Psi_{\alpha}$.

Let us now prove \eqref{eq.exe.Psi.h.1}. Indeed, we will show
\eqref{eq.exe.Psi.h.1} by strong induction over $n$. So let $N\in\NN$
be arbitrary, and let us assume that \eqref{eq.exe.Psi.h.1} has been proven
for every $n<N$. We now need to prove \eqref{eq.exe.Psi.h.1} for $n=N$.

If $N=0$, then \eqref{eq.exe.Psi.h.1} obviously holds for $n=N$ (because both
sides of \eqref{eq.exe.Psi.h.1} are $1_{\Nsym }$ in this case).
We thus WLOG assume that $N\neq0$. Thus, every $\alpha\in\Comp_{N}$ can be
written uniquely in the form $\alpha=\beta\cdot\left(
q\right)  $ for some $q\in\left\{  1,2,...,N\right\}  $ and some $\beta
\in\Comp_{N-q}$ (indeed, the $q$ is just the last
entry of the composition $\alpha$, and $\beta$ is the composition obtained by
erasing this last entry). Hence,
\begin{align}
\sum_{\alpha\in\Comp_{N}}\frac{1}{\pi_{u}\left(  \alpha\right)
}\Psi_{\alpha}  & =\underbrace{\sum\limits_{\substack{q\in\left\{  1,2,...,N\right\}
;\\\beta\in\Comp_{N-q}}}}_{=\sum_{q=1}^{N}\sum
_{\beta\in\Comp_{N-q}}}\underbrace{\frac{1}{\pi
_{u}\left(  \beta\cdot\left(  q\right)  \right)  }}
_{\substack{=\dfrac{1}{\pi_{u}\left(  \beta\right)  \cdot N}\\
\text{(since it is easy to see
that}\\\pi_{u}\left(  \beta\cdot\left(  q\right)  \right)  =\pi_{u}\left(
\beta\right)  \cdot N\text{)}}}\underbrace{\Psi_{\beta\cdot\left(  q\right)
}}_{\substack{=\Psi_{\beta}\cdot\Psi_{\left(  q\right)  }\\\text{(since the
family}\\\left(  \Psi_{\alpha}\right)  _{\alpha\in\Nsym }\text{
is}\\\text{multiplicative)}}}\nonumber\\
& =\sum_{q=1}^{N}\sum_{\beta\in\Comp_{N-q}}
\underbrace{\dfrac{1}{\pi_{u}\left(  \beta\right)  \cdot N}}
_{=\dfrac{1}{N}\cdot\dfrac{1}{\pi_{u}\left(  \beta\right)  }}
\Psi_{\beta}\cdot
\underbrace{\Psi_{\left(  q\right)  }}_{=\Psi_{q}}\nonumber\\
& =\sum_{q=1}^{N}\sum_{\beta\in\Comp_{N-q}}\dfrac
{1}{N}\cdot\dfrac{1}{\pi_{u}\left(  \beta\right)  }\Psi_{\beta}\cdot\Psi
_{q}\nonumber\\
& =\dfrac{1}{N}\sum_{q=1}^{N}\left(  \sum_{\beta\in\Comp_{N-q}}
\dfrac{1}{\pi_{u}\left(  \beta\right)  }\Psi_{\beta}\right)
\cdot\Psi_{q}.\label{sol.Psi.h.1.1}
\end{align}
However, for every $q\in\left\{  1,2,...,N\right\}  $, we can apply
\eqref{eq.exe.Psi.h.1} to $n=N-q$ (because $N-q<N$, and because
\eqref{eq.exe.Psi.h.1} has been proven for every $n<N$). Thus, for every
$q\in\left\{  1,2,...,N\right\}  $, we obtain
\begin{equation}
H_{N-q}=\sum_{\alpha\in\Comp_{N-q}}\frac{1}{\pi_{u}\left(
\alpha\right)  }\Psi_{\alpha}=\sum_{\beta\in\Comp_{N-q}}
\dfrac{1}{\pi_{u}\left(  \beta\right)  }\Psi_{\beta
}\label{sol.Psi.h.1.2}
\end{equation}
(here, we renamed the summation index $\alpha$ as $\beta$). Thus,
\eqref{sol.Psi.h.1.1} becomes
\begin{align*}
\sum_{\alpha\in\Comp_{N}}\frac{1}{\pi_{u}\left(  \alpha\right)
}\Psi_{\alpha}  & =\dfrac{1}{N}\sum_{q=1}^{N}\underbrace{\left(  \sum
_{\beta\in\Comp_{N-q}}\dfrac{1}{\pi_{u}\left(
\beta\right)  }\Psi_{\beta}\right)  }_{\substack{=H_{N-q}\\\text{(by
\eqref{sol.Psi.h.1.2})}}}\cdot\Psi_{q}\\
& =\dfrac{1}{N}\sum_{q=1}^{N}H_{N-q}\Psi_{q}=\dfrac{1}{N}\underbrace{\sum
_{k=0}^{N-1}H_{k}\Psi_{N-k}}_{\substack{=NH_{N}\\
\text{(by Exercise}~\text{\ref{exe.Psi}(d),}\\
\text{applied to }n=N\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }N-k\text{ for
}q\text{ in the sum}\right)  \\
& =\dfrac{1}{N}NH_{N}=H_{N}.
\end{align*}
In other words, $H_{N}=\sum_{\alpha\in\Comp_{N}}\frac{1}{\pi
_{u}\left(  \alpha\right)  }\Psi_{\alpha}$. Thus, \eqref{eq.exe.Psi.h.1} is
proven for $n=N$. This completes the induction step, and therefore the proof
of \eqref{eq.exe.Psi.h.1} is complete.

We now need to prove that $\left(  \Psi_{\alpha}\right)  _{\alpha
\in\Comp_{n}}$ is a $\kk$-basis of $\Nsym_{n}$ for
every $n\in{\NN }$. It is clear that $\left(
\Psi_{\alpha}\right)  _{\alpha\in\Comp_{n}}$ is a
family of elements of $\Nsym_{n}$ (because for every
composition $\alpha$, the element $\Psi_{\alpha}$ of $\Nsym$ is
homogeneous of degree $\left\vert \alpha\right\vert $).

Let $\mathfrak{A}$ be the $\kk$-subalgebra of $\Nsym$
generated by the elements $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, $...$. Then,
$\mathfrak{A}$ contains $\Psi_{\alpha}$ for every composition $\alpha$ (by the
definition of $\Psi_{\alpha}$). Therefore, $\mathfrak{A}$ contains $H_{n}$ for
every $n\geq1$ (by \eqref{eq.exe.Psi.h.1}). Consequently, $\mathfrak{A}
=\Nsym$ (because $\Nsym$ is generated as a
$\kk$-algebra by $H_{1}$, $H_{2}$, $H_{3}$, $...$). In other words, the
$\kk$-algebra $\Nsym$ is generated by the elements
$\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, $...$ (since we defined $\mathfrak{A}$ as
the $\kk$-subalgebra of $\Nsym$ generated by the
elements $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, $...$). In other words, the
$\kk$-module $\Nsym$ is spanned by all possible products
of the elements $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, $...$. In other words, the
$\kk$-module $\Nsym$ is spanned by the elements
$\Psi_{\alpha}$ with $\alpha\in\Comp $ (because the elements
$\Psi_{\alpha}$ with $\alpha\in\Comp $ are precisely all
possible products of the elements $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, $...$).
In yet other words, the family $\left(  \Psi_{\alpha}\right)  _{\alpha
\in\Comp }$ spans the $\kk$-module $\Nsym$.

Now, fix $n\in\NN$. Every element of $\Nsym_{n}$ can be written
as a $\kk$-linear combination of the
elements $\Psi_{\alpha}$ with $\alpha\in\Comp $ (since the
family $\left(  \Psi_{\alpha}\right)  _{\alpha\in\Comp }$ spans
the $\kk$-module $\Nsym$). In this $\kk$-linear
combination, we can remove all terms $\Psi_{\alpha}$ with $\alpha\notin
\Comp_{n}$ without changing its value (by
gradedness, because $\Psi_{\alpha}$ is homogeneous of degree $\left\vert
\alpha\right\vert$), and so we conclude that every element of
$\Nsym_{n}$ can be written as a $\kk$-linear
combination of the elements $\Psi_{\alpha}$ with $\alpha\in
\Comp_{n}$. In other words,
the family $\left(  \Psi_{\alpha}\right)  _{\alpha\in\Comp_{n}}$ spans
the $\kk$-module $\Nsym_{n}$.

Now, we can apply Exercise~\ref{exe.linalg.sur-bij}(b) to $A=\Nsym_{n}$,
$I=\Comp_{n}$, $\left(  \gamma_{i}\right)  _{i\in
I}=\left(  H_{\alpha}\right)  _{\alpha\in\Comp_{n}}$
and $\left(  \beta_{i}\right)  _{i\in I}=\left(  \Psi_{\alpha}\right)
_{\alpha\in\Comp_{n}}$ (since we know that $\left(  H_{\alpha
}\right)  _{\alpha\in\Comp_{n}}$ is a $\kk$-basis
of $\Nsym_{n}$, whereas $\left(  \Psi_{\alpha
}\right)  _{\alpha\in\Comp_{n}}$ spans the $\kk$-module
$\Nsym_{n}$). We conclude that $\left(  \Psi_{\alpha
}\right)  _{\alpha\in\Comp_{n}}$ is a $\kk$-basis of
$\Nsym_{n}$. This completes the solution of part (i).

(j) For every composition $\alpha$, define an element $\mathfrak{b}_{\alpha}$
of $T\left(  V\right)  $ by $\mathfrak{b}_{\alpha}=\mathfrak{b}_{\alpha_{1}%
}\mathfrak{b}_{\alpha_{2}}\cdots\mathfrak{b}_{\alpha_{\ell}}$, where $\alpha$
is written in the form $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right)  $ with $\ell=\ell\left(  \alpha\right)  $. Then,
$\left(  \mathfrak{b}_{\alpha}\right)  _{\alpha\in \Comp }$ is a
$\kk$-module basis of $T\left(  V\right)  $ (by the basic properties of
tensor algebras, since $\left(  \mathfrak{b}_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }$ is a $\kk$-module basis of $V$). Notice that
the family $\left(  \mathfrak{b}_{n}\right)  _{n\in\left\{  1,2,3,\ldots
\right\}  }$ generates the $\kk$-algebra $T\left(  V\right)  $ (since
it is a basis of $V$).

For every composition $\alpha$, define an element $\Psi_{\alpha}$ of
$\operatorname*{NSym}$ as in Exercise \ref{exe.Psi}(i). We know from Exercise
\ref{exe.Psi}(i) that $\left(  \Psi_{\alpha}\right)  _{\alpha\in
 \Comp_{n}}$ is a $\kk$-basis of $\Nsym_{n}$
for every $n\in{ \NN }$. Hence, $\left(  \Psi_{\alpha}\right)
_{\alpha\in\operatorname{Comp}}$ is a $\kk$-basis of
$\Nsym$.

Every $\alpha\in \Comp $ satisfies $F\left(  \mathfrak{b}%
_{\alpha}\right)  =\Psi_{\alpha}$\ \ \ \ \footnote{\textit{Proof.} Let
$\alpha\in \Comp $. Write $\alpha$ in the form $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ with $\ell=\ell\left(
\alpha\right)  $. Then, $\mathfrak{b}_{\alpha}=\mathfrak{b}_{\alpha_{1}%
}\mathfrak{b}_{\alpha_{2}}\cdots\mathfrak{b}_{\alpha_{\ell}}$, so that%
\begin{align*}
F\left(  \mathfrak{b}_{\alpha}\right)    & =F\left(  \mathfrak{b}_{\alpha_{1}%
}\mathfrak{b}_{\alpha_{2}}\cdots\mathfrak{b}_{\alpha_{\ell}}\right)
=\underbrace{f\left(  \mathfrak{b}_{\alpha_{1}}\right)  }_{\substack{=\Psi
_{\alpha_{1}}\\\text{(by the definition of }f\text{)}}}\underbrace{f\left(
\mathfrak{b}_{\alpha_{2}}\right)  }_{\substack{=\Psi_{\alpha_{2}}\\\text{(by
the definition of }f\text{)}}}\cdots\underbrace{f\left(  \mathfrak{b}%
_{\alpha_{\ell}}\right)  }_{\substack{=\Psi_{\alpha_{\ell}}\\\text{(by the
definition of }f\text{)}}}\\
& \qquad \qquad \left(\text{by the definition of } F\right) \\
& =\Psi_{\alpha_{1}}\Psi_{\alpha_{2}}\cdots\Psi_{\alpha_{\ell}}=\Psi_{\alpha
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Psi_{\alpha}\text{ was defined to
be }\Psi_{\alpha_{1}}\Psi_{\alpha_{2}}\cdots\Psi_{\alpha_{\ell}}\right)  ,
\end{align*}
qed.}. The map $F$ thus maps the basis $\left(  \mathfrak{b}_{\alpha}\right)
_{\alpha\in \Comp }$ of the $\kk$-module $T\left(
V\right)  $ to the basis $\left(  \Psi_{\alpha}\right)  _{\alpha
\in\operatorname{Comp}}$ of the $\kk$-module $\operatorname*{NSym}$.
Hence, $F$ is a $\kk$-module isomorphism (since any $\kk$-linear
map mapping a basis to a basis is a $\kk$-module isomorphism).

Next, we are going to show the equality $\Delta_{\operatorname*{NSym}}\circ
F=\left(  F\otimes F\right)  \circ\Delta_{T\left(  V\right)  }$. Indeed, this
is an equality between $\kk$-algebra homomorphisms, and thus needs only
to be verified on a generating set of the $\kk$-algebra $T\left(
V\right)  $. Picking $\left(  \mathfrak{b}_{n}\right)  _{n\in\left\{
1,2,3,\ldots\right\}  }$ as this generating set, we thus only need to check
that $\left(  \Delta_{\operatorname*{NSym}}\circ F\right)  \left(
\mathfrak{b}_{n}\right)  =\left(  \left(  F\otimes F\right)  \circ
\Delta_{T\left(  V\right)  }\right)  \left(  \mathfrak{b}_{n}\right)  $ for
every positive integer $n$. This is straightforward: If $n$ is any positive
integer, then comparing the equalities%
\begin{align*}
\left(  \Delta_{\operatorname*{NSym}}\circ F\right)  \left(  \mathfrak{b}%
_{n}\right)    & =\Delta_{\operatorname*{NSym}}\left(  \underbrace{F\left(
\mathfrak{b}_{n}\right)  }_{\substack{=f\left(  \mathfrak{b}_{n}\right)
\\\text{(since }F\text{ is induced by }f\text{)}}}\right)  =\Delta
_{\operatorname*{NSym}}\left(  \underbrace{f\left(  \mathfrak{b}_{n}\right)
}_{\substack{=\Psi_{n}\\\text{(by the definition of }f\text{)}}}\right)
=\Delta_{\operatorname*{NSym}}\left(  \Psi_{n}\right)  \\
& =1\otimes\Psi_{n}+\Psi_{n}\otimes1\ \ \ \ \ \ \ \ \ \ \left(  \text{since
Exercise \ref{exe.Psi}(c) shows that }\Psi_{n}\text{ is primitive}\right)
\end{align*}
and%
\begin{align*}
\left(  \left(  F\otimes F\right)  \circ\Delta_{T\left(  V\right)  }\right)
\left(  \mathfrak{b}_{n}\right)    & =\left(  F\otimes F\right)  \left(
\underbrace{\Delta_{T\left(  V\right)  }\left(  \mathfrak{b}_{n}\right)
}_{\substack{=1\otimes\mathfrak{b}_{n}+\mathfrak{b}_{n}\otimes1\\\text{(by the
definition of the}\\\text{comultiplication on }T\left(  V\right)  \text{)}%
}}\right)  =\left(  F\otimes F\right)  \left(  1\otimes\mathfrak{b}%
_{n}+\mathfrak{b}_{n}\otimes1\right)  \\
& =\underbrace{F\left(  1\right)  }_{\substack{=1\\\text{(since }F\text{ is a
} \kk \text{-algebra}\\\text{homomorphism)}}}\otimes\underbrace{F\left(
\mathfrak{b}_{n}\right)  }_{\substack{=f\left(  \mathfrak{b}_{n}\right)
\\\text{(since }F\text{ is induced by }f\text{)}}}+\underbrace{F\left(
\mathfrak{b}_{n}\right)  }_{\substack{=f\left(  \mathfrak{b}_{n}\right)
\\\text{(since }F\text{ is induced by }f\text{)}}}\otimes\underbrace{F\left(
1\right)  }_{\substack{=1\\\text{(since }F\text{ is a } \kk %
\text{-algebra}\\\text{homomorphism)}}}\\
& =1\otimes\underbrace{f\left(  \mathfrak{b}_{n}\right)  }_{\substack{=\Psi
_{n}\\\text{(by the definition of }f\text{)}}}+\underbrace{f\left(
\mathfrak{b}_{n}\right)  }_{\substack{=\Psi_{n}\\\text{(by the definition of
}f\text{)}}}\otimes1
= 1 \otimes \Psi_n + \Psi_n \otimes 1
\end{align*}
yields $\left(  \Delta_{\operatorname*{NSym}}\circ F\right)  \left(
\mathfrak{b}_{n}\right)  =\left(  \left(  F\otimes F\right)  \circ
\Delta_{T\left(  V\right)  }\right)  \left(  \mathfrak{b}_{n}\right)  $. We
thus have shown that $\Delta_{\operatorname*{NSym}}\circ F=\left(  F\otimes
F\right)  \circ\Delta_{T\left(  V\right)  }$. Combining this with the equality
$\epsilon_{\operatorname*{NSym}}\circ F=\epsilon_{T\left(  V\right)  }$ (whose
proof is similar but yet simpler), we conclude that $F$ is a $\kk %
$-coalgebra homomorphism, thus a $\kk$-bialgebra homomorphism, hence a
$\kk$-Hopf algebra homomorphism (by
Corollary~\ref{cor.bialg-mor-is-Hopf}), therefore a $\kk$-Hopf
algebra isomorphism (since it is a $\kk$-module isomorphism). This
solves Exercise \ref{exe.Psi}(j).

(k) Define the map $F$ as in Exercise \ref{exe.Psi}(j). Then, Exercise
\ref{exe.Psi}(j) shows that $F$ is a Hopf algebra isomorphism, hence a
$\kk$-module isomorphism.

We can endow the $\kk$-module $V$ with a grading by assigning to each
basis vector $\mathfrak{b}_{n}$ the degree $n$. Then, $V$ is of finite type
and satisfies $V_{0}=0$, and thus $T\left(  V\right)  $ is a connected graded
$\kk$-Hopf algebra.

The element $\Psi_{n}$ of $\operatorname*{NSym}$ is homogeneous of degree $n$
for every positive integer $n$. (This follows from the definition of $\Psi
_{n}$.)

The map $f$ is graded (since it sends every basis vector $\mathfrak{b}_{n}$ of
$V$ to the vector $\Psi_{n}\in\operatorname*{NSym}$, which is homogeneous of
the same degree $n$ as $\mathfrak{b}_{n}$). Hence, the map $F$ (being the
$\kk$-algebra homomorphism $T\left(  V\right)  \rightarrow
\operatorname*{NSym}$ induced by $f$) is also graded. It is well-known that if
a $\kk$-module isomorphism is graded, then it is an isomorphism of
graded $\kk$-modules. Thus, $F$ is an isomorphism of graded
$\kk$-modules (since $F$ is a $\kk$-module isomorphism and is
graded), hence an isomorphism of graded Hopf algebras (since $F$ is a Hopf
algebra isomorphism). Thus, $T\left(  V\right)  \cong\Nsym$ as
graded Hopf algebras. Therefore, $T\left(  V\right)  ^{o}\cong%
\Nsym^{o}\cong\Qsym$ (since
$\operatorname*{NSym}=\Qsym^{o}$) as graded Hopf
algebras. Hence, $\Qsym\cong T\left(  V\right)  ^{o}$ as graded
Hopf algebras.

Remark \ref{rmk.shuffle-alg.dual}(b) shows that the Hopf algebra $T\left(
V\right)  ^{o}$ is naturally isomorphic to the shuffle algebra
$\operatorname*{Sh}\left(  V^{o}\right)  $ as Hopf algebras. But $V^{o}\cong
V$ as $\kk$-modules (since $V$ is of finite type), and thus
$\operatorname*{Sh}\left(  V^{o}\right)  \cong\operatorname*{Sh}\left(
V\right)  $ as Hopf algebras. Altogether, we obtain $\Qsym\cong
T\left(  V\right)  ^{o}\cong\operatorname*{Sh}\left(  V^{o}\right)
\cong\operatorname*{Sh}\left(  V\right)  $ as Hopf algebras. This solves
Exercise~\ref{exe.Psi}(k).

(l) We recall that
\begin{equation}
\pi\left(  R_{\left(  1^{i},n-i\right)  }\right)  =s_{\left(  n-i,1^{i}%
\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }n\text{ and
every }i\in\left\{  0,1,...,n-1\right\}  .
\label{sol.Psi.l.piR1in-i}
\end{equation}
(This has been proven during the solution to Exercise~\ref{exe.Psi}(g).) Also,
Theorem~\ref{Nsym-structure-on-ribbons-theorem}(b) (applied to $\alpha=\left(
1^{i}\right)  $) yields that
\begin{equation}
\pi\left(  R_{\left(  1^{i}\right)  }\right)  =s_{\left(  1^{i}\right)
}=e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every } i \in \NN.
\label{sol.Psi.l.piR1}
\end{equation}


In our solution to Exercise~\ref{exe.Psi}(a), we have shown that
\begin{equation}
R_{\left(  1^{i}\right)  }H_{n-i}=R_{\left(  1^{i},n-i\right)  }+R_{\left(
1^{i-1},n-i+1\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every positive integer
}n\text{ and every }i\in\left\{  1,2,...,n-1\right\}  .
\label{sol.Psi.l.RH}
\end{equation}


Now, we are ready to solve parts (a) and (b) of Exercise~\ref{exe.hookschur} anew.

\textit{Alternative solution to Exercise \ref{exe.hookschur}(a):} Let $n$ and
$m$ be positive integers. We need to prove that $e_{n}h_{m}=s_{\left(
m+1,1^{n-1}\right)  }+s_{\left(  m,1^{n}\right)  }$.

Applying (\ref{sol.Psi.l.RH}) to $n+m$ and $n$ instead of $n$ and $i$, we
obtain $R_{\left(  1^{n}\right)  }H_{\left(  n+m\right)  -n}=R_{\left(
1^{n},\left(  n+m\right)  -n\right)  }+R_{\left(  1^{n-1},\left(  n+m\right)
-n+1\right)  }=R_{\left(  1^{n},\left(  n+m\right)  -n\right)  }+R_{\left(
1^{n-1},\left(  n+m\right)  -\left(  n-1\right)  \right)  }$ (since $\left(
n+m\right)  -n+1=\left(  n+m\right)  -\left(  n-1\right)  $). Applying the map
$\pi$ to both sides of this equality, we obtain%
\begin{align*}
\pi\left(  R_{\left(  1^{n}\right)  }H_{\left(  n+m\right)  -n}\right)   &
=\pi\left(  R_{\left(  1^{n},\left(  n+m\right)  -n\right)  }+R_{\left(
1^{n-1},\left(  n+m\right)  -\left(  n-1\right)  \right)  }\right)  \\
&  =\underbrace{\pi\left(  R_{\left(  1^{n},\left(  n+m\right)  -n\right)
}\right)  }_{\substack{=s_{\left(  \left(  n+m\right)  -n,1^{n}\right)
}\\\text{(by (\ref{sol.Psi.l.piR1in-i}), applied to}\\n+m\text{ and }n\text{
instead of }n\text{ and }i\text{)}}}+\underbrace{\pi\left(  R_{\left(
1^{n-1},\left(  n+m\right)  -\left(  n-1\right)  \right)  }\right)
}_{\substack{=s_{\left(  \left(  n+m\right)  -\left(  n-1\right)
,1^{n-1}\right)  }\\\text{(by (\ref{sol.Psi.l.piR1in-i}), applied
to}\\n+m\text{ and }n-1\text{ instead of }n\text{ and }i\text{)}}}\\
&  =s_{\left(  \left(  n+m\right)  -n,1^{n}\right)  }+s_{\left(  \left(
n+m\right)  -\left(  n-1\right)  ,1^{n-1}\right)  }\\
&  =s_{\left(  m,1^{n}\right)  }+s_{\left(  m+1,1^{n-1}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  n+m\right)  -n = m
\text{ and }\left(  n+m\right)  -\left(  n-1\right)  =m+1\right)  \\
&  =s_{\left(  m+1,1^{n-1}\right)  }+s_{\left(  m,1^{n}\right)  }.
\end{align*}
Compared with%
\begin{align*}
\pi\left(  R_{\left(  1^{n}\right)  }H_{\left(  n+m\right)  -n}\right)   &
=\underbrace{\pi\left(  R_{\left(  1^{n}\right)  }\right)  }_{\substack{=e_{n}%
\\\text{(by (\ref{sol.Psi.l.piR1}), applied to}\\n\text{ instead of }%
i\text{)}}}\cdot\pi\left(  \underbrace{H_{\left(  n+m\right)  -n}}_{=H_{m}%
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }%
 \kk \text{-algebra morphism}\right)  \\
&  =e_{n}\cdot\underbrace{\pi\left(  H_{m}\right)  }_{\substack{=h_{m}%
\\\text{(by the definition of }\pi\text{)}}}=e_{n}h_{m},
\end{align*}
this yields $e_{n}h_{m}=s_{\left(  m+1,1^{n-1}\right)  }+s_{\left(
m,1^{n}\right)  }$. Thus, Exercise \ref{exe.hookschur}(a) is solved again.

\textit{Alternative solution to Exercise \ref{exe.hookschur}(b):} Let
$a\in \NN$ and $b\in \NN$. Applying Exercise \ref{exe.Psi}(b) to
$n=a+b+1$ and $i=b$, we obtain%
\[
\left(  -1\right)  ^{b}R_{\left(  1^{b},\left(  a+b+1\right)  -b\right)
}=\sum_{j=0}^{b}S\left(  H_{j}\right)  H_{\left(  a+b+1\right)  -j}=\sum
_{j=0}^{b}S\left(  H_{b-j}\right)  H_{\left(  a+b+1\right)  -\left(
b-j\right)  }%
\]
(here, we substituted $b-j$ for $j$ in the sum). Multiplying both sides of
this equality by $\left(  -1\right)  ^{b}$, we obtain%
\begin{align*}
R_{\left(  1^{b},\left(  a+b+1\right)  -b\right)  }  &  =\left(  -1\right)
^{b}\sum_{j=0}^{b}S\left(  H_{b-j}\right)  \underbrace{H_{\left(
a+b+1\right)  -\left(  b-j\right)  }}_{\substack{=H_{a+j+1}\\\text{(since
}\left(  a+b+1\right)  -\left(  b-j\right)  =a+j+1\text{)}}}=\left(
-1\right)  ^{b}\sum_{j=0}^{b}S\left(  H_{b-j}\right)  H_{a+j+1}\\
&  =\left(  -1\right)  ^{b}\sum_{i=0}^{b}S\left(  H_{b-i}\right)
H_{a+i+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation
index }j\text{ as }i\right)  .
\end{align*}


Applying the map $\pi$ to both sides of this equality, we obtain%
\begin{align*}
\pi\left(  R_{\left(  1^{b},\left(  a+b+1\right)  -b\right)  }\right)   &
=\pi\left(  \left(  -1\right)  ^{b}\sum_{i=0}^{b}S\left(  H_{b-i}\right)
H_{a+i+1}\right)  =\left(  -1\right)  ^{b}\sum_{i=0}^{b}S\left(
\underbrace{\pi\left(  H_{b-i}\right)  }_{\substack{=h_{b-i}\\\text{(by the
definition of }\pi\text{)}}}\right)  \underbrace{\pi\left(  H_{a+i+1}\right)
}_{\substack{=h_{a+i+1}\\\text{(by the definition of }\pi\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a Hopf algebra
homomorphism}\right)  \\
&  =\left(  -1\right)  ^{b}\sum_{i=0}^{b}\underbrace{S\left(  h_{b-i}\right)
}_{\substack{=\left(  -1\right)  ^{b-i}e_{b-i}\\\text{(by Proposition
\ref{symm-antipode-and-h-basis}(iii),}\\\text{applied to }n=b-i\text{)}%
}}h_{a+i+1}\\
&  =\left(  -1\right)  ^{b}\sum_{i=0}^{b}\left(  -1\right)  ^{b-i}%
e_{b-i}h_{a+i+1}=\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}.
\end{align*}


Thus,
\begin{align*}
\sum_{i=0}^{b}\left(  -1\right)  ^{i}h_{a+i+1}e_{b-i}  &  =\pi\left(
R_{\left(  1^{b},\left(  a+b+1\right)  -b\right)  }\right)  =s_{\left(
\left(  a+b+1\right)  -b,1^{b}\right)  }\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.Psi.l.piR1in-i}), applied to }n=a+b+1\text{ and }i=b\right) \\
&  =s_{\left(  a+1,1^{b}\right)  }.
\end{align*}
Thus, Exercise \ref{exe.hookschur}(b) is once again solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.QSym.frobenius}}

\textit{Solution to Exercise~\ref{exe.QSym.frobenius}.}

(a) The solution of Exercise \ref{exe.QSym.frobenius}(a) can be obtained from
the solution of Exercise \ref{exe.witt.ghost-app}(a) upon replacing
$\mathbf{f}_{n}$ by $\mathbf{F}_{n}$.

(b) The solution of Exercise \ref{exe.QSym.frobenius}(b) can be obtained from
the solution of Exercise \ref{exe.witt.ghost-app}(b) upon replacing
$\mathbf{f}_{n}$, $\mathbf{f}_{m}$, $\mathbf{f}_{nm}$ and $\Lambda$ by
$\mathbf{F}_{n}$, $\mathbf{F}_{m}$, $\mathbf{F}_{nm}$ and
$ \Qsym $.

(c) The solution of Exercise \ref{exe.QSym.frobenius}(c) can be obtained from
the solution of Exercise \ref{exe.witt.ghost-app}(c) upon replacing
$\mathbf{f}_{1}$ and $\Lambda$ by $\mathbf{F}_{1}$ and $ \Qsym $.

(d) Let $n\in\left\{  1,2,3,\ldots\right\}  $ and $\left(  \beta_{1},\beta
_{2},\ldots,\beta_{s}\right)  \in \Comp $. The definition of
$M_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)  }$ yields that
$M_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)  }=\sum_{i_{1}%
<i_{2}<\cdots<i_{s}}x_{i_{1}}^{\beta_{1}}x_{i_{2}}^{\beta_{2}}\cdots x_{i_{s}%
}^{\beta_{s}}$ (where the sum is over all $s$-tuples $\left(  i_{1}%
,i_{2},\ldots,i_{s}\right)  $ of positive integers satisfying $i_{1}%
<i_{2}<\cdots<i_{s}$). Applying the map $\mathbf{F}_{n}$ to both sides of this
equality, we obtain%
\begin{align*}
\mathbf{F}_{n}\left(  M_{\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)
}\right)   &  =\mathbf{F}_{n}\left(  \sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}%
}^{\beta_{1}}x_{i_{2}}^{\beta_{2}}\cdots x_{i_{s}}^{\beta_{s}}\right) \\
&  =\left(  \sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}^{\beta_{1}}x_{i_{2}%
}^{\beta_{2}}\cdots x_{i_{s}}^{\beta_{s}}\right)  \left(  x_{1}^{n},x_{2}%
^{n},x_{3}^{n},\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\mathbf{F}_{n}\right) \\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{s}}\left(  x_{i_{1}}^{n}\right)  ^{\beta_{1}%
}\left(  x_{i_{2}}^{n}\right)  ^{\beta_{2}}\cdots\left(  x_{i_{s}}^{n}\right)
^{\beta_{s}}=\sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}^{n\beta_{1}}x_{i_{2}%
}^{n\beta_{2}}\cdots x_{i_{s}}^{n\beta_{s}}.
\end{align*}
Compared with%
\[
M_{\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{s}\right)  }=\sum_{i_{1}%
<i_{2}<\cdots<i_{s}}x_{i_{1}}^{n\beta_{1}}x_{i_{2}}^{n\beta_{2}}\cdots
x_{i_{s}}^{n\beta_{s}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}M_{\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{s}\right)  }\right)  ,
\]
this yields $\mathbf{F}_{n}\left(  M_{\left(  \beta_{1},\beta_{2},\ldots
,\beta_{s}\right)  }\right)  =M_{\left(  n\beta_{1},n\beta_{2},\ldots
,n\beta_{s}\right)  }$. This solves Exercise \ref{exe.QSym.frobenius}(d).

(e) Fix $n\in\left\{  1,2,3,\ldots\right\}  $.
We now know that $\mathbf{F}_{n}$ is a $\kk$-algebra homomorphism (due
to Exercise \ref{exe.QSym.frobenius}(a)), thus a $\kk$-linear map.

Let $\alpha\in
 \Comp $. Write $\alpha$ in the form $\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $. Then,%
\begin{equation}
\Delta M_{\alpha}=\sum_{k=0}^{\ell}M_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{k}\right)  }\otimes M_{\left(  \alpha_{k+1},\alpha
_{k+2},\ldots,\alpha_{\ell}\right)  } \label{sol.QSym.frobenius.e.0}%
\end{equation}
(according to Proposition \ref{Qsym-coproduct-on-monomials}). Applying the map
$\mathbf{F}_{n}\otimes\mathbf{F}_{n}$ to both sides of this equality, we
obtain%
\begin{align}
\left(  \mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)  \left(  \Delta M_{\alpha
}\right)   &  =\left(  \mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)  \left(
\sum_{k=0}^{\ell}M_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k}\right)
}\otimes M_{\left(  \alpha_{k+1},\alpha_{k+2},\ldots,\alpha_{\ell}\right)
}\right) \nonumber\\
&  =\sum_{k=0}^{\ell}\underbrace{\mathbf{F}_{n}\left(  M_{\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{k}\right)  }\right)  }_{\substack{=M_{\left(
n\alpha_{1},n\alpha_{2},\ldots,n\alpha_{k}\right)  }\\\text{(by Exercise
\ref{exe.QSym.frobenius}(d), applied to}\\\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{k}\right)  \text{ instead of }\left(  \beta_{1},\beta
_{2},\ldots,\beta_{s}\right)  \text{)}}}\otimes\underbrace{\mathbf{F}%
_{n}\left(  M_{\left(  \alpha_{k+1},\alpha_{k+2},\ldots,\alpha_{\ell}\right)
}\right)  }_{\substack{=M_{\left(  n\alpha_{k+1},n\alpha_{k+2},\ldots
,n\alpha_{\ell}\right)  }\\\text{(by Exercise \ref{exe.QSym.frobenius}(d),
applied to}\\\left(  \alpha_{k+1},\alpha_{k+2},\ldots,\alpha_{\ell}\right)
\text{ instead of }\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)
\text{)}}}\nonumber\\
&  =\sum_{k=0}^{\ell}M_{\left(  n\alpha_{1},n\alpha_{2},\ldots,n\alpha
_{k}\right)  }\otimes M_{\left(  n\alpha_{k+1},n\alpha_{k+2},\ldots
,n\alpha_{\ell}\right)  }. \label{sol.QSym.frobenius.e.1}%
\end{align}


But
\begin{align*}
\left(  \Delta\circ\mathbf{F}_{n}\right)  \left(  M_{\alpha}\right)   &
=\Delta\left(  \mathbf{F}_{n}\left(  \underbrace{M_{\alpha}}%
_{\substack{=M_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
}\\\text{(since }\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  \text{)}}}\right)  \right)  =\Delta\left(  \underbrace{\mathbf{F}%
_{n}\left(  M_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
}\right)  }_{\substack{=M_{\left(  n\alpha_{1},n\alpha_{2},\ldots
,n\alpha_{\ell}\right)  }\\\text{(by Exercise \ref{exe.QSym.frobenius}(d),
applied to}\\\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  \text{
instead of }\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)  \text{)}%
}}\right) \\
&  =\Delta M_{\left(  n\alpha_{1},n\alpha_{2},\ldots,n\alpha_{\ell}\right)
}=\sum_{k=0}^{\ell}M_{\left(  n\alpha_{1},n\alpha_{2},\ldots,n\alpha
_{k}\right)  }\otimes M_{\left(  n\alpha_{k+1},n\alpha_{k+2},\ldots
,n\alpha_{\ell}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{sol.QSym.frobenius.e.0}), applied to }\left(  n\alpha
_{1},n\alpha_{2},\ldots,n\alpha_{\ell}\right)  \text{ and}\\
\left(  n\alpha_{1},n\alpha_{2},\ldots,n\alpha_{\ell}\right)  \text{ instead
of }\alpha\text{ and }\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)
\end{array}
\right) \\
&  =\left(  \mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)  \left(  \Delta
M_{\alpha}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.QSym.frobenius.e.1})}\right) \\
&  =\left(  \left(  \mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)  \circ
\Delta\right)  \left(  M_{\alpha}\right)  .
\end{align*}


Let us now forget that we fixed $\alpha$. We thus have shown that%
\begin{equation}
\left(  \Delta\circ\mathbf{F}_{n}\right)  \left(  M_{\alpha}\right)  =\left(
\left(  \mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)  \circ\Delta\right)
\left(  M_{\alpha}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\alpha
\in \Comp . \label{sol.QSym.frobenius.e.main}%
\end{equation}


Now, let us recall that the family $\left(  M_{\alpha}\right)  _{\alpha
\in \Comp }$ is a basis of the $\kk$-module
$ \Qsym $. The two $\kk$-linear maps $\Delta
\circ\mathbf{F}_{n}$ and $\left(  \mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)
\circ\Delta$ are equal to each other on this basis (according to
(\ref{sol.QSym.frobenius.e.main})), and therefore must be identical (because
if two $\kk$-linear maps from the same domain are equal to each other
on a basis of their domain, then these two maps must be identical). In other
words, $\Delta\circ\mathbf{F}_{n}=\left(  \mathbf{F}_{n}\otimes\mathbf{F}%
_{n}\right)  \circ\Delta$. We can prove (using a similar but simpler argument)
that $\epsilon\circ\mathbf{F}_{n}=\epsilon$. Thus, the map
$\mathbf{F}_{n}$ is a $\kk$-coalgebra homomorphism (since it is
$\kk$-linear and satisfies $\Delta\circ\mathbf{F}_{n}=\left(
\mathbf{F}_{n}\otimes\mathbf{F}_{n}\right)  \circ\Delta$ and $\epsilon
\circ\mathbf{F}_{n}=\epsilon$).

We now know that $\mathbf{F}_{n}$ is a $\kk$-algebra homomorphism
and a $\kk$-coalgebra
homomorphism. Hence, $\mathbf{F}_{n}$ is a $\kk$-bialgebra
homomorphism, thus a Hopf algebra homomorphism (due to
Corollary~\ref{cor.bialg-mor-is-Hopf}). This solves Exercise
\ref{exe.QSym.frobenius}(e).

(f) Let $n\in\left\{  1,2,3,\ldots\right\}  $. For every $a\in\Lambda$, we
have
\begin{align*}
\left(  \mathbf{F}_{n}\mid_{\Lambda}\right)  \left(  a\right)   &
=\mathbf{F}_{n}\left(  a\right)  =a\left(  x_{1}^{n},x_{2}^{n},x_{3}%
^{n},\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{F}_{n}\right) \\
&  =\mathbf{f}_{n}\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{f}_{n}\left(  a\right)  =a\left(  x_{1}^{n},x_{2}^{n},x_{3}%
^{n},\ldots\right)  \text{ (by the definition of }\mathbf{f}_{n}%
\text{)}\right)  .
\end{align*}
Thus, $\mathbf{F}_{n}\mid_{\Lambda}=\mathbf{f}_{n}$. This solves Exercise
\ref{exe.QSym.frobenius}(f).

(g) The solution of Exercise \ref{exe.QSym.frobenius}(g) can be obtained from
the solution of Exercise \ref{exe.witt.ghost-app}(f) upon replacing
$\mathbf{f}_{p}$ and $\Lambda$ by $\mathbf{F}_{p}$ and $\Qsym$,
and replacing the word ``symmetric'' by ``quasisymmetric''.

(h) \textit{Alternative solution of Exercise \ref{exe.witt.ghost-app}(d).} Fix
$n\in\left\{  1,2,3,\ldots\right\}  $. Exercise \ref{exe.QSym.frobenius}(f)
yields that $\mathbf{F}_{n}\mid_{\Lambda}=\mathbf{f}_{n}$. Hence,
$\mathbf{f}_{n}=\mathbf{F}_{n}\mid_{\Lambda}$ is the restriction of
$\mathbf{F}_{n}$ to the Hopf subalgebra $\Lambda$ of $ \Qsym $.
Thus, $\mathbf{f}_{n}$ is the restriction of a Hopf algebra homomorphism to
the Hopf subalgebra $\Lambda$ of $ \Qsym $ (since $\mathbf{F}%
_{n}$ is a Hopf algebra homomorphism (by Exercise \ref{exe.QSym.frobenius}%
(e))). Consequently, $\mathbf{f}_{n}$ is a Hopf algebra homomorphism itself
(since the restriction of a Hopf algebra homomorphism to a Hopf subalgebra
must always be a Hopf algebra homomorphism). This gives a new solution to
Exercise \ref{exe.witt.ghost-app}(d). Thus, Exercise \ref{exe.QSym.frobenius}%
(h) is solved.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.NSym.verschiebung}}

\textit{Solution to Exercise \ref{exe.NSym.verschiebung}.} Let us first notice
that every positive integer $n$ satisfies%
\begin{equation}
\mathbf{V}_{n} \left(  H_{m}\right)  =
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in \NN 
\label{sol.NSym.verschiebung.vnhm}%
\end{equation}
\footnote{\textit{Proof.} To obtain a proof of
(\ref{sol.NSym.verschiebung.vnhm}), it is enough to repeat the proof of
(\ref{sol.Lambda.verschiebung.vnhm}), making just the following changes:
\par
\begin{itemize}
\item replacing every $h_{i}$ by $H_{i}$;
\par
\item replacing $\mathbf{v}_{n}$ by $\mathbf{V}_{n} $.
\end{itemize}
}.

Recall that $\Nsym$ is (isomorphic to) the free associative
algebra with generators $H_{1},H_{2},H_{3},\ldots$ (according to
(\ref{eq.thm.NSym.free})). Hence, the elements $H_{1},H_{2},H_{3},\ldots$
generate the $\kk$-algebra $\Nsym$. In other words,%
\begin{equation}
\text{the family }\left(  H_{r}\right)  _{r\geq1}\text{ generates the }
 \kk \text{-algebra } \Nsym .
\label{sol.NSym.verschiebung.gen}
\end{equation}


(e) A solution to Exercise \ref{exe.NSym.verschiebung}(e) can be obtained by
copying the solution of Exercise \ref{exe.Lambda.verschiebung}(e) and making
the following changes:

\begin{itemize}
\item Replace every appearance of $\mathbf{v}_{n}$ by $\mathbf{V}_{n} $.

\item Replace every appearance of $\Lambda$ by $\Nsym$.

\item Replace every appearance of $h_{j}$ (for some $j\in \NN$) by
$H_{j}$.

\item Replace the reference to Proposition
\ref{symm-comultiplication-formulas-prop}(iii) by a reference to
(\ref{Nsym-coproduct-on-H}).

\item Replace the reference to Proposition \ref{symm-antipode-and-h-basis} by
a reference to (\ref{sol.NSym.verschiebung.gen}).

\item Replace every reference to (\ref{sol.Lambda.verschiebung.vnhm}) by a
reference to (\ref{sol.NSym.verschiebung.vnhm}).
\end{itemize}

(c) A solution to Exercise \ref{exe.NSym.verschiebung}(c) can be obtained by
copying the solution of Exercise \ref{exe.Lambda.verschiebung}(c) and making
the following changes:

\begin{itemize}
\item Replace every appearance of $\mathbf{v}_{j}$ (for some positive integer
$j$) by $\mathbf{V}_{j}$.

\item Replace every appearance of $\Lambda$ by $\Nsym$.

\item Replace every appearance of $h_{j}$ (for some $j\in \NN$) by
$H_{j}$.

\item Replace the reference to Proposition \ref{symm-antipode-and-h-basis} by
a reference to (\ref{sol.NSym.verschiebung.gen}).
\end{itemize}

(d) A solution to Exercise \ref{exe.NSym.verschiebung}(d) can be obtained by
copying the solution of Exercise \ref{exe.Lambda.verschiebung}(d) and making
the following changes:

\begin{itemize}
\item Replace every appearance of $\mathbf{v}_{1}$ by $\mathbf{V}_{1} $.

\item Replace every appearance of $\Lambda$ by $\Nsym$.

\item Replace every appearance of $h_{j}$ (for some $j\in \NN$) by
$H_{j}$.

\item Replace the reference to Proposition \ref{symm-antipode-and-h-basis} by
a reference to (\ref{sol.NSym.verschiebung.gen}).
\end{itemize}

(a) Here is one of several possible solutions of Exercise
\ref{exe.NSym.verschiebung}(a):\footnote{Another solution (which the reader
can easily find) proceeds by making some relatively straightforward
modifications to the solution of Exercise \ref{exe.Lambda.verschiebung}(a).
(One needs to keep in mind that $\Nsym$, unlike $\Lambda$, is
not commutative -- but this does not prevent us from doing things such as
substituting $t^n$ for $t$ in a power series over $\Nsym$
(since $t^n$ is a central element of $\Nsym \left[  \left[
t\right]  \right]  $).)}
Define a $\kk$-linear map $E : \Nsym \rightarrow \Nsym$ as in Exercise
\ref{exe.dynkin} (but with $\Nsym$ instead of $A$). Every
positive integer $n$ satisfies
\begin{equation}
\Psi_n = \left(  S\star E\right)  \left(  H_{n}\right)
\label{sol.NSym.verschiebung.e.Psin}%
\end{equation}
(according to Exercise \ref{exe.Psi}(c)).

Fix a positive integer $n$. Let us make some auxiliary observations first:

\begin{itemize}
\item Every composition $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $ satisfies%
\begin{equation}
H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }=H_{\alpha_{1}%
}H_{\alpha_{2}}\cdots H_{\alpha_{\ell}}. \label{sol.NSym.verschiebung.a.H}%
\end{equation}
\footnote{This is precisely the equality (\ref{eq.Nsym.Halpha}), and has been
proven before.}

\item If $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ is a
composition such that $\left(  \text{not every }i\in\left\{  1,2,\ldots
,\ell\right\}  \text{ satisfies }n\mid\alpha_{i}\right)  $, then%
\begin{equation}
\mathbf{V}_{n}\left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)  =0. \label{sol.NSym.verschiebung.a.H1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.a.H1}):} Let $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be a composition such
that $\left(  \text{not every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{
satisfies }n\mid\alpha_{i}\right)  $. Then, there exists an $i\in\left\{
1,2,\ldots,\ell\right\}  $ satisfying $n\nmid\alpha_{i}$ (since not every
$i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies $n\mid\alpha_{i}$).
Consider this $i$. We have
\begin{align*}
\mathbf{V}_{n}\left(  H_{\alpha_{i}}\right)   &  =
\begin{cases}
H_{\alpha_{i}/n},& \text{if }n\mid\alpha_{i};\\
0,& \text{if }n\nmid\alpha_{i}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{V}_{n}\left(  H_{\alpha_{i}}\right)  \right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\nmid\alpha_{i}\right)  .
\end{align*}
But (\ref{sol.NSym.verschiebung.a.H}) yields $H_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  }=H_{\alpha_{1}}H_{\alpha_{2}}\cdots
H_{\alpha_{\ell}}$. Applying $\mathbf{V}_{n}$ to both sides of this equality,
we obtain%
\begin{align*}
\mathbf{V}_{n}\left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)   &  =\mathbf{V}_{n}\left(  H_{\alpha_{1}}H_{\alpha_{2}%
}\cdots H_{\alpha_{\ell}}\right)  =\mathbf{V}_{n}\left(  H_{\alpha_{1}%
}\right)  \mathbf{V}_{n}\left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}%
_{n}\left(  H_{\alpha_{\ell}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{V}_{n}\text{ is a
} \kk \text{-algebra homomorphism}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since at least one factor of the product}\\
\mathbf{V}_{n}\left(  H_{\alpha_{1}}\right)  \mathbf{V}_{n}\left(
H_{\alpha_{2}}\right)  \cdots\mathbf{V}_{n}\left(  H_{\alpha_{\ell}}\right)
\text{ is }0\text{ (namely, the factor }\mathbf{V}_{n}\left(  H_{\alpha_{i}%
}\right)  \text{)}%
\end{array}
\right)  .
\end{align*}
This proves (\ref{sol.NSym.verschiebung.a.H1}).}

\item If $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ is a
composition such that $\left(  \text{every }i\in\left\{  1,2,\ldots
,\ell\right\}  \text{ satisfies }n\mid\alpha_{i}\right)  $, then%
\begin{equation}
\mathbf{V}_{n}\left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)  =H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }. \label{sol.NSym.verschiebung.a.H2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.a.H2}):} Let $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be a composition such
that $\left(  \text{every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{
satisfies }n\mid\alpha_{i}\right)  $. Then, $\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  $ is a composition. Therefore,
(\ref{sol.NSym.verschiebung.a.H}) (applied to $\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  $ instead of $\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $) yields $H_{\left(  \alpha
_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }=H_{\alpha_{1}/n}%
H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}$.
\par
But every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies%
\begin{align*}
\mathbf{V}_{n}\left(  H_{\alpha_{i}}\right)   &  =
\begin{cases}
H_{\alpha_{i}/n},& \text{if }n\mid\alpha_{i};\\
0,& \text{if }n\nmid\alpha_{i}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{V}_{n}\left(  H_{\alpha_{i}}\right)  \right) \\
&  =H_{\alpha_{i}/n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\mid\alpha
_{i}\right)  .
\end{align*}
Multiplying these equalities for all $i\in\left\{  1,2,\ldots,\ell\right\}  $,
we obtain
\[
\mathbf{V}_{n}\left(  H_{\alpha_{1}}\right)  \mathbf{V}_{n}\left(
H_{\alpha_{2}}\right)  \cdots\mathbf{V}_{n}\left(  H_{\alpha_{\ell}}\right)
=H_{\alpha_{1}/n}H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}.
\]
\par
But (\ref{sol.NSym.verschiebung.a.H}) yields $H_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  }=H_{\alpha_{1}}H_{\alpha_{2}}\cdots
H_{\alpha_{\ell}}$. Applying $\mathbf{V}_{n}$ to both sides of this equality,
we obtain%
\begin{align*}
\mathbf{V}_{n}\left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)   &  =\mathbf{V}_{n}\left(  H_{\alpha_{1}}H_{\alpha_{2}%
}\cdots H_{\alpha_{\ell}}\right)  =\mathbf{V}_{n}\left(  H_{\alpha_{1}%
}\right)  \mathbf{V}_{n}\left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}%
_{n}\left(  H_{\alpha_{\ell}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{V}_{n}\text{ is a
}{ \kk }\text{-algebra homomorphism}\right) \\
&  =H_{\alpha_{1}/n}H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}=H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }.
\end{align*}
This proves (\ref{sol.NSym.verschiebung.a.H2}).}

\item We have%
\begin{equation}
\mathbf{V}_{n}\circ E=n\cdot\left(  E\circ\mathbf{V}_{n}\right)
\label{sol.NSym.verschiebung.a.E}%
\end{equation}
(as endomorphisms of $\Nsym$).\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.NSym.verschiebung.a.E}):} Let us show that%
\begin{equation}
\left(  \mathbf{V}_{n}\circ E\right)  \left(  H_{\alpha}\right)  =\left(
n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \right)  \left(  H_{\alpha}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\alpha\in\operatorname{Comp}.
\label{sol.NSym.verschiebung.a.E.pf.1}
\end{equation}
\par
\textit{Proof of (\ref{sol.NSym.verschiebung.a.E.pf.1}):} Let $\alpha
\in\operatorname{Comp}$. Then, $H_{\alpha}$ is a homogeneous element of
$\Nsym$ of degree $\deg\left(  H_{\alpha}\right)  =\left\vert
\alpha\right\vert $. Thus, the definition of $E\left(  H_{\alpha}\right)  $
yields $E\left(  H_{\alpha}\right)  =\underbrace{\left(  \deg\left(
H_{\alpha}\right)  \right)  }_{=\left\vert \alpha\right\vert }\cdot H_{\alpha
}=\left\vert \alpha\right\vert \cdot H_{\alpha}$. Now,%
\[
\left(  \mathbf{V}_{n}\circ E\right)  \left(  H_{\alpha}\right)
=\mathbf{V}_{n}\left(  \underbrace{E\left(  H_{\alpha}\right)  }_{=\left\vert
\alpha\right\vert \cdot H_{\alpha}}\right)  =\mathbf{V}_{n}\left(  \left\vert
\alpha\right\vert \cdot H_{\alpha}\right)  =\left\vert \alpha\right\vert
\cdot\mathbf{V}_{n}\left(  H_{\alpha}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\mathbf{V}_{n}\text{ is }{ \kk }\text{-linear}%
\right)  .
\]
\par
Now, let us write the composition $\alpha$ in the form $\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. We distinguish between two
cases:
\par
\textit{Case 1:} Not every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies
$n\mid\alpha_{i}$.
\par
\textit{Case 2:} Every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies
$n\mid\alpha_{i}$.
\par
Let us first consider Case 1. In this case, not every $i\in\left\{
1,2,\ldots,\ell\right\}  $ satisfies $n\mid\alpha_{i}$. Thus,
(\ref{sol.NSym.verschiebung.a.H1}) yields $\mathbf{V}_{n}\left(  H_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }\right)  =0$. Since
$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  =\alpha$, this
rewrites as $\mathbf{V}_{n}\left(  H_{\alpha}\right)  =0$. Thus, $\left(
\mathbf{V}_{n}\circ E\right)  \left(  H_{\alpha}\right)  =\left\vert
\alpha\right\vert \cdot\underbrace{\mathbf{V}_{n}\left(  H_{\alpha}\right)
}_{=0}=0$. Compared with
\[
\left(  n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \right)  \left(  H_{\alpha
}\right)  =n\cdot\underbrace{\left(  E\circ\mathbf{V}_{n}\right)  \left(
H_{\alpha}\right)  }_{=E\left(  \mathbf{V}_{n}\left(  H_{\alpha}\right)
\right)  }=n\cdot E\left(  \underbrace{\mathbf{V}_{n}\left(  H_{\alpha
}\right)  }_{=0}\right)  =n\cdot\underbrace{E\left(  0\right)  }%
_{\substack{=0\\\text{(since }E\text{ is }{ \kk }\text{-linear)}}}=0,
\]
this yields $\left(  \mathbf{V}_{n}\circ E\right)  \left(  H_{\alpha}\right)
=\left(  n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \right)  \left(
H_{\alpha}\right)  $. Hence, (\ref{sol.NSym.verschiebung.a.E.pf.1}) is proven
in Case 1.
\par
Let us now consider Case 2. In this case, every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ satisfies $n\mid\alpha_{i}$. Thus,
(\ref{sol.NSym.verschiebung.a.H2}) yields $\mathbf{V}_{n}\left(  H_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }\right)  =H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }$. Since $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  =\alpha$, this rewrites as
$\mathbf{V}_{n}\left(  H_{\alpha}\right)  =H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }$. Thus,
\begin{equation}
\left(  \mathbf{V}_{n}\circ E\right)  \left(  H_{\alpha}\right)  =\left\vert
\alpha\right\vert \cdot\underbrace{\mathbf{V}_{n}\left(  H_{\alpha}\right)
}_{=H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }%
}=\left\vert \alpha\right\vert \cdot H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }.
\label{sol.NSym.verschiebung.a.E.pf.1.pf.1}%
\end{equation}
\par
On the other hand, $H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }$ is a homogeneous element of $\Nsym$ of degree
$\deg\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }\right)  =\left\vert \left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  \right\vert =\left\vert \underbrace{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }_{=\alpha}\right\vert
/n=\left\vert \alpha\right\vert /n$. Hence, the definition of $E\left(
H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }\right)
$ yields $E\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }\right)  =\underbrace{\left(  \deg\left(  H_{\left(  \alpha
_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }\right)  \right)
}_{=\left\vert \alpha\right\vert /n}\cdot H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }=\left(  \left\vert \alpha\right\vert
/n\right)  \cdot H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }$. Now,%
\begin{align*}
\left(  n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \right)  \left(  H_{\alpha
}\right)   &  =n\cdot\underbrace{\left(  E\circ\mathbf{V}_{n}\right)  \left(
H_{\alpha}\right)  }_{=E\left(  \mathbf{V}_{n}\left(  H_{\alpha}\right)
\right)  }=n\cdot E\left(  \underbrace{\mathbf{V}_{n}\left(  H_{\alpha
}\right)  }_{=H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }}\right) \\
&  =n\cdot\underbrace{E\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  }\right)  }_{=\left(  \left\vert
\alpha\right\vert /n\right)  \cdot H_{\left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  }}=n\cdot\left(  \left\vert \alpha
\right\vert /n\right)  \cdot H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots
,\alpha_{\ell}/n\right)  }\\
&  =\left\vert \alpha\right\vert \cdot H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }.
\end{align*}
Compared with (\ref{sol.NSym.verschiebung.a.E.pf.1.pf.1}), this yields
$\left(  \mathbf{V}_{n}\circ E\right)  \left(  H_{\alpha}\right)  =\left(
n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \right)  \left(  H_{\alpha}\right)
$. Hence, (\ref{sol.NSym.verschiebung.a.E.pf.1}) is proven in Case 2.
\par
Thus, (\ref{sol.NSym.verschiebung.a.E.pf.1}) is proven in both Cases 1 and 2.
Since these two Cases cover all possibilities, this shows that
(\ref{sol.NSym.verschiebung.a.E.pf.1}) always holds.
\par
Now, $\left(  H_{\alpha}\right)  _{\alpha\in\operatorname{Comp}}$ is a basis
of the $\kk$-module $\Nsym$. The equality
(\ref{sol.NSym.verschiebung.a.E.pf.1}) shows that the two ${ \kk }%
$-linear maps $\mathbf{V}_{n}\circ E$ and $n\cdot\left(  E\circ\mathbf{V}%
_{n}\right)  $ are equal to each other on every element of this basis. Hence,
these two maps $\mathbf{V}_{n}\circ E$ and $n\cdot\left(  E\circ\mathbf{V}%
_{n}\right)  $ are identical (because if two $\kk$-linear maps having
the same domain are equal to each other on every element of some basis of this
domain, then these two maps must be identical). In other words, $\mathbf{V}%
_{n}\circ E=n\cdot\left(  E\circ\mathbf{V}_{n}\right)  $. This proves
(\ref{sol.NSym.verschiebung.a.E}).}

\item We have%
\begin{equation}
\mathbf{V}_{n}\circ\left(  S\star E\right)  =n\cdot\left(  \left(  S\star
E\right)  \circ\mathbf{V}_{n}\right)  \label{sol.NSym.verschiebung.a.dyn}%
\end{equation}
(where $S$, as usual, denotes the antipode of $\Nsym$).
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.a.dyn}):}
We have $S\star E=m\circ\left(  S\otimes E\right)  \circ\Delta$ (by the
definition of convolution). But we know (from Exercise
\ref{exe.NSym.verschiebung}(e)) that $\mathbf{V}_{n}$ is a Hopf algebra
homomorphism; thus, $\mathbf{V}_{n}\circ S=S\circ\mathbf{V}_{n}$. Also,
$\mathbf{V}_{n}$ is a $\kk$-coalgebra homomorphism (since
$\mathbf{V}_{n}$ is a Hopf algebra homomorphism); therefore, $\left(
\mathbf{V}_{n}\otimes\mathbf{V}_{n}\right)  \circ\Delta=\Delta\circ
\mathbf{V}_{n}$. Finally, $\mathbf{V}_{n}$ is a $\kk$-algebra
homomorphism; thus, $\mathbf{V}_{n}\circ m=m\circ\left(  \mathbf{V}_{n}%
\otimes\mathbf{V}_{n}\right)  $. Now,%
\begin{align*}
\mathbf{V}_{n}\circ\left(  \underbrace{S\star E}_{=m\circ\left(  S\otimes
E\right)  \circ\Delta}\right)   &  =\underbrace{\mathbf{V}_{n}\circ
m}_{=m\circ\left(  \mathbf{V}_{n}\otimes\mathbf{V}_{n}\right)  }\circ\left(
S\otimes E\right)  \circ\Delta=m\circ\underbrace{\left(  \mathbf{V}_{n}%
\otimes\mathbf{V}_{n}\right)  \circ\left(  S\otimes E\right)  }_{=\left(
\mathbf{V}_{n}\circ S\right)  \otimes\left(  \mathbf{V}_{n}\circ E\right)
}\circ\Delta\\
&  =m\circ\left(  \underbrace{\left(  \mathbf{V}%
_{n}\circ S\right)  }_{=S\circ\mathbf{V}_{n}}
\otimes
\underbrace{\left(  \mathbf{V}_{n}\circ E\right)
}_{\substack{=n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \\\text{(by
(\ref{sol.NSym.verschiebung.a.E}))}}}
\right)  \circ\Delta
=m\circ\left(
\left(  S\circ\mathbf{V}_{n}\right)
\otimes
\left(  n\cdot\left(  E\circ\mathbf{V}_{n}\right)  \right)
\right)  \circ\Delta\\
&  =n\cdot\left(  m\circ\underbrace{\left(
\left(  S\circ\mathbf{V}_{n}\right)
\otimes
\left(  E\circ\mathbf{V}_{n}\right)  \right)  }_{=\left(
S\otimes E\right)  \circ\left(  \mathbf{V}_{n}\otimes\mathbf{V}_{n}\right)
}\circ\Delta\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\text{ is just
a scalar factor}\right) \\
&  =n\cdot\left(  m\circ\left(  S\otimes E\right)  \circ\underbrace{\left(
\mathbf{V}_{n}\otimes\mathbf{V}_{n}\right)  \circ\Delta}_{=\Delta
\circ\mathbf{V}_{n}}\right)  =n\cdot\left(  \underbrace{m\circ\left(  S\otimes
E\right)  \circ\Delta}_{\substack{=S\star E\\\text{(since }S\star E
=m\circ\left(  S\otimes E\right)  \circ\Delta\text{)}}}\circ\mathbf{V}%
_{n}\right)  =n\cdot\left(  \left(  S\star E\right)  \circ\mathbf{V}%
_{n}\right)  .
\end{align*}
This proves (\ref{sol.NSym.verschiebung.a.dyn}).}
\end{itemize}

But fix a positive integer $m$. We have $\Psi_{m}=\left(  S\star E\right)
\left(  H_{m}\right)  $ (by (\ref{sol.NSym.verschiebung.e.Psin}), applied to
$m$ instead of $n$) and thus%
\begin{align*}
\mathbf{V}_{n}\left(  \underbrace{\Psi_{m}}_{=\left(  S\star E\right)  \left(
H_{m}\right)  }\right)   &  =\mathbf{V}_{n}\left(  \left(  S\star E\right)
\left(  H_{m}\right)  \right)  =\underbrace{\left(  \mathbf{V}_{n}\circ\left(
S\star E\right)  \right)  }_{\substack{=n\cdot\left(  \left(  S\star E\right)
\circ\mathbf{V}_{n}\right)  \\\text{(by (\ref{sol.NSym.verschiebung.a.dyn}))}%
}}\left(  H_{m}\right) \\
&  =\left(  n\cdot\left(  \left(  S\star E\right)  \circ\mathbf{V}_{n}\right)
\right)  \left(  H_{m}\right)  =n\cdot\underbrace{\left(  \left(  S\star E
\right)  \circ\mathbf{V}_{n}\right)  \left(  H_{m}\right)  }_{=\left(  S\star
E\right)  \left(  \mathbf{V}_{n}\left(  H_{m}\right)  \right)  }\\
&  =n\cdot\left(  S\star E\right)  \left(  \underbrace{\mathbf{V}_{n}\left(
H_{m}\right)  }_{\substack{=
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\\\text{(by (\ref{sol.NSym.verschiebung.vnhm}))}}}\right)
=n\cdot\left(  S\star E\right)  \left(
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\right) \\
&  =
\begin{cases}
n\cdot\left(  S\star E\right)  \left(  H_{m/n}\right)
,& \text{if }n\mid m;\\
n\cdot\left(  S\star E\right)  \left(  0\right)  ,& \text{if }n\nmid m
\end{cases}
\\
&  =
\begin{cases}
n\cdot\left(  S\star E\right)  \left(  H_{m/n}\right)
,& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\cdot\underbrace{\left(
S\star E\right)  \left(  0\right)  }_{=0}=0\right) \\
&  =
\begin{cases}
n\Psi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\end{align*}
(because if $n\mid m$, then $\left(  S\star E\right)  \left(  H_{m/n}\right)
=\Psi_{m/n}$\ $\ \ \ $\footnote{\textit{Proof.} Assume that $n\mid m$. Then,
$m/n$ is a positive integer. Hence, (\ref{sol.NSym.verschiebung.e.Psin})
(applied to $m/n$ instead of $n$) yields $\Psi_{m/n}=\left(  S\star E\right)
\left(  H_{m/n}\right)  $, so that $\left(  S\star E\right)  \left(
H_{m/n}\right)  =\Psi_{m/n}$, qed.}). This solves Exercise
\ref{exe.NSym.verschiebung}(a).

(b) Recall that a ring of formal power series $R\left[  \left[  t\right]
\right]  $ over a (not necessarily commutative) $\kk$-algebra $R$ has a canonical
topology which makes it into a topological $\kk$-algebra\footnote{This is proven
just as in the case when $R$ is commutative.}. Thus, in particular,
$\Nsym \left[  \left[  t\right]  \right]$ becomes a
topological $\kk$-algebra. We shall be considering this
topology when we speak of continuity.

Assume that ${\QQ}$ is a subring of $\kk$.

Define two power series $\widetilde{H}\left(  t\right)  $ and $\xi\left(
t\right)  $ in $\Nsym \left[  \left[  t\right]  \right]  $ by%
\begin{align*}
\widetilde{H}\left(  t\right)   &  =\sum_{n\geq0}H_{n}t^{n};\\
\xi\left(  t\right)   &  =\sum_{n\geq1}\xi_{n}t^{n}=\log\left(  \widetilde{H}%
\left(  t\right)  \right)  .
\end{align*}
(Here, the equality $\sum_{n\geq1}\xi_{n}t^{n}=\log\left(  \widetilde{H}%
\left(  t\right)  \right)  $ follows from \eqref{relation-definining-pi's}.)

We have $\widetilde{H}\left(  t\right)  =\sum_{n\geq0}H_{n}t^{n}=\sum_{i\geq
0}H_{i}t^{i}$ (here, we renamed the summation index $n$ as $i$) and
$\xi\left(  t\right)  =\sum_{n\geq1}\xi_{n}t^{n}=\sum_{i\geq1}\xi_{i}t^{i}$
(here, we renamed the summation index $n$ as $i$).

Now, let $n$ be a positive integer. The $\kk$-algebra homomorphism
$\mathbf{V}_{n} : \Nsym \rightarrow\Nsym $ induces
a continuous $\kk$-algebra homomorphism $\mathbf{V}_{n} \left[
\left[  t\right]  \right]  : \Nsym \left[  \left[  t\right]
\right]  \rightarrow\Nsym \left[  \left[  t\right]  \right]  $
given by%
\[
\left(  \mathbf{V}_{n} \left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\in \NN }a_{i}t^{i}\right)  =\sum_{i\in \NN }\mathbf{V}%
_{n}\left(  a_{i}\right)  t^{i}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
a_{i}\right)  _{i\in \NN }\in\Nsym^{ \NN }.
\]
This $\kk$-algebra homomorphism $\mathbf{V}_{n} \left[  \left[
t\right]  \right]  $ commutes with taking logarithms (since it is continuous
and a $\kk$-algebra homomorphism) -- i.e., we have%
\begin{equation}
\left(  \mathbf{V}_{n} \left[  \left[  t\right]  \right]  \right)  \left(
\log Q\right)  =\log\left(  \left(  \mathbf{V}_{n} \left[  \left[  t\right]
\right]  \right)  \left(  Q\right)  \right)
\label{sol.NSym.verschiebung.b.deriv}
\end{equation}
for every $Q\in\Nsym \left[  \left[  t\right]  \right]$
having constant coefficient $1$.

We have $\xi\left(  t\right)  =\sum_{i\geq1}\xi_{i}t^{i}$, thus%
\[
\sum_{i\geq1}\xi_{i}t^{i}=\xi\left(  t\right)  =\log\left(
\underbrace{\widetilde{H}\left(  t\right)  }_{=\sum_{i\geq0}H_{i}t^{i}%
}\right)  =\log\left(  \sum_{i\geq0}H_{i}t^{i}\right)  .
\]
We can substitute $t^{n}$ for $t$ on both sides of this equality\footnote{This
is allowed because the element $t^{n}$ of $\Nsym \left[  \left[
t\right]  \right]  $ is central. (Generally, if $R$ is a ring (possibly not
commutative) and $Q\in R\left[  \left[  t\right]  \right]  $ is a power
series, then every central element $z$ of $R\left[  \left[  t\right]  \right]
$ can be substituted for $t$ in $Q$ as long as the constant coefficient of $z$
is $0$. The result of this substitution is denoted by $Q\left(  z\right)  $.
The map $R\left[  \left[  t\right]  \right]  \rightarrow R\left[  \left[
t\right]  \right]  $ which sends every $Q$ to $Q\left(  z\right)  $ (for a
fixed $z$) is a ring homomorphism $R\left[  \left[  t\right]  \right]
\rightarrow R\left[  \left[  t\right]  \right]  $.)}. As a result, we obtain%
\begin{equation}
\sum_{i\geq1}\xi_{i}\left(  t^{n}\right)  ^{i}=\log\left(  \sum_{i\geq0}%
H_{i}\left(  t^{n}\right)  ^{i}\right)  . \label{sol.NSym.verschiebung.b.3}%
\end{equation}


But $\widetilde{H}\left(  t\right)  =\sum_{i\geq0}H_{i}t^{i}$. Applying the
map $\mathbf{V}_{n}\left[  \left[  t\right]  \right]  $ to both sides of this
equality, we obtain
\begin{align}
\left(  \mathbf{V}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\widetilde{H}\left(  t\right)  \right)   &  =\left(  \mathbf{V}_{n}\left[
\left[  t\right]  \right]  \right)  \left(  \sum_{i\geq0}H_{i}t^{i}\right)
=\sum_{i\geq0}\underbrace{\mathbf{V}_{n}\left(  H_{i}\right)  }%
_{\substack{=
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
\\\text{(by (\ref{sol.NSym.verschiebung.vnhm}), applied to
}m=i\text{)}}}t^{i}\nonumber\\
&  =\sum_{i\geq0}
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} \cdot
t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}\underbrace{
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
}_{\substack{=H_{i/n}\\\text{(since }n\mid i\text{)}}}
\cdot t^{i}%
+\sum\limits_{\substack{i\geq0;\\n\nmid i}}\underbrace{
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}H_{i/n}t^{i}+\underbrace{\sum\limits
_{\substack{i\geq0;\\n\nmid i}}0t^{i}}_{=0}=\sum\limits_{\substack{i\geq0;\\n\mid
i}}H_{i/n}t^{i}\nonumber\\
&  =\sum_{i\geq0}\underbrace{H_{ni/n}}_{=H_{i}}\underbrace{t^{ni}}_{=\left(
t^{n}\right)  ^{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted
}ni\text{ for }i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\geq0}H_{i}\left(  t^{n}\right)  ^{i}.
\label{sol.NSym.verschiebung.b.4}%
\end{align}
Now,%
\begin{align}
\left(  \mathbf{V}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\underbrace{\xi\left(  t\right)  }_{=\log\left(  \widetilde{H}\left(
t\right)  \right)  }\right)   &  =\left(  \mathbf{V}_{n}\left[  \left[
t\right]  \right]  \right)  \left(  \log\left(  \widetilde{H}\left(  t\right)
\right)  \right)  =\log\left(  \underbrace{\left(  \mathbf{V}_{n}\left[
\left[  t\right]  \right]  \right)  \left(  \widetilde{H}\left(  t\right)
\right)  }_{\substack{=\sum_{i\geq0}H_{i}\left(  t^{n}\right)  ^{i}\\\text{(by
(\ref{sol.NSym.verschiebung.b.4}))}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.b.deriv}),
applied to }Q=\widetilde{H}\left(  t\right)  \right) \nonumber\\
&  =\log\left(  \sum_{i\geq0}H_{i}\left(  t^{n}\right)  ^{i}\right)
=\sum_{i\geq1}\xi_{i}\underbrace{\left(  t^{n}\right)  ^{i}}_{=t^{ni}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.b.3}%
)}\right) \nonumber\\
&  =\sum_{i\geq1}\xi_{i}t^{ni}. \label{sol.NSym.verschiebung.b.6}%
\end{align}
Comparing this with%
\begin{align}
\left(  \mathbf{V}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\underbrace{\xi\left(  t\right)  }_{=\sum_{i\geq1}\xi_{i}t^{i}}\right)   &
=\left(  \mathbf{V}_{n}\left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\geq1}\xi_{i}t^{i}\right)  =\sum_{i\geq1}\mathbf{V}_{n}\left(  \xi
_{i}\right)  t^{i}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{V}%
_{n}\left[  \left[  t\right]  \right]  \right) , \nonumber
\end{align}
we obtain%
\begin{equation}
\sum_{i\geq1}\mathbf{V}_{n}\left(  \xi_{i}\right)  t^{i}=\sum_{i\geq1}\xi
_{i}t^{ni}. \label{sol.NSym.verschiebung.b.8}%
\end{equation}


Now, let $m$ be a positive integer. Comparing coefficients before $t^{m}$ in
the equality (\ref{sol.NSym.verschiebung.b.8}), we obtain%
\[
\mathbf{V}_{n}\left(  \xi_{m}\right)  =
\begin{cases}
\xi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad
.
\]
This solves Exercise \ref{exe.NSym.verschiebung}(b).

(f) Recall first that
$\left( H_{\alpha} \right) _{\alpha \in \Comp}$ and
$\left( M_{\alpha} \right) _{\alpha \in \Comp}$ are
mutually dual bases with respect to the dual pairing
$\Nsym \otimes \Qsym
\overset{\left(\cdot,\cdot\right)  }{\longrightarrow}
\kk$.\ \ \ \ \footnote{This
follows from the definition of $\left(  H_{\alpha}\right)  _{\alpha\in
 \Comp }$.} Thus,
\begin{equation}
\left(  H_{\alpha},M_{\beta}\right)  = \delta_{\alpha,\beta}
\ \ \ \ \ \ \ \ \ \ \text{for any two compositions }\alpha\text{ and }\beta.
\label{sol.NSym.verschiebung.e.duality}
\end{equation}


Let us introduce a notation: For every composition $\alpha$ and every positive
integer $s$, let $\alpha\left\{  s\right\}  $ denote the $\ell$-tuple $\left(
s\alpha_{1},s\alpha_{2},\ldots,s\alpha_{\ell}\right)  $, where $\alpha$ is
written in the form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $. Notice that if $\alpha$ is a composition and $s$ is a positive
integer, then $\alpha\left\{  s\right\}  $ is a composition again.

We have
\begin{equation}
\mathbf{F}_{n}M_{\alpha} = M_{\alpha\left\{  n\right\}  }
\ \ \ \ \ \ \ \ \ \ \text{for every composition }\alpha.
\label{sol.NSym.verschiebung.e.f}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.e.f}):} Let $\alpha$ be
a composition. Write $\alpha$ in the form $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $. By the definition of $M_{\alpha}$,
we have $M_{\alpha}=\sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}}^{\alpha_{1}%
}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}$ (where the sum is
over all $\ell$-tuples $\left(  i_{1},i_{2},\ldots,i_{\ell}\right)  $ of
positive integers satisfying $i_{1}<i_{2}<\cdots<i_{\ell}$).
\par
On the other hand, the definition of $\alpha\left\{  n\right\}  $ yields
$\alpha\left\{  n\right\}  =\left(  n\alpha_{1},n\alpha_{2},\ldots
,n\alpha_{\ell}\right)  $ (since $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  $). Hence, the definition of $M_{\alpha\left\{
n\right\}  }$ yields
\begin{equation}
M_{\alpha\left\{  n\right\}  }=\sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}%
}^{n\alpha_{1}}x_{i_{2}}^{n\alpha_{2}}\cdots x_{i_{\ell}}^{n\alpha_{\ell}}
\label{sol.NSym.verschiebung.e.f.pf.1}%
\end{equation}
(where the sum is over all $\ell$-tuples $\left(  i_{1},i_{2},\ldots,i_{\ell
}\right)  $ of positive integers satisfying $i_{1}<i_{2}<\cdots<i_{\ell}$).
\par
Now, the definition of $\mathbf{F}_{n}$ yields%
\begin{align*}
\mathbf{F}_{n}M_{\alpha}  &  =\underbrace{M_{\alpha}}_{=\sum_{i_{1}%
<i_{2}<\cdots<i_{\ell}}x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots
x_{i_{\ell}}^{\alpha_{\ell}}}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n}%
,\ldots\right)  =\left(  \sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}}%
^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}\right)
\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right) \\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}}^{n\alpha_{1}}x_{i_{2}%
}^{n\alpha_{2}}\cdots x_{i_{\ell}}^{n\alpha_{\ell}}=M_{\alpha\left\{
n\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.NSym.verschiebung.e.f.pf.1})}\right)  .
\end{align*}
This proves (\ref{sol.NSym.verschiebung.e.f}).}.

We need to prove that the maps $\mathbf{F}_{n}: \Qsym
\rightarrow \Qsym $ and $\mathbf{V}_{n} : \Nsym
\rightarrow\Nsym $ are adjoint with respect to the dual pairing
$\Nsym \otimes \Qsym \overset{\left(  \cdot
,\cdot\right)  }{\longrightarrow} { \kk } $. In other words, we need to
prove that%
\begin{equation}
\left(  b,\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_{n} b,a\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in \Qsym  \text{ and }%
b\in\Nsym . \label{sol.NSym.verschiebung.e.goal}%
\end{equation}


\textit{Proof of (\ref{sol.NSym.verschiebung.e.goal}):} Fix $a\in
 \Qsym  $ and $b\in\Nsym $.

Recall that $\left(  M_{\alpha}\right)  _{\alpha\in\operatorname{Comp} }$ is a
basis of the $\kk$-module $\Nsym$. Hence, in proving
(\ref{sol.NSym.verschiebung.e.goal}), we can WLOG assume that $a$ is an
element of this basis $\left(  M_{\alpha}\right)  _{\alpha\in
\operatorname{Comp} }$ (because the equality
(\ref{sol.NSym.verschiebung.e.goal}) is $\kk$-linear in $a$). Assume
this. Thus, $a$ is an element of the basis $\left(  M_{\alpha}\right)
_{\alpha\in\operatorname{Comp} }$. In other words, there exists a $\beta
\in\operatorname{Comp} $ such that $a=M_{\beta}$. Consider this $\beta$. We
have $\mathbf{F}_{n}\underbrace{a}_{=M_{\beta}}=\mathbf{F}_{n}M_{\beta
}=M_{\beta\left\{  n\right\}  }$ (by (\ref{sol.NSym.verschiebung.e.f}),
applied to $\alpha=\beta$).

Recall that $\left(  H_{\alpha}\right)  _{\alpha\in\operatorname{Comp} }$ is a
basis of the $\kk$-module $\Lambda$. Hence, in proving
(\ref{sol.NSym.verschiebung.e.goal}), we can WLOG assume that $b$ is an
element of this basis $\left(  H_{\alpha}\right)  _{\alpha\in
\operatorname{Comp} }$ (because the equality
(\ref{sol.NSym.verschiebung.e.goal}) is $\kk$-linear in $b$). Assume
this. Thus, $b$ is an element of the basis $\left(  H_{\alpha}\right)
_{\alpha\in\operatorname{Comp} }$. In other words, there exists a $\gamma
\in\operatorname{Comp} $ such that $b=H_{\gamma}$. Consider this $\gamma$. We
have%
\begin{equation}
\left(  \underbrace{b}_{=H_{\gamma}},\underbrace{\mathbf{F}_{n}a}%
_{=M_{\beta\left\{  n\right\}  }}\right)  =\left(  H_{\gamma},M_{\beta\left\{
n\right\}  }\right)  =\delta_{\gamma,\beta\left\{  n\right\}  }
\label{sol.NSym.verschiebung.e.lhs}%
\end{equation}
(by (\ref{sol.NSym.verschiebung.e.duality}), applied to $\gamma$ and
$\beta\left\{  n\right\}  $ instead of $\alpha$ and $\beta$).

Let us write the composition $\gamma$ in the form $\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  $. Then, $\gamma=\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  $, so that $H_{\gamma}=H_{\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }=H_{\gamma_{1}}%
H_{\gamma_{2}}\cdots H_{\gamma_{\ell}}$ (by (\ref{sol.NSym.verschiebung.a.H}),
applied to $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  $
instead of $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $).

Let us first assume that%
\begin{equation}
\left(  \text{not every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{
satisfies }n\mid\gamma_{i}\right)  .
\label{sol.NSym.verschiebung.e.stupid-assumption}%
\end{equation}
Then, $\mathbf{V}_{n} \left(  \underbrace{H_{\gamma}}_{=H_{\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }}\right)  =\mathbf{V}_{n}
\left(  H_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)
}\right)  =0$ (by (\ref{sol.NSym.verschiebung.a.H1}), applied to $\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  $ instead of $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $). Also, $\beta\left\{
n\right\}  \neq\gamma$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary.
Thus, $\beta\left\{  n\right\}  =\gamma$. Let us write the composition $\beta$
in the form $\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  $. Then,
$\beta\left\{  n\right\}  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta
_{q}\right)  $ (by the definition of $\beta\left\{  n\right\}  $). Hence,
$\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  =\gamma
=\beta\left\{  n\right\}  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta
_{q}\right)  $. As a consequence, $\ell=q$, so that $q=\ell$ and thus $\left(
n\beta_{1},n\beta_{2},\ldots,n\beta_{q}\right)  =\left(  n\beta_{1},n\beta
_{2},\ldots,n\beta_{\ell}\right)  $. Hence, $\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{\ell}\right)  =\left(  n\beta_{1},n\beta_{2},\ldots
,n\beta_{q}\right)  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{\ell
}\right)  $. Consequently, every $i\in\left\{  1,2,\ldots,\ell\right\}  $
satisfies $\gamma_{i}=n\beta_{i}$. Thus, every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ satisfies $n\mid\gamma_{i}$. This contradicts
(\ref{sol.NSym.verschiebung.e.stupid-assumption}). This contradiction proves
that our assumption was wrong, qed.}, so that $\delta_{\gamma,\beta\left\{
n\right\}  }=0$. Thus, (\ref{sol.NSym.verschiebung.e.lhs}) becomes $\left(
b,\mathbf{F}_{n}a\right)  =\delta_{\gamma,\beta\left\{  n\right\}  }=0$.
Compared with $\left(  \mathbf{V}_{n} \underbrace{b}_{=H_{\gamma}},a\right)
=\left(  \underbrace{\mathbf{V}_{n} \left(  H_{\gamma}\right)  }%
_{=0},a\right)  =\left(  0,a\right)  =0$ (since the dual pairing
$\Nsym \otimes \Qsym \overset{\left(  \cdot
,\cdot\right)  }{\longrightarrow} { \kk } $ is $\kk$-bilinear),
this yields $\left(  b,\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_{n}
b,a\right)  $. Thus, (\ref{sol.NSym.verschiebung.e.goal}) holds.

Now, let us forget that we assumed that
(\ref{sol.NSym.verschiebung.e.stupid-assumption}) holds. We thus have proven
(\ref{sol.NSym.verschiebung.e.goal}) under the assumption that
(\ref{sol.NSym.verschiebung.e.stupid-assumption}) holds. Hence, for the rest
of our proof of (\ref{sol.NSym.verschiebung.e.goal}), we can WLOG assume that
(\ref{sol.NSym.verschiebung.e.stupid-assumption}) does not hold. Assume this.

We have assumed that (\ref{sol.NSym.verschiebung.e.stupid-assumption}) does
not hold. In other words,%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{ satisfies }%
n\mid\gamma_{i}. \label{sol.NSym.verschiebung.e.anti}%
\end{equation}
Thus, $\gamma_{i}/n$ is a positive integer for every $i\in\left\{
1,2,\ldots,\ell\right\}  $ (since $\gamma_{i}$ is a positive integer for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $). Thus, $\left(  \gamma_{1}%
/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)  $ is a composition. Let us
denote this composition by $\zeta$. We have $\zeta\left\{  n\right\}  =\gamma
$\ \ \ \ \footnote{\textit{Proof.} We have $\zeta=\left(  \gamma_{1}%
/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)  $. Hence, the definition of
$\zeta\left\{  n\right\}  $ yields $\zeta\left\{  n\right\}  =\left(
n\gamma_{1}/n,n\gamma_{2}/n,\ldots,n\gamma_{\ell}/n\right)  =\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  =\gamma$, qed.} and
$\mathbf{V}_{n} \left(  H_{\gamma}\right)  =H_{\zeta}$%
\ \ \ \ \footnote{\textit{Proof.} We have $\gamma=\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  $, so that%
\begin{align*}
\mathbf{V}_{n} \left(  H_{\gamma}\right)   &  =\mathbf{V}_{n} \left(
H_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }\right) \\
&  =H_{\left(  \gamma_{1}/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.a.H2}),
applied to }\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  \text{
instead of }\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  \right)
\\
&  =H_{\zeta}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \gamma
_{1}/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)  =\zeta\right)  ,
\end{align*}
qed.} and thus%
\begin{equation}
\left(  \mathbf{V}_{n} \underbrace{b}_{=H_{\gamma}},\underbrace{a}_{=M_{\beta
}}\right)  =\left(  \underbrace{\mathbf{V}_{n} \left(  H_{\gamma}\right)
}_{=H_{\zeta}},M_{\beta}\right)  =\left(  H_{\zeta},M_{\beta}\right)
=\delta_{\zeta,\beta} \label{sol.NSym.verschiebung.e.rhs}%
\end{equation}
(by (\ref{sol.NSym.verschiebung.e.duality}), applied to $\zeta$ instead of
$\alpha$).

Now, let us write the composition $\beta$ in the form $\left(  \beta_{1}%
,\beta_{2},\ldots,\beta_{s}\right)  $. Then, $\beta\left\{  n\right\}
=\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{s}\right)  $ (by the definition
of $\beta\left\{  n\right\}  $).

Now, we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \gamma=\beta\left\{  n\right\}  \right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \gamma_{1},\gamma_{2},\ldots
,\gamma_{\ell}\right)  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta
_{s}\right)  \right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma=\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  \text{ and }\beta\left\{  n\right\}
=\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{s}\right)  \right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{we have }\ell=s\text{, and every }%
i\in\left\{  1,2,\ldots,\ell\right\}  \text{ satisfies }\underbrace{\gamma
_{i}=n\beta_{i}}_{\substack{\text{this is equivalent to}\\\gamma_{i}%
/n=\beta_{i}}}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{we have }\ell=s\text{, and every }%
i\in\left\{  1,2,\ldots,\ell\right\}  \text{ satisfies }\gamma_{i}/n=\beta
_{i}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \gamma_{1}/n,\gamma_{2}/n,\ldots
,\gamma_{\ell}/n\right)  =\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)
\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \zeta=\beta\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \gamma_{1}/n,\gamma
_{2}/n,\ldots,\gamma_{\ell}/n\right)  =\zeta\text{ and }\left(  \beta
_{1},\beta_{2},\ldots,\beta_{s}\right)  =\beta\right)
.\label{sol.NSym.verschiebung.e.equiv}%
\end{align}
This equivalence shows that $\delta_{\gamma,\beta\left\{  n\right\}  }%
=\delta_{\zeta,\beta}$. But (\ref{sol.NSym.verschiebung.e.lhs}) becomes%
\[
\left(  b,\mathbf{F}_{n}a\right)  =\delta_{\gamma,\beta\left\{  n\right\}
}=\delta_{\zeta,\beta}=\left(  \mathbf{V}_{n}b,a\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.e.rhs}%
)}\right)  .
\]
Thus, (\ref{sol.NSym.verschiebung.e.goal}) is proven. As we know, this
completes the solution of Exercise \ref{exe.NSym.verschiebung}(f).

(g) We know that $\pi: \Nsym \rightarrow\Lambda$ is a
$\kk$-algebra homomorphism. Thus, $\pi\left(  0\right)  =0$. Also,
we know that
\begin{equation}
\pi\left(  H_{n}\right)  =h_{n}\ \ \ \ \ \ \ \ \ \ \text{for every positive
integer }n. \label{sol.NSym.verschiebung.h.pi}%
\end{equation}
Now, let $n$ be a positive integer. The maps $\mathbf{v}_{n}\circ\pi$ and
$\pi\circ\mathbf{V}_{n} $ are $\kk$-algebra homomorphisms (since
$\pi$ and $\mathbf{v}_{n}$ are $\kk$-algebra homomorphisms).

Let $r$ be a positive integer. Applying (\ref{sol.NSym.verschiebung.h.pi}) to
$r$ instead of $n$, we obtain $\pi\left(  H_{r}\right)  =h_{r}$.

If $n\mid r$, then $r/n$ is a positive integer. Hence,%
\begin{equation}
\text{if }n\mid r\text{, then }\pi\left(  H_{r/n}\right)  =h_{r/n}.
\label{sol.NSym.verschiebung.h.pi2}%
\end{equation}


Now,%
\[
\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  H_{r}\right)  =\mathbf{v}%
_{n}\left(  \underbrace{\pi\left(  H_{r}\right)  }_{=h_{r}}\right)
=\mathbf{v}_{n}\left(  h_{r}\right)  =
\begin{cases}
h_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{v}_{n}\right)  .
\]
Compared with%
\begin{align*}
\left(  \pi\circ\mathbf{V}_{n} \right)  \left(  H_{r}\right)   &  =\pi\left(
\underbrace{\mathbf{V}_{n} \left(  H_{r}\right)  }_{\substack{=
\begin{cases}
H_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\\\text{(by the definition of }\mathbf{V}_{n} \text{)}}}\right)
=\pi\left(
\begin{cases}
H_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\right) \\
&  =
\begin{cases}
\pi\left(  H_{r/n}\right)  ,& \text{if }n\mid r;\\
\pi\left(  0\right)  ,& \text{if }n\nmid r
\end{cases} \quad
=
\begin{cases}
\pi\left(  H_{r/n}\right)  ,& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\left(  0\right)  =0\text{ in
the case when }n\nmid r\right) \\
&  =
\begin{cases}
h_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\left(  H_{r/n}\right)
=h_{r/n}\text{ in the case when }n\mid r\text{ (according to
(\ref{sol.NSym.verschiebung.h.pi2}))}\right)  ,
\end{align*}
this yields $\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  H_{r}\right)
=\left(  \pi\circ\mathbf{V}_{n} \right)  \left(  H_{r}\right)  $.

Let us now forget that we fixed $r$. We thus have proven that%
\begin{equation}
\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  H_{r}\right)  =\left(  \pi
\circ\mathbf{V}_{n} \right)  \left(  H_{r}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }r\text{.}
\label{sol.NSym.verschiebung.h.main}%
\end{equation}


Now, recall that the family $\left(  H_{r}\right)  _{r\geq1}$ generates the
$\kk$-algebra $\Nsym$ (according to
(\ref{sol.NSym.verschiebung.gen})). In other words, $\left(  H_{r}\right)
_{r\geq1}$ is a generating set of the $\kk$-algebra
$\Nsym$. The two $\kk$-algebra homomorphisms
$\mathbf{v}_{n}\circ\pi$ and $\pi\circ\mathbf{V}_{n} $ are equal to each other
on this generating set (according to (\ref{sol.witt.ghost-app.d.main})), and
therefore must be identical (because if two $\kk$-algebra
homomorphisms from the same domain are equal to each other on a generating set
of their domain, then these two homomorphisms must be identical). In other
words, $\mathbf{v}_{n}\circ\pi=\pi\circ\mathbf{V}_{n} $. This completes the
solution of Exercise \ref{exe.NSym.verschiebung}(g).

(h) \textit{Alternative solution of Exercise~\ref{exe.Lambda.verschiebung}%
(f).} Let $i$ denote the inclusion map $\Lambda\rightarrow \Qsym 
$. Then, Corollary \ref{cor.NSym.pi} yields that the map $\pi$ is adjoint to
the map $i$ with respect to the dual pairing $\Nsym
\otimes \Qsym \overset{\left(  \cdot,\cdot\right)
}{\longrightarrow} { \kk } $. In other words,%
\begin{equation}
\left(  \pi\left(  b\right)  ,a\right)  =\left(  b,i\left(  a\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }b\in\Nsym \text{ and }%
a\in\Lambda. \label{sol.NSym.verschiebung.realh.adj}%
\end{equation}


But Exercise \ref{exe.NSym.verschiebung}(f) yields that the maps
$\mathbf{F}_{n} : \Qsym \rightarrow \Qsym$ and
$\mathbf{V}_{n} : \Nsym \rightarrow \Nsym$ are
adjoint with respect to the dual pairing $\Nsym \otimes
\Qsym \overset{\left(  \cdot,\cdot\right)  }{\longrightarrow}
{ \kk } $. In other words,%
\begin{equation}
\left(  b,\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_{n} b,a\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in \Qsym \text{ and }%
b\in\Nsym . \label{sol.NSym.verschiebung.realh.adj2}%
\end{equation}


Now, fix a positive integer $n$. Let $a\in\Lambda$ and $b\in\Lambda$ be
arbitrary. Then, $a\in\Lambda\subset \Qsym $. Exercise
\ref{exe.QSym.frobenius}(f) yields $\mathbf{F}_{n}\mid_{\Lambda}%
=\mathbf{f}_{n}$. Thus, $\mathbf{f}_{n}=\mathbf{F}_{n}\mid_{\Lambda}$, so that
$\mathbf{f}_{n}a=\left(  \mathbf{F}_{n}\mid_{\Lambda}\right)  a=\mathbf{F}%
_{n}a$.

On the other hand, the projection $\pi: \Nsym \rightarrow
\Lambda$ is surjective. Hence, there exists some $b^{\prime}\in
\Nsym $ satisfying $b=\pi\left(  b^{\prime}\right)  $. Consider
this $b^{\prime}$.

Since the Hall inner product on $\Lambda$ is symmetric, we have%
\begin{align*}
\left(  \mathbf{f}_{n}a,b\right)   &  =\left(  \underbrace{b}_{=\pi\left(
b^{\prime}\right)  },\mathbf{f}_{n}a\right)  =\left(  \pi\left(  b^{\prime
}\right)  ,\mathbf{f}_{n}a\right)  =\left(  b^{\prime},\underbrace{i\left(
\mathbf{f}_{n}a\right)  }_{\substack{=\mathbf{f}_{n}a\\\text{(since }i\text{
is an inclusion map)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.realh.adj}%
), applied to }b^{\prime}\text{ and }\mathbf{f}_{n}a\text{ instead of }b\text{
and }a\right) \\
&  =\left(  b^{\prime},\underbrace{\mathbf{f}_{n}a}_{=\mathbf{F}_{n}a}\right)
=\left(  b^{\prime},\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_{n} b^{\prime
},a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.NSym.verschiebung.realh.adj2}), applied to }b^{\prime}\text{ instead
of }b\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  a,\mathbf{v}_{n}b\right)   &  =\left(  \mathbf{v}_{n}\underbrace{b}%
_{=\pi\left(  b^{\prime}\right)  },a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since the Hall inner product on }\Lambda\text{ is symmetric}\right) \\
&  =\left(  \underbrace{\mathbf{v}_{n}\left(  \pi\left(  b^{\prime}\right)
\right)  }_{=\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  b^{\prime}\right)
},a\right)  =\left(  \underbrace{\left(  \mathbf{v}_{n}\circ\pi\right)
}_{\substack{=\pi\circ\mathbf{V}_{n} \\\text{(by Exercise
\ref{exe.NSym.verschiebung}(g))}}}\left(  b^{\prime}\right)  ,a\right)
=\left(  \underbrace{\left(  \pi\circ\mathbf{V}_{n} \right)  \left(
b^{\prime}\right)  }_{=\pi\left(  \mathbf{V}_{n} b^{\prime}\right)  },a\right)
\\
&  =\left(  \pi\left(  \mathbf{V}_{n} b^{\prime}\right)  ,a\right)  =\left(
\mathbf{V}_{n} b^{\prime},\underbrace{i\left(  a\right)  }%
_{\substack{=a\\\text{(since }i\text{ is an inclusion map)}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.realh.adj}),
applied to }\mathbf{V}_{n} b^{\prime}\text{ instead of }b\right) \\
&  =\left(  \mathbf{V}_{n} b^{\prime},a\right)  ,
\end{align*}
we obtain $\left(  \mathbf{f}_{n}a,b\right)  =\left(  a,\mathbf{v}%
_{n}b\right)  $.

Let us now forget that we fixed $a$ and $b$. We have thus shown that $\left(
\mathbf{f}_{n}a,b\right)  =\left(  a,\mathbf{v}_{n}b\right)  $ for every
$a\in\Lambda$ and $b\in\Lambda$. In other words, the maps $\mathbf{f}%
_{n}:\Lambda\rightarrow\Lambda$ and $\mathbf{v}_{n}:\Lambda\rightarrow\Lambda$
are adjoint with respect to the Hall inner product on $\Lambda$. Thus,
Exercise~\ref{exe.Lambda.verschiebung}(f) is solved once again. Hence,
Exercise \ref{exe.NSym.verschiebung}(h) is solved.
%\end{solutions}



\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.NSym.verschiebung}}

\textit{Solution to Exercise \ref{exe.NSym.verschiebung}.} Let us first notice
that every positive integer $n$ satisfies%
\begin{equation}
\mathbf{V}_n \left(  H_{m}\right)  =
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in \NN 
\label{sol.NSym.verschiebung.vnhm}
\end{equation}
\footnote{\textit{Proof.} To obtain a proof of
(\ref{sol.NSym.verschiebung.vnhm}), it is enough to repeat the proof of
(\ref{sol.Lambda.verschiebung.vnhm}), making just the following changes:
\par
\begin{itemize}
\item replacing every $h_{i}$ by $H_{i}$;
\par
\item replacing $\mathbf{v}_{n}$ by $\mathbf{V}_n $.
\end{itemize}
}.

Recall that $\Nsym$ is (isomorphic to) the free associative
algebra with generators $H_{1},H_{2},H_{3},\ldots$ (according to
(\ref{eq.thm.NSym.free})). Hence, the elements $H_{1},H_{2},H_{3},\ldots$
generate the $\kk$-algebra $\Nsym$. In other words,%
\begin{equation}
\text{the family }\left(  H_{r}\right)  _{r\geq1}\text{ generates the }
 \kk \text{-algebra } \Nsym .
\label{sol.NSym.verschiebung.gen}
\end{equation}


(e) A solution to Exercise \ref{exe.NSym.verschiebung}(e) can be obtained by
copying the solution of Exercise \ref{exe.Lambda.verschiebung}(e) and making
the following changes:

\begin{itemize}
\item Replace every appearance of $\mathbf{v}_{n}$ by $\mathbf{V}_n $.

\item Replace every appearance of $\Lambda$ by $\Nsym$.

\item Replace every appearance of $h_{j}$ (for some $j\in \NN$) by
$H_{j}$.

\item Replace the reference to Proposition
\ref{symm-comultiplication-formulas-prop}(iii) by a reference to
(\ref{Nsym-coproduct-on-H}).

\item Replace the reference to Proposition \ref{symm-antipode-and-h-basis} by
a reference to (\ref{sol.NSym.verschiebung.gen}).

\item Replace every reference to (\ref{sol.Lambda.verschiebung.vnhm}) by a
reference to (\ref{sol.NSym.verschiebung.vnhm}).
\end{itemize}

(c) A solution to Exercise \ref{exe.NSym.verschiebung}(c) can be obtained by
copying the solution of Exercise \ref{exe.Lambda.verschiebung}(c) and making
the following changes:

\begin{itemize}
\item Replace every appearance of $\mathbf{v}_{j}$ (for some positive integer
$j$) by $\mathbf{V}_{j}$.

\item Replace every appearance of $\Lambda$ by $\Nsym$.

\item Replace every appearance of $h_{j}$ (for some $j\in \NN$) by
$H_{j}$.

\item Replace the reference to Proposition \ref{symm-antipode-and-h-basis} by
a reference to (\ref{sol.NSym.verschiebung.gen}).
\end{itemize}

(d) A solution to Exercise \ref{exe.NSym.verschiebung}(d) can be obtained by
copying the solution of Exercise \ref{exe.Lambda.verschiebung}(d) and making
the following changes:

\begin{itemize}
\item Replace every appearance of $\mathbf{v}_{1}$ by $\mathbf{V}_1 $.

\item Replace every appearance of $\Lambda$ by $\Nsym$.

\item Replace every appearance of $h_{j}$ (for some $j\in \NN$) by
$H_{j}$.

\item Replace the reference to Proposition \ref{symm-antipode-and-h-basis} by
a reference to (\ref{sol.NSym.verschiebung.gen}).
\end{itemize}

(a) Here is one of several possible solutions of Exercise
\ref{exe.NSym.verschiebung}(a):\footnote{Another solution (which the reader
can easily find) proceeds by making some relatively straightforward
modifications to the solution of Exercise \ref{exe.Lambda.verschiebung}(a).
(One needs to keep in mind that $\Nsym$, unlike $\Lambda$, is
not commutative -- but this does not prevent us from doing things such as
substituting $t^{n}$ for $t$ in a power series over $\Nsym$
(since $t^{n}$ is a central element of $ \Nsym \left[  \left[
t\right]  \right]  $).)} Define a $\kk$-linear map
$E: \Nsym \rightarrow \Nsym $ as in Exercise
\ref{exe.dynkin} (but with $\Nsym$ instead of $A$). Every
positive integer $n$ satisfies%
\begin{equation}
\Psi_{n}=\left(  S\star E\right)  \left(  H_{n}\right)
\label{sol.NSym.verschiebung.e.Psin}%
\end{equation}
(according to Exercise \ref{exe.Psi}(c)).

Fix a positive integer $n$. Let us make some auxiliary observations first:

\begin{itemize}
\item Every composition $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $ satisfies%
\begin{equation}
H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }=H_{\alpha_{1}%
}H_{\alpha_{2}}\cdots H_{\alpha_{\ell}}. \label{sol.NSym.verschiebung.a.H}%
\end{equation}
\footnote{This is precisely the equality (\ref{eq.Nsym.Halpha}), and has been
proven before.}

\item If $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ is a
composition such that $\left(  \text{not every }i\in\left\{  1,2,\ldots
,\ell\right\}  \text{ satisfies }n\mid\alpha_{i}\right)  $, then%
\begin{equation}
\mathbf{V}_n \left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)  =0. \label{sol.NSym.verschiebung.a.H1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.a.H1}):} Let $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be a composition such
that $\left(  \text{not every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{
satisfies }n\mid\alpha_{i}\right)  $. Then, there exists an $i\in\left\{
1,2,\ldots,\ell\right\}  $ satisfying $n\nmid\alpha_{i}$ (since not every
$i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies $n\mid\alpha_{i}$).
Consider this $i$. We have
\begin{align*}
\mathbf{V}_n \left(  H_{\alpha_{i}}\right)   &  =
\begin{cases}
H_{\alpha_{i}/n},& \text{if }n\mid\alpha_{i};\\
0,& \text{if }n\nmid\alpha_{i}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{V}_n \left(  H_{\alpha_{i}}\right)  \right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\nmid\alpha_{i}\right)  .
\end{align*}
But (\ref{sol.NSym.verschiebung.a.H}) yields $H_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  }=H_{\alpha_{1}}H_{\alpha_{2}}\cdots
H_{\alpha_{\ell}}$. Applying $\mathbf{V}_n $ to both sides of this equality,
we obtain%
\[
\mathbf{V}_n \left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)  =\mathbf{V}_n \left(  H_{\alpha_{1}}H_{\alpha_{2}}\cdots
H_{\alpha_{\ell}}\right)  =\mathbf{V}_n \left(  H_{\alpha_{1}}\right)
\mathbf{V}_n \left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}_n \left(
H_{\alpha_{\ell}}\right)
\]
(since $\mathbf{V}_n $ is a $\kk$-algebra homomorphism).
\par
But $\mathbf{V}_n \left(  H_{\alpha_{i}}\right)  $ is a factor of the product
$\mathbf{V}_n \left(  H_{\alpha_{1}}\right)  \mathbf{V}_n \left(
H_{\alpha_{2}}\right)  \cdots\mathbf{V}_n \left(  H_{\alpha_{\ell}}\right)
$, and equals $0$ (since $\mathbf{V}_n \left(  H_{\alpha_{i}}\right)  =0$).
Hence, the product $\mathbf{V}_n \left(  H_{\alpha_{1}}\right)
\mathbf{V}_n \left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}_n \left(
H_{\alpha_{\ell}}\right)  $ has at least one factor which is equal to $0$. As
a consequence, this product $\mathbf{V}_n \left(  H_{\alpha_{1}}\right)
\mathbf{V}_n \left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}_n \left(
H_{\alpha_{\ell}}\right)  $ is $0$ (because if a product has one factor which
is equal to zero, then the product must be $0$). In other words,
$\mathbf{V}_n \left(  H_{\alpha_{1}}\right)  \mathbf{V}_n \left(
H_{\alpha_{2}}\right)  \cdots\mathbf{V}_n \left(  H_{\alpha_{\ell}}\right)
=0$, so that $\mathbf{V}_n \left(  H_{\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  }\right)  =\mathbf{V}_n \left(  H_{\alpha_{1}%
}\right)  \mathbf{V}_n \left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}%
_{n}\left(  H_{\alpha_{\ell}}\right)  =0$. This proves
(\ref{sol.NSym.verschiebung.a.H1}).}

\item If $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ is a
composition such that $\left(  \text{every }i\in\left\{  1,2,\ldots
,\ell\right\}  \text{ satisfies }n\mid\alpha_{i}\right)  $, then%
\begin{equation}
\mathbf{V}_n \left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)  =H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }. \label{sol.NSym.verschiebung.a.H2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.a.H2}):} Let $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ be a composition such
that $\left(  \text{every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{
satisfies }n\mid\alpha_{i}\right)  $. Then, $\alpha_{i}/n$ is a positive
integer for every $i\in\left\{  1,2,\ldots,\ell\right\}  $ (since $n\mid
\alpha_{i}$ for every $i\in\left\{  1,2,\ldots,\ell\right\}  $). Hence,
$\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  $ is a
composition. Therefore, (\ref{sol.NSym.verschiebung.a.H}) (applied to $\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  $ instead of $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $) yields $H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }=H_{\alpha_{1}%
/n}H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}$.
\par
But every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies%
\begin{align*}
\mathbf{V}_n \left(  H_{\alpha_{i}}\right)   &  =
\begin{cases}
H_{\alpha_{i}/n},& \text{if }n\mid\alpha_{i};\\
0,& \text{if }n\nmid\alpha_{i}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{V}_n \left(  H_{\alpha_{i}}\right)  \right) \\
&  =H_{\alpha_{i}/n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\mid\alpha
_{i}\right)  .
\end{align*}
Multiplying these equalities for all $i\in\left\{  1,2,\ldots,\ell\right\}  $,
we obtain
\[
\mathbf{V}_n \left(  H_{\alpha_{1}}\right)  \mathbf{V}_n \left(
H_{\alpha_{2}}\right)  \cdots\mathbf{V}_n \left(  H_{\alpha_{\ell}}\right)
=H_{\alpha_{1}/n}H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}.
\]
\par
But (\ref{sol.NSym.verschiebung.a.H}) yields $H_{\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  }=H_{\alpha_{1}}H_{\alpha_{2}}\cdots
H_{\alpha_{\ell}}$. Applying $\mathbf{V}_n $ to both sides of this equality,
we obtain%
\begin{align*}
\mathbf{V}_n \left(  H_{\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }\right)   &  =\mathbf{V}_n \left(  H_{\alpha_{1}}H_{\alpha_{2}%
}\cdots H_{\alpha_{\ell}}\right)  =\mathbf{V}_n \left(  H_{\alpha_{1}%
}\right)  \mathbf{V}_n \left(  H_{\alpha_{2}}\right)  \cdots\mathbf{V}%
_{n}\left(  H_{\alpha_{\ell}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{V}_n \text{ is a
} \kk \text{-algebra homomorphism}\right) \\
&  =H_{\alpha_{1}/n}H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}=H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }%
\end{align*}
(since $H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)
}=H_{\alpha_{1}/n}H_{\alpha_{2}/n}\cdots H_{\alpha_{\ell}/n}$). This proves
(\ref{sol.NSym.verschiebung.a.H2}).}

\item We have%
\begin{equation}
\mathbf{V}_n \circ E=n\cdot\left(  E\circ\mathbf{V}_n \right)
\label{sol.NSym.verschiebung.a.E}%
\end{equation}
(as endomorphisms of $\Nsym$).\ \ \ \ \footnote{\textit{Proof
of (\ref{sol.NSym.verschiebung.a.E}):} Let us show that%
\begin{equation}
\left(  \mathbf{V}_n \circ E\right)  \left(  H_{\alpha}\right)  =\left(
n\cdot\left(  E\circ\mathbf{V}_n \right)  \right)  \left(  H_{\alpha}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\alpha\in \Comp .
\label{sol.NSym.verschiebung.a.E.pf.1}%
\end{equation}
\par
\textit{Proof of (\ref{sol.NSym.verschiebung.a.E.pf.1}):} Let $\alpha
\in \Comp $. Then, $H_{\alpha}$ is a homogeneous element of
$\Nsym$ of degree $\deg\left(  H_{\alpha}\right)  =\left\vert
\alpha\right\vert $. Thus, the definition of $E\left(  H_{\alpha}\right)  $
yields $E\left(  H_{\alpha}\right)  =\underbrace{\left(  \deg\left(
H_{\alpha}\right)  \right)  }_{=\left\vert \alpha\right\vert }\cdot H_{\alpha
}=\left\vert \alpha\right\vert \cdot H_{\alpha}$. Now,%
\[
\left(  \mathbf{V}_n \circ E\right)  \left(  H_{\alpha}\right)
=\mathbf{V}_n \left(  \underbrace{E\left(  H_{\alpha}\right)  }_{=\left\vert
\alpha\right\vert \cdot H_{\alpha}}\right)  =\mathbf{V}_n \left(  \left\vert
\alpha\right\vert \cdot H_{\alpha}\right)  =\left\vert \alpha\right\vert
\cdot\mathbf{V}_n \left(  H_{\alpha}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\mathbf{V}_n \text{ is } \kk \text{-linear}\right)
.
\]
\par
Now, let us write the composition $\alpha$ in the form $\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. We distinguish between two
cases:
\par
\textit{Case 1:} Not every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies
$n\mid\alpha_{i}$.
\par
\textit{Case 2:} Every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies
$n\mid\alpha_{i}$.
\par
Let us first consider Case 1. In this case, not every $i\in\left\{
1,2,\ldots,\ell\right\}  $ satisfies $n\mid\alpha_{i}$. Thus,
(\ref{sol.NSym.verschiebung.a.H1}) yields $\mathbf{V}_n \left(  H_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }\right)  =0$. Since
$\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  =\alpha$, this
rewrites as $\mathbf{V}_n \left(  H_{\alpha}\right)  =0$. Thus, $\left(
\mathbf{V}_n \circ E\right)  \left(  H_{\alpha}\right)  =\left\vert
\alpha\right\vert \cdot\underbrace{\mathbf{V}_n \left(  H_{\alpha}\right)
}_{=0}=0$. Compared with
\[
\left(  n\cdot\left(  E\circ\mathbf{V}_n \right)  \right)  \left(  H_{\alpha
}\right)  =n\cdot\underbrace{\left(  E\circ\mathbf{V}_n \right)  \left(
H_{\alpha}\right)  }_{=E\left(  \mathbf{V}_n \left(  H_{\alpha}\right)
\right)  }=n\cdot E\left(  \underbrace{\mathbf{V}_n \left(  H_{\alpha
}\right)  }_{=0}\right)  =n\cdot\underbrace{E\left(  0\right)  }%
_{\substack{=0\\\text{(since }E\text{ is } \kk \text{-linear)}}}=0,
\]
this yields $\left(  \mathbf{V}_n \circ E\right)  \left(  H_{\alpha}\right)
=\left(  n\cdot\left(  E\circ\mathbf{V}_n \right)  \right)  \left(
H_{\alpha}\right)  $. Hence, (\ref{sol.NSym.verschiebung.a.E.pf.1}) is proven
in Case 1.
\par
Let us now consider Case 2. In this case, every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ satisfies $n\mid\alpha_{i}$. Thus,
(\ref{sol.NSym.verschiebung.a.H2}) yields $\mathbf{V}_n \left(  H_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }\right)  =H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }$. Since $\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  =\alpha$, this rewrites as
$\mathbf{V}_n \left(  H_{\alpha}\right)  =H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }$. Thus,
\begin{equation}
\left(  \mathbf{V}_n \circ E\right)  \left(  H_{\alpha}\right)  =\left\vert
\alpha\right\vert \cdot\underbrace{\mathbf{V}_n \left(  H_{\alpha}\right)
}_{=H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }%
}=\left\vert \alpha\right\vert \cdot H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }.
\label{sol.NSym.verschiebung.a.E.pf.1.pf.1}%
\end{equation}
\par
On the other hand, $\left\vert \alpha\right\vert =\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell}$ (since $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  $) and%
\[
\left\vert \left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)
\right\vert =\alpha_{1}/n+\alpha_{2}/n+\cdots+\alpha_{\ell}%
/n=\underbrace{\left(  \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}\right)
}_{=\left\vert \alpha\right\vert }/n=\left\vert \alpha\right\vert /n.
\]
Recall that $H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }$ is a homogeneous element of $\Nsym$ of degree
$\deg\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }\right)  =\left\vert \left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  \right\vert $. Hence, the definition of
$E\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)
}\right)  $ yields $E\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots
,\alpha_{\ell}/n\right)  }\right)  =\underbrace{\left(  \deg\left(  H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }\right)  \right)
}_{=\left\vert \left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  \right\vert =\left\vert \alpha\right\vert /n}\cdot H_{\left(
\alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell}/n\right)  }=\left(  \left\vert
\alpha\right\vert /n\right)  \cdot H_{\left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  }$. Now,%
\begin{align*}
\left(  n\cdot\left(  E\circ\mathbf{V}_n \right)  \right)  \left(  H_{\alpha
}\right)   &  =n\cdot\underbrace{\left(  E\circ\mathbf{V}_n \right)  \left(
H_{\alpha}\right)  }_{=E\left(  \mathbf{V}_n \left(  H_{\alpha}\right)
\right)  }=n\cdot E\left(  \underbrace{\mathbf{V}_n \left(  H_{\alpha
}\right)  }_{=H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots,\alpha_{\ell
}/n\right)  }}\right) \\
&  =n\cdot\underbrace{E\left(  H_{\left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  }\right)  }_{=\left(  \left\vert
\alpha\right\vert /n\right)  \cdot H_{\left(  \alpha_{1}/n,\alpha_{2}%
/n,\ldots,\alpha_{\ell}/n\right)  }}=n\cdot\left(  \left\vert \alpha
\right\vert /n\right)  \cdot H_{\left(  \alpha_{1}/n,\alpha_{2}/n,\ldots
,\alpha_{\ell}/n\right)  }\\
&  =\left\vert \alpha\right\vert \cdot H_{\left(  \alpha_{1}/n,\alpha
_{2}/n,\ldots,\alpha_{\ell}/n\right)  }.
\end{align*}
Compared with (\ref{sol.NSym.verschiebung.a.E.pf.1.pf.1}), this yields
$\left(  \mathbf{V}_n \circ E\right)  \left(  H_{\alpha}\right)  =\left(
n\cdot\left(  E\circ\mathbf{V}_n \right)  \right)  \left(  H_{\alpha}\right)
$. Hence, (\ref{sol.NSym.verschiebung.a.E.pf.1}) is proven in Case 2.
\par
Thus, (\ref{sol.NSym.verschiebung.a.E.pf.1}) is proven in both Cases 1 and 2.
Since these two Cases cover all possibilities, this shows that
(\ref{sol.NSym.verschiebung.a.E.pf.1}) always holds.
\par
Now, $\left(  H_{\alpha}\right)  _{\alpha\in \Comp }$ is a basis
of the $\kk$-module $\Nsym$. The equality
(\ref{sol.NSym.verschiebung.a.E.pf.1}) shows that the two $\kk$-linear
maps $\mathbf{V}_n \circ E$ and $n\cdot\left(  E\circ\mathbf{V}_n \right)  $
are equal to each other on every element of this basis. Hence, these two maps
$\mathbf{V}_n \circ E$ and $n\cdot\left(  E\circ\mathbf{V}_n \right)  $ are
identical (because if two $\kk$-linear maps having the same domain are
equal to each other on every element of some basis of this domain, then these
two maps must be identical). In other words, $\mathbf{V}_n \circ
E=n\cdot\left(  E\circ\mathbf{V}_n \right)  $. This proves
(\ref{sol.NSym.verschiebung.a.E}).}

\item We have%
\begin{equation}
\mathbf{V}_n \circ\left(  S\star E\right)  =n\cdot\left(  \left(  S\star
E\right)  \circ\mathbf{V}_n \right)
\label{sol.NSym.verschiebung.a.dyn}
\end{equation}
(where $S$, as usual, denotes the antipode of $\Nsym$).
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.a.dyn}):}
We have $S\star E=m\circ\left(  S\otimes E\right)  \circ\Delta$ (by the
definition of convolution). But we know (from Exercise
\ref{exe.NSym.verschiebung}(e)) that $\mathbf{V}_n$ is a Hopf algebra
homomorphism; thus, $\mathbf{V}_n \circ S=S\circ\mathbf{V}_n $. Also,
$\mathbf{V}_n $ is a $\kk$-coalgebra homomorphism (since
$\mathbf{V}_n $ is a Hopf algebra homomorphism); therefore, $\left(
\mathbf{V}_n \otimes\mathbf{V}_n \right)  \circ\Delta=\Delta\circ
\mathbf{V}_n $. Finally, $\mathbf{V}_n $ is a $\kk$-algebra
homomorphism; thus, $\mathbf{V}_n \circ m=m\circ\left(  \mathbf{V}_n %
\otimes\mathbf{V}_n \right)  $. Now,
\begin{align*}
\mathbf{V}_n \circ\left(  \underbrace{S\star E}_{=m\circ\left(  S\otimes
E\right)  \circ\Delta}\right)   &  =\underbrace{\mathbf{V}_n \circ
m}_{=m\circ\left(  \mathbf{V}_n \otimes\mathbf{V}_n \right)  }\circ\left(
S\otimes E\right)  \circ\Delta=m\circ\underbrace{\left(  \mathbf{V}_n %
\otimes\mathbf{V}_n \right)  \circ\left(  S\otimes E\right)  }_{=\left(
\mathbf{V}_n \circ S\right)  \otimes\left(  \mathbf{V}_n \circ E\right)
}\circ\Delta\\
&  =m\circ\left(  \underbrace{\left(  \mathbf{V}%
_{n}\circ S\right)  }_{=S\circ\mathbf{V}_n }
\otimes
\underbrace{\left(  \mathbf{V}_n \circ E\right)
}_{\substack{=n\cdot\left(  E\circ\mathbf{V}_n \right)  \\\text{(by
(\ref{sol.NSym.verschiebung.a.E}))}}}
\right)  \circ\Delta
=m\circ\underbrace{\left(
\left(  S\circ\mathbf{V}_n \right)
\otimes
\left(  n\cdot\left(  E\circ\mathbf{V}_n \right) \right)
\right)  }%
_{\substack{=n\cdot\left(  \left(  S\circ\mathbf{V}_n \right)  \otimes\left(
E\circ\mathbf{V}_n \right)  \right)  \\\text{(since tensoring }%
 \kk \text{-linear maps is a } \kk \text{-bilinear operation)}%
}}\circ\Delta\\
&  =m\circ\left(  n\cdot\underbrace{\left(  \left(  S\circ\mathbf{V}%
_{n}\right)  \otimes\left(  E\circ\mathbf{V}_n \right)  \right)  }_{=\left(
S\otimes E\right)  \circ\left(  \mathbf{V}_n \otimes\mathbf{V}_n \right)
}\right)  \circ\Delta=\underbrace{m\circ\left(  n\cdot\left(  S\otimes
E\right)  \circ\left(  \mathbf{V}_n \otimes\mathbf{V}_n \right)  \right)
}_{\substack{=n\cdot\left(  m\circ\left(  S\otimes E\right)  \circ\left(
\mathbf{V}_n \otimes\mathbf{V}_n \right)  \right)  \\\text{(since
composition of } \kk \text{-linear maps is a } \kk \text{-bilinear
operation)}}}\circ\Delta\\
&  =\left(  n\cdot\left(  m\circ\left(  S\otimes E\right)  \circ\left(
\mathbf{V}_n \otimes\mathbf{V}_n \right)  \right)  \right)  \circ
\Delta=n\cdot\left(  m\circ\left(  S\otimes E\right)  \circ\underbrace{\left(
\mathbf{V}_n \otimes\mathbf{V}_n \right)  \circ\Delta}_{=\Delta
\circ\mathbf{V}_n }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of } \kk %
\text{-linear maps is a } \kk \text{-bilinear operation}\right) \\
&  =n\cdot\left(  \underbrace{m\circ\left(  S\otimes E\right)  \circ\Delta
}_{\substack{=S\star E\\\text{(since }S\star E=m\circ\left(  S\otimes
E\right)  \circ\Delta\text{)}}}\circ\mathbf{V}_n \right)  =n\cdot\left(
\left(  S\star E\right)  \circ\mathbf{V}_n \right)  .
\end{align*}
This proves (\ref{sol.NSym.verschiebung.a.dyn}).}
\end{itemize}

But fix a positive integer $m$. We have $\Psi_{m}=\left(  S\star E\right)
\left(  H_{m}\right)  $ (by (\ref{sol.NSym.verschiebung.e.Psin}), applied to
$m$ instead of $n$) and thus%
\begin{align}
\mathbf{V}_n \left(  \underbrace{\Psi_{m}}_{=\left(  S\star E\right)  \left(
H_{m}\right)  }\right)   &  =\mathbf{V}_n \left(  \left(  S\star E\right)
\left(  H_{m}\right)  \right)  =\underbrace{\left(  \mathbf{V}_n \circ\left(
S\star E\right)  \right)  }_{\substack{=n\cdot\left(  \left(  S\star E\right)
\circ\mathbf{V}_n \right)  \\\text{(by (\ref{sol.NSym.verschiebung.a.dyn}))}%
}}\left(  H_{m}\right) \nonumber\\
&  =\left(  n\cdot\left(  \left(  S\star E\right)  \circ\mathbf{V}_n \right)
\right)  \left(  H_{m}\right)  =n\cdot\underbrace{\left(  \left(  S\star
E\right)  \circ\mathbf{V}_n \right)  \left(  H_{m}\right)  }_{=\left(  S\star
E\right)  \left(  \mathbf{V}_n \left(  H_{m}\right)  \right)  }\nonumber\\
&  =n\cdot\left(  S\star E\right)  \left(  \underbrace{\mathbf{V}_n \left(
H_{m}\right)  }_{\substack{=
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\\\text{(by (\ref{sol.NSym.verschiebung.vnhm}))}}}\right) \nonumber\\
&  =n\cdot\left(  S\star E\right)  \left(
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
\right)  . \label{sol.NSym.verschiebung.a.almost}%
\end{align}


Now, if $n\nmid m$, then Exercise \ref{exe.NSym.verschiebung}(a) is easily
seen to hold\footnote{\textit{Proof.} Assume that $n\nmid m$. Then,
(\ref{sol.NSym.verschiebung.a.almost}) becomes%
\[
\mathbf{V}_n \left(  \Psi_{m}\right)  =n\cdot\left(  S\star E\right)  \left(
\underbrace{
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid m\text{)}}}\right)
=n\cdot\underbrace{\left(  S\star E\right)  \left(  0\right)  }%
_{\substack{=0\\\text{(since }S\star E\text{ is } \kk \text{-linear)}%
}}=0.
\]
Compared with%
\[
\begin{cases}
n\Psi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\nmid m\right)  ,
\]
this yields $\mathbf{V}_n \left(  \Psi_{m}\right)  =
\begin{cases}
n\Psi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad$.
Hence, Exercise \ref{exe.NSym.verschiebung}(a) is solved (under
the assumption that $n\nmid m$), qed.}. Hence, for the rest of this solution,
we can WLOG assume that we don't have $n\nmid m$. Assume this.

We have $n\mid m$ (since we don't have $n\nmid m$). Hence, $m/n$ is a positive
integer. Thus, (\ref{sol.NSym.verschiebung.e.Psin}) (applied to $m/n$ instead
of $n$) yields $\Psi_{m/n}=\left(  S\star E\right)  \left(  H_{m/n}\right)  $.
Now, (\ref{sol.NSym.verschiebung.a.almost}) becomes%
\[
\mathbf{V}_n \left(  \Psi_{m}\right)  =n\cdot\left(  S\star E\right)  \left(
\underbrace{
\begin{cases}
H_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases}
}_{\substack{=H_{m/n}\\\text{(since }n\mid m\text{)}}}\right)
=n\cdot\underbrace{\left(  S\star E\right)  \left(  H_{m/n}\right)  }%
_{=\Psi_{m/n}}=n\Psi_{m/n}.
\]
Compared with%
\[
\begin{cases}
n\Psi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad
=n\Psi_{m/n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\mid m\right)
,
\]
this yields $\mathbf{V}_n \left(  \Psi_{m}\right)  =
\begin{cases}
n\Psi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad
$. This solves Exercise \ref{exe.NSym.verschiebung}(a).

(b) Recall that a ring of formal power series $R\left[  \left[  t\right]
\right]  $ over a (not necessarily commutative) $\kk$-algebra $R$ has a canonical
topology which makes it into a topological $\kk$-algebra\footnote{This is proven
just as in the case when $R$ is commutative.}. Thus, in particular,
$\Nsym \left[  \left[  t\right]  \right]$ becomes a
topological $\kk$-algebra. We shall be considering this
topology when we speak of continuity.

Assume that $\QQ$ is a subring of $\kk$.

Define two power series $\widetilde{H}\left(  t\right)  $ and $\xi\left(
t\right)  $ in $\Nsym \left[  \left[  t\right]  \right]  $ by%
\begin{align*}
\widetilde{H}\left(  t\right)   &  =\sum_{n\geq0}H_{n}t^{n};\\
\xi\left(  t\right)   &  =\sum_{n\geq1}\xi_{n}t^{n}=\log\left(  \widetilde{H}%
\left(  t\right)  \right)  .
\end{align*}
(Here, the equality $\sum_{n\geq1}\xi_{n}t^{n}=\log\left(  \widetilde{H}%
\left(  t\right)  \right)  $ follows from \eqref{relation-definining-pi's}.)

We have $\widetilde{H}\left(  t\right)  =\sum_{n\geq0}H_{n}t^{n}=\sum_{i\geq
0}H_{i}t^{i}$ (here, we renamed the summation index $n$ as $i$) and
$\xi\left(  t\right)  =\sum_{n\geq1}\xi_{n}t^{n}=\sum_{i\geq1}\xi_{i}t^{i}$
(here, we renamed the summation index $n$ as $i$).

Now, let $n$ be a positive integer. The $\kk$-algebra homomorphism
$\mathbf{V}_n : \Nsym \rightarrow \Nsym $ induces a
continuous $\kk$-algebra homomorphism $\mathbf{V}_n \left[  \left[
t\right]  \right]  : \Nsym \left[  \left[  t\right]  \right]
\rightarrow \Nsym \left[  \left[  t\right]  \right]  $ given by%
\[
\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\in \NN }a_{i}t^{i}\right)  =\sum_{i\in \NN }\mathbf{V}%
_{n}\left(  a_{i}\right)  t^{i}\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
a_{i}\right)  _{i\in \NN }\in \Nsym^{ \NN }.
\]
This $\kk$-algebra homomorphism $\mathbf{V}_n \left[  \left[
t\right]  \right]  $ commutes with taking logarithms (since it is continuous
and a $\kk$-algebra homomorphism) -- i.e., we have%
\begin{equation}
\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(  \log
Q\right)  =\log\left(  \left(  \mathbf{V}_n \left[  \left[  t\right]
\right]  \right)  \left(  Q\right)  \right)
\label{sol.NSym.verschiebung.b.deriv}
\end{equation}
for every $Q\in \Nsym \left[  \left[  t\right]  \right]$
having constant coefficient $1$.

We have $\xi\left(  t\right)  =\sum_{i\geq1}\xi_{i}t^{i}$, thus%
\[
\sum_{i\geq1}\xi_{i}t^{i}=\xi\left(  t\right)  =\log\left(
\underbrace{\widetilde{H}\left(  t\right)  }_{=\sum_{i\geq0}H_{i}t^{i}%
}\right)  =\log\left(  \sum_{i\geq0}H_{i}t^{i}\right)  .
\]
We can substitute $t^{n}$ for $t$ on both sides of this equality\footnote{This
is allowed because the element $t^{n}$ of $ \Nsym \left[  \left[
t\right]  \right]  $ is central. (Generally, if $R$ is a ring (possibly not
commutative) and $Q\in R\left[  \left[  t\right]  \right]  $ is a power
series, then every central element $z$ of $R\left[  \left[  t\right]  \right]
$ can be substituted for $t$ in $Q$ as long as the constant coefficient of $z$
is $0$. The result of this substitution is denoted by $Q\left(  z\right)  $.
The map $R\left[  \left[  t\right]  \right]  \rightarrow R\left[  \left[
t\right]  \right]  $ which sends every $Q$ to $Q\left(  z\right)  $ (for a
fixed $z$) is a ring homomorphism $R\left[  \left[  t\right]  \right]
\rightarrow R\left[  \left[  t\right]  \right]  $.)}. As a result, we obtain%
\begin{equation}
\sum_{i\geq1}\xi_{i}\left(  t^{n}\right)  ^{i}=\log\left(  \sum_{i\geq0}%
H_{i}\left(  t^{n}\right)  ^{i}\right)  . \label{sol.NSym.verschiebung.b.3}%
\end{equation}


But $\widetilde{H}\left(  t\right) = \sum_{i\geq0}H_{i}t^{i}$. Applying the
map $\mathbf{V}_n \left[  \left[  t\right]  \right]  $ to both sides of this
equality, we obtain%
\begin{align}
\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(
\widetilde{H}\left(  t\right)  \right)   &  =\left(  \mathbf{V}_n \left[
\left[  t\right]  \right]  \right)  \left(  \sum_{i\geq0}H_{i}t^{i}\right)
=\sum_{i\geq0}\underbrace{\mathbf{V}_n \left(  H_{i}\right)  }%
_{\substack{=
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
\\\text{(by (\ref{sol.NSym.verschiebung.vnhm}), applied to
}m=i\text{)}}}t^{i}\nonumber\\
&  =\sum_{i\geq0}
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} \cdot
t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}\underbrace{
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
}_{\substack{=H_{i/n}\\\text{(since }n\mid i\text{)}}}
\cdot t^{i}%
+\sum\limits_{\substack{i\geq0;\\n\nmid i}}\underbrace{
\begin{cases}
H_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq0;\\n\mid i}}H_{i/n}t^{i}+\underbrace{\sum\limits
_{\substack{i\geq0;\\n\nmid i}}0t^{i}}_{=0}=\sum\limits_{\substack{i\geq0;\\n\mid
i}}H_{i/n}t^{i}\nonumber\\
&  =\sum_{i\geq0}\underbrace{H_{ni/n}}_{=H_{i}}\underbrace{t^{ni}}_{=\left(
t^{n}\right)  ^{i}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted
}ni\text{ for }i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i\geq0}H_{i}\left(  t^{n}\right)  ^{i}.
\label{sol.NSym.verschiebung.b.4}%
\end{align}
Now,%
\begin{align}
\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(
\underbrace{\xi\left(  t\right)  }_{=\log\left(  \widetilde{H}\left(
t\right)  \right)  }\right)   &  =\left(  \mathbf{V}_n \left[  \left[
t\right]  \right]  \right)  \left(  \log\left(  \widetilde{H}\left(  t\right)
\right)  \right)  =\log\left(  \underbrace{\left(  \mathbf{V}_n \left[
\left[  t\right]  \right]  \right)  \left(  \widetilde{H}\left(  t\right)
\right)  }_{\substack{=\sum_{i\geq0}H_{i}\left(  t^{n}\right)  ^{i}\\\text{(by
(\ref{sol.NSym.verschiebung.b.4}))}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.b.deriv}),
applied to }Q=\widetilde{H}\left(  t\right)  \right) \nonumber\\
&  =\log\left(  \sum_{i\geq0}H_{i}\left(  t^{n}\right)  ^{i}\right)
=\sum_{i\geq1}\xi_{i}\underbrace{\left(  t^{n}\right)  ^{i}}_{=t^{ni}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.b.3}%
)}\right) \nonumber\\
&  =\sum_{i\geq1}\xi_{i}t^{ni}. \label{sol.NSym.verschiebung.b.6}%
\end{align}


Comparing this with%
\begin{align}
\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(
\underbrace{\xi\left(  t\right)  }_{=\sum_{i\geq1}\xi_{i}t^{i}}\right)   &
=\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(
\sum_{i\geq1}\xi_{i}t^{i}\right)  =\sum_{i\geq1}\underbrace{\left(
\mathbf{V}_n \left[  \left[  t\right]  \right]  \right)  \left(  \xi
_{i}\right)  }_{\substack{=\mathbf{V}_n \left(  \xi_{i}\right)  \\\text{(by
the definition of }\mathbf{V}_n \left[  \left[  t\right]  \right]  \text{)}%
}}\underbrace{\left(  \mathbf{V}_n \left[  \left[  t\right]  \right]
\right)  \left(  t^{i}\right)  }_{\substack{=t^{i}\\\text{(by the definition
of }\mathbf{V}_n \left[  \left[  t\right]  \right]  \text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{V}_n \left[  \left[
t\right]  \right]  \text{ is a continuous } \kk \text{-algebra
homomorphism}\right) \nonumber\\
&  =\sum_{i\geq1}\mathbf{V}_n \left(  \xi_{i}\right)  t^{i},\nonumber
\end{align}
we obtain%
\begin{equation}
\sum_{i\geq1}\mathbf{V}_n \left(  \xi_{i}\right)  t^{i}=\sum_{i\geq1}\xi
_{i}t^{ni}. \label{sol.NSym.verschiebung.b.8}%
\end{equation}


Now,%
\begin{align}
&  \sum_{i\geq1}
\begin{cases}
\xi_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases} \cdot
t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq1;\\n\mid i}}\underbrace{
\begin{cases}
\xi_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
}_{\substack{=np_{i/n}\\\text{(since }n\mid i\text{)}}}
\cdot t^{i}%
+\sum\limits_{\substack{i\geq1;\\n\nmid i}}\underbrace{
\begin{cases}
\xi_{i/n},& \text{if }n\mid i;\\
0,& \text{if }n\nmid i
\end{cases}
}_{\substack{=0\\\text{(since }n\nmid i\text{)}}}
\cdot t^{i}\nonumber\\
&  =\sum\limits_{\substack{i\geq1;\\n\mid i}}\xi_{i/n}t^{i}+\underbrace{\sum\limits
_{\substack{i\geq1;\\n\nmid i}}0t^{i}}_{=0}=\sum\limits_{\substack{i\geq1;\\n\mid
i}}\xi_{i/n}t^{i}=\sum_{i\geq1}\xi_{i}t^{ni}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have substituted }ni\text{ for }i\text{ in the sum}\right)
\nonumber\\
&  =\sum_{i\geq1}\mathbf{V}_n \left(  \xi_{i}\right)  t^{i}\qquad
\qquad\left(  \text{by \eqref{sol.NSym.verschiebung.b.8}}\right)  .
\label{sol.NSym.verschiebung.b.almost}%
\end{align}
Now, let $m$ be a positive integer. Comparing coefficients before $t^{m}$ in
the equality (\ref{sol.NSym.verschiebung.b.almost}), we obtain%
\[
\begin{cases}
\xi_{m/n},& \text{if }n\mid m;\\
0,& \text{if }n\nmid m
\end{cases} \quad
=\mathbf{V}_n \left(  \xi_{m}\right)  .
\]
This solves Exercise \ref{exe.NSym.verschiebung}(b).

(f) Recall first that
$\left( H_{\alpha} \right) _{\alpha \in \Comp}$ and
$\left( M_{\alpha} \right) _{\alpha \in \Comp}$ are
mutually dual bases with respect to the dual pairing
$\Nsym \otimes \Qsym
\overset{\left(\cdot,\cdot\right)  }{\longrightarrow}
\kk$.\ \ \ \ \footnote{This
follows from the definition of $\left(  H_{\alpha}\right)  _{\alpha\in
 \Comp }$.} Thus,
\begin{equation}
\left(  H_{\alpha},M_{\beta}\right)  = \delta_{\alpha,\beta}
\ \ \ \ \ \ \ \ \ \ \text{for any two compositions }\alpha\text{ and }\beta.
\label{sol.NSym.verschiebung.e.duality}
\end{equation}


Let us introduce a notation: For every composition $\alpha$ and every positive
integer $s$, let $\alpha\left\{  s\right\}  $ denote the $\ell$-tuple $\left(
s\alpha_{1},s\alpha_{2},\ldots,s\alpha_{\ell}\right)  $, where $\alpha$ is
written in the form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  $. Notice that if $\alpha$ is a composition and $s$ is a positive
integer, then $\alpha\left\{  s\right\}  $ is a composition again and
satisfies
\begin{equation}
\ell\left(  \alpha\left\{  s\right\}  \right)  =\ell\left(  \alpha\right)  .
\label{sol.NSym.verschiebung.e.llambdas}
\end{equation}


We have%
\begin{equation}
\mathbf{F}_{n}M_{\alpha}=M_{\alpha\left\{  n\right\}  }%
\ \ \ \ \ \ \ \ \ \ \text{for every composition }\alpha.
\label{sol.NSym.verschiebung.e.f}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.NSym.verschiebung.e.f}):} Let $\alpha$ be
a composition. Write $\alpha$ in the form $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $. By the definition of $M_{\alpha}$,
we have $M_{\alpha}=\sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}}^{\alpha_{1}%
}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}$ (where the sum is
over all $\ell$-tuples $\left(  i_{1},i_{2},\ldots,i_{\ell}\right)  $ of
positive integers satisfying $i_{1}<i_{2}<\cdots<i_{\ell}$).
\par
On the other hand, the definition of $\alpha\left\{  n\right\}  $ yields
$\alpha\left\{  n\right\}  =\left(  n\alpha_{1},n\alpha_{2},\ldots
,n\alpha_{\ell}\right)  $ (since $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  $). Hence, the definition of $M_{\alpha\left\{
n\right\}  }$ yields
\begin{equation}
M_{\alpha\left\{  n\right\}  }=\sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}%
}^{n\alpha_{1}}x_{i_{2}}^{n\alpha_{2}}\cdots x_{i_{\ell}}^{n\alpha_{\ell}}
\label{sol.NSym.verschiebung.e.f.pf.1}%
\end{equation}
(where the sum is over all $\ell$-tuples $\left(  i_{1},i_{2},\ldots,i_{\ell
}\right)  $ of positive integers satisfying $i_{1}<i_{2}<\cdots<i_{\ell}$).
\par
Now, the definition of $\mathbf{F}_{n}$ yields%
\begin{align*}
\mathbf{F}_{n}M_{\alpha}  &  =\underbrace{M_{\alpha}}_{=\sum_{i_{1}%
<i_{2}<\cdots<i_{\ell}}x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots
x_{i_{\ell}}^{\alpha_{\ell}}}\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n}%
,\ldots\right)  =\left(  \sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}}%
^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}\right)
\left(  x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right) \\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{\ell}}x_{i_{1}}^{n\alpha_{1}}x_{i_{2}%
}^{n\alpha_{2}}\cdots x_{i_{\ell}}^{n\alpha_{\ell}}=M_{\alpha\left\{
n\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.NSym.verschiebung.e.f.pf.1})}\right)  .
\end{align*}
This proves (\ref{sol.NSym.verschiebung.e.f}).}.

We need to prove that the maps $\mathbf{F}_{n}: \Qsym %
\rightarrow \Qsym $ and $\mathbf{V}_n : \Nsym %
\rightarrow \Nsym $ are adjoint with respect to the dual pairing
$\Nsym \otimes \Qsym \overset{\left(  \cdot
,\cdot\right)  }{\longrightarrow} \kk$. In other words, we need to prove
that%
\begin{equation}
\left(  b,\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_n b,a\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in \Qsym \text{ and }%
b\in \Nsym .\label{sol.NSym.verschiebung.e.goal}%
\end{equation}


\textit{Proof of (\ref{sol.NSym.verschiebung.e.goal}):} Fix $a\in
 \Qsym $ and $b\in \Nsym $.

Recall that $\left(  M_{\alpha}\right)  _{\alpha\in \Comp }$ is a
basis of the $\kk$-module $\Nsym$. Hence, in proving
(\ref{sol.NSym.verschiebung.e.goal}), we can WLOG assume that $a$ is an
element of this basis $\left(  M_{\alpha}\right)  _{\alpha\in
 \Comp }$ (because the equality
(\ref{sol.NSym.verschiebung.e.goal}) is $\kk$-linear in $a$). Assume
this. Thus, $a$ is an element of the basis $\left(  M_{\alpha}\right)
_{\alpha\in \Comp }$. In other words, there exists a $\beta
\in \Comp $ such that $a=M_{\beta}$. Consider this $\beta$. We
have $\mathbf{F}_{n}\underbrace{a}_{=M_{\beta}}=\mathbf{F}_{n}M_{\beta
}=M_{\beta\left\{  n\right\}  }$ (by (\ref{sol.NSym.verschiebung.e.f}),
applied to $\alpha=\beta$).

Recall that $\left(  H_{\alpha}\right)  _{\alpha\in \Comp }$ is a
basis of the $\kk$-module $\Lambda$. Hence, in proving
(\ref{sol.NSym.verschiebung.e.goal}), we can WLOG assume that $b$ is an
element of this basis $\left(  H_{\alpha}\right)  _{\alpha\in
 \Comp }$ (because the equality
(\ref{sol.NSym.verschiebung.e.goal}) is $\kk$-linear in $b$). Assume
this. Thus, $b$ is an element of the basis $\left(  H_{\alpha}\right)
_{\alpha\in \Comp }$. In other words, there exists a $\gamma
\in \Comp $ such that $b=H_{\gamma}$. Consider this $\gamma$. We
have%
\begin{equation}
\left(  \underbrace{b}_{=H_{\gamma}},\underbrace{\mathbf{F}_{n}a}%
_{=M_{\beta\left\{  n\right\}  }}\right)  =\left(  H_{\gamma},M_{\beta\left\{
n\right\}  }\right)  =\delta_{\gamma,\beta\left\{  n\right\}  }
\label{sol.NSym.verschiebung.e.lhs}%
\end{equation}
(by (\ref{sol.NSym.verschiebung.e.duality}), applied to $\gamma$ and
$\beta\left\{  n\right\}  $ instead of $\alpha$ and $\beta$).

Let us write the composition $\gamma$ in the form $\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  $. Then, $\gamma=\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  $, so that $H_{\gamma}=H_{\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }=H_{\gamma_{1}}%
H_{\gamma_{2}}\cdots H_{\gamma_{\ell}}$ (by (\ref{sol.NSym.verschiebung.a.H}),
applied to $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  $
instead of $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $).

Let us first assume that%
\begin{equation}
\left(  \text{not every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{
satisfies }n\mid\gamma_{i}\right)
.\label{sol.NSym.verschiebung.e.stupid-assumption}%
\end{equation}
Then, $\mathbf{V}_n \left(  \underbrace{H_{\gamma}}_{=H_{\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }}\right)  =\mathbf{V}_n \left(
H_{\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }\right)  =0$
(by (\ref{sol.NSym.verschiebung.a.H1}), applied to $\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  $ instead of $\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $). Also, $\beta\left\{
n\right\}  \neq\gamma$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary.
Thus, $\beta\left\{  n\right\}  =\gamma$. Let us write the composition $\beta$
in the form $\left(  \beta_{1},\beta_{2},\ldots,\beta_{q}\right)  $. Then,
$\beta\left\{  n\right\}  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta
_{q}\right)  $ (by the definition of $\beta\left\{  n\right\}  $). Hence,
$\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  =\gamma
=\beta\left\{  n\right\}  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta
_{q}\right)  $. As a consequence, $\ell=q$, so that $q=\ell$ and thus $\left(
n\beta_{1},n\beta_{2},\ldots,n\beta_{q}\right)  =\left(  n\beta_{1},n\beta
_{2},\ldots,n\beta_{\ell}\right)  $. Hence, $\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{\ell}\right)  =\left(  n\beta_{1},n\beta_{2},\ldots
,n\beta_{q}\right)  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{\ell
}\right)  $. Consequently, every $i\in\left\{  1,2,\ldots,\ell\right\}  $
satisfies $\gamma_{i}=n\beta_{i}$. Thus, every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ satisfies $n\mid\gamma_{i}$. This contradicts
(\ref{sol.NSym.verschiebung.e.stupid-assumption}). This contradiction proves
that our assumption was wrong, qed.}, so that $\delta_{\gamma,\beta\left\{
n\right\}  }=0$. Thus, (\ref{sol.NSym.verschiebung.e.lhs}) becomes $\left(
b,\mathbf{F}_{n}a\right)  =\delta_{\gamma,\beta\left\{  n\right\}  }=0$.
Compared with $\left(  \mathbf{V}_n \underbrace{b}_{=H_{\gamma}},a\right)
=\left(  \underbrace{\mathbf{V}_n \left(  H_{\gamma}\right)  }_{=0},a\right)
=\left(  0,a\right)  =0$ (since the dual pairing $\Nsym %
\otimes \Qsym \overset{\left(  \cdot,\cdot\right)
}{\longrightarrow} \kk$ is $\kk$-bilinear), this yields $\left(
b,\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_n b,a\right)  $. Thus,
(\ref{sol.NSym.verschiebung.e.goal}) holds.

Now, let us forget that we assumed that
(\ref{sol.NSym.verschiebung.e.stupid-assumption}) holds. We thus have proven
(\ref{sol.NSym.verschiebung.e.goal}) under the assumption that
(\ref{sol.NSym.verschiebung.e.stupid-assumption}) holds. Hence, for the rest
of our proof of (\ref{sol.NSym.verschiebung.e.goal}), we can WLOG assume that
(\ref{sol.NSym.verschiebung.e.stupid-assumption}) does not hold. Assume this.

We have assumed that (\ref{sol.NSym.verschiebung.e.stupid-assumption}) does
not hold. In other words,%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,\ell\right\}  \text{ satisfies }%
n\mid\gamma_{i}. \label{sol.NSym.verschiebung.e.anti}%
\end{equation}
Thus, $\gamma_{i}/n$ is a positive integer for every $i\in\left\{
1,2,\ldots,\ell\right\}  $ (since $\gamma_{i}$ is a positive integer for every
$i\in\left\{  1,2,\ldots,\ell\right\}  $). Thus, $\left(  \gamma_{1}%
/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)  $ is a composition. Let us
denote this composition by $\zeta$. We have $\zeta\left\{  n\right\}  =\gamma
$\ \ \ \ \footnote{\textit{Proof.} We have $\zeta=\left(  \gamma_{1}%
/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)  $. Hence, the definition of
$\zeta\left\{  n\right\}  $ yields $\zeta\left\{  n\right\}  =\left(
n\gamma_{1}/n,n\gamma_{2}/n,\ldots,n\gamma_{\ell}/n\right)  =\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  =\gamma$, qed.} and
$\mathbf{V}_n \left(  H_{\gamma}\right)  =H_{\zeta}$%
\ \ \ \ \footnote{\textit{Proof.} We have $\gamma=\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  $, so that%
\begin{align*}
\mathbf{V}_n \left(  H_{\gamma}\right)   &  =\mathbf{V}_n \left(  H_{\left(
\gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  }\right) \\
&  =H_{\left(  \gamma_{1}/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.a.H2}),
applied to }\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right)  \text{
instead of }\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  \right)
\\
&  =H_{\zeta}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \gamma
_{1}/n,\gamma_{2}/n,\ldots,\gamma_{\ell}/n\right)  =\zeta\right)  ,
\end{align*}
qed.} and thus%
\begin{equation}
\left(  \mathbf{V}_n \underbrace{b}_{=H_{\gamma}},\underbrace{a}_{=M_{\beta}%
}\right)  =\left(  \underbrace{\mathbf{V}_n \left(  H_{\gamma}\right)
}_{=H_{\zeta}},M_{\beta}\right)  =\left(  H_{\zeta},M_{\beta}\right)
=\delta_{\zeta,\beta} \label{sol.NSym.verschiebung.e.rhs}%
\end{equation}
(by (\ref{sol.NSym.verschiebung.e.duality}), applied to $\zeta$ instead of
$\alpha$).

Now, let us write the composition $\beta$ in the form $\left(  \beta_{1}%
,\beta_{2},\ldots,\beta_{s}\right)  $. Then, $\beta\left\{  n\right\}
=\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{s}\right)  $ (by the definition
of $\beta\left\{  n\right\}  $).

Now, we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \gamma=\beta\left\{  n\right\}  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \gamma_{1},\gamma_{2},\ldots
,\gamma_{\ell}\right)  =\left(  n\beta_{1},n\beta_{2},\ldots,n\beta
_{s}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma=\left(  \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right)  \text{ and }\beta\left\{  n\right\}
=\left(  n\beta_{1},n\beta_{2},\ldots,n\beta_{s}\right)  \right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{we have }\ell=s\text{, and every }%
i\in\left\{  1,2,\ldots,\ell\right\}  \text{ satisfies }\underbrace{\gamma
_{i}=n\beta_{i}}_{\substack{\text{this is equivalent to}\\\gamma_{i}%
/n=\beta_{i}}}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{we have }\ell=s\text{, and every }%
i\in\left\{  1,2,\ldots,\ell\right\}  \text{ satisfies }\gamma_{i}/n=\beta
_{i}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \left(  \gamma_{1}/n,\gamma_{2}/n,\ldots
,\gamma_{\ell}/n\right)  =\left(  \beta_{1},\beta_{2},\ldots,\beta_{s}\right)
\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \zeta=\beta\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \gamma_{1}/n,\gamma
_{2}/n,\ldots,\gamma_{\ell}/n\right)  =\zeta\text{ and }\left(  \beta
_{1},\beta_{2},\ldots,\beta_{s}\right)  =\beta\right)  .
\label{sol.NSym.verschiebung.e.equiv}%
\end{align}
But (\ref{sol.NSym.verschiebung.e.lhs}) becomes%
\begin{align*}
\left(  b,\mathbf{F}_{n}a\right)   &  =\delta_{\gamma,\beta\left\{  n\right\}
}=
\begin{cases}
1,& \text{if }\gamma=\beta\left\{  n\right\}  ;\\
0,& \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\delta_{\gamma,\beta\left\{  n\right\}  }\right) \\
&  =
\begin{cases}
1,& \text{if }\zeta=\beta;\\
0,& \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma=\beta\left\{
n\right\}  \text{ is equivalent to }\zeta=\beta\text{ (by
(\ref{sol.NSym.verschiebung.e.equiv}))}\right) \\
&  =\delta_{\zeta,\beta}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\delta
_{\zeta,\beta}=
\begin{cases}
1,& \text{if }\zeta=\beta;\\
0,& \text{otherwise}
\end{cases}
\text{ (by the definition of }\delta_{\mu,\kappa}\text{)}\right) \\
&  =\left(  \mathbf{V}_n b,a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.NSym.verschiebung.e.rhs})}\right)  .
\end{align*}
Thus, (\ref{sol.NSym.verschiebung.e.goal}) is proven. As we know, this
completes the solution of Exercise \ref{exe.NSym.verschiebung}(f).

(g) We know that $\pi: \Nsym \rightarrow\Lambda$ is a
$\kk$-algebra homomorphism. Thus, $\pi\left(  0\right)  =0$. Also, we
know that
\begin{equation}
\pi\left(  H_{n}\right)  =h_{n}\ \ \ \ \ \ \ \ \ \ \text{for every positive
integer }n. \label{sol.NSym.verschiebung.h.pi}%
\end{equation}
Now, let $n$ be a positive integer. The maps $\mathbf{v}_{n}\circ\pi$ and
$\pi\circ\mathbf{V}_n $ are $\kk$-algebra homomorphisms (since $\pi$
and $\mathbf{v}_{n}$ are $\kk$-algebra homomorphisms).

Let $r$ be a positive integer. Applying (\ref{sol.NSym.verschiebung.h.pi}) to
$r$ instead of $n$, we obtain $\pi\left(  H_{r}\right)  =h_{r}$.

If $n\mid r$, then $r/n$ is a positive integer. Hence,%
\begin{equation}
\text{if }n\mid r\text{, then }\pi\left(  H_{r/n}\right)  =h_{r/n}.
\label{sol.NSym.verschiebung.h.pi2}%
\end{equation}


Now,%
\[
\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  H_{r}\right)  =\mathbf{v}%
_{n}\left(  \underbrace{\pi\left(  H_{r}\right)  }_{=h_{r}}\right)
=\mathbf{v}_{n}\left(  h_{r}\right)  =
\begin{cases}
h_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\mathbf{v}_{n}\right)  .
\]
Compared with%
\begin{align*}
\left(  \pi\circ\mathbf{V}_n \right)  \left(  H_{r}\right)   &  =\pi\left(
\underbrace{\mathbf{V}_n \left(  H_{r}\right)  }_{\substack{=
\begin{cases}
H_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\\\text{(by the definition of }\mathbf{V}_n \text{)}}}\right)
=\pi\left(
\begin{cases}
H_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\right) \\
&  =
\begin{cases}
\pi\left(  H_{r/n}\right)  ,& \text{if }n\mid r;\\
\pi\left(  0\right)  ,& \text{if }n\nmid r
\end{cases}  \quad
=
\begin{cases}
\pi\left(  H_{r/n}\right)  ,& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\left(  0\right)  =0\text{ in
the case when }n\nmid r\right) \\
&  =
\begin{cases}
h_{r/n},& \text{if }n\mid r;\\
0,& \text{if }n\nmid r
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\left(  H_{r/n}\right)
=h_{r/n}\text{ in the case when }n\mid r\text{ (according to
(\ref{sol.NSym.verschiebung.h.pi2}))}\right)  ,
\end{align*}
this yields $\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  H_{r}\right)
=\left(  \pi\circ\mathbf{V}_n \right)  \left(  H_{r}\right)  $.

Let us now forget that we fixed $r$. We thus have proven that%
\begin{equation}
\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  H_{r}\right)  =\left(  \pi
\circ\mathbf{V}_n \right)  \left(  H_{r}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }r\text{.}
\label{sol.NSym.verschiebung.h.main}%
\end{equation}


Now, recall that the family $\left(  H_{r}\right)  _{r\geq1}$ generates the
$\kk$-algebra $\Nsym$ (according to
(\ref{sol.NSym.verschiebung.gen})). In other words, $\left(  H_{r}\right)
_{r\geq1}$ is a generating set of the $\kk$-algebra
$\Nsym$. The two $\kk$-algebra homomorphisms
$\mathbf{v}_{n}\circ\pi$ and $\pi\circ\mathbf{V}_n $ are equal to each other
on this generating set (according to (\ref{sol.witt.ghost-app.d.main})), and
therefore must be identical (because if two $\kk$-algebra homomorphisms
from the same domain are equal to each other on a generating set of their
domain, then these two homomorphisms must be identical). In other words,
$\mathbf{v}_{n}\circ\pi=\pi\circ\mathbf{V}_n $. This completes the solution
of Exercise \ref{exe.NSym.verschiebung}(g).

(h) \textit{Alternative solution of Exercise~\ref{exe.Lambda.verschiebung}%
(f).} Let $i$ denote the inclusion map $\Lambda\rightarrow \Qsym %
$. Then, Corollary \ref{cor.NSym.pi} yields that the map $\pi$ is adjoint to
the map $i$ with respect to the dual pairing $\Nsym %
\otimes \Qsym \overset{\left(  \cdot,\cdot\right)
}{\longrightarrow} \kk$. In other words,%
\begin{equation}
\left(  \pi\left(  b\right)  ,a\right)  =\left(  b,i\left(  a\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }b\in \Nsym \text{ and }%
a\in\Lambda.\label{sol.NSym.verschiebung.realh.adj}%
\end{equation}


But Exercise \ref{exe.NSym.verschiebung}(f) yields that the maps
$\mathbf{F}_{n}: \Qsym \rightarrow \Qsym $ and
$\mathbf{V}_n : \Nsym \rightarrow \Nsym $ are
adjoint with respect to the dual pairing $\Nsym \otimes
 \Qsym \overset{\left(  \cdot,\cdot\right)  }{\longrightarrow
} \kk$. In other words,%
\begin{equation}
\left(  b,\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_n b,a\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }a\in \Qsym \text{ and }%
b\in \Nsym .\label{sol.NSym.verschiebung.realh.adj2}%
\end{equation}


Now, fix a positive integer $n$. Let $a\in\Lambda$ and $b\in\Lambda$ be
arbitrary. Then, $a\in\Lambda\subset \Qsym $. Exercise
\ref{exe.QSym.frobenius}(f) yields $\mathbf{F}_{n}\mid_{\Lambda}%
=\mathbf{f}_{n}$. Thus, $\mathbf{f}_{n}=\mathbf{F}_{n}\mid_{\Lambda}$, so that
$\mathbf{f}_{n}a=\left(  \mathbf{F}_{n}\mid_{\Lambda}\right)  a=\mathbf{F}%
_{n}a$.

On the other hand, the projection $\pi: \Nsym \rightarrow\Lambda$
is surjective. Hence, there exists some $b^{\prime}\in \Nsym $
satisfying $b=\pi\left(  b^{\prime}\right)  $. Consider this $b^{\prime}$.

Since the Hall inner product on $\Lambda$ is symmetric, we have%
\begin{align*}
\left(  \mathbf{f}_{n}a,b\right)    & =\left(  \underbrace{b}_{=\pi\left(
b^{\prime}\right)  },\mathbf{f}_{n}a\right)  =\left(  \pi\left(  b^{\prime
}\right)  ,\mathbf{f}_{n}a\right)  =\left(  b^{\prime},\underbrace{i\left(
\mathbf{f}_{n}a\right)  }_{\substack{=\mathbf{f}_{n}a\\\text{(since }i\text{
is an inclusion map)}}}\right)  \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.realh.adj}%
), applied to }b^{\prime}\text{ and }\mathbf{f}_{n}a\text{ instead of }b\text{
and }a\right)  \\
& =\left(  b^{\prime},\underbrace{\mathbf{f}_{n}a}_{=\mathbf{F}_{n}a}\right)
=\left(  b^{\prime},\mathbf{F}_{n}a\right)  =\left(  \mathbf{V}_n b^{\prime
},a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.NSym.verschiebung.realh.adj2}), applied to }b^{\prime}\text{ instead
of }b\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  a,\mathbf{v}_{n}b\right)    & =\left(  \mathbf{v}_{n}\underbrace{b}%
_{=\pi\left(  b^{\prime}\right)  },a\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since the Hall inner product on }\Lambda\text{ is symmetric}\right)  \\
& =\left(  \underbrace{\mathbf{v}_{n}\left(  \pi\left(  b^{\prime}\right)
\right)  }_{=\left(  \mathbf{v}_{n}\circ\pi\right)  \left(  b^{\prime}\right)
},a\right)  =\left(  \underbrace{\left(  \mathbf{v}_{n}\circ\pi\right)
}_{\substack{=\pi\circ\mathbf{V}_n \\\text{(by Exercise
\ref{exe.NSym.verschiebung}(g))}}}\left(  b^{\prime}\right)  ,a\right)
=\left(  \underbrace{\left(  \pi\circ\mathbf{V}_n \right)  \left(  b^{\prime
}\right)  }_{=\pi\left(  \mathbf{V}_n b^{\prime}\right)  },a\right)  \\
& =\left(  \pi\left(  \mathbf{V}_n b^{\prime}\right)  ,a\right)  =\left(
\mathbf{V}_n b^{\prime},\underbrace{i\left(  a\right)  }%
_{\substack{=a\\\text{(since }i\text{ is an inclusion map)}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.NSym.verschiebung.realh.adj}),
applied to }\mathbf{V}_n b^{\prime}\text{ instead of }b\right)  \\
& =\left(  \mathbf{V}_n b^{\prime},a\right)  ,
\end{align*}
we obtain $\left(  \mathbf{f}_{n}a,b\right)  =\left(  a,\mathbf{v}%
_{n}b\right)  $.

Let us now forget that we fixed $a$ and $b$. We have thus shown that $\left(
\mathbf{f}_{n}a,b\right)  =\left(  a,\mathbf{v}_{n}b\right)  $ for every
$a\in\Lambda$ and $b\in\Lambda$. In other words, the maps $\mathbf{f}%
_{n}:\Lambda\rightarrow\Lambda$ and $\mathbf{v}_{n}:\Lambda\rightarrow\Lambda$
are adjoint with respect to the Hall inner product on $\Lambda$. Thus,
Exercise~\ref{exe.Lambda.verschiebung}(f) is solved once again. Hence,
Exercise \ref{exe.NSym.verschiebung}(h) is solved.
%\end{solutions}



\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lex}}

\textit{Solution to Exercise \ref{exe.words.lex}.}

\begin{proof}
[Proof of Proposition~\ref{prop.words.lex}.](a) This can be easily verified by
hand, but here is a slicker way to see it: Let $\mathfrak{B}$ be the set
$\mathfrak{A}\sqcup\left\{  -\infty\right\}  $, where $-\infty$ is a symbol.
We define a total order on $\mathfrak{B}$ by extending the given total order
on $\mathfrak{A}$ in such a way that every $a\in\mathfrak{A}$ satisfies
$-\infty<a$. Consider the set $\mathfrak{B}^{\infty}$ of all infinite
sequences of elements of $\mathfrak{B}$. Then, $\mathfrak{A}^{\ast}$ embeds
into $\mathfrak{B}^{\infty}$ by identifying every word $\left(  a_{1}%
,a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{\ast}$ with the sequence $\left(
a_{1},a_{2},\ldots,a_{n},-\infty,-\infty,-\infty,\ldots\right)  \in
\mathfrak{B}^{\infty}$. It is easy to check that our order relation $\leq$ on
$\mathfrak{A}^{\ast}$ thus becomes the restriction to $\mathfrak{A}^{\ast}$ of
the smaller-or-equal relation of the lexicographic order on $\mathfrak{B}%
^{\infty}$. Consequently, it is a total order. Proposition
\ref{prop.words.lex}(a) is thus proven.

(h) follows immediately from the definition of $\leq$.

(i) follows from common sense.

(k) Let $a\in\mathfrak{A}^{\ast}$ and $b\in\mathfrak{A}^{\ast}$ be such that
$b$ is nonempty. Then, $a$ is a prefix of $ab$. Thus, $a\leq ab$ (by
Proposition \ref{prop.words.lex}(h), applied to $ab$ instead of $b$). But
$a\neq ab$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $a=ab$.
Hence, $a\varnothing=a=ab$. Cancelling $a$ from this equality, we obtain
$\varnothing=b$, so that the word $b$ is empty. This contradicts the fact that
$b$ is nonempty. This contradiction proves that our assumption was wrong,
qed.}. Combined with $a\leq ab$, this yields $a<ab$. Proposition
\ref{prop.words.lex}(k) is thus proven.

(b) Proposition \ref{prop.words.lex}(b) is an almost immediate consequence of
the definition of $\leq$; its proof is thus left to the reader.

(c) Let $a,c,d\in\mathfrak{A}^{\ast}$ satisfy $ac\leq ad$. We have $ac=\left(
a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  },c_{1},c_{2},\ldots
,c_{\ell\left(  c\right)  }\right)  $ and \newline
$ad=\left(  a_{1},a_{2}%
,\ldots,a_{\ell\left(  a\right)  },d_{1},d_{2},\ldots,d_{\ell\left(  d\right)
}\right)  $, and we have $ac\leq ad$. Due to the definition of $\leq$, this
means that we must be in one of the following two situations:

\begin{itemize}
\item There exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
a\right)  +\ell\left(  c\right)  ,\ell\left(  a\right)  +\ell\left(  d\right)
\right\}  \right\}  $ such that%
\begin{equation}
\left(  \left(  ac\right)  _{i}<\left(  ad\right)  _{i}\text{, and every }%
j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  ac\right)
_{j}=\left(  ad\right)  _{j}\right)  . \label{pf.prop.words.lex.c.1}%
\end{equation}


\item The word $ac$ is a prefix of $ad$.
\end{itemize}

If we are in the first of these two situations, then we clearly must have
$i>\ell\left(  a\right)  $ (since otherwise, both $\left(  ac\right)  _{i}$
and $\left(  ad\right)  _{i}$ would equal $a_{i}$, and then $\left(
ac\right)  _{i}<\left(  ad\right)  _{i}$ would contradict $\left(  ac\right)
_{i}=a_{i}=\left(  ad\right)  _{i}$), and therefore $i$ has the form
$i=\ell\left(  a\right)  +i^{\prime}$ for some $i^{\prime}\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}
\right\}  $. Then, (\ref{pf.prop.words.lex.c.1}) yields $\left(  c_{i^{\prime
}}<d_{i^{\prime}}\text{, and every }j\in\left\{  1,2,\ldots,i^{\prime
}-1\right\}  \text{ satisfies }c_{j}=d_{j}\right)  $. But this shows that
$c\leq d$, and so we are done in the first situation. In the second situation,
we also have $c\leq d$\ \ \ \ \footnote{\textit{Proof.} Let us consider the
second situation. In this situation, the word $ac$ is a prefix of $ad$. In
other words, there exists a word $z$ such that $ad=acz$. Consider this $z$.
Cancelling $a$ from the equality $ad=acz$, we obtain $d=cz$. Hence, the word
$c$ is a prefix of $d$. Thus, $c\leq d$, qed.}, and again we are done. Thus,
Proposition \ref{prop.words.lex}(c) is proven.

(d) Let $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $a\leq c$. We have $a=\left(
a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  }\right)  $ and $c=\left(
c_{1},c_{2},\ldots,c_{\ell\left(  c\right)  }\right)  $, and we have $a\leq
c$. Due to the definition of $\leq$, this means that we must be in one of the
following two situations:

\begin{itemize}
\item There exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
a\right)  ,\ell\left(  c\right)  \right\}  \right\}  $ such that%
\[
\left(  a_{i}<c_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}
\text{ satisfies }a_{j}=c_{j}\right)  .
\]


\item The word $a$ is a prefix of $c$.
\end{itemize}

The first of these situations entails $ab\leq cd$ (because every $k\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}
\right\}  $ satisfies $\left(  ab\right)  _{k}=a_{k}$ and $\left(  cd\right)
_{k}=c_{k}$). Hence, in both situations, Proposition \ref{prop.words.lex}(d)
is proven.

(e) Let $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $ab\leq cd$. We have
$a=\left(  a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  }\right)  $, $b=\left(
b_{1},b_{2},\ldots,b_{\ell\left(  b\right)  }\right)  $, $c=\left(
c_{1},c_{2},\ldots,c_{\ell\left(  c\right)  }\right)  $ and $d=\left(
d_{1},d_{2},\ldots,d_{\ell\left(  d\right)  }\right)  $, and we have $ab\leq
cd$. Due to the definition of $\leq$, this means that we must be in one of the
following two situations:

\begin{itemize}
\item There exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
a\right)  +\ell\left(  b\right)  ,\ell\left(  c\right)  +\ell\left(  d\right)
\right\}  \right\}  $ such that%
\begin{equation}
\left(  \left(  ab\right)  _{i}<\left(  cd\right)  _{i}\text{, and every }%
j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  ab\right)
_{j}=\left(  cd\right)  _{j}\right)  . \label{pf.prop.words.lex.e.1}%
\end{equation}


\item The word $ab$ is a prefix of $cd$.
\end{itemize}

It is easy to see that if we are in the second situation, then either we have
$a\leq c$ or the word $c$ is a prefix of $a$\ \ \ \ \footnote{\textit{Proof.}
Assume that we are in the second situation. Then, the word $ab$ is a prefix of
$cd$. In other words, there exists a word $z\in\mathfrak{A}^{\ast}$ such that
$cd=abz$. Consider this $z$.
\par
Now, $a$ is a prefix of the word $cd$ (since $cd=abz=a\left(  bz\right)  $).
Also, $c$ is a prefix of the word $cd$. Hence, $a$ and $c$ are two prefixes of
the word $cd$. Thus, Proposition \ref{prop.words.lex}(i) (applied to $c$ and
$cd$ instead of $b$ and $c$) yields that either $a$ is a prefix of $c$, or $c$
is a prefix of $a$. Hence, either we have $a\leq c$ or $c$ is a prefix of $a$
(because if $a$ is a prefix of $c$, then $a\leq c$). Qed.}. We are therefore
done in the second situation. We can thus WLOG assume that we are in the first
situation. Assume this. We thus have an $i\in\left\{  1,2,\ldots,\min\left\{
\ell\left(  a\right)  +\ell\left(  b\right)  ,\ell\left(  c\right)
+\ell\left(  d\right)  \right\}  \right\}  $ satisfying
(\ref{pf.prop.words.lex.e.1}). If $i\leq\min\left\{  \ell\left(  a\right)
,\ell\left(  c\right)  \right\}  $, then this yields $a\leq c$ (because every
$k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \right\}  $ satisfies $\left(  ab\right)  _{k}=a_{k}$ and
$\left(  cd\right)  _{k}=c_{k}$), and we are done. If $i>\min\left\{
\ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $, then every
$j\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \right\}  $ satisfies $a_{j}=c_{j}$ (because every
$j\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \right\}  $ satisfies $j\in\left\{  1,2,\ldots
,i-1\right\}  $ and thus%
\begin{align*}
a_{j}  &  =\left(  ab\right)  _{j}=\left(  cd\right)  _{j}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.words.lex.e.1})}\right) \\
&  =c_{j}%
\end{align*}
), and thus we conclude that either $a$ is a prefix of $c$, or $c$ is a prefix
of $a$. Again, this means that we are done. Thus, Proposition
\ref{prop.words.lex}(e) is proven.

(f) follows from (e), because if $c$ is a prefix of $a$ satisfying
$\ell\left(  a\right)  \leq\ell\left(  c\right)  $, then $c=a$.

(g) Let $a,b,c\in\mathfrak{A}^{\ast}$ satisfy $a\leq b\leq ac$. Then,
$b\varnothing=b\leq ac$. Hence, Proposition \ref{prop.words.lex}(e) (applied
to $b,\varnothing,a,c$ instead of $a,b,c,d$) yields that either we have $b\leq
a$ or the word $a$ is a prefix of $b$. Since $b\leq a$ leads to $a=b$ (in view
of $a\leq b$), we get in both of these cases that $a$ is a prefix of $b$. This
proves Proposition \ref{prop.words.lex}(g).

(j) Let $a,b,c\in\mathfrak{A}^{\ast}$ satisfy $a\leq b$ and $\ell\left(
a\right)  \geq\ell\left(  b\right)  $. Proposition \ref{prop.words.lex}(d)
(applied to $a$, $c$, $b$ and $c$) yields that either we have $ac\leq bc$ or
the word $a$ is a prefix of $b$. In the first of these two cases, we are
obviously done. Hence, we WLOG assume that we are in the second of these two
cases. Thus, the word $a$ is a prefix of $b$. Since the word $a$ is at least
as long as $b$ (in fact, $\ell\left(  a\right)  \geq\ell\left(  b\right)  $),
this yields that $a=b$, so that $\underbrace{a}_{=b}c=bc\leq bc$. This proves
Proposition \ref{prop.words.lex}(j).
\end{proof}
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lex}}

\textit{Solution to Exercise \ref{exe.words.lex}.}

\begin{proof}
[Proof of Proposition~\ref{prop.words.lex}.]
(a) Let $\omega$ be an object not
in the alphabet $\mathfrak{A}$. Let us equip the set $\mathfrak{A}\cup\left\{
\omega\right\}  $ with a total order which extends the total order on the
alphabet $\mathfrak{A}$ and which satisfies
\begin{equation}
\left(  a>\omega\text{ for every }a\in\mathfrak{A}\right)  .
\label{pf.prop.words.lex.a.long.omega}%
\end{equation}


Let $\left(  \mathfrak{A}\cup\left\{  \omega\right\}  \right)  ^{\infty}$
denote the set of all infinite sequences $\left(  b_{1},b_{2},b_{3}%
,\ldots\right)  $ of elements of $\mathfrak{A}\cup\left\{  \omega\right\}  $.
The total order on the set $\mathfrak{A}\cup\left\{  \omega\right\}  $ gives
rise to a lexicographic order on the set $\left(  \mathfrak{A}\cup\left\{
\omega\right\}  \right)  ^{\infty}$; this latter lexicographic order is a
partial order on the set $\left(  \mathfrak{A}\cup\left\{  \omega\right\}
\right)  ^{\infty}$\ \ \ \ \footnote{In classical logic, this lexicographic
order on the set $\left(  \mathfrak{A}\cup\left\{  \omega\right\}  \right)
^{\infty}$ is actually a total order. But we will not use this, since we
prefer our proof to stay constructive.}.

For every sequence $g\in\left(  \mathfrak{A}\cup\left\{  \omega\right\}
\right)  ^{\infty}$ and every positive integer $i$, we shall denote by $g_{i}$
the $i$-th entry of the sequence $g$. Thus,%
\[
g=\left(  g_{1},g_{2},g_{3},\ldots\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
sequence }g\in\left(  \mathfrak{A}\cup\left\{  \omega\right\}  \right)
^{\infty}.
\]


Let us define a map $\mathbf{i}:\mathfrak{A}^{\ast}\rightarrow\left(
\mathfrak{A}\cup\left\{  \omega\right\}  \right)  ^{\infty}$ by setting%
\begin{equation}
\left(  \mathbf{i}\left(  \left(  w_{1},w_{2},\ldots,w_{\ell}\right)  \right)
=\left(  w_{1},w_{2},\ldots,w_{\ell},\omega,\omega,\omega,\ldots\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  w_{1},w_{2},\ldots,w_{\ell
}\right)  \in\mathfrak{A}^{\ast}\right)  . \label{pf.prop.words.lex.a.long.i}%
\end{equation}


We notice the following facts:

\begin{itemize}
\item Every word $w\in\mathfrak{A}^{\ast}$ and every $i\in\left\{
1,2,\ldots,\ell\left(  w\right)  \right\}  $ satisfy%
\begin{equation}
\left(  \mathbf{i}\left(  w\right)  \right)  _{i}=w_{i}.
\label{pf.prop.words.lex.a.long.lem1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.words.lex.a.long.lem1}):} Let
$w\in\mathfrak{A}^{\ast}$ and $i\in\left\{  1,2,\ldots,\ell\left(  w\right)
\right\}  $. Then, $i\leq\ell\left(  w\right)  $ (since $i\in\left\{
1,2,\ldots,\ell\left(  w\right)  \right\}  $). We have $w=\left(  w_{1}%
,w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  $. Applying the map
$\mathbf{i}$ to both sides of this equality, we obtain%
\begin{align*}
\mathbf{i}\left(  w\right)   &  =\mathbf{i}\left(  \left(  w_{1},w_{2}%
,\ldots,w_{\ell\left(  w\right)  }\right)  \right)  =\left(  w_{1}%
,w_{2},\ldots,w_{\ell\left(  w\right)  },\omega,\omega,\omega,\ldots\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{i}\left(
w\right)  \right)  .
\end{align*}
Hence,%
\[
\left(  \underbrace{\mathbf{i}\left(  w\right)  }_{=\left(  w_{1},w_{2}%
,\ldots,w_{\ell\left(  w\right)  },\omega,\omega,\omega,\ldots\right)
}\right)  _{i}=\left(  \left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
},\omega,\omega,\omega,\ldots\right)  \right)  _{i}=w_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq\ell\left(  w\right)  \right)
.
\]
This proves (\ref{pf.prop.words.lex.a.long.lem1}).}

\item Every word $w\in\mathfrak{A}^{\ast}$ and every positive integer $i$
satisfying $i>\ell\left(  w\right)  $ satisfy%
\begin{equation}
\left(  \mathbf{i}\left(  w\right)  \right)  _{i}=\omega.
\label{pf.prop.words.lex.a.long.lem2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.words.lex.a.long.lem2}):} Let
$w\in\mathfrak{A}^{\ast}$ and let $i$ be a positive integer satisfying
$i>\ell\left(  w\right)  $. We have $w=\left(  w_{1},w_{2},\ldots
,w_{\ell\left(  w\right)  }\right)  $. Applying the map $\mathbf{i}$ to both
sides of this equality, we obtain%
\begin{align*}
\mathbf{i}\left(  w\right)   &  =\mathbf{i}\left(  \left(  w_{1},w_{2}%
,\ldots,w_{\ell\left(  w\right)  }\right)  \right)  =\left(  w_{1}%
,w_{2},\ldots,w_{\ell\left(  w\right)  },\omega,\omega,\omega,\ldots\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{i}\left(
w\right)  \right)  .
\end{align*}
Hence,%
\[
\left(  \underbrace{\mathbf{i}\left(  w\right)  }_{=\left(  w_{1},w_{2}%
,\ldots,w_{\ell\left(  w\right)  },\omega,\omega,\omega,\ldots\right)
}\right)  _{i}=\left(  \left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
},\omega,\omega,\omega,\ldots\right)  \right)  _{i}=\omega
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>\ell\left(  w\right)  \right)  .
\]
This proves (\ref{pf.prop.words.lex.a.long.lem2}).}

\item If two words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$
satisfy $u\leq v$, then%
\begin{equation}
\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)  .
\label{pf.prop.words.lex.a.long.lem5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.words.lex.a.long.lem5}):} Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words which
satisfy $u\leq v$. We need to prove that $\mathbf{i}\left(  u\right)
\leq\mathbf{i}\left(  v\right)  $.
\par
The definition of ``$u\leq v$'' shows that the
following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $u\leq v$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
\end{statement}
\par
On the other hand, recall that the order on $\left(  \mathfrak{A}\cup\left\{
\omega\right\}  \right)  ^{\infty}$ is the lexicographic order. Hence, the
definition of ``$\mathbf{i}\left(  u\right)  \leq
\mathbf{i}\left(  v\right)  $'' shows that the following fact
holds:
\par
\begin{statement}
\textit{Fact B:} The relation $\mathbf{i}\left(  u\right)  \leq\mathbf{i}%
\left(  v\right)  $ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
\end{statement}
\par
We need to prove that $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(
v\right)  $. If $u=v$, then this is obvious (because if $u=v$, then
$\mathbf{i}\left(  \underbrace{u}_{=v}\right)  =\mathbf{i}\left(  v\right)
$). Hence, for the rest of this proof, we can WLOG assume that we don't have
$u=v$. Assume this. Thus, $u\neq v$.
\par
Now, let us recall that the relation $u\leq v$ holds (by assumption). Hence,
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
(according to Fact A). Thus, we must be in one of the following two cases:
\par
\textit{Case 1:} There exists an $i\in\left\{  1,2,\ldots,\min\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  u_{i}<v_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $.
\par
\textit{Case 2:} The word $u$ is a prefix of $v$.
\par
Let us first consider Case 1. In this case, there exists an $i\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  u_{i}<v_{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $. Let us denote
this $i$ by $k$. Thus, $k$ is an element of $\left\{  1,2,\ldots,\min\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  u_{k}<v_{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $.
\par
We have $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  \right\}  $, hence $k\geq1$ and $k\leq
\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  $. Hence,
$k\leq\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\leq\ell\left(  u\right)  $ and thus $k\in\left\{  1,2,\ldots,\ell\left(
u\right)  \right\}  $ (since $k\geq1$). Also, $k\leq\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  \leq\ell\left(  v\right)  $, so
that $k\in\left\{  1,2,\ldots,\ell\left(  v\right)  \right\}  $ (since
$k\geq1$).
\par
Applying (\ref{pf.prop.words.lex.a.long.lem1}) to $w=u$ and $i=k$, we obtain
$\left(  \mathbf{i}\left(  u\right)  \right)  _{k}=u_{k}$. Applying
(\ref{pf.prop.words.lex.a.long.lem1}) to $w=v$ and $i=k$, we obtain $\left(
\mathbf{i}\left(  v\right)  \right)  _{k}=v_{k}$. Thus, $v_{k}=\left(
\mathbf{i}\left(  v\right)  \right)  _{k}$, so that $\left(  \mathbf{i}\left(
u\right)  \right)  _{k}=u_{k}<v_{k}=\left(  \mathbf{i}\left(  v\right)
\right)  _{k}$.
\par
Let $j\in\left\{  1,2,\ldots,k-1\right\}  $. Then, $1\leq j\leq k-1$. Thus,
$j\leq k-1<k\leq\ell\left(  u\right)  $, so that $j\in\left\{  1,2,\ldots
,\ell\left(  u\right)  \right\}  $ (since $1\leq j$). Hence, applying
(\ref{pf.prop.words.lex.a.long.lem1}) to $w=u$ and $i=j$, we obtain $\left(
\mathbf{i}\left(  u\right)  \right)  _{j}=u_{j}$. The same argument (but
applied to $v$ instead of $u$) yields $\left(  \mathbf{i}\left(  v\right)
\right)  _{j}=v_{j}$. Thus, $v_{j}=\left(  \mathbf{i}\left(  v\right)
\right)  _{j}$.
\par
But $u_{j}=v_{j}$ (since we know that every $j\in\left\{  1,2,\ldots
,k-1\right\}  $ satisfies $u_{j}=v_{j}$). Hence, $\left(  \mathbf{i}\left(
u\right)  \right)  _{j}=u_{j}=v_{j}=\left(  \mathbf{i}\left(  v\right)
\right)  _{j}$.
\par
Let us now forget that we fixed $j$. We thus have shown that every
$j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $\left(  \mathbf{i}\left(
u\right)  \right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}$. So
we know that%
\[
\left(  \left(  \mathbf{i}\left(  u\right)  \right)  _{k}<\left(
\mathbf{i}\left(  v\right)  \right)  _{k}\text{, and every }j\in\left\{
1,2,\ldots,k-1\right\}  \text{ satisfies }\left(  \mathbf{i}\left(  u\right)
\right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}\right)  .
\]
Hence, there exists an $i\in\left\{  1,2,3,\ldots\right\}  $ such that
$\left(  \left(  \mathbf{i}\left(  u\right)  \right)  _{i}<\left(
\mathbf{i}\left(  v\right)  \right)  _{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  \mathbf{i}\left(  u\right)
\right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}\right)  $
(namely, $i=k$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
Therefore, the relation $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(
v\right)  $ holds (according to Fact B). Thus, $\mathbf{i}\left(  u\right)
\leq\mathbf{i}\left(  v\right)  $ is proven in Case 1.
\par
Let us now consider Case 2. In this case, the word $u$ is a prefix of $v$. In
other words, there exists a word $q\in\mathfrak{A}^{\ast}$ such that $v=uq$.
Consider this $q$. If we had $q=\varnothing$, then we would have
$v=u\underbrace{q}_{=\varnothing}=u\neq v$, which is absurd. Hence, we cannot
have $q=\varnothing$. Thus, $q\neq\varnothing$, so that $\ell\left(  q\right)
>0$. Now, $\ell\left(  \underbrace{v}_{=uq}\right)  =\ell\left(  uq\right)
=\ell\left(  u\right)  +\underbrace{\ell\left(  q\right)  }_{>0}>\ell\left(
u\right)  $, so that $\ell\left(  v\right)  \geq\ell\left(  u\right)  +1$
(since $\ell\left(  v\right)  $ and $\ell\left(  u\right)  $ are integers).
Thus, $\ell\left(  u\right)  +1\leq\ell\left(  v\right)  $. Combined with
$\ell\left(  u\right)  +1\geq1$, this yields $\ell\left(  u\right)
+1\in\left\{  1,2,\ldots,\ell\left(  v\right)  \right\}  $. Thus, applying
(\ref{pf.prop.words.lex.a.long.lem1}) to $w=v$ and $i=\ell\left(  u\right)
+1$, we obtain $\left(  \mathbf{i}\left(  v\right)  \right)  _{\ell\left(
u\right)  +1}=v_{\ell\left(  u\right)  +1}\in\mathfrak{A}$ (since
$v\in\mathfrak{A}^{\ast}$). But (\ref{pf.prop.words.lex.a.long.lem2}) (applied
to $w=u$ and $i=\ell\left(  u\right)  +1$) yields $\left(  \mathbf{i}\left(
u\right)  \right)  _{\ell\left(  u\right)  +1}=\omega$ (since $\ell\left(
u\right)  +1>\ell\left(  u\right)  $).
\par
But (\ref{pf.prop.words.lex.a.long.omega}) (applied to $a=\left(
\mathbf{i}\left(  v\right)  \right)  _{\ell\left(  u\right)  +1}$) yields
$\left(  \mathbf{i}\left(  v\right)  \right)  _{\ell\left(  u\right)
+1}>\omega$ (since $\left(  \mathbf{i}\left(  v\right)  \right)  _{\ell\left(
u\right)  +1}\in\mathfrak{A}$). Thus, $\omega<\left(  \mathbf{i}\left(
v\right)  \right)  _{\ell\left(  u\right)  +1}$, so that $\left(
\mathbf{i}\left(  u\right)  \right)  _{\ell\left(  u\right)  +1}%
=\omega<\left(  \mathbf{i}\left(  v\right)  \right)  _{\ell\left(  u\right)
+1}$.
\par
Now, let $j\in\left\{  1,2,\ldots,\left(  \ell\left(  u\right)  +1\right)
-1\right\}  $. Then, $j\geq1$ and $j\leq\left(  \ell\left(  u\right)
+1\right)  -1=\ell\left(  u\right)  $. Now,%
\[
v=\underbrace{u}_{=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)
}\right)  }\underbrace{q}_{=\left(  q_{1},q_{2},\ldots,q_{\ell\left(
q\right)  }\right)  }=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)
}\right)  \left(  q_{1},q_{2},\ldots,q_{\ell\left(  q\right)  }\right)
=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  },q_{1},q_{2}%
,\ldots,q_{\ell\left(  q\right)  }\right)  .
\]
Hence, $v_{j}=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }%
,q_{1},q_{2},\ldots,q_{\ell\left(  q\right)  }\right)  _{j}=u_{j}$ (since
$j\leq\ell\left(  u\right)  $), so that $u_{j}=v_{j}$.
\par
Let us now forget that we fixed $j$. We thus have shown that every
$j\in\left\{  1,2,\ldots,\left(  \ell\left(  u\right)  +1\right)  -1\right\}
$ satisfies $u_{j}=v_{j}$. So we know that%
\[
\left(  \left(  \mathbf{i}\left(  u\right)  \right)  _{\ell\left(  u\right)
+1}<\left(  \mathbf{i}\left(  v\right)  \right)  _{\ell\left(  u\right)
+1}\text{, and every }j\in\left\{  1,2,\ldots,\left(  \ell\left(  u\right)
+1\right)  -1\right\}  \text{ satisfies }\left(  \mathbf{i}\left(  u\right)
\right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}\right)  .
\]
Hence, there exists an $i\in\left\{  1,2,3,\ldots\right\}  $ such that
$\left(  \left(  \mathbf{i}\left(  u\right)  \right)  _{i}<\left(
\mathbf{i}\left(  v\right)  \right)  _{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  \mathbf{i}\left(  u\right)
\right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}\right)  $
(namely, $i=\ell\left(  u\right)  +1$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
Therefore, the relation $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(
v\right)  $ holds (according to Fact B). Thus, $\mathbf{i}\left(  u\right)
\leq\mathbf{i}\left(  v\right)  $ is proven in Case 2.
\par
We have thus proven that $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(
v\right)  $ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this yields that $\mathbf{i}\left(  u\right)  \leq
\mathbf{i}\left(  v\right)  $ always holds. This proves
(\ref{pf.prop.words.lex.a.long.lem5}).}

\item If two words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$
satisfy $\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)  $, then%
\begin{equation}
u=v. \label{pf.prop.words.lex.a.long.lem6}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.words.lex.a.long.lem6}):} Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words which
satisfy $\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)  $. We need
to prove that $u=v$.
\par
We can WLOG assume that $\ell\left(  u\right)  \leq\ell\left(  v\right)  $
(because otherwise, we can simply interchange $u$ with $v$). Assume this.
\par
Let us assume (for the sake of contradiction) that $\ell\left(  u\right)
\neq\ell\left(  v\right)  $. Then, $\ell\left(  u\right)  <\ell\left(
v\right)  $ (since $\ell\left(  u\right)  \leq\ell\left(  v\right)  $). Hence,
$\ell\left(  u\right)  \leq\ell\left(  v\right)  -1$ (since $\ell\left(
u\right)  $ and $\ell\left(  v\right)  $ are integers), so that $\ell\left(
u\right)  +1\leq\ell\left(  v\right)  $. Combined with $\ell\left(  u\right)
+1\geq1$, this yields $\ell\left(  u\right)  +1\in\left\{  1,2,\ldots
,\ell\left(  v\right)  \right\}  $. Hence,
(\ref{pf.prop.words.lex.a.long.lem1}) (applied to $w=v$ and $i=\ell\left(
u\right)  +1$) yields $\left(  \mathbf{i}\left(  v\right)  \right)
_{\ell\left(  u\right)  +1}=v_{\ell\left(  u\right)  +1}\in\mathfrak{A}$
(since $v\in\mathfrak{A}^{\ast}$). But (\ref{pf.prop.words.lex.a.long.lem2})
(applied to $w=u$ and $i=\ell\left(  u\right)  +1$) yields $\left(
\mathbf{i}\left(  u\right)  \right)  _{\ell\left(  u\right)  +1}=\omega$
(since $\ell\left(  u\right)  +1>\ell\left(  u\right)  $). Hence,
$\omega=\left(  \underbrace{\mathbf{i}\left(  u\right)  }_{=\mathbf{i}\left(
v\right)  }\right)  _{\ell\left(  u\right)  +1}=\left(  \mathbf{i}\left(
v\right)  \right)  _{\ell\left(  u\right)  +1}\in\mathfrak{A}$. This
contradicts the fact that $\omega\notin\mathfrak{A}$. This contradiction
proves that our assumption (that $\ell\left(  u\right)  \neq\ell\left(
v\right)  $) is wrong. Hence, we cannot have $\ell\left(  u\right)  \neq
\ell\left(  v\right)  $. We thus must have $\ell\left(  u\right)  =\ell\left(
v\right)  $. Hence, $\ell\left(  v\right)  =\ell\left(  u\right)  $.
\par
We have $u=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $ and
$v=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $ (since $\ell\left(
v\right)  =\ell\left(  u\right)  $).
\par
Now, let $i\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $. Then,
$\left(  \mathbf{i}\left(  u\right)  \right)  _{i}=u_{i}$ (by
(\ref{pf.prop.words.lex.a.long.lem1}), applied to $w=u$), so that
$u_{i}=\left(  \mathbf{i}\left(  u\right)  \right)  _{i}$. Also, $i\in\left\{
1,2,\ldots,\ell\left(  u\right)  \right\}  =\left\{  1,2,\ldots,\ell\left(
v\right)  \right\}  $ (since $\ell\left(  u\right)  =\ell\left(  v\right)  $),
and therefore $\left(  \mathbf{i}\left(  v\right)  \right)  _{i}=v_{i}$ (by
(\ref{pf.prop.words.lex.a.long.lem1}), applied to $w=v$). Now, $u_{i}=\left(
\underbrace{\mathbf{i}\left(  u\right)  }_{=\mathbf{i}\left(  v\right)
}\right)  _{i}=\left(  \mathbf{i}\left(  v\right)  \right)  _{i}=v_{i}$.
\par
Let us now forget that we fixed $i$. We thus have shown that every
$i\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $ satisfies
$u_{i}=v_{i}$. In other words, corresponding entries of the two sequences
$\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $ and $\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $ are equal. Thus,
$\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  =\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $. Hence, $u=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  =v$ (since $v=\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $). This proves
(\ref{pf.prop.words.lex.a.long.lem6}).}

\item If two words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$
satisfy $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)  $, then%
\begin{equation}
u\leq v. \label{pf.prop.words.lex.a.long.lem7}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.words.lex.a.long.lem7}):} Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words which
satisfy $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)  $. We
need to prove that $u\leq v$.
\par
The definition of ``$u\leq v$'' shows that the
following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $u\leq v$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
\par
\end{statement}
\par
On the other hand, recall that the order on $\left(  \mathfrak{A}\cup\left\{
\omega\right\}  \right)  ^{\infty}$ is the lexicographic order. Hence, the
definition of ``$\mathbf{i}\left(  u\right)  \leq
\mathbf{i}\left(  v\right)  $'' shows that the following fact
holds:
\par
\begin{statement}
\textit{Fact B:} The relation $\mathbf{i}\left(  u\right)  \leq\mathbf{i}%
\left(  v\right)  $ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
\par
\end{statement}
\par
But we know that the relation $\mathbf{i}\left(  u\right)  \leq\mathbf{i}%
\left(  v\right)  $ holds (by assumption). Hence,
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
(according to Fact B). Thus, we must be in one of the following two cases:
\par
\textit{Case 1:} There exists an $i\in\left\{  1,2,3,\ldots\right\}  $ such
that $\left(  \left(  \mathbf{i}\left(  u\right)  \right)  _{i}<\left(
\mathbf{i}\left(  v\right)  \right)  _{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  \mathbf{i}\left(  u\right)
\right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}\right)  $.
\par
\textit{Case 2:} We have $\mathbf{i}\left(  u\right)  =\mathbf{i}\left(
v\right)  $.
\par
Let us first consider Case 1. In this case, there exists an $i\in\left\{
1,2,3,\ldots\right\}  $ such that $\left(  \left(  \mathbf{i}\left(  u\right)
\right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)  _{i}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }\left(
\mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(  v\right)
\right)  _{j}\right)  $. Let us denote this $i$ by $k$. Thus, $k$ is an
element of $\left\{  1,2,3,\ldots\right\}  $ such that $\left(  \left(
\mathbf{i}\left(  u\right)  \right)  _{k}<\left(  \mathbf{i}\left(  v\right)
\right)  _{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{
satisfies }\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(
\mathbf{i}\left(  v\right)  \right)  _{j}\right)  $.
\par
We have $k\in\left\{  1,2,3,\ldots\right\}  $, hence $k\geq1$. Also, recall
that%
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }\left(
\mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(  v\right)
\right)  _{j}. \label{pf.prop.words.lex.a.long.lem7.pf.2}%
\end{equation}
\par
Let us first (for the sake of contradiction) assume that $k>\ell\left(
v\right)  $. Then, $\left(  \mathbf{i}\left(  v\right)  \right)  _{k}=\omega$
(by (\ref{pf.prop.words.lex.a.long.lem2}), applied to $w=v$ and $i=k$). Now,
$\left(  \mathbf{i}\left(  u\right)  \right)  _{k}<\left(  \mathbf{i}\left(
v\right)  \right)  _{k}=\omega$, hence $\left(  \mathbf{i}\left(  u\right)
\right)  _{k}\neq\omega$. Combined with $\left(  \mathbf{i}\left(  u\right)
\right)  _{k}\in\mathfrak{A}\cup\left\{  \omega\right\}  $, this yields
$\left(  \mathbf{i}\left(  u\right)  \right)  _{k}\in\left(  \mathfrak{A}%
\cup\left\{  \omega\right\}  \right)  \setminus\left\{  \omega\right\}
\subset\mathfrak{A}$. Hence, (\ref{pf.prop.words.lex.a.long.omega}) (applied
to $a=\left(  \mathbf{i}\left(  u\right)  \right)  _{k}$) yields $\left(
\mathbf{i}\left(  u\right)  \right)  _{k}>\omega$. This contradicts $\left(
\mathbf{i}\left(  u\right)  \right)  _{k}<\omega$. This contradiction proves
that our assumption (that $k>\ell\left(  v\right)  $) was wrong. Hence, we
cannot have $k>\ell\left(  v\right)  $. Thus, we have $k\leq\ell\left(
v\right)  $. Hence, $k\in\left\{  1,2,\ldots,\ell\left(  v\right)  \right\}  $
(since $k\geq1$). Thus, applying (\ref{pf.prop.words.lex.a.long.lem1}) to
$w=v$ and $i=k$ yields $\left(  \mathbf{i}\left(  v\right)  \right)
_{k}=v_{k}$.
\par
We now must be in one of the following two subcases:
\par
\textit{Subcase 1.1:} We have $k\leq\ell\left(  u\right)  $.
\par
\textit{Subcase 1.2:} We have $k>\ell\left(  u\right)  $.
\par
Let us first consider Subcase 1.1. In this subcase, we have $k\leq\ell\left(
u\right)  $. Combined with $k\leq\ell\left(  v\right)  $, this yields
$k\leq\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  $,
so that $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  \right\}  $ (since $k\geq1$).
\par
Applying (\ref{pf.prop.words.lex.a.long.lem1}) to $w=u$ and $i=k$ yields
$\left(  \mathbf{i}\left(  u\right)  \right)  _{k}=u_{k}$ (since $k\in\left\{
1,2,\ldots,\ell\left(  u\right)  \right\}  $ (since $k\leq\ell\left(
u\right)  $ and $k\geq1$)). Thus, $u_{k}=\left(  \mathbf{i}\left(  u\right)
\right)  _{k}<\left(  \mathbf{i}\left(  v\right)  \right)  _{k}=v_{k}$.
\par
Now, let $j\in\left\{  1,2,\ldots,k-1\right\}  $ be arbitrary. Then, $j\geq1$
and $j\leq k-1$. Since $j\leq k-1<k\leq\ell\left(  u\right)  $, we have
$j\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $ (since $j\geq1$).
Thus, (\ref{pf.prop.words.lex.a.long.lem1}) (applied to $w=u$ and $i=j$)
yields $\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=u_{j}$, so that
$u_{j}=\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(
\mathbf{i}\left(  v\right)  \right)  _{j}$ (by
(\ref{pf.prop.words.lex.a.long.lem7.pf.2})).
\par
But since $j\leq k-1<k\leq\ell\left(  v\right)  $, we have $j\in\left\{
1,2,\ldots,\ell\left(  v\right)  \right\}  $ (since $j\geq1$). Thus,
(\ref{pf.prop.words.lex.a.long.lem1}) (applied to $w=v$ and $i=j$) yields
$\left(  \mathbf{i}\left(  v\right)  \right)  _{j}=v_{j}$, so that
$u_{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}=v_{j}$.
\par
Let us now forget that we fixed $j$. We thus have proven that every
$j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $u_{j}=v_{j}$.
\par
Altogether, we now know that $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $, and that $\left(
u_{k}<v_{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  $. Hence, there exists an $i\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  u_{i}<v_{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $ (namely,
$i=k$). Hence,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
Therefore, the relation $u\leq v$ holds (according to Fact A). Thus, $u\leq v$
is proven in Subcase 1.1.
\par
Let us now consider Subcase 1.2. In this subcase, we have $k>\ell\left(
u\right)  $. Thus, $\ell\left(  u\right)  <k\leq\ell\left(  v\right)  $. Thus,
the sequence $\left(  v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $
is a well-defined subsequence of the sequence $\left(  v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right)  $.
\par
Now, let $j\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $. Then,
$j\geq1$ and $j\leq\ell\left(  u\right)  $. Since $j\leq\ell\left(  u\right)
<k$, we have $j\leq k-1$ (since $j$ and $k$ are integers). Thus, $j\in\left\{
1,2,\ldots,k-1\right\}  $ (since $j\geq1$), so that $\left(  \mathbf{i}\left(
u\right)  \right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}$ (by
(\ref{pf.prop.words.lex.a.long.lem7.pf.2})).
\par
But (\ref{pf.prop.words.lex.a.long.lem1}) (applied to $w=u$ and $i=j$) yields
$\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=u_{j}$, so that
$u_{j}=\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(
\mathbf{i}\left(  v\right)  \right)  _{j}$.
\par
Since $j\leq k-1<k\leq\ell\left(  v\right)  $, we have $j\in\left\{
1,2,\ldots,\ell\left(  v\right)  \right\}  $ (since $j\geq1$). Thus,
(\ref{pf.prop.words.lex.a.long.lem1}) (applied to $w=v$ and $i=j$) yields
$\left(  \mathbf{i}\left(  v\right)  \right)  _{j}=v_{j}$, so that
$u_{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}=v_{j}$.
\par
Let us now forget that we fixed $j$. We thus have proven that every
$j\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $ satisfies
$u_{j}=v_{j}$. Hence, every entry of the sequence $\left(  u_{1},u_{2}%
,\ldots,u_{\ell\left(  u\right)  }\right)  $ equals the corresponding term of
the sequence $\left(  v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $.
Thus, $\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  =\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $. Now, $u=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $, so that $\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  =u$. But%
\begin{align*}
v  &  =\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  },v_{\ell\left(  u\right)
+1},v_{\ell\left(  u\right)  +2},\ldots,v_{\ell\left(  v\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  u\right)  <\ell\left(
v\right)  \right) \\
&  =\underbrace{\left(  v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)
}_{=u}\left(  v_{\ell\left(  u\right)  +1},v_{\ell\left(  u\right)  +2}%
,\ldots,v_{\ell\left(  v\right)  }\right)  =u\left(  v_{\ell\left(  u\right)
+1},v_{\ell\left(  u\right)  +2},\ldots,v_{\ell\left(  v\right)  }\right)  .
\end{align*}
Thus, $u$ is a prefix of $v$. Hence,
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
Therefore, the relation $u\leq v$ holds (according to Fact A). Thus, $u\leq v$
is proven in Subcase 1.2.
\par
Hence, $u\leq v$ is proven in each of the two Subcases 1.1 and 1.2. Since
these two Subcases cover all of Case 1, this yields that $u\leq v$ holds in
Case 1.
\par
Let us now consider Case 2. In this case, we have $\mathbf{i}\left(  u\right)
=\mathbf{i}\left(  v\right)  $. Thus, $u=v$ (by
(\ref{pf.prop.words.lex.a.long.lem6})). Hence, $u$ is a prefix of $v$. Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
Therefore, the relation $u\leq v$ holds (according to Fact A). Thus, $u\leq v$
is proven in Case 2.
\par
Now, $u\leq v$ is proven in each of the two Cases 1 and 2. Since these two
Cases cover all possibilities, this shows that $u\leq v$ always holds. This
proves (\ref{pf.prop.words.lex.a.long.lem7}).}
\end{itemize}

Let us now observe that the relation $\leq$ on $\mathfrak{A}^{\ast}$ is
reflexive\footnote{\textit{Proof.} Let $u\in\mathfrak{A}^{\ast}$. The relation
$\leq$ on $\left(  \mathfrak{A}\cup\left\{  \omega\right\}  \right)  ^{\infty
}$ is reflexive (since the lexicographic order on $\left(  \mathfrak{A}%
\cup\left\{  \omega\right\}  \right)  ^{\infty}$ is a partial order). Hence,
$\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  u\right)  $. Thus, $u\leq
u$ (by (\ref{pf.prop.words.lex.a.long.lem7}), applied to $v=u$).
\par
Now, let us forget that we fixed $u$. We thus have shown that $u\leq u$ for
every $u\in\mathfrak{A}^{\ast}$. In other words, the relation $\leq$ on
$\mathfrak{A}^{\ast}$ is reflexive. Qed.}, transitive\footnote{\textit{Proof.}
Let $u$, $v$ and $w$ be three elements of $\mathfrak{A}^{\ast}$ satisfying
$u\leq v$ and $v\leq w$. The relation $\leq$ on $\left(  \mathfrak{A}%
\cup\left\{  \omega\right\}  \right)  ^{\infty}$ is transitive (since the
lexicographic order on $\left(  \mathfrak{A}\cup\left\{  \omega\right\}
\right)  ^{\infty}$ is a partial order).
\par
We have $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)  $ (by
(\ref{pf.prop.words.lex.a.long.lem5})) and $\mathbf{i}\left(  v\right)
\leq\mathbf{i}\left(  w\right)  $ (by (\ref{pf.prop.words.lex.a.long.lem5}),
applied to $v$ and $w$ instead of $u$ and $v$). Combining these two
inequalities, we obtain $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(
v\right)  \leq\mathbf{i}\left(  w\right)  $, thus $\mathbf{i}\left(  u\right)
\leq\mathbf{i}\left(  w\right)  $ (since the relation $\leq$ on $\left(
\mathfrak{A}\cup\left\{  \omega\right\}  \right)  ^{\infty}$ is transitive).
Thus, $u\leq w$ (by (\ref{pf.prop.words.lex.a.long.lem7}), applied to $w$
instead of $v$).
\par
Now, let us forget that we fixed $u$, $v$ and $w$. We thus have shown that any
three elements $u$, $v$ and $w$ of $\mathfrak{A}^{\ast}$ satisfying $u\leq v$
and $v\leq w$ must satisfy $u\leq w$. In other words, the relation $\leq$ on
$\mathfrak{A}^{\ast}$ is transitive. Qed.} and
antisymmetric\footnote{\textit{Proof.} Let $u$ and $v$ be two elements of
$\mathfrak{A}^{\ast}$ satisfying $u\leq v$ and $v\leq u$. The relation $\leq$
on $\left(  \mathfrak{A}\cup\left\{  \omega\right\}  \right)  ^{\infty}$ is
antisymmetric (since the lexicographic order on $\left(  \mathfrak{A}%
\cup\left\{  \omega\right\}  \right)  ^{\infty}$ is a partial order).
\par
We have $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)  $ (by
(\ref{pf.prop.words.lex.a.long.lem5})) and $\mathbf{i}\left(  v\right)
\leq\mathbf{i}\left(  u\right)  $ (by (\ref{pf.prop.words.lex.a.long.lem5}),
applied to $v$ and $u$ instead of $u$ and $v$). Combining these two
inequalities, we obtain $\mathbf{i}\left(  u\right)  =\mathbf{i}\left(
v\right)  $ (since the relation $\leq$ on $\left(  \mathfrak{A}\cup\left\{
\omega\right\}  \right)  ^{\infty}$ is antisymmetric). Thus, $u=v$ (by
(\ref{pf.prop.words.lex.a.long.lem6})).
\par
Now, let us forget that we fixed $u$ and $v$. We thus have shown that any two
elements $u$ and $v$ of $\mathfrak{A}^{\ast}$ satisfying $u\leq v$ and $v\leq
u$ must satisfy $u=v$. In other words, the relation $\leq$ on $\mathfrak{A}%
^{\ast}$ is antisymmetric. Qed.}. In other words, the relation $\leq$ on
$\mathfrak{A}^{\ast}$ is (the smaller-or-equal relation of) a partial order on
the set $\mathfrak{A}^{\ast}$. In order to complete the proof Proposition
\ref{prop.words.lex}(a), we thus only need to prove that this partial order is
a total order.

Let $u$ and $v$ be two elements of $\mathfrak{A}^{\ast}$. We shall show that
either $u\leq v$ or $v\leq u$.

We distinguish between two cases:

\textit{Case 1:} There exists a $g\in\left\{  1,2,\ldots,\max\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  \mathbf{i}\left(  u\right)  \right)  _{g}\neq\left(  \mathbf{i}%
\left(  v\right)  \right)  _{g}$.

\textit{Case 2:} There exists no $g\in\left\{  1,2,\ldots,\max\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  \mathbf{i}\left(  u\right)  \right)  _{g}\neq\left(  \mathbf{i}%
\left(  v\right)  \right)  _{g}$.

Let us consider Case 1 first. In this case, there exists a $g\in\left\{
1,2,\ldots,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  \mathbf{i}\left(  u\right)  \right)  _{g}%
\neq\left(  \mathbf{i}\left(  v\right)  \right)  _{g}$. Let $k$ be the
\textbf{smallest} such $g$. Then, $k$ is an element $g\in\left\{
1,2,\ldots,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  \mathbf{i}\left(  u\right)  \right)  _{g}%
\neq\left(  \mathbf{i}\left(  v\right)  \right)  _{g}$. In other words, $k$ is
an element of $\left\{  1,2,\ldots,\max\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  \right\}  $ and satisfies $\left(
\mathbf{i}\left(  u\right)  \right)  _{k}\neq\left(  \mathbf{i}\left(
v\right)  \right)  _{k}$.

We can WLOG assume that $\left(  \mathbf{i}\left(  u\right)  \right)  _{k}%
\leq\left(  \mathbf{i}\left(  v\right)  \right)  _{k}$ (because otherwise, we
can simply interchange $u$ with $v$). Assume this. Then, $\left(
\mathbf{i}\left(  u\right)  \right)  _{k}<\left(  \mathbf{i}\left(  v\right)
\right)  _{k}$ (since $\left(  \mathbf{i}\left(  u\right)  \right)  _{k}%
\leq\left(  \mathbf{i}\left(  v\right)  \right)  _{k}$ and $\left(
\mathbf{i}\left(  u\right)  \right)  _{k}\neq\left(  \mathbf{i}\left(
v\right)  \right)  _{k}$).

Let $j\in\left\{  1,2,\ldots,k-1\right\}  $ be arbitrary. Then, $j\geq1$ and
$j\leq k-1$. Hence, $j\leq k-1<k\leq\max\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  $ (since $k\in\left\{  1,2,\ldots
,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
$), so that $j\in\left\{  1,2,\ldots,\max\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  \right\}  $ (since $j\geq1$).

Assume (for the sake of contradiction) that $\left(  \mathbf{i}\left(
u\right)  \right)  _{j}\neq\left(  \mathbf{i}\left(  v\right)  \right)  _{j}$.
Then, $j$ is an element $g\in\left\{  1,2,\ldots,\max\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that $\left(
\mathbf{i}\left(  u\right)  \right)  _{g}\neq\left(  \mathbf{i}\left(
v\right)  \right)  _{g}$ (since $j\in\left\{  1,2,\ldots,\max\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ and
$\left(  \mathbf{i}\left(  u\right)  \right)  _{j}\neq\left(  \mathbf{i}%
\left(  v\right)  \right)  _{j}$). Since $k$ is the \textbf{smallest} such $g$
(because this is how $k$ was defined), this shows that $j\geq k$. But this
contradicts $j<k$. This contradiction proves that our assumption (that
$\left(  \mathbf{i}\left(  u\right)  \right)  _{j}\neq\left(  \mathbf{i}%
\left(  v\right)  \right)  _{j}$) was wrong. Hence, we cannot have $\left(
\mathbf{i}\left(  u\right)  \right)  _{j}\neq\left(  \mathbf{i}\left(
v\right)  \right)  _{j}$. In other words, we must have $\left(  \mathbf{i}%
\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)
_{j}$.

Let us now forget that we fixed $j$. We thus have proven that every
$j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $\left(  \mathbf{i}\left(
u\right)  \right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}$. We
also know that $\left(  \mathbf{i}\left(  u\right)  \right)  _{k}<\left(
\mathbf{i}\left(  v\right)  \right)  _{k}$. From this, it is easy to conclude
that $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)
$\ \ \ \ \footnote{\textit{Proof.} Recall that the order on $\left(
\mathfrak{A}\cup\left\{  \omega\right\}  \right)  ^{\infty}$ is the
lexicographic order. Hence, the definition of ``$\mathbf{i}%
\left(  u\right)  \leq\mathbf{i}\left(  v\right)  $'' shows
that the following fact holds:
\par
\begin{statement}
\textit{Fact B:} The relation $\mathbf{i}\left(  u\right)  \leq\mathbf{i}%
\left(  v\right)  $ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
\end{statement}
\par
But we know that $\left(  \mathbf{i}\left(  u\right)  \right)  _{k}<\left(
\mathbf{i}\left(  v\right)  \right)  _{k}$, and that every $j\in\left\{
1,2,\ldots,k-1\right\}  $ satisfies $\left(  \mathbf{i}\left(  u\right)
\right)  _{j}=\left(  \mathbf{i}\left(  v\right)  \right)  _{j}$. Hence, there
exists an $i\in\left\{  1,2,3,\ldots\right\}  $ such that $\left(  \left(
\mathbf{i}\left(  u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)
\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(
\mathbf{i}\left(  v\right)  \right)  _{j}\right)  $ (namely, $i=k$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,3,\ldots
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \mathbf{i}\left(
u\right)  \right)  _{i}<\left(  \mathbf{i}\left(  v\right)  \right)
_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies
}\left(  \mathbf{i}\left(  u\right)  \right)  _{j}=\left(  \mathbf{i}\left(
v\right)  \right)  _{j}\right)  ,\\
&  \text{\textbf{or} }\mathbf{i}\left(  u\right)  =\mathbf{i}\left(  v\right)
.
\end{align*}
Therefore, $\mathbf{i}\left(  u\right)  \leq\mathbf{i}\left(  v\right)  $
(according to Fact B). Qed.}. Thus, $u\leq v$ (according to
(\ref{pf.prop.words.lex.a.long.lem7})). Hence, $u\leq v$ or $v\leq u$. Thus,
we have shown that $\left(  \text{either }u\leq v\text{ or }v\leq u\right)  $
in Case 1.

Let us now consider Case 2. In this case, there exists no $g\in\left\{
1,2,\ldots,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  \mathbf{i}\left(  u\right)  \right)  _{g}%
\neq\left(  \mathbf{i}\left(  v\right)  \right)  _{g}$. In other words,%
\begin{equation}
\text{every }g\in\left\{  1,2,\ldots,\max\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  \right\}  \text{ satisfies }\left(
\mathbf{i}\left(  u\right)  \right)  _{g}=\left(  \mathbf{i}\left(  v\right)
\right)  _{g}. \label{pf.prop.words.lex.a.long.12}%
\end{equation}


Now, let $g\in\left\{  1,2,3,\ldots\right\}  $. We shall show that $\left(
\mathbf{i}\left(  u\right)  \right)  _{g}=\left(  \mathbf{i}\left(  v\right)
\right)  _{g}$. If $g\in\left\{  1,2,\ldots,\max\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  \right\}  $, then this follows immediately
from (\ref{pf.prop.words.lex.a.long.12}). Hence, for the rest of this proof,
we can WLOG assume that we don't have $g\in\left\{  1,2,\ldots,\max\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $. Assume this.

We have $g\in\left\{  1,2,3,\ldots\right\}  $ but we don't have $g\in\left\{
1,2,\ldots,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $. Thus,%
\begin{align*}
g
&\in\left\{  1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots,\max\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  \\
&=\left\{
\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
+1,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
+2,\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
+3,\ldots\right\}  .
\end{align*}
In other words, $g>\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)
\right\}  $. Hence, $g>\max\left\{  \ell\left(  u\right)  ,\ell\left(
v\right)  \right\}  \geq\ell\left(  u\right)  $. Thus,
(\ref{pf.prop.words.lex.a.long.lem2}) (applied to $w=u$ and $i=g$) yields
$\left(  \mathbf{i}\left(  u\right)  \right)  _{g}=\omega$. Also,
$g>\max\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\geq\ell\left(  v\right)  $. Thus, (\ref{pf.prop.words.lex.a.long.lem2})
(applied to $w=v$ and $i=g$) yields $\left(  \mathbf{i}\left(  v\right)
\right)  _{g}=\omega$. Hence, $\omega=\left(  \mathbf{i}\left(  v\right)
\right)  _{g}$, so that $\left(  \mathbf{i}\left(  u\right)  \right)
_{g}=\omega=\left(  \mathbf{i}\left(  v\right)  \right)  _{g}$. Thus, $\left(
\mathbf{i}\left(  u\right)  \right)  _{g}=\left(  \mathbf{i}\left(  v\right)
\right)  _{g}$ is proven.

Now let us forget that we fixed $g$. We thus have shown that $\left(
\mathbf{i}\left(  u\right)  \right)  _{g}=\left(  \mathbf{i}\left(  v\right)
\right)  _{g}$ for every $g\in\left\{  1,2,3,\ldots\right\}  $. In other
words, each entry of the sequence $\left(  \left(  \mathbf{i}\left(  u\right)
\right)  _{1},\left(  \mathbf{i}\left(  u\right)  \right)  _{2},\left(
\mathbf{i}\left(  u\right)  \right)  _{3},\ldots\right)  $ equals the
corresponding entry of the sequence $\left(  \left(  \mathbf{i}\left(
v\right)  \right)  _{1},\left(  \mathbf{i}\left(  v\right)  \right)
_{2},\left(  \mathbf{i}\left(  v\right)  \right)  _{3},\ldots\right)  $. Thus,%
\[
\left(  \left(  \mathbf{i}\left(  u\right)  \right)  _{1},\left(
\mathbf{i}\left(  u\right)  \right)  _{2},\left(  \mathbf{i}\left(  u\right)
\right)  _{3},\ldots\right)  =\left(  \left(  \mathbf{i}\left(  v\right)
\right)  _{1},\left(  \mathbf{i}\left(  v\right)  \right)  _{2},\left(
\mathbf{i}\left(  v\right)  \right)  _{3},\ldots\right)  .
\]
Hence,%
\[
\mathbf{i}\left(  u\right)  =\left(  \left(  \mathbf{i}\left(  u\right)
\right)  _{1},\left(  \mathbf{i}\left(  u\right)  \right)  _{2},\left(
\mathbf{i}\left(  u\right)  \right)  _{3},\ldots\right)  =\left(  \left(
\mathbf{i}\left(  v\right)  \right)  _{1},\left(  \mathbf{i}\left(  v\right)
\right)  _{2},\left(  \mathbf{i}\left(  v\right)  \right)  _{3},\ldots\right)
=\mathbf{i}\left(  v\right)  ,
\]
so that $u=v$ (according to (\ref{pf.prop.words.lex.a.long.lem6})). Thus,
$u\leq v$. Hence, $u\leq v$ or $v\leq u$. Thus, we have shown that $\left(
\text{either }u\leq v\text{ or }v\leq u\right)  $ in Case 2.

We have thus shown that $\left(  \text{either }u\leq v\text{ or }v\leq
u\right)  $ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this yields that $\left(  \text{either }u\leq v\text{ or }v\leq
u\right)  $ always holds.

Let us now forget that we fixed $u$ and $v$. We thus have proven that if $u$
and $v$ are two elements of $\mathfrak{A}^{\ast}$, then either $u\leq v$ or
$v\leq u$. In other words, the relation $\leq$ on $\mathfrak{A}^{\ast}$ is
(the smaller-or-equal relation of) a total order on the set $\mathfrak{A}%
^{\ast}$ (since we already know that this relation is (the smaller-or-equal
relation of) a partial order on the set $\mathfrak{A}^{\ast}$). This proves
Proposition \ref{prop.words.lex}(a).

(h) Let $b\in\mathfrak{A}^{\ast}$. Let $a\in\mathfrak{A}^{\ast}$ be a prefix
of $b$. Then, $a\leq b$\ \ \ \ \footnote{\textit{Proof.} The definition of
``$a\leq b$'' shows that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $a\leq b$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  b\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  a_{i}<b_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}=b_{j}\right)
,\\
&  \text{\textbf{or} the word }a\text{ is a prefix of }b.
\end{align*}
\end{statement}
\par
But $a$ is a prefix of $b$. Hence,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  b\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  a_{i}<b_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}=b_{j}\right)
,\\
&  \text{\textbf{or} the word }a\text{ is a prefix of }b.
\end{align*}
Thus, $a\leq b$ (according to Fact A), qed.}. This proves Proposition
\ref{prop.words.lex}(h).

(i) Let $c\in\mathfrak{A}^{\ast}$. Let $a$ and $b$ be two prefixes of $c$. We
need to prove that either $a$ is a prefix of $b$, or $b$ is a prefix of $a$.
Our situation is symmetric in $a$ and $b$. Hence, we can WLOG assume that
$\ell\left(  a\right)  \leq\ell\left(  b\right)  $ (since otherwise, we can
switch $a$ with $b$). Assume this.

We have $a=\left(  c_{1},c_{2},\ldots,c_{\ell\left(  a\right)  }\right)
$\ \ \ \ \footnote{\textit{Proof.} We know that $a$ is a prefix of $c$. But a
prefix of $c$ means a word of the form $\left(  c_{1},c_{2},\ldots
,c_{i}\right)  $ for some $i\in\left\{  0,1,\ldots,\ell\left(  c\right)
\right\}  $ (according to the definition of ``a prefix of
$c$''). Thus, $a$ is a word of the form $\left(  c_{1}%
,c_{2},\ldots,c_{i}\right)  $ for some $i\in\left\{  0,1,\ldots,\ell\left(
c\right)  \right\}  $ (since $a$ is a prefix of $c$). In other words,
$a=\left(  c_{1},c_{2},\ldots,c_{i}\right)  $ for some $i\in\left\{
0,1,\ldots,\ell\left(  c\right)  \right\}  $. Consider this $i$. Then,
$\ell\left(  \underbrace{a}_{=\left(  c_{1},c_{2},\ldots,c_{i}\right)
}\right)  =\ell\left(  \left(  c_{1},c_{2},\ldots,c_{i}\right)  \right)  =i$,
so that $i=\ell\left(  a\right)  $. Now, $a=\left(  c_{1},c_{2},\ldots
,c_{i}\right)  =\left(  c_{1},c_{2},\ldots,c_{\ell\left(  a\right)  }\right)
$ (since $i=\ell\left(  a\right)  $), qed.} and $b=\left(  c_{1},c_{2}%
,\ldots,c_{\ell\left(  b\right)  }\right)  $\ \ \ \ \footnote{This can be
proved using the same argument that we have used to show that $a=\left(
c_{1},c_{2},\ldots,c_{\ell\left(  a\right)  }\right)  $ (but with $a$ replaced
by $b$).}. From $b=\left(  c_{1},c_{2},\ldots,c_{\ell\left(  b\right)
}\right)  $, we obtain $\left(  c_{1},c_{2},\ldots,c_{\ell\left(  b\right)
}\right)  =b=\left(  b_{1},b_{2},\ldots,b_{\ell\left(  b\right)  }\right)  $.
In other words,%
\begin{equation}
c_{i}=b_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{  1,2,\ldots
,\ell\left(  b\right)  \right\}  . \label{pf.prop.words.lex.i.cb}%
\end{equation}


Notice that $\ell\left(  a\right)  \in\left\{  0,1,\ldots,\ell\left(
b\right)  \right\}  $ (since $\ell\left(  a\right)  \leq\ell\left(  b\right)
$). We have $a=\left(  a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  }\right)
$, so that $\left(  a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  }\right)
=a=\left(  c_{1},c_{2},\ldots,c_{\ell\left(  a\right)  }\right)  $. In other
words,
\begin{equation}
a_{i}=c_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{  1,2,\ldots
,\ell\left(  a\right)  \right\}  . \label{pf.prop.words.lex.i.ac}%
\end{equation}
Now, fix any $i\in\left\{  1,2,\ldots,\ell\left(  a\right)  \right\}  $. Then,
$1\leq i$ and $i\leq\ell\left(  a\right)  \leq\ell\left(  b\right)  $, so that
$i\in\left\{  1,2,\ldots,\ell\left(  b\right)  \right\}  $, thus $c_{i}=b_{i}$
(by (\ref{pf.prop.words.lex.i.cb})). Hence, (\ref{pf.prop.words.lex.i.ac})
yields $a_{i}=c_{i}=b_{i}$.

Let us now forget that we fixed $i$. We thus have shown that $a_{i}=b_{i}$ for
every $i\in\left\{  1,2,\ldots,\ell\left(  a\right)  \right\}  $. In other
words, every entry of the sequence $\left(  a_{1},a_{2},\ldots,a_{\ell\left(
a\right)  }\right)  $ equals the corresponding entry of the sequence $\left(
b_{1},b_{2},\ldots,b_{\ell\left(  a\right)  }\right)  $. Hence, $\left(
a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  }\right)  =\left(  b_{1}%
,b_{2},\ldots,b_{\ell\left(  a\right)  }\right)  $. Thus, $a=\left(
a_{1},a_{2},\ldots,a_{\ell\left(  a\right)  }\right)  =\left(  b_{1}%
,b_{2},\ldots,b_{\ell\left(  a\right)  }\right)  $. Thus, $a$ is a word of the
form $\left(  b_{1},b_{2},\ldots,b_{i}\right)  $ for some $i\in\left\{
0,1,\ldots,\ell\left(  b\right)  \right\}  $ (namely, for $i=\ell\left(
a\right)  $). In other words, $a$ is a prefix of $b$ (since a prefix of $b$
means a word of the form $\left(  b_{1},b_{2},\ldots,b_{i}\right)  $ for some
$i\in\left\{  0,1,\ldots,\ell\left(  b\right)  \right\}  $ (according to the
definition of ``a prefix of $b$'')). Thus,
either $a$ is a prefix of $b$, or $b$ is a prefix of $a$. This proves
Proposition \ref{prop.words.lex}(i).

(k) Let $a\in\mathfrak{A}^{\ast}$ and $b\in\mathfrak{A}^{\ast}$ be such that
$b$ is nonempty. Then, $a$ is a prefix of $ab$. Thus, $a\leq ab$ (by
Proposition \ref{prop.words.lex}(h), applied to $ab$ instead of $b$). But
$a\neq ab$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $a=ab$.
Hence, $a\varnothing=a=ab$. Cancelling $a$ from this equality, we obtain
$\varnothing=b$, so that the word $b$ is empty. This contradicts the fact that
$b$ is nonempty. This contradiction proves that our assumption was wrong,
qed.}. Combined with $a\leq ab$, this yields $a<ab$. Proposition
\ref{prop.words.lex}(k) is thus proven.

Before we move on to the proofs of further parts of Proposition
\ref{prop.words.lex}, let us state some very simple properties of words:

\begin{lemma}
\label{lem.pf.prop.words.lex.uv}Let $u\in\mathfrak{A}^{\ast}$ and
$v\in\mathfrak{A}^{\ast}$.

\begin{enumerate}
\item[(a)] We have%
\begin{equation}
uv=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right)  .
\label{eq.lem.pf.prop.words.lex.uv.a}%
\end{equation}


\item[(b)] Every $j\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $
satisfies%
\begin{equation}
\left(  uv\right)  _{j}=u_{j}. \label{eq.lem.pf.prop.words.lex.uv.b}%
\end{equation}


\item[(c)] Every $j\in\left\{  \ell\left(  u\right)  +1,\ell\left(  u\right)
+2,\ldots,\ell\left(  uv\right)  \right\}  $ satisfies%
\begin{equation}
\left(  uv\right)  _{j}=v_{j-\ell\left(  u\right)  }.
\label{eq.lem.pf.prop.words.lex.uv.c}%
\end{equation}


\item[(d)] Assume that every $j\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ satisfies $u_{j}=v_{j}%
$. Then, either $u$ is a prefix of $v$, or $v$ is a prefix of $u$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.pf.prop.words.lex.uv}.](a) We have%
\[
\underbrace{u}_{=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)
}\underbrace{v}_{=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)
}\right)  }=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)
\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2},\ldots
,v_{\ell\left(  v\right)  }\right)  .
\]
This proves Lemma \ref{lem.pf.prop.words.lex.uv}(a).

(b) Let $j\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $. Taking the
$j$-th entry on both sides of the equality
(\ref{eq.lem.pf.prop.words.lex.uv.a}), we obtain%
\[
\left(  uv\right)  _{j}=\left(  \left(  u_{1},u_{2},\ldots,u_{\ell\left(
u\right)  },v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  \right)
_{j}=u_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j\leq\ell\left(  u\right)
\right)  .
\]
This proves Lemma \ref{lem.pf.prop.words.lex.uv}(b).

(c) Let $j\in\left\{  \ell\left(  u\right)  +1,\ell\left(  u\right)
+2,\ldots,\ell\left(  uv\right)  \right\}  $. Then, \newline$j\in\left\{
\ell\left(  u\right)  +1,\ell\left(  u\right)  +2,\ldots,\ell\left(
uv\right)  \right\}  =\left\{  \ell\left(  u\right)  +1,\ell\left(  u\right)
+2,\ldots,\ell\left(  u\right)  +\ell\left(  v\right)  \right\}  $ (since
$\ell\left(  uv\right)  =\ell\left(  u\right)  +\ell\left(  v\right)  $), so
that $j-\ell\left(  u\right)  \in\left\{  1,2,\ldots,\ell\left(  v\right)
\right\}  $. Hence, $v_{j-\ell\left(  u\right)  }$ is well-defined.

Taking the $j$-th entry on both sides of the equality
(\ref{eq.lem.pf.prop.words.lex.uv.a}), we obtain%
\[
\left(  uv\right)  _{j}=\left(  \left(  u_{1},u_{2},\ldots,u_{\ell\left(
u\right)  },v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  \right)
_{j}=v_{j-\ell\left(  u\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}j>\ell\left(  u\right)  \right)  .
\]
This proves Lemma \ref{lem.pf.prop.words.lex.uv}(c).

(d) Our situation is symmetric in $u$ and $v$. Hence, we can WLOG assume that
$\ell\left(  u\right)  \leq\ell\left(  v\right)  $ (since otherwise, we can
switch $u$ with $v$). Assume this.

Let $i\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $. Then, $i\geq1$
and $i\leq\ell\left(  u\right)  $. Combining $i\geq1$ with $i\leq\ell\left(
u\right)  \leq\ell\left(  v\right)  $, we see that $i\in\left\{
1,2,\ldots,\ell\left(  v\right)  \right\}  $. Hence, $v_{i}$ is well-defined.
Also, $u_{i}$ is well-defined (since $i\in\left\{  1,2,\ldots,\ell\left(
u\right)  \right\}  $). Also, $\min\left\{  \ell\left(  u\right)  ,\ell\left(
v\right)  \right\}  =\ell\left(  u\right)  $ (since $\ell\left(  u\right)
\leq\ell\left(  v\right)  $), so that $\ell\left(  u\right)  =\min\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  $.

We now have $i\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  =\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ (since $\ell\left(  u\right)  =\min\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  $). But we assumed that every $j\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ satisfies $u_{j}=v_{j}$. Applying this to $j=i$, we obtain
$u_{i}=v_{i}$.

Let us now forget that we fixed $i$. We thus have shown that for every
$i\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $, we have
$u_{i}=v_{i}$ (and both $u_{i}$ and $v_{i}$ are well-defined). In other words,
each entry of the sequence $\left(  u_{1},u_{2},\ldots,u_{\ell\left(
u\right)  }\right)  $ equals the corresponding entry of the sequence $\left(
v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $. Thus, $\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $, so that $u=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  $. Thus, $\left(  v_{1}%
,v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  =u$.

Now, $\ell\left(  u\right)  \leq\ell\left(  v\right)  $, so that the sequence
$\left(  v_{\ell\left(  u\right)  +1},v_{\ell\left(  u\right)  +2}%
,\ldots,v_{\ell\left(  v\right)  }\right)  $ makes sense and we have%
\begin{align*}
&  \left(  v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  }\right)  \left(
v_{\ell\left(  u\right)  +1},v_{\ell\left(  u\right)  +2},\ldots
,v_{\ell\left(  v\right)  }\right) \\
&  =\left(  v_{1},v_{2},\ldots,v_{\ell\left(  u\right)  },v_{\ell\left(
u\right)  +1},v_{\ell\left(  u\right)  +2},\ldots,v_{\ell\left(  v\right)
}\right)  =\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =v
\end{align*}
(since $v=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $), so
that $v=\underbrace{\left(  v_{1},v_{2},\ldots,v_{\ell\left(  u\right)
}\right)  }_{=u}\left(  v_{\ell\left(  u\right)  +1},v_{\ell\left(  u\right)
+2},\ldots,v_{\ell\left(  v\right)  }\right)  =u\left(  v_{\ell\left(
u\right)  +1},v_{\ell\left(  u\right)  +2},\ldots,v_{\ell\left(  v\right)
}\right)  $. Thus, $u$ is a prefix of $v$. Hence, either $u$ is a prefix of
$v$, or $v$ is a prefix of $u$. This proves Lemma
\ref{lem.pf.prop.words.lex.uv}(d).
\end{proof}

We shall now return to our proof of Proposition \ref{prop.words.lex}.

(b) Let $a,c,d\in\mathfrak{A}^{\ast}$ be such that $c\leq d$. We need to prove
that $ac\leq ad$.

If $c$ is a prefix of $d$, then the claim $ac\leq ad$
holds\footnote{\textit{Proof.} Assume that $c$ is a prefix of $d$. Then, there
exists a word $v\in\mathfrak{A}^{\ast}$ such that $d=cv$. Consider this $v$.
We thus have $a\underbrace{d}_{=cv}=acv=\left(  ac\right)  v$. Hence, $ac$ is
a prefix of $ad$. Thus, $ac\leq ad$ (by Proposition \ref{prop.words.lex}(h),
applied to $ac$ and $ad$ instead of $a$ and $b$), qed.}. Hence, for the rest
of this proof, we can WLOG assume that $c$ is not a prefix of $d$.

There exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  c\right)
,\ell\left(  d\right)  \right\}  \right\}  $ such that $\left(  c_{k}%
<d_{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies
}c_{j}=d_{j}\right)  $\ \ \ \ \footnote{\textit{Proof.} The definition of
``$c\leq d$'' shows that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $c\leq d$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  c_{i}<d_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }c_{j}=d_{j}\right)
,\\
&  \text{\textbf{or} the word }c\text{ is a prefix of }d.
\end{align*}
\end{statement}
\par
But $c\leq d$. Thus, (according to Fact A) we can conclude that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  c_{i}<d_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }c_{j}=d_{j}\right)
,\\
&  \text{\textbf{or} the word }c\text{ is a prefix of }d.
\end{align*}
Since $c$ is not a prefix of $d$, this yields that there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  c\right)  ,\ell\left(
d\right)  \right\}  \right\}  $ such that $\left(  c_{i}<d_{i}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }c_{j}%
=d_{j}\right)  $. Renaming the index $i$ as $k$ in this statement, we conclude
that there exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
c\right)  ,\ell\left(  d\right)  \right\}  \right\}  $ such that $\left(
c_{k}<d_{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{
satisfies }c_{j}=d_{j}\right)  $. Qed.}. Consider this $k$.

We have $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  c\right)
,\ell\left(  d\right)  \right\}  \right\}  $. Thus, $k\geq1$ and $k\leq
\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  $. From
this, it is easy to obtain $\ell\left(  a\right)  +k\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\right\}  $\ \ \ \ \footnote{\textit{Proof.} We have $\min\left\{
\underbrace{\ell\left(  ac\right)  }_{=\ell\left(  a\right)  +\ell\left(
c\right)  },\underbrace{\ell\left(  ad\right)  }_{=\ell\left(  a\right)
+\ell\left(  d\right)  }\right\}  =\min\left\{  \ell\left(  a\right)
+\ell\left(  c\right)  ,\ell\left(  a\right)  +\ell\left(  d\right)  \right\}
=\ell\left(  a\right)  +\min\left\{  \ell\left(  c\right)  ,\ell\left(
d\right)  \right\}  $. Hence, $\ell\left(  a\right)  +\min\left\{  \ell\left(
c\right)  ,\ell\left(  d\right)  \right\}  =\min\left\{  \ell\left(
ac\right)  ,\ell\left(  ad\right)  \right\}  $.
\par
But $\underbrace{k}_{\geq1}+\underbrace{\ell\left(  a\right)  }_{\geq0}\geq1$
and $\underbrace{k}_{\leq\min\left\{  \ell\left(  c\right)  ,\ell\left(
d\right)  \right\}  }+\ell\left(  a\right)  \leq\min\left\{  \ell\left(
c\right)  ,\ell\left(  d\right)  \right\}  +\ell\left(  a\right)  =\ell\left(
a\right)  +\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}
=\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}  $.
Combining these two inequalities, we obtain $\ell\left(  a\right)
+k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ac\right)  ,\ell\left(
ad\right)  \right\}  \right\}  $, qed.}. Thus, $\ell\left(  a\right)  +k\geq1$
and $\ell\left(  a\right)  +k\leq\min\left\{  \ell\left(  ac\right)
,\ell\left(  ad\right)  \right\}  \leq\ell\left(  ac\right)  $, so that
$\ell\left(  a\right)  +k\in\left\{  1,2,\ldots,\ell\left(  ac\right)
\right\}  $. Hence, $\left(  ac\right)  _{\ell\left(  a\right)  +k}$ is
well-defined. Similarly, $\left(  ad\right)  _{\ell\left(  a\right)  +k}$ is well-defined.

Recall that $c_{k}<d_{k}$, and
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }%
c_{j}=d_{j}. \label{pf.prop.words.lex.b.ass}%
\end{equation}


Now, $\left(  ac\right)  _{\ell\left(  a\right)  +k}<\left(  ad\right)
_{\ell\left(  a\right)  +k}$\ \ \ \ \footnote{\textit{Proof.} Since
$\ell\left(  a\right)  +\underbrace{k}_{\geq1}\geq\ell\left(  a\right)  +1$
and $\ell\left(  a\right)  +k\leq\min\left\{  \ell\left(  ac\right)
,\ell\left(  ad\right)  \right\}  \leq\ell\left(  ac\right)  $, we have
$\ell\left(  a\right)  +k\in\left\{  \ell\left(  a\right)  +1,\ell\left(
a\right)  +2,\ldots,\ell\left(  ac\right)  \right\}  $. Hence, Lemma
\ref{lem.pf.prop.words.lex.uv}(c) (applied to $u=a$, $v=c$ and $j=\ell\left(
a\right)  +k$) yields $\left(  ac\right)  _{\ell\left(  a\right)
+k}=c_{\left(  \ell\left(  a\right)  +k\right)  -\ell\left(  a\right)  }%
=c_{k}$.
\par
Since $\ell\left(  a\right)  +\underbrace{k}_{\geq1}\geq\ell\left(  a\right)
+1$ and $\ell\left(  a\right)  +k\leq\min\left\{  \ell\left(  ac\right)
,\ell\left(  ad\right)  \right\}  \leq\ell\left(  ad\right)  $, we have
$\ell\left(  a\right)  +k\in\left\{  \ell\left(  a\right)  +1,\ell\left(
a\right)  +2,\ldots,\ell\left(  ad\right)  \right\}  $. Hence, Lemma
\ref{lem.pf.prop.words.lex.uv}(c) (applied to $u=a$, $v=d$ and $j=\ell\left(
a\right)  +k$) yields $\left(  ad\right)  _{\ell\left(  a\right)
+k}=d_{\left(  \ell\left(  a\right)  +k\right)  -\ell\left(  a\right)  }%
=d_{k}$. Thus, $d_{k}=\left(  ad\right)  _{\ell\left(  a\right)  +k}$.
\par
Now, $\left(  ac\right)  _{\ell\left(  a\right)  +k}=c_{k}<d_{k}=\left(
ad\right)  _{\ell\left(  a\right)  +k}$, qed.}. Also, every $j\in\left\{
1,2,\ldots,\left(  \ell\left(  a\right)  +k\right)  -1\right\}  $ satisfies
$\left(  ac\right)  _{j}=\left(  ad\right)  _{j}$%
\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,\left(
\ell\left(  a\right)  +k\right)  -1\right\}  $. We need to prove that $\left(
ac\right)  _{j}=\left(  ad\right)  _{j}$.
\par
If $j\in\left\{  1,2,\ldots,\ell\left(  a\right)  \right\}  $, then%
\begin{align*}
\left(  ac\right)  _{j}  &  =a_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.pf.prop.words.lex.uv}(b), applied to }u=a\text{ and }v=c\right) \\
&  =\left(  ad\right)  _{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
ad\right)  _{j}=a_{j}\text{ (by Lemma \ref{lem.pf.prop.words.lex.uv}(b),
applied to }u=a\text{ and }v=d\text{)}\right)  .
\end{align*}
Hence, $\left(  ac\right)  _{j}=\left(  ad\right)  _{j}$ is proven in the case
when $j\in\left\{  1,2,\ldots,\ell\left(  a\right)  \right\}  $. Thus, for the
rest of this proof, we can WLOG assume that we don't have $j\in\left\{
1,2,\ldots,\ell\left(  a\right)  \right\}  $. Assume this.
\par
We have $j\in\left\{  1,2,\ldots,\left(  \ell\left(  a\right)  +k\right)
-1\right\}  $ but we don't have $j\in\left\{  1,2,\ldots,\ell\left(  a\right)
\right\}  $. Thus,
\begin{align*}
j  &  \in\left\{  1,2,\ldots,\left(  \ell\left(  a\right)  +k\right)
-1\right\}  \setminus\left\{  1,2,\ldots,\ell\left(  a\right)  \right\}
\subset\left\{  \ell\left(  a\right)  +1,\ell\left(  a\right)  +2,\ldots
,\left(  \ell\left(  a\right)  +k\right)  -1\right\} \\
&  \subset\left\{  \ell\left(  a\right)  +1,\ell\left(  a\right)
+2,\ldots,\ell\left(  ac\right)  \right\}
\end{align*}
(since $\left(  \ell\left(  a\right)  +k\right)  -1<\ell\left(  a\right)
+k\leq\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\leq\ell\left(  ac\right)  $). Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(c)
(applied to $u=a$ and $v=c$) yields $\left(  ac\right)  _{j}=c_{j-\ell\left(
a\right)  }$. Also,%
\[
j\in\left\{  \ell\left(  a\right)  +1,\ell\left(  a\right)  +2,\ldots,\left(
\ell\left(  a\right)  +k\right)  -1\right\}  \subset\left\{  \ell\left(
a\right)  +1,\ell\left(  a\right)  +2,\ldots,\ell\left(  ad\right)  \right\}
\]
(since $\left(  \ell\left(  a\right)  +k\right)  -1<\ell\left(  a\right)
+k\leq\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\leq\ell\left(  ad\right)  $). Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(c)
(applied to $u=a$ and $v=d$) yields $\left(  ad\right)  _{j}=d_{j-\ell\left(
a\right)  }$. Thus, $d_{j-\ell\left(  a\right)  }=\left(  ad\right)  _{j}$.
\par
But $j\in\left\{  \ell\left(  a\right)  +1,\ell\left(  a\right)
+2,\ldots,\underbrace{\left(  \ell\left(  a\right)  +k\right)  -1}%
_{=\ell\left(  a\right)  +\left(  k-1\right)  }\right\}  =\left\{  \ell\left(
a\right)  +1,\ell\left(  a\right)  +2,\ldots,\ell\left(  a\right)  +\left(
k-1\right)  \right\}  $, so that $j-\ell\left(  a\right)  \in\left\{
1,2,\ldots,k-1\right\}  $. Hence, (\ref{pf.prop.words.lex.b.ass}) (applied to
$j-\ell\left(  a\right)  $ instead of $j$) yields $c_{j-\ell\left(  a\right)
}=d_{j-\ell\left(  a\right)  }$. Altogether, we now obtain $\left(  ac\right)
_{j}=c_{j-\ell\left(  a\right)  }=d_{j-\ell\left(  a\right)  }=\left(
ad\right)  _{j}$, qed.}. From this, it is easy to obtain $ac\leq
ad$\ \ \ \ \footnote{\textit{Proof.} The definition of
``$ac\leq ad$'' shows that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $ac\leq ad$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ac\right)
_{i}<\left(  ad\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(  ad\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ac\text{ is a prefix of }ad.
\end{align*}
\end{statement}
\par
But $\ell\left(  a\right)  +k$ is an element of $\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\right\}  $ and has the property that $\left(  ac\right)  _{\ell\left(
a\right)  +k}<\left(  ad\right)  _{\ell\left(  a\right)  +k}$, while every
$j\in\left\{  1,2,\ldots,\left(  \ell\left(  a\right)  +k\right)  -1\right\}
$ satisfies $\left(  ac\right)  _{j}=\left(  ad\right)  _{j}$. Hence, there
exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ac\right)
,\ell\left(  ad\right)  \right\}  \right\}  $ such that $\left(  \left(
ac\right)  _{i}<\left(  ad\right)  _{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(
ad\right)  _{j}\right)  $ (namely, $i=\ell\left(  a\right)  +k$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ac\right)
_{i}<\left(  ad\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(  ad\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ac\text{ is a prefix of }ad.
\end{align*}
In other words, $ac\leq ad$ (according to Fact A). Qed.}. This completes the
proof of Proposition \ref{prop.words.lex}(b).

(c) Let $a,c,d\in\mathfrak{A}^{\ast}$ satisfy $ac\leq ad$. We need to prove
that $c\leq d$.

If $ac$ is a prefix of $ad$, then the claim $c\leq d$
holds\footnote{\textit{Proof.} Assume that $ac$ is a prefix of $ad$. Then,
there exists a word $v\in\mathfrak{A}^{\ast}$ such that $ad=\left(  ac\right)
v$. Consider this $v$. We thus have $ad=\left(  ac\right)  v=a\left(
cv\right)  $. Cancelling $a$ from this equality, we obtain $d=cv$. Hence, $c$
is a prefix of $d$. Thus, $c\leq d$ (by Proposition \ref{prop.words.lex}(h),
applied to $c$ and $d$ instead of $a$ and $b$), qed.}. Hence, for the rest of
this proof, we can WLOG assume that $ac$ is not a prefix of $ad$.

There exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ac\right)
,\ell\left(  ad\right)  \right\}  \right\}  $ such that \newline$\left(
\left(  ac\right)  _{k}<\left(  ad\right)  _{k}\text{, and every }j\in\left\{
1,2,\ldots,k-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(
ad\right)  _{j}\right)  $\ \ \ \ \footnote{\textit{Proof.} The definition of
``$ac\leq ad$'' shows that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $ac\leq ad$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ac\right)
_{i}<\left(  ad\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(  ad\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ac\text{ is a prefix of }ad.
\end{align*}
\end{statement}
\par
But $ac\leq ad$. Thus, (according to Fact A) we can conclude that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ac\right)
_{i}<\left(  ad\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(  ad\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ac\text{ is a prefix of }ad.
\end{align*}
Since $ac$ is not a prefix of $ad$, this yields that there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ac\right)  ,\ell\left(
ad\right)  \right\}  \right\}  $ such that $\left(  \left(  ac\right)
_{i}<\left(  ad\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(  ad\right)
_{j}\right)  $. Renaming the index $i$ as $k$ in this statement, we conclude
that there exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
ac\right)  ,\ell\left(  ad\right)  \right\}  \right\}  $ such that $\left(
\left(  ac\right)  _{k}<\left(  ad\right)  _{k}\text{, and every }j\in\left\{
1,2,\ldots,k-1\right\}  \text{ satisfies }\left(  ac\right)  _{j}=\left(
ad\right)  _{j}\right)  $. Qed.}. Consider this $k$.

We have $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ac\right)
,\ell\left(  ad\right)  \right\}  \right\}  $. Thus, $k\geq1$ and $k\leq
\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}  $. From
this, it is easy to obtain $k-\ell\left(  a\right)  \leq\min\left\{
\ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  $%
\ \ \ \ \footnote{\textit{Proof.} We have $k\leq\min\left\{  \underbrace{\ell
\left(  ac\right)  }_{=\ell\left(  a\right)  +\ell\left(  c\right)
},\underbrace{\ell\left(  ad\right)  }_{=\ell\left(  a\right)  +\ell\left(
d\right)  }\right\}  =\min\left\{  \ell\left(  a\right)  +\ell\left(
c\right)  ,\ell\left(  a\right)  +\ell\left(  d\right)  \right\}  =\ell\left(
a\right)  +\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}
$, so that $k-\ell\left(  a\right)  \leq\min\left\{  \ell\left(  c\right)
,\ell\left(  d\right)  \right\}  $, qed.}.

Recall that $\left(  ac\right)  _{k}<\left(  ad\right)  _{k}$, and
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }\left(
ac\right)  _{j}=\left(  ad\right)  _{j}. \label{pf.prop.words.lex.c.ass}%
\end{equation}


Now, we have $k-\ell\left(  a\right)  \geq1$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Then, we don't have $k-\ell\left(  a\right)  \geq1$.
Hence, $k-\ell\left(  a\right)  <1$. Since $k-\ell\left(  a\right)  $ and $1$
are integers, this yields $k-\ell\left(  a\right)  \leq0$, so that $k\leq
\ell\left(  a\right)  $. Combined with $k\geq1$, this yields $k\in\left\{
1,2,\ldots,\ell\left(  a\right)  \right\}  $. Thus, Lemma
\ref{lem.pf.prop.words.lex.uv}(b) (applied to $u=a$, $v=c$ and $j=k$) yields
$\left(  ac\right)  _{k}=a_{k}$. But Lemma \ref{lem.pf.prop.words.lex.uv}(b)
(applied to $u=a$, $v=d$ and $j=k$) yields $\left(  ad\right)  _{k}=a_{k}$.
Thus, $a_{k}=\left(  ad\right)  _{k}$, so that $\left(  ac\right)  _{k}%
=a_{k}=\left(  ad\right)  _{k}$. This contradicts $\left(  ac\right)
_{k}<\left(  ad\right)  _{k}$. This contradiction proves that our assumption
was wrong, qed.}. Combined with $k-\ell\left(  a\right)  \leq\min\left\{
\ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  $, this yields
$k-\ell\left(  a\right)  \in\left\{  1,2,\ldots,\min\left\{  \ell\left(
c\right)  ,\ell\left(  d\right)  \right\}  \right\}  $.

Combining $k-\ell\left(  a\right)  \leq\min\left\{  \ell\left(  c\right)
,\ell\left(  d\right)  \right\}  \leq\ell\left(  c\right)  $ with
$k-\ell\left(  a\right)  \geq1$, we see that $k-\ell\left(  a\right)
\in\left\{  1,2,\ldots,\ell\left(  c\right)  \right\}  $, so that
$c_{k-\ell\left(  a\right)  }$ is well-defined. Similarly, $d_{k-\ell\left(
a\right)  }$ is well-defined. It is easy to see that $c_{k-\ell\left(
a\right)  }<d_{k-\ell\left(  a\right)  }$\ \ \ \ \footnote{\textit{Proof.} We
have $k-\ell\left(  a\right)  \geq1$, thus $k\geq\ell\left(  a\right)  +1$.
Combined with $k\leq\min\left\{  \ell\left(  ac\right)  ,\ell\left(
ad\right)  \right\}  \leq\ell\left(  ac\right)  $, this yields $k\in\left\{
\ell\left(  a\right)  +1,\ell\left(  a\right)  +2,\ldots,\ell\left(
ac\right)  \right\}  $. Thus, Lemma \ref{lem.pf.prop.words.lex.uv}(c) (applied
to $u=a$, $v=c$ and $j=k$) yields $\left(  ac\right)  _{k}=c_{k-\ell\left(
a\right)  }$. Thus, $c_{k-\ell\left(  a\right)  }=\left(  ac\right)  _{k}$.
\par
But combining $k\geq\ell\left(  a\right)  +1$ with $k\leq\min\left\{
\ell\left(  ac\right)  ,\ell\left(  ad\right)  \right\}  \leq\ell\left(
ad\right)  $, we obtain $k\in\left\{  \ell\left(  a\right)  +1,\ell\left(
a\right)  +2,\ldots,\ell\left(  ad\right)  \right\}  $. Thus, Lemma
\ref{lem.pf.prop.words.lex.uv}(c) (applied to $u=a$, $v=d$ and $j=k$) yields
$\left(  ad\right)  _{k}=d_{k-\ell\left(  a\right)  }$.
\par
Now, $c_{k-\ell\left(  a\right)  }=\left(  ac\right)  _{k}<\left(  ad\right)
_{k}=d_{k-\ell\left(  a\right)  }$, qed.}. Also, every $j\in\left\{
1,2,\ldots,\left(  k-\ell\left(  a\right)  \right)  -1\right\}  $ satisfies
$c_{j}=d_{j}$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{
1,2,\ldots,\left(  k-\ell\left(  a\right)  \right)  -1\right\}  $. We need to
prove that $c_{j}=d_{j}$.
\par
We have $j\in\left\{  1,2,\ldots,\left(  k-\ell\left(  a\right)  \right)
-1\right\}  $. Thus, $j\geq1$ and $j\leq\left(  k-\ell\left(  a\right)
\right)  -1$. From $j\leq\left(  k-\ell\left(  a\right)  \right)  -1=\left(
k-1\right)  -\ell\left(  a\right)  $, we obtain $j+\ell\left(  a\right)  \leq
k-1$. Combined with $j+\underbrace{\ell\left(  a\right)  }_{\geq0}\geq j\geq
1$, this yields $j+\ell\left(  a\right)  \in\left\{  1,2,\ldots,k-1\right\}
$. Hence, (\ref{pf.prop.words.lex.c.ass}) (applied to $j+\ell\left(  a\right)
$ instead of $j$) yields $\left(  ac\right)  _{j+\ell\left(  a\right)
}=\left(  ad\right)  _{j+\ell\left(  a\right)  }$.
\par
But $\underbrace{j}_{\geq1}+\ell\left(  a\right)  \geq1+\ell\left(  a\right)
=\ell\left(  a\right)  +1$. Combined with $j+\ell\left(  a\right)  \leq
k-1\leq k\leq\min\left\{  \ell\left(  ac\right)  ,\ell\left(  ad\right)
\right\}  \leq\ell\left(  ac\right)  $, this yields $j+\ell\left(  a\right)
\in\left\{  \ell\left(  a\right)  +1,\ell\left(  a\right)  +2,\ldots
,\ell\left(  ac\right)  \right\}  $. Thus, Lemma
\ref{lem.pf.prop.words.lex.uv}(c) (applied to $a$, $c$ and $j+\ell\left(
a\right)  $ instead of $u$, $v$ and $j$) yields $\left(  ac\right)
_{j+\ell\left(  a\right)  }=c_{\left(  j+\ell\left(  a\right)  \right)
-\ell\left(  a\right)  }=c_{j}$. Thus, $c_{j}=\left(  ac\right)
_{j+\ell\left(  a\right)  }$.
\par
Also, combining $j+\ell\left(  a\right)  \geq\ell\left(  a\right)  +1$ with
$j+\ell\left(  a\right)  \leq\min\left\{  \ell\left(  ac\right)  ,\ell\left(
ad\right)  \right\}  \leq\ell\left(  ad\right)  $, we obtain $j+\ell\left(
a\right)  \in\left\{  \ell\left(  a\right)  +1,\ell\left(  a\right)
+2,\ldots,\ell\left(  ad\right)  \right\}  $. Thus, Lemma
\ref{lem.pf.prop.words.lex.uv}(c) (applied to $a$, $d$ and $j+\ell\left(
a\right)  $ instead of $u$, $v$ and $j$) yields $\left(  ad\right)
_{j+\ell\left(  a\right)  }=d_{\left(  j+\ell\left(  a\right)  \right)
-\ell\left(  a\right)  }=d_{j}$. Altogether, $c_{j}=\left(  ac\right)
_{j+\ell\left(  a\right)  }=\left(  ad\right)  _{j+\ell\left(  a\right)
}=d_{j}$, qed.}. From this, it is easy to obtain $c\leq d$%
\ \ \ \ \footnote{\textit{Proof.} The definition of ``$c\leq d$''
shows that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $c\leq d$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  c_{i}<d_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }c_{j}=d_{j}\right)
,\\
&  \text{\textbf{or} the word }c\text{ is a prefix of }d.
\end{align*}
\end{statement}
\par
But $k-\ell\left(  a\right)  $ is an element of $\left\{  1,2,\ldots
,\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  \right\}
$ and has the property that $c_{k-\ell\left(  a\right)  }<d_{k-\ell\left(
a\right)  }$, while every $j\in\left\{  1,2,\ldots,\left(  k-\ell\left(
a\right)  \right)  -1\right\}  $ satisfies $c_{j}=d_{j}$. Hence, there exists
an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  c\right)  ,\ell\left(
d\right)  \right\}  \right\}  $ such that $\left(  c_{i}<d_{i}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }c_{j}%
=d_{j}\right)  $ (namely, $i=k-\ell\left(  a\right)  $). Thus,
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  c\right)  ,\ell\left(  d\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  c_{i}<d_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }c_{j}=d_{j}\right)
,\\
&  \text{\textbf{or} the word }c\text{ is a prefix of }d.
\end{align*}
In other words, $c\leq d$ (according to Fact A). Qed.}. This completes the
proof of Proposition \ref{prop.words.lex}(c).

(d) Let $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $a\leq c$. We need to prove
that either we have $ab\leq cd$ or the word $a$ is a prefix of $c$.

If $a$ is a prefix of $c$, then the claim holds\footnote{\textit{Proof.}
Assume that $a$ is a prefix of $c$. Then, either we have $ab\leq cd$ or the
word $a$ is a prefix of $c$. Qed.}. Hence, for the rest of this proof, we can
WLOG assume that $a$ is not a prefix of $c$. Assume this.

There exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)
,\ell\left(  c\right)  \right\}  \right\}  $ such that $\left(  a_{k}%
<c_{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies
}a_{j}=c_{j}\right)  $\ \ \ \ \footnote{\textit{Proof.} The definition of
``$a\leq c$'' shows that the following fact
holds:
\par
\begin{statement}
\textit{Fact A:} The relation $a\leq c$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  a_{i}<c_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}=c_{j}\right)
,\\
&  \text{\textbf{or} the word }a\text{ is a prefix of }c.
\end{align*}
\end{statement}
\par
But $a\leq c$. Thus, (according to Fact A) we can conclude that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  a_{i}<c_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}=c_{j}\right)
,\\
&  \text{\textbf{or} the word }a\text{ is a prefix of }c.
\end{align*}
Since $a$ is not a prefix of $c$, this yields that there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \right\}  $ such that $\left(  a_{i}<c_{i}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}%
=c_{j}\right)  $. Renaming the index $i$ as $k$ in this statement, we conclude
that there exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
a\right)  ,\ell\left(  c\right)  \right\}  \right\}  $ such that $\left(
a_{k}<c_{k}\text{, and every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{
satisfies }a_{j}=c_{j}\right)  $. Qed.}. Consider this $k$.

We have $\ell\left(  ab\right)  =\ell\left(  a\right)  +\ell\left(  b\right)
\geq\ell\left(  a\right)  $ and $\ell\left(  cd\right)  =\ell\left(  c\right)
+\ell\left(  d\right)  \geq\ell\left(  c\right)  $, so that $\min\left\{
\underbrace{\ell\left(  ab\right)  }_{\geq\ell\left(  a\right)  }%
,\underbrace{\ell\left(  cd\right)  }_{\geq\ell\left(  c\right)  }\right\}
\geq\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $,
thus $\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}
\leq\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}  $.
Now, $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \right\}  \subset\left\{  1,2,\ldots,\min\left\{
\ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}  \right\}  $ (since
$\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \leq
\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}  $).

Now, recall that $a_{k}<c_{k}$, and
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }%
a_{j}=c_{j}. \label{pf.prop.words.lex.d.ass}%
\end{equation}


We have $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)
,\ell\left(  c\right)  \right\}  \right\}  $, so that $k\geq1$ and $k\leq
\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $.
Combining $k\geq1$ with $k\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \leq\ell\left(  a\right)  $, we obtain $k\in\left\{
1,2,\ldots,\ell\left(  a\right)  \right\}  $. Hence, Lemma
\ref{lem.pf.prop.words.lex.uv}(b) (applied to $u=a$, $v=b$ and $j=k$) yields
$\left(  ab\right)  _{k}=a_{k}$. But combining $k\geq1$ with $k\leq
\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \leq
\ell\left(  c\right)  $, we obtain $k\in\left\{  1,2,\ldots,\ell\left(
c\right)  \right\}  $. Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(b) (applied
to $u=c$, $v=d$ and $j=k$) yields $\left(  cd\right)  _{k}=c_{k}$. Hence,
$c_{k}=\left(  cd\right)  _{k}$. So we have $\left(  ab\right)  _{k}%
=a_{k}<c_{k}=\left(  cd\right)  _{k}$. Also, every $j\in\left\{
1,2,\ldots,k-1\right\}  $ satisfies $\left(  ab\right)  _{j}=\left(
cd\right)  _{j}$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{
1,2,\ldots,k-1\right\}  $. Then, $j\geq1$ and $j\leq k-1$. Combining $j\geq1$
with $j\leq k-1\leq k\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \leq\ell\left(  a\right)  $, we obtain $j\in\left\{
1,2,\ldots,\ell\left(  a\right)  \right\}  $. Hence, Lemma
\ref{lem.pf.prop.words.lex.uv}(b) (applied to $u=a$ and $v=b$) yields $\left(
ab\right)  _{j}=a_{j}$. But combining $j\geq1$ with $j\leq\min\left\{
\ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \leq\ell\left(
c\right)  $, we obtain $j\in\left\{  1,2,\ldots,\ell\left(  c\right)
\right\}  $. Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(b) (applied to $u=c$
and $v=d$) yields $\left(  cd\right)  _{j}=c_{j}$. Hence, $c_{j}=\left(
cd\right)  _{j}$. But (\ref{pf.prop.words.lex.d.ass}) yields $a_{j}=c_{j}$. So
we have $\left(  ab\right)  _{j}=a_{j}=c_{j}=\left(  cd\right)  _{j}$, qed.}.
From this, it is easy to obtain $ab\leq cd$\ \ \ \ \footnote{\textit{Proof.}
The definition of ``$ab\leq cd$'' shows that
the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $ab\leq cd$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ab\right)
_{i}<\left(  cd\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(  cd\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ab\text{ is a prefix of }cd.
\end{align*}
\end{statement}
\par
But $k$ is an element of $\left\{  1,2,\ldots,\min\left\{  \ell\left(
ab\right)  ,\ell\left(  cd\right)  \right\}  \right\}  $ and has the property
that $\left(  ab\right)  _{k}<\left(  cd\right)  _{k}$, while every
$j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $\left(  ab\right)
_{j}=\left(  cd\right)  _{j}$. Hence, there exists an $i\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)
\right\}  \right\}  $ such that $\left(  \left(  ab\right)  _{i}<\left(
cd\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }\left(  ab\right)  _{j}=\left(  cd\right)  _{j}\right)  $ (namely,
$i=k$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ab\right)
_{i}<\left(  cd\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(  cd\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ab\text{ is a prefix of }cd.
\end{align*}
In other words, $ab\leq cd$ (according to Fact A). Qed.}. This completes the
proof of Proposition \ref{prop.words.lex}(d).

(e) Let $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $ab\leq cd$. We need to show
that either we have $a\leq c$ or the word $c$ is a prefix of $a$.

If $c$ is a prefix of $a$, then the claim holds\footnote{\textit{Proof.}
Assume that $c$ is a prefix of $a$. Then, either we have $a\leq c$ or the word
$c$ is a prefix of $a$. Qed.}. Hence, for the rest of this proof, we can WLOG
assume that $c$ is not a prefix of $a$. Assume this.

If $a$ is a prefix of $c$, then the claim (that either we have $a\leq c$ or
the word $c$ is a prefix of $a$) holds\footnote{\textit{Proof.} Assume that
$a$ is a prefix of $c$. Then, $a\leq c$ (according to Proposition
\ref{prop.words.lex}(h), applied to $c$ instead of $b$). Thus, either we have
$a\leq c$ or the word $c$ is a prefix of $a$. Qed.}. Hence, for the rest of
this proof, we can WLOG assume that $a$ is not a prefix of $c$. Assume this.

It is now easy to see that $ab$ is not a prefix of $cd$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $ab$ is a prefix
of $cd$. Then, there exists a word $v\in\mathfrak{A}^{\ast}$ such that
$cd=\left(  ab\right)  v$. Consider this $v$. We have $cd=\left(  ab\right)
v=abv$. Hence, the word $c$ is a prefix of $abv$. But $a$ is also a prefix of
$abv$ (since $abv=a\left(  bv\right)  $). Thus, Proposition
\ref{prop.words.lex}(i) (applied to $a$, $c$ and $abv$ instead of $a$, $b$ and
$c$) yields that either $a$ is a prefix of $c$, or $c$ is a prefix of $a$.
Since $c$ is not a prefix of $a$, we can thus conclude that $a$ is a prefix of
$c$. This contradicts the fact that $a$ is not a prefix of $c$. This
contradiction proves that our assumption was wrong, qed.}.

There exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ab\right)
,\ell\left(  cd\right)  \right\}  \right\}  $ such that \newline$\left(
\left(  ab\right)  _{k}<\left(  cd\right)  _{k}\text{, and every }j\in\left\{
1,2,\ldots,k-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(
cd\right)  _{j}\right)  $\ \ \ \ \footnote{\textit{Proof.} The definition of
``$ab\leq cd$'' shows that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $ab\leq cd$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}
\right\}  \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ab\right)
_{i}<\left(  cd\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(  cd\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ab\text{ is a prefix of }cd.
\end{align*}
\end{statement}
\par
But $ab\leq cd$. Thus, (according to Fact A) we can conclude that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}
\right\}  \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  ab\right)
_{i}<\left(  cd\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(  cd\right)
_{j}\right)  ,\\
&  \text{\textbf{or} the word }ab\text{ is a prefix of }cd.
\end{align*}
Since $ab$ is not a prefix of $cd$, this yields that there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ab\right)  ,\ell\left(
cd\right)  \right\}  \right\}  $ such that $\left(  \left(  ab\right)
_{i}<\left(  cd\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(  cd\right)
_{j}\right)  $. Renaming the index $i$ as $k$ in this statement, we conclude
that there exists a $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
ab\right)  ,\ell\left(  cd\right)  \right\}  \right\}  $ such that $\left(
\left(  ab\right)  _{k}<\left(  cd\right)  _{k}\text{, and every }j\in\left\{
1,2,\ldots,k-1\right\}  \text{ satisfies }\left(  ab\right)  _{j}=\left(
cd\right)  _{j}\right)  $. Qed.}. Consider this $k$.

We have $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  ab\right)
,\ell\left(  cd\right)  \right\}  \right\}  $. Thus, $k\geq1$ and $k\leq
\min\left\{  \ell\left(  ab\right)  ,\ell\left(  cd\right)  \right\}  $. Also,
we know that $\left(  ab\right)  _{k}<\left(  cd\right)  _{k}$, and that%
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }\left(
ab\right)  _{j}=\left(  cd\right)  _{j}.\label{pf.prop.words.lex.e.ass}%
\end{equation}


We shall now show that $a\leq c$. We distinguish between two cases:

\textit{Case 1:} We have $k\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  $.

\textit{Case 2:} We don't have $k\leq\min\left\{  \ell\left(  a\right)
,\ell\left(  c\right)  \right\}  $.

Let us first consider Case 1. In this case, we have $k\leq\min\left\{
\ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $. Combined with
$k\geq1$, this yields $k\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
a\right)  ,\ell\left(  c\right)  \right\}  \right\}  $.

Combining $k\geq1$ with $k\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \leq\ell\left(  a\right)  $, we obtain $k\in\left\{
1,2,\ldots,\ell\left(  a\right)  \right\}  $. Hence, Lemma
\ref{lem.pf.prop.words.lex.uv}(b) (applied to $u=a$, $v=b$ and $j=k$) yields
$\left(  ab\right)  _{k}=a_{k}$. Thus, $a_{k}=\left(  ab\right)  _{k}$. But
combining $k\geq1$ with $k\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \leq\ell\left(  c\right)  $, we obtain $k\in\left\{
1,2,\ldots,\ell\left(  c\right)  \right\}  $. Hence, Lemma
\ref{lem.pf.prop.words.lex.uv}(b) (applied to $u=c$, $v=d$ and $j=k$) yields
$\left(  cd\right)  _{k}=c_{k}$. Hence, $c_{k}=\left(  cd\right)  _{k}$. So we
have $a_{k}=\left(  ab\right)  _{k}<\left(  cd\right)  _{k}=c_{k}$. Also,
every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $a_{j}=c_{j}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j\in\left\{  1,2,\ldots,k-1\right\}
$. Then, $j\geq1$ and $j\leq k-1$. Combining $j\geq1$ with $j\leq k-1\leq
k\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}
\leq\ell\left(  a\right)  $, we obtain $j\in\left\{  1,2,\ldots,\ell\left(
a\right)  \right\}  $. Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(b) (applied
to $u=a$ and $v=b$) yields $\left(  ab\right)  _{j}=a_{j}$. Thus,
$a_{j}=\left(  ab\right)  _{j}$. But combining $j\geq1$ with $j\leq
\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \leq
\ell\left(  c\right)  $, we obtain $j\in\left\{  1,2,\ldots,\ell\left(
c\right)  \right\}  $. Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(b) (applied
to $u=c$ and $v=d$) yields $\left(  cd\right)  _{j}=c_{j}$. But
(\ref{pf.prop.words.lex.e.ass}) yields $\left(  ab\right)  _{j}=\left(
cd\right)  _{j}$. So we have $a_{j}=\left(  ab\right)  _{j}=\left(  cd\right)
_{j}=c_{j}$, qed.}. From this, it is easy to obtain $a\leq c$%
\ \ \ \ \footnote{\textit{Proof.} The definition of ``$a\leq c$'' shows
that the following fact holds:
\par
\begin{statement}
\textit{Fact A:} The relation $a\leq c$ holds if and only if%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  a_{i}<c_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}=c_{j}\right)
,\\
&  \text{\textbf{or} the word }a\text{ is a prefix of }c.
\end{align*}
\end{statement}
\par
But $k$ is an element of $\left\{  1,2,\ldots,\min\left\{  \ell\left(
a\right)  ,\ell\left(  c\right)  \right\}  \right\}  $ and has the property
that $a_{k}<c_{k}$, while every $j\in\left\{  1,2,\ldots,k-1\right\}  $
satisfies $a_{j}=c_{j}$. Hence, there exists an $i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \right\}
$ such that $\left(  a_{i}<c_{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }a_{j}=c_{j}\right)  $ (namely, $i=k$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  a_{i}<c_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }a_{j}=c_{j}\right)
,\\
&  \text{\textbf{or} the word }a\text{ is a prefix of }c.
\end{align*}
In other words, $a\leq c$ (according to Fact A). Qed.}. Thus, $a\leq c$ is
proven in Case 1.

Let us now consider Case 2. In this case, we don't have $k\leq\min\left\{
\ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $. Thus, we have
$k>\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $.
Since $k$ and $\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)
\right\}  $ are integers, this yields $k\geq\min\left\{  \ell\left(  a\right)
,\ell\left(  c\right)  \right\}  +1$, so that $k-1\geq\min\left\{  \ell\left(
a\right)  ,\ell\left(  c\right)  \right\}  $. Thus, $\min\left\{  \ell\left(
a\right)  ,\ell\left(  c\right)  \right\}  \leq k-1$.

Now, let $j\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)
,\ell\left(  c\right)  \right\}  \right\}  $ be arbitrary. Then, $j\geq1$ and
$j\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  $.
Combining $j\geq1$ with $j\leq\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \leq k-1$, we obtain $j\in\left\{  1,2,\ldots,k-1\right\}
$, and therefore $\left(  ab\right)  _{j}=\left(  cd\right)  _{j}$ (by
(\ref{pf.prop.words.lex.e.ass})). But combining $j\geq1$ with $j\leq
\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \leq
\ell\left(  a\right)  $, we obtain $j\in\left\{  1,2,\ldots,\ell\left(
a\right)  \right\}  $. Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(b) (applied
to $u=a$ and $v=b$) yields $\left(  ab\right)  _{j}=a_{j}$. Thus,
$a_{j}=\left(  ab\right)  _{j}$. But combining $j\geq1$ with $j\leq
\min\left\{  \ell\left(  a\right)  ,\ell\left(  c\right)  \right\}  \leq
\ell\left(  c\right)  $, we obtain $j\in\left\{  1,2,\ldots,\ell\left(
c\right)  \right\}  $. Hence, Lemma \ref{lem.pf.prop.words.lex.uv}(b) (applied
to $u=c$ and $v=d$) yields $\left(  cd\right)  _{j}=c_{j}$. Thus,
$a_{j}=\left(  ab\right)  _{j}=\left(  cd\right)  _{j}=c_{j}$.

Now, let us forget that we fixed $j$. We thus have proven that every
$j\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  a\right)  ,\ell\left(
c\right)  \right\}  \right\}  $ satisfies $a_{j}=c_{j}$. Thus, Lemma
\ref{lem.pf.prop.words.lex.uv}(d) (applied to $u=a$ and $v=c$) yields that
either $a$ is a prefix of $c$, or $c$ is a prefix of $a$. Since $a$ is not a
prefix of $c$, this shows that $c$ is a prefix of $a$. But this contradicts
the fact that $c$ is not a prefix of $a$. We thus have obtained a
contradiction. Hence, $a\leq c$ (since ex falso quod libet). We thus have
proven that $a\leq c$ in Case 2.

We now have shown that $a\leq c$ holds in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, we thus conclude that $a\leq c$
always holds. Hence, either we have $a\leq c$ or the word $c$ is a prefix of
$a$. This completes the proof of Proposition \ref{prop.words.lex}(e).

(f) Let $a,b,c,d\in\mathfrak{A}^{\ast}$ satisfy $ab\leq cd$ and $\ell\left(
a\right)  \leq\ell\left(  c\right)  $. We have to prove that $a\leq c$.
Proposition \ref{prop.words.lex}(e) yields that either we have $a\leq c$ or
the word $c$ is a prefix of $a$. In other words, we are in one of the
following two Cases:

\textit{Case 1:} We have $a\leq c$.

\textit{Case 2:} The word $c$ is a prefix of $a$.

Let us first consider Case 1. In this case, we have $a\leq c$. Hence, $a\leq
c$ is proven in Case 1.

Now, let us consider Case 2. In this case, the word $c$ is a prefix of $a$.
Thus, there exists a word $v\in\mathfrak{A}^{\ast}$ such that $a=cv$. Consider
this $v$. We have $\ell\left(  \underbrace{a}_{=cv}\right)  =\ell\left(
cv\right)  =\ell\left(  c\right)  +\ell\left(  v\right)  $, so that
$\ell\left(  c\right)  +\ell\left(  v\right)  =\ell\left(  a\right)  \leq
\ell\left(  c\right)  $. Subtracting $\ell\left(  c\right)  $ from this
inequality, we obtain $\ell\left(  v\right)  \leq0$. Combined with
$\ell\left(  v\right)  \geq0$ (which is obvious), this yields $\ell\left(
v\right)  =0$. Thus, $v$ is the empty word; that is, we have $v=\varnothing$.
Thus, $a=c\underbrace{v}_{=\varnothing}=c$, so that $a\leq c$. Hence, $a\leq
c$ is proven in Case 2.

We now have shown that $a\leq c$ holds in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, we thus conclude that $a\leq c$
always holds. This proves Proposition \ref{prop.words.lex}(f).

(g) Let $a,b,c\in\mathfrak{A}^{\ast}$ satisfy $a\leq b\leq ac$. Then,
$b\varnothing=b\leq ac$. Hence, Proposition \ref{prop.words.lex}(e) (applied
to $b,\varnothing,a,c$ instead of $a,b,c,d$) yields that either we have $b\leq
a$ or the word $a$ is a prefix of $b$. In other words, we must be in one of
the following two cases:

\textit{Case 1:} We have $b\leq a$.

\textit{Case 2:} The word $a$ is a prefix of $b$.

Let us first consider Case 1. In this case, we have $b\leq a$. Combined with
$a\leq b$, this yields $a=b$ (since the order relation $\leq$ is antisymmetric
(because it is the smaller-or-equal relation of a partial order on the set
$\mathfrak{A}^{\ast}$)). Thus, $a$ is a prefix of $b$ (since $b$ is a prefix
of $b$). We have thus shown in Case 1 that $a$ is a prefix of $b$.

Let us now consider Case 2. In this case, the word $a$ is a prefix of $b$.
Thus, we have shown in Case 2 that $a$ is a prefix of $b$.

We now have shown that $a$ is a prefix of $b$ in each of the two Cases 1 and
2. Since these two Cases cover all possibilities, we thus conclude that $a$ is
always a prefix of $b$. This completes the proof of Proposition
\ref{prop.words.lex}(g).

(j) Let $a,b,c\in\mathfrak{A}^{\ast}$ be such that $a\leq b$ and $\ell\left(
a\right)  \geq\ell\left(  b\right)  $. We need to prove that $ac\leq bc$.

Proposition \ref{prop.words.lex}(d) (applied to $a$, $c$, $b$ and $c$) yields
that either we have $ac\leq bc$ or the word $a$ is a prefix of $b$. In other
words, we must be in one of the following two cases:

\textit{Case 1:} We have $ac\leq bc$.

\textit{Case 2:} The word $a$ is a prefix of $b$.

Let us first consider Case 1. In this case, we have $ac\leq bc$. Hence,
$ac\leq bc$ is proven in Case 1.

Let us now consider Case 2. In this case, the word $a$ is a prefix of $b$.
Thus, there exists a word $v\in\mathfrak{A}^{\ast}$ such that $b=av$. Consider
this $v$. We have $\ell\left(  \underbrace{b}_{=av}\right)  =\ell\left(
av\right)  =\ell\left(  a\right)  +\ell\left(  v\right)  $, hence $\ell\left(
a\right)  +\ell\left(  v\right)  =\ell\left(  b\right)  \leq\ell\left(
a\right)  $ (since $\ell\left(  a\right)  \geq\ell\left(  b\right)  $).
Subtracting $\ell\left(  a\right)  $ from this inequality, we obtain
$\ell\left(  v\right)  \leq0$. Combined with $\ell\left(  v\right)  \geq0$
(which is obvious), this yields $\ell\left(  v\right)  =0$. Thus, $v$ is the
empty word; that is, we have $v=\varnothing$. Thus, $b=a\underbrace{v}%
_{=\varnothing}=a$, so that $a=b$. Hence, $\underbrace{a}_{=b}c=bc\leq bc$.
Thus, $ac\leq bc$ is proven in Case 2.

We now have shown that $ac\leq bc$ holds in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, we thus conclude that $ac\leq
bc$ always holds. This proves Proposition \ref{prop.words.lex}(j).
\end{proof}
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.uw-wu}}

\textit{Solution to Exercise \ref{exe.words.uw-wu}.}

\begin{proof}
[Alternative proof of Corollary \ref{cor.words.uv=vu.3}.]We shall prove
Corollary \ref{cor.words.uv=vu.3} by strong induction on $\ell\left(
u\right)  +\ell\left(  v\right)  +\ell\left(  w\right)  $. That is, we fix an
$N\in \NN$, and we assume (as the induction hypothesis) that Corollary
\ref{cor.words.uv=vu.3} holds in the case when $\ell\left(  u\right)
+\ell\left(  v\right)  +\ell\left(  w\right)  <N$. We then need to prove that
Corollary \ref{cor.words.uv=vu.3} holds in the case when $\ell\left(
u\right)  +\ell\left(  v\right)  +\ell\left(  w\right)  =N$.

So let $u$, $v$ and $w$ be words satisfying $uv\geq vu$ and $vw\geq wv$ and
$\ell\left(  u\right)  +\ell\left(  v\right)  +\ell\left(  w\right)  =N$.
Assume that $v$ is nonempty. We want to prove that $uw\geq wu$.

If $u=\varnothing$, then $uw\geq wu$ holds obviously (because if
$u=\varnothing$, then $\underbrace{u}_{=\varnothing}%
w=w=w\underbrace{\varnothing}_{=u}=wu$). Hence, for the rest of this proof, we
can WLOG assume that $u\neq\varnothing$. Assume this. For a similar reason, we
WLOG assume that $w\neq\varnothing$. Recall also that $v$ is nonempty; that
is, $v\neq\varnothing$.

We have $\ell\left(  u\right)  >0$ (since $u\neq\varnothing$) and $\ell\left(
v\right)  >0$ (since $v\neq\varnothing$) and $\ell\left(  w\right)  >0$ (since
$w\neq\varnothing$).

Notice that%
\begin{equation}
uvc\geq vuc\ \ \ \ \ \ \ \ \ \ \text{for every }c\in\mathfrak{A}^{\ast}.
\label{sol.words.uw-wu.uvt}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.uw-wu.uvt}):} Let $c\in
\mathfrak{A}^{\ast}$. We have $uv\geq vu$ and thus $vu\leq uv$. Also,
$\ell\left(  vu\right)  =\ell\left(  v\right)  +\ell\left(  u\right)
=\ell\left(  u\right)  +\ell\left(  v\right)  =\ell\left(  uv\right)  $.
Hence, Proposition \ref{prop.words.lex}(j) (applied to $a=vu$ and $b=uv$)
yields $vuc\leq uvc$, so that $uvc\geq vuc$. This proves
(\ref{sol.words.uw-wu.uvt}).} Also,%
\begin{equation}
vwc\geq wvc\ \ \ \ \ \ \ \ \ \ \text{for every }c\in\mathfrak{A}^{\ast}.
\label{sol.words.uw-wu.vwt}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.uw-wu.vwt}):} Let $c\in
\mathfrak{A}^{\ast}$. We have $vw\geq wv$ and thus $wv\leq vw$. Also,
$\ell\left(  wv\right)  =\ell\left(  w\right)  +\ell\left(  v\right)
=\ell\left(  v\right)  +\ell\left(  w\right)  =\ell\left(  vw\right)  $.
Hence, Proposition \ref{prop.words.lex}(j) (applied to $a=wv$ and $b=vw$)
yields $wvc\leq vwc$, so that $vwc\geq wvc$. This proves
(\ref{sol.words.uw-wu.vwt}).}

Let us first assume that $u$ is a prefix of $w$. Then, there exists a
$w^{\prime}\in\mathfrak{A}^{\ast}$ satisfying $w=uw^{\prime}$ (since $u$ is a
prefix of $w$). Consider this $w^{\prime}$. Applying
(\ref{sol.words.uw-wu.uvt}) to $c=w^{\prime}$, we obtain $uvw^{\prime}\geq
v\underbrace{uw^{\prime}}_{=w}=vw\geq\underbrace{w}_{=uw^{\prime}}%
v=uw^{\prime}v$. That is, $uw^{\prime}v\leq uvw^{\prime}$. Hence, Proposition
\ref{prop.words.lex}(c) (applied to $a=u$, $c=w^{\prime}v$ and $d=vw^{\prime}%
$) yields $w^{\prime}v\leq vw^{\prime}$, so that $vw^{\prime}\geq w^{\prime}%
v$. Furthermore, $\ell\left(  \underbrace{w}_{=uw^{\prime}}\right)
=\underbrace{\ell\left(  u\right)  }_{>0}+\ell\left(  w^{\prime}\right)
>\ell\left(  w^{\prime}\right)  $, thus $\ell\left(  w^{\prime}\right)
<\ell\left(  w\right)  $ and therefore $\ell\left(  u\right)  +\ell\left(
v\right)  +\underbrace{\ell\left(  w^{\prime}\right)  }_{<\ell\left(
w\right)  }<\ell\left(  u\right)  +\ell\left(  v\right)  +\ell\left(
w\right)  =N$. Hence, Corollary \ref{cor.words.uv=vu.3} holds for $w^{\prime}$
instead of $w$ (by the induction hypothesis). Consequently, we obtain
$uw^{\prime}\geq w^{\prime}u$ (since $vw^{\prime}\geq w^{\prime}v$). Thus,
$w^{\prime}u\leq uw^{\prime}$, so that $uw^{\prime}u\leq uuw^{\prime}$ (by
Proposition \ref{prop.words.lex}(b), applied to $a=u$, $c=w^{\prime}u$ and
$d=uw^{\prime}$). Thus, $uuw^{\prime}\geq\underbrace{uw^{\prime}}_{=w}u=wu$,
so that $u\underbrace{w}_{=uw^{\prime}}=uuw^{\prime}\geq wu$.

Now, let us forget that we assumed that $u$ is a prefix of $w$. We thus have
proven that $uw\geq wu$ under the assumption that $u$ is a prefix of $w$.
Hence, for the rest of this proof of $uw\geq wu$, we can WLOG assume that
\[
u\text{ is not a prefix of }w.
\]
Assume this.

Let us next assume that $v$ is a prefix of $w$. Then, there exists a
$w^{\prime}\in\mathfrak{A}^{\ast}$ satisfying $w=vw^{\prime}$ (since $v$ is a
prefix of $w$). Consider this $w^{\prime}$. We have $v\underbrace{vw^{\prime}%
}_{=w}=vw\geq\underbrace{w}_{=vw^{\prime}}v=vw^{\prime}v$, thus $vw^{\prime
}v\leq vvw^{\prime}$. Thus, Proposition \ref{prop.words.lex}(c) (applied to
$a=v$, $c=w^{\prime}v$ and $d=vw^{\prime}$) yields $w^{\prime}v\leq
vw^{\prime}$, so that $vw^{\prime}\geq w^{\prime}v$. Furthermore, $\ell\left(
\underbrace{w}_{=vw^{\prime}}\right)  =\underbrace{\ell\left(  v\right)
}_{>0}+\ell\left(  w^{\prime}\right)  >\ell\left(  w^{\prime}\right)  $, thus
$\ell\left(  w^{\prime}\right)  <\ell\left(  w\right)  $ and therefore
$\ell\left(  u\right)  +\ell\left(  v\right)  +\underbrace{\ell\left(
w^{\prime}\right)  }_{<\ell\left(  w\right)  }<\ell\left(  u\right)
+\ell\left(  v\right)  +\ell\left(  w\right)  =N$. Hence, Corollary
\ref{cor.words.uv=vu.3} holds for $w^{\prime}$ instead of $w$ (by the
induction hypothesis). Consequently, we obtain $uw^{\prime}\geq w^{\prime}u$
(since $vw^{\prime}\geq w^{\prime}v$). Thus, $w^{\prime}u\leq uw^{\prime}$, so
that $vw^{\prime}u\leq vuw^{\prime}$ (by Proposition \ref{prop.words.lex}(b),
applied to $a=v$, $c=w^{\prime}u$ and $d=uw^{\prime}$). Thus, $vuw^{\prime
}\geq\underbrace{vw^{\prime}}_{=w}u=wu$. But (\ref{sol.words.uw-wu.uvt})
(applied to $c=w^{\prime}$) yields $uvw^{\prime}\geq vuw^{\prime}\geq wu$,
thus $u\underbrace{w}_{=vw^{\prime}}=uvw^{\prime}\geq wu$.

Now, let us forget that we assumed that $v$ is a prefix of $w$. We thus have
proven that $uw\geq wu$ under the assumption that $v$ is a prefix of $w$.
Hence, for the rest of this proof of $uw\geq wu$, we can WLOG assume that%
\[
v\text{ is not a prefix of }w.
\]
Assume this.

We have $vw\geq wv$, thus $wv\leq vw$. Hence, Proposition \ref{prop.words.lex}%
(e) (applied to $a=w$, $b=v$, $c=v$ and $d=w$) yields that either we have
$w\leq v$ or the word $v$ is a prefix of $w$. Since the word $v$ is not a
prefix of $w$, this yields that we have $w\leq v$. Thus,%
\[
v\geq w.
\]


Let us next assume that $w$ is a prefix of $u$. Then, there exists a
$u^{\prime}\in\mathfrak{A}^{\ast}$ satisfying $u=wu^{\prime}$ (since $w$ is a
prefix of $u$). Consider this $u^{\prime}$. We have $uv\geq vu$ and thus
$\underbrace{wu^{\prime}}_{=u}v=uv\geq v\underbrace{u}_{=wu^{\prime}%
}=vwu^{\prime}\geq wvu^{\prime}$ (by (\ref{sol.words.uw-wu.vwt}), applied to
$c=u^{\prime}$). In other words, $wvu^{\prime}\leq wu^{\prime}v$. Hence,
Proposition \ref{prop.words.lex}(c) (applied to $a=w$, $c=vu^{\prime}$ and
$d=u^{\prime}v$) yields $vu^{\prime}\leq u^{\prime}v$, so that $u^{\prime
}v\geq vu^{\prime}$. Furthermore, $\ell\left(  \underbrace{u}_{=wu^{\prime}%
}\right)  =\underbrace{\ell\left(  w\right)  }_{>0}+\ell\left(  u^{\prime
}\right)  >\ell\left(  u^{\prime}\right)  $, thus $\ell\left(  u^{\prime
}\right)  <\ell\left(  u\right)  $ and therefore $\underbrace{\ell\left(
u^{\prime}\right)  }_{<\ell\left(  u\right)  }+\ell\left(  v\right)
+\ell\left(  w\right)  <\ell\left(  u\right)  +\ell\left(  v\right)
+\ell\left(  w\right)  =N$. Hence, Corollary \ref{cor.words.uv=vu.3} holds for
$u^{\prime}$ instead of $u$ (by the induction hypothesis). Consequently, we
obtain $u^{\prime}w\geq wu^{\prime}$ (since $u^{\prime}v\geq vu^{\prime}$).
Thus, $wu^{\prime}\leq u^{\prime}w$, so that $wwu^{\prime}\leq wu^{\prime}w$
(by Proposition \ref{prop.words.lex}(b), applied to $a=w$, $c=wu^{\prime}$ and
$d=u^{\prime}w$). Thus, $wu^{\prime}w\geq w\underbrace{wu^{\prime}}_{=u}=wu$,
so that $\underbrace{u}_{=wu^{\prime}}w=wu^{\prime}w\geq wu$.

Now, let us forget that we assumed that $w$ is a prefix of $u$. We thus have
proven that $uw\geq wu$ under the assumption that $w$ is a prefix of $u$.
Hence, for the rest of this proof of $uw\geq wu$, we can WLOG assume that%
\[
w\text{ is not a prefix of }u.
\]
Assume this.

Let us now assume that $u$ is a prefix of $v$. Then, there exists a
$v^{\prime}\in\mathfrak{A}^{\ast}$ satisfying $v=uv^{\prime}$ (since $u$ is a
prefix of $v$). Consider this $v^{\prime}$. We have $uv^{\prime}=v\geq
w=w\varnothing$. Hence, $w\varnothing\leq uv^{\prime}$. Thus, Proposition
\ref{prop.words.lex}(e) (applied to $a=w$, $b=\varnothing$, $c=u$ and
$d=v^{\prime}$) yields that either we have $w\leq u$ or the word $u$ is a
prefix of $w$. Since the word $u$ is not a prefix of $w$, we can conclude from
this that $w\leq u$. Thus, Proposition \ref{prop.words.lex}(d) (applied to
$a=w$, $b=u$, $c=u$ and $d=w$) shows that either we have $wu\leq uw$ or the
word $w$ is a prefix of $u$. Since the word $w$ is not a prefix of $u$, this
yields that $wu\leq uw$, so that $uw\geq wu$.

Now, let us forget that we assumed that $u$ is a prefix of $v$. We thus have
proven that $uw\geq wu$ under the assumption that $u$ is a prefix of $v$.
Hence, for the rest of this proof of $uw\geq wu$, we can WLOG assume that%
\[
u\text{ is not a prefix of }v.
\]
Assume this.

We have $uv\geq vu$, thus $vu\leq uv$. Hence, Proposition \ref{prop.words.lex}%
(e) (applied to $a=v$, $b=u$, $c=u$ and $d=v$) yields that either we have
$v\leq u$ or the word $u$ is a prefix of $v$. Since the word $u$ is not a
prefix of $v$, this yields that we have $v\leq u$. Thus, $u\geq v$. Combined
with $v\geq w$, this yields $u\geq w$, so that $w\leq u$. Thus, Proposition
\ref{prop.words.lex}(d) (applied to $a=w$, $b=u$, $c=u$ and $d=w$) shows that
either we have $wu\leq uw$ or the word $w$ is a prefix of $u$. Since the word
$w$ is not a prefix of $u$, this yields that $wu\leq uw$, so that $uw\geq wu$.
Our proof of $uw\geq wu$ is thus complete.

Now, let us forget that we fixed $u$, $v$ and $w$. We have thus shown that if
$u$, $v$ and $w$ are words satisfying $uv\geq vu$ and $vw\geq wv$ and
$\ell\left(  u\right)  +\ell\left(  v\right)  +\ell\left(  w\right)  =N$, and
if $v$ is nonempty, then $uw\geq wu$. In other words, Corollary
\ref{cor.words.uv=vu.3} holds in the case when $\ell\left(  u\right)
+\ell\left(  v\right)  +\ell\left(  w\right)  =N$. This completes the
induction step. We thus have proven Corollary \ref{cor.words.uv=vu.3} (again).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.u=pn=qm}}

\textit{Solution to Exercise \ref{exe.words.u=pn=qm}.} If the word $u$
is empty, then Exercise \ref{exe.words.u=pn=qm} is easy to
solve\footnote{\textit{Proof.} Assume that the word $u$ is empty. Thus,
$u=\varnothing$. Hence, $u^{n}=\varnothing^{n}=\varnothing$, so that
$\varnothing=u^{n}=v^{m}$ and thus $v^{m}=\varnothing$. Consequently,
$v=\varnothing$ (since $m$ is positive). Thus, there exists a word $t$ and
positive integers $i$ and $j$ such that $u=t^{i}$ and $v=t^{j}$ (namely, we
can take $t=\varnothing$ and $i=1$ and $j=1$). In other words, Exercise
\ref{exe.words.u=pn=qm} is solved.}. Hence, for the rest of this solution, we
can WLOG assume that the word $u$ is nonempty. Assume this. The word $u^{n}$
is nonempty (since $u$ is nonempty and $n$ is a positive integer). In other
words, the word $v^{m}$ is nonempty (since $u^{n}=v^{m}$). Thus, $v$ is nonempty.

We have $u\underbrace{v^{m}}_{=u^{n}}=uu^{n}=u^{n+1}=\underbrace{u^{n}%
}_{=v^{m}}u=v^{m}u$ and $v^{m}v=v^{m+1}=vv^{m}$. Hence, Corollary
\ref{cor.words.uv=vu.3} (applied to $u$, $v^{m}$ and $v$ instead of $u$, $v$
and $w$) yields $uv\geq vu$ (since $v^{m}$ is nonempty).

But we also have $vv^{m}=v^{m}v$ (since $v^{m}v=vv^{m}$) and $v^{m}u=uv^{m}$
(since $uv^{m}=v^{m}u$). Thus, Corollary \ref{cor.words.uv=vu.3} (applied to
$v$, $v^{m}$ and $u$ instead of $u$, $v$ and $w$) yields $vu\geq uv$ (since
$v^{m}$ is nonempty). Combined with $uv\geq vu$, this yields $uv=vu$. Hence,
Proposition \ref{prop.words.uv=vu.2} yields that there exist a $t\in
\mathfrak{A}^{\ast}$ and two nonnegative integers $i$ and $j$ such that
$u=t^{i}$ and $v=t^{j}$\ \ \ \ \footnote{The $i$ and $j$ here are the
variables called $n$ and $m$ in Proposition \ref{prop.words.uv=vu.2}.}.
Consider this $t$ and these $i$ and $j$. We have $i\neq0$ (since $t^{i}=u$ is
nonempty), so that $i$ is a positive integer. Also, $j\neq0$ (since $t^{j}=v$
is nonempty), and therefore $j$ is a positive integer.

We have thus shown that there exists a word $t$ and positive integers $i$ and
$j$ such that $u=t^{i}$ and $v=t^{j}$. Exercise \ref{exe.words.u=pn=qm} is
thus solved.

\begin{commentedout}
[Here comes an \textit{alternative solution to Exercise
\ref{exe.words.u=pn=qm}}, which is a bit shorter:]

We assume WLOG that $vu\geq uv$ (otherwise, it is enough to switch $u$ and
$v$). We also WLOG assume that $u$ and $v$ are nonempty (otherwise, the
solution is easy).

We have $u\underbrace{v^{m}}_{=u^{n}}=uu^{n}=u^{n+1}=\underbrace{u^{n}%
}_{=v^{m}}u=v^{m}u$ and $v^{m}v=v^{m+1}=vv^{m}$. Hence, Proposition
\ref{prop.words.uv=vu.3} (applied to $u$, $v^{m}$ and $v$ instead of $u$, $v$
and $w$) yields that there exist a $t\in\mathfrak{A}^{\ast}$ and three
nonnegative integers $n^{\prime}$, $m^{\prime}$ and $p^{\prime}$ such that
$u=t^{n^{\prime}}$, $v^{m}=t^{m^{\prime}}$ and $v=t^{p^{\prime}}$. Discarding
$m^{\prime}$, we thus see that the exercise is solved.]
\end{commentedout}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.unvm.vmun}}

\textit{Solution to Exercise \ref{exe.words.unvm.vmun}.} We WLOG assume
that $u$ is nonempty (because otherwise, both $uv\geq vu$ and $u^{n}v^{m}\geq
v^{m}u^{n}$ hold for trivial reasons). Similarly, we WLOG assume that $v$ is nonempty.

The exercise asks us to prove the equivalence $\left(  uv\geq vu\right)
\Longleftrightarrow\left(  u^{n}v^{m}\geq v^{m}u^{n}\right)  $. We shall
verify its $\Longrightarrow$ and $\Longleftarrow$ parts separately:

$\Longrightarrow:$ Assume that $uv\geq vu$ holds. Then, Corollary
\ref{cor.words.uv=vu.3} (applied to $w=v^{m}$) yields $uv^{m}\geq v^{m}u$
(since $vv^{m}=v^{m+1}=v^{m}v$). Thus, Corollary \ref{cor.words.uv=vu.3}
(applied to $u^{n}$, $u$ and $v^{m}$ instead of $u$, $v$ and $w$) yields
$u^{n}v^{m}\geq v^{m}u^{n}$ (since $u^{n}u=u^{n+1}=uu^{n}$). This proves the
$\Longrightarrow$ part of the equivalence $\left(  uv\geq vu\right)
\Longleftrightarrow\left(  u^{n}v^{m}\geq v^{m}u^{n}\right)  $.

$\Longleftarrow:$ Assume that $u^{n}v^{m}\geq v^{m}u^{n}$ holds. Then,
Corollary \ref{cor.words.uv=vu.3} (applied to $u$, $u^{n}$ and $v^{m}$ instead
of $u$, $v$ and $w$) yields $uv^{m}\geq v^{m}u$ (since $uu^{n}=u^{n+1}=u^{n}%
u$, and since the word $u^{n}$ is nonempty\footnote{because $u$ is nonempty
and $n$ is positive}). Hence, Corollary \ref{cor.words.uv=vu.3} (applied to
$u$, $v^{m}$ and $v$ instead of $u$, $v$ and $w$) yields $uv\geq vu$ (since
$v^{m}v=v^{m+1}=vv^{m}$, and since the word $v^{m}$ is
nonempty\footnote{because $v$ is nonempty and $m$ is positive}). This proves
the $\Longleftarrow$ part of the equivalence $\left(  uv\geq vu\right)
\Longleftrightarrow\left(  u^{n}v^{m}\geq v^{m}u^{n}\right)  $.

Thus, both the $\Longrightarrow$ and the $\Longleftarrow$ part of the
equivalence $\left(  uv\geq vu\right)  \Longleftrightarrow\left(  u^{n}%
v^{m}\geq v^{m}u^{n}\right)  $ are proven, and Exercise
\ref{exe.words.unvm.vmun} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.un.vm}}

\textit{Solution to Exercise \ref{exe.words.un.vm}.}

Notice that $\ell\left(  u^{n}\right)  =n\ell\left(  u\right)  =m\ell\left(
v\right)  =\ell\left(  v^{m}\right)  $ (since $\ell\left(  v^{m}\right)
=m\ell\left(  v\right)  $), thus $\ell\left(  v^{m}\right)  =\ell\left(
u^{n}\right)  $. Now, we have the following two logical implications:%
\[
\left(  u^{n}v^{m}\geq v^{m}u^{n}\right)  \Longrightarrow\left(  u^{n}\geq
v^{m}\right)
\]
\footnote{\textit{Proof.} Assume that $u^{n}v^{m}\geq v^{m}u^{n}$. We need to
prove that $u^{n}\geq v^{m}$.
\par
We have $v^{m}u^{n}\leq u^{n}v^{m}$ (since $u^{n}v^{m}\geq v^{m}u^{n}$) and
$\ell\left(  v^{m}\right)  \leq\ell\left(  u^{n}\right)  $ (since $\ell\left(
v^{m}\right)  =\ell\left(  u^{n}\right)  $). Hence, $v^{m}\leq u^{n}$ (by
Proposition \ref{prop.words.lex}(f), applied to $a=v^{m}$, $b=u^{n}$,
$c=u^{n}$ and $d=v^{m}$). Thus, $u^{n}\geq v^{m}$, qed.} and%
\[
\left(  u^{n}\geq v^{m}\right)  \Longrightarrow\left(  u^{n}v^{m}\geq
v^{m}u^{n}\right)
\]
\footnote{\textit{Proof.} Assume that $u^{n}\geq v^{m}$. We need to prove that
$u^{n}v^{m}\geq v^{m}u^{n}$.
\par
If $v^{m}u^{n}\leq u^{n}v^{m}$, then $u^{n}v^{m}\geq v^{m}u^{n}$ is obviously
true. Hence, for the rest of this proof of $u^{n}v^{m}\geq v^{m}u^{n}$, we can
WLOG assume that we don't have $v^{m}u^{n}\leq u^{n}v^{m}$. Assume this.
\par
We have $v^{m}\leq u^{n}$ (since $u^{n}\geq v^{m}$). Hence, Proposition
\ref{prop.words.lex}(d) (applied to $a=v^{m}$, $b=u^{n}$, $c=u^{n}$ and
$d=v^{m}$) yields that either we have $v^{m}u^{n}\leq u^{n}v^{m}$ or the word
$v^{m}$ is a prefix of $u^{n}$. Since we don't have $v^{m}u^{n}\leq u^{n}%
v^{m}$, we therefore conclude that the word $v^{m}$ is a prefix of $u^{n}$. In
other words, there exists a $t\in\mathfrak{A}^{\ast}$ such that $u^{n}=v^{m}%
t$. Consider this $t$.
\par
We have $\ell\left(  v^{m}\right)  =\ell\left(  \underbrace{u^{n}}_{=v^{m}%
t}\right)  =\ell\left(  v^{m}t\right)  =\ell\left(  v^{m}\right)  +\ell\left(
t\right)  $, thus $0=\ell\left(  t\right)  $. Hence, the word $t$ is empty,
i.e., we have $t=\varnothing$. Thus, $u^{n}=v^{m}\underbrace{t}_{=\varnothing
}=v^{m}$, so that $\underbrace{u^{n}}_{=v^{m}}v^{m}=v^{m}\underbrace{v^{m}%
}_{=u^{n}}=v^{m}u^{n}\geq v^{m}u^{n}$, qed.}. Combining these two
implications, we obtain the equivalence $\left(  u^{n}v^{m}\geq v^{m}%
u^{n}\right)  \Longleftrightarrow\left(  u^{n}\geq v^{m}\right)  $. But
Exercise \ref{exe.words.unvm.vmun} shows that we have the equivalence $\left(
uv\geq vu\right)  \Longleftrightarrow\left(  u^{n}v^{m}\geq v^{m}u^{n}\right)
$. Altogether, we thus obtain the following chain of equivalences:%
\[
\left(  uv\geq vu\right)  \Longleftrightarrow\left(  u^{n}v^{m}\geq v^{m}%
u^{n}\right)  \Longleftrightarrow\left(  u^{n}\geq v^{m}\right)  .
\]
This solves Exercise \ref{exe.words.un.vm}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.uv=vu.k}}

\textit{Solution to Exercise \ref{exe.words.uv=vu.k}.} If $w$ is a
nonempty word, then let us denote by $\operatorname*{rad}\left(  w\right)  $
the shortest word $p$ such that $w$ is a power of $p$. (This is clearly
well-defined because $w$ is a power of itself, and because for every given
integer $\lambda$ there exists at most one word $v$ of length $\lambda$ such
that $w$ is a power of $v$.)

Every nonempty word $w$ and every positive integer $n$ satisfy%
\begin{equation}
\operatorname*{rad}\left(  w^{n}\right)  =\operatorname*{rad}\left(  w\right)
. \label{sol.words.uv=vu.k.rad1}%
\end{equation}


\textit{Proof of (\ref{sol.words.uv=vu.k.rad1}):} Let $w$ be a nonempty word,
and let $n$ be a positive integer. Let $q=\operatorname*{rad}\left(
w^{n}\right)  $. Thus, $q$ is the shortest word $p$ such that $w^{n}$ is a
power of $p$. Consequently, $w^{n}$ is a power of $q$, so that there exists a
positive integer $N$ such that $w^{n}=q^{N}$. Consider this $N$.

The word $w^{n}$ is nonempty (since $w$ is nonempty and $n$ is positive). That
is, the word $q^{N}$ is nonempty (since $w^{n}=q^{N}$). Hence, the word $q$ is
nonempty, so that $\ell\left(  q\right)  $ is nonzero.

Exercise \ref{exe.words.u=pn=qm} (applied to $w$, $q$, $n$ and $N$ instead of
$u$, $v$, $n$ and $m$) yields that there exists a word $t$ and positive
integers $i$ and $j$ such that $w=t^{i}$ and $q=t^{j}$. Consider these $t$,
$i$ and $j$. We have $q=t^{j}$ and thus $q^{N}=\left(  t^{j}\right)
^{N}=t^{jN} $, so that $t^{jN}=q^{N}=w^{n}$. Hence, $w^{n}$ is a power of $t$.
The word $t$ thus cannot be shorter than $q$ (since $q$ is the shortest word
$p$ such that $w^{n}$ is a power of $p$). Consequently, $j=1$ (because
otherwise, $t$ would be shorter than $t^{j}=q$). Hence, $q=t^{j}=t$ (since
$j=1$) and $w=t^{i}=q^{i}$ (since $t=q$). Thus, $w$ is a power of $q$.
Consequently, $\ell\left(  q\right)  \geq\ell\left(  \operatorname*{rad}%
\left(  w\right)  \right)  $ (since $\operatorname*{rad}\left(  w\right)  $ is
the shortest word $p$ such that $w$ is a power of $p$).

On the other hand, $\operatorname*{rad}\left(  w\right)  $ is the shortest
word $p$ such that $w$ is a power of $p$. Thus, $w$ is a power of
$\operatorname*{rad}\left(  w\right)  $. That is, there exists a
$P\in \NN$ such that $w=\left(  \operatorname*{rad}\left(  w\right)
\right)  ^{P}$. Consider this $P$. We have $P>0$ (since $\left(
\operatorname*{rad}\left(  w\right)  \right)  ^{P}=w$ is nonempty), so that
$Pn>0$ (since $n>0$). Also, taking both sides of the equality $w=\left(
\operatorname*{rad}\left(  w\right)  \right)  ^{P}$ to the $n$-th power, we
obtain $w^{n}=\left(  \left(  \operatorname*{rad}\left(  w\right)  \right)
^{P}\right)  ^{n}=\left(  \operatorname*{rad}\left(  w\right)  \right)  ^{Pn}%
$, so that $w^{n}$ is a power of $\operatorname*{rad}\left(  w\right)  $.

Recall that $q$ is the shortest word $p$ such that $w^{n}$ is a power of $p$.
Since $w^{n}$ is a power of $\operatorname*{rad}\left(  w\right)  $, this
yields that $\ell\left(  \operatorname*{rad}\left(  w\right)  \right)
\geq\ell\left(  q\right)  $. Combined with $\ell\left(  q\right)  \geq
\ell\left(  \operatorname*{rad}\left(  w\right)  \right)  $, this yields
$\ell\left(  \operatorname*{rad}\left(  w\right)  \right)  =\ell\left(
q\right)  $. Now,%
\[
\ell\left(  \underbrace{w^{n}}_{=\left(  \operatorname*{rad}\left(  w\right)
\right)  ^{Pn}}\right)  =\ell\left(  \left(  \operatorname*{rad}\left(
w\right)  \right)  ^{Pn}\right)  =Pn\cdot\underbrace{\ell\left(
\operatorname*{rad}\left(  w\right)  \right)  }_{=\ell\left(  q\right)
}=Pn\cdot\ell\left(  q\right)  .
\]
Compared with $\ell\left(  \underbrace{w^{n}}_{=q^{N}}\right)  =\ell\left(
q^{N}\right)  =N\cdot\ell\left(  q\right)  $, this yields $Pn\cdot\ell\left(
q\right)  =N\cdot\ell\left(  q\right)  $. Division by $\ell\left(  q\right)  $
(which is nonzero) yields $Pn=N$. Thus,
\[
q^{Pn}=q^{N}=w^{n}=\left(  \operatorname*{rad}\left(  w\right)  \right)
^{Pn}.
\]
Since taking the $Pn$-th root of a word is unambiguous (when said root
exists)\footnote{because $Pn>0$}, this yields $q=\operatorname*{rad}\left(
w\right)  $. Because of $q=\operatorname*{rad}\left(  w^{n}\right)  $, this
rewrites as $\operatorname*{rad}\left(  w^{n}\right)  =\operatorname*{rad}%
\left(  w\right)  $. This proves (\ref{sol.words.uv=vu.k.rad1}).

Next, we notice that%
\begin{equation}
\text{any two nonempty words }u\text{ and }v\text{ satisfying }uv=vu\text{
satisfy }\operatorname*{rad}\left(  u\right)  =\operatorname*{rad}\left(
v\right)  . \label{sol.words.uv=vu.k.rad2}%
\end{equation}


\textit{Proof of (\ref{sol.words.uv=vu.k.rad2}):} Let $u$ and $v$ be two
nonempty words satisfying $uv=vu$. Proposition \ref{prop.words.uv=vu.2} yields
that there exist a $t\in\mathfrak{A}^{\ast}$ and two nonnegative integers $n$
and $m$ such that $u=t^{n}$ and $v=t^{m}$. Consider this $t$ and these $n$ and
$m$. Since $u=t^{n}$, we have $\operatorname*{rad}\left(  u\right)
=\operatorname*{rad}\left(  t^{n}\right)  =\operatorname*{rad}\left(
t\right)  $ (by (\ref{sol.words.uv=vu.k.rad1}), applied to $t$ instead of
$w$). Similarly, $\operatorname*{rad}\left(  v\right)  =\operatorname*{rad}%
\left(  t\right)  $. Thus, $\operatorname*{rad}\left(  u\right)
=\operatorname*{rad}\left(  t\right)  =\operatorname*{rad}\left(  v\right)  $,
and this proves (\ref{sol.words.uv=vu.k.rad2}).

Now, we can finally solve the exercise. Let us show that%
\begin{equation}
u_{1}u_{i}\geq u_{i}u_{1}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{
1,2,\ldots,k\right\}  . \label{sol.words.uv=vu.k.rad3}%
\end{equation}


Indeed, (\ref{sol.words.uv=vu.k.rad3}) can be proven by induction over $i $:
The \textit{base case} ($i=1$) is obvious, whereas the \textit{induction step}
(proving $u_{1}u_{i+1}\geq u_{i+1}u_{1}$ using $u_{1}u_{i}\geq u_{i}u_{1}$)
results from applying Corollary \ref{cor.words.uv=vu.3} to $u=u_{1}$,
$v=u_{i}$ and $w=u_{i+1}$ (because we have $u_{i}u_{i+1}\geq u_{i+1}u_{i}$ by
assumption). We thus have shown (\ref{sol.words.uv=vu.k.rad3}).

Now, we can apply (\ref{sol.words.uv=vu.k.rad3}) to $i=k$, and obtain
$u_{1}u_{k}\geq u_{k}u_{1}$. But on the other hand, we can apply $u_{i}%
u_{i+1}\geq u_{i+1}u_{i}$ to $i=k$, and obtain $u_{k}u_{k+1}\geq u_{k+1}u_{k}%
$. Since $u_{k+1}=u_{1}$, this rewrites as $u_{k}u_{1}\geq u_{1}u_{k}$.
Contrasting this with $u_{1}u_{k}\geq u_{k}u_{1}$, we obtain $u_{1}u_{k}%
=u_{k}u_{1}$. Hence, (\ref{sol.words.uv=vu.k.rad2}) (applied to $u=u_{1}$ and
$v=u_{k}$) yields $\operatorname*{rad}\left(  u_{1}\right)
=\operatorname*{rad}\left(  u_{k}\right)  $. Similarly, we can show that
$\operatorname*{rad}\left(  u_{i}\right)  =\operatorname*{rad}\left(
u_{i-1}\right)  $ for every $i\in\left\{  1,2,\ldots,k\right\}  $, where
$u_{0}$ means $u_{k}$ (in fact, our situation is invariant under cyclically
shifting the $k$-tuple $\left(  u_{1},u_{2},\ldots,u_{k}\right)  $). Hence,
$\operatorname*{rad}\left(  u_{k}\right)  =\operatorname*{rad}\left(
u_{k-1}\right)  = \cdots=\operatorname*{rad}\left(  u_{1}\right)  $. Denote
this common value $\operatorname*{rad}\left(  u_{k}\right)
=\operatorname*{rad}\left(  u_{k-1}\right)  = \cdots=\operatorname*{rad}%
\left(  u_{1}\right)  $ by $t$. Then, for every $i\in\left\{  1,2,\ldots
,k\right\}  $, we have $t=\operatorname*{rad}\left(  u_{i}\right)  $, whence
$u_{i}$ is a power of $t$ (because $\operatorname*{rad}\left(  u_{i}\right)  $
is defined as the shortest word $p$ such that $u_{i}$ is a power of $p$). This
solves Exercise \ref{exe.words.uv=vu.k}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.equiv2}}

\textit{Solution to Exercise \ref{exe.words.lyndon.equiv2}.} (a) We
shall solve Exercise \ref{exe.words.lyndon.equiv2}(a) by strong induction on
$\ell\left(  u\right)  +\ell\left(  v\right)  $.

\textit{Induction step:} Let $N\in \NN$. Assume that Exercise
\ref{exe.words.lyndon.equiv2}(a) is already solved in the case when
$\ell\left(  u\right)  +\ell\left(  v\right)  <N$. We now need to solve
Exercise \ref{exe.words.lyndon.equiv2}(a) in the case when $\ell\left(
u\right)  +\ell\left(  v\right)  =N$.

We have assumed that Exercise \ref{exe.words.lyndon.equiv2}(a) is already
solved in the case when $\ell\left(  u\right)  +\ell\left(  v\right)  <N$. In
other words, we know that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }u\in\mathfrak{A}^{\ast}\text{ and }v\in\mathfrak{A}^{\ast}\text{ are
two words satisfying }uv<vu\text{ and }\ell\left(  u\right)  +\ell\left(
v\right)  <N\text{,}\\
\text{then there exists a nonempty suffix }s\text{ of }u\text{ satisfying
}sv<v
\end{array}
\right)  . \label{sol.words.lyndon.equiv2.a.indhyp}%
\end{equation}


So let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two words
satisfying $uv<vu$ and $\ell\left(  u\right)  +\ell\left(  v\right)  =N$. We
are going to prove that there exists a nonempty suffix $s$ of $u$ satisfying
$sv<v$.

Indeed, let us assume the contrary. Thus, there exists no nonempty suffix $s$
of $u$ satisfying $sv<v$. Hence,%
\begin{equation}
\text{whenever }s\text{ is a nonempty suffix of }u\text{, we do \textbf{not}
have }sv<v. \label{sol.words.lyndon.equiv2.a.1}%
\end{equation}


We have $u\neq\varnothing$ (since otherwise, we would have $u=\varnothing$ and
thus $\underbrace{u}_{=\varnothing}v=v=v\underbrace{\varnothing}_{=u}=vu$,
which would contradict $uv<vu$). In other words, the word $u$ is nonempty.
Similarly, the word $v$ is nonempty. We have $\ell\left(  u\right)  >0$ (since
$u$ is nonempty) and $\ell\left(  v\right)  >0$ (since $v$ is nonempty).

We have $\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)
}_{\substack{>0}}+\ell\left(  v\right)  >\ell\left(  v\right)  $, thus
$\ell\left(  uv\right)  \neq\ell\left(  v\right)  $, hence $uv\neq v$.

Since $u$ is a nonempty suffix of $u$, we can apply
(\ref{sol.words.lyndon.equiv2.a.1}) to $s=u$. As a result, we conclude that we
do \textbf{not} have $uv<v$. Hence, we do \textbf{not} have $uv\leq
v$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $uv\leq v$.
Since $uv\neq v$, this yields $uv<v$. This contradicts the fact that we do
\textbf{not} have $uv<v$. This contradiction proves that our assumption was
wrong, qed.}. In other words, we do \textbf{not} have $uv\leq v\varnothing$
(since $v\varnothing=v$).

But $uv\varnothing=uv<vu$. Thus, Proposition \ref{prop.words.lex}(e) (applied
to $a=uv$, $b=\varnothing$, $c=v$ and $d=u$) yields that either we have
$uv\leq v$ or the word $v$ is a prefix of $uv$. Since we do not have $uv\leq
v$, we can therefore conclude that the word $v$ is a prefix of the word $uv$.
In other words, there exists a word $g\in\mathfrak{A}^{\ast}$ such that
$uv=vg$. Consider this $g$.

Since $u$ is a prefix of $uv$, we have $u\leq uv<vu$. Thus, $u\varnothing
=u<vu$. Hence, Proposition \ref{prop.words.lex}(e) (applied to $a=u$,
$b=\varnothing$, $c=v$ and $d=u$) yields that either we have $u\leq v$ or the
word $v$ is a prefix of $u$. In other words, we are in one of the following
two cases:

\textit{Case 1:} We have $u\leq v$.

\textit{Case 2:} The word $v$ is a prefix of $u$.

Let us consider Case 1 first. In this case, we have $u\leq v$. Thus,
Proposition \ref{prop.words.lex}(d) (applied to $a=u$, $b=v$, $c=v$ and
$d=\varnothing$) yields that either we have $uv\leq v\varnothing$ or the word
$u$ is a prefix of $v$. Since we do not have $uv\leq v\varnothing$, this
yields that the word $u$ is a prefix of $v$. In other words, there exists a
word $q\in\mathfrak{A}^{\ast}$ such that $v=uq$. Consider this $q$. We have
$\ell\left(  \underbrace{v}_{=uq}\right)  =\ell\left(  uq\right)
=\underbrace{\ell\left(  u\right)  }_{>0}+\ell\left(  q\right)  >\ell\left(
q\right)  $, so that $\ell\left(  q\right)  <\ell\left(  v\right)  $. Also,
$u\underbrace{uq}_{=v}=uv<\underbrace{v}_{=uq}u=uqu$. Hence, Proposition
\ref{prop.words.lex}(c) (applied to $a=u$, $c=uq$ and $d=qu$) yields that
$uq\leq qu$. Hence, $uq<qu$\ \ \ \ \footnote{\textit{Proof.} If we had
$uq=qu$, then we would have $u\underbrace{uq}_{=qu}=uqu$, which would
contradict $uuq<uqu$. Hence, we cannot have $uq=qu$. Thus, we have $uq\neq
qu$. Combined with $uq\leq qu$, this yields $uq<qu$, qed.}. Since $\ell\left(
u\right)  +\underbrace{\ell\left(  q\right)  }_{<\ell\left(  v\right)  }%
<\ell\left(  u\right)  +\ell\left(  v\right)  =N$, we can therefore apply
(\ref{sol.words.lyndon.equiv2.a.indhyp}) to $q$ instead of $v$. As a result,
we obtain that there exists a nonempty suffix $s$ of $u$ satisfying $sq<q$.
Let us denote this $s$ by $t$. Thus, $t$ is a nonempty suffix of $u$
satisfying $tq<q$. Thus, Proposition \ref{prop.words.lex}(j) (applied to
$a=tq$, $b=q$ and $c=g$) yields that $tqg\leq qg$ (because $\ell\left(
tq\right)  =\underbrace{\ell\left(  t\right)  }_{\substack{>0\\\text{(since
}t\text{ is nonempty)}}}+\ell\left(  q\right)  >\ell\left(  q\right)  $).

But $\underbrace{uq}_{=v}g=vg=uv$ (since $uv=vg$). Cancelling $u$ from this
equality, we obtain $qg=v$. Hence, $t\underbrace{v}_{=qg}=tqg\leq qg=v$. But
we have $\ell\left(  tv\right)  =\underbrace{\ell\left(  t\right)
}_{\substack{>0\\\text{(since }t\text{ is nonempty)}}}+\ell\left(  v\right)
>\ell\left(  v\right)  $, thus $\ell\left(  tv\right)  \neq\ell\left(
v\right)  $, hence $tv\neq v$. Combined with $tv\leq v$, this yields $tv<v$.

On the other hand, $t$ is a nonempty suffix of $u$. Hence,
(\ref{sol.words.lyndon.equiv2.a.1}) (applied to $s=t$) yields that we do
\textbf{not} have $tv<v$. This contradicts the fact that $tv<v$. Hence, we
have obtained a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $v$ is a prefix of $u$.
Hence, there exists a word $r\in\mathfrak{A}^{\ast}$ satisfying $u=vr$.
Consider this $r$. Clearly, $r$ is a suffix of $u$ (since $u=vr$).

We have $\underbrace{vr}_{=u}v=uv<v\underbrace{u}_{=vr}=vvr$. Thus,
Proposition \ref{prop.words.lex}(c) (applied to $a=v$, $c=rv$ and $d=vr$)
yields $rv\leq vr$. Hence, $rv<vr$\ \ \ \ \footnote{\textit{Proof.} If we had
$rv=vr$, then we would have $v\underbrace{rv}_{=vr}=vvr$, which would
contradict $vrv<vvr$. Hence, we cannot have $rv=vr$. Thus, we have $rv\neq
vr$. Combined with $rv\leq vr$, this yields $rv<vr$, qed.}. Also, $\ell\left(
\underbrace{u}_{=vr}\right)  =\ell\left(  vr\right)  =\underbrace{\ell\left(
v\right)  }_{>0}+\ell\left(  r\right)  >\ell\left(  r\right)  $, so that
$\ell\left(  r\right)  <\ell\left(  u\right)  $ and thus $\underbrace{\ell
\left(  r\right)  }_{<\ell\left(  u\right)  }+\ell\left(  v\right)
<\ell\left(  u\right)  +\ell\left(  v\right)  =N$. Therefore,
(\ref{sol.words.lyndon.equiv2.a.indhyp}) can be applied to $r$ instead of $u$.
As a result, we obtain that there exists a nonempty suffix $s$ of $r$
satisfying $sv<v$. Let us denote this by $t$. Thus, $t$ is a nonempty suffix
of $r$ satisfying $tv<v$.

We know that $t$ is a suffix of the word $r$, which (in turn) is a suffix of
$u$. Hence, $t$ is a suffix of $u$. Thus, (\ref{sol.words.lyndon.equiv2.a.1})
(applied to $s=t$) yields that we do \textbf{not} have $tv<v$. This
contradicts the fact that $tv<v$. Hence, we have obtained a contradiction in
Case 2.

We thus have obtained a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this yields that we always have a
contradiction. Therefore, our assumption was wrong. So we have proven that
there exists a nonempty suffix $s$ of $u$ satisfying $sv<v$.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that if
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ are two words
satisfying $uv<vu$ and $\ell\left(  u\right)  +\ell\left(  v\right)  =N$, then
there exists a nonempty suffix $s$ of $u$ satisfying $sv<v$. In other words,
we have solved Exercise \ref{exe.words.lyndon.equiv2}(a) in the case when
$\ell\left(  u\right)  +\ell\left(  v\right)  =N$. This completes the
induction step. Hence, Exercise \ref{exe.words.lyndon.equiv2}(a) is solved by
strong induction.

(b)

\begin{proof}
[Alternative proof of Theorem \ref{thm.words.lyndon.equiv}.]\textit{Proof of
the implication $\mathcal{A}\Longrightarrow\mathcal{B}$:} Assume that
Assertion $\mathcal{A}$ holds. Thus, the word $w$ is Lyndon. Hence, every
nonempty proper suffix of $w$ is $>w$ (by the definition of a Lyndon word).
Now, let $u$ and $v$ be any nonempty words satisfying $w=uv$. Then, $v$ is a
nonempty proper suffix of $w$, and therefore $>w$ (since every nonempty proper
suffix of $w$ is $>w$). That is, we have $v>w$.

Now, let us forget that we fixed $u$ and $v$. We thus have shown that any
nonempty words $u$ and $v$ satisfying $w=uv$ satisfy $v>w$. In other words,
Assertion $\mathcal{B}$ holds. Thus, the implication $\mathcal{A}%
\Longrightarrow\mathcal{B}$ is proven.

\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{C}%
$:} This implication follows from Proposition
\ref{prop.words.lyndon.trivia}(b).

\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{D}%
$:} This implication follows from Proposition
\ref{prop.words.lyndon.trivia}(c).

\textit{Proof of the implication $\mathcal{B}\Longrightarrow\mathcal{A}%
$:} Assume that Assertion $\mathcal{B}$ holds. Let $v$ be a nonempty
proper suffix of $w$. Then, there exists a nonempty $u\in\mathfrak{A}^{\ast}$
satisfying $w=uv$ (since $v$ is a proper suffix of $w$). Consider this $u$.
Assertion $\mathcal{B}$ yields $v>w$.

Now, let us forget that we fixed $v$. We thus have shown that every nonempty
proper suffix $v$ of $w$ satisfies $v>w$. By the definition of a Lyndon word,
this yields that $w$ is Lyndon (since $w$ is nonempty), so that Assertion
$\mathcal{A}$ holds. Hence, the implication $\mathcal{B}\Longrightarrow
\mathcal{A}$ is proven.

\textit{Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{B}%
$:} Assume that Assertion $\mathcal{C}$ holds. Let us prove that
Assertion $\mathcal{B}$ holds.

Indeed, let $u$ and $v$ be two nonempty words satisfying $w=uv$. We will show
that $v>w$.

Let us first notice that Assertion $\mathcal{C}$ yields $v>u$.

Let $\mathfrak{K}$ be the set $\left\{  k\in \NN \ \mid\ u^{k}\text{ is a
prefix of }v\right\}  $. The integer $0$ belongs to this set $\mathfrak{K}$
(since $u^{0}=\varnothing$ is a prefix of $v$), and therefore $\mathfrak{K}$
is nonempty. Also, this set $\mathfrak{K}$ is finite\footnote{\textit{Proof.}
Let $i$ be an element of $\mathfrak{K}$. Thus, $i\in\mathfrak{K}=\left\{
k\in \NN \ \mid\ u^{k}\text{ is a prefix of }v\right\}  $. In other words
$i$ is an element of $\NN$ such that $u^{i}$ is a prefix of $v$. The
word $u^{i}$ is not longer than $v$ (since it is a prefix of $v$); thus
$\ell\left(  u^{i}\right)  \leq\ell\left(  v\right)  $. But $u$ is nonempty,
and thus $\ell\left(  u\right)  \geq1$. Hence, $\ell\left(  u^{i}\right)
=i\underbrace{\ell\left(  u\right)  }_{\geq1}\geq i$, so that $i\leq
\ell\left(  u^{i}\right)  \leq\ell\left(  v\right)  $.
\par
Now, let us forget that we fixed $i$. We thus have proven that every element
$i$ of $\mathfrak{K}$ satisfies $i\leq\ell\left(  v\right)  $. Thus, there are
only finitely many elements of $\mathfrak{K}$ (since there are only finitely
many $i\in \NN$ satisfying $i\leq\ell\left(  v\right)  $). In other
words, the set $\mathfrak{K}$ is finite.}. Hence, the set $\mathfrak{K}$ has a
maximum element (since it is nonempty and finite). Let $m$ be this maximum
element. Then, $m\in\mathfrak{K}$ but $m+1\notin\mathfrak{K}$.

We have $m\in\mathfrak{K}=\left\{  k\in \NN \ \mid\ u^{k}\text{ is a
prefix of }v\right\}  $. Thus, $m$ is an element of $\NN$ such that
$u^{m}$ is a prefix of $v$. In other words, there exists a word $v^{\prime}%
\in\mathfrak{A}^{\ast}$ such that $v=u^{m}v^{\prime}$. Consider this
$v^{\prime}$. Using $m+1\notin\mathfrak{K}$, it is easy to see that $u$ is not
a prefix of $v^{\prime}$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary.
Thus, $u$ is a prefix of $v^{\prime}$. In other words, there exists a word
$t\in\mathfrak{A}^{\ast}$ such that $v^{\prime}=ut$. Consider this $t$. We
have $v=u^{m}\underbrace{v^{\prime}}_{=ut}=\underbrace{u^{m}u}_{=u^{m+1}%
}t=u^{m+1}t$, and thus the word $u^{m+1}$ is a prefix of $v$. Hence, $m+1$ is
an element of $\NN$ such that $u^{m+1}$ is a prefix of $v$. In other
words, $m+1\in\left\{  k\in \NN \ \mid\ u^{k}\text{ is a prefix of
}v\right\}  =\mathfrak{K}$. But this contradicts $m+1\notin\mathfrak{K}$. This
contradiction proves that our assumption was wrong, qed.}.

It is easy to see that the word $v^{\prime}$ is
nonempty\footnote{\textit{Proof.} Assume the contrary. Thus, the word
$v^{\prime}$ is empty; that is, we have $v^{\prime}=\varnothing$. Now,
$v=u^{m}\underbrace{v^{\prime}}_{=\varnothing}=u^{m}$ and $w=u\underbrace{v}%
_{=u^{m}}=uu^{m}=u^{m+1}=\underbrace{u^{m}}_{=v}u=vu$. Hence, Assertion
$\mathfrak{C}$ (applied to $v$ and $u$ instead of $u$ and $v$) yields $u>v$.
This contradicts $v>u$. This contradiction proves that our assumption was
wrong, qed.}. Also, the word $uu^{m}$ is nonempty (since $u$ is nonempty). We
have $w=u\underbrace{v}_{=u^{m}v^{\prime}}=uu^{m}v^{\prime}$. Therefore,
Assertion $\mathcal{C}$ (applied to $uu^{m}$ and $v^{\prime}$ instead of $u$
and $v$) yields $v^{\prime}>uu^{m}\geq u$. Thus, $u\leq v^{\prime}$. Hence,
Proposition \ref{prop.words.lex}(d) (applied to $a=u$, $b=v^{\prime}$,
$c=v^{\prime}$ and $d=\varnothing$) yields that either we have $uv^{\prime
}\leq v^{\prime}\varnothing$ or the word $u$ is a prefix of $v^{\prime}$.
Since we know that the word $u$ is not a prefix of $v^{\prime}$, we can thus
conclude that $uv^{\prime}\leq v^{\prime}\varnothing$. Thus, $uv^{\prime}\leq
v^{\prime}\varnothing=v^{\prime}$. Hence, Proposition \ref{prop.words.lex}(b)
(applied to $a=u^{m}$, $c=uv^{\prime}$ and $d=v^{\prime}$) yields
$u^{m}uv^{\prime}\leq u^{m}v^{\prime}$. Thus, $u^{m}v^{\prime}\geq
\underbrace{u^{m}u}_{=u^{m+1}=uu^{m}}v^{\prime}=uu^{m}v^{\prime}=w$. Since
$u^{m}v^{\prime}=v$, this rewrites as $v\geq w$. Hence, $v>w$%
\ \ \ \ \footnote{\textit{Proof.} We have $\ell\left(  \underbrace{w}%
_{=uv}\right)  =\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)
}_{\substack{>0\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  v\right)
>\ell\left(  v\right)  $, so that $\ell\left(  w\right)  \neq\ell\left(
v\right)  $ and thus $w\neq v$. Combined with $v\geq w$, this yields $v>w$,
qed.}.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that any
nonempty words $u$ and $v$ satisfying $w=uv$ satisfy $v>w$. In other words,
Assertion $\mathcal{B}$ holds. Hence, the implication $\mathcal{C}%
\Longrightarrow\mathcal{B}$ is proven.

\textit{Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{B}%
$:} Assume that Assertion $\mathcal{D}$ holds. Let us prove that
Assertion $\mathcal{B}$ holds.

We shall prove Assertion $\mathcal{B}$ by strong induction over $\ell\left(
u\right)  $:

\textit{Induction step:} Let $N\in \NN$. Assume that Assertion
$\mathcal{B}$ holds in the case when $\ell\left(  u\right)  <N$. We now need
to prove Assertion $\mathcal{B}$ in the case when $\ell\left(  u\right)  =N$.

We have assumed that Assertion $\mathcal{B}$ holds in the case when
$\ell\left(  u\right)  <N$. In other words,%
\begin{equation}
\text{any nonempty words }u\text{ and }v\text{ satisfying }w=uv\text{ and
}\ell\left(  u\right)  <N\text{ satisfy }v>w.
\label{sol.words.lyndon.equiv2.b.DA.a}%
\end{equation}


Now, let $u$ and $v$ be two nonempty words satisfying $w=uv$ and $\ell\left(
u\right)  =N$. We shall prove that $v>w$.

From Assertion $\mathcal{D}$, we obtain $vu>uv$. Thus, $uv<vu$. Hence,
Exercise \ref{exe.words.lyndon.equiv2}(a) yields that there exists a nonempty
suffix $s$ of $u$ satisfying $sv<v$. Consider this $s$. There exists a
$p\in\mathfrak{A}^{\ast}$ such that $u=ps$ (since $s$ is a suffix of $u$).
Consider this $p$. Since $s$ is nonempty, we have $\ell\left(  s\right)  >0$.
Now, $\ell\left(  \underbrace{u}_{=ps}\right)  =\ell\left(  ps\right)
=\ell\left(  p\right)  +\underbrace{\ell\left(  s\right)  }_{>0}>\ell\left(
p\right)  $, so that $\ell\left(  p\right)  <\ell\left(  u\right)  =N$. Also,
$w=\underbrace{u}_{=ps}v=psv$. Furthermore, if $p=\varnothing$, then $v>w$ is
true\footnote{\textit{Proof.} Assume that $p=\varnothing$. Then,
$u=\underbrace{p}_{=\varnothing}s=s$, so that $s=u$. Then, $\underbrace{s}%
_{=u}v=uv=w$, so that $w=sv<v$ and thus $v>w$, qed.}. Hence, for the rest of
our proof of $v>w$, we can WLOG assume that we don't have $p=\varnothing$.
Assume this. Thus, $p\neq\varnothing$, so that the word $p$ is nonempty. Also,
the word $sv$ is nonempty (since $v$ is nonempty).

Now, the words $p$ and $sv$ are nonempty and satisfy $w=psv$ and $\ell\left(
p\right)  <N$. Hence, we can apply (\ref{sol.words.lyndon.equiv2.b.DA.a}) to
$p$ and $sv$ instead of $u$ and $v$. As a result, we obtain $sv>w$. But
$sv<v$, so that $v>sv>w$. Hence, we have proven that $v>w$.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that any
nonempty words $u$ and $v$ satisfying $w=uv$ and $\ell\left(  u\right)  =N$
satisfy $v>w$. In other words, we have proven Assertion $\mathcal{B}$ in the
case when $\ell\left(  u\right)  =N$. Hence, the induction step is complete.
Assertion $\mathcal{B}$ is thus proven by strong induction. And so, we have
established the implication $\mathcal{D}\Longrightarrow\mathcal{B}$.

Combining the implications $\mathcal{A}\Longrightarrow\mathcal{B}$,
$\mathcal{A}\Longrightarrow\mathcal{C}$, $\mathcal{A}\Longrightarrow
\mathcal{D}$, $\mathcal{B}\Longrightarrow\mathcal{A}$, $\mathcal{C}%
\Longrightarrow\mathcal{B}$ and $\mathcal{D}\Longrightarrow\mathcal{B}$ that
we have proven, we obtain the equivalence $\mathcal{A}\Longleftrightarrow
\mathcal{B}\Longleftrightarrow\mathcal{C}\Longleftrightarrow\mathcal{D}$.
Hence, Theorem \ref{thm.words.lyndon.equiv} is proven again.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.power-criterion}}

\textit{Solution to Exercise \ref{exe.words.lyndon.power-criterion}.}
Let us first assume that $w$ is Lyndon. We shall prove that%
\begin{equation}
\left(  \text{every nonempty word }t\text{ and every positive integer }n\text{
satisfy }\left(  \text{if }w\leq t^{n}\text{, then }w\leq t\right)  \right)  .
\label{sol.words.lyndon.power-criterion.this}%
\end{equation}


Let $t$ be a nonempty word, and let $n$ be a positive integer. Assume that
$w\leq t^{n}$. We need to prove that $w\leq t$.

Assume the contrary. Thus, we don't have $w\leq t$. In other words, we don't
have $w\leq t^{1}$.

Let $m$ be the minimal $i\in\left\{  1,2,\ldots,n\right\}  $ satisfying $w\leq
t^{i}$.\ \ \ \ \footnote{Such an $i$ exists, because $n\in\left\{
1,2,\ldots,n\right\}  $ satisfies $w\leq t^{n}$.} Then, $w\leq t^{m}$. Hence,
$m\neq1$ (because $w\leq t^{m}$, but we don't have $w\leq t^{1}$). Thus,
$m\geq2$, so that $m-1$ is also an element of $\left\{  1,2,\ldots,n\right\}
$. If we had $w\leq t^{m-1}$, then $m-1$ would be an $i\in\left\{
1,2,\ldots,n\right\}  $ satisfying $w\leq t^{i}$, which would contradict the
fact that $m$ is the \textbf{minimal} such $i$. Thus, we cannot have $w\leq
t^{m-1}$.

But $w\varnothing=w\leq t^{m-1}t$. Hence, Proposition \ref{prop.words.lex}(e)
(applied to $a=w$, $b=\varnothing$, $c=t^{m-1}$ and $d=t$) yields that either
we have $w\leq t^{m-1}$ or the word $t^{m-1}$ is a prefix of $w$. Since we
cannot have $w\leq t^{m-1}$, this shows that the word $t^{m-1}$ is a prefix of
$w$. In other words, there exists a $v\in\mathfrak{A}^{\ast}$ such that
$w=t^{m-1}v$. Consider this $v$. We have $v\neq\varnothing$ (because
otherwise, we would have $v=\varnothing$ and thus $w=t^{m-1}\underbrace{v}%
_{=\varnothing}=t^{m-1}\leq t^{m-1}$, contradicting the fact that we cannot
have $w\leq t^{m-1}$), so that $v$ is nonempty. Hence, Proposition
\ref{prop.words.lyndon.trivia}(b) (applied to $u=t^{m-1}$) now yields
$v>t^{m-1}$. Hence, $t^{m-1}<v$. Thus, Proposition \ref{prop.words.lex}(b)
(applied to $a=t^{m-1}$, $c=t^{m-1}$ and $d=v$) yields $t^{m-1}t^{m-1}\leq
t^{m-1}v=w$. Hence,
\[
w\geq t^{m-1}t^{m-1}=t^{2\left(  m-1\right)  }.
\]
Combined with
\begin{align*}
w  &  \leq t^{m}\leq t^{m}t^{m-2}\ \ \ \ \ \ \ \ \ \ \left(  \text{this makes
sense since }m\geq2\right) \\
&  =t^{m+\left(  m-2\right)  }=t^{2\left(  m-1\right)  },
\end{align*}
this yields $w=t^{2\left(  m-1\right)  }=t^{m-1}t^{m-1}$, so that
$t^{m-1}t^{m-1}=w=t^{m-1}v$. Cancelling $t^{m-1}$ in this, we obtain
$t^{m-1}=v$, which contradicts $t^{m-1}<v$. This contradiction shows that our
assumption (that we don't have $w\leq t$) was false. Hence, $w\leq t$.

Forget now that we assumed that $w\leq t^{n}$. We thus have proven that if
$w\leq t^{n}$, then $w\leq t$.

Now, forget that we fixed $t$ and assumed that $w$ is Lyndon. We thus have
shown that
\begin{equation}
\left(  \text{if }w\text{ is Lyndon, then
(\ref{sol.words.lyndon.power-criterion.this}) holds}\right)  .
\label{sol.words.lyndon.power-criterion.1}%
\end{equation}


Now, conversely, assume that (\ref{sol.words.lyndon.power-criterion.this})
holds. We will prove that $w$ is Lyndon.

In fact, assume the contrary. Then, $w$ is not Lyndon. Let $v$ be the
(lexicographically) smallest nonempty suffix of $w$. Proposition
\ref{prop.words.lyndon.cutoff}(b) yields that there exists a nonempty
$u\in\mathfrak{A}^{\ast}$ such that $w=uv$, $u\geq v$ and $uv\geq vu$.
Consider this $u$. We have $u\geq v$, thus $v\leq u$, and therefore
Proposition \ref{prop.words.lex}(b) (applied to $a=u$, $c=v$ and $d=u$) yields
$uv\leq uu$. Now, $w=uv\leq uu=u^{2}$. Thus,
(\ref{sol.words.lyndon.power-criterion.this}) (applied to $t=u$ and $n=2$)
yields $w\leq u$. But this contradicts the fact that $w=uv>u$ (since $v$ is
nonempty). As this contradiction shows, our assumption was wrong. Thus, we
have shown that $w$ is Lyndon.

Now, forget that we fixed $w$. We thus have proven that
\[
\left(  \text{if (\ref{sol.words.lyndon.power-criterion.this}) holds, then
}w\text{ is Lyndon}\right)  .
\]
Combined with (\ref{sol.words.lyndon.power-criterion.1}), this yields that $w
$ is Lyndon if and only if (\ref{sol.words.lyndon.power-criterion.this})
holds. This solves the exercise.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.concatn}}

\textit{Solution to Exercise \ref{exe.words.lyndon.concatn}.} We will
solve Exercise \ref{exe.words.lyndon.concatn} by induction over $n$:

The \textit{induction base} is the case $n=1$; this case is vacuously true
(since $w_{1}<w_{n}$ is impossible for $n=1$).

For the \textit{induction step}, we fix a positive integer $N>1$, and we
assume that Exercise \ref{exe.words.lyndon.concatn} has been solved for
$n=N-1$. We now must solve Exercise \ref{exe.words.lyndon.concatn} for $n=N$.

So let $w_{1}$, $w_{2}$, $\ldots$, $w_{N}$ be $N$ Lyndon words. Assume that
$w_{1}\leq w_{2}\leq\cdots\leq w_{N}$ and $w_{1}<w_{N}$. We need to show that
$w_{1}w_{2} \cdots w_{N}$ is a Lyndon word.

Proposition \ref{prop.words.lyndon.concat}(a) (applied to $u=w_{1}$ and
$v=w_{N}$) yields that the word $w_{1}w_{N}$ is Lyndon. If $N=2$, then this
yields that $w_{1}w_{2}\cdots w_{N}$ is Lyndon (because $w_{1}w_{2}\cdots
w_{N}=w_{1}w_{N}$ if $N=2$). Thus, if $N=2$, then we are done. We therefore
WLOG assume that we don't have $N=2$. Hence, $N\geq3$. Thus, $w_{3}w_{4}\cdots
w_{N}$ is a nonempty product. More precisely, $w_{3}w_{4}\cdots w_{N}$ is a
nonempty product of nonempty words (since the words $w_{3}$, $w_{4}$, $\ldots
$, $w_{N}$ are nonempty (because they are Lyndon)), and therefore a nonempty
word itself. Hence, $w_{2}<w_{2}\left(  w_{3}w_{4}\cdots w_{N}\right)
=w_{2}w_{3}\cdots w_{N}$.

We have $w_{1}\leq w_{2}\leq\cdots\leq w_{N}$. Hence, $w_{2}\leq w_{3}%
\leq\cdots\leq w_{N}$ and, in particular, $w_{2}\leq w_{N}$. We distinguish
between two cases:

\textit{Case 1:} We have $w_{2}=w_{N}$.

\textit{Case 2:} We have $w_{2}\neq w_{N}$.

Let us first consider Case 1. In this case, we have $w_{2}=w_{N}$. Hence,
$w_{1}<w_{N}=w_{2}$. Thus, Proposition \ref{prop.words.lyndon.concat}(a)
(applied to $u=w_{1}$ and $v=w_{2}$) yields that the word $w_{1}w_{2}$ is
Lyndon. Moreover, $w_{1}w_{2}<w_{2}$ (by Proposition
\ref{prop.words.lyndon.concat}(b), applied to $u=w_{1}$ and $v=w_{2}$). Hence,
$w_{1}w_{2}\leq w_{3}\leq w_{4}\leq\cdots\leq w_{N}$ (this follows by
combining $w_{1}w_{2}<w_{2}$ and $w_{2}\leq w_{3}\leq\cdots\leq w_{N}$) and
$w_{1}w_{2}<w_{2}=w_{N}$. Thus, we can apply Exercise
\ref{exe.words.lyndon.concatn} to $N-1$ and $\left(  w_{1}w_{2},w_{3}%
,w_{4},\ldots,w_{N}\right)  $ instead of $n$ and $\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  $ (because we have assumed that Exercise
\ref{exe.words.lyndon.concatn} has been solved for $n=N-1$). We thus obtain
that $w_{1}w_{2}w_{3}w_{4} \cdots w_{N}$ is a Lyndon word. In other words,
$w_{1}w_{2} \cdots w_{N}$ is a Lyndon word. So we have shown that $w_{1}w_{2}
\cdots w_{N}$ is a Lyndon word in Case 1.

Let us now consider Case 2. In this case, we have $w_{2}\neq w_{N}$. Since
$w_{2}\leq w_{N}$, this yields $w_{2}<w_{N}$. Thus, we can apply Exercise
\ref{exe.words.lyndon.concatn} to $N-1$ and $\left(  w_{2},w_{3},\ldots
,w_{N}\right)  $ instead of $n$ and $\left(  w_{1},w_{2},\ldots,w_{n}\right)
$ (because we have assumed that Exercise \ref{exe.words.lyndon.concatn} has
been solved for $n=N-1$). We thus obtain that $w_{2}w_{3}\cdots w_{N}$ is a
Lyndon word. Now, $w_{1}$ and $w_{2}w_{3}\cdots w_{N}$ are two Lyndon words
satisfying $w_{1}\leq w_{2}<w_{2}w_{3}\cdots w_{N}$. Therefore, Proposition
\ref{prop.words.lyndon.concat}(a) (applied to $u=w_{1}$ and $v=w_{2}%
w_{3}\cdots w_{N}$) yields that the word $w_{1}w_{2}w_{3}\cdots w_{N}$ is
Lyndon. In other words, the word $w_{1}w_{2}\cdots w_{N}$ is Lyndon. So we
have shown that $w_{1}w_{2}\cdots w_{N}$ is a Lyndon word in Case 2.

Now, we have proven that $w_{1}w_{2} \cdots w_{N}$ is a Lyndon word in both
possible Cases 1 and 2. Hence, $w_{1}w_{2} \cdots w_{N}$ always is a Lyndon
word. In other words, Exercise \ref{exe.words.lyndon.concatn} is solved for
$n=N$. This completes the induction, and therefore Exercise
\ref{exe.words.lyndon.concatn} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.concatn-lyndon}}

\textit{Solution to Exercise \ref{exe.words.lyndon.concatn-lyndon}.} We
have assumed that%
\begin{equation}
w_{i}w_{i+1}\cdots w_{n}\geq w_{1}w_{2}\cdots w_{n}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}  .
\label{sol.words.lyndon.concatn-lyndon.ass}%
\end{equation}
As a consequence,%
\begin{equation}
w_{i}w_{i+1}\cdots w_{n}>w_{1}w_{2}\cdots w_{n}\ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  2,3,\ldots,n\right\}  .
\label{sol.words.lyndon.concatn-lyndon.ass2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.lyndon.concatn-lyndon.ass2}):} Let
$i\in\left\{  2,3,\ldots,n\right\}  $. Then, $i-1\geq1$. The words $w_{1}$,
$w_{2}$, $\ldots$, $w_{i-1}$ are Lyndon (since the words $w_{1}$, $w_{2}$,
$\ldots$, $w_{n}$ are Lyndon), and thus nonempty. Now, $i-1\geq1>0$. Hence,
the product $w_{1}w_{2}\cdots w_{i-1}$ is a nonempty product of nonempty words
(since $w_{1}$, $w_{2}$, $\ldots$, $w_{i-1}$ are nonempty words), and thus a
nonempty word itself. Since $w_{1}w_{2}\cdots w_{n}=\left(  w_{1}w_{2}\cdots
w_{i-1}\right)  \left(  w_{i}w_{i+1}\cdots w_{n}\right)  $, this yields that
$w_{i}w_{i+1}\cdots w_{n}$ is a \textbf{proper} suffix of the word $w_{1}%
w_{2}\cdots w_{n}$. As a consequence, $w_{i}w_{i+1}\cdots w_{n}\neq w_{1}%
w_{2}\cdots w_{n}$. Combining this with $w_{i}w_{i+1}\cdots w_{n}\geq
w_{1}w_{2}\cdots w_{n}$ (by (\ref{sol.words.lyndon.concatn-lyndon.ass})), we
obtain $w_{i}w_{i+1}\cdots w_{n}>w_{1}w_{2}\cdots w_{n}$. This proves
(\ref{sol.words.lyndon.concatn-lyndon.ass2}).}

Now, we claim that if $j$ is any element of $\left\{  0,1,\ldots,n\right\}  $,
then%
\begin{equation}
\text{every nonempty proper suffix }v\text{ of }w_{n-j+1}w_{n-j+2}\cdots
w_{n}\text{ satisfies }v>w_{1}w_{2}\cdots w_{n}.
\label{sol.words.lyndon.concatn-lyndon.hook}%
\end{equation}


\textit{Proof of (\ref{sol.words.lyndon.concatn-lyndon.hook}):} We will prove
(\ref{sol.words.lyndon.concatn-lyndon.hook}) by induction over $j$:

\textit{Induction base:} For $j=0$, we have $w_{n-j+1}w_{n-j+2}\cdots
w_{n}=w_{n-0+1}w_{n-0+2}\cdots w_{n}=\left(  \text{empty product}\right)
=\varnothing$. Hence, for $j=0$, the word $w_{n-j+1}w_{n-j+2}\cdots w_{n}$ has
no nonempty proper suffix. Thus, (\ref{sol.words.lyndon.concatn-lyndon.hook})
is vacuously true for $j=0$. The induction base is thus complete.

\textit{Induction step:} Let $J\in\left\{  0,1,\ldots,n-1\right\}  $. We
assume that (\ref{sol.words.lyndon.concatn-lyndon.hook}) holds for $j=J$. We
now need to prove that (\ref{sol.words.lyndon.concatn-lyndon.hook}) holds for
$j=J+1$.

From $J\in\left\{  0,1,\ldots,n-1\right\}  $, we obtain $n-J\in\left\{
1,2,\ldots,n\right\}  $.

Let $g=w_{n-J+1}w_{n-J+2}\cdots w_{n}$. We have assumed that
(\ref{sol.words.lyndon.concatn-lyndon.hook}) holds for $j=J$. In other words,
every nonempty proper suffix $v$ of $w_{n-J+1}w_{n-J+2}\cdots w_{n}$ satisfies
$v>w_{1}w_{2}\cdots w_{n}$. Since $w_{n-J+1}w_{n-J+2}\cdots w_{n}=g$, this
rewrites as follows:%
\begin{equation}
\text{Every nonempty proper suffix }v\text{ of }g\text{ satisfies }%
v>w_{1}w_{2}\cdots w_{n}\text{.}
\label{sol.words.lyndon.concatn-lyndon.hook.pf.indass}%
\end{equation}


Now, let us notice that%
\begin{align}
w_{n-\left(  J+1\right)  +1}w_{n-\left(  J+1\right)  +2}\cdots w_{n}%
&=w_{n-J}w_{n-J+1}\cdots w_{n}=w_{n-J}\underbrace{\left(  w_{n-J+1}%
w_{n-J+2}\cdots w_{n}\right)  }_{=g} \nonumber\\
&=w_{n-J}g.
\label{sol.words.lyndon.concatn-lyndon.hook.pf.2}%
\end{align}


Let now $v$ be a nonempty proper suffix of $w_{n-J}g$. We are going to prove
that $v>w_{1}w_{2}\cdots w_{n}$.

We have $v\neq\varnothing$ (since $v$ is nonempty). Since $v$ is a nonempty
suffix of $w_{n-J}g$, we must be in one of the following two cases (depending
on whether this suffix begins before the suffix $g$ of $w_{n-J}g$ begins or afterwards):

\textit{Case 1:} The word $v$ is a nonempty suffix of $g$. (Note that $v$ is
allowed to be $g$.)

\textit{Case 2:} The word $v$ has the form $hg$ where $h$ is a nonempty suffix
of $w_{n-J}$.

Let us first consider Case 1. In this case, the word $v$ is a nonempty suffix
of $g$. If $v$ is a proper suffix of $g$, then we immediately obtain
$v>w_{1}w_{2}\cdots w_{n}$ (from
(\ref{sol.words.lyndon.concatn-lyndon.hook.pf.indass})). Thus, for the rest of
the proof of $v>w_{1}w_{2}\cdots w_{n}$ in Case 1, we can WLOG assume that $v$
is not a proper suffix of $g$. Assume this.

So we know that $v$ is a suffix of $g$, but not a proper suffix of $g$. Hence,
$v$ must be $g$ itself. That is, we have $v=g$. Hence, $v=g=w_{n-J+1}%
w_{n-J+2}\cdots w_{n}$. Consequently, $J\neq0$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $J=0$, so that
\begin{align*}
v  &  =w_{n-J+1}w_{n-J+2}\cdots w_{n}=w_{n-0+1}w_{n-0+2}\cdots w_{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }J=0\right) \\
&  =\left(  \text{empty product}\right)  =\varnothing,
\end{align*}
contradicting the fact that $v\neq\varnothing$. This contradiction shows that
our assumption was wrong, qed.}. Combined with $J\in\left\{  0,1,\ldots
,n-1\right\}  $, this yields $J\in\left\{  0,1,\ldots,n-1\right\}
\setminus\left\{  0\right\}  =\left\{  1,2,\ldots,n-1\right\}  $, so that
$n-J+1\in\left\{  2,3,\ldots,n\right\}  $. Now,%
\[
v=g=w_{n-J+1}w_{n-J+2}\cdots w_{n}=w_{n-J+1}w_{\left(  n-J+1\right)  +1}\cdots
w_{n}>w_{1}w_{2}\cdots w_{n}%
\]
(by (\ref{sol.words.lyndon.concatn-lyndon.ass2}), applied to $i=n-J+1$). Thus,
$v>w_{1}w_{2}\cdots w_{n}$ is proven in Case 1.

Let us now consider Case 2. In this case, the word $v$ has the form $hg$ where
$h$ is a nonempty suffix of $w_{n-J}$. Consider this $h$. Since $h$ is a
suffix of $w_{n-J}$, we have $\ell\left(  h\right)  \leq\ell\left(
w_{n-J}\right)  $, so that $\ell\left(  w_{n-J}\right)  \geq\ell\left(
h\right)  $. But $w_{n-J}$ is a Lyndon word (since $w_{1}$, $w_{2}$, $\ldots$,
$w_{n}$ are Lyndon words), and thus $h\geq w_{n-J}$ (by Corollary
\ref{cor.words.lyndon.trivia}, applied to $w_{n-J}$ and $h$ instead of $w$ and
$v$). Thus, $w_{n-J}\leq h$. Thus, Proposition \ref{prop.words.lex}(j)
(applied to $a=w_{n-J}$, $b=h$ and $c=g$) yields $w_{n-J}g\leq hg=v$. Thus,
$v\geq w_{n-J}g$.

But we also have $v\neq w_{n-J}g$ (since $v$ is a proper suffix of $w_{n-J}%
g$). Combined with $v\geq w_{n-J}g$, this yields $v>w_{n-J}g$, so that
\begin{align*}
v  &  >w_{n-J}g=w_{n-\left(  J+1\right)  +1}w_{n-\left(  J+1\right)  +2}\cdots
w_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.words.lyndon.concatn-lyndon.hook.pf.2})}\right) \\
&  =w_{n-J}w_{n-J+1}\cdots w_{n}\geq w_{1}w_{2}\cdots w_{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.words.lyndon.concatn-lyndon.ass}), applied to }i=n-J\right)  .
\end{align*}
Thus, $v>w_{1}w_{2}\cdots w_{n}$ is proven in Case 2.

Now, $v>w_{1}w_{2}\cdots w_{n}$ is proven in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this yields that $v>w_{1}%
w_{2}\cdots w_{n}$ always holds.

Now, let us forget that we fixed $v$. We thus have proven that every nonempty
proper suffix $v$ of $w_{n-J}g$ satisfies $v>w_{1}w_{2}\cdots w_{n}$. Since
$w_{n-J}g=w_{n-\left(  J+1\right)  +1}w_{n-\left(  J+1\right)  +2}\cdots
w_{n}$ (by (\ref{sol.words.lyndon.concatn-lyndon.hook.pf.2})), this rewrites
as follows: Every nonempty proper suffix $v$ of $w_{n-\left(  J+1\right)
+1}w_{n-\left(  J+1\right)  +2}\cdots w_{n}$ satisfies $v>w_{1}w_{2}\cdots
w_{n}$. In other words, (\ref{sol.words.lyndon.concatn-lyndon.hook}) holds for
$j=J+1$. This completes the induction step. Thus,
(\ref{sol.words.lyndon.concatn-lyndon.hook}) is proven by induction.

Now, we can apply (\ref{sol.words.lyndon.concatn-lyndon.hook}) to $j=n$. As a
result, we obtain that every nonempty proper suffix $v$ of $w_{n-n+1}%
w_{n-n+2}\cdots w_{n}$ satisfies $v>w_{1}w_{2}\cdots w_{n}$. In other words,
every nonempty proper suffix $v$ of $w_{1}w_{2}\cdots w_{n}$ satisfies
$v>w_{1}w_{2}\cdots w_{n}$ (since $w_{n-n+1}w_{n-n+2}\cdots w_{n}=w_{1}%
w_{2}\cdots w_{n}$). Since the word $w_{1}w_{2}\cdots w_{n}$ is also
nonempty\footnote{\textit{Proof.} The product $w_{1}w_{2}\cdots w_{n}$ is
nonempty (since $n$ is a positive integer), and the words $w_{1}$, $w_{2}$,
$\ldots$, $w_{n}$ are nonempty (since they are Lyndon words). Hence,
$w_{1}w_{2}\cdots w_{n}$ is a nonempty product of nonempty words. Thus,
$w_{1}w_{2}\cdots w_{n}$ is a nonempty word, qed.}, this shows that the word
$w_{1}w_{2}\cdots w_{n}$ is Lyndon (by the definition of a Lyndon word). This
solves Exercise \ref{exe.words.lyndon.concatn-lyndon}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.count}}

\textit{Solution to Exercise \ref{exe.words.lyndon.count}.}
Let us first forget about the setting of
Exercise \ref{exe.words.lyndon.count} (so $\mathfrak{A}$ can be any
alphabet, not necessarily finite).

Our solution to Exercise \ref{exe.words.lyndon.count} will rely on two
basic propositions:

\begin{proposition}
\label{prop.lyndon.CFL.bij}Let $\mathfrak{M}$ denote the set of all finite
multisets of Lyndon words.

Define a map $\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$ as
follows: Given an $M\in\mathfrak{M}$, we set $\mathbf{m}\left(  M\right)
=a_{1}a_{2}\cdots a_{k}$, where $a_{1},a_{2},\ldots,a_{k}$ denote the elements
of $M$ listed in decreasing order\footnote{When we say ``the
elements of $M$ listed in decreasing order'', we mean that
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ should be the unique tuple
satisfying $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ and $M=\left\{  a_{1}%
,a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}$. (The existence and
the uniqueness of this tuple follow from basic properties of finite multisets,
since $\geq$ is a total order.) Note that each element appears in the tuple
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ with the same multiplicity with
which it appears in the multiset $M$.
\par
For example, if $M=\left\{  2,2,3536,24\right\}  _{\operatorname*{multiset}}$,
then $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(  3536,24,2,2\right)
$.}. Thus, a map $\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$ is defined.

Let $\mathbf{n}:\mathfrak{A}^{\ast}\rightarrow\mathfrak{M}$ be the map that
sends each word $w\in\mathfrak{A}^{\ast}$ to the multiset $\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}$, where $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of $w$.

Then, these two maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse bijections.
\end{proposition}

Note that $\mathfrak{A}$ is \textbf{not} assumed to be finite in
Proposition~\ref{prop.lyndon.CFL.bij}.

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.lyndon.CFL.bij}.]This follows
from the definition of the CFL factorization (and from the fact that it exists
and is unique).\footnote{Here are a few details:
\begin{itemize}
\item The equality $\mathbf{m} \circ \mathbf{n} = \id$ follows immediately
      from the definition of the CFL factorization.
\item In order to prove the equality $\mathbf{n} \circ \mathbf{m} = \id$,
      we fix some $M\in\mathfrak{M}$; our goal is thus to show that
      $\left(\mathbf{n} \circ \mathbf{m}\right) \left(M\right) = M$.
      Let $a_{1},a_{2},\ldots,a_{k}$ denote the elements
      of $M$ listed in decreasing order (so that
      $M = \left\{a_1, a_2, \ldots, a_k\right\}_{\operatorname{multiset}}$
      and $a_1 \geq a_2 \geq \cdots \geq a_k$).
      The definition of $\mathbf{m}$
      then yields $\mathbf{m}\left(M\right) = a_1 a_2 \cdots a_k$.
      Hence, $a_1, a_2, \ldots, a_k$ are the elements of $M$, and thus
      are Lyndon words (since $M$ is a multiset of Lyndon words).
      Therefore, $\left(a_1, a_2, \ldots, a_k\right)$ is a tuple of
      Lyndon words satisfying $\mathbf{m}\left(M\right) = a_1 a_2 \cdots a_k$
      and $a_1 \geq a_2 \geq \cdots \geq a_k$.
      In other words, $\left(a_1, a_2, \ldots, a_k\right)$ is a CFL
      factorization of $\mathbf{m}\left(M\right)$ (by the definition
      of a CFL factorization).
      Since the CFL factorization of a word is unique, this shows that
      $\left(a_1, a_2, \ldots, a_k\right)$ is \textbf{the} CFL
      factorization of $\mathbf{m}\left(M\right)$.
      Hence, the definition of $\mathbf{n}$ yields
      $\mathbf{n}\left(\mathbf{m}\left(M\right)\right)
      = \left\{a_1, a_2, \ldots, a_k\right\}_{\operatorname{multiset}}
      = M$.
      In other words,
      $\left(\mathbf{n} \circ \mathbf{m}\right) \left(M\right) = M$.
      This concludes the proof of $\mathbf{n} \circ \mathbf{m} = \id$.
\end{itemize}
Combining the equalities $\mathbf{m} \circ \mathbf{n} = \id$
and $\mathbf{n} \circ \mathbf{m} = \id$, we conclude that the
maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse, hence
bijections.}
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.lyndon.CFL.bij}.]Let us first convince
ourselves that the maps $\mathbf{m}$ and $\mathbf{n}$ are well-defined. For
this, we shall need the following two claims:

\begin{statement}
\textit{Claim 1:} Let $M\in\mathfrak{M}$. Let $a_{1},a_{2},\ldots,a_{k}$
denote the elements of $M$ listed in decreasing order. Then, $a_{1}a_{2}\cdots
a_{k}\in\mathfrak{A}^{\ast}$.
\end{statement}

[\textit{Proof of Claim 1:} We have $M\in\mathfrak{M}$. In other words, $M$ is
a finite multiset of Lyndon words (since $\mathfrak{M}$ is the set of all
finite multisets of Lyndon words). Hence, in particular, $M$ is a finite
multiset of words.

Recall that $a_{1},a_{2},\ldots,a_{k}$ are the elements of $M$ listed in
decreasing order. In other words, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $
is the unique tuple satisfying $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ and
$M=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}$.

Clearly, $a_{1},a_{2},\ldots,a_{k}$ are elements of $M$ (since $M=\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}$). Hence,
$a_{1},a_{2},\ldots,a_{k}$ are words (since $M$ is a multiset of words). Thus,
$a_{1}a_{2}\cdots a_{k}$ is a word. In other words, $a_{1}a_{2}\cdots a_{k}%
\in\mathfrak{A}^{\ast}$. This proves Claim 1.]

\begin{statement}
\textit{Claim 2:} Let $w\in\mathfrak{A}^{\ast}$. Let $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of $w$. Then, $\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}\in\mathfrak{M}$.
\end{statement}

[\textit{Proof of Claim 2:} We know that $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is the CFL factorization of $w$. In other words, $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words satisfying
$w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (by the
definition of a ``CFL factorization''). Thus,
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words. In
other words, $a_{1},a_{2},\ldots,a_{k}$ are Lyndon words. Hence, $\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}$ is a finite
multiset of Lyndon words. In other words, $\left\{  a_{1},a_{2},\ldots
,a_{k}\right\}  _{\operatorname{multiset}}\in\mathfrak{M}$ (since
$\mathfrak{M}$ is the set of all finite multisets of Lyndon words). This
proves Claim 2.]

Claim 1 shows that the map $\mathbf{m}$ is well-defined (because the
definition of $\mathbf{m}$ says that $\mathbf{m}\left(  M\right)  =a_{1}%
a_{2}\cdots a_{k}$ for each $M\in\mathfrak{M}$, where $a_{1},a_{2}%
,\ldots,a_{k}$ are defined as in Claim 1). Claim 2 shows that the map
$\mathbf{n}$ is well-defined (since the definition of $\mathbf{n}$ says that
$\mathbf{n}\left(  w\right)  =\left\{  a_{1},a_{2},\ldots,a_{k}\right\}
_{\operatorname{multiset}}$ for every $w\in\mathfrak{A}^{\ast}$, where
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is defined as in Claim 2).

We shall now show that the maps $\mathbf{m}$ and $\mathbf{n}$ are mutually
inverse. Indeed, we have the following two claims:

\begin{statement}
\textit{Claim 3:} We have $\mathbf{m}\circ\mathbf{n}=\operatorname*{id}$.
\end{statement}

[\textit{Proof of Claim 3:} Let $w\in\mathfrak{A}^{\ast}$.

The definition of $\mathbf{m}$ yields $\mathbf{m}\left(  M\right)  =a_{1}%
a_{2}\cdots a_{k}$, where $a_{1},a_{2},\ldots,a_{k}$ denote the elements of
$M$ listed in decreasing order. In other words, if $a_{1},a_{2},\ldots,a_{k}$
are the elements of $M$ listed in decreasing order, then%
\begin{equation}
\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}%
.\label{pf.prop.lyndon.CFL.bij.c3.pf.1}%
\end{equation}


Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Then, the definition of $\mathbf{n}$ yields $\mathbf{n}\left(  w\right)
=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}$.

We know that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (by the definition of a
``CFL factorization'').

Let $M=\mathbf{n}\left(  w\right)  $. Thus, $M=\mathbf{n}\left(  w\right)
=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}$. Thus,
we know that $M=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}
_{\operatorname{multiset}}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$. In
other words, $a_{1},a_{2},\ldots,a_{k}$ are the elements of $M$ listed in
decreasing order. Hence, (\ref{pf.prop.lyndon.CFL.bij.c3.pf.1}) yields
$\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$. Comparing this with
$w=a_{1}a_{2}\cdots a_{k}$, we obtain $\mathbf{m}\left(  M\right)  =w$. In
view of $M=\mathbf{n}\left(  w\right)  $, this rewrites as $\mathbf{m}\left(
\mathbf{n}\left(  w\right)  \right)  =w$. Hence, $\left(  \mathbf{m}%
\circ\mathbf{n}\right)  \left(  w\right)  =\mathbf{m}\left(  \mathbf{n}\left(
w\right)  \right)  =w=\operatorname*{id}\left(  w\right)  $.

Forget that we fixed $w$. We thus have proved that $\left(  \mathbf{m}%
\circ\mathbf{n}\right)  \left(  w\right)  =\operatorname*{id}\left(  w\right)
$ for each $w\in\mathfrak{A}^{\ast}$. In other words, $\mathbf{m}%
\circ\mathbf{n}=\operatorname*{id}$. This proves Claim 3.]

\begin{statement}
\textit{Claim 4:} We have $\mathbf{n}\circ\mathbf{m}=\operatorname*{id}$.
\end{statement}

[\textit{Proof of Claim 4:} Let $M\in\mathfrak{M}$. Thus, $M$ is a finite
multiset of Lyndon words (since $\mathfrak{M}$ is the set of all finite
multisets of Lyndon words). Hence, in particular, $M$ is a finite multiset of words.

Let $w=\mathbf{m}\left(  M\right)  $. The definition of $\mathbf{n}$ shows
that $\mathbf{n}\left(  w\right)  =\left\{  a_{1},a_{2},\ldots,a_{k}\right\}
_{\operatorname{multiset}}$, where $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $
is the CFL factorization of $w$. In other words, if $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of $w$, then%
\begin{equation}
\mathbf{n}\left(  w\right)  =\left\{  a_{1},a_{2},\ldots,a_{k}\right\}
_{\operatorname{multiset}}. \label{pf.prop.lyndon.CFL.bij.c4.pf.1}%
\end{equation}


Now, let $a_{1},a_{2},\ldots,a_{k}$ denote the elements of $M$ listed in
decreasing order. Then, the definition of $\mathbf{m}$ shows that
$\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$. Hence, $w=\mathbf{m}%
\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$.

Recall that $a_{1},a_{2},\ldots,a_{k}$ are the elements of $M$ listed in
decreasing order. In other words, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $
is the unique tuple satisfying $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ and
$M=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}$.
Also, $a_{1},a_{2},\ldots,a_{k}$ are elements of $M$ (since $M=\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}$). Hence,
$a_{1},a_{2},\ldots,a_{k}$ are Lyndon words (since $M$ is a multiset of Lyndon
words). Hence, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of
Lyndon words.

Now, we know that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of
Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}%
\geq\cdots\geq a_{k}$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a CFL factorization of $w$ (by the definition of a
``CFL factorization''). In other words,
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is \textbf{the} CFL factorization
of $w$ (since the CFL factorization of $w$ is unique\footnote{by Theorem
\ref{thm.words.CFL}}). Hence, (\ref{pf.prop.lyndon.CFL.bij.c4.pf.1}) shows
that $\mathbf{n}\left(  w\right)  =\left\{  a_{1},a_{2},\ldots,a_{k}\right\}
_{\operatorname{multiset}}=M$ (since $M=\left\{  a_{1},a_{2},\ldots
,a_{k}\right\}  _{\operatorname{multiset}}$). In view of $w=\mathbf{m}\left(
M\right)  $, this rewrites as $\mathbf{n}\left(  \mathbf{m}\left(  M\right)
\right)  =M$.

Thus, $\left(  \mathbf{n}\circ\mathbf{m}\right)  \left(  M\right)
=\mathbf{n}\left(  \mathbf{m}\left(  M\right)  \right)  =M=\operatorname*{id}%
\left(  M\right)  $.

Forget that we fixed $M$. We thus have showed that $\left(  \mathbf{n}%
\circ\mathbf{m}\right)  \left(  M\right)  =\operatorname*{id}\left(  M\right)
$ for each $M\in\mathfrak{M}$. In other words, $\mathbf{n}\circ\mathbf{m}%
=\operatorname*{id}$. This proves Claim 4.]

Now, we have $\mathbf{m}\circ\mathbf{n}=\operatorname*{id}$ (by Claim 3) and
$\mathbf{n}\circ\mathbf{m}=\operatorname*{id}$ (by Claim 4). In other words,
the two maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse. Hence, these
two maps are bijections. This proves Proposition \ref{prop.lyndon.CFL.bij}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.multiset.bij.seqs}Let $S$ be any set.

Let $\mathfrak{M}$ be the set of all finite multisets of elements of $S$.

Let $\mathfrak{N}$ be the set of all families $\left(  k_{w}\right)  _{w\in
S}\in{\NN}^{S}$ of nonnegative integers (indexed by the elements of
$S$) such that all but finitely many $w\in S$ satisfy $k_{w}=0$.

Then, the map $\operatorname*{mult}:\mathfrak{M}\rightarrow\mathfrak{N}$ that
sends each multiset $M\in\mathfrak{M}$ to the family
\[
\left(  \left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\right)  _{w\in S}\in\mathfrak{N}%
\]
is well-defined and is a bijection.
\end{proposition}

\begin{example}
If $S=\mathbb{N}$, then the map $\operatorname*{mult}$ defined in Proposition
\ref{prop.multiset.bij.seqs} sends the finite multiset $\left\{
1,4,4,5\right\}  _{\operatorname*{multiset}}\in\mathfrak{M}$ to the family
$\left(  k_{w}\right)  _{w\in S}\in \NN^{S}$, where%
\[
k_{1}=1,\ \ \ \ \ \ \ \ \ \ k_{4}=2,\ \ \ \ \ \ \ \ \ \ k_{5}%
=1,\ \ \ \ \ \ \ \ \ \ \text{and }k_{w}=0\text{ for all }w\notin\left\{
1,4,5\right\}  .
\]

\end{example}

\begin{proof}
[Proof of Proposition \ref{prop.multiset.bij.seqs}.]It is straightforward to
see that the map $\operatorname*{mult}$ is well-defined. On the other hand,
the map%
\begin{align*}
\mathfrak{N}  &  \rightarrow\mathfrak{M},\\
\left(  k_{w}\right)  _{w\in S}  &  \mapsto\left(  \text{the multiset that
contains each }w\in S\text{ with multiplicity }k_{w}\right)
\end{align*}
is also well-defined. These two maps are clearly mutually inverse. Thus, the
map $\operatorname*{mult}$ is invertible, i.e., is a bijection. This proves
Proposition \ref{prop.multiset.bij.seqs}.
\end{proof}

Let us now solve Exercise~\ref{exe.words.lyndon.count}.
Let $\mathfrak{A}$ and $q$ be as in Exercise~\ref{exe.words.lyndon.count}.

For every
positive integer $n$, let $\operatorname*{lyn}n$ denote the number of Lyndon
words of length $n$. We need to prove that%
\begin{equation}
\operatorname*{lyn}n=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)
q^{n/d}\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }n.
\label{sol.words.lyndon.count.goal}%
\end{equation}


Let $\mathfrak{M}$ denote the set of all finite multisets of Lyndon words.
Define two maps $\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$
and $\mathbf{n} : \mathfrak{A}^{\ast} \to \mathfrak{M}$ as in
Proposition \ref{prop.lyndon.CFL.bij}.
Then, Proposition \ref{prop.lyndon.CFL.bij} shows that
these two maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse bijections.
Hence, the map $\mathbf{m}$ is a bijection.

On the other hand, let $\mathfrak{L}$ be the set of all Lyndon words. Thus,
the Lyndon words are precisely the elements of $\mathfrak{L}$. But the
definition of $\mathfrak{M}$ says that $\mathfrak{M}$ is the set of all finite
multisets of Lyndon words. In other words, $\mathfrak{M}$ is the set of all
finite multisets of elements of $\mathfrak{L}$ (since the Lyndon words are
precisely the elements of $\mathfrak{L}$).

The definition of $\operatorname*{lyn}n$ now rewrites as
\begin{equation}
\operatorname*{lyn}n=\left\vert \left\{  w\in\mathfrak{L}\ \mid\ \ell\left(
w\right)  =n\right\}  \right\vert \ \ \ \ \ \ \ \ \ \ \text{for every positive
integer }n
\label{sol.words.lyndon.count.lynn}
\end{equation}
(since the Lyndon words are precisely the elements of $\mathfrak{L}$).


Let $\mathfrak{N}$ be the set of all families $\left(  k_{w}\right)
_{w\in\mathfrak{L}}\in{\NN}^{\mathfrak{L}}$ of nonnegative integers
(indexed by the Lyndon words) such that all but finitely many $w\in
\mathfrak{L}$ satisfy $k_{w}=0$. Proposition \ref{prop.multiset.bij.seqs}
(applied to $S=\mathfrak{L}$) shows that the map $\operatorname*{mult}%
:\mathfrak{M}\rightarrow\mathfrak{N}$ that sends each multiset $M\in\mathfrak{M}$ to
the family
\[
\left(  \left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\right)  _{w\in S}\in\mathfrak{N}%
\]
is well-defined and is a bijection. Consider this map $\operatorname*{mult}$.

The composition $\mathbf{m}\circ\operatorname{mult}^{-1}
:\mathfrak{N}\rightarrow\mathfrak{A}^{\ast}$ of the bijections $\mathbf{m}$
and $\operatorname{mult}^{-1}$ is clearly a bijection. It can easily
be seen to satisfy
\begin{equation}
\ell\left(  \left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right)
=\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\ell\left(  w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  k_{w}\right)  _{w\in\mathfrak{L}%
}\in\mathfrak{N}.
\label{sol.words.lyndon.count.1}
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.lyndon.count.1}):} Let $\left(
k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}$. Let
$M = \operatorname{mult}^{-1}
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  $.
Then, $M$ is a multiset of elements of $\mathfrak{L}$ and satisfies $\left(
k_{w}\right)  _{w\in\mathfrak{L}}=\operatorname*{mult}M=\left(  \left(
\text{multiplicity of }w\text{ in the multiset }M\right)  \right)
_{w\in\mathfrak{L}}$. In other words, every $w\in\mathfrak{L}$ satisfies%
\begin{equation}
k_{w}=\left(  \text{multiplicity of }w\text{ in the multiset }M\right)  .
\label{sol.words.lyndon.count.2}
\end{equation}
\par
Let $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ denote the elements of this multiset
$M$ listed in decreasing order. Then, the definition of $\mathbf{m}$ yields
$\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$, so that%
\begin{align*}
\ell\left(  \mathbf{m}\left(  M\right)  \right)   &  =\ell\left(  a_{1}%
a_{2}\cdots a_{k}\right)  =\sum_{i\in\left\{  1,2,\ldots,k\right\}  }%
\ell\left(  a_{i}\right)  =\sum\limits_{w\in\mathfrak{L}}\sum\limits_{\substack{i\in
\left\{  1,2,\ldots,k\right\}  ;\\a_{i}=w}}\ell\left(  \underbrace{a_{i}%
}_{\substack{=w\\\text{(since }a_{i}=w\text{)}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since every }a_{i}\text{ belongs to
}\mathfrak{L}\right) \\
&  =\sum\limits_{w\in\mathfrak{L}}\underbrace{\sum\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=w}}\ell\left(  w\right)  }_{=\left(
\text{number of }i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfying }%
a_{i}=w\right)  \cdot\ell\left(  w\right)  }=\sum\limits_{w\in\mathfrak{L}%
}\underbrace{\left(  \text{number of }i\in\left\{  1,2,\ldots,k\right\}
\text{ satisfying }a_{i}=w\right)  }_{\substack{=\left(  \text{multiplicity of
}w\text{ in the multiset }M\right)  =k_{w}\\\text{(by
(\ref{sol.words.lyndon.count.2}))}}}\cdot\ell\left(  w\right) \\
&  =\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\ell\left(  w\right)  .
\end{align*}
Now,%
\[
\ell\left(  \left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right)
=\ell\left(  \mathbf{m}\left(  \underbrace{\operatorname{mult}^{-1}%
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  }_{=M}\right)
\right)  =\ell\left(  \mathbf{m}\left(  M\right)  \right)  =\sum
\limits_{w\in\mathfrak{L}}k_{w}\cdot\ell\left(  w\right)  ,
\]
which proves (\ref{sol.words.lyndon.count.1}).} Now, in the ring
$\QQ \left[  \left[  t\right]  \right]  $ of formal power series, we
have%
\[
\sum_{w\in\mathfrak{A}^{\ast}}t^{\ell\left(  w\right)  }=\sum_{n\in \NN %
}\underbrace{\left\vert \left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \ell\left(
w\right)  =n\right\}  \right\vert }_{\substack{=\left\vert \mathfrak{A}%
^{n}\right\vert =\left\vert \mathfrak{A}\right\vert ^{n}=q^{n}\\\text{(since
}\left\vert \mathfrak{A}\right\vert =q\text{)}}}t^{n}=\sum_{n\in \NN %
}q^{n}t^{n}=\dfrac{1}{1-qt}.
\]
Hence,%
\begin{align*}
\dfrac{1}{1-qt}  &  =\sum_{w\in\mathfrak{A}^{\ast}}t^{\ell\left(  w\right)
}=\sum_{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}}%
t^{\ell\left(  \left(  \mathbf{m}\circ\operatorname{mult}^{-1}%
\right)  \left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right)
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  \mathbf{m}\circ
\operatorname{mult}^{-1}\right) 
 \left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}%
}\right)  \text{ for }w\text{ in the sum,}\\
\text{since the map }\mathbf{m}\circ\operatorname{mult}^{-1}%
:\mathfrak{N}\rightarrow\mathfrak{A}^{\ast}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum_{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}}%
t^{\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\ell\left(  w\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.words.lyndon.count.1})}\right)
\\
&  =\sum_{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}}\prod
_{w\in\mathfrak{L}}t^{k_{w}\cdot\ell\left(  w\right)  }=\prod_{w\in
\mathfrak{L}}\underbrace{\sum_{k\in \NN }t^{k\cdot\ell\left(  w\right)
}}_{=\dfrac{1}{1-t^{\ell\left(  w\right)  }}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the product rule}\right) \\
&  =\prod_{w\in\mathfrak{L}}\dfrac{1}{1-t^{\ell\left(  w\right)  }}%
=\prod_{n\geq1}\prod\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=n}}\underbrace{\dfrac{1}{1-t^{\ell\left(  w\right)  }}}_{\substack{=\dfrac
{1}{1-t^{n}}\\\text{(since }\ell\left(  w\right)  =n\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  w\right)  \geq1\text{
for every }w\in\mathfrak{L}\right) \\
&  =\prod_{n\geq1}\underbrace{\prod\limits_{\substack{w\in\mathfrak{L};\\\ell\left(
w\right)  =n}}\dfrac{1}{1-t^{n}}}_{\substack{=\left(  \dfrac{1}{1-t^{n}%
}\right)  ^{\operatorname*{lyn}n}\\\text{(by
(\ref{sol.words.lyndon.count.lynn}))}}}=\prod_{n\geq1}\left(  \dfrac
{1}{1-t^{n}}\right)  ^{\operatorname*{lyn}n}.
\end{align*}
Taking the logarithm of both sides of this identity, we obtain%
\begin{align*}
\log\dfrac{1}{1-qt}  &  =\log\left(  \prod_{n\geq1}\left(  \dfrac{1}{1-t^{n}%
}\right)  ^{\operatorname*{lyn}n}\right)  =\sum_{n\geq1}\left(
\operatorname*{lyn}n\right)  \cdot\underbrace{\log\left(  \dfrac{1}{1-t^{n}%
}\right)  }_{\substack{=-\log\left(  1-t^{n}\right)  =\sum_{u\geq1}\dfrac
{1}{u}\left(  t^{n}\right)  ^{u}\\\text{(by the Mercator series for the
logarithm)}}}\\
&  =\sum_{n\geq1}\left(  \operatorname*{lyn}n\right)  \cdot\sum_{u\geq1}%
\dfrac{1}{u}\left(  t^{n}\right)  ^{u}=\sum_{n\geq1}\sum_{u\geq1}\left(
\operatorname*{lyn}n\right)  \dfrac{1}{u}\underbrace{\left(  t^{n}\right)
^{u}}_{=t^{nu}}=\sum_{n\geq1}\sum_{u\geq1}\left(  \operatorname*{lyn}n\right)
\dfrac{1}{u}t^{nu}\\
&  =\underbrace{\sum_{n\geq1}\sum\limits_{\substack{v\geq1;\\n\mid v}}}_{=\sum
_{v\geq1}\sum_{n\mid v}}\left(  \operatorname*{lyn}n\right)
\underbrace{\dfrac{1}{v/n}}_{=\dfrac{n}{v}}t^{v}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we substituted }v/n\text{ for }u\text{ in the second sum}\right)
\\
&  =\sum_{v\geq1}\sum_{n\mid v}\left(  \operatorname*{lyn}n\right)  \dfrac
{n}{v}t^{v}=\sum_{n\geq1}\sum_{d\mid n}\left(  \operatorname*{lyn}d\right)
\dfrac{d}{n}t^{n}%
\end{align*}
(here, we renamed the summation indices $v$ and $n$ as $n$ and $d$). Since%
\begin{align*}
\log\dfrac{1}{1-qt}  &  =-\log\left(  1-qt\right)  =\sum_{n\geq1}\dfrac{1}%
{n}\left(  qt\right)  ^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the Mercator
series for the logarithm}\right) \\
&  =\sum_{n\geq1}\dfrac{1}{n}q^{n}t^{n},
\end{align*}
this rewrites as%
\[
\sum_{n\geq1}\dfrac{1}{n}q^{n}t^{n}=\sum_{n\geq1}\sum_{d\mid n}\left(
\operatorname*{lyn}d\right)  \dfrac{d}{n}t^{n}.
\]
Comparing coefficients, we conclude that every positive integer $n$ satisfies%
\[
\dfrac{1}{n}q^{n}=\sum_{d\mid n}\left(  \operatorname*{lyn}d\right)  \dfrac
{d}{n}.
\]
Multiplying this with $n$, we obtain%
\begin{equation}
q^{n}=\sum_{d\mid n}\left(  \operatorname*{lyn}d\right)  d.
\label{sol.words.lyndon.count.6}%
\end{equation}


Now, recall that every positive integer $N$ satisfies%
\begin{equation}
\sum_{d\mid N}\mu\left(  d\right)  =\delta_{N,1}.
\label{sol.words.lyndon.count.mu}%
\end{equation}
\footnote{This is one of the most fundamental properties of the
number-theoretic M\"{o}bius function. For a proof of
\eqref{sol.words.lyndon.count.mu}, see the solution of
Exercise~\ref{exe.witt.ghost-equiv}. (More precisely,
the equality \eqref{sol.words.lyndon.count.mu} is obtained
from \eqref{sol.exe.witt.ghost-equiv.mu*1} by renaming $n$
as $N$.)} Now, every positive integer $n$ satisfies%
\begin{align*}
\sum\limits_{d\mid n}\mu\left(  d\right)  q^{n/d}  &  =\sum\limits_{e\mid
n}\mu\left(  e\right)  \underbrace{q^{n/e}}_{\substack{=\sum_{d\mid
n/e}\left(  \operatorname*{lyn}d\right)  d\\\text{(by
(\ref{sol.words.lyndon.count.6}), applied}\\\text{to }n/e\text{ instead of
}n\text{)}}}=\sum\limits_{e\mid n}\mu\left(  e\right)  \sum_{d\mid n/e}\left(
\operatorname*{lyn}d\right)  d\\
&  =\underbrace{\sum\limits_{e\mid n}\sum_{d\mid n/e}}_{=\sum_{d\mid n}%
\sum_{e\mid n/d}}\mu\left(  e\right)  \left(  \operatorname*{lyn}d\right)
d=\sum_{d\mid n}\underbrace{\sum_{e\mid n/d}\mu\left(  e\right)
}_{\substack{=\delta_{n/d,1}\\\text{(by (\ref{sol.words.lyndon.count.mu}),
applied}\\\text{to }N=n/d\text{)}}}\left(  \operatorname*{lyn}d\right)  d\\
&  =\sum_{d\mid n}\underbrace{\delta_{n/d,1}}_{=\delta_{n,d}}\left(
\operatorname*{lyn}d\right)  d=\sum_{d\mid n}\delta_{n,d}\left(
\operatorname*{lyn}d\right)  d=\left(  \operatorname*{lyn}n\right)  n.
\end{align*}
Dividing this by $n$, we obtain $\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  q^{n/d}=\operatorname*{lyn}n$. This proves
(\ref{sol.words.lyndon.count.goal}). Thus, Exercise
\ref{exe.words.lyndon.count} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.std2}}

\textit{Solution to Exercise \ref{exe.words.lyndon.std2}.} The word $v$
is nonempty (since it is Lyndon). Thus, $u\neq w$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $u=w$. Hence,
$u\varnothing=w=uv$. Cancelling $u$ from this equality, we obtain
$\varnothing=v$. Thus, the word $v$ is empty; this contradicts the fact that
$v$ is nonempty. This contradiction proves that our assumption was wrong,
qed.}. Combined with $u\leq uv=w$, this yields $u<w$.

(b) We have $w=uv$; thus, $v$ is a proper suffix of $w$ (since $u$ is
nonempty). Also, the word $w$ is Lyndon, and therefore%
\begin{equation}
\text{every nonempty proper suffix of }w\text{ is }>w
\label{sol.words.lyndon.std2.a.axi}%
\end{equation}
(by the definition of a Lyndon word). Applying
(\ref{sol.words.lyndon.std2.a.axi}) to the nonempty proper suffix $v$ of $w$,
we obtain $v>w$. Thus, $w<v$. Hence, $u<w<v$, and this solves Exercise
\ref{exe.words.lyndon.std2}(b).

(a) The word $v$ is Lyndon (by its definition). It remains to prove that the
word $u$ is Lyndon.

Recall that a word $x$ is Lyndon if and only if it is nonempty and satisfies
the property that every nonempty proper suffix $y$ of $x$ satisfies
$y>x$.\ \ \ \ \footnote{This is just a restatement of the definition of a
Lyndon word.} Applied to $x=u$, this shows that the word $u$ is Lyndon if and
only if it is nonempty and satisfies the property that every nonempty proper
suffix $y$ of $u$ satisfies $y>u$. Let us now prove that%
\begin{equation}
\text{every nonempty proper suffix }y\text{ of }u\text{ satisfies }y>u\text{.}
\label{sol.words.lyndon.std2.a.1}%
\end{equation}


\textit{Proof of (\ref{sol.words.lyndon.std2.a.1}):} Assume the contrary.
Then, there exists a nonempty proper suffix $y$ of $u$ which does not satisfy
$y>u$. Let $p$ be the \textbf{shortest} such suffix. Thus, $p$ is a nonempty
proper suffix of $u$ which does not satisfy $p>u$.

Notice that $p\neq u$ (since $p$ is a \textbf{proper} suffix of $u$).

There exists a nonempty $t\in\mathfrak{A}^{\ast}$ such that $u=tp$ (since $p$
is a proper suffix of $u$). Consider this $t$. We have $w=\underbrace{u}%
_{=tp}v=tpv=t\left(  pv\right)  $, and thus the word $pv$ is a proper suffix
of $w$ (since $t$ is nonempty). Also, the word $pv$ is nonempty (since $v$ is
nonempty). Hence, $pv$ is a nonempty proper suffix of $w$. But applying
(\ref{sol.words.lyndon.std2.a.axi}) to the nonempty proper suffix $pv$ of $w$,
we obtain $pv>w=uv$. Thus, $uv<pv$. Hence, Proposition \ref{prop.words.lex}(e)
(applied to $a=u$, $b=v$, $c=p$ and $d=v$) yields that either we have $u\leq
p$ or the word $p$ is a prefix of $u$. Since we don't have $u\leq
p$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $u\leq p$.
Thus, $p\geq u$. Combined with $p\neq u$, this yields $p>u$. This contradicts
the fact that $p$ does not satisfy $p>u$. This contradiction proves that our
assumption was wrong, qed.}, we therefore conclude that the word $p$ is a
prefix of $u$. Hence, $p\leq u$. Combined with $p\neq u$, this yields $p<u$.
Combined with $u<v$ (by Exercise \ref{exe.words.lyndon.std2}(b)), this yields
$p<u<v$.

Let us now show that the word $p$ is Lyndon. Indeed, let $r$ be a nonempty
proper suffix of $p$. Then, $r$ is shorter than $p$ (being a proper suffix of
$p$); in other words, $\ell\left(  r\right)  <\ell\left(  p\right)  $. On the
other hand, there exists some nonempty $q\in\mathfrak{A}^{\ast}$ satisfying
$p=qr$ (since $r$ is a proper suffix of $p$). Consider this $q$. We have
$u=t\underbrace{p}_{=qr}=tqr=\left(  tq\right)  r$. Thus, $r$ is a proper
suffix of $u$ (since $tq$ is nonempty (because $q$ is nonempty)). Hence,
$r>u$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, we don't
have $r>u$. Thus, $r$ is a nonempty proper suffix of $u$ which does not
satisfy $r>u$. In other words, $r$ is a nonempty proper suffix $y$ of $u$
which does not satisfy $y>u$. Since the shortest such suffix is $p$ (by the
definition of $p$), this yields that $r$ is not shorter than $p$. That is,
$\ell\left(  r\right)  \geq\ell\left(  p\right)  $. This contradicts
$\ell\left(  r\right)  <\ell\left(  p\right)  $. This contradiction proves
that our assumption was wrong, qed.}. Hence, $r>u>p$ (since $p<u$). Now, let
us forget that we fixed $r$. We thus have shown that
\[
\text{every nonempty proper suffix }r\text{ of }p\text{ satisfies }r>p\text{.}%
\]
But recall that the word $p$ is Lyndon if and only if it is nonempty and
satisfies the property that every nonempty proper suffix $r$ of $p$ satisfies
$r>p$.\ \ \ \ \footnote{This is just a restatement of the definition of a
Lyndon word (applied to $p$).} Hence, we conclude that the word $p$ is Lyndon
(since we have shown that $p$ is nonempty and satisfies the property that
every nonempty proper suffix $r$ of $p$ satisfies $r>p$).

Now, the words $p$ and $v$ are Lyndon and satisfy $p<v$. Thus, Proposition
\ref{prop.words.lyndon.concat}(a) (applied to $p$ instead of $u$) yields that
the word $pv$ is Lyndon. Notice also that $w=\underbrace{u}_{=tp}%
v=tpv=t\left(  pv\right)  $, and thus $pv$ is a proper suffix of $w$ (since
$t$ is nonempty). Hence, $pv$ is a proper suffix of $w$ such that $pv$ is
Lyndon. In other words, $pv$ is a proper suffix $z$ of $w$ such that $z$ is
Lyndon. Since $v$ is the longest such suffix\footnote{\textit{Proof.} We
defined $v$ as the longest proper suffix of $w$ such that $v$ is Lyndon. In
other words, $v$ is the longest proper suffix $z$ of $w$ such that $z$ is
Lyndon, qed.}, this yields that $pv$ is not longer than $v$. In other words,
$\ell\left(  pv\right)  \leq\ell\left(  v\right)  $. This contradicts
$\ell\left(  pv\right)  =\underbrace{\ell\left(  p\right)  }%
_{\substack{>0\\\text{(since }p\text{ is nonempty)}}}+\ell\left(  v\right)
>\ell\left(  v\right)  $. This contradiction proves that our assumption was
wrong. Hence, (\ref{sol.words.lyndon.std2.a.1}) is proven.

Now, recall that the word $u$ is Lyndon if and only if it is nonempty and
satisfies the property that every nonempty proper suffix $y$ of $u$ satisfies
$y>u$. Hence, the word $u$ is Lyndon (because we have shown that $u$ is
nonempty and satisfies the property that every nonempty proper suffix $y$ of
$u$ satisfies $y>u$). The solution of Exercise \ref{exe.words.lyndon.std2}(a)
is thus complete.

(c) Let us denote by $u^{\prime}$ and $v^{\prime}$ the words $u$ and $v$
constructed in Theorem \ref{thm.words.lyndon.std} (to avoid confusing them
with the words $u$ and $v$ defined in Exercise \ref{exe.words.lyndon.std2}).
We must then show that $u=u^{\prime}$ and $v=v^{\prime}$.

From Theorem \ref{thm.words.lyndon.std}, we see that $v^{\prime}$ is the
(lexicographically) smallest nonempty \textbf{proper} suffix of $w$, and that
$u^{\prime}$ is a nonempty word such that $w=u^{\prime}v^{\prime}$.

The word $v$ is a nonempty proper suffix of $w$. Since $v^{\prime}$ is the
smallest such suffix, we thus conclude that $v^{\prime}\leq v$. We shall now
prove that $v^{\prime}=v$.

Indeed, $v$ is a proper suffix of $w$. In other words, $v$ is a
proper suffix of $u^{\prime}v^{\prime}$ (since $w=u^{\prime}v^{\prime}$).
Hence, we must be in one of the following two cases (depending on whether this
suffix begins before the suffix $v^{\prime}$ of $u^{\prime}v^{\prime}$ begins
or afterwards):

\textit{Case 1:} The word $v$ is a proper suffix of $v^{\prime}$.

\textit{Case 2:} The word $v$ has the form $qv^{\prime}$ where $q$ is a proper
suffix of $u^{\prime}$. (This suffix $q$ may be empty.)

Let us first consider Case 1. In this case, the word $v$ is a proper suffix of
$v^{\prime}$. Now, every nonempty proper suffix $s$ of $v^{\prime}$ satisfies
$s>v^{\prime}$\ \ \ \ \footnote{\textit{Proof.} Let $s$ be a nonempty proper
suffix of $v^{\prime}$. The word $s$ is a proper suffix of $v^{\prime}$, which
(in turn) is a suffix of $w$. Hence, $s$ is a proper suffix of $w$. Thus, $s$
is a nonempty proper suffix of $w$. Since the smallest such suffix is
$v^{\prime}$ (by the definition of $v^{\prime}$), this yields $s\geq
v^{\prime}$. Since $s\neq v^{\prime}$ (because $s$ is a \textbf{proper} suffix
of $v^{\prime}$), this yields $s>v^{\prime}$, qed.}. Thus, the word
$v^{\prime}$ is nonempty and satisfies the property that every nonempty proper
suffix $s$ of $v^{\prime}$ satisfies $s>v^{\prime}$. In other words, the word
$v^{\prime}$ is Lyndon (according to the definition of a Lyndon word).

Now, $v$ is the longest proper suffix of $w$ such that $v$ is Lyndon. In other
words, $v$ is the longest proper suffix $z$ of $w$ such that $z$ is Lyndon.
But $v^{\prime}$ also is a proper suffix $z$ of $w$ such that $z$ is Lyndon
(since $v^{\prime}$ is Lyndon). Since $v$ is the longest such suffix, this
yields that $v^{\prime}$ is not longer than $v$. In other words, $\ell\left(
v^{\prime}\right)  \leq\ell\left(  v\right)  $. But since $v$ is a proper
suffix of $v^{\prime}$, we have $\ell\left(  v\right)  <\ell\left(  v^{\prime
}\right)  \leq\ell\left(  v\right)  $. This is absurd. Hence, $v^{\prime}=v$
(since ex falso quodlibet). Thus, $v^{\prime}=v$ is proven in Case 1.

Let us now consider Case 2. In this case, the word $v$ has the form
$qv^{\prime}$ where $q$ is a proper suffix of $u^{\prime}$. Consider this $q$.

We need to prove that $v^{\prime}=v$. If $q=\varnothing$, then this is obvious
(because if $q=\varnothing$, then $v=\underbrace{q}_{=\varnothing}v^{\prime
}=v^{\prime}$ and thus $v^{\prime}=v$). Hence, for the rest of this proof, we
can WLOG assume that we don't have $q=\varnothing$. Assume this. The word $q$
is nonempty (since we don't have $q=\varnothing$), and thus $v^{\prime}$ is a
proper suffix of $v$ (since $v=qv^{\prime}$).

The word $v$ is Lyndon. Hence, every nonempty proper suffix of $v$ is $>v$ (by
the definition of a Lyndon word). Applying this to the nonempty proper suffix
$v^{\prime}$ of $v$, we conclude that $v^{\prime}>v$. This contradicts
$v^{\prime}\leq v$. This contradiction shows that we have $v^{\prime}=v$
(since ex falso quodlibet). Thus, $v^{\prime}=v$ is proven in Case 2.

Now, $v^{\prime}=v$ is proven in each of the two Cases 1 and 2. Since these
two Cases cover all possibilities, this yields that $v^{\prime}=v$ always holds.

Now we have proven that $v^{\prime}=v$. Cancelling $v$ from the equality
$uv=w=u^{\prime}\underbrace{v^{\prime}}_{=v}=u^{\prime}v$, we obtain
$u=u^{\prime}$. Thus, we have $u=u^{\prime}$ and $v=v^{\prime}$. This solves
Exercise \ref{exe.words.lyndon.std2}(c).

[\textit{Remark:} We have not used any statement of Theorem
\ref{thm.words.lyndon.std} in our above solution. Thus, parts (a) and (b) of
Exercise \ref{exe.words.lyndon.std2} provide an alternative proof of Theorem
\ref{thm.words.lyndon.std} (because Exercise \ref{exe.words.lyndon.std2}(c)
shows that the words $u$ and $v$ defined in Exercise
\ref{exe.words.lyndon.std2} are precisely the words $u$ and $v$ constructed in
Theorem \ref{thm.words.lyndon.std}).]
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.powerlyndon}}

\textit{Solution to Exercise \ref{exe.words.powerlyndon}.} (a) Let us
prove the implications $\mathcal{A}^{\prime}\Longrightarrow\mathcal{D}%
^{\prime}$ and $\mathcal{D}^{\prime}\Longrightarrow\mathcal{A}^{\prime}$.

\textit{Proof of the implication $\mathcal{A}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$:} Assume that Assertion $\mathcal{A}^{\prime}$
holds. Thus, $w$ is a power of a Lyndon word. Let said Lyndon word be $t$.
Thus, $w=t^{n}$ for some $n\in \NN$. Consider this $n$. Since $w$ is
nonempty, we have $n\geq1$. Hence, $t^{n-1}$ is well-defined.

We will first show the following simple lemma:

\textit{Lemma A:} Let $u^{\prime}$, $v^{\prime}$ and $p$ be three words, and
$N\in \NN$ be such that $u^{\prime}v^{\prime}=p^{N}$. Assume that $p$ is
not a prefix of $u^{\prime}$, and that $p$ is not a suffix of $v^{\prime}$.
Then, $N\leq1$.

\textit{Proof of Lemma A:} Assume the contrary. Thus, $N>1$, so that $N\geq2$.
The word $v^{\prime}$ is a suffix of $p^{N-1}p$ (since $u^{\prime}v^{\prime
}=p^{N}=p^{N-1}p$). Thus, we must be in one of the following two cases
(depending on whether this suffix begins before the suffix $p$ of $p^{N-1}p$
begins, or afterwards):

\textit{Case 1:} The word $v^{\prime}$ has the form $rp$ where $r$ is a suffix
of $p^{N-1}$.

\textit{Case 2:} The word $v^{\prime}$ is a suffix of $p$.

Let us first consider Case 1. In this case, the word $v^{\prime}$ has the form
$rp$ where $r$ is a suffix of $p^{N-1}$. Hence, $p$ is a suffix of $v^{\prime
}$, contradicting the fact that $p$ is not a suffix of $v^{\prime}$. Hence,
Case 1 leads to a contradiction.

Let us now consider Case 2. In this case, the word $v^{\prime}$ is a suffix of
$p$. In other words, there exists a $q\in\mathfrak{A}^{\ast}$ such that
$p=qv^{\prime}$. Consider this $q$. We have $u^{\prime}v^{\prime}%
=p^{N}=p^{N-1}\underbrace{p}_{=qv^{\prime}}=p^{N-1}qv^{\prime}$. Cancelling
$v^{\prime}$ from this equality, we obtain $u^{\prime}=p^{N-1}q$. Since
$p^{N-1}=pp^{N-2}$ (this is well-defined because $N\geq2$), this further
becomes $u^{\prime}=\underbrace{p^{N-1}}_{=pp^{N-2}}q=pp^{N-2}q=p\left(
p^{N-2}q\right)  $. Hence, $p$ is a prefix of $u^{\prime}$, contradicting the
fact that $p$ is not a prefix of $u^{\prime}$. We have thus obtained a
contradiction in Case 2.

We have thus obtained a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that we always get a
contradiction. This completes the proof of Lemma A.

Now, let us return to the proof of the implication $\mathcal{A}^{\prime
}\Longrightarrow\mathcal{D}^{\prime}$. Let $u$ and $v$ be nonempty words
satisfying $w=uv$. We want to show that $vu\geq uv$.

The word $t$ is nonempty (since it is Lyndon); thus, $\ell\left(  t\right)
\geq1$. Hence, for every sufficiently large $a\in \NN$, we have that the
word $t^{a}$ is not a prefix of $u$\ \ \ \ \footnote{\textit{Proof.} Let
$a\in \NN$ be such that $a>\ell\left(  u\right)  $. Then, $\ell\left(
t^{a}\right)  =\underbrace{a}_{>\ell\left(  u\right)  }\underbrace{\ell\left(
t\right)  }_{\geq1}>\ell\left(  u\right)  $, so that the word $t^{a}$ is
longer than $u$. Hence, the word $t^{a}$ is not a prefix of $u$.
\par
Now, let us forget that we have fixed $a$. We thus have shown that for every
$a\in \NN$ satisfying $a>\ell\left(  u\right)  $, we have that the word
$t^{a}$ is not a prefix of $u$. Consequently, for every sufficiently large
$a\in \NN$, we have that the word $t^{a}$ is not a prefix of $u$, qed.}.
Similarly, for every sufficiently large $b\in \NN$, we have that the
word $t^{b}$ is not a suffix of $v$.

Let $A$ be the largest nonnegative integer $a$ such that $t^{a}$ is a prefix
of $u\ \ \ \ $\footnote{This is well-defined because of the following two
facts:
\par
\begin{itemize}
\item There exists an $a\in \NN$ such that the word $t^{a}$ is a prefix
of $u$ (namely, $a=0$).
\par
\item For every sufficiently large $a\in \NN$, we have that the word
$t^{a}$ is not a prefix of $u$.
\end{itemize}
}. Thus, $t^{A}$ is a prefix of $u$, but $t^{A+1}$ is not a prefix of $u$. Let
$B$ be the largest nonnegative integer $b$ such that $t^{b}$ is a suffix of
$v$\ \ \ \ \footnote{This is well-defined because of the following two facts:
\par
\begin{itemize}
\item There exists a $b\in \NN$ such that the word $t^{b}$ is a suffix
of $v$ (namely, $b=0$).
\par
\item For every sufficiently large $b\in \NN$, we have that the word
$t^{b}$ is not a suffix of $v$.
\end{itemize}
}. Thus, $t^{B}$ is a suffix of $v$, but $t^{B+1}$ is not a suffix of $v$.

There exists a $u^{\prime}\in\mathfrak{A}^{\ast}$ such that $u=t^{A}u^{\prime
}$ (since $t^{A}$ is a prefix of $u$). Consider this $u^{\prime}$. Recall that
$t^{A+1}$ is not a prefix of $u$. In other words, $t^{A}t$ is not a prefix of
$t^{A}u^{\prime}$ (since $t^{A+1}=t^{A}t$ and $u=t^{A}u^{\prime}$).

There exists a $v^{\prime}\in\mathfrak{A}^{\ast}$ such that $v=v^{\prime}%
t^{B}$ (since $t^{B}$ is a suffix of $v$). Consider this $v^{\prime}$. Recall
that $t^{B+1}$ is not a suffix of $v$. In other words, $tt^{B}$ is not a
suffix of $v^{\prime}t^{B}$ (since $t^{B+1}=tt^{B}$ and $v=v^{\prime}t^{B}$).

If the word $t$ was a prefix of $u^{\prime}$, then $t^{A}t$ would be a prefix
of $t^{A}u^{\prime}$, which would contradict the fact that $t^{A}t$ is not a
prefix of $t^{A}u^{\prime}$. Thus, $t$ is not a prefix of $u^{\prime}$. In
particular, $t\neq u^{\prime}$.

If the word $t$ was a suffix of $v^{\prime}$, then $tt^{B}$ would be a suffix
of $v^{\prime}t^{B}$, which would contradict the fact that $tt^{B}$ is not a
suffix of $v^{\prime}t^{B}$. Thus, $t$ is not a suffix of $v^{\prime}$. In
particular, $t\neq v^{\prime}$.

But we have $t^{n}=w=\underbrace{u}_{=t^{A}u^{\prime}}\underbrace{v}%
_{=v^{\prime}t^{B}}=t^A u^\prime v^\prime t^B
=t^{A}\left(  u^{\prime}v^{\prime}t^{B}\right)  $. Hence,
$t^{A}$ is a prefix of $t^{n}$; therefore, $A\leq n$ and thus $n-A\geq0$.
Hence, $t^{A}t^{n-A}=t^{n}=t^{A}\left(  u^{\prime}v^{\prime}t^{B}\right)  $.
Cancelling $t^{A}$ in this equality, we obtain $t^{n-A}=u^{\prime}v^{\prime
}t^{B}=\left(  u^{\prime}v^{\prime}\right)  t^{B}$, which shows that $t^{B}$
is a suffix of $t^{n-A}$. Thus, $B\leq n-A$ and therefore $n-A-B\geq0$. Hence,
$t^{n-A-B}t^{B}=t^{n-A}=\left(  u^{\prime}v^{\prime}\right)  t^{B}$.
Cancelling $t^{B}$ in this yields $t^{n-A-B}=u^{\prime}v^{\prime}$. Now, Lemma
A (applied to $p=t$ and $N=n-A-B$) yields $n-A-B\leq1$. Since $n-A-B\geq0$,
this shows that we have either $n-A-B=0$ or $n-A-B=1$. We thus must be in one
of the following two cases:

\textit{Case 1:} We have $n-A-B=0$.

\textit{Case 2:} We have $n-A-B=1$.

Let us first consider Case 1. In this case, we have $n-A-B=0$. Thus,
$t^{n-A-B}=u^{\prime}v^{\prime}$ rewrites as $t^{0}=u^{\prime}v^{\prime}$, so
that $u^{\prime}v^{\prime}=t^{0}=\varnothing$. Consequently, $u^{\prime
}=\varnothing$ and $v^{\prime}=\varnothing$. Now, $u=t^{A}%
\underbrace{u^{\prime}}_{=\varnothing}=t^{A}$ and $v=\underbrace{v^{\prime}%
}_{=\varnothing}t^{B}=t^{B}$, so that $\underbrace{v}_{=t^{B}}\underbrace{u}%
_{=t^{A}}=t^{B}t^{A}=t^{B+A}=t^{A+B}=\underbrace{t^{A}}_{=u}\underbrace{t^{B}%
}_{=v}=uv\geq uv$. Hence, $vu\geq uv$ is proven in Case 1.

Let us now consider Case 2. In this case, we have $n-A-B=1$. Thus,
$t^{n-A-B}=u^{\prime}v^{\prime}$ rewrites as $t^{1}=u^{\prime}v^{\prime}$, so
that $u^{\prime}v^{\prime}=t^{1}=t$. As a consequence, $u^{\prime}$ is
nonempty (because if $u^{\prime}$ was empty, then we would have $u^{\prime
}v^{\prime}=v^{\prime}$ and thus $v^{\prime}=u^{\prime}v^{\prime}=t$,
contradicting $t\neq v^{\prime}$), so that $\ell\left(  u^{\prime}\right)
>0$. Now, $\ell\left(  \underbrace{t}_{=u^{\prime}v^{\prime}}\right)
=\underbrace{\ell\left(  u^{\prime}\right)  }_{>0}+\ell\left(  v^{\prime
}\right)  >\ell\left(  v^{\prime}\right)  $, and therefore $t$ is not a prefix
of $v^{\prime}$. Also, $v^{\prime}$ is nonempty (because if $v^{\prime}$ was
empty, then we would have $u^{\prime}v^{\prime}=u^{\prime}$ and thus
$u^{\prime}=u^{\prime}v^{\prime}=t$, contradicting $t\neq u^{\prime}$). But
$t$ is Lyndon, and therefore Proposition \ref{prop.words.lyndon.trivia}(a)
(applied to $t$, $u^{\prime}$ and $v^{\prime}$ instead of $w$, $u$ and $v$)
yields $v^{\prime}\geq t$. That is, $t\leq v^{\prime}$. Thus, Proposition
\ref{prop.words.lex}(d) (applied to $a=t$, $b=t^{n-1}$, $c=v^{\prime}$ and
$d=t^{B}u$) yields that either we have $tt^{n-1}\leq v^{\prime}t^{B}u$ or the
word $t$ is a prefix of $v^{\prime}$. Since $t$ is not a prefix of $v^{\prime
}$, this yields $tt^{n-1}\leq v^{\prime}t^{B}u$. Since $tt^{n-1}=t^{n}=w=uv$
and $\underbrace{v^{\prime}t^{B}}_{=v}u=vu$, this rewrites as $uv\leq vu$.
Hence, $vu\geq uv$ is proven in Case 2.

We thus have shown that $vu\geq uv$ in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this yields that we always have
$vu\geq uv$. Thus, Assertion $\mathcal{D}^{\prime}$ is satisfied, so we have
proven the implication $\mathcal{A}^{\prime}\Longrightarrow\mathcal{D}%
^{\prime}$.

\textit{Proof of the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{A}^{\prime}$:} Assume that Assertion $\mathcal{D}^{\prime}$
holds. Thus, if $u$ and $v$ are nonempty words satisfying $w=uv$, then we have
$vu\geq uv$.

We need to prove that Assertion $\mathcal{A}^{\prime}$ holds, i.e., that $w$
is a power of a Lyndon word. Assume the contrary. Thus, $w$ is not a power of
a Lyndon word; hence, $w$ is not a Lyndon word itself. Consequently, there
exist nonempty words $u$ and $v$ such that $w=uv$ and $vu\leq uv$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Hence, any nonempty
words $u$ and $v$ satisfying $w=uv$ satisfy $vu>uv$. In other words, $w$
satisfies Assertion $\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv}.
Hence, $w$ also satisfies Assertion $\mathcal{A}$ of Theorem
\ref{thm.words.lyndon.equiv} (since Theorem \ref{thm.words.lyndon.equiv}
yields the equivalence of these two assertions $\mathcal{D}$ and $\mathcal{A}%
$). In other words, $w$ is a Lyndon word. This contradicts the knowledge that
$w$ is not a Lyndon word, qed.}. Consider such a pair of nonempty words $u$
and $v$ with \textbf{minimum }$\ell\left(  u\right)  $. The minimality of
$\ell\left(  u\right)  $ shows that%
\begin{equation}
\left(  \text{if }u^{\prime}\text{ and }v^{\prime}\text{ are nonempty words
such that }w=u^{\prime}v^{\prime}\text{ and }v^{\prime}u^{\prime}\leq
u^{\prime}v^{\prime}\text{, then }\ell\left(  u^{\prime}\right)  \geq
\ell\left(  u\right)  \right)  . \label{sol.words.powerlyndon.DA.1}%
\end{equation}


We have $vu\leq uv$. In combination with $vu\geq uv$ (which follows from
Assertion $\mathcal{D}^{\prime}$), this yields $vu=uv$. Therefore, Proposition
\ref{prop.words.uv=vu.2} yields that there exist a $t\in\mathfrak{A}^{\ast}$
and two nonnegative integers $n$ and $m$ such that $u=t^{n}$ and $v=t^{m}$.
Consider this $t$ and these $n$ and $m$. We have $n\neq0$ (since $t^{n}=u$ is
nonempty) and $m\neq0$ (since $t^{m}=v$ is nonempty), and the word $t$ is
nonempty (since $t^{n}=u$ is nonempty). Moreover, we have $n=1$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Hence, $n\neq1$.
Combined with $n\neq0$, this leads to $n\geq2$. As a consequence, $u=t^{n}$
can be rewritten as $u=tt^{n-1}$. The word $t^{n-1+m}$ is nonempty (since $t$
is nonempty and since $\underbrace{n}_{\geq2}-1+\underbrace{m}_{\geq0}%
\geq2-1+0=1$). Now, $w=\underbrace{u}_{=tt^{n-1}}\underbrace{v}_{=t^{m}%
}=tt^{n-1}t^{m}=tt^{n-1+m}$. Also, $t^{n-1+m}t=t^{n-1+m+1}=tt^{n-1+m}$. Hence,
(\ref{sol.words.powerlyndon.DA.1}) (applied to $u^{\prime}=t$ and $v^{\prime
}=t^{n-1+m}$) yields $\ell\left(  t\right)  \geq\ell\left(  u\right)  $ (since
$t$ and $t^{n-1+m}$ are nonempty). Since $u=t^{n}$, this rewrites as
$\ell\left(  t\right)  \geq\ell\left(  t^{n}\right)  =\underbrace{n}_{\geq
2}\ell\left(  t\right)  \geq2\ell\left(  t\right)  $, whence $\ell\left(
t\right)  =0$, which contradicts the fact that $t$ is nonempty. This
contradiction shows that our assumption was wrong, qed.}. Hence, $t^{n}%
=t^{1}=t$, so that $u=t^{n}=t$ and $w=\underbrace{u}_{=t}\underbrace{v}%
_{=t^{m}}=tt^{m}=t^{m+1}$. We shall now prove that the word $t$ is Lyndon.

Assume the contrary. Then, $t$ is not Lyndon. Let $q$ be the
(lexicographically) smallest nonempty suffix of $t$. Then, Proposition
\ref{prop.words.lyndon.cutoff}(b) (applied to $t$ and $q$ instead of $w$ and
$v$) yields that there exists a nonempty $p\in\mathfrak{A}^{\ast}$ such that
$t=pq$, $p\geq q$ and $pq\geq qp$\ \ \ \ \footnote{Notice that the variable
$p$ here is what has been called $u$ in Proposition
\ref{prop.words.lyndon.cutoff}(b). (We had to rename it since the letter $u$
is already in use.)}. Consider this $p$. Since $q$ is nonempty, we have
$\ell\left(  pq\right)  >\ell\left(  p\right)  $, so that $\ell\left(
\underbrace{u}_{=t=pq}\right)  =\ell\left(  pq\right)  >\ell\left(  p\right)
$. From $pq\geq qp$, we obtain $qp\leq pq$. Thus, it is easy to see that
$\left(  qp\right)  ^{m+1}\leq\left(  pq\right)  ^{m+1}$%
\ \ \ \ \footnote{\textit{Proof.} We have $qp\leq pq$. Hence, Proposition
\ref{prop.words.lex}(d) (applied to $qp$, $\left(  qp\right)  ^{m}$, $pq$ and
$\left(  pq\right)  ^{m}$ instead of $a$, $b$, $c$ and $d$) yields that either
we have $\left(  qp\right)  \left(  qp\right)  ^{m}\leq\left(  pq\right)
\left(  pq\right)  ^{m}$ or the word $qp$ is a prefix of $pq$. In the first of
these two cases, we are done (because in the first of these two cases, we have
$\left(  qp\right)  \left(  qp\right)  ^{m}\leq\left(  pq\right)  \left(
pq\right)  ^{m}$ and thus $\left(  qp\right)  ^{m+1}=\left(  qp\right)
\left(  qp\right)  ^{m}\leq\left(  pq\right)  \left(  pq\right)  ^{m}=\left(
pq\right)  ^{m+1}$). Hence, we can WLOG assume that we are in the second of
these two cases. Assume this. Then, the word $qp$ is a prefix of $pq$. Since
the word $qp$ has the same length as $pq$ (in fact, $\ell\left(  qp\right)
=\ell\left(  q\right)  +\ell\left(  p\right)  =\ell\left(  p\right)
+\ell\left(  q\right)  =\ell\left(  pq\right)  $), this yields that $qp=pq$,
so that $\left(  \underbrace{qp}_{=pq}\right)  ^{m+1}=\left(  pq\right)
^{m+1}\leq\left(  pq\right)  ^{m+1}$, qed.}.

We have $v=t^{m}$. Since $t=pq$, this rewrites as $v=\left(  pq\right)  ^{m}$.
Now,%
\[
q\underbrace{v}_{=\left(  pq\right)  ^{m}}p=\underbrace{q\left(  pq\right)
^{m}}_{=\left(  qp\right)  ^{m}q}p=\left(  qp\right)  ^{m}qp=\left(
qp\right)  ^{m+1}\leq\left(  pq\right)  ^{m+1}=\left(  pq\right)  \left(
\underbrace{pq}_{=t}\right)  ^{m}=\left(  pq\right)  \underbrace{t^{m}}%
_{=v}=\left(  pq\right)  v=pqv.
\]
Since $w=\underbrace{u}_{=t=pq}v=pqv$, we can thus apply
(\ref{sol.words.powerlyndon.DA.1}) to $u^{\prime}=p$ and $v^{\prime}=qv$. As a
result, we obtain $\ell\left(  p\right)  \geq\ell\left(  u\right)  $. This
contradicts $\ell\left(  u\right)  >\ell\left(  p\right)  $. This
contradiction shows that our assumption (that $t$ is not Lyndon) was wrong.
Hence, $t$ is Lyndon. Thus, $w$ is a power of a Lyndon word (since $w=t^{m+1}$
is a power of $t$). Thus, Assertion $\mathcal{A}^{\prime}$ is satisfied, so we
have proven the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{A}^{\prime}$.

Now we have proven both implications $\mathcal{A}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$ and $\mathcal{D}^{\prime}\Longrightarrow\mathcal{A}%
^{\prime}$. Therefore, the equivalence $\mathcal{A}^{\prime}%
\Longleftrightarrow\mathcal{D}^{\prime}$ follows. Thus, Exercise
\ref{exe.words.powerlyndon}(a) is solved.

(b) Consider the letter $m$ and the alphabet $\mathfrak{A}\cup\left\{
m\right\}  $ defined in Assertion $\mathcal{F}^{\prime\prime}$. We notice that
the lexicographic order on $\mathfrak{A}^{\ast}$ is the restriction of the
lexicographic order on $\left(  \mathfrak{A}\cup\left\{  m\right\}  \right)
^{\ast}$ to $\mathfrak{A}^{\ast}$. Therefore, when we have two words $p$ and
$q$ in $\mathfrak{A}^{\ast}$, statements like ``$p<q$'' do not depend on
whether we are regarding $p$ and $q$ as
elements of $\mathfrak{A}^{\ast}$ or as elements of $\left(  \mathfrak{A}%
\cup\left\{  m\right\}  \right)  ^{\ast}$. It is easy to see that the
one-letter word $m$ satisfies%
\begin{equation}
m>p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathfrak{A}^{\ast}.
\label{sol.words.powerlyndon.b.mp}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.powerlyndon.b.mp}):} Let
$p\in\mathfrak{A}^{\ast}$. If $p$ is empty, then
(\ref{sol.words.powerlyndon.b.mp}) is obviously satisfied. Hence, for the rest
of the proof of (\ref{sol.words.powerlyndon.b.mp}), we WLOG assume that $p$ is
nonempty. Then, $p$ has a first letter, and we have $\left(  \text{the first
letter of the word }p\right)  \in\mathfrak{A}$ (since $p\in\mathfrak{A}^{\ast
}$). Thus,%
\begin{align*}
\left(  \text{the first letter of the word }p\right)   &
<m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a<m\text{ for every }%
a\in\mathfrak{A}\right) \\
&  =\left(  \text{the first letter of the word }m\right)  .
\end{align*}
Hence, $p<m$ (by the definition of the lexicographic order), that is, $m>p$.
This proves (\ref{sol.words.powerlyndon.b.mp}).}

We shall prove the implications $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{E}^{\prime}$, $\mathcal{C}^{\prime}\Longrightarrow\mathcal{E}%
^{\prime}$, $\mathcal{G}^{\prime}\Longrightarrow\mathcal{H}^{\prime}$,
$\mathcal{E}^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$,
$\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$,
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{C}^{\prime}$, $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$ and $\mathcal{H}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$.

First of all, the implication $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{E}^{\prime}$ holds for obvious reasons (in fact, if two words $u$ and
$v$ satisfying $w=uv$ satisfy $v\geq w$, then $v\geq u$ (because $v\geq
w=uv\geq u$)). Also, the implication $\mathcal{C}^{\prime}\Longrightarrow
\mathcal{E}^{\prime}$ holds for obvious reasons (in fact, if two words $u$ and
$v$ satisfying $w=uv$ satisfy $\left(  v\text{ is a prefix of }u\right)  $,
then $\left(  v\text{ is a prefix of }w\right)  $ (because $v$ is a prefix of
$u$, and $u$ in turn is a prefix of $uv=w$)). The implication $\mathcal{G}%
^{\prime}\Longrightarrow\mathcal{H}^{\prime}$ is also trivially true.

\textit{Proof of the implication $\mathcal{E}^{\prime}\Longrightarrow
\mathcal{F}^{\prime\prime}$:} Assume that Assertion $\mathcal{E}%
^{\prime}$ holds.

Assume (for the sake of contradiction) that the word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ is not a Lyndon word.
Clearly, this word $wm$ is nonempty. Let $v^{\prime}$ be the
(lexicographically) smallest nonempty suffix of this word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$. Since $wm$ is not a
Lyndon word, we can apply Proposition \ref{prop.words.lyndon.cutoff}(b) to
$\mathfrak{A}\cup\left\{  m\right\}  $, $wm$ and $v^{\prime}$ instead of
$\mathfrak{A}$, $w$ and $v$. As a result, we conclude that there exists a
nonempty $u\in\left(  \mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$
such that $wm=uv^{\prime}$, $u\geq v^{\prime}$ and $uv^{\prime}\geq v^{\prime
}u$. Consider this $u$.

We know that $v^{\prime}$ is a suffix of $wm$. Thus, we must be in one of the
following two cases (depending on whether this suffix begins before the suffix
$m$ of $wm$ begins or afterwards):

\textit{Case 1:} The word $v^{\prime}$ is a nonempty suffix of $m$. (Note that
$v^{\prime}=m$ is allowed.)

\textit{Case 2:} The word $v^{\prime}$ has the form $vm$ where $v$ is a
nonempty suffix of $w$.

Let us consider Case 1 first. In this case, the word $v^{\prime}$ is a
nonempty suffix of $m$. Since the only nonempty suffix of $m$ is $m$ itself
(because $m$ is a one-letter word), this yields $v^{\prime}=m$. Now,
$wm=u\underbrace{v^{\prime}}_{=m}=um$. Cancelling $m$ from this equality, we
obtain $w=u$, so that $u=w\in\mathfrak{A}^{\ast}$. Hence, $m>u$ (by
(\ref{sol.words.powerlyndon.b.mp}), applied to $p=u$). This contradicts $u\geq
v^{\prime}=m$. Thus, we have obtained a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $v^{\prime}$ has the form
$vm$ where $v$ is a nonempty suffix of $w$. Consider this $v$. We have
$wm=u\underbrace{v^{\prime}}_{=vm}=uvm$. By cancelling $m$ from this equality,
we obtain $w=uv$. Thus, $u$ and $v$ are subwords of $w$, and therefore belong
to $\mathfrak{A}^{\ast}$ (since $w\in\mathfrak{A}^{\ast}$). Moreover, $u$ and
$v$ are nonempty. Hence, Assertion $\mathcal{E}^{\prime}$ yields that either
we have $v\geq u$ or the word $v$ is a prefix of $w$. Since we cannot have
$v\geq u$ (because if we had $v\geq u$, then we would have $v^{\prime
}=vm>v\geq u\geq v^{\prime}$, which is absurd), we therefore must have that
$v$ is a prefix of $w$. In other words, there exists a $q\in\mathfrak{A}%
^{\ast}$ such that $w=vq$. Consider this $q$. We have $m>q$ (by
(\ref{sol.words.powerlyndon.b.mp}), applied to $p=q$). Thus, $q\leq m$. Hence,
Proposition \ref{prop.words.lex}(b) (applied to $\mathfrak{A}\cup\left\{
m\right\}  $, $v$, $q$ and $m$ instead of $\mathfrak{A}$, $a$, $c$ and $d$)
yields $vq\leq vm$. Therefore, $vm\geq vq=w$ (since $w=vq$), so that
$v^{\prime}=vm\geq w=uv>u$ (since $v$ is nonempty). This contradicts $u\geq
v^{\prime}$. Thus, we have found a contradiction in Case 2.

We have therefore obtained a contradiction in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this shows that we always get a
contradiction. Hence, our assumption (that the word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ is not a Lyndon word)
was false. Hence, the word $wm\in\left(  \mathfrak{A}\cup\left\{  m\right\}
\right)  ^{\ast}$ is a Lyndon word. That is, Assertion $\mathcal{F}%
^{\prime\prime}$ holds. Hence, we have proven the implication $\mathcal{E}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$.

\textit{Proof of the implication $\mathcal{F}^{\prime\prime}\Longrightarrow
\mathcal{B}^{\prime}$:} Assume that Assertion $\mathcal{F}%
^{\prime\prime}$ holds. Thus, the word $wm\in\left(  \mathfrak{A}\cup\left\{
m\right\}  \right)  ^{\ast}$ is a Lyndon word.

Let $u$ and $v$ be nonempty words satisfying $w=uv$. We shall prove that
either we have $v\geq w$ or the word $v$ is a prefix of $w$.

Indeed, $v$ is a suffix of $w$ (since $w=uv$), so that $vm$ is a suffix of
$wm$. Clearly, $vm$ is nonempty. Thus, Corollary \ref{cor.words.lyndon.trivia}
(applied to $\mathfrak{A}\cup\left\{  m\right\}  $, $wm$ and $vm$ instead of
$\mathfrak{A}$, $w$ and $v$) yields $vm\geq wm$. In other words, $wm\leq vm$.
Hence, Proposition \ref{prop.words.lex}(e) (applied to $\mathfrak{A}%
\cup\left\{  m\right\}  $, $w$, $m$, $v$ and $m$ instead of $\mathfrak{A}$,
$a$, $b$, $c$ and $d$) yields that either we have $w\leq v$ or the word $v$ is
a prefix of $w$. In other words, either we have $v\geq w$ or the word $v$ is a
prefix of $w$.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that if $u$
and $v$ are nonempty words satisfying $w=uv$, then either we have $v\geq w$ or
the word $v$ is a prefix of $w$. In other words, Assertion $\mathcal{B}%
^{\prime}$ holds. Hence, we have proven the implication $\mathcal{F}%
^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$.

\textit{Proof of the implication $\mathcal{F}^{\prime\prime}\Longrightarrow
\mathcal{C}^{\prime}$:} Assume that Assertion $\mathcal{F}%
^{\prime\prime}$ holds. Thus, the word $wm\in\left(  \mathfrak{A}\cup\left\{
m\right\}  \right)  ^{\ast}$ is a Lyndon word.

Let $u$ and $v$ be nonempty words satisfying $w=uv$. We shall prove that
either we have $v\geq u$ or the word $v$ is a prefix of $u$.

Indeed, Proposition \ref{prop.words.lyndon.trivia}(b) (applied to
$\mathfrak{A}\cup\left\{  m\right\}  $, $wm$ and $vm$ instead of
$\mathfrak{A}$, $w$ and $v$) yields $vm>u$ (since $\underbrace{w}_{=uv}m=uvm$,
and since $vm$ is nonempty). Hence, $vm\geq u=u\varnothing$, so that
$u\varnothing\leq vm$. Thus, Proposition \ref{prop.words.lex}(e) (applied to
$\mathfrak{A}\cup\left\{  m\right\}  $, $u$, $\varnothing$, $v$ and $m$
instead of $\mathfrak{A}$, $a$, $b$, $c$ and $d$) yields that either we have
$u\leq v$ or the word $v$ is a prefix of $u$. In other words, either we have
$v\geq u$ or the word $v$ is a prefix of $u$.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that if $u$
and $v$ are nonempty words satisfying $w=uv$, then either we have $v\geq u$ or
the word $v$ is a prefix of $u$. In other words, Assertion $\mathcal{C}%
^{\prime}$ holds. Hence, we have proven the implication $\mathcal{F}%
^{\prime\prime}\Longrightarrow\mathcal{C}^{\prime}$.

\textit{Proof of the implication $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{G}^{\prime}$:} Assume that Assertion $\mathcal{B}^{\prime}$ holds.

Let $s$ be the longest suffix $v$ of $w$ satisfying $v<w$. (This is
well-defined, because there exists a suffix $v$ of $w$ satisfying $v<w$ --
namely, the empty word.) So we know that $s$ is a suffix $v$ of $w$ satisfying
$v<w$. In other words, $s$ is a suffix of $w$ and satisfies $s<w$. As a
consequence, $s$ is a proper suffix of $w$ (because otherwise, $s$ would be
$w$, and this would contradict $s<w$). Hence, there exists a nonempty word
$h\in\mathfrak{A}^{\ast}$ satisfying $w=hs$. Consider this $h$. Using
Assertion $\mathcal{B}^{\prime}$, it is easy to see that $s$ is a prefix of
$w$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $s$ is not a
prefix of $w$. Hence, $s$ is nonempty. Therefore, Assertion $\mathcal{B}%
^{\prime}$ (applied to $u=h$ and $v=s$) yields that either we have $s\geq w$
or the word $s$ is a prefix of $w$. Since $s$ is not a prefix of $w$, we must
thus have $s\geq w$. But this contradicts $s<w$. This contradiction shows that
our assumption was wrong, qed.}. In other words, there exists a $g\in
\mathfrak{A}^{\ast}$ such that $w=sg$. Consider this $g$.

We know that $s$ is the longest suffix $v$ of $w$ satisfying $v<w$. Hence,%
\begin{equation}
\left(  \text{if }v\text{ is a suffix of }w\text{ satisfying }v<w\text{, then
}\ell\left(  v\right)  \leq\ell\left(  s\right)  \right)  .
\label{sol.words.powerlyndon.BG.1}%
\end{equation}


There exists a nonnegative integer $m$ such that $h^{m}$ is a prefix of $s$
(for example, the nonnegative integer $m=0$). Consider the \textbf{maximal}
such integer $m$\ \ \ \ \footnote{This is well-defined, because of the
following reason:
\par
We have $\ell\left(  h\right)  \geq1$ (since the word $h$ is nonempty). Thus,
for every $m\in \NN$ satisfying $m>\ell\left(  s\right)  $, we have
$\ell\left(  h^{m}\right)  =m\underbrace{\ell\left(  h\right)  }_{\geq1}\geq
m>\ell\left(  s\right)  $. In other words, for every $m\in \NN$
satisfying $m>\ell\left(  s\right)  $, the word $h^{m}$ is longer than $s$.
Hence, for every $m\in \NN$ satisfying $m>\ell\left(  s\right)  $, the
word $h^{m}$ cannot be a prefix of $s$. Thus, for every sufficiently high
$m\in \NN$, the word $h^{m}$ cannot be a prefix of $s$.
\par
We thus know the following:
\par
\begin{itemize}
\item There exists a nonnegative integer $m$ such that $h^{m}$ is a prefix of
$s$.
\par
\item For every sufficiently high $m\in \NN$, the word $h^{m}$ cannot be
a prefix of $s$.
\end{itemize}
\par
Consequently, there exists a \textbf{maximal} nonnegative integer $m$ such
that $h^{m}$ is a prefix of $s$, qed.}. Then, $h^{m}$ is a prefix of $s$, but
$h^{m+1}$ is not a prefix of $s$. Since $h^{m}$ is a prefix of $s$, there
exists a word $q\in\mathfrak{A}^{\ast}$ such that $s=h^{m}q$. Consider this
$q$. Clearly, $w=h\underbrace{s}_{=h^{m}q}=\underbrace{hh^{m}}_{=h^{m+1}%
}q=\underbrace{h^{m+1}}_{=h^{m}h}q=h^{m}hq$. Hence, $h^{m}hq=w=\underbrace{s}%
_{=h^{m}q}g=h^{m}qg$. Cancelling $h^{m}$ from this equality, we obtain
$hq=qg$. It is now easy to see that $h>q$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Then, $h\leq q$. Hence, $h\leq q\leq qg=hq$ (since
$hq=qg$). Therefore, Proposition \ref{prop.words.lex}(g) (applied to $h$, $q$
and $q$ instead of $a$, $b$ and $c$) yields that $h$ is a prefix of $q$. In
other words, there exists a word $r\in\mathfrak{A}^{\ast}$ such that $q=hr$.
Consider this $r$. Now, $s=h^{m}\underbrace{q}_{=hr}=\underbrace{h^{m}%
h}_{=h^{m+1}}r=h^{m+1}r$, so that $h^{m+1}$ is a prefix of $s$. This
contradicts the fact that $h^{m+1}$ is not a prefix of $s$. This contradiction
proves that our assumption was wrong, qed.}. Hence, $q\leq h\leq hq=qg$. Thus,
Proposition \ref{prop.words.lex}(g) (applied to $q$, $h$ and $g$ instead of
$a$, $b$ and $c$) yields that $q$ is a prefix of $h$.

Next, we shall prove that the word $h$ is Lyndon.

In fact, assume the contrary. Then, $h$ is not Lyndon. Let $v$ be the
(lexicographically) smallest nonempty suffix of $h$. Then, Proposition
\ref{prop.words.lyndon.cutoff}(b) (applied to $h$ instead of $w$) yields that
there exists a nonempty $u\in\mathfrak{A}^{\ast}$ such that $h=uv$, $u\geq v$
and $uv\geq vu$. Consider this $u$. Since $w=\underbrace{h}_{=uv}s=uvs$, it is
clear that the word $vs$ is a suffix of $w$. If we had $vs<w$, then we could
therefore obtain $\ell\left(  vs\right)  \leq\ell\left(  s\right)  $ (by
(\ref{sol.words.powerlyndon.BG.1}), applied to $vs$ instead of $v$), which
would contradict $\ell\left(  vs\right)  =\underbrace{\ell\left(  v\right)
}_{\substack{>0\\\text{(since }v\text{ is nonempty)}}}+\ell\left(  s\right)
>\ell\left(  s\right)  $. Thus, we cannot have $vs<w$. We thus have $vs\geq
w=hs\geq h=uv$, so that $uv\leq vs$.

Recall that $s<w$. Hence, $vs\leq vw$ (by Proposition \ref{prop.words.lex}(b),
applied to $v$, $s$ and $w$ instead of $a$, $c$ and $d$). Now, $uv\geq vu$, so
that $vu\leq uv\leq vs\leq v\underbrace{w}_{=hs}=v\underbrace{h}_{=uv}s=vuvs$.
Hence, Proposition \ref{prop.words.lex}(g) (applied to $vu$, $uv$ and $vs$
instead of $a$, $b$ and $c$) yields that $vu$ is a prefix of $uv$. Since $vu$
has the same length as $uv$ (because $\ell\left(  vu\right)  =\ell\left(
v\right)  +\ell\left(  u\right)  =\ell\left(  u\right)  +\ell\left(  v\right)
=\ell\left(  uv\right)  $), this yields that $vu=uv$. Thus, the elements $u$
and $v$ of the monoid $\mathfrak{A}^{\ast}$ commute. Thus, the submonoid of
$\mathfrak{A}^{\ast}$ generated by $u$ and $v$ is commutative. Since $h=uv$,
the element $h$ lies in this submonoid, and therefore the element $h^{m}$ lies
in it as well. Thus, $h^{m}$ commutes with $v$ (since this submonoid is
commutative), i.e., we have $vh^{m}=h^{m}v$. Thus, $v\underbrace{s}_{=h^{m}%
q}=\underbrace{vh^{m}}_{=h^{m}v}q=h^{m}vq$. Thus, $h^{m}vq=vs\geq
w=h\underbrace{s}_{=h^{m}q}=\underbrace{hh^{m}}_{=h^{m+1}=h^{m}h}q=h^{m}hq$,
so that $h^{m}hq\leq h^{m}vq$. Hence, Proposition \ref{prop.words.lex}(c)
(applied to $h^{m}$, $hq$ and $vq$ instead of $a$, $c$ and $d$) yields $hq\leq
vq$. But since $q$ is a prefix of $h$, there exists a word $z\in
\mathfrak{A}^{\ast}$ such that $h=qz$. Consider this $z$. We have%
\begin{align*}
v\underbrace{qz}_{=h=uv}  &  =\underbrace{vu}_{=uv=h}v=hv\leq h\underbrace{vu}%
_{=uv=h=qz}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }hv\text{ is a prefix of
}hvu\right) \\
&  =hqz.
\end{align*}


Also, $\ell\left(  \underbrace{h}_{=uv}q\right)  =\ell\left(  uvq\right)
=\ell\left(  u\left(  vq\right)  \right)  =\underbrace{\ell\left(  u\right)
}_{\substack{>0\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  vq\right)
>\ell\left(  vq\right)  $, so that $\ell\left(  vq\right)  \leq\ell\left(
hq\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied to $vq$,
$z$, $hq$ and $z$ instead of $a$, $b$, $c$ and $d$) yields $vq\leq hq$.
Combined with $hq\leq vq$, this yields $vq=hq$. Hence, $\ell\left(
\underbrace{vq}_{=hq}\right)  =\ell\left(  hq\right)  >\ell\left(  vq\right)
$, which is absurd. This contradiction proves that our assumption is wrong.
Thus, we have shown that the word $h$ is Lyndon.

We now know that $h\in\mathfrak{A}^{\ast}$ is a Lyndon word, $m+1$ is a
positive integer, and $q$ is a prefix of $h$, and we have $w=h^{m+1}q$. Hence,
there exists a Lyndon word $t\in\mathfrak{A}^{\ast}$, a positive integer
$\ell$ and a prefix $p$ of $t$ (possibly empty) such that $w=t^{\ell}p$
(namely, $t=h$, $\ell=m+1$ and $p=q$). In other words, Assertion
$\mathcal{G}^{\prime}$ holds. This proves the implication $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$.

We are now going to prove the implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$; this implication will later be used in the proof of the
implication $\mathcal{H}^{\prime}\Longrightarrow\mathcal{B}^{\prime}$.

\textit{Proof of the implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$:} Assume that Assertion $\mathcal{F}^{\prime}$
holds. In other words, the word $w$ is a prefix of a Lyndon word in
$\mathfrak{A}^{\ast}$. Let $z$ be this Lyndon word. Thus, $w$ is a prefix of
$z$. In other words, there exists a word $q\in\mathfrak{A}^{\ast}$ such that
$z=wq$. Consider this $q$.

Let $u$ and $v$ be nonempty words satisfying $w=uv$. We are going to prove
that either we have $v\geq w$ or the word $v$ is a prefix of $w$.

We have $z=\underbrace{w}_{=uv}q=uvq$. Thus, Proposition
\ref{prop.words.lyndon.trivia}(a) (applied to $z$ and $vq$ instead of $w$ and
$v$) yields $vq\geq z$ (since $vq$ is nonempty (since $v$ is nonempty)).
Hence, $vq\geq z=wq$. In other words, $wq\leq vq$. Thus, Proposition
\ref{prop.words.lex}(e) (applied to $w$, $q$, $v$ and $q$ instead of $a$, $b$,
$c$ and $d$) yields that either we have $w\leq v$ or the word $v$ is a prefix
of $w$. In other words, either we have $v\geq w$ or the word $v$ is a prefix
of $w$.

Now, forget that we fixed $u$ and $v$. We thus have shown that if $u$ and $v$
are nonempty words satisfying $w=uv$, then either we have $v\geq w$ or the
word $v$ is a prefix of $w$. In other words, Assertion $\mathcal{B}^{\prime}$
holds. Thus, the implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ is proven.

\textit{Proof of the implication $\mathcal{H}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$:} Assume that Assertion $\mathcal{H}^{\prime}$
holds. In other words, there exists a Lyndon word $t\in\mathfrak{A}^{\ast}$, a
nonnegative integer $\ell$ and a prefix $p$ of $t$ (possibly empty) such that
$w=t^{\ell}p$. Consider this $t$, this $\ell$ and this $p$.

We are going to prove that for every $m\in \NN$,%
\begin{equation}
\left(  \text{every suffix }s\text{ of }t^{m}p\text{ satisfies either }s\geq
t^{m}p\text{ or }\left(  \text{the word }s\text{ is a prefix of }%
t^{m}p\right)  \right)  . \label{sol.words.powerlyndon.HB.1}%
\end{equation}


\textit{Proof of (\ref{sol.words.powerlyndon.HB.1}):} We will prove
(\ref{sol.words.powerlyndon.HB.1}) by induction over $m$:

\textit{Induction base:} Using the implication $\mathcal{F}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$, it is easy to see that
(\ref{sol.words.powerlyndon.HB.1}) holds for $m=0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $m=0$. Then, $t^{m}%
p=\underbrace{t^{0}}_{=\varnothing}p=\varnothing p=p$.
\par
Let $s$ be a suffix of $t^{m}p$. Then, $s$ is a suffix of $t^{m}p=p$. In other
words, there exists a word $g\in\mathfrak{A}^{\ast}$ satisfying $p=gs$.
Consider this $g$.
\par
We are going to prove that either $s\geq t^{m}p$ or $\left(  \text{the word
}s\text{ is a prefix of }t^{m}p\right)  $. If the word $s$ is empty, then this
is obvious (because if the word $s$ is empty, then the word $s$ is a prefix of
$t^{m}p$). Hence, we WLOG assume that the word $s$ is nonempty. If the word
$g$ is empty, then it is also clear that either $s\geq t^{m}p$ or $\left(
\text{the word }s\text{ is a prefix of }t^{m}p\right)  $ (because if the word
$g$ is empty, then $g=\varnothing$ and thus $t^{m}p=p=\underbrace{g}%
_{=\varnothing}s=s$, so that the word $s$ is a prefix of $t^{m}p$). Hence, we
WLOG assume that the word $g$ is nonempty.
\par
But the word $p$ is a prefix of a Lyndon word in $\mathfrak{A}^{\ast}$ (since
$p$ is a prefix of $t$, and since $t$ is a Lyndon word in $\mathfrak{A}^{\ast
}$). In other words, Assertion $\mathcal{F}^{\prime}$ with $w$ replaced by $p$
is satisfied. Hence, Assertion $\mathcal{B}^{\prime}$ with $w$ replaced by $p$
is satisfied as well (since we have already proven the implication
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}^{\prime}$). In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }u\text{ and }v\text{ are nonempty words satisfying }p=uv\text{,
then}\\
\text{either we have }v\geq p\text{ or the word }v\text{ is a prefix of }p
\end{array}
\right)  . \label{sol.words.powerlyndon.HB.base}%
\end{equation}
Since the words $g$ and $s$ are nonempty, we can apply
(\ref{sol.words.powerlyndon.HB.base}) to $u=g$ and $v=s$. As a result, we
obtain that either we have $s\geq p$ or the word $s$ is a prefix of $p$. In
other words, either $s\geq p$ or $\left(  \text{the word }s\text{ is a prefix
of }p\right)  $. In other words, either $s\geq t^{m}p$ or $\left(  \text{the
word }s\text{ is a prefix of }t^{m}p\right)  $ (since $t^{m}p=p$). This proves
(\ref{sol.words.powerlyndon.HB.1}).}. This completes the induction base.

\textit{Induction step:} Let $M$ be a positive integer. Assume that
(\ref{sol.words.powerlyndon.HB.1}) is proven for $m=M-1$. We will now show
that (\ref{sol.words.powerlyndon.HB.1}) holds for $m=M$.

Let $r$ denote the word $t^{M-1}p$. It is easy to see that $r$ is a prefix of
$t^{M}p$\ \ \ \ \footnote{\textit{Proof.} There exists a word $q\in
\mathfrak{A}^{\ast}$ such that $t=pq$ (since $p$ is a prefix of $t$). Consider
this $q$. We have $\underbrace{t^{M}}_{=t^{M-1}t}p=t^{M-1}\underbrace{t}%
_{=pq}p=\underbrace{t^{M-1}p}_{=r}qp=rqp=r\left(  qp\right)  $. Hence, $r$ is
a prefix of $t^{M}p$, qed.}. In other words, there exists a word
$g\in\mathfrak{A}^{\ast}$ such that $t^{M}p=rg$. Consider this $g$.

Let $s$ be a suffix of $t^{M}p$. We shall show that either $s\geq t^{M}p$ or
$\left(  \text{the word }s\text{ is a prefix of }t^{M}p\right)  $.

In order to prove this, let us assume the contrary (for the sake of
contradiction). Then, neither $s\geq t^{M}p$ nor $\left(  \text{the word
}s\text{ is a prefix of }t^{M}p\right)  $. In other words, we have $s<t^{M}p$,
and the word $s$ is not a prefix of $t^{M}p$. If the word $s$ was a prefix of
$r$, then the word $s$ would be a prefix of $t^{M}p$ (since $r$ is a prefix of
$t^{M}p$), which would contradict the fact that the word $s$ is not a prefix
of $t^{M}p$. Hence, the word $s$ cannot be a prefix of $r$. In other words,
the word $s$ cannot be a prefix of $t^{M-1}p$ (since $r=t^{M-1}p$).

The word $s$ is a suffix of $\underbrace{t^{M}}_{=tt^{M-1}}%
p=t\underbrace{t^{M-1}p}_{=r}=tr$. Therefore, we must be in one of the
following two cases (depending on whether this suffix begins before the suffix
$r$ of $tr$ begins or afterwards):

\textit{Case 1:} The word $s$ is a suffix of $r$. (Note that $s=r$ is allowed.)

\textit{Case 2:} The word $s$ has the form $s^{\prime}r$ where $s^{\prime}$ is
a nonempty suffix of $t$.

Let us consider Case 1 first. In this case, the word $s$ is a suffix of $r$.
In other words, the word $s$ is a suffix of $t^{M-1}p$ (since $r=t^{M-1}p$).
Hence, (\ref{sol.words.powerlyndon.HB.1}) (applied to $m=M-1$) yields that
either $s\geq t^{M-1}p$ or $\left(  \text{the word }s\text{ is a prefix of
}t^{M-1}p\right)  $ (since (\ref{sol.words.powerlyndon.HB.1}) holds for
$m = M-1$). Since the word $s$ cannot be a prefix of $t^{M-1}p$, we
thus must have $s\geq t^{M-1}p$. Thus, $t^{M-1}p\leq s<t^{M}p=rg$. Hence,
$r=t^{M-1}p\leq s<rg$. Therefore, Proposition \ref{prop.words.lex}(g) (applied
to $r$, $s$ and $g$ instead of $a$, $b$ and $c$) yields that $r$ is a prefix
of $s$. Since $\ell\left(  r\right)  \geq\ell\left(  s\right)  $ (because $s$
is a suffix of $r$), this can only hold if $r=s$. We thus have $r=s$. Thus,
$s=r$, so that $s$ is a prefix of $s=r$. This contradicts the fact that the
word $s$ cannot be a prefix of $r$. Thus, we have found a contradiction in
Case 1.

Let us now consider Case 2. In this case, the word $s$ has the form
$s^{\prime}r$ where $s^{\prime}$ is a nonempty suffix of $t$. Consider this
$s^{\prime}$. Corollary \ref{cor.words.lyndon.trivia} (applied to $t$ and
$s^{\prime}$ instead of $w$ and $v$) yields $s^{\prime}\geq t$. But
$s^{\prime}$ is a suffix of $t$, so that $\ell\left(  s^\prime\right)  \leq
\ell\left(  t\right)  $. Also, $s=s^{\prime}r$, so that $s^{\prime}%
r=s<t^{M}p=tr$. Hence, Proposition \ref{prop.words.lex}(f) (applied to
$s^{\prime}$, $r$, $t$ and $r$ instead of $a$, $b$, $c$ and $d$) yields that
$s^{\prime}\leq t$. Combined with $s^{\prime}\geq t$, this yields $s^{\prime
}=t$. Hence, $s=\underbrace{s^{\prime}}_{=t}r=t^{M}p$ (since $t^{M}p=tr$),
which contradicts the fact that the word $s$ is not a prefix of $t^{M}p$.
Thus, we have found a contradiction in Case 2.

We have thus obtained a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that we always get a
contradiction. This completes the proof that either $s\geq t^{M}p$ or $\left(
\text{the word }s\text{ is a prefix of }t^{M}p\right)  $.

Now, forget that we fixed $s$. We thus have shown that every suffix $s$ of
$t^{M}p$ satisfies either $s\geq t^{M}p$ or $\left(  \text{the word }s\text{
is a prefix of }t^{M}p\right)  $. In other words,
(\ref{sol.words.powerlyndon.HB.1}) holds for $m=M$. This completes the
induction step, and thus (\ref{sol.words.powerlyndon.HB.1}) is proven by induction.

Now, let $u$ and $v$ be nonempty words satisfying $w=uv$. Then, $v$ is a
suffix of $w=t^{\ell}p$. Hence, (\ref{sol.words.powerlyndon.HB.1}) (applied to
$m=\ell$ and $s=v$) yields that either $v\geq t^{\ell}p$ or $\left(  \text{the
word }v\text{ is a prefix of }t^{\ell}p\right)  $. In other words, either
$v\geq w$ or $\left(  \text{the word }v\text{ is a prefix of }w\right)  $
(since $w=t^{\ell}p$). In other words, either we have $v\geq w$ or the word
$v$ is a prefix of $w$.

Now, forget that we fixed $u$ and $v$. We thus have shown that if $u$ and $v$
are nonempty words satisfying $w=uv$, then either we have $v\geq w$ or the
word $v$ is a prefix of $w$. In other words, Assertion $\mathcal{B}^{\prime}$
holds. Thus, the implication $\mathcal{H}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ is proven.

We have thus proven the implications $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{E}^{\prime}$, $\mathcal{C}^{\prime}\Longrightarrow\mathcal{E}%
^{\prime}$, $\mathcal{G}^{\prime}\Longrightarrow\mathcal{H}^{\prime}$,
$\mathcal{E}^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$,
$\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$,
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{C}^{\prime}$, $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$ and $\mathcal{H}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$. Combined, these yield the equivalence
$\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime}%
\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime}\Longleftrightarrow
\mathcal{H}^{\prime}$. This solves Exercise \ref{exe.words.powerlyndon}(b).

(c) The implication $\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}^{\prime}$
has already been proven in our solution of Exercise
\ref{exe.words.powerlyndon}(b). Hence, Exercise \ref{exe.words.powerlyndon}(c)
is solved.

(d) Assume that Assertion $\mathcal{D}^{\prime}$ holds. Then, Assertion
$\mathcal{A}^{\prime}$ holds as well (because of the equivalence
$\mathcal{A}^{\prime}\Longleftrightarrow\mathcal{D}^{\prime}$). In other
words, the word $w$ is a power of a Lyndon word. In other words, there exist a
Lyndon word $z\in\mathfrak{A}^{\ast}$ and a nonnegative integer $m$ such that
$w=z^{m}$. Consider these $z$ and $m$. There exists a Lyndon word
$t\in\mathfrak{A}^{\ast}$, a nonnegative integer $\ell$ and a prefix $p$ of
$t$ (possibly empty) such that $w=t^{\ell}p$ (namely, $t=z$, $\ell=m$ and
$p=\varnothing$). In other words, Assertion $\mathcal{H}^{\prime}$ holds.
Since Assertion $\mathcal{H}^{\prime}$ is equivalent to Assertion
$\mathcal{B}^{\prime}$ (because of the equivalence $\mathcal{B}^{\prime
}\Longleftrightarrow\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}%
^{\prime}\Longleftrightarrow\mathcal{F}^{\prime\prime}\Longleftrightarrow
\mathcal{G}^{\prime}\Longleftrightarrow\mathcal{H}^{\prime}$), this yields
that Assertion $\mathcal{B}^{\prime}$ holds. Thus, the implication
$\mathcal{D}^{\prime}\Longrightarrow\mathcal{B}^{\prime}$ is proven.

(e) Assume that there exists a letter $\mu\in\mathfrak{A}$ such that $\left(
\mu>a\text{ for every letter }a\text{ of }w\right)  $. Consider this $\mu$. We
need to prove that the equivalence $\mathcal{F}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}$ holds.

Combining the implication $\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}%
^{\prime}$ (which has already been proven) and the implication $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$ (which follows from the
equivalence $\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime
}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime}\Longleftrightarrow
\mathcal{H}^{\prime}$ proven above), we obtain the implication $\mathcal{F}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$. Thus, in order to prove
the equivalence $\mathcal{F}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}$, it is enough to verify the implication $\mathcal{F}%
^{\prime\prime}\Longrightarrow\mathcal{F}^{\prime}$. Let us do this now.

Assume that Assertion $\mathcal{F}^{\prime\prime}$ holds. Let $\mathfrak{B}$
denote the alphabet consisting of all letters that appear in $w$. Clearly,
$\mathfrak{B}$ is a subalphabet of $\mathfrak{A}$, and we have $w\in
\mathfrak{B}^{\ast}$. Moreover, we have $\mu>a$ for every letter $a$ of $w$.
Therefore, $\mu>a$ for every $a\in\mathfrak{B}$ (because the elements of
$\mathfrak{B}$ are precisely the letters of $w$). In other words, $a<\mu$ for
every $a\in\mathfrak{B}$. As a consequence, $\mu\notin\mathfrak{B}$; that is,
$\mu$ is an object not in the alphabet $\mathfrak{B}$.

We know that Assertion $\mathcal{F}^{\prime\prime}$ holds. Due to the
implication $\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$
(which follows from the equivalence $\mathcal{B}^{\prime}\Longleftrightarrow
\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime
}\Longleftrightarrow\mathcal{H}^{\prime}$ proven above), this yields that
Assertion $\mathcal{B}^{\prime}$ holds. Thus, Assertion $\mathcal{B}^{\prime}$
with $\mathfrak{A}$ replaced by $\mathfrak{B}$ holds as well (since the
Assertion $\mathcal{B}^{\prime}$ does not change if we extend our alphabet).
Due to the implication $\mathcal{B}^{\prime}\Longrightarrow\mathcal{F}%
^{\prime\prime}$ (which follows from the equivalence $\mathcal{B}^{\prime
}\Longleftrightarrow\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}%
^{\prime}\Longleftrightarrow\mathcal{F}^{\prime\prime}\Longleftrightarrow
\mathcal{G}^{\prime}\Longleftrightarrow\mathcal{H}^{\prime}$ proven above),
this yields that Assertion $\mathcal{F}^{\prime\prime}$ with $\mathfrak{A}$
replaced by $\mathfrak{B}$ holds as well. Thus, we can apply Assertion
$\mathcal{F}^{\prime\prime}$ with $\mathfrak{A}$ replaced by $\mathfrak{B}$ to
$m=\mu$ (because $\mu$ is an object not in the alphabet $\mathfrak{B}$, and
the total order on $\mathfrak{A}$ satisfies $\left(  a<\mu\text{ for every
}a\in\mathfrak{B}\right)  $). As a result, we conclude that the word $w\mu
\in\left(  \mathfrak{B}\cup\left\{  \mu\right\}  \right)  ^{\ast}$ is a Lyndon
word. Since $\mathfrak{B}\cup\left\{  \mu\right\}  \subset\mathfrak{A}$
(because $\mathfrak{B}\subset\mathfrak{A}$ and $\mu\in\mathfrak{A}$), we have
$w\mu\in\left(  \underbrace{\mathfrak{B}\cup\left\{  \mu\right\}  }%
_{\subset\mathfrak{A}}\right)  ^{\ast}\subset\mathfrak{A}^{\ast}$, and thus
$w\mu\in\mathfrak{A}^{\ast}$ is a Lyndon word. Of course, the word $w$ is a
prefix of $w\mu$. As a consequence, the word $w$ is a prefix of a Lyndon word
in $\mathfrak{A}^{\ast}$ (namely, of the word $w\mu$). In other words,
Assertion $\mathcal{F}^{\prime}$ holds. This proves the implication
$\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{F}^{\prime}$. Thus, the
solution of Exercise \ref{exe.words.powerlyndon}(e) is complete.

(f) Assume that there exists a letter $\mu\in\mathfrak{A}$ such that $\left(
\mu>a\text{ for some letter }a\text{ of }w\right)  $. Consider this $\mu$. We
need to prove that the equivalence $\mathcal{F}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}$ holds.

Just as in our solution of Exercise \ref{exe.words.powerlyndon}(e) above, we
can see that it is enough to verify the implication $\mathcal{F}^{\prime
\prime}\Longrightarrow\mathcal{F}^{\prime}$. Let us do this now.

Assume that Assertion $\mathcal{F}^{\prime\prime}$ holds. Due to the
implication $\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$
(which follows from the equivalence $\mathcal{B}^{\prime}\Longleftrightarrow
\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime
}\Longleftrightarrow\mathcal{H}^{\prime}$ proven above), this yields that
Assertion $\mathcal{B}^{\prime}$ holds.

Let $\beta$ be the highest letter of the word $w\mu\in\mathfrak{A}^{\ast}$
(the concatenation of the word $w$ with the one-letter word $\mu$).
The word $w\beta^{\ell\left(  w\right)  }$ is
clearly nonempty (since $w$ is nonempty). We shall now prove that the word
$w\beta^{\ell\left(  w\right)  }$ is Lyndon.

Indeed, assume the contrary. Thus, $w\beta^{\ell\left(  w\right)  }$ is not
Lyndon. Let $v$ denote the (lexicographically) smallest nonempty suffix of
$w\beta^{\ell\left(  w\right)  }$. Then, Proposition
\ref{prop.words.lyndon.cutoff}(b) (applied to $w\beta^{\ell\left(  w\right)
}$ instead of $w$) yields that there exists a nonempty $u\in\mathfrak{A}%
^{\ast}$ such that $w\beta^{\ell\left(  w\right)  }=uv$, $u\geq v$ and $uv\geq
vu$.

Let $f$ be the first letter of the word $w$ (this is well-defined since $w$ is
nonempty). Then, we can write $w$ in the form $w=fs$ for some word
$s\in\mathfrak{A}^{\ast}$. Consider this $s$.

We know that $\mu>a$ for some letter $a$ of $w$. Consider this $a$. We also
know that $\beta$ is the highest letter of the word $w\mu$. Thus, $\beta$ is
$\geq$ to every letter of the word $w\mu$. In particular, this yields that
$\beta\geq\mu$ (since $\mu$ is a letter of the word $w\mu$), so that
$\beta\geq\mu>a$. But it is fairly easy to see (using the fact that Assertion
$\mathcal{B}^{\prime}$ holds) that $a\geq f$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, $a<f$.
\par
Since $a$ is a letter of $w$, the word $w$ must have a suffix which begins
with the letter $a$. Let $p$ be this suffix. Then, there exists a word
$q\in\mathfrak{A}^{\ast}$ such that $w=qp$ (since $p$ is a suffix of $w$).
Consider this $q$. The word $p$ is nonempty (since it begins with $a$). Since
the word $p$ begins with the letter $a$, we have%
\begin{equation}
\left(  \text{the first letter of }p\right)  =a<f=\left(  \text{the first
letter of }w\right)  . \label{sol.words.powerlyndon.f.fn1}%
\end{equation}
By the definition of the lexicographic order, this shows that $p<w$. Hence,
$p\neq w$. Now, the word $q$ is nonempty (since otherwise, we would have
$q=\varnothing$ and thus $w=\underbrace{q}_{=\varnothing}p=p$, contradicting
$p\neq w$). Hence, applying Assertion $\mathcal{B}^{\prime}$ to $q$ and $p$
instead of $u$ and $v$, we conclude that either we have $p\geq w$ or the word
$p$ is a prefix of $w$. Since $p\geq w$ is impossible (because $p<w$), this
yields that the word $p$ is a prefix of $w$. Since $p$ is nonempty, this shows
that $p$ is a nonempty prefix of $w$. But this yields that%
\[
\left(  \text{the first letter of }p\right)  =\left(  \text{the first letter
of }w\right)  ;
\]
this contradicts (\ref{sol.words.powerlyndon.f.fn1}). This contradiction
proves that our assumption was wrong, qed.}. Hence, $\beta>a\geq f$.

The word $v$ is a proper suffix of $w\beta^{\ell\left(  w\right)  }$ (since
$w\beta^{\ell\left(  w\right)  }=uv$ and since $u$ is nonempty) and is
nonempty. Hence, $v$ is a nonempty proper suffix of $w\beta^{\ell\left(
w\right)  }$. Therefore, we must be in one of the following two cases
(depending on whether this suffix begins before the suffix $\beta^{\ell\left(
w\right)  }$ of $w\beta^{\ell\left(  w\right)  }$ begins or afterwards):

\textit{Case 1:} The word $v$ is a nonempty suffix of $\beta^{\ell\left(
w\right)  }$. (Note that $v=\beta^{\ell\left(  w\right)  }$ is allowed.)

\textit{Case 2:} The word $v$ has the form $q\beta^{\ell\left(  w\right)  }$
where $q$ is a nonempty proper suffix of $w$.

Let us first consider Case 1. In this case, the word $v$ is a nonempty suffix
of $\beta^{\ell\left(  w\right)  }$. Thus, $v$ has the form $\beta^{i}$ for
some $i\in\left\{  0,1,\ldots,\ell\left(  w\right)  \right\}  $. Consider this
$i$. We have $v=\beta^{i}$. Thus, $i\neq0$ (since $v$ is nonempty). Hence,
$\left(  \text{the first letter of }v\right)  =\beta$. But since the word $w$
is nonempty, we have%
\[
\left(  \text{the first letter of }w\beta^{\ell\left(  w\right)  }\right)
=\left(  \text{the first letter of }w\right)  =f
\]
(since $f$ was defined to be the first letter of $w$). Now,%
\[
\left(  \text{the first letter of }v\right)  =\beta>f=\left(  \text{the first
letter of }w\beta^{\ell\left(  w\right)  }\right)  .
\]
By the definition of lexicographic order, this shows that $v>w\beta
^{\ell\left(  w\right)  }$. But this contradicts $w\beta^{\ell\left(
w\right)  }=uv\geq vu\geq v$. Hence, we have found a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $v$ has the form
$q\beta^{\ell\left(  w\right)  }$ where $q$ is a nonempty proper suffix of
$w$. Consider this $q$. Notice that $\ell\left(  q\right)  <\ell\left(
w\right)  $ (since $q$ is a proper suffix of $w$), so that $\ell\left(
w\right)  >\ell\left(  q\right)  $.

Since $q$ is a proper suffix of $w$, there exists a nonempty word
$u\in\mathfrak{A}^{\ast}$ such that $w=uq$. Consider this $u$. Recall that
Assertion $\mathcal{B}^{\prime}$ holds. Applying this Assertion $\mathcal{B}%
^{\prime}$ to $q$ instead of $v$, we conclude that either we have $q\geq w$ or
the word $q$ is a prefix of $w$. If the word $q$ is not a prefix of $w$, then
it is very easy to derive a contradiction\footnote{\textit{Proof.} Assume that
the word $q$ is not a prefix of $w$. Then, we have $q\geq w$ (since we know
that either we have $q\geq w$ or the word $q$ is a prefix of $w$). In other
words, $w\leq q$. Hence, Proposition \ref{prop.words.lex}(d) (applied to $w$,
$\beta^{\ell\left(  w\right)  }$, $q$ and $\beta^{\ell\left(  w\right)  }$
instead of $a$, $b$, $c$ and $d$) yields that either we have $w\beta
^{\ell\left(  w\right)  }\leq q\beta^{\ell\left(  w\right)  }$ or the word $w$
is a prefix of $q$. Since the word $w$ is not a prefix of $q$ (because
$\ell\left(  w\right)  >\ell\left(  q\right)  $), this yields that we have
$w\beta^{\ell\left(  w\right)  }\leq q\beta^{\ell\left(  w\right)  }$. Since
$\ell\left(  w\beta^{\ell\left(  w\right)  }\right)  =\underbrace{\ell\left(
w\right)  }_{>\ell\left(  q\right)  }+\ell\left(  \beta^{\ell\left(  w\right)
}\right)  >\ell\left(  q\right)  +\ell\left(  \beta^{\ell\left(  w\right)
}\right)  =\ell\left(  q\beta^{\ell\left(  w\right)  }\right)  $, this shows
that $w\beta^{\ell\left(  w\right)  }<q\beta^{\ell\left(  w\right)  }=v$
(because $v=q\beta^{\ell\left(  w\right)  }$). Hence, $v>w\beta^{\ell\left(
w\right)  }=uv\geq vu\geq v$, which is a contradiction, qed.}. Hence, the word
$q$ must be a prefix of $w$. In other words, there exists a word
$g\in\mathfrak{A}^{\ast}$ such that $w=qg$. Consider this $g$. Notice that
$\ell\left(  \underbrace{w}_{=qg}\right)
= \ell \left( q g \right)
=\underbrace{\ell\left(  q\right)
}_{\substack{>0\\\text{(since }q\text{ is nonempty)}}}+\ell\left(  g\right)
>\ell\left(  g\right)  $, so that $\ell\left(  \beta^{\ell\left(  w\right)
}\right)  =\ell\left(  w\right)  >\ell\left(  g\right)  $. In other words, the
word $\beta^{\ell\left(  w\right)  }$ is longer than $g$. It is now easy to
see that $\beta^{\ell\left(  w\right)  }\geq g$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $\beta
^{\ell\left(  w\right)  }<g$. By the definition of the lexicographic order,
this yields that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  \beta^{\ell\left(  w\right)  }\right)  ,\ell\left(
g\right)  \right\}  \right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  \beta^{\ell\left(
w\right)  }\right)  _{i}<g_{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  \beta^{\ell\left(  w\right)  }\right)
_{j}=g_{j}\right)  ,\\
&  \text{\textbf{or} the word }\beta^{\ell\left(  w\right)  }\text{ is a
prefix of }g.
\end{align*}
Since the word $\beta^{\ell\left(  w\right)  }$ cannot be a prefix of $g$
(because the word $\beta^{\ell\left(  w\right)  }$ is longer than $g$), this
yields that there exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
\beta^{\ell\left(  w\right)  }\right)  ,\ell\left(  g\right)  \right\}
\right\}  $ such that $\left(  \left(  \beta^{\ell\left(  w\right)  }\right)
_{i}<g_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }\left(  \beta^{\ell\left(  w\right)  }\right)  _{j}=g_{j}\right)
$. Consider this $i$. We have $\left(  \beta^{\ell\left(  w\right)  }\right)
_{i}=\beta$, so that $\beta=\left(  \beta^{\ell\left(  w\right)  }\right)
_{i}<g_{i}$.
\par
But $g$ is a suffix of $w$ (since $w=qg$), and thus every letter of $g$ is a
letter of $w$. In particular, $g_{i}$ is a letter of $w$. Hence, $g_{i}$ is a
letter of $w\mu$ as well (since every letter of $w$ is a letter of $w\mu$
(because $w$ is a prefix of $w\mu$)). Since $\beta$ is the highest letter of
$w\mu$, this yields $\beta\geq g_{i}$. But this contradicts $\beta<g_{i}$.
This contradiction proves that our assumption was false, qed.}. In other
words, $g\leq\beta^{\ell\left(  w\right)  }$. Hence, Proposition
\ref{prop.words.lex}(b) (applied to $q$, $g$ and $\beta^{\ell\left(  w\right)
}$ instead of $a$, $c$ and $d$) yields $qg\leq q\beta^{\ell\left(  w\right)
}$, so that $w=qg\leq q\beta^{\ell\left(  w\right)  }=v$ (since $v=q\beta
^{\ell\left(  w\right)  }$). Thus, $v\geq w$, so that $u\geq v\geq w$.

But $v=q\beta^{\ell\left(  w\right)  }$, so that $w\beta^{\ell\left(
w\right)  }=u\underbrace{v}_{=q\beta^{\ell\left(  w\right)  }}=uq\beta
^{\ell\left(  w\right)  }$. Cancelling $\beta^{\ell\left(  w\right)  }$ from
this equation, we obtain $w=uq$. Thus, $u\leq uq=w$. Since $u\neq
w$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $u=w$. Hence,
$u\varnothing=u=w=uq$. Cancelling $u$ from this equation, we obtain
$\varnothing=q$. Hence, the word $q$ is empty. This contradicts the fact that
$q$ is nonempty. This contradiction shows that our assumption was wrong,
qed.}, this becomes $u<w$. This contradicts $u\geq w$. Hence, we have found a
contradiction in Case 2.

We have now obtained a contradiction in each of our two Cases 1 and 2. Since
these two Cases cover all possibilities, this yields that we always obtain a
contradiction. Thus, our assumption was wrong, and we conclude that the word
$w\beta^{\ell\left(  w\right)  }$ is Lyndon. Hence, $w$ is a prefix of a
Lyndon word in $\mathfrak{A}^{\ast}$ (because $w$ is a prefix of the word
$w\beta^{\ell\left(  w\right)  }$). In other words, Assertion $\mathcal{F}%
^{\prime}$ holds. This proves the implication $\mathcal{F}^{\prime\prime
}\Longrightarrow\mathcal{F}^{\prime}$. This solves Exercise
\ref{exe.words.powerlyndon}(f).

\textit{Remark:} Of course, for a letter $\mu\in\mathfrak{A}$, if we have
$\left(  \mu>a\text{ for every letter }a\text{ of }w\right)  $, then we also
have $\left(  \mu>a\text{ for some letter }a\text{ of }w\right)  $ (since $w$
is nonempty and thus has at least one letter). Hence, Exercise
\ref{exe.words.powerlyndon}(e) is a particular case of Exercise
\ref{exe.words.powerlyndon}(f).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.poset}}

\textit{Solution to Exercise \ref{exe.words.poset}.}

(a) If we replace the word ``total'' by
``partial'' throughout the proof of
Proposition \ref{prop.words.lex}, then the resulting argument can be used in
the partial-order setting to prove Proposition \ref{prop.words.lex} with
``a total order'' replaced by
``a partial order''. Hence, Proposition
\ref{prop.words.lex} holds in the partial-order setting, as long as one
replaces ``a total order'' by
``a partial order'' in part (a) of this
Proposition. This solves Exercise \ref{exe.words.poset}(a).

(b) Let us work in the partial-order setting. Let $a,b,c,d\in\mathfrak{A}%
^{\ast}$ be four words such that the words $ab$ and $cd$ are comparable. We
need to prove that the words $a$ and $c$ are comparable.

The words $ab$ and $cd$ are comparable. In other words, we have either $ab\leq
cd$ or $ab\geq cd$. In other words, we have either $ab\leq cd$ or $cd\leq ab$.
We can WLOG assume that $ab\leq cd$ (since otherwise, we can achieve $ab\leq
cd$ by switching $a$ and $b$ with $c$ and $d$, respectively). Assume this. We
know (from Exercise \ref{exe.words.poset}(a)) that Proposition
\ref{prop.words.lex} holds in the partial-order setting, as long as one
replaces ``a total order'' by ``a partial order'' in part (a) of this
Proposition. In particular, Proposition \ref{prop.words.lex}(e) holds in the
partial-order setting. Applying Proposition \ref{prop.words.lex}(e), we thus
conclude that either we have $a\leq c$ or the word $c$ is a prefix of $a$.
Thus, either we have $a\leq c$ or we have $c\leq a$ (because if $c$ is a
prefix of $a$, then $c\leq a$). In other words, either we have $a\leq c$ or we
have $a\geq c$. In other words, the words $a$ and $c$ are comparable. This
solves Exercise \ref{exe.words.poset}(b).

(c) Recall (from Exercise \ref{exe.words.poset}(a)) that Proposition
\ref{prop.words.lex} holds in the partial-order setting, as long as one
replaces ``a total order'' by
``a partial order'' in part (a) of this Proposition.

Proposition \ref{prop.words.uv=vu.2} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Proposition \ref{prop.words.uv=vu.3} holds in the partial-order setting,
because its proof applies verbatim in this setting.

The proof of Corollary \ref{cor.words.uv=vu.3} given above is no longer
applicable in the partial-order setting. However, the alternative proof of
Corollary \ref{cor.words.uv=vu.3} given in the solution to Exercise
\ref{exe.words.uw-wu} does apply verbatim in the partial-order setting. Thus,
Corollary \ref{cor.words.uv=vu.3} holds in the partial-order setting.

Corollary \ref{cor.words.uv=vu.partord} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Exercise \ref{exe.words.u=pn=qm} and Exercise \ref{exe.words.unvm.vmun} hold
in the partial-order setting, because their solutions apply verbatim in this setting.

Exercise \ref{exe.words.un.vm} and Exercise \ref{exe.words.uv=vu.k} hold in
the partial-order setting, because their solutions apply verbatim in this setting.

Proposition \ref{prop.words.lyndon.trivia} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Corollary \ref{cor.words.lyndon.trivia} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Proposition \ref{prop.words.lyndon.concat} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Corollary \ref{cor.words.lyndon.uv=vu} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Our proof of Proposition \ref{prop.words.lyndon.preorder} does not directly
apply in the partial-order setting; however, it can be tweaked so that it does:

\begin{proof}
[Proof of Proposition \ref{prop.words.lyndon.preorder} in the partial-order
setting.]If the words $u$ and $v$ are incomparable, then Proposition
\ref{prop.words.lyndon.preorder} is easily seen to
hold\footnote{\textit{Proof.} Assume that $u$ and $v$ are incomparable. If the
words $uv$ and $vu$ were comparable, then the words $u$ and $v$ would also be
comparable (by Exercise \ref{exe.words.poset}(b), applied to $a=u$, $b=v$,
$c=v$ and $d=u$), which would contradict the fact that $u$ and $v$ are
incomparable. Hence, the words $uv$ and $vu$ are incomparable. Thus, we cannot
have $uv\geq vu$. But we cannot have $u\geq v$ either (since $u$ and $v$ are
incomparable). Thus, neither $u\geq v$ nor $uv\geq vu$ holds. Hence, $u\geq v$
if and only if $uv\geq vu$; therefore, Proposition
\ref{prop.words.lyndon.preorder} holds, qed.}. Hence, for the rest of this
proof, we can WLOG assume that the words $u$ and $v$ are comparable. Assume this.

The words $u$ and $v$ are comparable. In other words, we have either $u<v$ or
$u=v$ or $u>v$. From here, we can proceed as in our proof of Proposition
\ref{prop.words.lyndon.preorder} in the total-order setting. Proposition
\ref{prop.words.lyndon.preorder} is thus proven in the partial-order setting.
\end{proof}

Exercise \ref{exe.words.lyndon.equiv2}(a) holds in the partial-order setting,
because its solution applies verbatim in this setting.

The proof of Theorem \ref{thm.words.lyndon.equiv} we gave (using Proposition
\ref{prop.words.lyndon.cutoff}) does not work in the partial-order
setting\footnote{One might try tweaking Proposition
\ref{prop.words.lyndon.cutoff} for the partial-order setting by replacing
``the (lexicographically) smallest nonempty
suffix'' by ``a nonempty suffix which is
(lexicographically) minimal among the nonempty suffices'', and
by replacing ``$u\geq v$ and $uv\geq vu$'' by
``neither $u<v$ nor $uv<vu$''. But this still
would not hold. For example, if $w$ is the word $XX12$ over the partially
ordered alphabet $\left\{  X,1,2\right\}  $ with relation $1<2$, then the
nonempty suffix $X12$ is lexicographically minimal among such suffixes, but we
do have $X<X12$. (At least $uv<vu$ does not hold indeed.)}. However, the
alternative proof of Theorem \ref{thm.words.lyndon.equiv} given in Exercise
\ref{exe.words.lyndon.equiv2}(b) does apply verbatim in the partial-order
setting. Thus, Theorem \ref{thm.words.lyndon.equiv} holds in the partial-order setting.

Exercise \ref{exe.words.lyndon.concatn} and Exercise
\ref{exe.words.lyndon.concatn-lyndon} hold in the partial-order setting,
because their solutions apply verbatim in this setting.

Exercise \ref{exe.words.lyndon.std2}(a) and Exercise
\ref{exe.words.lyndon.std2}(b) hold in the partial-order setting, because
their solutions apply verbatim in this setting.

Exercise \ref{exe.words.poset}(c) is thus solved.

[\textit{Remark:} In the above solution of Exercise \ref{exe.words.poset}(c),
we have used the results of Exercise \ref{exe.words.lyndon.equiv2}(b) and
Exercise \ref{exe.words.uw-wu}. This is the main reason why the latter two
exercises have been written. However, there is a way to avoid them and still
prove that Theorem \ref{thm.words.lyndon.equiv} and Corollary
\ref{cor.words.uv=vu.3} hold in the partial-order setting. This uses a trick,
which we shall now explain.

First, a definition:

\begin{definition}
Let $P$ be a poset.

\begin{enumerate}
\item[(a)] A poset $Q$ is said to be an
\emph{extension}\index{extension of a poset}
of $P$ if and only if the following two statements hold:

\begin{itemize}
\item We have $Q=P$ as sets.

\item Any two elements $a$ and $b$ of $P$ satisfying $a\leq b$ in $P$ satisfy
$a\leq b$ in $Q$.
\end{itemize}

\item[(b)] An extension $Q$ of $P$ is called a
\emph{linear extension}\index{linear extension of a poset} of
$P$ if and only if the poset $Q$ is totally ordered.\footnote{This notion of a
linear extension is identical to the one used in
Theorem~\ref{Stanley's-P-partition-theorem}, except that we don't require $P$
to be finite here.}
\end{enumerate}
\end{definition}

The following fact about extensions of posets is well-known:

\begin{proposition}
\label{prop.sol.words.poset.linex}Let $P$ be a finite poset.

\begin{enumerate}
\item[(a)] If $a$ and $b$ are two incomparable elements of $P$, then there
exists an extension $Q$ of $P$ such that we have $a<b$ in $Q$.

\item[(b)] If $a$ and $b$ are two elements of $P$ which don't satisfy $a\geq
b$, then there exists an extension $Q$ of $P$ such that we have $a<b$ in $Q$.

\item[(c)] There exists a linear extension of $P$.

\item[(d)] If $a$ and $b$ are two elements of $P$ which don't satisfy $a\geq
b$, then there exists a linear extension $Q$ of $P$ such that we have $a<b$ in
$Q$.
\end{enumerate}
\end{proposition}

(Actually, the requirement that $P$ be finite in Proposition
\ref{prop.sol.words.poset.linex} can be dropped if you accept Zorn's lemma,
but we do not need this generality.)

We can now apply this all to alphabets. In the partial-order setting,
alphabets are posets, and so it makes sense to speak of an extension of an
alphabet. We notice the following fact:

\begin{proposition}
\label{prop.sol.words.poset.reduce-to-linex}Let $\mathfrak{A}$ be a finite poset.

\begin{enumerate}
\item[(a)] If $\mathfrak{B}$ is an extension of the poset $\mathfrak{A}$, then
$\mathfrak{B}^{\ast}$ is an extension of the poset $\mathfrak{A}^{\ast}$.

\item[(b)] Let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two
words. Then, $u<v$ holds in $\mathfrak{A}^{\ast}$ if and only if we have%
\[
\left(  u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]

\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.reduce-to-linex}.](a) Let
$\mathfrak{B}$ be an extension of the poset $\mathfrak{A}$. Then, by the
definition of an ``extension'', we see that:

\begin{itemize}
\item We have $\mathfrak{B}=\mathfrak{A}$ as sets.

\item Any two elements $a$ and $b$ of $\mathfrak{A}$ satisfying $a\leq b$ in
$\mathfrak{A}$ satisfy $a\leq b$ in $\mathfrak{B}$.
\end{itemize}

Now, it is clear that $\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ as sets (since
$\mathfrak{B}=\mathfrak{A}$ as sets). Also, any two elements $a$ and $b$ of
$\mathfrak{A}^{\ast}$ satisfying $a\leq b$ in $\mathfrak{A}^{\ast}$ satisfy
$a\leq b$ in $\mathfrak{B}^{\ast}$\ \ \ \ \footnote{\textit{Proof.} Let $u$
and $v$ be two elements of $\mathfrak{A}^{\ast}$ satisfying $u\leq v$ in
$\mathfrak{A}^{\ast}$. We are going to prove that $u\leq v$ in $\mathfrak{B}%
^{\ast}$.
\par
According to the definition of the relation $\leq$ on $\mathfrak{A}^{\ast}$,
we have that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{ in
}\mathfrak{A}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v
\end{align*}
(because $u\leq v$ in $\mathfrak{A}$).
In other words, we must be in one of the following two cases:
\par
\textit{Case 1:} There exists an $i\in\left\{  1,2,\ldots,\min\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  u_{i}<v_{i}\text{ in }\mathfrak{A}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $.
\par
\textit{Case 2:} The word $u$ is a prefix of $v$.
\par
Let us first consider Case 1. In this case, there exists an $i\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  u_{i}<v_{i}\text{ in }\mathfrak{A}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}%
=v_{j}\right)  $. Let $i^{\prime}$ be this $i$. Thus, $i^{\prime}\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ satisfies $\left(  u_{i^{\prime}}<v_{i^{\prime}}\text{ in
}\mathfrak{A}\text{, and every }j\in\left\{  1,2,\ldots,i^{\prime}-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $.
\par
We have $u_{i^{\prime}}<v_{i^{\prime}}$ in $\mathfrak{A}$; thus,
$u_{i^{\prime}}\leq v_{i^{\prime}}$ in $\mathfrak{A}$ and $u_{i^{\prime}}\neq
v_{i^{\prime}}$.
\par
Recall that any two elements $a$ and $b$ of $\mathfrak{A}$ satisfying $a\leq
b$ in $\mathfrak{A}$ satisfy $a\leq b$ in $\mathfrak{B}$. Applied to
$a=u_{i^{\prime}}$ and $b=v_{i^{\prime}}$, this yields that $u_{i^{\prime}%
}\leq v_{i^{\prime}}$ in $\mathfrak{B}$ (since $u_{i^{\prime}}\leq
v_{i^{\prime}}$ in $\mathfrak{A}$). Combined with $u_{i^{\prime}}\neq
v_{i^{\prime}}$, this yields $u_{i^{\prime}}<v_{i^{\prime}}$ in $\mathfrak{B}%
$. Thus, we have $\left(  u_{i^{\prime}}<v_{i^{\prime}}\text{ in }%
\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i^{\prime}-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $. Consequently, there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(
v\right)  \right\}  \right\}  $ such that $\left(  u_{i}<v_{i}\text{ in
}\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  $ (namely, $i=i^{\prime}$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{ in
}\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
In other words, $u\leq v$ in $\mathfrak{B}^{\ast}$. Thus, $u\leq v$ in
$\mathfrak{B}^{\ast}$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, the word $u$ is a prefix of $v$.
Hence, $u\leq v$ in $\mathfrak{B}^{\ast}$. Thus, $u\leq v$ in $\mathfrak{B}%
^{\ast}$ is proven in Case 2.
\par
Now, $u\leq v$ in $\mathfrak{B}^{\ast}$ is proven in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this yields that $u\leq
v$ in $\mathfrak{B}^{\ast}$ always holds.
\par
Now, let us forget that we fixed $u$ and $v$. We have thus shown that any two
elements $u$ and $v$ of $\mathfrak{A}^{\ast}$ satisfying $u\leq v$ in
$\mathfrak{A}^{\ast}$ satisfy $u\leq v$ in $\mathfrak{B}^{\ast}$. Renaming the
variables $u$ and $v$ as $a$ and $b$ in this statement, we conclude the
following: Any two elements $a$ and $b$ of $\mathfrak{A}^{\ast}$ satisfying
$a\leq b$ in $\mathfrak{A}^{\ast}$ satisfy $a\leq b$ in $\mathfrak{B}^{\ast}$.
Qed.}. Consequently, $\mathfrak{B}^{\ast}$ is an extension of the poset
$\mathfrak{A}^{\ast}$. This proves Proposition
\ref{prop.sol.words.poset.reduce-to-linex}(a).

(b) It is easy to prove the logical implication%
\begin{equation}
\left(  u<v\text{ in }\mathfrak{A}^{\ast}\right)  \Longrightarrow\left(
u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)
\label{pf.prop.sol.words.poset.reduce-to-linex.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.words.poset.reduce-to-linex.1}):}
Assume that $u<v$ in $\mathfrak{A}^{\ast}$. Thus, $u\neq v$ and $u\leq v$ in
$\mathfrak{A}^{\ast}$. Let $\mathfrak{B}$ be a linear extension of
$\mathfrak{A}$. Then, $\mathfrak{B}^{\ast}$ is an extension of the poset
$\mathfrak{A}^{\ast}$ (by Proposition
\ref{prop.sol.words.poset.reduce-to-linex}(a)). Hence, $u\leq v$ in
$\mathfrak{B}^{\ast}$ (since $u\leq v$ in $\mathfrak{A}^{\ast}$). Combined
with $u\neq v$, this yields $u<v$ in $\mathfrak{B}^{\ast}$.
\par
Now, let us forget that we fixed $\mathfrak{B}$. We thus have proven that
$u<v$ in $\mathfrak{B}^{\ast}$ for every linear extension $\mathfrak{B}$ of
$\mathfrak{A}$. This proves (\ref{pf.prop.sol.words.poset.reduce-to-linex.1}%
).}. We shall now focus on proving the implication%
\begin{equation}
\left(  u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  \Longrightarrow\left(  u<v\text{
in }\mathfrak{A}^{\ast}\right)  .
\label{pf.prop.sol.words.poset.reduce-to-linex.2}%
\end{equation}


\textit{Proof of (\ref{pf.prop.sol.words.poset.reduce-to-linex.1}):} Assume
that%
\begin{equation}
u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}.
\label{pf.prop.sol.words.poset.reduce-to-linex.2.pf.1}%
\end{equation}


We need to show that $u<v$ in $\mathfrak{A}^{\ast}$.

First of all, it is impossible that $u\geq v$ in $\mathfrak{A}^{\ast}%
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $u\geq v$ in
$\mathfrak{A}^{\ast}$. Hence, $v\leq u$ in $\mathfrak{A}^{\ast}$.
\par
Proposition \ref{prop.sol.words.poset.linex}(c) (applied to $P=\mathfrak{A}%
^{\ast}$) yields that there exists a linear extension of $\mathfrak{A}$. Let
$\mathfrak{B}$ be such a linear extension. Then, $u<v$ in $\mathfrak{B}^{\ast
}$ (by (\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.1})). But
$\mathfrak{B}^{\ast}$ is an extension of the poset $\mathfrak{A}^{\ast}$ (by
Proposition \ref{prop.sol.words.poset.linex}(a)). Thus, $v\leq u$ in
$\mathfrak{B}^{\ast}$ (since $v\leq u$ in $\mathfrak{A}^{\ast}$). This
contradicts $u<v$ in $\mathfrak{B}^{\ast}$. This contradiction proves that our
assumption was wrong, qed.}. Consequently, we have $u\neq v$ (because
otherwise, we would have $u=v$ and thus $u\geq v$ in $\mathfrak{A}^{\ast}$,
which would contradict the fact that it is impossible that $u\geq v$ in
$\mathfrak{A}^{\ast}$).

Let $m=\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  $.
Then, $m\leq\ell\left(  u\right)  $ and $m\leq\ell\left(  v\right)  $. Recall
that $u=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $ and
$v=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $. If every
$i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}=v_{i}$, then it is easy
to see that $u<v$ in $\mathfrak{A}^{\ast}$\ \ \ \ \footnote{\textit{Proof.}
Assume that every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}%
=v_{i}$. Thus, $\left(  u_{1},u_{2},\ldots,u_{m}\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{m}\right)  $.
\par
Let us first assume (for the sake of contradiction) that $\ell\left(
u\right)  \geq\ell\left(  v\right)  $. Then, $m=\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  =\ell\left(  v\right)  $ (since
$\ell\left(  u\right)  \geq\ell\left(  v\right)  $). Now,%
\begin{align*}
v  &  =\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(
v_{1},v_{2},\ldots,v_{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  v\right)  =m\right) \\
&  =\left(  u_{1},u_{2},\ldots,u_{m}\right)  .
\end{align*}
But the word $\left(  u_{1},u_{2},\ldots,u_{m}\right)  $ is clearly a prefix
of $\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $ (since
$m\leq\ell\left(  u\right)  $). In other words, the word $v$ is a prefix of
$u$ (since $v=\left(  u_{1},u_{2},\ldots,u_{m}\right)  $ and $u=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $). Hence, $v\leq u$ in
$\mathfrak{A}^{\ast}$. In other words, $u\geq v$ in $\mathfrak{A}^{\ast}$.
This contradicts the fact that it is impossible that $u\geq v$ in
$\mathfrak{A}^{\ast}$.
\par
This contradiction proves that our assumption (that $\ell\left(  u\right)
\geq\ell\left(  v\right)  $) was wrong. Hence, we cannot have $\ell\left(
u\right)  \geq\ell\left(  v\right)  $. We thus have $\ell\left(  u\right)
<\ell\left(  v\right)  $. Thus, $m=\min\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  =\ell\left(  u\right)  $ (since $\ell\left(
u\right)  <\ell\left(  v\right)  $). Now,%
\begin{align*}
u  &  =\left(  u_{1},u_{2},\ldots,u_{\ell\left(  v\right)  }\right)  =\left(
u_{1},u_{2},\ldots,u_{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  u\right)  =m\right) \\
&  =\left(  v_{1},v_{2},\ldots,v_{m}\right)  .
\end{align*}
But the word $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ is clearly a prefix
of $\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $ (since
$m\leq\ell\left(  v\right)  $). In other words, the word $u$ is a prefix of
$v$ (since $u=\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ and $v=\left(
v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $). Hence, $u\leq v$ in
$\mathfrak{A}^{\ast}$. Combined with $u\neq v$, this yields that $u<v$ in
$\mathfrak{A}^{\ast}$, qed.}. Hence, for the rest of the proof of
(\ref{pf.prop.sol.words.poset.reduce-to-linex.1}), we can WLOG assume that not
every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}=v_{i}$. Assume this.

Not every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}=v_{i}$. In
other words, there exists an $i\in\left\{  1,2,\ldots,m\right\}  $ which does
not satisfy $u_{i}=v_{i}$. Let $k$ be the smallest such $i$. Thus,
$k\in\left\{  1,2,\ldots,m\right\}  $ does not satisfy $u_{k}=v_{k}$, whereas%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,m\right\}  \text{ satisfying }i<k\text{
satisfies }u_{i}=v_{i}. \label{pf.prop.sol.words.poset.reduce-to-linex.2.pf.4}%
\end{equation}


We have $u_{k}\neq v_{k}$ (since $k$ does not satisfy $u_{k}=v_{k}$). We must
have $v_{k}\geq u_{k}$ in $\mathfrak{A}$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, we don't have $v_{k}\geq u_{k}$ in $\mathfrak{A}$.
Hence, Proposition \ref{prop.sol.words.poset.linex}(d) (applied to
$P=\mathfrak{A}^{\ast}$, $a=v_{k}$ and $b=u_{k}$) yields that there exists a
linear extension $Q$ of $\mathfrak{A}$ such that we have $v_{k}<u_{k}$ in $Q$.
Let $\mathfrak{B}$ be such a linear extension. Thus, $\mathfrak{B}$ is a
linear extension of $\mathfrak{A}$ such that we have $v_{k}<u_{k}$ in
$\mathfrak{B}$. We have $u<v$ in $\mathfrak{B}^{\ast}$ (by
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.1})).
\par
Every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $u_{j}=v_{j}$ (by
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.4}), applied to $i=j$). In
other words, every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies
$v_{j}=u_{j}$. Also, we have $k\in\left\{  1,2,\ldots,m\right\}  =\left\{
1,2,\ldots,\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)  \right\}
\right\}  $ (since $m=\min\left\{  \ell\left(  u\right)  ,\ell\left(
v\right)  \right\}  =\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)
\right\}  $). Altogether, we thus have shown that $k\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)  \right\}  \right\}
$ satisfies $\left(  v_{k}<u_{k}\text{ in }\mathfrak{B}\text{, and every }%
j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }v_{j}=u_{j}\right)  $.
Thus, there exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
v\right)  ,\ell\left(  u\right)  \right\}  \right\}  $ such that \newline
$\left(
v_{i}<u_{i}\text{ in }\mathfrak{B}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }v_{j}=u_{j}\right)  $ (namely,
$i=k$). Hence,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  v_{i}<u_{i}\text{ in
}\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }v_{j}=u_{j}\right)  ,\\
&  \text{\textbf{or} the word }v\text{ is a prefix of }u.
\end{align*}
In other words, $v<u$ in $\mathfrak{B}^{\ast}$ (by the definition of the
relation $<$ on $\mathfrak{B}^{\ast}$). This contradicts the fact that $u<v$
in $\mathfrak{B}^{\ast}$. This contradiction proves that our assumption was
wrong, qed.}. That is, $u_{k}\leq v_{k}$ in $\mathfrak{A}$. Combined with
$u_{k}\neq v_{k}$, this yields that $u_{k}<v_{k}$ in $\mathfrak{A}$.

Every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $u_{j}=v_{j}$ (by
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.4}), applied to $i=j$).
Also, we have $k\in\left\{  1,2,\ldots,m\right\}  =\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
$ (since $m=\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)
\right\}  $). Altogether, we thus have shown that $k\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
$ satisfies $\left(  u_{k}<v_{k}\text{ in }\mathfrak{A}\text{, and every }%
j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $.
Thus, there exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that \newline
$\left(
u_{i}<v_{i}\text{ in }\mathfrak{A}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $ (namely,
$i=k$). Hence,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{ in
}\mathfrak{A}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
In other words, $u<v$ in $\mathfrak{A}^{\ast}$ (by the definition of the
relation $<$ on $\mathfrak{A}^{\ast}$). This proves
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2}).

Combining the implications (\ref{pf.prop.sol.words.poset.reduce-to-linex.1})
and (\ref{pf.prop.sol.words.poset.reduce-to-linex.2}), we obtain the following
equivalence of statements:%
\[
\left(  u<v\text{ in }\mathfrak{A}^{\ast}\right)  \Longleftrightarrow\left(
u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
Thus, Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b) is proven.
\end{proof}

Let us now give a new proof of the fact that Corollary \ref{cor.words.uv=vu.3}
holds in the partial-order setting:

\begin{proof}
[Alternative proof of Corollary \ref{cor.words.uv=vu.3} in the partial-order
setting.]We need to prove that $uw\geq wu$. If $uw=wu$, then this is obvious.
Hence, for the rest of this proof, we can WLOG assume that $uw\neq wu$. Assume this.

Let $\mathfrak{B}$ be any linear extension of the alphabet $\mathfrak{A}$.
Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)).
Hence, $uv\geq vu$ in $\mathfrak{B}^{\ast}$ (since $uv\geq vu$ in
$\mathfrak{A}^{\ast}$) and $vw\geq wv$ in $\mathfrak{B}^{\ast}$ (since $vw\geq
wv$ in $\mathfrak{A}^{\ast}$). But the alphabet $\mathfrak{B}$ is totally
ordered, and thus Corollary \ref{cor.words.uv=vu.3} (applied to $\mathfrak{B}$
instead of $\mathfrak{A}$) yields that $uw\geq wu$ in $\mathfrak{B}^{\ast}$
(since we know that Corollary \ref{cor.words.uv=vu.3} holds in the total-order
setting). Thus, $uw>wu$ in $\mathfrak{B}^{\ast}$ (since $uw\neq wu$). In other
words, $wu<uw$ in $\mathfrak{B}^{\ast}$.

Now, let us forget that we fixed $\mathfrak{B}$. We thus have shown that
$wu<uw$ in $\mathfrak{B}^{\ast}$ for every linear extension $\mathfrak{B}$ of
$\mathfrak{A}$. But Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b)
(applied to $wu$ and $uw$ instead of $u$ and $v$) yields that $wu<uw$ holds in
$\mathfrak{A}^{\ast}$ if and only if%
\[
\left(  wu<uw\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
Thus, we conclude that $wu<uw$ holds in $\mathfrak{A}^{\ast}$ (since we
already know that $wu<uw$ in $\mathfrak{B}^{\ast}$ for every linear extension
$\mathfrak{B}$ of $\mathfrak{A}$). Thus, $uw>wu$ in $\mathfrak{A}^{\ast}$;
hence, $uw\geq wu$ in $\mathfrak{A}^{\ast}$. This proves Corollary
\ref{cor.words.uv=vu.3} in the partial-order setting.
\end{proof}

Next, let us give a new proof of the fact that Theorem
\ref{thm.words.lyndon.equiv} holds in the partial-order setting:

\begin{proof}
[Alternative proof of Theorem \ref{thm.words.lyndon.equiv} in the
partial-order setting.]We need to prove the equivalence $\mathcal{A}%
\Longleftrightarrow\mathcal{B}\Longleftrightarrow\mathcal{C}%
\Longleftrightarrow\mathcal{D}$. We can prove the implications $\mathcal{A}%
\Longrightarrow\mathcal{B}$, $\mathcal{A}\Longrightarrow\mathcal{C}$,
$\mathcal{A}\Longrightarrow\mathcal{D}$ and $\mathcal{B}\Longrightarrow
\mathcal{A}$ in the same way as we did in the total-order setting. Hence, in
order to prove Theorem \ref{thm.words.lyndon.equiv}, it will be enough to
prove the implications $\mathcal{C}\Longrightarrow\mathcal{B}$ and
$\mathcal{D}\Longrightarrow\mathcal{B}$.

\textit{Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{B}%
$:} Assume that Assertion $\mathcal{C}$ holds.

Let $\mathfrak{B}$ be any linear extension of the alphabet $\mathfrak{A}$.
Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)). It
is easy to see that Assertion $\mathcal{C}$ (with $\mathfrak{A}$ replaced by
$\mathfrak{B}$) holds\footnote{\textit{Proof.} Let $u\in\mathfrak{B}^{\ast}$
and $v\in\mathfrak{B}^{\ast}$ be two nonempty words satisfying $w=uv$. We have
$u\in\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ and $v\in\mathfrak{B}^{\ast
}=\mathfrak{A}^{\ast}$, and thus $u\in\mathfrak{A}^{\ast}$ and $v\in
\mathfrak{A}^{\ast}$ are two nonempty words satisfying $w=uv$. Hence, we have
$v>u$ in $\mathfrak{A}^{\ast}$ (since we assumed that Assertion $\mathcal{C}$
holds). Thus, $u<v$ in $\mathfrak{A}^{\ast}$, so that $u\neq v$ and $u\leq v$
in $\mathfrak{A}^{\ast}$.
\par
So we have $u\leq v$ in $\mathfrak{A}^{\ast}$. Hence, $u\leq v$ in
$\mathfrak{B}^{\ast}$ (since $\mathfrak{B}^{\ast}$ is an extension of
$\mathfrak{A}^{\ast}$). Since $u\neq v$, this becomes $u<v$ in $\mathfrak{B}%
^{\ast}$. In other words, $v>u$ in $\mathfrak{B}^{\ast}$.
\par
Now, let us forget that we fixed $u$ and $v$. We thus have shown that any
nonempty words $u\in\mathfrak{B}^{\ast}$ and $v\in\mathfrak{B}^{\ast}$
satisfying $w=uv$ satisfy $v>u$ in $\mathfrak{B}^{\ast}$. In other words,
Assertion $\mathcal{C}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$)
holds}. But we can apply Theorem \ref{thm.words.lyndon.equiv} to
$\mathfrak{B}$ instead of $\mathfrak{A}$ (since $\mathfrak{B}$ is totally
ordered). As a consequence, we see that the four assertions $\mathcal{A}$,
$\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ (all with $\mathfrak{A}$
replaced by $\mathfrak{B}$) are equivalent. In particular, Assertion
$\mathcal{C}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$) is equivalent
to Assertion $\mathcal{B}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$).
Since Assertion $\mathcal{C}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$)
holds, we thus conclude that Assertion $\mathcal{B}$ (with $\mathfrak{A}$
replaced by $\mathfrak{B}$) holds. In other words,%
\begin{equation}
\text{any two nonempty words }u\in\mathfrak{B}^{\ast}\text{ and }%
v\in\mathfrak{B}^{\ast}\text{ satisfying }w=uv\text{ satisfy }v>w\text{ in
}\mathfrak{B}^{\ast}. \label{pf.thm.words.lyndon.equiv.poset.CB.1}%
\end{equation}


Now, let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two
nonempty words satisfying $w=uv$. It is easy to see that%
\begin{equation}
\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)
\label{pf.thm.words.lyndon.equiv.poset.CB.2}
\end{equation}
\footnote{\textit{Proof.} Let $\mathfrak{B}$ be a linear extension of
$\mathfrak{A}$. Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)).
Thus, $\mathfrak{B}^{\ast} = \mathfrak{A}^{\ast}$ as sets, so that
$u \in \mathfrak{A}^{\ast} = \mathfrak{B}^{\ast}$ and
$v \in \mathfrak{A}^{\ast} = \mathfrak{B}^{\ast}$. Now,
\eqref{pf.thm.words.lyndon.equiv.poset.CB.1} yields $v > w$ in
$\mathfrak{B}^\ast$. In other words, $w < v$ in $\mathfrak{B}^\ast$.
This proves \eqref{pf.thm.words.lyndon.equiv.poset.CB.2}.}.

But Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b) (applied to $w$
instead of $u$) yields that $w<v$ holds in $\mathfrak{A}^{\ast}$ if and only
if we have%
\[
\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
Consequently, we conclude that $w<v$ holds in $\mathfrak{A}^{\ast}$ (since we
know that we have \newline
$\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every
linear extension }\mathfrak{B}\text{ of }\mathfrak{A}\right)  $). In other
words, $v>w$ in $\mathfrak{A}^{\ast}$.

Now, let us forget that we fixed $u$ and $v$. We have thus shown that any
nonempty words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$
satisfying $w=uv$ satisfy $v>w$ in $\mathfrak{A}^{\ast}$. In other words,
Assertion $\mathcal{B}$ holds. Thus, we have proven the implication
$\mathcal{C}\Longrightarrow\mathcal{B}$.

\textit{Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{B}%
$:} The proof of the implication $\mathcal{D}\Longrightarrow
\mathcal{B}$ is analogous to our above proof of the implication $\mathcal{C}%
\Longrightarrow\mathcal{B}$, and thus left to the reader.

The proof of Theorem \ref{thm.words.lyndon.equiv} in the partial-order setting
is complete.
\end{proof}

The examples of Theorem \ref{thm.words.lyndon.equiv} and Corollary
\ref{cor.words.uv=vu.3} should have illustrated how Proposition
\ref{prop.sol.words.poset.reduce-to-linex} allows deriving certain facts about
the partial-order setting from the corresponding facts about the total-order
setting. This trick, however, has its limits. For example, in the total-order
setting, the fact that any Lyndon word $w$ of length $>1$ can be written in
the form $w=uv$ for two Lyndon words $u$ and $v$ satisfying $u<w<v$ is a
consequence of Theorem \ref{thm.words.lyndon.std}. But in the partial-order
setting, it is not clear how to derive it from the total-order setting,
although it \textbf{is} true in the partial-order setting (and follows from
Exercise \ref{exe.words.lyndon.std2}).]

(d) If $\mathfrak{A}$ is the partially ordered alphabet $\left\{  1,2\right\}
$ with no relations whatsoever (i.e., a $2$-element antichain), and $w$ is the
word $12$, then $w$ does satisfy $\left(  \text{if }w\leq t^{n}\text{, then
}w\leq t\right)  $ for every nonempty word $t$ and every positive integer $n$,
but $w$ is not Lyndon.

[\textit{Remark:} One direction of Exercise
\ref{exe.words.lyndon.power-criterion} does hold in the partial-order setting:
Namely, if $w$ is Lyndon, then every nonempty word $t$ and every positive
integer $n$ satisfy $\left(  \text{if }w\leq t^{n}\text{, then }w\leq
t\right)  $. The proof of this is the same as in the total-order setting.]

(e) The following statement is clearly equivalent to Exercise
\ref{exe.words.lyndon.power-criterion} in the total-order setting, while still
being valid in the partial-order setting:

\begin{proposition}
\label{prop.sol.words.poset.e}Let $w$ be a nonempty word. Then, $w$ is Lyndon
if and only if every nonempty word $t$ and every positive integer $n$ satisfy
$\left(  \text{if }w>t\text{, then }w>t^{n}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.e} in the partial-order
setting.]Let us first assume that $w$ is Lyndon. We shall prove that%
\begin{equation}
\left(  \text{every nonempty word }t\text{ and every positive integer }n\text{
satisfy }\left(  \text{if }w>t\text{, then }w>t^{n}\right)  \right)  .
\label{pf.prop.sol.words.poset.e.this}%
\end{equation}


Let $t$ be a nonempty word, and let $n$ be a positive integer. Assume that
$w>t$. We need to prove that $w>t^{n}$.

Assume the contrary. Thus, we don't have $w>t^{n}$. Thus, there exists an
$i\in\left\{  1,2,\ldots,n\right\}  $ such that we don't have $w>t^{i}$
(namely, $i=n$). Let $m$ be the \textbf{minimal} such $i$. Thus, $m\in\left\{
1,2,\ldots,n\right\}  $, and we don't have $w>t^{m}$. Hence, $m\neq1$ (since
we don't have $w>t^{m}$, but we do have $w>t=t^{1}$). Thus, $m\geq2$, so that
$m-1$ is also an element of $\left\{  1,2,\ldots,n\right\}  $. If we did not
have $w>t^{m-1}$, then $m-1$ would therefore be an $i\in\left\{
1,2,\ldots,n\right\}  $ such that we don't have $w>t^{i}$. But this would
contradict the fact that $m$ is the \textbf{minimal} such $i$. Thus, we must
have $w>t^{m-1}$. In other words, $t^{m-1}<w$.

Notice that $m-1\geq1$ (since $m\geq2$). Hence, the word $t^{m-1}$ is nonempty
(since $t$ is nonempty). In other words, $t^{m-1}\neq\varnothing$.

Recall that we don't have $w>t^{m}$. It is now easy to see that we don't have
$t^{m}\leq w$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we
have $t^{m}\leq w$. Hence, $w\geq t^{m}$, so that $w=t^{m}$ (since we don't
have $w>t^{m}$). Thus, $w=t^{m}=t^{m-1}t$. Thus, $t$ is a suffix of $w$. Thus,
Corollary \ref{cor.words.lyndon.trivia} (applied to $t$ instead of $v$) yields
$t\geq w$. Thus, $t\geq w=t^{m}=tt^{m-1}$. Combined with $tt^{m-1}\geq t$
(since $t$ is a prefix of $tt^{m-1}$), this yields $t=tt^{m-1}$. Thus,
$tt^{m-1}=t=t\varnothing$. Cancelling $t$ from this equality, we obtain
$t^{m-1}=\varnothing$. This contradicts $t^{m-1}\neq\varnothing$. This
contradiction proves that our assumption was wrong, qed.}. In other words, we
don't have $t^{m-1}t\leq w\varnothing$ (since $t^{m}=t^{m-1}t$ and
$w=w\varnothing$).

But recall that $t^{m-1}<w$. Hence, Proposition \ref{prop.words.lex}(d)
(applied to $a=t^{m-1}$, $b=t$, $c=w$ and $d=\varnothing$) yields that either
we have $t^{m-1}t\leq w\varnothing$ or the word $t^{m-1}$ is a prefix of $w$.
Thus, the word $t^{m-1}$ is a prefix of $w$ (since we don't have $t^{m-1}t\leq
w\varnothing$). In other words, there exists a $v\in\mathfrak{A}^{\ast}$ such
that $w=t^{m-1}v$. Consider this $v$. We have $v\neq\varnothing$ (because
otherwise, we would have $v=\varnothing$ and thus $w=t^{m-1}\underbrace{v}%
_{=\varnothing}=t^{m-1}$, contradicting the fact that $w>t^{m-1}$), so that
$v$ is nonempty. Hence, Proposition \ref{prop.words.lyndon.trivia}(b) (applied
to $u=t^{m-1}$) now yields $v>t^{m-1}$. Hence, $t^{m-1}<v$. Thus, Proposition
\ref{prop.words.lex}(b) (applied to $a=t^{m-1}$, $c=t^{m-1}$ and $d=v$) yields
$t^{m-1}t^{m-1}\leq t^{m-1}v=w$. Hence,
\begin{align*}
w  &  \geq t^{m-1}t^{m-1}=t^{2\left(  m-1\right)  }=t^{m+\left(  m-2\right)
}=t^{m}t^{m-2}\ \ \ \ \ \ \ \ \ \ \left(  \text{this makes sense since }%
m\geq2\right) \\
&  \geq t^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t^{m}\text{ is a prefix
of }t^{m}t^{m-2}\right)  .
\end{align*}
Hence, $t^m \leq w$.
This contradicts the fact that we don't have $t^m \leq w$. This contradiction
proves that our assumption (that we don't have $w>t^{n}$) was false. Hence,
$w>t^{n}$.

Forget now that we assumed that $w>t$. We thus have proven that if $w>t$, then
$w>t^{n}$.

Now, forget that we fixed $t$ and assumed that $w$ is Lyndon. We thus have
shown that
\begin{equation}
\left(  \text{if }w\text{ is Lyndon, then
(\ref{pf.prop.sol.words.poset.e.this}) holds}\right)  .
\label{pf.prop.sol.words.poset.e.1}%
\end{equation}


Now, conversely, assume that (\ref{pf.prop.sol.words.poset.e.this}) holds. We
will prove that $w$ is Lyndon.

Let $u$ and $v$ be any nonempty words satisfying $w=uv$. We have $w\neq
u$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $w=u$. Hence,
$uv=w=u=u\varnothing$. Cancelling $u$ from this equality, we obtain
$v=\varnothing$, so that $v$ is empty. This contradicts the fact that $v$ is
nonempty. This contradiction proves that our assumption was wrong, qed.}.
Combined with $w=uv\geq u$ (since $u$ is a prefix of $uv$), this yields $w>u$.
But (\ref{pf.prop.sol.words.poset.e.this}) (applied to $t=u$ and $n=2$) yields
that if $w>u$, then $w>u^{2}$. Hence, $w>u^{2}$ (since we know that $w>u$).
Thus, $uv=w>u^{2}=uu$, so that $uu<uv$. Thus, Proposition \ref{prop.words.lex}%
(c) (applied to $a=u$, $c=u$ and $d=v$) yields $u\leq v$. Since $u\neq
v$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $u=v$. Hence,
$w=u\underbrace{v}_{=u}=uu=u^{2}$. This contradicts $w>u^{2}$. This
contradiction proves that our assumption was wrong, qed.}, this becomes $u<v$,
so that $v>u$.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that any
nonempty words $u$ and $v$ satisfying $w=uv$ satisfy $v>u$. In other words,
Assertion $\mathcal{C}$ of Theorem \ref{thm.words.lyndon.equiv} holds. Hence,
Assertion $\mathcal{A}$ of Theorem \ref{thm.words.lyndon.equiv} holds as well
(since Theorem \ref{thm.words.lyndon.equiv} yields that these Assertions
$\mathcal{C}$ and $\mathcal{A}$ are equivalent). In other words, the word $w$
is Lyndon.

Now, forget that we fixed $w$. We thus have proven that
\[
\left(  \text{if (\ref{pf.prop.sol.words.poset.e.this}) holds, then }w\text{
is Lyndon}\right)  .
\]
Combined with (\ref{pf.prop.sol.words.poset.e.1}), this yields that $w$ is
Lyndon if and only if (\ref{pf.prop.sol.words.poset.e.this}) holds. This
proves Proposition \ref{prop.sol.words.poset.e}.
\end{proof}

Thus, we have salvaged Exercise \ref{exe.words.lyndon.power-criterion} in the
partial-order setting. (And, as a side effect, we have obtained an alternative
solution for Exercise \ref{exe.words.lyndon.power-criterion} in the
total-order setting.)

[\textit{Remark:} Speaking of salvaging, there is a little piece of
Proposition \ref{prop.words.lyndon.cutoff} which can be salvaged for the
partial-order case:

\begin{proposition}
\label{prop.sol.words.poset.prop.words.lyndon.cutoff}Let $w$ be a nonempty
word. Let $v$ be a (lexicographically) minimal nonempty suffix of
$w$\ \ \ \ \footnote{By this, we mean that:
\par
\begin{itemize}
\item $v$ is a nonempty suffix of $w$;
\par
\item no nonempty suffix $s$ of $w$ satisfies $s<v$.
\end{itemize}
\par
Such a $v$ always exists (since $w$ is nonempty), but is not always unique.}.
Then, there exists a $u\in\mathfrak{A}^{\ast}$ such that $w = uv$ but
we don't have $uv<vu$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.prop.words.lyndon.cutoff} in
the partial-order case.] We know that $v$ is a suffix of $w$. Hence, there
exists a $u\in\mathfrak{A}^{\ast}$ such that $w=uv$. Consider this $u$. It
will clearly be enough to show that we don't have $uv<vu$.

So let us prove that we don't have $uv<vu$. Indeed, assume the contrary. Then,
$uv<vu$. Thus, there exists at least one suffix $t$ of $u$ such that $tv<vt$
(namely, $t=u$). Let $p$ be the \textbf{minimum-length} such suffix. Then,
$pv<vp$. Thus, $p$ is nonempty. In other words, $p\neq\varnothing$.

Since $p$ is a suffix of $u$, it is clear that $pv$ is a suffix of $uv=w$. So
we know that $pv$ is a nonempty suffix of $w$. Since $v$ is a minimal such
suffix, this yields that we don't have $pv<v$. Hence, we don't have $pv\leq
v$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $pv\leq v$.
This yields that $pv=v$ (since we don't have $pv<v$). Thus, $pv=v=\varnothing
v$. Cancelling $v$ from this equation, we obtain $p=\varnothing$. This
contradicts $p\neq\varnothing$. This contradiction proves that our assumption
was wrong, qed.}.

But $pv\varnothing=pv<vp$. Hence, Proposition \ref{prop.words.lex}(e) (applied
to $a=pv$, $b=\varnothing$, $c=v$ and $d=p$) yields that either we have
$pv\leq v$ or the word $v$ is a prefix of $pv$. Thus, the word $v$ is a prefix
of $pv$ (since we don't have $pv\leq v$). In other words, there exists a
$q\in\mathfrak{A}^{\ast}$ such that $pv=vq$. Consider this $q$. This $q$ is
nonempty (because otherwise we would have $pv=v\underbrace{q}_{=\varnothing
}=v$, contradicting the fact that $p$ is nonempty). From $vq=pv<vp$, we obtain
$q\leq p$ (by Proposition \ref{prop.words.lex}(c), applied to $a=v$, $c=q$ and
$d=p$).

We know that $q$ is a suffix of $pv$ (since $vq=pv$), whereas $pv$ is a suffix
of $w$. Thus, $q$ is a suffix of $w$. So $q$ is a nonempty suffix of $w$.
Since $v$ is a minimal such suffix, this yields that we don't have $q<v$.
But we have $p\varnothing=p\leq pv<vp$. Hence, Proposition
\ref{prop.words.lex}(e) (applied to $a=p$, $b=\varnothing$, $c=v$ and $d=p$)
yields that either we have $p\leq v$ or the word $v$ is a prefix of $p$. From
this, it is easy to obtain that $v$ is a prefix of $p$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $v$ is not a
prefix of $p$. Hence, $p\leq v$ (since either we have $p\leq v$ or the word
$v$ is a prefix of $p$). Now, $q\leq p\leq v$. Now, $q=v$ (since $q\leq v$ but
not $q<v$). Hence, $v=q\leq p$. Combined with $p\leq v$, this yields $v=p$.
Hence, $p\underbrace{v}_{=p}=\underbrace{p}_{=v}p=vp$, which contradicts
$pv<vp$. This contradiction proves that our assumption was wrong, qed.}. In
other words, there exists an $r\in\mathfrak{A}^{\ast}$ such that $p=vr$.
Consider this $r$. Clearly, $r$ is a suffix of $p$, while $p$ is a suffix of
$u$; therefore, $r$ is a suffix of $u$. Also, $pv<vp$ rewrites as $vrv<vvr$
(because $p=vr$). Thus, Proposition \ref{prop.words.lex}(c) (applied to $a=v$,
$c=rv$ and $d=vr$) yields $rv\leq vr$. Since $rv\neq vr$ (because otherwise,
we would have $rv=vr$, thus $v\underbrace{rv}_{=vr}=vvr$, contradicting
$vrv<vvr$), this becomes $rv<vr$.

The word $v$ is nonempty; thus, $\ell\left(v\right) > 0$.

Now, $r$ is a suffix of $u$ such that $rv<vr$. Since $p$ is the minimum-length
such suffix, this yields $\ell\left(  r\right)  \geq\ell\left(  p\right)  $.
But this contradicts the fact that $\ell\left(  \underbrace{p}_{=vr}\right)
=\ell\left(  vr\right)  =\underbrace{\ell\left(  v\right)  }_{>0}+\ell\left(
r\right)  >\ell\left(  r\right)  $. This contradiction proves our assumption
wrong. Thus, we have proven that we don't have $uv<vu$.
This completes the proof of
Proposition \ref{prop.sol.words.poset.prop.words.lyndon.cutoff} in the
partial-order case.
\end{proof}

]

(f) Let us first establish a lemma which plays a role similar to that of
Lemma~\ref{lem.words.CFLlemma} in the total-order setting:

\begin{lemma}
\label{lem.sol.words.poset.f.CFLlemma} Let $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ be a Hazewinkel-CFL factorization of a nonempty word $w$ (in
the partial-order setting). Let $p$ be a suffix of $w$ such that $p$ is
Lyndon. Then, $p\geq a_{k}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.words.poset.f.CFLlemma}.]We will prove Lemma
\ref{lem.sol.words.poset.f.CFLlemma} by induction over the (obviously)
positive integer $k$.

\textit{Induction base:} Assume that $k=1$. By the definition of a
Hazewinkel-CFL factorization, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a
tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ (since $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a Hazewinkel-CFL factorization of $w$).
Thus, $w=a_{1}a_{2}\cdots a_{k}=a_{1}$ (since $k=1$), so that $w$ is a Lyndon
word (since $a_{1}$ is a Lyndon word). But $p$ is nonempty (since $p$ is
Lyndon). Thus, Corollary \ref{cor.words.lyndon.trivia} (applied to $v=p$)
yields $p\geq w=a_{1}=a_{k}$ (since $1=k$). Thus, Lemma
\ref{lem.sol.words.poset.f.CFLlemma} is proven in the case $k=1$. The
induction base is complete.

\textit{Induction step:} Let $K$ be a positive integer. Assume (as the
induction hypothesis) that Lemma \ref{lem.sol.words.poset.f.CFLlemma} is
proven for $k=K$. We now need to show that Lemma
\ref{lem.sol.words.poset.f.CFLlemma} holds for $k=K+1$.

So let $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ be a Hazewinkel-CFL
factorization of a nonempty word $w$. Let $p$ be a nonempty suffix of $w$ such
that $p$ is Lyndon. We need to prove that $p\geq a_{K+1}$.

The tuple $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ is a Hazewinkel-CFL
factorization of $w$. By the definition of a Hazewinkel-CFL factorization,
this yields that $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ is a tuple of
Lyndon words such that $w=a_{1}a_{2}\cdots a_{K+1}$ and such that no
$i\in\left\{  1,2,\ldots,K\right\}  $ satisfies $a_{i}<a_{i+1}$.

Let $w^{\prime}=a_{2}a_{3}\cdots a_{K+1}$; then, $w=a_{1}a_{2}\cdots
a_{K+1}=a_{1}\underbrace{\left(  a_{2}a_{3}\cdots a_{K+1}\right)
}_{=w^{\prime}}=a_{1}w^{\prime}$. Hence, every nonempty suffix of $w$ is
either a nonempty suffix of $w^{\prime}$, or has the form $qw^{\prime}$ for a
nonempty suffix $q$ of $a_{1}$. Since $p$ is a nonempty suffix of $w$, we thus
must be in one of the following two cases:

\textit{Case 1:} The word $p$ is a nonempty suffix of $w^{\prime}$.

\textit{Case 2:} The word $p$ has the form $qw^{\prime}$ for a nonempty suffix
$q$ of $a_{1}$.

Let us first consider Case 1. In this case, $p$ is a nonempty suffix of
$w^{\prime}$. The $K$-tuple $\left(  a_{2},a_{3},\ldots,a_{K+1}\right)  $ of
Lyndon words satisfies $w^{\prime}=a_{2}a_{3}\cdots a_{K+1}$ and has the
property that no $i\in\left\{  1,2,\ldots,K-1\right\}  $ satisfies
$a_{i+1}<a_{\left(  i+1\right)  +1}$\ \ \ \ \footnote{\textit{Proof.} We
already know that $w^{\prime}=a_{2}a_{3}\cdots a_{K+1}$. It remains to show
that no $i\in\left\{  1,2,\ldots,K-1\right\}  $ satisfies $a_{i+1}<a_{\left(
i+1\right)  +1}$.
\par
To prove this, let us assume this contrary. Thus, there exists an
$i\in\left\{  1,2,\ldots,K-1\right\}  $ satisfying $a_{i+1}<a_{\left(
i+1\right)  +1}$. Let $j$ be such an $i$. Then, $j\in\left\{  1,2,\ldots
,K-1\right\}  $ satisfies $a_{j+1}<a_{\left(  j+1\right)  +1}$. Hence, we have
$a_{i}<a_{i+1}$ for $i=j+1$. This contradicts the fact that no $i\in\left\{
1,2,\ldots,K\right\}  $ satisfies $a_{i}<a_{i+1}$. This contradiction shows
that our assumption was wrong, qed.}. Therefore, $\left(  a_{2},a_{3}%
,\ldots,a_{K+1}\right)  $ is a Hazewinkel-CFL factorization of $w^{\prime}$.
We can thus apply Lemma \ref{lem.sol.words.poset.f.CFLlemma} to $K$,
$w^{\prime}$ and $\left(  a_{2},a_{3},\ldots,a_{K+1}\right)  $ instead of $k$,
$w$ and $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ (because we assumed that
Lemma \ref{lem.sol.words.poset.f.CFLlemma} is proven for $k=K$). As a result,
we obtain that $p\geq a_{K+1}$. Thus, $p\geq a_{K+1}$ is proven in Case 1.

Let us now consider Case 2. In this case, $p$ has the form $qw^{\prime}$ for a
nonempty suffix $q$ of $a_{1}$. Consider this $q$. Since $a_{1}$ is a Lyndon
word, we have $q\geq a_{1}$ (by Corollary \ref{cor.words.lyndon.trivia},
applied to $a_{1}$ and $q$ instead of $w$ and $v$). Our goal, however, is to
prove that $q\geq a_{K+1}$.

We will show that%
\begin{equation}
q\geq a_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,K+1\right\}  .\label{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}%
\end{equation}


\textit{Proof of (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}):} We will
prove (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) by induction over
$i$:

\textit{Induction base:} We know that $q\geq a_{1}$. In other words,
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=1$. This
completes the induction base.

\textit{Induction step:} Let $j\in\left\{  1,2,\ldots,K\right\}  $. Assume
that (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=j$. We
must prove that (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for
$i=j+1$.

We have $j\leq K$. Hence, the product $a_{j+1}a_{j+2}\cdots a_{K+1}$ contains
at least one factor. Also, the factors of this product are nonempty words
(because the words $a_{j+1}$, $a_{j+2}$, $\ldots$, $a_{K+1}$ are nonempty
(since these words are Lyndon)). Hence, $a_{j+1}a_{j+2}\cdots a_{K+1}$ is a
nonempty product of nonempty words. Consequently, $a_{j+1}a_{j+2}\cdots
a_{K+1}$ is a nonempty word.

The product $a_{j+2}a_{j+3}\cdots a_{K+1}$ is well-defined (because $j\leq
K$). Denote this product by $g$. Thus, $g=a_{j+2}a_{j+3}\cdots a_{K+1}$.

We have $q\geq a_{j}$ (since
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=j$). But%
\[
p=q\underbrace{w^{\prime}}_{\substack{=a_{2}a_{3}\cdots a_{K+1}\\=\left(
a_{2}a_{3}\cdots a_{j}\right)  \left(  a_{j+1}a_{j+2}\cdots a_{K+1}\right)
}}=q\left(  a_{2}a_{3}\cdots a_{j}\right)  \left(  a_{j+1}a_{j+2}\cdots
a_{K+1}\right)  =\left(  q\left(  a_{2}a_{3}\cdots a_{j}\right)  \right)
\left(  a_{j+1}a_{j+2}\cdots a_{K+1}\right)  .
\]
Consequently, $a_{j+1}a_{j+2}\cdots a_{K+1}$ is a suffix of $p$. Thus,
Corollary \ref{cor.words.lyndon.trivia} (applied to $p$ and $a_{j+1}%
a_{j+2}\cdots a_{K+1}$ instead of $w$ and $v$) yields $a_{j+1}a_{j+2}\cdots
a_{K+1}\geq p$ (since $p$ is Lyndon). Thus,%
\[
a_{j+1}a_{j+2}\cdots a_{K+1}\geq p=qw^{\prime}\geq q\geq a_{j}.
\]
In other words,%
\[
a_{j}\leq a_{j+1}a_{j+2}\cdots a_{K+1}=a_{j+1}\underbrace{\left(
a_{j+2}a_{j+3}\cdots a_{K+1}\right)  }_{=g}=a_{j+1}g.
\]
Hence, $a_{j}\varnothing=a_{j}\leq a_{j+1}g$. Proposition \ref{prop.words.lex}%
(e) (applied to $a_{j}$, $\varnothing$, $a_{j+1}$ and $g$ instead of $a$, $b$,
$c$ and $d$) thus yields that either we have $a_{j}\leq a_{j+1}$ or the word
$a_{j+1}$ is a prefix of $a_{j}$. From this, it is easy to conclude that the
word $a_{j+1}$ is a prefix of $a_{j}$\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Thus, the word $a_{j+1}$ is not a prefix of $a_{j}$. Hence, we
have $a_{j}\leq a_{j+1}$ (because we know that either we have $a_{j}\leq
a_{j+1}$ or the word $a_{j+1}$ is a prefix of $a_{j}$). Since we cannot have
$a_{j}<a_{j+1}$ (because no $i\in\left\{  1,2,\ldots,K\right\}  $ satisfies
$a_{i}<a_{i+1}$), this yields that we have $a_{j}=a_{j+1}$. Therefore,
$a_{j+1}$ is a prefix of $a_{j}$; this contradicts our assumption that the
word $a_{j+1}$ is not a prefix of $a_{j}$. This contradiction proves that our
assumption was wrong, qed.}. Consequently, $a_{j+1}\leq a_{j}$, so that
$a_{j}\geq a_{j+1}$ and $q\geq a_{j}\geq a_{j+1}$. In other words,
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=j+1$. This
completes the induction step. Thus,
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) is proven by induction.

Now, (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) (applied to $i=K+1$)
yields $q\geq a_{K+1}$. Hence, $p=qw^{\prime}\geq q\geq a_{K+1}$. Thus, $p\geq
a_{K+1}$ is proven in Case 2.

We have now proven $p\geq a_{K+1}$ in all cases. This proves that Lemma
\ref{lem.sol.words.poset.f.CFLlemma} holds for $k=K+1$. The induction step is
thus finished, and with it the proof of Lemma
\ref{lem.sol.words.poset.f.CFLlemma}.
\end{proof}

We can now conclude the solution of Exercise \ref{exe.words.poset}(f) by
proving the following proposition:

\begin{proposition}
\label{prop.sol.words.poset.f.CFL}Let $w$ be a word (in the partial-order
setting). Then, there exists a unique Hazewinkel-CFL factorization of $w$.
\end{proposition}

The proof is an almost literal adaptation of the proof of Theorem
\ref{thm.words.CFL}:

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.f.CFL}.]Let us first prove
that there exists a Hazewinkel-CFL factorization of $w$.

Indeed, there clearly exists a tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
$ of Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$\ \ \ \ \footnote{For
instance, the tuple $\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
}\right)  $ of one-letter words is a valid example (recall that one-letter
words are always Lyndon).}. Fix such a tuple with \textbf{minimum} $k$. We
claim that no $i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $a_{i}%
<a_{i+1}$.

Indeed, if some $i\in\left\{  1,2,\ldots,k-1\right\}  $ would satisfy
$a_{i}<a_{i+1}$, then the word $a_{i}a_{i+1}$ would be Lyndon (by Proposition
\ref{prop.words.lyndon.concat}(a), applied to $u=a_{i}$ and $v=a_{i+1}$),
whence $\left(  a_{1},a_{2},\ldots,a_{i-1},a_{i}a_{i+1},a_{i+2},a_{i+3}%
,\ldots,a_{k}\right)  $ would also be a tuple of Lyndon words satisfying
$w=a_{1}a_{2}\cdots a_{i-1}\left(  a_{i}a_{i+1}\right)  a_{i+2}a_{i+3}\cdots
a_{k}$ but having length $k-1<k$, contradicting the fact that $k$ is the
minimum length of such a tuple. Hence, no $i\in\left\{  1,2,\ldots
,k-1\right\}  $ can satisfy $a_{i}<a_{i+1}$. Thus, $\left(  a_{1},a_{2}%
,\ldots,a_{k}\right)  $ is a Hazewinkel-CFL factorization of $w$, so we have
shown that such a Hazewinkel-CFL factorization exists.

It remains to show that there exists at most one Hazewinkel-CFL factorization
of $w$. We shall prove this by induction over $\ell\left(  w\right)  $. Thus,
we fix a word $w$ and assume that%
\begin{equation}
\text{for every word }v\text{ with }\ell\left(  v\right)  <\ell\left(
w\right)  \text{, there exists at most one Hazewinkel-CFL factorization of }v.
\label{pf.thm.words.Hazewinkel-CFL.uni.1}%
\end{equation}
We now have to prove that there exists at most one Hazewinkel-CFL
factorization of $w$.

Indeed, let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(
b_{1},b_{2},\ldots,b_{m}\right)  $ be two Hazewinkel-CFL factorizations of
$w$. We need to prove that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(
b_{1},b_{2},\ldots,b_{m}\right)  $. If $w$ is empty, then this is obvious, so
we WLOG assume that it is not; thus, $k>0$ and $m>0$.

The tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a Hazewinkel-CFL
factorization of $w$. Thus, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a
tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and such that no
$i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $a_{i}<a_{i+1}$.

The tuple $\left(  b_{1},b_{2},\ldots,b_{m}\right)  $ is a Hazewinkel-CFL
factorization of $w$. Thus, $\left(  b_{1},b_{2},\ldots,b_{m}\right)  $ is a
tuple of Lyndon words satisfying $w=b_{1}b_{2}\cdots b_{m}$ and such that no
$i\in\left\{  1,2,\ldots,m-1\right\}  $ satisfies $b_{i}<b_{i+1}$. Now,
$b_{m}$ is a suffix of $w$ (since $w=b_{1}b_{2}\cdots b_{m}$). Also, $b_{m}$
is Lyndon. Thus, Lemma \ref{lem.sol.words.poset.f.CFLlemma} (applied to
$p=b_{m}$) yields $b_{m}\geq a_{k}$. The same argument (but with the roles of
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(  b_{1},b_{2}%
,\ldots,b_{m}\right)  $ switched) shows that $a_{k}\geq b_{m}$. Combined with
$b_{m}\geq a_{k}$, this yields $a_{k}=b_{m}$. Now let $v=a_{1}a_{2}\cdots
a_{k-1}$. Then, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)  $ is a tuple of
Lyndon words satisfying $v=a_{1}a_{2}\cdots a_{k-1}$ and such that no
$i\in\left\{  1,2,\ldots,\left(  k-1\right)  -1\right\}  $ satisfies
$a_{i}<a_{i+1}$ (because no $i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies
$a_{i}<a_{i+1}$). In other words, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)
$ is a Hazewinkel-CFL factorization of $v$.

But $w=a_{1}a_{2}\cdots a_{k}=\underbrace{a_{1}a_{2}\cdots a_{k-1}}%
_{=v}\underbrace{a_{k}}_{=b_{m}}=vb_{m}$, so that%
\[
vb_{m}=w=b_{1}b_{2}\cdots b_{m}=b_{1}b_{2}\cdots b_{m-1}b_{m}.
\]
Cancelling $b_{m}$ from this equality yields $v=b_{1}b_{2}\cdots b_{m-1}$.
Thus, $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $ is a tuple of Lyndon words
satisfying $v=b_{1}b_{2}\cdots b_{m-1}$ and such that no $i\in\left\{
1,2,\ldots,\left(  m-1\right)  -1\right\}  $ satisfies $b_{i}<b_{i+1}$ (since
no $i\in\left\{  1,2,\ldots,m-1\right\}  $ satisfies $b_{i}<b_{i+1}$). In
other words, $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $ is a Hazewinkel-CFL
factorization of $v$. Since $\ell\left(  v\right)  <\ell\left(  w\right)  $
(because $v=a_{1}a_{2}\cdots a_{k-1}$ is shorter than $w=a_{1}a_{2}\cdots
a_{k}$), we can apply (\ref{pf.thm.words.Hazewinkel-CFL.uni.1}) to obtain that
there exists at most one Hazewinkel-CFL factorization of $v$. But we already
know two such Hazewinkel-CFL factorizations: $\left(  a_{1},a_{2}%
,\ldots,a_{k-1}\right)  $ and $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $.
Thus, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)  =\left(  b_{1},b_{2}%
,\ldots,b_{m-1}\right)  $. Combining this with $a_{k}=b_{m}$, we obtain
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(  b_{1},b_{2},\ldots
,b_{m}\right)  $. This is exactly what we needed to prove. So we have shown
(by induction) that there exists at most one Hazewinkel-CFL factorization of
$w$. This completes the proof of Proposition \ref{prop.sol.words.poset.f.CFL}.
\end{proof}

(g) \textit{Solution to Exercise \ref{exe.words.powerlyndon} in the
partial-order setting.} Let us first observe the following fact (in the
partial-order setting): If $h$ is a nonempty word which is not a Lyndon word,
then%
\begin{equation}
\text{there exist nonempty words }u\text{ and }v\text{ such that }h=uv\text{
and not }vu>uv\text{.} \label{sol.words.poset.g.lem1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.poset.g.lem1}):} Let $h$ be a
nonempty word which is not a Lyndon word. We need to prove that
(\ref{sol.words.poset.g.lem1}) holds.
\par
In fact, assume the contrary. Then, there exist no nonempty words $u$ and $v$
such that $h=uv$ and not $vu>uv$. In other words, any nonempty words $u$ and
$v$ satisfying $h=uv$ must satisfy $vu>uv$. In other words, Assertion
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) holds.
\par
But we know (from Exercise \ref{exe.words.poset}(c)) that Theorem
\ref{thm.words.lyndon.equiv} holds in the partial-order setting. Hence, we can
apply Theorem \ref{thm.words.lyndon.equiv} to $h$ instead of $w$. We thus
conclude that Assertions $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) are equivalent. Hence, Assertion $\mathcal{A}$ of Theorem
\ref{thm.words.lyndon.equiv} (with $h$ instead of $w$) holds (since Assertion
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) holds). In other words, the word $h$ is Lyndon. This contradicts the fact
that the word $h$ is not Lyndon. This contradiction shows that our assumption
was wrong. Hence, (\ref{sol.words.poset.g.lem1}) is proven, qed.} Renaming $u$
and $v$ as $p$ and $q$ in this result, we obtain the following: If $h$ is a
nonempty word which is not a Lyndon word, then%
\begin{equation}
\text{there exist nonempty words }p\text{ and }q\text{ such that }h=pq\text{
and not }qp>pq\text{.} \label{sol.words.poset.g.lem1a}%
\end{equation}


Furthermore, if $h$ is a nonempty word which is not a Lyndon word, then%
\begin{equation}
\text{there exist nonempty words }u\text{ and }v\text{ such that }h=uv\text{
and not }v>u\text{.} \label{sol.words.poset.g.lem2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.poset.g.lem2}):} Let $h$ be a
nonempty word which is not a Lyndon word. We need to prove that
(\ref{sol.words.poset.g.lem2}) holds.
\par
In fact, assume the contrary. Then, there exist no nonempty words $u$ and $v$
such that $h=uv$ and not $v>u$. In other words, any nonempty words $u$ and $v$
satisfying $h=uv$ must satisfy $v>u$. In other words, Assertion $\mathcal{C}$
of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of $w$) holds.
\par
But we know (from Exercise \ref{exe.words.poset}(c)) that Theorem
\ref{thm.words.lyndon.equiv} holds in the partial-order setting. Hence, we can
apply Theorem \ref{thm.words.lyndon.equiv} to $h$ instead of $w$. We thus
conclude that Assertions $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) are equivalent. Hence, Assertion $\mathcal{A}$ of Theorem
\ref{thm.words.lyndon.equiv} (with $h$ instead of $w$) holds (since Assertion
$\mathcal{C}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) holds). In other words, the word $h$ is Lyndon. This contradicts the fact
that the word $h$ is not Lyndon. This contradiction shows that our assumption
was wrong. Hence, (\ref{sol.words.poset.g.lem2}) is proven, qed.}

Now, let us come to the solution of Exercise \ref{exe.words.powerlyndon} in
the partial-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(a) in the
partial-order setting.} The implication $\mathcal{A}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$ can be proven in the same way as it was proven in the
total-order setting.

Let us now prove the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{A}^{\prime}$:

\textit{Proof of the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{A}^{\prime}$:} Assume that Assertion $\mathcal{D}^{\prime}$
holds. Thus, if $u$ and $v$ are nonempty words satisfying $w=uv$, then we have
$vu\geq uv$.

We need to prove that Assertion $\mathcal{A}^{\prime}$ holds, i.e., that $w$
is a power of a Lyndon word. Assume the contrary. Thus, $w$ is not a power of
a Lyndon word; hence, $w$ is not a Lyndon word itself. Consequently,
(\ref{sol.words.poset.g.lem1}) (applied to $h=w$) shows that there exist
nonempty words $u$ and $v$ such that $w=uv$ and not $vu>uv$. Consider such a
pair of nonempty words $u$ and $v$ with \textbf{minimum }$\ell\left(
u\right)  $. The minimality of $\ell\left(  u\right)  $ shows that%
\begin{equation}
\left( \begin{array}{c}
 \text{if }u^{\prime}\text{ and }v^{\prime}\text{ are nonempty words
such that }w=u^{\prime}v^{\prime}\text{ and not }v^{\prime}u^{\prime
}>u^{\prime}v^{\prime}\text{,} \\
\text{then }\ell\left(  u^{\prime}\right)  \geq
\ell\left(  u\right) 
\end{array}
\right)  .
\label{sol.words.poset.g.DA.1}
\end{equation}


We have $vu\geq uv$ (according to Assertion $\mathcal{D}^{\prime}$) but not
$vu>uv$. Thus, $vu=uv$. Therefore, Proposition \ref{prop.words.uv=vu.2} yields
that there exist a $t\in\mathfrak{A}^{\ast}$ and two nonnegative integers $n$
and $m$ such that $u=t^{n}$ and $v=t^{m}$. Consider this $t$ and these $n$ and
$m$. We have $n\neq0$ (since $t^{n}=u$ is nonempty) and $m\neq0$ (since
$t^{m}=v$ is nonempty), and the word $t$ is nonempty (since $t^{n}=u$ is
nonempty). Moreover, we have $n=1$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Hence, $n\neq1$. Combined with $n\neq0$, this leads to $n\geq2$. As
a consequence, $u=t^{n}$ can be rewritten as $u=tt^{n-1}$. The word
$t^{n-1+m}$ is nonempty (since $t$ is nonempty and since $\underbrace{n}%
_{\geq2}-1+\underbrace{m}_{\geq0}\geq2-1+0=1$). Now, $w=\underbrace{u}%
_{=tt^{n-1}}\underbrace{v}_{=t^{m}}=tt^{n-1}t^{m}=tt^{n-1+m}$. Also,
$t^{n-1+m}t=t^{n-1+m+1}=tt^{n-1+m}$. Thus, we do not have $t^{n-1+m}%
t>tt^{n-1+m}$. Hence, (\ref{sol.words.poset.g.DA.1}) (applied to $u^{\prime
}=t$ and $v^{\prime}=t^{n-1+m}$) yields $\ell\left(  t\right)  \geq\ell\left(
u\right)  $ (since $t$ and $t^{n-1+m}$ are nonempty). Since $u=t^{n}$, this
rewrites as $\ell\left(  t\right)  \geq\ell\left(  t^{n}\right)
=\underbrace{n}_{\geq2}\ell\left(  t\right)  \geq2\ell\left(  t\right)  $,
whence $\ell\left(  t\right)  =0$, which contradicts the fact that $t$ is
nonempty. This contradiction shows that our assumption was wrong, qed.}.
Hence, $t^{n}=t^{1}=t$, so that $u=t^{n}=t$ and $w=\underbrace{u}%
_{=t}\underbrace{v}_{=t^{m}}=tt^{m}=t^{m+1}$. We shall now prove that the word
$t$ is Lyndon.

Assume the contrary. Then, $t$ is not Lyndon. Hence,
(\ref{sol.words.poset.g.lem1a}) (applied to $h=t$) shows that there exist
nonempty words $p$ and $q$ such that $t=pq$ and not $qp>pq$. Consider these
$p$ and $q$. Since $q$ is nonempty, we have $\ell\left(  q\right)  >0$, so
that $\ell\left(  \underbrace{u}_{=t=pq}\right)  =\ell\left(  pq\right)
=\ell\left(  p\right)  +\underbrace{\ell\left(  q\right)  }_{>0}>\ell\left(
p\right)  $.

We have $w=\underbrace{u}_{=t=pq}v=pqv$, and the words $p$ and $qv$ are
nonempty\footnote{For $qv$, this follows from the nonemptiness of $q$.}. Now,
using (\ref{sol.words.poset.g.DA.1}), it is easy to see that $qvp>pqv$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, we do not have
$qvp>pqv$. We can thus apply (\ref{sol.words.poset.g.DA.1}) to $u^{\prime}=p$
and $v^{\prime}=qv$. As a result, we obtain $\ell\left(  p\right)  \geq
\ell\left(  u\right)  $. This contradicts $\ell\left(  u\right)  >\ell\left(
p\right)  $. This contradiction shows that our assumption was wrong, qed.}.

Notice that $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ for every
$i\in \NN$\ \ \ \ \footnote{\textit{Proof.} We shall prove the equality
$q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ by induction over $i$:
\par
\textit{Induction base:} We have $q\underbrace{\left(  pq\right)  ^{0}%
}_{=\varnothing}=q\varnothing=q=\underbrace{\varnothing}_{=\left(  qp\right)
^{0}}q=\left(  qp\right)  ^{0}q$. In other words, the equality $q\left(
pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ holds for $i=0$. This completes the
induction base.
\par
\textit{Induction step:} Let $I\in \NN$ be such that the equality
$q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ holds for $i=I$. We need
to show that the equality $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$
also holds for $i=I+1$.
\par
We have $q\left(  pq\right)  ^{I}=\left(  qp\right)  ^{I}q$ (since the
equality $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ holds for $i=I$).
Thus,
\[
q\underbrace{\left(  pq\right)  ^{I+1}}_{=\left(  pq\right)  ^{I}\left(
pq\right)  }=\underbrace{q\left(  pq\right)  ^{I}}_{=\left(  qp\right)  ^{I}%
q}\left(  pq\right)  =\left(  qp\right)  ^{I}\underbrace{q\left(  pq\right)
}_{=\left(  qp\right)  q}=\underbrace{\left(  qp\right)  ^{I}\left(
qp\right)  }_{=\left(  qp\right)  ^{I+1}}q=\left(  qp\right)  ^{I+1}q.
\]
In other words, the equality $q\left(  pq\right)  ^{i}=\left(  qp\right)
^{i}q$ holds for $i=I+1$. This completes the induction step. The induction
proof of the equality $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ is
thus complete.}. Applied to $i=m$, this yields $q\left(  pq\right)
^{m}=\left(  qp\right)  ^{m}q$.

But $v=t^{m}$. Since $t=pq$, this rewrites as $v=\left(  pq\right)  ^{m}$.
Hence,%
\[
q\underbrace{v}_{=\left(  pq\right)  ^{m}}p=\underbrace{q\left(  pq\right)
^{m}}_{=\left(  qp\right)  ^{m}q}p=\left(  qp\right)  ^{m}qp=\left(
qp\right)  ^{m}\left(  qp\right)  =\left(  qp\right)  ^{m+1},
\]
so that $\left(  qp\right)  ^{m+1}=qvp>pqv$. But%
\[
\left(  pq\right)  ^{m+1}=\left(  pq\right)  \left(  \underbrace{pq}%
_{=t}\right)  ^{m}=\left(  pq\right)  \underbrace{t^{m}}_{=v}=\left(
pq\right)  v=pqv<\left(  qp\right)  ^{m+1}%
\]
(since $\left(  qp\right)  ^{m+1}>pqv$). From this, it is easy to see that
$pq<qp$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we don't
have $pq<qp$. But $\left(  pq\right)  \left(  pq\right)  ^{m}=\left(
pq\right)  ^{m+1}<\left(  qp\right)  ^{m+1}=\left(  qp\right)  \left(
qp\right)  ^{m+1}$ and $\ell\left(  pq\right)  =\ell\left(  p\right)
+\ell\left(  q\right)  =\ell\left(  q\right)  +\ell\left(  p\right)
=\ell\left(  qp\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied
to $a=pq$, $b=\left(  pq\right)  ^{m}$, $c=qp$ and $d=\left(  qp\right)  ^{m}%
$) yields $pq\leq qp$. Thus, $pq=qp$ (since we don't have $pq<qp$). Taking
both sides of this equality to the $\left(  m+1\right)  $-th power, we obtain
$\left(  pq\right)  ^{m+1}=\left(  qp\right)  ^{m+1}$, which contradicts
$\left(  pq\right)  ^{m+1}<\left(  qp\right)  ^{m+1}$. This contradiction
proves that our assumption was wrong, qed.}, so that $qp>pq$. This contradicts
the fact that we do not have $qp>pq$. This contradiction shows that our
assumption (that $t$ is not Lyndon) was wrong. Hence, $t$ is Lyndon. Thus, $w$
is a power of a Lyndon word (since $w=t^{m+1}$ is a power of $t$). Thus,
Assertion $\mathcal{A}^{\prime}$ is satisfied, so we have proven the
implication $\mathcal{D}^{\prime}\Longrightarrow\mathcal{A}^{\prime}$.

Now we have proven both implications $\mathcal{A}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$ and $\mathcal{D}^{\prime}\Longrightarrow\mathcal{A}%
^{\prime}$. Therefore, the equivalence $\mathcal{A}^{\prime}%
\Longleftrightarrow\mathcal{D}^{\prime}$ follows. Thus, Exercise
\ref{exe.words.powerlyndon}(a) is solved in the partial-order case.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(b) in the
partial-order setting.} Consider the letter $m$ and the alphabet
$\mathfrak{A}\cup\left\{  m\right\}  $ defined in Assertion $\mathcal{F}%
^{\prime\prime}$. We notice that the lexicographic order on $\mathfrak{A}%
^{\ast}$ is the restriction of the lexicographic order on $\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ to $\mathfrak{A}^{\ast}%
$. Therefore, when we have two words $p$ and $q$ in $\mathfrak{A}^{\ast}$,
statements like ``$p<q$'' do not depend on
whether we are regarding $p$ and $q$ as elements of $\mathfrak{A}^{\ast}$ or
as elements of $\left(  \mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$.
It is easy to see that the one-letter word $m$ satisfies%
\begin{equation}
m>p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathfrak{A}^{\ast}.
\label{sol.words.poset.g.b.mp}%
\end{equation}
\footnote{This can be proven in the same way as it was proven in the
total-order setting.}
% Also,
%\begin{equation}
%m>pm\ \ \ \ \ \ \ \ \ \ \text{for every nonempty }p\in\mathfrak{A}^{\ast}.
%\label{sol.words.poset.g.b.mpm}
%\end{equation}
%
%
%\footnote{\textit{Proof of (\ref{sol.words.poset.g.b.mpm}):} Let
%$p\in\mathfrak{A}^{\ast}$ be nonempty. Then, $p$ has a first letter, and we
%have $\left(  \text{the first letter of the word }p\right)  \in\mathfrak{A}$
%(since $p\in\mathfrak{A}^{\ast}$). Thus,
%\begin{align*}
%\left(  \text{the first letter of the word }p\right)   &
%<m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a<m\text{ for every }%
%a\in\mathfrak{A}\right) \\
%&  =\left(  \text{the first letter of the word }m\right)  .
%\end{align*}
%Now, the word $p$ is nonempty, and thus
%\begin{align*}
%\left(  \text{the first letter of the word }pm\right)   &  =\left(  \text{the
%first letter of the word }pm\right) \\
%&  <\left(  \text{the first letter of the word }m\right)  .
%\end{align*}
%Hence, $pm<m$ (by the definition of the lexicographic order), that is, $m>pm$.
%This proves (\ref{sol.words.poset.g.b.mpm}).}

The implications $\mathcal{B}^{\prime}\Longrightarrow\mathcal{E}^{\prime}$,
$\mathcal{C}^{\prime}\Longrightarrow\mathcal{E}^{\prime}$, $\mathcal{G}%
^{\prime}\Longrightarrow\mathcal{H}^{\prime}$, $\mathcal{F}^{\prime\prime
}\Longrightarrow\mathcal{B}^{\prime}$, $\mathcal{F}^{\prime\prime
}\Longrightarrow\mathcal{C}^{\prime}$ and $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ can be proven in the same way as they were proven in the
total-order setting. We shall now prove some further implications.

\textit{Proof of the implication $\mathcal{E}^{\prime}\Longrightarrow
\mathcal{F}^{\prime\prime}$:} Assume that Assertion $\mathcal{E}%
^{\prime}$ holds.

Assume (for the sake of contradiction) that the word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ is not a Lyndon word.
Clearly, this word $wm$ is nonempty. Thus, (\ref{sol.words.poset.g.lem2})
(applied to $h=wm$) shows that there exist nonempty words $u$ and $v$ in
$\left(  \mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ such that
$wm=uv$ and not $v>u$. Denote these two nonempty words $u$ and $v$ by $u$ and
$v^{\prime}$. Then, $u$ and $v^{\prime}$ are nonempty words in $\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ such that $wm=uv^{\prime
}$ and not $v^{\prime}>u$.

The word $v^{\prime}$ is a proper suffix of $wm$ (since $wm=uv^{\prime}$ and
since $w$ is nonempty). Hence, $v^{\prime}$ is a nonempty suffix of $wm$.
Thus, we must be in one of the following two cases (depending on whether this
suffix begins before the suffix $m$ of $wm$ begins or afterwards):

\textit{Case 1:} The word $v^{\prime}$ is a nonempty suffix of $m$. (Note that
$v^{\prime}=m$ is allowed.)

\textit{Case 2:} The word $v^{\prime}$ has the form $vm$ where $v$ is a
nonempty proper suffix of $w$.

Let us consider Case 1 first. In this case, the word $v^{\prime}$ is a
nonempty suffix of $m$. Since the only nonempty suffix of $m$ is $m$ itself
(because $m$ is a one-letter word), this yields $v^{\prime}=m$. Now,
$wm=u\underbrace{v^{\prime}}_{=m}=um$. Cancelling $m$ from this equality, we
obtain $w=u$. But $v^{\prime}=m>w$ (by (\ref{sol.words.poset.g.b.mp}), applied
to $p=w$), thus $v^{\prime}>w=u$. This contradicts the fact that we don't have
$v^{\prime}>u$. Thus, we have obtained a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $v^{\prime}$ has the form
$vm$ where $v$ is a nonempty proper suffix of $w$. Consider this $v$. We have
$wm=u\underbrace{v^{\prime}}_{=vm}=uvm$. By cancelling $m$ from this equality,
we obtain $w=uv$. Thus, $u$ and $v$ are subwords of $w$, and therefore belong
to $\mathfrak{A}^{\ast}$ (since $w\in\mathfrak{A}^{\ast}$). Moreover, $u$ and
$v$ are nonempty. Hence, Assertion $\mathcal{E}^{\prime}$ yields that either
we have $v\geq u$ or the word $v$ is a prefix of $w$. Since we cannot have
$v\geq u$ (because if we had $v\geq u$, then we would have $v^{\prime
}=vm>v\geq u$, which would contradict the fact that we don't have $v^{\prime
}>u$), we therefore must have that $v$ is a prefix of $w$. In other words,
there exists a $q\in\mathfrak{A}^{\ast}$ such that $w=vq$. Consider this $q$.
We have $m>q$ (by (\ref{sol.words.poset.g.b.mp}), applied to $p=q$). Thus,
$q\leq m$. Hence, Proposition \ref{prop.words.lex}(b) (applied to
$\mathfrak{A}\cup\left\{  m\right\}  $, $v$, $q$ and $m$ instead of
$\mathfrak{A}$, $a$, $c$ and $d$) yields $vq\leq vm$. Therefore, $vm\geq vq=w$
(since $w=vq$), so that $v^{\prime}=vm\geq w=uv>u$ (since $v$ is nonempty).
This contradicts the fact that we don't have $v^{\prime}>u$. Thus, we have
found a contradiction in Case 2.

We have therefore obtained a contradiction in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this shows that we always get a
contradiction. Hence, our assumption (that the word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ is not a Lyndon word)
was false. Hence, the word $wm\in\left(  \mathfrak{A}\cup\left\{  m\right\}
\right)  ^{\ast}$ is a Lyndon word. That is, Assertion $\mathcal{F}%
^{\prime\prime}$ holds. Hence, we have proven the implication $\mathcal{E}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$.

\textit{Proof of the implication $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{G}^{\prime}$:} Assume that Assertion $\mathcal{B}^{\prime}$ holds.

Let $s$ be the longest suffix $v$ of $w$ which does not satisfy $v\geq w$.
(This is well-defined, because there exists a suffix $v$ of $w$ which does not
satisfy $v\geq w$ -- namely, the empty word.) So we know that $s$ is a suffix
$v$ of $w$ which does not satisfy $v\geq w$. In other words, $s$ is a suffix
of $w$ and does not satisfy $s\geq w$. Hence, $s\neq w$ (because otherwise, we
would have $s=w$ and thus $s\geq w$; but this would contradict the fact that
$s$ does not satisfy $s\geq w$). Thus, $s$ is a proper suffix of $w$ (because
$s$ is a suffix of $w$ and satisfies $s\neq w$). Hence, there exists a
nonempty word $h\in\mathfrak{A}^{\ast}$ satisfying $w=hs$. Consider this $h$.
Using Assertion $\mathcal{B}^{\prime}$, it is easy to see that $s$ is a prefix
of $w$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $s$ is not
a prefix of $w$. Hence, $s$ is nonempty. Therefore, Assertion $\mathcal{B}%
^{\prime}$ (applied to $u=h$ and $v=s$) yields that either we have $s\geq w$
or the word $s$ is a prefix of $w$. Since $s$ is not a prefix of $w$, we must
thus have $s\geq w$. But this contradicts the fact that $s$ does not satisfy
$s\geq w$. This contradiction shows that our assumption was wrong, qed.}. In
other words, there exists a $g\in\mathfrak{A}^{\ast}$ such that $w=sg$.
Consider this $g$. We have $g\neq\varnothing$ (because otherwise, we would
have $g=\varnothing$ and thus $w=s\underbrace{g}_{=\varnothing}=s\neq w$,
which would be absurd). In other words, the word $g$ is nonempty. Thus,
$s<sg=w$.

We know that $s$ is the longest suffix $v$ of $w$ which does not satisfy
$v\geq w$. Hence,%
\begin{equation}
\left(  \text{if }v\text{ is a suffix of }w\text{ which does not satisfy
}v\geq w\text{, then }\ell\left(  v\right)  \leq\ell\left(  s\right)  \right)
. \label{sol.words.poset.g.BG.1}%
\end{equation}


There exists a nonnegative integer $m$ such that $h^{m}$ is a prefix of $s$
(for example, the nonnegative integer $m=0$). Consider the \textbf{maximal}
such integer $m$\ \ \ \ \footnote{This is well-defined, because of the
following reason:
\par
We have $\ell\left(  h\right)  \geq1$ (since the word $h$ is nonempty). Thus,
for every $m\in \NN$ satisfying $m>\ell\left(  s\right)  $, we have
$\ell\left(  h^{m}\right)  =m\underbrace{\ell\left(  h\right)  }_{\geq1}\geq
m>\ell\left(  s\right)  $. In other words, for every $m\in \NN$
satisfying $m>\ell\left(  s\right)  $, the word $h^{m}$ is longer than $s$.
Hence, for every $m\in \NN$ satisfying $m>\ell\left(  s\right)  $, the
word $h^{m}$ cannot be a prefix of $s$. Thus, for every sufficiently high
$m\in \NN$, the word $h^{m}$ cannot be a prefix of $s$.
\par
We thus know the following:
\par
\begin{itemize}
\item There exists a nonnegative integer $m$ such that $h^{m}$ is a prefix of
$s$.
\par
\item For every sufficiently high $m\in \NN$, the word $h^{m}$ cannot
be a prefix of $s$.
\end{itemize}
\par
Consequently, there exists a \textbf{maximal} nonnegative integer $m$ such
that $h^{m}$ is a prefix of $s$, qed.}. Then, $h^{m}$ is a prefix of $s$, but
$h^{m+1}$ is not a prefix of $s$. Since $h^{m}$ is a prefix of $s$, there
exists a word $q\in\mathfrak{A}^{\ast}$ such that $s=h^{m}q$. Consider this
$q$. Clearly, $w=h\underbrace{s}_{=h^{m}q}=\underbrace{hh^{m}}_{=h^{m+1}%
}q=\underbrace{h^{m+1}}_{=h^{m}h}q=h^{m}hq$. Hence, $h^{m}hq=w=\underbrace{s}%
_{=h^{m}q}g=h^{m}qg$. Cancelling $h^{m}$ from this equality, we obtain
$hq=qg$. It is now easy to see that we don't have $h\leq q$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $h\leq q$. Hence,
$h\leq q\leq qg=hq$ (since $hq=qg$). Therefore, Proposition
\ref{prop.words.lex}(g) (applied to $h$, $q$ and $q$ instead of $a$, $b$ and
$c$) yields that $h$ is a prefix of $q$. In other words, there exists a word
$r\in\mathfrak{A}^{\ast}$ such that $q=hr$. Consider this $r$. Now,
$s=h^{m}\underbrace{q}_{=hr}=\underbrace{h^{m}h}_{=h^{m+1}}r=h^{m+1}r$, so
that $h^{m+1}$ is a prefix of $s$. This contradicts the fact that $h^{m+1}$ is
not a prefix of $s$. This contradiction proves that our assumption was wrong,
qed.}. But $h\varnothing=h\leq hq=qg$. Thus, Proposition \ref{prop.words.lex}%
(e) (applied to $h$, $\varnothing$, $q$ and $g$ instead of $a$, $b$, $c$ and
$d$) yields that either we have $h\leq q$ or the word $q$ is a prefix of $h$.
Thus, the word $q$ is a prefix of $h$ (since we don't have $h\leq q$).

Next, we shall prove that the word $h$ is Lyndon.

In fact, assume the contrary. Then, $h$ is not Lyndon. Hence,
(\ref{sol.words.poset.g.lem1}) shows that there exist nonempty words $u$ and
$v$ such that $h=uv$ and not $vu>uv$. Consider these $u$ and $v$. Since
$w=\underbrace{h}_{=uv}s=uvs=u\left(  vs\right)  $, it is clear that the word
$vs$ is a suffix of $w$. If this suffix $vs$ would not satisfy $vs\geq w$,
then we could therefore obtain $\ell\left(  vs\right)  \leq\ell\left(
s\right)  $ (by (\ref{sol.words.poset.g.BG.1}), applied to $vs$ instead of
$v$), which would contradict $\ell\left(  vs\right)  =\underbrace{\ell\left(
v\right)  }_{\substack{>0\\\text{(since }v\text{ is nonempty)}}}+\ell\left(
s\right)  >\ell\left(  s\right)  $. Thus, the suffix $vs$ must satisfy $vs\geq
w$. We thus have $vs\geq w=hs\geq h=uv$, so that $uv\leq vs$.

Recall that $s<w$. Hence, $vs\leq vw$ (by Proposition \ref{prop.words.lex}(b),
applied to $v$, $s$ and $w$ instead of $a$, $c$ and $d$). Now, $uv\varnothing
=uv\leq vs\leq v\underbrace{w}_{=hs}=v\underbrace{h}_{=uv}s=vuvs$ and
$\ell\left(  uv\right)  =\ell\left(  u\right)  +\ell\left(  v\right)
=\ell\left(  v\right)  +\ell\left(  u\right)  =\ell\left(  vu\right)  \leq
\ell\left(  vu\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied
to $uv$, $\varnothing$, $vu$ and $vs$ instead of $a$, $b$, $c$ and $d$) yields
that $uv\leq vu$. In other words, $vu\geq uv$. Since we don't have $vu>uv$, we
therefore must have $vu=uv$. Thus, the elements $u$ and $v$ of the monoid
$\mathfrak{A}^{\ast}$ commute. Thus, the submonoid of $\mathfrak{A}^{\ast}$
generated by $u$ and $v$ is commutative. Since $h=uv$, the element $h$ lies in
this submonoid, and therefore the element $h^{m}$ lies in it as well. Thus,
$h^{m}$ commutes with $v$ (since this submonoid is commutative), i.e., we have
$vh^{m}=h^{m}v$. Thus, $v\underbrace{s}_{=h^{m}q}=\underbrace{vh^{m}}%
_{=h^{m}v}q=h^{m}vq$. Thus, $h^{m}vq=vs\geq w=h\underbrace{s}_{=h^{m}%
q}=\underbrace{hh^{m}}_{=h^{m+1}=h^{m}h}q=h^{m}hq$, so that $h^{m}hq\leq
h^{m}vq$. Hence, Proposition \ref{prop.words.lex}(c) (applied to $h^{m}$, $hq$
and $vq$ instead of $a$, $c$ and $d$) yields $hq\leq vq$. But since $q$ is a
prefix of $h$, there exists a word $z\in\mathfrak{A}^{\ast}$ such that $h=qz$.
Consider this $z$. We have%
\begin{align*}
v\underbrace{qz}_{=h=uv}  &  =\underbrace{vu}_{=uv=h}v=hv\leq h\underbrace{vu}%
_{=uv=h=qz}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }hv\text{ is a prefix of
}hvu\right) \\
&  =hqz.
\end{align*}


Also, $\ell\left(  \underbrace{h}_{=uv}q\right)  =\ell\left(  uvq\right)
=\ell\left(  u\left(  vq\right)  \right)  =\underbrace{\ell\left(  u\right)
}_{\substack{>0\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  vq\right)
>\ell\left(  vq\right)  $, so that $\ell\left(  vq\right)  \leq\ell\left(
hq\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied to $vq$,
$z$, $hq$ and $z$ instead of $a$, $b$, $c$ and $d$) yields $vq\leq hq$.
Combined with $hq\leq vq$, this yields $vq=hq$. Hence, $\ell\left(
\underbrace{vq}_{=hq}\right)  =\ell\left(  hq\right)  >\ell\left(  vq\right)
$, which is absurd. This contradiction proves that our assumption is wrong.
Thus, we have shown that the word $h$ is Lyndon.

We now know that $h\in\mathfrak{A}^{\ast}$ is a Lyndon word, $m+1$ is a
positive integer, and $q$ is a prefix of $h$, and we have $w=h^{m+1}q$. Hence,
there exists a Lyndon word $t\in\mathfrak{A}^{\ast}$, a positive integer
$\ell$ and a prefix $p$ of $t$ (possibly empty) such that $w=t^{\ell}p$
(namely, $t=h$, $\ell=m+1$ and $p=q$). In other words, Assertion
$\mathcal{G}^{\prime}$ holds. This proves the implication $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$.

Furthermore, the implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ holds. (In fact, it can be proven in the same way
as it was proven in the total-order setting.)

\textit{Proof of the implication $\mathcal{H}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$:} Assume that Assertion $\mathcal{H}^{\prime}$
holds. In other words, there exists a Lyndon word $t\in\mathfrak{A}^{\ast}$, a
nonnegative integer $\ell$ and a prefix $p$ of $t$ (possibly empty) such that
$w=t^{\ell}p$. Consider this $t$, this $\ell$ and this $p$.

We are going to prove that for every $m \in \NN$,
\begin{equation}
\left(  \text{every suffix }s\text{ of }t^{m}p\text{ satisfies either }s\geq
t^{m}p\text{ or }\left(  \text{the word }s\text{ is a prefix of }%
t^{m}p\right)  \right)  .
\label{sol.words.poset.g.HB.1}
\end{equation}


\textit{Proof of (\ref{sol.words.poset.g.HB.1}):} We will prove
(\ref{sol.words.poset.g.HB.1}) by induction over $m$:

\textit{Induction base:} Using the implication $\mathcal{F}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$, it is easy to see that
(\ref{sol.words.poset.g.HB.1}) holds for $m=0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $m=0$. Then, $t^{m}%
p=\underbrace{t^{0}}_{=\varnothing}p=\varnothing p=p$.
\par
Let $s$ be a suffix of $t^{m}p$. Then, $s$ is a suffix of $t^{m}p=p$. In other
words, there exists a word $g\in\mathfrak{A}^{\ast}$ satisfying $p=gs$.
Consider this $g$.
\par
We are going to prove that either $s\geq t^{m}p$ or $\left(  \text{the word
}s\text{ is a prefix of }t^{m}p\right)  $. If the word $s$ is empty, then this
is obvious (because if the word $s$ is empty, then the word $s$ is a prefix of
$t^{m}p$). Hence, we WLOG assume that the word $s$ is nonempty. If the word
$g$ is empty, then it is also clear that either $s\geq t^{m}p$ or $\left(
\text{the word }s\text{ is a prefix of }t^{m}p\right)  $ (because if the word
$g$ is empty, then $g=\varnothing$ and thus $t^{m}p=p=\underbrace{g}%
_{=\varnothing}s=s$, so that the word $s$ is a prefix of $t^{m}p$). Hence, we
WLOG assume that the word $g$ is nonempty.
\par
But the word $p$ is a prefix of a Lyndon word in $\mathfrak{A}^{\ast}$ (since
$p$ is a prefix of $t$, and since $t$ is a Lyndon word in $\mathfrak{A}^{\ast
}$). In other words, Assertion $\mathcal{F}^{\prime}$ with $w$ replaced by $p$
is satisfied. Hence, Assertion $\mathcal{B}^{\prime}$ with $w$ replaced by $p$
is satisfied as well (since we have already proven the implication
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}^{\prime}$). In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }u\text{ and }v\text{ are nonempty words satisfying }p=uv\text{,
then}\\
\text{either we have }v\geq p\text{ or the word }v\text{ is a prefix of }p
\end{array}
\right)  .
\label{sol.words.poset.g.HB.base}
\end{equation}
Since the words $g$ and $s$ are nonempty, we can apply
(\ref{sol.words.poset.g.HB.base}) to $u=g$ and $v=s$. As a result, we obtain
that either we have $s\geq p$ or the word $s$ is a prefix of $p$. In other
words, either $s\geq p$ or $\left(  \text{the word }s\text{ is a prefix of
}p\right)  $. In other words, either $s\geq t^{m}p$ or $\left(  \text{the word
}s\text{ is a prefix of }t^{m}p\right)  $ (since $t^{m}p=p$). This proves
(\ref{sol.words.poset.g.HB.1}).}. This completes the induction base.

\textit{Induction step:} Let $M$ be a positive integer. Assume that
(\ref{sol.words.poset.g.HB.1}) is proven for $m=M-1$. We will now show that
(\ref{sol.words.poset.g.HB.1}) holds for $m=M$.

Let $r$ denote the word $t^{M-1}p$. It is easy to see that $r$ is a prefix of
$t^{M}p$\ \ \ \ \footnote{\textit{Proof.} There exists a word $q\in
\mathfrak{A}^{\ast}$ such that $t=pq$ (since $p$ is a prefix of $t$). Consider
this $q$. We have $\underbrace{t^{M}}_{=t^{M-1}t}p=t^{M-1}\underbrace{t}%
_{=pq}p=\underbrace{t^{M-1}p}_{=r}qp=rqp=r\left(  qp\right)  $. Hence, $r$ is
a prefix of $t^{M}p$, qed.}. In other words, there exists a word
$g\in\mathfrak{A}^{\ast}$ such that $t^{M}p=rg$. Consider this $g$.

Let $s$ be a suffix of $t^{M}p$. We shall show that either $s\geq t^{M}p$ or
$\left(  \text{the word }s\text{ is a prefix of }t^{M}p\right)  $.

In order to prove this, let us assume the contrary (for the sake of
contradiction). Then, neither $s\geq t^{M}p$ nor $\left(  \text{the word
}s\text{ is a prefix of }t^{M}p\right)  $. In other words, we don't have
$s\geq t^{M}p$, and the word $s$ is not a prefix of $t^{M}p$. If the word $s$
was a prefix of $r$, then the word $s$ would be a prefix of $t^{M}p$ (since
$r$ is a prefix of $t^{M}p$), which would contradict the fact that the word
$s$ is not a prefix of $t^{M}p$. Hence, the word $s$ cannot be a prefix of
$r$. In other words, the word $s$ cannot be a prefix of $t^{M-1}p$ (since
$r=t^{M-1}p$).

We don't have $s\geq t^{M}p$. Since $s=s\varnothing$ and $t^{M}p=rg$, this
rewrites as follows: We don't have $s\varnothing\geq rg$. In other words, we
don't have $rg\leq s\varnothing$.

The word $s$ is a suffix of $\underbrace{t^{M}}_{=tt^{M-1}}%
p=t\underbrace{t^{M-1}p}_{=r}=tr$. Therefore, we must be in one of the
following two cases (depending on whether this suffix begins before the suffix
$r$ of $tr$ begins or afterwards):

\textit{Case 1:} The word $s$ is a suffix of $r$. (Note that $s=r$ is allowed.)

\textit{Case 2:} The word $s$ has the form $s^{\prime}r$ where $s^{\prime}$ is
a nonempty suffix of $t$.

Let us consider Case 1 first. In this case, the word $s$ is a suffix of $r$.
In other words, the word $s$ is a suffix of $t^{M-1}p$ (since $r=t^{M-1}p$).
Hence, (\ref{sol.words.poset.g.HB.1}) (applied to $m=M-1$) yields that either
$s\geq t^{M-1}p$ or $\left(  \text{the word }s\text{ is a prefix of }%
t^{M-1}p\right)  $ (since (\ref{sol.words.poset.g.HB.1}) holds for $m=M-1$).
Since the word $s$ cannot be a prefix of $t^{M-1}p$, we thus must have $s\geq
t^{M-1}p$. Thus, $t^{M-1}p\leq s$, so that $r=t^{M-1}p\leq s$. Therefore,
Proposition \ref{prop.words.lex}(d) (applied to $r$, $g$, $s$ and
$\varnothing$ instead of $a$, $b$, $c$ and $d$) yields that either we have
$rg\leq s\varnothing$ or the word $r$ is a prefix of $s$. Thus, the word $r$
is a prefix of $s$ (since we don't have $rg\leq s\varnothing$). But
$\ell\left(  r\right)  \geq\ell\left(  s\right)  $ (since $s$ is a suffix of
$r$). Hence, $r$ is a prefix of $s$ which is at least as long as $s$ itself.
Consequently, $r=s$. Hence, $s=r$, so that $s$ is a prefix of $s=r$. This
contradicts the fact that the word $s$ cannot be a prefix of $r$. Thus, we
have found a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $s$ has the form
$s^{\prime}r$ where $s^{\prime}$ is a nonempty suffix of $t$. Consider this
$s^{\prime}$. Corollary \ref{cor.words.lyndon.trivia} (applied to $t$ and
$s^{\prime}$ instead of $w$ and $v$) yields $s^{\prime}\geq t$. That is,
$t\leq s^{\prime}$. But $s^{\prime}$ is a suffix of $t$, so that $\ell\left(
s^{\prime}\right)  \leq\ell\left(  t\right)  $. Hence, $\ell\left(  t\right)
\geq\ell\left(  s^{\prime}\right)  $. Thus, Proposition \ref{prop.words.lex}%
(j) (applied to $t$, $s^{\prime}$ and $r$ instead of $a$, $b$ and $c$) yields
$tr\leq s^{\prime}r$. Hence, $s^{\prime}r\geq tr$, so that $s=s^{\prime}r\geq
tr=t^{M}p$. This contradicts the fact that we don't have $s\geq t^{M}p$. Thus,
we have found a contradiction in Case 2.

We have thus obtained a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that we always get a
contradiction. This completes the proof that either $s\geq t^{M}p$ or $\left(
\text{the word }s\text{ is a prefix of }t^{M}p\right)  $.

Now, forget that we fixed $s$. We thus have shown that every suffix $s$ of
$t^{M}p$ satisfies either $s\geq t^{M}p$ or $\left(  \text{the word }s\text{
is a prefix of }t^{M}p\right)  $. In other words,
(\ref{sol.words.poset.g.HB.1}) holds for $m=M$. This completes the induction
step, and thus (\ref{sol.words.poset.g.HB.1}) is proven by induction.

Now, let $u$ and $v$ be nonempty words satisfying $w=uv$. Then, $v$ is a
suffix of $w=t^{\ell}p$. Hence, (\ref{sol.words.poset.g.HB.1}) (applied to
$m=\ell$ and $s=v$) yields that either $v\geq t^{\ell}p$ or $\left(  \text{the
word }v\text{ is a prefix of }t^{\ell}p\right)  $. In other words, either
$v\geq w$ or $\left(  \text{the word }v\text{ is a prefix of }w\right)  $
(since $w=t^{\ell}p$). In other words, either we have $v\geq w$ or the word
$v$ is a prefix of $w$.

Now, forget that we fixed $u$ and $v$. We thus have shown that if $u$ and $v$
are nonempty words satisfying $w=uv$, then either we have $v\geq w$ or the
word $v$ is a prefix of $w$. In other words, Assertion $\mathcal{B}^{\prime}$
holds. Thus, the implication $\mathcal{H}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ is proven.

We have thus proven the implications $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{E}^{\prime}$, $\mathcal{C}^{\prime}\Longrightarrow\mathcal{E}%
^{\prime}$, $\mathcal{G}^{\prime}\Longrightarrow\mathcal{H}^{\prime}$,
$\mathcal{E}^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$,
$\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$,
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{C}^{\prime}$, $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$ and $\mathcal{H}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$. Combined, these yield the equivalence
$\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime}%
\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime}\Longleftrightarrow
\mathcal{H}^{\prime}$. This solves Exercise \ref{exe.words.powerlyndon}(b) in
the partial-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(c) in the
partial-order setting.} The implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ has already been proven in our solution of Exercise
\ref{exe.words.powerlyndon}(b). Hence, Exercise \ref{exe.words.powerlyndon}(c)
is solved in the partial-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(d) in the
partial-order setting.} The solution of Exercise \ref{exe.words.powerlyndon}%
(d) in the partial-order setting proceeds precisely as in the total-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(e) in the
partial-order setting.} The solution of Exercise \ref{exe.words.powerlyndon}%
(e) in the partial-order setting proceeds precisely as in the total-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(f) in the
partial-order setting.} Assume that there exists a letter $\mu\in\mathfrak{A}$
such that $\left(  \mu>a\text{ for some letter }a\text{ of }w\right)  $.
Consider this $\mu$. We need to prove that the equivalence $\mathcal{F}%
^{\prime}\Longleftrightarrow\mathcal{F}^{\prime\prime}$ holds.

Combining the implication $\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}%
^{\prime}$ (which has already been proven) and the implication $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$ (which follows from the
equivalence $\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime
}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime}\Longleftrightarrow
\mathcal{H}^{\prime}$ proven above), we obtain the implication $\mathcal{F}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$. Thus, in order to prove
the equivalence $\mathcal{F}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}$, it is enough to verify the implication $\mathcal{F}%
^{\prime\prime}\Longrightarrow\mathcal{F}^{\prime}$. Let us do this now.

Assume that Assertion $\mathcal{F}^{\prime\prime}$ holds. Due to the
implication $\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{G}^{\prime}$
(which follows from the equivalence $\mathcal{B}^{\prime}\Longleftrightarrow
\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime
}\Longleftrightarrow\mathcal{H}^{\prime}$ proven above), this yields that
Assertion $\mathcal{G}^{\prime}$ holds. In other words, there exists a Lyndon
word $t\in\mathfrak{A}^{\ast}$, a positive integer $\ell$ and a prefix $p$ of
$t$ (possibly empty) such that $w=t^{\ell}p$. Consider these $t$, $\ell$ and
$p$.

We shall first prove that
\begin{equation}
\text{there exists a letter }z\in\mathfrak{A}\text{ such that }%
z>t.
\label{sol.words.poset.g.f.z}
\end{equation}


\textit{Proof of (\ref{sol.words.poset.g.f.z}):} The word $t$ is Lyndon and
thus nonempty. Hence, $\ell\left(  t\right)  \geq1$. We thus are in one of the
following two cases:

\textit{Case 1:} We have $\ell\left(  t\right)  =1$.

\textit{Case 2:} We have $\ell\left(  t\right)  >1$.

Let us consider Case 1 first. In this case, we have $\ell\left(  t\right)
=1$. Thus, $t$ is a one-letter word. In other words, $t=b$ for some letter
$b$. Let us consider this $b$.

Recall that $\mu>a$ for some letter $a$ of $w$. Consider this letter $a$.

The word $t^\ell p$ is a prefix of $t^\ell t$ (since $p$ is a prefix of
$t$). In other words, the word $w$ is a prefix of $t^{\ell+1}$ (since
$w = t^\ell p$ and $t^{\ell+1} = t^\ell t$). Hence, each letter of $w$
is a letter of $t^{\ell+1}$. Since each letter of $t^{\ell+1}$ is a
letter of $t$, this shows that each letter of $w$ is a letter of $t$.
Applying this to the letter $a$, we conclude that $a$ is a letter of $t$
(since $a$ is a letter of $w$).

Now, both
$a$ and $b$ are letters of $t$. Since the word $t$ has only one letter
(because $\ell\left(  t\right)  =1$), this yields that $a=b$. Hence,
$\mu>a=b=t$ (since $t=b$). Hence, there exists a letter $z\in\mathfrak{A}$
such that $z>t$ (namely, $z=\mu$). Thus, (\ref{sol.words.poset.g.f.z}) is
proven in Case 1.

Let us now consider Case 2. In this case, we have $\ell\left(  t\right)  >1$.
Let $g$ be the last letter of the word $t$ (this is well-defined since $t$ is
nonempty). Then, the one-letter word $g$ is a suffix of the word $t$, and
therefore there exists a word $t^{\prime}\in\mathfrak{A}^{\ast}$ such that
$t=t^{\prime}g$. Consider this $t^{\prime}$. Since $g$ is a one-letter word,
we have $\ell\left(  g\right)  =1$. Thus, $\ell\left(  \underbrace{t}%
_{=t^{\prime}g}\right)  =\ell\left(  t^{\prime}\right)  +\underbrace{\ell
\left(  g\right)  }_{=1}=\ell\left(  t^{\prime}\right)  +1$, so that
$\ell\left(  t^{\prime}\right)  =\underbrace{\ell\left(  t\right)  }%
_{>1}-1>1-1=0$. Hence, the word $t^{\prime}$ is nonempty. Thus, $g$ is a
proper suffix of $t$ (since $t=t^{\prime}g$). Also, $g$ is nonempty (since
$\ell\left(  g\right)  =1$). But recall that the word $t$ is Lyndon. By the
definition of a Lyndon word, this yields that every nonempty proper suffix $v$
of $t$ satisfies $v>t$. Applying this to $v=g$, we obtain $g>t$. Thus, there
exists a letter $z\in\mathfrak{A}$ such that $z>t$ (namely, $z=g$). Thus,
(\ref{sol.words.poset.g.f.z}) is proven in Case 2.

We have thus proven (\ref{sol.words.poset.g.f.z}) in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
(\ref{sol.words.poset.g.f.z}) always holds.

Now, (\ref{sol.words.poset.g.f.z}) shows that there exists a letter
$z\in\mathfrak{A}$ such that $z>t$. Consider this letter $z$. We have $z>t$,
thus $t<z$, hence $t\leq z$. The one-letter word $z$ is Lyndon (since every
one-letter word is Lyndon). Thus, $\underbrace{t\text{, }t\text{, }%
\ldots\text{, }t}_{\ell+1\text{ times }t}$, $z$ are $\ell+2$ Lyndon words
(since both $t$ and $z$ are Lyndon) satisfying $t\leq t\leq\cdots\leq t\leq z$
(since $t\leq z$) and $t<z$. Hence, Exercise \ref{exe.words.lyndon.concatn}
(applied to $\ell+2$ and $\left(
\underbrace{t,t,\ldots,t}_{\ell+1\text{ times }t},z\right)  $
instead of $n$ and $\left(  w_{1},w_{2},\ldots,w_{n}\right)$) yields that
$\underbrace{tt\cdots t}_{\ell+1\text{ times }t}z$ is a Lyndon word. In other
words, $t^{\ell+1}z$ is a Lyndon word (since $\underbrace{tt\cdots t}%
_{\ell+1\text{ times }t}=t^{\ell+1}$).

But $p$ is a prefix of $t$. In other words, there exists a $q\in
\mathfrak{A}^{\ast}$ such that $t=pq$. Consider this $q$. We then have
$\underbrace{t^{\ell+1}}_{=t^{\ell}t}z=t^{\ell}\underbrace{t}_{=pq}%
z=\underbrace{t^{\ell}p}_{=w}qz=wqz=w\left(  qz\right)  $. Hence, $w$ is a
prefix of the word $t^{\ell+1}z$. Thus, $w$ is a prefix of a Lyndon word in
$\mathfrak{A}^{\ast}$ (because $t^{\ell+1}z$ is a Lyndon word in
$\mathfrak{A}^{\ast}$). In other words, Assertion $\mathcal{F}^{\prime}$
holds. This proves the implication $\mathcal{F}^{\prime\prime}\Longrightarrow
\mathcal{F}^{\prime}$. This solves Exercise \ref{exe.words.powerlyndon}(f) in
the partial-order case.

\textit{Remark:} Of course, for a letter $\mu\in\mathfrak{A}$, if we have
$\left(  \mu>a\text{ for every letter }a\text{ of }w\right)  $, then we also
have $\left(  \mu>a\text{ for some letter }a\text{ of }w\right)  $ (since $w$
is nonempty and thus has at least one letter). Hence, Exercise
\ref{exe.words.powerlyndon}(e) is a particular case of Exercise
\ref{exe.words.powerlyndon}(f).

Altogether, we have now solved all parts of Exercise
\ref{exe.words.powerlyndon} in the partial-order case. Exercise
\ref{exe.words.poset}(g) is thus solved.

[\textit{Remark:} The validity of some parts of Exercise
\ref{exe.words.powerlyndon} in the partial-order case can also be deduced from
their validity in the total-order case using Proposition
\ref{prop.sol.words.poset.reduce-to-linex}. For example, the equivalence
$\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime}%
\Longleftrightarrow\mathcal{E}^{\prime}$ can be treated this way. We shall not
give any details of this alternative approach, however.]
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.poset}}

\textit{Solution to Exercise \ref{exe.words.poset}.}

(a) If we replace the word ``total'' by
``partial'' throughout the proof of
Proposition \ref{prop.words.lex}, then the resulting argument can be used in
the partial-order setting to prove Proposition \ref{prop.words.lex} with
``a total order'' replaced by
``a partial order''. Hence, Proposition
\ref{prop.words.lex} holds in the partial-order setting, as long as one
replaces ``a total order'' by
``a partial order'' in part (a) of this
Proposition. This solves Exercise \ref{exe.words.poset}(a).

(b) Let us work in the partial-order setting. Let $a,b,c,d\in\mathfrak{A}%
^{\ast}$ be four words such that the words $ab$ and $cd$ are comparable. We
need to prove that the words $a$ and $c$ are comparable.

The words $ab$ and $cd$ are comparable. In other words, we have either $ab\leq
cd$ or $ab\geq cd$. In other words, we have either $ab\leq cd$ or $cd\leq ab$.
We can WLOG assume that $ab\leq cd$ (since otherwise, we can achieve $ab\leq
cd$ by switching $a$ and $b$ with $c$ and $d$, respectively). Assume this. We
know (from Exercise \ref{exe.words.poset}(a)) that Proposition
\ref{prop.words.lex} holds in the partial-order setting, as long as one
replaces ``a total order'' by ``a partial order'' in part (a) of this
Proposition. In particular, Proposition \ref{prop.words.lex}(e) holds in the
partial-order setting. Applying Proposition \ref{prop.words.lex}(e), we thus
conclude that either we have $a\leq c$ or the word $c$ is a prefix of $a$.
Thus, either we have $a\leq c$ or we have $c\leq a$ (because if $c$ is a
prefix of $a$, then $c\leq a$). In other words, either we have $a\leq c$ or we
have $a\geq c$. In other words, the words $a$ and $c$ are comparable. This
solves Exercise \ref{exe.words.poset}(b).

(c) Recall (from Exercise \ref{exe.words.poset}(a)) that Proposition
\ref{prop.words.lex} holds in the partial-order setting, as long as one
replaces ``a total order'' by
``a partial order'' in part (a) of this Proposition.

Proposition \ref{prop.words.uv=vu.2} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Proposition \ref{prop.words.uv=vu.3} holds in the partial-order setting,
because its proof applies verbatim in this setting.

The proof of Corollary \ref{cor.words.uv=vu.3} given above is no longer
applicable in the partial-order setting. However, the alternative proof of
Corollary \ref{cor.words.uv=vu.3} given in the solution to Exercise
\ref{exe.words.uw-wu} does apply verbatim in the partial-order setting. Thus,
Corollary \ref{cor.words.uv=vu.3} holds in the partial-order setting.

Corollary \ref{cor.words.uv=vu.partord} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Exercise \ref{exe.words.u=pn=qm} and Exercise \ref{exe.words.unvm.vmun} hold
in the partial-order setting, because their solutions apply verbatim in this setting.

Exercise \ref{exe.words.un.vm} and Exercise \ref{exe.words.uv=vu.k} hold in
the partial-order setting, because their solutions apply verbatim in this setting.

Proposition \ref{prop.words.lyndon.trivia} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Corollary \ref{cor.words.lyndon.trivia} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Proposition \ref{prop.words.lyndon.concat} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Corollary \ref{cor.words.lyndon.uv=vu} holds in the partial-order setting,
because its proof applies verbatim in this setting.

Our proof of Proposition \ref{prop.words.lyndon.preorder} does not directly
apply in the partial-order setting; however, it can be tweaked so that it does:

\begin{proof}
[Proof of Proposition \ref{prop.words.lyndon.preorder} in the partial-order
setting.]If the words $u$ and $v$ are incomparable, then Proposition
\ref{prop.words.lyndon.preorder} is easily seen to
hold\footnote{\textit{Proof.} Assume that $u$ and $v$ are incomparable. If the
words $uv$ and $vu$ were comparable, then the words $u$ and $v$ would also be
comparable (by Exercise \ref{exe.words.poset}(b), applied to $a=u$, $b=v$,
$c=v$ and $d=u$), which would contradict the fact that $u$ and $v$ are
incomparable. Hence, the words $uv$ and $vu$ are incomparable. Thus, we cannot
have $uv\geq vu$. But we cannot have $u\geq v$ either (since $u$ and $v$ are
incomparable). Thus, neither $u\geq v$ nor $uv\geq vu$ holds. Hence, $u\geq v$
if and only if $uv\geq vu$; therefore, Proposition
\ref{prop.words.lyndon.preorder} holds, qed.}. Hence, for the rest of this
proof, we can WLOG assume that the words $u$ and $v$ are comparable. Assume this.

The words $u$ and $v$ are comparable. In other words, we have either $u<v$ or
$u=v$ or $u>v$. From here, we can proceed as in our proof of Proposition
\ref{prop.words.lyndon.preorder} in the total-order setting. Proposition
\ref{prop.words.lyndon.preorder} is thus proven in the partial-order setting.
\end{proof}

Exercise \ref{exe.words.lyndon.equiv2}(a) holds in the partial-order setting,
because its solution applies verbatim in this setting.

The proof of Theorem \ref{thm.words.lyndon.equiv} we gave (using Proposition
\ref{prop.words.lyndon.cutoff}) does not work in the partial-order
setting\footnote{One might try tweaking Proposition
\ref{prop.words.lyndon.cutoff} for the partial-order setting by replacing
``the (lexicographically) smallest nonempty
suffix'' by ``a nonempty suffix which is
(lexicographically) minimal among the nonempty suffices'', and
by replacing ``$u\geq v$ and $uv\geq vu$'' by
``neither $u<v$ nor $uv<vu$''. But this still
would not hold. For example, if $w$ is the word $XX12$ over the partially
ordered alphabet $\left\{  X,1,2\right\}  $ with relation $1<2$, then the
nonempty suffix $X12$ is lexicographically minimal among such suffixes, but we
do have $X<X12$. (At least $uv<vu$ does not hold indeed.)}. However, the
alternative proof of Theorem \ref{thm.words.lyndon.equiv} given in Exercise
\ref{exe.words.lyndon.equiv2}(b) does apply verbatim in the partial-order
setting. Thus, Theorem \ref{thm.words.lyndon.equiv} holds in the partial-order setting.

Exercise \ref{exe.words.lyndon.concatn} and Exercise
\ref{exe.words.lyndon.concatn-lyndon} hold in the partial-order setting,
because their solutions apply verbatim in this setting.

Exercise \ref{exe.words.lyndon.std2}(a) and Exercise
\ref{exe.words.lyndon.std2}(b) hold in the partial-order setting, because
their solutions apply verbatim in this setting.

Exercise \ref{exe.words.poset}(c) is thus solved.

[\textit{Remark:} In the above solution of Exercise \ref{exe.words.poset}(c),
we have used the results of Exercise \ref{exe.words.lyndon.equiv2}(b) and
Exercise \ref{exe.words.uw-wu}. This is the main reason why the latter two
exercises have been written. However, there is a way to avoid them and still
prove that Theorem \ref{thm.words.lyndon.equiv} and Corollary
\ref{cor.words.uv=vu.3} hold in the partial-order setting. This uses a trick,
which we shall now explain.

First, a definition:

\begin{definition}
Let $P$ be a poset.

\begin{enumerate}
\item[(a)] A poset $Q$ is said to be an
\emph{extension}\index{extension of a poset}
of $P$ if and only if the following two statements hold:

\begin{itemize}
\item We have $Q=P$ as sets.

\item Any two elements $a$ and $b$ of $P$ satisfying $a\leq b$ in $P$ satisfy
$a\leq b$ in $Q$.
\end{itemize}

\item[(b)] An extension $Q$ of $P$ is called a
\emph{linear extension}\index{linear extension of a poset} of
$P$ if and only if the poset $Q$ is totally ordered.\footnote{This notion of a
linear extension is identical to the one used in
Theorem~\ref{Stanley's-P-partition-theorem}, except that we don't require $P$
to be finite here.}
\end{enumerate}
\end{definition}

The following fact about extensions of posets is well-known:

\begin{proposition}
\label{prop.sol.words.poset.linex}Let $P$ be a finite poset.

\begin{enumerate}
\item[(a)] If $a$ and $b$ are two incomparable elements of $P$, then there
exists an extension $Q$ of $P$ such that we have $a<b$ in $Q$.

\item[(b)] If $a$ and $b$ are two elements of $P$ which don't satisfy $a\geq
b$, then there exists an extension $Q$ of $P$ such that we have $a<b$ in $Q$.

\item[(c)] There exists a linear extension of $P$.

\item[(d)] If $a$ and $b$ are two elements of $P$ which don't satisfy $a\geq
b$, then there exists a linear extension $Q$ of $P$ such that we have $a<b$ in
$Q$.
\end{enumerate}
\end{proposition}

(Actually, the requirement that $P$ be finite in Proposition
\ref{prop.sol.words.poset.linex} can be dropped if you accept Zorn's lemma,
but we do not need this generality.)

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.linex}.](a) Proposition
\ref{prop.sol.words.poset.linex}(a) is a well-known fact (and was used in the
proof of Theorem \ref{Stanley's-P-partition-theorem}). Let us nevertheless
give a proof:

Let $a$ and $b$ be two incomparable elements of $P$. Let $\leq_{P}$ denote the
smaller-or-equal relation of the poset $P$. Define a new binary relation
$\leq_{P}^{\prime}$ on the set $P$ by saying that two elements $u$ and $v$ of
$P$ satisfy $u\leq_{P}^{\prime}v$ if and only if we have%
\[
\left(  \text{either }u\leq_{P}v\text{ or }\left(  u,v\right)  =\left(
a,b\right)  \right)  .
\]
Then, the relation $\leq_{P}^{\prime}$ is weaker than the relation $\leq_{P}$
(in the sense that every $\left(  u,v\right)  \in P^{2}$ satisfying $u\leq
_{P}v$ satisfies $u\leq_{P}^{\prime}v$ as well)\footnote{\textit{Proof:} Let
$\left(  u,v\right)  \in P^{2}$ be such that $u\leq_{P}v$. Then, $\left(
\text{either }u\leq_{P}v\text{ or }\left(  u,v\right)  =\left(  a,b\right)
\right)  $ (since $u\leq_{P}v$). In other words, $u\leq_{P}^{\prime}v$ (by the
definition of $\leq_{P}^{\prime}$), qed.} and thus reflexive (since $\leq_{P}$
is reflexive). Let $\leq_{P}^{\prime\prime}$ denote the
reflexive-and-transitive closure\footnote{We could have just as well taken the
transitive closure, but the reflexive-and-transitive closure seems like a more
sensible notion to use.} of the relation $\leq_{P}^{\prime}$. Then, $\leq
_{P}^{\prime\prime}$ is clearly a reflexive and transitive relation on the set
$P$. If we can show that this relation $\leq_{P}^{\prime\prime}$ is also
antisymmetric, then it will follow that this relation $\leq_{P}^{\prime\prime
}$ is a partial order on the set $P$.

Before we prove that the relation $\leq_{P}^{\prime\prime}$ is antisymmetric,
let us first show that if $u$ and $v$ are two elements of $P$ satisfying
$u\leq_{P}^{\prime\prime}v$, then%
\begin{equation}
\left(  \text{either }u\leq_{P}v\text{ or }\left(  u\leq_{P}a\text{ and }%
b\leq_{P}v\right)  \right)  \label{pf.prop.sol.words.poset.linex.0}%
\end{equation}
(where the word ``either'' is used
non-exclusively).\footnote{\textit{Proof of
(\ref{pf.prop.sol.words.poset.linex.0}):} Let $u$ and $v$ be two elements of
$P$ satisfying $u\leq_{P}^{\prime\prime}v$.
\par
Recall that the relation $\leq_{P}^{\prime\prime}$ is the
reflexive-and-transitive closure of $\leq_{P}^{\prime}$. Thus, if $\alpha$ and
$\beta$ are two elements of $P$ satisfying $\alpha\leq_{P}^{\prime\prime}%
\beta$, then there exists a sequence $\left(  \gamma_{0},\gamma_{1}%
,\ldots,\gamma_{\ell}\right)  $ of elements of $P$ such that $\gamma
_{0}=\alpha$ and $\gamma_{\ell}=\beta$ and such that every $i\in\left\{
1,2,\ldots,\ell\right\}  $ satisfies $\gamma_{i-1}\leq_{P}^{\prime}\gamma_{i}%
$. Applying this to $\alpha=u$ and $\beta=v$, we conclude that there exists a
sequence $\left(  \gamma_{0},\gamma_{1},\ldots,\gamma_{\ell}\right)  $ of
elements of $P$ such that $\gamma_{0}=u$ and $\gamma_{\ell}=v$ and such that
every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies $\gamma_{i-1}\leq
_{P}^{\prime}\gamma_{i}$. Consider this sequence $\left(  \gamma_{0}%
,\gamma_{1},\ldots,\gamma_{\ell}\right)  $. We shall prove that every
$k\in\left\{  0,1,\ldots,\ell\right\}  $ satisfies%
\begin{equation}
\left(  \text{either }u\leq_{P}\gamma_{k}\text{ or }\left(  u\leq_{P}a\text{
and }b\leq_{P}\gamma_{k}\right)  \right)
.\label{pf.prop.sol.words.poset.linex.0.pf.1}%
\end{equation}
\par
\textit{Proof of (\ref{pf.prop.sol.words.poset.linex.0.pf.1}):} We will prove
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) by induction over $k$:
\par
\textit{Induction base:} We have $\gamma_{0}=u$, thus $u=\gamma_{0}$, hence
$u\leq_{P}\gamma_{0}$. Hence, either $u\leq_{P}\gamma_{0}$ or $\left(
u\leq_{P}a\text{ and }b\leq_{P}\gamma_{0}\right)  $. In other words,
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=0$. This completes
the induction base.
\par
\textit{Induction step:} Let $K\in\left\{  0,1,\ldots,\ell-1\right\}  $.
Assume that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K$. We
need to prove that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for
$k=K+1$.
\par
Recall that every $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies
$\gamma_{i-1}\leq_{P}^{\prime}\gamma_{i}$. Applied to $i=K+1$, this yields
$\gamma_{\left(  K+1\right)  -1}\leq_{P}^{\prime}\gamma_{K+1}$. Since $\left(
K+1\right)  -1=K$, this rewrites as $\gamma_{K}\leq_{P}^{\prime}\gamma_{K+1}$.
According to the definition of $\leq_{P}^{\prime}$, this is equivalent to
saying that either $\gamma_{K}\leq_{P}\gamma_{K+1}$ or $\left(  \gamma
_{K},\gamma_{K+1}\right)  =\left(  a,b\right)  $. Thus, we see that either
$\gamma_{K}\leq_{P}\gamma_{K+1}$ or $\left(  \gamma_{K},\gamma_{K+1}\right)
=\left(  a,b\right)  $.
\par
We have assumed that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for
$k=K$. In other words, either $u\leq_{P}\gamma_{K}$ or $\left(  u\leq
_{P}a\text{ and }b\leq_{P}\gamma_{K}\right)  $. Thus, we must be in one of the
following two cases:
\par
\textit{Case 1:} We have $u\leq_{P}\gamma_{K}$.
\par
\textit{Case 2:} We have $u\leq_{P}a$ and $b\leq_{P}\gamma_{K}$.
\par
Let us first consider Case 1. In this case, we have $u\leq_{P}\gamma_{K}$.
Recall that we have either $\gamma_{K}\leq_{P}\gamma_{K+1}$ or $\left(
\gamma_{K},\gamma_{K+1}\right)  =\left(  a,b\right)  $. Thus, we must be in
one of the following two subcases:
\par
\textit{Subcase 1.1:} We have $\gamma_{K}\leq_{P}\gamma_{K+1}$.
\par
\textit{Subcase 1.2:} We have $\left(  \gamma_{K},\gamma_{K+1}\right)
=\left(  a,b\right)  $.
\par
Let us consider Subcase 1.1 first. In this subcase, we have $\gamma_{K}%
\leq_{P}\gamma_{K+1}$. Thus, $u\leq_{P}\gamma_{K}\leq_{P}\gamma_{K+1}$. Hence,
$u\leq_{P}\gamma_{K+1}$. Thus, either $u\leq_{P}\gamma_{K+1}$ or $\left(
u\leq_{P}a\text{ and }b\leq_{P}\gamma_{K+1}\right)  $. In other words,
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$. We have thus
shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$ in
Subcase 1.1.
\par
Let us now consider Subcase 1.2. In this subcase, we have $\left(  \gamma
_{K},\gamma_{K+1}\right)  =\left(  a,b\right)  $. Thus, $\gamma_{K}=a$ and
$\gamma_{K+1}=b$. We have $u\leq_{P}\gamma_{K}=a$ and $b=\gamma_{K+1}\leq
_{P}\gamma_{K+1}$. Hence, $\left(  u\leq_{P}a\text{ and }b\leq_{P}\gamma
_{K+1}\right)  $. Thus, either $u\leq_{P}\gamma_{K+1}$ or $\left(  u\leq
_{P}a\text{ and }b\leq_{P}\gamma_{K+1}\right)  $. In other words,
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$. We have thus
shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$ in
Subcase 1.2.
\par
We have now shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for
$k=K+1$ in each of the two Subcases 1.1 and 1.2. Since these two Subcases
cover all of Case 1, this yields that
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$ in Case 1.
\par
Let us now consider Case 2. In this case, we have $u\leq_{P}a$ and $b\leq
_{P}\gamma_{K}$. Recall that we have either $\gamma_{K}\leq_{P}\gamma_{K+1}$
or $\left(  \gamma_{K},\gamma_{K+1}\right)  =\left(  a,b\right)  $. Thus, we
must be in one of the following two subcases:
\par
\textit{Subcase 2.1:} We have $\gamma_{K}\leq_{P}\gamma_{K+1}$.
\par
\textit{Subcase 2.2:} We have $\left(  \gamma_{K},\gamma_{K+1}\right)
=\left(  a,b\right)  $.
\par
Let us consider Subcase 2.1 first. In this subcase, we have $\gamma_{K}%
\leq_{P}\gamma_{K+1}$. Thus, $b\leq_{P}\gamma_{K}\leq_{P}\gamma_{K+1}$. Hence,
$\left(  u\leq_{P}a\text{ and }b\leq_{P}\gamma_{K+1}\right)  $. Thus, either
$u\leq_{P}\gamma_{K+1}$ or $\left(  u\leq_{P}a\text{ and }b\leq_{P}%
\gamma_{K+1}\right)  $. In other words,
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$. We have thus
shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$ in
Subcase 2.1.
\par
Let us now consider Subcase 2.2. In this subcase, we have $\left(  \gamma
_{K},\gamma_{K+1}\right)  =\left(  a,b\right)  $. Thus, $\gamma_{K}=a$ and
$\gamma_{K+1}=b$. We have $u\leq_{P}a$ and $b=\gamma_{K+1}\leq_{P}\gamma
_{K+1}$. Hence, $\left(  u\leq_{P}a\text{ and }b\leq_{P}\gamma_{K+1}\right)
$. Thus, either $u\leq_{P}\gamma_{K+1}$ or $\left(  u\leq_{P}a\text{ and
}b\leq_{P}\gamma_{K+1}\right)  $. In other words,
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$. We have thus
shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$ in
Subcase 2.2.
\par
We have now shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for
$k=K+1$ in each of the two Subcases 2.1 and 2.2. Since these two Subcases
cover all of Case 2, this yields that
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for $k=K+1$ in Case 2.
\par
We have now shown that (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) holds for
$k=K+1$ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this yields that (\ref{pf.prop.sol.words.poset.linex.0.pf.1})
always holds.
\par
Now, (\ref{pf.prop.sol.words.poset.linex.0.pf.1}) is proven. Applying
(\ref{pf.prop.sol.words.poset.linex.0.pf.1}) to $k=\ell$, we conclude that%
\[
\left(  \text{either }u\leq_{P}\gamma_{\ell}\text{ or }\left(  u\leq
_{P}a\text{ and }b\leq_{P}\gamma_{\ell}\right)  \right)  .
\]
Since $\gamma_{\ell}=v$, this rewrites as follows:%
\[
\left(  \text{either }u\leq_{P}v\text{ or }\left(  u\leq_{P}a\text{ and }%
b\leq_{P}v\right)  \right)  .
\]
This proves (\ref{pf.prop.sol.words.poset.linex.0}).}

Let us now prove that the relation $\leq_{P}^{\prime\prime}$ is antisymmetric.
Indeed, let $u$ and $v$ be two elements of $P$ satisfying $u\leq_{P}%
^{\prime\prime}v$ and $v\leq_{P}^{\prime\prime}u$. We shall prove that $u=v$.

From (\ref{pf.prop.sol.words.poset.linex.0}), we see that either $u\leq_{P}v$
or $\left(  u\leq_{P}a\text{ and }b\leq_{P}v\right)  $. In other words, we
must be in one of the following two cases:

\textit{Case 1:} We have $u\leq_{P}v$.

\textit{Case 2:} We have $u\leq_{P}a$ and $b\leq_{P}v$.

Let us first consider Case 1. In this case, we have $u\leq_{P}v$. But we can
also apply (\ref{pf.prop.sol.words.poset.linex.0}) to $v$ and $u$ instead of
$u$ and $v$ (since $v\leq_{P}^{\prime\prime}u$). As a consequence, we see that
either $v\leq_{P}u$ or $\left(  v\leq_{P}a\text{ and }b\leq_{P}u\right)  $.
But we cannot have $\left(  v\leq_{P}a\text{ and }b\leq_{P}u\right)
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we have
$v\leq_{P}a$ and $b\leq_{P}u$. Hence, $b\leq_{P}u\leq_{P}v\leq_{P}a$, which
contradicts the fact that $a$ and $b$ are incomparable elements of $P$. This
contradiction proves that our assumption was wrong, qed.}. Hence, we must have
$v\leq_{P}u$ (since we have either $v\leq_{P}u$ or $\left(  v\leq_{P}a\text{
and }b\leq_{P}u\right)  $). Combined with $u\leq_{P}v$, this yields $u=v$.
Thus, $u=v$ holds in Case 1.

Let us now consider Case 2. In this case, we have $u\leq_{P}a$ and $b\leq
_{P}v$. But we can also apply (\ref{pf.prop.sol.words.poset.linex.0}) to $v$
and $u$ instead of $u$ and $v$ (since $v\leq_{P}^{\prime\prime}u$). As a
consequence, we see that either $v\leq_{P}u$ or $\left(  v\leq_{P}a\text{ and
}b\leq_{P}u\right)  $. But we cannot have $v\leq_{P}u$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we have
$v\leq_{P}u$. Hence, $b\leq_{P}v\leq_{P}u\leq_{P}a$, which contradicts the
fact that $a$ and $b$ are incomparable elements of $P$. This contradiction
proves that our assumption was wrong, qed.}. Hence, we must have $\left(
v\leq_{P}a\text{ and }b\leq_{P}u\right)  $ (since we have either $v\leq_{P}u$
or $\left(  v\leq_{P}a\text{ and }b\leq_{P}u\right)  $). Now, $b\leq_{P}%
u\leq_{P}a$ contradicts the fact that $a$ and $b$ are incomparable elements of
$P$. This contradiction yields that $u=v$ (since ex falso quodlibet). Hence,
$u=v$ holds in Case 2.

We have now shown that $u=v$ holds in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this yields that $u=v$ always holds.

Now, let us forget that we fixed $u$ and $v$. We thus have seen that if $u$
and $v$ are two elements of $P$ satisfying $u\leq_{P}^{\prime\prime}v$ and
$v\leq_{P}^{\prime\prime}u$, then $u=v$. In other words, the relation
$\leq_{P}^{\prime\prime}$ is antisymmetric. Since this relation is also
reflexive and transitive, this shows that the set $P$, endowed with this
relation $\leq_{P}^{\prime\prime}$ (as a smaller-or-equal relation), becomes a
poset. Let us denote this poset by $R$. Thus, $R=P$ as sets, and the
smaller-or-equal relation of the poset $R$ is $\leq_{P}^{\prime\prime}$.

The relation $\leq_{P}^{\prime\prime}$ is weaker than the relation $\leq
_{P}^{\prime}$ (because it is the reflexive-and-transitive closure of
$\leq_{P}^{\prime}$), which in turn is weaker than the relation $\leq_{P}$.
Thus, the relation $\leq_{P}^{\prime\prime}$ is weaker than the relation
$\leq_{P}$. Hence, $R$ is an extension of
$P$\ \ \ \ \footnote{\textit{Proof.} Let $u$ and $v$ be two elements of $P$
satisfying $u \leq v$ in $P$. We have $u\leq v$ in $P$. In other words,
$u \leq_P v$ (since the smaller-or-equal relation of $P$ is $\leq_P$).
Hence, $u \leq_P^{\prime\prime} v$ (since the relation
$\leq_{P}^{\prime\prime}$ is weaker than the relation $\leq_{P}$). In
other words, $u \leq v$ in $R$ (since the smaller-or-equal
relation of $R$ is $\leq_{P}^{\prime\prime}$).

Let us now forget that we fixed $u$ and $v$. We thus have proven that
any two elements $u$ and $v$ of $P$ satisfying $u \leq v$ in $P$ satisfy
$u \leq v$ in $R$. In other words, $R$ is an extension of $P$, qed.}.

Clearly, $\left(  a,b\right)  =\left(  a,b\right)  $. Thus, $\left(
\text{either }a\leq_{P}b\text{ or }\left(  a,b\right)  =\left(  a,b\right)
\right)  $. But by the definition of $\leq_{P}^{\prime}$, we have $a\leq
_{P}^{\prime}b$ if and only if $\left(  \text{either }a\leq_{P}b\text{ or
}\left(  a,b\right)  =\left(  a,b\right)  \right)  $. Since $\left(
\text{either }a\leq_{P}b\text{ or }\left(  a,b\right)  =\left(  a,b\right)
\right)  $ is true, we thus conclude that $a\leq_{P}^{\prime}b$. Hence,
$a\leq_{P}^{\prime\prime}b$ (since the relation $\leq_{P}^{\prime\prime}$ is
weaker than the relation $\leq_{P}^{\prime}$). In other words, we have $a\leq
b$ in $R$ (since the smaller-or-equal relation of $R$ is $\leq_{P}%
^{\prime\prime}$). Combined with $a\neq b$ (since the elements $a$ and $b$ of
$P$ are incomparable), this yields $a<b$ in $R$.

We have now shown that $R$ is an extension of $P$ such that we have $a<b$ in
$R$. Thus, there exists an extension $Q$ of $P$ such that we have $a<b$ in $Q$
(namely, $Q=R$). This proves Proposition \ref{prop.sol.words.poset.linex}(a).

(b) Let $a$ and $b$ be two elements of $P$ which don't satisfy $a\geq b$. We
need to prove that there exists an extension $Q$ of $P$ such that we have
$a<b$ in $Q$.

This follows immediately from Proposition \ref{prop.sol.words.poset.linex}(a)
if the elements $a$ and $b$ of $P$ are incomparable. Hence, for the rest of
this proof, we can WLOG assume that the elements $a$ and $b$ of $P$ are not
incomparable. Assume this.

The elements $a$ and $b$ of $P$ are comparable (since they are not
incomparable). In other words, either $a\leq b$ in $P$, or $a\geq b$ in $P$.
Hence, $a\leq b$ in $P$ (since we don't have $a\geq b$ in $P$). But $a\neq b$
(because otherwise, we would have $a=b$ and thus $a\geq b$ in $P$, which would
contradict the fact that we don't have $a\geq b$ in $P$). Combined with $a\leq
b$ in $P$, this yields $a<b$ in $P$. Hence, there exists an extension $Q$ of
$P$ such that we have $a<b$ in $Q$ (namely, $Q=P$). This proves Proposition
\ref{prop.sol.words.poset.linex}(b).

[For an alternative proof of Proposition \ref{prop.sol.words.poset.linex}(b),
we could simply repeat our above proof of Proposition
\ref{prop.sol.words.poset.linex}(a) with trivial modifications -- because
every time we claimed to use the incomparability of $a$ and $b$, the only
thing that we really used was the fact that we don't have $a\geq b$ in $P$.]

(c) Proposition \ref{prop.sol.words.poset.linex}(c) is a well-known fact; its
proof is easy and can be recovered from the proof of Theorem
\ref{Stanley's-P-partition-theorem}. Let us just give the main idea: We define
$\operatorname*{inc}\left(  P\right)  $ to be the number of all $\left(
u,v\right)  \in P^{2}$ such that $u$ and $v$ are incomparable elements of $P$.
We perform a strong induction over the number $\operatorname*{inc}\left(
P\right)  $: If this number is $0$, then $P$ is already a total order, and
thus $P$ itself is a linear extension of $P$. If this number is positive, then
there exist two incomparable elements $a$ and $b$ of $P$, and thus Proposition
\ref{prop.sol.words.poset.linex}(a) yields the existence of an extension $Q$
of $P$ such that we have $a<b$ in $Q$ for these two elements $a$ and $b$. This
poset $Q$ has $\operatorname*{inc}\left(  Q\right)  <\operatorname*{inc}%
\left(  P\right)  $, and so, by the induction hypothesis, there exists a
linear extension of $Q$. But such an extension must also be a linear extension
of $P$, and thus the induction step is finished. The details of this proof are
left to the reader.

(d) Let $a$ and $b$ be two elements of $P$ which don't satisfy $a\geq b$.
Proposition \ref{prop.sol.words.poset.linex}(b) yields that there exists an
extension $Q$ of $P$ such that we have $a<b$ in $Q$. Denote this extension $Q$
by $R$. Thus, $R$ is an extension of $P$ such that we have $a<b$ in $R$.

Proposition \ref{prop.sol.words.poset.linex}(c) (applied to $R$ instead of
$P$) yields that there exists a linear extension of $R$. Denote this linear
extension by $S$. Hence, $S$ is a linear extension of $R$. In other words, $S$
is an extension of $R$ and is totally ordered (according to the definition of
a linear extension).

The poset $S$ is an extension of $R$, which in turn is an extension of $P$.
Hence, $S$ is an extension of $P$. Thus, $S$ is a linear extension of $P$
(because $S$ is totally ordered).

We have $a\leq b$ in $R$ (since $a<b$ in $Q$) and thus $a\leq b$ in $S$ (since
$S$ is an extension of $R$). Since $a\neq b$ (because $a<b$ in $R$), this
becomes $a<b$ in $R$.

Now, we have shown that there exists a linear extension $Q$ of $P$ such that
we have $a<b$ in $Q$ (namely, $Q=S$). This proves Proposition
\ref{prop.sol.words.poset.linex}(d).
\end{proof}

We can now apply this all to alphabets. In the partial-order setting,
alphabets are posets, and so it makes sense to speak of an extension of an
alphabet. We notice the following fact:

\begin{proposition}
\label{prop.sol.words.poset.reduce-to-linex}Let $\mathfrak{A}$ be a finite poset.

\begin{enumerate}
\item[(a)] If $\mathfrak{B}$ is an extension of the poset $\mathfrak{A}$, then
$\mathfrak{B}^{\ast}$ is an extension of the poset $\mathfrak{A}^{\ast}$.

\item[(b)] Let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two
words. Then, $u<v$ holds in $\mathfrak{A}^{\ast}$ if and only if we have%
\[
\left(  u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]

\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.reduce-to-linex}.](a) Let
$\mathfrak{B}$ be an extension of the poset $\mathfrak{A}$. Then, by the
definition of an ``extension'', we see that:

\begin{itemize}
\item We have $\mathfrak{B}=\mathfrak{A}$ as sets.

\item Any two elements $a$ and $b$ of $\mathfrak{A}$ satisfying $a\leq b$ in
$\mathfrak{A}$ satisfy $a\leq b$ in $\mathfrak{B}$.
\end{itemize}

Now, it is clear that $\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ as sets (since
$\mathfrak{B}=\mathfrak{A}$ as sets). Also, any two elements $a$ and $b$ of
$\mathfrak{A}^{\ast}$ satisfying $a\leq b$ in $\mathfrak{A}^{\ast}$ satisfy
$a\leq b$ in $\mathfrak{B}^{\ast}$\ \ \ \ \footnote{\textit{Proof.} Let $u$
and $v$ be two elements of $\mathfrak{A}^{\ast}$ satisfying $u\leq v$ in
$\mathfrak{A}^{\ast}$. We are going to prove that $u\leq v$ in $\mathfrak{B}%
^{\ast}$.
\par
According to the definition of the relation $\leq$ on $\mathfrak{A}^{\ast}$,
we have that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{ in
}\mathfrak{A}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v
\end{align*}
(because $u\leq v$ in $\mathfrak{A}$).
In other words, we must be in one of the following two cases:
\par
\textit{Case 1:} There exists an $i\in\left\{  1,2,\ldots,\min\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  u_{i}<v_{i}\text{ in }\mathfrak{A}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $.
\par
\textit{Case 2:} The word $u$ is a prefix of $v$.
\par
Let us first consider Case 1. In this case, there exists an $i\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  u_{i}<v_{i}\text{ in }\mathfrak{A}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}%
=v_{j}\right)  $. Let $i^{\prime}$ be this $i$. Thus, $i^{\prime}\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ satisfies $\left(  u_{i^{\prime}}<v_{i^{\prime}}\text{ in
}\mathfrak{A}\text{, and every }j\in\left\{  1,2,\ldots,i^{\prime}-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $.
\par
We have $u_{i^{\prime}}<v_{i^{\prime}}$ in $\mathfrak{A}$; thus,
$u_{i^{\prime}}\leq v_{i^{\prime}}$ in $\mathfrak{A}$ and $u_{i^{\prime}}\neq
v_{i^{\prime}}$.
\par
Recall that any two elements $a$ and $b$ of $\mathfrak{A}$ satisfying $a\leq
b$ in $\mathfrak{A}$ satisfy $a\leq b$ in $\mathfrak{B}$. Applied to
$a=u_{i^{\prime}}$ and $b=v_{i^{\prime}}$, this yields that $u_{i^{\prime}%
}\leq v_{i^{\prime}}$ in $\mathfrak{B}$ (since $u_{i^{\prime}}\leq
v_{i^{\prime}}$ in $\mathfrak{A}$). Combined with $u_{i^{\prime}}\neq
v_{i^{\prime}}$, this yields $u_{i^{\prime}}<v_{i^{\prime}}$ in $\mathfrak{B}%
$. Thus, we have $\left(  u_{i^{\prime}}<v_{i^{\prime}}\text{ in }%
\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i^{\prime}-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $. Consequently, there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(
v\right)  \right\}  \right\}  $ such that $\left(  u_{i}<v_{i}\text{ in
}\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  $ (namely, $i=i^{\prime}$). Thus,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{ in
}\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
In other words, $u\leq v$ in $\mathfrak{B}^{\ast}$. Thus, $u\leq v$ in
$\mathfrak{B}^{\ast}$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, the word $u$ is a prefix of $v$.
Hence, $u\leq v$ in $\mathfrak{B}^{\ast}$. Thus, $u\leq v$ in $\mathfrak{B}%
^{\ast}$ is proven in Case 2.
\par
Now, $u\leq v$ in $\mathfrak{B}^{\ast}$ is proven in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this yields that $u\leq
v$ in $\mathfrak{B}^{\ast}$ always holds.
\par
Now, let us forget that we fixed $u$ and $v$. We have thus shown that any two
elements $u$ and $v$ of $\mathfrak{A}^{\ast}$ satisfying $u\leq v$ in
$\mathfrak{A}^{\ast}$ satisfy $u\leq v$ in $\mathfrak{B}^{\ast}$. Renaming the
variables $u$ and $v$ as $a$ and $b$ in this statement, we conclude the
following: Any two elements $a$ and $b$ of $\mathfrak{A}^{\ast}$ satisfying
$a\leq b$ in $\mathfrak{A}^{\ast}$ satisfy $a\leq b$ in $\mathfrak{B}^{\ast}$.
Qed.}. Consequently, $\mathfrak{B}^{\ast}$ is an extension of the poset
$\mathfrak{A}^{\ast}$. This proves Proposition
\ref{prop.sol.words.poset.reduce-to-linex}(a).

(b) It is easy to prove the logical implication%
\begin{equation}
\left(  u<v\text{ in }\mathfrak{A}^{\ast}\right)  \Longrightarrow\left(
u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)
\label{pf.prop.sol.words.poset.reduce-to-linex.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.words.poset.reduce-to-linex.1}):}
Assume that $u<v$ in $\mathfrak{A}^{\ast}$. Thus, $u\neq v$ and $u\leq v$ in
$\mathfrak{A}^{\ast}$. Let $\mathfrak{B}$ be a linear extension of
$\mathfrak{A}$. Then, $\mathfrak{B}^{\ast}$ is an extension of the poset
$\mathfrak{A}^{\ast}$ (by Proposition
\ref{prop.sol.words.poset.reduce-to-linex}(a)). Hence, $u\leq v$ in
$\mathfrak{B}^{\ast}$ (since $u\leq v$ in $\mathfrak{A}^{\ast}$). Combined
with $u\neq v$, this yields $u<v$ in $\mathfrak{B}^{\ast}$.
\par
Now, let us forget that we fixed $\mathfrak{B}$. We thus have proven that
$u<v$ in $\mathfrak{B}^{\ast}$ for every linear extension $\mathfrak{B}$ of
$\mathfrak{A}$. This proves (\ref{pf.prop.sol.words.poset.reduce-to-linex.1}%
).}. We shall now focus on proving the implication%
\begin{equation}
\left(  u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  \Longrightarrow\left(  u<v\text{
in }\mathfrak{A}^{\ast}\right)  .
\label{pf.prop.sol.words.poset.reduce-to-linex.2}%
\end{equation}


\textit{Proof of (\ref{pf.prop.sol.words.poset.reduce-to-linex.1}):} Assume
that%
\begin{equation}
u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}.
\label{pf.prop.sol.words.poset.reduce-to-linex.2.pf.1}%
\end{equation}


We need to show that $u<v$ in $\mathfrak{A}^{\ast}$.

First of all, it is impossible that $u\geq v$ in $\mathfrak{A}^{\ast}%
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $u\geq v$ in
$\mathfrak{A}^{\ast}$. Hence, $v\leq u$ in $\mathfrak{A}^{\ast}$.
\par
Proposition \ref{prop.sol.words.poset.linex}(c) (applied to $P=\mathfrak{A}%
^{\ast}$) yields that there exists a linear extension of $\mathfrak{A}$. Let
$\mathfrak{B}$ be such a linear extension. Then, $u<v$ in $\mathfrak{B}^{\ast
}$ (by (\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.1})). But
$\mathfrak{B}^{\ast}$ is an extension of the poset $\mathfrak{A}^{\ast}$ (by
Proposition \ref{prop.sol.words.poset.linex}(a)). Thus, $v\leq u$ in
$\mathfrak{B}^{\ast}$ (since $v\leq u$ in $\mathfrak{A}^{\ast}$). This
contradicts $u<v$ in $\mathfrak{B}^{\ast}$. This contradiction proves that our
assumption was wrong, qed.}. Consequently, we have $u\neq v$ (because
otherwise, we would have $u=v$ and thus $u\geq v$ in $\mathfrak{A}^{\ast}$,
which would contradict the fact that it is impossible that $u\geq v$ in
$\mathfrak{A}^{\ast}$).

Let $m=\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  $.
Then, $m\leq\ell\left(  u\right)  $ and $m\leq\ell\left(  v\right)  $. Recall
that $u=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $ and
$v=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $. If every
$i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}=v_{i}$, then it is easy
to see that $u<v$ in $\mathfrak{A}^{\ast}$\ \ \ \ \footnote{\textit{Proof.}
Assume that every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}%
=v_{i}$. Thus, $\left(  u_{1},u_{2},\ldots,u_{m}\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{m}\right)  $.
\par
Let us first assume (for the sake of contradiction) that $\ell\left(
u\right)  \geq\ell\left(  v\right)  $. Then, $m=\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  =\ell\left(  v\right)  $ (since
$\ell\left(  u\right)  \geq\ell\left(  v\right)  $). Now,%
\begin{align*}
v  &  =\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(
v_{1},v_{2},\ldots,v_{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  v\right)  =m\right) \\
&  =\left(  u_{1},u_{2},\ldots,u_{m}\right)  .
\end{align*}
But the word $\left(  u_{1},u_{2},\ldots,u_{m}\right)  $ is clearly a prefix
of $\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $ (since
$m\leq\ell\left(  u\right)  $). In other words, the word $v$ is a prefix of
$u$ (since $v=\left(  u_{1},u_{2},\ldots,u_{m}\right)  $ and $u=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $). Hence, $v\leq u$ in
$\mathfrak{A}^{\ast}$. In other words, $u\geq v$ in $\mathfrak{A}^{\ast}$.
This contradicts the fact that it is impossible that $u\geq v$ in
$\mathfrak{A}^{\ast}$.
\par
This contradiction proves that our assumption (that $\ell\left(  u\right)
\geq\ell\left(  v\right)  $) was wrong. Hence, we cannot have $\ell\left(
u\right)  \geq\ell\left(  v\right)  $. We thus have $\ell\left(  u\right)
<\ell\left(  v\right)  $. Thus, $m=\min\left\{  \ell\left(  u\right)
,\ell\left(  v\right)  \right\}  =\ell\left(  u\right)  $ (since $\ell\left(
u\right)  <\ell\left(  v\right)  $). Now,%
\begin{align*}
u  &  =\left(  u_{1},u_{2},\ldots,u_{\ell\left(  v\right)  }\right)  =\left(
u_{1},u_{2},\ldots,u_{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  u\right)  =m\right) \\
&  =\left(  v_{1},v_{2},\ldots,v_{m}\right)  .
\end{align*}
But the word $\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ is clearly a prefix
of $\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $ (since
$m\leq\ell\left(  v\right)  $). In other words, the word $u$ is a prefix of
$v$ (since $u=\left(  v_{1},v_{2},\ldots,v_{m}\right)  $ and $v=\left(
v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  $). Hence, $u\leq v$ in
$\mathfrak{A}^{\ast}$. Combined with $u\neq v$, this yields that $u<v$ in
$\mathfrak{A}^{\ast}$, qed.}. Hence, for the rest of the proof of
(\ref{pf.prop.sol.words.poset.reduce-to-linex.1}), we can WLOG assume that not
every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}=v_{i}$. Assume this.

Not every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies $u_{i}=v_{i}$. In
other words, there exists an $i\in\left\{  1,2,\ldots,m\right\}  $ which does
not satisfy $u_{i}=v_{i}$. Let $k$ be the smallest such $i$. Thus,
$k\in\left\{  1,2,\ldots,m\right\}  $ does not satisfy $u_{k}=v_{k}$, whereas%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,m\right\}  \text{ satisfying }i<k\text{
satisfies }u_{i}=v_{i}. \label{pf.prop.sol.words.poset.reduce-to-linex.2.pf.4}%
\end{equation}


We have $u_{k}\neq v_{k}$ (since $k$ does not satisfy $u_{k}=v_{k}$). We must
have $v_{k}\geq u_{k}$ in $\mathfrak{A}$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, we don't have $v_{k}\geq u_{k}$ in $\mathfrak{A}$.
Hence, Proposition \ref{prop.sol.words.poset.linex}(d) (applied to
$P=\mathfrak{A}^{\ast}$, $a=v_{k}$ and $b=u_{k}$) yields that there exists a
linear extension $Q$ of $\mathfrak{A}$ such that we have $v_{k}<u_{k}$ in $Q$.
Let $\mathfrak{B}$ be such a linear extension. Thus, $\mathfrak{B}$ is a
linear extension of $\mathfrak{A}$ such that we have $v_{k}<u_{k}$ in
$\mathfrak{B}$. We have $u<v$ in $\mathfrak{B}^{\ast}$ (by
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.1})).
\par
Every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $u_{j}=v_{j}$ (by
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.4}), applied to $i=j$). In
other words, every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies
$v_{j}=u_{j}$. Also, we have $k\in\left\{  1,2,\ldots,m\right\}  =\left\{
1,2,\ldots,\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)  \right\}
\right\}  $ (since $m=\min\left\{  \ell\left(  u\right)  ,\ell\left(
v\right)  \right\}  =\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)
\right\}  $). Altogether, we thus have shown that $k\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)  \right\}  \right\}
$ satisfies $\left(  v_{k}<u_{k}\text{ in }\mathfrak{B}\text{, and every }%
j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }v_{j}=u_{j}\right)  $.
Thus, there exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
v\right)  ,\ell\left(  u\right)  \right\}  \right\}  $ such that \newline
$\left(
v_{i}<u_{i}\text{ in }\mathfrak{B}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }v_{j}=u_{j}\right)  $ (namely,
$i=k$). Hence,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  v\right)  ,\ell\left(  u\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  v_{i}<u_{i}\text{ in
}\mathfrak{B}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }v_{j}=u_{j}\right)  ,\\
&  \text{\textbf{or} the word }v\text{ is a prefix of }u.
\end{align*}
In other words, $v<u$ in $\mathfrak{B}^{\ast}$ (by the definition of the
relation $<$ on $\mathfrak{B}^{\ast}$). This contradicts the fact that $u<v$
in $\mathfrak{B}^{\ast}$. This contradiction proves that our assumption was
wrong, qed.}. That is, $u_{k}\leq v_{k}$ in $\mathfrak{A}$. Combined with
$u_{k}\neq v_{k}$, this yields that $u_{k}<v_{k}$ in $\mathfrak{A}$.

Every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $u_{j}=v_{j}$ (by
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2.pf.4}), applied to $i=j$).
Also, we have $k\in\left\{  1,2,\ldots,m\right\}  =\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
$ (since $m=\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)
\right\}  $). Altogether, we thus have shown that $k\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
$ satisfies $\left(  u_{k}<v_{k}\text{ in }\mathfrak{A}\text{, and every }%
j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $.
Thus, there exists an $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that \newline
$\left(
u_{i}<v_{i}\text{ in }\mathfrak{A}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $ (namely,
$i=k$). Hence,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{ in
}\mathfrak{A}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }u_{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
In other words, $u<v$ in $\mathfrak{A}^{\ast}$ (by the definition of the
relation $<$ on $\mathfrak{A}^{\ast}$). This proves
(\ref{pf.prop.sol.words.poset.reduce-to-linex.2}).

Combining the implications (\ref{pf.prop.sol.words.poset.reduce-to-linex.1})
and (\ref{pf.prop.sol.words.poset.reduce-to-linex.2}), we obtain the following
equivalence of statements:%
\[
\left(  u<v\text{ in }\mathfrak{A}^{\ast}\right)  \Longleftrightarrow\left(
u<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
Thus, Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b) is proven.
\end{proof}

Let us now give a new proof of the fact that Corollary \ref{cor.words.uv=vu.3}
holds in the partial-order setting:

\begin{proof}
[Alternative proof of Corollary \ref{cor.words.uv=vu.3} in the partial-order
setting.]We need to prove that $uw\geq wu$. If $uw=wu$, then this is obvious.
Hence, for the rest of this proof, we can WLOG assume that $uw\neq wu$. Assume this.

Let $\mathfrak{B}$ be any linear extension of the alphabet $\mathfrak{A}$.
Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)).
Hence, $uv\geq vu$ in $\mathfrak{B}^{\ast}$ (since $uv\geq vu$ in
$\mathfrak{A}^{\ast}$) and $vw\geq wv$ in $\mathfrak{B}^{\ast}$ (since $vw\geq
wv$ in $\mathfrak{A}^{\ast}$). But the alphabet $\mathfrak{B}$ is totally
ordered, and thus Corollary \ref{cor.words.uv=vu.3} (applied to $\mathfrak{B}$
instead of $\mathfrak{A}$) yields that $uw\geq wu$ in $\mathfrak{B}^{\ast}$
(since we know that Corollary \ref{cor.words.uv=vu.3} holds in the total-order
setting). Thus, $uw>wu$ in $\mathfrak{B}^{\ast}$ (since $uw\neq wu$). In other
words, $wu<uw$ in $\mathfrak{B}^{\ast}$.

Now, let us forget that we fixed $\mathfrak{B}$. We thus have shown that
$wu<uw$ in $\mathfrak{B}^{\ast}$ for every linear extension $\mathfrak{B}$ of
$\mathfrak{A}$. But Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b)
(applied to $wu$ and $uw$ instead of $u$ and $v$) yields that $wu<uw$ holds in
$\mathfrak{A}^{\ast}$ if and only if%
\[
\left(  wu<uw\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
Thus, we conclude that $wu<uw$ holds in $\mathfrak{A}^{\ast}$ (since we
already know that $wu<uw$ in $\mathfrak{B}^{\ast}$ for every linear extension
$\mathfrak{B}$ of $\mathfrak{A}$). Thus, $uw>wu$ in $\mathfrak{A}^{\ast}$;
hence, $uw\geq wu$ in $\mathfrak{A}^{\ast}$. This proves Corollary
\ref{cor.words.uv=vu.3} in the partial-order setting.
\end{proof}

Next, let us give a new proof of the fact that Theorem
\ref{thm.words.lyndon.equiv} holds in the partial-order setting:

\begin{proof}
[Alternative proof of Theorem \ref{thm.words.lyndon.equiv} in the
partial-order setting.]We need to prove the equivalence $\mathcal{A}%
\Longleftrightarrow\mathcal{B}\Longleftrightarrow\mathcal{C}%
\Longleftrightarrow\mathcal{D}$. We can prove the implications $\mathcal{A}%
\Longrightarrow\mathcal{B}$, $\mathcal{A}\Longrightarrow\mathcal{C}$,
$\mathcal{A}\Longrightarrow\mathcal{D}$ and $\mathcal{B}\Longrightarrow
\mathcal{A}$ in the same way as we did in the total-order setting. Hence, in
order to prove Theorem \ref{thm.words.lyndon.equiv}, it will be enough to
prove the implications $\mathcal{C}\Longrightarrow\mathcal{B}$ and
$\mathcal{D}\Longrightarrow\mathcal{B}$.

\textit{Proof of the implication $\mathcal{C}\Longrightarrow\mathcal{B}%
$:} Assume that Assertion $\mathcal{C}$ holds.

Let $\mathfrak{B}$ be any linear extension of the alphabet $\mathfrak{A}$.
Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)). It
is easy to see that Assertion $\mathcal{C}$ (with $\mathfrak{A}$ replaced by
$\mathfrak{B}$) holds\footnote{\textit{Proof.} Let $u\in\mathfrak{B}^{\ast}$
and $v\in\mathfrak{B}^{\ast}$ be two nonempty words satisfying $w=uv$. We have
$u\in\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ and $v\in\mathfrak{B}^{\ast
}=\mathfrak{A}^{\ast}$, and thus $u\in\mathfrak{A}^{\ast}$ and $v\in
\mathfrak{A}^{\ast}$ are two nonempty words satisfying $w=uv$. Hence, we have
$v>u$ in $\mathfrak{A}^{\ast}$ (since we assumed that Assertion $\mathcal{C}$
holds). Thus, $u<v$ in $\mathfrak{A}^{\ast}$, so that $u\neq v$ and $u\leq v$
in $\mathfrak{A}^{\ast}$.
\par
So we have $u\leq v$ in $\mathfrak{A}^{\ast}$. Hence, $u\leq v$ in
$\mathfrak{B}^{\ast}$ (since $\mathfrak{B}^{\ast}$ is an extension of
$\mathfrak{A}^{\ast}$). Since $u\neq v$, this becomes $u<v$ in $\mathfrak{B}%
^{\ast}$. In other words, $v>u$ in $\mathfrak{B}^{\ast}$.
\par
Now, let us forget that we fixed $u$ and $v$. We thus have shown that any
nonempty words $u\in\mathfrak{B}^{\ast}$ and $v\in\mathfrak{B}^{\ast}$
satisfying $w=uv$ satisfy $v>u$ in $\mathfrak{B}^{\ast}$. In other words,
Assertion $\mathcal{C}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$)
holds}. But we can apply Theorem \ref{thm.words.lyndon.equiv} to
$\mathfrak{B}$ instead of $\mathfrak{A}$ (since $\mathfrak{B}$ is totally
ordered). As a consequence, we see that the four assertions $\mathcal{A}$,
$\mathcal{B}$, $\mathcal{C}$ and $\mathcal{D}$ (all with $\mathfrak{A}$
replaced by $\mathfrak{B}$) are equivalent. In particular, Assertion
$\mathcal{C}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$) is equivalent
to Assertion $\mathcal{B}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$).
Since Assertion $\mathcal{C}$ (with $\mathfrak{A}$ replaced by $\mathfrak{B}$)
holds, we thus conclude that Assertion $\mathcal{B}$ (with $\mathfrak{A}$
replaced by $\mathfrak{B}$) holds. In other words,%
\begin{equation}
\text{any two nonempty words }u\in\mathfrak{B}^{\ast}\text{ and }%
v\in\mathfrak{B}^{\ast}\text{ satisfying }w=uv\text{ satisfy }v>w\text{ in
}\mathfrak{B}^{\ast}. \label{pf.thm.words.lyndon.equiv.poset.CB.1}%
\end{equation}


Now, let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two
nonempty words satisfying $w=uv$. It is easy to see that%
\begin{equation}
\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)
\label{pf.thm.words.lyndon.equiv.poset.CB.2}
\end{equation}
\footnote{\textit{Proof.} Let $\mathfrak{B}$ be a linear extension of
$\mathfrak{A}$. Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)).
Thus, $\mathfrak{B}^{\ast} = \mathfrak{A}^{\ast}$ as sets, so that
$u \in \mathfrak{A}^{\ast} = \mathfrak{B}^{\ast}$ and
$v \in \mathfrak{A}^{\ast} = \mathfrak{B}^{\ast}$. Now,
\eqref{pf.thm.words.lyndon.equiv.poset.CB.1} yields $v > w$ in
$\mathfrak{B}^\ast$. In other words, $w < v$ in $\mathfrak{B}^\ast$.
This proves \eqref{pf.thm.words.lyndon.equiv.poset.CB.2}.}.

But Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b) (applied to $w$
instead of $u$) yields that $w<v$ holds in $\mathfrak{A}^{\ast}$ if and only
if we have%
\[
\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
Consequently, we conclude that $w<v$ holds in $\mathfrak{A}^{\ast}$ (since we
know that we have \newline
$\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every
linear extension }\mathfrak{B}\text{ of }\mathfrak{A}\right)  $). In other
words, $v>w$ in $\mathfrak{A}^{\ast}$.

Now, let us forget that we fixed $u$ and $v$. We have thus shown that any
nonempty words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$
satisfying $w=uv$ satisfy $v>w$ in $\mathfrak{A}^{\ast}$. In other words,
Assertion $\mathcal{B}$ holds. Thus, we have proven the implication
$\mathcal{C}\Longrightarrow\mathcal{B}$.

\textit{Proof of the implication $\mathcal{D}\Longrightarrow\mathcal{B}%
$:} The proof of the implication $\mathcal{D}\Longrightarrow
\mathcal{B}$ is analogous to our above proof of the implication $\mathcal{C}%
\Longrightarrow\mathcal{B}$, and thus left to the reader.

The proof of Theorem \ref{thm.words.lyndon.equiv} in the partial-order setting
is complete.
\end{proof}

The examples of Theorem \ref{thm.words.lyndon.equiv} and Corollary
\ref{cor.words.uv=vu.3} should have illustrated how Proposition
\ref{prop.sol.words.poset.reduce-to-linex} allows deriving certain facts about
the partial-order setting from the corresponding facts about the total-order
setting. This trick, however, has its limits. For example, in the total-order
setting, the fact that any Lyndon word $w$ of length $>1$ can be written in
the form $w=uv$ for two Lyndon words $u$ and $v$ satisfying $u<w<v$ is a
consequence of Theorem \ref{thm.words.lyndon.std}. But in the partial-order
setting, it is not clear how to derive it from the total-order setting,
although it \textbf{is} true in the partial-order setting (and follows from
Exercise \ref{exe.words.lyndon.std2}).]

(d) If $\mathfrak{A}$ is the partially ordered alphabet $\left\{  1,2\right\}
$ with no relations whatsoever (i.e., a $2$-element antichain), and $w$ is the
word $12$, then $w$ does satisfy $\left(  \text{if }w\leq t^{n}\text{, then
}w\leq t\right)  $ for every nonempty word $t$ and every positive integer $n$,
but $w$ is not Lyndon.

[\textit{Remark:} One direction of Exercise
\ref{exe.words.lyndon.power-criterion} does hold in the partial-order setting:
Namely, if $w$ is Lyndon, then every nonempty word $t$ and every positive
integer $n$ satisfy $\left(  \text{if }w\leq t^{n}\text{, then }w\leq
t\right)  $. The proof of this is the same as in the total-order setting.]

(e) The following statement is clearly equivalent to Exercise
\ref{exe.words.lyndon.power-criterion} in the total-order setting, while still
being valid in the partial-order setting:

\begin{proposition}
\label{prop.sol.words.poset.e}Let $w$ be a nonempty word. Then, $w$ is Lyndon
if and only if every nonempty word $t$ and every positive integer $n$ satisfy
$\left(  \text{if }w>t\text{, then }w>t^{n}\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.e} in the partial-order
setting.]Let us first assume that $w$ is Lyndon. We shall prove that%
\begin{equation}
\left(  \text{every nonempty word }t\text{ and every positive integer }n\text{
satisfy }\left(  \text{if }w>t\text{, then }w>t^{n}\right)  \right)  .
\label{pf.prop.sol.words.poset.e.this}%
\end{equation}


Let $t$ be a nonempty word, and let $n$ be a positive integer. Assume that
$w>t$. We need to prove that $w>t^{n}$.

Assume the contrary. Thus, we don't have $w>t^{n}$. Thus, there exists an
$i\in\left\{  1,2,\ldots,n\right\}  $ such that we don't have $w>t^{i}$
(namely, $i=n$). Let $m$ be the \textbf{minimal} such $i$. Thus, $m\in\left\{
1,2,\ldots,n\right\}  $, and we don't have $w>t^{m}$. Hence, $m\neq1$ (since
we don't have $w>t^{m}$, but we do have $w>t=t^{1}$). Thus, $m\geq2$, so that
$m-1$ is also an element of $\left\{  1,2,\ldots,n\right\}  $. If we did not
have $w>t^{m-1}$, then $m-1$ would therefore be an $i\in\left\{
1,2,\ldots,n\right\}  $ such that we don't have $w>t^{i}$. But this would
contradict the fact that $m$ is the \textbf{minimal} such $i$. Thus, we must
have $w>t^{m-1}$. In other words, $t^{m-1}<w$.

Notice that $m-1\geq1$ (since $m\geq2$). Hence, the word $t^{m-1}$ is nonempty
(since $t$ is nonempty). In other words, $t^{m-1}\neq\varnothing$.

Recall that we don't have $w>t^{m}$. It is now easy to see that we don't have
$t^{m}\leq w$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we
have $t^{m}\leq w$. Hence, $w\geq t^{m}$, so that $w=t^{m}$ (since we don't
have $w>t^{m}$). Thus, $w=t^{m}=t^{m-1}t$. Thus, $t$ is a suffix of $w$. Thus,
Corollary \ref{cor.words.lyndon.trivia} (applied to $t$ instead of $v$) yields
$t\geq w$. Thus, $t\geq w=t^{m}=tt^{m-1}$. Combined with $tt^{m-1}\geq t$
(since $t$ is a prefix of $tt^{m-1}$), this yields $t=tt^{m-1}$. Thus,
$tt^{m-1}=t=t\varnothing$. Cancelling $t$ from this equality, we obtain
$t^{m-1}=\varnothing$. This contradicts $t^{m-1}\neq\varnothing$. This
contradiction proves that our assumption was wrong, qed.}. In other words, we
don't have $t^{m-1}t\leq w\varnothing$ (since $t^{m}=t^{m-1}t$ and
$w=w\varnothing$).

But recall that $t^{m-1}<w$. Hence, Proposition \ref{prop.words.lex}(d)
(applied to $a=t^{m-1}$, $b=t$, $c=w$ and $d=\varnothing$) yields that either
we have $t^{m-1}t\leq w\varnothing$ or the word $t^{m-1}$ is a prefix of $w$.
Thus, the word $t^{m-1}$ is a prefix of $w$ (since we don't have $t^{m-1}t\leq
w\varnothing$). In other words, there exists a $v\in\mathfrak{A}^{\ast}$ such
that $w=t^{m-1}v$. Consider this $v$. We have $v\neq\varnothing$ (because
otherwise, we would have $v=\varnothing$ and thus $w=t^{m-1}\underbrace{v}%
_{=\varnothing}=t^{m-1}$, contradicting the fact that $w>t^{m-1}$), so that
$v$ is nonempty. Hence, Proposition \ref{prop.words.lyndon.trivia}(b) (applied
to $u=t^{m-1}$) now yields $v>t^{m-1}$. Hence, $t^{m-1}<v$. Thus, Proposition
\ref{prop.words.lex}(b) (applied to $a=t^{m-1}$, $c=t^{m-1}$ and $d=v$) yields
$t^{m-1}t^{m-1}\leq t^{m-1}v=w$. Hence,
\begin{align*}
w  &  \geq t^{m-1}t^{m-1}=t^{2\left(  m-1\right)  }=t^{m+\left(  m-2\right)
}=t^{m}t^{m-2}\ \ \ \ \ \ \ \ \ \ \left(  \text{this makes sense since }%
m\geq2\right) \\
&  \geq t^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t^{m}\text{ is a prefix
of }t^{m}t^{m-2}\right)  .
\end{align*}
Hence, $t^m \leq w$.
This contradicts the fact that we don't have $t^m \leq w$. This contradiction
proves that our assumption (that we don't have $w>t^{n}$) was false. Hence,
$w>t^{n}$.

Forget now that we assumed that $w>t$. We thus have proven that if $w>t$, then
$w>t^{n}$.

Now, forget that we fixed $t$ and assumed that $w$ is Lyndon. We thus have
shown that
\begin{equation}
\left(  \text{if }w\text{ is Lyndon, then
(\ref{pf.prop.sol.words.poset.e.this}) holds}\right)  .
\label{pf.prop.sol.words.poset.e.1}%
\end{equation}


Now, conversely, assume that (\ref{pf.prop.sol.words.poset.e.this}) holds. We
will prove that $w$ is Lyndon.

Let $u$ and $v$ be any nonempty words satisfying $w=uv$. We have $w\neq
u$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $w=u$. Hence,
$uv=w=u=u\varnothing$. Cancelling $u$ from this equality, we obtain
$v=\varnothing$, so that $v$ is empty. This contradicts the fact that $v$ is
nonempty. This contradiction proves that our assumption was wrong, qed.}.
Combined with $w=uv\geq u$ (since $u$ is a prefix of $uv$), this yields $w>u$.
But (\ref{pf.prop.sol.words.poset.e.this}) (applied to $t=u$ and $n=2$) yields
that if $w>u$, then $w>u^{2}$. Hence, $w>u^{2}$ (since we know that $w>u$).
Thus, $uv=w>u^{2}=uu$, so that $uu<uv$. Thus, Proposition \ref{prop.words.lex}%
(c) (applied to $a=u$, $c=u$ and $d=v$) yields $u\leq v$. Since $u\neq
v$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $u=v$. Hence,
$w=u\underbrace{v}_{=u}=uu=u^{2}$. This contradicts $w>u^{2}$. This
contradiction proves that our assumption was wrong, qed.}, this becomes $u<v$,
so that $v>u$.

Now, let us forget that we fixed $u$ and $v$. We thus have proven that any
nonempty words $u$ and $v$ satisfying $w=uv$ satisfy $v>u$. In other words,
Assertion $\mathcal{C}$ of Theorem \ref{thm.words.lyndon.equiv} holds. Hence,
Assertion $\mathcal{A}$ of Theorem \ref{thm.words.lyndon.equiv} holds as well
(since Theorem \ref{thm.words.lyndon.equiv} yields that these Assertions
$\mathcal{C}$ and $\mathcal{A}$ are equivalent). In other words, the word $w$
is Lyndon.

Now, forget that we fixed $w$. We thus have proven that
\[
\left(  \text{if (\ref{pf.prop.sol.words.poset.e.this}) holds, then }w\text{
is Lyndon}\right)  .
\]
Combined with (\ref{pf.prop.sol.words.poset.e.1}), this yields that $w$ is
Lyndon if and only if (\ref{pf.prop.sol.words.poset.e.this}) holds. This
proves Proposition \ref{prop.sol.words.poset.e}.
\end{proof}

Thus, we have salvaged Exercise \ref{exe.words.lyndon.power-criterion} in the
partial-order setting. (And, as a side effect, we have obtained an alternative
solution for Exercise \ref{exe.words.lyndon.power-criterion} in the
total-order setting.)

[\textit{Remark:} Speaking of salvaging, there is a little piece of
Proposition \ref{prop.words.lyndon.cutoff} which can be salvaged for the
partial-order case:

\begin{proposition}
\label{prop.sol.words.poset.prop.words.lyndon.cutoff}Let $w$ be a nonempty
word. Let $v$ be a (lexicographically) minimal nonempty suffix of
$w$\ \ \ \ \footnote{By this, we mean that:
\par
\begin{itemize}
\item $v$ is a nonempty suffix of $w$;
\par
\item no nonempty suffix $s$ of $w$ satisfies $s<v$.
\end{itemize}
\par
Such a $v$ always exists (since $w$ is nonempty), but is not always unique.}.
Then, there exists a $u\in\mathfrak{A}^{\ast}$ such that $w = uv$ but
we don't have $uv<vu$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.prop.words.lyndon.cutoff} in
the partial-order case.] We know that $v$ is a suffix of $w$. Hence, there
exists a $u\in\mathfrak{A}^{\ast}$ such that $w=uv$. Consider this $u$. It
will clearly be enough to show that we don't have $uv<vu$.

So let us prove that we don't have $uv<vu$. Indeed, assume the contrary. Then,
$uv<vu$. Thus, there exists at least one suffix $t$ of $u$ such that $tv<vt$
(namely, $t=u$). Let $p$ be the \textbf{minimum-length} such suffix. Then,
$pv<vp$. Thus, $p$ is nonempty. In other words, $p\neq\varnothing$.

Since $p$ is a suffix of $u$, it is clear that $pv$ is a suffix of $uv=w$. So
we know that $pv$ is a nonempty suffix of $w$. Since $v$ is a minimal such
suffix, this yields that we don't have $pv<v$. Hence, we don't have $pv\leq
v$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $pv\leq v$.
This yields that $pv=v$ (since we don't have $pv<v$). Thus, $pv=v=\varnothing
v$. Cancelling $v$ from this equation, we obtain $p=\varnothing$. This
contradicts $p\neq\varnothing$. This contradiction proves that our assumption
was wrong, qed.}.

But $pv\varnothing=pv<vp$. Hence, Proposition \ref{prop.words.lex}(e) (applied
to $a=pv$, $b=\varnothing$, $c=v$ and $d=p$) yields that either we have
$pv\leq v$ or the word $v$ is a prefix of $pv$. Thus, the word $v$ is a prefix
of $pv$ (since we don't have $pv\leq v$). In other words, there exists a
$q\in\mathfrak{A}^{\ast}$ such that $pv=vq$. Consider this $q$. This $q$ is
nonempty (because otherwise we would have $pv=v\underbrace{q}_{=\varnothing
}=v$, contradicting the fact that $p$ is nonempty). From $vq=pv<vp$, we obtain
$q\leq p$ (by Proposition \ref{prop.words.lex}(c), applied to $a=v$, $c=q$ and
$d=p$).

We know that $q$ is a suffix of $pv$ (since $vq=pv$), whereas $pv$ is a suffix
of $w$. Thus, $q$ is a suffix of $w$. So $q$ is a nonempty suffix of $w$.
Since $v$ is a minimal such suffix, this yields that we don't have $q<v$.
But we have $p\varnothing=p\leq pv<vp$. Hence, Proposition
\ref{prop.words.lex}(e) (applied to $a=p$, $b=\varnothing$, $c=v$ and $d=p$)
yields that either we have $p\leq v$ or the word $v$ is a prefix of $p$. From
this, it is easy to obtain that $v$ is a prefix of $p$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $v$ is not a
prefix of $p$. Hence, $p\leq v$ (since either we have $p\leq v$ or the word
$v$ is a prefix of $p$). Now, $q\leq p\leq v$. Now, $q=v$ (since $q\leq v$ but
not $q<v$). Hence, $v=q\leq p$. Combined with $p\leq v$, this yields $v=p$.
Hence, $p\underbrace{v}_{=p}=\underbrace{p}_{=v}p=vp$, which contradicts
$pv<vp$. This contradiction proves that our assumption was wrong, qed.}. In
other words, there exists an $r\in\mathfrak{A}^{\ast}$ such that $p=vr$.
Consider this $r$. Clearly, $r$ is a suffix of $p$, while $p$ is a suffix of
$u$; therefore, $r$ is a suffix of $u$. Also, $pv<vp$ rewrites as $vrv<vvr$
(because $p=vr$). Thus, Proposition \ref{prop.words.lex}(c) (applied to $a=v$,
$c=rv$ and $d=vr$) yields $rv\leq vr$. Since $rv\neq vr$ (because otherwise,
we would have $rv=vr$, thus $v\underbrace{rv}_{=vr}=vvr$, contradicting
$vrv<vvr$), this becomes $rv<vr$.

The word $v$ is nonempty; thus, $\ell\left(v\right) > 0$.

Now, $r$ is a suffix of $u$ such that $rv<vr$. Since $p$ is the minimum-length
such suffix, this yields $\ell\left(  r\right)  \geq\ell\left(  p\right)  $.
But this contradicts the fact that $\ell\left(  \underbrace{p}_{=vr}\right)
=\ell\left(  vr\right)  =\underbrace{\ell\left(  v\right)  }_{>0}+\ell\left(
r\right)  >\ell\left(  r\right)  $. This contradiction proves our assumption
wrong. Thus, we have proven that we don't have $uv<vu$.
This completes the proof of
Proposition \ref{prop.sol.words.poset.prop.words.lyndon.cutoff} in the
partial-order case.
\end{proof}

]

(f) Let us first establish a lemma which plays a role similar to that of
Lemma~\ref{lem.words.CFLlemma} in the total-order setting:

\begin{lemma}
\label{lem.sol.words.poset.f.CFLlemma} Let $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ be a Hazewinkel-CFL factorization of a nonempty word $w$ (in
the partial-order setting). Let $p$ be a suffix of $w$ such that $p$ is
Lyndon. Then, $p\geq a_{k}$.
\end{lemma}

\begin{proof}
[Proof of Lemma~\ref{lem.sol.words.poset.f.CFLlemma}.]We will prove Lemma
\ref{lem.sol.words.poset.f.CFLlemma} by induction over the (obviously)
positive integer $k$.

\textit{Induction base:} Assume that $k=1$. By the definition of a
Hazewinkel-CFL factorization, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a
tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ (since $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a Hazewinkel-CFL factorization of $w$).
Thus, $w=a_{1}a_{2}\cdots a_{k}=a_{1}$ (since $k=1$), so that $w$ is a Lyndon
word (since $a_{1}$ is a Lyndon word). But $p$ is nonempty (since $p$ is
Lyndon). Thus, Corollary \ref{cor.words.lyndon.trivia} (applied to $v=p$)
yields $p\geq w=a_{1}=a_{k}$ (since $1=k$). Thus, Lemma
\ref{lem.sol.words.poset.f.CFLlemma} is proven in the case $k=1$. The
induction base is complete.

\textit{Induction step:} Let $K$ be a positive integer. Assume (as the
induction hypothesis) that Lemma \ref{lem.sol.words.poset.f.CFLlemma} is
proven for $k=K$. We now need to show that Lemma
\ref{lem.sol.words.poset.f.CFLlemma} holds for $k=K+1$.

So let $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ be a Hazewinkel-CFL
factorization of a nonempty word $w$. Let $p$ be a nonempty suffix of $w$ such
that $p$ is Lyndon. We need to prove that $p\geq a_{K+1}$.

The tuple $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ is a Hazewinkel-CFL
factorization of $w$. By the definition of a Hazewinkel-CFL factorization,
this yields that $\left(  a_{1},a_{2},\ldots,a_{K+1}\right)  $ is a tuple of
Lyndon words such that $w=a_{1}a_{2}\cdots a_{K+1}$ and such that no
$i\in\left\{  1,2,\ldots,K\right\}  $ satisfies $a_{i}<a_{i+1}$.

Let $w^{\prime}=a_{2}a_{3}\cdots a_{K+1}$; then, $w=a_{1}a_{2}\cdots
a_{K+1}=a_{1}\underbrace{\left(  a_{2}a_{3}\cdots a_{K+1}\right)
}_{=w^{\prime}}=a_{1}w^{\prime}$. Hence, every nonempty suffix of $w$ is
either a nonempty suffix of $w^{\prime}$, or has the form $qw^{\prime}$ for a
nonempty suffix $q$ of $a_{1}$. Since $p$ is a nonempty suffix of $w$, we thus
must be in one of the following two cases:

\textit{Case 1:} The word $p$ is a nonempty suffix of $w^{\prime}$.

\textit{Case 2:} The word $p$ has the form $qw^{\prime}$ for a nonempty suffix
$q$ of $a_{1}$.

Let us first consider Case 1. In this case, $p$ is a nonempty suffix of
$w^{\prime}$. The $K$-tuple $\left(  a_{2},a_{3},\ldots,a_{K+1}\right)  $ of
Lyndon words satisfies $w^{\prime}=a_{2}a_{3}\cdots a_{K+1}$ and has the
property that no $i\in\left\{  1,2,\ldots,K-1\right\}  $ satisfies
$a_{i+1}<a_{\left(  i+1\right)  +1}$\ \ \ \ \footnote{\textit{Proof.} We
already know that $w^{\prime}=a_{2}a_{3}\cdots a_{K+1}$. It remains to show
that no $i\in\left\{  1,2,\ldots,K-1\right\}  $ satisfies $a_{i+1}<a_{\left(
i+1\right)  +1}$.
\par
To prove this, let us assume this contrary. Thus, there exists an
$i\in\left\{  1,2,\ldots,K-1\right\}  $ satisfying $a_{i+1}<a_{\left(
i+1\right)  +1}$. Let $j$ be such an $i$. Then, $j\in\left\{  1,2,\ldots
,K-1\right\}  $ satisfies $a_{j+1}<a_{\left(  j+1\right)  +1}$. Hence, we have
$a_{i}<a_{i+1}$ for $i=j+1$. This contradicts the fact that no $i\in\left\{
1,2,\ldots,K\right\}  $ satisfies $a_{i}<a_{i+1}$. This contradiction shows
that our assumption was wrong, qed.}. Therefore, $\left(  a_{2},a_{3}%
,\ldots,a_{K+1}\right)  $ is a Hazewinkel-CFL factorization of $w^{\prime}$.
We can thus apply Lemma \ref{lem.sol.words.poset.f.CFLlemma} to $K$,
$w^{\prime}$ and $\left(  a_{2},a_{3},\ldots,a_{K+1}\right)  $ instead of $k$,
$w$ and $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ (because we assumed that
Lemma \ref{lem.sol.words.poset.f.CFLlemma} is proven for $k=K$). As a result,
we obtain that $p\geq a_{K+1}$. Thus, $p\geq a_{K+1}$ is proven in Case 1.

Let us now consider Case 2. In this case, $p$ has the form $qw^{\prime}$ for a
nonempty suffix $q$ of $a_{1}$. Consider this $q$. Since $a_{1}$ is a Lyndon
word, we have $q\geq a_{1}$ (by Corollary \ref{cor.words.lyndon.trivia},
applied to $a_{1}$ and $q$ instead of $w$ and $v$). Our goal, however, is to
prove that $q\geq a_{K+1}$.

We will show that%
\begin{equation}
q\geq a_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots
,K+1\right\}  .\label{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}%
\end{equation}


\textit{Proof of (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}):} We will
prove (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) by induction over
$i$:

\textit{Induction base:} We know that $q\geq a_{1}$. In other words,
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=1$. This
completes the induction base.

\textit{Induction step:} Let $j\in\left\{  1,2,\ldots,K\right\}  $. Assume
that (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=j$. We
must prove that (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for
$i=j+1$.

We have $j\leq K$. Hence, the product $a_{j+1}a_{j+2}\cdots a_{K+1}$ contains
at least one factor. Also, the factors of this product are nonempty words
(because the words $a_{j+1}$, $a_{j+2}$, $\ldots$, $a_{K+1}$ are nonempty
(since these words are Lyndon)). Hence, $a_{j+1}a_{j+2}\cdots a_{K+1}$ is a
nonempty product of nonempty words. Consequently, $a_{j+1}a_{j+2}\cdots
a_{K+1}$ is a nonempty word.

The product $a_{j+2}a_{j+3}\cdots a_{K+1}$ is well-defined (because $j\leq
K$). Denote this product by $g$. Thus, $g=a_{j+2}a_{j+3}\cdots a_{K+1}$.

We have $q\geq a_{j}$ (since
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=j$). But%
\[
p=q\underbrace{w^{\prime}}_{\substack{=a_{2}a_{3}\cdots a_{K+1}\\=\left(
a_{2}a_{3}\cdots a_{j}\right)  \left(  a_{j+1}a_{j+2}\cdots a_{K+1}\right)
}}=q\left(  a_{2}a_{3}\cdots a_{j}\right)  \left(  a_{j+1}a_{j+2}\cdots
a_{K+1}\right)  =\left(  q\left(  a_{2}a_{3}\cdots a_{j}\right)  \right)
\left(  a_{j+1}a_{j+2}\cdots a_{K+1}\right)  .
\]
Consequently, $a_{j+1}a_{j+2}\cdots a_{K+1}$ is a suffix of $p$. Thus,
Corollary \ref{cor.words.lyndon.trivia} (applied to $p$ and $a_{j+1}%
a_{j+2}\cdots a_{K+1}$ instead of $w$ and $v$) yields $a_{j+1}a_{j+2}\cdots
a_{K+1}\geq p$ (since $p$ is Lyndon). Thus,%
\[
a_{j+1}a_{j+2}\cdots a_{K+1}\geq p=qw^{\prime}\geq q\geq a_{j}.
\]
In other words,%
\[
a_{j}\leq a_{j+1}a_{j+2}\cdots a_{K+1}=a_{j+1}\underbrace{\left(
a_{j+2}a_{j+3}\cdots a_{K+1}\right)  }_{=g}=a_{j+1}g.
\]
Hence, $a_{j}\varnothing=a_{j}\leq a_{j+1}g$. Proposition \ref{prop.words.lex}%
(e) (applied to $a_{j}$, $\varnothing$, $a_{j+1}$ and $g$ instead of $a$, $b$,
$c$ and $d$) thus yields that either we have $a_{j}\leq a_{j+1}$ or the word
$a_{j+1}$ is a prefix of $a_{j}$. From this, it is easy to conclude that the
word $a_{j+1}$ is a prefix of $a_{j}$\ \ \ \ \footnote{\textit{Proof.} Assume
the contrary. Thus, the word $a_{j+1}$ is not a prefix of $a_{j}$. Hence, we
have $a_{j}\leq a_{j+1}$ (because we know that either we have $a_{j}\leq
a_{j+1}$ or the word $a_{j+1}$ is a prefix of $a_{j}$). Since we cannot have
$a_{j}<a_{j+1}$ (because no $i\in\left\{  1,2,\ldots,K\right\}  $ satisfies
$a_{i}<a_{i+1}$), this yields that we have $a_{j}=a_{j+1}$. Therefore,
$a_{j+1}$ is a prefix of $a_{j}$; this contradicts our assumption that the
word $a_{j+1}$ is not a prefix of $a_{j}$. This contradiction proves that our
assumption was wrong, qed.}. Consequently, $a_{j+1}\leq a_{j}$, so that
$a_{j}\geq a_{j+1}$ and $q\geq a_{j}\geq a_{j+1}$. In other words,
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) holds for $i=j+1$. This
completes the induction step. Thus,
(\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) is proven by induction.

Now, (\ref{pf.lem.sol.words.poset.f.CFLlemma.c2indhook}) (applied to $i=K+1$)
yields $q\geq a_{K+1}$. Hence, $p=qw^{\prime}\geq q\geq a_{K+1}$. Thus, $p\geq
a_{K+1}$ is proven in Case 2.

We have now proven $p\geq a_{K+1}$ in all cases. This proves that Lemma
\ref{lem.sol.words.poset.f.CFLlemma} holds for $k=K+1$. The induction step is
thus finished, and with it the proof of Lemma
\ref{lem.sol.words.poset.f.CFLlemma}.
\end{proof}

We can now conclude the solution of Exercise \ref{exe.words.poset}(f) by
proving the following proposition:

\begin{proposition}
\label{prop.sol.words.poset.f.CFL}Let $w$ be a word (in the partial-order
setting). Then, there exists a unique Hazewinkel-CFL factorization of $w$.
\end{proposition}

The proof is an almost literal adaptation of the proof of Theorem
\ref{thm.words.CFL}:

\begin{proof}
[Proof of Proposition \ref{prop.sol.words.poset.f.CFL}.]Let us first prove
that there exists a Hazewinkel-CFL factorization of $w$.

Indeed, there clearly exists a tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)
$ of Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$\ \ \ \ \footnote{For
instance, the tuple $\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
}\right)  $ of one-letter words is a valid example (recall that one-letter
words are always Lyndon).}. Fix such a tuple with \textbf{minimum} $k$. We
claim that no $i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $a_{i}%
<a_{i+1}$.

Indeed, if some $i\in\left\{  1,2,\ldots,k-1\right\}  $ would satisfy
$a_{i}<a_{i+1}$, then the word $a_{i}a_{i+1}$ would be Lyndon (by Proposition
\ref{prop.words.lyndon.concat}(a), applied to $u=a_{i}$ and $v=a_{i+1}$),
whence $\left(  a_{1},a_{2},\ldots,a_{i-1},a_{i}a_{i+1},a_{i+2},a_{i+3}%
,\ldots,a_{k}\right)  $ would also be a tuple of Lyndon words satisfying
$w=a_{1}a_{2}\cdots a_{i-1}\left(  a_{i}a_{i+1}\right)  a_{i+2}a_{i+3}\cdots
a_{k}$ but having length $k-1<k$, contradicting the fact that $k$ is the
minimum length of such a tuple. Hence, no $i\in\left\{  1,2,\ldots
,k-1\right\}  $ can satisfy $a_{i}<a_{i+1}$. Thus, $\left(  a_{1},a_{2}%
,\ldots,a_{k}\right)  $ is a Hazewinkel-CFL factorization of $w$, so we have
shown that such a Hazewinkel-CFL factorization exists.

It remains to show that there exists at most one Hazewinkel-CFL factorization
of $w$. We shall prove this by induction over $\ell\left(  w\right)  $. Thus,
we fix a word $w$ and assume that%
\begin{equation}
\text{for every word }v\text{ with }\ell\left(  v\right)  <\ell\left(
w\right)  \text{, there exists at most one Hazewinkel-CFL factorization of }v.
\label{pf.thm.words.Hazewinkel-CFL.uni.1}%
\end{equation}
We now have to prove that there exists at most one Hazewinkel-CFL
factorization of $w$.

Indeed, let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(
b_{1},b_{2},\ldots,b_{m}\right)  $ be two Hazewinkel-CFL factorizations of
$w$. We need to prove that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(
b_{1},b_{2},\ldots,b_{m}\right)  $. If $w$ is empty, then this is obvious, so
we WLOG assume that it is not; thus, $k>0$ and $m>0$.

The tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a Hazewinkel-CFL
factorization of $w$. Thus, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a
tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and such that no
$i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $a_{i}<a_{i+1}$.

The tuple $\left(  b_{1},b_{2},\ldots,b_{m}\right)  $ is a Hazewinkel-CFL
factorization of $w$. Thus, $\left(  b_{1},b_{2},\ldots,b_{m}\right)  $ is a
tuple of Lyndon words satisfying $w=b_{1}b_{2}\cdots b_{m}$ and such that no
$i\in\left\{  1,2,\ldots,m-1\right\}  $ satisfies $b_{i}<b_{i+1}$. Now,
$b_{m}$ is a suffix of $w$ (since $w=b_{1}b_{2}\cdots b_{m}$). Also, $b_{m}$
is Lyndon. Thus, Lemma \ref{lem.sol.words.poset.f.CFLlemma} (applied to
$p=b_{m}$) yields $b_{m}\geq a_{k}$. The same argument (but with the roles of
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(  b_{1},b_{2}%
,\ldots,b_{m}\right)  $ switched) shows that $a_{k}\geq b_{m}$. Combined with
$b_{m}\geq a_{k}$, this yields $a_{k}=b_{m}$. Now let $v=a_{1}a_{2}\cdots
a_{k-1}$. Then, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)  $ is a tuple of
Lyndon words satisfying $v=a_{1}a_{2}\cdots a_{k-1}$ and such that no
$i\in\left\{  1,2,\ldots,\left(  k-1\right)  -1\right\}  $ satisfies
$a_{i}<a_{i+1}$ (because no $i\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies
$a_{i}<a_{i+1}$). In other words, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)
$ is a Hazewinkel-CFL factorization of $v$.

But $w=a_{1}a_{2}\cdots a_{k}=\underbrace{a_{1}a_{2}\cdots a_{k-1}}%
_{=v}\underbrace{a_{k}}_{=b_{m}}=vb_{m}$, so that%
\[
vb_{m}=w=b_{1}b_{2}\cdots b_{m}=b_{1}b_{2}\cdots b_{m-1}b_{m}.
\]
Cancelling $b_{m}$ from this equality yields $v=b_{1}b_{2}\cdots b_{m-1}$.
Thus, $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $ is a tuple of Lyndon words
satisfying $v=b_{1}b_{2}\cdots b_{m-1}$ and such that no $i\in\left\{
1,2,\ldots,\left(  m-1\right)  -1\right\}  $ satisfies $b_{i}<b_{i+1}$ (since
no $i\in\left\{  1,2,\ldots,m-1\right\}  $ satisfies $b_{i}<b_{i+1}$). In
other words, $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $ is a Hazewinkel-CFL
factorization of $v$. Since $\ell\left(  v\right)  <\ell\left(  w\right)  $
(because $v=a_{1}a_{2}\cdots a_{k-1}$ is shorter than $w=a_{1}a_{2}\cdots
a_{k}$), we can apply (\ref{pf.thm.words.Hazewinkel-CFL.uni.1}) to obtain that
there exists at most one Hazewinkel-CFL factorization of $v$. But we already
know two such Hazewinkel-CFL factorizations: $\left(  a_{1},a_{2}%
,\ldots,a_{k-1}\right)  $ and $\left(  b_{1},b_{2},\ldots,b_{m-1}\right)  $.
Thus, $\left(  a_{1},a_{2},\ldots,a_{k-1}\right)  =\left(  b_{1},b_{2}%
,\ldots,b_{m-1}\right)  $. Combining this with $a_{k}=b_{m}$, we obtain
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(  b_{1},b_{2},\ldots
,b_{m}\right)  $. This is exactly what we needed to prove. So we have shown
(by induction) that there exists at most one Hazewinkel-CFL factorization of
$w$. This completes the proof of Proposition \ref{prop.sol.words.poset.f.CFL}.
\end{proof}

(g) \textit{Solution to Exercise \ref{exe.words.powerlyndon} in the
partial-order setting.} Let us first observe the following fact (in the
partial-order setting): If $h$ is a nonempty word which is not a Lyndon word,
then%
\begin{equation}
\text{there exist nonempty words }u\text{ and }v\text{ such that }h=uv\text{
and not }vu>uv\text{.} \label{sol.words.poset.g.lem1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.poset.g.lem1}):} Let $h$ be a
nonempty word which is not a Lyndon word. We need to prove that
(\ref{sol.words.poset.g.lem1}) holds.
\par
In fact, assume the contrary. Then, there exist no nonempty words $u$ and $v$
such that $h=uv$ and not $vu>uv$. In other words, any nonempty words $u$ and
$v$ satisfying $h=uv$ must satisfy $vu>uv$. In other words, Assertion
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) holds.
\par
But we know (from Exercise \ref{exe.words.poset}(c)) that Theorem
\ref{thm.words.lyndon.equiv} holds in the partial-order setting. Hence, we can
apply Theorem \ref{thm.words.lyndon.equiv} to $h$ instead of $w$. We thus
conclude that Assertions $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) are equivalent. Hence, Assertion $\mathcal{A}$ of Theorem
\ref{thm.words.lyndon.equiv} (with $h$ instead of $w$) holds (since Assertion
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) holds). In other words, the word $h$ is Lyndon. This contradicts the fact
that the word $h$ is not Lyndon. This contradiction shows that our assumption
was wrong. Hence, (\ref{sol.words.poset.g.lem1}) is proven, qed.} Renaming $u$
and $v$ as $p$ and $q$ in this result, we obtain the following: If $h$ is a
nonempty word which is not a Lyndon word, then%
\begin{equation}
\text{there exist nonempty words }p\text{ and }q\text{ such that }h=pq\text{
and not }qp>pq\text{.} \label{sol.words.poset.g.lem1a}%
\end{equation}


Furthermore, if $h$ is a nonempty word which is not a Lyndon word, then%
\begin{equation}
\text{there exist nonempty words }u\text{ and }v\text{ such that }h=uv\text{
and not }v>u\text{.} \label{sol.words.poset.g.lem2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.poset.g.lem2}):} Let $h$ be a
nonempty word which is not a Lyndon word. We need to prove that
(\ref{sol.words.poset.g.lem2}) holds.
\par
In fact, assume the contrary. Then, there exist no nonempty words $u$ and $v$
such that $h=uv$ and not $v>u$. In other words, any nonempty words $u$ and $v$
satisfying $h=uv$ must satisfy $v>u$. In other words, Assertion $\mathcal{C}$
of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of $w$) holds.
\par
But we know (from Exercise \ref{exe.words.poset}(c)) that Theorem
\ref{thm.words.lyndon.equiv} holds in the partial-order setting. Hence, we can
apply Theorem \ref{thm.words.lyndon.equiv} to $h$ instead of $w$. We thus
conclude that Assertions $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ and
$\mathcal{D}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) are equivalent. Hence, Assertion $\mathcal{A}$ of Theorem
\ref{thm.words.lyndon.equiv} (with $h$ instead of $w$) holds (since Assertion
$\mathcal{C}$ of Theorem \ref{thm.words.lyndon.equiv} (with $h$ instead of
$w$) holds). In other words, the word $h$ is Lyndon. This contradicts the fact
that the word $h$ is not Lyndon. This contradiction shows that our assumption
was wrong. Hence, (\ref{sol.words.poset.g.lem2}) is proven, qed.}

Now, let us come to the solution of Exercise \ref{exe.words.powerlyndon} in
the partial-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(a) in the
partial-order setting.} The implication $\mathcal{A}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$ can be proven in the same way as it was proven in the
total-order setting.

Let us now prove the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{A}^{\prime}$:

\textit{Proof of the implication $\mathcal{D}^{\prime}\Longrightarrow
\mathcal{A}^{\prime}$:} Assume that Assertion $\mathcal{D}^{\prime}$
holds. Thus, if $u$ and $v$ are nonempty words satisfying $w=uv$, then we have
$vu\geq uv$.

We need to prove that Assertion $\mathcal{A}^{\prime}$ holds, i.e., that $w$
is a power of a Lyndon word. Assume the contrary. Thus, $w$ is not a power of
a Lyndon word; hence, $w$ is not a Lyndon word itself. Consequently,
(\ref{sol.words.poset.g.lem1}) (applied to $h=w$) shows that there exist
nonempty words $u$ and $v$ such that $w=uv$ and not $vu>uv$. Consider such a
pair of nonempty words $u$ and $v$ with \textbf{minimum }$\ell\left(
u\right)  $. The minimality of $\ell\left(  u\right)  $ shows that
\begin{equation}
\left( \begin{array}{c}
 \text{if }u^{\prime}\text{ and }v^{\prime}\text{ are nonempty words
such that }w=u^{\prime}v^{\prime}\text{ and not }v^{\prime}u^{\prime
}>u^{\prime}v^{\prime}\text{,} \\
\text{then }\ell\left(  u^{\prime}\right)  \geq
\ell\left(  u\right)
\end{array}
\right)  .
\label{sol.words.poset.g.DA.1}
\end{equation}


We have $vu\geq uv$ (according to Assertion $\mathcal{D}^{\prime}$) but not
$vu>uv$. Thus, $vu=uv$. Therefore, Proposition \ref{prop.words.uv=vu.2} yields
that there exist a $t\in\mathfrak{A}^{\ast}$ and two nonnegative integers $n$
and $m$ such that $u=t^{n}$ and $v=t^{m}$. Consider this $t$ and these $n$ and
$m$. We have $n\neq0$ (since $t^{n}=u$ is nonempty) and $m\neq0$ (since
$t^{m}=v$ is nonempty), and the word $t$ is nonempty (since $t^{n}=u$ is
nonempty). Moreover, we have $n=1$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Hence, $n\neq1$. Combined with $n\neq0$, this leads to $n\geq2$. As
a consequence, $u=t^{n}$ can be rewritten as $u=tt^{n-1}$. The word
$t^{n-1+m}$ is nonempty (since $t$ is nonempty and since $\underbrace{n}%
_{\geq2}-1+\underbrace{m}_{\geq0}\geq2-1+0=1$). Now, $w=\underbrace{u}%
_{=tt^{n-1}}\underbrace{v}_{=t^{m}}=tt^{n-1}t^{m}=tt^{n-1+m}$. Also,
$t^{n-1+m}t=t^{n-1+m+1}=tt^{n-1+m}$. Thus, we do not have $t^{n-1+m}%
t>tt^{n-1+m}$. Hence, (\ref{sol.words.poset.g.DA.1}) (applied to $u^{\prime
}=t$ and $v^{\prime}=t^{n-1+m}$) yields $\ell\left(  t\right)  \geq\ell\left(
u\right)  $ (since $t$ and $t^{n-1+m}$ are nonempty). Since $u=t^{n}$, this
rewrites as $\ell\left(  t\right)  \geq\ell\left(  t^{n}\right)
=\underbrace{n}_{\geq2}\ell\left(  t\right)  \geq2\ell\left(  t\right)  $,
whence $\ell\left(  t\right)  =0$, which contradicts the fact that $t$ is
nonempty. This contradiction shows that our assumption was wrong, qed.}.
Hence, $t^{n}=t^{1}=t$, so that $u=t^{n}=t$ and $w=\underbrace{u}%
_{=t}\underbrace{v}_{=t^{m}}=tt^{m}=t^{m+1}$. We shall now prove that the word
$t$ is Lyndon.

Assume the contrary. Then, $t$ is not Lyndon. Hence,
(\ref{sol.words.poset.g.lem1a}) (applied to $h=t$) shows that there exist
nonempty words $p$ and $q$ such that $t=pq$ and not $qp>pq$. Consider these
$p$ and $q$. Since $q$ is nonempty, we have $\ell\left(  q\right)  >0$, so
that $\ell\left(  \underbrace{u}_{=t=pq}\right)  =\ell\left(  pq\right)
=\ell\left(  p\right)  +\underbrace{\ell\left(  q\right)  }_{>0}>\ell\left(
p\right)  $.

We have $w=\underbrace{u}_{=t=pq}v=pqv$, and the words $p$ and $qv$ are
nonempty\footnote{For $qv$, this follows from the nonemptiness of $q$.}. Now,
using (\ref{sol.words.poset.g.DA.1}), it is easy to see that $qvp>pqv$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, we do not have
$qvp>pqv$. We can thus apply (\ref{sol.words.poset.g.DA.1}) to $u^{\prime}=p$
and $v^{\prime}=qv$. As a result, we obtain $\ell\left(  p\right)  \geq
\ell\left(  u\right)  $. This contradicts $\ell\left(  u\right)  >\ell\left(
p\right)  $. This contradiction shows that our assumption was wrong, qed.}.

Notice that $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ for every
$i\in \NN$\ \ \ \ \footnote{\textit{Proof.} We shall prove the equality
$q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ by induction over $i$:
\par
\textit{Induction base:} We have $q\underbrace{\left(  pq\right)  ^{0}%
}_{=\varnothing}=q\varnothing=q=\underbrace{\varnothing}_{=\left(  qp\right)
^{0}}q=\left(  qp\right)  ^{0}q$. In other words, the equality $q\left(
pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ holds for $i=0$. This completes the
induction base.
\par
\textit{Induction step:} Let $I\in \NN$ be such that the equality
$q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ holds for $i=I$. We need
to show that the equality $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$
also holds for $i=I+1$.
\par
We have $q\left(  pq\right)  ^{I}=\left(  qp\right)  ^{I}q$ (since the
equality $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ holds for $i=I$).
Thus,
\[
q\underbrace{\left(  pq\right)  ^{I+1}}_{=\left(  pq\right)  ^{I}\left(
pq\right)  }=\underbrace{q\left(  pq\right)  ^{I}}_{=\left(  qp\right)  ^{I}%
q}\left(  pq\right)  =\left(  qp\right)  ^{I}\underbrace{q\left(  pq\right)
}_{=\left(  qp\right)  q}=\underbrace{\left(  qp\right)  ^{I}\left(
qp\right)  }_{=\left(  qp\right)  ^{I+1}}q=\left(  qp\right)  ^{I+1}q.
\]
In other words, the equality $q\left(  pq\right)  ^{i}=\left(  qp\right)
^{i}q$ holds for $i=I+1$. This completes the induction step. The induction
proof of the equality $q\left(  pq\right)  ^{i}=\left(  qp\right)  ^{i}q$ is
thus complete.}. Applied to $i=m$, this yields $q\left(  pq\right)
^{m}=\left(  qp\right)  ^{m}q$.

But $v=t^{m}$. Since $t=pq$, this rewrites as $v=\left(  pq\right)  ^{m}$.
Hence,%
\[
q\underbrace{v}_{=\left(  pq\right)  ^{m}}p=\underbrace{q\left(  pq\right)
^{m}}_{=\left(  qp\right)  ^{m}q}p=\left(  qp\right)  ^{m}qp=\left(
qp\right)  ^{m}\left(  qp\right)  =\left(  qp\right)  ^{m+1},
\]
so that $\left(  qp\right)  ^{m+1}=qvp>pqv$. But%
\[
\left(  pq\right)  ^{m+1}=\left(  pq\right)  \left(  \underbrace{pq}%
_{=t}\right)  ^{m}=\left(  pq\right)  \underbrace{t^{m}}_{=v}=\left(
pq\right)  v=pqv<\left(  qp\right)  ^{m+1}%
\]
(since $\left(  qp\right)  ^{m+1}>pqv$). From this, it is easy to see that
$pq<qp$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we don't
have $pq<qp$. But $\left(  pq\right)  \left(  pq\right)  ^{m}=\left(
pq\right)  ^{m+1}<\left(  qp\right)  ^{m+1}=\left(  qp\right)  \left(
qp\right)  ^{m+1}$ and $\ell\left(  pq\right)  =\ell\left(  p\right)
+\ell\left(  q\right)  =\ell\left(  q\right)  +\ell\left(  p\right)
=\ell\left(  qp\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied
to $a=pq$, $b=\left(  pq\right)  ^{m}$, $c=qp$ and $d=\left(  qp\right)  ^{m}%
$) yields $pq\leq qp$. Thus, $pq=qp$ (since we don't have $pq<qp$). Taking
both sides of this equality to the $\left(  m+1\right)  $-th power, we obtain
$\left(  pq\right)  ^{m+1}=\left(  qp\right)  ^{m+1}$, which contradicts
$\left(  pq\right)  ^{m+1}<\left(  qp\right)  ^{m+1}$. This contradiction
proves that our assumption was wrong, qed.}, so that $qp>pq$. This contradicts
the fact that we do not have $qp>pq$. This contradiction shows that our
assumption (that $t$ is not Lyndon) was wrong. Hence, $t$ is Lyndon. Thus, $w$
is a power of a Lyndon word (since $w=t^{m+1}$ is a power of $t$). Thus,
Assertion $\mathcal{A}^{\prime}$ is satisfied, so we have proven the
implication $\mathcal{D}^{\prime}\Longrightarrow\mathcal{A}^{\prime}$.

Now we have proven both implications $\mathcal{A}^{\prime}\Longrightarrow
\mathcal{D}^{\prime}$ and $\mathcal{D}^{\prime}\Longrightarrow\mathcal{A}%
^{\prime}$. Therefore, the equivalence $\mathcal{A}^{\prime}%
\Longleftrightarrow\mathcal{D}^{\prime}$ follows. Thus, Exercise
\ref{exe.words.powerlyndon}(a) is solved in the partial-order case.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(b) in the
partial-order setting.} Consider the letter $m$ and the alphabet
$\mathfrak{A}\cup\left\{  m\right\}  $ defined in Assertion $\mathcal{F}%
^{\prime\prime}$. We notice that the lexicographic order on $\mathfrak{A}%
^{\ast}$ is the restriction of the lexicographic order on $\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ to $\mathfrak{A}^{\ast}%
$. Therefore, when we have two words $p$ and $q$ in $\mathfrak{A}^{\ast}$,
statements like ``$p<q$'' do not depend on
whether we are regarding $p$ and $q$ as elements of $\mathfrak{A}^{\ast}$ or
as elements of $\left(  \mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$.
It is easy to see that the one-letter word $m$ satisfies%
\begin{equation}
m>p\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathfrak{A}^{\ast}.
\label{sol.words.poset.g.b.mp}%
\end{equation}
\footnote{This can be proven in the same way as it was proven in the
total-order setting.}
% Also,
%\begin{equation}
%m>pm\ \ \ \ \ \ \ \ \ \ \text{for every nonempty }p\in\mathfrak{A}^{\ast}.
%\label{sol.words.poset.g.b.mpm}
%\end{equation}
%
%
%\footnote{\textit{Proof of (\ref{sol.words.poset.g.b.mpm}):} Let
%$p\in\mathfrak{A}^{\ast}$ be nonempty. Then, $p$ has a first letter, and we
%have $\left(  \text{the first letter of the word }p\right)  \in\mathfrak{A}$
%(since $p\in\mathfrak{A}^{\ast}$). Thus,
%\begin{align*}
%\left(  \text{the first letter of the word }p\right)   &
%<m\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a<m\text{ for every }%
%a\in\mathfrak{A}\right) \\
%&  =\left(  \text{the first letter of the word }m\right)  .
%\end{align*}
%Now, the word $p$ is nonempty, and thus
%\begin{align*}
%\left(  \text{the first letter of the word }pm\right)   &  =\left(  \text{the
%first letter of the word }pm\right) \\
%&  <\left(  \text{the first letter of the word }m\right)  .
%\end{align*}
%Hence, $pm<m$ (by the definition of the lexicographic order), that is, $m>pm$.
%This proves (\ref{sol.words.poset.g.b.mpm}).}

The implications $\mathcal{B}^{\prime}\Longrightarrow\mathcal{E}^{\prime}$,
$\mathcal{C}^{\prime}\Longrightarrow\mathcal{E}^{\prime}$, $\mathcal{G}%
^{\prime}\Longrightarrow\mathcal{H}^{\prime}$, $\mathcal{F}^{\prime\prime
}\Longrightarrow\mathcal{B}^{\prime}$, $\mathcal{F}^{\prime\prime
}\Longrightarrow\mathcal{C}^{\prime}$ and $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ can be proven in the same way as they were proven in the
total-order setting. We shall now prove some further implications.

\textit{Proof of the implication $\mathcal{E}^{\prime}\Longrightarrow
\mathcal{F}^{\prime\prime}$:} Assume that Assertion $\mathcal{E}%
^{\prime}$ holds.

Assume (for the sake of contradiction) that the word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ is not a Lyndon word.
Clearly, this word $wm$ is nonempty. Thus, (\ref{sol.words.poset.g.lem2})
(applied to $h=wm$) shows that there exist nonempty words $u$ and $v$ in
$\left(  \mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ such that
$wm=uv$ and not $v>u$. Denote these two nonempty words $u$ and $v$ by $u$ and
$v^{\prime}$. Then, $u$ and $v^{\prime}$ are nonempty words in $\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ such that $wm=uv^{\prime
}$ and not $v^{\prime}>u$.

The word $v^{\prime}$ is a proper suffix of $wm$ (since $wm=uv^{\prime}$ and
since $w$ is nonempty). Hence, $v^{\prime}$ is a nonempty suffix of $wm$.
Thus, we must be in one of the following two cases (depending on whether this
suffix begins before the suffix $m$ of $wm$ begins or afterwards):

\textit{Case 1:} The word $v^{\prime}$ is a nonempty suffix of $m$. (Note that
$v^{\prime}=m$ is allowed.)

\textit{Case 2:} The word $v^{\prime}$ has the form $vm$ where $v$ is a
nonempty proper suffix of $w$.

Let us consider Case 1 first. In this case, the word $v^{\prime}$ is a
nonempty suffix of $m$. Since the only nonempty suffix of $m$ is $m$ itself
(because $m$ is a one-letter word), this yields $v^{\prime}=m$. Now,
$wm=u\underbrace{v^{\prime}}_{=m}=um$. Cancelling $m$ from this equality, we
obtain $w=u$. But $v^{\prime}=m>w$ (by (\ref{sol.words.poset.g.b.mp}), applied
to $p=w$), thus $v^{\prime}>w=u$. This contradicts the fact that we don't have
$v^{\prime}>u$. Thus, we have obtained a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $v^{\prime}$ has the form
$vm$ where $v$ is a nonempty proper suffix of $w$. Consider this $v$. We have
$wm=u\underbrace{v^{\prime}}_{=vm}=uvm$. By cancelling $m$ from this equality,
we obtain $w=uv$. Thus, $u$ and $v$ are subwords of $w$, and therefore belong
to $\mathfrak{A}^{\ast}$ (since $w\in\mathfrak{A}^{\ast}$). Moreover, $u$ and
$v$ are nonempty. Hence, Assertion $\mathcal{E}^{\prime}$ yields that either
we have $v\geq u$ or the word $v$ is a prefix of $w$. Since we cannot have
$v\geq u$ (because if we had $v\geq u$, then we would have $v^{\prime
}=vm>v\geq u$, which would contradict the fact that we don't have $v^{\prime
}>u$), we therefore must have that $v$ is a prefix of $w$. In other words,
there exists a $q\in\mathfrak{A}^{\ast}$ such that $w=vq$. Consider this $q$.
We have $m>q$ (by (\ref{sol.words.poset.g.b.mp}), applied to $p=q$). Thus,
$q\leq m$. Hence, Proposition \ref{prop.words.lex}(b) (applied to
$\mathfrak{A}\cup\left\{  m\right\}  $, $v$, $q$ and $m$ instead of
$\mathfrak{A}$, $a$, $c$ and $d$) yields $vq\leq vm$. Therefore, $vm\geq vq=w$
(since $w=vq$), so that $v^{\prime}=vm\geq w=uv>u$ (since $v$ is nonempty).
This contradicts the fact that we don't have $v^{\prime}>u$. Thus, we have
found a contradiction in Case 2.

We have therefore obtained a contradiction in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this shows that we always get a
contradiction. Hence, our assumption (that the word $wm\in\left(
\mathfrak{A}\cup\left\{  m\right\}  \right)  ^{\ast}$ is not a Lyndon word)
was false. Hence, the word $wm\in\left(  \mathfrak{A}\cup\left\{  m\right\}
\right)  ^{\ast}$ is a Lyndon word. That is, Assertion $\mathcal{F}%
^{\prime\prime}$ holds. Hence, we have proven the implication $\mathcal{E}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$.

\textit{Proof of the implication $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{G}^{\prime}$:} Assume that Assertion $\mathcal{B}^{\prime}$ holds.

Let $s$ be the longest suffix $v$ of $w$ which does not satisfy $v\geq w$.
(This is well-defined, because there exists a suffix $v$ of $w$ which does not
satisfy $v\geq w$ -- namely, the empty word.) So we know that $s$ is a suffix
$v$ of $w$ which does not satisfy $v\geq w$. In other words, $s$ is a suffix
of $w$ and does not satisfy $s\geq w$. Hence, $s\neq w$ (because otherwise, we
would have $s=w$ and thus $s\geq w$; but this would contradict the fact that
$s$ does not satisfy $s\geq w$). Thus, $s$ is a proper suffix of $w$ (because
$s$ is a suffix of $w$ and satisfies $s\neq w$). Hence, there exists a
nonempty word $h\in\mathfrak{A}^{\ast}$ satisfying $w=hs$. Consider this $h$.
Using Assertion $\mathcal{B}^{\prime}$, it is easy to see that $s$ is a prefix
of $w$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $s$ is not
a prefix of $w$. Hence, $s$ is nonempty. Therefore, Assertion $\mathcal{B}%
^{\prime}$ (applied to $u=h$ and $v=s$) yields that either we have $s\geq w$
or the word $s$ is a prefix of $w$. Since $s$ is not a prefix of $w$, we must
thus have $s\geq w$. But this contradicts the fact that $s$ does not satisfy
$s\geq w$. This contradiction shows that our assumption was wrong, qed.}. In
other words, there exists a $g\in\mathfrak{A}^{\ast}$ such that $w=sg$.
Consider this $g$. We have $g\neq\varnothing$ (because otherwise, we would
have $g=\varnothing$ and thus $w=s\underbrace{g}_{=\varnothing}=s\neq w$,
which would be absurd). In other words, the word $g$ is nonempty. Thus,
$s<sg=w$.

We know that $s$ is the longest suffix $v$ of $w$ which does not satisfy
$v\geq w$. Hence,%
\begin{equation}
\left(  \text{if }v\text{ is a suffix of }w\text{ which does not satisfy
}v\geq w\text{, then }\ell\left(  v\right)  \leq\ell\left(  s\right)  \right)
. \label{sol.words.poset.g.BG.1}%
\end{equation}


There exists a nonnegative integer $m$ such that $h^{m}$ is a prefix of $s$
(for example, the nonnegative integer $m=0$). Consider the \textbf{maximal}
such integer $m$\ \ \ \ \footnote{This is well-defined, because of the
following reason:
\par
We have $\ell\left(  h\right)  \geq1$ (since the word $h$ is nonempty). Thus,
for every $m\in \NN$ satisfying $m>\ell\left(  s\right)  $, we have
$\ell\left(  h^{m}\right)  =m\underbrace{\ell\left(  h\right)  }_{\geq1}\geq
m>\ell\left(  s\right)  $. In other words, for every $m\in \NN$
satisfying $m>\ell\left(  s\right)  $, the word $h^{m}$ is longer than $s$.
Hence, for every $m\in \NN$ satisfying $m>\ell\left(  s\right)  $, the
word $h^{m}$ cannot be a prefix of $s$. Thus, for every sufficiently high
$m\in \NN$, the word $h^{m}$ cannot be a prefix of $s$.
\par
We thus know the following:
\par
\begin{itemize}
\item There exists a nonnegative integer $m$ such that $h^{m}$ is a prefix of
$s$.
\par
\item For every sufficiently high $m\in \NN$, the word $h^{m}$ cannot
be a prefix of $s$.
\end{itemize}
\par
Consequently, there exists a \textbf{maximal} nonnegative integer $m$ such
that $h^{m}$ is a prefix of $s$, qed.}. Then, $h^{m}$ is a prefix of $s$, but
$h^{m+1}$ is not a prefix of $s$. Since $h^{m}$ is a prefix of $s$, there
exists a word $q\in\mathfrak{A}^{\ast}$ such that $s=h^{m}q$. Consider this
$q$. Clearly, $w=h\underbrace{s}_{=h^{m}q}=\underbrace{hh^{m}}_{=h^{m+1}%
}q=\underbrace{h^{m+1}}_{=h^{m}h}q=h^{m}hq$. Hence, $h^{m}hq=w=\underbrace{s}%
_{=h^{m}q}g=h^{m}qg$. Cancelling $h^{m}$ from this equality, we obtain
$hq=qg$. It is now easy to see that we don't have $h\leq q$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then, $h\leq q$. Hence,
$h\leq q\leq qg=hq$ (since $hq=qg$). Therefore, Proposition
\ref{prop.words.lex}(g) (applied to $h$, $q$ and $q$ instead of $a$, $b$ and
$c$) yields that $h$ is a prefix of $q$. In other words, there exists a word
$r\in\mathfrak{A}^{\ast}$ such that $q=hr$. Consider this $r$. Now,
$s=h^{m}\underbrace{q}_{=hr}=\underbrace{h^{m}h}_{=h^{m+1}}r=h^{m+1}r$, so
that $h^{m+1}$ is a prefix of $s$. This contradicts the fact that $h^{m+1}$ is
not a prefix of $s$. This contradiction proves that our assumption was wrong,
qed.}. But $h\varnothing=h\leq hq=qg$. Thus, Proposition \ref{prop.words.lex}%
(e) (applied to $h$, $\varnothing$, $q$ and $g$ instead of $a$, $b$, $c$ and
$d$) yields that either we have $h\leq q$ or the word $q$ is a prefix of $h$.
Thus, the word $q$ is a prefix of $h$ (since we don't have $h\leq q$).

Next, we shall prove that the word $h$ is Lyndon.

In fact, assume the contrary. Then, $h$ is not Lyndon. Hence,
(\ref{sol.words.poset.g.lem1}) shows that there exist nonempty words $u$ and
$v$ such that $h=uv$ and not $vu>uv$. Consider these $u$ and $v$. Since
$w=\underbrace{h}_{=uv}s=uvs=u\left(  vs\right)  $, it is clear that the word
$vs$ is a suffix of $w$. If this suffix $vs$ would not satisfy $vs\geq w$,
then we could therefore obtain $\ell\left(  vs\right)  \leq\ell\left(
s\right)  $ (by (\ref{sol.words.poset.g.BG.1}), applied to $vs$ instead of
$v$), which would contradict $\ell\left(  vs\right)  =\underbrace{\ell\left(
v\right)  }_{\substack{>0\\\text{(since }v\text{ is nonempty)}}}+\ell\left(
s\right)  >\ell\left(  s\right)  $. Thus, the suffix $vs$ must satisfy $vs\geq
w$. We thus have $vs\geq w=hs\geq h=uv$, so that $uv\leq vs$.

Recall that $s<w$. Hence, $vs\leq vw$ (by Proposition \ref{prop.words.lex}(b),
applied to $v$, $s$ and $w$ instead of $a$, $c$ and $d$). Now, $uv\varnothing
=uv\leq vs\leq v\underbrace{w}_{=hs}=v\underbrace{h}_{=uv}s=vuvs$ and
$\ell\left(  uv\right)  =\ell\left(  u\right)  +\ell\left(  v\right)
=\ell\left(  v\right)  +\ell\left(  u\right)  =\ell\left(  vu\right)  \leq
\ell\left(  vu\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied
to $uv$, $\varnothing$, $vu$ and $vs$ instead of $a$, $b$, $c$ and $d$) yields
that $uv\leq vu$. In other words, $vu\geq uv$. Since we don't have $vu>uv$, we
therefore must have $vu=uv$. Thus, the elements $u$ and $v$ of the monoid
$\mathfrak{A}^{\ast}$ commute. Thus, the submonoid of $\mathfrak{A}^{\ast}$
generated by $u$ and $v$ is commutative. Since $h=uv$, the element $h$ lies in
this submonoid, and therefore the element $h^{m}$ lies in it as well. Thus,
$h^{m}$ commutes with $v$ (since this submonoid is commutative), i.e., we have
$vh^{m}=h^{m}v$. Thus, $v\underbrace{s}_{=h^{m}q}=\underbrace{vh^{m}}%
_{=h^{m}v}q=h^{m}vq$. Thus, $h^{m}vq=vs\geq w=h\underbrace{s}_{=h^{m}%
q}=\underbrace{hh^{m}}_{=h^{m+1}=h^{m}h}q=h^{m}hq$, so that $h^{m}hq\leq
h^{m}vq$. Hence, Proposition \ref{prop.words.lex}(c) (applied to $h^{m}$, $hq$
and $vq$ instead of $a$, $c$ and $d$) yields $hq\leq vq$. But since $q$ is a
prefix of $h$, there exists a word $z\in\mathfrak{A}^{\ast}$ such that $h=qz$.
Consider this $z$. We have%
\begin{align*}
v\underbrace{qz}_{=h=uv}  &  =\underbrace{vu}_{=uv=h}v=hv\leq h\underbrace{vu}%
_{=uv=h=qz}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }hv\text{ is a prefix of
}hvu\right) \\
&  =hqz.
\end{align*}


Also, $\ell\left(  \underbrace{h}_{=uv}q\right)  =\ell\left(  uvq\right)
=\ell\left(  u\left(  vq\right)  \right)  =\underbrace{\ell\left(  u\right)
}_{\substack{>0\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  vq\right)
>\ell\left(  vq\right)  $, so that $\ell\left(  vq\right)  \leq\ell\left(
hq\right)  $. Hence, Proposition \ref{prop.words.lex}(f) (applied to $vq$,
$z$, $hq$ and $z$ instead of $a$, $b$, $c$ and $d$) yields $vq\leq hq$.
Combined with $hq\leq vq$, this yields $vq=hq$. Hence, $\ell\left(
\underbrace{vq}_{=hq}\right)  =\ell\left(  hq\right)  >\ell\left(  vq\right)
$, which is absurd. This contradiction proves that our assumption is wrong.
Thus, we have shown that the word $h$ is Lyndon.

We now know that $h\in\mathfrak{A}^{\ast}$ is a Lyndon word, $m+1$ is a
positive integer, and $q$ is a prefix of $h$, and we have $w=h^{m+1}q$. Hence,
there exists a Lyndon word $t\in\mathfrak{A}^{\ast}$, a positive integer
$\ell$ and a prefix $p$ of $t$ (possibly empty) such that $w=t^{\ell}p$
(namely, $t=h$, $\ell=m+1$ and $p=q$). In other words, Assertion
$\mathcal{G}^{\prime}$ holds. This proves the implication $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$.

Furthermore, the implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ holds. (In fact, it can be proven in the same way
as it was proven in the total-order setting.)

\textit{Proof of the implication $\mathcal{H}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$:} Assume that Assertion $\mathcal{H}^{\prime}$
holds. In other words, there exists a Lyndon word $t\in\mathfrak{A}^{\ast}$, a
nonnegative integer $\ell$ and a prefix $p$ of $t$ (possibly empty) such that
$w=t^{\ell}p$. Consider this $t$, this $\ell$ and this $p$.

We are going to prove that for every $m \in \NN$,
\begin{equation}
\left(  \text{every suffix }s\text{ of }t^{m}p\text{ satisfies either }s\geq
t^{m}p\text{ or }\left(  \text{the word }s\text{ is a prefix of }%
t^{m}p\right)  \right)  .
\label{sol.words.poset.g.HB.1}
\end{equation}


\textit{Proof of (\ref{sol.words.poset.g.HB.1}):} We will prove
(\ref{sol.words.poset.g.HB.1}) by induction over $m$:

\textit{Induction base:} Using the implication $\mathcal{F}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$, it is easy to see that
(\ref{sol.words.poset.g.HB.1}) holds for $m=0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $m=0$. Then, $t^{m}%
p=\underbrace{t^{0}}_{=\varnothing}p=\varnothing p=p$.
\par
Let $s$ be a suffix of $t^{m}p$. Then, $s$ is a suffix of $t^{m}p=p$. In other
words, there exists a word $g\in\mathfrak{A}^{\ast}$ satisfying $p=gs$.
Consider this $g$.
\par
We are going to prove that either $s\geq t^{m}p$ or $\left(  \text{the word
}s\text{ is a prefix of }t^{m}p\right)  $. If the word $s$ is empty, then this
is obvious (because if the word $s$ is empty, then the word $s$ is a prefix of
$t^{m}p$). Hence, we WLOG assume that the word $s$ is nonempty. If the word
$g$ is empty, then it is also clear that either $s\geq t^{m}p$ or $\left(
\text{the word }s\text{ is a prefix of }t^{m}p\right)  $ (because if the word
$g$ is empty, then $g=\varnothing$ and thus $t^{m}p=p=\underbrace{g}%
_{=\varnothing}s=s$, so that the word $s$ is a prefix of $t^{m}p$). Hence, we
WLOG assume that the word $g$ is nonempty.
\par
But the word $p$ is a prefix of a Lyndon word in $\mathfrak{A}^{\ast}$ (since
$p$ is a prefix of $t$, and since $t$ is a Lyndon word in $\mathfrak{A}^{\ast
}$). In other words, Assertion $\mathcal{F}^{\prime}$ with $w$ replaced by $p$
is satisfied. Hence, Assertion $\mathcal{B}^{\prime}$ with $w$ replaced by $p$
is satisfied as well (since we have already proven the implication
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}^{\prime}$). In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }u\text{ and }v\text{ are nonempty words satisfying }p=uv\text{,
then}\\
\text{either we have }v\geq p\text{ or the word }v\text{ is a prefix of }p
\end{array}
\right)  .
\label{sol.words.poset.g.HB.base}
\end{equation}
Since the words $g$ and $s$ are nonempty, we can apply
(\ref{sol.words.poset.g.HB.base}) to $u=g$ and $v=s$. As a result, we obtain
that either we have $s\geq p$ or the word $s$ is a prefix of $p$. In other
words, either $s\geq p$ or $\left(  \text{the word }s\text{ is a prefix of
}p\right)  $. In other words, either $s\geq t^{m}p$ or $\left(  \text{the word
}s\text{ is a prefix of }t^{m}p\right)  $ (since $t^{m}p=p$). This proves
(\ref{sol.words.poset.g.HB.1}).}. This completes the induction base.

\textit{Induction step:} Let $M$ be a positive integer. Assume that
(\ref{sol.words.poset.g.HB.1}) is proven for $m=M-1$. We will now show that
(\ref{sol.words.poset.g.HB.1}) holds for $m=M$.

Let $r$ denote the word $t^{M-1}p$. It is easy to see that $r$ is a prefix of
$t^{M}p$\ \ \ \ \footnote{\textit{Proof.} There exists a word $q\in
\mathfrak{A}^{\ast}$ such that $t=pq$ (since $p$ is a prefix of $t$). Consider
this $q$. We have $\underbrace{t^{M}}_{=t^{M-1}t}p=t^{M-1}\underbrace{t}%
_{=pq}p=\underbrace{t^{M-1}p}_{=r}qp=rqp=r\left(  qp\right)  $. Hence, $r$ is
a prefix of $t^{M}p$, qed.}. In other words, there exists a word
$g\in\mathfrak{A}^{\ast}$ such that $t^{M}p=rg$. Consider this $g$.

Let $s$ be a suffix of $t^{M}p$. We shall show that either $s\geq t^{M}p$ or
$\left(  \text{the word }s\text{ is a prefix of }t^{M}p\right)  $.

In order to prove this, let us assume the contrary (for the sake of
contradiction). Then, neither $s\geq t^{M}p$ nor $\left(  \text{the word
}s\text{ is a prefix of }t^{M}p\right)  $. In other words, we don't have
$s\geq t^{M}p$, and the word $s$ is not a prefix of $t^{M}p$. If the word $s$
was a prefix of $r$, then the word $s$ would be a prefix of $t^{M}p$ (since
$r$ is a prefix of $t^{M}p$), which would contradict the fact that the word
$s$ is not a prefix of $t^{M}p$. Hence, the word $s$ cannot be a prefix of
$r$. In other words, the word $s$ cannot be a prefix of $t^{M-1}p$ (since
$r=t^{M-1}p$).

We don't have $s\geq t^{M}p$. Since $s=s\varnothing$ and $t^{M}p=rg$, this
rewrites as follows: We don't have $s\varnothing\geq rg$. In other words, we
don't have $rg\leq s\varnothing$.

The word $s$ is a suffix of $\underbrace{t^{M}}_{=tt^{M-1}}%
p=t\underbrace{t^{M-1}p}_{=r}=tr$. Therefore, we must be in one of the
following two cases (depending on whether this suffix begins before the suffix
$r$ of $tr$ begins or afterwards):

\textit{Case 1:} The word $s$ is a suffix of $r$. (Note that $s=r$ is allowed.)

\textit{Case 2:} The word $s$ has the form $s^{\prime}r$ where $s^{\prime}$ is
a nonempty suffix of $t$.

Let us consider Case 1 first. In this case, the word $s$ is a suffix of $r$.
In other words, the word $s$ is a suffix of $t^{M-1}p$ (since $r=t^{M-1}p$).
Hence, (\ref{sol.words.poset.g.HB.1}) (applied to $m=M-1$) yields that either
$s\geq t^{M-1}p$ or $\left(  \text{the word }s\text{ is a prefix of }%
t^{M-1}p\right)  $ (since (\ref{sol.words.poset.g.HB.1}) holds for $m=M-1$).
Since the word $s$ cannot be a prefix of $t^{M-1}p$, we thus must have $s\geq
t^{M-1}p$. Thus, $t^{M-1}p\leq s$, so that $r=t^{M-1}p\leq s$. Therefore,
Proposition \ref{prop.words.lex}(d) (applied to $r$, $g$, $s$ and
$\varnothing$ instead of $a$, $b$, $c$ and $d$) yields that either we have
$rg\leq s\varnothing$ or the word $r$ is a prefix of $s$. Thus, the word $r$
is a prefix of $s$ (since we don't have $rg\leq s\varnothing$). But
$\ell\left(  r\right)  \geq\ell\left(  s\right)  $ (since $s$ is a suffix of
$r$). Hence, $r$ is a prefix of $s$ which is at least as long as $s$ itself.
Consequently, $r=s$. Hence, $s=r$, so that $s$ is a prefix of $s=r$. This
contradicts the fact that the word $s$ cannot be a prefix of $r$. Thus, we
have found a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $s$ has the form
$s^{\prime}r$ where $s^{\prime}$ is a nonempty suffix of $t$. Consider this
$s^{\prime}$. Corollary \ref{cor.words.lyndon.trivia} (applied to $t$ and
$s^{\prime}$ instead of $w$ and $v$) yields $s^{\prime}\geq t$. That is,
$t\leq s^{\prime}$. But $s^{\prime}$ is a suffix of $t$, so that $\ell\left(
s^{\prime}\right)  \leq\ell\left(  t\right)  $. Hence, $\ell\left(  t\right)
\geq\ell\left(  s^{\prime}\right)  $. Thus, Proposition \ref{prop.words.lex}%
(j) (applied to $t$, $s^{\prime}$ and $r$ instead of $a$, $b$ and $c$) yields
$tr\leq s^{\prime}r$. Hence, $s^{\prime}r\geq tr$, so that $s=s^{\prime}r\geq
tr=t^{M}p$. This contradicts the fact that we don't have $s\geq t^{M}p$. Thus,
we have found a contradiction in Case 2.

We have thus obtained a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that we always get a
contradiction. This completes the proof that either $s\geq t^{M}p$ or $\left(
\text{the word }s\text{ is a prefix of }t^{M}p\right)  $.

Now, forget that we fixed $s$. We thus have shown that every suffix $s$ of
$t^{M}p$ satisfies either $s\geq t^{M}p$ or $\left(  \text{the word }s\text{
is a prefix of }t^{M}p\right)  $. In other words,
(\ref{sol.words.poset.g.HB.1}) holds for $m=M$. This completes the induction
step, and thus (\ref{sol.words.poset.g.HB.1}) is proven by induction.

Now, let $u$ and $v$ be nonempty words satisfying $w=uv$. Then, $v$ is a
suffix of $w=t^{\ell}p$. Hence, (\ref{sol.words.poset.g.HB.1}) (applied to
$m=\ell$ and $s=v$) yields that either $v\geq t^{\ell}p$ or $\left(  \text{the
word }v\text{ is a prefix of }t^{\ell}p\right)  $. In other words, either
$v\geq w$ or $\left(  \text{the word }v\text{ is a prefix of }w\right)  $
(since $w=t^{\ell}p$). In other words, either we have $v\geq w$ or the word
$v$ is a prefix of $w$.

Now, forget that we fixed $u$ and $v$. We thus have shown that if $u$ and $v$
are nonempty words satisfying $w=uv$, then either we have $v\geq w$ or the
word $v$ is a prefix of $w$. In other words, Assertion $\mathcal{B}^{\prime}$
holds. Thus, the implication $\mathcal{H}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ is proven.

We have thus proven the implications $\mathcal{B}^{\prime}\Longrightarrow
\mathcal{E}^{\prime}$, $\mathcal{C}^{\prime}\Longrightarrow\mathcal{E}%
^{\prime}$, $\mathcal{G}^{\prime}\Longrightarrow\mathcal{H}^{\prime}$,
$\mathcal{E}^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$,
$\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{B}^{\prime}$,
$\mathcal{F}^{\prime}\Longrightarrow\mathcal{C}^{\prime}$, $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{G}^{\prime}$ and $\mathcal{H}^{\prime
}\Longrightarrow\mathcal{B}^{\prime}$. Combined, these yield the equivalence
$\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime}%
\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime}\Longleftrightarrow
\mathcal{H}^{\prime}$. This solves Exercise \ref{exe.words.powerlyndon}(b) in
the partial-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(c) in the
partial-order setting.} The implication $\mathcal{F}^{\prime}\Longrightarrow
\mathcal{B}^{\prime}$ has already been proven in our solution of Exercise
\ref{exe.words.powerlyndon}(b). Hence, Exercise \ref{exe.words.powerlyndon}(c)
is solved in the partial-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(d) in the
partial-order setting.} The solution of Exercise \ref{exe.words.powerlyndon}%
(d) in the partial-order setting proceeds precisely as in the total-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(e) in the
partial-order setting.} The solution of Exercise \ref{exe.words.powerlyndon}%
(e) in the partial-order setting proceeds precisely as in the total-order setting.

\textit{Solution to Exercise \ref{exe.words.powerlyndon}(f) in the
partial-order setting.} Assume that there exists a letter $\mu\in\mathfrak{A}$
such that $\left(  \mu>a\text{ for some letter }a\text{ of }w\right)  $.
Consider this $\mu$. We need to prove that the equivalence $\mathcal{F}%
^{\prime}\Longleftrightarrow\mathcal{F}^{\prime\prime}$ holds.

Combining the implication $\mathcal{F}^{\prime}\Longrightarrow\mathcal{B}%
^{\prime}$ (which has already been proven) and the implication $\mathcal{B}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$ (which follows from the
equivalence $\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime
}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime}\Longleftrightarrow
\mathcal{H}^{\prime}$ proven above), we obtain the implication $\mathcal{F}%
^{\prime}\Longrightarrow\mathcal{F}^{\prime\prime}$. Thus, in order to prove
the equivalence $\mathcal{F}^{\prime}\Longleftrightarrow\mathcal{F}%
^{\prime\prime}$, it is enough to verify the implication $\mathcal{F}%
^{\prime\prime}\Longrightarrow\mathcal{F}^{\prime}$. Let us do this now.

Assume that Assertion $\mathcal{F}^{\prime\prime}$ holds. Due to the
implication $\mathcal{F}^{\prime\prime}\Longrightarrow\mathcal{G}^{\prime}$
(which follows from the equivalence $\mathcal{B}^{\prime}\Longleftrightarrow
\mathcal{C}^{\prime}\Longleftrightarrow\mathcal{E}^{\prime}\Longleftrightarrow
\mathcal{F}^{\prime\prime}\Longleftrightarrow\mathcal{G}^{\prime
}\Longleftrightarrow\mathcal{H}^{\prime}$ proven above), this yields that
Assertion $\mathcal{G}^{\prime}$ holds. In other words, there exists a Lyndon
word $t\in\mathfrak{A}^{\ast}$, a positive integer $\ell$ and a prefix $p$ of
$t$ (possibly empty) such that $w=t^{\ell}p$. Consider these $t$, $\ell$ and
$p$.

We shall first prove that
\begin{equation}
\text{there exists a letter }z\in\mathfrak{A}\text{ such that }%
z>t.
\label{sol.words.poset.g.f.z}
\end{equation}


\textit{Proof of (\ref{sol.words.poset.g.f.z}):} The word $t$ is Lyndon and
thus nonempty. Hence, $\ell\left(  t\right)  \geq1$. We thus are in one of the
following two cases:

\textit{Case 1:} We have $\ell\left(  t\right)  =1$.

\textit{Case 2:} We have $\ell\left(  t\right)  >1$.

Let us consider Case 1 first. In this case, we have $\ell\left(  t\right)
=1$. Thus, $t$ is a one-letter word. In other words, $t=b$ for some letter
$b$. Let us consider this $b$.

Recall that $\mu>a$ for some letter $a$ of $w$. Consider this letter $a$.

The word $t^\ell p$ is a prefix of $t^\ell t$ (since $p$ is a prefix of
$t$). In other words, the word $w$ is a prefix of $t^{\ell+1}$ (since
$w = t^\ell p$ and $t^{\ell+1} = t^\ell t$). Hence, each letter of $w$
is a letter of $t^{\ell+1}$. Since each letter of $t^{\ell+1}$ is a
letter of $t$, this shows that each letter of $w$ is a letter of $t$.
Applying this to the letter $a$, we conclude that $a$ is a letter of $t$
(since $a$ is a letter of $w$).

Now, both
$a$ and $b$ are letters of $t$. Since the word $t$ has only one letter
(because $\ell\left(  t\right)  =1$), this yields that $a=b$. Hence,
$\mu>a=b=t$ (since $t=b$). Hence, there exists a letter $z\in\mathfrak{A}$
such that $z>t$ (namely, $z=\mu$). Thus, (\ref{sol.words.poset.g.f.z}) is
proven in Case 1.

Let us now consider Case 2. In this case, we have $\ell\left(  t\right)  >1$.
Let $g$ be the last letter of the word $t$ (this is well-defined since $t$ is
nonempty). Then, the one-letter word $g$ is a suffix of the word $t$, and
therefore there exists a word $t^{\prime}\in\mathfrak{A}^{\ast}$ such that
$t=t^{\prime}g$. Consider this $t^{\prime}$. Since $g$ is a one-letter word,
we have $\ell\left(  g\right)  =1$. Thus, $\ell\left(  \underbrace{t}%
_{=t^{\prime}g}\right)  =\ell\left(  t^{\prime}\right)  +\underbrace{\ell
\left(  g\right)  }_{=1}=\ell\left(  t^{\prime}\right)  +1$, so that
$\ell\left(  t^{\prime}\right)  =\underbrace{\ell\left(  t\right)  }%
_{>1}-1>1-1=0$. Hence, the word $t^{\prime}$ is nonempty. Thus, $g$ is a
proper suffix of $t$ (since $t=t^{\prime}g$). Also, $g$ is nonempty (since
$\ell\left(  g\right)  =1$). But recall that the word $t$ is Lyndon. By the
definition of a Lyndon word, this yields that every nonempty proper suffix $v$
of $t$ satisfies $v>t$. Applying this to $v=g$, we obtain $g>t$. Thus, there
exists a letter $z\in\mathfrak{A}$ such that $z>t$ (namely, $z=g$). Thus,
(\ref{sol.words.poset.g.f.z}) is proven in Case 2.

We have thus proven (\ref{sol.words.poset.g.f.z}) in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this yields that
(\ref{sol.words.poset.g.f.z}) always holds.

Now, (\ref{sol.words.poset.g.f.z}) shows that there exists a letter
$z\in\mathfrak{A}$ such that $z>t$. Consider this letter $z$. We have $z>t$,
thus $t<z$, hence $t\leq z$. The one-letter word $z$ is Lyndon (since every
one-letter word is Lyndon). Thus, $\underbrace{t\text{, }t\text{, }%
\ldots\text{, }t}_{\ell+1\text{ times }t}$, $z$ are $\ell+2$ Lyndon words
(since both $t$ and $z$ are Lyndon) satisfying $t\leq t\leq\cdots\leq t\leq z$
(since $t\leq z$) and $t<z$. Hence, Exercise \ref{exe.words.lyndon.concatn}
(applied to $\ell+2$ and $\left(
\underbrace{t,t,\ldots,t}_{\ell+1\text{ times }t},z\right)  $
instead of $n$ and $\left(  w_{1},w_{2},\ldots,w_{n}\right)$) yields that
$\underbrace{tt\cdots t}_{\ell+1\text{ times }t}z$ is a Lyndon word. In other
words, $t^{\ell+1}z$ is a Lyndon word (since $\underbrace{tt\cdots t}%
_{\ell+1\text{ times }t}=t^{\ell+1}$).

But $p$ is a prefix of $t$. In other words, there exists a $q\in
\mathfrak{A}^{\ast}$ such that $t=pq$. Consider this $q$. We then have
$\underbrace{t^{\ell+1}}_{=t^{\ell}t}z=t^{\ell}\underbrace{t}_{=pq}%
z=\underbrace{t^{\ell}p}_{=w}qz=wqz=w\left(  qz\right)  $. Hence, $w$ is a
prefix of the word $t^{\ell+1}z$. Thus, $w$ is a prefix of a Lyndon word in
$\mathfrak{A}^{\ast}$ (because $t^{\ell+1}z$ is a Lyndon word in
$\mathfrak{A}^{\ast}$). In other words, Assertion $\mathcal{F}^{\prime}$
holds. This proves the implication $\mathcal{F}^{\prime\prime}\Longrightarrow
\mathcal{F}^{\prime}$. This solves Exercise \ref{exe.words.powerlyndon}(f) in
the partial-order case.

\textit{Remark:} Of course, for a letter $\mu\in\mathfrak{A}$, if we have
$\left(  \mu>a\text{ for every letter }a\text{ of }w\right)  $, then we also
have $\left(  \mu>a\text{ for some letter }a\text{ of }w\right)  $ (since $w$
is nonempty and thus has at least one letter). Hence, Exercise
\ref{exe.words.powerlyndon}(e) is a particular case of Exercise
\ref{exe.words.powerlyndon}(f).

Altogether, we have now solved all parts of Exercise
\ref{exe.words.powerlyndon} in the partial-order case. Exercise
\ref{exe.words.poset}(g) is thus solved.

[\textit{Remark:} The validity of some parts of Exercise
\ref{exe.words.powerlyndon} in the partial-order case can also be deduced from
their validity in the total-order case using Proposition
\ref{prop.sol.words.poset.reduce-to-linex}. For example, the equivalence
$\mathcal{B}^{\prime}\Longleftrightarrow\mathcal{C}^{\prime}%
\Longleftrightarrow\mathcal{E}^{\prime}$ can be treated this way. We shall not
give any details of this alternative approach, however.]
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.necklaces}}

\textit{Solution to Exercise \ref{exe.words.necklaces}.} In the following,
whenever $G$ is a group and $P$ is a (left) $G$-set, we denote by $G_{u}$ the
stabilizer of $u$ in $G$, that is, the subgroup $\left\{  g\in G\ \mid
\ gu=u\right\}  $ of $G$.

Let $C^{\left(  n\right)  }$ denote the subgroup $\left\langle c^{n}%
\right\rangle $ of the infinite cyclic group $C$. Then, $C/C^{\left(
n\right)  }$ is a cyclic group with $n$ elements. Hence, $\left\vert
C/C^{\left(  n\right)  }\right\vert =n$.

Also, recall that $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating
$n$-tuples one step to the left ($c\cdot\left(  a_{1},a_{2},\ldots
,a_{n}\right)  =\left(  a_{2},a_{3},\ldots,a_{n},a_{1}\right)  $). Thus,
$c^{n}$ acts trivially on $\mathfrak{A}^{n}$ (since cyclically rotating an
$n$-tuple $n$ steps to the left does nothing). Therefore, the whole subgroup
$C^{\left(  n\right)  }$ acts trivially on $\mathfrak{A}^{n}$ (since
$C^{\left(  n\right)  }=\left\langle c^{n}\right\rangle $).

The word ``divisor'' shall mean
``positive divisor'' throughout this solution.

(a) Let $N$ be any $n$-necklace. Then, $N$ is an orbit of the $C$-action, thus
a nonempty set. Fix an element $w$ of $N$ (such a $w$ exists since $N$ is nonempty).

Recall that the subgroup $C^{\left(  n\right)  }$ acts trivially on
$\mathfrak{A}^{n}$. In particular, $C^{\left(  n\right)  }$ stabilizes $w$.
Hence, $C^{\left(  n\right)  }\subset C_{w}$. Thus, $\left[  C:C^{\left(
n\right)  }\right]  =\left[  C:C_{w}\right]  \cdot\left[  C_{w}:C^{\left(
n\right)  }\right]  $. As a consequence, $\left[  C:C_{w}\right]  \mid\left[
C:C_{w}\right]  \cdot\left[  C_{w}:C^{\left(  n\right)  }\right]  =\left[
C:C^{\left(  n\right)  }\right]  =\left\vert C/C^{\left(  n\right)
}\right\vert =n$, so that $\left[  C:C_{w}\right]  <\infty$.

But $N$ is an orbit of the $C$-action containing $w$. Hence, $N$ is the
$C$-orbit of $w$. Thus, by the orbit-stabilizer theorem, we have $\left\vert
N\right\vert =\left[  C:C_{w}\right]  $. This yields $\left\vert N\right\vert
=\left[  C:C_{w}\right]  <\infty$, so that $N$ is a finite set.

Also, $\left\vert N\right\vert =\left[  C:C_{w}\right]  \mid n$. This solves
Exercise \ref{exe.words.necklaces}(a).

(b) Let $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n}$ be an
$n$-tuple. We need to prove the equivalence between the following two assertions:

\begin{statement}
\textit{Assertion $\mathcal{A}$:} The $n$-necklace $\left[  w\right]
$ is aperiodic.
\end{statement}

\begin{statement}
\textit{Assertion $\mathcal{B}$:} Every $k\in\left\{  1,2,\ldots
,n-1\right\}  $ satisfies $\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1}%
,w_{2},\ldots,w_{k}\right)  \neq w$.
\end{statement}

We will achieve this by proving the implications $\mathcal{A}\Longrightarrow
\mathcal{B}$ and $\mathcal{B}\Longrightarrow\mathcal{A}$. But let us first do
some preparatory work.

Let $N$ denote the $n$-necklace $\left[  w\right]  $. Thus, $w\in N$ and
therefore (as we have shown in the solution of Exercise
\ref{exe.words.necklaces}(a)) we have $C^{\left(  n\right)  }\subset C_{w}$
and $\left\vert N\right\vert =\left[  C:C_{w}\right]  $.

We are now ready to prove implications $\mathcal{A}\Longrightarrow\mathcal{B}$
and $\mathcal{B}\Longrightarrow\mathcal{A}$.

\textit{Proof of the implication $\mathcal{A}\Longrightarrow\mathcal{B}%
$:} Assume that Assertion $\mathcal{A}$ holds. In other words, the
$n$-necklace $\left[  w\right]  $ is aperiodic.

In other words, $N$ is aperiodic (since $N = \left[  w\right]  $). In other
words, the period of $N$ is $n$ (by the definition of ``aperiodic'').
In other words, $\left\vert N\right\vert $ is
$n$ (since the period of $N$ is defined as $\left\vert N\right\vert $). In
other words, $\left\vert N\right\vert =n$. Thus, $n=\left\vert N\right\vert
=\left[  C:C_{w}\right]  $. But%
\begin{align*}
n  & =\left[  C:C^{\left(  n\right)  }\right]  =\underbrace{\left[
C:C_{w}\right]  }_{=n}\cdot\left[  C_{w}:C^{\left(  n\right)  }\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }C^{\left(  n\right)  }\subset
C_{w}\right)  \\
& =n\cdot\left[  C_{w}:C^{\left(  n\right)  }\right]  .
\end{align*}
Solving this for $\left[  C_{w}:C^{\left(  n\right)  }\right]  $, we obtain
$\left[  C_{w}:C^{\left(  n\right)  }\right]  =1$, so that $C^{\left(
n\right)  }=C_{w}$.

Let now $k\in\left\{  1,2,\ldots,n-1\right\}  $. We are going to prove that
$\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  \neq
w$. Indeed, assume the contrary. Thus, $\left(  w_{k+1},w_{k+2},\ldots
,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  =w$.

Recall that $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples
one step to the left. Hence, $c^{k}$ acts on $\mathfrak{A}^{n}$ by cyclically
rotating $n$-tuples $k$ steps to the left. Hence,%
\[
c^{k}w=\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)
=w.
\]
Thus, $c^{k}\in C_{w}=C^{\left(  n\right)  }=\left\langle c^{n}\right\rangle
$. But if an integer $a\in \ZZ$ satisfies $c^{a}\in\left\langle
c^{n}\right\rangle $, then we must have $n\mid a$ (this follows from the
structure of the infinite cyclic group $C$). Applied to $a=k$, this yields
$n\mid k$. But this is absurd, since $k\in\left\{  1,2,\ldots,n-1\right\}  $.
This contradiction proves that our assumption was wrong. Hence, we have proven
$\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  \neq w$.

Now, let us forget that we fixed $k$. We thus have shown that every
$k\in\left\{  1,2,\ldots,n-1\right\}  $ satisfies $\left(  w_{k+1}%
,w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  \neq w$. In other
words, Assertion $\mathcal{B}$ holds. This proves the implication
$\mathcal{A}\Longrightarrow\mathcal{B}$.

\textit{Proof of the implication $\mathcal{B}\Longrightarrow\mathcal{A}%
$:} Assume that Assertion $\mathcal{B}$ holds. In other words, every
$k\in\left\{  1,2,\ldots,n-1\right\}  $ satisfies $\left(  w_{k+1}%
,w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  \neq w$.

We will prove that $C^{\left(  n\right)  }=C_{w}$. Indeed, assume the
contrary. Thus, $C^{\left(  n\right)  }\neq C_{w}$. Since $C^{\left(
n\right)  }\subset C_{w}$, this yields that $C^{\left(  n\right)  }$ is a
proper subset of $C_{w}$. Hence, there exists some $d\in C_{w}$ such that
$d\notin C^{\left(  n\right)  }$. Consider this $d$. Write $d$ in the form
$d=c^{h}$ for some $h\in \ZZ$ (this is possible since $c$ generates
$C$). Let $k$ denote the remainder of $h$ modulo $n$. Then, $h-k$ is divisible
by $n$, and thus $c^{h-k}\in\left\langle c^{n}\right\rangle =C^{\left(
n \right)} \subset C_{w}$, so that $c^{h-k}w=w$. But also, $c^{h}w=w$ (since
$c^{h}=d\in C_{w}$). Hence, $w=\underbrace{c^{h}}_{=c^{k}c^{h-k}}
w=c^{k}\underbrace{c^{h-k}w}_{=w}=c^{k}w$.

But $k\in\left\{  0,1,\ldots,n-1\right\}  $ (since $k$ is a remainder modulo
$n$). We have $c^{h-k}\neq c^{h}$ (since $c^{h-k}\in C^{\left(  n\right)  }$
whereas $c^{h}=d\notin C^{\left(  n\right)  }$). Hence, $h-k\neq h$, so that
$k\neq0$. Combined with $k\in\left\{  0,1,\ldots,n-1\right\}  $, this yields
$k\in\left\{  0,1,\ldots,n-1\right\}  \setminus\left\{  0\right\}  =\left\{
1,2,\ldots,n-1\right\}  $. Thus, Assertion $\mathcal{B}$ yields that $\left(
w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  \neq w$.

Recall that $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples
one step to the left. Hence, $c^{k}$ acts on $\mathfrak{A}^{n}$ by cyclically
rotating $n$-tuples $k$ steps to the left. Hence,
\[
c^{k}w=\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)
\neq w=c^{k}w.
\]
This is absurd. This contradiction shows that our assumption was wrong. Hence,
we have shown that $C^{\left(  n\right)  }=C_{w}$. Now, $\left\vert
N\right\vert =\left[  C:\underbrace{C_{w}}_{=C^{\left(  n\right)  }}\right]
=\left[  C:C^{\left(  n\right)  }\right]  =n$. In other words, the period of
$N$ is $n$ (since the period of $N$ is defined as $\left\vert N\right\vert $).
In other words, $N$ is aperiodic (by the definition of ``aperiodic'').
In other words, $\left[  w\right]  $ is
aperiodic (since $N=\left[  w\right]  $). In other words, Assertion
$\mathcal{A}$ holds. This proves the implication $\mathcal{B}\Longrightarrow
\mathcal{A}$.

Now we have proven both implications $\mathcal{A}\Longrightarrow\mathcal{B}$
and $\mathcal{B}\Longrightarrow\mathcal{A}$. Hence, the Assertions
$\mathcal{A}$ and $\mathcal{B}$ are equivalent. Exercise
\ref{exe.words.necklaces}(b) is solved.

Before we come to the solution of Exercise \ref{exe.words.necklaces}(c), let
us state a simple lemma:

\begin{lemma}
\label{lem.sol.words.necklaces.lyndon-rot}Let $n$ be a positive integer.
Assume that the set $\mathfrak{A}$ is totally ordered. Let $w\in
\mathfrak{A}^{n}$ be a Lyndon word. Let $k\in\left\{  1,2,\ldots,n-1\right\}
$. Then, $c^{k}w>w$ in the lexicographic order.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.words.necklaces.lyndon-rot}.]We have $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ (since $w\in\mathfrak{A}^{n}$). Let
$u=\left(  w_{1},w_{2},\ldots,w_{k}\right)  $ and $v=\left(  w_{k+1}
,w_{k+2},\ldots,w_{n}\right)  $. These words $u$ and $v$ are well-defined and
nonempty (since $k\in\left\{  1,2,\ldots,n-1\right\}  $) and satisfy
\begin{align*}
uv & = \left(w_1, w_2, \ldots, w_k\right) \left(w_{k+1}, w_{k+2}, \ldots,
w_n\right) = \left(  w_{1},w_{2},\ldots,w_{k},w_{k+1},w_{k+2},\ldots
,w_{n}\right) \\
& =  \left(  w_{1},w_{2},\ldots,w_{n}\right)  =w .
\end{align*}

But the $n$-tuple $c^{k}w$ is obtained from $w$ by $k$-fold cyclic rotation to
the left (since $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating
$n$-tuples one step to the left). In other words,
\begin{align*}
c^{k}w  &  =\left(  w_{k+1},w_{k+2},\ldots,w_{n},w_{1},w_{2},\ldots
,w_{k}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  \text{ and }k\in\left\{  1,2,\ldots,n-1\right\}
\right) \\
&  =\underbrace{\left(  w_{k+1},w_{k+2},\ldots,w_{n}\right)  }_{=v}%
\underbrace{\left(  w_{1},w_{2},\ldots,w_{k}\right)  }_{=u}%
=vu>uv\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.words.lyndon.trivia}(c)}\right) \\
&  =w.
\end{align*}
This proves Lemma \ref{lem.sol.words.necklaces.lyndon-rot}.
\end{proof}

(c) Let $N$ be any aperiodic $n$-necklace. We need to show that $N$ contains
exactly one Lyndon word.

Since the necklace $N$ is aperiodic, we know that the period of $N$ is $n$. In
other words, $\left\vert N\right\vert =n$.

Let $w$ be the lexicographically smallest word contained in $N$. The word $w$
has $n$ letters (since $w\in N\subset\mathfrak{A}^{n}$), and thus is nonempty.
We will prove that the word $w$ is Lyndon.

Clearly, $N$ is a $C$-orbit (since $N$ is an $n$-necklace), and thus $N$ is
the orbit of the word $w$ (since $w\in N$). In other words, $N=Cw$.

We can see (as in the solution of Exercise \ref{exe.words.necklaces}(a)) that
$C^{\left(n\right)} \subset C_w$ and
$\left\vert N\right\vert =\left[  C:C_{w}\right]  $. Now, $n=\left[
C:C^{\left(  n\right)  }\right]  =\underbrace{\left[  C:C_{w}\right]
}_{=\left\vert N\right\vert =n}\cdot\left[  C_{w}:C^{\left(  n\right)
}\right]  =n\cdot\left[  C_{w}:C^{\left(  n\right)  }\right]  $. Thus,
$\left[  C_{w}:C^{\left(  n\right)  }\right]  =1$, so that $C_{w}=C^{\left(
n\right)  }$.

Now, let $u$ and $v$ be two nonempty words satisfying $w=uv$. We will prove
that $vu>uv$.

Assume the contrary. Thus, $vu\leq uv$.

We have $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $ (since $w\in
N\subset\mathfrak{A}^{n}$). Thus, there exists some $k\in\left\{
0,1,\ldots,n\right\}  $ such that $u=\left(  w_{1},w_{2},\ldots,w_{k}\right)
$ and $v=\left(  w_{k+1},w_{k+2},\ldots,w_{n}\right)  $ (since $w=uv$).
Consider this $k$. We have $0<k<n$ (since $u$ and $v$ are nonempty). Since
$v=\left(  w_{k+1},w_{k+2},\ldots,w_{n}\right)  $ and $u=\left(  w_{1}%
,w_{2},\ldots,w_{k}\right)  $, we have $vu=\left(  w_{k+1},w_{k+2}%
,\ldots,w_{n},w_{1},w_{2},\ldots,w_{k}\right)  $. In other words, the
$n$-tuple $vu$ is obtained from $w$ by $k$-fold cyclic rotation to the left.
In yet other words, $vu=c^{k}w$ (since $c$ acts on $\mathfrak{A}^{n}$ by
cyclically rotating $n$-tuples one step to the left). Hence,
$vu=\underbrace{c^{k}}_{\in C}w\in Cw=N$. Thus, $vu\geq w$ (since $w$ is the
lexicographically smallest word contained in $N$). Combined with $vu\leq
uv=w$, this yields $vu=w$. Hence, $c^{k}w=vu=w$, so that $c^{k}$ stabilizes
$w$. In other words, $c^{k}\in C_{w}=C^{\left(  n\right)  }$. But this is
impossible, since $0<k<n$ (and since $C^{\left(  n\right)  }$ is the subgroup
$\left\langle c^{n}\right\rangle $ of $C$). This contradiction proves that our
assumption was wrong. Thus, we have proven that $vu>uv$.

Let us now forget that we fixed $u$ and $v$. We thus have proven that any
nonempty words $u$ and $v$ satisfying $w=uv$ satisfy $vu>uv$. In other words,
the word $w$ satisfies Assertion $\mathcal{D}$ of Theorem
\ref{thm.words.lyndon.equiv}. Consequently, the word $w$ satisfies Assertion
$\mathcal{A}$ of Theorem \ref{thm.words.lyndon.equiv} as well (since Theorem
\ref{thm.words.lyndon.equiv} yields that these two assertions are equivalent);
in other words, $w$ is Lyndon. The orbit $N$ thus contains at least one Lyndon
word (namely, $w$).

We shall next prove that $w$ is the only Lyndon word in $N$. Indeed, let $p$
be any Lyndon word in $N$ distinct from $w$. We will derive a contradiction.

It is easy to see that there exists a $k\in\left\{  1,2,\ldots,n-1\right\}  $
satisfying $p=c^{k}w$\ \ \ \ \footnote{\textit{Proof.} We have $p\in N=Cw$.
Hence, there exists some $e\in C$ satisfying $p=ew$. Consider this $e$.
\par
We know that $C^{\left(  n\right)  }$ acts trivially on $\mathfrak{A}^{n}$,
and thus stabilizes $w$. Hence, $C^{\left(  n\right)  }w=\left\{  w\right\}
$.
\par
By the well-known properties of the infinite cyclic group, we know that the
elements $c^{0}$, $c^{1}$, $\ldots$, $c^{n-1}$ of $C$ form a system of unique
representatives for the cosets of $C^{\left(  n\right)  }$ in $C$. Hence,
every $d\in C$ belongs to the coset $c^{\ell}C^{\left(  n\right)  }$
for some $\ell\in\left\{  0,1,\ldots,n-1\right\}  $. Applied to $d=e$, this
yields that $e$ belongs to the coset $c^{\ell}C^{\left(  n\right)  }$ for some
$\ell\in\left\{  0,1,\ldots,n-1\right\}  $. Consider this $\ell$. We have
$e\in c^{\ell}C^{\left(  n\right)  }$, thus $p=\underbrace{e}_{\in c^{\ell
}C^{\left(  n\right)  }}w\in c^{\ell}\underbrace{C^{\left(  n\right)  }%
w}_{=\left\{  w\right\}  }=c^{\ell}\left\{  w\right\}  =\left\{  c^{\ell
}w\right\}  $, so that $p=c^{\ell}w$.
\par
If $\ell=0$, then $p=\underbrace{c^{\ell}}_{=c^{0}=1_{C}}w=1_{C}w=w$, which
contradicts the fact that $p$ is distinct from $w$. Hence, we cannot have
$\ell=0$. Thus, $\ell\in\left\{  0,1,\ldots,n-1\right\}  \setminus\left\{
0\right\}  =\left\{  1,2,\ldots,n-1\right\}  $. Hence, there exists a
$k\in\left\{  1,2,\ldots,n-1\right\}  $ satisfying $p=c^{k}w$ (namely,
$k=\ell$), qed.}. Consider this $k$. Then, $w=c^{n-k}p$%
\ \ \ \ \footnote{\textit{Proof.} We know that $c^{n}$ acts trivially on
$\mathfrak{A}^{n}$. Thus, $c^{n}w=w$. Hence, $w=\underbrace{c^{n}}%
_{=c^{n-k}c^{k}}w=c^{n-k}\underbrace{c^{k}w}_{=p}=c^{n-k}p$, qed.}. Notice
that $n-k\in\left\{  1,2,\ldots,n-1\right\}  $ (since $k\in\left\{
1,2,\ldots,n-1\right\}  $) and $p\in N\subset\mathfrak{A}^{n}$. Hence, Lemma
\ref{lem.sol.words.necklaces.lyndon-rot} (applied to $p$ and $n-k$ instead of
$w$ and $k$) yields that $c^{n-k}p>p$. Hence, $p<c^{n-k}p=w$. Since $p\in N$,
this shows that there exists an element of $N$ which is lexicographically
smaller than $w$ (namely, $p$). This contradicts the fact that $w$ is the
lexicographically smallest word contained in $N$.

Now, let us forget that we fixed $w$. We thus have obtained a contradiction
for every Lyndon word $p$ in $N$ distinct from $w$. Thus, there exists no
Lyndon word $p$ in $N$ distinct from $w$. Hence, $w$ is the only Lyndon word
in $N$. Since we already know that $w$ is a Lyndon word in $N$, this yields
that $N$ contains exactly one Lyndon word (namely, $w$). This solves Exercise
\ref{exe.words.necklaces}(c).

(d) Let $N$ be an $n$-necklace which is not aperiodic. We shall prove that $N$
contains no Lyndon word.

Indeed, assume the contrary. Then, $N$ contains a Lyndon word. Let $w$ be this
Lyndon word.

But the $n$-necklace $N$ is not aperiodic. Thus, $\left\vert N\right\vert \neq
n$ (since $N$ is aperiodic if and only if $\left\vert N\right\vert =n$). We
can see (as in the solution of Exercise \ref{exe.words.necklaces}(a)) that
$\left\vert N\right\vert =\left[  C:C_{w}\right]  $. Thus, $\left[
C:C_{w}\right]  =\left\vert N\right\vert \neq n=\left[  C:C^{\left(  n\right)
}\right]  $, so that $C_{w}\neq C^{\left(  n\right)  }$. But $C^{\left(
n\right)  }\subset C_{w}$ (this can be proven just as in the solution of
Exercise \ref{exe.words.necklaces}(a)). Hence, $C^{\left(  n\right)  }$ is a
proper subset of $C_{w}$. Hence, there exists some $e\in C_{w}$ such that
$e\notin C^{\left(  n\right)  }$. Consider this $e$.

We have $ew=w$ (since $e\in C_{w}$). It is now easy to see that there exists a
$k\in\left\{  1,2,\ldots,n-1\right\}  $ such that $c^{k}w=w$%
\ \ \ \ \footnote{\textit{Proof.} We know that $C^{\left(  n\right)  }$ acts
trivially on $\mathfrak{A}^{n}$, and thus stabilizes $w$. Hence, $C^{\left(
n\right)  }w=\left\{  w\right\}  $.
\par
By the well-known properties of the infinite cyclic group, we know that the
elements $c^{0}$, $c^{1}$, $\ldots$, $c^{n-1}$ of $C$ form a system of unique
representatives for the cosets of $C^{\left(  n\right)  }$ in $C$. Hence,
every $d\in C$ belongs to the coset $c^{\ell}C^{\left(  n\right)  }$
for some $\ell\in\left\{  0,1,\ldots,n-1\right\}  $. Applied to $d=e$, this
yields that $e$ belongs to the coset $c^{\ell}C^{\left(  n\right)  }$ for some
$\ell\in\left\{  0,1,\ldots,n-1\right\}  $. Consider this $\ell$. We have
$e\in c^{\ell}C^{\left(  n\right)  }$, thus $w=\underbrace{e}_{\in c^{\ell
}C^{\left(  n\right)  }}w\in c^{\ell}\underbrace{C^{\left(  n\right)  }%
w}_{=\left\{  w\right\}  }=c^{\ell}\left\{  w\right\}  =\left\{  c^{\ell
}w\right\}  $, so that $w=c^{\ell}w$. Thus, $c^{\ell}w=w$.
\par
If $\ell=0$, then $e\in\underbrace{c^{\ell}}_{=c^{0}=1_{C}}C^{\left(
n\right)  }=1^{C}C^{\left(  n\right)  }=C^{\left(  n\right)  }$, which
contradicts $e\notin C^{\left(  n\right)  }$. Hence, we cannot have $\ell=0$.
Thus, $\ell\in\left\{  0,1,\ldots,n-1\right\}  \setminus\left\{  0\right\}
=\left\{  1,2,\ldots,n-1\right\}  $. Hence, there exists a $k\in\left\{
1,2,\ldots,n-1\right\}  $ satisfying $c^{k}w=w$ (namely, $k=\ell$), qed.}.
Consider this $k$. Lemma \ref{lem.sol.words.necklaces.lyndon-rot} yields
$c^{k}w>w$ (since $w\in N\subset\mathfrak{A}^{n}$), which contradicts
$c^{k}w=w$. This contradiction proves that our assumption was false. Thus, $N$
contains no Lyndon word. This solves Exercise \ref{exe.words.necklaces}(d).

(e) If $w$ is a Lyndon word of length $n$, then $\left[  w\right]  $ is an
aperiodic $n$-necklace\footnote{\textit{Proof.} Let $w$ be a Lyndon word of
length $n$. Then, $\left[  w\right]  $ is an $n$-necklace. If $\left[
w\right]  $ was not aperiodic, then the necklace $\left[  w\right]  $ would
contain no Lyndon word (by Exercise \ref{exe.words.necklaces}(d), applied to
$N=\left[  w\right]  $), which would contradict the fact that this necklace
$\left[  w\right]  $ contains the Lyndon word $w$. Hence, $\left[  w\right]  $
must be aperiodic, qed.}. Hence, the map%
\begin{align*}
\left(  \text{the set of all Lyndon words of length }n\right)   &
\rightarrow\left(  \text{the set of all aperiodic }n\text{-necklaces}\right)
,\\
w  &  \mapsto\left[  w\right]
\end{align*}
is well-defined. Denote this map by $\Phi$. This map $\Phi$ is
injective\footnote{\textit{Proof.} Let $w$ and $w^{\prime}$ be two Lyndon
words of length $n$ such that $\Phi\left(  w\right)  =\Phi\left(  w^{\prime
}\right)  $. We shall prove that $w=w^{\prime}$.
\par
We have $\Phi\left(  w\right)  =\left[  w\right]  $ (by the definition of
$\Phi$) and $\Phi\left(  w^{\prime}\right)  =\left[  w^{\prime}\right]  $
(similarly). Thus, $\left[  w^{\prime}\right]  =\Phi\left(  w^{\prime}\right)
=\Phi\left(  w\right)  =\left[  w\right]  $.
\par
The word $w$ is contained in $\left[  w\right]  $ (obviously), and the word
$w^{\prime}$ is contained in $\left[  w^{\prime}\right]  =\left[  w\right]  $.
Thus, both $w$ and $w^{\prime}$ are contained in $\left[  w\right]  $.
\par
We know that $\left[  w\right]  $ is an aperiodic $n$-necklace. Hence,
$\left[  w\right]  $ contains exactly one Lyndon word (by Exercise
\ref{exe.words.necklaces}(c)). Hence, any two Lyndon words contained in
$\left[  w\right]  $ must be identical. Applying this to the two Lyndon words
$w$ and $w^{\prime}$ both contained in $\left[  w\right]  $, we obtain that
$w$ and $w^{\prime}$ are identical, i.e., we have $w=w^{\prime}$.
\par
Let us now forget that we fixed $w$ and $w^{\prime}$. We have thus proven that
any two Lyndon words $w$ and $w^{\prime}$ of length $n$ satisfying
$\Phi\left(  w\right)  =\Phi\left(  w^{\prime}\right)  $ must satisfy
$w=w^{\prime}$. In other words, the map $\Phi$ is injective, qed.} and
surjective\footnote{\textit{Proof.} Let $N$ be an aperiodic $n$-necklace.
Hence, $N$ contains exactly one Lyndon word (by Exercise
\ref{exe.words.necklaces}(c)). Let $w$ be this Lyndon word. The definition of
$\Phi$ yields $\Phi\left(  w\right)  =\left[  w\right]  =N$ (since $N$ is the
aperiodic $n$-necklace containing $w$). Thus, $N=\Phi\left(  w\right)
\in\operatorname{Im}\Phi$.
\par
Now, let us forget that we fixed $N$. We have thus shown that $N\in
\operatorname{Im}\Phi$ for every aperiodic $n$-necklace $N$. In other words,
the map $\Phi$ is surjective, qed.}. Hence, $\Phi$ is bijective. In other
words, $\Phi$ is a bijection between the set of all Lyndon words of length $n$
and the set of all aperiodic $n$-necklaces. Thus, the aperiodic $n$-necklaces
are in bijection with Lyndon words of length $n$. This solves Exercise
\ref{exe.words.necklaces}(e).

Before we start solving Exercise \ref{exe.words.necklaces}(f), we state a
lemma about words:

\begin{lemma}
\label{lem.sol.words.necklaces.rooting}Let $N$ be a positive integer, and let
$w\in\mathfrak{A}^{N}$. Let $p$ be a positive divisor of $N$. Assume that
$c^{p}w=w$. Then, there exists a word $q\in\mathfrak{A}^{p}$ such that
$w=q^{N/p}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.words.necklaces.rooting}.](The following proof is
overkill, but it is the simplest proof to formalize.)

Notice that $N/p$ is a positive integer (since $p$ is a positive divisor of
$N$), so that $N/p-1\in \NN$.

We have $w=\left(  w_{1},w_{2},\ldots,w_{N}\right)  $ (since $w\in
\mathfrak{A}^{N}$). Let $u=\left(  w_{1},w_{2},\ldots,w_{p}\right)  $ and
$v=\left(  w_{p+1},w_{p+2},\ldots,w_{N}\right)  $. (These $u$ and $v$ are
well-defined since $p\in\left\{  0,1,\ldots,N\right\}  $.) Then,
\begin{align*}
\underbrace{u}_{=\left(  w_{1},w_{2},\ldots,w_{p}\right)  }\underbrace{v}%
_{=\left(  w_{p+1},w_{p+2},\ldots,w_{N}\right)  }  &  =\left(  w_{1}%
,w_{2},\ldots,w_{p}\right)  \left(  w_{p+1},w_{p+2},\ldots,w_{N}\right) \\
&  =\left(  w_{1},w_{2},\ldots,w_{p},w_{p+1},w_{p+2},\ldots,w_{N}\right) \\
&  =\left(  w_{1},w_{2},\ldots,w_{N}\right)  =w.
\end{align*}
On the other hand, $c$ acts on $\mathfrak{A}^{N}$ by cyclically rotating
$N$-tuples one step to the left. Hence, $c^{p}w$ is the result of cyclically
rotating the $N$-tuple $w$ to the left $p$ times. In other words,%
\[
c^{p}w=\left(  w_{p+1},w_{p+2},\ldots,w_{N},w_{1},w_{2},\ldots,w_{p}\right)
=\underbrace{\left(  w_{p+1},w_{p+2},\ldots,w_{N}\right)  }_{=v}%
\underbrace{\left(  w_{1},w_{2},\ldots,w_{p}\right)  }_{=u}=vu.
\]
Compared with $c^{p}w=w=uv$, this yields $uv=vu$. Hence, Proposition
\ref{prop.words.uv=vu.2} yields that there exist a $t\in\mathfrak{A}^{\ast}$
and two nonnegative integers $n$ and $m$ such that $u=t^{n}$ and $v=t^{m}$.
Consider this $t$ and these $n$ and $m$. We have $u=\left(  w_{1},w_{2}%
,\ldots,w_{p}\right)  $, so that $u\in\mathfrak{A}^{p}$ and thus $\ell\left(
u\right)  =p$. Hence, $p=\ell\left(  \underbrace{u}_{=t^{n}}\right)
=\ell\left(  t^{n}\right)  =n\ell\left(  t\right)  $, so that $n\ell\left(
t\right)  =p$. On the other hand, $v=\left(  w_{p+1},w_{p+2},\ldots
,w_{N}\right)  $, so that $\ell\left(  v\right)  =N-p$ and thus $N-p=\ell
\left(  \underbrace{v}_{=t^{m}}\right)  =\ell\left(  t^{m}\right)
=m\ell\left(  t\right)  $ and thus $m\ell\left(  t\right)  =N-p$. Now,
$n\ell\left(  t\right)  =p\neq0$. Hence, $n\neq0$ and $\ell\left(  t\right)
\neq0$. Now,
\begin{align*}
\dfrac{m}{n}  &  =\dfrac{m\ell\left(  t\right)  }{n\ell\left(  t\right)
}=\dfrac{N-p}{p}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m\ell\left(
t\right)  =N-p\text{ and }n\ell\left(  t\right)  =p\right) \\
&  =N/p-1,
\end{align*}
so that $m=n\cdot\left(  N/p-1\right)  $. Hence, $t^{m}=t^{n\cdot\left(
N/p-1\right)  }=\left(  \underbrace{t^{n}}_{=u}\right)  ^{N/p-1}=u^{N/p-1}$.
Now, $w=u\underbrace{v}_{=t^{m}=u^{N/p-1}}=uu^{N/p-1}=u^{N/p}$. Hence, there
exists a word $q\in\mathfrak{A}^{p}$ such that $w=q^{N/p}$ (namely, $q=u$).
Lemma \ref{lem.sol.words.necklaces.rooting} is proven.
\end{proof}

We also recall a lemma about the functions $\mu$ and $\phi$:

\begin{lemma}
\label{lem.sol.words.necklaces.muphi}Every positive integer $n$ satisfies%
\begin{align}
\sum_{d\mid n}\mu\left(  d\right)   &  =\delta_{n,1}%
;\label{pf.lem.sol.words.necklaces.muphi.mu*1}\\
\sum_{d\mid n}\mu\left(  d\right)  \dfrac{n}{d}  &  =\phi\left(  n\right)  .
\label{pf.lem.sol.words.necklaces.muphi.mu*n}%
\end{align}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.words.necklaces.muphi}.]Both equalities
(\ref{pf.lem.sol.words.necklaces.muphi.mu*1}) and
(\ref{pf.lem.sol.words.necklaces.muphi.mu*n}) have been proven in the solution
of Exercise \ref{exe.witt.ghost-equiv}. (Indeed,
(\ref{pf.lem.sol.words.necklaces.muphi.mu*1}) is
(\ref{sol.exe.witt.ghost-equiv.mu*1}), and
(\ref{pf.lem.sol.words.necklaces.muphi.mu*n}) is
(\ref{sol.exe.witt.ghost-equiv.mu*n}).)
\end{proof}

In Exercise \ref{exe.words.necklaces}, an action of the cyclic group $C$
on $\mathfrak{A}^n$ was defined.
In the same way, we define an action of the cyclic group $C$ on
$\mathfrak{A}^m$ for any positive integer $m$.
The orbits of this latter $C$-action will be called the
\emph{$m$-necklaces}.
(The notion of ``$n$-necklaces'' defined in Exercise \ref{exe.words.necklaces}
is clearly a particular case of this.)
We define the notions of ``period'' and ``aperiodic'' for $m$-necklaces
in the same way as they were defined for $n$-necklaces in
Exercise \ref{exe.words.necklaces}.
We will use the same notations for $m$-necklaces as we do for $n$-necklaces
(i.e., the $m$-necklace containing a given $w \in \mathfrak{A}^m$ will
be denoted by $\left[w\right]$).

Let us now state a few lemmas about the actions of $C$
on words:

\begin{lemma}
\label{lem.sol.words.necklaces.f1}
Let $m$ be a positive integer.
Let $u \in \mathfrak{A}$ be a letter.
Let $v\in\mathfrak{A}^{m-1}$ be a word.
Then,
\[
c\left(  uv\right)  = vu
\]
(where we identify the letter $u$ with the one-letter word $\left(  u\right)
$).
\end{lemma}

\begin{proof}[Proof of Lemma \ref{lem.sol.words.necklaces.f1}.]
From $v\in \mathfrak{A}^{m-1}$, we obtain $v=\left(  v_{1},v_{2},\ldots
,v_{m-1}\right)  $, so that
\[
\underbrace{u}_{=\left(  u\right)  }\underbrace{v}_{=\left(  v_{1}%
,v_{2},\ldots,v_{m-1}\right)  }=\left(  u\right)  \left(  v_{1},v_{2}%
,\ldots,v_{m-1}\right)  =\left(  u,v_{1},v_{2},\ldots,v_{m-1}\right)  .
\]
\par
But recall that $c$ acts on $\mathfrak{A}^m$ by cyclically rotating
$m$-tuples one step to the left. Thus,
\[
c\left(  u,v_{1},v_{2},\ldots,v_{m-1}\right)  =\left(  v_{1},v_{2}%
,\ldots,v_{m-1},u\right)  =\underbrace{\left(  v_{1},v_{2},\ldots
,v_{m-1}\right)  }_{=v}\underbrace{\left(  u\right)  }_{=u}=vu.
\]
Hence,
\[
c\underbrace{\left(  uv\right)  }_{=\left(  u,v_{1},v_{2},\ldots
,v_{m-1}\right)  }=c\left(  u,v_{1},v_{2},\ldots,v_{m-1}\right)  =vu,
\]
and thus Lemma \ref{lem.sol.words.necklaces.f1} is proven.
\end{proof}

\begin{lemma}
\label{lem.sol.words.necklaces.f2}
Let $n$ be a positive integer.
Let $d$ be a positive divisor of $n$.
Thus, $n/d$ is a positive integer.

Define a map
\begin{align*}
\Delta:\mathfrak{A}^{n/d}  &  \rightarrow\mathfrak{A}^{n},\\
w  &  \mapsto w^{d}
\end{align*}
(where, as we recall, $w^{d}$ means the $d$-fold concatenation
$\underbrace{ww\cdots w}_{d\text{ times}}$ of $w$ with itself).

Then:

\begin{enumerate}
\item[(a)] This map $\Delta$ is well-defined.

\item[(b)] The map $\Delta$ is $C$-equivariant
(meaning that $\Delta\left(
gw\right)  =g\cdot\Delta\left(  w\right)  $ for every $g\in C$ and
$w\in\mathfrak{A}^{n/d}$).

\item[(c)] The map $\Delta$ is injective.
\end{enumerate}
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem.sol.words.necklaces.f2}.]
(a) If $w \in \mathfrak{A}^{n/d}$, then the word $w^d$ has length
\[
\ell\left(w^d\right)
= d \underbrace{\ell\left(w\right)}_{\substack{= n/d\\ \text{(since $w \in \mathfrak{A}^{n/d}$)}}}
= d \left(n/d\right) = n
\]
and thus satisfies $w^d \in \mathfrak{A}^n$.
Hence, the map $\Delta$ is well-defined.
This proves Lemma~\ref{lem.sol.words.necklaces.f2}(a).

(b) We need to show
that $\Delta$ is $C$-equivariant. Clearly, it is enough to prove that
$\Delta\left(  cw\right)  =c\cdot\Delta\left(  w\right)  $ for every
$w\in\mathfrak{A}^{n/d}$ (since $c$ generates the group $C$). So let us prove
this.
\par
Let $w\in\mathfrak{A}^{n/d}$. Thus, $w=\left(  w_{1},w_{2},\ldots
,w_{n/d}\right)  $. Let $\overline{w}$ denote the word $\left(  w_{2}%
,w_{3},\ldots,w_{n/d}\right)  $. Then, (identifying the letter $w_{1}$ with
the one-letter word $\left(  w_{1}\right)  $) we have%
\[
\underbrace{w_{1}}_{=\left(  w_{1}\right)  }\underbrace{\overline{w}%
}_{=\left(  w_{2},w_{3},\ldots,w_{n/d}\right)  }=\left(  w_{1}\right)  \left(
w_{2},w_{3},\ldots,w_{n/d}\right)  =\left(  w_{1},w_{2},w_{3},\ldots
,w_{n/d}\right)  =\left(  w_{1},w_{2},\ldots,w_{n/d}\right)  =w.
\]
Thus, $w=w_{1}\overline{w}$, so that%
\[
cw=c\left(  w_{1}\overline{w}\right)  =\overline{w}w_{1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.sol.words.necklaces.f1},
applied to }m=n/d\text{, }u=w_{1}\text{ and }v=\overline{w}\right)  .
\]
\par
Also, it is easy to see that
\begin{align}
\left(pq\right)^d = p\left(qp\right)^{d-1}q
\ \ \ \ \ \ \ \ \ \ \text{for any $p \in \mathfrak{A}^*$ and $q \in \mathfrak{A}^*$}.
\label{pf.lem.sol.words.necklaces.f2.pqd}
\end{align}
\footnote{\textit{Proof.} Let
$p \in \mathfrak{A}^*$ and $q \in \mathfrak{A}^*$.
Then,
\[
\left(pq\right)^d =\underbrace{\left(
pq\right)  \left(  pq\right)  \cdots\left(
pq\right)  }_{d\text{ times}}
= p
\underbrace{\underbrace{\left(  qp\right)  \left( qp\right)
\cdots\left(  qp\right)  }_{d-1\text{ times}}%
}_{=\left(  qp\right)  ^{d-1}} q
= p\left( qp\right)  ^{d-1} q,
\]
qed.}.

Taking both sides of the equality $w=w_{1}\overline{w}$ to the $d$-th
power, we obtain%
\[
w^{d}=\left(  w_{1}\overline{w}\right)  ^{d}
= w_{1}\left( \overline{w}w_{1}\right)  ^{d-1}\overline{w}
\ \ \ \ \ \ \ \ \ \ \ \left(\text{by \eqref{pf.lem.sol.words.necklaces.f2.pqd},
applied to $p = w_1$ and $q = \overline{w}$}\right) .
\]
The definition of $\Delta$ yields $\Delta\left(  w\right)
=w^{d}=w_{1}\left(  \overline{w}w_{1}\right)  ^{d-1}\overline{w}$. Applying
$c$ to both sides of this identity, we obtain%
\begin{align*}
c\cdot\Delta\left(  w\right)
&  =c\cdot\left(  w_{1}\left(  \overline{w}w_{1}\right)  ^{d-1}
\overline{w}\right)  =\left(  \overline{w}w_{1}\right)
^{d-1}\overline{w}w_{1}
\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Lemma \ref{lem.sol.words.necklaces.f1}, applied to }m=n\text{, }%
u=w_{1}\text{ and }v=\left(  \overline{w}w_{1}\right)  ^{d-1}
\overline{w}\right) \\
&  =\left(  \overline{w}w_{1}\right)  ^{d-1}\left(  \overline{w}w_{1}\right)
=\left(  \underbrace{\overline{w}w_{1}}_{=cw}\right)  ^{d}=\left(  cw\right)
^{d}.
\end{align*}
Compared with $\Delta\left(  cw\right)  =\left(  cw\right)  ^{d}$ (by the
definition of $\Delta$), this yields $\Delta\left(  cw\right)  =c\cdot
\Delta\left(  w\right)  $. We thus have shown that $\Delta$ is $C$%
-equivariant.
This proves Lemma~\ref{lem.sol.words.necklaces.f2}(b).

(c) We need to prove
that $\Delta$ is injective. In other words, we need to prove that every
$w\in\mathfrak{A}^{n/d}$ can be reconstructed from $\Delta\left(  w\right)  $.
But this is easy: Since $\Delta\left(  w\right)  =w^{d}$ (by the definition of
$\Delta$), the word $w$ can be obtained from $\Delta\left(  w\right)  $ by
taking the first $n/d$ letters of $\Delta\left(  w\right)  $. Hence, $w$ can
be reconstructed from $\Delta\left(  w\right)  $.
This proves Lemma~\ref{lem.sol.words.necklaces.f2}(c).

\end{proof}

We now step to the solution of Exercise \ref{exe.words.necklaces}(f):

(f) Exercise \ref{exe.words.necklaces}(a) yields that for any $n$-necklace
$N$, we have $\left\vert N\right\vert \mid n$. Hence, for any $n$-necklace
$N$, the cardinality $\left\vert N\right\vert $ is a divisor of $n$ (since
$\left\vert N\right\vert \in \NN$). Now, recall that the $n$-necklaces
are the orbits of the $C$-action on $\mathfrak{A}^{n}$, and therefore form a
set partition of the set $\mathfrak{A}^{n}$. Hence,%
\[
\left\vert \mathfrak{A}^{n}\right\vert =\sum_{N\text{ is an }n\text{-necklace}%
}\left\vert N\right\vert =\sum_{d\mid n}\sum\limits_{\substack{N\text{ is an
}n\text{-necklace;}\\\left\vert N\right\vert =d}}d
\]
(because for any $n$-necklace $N$, the cardinality $\left\vert N\right\vert $
is a divisor of $n$). Hence,%
\begin{equation}
\left\vert \mathfrak{A}\right\vert ^{n}=\left\vert \mathfrak{A}^{n}\right\vert
=\sum_{d\mid n}\underbrace{\sum\limits_{\substack{N\text{ is an }n\text{-necklace;}%
\\\left\vert N\right\vert =d}}d}_{=d\left\vert \left\{  N\text{ is an
}n\text{-necklace }\mid\ \left\vert N\right\vert =d\right\}  \right\vert
}=\sum_{d\mid n}d\left\vert \left\{  N\text{ is an }n\text{-necklace }%
\mid\ \left\vert N\right\vert =d\right\}  \right\vert .
\label{sol.words.necklaces.e.1}%
\end{equation}


Now, for every positive integer $e$, let $\operatorname*{Aper}\left(
e\right)  $ denote the set of all aperiodic $e$-necklaces. We shall now prove
that%
\begin{equation}
\left\vert \left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert
N\right\vert =n/d\right\}  \right\vert =\left\vert \operatorname*{Aper}\left(
n/d\right)  \right\vert \label{sol.words.necklaces.e.2}%
\end{equation}
for every divisor $d$ of $n$.

\textit{Proof of (\ref{sol.words.necklaces.e.2}):}
Let $d$ be a divisor of $n$. Then, $n/d$ is a positive integer. We can thus
define a map
\begin{align*}
\Delta:\mathfrak{A}^{n/d}  &  \rightarrow\mathfrak{A}^{n},\\
w  &  \mapsto w^{d}%
\end{align*}
(where, as we recall, $w^{d}$ means the $d$-fold concatenation
$\underbrace{ww\cdots w}_{d\text{ times}}$ of $w$ with itself).
Lemma~\ref{lem.sol.words.necklaces.f2}(a) shows that this map
$\Delta$ is well-defined.
Lemma~\ref{lem.sol.words.necklaces.f2}(b) shows that this map
$\Delta$ is $C$-equivariant (meaning that $\Delta\left(
gw\right)  =g\cdot\Delta\left(  w\right)  $ for every $g\in C$ and
$w\in\mathfrak{A}^{n/d}$).
Lemma~\ref{lem.sol.words.necklaces.f2}(c) shows that this
map $\Delta$ is injective.

Since the map $\Delta$ is $C$-equivariant, it gives rise to a map%
\begin{align*}
\overline{\Delta}:\left(  \text{the set of all }C\text{-orbits on
}\mathfrak{A}^{n/d}\right)   &  \rightarrow\left(  \text{the set of all
}C\text{-orbits on }\mathfrak{A}^{n}\right)  ,\\
N  &  \mapsto\Delta\left(  N\right)  .
\end{align*}
Consider this map $\overline{\Delta}$. The map $\overline{\Delta}$ is
injective\footnote{\textit{Proof.} Let $P$ and $Q$ be two $C$-orbits on
$\mathfrak{A}^{n/d}$ such that $\overline{\Delta}\left(  P\right)
=\overline{\Delta}\left(  Q\right)  $. We want to show that $P=Q$.
\par
By the definition of $\overline{\Delta}$, we have $\overline{\Delta}\left(
P\right)  =\Delta\left(  P\right)  $ and $\overline{\Delta}\left(  Q\right)
=\Delta\left(  Q\right)  $. Thus, $\Delta\left(  P\right)  =\overline{\Delta
}\left(  P\right)  =\overline{\Delta}\left(  Q\right)  =\Delta\left(
Q\right)  $.
\par
The orbit $P$ is nonempty, and thus contains an element. Let $p$ be such an
element.
\par
We have $\Delta\left(  \underbrace{p}_{\in P}\right)  \in\Delta\left(
P\right)  =\Delta\left(  Q\right)  $. Hence, there exists some $q\in Q$ such
that $\Delta\left(  p\right)  =\Delta\left(  q\right)  $. Consider this $q$.
We have $\Delta\left(  p\right)  =\Delta\left(  q\right)  $, and thus $p=q$
(since $\Delta$ is injective). The element $p$ belongs to $P\cap Q$ (since
$p\in P$ and $p=q\in Q$), and thus the two orbits $P$ and $Q$ have an element
in common (namely, $p$). But any two orbits which have an element in common
must be identical. Thus, $P$ and $Q$ are identical, i.e., we have $P=Q$.
\par
Let us forget that we fixed $P$ and $Q$. We thus have shown that any two
$C$-orbits $P$ and $Q$ on $\mathfrak{A}^{n/d}$ which satisfy
$\overline{\Delta}\left(  P\right)  =\overline{\Delta}\left(  Q\right)  $
must satisfy
$P=Q$. In other words, the map $\overline{\Delta}$ is injective.}.
Furthermore, $\overline{\Delta}$ is a map from the set of all $C$-orbits on
$\mathfrak{A}^{n/d}$ to the set of all $C$-orbits on $\mathfrak{A}^{n}$. In
other words, $\overline{\Delta}$ is a map from the set of all $\left(
n/d\right)  $-necklaces to the set of all $n$-necklaces (since the $C$-orbits
on $\mathfrak{A}^{n/d}$ are the $\left(  n/d\right)  $-necklaces, and the
$C$-orbits on $\mathfrak{A}^{n}$ are the $n$-necklaces). It is now easy to see
that
\begin{equation}
\overline{\Delta}\left(  \operatorname*{Aper}\left(  n/d\right)  \right)
\subset\left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert
=n/d\right\}  . \label{sol.words.necklaces.e.2.pf.oneside}%
\end{equation}
\footnote{\textit{Proof.} Let $M\in\operatorname*{Aper}\left(  n/d\right)  $.
Then, $M$ is an aperiodic $\left(  n/d\right)  $-necklace (since
$\operatorname*{Aper}\left(  n/d\right)  $ is the set of all aperiodic
$\left(  n/d\right)  $-necklaces). In other words, $M$ is an $\left(
n/d\right)  $-necklace whose period is $n/d$.
\par
The period of $M$ is defined to be $\left\vert M\right\vert $. Thus,
$\left\vert M\right\vert $ is $n/d$ (since the period of $M$ is $n/d$). In
other words, $\left\vert M\right\vert =n/d$. But the definition of
$\overline{\Delta}$ yields $\overline{\Delta}\left(  M\right)  =\Delta\left(
M\right)  $, thus
\begin{align*}
\left\vert \overline{\Delta}\left(  M\right)  \right\vert  &  =\left\vert
\Delta\left(  M\right)  \right\vert =\left\vert M\right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is injective}\right) \\
&  =n/d.
\end{align*}
Hence, $\overline{\Delta}\left(  M\right)  \in\left\{  N\text{ is an
}n\text{-necklace }\mid\ \left\vert N\right\vert =n/d\right\}  $
(since $\overline{\Delta}\left(  M\right)$ is an $n$-necklace (because
$\overline{\Delta}$ is a map from the set of all $\left(
n/d\right)  $-necklaces to the set of all $n$-necklaces)).
\par
Now, let us forget that we fixed $M$. We thus have proven that
$\overline{\Delta}\left(  M\right)  \in
\left\{  N\text{ is an }n\text{-necklace }
\mid\ \left\vert N\right\vert =n/d\right\}  $ for every $M\in
\operatorname*{Aper}\left(  n/d\right)  $. In other words, $\overline{\Delta
}\left(  \operatorname*{Aper}\left(  n/d\right)  \right)  \subset\left\{
N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert =n/d\right\}
$. This proves (\ref{sol.words.necklaces.e.2.pf.oneside}).} But we also have
\begin{equation}
\left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert
=n/d\right\}  \subset\overline{\Delta}\left(  \operatorname*{Aper}\left(
n/d\right)  \right)  .
\label{sol.words.necklaces.e.2.pf.otherside}
\end{equation}


\textit{Proof of (\ref{sol.words.necklaces.e.2.pf.otherside}):} Let
$P\in\left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert
=n/d\right\}  $. Thus, $P$ is an $n$-necklace such that $\left\vert
P\right\vert =n/d$. We shall now prove that $P\in\overline{\Delta}\left(
\operatorname*{Aper}\left(  n/d\right)  \right)  $.

The set $P$ is an $n$-necklace, thus a $C$-orbit on $\mathfrak{A}^{n}$, thus
nonempty. Pick some $w\in P$ (this clearly exists since $P$ is nonempty). Then
$P$ is the $C$-orbit of $w$. Hence, by the orbit-stabilizer theorem, we have
$\left\vert P\right\vert =\left[  C:C_{w}\right]  $. Hence, $\left[
C:C_{w}\right]  =\left\vert P\right\vert =n/d$. Hence, $C_{w}$ is a subgroup
of $C$ having index $n/d$. Since the only subgroup of $C$ having index $n/d$
is $\left\langle c^{n/d}\right\rangle $\ \ \ \ \footnote{This is a particular
case of the following elementary fact: If $e$ is a positive integer, then the
only subgroup of $C$ having index $e$ is $\left\langle c^{e}\right\rangle $.
(This is because $C\cong\left(   \ZZ ,+\right)  $, and because the only
subgroup of $\left(   \ZZ ,+\right)  $ having index $e$ is $e \ZZ %
$.)}, this yields that $C_{w}$ is $\left\langle c^{n/d}\right\rangle $. Hence,
$C_{w}=\left\langle c^{n/d}\right\rangle $, so that $c^{n/d}\in\left\langle
c^{n/d}\right\rangle =C_{w}$, and thus $c^{n/d}w=w$. Now, Lemma
\ref{lem.sol.words.necklaces.rooting} (applied to $N=n$ and $p=n/d$) yields
that there exists a word $q\in\mathfrak{A}^{n/d}$ such that $w=q^{n/\left(
n/d\right)  }$. Consider this $q$. We have $w=q^{n/\left(  n/d\right)  }%
=q^{d}$, whereas the definition of $\Delta$ yields $\Delta\left(  q\right)
=q^{d}$. Thus, $w=q^{d}=\Delta\left(  \underbrace{q}_{\in\left[  q\right]
}\right)  \in\Delta\left(  \left[  q\right]  \right)  $. The definition of
$\overline{\Delta}$ yields $\overline{\Delta}\left(  \left[  q\right]
\right)  =\Delta\left(  \left[  q\right]  \right)  $. Hence, $w\in
\Delta\left(  \left[  q\right]  \right)  =\overline{\Delta}\left(  \left[
q\right]  \right)  $. Thus, $\overline{\Delta}\left(  \left[  q\right]
\right)  $ is the $C$-orbit on $\mathfrak{A}^{n}$ containing $w$ (since
$\overline{\Delta}\left(  \left[  q\right]  \right)  $ is a $C$-orbit on
$\mathfrak{A}^{n}$ (since $\overline{\Delta}\left(  \left[  q\right]  \right)
$ is an $n$-necklace)). In other words, $\overline{\Delta}\left(  \left[
q\right]  \right)  $ is the $C$-orbit of $w$. Hence, $\overline{\Delta}\left(
\left[  q\right]  \right)  =P$ (since $P$ is the $C$-orbit of $w$).

But $P=\overline{\Delta}\left(  \left[  q\right]  \right)  =\Delta\left(
\left[  q\right]  \right)  $ and thus $\left\vert P\right\vert =\left\vert
\Delta\left(  \left[  q\right]  \right)  \right\vert =\left\vert \left[
q\right]  \right\vert $ (since $\Delta$ is injective), so that $\left\vert
\left[  q\right]  \right\vert =\left\vert P\right\vert =n/d$. By the
definition of the period of an $\left(  n/d\right)  $-necklace, we see that
the period of the $\left(  n/d\right)  $-necklace $\left[  q\right]  $ is
$\left\vert \left[  q\right]  \right\vert =n/d$. In other words, the $\left(
n/d\right)  $-necklace $\left[  q\right]  $ is aperiodic (by the definition of
``aperiodic''). In other words, $\left[
q\right]  \in\operatorname*{Aper}\left(  n/d\right)  $ (since
$\operatorname*{Aper}\left(  n/d\right)  $ is the set of all aperiodic
$\left(  n/d\right)  $-necklaces). Now, $P = \overline{\Delta}\left(
\underbrace{\left[  q\right]  }_{\in \operatorname*{Aper}\left(  n/d\right)
}\right)  \in\overline{\Delta}\left(
\operatorname*{Aper}\left(  n/d\right)  \right)  $.

Let us now forget that we fixed $P$. We thus have proven that every
$P\in\left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert
=n/d\right\}  $ satisfies $P \in \overline{\Delta}\left(  \operatorname*{Aper}%
\left(  n/d\right)  \right)  $. In other words, $\left\{  N\text{ is an
}n\text{-necklace }\mid\ \left\vert N\right\vert =n/d\right\}  \subset
\overline{\Delta}\left(  \operatorname*{Aper}\left(  n/d\right)  \right)  $.
This proves (\ref{sol.words.necklaces.e.2.pf.otherside}).

Combining (\ref{sol.words.necklaces.e.2.pf.oneside}) with
(\ref{sol.words.necklaces.e.2.pf.otherside}), we obtain%
\[
\overline{\Delta}\left(  \operatorname*{Aper}\left(  n/d\right)  \right)
=\left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert
=n/d\right\}  .
\]
Hence,%
\[
\left\vert \overline{\Delta}\left(  \operatorname*{Aper}\left(  n/d\right)
\right)  \right\vert =\left\vert \left\{  N\text{ is an }n\text{-necklace
}\mid\ \left\vert N\right\vert =n/d\right\}  \right\vert ,
\]
so that%
\[
\left\vert \left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert
N\right\vert =n/d\right\}  \right\vert =\left\vert \overline{\Delta}\left(
\operatorname*{Aper}\left(  n/d\right)  \right)  \right\vert =\left\vert
\operatorname*{Aper}\left(  n/d\right)  \right\vert
\]
(since $\overline{\Delta}$ is injective). This proves
(\ref{sol.words.necklaces.e.2}).

Now, every divisor $d$ of $n$ satisfies%
\begin{align}
\left\vert \left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert
N\right\vert =\underbrace{d}_{=n/\left(  n/d\right)  }\right\}  \right\vert
&  =\left\vert \left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert
N\right\vert =n/\left(  n/d\right)  \right\}  \right\vert \nonumber\\
&  =\left\vert \operatorname*{Aper}\left(  \underbrace{n/\left(  n/d\right)
}_{=d}\right)  \right\vert \nonumber\\
&\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.words.necklaces.e.2}), applied to }n/d\text{ instead of }d\right)
\nonumber\\
&  =\left\vert \operatorname*{Aper}\left(  d\right)  \right\vert .
\label{sol.words.necklaces.e.3}%
\end{align}
Now, (\ref{sol.words.necklaces.e.1}) becomes%
\begin{equation}
\left\vert \mathfrak{A}\right\vert ^{n}=\sum_{d\mid n}d\underbrace{\left\vert
\left\{  N\text{ is an }n\text{-necklace }\mid\ \left\vert N\right\vert
=d\right\}  \right\vert }_{\substack{=\left\vert \operatorname*{Aper}\left(
d\right)  \right\vert \\\text{(by (\ref{sol.words.necklaces.e.3}))}}%
}=\sum_{d\mid n}d\left\vert \operatorname*{Aper}\left(  d\right)  \right\vert
=\sum_{e\mid n}e\left\vert \operatorname*{Aper}\left(  e\right)  \right\vert
\label{sol.words.necklaces.e.4}%
\end{equation}
(here, we renamed the summation index $d$ as $e$). Now,%
\begin{align*}
\sum_{d\mid n}\mu\left(  d\right)  \underbrace{\left\vert \mathfrak{A}%
\right\vert ^{n/d}}_{\substack{=\sum_{e\mid n/d}e\left\vert
\operatorname*{Aper}\left(  e\right)  \right\vert \\\text{(by
(\ref{sol.words.necklaces.e.4}), applied to}\\n/d\text{ instead of }n\text{)}%
}}  &  =\sum_{d\mid n}\mu\left(  d\right)  \sum_{e\mid n/d}e\left\vert
\operatorname*{Aper}\left(  e\right)  \right\vert =\underbrace{\sum_{d\mid
n}\sum_{e\mid n/d}}_{=\sum_{e\mid n}\sum_{d\mid n/e}}\mu\left(  d\right)
e\left\vert \operatorname*{Aper}\left(  e\right)  \right\vert \\
&  =\sum_{e\mid n}\sum_{d\mid n/e}\mu\left(  d\right)  e\left\vert
\operatorname*{Aper}\left(  e\right)  \right\vert =\sum_{e\mid n}%
\underbrace{\left(  \sum_{d\mid n/e}\mu\left(  d\right)  \right)
}_{\substack{=\delta_{n/e,1}\\\text{(by
(\ref{pf.lem.sol.words.necklaces.muphi.mu*1}), applied}\\\text{to }n/e\text{
instead of }n\text{)}}}e\left\vert \operatorname*{Aper}\left(  e\right)
\right\vert \\
&  =\sum_{e\mid n}\underbrace{\delta_{n/e,1}}_{=\delta_{n,e}}e\left\vert
\operatorname*{Aper}\left(  e\right)  \right\vert =\sum_{e\mid n}\delta
_{n,e}e\left\vert \operatorname*{Aper}\left(  e\right)  \right\vert
=n\left\vert \operatorname*{Aper}\left(  n\right)  \right\vert .
\end{align*}
Solving this for $\left\vert \operatorname*{Aper}\left(  n\right)  \right\vert
$, we obtain
\begin{equation}
\left\vert \operatorname*{Aper}\left(  n\right)  \right\vert =\dfrac{1}{n}%
\sum_{d\mid n}\mu\left(  d\right)  \left\vert \mathfrak{A}\right\vert ^{n/d}.
\label{sol.words.necklaces.e.7}%
\end{equation}


Now, let us recall that $\operatorname*{Aper}\left(  n\right)  $ is the set of
all aperiodic $n$-necklaces. Thus,%
\[
\left\vert \operatorname*{Aper}\left(  n\right)  \right\vert =\left(
\text{the number of all aperiodic }n\text{-necklaces}\right)  .
\]
Hence,%
\[
\left(  \text{the number of all aperiodic }n\text{-necklaces}\right)
=\left\vert \operatorname*{Aper}\left(  n\right)  \right\vert =\dfrac{1}%
{n}\sum_{d\mid n}\mu\left(  d\right)  \left\vert \mathfrak{A}\right\vert
^{n/d}.
\]
This solves Exercise \ref{exe.words.necklaces}(f).

(g) We shall use the notations we introduced in the solution of Exercise
\ref{exe.words.necklaces}(f).

We have seen (in the solution of Exercise \ref{exe.words.necklaces}(f)) that
for any $n$-necklace $N$, the cardinality $\left\vert N\right\vert $ is a
divisor of $n$. Hence,%
\begin{align}
&  \left(  \text{the number of all }n\text{-necklaces}\right) \nonumber\\
&  =\sum_{d\mid n}\underbrace{\left(  \text{the number of all }%
n\text{-necklaces }N\text{ such that }\left\vert N\right\vert =d\right)
}_{\substack{=\left\vert \left\{  N\text{ is an }n\text{-necklace }%
\mid\ \left\vert N\right\vert =d\right\}  \right\vert =\left\vert
\operatorname*{Aper}\left(  d\right)  \right\vert \\\text{(by
(\ref{sol.words.necklaces.e.3}))}}}\nonumber\\
&  =\sum_{d\mid n}\left\vert \operatorname*{Aper}\left(  d\right)  \right\vert
=\sum_{e\mid n}\underbrace{\left\vert \operatorname*{Aper}\left(  e\right)
\right\vert }_{\substack{=\dfrac{1}{e}\sum_{d\mid e}\mu\left(  d\right)
\left\vert \mathfrak{A}\right\vert ^{e/d}\\\text{(by
(\ref{sol.words.necklaces.e.7}), applied to }e\\\text{instead of }n\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}d\text{ as }e\right) \nonumber\\
&  =\sum_{e\mid n}\dfrac{1}{e}\sum_{d\mid e}\mu\left(  d\right)  \left\vert
\mathfrak{A}\right\vert ^{e/d}=\underbrace{\sum_{e\mid n}\sum_{d\mid e}%
}_{=\sum_{d\mid n}\sum\limits_{\substack{e\mid n;\\d\mid e}}}\dfrac{1}{e}\mu\left(
d\right)  \left\vert \mathfrak{A}\right\vert ^{e/d}=\sum_{d\mid n}%
\sum\limits_{\substack{e\mid n;\\d\mid e}}\dfrac{1}{e}\mu\left(  d\right)  \left\vert
\mathfrak{A}\right\vert ^{e/d}\nonumber\\
&  =\underbrace{\sum_{d\mid n}\sum_{e\mid n/d}}_{=\sum_{e\mid n}\sum_{d\mid
n/e}}\dfrac{1}{de}\mu\left(  d\right)  \left\vert \mathfrak{A}\right\vert
^{e}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }de\text{ for
}e\text{ in the inner sum}\right) \nonumber\\
&  =\sum_{e\mid n}\sum_{d\mid n/e}\dfrac{1}{de}\mu\left(  d\right)  \left\vert
\mathfrak{A}\right\vert ^{e}. \label{sol.words.necklaces.f.0}%
\end{align}


From (\ref{pf.lem.sol.words.necklaces.muphi.mu*n}), we have%
\begin{equation}
\phi\left(  n\right)  =\sum_{d\mid n}\mu\left(  d\right)  \dfrac{n}{d}.
\label{sol.words.necklaces.f.1}%
\end{equation}
Now,%
\begin{align*}
\dfrac{1}{n}\sum_{d\mid n}\phi\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}  &  =\dfrac{1}{n}\sum_{d\mid n}\phi\left(  n/d\right)
\left\vert \mathfrak{A}\right\vert ^{d}\ \ \ \ \ \ \ \ \ \ \left(  \text{here,
we have substituted }n/d\text{ for }d\text{ in the sum}\right) \\
&  =\dfrac{1}{n}\sum_{e\mid n}\underbrace{\phi\left(  n/e\right)
}_{\substack{=\sum_{d\mid n/e}\mu\left(  d\right)  \dfrac{n/e}{d}\\\text{(by
(\ref{sol.words.necklaces.f.1}), applied to }n/e\\\text{instead of }n\text{)}%
}}\left\vert \mathfrak{A}\right\vert ^{e}\\
&\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have renamed the summation index }d\text{ as }e\right) \\
&  =\dfrac{1}{n}\sum_{e\mid n}\left(  \sum_{d\mid n/e}\mu\left(  d\right)
\dfrac{n/e}{d}\right)  \left\vert \mathfrak{A}\right\vert ^{e}=\dfrac{1}%
{n}\sum_{e\mid n}\sum_{d\mid n/e}\mu\left(  d\right)  \dfrac{n/e}{d}\left\vert
\mathfrak{A}\right\vert ^{e}=\dfrac{1}{n}\sum_{e\mid n}\sum_{d\mid n/e}%
\mu\left(  d\right)  \dfrac{n}{de}\left\vert \mathfrak{A}\right\vert ^{e}\\
&  =\sum_{e\mid n}\sum_{d\mid n/e}\dfrac{1}{de}\mu\left(  d\right)  \left\vert
\mathfrak{A}\right\vert ^{e}.
\end{align*}
Compared with (\ref{sol.words.necklaces.f.0}), this yields%
\[
\left(  \text{the number of all }n\text{-necklaces}\right)  =\dfrac{1}{n}%
\sum_{d\mid n}\phi\left(  d\right)  \left\vert \mathfrak{A}\right\vert
^{n/d}.
\]
This solves Exercise \ref{exe.words.necklaces}(g).

(h) \textit{Alternative solution of Exercise \ref{exe.words.lyndon.count}:}
Let $n$ be a positive integer. Exercise \ref{exe.words.necklaces}(e) yields
that the aperiodic $n$-necklaces are in bijection with Lyndon words of length
$n$. Hence,%
\[
\left(  \text{the number of all aperiodic }n\text{-necklaces}\right)  =\left(
\text{the number of all Lyndon words of length }n\right)  ,
\]
so that%
\begin{align*}
&  \left(  \text{the number of all Lyndon words of length }n\right) \\
&  =\left(  \text{the number of all aperiodic }n\text{-necklaces}\right)
=\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}%
\end{align*}
(by Exercise \ref{exe.words.necklaces}(f)). Thus,%
\[
\left(  \text{the number of all Lyndon words of length }n\right)  =\dfrac
{1}{n}\sum_{d\mid n}\mu\left(  d\right)  \underbrace{\left\vert \mathfrak{A}%
\right\vert ^{n/d}}_{\substack{=q^{n/d}\\\text{(since }\left\vert
\mathfrak{A}\right\vert =q\text{)}}}=\dfrac{1}{n}\sum\limits_{d\mid n}%
\mu\left(  d\right)  q^{n/d}.
\]
This solves Exercise \ref{exe.words.lyndon.count} again. Thus, Exercise
\ref{exe.words.necklaces}(h) is solved.

(i) \textit{First solution of Exercise \ref{exe.words.necklaces}(i):} Let
$q\in \ZZ$. We need to show that $n\mid\sum_{d\mid n}\mu\left(
d\right)  q^{n/d}$ and $n\mid\sum_{d\mid n}\phi\left(  d\right)  q^{n/d}$.

Let $r$ be the remainder of $q$ modulo $n$. Then, $r\in\left\{  0,1,\ldots
,n-1\right\}  \subset \NN$.
Fix a finite totally ordered set $\mathfrak{A}$ containing
$r$ elements. (Such an $\mathfrak{A}$ exists, since $r\in \NN$.
For example, we can set $\mathfrak{A} = \left\{1,2,\ldots,r\right\}$.)
Exercise \ref{exe.words.necklaces}(f) shows
that the number of all aperiodic $n$-necklaces is $\dfrac{1}{n}\sum_{d\mid
n}\mu\left(  d\right)  \left\vert \mathfrak{A}\right\vert ^{n/d}$. Hence,
$\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}$ is an integer (since the number of all aperiodic
$n$-necklaces is an integer). In other words, $n\mid\sum_{d\mid n}\mu\left(
d\right)  \left\vert \mathfrak{A}\right\vert ^{n/d}$, so that $\sum_{d\mid
n}\mu\left(  d\right)  \left\vert \mathfrak{A}\right\vert ^{n/d}
\equiv0\operatorname{mod}n$. But $r$ is the remainder of $q$ modulo $n$. Thus,
$r\equiv q\operatorname{mod}n$, so that $q\equiv r=\left\vert \mathfrak{A}%
\right\vert \operatorname{mod}n$ (since the set $\mathfrak{A}$ has $r$
elements). Now, $\sum_{d\mid n}\mu\left(  d\right)  \underbrace{q^{n/d}%
}_{\substack{\equiv\left\vert \mathfrak{A}\right\vert ^{n/d}\operatorname{mod}%
n\\\text{(since }q\equiv\left\vert \mathfrak{A}\right\vert \operatorname{mod}%
n\text{)}}}\equiv\sum_{d\mid n}\mu\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}\equiv0\operatorname{mod}n$, so that $n\mid\sum_{d\mid n}%
\mu\left(  d\right)  q^{n/d}$.

Also, Exercise \ref{exe.words.necklaces}(g)
shows that the number of all $n$-necklaces is $\dfrac{1}{n}\sum_{d\mid n}%
\phi\left(  d\right)  \left\vert \mathfrak{A}\right\vert ^{n/d}$. Hence,
\newline
$\dfrac{1}{n}\sum_{d\mid n}\phi\left(  d\right)  \left\vert \mathfrak{A}%
\right\vert ^{n/d}$ is an integer (since the number of all $n$-necklaces is an
integer). In other words, $n\mid\sum_{d\mid n}\phi\left(  d\right)  \left\vert
\mathfrak{A}\right\vert ^{n/d}$, so that $\sum_{d\mid n}\phi\left(  d\right)
\left\vert \mathfrak{A}\right\vert ^{n/d}\equiv0\operatorname{mod}n$. Now,
$\sum_{d\mid n}\phi\left(  d\right)  \underbrace{q^{n/d}}_{\substack{\equiv
\left\vert \mathfrak{A}\right\vert ^{n/d}\operatorname{mod}n\\\text{(since
}q\equiv\left\vert \mathfrak{A}\right\vert \operatorname{mod}n\text{)}}%
}\equiv\sum_{d\mid n}\phi\left(  d\right)  \left\vert \mathfrak{A}\right\vert
^{n/d}\equiv0\operatorname{mod}n$, so that $n\mid\sum_{d\mid n}\phi\left(
d\right)  q^{n/d}$. The solution of Exercise \ref{exe.words.necklaces}(i) is
thus complete.

\textit{Second solution of Exercise \ref{exe.words.necklaces}(i):} Forget that
we fixed $n$. Let $q\in \ZZ$. Let $A$ denote the ring $\ZZ$. For
every $n\in\left\{  1,2,3,\ldots\right\}  $, let $\varphi_{n}$ denote the
identity endomorphism $\operatorname{id}$ of $A$. Exercise
\ref{exe.witt.ghost-exa} yields (among other things) that the seven equivalent
assertions $\mathcal{C}$, $\mathcal{D}$, $\mathcal{E}$, $\mathcal{F}$,
$\mathcal{G}$, $\mathcal{H}$ and $\mathcal{J}$ of Exercise
\ref{exe.witt.ghost-equiv} are satisfied for the family $\left(  b_{n}\right)
_{n\geq1}=\left(  q^{n}\right)  _{n\geq1}$. In particular, Assertion
$\mathcal{F}$ of Exercise \ref{exe.witt.ghost-equiv} is satisfied for the
family $\left(  b_{n}\right)  _{n\geq1}=\left(  q^{n}\right)  _{n\geq1}$. In
other words, every positive integer $n$ satisfies
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(  q^{n/d}\right)  \in
n \ZZ . \label{sol.words.necklaces.h.4}%
\end{equation}
Also, Assertion $\mathcal{G}$ of Exercise \ref{exe.witt.ghost-equiv} is
satisfied for the family $\left(  b_{n}\right)  _{n\geq1}=\left(
q^{n}\right)  _{n\geq1}$ (since the seven equivalent assertions $\mathcal{C}$,
$\mathcal{D}$, $\mathcal{E}$, $\mathcal{F}$, $\mathcal{G}$, $\mathcal{H}$ and
$\mathcal{J}$ of Exercise \ref{exe.witt.ghost-equiv} are satisfied for the
family $\left(  b_{n}\right)  _{n\geq1}=\left(  q^{n}\right)  _{n\geq1}$). In
other words, every positive integer $n$ satisfies
\begin{equation}
\sum_{d\mid n}\phi\left(  d\right)  \varphi_{d}\left(  q^{n/d}\right)  \in
n \ZZ .\label{sol.words.necklaces.h.5}%
\end{equation}


Now, fix a positive integer $n$. We can rewrite (\ref{sol.words.necklaces.h.4}%
) as $n\mid\sum_{d\mid n}\mu\left(  d\right)  \varphi_{d}\left(
q^{n/d}\right)  $. Since
\[
\sum_{d\mid n}\mu\left(  d\right)
\underbrace{\varphi_{d}}_{=\id}\left(  q^{n/d}\right)
=\sum_{d\mid n}\mu\left(  d\right)  q^{n/d} ,
\]
this simplifies to $n\mid
\sum_{d\mid n}\mu\left(  d\right)  q^{n/d}$. Also, we can rewrite
(\ref{sol.words.necklaces.h.5}) as $n\mid\sum_{d\mid n}\phi\left(  d\right)
\varphi_{d}\left(  q^{n/d}\right)  $. Since
\[
\sum_{d\mid n}\phi\left(
d\right)  \underbrace{\varphi_{d}}_{=\id}\left(  q^{n/d}%
\right)  =\sum_{d\mid n}\phi\left(  d\right)  q^{n/d} ,
\]
this simplifies to
$n\mid\sum_{d\mid n}\phi\left(  d\right)  q^{n/d}$.
Exercise~\ref{exe.words.necklaces}(i) is thus solved again.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.stf.props1}}

\textit{Solution to Exercise \ref{exe.words.lyndon.stf.props1}.} There exists
a $u\in\mathfrak{A}^{\ast}$ such that $w=uv$ (since $v$ is a suffix of $w$).
Consider this $u$.

The word $v$ is Lyndon and thus nonempty. Hence, $\ell\left(  v\right)  >0$.

The exercise asks us to prove the logical equivalence%
\begin{equation}
\left(  t\text{ is the longest Lyndon suffix of }wt\right)
\ \Longleftrightarrow\ \left(  \text{we do not have }v<t\right)  .
\label{sol.words.lyndon.stf.props1.equiv}%
\end{equation}
We shall prove the $\Longrightarrow$ and $\Longleftarrow$ parts of this
equivalence separately:

$\Longrightarrow:$ Assume that $t$ is the longest Lyndon suffix of $wt$. We
need to prove that we do not have $v<t$.

Assume the contrary. Thus, $v<t$. Now, both $v$ and $t$ are Lyndon words.
Hence, Proposition \ref{prop.words.lyndon.concat}(a) (applied to $v$ and $t$
instead of $u$ and $v$) yields that the word $vt$ is Lyndon. But $vt$ is a
suffix of $wt$ (since $v$ is a suffix of $w$). Hence, $vt$ is a Lyndon suffix
of $wt$. This Lyndon suffix is clearly longer than $t$ (since $\ell\left(
vt\right)  =\underbrace{\ell\left(  v\right)  }_{>0}+\ell\left(  t\right)
>\ell\left(  t\right)  $), which flies in the face of the fact that $t$ is the
longest Lyndon suffix of $wt$. This contradiction shows that our assumption
was wrong. Hence, we have proven that we do not have $v<t$. This proves the
$\Longrightarrow$ part of the equivalence
(\ref{sol.words.lyndon.stf.props1.equiv}).

$\Longleftarrow:$ Assume that we do not have $v<t$. We must show that $t$ is
the longest Lyndon suffix of $wt$.

Assume the contrary. Then, $t$ is not the longest Lyndon suffix of $wt$. Since
$t$ is a Lyndon suffix of $wt$, this means that there exists a Lyndon suffix
$r$ of $wt$ which satisfies $\ell\left(  r\right)  >\ell\left(  t\right)  $.
Let $q$ be the shortest such suffix.

Thus, $q$ is a Lyndon suffix of $wt$ and satisfies $\ell\left(  q\right)
>\ell\left(  t\right)  $. Moreover, $q$ is the shortest such Lyndon suffix.
Hence, if $r$ is a Lyndon suffix of $wt$ which satisfies $\ell\left(
r\right)  >\ell\left(  t\right)  $, then%
\begin{equation}
\ell\left(  r\right)  \geq\ell\left(  q\right)  .
\label{sol.words.lyndon.stf.props1.1.pf.1}%
\end{equation}


Both $q$ and $t$ are suffixes of $wt$, and the suffix $q$ begins earlier
(since $\ell\left(  q\right)  >\ell\left(  t\right)  $). Thus, there exists a
nonempty suffix $g$ of $w$ such that $q=gt$. Consider this $g$.

The word $t$ is nonempty (since it is Lyndon) and a suffix of $q$ (since
$q=gt$). Thus, Corollary \ref{cor.words.lyndon.trivia} (applied to $q$ and $t$
instead of $w$ and $v$) yields $t\geq q$ (since $q$ is Lyndon).

The word $t$ is Lyndon (by assumption) and a proper suffix of $q$ (since
$q=gt$ and since $g$ is nonempty). In other words, $t$ is a Lyndon proper
suffix\footnote{Of course, a Lyndon proper suffix of $q$ just means a proper
suffix $z$ of $q$ such that $z$ is Lyndon.} of $q$. Moreover, the word $t$ is
the longest Lyndon proper suffix of $q$\ \ \ \ \footnote{\textit{Proof.} Let
$r$ be a Lyndon proper suffix of $q$. We shall prove that $\ell\left(
r\right)  \leq\ell\left(  t\right)  $.
\par
Indeed, assume the contrary (for the sake of contradiction). Thus,
$\ell\left(  r\right)  >\ell\left(  t\right)  $. Now, since $r$ is a suffix of
$q$, and since $q$ (in turn) is a suffix of $wt$, we see that $r$ is a suffix
of $wt$. Thus, $r$ is a Lyndon suffix of $wt$ (since $r$ is Lyndon). Thus,
(\ref{sol.words.lyndon.stf.props1.1.pf.1}) shows that $\ell\left(  r\right)
\geq\ell\left(  q\right)  $. But since $r$ is a proper suffix of $q$, we have
$\ell\left(  r\right)  <\ell\left(  q\right)  $. This contradicts $\ell\left(
r\right)  \geq\ell\left(  q\right)  $. This contradiction proves that our
assumption was wrong. Hence, we have shown that $\ell\left(  r\right)
\leq\ell\left(  t\right)  $.
\par
Now, let us forget that we fixed $r$. We thus have shown that any Lyndon
proper suffix $r$ of $q$ satisfies $\ell\left(  r\right)  \leq\ell\left(
t\right)  $. In other words, any Lyndon proper suffix of $q$ is at most as
long as $t$. Since $t$ is a Lyndon proper suffix of $q$, this shows that the
word $t$ is the \textbf{longest} Lyndon proper suffix of $q$. Qed.}. In other
words, the word $t$ is the longest proper suffix of $q$ such that $t$ is
Lyndon. Also, $\ell\left(  q\right)  >\ell\left(  t\right)  \geq1$ (since $t$
is nonempty). Thus, $q$ is a Lyndon word of length $>1$. Hence, Exercise
\ref{exe.words.lyndon.std2}(a) (applied to $q$, $g$ and $t$ instead of $w$,
$u$ and $v$) shows that the words $g$ and $t$ are Lyndon. In particular, $g$
is a Lyndon suffix of $w$ (since $g$ is Lyndon and a suffix of $w$). Since $v$
is the longest Lyndon suffix of $w$, this shows that $g$ is at most as long as
$v$. In other words, $\ell\left(  g\right)  \leq\ell\left(  v\right)  $.

Since $t$ is nonempty, we have $g<gt$.

Both $g$ and $v$ are suffixes of $w$, and the suffix $g$ begins no earlier
than $v$ (since $\ell\left(  g\right)  \leq\ell\left(  v\right)  $).
Therefore, $g$ is a suffix of $v$. Hence, Corollary
\ref{cor.words.lyndon.trivia} (applied to $v$ and $g$ instead of $w$ and $v$)
yields $g\geq v$. Hence, $v\leq g<gt=q\leq t$ (since $t\geq q$). This
contradicts the fact that we do not have $v<t$. Thus, we have obtained a
contradiction. Our assumption was therefore wrong, and we have shown that $t$
is the longest Lyndon suffix of $wt$. This proves the $\Longleftarrow$ part of
the equivalence (\ref{sol.words.lyndon.stf.props1.equiv}).

We have now proven both parts of the equivalence
(\ref{sol.words.lyndon.stf.props1.equiv}), and thus solved Exercise
\ref{exe.words.lyndon.stf.props1}.

[\textit{Remark:} Exercise \ref{exe.words.lyndon.stf.props1} still holds in
the partial-order setting\footnote{See Exercise \ref{exe.words.poset} for an
explanation of what the partial-order setting is.}. In fact, the solution we
have given above still applies in this setting.]
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.stf.props1}}

\textit{Solution to Exercise \ref{exe.words.lyndon.stf.props1}.} The word $v$
is a Lyndon suffix of $w$. In other words, the word $v$ is Lyndon and is a
suffix of $w$.

There exists a $u\in\mathfrak{A}^{\ast}$ such that $w=uv$ (since $v$ is a
suffix of $w$). Consider this $u$.

The word $v$ is Lyndon and thus nonempty. Hence, $\ell\left(  v\right)  >0$.

The exercise asks us to prove the logical equivalence%
\begin{equation}
\left(  t\text{ is the longest Lyndon suffix of }wt\right)
\ \Longleftrightarrow\ \left(  \text{we do not have }v<t\right)  .
\label{sol.words.lyndon.stf.props1.equiv}%
\end{equation}
We shall prove the $\Longrightarrow$ and $\Longleftarrow$ parts of this
equivalence separately:

$\Longrightarrow:$ Assume that $t$ is the longest Lyndon suffix of $wt$. We
need to prove that we do not have $v<t$.

Assume the contrary (for the sake of contradiction). Thus, $v<t$. Now, both
$v$ and $t$ are Lyndon words. Hence, Proposition
\ref{prop.words.lyndon.concat}(a) (applied to $v$ and $t$ instead of $u$ and
$v$) yields that the word $vt$ is Lyndon. But $vt$ is a suffix of $wt$ (since
$v$ is a suffix of $w$). Thus, $vt$ is a Lyndon suffix of $wt$ (since $vt$ is Lyndon).

Now, recall that $t$ is the longest Lyndon suffix of $wt$. Thus, every Lyndon
suffix $q$ of $wt$ satisfies $\ell\left(  q\right)  \leq\ell\left(  t\right)
$. Applying this to $q=vt$, we conclude that $\ell\left(  vt\right)  \leq
\ell\left(  t\right)  $ (since $vt$ is a Lyndon suffix of $wt$). Thus,
$\ell\left(  t\right)  \geq\ell\left(  vt\right)  =\underbrace{\ell\left(
v\right)  }_{>0}+\ell\left(  t\right)  >\ell\left(  t\right)  $, which is
absurd. This contradiction shows that our assumption was wrong. Hence, we have
proven that we do not have $v<t$. This proves the $\Longrightarrow$ part of
the equivalence (\ref{sol.words.lyndon.stf.props1.equiv}).

$\Longleftarrow:$ Assume that we do not have $v<t$. We must show that $t$ is
the longest Lyndon suffix of $wt$.

The word $t$ is Lyndon and a suffix of $wt$. Thus, $t$ is a Lyndon suffix of
$wt$.

Now, we shall show that%
\begin{equation}
\text{every Lyndon suffix }r\text{ of }wt\text{ satisfies }\ell\left(
r\right)  \leq\ell\left(  t\right)  . \label{sol.words.lyndon.stf.props1.1}%
\end{equation}


\textit{Proof of (\ref{sol.words.lyndon.stf.props1.1}):} Assume the contrary
(for the sake of contradiction). Thus, not every Lyndon suffix $r$ of $wt$
satisfies $\ell\left(  r\right)  \leq\ell\left(  t\right)  $. In other words,
there exists a Lyndon suffix $r$ of $wt$ which satisfies $\ell\left(
r\right)  >\ell\left(  t\right)  $. Let $q$ be the shortest such suffix.

Thus, $q$ is the shortest Lyndon suffix $r$ of $wt$ which satisfies
$\ell\left(  r\right)  >\ell\left(  t\right)  $. In particular, $q$ is a
Lyndon suffix $r$ of $wt$ which satisfies $\ell\left(  r\right)  >\ell\left(
t\right)  $. In other words, $q$ is a Lyndon suffix of $wt$ and satisfies
$\ell\left(  q\right)  >\ell\left(  t\right)  $. Moreover, every Lyndon suffix
$r$ of $wt$ which satisfies $\ell\left(  r\right)  >\ell\left(  t\right)  $
must be at least as long as $q$ (since $q$ is the \textbf{shortest} Lyndon
suffix $r$ of $wt$ which satisfies $\ell\left(  r\right)  >\ell\left(
t\right)  $). In other words, if $r$ is a Lyndon suffix of $wt$ which
satisfies $\ell\left(  r\right)  >\ell\left(  t\right)  $, then%
\begin{equation}
\ell\left(  r\right)  \geq\ell\left(  q\right)  .
\label{sol.words.lyndon.stf.props1.1.pf.1}%
\end{equation}


The word $q$ is a Lyndon suffix of $wt$. In other words, $q$ is Lyndon and is
a suffix of $wt$.

We have $\ell\left(  q\right)  >\ell\left(  t\right)  $. Therefore, there
exists a nonempty suffix $g$ of $w$ such that $q=gt$%
\ \ \ \ \footnote{\textit{Proof.} If the word $q$ was a suffix of $t$, then we
would have $\ell\left(  q\right)  \leq\ell\left(  t\right)  $, which would
contradict $\ell\left(  q\right)  >\ell\left(  t\right)  $. Hence, the word
$q$ is not a suffix of $t$.
\par
We know that $q$ is a suffix of $wt$. Hence, \textbf{either} the word $q$ is a
suffix of $t$, \textbf{or} the word $q$ has the form $gt$ for some nonempty
suffix $g$ of $w$ (depending on whether the suffix $q$ of $wt$ begins after
the suffix $t$ of $wt$ does, or before that). Therefore, the word $q$ has the
form $gt$ for some nonempty suffix $g$ of $w$ (because we know that the word
$q$ is not a suffix of $t$). In other words, there exists a nonempty suffix
$g$ of $w$ such that $q=gt$. Qed.}. Consider this $g$.

The word $t$ is nonempty (since it is Lyndon) and a suffix of $q$ (since
$q=gt$). Thus, Corollary \ref{cor.words.lyndon.trivia} (applied to $q$ and $t$
instead of $w$ and $v$) yields $t\geq q$ (since $q$ is Lyndon).

The word $t$ is Lyndon (by assumption) and a proper suffix of $q$ (since
$q=gt$ and since $g$ is nonempty). In other words, $t$ is a Lyndon proper
suffix\footnote{Of course, a Lyndon proper suffix of $q$ just means a proper
suffix $z$ of $q$ such that $z$ is Lyndon.} of $q$. Moreover, the word $t$ is
the longest Lyndon proper suffix of $q$\ \ \ \ \footnote{\textit{Proof.} Let
$r$ be a Lyndon proper suffix of $q$. We shall prove that $\ell\left(
r\right)  \leq\ell\left(  t\right)  $.
\par
Indeed, assume the contrary (for the sake of contradiction). Thus,
$\ell\left(  r\right)  >\ell\left(  t\right)  $. Now, $r$ is a Lyndon proper
suffix of $q$. In other words, $r$ is Lyndon and a proper suffix of $q$. Since
$r$ is a suffix of $q$, and since $q$ (in turn) is a suffix of $wt$, we see
that $r$ is a suffix of $wt$. Thus, $r$ is a Lyndon suffix of $wt$ (since $r$
is Lyndon). Thus, (\ref{sol.words.lyndon.stf.props1.1.pf.1}) shows that
$\ell\left(  r\right)  \geq\ell\left(  q\right)  $. But since $r$ is a proper
suffix of $q$, we have $\ell\left(  r\right)  <\ell\left(  q\right)  $. This
contradicts $\ell\left(  r\right)  \geq\ell\left(  q\right)  $. This
contradiction proves that our assumption was wrong. Hence, we have shown that
$\ell\left(  r\right)  \leq\ell\left(  t\right)  $.
\par
Now, let us forget that we fixed $r$. We thus have shown that any Lyndon
proper suffix $r$ of $q$ satisfies $\ell\left(  r\right)  \leq\ell\left(
t\right)  $. In other words, any Lyndon proper suffix of $q$ is at most as
long as $t$. Since $t$ is a Lyndon proper suffix of $q$, this shows that the
word $t$ is the \textbf{longest} Lyndon proper suffix of $q$. Qed.}. In other
words, the word $t$ is the longest proper suffix of $q$ such that $t$ is
Lyndon. Also, $\ell\left(  q\right)  >\ell\left(  t\right)  \geq1$ (since $t$
is nonempty). Thus, $q$ is a Lyndon word of length $>1$. Hence, Exercise
\ref{exe.words.lyndon.std2}(a) (applied to $q$, $g$ and $t$ instead of $w$,
$u$ and $v$) shows that the words $g$ and $t$ are Lyndon.

The word $g$ is Lyndon and a suffix of $w$. In other words, $g$ is a Lyndon
suffix of $w$.

We have $g\neq gt$ (since $t$ is nonempty). Combined with $g\leq gt$, this
yields $g<gt$.

The word $g$ is a suffix of $w$. In other words, the word $g$ is a suffix of
$uv$ (since $w=uv$). Hence, we must be in one of the following two cases
(depending on whether this suffix $g$ begins after the suffix $v$ begins or before):

\textit{Case 1:} The word $g$ is a suffix of $v$.

\textit{Case 2:} The word $g$ has the form $pv$, where $p$ is a nonempty
suffix of $u$.

Let us first consider Case 1. In this case, the word $g$ is a suffix of $v$.
Hence, Corollary \ref{cor.words.lyndon.trivia} (applied to $v$ and $g$ instead
of $w$ and $v$) yields $g\geq v$. Hence, $v\leq g<gt=q\leq t$ (since $t\geq
q$). This contradicts the fact that we do not have $v<t$. Thus, we have
obtained a contradiction in Case 1.

Let us now consider Case 2. In this case, the word $g$ has the form $pv$,
where $p$ is a nonempty suffix of $u$. Consider this $p$.

Recall that $v$ is the longest Lyndon suffix of $w$. Thus, any Lyndon suffix
of $w$ is at most as long as $v$. In other words, if $x$ is a Lyndon suffix of
$w$, then $\ell\left(  x\right)  \leq\ell\left(  v\right)  $. We can apply
this to $x=g$ (since $g$ is a Lyndon suffix of $w$), and thus obtain
$\ell\left(  g\right)  \leq\ell\left(  v\right)  $. Hence, $\ell\left(
v\right)  \geq\ell\left(  \underbrace{g}_{=pv}\right)  =\ell\left(  pv\right)
=\underbrace{\ell\left(  p\right)  }_{\substack{>0\\\text{(since }p\text{ is
nonempty)}}}+\ell\left(  v\right)  >\ell\left(  v\right)  $. This is absurd.
Thus, we have obtained a contradiction in Case 2.

We have now obtained a contradiction in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this shows that we always obtain a
contradiction. Therefore, our assumption was wrong. This proves (\ref{sol.words.lyndon.stf.props1.1}).

We have now proven (\ref{sol.words.lyndon.stf.props1.1}). In other words, any
Lyndon suffix of $wt$ is at most as long as $t$. Since we also know that $t$
is a Lyndon suffix of $wt$, we thus conclude that $t$ is the longest Lyndon
suffix of $wt$. This proves the $\Longleftarrow$ part of the equivalence
(\ref{sol.words.lyndon.stf.props1.equiv}).

We have now proven both parts of the equivalence
(\ref{sol.words.lyndon.stf.props1.equiv}). Thus, the equivalence
(\ref{sol.words.lyndon.stf.props1.equiv}) holds. Exercise
\ref{exe.words.lyndon.stf.props1} is thus solved.

[\textit{Remark:} Exercise \ref{exe.words.lyndon.stf.props1} still holds in
the partial-order setting\footnote{See Exercise \ref{exe.words.poset} for an
explanation of what the partial-order setting is.}. In fact, the solution we
have given above still applies in this setting.]
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.stf.props2}}

\textit{Solution to Exercise \ref{exe.words.lyndon.stf.props2}.} Let $a$ be
the first letter of the word $w$. (This is well-defined, since $w$ has length
$>1>0$.) We consider $a$ as a one-letter word; thus, $\ell\left(  a\right)
=1$. Clearly, $a$ is a prefix of $w$ (since $a$ is the first letter of $w$).
Hence, there exists a word $w^{\prime}$ such that $w=aw^{\prime}$. Consider
this $w^{\prime}$. The word $w^{\prime}$ is nonempty (since $w$ has length
$>1$).

Now, if $h$ is any word, then%
\begin{equation}
\text{the suffixes of }h\text{ are precisely the proper suffixes of }ah
\label{sol.words.lyndon.stf.props2.translator}%
\end{equation}
(since $a$ is a
single letter). Applying this to $h=w^{\prime}$, we see that the suffixes of
$w^{\prime}$ are precisely the proper suffixes of $aw^{\prime}$. In other
words, the suffixes of $w^{\prime}$ are precisely the proper suffixes of $w$
(since $aw^{\prime}=w$). Thus, $v$ is the longest Lyndon suffix of $w^{\prime
}$ (since $v$ is the longest Lyndon proper suffix of $w$).

On the other hand, applying (\ref{sol.words.lyndon.stf.props2.translator}) to
$h=w^{\prime}t$, we see that the suffixes of $w^{\prime}t$ are precisely the
proper suffixes of $aw^{\prime}t$. In other words, the suffixes of $w^{\prime
}t$ are precisely the proper suffixes of $wt$ (since $aw^{\prime}=w$).

But Exercise \ref{exe.words.lyndon.stf.props1} (applied to $w^{\prime}$
instead of $w$) shows that%
\[
\left(  t\text{ is the longest Lyndon suffix of }w^{\prime}t\text{ if and only
if we do not have }v<t\right)  .
\]
Since the suffixes of $w^{\prime}t$ are precisely the proper suffixes of $wt$,
this result rewrites as follows:
\[
\left(  t\text{ is the longest Lyndon proper suffix of }wt\text{ if and only
if we do not have }v<t\right)  .
\]
This solves Exercise \ref{exe.words.lyndon.stf.props2}.

[\textit{Remark:} Exercise \ref{exe.words.lyndon.stf.props2} still holds in
the partial-order setting\footnote{See Exercise \ref{exe.words.poset} for an
explanation of what the partial-order setting is.}. In fact, the solution we
have given above still applies in this setting.]
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.stf.props2}}

\textit{Solution to Exercise \ref{exe.words.lyndon.stf.props2}.} The word $w$
has length $>1$. In other words, $\ell\left(  w\right)  >1$.

Let $a$ be the first letter of the word $w$.\ \ \ \ \footnote{This is
well-defined, since $\ell\left(  w\right)  >1\geq0$.} We consider $a$ as a
one-letter word; thus, $\ell\left(  a\right)  =1$. Clearly, $a$ is a prefix of
$w$ (since $a$ is the first letter of $w$). Hence, there exists a word
$w^{\prime}$ such that $w=aw^{\prime}$. Consider this $w^{\prime}$. We have
$\ell\left(  \underbrace{w}_{=aw^{\prime}}\right)  =\ell\left(  aw^{\prime
}\right)  =\underbrace{\ell\left(  a\right)  }_{=1}+\ell\left(  w^{\prime
}\right)  =1+\ell\left(  w^{\prime}\right)  $, so that $\ell\left(  w^{\prime
}\right)  =\underbrace{\ell\left(  w\right)  }_{>1}-1>1-1=0$. In other words,
the word $w^{\prime}$ is nonempty.

Now, for every words $p$ and $h$, we have the following logical equivalence:%
\begin{equation}
\left(  p\text{ is a Lyndon proper suffix of }ah\right)  \ \Longleftrightarrow
\ \left(  p\text{ is a Lyndon suffix of }h\right)
\label{sol.words.lyndon.stf.props2.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.lyndon.stf.props2.1}):} Let $p$ and
$h$ be words. We shall prove the $\Longrightarrow$ and $\Longleftarrow$ parts
of the equivalence (\ref{sol.words.lyndon.stf.props2.1}) separately:
\par
$\Longrightarrow:$ Assume that $p$ is a Lyndon proper suffix of $ah$. In other
words, $p$ is Lyndon and a proper suffix of $ah$.
\par
The word $p$ is a proper suffix of $ah$. In other words, there exists a
nonempty word $q$ such that $ah=qp$. Consider this $q$. The word $q$ is
nonempty, and thus its first letter is well-defined. Moreover, $q$ is a prefix
of $ah$ (since $ah=qp$), and thus we have%
\[
\left(  \text{the first letter of }q\right)  =\left(  \text{the first letter
of }ah\right)  =a.
\]
Therefore, $a$ is a prefix of $q$. In other words, there exists a word
$q^{\prime}$ such that $q=aq^{\prime}$. Consider this $q^{\prime}$. We have
$ah=\underbrace{q}_{=aq^{\prime}}p=aq^{\prime}p=a\left(  q^{\prime}p\right)
$. Cancelling $a$ from this equality, we obtain $h=q^{\prime}p$. Hence, $p$ is
a suffix of $h$. Since $p$ is Lyndon, this shows that $p$ is a Lyndon suffix
of $h$. This proves the $\Longrightarrow$ part of the equivalence
(\ref{sol.words.lyndon.stf.props2.1}).
\par
$\Longleftarrow:$ Assume that $p$ is a Lyndon suffix of $h$. In other words,
$p$ is Lyndon and a suffix of $h$.
\par
The word $p$ is a suffix of $h$. In other words, there exists a word $q$ such
that $h=qp$. Consider this $q$. Now, the word $aq$ is nonempty (since
$\ell\left(  aq\right)  =\underbrace{\ell\left(  a\right)  }_{=1}%
+\underbrace{\ell\left(  q\right)  }_{\geq0}\geq1$), and satisfies
$a\underbrace{h}_{=qp}=aqp=\left(  aq\right)  p$. Therefore, $p$ is a proper
suffix of $ah$ (since $ah=\left(  aq\right)  p$ and since $aq$ is nonempty).
Since $p$ is Lyndon, this shows that $p$ is a Lyndon proper suffix of $ah$.
This proves the $\Longleftarrow$ part of the equivalence
(\ref{sol.words.lyndon.stf.props2.1}).
\par
We have now proven both parts of the equivalence
(\ref{sol.words.lyndon.stf.props2.1}). Hence, the equivalence
(\ref{sol.words.lyndon.stf.props2.1}) holds. Qed.}. Thus, for every word $p$,
we have the following logical equivalence:%
\[
\left(  p\text{ is a Lyndon proper suffix of }aw^{\prime}\right)
\ \Longleftrightarrow\ \left(  p\text{ is a Lyndon suffix of }w^{\prime
}\right)
\]
(because of (\ref{sol.words.lyndon.stf.props2.1}), applied to $h=w^{\prime}$).
In other words, the Lyndon proper suffixes of $aw^{\prime}$ are precisely the
Lyndon suffixes of $w^{\prime}$. In other words, the Lyndon proper suffixes of
$w$ are precisely the Lyndon suffixes of $w^{\prime}$ (since $aw^{\prime}=w$).
Therefore, the longest Lyndon proper suffix of $w$ is precisely the longest
Lyndon suffix of $w^{\prime}$. Thus, $v$ is the longest Lyndon suffix of
$w^{\prime}$ (since we know that $v$ is the longest Lyndon proper suffix of
$w$). Therefore, Exercise \ref{exe.words.lyndon.stf.props1} (applied to
$w^{\prime}$ instead of $w$) shows that%
\begin{equation}
\left(  t\text{ is the longest Lyndon suffix of }w^{\prime}t\text{ if and only
if we do not have }v<t\right)  . \label{sol.words.lyndon.stf.props2.4}%
\end{equation}


However, for every word $p$, we have the following logical equivalence:
\[
\left(  p\text{ is a Lyndon proper suffix of }aw^{\prime}t\right)
\ \Longleftrightarrow\ \left(  p\text{ is a Lyndon suffix of }w^{\prime
}t\right)
\]
(because of (\ref{sol.words.lyndon.stf.props2.1}), applied to $h=w^{\prime}%
t$). In other words, the Lyndon proper suffixes of $aw^{\prime}t$ are
precisely the Lyndon suffixes of $w^{\prime}t$. In other words, the Lyndon
proper suffixes of $wt$ are precisely the Lyndon suffixes of $w^{\prime}t$
(since $aw^{\prime}=w$). Therefore, the longest Lyndon proper suffix of $wt$
is precisely the longest Lyndon suffix of $w^{\prime}t$. Therefore,
(\ref{sol.words.lyndon.stf.props2.4}) rewrites as follows:%
\[
\left(  t\text{ is the longest Lyndon proper suffix of }wt\text{ if and only
if we do not have }v<t\right)  .
\]
This solves Exercise \ref{exe.words.lyndon.stf.props2}.

[\textit{Remark:} Exercise \ref{exe.words.lyndon.stf.props2} still holds in
the partial-order setting\footnote{See Exercise \ref{exe.words.poset} for an
explanation of what the partial-order setting is.}. In fact, the solution we
have given above still applies in this setting.]
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.stf.basic}}

\textit{Solution to Exercise \ref{exe.words.lyndon.stf.basic}.} Recall the
definition of $\operatorname*{stf}w$. It says that $\operatorname*{stf}%
w=\left(  u,v\right)  $, where $u$ and $v$ are defined as follows:

\begin{itemize}
\item The word $v$ is defined as the longest proper suffix of $w$ such that
$v$ is Lyndon.

\item The word $u$ is defined as the nonempty word such that $w=uv$.
\end{itemize}

Consider these $u$ and $v$. Thus, $\left(  u,v\right)  =\operatorname*{stf}%
w=\left(  g,h\right)  $. In other words, $u=g$ and $v=h$.

We know that $v$ is the longest proper suffix of $w$ such that $v$ is Lyndon.
In other words, $v$ is the longest Lyndon proper suffix of $w$. In other
words, $h$ is the longest Lyndon proper suffix of $w$ (since $v=h$). This
solves Exercise \ref{exe.words.lyndon.stf.basic}(a).

We have $w=\underbrace{u}_{=g}\underbrace{v}_{=h}=gh$. This solves Exercise
\ref{exe.words.lyndon.stf.basic}(b).

From Exercise \ref{exe.words.lyndon.std2}(b), we conclude that $u<w<v$. Since
$u=g$, $w=gh$ and $v=h$, this rewrites as $g<gh<h$. This solves Exercise
\ref{exe.words.lyndon.stf.basic}(c).

From Exercise \ref{exe.words.lyndon.std2}(a), we conclude that the words $u$
and $v$ are Lyndon. Thus, the word $u$ is Lyndon. In other words, the word $g$
is Lyndon (since $u=g$). This solves solves Exercise
\ref{exe.words.lyndon.stf.basic}(d).

The word $v$ is the longest proper suffix of $w$ such that $v$ is Lyndon. In
particular, the word $v$ is Lyndon. In other words, $v\in\mathfrak{L}$.

We have $g\in\mathfrak{L}$ (since the word $g$ is Lyndon) and $h=v\in
\mathfrak{L}$. Also, the word $g$ is nonempty (since $g$ is Lyndon), and thus
we have $\ell\left(  g\right)  \geq1$. Furthermore, the word $h$ is Lyndon
(since $h\in\mathfrak{L}$) and thus nonempty. Hence, $\ell\left(  h\right)
\geq1$. Now, $\ell\left(  \underbrace{w}_{=gh}\right)  =\ell\left(  gh\right)
=\ell\left(  g\right)  +\underbrace{\ell\left(  h\right)  }_{\geq1}\geq
\ell\left(  g\right)  +1>\ell\left(  g\right)  $, so that $\ell\left(
g\right)  <\ell\left(  w\right)  $. Also, $\ell\left(  \underbrace{w}%
_{=gh}\right)  =\ell\left(  gh\right)  =\underbrace{\ell\left(  g\right)
}_{\geq1}+\ell\left(  h\right)  \geq1+\ell\left(  h\right)  >\ell\left(
h\right)  $, so that $\ell\left(  g\right)  <\ell\left(  w\right)  $. Thus,
Exercise \ref{exe.words.lyndon.stf.basic}(e) is solved.

(f) Exercise \ref{exe.words.lyndon.stf.props2} (applied to $v=h$) shows that
$t$ is the longest Lyndon proper suffix of $wt$ if and only if we do not have
$h<t$ (since $h$ is the longest Lyndon proper suffix of $w$). This solves
Exercise \ref{exe.words.lyndon.stf.basic}(f).
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.freelie.1}}

\textit{Solution to Exercise \ref{exe.freelie.1}.} We define a binary relation
$\sim$ on the set $\mathfrak{A}^{\ast}$ as follows: If $w$ and $w^{\prime}$
are two words, then we write $w\sim w^{\prime}$ if and only if $w^{\prime}$ is
a permutation of the word $w$ (that is, if and only if there exists a
permutation $\sigma\in\Symm_k$ satisfying $w^{\prime}=\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  k\right)  }\right)  $, where $k=\ell\left(  w\right)  $). We
notice the following properties of the relation $\sim$:

\begin{itemize}
\item The relation $\sim$ is an equivalence relation; in other words, it is
reflexive, symmetric and transitive.

\item If $w$ and $w^{\prime}$ are two words satisfying $w\sim w^{\prime}$,
then $\ell\left(  w\right)  =\ell\left(  w^{\prime}\right)  $.

\item If $u$, $v$, $u^{\prime}$ and $v^{\prime}$ are four words satisfying
$u\sim u^{\prime}$ and $v\sim v^{\prime}$, then $uv\sim u^{\prime}v^{\prime}$.
We shall refer to this fact as the \emph{monoidality} of the relation $\sim$.

\item If $u$ and $v$ are any two words, then $uv\sim vu$.
\end{itemize}

We also recall a fundamental property of Lie algebras (one of the forms of the
Jacobi identity):

\begin{itemize}
\item Every three elements $x$, $y$ and $z$ of a Lie algebra $\mathfrak{k}$
satisfy%
\begin{equation}
\left[  \left[  x,y\right]  ,z\right]  =\left[  \left[  x,z\right]  ,y\right]
-\left[  \left[  y,z\right]  ,x\right]  . \label{sol.freelie.1.jacobi}%
\end{equation}

\end{itemize}

We can now finally come to the solution of Exercise \ref{exe.freelie.1}.

For every $h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}^{\ast}$, we define a
subset $\mathfrak{L}_{h,s}$ of $\mathfrak{L}$ by%
\[
\mathfrak{L}_{h,s}=\left\{  w\in\mathfrak{L}\ \mid\ w\sim h\text{ and
}w<s\right\}  .
\]
For every $h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}^{\ast}$, we define a
$\kk$-submodule $B_{h,s}$ of $B$ by%
\[
B_{h,s}=\sum_{w\in\mathfrak{L}_{h,s}}\kk b_{w}.
\]
(In other words, for every $h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}%
^{\ast}$, we define $B_{h,s}$ as the $\kk$-linear span of the elements
$b_{w}$ with $w\in\mathfrak{L}_{h,s}$.) If $h$, $s$, $g$ and $t$ are four
words satisfying $h\sim g$ and $s<t$, then%
\begin{equation}
B_{h,s}\subset B_{g,t} \label{sol.freelie.1.a.BhsBgt}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.BhsBgt}):} Let $h$, $s$, $g$
and $t$ be four words satisfying $h\sim g$ and $s<t$.
\par
Let $v\in\mathfrak{L}_{h,s}$. Thus, $v\in\mathfrak{L}_{h,s}=\left\{
w\in\mathfrak{L}\ \mid\ w\sim h\text{ and }w<s\right\}  $ (by the definition
of $\mathfrak{L}_{h,s}$). In other words, $v$ is an element of $\mathfrak{L}$
and satisfies $v\sim h$ and $v<s$. From $v\sim h$ and $h\sim g$, we obtain
$v\sim g$ (since the relation $\sim$ is transitive). Also, $v<s<t$. Thus, $v$
is an element of $\mathfrak{L}$ and satisfies $v\sim g$ and $v<t$. In other
words, $v\in\left\{  w\in\mathfrak{L}\ \mid\ w\sim g\text{ and }w<t\right\}
=\mathfrak{L}_{g,t}$ (since $\mathfrak{L}_{g,t}$ is defined to be $\left\{
w\in\mathfrak{L}\ \mid\ w\sim g\text{ and }w<t\right\}  $).
\par
Now, let us forget that we fixed $v$. We thus have proven that every
$v\in\mathfrak{L}_{h,s}$ satisfies $v\in\mathfrak{L}_{g,t}$. In other words,
$\mathfrak{L}_{h,s}\subset\mathfrak{L}_{g,t}$. Now, the definition of
$B_{h,s}$ yields $B_{h,s}=\sum_{w\in\mathfrak{L}_{h,s}}\kk b_{w}%
\subset\sum_{w\in\mathfrak{L}_{g,t}}\kk b_{w}$ (since $\mathfrak{L}%
_{h,s}\subset\mathfrak{L}_{g,t}$). Since $B_{g,t}=\sum_{w\in\mathfrak{L}%
_{g,t}}\kk b_{w}$ (by the definition of $B_{g,t}$), this rewrites as
$B_{h,s}\subset B_{g,t}$. This proves (\ref{sol.freelie.1.a.BhsBgt}).}.

(a) We claim that%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q.
\label{sol.freelie.1.a.mainclaim}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.1.a.mainclaim}):} We can WLOG assume that
the alphabet $\mathfrak{A}$ is finite\footnote{In fact, assume that
(\ref{sol.freelie.1.a.mainclaim}) is proven in the case when the alphabet
$\mathfrak{A}$ is finite. Now, let $\mathfrak{A}$ be arbitrary. We must prove
(\ref{sol.freelie.1.a.mainclaim}) for this $\mathfrak{A}$.
\par
Fix $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ satisfying $p<q$.
We need to prove that $\left[  b_{p},b_{q}\right]  \in B_{pq,q}$. Let
$\mathfrak{B}$ denote the set of all letters that appear in (at least) one of
the words $p$ and $q$. Then, $\mathfrak{B}$ is a finite subset of
$\mathfrak{A}$, and the words $p$ and $q$ belong to $\mathfrak{B}^{\ast}$.
\par
Let $\mathfrak{L}^{\prime}$ denote the set of all Lyndon words over the
alphabet $\mathfrak{B}$. Clearly, a word $w\in\mathfrak{B}^{\ast}$ is Lyndon
as a word over the alphabet $\mathfrak{B}$ if and only if it is Lyndon as a
word over the alphabet $\mathfrak{A}$. Thus, $\mathfrak{L}^{\prime
}=\mathfrak{L}\cap\mathfrak{B}^{\ast}$, so that $p$ and $q$ belong to
$\mathfrak{L}^{\prime}$. Also, for every given $w\in\mathfrak{L}^{\prime}$ of
length $>1$, the pair $\operatorname*{stf}w$ does not depend on whether $w$ is
considered as a Lyndon word over the alphabet $\mathfrak{A}$ or as a Lyndon
word over the alphabet $\mathfrak{B}$ (because the definition of
$\operatorname*{stf}w$ involves only suffixes of $w$, and all of these
suffixes belong to $\mathfrak{B}^{\ast}$).
\par
For every $h\in\mathfrak{B}^{\ast}$ and $s\in\mathfrak{B}^{\ast}$, let us
define the set $\mathfrak{L}_{h,s}^{\prime}$, the $\kk$-module
$B^{\prime}$ and the $\kk$-module $B_{h,s}^{\prime}$ in the same way as
we have defined the set $\mathfrak{L}_{h,s}$, the $\kk$-module $B$ and
the $\kk$-module $B_{h,s}$, but using the alphabet $\mathfrak{B}$
instead of $\mathfrak{A}$. (Thus, $\mathfrak{L}_{h,s}^{\prime}=\left\{
w\in\mathfrak{L}^{\prime}\ \mid\ w\sim h\text{ and }w<s\right\}  $,
$B^{\prime}=\left(  \text{the }\kk\text{-submodule of }\mathfrak{g}%
\text{ spanned by the family }\left(  b_{w}\right)  _{w\in\mathfrak{L}%
^{\prime}}\right)  $ and $B_{h,s}^{\prime}=\sum_{w\in\mathfrak{L}%
_{h,s}^{\prime}}\kk b_{w}$.)
\par
Now, whenever $w$ is a Lyndon word over $\mathfrak{B}$ of length $>1$, we must
have $w\in\mathfrak{L}^{\prime}=\mathfrak{L}\cap\mathfrak{B}^{\ast}%
\subset\mathfrak{L}$ and thus $b_{w}=\left[  b_{u},b_{v}\right]  $, where
$\left(  u,v\right)  =\operatorname*{stf}w$ (according to
(\ref{eq.exe.freelie.1.bw})). Here, when we speak of $\operatorname*{stf}w$,
we are regarding $w$ as a Lyndon word over $\mathfrak{A}$, but as we have
already explained, the result is the same if we regard $w$ as a Lyndon word
over $\mathfrak{B}$ instead. Thus, we can apply
(\ref{sol.freelie.1.a.mainclaim}) to $\mathfrak{B}$ and $\mathfrak{L}^{\prime
}$ instead of $\mathfrak{A}$ and $\mathfrak{L}$ (since we assumed that
(\ref{sol.freelie.1.a.mainclaim}) is proven in the case when the alphabet
$\mathfrak{A}$ is finite), and obtain $\left[  b_{p},b_{q}\right]  \in
B_{pq,q}^{\prime}$.
\par
But we have $\mathfrak{L}^{\prime}=\mathfrak{L}\cap\mathfrak{B}^{\ast}%
\subset\mathfrak{L}$, thus $\mathfrak{L}_{pq,q}^{\prime}\subset\mathfrak{L}%
_{pq,q}$ and therefore $B_{pq,q}^{\prime}\subset B_{pq,q}$ (actually, a little
thought shows that $B_{pq,q}^{\prime}=B_{pq,q}$), so that we have $\left[
b_{p},b_{q}\right]  \in B_{pq,q}^{\prime}\subset B_{pq,q}$, and thus
(\ref{sol.freelie.1.a.mainclaim}) is proven.}. Assume this.

We shall prove (\ref{sol.freelie.1.a.mainclaim}) by strong induction over
$\ell\left(  pq\right)  $:

\textit{Induction step:} Let $N$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.a.mainclaim}) holds in the case when $\ell\left(
pq\right)  <N$. We need to prove that (\ref{sol.freelie.1.a.mainclaim}) holds
in the case when $\ell\left(  pq\right)  =N$.

We have assumed that (\ref{sol.freelie.1.a.mainclaim}) holds in the case when
$\ell\left(  pq\right)  <N$. In other words, we have%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  <N. \label{sol.freelie.1.a.indhyp1}%
\end{equation}


We now must prove that (\ref{sol.freelie.1.a.mainclaim}) holds in the case
when $\ell\left(  pq\right)  =N$. In other words, we must prove that
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  =N. \label{sol.freelie.1.a.indgoal1}%
\end{equation}


Let $\mathfrak{G}$ be the set $\bigcup_{i=0}^{N}\mathfrak{A}^{i}$. This set
$\mathfrak{G}$ is finite (since the set $\mathfrak{A}$ is finite) and totally
ordered (by the lexicographic order). Thus, for every $w\in\mathfrak{G}$, we
can define a nonnegative integer $\rho\left(  w\right)  $ by%
\[
\rho\left(  w\right)  =\left\vert \left\{  g\in\mathfrak{G}\ \mid
\ g<w\right\}  \right\vert .
\]
In other words, for every $w\in\mathfrak{G}$, we define $\rho\left(  w\right)
$ to be the number of all $g\in\mathfrak{G}$ which are smaller than $w$. It is
clear that if $w$ and $w^{\prime}$ are two elements of $\mathfrak{G}$
satisfying $w<w^{\prime}$, then%
\begin{equation}
\rho\left(  w\right)  <\rho\left(  w^{\prime}\right)
\label{sol.freelie.1.a.rank}%
\end{equation}
\footnote{\textit{Proof.} Let $w$ and $w^{\prime}$ be two elements of
$\mathfrak{G}$ satisfying $w<w^{\prime}$. Then, $\left\{  g\in\mathfrak{G}%
\ \mid\ g<w\right\}  $ is a proper subset of $\left\{  g\in\mathfrak{G}%
\ \mid\ g<w^{\prime}\right\}  $ (proper because $w$ belongs to the latter set
but not to the former set). Hence, $\left\vert \left\{  g\in\mathfrak{G}%
\ \mid\ g<w\right\}  \right\vert <\left\vert \left\{  g\in\mathfrak{G}%
\ \mid\ g<w^{\prime}\right\}  \right\vert $. Since $\rho\left(  w\right)
=\left\vert \left\{  g\in\mathfrak{G}\ \mid\ g<w\right\}  \right\vert $ and
$\rho\left(  w^{\prime}\right)  =\left\vert \left\{  g\in\mathfrak{G}%
\ \mid\ g<w^{\prime}\right\}  \right\vert $ (similarly), this rewrites as
$\rho\left(  w\right)  <\rho\left(  w^{\prime}\right)  $, qed.}.

For every $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ satisfying
$p<q$ and $\ell\left(  pq\right)  =N$, we have $q\in\mathfrak{G}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  p,q\right)  \in\mathfrak{L}%
\times\mathfrak{L}$ be such that $p<q$ and $\ell\left(  pq\right)  =N$. Then,
$N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{\geq0}%
+\ell\left(  q\right)  \geq\ell\left(  q\right)  $, so that $\ell\left(
q\right)  \leq N$ and thus $q\in\bigcup_{i=0}^{N}\mathfrak{A}^{i}%
=\mathfrak{G}$, qed.}, and thus $\rho\left(  q\right)  $ is well-defined. We
are thus going to prove (\ref{sol.freelie.1.a.indgoal1}) by strong induction
over $\rho\left(  q\right)  $\ \ \ \ \footnote{Of course, this will be an
induction within our current induction step, so the reader should try not to
confuse the two inductions going on.}:

\textit{Induction step:} Let $K$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.a.indgoal1}) holds in the case when $\rho\left(  q\right)
<K$. We need to prove that (\ref{sol.freelie.1.a.indgoal1}) holds in the case
when $\rho\left(  q\right)  =K$.

We have assumed that (\ref{sol.freelie.1.a.indgoal1}) holds in the case when
$\rho\left(  q\right)  <K$. In other words, we have%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  =N\text{ and }\rho\left(  q\right)  <K.
\label{sol.freelie.1.a.indhyp2}%
\end{equation}


Now, let $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ be such that
$p<q$ and $\ell\left(  pq\right)  =N$ and $\rho\left(  q\right)  =K$. We are
going to prove that $\left[  b_{p},b_{q}\right]  \in B_{pq,q}$.

We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$. In other
words, the words $p$ and $q$ are Lyndon. Proposition
\ref{prop.words.lyndon.concat}(a) (applied to $u=p$ and $v=q$) thus shows that
the word $pq$ is Lyndon. In other words, $pq\in\mathfrak{L}$.

Furthermore, Proposition \ref{prop.words.lyndon.concat}(b) (applied to $u=p$
and $v=q$) shows that $pq<q$.

The definition of $\mathfrak{L}_{pq,q}$ yields $\mathfrak{L}_{pq,q}=\left\{
w\in\mathfrak{L}\ \mid\ w\sim pq\text{ and }w<q\right\}  $. The definition of
$B_{pq,q}$ yields $B_{pq,q}=\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$.

The words $p$ and $q$ are nonempty (since they are Lyndon), and thus have
length $\geq1$ each. Hence, the word $pq$ has length $\geq1+1>1$. Hence, $pq$
is a Lyndon word of length $>1$ (since $pq$ is Lyndon and since $pq$ has
length $\ell\left(  pq\right)  >1$). Therefore, $\operatorname*{stf}\left(
pq\right)  $ is well-defined. If $\operatorname*{stf}\left(  pq\right)
=\left(  p,q\right)  $, then it is easy to see that $\left[  b_{p}%
,b_{q}\right]  \in B_{pq,q}$\ \ \ \ \footnote{\textit{Proof.} Assume that
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $. Thus,
(\ref{eq.exe.freelie.1.bw}) (applied to $w=pq$, $u=p$ and $v=q$) shows that
$b_{pq}=\left[  b_{p},b_{q}\right]  $. But $pq$ is an element of
$\mathfrak{L}$ and satisfies $pq\sim pq$ (since the relation $\sim$ is
reflexive) and $pq<q$. In other words, $pq\in\left\{  w\in\mathfrak{L}%
\ \mid\ w\sim pq\text{ and }w<q\right\}  =\mathfrak{L}_{pq,q}$. Thus,
$\kk b_{pq}\subset\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$.
\par
Now, from $b_{pq}=\left[  b_{p},b_{q}\right]  $, we obtain%
\[
\left[  b_{p},b_{q}\right]  =b_{pq}\in\kk b_{pq}\subset\sum
_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}=B_{pq,q}%
\]
(since $B_{pq,q}=\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$), qed.}.
Hence, for the rest of the proof of $\left[  b_{p},b_{q}\right]  \in B_{pq,q}%
$, we WLOG assume that we don't have $\operatorname*{stf}\left(  pq\right)
=\left(  p,q\right)  $. Thus, $q$ is not the longest Lyndon proper suffix of
$pq$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $q$ is the
longest Lyndon proper suffix of $pq$.
\par
Let $w=pq$. Recall that $q$ is the longest Lyndon proper suffix of $pq$. In
other words, $q$ is the longest Lyndon proper suffix of $w$ (since $w=pq$).
\par
Let $\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Then,
$\left(  g,h\right)  =\operatorname*{stf}\left(  \underbrace{pq}_{=w}\right)
=\operatorname*{stf}w$. Thus, $h$ is the longest Lyndon proper suffix of $w$
(by Exercise \ref{exe.words.lyndon.stf.basic}(a)). Comparing this with the
fact that $q$ is the longest Lyndon proper suffix of $w$, we obtain that
$h=q$.
\par
But Exercise \ref{exe.words.lyndon.stf.basic}(b) shows that $w=g\underbrace{h}%
_{=q}=gq$. Thus, $gq=w=pq$. Cancelling $q$ from this equality, we obtain
$g=p$. Now, from $\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)
$, we obtain $\operatorname*{stf}\left(  pq\right)  =\left(  \underbrace{g}%
_{=p},\underbrace{h}_{=q}\right)  =\left(  p,q\right)  $. This contradicts the
fact that we don't have $\operatorname*{stf}\left(  pq\right)  =\left(
p,q\right)  $. This contradiction proves that our assumption was wrong, qed.}.
In other words, the statement that $q$ is the longest Lyndon proper suffix of
$pq$ is \textbf{false}.

We have $\ell\left(  p\right)  >1$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, $\ell\left(  p\right)  \leq1$. Since $p$ is nonempty, this
shows that $\ell\left(  p\right)  =1$. Therefore, $q$ is the longest proper
suffix of $pq$. Thus, $q$ is the longest Lyndon proper suffix of $pq$ (since
$q$ is Lyndon). This contradicts the fact that $q$ is not the longest Lyndon
proper suffix of $pq$. This contradiction shows that our assumption was wrong,
qed.}. Thus, $p$ is a Lyndon word of length $>1$ (since $p$ is Lyndon and
since $p$ has length $\ell\left(  p\right)  >1$). Therefore,
$\operatorname*{stf}p$ is well-defined. Set $\left(  u,v\right)
=\operatorname*{stf}p$. Then, Exercise \ref{exe.words.lyndon.stf.basic}(a)
(applied to $w=p$, $g=u$ and $h=v$) says that $v$ is the longest Lyndon proper
suffix of $p$. In particular, $v$ is a Lyndon proper suffix of $p$. Moreover,
Exercise \ref{exe.words.lyndon.stf.basic}(b) (applied to $w=p$, $g=u$ and
$h=v$) says that we have $p=uv$. Also, Exercise
\ref{exe.words.lyndon.stf.basic}(c) (applied to $w=p$, $g=u$ and $h=v$) says
that we have $u<uv<v$. Finally, Exercise \ref{exe.words.lyndon.stf.basic}(d)
(applied to $w=p$, $g=u$ and $h=v$) says that the word $u$ is Lyndon. In other
words, $u\in\mathfrak{L}$. The words $u$ and $v$ are nonempty (since they are Lyndon).

Exercise \ref{exe.words.lyndon.stf.props2} (applied to $w=p$ and $t=q$) now
shows that $q$ is the longest Lyndon proper suffix of $pq$ if and only if we
do not have $v<q$. Thus, the statement that we do not have $v<q$ is
\textbf{false} (because the statement that $q$ is the longest Lyndon proper
suffix of $pq$ is \textbf{false}). Thus, we have $v<q$. Thus, $u<v<q$.

Recall that $\left(  u,v\right)  =\operatorname*{stf}p$. Thus,
(\ref{eq.exe.freelie.1.bw}) (applied to $w=p$) shows that $b_{p}=\left[
b_{u},b_{v}\right]  $. Thus,%
\begin{equation}
\left[  \underbrace{b_{p}}_{=\left[  b_{u},b_{v}\right]  },b_{q}\right]
=\left[  \left[  b_{u},b_{v}\right]  ,b_{q}\right]  =\left[  \left[
b_{u},b_{q}\right]  ,b_{v}\right]  -\left[  \left[  b_{v},b_{q}\right]
,b_{u}\right]  \label{sol.freelie.1.a.jacobi}%
\end{equation}
(by (\ref{sol.freelie.1.jacobi}), applied to $\mathfrak{k}=\mathfrak{g}$,
$x=b_{u}$, $y=b_{v}$ and $z=b_{q}$).

Now,%
\begin{equation}
\left[  b_{u},b_{q}\right]  \in B_{uq,q} \label{sol.freelie.1.a.in1}%
\end{equation}


\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in1}):} We have $\left(
u,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $u\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $u<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\ell\left(  u\right)  +\underbrace{\ell\left(
v\right)  }_{\substack{>0\\\text{(since }v\text{ is nonempty)}}}>\ell\left(
u\right)  $. But $\ell\left(  pq\right)  =N$, so that $N=\ell\left(
pq\right)  =\underbrace{\ell\left(  p\right)  }_{>\ell\left(  u\right)  }%
+\ell\left(  q\right)  >\ell\left(  u\right)  +\ell\left(  q\right)
=\ell\left(  uq\right)  $. Thus, $\ell\left(  uq\right)  <N$. Therefore,
(\ref{sol.freelie.1.a.indhyp1}) (applied to $\left(  u,q\right)  $ instead of
$\left(  p,q\right)  $) yields $\left[  b_{u},b_{q}\right]  \in B_{uq,q}$.
This proves (\ref{sol.freelie.1.a.in1}).} and%
\begin{equation}
\left[  b_{v},b_{q}\right]  \in B_{vq,q} \label{sol.freelie.1.a.in2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in2}):} We have $\left(
v,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $v\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $v<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)  }_{\substack{
>0\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  v\right)  >\ell\left(
v\right)  $. But $\ell\left(  pq\right)  =N$, so that $N=\ell\left(
pq\right)  =\underbrace{\ell\left(  p\right)  }_{>\ell\left(  v\right)  }%
+\ell\left(  q\right)  >\ell\left(  v\right)  +\ell\left(  q\right)
=\ell\left(  vq\right)  $. Thus, $\ell\left(  vq\right)  <N$. Therefore,
(\ref{sol.freelie.1.a.indhyp1}) (applied to $\left(  v,q\right)  $ instead of
$\left(  p,q\right)  $) yields $\left[  b_{v},b_{q}\right]  \in B_{vq,q}$.
This proves (\ref{sol.freelie.1.a.in2}).}. Hence,
(\ref{sol.freelie.1.a.jacobi}) becomes%
\begin{equation}
\left[  b_{p},b_{q}\right]  =\left[  \underbrace{\left[  b_{u},b_{q}\right]
}_{\substack{\in B_{uq,q}\\\text{(by (\ref{sol.freelie.1.a.in1}))}}%
},b_{v}\right]  -\left[  \underbrace{\left[  b_{v},b_{q}\right]
}_{\substack{\in B_{vq,q}\\\text{(by (\ref{sol.freelie.1.a.in2}))}}%
},b_{u}\right]  \in\left[  B_{uq,q},b_{v}\right]  -\left[  B_{vq,q}%
,b_{u}\right]  . \label{sol.freelie.1.a.jacobi2}%
\end{equation}


On the other hand, if $g$ and $h$ are two Lyndon words satisfying $g<q$, $h<q$
and $gh\sim p$, then%
\begin{equation}
\left(  \text{every }r\in\mathfrak{L}_{gq,q}\text{ satisfies }\left[
b_{r},b_{h}\right]  \in B_{pq,q}\right)  \label{sol.freelie.1.a.in3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in3a}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. Let $r\in
\mathfrak{L}_{gq,q}$. We must prove that $\left[  b_{r},b_{h}\right]  \in
B_{pq,q}$.
\par
The words $g$ and $h$ are Lyndon. In other words, $g\in\mathfrak{L}$ and
$h\in\mathfrak{L}$.
\par
We have $gh\sim p$ and thus $\ell\left(  gh\right)  =\ell\left(  p\right)  $.
Hence, $\ell\left(  p\right)  =\ell\left(  gh\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  $.
\par
We have $gh\sim p$ and $q\sim q$ (since the relation $\sim$ is reflexive).
Hence, $ghq\sim pq$ (by the monoidality of the relation $\sim$).
\par
We have $N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{\geq
0}+\ell\left(  q\right)  \geq\ell\left(  q\right)  $, whence $\ell\left(
q\right)  \leq N$ and thus $q\in\bigcup_{i=0}^{N}\mathfrak{A}^{i}%
=\mathfrak{G}$.
\par
We have $r\in\mathfrak{L}_{gq,q}=\left\{  w\in\mathfrak{L}\ \mid\ w\sim
gq\text{ and }w<q\right\}  $ (by the definition of $\mathfrak{L}_{gq,q}$). In
other words, $r$ is an element of $\mathfrak{L}$ and satisfies $r\sim gq$ and
$r<q$. From $r\sim gq$, we obtain $\ell\left(  r\right)  =\ell\left(
gq\right)  =\ell\left(  g\right)  +\ell\left(  q\right)  $.
\par
If $r=h$, then $\left[  \underbrace{b_{r}}_{\substack{=b_{h}\\\text{(since
}r=h\text{)}}},b_{h}\right]  =\left[  b_{h},b_{h}\right]  =0\in B_{pq,q}$
(since $B_{pq,q}$ is a $\kk$-module). Hence, for the rest of the proof
of $\left[  b_{r},b_{h}\right]  \in B_{pq,q}$, we WLOG assume that we don't
have $r=h$.
\par
Thus, we have $r\neq h$. Hence, we have either $r<h$ or $r>h$ (since the
lexicographic order on $\mathfrak{A}^{\ast}$ is a total order). In other
words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $r<h$.
\par
\textit{Case 2:} We have $r>h$.
\par
Let us first consider Case 1. In this case, we have $r<h$. We have
$\ell\left(  rh\right)  =\underbrace{\ell\left(  r\right)  }_{=\ell\left(
g\right)  +\ell\left(  q\right)  }+\ell\left(  h\right)  =\ell\left(
g\right)  +\ell\left(  q\right)  +\ell\left(  h\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  +\ell\left(  q\right)  $. Comparing this with
$\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=\ell\left(
g\right)  +\ell\left(  h\right)  }+\ell\left(  q\right)  =\ell\left(
g\right)  +\ell\left(  h\right)  +\ell\left(  q\right)  $, we obtain
$\ell\left(  rh\right)  =\ell\left(  pq\right)  =N$.
\par
We have $r\sim gq$ and $h\sim h$ (since the relation $\sim$ is reflexive).
Thus, $rh\sim gqh$ (by the monoidality of the relation $\sim$). But $gqh\sim
ghq$ (by the monoidality of the relation $\sim$ again, since $g\sim g$ and
$qh\sim hq$). From $rh\sim gqh$ and $gqh\sim ghq$, we obtain $rh\sim ghq$
(since the relation $\sim$ is transitive). From $rh\sim ghq$ and $ghq\sim pq$,
we obtain $rh\sim pq$ (since the relation $\sim$ is transitive). Now,
(\ref{sol.freelie.1.a.BhsBgt}) (applied to $rh$, $h$, $pq$ and $q$ instead of
$h$, $s$, $g$ and $t$) yields $B_{rh,h}\subset B_{pq,q}$ (since $h<q$).
\par
On the other hand, $N=\ell\left(  rh\right)  =\underbrace{\ell\left(
r\right)  }_{\geq0}+\ell\left(  h\right)  \geq\ell\left(  h\right)  $, whence
$\ell\left(  h\right)  \leq N$ and thus $h\in\bigcup_{i=0}^{N}\mathfrak{A}%
^{i}=\mathfrak{G}$. Since $h<q$, we have $\rho\left(  h\right)  <\rho\left(
q\right)  $ (by (\ref{sol.freelie.1.a.rank}), applied to $h$ and $q$ instead
of $w$ and $w^{\prime}$), so that $\rho\left(  h\right)  <\rho\left(
q\right)  =K$.
\par
Now, we have $\left(  r,h\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$r\in\mathfrak{L}$ and $h\in\mathfrak{L}$) and $r<h$ and $\ell\left(
rh\right)  =N$ and $\rho\left(  h\right)  <K$. Therefore,
(\ref{sol.freelie.1.a.indhyp2}) (applied to $r$ and $h$ instead of $p$ and
$q$) shows that $\left[  b_{r},b_{h}\right]  \in B_{rh,h}\subset B_{pq,q}$.
Thus, $\left[  b_{r},b_{h}\right]  \in B_{pq,q}$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $h<r$. We have $\ell\left(
hr\right)  =\ell\left(  h\right)  +\underbrace{\ell\left(  r\right)  }%
_{=\ell\left(  g\right)  +\ell\left(  q\right)  }=\ell\left(  h\right)
+\ell\left(  g\right)  +\ell\left(  q\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  +\ell\left(  q\right)  $. Comparing this with
$\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=\ell\left(
g\right)  +\ell\left(  h\right)  }+\ell\left(  q\right)  =\ell\left(
g\right)  +\ell\left(  h\right)  +\ell\left(  q\right)  $, we obtain
$\ell\left(  hr\right)  =\ell\left(  pq\right)  =N$.
\par
We have $h\sim h$ (since the relation $\sim$ is reflexive) and $r\sim gq$.
Thus, $hr\sim hgq$ (by the monoidality of the relation $\sim$). But $hgq\sim
ghq$ (by the monoidality of the relation $\sim$, since $hg\sim gh$ and $q\sim
q$). From $hr\sim hgq$ and $hgq\sim ghq$, we obtain $hr\sim ghq$ (since the
relation $\sim$ is transitive). From $hr\sim ghq$ and $ghq\sim pq$, we obtain
$hr\sim pq$ (since the relation $\sim$ is transitive). Now,
(\ref{sol.freelie.1.a.BhsBgt}) (applied to $hr$, $r$, $pq$ and $q$ instead of
$h$, $s$, $g$ and $t$) yields $B_{hr,r}\subset B_{pq,q}$ (since $r<q$).
\par
On the other hand, $N=\ell\left(  hr\right)  =\underbrace{\ell\left(
h\right)  }_{\geq0}+\ell\left(  r\right)  \geq\ell\left(  r\right)  $, whence
$\ell\left(  r\right)  \leq N$ and thus $r\in\bigcup_{i=0}^{N}\mathfrak{A}%
^{i}=\mathfrak{G}$. Since $r<q$, we have $\rho\left(  r\right)  <\rho\left(
q\right)  $ (by (\ref{sol.freelie.1.a.rank}), applied to $r$ and $q$ instead
of $w$ and $w^{\prime}$), so that $\rho\left(  r\right)  <\rho\left(
q\right)  =K$.
\par
Now, we have $\left(  h,r\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$h\in\mathfrak{L}$ and $r\in\mathfrak{L}$) and $h<r$ and $\ell\left(
hr\right)  =N$ and $\rho\left(  r\right)  <K$. Therefore,
(\ref{sol.freelie.1.a.indhyp2}) (applied to $h$ and $r$ instead of $p$ and
$q$) shows that $\left[  b_{h},b_{r}\right]  \in B_{hr,r}\subset B_{pq,q}$.
Thus, $\left[  b_{r},b_{h}\right]  =-\underbrace{\left[  b_{h},b_{r}\right]
}_{\in B_{pq,q}}\in-B_{pq,q}\subset B_{pq,q}$ (since $B_{pq,q}$ is a
$\kk$-module). Hence, $\left[  b_{r},b_{h}\right]  \in B_{pq,q}$ is
proven in Case 2.
\par
Thus, $\left[  b_{r},b_{h}\right]  \in B_{pq,q}$ is proven in each of the two
Cases 1 and 2. This completes the proof of (\ref{sol.freelie.1.a.in3a}).} and
therefore%
\begin{equation}
\left[  B_{gq,q},b_{h}\right]  \subset B_{pq,q} \label{sol.freelie.1.a.in3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in3b}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. The definition of
$B_{gq,q}$ shows that $B_{gq,q}=\sum_{w\in\mathfrak{L}_{gq,q}}\kk b_{w}%
$. In other words, $B_{gq,q}$ is the $\kk$-linear span of the elements
$b_{w}$ with $w\in\mathfrak{L}_{gq,q}$. Hence, in order to prove the relation
(\ref{sol.freelie.1.a.in3b}), it suffices to show that $\left[  b_{w}%
,b_{h}\right]  \in B_{pq,q}$ for every $w\in\mathfrak{L}_{gq,q}$. But this
follows from (\ref{sol.freelie.1.a.in3a}) (applied to $r=w$). This proves
(\ref{sol.freelie.1.a.in3b}).}.

We have $p\sim p$ (since the relation $\sim$ is reflexive). In other words,
$uv\sim p$ (since $p=uv$). Also, $vu\sim uv$. Since $p=uv$, this rewrites as
$vu\sim p$.

Now, the words $u$ and $v$ are Lyndon and satisfy $u<q$, $v<q$ and $uv\sim p$.
Thus, we can apply (\ref{sol.freelie.1.a.in3b}) to $g=u$ and $h=v$. As a
result, we obtain $\left[  B_{uq,q},b_{v}\right]  \subset B_{pq,q}$.

Furthermore, the words $v$ and $u$ are Lyndon and satisfy $v<q$, $u<q$ and
$vu\sim p$. Thus, we can apply (\ref{sol.freelie.1.a.in3b}) to $g=v$ and
$h=u$. As a result, we obtain $\left[  B_{vq,q},b_{u}\right]  \subset
B_{pq,q}$.

Now, (\ref{sol.freelie.1.a.jacobi2}) becomes%
\[
\left[  b_{p},b_{q}\right]  \in\underbrace{\left[  B_{uq,q},b_{v}\right]
}_{\subset B_{pq,q}}-\underbrace{\left[  B_{vq,q},b_{u}\right]  }_{\subset
B_{pq,q}}\subset B_{pq,q}-B_{pq,q}\subset B_{pq,q}%
\]
(since $B_{pq,q}$ is a $\kk$-module).

Let us now forget that we fixed $\left(  p,q\right)  $. We thus have proven
that%
\[
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  =N\text{ and }\rho\left(  q\right)
=K.
\]
In other words, (\ref{sol.freelie.1.a.indgoal1}) holds in the case when
$\rho\left(  q\right)  =K$. This completes the induction step (in the
induction proof of (\ref{sol.freelie.1.a.indgoal1})). The induction proof of
(\ref{sol.freelie.1.a.indgoal1}) is thus finished.

We thus have proven (\ref{sol.freelie.1.a.indgoal1}). In other words,
(\ref{sol.freelie.1.a.mainclaim}) holds in the case when $\ell\left(
pq\right)  =N$. This completes the induction step (in the induction proof of
(\ref{sol.freelie.1.a.mainclaim})). The induction proof of
(\ref{sol.freelie.1.a.mainclaim}) is thus complete.

Now, we recall that $B$ is the $\kk$-submodule of $\mathfrak{g}$
spanned by the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$. In other
words,%
\begin{equation}
B=\sum_{w\in\mathfrak{L}}\kk b_{w}. \label{sol.freelie.1.a.endgame.B}%
\end{equation}
Thus,%
\begin{equation}
B_{h,s}\subset B\ \ \ \ \ \ \ \ \ \ \text{for every }h\in\mathfrak{A}^{\ast
}\text{ and }s\in\mathfrak{A}^{\ast} \label{sol.freelie.1.a.endgame1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.endgame1}):} Let
$h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}^{\ast}$.
\par
Clearly, $\mathfrak{L}_{h,s}\subset\mathfrak{L}$. Thus, $\sum_{w\in
\mathfrak{L}_{h,s}}\kk b_{w}\subset\sum_{w\in\mathfrak{L}}%
\kk b_{w}$. Since $B_{h,s}=\sum_{w\in\mathfrak{L}_{h,s}}\kk b_{w}$
and $B=\sum_{w\in\mathfrak{L}}\kk b_{w}$ (by
(\ref{sol.freelie.1.a.endgame.B})), this rewrites as $B_{h,s}\subset B$. This
proves (\ref{sol.freelie.1.a.endgame1}).}.

Now, using (\ref{sol.freelie.1.a.mainclaim}), we can easily see the following
fact: For any $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$, we have%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B \label{sol.freelie.1.a.endgame3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.endgame3}):} Let
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$. We need to prove that $\left[
b_{p},b_{q}\right]  \in B$. If $p=q$, then this is clear (because if $p=q$,
then $\left[  \underbrace{b_{p}}_{\substack{=b_{q}\\\text{(since }p=q\text{)}%
}},b_{q}\right]  =\left[  b_{q},b_{q}\right]  =0\in B$ (since $B$ is a
$\kk$-module)). Thus, for the rest of this proof, we WLOG assume that
we don't have $p=q$.
\par
We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$) and $\left(  q,p\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $q\in\mathfrak{L}$ and
$p\in\mathfrak{L}$). We have $p\neq q$ (since we don't have $p=q$). Thus, we
have either $p<q$ or $p>q$ (since the lexicographic order on $\mathfrak{A}%
^{\ast}$ is a total order). In other words, we are in one of the following two
Cases:
\par
\textit{Case 1:} We have $p<q$.
\par
\textit{Case 2:} We have $p>q$.
\par
Let us first consider Case 1. In this case, we have $p<q$. Thus, from
(\ref{sol.freelie.1.a.mainclaim}), we obtain $\left[  b_{p},b_{q}\right]  \in
B_{pq,q}\subset B$ (by (\ref{sol.freelie.1.a.endgame1}) (applied to $h=pq$ and
$s=q$)). Thus, $\left[  b_{p},b_{q}\right]  \in B$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $p>q$. In other words,
$q<p$. Thus, (\ref{sol.freelie.1.a.mainclaim}) (applied to $\left(
q,p\right)  $ instead of $\left(  p,q\right)  $) shows that $\left[
b_{q},b_{p}\right]  \in B_{qp,p}\subset B$ (by (\ref{sol.freelie.1.a.endgame1}%
) (applied to $h=qp$ and $s=p$)). Now, $\left[  b_{p},b_{q}\right]
=-\underbrace{\left[  b_{q},b_{p}\right]  }_{\in B}\in-B\subset B$ (since $B$
is a $\kk$-module). Thus, $\left[  b_{p},b_{q}\right]  \in B$ is proven
in Case 2.
\par
Now, $\left[  b_{p},b_{q}\right]  \in B$ is proven in each of the two Cases 1
and 2. Thus, (\ref{sol.freelie.1.a.endgame3}) is proven.}. Now,
(\ref{sol.freelie.1.a.endgame.B}) yields $B=\sum_{w\in\mathfrak{L}}%
\kk b_{w}=\sum_{p\in\mathfrak{L}}\kk b_{p}$ (here, we renamed the
summation index $w$ as $p$) and $B=\sum_{w\in\mathfrak{L}}\kk b_{w}%
=\sum_{q\in\mathfrak{L}}\kk b_{q}$ (here, we renamed the summation index
$w$ as $q$). Thus,%
\begin{align*}
&  \left[  \underbrace{B}_{=\sum_{p\in\mathfrak{L}}\kk b_{p}%
},\underbrace{B}_{=\sum_{q\in\mathfrak{L}}\kk b_{q}}\right] \\
&  =\left[  \sum_{p\in\mathfrak{L}}\kk b_{p},\sum_{q\in\mathfrak{L}%
}\kk b_{q}\right]  =\sum_{p\in\mathfrak{L}}\sum_{q\in\mathfrak{L}%
}\underbrace{\kk\kk}_{\subset\kk}\underbrace{\left[  b_{p}%
,b_{q}\right]  }_{\substack{\in B\\\text{(by (\ref{sol.freelie.1.a.endgame3}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie bracket on }%
\mathfrak{g}\text{ is }\kk\text{-bilinear}\right) \\
&  \subset\sum_{p\in\mathfrak{L}}\sum_{q\in\mathfrak{L}}\kk B\subset
B\ \ \ \ \ \ \ \ \ \ \left(  \text{since }B\text{ is a }\kk%
\text{-module}\right)  .
\end{align*}
In other words, $B$ is a Lie subalgebra of $\mathfrak{g}$. This solves
Exercise \ref{exe.freelie.1}(a).

(b) We shall first prove that
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q.
\label{sol.freelie.1.b.mainclaim}%
\end{equation}


The proof of (\ref{sol.freelie.1.b.mainclaim}) is very similar to our above
proof of (\ref{sol.freelie.1.a.mainclaim}); it proceeds using the same kind of
double induction, with almost the same computations. Here are the details of
this proof:

\textit{Proof of (\ref{sol.freelie.1.b.mainclaim}):} We can WLOG assume that
the alphabet $\mathfrak{A}$ is finite\footnote{The reasons why this is
legitimate are similar to the analogous reasons in the proof of
(\ref{sol.freelie.1.a.mainclaim}).}. Assume this.

We shall prove (\ref{sol.freelie.1.b.mainclaim}) by strong induction over
$\ell\left(  pq\right)  $:

\textit{Induction step:} Let $N$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.b.mainclaim}) holds in the case when $\ell\left(
pq\right)  <N$. We need to prove that (\ref{sol.freelie.1.b.mainclaim}) holds
in the case when $\ell\left(  pq\right)  =N$.

We have assumed that (\ref{sol.freelie.1.b.mainclaim}) holds in the case when
$\ell\left(  pq\right)  <N$. In other words, we have%
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  <N. \label{sol.freelie.1.b.indhyp1}%
\end{equation}


We now must prove that (\ref{sol.freelie.1.b.mainclaim}) holds in the case
when $\ell\left(  pq\right)  =N$. In other words, we must prove that
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  =N. \label{sol.freelie.1.b.indgoal1}%
\end{equation}


We define a set $\mathfrak{G}$ in the same fashion as in our proof of
(\ref{sol.freelie.1.a.mainclaim}). Likewise, we define a nonnegative integer
$\rho\left(  w\right)  $ for every $w\in\mathfrak{G}$ in the same way as we
did in our proof of (\ref{sol.freelie.1.a.mainclaim}).

For every $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ satisfying
$p<q$ and $\ell\left(  pq\right)  =N$, we have $q\in\mathfrak{G}%
$\ \ \ \ \footnote{This can be proven in the same way as in our proof of
(\ref{sol.freelie.1.a.mainclaim}).}, and thus $\rho\left(  q\right)  $ is
well-defined. We are thus going to prove (\ref{sol.freelie.1.b.indgoal1}) by
strong induction over $\rho\left(  q\right)  $\ \ \ \ \footnote{Of course,
this will be an induction within our current induction step, so the reader
should try not to confuse the two inductions going on.}:

\textit{Induction step:} Let $K$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.b.indgoal1}) holds in the case when $\rho\left(  q\right)
<K$. We need to prove that (\ref{sol.freelie.1.b.indgoal1}) holds in the case
when $\rho\left(  q\right)  =K$.

We have assumed that (\ref{sol.freelie.1.b.indgoal1}) holds in the case when
$\rho\left(  q\right)  <K$. In other words, we have%
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  =N\text{ and }\rho\left(  q\right)  <K.
\label{sol.freelie.1.b.indhyp2}%
\end{equation}


Now, let $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ be such that
$p<q$ and $\ell\left(  pq\right)  =N$ and $\rho\left(  q\right)  =K$. We are
going to prove that $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[
f\left(  b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $.

As in our proof of (\ref{sol.freelie.1.a.indgoal1}), we can prove the
following facts:

\begin{itemize}
\item The words $p$ and $q$ are Lyndon.

\item The word $pq$ is Lyndon. In other words, $pq\in\mathfrak{L}$.

\item We have $pq<q$.

\item The word $pq$ is a Lyndon word of length $>1$. Therefore,
$\operatorname*{stf}\left(  pq\right)  $ is well-defined.
\end{itemize}

If $\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $, then it is
easy to see that $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[
f\left(  b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $%
\ \ \ \ \footnote{\textit{Proof.} Assume that $\operatorname*{stf}\left(
pq\right)  =\left(  p,q\right)  $. Thus, $\left(  p,q\right)
=\operatorname*{stf}\left(  pq\right)  $. Hence, (\ref{eq.exe.freelie.1.bcond}%
) (applied to $w=pq$, $u=p$ and $v=q$) shows that $f\left(  \left[
b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $, qed.}. Hence, for the rest of the proof of $f\left(
\left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $, we WLOG assume that we don't have
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $. As in our proof
of (\ref{sol.freelie.1.a.indgoal1}), we can see that $p$ is a Lyndon word of
length $>1$. Therefore, $\operatorname*{stf}p$ is well-defined. Set $\left(
u,v\right)  =\operatorname*{stf}p$.

As in our proof of (\ref{sol.freelie.1.a.indgoal1}), we can see the following facts:

\begin{itemize}
\item The word $v$ is a Lyndon proper suffix of $p$.

\item We have $p=uv$.

\item We have $u<uv<v$.

\item The word $u$ is Lyndon. In other words, $u\in\mathfrak{L}$.

\item The words $u$ and $v$ are nonempty.

\item We have $u<v<q$.

\item We have $b_{p}=\left[  b_{u},b_{v}\right]  $.

\item The equality (\ref{sol.freelie.1.a.jacobi}) holds.
\end{itemize}

On the other hand, (\ref{eq.exe.freelie.1.bcond}) (applied to $w=p$) shows
that $f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(
b_{u}\right)  ,f\left(  b_{v}\right)  \right]  $ (since $p$ is a Lyndon word
of length $>1$, and since $\left(  u,v\right)  =\operatorname*{stf}p$). Now,
applying the map $f$ to both sides of the equality $b_{p}=\left[  b_{u}%
,b_{v}\right]  $, we obtain%
\begin{equation}
f\left(  b_{p}\right)  =f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[
f\left(  b_{u}\right)  ,f\left(  b_{v}\right)  \right]  .
\label{sol.freelie.1.b.in0}%
\end{equation}


Now,%
\begin{equation}
f\left(  \left[  b_{u},b_{q}\right]  \right)  =\left[  f\left(  b_{u}\right)
,f\left(  b_{q}\right)  \right]  \label{sol.freelie.1.b.in1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in1}):} We have $\left(
u,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $u\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $u<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\ell\left(  u\right)  +\underbrace{\ell\left(
v\right)  }_{\substack{>0\\\text{(since }v\text{ is nonempty)}}}>\ell\left(
u\right)  $. But $\ell\left(  pq\right)  =N$, so that $N=\ell\left(
pq\right)  =\underbrace{\ell\left(  p\right)  }_{>\ell\left(  u\right)  }%
+\ell\left(  q\right)  >\ell\left(  u\right)  +\ell\left(  q\right)
=\ell\left(  uq\right)  $. Thus, $\ell\left(  uq\right)  <N$. Therefore,
(\ref{sol.freelie.1.b.indhyp1}) (applied to $\left(  u,q\right)  $ instead of
$\left(  p,q\right)  $) yields $f\left(  \left[  b_{u},b_{q}\right]  \right)
=\left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]  $. This
proves (\ref{sol.freelie.1.b.in1}).} and%
\begin{equation}
f\left(  \left[  b_{v},b_{q}\right]  \right)  =\left[  f\left(  b_{v}\right)
,f\left(  b_{q}\right)  \right]  \label{sol.freelie.1.b.in2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in2}):} We have $\left(
v,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $v\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $v<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)  }%
_{\substack{>0\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  v\right)
>\ell\left(  v\right)  $. But $\ell\left(  pq\right)  =N$, so that
$N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{>\ell\left(
v\right)  }+\ell\left(  q\right)  >\ell\left(  v\right)  +\ell\left(
q\right)  =\ell\left(  vq\right)  $. Thus, $\ell\left(  vq\right)  <N$.
Therefore, (\ref{sol.freelie.1.b.indhyp1}) (applied to $\left(  v,q\right)  $
instead of $\left(  p,q\right)  $) yields $f\left(  \left[  b_{v}%
,b_{q}\right]  \right)  =\left[  f\left(  b_{v}\right)  ,f\left(
b_{q}\right)  \right]  $. This proves (\ref{sol.freelie.1.b.in2}).}.

On the other hand, if $g$ and $h$ are two Lyndon words satisfying $g<q$, $h<q$
and $gh\sim p$, then%
\begin{equation}
\left(  \text{every }r\in\mathfrak{L}_{gq,q}\text{ satisfies }f\left(  \left[
b_{r},b_{h}\right]  \right)  =\left[  f\left(  b_{r}\right)  ,f\left(
b_{h}\right)  \right]  \right)  \label{sol.freelie.1.b.in3a}%
\end{equation}
\footnote{The proof of (\ref{sol.freelie.1.b.in3a}) can be obtained by
modifying our proof of (\ref{sol.freelie.1.a.in3a}) in a straightforward way.
(We now must use (\ref{sol.freelie.1.b.indhyp2}) instead of
(\ref{sol.freelie.1.a.indhyp2}).)} and therefore%
\begin{equation}
\left(  \text{every }x\in B_{gq,q}\text{ satisfies }f\left(  \left[
x,b_{h}\right]  \right)  =\left[  f\left(  x\right)  ,f\left(  b_{h}\right)
\right]  \right)  \label{sol.freelie.1.b.in3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in3b}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. The definition of
$B_{gq,q}$ shows that $B_{gq,q}=\sum_{w\in\mathfrak{L}_{gq,q}}\kk b_{w}%
$. In other words, $B_{gq,q}$ is the $\kk$-linear span of the elements
$b_{w}$ with $w\in\mathfrak{L}_{gq,q}$. Hence, in order to prove the relation
(\ref{sol.freelie.1.b.in3b}), it suffices to show that $f\left(  \left[
b_{w},b_{h}\right]  \right)  =\left[  f\left(  b_{w}\right)  ,f\left(
b_{h}\right)  \right]  $ for every $w\in\mathfrak{L}_{gq,q}$. But this follows
from (\ref{sol.freelie.1.b.in3a}) (applied to $r=w$). This proves
(\ref{sol.freelie.1.b.in3b}).}.

Recall that (\ref{sol.freelie.1.a.mainclaim}) holds. (This was proven in our
solution to Exercise \ref{exe.freelie.1}(a).) Now, $\left(  u,q\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $u\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $u<q$. Thus, (\ref{sol.freelie.1.a.mainclaim})
(applied to $\left(  u,q\right)  $ instead of $\left(  p,q\right)  $) shows
that $\left[  b_{u},b_{q}\right]  \in B_{uq,q}$. Also, $\left(  v,q\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $v\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $v<q$. Thus, (\ref{sol.freelie.1.a.mainclaim})
(applied to $\left(  v,q\right)  $ instead of $\left(  p,q\right)  $) shows
that $\left[  b_{v},b_{q}\right]  \in B_{vq,q}$.

As in our proof of (\ref{sol.freelie.1.a.indgoal1}), we can show that $uv\sim
p$ and $vu\sim p$.

Now, the words $u$ and $v$ are Lyndon and satisfy $u<q$, $v<q$, $uv\sim p$ and
$\left[  b_{u},b_{q}\right]  \in B_{uq,q}$. Thus, we can apply
(\ref{sol.freelie.1.b.in3b}) to $g=u$, $h=v$ and $x=\left[  b_{u}%
,b_{q}\right]  $. As a result, we obtain%
\begin{equation}
f\left(  \left[  \left[  b_{u},b_{q}\right]  ,b_{v}\right]  \right)  =\left[
\underbrace{f\left(  \left[  b_{u},b_{q}\right]  \right)  }%
_{\substack{=\left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
\\\text{(by (\ref{sol.freelie.1.b.in1}))}}},f\left(  b_{v}\right)  \right]
=\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{v}\right)  \right]  . \label{sol.freelie.1.b.in4}%
\end{equation}


Furthermore, the words $v$ and $u$ are Lyndon and satisfy $v<q$, $u<q$,
$vu\sim p$ and $\left[  b_{v},b_{q}\right]  \in B_{vq,q}$. Thus, we can apply
(\ref{sol.freelie.1.b.in3b}) to $g=v$, $h=u$ and $x=\left[  b_{v}%
,b_{q}\right]  $. As a result, we obtain%
\begin{equation}
f\left(  \left[  \left[  b_{v},b_{q}\right]  ,b_{u}\right]  \right)  =\left[
\underbrace{f\left(  \left[  b_{v},b_{q}\right]  \right)  }%
_{\substack{=\left[  f\left(  b_{v}\right)  ,f\left(  b_{q}\right)  \right]
\\\text{(by (\ref{sol.freelie.1.b.in2}))}}},f\left(  b_{u}\right)  \right]
=\left[  \left[  f\left(  b_{v}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{u}\right)  \right]  . \label{sol.freelie.1.b.in5}%
\end{equation}


Now, applying the map $f$ to both sides of the equality
(\ref{sol.freelie.1.a.jacobi}), we obtain%
\begin{align*}
f\left(  \left[  b_{p},b_{q}\right]  \right)   &  =f\left(  \left[  \left[
b_{u},b_{q}\right]  ,b_{v}\right]  -\left[  \left[  b_{v},b_{q}\right]
,b_{u}\right]  \right) \\
&  =\underbrace{f\left(  \left[  \left[  b_{u},b_{q}\right]  ,b_{v}\right]
\right)  }_{\substack{=\left[  \left[  f\left(  b_{u}\right)  ,f\left(
b_{q}\right)  \right]  ,f\left(  b_{v}\right)  \right]  \\\text{(by
(\ref{sol.freelie.1.b.in4}))}}}-\underbrace{f\left(  \left[  \left[
b_{v},b_{q}\right]  ,b_{u}\right]  \right)  }_{\substack{=\left[  \left[
f\left(  b_{v}\right)  ,f\left(  b_{q}\right)  \right]  ,f\left(
b_{u}\right)  \right]  \\\text{(by (\ref{sol.freelie.1.b.in5}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{v}\right)  \right]  -\left[  \left[  f\left(  b_{v}\right)
,f\left(  b_{q}\right)  \right]  ,f\left(  b_{u}\right)  \right]  .
\end{align*}
Comparing this with%
\begin{align*}
\left[  \underbrace{f\left(  b_{p}\right)  }_{\substack{=\left[  f\left(
b_{u}\right)  ,f\left(  b_{v}\right)  \right]  \\\text{(by
(\ref{sol.freelie.1.b.in0}))}}},f\left(  b_{q}\right)  \right]
&  =\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{v}\right)  \right]
,f\left(  b_{q}\right)  \right] \\
&  =\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{v}\right)  \right]  -\left[  \left[  f\left(  b_{v}\right)
,f\left(  b_{q}\right)  \right]  ,f\left(  b_{u}\right)  \right]  .\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.1.jacobi}), applied
to }\mathfrak{k}=\mathfrak{h}\text{, }x=f\left(  b_{u}\right)  \text{,
}y=f\left(  b_{v}\right)  \text{ and }z=f\left(  b_{q}\right)  \right)  ,
\end{align*}
we obtain $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $.

Let us now forget that we fixed $\left(  p,q\right)  $. We thus have proven
that%
\[
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  =N\text{ and } \rho\left(
q\right) =K.
\]
In other words, (\ref{sol.freelie.1.b.indgoal1}) holds in the case when
$\rho\left(  q\right)  =K$. This completes the induction step (in the
induction proof of (\ref{sol.freelie.1.b.indgoal1})). The induction proof of
(\ref{sol.freelie.1.b.indgoal1}) is thus finished.

We thus have proven (\ref{sol.freelie.1.b.indgoal1}). In other words,
(\ref{sol.freelie.1.b.mainclaim}) holds in the case when $\ell\left(
pq\right)  =N$. This completes the induction step (in the induction proof of
(\ref{sol.freelie.1.b.mainclaim})). The induction proof of
(\ref{sol.freelie.1.b.mainclaim}) is thus complete.

Using (\ref{sol.freelie.1.b.mainclaim}), we can easily see the following fact:
For any $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$, we have%
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \label{sol.freelie.1.b.endgame3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.endgame3}):} Let
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$. We need to prove that $f\left(
\left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $. If $p=q$, then this is clear (because if $p=q$, then
$f\left(  \left[  \underbrace{b_{p}}_{\substack{=b_{q}\\\text{(since
}p=q\text{)}}},b_{q}\right]  \right)  =f\left(  \underbrace{\left[
b_{q},b_{q}\right]  }_{=0}\right)  =f\left(  0\right)  =0$ and $\left[
f\left(  \underbrace{b_{p}}_{\substack{=b_{q}\\\text{(since }p=q\text{)}%
}}\right)  ,f\left(  b_{q}\right)  \right]  =\left[  f\left(  b_{q}\right)
,f\left(  b_{q}\right)  \right]  =0$, so that both sides of the equality
$f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  $ vanish). Thus, for the rest of this proof,
we WLOG assume that we don't have $p=q$.
\par
We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$) and $\left(  q,p\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $q\in\mathfrak{L}$ and
$p\in\mathfrak{L}$). We have $p\neq q$ (since we don't have $p=q$). Thus, we
have either $p<q$ or $p>q$ (since the lexicographic order on $\mathfrak{A}%
^{\ast}$ is a total order). In other words, we are in one of the following two
Cases:
\par
\textit{Case 1:} We have $p<q$.
\par
\textit{Case 2:} We have $p>q$.
\par
Let us first consider Case 1. In this case, we have $p<q$. Thus, from
(\ref{sol.freelie.1.b.mainclaim}), we obtain $f\left(  \left[  b_{p}%
,b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $. Thus, $f\left(  \left[  b_{p},b_{q}\right]  \right)
=\left[  f\left(  b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $ is proven
in Case 1.
\par
Let us now consider Case 2. In this case, we have $p>q$. In other words,
$q<p$. Thus, (\ref{sol.freelie.1.b.mainclaim}) (applied to $\left(
q,p\right)  $ instead of $\left(  p,q\right)  $) shows that $f\left(  \left[
b_{q},b_{p}\right]  \right)  =\left[  f\left(  b_{q}\right)  ,f\left(
b_{p}\right)  \right]  $. Now,
\[
f\left(  \underbrace{\left[  b_{p},b_{q}\right]  }_{=-\left[  b_{q}%
,b_{p}\right]  }\right)  =f\left(  -\left[  b_{q},b_{p}\right]  \right)
=-\underbrace{f\left(  \left[  b_{q},b_{p}\right]  \right)  }_{=\left[
f\left(  b_{q}\right)  ,f\left(  b_{p}\right)  \right]  }=-\left[  f\left(
b_{q}\right)  ,f\left(  b_{p}\right)  \right]  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  .
\]
Thus, $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $ is proven in Case 2.
\par
Now, $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $ is proven in each of the two
Cases 1 and 2. This completes the proof of (\ref{sol.freelie.1.b.endgame3}).}.

Now, we recall that $B$ is the $\kk$-submodule of $\mathfrak{g}$
spanned by the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$. Thus, the
family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk%
$-module $B$.

Now, we have%
\begin{equation}
f\left(  \left[  x,y\right]  \right)  =\left[  f\left(  x\right)  ,f\left(
y\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in B\text{ and }y\in
B \label{sol.freelie.1.b.endgame}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.endgame}):} Let $x\in B$ and
$y\in B$. We must prove the equality $f\left(  \left[  x,y\right]  \right)
=\left[  f\left(  x\right)  ,f\left(  y\right)  \right]  $. Both sides of this
equality are $\kk$-linear in each of $x$ and $y$. Hence, we can WLOG
assume that both $x$ and $y$ belong to the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ (since the family $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$ spans the $\kk$-module $B$). In other words, we can WLOG
assume that there exist $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$ satisfying
$x=b_{p}$ and $y=b_{q}$. Assume this, and consider these $p$ and $q$.
\par
From (\ref{sol.freelie.1.b.endgame3}), we obtain $f\left(  \left[  b_{p}%
,b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $. This rewrites as $f\left(  \left[  x,y\right]
\right)  =\left[  f\left(  x\right)  ,f\left(  y\right)  \right]  $ (since
$x=b_{p}$ and $y=b_{q}$). This proves (\ref{sol.freelie.1.b.endgame}).}. In
other words, the map $f:B\rightarrow\mathfrak{h}$ is a Lie algebra
homomorphism. This solves Exercise \ref{exe.freelie.1}(b).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.freelie.1}}

\textit{Solution to Exercise \ref{exe.freelie.1}.} We begin with the following
fundamental fact:

\begin{proposition}
\label{prop.sol.freelie.1.permutation-conc}Let $k\in \NN$ and $\ell
\in \NN$.

\begin{enumerate}
\item[(a)] For every $\tau\in\Symm_k$ and $\rho\in\Symm_{\ell
}$, there exists a permutation $\eta\in\Symm_{k+\ell}$ such that%
\[
\left(  \eta\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)  .
\]


\item[(b)] There exists a permutation $\zeta\in\Symm_{k+\ell}$ such
that
\[
\left(  \zeta\left(  i\right)  =%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)
\]

\end{enumerate}
\end{proposition}

Proposition \ref{prop.sol.freelie.1.permutation-conc} is a very basic result
that has been tacitly used many times already\footnote{In particular,
Proposition \ref{prop.sol.freelie.1.permutation-conc}(a) can be used to define
an embedding $\Symm_k\times\Symm_{\ell}\rightarrow
\Symm_{k+\ell}$ (which sends $\left(  \tau,\rho\right)  $ to $\eta$).
If we rename $k$ and $\ell$ as $i$ and $j$, this becomes precisely the
embedding $\Symm_i\times\Symm_j\rightarrow\Symm_{i+j}$
used in Definition \ref{ind-res-defn}.}. We give a proof here just for
the sake of completeness.

\begin{proof}
[Proof of Proposition \ref{prop.sol.freelie.1.permutation-conc}.](a) Let
$\tau\in\Symm_k$ and $\rho\in\Symm_{\ell}$. For every
$i\in\left\{  1,2,\ldots,k+\ell\right\}  $, the expression $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k+\ell
\right\}  $. We need to show that $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $i\leq k$.
\par
\textit{Case 2:} We have $i>k$.
\par
Let us first consider Case 1. In this case, we have $i\leq k$. Thus,
$i\in\left\{  1,2,\ldots,k\right\}  $ (since $i\in\left\{  1,2,\ldots
,k+\ell\right\}  $ and $i\leq k$). Therefore, $\tau\left(  i\right)  $ is a
well-defined element of $\left\{  1,2,\ldots,k\right\}  $ (since $\tau
\in\Symm_k$). Hence, $\tau\left(  i\right)  \in\left\{
1,2,\ldots,k\right\}  \subset\left\{  1,2,\ldots,k+\ell\right\}  $. Now,
\begin{align*}%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}  \quad
&  =\tau\left(  i\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq
k\right) \\
&  \in\left\{  1,2,\ldots,k+\ell\right\}  .
\end{align*}
Hence, $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $. We thus
have shown that $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $ in Case
1.
\par
Let us now consider Case 2. In this case, we have $i>k$. Thus, $i\in\left\{
k+1,k+2,\ldots,k+\ell\right\}  $ (since $i\in\left\{  1,2,\ldots
,k+\ell\right\}  $ and $i>k$). Therefore, $i-k\in\left\{  1,2,\ldots
,\ell\right\}  $. Hence, $\rho\left(  i-k\right)  $ is a well-defined element
of $\left\{  1,2,\ldots,\ell\right\}  $ (since $\rho\in\Symm_{\ell}$).
Hence, $\rho\left(  i-k\right)  \in\left\{  1,2,\ldots,\ell\right\}  $, so
that $\rho\left(  i-k\right)  +k\in\left\{  k+1,k+2,\ldots,k+\ell\right\}
\subset\left\{  1,2,\ldots,k+\ell\right\}  $. Now,
\begin{align*}%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}  \quad
&  =\rho\left(  i-k\right)  +k\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}i>k\right) \\
&  \in\left\{  1,2,\ldots,k+\ell\right\}  .
\end{align*}
Hence, $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $. We thus
have shown that $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $ in Case
2.
\par
Now, in each of the two Cases 1 and 2, we have shown that $%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $. Hence,
$%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases} \quad
$ is always a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $
(because the two Cases 1 and 2 cover all possibilities). Qed.}. Thus, we can
define a map $g_{\tau,\rho}:\left\{  1,2,\ldots,k+\ell\right\}  \rightarrow
\left\{  1,2,\ldots,k+\ell\right\}  $ by%
\begin{equation}
\left(  g_{\tau,\rho}\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)  . \label{pf.prop.sol.freelie.1.permutation-conc.1}%
\end{equation}
Consider this map $g_{\tau,\rho}$.

Now, let us forget that we fixed $\tau$ and $\rho$. Thus, for every $\tau
\in\Symm_k$ and $\rho\in\Symm_{\ell}$, we have constructed a
map $g_{\tau,\rho}:\left\{  1,2,\ldots,k+\ell\right\}  \rightarrow\left\{
1,2,\ldots,k+\ell\right\}  $.

Now, let $\tau\in\Symm_k$ and $\rho\in\Symm_{\ell}$. Then,
$\tau^{-1}\in\Symm_k$ and $\rho^{-1}\in\Symm_{\ell}$. Hence,
the map $g_{\tau^{-1},\rho^{-1}}:\left\{  1,2,\ldots,k+\ell\right\}
\rightarrow\left\{  1,2,\ldots,k+\ell\right\}  $ is well-defined.

We have%
\begin{equation}
g_{\tau,\rho}\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\label{pf.prop.sol.freelie.1.permutation-conc.2}%
\end{equation}
(by the definition of $g_{\tau,\rho}$). Also, we have%
\begin{equation}
g_{\tau^{-1},\rho^{-1}}\left(  i\right)  =%
\begin{cases}
\tau^{-1}\left(  i\right)  , & \text{if }i\leq k;\\
\rho^{-1}\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\label{pf.prop.sol.freelie.1.permutation-conc.3}%
\end{equation}
(by the definition of $g_{\tau^{-1},\rho^{-1}}$).

We have $g_{\tau^{-1},\rho^{-1}}\circ g_{\tau,\rho}=\operatorname*{id}%
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k+\ell
\right\}  $. We shall show that $g_{\tau^{-1},\rho^{-1}}\left(  g_{\tau,\rho
}\left(  i\right)  \right)  =i$.
\par
Indeed, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $i\leq k$.
\par
\textit{Case 2:} We have $i>k$.
\par
Let us first consider Case 1. In this case, we have $i\leq k$. Now,
(\ref{pf.prop.sol.freelie.1.permutation-conc.1}) shows that $g_{\tau,\rho
}\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}  \quad
=\tau\left(  i\right)  $ (since $i\leq k$). But from $i\leq k$ and
$i\in\left\{  1,2,\ldots,k+\ell\right\}  $, we obtain $i\in\left\{
1,2,\ldots,k\right\}  $, and thus $\tau\left(  i\right)  \in\left\{
1,2,\ldots,k\right\}  $ (since $\tau\in\Symm_k$). Thus, $\tau\left(
i\right)  \leq k$. Now,
\begin{align*}
g_{\tau^{-1},\rho^{-1}}\left(  \underbrace{g_{\tau,\rho}\left(  i\right)
}_{=\tau\left(  i\right)  }\right)   &  =g_{\tau^{-1},\rho^{-1}}\left(
\tau\left(  i\right)  \right)  =%
\begin{cases}
\tau^{-1}\left(  \tau\left(  i\right)  \right)  , & \text{if }\tau\left(
i\right)  \leq k;\\
\rho^{-1}\left(  \tau\left(  i\right)  -k\right)  +k, & \text{if }\tau\left(
i\right)  >k
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.freelie.1.permutation-conc.2}), applied to }\tau\left(
i\right)  \text{ instead of }i\right) \\
&  =\tau^{-1}\left(  \tau\left(  i\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau\left(  i\right)  \leq k\right)
\\
&  =i.
\end{align*}
Thus, $g_{\tau^{-1},\rho^{-1}}\left(  g_{\tau,\rho}\left(  i\right)  \right)
=i$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i>k$. Now,
(\ref{pf.prop.sol.freelie.1.permutation-conc.1}) shows that $g_{\tau,\rho
}\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}  \quad
=\rho\left(  i-k\right)  +k$ (since $i>k$). But from $i\leq k$ and
$i\in\left\{  1,2,\ldots,k+\ell\right\}  $, we obtain $i\in\left\{
k+1,k+2,\ldots,k+\ell\right\}  $, so that $i-k\in\left\{  1,2,\ldots
,\ell\right\}  $. Hence, $\rho\left(  i-k\right)  \in\left\{  1,2,\ldots
,\ell\right\}  $ (since $\rho\in\Symm_{\ell}$), so that $\rho\left(
i-k\right)  +k\in\left\{  k+1,k+2,\ldots,k+\ell\right\}  $. Thus, $\rho\left(
i-k\right)  +k>k$. Now,
\begin{align*}
g_{\tau^{-1},\rho^{-1}}\left(  \underbrace{g_{\tau,\rho}\left(  i\right)
}_{=\rho\left(  i-k\right)  +k}\right)   &  =g_{\tau^{-1},\rho^{-1}}\left(
\rho\left(  i-k\right)  +k\right)  =%
\begin{cases}
\tau^{-1}\left(  \rho\left(  i-k\right)  +k\right)  , & \text{if }\rho\left(
i-k\right)  +k\leq k;\\
\rho^{-1}\left(  \left(  \rho\left(  i-k\right)  +k\right)  -k\right)  +k, &
\text{if }\rho\left(  i-k\right)  +k>k
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.freelie.1.permutation-conc.2}), applied to }\rho\left(
i-k\right)  +k\text{ instead of }i\right) \\
&  =\rho^{-1}\left(  \underbrace{\left(  \rho\left(  i-k\right)  +k\right)
-k}_{=\rho\left(  i-k\right)  }\right)  +k\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\rho\left(  i-k\right)  +k>k\right) \\
&  =\underbrace{\rho^{-1}\left(  \rho\left(  i-k\right)  \right)  }%
_{=i-k}+k=\left(  i-k\right)  +k=i.
\end{align*}
Thus, $g_{\tau^{-1},\rho^{-1}}\left(  g_{\tau,\rho}\left(  i\right)  \right)
=i$ is proven in Case 2.
\par
We have now proven $g_{\tau^{-1},\rho^{-1}}\left(  g_{\tau,\rho}\left(
i\right)  \right)  =i$ in each of the two Cases 1 and 2. Since these two Cases
cover all possibilities, this shows that $g_{\tau^{-1},\rho^{-1}}\left(
g_{\tau,\rho}\left(  i\right)  \right)  =i$ always holds.
\par
Hence, $\left(  g_{\tau^{-1},\rho^{-1}}\circ g_{\tau,\rho}\right)  \left(
i\right)  =g_{\tau^{-1},\rho^{-1}}\left(  g_{\tau,\rho}\left(  i\right)
\right)  =i=\operatorname*{id}\left(  i\right)  $.
\par
Now, let us forget that we fixed $i$. We thus have proven that $\left(
g_{\tau^{-1},\rho^{-1}}\circ g_{\tau,\rho}\right)  \left(  i\right)
=\operatorname*{id}\left(  i\right)  $ for every $i\in\left\{  1,2,\ldots
,k+\ell\right\}  $. In other words, $g_{\tau^{-1},\rho^{-1}}\circ g_{\tau
,\rho}=\operatorname*{id}$, qed.}. Applying this to $\tau^{-1}$ and $\rho
^{-1}$ instead of $\tau$ and $\rho$, we obtain $g_{\left(  \tau^{-1}\right)
^{-1},\left(  \rho^{-1}\right)  ^{-1}}\circ g_{\tau^{-1},\rho^{-1}%
}=\operatorname*{id}$. Since $\left(  \tau^{-1}\right)  ^{-1}=\tau$ and
$\left(  \rho^{-1}\right)  ^{-1}=\rho$, this rewrites as $g_{\tau,\rho}\circ
g_{\tau^{-1},\rho^{-1}}=\operatorname*{id}$.

The equalities $g_{\tau^{-1},\rho^{-1}}\circ g_{\tau,\rho}=\operatorname*{id}$
and $g_{\tau,\rho}\circ g_{\tau^{-1},\rho^{-1}}=\operatorname*{id}$ show that
the maps $g_{\tau^{-1},\rho^{-1}}$ and $g_{\tau,\rho}$ are mutually inverse.
Thus, the map $g_{\tau,\rho}$ is invertible. In other words, $g_{\tau,\rho}$
is a bijection. In other words, $g_{\tau,\rho}$ is a permutation of $\left\{
1,2,\ldots,k+\ell\right\}  $. In other words, $g_{\tau,\rho}$ is a permutation
in $\Symm_{k+\ell}$. As we know, this permutation satisfies
(\ref{pf.prop.sol.freelie.1.permutation-conc.1}). Hence, there exists a
permutation $\eta\in\Symm_{k+\ell}$ such that%
\[
\left(  \eta\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)
\]
(namely, $\eta=g_{\tau,\rho}$). This proves Proposition
\ref{prop.sol.freelie.1.permutation-conc}(a).

(b) For every $i\in\left\{  1,2,\ldots,k+\ell\right\}  $, the expression $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}
$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k+\ell
\right\}  $. We need to show that $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $.
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $i\leq\ell$.
\par
\textit{Case 2:} We have $i>\ell$.
\par
Let us first consider Case 1. In this case, we have $i\leq\ell$. Thus,
$i\in\left\{  1,2,\ldots,\ell\right\}  $ (since $i\in\left\{  1,2,\ldots
,k+\ell\right\}  $ and $i\leq\ell$) and thus $i+k\in\left\{  k+1,k+2,\ldots
,k+\ell\right\}  \subset\left\{  1,2,\ldots,k+\ell\right\}  $. Now,
\begin{align*}%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}  \quad
&  =i+k\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq\ell\right) \\
&  \in\left\{  1,2,\ldots,k+\ell\right\}  .
\end{align*}
Hence, $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $. We thus
have shown that $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $ in Case
1.
\par
Let us now consider Case 2. In this case, we have $i>\ell$. Thus,
$i\in\left\{  \ell+1,\ell+2,\ldots,k+\ell\right\}  $ (since $i\in\left\{
1,2,\ldots,k+\ell\right\}  $ and $i>\ell$). Therefore, $i-\ell\in\left\{
1,2,\ldots,\left(  k+\ell\right)  -\ell\right\}  =\left\{  1,2,\ldots
,k\right\}  \subset\left\{  1,2,\ldots,k+\ell\right\}  $. Hence,
\begin{align*}%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}  \quad
&  =i-\ell\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>\ell\right) \\
&  \in\left\{  1,2,\ldots,k+\ell\right\}  .
\end{align*}
Hence, $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $. We thus
have shown that $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $ in Case
2.
\par
Now, in each of the two Cases 1 and 2, we have shown that $%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $. Hence,
$%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases} \quad
$ is always a well-defined element of $\left\{  1,2,\ldots,k+\ell\right\}  $
(because the two Cases 1 and 2 cover all possibilities). Qed.}. Hence, we can
define a map $\omega:\left\{  1,2,\ldots,k+\ell\right\}  \rightarrow\left\{
1,2,\ldots,k+\ell\right\}  $ by
\begin{equation}
\left(  \omega\left(  i\right)  =%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)  . \label{pf.prop.sol.freelie.1.permutation-conc.b.def}%
\end{equation}


Consider this map $\omega$. The map $\omega$ is
injective\footnote{\textit{Proof.} Let $i_{1}$ and $i_{2}$ be two elements of
$\left\{  1,2,\ldots,k+\ell\right\}  $ such that $\omega\left(  i_{1}\right)
=\omega\left(  i_{2}\right)  $. We will show that $i_{1}=i_{2}$.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $i_{1}\leq\ell$.
\par
\textit{Case 2:} We have $i_{1}>\ell$.
\par
Let us first consider Case 1. In this case, we have $i_{1}\leq\ell$. Now, the
definition of $\omega$ yields $\omega\left(  i_{1}\right)  =%
\begin{cases}
i_{1}+k, & \text{if }i_{1}\leq\ell;\\
i_{1}-\ell, & \text{if }i_{1}>\ell
\end{cases}  \quad
=i_{1}+k$ (since $i_{1}\leq\ell$). Thus, $\omega\left(  i_{1}\right)
=\underbrace{i_{1}}_{\substack{>0\\\text{(since }i_{1}\in\left\{
1,2,\ldots,k+\ell\right\}  \text{)}}}+k>k$. Hence, $k<\omega\left(
i_{1}\right)  $. But $i_{2}\in\left\{  1,2,\ldots,k+\ell\right\}  $, so that
$i_{2}\leq k+\ell$ and thus $i_{2}-\ell\leq k$.
\par
Now, if we had $i_{2}>\ell$, then we would have%
\begin{align*}
k  &  <\omega\left(  i_{1}\right)  =\omega\left(  i_{2}\right)  =%
\begin{cases}
i_{2}+k, & \text{if }i_{2}\leq\ell;\\
i_{2}-\ell, & \text{if }i_{2}>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right) \\
&  =i_{2}-\ell\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i_{2}>\ell\right) \\
&  \leq k,
\end{align*}
which is absurd. Hence, we cannot have $i_{2}>\ell$. Thus, we must have
$i_{2}\leq\ell$. Hence,
\begin{align*}
i_{1}+k  &  =\omega\left(  i_{1}\right)  =\omega\left(  i_{2}\right)  =%
\begin{cases}
i_{2}+k, & \text{if }i_{2}\leq\ell;\\
i_{2}-\ell, & \text{if }i_{2}>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right) \\
&  =i_{2}+k\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i_{2}\leq\ell\right)  .
\end{align*}
Subtracting $k$ from both sides of this equality, we obtain $i_{1}=i_{2}$.
Thus, $i_{1}=i_{2}$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $i_{1}>\ell$. Now, the
definition of $\omega$ yields $\omega\left(  i_{1}\right)  =%
\begin{cases}
i_{1}+k, & \text{if }i_{1}\leq\ell;\\
i_{1}-\ell, & \text{if }i_{1}>\ell
\end{cases}  \quad
=i_{1}-\ell$ (since $i_{1}>\ell$). Thus, $\omega\left(  i_{1}\right)
=\underbrace{i_{1}}_{\substack{\leq k+\ell\\\text{(since }i_{1}\in\left\{
1,2,\ldots,k+\ell\right\}  \text{)}}}-\ell\leq\left(  k+\ell\right)  -\ell=k$.
Hence, $k\geq\omega\left(  i_{1}\right)  $. But $i_{2}\in\left\{
1,2,\ldots,k+\ell\right\}  $, so that $i_{2}>0$ and thus $\underbrace{i_{2}%
}_{>0}+k>k$.
\par
Now, if we had $i_{2}\leq\ell$, then we would have%
\begin{align*}
k  &  \geq\omega\left(  i_{1}\right)  =\omega\left(  i_{2}\right)  =%
\begin{cases}
i_{2}+k, & \text{if }i_{2}\leq\ell;\\
i_{2}-\ell, & \text{if }i_{2}>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right) \\
&  =i_{2}+k\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i_{2}\leq\ell\right) \\
&  >k,
\end{align*}
which is absurd. Hence, we cannot have $i_{2}\leq\ell$. Thus, we must have
$i_{2}>\ell$. Hence,
\begin{align*}
i_{1}-\ell &  =\omega\left(  i_{1}\right)  =\omega\left(  i_{2}\right)  =%
\begin{cases}
i_{2}+k, & \text{if }i_{2}\leq\ell;\\
i_{2}-\ell, & \text{if }i_{2}>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right) \\
&  =i_{2}-\ell\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i_{2}>\ell\right)  .
\end{align*}
Adding $\ell$ to both sides of this equality, we obtain $i_{1}=i_{2}$. Thus,
$i_{1}=i_{2}$ is proven in Case 2.
\par
We have now proven $i_{1}=i_{2}$ in each of the two Cases 1 and 2. Since these
two Cases cover all possibilities, this shows that $i_{1}=i_{2}$ always holds.
\par
Now, let us forget that we fixed $i_{1}$ and $i_{2}$. We thus have shown that
if $i_{1}$ and $i_{2}$ are two elements of $\left\{  1,2,\ldots,k+\ell
\right\}  $ such that $\omega\left(  i_{1}\right)  =\omega\left(
i_{2}\right)  $, then $i_{1}=i_{2}$. In other words, the map $\omega$ is
injective. Qed.} and surjective\footnote{\textit{Proof.} Let $j\in\left\{
1,2,\ldots,k+\ell\right\}  $. We will show that $j\in\omega\left(  \left\{
1,2,\ldots,k+\ell\right\}  \right)  $.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $j\leq k$.
\par
\textit{Case 2:} We have $j>k$.
\par
Let us first consider Case 1. In this case, we have $j\leq k$. Thus,
$j\in\left\{  1,2,\ldots,k\right\}  $ (since $j\in\left\{  1,2,\ldots
,k+\ell\right\}  $). Hence, $j+\ell\in\left\{  \ell+1,\ell+2,\ldots
,\ell+k\right\}  \subset\left\{  1,2,\ldots,\ell+k\right\}  =\left\{
1,2,\ldots,k+\ell\right\}  $ (since $\ell+k=k+\ell$). Thus, $\omega\left(
j+\ell\right)  $ is well-defined. We have $j+\ell\in\left\{  \ell
+1,\ell+2,\ldots,\ell+k\right\}  $, and thus $j+\ell>\ell$. The definition of
$\omega$ yields
\begin{align*}
\omega\left(  j+\ell\right)   &  =%
\begin{cases}
\left(  j+\ell\right)  +k, & \text{if }j+\ell\leq\ell;\\
\left(  j+\ell\right)  -\ell, & \text{if }j+\ell>\ell
\end{cases}  \quad
=\left(  j+\ell\right)  -\ell\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
j+\ell>\ell\right) \\
&  =j,
\end{align*}
and thus $j=\omega\left(  \underbrace{j+\ell}_{\in\left\{  1,2,\ldots
,k+\ell\right\}  }\right)  \in\omega\left(  \left\{  1,2,\ldots,k+\ell
\right\}  \right)  $. Thus, $j\in\omega\left(  \left\{  1,2,\ldots
,k+\ell\right\}  \right)  $ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $j>k$. Thus, $j\in\left\{
k+1,k+2,\ldots,k+\ell\right\}  $ (since $j\in\left\{  1,2,\ldots
,k+\ell\right\}  $). Hence, $j-k\in\left\{  1,2,\ldots,\ell\right\}
\subset\left\{  1,2,\ldots,\ell+k\right\}  =\left\{  1,2,\ldots
,k+\ell\right\}  $ (since $\ell+k=k+\ell$). Thus, $\omega\left(  j-k\right)  $
is well-defined. We have $j-k\in\left\{  1,2,\ldots,\ell\right\}  $ and thus
$j-k\leq\ell$. The definition of $\omega$ yields
\begin{align*}
\omega\left(  j-k\right)   &  =%
\begin{cases}
\left(  j-k\right)  +k, & \text{if }j-k\leq\ell;\\
\left(  j-k\right)  -\ell, & \text{if }j-k>\ell
\end{cases}  \quad
=\left(  j-k\right)  +k\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j-k\leq
\ell\right) \\
&  =j,
\end{align*}
and thus $j=\omega\left(  \underbrace{j-k}_{\in\left\{  1,2,\ldots
,k+\ell\right\}  }\right)  \in\omega\left(  \left\{  1,2,\ldots,k+\ell
\right\}  \right)  $. Thus, $j\in\omega\left(  \left\{  1,2,\ldots
,k+\ell\right\}  \right)  $ is proven in Case 2.
\par
We have now proven $j\in\omega\left(  \left\{  1,2,\ldots,k+\ell\right\}
\right)  $ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this shows that $j\in\omega\left(  \left\{  1,2,\ldots
,k+\ell\right\}  \right)  $ always holds.
\par
Now, let us forget that we fixed $j$. We thus have shown that $j\in
\omega\left(  \left\{  1,2,\ldots,k+\ell\right\}  \right)  $ for every
$j\in\left\{  1,2,\ldots,k+\ell\right\}  $. In other words, $\left\{
1,2,\ldots,k+\ell\right\}  \subset\omega\left(  \left\{  1,2,\ldots
,k+\ell\right\}  \right)  $. In other words, the map $\omega$ is surjective.
Qed.}. Thus, the map $\omega$ is bijective. Hence, $\omega$ is a bijection
$\left\{  1,2,\ldots,k+\ell\right\}  \rightarrow\left\{  1,2,\ldots
,k+\ell\right\}  $. In other words, $\omega$ is a permutation of the set
$\left\{  1,2,\ldots,k+\ell\right\}  $. In other words, $\omega\in
\Symm_{k+\ell}$. Hence, there exists a permutation $\zeta
\in\Symm_{k+\ell}$ such that
\[
\left(  \zeta\left(  i\right)  =%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)
\]
(namely, $\zeta=\omega$). This proves Proposition
\ref{prop.sol.freelie.1.permutation-conc}(b).
\end{proof}

Now, let us introduce some notations.

For every word $w\in\mathfrak{A}^{\ast}$, we define a subset
$\operatorname*{Alph}w$ of $\mathfrak{A}$ by%
\begin{equation}
\operatorname*{Alph}w=\left\{  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
}\right\}  . \label{sol.freelie.1.Alph.def}%
\end{equation}
We notice the following facts:

\begin{itemize}
\item If $w$ is any word, then $\operatorname*{Alph}w$ is the set of all
letters of $w$\ \ \ \ \footnote{\textit{Proof.} Let $w$ be any word. Thus,
$w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  $. Therefore,
the letters of $w$ (in the order in which they appear in $w$) are $w_{1}$,
$w_{2}$, $\ldots$, $w_{\ell\left(  w\right)  }$. Hence, the set of all letters
of $w$ is $\left\{  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right\}  $.
Since $\operatorname*{Alph}w=\left\{  w_{1},w_{2},\ldots,w_{\ell\left(
w\right)  }\right\}  $ (by the definition of $\operatorname*{Alph}w$), this
rewrites as follows: The set of all letters of $w$ is $\operatorname*{Alph}w$.
Qed.}.

\item If $w$ is any word, then%
\begin{equation}
\text{the set }\operatorname*{Alph}w\text{ is finite}
\label{sol.freelie.1.Alph.fin}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Alph.fin}):} Let $w$ be any
word. Then, the definition of $\operatorname*{Alph}w$ shows that
$\operatorname*{Alph}w=\left\{  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
}\right\}  $. Thus, the set $\operatorname*{Alph}w$ has at most $\ell\left(
w\right)  $ elements. In particular, the set $\operatorname*{Alph}w$ is
finite. This proves (\ref{sol.freelie.1.Alph.fin}).}.

\item If $w$ is any word, then%
\begin{equation}
w\in\left(  \operatorname*{Alph}w\right)  ^{\ell\left(  w\right)  }
\label{sol.freelie.1.Alph.w}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Alph.w}):} Let $w$ be any word.
Then, $w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  $. Now,
every $i\in\left\{  1,2,\ldots,\ell\left(  w\right)  \right\}  $ satisfies%
\begin{align*}
w_{i}  &  \in\left\{  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\in\left\{  1,2,\ldots,\ell\left(
w\right)  \right\}  \right) \\
&  =\operatorname*{Alph}w\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\operatorname*{Alph}w=\left\{  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)
}\right\}  \text{ (by the definition of }\operatorname*{Alph}w\text{)}\right)
.
\end{align*}
Thus, $\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)
\in\left(  \operatorname*{Alph}w\right)  ^{\ell\left(  w\right)  }$. Hence,
$w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  \in\left(
\operatorname*{Alph}w\right)  ^{\ell\left(  w\right)  }$. This proves
(\ref{sol.freelie.1.Alph.w}).}.

\item If $u$ and $v$ are any two words, then%
\begin{equation}
\operatorname*{Alph}\left(  uv\right)  =\left(  \operatorname*{Alph}u\right)
\cup\left(  \operatorname*{Alph}v\right)  \label{sol.freelie.1.Alph.uv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Alph.uv1}):} Let $u$ and $v$ be
two words. We have $uv=\left(  \left(  uv\right)  _{1},\left(  uv\right)
_{2},\ldots,\left(  uv\right)  _{\ell\left(  uv\right)  }\right)  $, and thus
\begin{align*}
\left(  \left(  uv\right)  _{1},\left(  uv\right)  _{2},\ldots,\left(
uv\right)  _{\ell\left(  uv\right)  }\right)   &  =\underbrace{u}_{=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  }\underbrace{v}%
_{=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  }=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  \left(  v_{1}%
,v_{2},\ldots,v_{\ell\left(  v\right)  }\right) \\
&  =\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right)  .
\end{align*}
Now, the definition of $\operatorname*{Alph}\left(  uv\right)  $ yields%
\begin{align*}
\operatorname*{Alph}\left(  uv\right)   &  =\left\{  \left(  uv\right)
_{1},\left(  uv\right)  _{2},\ldots,\left(  uv\right)  _{\ell\left(
uv\right)  }\right\} \\
&  =\left(  \text{the set of all entries of the list }\underbrace{\left(
\left(  uv\right)  _{1},\left(  uv\right)  _{2},\ldots,\left(  uv\right)
_{\ell\left(  uv\right)  }\right)  }_{=\left(  u_{1},u_{2},\ldots
,u_{\ell\left(  u\right)  },v_{1},v_{2},\ldots,v_{\ell\left(  v\right)
}\right)  }\right) \\
&  =\left(  \text{the set of all entries of the list }\left(  u_{1}%
,u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2},\ldots,v_{\ell\left(
v\right)  }\right)  \right) \\
&  =\left\{  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right\}  =\left\{  u_{1},u_{2}%
,\ldots,u_{\ell\left(  u\right)  }\right\}  \cup\left\{  v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right\}  .
\end{align*}
Compared with%
\[
\underbrace{\left(  \operatorname*{Alph}u\right)  }_{\substack{=\left\{
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right\}  \\\text{(by the
definition of }\operatorname*{Alph}u\text{)}}}\cup\underbrace{\left(
\operatorname*{Alph}v\right)  }_{\substack{=\left\{  v_{1},v_{2}%
,\ldots,v_{\ell\left(  v\right)  }\right\}  \\\text{(by the definition of
}\operatorname*{Alph}v\text{)}}}=\left\{  u_{1},u_{2},\ldots,u_{\ell\left(
u\right)  }\right\}  \cup\left\{  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)
}\right\}  ,
\]
this yields $\operatorname*{Alph}\left(  uv\right)  =\left(
\operatorname*{Alph}u\right)  \cup\left(  \operatorname*{Alph}v\right)  $.
This proves (\ref{sol.freelie.1.Alph.uv1}).}.
\end{itemize}

Next, we make another definition: For every $w\in\mathfrak{A}^{\ast}$, we
define a subset $\operatorname*{Und}w$ of $\mathfrak{A}^{\ast}$ by%
\begin{equation}
\operatorname*{Und}w=\bigcup\limits_{j=0}^{\ell\left(  w\right)  }\left(
\operatorname*{Alph}w\right)  ^{j}. \label{sol.freelie.1.Und.def}%
\end{equation}


We notice the following fact:

\begin{itemize}
\item If $w$ is any word, then%
\begin{equation}
\text{the set }\operatorname*{Und}w\text{ is finite}
\label{sol.freelie.1.Und.fin}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Und.fin}):} Let $w$ be any
word. Then, $\ell\left(  w\right)  \in \NN$. Also, the set
$\operatorname*{Alph}w$ is finite (by (\ref{sol.freelie.1.Alph.fin})). Thus,
for every $j\in\left\{  0,1,\ldots,\ell\left(  w\right)  \right\}  $, the set
$\left(  \operatorname*{Alph}w\right)  ^{j}$ is finite. Hence, $\bigcup
\limits_{j=0}^{\ell\left(  w\right)  }\left(  \operatorname*{Alph}w\right)
^{j}$ is a finite union of finite sets, and therefore itself a finite set
(since any finite union of finite sets must itself be a finite set). In other
words, $\operatorname*{Und}w$ is a finite set (since $\operatorname*{Und}%
w=\bigcup\limits_{j=0}^{\ell\left(  w\right)  }\left(  \operatorname*{Alph}%
w\right)  ^{j}$). This proves (\ref{sol.freelie.1.Und.fin}).} and we have%
\begin{equation}
w\in\operatorname*{Und}w \label{sol.freelie.1.Und.w}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Und.w}):} Let $w$ be any word.
Then, (\ref{sol.freelie.1.Alph.w}) shows that $w\in\left(
\operatorname*{Alph}w\right)  ^{\ell\left(  w\right)  }$. But $\ell\left(
w\right)  \in\left\{  0,1,\ldots,\ell\left(  w\right)  \right\}  $. Hence,
$\left(  \operatorname*{Alph}w\right)  ^{\ell\left(  w\right)  }$ is a term in
the union $\bigcup\limits_{j=0}^{\ell\left(  w\right)  }\left(
\operatorname*{Alph}w\right)  ^{j}$. Therefore, $\left(  \operatorname*{Alph}%
w\right)  ^{\ell\left(  w\right)  }\subset\bigcup\limits_{j=0}^{\ell\left(
w\right)  }\left(  \operatorname*{Alph}w\right)  ^{j}$. Now, $w\in\left(
\operatorname*{Alph}w\right)  ^{\ell\left(  w\right)  }\subset\bigcup
\limits_{j=0}^{\ell\left(  w\right)  }\left(  \operatorname*{Alph}w\right)
^{j}=\operatorname*{Und}w$ (by (\ref{sol.freelie.1.Und.def})). This proves
(\ref{sol.freelie.1.Und.w}).}.

\item If $u$ and $v$ are any two words, then%
\begin{equation}
\operatorname*{Und}u\subset\operatorname*{Und}\left(  uv\right)
\label{sol.freelie.1.Und.uv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Und.uv1}):} Let $u$ and $v$ be
two words. Then, (\ref{sol.freelie.1.Alph.uv1}) shows that
$\operatorname*{Alph}\left(  uv\right)  =\left(  \operatorname*{Alph}u\right)
\cup\left(  \operatorname*{Alph}v\right)  \supset \operatorname*{Alph}u$.
Thus, $\operatorname*{Alph}u\subset\operatorname*{Alph}\left(  uv\right)  $.
Furthermore, $\ell\left(  uv\right)  =\ell\left(  u\right)  +\underbrace{\ell
\left(  v\right)  }_{\geq0}\geq\ell\left(  u\right)  $. Thus, $\ell\left(
u\right)  \leq\ell\left(  uv\right)  $. Therefore, the union $\bigcup
\limits_{j=0}^{\ell\left(  u\right)  }\left(  \operatorname*{Alph}\left(
uv\right)  \right)  ^{j}$ is contained in the union $\bigcup\limits_{j=0}%
^{\ell\left(  uv\right)  }\left(  \operatorname*{Alph}\left(  uv\right)
\right)  ^{j}$. Thus, we have $\bigcup\limits_{j=0}^{\ell\left(  u\right)
}\left(  \operatorname*{Alph}\left(  uv\right)  \right)  ^{j}\subset
\bigcup\limits_{j=0}^{\ell\left(  uv\right)  }\left(  \operatorname*{Alph}%
\left(  uv\right)  \right)  ^{j}$.
\par
Now, the definition of $\operatorname*{Und}\left(  uv\right)  $ yields
$\operatorname*{Und}\left(  uv\right)  =\bigcup\limits_{j=0}^{\ell\left(
uv\right)  }\left(  \operatorname*{Alph}\left(  uv\right)  \right)  ^{j}$. But
the definition of $\operatorname*{Und}u$ yields%
\[
\operatorname*{Und}u=\bigcup\limits_{j=0}^{\ell\left(  u\right)  }\left(
\underbrace{\operatorname*{Alph}u}_{\subset\operatorname*{Alph}\left(
uv\right)  }\right)  ^{j}\subset\bigcup\limits_{j=0}^{\ell\left(  u\right)
}\left(  \operatorname*{Alph}\left(  uv\right)  \right)  ^{j}\subset
\bigcup\limits_{j=0}^{\ell\left(  uv\right)  }\left(  \operatorname*{Alph}%
\left(  uv\right)  \right)  ^{j}=\operatorname*{Und}\left(  uv\right)  .
\]
This proves (\ref{sol.freelie.1.Und.uv1}).} and%
\begin{equation}
\operatorname*{Und}v\subset\operatorname*{Und}\left(  uv\right)
\label{sol.freelie.1.Und.uv2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Und.uv2}):} Let $u$ and $v$ be
two words. Then, (\ref{sol.freelie.1.Alph.uv1}) shows that
$\operatorname*{Alph}\left(  uv\right)  =\left(  \operatorname*{Alph}u\right)
\cup\left(  \operatorname*{Alph}v\right)  \supset \operatorname*{Alph}v$.
Thus, $\operatorname*{Alph}v\subset\operatorname*{Alph}\left(  uv\right)  $.
Furthermore, $\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)
}_{\geq0}+\ell\left(  v\right)  \geq\ell\left(  v\right)  $. Thus,
$\ell\left(  v\right)  \leq\ell\left(  uv\right)  $. Therefore, the union
$\bigcup\limits_{j=0}^{\ell\left(  v\right)  }\left(  \operatorname*{Alph}%
\left(  uv\right)  \right)  ^{j}$ is contained in the union $\bigcup
\limits_{j=0}^{\ell\left(  uv\right)  }\left(  \operatorname*{Alph}\left(
uv\right)  \right)  ^{j}$. Thus, we have $\bigcup\limits_{j=0}^{\ell\left(
v\right)  }\left(  \operatorname*{Alph}\left(  uv\right)  \right)  ^{j}%
\subset\bigcup\limits_{j=0}^{\ell\left(  uv\right)  }\left(
\operatorname*{Alph}\left(  uv\right)  \right)  ^{j}$.
\par
Now, the definition of $\operatorname*{Und}\left(  uv\right)  $ yields
$\operatorname*{Und}\left(  uv\right)  =\bigcup\limits_{j=0}^{\ell\left(
uv\right)  }\left(  \operatorname*{Alph}\left(  uv\right)  \right)  ^{j}$. But
the definition of $\operatorname*{Und}u$ yields%
\[
\operatorname*{Und}v=\bigcup\limits_{j=0}^{\ell\left(  v\right)  }\left(
\underbrace{\operatorname*{Alph}v}_{\subset\operatorname*{Alph}\left(
uv\right)  }\right)  ^{j}\subset\bigcup\limits_{j=0}^{\ell\left(  v\right)
}\left(  \operatorname*{Alph}\left(  uv\right)  \right)  ^{j}\subset
\bigcup\limits_{j=0}^{\ell\left(  uv\right)  }\left(  \operatorname*{Alph}%
\left(  uv\right)  \right)  ^{j}=\operatorname*{Und}\left(  uv\right)  .
\]
This proves (\ref{sol.freelie.1.Und.uv2}).}.
\end{itemize}

We define a binary relation $\sim$ on the set $\mathfrak{A}^{\ast}$ as
follows: If $w$ and $w^{\prime}$ are two words, then we write $w\sim
w^{\prime}$ if and only if there exists a permutation $\sigma\in
\Symm_k$ satisfying $w^{\prime}=\left(  w_{\sigma\left(  1\right)
},w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  k\right)  }\right)  $,
where $k=\ell\left(  w\right)  $. Thus, if $w$ and $w^{\prime}$ are two words,
then $w\sim w^{\prime}$ holds if and only if $w^{\prime}$ is a permutation of
the word $w$. We notice the following properties of the relation $\sim$:

\begin{itemize}
\item The relation $\sim$ is reflexive\footnote{\textit{Proof.} Let
$w\in\mathfrak{A}^{\ast}$. Let $k=\ell\left(  w\right)  $. Now, $w=\left(
w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  =\left(  w_{1}%
,w_{2},\ldots,w_{k}\right)  $ (since $\ell\left(  w\right)  =k$).
\par
Now, consider the permutation $\operatorname*{id}\in\Symm_k$. We have
$w_{\operatorname*{id}\left(  i\right)  }=w_{i}$ for every $i\in\left\{
1,2,\ldots,k\right\}  $. Thus, $\left(  w_{\operatorname*{id}\left(  1\right)
},w_{\operatorname*{id}\left(  2\right)  },\ldots,w_{\operatorname*{id}\left(
k\right)  }\right)  =\left(  w_{1},w_{2},\ldots,w_{k}\right)  =w$, so that
$w=\left(  w_{\operatorname*{id}\left(  1\right)  },w_{\operatorname*{id}%
\left(  2\right)  },\ldots,w_{\operatorname*{id}\left(  k\right)  }\right)  $.
Hence, there exists a permutation $\sigma\in\Symm_k$ satisfying
$w=\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  }%
,\ldots,w_{\sigma\left(  k\right)  }\right)  $ (namely, $\sigma
=\operatorname*{id}$).
\par
But recall that $w\sim w$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $w=\left(  w_{\sigma\left(  1\right)
},w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  k\right)  }\right)  $
(by the definition of the relation $\sim$, since $k=\ell\left(  w\right)  $).
Thus, $w\sim w$ holds (since there exists a permutation $\sigma\in
\Symm_k$ satisfying $w=\left(  w_{\sigma\left(  1\right)  }%
,w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  k\right)  }\right)  $).
\par
Let us now forget that we fixed $w$. We thus have shown that $w\sim w$ for
every $w\in\mathfrak{A}^{\ast}$. In other words, the relation $\sim$ is
reflexive. Qed.}, symmetric\footnote{\textit{Proof.} Let $u\in\mathfrak{A}%
^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be such that $u\sim v$. Let
$k=\ell\left(  u\right)  $.
\par
Recall that $u\sim v$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $v=\left(  u_{\sigma\left(  1\right)
},u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $
(by the definition of the relation $\sim$, since $k=\ell\left(  u\right)  $).
Thus, there exists a permutation $\sigma\in\Symm_k$ satisfying
$v=\left(  u_{\sigma\left(  1\right)  },u_{\sigma\left(  2\right)  }%
,\ldots,u_{\sigma\left(  k\right)  }\right)  $ (since $u\sim v$ holds). Let
$\tau$ be this permutation $\sigma$. Thus, $\tau$ is a permutation $\sigma
\in\Symm_k$ satisfying $v=\left(  u_{\sigma\left(  1\right)
},u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $.
In other words, $\tau$ is a permutation in $\Symm_k$ and satisfies
$v=\left(  u_{\tau\left(  1\right)  },u_{\tau\left(  2\right)  }%
,\ldots,u_{\tau\left(  k\right)  }\right)  $. From $v=\left(  u_{\tau\left(
1\right)  },u_{\tau\left(  2\right)  },\ldots,u_{\tau\left(  k\right)
}\right)  $, we obtain $\ell\left(  v\right)  =k$ and thus $v=\left(
v_{1},v_{2},\ldots,v_{k}\right)  $. Hence, $\left(  v_{1},v_{2},\ldots
,v_{k}\right)  =v=\left(  u_{\tau\left(  1\right)  },u_{\tau\left(  2\right)
},\ldots,u_{\tau\left(  k\right)  }\right)  $.
In other words,
$\left(  u_{\tau\left(  1\right)  },u_{\tau\left(  2\right)
},\ldots,u_{\tau\left(  k\right)  }\right)
= \left(  v_{1},v_{2},\ldots ,v_{k}\right) $.
In other words,
\begin{equation}
u_{\tau\left(  i\right)  } = v_{i}
\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  1,2,\ldots,k\right\}  . \label{sol.freelie.1.rel.sym.pf.1}%
\end{equation}
\par
Now, for every $j\in\left\{  1,2,\ldots,k\right\}  $, we have%
\begin{align*}
u_{j}  &  =u_{\tau\left(  \tau^{-1}\left(  j\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j=\tau\left(  \tau^{-1}\left(
j\right)  \right)  \text{ (since }\tau\left(  \tau^{-1}\left(  j\right)
\right)  =j\text{)}\right) \\
&  =v_{\tau^{-1}\left(  j\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.1.rel.sym.pf.1}), applied to }i=\tau^{-1}\left(  j\right)
\right)  .
\end{align*}
In other words, $\left(  u_{1},u_{2},\ldots,u_{k}\right)  =\left(
v_{\tau^{-1}\left(  1\right)  },v_{\tau^{-1}\left(  2\right)  },\ldots
,v_{\tau^{-1}\left(  k\right)  }\right)  $. Now, from $\ell\left(  u\right)
=k$, we obtain $u=\left(  u_{1},u_{2},\ldots,u_{k}\right)  =\left(
v_{\tau^{-1}\left(  1\right)  },v_{\tau^{-1}\left(  2\right)  },\ldots
,v_{\tau^{-1}\left(  k\right)  }\right)  $. In other words, there exists a
permutation $\sigma\in\Symm_k$ satisfying $u=\left(  v_{\sigma\left(
1\right)  },v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  k\right)
}\right)  $ (namely, $\sigma=\tau^{-1}$).
\par
But recall that $v\sim u$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $u=\left(  v_{\sigma\left(  1\right)
},v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  k\right)  }\right)  $
(by the definition of the relation $\sim$, since $k=\ell\left(  v\right)  $).
Thus, $v\sim u$ holds (since there exists a permutation $\sigma\in
\Symm_k$ satisfying $u=\left(  v_{\sigma\left(  1\right)  }%
,v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  k\right)  }\right)  $).
\par
Let us now forget that we fixed $u$ and $v$. We thus have shown that if
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ are such that $u\sim
v$, then $v\sim u$. In other words, the relation $\sim$ is symmetric. Qed.}
and transitive\footnote{\textit{Proof.} Let $u\in\mathfrak{A}^{\ast}$,
$v\in\mathfrak{A}^{\ast}$ and $w\in\mathfrak{A}^{\ast}$ be such that $u\sim v$
and $v\sim w$. Let $k=\ell\left(  u\right)  $.
\par
Recall that $u\sim v$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $v=\left(  u_{\sigma\left(  1\right)
},u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $
(by the definition of the relation $\sim$, since $k=\ell\left(  u\right)  $).
Thus, there exists a permutation $\sigma\in\Symm_k$ satisfying
$v=\left(  u_{\sigma\left(  1\right)  },u_{\sigma\left(  2\right)  }%
,\ldots,u_{\sigma\left(  k\right)  }\right)  $ (since $u\sim v$ holds). Let
$\tau$ be this permutation $\sigma$. Thus, $\tau$ is a permutation $\sigma
\in\Symm_k$ satisfying $v=\left(  u_{\sigma\left(  1\right)
},u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $.
In other words, $\tau$ is a permutation in $\Symm_k$ and satisfies
$v=\left(  u_{\tau\left(  1\right)  },u_{\tau\left(  2\right)  }%
,\ldots,u_{\tau\left(  k\right)  }\right)  $. From $v=\left(  u_{\tau\left(
1\right)  },u_{\tau\left(  2\right)  },\ldots,u_{\tau\left(  k\right)
}\right)  $, we obtain $\ell\left(  v\right)  =k$ and thus $v=\left(
v_{1},v_{2},\ldots,v_{k}\right)  $. Hence, $\left(  v_{1},v_{2},\ldots
,v_{k}\right)  =v=\left(  u_{\tau\left(  1\right)  },u_{\tau\left(  2\right)
},\ldots,u_{\tau\left(  k\right)  }\right)  $. Hence,
\begin{equation}
v_{i}=u_{\tau\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  1,2,\ldots,k\right\}  . \label{sol.freelie.1.rel.tran.pf.1}%
\end{equation}
\par
Recall that $v\sim w$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $w=\left(  v_{\sigma\left(  1\right)
},v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  k\right)  }\right)  $
(by the definition of the relation $\sim$, since $k=\ell\left(  v\right)  $).
Thus, there exists a permutation $\sigma\in\Symm_k$ satisfying
$w=\left(  v_{\sigma\left(  1\right)  },v_{\sigma\left(  2\right)  }%
,\ldots,v_{\sigma\left(  k\right)  }\right)  $ (since $v\sim w$ holds). Let
$\rho$ be this permutation $\sigma$. Thus, $\rho$ is a permutation $\sigma
\in\Symm_k$ satisfying $w=\left(  v_{\sigma\left(  1\right)
},v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  k\right)  }\right)  $.
In other words, $\rho$ is a permutation in $\Symm_k$ and satisfies
$w=\left(  v_{\rho\left(  1\right)  },v_{\rho\left(  2\right)  }%
,\ldots,v_{\rho\left(  k\right)  }\right)  $. From $w=\left(  v_{\rho\left(
1\right)  },v_{\rho\left(  2\right)  },\ldots,v_{\rho\left(  k\right)
}\right)  $, we obtain $\ell\left(  w\right)  =k$ and thus $w=\left(
w_{1},w_{2},\ldots,w_{k}\right)  $. Hence, $\left(  w_{1},w_{2},\ldots
,w_{k}\right)  =w=\left(  v_{\rho\left(  1\right)  },v_{\rho\left(  2\right)
},\ldots,v_{\rho\left(  k\right)  }\right)  $. Hence,
\begin{equation}
w_{i}=v_{\rho\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  1,2,\ldots,k\right\}  . \label{sol.freelie.1.rel.tran.pf.2}%
\end{equation}
\par
Now, for every $j\in\left\{  1,2,\ldots,k\right\}  $, we have%
\begin{align*}
w_{j}  &  =v_{\rho\left(  j\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.1.rel.tran.pf.2}), applied to }i=j\right) \\
&  =u_{\tau\left(  \rho\left(  j\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{sol.freelie.1.rel.tran.pf.1}), applied to }i=\rho\left(
j\right)  \right) \\
&  =u_{\left(  \tau\circ\rho\right)  \left(  j\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau\left(  \rho\left(  j\right)
\right)  =\left(  \tau\circ\rho\right)  \left(  j\right)  \right)  .
\end{align*}
In other words, $\left(  w_{1},w_{2},\ldots,w_{k}\right)  =\left(  u_{\left(
\tau\circ\rho\right)  \left(  1\right)  },u_{\left(  \tau\circ\rho\right)
\left(  2\right)  },\ldots,u_{\left(  \tau\circ\rho\right)  \left(  k\right)
}\right)  $. Now, from $\ell\left(  w\right)  =k$, we obtain $w=\left(
w_{1},w_{2},\ldots,w_{k}\right)  =\left(  u_{\left(  \tau\circ\rho\right)
\left(  1\right)  },u_{\left(  \tau\circ\rho\right)  \left(  2\right)
},\ldots,u_{\left(  \tau\circ\rho\right)  \left(  k\right)  }\right)  $. In
other words, there exists a permutation $\sigma\in\Symm_k$ satisfying
$w=\left(  u_{\sigma\left(  1\right)  },u_{\sigma\left(  2\right)  }%
,\ldots,u_{\sigma\left(  k\right)  }\right)  $ (namely, $\sigma=\tau\circ\rho
$).
\par
But recall that $u\sim w$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $w=\left(  u_{\sigma\left(  1\right)
},u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $
(by the definition of the relation $\sim$, since $k=\ell\left(  u\right)  $).
Thus, $u\sim w$ holds (since there exists a permutation $\sigma\in
\Symm_k$ satisfying $w=\left(  u_{\sigma\left(  1\right)  }%
,u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $).
\par
Let us now forget that we fixed $u$, $v$ and $w$. We thus have shown that if
$u\in\mathfrak{A}^{\ast}$, $v\in\mathfrak{A}^{\ast}$ and $w\in\mathfrak{A}%
^{\ast}$ are such that $u\sim v$ and $v\sim w$, then $u\sim w$. In other
words, the relation $\sim$ is transitive. Qed.}. In other words, the relation
$\sim$ is an equivalence relation.

\item If $w$ and $w^{\prime}$ are two words satisfying $w\sim w^{\prime}$,
then%
\begin{equation}
\ell\left(  w\right)  =\ell\left(  w^{\prime}\right)
\label{sol.freelie.1.rel.len}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.len}):} Let $w$ and
$w^{\prime}$ be two words satisfying $w\sim w^{\prime}$.
\par
Let $k=\ell\left(  w\right)  $.
\par
Recall that $w\sim w^{\prime}$ if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $w^{\prime}=\left(  w_{\sigma\left(
1\right)  },w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  k\right)
}\right)  $ (by the definition of the relation $\sim$, since $k=\ell\left(
w\right)  $). Thus, there exists a permutation $\sigma\in\Symm_k$
satisfying $w^{\prime}=\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(
2\right)  },\ldots,w_{\sigma\left(  k\right)  }\right)  $ (since we have
$w\sim w^{\prime}$). Consider this $\sigma$. We have $w^{\prime}=\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  k\right)  }\right)  $ and thus $\ell\left(  w^{\prime
}\right)  =k=\ell\left(  w\right)  $. This proves (\ref{sol.freelie.1.rel.len}%
).}.

\item If $w$ and $w^{\prime}$ are two words satisfying $w\sim w^{\prime}$,
then%
\begin{equation}
\operatorname*{Alph}w=\operatorname*{Alph}\left(  w^{\prime}\right)
\label{sol.freelie.1.rel.Alph}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.Alph}):} Let $w$ and
$w^{\prime}$ be two words satisfying $w\sim w^{\prime}$.
\par
Let $k=\ell\left(  w\right)  $.
\par
Recall that $w\sim w^{\prime}$ if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $w^{\prime}=\left(  w_{\sigma\left(
1\right)  },w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  k\right)
}\right)  $ (by the definition of the relation $\sim$, since $k=\ell\left(
w\right)  $). Thus, there exists a permutation $\sigma\in\Symm_k$
satisfying $w^{\prime}=\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(
2\right)  },\ldots,w_{\sigma\left(  k\right)  }\right)  $ (since we have
$w\sim w^{\prime}$). Consider this $\sigma$. The map $\sigma$ is a permutation
in $\Symm_k$, thus a bijection $\left\{  1,2,\ldots,k\right\}
\rightarrow\left\{  1,2,\ldots,k\right\}  $. Now, we have $w^{\prime}=\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  k\right)  }\right)  $ and thus $\ell\left(  w^{\prime
}\right)  =k$. Hence, $w^{\prime}=\left(  \left(  w^{\prime}\right)
_{1},\left(  w^{\prime}\right)  _{2},\ldots,\left(  w^{\prime}\right)
_{k}\right)  $. Thus, $\left(  \left(  w^{\prime}\right)  _{1},\left(
w^{\prime}\right)  _{2},\ldots,\left(  w^{\prime}\right)  _{k}\right)
=w^{\prime}=\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)
},\ldots,w_{\sigma\left(  k\right)  }\right)  $. In other words, we have%
\begin{equation}
\left(  w^{\prime}\right)  _{i}=w_{\sigma\left(  i\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k\right\}  .
\label{sol.freelie.1.rel.Alph.pf.1}%
\end{equation}
\par
Now, the definition of $\operatorname*{Alph}\left(  w^{\prime}\right)  $ shows%
\begin{align*}
\operatorname*{Alph}\left(  w^{\prime}\right)   &  =\left\{  \left(
w^{\prime}\right)  _{1},\left(  w^{\prime}\right)  _{2},\ldots,\left(
w^{\prime}\right)  _{k}\right\}  =\left\{  \underbrace{\left(  w^{\prime
}\right)  _{i}}_{\substack{=w_{\sigma\left(  i\right)  }\\\text{(by
(\ref{sol.freelie.1.rel.Alph.pf.1}))}}}\mid i\in\left\{  1,2,\ldots,k\right\}
\right\} \\
&  =\left\{  w_{\sigma\left(  i\right)  }\mid i\in\left\{  1,2,\ldots
,k\right\}  \right\}  =\left\{  w_{i}\mid i\in\left\{  1,2,\ldots,k\right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }i\text{ for }\sigma\left(  i\right)  \text{
in the set, since}\\
\text{the map }\sigma:\left\{  1,2,\ldots,k\right\}  \rightarrow\left\{
1,2,\ldots,k\right\}  \text{ is a bijection}%
\end{array}
\right) \\
&  =\left\{  w_{1},w_{2},\ldots,w_{k}\right\}  
=\left\{  w_{1},w_{2},\ldots,w_{\ell\left(w\right)}\right\}
\ \ \ \ \ \ \ \ \ \ \left(\text{since } k = \ell\left(w\right)\right) \\
&=\operatorname*{Alph}%
w\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.1.Alph.def})}\right)
.
\end{align*}
This proves (\ref{sol.freelie.1.rel.Alph}).}.

\item If $w$ and $w^{\prime}$ are two words satisfying $w\sim w^{\prime}$,
then%
\begin{equation}
\operatorname*{Und}w=\operatorname*{Und}\left(  w^{\prime}\right)
\label{sol.freelie.1.rel.Und}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.Und}):} Let $w$ and
$w^{\prime}$ be two words satisfying $w\sim w^{\prime}$. From
(\ref{sol.freelie.1.rel.len}), we obtain $\ell\left(  w\right)  =\ell\left(
w^{\prime}\right)  $. From (\ref{sol.freelie.1.rel.Alph}), we obtain
$\operatorname*{Alph}w=\operatorname*{Alph}\left(  w^{\prime}\right)  $. Now,
the definition of $\operatorname*{Und}w$ yields
\[
\operatorname*{Und}w=\bigcup\limits_{j=0}^{\ell\left(  w\right)  }\left(
\operatorname*{Alph}w\right)  ^{j}=\bigcup\limits_{j=0}^{\ell\left(
w^{\prime}\right)  }\left(  \operatorname*{Alph}\left(  w^{\prime}\right)
\right)  ^{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  w\right)
=\ell\left(  w^{\prime}\right)  \text{ and }\operatorname*{Alph}%
w=\operatorname*{Alph}\left(  w^{\prime}\right)  \right)  .
\]
Compared with%
\[
\operatorname*{Und}\left(  w^{\prime}\right)  =\bigcup\limits_{j=0}%
^{\ell\left(  w^{\prime}\right)  }\left(  \operatorname*{Alph}\left(
w^{\prime}\right)  \right)  ^{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\operatorname*{Und}\left(  w^{\prime}\right)  \right)  ,
\]
this yields $\operatorname*{Und}w=\operatorname*{Und}\left(  w^{\prime
}\right)  $. This proves (\ref{sol.freelie.1.rel.Und}).}.

\item If $u$, $v$, $u^{\prime}$ and $v^{\prime}$ are four words satisfying
$u\sim u^{\prime}$ and $v\sim v^{\prime}$, then%
\begin{equation}
uv\sim u^{\prime}v^{\prime} \label{sol.freelie.1.rel.uv}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.uv}):} Let $u$, $v$,
$u^{\prime}$ and $v^{\prime}$ be four words satisfying $u\sim u^{\prime}$ and
$v\sim v^{\prime}$. We must prove (\ref{sol.freelie.1.rel.uv}).
\par
Let $k=\ell\left(  u\right)  $ and $\ell=\ell\left(  v\right)  $.
\par
Recall that $u\sim u^{\prime}$ holds if and only if there exists a permutation
$\sigma\in\Symm_k$ satisfying $u^{\prime}=\left(  u_{\sigma\left(
1\right)  },u_{\sigma\left(  2\right)  },\ldots,u_{\sigma\left(  k\right)
}\right)  $ (by the definition of the relation $\sim$, since $k=\ell\left(
u\right)  $). Thus, there exists a permutation $\sigma\in\Symm_k$
satisfying $u^{\prime}=\left(  u_{\sigma\left(  1\right)  },u_{\sigma\left(
2\right)  },\ldots,u_{\sigma\left(  k\right)  }\right)  $ (since $u\sim
u^{\prime}$ holds). Let $\tau$ be this permutation $\sigma$. Thus, $\tau$ is a
permutation $\sigma\in\Symm_k$ satisfying $u^{\prime}=\left(
u_{\sigma\left(  1\right)  },u_{\sigma\left(  2\right)  },\ldots
,u_{\sigma\left(  k\right)  }\right)  $. In other words, $\tau$ is a
permutation in $\Symm_k$ and satisfies $u^{\prime}=\left(
u_{\tau\left(  1\right)  },u_{\tau\left(  2\right)  },\ldots,u_{\tau\left(
k\right)  }\right)  $. From $u^{\prime}=\left(  u_{\tau\left(  1\right)
},u_{\tau\left(  2\right)  },\ldots,u_{\tau\left(  k\right)  }\right)  $, we
obtain $\ell\left(  u^{\prime}\right)  =k$ and thus $u^{\prime}=\left(
\left(  u^{\prime}\right)  _{1},\left(  u^{\prime}\right)  _{2},\ldots,\left(
u^{\prime}\right)  _{k}\right)  $. Hence, $\left(  \left(  u^{\prime}\right)
_{1},\left(  u^{\prime}\right)  _{2},\ldots,\left(  u^{\prime}\right)
_{k}\right)  =u^{\prime}=\left(  u_{\tau\left(  1\right)  },u_{\tau\left(
2\right)  },\ldots,u_{\tau\left(  k\right)  }\right)  $. Hence,
\begin{equation}
\left(  u^{\prime}\right)  _{i}=u_{\tau\left(  i\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k\right\}  .
\label{sol.freelie.1.rel.uv.pf.1}%
\end{equation}
\par
Recall that $v\sim v^{\prime}$ holds if and only if there exists a permutation
$\sigma\in\Symm_{\ell}$ satisfying $v^{\prime}=\left(  v_{\sigma\left(
1\right)  },v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  \ell\right)
}\right)  $ (by the definition of the relation $\sim$, since $\ell=\ell\left(
v\right)  $). Thus, there exists a permutation $\sigma\in\Symm_{\ell}$
satisfying $v^{\prime}=\left(  v_{\sigma\left(  1\right)  },v_{\sigma\left(
2\right)  },\ldots,v_{\sigma\left(  \ell\right)  }\right)  $ (since $v\sim
v^{\prime}$ holds). Let $\rho$ be this permutation $\sigma$. Thus, $\rho$ is a
permutation $\sigma\in\Symm_{\ell}$ satisfying $v^{\prime}=\left(
v_{\sigma\left(  1\right)  },v_{\sigma\left(  2\right)  },\ldots
,v_{\sigma\left(  \ell\right)  }\right)  $. In other words, $\rho$ is a
permutation in $\Symm_{\ell}$ and satisfies $v^{\prime}=\left(
v_{\rho\left(  1\right)  },v_{\rho\left(  2\right)  },\ldots,v_{\rho\left(
\ell\right)  }\right)  $. From $v^{\prime}=\left(  v_{\rho\left(  1\right)
},v_{\rho\left(  2\right)  },\ldots,v_{\rho\left(  \ell\right)  }\right)  $,
we obtain $\ell\left(  v^{\prime}\right)  =\ell$ and thus $v^{\prime}=\left(
\left(  v^{\prime}\right)  _{1},\left(  v^{\prime}\right)  _{2},\ldots,\left(
v^{\prime}\right)  _{\ell}\right)  $. Hence, $\left(  \left(  v^{\prime
}\right)  _{1},\left(  v^{\prime}\right)  _{2},\ldots,\left(  v^{\prime
}\right)  _{\ell}\right)  =v^{\prime}=\left(  v_{\rho\left(  1\right)
},v_{\rho\left(  2\right)  },\ldots,v_{\rho\left(  \ell\right)  }\right)  $.
Hence,
\begin{equation}
\left(  v^{\prime}\right)  _{i}=v_{\rho\left(  i\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,\ell\right\}  .
\label{sol.freelie.1.rel.uv.pf.2}%
\end{equation}
\par
Now, Proposition \ref{prop.sol.freelie.1.permutation-conc}(a) shows that there
exists a permutation $\eta\in\Symm_{k+\ell}$ such that%
\begin{equation}
\left(  \eta\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)  . \label{sol.freelie.1.rel.uv.pf.3}%
\end{equation}
Consider this $\eta$.
\par
We have $\ell\left(  u^{\prime}v^{\prime}\right)  =\underbrace{\ell\left(
u^{\prime}\right)  }_{=k}+\underbrace{\ell\left(  v^{\prime}\right)  }_{=\ell
}=k+\ell$ and $\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)
}_{=k}+\underbrace{\ell\left(  v\right)  }_{=\ell}=k+\ell$.
\par
Now,
\[
\underbrace{u}_{\substack{=\left(  u_{1},u_{2},\ldots,u_{k}\right)
\\\text{(since }\ell\left(  u\right)  =k\text{)}}}\underbrace{v}%
_{\substack{=\left(  v_{1},v_{2},\ldots,v_{\ell}\right)  \\\text{(since }%
\ell\left(  v\right)  =\ell\text{)}}}=\left(  u_{1},u_{2},\ldots,u_{k}\right)
\left(  v_{1},v_{2},\ldots,v_{\ell}\right)  =\left(  u_{1},u_{2},\ldots
,u_{k},v_{1},v_{2},\ldots,v_{\ell}\right)  .
\]
Therefore, we have $\ell\left(  uv\right)  =k+\ell$, and every $i\in\left\{
1,2,\ldots,k+\ell\right\}  $ satisfies%
\begin{equation}
\left(  uv\right)  _{i}=%
\begin{cases}
u_{i}, & \text{if }i\leq k;\\
v_{i-k}, & \text{if }i>k
\end{cases} \quad
. \label{sol.freelie.1.rel.uv.pf.4}
\end{equation}
\par
Let $i\in\left\{  1,2,\ldots,k\right\}  $. Thus, $i\leq k$ and $i\in\left\{
1,2,\ldots,k\right\}  \subset\left\{  1,2,\ldots,k+\ell\right\}  $. But
(\ref{sol.freelie.1.rel.uv.pf.3}) shows that $\eta\left(  i\right)  =%
\begin{cases}
\tau\left(  i\right)  , & \text{if }i\leq k;\\
\rho\left(  i-k\right)  +k, & \text{if }i>k
\end{cases}  \quad
=\tau\left(  i\right)  $ (since $i\leq k$). Thus, $\eta\left(  i\right)
=\tau\left(  i\right)  \in\left\{  1,2,\ldots,k\right\}  $ (since $\tau
\in\Symm_k$), so that $\eta\left(  i\right)  \leq k$. Clearly,
$\eta\left(  i\right)  \in\left\{  1,2,\ldots,k\right\}  \subset\left\{
1,2,\ldots,k+\ell\right\}  $. Applying (\ref{sol.freelie.1.rel.uv.pf.4}) to
$\eta\left(  i\right)  $ instead of $i$, we obtain $\left(  uv\right)
_{\eta\left(  i\right)  }=%
\begin{cases}
u_{\eta\left(  i\right)  }, & \text{if }\eta\left(  i\right)  \leq k;\\
v_{\eta\left(  i\right)  -k}, & \text{if }\eta\left(  i\right)  >k
\end{cases}  \quad
=u_{\eta\left(  i\right)  }$ (since $\eta\left(  i\right)  \leq k$), so that
$\left(  uv\right)  _{\eta\left(  i\right)  }=u_{\eta\left(  i\right)
}=u_{\tau\left(  i\right)  }$ (since $\eta\left(  i\right)  =\tau\left(
i\right)  $). Thus, $\left(  uv\right)  _{\eta\left(  i\right)  }%
=u_{\tau\left(  i\right)  }=\left(  u^{\prime}\right)  _{i}$ (by
(\ref{sol.freelie.1.rel.uv.pf.1})).
\par
Now, let us forget that we fixed $i$. We thus have shown that $\left(
uv\right)  _{\eta\left(  i\right)  }=\left(  u^{\prime}\right)  _{i}$ for
every $i\in\left\{  1,2,\ldots,k\right\}  $. In other words,%
\begin{equation}
\left(  \left(  uv\right)  _{\eta\left(  1\right)  },\left(  uv\right)
_{\eta\left(  2\right)  },\ldots,\left(  uv\right)  _{\eta\left(  k\right)
}\right)  =\left(  \left(  u^{\prime}\right)  _{1},\left(  u^{\prime}\right)
_{2},\ldots,\left(  u^{\prime}\right)  _{k}\right)  =u^{\prime}.
\label{sol.freelie.1.rel.uv.pf.6}%
\end{equation}
\par
On the other hand, let $j\in\left\{  1,2,\ldots,\ell\right\}  $. Thus,
$j+k\in\left\{  k+1,k+2,\ldots,k+\ell\right\}  \subset\left\{
1,2,\ldots,k+\ell\right\}  $. Also, $j+k>k$ (since $j+k\in\left\{
k+1,k+2,\ldots,k+\ell\right\}  $). But (\ref{sol.freelie.1.rel.uv.pf.3})
(applied to $i=j+k$) shows that
\begin{align*}
\eta\left(  j+k\right)   &  =%
\begin{cases}
\tau\left(  j+k\right)  , & \text{if }j+k\leq k;\\
\rho\left(  \left(  j+k\right)  -k\right)  +k, & \text{if }j+k>k
\end{cases}  \quad
=\rho\left(  \underbrace{\left(  j+k\right)  -k}_{=j}\right)
+k\ \ \ \ \ \ \ \ \ \ \left(  \text{since }j+k>k\right) \\
&  =\rho\left(  j\right)  +k.
\end{align*}
Hence, $\eta\left(  j+k\right)  -k=\rho\left(  j\right)  $. But $\rho\left(
j\right)  \in\left\{  1,2,\ldots,\ell\right\}  $ (since $\rho\in
\Symm_{\ell}$), so that $\rho\left(  j\right)  +k\in\left\{
k+1,k+2,\ldots,k+\ell\right\}  $. Thus, $\eta\left(  j+k\right)  =\rho\left(
j\right)  +k\in\left\{  k+1,k+2,\ldots,k+\ell\right\}  \subset\left\{
1,2,\ldots,k+\ell\right\}  $. Thus, $\eta\left(  j+k\right)  >k$ (since
$\eta\left(  j+k\right)  \in\left\{  k+1,k+2,\ldots,k+\ell\right\}  $).
Applying (\ref{sol.freelie.1.rel.uv.pf.4}) to $\eta\left(  j+k\right)  $
instead of $i$, we obtain $\left(  uv\right)  _{\eta\left(  j+k\right)  }=%
\begin{cases}
u_{\eta\left(  j+k\right)  }, & \text{if }\eta\left(  j+k\right)  \leq k;\\
v_{\eta\left(  j+k\right)  -k}, & \text{if }\eta\left(  j+k\right)  >k
\end{cases}  \quad
=v_{\eta\left(  j+k\right)  -k}$ (since $\eta\left(  j+k\right)  >k$), so that
$\left(  uv\right)  _{\eta\left(  j+k\right)  }=v_{\eta\left(  j+k\right)
-k}=v_{\rho\left(  j\right)  }$ (since $\eta\left(  j+k\right)  -k=\rho\left(
j\right)  $). Thus, $\left(  uv\right)  _{\eta\left(  j+k\right)  }%
=v_{\rho\left(  j\right)  }=\left(  v^{\prime}\right)  _{j}$ (since $\left(
v^{\prime}\right)  _{j}=v_{\rho\left(  j\right)  }$ (by
(\ref{sol.freelie.1.rel.uv.pf.2}), applied to $j=i$)). Thus, $\left(
uv\right)  _{\eta\left(  k+j\right)  }=\left(  uv\right)  _{\eta\left(
j+k\right)  }=\left(  v^{\prime}\right)  _{j}$.
\par
Now, let us forget that we fixed $j$. We thus have shown that $\left(
uv\right)  _{\eta\left(  k+j\right)  }=\left(  v^{\prime}\right)  _{j}$ for
every $j\in\left\{  1,2,\ldots,\ell\right\}  $. In other words,%
\begin{equation}
\left(  \left(  uv\right)  _{\eta\left(  k+1\right)  },\left(  uv\right)
_{\eta\left(  k+2\right)  },\ldots,\left(  uv\right)  _{\eta\left(
k+\ell\right)  }\right)  =\left(  \left(  v^{\prime}\right)  _{1},\left(
v^{\prime}\right)  _{2},\ldots,\left(  v^{\prime}\right)  _{\ell}\right)
=v^{\prime}. \label{sol.freelie.1.rel.uv.pf.7}%
\end{equation}
\par
Now,
\begin{align*}
&  \underbrace{u^{\prime}}_{\substack{=\left(  \left(  uv\right)
_{\eta\left(  1\right)  },\left(  uv\right)  _{\eta\left(  2\right)  }%
,\ldots,\left(  uv\right)  _{\eta\left(  k\right)  }\right)  \\\text{(by
(\ref{sol.freelie.1.rel.uv.pf.6}))}}}\underbrace{v^{\prime}}%
_{\substack{=\left(  \left(  uv\right)  _{\eta\left(  k+1\right)  },\left(
uv\right)  _{\eta\left(  k+2\right)  },\ldots,\left(  uv\right)  _{\eta\left(
k+\ell\right)  }\right)  \\\text{(by (\ref{sol.freelie.1.rel.uv.pf.7}))}}}\\
&  =\left(  \left(  uv\right)  _{\eta\left(  1\right)  },\left(  uv\right)
_{\eta\left(  2\right)  },\ldots,\left(  uv\right)  _{\eta\left(  k\right)
}\right)  \left(  \left(  uv\right)  _{\eta\left(  k+1\right)  },\left(
uv\right)  _{\eta\left(  k+2\right)  },\ldots,\left(  uv\right)  _{\eta\left(
k+\ell\right)  }\right) \\
&  =\left(  \left(  uv\right)  _{\eta\left(  1\right)  },\left(  uv\right)
_{\eta\left(  2\right)  },\ldots,\left(  uv\right)  _{\eta\left(  k\right)
},\left(  uv\right)  _{\eta\left(  k+1\right)  },\left(  uv\right)
_{\eta\left(  k+2\right)  },\ldots,\left(  uv\right)  _{\eta\left(
k+\ell\right)  }\right) \\
&  =\left(  \left(  uv\right)  _{\eta\left(  1\right)  },\left(  uv\right)
_{\eta\left(  2\right)  },\ldots,\left(  uv\right)  _{\eta\left(
k+\ell\right)  }\right)  .
\end{align*}
Thus, there exists a permutation $\sigma\in\Symm_{k+\ell}$ satisfying
$u^{\prime}v^{\prime}=\left(  \left(  uv\right)  _{\sigma\left(  1\right)
},\left(  uv\right)  _{\sigma\left(  2\right)  },\ldots,\left(  uv\right)
_{\sigma\left(  k+\ell\right)  }\right)  $ (namely, $\sigma=\eta$).
\par
Recall that $uv\sim u^{\prime}v^{\prime}$ holds if and only if there exists a
permutation $\sigma\in\Symm_{k+\ell}$ satisfying $u^{\prime}v^{\prime
}=\left(  \left(  uv\right)  _{\sigma\left(  1\right)  },\left(  uv\right)
_{\sigma\left(  2\right)  },\ldots,\left(  uv\right)  _{\sigma\left(
k+\ell\right)  }\right)  $ (by the definition of the relation $\sim$, since
$k+\ell=\ell\left(  uv\right)  $). Thus, $uv\sim u^{\prime}v^{\prime}$ holds
(since there exists a permutation $\sigma\in\Symm_{k+\ell}$ satisfying
$u^{\prime}v^{\prime}=\left(  \left(  uv\right)  _{\sigma\left(  1\right)
},\left(  uv\right)  _{\sigma\left(  2\right)  },\ldots,\left(  uv\right)
_{\sigma\left(  k+\ell\right)  }\right)  $). This proves
(\ref{sol.freelie.1.rel.uv}).}.

\item If $u$ and $v$ are any two words, then%
\begin{equation}
uv\sim vu \label{sol.freelie.1.rel.vu}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.vu}):} Let $u$ and $v$ be
two words. We must prove that $uv\sim vu$.
\par
Let $k=\ell\left(  u\right)  $ and $\ell=\ell\left(  v\right)  $.
\par
Proposition \ref{prop.sol.freelie.1.permutation-conc}(b) shows that there
exists a permutation $\zeta\in\Symm_{k+\ell}$ such that
\begin{equation}
\left(  \zeta\left(  i\right)  =%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,k+\ell\right\}
\right)  \label{sol.freelie.1.rel.vu.pf.zeta}%
\end{equation}
Consider this $\zeta$.
\par
We have $\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)  }%
_{=k}+\underbrace{\ell\left(  v\right)  }_{=\ell}=k+\ell$.
\par
Now,
\[
\underbrace{u}_{\substack{=\left(  u_{1},u_{2},\ldots,u_{k}\right)
\\\text{(since }\ell\left(  u\right)  =k\text{)}}}\underbrace{v}%
_{\substack{=\left(  v_{1},v_{2},\ldots,v_{\ell}\right)  \\\text{(since }%
\ell\left(  v\right)  =\ell\text{)}}}=\left(  u_{1},u_{2},\ldots,u_{k}\right)
\left(  v_{1},v_{2},\ldots,v_{\ell}\right)  =\left(  u_{1},u_{2},\ldots
,u_{k},v_{1},v_{2},\ldots,v_{\ell}\right)  .
\]
Therefore, we have $\ell\left(  uv\right)  =k+\ell$, and every $i\in\left\{
1,2,\ldots,k+\ell\right\}  $ satisfies%
\begin{equation}
\left(  uv\right)  _{i}=%
\begin{cases}
u_{i}, & \text{if }i\leq k;\\
v_{i-k}, & \text{if }i>k
\end{cases} \quad
. \label{sol.freelie.1.rel.vu.pf.uvi}
\end{equation}
\par
On the other hand,
\[
\underbrace{v}_{\substack{=\left(  v_{1},v_{2},\ldots,v_{\ell}\right)
\\\text{(since }\ell\left(  v\right)  =\ell\text{)}}}\underbrace{u}%
_{\substack{=\left(  u_{1},u_{2},\ldots,u_{k}\right)  \\\text{(since }%
\ell\left(  u\right)  =k\text{)}}}=\left(  v_{1},v_{2},\ldots,v_{\ell}\right)
\left(  u_{1},u_{2},\ldots,u_{k}\right)  =\left(  v_{1},v_{2},\ldots,v_{\ell
},u_{1},u_{2},\ldots,u_{k}\right)  .
\]
Therefore, we have $\ell\left(  vu\right)  =\ell+k$, and every $i\in\left\{
1,2,\ldots,\ell+k\right\}  $ satisfies%
\begin{equation}
\left(  vu\right)  _{i}=%
\begin{cases}
v_{i}, & \text{if }i\leq\ell;\\
u_{i-\ell}, & \text{if }i>\ell
\end{cases} \quad
. \label{sol.freelie.1.rel.vu.pf.vui}
\end{equation}
\par
Now, let $i\in\left\{  1,2,\ldots,k+\ell\right\}  $. We shall show that
$\left(  vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(  i\right)  }$.
\par
First of all, from $i\in\left\{  1,2,\ldots,k+\ell\right\}  $, we obtain
$1\leq i\leq k+\ell$. From $1\leq i$, we obtain $i\geq1>0$.
\par
But $i\in\left\{  1,2,\ldots,k+\ell\right\}  =\left\{  1,2,\ldots
,\ell+k\right\}  $. Hence, $\left(  vu\right)  _{i}$ is well-defined (since
$\ell\left(  vu\right)  =\ell+k$). Moreover,
(\ref{sol.freelie.1.rel.vu.pf.vui}) yields $\left(  vu\right)  _{i}=%
\begin{cases}
v_{i}, & \text{if }i\leq\ell;\\
u_{i-\ell}, & \text{if }i>\ell
\end{cases} \quad
$. On the other hand, $\zeta\left(  i\right)  \in\left\{  1,2,\ldots
,k+\ell\right\}  $ (since $i\in\left\{  1,2,\ldots,k+\ell\right\}  $), and
thus $\left(  uv\right)  _{\zeta\left(  i\right)  }$ is well-defined (since
$\ell\left(  uv\right)  =k+\ell$).
\par
Now, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $i\leq\ell$.
\par
\textit{Case 2:} We have $i>\ell$.
\par
Let us consider Case 1 first. In this case, we have $i\leq\ell$. Now, $\left(
vu\right)  _{i}=%
\begin{cases}
v_{i}, & \text{if }i\leq\ell;\\
u_{i-\ell}, & \text{if }i>\ell
\end{cases}  \quad
=v_{i}$ (since $i\leq\ell$). On the other hand,
(\ref{sol.freelie.1.rel.vu.pf.zeta}) yields $\zeta\left(  i\right)  =%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}  \quad
=i+k$ (since $i\leq\ell$), so that
\begin{align*}
\left(  uv\right)  _{\zeta\left(  i\right)  }  &  =\left(  uv\right)  _{i+k}=%
\begin{cases}
u_{i+k}, & \text{if }i+k\leq k;\\
v_{\left(  i+k\right)  -k}, & \text{if }i+k>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.1.rel.vu.pf.uvi}),
applied to }i+k\text{ instead of }i\right) \\
&  =v_{\left(  i+k\right)  -k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\underbrace{i}_{>0}+k>k\right) \\
&  =v_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  i+k\right)
-k=i\right)  .
\end{align*}
Comparing this with $\left(  vu\right)  _{i}=v_{i}$, we obtain $\left(
vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(  i\right)  }$. Thus, $\left(
vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(  i\right)  }$ is proven in
Case 1.
\par
Let us now consider Case 2. In this case, we have $i>\ell$. Now, $\left(
vu\right)  _{i}=%
\begin{cases}
v_{i}, & \text{if }i\leq\ell;\\
u_{i-\ell}, & \text{if }i>\ell
\end{cases}  \quad
=u_{i-\ell}$ (since $i>\ell$). On the other hand,
(\ref{sol.freelie.1.rel.vu.pf.zeta}) yields $\zeta\left(  i\right)  =%
\begin{cases}
i+k, & \text{if }i\leq\ell;\\
i-\ell, & \text{if }i>\ell
\end{cases}  \quad
=i-\ell$ (since $i>\ell$), so that
\begin{align*}
\left(  uv\right)  _{\zeta\left(  i\right)  }  &  =\left(  uv\right)
_{i-\ell}=%
\begin{cases}
u_{i-\ell}, & \text{if }i-\ell\leq k;\\
v_{\left(  i-\ell\right)  -k}, & \text{if }i-\ell>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.1.rel.vu.pf.uvi}),
applied to }i-\ell\text{ instead of }i\right) \\
&  =u_{i-\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{i}_{\leq
k+\ell}-\ell\leq\left(  k+\ell\right)  -\ell=k\right)  .
\end{align*}
Comparing this with $\left(  vu\right)  _{i}=u_{i-\ell}$, we obtain $\left(
vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(  i\right)  }$. Thus, $\left(
vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(  i\right)  }$ is proven in
Case 2.
\par
Now, we have proven $\left(  vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(
i\right)  }$ in each of the two Cases 1 and 2. Since these two Cases cover all
possibilities, this shows that $\left(  vu\right)  _{i}=\left(  uv\right)
_{\zeta\left(  i\right)  }$ always holds.
\par
Now, let us forget that we fixed $i$. We thus have shown that $\left(
vu\right)  _{i}=\left(  uv\right)  _{\zeta\left(  i\right)  }$ for every
$i\in\left\{  1,2,\ldots,k+\ell\right\}  $. In other words,%
\[
\left(  \left(  vu\right)  _{1},\left(  vu\right)  _{2},\ldots,\left(
vu\right)  _{k+\ell}\right)  =\left(  \left(  uv\right)  _{\zeta\left(
1\right)  },\left(  uv\right)  _{\zeta\left(  2\right)  },\ldots,\left(
uv\right)  _{\zeta\left(  k+\ell\right)  }\right)  .
\]
Now, $\ell\left(  vu\right)  =\ell+k=k+\ell$, so that
\[
vu=\left(  \left(  vu\right)  _{1},\left(  vu\right)  _{2},\ldots,\left(
vu\right)  _{k+\ell}\right)  =\left(  \left(  uv\right)  _{\zeta\left(
1\right)  },\left(  uv\right)  _{\zeta\left(  2\right)  },\ldots,\left(
uv\right)  _{\zeta\left(  k+\ell\right)  }\right)  .
\]
Thus, there exists a permutation $\sigma\in\Symm_{k+\ell}$ satisfying
$vu=\left(  \left(  uv\right)  _{\sigma\left(  1\right)  },\left(  uv\right)
_{\sigma\left(  2\right)  },\ldots,\left(  uv\right)  _{\sigma\left(
k+\ell\right)  }\right)  $ (namely, $\sigma=\zeta$).
\par
But recall that $uv\sim vu$ holds if and only if there exists a permutation
$\sigma\in\Symm_{k+\ell}$ satisfying $vu=\left(  \left(  uv\right)
_{\sigma\left(  1\right)  },\left(  uv\right)  _{\sigma\left(  2\right)
},\ldots,\left(  uv\right)  _{\sigma\left(  k+\ell\right)  }\right)  $ (by the
definition of the relation $\sim$, since $k+\ell=\ell\left(  uv\right)  $).
Thus, $uv\sim vu$ holds (since there exists a permutation $\sigma
\in\Symm_{k+\ell}$ satisfying $vu=\left(  \left(  uv\right)
_{\sigma\left(  1\right)  },\left(  uv\right)  _{\sigma\left(  2\right)
},\ldots,\left(  uv\right)  _{\sigma\left(  k+\ell\right)  }\right)  $). This
proves (\ref{sol.freelie.1.rel.vu}).}.

\item If $u$, $v$ and $w$ are any three words, then%
\begin{equation}
wuv\sim wvu \label{sol.freelie.1.rel.wuv}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.wuv}):} Let $u$, $v$ and
$w$ be any three words. From (\ref{sol.freelie.1.rel.vu}), we have $uv\sim
vu$. Also, the relation $\sim$ is reflexive, and thus we have $w\sim w$. Now,
(\ref{sol.freelie.1.rel.uv}) (applied to $w$, $uv$, $w$ and $vu$ instead of
$u$, $v$, $u^{\prime}$ and $v^{\prime}$) shows that $wuv\sim wvu$ (since
$w\sim w$ and $uv\sim vu$). This proves (\ref{sol.freelie.1.rel.wuv}).} and%
\begin{equation}
uvw\sim vuw \label{sol.freelie.1.rel.uvw}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.rel.uvw}):} Let $u$, $v$ and
$w$ be any three words. From (\ref{sol.freelie.1.rel.vu}), we have $uv\sim
vu$. Also, the relation $\sim$ is reflexive, and thus we have $w\sim w$. Now,
(\ref{sol.freelie.1.rel.uv}) (applied to $uv$, $w$, $vu$ and $w$ instead of
$u$, $v$, $u^{\prime}$ and $v^{\prime}$) shows that $uvw\sim vuw$ (since
$uv\sim vu$ and $w\sim w$). This proves (\ref{sol.freelie.1.rel.uvw}).}.
\end{itemize}

Finally, here is one more definition:

For every words $w$ and $q$, we define a subset $\mathfrak{Z}_{w}\left(
q\right)  $ of $\operatorname*{Und}w$ by%
\begin{equation}
\mathfrak{Z}_{w}\left(  q\right)  =\left\{  t\in\operatorname*{Und}%
w\ \mid\ t<q\right\}  . \label{sol.freelie.1.Z.def}%
\end{equation}
We notice the following facts:

\begin{itemize}
\item If $w$ and $q$ are any two words, then%
\begin{equation}
\text{the set }\mathfrak{Z}_{w}\left(  q\right)  \text{ is finite}
\label{sol.freelie.1.Z.fin}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Z.fin}):} Let $w$ and $q$ be
two words. Then, $\mathfrak{Z}_{w}\left(  q\right)  $ is a subset of
$\operatorname*{Und}w$ (by the definition of $\mathfrak{Z}_{w}\left(
q\right)  $). Therefore, the set $\mathfrak{Z}_{w}\left(  q\right)  $ is
finite (since the set $\operatorname*{Und}w$ is finite (by
\eqref{sol.freelie.1.Und.fin})). This proves \eqref{sol.freelie.1.Z.fin}.}.
Therefore, if $w$ and $q$ are any two words, then $\left\vert \mathfrak{Z}%
_{w}\left(  q\right)  \right\vert $ is a well-defined nonnegative integer. We
will use this rather heavily.

\item If $w$, $v$ and $q$ are any three words such that $v\sim w$, and if
$h\in\operatorname*{Und}w$ is such that $h<q$, then%
\begin{equation}
\left\vert \mathfrak{Z}_{v}\left(  h\right)  \right\vert <\left\vert
\mathfrak{Z}_{w}\left(  q\right)  \right\vert \label{sol.freelie.1.Z.dec}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.Z.dec}):} Let $w$, $v$ and $q$
be three words such that $v\sim w$. Let $h\in\operatorname*{Und}w$ be such
that $h<q$.
\par
Let $z\in\mathfrak{Z}_{v}\left(  h\right)  $. Thus, $z\in\mathfrak{Z}%
_{v}\left(  h\right)  =\left\{  t\in\operatorname*{Und}v\ \mid\ t<h\right\}  $
(by the definition of $\mathfrak{Z}_{v}\left(  h\right)  $). In other words,
$z$ is an element $t$ of $\operatorname*{Und}v$ satisfying $t<h$. In other
words, $z$ is an element of $\operatorname*{Und}v$ and satisfies $z<h$. We
have $v\sim w$ and thus $\operatorname*{Und}v=\operatorname*{Und}w$ (by
(\ref{sol.freelie.1.rel.Und}), applied to $v$ and $w$ instead of $w$ and
$w^{\prime}$). Hence, $z\in\operatorname*{Und}v=\operatorname*{Und}w$. Thus,
$z$ is an element of $\operatorname*{Und}w$ and satisfies $z<q$ (since
$z<q<h$). In other words, $z$ is an element $t$ of $\operatorname*{Und}w$
satisfying $t<q$. In other words, $z\in\left\{  t\in\operatorname*{Und}%
w\ \mid\ t<q\right\}  =\mathfrak{Z}_{w}\left(  q\right)  $ (by
(\ref{sol.freelie.1.Z.def})).
\par
Let us now forget that we fixed $z$. We thus have proven that $z\in
\mathfrak{Z}_{w}\left(  q\right)  $ for every $z\in\mathfrak{Z}_{v}\left(
h\right)  $. In other words, $\mathfrak{Z}_{v}\left(  h\right)  \subset
\mathfrak{Z}_{w}\left(  q\right)  $.
\par
On the other hand, $h$ is an element of $\operatorname*{Und}w$ and satisfies
$h<q$. In other words, $h$ is an element $t$ of $\operatorname*{Und}w$
satisfying $t<q$. In other words, $h\in\left\{  t\in\operatorname*{Und}%
w\ \mid\ t<q\right\}  =\mathfrak{Z}_{w}\left(  q\right)  $ (by
(\ref{sol.freelie.1.Z.def})).
\par
On the other hand, let us prove that $h\notin\mathfrak{Z}_{v}\left(  h\right)
$. Indeed, assume the contrary. Thus, $h\in\mathfrak{Z}_{v}\left(  h\right)
=\left\{  t\in\operatorname*{Und}v\ \mid\ t<h\right\}  $ (by the definition of
$\mathfrak{Z}_{v}\left(  h\right)  $). In other words, $h$ is an element $t$
of $\operatorname*{Und}v$ satisfying $t<h$. In other words, $h$ is an element
of $\operatorname*{Und}v$ and satisfies $h<h$. This is a contradiction (since
$h<h$ is clearly absurd). This contradiction proves that our assumption was
false. Thus, $h\notin\mathfrak{Z}_{v}\left(  h\right)  $ is proven. Since
$h\notin\mathfrak{Z}_{v}\left(  h\right)  $ but $h\in\mathfrak{Z}_{w}\left(
q\right)  $, we must have $\mathfrak{Z}_{v}\left(  h\right)  \neq
\mathfrak{Z}_{w}\left(  q\right)  $. Combining this with $\mathfrak{Z}%
_{v}\left(  h\right)  \subset\mathfrak{Z}_{w}\left(  q\right)  $, we see
that $\mathfrak{Z}_{v}\left(  h\right)  $ is a proper subset of $\mathfrak{Z}%
_{w}\left(  q\right)  $. Hence, $\left\vert \mathfrak{Z}_{v}\left(  h\right)
\right\vert <\left\vert \mathfrak{Z}_{w}\left(  q\right)  \right\vert $ (since
the set $\mathfrak{Z}_{w}\left(  q\right)  $ is finite). This proves
(\ref{sol.freelie.1.Z.dec}).}
\end{itemize}

We also recall a fundamental property of Lie algebras (one of the forms of the
Jacobi identity):

\begin{itemize}
\item Every three elements $x$, $y$ and $z$ of a Lie algebra $\mathfrak{k}$
satisfy%
\begin{equation}
\left[  \left[  x,y\right]  ,z\right]  =\left[  \left[  x,z\right]  ,y\right]
-\left[  \left[  y,z\right]  ,x\right]  . \label{sol.freelie.1.jacobi}%
\end{equation}

\end{itemize}

We can now finally come to the solution of Exercise \ref{exe.freelie.1}.

For every $h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}^{\ast}$, we define a
subset $\mathfrak{L}_{h,s}$ of $\mathfrak{L}$ by%
\[
\mathfrak{L}_{h,s}=\left\{  w\in\mathfrak{L}\ \mid\ w\sim h\text{ and
}w<s\right\}  .
\]
For every $h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}^{\ast}$, we define a
$\kk$-submodule $B_{h,s}$ of $B$ by%
\[
B_{h,s}=\sum_{w\in\mathfrak{L}_{h,s}}\kk b_{w}.
\]
(In other words, for every $h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}%
^{\ast}$, we define $B_{h,s}$ as the $\kk$-linear span of the elements
$b_{w}$ with $w\in\mathfrak{L}_{h,s}$.) If $h$, $s$, $g$ and $t$ are four
words satisfying $h\sim g$ and $s<t$, then%
\begin{equation}
B_{h,s}\subset B_{g,t} \label{sol.freelie.1.a.BhsBgt}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.BhsBgt}):} Let $h$, $s$, $g$
and $t$ be four words satisfying $h\sim g$ and $s<t$.
\par
Let $v\in\mathfrak{L}_{h,s}$. Thus, $v\in\mathfrak{L}_{h,s}=\left\{
w\in\mathfrak{L}\ \mid\ w\sim h\text{ and }w<s\right\}  $ (by the definition
of $\mathfrak{L}_{h,s}$). In other words, $v$ is an element $w$ of
$\mathfrak{L}$ such that $w\sim h$ and $w<s$. In other words, $v$ is an
element of $\mathfrak{L}$ and satisfies $v\sim h$ and $v<s$. From $v\sim h$
and $h\sim g$, we obtain $v\sim g$ (since the relation $\sim$ is transitive).
Also, $v<s<t$. Thus, $v$ is an element of $\mathfrak{L}$ and satisfies $v\sim
g$ and $v<t$. In other words, an element $w$ of $\mathfrak{L}$ such that
$w\sim g$ and $w<t$. In other words, $v\in\left\{  w\in\mathfrak{L}%
\ \mid\ w\sim g\text{ and }w<t\right\}  $. But the definition of
$\mathfrak{L}_{g,t}$ shows that $\mathfrak{L}_{g,t}=\left\{  w\in
\mathfrak{L}\ \mid\ w\sim g\text{ and }w<t\right\}  $, so that $\left\{
w\in\mathfrak{L}\ \mid\ w\sim g\text{ and }w<t\right\}  =\mathfrak{L}_{g,t}$.
Thus, $v\in\left\{  w\in\mathfrak{L}\ \mid\ w\sim g\text{ and }w<t\right\}
=\mathfrak{L}_{g,t}$.
\par
Now, let us forget that we fixed $v$. We thus have proven that every
$v\in\mathfrak{L}_{h,s}$ satisfies $v\in\mathfrak{L}_{g,t}$. In other words,
$\mathfrak{L}_{h,s}\subset\mathfrak{L}_{g,t}$. Thus, the sum $\sum
_{w\in\mathfrak{L}_{h,s}}\kk b_{w}$ is a subsum of the sum $\sum
_{w\in\mathfrak{L}_{g,t}}\kk b_{w}$. Thus, $\sum_{w\in\mathfrak{L}%
_{h,s}}\kk b_{w}\subset\sum_{w\in\mathfrak{L}_{g,t}}\kk b_{w}$.
But the definition of $B_{h,s}$ yields $B_{h,s}=\sum_{w\in\mathfrak{L}_{h,s}%
}\kk b_{w}\subset\sum_{w\in\mathfrak{L}_{g,t}}\kk b_{w}=B_{g,t}$
(since $B_{g,t}=\sum_{w\in\mathfrak{L}_{g,t}}\kk b_{w}$ (by the
definition of $B_{g,t}$)). This proves (\ref{sol.freelie.1.a.BhsBgt}).}.

(a) We claim that%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q.
\label{sol.freelie.1.a.mainclaim}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.1.a.mainclaim}):} We shall prove
(\ref{sol.freelie.1.a.mainclaim}) by strong induction over $\ell\left(
pq\right)  $:

\textit{Induction step:} Let $N$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.a.mainclaim}) holds in the case when $\ell\left(
pq\right)  <N$. We need to prove that (\ref{sol.freelie.1.a.mainclaim}) holds
in the case when $\ell\left(  pq\right)  =N$.

We have assumed that (\ref{sol.freelie.1.a.mainclaim}) holds in the case when
$\ell\left(  pq\right)  <N$. In other words, we have%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  <N. \label{sol.freelie.1.a.indhyp1}%
\end{equation}


We now must prove that (\ref{sol.freelie.1.a.mainclaim}) holds in the case
when $\ell\left(  pq\right)  =N$. In other words, we must prove that
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  =N. \label{sol.freelie.1.a.indgoal1}%
\end{equation}


We shall prove (\ref{sol.freelie.1.a.indgoal1}) by strong induction on
$\left\vert \mathfrak{Z}_{pq}\left(  q\right)  \right\vert $%
\ \ \ \ \footnote{This is legitimate, because $\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert $ is a well-defined nonnegative integer for
every $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (because
(\ref{sol.freelie.1.Z.fin}) (applied to $w=pq$) shows that the set
$\mathfrak{Z}_{pq}\left(  q\right)  $ is finite).
\par
Of course, this will be an induction within our current induction step, so the
reader should try not to confuse the two inductions going on.}:

\textit{Induction step:} Let $K$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.a.indgoal1}) holds in the case when $\left\vert
\mathfrak{Z}_{pq}\left(  q\right)  \right\vert <K$. We need to prove that
(\ref{sol.freelie.1.a.indgoal1}) holds in the case when $\left\vert
\mathfrak{Z}_{pq}\left(  q\right)  \right\vert =K$.

We have assumed that (\ref{sol.freelie.1.a.indgoal1}) holds in the case when
$\left\vert \mathfrak{Z}_{pq}\left(  q\right)  \right\vert <K$. In other
words, we have%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  =N\text{ and }\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert <K. \label{sol.freelie.1.a.indhyp2}%
\end{equation}


Now, let $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ be such that
$p<q$ and $\ell\left(  pq\right)  =N$ and $\left\vert \mathfrak{Z}_{pq}\left(
q\right)  \right\vert =K$. We are going to prove that $\left[  b_{p}%
,b_{q}\right]  \in B_{pq,q}$.

We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$. In other
words, $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$. In other words, the words
$p$ and $q$ are Lyndon. Proposition \ref{prop.words.lyndon.concat}(a) (applied
to $u=p$ and $v=q$) thus shows that the word $pq$ is Lyndon. In other words,
$pq\in\mathfrak{L}$.

Furthermore, Proposition \ref{prop.words.lyndon.concat}(b) (applied to $u=p$
and $v=q$) shows that $pq<q$.

The definition of $\mathfrak{L}_{pq,q}$ yields $\mathfrak{L}_{pq,q}=\left\{
w\in\mathfrak{L}\ \mid\ w\sim pq\text{ and }w<q\right\}  $. The definition of
$B_{pq,q}$ yields $B_{pq,q}=\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$.

Moreover, the word $p$ is nonempty (since it is Lyndon); thus, $\ell\left(
p\right)  \geq1$. Also, the word $q$ is nonempty (since it is Lyndon); thus,
$\ell\left(  q\right)  \geq1$. Now, $\ell\left(  pq\right)  =\underbrace{\ell
\left(  p\right)  }_{\geq1}+\underbrace{\ell\left(  q\right)  }_{\geq1}%
\geq1+1=2>1$. Therefore, $pq$ is a Lyndon word of length $>1$ (since $pq$ is
Lyndon and since $pq$ has length $\ell\left(  pq\right)  >1$). Therefore,
$\operatorname*{stf}\left(  pq\right)  $ is well-defined. If
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $, then it is easy
to see that $\left[  b_{p},b_{q}\right]  \in B_{pq,q}$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $\operatorname*{stf}\left(
pq\right)  =\left(  p,q\right)  $. Thus, $\left(  p,q\right)
=\operatorname*{stf}\left(  pq\right)  $. Hence, (\ref{eq.exe.freelie.1.bw})
(applied to $w=pq$, $u=p$ and $v=q$) shows that $b_{pq}=\left[  b_{p}%
,b_{q}\right]  $. But $pq$ is an element of $\mathfrak{L}$ and satisfies
$pq\sim pq$ (since the relation $\sim$ is reflexive) and $pq<q$. In other
words, $pq$ is an element $w$ of $\mathfrak{L}$ satisfying $w\sim pq$ and
$w<q$. In other words, $pq\in\left\{  w\in\mathfrak{L}\ \mid\ w\sim pq\text{
and }w<q\right\}  =\mathfrak{L}_{pq,q}$ (since $\mathfrak{L}_{pq,q}=\left\{
w\in\mathfrak{L}\ \mid\ w\sim pq\text{ and }w<q\right\}  $ (by the definition
of $\mathfrak{L}_{pq,q}$)). Therefore, $\kk b_{pq}$ is an addend of the
sum $\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$. Thus, $\kk%
b_{pq}\subset\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$.
\par
Now, from $b_{pq}=\left[  b_{p},b_{q}\right]  $, we obtain%
\[
\left[  b_{p},b_{q}\right]  =b_{pq}=\underbrace{1}_{\in\kk}b_{pq}%
\in\kk b_{pq}\subset\sum_{w\in\mathfrak{L}_{pq,q}}\kk%
b_{w}=B_{pq,q}%
\]
(since $B_{pq,q}=\sum_{w\in\mathfrak{L}_{pq,q}}\kk b_{w}$), qed.}.
Hence, for the rest of the proof of $\left[  b_{p},b_{q}\right]  \in B_{pq,q}%
$, we can WLOG assume that we don't have $\operatorname*{stf}\left(
pq\right)  =\left(  p,q\right)  $. Assume this. Thus, $q$ is not the longest
Lyndon proper suffix of $pq$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, $q$ is the longest Lyndon proper suffix of $pq$.
\par
Let $w=pq$. Recall that $q$ is the longest Lyndon proper suffix of $pq$. In
other words, $q$ is the longest Lyndon proper suffix of $w$ (since $w=pq$).
\par
Let $\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Then,
$\left(  g,h\right)  =\operatorname*{stf}\left(  \underbrace{pq}_{=w}\right)
=\operatorname*{stf}w$. Thus, $h$ is the longest Lyndon proper suffix of $w$
(by Exercise \ref{exe.words.lyndon.stf.basic}(a)). Comparing this with the
fact that $q$ is the longest Lyndon proper suffix of $w$, we obtain that
$h=q$.
\par
But Exercise \ref{exe.words.lyndon.stf.basic}(b) shows that we have
$w=g\underbrace{h}_{=q}=gq$. Thus, $gq=w=pq$. Cancelling $q$ from this
equality, we obtain $g=p$. Now, from $\left(  g,h\right)  =\operatorname*{stf}%
\left(  pq\right)  $, we obtain $\operatorname*{stf}\left(  pq\right)
=\left(  \underbrace{g}_{=p},\underbrace{h}_{=q}\right)  =\left(  p,q\right)
$. This contradicts the fact that we don't have $\operatorname*{stf}\left(
pq\right)  =\left(  p,q\right)  $. This contradiction proves that our
assumption was wrong, qed.}. In other words, the statement that $q$ is the
longest Lyndon proper suffix of $pq$ is \textbf{false}.

We have $\ell\left(  p\right)  >1$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, $\ell\left(  p\right)  \leq1$. Combined with $\ell\left(
p\right)  \geq1$, this shows that $\ell\left(  p\right)  =1$.
\par
Let $w=pq$. The word $q$ is Lyndon, and is a proper suffix of $w$ (since
$w=pq$, and since the word $p$ is nonempty). In other words, the word $q$ is a
Lyndon proper suffix of $w$.
\par
Let $\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Then,
$\left(  g,h\right)  =\operatorname*{stf}\left(  \underbrace{pq}_{=w}\right)
=\operatorname*{stf}w$. Thus, $h$ is the longest Lyndon proper suffix of $w$
(by Exercise \ref{exe.words.lyndon.stf.basic}(a)). Thus, every Lyndon proper
suffix of $w$ is at most as long as $h$. In other words, if $h^{\prime}$ is
any Lyndon proper suffix of $w$, then $\ell\left(  h^{\prime}\right)  \leq
\ell\left(  h\right)  $. Applying this to $h^{\prime}=q$, we obtain
$\ell\left(  q\right)  \leq\ell\left(  h\right)  $ (since $q$ is a Lyndon
proper suffix of $w$). Hence, $\ell\left(  h\right)  \geq\ell\left(  q\right)
$.
\par
From Exercise \ref{exe.words.lyndon.stf.basic}(d), we know that the word $g$
is Lyndon. Thus, $g$ is nonempty, so that $\ell\left(  g\right)  \geq1$.
\par
But Exercise \ref{exe.words.lyndon.stf.basic}(b) shows that we have $w=gh$, so
that $\ell\left(  \underbrace{w}_{=gh}\right)  =\ell\left(  gh\right)
=\ell\left(  g\right)  +\ell\left(  h\right)  $ and therefore $\ell\left(
g\right)  +\ell\left(  h\right)  =\ell\left(  \underbrace{w}_{=pq}\right)
=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=1}%
+\underbrace{\ell\left(  q\right)  }_{\leq\ell\left(  h\right)  }\leq
1+\ell\left(  h\right)  $. Subtracting $\ell\left(  h\right)  $ from both
sides of this inequality, we obtain $\ell\left(  g\right)  \leq1$. Combined
with $\ell\left(  g\right)  \geq1$, this shows that $\ell\left(  g\right)
=1$.
\par
Now, clearly, if $x$ is a word and $y$ is a prefix of $x$, then%
\begin{equation}
y=\left(  x_{1},x_{2},\ldots,x_{\ell\left(  y\right)  }\right)  .
\label{sol.freelie.1.a.prefix-lem}%
\end{equation}
\par
But $g$ is a prefix of $w$ (since $w=gh$). Thus,
(\ref{sol.freelie.1.a.prefix-lem}) (applied to $x=w$ and $y=g$) shows that%
\begin{align*}
g  &  =\left(  w_{1},w_{2},\ldots,w_{\ell\left(  g\right)  }\right)  =\left(
w_{1},w_{2},\ldots,w_{1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  g\right)  =1\right) \\
&  =\left(  w_{1}\right)  .
\end{align*}
On the other hand, $p$ is a prefix of $w$ (since $w=pq$). Thus,
(\ref{sol.freelie.1.a.prefix-lem}) (applied to $x=w$ and $y=p$) shows that%
\begin{align*}
p  &  =\left(  w_{1},w_{2},\ldots,w_{\ell\left(  p\right)  }\right)  =\left(
w_{1},w_{2},\ldots,w_{1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  p\right)  =1\right) \\
&  =\left(  w_{1}\right)  .
\end{align*}
Comparing this with $g=\left(  w_{1}\right)  $, we obtain $p=g$. Hence, $g=p$,
so that $\underbrace{g}_{=p}q=pq=w=gh$. Cancelling $g$ from this equality, we
obtain $q=h$. Now, $\left(  \underbrace{p}_{=g},\underbrace{q}_{=h}\right)
=\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Hence,
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $. This
contradicts the fact that we don't have $\operatorname*{stf}\left(  pq\right)
=\left(  p,q\right)  $. This contradiction proves that our assumption was
wrong, qed.}. Thus, $p$ is a Lyndon word of length $>1$ (since $p$ is Lyndon
and since $p$ has length $\ell\left(  p\right)  >1$). Therefore,
$\operatorname*{stf}p$ is well-defined. Set $\left(  u,v\right)
=\operatorname*{stf}p$. Then, Exercise \ref{exe.words.lyndon.stf.basic}(a)
(applied to $w=p$, $g=u$ and $h=v$) says that $v$ is the longest Lyndon proper
suffix of $p$. In particular, $v$ is a Lyndon proper suffix of $p$. In other
words, $v$ is Lyndon and is a proper suffix of $p$. Moreover, Exercise
\ref{exe.words.lyndon.stf.basic}(b) (applied to $w=p$, $g=u$ and $h=v$) says
that we have $p=uv$. Also, Exercise \ref{exe.words.lyndon.stf.basic}(c)
(applied to $w=p$, $g=u$ and $h=v$) says that we have $u<uv<v$. Finally,
Exercise \ref{exe.words.lyndon.stf.basic}(d) (applied to $w=p$, $g=u$ and
$h=v$) says that the word $u$ is Lyndon. In other words, $u\in\mathfrak{L}$.
The words $u$ and $v$ are nonempty (since they are Lyndon).

Exercise \ref{exe.words.lyndon.stf.props2} (applied to $w=p$ and $t=q$) now
shows that $q$ is the longest Lyndon proper suffix of $pq$ if and only if we
do not have $v<q$. Thus, the statement that we do not have $v<q$ is
\textbf{false} (because the statement that $q$ is the longest Lyndon proper
suffix of $pq$ is \textbf{false}). Thus, we have $v<q$. Thus, $u<v<q$.

Recall that $\left(  u,v\right)  =\operatorname*{stf}p$. Thus,
(\ref{eq.exe.freelie.1.bw}) (applied to $w=p$) shows that $b_{p}=\left[
b_{u},b_{v}\right]  $. Thus,%
\begin{equation}
\left[  \underbrace{b_{p}}_{=\left[  b_{u},b_{v}\right]  },b_{q}\right]
=\left[  \left[  b_{u},b_{v}\right]  ,b_{q}\right]  =\left[  \left[
b_{u},b_{q}\right]  ,b_{v}\right]  -\left[  \left[  b_{v},b_{q}\right]
,b_{u}\right]  \label{sol.freelie.1.a.jacobi}%
\end{equation}
(by (\ref{sol.freelie.1.jacobi}), applied to $\mathfrak{k}=\mathfrak{g}$,
$x=b_{u}$, $y=b_{v}$ and $z=b_{q}$).

Now,%
\begin{equation}
\left[  b_{u},b_{q}\right]  \in B_{uq,q} \label{sol.freelie.1.a.in1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in1}):} We have $\left(
u,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $u\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $u<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\ell\left(  u\right)  +\underbrace{\ell\left(
v\right)  }_{\substack{\geq1\\\text{(since }v\text{ is nonempty)}}}\geq
\ell\left(  u\right)  +1>\ell\left(  u\right)  $. But $\ell\left(  pq\right)
=N$, so that $N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)
}_{>\ell\left(  u\right)  }+\ell\left(  q\right)  >\ell\left(  u\right)
+\ell\left(  q\right)  =\ell\left(  uq\right)  $. Thus, $\ell\left(
uq\right)  <N$. Therefore, (\ref{sol.freelie.1.a.indhyp1}) (applied to
$\left(  u,q\right)  $ instead of $\left(  p,q\right)  $) yields $\left[
b_{u},b_{q}\right]  \in B_{uq,q}$. This proves (\ref{sol.freelie.1.a.in1}).}
and%
\begin{equation}
\left[  b_{v},b_{q}\right]  \in B_{vq,q} \label{sol.freelie.1.a.in2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in2}):} We have $\left(
v,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $v\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $v<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)  }_{\substack{\geq
1\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  v\right)  \geq
1+\ell\left(  v\right)  >\ell\left(  v\right)  $. But $\ell\left(  pq\right)
=N$, so that $N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)
}_{>\ell\left(  v\right)  }+\ell\left(  q\right)  >\ell\left(  v\right)
+\ell\left(  q\right)  =\ell\left(  vq\right)  $. Thus, $\ell\left(
vq\right)  <N$. Therefore, (\ref{sol.freelie.1.a.indhyp1}) (applied to
$\left(  v,q\right)  $ instead of $\left(  p,q\right)  $) yields $\left[
b_{v},b_{q}\right]  \in B_{vq,q}$. This proves (\ref{sol.freelie.1.a.in2}).}.
Hence, (\ref{sol.freelie.1.a.jacobi}) becomes%
\begin{equation}
\left[  b_{p},b_{q}\right]  =\left[  \underbrace{\left[  b_{u},b_{q}\right]
}_{\substack{\in B_{uq,q}\\\text{(by (\ref{sol.freelie.1.a.in1}))}}%
},b_{v}\right]  -\left[  \underbrace{\left[  b_{v},b_{q}\right]
}_{\substack{\in B_{vq,q}\\\text{(by (\ref{sol.freelie.1.a.in2}))}}%
},b_{u}\right]  \in\left[  B_{uq,q},b_{v}\right]  -\left[  B_{vq,q}%
,b_{u}\right]  . \label{sol.freelie.1.a.jacobi2}%
\end{equation}


On the other hand, if $g$ and $h$ are two Lyndon words satisfying $g<q$, $h<q$
and $gh\sim p$, then%
\begin{equation}
\left(  \text{every }r\in\mathfrak{L}_{gq,q}\text{ satisfies }\left[
b_{r},b_{h}\right]  \in B_{pq,q}\right)  \label{sol.freelie.1.a.in3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in3a}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. Let $r\in
\mathfrak{L}_{gq,q}$. We must prove that $\left[  b_{r},b_{h}\right]  \in
B_{pq,q}$.
\par
The words $g$ and $h$ are Lyndon. In other words, $g\in\mathfrak{L}$ and
$h\in\mathfrak{L}$.
\par
We have $gh\sim p$ and thus $\ell\left(  gh\right)  =\ell\left(  p\right)  $
(by (\ref{sol.freelie.1.rel.len}), applied to $w=gh$ and $w^{\prime}=p$).
Hence, $\ell\left(  p\right)  =\ell\left(  gh\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  $.
\par
We have $gh\sim p$ and $q\sim q$ (since the relation $\sim$ is reflexive).
Hence, $ghq\sim pq$ (by (\ref{sol.freelie.1.rel.uv}), applied to $gh$, $q$,
$p$ and $q$ instead of $u$, $v$, $u^{\prime}$ and $v^{\prime}$).
\par
We have $r\in\mathfrak{L}_{gq,q}=\left\{  w\in\mathfrak{L}\ \mid\ w\sim
gq\text{ and }w<q\right\}  $ (by the definition of $\mathfrak{L}_{gq,q}$). In
other words, $r$ is an element $w\in\mathfrak{L}$ satisfying $w\sim gq$ and
$w<q$. In other words, $r$ is an element of $\mathfrak{L}$ and satisfies
$r\sim gq$ and $r<q$. From $r\sim gq$, we obtain $\ell\left(  r\right)
=\ell\left(  gq\right)  $ (by (\ref{sol.freelie.1.rel.len}), applied to $w=r$
and $w^{\prime}=gq$). Thus, $\ell\left(  r\right)  =\ell\left(  gq\right)
=\ell\left(  g\right)  +\ell\left(  q\right)  $.
\par
If $r=h$, then $\left[  \underbrace{b_{r}}_{\substack{=b_{h}\\\text{(since
}r=h\text{)}}},b_{h}\right]  =\left[  b_{h},b_{h}\right]  =0\in B_{pq,q}$
(since $B_{pq,q}$ is a $\kk$-module). Thus, if $r=h$, then $\left[
b_{r},b_{h}\right]  \in B_{pq,q}$ holds. Hence, for the rest of the proof of
$\left[  b_{r},b_{h}\right]  \in B_{pq,q}$, we can WLOG assume that we don't
have $r=h$. Assume this.
\par
We have $r\neq h$ (since we don't have $r=h$). Thus, we have either $r<h$ or
$r>h$ (since the lexicographic order on $\mathfrak{A}^{\ast}$ is a total
order). In other words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $r<h$.
\par
\textit{Case 2:} We have $r>h$.
\par
Let us first consider Case 1. In this case, we have $r<h$. We have
$\ell\left(  rh\right)  =\underbrace{\ell\left(  r\right)  }_{=\ell\left(
g\right)  +\ell\left(  q\right)  }+\ell\left(  h\right)  =\ell\left(
g\right)  +\ell\left(  q\right)  +\ell\left(  h\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  +\ell\left(  q\right)  $. Comparing this with
$\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=\ell\left(
g\right)  +\ell\left(  h\right)  }+\ell\left(  q\right)  =\ell\left(
g\right)  +\ell\left(  h\right)  +\ell\left(  q\right)  $, we obtain
$\ell\left(  rh\right)  =\ell\left(  pq\right)  =N$.
\par
We have $r\sim gq$ and $h\sim h$ (since the relation $\sim$ is reflexive).
Thus, (\ref{sol.freelie.1.rel.uv}) (applied to $r$, $h$, $gq$ and $h$ instead
of $u$, $v$, $u^{\prime}$ and $v^{\prime}$) shows that $rh\sim gqh$. But
$gqh\sim ghq$ (by (\ref{sol.freelie.1.rel.wuv}), applied to $q$, $h$ and $g$
instead of $u$, $v$ and $w$). From $rh\sim gqh$ and $gqh\sim ghq$, we obtain
$rh\sim ghq$ (since the relation $\sim$ is transitive). From $rh\sim ghq$ and
$ghq\sim pq$, we obtain $rh\sim pq$ (since the relation $\sim$ is transitive).
Now, (\ref{sol.freelie.1.a.BhsBgt}) (applied to $rh$, $h$, $pq$ and $q$
instead of $h$, $s$, $g$ and $t$) yields $B_{rh,h}\subset B_{pq,q}$ (since
$h<q$).
\par
On the other hand, $rh\sim pq$ and thus $\operatorname*{Und}\left(  rh\right)
=\operatorname*{Und}\left(  pq\right)  $ (by (\ref{sol.freelie.1.rel.Und}),
applied to $w=rh$ and $w^{\prime}=pq$). But $\operatorname*{Und}%
h\subset\operatorname*{Und}\left(  rh\right)  $ (by
(\ref{sol.freelie.1.Und.uv2}), applied to $r$ and $h$ instead of $u$ and $v$).
Thus, $\operatorname*{Und}h\subset\operatorname*{Und}\left(  rh\right)
=\operatorname*{Und}\left(  pq\right)  $. Now, (\ref{sol.freelie.1.Und.w})
(applied to $w=h$) shows that $h\in\operatorname*{Und}h\subset
\operatorname*{Und}\left(  pq\right)  $.
\par
So we know that $rh\sim pq$, $h\in\operatorname*{Und}\left(  pq\right)  $ and
$h<q$. Thus, we can apply (\ref{sol.freelie.1.Z.dec}) to $pq$, $rh$, $q$ and
$h$ instead of $w$, $v$, $q$ and $h$. As a result, we obtain $\left\vert
\mathfrak{Z}_{rh}\left(  h\right)  \right\vert <\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert =K$.
\par
Now, we have $\left(  r,h\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$r\in\mathfrak{L}$ and $h\in\mathfrak{L}$) and $r<h$ and $\ell\left(
rh\right)  =N$ and $\left\vert \mathfrak{Z}_{rh}\left(  h\right)  \right\vert
<K$. Therefore, (\ref{sol.freelie.1.a.indhyp2}) (applied to $r$ and $h$
instead of $p$ and $q$) shows that $\left[  b_{r},b_{h}\right]  \in
B_{rh,h}\subset B_{pq,q}$. Thus, $\left[  b_{r},b_{h}\right]  \in B_{pq,q}$ is
proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $h<r$. We have $\ell\left(
hr\right)  =\ell\left(  h\right)  +\underbrace{\ell\left(  r\right)  }%
_{=\ell\left(  g\right)  +\ell\left(  q\right)  }=\ell\left(  h\right)
+\ell\left(  g\right)  +\ell\left(  q\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  +\ell\left(  q\right)  $. Comparing this with
$\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=\ell\left(
g\right)  +\ell\left(  h\right)  }+\ell\left(  q\right)  =\ell\left(
g\right)  +\ell\left(  h\right)  +\ell\left(  q\right)  $, we obtain
$\ell\left(  hr\right)  =\ell\left(  pq\right)  =N$.
\par
We have $h\sim h$ (since the relation $\sim$ is reflexive) and $r\sim gq$.
Thus, (\ref{sol.freelie.1.rel.uv}) (applied to $h$, $r$, $h$ and $gq$ instead
of $u$, $v$, $u^{\prime}$ and $v^{\prime}$) shows that $hr\sim hgq$. But
$hgq\sim ghq$ (by (\ref{sol.freelie.1.rel.uvw}), applied to $h$, $g$ and $q$
instead of $u$, $v$ and $w$). From $hr\sim hgq$ and $hgq\sim ghq$, we obtain
$hr\sim ghq$ (since the relation $\sim$ is transitive). From $hr\sim ghq$ and
$ghq\sim pq$, we obtain $hr\sim pq$ (since the relation $\sim$ is transitive).
Now, (\ref{sol.freelie.1.a.BhsBgt}) (applied to $hr$, $r$, $pq$ and $q$
instead of $h$, $s$, $g$ and $t$) yields $B_{hr,r}\subset B_{pq,q}$ (since
$r<q$).
\par
On the other hand, $hr\sim pq$ and thus $\operatorname*{Und}\left(  hr\right)
=\operatorname*{Und}\left(  pq\right)  $ (by (\ref{sol.freelie.1.rel.Und}),
applied to $w=hr$ and $w^{\prime}=pq$). But $\operatorname*{Und}%
r\subset\operatorname*{Und}\left(  hr\right)  $ (by
(\ref{sol.freelie.1.Und.uv2}), applied to $h$ and $r$ instead of $u$ and $v$).
Thus, $\operatorname*{Und}r\subset\operatorname*{Und}\left(  hr\right)
=\operatorname*{Und}\left(  pq\right)  $. Now, (\ref{sol.freelie.1.Und.w})
(applied to $w=r$) shows that $r\in\operatorname*{Und}r\subset
\operatorname*{Und}\left(  pq\right)  $.
\par
So we know that $hr\sim pq$, $r\in\operatorname*{Und}\left(  pq\right)  $ and
$r<q$. Thus, we can apply (\ref{sol.freelie.1.Z.dec}) to $pq$, $hr$, $q$ and
$r$ instead of $w$, $v$, $q$ and $h$. As a result, we obtain $\left\vert
\mathfrak{Z}_{hr}\left(  r\right)  \right\vert <\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert =K$.
\par
Now, we have $\left(  h,r\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$h\in\mathfrak{L}$ and $r\in\mathfrak{L}$) and $h<r$ and $\ell\left(
hr\right)  =N$ and $\left\vert \mathfrak{Z}_{hr}\left(  r\right)  \right\vert
<K$. Therefore, (\ref{sol.freelie.1.a.indhyp2}) (applied to $h$ and $r$
instead of $p$ and $q$) shows that $\left[  b_{h},b_{r}\right]  \in
B_{hr,r}\subset B_{pq,q}$. Thus, $\left[  b_{r},b_{h}\right]
=-\underbrace{\left[  b_{h},b_{r}\right]  }_{\in B_{pq,q}}\in-B_{pq,q}\subset
B_{pq,q}$ (since $B_{pq,q}$ is a $\kk$-module). Hence, $\left[
b_{r},b_{h}\right]  \in B_{pq,q}$ is proven in Case 2.
\par
Thus, $\left[  b_{r},b_{h}\right]  \in B_{pq,q}$ is proven in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, this shows that
$\left[  b_{r},b_{h}\right]  \in B_{pq,q}$ always holds. Hence,
(\ref{sol.freelie.1.a.in3a}) is proven.} and therefore%
\begin{equation}
\left[  B_{gq,q},b_{h}\right]  \subset B_{pq,q} \label{sol.freelie.1.a.in3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.in3b}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. The definition of
$B_{gq,q}$ shows that $B_{gq,q}=\sum_{w\in\mathfrak{L}_{gq,q}}\kk b_{w}%
$. Hence,%
\begin{align*}
\left[  \underbrace{B_{gq,q}}_{=\sum_{w\in\mathfrak{L}_{gq,q}}\kk b_{w}%
},b_{h}\right]   &  =\left[  \sum_{w\in\mathfrak{L}_{gq,q}}\kk%
b_{w},b_{h}\right]  =\sum_{w\in\mathfrak{L}_{gq,q}}\kk%
\underbrace{\left[  b_{w},b_{h}\right]  }_{\substack{\in B_{pq,q}\\\text{(by
(\ref{sol.freelie.1.a.in3a}), applied to }r=w\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie bracket of }\mathfrak{g}%
\text{ is }\kk\text{-bilinear}\right) \\
&  \subset\sum_{w\in\mathfrak{L}_{gq,q}}\kk B_{pq,q}\subset
B_{pq,q}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }B_{pq,q}\text{ is a
}\kk\text{-module}\right)  .
\end{align*}
This proves (\ref{sol.freelie.1.a.in3b}).}.

We have $p\sim p$ (since the relation $\sim$ is reflexive). In other words,
$uv\sim p$ (since $p=uv$). Also, (\ref{sol.freelie.1.rel.vu}) yields $uv\sim
vu$, so that $vu\sim uv$ (since the relation $\sim$ is symmetric). Since
$p=uv$, this rewrites as $vu\sim p$.

Now, the words $u$ and $v$ are Lyndon and satisfy $u<q$, $v<q$ and $uv\sim p$.
Thus, we can apply (\ref{sol.freelie.1.a.in3b}) to $g=u$ and $h=v$. As a
result, we obtain $\left[  B_{uq,q},b_{v}\right]  \subset B_{pq,q}$.

Furthermore, the words $v$ and $u$ are Lyndon and satisfy $v<q$, $u<q$ and
$vu\sim p$. Thus, we can apply (\ref{sol.freelie.1.a.in3b}) to $g=v$ and
$h=u$. As a result, we obtain $\left[  B_{vq,q},b_{u}\right]  \subset
B_{pq,q}$.

Now, (\ref{sol.freelie.1.a.jacobi2}) becomes%
\[
\left[  b_{p},b_{q}\right]  \in\underbrace{\left[  B_{uq,q},b_{v}\right]
}_{\subset B_{pq,q}}-\underbrace{\left[  B_{vq,q},b_{u}\right]  }_{\subset
B_{pq,q}}\subset B_{pq,q}-B_{pq,q}\subset B_{pq,q}%
\]
(since $B_{pq,q}$ is a $\kk$-module).

Let us now forget that we fixed $\left(  p,q\right)  $. We thus have proven
that%
\[
\left[  b_{p},b_{q}\right]  \in B_{pq,q}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying
}p<q\text{ and }\ell\left(  pq\right)  =N\text{ and }\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert =K.
\]
In other words, (\ref{sol.freelie.1.a.indgoal1}) holds in the case when
$\left\vert \mathfrak{Z}_{pq}\left(  q\right)  \right\vert =K$. This completes
the induction step (in the induction proof of (\ref{sol.freelie.1.a.indgoal1}%
)). The induction proof of (\ref{sol.freelie.1.a.indgoal1}) is thus finished.

We thus have proven (\ref{sol.freelie.1.a.indgoal1}). In other words,
(\ref{sol.freelie.1.a.mainclaim}) holds in the case when $\ell\left(
pq\right)  =N$. This completes the induction step (in the induction proof of
(\ref{sol.freelie.1.a.mainclaim})). The induction proof of
(\ref{sol.freelie.1.a.mainclaim}) is thus complete.

Now, we recall that $B$ is the $\kk$-submodule of $\mathfrak{g}$
spanned by the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$. In other
words,%
\begin{equation}
B=\left(  \text{the }\kk\text{-submodule of }\mathfrak{g}\text{ spanned
by the family }\left(  b_{w}\right)  _{w\in\mathfrak{L}}\right)  =\sum
_{w\in\mathfrak{L}}\kk b_{w}. \label{sol.freelie.1.a.endgame.B}%
\end{equation}
Thus,%
\begin{equation}
B_{h,s}\subset B\ \ \ \ \ \ \ \ \ \ \text{for every }h\in\mathfrak{A}^{\ast
}\text{ and }s\in\mathfrak{A}^{\ast} \label{sol.freelie.1.a.endgame1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.endgame1}):} Let
$h\in\mathfrak{A}^{\ast}$ and $s\in\mathfrak{A}^{\ast}$.
\par
Clearly, $\mathfrak{L}_{h,s}\subset\mathfrak{L}$. Thus, the sum $\sum
_{w\in\mathfrak{L}_{h,s}}\kk b_{w}$ is a subsum of the sum $\sum
_{w\in\mathfrak{L}}\kk b_{w}$. Thus, $\sum_{w\in\mathfrak{L}_{h,s}%
}\kk b_{w}\subset\sum_{w\in\mathfrak{L}}\kk b_{w}$.
\par
The definition of $B_{h,s}$ shows that $B_{h,s}=\sum_{w\in\mathfrak{L}_{h,s}%
}\kk b_{w}\subset\sum_{w\in\mathfrak{L}}\kk b_{w}=B$ (by
(\ref{sol.freelie.1.a.endgame.B})). This proves
(\ref{sol.freelie.1.a.endgame1}).}.

Now, using (\ref{sol.freelie.1.a.mainclaim}), we can easily see the following
fact: For any $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$, we have%
\begin{equation}
\left[  b_{p},b_{q}\right]  \in B \label{sol.freelie.1.a.endgame3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.a.endgame3}):} Let
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$. We need to prove that $\left[
b_{p},b_{q}\right]  \in B$. If $p=q$, then this is clear (because if $p=q$,
then $\left[  \underbrace{b_{p}}_{\substack{=b_{q}\\\text{(since }p=q\text{)}%
}},b_{q}\right]  =\left[  b_{q},b_{q}\right]  =0\in B$ (since $B$ is a
$\kk$-module)). Thus, for the rest of this proof, we can WLOG assume
that we don't have $p=q$. Assume this.
\par
We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$) and $\left(  q,p\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $q\in\mathfrak{L}$ and
$p\in\mathfrak{L}$). We have $p\neq q$ (since we don't have $p=q$). Thus, we
have either $p<q$ or $p>q$ (since the lexicographic order on $\mathfrak{A}%
^{\ast}$ is a total order). In other words, we are in one of the following two
Cases:
\par
\textit{Case 1:} We have $p<q$.
\par
\textit{Case 2:} We have $p>q$.
\par
Let us first consider Case 1. In this case, we have $p<q$. Thus, from
(\ref{sol.freelie.1.a.mainclaim}), we obtain $\left[  b_{p},b_{q}\right]  \in
B_{pq,q}\subset B$ (by (\ref{sol.freelie.1.a.endgame1}) (applied to $h=pq$ and
$s=q$)). Thus, $\left[  b_{p},b_{q}\right]  \in B$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $p>q$. In other words,
$q<p$. Thus, (\ref{sol.freelie.1.a.mainclaim}) (applied to $\left(
q,p\right)  $ instead of $\left(  p,q\right)  $) shows that $\left[
b_{q},b_{p}\right]  \in B_{qp,p}\subset B$ (by (\ref{sol.freelie.1.a.endgame1}%
) (applied to $h=qp$ and $s=p$)). Now, $\left[  b_{p},b_{q}\right]
=-\underbrace{\left[  b_{q},b_{p}\right]  }_{\in B}\in-B\subset B$ (since $B$
is a $\kk$-module). Thus, $\left[  b_{p},b_{q}\right]  \in B$ is proven
in Case 2.
\par
Now, $\left[  b_{p},b_{q}\right]  \in B$ is proven in each of the two Cases 1
and 2. Since these two Cases cover all possibilities, this shows that $\left[
b_{p},b_{q}\right]  \in B$ always holds. Thus, (\ref{sol.freelie.1.a.endgame3}%
) is proven.}. Now, (\ref{sol.freelie.1.a.endgame.B}) yields $B=\sum
_{w\in\mathfrak{L}}\kk b_{w}=\sum_{p\in\mathfrak{L}}\kk b_{p}$
(here, we renamed the summation index $w$ as $p$) and $B=\sum_{w\in
\mathfrak{L}}\kk b_{w}=\sum_{q\in\mathfrak{L}}\kk b_{q}$ (here, we
renamed the summation index $w$ as $q$). Thus,%
\begin{align*}
&  \left[  \underbrace{B}_{=\sum_{p\in\mathfrak{L}}\kk b_{p}%
},\underbrace{B}_{=\sum_{q\in\mathfrak{L}}\kk b_{q}}\right] \\
&  =\left[  \sum_{p\in\mathfrak{L}}\kk b_{p},\sum_{q\in\mathfrak{L}%
}\kk b_{q}\right]  =\sum_{p\in\mathfrak{L}}\sum_{q\in\mathfrak{L}%
}\underbrace{\kk\kk}_{\subset\kk}\underbrace{\left[  b_{p}%
,b_{q}\right]  }_{\substack{\in B\\\text{(by (\ref{sol.freelie.1.a.endgame3}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie bracket on }%
\mathfrak{g}\text{ is }\kk\text{-bilinear}\right) \\
&  \subset\sum_{p\in\mathfrak{L}}\sum_{q\in\mathfrak{L}}\kk B\subset
B\ \ \ \ \ \ \ \ \ \ \left(  \text{since }B\text{ is a }\kk%
\text{-module}\right)  .
\end{align*}
In other words, $B$ is a Lie subalgebra of $\mathfrak{g}$. This solves
Exercise \ref{exe.freelie.1}(a).

(b) We shall first prove that
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q.
\label{sol.freelie.1.b.mainclaim}%
\end{equation}


The proof of (\ref{sol.freelie.1.b.mainclaim}) is very similar to our above
proof of (\ref{sol.freelie.1.a.mainclaim}); it proceeds using the same kind of
double induction, with almost the same computations. Here are the details of
this proof:

\textit{Proof of (\ref{sol.freelie.1.b.mainclaim}):} We shall prove
(\ref{sol.freelie.1.b.mainclaim}) by strong induction over $\ell\left(
pq\right)  $:

\textit{Induction step:} Let $N$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.b.mainclaim}) holds in the case when $\ell\left(
pq\right)  <N$. We need to prove that (\ref{sol.freelie.1.b.mainclaim}) holds
in the case when $\ell\left(  pq\right)  =N$.

We have assumed that (\ref{sol.freelie.1.b.mainclaim}) holds in the case when
$\ell\left(  pq\right)  <N$. In other words, we have%
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  <N. \label{sol.freelie.1.b.indhyp1}%
\end{equation}


We now must prove that (\ref{sol.freelie.1.b.mainclaim}) holds in the case
when $\ell\left(  pq\right)  =N$. In other words, we must prove that
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  =N. \label{sol.freelie.1.b.indgoal1}%
\end{equation}


We shall prove (\ref{sol.freelie.1.b.indgoal1}) by strong induction on
$\left\vert \mathfrak{Z}_{pq}\left(  q\right)  \right\vert $%
\ \ \ \ \footnote{This is legitimate, because $\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert $ is a well-defined nonnegative integer for
every $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (because
(\ref{sol.freelie.1.Z.fin}) (applied to $w=pq$) shows that the set
$\mathfrak{Z}_{pq}\left(  q\right)  $ is finite).
\par
Of course, this will be an induction within our current induction step, so the
reader should try not to confuse the two inductions going on.}:

\textit{Induction step:} Let $K$ be a nonnegative integer. Assume that
(\ref{sol.freelie.1.b.indgoal1}) holds in the case when $\left\vert
\mathfrak{Z}_{pq}\left(  q\right)  \right\vert <K$. We need to prove that
(\ref{sol.freelie.1.b.indgoal1}) holds in the case when $\left\vert
\mathfrak{Z}_{pq}\left(  q\right)  \right\vert =K$.

We have assumed that (\ref{sol.freelie.1.b.indgoal1}) holds in the case when
$\left\vert \mathfrak{Z}_{pq}\left(  q\right)  \right\vert <K$. In other
words, we have%
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  =N\text{ and }\left\vert \mathfrak{Z}_{pq}\left(
q\right)  \right\vert <K. \label{sol.freelie.1.b.indhyp2}%
\end{equation}


Now, let $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ be such that
$p<q$ and $\ell\left(  pq\right)  =N$ and $\left\vert \mathfrak{Z}_{pq}\left(
q\right)  \right\vert =K$. We are going to prove that $f\left(  \left[
b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $.

We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$. In other
words, $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$. In other words, the words
$p$ and $q$ are Lyndon. Proposition \ref{prop.words.lyndon.concat}(a) (applied
to $u=p$ and $v=q$) thus shows that the word $pq$ is Lyndon. In other words,
$pq\in\mathfrak{L}$.

Furthermore, Proposition \ref{prop.words.lyndon.concat}(b) (applied to $u=p$
and $v=q$) shows that $pq<q$.

Moreover, the word $p$ is nonempty (since it is Lyndon); thus, $\ell\left(
p\right)  \geq1$. Also, the word $q$ is nonempty (since it is Lyndon); thus,
$\ell\left(  q\right)  \geq1$. Now, $\ell\left(  pq\right)  =\underbrace{\ell
\left(  p\right)  }_{\geq1}+\underbrace{\ell\left(  q\right)  }_{\geq1}%
\geq1+1=2>1$. Therefore, $pq$ is a Lyndon word of length $>1$ (since $pq$ is
Lyndon and since $pq$ has length $\ell\left(  pq\right)  >1$). Therefore,
$\operatorname*{stf}\left(  pq\right)  $ is well-defined. If
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $, then it is easy
to see that $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $%
\ \ \ \ \footnote{\textit{Proof.} Assume that $\operatorname*{stf}\left(
pq\right)  =\left(  p,q\right)  $. Thus, $\left(  p,q\right)
=\operatorname*{stf}\left(  pq\right)  $. Hence, (\ref{eq.exe.freelie.1.bcond}%
) (applied to $w=pq$, $u=p$ and $v=q$) shows that $f\left(  \left[
b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $, qed.}. Hence, for the rest of the proof of $f\left(
\left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $, we can WLOG assume that we don't have
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $. Assume this.
Thus, $q$ is not the longest Lyndon proper suffix of $pq$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $q$ is the
longest Lyndon proper suffix of $pq$.
\par
Let $w=pq$. Recall that $q$ is the longest Lyndon proper suffix of $pq$. In
other words, $q$ is the longest Lyndon proper suffix of $w$ (since $w=pq$).
\par
Let $\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Then,
$\left(  g,h\right)  =\operatorname*{stf}\left(  \underbrace{pq}_{=w}\right)
=\operatorname*{stf}w$. Thus, $h$ is the longest Lyndon proper suffix of $w$
(by Exercise \ref{exe.words.lyndon.stf.basic}(a)). Comparing this with the
fact that $q$ is the longest Lyndon proper suffix of $w$, we obtain that
$h=q$.
\par
But Exercise \ref{exe.words.lyndon.stf.basic}(b) shows that we have
$w=g\underbrace{h}_{=q}=gq$. Thus, $gq=w=pq$. Cancelling $q$ from this
equality, we obtain $g=p$. Now, from $\left(  g,h\right)  =\operatorname*{stf}%
\left(  pq\right)  $, we obtain $\operatorname*{stf}\left(  pq\right)
=\left(  \underbrace{g}_{=p},\underbrace{h}_{=q}\right)  =\left(  p,q\right)
$. This contradicts the fact that we don't have $\operatorname*{stf}\left(
pq\right)  =\left(  p,q\right)  $. This contradiction proves that our
assumption was wrong, qed.}. In other words, the statement that $q$ is the
longest Lyndon proper suffix of $pq$ is \textbf{false}.

We have $\ell\left(  p\right)  >1$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, $\ell\left(  p\right)  \leq1$. Combined with $\ell\left(
p\right)  \geq1$, this shows that $\ell\left(  p\right)  =1$.
\par
Let $w=pq$. The word $q$ is Lyndon, and is a proper suffix of $w$ (since
$w=pq$, and since the word $p$ is nonempty). In other words, the word $q$ is a
Lyndon proper suffix of $w$.
\par
Let $\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Then,
$\left(  g,h\right)  =\operatorname*{stf}\left(  \underbrace{pq}_{=w}\right)
=\operatorname*{stf}w$. Thus, $h$ is the longest Lyndon proper suffix of $w$
(by Exercise \ref{exe.words.lyndon.stf.basic}(a)). Thus, every Lyndon proper
suffix of $w$ is at most as long as $h$. In other words, if $h^{\prime}$ is
any Lyndon proper suffix of $w$, then $\ell\left(  h^{\prime}\right)  \leq
\ell\left(  h\right)  $. Applying this to $h^{\prime}=q$, we obtain
$\ell\left(  q\right)  \leq\ell\left(  h\right)  $ (since $q$ is a Lyndon
proper suffix of $w$). Hence, $\ell\left(  h\right)  \geq\ell\left(  q\right)
$.
\par
From Exercise \ref{exe.words.lyndon.stf.basic}(d), we know that the word $g$
is Lyndon. Thus, $g$ is nonempty, so that $\ell\left(  g\right)  \geq1$.
\par
But Exercise \ref{exe.words.lyndon.stf.basic}(b) shows that we have $w=gh$, so
that $\ell\left(  \underbrace{w}_{=gh}\right)  =\ell\left(  gh\right)
=\ell\left(  g\right)  +\ell\left(  h\right)  $ and therefore $\ell\left(
g\right)  +\ell\left(  h\right)  =\ell\left(  \underbrace{w}_{=pq}\right)
=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=1}%
+\underbrace{\ell\left(  q\right)  }_{\leq\ell\left(  h\right)  }\leq
1+\ell\left(  h\right)  $. Subtracting $\ell\left(  h\right)  $ from both
sides of this inequality, we obtain $\ell\left(  g\right)  \leq1$. Combined
with $\ell\left(  g\right)  \geq1$, this shows that $\ell\left(  g\right)
=1$.
\par
Now, clearly, if $x$ is a word and $y$ is a prefix of $x$, then%
\begin{equation}
y=\left(  x_{1},x_{2},\ldots,x_{\ell\left(  y\right)  }\right)  .
\label{sol.freelie.1.b.prefix-lem}%
\end{equation}
\par
But $g$ is a prefix of $w$ (since $w=gh$). Thus,
(\ref{sol.freelie.1.b.prefix-lem}) (applied to $x=w$ and $y=g$) shows that%
\begin{align*}
g  &  =\left(  w_{1},w_{2},\ldots,w_{\ell\left(  g\right)  }\right)  =\left(
w_{1},w_{2},\ldots,w_{1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  g\right)  =1\right) \\
&  =\left(  w_{1}\right)  .
\end{align*}
On the other hand, $p$ is a prefix of $w$ (since $w=pq$). Thus,
(\ref{sol.freelie.1.b.prefix-lem}) (applied to $x=w$ and $y=p$) shows that%
\begin{align*}
p  &  =\left(  w_{1},w_{2},\ldots,w_{\ell\left(  p\right)  }\right)  =\left(
w_{1},w_{2},\ldots,w_{1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\ell\left(  p\right)  =1\right) \\
&  =\left(  w_{1}\right)  .
\end{align*}
Comparing this with $g=\left(  w_{1}\right)  $, we obtain $p=g$. Hence, $g=p$,
so that $\underbrace{g}_{=p}q=pq=w=gh$. Cancelling $g$ from this equality, we
obtain $q=h$. Now, $\left(  \underbrace{p}_{=g},\underbrace{q}_{=h}\right)
=\left(  g,h\right)  =\operatorname*{stf}\left(  pq\right)  $. Hence,
$\operatorname*{stf}\left(  pq\right)  =\left(  p,q\right)  $. This
contradicts the fact that we don't have $\operatorname*{stf}\left(  pq\right)
=\left(  p,q\right)  $. This contradiction proves that our assumption was
wrong, qed.}. Thus, $p$ is a Lyndon word of length $>1$ (since $p$ is Lyndon
and since $p$ has length $\ell\left(  p\right)  >1$). Therefore,
$\operatorname*{stf}p$ is well-defined. Set $\left(  u,v\right)
=\operatorname*{stf}p$. Then, Exercise \ref{exe.words.lyndon.stf.basic}(a)
(applied to $w=p$, $g=u$ and $h=v$) says that $v$ is the longest Lyndon proper
suffix of $p$. In particular, $v$ is a Lyndon proper suffix of $p$. In other
words, $v$ is Lyndon and is a proper suffix of $p$. Moreover, Exercise
\ref{exe.words.lyndon.stf.basic}(b) (applied to $w=p$, $g=u$ and $h=v$) says
that we have $p=uv$. Also, Exercise \ref{exe.words.lyndon.stf.basic}(c)
(applied to $w=p$, $g=u$ and $h=v$) says that we have $u<uv<v$. Finally,
Exercise \ref{exe.words.lyndon.stf.basic}(d) (applied to $w=p$, $g=u$ and
$h=v$) says that the word $u$ is Lyndon. In other words, $u\in\mathfrak{L}$.
The words $u$ and $v$ are nonempty (since they are Lyndon).

Exercise \ref{exe.words.lyndon.stf.props2} (applied to $w=p$ and $t=q$) now
shows that $q$ is the longest Lyndon proper suffix of $pq$ if and only if we
do not have $v<q$. Thus, the statement that we do not have $v<q$ is
\textbf{false} (because the statement that $q$ is the longest Lyndon proper
suffix of $pq$ is \textbf{false}). Thus, we have $v<q$. Thus, $u<v<q$.

Recall that $\left(  u,v\right)  =\operatorname*{stf}p$. Thus,
(\ref{eq.exe.freelie.1.bw}) (applied to $w=p$) shows that $b_{p}=\left[
b_{u},b_{v}\right]  $. Thus,%
\begin{equation}
\left[  \underbrace{b_{p}}_{=\left[  b_{u},b_{v}\right]  },b_{q}\right]
=\left[  \left[  b_{u},b_{v}\right]  ,b_{q}\right]  =\left[  \left[
b_{u},b_{q}\right]  ,b_{v}\right]  -\left[  \left[  b_{v},b_{q}\right]
,b_{u}\right]  \label{sol.freelie.1.b.jacobi}%
\end{equation}
(by (\ref{sol.freelie.1.jacobi}), applied to $\mathfrak{k}=\mathfrak{g}$,
$x=b_{u}$, $y=b_{v}$ and $z=b_{q}$).

On the other hand, (\ref{eq.exe.freelie.1.bcond}) (applied to $w=p$) shows
that $f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(
b_{u}\right)  ,f\left(  b_{v}\right)  \right]  $ (since $p$ is a Lyndon word
of length $>1$, and since $\left(  u,v\right)  =\operatorname*{stf}p$). Now,
applying the map $f$ to both sides of the equality $b_{p}=\left[  b_{u}%
,b_{v}\right]  $, we obtain%
\begin{equation}
f\left(  b_{p}\right)  =f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[
f\left(  b_{u}\right)  ,f\left(  b_{v}\right)  \right]  .
\label{sol.freelie.1.b.in0}%
\end{equation}


Now,%
\begin{equation}
f\left(  \left[  b_{u},b_{q}\right]  \right)  =\left[  f\left(  b_{u}\right)
,f\left(  b_{q}\right)  \right]  \label{sol.freelie.1.b.in1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in1}):} We have $\left(
u,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $u\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $u<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\ell\left(  u\right)  +\underbrace{\ell\left(
v\right)  }_{\substack{\geq1\\\text{(since }v\text{ is nonempty)}}}\geq
\ell\left(  u\right)  +1>\ell\left(  u\right)  $. But $\ell\left(  pq\right)
=N$, so that $N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)
}_{>\ell\left(  u\right)  }+\ell\left(  q\right)  >\ell\left(  u\right)
+\ell\left(  q\right)  =\ell\left(  uq\right)  $. Thus, $\ell\left(
uq\right)  <N$. Therefore, (\ref{sol.freelie.1.b.indhyp1}) (applied to
$\left(  u,q\right)  $ instead of $\left(  p,q\right)  $) yields $f\left(
\left[  b_{u},b_{q}\right]  \right)  =\left[  f\left(  b_{u}\right)  ,f\left(
b_{q}\right)  \right]  $. This proves (\ref{sol.freelie.1.b.in1}).} and%
\begin{equation}
f\left(  \left[  b_{v},b_{q}\right]  \right)  =\left[  f\left(  b_{v}\right)
,f\left(  b_{q}\right)  \right]  \label{sol.freelie.1.b.in2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in2}):} We have $\left(
v,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since $v\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $v<q$. Also, $\ell\left(  \underbrace{p}_{=uv}\right)
=\ell\left(  uv\right)  =\underbrace{\ell\left(  u\right)  }_{\substack{\geq
1\\\text{(since }u\text{ is nonempty)}}}+\ell\left(  v\right)  \geq
1+\ell\left(  v\right)  >\ell\left(  v\right)  $. But $\ell\left(  pq\right)
=N$, so that $N=\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)
}_{>\ell\left(  v\right)  }+\ell\left(  q\right)  >\ell\left(  v\right)
+\ell\left(  q\right)  =\ell\left(  vq\right)  $. Thus, $\ell\left(
vq\right)  <N$. Therefore, (\ref{sol.freelie.1.b.indhyp1}) (applied to
$\left(  v,q\right)  $ instead of $\left(  p,q\right)  $) yields $f\left(
\left[  b_{v},b_{q}\right]  \right)  =\left[  f\left(  b_{v}\right)  ,f\left(
b_{q}\right)  \right]  $. This proves (\ref{sol.freelie.1.b.in2}).}.

On the other hand, if $g$ and $h$ are two Lyndon words satisfying $g<q$, $h<q$
and $gh\sim p$, then%
\begin{equation}
\left(  \text{every }r\in\mathfrak{L}_{gq,q}\text{ satisfies }f\left(  \left[
b_{r},b_{h}\right]  \right)  =\left[  f\left(  b_{r}\right)  ,f\left(
b_{h}\right)  \right]  \right)  \label{sol.freelie.1.b.in3a}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in3a}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. Let $r\in
\mathfrak{L}_{gq,q}$. We must prove that $\left[  f\left(  b_{r}\right)
,f\left(  b_{h}\right)  \right]  =f\left(  \left[  b_{r},b_{h}\right]
\right)  $.
\par
The words $g$ and $h$ are Lyndon. In other words, $g\in\mathfrak{L}$ and
$h\in\mathfrak{L}$.
\par
We have $gh\sim p$ and thus $\ell\left(  gh\right)  =\ell\left(  p\right)  $
(by (\ref{sol.freelie.1.rel.len}), applied to $w=gh$ and $w^{\prime}=p$).
Hence, $\ell\left(  p\right)  =\ell\left(  gh\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  $.
\par
We have $gh\sim p$ and $q\sim q$ (since the relation $\sim$ is reflexive).
Hence, $ghq\sim pq$ (by (\ref{sol.freelie.1.rel.uv}), applied to $gh$, $q$,
$p$ and $q$ instead of $u$, $v$, $u^{\prime}$ and $v^{\prime}$).
\par
We have $r\in\mathfrak{L}_{gq,q}=\left\{  w\in\mathfrak{L}\ \mid\ w\sim
gq\text{ and }w<q\right\}  $ (by the definition of $\mathfrak{L}_{gq,q}$). In
other words, $r$ is an element $w\in\mathfrak{L}$ satisfying $w\sim gq$ and
$w<q$. In other words, $r$ is an element of $\mathfrak{L}$ and satisfies
$r\sim gq$ and $r<q$. From $r\sim gq$, we obtain $\ell\left(  r\right)
=\ell\left(  gq\right)  $ (by (\ref{sol.freelie.1.rel.len}), applied to $w=r$
and $w^{\prime}=gq$). Thus, $\ell\left(  r\right)  =\ell\left(  gq\right)
=\ell\left(  g\right)  +\ell\left(  q\right)  $.
\par
If $r=h$, then $\left[  \underbrace{b_{r}}_{\substack{=b_{h}\\\text{(since
}r=h\text{)}}},b_{h}\right]  =\left[  b_{h},b_{h}\right]  =0\in B_{pq,q}$
(since $B_{pq,q}$ is a $\kk$-module). Thus, if $r=h$, then $\left[
b_{r},b_{h}\right]  \in B_{pq,q}$ holds. Hence, for the rest of the proof of
$\left[  b_{r},b_{h}\right]  \in B_{pq,q}$, we can WLOG assume that we don't
have $r=h$. Assume this.
\par
We have $r\neq h$ (since we don't have $r=h$). Thus, we have either $r<h$ or
$r>h$ (since the lexicographic order on $\mathfrak{A}^{\ast}$ is a total
order). In other words, we are in one of the following two Cases:
\par
\textit{Case 1:} We have $r<h$.
\par
\textit{Case 2:} We have $r>h$.
\par
Let us first consider Case 1. In this case, we have $r<h$. We have
$\ell\left(  rh\right)  =\underbrace{\ell\left(  r\right)  }_{=\ell\left(
g\right)  +\ell\left(  q\right)  }+\ell\left(  h\right)  =\ell\left(
g\right)  +\ell\left(  q\right)  +\ell\left(  h\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  +\ell\left(  q\right)  $. Comparing this with
$\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=\ell\left(
g\right)  +\ell\left(  h\right)  }+\ell\left(  q\right)  =\ell\left(
g\right)  +\ell\left(  h\right)  +\ell\left(  q\right)  $, we obtain
$\ell\left(  rh\right)  =\ell\left(  pq\right)  =N$.
\par
We have $r\sim gq$ and $h\sim h$ (since the relation $\sim$ is reflexive).
Thus, (\ref{sol.freelie.1.rel.uv}) (applied to $r$, $h$, $gq$ and $h$ instead
of $u$, $v$, $u^{\prime}$ and $v^{\prime}$) shows that $rh\sim gqh$. But
$gqh\sim ghq$ (by (\ref{sol.freelie.1.rel.wuv}), applied to $q$, $h$ and $g$
instead of $u$, $v$ and $w$). From $rh\sim gqh$ and $gqh\sim ghq$, we obtain
$rh\sim ghq$ (since the relation $\sim$ is transitive). From $rh\sim ghq$ and
$ghq\sim pq$, we obtain $rh\sim pq$ (since the relation $\sim$ is transitive).
\par
Thus, $\operatorname*{Und}\left(  rh\right)  =\operatorname*{Und}\left(
pq\right)  $ (by (\ref{sol.freelie.1.rel.Und}), applied to $w=rh$ and
$w^{\prime}=pq$). But $\operatorname*{Und}h\subset\operatorname*{Und}\left(
rh\right)  $ (by (\ref{sol.freelie.1.Und.uv2}), applied to $r$ and $h$ instead
of $u$ and $v$). Thus, $\operatorname*{Und}h\subset\operatorname*{Und}\left(
rh\right)  =\operatorname*{Und}\left(  pq\right)  $. Now,
(\ref{sol.freelie.1.Und.w}) (applied to $w=h$) shows that $h\in
\operatorname*{Und}h\subset\operatorname*{Und}\left(  pq\right)  $.
\par
So we know that $rh\sim pq$, $h\in\operatorname*{Und}\left(  pq\right)  $ and
$h<q$. Thus, we can apply (\ref{sol.freelie.1.Z.dec}) to $pq$, $rh$, $q$ and
$h$ instead of $w$, $v$, $q$ and $h$. As a result, we obtain $\left\vert
\mathfrak{Z}_{rh}\left(  h\right)  \right\vert <\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert =K$.
\par
Now, we have $\left(  r,h\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$r\in\mathfrak{L}$ and $h\in\mathfrak{L}$) and $r<h$ and $\ell\left(
rh\right)  =N$ and $\left\vert \mathfrak{Z}_{rh}\left(  h\right)  \right\vert
<K$. Therefore, (\ref{sol.freelie.1.b.indhyp2}) (applied to $r$ and $h$
instead of $p$ and $q$) shows that $f\left(  \left[  b_{r},b_{h}\right]
\right)  =\left[  f\left(  b_{r}\right)  ,f\left(  b_{h}\right)  \right]  $.
Thus, $f\left(  \left[  b_{r},b_{h}\right]  \right)  =\left[  f\left(
b_{r}\right)  ,f\left(  b_{h}\right)  \right]  $ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $h<r$. We have $\ell\left(
hr\right)  =\ell\left(  h\right)  +\underbrace{\ell\left(  r\right)  }%
_{=\ell\left(  g\right)  +\ell\left(  q\right)  }=\ell\left(  h\right)
+\ell\left(  g\right)  +\ell\left(  q\right)  =\ell\left(  g\right)
+\ell\left(  h\right)  +\ell\left(  q\right)  $. Comparing this with
$\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)  }_{=\ell\left(
g\right)  +\ell\left(  h\right)  }+\ell\left(  q\right)  =\ell\left(
g\right)  +\ell\left(  h\right)  +\ell\left(  q\right)  $, we obtain
$\ell\left(  hr\right)  =\ell\left(  pq\right)  =N$.
\par
We have $h\sim h$ (since the relation $\sim$ is reflexive) and $r\sim gq$.
Thus, (\ref{sol.freelie.1.rel.uv}) (applied to $h$, $r$, $h$ and $gq$ instead
of $u$, $v$, $u^{\prime}$ and $v^{\prime}$) shows that $hr\sim hgq$. But
$hgq\sim ghq$ (by (\ref{sol.freelie.1.rel.uvw}), applied to $h$, $g$ and $q$
instead of $u$, $v$ and $w$). From $hr\sim hgq$ and $hgq\sim ghq$, we obtain
$hr\sim ghq$ (since the relation $\sim$ is transitive). From $hr\sim ghq$ and
$ghq\sim pq$, we obtain $hr\sim pq$ (since the relation $\sim$ is transitive).
\par
Hence, $\operatorname*{Und}\left(  hr\right)  =\operatorname*{Und}\left(
pq\right)  $ (by (\ref{sol.freelie.1.rel.Und}), applied to $w=hr$ and
$w^{\prime}=pq$). But $\operatorname*{Und}r\subset\operatorname*{Und}\left(
hr\right)  $ (by (\ref{sol.freelie.1.Und.uv2}), applied to $h$ and $r$ instead
of $u$ and $v$). Thus, $\operatorname*{Und}r\subset\operatorname*{Und}\left(
hr\right)  =\operatorname*{Und}\left(  pq\right)  $. Now,
(\ref{sol.freelie.1.Und.w}) (applied to $w=r$) shows that $r\in
\operatorname*{Und}r\subset\operatorname*{Und}\left(  pq\right)  $.
\par
So we know that $hr\sim pq$, $r\in\operatorname*{Und}\left(  pq\right)  $ and
$r<q$. Thus, we can apply (\ref{sol.freelie.1.Z.dec}) to $pq$, $hr$, $q$ and
$r$ instead of $w$, $v$, $q$ and $h$. As a result, we obtain $\left\vert
\mathfrak{Z}_{hr}\left(  r\right)  \right\vert <\left\vert \mathfrak{Z}%
_{pq}\left(  q\right)  \right\vert =K$.
\par
Now, we have $\left(  h,r\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$h\in\mathfrak{L}$ and $r\in\mathfrak{L}$) and $h<r$ and $\ell\left(
hr\right)  =N$ and $\left\vert \mathfrak{Z}_{hr}\left(  r\right)  \right\vert
<K$. Therefore, (\ref{sol.freelie.1.b.indhyp2}) (applied to $h$ and $r$
instead of $p$ and $q$) shows that $f\left(  \left[  b_{h},b_{r}\right]
\right)  =\left[  f\left(  b_{h}\right)  ,f\left(  b_{r}\right)  \right]  $.
Thus,%
\begin{align*}
f\left(  \underbrace{\left[  b_{r},b_{h}\right]  }_{=-\left[  b_{h}%
,b_{r}\right]  }\right)   &  =f\left(  -\left[  b_{h},b_{r}\right]  \right)
=-\underbrace{f\left(  \left[  b_{h},b_{r}\right]  \right)  }_{=\left[
f\left(  b_{h}\right)  ,f\left(  b_{r}\right)  \right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =-\left[  f\left(  b_{h}\right)  ,f\left(  b_{r}\right)  \right]  =\left[
f\left(  b_{r}\right)  ,f\left(  b_{h}\right)  \right]  .
\end{align*}
Thus, $f\left(  \left[  b_{r},b_{h}\right]  \right)  =\left[  f\left(
b_{r}\right)  ,f\left(  b_{h}\right)  \right]  $ is proven in Case 2.
\par
Thus, $f\left(  \left[  b_{r},b_{h}\right]  \right)  =\left[  f\left(
b_{r}\right)  ,f\left(  b_{h}\right)  \right]  $ is proven in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, this shows that
$f\left(  \left[  b_{r},b_{h}\right]  \right)  =\left[  f\left(  b_{r}\right)
,f\left(  b_{h}\right)  \right]  $ always holds. Hence,
(\ref{sol.freelie.1.b.in3a}) is proven.} and therefore%
\begin{equation}
\left(  \text{every }x\in B_{gq,q}\text{ satisfies }f\left(  \left[
x,b_{h}\right]  \right)  =\left[  f\left(  x\right)  ,f\left(  b_{h}\right)
\right]  \right)  \label{sol.freelie.1.b.in3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.in3b}):} Let $g$ and $h$ be
two Lyndon words satisfying $g<q$, $h<q$ and $gh\sim p$. Let $x\in B_{gq,q}$.
The definition of $B_{gq,q}$ shows that $B_{gq,q}=\sum_{w\in\mathfrak{L}%
_{gq,q}}\kk b_{w}$. In other words, $B_{gq,q}$ is the $\kk%
$-linear span of the elements $b_{w}$ for $w\in\mathfrak{L}_{gq,q}$. Thus, the
family $\left(  b_{w}\right)  _{w\in\mathfrak{L}_{gq,q}}$ spans the
$\kk$-module $B_{gq,q}$.
\par
Now, fix $x \in B_{gq,q}$. We must prove the equality $f\left(  \left[
x,b_{h}\right]  \right)  =\left[  f\left(  x\right)  ,f\left(  b_{h}\right)
\right]  $. Both sides of this equality are $\kk$-linear in $x$ (since
the map $f$ is $\kk$-linear, and since the Lie brackets of the Lie
algebras $\mathfrak{g}$ and $\mathfrak{h}$ are $\kk$-bilinear). Hence,
we can WLOG assume that $x$ belongs to the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}_{gq,q}}$ (since this family $\left(  b_{w}\right)
_{w\in\mathfrak{L}_{gq,q}}$ spans the $\kk$-module $B_{gq,q}$, and
since $x \in B_{gq,q}$). Assume this.
\par
We know that $x$ belongs to the family $\left(  b_{w}\right)  _{w\in
\mathfrak{L}_{gq,q}}$. In other words, there exists a $w\in\mathfrak{L}%
_{gq,q}$ such that $x=b_{w}$. Consider this $w$. Now,
(\ref{sol.freelie.1.b.in3a}) (applied to $r=w$) yields $f\left(  \left[
b_{w},b_{h}\right]  \right)  =\left[  f\left(  b_{w}\right)  ,f\left(
b_{h}\right)  \right]  $. Since $x=b_{w}$, this rewrites as $f\left(  \left[
x,b_{h}\right]  \right)  =\left[  f\left(  x\right)  ,f\left(  b_{h}\right)
\right]  $. This proves (\ref{sol.freelie.1.b.in3b}).}.

Recall that (\ref{sol.freelie.1.a.mainclaim}) holds. (This was proven in our
solution to Exercise \ref{exe.freelie.1}(a).) Now, $\left(  u,q\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $u\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $u<q$. Thus, (\ref{sol.freelie.1.a.mainclaim})
(applied to $\left(  u,q\right)  $ instead of $\left(  p,q\right)  $) shows
that $\left[  b_{u},b_{q}\right]  \in B_{uq,q}$. Also, $\left(  v,q\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $v\in\mathfrak{L}$ and
$q\in\mathfrak{L}$) and $v<q$. Thus, (\ref{sol.freelie.1.a.mainclaim})
(applied to $\left(  v,q\right)  $ instead of $\left(  p,q\right)  $) shows
that $\left[  b_{v},b_{q}\right]  \in B_{vq,q}$.

We have $p\sim p$ (since the relation $\sim$ is reflexive). In other words,
$uv\sim p$ (since $p=uv$). Also, (\ref{sol.freelie.1.rel.vu}) yields $uv\sim
vu$, so that $vu\sim uv$ (since the relation $\sim$ is symmetric). Since
$p=uv$, this rewrites as $vu\sim p$.

Now, the words $u$ and $v$ are Lyndon and satisfy $u<q$, $v<q$, $uv\sim p$ and
$\left[  b_{u},b_{q}\right]  \in B_{uq,q}$. Thus, we can apply
(\ref{sol.freelie.1.b.in3b}) to $g=u$, $h=v$ and $x=\left[  b_{u}%
,b_{q}\right]  $. As a result, we obtain%
\begin{equation}
f\left(  \left[  \left[  b_{u},b_{q}\right]  ,b_{v}\right]  \right)  =\left[
\underbrace{f\left(  \left[  b_{u},b_{q}\right]  \right)  }%
_{\substack{=\left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
\\\text{(by (\ref{sol.freelie.1.b.in1}))}}},f\left(  b_{v}\right)  \right]
=\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{v}\right)  \right]  . \label{sol.freelie.1.b.in4}%
\end{equation}


Furthermore, the words $v$ and $u$ are Lyndon and satisfy $v<q$, $u<q$,
$vu\sim p$ and $\left[  b_{v},b_{q}\right]  \in B_{vq,q}$. Thus, we can apply
(\ref{sol.freelie.1.b.in3b}) to $g=v$, $h=u$ and $x=\left[  b_{v}%
,b_{q}\right]  $. As a result, we obtain%
\begin{equation}
f\left(  \left[  \left[  b_{v},b_{q}\right]  ,b_{u}\right]  \right)  =\left[
\underbrace{f\left(  \left[  b_{v},b_{q}\right]  \right)  }%
_{\substack{=\left[  f\left(  b_{v}\right)  ,f\left(  b_{q}\right)  \right]
\\\text{(by (\ref{sol.freelie.1.b.in2}))}}},f\left(  b_{u}\right)  \right]
=\left[  \left[  f\left(  b_{v}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{u}\right)  \right]  . \label{sol.freelie.1.b.in5}%
\end{equation}


Now, applying the map $f$ to both sides of the equality
(\ref{sol.freelie.1.b.jacobi}), we obtain%
\begin{align*}
f\left(  \left[  b_{p},b_{q}\right]  \right)   &  =f\left(  \left[  \left[
b_{u},b_{q}\right]  ,b_{v}\right]  -\left[  \left[  b_{v},b_{q}\right]
,b_{u}\right]  \right) \\
&  =\underbrace{f\left(  \left[  \left[  b_{u},b_{q}\right]  ,b_{v}\right]
\right)  }_{\substack{=\left[  \left[  f\left(  b_{u}\right)  ,f\left(
b_{q}\right)  \right]  ,f\left(  b_{v}\right)  \right]  \\\text{(by
(\ref{sol.freelie.1.b.in4}))}}}-\underbrace{f\left(  \left[  \left[
b_{v},b_{q}\right]  ,b_{u}\right]  \right)  }_{\substack{=\left[  \left[
f\left(  b_{v}\right)  ,f\left(  b_{q}\right)  \right]  ,f\left(
b_{u}\right)  \right]  \\\text{(by (\ref{sol.freelie.1.b.in5}))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{v}\right)  \right]  -\left[  \left[  f\left(  b_{v}\right)
,f\left(  b_{q}\right)  \right]  ,f\left(  b_{u}\right)  \right]  .
\end{align*}
Comparing this with%
\begin{align*}
&  \left[  \underbrace{f\left(  b_{p}\right)  }_{\substack{=\left[  f\left(
b_{u}\right)  ,f\left(  b_{v}\right)  \right]  \\\text{(by
(\ref{sol.freelie.1.b.in0}))}}},f\left(  b_{q}\right)  \right] \\
&  =\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{v}\right)  \right]
,f\left(  b_{q}\right)  \right] \\
&  =\left[  \left[  f\left(  b_{u}\right)  ,f\left(  b_{q}\right)  \right]
,f\left(  b_{v}\right)  \right]  -\left[  \left[  f\left(  b_{v}\right)
,f\left(  b_{q}\right)  \right]  ,f\left(  b_{u}\right)  \right]  .\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.1.jacobi}), applied
to }\mathfrak{k}=\mathfrak{h}\text{, }x=f\left(  b_{u}\right)  \text{,
}y=f\left(  b_{v}\right)  \text{ and }z=f\left(  b_{q}\right)  \right)  ,
\end{align*}
we obtain $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $.

Let us now forget that we fixed $\left(  p,q\right)  $. We thus have proven
that%
\[
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
p,q\right)  \in\mathfrak{L}\times\mathfrak{L}\text{ satisfying }p<q\text{ and
}\ell\left(  pq\right)  =N\text{ and }\left\vert \mathfrak{Z}_{pq}\left(
q\right)  \right\vert =K.
\]
In other words, (\ref{sol.freelie.1.b.indgoal1}) holds in the case when
$\left\vert \mathfrak{Z}_{pq}\left(  q\right)  \right\vert =K$. This completes
the induction step (in the induction proof of (\ref{sol.freelie.1.b.indgoal1}%
)). The induction proof of (\ref{sol.freelie.1.b.indgoal1}) is thus finished.

We thus have proven (\ref{sol.freelie.1.b.indgoal1}). In other words,
(\ref{sol.freelie.1.b.mainclaim}) holds in the case when $\ell\left(
pq\right)  =N$. This completes the induction step (in the induction proof of
(\ref{sol.freelie.1.b.mainclaim})). The induction proof of
(\ref{sol.freelie.1.b.mainclaim}) is thus complete.

Using (\ref{sol.freelie.1.b.mainclaim}), we can easily see the following fact:
For any $p\in\mathfrak{L}$ and $q\in\mathfrak{L}$, we have%
\begin{equation}
f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  \label{sol.freelie.1.b.endgame3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.endgame3}):} Let
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$. We need to prove that $f\left(
\left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $. If $p=q$, then this is clear (because if $p=q$,
then
\begin{align*}
f\left(  \left[  \underbrace{b_{p}}_{\substack{=b_{q}\\\text{(since
}p=q\text{)}}},b_{q}\right]  \right)   &  =f\left(  \underbrace{\left[
b_{q},b_{q}\right]  }_{=0}\right)  =f\left(  0\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =\left[  f\left(  b_{p}\right)  ,f\left(  \underbrace{b_{p}}%
_{\substack{=b_{q}\\\text{(since }p=q\text{)}}}\right)  \right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[  f\left(  b_{p}\right)
,f\left(  b_{p}\right)  \right]  =0\right) \\
&  =\left[  f\left(  b_{p}\right)  ,f\left(  b_{q}\right)  \right]
\end{align*}
). Thus, for the rest of this proof, we can WLOG assume that we don't have
$p=q$. Assume this.
\par
We have $\left(  p,q\right)  \in\mathfrak{L}\times\mathfrak{L}$ (since
$p\in\mathfrak{L}$ and $q\in\mathfrak{L}$) and $\left(  q,p\right)
\in\mathfrak{L}\times\mathfrak{L}$ (since $q\in\mathfrak{L}$ and
$p\in\mathfrak{L}$). We have $p\neq q$ (since we don't have $p=q$). Thus, we
have either $p<q$ or $p>q$ (since the lexicographic order on $\mathfrak{A}%
^{\ast}$ is a total order). In other words, we are in one of the following two
Cases:
\par
\textit{Case 1:} We have $p<q$.
\par
\textit{Case 2:} We have $p>q$.
\par
Let us first consider Case 1. In this case, we have $p<q$. Thus, from
(\ref{sol.freelie.1.b.mainclaim}), we obtain $f\left(  \left[  b_{p}%
,b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $. Thus, $f\left(  \left[  b_{p},b_{q}\right]  \right)
=\left[  f\left(  b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $ is proven
in Case 1.
\par
Let us now consider Case 2. In this case, we have $p>q$. In other words,
$q<p$. Thus, (\ref{sol.freelie.1.b.mainclaim}) (applied to $\left(
q,p\right)  $ instead of $\left(  p,q\right)  $) shows that $f\left(  \left[
b_{q},b_{p}\right]  \right)  =\left[  f\left(  b_{q}\right)  ,f\left(
b_{p}\right)  \right]  $. Now,
\begin{align*}
f\left(  \underbrace{\left[  b_{p},b_{q}\right]  }_{=-\left[  b_{q}%
,b_{p}\right]  }\right)   &  =f\left(  -\left[  b_{q},b_{p}\right]  \right)
=-\underbrace{f\left(  \left[  b_{q},b_{p}\right]  \right)  }_{=\left[
f\left(  b_{q}\right)  ,f\left(  b_{p}\right)  \right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\kk%
\text{-linear}\right) \\
&  =-\left[  f\left(  b_{q}\right)  ,f\left(  b_{p}\right)  \right]  =\left[
f\left(  b_{p}\right)  ,f\left(  b_{q}\right)  \right]  .
\end{align*}
Thus, $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $ is proven in Case 2.
\par
Now, $f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(
b_{p}\right)  ,f\left(  b_{q}\right)  \right]  $ is proven in each of the two
Cases 1 and 2. Since these two Cases cover all possibilities, this shows that
$f\left(  \left[  b_{p},b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)
,f\left(  b_{q}\right)  \right]  $ always holds. Thus,
(\ref{sol.freelie.1.b.endgame3}) is proven.}.

Now, we recall that $B$ is the $\kk$-submodule of $\mathfrak{g}$
spanned by the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$. Thus, the
family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk%
$-module $B$.

Now, we have%
\begin{equation}
f\left(  \left[  x,y\right]  \right)  =\left[  f\left(  x\right)  ,f\left(
y\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }x\in B\text{ and }y\in
B \label{sol.freelie.1.b.endgame}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.1.b.endgame}):} Let $x\in B$ and
$y\in B$. We must prove the equality $f\left(  \left[  x,y\right]  \right)
=\left[  f\left(  x\right)  ,f\left(  y\right)  \right]  $. Both sides of this
equality are $\kk$-linear in each of $x$ and $y$ (since the Lie
brackets of the Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ are
$\kk$-bilinear, and since the map $f$ is $\kk$-linear). Hence,
we can WLOG assume that both $x$ and $y$ belong to the family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ (since the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ spans the $\kk$-module $B$). Assume this.
\par
We know that $x$ belongs to the family $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$. In other words, there exists a $p\in\mathfrak{L}$ such that
$x=b_{p}$. Consider this $p$.
\par
We know that $y$ belongs to the family $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$. In other words, there exists a $q\in\mathfrak{L}$ such that
$y=b_{q}$. Consider this $q$.
\par
From (\ref{sol.freelie.1.b.endgame3}), we obtain $f\left(  \left[  b_{p}%
,b_{q}\right]  \right)  =\left[  f\left(  b_{p}\right)  ,f\left(
b_{q}\right)  \right]  $. This rewrites as $f\left(  \left[  x,b_{q}\right]
\right)  =\left[  f\left(  x\right)  ,f\left(  b_{q}\right)  \right]  $ (since
$x=b_{p}$). This rewrites as $f\left(  \left[  x,y\right]  \right)  =\left[
f\left(  x\right)  ,f\left(  y\right)  \right]  $ (since $y=b_{q}$). This
proves (\ref{sol.freelie.1.b.endgame}).}. In other words, the map
$f:B\rightarrow\mathfrak{h}$ is a Lie algebra homomorphism. This solves
Exercise \ref{exe.freelie.1}(b).
%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.freelie.2}}

\textit{Solution to Exercise \ref{exe.freelie.2}.} The definition of
$\mathfrak{g}$ shows that $\mathfrak{g}=\mathfrak{g}_{1}+\mathfrak{g}%
_{2}+\mathfrak{g}_{3}+\cdots=\sum_{i\geq1}\mathfrak{g}_{i}$. Thus, for every
positive integer $k$, we have%
\begin{equation}
\mathfrak{g}_{k}\subset\mathfrak{g}. \label{sol.freelie.2.gk}%
\end{equation}
Notice that $\mathfrak{g}_{1}=V$, so that%
\begin{equation}
V=\mathfrak{g}_{1}\subset\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.gk}), applied to }k=1\right)  . \label{sol.freelie.2.V}%
\end{equation}


We also recall a fundamental property of Lie algebras (one of the forms of the
Jacobi identity):

\begin{itemize}
\item Every three elements $x$, $y$ and $z$ of a Lie algebra $\mathfrak{k}$
satisfy%
\begin{equation}
\left[  \left[  x,y\right]  ,z\right]  =\left[  \left[  x,z\right]  ,y\right]
-\left[  \left[  y,z\right]  ,x\right]  . \label{sol.freelie.2.jacobi}%
\end{equation}

\end{itemize}

(a) For every two positive integers $i$ and $j$, we have%
\begin{equation}
\left[  \mathfrak{g}_{i},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{i+j}.
\label{sol.freelie.2.a.claim}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.2.a.claim}):} We shall prove
(\ref{sol.freelie.2.a.claim}) by induction over $i$:

\textit{Induction base:} For every positive integer $j$, we have $\left[
\mathfrak{g}_{1},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{1+j}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j$ be a positive integer. Thus, the
recursive definition of $\mathfrak{g}_{1+j}$ yields $\mathfrak{g}%
_{1+j}=\left[  V,\underbrace{\mathfrak{g}_{\left(  1+j\right)  -1}%
}_{=\mathfrak{g}_{j}}\right]  =\left[  V,\mathfrak{g}_{j}\right]  $, so that
$\left[  V,\mathfrak{g}_{j}\right]  =\mathfrak{g}_{1+j}$. Now, $\left[
\underbrace{\mathfrak{g}_{1}}_{=V},\mathfrak{g}_{j}\right]  =\left[
V,\mathfrak{g}_{j}\right]  =\mathfrak{g}_{1+j}\subset\mathfrak{g}_{1+j}$,
qed.}. In other words, (\ref{sol.freelie.2.a.claim}) holds for $i=1$. This
completes the induction base.

\textit{Induction step:} Let $I$ be a positive integer. Assume that
(\ref{sol.freelie.2.a.claim}) holds for $i=I$. We must prove that
(\ref{sol.freelie.2.a.claim}) holds for $i=I+1$.

We have assumed that (\ref{sol.freelie.2.a.claim}) holds for $i=I$. In other
words, for every positive integer $j$, we have%
\begin{equation}
\left[  \mathfrak{g}_{I},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{I+j}.
\label{sol.freelie.2.a.claim.pf.hyp}%
\end{equation}


Now, let $j$ be a positive integer. Thus, the recursive definition of
$\mathfrak{g}_{j+1}$ yields $\mathfrak{g}_{j+1}=\left[
V,\underbrace{\mathfrak{g}_{\left(  j+1\right)  -1}}_{=\mathfrak{g}_{j}%
}\right]  =\left[  V,\mathfrak{g}_{j}\right]  $. The same argument (applied to
$I$ instead of $j$) shows that $\mathfrak{g}_{I+1}=\left[  V,\mathfrak{g}%
_{I}\right]  $.

Also, the recursive definition of $\mathfrak{g}_{I+j+1}$ yields $\mathfrak{g}%
_{I+j+1}=\left[  V,\underbrace{\mathfrak{g}_{\left(  I+j+1\right)  -1}%
}_{=\mathfrak{g}_{I+j}}\right]  =\left[  V,\mathfrak{g}_{I+j}\right]  $.

Now, let $z\in\mathfrak{g}_{j}$. Then, $z\in\mathfrak{g}_{j}\subset
\mathfrak{g}$ (by (\ref{sol.freelie.2.gk}), applied to $k=j$).

Also, let $p\in\mathfrak{g}_{I+1}$. We are going to prove that $\left[
p,z\right]  \in\mathfrak{g}_{I+1+j}$.

Recall that $\left[  V,\mathfrak{g}_{I}\right]  $ is the $\kk$-linear
span of all elements of the form $\left[  x,y\right]  $ with $\left(
x,y\right)  \in V\times\mathfrak{g}_{I}$ (indeed, this is how $\left[
V,\mathfrak{g}_{I}\right]  $ is defined). Thus, $p$ is a $\kk$-linear
combination of elements of the form $\left[  x,y\right]  $ with $\left(
x,y\right)  \in V\times\mathfrak{g}_{I}$ (since $p\in\mathfrak{g}%
_{I+1}=\left[  V,\mathfrak{g}_{I}\right]  $).

Now, we must prove the relation $\left[  p,z\right]  \in\mathfrak{g}_{I+1+j}$.
But this relation is $\kk$-linear in $p$. Thus, we WLOG assume that $p$
is an element of the form $\left[  x,y\right]  $ for with $\left(  x,y\right)
\in V\times\mathfrak{g}_{I}$ (since $p$ is a $\kk$-linear combination
of elements of the form $\left[  x,y\right]  $ with $\left(  x,y\right)  \in
V\times\mathfrak{g}_{I}$). In other words, there exists an $\left(
x,y\right)  \in V\times\mathfrak{g}_{I}$ such that $p=\left[  x,y\right]  $.
Consider this $\left(  x,y\right)  $.

We have $\left(  x,y\right)  \in V\times\mathfrak{g}_{I}$. In other words,
$x\in V$ and $y\in\mathfrak{g}_{I}$. Thus, $y\in\mathfrak{g}_{I}%
\subset\mathfrak{g}$ (by (\ref{sol.freelie.2.gk}), applied to $k=I$). Also,
$x\in V\subset\mathfrak{g}$ (by (\ref{sol.freelie.2.V})). Now,%
\begin{align*}
\left[  \underbrace{p}_{=\left[  x,y\right]  },z\right]   &  =\left[  \left[
x,y\right]  ,z\right]  =\underbrace{\left[  \left[  x,z\right]  ,y\right]
}_{=-\left[  y,\left[  x,z\right]  \right]  }-\underbrace{\left[  \left[
y,z\right]  ,x\right]  }_{=-\left[  x,\left[  y,z\right]  \right]
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.2.jacobi})}\right) \\
&  =-\left[  y,\left[  x,z\right]  \right]  -\left(  -\left[  x,\left[
y,z\right]  \right]  \right)  =\left[  \underbrace{x}_{\in V},\left[
\underbrace{y}_{\in\mathfrak{g}_{I}},\underbrace{z}_{\in\mathfrak{g}_{j}%
}\right]  \right]  -\left[  \underbrace{y}_{\in\mathfrak{g}_{I}},\left[
\underbrace{x}_{\in V},\underbrace{z}_{\in\mathfrak{g}_{j}}\right]  \right] \\
&  \in\left[  V,\underbrace{\left[  \mathfrak{g}_{I},\mathfrak{g}_{j}\right]
}_{\substack{\subset\mathfrak{g}_{I+j}\\\text{(by
(\ref{sol.freelie.2.a.claim.pf.hyp}))}}}\right]  -\left[  \mathfrak{g}%
_{I},\underbrace{\left[  V,\mathfrak{g}_{j}\right]  }_{\substack{=\mathfrak{g}%
_{j+1}\\\text{(since }\mathfrak{g}_{j+1}=\left[  V,\mathfrak{g}_{j}\right]
\text{)}}}\right] \\
&  \subset\underbrace{\left[  V,\mathfrak{g}_{I+j}\right]  }%
_{\substack{=\mathfrak{g}_{I+j+1}\\\text{(since }\mathfrak{g}_{I+j+1}=\left[
V,\mathfrak{g}_{I+j}\right]  \text{)}}}-\underbrace{\left[  \mathfrak{g}%
_{I},\mathfrak{g}_{j+1}\right]  }_{\substack{\subset\mathfrak{g}%
_{I+j+1}\\\text{(by (\ref{sol.freelie.2.a.claim.pf.hyp}), applied to
}j+1\text{ instead of }j\text{)}}}\\
&  \subset\mathfrak{g}_{I+j+1}-\mathfrak{g}_{I+j+1}\subset\mathfrak{g}%
_{I+j+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{g}_{I+j+1}\text{ is
a }\kk\text{-module}\right) \\
&  =\mathfrak{g}_{I+1+j}.
\end{align*}


Now, let us forget that we fixed $z$ and $p$. We thus have proven that
$\left[  p,z\right]  \in\mathfrak{g}_{I+1+j}$ for any $p\in\mathfrak{g}_{I+1}$
and $z\in\mathfrak{g}_{j}$. Thus, $\left[  \mathfrak{g}_{I+1},\mathfrak{g}%
_{j}\right]  \subset\mathfrak{g}_{I+1+j}$ (since $\mathfrak{g}_{I+1+j}$ is a
$\kk$-module).

Let us now forget that we fixed $j$. We thus have proven that, for every
positive integer $j$, we have $\left[  \mathfrak{g}_{I+1},\mathfrak{g}%
_{j}\right]  \subset\mathfrak{g}_{I+1+j}$. In other words,
(\ref{sol.freelie.2.a.claim}) holds for $i=I+1$. This completes the induction
step. The induction proof of (\ref{sol.freelie.2.a.claim}) is thus complete.

Now, for every two positive integers $i$ and $j$, we have%
\begin{align}
\left[  \mathfrak{g}_{i},\mathfrak{g}_{j}\right]   &  \subset\mathfrak{g}%
_{i+j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.2.a.claim}%
)}\right) \nonumber\\
&  \subset\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.gk}), applied to }k=i+j\right)  .
\label{sol.freelie.2.a.alm}%
\end{align}


But $\mathfrak{g}=\sum_{i\geq1}\mathfrak{g}_{i}=\sum_{j\geq1}\mathfrak{g}_{j}$
(here, we have renamed the summation index $i$ as $j$). Thus,%
\begin{align*}
\left[  \underbrace{\mathfrak{g}}_{=\sum_{i\geq1}\mathfrak{g}_{i}%
},\underbrace{\mathfrak{g}}_{=\sum_{j\geq1}\mathfrak{g}_{j}}\right]   &
=\left[  \sum_{i\geq1}\mathfrak{g}_{i},\sum_{j\geq1}\mathfrak{g}_{j}\right] \\
&  \subset\sum_{i\geq1}\sum_{j\geq1}\underbrace{\left[  \mathfrak{g}%
_{i},\mathfrak{g}_{j}\right]  }_{\substack{\subset\mathfrak{g}\\\text{(by
(\ref{sol.freelie.2.a.alm}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie
bracket on }T\left(  V\right)  \text{ is }\kk\text{-bilinear}\right) \\
&  \subset\sum_{i\geq1}\sum_{j\geq1}\mathfrak{g}\subset\mathfrak{g}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{g}\text{ is a }%
\kk\text{-module}\right)  .
\end{align*}
Thus, $\mathfrak{g}$ is a Lie subalgebra of $T\left(  V\right)  $. This solves
Exercise \ref{exe.freelie.2}(a).

(b) Let $\mathfrak{k}$ be any Lie subalgebra of $T\left(  V\right)  $
satisfying $V\subset\mathfrak{k}$. We must prove that $\mathfrak{g}%
\subset\mathfrak{k}$.

We claim that%
\begin{equation}
\mathfrak{g}_{k}\subset\mathfrak{k}\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }k. \label{sol.freelie.2.b.claim}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.2.b.claim}):} We shall prove
(\ref{sol.freelie.2.b.claim}) by induction over $k$:

\textit{Induction base:} We have $\mathfrak{g}_{1}=V\subset\mathfrak{k}$. In
other words, (\ref{sol.freelie.2.b.claim}) holds for $i=1$. This completes the
induction base.

\textit{Induction step:} Let $K$ be a positive integer. Assume that
(\ref{sol.freelie.2.b.claim}) holds for $k=K$. We must prove that
(\ref{sol.freelie.2.b.claim}) holds for $k=K+1$.

We have assumed that (\ref{sol.freelie.2.b.claim}) holds for $k=K$. In other
words, we have $\mathfrak{g}_{K}\subset\mathfrak{k}$.

But $\mathfrak{k}$ is a Lie subalgebra of $T\left(  V\right)  $. Hence,
$\left[  \mathfrak{k},\mathfrak{k}\right]  \subset\mathfrak{k}$.

Now, the recursive definition of $\mathfrak{g}_{K+1}$ yields $\mathfrak{g}%
_{K+1}=\left[  \underbrace{V}_{\subset\mathfrak{k}},\underbrace{\mathfrak{g}%
_{\left(  K+1\right)  -1}}_{=\mathfrak{g}_{K}\subset\mathfrak{k}}\right]
\subset\left[  \mathfrak{k},\mathfrak{k}\right]  \subset\mathfrak{k}$. In
other words, (\ref{sol.freelie.2.b.claim}) holds for $k=K+1$. This completes
the induction step. The induction proof of (\ref{sol.freelie.2.b.claim}) is
thus complete.

Now,%
\[
\mathfrak{g}=\sum_{i\geq1}\underbrace{\mathfrak{g}_{i}}_{\substack{\subset
\mathfrak{k}\\\text{(by (\ref{sol.freelie.2.b.claim}), applied to }%
k=i\text{)}}}\subset\sum_{i\geq1}\mathfrak{k}\subset\mathfrak{k}%
\]
(since $\mathfrak{k}$ is a $\kk$-module (since $\mathfrak{k}$ is a Lie
algebra)). This solves Exercise \ref{exe.freelie.2}(b).

Before we step to the solution of Exercise \ref{exe.freelie.2}(c), we record a
simple fact: If $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$, then%
\begin{equation}
x_{u}x_{v}=x_{uv} \label{sol.freelie.2.c.xuxv}%
\end{equation}
in the $\kk$-algebra $T\left(  V\right)  $%
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.freelie.2.c.xuxv}):} Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$. We have $uv=\left(
\left(  uv\right)  _{1},\left(  uv\right)  _{2},\ldots,\left(  uv\right)
_{\ell\left(  uv\right)  }\right)  $ and thus%
\begin{align*}
\left(  \left(  uv\right)  _{1},\left(  uv\right)  _{2},\ldots,\left(
uv\right)  _{\ell\left(  uv\right)  }\right)   &  =\underbrace{u}_{=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  }\underbrace{v}%
_{=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  }\\
&  =\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  \left(
v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(  u_{1}%
,u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2},\ldots,v_{\ell\left(
v\right)  }\right)
\end{align*}
and therefore%
\[
x_{\left(  uv\right)  _{1}}\otimes x_{\left(  uv\right)  _{2}}\otimes
\cdots\otimes x_{\left(  uv\right)  _{\ell\left(  uv\right)  }}=x_{u_{1}%
}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(  u\right)  }}\otimes
x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes x_{v_{\ell\left(  v\right)  }%
}.
\]
The definition of $x_{uv}$ now yields%
\begin{equation}
x_{uv}=x_{\left(  uv\right)  _{1}}\otimes x_{\left(  uv\right)  _{2}}%
\otimes\cdots\otimes x_{\left(  uv\right)  _{\ell\left(  uv\right)  }%
}=x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(  u\right)
}}\otimes x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes x_{v_{\ell\left(
v\right)  }}. \label{sol.freelie.2.c.xuxv.pf.1}%
\end{equation}
But the definition of $x_{u}$ yields $x_{u}=x_{u_{1}}\otimes x_{u_{2}}%
\otimes\cdots\otimes x_{u_{\ell\left(  u\right)  }}$, and the definition of
$x_{v}$ yields $x_{v}=x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes
x_{v_{\ell\left(  v\right)  }}$. Thus,%
\begin{align*}
&  \underbrace{x_{u}}_{=x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes
x_{u_{\ell\left(  u\right)  }}}\underbrace{x_{v}}_{=x_{v_{1}}\otimes x_{v_{2}%
}\otimes\cdots\otimes x_{v_{\ell\left(  v\right)  }}}\\
&  =\left(  x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(
u\right)  }}\right)  \left(  x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes
x_{v_{\ell\left(  v\right)  }}\right) \\
&  =x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(  u\right)
}}\otimes x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes x_{v_{\ell\left(
v\right)  }}\\
&  =x_{uv}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.c.xuxv.pf.1})}\right)  .
\end{align*}
This proves (\ref{sol.freelie.2.c.xuxv}).}.

(c) We shall solve Exercise \ref{exe.freelie.2}(c) by strong induction on
$\ell\left(  w\right)  $:

\textit{Induction step:} Let $N\in \NN$. Assume that Exercise
\ref{exe.freelie.2}(c) holds under the condition that $\ell\left(  w\right)
<N$. We need to show that Exercise \ref{exe.freelie.2}(c) also holds under the
condition that $\ell\left(  w\right)  =N$.

We have assumed that Exercise \ref{exe.freelie.2}(c) holds under the condition
that $\ell\left(  w\right)  <N$. In other words, we have%
\begin{equation}
b_{w}\in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in
\mathfrak{L}\text{ satisfying }\ell\left(  w\right)  <N.
\label{sol.freelie.2.c.hyp}%
\end{equation}


For every $w\in\mathfrak{L}$ satisfying $\ell\left(  w\right)  <N$, we have%
\begin{align}
b_{w}  &  \in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}\nonumber\\
&  =x_{w}+\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\p>w}}\kk x_{p}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }v\text{ as }p\right) \label{sol.freelie.2.c.hyp2}\\
&  =x_{w}+\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\q>w}}\kk x_{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }p\text{ as }q\right)  . \label{sol.freelie.2.c.hyp3}%
\end{align}
From this, it is easy to conclude the following: For every $w\in\mathfrak{L}$
satisfying $\ell\left(  w\right)  <N$, we have%
\begin{equation}
b_{w}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q} \label{sol.freelie.2.c.hyp5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.hyp5}):} Let $w\in
\mathfrak{L}$ be such that $\ell\left(  w\right)  <N$.
\par
Every $q\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying $q>w$ must also
satisfy $q\geq w$. Thus, $\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
w\right)  };\\q>w}}\kk x_{q}\subset\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\q\geq w}}\kk x_{q}$.
\par
On the other hand, $w$ is an element of $\mathfrak{A}^{\ell\left(  w\right)
}$ (since $\ell\left(  w\right)  =\ell\left(  w\right)  $) and satisfies
$w\geq w$. In other words, $w$ is a $q\in\mathfrak{A}^{\ell\left(  w\right)
}$ satisfying $q\geq w$. Thus, $\kk x_{w}$ is an addend of the sum
$\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$. Hence, $\kk x_{w}\subset\sum\limits_{\substack{q\in
\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq w}}\kk x_{q}$.
\par
Now, $x_{w}\in\kk x_{w}\subset\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\q\geq w}}\kk x_{q}$. But
(\ref{sol.freelie.2.c.hyp3}) becomes%
\[
b_{w}\in\underbrace{x_{w}}_{\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
w\right)  };\\q\geq w}}\kk x_{q}}+\underbrace{\sum\limits_{\substack{q\in
\mathfrak{A}^{\ell\left(  w\right)  };\\q>w}}\kk x_{q}}_{\subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}}\subset\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
w\right)  };\\q\geq w}}\kk x_{q}+\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\q\geq w}}\kk x_{q}\subset\sum\limits
_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq w}%
}\kk x_{q}%
\]
(since $\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.hyp5}).}.

Now, let $w\in\mathfrak{L}$ be such that $\ell\left(  w\right)  =N$. We shall
prove that $b_{w}\in x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(
w\right)  };\\r>w}}\kk x_{r}$.

If $\ell\left(  w\right)  \leq1$, then $b_{w}\in x_{w}+\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$
holds\footnote{\textit{Proof.} Assume that $\ell\left(  w\right)  \leq1$. The
word $w$ is Lyndon (since $w\in\mathfrak{L}$) and thus nonempty. Hence,
$\ell\left(  w\right)  \geq1$. Combined with $\ell\left(  w\right)  \leq1$,
this yields $\ell\left(  w\right)  =1$. In other words, the word $w$ consists
of a single letter. In other words, $w=\left(  a\right)  $ for some
$a\in\mathfrak{A}$. Consider this $a$. The definition of $b_{w}$ then yields
$b_{w}=x_{a}$ (since $\ell\left(  w\right)  =1$ and $w=\left(  a\right)  $).
\par
On the other hand, the definition of $x_{w}$ yields $x_{w}=x_{a}$ (since
$w=\left(  a\right)  $). Compared with $b_{w}=x_{a}$, this yields
\[
b_{w}=x_{w}=x_{w}+\underbrace{0}_{\substack{\in\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}\\\text{(since
}\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r}\text{ is a }\kk\text{-module)}}}\in x_{w}%
+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r},
\]
qed.}. Hence, for the rest of this proof, we can WLOG assume that we don't
have $\ell\left(  w\right)  \leq1$. Assume this.

The word $w$ is Lyndon (since $w\in\mathfrak{L}$) and satisfies $\ell\left(
w\right)  >1$ (since we don't have $\ell\left(  w\right)  \leq1$). Thus, $w$
is a Lyndon word of length $>1$. Therefore, $\operatorname*{stf}w$ is
well-defined. Let $\left(  u,v\right)  =\operatorname*{stf}w$. The recursive
definition of $b_{w}$ yields $b_{w}=\left[  b_{u},b_{v}\right]  $ (since
$\ell\left(  w\right)  >1$ and $\left(  u,v\right)  =\operatorname*{stf}w$).

From Exercise \ref{exe.words.lyndon.stf.basic}(a) (applied to $\left(
g,h\right)  =\left(  u,v\right)  $), we see that $v$ is the longest Lyndon
proper suffix of $w$. In particular, $v$ is a Lyndon proper suffix of $w$.
Also, Exercise \ref{exe.words.lyndon.stf.basic}(d) (applied to $\left(
g,h\right)  =\left(  u,v\right)  $) shows that the word $u$ is Lyndon. From
Exercise \ref{exe.words.lyndon.stf.basic}(c) (applied to $\left(  g,h\right)
=\left(  u,v\right)  $), we obtain $u<uv<v$. Finally, Exercise
\ref{exe.words.lyndon.stf.basic}(b) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that $w=uv$. Exercise \ref{exe.words.lyndon.stf.basic}(e)
(applied to $\left(  g,h\right)  =\left(  u,v\right)  $) shows that
$u\in\mathfrak{L}$, $v\in\mathfrak{L}$, $\ell\left(  u\right)  <\ell\left(
w\right)  $ and $\ell\left(  v\right)  <\ell\left(  w\right)  $.

The word $u$ is Lyndon and thus nonempty. Hence, $\ell\left(  u\right)  \geq
1$. Also, the word $v$ is Lyndon and thus nonempty. Thus, $\ell\left(
v\right)  \geq1$.

Now, $\ell\left(  u\right)  <\ell\left(  w\right)  =N$. Thus,
(\ref{sol.freelie.2.c.hyp2}) (applied to $u$ instead of $w$) shows that%
\begin{equation}
b_{u}\in x_{u}+\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)
};\\p>u}}\kk x_{p}. \label{sol.freelie.2.c.ineq1a}%
\end{equation}
Also, (\ref{sol.freelie.2.c.hyp5}) (applied to $u$ instead of $w$) shows that%
\begin{equation}
b_{u}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  u\right)  };\\q\geq
u}}\kk x_{q}=\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)
};\\p\geq u}}\kk x_{p} \label{sol.freelie.2.c.ineq1b}%
\end{equation}
(here, we renamed the summation index $q$ as $p$).

Also, $\ell\left(  v\right)  <\ell\left(  w\right)  =N$. Thus,
(\ref{sol.freelie.2.c.hyp3}) (applied to $v$ instead of $w$) shows that%
\begin{equation}
b_{v}\in x_{v}+\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q>v}}\kk x_{q}. \label{sol.freelie.2.c.ineq2a}%
\end{equation}
Also, (\ref{sol.freelie.2.c.hyp5}) (applied to $v$ instead of $w$) shows that%
\begin{equation}
b_{v}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\kk x_{q}. \label{sol.freelie.2.c.ineq2b}%
\end{equation}


Now, let $G=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\r>w}}\kk x_{r}$. Thus, $G=\sum\limits_{\substack{r\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$ is a $\kk$-submodule
of $T\left(  V\right)  $.

If $p\in\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in\mathfrak{A}%
^{\ell\left(  v\right)  }$ are such that $q\geq v$, then%
\begin{equation}
x_{q}x_{p}\in G \label{sol.freelie.2.c.bound1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound1}):} Let $p\in
\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in\mathfrak{A}^{\ell\left(
v\right)  }$ be such that $q\geq v$.
\par
We have $\ell\left(  qp\right)  =\underbrace{\ell\left(  q\right)
}_{\substack{=\ell\left(  v\right)  \\\text{(since }q\in\mathfrak{A}%
^{\ell\left(  v\right)  }\text{)}}}+\underbrace{\ell\left(  p\right)
}_{\substack{=\ell\left(  u\right)  \\\text{(since }p\in\mathfrak{A}%
^{\ell\left(  u\right)  }\text{)}}}=\ell\left(  v\right)  +\ell\left(
u\right)  =\ell\left(  u\right)  +\ell\left(  v\right)  $. Compared with
$\ell\left(  w\right)  =\ell\left(  u\right)  +\ell\left(  v\right)  $, this
yields $\ell\left(  qp\right)  =\ell\left(  w\right)  $. In other words,
$qp\in\mathfrak{A}^{\ell\left(  w\right)  }$.
\par
But the word $w$ is Lyndon and satisfies $w=uv$. Thus, Proposition
\ref{prop.words.lyndon.trivia}(a) shows that $v\geq w$ (since $v$ is
nonempty). Since $\ell\left(  v\right)  \neq\ell\left(  w\right)  $ (because
$\ell\left(  v\right)  <\ell\left(  w\right)  $), we have $v\neq w$. Combined
with $v\geq w$, this yields $v>w$. Now, $qp\geq q\geq v>w$.
\par
So we know that $qp\in\mathfrak{A}^{\ell\left(  w\right)  }$ and $qp>w$. In
other words, $qp$ is an $r\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying
$r>w$. Thus, $\kk x_{qp}$ is an addend of the sum $\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$. Therefore,
$\kk x_{qp}\subset\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(
w\right)  };\\r>w}}\kk x_{r}=G$ (since $G=\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$).
\par
But (\ref{sol.freelie.2.c.xuxv}) (applied to $q$ and $p$ instead of $u$ and
$v$) shows that $x_{q}x_{p}=x_{qp}\in\kk x_{qp}\subset G$. This proves
(\ref{sol.freelie.2.c.bound1}).}. Hence,%
\begin{equation}
b_{v}b_{u}\in G \label{sol.freelie.2.c.bound1b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound1b}):} We have%
\begin{align*}
&  \underbrace{b_{v}}_{\substack{\in\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  v\right)  };\\q\geq v}}\kk x_{q}\\\text{(by
(\ref{sol.freelie.2.c.ineq2b}))}}}\underbrace{b_{u}}_{\substack{\in
\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq
u}}\kk x_{p}\\\text{(by (\ref{sol.freelie.2.c.ineq1b}))}}}\\
&  =\left(  \sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\kk x_{q}\right)  \left(  \sum\limits_{\substack{p\in\mathfrak{A}%
^{\ell\left(  u\right)  };\\p\geq u}}\kk x_{p}\right)  \subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq v}%
}\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq
u}}\underbrace{\kk\kk}_{\subset\kk}\underbrace{x_{q}x_{p}%
}_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound1}))}}}\subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq v}%
}\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq
u}}\kk G\subset G
\end{align*}
(since $G$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.bound1b}).}.

Furthermore, if $p\in\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in
\mathfrak{A}^{\ell\left(  v\right)  }$ are such that $p>u$, then
\begin{equation}
x_{p}x_{q}\in G \label{sol.freelie.2.c.bound2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound2}):} Let $p\in
\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in\mathfrak{A}^{\ell\left(
v\right)  }$ be such that $p>u$.
\par
We have $\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)
}_{\substack{=\ell\left(  u\right)  \\\text{(since }p\in\mathfrak{A}%
^{\ell\left(  u\right)  }\text{)}}}+\underbrace{\ell\left(  q\right)
}_{\substack{=\ell\left(  v\right)  \\\text{(since }q\in\mathfrak{A}%
^{\ell\left(  v\right)  }\text{)}}}=\ell\left(  u\right)  +\ell\left(
v\right)  $. Compared with $\ell\left(  w\right)  =\ell\left(  u\right)
+\ell\left(  v\right)  $, this yields $\ell\left(  pq\right)  =\ell\left(
w\right)  $. In other words, $pq\in\mathfrak{A}^{\ell\left(  w\right)  }$.
\par
We have $\ell\left(  p\right)  =\ell\left(  u\right)  $ (since $p\in
\mathfrak{A}^{\ell\left(  u\right)  }$), so that $\ell\left(  u\right)
=\ell\left(  p\right)  $.
\par
Assume (for the sake of contradiction) that the word $u$ is a prefix of $p$.
Since the word $u$ has the same length as $p$ (since $\ell\left(  u\right)
=\ell\left(  p\right)  $), this shows that $u=p$. Thus, $u=p>u$, which is
absurd. This contradiction shows that our assumption (that the word $u$ is a
prefix of $p$) was false. In other words, the word $u$ is not a prefix of $p$.
\par
We have $u<p$ (since $p>u$), thus $u\leq p$. Thus, Proposition
\ref{prop.words.lex}(d) (applied to $u$, $v$, $p$ and $\varnothing$ instead of
$a$, $b$, $c$ and $d$) shows that either we have $uv\leq p\varnothing$ or the
word $u$ is a prefix of $p$. Since the word $u$ is not a prefix of $p$, we
thus conclude that $uv\leq p\varnothing$. Hence, $w=uv\leq p\varnothing=p$.
\par
Now, $q\in\mathfrak{A}^{\ell\left(  v\right)  }$, so that $\ell\left(
q\right) = \ell\left(  v\right)  \geq1$. The word $q$ is thus nonempty.
Hence, $p<pq$. Thus, $w\leq p<pq$, so that $pq>w$.
\par
So we know that $pq\in\mathfrak{A}^{\ell\left(  w\right)  }$ and $pq>w$. In
other words, $pq$ is an $r\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying
$r>w$. Thus, $\kk x_{pq}$ is an addend of the sum $\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$. Therefore,
$\kk x_{pq}\subset\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(
w\right)  };\\r>w}}\kk x_{r}=G$ (since $G=\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$).
\par
But (\ref{sol.freelie.2.c.xuxv}) (applied to $p$ and $q$ instead of $u$ and
$v$) shows that $x_{p}x_{q}=x_{pq}\in\kk x_{pq}\subset G$. This proves
(\ref{sol.freelie.2.c.bound2}).}. Hence,
\begin{equation}
\left(  b_{u}-x_{u}\right)  b_{v}\in G \label{sol.freelie.2.c.bound2b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound2b}):} Subtracting
$x_{u}$ from both sides of the relation (\ref{sol.freelie.2.c.ineq1a}), we
obtain%
\[
b_{u}-x_{u}\in\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  }%
;\\p>u}}\kk x_{p}.
\]
Now,%
\begin{align*}
&  \underbrace{\left(  b_{u}-x_{u}\right)  }_{\in\sum\limits_{\substack{p\in
\mathfrak{A}^{\ell\left(  u\right)  };\\p>u}}\kk x_{p}}\underbrace{b_{v}%
}_{\substack{\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q\geq v}}\kk x_{q}\\\text{(by (\ref{sol.freelie.2.c.ineq2b}))}}}\\
&  \in\left(  \sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  }%
;\\p>u}}\kk x_{p}\right)  \left(  \sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  v\right)  };\\q\geq v}}\kk x_{q}\right)  \subset
\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p>u}%
}\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\underbrace{\kk\kk}_{\subset\kk}\underbrace{x_{p}x_{q}%
}_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound2}))}}}\subset
\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p>u}%
}\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\kk G\subset G
\end{align*}
(since $G$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.bound2b}).}.

Furthermore, if $q\in\mathfrak{A}^{\ell\left(  v\right)  }$ is such that
$q>v$, then%
\begin{equation}
x_{u}x_{q}\in G \label{sol.freelie.2.c.bound3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound3}):} Let $q\in
\mathfrak{A}^{\ell\left(  v\right)  }$ be such that $q>v$.
\par
We have $\ell\left(  uq\right)  =\ell\left(  u\right)  +\underbrace{\ell
\left(  q\right)  }_{\substack{=\ell\left(  v\right)  \\\text{(since }%
q\in\mathfrak{A}^{\ell\left(  v\right)  }\text{)}}}=\ell\left(  u\right)
+\ell\left(  v\right)  $. Compared with $\ell\left(  w\right)  =\ell\left(
u\right)  +\ell\left(  v\right)  $, this yields $\ell\left(  uq\right)
=\ell\left(  w\right)  $. In other words, $uq\in\mathfrak{A}^{\ell\left(
w\right)  }$.
\par
We have $q>v$, thus $v<q$, thus $v\leq q$. Hence, Proposition
\ref{prop.words.lex}(b) (applied to $u$, $v$ and $q$ instead of $a$, $c$ and
$d$) shows that $uv\leq uq$. If we had $uv=uq$, then we would have $v=q$
(since we could cancel $u$ from the equality $uv=uq$), which would contradict
$v<q$. Thus, we cannot have $uv=uq$. Hence, we have $uv\neq uq$. Combined with
$uv\leq uq$, this shows that $uv<uq$. In other words, $uq>uv=w$ (since
$w=uv$).
\par
So we know that $uq\in\mathfrak{A}^{\ell\left(  w\right)  }$ and $uq>w$. In
other words, $uq$ is an $r\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying
$r>w$. Thus, $\kk x_{uq}$ is an addend of the sum $\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$. Therefore,
$\kk x_{uq}\subset\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(
w\right)  };\\r>w}}\kk x_{r}=G$ (since $G=\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$).
\par
But (\ref{sol.freelie.2.c.xuxv}) (applied to $u$ and $q$ instead of $u$ and
$v$) shows that $x_{u}x_{q}=x_{uq}\in\kk x_{uq}\subset G$.
This proves (\ref{sol.freelie.2.c.bound3}).}. Hence,%
\begin{equation}
x_{u}\left(  b_{v}-x_{v}\right)  \in G \label{sol.freelie.2.c.bound3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound3b}):} Subtracting
$x_{v}$ from both sides of the relation (\ref{sol.freelie.2.c.ineq2a}), we
obtain%
\[
b_{v}-x_{v}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  }%
;\\q>v}}\kk x_{q}.
\]
Now,%
\[
x_{u}\underbrace{\left(  b_{v}-x_{v}\right)  }_{\in\sum\limits_{\substack{q\in
\mathfrak{A}^{\ell\left(  v\right)  };\\q>v}}\kk x_{q}}\in x_{u}\left(
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q>v}%
}\kk x_{q}\right)  \subset\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
v\right)  };\\q>v}}\kk\underbrace{x_{u}x_{q}}_{\substack{\in
G\\\text{(by (\ref{sol.freelie.2.c.bound3}))}}}\subset\sum\limits_{\substack{q\in
\mathfrak{A}^{\ell\left(  v\right)  };\\q>v}}\kk G\subset G
\]
(since $G$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.bound3b}).}.

Now,
\begin{align*}
b_{w}  &  =\left[  b_{u},b_{v}\right]  =b_{u}b_{v}-b_{v}b_{u}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the Lie bracket on
}T\left(  V\right)  \right) \\
&  =\underbrace{\left(  b_{u}-x_{u}\right)  b_{v}}_{\substack{\in G\\\text{(by
(\ref{sol.freelie.2.c.bound2b}))}}}+\underbrace{x_{u}\left(  b_{v}%
-x_{v}\right)  }_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound3b}%
))}}}+x_{u}x_{v}-\underbrace{b_{v}b_{u}}_{\substack{\in G\\\text{(by
(\ref{sol.freelie.2.c.bound1b}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by straightforward computation}\right) \\
&  \in G+G+x_{u}x_{v}-G=\underbrace{x_{u}x_{v}}_{\substack{=x_{uv}\\\text{(by
(\ref{sol.freelie.2.c.xuxv}))}}}+\underbrace{G+G-G}_{\substack{\subset
G\\\text{(since }G\text{ is a }\kk\text{-module)}}}\subset
\underbrace{x_{uv}}_{\substack{=x_{w}\\\text{(since }uv=w\text{)}%
}}+\underbrace{G}_{=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)
};\\r>w}}\kk x_{r}}\\
&  =x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\r>w}}\kk x_{r}.
\end{align*}


Thus, $b_{w}\in x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)
};\\r>w}}\kk x_{r}$ is proven.

Let us now forget that we defined $\left(  u,v\right)  $. We have%
\[
b_{w}\in x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)
};\\r>w}}\kk x_{r}=x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(
w\right)  };\\v>w}}\kk x_{v}%
\]
(here, we have renamed the summation index $r$ as $v$).

Now, let us forget that we fixed $w$. We thus have proven that%
\[
b_{w}\in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in
\mathfrak{L}\text{ satisfying }\ell\left(  w\right)  =N.
\]
In other words, we have proven that Exercise \ref{exe.freelie.2}(c) holds
under the condition that $\ell\left(  w\right)  =N$. Thus, our induction is
complete, and Exercise \ref{exe.freelie.2}(c) is solved.

(d) We know that $\mathfrak{g}$ is a Lie subalgebra of $T\left(  V\right)  $
(by Exercise \ref{exe.freelie.2}(a)). Thus, $\left[  \mathfrak{g}%
,\mathfrak{g}\right]  \subset\mathfrak{g}$.

We first notice that
\begin{equation}
b_{w}\in\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}.
\label{sol.freelie.2.d.1}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.2.d.1}):} We shall prove
(\ref{sol.freelie.2.d.1}) by strong induction on $\ell\left(  w\right)  $:

\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{sol.freelie.2.d.1}) holds under the condition that $\ell\left(
w\right)  <N$. We need to show that (\ref{sol.freelie.2.d.1}) also holds under
the condition that $\ell\left(  w\right)  =N$.

We have assumed that (\ref{sol.freelie.2.d.1}) holds under the condition that
$\ell\left(  w\right)  <N$. In other words, we have%
\begin{equation}
b_{w}\in\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}%
\text{ satisfying }\ell\left(  w\right)  <N. \label{sol.freelie.2.d.1.pf.hyp}%
\end{equation}


Now, let $w\in\mathfrak{L}$ be such that $\ell\left(  w\right)  =N$. We shall
prove that $b_{w}\in\mathfrak{g}$.

If $\ell\left(  w\right)  \leq1$, then $b_{w}\in\mathfrak{g}$
holds\footnote{\textit{Proof.} Assume that $\ell\left(  w\right)  \leq1$. The
word $w$ is Lyndon (since $w\in\mathfrak{L}$) and thus nonempty. Hence,
$\ell\left(  w\right)  \geq1$. Combined with $\ell\left(  w\right)  \leq1$,
this yields $\ell\left(  w\right)  =1$. In other words, the word $w$ consists
of a single letter. In other words, $w=\left(  a\right)  $ for some
$a\in\mathfrak{A}$. Consider this $a$. The definition of $b_{w}$ then yields
$b_{w}=x_{a}$ (since $\ell\left(  w\right)  =1$ and $w=\left(  a\right)  $).
\par
Now, $b_{w}=x_{a}\in V\subset\mathfrak{g}$, qed.}. Hence, for the rest of this
proof, we can WLOG assume that we don't have $\ell\left(  w\right)  \leq1$.
Assume this.

The word $w$ is Lyndon (since $w\in\mathfrak{L}$) and satisfies $\ell\left(
w\right)  >1$ (since we don't have $\ell\left(  w\right)  \leq1$). Thus, $w$
is a Lyndon word of length $>1$. Therefore, $\operatorname*{stf}w$ is
well-defined. Let $\left(  u,v\right)  =\operatorname*{stf}w$. The recursive
definition of $b_{w}$ yields $b_{w}=\left[  b_{u},b_{v}\right]  $ (since
$\ell\left(  w\right)  >1$ and $\left(  u,v\right)  =\operatorname*{stf}w$).

Exercise \ref{exe.words.lyndon.stf.basic}(e) (applied to $\left(  g,h\right)
=\left(  u,v\right)  $) shows that $u\in\mathfrak{L}$, $v\in\mathfrak{L}$,
$\ell\left(  u\right)  <\ell\left(  w\right)  $ and $\ell\left(  v\right)
<\ell\left(  w\right)  $.

Now, $\ell\left(  u\right)  <\ell\left(  w\right)  =N$. Thus,
(\ref{sol.freelie.2.d.1.pf.hyp}) (applied to $u$ instead of $w$) shows that
$b_{u}\in\mathfrak{g}$.

Also, $\ell\left(  v\right)  <\ell\left(  w\right)  =N$. Thus,
(\ref{sol.freelie.2.d.1.pf.hyp}) (applied to $v$ instead of $w$) shows that
$b_{v}\in\mathfrak{g}$.

Now, $b_{w}=\left[  \underbrace{b_{u}}_{\in\mathfrak{g}},\underbrace{b_{v}%
}_{\in\mathfrak{g}}\right]  \in\left[  \mathfrak{g},\mathfrak{g}\right]
\subset\mathfrak{g}$.

Let us now forget that we fixed $w$. We thus have shown that $b_{w}%
\in\mathfrak{g}$ for every $w\in\mathfrak{L}$ satisfying $\ell\left(
w\right)  =N$. In other words, (\ref{sol.freelie.2.d.1}) holds under the
condition that $\ell\left(  w\right)  =N$. This completes the induction step.
The induction proof of (\ref{sol.freelie.2.d.1}) is thus complete.

Now, (\ref{sol.freelie.2.d.1}) shows that $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a family of elements of $\mathfrak{g}$. Using Exercise
\ref{exe.freelie.1}(a), it is easy to conclude that this family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk$-module
$\mathfrak{g}$\ \ \ \ \footnote{\textit{Proof.} Let $B$ be the $\kk%
$-submodule of $\mathfrak{g}$ spanned by the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ (this is well-defined since $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a family of elements of $\mathfrak{g}$). Then,
clearly,%
\begin{equation}
b_{w}\in B\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}.
\label{sol.freelie.2.d.span.pf.1}%
\end{equation}
\par
Whenever $w$ is a Lyndon word of length $>1$, we have%
\[
b_{w}=\left[  b_{u},b_{v}\right]  ,\ \ \ \ \ \ \ \ \ \ \text{where }\left(
u,v\right)  =\operatorname*{stf}w
\]
(due to the recursive definition of $b_{w}$). Thus, Exercise
\ref{exe.freelie.1}(a) shows that $B$ is a Lie subalgebra of $\mathfrak{g}$.
Hence, $B$ is a Lie subalgebra of $T\left(  V\right)  $.
\par
Recall that $\left(  x_{a}\right)  _{a\in\mathfrak{A}}$ is a basis of the
$\kk$-module $V$. Hence,%
\begin{equation}
\left(  \text{the }\kk\text{-linear span of the family }\left(
x_{a}\right)  _{a\in\mathfrak{A}}\right)  =V.
\label{sol.freelie.2.d.span.pf.2}%
\end{equation}
\par
Now, let $a\in\mathfrak{A}$. We shall show that $x_{a}\in B$.
\par
Indeed, let $w$ be the one-letter word $\left(  a\right)  $. Then, $w$ is
Lyndon (since $w$ is a one-letter word) and has length $1$. Thus, $b_{w}%
=x_{a}$ (by the definition of $b_{w}$, since $w=\left(  a\right)  $). Hence,
$x_{a}=b_{w}\in B$ (by (\ref{sol.freelie.2.d.span.pf.1})).
\par
Now, let us forget that we fixed $a$. We thus have shown that $x_{a}\in B$ for
every $a\in\mathfrak{A}$. Since $B$ is a $\kk$-module, this entails
that $\left(  \text{the }\kk\text{-linear span of the family }\left(
x_{a}\right)  _{a\in\mathfrak{A}}\right)  \subset B$. Because of
(\ref{sol.freelie.2.d.span.pf.2}), this rewrites as $V\subset B$.
\par
Now, we know that $B$ is a Lie subalgebra of $T\left(  V\right)  $ satisfying
$V\subset B$. Thus, $\mathfrak{g}\subset B$ (by Exercise \ref{exe.freelie.2}%
(b), applied to $\mathfrak{k}=B$). Combined with $B\subset\mathfrak{g}$ (since
$B$ is a $\kk$-submodule of $\mathfrak{g}$), this yields $\mathfrak{g}%
=B$. But the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the
$\kk$-module $B$ (since $B$ is the $\kk$-submodule of
$\mathfrak{g}$ spanned by the family $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$). Since $\mathfrak{g}=B$, this rewrites as follows: The family
$\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk$-module
$\mathfrak{g}$. Qed.}.

We shall now show that the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$
is $\kk$-linearly independent. In order to do so, we will prove a more
general result:

\begin{proposition}
\label{prop.sol.freelie.2.d.general}Let $\mathfrak{V}$ be a $\kk%
$-module, and let $W$ be a totally ordered set. Let $\left(  p_{w}\right)
_{w\in W}$ be a $\kk$-linearly independent family of elements of
$\mathfrak{V}$. Let $L$ be a subset of $W$. Let $\left(  s_{w}\right)  _{w\in
L}$ be a family of elements of $\mathfrak{V}$. Assume that every $w\in L$
satisfies%
\begin{equation}
s_{w}\in p_{w}+\sum\limits_{\substack{v\in W;\\v>w}}\kk p_{v}.
\label{pf.prop.sol.freelie.2.d.general.premise}%
\end{equation}
(Here, the ``$>$'' sign under the sum refers
to the total order on $W$.) Then, the family $\left(  s_{w}\right)  _{w\in L}$
is $\kk$-linearly independent.
\end{proposition}

Proposition \ref{prop.sol.freelie.2.d.general} is actually a standard
criterion for linear independence. It is often summarized by the motto
``vectors with distinct leading coordinates are linearly
independent''. For the sake of completeness, we shall
nevertheless prove it. First, let us state an obvious lemma:

\begin{lemma}
\label{lem.sol.freelie.2.d.general.lem2}Let $\mathfrak{V}$ be a $\kk%
$-module, and let $W$ be a set. Let $\left(  p_{w}\right)  _{w\in W}$ be a
$\kk$-linearly independent family of elements of $\mathfrak{V}$. Let
$x\in W$ and $\lambda\in\kk$ be such that%
\begin{equation}
0\in\lambda p_{x}+\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}.
\label{pf.prop.sol.freelie.2.d.general.lem.premise}%
\end{equation}
Then, $\lambda=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.freelie.2.d.general.lem2}.]The equality
(\ref{pf.prop.sol.freelie.2.d.general.lem.premise}) shows that we can write
$0$ as a $\kk$-linear combination of the elements $p_{v}$ with $v\in W$
in such a way that the coefficient in front of $p_{x}$ is $\lambda$. But since
$\left(  p_{w}\right)  _{w\in W}$ is $\kk$-linearly independent, the
only such linear combination which gives $0$ is the trivial one (i.e., the one
where all the coefficients are $0$). Hence, all coefficients in our linear
combination are $0$. In particular, the coefficient in front of $p_{x}$ is
$0$. Since this coefficient is $\lambda$, this means that $\lambda=0$. This
proves Lemma \ref{lem.sol.freelie.2.d.general.lem2}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.sol.freelie.2.d.general}.]Let us first prove
the following fact: If $S$ is a finite subset of $L$, then
\begin{equation}
\text{the family }\left(  s_{w}\right)  _{w\in S}\text{ is }\kk%
\text{-linearly independent.} \label{pf.prop.sol.freelie.2.d.general.1}%
\end{equation}


\textit{Proof of (\ref{pf.prop.sol.freelie.2.d.general.1}):} We shall prove
(\ref{pf.prop.sol.freelie.2.d.general.1}) by induction over $\left\vert
S\right\vert $:

\textit{Induction base:} If $S$ is a finite subset of $L$ satisfying
$\left\vert S\right\vert =0$, then the family $\left(  s_{w}\right)  _{w\in
S}$ is $\kk$-linearly independent (because $\left\vert S\right\vert =0$
entails $S=\varnothing$, and thus the family $\left(  s_{w}\right)  _{w\in S}$
is empty). In other words, (\ref{pf.prop.sol.freelie.2.d.general.1}) holds in
the case when $\left\vert S\right\vert =0$. This completes the induction base.

\textit{Induction step:} Let $K$ be a positive integer. Assume that
(\ref{pf.prop.sol.freelie.2.d.general.1}) is proven in the case when
$\left\vert S\right\vert =K-1$. We must prove that
(\ref{pf.prop.sol.freelie.2.d.general.1}) holds in the case when $\left\vert
S\right\vert =K$.

We have assumed that (\ref{pf.prop.sol.freelie.2.d.general.1}) is proven in
the case when $\left\vert S\right\vert =K-1$. In other words, if $S$ is a
finite subset of $L$ satisfying $\left\vert S\right\vert =K-1$, then
\begin{equation}
\text{the family }\left(  s_{w}\right)  _{w\in S}\text{ is }\kk%
\text{-linearly independent.} \label{pf.prop.sol.freelie.2.d.general.1.pf.hyp}%
\end{equation}


Now, let $S$ be a finite subset of $L$ satisfying $\left\vert S\right\vert
=K$. Let $\left(  \lambda_{w}\right)  _{w\in S}\in\kk^{S}$ be a family
of elements of $\kk$ such that $\sum_{w\in S}\lambda_{w}s_{w}=0$. We
shall prove that $\left(  \lambda_{w}\right)  _{w\in S}=\left(  0\right)
_{w\in S}$.

We have $\left\vert S\right\vert =K>0$. Thus, the set $S$ is nonempty. Also,
$S\subset L\subset W$, and thus the set $S$ is totally ordered (since it is a
subset of the totally ordered set $W$). Therefore, this set $S$ has a smallest
element (since every nonempty finite totally ordered set has a smallest
element). Let $x$ be this smallest element. Thus,%
\begin{equation}
x\leq w\ \ \ \ \ \ \ \ \ \ \text{for every }w\in S
\label{pf.prop.sol.freelie.2.d.general.1.pf.1}%
\end{equation}
(by the definition of the smallest element). Also, $x\in S$ (since $x$ is the
smallest element of $S$), and thus $\left\vert S\setminus\left\{  x\right\}
\right\vert =\underbrace{\left\vert S\right\vert }_{=K}-1=K-1$. Hence, we can
apply (\ref{pf.prop.sol.freelie.2.d.general.1.pf.hyp}) to $S\setminus\left\{
x\right\}  $ instead of $S$. As a result, we see that the family $\left(
s_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }$ is $\kk$-linearly
independent. In other words, if $\left(  \mu_{w}\right)  _{w\in S\setminus
\left\{  x\right\}  }\in\kk^{S\setminus\left\{  x\right\}  }$ is a
family of elements of $\kk$ such that $\sum_{w\in S\setminus\left\{
x\right\}  }\mu_{w}s_{w}=0$, then
\begin{equation}
\left(  \mu_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }=\left(
0\right)  _{w\in S\setminus\left\{  x\right\}  }.
\label{pf.prop.sol.freelie.2.d.general.1.pf.hyp-used}%
\end{equation}


On the other hand, every $w\in S\setminus\left\{  x\right\}  $ satisfies%
\begin{equation}
s_{w}\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}
\label{pf.prop.sol.freelie.2.d.general.1.pf.bound1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound1}%
):} Let $w\in S\setminus\left\{  x\right\}  $. Thus, $w\in S$ and $w\neq x$.
\par
Let $v\in\left\{  q\in W\ \mid\ q>w\right\}  $. We shall show that $v\in
W\setminus\left\{  x\right\}  $.
\par
Indeed, we have $v\in\left\{  q\in W\ \mid\ q>w\right\}  $. In other words,
$v$ is an element of $W$ and satisfies $v>w$. But $w\in S\setminus\left\{
x\right\}  \subset S$, so that (\ref{pf.prop.sol.freelie.2.d.general.1.pf.1})
shows that $x\leq w$. Hence, $w\geq x$, so that $v>w\geq x$. Thus, $v\neq x$.
Combining $v\in W$ with $v\neq x$, we obtain $v\in W\setminus\left\{
x\right\}  $.
\par
Let us now forget that we fixed $v$. We thus have proven that every
$v\in\left\{  q\in W\ \mid\ q>w\right\}  $ satisfies $v\in W\setminus\left\{
x\right\}  $. In other words, $\left\{  q\in W\ \mid\ q>w\right\}  \subset
W\setminus\left\{  x\right\}  $. Therefore, $\sum_{v\in\left\{  q\in
W\ \mid\ q>w\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus\left\{
x\right\}  }\kk p_{v}$.
\par
Also, combining $w\in W$ with $w\neq x$, we obtain $w\in W\setminus\left\{
x\right\}  $. Hence, $\kk p_{w}$ is an addend of the sum $\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}$. Thus, $\kk p_{w}%
\subset\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}$. Now,
$p_{w}\in\kk p_{w}\subset\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}$.
\par
Now, $w\in S\subset L\subset W$. Hence,
(\ref{pf.prop.sol.freelie.2.d.general.premise}) yields%
\begin{align*}
s_{w}  &  \in\underbrace{p_{w}}_{\in\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}}+\underbrace{\sum\limits_{\substack{v\in W;\\v>w}}}_{=\sum
_{v\in\left\{  q\in W\ \mid\ q>w\right\}  }}\kk p_{v}\subset\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}+\underbrace{\sum_{v\in\left\{
q\in W\ \mid\ q>w\right\}  }\kk p_{v}}_{\subset\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}}\\
&  \subset\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}+\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}%
\end{align*}
(since $\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}$ is a
$\kk$-module). This proves
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound1}).}. Also,
\begin{equation}
s_{x}-p_{x}\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}
\label{pf.prop.sol.freelie.2.d.general.1.pf.bound2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound2}%
):} Let $v\in\left\{  q\in W\ \mid\ q>x\right\}  $. We shall show that $v\in
W\setminus\left\{  x\right\}  $.
\par
Indeed, we have $v\in\left\{  q\in W\ \mid\ q>x\right\}  $. In other words,
$v$ is an element of $W$ and satisfies $v>x$. Hence, $v\neq x$ (since $v>x$).
Combining $v\in W$ with $v\neq x$, we obtain $v\in W\setminus\left\{
x\right\}  $.
\par
Let us now forget that we fixed $v$. We thus have proven that every
$v\in\left\{  q\in W\ \mid\ q>x\right\}  $ satisfies $v\in W\setminus\left\{
x\right\}  $. In other words, $\left\{  q\in W\ \mid\ q>x\right\}  \subset
W\setminus\left\{  x\right\}  $. Therefore, $\sum_{v\in\left\{  q\in
W\ \mid\ q>x\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus\left\{
x\right\}  }\kk p_{v}$.
\par
Now, $x\in S\subset L\subset W$. Hence,
(\ref{pf.prop.sol.freelie.2.d.general.premise}) (applied to $w=x$) yields%
\[
s_{x}\in p_{x}+\sum\limits_{\substack{v\in W;\\v>x}}\kk p_{v}.
\]
Subtracting $p_{x}$ from both sides of this relation, we obtain%
\[
s_{x}-p_{x}\in\underbrace{\sum\limits_{\substack{v\in W;\\v>x}}}_{=\sum_{v\in\left\{
q\in W\ \mid\ q>x\right\}  }}\kk p_{v}=\sum_{v\in\left\{  q\in
W\ \mid\ q>x\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus\left\{
x\right\}  }\kk p_{v}.
\]
This proves (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound2}).}. Now,
$\sum_{w\in S}\lambda_{w}s_{w}=0$, so that%
\begin{align*}
0  &  =\sum_{w\in S}\lambda_{w}s_{w}=\lambda_{x}\underbrace{s_{x}}%
_{=p_{x}+\left(  s_{x}-p_{x}\right)  }+\underbrace{\sum\limits_{\substack{w\in
S;\\w\neq x}}}_{=\sum_{w\in S\setminus\left\{  x\right\}  }}\lambda_{w}%
s_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }x\in S\right) \\
&  =\underbrace{\lambda_{x}\left(  p_{x}+\left(  s_{x}-p_{x}\right)  \right)
}_{=\lambda_{x}p_{x}+\lambda_{x}\left(  s_{x}-p_{x}\right)  }+\sum_{w\in
S\setminus\left\{  x\right\}  }\lambda_{w}\underbrace{s_{w}}_{\substack{\in
\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}\\\text{(by
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound1}))}}}\\
&  \in\lambda_{x}p_{x}+\lambda_{x}\underbrace{\left(  s_{x}-p_{x}\right)
}_{\substack{\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk%
p_{v}\\\text{(by (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound2}))}%
}}+\underbrace{\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}}_{\substack{\subset\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\\\text{(since }\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\text{ is a }\kk%
\text{-module)}}}\\
&  \subset\lambda_{x}p_{x}+\underbrace{\lambda_{x}\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}+\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}}_{\substack{\subset\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}\\\text{(since }\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}\text{ is a }\kk\text{-module)}}}\subset\lambda
_{x}p_{x}+\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}.
\end{align*}
Lemma \ref{lem.sol.freelie.2.d.general.lem2} (applied to $\lambda=\lambda_{x}%
$) thus shows that $\lambda_{x}=0$. Thus,%
\[
0=\underbrace{\lambda_{x}}_{=0}s_{x}+\underbrace{\sum\limits_{\substack{w\in
S;\\w\neq x}}}_{=\sum_{w\in S\setminus\left\{  x\right\}  }}\lambda_{w}%
s_{w}=\underbrace{0s_{x}}_{=0}+\sum_{w\in S\setminus\left\{  x\right\}
}\lambda_{w}s_{w}=\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}s_{w}.
\]
Thus, the family $\left(  \lambda_{w}\right)  _{w\in S\setminus\left\{
x\right\}  }\in\kk^{S\setminus\left\{  x\right\}  }$ satisfies
$\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}s_{w}=0$. Hence,
$\left(  \lambda_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }=\left(
0\right)  _{w\in S\setminus\left\{  x\right\}  }$ (according to
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.hyp-used}), applied to $\left(
\mu_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }=\left(  \lambda
_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }$). In other words,%
\begin{equation}
\lambda_{w}=0\ \ \ \ \ \ \ \ \ \ \text{for every }w\in S\setminus\left\{
x\right\}  . \label{pf.prop.sol.freelie.2.d.general.1.pf.alm1}%
\end{equation}
Combining this with the fact that $\lambda_{x}=0$, we conclude that
$\lambda_{w}=0$ for every $w\in S$. In other words, $\left(  \lambda
_{w}\right)  _{w\in S}=\left(  0\right)  _{w\in S}$.

Now, let us forget that we fixed $\left(  \lambda_{w}\right)  _{w\in S}$. We
thus have shown that if $\left(  \lambda_{w}\right)  _{w\in S}\in
\kk^{S}$ is a family of elements of $\kk$ such that $\sum_{w\in
S}\lambda_{w}s_{w}=0$, then $\left(  \lambda_{w}\right)  _{w\in S}=\left(
0\right)  _{w\in S}$. In other words, the family $\left(  s_{w}\right)  _{w\in
S}$ is $\kk$-linearly independent.

Let us now forget that we fixed $S$. We thus have proven that if $S$ is a
finite subset of $L$ satisfying $\left\vert S\right\vert =K$, then the family
$\left(  s_{w}\right)  _{w\in S}$ is $\kk$-linearly independent. In
other words, (\ref{pf.prop.sol.freelie.2.d.general.1}) holds in the case when
$\left\vert S\right\vert =K$. This completes the induction step. Thus, we have
proven (\ref{pf.prop.sol.freelie.2.d.general.1}) by induction.

Now, a family $\mathfrak{f}$ of vectors in a $\kk$-module is
$\kk$-linearly independent if every finite subfamily of $\mathfrak{f}$
is $\kk$-linearly independent (because every linear dependence relation
for $\mathfrak{f}$ has only finitely many nonzero coefficients, and thus can
be recast as a linear dependence relation for some finite subfamily of
$\mathfrak{f}$). Now, (\ref{pf.prop.sol.freelie.2.d.general.1}) shows that
every finite subfamily of the family $\left(  s_{w}\right)  _{w\in L}$ is
$\kk$-linearly independent; therefore, the previous sentence shows that
the family $\left(  s_{w}\right)  _{w\in L}$ is $\kk$-linearly
independent. This proves Proposition \ref{prop.sol.freelie.2.d.general}.
\end{proof}

Let us now return to the solution of Exercise \ref{exe.freelie.2}(d). The set
$\mathfrak{A}^{\ast}$ is totally ordered (by the lexicographic order), and the
set $\mathfrak{L}$ is a subset of $\mathfrak{A}^{\ast}$. The family $\left(
x_{w}\right)  _{w\in\mathfrak{A}^{\ast}}$ is a basis of the $\kk%
$-module $T\left(  V\right)  $, and thus is a $\kk$-linearly
independent family of elements of this $\kk$-module. The family
$\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is a family of elements of
$T\left(  V\right)  $. Every $w\in\mathfrak{L}$ satisfies%
\begin{align*}
b_{w}  &  \in x_{w}+\underbrace{\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(
w\right)  };\\v>w}}\kk x_{v}}_{\substack{\subset\sum\limits_{\substack{v\in
\mathfrak{A}^{\ast};\\v>w}}\kk x_{v}\\\text{(since }\mathfrak{A}%
^{\ell\left(  w\right)  }\subset\mathfrak{A}^{\ast}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.freelie.2}(c)}\right)
\\
&  \subset x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ast};\\v>w}}\kk%
x_{v}.
\end{align*}
Hence, Proposition \ref{prop.sol.freelie.2.d.general} (applied to
$\mathfrak{V}=T\left(  V\right)  $, $W=\mathfrak{A}^{\ast}$, $\left(
p_{w}\right)  _{w\in\mathfrak{A}^{\ast}}=\left(  x_{w}\right)  _{w\in
\mathfrak{A}^{\ast}}$, $L=\mathfrak{L}$ and $\left(  s_{w}\right)
_{w\in\mathfrak{L}}=\left(  b_{w}\right)  _{w\in\mathfrak{L}}$) shows that the
family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is $\kk$-linearly
independent. Combining this with the fact that this family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk$-module
$\mathfrak{g}$, we therefore conclude that the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a basis of the $\kk$-module $\mathfrak{g}$.
This solves Exercise \ref{exe.freelie.2}(d).

(e) Let us first show a simple lemma:

\begin{lemma}
\label{lem.sol.freelie.2.e.generators-determine}Let $\mathfrak{a}$ and
$\mathfrak{b}$ be two $\kk$-Lie algebras. Let $p:\mathfrak{a}%
\rightarrow\mathfrak{b}$ and $q:\mathfrak{a}\rightarrow\mathfrak{b}$ be two
Lie algebra homomorphisms. Then, $\ker\left(  p-q\right)  $ is
a Lie subalgebra of $\mathfrak{a}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.freelie.2.e.generators-determine}.]Let
$x\in\ker\left(  p-q\right)  $ and $y\in\ker%
\left(  p-q\right)  $. Then, $\left(  p-q\right)  \left(  x\right)  =0$ (since
$x\in\ker\left(  p-q\right)  $), so that $0=\left(  p-q\right)
\left(  x\right)  =p\left(  x\right)  -q\left(  x\right)  $, and thus
$p\left(  x\right)  =q\left(  x\right)  $. The same argument (applied to $y$
instead of $x$) shows that $p\left(  y\right)  =q\left(  y\right)  $.

But $p$ is a Lie algebra homomorphism, and thus satisfies $p\left(  \left[
x,y\right]  \right)  =\left[  \underbrace{p\left(  x\right)  }_{=q\left(
x\right)  },\underbrace{p\left(  y\right)  }_{=q\left(  y\right)  }\right]
=\left[  q\left(  x\right)  ,q\left(  y\right)  \right]  =q\left(  \left[
x,y\right]  \right)  $ (since $q$ is a Lie algebra homomorphism). Now,
$\left(  p-q\right)  \left(  \left[  x,y\right]  \right)
=\underbrace{p\left(  \left[  x,y\right]  \right)  }_{=q\left(  \left[
x,y\right]  \right)  }-q\left(  \left[  x,y\right]  \right)  =q\left(  \left[
x,y\right]  \right)  -q\left(  \left[  x,y\right]  \right)  =0$. In other
words, $\left[  x,y\right]  \in\ker\left(  p-q\right)  $.

Let us now forget that we fixed $x$ and $y$. We thus have proven that $\left[
x,y\right]  \in\ker\left(  p-q\right)  $ whenever
$x\in\ker\left(  p-q\right)  $ and $y\in\ker%
\left(  p-q\right)  $. In other words, the set $\ker\left(
p-q\right)  $ is closed under the Lie bracket. Also, clearly, $p-q$ is a
$\kk$-linear map (since both maps $p$ and $q$ are $\kk$-linear),
and therefore its kernel $\ker\left(  p-q\right)  $ is a
$\kk$-submodule of $\mathfrak{a}$. Thus, $\ker\left(
p-q\right)  $ is a $\kk$-submodule of $\mathfrak{a}$ which is closed
under the Lie bracket. In other words, $\ker\left(  p-q\right)
$ is a Lie subalgebra of $\mathfrak{a}$. This proves Lemma
\ref{lem.sol.freelie.2.e.generators-determine}.
\end{proof}

Now, it is easy to see that if $\Xi_{1}$ and $\Xi_{2}$ are two Lie algebra
homomorphisms $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that every
$a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(  a\right)
$, then $\Xi_{1}=\Xi_{2}$\ \ \ \ \footnote{\textit{Proof.} Let $\Xi_{1}$ and
$\Xi_{2}$ be two Lie algebra homomorphisms $\Xi:\mathfrak{g}\rightarrow
\mathfrak{h}$ such that every $a\in\mathfrak{A}$ satisfies $\Xi\left(
x_{a}\right)  =\xi\left(  a\right)  $. We must prove that $\Xi_{1}=\Xi_{2}$.
\par
Both $\Xi_{1}$ and $\Xi_{2}$ are $\kk$-linear maps, and therefore
$\Xi_{1}-\Xi_{2}$ is a $\kk$-linear map as well. Hence, its kernel
$\ker\left(  \Xi_{1}-\Xi_{2}\right)  $ is a $\kk%
$-submodule of $\mathfrak{g}$.
\par
We know that $\Xi_{1}$ is a Lie algebra homomorphism $\Xi:\mathfrak{g}%
\rightarrow\mathfrak{h}$ such that every $a\in\mathfrak{A}$ satisfies
$\Xi\left(  x_{a}\right)  =\xi\left(  a\right)  $. Thus, every $a\in
\mathfrak{A}$ satisfies $\Xi_{1}\left(  x_{a}\right)  =\xi\left(  a\right)  $.
Similarly, every $a\in\mathfrak{A}$ satisfies $\Xi_{2}\left(  x_{a}\right)
=\xi\left(  a\right)  $. Hence, every $a\in\mathfrak{A}$ satisfies $\left(
\Xi_{1}-\Xi_{2}\right)  \left(  x_{a}\right)  =\underbrace{\Xi_{1}\left(
x_{a}\right)  }_{=\xi\left(  a\right)  }-\underbrace{\Xi_{2}\left(
x_{a}\right)  }_{=\xi\left(  a\right)  }=\xi\left(  a\right)  -\xi\left(
a\right)  =0$, so that $x_{a}\in\ker\left(  \Xi_{1}-\Xi
_{2}\right)  $. In other words, the set $\ker\left(  \Xi
_{1}-\Xi_{2}\right)  $ contains $x_{a}$ for every $a\in\mathfrak{A}$. Since
$\ker\left(  \Xi_{1}-\Xi_{2}\right)  $ is a $\kk%
$-submodule of $\mathfrak{g}$, this shows that%
\[
\ker\left(  \Xi_{1}-\Xi_{2}\right)  \supset\left(  \text{the
}\kk\text{-linear span of the family }\left(  x_{a}\right)
_{a\in\mathfrak{A}}\right)  .
\]
Since $\left(  \text{the }\kk\text{-linear span of the family }\left(
x_{a}\right)  _{a\in\mathfrak{A}}\right)  =V$ (because the family $\left(
x_{a}\right)  _{a\in\mathfrak{A}}$ is a basis of the $\kk$-module $V$),
this rewrites as $\ker\left(  \Xi_{1}-\Xi_{2}\right)  \supset
V$. In other words, $V\subset\ker\left(  \Xi_{1}-\Xi
_{2}\right)  $.
\par
From Lemma \ref{lem.sol.freelie.2.e.generators-determine} (applied to
$\mathfrak{a}=\mathfrak{g}$, $\mathfrak{b}=\mathfrak{h}$, $p=\Xi_{1}$ and
$q=\Xi_{2}$), we see that $\ker\left(  \Xi_{1}-\Xi_{2}\right)
$ is a Lie subalgebra of $\mathfrak{g}$. Since $\mathfrak{g}$ (in turn) is a
Lie subalgebra of $T\left(  V\right)  $, this shows that $\ker%
\left(  \Xi_{1}-\Xi_{2}\right)  $ is a Lie subalgebra of $T\left(  V\right)
$.
\par
Thus, we know that $\ker\left(  \Xi_{1}-\Xi_{2}\right)  $ is a
Lie subalgebra of $T\left(  V\right)  $ satisfying $V\subset
\ker\left(  \Xi_{1}-\Xi_{2}\right)  $. Exercise
\ref{exe.freelie.2}(b)
(applied to $\mathfrak{k} = \ker\left(  \Xi_{1}-\Xi_{2}\right)$)
thus shows that $\mathfrak{g}\subset\ker%
\left(  \Xi_{1}-\Xi_{2}\right)  $. Hence, $\Xi_{1}-\Xi_{2}=0$, so that
$\Xi_{1}=\Xi_{2}$. Qed.}. In other words, there exists \textbf{at most one}
Lie algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that
every $a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(
a\right)  $. We shall now show that there actually exists such a homomorphism.

For every $w\in\mathfrak{L}$, we define an element $z_{w}$ of $\mathfrak{h}$
as follows: We define $z_{w}$ by recursion on the length of $w$. If the length
of $w$ is $1$\ \ \ \ \footnote{The length of any $w\in\mathfrak{L}$ must be at
least $1$. (Indeed, if $w\in\mathfrak{L}$, then the word $w$ is Lyndon and
thus nonempty, and hence its length must be at least $1$.)}, then we have
$w=\left(  a\right)  $ for some letter $a\in\mathfrak{A}$, and we set
$z_{w}=\xi\left(  a\right)  $ for this letter $a$. If the length of $w$ is
$>1$, then we set $z_{w}=\left[  z_{u},z_{v}\right]  $, where $\left(
u,v\right)  =\operatorname*{stf}w$\ \ \ \ \footnote{This is well-defined,
because $z_{u}$ and $z_{v}$ have already been defined. [\textit{Proof.} Let
$\left(  u,v\right)  =\operatorname*{stf}w$. Then, Exercise
\ref{exe.words.lyndon.stf.basic}(e) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that $u\in\mathfrak{L}$, $v\in\mathfrak{L}$, $\ell\left(
u\right)  <\ell\left(  w\right)  $ and $\ell\left(  v\right)  <\ell\left(
w\right)  $. Recall that we are defining $z_{w}$ by recursion on the length of
$w$. Hence, $z_{p}$ is already defined for every $p\in\mathfrak{L}$ satisfying
$\ell\left(  p\right)  <\ell\left(  w\right)  $. Applying this to $p=u$, we
see that $z_{u}$ is already defined (since $u\in\mathfrak{L}$ and $\ell\left(
u\right)  <\ell\left(  w\right)  $). The same argument (but applied to $v$
instead of $u$) shows that $z_{v}$ is already defined. Hence, $z_{u}$ and
$z_{v}$ have already been defined. Thus, $z_{w}$ is well-defined by
$z_{w}=\left[  z_{u},z_{v}\right]  $, qed.]}.

Thus, we have defined a $z_{w}\in\mathfrak{h}$ for every $w\in\mathfrak{L}$.
Now, Exercise \ref{exe.freelie.2}(d) shows that the family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ is a basis of the $\kk$-module
$\mathfrak{g}$. Thus, we can define a $\kk$-module homomorphism
$f:\mathfrak{g}\rightarrow\mathfrak{h}$ by requiring that%
\begin{equation}
\left(  f\left(  b_{w}\right)  =z_{w}\ \ \ \ \ \ \ \ \ \ \text{for every }%
w\in\mathfrak{L}\right)  .\label{sol.freelie.2.e.1}%
\end{equation}
Consider this $f$. Whenever $w$ is a Lyndon word of length $>1$, we have%
\begin{equation}
f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(  b_{u}\right)
,f\left(  b_{v}\right)  \right]  ,\ \ \ \ \ \ \ \ \ \ \text{where }\left(
u,v\right)  =\operatorname*{stf}w\label{sol.freelie.2.e.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.e.3}):} Let $w$ be a Lyndon
word of length $>1$. Let $\left(  u,v\right)  =\operatorname*{stf}w$. Thus,
$\left(  u,v\right)  =\operatorname*{stf}w\in\mathfrak{L}\times\mathfrak{L}$.
\par
The recursive definition of $z_{w}$ shows that $z_{w}=\left[  z_{u}%
,z_{v}\right]  $ (since $w$ is a Lyndon word of length $>1$, and since
$\left(  u,v\right)  =\operatorname*{stf}w$). The recursive definition of
$b_{w}$ shows that $b_{w}=\left[  b_{u},b_{v}\right]  $ (for the same
reasons). Thus, $\left[  b_{u},b_{v}\right]  =b_{w}$, so that%
\begin{align*}
f\left(  \underbrace{\left[  b_{u},b_{v}\right]  }_{=b_{w}}\right)   &
=f\left(  b_{w}\right)  =z_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.e.1})}\right)  \\
&  =\left[  z_{u},z_{v}\right]  .
\end{align*}
Comparing this with%
\[
\left[  \underbrace{f\left(  b_{u}\right)  }_{\substack{=z_{u}\\\text{(by
(\ref{sol.freelie.2.e.1}), applied to }u\text{ instead of }w\text{)}%
}},\underbrace{f\left(  b_{v}\right)  }_{\substack{=z_{v}\\\text{(by
(\ref{sol.freelie.2.e.1}), applied to }v\text{ instead of }w\text{)}}}\right]
=\left[  z_{u},z_{v}\right]  ,
\]
we obtain $f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(
b_{u}\right)  ,f\left(  b_{v}\right)  \right]  $. This proves
(\ref{sol.freelie.2.e.3}).}. But recall that the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a basis of the $\kk$-module $\mathfrak{g}$.
Thus, this family spans the $\kk$-module $\mathfrak{g}$. In other
words, $\mathfrak{g}$ is the $\kk$-submodule of $\mathfrak{g}$ spanned
by the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$. Thus, Exercise
\ref{exe.freelie.1}(b) (applied to $B=\mathfrak{g}$) shows that $f$ is a Lie
algebra homomorphism.

We notice that%
\[
\text{every }a\in\mathfrak{A}\text{ satisfies }f\left(  x_{a}\right)
=\xi\left(  a\right)
\]
\footnote{\textit{Proof.} Let $a\in\mathfrak{A}$. Let $w$ be the one-letter
word $\left(  a\right)  $. Then, $w$ is a Lyndon word (since $w$ is a
one-letter word) and has length $1$. Thus, the definition of $z_{w}$ shows
that $z_{w}=\xi\left(  a\right)  $ (since $w=\left(  a\right)  $).
\par
But the definition of $b_{w}$ yields $b_{w}=x_{a}$ (since $\ell\left(
w\right)  =1$ and $w=\left(  a\right)  $). Hence, $f\left(  \underbrace{b_{w}%
}_{=x_{a}}\right)  =f\left(  x_{a}\right)  $, so that $f\left(  x_{a}\right)
=f\left(  b_{w}\right)  =z_{w}$ (by (\ref{sol.freelie.2.e.1})) and thus
$f\left(  x_{a}\right)  =z_{w}=\xi\left(  a\right)  $, qed.}. Thus, $f$ is a
Lie algebra homomorphism $\mathfrak{g}\rightarrow\mathfrak{h}$ having the
property that every $a\in\mathfrak{A}$ satisfies $f\left(  x_{a}\right)
=\xi\left(  a\right)  $. Therefore, there exists \textbf{at least one} Lie
algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that every
$a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(  a\right)  $
(namely, $\Xi=f$). Combining this with the fact that there exists \textbf{at
most one} Lie algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$
such that every $a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)
=\xi\left(  a\right)  $, we conclude that there exists a \textbf{unique} Lie
algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that every
$a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(  a\right)
$. This solves Exercise \ref{exe.freelie.2}(e).
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.freelie.2}}

\textit{Solution to Exercise \ref{exe.freelie.2}.} The definition of
$\mathfrak{g}$ shows that $\mathfrak{g}=\mathfrak{g}_{1}+\mathfrak{g}%
_{2}+\mathfrak{g}_{3}+\cdots=\sum_{i\geq1}\mathfrak{g}_{i}$. For every
positive integer $k$, we have%
\begin{align}
\mathfrak{g}_{k}  &  \subset\sum_{i\geq1}\mathfrak{g}_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{g}_{k}\text{ is an addend
of the sum }\sum_{i\geq1}\mathfrak{g}_{i}\text{ (since }k\text{ is a positive
integer)}\right) \nonumber\\
&  =\mathfrak{g}. \label{sol.freelie.2.gk}%
\end{align}
Notice that $\mathfrak{g}_{1}=V$, so that%
\begin{equation}
V=\mathfrak{g}_{1}\subset\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.gk}), applied to }k=1\right)  . \label{sol.freelie.2.V}%
\end{equation}


We also recall a fundamental property of Lie algebras (one of the forms of the
Jacobi identity):

\begin{itemize}
\item Every three elements $x$, $y$ and $z$ of a Lie algebra $\mathfrak{k}$
satisfy%
\begin{equation}
\left[  \left[  x,y\right]  ,z\right]  =\left[  \left[  x,z\right]  ,y\right]
-\left[  \left[  y,z\right]  ,x\right]  . \label{sol.freelie.2.jacobi}%
\end{equation}

\end{itemize}

(a) For every two positive integers $i$ and $j$, we have%
\begin{equation}
\left[  \mathfrak{g}_{i},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{i+j}.
\label{sol.freelie.2.a.claim}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.2.a.claim}):} We shall prove
(\ref{sol.freelie.2.a.claim}) by induction over $i$:

\textit{Induction base:} For every positive integer $j$, we have $\left[
\mathfrak{g}_{1},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{1+j}%
$\ \ \ \ \footnote{\textit{Proof.} Let $j$ be a positive integer. Thus,
$j\geq1$, so that $1+\underbrace{j}_{\geq1}\geq1+1\geq2$. The recursive
definition of $\mathfrak{g}_{1+j}$ thus yields $\mathfrak{g}_{1+j}=\left[
V,\underbrace{\mathfrak{g}_{\left(  1+j\right)  -1}}_{=\mathfrak{g}_{j}%
}\right]  =\left[  V,\mathfrak{g}_{j}\right]  $, so that $\left[
V,\mathfrak{g}_{j}\right]  =\mathfrak{g}_{1+j}$. Now, $\left[
\underbrace{\mathfrak{g}_{1}}_{=V},\mathfrak{g}_{j}\right]  =\left[
V,\mathfrak{g}_{j}\right]  =\mathfrak{g}_{1+j}\subset\mathfrak{g}_{1+j}$,
qed.}. In other words, (\ref{sol.freelie.2.a.claim}) holds for $i=1$. This
completes the induction base.

\textit{Induction step:} Let $I$ be a positive integer. Assume that
(\ref{sol.freelie.2.a.claim}) holds for $i=I$. We must prove that
(\ref{sol.freelie.2.a.claim}) holds for $i=I+1$.

We have assumed that (\ref{sol.freelie.2.a.claim}) holds for $i=I$. In other
words, for every positive integer $j$, we have%
\begin{equation}
\left[  \mathfrak{g}_{I},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{I+j}.
\label{sol.freelie.2.a.claim.pf.hyp}%
\end{equation}


Now, let $j$ be a positive integer. Thus, $j\geq1$ and thus $j+1\geq2$. The
recursive definition of $\mathfrak{g}_{j+1}$ thus yields $\mathfrak{g}%
_{j+1}=\left[  V,\underbrace{\mathfrak{g}_{\left(  j+1\right)  -1}%
}_{=\mathfrak{g}_{j}}\right]  =\left[  V,\mathfrak{g}_{j}\right]  $.

On the other hand, $I\geq1$ (since $I$ is a positive integer) and thus
$I+1\geq2$. The recursive definition of $\mathfrak{g}_{I+1}$ thus yields
$\mathfrak{g}_{I+1}=\left[  V,\underbrace{\mathfrak{g}_{\left(  I+1\right)
-1}}_{=\mathfrak{g}_{I}}\right]  =\left[  V,\mathfrak{g}_{I}\right]  $.

Also, $\underbrace{I}_{\geq1}+\underbrace{j}_{\geq1\geq0}+1\geq1+1=2$. The
recursive definition of $\mathfrak{g}_{I+j+1}$ thus yields $\mathfrak{g}%
_{I+j+1}=\left[  V,\underbrace{\mathfrak{g}_{\left(  I+j+1\right)  -1}%
}_{=\mathfrak{g}_{I+j}}\right]  =\left[  V,\mathfrak{g}_{I+j}\right]  $.

Now, let $z\in\mathfrak{g}_{j}$. Then, $z\in\mathfrak{g}_{j}\subset
\mathfrak{g}$ (by (\ref{sol.freelie.2.gk}), applied to $k=j$).

Also, let $p\in\mathfrak{g}_{I+1}$. We are going to prove that $\left[
p,z\right]  \in\mathfrak{g}_{I+j+1}$.

Recall that $\left[  V,\mathfrak{g}_{I}\right]  $ is the $\kk$-linear
span of all elements of the form $\left[  x,y\right]  $ with $\left(
x,y\right)  \in V\times\mathfrak{g}_{I}$ (indeed, this is how $\left[
V,\mathfrak{g}_{I}\right]  $ is defined). Thus, $p$ is a $\kk$-linear
combination of elements of the form $\left[  x,y\right]  $ with $\left(
x,y\right)  \in V\times\mathfrak{g}_{I}$ (since $p\in\mathfrak{g}%
_{I+1}=\left[  V,\mathfrak{g}_{I}\right]  $).

Now, we must prove the relation $\left[  p,z\right]  \in\mathfrak{g}_{I+j+1}$.
But this relation is $\kk$-linear in $p$ (since the Lie bracket of
$T\left(  V\right)  $ is $\kk$-bilinear, and since $\mathfrak{g}%
_{I+j+1}$ is a $\kk$-module). Thus, we can WLOG assume that $p$ is an
element of the form $\left[  x,y\right]  $ for with $\left(  x,y\right)  \in
V\times\mathfrak{g}_{I}$ (since $p$ is a $\kk$-linear combination of
elements of the form $\left[  x,y\right]  $ with $\left(  x,y\right)  \in
V\times\mathfrak{g}_{I}$). Assume this.

We know that $p$ is an element of the form $\left[  x,y\right]  $ for with
$\left(  x,y\right)  \in V\times\mathfrak{g}_{I}$. In other words, there
exists an $\left(  x,y\right)  \in V\times\mathfrak{g}_{I}$ such that
$p=\left[  x,y\right]  $. Consider this $\left(  x,y\right)  $.

We have $\left(  x,y\right)  \in V\times\mathfrak{g}_{I}$. In other words,
$x\in V$ and $y\in\mathfrak{g}_{I}$. Thus, $y\in\mathfrak{g}_{I}%
\subset\mathfrak{g}$ (by (\ref{sol.freelie.2.gk}), applied to $k=I$). Also,
$x\in V\subset\mathfrak{g}$ (by (\ref{sol.freelie.2.V})). Now,%
\begin{align*}
\left[  \underbrace{p}_{=\left[  x,y\right]  },z\right]   &  =\left[  \left[
x,y\right]  ,z\right]  =\underbrace{\left[  \left[  x,z\right]  ,y\right]
}_{=-\left[  y,\left[  x,z\right]  \right]  }-\underbrace{\left[  \left[
y,z\right]  ,x\right]  }_{=-\left[  x,\left[  y,z\right]  \right]
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.2.jacobi})}\right) \\
&  =-\left[  y,\left[  x,z\right]  \right]  -\left(  -\left[  x,\left[
y,z\right]  \right]  \right)  =\left[  \underbrace{x}_{\in V},\left[
\underbrace{y}_{\in\mathfrak{g}_{I}},\underbrace{z}_{\in\mathfrak{g}_{j}%
}\right]  \right]  -\left[  \underbrace{y}_{\in\mathfrak{g}_{I}},\left[
\underbrace{x}_{\in V},\underbrace{z}_{\in\mathfrak{g}_{j}}\right]  \right] \\
&  \in\left[  V,\underbrace{\left[  \mathfrak{g}_{I},\mathfrak{g}_{j}\right]
}_{\substack{\subset\mathfrak{g}_{I+j}\\\text{(by
(\ref{sol.freelie.2.a.claim.pf.hyp}))}}}\right]  -\left[  \mathfrak{g}%
_{I},\underbrace{\left[  V,\mathfrak{g}_{j}\right]  }_{\substack{=\mathfrak{g}%
_{j+1}\\\text{(since }\mathfrak{g}_{j+1}=\left[  V,\mathfrak{g}_{j}\right]
\text{)}}}\right] \\
&  \subset\underbrace{\left[  V,\mathfrak{g}_{I+j}\right]  }%
_{\substack{=\mathfrak{g}_{I+j+1}\\\text{(since }\mathfrak{g}_{I+j+1}=\left[
V,\mathfrak{g}_{I+j}\right]  \text{)}}}-\underbrace{\left[  \mathfrak{g}%
_{I},\mathfrak{g}_{j+1}\right]  }_{\substack{\subset\mathfrak{g}%
_{I+j+1}\\\text{(by (\ref{sol.freelie.2.a.claim.pf.hyp}), applied to
}j+1\text{ instead of }j\text{)}}}\\
&  \subset\mathfrak{g}_{I+j+1}-\mathfrak{g}_{I+j+1}\subset\mathfrak{g}%
_{I+j+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{g}_{I+j+1}\text{ is
a }\kk\text{-module}\right)  .
\end{align*}


Now, let us forget that we fixed $z$ and $p$. We thus have proven that%
\begin{equation}
\left[  p,z\right]  \in\mathfrak{g}_{I+j+1}\ \ \ \ \ \ \ \ \ \ \text{for any
}z\in\mathfrak{g}_{j}\text{ and }p\in\mathfrak{g}_{I+1}.
\label{sol.freelie.2.a.claim.pf.alm}%
\end{equation}


Now, let $q\in\left[  \mathfrak{g}_{I+1},\mathfrak{g}_{j}\right]  $. We will
show that $q\in\mathfrak{g}_{I+j+1}$.

Recall that $\left[  \mathfrak{g}_{I+1},\mathfrak{g}_{j}\right]  $ is the
$\kk$-linear span of all elements of the form $\left[  x,y\right]  $
with $\left(  x,y\right)  \in\mathfrak{g}_{I+1}\times\mathfrak{g}_{j}$
(indeed, this is how $\left[  \mathfrak{g}_{I+1},\mathfrak{g}_{j}\right]  $ is
defined). Thus, $q$ is a $\kk$-linear combination of elements of the
form $\left[  x,y\right]  $ with $\left(  x,y\right)  \in\mathfrak{g}%
_{I+1}\times\mathfrak{g}_{j}$ (since $q\in\left[  \mathfrak{g}_{I+1}%
,\mathfrak{g}_{j}\right]  $).

Now, we must prove the relation $q\in\mathfrak{g}_{I+j+1}$. But this relation
is $\kk$-linear in $q$ (since $\mathfrak{g}_{I+j+1}$ is a $\kk%
$-module). Thus, we can WLOG assume that $q$ is an element of the form
$\left[  x,y\right]  $ with $\left(  x,y\right)  \in\mathfrak{g}_{I+1}%
\times\mathfrak{g}_{j}$ (since $q$ is a $\kk$-linear combination of
elements of the form $\left[  x,y\right]  $ with $\left(  x,y\right)
\in\mathfrak{g}_{I+1}\times\mathfrak{g}_{j}$). Assume this.

We know that $q$ is an element of the form $\left[  x,y\right]  $ with
$\left(  x,y\right)  \in\mathfrak{g}_{I+1}\times\mathfrak{g}_{j}$. In other
words, there exists an $\left(  x,y\right)  \in\mathfrak{g}_{I+1}%
\times\mathfrak{g}_{j}$ such that $q=\left[  x,y\right]  $. Consider this
$\left(  x,y\right)  $.

From $\left(  x,y\right)  \in\mathfrak{g}_{I+1}\times\mathfrak{g}_{j}$, we
obtain $x\in\mathfrak{g}_{I+1}$ and $y\in\mathfrak{g}_{j}$. Now,
(\ref{sol.freelie.2.a.claim.pf.alm}) (applied to $z=y$ and $p=x$) shows that
$\left[  x,y\right]  \in\mathfrak{g}_{I+j+1}$. Thus, $q=\left[  x,y\right]
\in\mathfrak{g}_{I+j+1}$.

Now, let us forget that we fixed $q$. We thus have proven that every
$q\in\left[  \mathfrak{g}_{I+1},\mathfrak{g}_{j}\right]  $ satisfies
$q\in\mathfrak{g}_{I+j+1}$. In other words, $\left[  \mathfrak{g}%
_{I+1},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{I+j+1}$.
In other words, $\left[  \mathfrak{g}%
_{I+1},\mathfrak{g}_{j}\right]  \subset\mathfrak{g}_{I+1+j}$
(since $I+j+1 = I+1+j$).

Let us now forget that we fixed $j$. We thus have proven that, for every
positive integer $j$, we have $\left[  \mathfrak{g}_{I+1},\mathfrak{g}%
_{j}\right]  \subset\mathfrak{g}_{I+1+j}$. In other words,
(\ref{sol.freelie.2.a.claim}) holds for $i=I+1$. This completes the induction
step. The induction proof of (\ref{sol.freelie.2.a.claim}) is thus complete.

Now, for every two positive integers $i$ and $j$, we have%
\begin{align}
\left[  \mathfrak{g}_{i},\mathfrak{g}_{j}\right]   &  \subset\mathfrak{g}%
_{i+j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{sol.freelie.2.a.claim}%
)}\right) \nonumber\\
&  \subset\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.gk}), applied to }k=i+j\right)  .
\label{sol.freelie.2.a.alm}%
\end{align}


But $\mathfrak{g}=\sum_{i\geq1}\mathfrak{g}_{i}=\sum_{j\geq1}\mathfrak{g}_{j}$
(here, we have renamed the summation index $i$ as $j$). Thus,%
\begin{align*}
\left[  \underbrace{\mathfrak{g}}_{=\sum_{i\geq1}\mathfrak{g}_{i}%
},\underbrace{\mathfrak{g}}_{=\sum_{j\geq1}\mathfrak{g}_{j}}\right]   &
=\left[  \sum_{i\geq1}\mathfrak{g}_{i},\sum_{j\geq1}\mathfrak{g}_{j}\right] \\
&  \subset\sum_{i\geq1}\sum_{j\geq1}\underbrace{\left[  \mathfrak{g}%
_{i},\mathfrak{g}_{j}\right]  }_{\substack{\subset\mathfrak{g}\\\text{(by
(\ref{sol.freelie.2.a.alm}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie
bracket on }T\left(  V\right)  \text{ is }\kk\text{-bilinear}\right) \\
&  \subset\sum_{i\geq1}\sum_{j\geq1}\mathfrak{g}\subset\mathfrak{g}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{g}\text{ is a }%
\kk\text{-module}\right)  .
\end{align*}
Thus, $\mathfrak{g}$ is a Lie subalgebra of $T\left(  V\right)  $. This solves
Exercise \ref{exe.freelie.2}(a).

(b) Let $\mathfrak{k}$ be any Lie subalgebra of $T\left(  V\right)  $
satisfying $V\subset\mathfrak{k}$. We must prove that $\mathfrak{g}%
\subset\mathfrak{k}$.

We claim that%
\begin{equation}
\mathfrak{g}_{k}\subset\mathfrak{k}\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }k. \label{sol.freelie.2.b.claim}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.2.b.claim}):} We shall prove
(\ref{sol.freelie.2.b.claim}) by induction over $k$:

\textit{Induction base:} We have $\mathfrak{g}_{1}=V\subset\mathfrak{k}$. In
other words, (\ref{sol.freelie.2.b.claim}) holds for $i=1$. This completes the
induction base.

\textit{Induction step:} Let $K$ be a positive integer. Assume that
(\ref{sol.freelie.2.b.claim}) holds for $k=K$. We must prove that
(\ref{sol.freelie.2.b.claim}) holds for $k=K+1$.

We have assumed that (\ref{sol.freelie.2.b.claim}) holds for $k=K$. In other
words, we have $\mathfrak{g}_{K}\subset\mathfrak{k}$.

But $\mathfrak{k}$ is a Lie subalgebra of $T\left(  V\right)  $. Hence,
$\left[  \mathfrak{k},\mathfrak{k}\right]  \subset\mathfrak{k}$.

Now, $K\geq1$ (since $K$ is a positive integer) and thus $K+1\geq2$. The
recursive definition of $\mathfrak{g}_{K+1}$ thus yields%
\[
\mathfrak{g}_{K+1}=\left[  \underbrace{V}_{\subset\mathfrak{k}}%
,\underbrace{\mathfrak{g}_{\left(  K+1\right)  -1}}_{=\mathfrak{g}_{K}%
\subset\mathfrak{k}}\right]  \subset\left[  \mathfrak{k},\mathfrak{k}\right]
\subset\mathfrak{k}.
\]
In other words, (\ref{sol.freelie.2.b.claim}) holds for $k=K+1$. This
completes the induction step. The induction proof of
(\ref{sol.freelie.2.b.claim}) is thus complete.

Now,%
\[
\mathfrak{g}=\sum_{i\geq1}\underbrace{\mathfrak{g}_{i}}_{\substack{\subset
\mathfrak{k}\\\text{(by (\ref{sol.freelie.2.b.claim}), applied to }%
k=i\text{)}}}\subset\sum_{i\geq1}\mathfrak{k}\subset\mathfrak{k}%
\]
(since $\mathfrak{k}$ is a $\kk$-module (since $\mathfrak{k}$ is a Lie
algebra)). This solves Exercise \ref{exe.freelie.2}(b).

Before we step to the solution of Exercise \ref{exe.freelie.2}(c), we record a
simple fact: If $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$, then%
\begin{equation}
x_{u}x_{v}=x_{uv} \label{sol.freelie.2.c.xuxv}%
\end{equation}
in the $\kk$-algebra $T\left(  V\right)  $%
\ \ \ \ \footnote{\textit{Proof of (\ref{sol.freelie.2.c.xuxv}):} Let
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$. We have $uv=\left(
\left(  uv\right)  _{1},\left(  uv\right)  _{2},\ldots,\left(  uv\right)
_{\ell\left(  uv\right)  }\right)  $ and thus%
\begin{align*}
\left(  \left(  uv\right)  _{1},\left(  uv\right)  _{2},\ldots,\left(
uv\right)  _{\ell\left(  uv\right)  }\right)   &  =\underbrace{u}_{=\left(
u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  }\underbrace{v}%
_{=\left(  v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  }\\
&  =\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  \left(
v_{1},v_{2},\ldots,v_{\ell\left(  v\right)  }\right)  =\left(  u_{1}%
,u_{2},\ldots,u_{\ell\left(  u\right)  },v_{1},v_{2},\ldots,v_{\ell\left(
v\right)  }\right)
\end{align*}
and therefore%
\[
x_{\left(  uv\right)  _{1}}\otimes x_{\left(  uv\right)  _{2}}\otimes
\cdots\otimes x_{\left(  uv\right)  _{\ell\left(  uv\right)  }}=x_{u_{1}%
}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(  u\right)  }}\otimes
x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes x_{v_{\ell\left(  v\right)  }%
}.
\]
The definition of $x_{uv}$ now yields%
\begin{equation}
x_{uv}=x_{\left(  uv\right)  _{1}}\otimes x_{\left(  uv\right)  _{2}}%
\otimes\cdots\otimes x_{\left(  uv\right)  _{\ell\left(  uv\right)  }%
}=x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(  u\right)
}}\otimes x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes x_{v_{\ell\left(
v\right)  }}. \label{sol.freelie.2.c.xuxv.pf.1}%
\end{equation}
But the definition of $x_{u}$ yields $x_{u}=x_{u_{1}}\otimes x_{u_{2}}%
\otimes\cdots\otimes x_{u_{\ell\left(  u\right)  }}$, and the definition of
$x_{v}$ yields $x_{v}=x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes
x_{v_{\ell\left(  v\right)  }}$. Thus,%
\begin{align*}
&  \underbrace{x_{u}}_{=x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes
x_{u_{\ell\left(  u\right)  }}}\underbrace{x_{v}}_{=x_{v_{1}}\otimes x_{v_{2}%
}\otimes\cdots\otimes x_{v_{\ell\left(  v\right)  }}}\\
&  =\left(  x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(
u\right)  }}\right)  \left(  x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes
x_{v_{\ell\left(  v\right)  }}\right) \\
&  =x_{u_{1}}\otimes x_{u_{2}}\otimes\cdots\otimes x_{u_{\ell\left(  u\right)
}}\otimes x_{v_{1}}\otimes x_{v_{2}}\otimes\cdots\otimes x_{v_{\ell\left(
v\right)  }}\\
&  =x_{uv}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.c.xuxv.pf.1})}\right)  .
\end{align*}
This proves (\ref{sol.freelie.2.c.xuxv}).}.

(c) We shall solve Exercise \ref{exe.freelie.2}(c) by strong induction on
$\ell\left(  w\right)  $:

\textit{Induction step:} Let $N\in \NN$. Assume that Exercise
\ref{exe.freelie.2}(c) holds under the condition that $\ell\left(  w\right)
<N$. We need to show that Exercise \ref{exe.freelie.2}(c) also holds under the
condition that $\ell\left(  w\right)  =N$.

We have assumed that Exercise \ref{exe.freelie.2}(c) holds under the condition
that $\ell\left(  w\right)  <N$. In other words, we have%
\begin{equation}
b_{w}\in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in
\mathfrak{L}\text{ satisfying }\ell\left(  w\right)  <N.
\label{sol.freelie.2.c.hyp}%
\end{equation}


For every $w\in\mathfrak{L}$ satisfying $\ell\left(  w\right)  <N$, we have%
\begin{align}
b_{w}  &  \in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}\nonumber\\
&  =x_{w}+\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\p>w}}\kk x_{p}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }v\text{ as }p\right) \label{sol.freelie.2.c.hyp2}\\
&  =x_{w}+\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\q>w}}\kk x_{q}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation index }p\text{ as }q\right)  . \label{sol.freelie.2.c.hyp3}%
\end{align}
From this, it is easy to conclude the following: For every $w\in\mathfrak{L}$
satisfying $\ell\left(  w\right)  <N$, we have%
\begin{equation}
b_{w}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q} \label{sol.freelie.2.c.hyp5}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.hyp5}):} Let $w\in
\mathfrak{L}$ be such that $\ell\left(  w\right)  <N$.
\par
Every $q\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying $q>w$ must also
satisfy $q\geq w$ (since it satisfies $q>w\geq w$). Thus, the sum
$\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q>w}%
}\kk x_{q}$ is a subsum of $\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\q\geq w}}\kk x_{q}$. Hence, $\sum\limits
_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q>w}}\kk%
x_{q}\subset\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$.
\par
On the other hand, $w$ is an element of $\mathfrak{A}^{\ell\left(  w\right)
}$ (since $\ell\left(  w\right)  =\ell\left(  w\right)  $) and satisfies
$w\geq w$. In other words, $w$ is a $q\in\mathfrak{A}^{\ell\left(  w\right)
}$ satisfying $q\geq w$. Thus, $\kk x_{w}$ is an addend of the sum
$\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$ (namely, the addend for $q=w$). Hence, $\kk%
x_{w}\subset\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$.
\par
Now, $x_{w}=\underbrace{1}_{\in\kk}x_{w}\in\kk x_{w}\subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$. But (\ref{sol.freelie.2.c.hyp3}) becomes%
\[
b_{w}\in\underbrace{x_{w}}_{\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
w\right)  };\\q\geq w}}\kk x_{q}}+\underbrace{\sum\limits_{\substack{q\in
\mathfrak{A}^{\ell\left(  w\right)  };\\q>w}}\kk x_{q}}_{\subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}}\subset\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
w\right)  };\\q\geq w}}\kk x_{q}+\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\q\geq w}}\kk x_{q}\subset\sum\limits
_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq w}%
}\kk x_{q}%
\]
(since $\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  w\right)  };\\q\geq
w}}\kk x_{q}$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.hyp5}).}.

Now, let $w\in\mathfrak{L}$ be such that $\ell\left(  w\right)  =N$. We shall
prove that $b_{w}\in x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(
w\right)  };\\r>w}}\kk x_{r}$.

If $\ell\left(  w\right)  \leq1$, then $b_{w}\in x_{w}+\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$
holds\footnote{\textit{Proof.} Assume that $\ell\left(  w\right)  \leq1$. The
word $w$ is Lyndon (since $w\in\mathfrak{L}$) and thus nonempty. Hence,
$\ell\left(  w\right)  \geq1$. Combined with $\ell\left(  w\right)  \leq1$,
this yields $\ell\left(  w\right)  =1$. In other words, the word $w$ consists
of a single letter. In other words, $w=\left(  a\right)  $ for some
$a\in\mathfrak{A}$. Consider this $a$. The definition of $b_{w}$ then yields
$b_{w}=x_{a}$ (since $\ell\left(  w\right)  =1$ and $w=\left(  a\right)  $).
\par
On the other hand, the definition of $x_{w}$ yields
\begin{align*}
x_{w}  &  =x_{w_{1}}\otimes x_{w_{2}}\otimes\cdots\otimes x_{w_{\ell\left(
w\right)  }}=x_{w_{1}}\otimes x_{w_{2}}\otimes\cdots\otimes x_{w_{1}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  w\right)  =1\right) \\
&  =x_{w_{1}}=x_{a}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }w_{1}=a\text{
(since }w=\left(  a\right)  \text{)}\right)  .
\end{align*}
Compared with $b_{w}=x_{a}$, this yields
\[
b_{w}=x_{w}=x_{w}+\underbrace{0}_{\substack{\in\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}\\\text{(since
}\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r}\text{ is a }\kk\text{-module)}}}\in x_{w}%
+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r},
\]
qed.}. Hence, for the rest of this proof, we can WLOG assume that we don't
have $\ell\left(  w\right)  \leq1$. Assume this.

The word $w$ is Lyndon (since $w\in\mathfrak{L}$) and satisfies $\ell\left(
w\right)  >1$ (since we don't have $\ell\left(  w\right)  \leq1$). Thus, $w$
is a Lyndon word of length $>1$. Therefore, $\operatorname*{stf}w$ is
well-defined. Let $\left(  u,v\right)  =\operatorname*{stf}w$. The recursive
definition of $b_{w}$ yields $b_{w}=\left[  b_{u},b_{v}\right]  $ (since
$\ell\left(  w\right)  >1$ and $\left(  u,v\right)  =\operatorname*{stf}w$).

From Exercise \ref{exe.words.lyndon.stf.basic}(a) (applied to $\left(
g,h\right)  =\left(  u,v\right)  $), we see that $v$ is the longest Lyndon
proper suffix of $w$. In particular, $v$ is a Lyndon proper suffix of $w$. In
other words, $v$ is a Lyndon word and a proper suffix of $w$. Also, Exercise
\ref{exe.words.lyndon.stf.basic}(d) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that the word $u$ is Lyndon. From Exercise
\ref{exe.words.lyndon.stf.basic}(c) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $), we see that we have $u<uv<v$. Finally, Exercise
\ref{exe.words.lyndon.stf.basic}(b) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that we have $w=uv$.

The word $u$ is Lyndon and thus nonempty. Hence, $\ell\left(  u\right)  \geq
1$. Also, the word $v$ is Lyndon and thus nonempty. Thus, $\ell\left(
v\right)  \geq1$. We have $u\in\mathfrak{L}$ (since $u$ is Lyndon) and
$v\in\mathfrak{L}$ (since $v$ is Lyndon).

Now, $\ell\left(  \underbrace{w}_{=uv}\right)  =\ell\left(  uv\right)
=\ell\left(  u\right)  +\underbrace{\ell\left(  v\right)  }_{\geq1>0}%
>\ell\left(  u\right)  $, so that $\ell\left(  u\right)  <\ell\left(
w\right)  =N$. Thus, (\ref{sol.freelie.2.c.hyp2}) (applied to $u$ instead of
$w$) shows that%
\begin{equation}
b_{u}\in x_{u}+\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)
};\\p>u}}\kk x_{p}. \label{sol.freelie.2.c.ineq1a}%
\end{equation}
Also, (\ref{sol.freelie.2.c.hyp5}) (applied to $u$ instead of $w$) shows that%
\begin{equation}
b_{u}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  u\right)  };\\q\geq
u}}\kk x_{q}=\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)
};\\p\geq u}}\kk x_{p} \label{sol.freelie.2.c.ineq1b}%
\end{equation}
(here, we renamed the summation index $q$ as $p$).

Also, $\ell\left(  \underbrace{w}_{=uv}\right)  =\ell\left(  uv\right)
=\underbrace{\ell\left(  u\right)  }_{\geq1>0}+\ell\left(  v\right)
>\ell\left(  v\right)  $, so that $\ell\left(  v\right)  <\ell\left(
w\right)  =N$. Thus, (\ref{sol.freelie.2.c.hyp3}) (applied to $v$ instead of
$w$) shows that%
\begin{equation}
b_{v}\in x_{v}+\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q>v}}\kk x_{q}. \label{sol.freelie.2.c.ineq2a}%
\end{equation}
Also, (\ref{sol.freelie.2.c.hyp5}) (applied to $v$ instead of $w$) shows that%
\begin{equation}
b_{v}\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\kk x_{q}. \label{sol.freelie.2.c.ineq2b}%
\end{equation}


Now, let $G=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\r>w}}\kk x_{r}$. Thus, $G=\sum\limits_{\substack{r\in\mathfrak{A}%
^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$ is a $\kk$-submodule
of $T\left(  V\right)  $.

If $p\in\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in\mathfrak{A}%
^{\ell\left(  v\right)  }$ are such that $q\geq v$, then%
\begin{equation}
x_{q}x_{p}\in G \label{sol.freelie.2.c.bound1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound1}):} Let $p\in
\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in\mathfrak{A}^{\ell\left(
v\right)  }$ be such that $q\geq v$.
\par
We have $\ell\left(  qp\right)  =\underbrace{\ell\left(  q\right)
}_{\substack{=\ell\left(  v\right)  \\\text{(since }q\in\mathfrak{A}%
^{\ell\left(  v\right)  }\text{)}}}+\underbrace{\ell\left(  p\right)
}_{\substack{=\ell\left(  u\right)  \\\text{(since }p\in\mathfrak{A}%
^{\ell\left(  u\right)  }\text{)}}}=\ell\left(  v\right)  +\ell\left(
u\right)  =\ell\left(  u\right)  +\ell\left(  v\right)  $. Compared with
$\ell\left(  w\right)  =\ell\left(  u\right)  +\ell\left(  v\right)  $, this
yields $\ell\left(  qp\right)  =\ell\left(  w\right)  $. In other words,
$qp\in\mathfrak{A}^{\ell\left(  w\right)  }$.
\par
But the word $w$ is Lyndon and satisfies $w=uv$. Thus, Proposition
\ref{prop.words.lyndon.trivia}(a) shows that $v\geq w$ (since $v$ is
nonempty). Since $\ell\left(  v\right)  \neq\ell\left(  w\right)  $ (because
$\ell\left(  v\right)  <\ell\left(  w\right)  $), we have $v\neq w$. Combined
with $v\geq w$, this yields $v>w$. Now, $qp\geq q\geq v>w$.
\par
So we know that $qp\in\mathfrak{A}^{\ell\left(  w\right)  }$ and $qp>w$. In
other words, $qp$ is an $r\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying
$r>w$. Thus, $\kk x_{qp}$ is an addend of the sum $\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$ (namely, the
addend for $r=qp$). Therefore, $\kk x_{qp}\subset\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}=G$ (since
$G=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r}$).
\par
But (\ref{sol.freelie.2.c.xuxv}) (applied to $q$ and $p$ instead of $u$ and
$v$) shows that $x_{q}x_{p}=x_{qp}=\underbrace{1}_{\in\kk}x_{qp}%
\in\kk x_{qp}\subset G$. This proves (\ref{sol.freelie.2.c.bound1}).}.
Hence,%
\begin{equation}
b_{v}b_{u}\in G \label{sol.freelie.2.c.bound1b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound1b}):} We have%
\begin{align*}
&  \underbrace{b_{v}}_{\substack{\in\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  v\right)  };\\q\geq v}}\kk x_{q}\\\text{(by
(\ref{sol.freelie.2.c.ineq2b}))}}}\underbrace{b_{u}}_{\substack{\in
\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq
u}}\kk x_{p}\\\text{(by (\ref{sol.freelie.2.c.ineq1b}))}}}\\
&  \in\left(  \sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q\geq v}}\kk x_{q}\right)  \left(  \sum\limits_{\substack{p\in
\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq u}}\kk x_{p}\right)
\subset\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq
u}}\underbrace{\kk\kk}_{\subset\kk}\underbrace{x_{q}x_{p}%
}_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound1}))}}}\subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq v}%
}\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p\geq
u}}\kk G\subset G
\end{align*}
(since $G$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.bound1b}).}.

Furthermore, if $p\in\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in
\mathfrak{A}^{\ell\left(  v\right)  }$ are such that $p>u$, then
\begin{equation}
x_{p}x_{q}\in G \label{sol.freelie.2.c.bound2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound2}):} Let $p\in
\mathfrak{A}^{\ell\left(  u\right)  }$ and $q\in\mathfrak{A}^{\ell\left(
v\right)  }$ be such that $p>u$.
\par
We have $\ell\left(  pq\right)  =\underbrace{\ell\left(  p\right)
}_{\substack{=\ell\left(  u\right)  \\\text{(since }p\in\mathfrak{A}%
^{\ell\left(  u\right)  }\text{)}}}+\underbrace{\ell\left(  q\right)
}_{\substack{=\ell\left(  v\right)  \\\text{(since }q\in\mathfrak{A}%
^{\ell\left(  v\right)  }\text{)}}}=\ell\left(  u\right)  +\ell\left(
v\right)  $. Compared with $\ell\left(  w\right)  =\ell\left(  u\right)
+\ell\left(  v\right)  $, this yields $\ell\left(  pq\right)  =\ell\left(
w\right)  $. In other words, $pq\in\mathfrak{A}^{\ell\left(  w\right)  }$.
\par
We have $\ell\left(  p\right)  =\ell\left(  u\right)  $ (since $p\in
\mathfrak{A}^{\ell\left(  u\right)  }$), so that $\ell\left(  u\right)
=\ell\left(  p\right)  $.
\par
Assume (for the sake of contradiction) that the word $u$ is a prefix of $p$.
Since the word $u$ has the same length as $p$ (since $\ell\left(  u\right)
=\ell\left(  p\right)  $), this shows that $u=p$. Thus, $u=p>u$, which is
absurd. This contradiction shows that our assumption (that the word $u$ is a
prefix of $p$) was false. In other words, the word $u$ is not a prefix of $p$.
\par
We have $u<p$ (since $p>u$), thus $u\leq p$. Thus, Proposition
\ref{prop.words.lex}(d) (applied to $u$, $v$, $p$ and $\varnothing$ instead of
$a$, $b$, $c$ and $d$) shows that either we have $uv\leq p\varnothing$ or the
word $u$ is a prefix of $p$. Since the word $u$ is not a prefix of $p$, we
thus conclude that $uv\leq p\varnothing$.
\par
Now, $q\in\mathfrak{A}^{\ell\left(  v\right)  }$, so that $\ell\left(
q\right) = \ell\left(  v\right)  \geq1$. The word $q$ is thus nonempty. In
other words, $q\neq\varnothing$. If we had $pq=p\varnothing$, then we would
have $q=\varnothing$ (since we could cancel $p$ from the equality
$pq=p\varnothing$), which would contradict $q\neq\varnothing$. Thus, we cannot
have $pq=p\varnothing$. Hence, $pq\neq p\varnothing$. Combining $pq\neq
p\varnothing=p$ with $pq\geq p$, we obtain $pq>p$. Hence, $pq>p=p\varnothing
\geq uv$ (since $uv\leq p\varnothing$), so that $pq>uv=w$ (since $w=uv$).
\par
So we know that $pq\in\mathfrak{A}^{\ell\left(  w\right)  }$ and $pq>w$. In
other words, $pq$ is an $r\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying
$r>w$. Thus, $\kk x_{pq}$ is an addend of the sum $\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$ (namely, the
addend for $r=pq$). Therefore, $\kk x_{pq}\subset\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}=G$ (since
$G=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r}$).
\par
But (\ref{sol.freelie.2.c.xuxv}) (applied to $p$ and $q$ instead of $u$ and
$v$) shows that $x_{p}x_{q}=x_{pq}=\underbrace{1}_{\in\kk}x_{pq}%
\in\kk x_{pq}\subset G$. This proves (\ref{sol.freelie.2.c.bound2}).}.
Hence,
\begin{equation}
\left(  b_{u}-x_{u}\right)  b_{v}\in G \label{sol.freelie.2.c.bound2b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound2b}):} Subtracting
$x_{u}$ from both sides of the relation (\ref{sol.freelie.2.c.ineq1a}), we
obtain%
\[
b_{u}-x_{u}\in x_{u}+\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)
};\\p>u}}\kk x_{p}-x_{u}=\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(
u\right)  };\\p>u}}\kk x_{p}.
\]
Now,%
\begin{align*}
&  \underbrace{\left(  b_{u}-x_{u}\right)  }_{\in\sum\limits_{\substack{p\in
\mathfrak{A}^{\ell\left(  u\right)  };\\p>u}}\kk x_{p}}\underbrace{b_{v}%
}_{\substack{\in\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q\geq v}}\kk x_{q}\\\text{(by (\ref{sol.freelie.2.c.ineq2b}))}}}\\
&  \in\left(  \sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  }%
;\\p>u}}\kk x_{p}\right)  \left(  \sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  v\right)  };\\q\geq v}}\kk x_{q}\right)  \subset
\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p>u}%
}\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\underbrace{\kk\kk}_{\subset\kk}\underbrace{x_{p}x_{q}%
}_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound2}))}}}\subset
\sum\limits_{\substack{p\in\mathfrak{A}^{\ell\left(  u\right)  };\\p>u}%
}\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q\geq
v}}\kk G\subset G
\end{align*}
(since $G$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.bound2b}).}.

Furthermore, if $q\in\mathfrak{A}^{\ell\left(  v\right)  }$ is such that
$q>v$, then%
\begin{equation}
x_{u}x_{q}\in G \label{sol.freelie.2.c.bound3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound3}):} Let $q\in
\mathfrak{A}^{\ell\left(  v\right)  }$ be such that $q>v$.
\par
We have $\ell\left(  uq\right)  =\ell\left(  u\right)  +\underbrace{\ell
\left(  q\right)  }_{\substack{=\ell\left(  v\right)  \\\text{(since }%
q\in\mathfrak{A}^{\ell\left(  v\right)  }\text{)}}}=\ell\left(  u\right)
+\ell\left(  v\right)  $. Compared with $\ell\left(  w\right)  =\ell\left(
u\right)  +\ell\left(  v\right)  $, this yields $\ell\left(  uq\right)
=\ell\left(  w\right)  $. In other words, $uq\in\mathfrak{A}^{\ell\left(
w\right)  }$.
\par
We have $q>v$, thus $v<q$, thus $v\leq q$. Hence, Proposition
\ref{prop.words.lex}(b) (applied to $u$, $v$ and $q$ instead of $a$, $c$ and
$d$) shows that $uv\leq uq$. If we had $uv=uq$, then we would have $v=q$
(since we could cancel $u$ from the equality $uv=uq$), which would contradict
$v<q$. Thus, we cannot have $uv=uq$. Hence, we have $uv\neq uq$. Combined with
$uv\leq uq$, this shows that $uv<uq$. In other words, $uq>uv=w$ (since
$w=uv$).
\par
So we know that $uq\in\mathfrak{A}^{\ell\left(  w\right)  }$ and $uq>w$. In
other words, $uq$ is an $r\in\mathfrak{A}^{\ell\left(  w\right)  }$ satisfying
$r>w$. Thus, $\kk x_{uq}$ is an addend of the sum $\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}$ (namely, the
addend for $r=uq$). Therefore, $\kk x_{uq}\subset\sum\limits_{\substack{r\in
\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}}\kk x_{r}=G$ (since
$G=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  };\\r>w}%
}\kk x_{r}$).
\par
But (\ref{sol.freelie.2.c.xuxv}) (applied to $u$ and $q$ instead of $u$ and
$v$) shows that $x_{u}x_{q}=x_{uq}=\underbrace{1}_{\in\kk}x_{uq}%
\in\kk x_{uq}\subset G$. This proves (\ref{sol.freelie.2.c.bound3}).}.
Hence,%
\begin{equation}
x_{u}\left(  b_{v}-x_{v}\right)  \in G \label{sol.freelie.2.c.bound3b}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.c.bound3b}):} Subtracting
$x_{v}$ from both sides of the relation (\ref{sol.freelie.2.c.ineq2a}), we
obtain%
\[
b_{v}-x_{v}\in x_{v}+\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q>v}}\kk x_{q}-x_{v}=\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(
v\right)  };\\q>v}}\kk x_{q}.
\]
Now,%
\begin{align*}
&  x_{u}\underbrace{\left(  b_{v}-x_{v}\right)  }_{\in\sum\limits_{\substack{q\in
\mathfrak{A}^{\ell\left(  v\right)  };\\q>v}}\kk x_{q}}\\
&  \in x_{u}\left(  \sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)
};\\q>v}}\kk x_{q}\right)  \subset\sum\limits_{\substack{q\in\mathfrak{A}%
^{\ell\left(  v\right)  };\\q>v}}\kk\underbrace{x_{u}x_{q}%
}_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound3}))}}}\subset
\sum\limits_{\substack{q\in\mathfrak{A}^{\ell\left(  v\right)  };\\q>v}%
}\kk G\subset G
\end{align*}
(since $G$ is a $\kk$-module). This proves
(\ref{sol.freelie.2.c.bound3b}).}.

Now,
\begin{align*}
b_{w}  &  =\left[  b_{u},b_{v}\right]  =\underbrace{b_{u}}_{=\left(
b_{u}-x_{u}\right)  +x_{u}}b_{v}-b_{v}b_{u}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of the Lie bracket on }T\left(  V\right)  \right) \\
&  =\underbrace{\left(  \left(  b_{u}-x_{u}\right)  +x_{u}\right)  b_{v}%
}_{=\left(  b_{u}-x_{u}\right)  b_{v}+x_{u}b_{v}}-b_{v}b_{u}=\left(
b_{u}-x_{u}\right)  b_{v}+x_{u}\underbrace{b_{v}}_{=\left(  b_{v}%
-x_{v}\right)  +x_{v}}-b_{v}b_{u}\\
&  =\left(  b_{u}-x_{u}\right)  b_{v}+\underbrace{x_{u}\left(  \left(
b_{v}-x_{v}\right)  +x_{v}\right)  }_{=x_{u}\left(  b_{v}-x_{v}\right)
+x_{u}x_{v}}-b_{v}b_{u}\\
&  =\underbrace{\left(  b_{u}-x_{u}\right)  b_{v}}_{\substack{\in G\\\text{(by
(\ref{sol.freelie.2.c.bound2b}))}}}+\underbrace{x_{u}\left(  b_{v}%
-x_{v}\right)  }_{\substack{\in G\\\text{(by (\ref{sol.freelie.2.c.bound3b}%
))}}}+x_{u}x_{v}-\underbrace{b_{v}b_{u}}_{\substack{\in G\\\text{(by
(\ref{sol.freelie.2.c.bound1b}))}}}\\
&  \in G+G+x_{u}x_{v}-G=\underbrace{x_{u}x_{v}}_{\substack{=x_{uv}\\\text{(by
(\ref{sol.freelie.2.c.xuxv}))}}}+\underbrace{G+G-G}_{\substack{\subset
G\\\text{(since }G\text{ is a }\kk\text{-module)}}}\subset
\underbrace{x_{uv}}_{\substack{=x_{w}\\\text{(since }uv=w\text{)}%
}}+\underbrace{G}_{=\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)
};\\r>w}}\kk x_{r}}\\
&  =x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)  }%
;\\r>w}}\kk x_{r}.
\end{align*}


Thus, $b_{w}\in x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)
};\\r>w}}\kk x_{r}$ is proven.

Let us now forget that we defined $\left(  u,v\right)  $. We have%
\[
b_{w}\in x_{w}+\sum\limits_{\substack{r\in\mathfrak{A}^{\ell\left(  w\right)
};\\r>w}}\kk x_{r}=x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(
w\right)  };\\v>w}}\kk x_{v}%
\]
(here, we have renamed the summation index $r$ as $v$).

Now, let us forget that we fixed $w$. We thus have proven that%
\[
b_{w}\in x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(  w\right)
};\\v>w}}\kk x_{v}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in
\mathfrak{L}\text{ satisfying }\ell\left(  w\right)  =N.
\]
In other words, we have proven that Exercise \ref{exe.freelie.2}(c) holds
under the condition that $\ell\left(  w\right)  =N$. Thus, our induction is
complete, and Exercise \ref{exe.freelie.2}(c) is solved.

(d) We know that $\mathfrak{g}$ is a Lie subalgebra of $T\left(  V\right)  $
(by Exercise \ref{exe.freelie.2}(a)). Thus, $\left[  \mathfrak{g}%
,\mathfrak{g}\right]  \subset\mathfrak{g}$.

We first notice that
\begin{equation}
b_{w}\in\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}.
\label{sol.freelie.2.d.1}%
\end{equation}


\textit{Proof of (\ref{sol.freelie.2.d.1}):} We shall prove
(\ref{sol.freelie.2.d.1}) by strong induction on $\ell\left(  w\right)  $:

\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{sol.freelie.2.d.1}) holds under the condition that $\ell\left(
w\right)  <N$. We need to show that (\ref{sol.freelie.2.d.1}) also holds under
the condition that $\ell\left(  w\right)  =N$.

We have assumed that (\ref{sol.freelie.2.d.1}) holds under the condition that
$\ell\left(  w\right)  <N$. In other words, we have%
\begin{equation}
b_{w}\in\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}%
\text{ satisfying }\ell\left(  w\right)  <N. \label{sol.freelie.2.d.1.pf.hyp}%
\end{equation}


Now, let $w\in\mathfrak{L}$ be such that $\ell\left(  w\right)  =N$. We shall
prove that $b_{w}\in\mathfrak{g}$.

If $\ell\left(  w\right)  \leq1$, then $b_{w}\in\mathfrak{g}$
holds\footnote{\textit{Proof.} Assume that $\ell\left(  w\right)  \leq1$. The
word $w$ is Lyndon (since $w\in\mathfrak{L}$) and thus nonempty. Hence,
$\ell\left(  w\right)  \geq1$. Combined with $\ell\left(  w\right)  \leq1$,
this yields $\ell\left(  w\right)  =1$. In other words, the word $w$ consists
of a single letter. In other words, $w=\left(  a\right)  $ for some
$a\in\mathfrak{A}$. Consider this $a$. The definition of $b_{w}$ then yields
$b_{w}=x_{a}$ (since $\ell\left(  w\right)  =1$ and $w=\left(  a\right)  $).
\par
Now, $b_{w}=x_{a}\in V\subset\mathfrak{g}$, qed.}. Hence, for the rest of this
proof, we can WLOG assume that we don't have $\ell\left(  w\right)  \leq1$.
Assume this.

The word $w$ is Lyndon (since $w\in\mathfrak{L}$) and satisfies $\ell\left(
w\right)  >1$ (since we don't have $\ell\left(  w\right)  \leq1$). Thus, $w$
is a Lyndon word of length $>1$. Therefore, $\operatorname*{stf}w$ is
well-defined. Let $\left(  u,v\right)  =\operatorname*{stf}w$. The recursive
definition of $b_{w}$ yields $b_{w}=\left[  b_{u},b_{v}\right]  $ (since
$\ell\left(  w\right)  >1$ and $\left(  u,v\right)  =\operatorname*{stf}w$).

From Exercise \ref{exe.words.lyndon.stf.basic}(a) (applied to $\left(
g,h\right)  =\left(  u,v\right)  $), we see that $v$ is the longest Lyndon
proper suffix of $w$. In particular, $v$ is a Lyndon proper suffix of $w$. In
other words, $v$ is a Lyndon word and a proper suffix of $w$. Also, Exercise
\ref{exe.words.lyndon.stf.basic}(d) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that the word $u$ is Lyndon. From Exercise
\ref{exe.words.lyndon.stf.basic}(b) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $), we conclude that we have $w=uv$.

The word $u$ is Lyndon and thus nonempty. Hence, $\ell\left(  u\right)  \geq
1$. Also, the word $v$ is Lyndon and thus nonempty. Thus, $\ell\left(
v\right)  \geq1$. We have $u\in\mathfrak{L}$ (since $u$ is Lyndon) and
$v\in\mathfrak{L}$ (since $v$ is Lyndon).

Now, $\ell\left(  \underbrace{w}_{=uv}\right)  =\ell\left(  uv\right)
=\ell\left(  u\right)  +\underbrace{\ell\left(  v\right)  }_{\geq1>0}%
>\ell\left(  u\right)  $, so that $\ell\left(  u\right)  <\ell\left(
w\right)  =N$. Thus, (\ref{sol.freelie.2.d.1.pf.hyp}) (applied to $u$ instead
of $w$) shows that $b_{u}\in\mathfrak{g}$.

Also, $\ell\left(  \underbrace{w}_{=uv}\right)  =\ell\left(  uv\right)
=\underbrace{\ell\left(  u\right)  }_{\geq1>0}+\ell\left(  v\right)
>\ell\left(  v\right)  $, so that $\ell\left(  v\right)  <\ell\left(
w\right)  =N$. Thus, (\ref{sol.freelie.2.d.1.pf.hyp}) (applied to $v$ instead
of $w$) shows that $b_{v}\in\mathfrak{g}$.

Now, $b_{w}=\left[  \underbrace{b_{u}}_{\in\mathfrak{g}},\underbrace{b_{v}%
}_{\in\mathfrak{g}}\right]  \in\left[  \mathfrak{g},\mathfrak{g}\right]
\subset\mathfrak{g}$.

Let us now forget that we fixed $w$. We thus have shown that $b_{w}%
\in\mathfrak{g}$ for every $w\in\mathfrak{L}$ satisfying $\ell\left(
w\right)  =N$. In other words, (\ref{sol.freelie.2.d.1}) holds under the
condition that $\ell\left(  w\right)  =N$. This completes the induction step.
The induction proof of (\ref{sol.freelie.2.d.1}) is thus complete.

Now, (\ref{sol.freelie.2.d.1}) shows that $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a family of elements of $\mathfrak{g}$. Now, using
Exercise \ref{exe.freelie.1}(a), it is easy to conclude that this family
$\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk$-module
$\mathfrak{g}$\ \ \ \ \footnote{\textit{Proof.} Let $B$ be the $\kk%
$-submodule of $\mathfrak{g}$ spanned by the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ (this is well-defined since $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a family of elements of $\mathfrak{g}$). Then,
clearly,%
\begin{equation}
b_{w}\in B\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}.
\label{sol.freelie.2.d.span.pf.1}%
\end{equation}
\par
Whenever $w$ is a Lyndon word of length $>1$, we have%
\[
b_{w}=\left[  b_{u},b_{v}\right]  ,\ \ \ \ \ \ \ \ \ \ \text{where }\left(
u,v\right)  =\operatorname*{stf}w
\]
(due to the recursive definition of $b_{w}$). Thus, Exercise
\ref{exe.freelie.1}(a) shows that $B$ is a Lie subalgebra of $\mathfrak{g}$.
\par
Now, fix $v\in V$. We shall show that $v\in B$.
\par
Indeed, recall that $\left(  x_{a}\right)  _{a\in\mathfrak{A}}$ is a basis of
the $\kk$-module $V$. Hence, the family $\left(  x_{a}\right)
_{a\in\mathfrak{A}}$ spans the $\kk$-module $V$.
\par
But we want to prove the relation $v\in B$. This relation is $\kk%
$-linear in $v$ (since $B$ is a $\kk$-module). Hence, we can WLOG
assume that $v$ belongs to the family $\left(  x_{a}\right)  _{a\in
\mathfrak{A}}$ (since this family $\left(  x_{a}\right)  _{a\in\mathfrak{A}}$
spans the $\kk$-module $V$). Assume this.
\par
We know that $v$ belongs to the family $\left(  x_{a}\right)  _{a\in
\mathfrak{A}}$. In other words, there exists some $a\in\mathfrak{A}$ such that
$v=x_{a}$. Consider this $a$. Then, $v=x_{a}$.
\par
But let $w$ be the one-letter word $\left(  a\right)  $. Then, $w$ is a
one-letter word, and thus is Lyndon (since every one-letter word is Lyndon).
The word $w$ has length $1$ (since it is a one-letter word), and thus
satisfies $b_{w}=x_{a}$ (by the definition of $b_{w}$, since $w=\left(
a\right)  $). Compared with $v=x_{a}$, this yields $v=b_{w}\in B$ (by
(\ref{sol.freelie.2.d.span.pf.1})).
\par
Now, let us forget that we have fixed $v$. We thus have proven that $v\in B$
for every $v\in V$. In other words, $V\subset B$.
\par
Now, we know that $B$ is a Lie subalgebra of $T\left(  V\right)  $ satisfying
$V\subset B$. Thus, $\mathfrak{g}\subset B$ (by Exercise \ref{exe.freelie.2}%
(b), applied to $\mathfrak{k}=B$). Combined with $B\subset\mathfrak{g}$ (since
$B$ is a $\kk$-submodule of $\mathfrak{g}$), this yields $\mathfrak{g}%
=B$. But the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the
$\kk$-module $B$ (since $B$ is the $\kk$-submodule of
$\mathfrak{g}$ spanned by the family $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$). Since $\mathfrak{g}=B$, this rewrites as follows: The family
$\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk$-module
$\mathfrak{g}$. Qed.}.

We shall now show that the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$
is $\kk$-linearly independent. In order to do so, we will prove a more
general result:

\begin{proposition}
\label{prop.sol.freelie.2.d.general}Let $\mathfrak{V}$ be a $\kk%
$-module, and let $W$ be a totally ordered set. Let $\left(  p_{w}\right)
_{w\in W}$ be a $\kk$-linearly independent family of elements of
$\mathfrak{V}$. Let $L$ be a subset of $W$. Let $\left(  s_{w}\right)  _{w\in
L}$ be a family of elements of $\mathfrak{V}$. Assume that every $w\in L$
satisfies%
\begin{equation}
s_{w}\in p_{w}+\sum\limits_{\substack{v\in W;\\v>w}}\kk p_{v}.
\label{pf.prop.sol.freelie.2.d.general.premise}%
\end{equation}
(Here, the ``$>$'' sign under the sum refers
to the total order on $W$.) Then, the family $\left(  s_{w}\right)  _{w\in L}$
is $\kk$-linearly independent.
\end{proposition}

Before we prove this, let us show a simple lemma:

\begin{lemma}
\label{lem.sol.freelie.2.d.general.lem2}Let $\mathfrak{V}$ be a $\kk%
$-module, and let $W$ be a set. Let $\left(  p_{w}\right)  _{w\in W}$ be a
$\kk$-linearly independent family of elements of $\mathfrak{V}$. Let
$x\in W$ and $\lambda\in\kk$ be such that%
\begin{equation}
0\in\lambda p_{x}+\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}.
\label{pf.prop.sol.freelie.2.d.general.lem.premise}%
\end{equation}
Then, $\lambda=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.freelie.2.d.general.lem2}.]The family $\left(
p_{w}\right)  _{w\in W}$ is $\kk$-linearly independent. In other words,
if $\left(  \mu_{w}\right)  _{w\in W}\in\kk^{W}$ is a family of
elements of $\kk$ such that $\left(  \text{all but finitely many }w\in
W\text{ satisfy }\mu_{w}=0\right)  $ and $\sum_{w\in W}\mu_{w}p_{w}=0$, then
\begin{equation}
\left(  \mu_{w}\right)  _{w\in W}=\left(  0\right)  _{w\in W}.
\label{pf.prop.sol.freelie.2.d.general.lem.pf.lind}%
\end{equation}


Now, we have $0\in\lambda p_{x}+\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}$. In other words, there exists some $\alpha\in\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}$ such that $0=\lambda
p_{x}+\alpha$. Consider this $\alpha$.

We have $\alpha\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}$.
In other words, there exists some family $\left(  \kappa_{v}\right)  _{v\in
W\setminus\left\{  x\right\}  }\in\kk^{W\setminus\left\{  x\right\}  }$
such that $\left(  \text{all but finitely many }v\in W\setminus\left\{
x\right\}  \text{ satisfy }\kappa_{v}=0\right)  $ and $\alpha=\sum_{v\in
W\setminus\left\{  x\right\}  }\kappa_{v}p_{v}$. Consider this family $\left(
\kappa_{v}\right)  _{v\in W\setminus\left\{  x\right\}  }$. Clearly,%
\begin{equation}
\alpha=\sum_{v\in W\setminus\left\{  x\right\}  }\kappa_{v}p_{v}=\sum_{w\in
W\setminus\left\{  x\right\}  }\kappa_{w}p_{w}
\label{pf.prop.sol.freelie.2.d.general.lem.pf.alpha}%
\end{equation}
(here, we have renamed the summation index $v$ as $w$).

Define a family $\left(  \mu_{w}\right)  _{w\in W}\in\kk^{W}$ by
setting%
\[
\left(  \mu_{w}=%
\begin{cases}
\kappa_{w}, & \text{if }w\in W\setminus\left\{  x\right\}  ;\\
\lambda, & \text{if }w\notin W\setminus\left\{  x\right\}
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }w\in W\right)  .
\]
(This is clearly well-defined, because for every $w\in W\setminus\left\{
x\right\}  $, the element $\kappa_{w}\in\kk$ is well-defined (since
$\left(  \kappa_{v}\right)  _{v\in W\setminus\left\{  x\right\}  }%
\in\kk^{W\setminus\left\{  x\right\}  }$).) Clearly, every $w\in
W\setminus\left\{  x\right\}  $ satisfies%
\begin{align}
\mu_{w}  &  =%
\begin{cases}
\kappa_{w}, & \text{if }w\in W\setminus\left\{  x\right\}  ;\\
\lambda, & \text{if }w\notin W\setminus\left\{  x\right\}
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu_{w}\text{ (since
}w\in W\setminus\left\{  x\right\}  \subset W\text{)}\right) \nonumber\\
&  =\kappa_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }w\in W\setminus\left\{
x\right\}  \right)  . \label{pf.prop.sol.freelie.2.d.general.lem.14}%
\end{align}


If we had $x\in W\setminus\left\{  x\right\}  $, then we would have $x\neq x$
(since $x\in W\setminus\left\{  x\right\}  $), which would contradict $x=x$.
Hence, we cannot have $x\in W\setminus\left\{  x\right\}  $. In other words,
we have $x\notin W\setminus\left\{  x\right\}  $. Now, the definition of
$\mu_{x}$ yields $\mu_{x}=%
\begin{cases}
\kappa_{x}, & \text{if }x\in W\setminus\left\{  x\right\}  ;\\
\lambda, & \text{if }x\notin W\setminus\left\{  x\right\}
\end{cases}  \quad
=\lambda$ (since $x\notin W\setminus\left\{  x\right\}  $).

We have $\left(  \text{all but finitely many }w\in W\text{ satisfy }\mu
_{w}=0\right)  $\ \ \ \ \footnote{\textit{Proof.} We know that all but
finitely many $v\in W\setminus\left\{  x\right\}  $ satisfy $\kappa_{v}=0$. In
other words, there exists a finite subset $S$ of $W\setminus\left\{
x\right\}  $ such that every $v\in\left(  W\setminus\left\{  x\right\}
\right)  \setminus S$ satisfies $\kappa_{v}=0$. Consider this $S$.
\par
Let $T$ be the set $S\cup\left\{  x\right\}  $. Then, $T=\underbrace{S}%
_{\subset W\setminus\left\{  x\right\}  \subset W}\cup\left\{  x\right\}
\subset W\cup\left\{  x\right\}  =W$ (since $x\in W$). Also, $T=S\cup\left\{
x\right\}  $; thus, the set $T$ is finite (since it is the union of the two
finite sets $S$ and $\left\{  x\right\}  $). Thus, $T$ is a finite subset of
$W$. Notice that $x\in\left\{  x\right\}  \subset S\cup\left\{  x\right\}
=T$.
\par
Let $w\in W\setminus T$. We will show that $\mu_{w}=0$.
\par
We have $w\in W\setminus T$. In other words, $w\in W$ and $w\notin T$. If we
had $w=x$, then we would have $w=x\in T$, which would contradict $w\notin T$.
Hence, we cannot have $w=x$. In other words, we have $w\neq x$. Also, if we
had $w\in S$, then we would have $w\in S\subset S\cup\left\{  x\right\}  =T$,
which would contradict $w\notin T$. Hence, we cannot have $w\in S$. In other
words, we have $w\notin S$. Now, $w\in W\setminus\left\{  x\right\}  $ (since
$w\in W$ and $w\neq x$). Combining this with $w\notin S$, we obtain
$w\in\left(  W\setminus\left\{  x\right\}  \right)  \setminus S$.
\par
Now, recall that every $v\in\left(  W\setminus\left\{  x\right\}  \right)
\setminus S$ satisfies $\kappa_{v}=0$. Applying this to $v=w$, we obtain
$\kappa_{w}=0$ (since $w\in\left(  W\setminus\left\{  x\right\}  \right)
\setminus S$). But the definition of $\mu_{w}$ yields $\mu_{w}=%
\begin{cases}
\kappa_{w}, & \text{if }w\in W\setminus\left\{  x\right\}  ;\\
\lambda, & \text{if }w\notin W\setminus\left\{  x\right\}
\end{cases}  \quad
=\kappa_{w}$ (since $w\in W\setminus\left\{  x\right\}  $) and thus $\mu
_{w}=\kappa_{w}=0$.
\par
Let us now forget that we fixed $w$. We thus have proven that every $w\in
W\setminus T$ satisfies $\mu_{w}=0$.
\par
Thus, there exists a finite subset $Q$ of $W$ such that every $w\in W\setminus
Q$ satisfies $\mu_{w}=0$ (namely, $Q=T$). In other words, all but finitely
many $w\in W$ satisfy $\mu_{w}=0$. Qed.} and
\begin{align*}
\sum_{w\in W}\mu_{w}p_{w}  &  =\underbrace{\mu_{x}}_{=\lambda}p_{x}%
+\underbrace{\sum\limits_{\substack{w\in W;\\w\neq x}}}_{=\sum_{w\in W\setminus
\left\{  x\right\}  }}\mu_{w}p_{w}=\lambda p_{x}+\sum_{w\in W\setminus\left\{
x\right\}  }\underbrace{\mu_{w}}_{\substack{=\kappa_{w}\\\text{(by
(\ref{pf.prop.sol.freelie.2.d.general.lem.14}))}}}p_{w}\\
&  =\lambda p_{x}+\underbrace{\sum_{w\in W\setminus\left\{  x\right\}  }%
\kappa_{w}p_{w}}_{\substack{=\alpha\\\text{(by
(\ref{pf.prop.sol.freelie.2.d.general.lem.pf.alpha}))}}}=\lambda p_{x}%
+\alpha=0
\end{align*}
(since $0=\lambda p_{x}+\alpha$). Thus,
(\ref{pf.prop.sol.freelie.2.d.general.lem.pf.lind}) shows that $\left(
\mu_{w}\right)  _{w\in W}=\left(  0\right)  _{w\in W}$. In other words,
$\mu_{w}=0$ for every $w\in W$. Applying this to $w=x$, we obtain $\mu_{x}=0$.
Compared with $\mu_{x}=\lambda$, this yields $\lambda=0$. This proves Lemma
\ref{lem.sol.freelie.2.d.general.lem2}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.sol.freelie.2.d.general}.]We shall first show
the following fact: If $S$ is a finite subset of $L$, then
\begin{equation}
\text{the family }\left(  s_{w}\right)  _{w\in S}\text{ is }\kk%
\text{-linearly independent.} \label{pf.prop.sol.freelie.2.d.general.1}%
\end{equation}


\textit{Proof of (\ref{pf.prop.sol.freelie.2.d.general.1}):} We shall prove
(\ref{pf.prop.sol.freelie.2.d.general.1}) by induction over $\left\vert
S\right\vert $:

\textit{Induction base:} If $S$ is a finite subset of $L$ satisfying
$\left\vert S\right\vert =0$, then the family $\left(  s_{w}\right)  _{w\in
S}$ is $\kk$-linearly independent\footnote{\textit{Proof.} Let $S$ be a
finite subset of $L$ satisfying $\left\vert S\right\vert =0$. Then,
$S=\varnothing$ (since $\left\vert S\right\vert =0$), and thus the family
$\left(  s_{w}\right)  _{w\in S}$ is the empty family. Hence, this family
$\left(  s_{w}\right)  _{w\in S}$ is $\kk$-linearly independent (since
the empty family is always $\kk$-linearly independent). Qed.}. In other
words, (\ref{pf.prop.sol.freelie.2.d.general.1}) holds in the case when
$\left\vert S\right\vert =0$. This completes the induction base.

\textit{Induction step:} Let $K$ be a positive integer. Assume that
(\ref{pf.prop.sol.freelie.2.d.general.1}) is proven in the case when
$\left\vert S\right\vert =K-1$. We must prove that
(\ref{pf.prop.sol.freelie.2.d.general.1}) holds in the case when $\left\vert
S\right\vert =K$.

We have assumed that (\ref{pf.prop.sol.freelie.2.d.general.1}) is proven in
the case when $\left\vert S\right\vert =K-1$. In other words, if $S$ is a
finite subset of $L$ satisfying $\left\vert S\right\vert =K-1$, then
\begin{equation}
\text{the family }\left(  s_{w}\right)  _{w\in S}\text{ is }\kk%
\text{-linearly independent.} \label{pf.prop.sol.freelie.2.d.general.1.pf.hyp}%
\end{equation}


Now, let $S$ be a finite subset of $L$ satisfying $\left\vert S\right\vert
=K$. Let $\left(  \lambda_{w}\right)  _{w\in S}\in\kk^{S}$ be a family
of elements of $\kk$ such that $\sum_{w\in S}\lambda_{w}s_{w}=0$. We
shall prove that $\left(  \lambda_{w}\right)  _{w\in S}=\left(  0\right)
_{w\in S}$.

We have $\left\vert S\right\vert =K>0$. Thus, the set $S$ is nonempty. Also,
$S\subset L\subset W$, and thus the set $S$ is totally ordered (since it is a
subset of the totally ordered set $W$). Thus, the set $S$ is a nonempty finite
totally ordered set. Therefore, this set $S$ has a smallest element (since
every nonempty finite totally ordered set has a smallest element). Let $x$ be
this smallest element.

We know that $x$ is the smallest element of $S$. In other words,%
\begin{equation}
x\leq w\ \ \ \ \ \ \ \ \ \ \text{for every }w\in S.
\label{pf.prop.sol.freelie.2.d.general.1.pf.1}%
\end{equation}
Also, $x\in S$ (since $x$ is the smallest element of $S$), and thus
$\left\vert S\setminus\left\{  x\right\}  \right\vert =\underbrace{\left\vert
S\right\vert }_{=K}-1=K-1$. Hence, we can apply
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.hyp}) to $S\setminus\left\{
x\right\}  $ instead of $S$. As a result, we see that the family $\left(
s_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }$ is $\kk$-linearly
independent. In other words, if $\left(  \mu_{w}\right)  _{w\in S\setminus
\left\{  x\right\}  }\in\kk^{S\setminus\left\{  x\right\}  }$ is a
family of elements of $\kk$ such that $\sum_{w\in S\setminus\left\{
x\right\}  }\mu_{w}s_{w}=0$, then
\begin{equation}
\left(  \mu_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }=\left(
0\right)  _{w\in S\setminus\left\{  x\right\}  }.
\label{pf.prop.sol.freelie.2.d.general.1.pf.hyp-used}%
\end{equation}


On the other hand, every $w\in S\setminus\left\{  x\right\}  $ satisfies%
\begin{equation}
s_{w}\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}
\label{pf.prop.sol.freelie.2.d.general.1.pf.bound1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound1}%
):} Let $w\in S\setminus\left\{  x\right\}  $. Thus, $w\in S$ and $w\neq x$.
\par
Let $v\in\left\{  q\in W\ \mid\ q>w\right\}  $. We shall show that $v\in
W\setminus\left\{  x\right\}  $.
\par
Indeed, we have $v\in\left\{  q\in W\ \mid\ q>w\right\}  $. In other words,
$v$ is an element $q\in W$ satisfying $q>w$. In other words, $v$ is an element
of $W$ and satisfies $v>w$. But $w\in S\setminus\left\{  x\right\}  \subset
S$, so that (\ref{pf.prop.sol.freelie.2.d.general.1.pf.1}) shows that $x\leq
w$. Hence, $w\geq x$, so that $v>w\geq x$. Thus, $v\neq x$. Combining $v\in W$
with $v\neq x$, we obtain $v\in W\setminus\left\{  x\right\}  $.
\par
Let us now forget that we fixed $v$. We thus have proven that every
$v\in\left\{  q\in W\ \mid\ q>w\right\}  $ satisfies $v\in W\setminus\left\{
x\right\}  $. In other words, $\left\{  q\in W\ \mid\ q>w\right\}  \subset
W\setminus\left\{  x\right\}  $. Therefore, the sum $\sum_{v\in\left\{  q\in
W\ \mid\ q>w\right\}  }\kk p_{v}$ is a subsum of the sum $\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}$. Hence, $\sum_{v\in\left\{
q\in W\ \mid\ q>w\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}$.
\par
Also, combining $w\in W$ with $w\neq x$, we obtain $w\in W\setminus\left\{
x\right\}  $. Hence, $\kk p_{w}$ is an addend of the sum $\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}$ (namely, the addend for
$v=w$). Thus, $\kk p_{w}\subset\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}$. Now, $p_{w}=\underbrace{1}_{\in\kk}p_{w}%
\in\kk p_{w}\subset\sum_{v\in W\setminus\left\{  x\right\}  }%
\kk p_{v}$.
\par
Now, $w\in S\subset L\subset W$. Hence,
(\ref{pf.prop.sol.freelie.2.d.general.premise}) yields%
\begin{align*}
s_{w}  &  \in\underbrace{p_{w}}_{\in\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}}+\underbrace{\sum\limits_{\substack{v\in W;\\v>w}}}_{=\sum
_{v\in\left\{  q\in W\ \mid\ q>w\right\}  }}\kk p_{v}\subset\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}+\underbrace{\sum_{v\in\left\{
q\in W\ \mid\ q>w\right\}  }\kk p_{v}}_{\subset\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}}\\
&  \subset\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}+\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}%
\end{align*}
(since $\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}$ is a
$\kk$-module). This proves
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound1}).}. Also,
\begin{equation}
s_{x}-p_{x}\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}
\label{pf.prop.sol.freelie.2.d.general.1.pf.bound2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound2}%
):} Let $v\in\left\{  q\in W\ \mid\ q>x\right\}  $. We shall show that $v\in
W\setminus\left\{  x\right\}  $.
\par
Indeed, we have $v\in\left\{  q\in W\ \mid\ q>x\right\}  $. In other words,
$v$ is an element $q\in W$ satisfying $q>x$. In other words, $v$ is an element
of $W$ and satisfies $v>x$. Hence, $v\neq x$ (since $v>x$). Combining $v\in W$
with $v\neq x$, we obtain $v\in W\setminus\left\{  x\right\}  $.
\par
Let us now forget that we fixed $v$. We thus have proven that every
$v\in\left\{  q\in W\ \mid\ q>x\right\}  $ satisfies $v\in W\setminus\left\{
x\right\}  $. In other words, $\left\{  q\in W\ \mid\ q>x\right\}  \subset
W\setminus\left\{  x\right\}  $. Therefore, the sum $\sum_{v\in\left\{  q\in
W\ \mid\ q>x\right\}  }\kk p_{v}$ is a subsum of the sum $\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}$. Hence, $\sum_{v\in\left\{
q\in W\ \mid\ q>x\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}$.
\par
Now, $x\in S\subset L\subset W$. Hence,
(\ref{pf.prop.sol.freelie.2.d.general.premise}) (applied to $w=x$) yields%
\[
s_{x}\in p_{x}+\sum\limits_{\substack{v\in W;\\v>x}}\kk p_{v}.
\]
Subtracting $p_{x}$ from both sides of this relation, we obtain%
\[
s_{x}-p_{x}\in p_{x}+\sum\limits_{\substack{v\in W;\\v>x}}\kk p_{v}%
-p_{x}=\underbrace{\sum\limits_{\substack{v\in W;\\v>x}}}_{=\sum_{v\in\left\{  q\in
W\ \mid\ q>x\right\}  }}\kk p_{v}=\sum_{v\in\left\{  q\in W\ \mid
\ q>x\right\}  }\kk p_{v}\subset\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}.
\]
This proves (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound2}).}. Now,
$\sum_{w\in S}\lambda_{w}s_{w}=0$, so that%
\begin{align*}
0  &  =\sum_{w\in S}\lambda_{w}s_{w}=\lambda_{x}\underbrace{s_{x}}%
_{=p_{x}+\left(  s_{x}-p_{x}\right)  }+\underbrace{\sum\limits_{\substack{w\in
S;\\w\neq x}}}_{=\sum_{w\in S\setminus\left\{  x\right\}  }}\lambda_{w}%
s_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }x\in S\right) \\
&  =\underbrace{\lambda_{x}\left(  p_{x}+\left(  s_{x}-p_{x}\right)  \right)
}_{=\lambda_{x}p_{x}+\lambda_{x}\left(  s_{x}-p_{x}\right)  }+\sum_{w\in
S\setminus\left\{  x\right\}  }\lambda_{w}\underbrace{s_{w}}_{\substack{\in
\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}\\\text{(by
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound1}))}}}\\
&  \in\lambda_{x}p_{x}+\lambda_{x}\underbrace{\left(  s_{x}-p_{x}\right)
}_{\substack{\in\sum_{v\in W\setminus\left\{  x\right\}  }\kk%
p_{v}\\\text{(by (\ref{pf.prop.sol.freelie.2.d.general.1.pf.bound2}))}%
}}+\underbrace{\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}}_{\substack{\subset\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\\\text{(since }\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\text{ is a }\kk%
\text{-module)}}}\\
&  \subset\lambda_{x}p_{x}+\underbrace{\lambda_{x}\sum_{v\in W\setminus
\left\{  x\right\}  }\kk p_{v}}_{\substack{\subset\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\\\text{(since }\sum_{v\in
W\setminus\left\{  x\right\}  }\kk p_{v}\text{ is a }\kk%
\text{-module)}}}+\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}\\
&  \subset\lambda_{x}p_{x}+\underbrace{\sum_{v\in W\setminus\left\{
x\right\}  }\kk p_{v}+\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}}_{\substack{\subset\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}\\\text{(since }\sum_{v\in W\setminus\left\{  x\right\}
}\kk p_{v}\text{ is a }\kk\text{-module)}}}\subset\lambda
_{x}p_{x}+\sum_{v\in W\setminus\left\{  x\right\}  }\kk p_{v}.
\end{align*}
Lemma \ref{lem.sol.freelie.2.d.general.lem2} (applied to $\lambda=\lambda_{x}%
$) thus shows that $\lambda_{x}=0$. Thus,%
\[
0=\underbrace{\lambda_{x}}_{=0}s_{x}+\underbrace{\sum\limits_{\substack{w\in
S;\\w\neq x}}}_{=\sum_{w\in S\setminus\left\{  x\right\}  }}\lambda_{w}%
s_{w}=\underbrace{0s_{x}}_{=0}+\sum_{w\in S\setminus\left\{  x\right\}
}\lambda_{w}s_{w}=\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}s_{w}.
\]
In other words, $\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}s_{w}%
=0$. Thus, the family $\left(  \lambda_{w}\right)  _{w\in S\setminus\left\{
x\right\}  }\in\kk^{S\setminus\left\{  x\right\}  }$ satisfies
$\sum_{w\in S\setminus\left\{  x\right\}  }\lambda_{w}s_{w}=0$. Hence,
$\left(  \lambda_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }=\left(
0\right)  _{w\in S\setminus\left\{  x\right\}  }$ (according to
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.hyp-used}), applied to $\left(
\mu_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }=\left(  \lambda
_{w}\right)  _{w\in S\setminus\left\{  x\right\}  }$). In other words,%
\begin{equation}
\lambda_{w}=0\ \ \ \ \ \ \ \ \ \ \text{for every }w\in S\setminus\left\{
x\right\}  . \label{pf.prop.sol.freelie.2.d.general.1.pf.alm1}%
\end{equation}
Hence, $\lambda_{w}=0$ for every $w\in S$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in S$. We must prove that $\lambda_{w}=0$.
\par
If $w\in S\setminus\left\{  x\right\}  $, then $\lambda_{w}=0$ follows from
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.alm1}). Hence, for the rest of the
proof of $\lambda_{w}=0$, we can WLOG assume that we don't have $w\in
S\setminus\left\{  x\right\}  $. Assume this.
\par
We don't have $w\in S\setminus\left\{  x\right\}  $. In other words, we have
$w\notin S\setminus\left\{  x\right\}  $. Combining $w\in S$ with $w\notin
S\setminus\left\{  x\right\}  $, we obtain $w\in S\setminus\left(
S\setminus\left\{  x\right\}  \right)  \subset\left\{  x\right\}  $. Hence,
$w=x$, so that $\lambda_{w}=\lambda_{x}=0$, qed.}. In other words, $\left(
\lambda_{w}\right)  _{w\in S}=\left(  0\right)  _{w\in S}$.

Now, let us forget that we fixed $\left(  \lambda_{w}\right)  _{w\in S}$. We
thus have shown that if $\left(  \lambda_{w}\right)  _{w\in S}\in
\kk^{S}$ is a family of elements of $\kk$ such that $\sum_{w\in
S}\lambda_{w}s_{w}=0$, then $\left(  \lambda_{w}\right)  _{w\in S}=\left(
0\right)  _{w\in S}$. In other words, the family $\left(  s_{w}\right)  _{w\in
S}$ is $\kk$-linearly independent.

Let us now forget that we fixed $S$. We thus have proven that if $S$ is a
finite subset of $L$ satisfying $\left\vert S\right\vert =K$, then the family
$\left(  s_{w}\right)  _{w\in S}$ is $\kk$-linearly independent. In
other words, (\ref{pf.prop.sol.freelie.2.d.general.1}) holds in the case when
$\left\vert S\right\vert =K$. This completes the induction step. Thus, we have
proven (\ref{pf.prop.sol.freelie.2.d.general.1}) by induction.

Now, let us recall that our goal is to show that the family $\left(
s_{w}\right)  _{w\in L}$ is $\kk$-linearly independent.

Let $\left(  \lambda_{w}\right)  _{w\in L}\in\kk^{L}$ be a family of
elements of $\kk$ such that $\left(  \text{all but finitely many }w\in
L\text{ satisfy }\lambda_{w}=0\right)  $ and $\sum_{w\in L}\lambda_{w}s_{w}%
=0$. We shall show that $\left(  \lambda_{w}\right)  _{w\in L}=\left(
0\right)  _{w\in L}$.

We know that all but finitely many $w\in L$ satisfy $\lambda_{w}=0$. In other
words, there exists a finite subset $S$ of $L$ such that
\begin{equation}
\text{every }w\in L\setminus S\text{ satisfies }\lambda_{w}=0.
\label{pf.prop.sol.freelie.2.d.general.1.pf.alm2}%
\end{equation}
Consider this $S$. From (\ref{pf.prop.sol.freelie.2.d.general.1}), we see that
the family $\left(  s_{w}\right)  _{w\in S}$ is $\kk$-linearly
independent. In other words, if $\left(  \mu_{w}\right)  _{w\in S}%
\in\kk^{S}$ is a family of elements of $S$ satisfying $\sum_{w\in S}%
\mu_{w}s_{w}=0$, then
\begin{equation}
\left(  \mu_{w}\right)  _{w\in S}=\left(  0\right)  _{w\in S}.
\label{pf.prop.sol.freelie.2.d.general.1.pf.alm3}%
\end{equation}


Since $\sum_{w\in L}\lambda_{w}s_{w}=0$, we have%
\begin{align*}
0  &  =\sum_{w\in L}\lambda_{w}s_{w}=\underbrace{\sum\limits_{\substack{w\in L;\\w\in
S}}}_{\substack{=\sum_{w\in S}\\\text{(since }S\subset L\text{)}}}\lambda
_{w}s_{w}+\underbrace{\sum\limits_{\substack{w\in L;\\w\notin S}}}_{=\sum_{w\in
L\setminus S}}\lambda_{w}s_{w}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }w\in L\text{ either
satisfies }w\in S\text{ or satisfies }w\notin S\text{ (but never both)}\right)
\\
&  =\sum_{w\in S}\lambda_{w}s_{w}+\sum_{w\in L\setminus S}\underbrace{\lambda
_{w}}_{\substack{=0\\\text{(by
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.alm2}))}}}s_{w}=\sum_{w\in
S}\lambda_{w}s_{w}+\underbrace{\sum_{w\in L\setminus S}0s_{w}}_{=0}=\sum_{w\in
S}\lambda_{w}s_{w}.
\end{align*}
In other words, $\sum_{w\in S}\lambda_{w}s_{w}=0$. Hence,
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.alm3}) (applied to $\left(  \mu
_{w}\right)  _{w\in S}=\left(  \lambda_{w}\right)  _{w\in S}$) shows that
$\left(  \lambda_{w}\right)  _{w\in S}=\left(  0\right)  _{w\in S}$. In other
words,%
\begin{equation}
\text{every }w\in S\text{ satisfies }\lambda_{w}=0.
\label{pf.prop.sol.freelie.2.d.general.1.pf.alm4}%
\end{equation}


Now, every $w\in L$ satisfies $\lambda_{w}=0$\ \ \ \ \footnote{\textit{Proof.}
Let $w\in L$. We must prove that $\lambda_{w}=0$.
\par
If $w\in S$, then $\lambda_{w}=0$ follows from
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.alm4}). Hence, for the rest of the
proof of $\lambda_{w}=0$, we can WLOG assume that we don't have $w\in S$.
Assume this.
\par
We don't have $w\in S$. In other words, we have $w\notin S$. Combining $w\in
L$ with $w\notin S$, we obtain $w\in L\setminus S$. Hence,
(\ref{pf.prop.sol.freelie.2.d.general.1.pf.alm2}) shows that $\lambda_{w}=0$,
qed.}. In other words, $\left(  \lambda_{w}\right)  _{w\in L}=\left(
0\right)  _{w\in L}$.

Now, let us forget that we fixed $\left(  \lambda_{w}\right)  _{w\in L}$. We
thus have shown that if $\left(  \lambda_{w}\right)  _{w\in L}\in
\kk^{L}$ is a family of elements of $\kk$ such that $\left(
\text{all but finitely many }w\in L\text{ satisfy }\lambda_{w}=0\right)  $ and
$\sum_{w\in L}\lambda_{w}s_{w}=0$, then $\left(  \lambda_{w}\right)  _{w\in
L}=\left(  0\right)  _{w\in L}$. In other words, the family $\left(
s_{w}\right)  _{w\in L}$ is $\kk$-linearly independent. This proves
Proposition \ref{prop.sol.freelie.2.d.general}.
\end{proof}

[\textit{Remark:} Proposition \ref{prop.sol.freelie.2.d.general} is actually a
standard criterion for linear independence. It is often summarized by the
motto ``vectors with distinct leading coordinates are linearly
independent''.]

Let us now return to the solution of Exercise \ref{exe.freelie.2}(d). The set
$\mathfrak{A}^{\ast}$ is totally ordered (by the lexicographic order), and the
set $\mathfrak{L}$ is a subset of $\mathfrak{A}^{\ast}$. The family $\left(
x_{w}\right)  _{w\in\mathfrak{A}^{\ast}}$ is a basis of the $\kk%
$-module $T\left(  V\right)  $, and thus is a $\kk$-linearly
independent family of elements of this $\kk$-module. The family
$\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is a family of elements of
$T\left(  V\right)  $. Every $w\in\mathfrak{L}$ satisfies%
\begin{align*}
b_{w}  &  \in x_{w}+\underbrace{\sum\limits_{\substack{v\in\mathfrak{A}^{\ell\left(
w\right)  };\\v>w}}\kk x_{v}}_{\substack{\subset\sum\limits_{\substack{v\in
\mathfrak{A}^{\ast};\\v>w}}\kk x_{v}\\\text{(since }\sum\limits_{\substack{v\in
\mathfrak{A}^{\ell\left(  w\right)  };\\v>w}}\kk x_{v}\text{ is a subsum
of}\\\text{the sum }\sum\limits_{\substack{v\in\mathfrak{A}^{\ast};\\v>w}%
}\kk x_{v}\\\text{(since }\mathfrak{A}^{\ell\left(  w\right)  }%
\subset\mathfrak{A}^{\ast}\text{))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.freelie.2}(c)}\right) \\
&  \subset x_{w}+\sum\limits_{\substack{v\in\mathfrak{A}^{\ast};\\v>w}}\kk%
x_{v}.
\end{align*}
Hence, Proposition \ref{prop.sol.freelie.2.d.general} (applied to
$\mathfrak{V}=T\left(  V\right)  $, $W=\mathfrak{A}^{\ast}$, $\left(
p_{w}\right)  _{w\in\mathfrak{A}^{\ast}}=\left(  x_{w}\right)  _{w\in
\mathfrak{A}^{\ast}}$, $L=\mathfrak{L}$ and $\left(  s_{w}\right)
_{w\in\mathfrak{L}}=\left(  b_{w}\right)  _{w\in\mathfrak{L}}$) shows that the
family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is $\kk$-linearly
independent. Combining this with the fact that this family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ spans the $\kk$-module
$\mathfrak{g}$, we therefore conclude that the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a basis of the $\kk$-module $\mathfrak{g}$.
This solves Exercise \ref{exe.freelie.2}(d).

(e) Let us first show a simple lemma:

\begin{lemma}
\label{lem.sol.freelie.2.e.generators-determine}Let $\mathfrak{a}$ and
$\mathfrak{b}$ be two $\kk$-Lie algebras. Let $p:\mathfrak{a}%
\rightarrow\mathfrak{b}$ and $q:\mathfrak{a}\rightarrow\mathfrak{b}$ be two
Lie algebra homomorphisms. Then, $\ker\left(  p-q\right)  $ is
a Lie subalgebra of $\mathfrak{a}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.freelie.2.e.generators-determine}.]Let
$x\in\ker\left(  p-q\right)  $ and $y\in\ker%
\left(  p-q\right)  $. Then, $\left(  p-q\right)  \left(  x\right)  =0$ (since
$x\in\ker\left(  p-q\right)  $), so that $0=\left(  p-q\right)
\left(  x\right)  =p\left(  x\right)  -q\left(  x\right)  $, and thus
$p\left(  x\right)  =q\left(  x\right)  $. The same argument (applied to $y$
instead of $x$) shows that $p\left(  y\right)  =q\left(  y\right)  $.

But $p$ is a Lie algebra homomorphism, and thus satisfies $p\left(  \left[
x,y\right]  \right)  =\left[  \underbrace{p\left(  x\right)  }_{=q\left(
x\right)  },\underbrace{p\left(  y\right)  }_{=q\left(  y\right)  }\right]
=\left[  q\left(  x\right)  ,q\left(  y\right)  \right]  =q\left(  \left[
x,y\right]  \right)  $ (since $q$ is a Lie algebra homomorphism). Now,
$\left(  p-q\right)  \left(  \left[  x,y\right]  \right)
=\underbrace{p\left(  \left[  x,y\right]  \right)  }_{=q\left(  \left[
x,y\right]  \right)  }-q\left(  \left[  x,y\right]  \right)  =q\left(  \left[
x,y\right]  \right)  -q\left(  \left[  x,y\right]  \right)  =0$. In other
words, $\left[  x,y\right]  \in\ker\left(  p-q\right)  $.

Let us now forget that we fixed $x$ and $y$. We thus have proven that $\left[
x,y\right]  \in\ker\left(  p-q\right)  $ whenever
$x\in\ker\left(  p-q\right)  $ and $y\in\ker%
\left(  p-q\right)  $. In other words, the set $\ker\left(
p-q\right)  $ is closed under the Lie bracket. Also, clearly, $p-q$ is a
$\kk$-linear map (since both maps $p$ and $q$ are $\kk$-linear),
and therefore its kernel $\ker\left(  p-q\right)  $ is a
$\kk$-submodule of $\mathfrak{a}$. Thus, $\ker\left(
p-q\right)  $ is a $\kk$-submodule of $\mathfrak{a}$ which is closed
under the Lie bracket. In other words, $\ker\left(  p-q\right)
$ is a Lie subalgebra of $\mathfrak{a}$. This proves Lemma
\ref{lem.sol.freelie.2.e.generators-determine}.
\end{proof}

Now, it is easy to see that if $\Xi_{1}$ and $\Xi_{2}$ are two Lie algebra
homomorphisms $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that every
$a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(  a\right)
$, then $\Xi_{1}=\Xi_{2}$\ \ \ \ \footnote{\textit{Proof.} Let $\Xi_{1}$ and
$\Xi_{2}$ be two Lie algebra homomorphisms $\Xi:\mathfrak{g}\rightarrow
\mathfrak{h}$ such that every $a\in\mathfrak{A}$ satisfies $\Xi\left(
x_{a}\right)  =\xi\left(  a\right)  $. We must prove that $\Xi_{1}=\Xi_{2}$.
\par
Let $v\in V$. We shall prove that $v\in\ker\left(  \Xi_{1}%
-\Xi_{2}\right)  $.
\par
Indeed, $\Xi_{1}$ and $\Xi_{2}$ are $\kk$-linear maps, and therefore
$\Xi_{1}-\Xi_{2}$ is a $\kk$-linear map as well. Hence, its kernel
$\ker\left(  \Xi_{1}-\Xi_{2}\right)  $ is a $\kk%
$-submodule of $\mathfrak{g}$.
\par
We know that $\Xi_{1}$ is a Lie algebra homomorphism $\Xi:\mathfrak{g}%
\rightarrow\mathfrak{h}$ such that every $a\in\mathfrak{A}$ satisfies
$\Xi\left(  x_{a}\right)  =\xi\left(  a\right)  $. In other words, $\Xi_{1}$
is a Lie algebra homomorphism $\mathfrak{g}\rightarrow\mathfrak{h}$, and has
the property that%
\begin{equation}
\text{every }a\in\mathfrak{A}\text{ satisfies }\Xi_{1}\left(  x_{a}\right)
=\xi\left(  a\right)  . \label{sol.freelie.2.e.uniq.pf.1}%
\end{equation}
\par
We know that $\Xi_{2}$ is a Lie algebra homomorphism $\Xi:\mathfrak{g}%
\rightarrow\mathfrak{h}$ such that every $a\in\mathfrak{A}$ satisfies
$\Xi\left(  x_{a}\right)  =\xi\left(  a\right)  $. In other words, $\Xi_{2}$
is a Lie algebra homomorphism $\mathfrak{g}\rightarrow\mathfrak{h}$, and has
the property that%
\begin{equation}
\text{every }a\in\mathfrak{A}\text{ satisfies }\Xi_{2}\left(  x_{a}\right)
=\xi\left(  a\right)  . \label{sol.freelie.2.e.uniq.pf.3}%
\end{equation}
\par
The family $\left(  x_{a}\right)  _{a\in\mathfrak{A}}$ is a basis of the
$\kk$-module $V$, and thus spans this $\kk$-module.
\par
Now, we must prove the relation $v\in\ker\left(  \Xi_{1}%
-\Xi_{2}\right)  $. Since this relation is $\kk$-linear in $v$ (because
$\ker\left(  \Xi_{1}-\Xi_{2}\right)  $ is a $\kk%
$-submodule of $\mathfrak{g}$), we can WLOG assume that $v$ belongs to the
family $\left(  x_{a}\right)  _{a\in\mathfrak{A}}$ (because the family
$\left(  x_{a}\right)  _{a\in\mathfrak{A}}$ spans the $\kk$-module $V$,
and because $v\in V$). Assume this.
\par
We have assumed that $v$ belongs to the family $\left(  x_{a}\right)
_{a\in\mathfrak{A}}$. In other words, there exists an $a\in\mathfrak{A}$ such
that $v=x_{a}$. Consider this $a$.
\par
Now, $\left(  \Xi_{1}-\Xi_{2}\right)  \left(  \underbrace{v}_{=x_{a}}\right)
=\left(  \Xi_{1}-\Xi_{2}\right)  \left(  x_{a}\right)  =\underbrace{\Xi
_{1}\left(  x_{a}\right)  }_{\substack{=\xi\left(  x_{a}\right)  \\\text{(by
(\ref{sol.freelie.2.e.uniq.pf.1}))}}}-\underbrace{\Xi_{2}\left(  x_{a}\right)
}_{\substack{=\xi\left(  x_{a}\right)  \\\text{(by
(\ref{sol.freelie.2.e.uniq.pf.3}))}}}=\xi\left(  x_{a}\right)  -\xi\left(
x_{a}\right)  =0$. In other words, $v\in\ker\left(  \Xi_{1}%
-\Xi_{2}\right)  $. Thus, $v\in\ker\left(  \Xi_{1}-\Xi
_{2}\right)  $ is proven.
\par
Let us now forget that we fixed $v$. We thus have proven that every $v\in V$
satisfies $v\in\ker\left(  \Xi_{1}-\Xi_{2}\right)  $. In other
words, $V\subset\ker\left(  \Xi_{1}-\Xi_{2}\right)  $.
\par
From Lemma \ref{lem.sol.freelie.2.e.generators-determine} (applied to
$\mathfrak{a}=\mathfrak{g}$, $\mathfrak{b}=\mathfrak{h}$, $p=\Xi_{1}$ and
$q=\Xi_{2}$), we see that $\ker\left(  \Xi_{1}-\Xi_{2}\right)
$ is a Lie subalgebra of $\mathfrak{g}$. Since $\mathfrak{g}$ (in turn) is a
Lie subalgebra of $T\left(  V\right)  $, this shows that $\ker%
\left(  \Xi_{1}-\Xi_{2}\right)  $ is a Lie subalgebra of $T\left(  V\right)
$.
\par
Thus, we know that $\ker\left(  \Xi_{1}-\Xi_{2}\right)  $ is a
Lie subalgebra of $T\left(  V\right)  $ satisfying $V\subset
\ker\left(  \Xi_{1}-\Xi_{2}\right)  $. Exercise
\ref{exe.freelie.2}(b)
(applied to $\mathfrak{k} = \ker\left(  \Xi_{1}-\Xi_{2}\right)$)
thus shows that $\mathfrak{g}\subset\ker%
\left(  \Xi_{1}-\Xi_{2}\right)  $. Hence, $\Xi_{1}-\Xi_{2}=0$, so that
$\Xi_{1}=\Xi_{2}$. Qed.}. In other words, there exists \textbf{at most one}
Lie algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that
every $a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(
a\right)  $. We shall now show that there actually exists such a homomorphism.

For every $w\in\mathfrak{L}$, we define an element $z_{w}$ of $\mathfrak{h}$
as follows: We define $z_{w}$ by recursion on the length of $w$. If the length
of $w$ is $1$\ \ \ \ \footnote{The length of any $w\in\mathfrak{L}$ must be at
least $1$. (Indeed, if $w\in\mathfrak{L}$, then the word $w$ is Lyndon and
thus nonempty, and hence its length must be at least $1$.)}, then we have
$w=\left(  a\right)  $ for some letter $a\in\mathfrak{A}$, and we set
$z_{w}=\xi\left(  a\right)  $ for this letter $a$. If the length of $w$ is
$>1$, then we set $z_{w}=\left[  z_{u},z_{v}\right]  $, where $\left(
u,v\right)  =\operatorname*{stf}w$\ \ \ \ \footnote{This is well-defined,
because $z_{u}$ and $z_{v}$ have already been defined. [\textit{Proof.} Let
$\left(  u,v\right)  =\operatorname*{stf}w$. Then, Exercise
\ref{exe.words.lyndon.stf.basic}(e) (applied to $\left(  g,h\right)  =\left(
u,v\right)  $) shows that $u\in\mathfrak{L}$, $v\in\mathfrak{L}$, $\ell\left(
u\right)  <\ell\left(  w\right)  $ and $\ell\left(  v\right)  <\ell\left(
w\right)  $. Recall that we are defining $z_{w}$ by recursion on the length of
$w$. Hence, $z_{p}$ is already defined for every $p\in\mathfrak{L}$ satisfying
$\ell\left(  p\right)  <\ell\left(  w\right)  $. Applying this to $p=u$, we
see that $z_{u}$ is already defined (since $u\in\mathfrak{L}$ and $\ell\left(
u\right)  <\ell\left(  w\right)  $). The same argument (but applied to $v$
instead of $u$) shows that $z_{v}$ is already defined. Hence, $z_{u}$ and
$z_{v}$ have already been defined. Thus, $z_{w}$ is well-defined by
$z_{w}=\left[  z_{u},z_{v}\right]  $, qed.]}.

Thus, we have defined a $z_{w}\in\mathfrak{h}$ for every $w\in\mathfrak{L}$.
Now, Exercise \ref{exe.freelie.2}(d) shows that the family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ is a basis of the $\kk$-module
$\mathfrak{g}$. Thus, we can define a $\kk$-module homomorphism
$f:\mathfrak{g}\rightarrow\mathfrak{h}$ by requiring that%
\begin{equation}
\left(  f\left(  b_{w}\right)  =z_{w}\ \ \ \ \ \ \ \ \ \ \text{for every }%
w\in\mathfrak{L}\right)  . \label{sol.freelie.2.e.1}%
\end{equation}
Consider this $f$. Whenever $w$ is a Lyndon word of length $>1$, we have%
\begin{equation}
f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(  b_{u}\right)
,f\left(  b_{v}\right)  \right]  ,\ \ \ \ \ \ \ \ \ \ \text{where }\left(
u,v\right)  =\operatorname*{stf}w \label{sol.freelie.2.e.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.freelie.2.e.3}):} Let $w$ be a Lyndon
word of length $>1$. Let $\left(  u,v\right)  =\operatorname*{stf}w$. Thus,
$\left(  u,v\right)  =\operatorname*{stf}w\in\mathfrak{L}\times\mathfrak{L}$.
\par
The recursive definition of $z_{w}$ shows that $z_{w}=\left[  z_{u}%
,z_{v}\right]  $ (since $w$ is a Lyndon word of length $>1$, and since
$\left(  u,v\right)  =\operatorname*{stf}w$). The recursive definition of
$b_{w}$ shows that $b_{w}=\left[  b_{u},b_{v}\right]  $ (since $w$ is a Lyndon
word of length $>1$, and since $\left(  u,v\right)  =\operatorname*{stf}w$).
Thus, $\left[  b_{u},b_{v}\right]  =b_{w}$, so that%
\begin{align*}
f\left(  \underbrace{\left[  b_{u},b_{v}\right]  }_{=b_{w}}\right)   &
=f\left(  b_{w}\right)  =z_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.freelie.2.e.1})}\right) \\
&  =\left[  z_{u},z_{v}\right]  .
\end{align*}
Comparing this with%
\[
\left[  \underbrace{f\left(  b_{u}\right)  }_{\substack{=z_{u}\\\text{(by
(\ref{sol.freelie.2.e.1}), applied to }u\text{ instead of }w\text{)}%
}},\underbrace{f\left(  b_{v}\right)  }_{\substack{=z_{v}\\\text{(by
(\ref{sol.freelie.2.e.1}), applied to }v\text{ instead of }w\text{)}}}\right]
=\left[  z_{u},z_{v}\right]  ,
\]
we obtain $f\left(  \left[  b_{u},b_{v}\right]  \right)  =\left[  f\left(
b_{u}\right)  ,f\left(  b_{v}\right)  \right]  $. This proves
(\ref{sol.freelie.2.e.3}).}. But recall that the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$ is a basis of the $\kk$-module $\mathfrak{g}$.
Thus, this family spans the $\kk$-module $\mathfrak{g}$. In other
words, $\mathfrak{g}$ is the $\kk$-submodule of $\mathfrak{g}$ spanned
by the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$. Thus, Exercise
\ref{exe.freelie.1}(b) (applied to $B=\mathfrak{g}$) shows that $f$ is a Lie
algebra homomorphism.

We notice that%
\[
\text{every }a\in\mathfrak{A}\text{ satisfies }f\left(  x_{a}\right)
=\xi\left(  a\right)
\]
\footnote{\textit{Proof.} Let $a\in\mathfrak{A}$. Let $w$ be the one-letter
word $\left(  a\right)  $. Then, $w$ is a Lyndon word (since $w$ is a
one-letter word, and since every one-letter word is a Lyndon word) and has
length $1$ (since $w$ is a one-letter word). Thus, the definition of $z_{w}$
shows that $z_{w}=\xi\left(  a\right)  $ (since $w=\left(  a\right)  $).
\par
But the definition of $b_{w}$ yields $b_{w}=x_{a}$ (since $\ell\left(
w\right)  =1$ and $w=\left(  a\right)  $). Hence, $f\left(  \underbrace{b_{w}%
}_{=x_{a}}\right)  =f\left(  x_{a}\right)  $, so that $f\left(  x_{a}\right)
=f\left(  b_{w}\right)  =z_{w}$ (by (\ref{sol.freelie.2.e.1})) and thus
$f\left(  x_{a}\right)  =z_{w}=\xi\left(  a\right)  $, qed.}. Thus, $f$ is a
Lie algebra homomorphism $\mathfrak{g}\rightarrow\mathfrak{h}$ having the
property that every $a\in\mathfrak{A}$ satisfies $f\left(  x_{a}\right)
=\xi\left(  a\right)  $. Therefore, there exists \textbf{at least one} Lie
algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that every
$a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(  a\right)  $
(namely, $\Xi=f$). Combining this with the fact that there exists \textbf{at
most one} Lie algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$
such that every $a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)
=\xi\left(  a\right)  $, we conclude that there exists a \textbf{unique} Lie
algebra homomorphism $\Xi:\mathfrak{g}\rightarrow\mathfrak{h}$ such that every
$a\in\mathfrak{A}$ satisfies $\Xi\left(  x_{a}\right)  =\xi\left(  a\right)
$. This solves Exercise \ref{exe.freelie.2}(e).
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.rmk.intsys.trivia}}

\textit{Solution to Exercise \ref{exe.rmk.intsys.trivia}.}

\begin{proof}
[Proof of Remark \ref{rmk.intsys.trivia}.](a) Let $I$ and $J$ be two nonempty
intervals of $\ZZ$ satisfying $I<J$. Then,
\begin{equation}
\text{every }i\in I\text{ and }j\in J\text{ satisfy }i<j
\label{pf.rmk.intsys.trivia.a.1}%
\end{equation}
(by the definition of $I<J$), since $I<J$.

Let $p\in I\cap J$. Then, $p\in I\cap J\subset I$ and $p\in I\cap J\subset J$.
Hence, $p<p$ (by (\ref{pf.rmk.intsys.trivia.a.1})), which is absurd. Now, let
us forget that we fixed $p$. Thus, we have obtained a contradiction for every
$p\in I\cap J$. Hence, there exists no $p\in I\cap J$. In other words, $I\cap
J=\varnothing$, so that the sets $I$ and $J$ are disjoint. This proves Remark
\ref{rmk.intsys.trivia}(a).

(b) Let $I$ and $J$ be two disjoint nonempty intervals of $\ZZ$. We
need to prove that $I<J$ or $J<I$.

Let $i_{0}$ be the smallest element of $I$ (this exists, since $I$ is
nonempty), and let $j_{0}$ be the smallest element of $J$ (this exists, since
$J$ is nonempty). We WLOG assume that $i_{0}\leq j_{0}$ (since otherwise, we
can simply interchange $I$ with $J$).

The intervals $I$ and $J$ are disjoint, and thus $I\cap J=\varnothing$.
%%The following argument has been commented out because it is trivial and
%%useless:
%Hence,
%\begin{equation}
%\text{every }i\in I\text{ and }j\in J\text{ satisfy }i\neq j
%\label{pf.rmk.intsys.trivia.b.1}
%\end{equation}
%(because otherwise, there would be two elements $i\in I$ and $j\in J$
%satisfying $i=j$, and thus these two elements would satisfy $i\in I\cap J$
%(since $i\in I$ and $i=j\in J$), which would contradict $I\cap J=\varnothing$).

Now, let $i\in I$ and $j\in J$ be arbitrary. Assume (for the sake of
contradiction) that $i\geq j$. Notice that $j\geq j_{0}$ (since $j$ is an
element of $J$, whereas $j_{0}$ is the smallest element of $J$), so that
$i\geq j\geq j_{0}$, so that $j_{0}\leq i$.

Write the interval $I$ in the form $\left[  p:q\right]  ^{+}$ for some
$p\in \ZZ$ and $q\in \ZZ$. Since $i\in I=\left[  p:q\right]
^{+}=\left\{  p+1,p+2,\ldots,q\right\}  $, we have $p<i\leq q$. Since
$i_{0}\in I=\left[  p:q\right]  ^{+}=\left\{  p+1,p+2,\ldots,q\right\}  $, we
have $p<i_{0}\leq q$. Now, $p<i_{0}\leq j_{0}$ and $j_{0}\leq i\leq q$. Hence,
$p<j_{0}\leq q$ and thus $j_{0}\in\left\{  p+1,p+2,\ldots,q\right\}  =\left[
p:q\right]  ^{+}=I$. Combined with $j_{0}\in J$ (since $j_{0}$ is the smallest
element of $J$), this yields $j_{0}\in I\cap J=\varnothing$, which is absurd.
Hence, our assumption (that $i\geq j$) was wrong. We thus have $i<j$.

Now, forget that we have fixed $i$ and $j$. We thus have shown that every
$i\in I$ and $j\in J$ satisfy $i<j$. In other words, $I<J$ (by the definition
of $I<J$). Hence, $I<J$ or $J<I$. This proves Remark \ref{rmk.intsys.trivia}(b).

(c) For any $\ell$-tuple $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $ of
nonempty intervals of $\ZZ$, we can state the following three
properties (which might and might not be satisfied):

\begin{itemize}
\item \textit{Property C1:} The intervals $I_{1}$, $I_{2}$, $\ldots$,
$I_{\ell}$ form a set partition of the set $\left[  0:n\right]  ^{+}$, where
$n=\left\vert \alpha\right\vert $.

\item \textit{Property C2:} We have $I_{1}<I_{2}< \cdots<I_{\ell}$.

\item \textit{Property C3:} We have $\left\vert I_{i}\right\vert =\alpha_{i}$
for every $i\in\left\{  1,2,\ldots,\ell\right\}  $.
\end{itemize}

We have to prove that the interval system $\operatorname{intsys}\alpha$ is the
unique $\ell$-tuple $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $ of nonempty
intervals of $\ZZ$ satisfying these three properties C1, C2 and C3.

First, it is easy to check that the interval system $\operatorname{intsys}%
\alpha$ is an $\ell$-tuple $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $ of
nonempty intervals of $\ZZ$ satisfying these three properties C1, C2
and C3.\footnote{\textit{Proof.} Let $\left(  I_{1},I_{2},\ldots,I_{\ell
}\right)  $ denote the interval system $\operatorname{intsys}\alpha$. We need
to prove that the Properties C1, C2 and C3 are satisfied.
\par
Let us first recall that%
\begin{equation}
I_{i}=\left[  \sum_{k=1}^{i-1}\alpha_{k}:\sum_{k=1}^{i}\alpha_{k}\right]
^{+}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,\ell\right\}
\label{pf.rmk.intsys.trivia.c.1}%
\end{equation}
(by the definition of $\operatorname{intsys}\alpha$).
\par
\textit{Proof that Property C1 is satisfied:} Let $n=\left\vert \alpha
\right\vert $. Then, $\sum_{k=1}^{\ell}\alpha_{k}=\left\vert \alpha\right\vert
=n$. Clearly, $I_{i}\subset\left[  0:n\right]  ^{+}$ for every $i\in\left\{
1,2,\ldots,\ell\right\}  $. We have%
\[
0=\sum_{k=1}^{0}\alpha_{k}\leq\sum_{k=1}^{1}\alpha_{k}\leq\cdots\leq\sum
_{k=1}^{\ell}\alpha_{k}=n
\]
(since $\alpha_{1}$, $\alpha_{2}$, $\ldots$, $\alpha_{\ell}$ are positive
integers). Hence, for every $x\in\left[  0:n\right]  ^{+}$, there exists
precisely one $i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfying $\sum
_{k=1}^{i-1}\alpha_{k}<x\leq\sum_{k=1}^{i}\alpha_{k}$. In other words, for
every $x\in\left[  0:n\right]  ^{+}$, there exists precisely one $i\in\left\{
1,2,\ldots,\ell\right\}  $ satisfying $x\in\left[  \sum_{k=1}^{i-1}\alpha
_{k}:\sum_{k=1}^{i}\alpha_{k}\right]  ^{+}$ (since $x\in\left[  \sum
_{k=1}^{i-1}\alpha_{k}:\sum_{k=1}^{i}\alpha_{k}\right]  ^{+}$ is equivalent to
$\sum_{k=1}^{i-1}\alpha_{k}<x\leq\sum_{k=1}^{i}\alpha_{k}$). In other words,
for every $x\in\left[  0:n\right]  ^{+}$, there exists precisely one
$i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfying $x\in I_{i}$ (since
$I_{i}=\left[  \sum_{k=1}^{i-1}\alpha_{k}:\sum_{k=1}^{i}\alpha_{k}\right]
^{+}$). In other words, the subsets $I_{1}$, $I_{2}$, $\ldots$, $I_{\ell}$ of
$\left[  0:n\right]  ^{+}$ are disjoint, and their union is $\left[
0:n\right]  ^{+}$. In other words, the subsets $I_{1}$, $I_{2}$, $\ldots$,
$I_{\ell}$ of $\left[  0:n\right]  ^{+}$ form a set partition of the set
$\left[  0:n\right]  ^{+}$. This proves that Property C1 is satisfied.
\par
\textit{Proof that Property C2 is satisfied:} Let $u\in\left\{  1,2,\ldots
,\ell-1\right\}  $. Let $i\in I_{u}$ and $j\in I_{u+1}$. Since $i\in
I_{u}=\left[  \sum_{k=1}^{u-1}\alpha_{k}:\sum_{k=1}^{u}\alpha_{k}\right]
^{+}$ (by (\ref{pf.rmk.intsys.trivia.c.1}), applied to $u$ instead of $i$), we
have $\sum_{k=1}^{u-1}\alpha_{k}<i\leq\sum_{k=1}^{u}\alpha_{k}$. Since $j\in
I_{u+1}=\left[  \sum_{k=1}^{\left(  u+1\right)  -1}\alpha_{k}:\sum_{k=1}%
^{u+1}\alpha_{k}\right]  ^{+}$ (by (\ref{pf.rmk.intsys.trivia.c.1}), applied
to $u+1$ instead of $i$), we have $\sum_{k=1}^{\left(  u+1\right)  -1}%
\alpha_{k}<j\leq\sum_{k=1}^{u+1}\alpha_{k}$. Now, $i\leq\sum_{k=1}^{u}%
\alpha_{k}=\sum_{k=1}^{\left(  u+1\right)  -1}\alpha_{k}<j$.
\par
Now, let us forget that we fixed $i$ and $j$. We thus have proven that every
$i\in I_{u}$ and $j\in I_{u+1}$ satisfy $i<j$. In other words, $I_{u}<I_{u+1}$
(by the definition of $I_{u}<I_{u+1}$).
\par
Now, let us forget that we fixed $u$. We thus have shown that $I_{u}<I_{u+1}$
for every $u\in\left\{  1,2,\ldots,\ell-1\right\}  $. In other words,
$I_{1}<I_{2}< \cdots<I_{\ell}$. This proves that Property C2 is satisfied.
\par
\textit{Proof that Property C3 is satisfied:} Let $i\in\left\{  1,2,\ldots
,\ell\right\}  $. We have $\sum_{k=1}^{i}\alpha_{k}=\sum_{k=1}^{i-1}\alpha
_{k}+\underbrace{\alpha_{i}}_{>0}>\sum_{k=1}^{i-1}\alpha_{k}$. Now,
(\ref{pf.rmk.intsys.trivia.c.1}) yields%
\begin{align*}
\left\vert I_{i}\right\vert  &  =\left\vert \left[  \sum_{k=1}^{i-1}\alpha
_{k}:\sum_{k=1}^{i}\alpha_{k}\right]  ^{+}\right\vert =\sum_{k=1}^{i}%
\alpha_{k}-\sum_{k=1}^{i-1}\alpha_{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\sum_{k=1}^{i}\alpha_{k}>\sum_{k=1}^{i-1}\alpha_{k}\right) \\
&  =\alpha_{i}.
\end{align*}
This proves that Property C3 is satisfied.
\par
We thus have shown that the Properties C1, C2 and C3 are satisfied, qed.} It
remains to prove that it is the only such $\ell$-tuple.

So, let us fix any $\ell$-tuple $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $
of nonempty intervals of $\ZZ$ satisfying the three properties C1, C2
and C3. We need to prove that this $\ell$-tuple $\left(  I_{1},I_{2}%
,\ldots,I_{\ell}\right)  $ is $\operatorname*{intsys}\alpha$. In order to
prove this, it is enough to show that%
\begin{equation}
I_{i}=\left[  \sum_{k=1}^{i-1}\alpha_{k}:\sum_{k=1}^{i}\alpha_{k}\right]
^{+}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,\ell\right\}
\label{pf.rmk.intsys.trivia.c.5}%
\end{equation}
(according to the definition of $\operatorname*{intsys}\alpha$). So it remains
to prove (\ref{pf.rmk.intsys.trivia.c.5}).

Set $n=\left\vert \alpha\right\vert $. The intervals $I_{1}$, $I_{2}$,
$\ldots$, $I_{\ell}$ form a set partition of the set $\left[  0:n\right]
^{+}$ (according to Property C1). Thus, the intervals $I_{1}$, $I_{2}$,
$\ldots$, $I_{\ell}$ are disjoint, and satisfy $I_{1}\cup I_{2}%
\cup\cdots\cup I_{\ell}=\left[  0:n\right]  ^{+}$.

Now, let us show that%
\begin{equation}
I_{1}\cup I_{2}\cup\cdots\cup I_{u}=\left[  0:\sum_{k=1}^{u}\alpha_{k}\right]
^{+}\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\left\{  0,1,\ldots,\ell\right\}
. \label{pf.rmk.intsys.trivia.c.6}%
\end{equation}


\textit{Proof of (\ref{pf.rmk.intsys.trivia.c.6}):} We will prove
(\ref{pf.rmk.intsys.trivia.c.6}) by induction over $u$.

\textit{Induction base:} For $u=0$, we have $I_{1}\cup I_{2}\cup\cdots\cup
I_{u}=I_{1}\cup I_{2}\cup\cdots\cup I_{0}=\left(  \text{empty union}\right)
=\varnothing$ and $\left[  0:\sum_{k=1}^{u}\alpha_{k}\right]  ^{+}=\left[
0:\underbrace{\sum_{k=1}^{0}\alpha_{k}}_{=0}\right]  ^{+}=\left[  0:0\right]
^{+}=\varnothing$. Hence, for $u=0$, both sides of the equality
(\ref{pf.rmk.intsys.trivia.c.6}) are $\varnothing$. Thus, for $u=0$, the
equality (\ref{pf.rmk.intsys.trivia.c.6}) holds. This completes the induction base.

\textit{Induction step:} Let $U\in\left\{  1,2,\ldots,\ell\right\}  $. Assume
that (\ref{pf.rmk.intsys.trivia.c.6}) holds for $u=U-1$. We now need to prove
that (\ref{pf.rmk.intsys.trivia.c.6}) holds for $u=U$ as well.

We have assumed that (\ref{pf.rmk.intsys.trivia.c.6}) holds for $u=U-1$. In
other words,
\begin{equation}
I_{1}\cup I_{2}\cup\cdots\cup I_{U-1}=\left[  0:\sum_{k=1}^{U-1}\alpha
_{k}\right]  ^{+}. \label{pf.rmk.intsys.trivia.c.6.pf.indhyp}%
\end{equation}


Let $\xi=\sum_{k=1}^{U-1}\alpha_{k}$. Then, $\xi\geq0$, so that $\xi+1\geq1$.
Then,
\[
n=\left\vert \alpha\right\vert =\sum_{k=1}^{\ell}\alpha_{k}=\sum_{k=1}%
^{U-1}\alpha_{k}+\underbrace{\sum_{k=U}^{\ell}\alpha_{k}}%
_{\substack{>0\\\text{(since all }\alpha_{k}\text{ are }>0\text{,}\\\text{and
since }U\leq\ell\text{)}}}>\sum_{k=1}^{U-1}\alpha_{k}=\xi.
\]
Hence, $n\geq\xi+1$ (since $\xi$ and $n$ are integers), so that $1\leq
\xi+1\leq n$. In other words, $\xi+1\in\left\{  1,2,\ldots,n\right\}  =\left[
0:n\right]  ^{+}=I_{1}\cup I_{2}\cup\cdots\cup I_{\ell}$ (since $I_{1}\cup
I_{2}\cup\cdots\cup I_{\ell}=\left[  0:n\right]  ^{+}$). In other words, there
exists some $v\in\left\{  1,2,\ldots,\ell\right\}  $ such that $\xi+1\in
I_{v}$. Consider this $v$. If we had $v<U$, then we would have%
\begin{align*}
\xi+1  &  \in I_{v}\subset I_{1}\cup I_{2}\cup\cdots\cup I_{U-1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v\in\left\{  1,2,\ldots,U-1\right\}
\text{ (because }v<U\text{)}\right) \\
&  =\left[  0:\underbrace{\sum_{k=1}^{U-1}\alpha_{k}}_{=\xi}\right]
^{+}=\left[  0:\xi\right]  ^{+}=\left\{  1,2,\ldots,\xi\right\}  ,
\end{align*}
which is absurd. Hence, we cannot have $v<U$. Thus, we have $v\geq U$.

Recall that
\begin{equation}
\left[  0:\underbrace{\xi}_{=\sum_{k=1}^{U-1}\alpha_{k}}\right]  ^{+}=\left[
0:\sum_{k=1}^{U-1}\alpha_{k}\right]  ^{+}=I_{1}\cup I_{2}\cup\cdots\cup
I_{U-1} \label{pf.rmk.intsys.trivia.c.6.pf.indhyp-by-other-name}%
\end{equation}
(by (\ref{pf.rmk.intsys.trivia.c.6.pf.indhyp})). Using this, it is easy to see
that%
\begin{equation}
\text{every }p\in I_{U}\text{ satisfies }p\geq\xi+1.
\label{pf.rmk.intsys.trivia.c.6.pf.pxi}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.rmk.intsys.trivia.c.6.pf.pxi}):} Let $p\in
I_{U}$. Assume (for the sake of contradiction) that $p\leq\xi$. But $p\in
I_{U}\subset I_{1}\cup I_{2}\cup\cdots\cup I_{\ell}=\left[  0:n\right]
^{+}=\left\{  1,2,\ldots,n\right\}  $, so that $0<p\leq n$. Since $0<p\leq\xi
$, we have $p\in\left\{  1,2,\ldots,\xi\right\}  =\left[  0:\xi\right]
^{+}=I_{1}\cup I_{2}\cup\cdots\cup I_{U-1}$, which shows that there exists
some $r\in\left\{  1,2,\ldots,U-1\right\}  $ such that $p\in I_{r}$. Consider
this $r$. Since $r\neq U$ (because $r\in\left\{  1,2,\ldots,U-1\right\}  $),
the intervals $I_{r}$ and $I_{U}$ are disjoint (since the intervals
$I_{1}$, $I_{2}$, $\ldots$, $I_{\ell}$ are disjoint). In other words,
$I_{r}\cap I_{U}=\varnothing$. But combining $p\in I_{r}$ with $p\in I_{U}$,
we obtain $p\in I_{r}\cap I_U =\varnothing$, which is absurd. This
contradiction proves that our assumption (that $p\leq\xi$) was wrong. Hence,
we have $p>\xi$. Thus, $p\geq\xi+1$ (since $p$ and $\xi$ are integers). This
proves (\ref{pf.rmk.intsys.trivia.c.6.pf.pxi}).}

We now assume (for the sake of contradiction) that $v\neq U$. Then, $v>U$
(since $v\geq U$ and $v\neq U$), whence $I_{U}<I_{v}$ (since Property C2
yields $I_{1}<I_{2}< \cdots<I_{\ell}$).

The interval $I_{U}$ is nonempty, and thus there exists some $p\in I_{U}$.
Consider such a $p$. Recall that $I_{U}<I_{v}$. Thus, every $i\in I_{U}$ and
$j\in I_{v}$ satisfy $i<j$ (by the definition of $I_{U}<I_{v}$). Applying this
to $i=p$ and $j=\xi+1$, we obtain $p<\xi+1$. But
(\ref{pf.rmk.intsys.trivia.c.6.pf.pxi}) yields $p\geq\xi+1$, which contradicts
$p<\xi+1$. This contradiction proves that our assumption (that $v\neq U$) was
wrong. Hence, we have $v=U$.

Now, $\xi+1\in I_{v}=I_{U}$ (since $v=U$). Hence, $\xi+1$ is an element of
$I_{U}$. Due to (\ref{pf.rmk.intsys.trivia.c.6.pf.pxi}), this element $\xi+1$
is the smallest element of $I_{U}$.

But since property C3 is satisfied, we have $\left\vert I_{U}\right\vert
=\alpha_{U}$ (by Property C3, applied to $i=U$). So we know that the interval
$I_{U}$ has length $\alpha_{U}$ (since $\left\vert I_{U}\right\vert
=\alpha_{U}$) and smallest element $\xi+1$. Therefore, $I_{U}=\left[  \xi
:\xi+\alpha_{U}\right]  ^{+}$ (because the only interval having length
$\alpha_{U}$ and smallest element $\xi+1$ is the interval $\left[  \xi
:\xi+\alpha_{U}\right]  ^{+}$). Now,%
\begin{align*}
I_{1}\cup I_{2}\cup\cdots\cup I_{U}  &  =\underbrace{\left(  I_{1}\cup
I_{2}\cup\cdots\cup I_{U-1}\right)  }_{\substack{=\left[  0:\xi\right]
^{+}\\\text{(by (\ref{pf.rmk.intsys.trivia.c.6.pf.indhyp-by-other-name}))}%
}}\cup\underbrace{I_{U}}_{=\left[  \xi:\xi+\alpha_{U}\right]  ^{+}}\\
&  =\left[  0:\xi\right]  ^{+}\cup\left[  \xi:\xi+\alpha_{U}\right]
^{+}=\left[  0:\xi+\alpha_{U}\right]  ^{+}%
\end{align*}
(since $0\leq\xi\leq\xi+\alpha_{U}$ (since $\xi\geq0$ and $\alpha_{U}\geq0$)).
Since
\[
\underbrace{\xi}_{=\sum_{k=1}^{U-1}\alpha_{k}}+\alpha_{U}=\sum_{k=1}%
^{U-1}\alpha_{k}+\alpha_{U}=\sum_{k=1}^{U}\alpha_{k},
\]
this rewrites as $I_{1}\cup I_{2}\cup\cdots\cup I_{U}=\left[  0:\sum_{k=1}%
^{U}\alpha_{k}\right]  ^{+}$. In other words, (\ref{pf.rmk.intsys.trivia.c.6})
holds for $u=U$. This completes the induction step. The induction proof of
(\ref{pf.rmk.intsys.trivia.c.6}) is thus complete.

\textit{Proof of (\ref{pf.rmk.intsys.trivia.c.5}):} Let $i\in\left\{
1,2,\ldots,\ell\right\}  $. Notice that $\sum_{k=1}^{i-1}\alpha_{k}\geq0$
(since $\alpha_{k}>0$ for every $k$) and $\sum_{k=1}^{i}\alpha_{k}=\sum
_{k=1}^{i-1}\alpha_{k}+\underbrace{\alpha_{i}}_{>0}>\sum_{k=1}^{i-1}\alpha
_{k}$. Hence, $0\leq\sum_{k=1}^{i-1}\alpha_{k}\leq\sum_{k=1}^{i}\alpha_{k}$.

The intervals $I_{1}$, $I_{2}$, $\ldots$, $I_{\ell}$ are disjoint.
Hence, $I_{i}$ is disjoint from $I_{1}\cup I_{2}\cup\cdots\cup I_{i-1}$. Thus,%
\begin{align*}
I_{i}  &  =\underbrace{\left(  \left(  I_{1}\cup I_{2}\cup\cdots\cup
I_{i-1}\right)  \cup I_{i}\right)  }_{=I_{1}\cup I_{2}\cup\cdots\cup I_{i}%
}\setminus\left(  I_{1}\cup I_{2}\cup\cdots\cup I_{i-1}\right) \\
&  =\underbrace{\left(  I_{1}\cup I_{2}\cup\cdots\cup I_{i}\right)
}_{\substack{=\left[  0:\sum_{k=1}^{i}\alpha_{k}\right]  ^{+}\\\text{(by
(\ref{pf.rmk.intsys.trivia.c.6}), applied to }u=i\text{)}}}\setminus
\underbrace{\left(  I_{1}\cup I_{2}\cup\cdots\cup I_{i-1}\right)
}_{\substack{=\left[  0:\sum_{k=1}^{i-1}\alpha_{k}\right]  ^{+}\\\text{(by
(\ref{pf.rmk.intsys.trivia.c.6}), applied to }u=i-1\text{)}}}\\
&  =\left[  0:\sum_{k=1}^{i}\alpha_{k}\right]  ^{+}\setminus\left[
0:\sum_{k=1}^{i-1}\alpha_{k}\right]  ^{+}=\left[  \sum_{k=1}^{i-1}\alpha
_{k}:\sum_{k=1}^{i}\alpha_{k}\right]  ^{+}%
\end{align*}
(since $0\leq\sum_{k=1}^{i-1}\alpha_{k}\leq\sum_{k=1}^{i}\alpha_{k}$). This
proves (\ref{pf.rmk.intsys.trivia.c.5}). As explained above, this completes
our proof of Remark \ref{rmk.intsys.trivia}(c).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.shuffle.lyndon.easystep}}

\textit{Solution to Exercise \ref{exe.lem.shuffle.lyndon.easystep}.}

\begin{proof}
[Proof of Lemma \ref{lem.shuffle.lyndon.easystep}.]We have $\sigma
\in\operatorname{Sh}_{n,m}$, so that $\sigma^{-1}\left(  1\right)
<\sigma^{-1}\left(  2\right)  <\cdots<\sigma^{-1}\left(  n\right)  $ and
$\sigma^{-1}\left(  n+1\right)  <\sigma^{-1}\left(  n+2\right)  <\cdots
<\sigma^{-1}\left(  n+m\right)  $. In other words, the restriction of the map
$\sigma^{-1}$ to the interval $\left[  0:n\right]  ^{+}$ is strictly
increasing, and so is the restriction of the map $\sigma^{-1}$ to the interval
$\left[  n:n+m\right]  ^{+}$.

(a) Let $I$ be an interval of $\ZZ$ such that $I\subset\left[
0:n+m\right]  ^{+}$. We will only show that $\sigma\left(  I\right)
\cap\left[  0:n\right]  ^{+}$ is an interval; the proof that $\sigma\left(
I\right)  \cap\left[  n:n+m\right]  ^{+}$ is an interval is completely analogous.

It is known that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }\mathfrak{R}\text{ is a finite subset of } \ZZ \text{ such that
every }\alpha\in\mathfrak{R}\text{, }\gamma\in\mathfrak{R}\text{ and }\beta
\in \ZZ \\
\text{satisfying }\alpha<\beta<\gamma\text{ satisfy }\beta\in\mathfrak{R}%
\text{, then }\mathfrak{R}\text{ is an interval of } \ZZ %
\end{array}
\right)  . \label{pf.lem.shuffle.lyndon.easystep.a.1}%
\end{equation}
(Indeed, it is clear that $\mathfrak{R}=\left[  \min\mathfrak{R}%
-1:\max\mathfrak{R}\right]  ^{+}$ in this case, unless $\mathfrak{R}$ is empty
in which case the statement is obvious anyway.)

We now denote $\mathfrak{R}=\sigma\left(  I\right)  \cap\left[  0:n\right]
^{+}$. Our next goal is to use (\ref{pf.lem.shuffle.lyndon.easystep.a.1}) to
show that $\mathfrak{R}$ is an interval.

Indeed, let $\alpha\in\mathfrak{R}$, $\gamma\in\mathfrak{R}$ and $\beta
\in \ZZ$ be such that $\alpha<\beta<\gamma$. Then, $\alpha
\in\mathfrak{R}=\sigma\left(  I\right)  \cap\left[  0:n\right]  ^{+}%
\subset\left[  0:n\right]  ^{+}$ and similarly $\gamma\in\left[  0:n\right]
^{+}$. Combined with $\alpha<\beta<\gamma$, these yield $\beta\in\left[
0:n\right]  ^{+}$. But recall that the restriction of the map $\sigma^{-1}$ to
the interval $\left[  0:n\right]  ^{+}$ is strictly increasing. Hence, from
$\alpha<\beta<\gamma$, we obtain $\sigma^{-1}\left(  \alpha\right)
<\sigma^{-1}\left(  \beta\right)  <\sigma^{-1}\left(  \gamma\right)  $ (since
$\alpha$, $\beta$ and $\gamma$ belong to $\left[  0:n\right]  ^{+}$). In other
words, the integer $\sigma^{-1}\left(  \beta\right)  $ lies strictly between
the integers $\sigma^{-1}\left(  \alpha\right)  $ and $\sigma^{-1}\left(
\gamma\right)  $. Since $\sigma^{-1}\left(  \alpha\right)  \in I$ (because
$\alpha\in\sigma\left(  I\right)  \cap\left[  0:n\right]  ^{+}\subset
\sigma\left(  I\right)  $) and $\sigma^{-1}\left(  \gamma\right)  \in I$ (for
similar reasons), this entails $\sigma^{-1}\left(  \beta\right)  \in I$
(because $I$ is an interval, and thus any integer lying between two elements
of $I$ must also belong to $I$). Hence, $\beta\in\sigma\left(  I\right)  $.
Combined with $\beta\in\left[  0:n\right]  ^{+}$, this yields $\beta\in
\sigma\left(  I\right)  \cap\left[  0:n\right]  ^{+}=\mathfrak{R}$.

Now, forget that we fixed $\alpha$, $\gamma$ and $\beta$. We thus have shown
that every $\alpha\in\mathfrak{R}$, $\gamma\in\mathfrak{R}$ and $\beta
\in \ZZ$ satisfying $\alpha<\beta<\gamma$ satisfy $\beta\in\mathfrak{R}%
$. Thus, (\ref{pf.lem.shuffle.lyndon.easystep.a.1}) shows that $\mathfrak{R}$
is an interval of $\ZZ$. In other words, $\sigma\left(  I\right)
\cap\left[  0:n\right]  ^{+}$ is an interval of $\ZZ$ (since
$\mathfrak{R}=\sigma\left(  I\right)  \cap\left[  0:n\right]  ^{+}$). This
completes the proof of Lemma \ref{lem.shuffle.lyndon.easystep}(a).

(b) The intervals $K$ and $L$ both are subsets of $\left[  0:n\right]  ^{+}$.
Therefore, from $K<L$, we obtain $\sigma^{-1}\left(  K\right)  <\sigma
^{-1}\left(  L\right)  $ (because the restriction of the map $\sigma^{-1}$ to
the interval $\left[  0:n\right]  ^{+}$ is strictly increasing).

Set $\mathcal{K}=\sigma^{-1}\left(  K\right)  $ and $\mathcal{L}=\sigma
^{-1}\left(  L\right)  $. Then, $\mathcal{K}$ is a nonempty
interval\footnote{because $\mathcal{K}=\sigma^{-1}\left(  K\right)  $ and
because we know that $\sigma^{-1}\left(  K\right)  $ is an interval and $K$ is
nonempty}, and thus can be written in the form $\mathcal{K}=\left[
x_{\mathcal{K}}:y_{\mathcal{K}}\right]  ^{+}$ for two elements $x_{\mathcal{K}%
}$ and $y_{\mathcal{K}}$ of $\left[  0:n+m\right]  ^{+}$ satisfying
$x_{\mathcal{K}}<y_{\mathcal{K}}$. Consider these $x_{\mathcal{K}}$ and
$y_{\mathcal{K}}$. Also, $\mathcal{L}$ is a nonempty interval\footnote{because
$\mathcal{L}=\sigma^{-1}\left(  L\right)  $ and because we know that
$\sigma^{-1}\left(  L\right)  $ is an interval and $L$ is nonempty}, and thus
can be written in the form $\mathcal{L}=\left[  x_{\mathcal{L}}:y_{\mathcal{L}%
}\right]  ^{+}$ for two elements $x_{\mathcal{L}}$ and $y_{\mathcal{L}}$ of
$\left[  0:n+m\right]  ^{+}$ satisfying $x_{\mathcal{L}}<y_{\mathcal{L}}$.
Consider these $x_{\mathcal{L}}$ and $y_{\mathcal{L}}$. Since $\left[
x_{\mathcal{K}}:y_{\mathcal{K}}\right]  ^{+}=\mathcal{K}=\sigma^{-1}\left(
K\right)  <\sigma^{-1}\left(  L\right)  =\mathcal{L}=\left[  x_{\mathcal{L}%
}:y_{\mathcal{L}}\right]  ^{+}$, we have $y_{\mathcal{K}}\leq x_{\mathcal{L}}$.

Notice that $\sigma\left(  \mathcal{K}\right)  =K$ (since $\mathcal{K}%
=\sigma^{-1}\left(  K\right)  $) and $\sigma\left(  \mathcal{L}\right)  =L$
(since $\mathcal{L}=\sigma^{-1}\left(  L\right)  $).

If we had $y_{\mathcal{K}}=x_{\mathcal{L}}$, then $\underbrace{\sigma
^{-1}\left(  K\right)  }_{\substack{=\mathcal{K}=\left[  x_{\mathcal{K}%
}:y_{\mathcal{K}}\right]  ^{+}=\left[  x_{\mathcal{K}}:x_{\mathcal{L}}\right]
^{+}\\\text{(since }y_{\mathcal{K}}=x_{\mathcal{L}}\text{)}}}\cup
\underbrace{\sigma^{-1}\left(  L\right)  }_{=\mathcal{L}=\left[
x_{\mathcal{L}}:y_{\mathcal{L}}\right]  ^{+}}=\left[  x_{\mathcal{K}%
}:x_{\mathcal{L}}\right]  ^{+}\cup\left[  x_{\mathcal{L}}:y_{\mathcal{L}%
}\right]  ^{+}=\left[  x_{\mathcal{K}}:y_{\mathcal{L}}\right]  ^{+}$ would be
an interval, which would contradict the assumption that $\sigma^{-1}\left(
K\right)  \cup\sigma^{-1}\left(  L\right)  $ is not an interval. Hence, we
cannot have $y_{\mathcal{K}}=x_{\mathcal{L}}$. Thus, $y_{\mathcal{K}}\neq
x_{\mathcal{L}}$. Therefore, $y_{\mathcal{K}}<x_{\mathcal{L}}$ (since
$y_{\mathcal{K}}\leq x_{\mathcal{L}}$). Hence, we can define a nonempty
interval $\mathcal{P}\subset\left[  0:n+m\right]  ^{+}$ by $\mathcal{P}%
=\left[  y_{\mathcal{K}}:x_{\mathcal{L}}\right]  ^{+}$. Consider this
$\mathcal{P}$. It satisfies $\left\vert \mathcal{P}\right\vert \neq0$ (since
it is nonempty) and $\mathcal{K}<\mathcal{P}<\mathcal{L}$ (since
$\mathcal{K}=\left[  x_{\mathcal{K}}:y_{\mathcal{K}}\right]  ^{+}$,
$\mathcal{P}=\left[  y_{\mathcal{K}}:x_{\mathcal{L}}\right]  ^{+}$ and
$\mathcal{L}=\left[  x_{\mathcal{L}}:y_{\mathcal{L}}\right]  ^{+}$), so that
the sets $\mathcal{K}$, $\mathcal{P}$ and $\mathcal{L}$ are disjoint.
As a consequence, the sets $\sigma\left(  \mathcal{K}\right)  $,
$\sigma\left(  \mathcal{P}\right)  $ and $\sigma\left(  \mathcal{L}\right)  $
are disjoint. In other words, the sets $K$, $\sigma\left(
\mathcal{P}\right)  $ and $L$ are disjoint (since $\sigma\left(
\mathcal{K}\right)  =K$ and $\sigma\left(  \mathcal{L}\right)  =L$).

It is easy to see that $\sigma\left(  \mathcal{P}\right)  \subset\left[
n:n+m\right]  ^{+}$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary.
Then, there exists some $q\in\sigma\left(  \mathcal{P}\right)  $ such that
$q\notin\left[  n:n+m\right]  ^{+}$. Consider this $q$. Since $q\notin\left[
n:n+m\right]  ^{+}$, we must have $q\in\left[  0:n\right]  ^{+}$.
\par
Notice that $q\in\sigma\left(  \mathcal{P}\right)  $, so that $\sigma
^{-1}\left(  q\right)  \in\mathcal{P}$.
\par
The elements $\max K$ and $\min L$ of $K$ and $L$ are well-defined, since $K$
and $L$ are nonempty.
\par
Now, $\max K\leq q-1$. (To prove this, assume the contrary. Thus, $\max
K>q-1$, so that $\max K\geq q$. But $\max K\in K\subset\left[  0:n\right]
^{+}$ and $q\in\left[  0:n\right]  ^{+}$. Therefore, $\sigma^{-1}\left(  \max
K\right)  \geq\sigma^{-1}\left(  q\right)  $ (because $\max K\geq q$, and
since the restriction of the map $\sigma^{-1}$ to the interval $\left[
0:n\right]  ^{+}$ is strictly increasing). But since $\mathcal{K}<\mathcal{P}%
$, we have $\sigma^{-1}\left(  \max K\right)  <\sigma^{-1}\left(  q\right)  $
(since $\sigma^{-1}\left(  \underbrace{\max K}_{\in K}\right)  \in\sigma
^{-1}\left(  K\right)  =\mathcal{K}$ and $\sigma^{-1}\left(  q\right)
\in\mathcal{P}$), which contradicts $\sigma^{-1}\left(  \max K\right)
\geq\sigma^{-1}\left(  q\right)  $. This contradiction shows that our
assumption was wrong, and we have shown that $\max K\leq q-1$.)
\par
Furthermore, $q\leq\min L-1$. (To prove this, assume the contrary. Thus,
$q>\min L-1$. Hence, $q\geq\min L$. We have $\min L\in L\subset\left[
0:n\right]  ^{+}$ and $q\in\left[  0:n\right]  ^{+}$. Therefore, $\sigma
^{-1}\left(  q\right)  \geq\sigma^{-1}\left(  \min L\right)  $ (because
$q\geq\min L$, and since the restriction of the map $\sigma^{-1}$ to the
interval $\left[  0:n\right]  ^{+}$ is strictly increasing). But since
$\mathcal{P}<\mathcal{L}$, we have $\sigma^{-1}\left(  q\right)  <\sigma
^{-1}\left(  \min L\right)  $ (since $\sigma^{-1}\left(  q\right)
\in\mathcal{P}$ and $\sigma^{-1}\left(  \underbrace{\min L}_{\in L}\right)
\in\sigma^{-1}\left(  L\right)  =\mathcal{L}$), which contradicts $\sigma
^{-1}\left(  q\right)  \geq\sigma^{-1}\left(  \min L\right)  $. This
contradiction shows that our assumption was wrong, and we have shown that
$q\leq\min L-1$.)
\par
So we have $\max K\leq\underbrace{q}_{\leq\min L-1}-1\leq\min L-2$. But $K$
and $L$ are disjoint nonempty intervals satisfying $K<L$, and their union is
an interval again (because $K\cup L$ is an interval). Hence, the interval $L$
must begin immediately after the end of the interval $K$; in other words, we
must have $\max K=\min L-1>\min L-2$. This contradicts $\max K\leq\min L-2$.
This contradiction completes our proof.}. Thus, $\sigma\left(  \mathcal{P}%
\right)  \cap\left[  n:n+m\right]  ^{+}=\sigma\left(  \mathcal{P}\right)  $.
But Lemma \ref{lem.shuffle.lyndon.easystep}(a) (applied to $\mathcal{P}$
instead of $I$) shows that $\sigma\left(  \mathcal{P}\right)  \cap\left[
0:n\right]  ^{+}$ and $\sigma\left(  \mathcal{P}\right)  \cap\left[
n:n+m\right]  ^{+}$ are intervals. In particular, $\sigma\left(
\mathcal{P}\right)  \cap\left[  n:n+m\right]  ^{+}$ is an interval. In other
words, $\sigma\left(  \mathcal{P}\right)  $ is an interval (since
$\sigma\left(  \mathcal{P}\right)  \cap\left[  n:n+m\right]  ^{+}%
=\sigma\left(  \mathcal{P}\right)  $).

So we know that $\sigma\left(  \mathcal{P}\right)  \subset\left[
n:n+m\right]  ^{+}$ is a nonempty interval\footnote{nonempty because
$\mathcal{P}$ is nonempty}, and $\sigma^{-1}\left(  \sigma\left(
\mathcal{P}\right)  \right)  $ is also an interval (since $\sigma^{-1}\left(
\sigma\left(  \mathcal{P}\right)  \right)  =\mathcal{P}$). Moreover, we have
$\sigma^{-1}\left(  K\right)  <\sigma^{-1}\left(  \sigma\left(  \mathcal{P}%
\right)  \right)  $ (since $\sigma^{-1}\left(  K\right)  =\mathcal{K}%
<\mathcal{P}=\sigma^{-1}\left(  \sigma\left(  \mathcal{P}\right)  \right)  $).
Also, $\underbrace{\sigma^{-1}\left(  K\right)  }_{=\mathcal{K}=\left[
x_{\mathcal{K}}:y_{\mathcal{K}}\right]  ^{+}}\cup\underbrace{\sigma
^{-1}\left(  \sigma\left(  \mathcal{P}\right)  \right)  }_{=\mathcal{P}%
=\left[  y_{\mathcal{K}}:x_{\mathcal{L}}\right]  ^{+}}=\left[  x_{\mathcal{K}%
}:y_{\mathcal{K}}\right]  ^{+}\cup\left[  y_{\mathcal{K}}:x_{\mathcal{L}%
}\right]  ^{+}=\left[  x_{\mathcal{K}}:x_{\mathcal{L}}\right]  ^{+}$ is an
interval. Furthermore, $\sigma^{-1}\left(  \sigma\left(  \mathcal{P}\right)
\right)  <\sigma^{-1}\left(  L\right)  $ (since $\sigma^{-1}\left(
\sigma\left(  \mathcal{P}\right)  \right)  =\mathcal{P}<\mathcal{L}%
=\sigma^{-1}\left(  L\right)  $), and the set $\underbrace{\sigma^{-1}\left(
\sigma\left(  \mathcal{P}\right)  \right)  }_{=\mathcal{P}=\left[
y_{\mathcal{K}}:x_{\mathcal{L}}\right]  ^{+}}\cup\underbrace{\sigma
^{-1}\left(  L\right)  }_{=\mathcal{L}=\left[  x_{\mathcal{L}}:y_{\mathcal{L}%
}\right]  ^{+}}=\left[  y_{\mathcal{K}}:x_{\mathcal{L}}\right]  ^{+}%
\cup\left[  x_{\mathcal{L}}:y_{\mathcal{L}}\right]  ^{+}=\left[
y_{\mathcal{K}}:y_{\mathcal{L}}\right]  ^{+}$ is an interval.

Altogether, we now know that $\sigma\left(  \mathcal{P}\right)  \subset\left[
n:n+m\right]  ^{+}$ is a nonempty interval such that $\sigma^{-1}\left(
\sigma\left(  \mathcal{P}\right)  \right)  $, $\sigma^{-1}\left(  K\right)
\cup\sigma^{-1}\left(  \sigma\left(  \mathcal{P}\right)  \right)  $ and
$\sigma^{-1}\left(  \sigma\left(  \mathcal{P}\right)  \right)  \cup\sigma
^{-1}\left(  L\right)  $ are intervals and such that $\sigma^{-1}\left(
K\right)  <\sigma^{-1}\left(  \sigma\left(  \mathcal{P}\right)  \right)
<\sigma^{-1}\left(  L\right)  $. Thus, there exists a nonempty interval
$P\subset\left[  n:n+m\right]  ^{+}$ such that $\sigma^{-1}\left(  P\right)
$, $\sigma^{-1}\left(  K\right)  \cup\sigma^{-1}\left(  P\right)  $ and
$\sigma^{-1}\left(  P\right)  \cup\sigma^{-1}\left(  L\right)  $ are intervals
and such that $\sigma^{-1}\left(  K\right)  <\sigma^{-1}\left(  P\right)
<\sigma^{-1}\left(  L\right)  $ (namely, $P=\sigma\left(  \mathcal{P}\right)
$). This proves Lemma \ref{lem.shuffle.lyndon.easystep}(b).

(c) The proof of Lemma \ref{lem.shuffle.lyndon.easystep}(c) is analogous to
the proof of Lemma \ref{lem.shuffle.lyndon.easystep}(b).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.shuffle.lyndon.step}}

\textit{Solution to Exercise \ref{exe.lem.shuffle.lyndon.step}.}

\begin{proof}
[Proof of Lemma \ref{lem.shuffle.lyndon.step}.]We have $\sigma\in
\operatorname{Sh}_{n,m}$, so that $\sigma^{-1}\left(  1\right)
<\sigma^{-1}\left(  2\right)  <\cdots<\sigma^{-1}\left(  n\right)  $ and
$\sigma^{-1}\left(  n+1\right)  <\sigma^{-1}\left(  n+2\right)  <\cdots
<\sigma^{-1}\left(  n+m\right)  $. In other words, the restriction of the map
$\sigma^{-1}$ to the interval $\left[  0:n\right]  ^{+}$ is strictly
increasing, and so is the restriction of the map $\sigma^{-1}$ to the interval
$\left[  n:n+m\right]  ^{+}$.

(a) Let $I$ be an interval of $\ZZ$ satisfying either $I\subset\left[
0:n\right]  ^{+}$ or $I\subset\left[  n:n+m\right]  ^{+}$. Assume that
$\sigma^{-1}\left(  I\right)  $ is an interval.

Recall that the restriction of the map $\sigma^{-1}$ to the interval $\left[
0:n\right]  ^{+}$ is strictly increasing, and so is the restriction of the map
$\sigma^{-1}$ to the interval $\left[  n:n+m\right]  ^{+}$. Hence, the
restriction of the map $\sigma^{-1}$ to the interval $I$ is strictly
increasing (since either $I\subset\left[  0:n\right]  ^{+}$ or $I\subset
\left[  n:n+m\right]  ^{+}$).

Write the interval $I$ in the form $I=\left[  \alpha:\beta\right]  ^{+}$ where
$0\leq\alpha\leq\beta\leq n+m$. Write the interval $\sigma^{-1}\left(
I\right)  $ in the form $\sigma^{-1}\left(  I\right)  =\left[  a:b\right]
^{+}$ where $0\leq a\leq b\leq n+m$. Since $\sigma$ is bijective, we have
$\left\vert \sigma^{-1}\left(  I\right)  \right\vert =\left\vert I\right\vert
=\beta-\alpha$ (since $I=\left[  \alpha:\beta\right]  ^{+}$). Compared with
$\left\vert \sigma^{-1}\left(  I\right)  \right\vert =b-a$ (since $\sigma
^{-1}\left(  I\right)  =\left[  a:b\right]  ^{+}$), this yields $\beta
-\alpha=b-a$.

The restriction of the map $\sigma^{-1}$ to the interval $I$ is injective;
hence, it can be viewed as a bijection $I\rightarrow\sigma^{-1}\left(
I\right)  $. This bijection must be strictly increasing (since the restriction
of the map $\sigma^{-1}$ to the interval $I$ is strictly increasing), and thus
is a strictly increasing bijection $\left[  \alpha:\beta\right]
^{+}\rightarrow\left[  a:b\right]  ^{+}$ (since it goes from $I=\left[
\alpha:\beta\right]  ^{+}$ to $\sigma^{-1}\left(  I\right)  =\left[
a:b\right]  ^{+}$). But the only such bijection is the one sending every
$x\in\left[  \alpha:\beta\right]  ^{+}$ to $x-\alpha+a$. Hence, our bijection
$I\rightarrow\sigma^{-1}\left(  I\right)  $ must be the map sending every
$x\in\left[  \alpha:\beta\right]  ^{+}$ to $x-\alpha+a$. Since our bijection
comes from restricting the map $\sigma^{-1}$, it thus follows that the map
$\sigma^{-1}$ sends every $x\in\left[  \alpha:\beta\right]  ^{+}$ to
$x-\alpha+a$. Thus, $\sigma^{-1}\left(  x\right)  =x-\alpha+a$ for every
$x\in\left[  \alpha:\beta\right]  ^{+}$. Substituting $y$ for $x-\alpha+a$ in
this fact, we conclude that $\sigma^{-1}\left(  y+\alpha-a\right)  =y$ for
every $y\in\left[  a:\beta-\alpha+a\right]  ^{+}$. In other words,
$\sigma^{-1}\left(  y+\alpha-a\right)  =y$ for every $y\in\left[  a:b\right]
^{+}$ (since $\beta-\alpha+a=b$ (because $\beta-\alpha=b-a$)). In other words,
$\sigma\left(  y\right)  =y+\alpha-a$ for every $y\in\left[  a:b\right]  ^{+}%
$. Thus,%
\[
\left(  \sigma\left(  a+1\right)  ,\sigma\left(  a+2\right)  ,\ldots
,\sigma\left(  b\right)  \right)  =\left(  \alpha+1,\alpha+2,\ldots
,\underbrace{b+\alpha-a}_{\substack{=\beta\\\text{(since }\beta-\alpha
=b-a\text{)}}}\right)  =\left(  \alpha+1,\alpha+2,\ldots,\beta\right)  .
\]


But let $w=\left(  w_{1},w_{2},\ldots,w_{n+m}\right)  $ denote the word $uv$.
The definition of $u\underset{\sigma}{\shuffle}v$ then yields \newline%
$u\underset{\sigma}{\shuffle}v=\left(  w_{\sigma\left(  1\right)  }%
,w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)  $,
so that%
\begin{align*}
&  \underbrace{\left(  u\underset{\sigma}{\shuffle}v\right)  }_{=\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  n+m\right)  }\right)  }\left[  \underbrace{\sigma
^{-1}\left(  I\right)  }_{=\left[  a:b\right]  ^{+}}\right] \\
&  =\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  }%
,\ldots,w_{\sigma\left(  n+m\right)  }\right)  \left[  \left[  a:b\right]
^{+}\right]  =\left(  w_{\sigma\left(  a+1\right)  },w_{\sigma\left(
a+2\right)  },\ldots,w_{\sigma\left(  b\right)  }\right) \\
&  =\left(  w_{\alpha+1},w_{\alpha+2},\ldots,w_{\beta}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \sigma\left(  a+1\right)
,\sigma\left(  a+2\right)  ,\ldots,\sigma\left(  b\right)  \right)  =\left(
\alpha+1,\alpha+2,\ldots,\beta\right)  \right) \\
&  =\underbrace{w}_{=uv}\left[  \underbrace{\left[  \alpha:\beta\right]  ^{+}%
}_{=I}\right]  =\left(  uv\right)  \left[  I\right]  .
\end{align*}
This proves Lemma \ref{lem.shuffle.lyndon.step}(a).

(b) Notice that $\sigma^{-1}\left(  I\right)  $ is a nonempty interval (since
$I$ is nonempty and $\sigma^{-1}\left(  I\right)  $ is an interval). Hence, we
can write the interval $\sigma^{-1}\left(  I\right)  $ in the form $\left[
a:b\right]  ^{+}$ for some elements $a$ and $b$ of $\left[  0:n+m\right]
^{+}$ satisfying $a<b$. Consider these $a$ and $b$. Similarly, write
$\sigma^{-1}\left(  J\right)  $ in the form $\left[  c:d\right]  ^{+}$ for
some elements $c$ and $d$ of $\left[  0:n+m\right]  ^{+}$ satisfying $c<d$. We
have $b\leq c$ (since $\sigma^{-1}\left(  I\right)  <\sigma^{-1}\left(
J\right)  $), but we cannot have $b<c$ (because $\underbrace{\left[
a:b\right]  ^{+}}_{=\sigma^{-1}\left(  I\right)  }\cup\underbrace{\left[
c:d\right]  ^{+}}_{=\sigma^{-1}\left(  J\right)  }=\sigma^{-1}\left(
I\right)  \cup\sigma^{-1}\left(  J\right)  $ is an interval). Thus, we have
$b=c$. Hence, $\left[  b:d\right]  ^{+}=\left[  c:d\right]  ^{+}=\sigma
^{-1}\left(  J\right)  $ and $b=c<d$.

Now, let $A=\left[  0:a\right]  ^{+}$ and $Z=\left[  d:n+m\right]  ^{+}$.
Then, the interval $\left[  0:n+m\right]  ^{+}$ is the union of the
disjoint intervals $A$, $\sigma^{-1}\left(  I\right)  $, $\sigma^{-1}\left(
J\right)  $ and $Z$ (because
\[
\left[  0:n+m\right]  ^{+}=\underbrace{\left[  0:a\right]  ^{+}}_{=A}%
\sqcup\underbrace{\left[  a:b\right]  ^{+}}_{=\sigma^{-1}\left(  I\right)
}\sqcup\underbrace{\left[  b:d\right]  ^{+}}_{=\sigma^{-1}\left(  J\right)
}\sqcup\underbrace{\left[  d:n+m\right]  ^{+}}_{=Z}=A\sqcup\sigma^{-1}\left(
I\right)  \sqcup\sigma^{-1}\left(  J\right)  \sqcup Z
\]
), and these intervals satisfy $A<\sigma^{-1}\left(  I\right)  <\sigma
^{-1}\left(  J\right)  <Z$ (since $A=\left[  0:a\right]  ^{+}$, $\sigma
^{-1}\left(  I\right)  =\left[  a:b\right]  ^{+}$, $\sigma^{-1}\left(
J\right)  =\left[  b:d\right]  ^{+}$ and $Z=\left[  d:n+m\right]  ^{+}$).
Therefore,%
\begin{equation}
p=p\left[  A\right]  \cdot p\left[  \sigma^{-1}\left(  I\right)  \right]
\cdot p\left[  \sigma^{-1}\left(  J\right)  \right]  \cdot p\left[  Z\right]
\ \ \ \ \ \ \ \ \ \ \text{for every word }p\in\mathfrak{B}^{n+m}
\label{pf.lem.shuffle.lyndon.step.p}%
\end{equation}
for any alphabet $\mathfrak{B}$.
Applying this to $p=\sigma$ and $\mathfrak{B}%
=\left\{  1,2,\ldots,n+m\right\}  $, we obtain%
\begin{equation}
\sigma=\sigma\left[  A\right]  \cdot\sigma\left[  \sigma^{-1}\left(  I\right)
\right]  \cdot\sigma\left[  \sigma^{-1}\left(  J\right)  \right]  \cdot
\sigma\left[  Z\right]  \label{pf.lem.shuffle.lyndon.step.sigma}%
\end{equation}
(where we consider the permutation $\sigma$ as a word in $\left\{
1,2,\ldots,n+m\right\}  ^{n+m}$ by writing it in one-line notation).

Now, define a word $\tau$ by%
\begin{equation}
\tau=\sigma\left[  A\right]  \cdot\sigma\left[  \sigma^{-1}\left(  J\right)
\right]  \cdot\sigma\left[  \sigma^{-1}\left(  I\right)  \right]  \cdot
\sigma\left[  Z\right]  . \label{pf.lem.shuffle.lyndon.step.tau}%
\end{equation}
This word $\tau$ is obtained from the word $\sigma$ by switching the two
factors $\sigma\left[  \sigma^{-1}\left(  I\right)  \right]  $ and
$\sigma\left[  \sigma^{-1}\left(  J\right)  \right]  $ (this is clear by
comparing (\ref{pf.lem.shuffle.lyndon.step.tau}) with
(\ref{pf.lem.shuffle.lyndon.step.sigma})), and thus is a permutation written
in one-line notation (because $\sigma$ is a permutation). In other words,
$\tau\in\Symm_{n+m}$.

From (\ref{pf.lem.shuffle.lyndon.step.tau}), we have%
\begin{align*}
\tau &  =\sigma\left[  \underbrace{A}_{=\left[  0:a\right]  ^{+}}\right]
\cdot\sigma\left[  \underbrace{\sigma^{-1}\left(  J\right)  }_{=\left[
b:d\right]  ^{+}}\right]  \cdot\sigma\left[  \underbrace{\sigma^{-1}\left(
I\right)  }_{=\left[  a:b\right]  ^{+}}\right]  \cdot\sigma\left[
\underbrace{Z}_{=\left[  d:n+m\right]  ^{+}}\right] \\
&  =\underbrace{\sigma\left[  \left[  0:a\right]  ^{+}\right]  }_{=\left(
\sigma\left(  1\right)  ,\sigma\left(  2\right)  ,\ldots,\sigma\left(
a\right)  \right)  }\cdot\underbrace{\sigma\left[  \left[  b:d\right]
^{+}\right]  }_{=\left(  \sigma\left(  b+1\right)  ,\sigma\left(  b+2\right)
,\ldots,\sigma\left(  d\right)  \right)  }\cdot\underbrace{\sigma\left[
\left[  a:b\right]  ^{+}\right]  }_{=\left(  \sigma\left(  a+1\right)
,\sigma\left(  a+2\right)  ,\ldots,\sigma\left(  b\right)  \right)  }%
\cdot\underbrace{\sigma\left[  \left[  d:n+m\right]  ^{+}\right]  }_{=\left(
\sigma\left(  d+1\right)  ,\sigma\left(  d+2\right)  ,\ldots,\sigma\left(
n+m\right)  \right)  }\\
&  =\left(  \sigma\left(  1\right)  ,\sigma\left(  2\right)  ,\ldots
,\sigma\left(  a\right)  \right)  \cdot\left(  \sigma\left(  b+1\right)
,\sigma\left(  b+2\right)  ,\ldots,\sigma\left(  d\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \sigma\left(  a+1\right)  ,\sigma\left(
a+2\right)  ,\ldots,\sigma\left(  b\right)  \right)  \cdot\left(
\sigma\left(  d+1\right)  ,\sigma\left(  d+2\right)  ,\ldots,\sigma\left(
n+m\right)  \right) \\
&  =\left(  \sigma\left(  1\right)  ,\sigma\left(  2\right)  ,\ldots
,\sigma\left(  a\right)  ,\sigma\left(  b+1\right)  ,\sigma\left(  b+2\right)
,\ldots,\sigma\left(  d\right)  ,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \sigma\left(  a+1\right)  ,\sigma\left(
a+2\right)  ,\ldots,\sigma\left(  b\right)  ,\sigma\left(  d+1\right)
,\sigma\left(  d+2\right)  ,\ldots,\sigma\left(  n+m\right)  \right)  .
\end{align*}
Thus,%
\begin{align}
\left(  \tau\left(  1\right)  ,\tau\left(  2\right)  ,\ldots,\tau\left(
a\right)  \right)   &  =\left(  \sigma\left(  1\right)  ,\sigma\left(
2\right)  ,\ldots,\sigma\left(  a\right)  \right)
;\label{pf.lem.shuffle.lyndon.step.tau.1}\\
\left(  \tau\left(  a+1\right)  ,\tau\left(  a+2\right)  ,\ldots,\tau\left(
a+d-b\right)  \right)   &  =\left(  \sigma\left(  b+1\right)  ,\sigma\left(
b+2\right)  ,\ldots,\sigma\left(  d\right)  \right)
;\label{pf.lem.shuffle.lyndon.step.tau.2}\\
\left(  \tau\left(  a+d-b+1\right)  ,\tau\left(  a+d-b+2\right)  ,\ldots
,\tau\left(  d\right)  \right)   &  =\left(  \sigma\left(  a+1\right)
,\sigma\left(  a+2\right)  ,\ldots,\sigma\left(  b\right)  \right)
;\label{pf.lem.shuffle.lyndon.step.tau.3}\\
\left(  \tau\left(  d+1\right)  ,\tau\left(  d+2\right)  ,\ldots,\tau\left(
n+m\right)  \right)   &  =\left(  \sigma\left(  d+1\right)  ,\sigma\left(
d+2\right)  ,\ldots,\sigma\left(  n+m\right)  \right)  .
\label{pf.lem.shuffle.lyndon.step.tau.4}%
\end{align}
From this, it is easy to see that $\tau^{-1}\left(  J\right)  =\left[
a:a+d-b\right]  ^{+}$\ \ \ \ \footnote{In fact, this follows from%
\begin{align*}
\tau\left(  \underbrace{\left[  a:a+d-b\right]  ^{+}}_{=\left\{
a+1,a+2,\ldots,a+d-b\right\}  }\right)   &  =\tau\left(  \left\{
a+1,a+2,\ldots,a+d-b\right\}  \right)  =\left\{  \tau\left(  a+1\right)
,\tau\left(  a+2\right)  ,\ldots,\tau\left(  a+d-b\right)  \right\} \\
&  =\left\{  \sigma\left(  b+1\right)  ,\sigma\left(  b+2\right)
,\ldots,\sigma\left(  d\right)  \right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.shuffle.lyndon.step.tau.2})}\right) \\
&  =\sigma\left(  \underbrace{\left\{  b+1,b+2,\ldots,d\right\}  }_{=\left[
b:d\right]  ^{+}=\sigma^{-1}\left(  J\right)  }\right)  =\sigma\left(
\sigma^{-1}\left(  J\right)  \right)  =J.
\end{align*}
} and $\tau^{-1}\left(  I\right)  =\left[  a+d-b:d\right]  ^{+}\ \ \ \ $%
\footnote{The proof of this is similar to that of $\tau^{-1}\left(  J\right)
=\left[  a:a+d-b\right]  ^{+}$, but we have to use
(\ref{pf.lem.shuffle.lyndon.step.tau.3}) this time.}. In particular,
$\tau^{-1}\left(  J\right)  $ and $\tau^{-1}\left(  I\right)  $ are intervals.
Now, obviously, the intervals $\left[  0:a\right]  ^{+}$, $\left[
a:a+d-b\right]  ^{+}$, $\left[  a+d-b:d\right]  ^{+}$ and $\left[
d:n+m\right]  ^{+}$ are disjoint intervals having union $\left[  0:n+m\right]
^{+}$ and satisfying $\left[  0:a\right]  ^{+}<\left[  a:a+d-b\right]
^{+}<\left[  a+d-b:d\right]  ^{+}<\left[  d:n+m\right]  ^{+}$. Since
$A=\left[  0:a\right]  ^{+}$, $\tau^{-1}\left(  J\right)  =\left[
a:a+d-b\right]  ^{+}$, $\tau^{-1}\left(  I\right)  =\left[  a+d-b:d\right]
^{+}$ and $Z=\left[  d:n+m\right]  ^{+}$, this rewrites as follows: The
intervals $A$, $\tau^{-1}\left(  J\right)  $, $\tau^{-1}\left(  I\right)  $
and $Z$ are disjoint intervals having union $\left[  0:n+m\right]  ^{+}$ and
satisfying $A<\tau^{-1}\left(  J\right)  <\tau^{-1}\left(  I\right)  <Z$.
Therefore,%
\begin{equation}
p=p\left[  A\right]  \cdot p\left[  \tau^{-1}\left(  J\right)  \right]  \cdot
p\left[  \tau^{-1}\left(  I\right)  \right]  \cdot p\left[  Z\right]
\ \ \ \ \ \ \ \ \ \ \text{for every word }p\in\mathfrak{B}^{n+m}
\label{pf.lem.shuffle.lyndon.step.p.tau}%
\end{equation}
for every alphabet $\mathfrak{B}$.

Next, we claim that $\tau$ belongs to $\operatorname{Sh}_{n,m}$. In
fact, let $i$ and $j$ be two elements of $\left\{  1,2,\ldots,n\right\}  $
such that $i<j$. Our next goal is to prove that $\tau^{-1}\left(  i\right)
<\tau^{-1}\left(  j\right)  $.

Indeed, assume the contrary. Thus, $\tau^{-1}\left(  i\right)  \geq\tau
^{-1}\left(  j\right)  $, so that $\tau^{-1}\left(  i\right)  >\tau
^{-1}\left(  j\right)  $ (since $\tau$ is a permutation). In other words, the
letter $i$ lies further right than the letter $j$ in the word $\tau$. But we
also have $\sigma^{-1}\left(  i\right)  <\sigma^{-1}\left(  j\right)  $ (since
$\sigma^{-1}\left(  1\right)  <\sigma^{-1}\left(  2\right)  <\cdots
<\sigma^{-1}\left(  n\right)  $ and $i<j$ and since $i,j\in\left\{
1,2,\ldots,n\right\}  $), which means that the letter $i$ lies further left
than the letter $j$ in the word $\sigma$.

So the order in which the letters $i$ and $j$ appear in the word $\tau$ is
different from that in $\sigma$. Since the word $\tau$ is obtained from the
word $\sigma$ by switching the two adjacent factors $\sigma\left[  \sigma
^{-1}\left(  I\right)  \right]  $ and $\sigma\left[  \sigma^{-1}\left(
J\right)  \right]  $, this is only possible if one of the letters $i$ and $j$
is contained in one of these two factors and the other is contained in the
other factor. Hence, this is what must be happening. In particular, one of the
letters $i$ and $j$ must be contained in the factor $\sigma\left[  \sigma
^{-1}\left(  J\right)  \right]  $. Since all letters of $\sigma\left[
\sigma^{-1}\left(  J\right)  \right]  $ belong to the set $\left[
n:n+m\right]  ^{+}$ (because the letters of $\sigma\left[  \sigma^{-1}\left(
J\right)  \right]  $ are precisely the elements of $J$, but we have
$J\subset\left[  n:n+m\right]  ^{+}$), this yields that one of the letters $i$
and $j$ must belong to the set $\left[  n:n+m\right]  ^{+}$. But this is
impossible, since the letters $i$ and $j$ belong to $\left\{  1,2,\ldots
,n\right\}  $ and thus neither of them can be an element of $\left[
n:n+m\right]  ^{+}$ (because $\left[  n:n+m\right]  ^{+}$ is disjoint with
$\left\{  1,2,\ldots,n\right\}  $). This contradiction shows that our
assumption was wrong, and so we do have $\tau^{-1}\left(  i\right)  <\tau
^{-1}\left(  j\right)  $.

Now, forget that we fixed $i$ and $j$. We have shown that any two elements $i$
and $j$ of $\left\{  1,2,\ldots,n\right\}  $ such that $i<j$ must satisfy
$\tau^{-1}\left(  i\right)  <\tau^{-1}\left(  j\right)  $. Thus, $\tau
^{-1}\left(  1\right)  <\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(
n\right)  $. Similarly, $\tau^{-1}\left(  n+1\right)  <\tau^{-1}\left(
n+2\right)  <\cdots<\tau^{-1}\left(  n+m\right)  $. These two chains of
inequalities, together, yield that $\tau\in\operatorname{Sh}_{n,m}$.
Hence, $u\underset{\tau}{\shuffle}v$ is a well-defined element of the multiset
$u\shuffle v$. Since $u\underset{\sigma}{\shuffle}v$ is the lexicographically
highest element of this multiset, this shows that
\begin{equation}
u\underset{\sigma}{\shuffle}v\geq u\underset{\tau}{\shuffle}v.
\label{pf.lem.shuffle.lyndon.step.lex}%
\end{equation}


Let $\mathfrak{a}$ denote the word $\left(  u\underset{\tau}{\shuffle}%
v\right)  \left[  A\right]  $, and let $\mathfrak{z}$ denote the word $\left(
u\underset{\tau}{\shuffle}v\right)  \left[  Z\right]  $.

We can now apply (\ref{pf.lem.shuffle.lyndon.step.p}) to $p=u\underset{\sigma
}{\shuffle}v$ and $\mathfrak{B }= \mathfrak{A}$. As a result, we obtain%
\begin{equation}
u\underset{\sigma}{\shuffle}v=\left(  u\underset{\sigma}{\shuffle}v\right)
\left[  A\right]  \cdot\left(  u\underset{\sigma}{\shuffle}v\right)  \left[
\sigma^{-1}\left(  I\right)  \right]  \cdot\left(  u\underset{\sigma
}{\shuffle}v\right)  \left[  \sigma^{-1}\left(  J\right)  \right]
\cdot\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  Z\right]  .
\label{pf.lem.shuffle.lyndon.step.sh1}%
\end{equation}


Let now $\left(  w_{1},w_{2},\ldots,w_{n+m}\right)  $ denote the word $uv$.
The definition of $u\underset{\sigma}{\shuffle}v$ then yields \newline%
$u\underset{\sigma}{\shuffle}v=\left(  w_{\sigma\left(  1\right)  }%
,w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)  $.
Now, it is easy to see that%
\[
\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  A\right]  =\mathfrak{a}%
\]
\footnote{\textit{Proof.} We have%
\[
\underbrace{\left(  u\underset{\sigma}{\shuffle}v\right)  }_{=\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  n+m\right)  }\right)  }\left[  \underbrace{A}_{=\left[
0:a\right]  ^{+}}\right]  =\left(  w_{\sigma\left(  1\right)  },w_{\sigma
\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)  \left[
\left[  0:a\right]  ^{+}\right]  =\left(  w_{\sigma\left(  1\right)
},w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  a\right)  }\right)  .
\]
The same argument, applied to $\tau$ instead of $\sigma$, shows that $\left(
u\underset{\tau}{\shuffle}v\right)  \left[  A\right]  =\left(  w_{\tau\left(
1\right)  },w_{\tau\left(  2\right)  },\ldots,w_{\tau\left(  a\right)
}\right)  $. But (\ref{pf.lem.shuffle.lyndon.step.tau.1}) yields $\left(
w_{\tau\left(  1\right)  },w_{\tau\left(  2\right)  },\ldots,w_{\tau\left(
a\right)  }\right)  =\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(
2\right)  },\ldots,w_{\sigma\left(  a\right)  }\right)  $, so that%
\[
\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  A\right]  =\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  a\right)  }\right)  =\left(  w_{\tau\left(  1\right)
},w_{\tau\left(  2\right)  },\ldots,w_{\tau\left(  a\right)  }\right)
=\left(  u\underset{\tau}{\shuffle}v\right)  \left[  A\right]  =\mathfrak{a},
\]
qed.} and%
\[
\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  Z\right]  =\mathfrak{z}%
\]
\footnote{\textit{Proof.} We have%
\[
\underbrace{\left(  u\underset{\sigma}{\shuffle}v\right)  }_{=\left(
w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)  },\ldots
,w_{\sigma\left(  n+m\right)  }\right)  }\left[  \underbrace{Z}_{=\left[
d:n+m\right]  ^{+}}\right]  =\left(  w_{\sigma\left(  1\right)  }%
,w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)
\left[  \left[  d:n+m\right]  ^{+}\right]  =\left(  w_{\sigma\left(
d+1\right)  },w_{\sigma\left(  d+2\right)  },\ldots,w_{\sigma\left(
n+m\right)  }\right)  .
\]
The same argument, applied to $\tau$ instead of $\sigma$, shows that $\left(
u\underset{\tau}{\shuffle}v\right)  \left[  Z\right]  =\left(  w_{\tau\left(
d+1\right)  },w_{\tau\left(  d+2\right)  },\ldots,w_{\tau\left(  n+m\right)
}\right)  $. But (\ref{pf.lem.shuffle.lyndon.step.tau.4}) yields $\left(
w_{\tau\left(  d+1\right)  },w_{\tau\left(  d+2\right)  },\ldots
,w_{\tau\left(  n+m\right)  }\right)  =\left(  w_{\sigma\left(  d+1\right)
},w_{\sigma\left(  d+2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)
$, so that%
\[
\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  Z\right]  =\left(
w_{\sigma\left(  d+1\right)  },w_{\sigma\left(  d+2\right)  },\ldots
,w_{\sigma\left(  n+m\right)  }\right)  =\left(  w_{\tau\left(  d+1\right)
},w_{\tau\left(  d+2\right)  },\ldots,w_{\tau\left(  n+m\right)  }\right)
=\left(  u\underset{\tau}{\shuffle}v\right)  \left[  Z\right]  =\mathfrak{z},
\]
qed.}. Thus, (\ref{pf.lem.shuffle.lyndon.step.sh1}) becomes%
\begin{align*}
u\underset{\sigma}{\shuffle}v  &  =\underbrace{\left(  u\underset{\sigma
}{\shuffle}v\right)  \left[  A\right]  }_{=\mathfrak{a}}\cdot
\underbrace{\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  \sigma
^{-1}\left(  I\right)  \right]  }_{\substack{=\left(  uv\right)  \left[
I\right]  \\\text{(by (\ref{pf.lem.shuffle.lyndon.step.gamma}))}}%
}\cdot\underbrace{\left(  u\underset{\sigma}{\shuffle}v\right)  \left[
\sigma^{-1}\left(  J\right)  \right]  }_{\substack{=\left(  uv\right)  \left[
J\right]  \\\text{(by (\ref{pf.lem.shuffle.lyndon.step.gamma}), applied to
}J\\\text{instead of }I\text{)}}}\cdot\underbrace{\left(  u\underset{\sigma
}{\shuffle}v\right)  \left[  Z\right]  }_{=\mathfrak{z}}\\
&  =\mathfrak{a}\cdot\left(  uv\right)  \left[  I\right]  \cdot\left(
uv\right)  \left[  J\right]  \cdot\mathfrak{z}.
\end{align*}


Thus,%
\begin{align*}
&  \mathfrak{a}\cdot\left(  uv\right)  \left[  I\right]  \cdot\left(
uv\right)  \left[  J\right]  \cdot\mathfrak{z}\\
&  =u\underset{\sigma}{\shuffle}v\geq u\underset{\tau}{\shuffle}%
v\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.shuffle.lyndon.step.lex}%
)}\right) \\
&  =\underbrace{\left(  u\underset{\tau}{\shuffle}v\right)  \left[  A\right]
}_{=\mathfrak{a}}\cdot\underbrace{\left(  u\underset{\tau}{\shuffle}v\right)
\left[  \tau^{-1}\left(  J\right)  \right]  }_{\substack{=\left(  uv\right)
\left[  J\right]  \\\text{(by (\ref{pf.lem.shuffle.lyndon.step.gamma}),
applied to }\tau\text{ and }J\\\text{instead of }\sigma\text{ and }I\text{)}%
}}\cdot\underbrace{\left(  u\underset{\tau}{\shuffle}v\right)  \left[
\tau^{-1}\left(  I\right)  \right]  }_{\substack{=\left(  uv\right)  \left[
I\right]  \\\text{(by (\ref{pf.lem.shuffle.lyndon.step.gamma}), applied to
}\tau\\\text{instead of }\sigma\text{)}}}\cdot\underbrace{\left(
u\underset{\tau}{\shuffle}v\right)  \left[  Z\right]  }_{=\mathfrak{z}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.shuffle.lyndon.step.p.tau}), applied to }p=u\underset{\tau
}{\shuffle}v\text{ and }\mathfrak{B}=\mathfrak{A}\right) \\
&  =\mathfrak{a}\cdot\left(  uv\right)  \left[  J\right]  \cdot\left(
uv\right)  \left[  I\right]  \cdot\mathfrak{z}.
\end{align*}
In other words, $\mathfrak{a}\cdot\left(  uv\right)  \left[  J\right]
\cdot\left(  uv\right)  \left[  I\right]  \cdot\mathfrak{z}\leq\mathfrak{a}%
\cdot\left(  uv\right)  \left[  I\right]  \cdot\left(  uv\right)  \left[
J\right]  \cdot\mathfrak{z}$. Hence, Proposition \ref{prop.words.lex}(c)
(applied to $\mathfrak{a}$, $\left(  uv\right)  \left[  J\right]  \cdot\left(
uv\right)  \left[  I\right]  \cdot\mathfrak{z}$ and $\left(  uv\right)
\left[  I\right]  \cdot\left(  uv\right)  \left[  J\right]  \cdot\mathfrak{z}$
instead of $a$, $c$ and $d$) yields $\left(  uv\right)  \left[  J\right]
\cdot\left(  uv\right)  \left[  I\right]  \cdot\mathfrak{z}\leq\left(
uv\right)  \left[  I\right]  \cdot\left(  uv\right)  \left[  J\right]
\cdot\mathfrak{z}$. Thus, Proposition \ref{prop.words.lex}(d) (applied to
$\left(  uv\right)  \left[  J\right]  \cdot\left(  uv\right)  \left[
I\right]  $, $\mathfrak{z}$, $\left(  uv\right)  \left[  I\right]
\cdot\left(  uv\right)  \left[  J\right]  $ and $\mathfrak{z}$ instead of $a$,
$b$, $c$ and $d$) yields $\left(  uv\right)  \left[  J\right]  \cdot\left(
uv\right)  \left[  I\right]  \leq\left(  uv\right)  \left[  I\right]
\cdot\left(  uv\right)  \left[  J\right]  $ (since $\ell\left(  \left(
uv\right)  \left[  J\right]  \cdot\left(  uv\right)  \left[  I\right]
\right)  =\ell\left(  \left(  uv\right)  \left[  J\right]  \right)
+\ell\left(  \left(  uv\right)  \left[  I\right]  \right)  =\ell\left(
\left(  uv\right)  \left[  I\right]  \right)  +\ell\left(  \left(  uv\right)
\left[  J\right]  \right)  =\ell\left(  \left(  uv\right)  \left[  I\right]
\cdot\left(  uv\right)  \left[  J\right]  \right)  $). In other words,
$\left(  uv\right)  \left[  I\right]  \cdot\left(  uv\right)  \left[
J\right]  \geq\left(  uv\right)  \left[  J\right]  \cdot\left(  uv\right)
\left[  I\right]  $. This proves Lemma \ref{lem.shuffle.lyndon.step}(b).

(c) The proof of Lemma \ref{lem.shuffle.lyndon.step}(c) is analogous to that
of Lemma \ref{lem.shuffle.lyndon.step}(b).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.iper.intinverse}}

\textit{Solution to Exercise \ref{exe.prop.iper.intinverse}.}

\begin{proof}
[Proof of Proposition \ref{prop.iper.intinverse}.](a) Let $j\in\left\{
1,2,\ldots,\ell\right\}  $. We have $\sigma=\operatorname*{iper}\left(
\alpha,\tau\right)  =\overrightarrow{I_{\tau\left(  1\right)  }}%
\overrightarrow{I_{\tau\left(  2\right)  }}\cdots\overrightarrow{I_{\tau
\left(  \ell\right)  }}$ (in one-line notation). Therefore, the word
$\overrightarrow{I_{\tau\left(  j\right)  }}$ appears as a factor in this word
$\sigma$, starting at position $\sum_{k=1}^{j-1}\alpha_{\tau\left(  k\right)
}+1$ and ending at position $\sum_{k=1}^{j}\alpha_{\tau\left(  k\right)  }$.
In other words, the letters of the word $\overrightarrow{I_{\tau\left(
j\right)  }}$ are the letters $\sigma_{j}$ of $\sigma$ for $j\in\left[
\sum_{k=1}^{j-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{j}\alpha
_{\tau\left(  k\right)  }\right]  ^{+}$. Since the letters of the word
$\overrightarrow{I_{\tau\left(  j\right)  }}$ are precisely the elements of
$I_{\tau\left(  j\right)  }$, this rewrites as follows: The elements of
$I_{\tau\left(  j\right)  }$ are the letters $\sigma_{j}$ of $\sigma$ for
$j\in\left[  \sum_{k=1}^{j-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}%
^{j}\alpha_{\tau\left(  k\right)  }\right]  ^{+}$. In other words,%
\[
I_{\tau\left(  j\right)  }=\left\{  \sigma_{j}\ \mid\ j\in\left[  \sum
_{k=1}^{j-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{j}\alpha_{\tau\left(
k\right)  }\right]  ^{+}\right\}  =\sigma\left(  \left[  \sum_{k=1}%
^{j-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{j}\alpha_{\tau\left(
k\right)  }\right]  ^{+}\right)  .
\]
Hence, $\sigma^{-1}\left(  I_{\tau\left(  j\right)  }\right)  =\left[
\sum_{k=1}^{j-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{j}\alpha
_{\tau\left(  k\right)  }\right]  ^{+}$, so that Proposition
\ref{prop.iper.intinverse}(a) is proven.

(b) Let $j\in\left\{  1,2,\ldots,\ell\right\}  $. We need to show that the
restriction of the map $\sigma^{-1}$ to the interval $I_{\tau\left(  j\right)
}$ is increasing. In other words, we need to prove that the elements of
$I_{\tau\left(  j\right)  }$ occur in increasing order in the word $\sigma$.
But this is clear, because these elements all occur in the factor
$\overrightarrow{I_{\tau\left(  j\right)  }}$ of the word $\sigma$, and this
factor has them in increasing order (by its definition). This proves
Proposition \ref{prop.iper.intinverse}(b).

(c) Let $i\in\left\{  1,2,\ldots,\ell\right\}  $. Then, $\sigma^{-1}\left(
I_{i}\right)  =\sigma^{-1}\left(  I_{\tau\left(  \tau^{-1}\left(  i\right)
\right)  }\right)  =\left[  \sum_{k=1}^{\tau^{-1}\left(  i\right)  -1}%
\alpha_{\tau\left(  k\right)  },\sum_{k=1}^{\tau^{-1}\left(  i\right)  }%
\alpha_{\tau\left(  k\right)  }\right]  ^{+}$ (according to Proposition
\ref{prop.iper.intinverse}(a), applied to $j=\tau^{-1}\left(  i\right)  $).
Hence, $\sigma^{-1}\left(  I_{i}\right)  $ is an interval. Furthermore, the
restriction of the map $\sigma^{-1}$ to the interval $I_{i}=I_{\tau\left(
\tau^{-1}\left(  i\right)  \right)  }$ is increasing (according to Proposition
\ref{prop.iper.intinverse}(b), applied to $j=\tau^{-1}\left(  i\right)  $).

Now, forget that we fixed $i$. We thus have proven that every $i\in\left\{
1,2,\ldots,\ell\right\}  $ has the two properties that:

\begin{itemize}
\item the set $\sigma^{-1}\left(  I_{i}\right)  $ is an interval;

\item the restriction of the map $\sigma^{-1}$ to the interval $I_{i}$ is increasing.
\end{itemize}

In other words, the permutation $\sigma$ is $\alpha$-clumping. Since
$\sigma=\operatorname*{iper}\left(  \alpha,\tau\right)  $, this shows that
$\operatorname*{iper}\left(  \alpha,\tau\right)  $ is $\alpha$-clumping.
Proposition \ref{prop.iper.intinverse}(c) is proven.

(d) Let $i\in\left\{  1,2,\ldots,\ell-1\right\}  $. Then,
\begin{equation}
\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  =\left[  \sum
_{k=1}^{i-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{i}\alpha_{\tau\left(
k\right)  }\right]  ^{+} \label{pf.prop.iper.intinverse.d.1}%
\end{equation}
(by Proposition \ref{prop.iper.intinverse}(a), applied to $j=i$), so that
$\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  $ is an interval. Thus,
$\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  $ is a nonempty
interval (nonempty because $I_{\tau\left(  i\right)  }$ is nonempty).
Similarly, $\sigma^{-1}\left(  I_{\tau\left(  i+1\right)  }\right)  $ is a
nonempty interval.

Also, Proposition \ref{prop.iper.intinverse}(a) (applied to $j=i+1$) yields%
\begin{equation}
\sigma^{-1}\left(  I_{\tau\left(  i+1\right)  }\right)  =\left[  \sum
_{k=1}^{i}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{i+1}\alpha_{\tau\left(
k\right)  }\right]  ^{+}. \label{pf.prop.iper.intinverse.d.2}%
\end{equation}
Now,%
\[
\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  =\left[  \sum
_{k=1}^{i-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{i}\alpha_{\tau\left(
k\right)  }\right]  ^{+}<\left[  \sum_{k=1}^{i}\alpha_{\tau\left(  k\right)
}:\sum_{k=1}^{i+1}\alpha_{\tau\left(  k\right)  }\right]  ^{+}=\sigma
^{-1}\left(  I_{\tau\left(  i+1\right)  }\right)  .
\]
Also,%
\begin{align*}
&  \underbrace{\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)
}_{\substack{=\left[  \sum_{k=1}^{i-1}\alpha_{\tau\left(  k\right)  }%
:\sum_{k=1}^{i}\alpha_{\tau\left(  k\right)  }\right]  ^{+}\\\text{(by
(\ref{pf.prop.iper.intinverse.d.1}))}}}\cup\underbrace{\sigma^{-1}\left(
I_{\tau\left(  i+1\right)  }\right)  }_{\substack{=\left[  \sum_{k=1}%
^{i}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{i+1}\alpha_{\tau\left(
k\right)  }\right]  ^{+}\\\text{(by (\ref{pf.prop.iper.intinverse.d.2}))}}}\\
&  =\left[  \sum_{k=1}^{i-1}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}%
^{i}\alpha_{\tau\left(  k\right)  }\right]  ^{+}\cup\left[  \sum_{k=1}%
^{i}\alpha_{\tau\left(  k\right)  }:\sum_{k=1}^{i+1}\alpha_{\tau\left(
k\right)  }\right]  ^{+}=\left[  \sum_{k=1}^{i-1}\alpha_{\tau\left(  k\right)
}:\sum_{k=1}^{i+1}\alpha_{\tau\left(  k\right)  }\right]  ^{+},
\end{align*}
which is obviously an interval. Proposition \ref{prop.iper.intinverse}(d) is proven.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.iper.biject}}

\textit{Solution to Exercise \ref{exe.prop.iper.biject}.}

\begin{proof}
[Proof of Proposition \ref{prop.iper.biject}.]
We shall use Definition~\ref{def.sol.Lalpha-std.proofs.Inv} and
Proposition~\ref{prop.sol.Lalpha-std.proofs.Inv=}.

(a) The interval system corresponding to $\alpha$ is an $\ell$-tuple of
intervals (since $\ell\left(  \alpha\right)  =\ell$); denote this $\ell$-tuple
by $\left(  I_{1},I_{2},\ldots,I_{\ell}\right)  $. Then, the intervals $I_{1}%
$, $I_{2}$, $\ldots$, $I_{\ell}$ form a set partition of $\left[  0:n\right]
^{+}$ (according to Remark \ref{rmk.intsys.trivia}(c)) and are nonempty (also
according to Remark \ref{rmk.intsys.trivia}(c)).

We define a map $\operatorname{iper}_{\alpha}^{\prime}:\left\{
\omega\in\Symm_n\mid\omega\text{ is }\alpha\text{-clumping}\right\}
\rightarrow\Symm_{\ell}$ as follows: Let $\omega$ be an $\alpha
$-clumping element of $\Symm_n$. Then, every $i\in\left\{
1,2,\ldots,\ell\right\}  $ has the property that $\omega^{-1}\left(
I_{i}\right)  $ is an interval (since $\omega$ is $\alpha$-clumping). These
intervals $\omega^{-1}\left(  I_{1}\right)  $, $\omega^{-1}\left(
I_{2}\right)  $, $\ldots$, $\omega^{-1}\left(  I_{\ell}\right)  $ form a set
partition of $\left[  0:n\right]  ^{+}$ (since the intervals $I_{1}$, $I_{2}$,
$\ldots$, $I_{\ell}$ form a set partition of $\left[  0:n\right]  ^{+}$), and
thus are disjoint (and nonempty\footnote{Their nonemptiness follows
from the fact that the intervals $I_{1}$, $I_{2}$, $\ldots$, $I_{\ell}$ are
nonempty.}). Hence, these intervals form a totally ordered set with respect to
the relation $<$ (by Remark \ref{rmk.intsys.trivia}(b)). Thus, there exists a
unique permutation $\tau\in\Symm_{\ell}$ such that $\omega^{-1}\left(
I_{\tau\left(  1\right)  }\right)  <\omega^{-1}\left(  I_{\tau\left(
2\right)  }\right)  < \cdots<\omega^{-1}\left(  I_{\tau\left(  \ell\right)
}\right)  $. We define $\operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  $ to be this permutation $\tau$.

Thus, we have defined a map $\operatorname{iper}_{\alpha}^{\prime
}:\left\{  \omega\in\Symm_n\mid\omega\text{ is }\alpha
\text{-clumping}\right\}  \rightarrow\Symm_{\ell}$. It is easy to see
that $\operatorname{iper}_{\alpha}\circ
\operatorname{iper}_{\alpha}^{\prime}=\id$%
\ \ \ \ \footnote{\textit{Proof.} Let $\omega\in\Symm_n$ be $\alpha
$-clumping. We are going to prove that
$\left(  \operatorname{iper}_{\alpha}\circ
\operatorname{iper}_{\alpha}^{\prime}\right)
\left(  \omega\right)  =\operatorname*{id}\left(  \omega\right)  $.
\par
Indeed, let us recall how $\operatorname{iper}_{\alpha}^{\prime
}\left(  \omega\right)  $ was defined: Every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ has the property that $\omega^{-1}\left(  I_{i}\right)  $ is
an interval. These intervals $\omega^{-1}\left(  I_{1}\right)  $, $\omega
^{-1}\left(  I_{2}\right)  $, $\ldots$, $\omega^{-1}\left(  I_{\ell}\right)  $
form a totally ordered set with respect to the relation $<$. Then,
$\operatorname{iper}_{\alpha}^{\prime}\left(  \omega\right)  $ is
defined as the unique permutation $\tau\in\Symm_{\ell}$ such that
$\omega^{-1}\left(  I_{\tau\left(  1\right)  }\right)  <\omega^{-1}\left(
I_{\tau\left(  2\right)  }\right)  < \cdots<\omega^{-1}\left(  I_{\tau\left(
\ell\right)  }\right)  $. Thus, we have $\operatorname{iper}_{\alpha
}^{\prime}\left(  \omega\right)  \in\Symm_{\ell}$ and%
\begin{equation}
\omega^{-1}\left(  I_{\left(  \operatorname{iper}_{\alpha}^{\prime
}\left(  \omega\right)  \right)  \left(  1\right)  }\right)  <\omega
^{-1}\left(  I_{\left(  \operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  \right)  \left(  2\right)  }\right)  < \cdots<\omega
^{-1}\left(  I_{\left(  \operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  \right)  \left(  \ell\right)  }\right)  .
\label{pf.prop.iper.biject.a.1}%
\end{equation}
\par
Denote the permutation $\operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  \in\Symm_{\ell}$ by $\tau$. Then,
(\ref{pf.prop.iper.biject.a.1}) rewrites as
\begin{equation}
\omega^{-1}\left(  I_{\tau\left(  1\right)  }\right)  <\omega^{-1}\left(
I_{\tau\left(  2\right)  }\right)  < \cdots<\omega^{-1}\left(  I_{\tau\left(
\ell\right)  }\right)  \label{pf.prop.iper.biject.a.1tau}%
\end{equation}
(since $\tau=\operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  $). The definition of $\operatorname*{iper}\left(  \alpha
,\tau\right)  $ shows that $\operatorname*{iper}\left(  \alpha,\tau\right)  $
is the permutation in $\Symm_n$ which (in one-line notation) is the
word $\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau
\left(  2\right)  }}\cdots\overrightarrow{I_{\tau\left(  \ell\right)  }}$ (a
concatenation of $\ell$ words). In other words, $\operatorname*{iper}\left(
\alpha,\tau\right)  =\overrightarrow{I_{\tau\left(  1\right)  }}%
\overrightarrow{I_{\tau\left(  2\right)  }}\cdots\overrightarrow{I_{\tau
\left(  \ell\right)  }}$. Denote this permutation $\operatorname*{iper}\left(
\alpha,\tau\right)  $ by $\eta$. Then, $\eta=\operatorname*{iper}\left(
\alpha,\tau\right)  =\overrightarrow{I_{\tau\left(  1\right)  }}%
\overrightarrow{I_{\tau\left(  2\right)  }}\cdots\overrightarrow{I_{\tau
\left(  \ell\right)  }}$ and%
\begin{align}
\left(  \operatorname{iper}_{\alpha}\circ
\operatorname{iper}_{\alpha}^{\prime}\right)  \left(  \omega\right)   &
=\operatorname{iper}_{\alpha}\left(
\underbrace{\operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  }_{=\tau}\right)  =\operatorname{iper}_{\alpha}%
\tau=\operatorname*{iper}\left(  \alpha,\tau\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }
\operatorname{iper}_{\alpha}\tau\right) \nonumber\\
&  =\eta.
\label{pf.prop.iper.biject.a.iiwn}
\end{align}
\par
Our next goal is to prove that $\eta=\omega$.
\par
We know that $\omega$ is $\alpha$-clumping. In other words, every
$i\in\left\{  1,2,\ldots,\ell\right\}  $ has the two properties that:%
\begin{equation}
\text{the set }\omega^{-1}\left(  I_{i}\right)  \text{ is an interval,}
\label{pf.prop.iper.biject.a.2}%
\end{equation}
and%
\begin{equation}
\text{the restriction of the map }\omega^{-1}\text{ to the interval }%
I_{i}\text{ is increasing} \label{pf.prop.iper.biject.a.3}%
\end{equation}
(by the definition of ``$\alpha$-clumping'').
\par
We shall now prove that $\operatorname*{Inv}\left(  \eta^{-1}\right)
\subset\operatorname*{Inv}\left(  \omega^{-1}\right)  $.
\par
Indeed, let $\left(  i,j\right)  $ be some element of $\operatorname*{Inv}%
\left(  \eta^{-1}\right)  $. We want to prove that $\left(  i,j\right)
\in\operatorname*{Inv}\left(  \omega^{-1}\right)  $.
\par
We have $\left(  i,j\right)  \in\operatorname*{Inv}\left(  \eta^{-1}\right)
$. By the definition of $\operatorname*{Inv}\left(  \eta^{-1}\right)  $, this
yields that $\left(  i,j\right)  $ is an element of $\left\{  1,2,\ldots
,n\right\}  ^{2}$ satisfying $i<j$ and $\eta^{-1}\left(  i\right)  >\eta
^{-1}\left(  j\right)  $. In particular, $\eta^{-1}\left(  i\right)
>\eta^{-1}\left(  j\right)  $. In other words, the letter $i$ must appear
after the letter $j$ in the word $\eta$ (where our use of the word
``after'' does not imply ``immediately after'').
Since $\eta=\overrightarrow{I_{\tau
\left(  1\right)  }}\overrightarrow{I_{\tau\left(  2\right)  }}\cdots
\overrightarrow{I_{\tau\left(  \ell\right)  }}$, this rewrites as follows: The
letter $i$ must appear after the letter $j$ in the word
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  \ell\right)  }}$. Thus, we
must be in one of the following two cases:
\par
\textit{Case 1:} There exist some elements $\mathbf{i}$ and $\mathbf{j}$ of
$\left\{  1,2,\ldots,\ell\right\}  $ such that $\mathbf{i}>\mathbf{j}$ and
such that the letter $i$ appears in the word $\overrightarrow{I_{\tau\left(
\mathbf{i}\right)  }}$, whereas the letter $j$ appears in the word
$\overrightarrow{I_{\tau\left(  \mathbf{j}\right)  }}$.
\par
\textit{Case 2:} There exists some element $\mathbf{i}$ of $\left\{
1,2,\ldots,\ell\right\}  $ such that both letters $i$ and $j$ appear in the
word $\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$, and the letter
$i$ appears after the letter $j$ in this word.
\par
Let us first consider Case 1. In this case, there exist some elements
$\mathbf{i}$ and $\mathbf{j}$ of $\left\{  1,2,\ldots,\ell\right\}  $ such
that $\mathbf{i}>\mathbf{j}$ and such that the letter $i$ appears in the word
$\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$, whereas the letter $j$
appears in the word $\overrightarrow{I_{\tau\left(  \mathbf{j}\right)  }}$.
Consider these $\mathbf{i}$ and $\mathbf{j}$. The letter $i$ appears in the
word $\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$, and thus is an
element of $I_{\tau\left(  \mathbf{i}\right)  }$ (since the letters of the
word $\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$ are precisely the
elements of $I_{\tau\left(  \mathbf{i}\right)  }$). In other words, $i\in
I_{\tau\left(  \mathbf{i}\right)  }$. Hence, $\omega^{-1}\left(  i\right)
\in\omega^{-1}\left(  I_{\tau\left(  \mathbf{i}\right)  }\right)  $.
Similarly, $\omega^{-1}\left(  j\right)  \in\omega^{-1}\left(  I_{\tau\left(
\mathbf{j}\right)  }\right)  $. Now, $\mathbf{i}>\mathbf{j}$, so that
$\mathbf{j}<\mathbf{i}$ and thus $\omega^{-1}\left(  I_{\tau\left(
\mathbf{j}\right)  }\right)  <\omega^{-1}\left(  I_{\tau\left(  \mathbf{i}%
\right)  }\right)  $ (due to (\ref{pf.prop.iper.biject.a.1tau})). In other
words, every $j^{\prime}\in\omega^{-1}\left(  I_{\tau\left(  \mathbf{j}%
\right)  }\right)  $ and $i^{\prime}\in\omega^{-1}\left(  I_{\tau\left(
\mathbf{i}\right)  }\right)  $ satisfy $j^{\prime}<i^{\prime}$ (by the
definition of
$\omega^{-1}\left(  I_{\tau\left(
\mathbf{j}\right)  }\right)  <\omega^{-1}\left(  I_{\tau\left(  \mathbf{i}%
\right)  }\right)  $). Applying this to $j^{\prime}=\omega^{-1}\left(
j\right)  $ and $i^{\prime}=\omega^{-1}\left(  i\right)  $, we obtain
$\omega^{-1}\left(  j\right)  <\omega^{-1}\left(  i\right)  $, so that
$\omega^{-1}\left(  i\right)  >\omega^{-1}\left(  j\right)  $.
\par
Now, we know that $\left(  i,j\right)  $ is an element of $\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$ and $\omega^{-1}\left(  i\right)
>\omega^{-1}\left(  j\right)  $. In other words, $\left(  i,j\right)
\in\operatorname*{Inv}\left(  \omega^{-1}\right)  $ (by the definition of
$\operatorname*{Inv}\left(  \omega^{-1}\right)  $). Hence, $\left(
i,j\right)  \in\operatorname*{Inv}\left(  \omega^{-1}\right)  $ is proven in
Case 1.
\par
Let us now consider Case 2. In this case, there exists some element
$\mathbf{i}$ of $\left\{  1,2,\ldots,\ell\right\}  $ such that both letters
$i$ and $j$ appear in the word $\overrightarrow{I_{\tau\left(  \mathbf{i}%
\right)  }}$, and the letter $i$ appears after the letter $j$ in this word.
Consider this $i$. The letters of the word $\overrightarrow{I_{\tau\left(
\mathbf{i}\right)  }}$ are in increasing order (since $\overrightarrow{I_{\tau
\left(  \mathbf{i}\right)  }}$ is defined as the list of all elements of
$I_{\tau\left(  \mathbf{i}\right)  }$ in increasing order). Since the letter
$i$ appears after the letter $j$ in this word, we must therefore have $i>j$.
But this contradicts $i<j$. This contradiction shows that Case 2 cannot occur.
Hence, the only possible case is Case 1. Since we have proven $\left(
i,j\right)  \in\operatorname*{Inv}\left(  \omega^{-1}\right)  $ in this case,
we therefore conclude that $\left(  i,j\right)  \in\operatorname*{Inv}\left(
\omega^{-1}\right)  $ always holds.
\par
Now, let us forget that we fixed $\left(  i,j\right)  $. We thus have proven
that $\left(  i,j\right)  \in\operatorname*{Inv}\left(  \omega^{-1}\right)  $
for every $\left(  i,j\right)  \in\operatorname*{Inv}\left(  \eta^{-1}\right)
$. In other words,%
\begin{equation}
\operatorname*{Inv}\left(  \eta^{-1}\right)  \subset\operatorname*{Inv}\left(
\omega^{-1}\right)  . \label{pf.prop.iper.biject.a.5}%
\end{equation}
\par
Next, let us prove that $\operatorname*{Inv}\left(  \omega^{-1}\right)
\subset\operatorname*{Inv}\left(  \eta^{-1}\right)  $.
\par
Indeed, let $\left(  i,j\right)  $ be some element of $\operatorname*{Inv}%
\left(  \omega^{-1}\right)  $. We want to prove that $\left(  i,j\right)
\in\operatorname*{Inv}\left(  \eta^{-1}\right)  $.
\par
We have $\left(  i,j\right)  \in\operatorname*{Inv}\left(  \omega^{-1}\right)
$. By the definition of $\operatorname*{Inv}\left(  \omega^{-1}\right)  $,
this yields that $\left(  i,j\right)  $ is an element of $\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfying $i<j$ and $\omega^{-1}\left(  i\right)
>\omega^{-1}\left(  j\right)  $.
\par
The intervals $I_{1}$, $I_{2}$, $\ldots$, $I_{\ell}$ form a set partition of
$\left[  0:n\right]  ^{+}$ (according to Remark \ref{rmk.intsys.trivia}(c)).
Hence, there exists some $u\in\left\{  1,2,\ldots,\ell\right\}  $ such that
$i\in I_{u}$ (because $i\in\left\{  1,2,\ldots,n\right\} =
\left[  0:n\right]  ^{+}$). Similarly, there
exists some $v\in\left\{  1,2,\ldots,\ell\right\}  $ such that $j\in I_{v}$.
Consider these $u$ and $v$.
\par
First, let us assume (for the sake of contradiction) that $u=v$. Then, $i\in
I_{u}=I_{v}$ (since $u=v$) and $j\in I_{v}$. But
(\ref{pf.prop.iper.biject.a.3}) (applied to $v$ instead of $i$) yields that
the restriction of the map $\omega^{-1}$ to the interval $I_{v}$ is
increasing. Hence, $\omega^{-1}\left(  i\right)  \leq\omega^{-1}\left(
j\right)  $ (since the elements $i$ and $j$ both lie in the interval $I_{v}$
and satisfy $i<j$), which contradicts $\omega^{-1}\left(  i\right)
>\omega^{-1}\left(  j\right)  $. This contradiction shows that our assumption
(that $u=v$) was wrong. Hence, we have $u\neq v$. Thus, $\tau^{-1}\left(
u\right)  \neq\tau^{-1}\left(  v\right)  $ (since $\tau$ is a permutation).
\par
Define two elements $\mathbf{i}$ and $\mathbf{j}$ of $\left\{  1,2,\ldots
,\ell\right\}  $ by $\mathbf{i}=\tau^{-1}\left(  u\right)  $ and
$\mathbf{j}=\tau^{-1}\left(  v\right)  $. Then, $\tau\left(  \mathbf{i}%
\right)  =u$ (since $\mathbf{i}=\tau^{-1}\left(  u\right)  $), so that
$I_{\tau\left(  \mathbf{i}\right)  }=I_{u}$ and thus $i\in I_{u}%
=I_{\tau\left(  \mathbf{i}\right)  }$ (since $I_{\tau\left(  \mathbf{i}%
\right)  }=I_{u}$). Similarly, $j\in I_{\tau\left(  \mathbf{j}\right)  }$. We
have $\mathbf{i}=\tau^{-1}\left(  u\right)  \neq\tau^{-1}\left(  v\right)
=\mathbf{j}$.
\par
Now, let us assume (for the sake of contradiction) that $\mathbf{i}%
\leq\mathbf{j}$. Combined with $\mathbf{i}\neq\mathbf{j}$, this yields
$\mathbf{i}<\mathbf{j}$. Hence, $\omega^{-1}\left(  I_{\tau\left(
\mathbf{i}\right)  }\right)  <\omega^{-1}\left(  I_{\tau\left(  \mathbf{j}%
\right)  }\right)  $ (due to (\ref{pf.prop.iper.biject.a.1tau})). Thus, every
$i^{\prime}\in\omega^{-1}\left(  I_{\tau\left(  \mathbf{i}\right)  }\right)  $
and $j^{\prime}\in\omega^{-1}\left(  I_{\tau\left(  \mathbf{j}\right)
}\right)  $ satisfy $i^{\prime}<j^{\prime}$ (by the definition of $\omega
^{-1}\left(  I_{\tau\left(  \mathbf{i}\right)  }\right)  <\omega^{-1}\left(
I_{\tau\left(  \mathbf{j}\right)  }\right)  $). Applying this to $i^{\prime
}=\omega^{-1}\left(  i\right)  $ and $j^{\prime}=\omega^{-1}\left(  j\right)
$, we obtain $\omega^{-1}\left(  i\right)  <\omega^{-1}\left(  j\right)  $
(since $\omega^{-1}\left(  \underbrace{i}_{\in I_{\tau\left(  \mathbf{i}%
\right)  }}\right)  \in\omega^{-1}\left(  I_{\tau\left(  \mathbf{i}\right)
}\right)  $ and $\omega^{-1}\left(  \underbrace{j}_{\in I_{\tau\left(
\mathbf{j}\right)  }}\right)  \in\omega^{-1}\left(  I_{\tau\left(
\mathbf{j}\right)  }\right)  $), which contradicts $\omega^{-1}\left(
i\right)  >\omega^{-1}\left(  j\right)  $. This contradiction proves that our
assumption (that $\mathbf{i}\leq\mathbf{j}$) was wrong. Hence, we have
$\mathbf{i}>\mathbf{j}$.
\par
Now, recall that the word $\overrightarrow{I_{\tau\left(  \mathbf{i}\right)
}}$ is defined as the list of all elements of $I_{\tau\left(  \mathbf{i}%
\right)  }$ in increasing order. Hence, $i$ is a letter of the word
$\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$ (since $i$ is an
element of $I_{\tau\left(  \mathbf{i}\right)  }$). Similarly, $j$ is a letter
of the word $\overrightarrow{I_{\tau\left(  \mathbf{j}\right)  }}$. Both words
$\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$ and
$\overrightarrow{I_{\tau\left(  \mathbf{j}\right)  }}$ are factors of the
concatenation $\overrightarrow{I_{\tau\left(  1\right)  }}%
\overrightarrow{I_{\tau\left(  2\right)  }}\cdots\overrightarrow{I_{\tau
\left(  \ell\right)  }}$, with the factor $\overrightarrow{I_{\tau\left(
\mathbf{i}\right)  }}$ appearing after the factor $\overrightarrow{I_{\tau
\left(  \mathbf{j}\right)  }}$ (since $\mathbf{i}>\mathbf{j}$). Thus, if
$i_{0}$ is any letter of the word $\overrightarrow{I_{\tau\left(
\mathbf{i}\right)  }}$, and if $j_{0}$ is any letter of the word
$\overrightarrow{I_{\tau\left(  \mathbf{j}\right)  }}$, then the letter
$i_{0}$ appears after the letter $j_{0}$ in the concatenation
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  \ell\right)  }}$. Applying
this to $i_{0}=i$ and $j_{0}=j$, we conclude that the letter $i$ appears after
the letter $j$ in the concatenation $\overrightarrow{I_{\tau\left(  1\right)
}}\overrightarrow{I_{\tau\left(  2\right)  }}\cdots\overrightarrow{I_{\tau
\left(  \ell\right)  }}$ (since $i$ is a letter of the word
$\overrightarrow{I_{\tau\left(  \mathbf{i}\right)  }}$, and since $j$ is a
letter of the word $\overrightarrow{I_{\tau\left(  \mathbf{j}\right)  }}$). In
other words, the letter $i$ appears after the letter $j$ in the word $\eta$
(since $\eta=\overrightarrow{I_{\tau\left(  1\right)  }}%
\overrightarrow{I_{\tau\left(  2\right)  }}\cdots\overrightarrow{I_{\tau
\left(  \ell\right)  }}$). In other words, $\eta^{-1}\left(  i\right)
>\eta^{-1}\left(  j\right)  $.
\par
So $\left(  i,j\right)  $ is an element of $\left\{  1,2,\ldots,n\right\}
^{2}$ satisfying $i<j$ and $\eta^{-1}\left(  i\right)  >\eta^{-1}\left(
j\right)  $. In other words, $\left(  i,j\right)  \in\operatorname*{Inv}%
\left(  \eta^{-1}\right)  $ (by the definition of $\operatorname*{Inv}\left(
\eta^{-1}\right)  $).
\par
Now, let us forget that we fixed $\left(  i,j\right)  $. We thus have proven
that $\left(  i,j\right)  \in\operatorname*{Inv}\left(  \eta^{-1}\right)  $
for every $\left(  i,j\right)  \in\operatorname*{Inv}\left(  \omega
^{-1}\right)  $. In other words,%
\[
\operatorname*{Inv}\left(  \omega^{-1}\right)  \subset\operatorname*{Inv}%
\left(  \eta^{-1}\right)  .
\]
Combined with (\ref{pf.prop.iper.biject.a.5}), this yields
$\operatorname*{Inv}\left(  \omega^{-1}\right)  =\operatorname*{Inv}\left(
\eta^{-1}\right)  $. Thus, Proposition~\ref{prop.sol.Lalpha-std.proofs.Inv=}
(applied to
$\varphi=\omega^{-1}$ and $\psi=\eta^{-1}$) yields $\omega^{-1}%
=\eta^{-1}$, whence $\omega=\eta=\left(  \operatorname{iper}_{\alpha
}\circ\operatorname{iper}_{\alpha}^{\prime}\right)  \left(
\omega\right)  $ (by (\ref{pf.prop.iper.biject.a.iiwn})), so that $\left(
\operatorname{iper}_{\alpha}\circ
\operatorname{iper}_{\alpha}^{\prime}\right)  \left(  \omega\right)
=\omega
=\operatorname*{id}\left(  \omega\right)  $.
\par
Now, let us forget that we fixed $\omega$. We thus have proven that $\left(
\operatorname{iper}_{\alpha}\circ\operatorname{iper}_{\alpha}^{\prime}
\right)  \left(  \omega\right)  =\operatorname*{id}%
\left(  \omega\right)  $ for every $\alpha$-clumping permutation $\omega
\in\Symm_n$. In other words, $\operatorname{iper}_{\alpha
}\circ\operatorname{iper}_{\alpha}^{\prime}=\operatorname*{id}$,
qed.} and that $\operatorname{iper}_{\alpha}^{\prime}\circ
\operatorname{iper}_{\alpha}=\operatorname*{id}$%
\ \ \ \ \footnote{\textit{Proof.} Let $\pi\in\Symm_{\ell}$. We shall
show that that $\left(  \operatorname{iper}_{\alpha}^{\prime}%
\circ\operatorname{iper}_{\alpha}\right)  \left(  \pi\right)
=\operatorname*{id}\left(  \pi\right)  $.
\par
Indeed, let $\omega=\operatorname{iper}_{\alpha}\pi$. Then,
$\omega=\operatorname{iper}_{\alpha}\pi=\operatorname*{iper}\left(
\alpha,\pi\right)  $ (by the definition of
$\operatorname{iper}_{\alpha}\pi$). The permutation
$\omega=\operatorname*{iper}\left(
\alpha,\pi\right)  $ is $\alpha$-clumping (according to Proposition
\ref{prop.iper.intinverse}(c), applied to $\tau=\pi$).
\par
Let us now recall how $\operatorname{iper}_{\alpha}^{\prime}\left(
\omega\right)  $ was defined: Every $i\in\left\{  1,2,\ldots,\ell\right\}  $
has the property that $\omega^{-1}\left(  I_{i}\right)  $ is an interval.
These intervals $\omega^{-1}\left(  I_{1}\right)  $, $\omega^{-1}\left(
I_{2}\right)  $, $\ldots$, $\omega^{-1}\left(  I_{\ell}\right)  $ form a
totally ordered set with respect to the relation $<$. Then,
$\operatorname{iper}_{\alpha}^{\prime}\left(  \omega\right)  $ is
defined as the unique permutation $\tau\in\Symm_{\ell}$ such that
$\omega^{-1}\left(  I_{\tau\left(  1\right)  }\right)  <\omega^{-1}\left(
I_{\tau\left(  2\right)  }\right)  < \cdots<\omega^{-1}\left(  I_{\tau\left(
\ell\right)  }\right)  $. Hence, if $\tau\in\Symm_{\ell}$ is a
permutation satisfying $\omega^{-1}\left(  I_{\tau\left(  1\right)  }\right)
<\omega^{-1}\left(  I_{\tau\left(  2\right)  }\right)  < \cdots<\omega
^{-1}\left(  I_{\tau\left(  \ell\right)  }\right)  $, then%
\begin{equation}
\tau=\operatorname{iper}_{\alpha}^{\prime}\left(  \omega\right)  .
\label{pf.prop.iper.biject.a.101}%
\end{equation}
\par
We shall now use this to prove that $\pi=\operatorname{iper}_{\alpha
}^{\prime}\left(  \omega\right)  $.
\par
We have $\omega=\operatorname*{iper}\left(  \alpha,\pi\right)
=\overrightarrow{I_{\pi\left(  1\right)  }}\overrightarrow{I_{\pi\left(
2\right)  }}\cdots\overrightarrow{I_{\pi\left(  \ell\right)  }}$ (in one-line
notation), according to the definition of $\operatorname*{iper}\left(
\alpha,\pi\right)  $.
\par
The intervals $\omega^{-1}\left(  I_{1}\right)  $, $\omega^{-1}\left(
I_{2}\right)  $, $\ldots$, $\omega^{-1}\left(  I_{\ell}\right)  $ form a set
partition of $\left[  0:n\right]  ^{+}$ (since the intervals $I_{1}$, $I_{2}$,
$\ldots$, $I_{\ell}$ form a set partition of $\left[  0:n\right]  ^{+}$), and
thus are disjoint. These intervals $\omega^{-1}\left(  I_{1}\right)
$, $\omega^{-1}\left(  I_{2}\right)  $, $\ldots$, $\omega^{-1}\left(  I_{\ell
}\right)  $ are also nonempty (since the intervals $I_{1}$, $I_{2}$, $\ldots$,
$I_{\ell}$ are nonempty).
\par
Now, let $u\in\left\{  1,2,\ldots,\ell-1\right\}  $. The intervals
$\omega^{-1}\left(  I_{\pi\left(  u\right)  }\right)  $ and $\omega
^{-1}\left(  I_{\pi\left(  u+1\right)  }\right)  $ are nonempty (since the
intervals $\omega^{-1}\left(  I_{1}\right)  $, $\omega^{-1}\left(
I_{2}\right)  $, $\ldots$, $\omega^{-1}\left(  I_{\ell}\right)  $ are
nonempty). We will show that $\omega^{-1}\left(  I_{\pi\left(  u\right)
}\right)  <\omega^{-1}\left(  I_{\pi\left(  u+1\right)  }\right)  $.
\par
Indeed, let $i\in\omega^{-1}\left(  I_{\pi\left(  u\right)  }\right)  $ and
$j\in\omega^{-1}\left(  I_{\pi\left(  u+1\right)  }\right)  $ be arbitrary. We
shall prove that $i<j$.
\par
We have $i\in\omega^{-1}\left(  I_{\pi\left(  u\right)  }\right)  $, so that
$\omega\left(  i\right)  \in I_{\pi\left(  u\right)  }$. Hence, $\omega\left(
i\right)  $ is a letter of the word $\overrightarrow{I_{\pi\left(  u\right)
}}$ (since the word $\overrightarrow{I_{\pi\left(  u\right)  }}$ is defined as
the list of all elements of $I_{\pi\left(  u\right)  }$ in increasing order).
Similarly, $\omega\left(  j\right)  $ is a letter of the word
$\overrightarrow{I_{\pi\left(  u+1\right)  }}$ (since $j\in\omega^{-1}\left(
I_{\pi\left(  u+1\right)  }\right)  $).
\par
Both words $\overrightarrow{I_{\pi\left(  u\right)  }}$ and
$\overrightarrow{I_{\pi\left(  u+1\right)  }}$ are factors of the
concatenation $\overrightarrow{I_{\pi\left(  1\right)  }}%
\overrightarrow{I_{\pi\left(  2\right)  }}\cdots\overrightarrow{I_{\pi\left(
\ell\right)  }}$, with the factor $\overrightarrow{I_{\pi\left(  u+1\right)
}}$ appearing after the factor $\overrightarrow{I_{\pi\left(  u\right)  }}$.
Thus, if $i_{0}$ is any letter of the word $\overrightarrow{I_{\pi\left(
u\right)  }}$, and if $j_{0}$ is any letter of the word
$\overrightarrow{I_{\pi\left(  u+1\right)  }}$, then the letter $j_{0}$
appears after the letter $i_{0}$ in the concatenation $\overrightarrow{I_{\pi
\left(  1\right)  }}\overrightarrow{I_{\pi\left(  2\right)  }}\cdots
\overrightarrow{I_{\pi\left(  \ell\right)  }}$. Applying this to $i_{0}%
=\omega\left(  i\right)  $ and $j_{0}=\omega\left(  j\right)  $, we conclude
that the letter $\omega\left(  j\right)  $ appears after the letter
$\omega\left(  i\right)  $ in the concatenation $\overrightarrow{I_{\pi\left(
1\right)  }}\overrightarrow{I_{\pi\left(  2\right)  }}\cdots
\overrightarrow{I_{\pi\left(  \ell\right)  }}$ (since $\omega\left(  i\right)
$ is a letter of the word $\overrightarrow{I_{\pi\left(  u\right)  }}$, and
since $\omega\left(  j\right)  $ is a letter of the word
$\overrightarrow{I_{\pi\left(  u+1\right)  }}$). In other words, the letter
$\omega\left(  j\right)  $ appears after the letter $\omega\left(  i\right)  $
in the word $\omega$ (since $\omega=\overrightarrow{I_{\pi\left(  1\right)  }%
}\overrightarrow{I_{\pi\left(  2\right)  }}\cdots\overrightarrow{I_{\pi\left(
\ell\right)  }}$). In other words, $\omega^{-1}\left(  \omega\left(  j\right)
\right)  >\omega^{-1}\left(  \omega\left(  i\right)  \right)  $. In other
words, $j>i$, so that $i<j$.
\par
Now, let us forget that we fixed $i$ and $j$. We thus have proven that every
$i\in\omega^{-1}\left(  I_{\pi\left(  u\right)  }\right)  $ and $j\in
\omega^{-1}\left(  I_{\pi\left(  u+1\right)  }\right)  $ satisfy $i<j$. In
other words, $\omega^{-1}\left(  I_{\pi\left(  u\right)  }\right)
<\omega^{-1}\left(  I_{\pi\left(  u+1\right)  }\right)  $ (by the definition
of $\omega^{-1}\left(  I_{\pi\left(  u\right)  }\right)  <\omega^{-1}\left(
I_{\pi\left(  u+1\right)  }\right)  $).
\par
Now, let us forget that we fixed $u$. We thus have proven that $\omega
^{-1}\left(  I_{\pi\left(  u\right)  }\right)  <\omega^{-1}\left(
I_{\pi\left(  u+1\right)  }\right)  $ for every $u\in\left\{  1,2,\ldots
,\ell-1\right\}  $. In other words, $\omega^{-1}\left(  I_{\pi\left(
1\right)  }\right)  <\omega^{-1}\left(  I_{\pi\left(  2\right)  }\right)  <
\cdots<\omega^{-1}\left(  I_{\pi\left(  \ell\right)  }\right)  $. Therefore,
(\ref{pf.prop.iper.biject.a.101}) (applied to $\tau=\pi$) yields
$\pi=\operatorname{iper}_{\alpha}^{\prime}\left(  \underbrace{\omega
}_{=\operatorname{iper}_{\alpha}\pi}\right)
=\operatorname{iper}_{\alpha}^{\prime}\left( 
 \operatorname{iper}_{\alpha}%
\pi\right)  =\left(  \operatorname{iper}_{\alpha}^{\prime}%
\circ\operatorname{iper}_{\alpha}\right)  \left(  \pi\right)  $.
Hence, $\left(  \operatorname{iper}_{\alpha}^{\prime}\circ
\operatorname{iper}_{\alpha}\right)  \left(  \pi\right)
=\pi=\operatorname*{id}\left(  \pi\right)  $.
\par
Now, let us forget that we fixed $\pi$. We thus have proven that $\left(
\operatorname{iper}_{\alpha}^{\prime}\circ\operatorname*{iper}%
\nolimits_{\alpha}\right)  \left(  \pi\right)  =\operatorname*{id}\left(
\pi\right)  $ for every $\pi\in\Symm_{\ell}$. In other words,
$\operatorname{iper}_{\alpha}^{\prime}\circ\operatorname*{iper}%
\nolimits_{\alpha}=\operatorname*{id}$, qed.}. Hence, the maps
$\operatorname{iper}_{\alpha}$ and $\operatorname*{iper}%
\nolimits_{\alpha}^{\prime}$ are mutually inverse, and thus the map
$\operatorname{iper}_{\alpha}$ is bijective. This completes the
proof of Proposition \ref{prop.iper.biject}(a).

(b) Consider the map $\operatorname{iper}_{\alpha}$ defined in Proposition
\ref{prop.iper.biject}(a). Since $\sigma$ is $\alpha$-clumping, we have
\newline
$\sigma\in\left\{  \omega\in\Symm_n\mid\omega\text{ is }%
\alpha\text{-clumping}\right\}  $. In other words, $\sigma$ belongs to the
target of the map $\operatorname{iper}_{\alpha}$. Thus, $\sigma$ has a unique
preimage under the map $\operatorname{iper}_{\alpha}$ (since the map
$\operatorname{iper}_{\alpha}$ is bijective (by Proposition
\ref{prop.iper.biject}(a))). In other words, there exists a unique $\tau
\in\Symm_{\ell}$ satisfying $\sigma=\operatorname*{iper}%
\nolimits_{\alpha}\tau$. Since $\operatorname{iper}_{\alpha}%
\tau=\operatorname*{iper}\left(  \alpha,\tau\right)  $ for every $\tau
\in\Symm_{p+q}$ (by the definition of $\operatorname*{iper}%
\nolimits_{\alpha}$), this rewrites as follows: There exists a unique $\tau
\in\Symm_{\ell}$ satisfying $\sigma=\operatorname{iper}\left(
\alpha,\tau\right)  $. This proves Proposition \ref{prop.iper.biject}(b).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.iper.shuffle}}

\textit{Solution to Exercise \ref{exe.prop.iper.shuffle}.}

\begin{proof}
[Proof of Proposition \ref{prop.iper.shuffle}.]We know that $\alpha\beta$ is a
composition of $n+m$ having length $\ell\left(  \alpha\beta\right)
=\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  =p+q$. Hence, the
interval system corresponding to $\alpha\beta$ is a $\left(  p+q\right)
$-tuple of intervals which covers $\left[  0:n+m\right]  ^{+}$. Denote this
$\left(  p+q\right)  $-tuple by $\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  $.
It is clear that $I_{1}\cup I_{2}\cup\cdots\cup I_{p}=\left[  0:n\right]
^{+}$ and $I_{p+1}\cup I_{p+2}\cup\cdots\cup I_{p+q}=\left[  n:n+m\right]
^{+}$ (since the first $p$ parts of the composition $\alpha\beta$ form the
composition $\alpha$ of $n$). Moreover, $I_{1}<I_{2}< \cdots<I_{p+q}$ (since
$\left(  I_{1},I_{2},\ldots,I_{p+q}\right)  $ is the interval system
corresponding to $\alpha\beta$).

The definition of $\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $
yields that $\operatorname*{iper}\left(  \alpha\beta,\tau\right)
=\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$ (in one-line
notation). Denote the permutation $\operatorname*{iper}\left(  \alpha
\beta,\tau\right)  $ by $\omega$; then, this becomes $\omega
=\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$. For every
$j\in\left\{  1,2,\ldots,p+q\right\}  $, the restriction of the map
$\omega^{-1}$ to the interval $I_{\tau\left(  j\right)  }$ is increasing (by
Proposition \ref{prop.iper.intinverse}(b), applied to $n+m$, $\alpha\beta$,
$p+q$ and $\omega$ instead of $n$, $\alpha$, $\ell$ and $\sigma$).
Substituting $k$ for $\tau\left(  j\right)  $ here, we obtain: For every
$k\in\left\{  1,2,\ldots,p+q\right\}  $, the restriction of the map
$\omega^{-1}$ to the interval $I_{k}$ is increasing. Of course, this yields
that for every $k\in\left\{  1,2,\ldots,p+q\right\}  $, the restriction of the
map $\omega^{-1}$ to the interval $I_{k}$ is strictly increasing (because
$\omega^{-1}$ is injective).

Now, we need to prove that $\tau\in\operatorname{Sh}_{p,q}$ if and
only if $\operatorname*{iper}\left(  \alpha\beta,\tau\right)  \in
\operatorname{Sh}_{n,m}$. In other words, we need to prove that
$\tau\in\operatorname{Sh}_{p,q}$ if and only if $\omega
\in\operatorname{Sh}_{n,m}$ (since $\omega=\operatorname*{iper}%
\left(  \alpha\beta,\tau\right)  $). We shall prove the $\Longrightarrow$ and
$\Longleftarrow$ directions of this statement separately:

$\Longrightarrow:$ Assume that $\tau\in\operatorname{Sh}_{p,q}$. We
need to show that $\omega\in\operatorname{Sh}_{n,m}$.

We have $\tau\in\operatorname{Sh}_{p,q}$. Thus, $\tau^{-1}\left(
1\right)  <\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(  p\right)  $ and
$\tau^{-1}\left(  p+1\right)  <\tau^{-1}\left(  p+2\right)  <\cdots<\tau
^{-1}\left(  p+q\right)  $.

Now, let $i$ and $j$ be two elements of $\left\{  1,2,\ldots,n\right\}  $ such
that $i<j$. We are going to prove that $\omega^{-1}\left(  i\right)
<\omega^{-1}\left(  j\right)  $.

Indeed, we have $i\in\left\{  1,2,\ldots,n\right\}  =\left[  0:n\right]
^{+}=I_{1}\cup I_{2}\cup\cdots\cup I_{p}$, so that $i\in I_{i^{\prime}}$ for
some $i^{\prime}\in\left\{  1,2,\ldots,p\right\}  $. Similarly, $j\in
I_{j^{\prime}}$ for some $j^{\prime}\in\left\{  1,2,\ldots,p\right\}  $.
Consider these $i^{\prime}$ and $j^{\prime}$. We cannot have $j^{\prime
}<i^{\prime}$ (because this would entail $I_{j^{\prime}}<I_{i^{\prime}}$ (due
to $I_{1}<I_{2}< \cdots<I_{p+q}$), which would lead to $j<i$ (since $j\in
I_{j^{\prime}}$ and $i\in I_{i^{\prime}}$), contradicting $i<j$). Hence, we
must have either $j^{\prime}=i^{\prime}$ or $j^{\prime}>i^{\prime}$. If
$j^{\prime}=i^{\prime}$, then $i$ and $j$ lie in one and the same $I_{k}$
(namely, the one with $k=j^{\prime}=i^{\prime}$), and so $\omega^{-1}\left(
i\right)  <\omega^{-1}\left(  j\right)  $ follows from the fact that the map
$\omega^{-1}$ restricted to $I_{k}$ is strictly increasing (and from the
inequality $i<j$). Hence, it only remains to consider the case when
$j^{\prime}>i^{\prime}$. Assume WLOG that we are in this case. Then,
$i^{\prime}<j^{\prime}$, and so $\tau^{-1}\left(  i^{\prime}\right)
<\tau^{-1}\left(  j^{\prime}\right)  $ (since $\tau^{-1}\left(  1\right)
<\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(  p\right)  $ and since
$i^{\prime}$ and $j^{\prime}$ belong to $\left\{  1,2,\ldots,p\right\}  $),
and thus the word $\overrightarrow{I_{i^{\prime}}}$ appears before the word
$\overrightarrow{I_{j^{\prime}}}$ in the concatenation
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$. Hence, the
letter $i$ appears before the letter $j$ in the concatenation
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$ (because the
letter $i$ appears in $\overrightarrow{I_{i^{\prime}}}$, while the letter $j$
appears in $\overrightarrow{I_{j^{\prime}}}$). Since $\overrightarrow{I_{\tau
\left(  1\right)  }}\overrightarrow{I_{\tau\left(  2\right)  }}\cdots
\overrightarrow{I_{\tau\left(  p+q\right)  }}=\omega$, this shows that the
letter $i$ appears before the letter $j$ in the word $\omega$. In other words,
$\omega^{-1}\left(  i\right)  <\omega^{-1}\left(  j\right)  $.

So we have shown that if two elements $i$ and $j$ of $\left\{  1,2,\ldots
,n\right\}  $ satisfy $i<j$, then $\omega^{-1}\left(  i\right)  <\omega
^{-1}\left(  j\right)  $. In other words, $\omega^{-1}\left(  1\right)
<\omega^{-1}\left(  2\right)  <\cdots<\omega^{-1}\left(  n\right)  $.
Similarly, $\omega^{-1}\left(  n+1\right)  <\omega^{-1}\left(  n+2\right)
<\cdots<\omega^{-1}\left(  n+m\right)  $. Combining these two chains of
inequalities, we conclude that $\omega\in\operatorname{Sh}_{n,m}$.
This proves the $\Longrightarrow$ direction.

$\Longleftarrow:$ Assume that $\omega\in\operatorname{Sh}_{n,m}$. We
need to prove that $\tau\in\operatorname{Sh}_{p,q}$.

We have $\omega\in\operatorname{Sh}_{n,m}$. Thus, $\omega
^{-1}\left(  1\right)  <\omega^{-1}\left(  2\right)  <\cdots<\omega
^{-1}\left(  n\right)  $ and $\omega^{-1}\left(  n+1\right)  <\omega
^{-1}\left(  n+2\right)  <\cdots<\omega^{-1}\left(  n+m\right)  $.

Let $i^{\prime}$ and $j^{\prime}$ be two elements of $\left\{  1,2,\ldots
,p\right\}  $ such that $i^{\prime}<j^{\prime}$. We are going to prove that
$\tau^{-1}\left(  i^{\prime}\right)  <\tau^{-1}\left(  j^{\prime}\right)  $.

Indeed, assume the contrary. Then, $\tau^{-1}\left(  i^{\prime}\right)
\geq\tau^{-1}\left(  j^{\prime}\right)  $, thus $\tau^{-1}\left(  i^{\prime
}\right)  >\tau^{-1}\left(  j^{\prime}\right)  $ (since $\tau$ is a
permutation). Hence, the word $\overrightarrow{I_{i^{\prime}}}$ appears
after\footnote{``after'' does not imply ``immediately after''.} the word
$\overrightarrow{I_{j^{\prime}}}$ in the concatenation
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$.

Now, fix any $i\in I_{i^{\prime}}$ (such an $i$ exists since $I_{i^{\prime}}$
is nonempty) and fix any $j\in I_{j^{\prime}}$ (this exists for similar
reasons). We have $I_{i^{\prime}}<I_{j^{\prime}}$ (since $I_{1}<I_{2}<
\cdots<I_{p+q}$ and $i^{\prime}<j^{\prime}$), so that $i<j$ (since $i\in
I_{i^{\prime}}$ and $j\in I_{j^{\prime}}$).

But the letter $i$ appears after the letter $j$ in the concatenation
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$ (because the
letter $i$ appears in the word $\overrightarrow{I_{i^{\prime}}}$, whereas the
letter $j$ appears in the word $\overrightarrow{I_{j^{\prime}}}$, and we know
that the word $\overrightarrow{I_{i^{\prime}}}$ appears after the word
$\overrightarrow{I_{j^{\prime}}}$ in the concatenation
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}$). In other
words, the letter $i$ appears after the letter $j$ in the word $\omega$ (since
$\overrightarrow{I_{\tau\left(  1\right)  }}\overrightarrow{I_{\tau\left(
2\right)  }}\cdots\overrightarrow{I_{\tau\left(  p+q\right)  }}=\omega$). In
other words, $\omega^{-1}\left(  i\right)  >\omega^{-1}\left(  j\right)  $.

But $i^{\prime}\in\left\{  1,2,\ldots,p\right\}  $, so that $I_{i^{\prime}%
}\subset I_{1}\cup I_{2}\cup\cdots\cup I_{p}=\left[  0:n\right]  ^{+}$ and
thus $i\in I_{i^{\prime}}\subset\left[  0:n\right]  ^{+}$. Similarly,
$j\in\left[  0:n\right]  ^{+}$. Since the map $\omega^{-1}$ restricted to
$\left[  0:n\right]  ^{+}$ is strictly increasing (since $\omega^{-1}\left(
1\right)  <\omega^{-1}\left(  2\right)  <\cdots<\omega^{-1}\left(  n\right)
$), we thus have $\omega^{-1}\left(  i\right)  <\omega^{-1}\left(  j\right)  $
(since $i<j$), contradicting $\omega^{-1}\left(  i\right)  >\omega^{-1}\left(
j\right)  $. This contradiction shows that our assumption was wrong, and so we
have shown that $\tau^{-1}\left(  i^{\prime}\right)  <\tau^{-1}\left(
j^{\prime}\right)  $.

Thus, we have proven that if two elements $i^{\prime}$ and $j^{\prime}$ of
$\left\{  1,2,\ldots,p\right\}  $ satisfy $i^{\prime}<j^{\prime}$, then
$\tau^{-1}\left(  i^{\prime}\right)  <\tau^{-1}\left(  j^{\prime}\right)  $.
In other words, $\tau^{-1}\left(  1\right)  <\tau^{-1}\left(  2\right)
<\cdots<\tau^{-1}\left(  p\right)  $. Similarly, we can show that $\tau
^{-1}\left(  p+1\right)  <\tau^{-1}\left(  p+2\right)  <\cdots<\tau
^{-1}\left(  p+q\right)  $. The combination of these two chains of
inequalities shows that $\tau\in\operatorname{Sh}_{p,q}$. Thus, the
$\Longleftarrow$ direction is proven. The proof of Proposition
\ref{prop.iper.shuffle} is thus complete.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.shuffle.lyndon.step2}}

\textit{Solution to Exercise \ref{exe.lem.shuffle.lyndon.step2}.}

\begin{proof}
[Proof of Lemma \ref{lem.shuffle.lyndon.step2}.]We have $\tau\in
\operatorname{Sh}_{p,q}$ if and only if $\operatorname*{iper}\left(
\alpha\beta,\tau\right)  \in\operatorname{Sh}_{n,m}$ (by Proposition
\ref{prop.iper.shuffle}). Since we know that $\tau\in
\operatorname{Sh}_{p,q}$, we thus conclude that $\operatorname*{iper}\left(
\alpha\beta,\tau\right)  \in\operatorname{Sh}_{n,m}$. In other
words, $\sigma\in\operatorname{Sh}_{n,m}$ (since $\sigma
=\operatorname*{iper}\left(  \alpha\beta,\tau\right)  $).

Write the composition $\alpha\beta$ in the form $\left(  \gamma_{1},\gamma
_{2},\ldots,\gamma_{p+q}\right)  $. For every $j\in\left\{  0,1,\ldots
,p+q\right\}  $, let $s_{j}$ denote the integer $\sum_{k=1}^{j}\gamma
_{\tau\left(  k\right)  }$. Then, $0=s_{0}<s_{1}<s_{2}< \cdots<s_{p+q}=n+m$.

Let $w$ denote the word $u\underset{\sigma}{\shuffle}v$.

The first $p$ parts of the composition $\alpha\beta$ form the composition
$\alpha$ of $n$. Hence, the interval system $\left(  I_{1},I_{2}%
,\ldots,I_{p+q}\right)  $ corresponding to $\alpha\beta$ satisfies $I_{1}\cup
I_{2}\cup\cdots\cup I_{p}=\left[  0:n\right]  ^{+}$ and $I_{p+1}\cup
I_{p+2}\cup\cdots\cup I_{p+q}=\left[  n:n+m\right]  ^{+}$.

Let $i\in\left\{  1,2,\ldots,p+q\right\}  $. We know that $I_{\tau\left(
i\right)  }$ is an interval of $\ZZ$ satisfying either $I_{\tau\left(
i\right)  }\subset\left[  0:n\right]  ^{+}$ or $I_{\tau\left(  i\right)
}\subset\left[  n:n+m\right]  ^{+}$ (in fact, if $\tau\left(  i\right)  \leq
p$, then $I_{\tau\left(  i\right)  }\subset I_{1}\cup I_{2}\cup\cdots\cup
I_{p}=\left[  0:n\right]  ^{+}$, whereas otherwise, $I_{\tau\left(  i\right)
}\subset I_{p+1}\cup I_{p+2}\cup\cdots\cup I_{p+q}=\left[  n:n+m\right]  ^{+}%
$). Also, Proposition \ref{prop.iper.intinverse}(a) (applied to $n+m$,
$\alpha\beta$, $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{p+q}\right)  $,
$p+q$ and $i$ instead of $n$, $\alpha$, $\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right)  $, $\ell$ and $j$) yields%
\begin{equation}
\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  =\left[
\underbrace{\sum_{k=1}^{i-1}\gamma_{\tau\left(  k\right)  }}%
_{\substack{=s_{i-1}\\\text{(by the definition of }s_{i-1}\text{)}%
}}:\underbrace{\sum_{k=1}^{i}\gamma_{\tau\left(  k\right)  }}%
_{\substack{=s_{i}\\\text{(by the definition of }s_{i}\text{)}}}\right]
^{+}=\left[  s_{i-1}:s_{i}\right]  ^{+}. \label{pf.lem.shuffle.lyndon.step2.2}%
\end{equation}
Consequently, $\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)  $ is an
interval. Therefore, we can apply Lemma \ref{lem.shuffle.lyndon.step}(a) to
$I_{\tau\left(  i\right)  }$ instead of $I$. As a result, we obtain%
\[
\left(  u\underset{\sigma}{\shuffle}v\right)  \left[  \sigma^{-1}\left(
I_{\tau\left(  i\right)  }\right)  \right]  =\left(  uv\right)  \left[
I_{\tau\left(  i\right)  }\right]  .
\]
Hence,%
\begin{equation}
\left(  uv\right)  \left[  I_{\tau\left(  i\right)  }\right]
=\underbrace{\left(  u\underset{\sigma}{\shuffle}v\right)  }_{=w}\left[
\underbrace{\sigma^{-1}\left(  I_{\tau\left(  i\right)  }\right)
}_{\substack{=\left[  s_{i-1}:s_{i}\right]  ^{+}\\\text{(by
(\ref{pf.lem.shuffle.lyndon.step2.2}))}}}\right]  =w\left[  \left[
s_{i-1}:s_{i}\right]  ^{+}\right]  . \label{pf.lem.shuffle.lyndon.step2.4}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven
(\ref{pf.lem.shuffle.lyndon.step2.4}) for every $i\in\left\{  1,2,\ldots
,p+q\right\}  $. Now,%
\begin{align*}
&  \underbrace{\left(  uv\right)  \left[  I_{\tau\left(  1\right)  }\right]
}_{\substack{=w\left[  \left[  s_{0}:s_{1}\right]  ^{+}\right]  \\\text{(by
(\ref{pf.lem.shuffle.lyndon.step2.4}))}}}\cdot\underbrace{\left(  uv\right)
\left[  I_{\tau\left(  2\right)  }\right]  }_{\substack{=w\left[  \left[
s_{1}:s_{2}\right]  ^{+}\right]  \\\text{(by
(\ref{pf.lem.shuffle.lyndon.step2.4}))}}}\cdot\cdots\cdot\underbrace{\left(
uv\right)  \left[  I_{\tau\left(  p+q\right)  }\right]  }_{\substack{=w\left[
\left[  s_{p+q-1}:s_{p+q}\right]  ^{+}\right]  \\\text{(by
(\ref{pf.lem.shuffle.lyndon.step2.4}))}}}\\
&  =w\left[  \left[  s_{0}:s_{1}\right]  ^{+}\right]  \cdot w\left[  \left[
s_{1}:s_{2}\right]  ^{+}\right]  \cdot\cdots\cdot w\left[  \left[
s_{p+q-1}:s_{p+q}\right]  ^{+}\right] \\
&  =w\left[  \left[  s_{0}:s_{p+q}\right]  ^{+}\right]  =w\left[  \left[
0:n+m\right]  ^{+}\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
s_{0}=0\text{ and }s_{p+q}=n+m\right) \\
&  =w\ \ \ \ \ \ \ \ \ \ \left(  \text{since the word }w=u\underset{\sigma
}{\shuffle}v\text{ has length }n+m\right) \\
&  =u\underset{\sigma}{\shuffle}v.
\end{align*}
This proves Lemma \ref{lem.shuffle.lyndon.step2}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.shuffle.decrmax}}

\textit{Solution to Exercise \ref{exe.prop.shuffle.decrmax}.}

\begin{proof}
[Proof of Proposition \ref{prop.shuffle.decrmax}.]We have $h\left(  1\right)
\geq h\left(  2\right)  \geq\cdots\geq h\left(  p\right)  $. Hence, for every
$w\in\mathfrak{W}$, the set \newline$\left\{  i\in\left\{  1,2,\ldots
,p\right\}  \ \mid\ h\left(  i\right)  =w\right\}  $ is a (possibly empty)
interval of $\left\{  1,2,\ldots,p\right\}  $, and will be denoted by $P_{w}$.
Similarly, for every $w\in\mathfrak{W}$, the set $\left\{  i\in\left\{
p+1,p+2,\ldots,p+q\right\}  \ \mid\ h\left(  i\right)  =w\right\}  $ is a
(possibly empty) interval of $\left\{  p+1,p+2,\ldots,p+q\right\}  $, and will
be denoted by $Q_{w}$. Every $w\in\mathfrak{W}$ satisfies $h^{-1}\left(
w\right)  =P_{w}\sqcup Q_{w}$. Notice that $\left\vert P_{w}\right\vert
=\mathfrak{a}\left(  w\right)  $ and $\left\vert Q_{w}\right\vert
=\mathfrak{b}\left(  w\right)  $ for all $w\in\mathfrak{W}$.

Let $\left(  g_{1},g_{2},\ldots,g_{p+q}\right)  $ be the result of sorting the
list $\left(  h\left(  1\right)  ,h\left(  2\right)  ,\ldots,h\left(
p+q\right)  \right)  $ in decreasing order. Then, $g_{1}\geq g_{2}\geq
\cdots\geq g_{p+q}$. Hence, for every $w\in\mathfrak{W}$, the set $\left\{
j\in\left\{  1,2,\ldots,p+q\right\}  \ \mid\ g_{j}=w\right\}  $ is a (possibly
empty) interval of $\left\{  1,2,\ldots,p+q\right\}  $. Denote this interval
by $I_{w}$. The size of this interval is%
\begin{align*}
\left\vert I_{w}\right\vert  &  =\left\vert \left\{  j\in\left\{
1,2,\ldots,p+q\right\}  \ \mid\ g_{j}=w\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }I_{w}=\left\{  j\in\left\{
1,2,\ldots,p+q\right\}  \ \mid\ g_{j}=w\right\}  \right) \\
&  =\left\vert \underbrace{\left\{  j\in\left\{  1,2,\ldots,p+q\right\}
\ \mid\ h\left(  j\right)  =w\right\}  }_{=h^{-1}\left(  w\right)
=P_{w}\sqcup Q_{w}}\right\vert \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  g_{1},g_{2},\ldots,g_{p+q}\right)  \text{ is the result
of}\\
\text{sorting the list }\left(  h\left(  1\right)  ,h\left(  2\right)
,\ldots,h\left(  p+q\right)  \right)
\end{array}
\right) \\
&  =\left\vert P_{w}\sqcup Q_{w}\right\vert =\left\vert P_{w}\right\vert
+\left\vert Q_{w}\right\vert .
\end{align*}
Notice that $\left\{  1,2,\ldots,p+q\right\}  =\bigsqcup_{w\in\mathfrak{W}%
}I_{w}$ (by the definition of the $I_{w}$).

Fix $w\in\mathfrak{W}$. Let us define an
\emph{$\left(  I_{w},P_{w},Q_{w}\right)$-shuffle}
to mean a bijection $\kappa:I_{w}\rightarrow P_{w}\sqcup
Q_{w}$ having the property that the maps $\kappa^{-1}\mid_{P_{w}}%
:P_{w}\rightarrow I_{w}$ and $\kappa^{-1}\mid_{Q_{w}}:Q_{w}\rightarrow I_{w}$
are strictly increasing. It is easy to see that such a $\left(  I_{w}%
,P_{w},Q_{w}\right)  $-shuffle $\kappa$ is uniquely determined by the subset
$\kappa^{-1}\left(  P_{w}\right)  $ of $I_{w}$, and that for a given subset
$U$ of $I_{w}$, such a $\left(  I_{w},P_{w},Q_{w}\right)  $-shuffle $\kappa$
satisfying $\kappa^{-1}\left(  P_{w}\right)  =U$ exists if and only if
$\left\vert U\right\vert =\left\vert P_{w}\right\vert $. Hence, there are as
many $\left(  I_{w},P_{w},Q_{w}\right)  $-shuffles as there are subsets $U$ of
$I_{w}$ satisfying $\left\vert U\right\vert =\left\vert P_{w}\right\vert $. In
other words,%
\begin{align}
&  \left(  \text{the number of }\left(  I_{w},P_{w},Q_{w}\right)
\text{-shuffles}\right) \nonumber\\
&  =\left(  \text{the number of subsets }U\text{ of }I_{w}\text{ satisfying
}\left\vert U\right\vert =\left\vert P_{w}\right\vert \right) \nonumber\\
&  =\dbinom{\left\vert I_{w}\right\vert }{\left\vert P_{w}\right\vert
}=\dbinom{\mathfrak{a}\left(  w\right)  +\mathfrak{b}\left(  w\right)
}{\mathfrak{a}\left(  w\right)  } \label{pf.prop.shuffle.decrmax.numshuf}%
\end{align}
(since $\left\vert I_{w}\right\vert =\underbrace{\left\vert P_{w}\right\vert
}_{=\mathfrak{a}\left(  w\right)  }+\underbrace{\left\vert Q_{w}\right\vert
}_{=\mathfrak{b}\left(  w\right)  }=\mathfrak{a}\left(  w\right)
+\mathfrak{b}\left(  w\right)  $ and $\left\vert P_{w}\right\vert
=\mathfrak{a}\left(  w\right)  $).

Now, forget that we fixed $w$.

Consider any $\tau\in\operatorname{Sh}_{p,q}$ satisfying $h\left(
\tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)
\geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $. Then,
\newline$\left(  h\left(  \tau\left(  1\right)  \right)  ,h\left(  \tau\left(
2\right)  \right)  ,\ldots,h\left(  \tau\left(  p+q\right)  \right)  \right)
=\left(  g_{1},g_{2},\ldots,g_{p+q}\right)  $\ \ \ \ \footnote{\textit{Proof.}
The list $\left(  h\left(  \tau\left(  1\right)  \right)  ,h\left(
\tau\left(  2\right)  \right)  ,\ldots,h\left(  \tau\left(  p+q\right)
\right)  \right)  $ is a permutation of the list $\left(  h\left(  1\right)
,h\left(  2\right)  ,\ldots,h\left(  p+q\right)  \right)  $ (since $\tau
\in\Symm_{p+q}$), but is weakly decreasing (since $h\left(  \tau\left(
1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)  \geq\cdots\geq
h\left(  \tau\left(  p+q\right)  \right)  $). Hence, the list $\left(
h\left(  \tau\left(  1\right)  \right)  ,h\left(  \tau\left(  2\right)
\right)  ,\ldots,h\left(  \tau\left(  p+q\right)  \right)  \right)  $ is the
result of sorting the list $\left(  h\left(  1\right)  ,h\left(  2\right)
,\ldots,h\left(  p+q\right)  \right)  $ in decreasing order. But this yields
that $\left(  h\left(  \tau\left(  1\right)  \right)  ,h\left(  \tau\left(
2\right)  \right)  ,\ldots,h\left(  \tau\left(  p+q\right)  \right)  \right)
=\left(  g_{1},g_{2},\ldots,g_{p+q}\right)  $ (since $\left(  g_{1}%
,g_{2},\ldots,g_{p+q}\right)  $, too, is the result of sorting the list
$\left(  h\left(  1\right)  ,h\left(  2\right)  ,\ldots,h\left(  p+q\right)
\right)  $ in decreasing order), qed.}. In other words, $h\left(  \tau\left(
j\right)  \right)  =g_{j}$ for every $j\in\left\{  1,2,\ldots,p+q\right\}  $.
Thus,%
\begin{equation}
\tau\left(  I_{w}\right)  =P_{w}\sqcup Q_{w}\ \ \ \ \ \ \ \ \ \ \text{for
every }w\in\mathfrak{W} \label{pf.prop.shuffle.decrmax.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.shuffle.decrmax.1}):} Let
$w\in\mathfrak{W}$. Then,%
\begin{align*}
I_{w}  &  =\left\{  j\in\left\{  1,2,\ldots,p+q\right\}  \ \mid
\ \underbrace{g_{j}}_{\substack{=h\left(  \tau\left(  j\right)  \right)
\\=\left(  h\circ\tau\right)  \left(  j\right)  }}=w\right\}  =\left\{
j\in\left\{  1,2,\ldots,p+q\right\}  \ \mid\ \left(  h\circ\tau\right)
\left(  j\right)  =w\right\} \\
&  =\left(  h\circ\tau\right)  ^{-1}\left(  w\right)  =\tau^{-1}\left(
h^{-1}\left(  w\right)  \right)  .
\end{align*}
Since $\tau$ is a permutation, this yields $\tau\left(  I_{w}\right)
=h^{-1}\left(  w\right)  =P_{w}\sqcup Q_{w}$, and thus
(\ref{pf.prop.shuffle.decrmax.1}) is proven.}. As a consequence, for every
$w\in\mathfrak{W}$, the bijection $\tau:\left\{  1,2,\ldots,p+q\right\}
\rightarrow\left\{  1,2,\ldots,p+q\right\}  $ restricts to a bijection
$\tau_{w}:I_{w}\rightarrow P_{w}\sqcup Q_{w}$. This bijection $\tau_{w}$ is an
$\left(  I_{w},P_{w},Q_{w}\right)  $-shuffle\footnote{\textit{Proof.} We have
$\tau\in\operatorname{Sh}_{p,q}$, so that $\tau^{-1}\left(
1\right)  <\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(  p\right)  $ and
$\tau^{-1}\left(  p+1\right)  <\tau^{-1}\left(  p+2\right)  <\cdots<\tau
^{-1}\left(  p+q\right)  $. In other words, the restriction of $\tau^{-1}$ to
$\left\{  1,2,\ldots,p\right\}  $ is strictly increasing, and the restriction
of $\tau^{-1}$ to $\left\{  p+1,p+2,\ldots,p+q\right\}  $ is strictly
increasing. Since the restriction of $\tau^{-1}$ to $\left\{  1,2,\ldots
,p\right\}  $ is strictly increasing, the restriction of $\tau^{-1}$ to
$P_{w}$ must also be strictly increasing (because $P_{w}\subset\left\{
1,2,\ldots,p\right\}  $). But this restriction is $\tau_{w}^{-1}\mid_{P_{w}%
}:P_{w}\rightarrow I_{w}$. Hence, $\tau_{w}^{-1}\mid_{P_{w}}:P_{w}\rightarrow
I_{w}$ is strictly increasing. Similarly, the same can be said about $\tau
_{w}^{-1}\mid_{Q_{w}}:Q_{w}\rightarrow I_{w}$. Since the maps $\tau_{w}%
^{-1}\mid_{P_{w}}:P_{w}\rightarrow I_{w}$ and $\tau_{w}^{-1}\mid_{Q_{w}}%
:Q_{w}\rightarrow I_{w}$ are strictly increasing, we conclude that $\tau_{w}$
is an $\left(  I_{w},P_{w},Q_{w}\right)  $-shuffle (by the definition of a
$\left(  I_{w},P_{w},Q_{w}\right)  $-shuffle), qed.}. Thus, we have obtained a
family $\left(  \tau_{w}\right)  _{w\in\mathfrak{W}}$ of $\left(  I_{w}%
,P_{w},Q_{w}\right)  $-shuffles parametrized over all $w\in\mathfrak{W}$.

Now, let us forget that we fixed $\tau$. We thus have constructed, for every
$\tau\in\operatorname{Sh}_{p,q}$ satisfying $h\left(  \tau\left(
1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)  \geq\cdots\geq
h\left(  \tau\left(  p+q\right)  \right)  $, a family $\left(  \tau
_{w}\right)  _{w\in\mathfrak{W}}$ of $\left(  I_{w},P_{w},Q_{w}\right)
$-shuffles parametrized over all $w\in\mathfrak{W}$. We thus obtain a map%
\begin{align*}
\left\{  \tau\in\operatorname{Sh}_{p,q}\ \mid\ h\left(  \tau\left(
1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)  \geq\cdots\geq
h\left(  \tau\left(  p+q\right)  \right)  \right\}   &  \rightarrow\prod
_{w\in\mathfrak{W}}\left(  \text{the set of all }\left(  I_{w},P_{w}%
,Q_{w}\right)  \text{-shuffles}\right)  ,\\
\tau &  \mapsto\left(  \tau_{w}\right)  _{w\in\mathfrak{W}}.
\end{align*}
This map is injective\footnote{In fact, any $\tau\in
\operatorname{Sh}_{p,q}$ is uniquely determined by $\left(  \tau_{w}\right)
_{w\in\mathfrak{W}}$ (because $\tau_{w}$ is the restriction of $\tau$ to
$I_{w}$ (with a restricted codomain, but this doesn't matter right now), and
so knowing $\left(  \tau_{w}\right)  _{w\in\mathfrak{W}}$ means knowing the
values $\tau$ on each of the intervals $I_{w}$; but this means knowing all
values of $\tau$, because $\left\{  1,2,\ldots,p+q\right\}  =\bigsqcup
_{w\in\mathfrak{W}}I_{w}$).} and surjective\footnote{\textit{Proof.} We need
to show that for every
\[
\left(  \sigma_{w}\right)  _{w\in\mathfrak{W}}\in\prod_{w\in\mathfrak{W}%
}\left(  \text{the set of all }\left(  I_{w},P_{w},Q_{w}\right)
\text{-shuffles}\right)  ,
\]
there exists a $\tau\in\operatorname{Sh}_{p,q}$ satisfying $h\left(
\tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)
\geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $ and $\left(
\tau_{w}\right)  _{w\in\mathfrak{W}}=\left(  \sigma_{w}\right)  _{w\in
\mathfrak{W}}$.
\par
So fix some $\left(  \sigma_{w}\right)  _{w\in\mathfrak{W}}\in\prod
_{w\in\mathfrak{W}}\left(  \text{the set of all }\left(  I_{w},P_{w}%
,Q_{w}\right)  \text{-shuffles}\right)  $. For every $w\in\mathfrak{W}$, the
map $\sigma_{w}$ is an $\left(  I_{w},P_{w},Q_{w}\right)  $-shuffle, hence a
bijection from $I_{w}$ to $P_{w}\sqcup Q_{w}$. Since $\bigsqcup_{w\in
\mathfrak{W}}I_{w}=\left\{  1,2,\ldots,p+q\right\}  $ and $\bigsqcup
_{w\in\mathfrak{W}}\underbrace{\left(  P_{w}\sqcup Q_{w}\right)  }%
_{=h^{-1}\left(  w\right)  }=\bigsqcup_{w\in\mathfrak{W}}h^{-1}\left(
w\right)  =\left\{  1,2,\ldots,p+q\right\}  $, we can piece these bijections
$\sigma_{w}$ together to a bijection%
\[
\bigsqcup_{w\in\mathfrak{W}}\sigma_{w}:\left\{  1,2,\ldots,p+q\right\}
\rightarrow\left\{  1,2,\ldots,p+q\right\}  ,
\]
whose restriction to each interval $I_{w}$ coincides with the respective
$\sigma_{w}$ (except that the codomains of the maps are different). Let $\tau$
be this bijection $\bigsqcup_{w\in\mathfrak{W}}\sigma_{w}$. Clearly, $\tau
\in\Symm_{p+q}$. We will now show that $\tau\in
\operatorname{Sh}_{p,q}$, $h\left(  \tau\left(  1\right)  \right)  \geq h\left(
\tau\left(  2\right)  \right)  \geq\cdots\geq h\left(  \tau\left(  p+q\right)
\right)  $ and $\left(  \tau_{w}\right)  _{w\in\mathfrak{W}}=\left(
\sigma_{w}\right)  _{w\in\mathfrak{W}}$. Once this is done, the required
surjectivity will clearly follow.
\par
Since $\tau=\bigsqcup_{w\in\mathfrak{W}}\sigma_{w}$, we have $\tau\mid_{I_{w}%
}=\sigma_{w}$ for every $w\in\mathfrak{W}$ (up to the fact that the maps
$\tau\mid_{I_{w}}$ and $\sigma_{w}$ have different codomains). More precisely,
$\tau_{w}=\sigma_{w}$ for every $w\in\mathfrak{W}$. Hence, for every
$w\in\mathfrak{W}$, the map $\sigma_{w}$ is a restriction of $\tau$ (with
appropriately restricted codomain).
\par
The map $\tau=\bigsqcup_{w\in\mathfrak{W}}\sigma_{w}$ is pieced together from
bijections $\sigma_{w}:I_{w}\rightarrow P_{w}\sqcup Q_{w}$. Thus, $\tau\left(
I_{w}\right)  =P_{w}\sqcup Q_{w}$ for every $w\in\mathfrak{W}$. In other
words, $\tau\left(  I_{w}\right)  =h^{-1}\left(  w\right)  $ for every
$w\in\mathfrak{W}$ (since every $w\in\mathfrak{W}$ satisfies $h^{-1}\left(
w\right)  =P_{w}\sqcup Q_{w}$). Hence,
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,p+q\right\}  \text{ satisfies }h\left(
\tau\left(  i\right)  \right)  =g_{i}.
\label{pf.prop.shuffle.decrmax.surj.pf.1}%
\end{equation}
\par
[\textit{Proof of (\ref{pf.prop.shuffle.decrmax.surj.pf.1}):} Let
$i\in\left\{  1,2,\ldots,p+q\right\}  $. Set $w=h\left(  \tau\left(  i\right)
\right)  $. Then, $\tau\left(  i\right)  \in h^{-1}\left(  w\right)
=\tau\left(  I_{w}\right)  $ (since $\tau\left(  I_{w}\right)  =h^{-1}\left(
w\right)  $) and thus $i\in I_{w}$ (since $\tau$ is a bijection), so that
$i\in I_{w}=\left\{  j\in\left\{  1,2,\ldots,p+q\right\}  \ \mid
\ g_{j}=w\right\}  $ and thus $g_{i}=w=h\left(  \tau\left(  i\right)  \right)
$. This proves (\ref{pf.prop.shuffle.decrmax.surj.pf.1}).]
\par
We have $g_{1}\geq g_{2}\geq\cdots\geq g_{p+q}$. Due to
(\ref{pf.prop.shuffle.decrmax.surj.pf.1}), this rewrites as $h\left(
\tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)
\geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $.
\par
We now are going to prove that $\tau\in\operatorname{Sh}_{p,q}$. In
order to prove this, it is enough to show that $\tau^{-1}\left(  1\right)
<\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(  p\right)  $ and
$\tau^{-1}\left(  p+1\right)  <\tau^{-1}\left(  p+2\right)  <\cdots<\tau
^{-1}\left(  p+q\right)  $. We shall only verify $\tau^{-1}\left(  1\right)
<\tau^{-1}\left(  2\right)  <\cdots<\tau^{-1}\left(  p\right)  $ (the proof of
$\tau^{-1}\left(  p+1\right)  <\tau^{-1}\left(  p+2\right)  <\cdots<\tau
^{-1}\left(  p+q\right)  $ being analogous).
\par
So let $i$ and $j$ be elements of $\left\{  1,2,\ldots,p\right\}  $ such that
$i<j$. We will prove that $\tau^{-1}\left(  i\right)  <\tau^{-1}\left(
j\right)  $. Indeed, assume the contrary. Then, $\tau^{-1}\left(  i\right)
\geq\tau^{-1}\left(  j\right)  $. In other words, $\tau^{-1}\left(  j\right)
\leq\tau^{-1}\left(  i\right)  $. Thus, $h\left(  \tau\left(  \tau^{-1}\left(
j\right)  \right)  \right)  \geq h\left(  \tau\left(  \tau^{-1}\left(
i\right)  \right)  \right)  $ (since $h\left(  \tau\left(  1\right)  \right)
\geq h\left(  \tau\left(  2\right)  \right)  \geq\cdots\geq h\left(
\tau\left(  p+q\right)  \right)  $). In other words, $h\left(  j\right)  \geq
h\left(  i\right)  $. Combined with $h\left(  i\right)  \geq h\left(
j\right)  $ (since $h\left(  1\right)  \geq h\left(  2\right)  \geq\cdots\geq
h\left(  p\right)  $ and $i<j$), this yields $h\left(  i\right)  =h\left(
j\right)  $. So we can define a $w\in\mathfrak{W}$ by $w=h\left(  i\right)
=h\left(  j\right)  $. Consider this $w$. We have $h\left(  i\right)  =w$, so
that $i\in h^{-1}\left(  w\right)  =P_{w}\sqcup Q_{w}$. But we cannot have
$i\in Q_{w}$ (because $i$ lies in the set $\left\{  1,2,\ldots,p\right\}  $,
which is disjoint to $Q_{w}$ (since $Q_{w}\subset\left\{  p+1,p+2,\ldots
,p+q\right\}  $)). Since we have $i\in P_{w}\sqcup Q_{w}$ but not $i\in Q_{w}%
$, we must have $i\in P_{w}$. Similarly, $j\in P_{w}$. But recall that
$\sigma_{w}$ is a $\left(  I_{w},P_{w},Q_{w}\right)  $-shuffle. In other
words, $\sigma_{w}:I_{w}\rightarrow P_{w}\sqcup Q_{w}$ is a bijection having
the property that the maps $\sigma_{w}^{-1}\mid_{P_{w}}:P_{w}\rightarrow
I_{w}$ and $\sigma_{w}^{-1}\mid_{Q_{w}}:Q_{w}\rightarrow I_{w}$ are strictly
increasing. Since $\sigma_{w}^{-1}\mid_{P_{w}}:P_{w}\rightarrow I_{w}$ is
strictly increasing, we have $\left(  \sigma_{w}^{-1}\mid_{P_{w}}\right)
\left(  i\right)  <\left(  \sigma_{w}^{-1}\mid_{P_{w}}\right)  \left(
j\right)  $ (since $i<j$ and $i\in P_{w}$ and $j\in P_{w}$). Since $\left(
\sigma_{w}^{-1}\mid_{P_{w}}\right)  \left(  i\right)  =\sigma_{w}^{-1}\left(
i\right)  =\tau^{-1}\left(  i\right)  $ (because $\sigma_{w}$ is a restriction
of $\tau$) and $\left(  \sigma_{w}^{-1}\mid_{P_{w}}\right)  \left(  j\right)
=\tau^{-1}\left(  j\right)  $ (similarly), this rewrites as $\tau^{-1}\left(
i\right)  <\tau^{-1}\left(  j\right)  $, which contradicts $\tau^{-1}\left(
i\right)  \geq\tau^{-1}\left(  j\right)  $. This contradiction proves our
assumption wrong, and so we have $\tau^{-1}\left(  i\right)  <\tau^{-1}\left(
j\right)  $.
\par
Let us forget that we fixed $i$ and $j$. We thus have seen that $\tau
^{-1}\left(  i\right)  <\tau^{-1}\left(  j\right)  $ for any elements $i$ and
$j$ of $\left\{  1,2,\ldots,p\right\}  $ such that $i<j$. In other words,
$\tau^{-1}\left(  1\right)  <\tau^{-1}\left(  2\right)  <\cdots<\tau
^{-1}\left(  p\right)  $. Similarly, $\tau^{-1}\left(  p+1\right)  <\tau
^{-1}\left(  p+2\right)  <\cdots<\tau^{-1}\left(  p+q\right)  $. Thus,
$\tau\in\operatorname{Sh}_{p,q}$.
\par
We now know that $\tau\in\operatorname{Sh}_{p,q}$ and $h\left(
\tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)
\geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  $. Finally, $\left(
\tau_{w}\right)  _{w\in\mathfrak{W}}=\left(  \sigma_{w}\right)  _{w\in
\mathfrak{W}}$ follows from the very definition of $\tau_{w}$ (since $\tau
_{w}=\sigma_{w}$ for every $w\in\mathfrak{W}$). This completes the proof.}.
Hence, it is bijective, and this yields that the sets
\[
\left\{  \tau\in\operatorname{Sh}_{p,q}\ \mid\ h\left(  \tau\left(
1\right)  \right)  \geq h\left(  \tau\left(  2\right)  \right)  \geq\cdots\geq
h\left(  \tau\left(  p+q\right)  \right)  \right\}
\]
and%
\[
\prod_{w\in\mathfrak{W}}\left(  \text{the set of all }\left(  I_{w}%
,P_{w},Q_{w}\right)  \text{-shuffles}\right)
\]
are in bijection. Thus,%
\begin{align*}
&  \left\vert \left\{  \tau\in\operatorname{Sh}_{p,q}\ \mid
\ h\left(  \tau\left(  1\right)  \right)  \geq h\left(  \tau\left(  2\right)
\right)  \geq\cdots\geq h\left(  \tau\left(  p+q\right)  \right)  \right\}
\right\vert \\
&  =\left\vert \prod_{w\in\mathfrak{W}}\left(  \text{the set of all }\left(
I_{w},P_{w},Q_{w}\right)  \text{-shuffles}\right)  \right\vert \\
&  =\prod_{w\in\mathfrak{W}}\underbrace{\left\vert \left(  \text{the set of
all }\left(  I_{w},P_{w},Q_{w}\right)  \text{-shuffles}\right)  \right\vert
}_{\substack{=\left(  \text{the number of }\left(  I_{w},P_{w},Q_{w}\right)
\text{-shuffles}\right)  =\dbinom{\mathfrak{a}\left(  w\right)  +\mathfrak{b}%
\left(  w\right)  }{\mathfrak{a}\left(  w\right)  }\\\text{(by
(\ref{pf.prop.shuffle.decrmax.numshuf}))}}}\\
&  =\prod_{w\in\mathfrak{W}}\dbinom{\mathfrak{a}\left(  w\right)
+\mathfrak{b}\left(  w\right)  }{\mathfrak{a}\left(  w\right)  }.
\end{align*}
This is precisely the statement of Proposition \ref{prop.shuffle.decrmax}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.shuffle.lyndon.cflcrit}}

\textit{Solution to Exercise \ref{exe.shuffle.lyndon.cflcrit}.}
$\Longrightarrow:$ Assume that $w$ is Lyndon. We need to prove that for any
two nonempty words $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$
satisfying $w=uv$, there exists at least one $s\in u\shuffle v$ satisfying
$s>w$.

Let $u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ be two nonempty
words satisfying $w=uv$. Then, $vu$ is an element of $u\shuffle v$%
\ \ \ \ \footnote{\textit{Proof.} Let $n=\ell\left(  u\right)  $ and
$m=\ell\left(  v\right)  $. Then, the permutation in $\Symm_{n+m}$
which is written as $\left(  n+1,n+2,\ldots,n+m,1,2,\ldots,n\right)  $ in
one-line notation belongs to $\operatorname{Sh}_{n,m}$. If we denote
this permutation by $\sigma$, then $vu=u\underset{\sigma}{\shuffle}v\in
u\shuffle
v$, qed.} and satisfies $vu>uv$ (by Proposition \ref{prop.words.lyndon.trivia}%
(c)). Thus, $vu>uv=w$. Hence, there exists at least one $s\in u\shuffle v$
satisfying $s>w$ (namely, $s=vu$). Thus, the $\Longrightarrow$ direction of
Exercise \ref{exe.shuffle.lyndon.cflcrit} is solved.

$\Longleftarrow:$ Assume that for any two nonempty words $u\in\mathfrak{A}%
^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ satisfying $w=uv$,%
\begin{equation}
\text{there exists at least one }s\in u\shuffle v\text{ satisfying }s>w.
\label{sol.shuffle.lyndon.cflcrit.1}%
\end{equation}
We need to prove that $w$ is Lyndon.

In fact, assume the contrary. Thus, $w$ is not Lyndon. Let $\left(
a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of $w$; then,
$a_{1}\geq a_{2}\geq\cdots\geq a_{p}$ and $a_{1}a_{2} \cdots a_{p}=w$. Also,
$p\neq0$ (since $a_{1}a_{2} \cdots a_{p}=w$ is nonempty). Since $a_{1}$ is
Lyndon but $w$ is not, we have $a_{1}\neq w$. Thus, $p\neq1$ (because
otherwise, we would have $a_{1}=a_{1}a_{2} \cdots a_{p}=w$, contradicting
$a_{1}\neq w$). Combined with $p\neq0$, this yields $p\geq2$.

Let $u=a_{1}$ and $v=a_{2}a_{3} \cdots a_{p}$. Then, $u$ is Lyndon (since
$u=a_{1}$), thus nonempty. Also, $v$ is a nonempty product of Lyndon words
(nonempty because $p\geq2$), and hence nonempty itself (since Lyndon words are
nonempty). Clearly, $\underbrace{u}_{=a_{1}}\underbrace{v}_{=a_{2}a_{3} \cdots
a_{p}}=a_{1}\left(  a_{2}a_{3} \cdots a_{p}\right)  =a_{1}a_{2} \cdots
a_{p}=w$. Thus, (\ref{sol.shuffle.lyndon.cflcrit.1}) yields that there exists
at least one $s\in u\shuffle v$ satisfying $s>w$. Thus, $w$ is not the
lexicographically highest element of the multiset $u\shuffle v$.

Now, notice that $a_{2}$, $a_{3}$, $\ldots$, $a_{p}$ are Lyndon words
satisfying $a_{2}\geq a_{3}\geq\cdots\geq a_{p}$ (since $a_{1}\geq a_{2}%
\geq\cdots\geq a_{p}$) and $a_{2}a_{3} \cdots a_{p}=v$. Hence, $\left(
a_{2},a_{3},\ldots,a_{p}\right)  $ is the CFL factorization of $v$. We have
$u=a_{1}\geq a_{j+1}$ for every $j\in\left\{  1,2,\ldots,p-1\right\}  $ (since
$a_{1}\geq a_{2}\geq\cdots\geq a_{p}$). Thus, Theorem
\ref{thm.shuffle.lyndon.used}(e) (applied to $p-1$ and $\left(  a_{2}%
,a_{3},\ldots,a_{p}\right)  $ instead of $q$ and $\left(  b_{1},b_{2}%
,\ldots,b_{q}\right)  $) yields that the lexicographically highest element of
the multiset $u\shuffle v$ is $uv$, and the multiplicity with which this word
$uv$ appears in the multiset $u\shuffle v$ is $\operatorname{mult}_{u}v+1$.

Now, we know that the lexicographically highest element of the multiset
$u\shuffle v$ is $uv=w$. This contradicts the fact that $w$ is not the
lexicographically highest element of the multiset $u\shuffle v$. This
contradiction shows that our assumption was wrong. Thus, $w$ is Lyndon.
This completes the solution
of the $\Longleftarrow$ direction of Exercise \ref{exe.shuffle.lyndon.cflcrit}.

Thus, both the $\Longrightarrow$ and $\Longleftarrow$ directions of
Exercise \ref{exe.shuffle.lyndon.cflcrit} are proven.
Exercise \ref{exe.shuffle.lyndon.cflcrit} is thus solved.

[\textit{Remark:} Exercise \ref{exe.shuffle.lyndon.cflcrit} still holds in the
partial-order setting\footnote{See Exercise \ref{exe.words.poset} for an
explanation of what the partial-order setting is.}. To prove this, we can use
Proposition \ref{prop.sol.words.poset.reduce-to-linex} along with the fact
that Exercise \ref{exe.shuffle.lyndon.cflcrit} holds in the total-order
setting. Here are the details:

\begin{proof}
[Solution to Exercise \ref{exe.shuffle.lyndon.cflcrit} in the partial-order
setting.]$\Longrightarrow:$ In the partial-order setting, the $\Longrightarrow
$ direction of Exercise \ref{exe.shuffle.lyndon.cflcrit} can be proved in the
same way as it was proven in the total-order setting.

$\Longleftarrow:$ Assume that for any two nonempty words $u\in\mathfrak{A}%
^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ satisfying $w=uv$,%
\begin{equation}
\text{there exists at least one }s\in u\shuffle v\text{ satisfying
}s>w.\label{sol.shuffle.lyndon.cflcrit.posetting.1}%
\end{equation}
We need to prove that $w$ is Lyndon.

Let $\mathfrak{B}$ be any linear extension of the alphabet $\mathfrak{A}$.
Then, $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$
(according to Proposition \ref{prop.sol.words.poset.reduce-to-linex}(a)).
Thus, $\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ as sets. It is easy to see
that for any two nonempty words $u\in\mathfrak{B}^{\ast}$ and $v\in
\mathfrak{B}^{\ast}$ satisfying $w=uv$,%
\begin{equation}
\text{there exists at least one }s\in u\shuffle v\text{ satisfying }s>w\text{
in }\mathfrak{B}^{\ast}.\label{sol.shuffle.lyndon.cflcrit.posetting.1B}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.shuffle.lyndon.cflcrit.posetting.1B}):}
Let $u\in\mathfrak{B}^{\ast}$ and $v\in\mathfrak{B}^{\ast}$. Then,
$u\in\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ and $v\in\mathfrak{B}^{\ast
}=\mathfrak{A}^{\ast}$. Hence, (\ref{sol.shuffle.lyndon.cflcrit.posetting.1})
shows that there exists at least one $s\in u\shuffle v$ satisfying $s>w$ in
$\mathfrak{A}^{\ast}$. Let $t$ be such an $s$. Then, $t$ is an element of
$u\shuffle v$ satisfying $t>w$ in $\mathfrak{A}^{\ast}$. Hence, $w<t$ in
$\mathfrak{A}^{\ast}$ (since $t>w$ in $\mathfrak{A}^{\ast}$), so that $w\leq
t$ in $\mathfrak{A}^{\ast}$.
\par
We know that $\mathfrak{B}^{\ast}$ is an extension of $\mathfrak{A}^{\ast}$.
Thus, any two elements $a$ and $b$ of $\mathfrak{A}^{\ast}$ satisfying $a\leq
b$ in $\mathfrak{A}^{\ast}$ satisfy $a\leq b$ in $\mathfrak{B}^{\ast}$
(according to the definition of an ``extension'').
Applying this to $a=w$ and $b=t$, we obtain
$w\leq t$ in $\mathfrak{B}^{\ast}$. Combined with $w\neq t$ (since $w<t$ in
$\mathfrak{A}^{\ast}$), this yields $w<t$ in $\mathfrak{B}^{\ast}$. In other
words, $t>w$ in $\mathfrak{B}^{\ast}$.
\par
Hence, there exists at least one $s\in u\shuffle v$ satisfying $s>w$ in
$\mathfrak{B}^{\ast}$ (namely, $s=t$). This proves
(\ref{sol.shuffle.lyndon.cflcrit.posetting.1B}).}

Now, we can apply Exercise \ref{exe.shuffle.lyndon.cflcrit} to $\mathfrak{B}$
instead of $\mathfrak{A}$ (since $\mathfrak{B}$ is totally ordered). As a
consequence, we see that $w$ is Lyndon as a word in $\mathfrak{B}^{\ast}$ if
and only if for any two nonempty words $u\in\mathfrak{B}^{\ast}$ and
$v\in\mathfrak{B}^{\ast}$ satisfying $w=uv$, there exists at least one $s\in
u\shuffle v$ satisfying $s>w$ in $\mathfrak{B}^{\ast}$. Thus, $w$ is Lyndon as
a word in $\mathfrak{B}^{\ast}$ (because we know that for any two nonempty
words $u\in\mathfrak{B}^{\ast}$ and $v\in\mathfrak{B}^{\ast}$ satisfying
$w=uv$, there exists at least one $s\in u\shuffle v$ satisfying $s>w$ in
$\mathfrak{B}^{\ast}$).

Now, the definition of a Lyndon word shows the following: The word $w$ (as a
word in $\mathfrak{B}^{\ast}$) is Lyndon if and only if it is nonempty and
satisfies the following property:%
\begin{equation}
\text{Every nonempty proper suffix }v\text{ of }w\text{ satisfies }v>w\text{
in }\mathfrak{B}^{\ast}\text{.}\label{sol.shuffle.lyndon.cflcrit.posetting.4}%
\end{equation}
Since the word $w$ is Lyndon, this shows that the word $w$ is nonempty and
satisfies the property (\ref{sol.shuffle.lyndon.cflcrit.posetting.4}).

Let now $v$ be a nonempty proper suffix of $w$ (as a word in $\mathfrak{A}%
^{\ast}$). Then, $v$ is a nonempty proper suffix of $w$ (as a word in
$\mathfrak{B}^{\ast}$). Thus, (\ref{sol.shuffle.lyndon.cflcrit.posetting.4})
shows that $v$ satisfies $v>w$ in $\mathfrak{B}^{\ast}$.

Now, let us forget that we fixed $\mathfrak{B}$ and $v$. We thus have shown
that if $v$ is any nonempty proper suffix of $w$ and if $\mathfrak{B}$ is any
linear extension of the alphabet $\mathfrak{A}$, then%
\begin{equation}
v>w\text{ in }\mathfrak{B}^{\ast}\text{.}%
\label{sol.shuffle.lyndon.cflcrit.posetting.2}%
\end{equation}


But the definition of a Lyndon word shows the following: The word $w$ (as a
word in $\mathfrak{A}^{\ast}$) is Lyndon if and only if it is nonempty and
satisfies the following property:%
\begin{equation}
\text{Every nonempty proper suffix }v\text{ of }w\text{ satisfies }v>w\text{
(in }\mathfrak{A}^{\ast}\text{).}%
\label{sol.shuffle.lyndon.cflcrit.posetting.3}%
\end{equation}


We already know that $w$ is nonempty. We are now going to prove
(\ref{sol.shuffle.lyndon.cflcrit.posetting.3}):

Let $v$ be a nonempty proper suffix of $w$. We have $v>w$ in $\mathfrak{B}%
^{\ast}$ for every linear extension $\mathfrak{B}$ of $\mathfrak{A}$
(according to (\ref{sol.shuffle.lyndon.cflcrit.posetting.2})). In other words,
$w<v$ in $\mathfrak{B}^{\ast}$ for every linear extension $\mathfrak{B}$ of
$\mathfrak{A}$.

But Proposition \ref{prop.sol.words.poset.reduce-to-linex}(b) (applied to $w$
instead of $u$) yields that $w<v$ holds in $\mathfrak{A}^{\ast}$ if and only
if we have
\[
\left(  w<v\text{ in }\mathfrak{B}^{\ast}\text{ for every linear extension
}\mathfrak{B}\text{ of }\mathfrak{A}\right)  .
\]
We thus conclude that $w<v$ holds in $\mathfrak{A}^{\ast}$ (since we know that
$w<v$ in $\mathfrak{B}^{\ast}$ for every linear extension $\mathfrak{B}$ of
$\mathfrak{A}$). In other words, $v>w$ in $\mathfrak{A}^{\ast}$. This proves
(\ref{sol.shuffle.lyndon.cflcrit.posetting.3}).

So we know that the word $w$ is nonempty and satisfies the property
(\ref{sol.shuffle.lyndon.cflcrit.posetting.3}). Thus, the word $w$ (as a word
in $\mathfrak{A}^{\ast}$) is Lyndon (since the word $w$ (as a word in
$\mathfrak{A}^{\ast}$) is Lyndon if and only if it is nonempty and satisfies
the property (\ref{sol.shuffle.lyndon.cflcrit.posetting.3})). Thus, the
$\Longleftarrow$ direction of Exercise \ref{exe.shuffle.lyndon.cflcrit} is
proven in the partial-order setting.

Thus, both the $\Longrightarrow$ and $\Longleftarrow$ directions of
Exercise \ref{exe.shuffle.lyndon.cflcrit} are proven in the
partial-order setting. Hence, Exercise
\ref{exe.shuffle.lyndon.cflcrit} is solved in the partial-order setting.
\end{proof}

]
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.rmk.shuffle.shufflealg}}

\textit{Solution to Exercise \ref{exe.rmk.shuffle.shufflealg}.}

\begin{proof}
[Proof of Remark \ref{rmk.shuffle.shufflealg}.]Let $\left(  w_{1},w_{2}%
,\ldots,w_{n+m}\right)  $ denote the concatenation $u\cdot v=\left(
u_{1},u_{2},\ldots,u_{n},v_{1},v_{2},\ldots,v_{m}\right)  $. Then, $\left(
u_{1},u_{2},\ldots,u_{n},v_{1},v_{2},\ldots,v_{m}\right)  =\left(  w_{1}%
,w_{2},\ldots,w_{n+m}\right)  $, so that $\left(  u_{1},u_{2},\ldots
,u_{n}\right)  =\left(  w_{1},w_{2},\ldots,w_{n}\right)  $ and $\left(
v_{1},v_{2},\ldots,v_{m}\right)  =\left(  w_{n+1},w_{n+2},\ldots
,w_{n+m}\right)  $.

By the definition of $b_{u}$, we have $b_{u}=b_{u_{1}}b_{u_{2}}\cdots
b_{u_{n}}=b_{w_{1}}b_{w_{2}}\cdots b_{w_{n}}$ (since $\left(  u_{1}%
,u_{2},\ldots,u_{n}\right)  =\left(  w_{1},w_{2},\ldots,w_{n}\right)  $). By
the definition of $b_{v}$, we have $b_{v}=b_{v_{1}}b_{v_{2}}\cdots b_{v_{m}%
}=b_{w_{n+1}}b_{w_{n+2}}\cdots b_{w_{n+m}}$ (since \newline
$\left(  v_{1},v_{2}%
,\ldots,v_{m}\right)  =\left(  w_{n+1},w_{n+2},\ldots,w_{n+m}\right)  $). Now,%
\begin{align}
\underbrace{b_{u}}_{=b_{w_{1}}b_{w_{2}}\cdots b_{w_{n}}} \shufmult%
\underbrace{b_{v}}_{=b_{w_{n+1}}b_{w_{n+2}}\cdots b_{w_{n+m}}}  &  =\left(
b_{w_{1}}b_{w_{2}}\cdots b_{w_{n}}\right)  \shufmult \left(
b_{w_{n+1}}b_{w_{n+2}}\cdots b_{w_{n+m}}\right) \nonumber\\
&  =\sum_{\sigma\in\operatorname{Sh}_{n,m}}b_{w_{\sigma\left(
1\right)  }}b_{w_{\sigma\left(  2\right)  }}\cdots b_{w_{\sigma\left(
n+m\right)  }} \label{pf.rmk.shuffle.shufflealg.2}%
\end{align}
(by the definition of $\shufmult$).

However, for every $\sigma\in\operatorname{Sh}_{n,m}$, we have
$u\underset{\sigma}{\shuffle}v=\left(  w_{\sigma\left(  1\right)  }%
,w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)  $
(by the definition of $u\underset{\sigma}{\shuffle}v$), and therefore
$b_{u\underset{\sigma}{\shuffle}v}=b_{w_{\sigma\left(  1\right)  }%
}b_{w_{\sigma\left(  2\right)  }}\cdots b_{w_{\sigma\left(  n+m\right)  }}$
(by the definition of $b_{u\underset{\sigma}{\shuffle}v}$). Hence,
$\sum_{\sigma\in\operatorname{Sh}_{n,m}}%
\underbrace{b_{u\underset{\sigma}{\shuffle}v}}_{=b_{w_{\sigma\left(  1\right)
}}b_{w_{\sigma\left(  2\right)  }}\cdots b_{w_{\sigma\left(  n+m\right)  }}%
}=\sum_{\sigma\in\operatorname{Sh}_{n,m}}b_{w_{\sigma\left(
1\right)  }}b_{w_{\sigma\left(  2\right)  }}\cdots b_{w_{\sigma\left(
n+m\right)  }}$. Compared with (\ref{pf.rmk.shuffle.shufflealg.2}), this
yields $b_{u} \shufmult b_{v}=\sum_{\sigma\in
\operatorname{Sh}_{n,m}}b_{u\underset{\sigma}{\shuffle}v}$. This proves Remark
\ref{rmk.shuffle.shufflealg}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.lyndonbases}}

\textit{Solution to Exercise \ref{exe.lem.lyndonbases}.}

\begin{proof}
[Proof of Lemma \ref{lem.lyndonbases}.](a) Let $\mathfrak{M}$ denote the set
of all finite multisets of Lyndon words over $\mathfrak{A}$. In other words,
$\mathfrak{M}$ is the set of all finite multisets of elements of
$\mathfrak{L}$ (since the elements of $\mathfrak{L}$ are the Lyndon words over
$\mathfrak{A}$).
Define two maps $\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$
and $\mathbf{n} : \mathfrak{A}^{\ast} \to \mathfrak{M}$ as in
Proposition \ref{prop.lyndon.CFL.bij}.
Then, Proposition \ref{prop.lyndon.CFL.bij} shows that
these two maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse bijections.
Hence, the map $\mathbf{m}$ is a bijection.

On the other hand, for every $M\in\mathfrak{M}$, let us define an element
$\mathbf{b}_{M}$ of $A$ by setting $\mathbf{b}_{M}=b_{a_{1}}b_{a_{2}} \cdots
b_{a_{k}}$, where $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ denote the elements of
$M$ listed in decreasing order. Then, $\left(  b_{w}\right)  _{w\in
\mathfrak{L}}$ is an algebraically independent generating set of the
$\kk$-algebra $A$ if and only if $\left(  \mathbf{b}_{M}\right)
_{M\in\mathfrak{M}}$ is a basis of the $\kk$-module $A$%
\ \ \ \ \footnote{\textit{Proof.} Let $\mathfrak{N}$ be the set of all
families $\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in \NN %
^{\mathfrak{L}}$ of nonnegative integers (indexed by the Lyndon words) such
that all but finitely many $w\in\mathfrak{L}$ satisfy $k_{w}=0$.
Proposition \ref{prop.multiset.bij.seqs}
(applied to $S=\mathfrak{L}$) shows that the map $\operatorname*{mult}%
:\mathfrak{M}\rightarrow\mathfrak{N}$ that sends each multiset $M\in\mathfrak{M}$ to
the family
\[
\left(  \left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\right)  _{w\in S}\in\mathfrak{N}%
\]
is well-defined and is a bijection. Consider this map $\operatorname*{mult}$.
\par
For every family $f=\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}$,
we can define an element $\mathbf{b}_{f}$ of $A$ by $\mathbf{b}_{f}%
=\prod_{w\in\mathfrak{L}}\left(  b_{w}\right)  ^{k_{w}}$. (This is
well-defined, since $A$ is commutative and since all but finitely many
$w\in\mathfrak{L}$ satisfy $k_{w}=0$). Then, $\left(  \mathbf{b}_{f}\right)
_{f\in\mathfrak{N}}$ is the family of all possible monomials in the
``variables'' $b_{w}$ (that is, of all
possible finite products of elements of the family $\left(  b_{w}\right)
_{w\in\mathfrak{L}}$, with multiplicities allowed). Hence,
\begin{equation}
\left(
\begin{array}
[c]{c}%
\left(  b_{w}\right)  _{w\in\mathfrak{L}}\text{ is an algebraically
independent generating set of the } \kk \text{-algebra }A\\
\text{if and only if }\left(  \mathbf{b}_{f}\right)  _{f\in\mathfrak{N}}\text{
is a basis of the } \kk \text{-module }A
\end{array}
\right)  . \label{pf.lem.lyndonbases.3}%
\end{equation}
\par
Now, we claim that the family $\left(  \mathbf{b}_{f}\right)  _{f\in
\mathfrak{N}}$ is a reindexing of the family $\left(  \mathbf{b}_{M}\right)
_{M\in\mathfrak{M}}$. Indeed, since $\operatorname*{mult}$ is a bijection, it
is clear that the family $\left(  \mathbf{b}_{f}\right)  _{f\in\mathfrak{N}}$
is a reindexing of the family $\left(  \mathbf{b}_{\operatorname*{mult}%
M}\right)  _{M\in\mathfrak{M}}$. We now will prove that every $M\in
\mathfrak{M}$ satisfies $\mathbf{b}_{\operatorname*{mult}M}=\mathbf{b}_{M}$.
\par
Indeed, let $M\in\mathfrak{M}$. Let $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ denote
the elements of $M$ listed in decreasing order. Then, every $w\in\mathfrak{L}$
satisfies%
\begin{equation}
\left(  \text{multiplicity of }w\text{ in the multiset }M\right)  =\left(
\text{the number of }i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfying
}a_{i}=w\right)  . \label{pf.lem.lyndonbases.5}%
\end{equation}
But by the definition of $\operatorname*{mult}M$, we have%
\begin{align*}
\operatorname*{mult}M  &  =\left(  \underbrace{\left(  \text{multiplicity of
}w\text{ in the multiset }M\right)  }_{\substack{=\left(  \text{the number of
}i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)
\\\text{(by (\ref{pf.lem.lyndonbases.5}))}}}\right)  _{w\in\mathfrak{L}}\\
&  =\left(  \left(  \text{the number of }i\in\left\{  1,2,\ldots,k\right\}
\text{ satisfying }a_{i}=w\right)  \right)  _{w\in\mathfrak{L}},
\end{align*}
so that the definition of $\mathbf{b}_{\operatorname*{mult}M}$ becomes%
\begin{align*}
\mathbf{b}_{\operatorname*{mult}M}  &  =\prod_{w\in\mathfrak{L}}%
\underbrace{\left(  b_{w}\right)  ^{\left(  \text{the number of }i\in\left\{
1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)  }}_{=\prod\limits
_{\substack{i\in\left\{  1,2,\ldots,k\right\}  ;\\a_{i}=w}}b_{w}}=\prod
_{w\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{  1,2,\ldots,k\right\}
;\\a_{i}=w}}\underbrace{b_{w}}_{\substack{=b_{a_{i}}\\\text{(since }%
w=a_{i}\text{)}}}=\prod_{w\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=w}}b_{a_{i}}\\
&  =\prod_{i\in\left\{  1,2,\ldots,k\right\}  }b_{a_{i}}=b_{a_{1}}b_{a_{2}}
\cdots b_{a_{k}}=\mathbf{b}_{M}%
\end{align*}
(since $\mathbf{b}_{M}$ was defined as $b_{a_{1}}b_{a_{2}} \cdots b_{a_{k}}$).
\par
Forget now that we fixed $M$. We thus have shown that every $M\in\mathfrak{M}$
satisfies $\mathbf{b}_{\operatorname*{mult}M}=\mathbf{b}_{M}$. Hence, $\left(
\mathbf{b}_{\operatorname*{mult}M}\right)  _{M\in\mathfrak{M}}=\left(
\mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$. But we know that the family
$\left(  \mathbf{b}_{f}\right)  _{f\in\mathfrak{N}}$ is a reindexing of the
family $\left(  \mathbf{b}_{\operatorname*{mult}M}\right)  _{M\in\mathfrak{M}%
}$. Since $\left(  \mathbf{b}_{\operatorname*{mult}M}\right)  _{M\in
\mathfrak{M}}=\left(  \mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$, this
rewrites as follows: The family $\left(  \mathbf{b}_{f}\right)  _{f\in
\mathfrak{N}}$ is a reindexing of the family $\left(  \mathbf{b}_{M}\right)
_{M\in\mathfrak{M}}$. Hence, $\left(  \mathbf{b}_{f}\right)  _{f\in
\mathfrak{N}}$ is a basis of the $\kk$-module $A$ if and only if
$\left(  \mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$ is a basis of the
$\kk$-module $A$. Combined with (\ref{pf.lem.lyndonbases.3}), this
yields that $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically
independent generating set of the $\kk$-algebra $A$ if and only if
$\left(  \mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$ is a basis of the
$\kk$-module $A$, qed.}. Hence, in order to prove Lemma
\ref{lem.lyndonbases}(a), it remains to prove that $\left(  \mathbf{b}%
_{M}\right)  _{M\in\mathfrak{M}}$ is a basis of the $\kk$-module $A$ if
and only if $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a
basis of the $\kk$-module $A$.

Let us show that
\begin{equation}
\mathbf{b}_{\mathbf{m}\left(  M\right)  }=\mathbf{b}_{M}%
\ \ \ \ \ \ \ \ \ \ \text{for every }M\in\mathfrak{M}.
\label{pf.lem.lyndonbases.9}%
\end{equation}


\textit{Proof of (\ref{pf.lem.lyndonbases.9}):} Let $M\in\mathfrak{M}$. Let
$a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ denote the elements of $M$ listed in
decreasing order. Then, $\mathbf{m}\left(  M\right)  =a_{1}a_{2} \cdots a_{k}$
(by the definition of $\mathbf{m}\left(  M\right)  $). Combining this with the
fact that $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ are Lyndon words (since they are
elements of $M$, which is a multiset of Lyndon words) and satisfy $a_{1}\geq
a_{2}\geq\cdots\geq a_{k}$ (since $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ are the
elements of $M$ listed in decreasing order), we thus conclude that $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of $\mathbf{m}%
\left(  M\right)  $. Hence, the definition of $\mathbf{b}_{\mathbf{m}\left(
M\right)  }$ says that
$\mathbf{b}_{\mathbf{m}\left(  M\right)  }
= b_{a_{1}}b_{a_{2}} \cdots b_{a_{k}}$.
Compared with $\mathbf{b}_{M}=b_{a_{1}}b_{a_{2}} \cdots b_{a_{k}}$
(which is just the definition of $\mathbf{b}_{M}$), this
yields $\mathbf{b}_{\mathbf{m}\left(  M\right)  }=\mathbf{b}_{M}$. This proves
(\ref{pf.lem.lyndonbases.9}).

Now, the family $\left(  \mathbf{b}_{\mathbf{m}\left(  M\right)  }\right)
_{M\in\mathfrak{M}}$ is a reindexing of the family $\left(  \mathbf{b}%
_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ (since $\mathbf{m}$ is a bijection).
Since $\left(  \mathbf{b}_{\mathbf{m}\left(  M\right)  }\right)
_{M\in\mathfrak{M}}=\left(  \mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$ (by
(\ref{pf.lem.lyndonbases.9})), this rewrites as follows: The family $\left(
\mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$ is a reindexing of the family
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$. Hence, $\left(
\mathbf{b}_{M}\right)  _{M\in\mathfrak{M}}$ is a basis of the $\kk %
$-module $A$ if and only if $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ is a basis of the $\kk$-module $A$. As we said
above, this completes our proof of Lemma \ref{lem.lyndonbases}(a).

(b) The proof of Lemma \ref{lem.lyndonbases}(b) is analogous to the above
proof of Lemma \ref{lem.lyndonbases}(a).

(c) We assumed that the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$
generates the $\kk$-algebra $A$. By Lemma \ref{lem.lyndonbases}(b),
this yields that the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ spans the $\kk$-module $A$. Recall also that the
family $\left(  g_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis of the
$\kk$-module $A$, and thus spans this $\kk$-module.

We need to prove that the family $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$
is an algebraically independent generating set of the $\kk$-algebra
$A$. According to Lemma \ref{lem.lyndonbases}(a), this is equivalent to
proving that the family $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}%
^{\ast}}$ is a basis of the $\kk$-module $A$. We are going to prove the
latter statement.

Let us first notice that $\mathbf{b}_{u}$ is a homogeneous element of $A$ of
degree $\operatorname*{Wt}\left(  u\right)  $ for every $u\in\mathfrak{A}%
^{\ast}$\ \ \ \ \footnote{\textit{Proof.} Let $u\in\mathfrak{A}^{\ast}$. Let
$\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of $u$.
Then, $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are Lyndon words and satisfy
$u=a_{1}a_{2} \cdots a_{p}$.
\par
The definition of $\operatorname*{Wt}$ easily yields that%
\[
\operatorname*{Wt}\left(  s_{1}s_{2} \cdots s_{k}\right)  =\operatorname*{Wt}%
\left(  s_{1}\right)  +\operatorname*{Wt}\left(  s_{2}\right)  +
\cdots+\operatorname*{Wt}\left(  s_{k}\right)  \ \ \ \ \ \ \ \ \ \ \text{for
any }k\in \NN \text{ and any }k\text{ words }s_{1}\text{, }s_{2}\text{,
}\ldots\text{, }s_{k}\text{ in }\mathfrak{A}^{\ast}.
\]
Applying this to $k=p$ and $s_{i}=a_{i}$, we obtain $\operatorname*{Wt}\left(
a_{1}a_{2} \cdots a_{p}\right)  =\operatorname*{Wt}\left(  a_{1}\right)
+\operatorname*{Wt}\left(  a_{2}\right)  + \cdots+\operatorname*{Wt}\left(
a_{p}\right)  $, so that%
\[
\operatorname*{Wt}\left(  a_{1}\right)  +\operatorname*{Wt}\left(
a_{2}\right)  + \cdots+\operatorname*{Wt}\left(  a_{p}\right)
=\operatorname*{Wt}\left(  \underbrace{a_{1}a_{2} \cdots a_{p}}_{=u}\right)
=\operatorname*{Wt}\left(  u\right)  .
\]
But the definition of $\mathbf{b}_{u}$ yields $\mathbf{b}_{u}=b_{a_{1}%
}b_{a_{2}} \cdots b_{a_{p}}$. Thus, the element $\mathbf{b}_{u}$ is
homogeneous of degree $\operatorname*{Wt}\left(  a_{1}\right)
+\operatorname*{Wt}\left(  a_{2}\right)  + \cdots+\operatorname*{Wt}\left(
a_{p}\right)  $ (since for every $w\in\mathfrak{L}$, the element $b_{w}$ of
$A$ is homogeneous of degree $\operatorname*{Wt}\left(  w\right)  $). Since
$\operatorname*{Wt}\left(  a_{1}\right)  +\operatorname*{Wt}\left(
a_{2}\right)  + \cdots+\operatorname*{Wt}\left(  a_{p}\right)
=\operatorname*{Wt}\left(  u\right)  $, this rewrites as follows: The element
$\mathbf{b}_{u}$ is homogeneous of degree $\operatorname*{Wt}\left(  u\right)
$, qed.}.

Now, let $n\in \NN$. It is easy to see that the family $\left(
\mathbf{b}_{u}\right)  _{u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)
}$ spans the $\kk$-module $A_{n}$ (that is, the $n$-th homogeneous
component of the $\kk$-module $A$)\ \ \ \ \footnote{\textit{Proof.} For
every $u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  $, the element
$\mathbf{b}_{u}$ is a homogeneous element of $A$ of degree $\operatorname*{Wt}%
\left(  u\right)  =n$ (since $u\in\operatorname*{Wt}\nolimits^{-1}\left(
n\right)  $). In other words, for every $u\in\operatorname*{Wt}\nolimits^{-1}%
\left(  n\right)  $, we have $\mathbf{b}_{u}\in A_{n}$.
\par
Let $\xi\in A_{n}$. Then, $\xi\in A_{n}\subset A$, so that $\xi$ is a
$\kk$-linear combination of the elements $\mathbf{b}_{u}$ for
$u\in\mathfrak{A}^{\ast}$ (since the family $\left(  \mathbf{b}_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ spans the $\kk$-module $A$). In other
words, $\xi=\sum_{u\in\mathfrak{A}^{\ast}}\lambda_{u}\mathbf{b}_{u}$ for some
family $\left(  \lambda_{u}\right)  _{u\in\mathfrak{A}^{\ast}}\in
 \kk ^{\mathfrak{A}^{\ast}}$ of elements of $\kk$ such that all
but finitely many $u\in\mathfrak{A}^{\ast}$ satisfy $\lambda_{u}=0$. Consider
this $\left(  \lambda_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$.
\par
What happens if we apply the canonical projection $A\rightarrow A_{n}$ (which
projects $A$ onto its $n$-th homogeneous component $A_{n}$, annihilating all
other components) to both sides of the equality $\xi=\sum_{u\in\mathfrak{A}%
^{\ast}}\lambda_{u}\mathbf{b}_{u}$ ? The left hand side remains $\xi$ (since
$\xi\in A_{n}$). On the right hand side, all addends of the sum in which the
$u$ satisfies $\operatorname*{Wt}\left(  u\right)  =n$ stay fixed (because we
know that for each such $u$, the element $\mathbf{b}_{u}$ is a homogeneous
element of $A$ of degree $\operatorname*{Wt}\left(  u\right)  =n$), whereas
all other addends become $0$ (for a similar reason); therefore, the sum on the
right hand side becomes $\sum\limits_{\substack{u\in\mathfrak{A}^{\ast}%
;\\\operatorname*{Wt}\left(  u\right)  =n}}\lambda_{u}\mathbf{b}_{u}$. Thus,
the equality becomes $\xi=\sum\limits_{\substack{u\in\mathfrak{A}^{\ast
};\\\operatorname*{Wt}\left(  u\right)  =n}}\lambda_{u}\mathbf{b}_{u}$. Thus,%
\[
\xi=\sum\limits_{\substack{u\in\mathfrak{A}^{\ast};\\\operatorname*{Wt}\left(
u\right)  =n}}\lambda_{u}\mathbf{b}_{u}=\sum_{u\in\operatorname*{Wt}%
\nolimits^{-1}\left(  n\right)  }\lambda_{u}\mathbf{b}_{u}.
\]
Hence, $\xi$ is a $\kk$-linear combination of the elements
$\mathbf{b}_{u}$ with $u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)
$.
\par
Now, let us forget that we fixed $\xi$. We thus have shown that every $\xi\in
A_{n}$ is a $\kk$-linear combination of the elements $\mathbf{b}_{u}$
with $u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  $. Since all these
elements $\mathbf{b}_{u}$ belong to $A_{n}$ (because for every $u\in
\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  $, we have $\mathbf{b}%
_{u}\in A_{n}$), this shows that the family $\left(  \mathbf{b}_{u}\right)
_{u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ spans the
$\kk$-module $A_{n}$, qed.}. The same argument (but with $\mathbf{b}%
_{u}$ replaced by $g_{u}$) shows that the family $\left(  g_{u}\right)
_{u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ spans the
$\kk$-module $A_{n}$. Since this family $\left(  g_{u}\right)
_{u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ is linearly
independent (because it is a subfamily of the basis $\left(  g_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ of $A$), this shows that the family $\left(
g_{u}\right)  _{u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ is a
basis of the $\kk$-module $A_{n}$.

But the set $\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  $ is
finite\footnote{\textit{Proof.} We have%
\begin{align*}
\operatorname*{Wt}\nolimits^{-1}\left(  n\right)   &  =\left\{  w\in
\mathfrak{A}^{\ast}\ \mid\ \operatorname*{Wt}\left(  w\right)  =n\right\} \\
&  =\bigcup_{k\in \NN }\left\{  \left(  w_{1},w_{2},\ldots,w_{k}\right)
\in\mathfrak{A}^{\ast}\ \mid\ \underbrace{\operatorname*{Wt}\left(  \left(
w_{1},w_{2},\ldots,w_{k}\right)  \right)  }_{\substack{=\operatorname*{wt}%
\left(  w_{1}\right)  +\operatorname*{wt}\left(  w_{2}\right)  +
\cdots+\operatorname*{wt}\left(  w_{k}\right)  \\\text{(by the definition of
}\operatorname*{Wt}\left(  \left(  w_{1},w_{2},\ldots,w_{k}\right)  \right)
\text{)}}}=n\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every word }w\in\mathfrak{A}^{\ast
}\text{ has the form }\left(  w_{1},w_{2},\ldots,w_{k}\right)  \text{ for some
}k\in \NN \right) \\
&  =\bigcup_{k\in \NN }\underbrace{\left\{  \left(  w_{1},w_{2}%
,\ldots,w_{k}\right)  \in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{wt}\left(
w_{1}\right)  +\operatorname*{wt}\left(  w_{2}\right)  + \cdots
+\operatorname*{wt}\left(  w_{k}\right)  =n\right\}  }_{=
\bigcup\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{k}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{k};\\i_{1}+i_{2}+ \cdots+i_{k}=n}}\left\{  \left(
w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{wt}\left(  w_{1}\right)  =i_{1},\ \operatorname*{wt}\left(
w_{2}\right)  =i_{2},\ \ldots,\ \operatorname*{wt}\left(  w_{k}\right)
=i_{k}\right\}  }\\
&  =\underbrace{\bigcup_{k\in \NN }\bigcup\limits_{\substack{\left(  i_{1}%
,i_{2},\ldots,i_{k}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{k}%
;\\i_{1}+i_{2}+ \cdots+i_{k}=n}}}_{=\bigcup_{\left(  i_{1},i_{2},\ldots
,i_{k}\right)  \in\Comp_{n}}}\underbrace{\left\{
\left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{wt}\left(  w_{1}\right)  =i_{1},\ \operatorname*{wt}\left(
w_{2}\right)  =i_{2},\ \ldots,\ \operatorname*{wt}\left(  w_{k}\right)
=i_{k}\right\}  }_{=\operatorname{wt}^{-1}\left(  i_{1}\right)
\times\operatorname{wt}^{-1}\left(  i_{2}\right)  \times\cdots
\times\operatorname{wt}^{-1}\left(  i_{k}\right)  }\\
&  =\bigcup_{\left(  i_{1},i_{2},\ldots,i_{k}\right)  \in
\Comp_{n}}\underbrace{\operatorname{wt}^{-1}\left(
i_{1}\right)  \times\operatorname{wt}^{-1}\left(  i_{2}\right)
\times\cdots\times\operatorname{wt}^{-1}\left(  i_{k}\right)
}_{\substack{\text{a finite set}\\\text{(since for every }N\in\left\{
1,2,3,\ldots\right\}  \text{, the set }\operatorname{wt}^{-1}\left(
N\right)  \text{ is finite)}}}.
\end{align*}
Hence, $\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  $ is a union of
finitely many finite sets, and thus itself finite, qed.}. Hence, $A_{n}$ is a
finite free $\kk$-module (since $\left(  g_{u}\right)  _{u\in
\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ is a basis of the
$\kk$-module $A_{n}$), and Exercise \ref{exe.linalg.sur-bij}(b)
(applied to $A_{n}$, $\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  $,
$u$, $g_{u}$ and $\mathbf{b}_{u}$ instead of $A$, $I$, $i$, $\gamma_{i}$ and
$\beta_{i}$) yields that $\left(  \mathbf{b}_{u}\right)  _{u\in
\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ is a $\kk$-basis
of $A_{n}$.

Now, let us forget that we fixed $n$. We thus have shown that for every
$n\in \NN$, the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\operatorname*{Wt}\nolimits^{-1}\left(  n\right)  }$ is a $\kk$-basis
of $A_{n}$. Hence, the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ (being the disjoint union of the families $\left(
\mathbf{b}_{u}\right)  _{u\in\operatorname*{Wt}\nolimits^{-1}\left(  n\right)
}$ over all $n\in \NN$) is a $\kk$-basis of $\bigoplus
_{n\in \NN }A_{n}=A$. In other words, the family $\left(  \mathbf{b}%
_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis of the $\kk$-module
$A$. As we know, this completes the proof of Lemma \ref{lem.lyndonbases}(c).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.shuffle.preserves-lex.pf}}

\textit{Solution to Exercise~\ref{exe.shuffle.preserves-lex.pf}.}

\begin{proof}
[Proof of Lemma \ref{lem.shuffle.preserves-lex}.](a) Let $u$, $v$ and
$v^{\prime}$ be three words satisfying $\ell\left(  u\right)  =n$,
$\ell\left(  v\right)  =m$, $\ell\left(  v^{\prime}\right)  =m$ and
$v^{\prime}<v$. We must show that $u\underset{\sigma}{\shuffle}v^{\prime
}<u\underset{\sigma}{\shuffle}v$.

We have $v^{\prime}<v$, thus $v^{\prime}\leq v$. By the definition of the
relation $\leq$, this means that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  v^{\prime}\right)  ,\ell\left(  v\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  v^{\prime}\right)
_{i}<v_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{
satisfies }\left(  v^{\prime}\right)  _{j}=v_{j}\right)  ,\\
&  \text{\textbf{or} the word }v^{\prime}\text{ is a prefix of }v.
\end{align*}
Since the word $v^{\prime}$ cannot be a prefix of $v$ (because this would
entail that $v^{\prime}=v$ (because $\ell\left(  v^{\prime}\right)
=m=\ell\left(  v\right)  $, so that $v^{\prime}$ has the same length as $v$),
which would contradict $v^{\prime}<v$), this shows that there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  v^{\prime}\right)
,\ell\left(  v\right)  \right\}  \right\}  $ such that $\left(  \left(
v^{\prime}\right)  _{i}<v_{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  v^{\prime}\right)  _{j}=v_{j}\right)
$. Denote this $i$ by $k$. Thus, $k\in\left\{  1,2,\ldots,\min\left\{
\ell\left(  v^{\prime}\right)  ,\ell\left(  v\right)  \right\}  \right\}  $
has the property that $\left(  v^{\prime}\right)  _{k}<v_{k}$, and
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,k-1\right\}  \text{ satisfies }\left(
v^{\prime}\right)  _{j}=v_{j}. \label{pf.lem.shuffle.preserves-lex.kj}%
\end{equation}
We have $k\in\left\{  1,2,\ldots,\min\left\{  \underbrace{\ell\left(
v^{\prime}\right)  }_{=m},\underbrace{\ell\left(  v\right)  }_{=m}\right\}
\right\}  =\left\{  1,2,\ldots,\underbrace{\min\left\{  m,m\right\}  }%
_{=m}\right\}  =\left\{  1,2,\ldots,m\right\}  $.

Now, let $\left(  w_{1},w_{2},\ldots,w_{n+m}\right)  $ denote the
concatenation $u\cdot v=\left(  u_{1},u_{2},\ldots,u_{n},v_{1},v_{2}%
,\ldots,v_{m}\right)  $, and let $\left(  w_{1}^{\prime},w_{2}^{\prime}%
,\ldots,w_{n+m}^{\prime}\right)  $ denote the concatenation $u\cdot v^{\prime
}=\left(  u_{1},u_{2},\ldots,u_{n},\left(  v^{\prime}\right)  _{1},\left(
v^{\prime}\right)  _{2},\ldots,\left(  v^{\prime}\right)  _{m}\right)  $. We
have $u\underset{\sigma}{\shuffle}v=\left(  w_{\sigma\left(  1\right)
},w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)  $
(by the definition of $u\underset{\sigma}{\shuffle}v$) and $u\underset{\sigma
}{\shuffle}v^{\prime}=\left(  w_{\sigma\left(  1\right)  }^{\prime}%
,w_{\sigma\left(  2\right)  }^{\prime},\ldots,w_{\sigma\left(  n+m\right)
}^{\prime}\right)  $ (by the definition of $u\underset{\sigma}{\shuffle}%
v^{\prime}$). Hence, we have $u\underset{\sigma}{\shuffle}v^{\prime}\neq
u\underset{\sigma}{\shuffle}v$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Then, $u\underset{\sigma}{\shuffle}v^{\prime}=u\underset{\sigma
}{\shuffle}v$, thus $\left(  w_{\sigma\left(  1\right)  }^{\prime}%
,w_{\sigma\left(  2\right)  }^{\prime},\ldots,w_{\sigma\left(  n+m\right)
}^{\prime}\right)  =u\underset{\sigma}{\shuffle}v^{\prime}=u\underset{\sigma
}{\shuffle}v=\left(  w_{\sigma\left(  1\right)  },w_{\sigma\left(  2\right)
},\ldots,w_{\sigma\left(  n+m\right)  }\right)  $. Hence, $\left(
w_{1}^{\prime},w_{2}^{\prime},\ldots,w_{n+m}^{\prime}\right)  =\left(
w_{1},w_{2},\ldots,w_{n+m}\right)  $ (since $\sigma$ is a permutation), and
thus $u\cdot v^{\prime}=\left(  w_{1}^{\prime},w_{2}^{\prime},\ldots
,w_{n+m}^{\prime}\right)  =\left(  w_{1},w_{2},\ldots,w_{n+m}\right)  =u\cdot
v$, so that $v^{\prime}=v$ (since $u$ can be cancelled from the equality
$u\cdot v^{\prime}=u\cdot v$). But this contradicts $v^{\prime}<v$. This
contradiction shows that our assumption was wrong, qed.}. But our goal is to
prove that $u\underset{\sigma}{\shuffle}v^{\prime}<u\underset{\sigma
}{\shuffle}v$. Hence, it remains to prove that $u\underset{\sigma
}{\shuffle}v^{\prime}\leq u\underset{\sigma}{\shuffle}v$ (since we have shown
that $u\underset{\sigma}{\shuffle}v^{\prime}\neq u\underset{\sigma
}{\shuffle}v$). Due to the definition of the relation $\leq$, this amounts to
proving that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\underset{\sigma}{\shuffle}v^{\prime}\right)
,\ell\left(  u\underset{\sigma}{\shuffle}v\right)  \right\}  \right\}  \text{
such that}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  _{i}<\left(  u\underset{\sigma}{\shuffle}v\right)  _{i}\text{, and
every }j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }\left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  _{j}=\left(  u\underset{\sigma
}{\shuffle}v\right)  _{j}\right)  ,\\
&  \text{\textbf{or} the word }u\underset{\sigma}{\shuffle}v^{\prime}\text{ is
a prefix of }u\underset{\sigma}{\shuffle}v.
\end{align*}
We shall prove that the first of these two alternatives holds, i.e., that
there exists an
\[
i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  ,\ell\left(  u\underset{\sigma
}{\shuffle}v\right)  \right\}  \right\}
\]
such that
\[
\left(  \left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  _{i}<\left(  u\underset{\sigma
}{\shuffle}v\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  _{j}=\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}\right)  .
\]

Indeed, we claim that $\sigma^{-1}\left(  n+k\right)  $ is such an $i$. In
order to conclude the proof, we then need to show that
\begin{equation}
\sigma^{-1}\left(  n+k\right)  \in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  ,\ell\left(  u\underset{\sigma
}{\shuffle}v\right)  \right\}  \right\}  ,
\label{pf.lem.shuffle.preserves-lex.goal1}%
\end{equation}
that we have
\begin{equation}
\left(  u\underset{\sigma}{\shuffle}v^{\prime}\right)  _{\sigma^{-1}\left(
n+k\right)  }<\left(  u\underset{\sigma}{\shuffle}v\right)  _{\sigma
^{-1}\left(  n+k\right)  }, \label{pf.lem.shuffle.preserves-lex.goal2}%
\end{equation}
and that
\begin{equation}
\text{every }j\in\left\{  1,2,\ldots,\sigma^{-1}\left(  n+k\right)
-1\right\}  \text{ satisfies }\left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  _{j}=\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}\text{.}
\label{pf.lem.shuffle.preserves-lex.goal3}%
\end{equation}


\textit{Proof of (\ref{pf.lem.shuffle.preserves-lex.goal1}):} We have
$\ell\left(  u\underset{\sigma}{\shuffle}v\right)  =\underbrace{\ell\left(
u\right)  }_{=n}+\underbrace{\ell\left(  v\right)  }_{=m}=n+m$ and similarly
$\ell\left(  u\underset{\sigma}{\shuffle}v^{\prime}\right)  =n+m$. Thus,
$\min\left\{  \underbrace{\ell\left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  }_{=n+m},\underbrace{\ell\left(  u\underset{\sigma}{\shuffle}%
v\right)  }_{=n+m}\right\}  =\min\left\{  n+m,n+m\right\}  =n+m$. But since
$\sigma\in\operatorname{Sh}_{n,m}\subset\Symm_{n+m}$, we have
$\sigma^{-1}\left(  n+k\right)  \in\left\{  1,2,\ldots,n+m\right\}  $. In
other words, $\sigma^{-1}\left(  n+k\right)  \in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\underset{\sigma}{\shuffle}v^{\prime}\right)
,\ell\left(  u\underset{\sigma}{\shuffle}v\right)  \right\}  \right\}  $
(since $\min\left\{  \ell\left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  ,\ell\left(  u\underset{\sigma}{\shuffle}v\right)  \right\}  =n+m$).
This proves (\ref{pf.lem.shuffle.preserves-lex.goal1}).

\textit{Proof of (\ref{pf.lem.shuffle.preserves-lex.goal2}):} Since
$u\underset{\sigma}{\shuffle}v=\left(  w_{\sigma\left(  1\right)  }%
,w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)  $,
we have $\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}=w_{\sigma\left(
j\right)  }$ for every $j\in\left\{  1,2,\ldots,n+m\right\}  $. Applying this
to $j=\sigma^{-1}\left(  n+k\right)  $, we obtain $\left(  u\underset{\sigma
}{\shuffle}v\right)  _{\sigma^{-1}\left(  n+k\right)  }=w_{\sigma\left(
\sigma^{-1}\left(  n+k\right)  \right)  }=w_{n+k}=v_{k}$ (since $k\in\left\{
1,2,\ldots,m\right\}  $ and $\left(  w_{1},w_{2},\ldots,w_{n+m}\right)
=\left(  u_{1},u_{2},\ldots,u_{n},v_{1},v_{2},\ldots,v_{m}\right)  $).
Similarly, $\left(  u\underset{\sigma}{\shuffle}v^{\prime}\right)
_{\sigma^{-1}\left(  n+k\right)  }=\left(  v^{\prime}\right)  _{k}$. Recalling
that $\left(  v^{\prime}\right)  _{k}<v_{k}$, we now see that $\left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  _{\sigma^{-1}\left(
n+k\right)  }=\left(  v^{\prime}\right)  _{k}<v_{k}=\left(  u\underset{\sigma
}{\shuffle}v\right)  _{\sigma^{-1}\left(  n+k\right)  }$. This proves
(\ref{pf.lem.shuffle.preserves-lex.goal2}).

\textit{Proof of (\ref{pf.lem.shuffle.preserves-lex.goal3}):} Let
$j\in\left\{  1,2,\ldots,\sigma^{-1}\left(  n+k\right)  -1\right\}  $. We need
to prove that $\left(  u\underset{\sigma}{\shuffle}v^{\prime}\right)
_{j}=\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}$.

Assume the contrary. Then, $\left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  _{j}\neq\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}$.

We have $j\in\left\{  1,2,\ldots,n+m\right\}  $ and $j<\sigma^{-1}\left(
n+k\right)  $ (since $j\in\left\{  1,2,\ldots,\sigma^{-1}\left(  n+k\right)
-1\right\}  $).

Since $u\underset{\sigma}{\shuffle}v=\left(  w_{\sigma\left(  1\right)
},w_{\sigma\left(  2\right)  },\ldots,w_{\sigma\left(  n+m\right)  }\right)
$, we have $\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}=w_{\sigma
\left(  j\right)  }$. Similarly, $\left(  u\underset{\sigma}{\shuffle}%
v^{\prime}\right)  _{j}=w_{\sigma\left(  j\right)  }^{\prime}$. Hence,
$w_{\sigma\left(  j\right)  }^{\prime}=\left(  u\underset{\sigma
}{\shuffle}v^{\prime}\right)  _{j}\neq\left(  u\underset{\sigma}{\shuffle}%
v\right)  _{j}=w_{\sigma\left(  j\right)  }$.

If we had $\sigma\left(  j\right)  \leq n$, then we would have $w_{\sigma
\left(  j\right)  }=u_{\sigma\left(  j\right)  }$ (since \newline
$\left(  w_{1}%
,w_{2},\ldots,w_{n+m}\right)  =\left(  u_{1},u_{2},\ldots,u_{n},v_{1}%
,v_{2},\ldots,v_{m}\right)  $) and $w_{\sigma\left(  j\right)  }^{\prime
}=u_{\sigma\left(  j\right)  }$ (for similar reasons), which would yield that
$w_{\sigma\left(  j\right)  }^{\prime}=u_{\sigma\left(  j\right)  }%
=w_{\sigma\left(  j\right)  }$, which would contradict $w_{\sigma\left(
j\right)  }^{\prime}\neq w_{\sigma\left(  j\right)  }$. Hence, we cannot have
$\sigma\left(  j\right)  \leq n$. We thus must have $\sigma\left(  j\right)
>n$, whence $\sigma\left(  j\right)  \in\left\{  n+1,n+2,\ldots,n+m\right\}
$. Consequently, $w_{\sigma\left(  j\right)  }=v_{\sigma\left(  j\right)  -n}$
(because $\left(  w_{1},w_{2},\ldots,w_{n+m}\right)  =\left(  u_{1}%
,u_{2},\ldots,u_{n},v_{1},v_{2},\ldots,v_{m}\right)  $) and $\left(
w^{\prime}\right)  _{\sigma\left(  j\right)  }=\left(  v^{\prime}\right)
_{\sigma\left(  j\right)  -n}$ (similarly).

If we had $\sigma\left(  j\right)  -n\in\left\{  1,2,\ldots,k-1\right\}  $,
then we would have $\left(  v^{\prime}\right)  _{\sigma\left(  j\right)
-n}=v_{\sigma\left(  j\right)  -n}$ (by (\ref{pf.lem.shuffle.preserves-lex.kj}%
), applied to $\sigma\left(  j\right)  -n$ instead of $j$). Hence, if we had
$\sigma\left(  j\right)  -n\in\left\{  1,2,\ldots,k-1\right\}  $, then we
would have $\left(  w^{\prime}\right)  _{\sigma\left(  j\right)  }=\left(
v^{\prime}\right)  _{\sigma\left(  j\right)  -n}=v_{\sigma\left(  j\right)
-n}=w_{\sigma\left(  j\right)  }$, which would contradict $w_{\sigma\left(
j\right)  }^{\prime}\neq w_{\sigma\left(  j\right)  }$. Hence, we cannot have
$\sigma\left(  j\right)  -n\in\left\{  1,2,\ldots,k-1\right\}  $. In other
words, we have $\sigma\left(  j\right)  -n\notin\left\{  1,2,\ldots
,k-1\right\}  $, so that $\sigma\left(  j\right)  -n\geq k$. Hence,
$\sigma\left(  j\right)  \geq n+k$.

We have $\sigma\in\operatorname{Sh}_{n,m}$, so that $\sigma
^{-1}\left(  1\right)  <\sigma^{-1}\left(  2\right)  <\cdots<\sigma
^{-1}\left(  n\right)  $ and $\sigma^{-1}\left(  n+1\right)  <\sigma
^{-1}\left(  n+2\right)  <\cdots<\sigma^{-1}\left(  n+m\right)  $. In
particular, the restriction of the map $\sigma^{-1}$ to the set $\left\{
n+1,n+2,\ldots,n+m\right\}  $ is strictly increasing (since $\sigma
^{-1}\left(  n+1\right)  <\sigma^{-1}\left(  n+2\right)  <\cdots<\sigma
^{-1}\left(  n+m\right)  $). Since $\sigma\left(  j\right)  $ and $n+k$ both
lie in this set $\left\{  n+1,n+2,\ldots,n+m\right\}  $, we thus have
$\sigma^{-1}\left(  \sigma\left(  j\right)  \right)  \geq\sigma^{-1}\left(
n+k\right)  $ (because $\sigma\left(  j\right)  \geq n+k$). Hence,
$j=\sigma^{-1}\left(  \sigma\left(  j\right)  \right)  \geq\sigma^{-1}\left(
n+k\right)  $, which contradicts $j<\sigma^{-1}\left(  n+k\right)  $. This
contradiction shows that our assumption was wrong. Hence,
(\ref{pf.lem.shuffle.preserves-lex.goal3}) is proven.

Now that (\ref{pf.lem.shuffle.preserves-lex.goal1}),
(\ref{pf.lem.shuffle.preserves-lex.goal2}) and
(\ref{pf.lem.shuffle.preserves-lex.goal3}) are all proven, we conclude that
there exists an
\[
i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  ,\ell\left(  u\underset{\sigma
}{\shuffle}v\right)  \right\}  \right\}
\]
such that
\[
\left(  \left(
u\underset{\sigma}{\shuffle}v^{\prime}\right)  _{i}<\left(  u\underset{\sigma
}{\shuffle}v\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  u\underset{\sigma}{\shuffle}v^{\prime
}\right)  _{j}=\left(  u\underset{\sigma}{\shuffle}v\right)  _{j}\right)
\]
(namely, $i=\sigma^{-1}\left(  n+k\right)  $). This concludes the proof of
Lemma \ref{lem.shuffle.preserves-lex}(a).

(b) The proof of Lemma \ref{lem.shuffle.preserves-lex}(b) is similar to the
proof of Lemma \ref{lem.shuffle.preserves-lex}(a) above. (One of the changes
necessary is to replace $\sigma^{-1}\left(  n+k\right)  $ by $\sigma
^{-1}\left(  k\right)  $.)

(Alternatively, it is not hard to derive Lemma
\ref{lem.shuffle.preserves-lex}(b) from Lemma \ref{lem.shuffle.preserves-lex}%
(a), because if $\tau\in\Symm_{n+m}$ denotes the permutation which is
written $\left(  m+1,m+2,\ldots,n+m,1,2,\ldots,m\right)  $ in one-line
notation, then the permutation $\tau\circ\sigma$ belongs to
$\operatorname{Sh}_{m,n}$ and satisfies $u\underset{\sigma
}{\shuffle}v=v\underset{\tau\circ\sigma}{\shuffle}u$ and $u^{\prime}
\underset{\sigma}{\shuffle}v=v\underset{\tau\circ\sigma}{\shuffle}u^{\prime}$, and
therefore we can obtain Lemma \ref{lem.shuffle.preserves-lex}(b) by applying
Lemma \ref{lem.shuffle.preserves-lex}(a) to $m$, $n$, $v$, $u$, $u^{\prime}$
and $\tau\circ\sigma$ instead of $n$, $m$, $u$, $v$, $v^{\prime}$ and $\sigma$.)

(c) Let $u$, $v$ and $v^{\prime}$ be three words satisfying $\ell\left(
u\right)  =n$, $\ell\left(  v\right)  =m$, $\ell\left(  v^{\prime}\right)  =m$
and $v^{\prime}\leq v$. We must show that $u\underset{\sigma}{\shuffle}%
v^{\prime}\leq u\underset{\sigma}{\shuffle}v$. This is obvious if $v^{\prime
}=v$ (in fact, if $v^{\prime}=v$, then $u\underset{\sigma}{\shuffle}v^{\prime
} = u\underset{\sigma}{\shuffle}v \leq u\underset{\sigma}{\shuffle}v$).
Hence, we can WLOG assume that
$v^{\prime}\neq v$. Assuming this, we immediately obtain $v^{\prime}<v$ (since
$v^{\prime}\neq v$ and $v^{\prime}\leq v$), so that $u\underset{\sigma
}{\shuffle}v^{\prime}<u\underset{\sigma}{\shuffle}v$ (by Lemma
\ref{lem.shuffle.preserves-lex}(a)). Thus, of course, $u\underset{\sigma
}{\shuffle}v^{\prime}\leq u\underset{\sigma}{\shuffle}v$, and Lemma
\ref{lem.shuffle.preserves-lex}(c) is proven.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.shuffle.radford.triang}}

\textit{Solution to Exercise \ref{exe.prop.shuffle.radford.triang}.}

\begin{proof}
[Proof of Proposition \ref{prop.shuffle.radford.triang}.]We shall prove
Proposition \ref{prop.shuffle.radford.triang} by strong induction over $\ell$.
So we fix some $L\in \NN$, and we assume that Proposition
\ref{prop.shuffle.radford.triang} holds whenever $\ell<L$. We now need to
prove that Proposition \ref{prop.shuffle.radford.triang} holds for $\ell=L$.
In other words, we need to prove that for every word $x\in\mathfrak{A}^{L}$,
there is a family $\left(  \eta_{x,y}\right)  _{y\in\mathfrak{A}^{L}}%
\in \NN ^{\mathfrak{A}^{L}}$ of elements of $\NN$ satisfying%
\begin{equation}
\mathbf{b}_{x}=\sum\limits_{\substack{y\in\mathfrak{A}^{L};\\y\leq x}}\eta_{x,y}b_{y}
\label{pf.prop.shuffle.radford.triang.goal}%
\end{equation}
and $\eta_{x,x}\neq0$ (in $\NN$).

Let $x\in\mathfrak{A}^{L}$ be a word. We need to prove that there is a family
$\left(  \eta_{x,y}\right)  _{y\in\mathfrak{A}^{L}}\in \NN %
^{\mathfrak{A}^{L}}$ of elements of $\NN$ satisfying
(\ref{pf.prop.shuffle.radford.triang.goal}) and $\eta_{x,x}\neq0$.

Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
$x$. Then, $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are Lyndon words satisfying
$x=a_{1}a_{2} \cdots a_{p}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$.

If $x$ is the empty word, then our claim is trivial (in fact, we can just set
$\eta_{x,x}=1$ in this case, and (\ref{pf.prop.shuffle.radford.triang.goal})
holds obviously). Hence, for the rest of this proof, we WLOG assume that $x$
is not the empty word. Thus, $p\neq0$ (because otherwise, $x=a_{1}a_{2} \cdots
a_{p}$ would be an empty product and thus the empty word, contradicting the
assumption that $x$ is not the empty word). Hence, we can define two words
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ by $u=a_{1}$ and
$v=a_{2}a_{3} \cdots a_{p}$. The word $u$ is Lyndon (since $u=a_{1}$ and since
$a_{1}$ is Lyndon), and thus has $\left(  u\right)  $ as its CFL
factorization. The CFL factorization of the word $v$ is $\left(  a_{2}%
,a_{3},\ldots,a_{p}\right)  $ (since the words $a_{2}$, $a_{3}$, $\ldots$,
$a_{p}$ are Lyndon and satisfy $v=a_{2}a_{3} \cdots a_{p}$ and $a_{2}\geq
a_{3}\geq\cdots\geq a_{p}$ (because $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$)).
Also, $u=a_{1}\geq a_{j+1}$ for every $i\in\left\{  1,2,\ldots,1\right\}  $
and $j\in\left\{  1,2,\ldots,p-1\right\}  $ (because $a_{1}\geq a_{2}%
\geq\cdots\geq a_{p}$). Hence, we can apply Theorem
\ref{thm.shuffle.lyndon.used}(c) to $1$, $\left(  u\right)  $, $p-1$ and
$\left(  a_{2},a_{3},\ldots,a_{p}\right)  $ instead of $p$, $\left(
a_{1},a_{2},\ldots,a_{p}\right)  $, $q$ and $\left(  b_{1},b_{2},\ldots
,b_{q}\right)  $. As a result, we conclude that the lexicographically highest
element of the multiset $u\shuffle v$ is $\underbrace{u}_{=a_{1}%
}\underbrace{v}_{=a_{2}a_{3} \cdots a_{p}}=a_{1}\left(  a_{2}a_{3} \cdots
a_{p}\right)  =a_{1}a_{2} \cdots a_{p}=x$.

But $\mathbf{b}_{u}=b_{u}$ (by the definition of $\mathbf{b}_{u}$, since $u$
has CFL factorization $\left(  u\right)  $), and $\mathbf{b}_{x}%
=\mathbf{b}_{u} \shufmult \mathbf{b}_{v}$%
\ \ \ \ \footnote{\textit{Proof.} Since the CFL factorization of $v$ is
$\left(  a_{2},a_{3},\ldots,a_{p}\right)  $, we have $\mathbf{b}_{v}=b_{a_{2}%
} \shufmult b_{a_{3}} \shufmult \cdots \shufmult%
b_{a_{p}}$ (by the definition of $\mathbf{b}_{v}$). But since the CFL
factorization of $x$ is $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $, we have
\begin{align*}
\mathbf{b}_{x}  &  =b_{a_{1}} \shufmult b_{a_{2}} \shufmult
\cdots \shufmult b_{a_{p}}
\ \ \ \ \ \ \ \ \ \ \left(\text{by the definition of }\mathbf{b}_{x}\right) \\
& =\underbrace{b_{a_{1}}}_{\substack{=b_{u}%
\\\text{(since }a_{1}=u\text{)}}} \shufmult \underbrace{\left(
b_{a_{2}} \shufmult b_{a_{3}} \shufmult \cdots
\shufmult b_{a_{p}}\right)  }_{=\mathbf{b}_{v}}
= \underbrace{b_{u}}_{\substack{=\mathbf{b}_{u}\\\text{(since }%
\mathbf{b}_{u}=b_{u}\text{)}}} \shufmult \mathbf{b}_{v}%
=\mathbf{b}_{u} \shufmult \mathbf{b}_{v},
\end{align*}
qed.}.

The word $u$ is Lyndon and thus nonempty, so that $\ell\left(  u\right)  >0$.
Now, $\ell\left(  \underbrace{x}_{=uv}\right)  =\ell\left(  uv\right)
=\underbrace{\ell\left(  u\right)  }_{>0}+\ell\left(  v\right)  >\ell\left(
v\right)  $, so that $\ell\left(  v\right)  <\ell\left(  x\right)  =L$ (since
$x\in\mathfrak{A}^{L}$). Hence, the induction hypothesis tells us that we can
apply Proposition \ref{prop.shuffle.radford.triang} to $\ell\left(  v\right)
$ and $v$ instead of $\ell$ and $x$ (since $v\in\mathfrak{A}^{\ell\left(
v\right)  }$). As a result, we see that there is a family $\left(  \eta
_{v,y}\right)  _{y\in\mathfrak{A}^{\ell\left(  v\right)  }}\in \NN %
^{\mathfrak{A}^{\ell\left(  v\right)  }}$ of elements of $\NN$
satisfying%
\[
\mathbf{b}_{v}=\sum\limits_{\substack{y\in\mathfrak{A}^{\ell\left(  v\right)
};\\y\leq v}}\eta_{v,y}b_{y}%
\]
and $\eta_{v,v}\neq0$ (in $\NN$). Consider this family $\left(
\eta_{v,y}\right)  _{y\in\mathfrak{A}^{\ell\left(  v\right)  }}$.

Now,%
\begin{align}
\mathbf{b}_{x}  &  =\underbrace{\mathbf{b}_{u}}_{=b_{u}} \shufmult%
\underbrace{\mathbf{b}_{v}}_{=\sum\limits_{\substack{y\in\mathfrak{A}^{\ell\left(
v\right)  };\\y\leq v}}\eta_{v,y}b_{y}}=b_{u} \shufmult \left(
\sum\limits_{\substack{y\in\mathfrak{A}^{\ell\left(  v\right)  };\\y\leq v}%
}\eta_{v,y}b_{y}\right)  =\sum\limits_{\substack{y\in\mathfrak{A}^{\ell\left(
v\right)  };\\y\leq v}}\eta_{v,y}\underbrace{b_{u} \shufmult b_{y}%
}_{\substack{=\sum_{\sigma\in\operatorname{Sh}_{\ell\left(
u\right)  ,\ell\left(  v\right)  }}b_{u\underset{\sigma}{\shuffle}%
y}\\\text{(by Remark \ref{rmk.shuffle.shufflealg}, applied to}\\y\text{, }%
\ell\left(  u\right)  \text{ and }\ell\left(  v\right)  \\\text{instead of
}v\text{, }n\text{ and }m\text{)}}}\nonumber\\
&  =\sum\limits_{\substack{y\in\mathfrak{A}^{\ell\left(  v\right)  };\\y\leq v}%
}\eta_{v,y}\sum_{\sigma\in\operatorname{Sh}_{\ell\left(  u\right)
,\ell\left(  v\right)  }}b_{u\underset{\sigma}{\shuffle}y}=\sum\limits
_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)  };\\z\leq v}}\eta
_{v,z}\sum_{\sigma\in\operatorname{Sh}_{\ell\left(  u\right)
,\ell\left(  v\right)  }}b_{u\underset{\sigma}{\shuffle}z}
\label{pf.prop.shuffle.radford.triang.bx}%
\end{align}
(here, we renamed the summation index $y$ as $z$).

Now, let $z\in\mathfrak{A}^{\ell\left(  v\right)  }$ and $\sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
}$ be such that $z\leq v$. We will prove that $u\underset{\sigma
}{\shuffle}z\in\mathfrak{A}^{L}$ and $u\underset{\sigma}{\shuffle}z\leq x$.
First of all, it is clear that $\ell\left(  u\underset{\sigma}{\shuffle}%
z\right)  =\ell\left(  u\right)  +\underbrace{\ell\left(  z\right)
}_{\substack{=\ell\left(  v\right)  \\\text{(since }z\in\mathfrak{A}%
^{\ell\left(  v\right)  }\text{)}}}=\ell\left(  u\right)  +\ell\left(
v\right)  =\ell\left(  \underbrace{uv}_{=x}\right)  =\ell\left(  x\right)
=L$, so that $u\underset{\sigma}{\shuffle}z\in\mathfrak{A}^{L}$. Applying
Lemma \ref{lem.shuffle.preserves-lex}(c) to $\ell\left(  u\right)  $,
$\ell\left(  v\right)  $ and $z$ instead of $n$, $m$ and $v^{\prime}$, we
obtain $u\underset{\sigma}{\shuffle}z\leq u\underset{\sigma}{\shuffle}v$. But
$u\underset{\sigma}{\shuffle}v$ is an element of the multiset $u\shuffle v$,
and thus is $\leq x$ (since the lexicographically highest element of the
multiset $u\shuffle v$ is $x$). Thus, $u\underset{\sigma}{\shuffle}v\leq x$,
so that $u\underset{\sigma}{\shuffle}z\leq u\underset{\sigma}{\shuffle}v\leq
x$.

Now, let us forget that we fixed $z$ and $\sigma$. We thus have shown that%
\begin{equation}
\text{any }z\in\mathfrak{A}^{\ell\left(  v\right)  }\text{ and }\sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
}\text{ satisfying }z\leq v\text{ satisfy }u\underset{\sigma}{\shuffle}%
z\in\mathfrak{A}^{L}\text{ and }u\underset{\sigma}{\shuffle}z\leq x.
\label{pf.prop.shuffle.radford.triang.leq}%
\end{equation}
A similar argument (but with some of the $\leq$ signs replaced by $<$ signs,
and with a reference to Lemma \ref{lem.shuffle.preserves-lex}(a) instead of a
reference to Lemma \ref{lem.shuffle.preserves-lex}(c)) shows that%
\begin{equation}
\text{any }z\in\mathfrak{A}^{\ell\left(  v\right)  }\text{ and }\sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
}\text{ satisfying }z<v\text{ satisfy }u\underset{\sigma}{\shuffle}%
z\in\mathfrak{A}^{L}\text{ and }u\underset{\sigma}{\shuffle}z<x.
\label{pf.prop.shuffle.radford.triang.less}%
\end{equation}


Now, (\ref{pf.prop.shuffle.radford.triang.bx}) becomes%
\begin{align}
\mathbf{b}_{x}  &  =\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)
};\\z\leq v}}\eta_{v,z}\underbrace{\sum_{\sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)  }}b_{u\underset{\sigma
}{\shuffle}z}}_{\substack{=\sum\limits_{\substack{y\in\mathfrak{A}^{L};\\y\leq
x}}\sum\limits_{\substack{\sigma\in\operatorname{Sh}_{\ell\left(  u\right)
,\ell\left(  v\right)  };\\u\underset{\sigma}{\shuffle}z=y}%
}b_{u\underset{\sigma}{\shuffle}z}\\\text{(due to
(\ref{pf.prop.shuffle.radford.triang.leq}))}}}=\sum\limits_{\substack{z\in
\mathfrak{A}^{\ell\left(  v\right)  };\\z\leq v}}\eta_{v,z}\sum\limits
_{\substack{y\in\mathfrak{A}^{L};\\y\leq x}}\sum\limits_{\substack{\sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
};\\u\underset{\sigma}{\shuffle}z=y}}b_{u\underset{\sigma}{\shuffle}%
z}\nonumber\\
&  =\sum\limits_{\substack{y\in\mathfrak{A}^{L};\\y\leq x}}\sum\limits_{\substack{z\in
\mathfrak{A}^{\ell\left(  v\right)  };\\z\leq v}}\eta_{v,z}\sum\limits
_{\substack{\sigma\in\operatorname{Sh}_{\ell\left(  u\right)
,\ell\left(  v\right)  };\\u\underset{\sigma}{\shuffle}z=y}%
}\underbrace{b_{u\underset{\sigma}{\shuffle}z}}_{\substack{=b_{y}%
\\\text{(since }u\underset{\sigma}{\shuffle}z=y\text{)}}}=\sum\limits_{\substack{y\in
\mathfrak{A}^{L};\\y\leq x}}\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(
v\right)  };\\z\leq v}}\eta_{v,z}\underbrace{\sum\limits_{\substack{\sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
};\\u\underset{\sigma}{\shuffle}z=y}}b_{y}}_{=\left\vert \left\{  \sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
}\ \mid\ u\underset{\sigma}{\shuffle}z=y\right\}  \right\vert \cdot b_{y}%
}\nonumber\\
&  =\sum\limits_{\substack{y\in\mathfrak{A}^{L};\\y\leq x}}\sum\limits_{\substack{z\in
\mathfrak{A}^{\ell\left(  v\right)  };\\z\leq v}}\eta_{v,z}\left\vert \left\{
\sigma\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(
v\right)  }\ \mid\ u\underset{\sigma}{\shuffle}z=y\right\}  \right\vert \cdot
b_{y}. \label{pf.prop.shuffle.radford.triang.bx.2}%
\end{align}


Recall that we must prove that there is a family $\left(  \eta_{x,y}\right)
_{y\in\mathfrak{A}^{L}}\in \NN ^{\mathfrak{A}^{L}}$ of elements of
$\NN$ satisfying (\ref{pf.prop.shuffle.radford.triang.goal}) and
$\eta_{x,x}\neq0$ (in $\NN$). In order to prove this, we define such a
family $\left(  \eta_{x,y}\right)  _{y\in\mathfrak{A}^{L}}\in \NN %
^{\mathfrak{A}^{L}}$ by setting
\[
\left(  \eta_{x,y}=\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)
};\\z\leq v}}\eta_{v,z}\left\vert \left\{  \sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)  }\ \mid
\ u\underset{\sigma}{\shuffle}z=y\right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }y\in\mathfrak{A}^{L}\right)  .
\]
Then, (\ref{pf.prop.shuffle.radford.triang.bx.2}) shows that this family
satisfies (\ref{pf.prop.shuffle.radford.triang.goal}). All that remains to be
proven is now to show that $\eta_{x,x}\neq0$ (in $\NN$).

Indeed, the definition of $\eta_{x,x}$ yields%
\begin{align*}
\eta_{x,x}  &  =\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)
};\\z\leq v}}\eta_{v,z}\left\vert \left\{  \sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)  }\ \mid
\ u\underset{\sigma}{\shuffle}z=x\right\}  \right\vert \\
&  =\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)  };\\z<v}%
}\eta_{v,z}\left\vert \underbrace{\left\{  \sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)  }\ \mid
\ u\underset{\sigma}{\shuffle}z=x\right\}  }_{\substack{=\varnothing
\\\text{(since every }\sigma\in\operatorname{Sh}_{\ell\left(
u\right)  ,\ell\left(  v\right)  }\text{ satisfies }u\underset{\sigma
}{\shuffle}z<x\\\text{(by (\ref{pf.prop.shuffle.radford.triang.less})),
so that it cannot satisfy } u\underset{\sigma}{\shuffle}z = x\text{)}
}}\right\vert +\eta_{v,v}\left\vert \left\{  \sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)  }\ \mid
\ u\underset{\sigma}{\shuffle}v=x\right\}  \right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}z=v\text{ from the sum}\right) \\
&  =\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)  };\\z<v}%
}\eta_{v,z}\underbrace{\left\vert \varnothing\right\vert }_{=0}+\eta
_{v,v}\left\vert \left\{  \sigma\in\operatorname{Sh}_{\ell\left(
u\right)  ,\ell\left(  v\right)  }\ \mid\ u\underset{\sigma}{\shuffle}%
v=x\right\}  \right\vert \\
&  =\underbrace{\sum\limits_{\substack{z\in\mathfrak{A}^{\ell\left(  v\right)
};\\z<v}}\eta_{v,z}0}_{=0}+\eta_{v,v}\left\vert \left\{  \sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
}\ \mid\ u\underset{\sigma}{\shuffle}v=x\right\}  \right\vert \\
&  =\underbrace{\eta_{v,v}}_{\neq0\text{ (in } \NN \text{)}%
}\underbrace{\left\vert \left\{  \sigma\in\operatorname{Sh}_{\ell
\left(  u\right)  ,\ell\left(  v\right)  }\ \mid\ u\underset{\sigma
}{\shuffle}v=x\right\}  \right\vert }_{\substack{\neq0\text{ (in }%
 \NN \text{)}\\\text{(since there exists some }\sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v\right)
}\text{ satisfying }u\underset{\sigma}{\shuffle}v=x\\\text{(because }x=uv\in
u\shuffle v\text{))}}}\\
&  \neq0\text{ (in } \NN \text{).}%
\end{align*}
This completes the proof that there is a family $\left(  \eta_{x,y}\right)
_{y\in\mathfrak{A}^{L}}\in \NN ^{\mathfrak{A}^{L}}$ of elements of
$\NN$ satisfying (\ref{pf.prop.shuffle.radford.triang.goal}) and
$\eta_{x,x}\neq0$ (in $\NN$). The induction step is thus complete, and
Proposition \ref{prop.shuffle.radford.triang} is proven by induction.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.thm.shuffle.radford}}

\textit{Solution to Exercise \ref{exe.thm.shuffle.radford}.}

\begin{proof}
[Proof of Theorem \ref{thm.shuffle.radford}.]It is clearly enough to prove
that $\left(  b_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically
independent generating set of the $\kk$-algebra $\operatorname*{Sh}%
\left(  V\right)  $.

For every word $u\in\mathfrak{A}^{\ast}$, define an element $\mathbf{b}_{u}$
by $\mathbf{b}_{u}=b_{a_{1}} \shufmult b_{a_{2}} \shufmult
\cdots \shufmult b_{a_{p}}$, where $\left(  a_{1},a_{2},\ldots
,a_{p}\right)  $ is the CFL factorization of $u$. According to Lemma
\ref{lem.lyndonbases}(a) (applied to $A=\operatorname*{Sh}\left(  V\right)
$)\ \ \ \ \footnote{Don't be confused by the fact that the multiplication of
the $\kk$-algebra $\operatorname*{Sh}\left(  V\right)  $ is denoted by
$\shufmult$, but the multiplication of the $\kk$-algebra $A$
is denoted by $\cdot$ in Lemma \ref{lem.lyndonbases}(a).}, the family $\left(
b_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent generating
set of the $\kk$-algebra $\operatorname*{Sh}\left(  V\right)  $ if and
only if the family $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$
is a basis of the $\kk$-module $\operatorname*{Sh}\left(  V\right)  $.
In order to prove the former statement (which is our goal), it is therefore
enough to prove the latter statement.

So we must prove the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ is a basis of the $\kk$-module
$\operatorname*{Sh}\left(  V\right)  $. We shall first prove a particular case
of this statement:

\begin{statement}
\textit{Assertion A:} If the set $\mathfrak{A}$ is finite, then the family
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis of the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $.
\end{statement}

\textit{Proof of Assertion A:} Assume that the set $\mathfrak{A}$ is finite.
Fix $\ell\in \NN$. Regard $V^{\otimes\ell}$ as a $\kk$-submodule
of $T\left(  V\right)  $. Then, $\left(  b_{u_{1}}\otimes b_{u_{2}}%
\otimes\cdots\otimes b_{u_{\ell}}\right)  _{u\in\mathfrak{A}^{\ell}}$ is a
basis of the $\kk$-module $V^{\otimes\ell}$ (since $\left(
b_{a}\right)  _{a\in\mathfrak{A}}$ is a basis of the $\kk$-module $V$).
In other words,
\begin{equation}
\left(  b_{u}\right)  _{u\in\mathfrak{A}^{\ell}}\text{ is a basis of the
} \kk \text{-module }V^{\otimes\ell}
\label{pf.thm.shuffle.radford.b-l-basis}
\end{equation}
(since $b_{u}=b_{u_{1}}b_{u_{2}} \cdots b_{u_{\ell}}=b_{u_{1}}\otimes
b_{u_{2}}\otimes\cdots\otimes b_{u_{\ell}}$ for every $u\in\mathfrak{A}^{\ell
}$).

We know that
$\QQ$ is a subring of $\kk$. Hence, every nonzero element of $\NN$
is an invertible element of $\kk$.

Now, consider the set $\mathfrak{A}^\ell$ as a poset, whose smaller
relation is $\leq$. (This poset is actually totally ordered, though
we will not need this.) The notion of an invertibly-triangular
$\mathfrak{A}^\ell \times \mathfrak{A}^\ell$-matrix is thus defined
(according to Definition~\ref{def.STmat.tria}(c)).

The set $\mathfrak{A}^\ell$ is finite (since $\mathfrak{A}$ is finite);
thus, it is a finite poset.

According to Proposition~\ref{prop.shuffle.radford.triang}, the family
$\left( \mathbf{b}_u \right) _{u \in \mathfrak{A}^{\ell}}$
expands invertibly triangularly in the basis
$\left( b_u \right) _{u \in \mathfrak{A}^{\ell}}$
\ \ \ \ \footnote{\textit{Proof.}
Proposition~\ref{prop.shuffle.radford.triang} shows that, for every
$x \in \mathfrak{A}^\ell$, there exists a family
$\left(\eta_{x,y}\right)_{y \in \mathfrak{A}^\ell} \in \NN^{\mathfrak{A}^\ell}$
of elements of $\NN$ such that
\begin{equation}
\mathbf{b}_x
= \sum\limits_{\substack{y \in \mathfrak{A}^\ell; \\ y \leq x}} \eta_{x,y} b_y
\label{pf.thm.shuffle.radford.fn1.1}
\end{equation}
and
\begin{equation}
\eta_{x,x}\neq 0 \qquad \text{(in }\NN \text{)} .
\label{pf.thm.shuffle.radford.fn1.2}
\end{equation}
Consider such a family
$\left(\eta_{x,y}\right)_{y \in \mathfrak{A}^\ell} \in \NN^{\mathfrak{A}^\ell}$ for each
$x \in \mathfrak{A}^\ell$. Thus, an integer $\eta_{x,y} \in \NN \subset
\QQ \subset \kk$ is defined for each
$\left(x, y\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell$.

We observe that the only elements $\eta_{s, t}$
(with $\left(s, t\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell$)
appearing
in the statements \eqref{pf.thm.shuffle.radford.fn1.1} and
\eqref{pf.thm.shuffle.radford.fn1.2} are those which
satisfy $t \leq s$. Hence,
if some $\left(s, t\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell$
does not satisfy
$t \leq s$, then the
corresponding element $\eta_{s, t}$ does not appear in any of
the statements \eqref{pf.thm.shuffle.radford.fn1.1} and
\eqref{pf.thm.shuffle.radford.fn1.2}; as a consequence,
we can arbitrarily change the value of this $\eta_{s, t}$ without
running the risk of invalidating
\eqref{pf.thm.shuffle.radford.fn1.1} and
\eqref{pf.thm.shuffle.radford.fn1.2}. Hence, we can
WLOG assume that
\begin{equation}
\text{every } \left(s, t\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell
\text{ which does not satisfy }
t \leq s
\text{ must satisfy }
\eta_{s,t} = 0
\label{pf.thm.shuffle.radford.fn1.3}
\end{equation}
(otherwise, we can just set all such $\eta_{s,t}$
to $0$). Assume this. Thus, the matrix
$\left(\eta_{x,y}\right)_{\left(x, y\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell}$
is triangular. The diagonal entries $\eta_{x,x}$
of this matrix are nonzero elements of $\NN$
(because of \eqref{pf.thm.shuffle.radford.fn1.2}) and
therefore invertible elements of $\kk$ (since
every nonzero element of $\NN$ is an invertible element of $\kk$).
Thus, the matrix
$\left(\eta_{x,y}\right)_{\left(x, y\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell}$
(regarded as a matrix in $\kk^{\mathfrak{A}^\ell \times \mathfrak{A}^\ell}$)
is invertibly triangular.

Now, every $x \in \mathfrak{A}^\ell$ satisfies
\begin{align*}
\sum_{y \in \mathfrak{A}^\ell} \eta_{x,y} b_y
&= \underbrace{\sum\limits_{\substack{y \in \mathfrak{A}^\ell; \\ y \leq x}} \eta_{x,y} b_y}_{\substack{= \mathbf{b}_x \\ \text{(by \eqref{pf.thm.shuffle.radford.fn1.1})}}}
+ \sum\limits_{\substack{y \in \mathfrak{A}^\ell; \\ \text{not } y \leq x}} \underbrace{\eta_{x,y}}_{\substack{=0 \\ \text{(by \eqref{pf.thm.shuffle.radford.fn1.3}, applied to } \left(s, t\right) = \left(x, y\right) \text{)}}} b_y
= \mathbf{b}_x + \underbrace{ \sum\limits_{\substack{y \in \mathfrak{A}^\ell; \\ \text{not } y \leq x}} 0 b_y}_{=0}
= \mathbf{b}_x .
\end{align*}
In other words, every $x \in \mathfrak{A}^\ell$ satisfies
$\mathbf{b}_x = \sum_{y \in \mathfrak{A}^\ell} \eta_{x,y} b_y$.
In other words, the family
$\left( \mathbf{b}_u \right) _{u \in \mathfrak{A}^\ell}$ expands
in the family
$\left( b_u \right) _{u \in \mathfrak{A}^\ell}$
through the matrix
$\left(\eta_{x,y}\right)_{\left(x, y\right) \in \mathfrak{A}^\ell \times \mathfrak{A}^\ell}$.
Since the latter matrix is invertibly triangular,
we thus conclude that the family
$\left( \mathbf{b}_u \right) _{u \in \mathfrak{A}^\ell}$ expands
invertibly triangularly in the family
$\left( b_u \right) _{u \in \mathfrak{A}^\ell}$.}.
Hence,
Corollary~\ref{cor.STmat.expansion-tria-inv}(e) (applied to
$V^{\otimes \ell}$, $\mathfrak{A}^\ell$,
$\left( \mathbf{b}_u \right) _{u \in \mathfrak{A}^\ell}$
and $\left( b_u \right) _{u \in \mathfrak{A}^\ell}$ instead of $M$, $S$,
$\left(e_s\right)_{s \in S}$ and $\left(f_s\right)_{s \in S}$)
yields that the family
$\left( \mathbf{b}_u \right) _{u \in \mathfrak{A}^\ell}$
is a basis of the $\kk$-module $V^{\otimes \ell}$ if and only if the
family
$\left( b_u \right) _{u \in \mathfrak{A}^\ell}$ is a basis of the
$\kk$-module $V^{\otimes \ell}$\ \ \ \ \footnote{Here, we have used the
fact that the set $\mathfrak{A}^\ell$ is finite.}. Hence, the family
$\left( \mathbf{b}_u \right) _{u \in \mathfrak{A}^\ell}$
is a basis of the $\kk$-module $V^{\otimes \ell}$ (since the family
$\left( b_u \right) _{u \in \mathfrak{A}^\ell}$
is a basis of the $\kk$-module $V^{\otimes \ell}$).

Now, let us forget that we fixed $\ell$. We thus have shown that, for every
$\ell\in \NN$, the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ell}}$ is a basis of the $\kk$-module $V^{\otimes\ell}$.
Hence, the disjoint union of the families $\left(  \mathbf{b}_{u}\right)
_{u\in\mathfrak{A}^{\ell}}$ over all $\ell\in \NN$ is a basis of the
direct sum $\bigoplus_{\ell\in \NN }V^{\otimes\ell}$. Since the former
disjoint union is the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$, while the latter direct sum is the $\kk$-module
$\bigoplus_{\ell\in \NN }V^{\otimes\ell}=T\left(  V\right)
=\operatorname*{Sh}\left(  V\right)  $, this rewrites as follows: The family
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis of the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $. This proves
Assertion A.

Now, we need to prove that $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ is a basis of the $\kk$-module
$\operatorname*{Sh}\left(  V\right)  $, without the assumption that
$\mathfrak{A}$ be finite. We will reduce this to Assertion A. First, we need
some preparations:

Let $\mathfrak{B}$ be any subset of $\mathfrak{A}$. Then, $\mathfrak{B}$
canonically becomes a totally ordered set (since $\mathfrak{A}$ is totally
ordered), so that the notion of a Lyndon word over $\mathfrak{B}$ is
well-defined. We view $\mathfrak{B}^{\ast}$ as a subset of $\mathfrak{A}%
^{\ast}$, and so the Lyndon words over $\mathfrak{B}$ are precisely the Lyndon
words over $\mathfrak{A}$ which lie in $\mathfrak{B}^{\ast}$.

We define a $\kk$-submodule $V_{\mathfrak{B}}$ of $V$ as the
$\kk$-linear span of the family $\left(  b_{a}\right)  _{a\in
\mathfrak{B}}$. Notice that the family $\left(  b_{a}\right)  _{a\in
\mathfrak{B}}$ is $\kk$-linearly independent (being a subfamily of the
basis $\left(  b_{a}\right)  _{a\in\mathfrak{A}}$ of $V$), and thus is a basis
of the $\kk$-submodule $V_{\mathfrak{B}}$.

The inclusion $V_{\mathfrak{B}}\rightarrow V$ gives rise to an injective
$\kk$-algebra homomorphism $\operatorname*{Sh}\left(  V_{\mathfrak{B}%
}\right)  \rightarrow\operatorname*{Sh}\left(  V\right)  $, which sends every
$b_{w}$ and every $\mathbf{b}_{u}$ (for $w\in\mathfrak{B}^{\ast}$ and
$u\in\mathfrak{B}^{\ast}$, respectively) to the corresponding elements $b_{w}$
and $\mathbf{b}_{u}$ of $\operatorname*{Sh}\left(  V\right)  $, respectively.
We regard this homomorphism as an inclusion, so that $\operatorname*{Sh}%
\left(  V_{\mathfrak{B}}\right)  $ is a $\kk$-subalgebra of
$\operatorname*{Sh}\left(  V\right)  $.

If $\mathfrak{B}$ is finite, then Assertion A (applied to $\mathfrak{B}$
instead of $\mathfrak{A}$)
yields that the family $\left(  \mathbf{b}_{u}\right)
_{u\in\mathfrak{B}^{\ast}}$ is a basis of the $\kk$-module
$\operatorname*{Sh}\left(  V_{\mathfrak{B}}\right)  $.

Now, let us forget that we fixed $\mathfrak{B}$. We thus have shown that for
every finite subset $\mathfrak{B}$ of $\mathfrak{A}$,%
\begin{equation}
\text{the family }\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{B}^{\ast}%
}\text{ is a basis of the } \kk \text{-module }\operatorname*{Sh}\left(
V_{\mathfrak{B}}\right)  \text{.} \label{pf.thm.shuffle.radford.B-basis}%
\end{equation}


Let us introduce one more notation: A family of elements of a
$\kk$-module is said to be \emph{finitely supported} if
all but finitely many elements of this family are $0$.

Now, let us show that the family $\left(  \mathbf{b}_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ is $\kk$-linearly independent and spans the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $.

\textit{Proof that the family }$\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ \textit{is }$\kk$\textit{-linearly independent:}
Let $\left(  \lambda_{u}\right)  _{u\in\mathfrak{A}^{\ast}}\in \kk %
^{\mathfrak{A}^{\ast}}$ be a finitely supported family of elements of
$\kk$ satisfying $\sum_{u\in\mathfrak{A}^{\ast}}\lambda_{u}%
\mathbf{b}_{u}=0$. We are going to prove that all $u\in\mathfrak{A}^{\ast}$
satisfy $\lambda_{u}=0$.

Indeed, the family $\left(  \lambda_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is
finitely supported, so that there exists a finite subset $Z$ of $\mathfrak{A}%
^{\ast}$ such that all $u\in\mathfrak{A}^{\ast}\setminus Z$ satisfy
$\lambda_{u}=0$. Consider this $Z$. Since $Z$ is finite, there exists a finite
subset $\mathfrak{B}$ of $\mathfrak{A}$ satisfying $Z\subset\mathfrak{B}%
^{\ast}$ (in fact, we can take $\mathfrak{B}$
to be the set of all letters occurring in
the words lying in $Z$). Consider this $\mathfrak{B}$. The family $\left(
\mathbf{b}_{u}\right)  _{u\in\mathfrak{B}^{\ast}}$ is a basis of the
$\kk$-module $\operatorname*{Sh}\left(  V_{\mathfrak{B}}\right)  $ (by
(\ref{pf.thm.shuffle.radford.B-basis})), and thus $\kk$-linearly
independent. We have $\mathfrak{A}^{\ast}\setminus\mathfrak{B}^{\ast}%
\subset\mathfrak{A}^{\ast}\setminus Z$ (since $Z\subset\mathfrak{B}^{\ast}$),
and therefore all $u\in\mathfrak{A}^{\ast}\setminus\mathfrak{B}^{\ast}$
satisfy $\lambda_{u}=0$ (since all
$u\in\mathfrak{A}^{\ast}\setminus Z$ satisfy
$\lambda_{u}=0$). Hence, $\sum_{u\in\mathfrak{A}^{\ast}\setminus
\mathfrak{B}^{\ast}}\underbrace{\lambda_{u}}_{=0}\mathbf{b}_{u}=\sum
_{u\in\mathfrak{A}^{\ast}\setminus\mathfrak{B}^{\ast}}0\mathbf{b}_{u}=0$. Now,%
\begin{align*}
0 &= \sum_{u\in\mathfrak{A}^{\ast}}\lambda_{u}\mathbf{b}_{u}=\sum_{u\in
\mathfrak{B}^{\ast}}\lambda_{u}\mathbf{b}_{u}+\underbrace{\sum_{u\in
\mathfrak{A}^{\ast}\setminus\mathfrak{B}^{\ast}}\lambda_{u}\mathbf{b}_{u}%
}_{=0} \ \ \ \ \ \ \ \ \ \ \left(\text{since } \mathfrak{B}^\ast
\text{ is a subset of } \mathfrak{A}^\ast \right)
\\
&= \sum_{u\in\mathfrak{B}^{\ast}}\lambda_{u}\mathbf{b}_{u}.
\end{align*}
So we have $\sum_{u\in\mathfrak{B}^{\ast}}\lambda_{u}\mathbf{b}_{u}=0$. Thus,
all $u\in\mathfrak{B}^{\ast}$ satisfy $\lambda_{u}=0$ (since the family
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{B}^{\ast}}$ is $\kk %
$-linearly independent). Combining this with the fact that all $u\in
\mathfrak{A}^{\ast}\setminus\mathfrak{B}^{\ast}$ satisfy $\lambda_{u}=0$, we
conclude that all $u\in\mathfrak{A}^{\ast}$ satisfy $\lambda_{u}=0$.

Now forget that we fixed $\left(  \lambda_{u}\right)  _{u\in\mathfrak{A}%
^{\ast}}$. We thus have shown that if $\left(  \lambda_{u}\right)
_{u\in\mathfrak{A}^{\ast}}\in \kk ^{\mathfrak{A}^{\ast}}$ is a finitely
supported family of elements of $\kk$ satisfying $\sum_{u\in
\mathfrak{A}^{\ast}}\lambda_{u}\mathbf{b}_{u}=0$, then all $u\in
\mathfrak{A}^{\ast}$ satisfy $\lambda_{u}=0$. In other words, the family
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is $\kk %
$-linearly independent.

\textit{Proof that the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{A}^{\ast}}$ spans the $\kk$-module
$\operatorname*{Sh}\left(  V\right)  $:} We are going to show that
the family $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ spans
the $\kk$-module $\operatorname*{Sh}\left(  V\right)  $. In order to
prove this, it is enough to show that $b_{w}$ lies in the $\kk$-linear
span of the family $\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$
for every $w\in\mathfrak{A}^{\ast}$ (because the family $\left(  b_{w}\right)
_{w\in\mathfrak{A}^{\ast}}$ is a basis of the $\kk$-module
$\operatorname*{Sh}\left(  V\right)  $\ \ \ \ \footnote{\textit{Proof.} For
every $\ell\in \NN$, the family $\left(  b_{u}\right)  _{u\in
\mathfrak{A}^{\ell}}$ is a basis of the $\kk$-module $V^{\otimes\ell}$
(by (\ref{pf.thm.shuffle.radford.b-l-basis})). Hence, the disjoint union of
the families $\left(  b_{u}\right)  _{u\in\mathfrak{A}^{\ell}}$ over all
$\ell\in \NN$ is a basis of the direct sum $\bigoplus_{\ell\in
 \NN }V^{\otimes\ell}$. Since the former disjoint union is the family
$\left(  b_{u}\right)  _{u\in\mathfrak{A}^{\ast}}=\left(  b_{w}\right)
_{w\in\mathfrak{A}^{\ast}}$, whereas the latter direct sum is the $\kk %
$-module $\bigoplus_{\ell\in \NN }V^{\otimes\ell}=T\left(  V\right)
=\operatorname*{Sh}\left(  V\right)  $, this rewrites as follows: The family
$\left(  b_{w}\right)  _{w\in\mathfrak{A}^{\ast}}$ is a basis of the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $, qed.}). So let us
show this now.

Let $w\in\mathfrak{A}^{\ast}$. Then, there exists a finite subset
$\mathfrak{B}$ of $\mathfrak{A}$ such that $w\in\mathfrak{B}^{\ast}$ (namely,
we can take $\mathfrak{B}$ to be the set of all letters of $w$). Consider this
$\mathfrak{B}$. The family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{B}^{\ast}}$ is a basis of the $\kk$-module
$\operatorname*{Sh}\left(  V_{\mathfrak{B}}\right)  $ (by
(\ref{pf.thm.shuffle.radford.B-basis})), and thus spans this $\kk %
$-module. But $b_{w}\in\operatorname*{Sh}\left(  V_{\mathfrak{B}}\right)  $
(since $w\in\mathfrak{B}^{\ast}$), and thus $b_{w}$ lies in the $\kk %
$-linear span of the family $\left(  \mathbf{b}_{u}\right)  _{u\in
\mathfrak{B}^{\ast}}$ (since this family is a basis of the $\kk$-module
$\operatorname*{Sh}\left(  V_{\mathfrak{B}}\right)  $). Hence, $b_{w}$ lies in
the $\kk$-linear span of the family $\left(  \mathbf{b}_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ as well (since this family $\left(  \mathbf{b}%
_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ includes the family $\left(
\mathbf{b}_{u}\right)  _{u\in\mathfrak{B}^{\ast}}$ as a subfamily). Thus, we
have proven that $b_{w}$ lies in the $\kk$-linear span of the family
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ for every
$w\in\mathfrak{A}^{\ast}$. This completes the proof that the family $\left(
\mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ spans the $\kk %
$-module $\operatorname*{Sh}\left(  V\right)  $.

Altogether, we now know that the family $\left(  \mathbf{b}_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ is $\kk$-linearly independent and spans the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $. In other words,
$\left(  \mathbf{b}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis of the
$\kk$-module $\operatorname*{Sh}\left(  V\right)  $. This completes our
proof of Theorem \ref{thm.shuffle.radford}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.words.lyndon.compositions.count}}

\textit{Solution to Exercise \ref{exe.words.lyndon.compositions.count}.}
We shall give two solutions to this exercise; but they both rest on the
following lemmas:

\begin{lemma}
\label{lem.sol.words.lyndon.compositions.count.moebinv}For every positive
integer $N$, we have%
\begin{equation}
\sum_{d\mid N}\mu\left(  d\right)  =\delta_{N,1}.
\label{sol.words.lyndon.compositions.count.mu}
\end{equation}

\end{lemma}

Lemma \ref{lem.sol.words.lyndon.compositions.count.moebinv} is one of the most
fundamental properties of the number-theoretic M\"{o}bius function; it will
not be proven here.\footnote{For a proof of
Lemma \ref{lem.sol.words.lyndon.compositions.count.moebinv}, see the solution of
Exercise~\ref{exe.witt.ghost-equiv}. (More precisely,
Lemma \ref{lem.sol.words.lyndon.compositions.count.moebinv} is obtained
from \eqref{sol.exe.witt.ghost-equiv.mu*1} by renaming $n$
as $N$.)}

\begin{lemma}
\label{lem.sol.words.lyndon.compositions.count.2}Every positive integer $n$
satisfies%
\[
\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  \left(  2^{n/d}%
-1\right)  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)
2^{n/d}-\delta_{n,1}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.words.lyndon.compositions.count.2}.]Let $n$ be a
positive integer. Then,%
\begin{align*}
&  \dfrac{1}{n}\underbrace{\sum\limits_{d\mid n}\mu\left(  d\right)  \left(
2^{n/d}-1\right)  }_{=\sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}%
-\sum\limits_{d\mid n}\mu\left(  d\right)  1}\\
&  =\dfrac{1}{n}\left(  \sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}%
-\sum\limits_{d\mid n}\mu\left(  d\right)  1\right)  =\dfrac{1}{n}\left(
\sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}-\sum\limits_{d\mid n}%
\mu\left(  d\right)  \right) \\
&  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}-\dfrac{1}%
{n}\underbrace{\sum\limits_{d\mid n}\mu\left(  d\right)  }_{\substack{=\delta
_{n,1}\\\text{(by (\ref{sol.words.lyndon.compositions.count.mu}),}%
\\\text{applied to }N=n\text{)}}}=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  2^{n/d}-\underbrace{\dfrac{1}{n}\delta_{n,1}}_{=\delta_{n,1}}%
=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}-\delta_{n,1}.
\end{align*}
This proves Lemma \ref{lem.sol.words.lyndon.compositions.count.2}.
\end{proof}

\textit{First solution to Exercise \ref{exe.words.lyndon.compositions.count}.}
[The following solution is similar to the solution of Exercise
\ref{exe.words.lyndon.count}.]

For every positive integer $n$, let $\operatorname*{lync}n$ denote the number
of Lyndon compositions of size $n$. We need to prove that%
\begin{equation}
\operatorname*{lync}n=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)
\left(  2^{n/d}-1\right)  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  2^{n/d}-\delta_{n,1}
\label{sol.words.lyndon.compositions.count.goal}%
\end{equation}
for every positive integer $n$.

We recall that $\mathfrak{A}^{\ast}= \Comp $; that is, the
elements of $\mathfrak{A}^{\ast}$ are the compositions. Hence, the notation
$\left\vert w\right\vert $ makes sense for any $w\in\mathfrak{A}^{\ast}$; it
denotes the size of the composition $w$.

For every $n\in \NN$, we have%
\[
\left\vert \Comp_{n}\right\vert =
\begin{cases}
2^{n-1},& \text{if }n\geq1;\\
1,& \text{if }n=0
\end{cases}
\]
\footnote{\textit{Proof.} Let $n\in \NN$. Recall that we have a
bijection $ \Comp_{n}\rightarrow2^{\left[  n-1\right]  }$, where
$\left[  n-1\right]  =\left\{  1,2,\ldots,n-1\right\}  $. Thus,
\begin{align*}
\left\vert  \Comp_{n}\right\vert  &  =\left\vert 2^{\left[
n-1\right]  }\right\vert =2^{\left\vert \left[  n-1\right]  \right\vert
}=
\begin{cases}
2^{n-1},& \text{if }n\geq1;\\
2^{0},& \text{if }n=0
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert \left[
n-1\right]  \right\vert =
\begin{cases}
n-1,& \text{if }n\geq1;\\
0,& \text{if }n=0
\end{cases}
\right) \\
&  =
\begin{cases}
2^{n-1},& \text{if }n\geq1;\\
1,& \text{if }n=0
\end{cases} \quad
,
\end{align*}
qed.}. Thus, for every $n\in \NN$, we have%
\begin{align}
\left\vert \left\{  w\in\underbrace{\mathfrak{A}^{\ast}}%
_{= \Comp }\ \mid\ \left\vert w\right\vert =n\right\}
\right\vert  &  =\left\vert \underbrace{\left\{  w\in \Comp %
\ \mid\ \left\vert w\right\vert =n\right\}  }_{=
\Comp_{n}}\right\vert =\left\vert \Comp_{n}%
\right\vert \nonumber\\
&  =
\begin{cases}
2^{n-1},& \text{if }n\geq1;\\
1,& \text{if }n=0
\end{cases} \quad
. \label{sol.words.lyndon.compositions.count.2power}
\end{align}


Let $\mathfrak{M}$ denote the set of all finite multisets of Lyndon
compositions.
Define two maps $\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$
and $\mathbf{n} : \mathfrak{A}^{\ast} \to \mathfrak{M}$ as in
Proposition \ref{prop.lyndon.CFL.bij}.
Then, Proposition \ref{prop.lyndon.CFL.bij} shows that
these two maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse bijections
(since a Lyndon composition is the same thing as a Lyndon word over
the alphabet $\mathfrak{A}$).
Hence, the map $\mathbf{m}$ is a bijection.

On the other hand, let $\mathfrak{L}$ be the set of all Lyndon compositions.
Thus, the definition of $\mathfrak{M}$ says that
$\mathfrak{M}$ is the set of all finite multisets of elements of
$\mathfrak{L}$. Also, the definition of $\operatorname*{lync}n$ now rewrites
as%
\begin{equation}
\operatorname*{lync}n=\left\vert \left\{  w\in\mathfrak{L}\ \mid\ \left\vert
w\right\vert =n\right\}  \right\vert \ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }n. \label{sol.words.lyndon.compositions.count.lynn}%
\end{equation}


Let $\mathfrak{N}$ be the set of all families $\left(  k_{w}\right)
_{w\in\mathfrak{L}}\in{\NN}^{\mathfrak{L}}$ of nonnegative integers
(indexed by the Lyndon compositions) such that all but finitely many $w\in
\mathfrak{L}$ satisfy $k_{w}=0$. Proposition \ref{prop.multiset.bij.seqs}
(applied to $S=\mathfrak{L}$) shows that the map $\operatorname*{mult}%
:\mathfrak{M}\rightarrow\mathfrak{N}$ that sends each multiset $M\in\mathfrak{M}$ to
the family
\[
\left(  \left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\right)  _{w\in S}\in\mathfrak{N}%
\]
is well-defined and is a bijection. Consider this map $\operatorname*{mult}$.

The composition $\mathbf{m}\circ\operatorname{mult}^{-1}%
:\mathfrak{N}\rightarrow\mathfrak{A}^{\ast}$ of the bijections $\mathbf{m}$
and $\operatorname{mult}^{-1}$ is clearly a bijection. It can easily
be seen to satisfy%
\begin{equation}
\left\vert \left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right\vert
=\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\left\vert w\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  k_{w}\right)  _{w\in\mathfrak{L}%
}\in\mathfrak{N}. \label{sol.words.lyndon.compositions.count.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.words.lyndon.compositions.count.1}):} Let
$\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}$. Let
$M=\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  $. Then, $M$ is a multiset of elements of
$\mathfrak{L}$ and satisfies $\left(  k_{w}\right)  _{w\in\mathfrak{L}%
}=\operatorname*{mult}M=\left(  \left(  \text{multiplicity of }w\text{ in the
multiset }M\right)  \right)  _{w\in\mathfrak{L}}$. In other words, every
$w\in\mathfrak{L}$ satisfies%
\begin{equation}
k_{w}=\left(  \text{multiplicity of }w\text{ in the multiset }M\right)  .
\label{sol.words.lyndon.compositions.count.2}%
\end{equation}
\par
Let $a_{1}$, $a_{2}$, $\ldots$, $a_{k}$ denote the elements of this multiset
$M$ listed in decreasing order. Then, the definition of $\mathbf{m}$ yields
$\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$, so that%
\begin{align*}
\left\vert \mathbf{m}\left(  M\right)  \right\vert  &  =\left\vert a_{1}%
a_{2}\cdots a_{k}\right\vert =\sum_{i\in\left\{  1,2,\ldots,k\right\}
}\left\vert a_{i}\right\vert =\sum\limits_{w\in\mathfrak{L}}\sum\limits
_{\substack{i\in\left\{  1,2,\ldots,k\right\}  ;\\a_{i}=w}}\left\vert
\underbrace{a_{i}}_{\substack{=w\\\text{(since }a_{i}=w\text{)}}}\right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since every }a_{i}\text{ belongs to
}\mathfrak{L}\right) \\
&  =\sum\limits_{w\in\mathfrak{L}}\underbrace{\sum\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=w}}\left\vert w\right\vert }_{=\left(
\text{number of }i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfying }%
a_{i}=w\right)  \cdot\left\vert w\right\vert }=\sum\limits_{w\in\mathfrak{L}%
}\underbrace{\left(  \text{number of }i\in\left\{  1,2,\ldots,k\right\}
\text{ satisfying }a_{i}=w\right)  }_{\substack{=\left(  \text{multiplicity of
}w\text{ in the multiset }M\right)  =k_{w}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.2}))}}}\cdot\left\vert w\right\vert
\\
&  =\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\left\vert w\right\vert .
\end{align*}
Now,%
\[
\left\vert \left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right\vert
=\left\vert \mathbf{m}\left(  \underbrace{\operatorname{mult}^{-1}%
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  }_{=M}\right)
\right\vert =\left\vert \mathbf{m}\left(  M\right)  \right\vert =\sum
\limits_{w\in\mathfrak{L}}k_{w}\cdot\left\vert w\right\vert ,
\]
which proves (\ref{sol.words.lyndon.compositions.count.1}).} Now, in the ring
$\QQ \left[  \left[  t\right]  \right]  $ of formal power series, we
have%
\begin{align*}
\sum_{w\in\mathfrak{A}^{\ast}}t^{\left\vert w\right\vert }  &  =\sum
_{n\in \NN }\underbrace{\left\vert \left\{  w\in\mathfrak{A}^{\ast}%
\ \mid\ \left\vert w\right\vert =n\right\}  \right\vert }_{\substack{=
\begin{cases}
2^{n-1},& \text{if }n\geq1;\\
1,& \text{if }n=0
\end{cases}
\\\text{(by (\ref{sol.words.lyndon.compositions.count.2power}))}%
}}t^{n}=\sum_{n\in \NN }
\begin{cases}
2^{n-1},& \text{if }n\geq1;\\
1,& \text{if }n=0
\end{cases} \cdot
t^{n}\\
&  =\underbrace{1t^{0}}_{=1}+\underbrace{\sum\limits_{\substack{n\in \NN %
;\\n\geq1}}2^{n-1}t^{n}}_{\substack{=\sum_{n\in \NN }2^{n}t^{n+1}%
\\\text{(here, we have substituted}\\n+1\text{ for }n\text{ in the sum)}%
}}=1+\sum_{n\in \NN }\underbrace{2^{n}t^{n+1}}_{=\left(  2t\right)
^{n}t}=1+\underbrace{\sum_{n\in \NN }\left(  2t\right)  ^{n}}_{=\dfrac
{1}{1-2t}}t=1+\dfrac{1}{1-2t}t=\dfrac{1-t}{1-2t}.
\end{align*}
Hence,%
\begin{align*}
\dfrac{1-t}{1-2t}  &  =\sum_{w\in\mathfrak{A}^{\ast}}t^{\left\vert
w\right\vert }=\sum_{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}%
}t^{\left\vert \left(  \mathbf{m}\circ\operatorname{mult}^{-1}%
\right)  \left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right\vert
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\left(  \mathbf{m}\circ\operatorname*{mult}%
\nolimits^{-1}\right)  \left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}%
}\right)  \text{ for }w\text{ in the sum,}\\
\text{since the map }\mathbf{m}\circ\operatorname{mult}^{-1}%
:\mathfrak{N}\rightarrow\mathfrak{A}^{\ast}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum_{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}}%
t^{\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\left\vert w\right\vert
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.words.lyndon.compositions.count.1})}\right) \\
&  =\sum_{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}}\prod
_{w\in\mathfrak{L}}t^{k_{w}\cdot\left\vert w\right\vert }=\prod_{w\in
\mathfrak{L}}\underbrace{\sum_{k\in \NN }t^{k\cdot\left\vert w\right\vert
}}_{=\dfrac{1}{1-t^{\left\vert w\right\vert }}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the product rule}\right) \\
&  =\prod_{w\in\mathfrak{L}}\dfrac{1}{1-t^{\left\vert w\right\vert }}%
=\prod_{n\geq1}\prod\limits_{\substack{w\in\mathfrak{L};\\\left\vert w\right\vert
=n}}\underbrace{\dfrac{1}{1-t^{\left\vert w\right\vert }}}_{\substack{=\dfrac
{1}{1-t^{n}}\\\text{(since }\left\vert w\right\vert =n\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert w\right\vert \geq1\text{
for every }w\in\mathfrak{L}\right) \\
&  =\prod_{n\geq1}\underbrace{\prod\limits_{\substack{w\in\mathfrak{L};\\\left\vert
w\right\vert =n}}\dfrac{1}{1-t^{n}}}_{\substack{=\left(  \dfrac{1}{1-t^{n}%
}\right)  ^{\operatorname*{lync}n}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.lynn}))}}}=\prod_{n\geq1}\left(
\dfrac{1}{1-t^{n}}\right)  ^{\operatorname*{lync}n}.
\end{align*}
Taking the logarithm of both sides of this identity, we obtain%
\begin{align*}
\log\dfrac{1-t}{1-2t}  &  =\log\left(  \prod_{n\geq1}\left(  \dfrac{1}%
{1-t^{n}}\right)  ^{\operatorname*{lync}n}\right)  =\sum_{n\geq1}\left(
\operatorname*{lync}n\right)  \cdot\underbrace{\log\left(  \dfrac{1}{1-t^{n}%
}\right)  }_{\substack{=-\log\left(  1-t^{n}\right)  =\sum_{u\geq1}\dfrac
{1}{u}\left(  t^{n}\right)  ^{u}\\\text{(by the Mercator series for the
logarithm)}}}\\
&  =\sum_{n\geq1}\left(  \operatorname*{lync}n\right)  \cdot\sum_{u\geq
1}\dfrac{1}{u}\left(  t^{n}\right)  ^{u}=\sum_{n\geq1}\sum_{u\geq1}\left(
\operatorname*{lync}n\right)  \dfrac{1}{u}\underbrace{\left(  t^{n}\right)
^{u}}_{=t^{nu}}=\sum_{n\geq1}\sum_{u\geq1}\left(  \operatorname*{lync}%
n\right)  \dfrac{1}{u}t^{nu}\\
&  =\underbrace{\sum_{n\geq1}\sum\limits_{\substack{v\geq1;\\n\mid v}}}_{=\sum
_{v\geq1}\sum_{n\mid v}}\left(  \operatorname*{lync}n\right)
\underbrace{\dfrac{1}{v/n}}_{=\dfrac{n}{v}}t^{v}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we substituted }v/n\text{ for }u\text{ in the second sum}\right)
\\
&  =\sum_{v\geq1}\sum_{n\mid v}\left(  \operatorname*{lync}n\right)  \dfrac
{n}{v}t^{v}=\sum_{n\geq1}\sum_{d\mid n}\left(  \operatorname*{lync}d\right)
\dfrac{d}{n}t^{n}%
\end{align*}
(here, we renamed the summation indices $v$ and $n$ as $n$ and $d$). Since%
\begin{align*}
\log\dfrac{1-t}{1-2t}  &  =\log\left(  1-t\right)  -\log\left(  1-2t\right) \\
&  =\underbrace{\left(  -\log\left(  1-2t\right)  \right)  }_{\substack{=\sum
_{n\geq1}\dfrac{1}{n}\left(  2t\right)  ^{n}\\\text{(by the Mercator series
for the logarithm)}}}-\underbrace{\left(  -\log\left(  1-t\right)  \right)
}_{\substack{=\sum_{n\geq1}\dfrac{1}{n}t^{n}\\\text{(by the Mercator series
for the logarithm)}}}\\
&  =\sum_{n\geq1}\dfrac{1}{n}\left(  2t\right)  ^{n}-\sum_{n\geq1}\dfrac{1}%
{n}t^{n}=\sum_{n\geq1}\dfrac{1}{n}\underbrace{\left(  \left(  2t\right)
^{n}-t^{n}\right)  }_{=\left(  2^{n}-1\right)  t^{n}}
=\sum_{n\geq1}\dfrac{1}{n}\left(  2^{n}-1\right)  t^{n} ,
\end{align*}
this rewrites as%
\[
\sum_{n\geq1}\dfrac{1}{n}\left(  2^{n}-1\right)  t^{n}=\sum_{n\geq1}%
\sum_{d\mid n}\left(  \operatorname*{lync}d\right)  \dfrac{d}{n}t^{n}.
\]
Comparing coefficients, we conclude that every positive integer $n$ satisfies%
\[
\dfrac{1}{n}\left(  2^{n}-1\right)  =\sum_{d\mid n}\left(
\operatorname*{lync}d\right)  \dfrac{d}{n}.
\]
Multiplying this with $n$, we obtain%
\begin{equation}
2^{n}-1=\sum_{d\mid n}\left(  \operatorname*{lync}d\right)  d.
\label{sol.words.lyndon.compositions.count.6}%
\end{equation}


Now, every positive integer $n$ satisfies%
\begin{align*}
\sum\limits_{d\mid n}\mu\left(  d\right)  \left(  2^{n/d}-1\right)   &
=\sum\limits_{e\mid n}\mu\left(  e\right)  \underbrace{\left(  2^{n/e}%
-1\right)  }_{\substack{=\sum_{d\mid n/e}\left(  \operatorname*{lync}d\right)
d\\\text{(by (\ref{sol.words.lyndon.compositions.count.6}), applied}\\\text{to
}n/e\text{ instead of }n\text{)}}}=\sum\limits_{e\mid n}\mu\left(  e\right)
\sum_{d\mid n/e}\left(  \operatorname*{lync}d\right)  d\\
&  =\underbrace{\sum\limits_{e\mid n}\sum_{d\mid n/e}}_{=\sum_{d\mid n}%
\sum_{e\mid n/d}}\mu\left(  e\right)  \left(  \operatorname*{lync}d\right)
d=\sum_{d\mid n}\underbrace{\sum_{e\mid n/d}\mu\left(  e\right)
}_{\substack{=\delta_{n/d,1}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.mu}), applied}\\\text{to
}N=n/d\text{)}}}\left(  \operatorname*{lync}d\right)  d\\
&  =\sum_{d\mid n}\underbrace{\delta_{n/d,1}}_{=\delta_{n,d}}\left(
\operatorname*{lync}d\right)  d=\sum_{d\mid n}\delta_{n,d}\left(
\operatorname*{lync}d\right)  d=\left(  \operatorname*{lync}n\right)  n.
\end{align*}
Dividing this by $n$, we obtain $\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  \left(  2^{n/d}-1\right)  =\operatorname*{lync}n$. Hence,%
\[
\operatorname*{lync}n=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)
\left(  2^{n/d}-1\right)  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  2^{n/d}-\delta_{n,1}%
\]
(by Lemma \ref{lem.sol.words.lyndon.compositions.count.2}). This proves
(\ref{sol.words.lyndon.compositions.count.goal}). Thus, Exercise
\ref{exe.words.lyndon.compositions.count} is solved.

\textit{Second solution to Exercise \ref{exe.words.lyndon.compositions.count}%
.} Let $\mathfrak{B}$ denote the two-element set $\left\{  \mathbf{0}%
,\mathbf{1}\right\}  $, where $\mathbf{0}$ and $\mathbf{1}$ are two new
objects. We make $\mathfrak{B}$ into a totally ordered set by setting
$\mathbf{0}<\mathbf{1}$. In the following, we will study not only words over
the alphabet $\mathfrak{A}=\left\{  1,2,3,\ldots\right\}  $, but also words
over the alphabet $\mathfrak{B}$. The latter words form the set $\mathfrak{B}%
^{\ast}$. Let $\mathfrak{L}_{\mathfrak{B}}$ denote the set of all Lyndon words
over the alphabet $\mathfrak{B}$.

For every $k\in\mathfrak{A}$, define an element $\widetilde{k}$ of
$\mathfrak{B}^{\ast}$ by $\widetilde{k}=\left(  \mathbf{0}%
,\underbrace{\mathbf{1},\mathbf{1},\ldots,\mathbf{1}}_{k-1\text{ times}%
}\right)  $. (This is well-defined, since $\mathfrak{A}=\left\{
1,2,3,\ldots\right\}  $.) We can rewrite the definition of $\widetilde{k}$ as
follows: We have\footnote{Here and in the following, expressions like
$\mathbf{01}^{n}$ (for $n\in \NN$) have to be understood as
$\mathbf{0}\left(  \mathbf{1}^{n}\right)  $ rather than as $\left(
\mathbf{01}\right)  ^{n}$. (The objects $\mathbf{0}$ and $\mathbf{1}$ are not
actually numbers; they don't form digital expansions.)}%
\begin{equation}
\widetilde{k}=\mathbf{01}^{k-1}\ \ \ \ \ \ \ \ \ \ \text{for every positive
integer }k \label{sol.words.lyndon.compositions.count.sol2.tilde}%
\end{equation}
(where ``$\mathbf{0}$'' and ``$\mathbf{1}$'' are regarded as
one-letter words). Thus,
\begin{equation}
\ell\left(  \widetilde{k}\right)  =k\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }k \label{sol.words.lyndon.compositions.count.sol2.lentilde}%
\end{equation}
\footnote{\textit{Proof.} Let $k$ be a positive integer. Then,
(\ref{sol.words.lyndon.compositions.count.sol2.tilde}) yields $\widetilde{k}%
=\mathbf{01}^{k-1}$. Now,%
\[
\ell\left(  \underbrace{\widetilde{k}}_{=\mathbf{01}^{k-1}}\right)
=\ell\left(  \mathbf{01}^{k-1}\right)  =\underbrace{\ell\left(  \mathbf{0}%
\right)  }_{=1}+\underbrace{\ell\left(  \mathbf{1}^{k-1}\right)  }_{=\left(
k-1\right)  \ell\left(  \mathbf{1}\right)  }=1+\left(  k-1\right)
\underbrace{\ell\left(  \mathbf{1}\right)  }_{=1}=1+\left(  k-1\right)  =k,
\]
qed.}. Also,%
\begin{equation}
\widetilde{k+1}=\widetilde{k}\mathbf{1}\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }k \label{sol.words.lyndon.compositions.count.sol2.k+1}%
\end{equation}
(where ``$\widetilde{k}\mathbf{1}$'' means the
concatenation of $\widetilde{k}$ with the one-letter word $\mathbf{1}%
$).\ \ \ \ \footnote{\textit{Proof.} Let $k$ be a positive integer. Then,
$k\geq1$, so that $k-1\geq0$. But
(\ref{sol.words.lyndon.compositions.count.sol2.tilde}) (applied to $k+1$
instead of $k$) yields%
\[
\widetilde{k+1}=\mathbf{0}\underbrace{\mathbf{1}^{\left(  k+1\right)  -1}%
}_{\substack{=\mathbf{1}^{k}=\mathbf{1}^{\left(  k-1\right)  +1}%
=\mathbf{1}^{k-1}\mathbf{1}\\\text{(since }k-1\geq0\text{) }}%
}=\underbrace{\mathbf{01}^{k-1}}_{\substack{=\widetilde{k}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.tilde}))}}}\mathbf{1}%
=\widetilde{k}\mathbf{1},
\]
qed.} As a consequence, $\widetilde{k}$ is a prefix of $\widetilde{k+1}$ for
every positive integer $k$. Thus, $\widetilde{k}<\widetilde{k+1}$ (in the
lexicographic order on $\mathfrak{B}^{\ast}$) for every positive integer
$k$\ \ \ \ \footnote{\textit{Proof.} Let $k$ be a positive integer. Then,
$\widetilde{k}\leq\widetilde{k+1}$ (since $\widetilde{k}$ is a prefix of
$\widetilde{k+1}$). Also,
(\ref{sol.words.lyndon.compositions.count.sol2.lentilde}) (applied to $k+1$
instead of $k$) yields $\ell\left(  \widetilde{k+1}\right)  =k+1$. Now,
(\ref{sol.words.lyndon.compositions.count.sol2.lentilde}) yields $\ell\left(
\widetilde{k}\right)  =k<k+1=\ell\left(  \widetilde{k+1}\right)  $, so that
$\ell\left(  \widetilde{k}\right)  \neq\ell\left(  \widetilde{k+1}\right)  $
and thus $\widetilde{k}\neq\widetilde{k+1}$. Combined with $\widetilde{k}%
\leq\widetilde{k+1}$, this yields $\widetilde{k}<\widetilde{k+1}$, qed.}. In
other words,%
\[
\widetilde{1}<\widetilde{2}<\widetilde{3}<\cdots\ \ \ \ \ \ \ \ \ \ \text{in
the lexicographic order on }\mathfrak{B}^{\ast}.
\]
We notice a slightly stronger property: For any $a\in\mathfrak{A}$ and
$b\in\mathfrak{A}$ satisfying $a<b$, we have%
\begin{equation}
\widetilde{a}\left(  \widetilde{c_{1}}\widetilde{c_{2}}\cdots\widetilde{c_{s}%
}\right)  <\widetilde{b}w \label{sol.words.lyndon.compositions.count.sol2.1}%
\end{equation}
for any $s\in \NN$, any $c_{1},c_{2},\ldots,c_{s}\in\mathfrak{A}$ and
any $w\in\mathfrak{B}^{\ast}$.\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.words.lyndon.compositions.count.sol2.1}):} Let $a\in\mathfrak{A}$
and $b\in\mathfrak{A}$ be such that $a<b$. Let $s\in \NN$, let
$c_{1},c_{2},\ldots,c_{s}\in\mathfrak{A}$ and let $w\in\mathfrak{B}^{\ast}$.
We need to prove that (\ref{sol.words.lyndon.compositions.count.sol2.1})
holds.
\par
Assume the contrary. Thus, $\widetilde{a}\left(  \widetilde{c_{1}%
}\widetilde{c_{2}}\cdots\widetilde{c_{s}}\right)  <\widetilde{b}w$ does not
hold. We thus have $\widetilde{a}\left(  \widetilde{c_{1}}\widetilde{c_{2}%
}\cdots\widetilde{c_{s}}\right)  \geq\widetilde{b}w$. Hence, $\widetilde{a}%
\left(  \widetilde{c_{1}}\widetilde{c_{2}}\cdots\widetilde{c_{s}}\right)
\geq\widetilde{b}w\geq\widetilde{b}$.
\par
We notice that $b-a>0$ (since $a<b$), so that the word $\mathbf{1}^{b-a}$ is
nonempty. Also, $b-a>0$, so that $b-a\geq1$ (since $b-a$ is an integer) and
thus $\left(  b-a\right)  -1\geq0$. Hence, the word $\mathbf{1}^{\left(
b-a\right)  -1}$ is well-defined, and we have $\mathbf{1}^{b-a}=\mathbf{11}%
^{\left(  b-a\right)  -1}$.
\par
Applying (\ref{sol.words.lyndon.compositions.count.sol2.tilde}) to $k=a$, we
obtain $\widetilde{a}=\mathbf{01}^{a-1}$. But $b-1=\left(  a-1\right)
+\left(  b-a\right)  $, so that $\mathbf{1}^{b-1}=\mathbf{1}^{\left(
a-1\right)  +\left(  b-a\right)  }=\mathbf{1}^{a-1}\mathbf{1}^{b-a}$ (since
$b-a>0$). Now, (\ref{sol.words.lyndon.compositions.count.sol2.tilde}) (applied
to $k=b$) yields
\[
\widetilde{b}=\mathbf{0}\underbrace{\mathbf{1}^{b-1}}_{=\mathbf{1}%
^{a-1}\mathbf{1}^{b-a}}=\underbrace{\mathbf{01}^{a-1}}_{=\widetilde{a}%
}\mathbf{1}^{b-a}=\widetilde{a}\mathbf{1}^{b-a}>\widetilde{a}%
\]
(since $\mathbf{1}^{b-a}$ is a nonempty word). Hence, $\widetilde{a}%
<\widetilde{b}$. Now, if we had $s=0$, then we would have $\widetilde{a}%
\underbrace{\left(  \widetilde{c_{1}}\widetilde{c_{2}}\cdots\widetilde{c_{s}%
}\right)  }_{=\left(  \text{empty product}\right)  =\varnothing}%
=\widetilde{a}<\widetilde{b}$, which would contradict $\widetilde{a}\left(
\widetilde{c_{1}}\widetilde{c_{2}}\cdots\widetilde{c_{s}}\right)
\geq\widetilde{b}$. Hence, we cannot have $s=0$. We thus have $s\geq1$.
Consequently, $c_{1}$ is well-defined. From
(\ref{sol.words.lyndon.compositions.count.sol2.tilde}) (applied to $k=c_{1}$),
we have $\widetilde{c_{1}}=\mathbf{01}^{c_{1}-1}$. Thus, $\mathbf{0}$ is a
prefix of $\widetilde{c_{1}}$. Since $\widetilde{c_{1}}$ is, in turn, a prefix
of $\widetilde{c_{1}}\widetilde{c_{2}}\cdots\widetilde{c_{s}}$, this yields
that $\mathbf{0}$ is a prefix of $\widetilde{c_{1}}\widetilde{c_{2}}%
\cdots\widetilde{c_{s}}$. In other words, there exists a $t\in\mathfrak{B}%
^{\ast}$ such that $\widetilde{c_{1}}\widetilde{c_{2}}\cdots\widetilde{c_{s}%
}=\mathbf{0}t$. Consider this $t$. Now,
\[
\widetilde{a}\underbrace{\mathbf{0}t}_{=\widetilde{c_{1}}
\widetilde{c_{2}}\cdots\widetilde{c_{s}}}
=\widetilde{a}\left(  \widetilde{c_{1}}
\widetilde{c_{2}}\cdots\widetilde{c_{s}}\right)  \geq\widetilde{b}
=\widetilde{a}\mathbf{1}^{b-a}.
\]
In other words, $\widetilde{a}\mathbf{1}^{b-a}\leq\widetilde{a}\mathbf{0}t$.
Thus, Proposition \ref{prop.words.lex}(c) (applied to $\mathfrak{B}$,
$\widetilde{a}$, $\mathbf{1}^{b-a}$ and $\mathbf{0}t$ instead of
$\mathfrak{A}$, $a$, $c$ and $d$) yields $\mathbf{1}^{b-a}\leq\mathbf{0}t$.
Now, $\mathbf{11}^{\left(  b-a\right)  -1}=\mathbf{1}^{b-a}\leq\mathbf{0}t$.
Therefore, Proposition \ref{prop.words.lex}(e) (applied to $\mathfrak{B}$,
$\mathbf{1}$, $\mathbf{1}^{\left(  b-a\right)  -1}$, $\mathbf{0}$ and $t$
instead of $\mathfrak{A}$, $a$, $b$, $c$ and $d$) yields that either we have
$\mathbf{1} \leq \mathbf{0}$ or the word $\mathbf{0}$ is a prefix of $\mathbf{1}$.
Since the word $\mathbf{0}$ is not a prefix of $\mathbf{1}$, this shows that
$\mathbf{1} \leq \mathbf{0}$. But this contradicts $\mathbf{0} < \mathbf{1}$. This
contradiction proves that our assumption was wrong, qed.}

Also,
\begin{equation}
\widetilde{k}\text{ is a Lyndon word over the alphabet }\mathfrak{B}\text{ for
every positive integer }k.
\label{sol.words.lyndon.compositions.count.sol2.klyn}
\end{equation}
\footnote{\textit{Proof.} Let $k$ be a positive integer. From
(\ref{sol.words.lyndon.compositions.count.sol2.tilde}), we obtain
$\widetilde{k}=\mathbf{01}^{k-1}$.
\par
From (\ref{sol.words.lyndon.compositions.count.sol2.lentilde}), we have
$\ell\left(  \widetilde{k}\right)  =k\geq1$. Thus, the word $\widetilde{k}$ is
nonempty.
\par
Let $v$ be a nonempty proper suffix of $\widetilde{k}$. We will show that
$v>\widetilde{k}$.
\par
Indeed, there exists a nonempty $u\in\mathfrak{B}^{\ast}$ satisfying
$\widetilde{k}=uv$ (since $v$ is a proper suffix of $\widetilde{k}$). Consider
this $u$. Since $u$ is nonempty, the first letter of $u$ is well-defined. We
have%
\begin{align*}
\left(  \text{the first letter of }u\right)   &  =\left(  \text{the first
letter of }\underbrace{\widetilde{k}}_{=\mathbf{01}^{k-1}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }u\text{ is a prefix of }%
\widetilde{k}\text{ (since }\widetilde{k}=uv\text{)}\right) \\
&  =\left(  \text{the first letter of }\mathbf{01}^{k-1}\right)  =\mathbf{0}.
\end{align*}
Thus, $\mathbf{0}$ is a prefix of $u$. In other words, there exists a word
$u^{\prime}\in\mathfrak{B}^{\ast}$ satisfying $u=\mathbf{0}u^{\prime}$.
Consider this $u^{\prime}$. We have $\underbrace{\mathbf{0}u^{\prime}}%
_{=u}v=uv=\widetilde{k}=\mathbf{01}^{k-1}$. Cancelling $\mathbf{0}$ from this
equality, we obtain $u^{\prime}v=\mathbf{1}^{k-1}$. Hence, $v$ is a suffix of
the word $\mathbf{1}^{k-1}$. Thus, $v$ has the form $\mathbf{1}^{p}$ for some
$p\in \NN$ (since every suffix of the word $\mathbf{1}^{k-1}$ has this
form). Consider this $p$. The word $v$ is nonempty; thus, $v\neq\varnothing$.
We have $p\neq0$ (since otherwise, we would have
\begin{align*}
v  &  =\mathbf{1}^{p}=\mathbf{1}^{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}p=0\right) \\
&  =\varnothing,
\end{align*}
contradicting $v\neq\varnothing$). Hence, $p\geq1$, so that $p-1\geq0$ and
thus $\mathbf{1}^{p}=\mathbf{11}^{p-1}$. Now, $\mathbf{0}<\mathbf{1}$. Hence,
Proposition \ref{prop.words.lex}(d) (applied to $\mathfrak{B}$, $\mathbf{0}$,
$\mathbf{1}^{k-1}$, $\mathbf{1}$ and $\mathbf{1}^{p-1}$ instead of
$\mathfrak{A}$, $a$, $b$, $c$ and $d$) yields that either we have
$\mathbf{01}^{k-1}\leq\mathbf{11}^{p-1}$ or the word $\mathbf{0}$ is a prefix
of $\mathbf{1}$. Since the word $\mathbf{0}$ is not a prefix of $\mathbf{1}$,
we thus obtain $\mathbf{01}^{k-1}\leq\mathbf{11}^{p-1}$, so that
$\widetilde{k}=\mathbf{01}^{k-1}\leq\mathbf{11}^{p-1}=\mathbf{1}^{p}=v$.
Hence, $v\geq\widetilde{k}$. Since $v\neq\widetilde{k}$ (because $v$ is a
\textbf{proper} suffix of $\widetilde{k}$), this yields $v>\widetilde{k}$.
\par
Now, let us forget that we fixed $v$. We thus have proven that every nonempty
proper suffix $v$ of $\widetilde{k}$ satisfies $v>\widetilde{k}$. Since the
word $\widetilde{k}$ is nonempty, this shows that the word $\widetilde{k}$ is
Lyndon (by the definition of a Lyndon word), qed.}

We can now define a map $\Phi:\mathfrak{A}^{\ast}\rightarrow\mathfrak{B}%
^{\ast}$ by setting%
\[
\left(  \Phi\left(  \left(  w_{1},w_{2},\ldots,w_{k}\right)  \right)
=\widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{k}}%
\ \ \ \ \ \ \ \ \ \ \text{for every word }\left(  w_{1},w_{2},\ldots
,w_{k}\right)  \in\mathfrak{A}^{\ast}\right)  .
\]
This map $\Phi$ is clearly a monoid homomorphism\footnote{Indeed, we could
just as well have defined $\Phi$ as the unique monoid homomorphism
$\mathfrak{A}^{\ast}\rightarrow\mathfrak{B}^{\ast}$ which sends every
$k\in\mathfrak{A}$ to $\widetilde{k}\in\mathfrak{B}^{\ast}$. This definition
makes sense since $\mathfrak{A}^{\ast}$ is the free monoid on $\mathfrak{A}$%
.}. Hence, $\Phi\left(  \varnothing\right)  =\varnothing$. Also, the map
$\Phi$ is strictly order-preserving (with respect to the lexicographical
orders on $\mathfrak{A}^{\ast}$ and $\mathfrak{B}^{\ast}$%
)\ \ \ \ \footnote{\textit{Proof.} Let $u$ and $v$ be two words in
$\mathfrak{A}^{\ast}$ satisfying $u<v$. We are going to prove that
$\Phi\left(  u\right)  <\Phi\left(  v\right)  $ in $\mathfrak{B}^{\ast}$.
\par
We have $u<v$. Thus, $u\leq v$. By the definition of the relation $\leq$, this
means that%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}
\\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  u_{i}<v_{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)
,\\
&  \text{\textbf{or} the word }u\text{ is a prefix of }v.
\end{align*}
We thus must be in one of the following two cases:
\par
\textit{Case 1:} There exists an $i\in\left\{  1,2,\ldots,\min\left\{
\ell\left(  u\right)  ,\ell\left(  v\right)  \right\}  \right\}  $ such that
$\left(  u_{i}<v_{i}\text{, and every }j\in\left\{  1,2,\ldots,i-1\right\}
\text{ satisfies }u_{j}=v_{j}\right)  $.
\par
\textit{Case 2:} The word $u$ is a prefix of $v$.
\par
Let us first consider Case 1. In this case, there exists an $i\in\left\{
1,2,\ldots,\min\left\{  \ell\left(  u\right)  ,\ell\left(  v\right)  \right\}
\right\}  $ such that $\left(  u_{i}<v_{i}\text{, and every }j\in\left\{
1,2,\ldots,i-1\right\}  \text{ satisfies }u_{j}=v_{j}\right)  $. Consider this
$i$. We have $\widetilde{a}<\widetilde{b}$ for any two elements $a$ and $b$ of
$\mathfrak{A}$ satisfying $a<b$ (because $\widetilde{1}<\widetilde{2}%
<\widetilde{3}<\cdots$). Applying this to $a=u_{i}$ and $b=v_{i}$, we obtain
$\widetilde{u_{i}}<\widetilde{v_{i}}$.
\par
But let $g=\widetilde{u_{1}}\widetilde{u_{2}}\cdots\widetilde{u_{i-1}}$. Every
$j\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies $u_{j}=v_{j}$. Thus, every
$j\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies $\widetilde{u_{j}%
}=\widetilde{v_{j}}$. Taking the product of these equalities over all
$j\in\left\{  1,2,\ldots,i-1\right\}  $, we obtain $\widetilde{u_{1}%
}\widetilde{u_{2}}\cdots\widetilde{u_{i-1}}=\widetilde{v_{1}}\widetilde{v_{2}%
}\cdots\widetilde{v_{i-1}}$, so that $g=\widetilde{u_{1}}\widetilde{u_{2}%
}\cdots\widetilde{u_{i-1}}=\widetilde{v_{1}}\widetilde{v_{2}}\cdots
\widetilde{v_{i-1}}$.
\par
We have $u_{i}<v_{i}$. Thus, $\widetilde{u_{i}}\left(  \widetilde{u_{i+1}%
}\widetilde{u_{i+2}}\cdots\widetilde{u_{\ell\left(  u\right)  }}\right)
<\widetilde{v_{i}}\left(  \widetilde{v_{i+1}}\widetilde{v_{i+2}}%
\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  $ (by
(\ref{sol.words.lyndon.compositions.count.sol2.1}), applied to $a=u_{i}$,
$b=v_{i}$, $s=\ell\left(  u\right)  -i$, $c_{k}=u_{i+k}$ and
$w=\widetilde{v_{i+1}}\widetilde{v_{i+2}}\cdots\widetilde{v_{\ell\left(
v\right)  }}$). Thus,%
\[
\widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(  u\right)
}}=\widetilde{u_{i}}\left(  \widetilde{u_{i+1}}\widetilde{u_{i+2}}%
\cdots\widetilde{u_{\ell\left(  u\right)  }}\right)  <\widetilde{v_{i}}\left(
\widetilde{v_{i+1}}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(
v\right)  }}\right)  =\widetilde{v_{i}}\widetilde{v_{i+1}}\cdots
\widetilde{v_{\ell\left(  v\right)  }}.
\]
Hence, Proposition \ref{prop.words.lex}(b) (applied to $\mathfrak{B}$, $g$,
$\widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(  u\right)
}}$ and $\widetilde{v_{i}}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(
v\right)  }}$ instead of $\mathfrak{A}$, $a$, $c$ and $d$) yields%
\[
g\left(  \widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(
u\right)  }}\right)  \leq g\left(  \widetilde{v_{i}}\widetilde{v_{i+1}}%
\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  .
\]
Moreover, $g\left(  \widetilde{u_{i}}\widetilde{u_{i+1}}\cdots
\widetilde{u_{\ell\left(  u\right)  }}\right)  \neq g\left(  \widetilde{v_{i}%
}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  $
(because otherwise, we would have $g\left(  \widetilde{u_{i}}%
\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(  u\right)  }}\right)
=g\left(  \widetilde{v_{i}}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(
v\right)  }}\right)  $, and thus (by cancelling $g$ from the equality
$g\left(  \widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(
u\right)  }}\right)  =g\left(  \widetilde{v_{i}}\widetilde{v_{i+1}}%
\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  $) we would obtain
$\widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(  u\right)
}}=\widetilde{v_{i}}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(
v\right)  }}$, which would contradict $\widetilde{u_{i}}\widetilde{u_{i+1}%
}\cdots\widetilde{u_{\ell\left(  u\right)  }}<\widetilde{v_{i}}%
\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(  v\right)  }}$). Combining
this with $g\left(  \widetilde{u_{i}}\widetilde{u_{i+1}}\cdots
\widetilde{u_{\ell\left(  u\right)  }}\right)  \leq g\left(  \widetilde{v_{i}%
}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  $, we
obtain $g\left(  \widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell
\left(  u\right)  }}\right)  <g\left(  \widetilde{v_{i}}\widetilde{v_{i+1}%
}\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  $.
\par
But $u=\left(  u_{1},u_{2},\ldots,u_{\ell\left(  u\right)  }\right)  $, so
that%
\begin{align*}
\Phi\left(  u\right)   &  =\Phi\left(  \left(  u_{1},u_{2},\ldots
,u_{\ell\left(  u\right)  }\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\Phi\left(  u\right)  \right) \\
&  =\widetilde{u_{1}}\widetilde{u_{2}}\cdots\widetilde{u_{\ell\left(
u\right)  }}=\underbrace{\left(  \widetilde{u_{1}}\widetilde{u_{2}}%
\cdots\widetilde{u_{i-1}}\right)  }_{=g}\left(  \widetilde{u_{i}%
}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(  u\right)  }}\right)
=g\left(  \widetilde{u_{i}}\widetilde{u_{i+1}}\cdots\widetilde{u_{\ell\left(
u\right)  }}\right) \\
&  <\underbrace{g}_{=\widetilde{v_{1}}\widetilde{v_{2}}\cdots
\widetilde{v_{i-1}}}\left(  \widetilde{v_{i}}\widetilde{v_{i+1}}%
\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)  =\left(  \widetilde{v_{1}%
}\widetilde{v_{2}}\cdots\widetilde{v_{i-1}}\right)  \left(  \widetilde{v_{i}%
}\widetilde{v_{i+1}}\cdots\widetilde{v_{\ell\left(  v\right)  }}\right)
=\widetilde{v_{1}}\widetilde{v_{2}}\cdots\widetilde{v_{\ell\left(  v\right)
}}=\Phi\left(  v\right)
\end{align*}
(since $\Phi\left(  v\right)  =\widetilde{v_{1}}\widetilde{v_{2}}%
\cdots\widetilde{v_{\ell\left(  v\right)  }}$ (by the definition of
$\Phi\left(  v\right)  $)). Thus, $\Phi\left(  u\right)  <\Phi\left(
v\right)  $ is proven in Case 1.
\par
Let us now consider Case 2. In this case, the word $u$ is a prefix of $v$.
That is, there exists a word $r\in\mathfrak{A}^{\ast}$ such that $v=ur$.
Consider this $r$. We have $\Phi\left(  \underbrace{v}_{=ur}\right)
=\Phi\left(  ur\right)  =\Phi\left(  u\right)  \Phi\left(  r\right)  $ (since
$\Phi$ is a monoid homomorphism). If we had $r=\varnothing$, then we would
have $v=u\underbrace{r}_{=\varnothing}=u<v$, which is absurd. Hence, we cannot
have $r=\varnothing$. In other words, $r$ is nonempty, so that $\ell\left(
r\right)  \geq1$.
\par
Now, let us check that $\Phi\left(  r\right)  $ is nonempty. In fact,
$r=\left(  r_{1},r_{2},\ldots,r_{\ell\left(  r\right)  }\right)  $, so that
$\Phi\left(  r\right)  =\widetilde{r_{1}}\widetilde{r_{2}}\cdots
\widetilde{r_{\ell\left(  r\right)  }}$ (by the definition of $\Phi\left(
r\right)  $). Notice that $r_{1}$ is well-defined, since $\ell\left(
r\right)  \geq1$. From (\ref{sol.words.lyndon.compositions.count.sol2.tilde})
(applied to $k=r_{1}$), we obtain $\widetilde{r_{1}}=\mathbf{01}^{r_{1}-1}$,
so that $\mathbf{0}$ is a prefix of the word $\widetilde{r_{1}}$. Since
$\widetilde{r_{1}}$ (in turn) is a prefix of the word $\widetilde{r_{1}%
}\widetilde{r_{2}}\cdots\widetilde{r_{\ell\left(  r\right)  }}$, this yields
that $\mathbf{0}$ is a prefix of the word $\widetilde{r_{1}}\widetilde{r_{2}%
}\cdots\widetilde{r_{\ell\left(  r\right)  }}$. In other words, $\mathbf{0}$
is a prefix of the word $\Phi\left(  r\right)  $ (since $\Phi\left(  r\right)
=\widetilde{r_{1}}\widetilde{r_{2}}\cdots\widetilde{r_{\ell\left(  r\right)
}}$). Hence, the word $\Phi\left(  r\right)  $ has a nonempty prefix (namely,
$\mathbf{0}$). Thus, the word $\Phi\left(  r\right)  $ is nonempty. In other
words, $\Phi\left(  r\right)  \neq\varnothing$. That is, $\varnothing\neq
\Phi\left(  r\right)  $.
\par
Since $\Phi\left(  v\right)  =\Phi\left(  u\right)  \Phi\left(  r\right)  $,
we now see that the word $\Phi\left(  u\right)  $ is a prefix of $\Phi\left(
v\right)  $ (since $\Phi\left(  r\right)  $ is nonempty). Hence, $\Phi\left(
u\right)  \leq\Phi\left(  v\right)  $. On the other hand, if we had
$\Phi\left(  u\right)  = \Phi\left(  v\right)  $, then we would have
$\Phi\left(  u\right)  \varnothing= \Phi\left(  u\right)  = \Phi\left(
v\right)  = \Phi\left(  u\right)  \Phi\left(  r\right)  $, and therefore we
would have $\varnothing= \Phi\left(  r\right)  $ (as a result of cancelling
$\Phi\left(  u\right)  $ from the equality $\Phi\left(  u\right)  \varnothing=
\Phi\left(  u\right)  \Phi\left(  r\right)  $), which would contradict
$\varnothing\neq\Phi\left(  r\right)  $. Hence, we cannot have $\Phi\left(
u\right)  = \Phi\left(  v\right)  $. We therefore must have $\Phi\left(
u\right)  \neq\Phi\left(  v\right)  $. Hence, $\Phi\left(  u\right)  \leq
\Phi\left(  v\right)  $ becomes $\Phi\left(  u\right)  <\Phi\left(  v\right)
$. We thus have proven $\Phi\left(  u\right)  <\Phi\left(  v\right)  $ in Case
2.
\par
Now, we have proven that $\Phi\left(  u\right)  <\Phi\left(  v\right)  $ in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
this shows that $\Phi\left(  u\right)  <\Phi\left(  v\right)  $ always holds.
\par
Now, let us forget that we fixed $u$ and $v$. We thus have proven that if $u$
and $v$ are two words in $\mathfrak{A}^{\ast}$ satisfying $u<v$, then
$\Phi\left(  u\right)  <\Phi\left(  v\right)  $ in $\mathfrak{B}^{\ast}$. In
other words, the map $\Phi$ is strictly order-preserving, qed.}. Consequently,
the map $\Phi$ is injective\footnote{because any strictly order-preserving map
from a totally ordered set to a poset must be injective}.

For every $w\in\mathfrak{B}^{\ast}$, we denote by $w\mathfrak{B}^{\ast}$ the
set $\left\{  wu\ \mid\ u\in\mathfrak{B}^{\ast}\right\}  $ (that is, the set
of all words in $\mathfrak{B}^{\ast}$ which have $w$ as a prefix). Every
nonempty word in $\mathfrak{B}^{\ast}$ starts with either the letter
$\mathbf{0}$ or the letter $\mathbf{1}$, but not both. In other words, every
nonempty word in $\mathfrak{B}^{\ast}$ belongs to either $\mathbf{0}%
\mathfrak{B}^{\ast}$ or $\mathbf{1}\mathfrak{B}^{\ast}$ (where $\mathbf{0}$
and $\mathbf{1}$ are regarded as one-letter words), but not both. In other
words, $\mathfrak{B}^{\ast}\setminus\left\{  \varnothing\right\}
=\mathbf{0}\mathfrak{B}^{\ast}\cup\mathbf{1}\mathfrak{B}^{\ast}$ and
$\mathbf{0}\mathfrak{B}^{\ast}\cap\mathbf{1}\mathfrak{B}^{\ast}=\varnothing$.
Thus, $\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}=\left\{
\varnothing\right\}  \cup\mathbf{0}\mathfrak{B}^{\ast}$%
\ \ \ \ \footnote{\textit{Proof.} We have $\mathfrak{B}^{\ast}\setminus
\left\{  \varnothing\right\}  =\mathbf{0}\mathfrak{B}^{\ast}\cup
\mathbf{1}\mathfrak{B}^{\ast}$ and $\mathbf{0}\mathfrak{B}^{\ast}%
\cap\mathbf{1}\mathfrak{B}^{\ast}=\varnothing$. Thus, the sets $\mathbf{0}%
\mathfrak{B}^{\ast}$ and $\mathbf{1}\mathfrak{B}^{\ast}$ are disjoint and have
union $\mathfrak{B}^{\ast}\setminus\left\{  \varnothing\right\}  $. In other
words, the sets $\mathbf{0}\mathfrak{B}^{\ast}$ and $\mathbf{1}\mathfrak{B}%
^{\ast}$ are complementary subsets of $\mathfrak{B}^{\ast}\setminus\left\{
\varnothing\right\}  $. Hence, $\left(  \mathbf{0}\mathfrak{B}^{\ast}%
\cup\mathbf{1}\mathfrak{B}^{\ast}\right)  \setminus\mathbf{1}\mathfrak{B}%
^{\ast}=\mathbf{0}\mathfrak{B}^{\ast}$. Now,%
\begin{align*}
\underbrace{\mathfrak{B}^{\ast}}_{=\left\{  \varnothing\right\}  \cup\left(
\mathfrak{B}^{\ast}\setminus\left\{  \varnothing\right\}  \right)  }%
\setminus\mathbf{1}\mathfrak{B}^{\ast}  &  =\left(  \left\{  \varnothing
\right\}  \cup\left(  \mathfrak{B}^{\ast}\setminus\left\{  \varnothing
\right\}  \right)  \right)  \setminus\mathbf{1}\mathfrak{B}^{\ast
}=\underbrace{\left(  \left\{  \varnothing\right\}  \setminus\mathbf{1}%
\mathfrak{B}^{\ast}\right)  }_{=\left\{  \varnothing\right\}  }\cup\left(
\underbrace{\left(  \mathfrak{B}^{\ast}\setminus\left\{  \varnothing\right\}
\right)  }_{=\left(  \mathbf{0}\mathfrak{B}^{\ast}\cup\mathbf{1}%
\mathfrak{B}^{\ast}\right)  }\setminus\mathbf{1}\mathfrak{B}^{\ast}\right) \\
&  =\left\{  \varnothing\right\}  \cup\underbrace{\left(  \left(
\mathbf{0}\mathfrak{B}^{\ast}\cup\mathbf{1}\mathfrak{B}^{\ast}\right)
\setminus\mathbf{1}\mathfrak{B}^{\ast}\right)  }_{=\mathbf{0}\mathfrak{B}%
^{\ast}}=\left\{  \varnothing\right\}  \cup\mathbf{0}\mathfrak{B}^{\ast},
\end{align*}
qed.}.

Furthermore, $\Phi\left(  \mathfrak{A}^{\ast}\right)  \subset\mathfrak{B}%
^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$%
\ \ \ \ \footnote{\textit{Proof.} Let $u\in\Phi\left(  \mathfrak{A}^{\ast
}\right)  $. We will prove that $u\in\mathfrak{B}^{\ast}\setminus
\mathbf{1}\mathfrak{B}^{\ast}$.
\par
Indeed, there exists some $w\in\mathfrak{A}^{\ast}$ satisfying $u=\Phi\left(
w\right)  $ (since $u\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $). Consider
this $w$. If $w=\varnothing$, then $u=\Phi\left(  \underbrace{w}%
_{=\varnothing}\right)  =\Phi\left(  \varnothing\right)  =\varnothing
\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$ (since clearly,
$\varnothing\notin\mathbf{1}\mathfrak{B}^{\ast}$). Hence, if $w=\varnothing$,
then $u\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$ is
proven. Thus, for the rest of this proof, we can WLOG assume that we don't
have $w=\varnothing$. Assume this.
\par
The word $w$ is nonempty (since $w\neq\varnothing$), so that $\ell\left(
w\right)  \geq1$. Hence, the letter $w_{1}$ is well-defined. We have
$w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  $ and thus%
\begin{align*}
\Phi\left(  w\right)   &  =\widetilde{w_{1}}\widetilde{w_{2}}\cdots
\widetilde{w_{\ell\left(  w\right)  }}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\Phi\right) \\
&  =\underbrace{\widetilde{w_{1}}}_{\substack{=\mathbf{01}^{w_{1}%
-1}\\\text{(by (\ref{sol.words.lyndon.compositions.count.sol2.tilde}),
applied}\\\text{to }k=w_{1}\text{)}}}\left(  \widetilde{w_{2}}\widetilde{w_{3}%
}\cdots\widetilde{w_{\ell\left(  w\right)  }}\right)  =\mathbf{0}%
\underbrace{\mathbf{1}^{w_{1}-1}\left(  \widetilde{w_{2}}\widetilde{w_{3}%
}\cdots\widetilde{w_{\ell\left(  w\right)  }}\right)  }_{\in\mathfrak{B}%
^{\ast}}
\in\mathbf{0}\mathfrak{B}^{\ast}\subset\left\{  \varnothing\right\}
\cup\mathbf{0}\mathfrak{B}^{\ast}=\mathfrak{B}^{\ast}\setminus\mathbf{1}%
\mathfrak{B}^{\ast}.
\end{align*}
Hence, $u=\Phi\left(  w\right)  \in\mathfrak{B}^{\ast}\setminus\mathbf{1}%
\mathfrak{B}^{\ast}$.
\par
Now, let us forget that we fixed $u$. We thus have proven that $u\in
\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$ for every $u\in
\Phi\left(  \mathfrak{A}^{\ast}\right)  $. In other words, $\Phi\left(
\mathfrak{A}^{\ast}\right)  \subset\mathfrak{B}^{\ast}\setminus\mathbf{1}%
\mathfrak{B}^{\ast}$, qed.} and $\mathfrak{B}^{\ast}\setminus\mathbf{1}%
\mathfrak{B}^{\ast}\subset\Phi\left(  \mathfrak{A}^{\ast}\right)
$\ \ \ \ \footnote{\textit{Proof.} We are going to show that
\begin{equation}
\text{every }u\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast
}\text{ satisfies }u\in\Phi\left(  \mathfrak{A}^{\ast}\right)  .
\label{sol.words.lyndon.compositions.count.sol2.revinc.1}%
\end{equation}
\par
\textit{Proof of (\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}):}
We will prove (\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}) by
strong induction over $\ell\left(  u\right)  $:
\par
\textit{Induction step:} Let $N\in \NN$. Assume that
(\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}) holds whenever
$\ell\left(  u\right)  <N$. We now will prove that
(\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}) holds whenever
$\ell\left(  u\right)  =N$.
\par
We know that (\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}) holds
whenever $\ell\left(  u\right)  <N$. In other words,%
\begin{equation}
\text{every }u\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast
}\text{ satisfying }\ell\left(  u\right)  <N\text{ satisfies }u\in\Phi\left(
\mathfrak{A}^{\ast}\right)  .
\label{sol.words.lyndon.compositions.count.sol2.revinc.1.pf.1}%
\end{equation}
\par
Now, fix an $u\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$
satisfying $\ell\left(  u\right)  =N$. We shall show that $u\in\Phi\left(
\mathfrak{A}^{\ast}\right)  $.
\par
If $u=\varnothing$, then $u=\varnothing=\Phi\left(  \varnothing\right)
\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $. Hence, for the rest of our proof
of $u\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $, we can WLOG assume that we
don't have $u=\varnothing$. Assume this.
\par
We have $u\neq\varnothing$ (since we don't have $u=\varnothing$), thus
$u\notin\left\{  \varnothing\right\}  $. We have $u\in\mathfrak{B}^{\ast
}\setminus\mathbf{1}\mathfrak{B}^{\ast}=\left\{  \varnothing\right\}
\cup\mathbf{0}\mathfrak{B}^{\ast}$ but $u\notin\left\{  \varnothing\right\}
$. Hence, $u\in\left(  \left\{  \varnothing\right\}  \cup\mathbf{0}%
\mathfrak{B}^{\ast}\right)  \setminus\left\{  \varnothing\right\}
\subset\mathbf{0}\mathfrak{B}^{\ast}$. In other words, there exists a
$p\in\mathfrak{B}^{\ast}$ such that $u=\mathbf{0}p$. Consider this $p$. The
one-letter word $\mathbf{0}$ is a prefix of $u$ (since $u=\mathbf{0}p$).
\par
Applying (\ref{sol.words.lyndon.compositions.count.sol2.tilde}) to $k=1$, we
obtain $\widetilde{1}=\mathbf{0}\underbrace{\mathbf{1}^{1-1}}_{=\mathbf{1}%
^{0}=\varnothing}=\mathbf{0}$, so that $\widetilde{1}$ is a prefix of $u$
(since $\mathbf{0}$ is a prefix of $u$). Hence, there exists a $k\in
\mathfrak{A}$ such that $\widetilde{k}$ is a prefix of $u$ (namely, $k=1$).
\par
On the other hand, if $k\in\mathfrak{A}$ is such that $\widetilde{k}$ is a
prefix of $u$, then $\ell\left(  \widetilde{k}\right)  \leq\ell\left(
u\right)  $. Hence, if $k\in\mathfrak{A}$ is such that $\widetilde{k}$ is a
prefix of $u$, then $\ell\left(  u\right)  \geq\ell\left(  \widetilde{k}%
\right)  =k$ (by (\ref{sol.words.lyndon.compositions.count.sol2.lentilde})).
Hence, if $k\in\mathfrak{A}$ is such that $\widetilde{k}$ is a prefix of $u$,
then $k\leq\ell\left(  u\right)  $. Hence, only finitely many $k\in
\mathfrak{A}$ have the property that $\widetilde{k}$ is a prefix of $u$
(because only finitely many $k\in\mathfrak{A}$ have the property that
$k\leq\ell\left(  u\right)  $).
\par
We now have made the following two observations:
\par
\begin{itemize}
\item There exists a $k\in\mathfrak{A}$ such that $\widetilde{k}$ is a prefix
of $u$.
\par
\item Only finitely many $k\in\mathfrak{A}$ have the property that
$\widetilde{k}$ is a prefix of $u$.
\end{itemize}
\par
Combining these two observations, we conclude that there exists a
\textbf{largest} $k\in\mathfrak{A}$ such that $\widetilde{k}$ is a prefix of
$u$. Consider this largest $k$. Then, $\widetilde{k}$ is a prefix of $u$, but
$\widetilde{k+1}$ is not a prefix of $u$.
\par
There exists a $v\in\mathfrak{B}^{\ast}$ such that $u=\widetilde{k}v$ (since
$\widetilde{k}$ is a prefix of $u$). Consider this $v$. We have $\ell\left(
\underbrace{u}_{=\widetilde{k}v}\right)  =\ell\left(  \widetilde{k}v\right)
=\underbrace{\ell\left(  \widetilde{k}\right)  }_{\substack{=k\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.lentilde}))}}}+\ell\left(
v\right)  =\underbrace{k}_{\substack{\geq1\\\text{(since }k\in\mathfrak{A}%
\text{)}}}+\ell\left(  v\right)  \geq1+\ell\left(  v\right)  >\ell\left(
v\right)  $, so that $\ell\left(  v\right)  <\ell\left(  u\right)  =N$.
\par
Let us now assume that $v\in\mathfrak{B}^{\ast}\setminus\mathbf{1}%
\mathfrak{B}^{\ast}$. Then, $v\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $ (by
(\ref{sol.words.lyndon.compositions.count.sol2.revinc.1.pf.1}), applied to $v$
instead of $u$). In other words, there exists a $v^{\prime}\in\mathfrak{A}%
^{\ast}$ such that $v=\Phi\left(  v^{\prime}\right)  $. Consider this
$v^{\prime}$. The definition of $\Phi\left(  k\right)  $ (where $k$ stands for
the one-letter word $\left(  k\right)  \in\mathfrak{A}^{\ast}$) yields
$\Phi\left(  k\right)  =\widetilde{k}$. Hence,
\begin{align*}
u  &  =\underbrace{\widetilde{k}}_{=\Phi\left(  k\right)  }\underbrace{v}%
_{=\Phi\left(  v^{\prime}\right)  }=\Phi\left(  k\right)  \Phi\left(
v^{\prime}\right)  =\Phi\left(  kv^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Phi\text{ is a monoid homomorphism}%
\right) \\
&  \in\Phi\left(  \mathfrak{A}^{\ast}\right)  .
\end{align*}
\par
Now, let us forget that we have assumed that $v\in\mathfrak{B}^{\ast}%
\setminus\mathbf{1}\mathfrak{B}^{\ast}$. We have thus proven that $u\in
\Phi\left(  \mathfrak{A}^{\ast}\right)  $ under the assumption that
$v\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$. Thus, for the
rest of the proof of $u\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $, we can
WLOG assume that we don't have $v\in\mathfrak{B}^{\ast}\setminus
\mathbf{1}\mathfrak{B}^{\ast}$. Assume this.
\par
We have $v\in\mathbf{1}\mathfrak{B}^{\ast}$ (since we don't have
$v\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$). Thus, there
exists a $v^{\prime}\in\mathfrak{B}^{\ast}$ such that $v=\mathbf{1}v^{\prime}%
$. Consider this $v^{\prime}$. We have%
\[
u=\widetilde{k}\underbrace{v}_{=\mathbf{1}v^{\prime}}%
=\underbrace{\widetilde{k}\mathbf{1}}_{\substack{=\widetilde{k+1}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.k+1}))}}}v^{\prime}=\left(
\widetilde{k+1}\right)  v^{\prime}.
\]
Hence, $\widetilde{k+1}$ is a prefix of $u$. This contradicts the fact that
$\widetilde{k+1}$ is not a prefix of $u$. From this contradiction, we conclude
that $u\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $ (since ex falso
quodlibet). Thus, $u\in\Phi\left(  \mathfrak{A}^{\ast}\right)  $ is proven.
\par
Now, let us forget that we fixed $u$. We thus have shown that every
$u\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$ satisfying
$\ell\left(  u\right)  =N$ satisfies $u\in\Phi\left(  \mathfrak{A}^{\ast
}\right)  $. In other words,
(\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}) holds whenever
$\ell\left(  u\right)  =N$. This completes the induction step. Thus,
(\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}) is proven by
induction.
\par
But from (\ref{sol.words.lyndon.compositions.count.sol2.revinc.1}), we
immediately obtain $\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast
}\subset\Phi\left(  \mathfrak{A}^{\ast}\right)  $, qed.}. Combining these two
relations, we obtain
\begin{equation}
\Phi\left(  \mathfrak{A}^{\ast}\right)  =\mathfrak{B}^{\ast}\setminus
\mathbf{1}\mathfrak{B}^{\ast}.
\label{sol.words.lyndon.compositions.count.sol2.2}%
\end{equation}


Next, we recall that $\mathfrak{L}_{\mathfrak{B}}$ is the set of all Lyndon
words over the alphabet $\mathfrak{B}$. We notice that $\Phi\left(
\mathfrak{L}\right)  \subset\mathfrak{L}_{\mathfrak{B}}\setminus\left\{
\mathbf{1}\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $u\in\Phi\left(
\mathfrak{L}\right)  $. We are going to prove that $u\in\mathfrak{L}%
_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}  $.
\par
There exists a $w\in\mathfrak{L}$ such that $u=\Phi\left(  w\right)  $ (since
$u\in\Phi\left(  \mathfrak{L}\right)  $). Consider this $w$. The word $w$ is
Lyndon (since $w\in\mathfrak{L}$), and thus nonempty. Hence, $\ell\left(
w\right)  \geq1$.
\par
Let $n=\ell\left(  w\right)  $. Thus, $n=\ell\left(  w\right)  \geq1$. We have
$w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  =\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ (since $\ell\left(  w\right)  =n$).
\par
We have $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $, so that $\Phi\left(
w\right)  =\widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{n}}$ (by the
definition of $\Phi\left(  w\right)  $). For every $i\in\left\{
1,2,\ldots,n\right\}  $, the word $\widetilde{w_{i}}$ is a Lyndon word over
the alphabet $\mathfrak{B}$ (by
(\ref{sol.words.lyndon.compositions.count.sol2.klyn}), applied to $k=w_{i}$).
Thus, $\widetilde{w}_1$, $\widetilde{w}_2$, $\ldots$, $\widetilde{w}_n$
are Lyndon words over the alphabet $\mathfrak{B}$.
\par
Let now $i\in\left\{  1,2,\ldots,n\right\}  $. Then, $\left(  w_{i}%
,w_{i+1},\ldots,w_{n}\right)  $ is a nonempty suffix of $w$ (since $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  $).
\par
Recall that $w$ is Lyndon. Hence, Corollary \ref{cor.words.lyndon.trivia}
(applied to $v=\left(  w_{i},w_{i+1},\ldots,w_{n}\right)  $) yields that
$\left(  w_{i},w_{i+1},\ldots,w_{n}\right)  \geq w$ (since $\left(
w_{i},w_{i+1},\ldots,w_{n}\right)  $ is a nonempty suffix of $w$). Hence,
$\Phi\left(  \left(  w_{i},w_{i+1},\ldots,w_{n}\right)  \right)  \geq
\Phi\left(  w\right)  $ (since the map $\Phi$ is strictly order-preserving).
But the definition of the map $\Phi$ yields $\Phi\left(  \left(  w_{i}%
,w_{i+1},\ldots,w_{n}\right)  \right)  =\widetilde{w_{i}}\widetilde{w_{i+1}%
}\cdots\widetilde{w_{n}}$. Thus, $\widetilde{w_{i}}\widetilde{w_{i+1}}%
\cdots\widetilde{w_{n}}=\Phi\left(  \left(  w_{i},w_{i+1},\ldots,w_{n}\right)
\right)  \geq\Phi\left(  w\right)  =\widetilde{w_{1}}\widetilde{w_{2}}%
\cdots\widetilde{w_{n}}$.
\par
Now, let us forget that we fixed $i$. We thus have shown that
$\widetilde{w_{i}}\widetilde{w_{i+1}}\cdots\widetilde{w_{n}}\geq
\widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{n}}$ for every
$i\in\left\{  1,2,\ldots,n\right\}  $. Hence, Exercise
\ref{exe.words.lyndon.concatn-lyndon} (applied to $\mathfrak{B}$ and
$\widetilde{w_{i}}$ instead of $\mathfrak{A}$ and $w_{i}$) yields that
$\widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{n}}$ is a Lyndon word.
In other words, $u$ is a Lyndon word (since $u=\Phi\left(  w\right)
=\widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{n}}$). More precisely,
$u$ is a Lyndon word over the alphabet $\mathfrak{B}$. In other words,
$u\in\mathfrak{L}_{\mathfrak{B}}$ (since $\mathfrak{L}_{\mathfrak{B}}$ is the
set of all Lyndon words over the alphabet $\mathfrak{B}$).
\par
On the other hand, $u\in\Phi\left(  \underbrace{\mathfrak{L}}_{\subset
\mathfrak{A}^{\ast}}\right)  \subset\Phi\left(  \mathfrak{A}^{\ast}\right)
=\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}$ (by
(\ref{sol.words.lyndon.compositions.count.sol2.2})), so that $u\notin%
\mathbf{1}\mathfrak{B}^{\ast}$. Hence, we cannot have $u=\mathbf{1}$ (because
if we had $u=\mathbf{1}$, we would have $u=\mathbf{1}=\mathbf{1}%
\underbrace{\varnothing}_{\in\mathfrak{B}^{\ast}}\in\mathbf{1}\mathfrak{B}%
^{\ast}$, which would contradict $u\notin\mathbf{1}\mathfrak{B}^{\ast}$).
Thus, we have $u\neq\mathbf{1}$, so that $u\notin\left\{  \mathbf{1}\right\}
$. Combined with $u\in\mathfrak{L}_{\mathfrak{B}}$, this yields $u\in
\mathfrak{L}_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}  $.
\par
Now, let us forget that we fixed $u$. We thus have proven that $u\in
\mathfrak{L}_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}  $ for every
$u\in\Phi\left(  \mathfrak{L}\right)  $. In other words, $\Phi\left(
\mathfrak{L}\right)  \subset\mathfrak{L}_{\mathfrak{B}}\setminus\left\{
\mathbf{1}\right\}  $, qed.} and $\mathfrak{L}_{\mathfrak{B}}\setminus\left\{
\mathbf{1}\right\}  \subset\Phi\left(  \mathfrak{L}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $u\in\mathfrak{L}_{\mathfrak{B}%
}\setminus\left\{  \mathbf{1}\right\}  $. We are going to prove that $u\in
\Phi\left(  \mathfrak{L}\right)  $.
\par
We have $u\in\mathfrak{L}_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}
\subset\mathfrak{L}_{\mathfrak{B}}$. Thus, $u$ is a Lyndon word over the
alphabet $\mathfrak{B}$ (since $\mathfrak{L}_{\mathfrak{B}}$ is the set of all
Lyndon words over the alphabet $\mathfrak{B}$). Also, $u\notin\left\{
\mathbf{1}\right\}  $ (since $u\in\mathfrak{L}_{\mathfrak{B}}\setminus\left\{
\mathbf{1}\right\}  $), so that $u\neq\mathbf{1}$.
\par
Let us now assume (for the sake of contradiction) that $u\in\mathbf{1}%
\mathfrak{B}^{\ast}$. Then, there exists a $u^{\prime}\in\mathfrak{B}^{\ast}$
such that $u=\mathbf{1}u^{\prime}$. Consider this $u^{\prime}$.
\par
The word $u$ is nonempty (since it is Lyndon), and thus the last letter of $u$
is well-defined. Let $g$ be this last letter of $u$. Then, $g$ is a suffix of
$u$. Clearly, $g$ is nonempty (when regarded as a word). Thus, Corollary
\ref{cor.words.lyndon.trivia} (applied to $\mathfrak{B}$, $u$ and $g$ instead
of $\mathfrak{A}$, $w$ and $v$) yields $g\geq u=\mathbf{1}u^{\prime}$, so that
$\mathbf{1}u^{\prime}\leq g=g\varnothing$. Also, $\ell\left(  g\right)  \geq1$
(since $g$ is nonempty), whence $\ell\left(  g\right)  \geq1=\ell\left(
\mathbf{1}\right)  $ and thus $\ell\left(  \mathbf{1}\right)  \leq\ell\left(
g\right)  $. Now, Proposition \ref{prop.words.lex}(f) (applied to
$\mathfrak{B}$, $\mathbf{1}$, $u^{\prime}$, $g$ and $\varnothing$ instead of
$\mathfrak{A}$, $a$, $b$, $c$ and $d$) yields $\mathbf{1}\leq g$. Thus,
$g\geq\mathbf{1}$. Since $g$ is a single letter, this yields that
$g=\mathbf{1}$ (because the only letter of $\mathfrak{B}$ which is
$\geq\mathbf{1}$ is $\mathbf{1}$ itself). Thus, $\mathbf{1}u^{\prime}\leq
g=\mathbf{1}=\mathbf{1}\varnothing$. Now, Proposition \ref{prop.words.lex}(f)
(applied to $\mathfrak{B}$, $\mathbf{1}$, $u^{\prime}$ and $\varnothing$
instead of $\mathfrak{A}$, $a$, $c$ and $d$) yields $u^{\prime}\leq
\varnothing$. This yields $u^{\prime}=\varnothing$ (since the only word which
is $\leq\varnothing$ is $\varnothing$ itself). Hence, $u=\mathbf{1}%
\underbrace{u^{\prime}}_{=\varnothing}=\mathbf{1}$, which contradicts
$u\neq\mathbf{1}$. This contradiction shows that our assumption (that
$u\in\mathbf{1}\mathfrak{B}^{\ast}$) was wrong. Hence, we have $u\notin%
\mathbf{1}\mathfrak{B}^{\ast}$.
\par
Since $u\in\mathfrak{B}^{\ast}$ and $u\notin\mathbf{1}\mathfrak{B}^{\ast}$, we
must have $u\in\mathfrak{B}^{\ast}\setminus\mathbf{1}\mathfrak{B}^{\ast}%
=\Phi\left(  \mathfrak{A}^{\ast}\right)  $ (by
(\ref{sol.words.lyndon.compositions.count.sol2.2})). In other words, there
exists a $w\in\mathfrak{A}^{\ast}$ such that $u=\Phi\left(  w\right)  $.
Consider this $w$.
\par
We have $u\neq\varnothing$ (since $u$ is nonempty) and thus $w\neq\varnothing$
(because otherwise, we would have $w=\varnothing$ and thus $u=\Phi\left(
\underbrace{w}_{=\varnothing}\right)  =\Phi\left(  \varnothing\right)
=\varnothing$, contradicting $u\neq\varnothing$). Hence, the word $w$ is
nonempty.
\par
Let now $v$ be a nonempty proper suffix of $w$. We assume (for the sake of
contradiction) that $v\leq w$. Since $v\neq w$ (because $v$ is a
\textbf{proper} suffix of $w$) and $v\leq w$, we have $v<w$. Thus,
$\Phi\left(  v\right)  <\Phi\left(  w\right)  $ (since the map $\Phi$ is
strictly order-preserving). Hence, $\Phi\left(  v\right)  <\Phi\left(
w\right)  =u$.
\par
Also, $v\neq\varnothing$ (since $v$ is nonempty). Thus, $\Phi\left(  v\right)
\neq\Phi\left(  \varnothing\right)  $ (because otherwise, we would have
$\Phi\left(  v\right)  =\Phi\left(  \varnothing\right)  $, so that
$v=\varnothing$ (since the map $\Phi$ is injective), which would contradict
$v\neq\varnothing$). Hence, $\Phi\left(  v\right)  \neq\Phi\left(
\varnothing\right)  =\varnothing$, so that the word $\Phi\left(  v\right)  $
is nonempty.
\par
But there exists a $p\in\mathfrak{A}^{\ast}$ such that $w=pv$ (since $v$ is a
suffix of $w$). Consider this $p$. We have $u=\Phi\left(  \underbrace{w}%
_{=pv}\right)  =\Phi\left(  pv\right)  =\Phi\left(  p\right)  \Phi\left(
v\right)  $ (since $\Phi$ is a monoid homomorphism). Thus, $\Phi\left(
v\right)  $ is a suffix of $u$. Now, Corollary \ref{cor.words.lyndon.trivia}
(applied to $\mathfrak{B}$, $u$ and $\Phi\left(  v\right)  $ instead of
$\mathfrak{A}$, $w$ and $v$) yields $\Phi\left(  v\right)  \geq u$. This
contradicts $\Phi\left(  v\right)  <u$. This contradiction shows that our
assumption (that $v\leq w$) was wrong. Hence, we cannot have $v\leq w$. We
thus have $v>w$.
\par
Now, let us forget that we fixed $v$. We thus have proven that every nonempty
proper suffix $v$ of $w$ satisfies $v>w$. Since the word $w$ is nonempty, this
yields that the word $w$ is Lyndon (by the definition of a Lyndon word). In
other words, $w\in\mathfrak{L}$ (since $\mathfrak{L}$ is the set of all Lyndon
words over the alphabet $\mathfrak{A}$). Now, $u=\Phi\left(  \underbrace{w}%
_{\in\mathfrak{L}}\right)  \in\Phi\left(  \mathfrak{L}\right)  $.
\par
Now, let us forget that we fixed $u$. We thus have shown that $u\in\Phi\left(
\mathfrak{L}\right)  $ for every $u\in\mathfrak{L}_{\mathfrak{B}}%
\setminus\left\{  \mathbf{1}\right\}  $. In other words, $\mathfrak{L}%
_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}  \subset\Phi\left(
\mathfrak{L}\right)  $, qed.}. Combining the latter two relations, we obtain%
\begin{equation}
\Phi\left(  \mathfrak{L}\right)  =\mathfrak{L}_{\mathfrak{B}}\setminus\left\{
\mathbf{1}\right\}  . \label{sol.words.lyndon.compositions.count.sol2.3}%
\end{equation}
Hence, the map $\Phi$ restricts to a bijection $\mathfrak{L}\rightarrow
\mathfrak{L}_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}  $.

Next, we notice that%
\begin{equation}
\ell\left(  \Phi\left(  w\right)  \right)  =\left\vert w\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{A}^{\ast}
\label{sol.words.lyndon.compositions.count.sol2.len}%
\end{equation}
(where the notation $\left\vert w\right\vert $ makes sense because every
$w\in\mathfrak{A}^{\ast}$ is a composition and thus has a
size)\footnote{\textit{Proof.} Let $w\in\mathfrak{A}^{\ast}$. Then, $w=\left(
w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  $, so that $\Phi\left(
w\right)  =\widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{\ell\left(
w\right)  }}$ (by the definition of $\Phi\left(  w\right)  $). Hence,
\begin{align*}
\ell\left(  \underbrace{\Phi\left(  w\right)  }_{=\widetilde{w_{1}%
}\widetilde{w_{2}}\cdots\widetilde{w_{\ell\left(  w\right)  }}}\right)   &
=\ell\left(  \widetilde{w_{1}}\widetilde{w_{2}}\cdots\widetilde{w_{\ell\left(
w\right)  }}\right)  =\ell\left(  \widetilde{w_{1}}\right)  +\ell\left(
\widetilde{w_{2}}\right)  +\cdots+\ell\left(  \widetilde{w_{\ell\left(
w\right)  }}\right)  =\sum_{i=1}^{\ell\left(  w\right)  }\underbrace{\ell
\left(  \widetilde{w_{i}}\right)  }_{\substack{=w_{i}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.lentilde}), applied to
}k=w_{i}\text{)}}}\\
&  =\sum_{i=1}^{\ell\left(  w\right)  }w_{i}=w_{1}+w_{2}+\cdots+w_{\ell\left(
w\right)  }=\left\vert w\right\vert
\end{align*}
(since $\left\vert w\right\vert =w_{1}+w_{2}+\cdots+w_{\ell\left(  w\right)
}$ (since $w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)
$)), qed.}.

[Incidentally, here is a similar identity which will not use:%
\begin{equation}
\left(  \text{the number of letters }\mathbf{0}\text{ in }\Phi\left(
w\right)  \right)  =\ell\left(  w\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}w\in\mathfrak{A}^{\ast} \label{sol.words.lyndon.compositions.count.sol2.num0}%
\end{equation}
\footnote{\textit{Proof.} Let $w\in\mathfrak{A}^{\ast}$.
\par
We notice that%
\begin{equation}
\left(  \text{the number of letters }\mathbf{0}\text{ in }%
\underbrace{\widetilde{k}}_{\substack{=\mathbf{01}^{k-1}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.tilde}))}}}\right)  =\left(
\text{the number of letters }\mathbf{0}\text{ in }\mathbf{01}^{k-1}\right)  =1
\label{sol.words.lyndon.compositions.count.sol2.num0.pf.1}%
\end{equation}
for every $k\in\mathfrak{A}$.
\par
We have $w=\left(  w_{1},w_{2},\ldots,w_{\ell\left(  w\right)  }\right)  $, so
that $\Phi\left(  w\right)  =\widetilde{w_{1}}\widetilde{w_{2}}\cdots
\widetilde{w_{\ell\left(  w\right)  }}$ (by the definition of $\Phi\left(
w\right)  $). Hence,
\begin{align*}
&  \left(  \text{the number of letters }\mathbf{0}\text{ in }\underbrace{\Phi
\left(  w\right)  }_{=\widetilde{w_{1}}\widetilde{w_{2}}\cdots
\widetilde{w_{\ell\left(  w\right)  }}}\right) \\
&  =\left(  \text{the number of letters }\mathbf{0}\text{ in }\widetilde{w_{1}%
}\widetilde{w_{2}}\cdots\widetilde{w_{\ell\left(  w\right)  }}\right) \\
&  =\left(  \text{the number of letters }\mathbf{0}\text{ in }\widetilde{w_{1}%
}\right)  +\left(  \text{the number of letters }\mathbf{0}\text{ in
}\widetilde{w_{2}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\cdots+\left(  \text{the number of letters }%
\mathbf{0}\text{ in }\widetilde{w_{\ell\left(  w\right)  }}\right) \\
&  =\sum_{i=1}^{\ell\left(  w\right)  }\underbrace{\left(  \text{the number of
letters }\mathbf{0}\text{ in }\widetilde{w_{i}}\right)  }%
_{\substack{=1\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.num0.pf.1}), applied to
}k=w_{i}\text{)}}}=\sum_{i=1}^{\ell\left(  w\right)  }1=\ell\left(  w\right)
,
\end{align*}
qed.}.]

Now, fix a positive integer $n$. The alphabet $\mathfrak{B}$ is finite and
satisfies $2=\left\vert \mathfrak{B}\right\vert $. Hence, Exercise
\ref{exe.words.lyndon.count} (applied to $\mathfrak{B}$ and $2$ instead of
$\mathfrak{A}$ and $q$) yields that the number of Lyndon words of length $n$
over the alphabet $\mathfrak{B}$ equals $\dfrac{1}{n}\sum\limits_{d\mid n}%
\mu\left(  d\right)  2^{n/d}$. That is,%
\begin{equation}
\left(  \text{the number of all Lyndon words of length }n\text{ over the
alphabet }\mathfrak{B}\right)  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  2^{n/d}. \label{sol.words.lyndon.compositions.count.sol2.import}%
\end{equation}


Now, let $\operatorname*{lync}n$ denote the number of Lyndon compositions of
size $n$. We need to prove that%
\begin{equation}
\operatorname*{lync}n=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)
\left(  2^{n/d}-1\right)  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(
d\right)  2^{n/d}-\delta_{n,1}.
\label{sol.words.lyndon.compositions.count.sol2.goal}%
\end{equation}


By the definition of $\operatorname*{lync}n$, we have%
\begin{align*}
\operatorname*{lync}n  &  =\left(  \text{the number of Lyndon compositions of
size }n\right) \\
&  =\left(  \text{the number of all }w\in\mathfrak{L}\text{ such that
}\left\vert w\right\vert =n\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the set of all Lyndon compositions
is }\mathfrak{L}\right) \\
&  =\left\vert \left\{  w\in\mathfrak{L}\ \mid\ \underbrace{\left\vert
w\right\vert }_{\substack{=\ell\left(  \Phi\left(  w\right)  \right)
\\\text{(by (\ref{sol.words.lyndon.compositions.count.sol2.len}))}%
}}=n\right\}  \right\vert \\
&  =\left\vert \left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  \Phi\left(
w\right)  \right)  =n\right\}  \right\vert =\left\vert \underbrace{\left\{
w\in\mathfrak{L}_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}
\ \mid\ \ell\left(  w\right)  =n\right\}  }_{=\left\{  w\in\mathfrak{L}%
_{\mathfrak{B}}\ \mid\ \ell\left(  w\right)  =n\right\}  \setminus\left\{
w\in\left\{  \mathbf{1}\right\}  \ \mid\ \ell\left(  w\right)  =n\right\}
}\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }w\text{ for }\Phi\left(  w\right)  \text{,
since the map}\\
\Phi\text{ restricts to a bijection }\mathfrak{L}\rightarrow\mathfrak{L}%
_{\mathfrak{B}}\setminus\left\{  \mathbf{1}\right\}
\end{array}
\right) \\
&  =\left\vert \left\{  w\in\mathfrak{L}_{\mathfrak{B}}\ \mid\ \ell\left(
w\right)  =n\right\}  \setminus\left\{  w\in\left\{  \mathbf{1}\right\}
\ \mid\ \ell\left(  w\right)  =n\right\}  \right\vert \\
&  =\underbrace{\left\vert \left\{  w\in\mathfrak{L}_{\mathfrak{B}}%
\ \mid\ \ell\left(  w\right)  =n\right\}  \right\vert }_{\substack{=\left(
\text{the number of all }w\in\mathfrak{L}_{\mathfrak{B}}\text{ such that }%
\ell\left(  w\right)  =n\right)  \\=\left(  \text{the number of all Lyndon
words of length }n\text{ over the alphabet }\mathfrak{B}\right)
\\\text{(since }\mathfrak{L}_{B}\text{ is the set of all Lyndon words over the
alphabet }\mathfrak{B}\text{)}}}-\underbrace{\left\vert \left\{  w\in\left\{
\mathbf{1}\right\}  \ \mid\ \ell\left(  w\right)  =n\right\}  \right\vert
}_{\substack{=\delta_{n,1}\\\text{(since }\ell\left(  \mathbf{1}\right)
=1\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\{  w\in\underbrace{\left\{
\mathbf{1}\right\}  }_{\subset\mathfrak{L}_{\mathfrak{B}}}\ \mid\ \ell\left(
w\right)  =n\right\}  \subset\left\{  w\in\mathfrak{L}_{\mathfrak{B}}%
\ \mid\ \ell\left(  w\right)  =n\right\}  \right) \\
&  =\underbrace{\left(  \text{the number of all Lyndon words of length
}n\text{ over the alphabet }\mathfrak{B}\right)  }_{\substack{=\dfrac{1}%
{n}\sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}\\\text{(by
(\ref{sol.words.lyndon.compositions.count.sol2.import}))}}}-\delta_{n,1}\\
&  =\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  2^{n/d}-\delta
_{n,1}=\dfrac{1}{n}\sum\limits_{d\mid n}\mu\left(  d\right)  \left(
2^{n/d}-1\right)
\end{align*}
(by Lemma \ref{lem.sol.words.lyndon.compositions.count.2}). This solves
Exercise \ref{exe.words.lyndon.compositions.count} again.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.QSym.lyndon.MaMb}}

\textit{Solution to Exercise \ref{exe.prop.QSym.lyndon.MaMb}.}

\begin{proof}
[Proof of Proposition \ref{prop.QSym.lyndon.MaMb}.]Write $\alpha$ and $\beta$
as $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ and
$\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{m}\right)  $, respectively;
then $\ell\left(  \alpha\right)  =\ell$ and $\ell\left(  \beta\right)  =m$.

Fix three disjoint chain posets $\left(  i_{1}<i_{2}<\cdots<i_{\ell
}\right)  $, $\left(  j_{1}<j_{2}<\cdots<j_{m}\right)  $ and $\left(
k_{1}<k_{2}<k_{3}<\cdots\right)  $. For any $p\in \NN$, we define a
\emph{$p$-shuffling map} to mean a map $f$ from the disjoint union of
two chains to a chain
\[
\left(  i_{1}<i_{2}<\cdots<i_{\ell}\right)  \sqcup\left(  j_{1}<j_{2}%
<\cdots<j_{m}\right)  \overset{f}{\longrightarrow}\left(  k_{1}<k_{2}%
<\cdots<k_{p}\right)
\]
which is both surjective and strictly order-preserving
(that is, if $x$ and $y$ are two elements in the domain of $f$
satisfying $x<y$, then
$f\left(  x\right)  <f\left(  y\right)  $). For every $p\in \NN$ and
every $p$-shuffling map $f$, we define a composition $\operatorname*{wt}%
\left(  f\right)  :=\left(  \operatorname{wt}_{1}\left(  f\right)
,\operatorname{wt}_{2}\left(  f\right) , \ldots,\operatorname{wt}%
_{p}\left(  f\right)  \right)  $ by $\operatorname{wt}_{s}\left(  f\right)
:=\sum_{i_{u}\in f^{-1}\left(  k_{s}\right)  }\alpha_{u}+\sum_{j_{v}\in
f^{-1}\left(  k_{s}\right)  }\beta_{v}$. Then, Proposition
\ref{QSym-multiplication} yields
\[
M_{\alpha}M_{\beta}=\sum_{f}M_{\operatorname{wt}\left(f\right)},
\]
where the sum is over all $p\in \NN$ and all $p$-shuffling maps $f$. In
other words,%
\begin{align}
M_{\alpha}M_{\beta}  &  =\sum\limits_{\substack{\left(  p,f\right)  ;\\p\in
 \NN \text{;}\\f\text{ is a }p\text{-shuffling map}}}M_{\operatorname{wt}%
\left(f\right)}
=\sum_{p\in \NN }\sum_{f\text{ is a }p\text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}\nonumber\\
&  =\sum\limits_{\substack{p\in \NN ;\\p\leq\ell+m}}\sum_{f\text{ is a
}p\text{-shuffling map}}M_{\operatorname{wt}\left(f\right)}
+\sum\limits_{\substack{p\in
 \NN ;\\p>\ell+m}}\underbrace{\sum_{f\text{ is a }p\text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}
}_{\substack{=\left(  \text{empty sum}\right)
\\\text{(because there exist no }p\text{-shuffling maps if }p>\ell
+m\\\text{(since a }p\text{-shuffling map has to be a surjective
map}\\\text{from an }\left(  \ell+m\right)  \text{-element set to a
}p\text{-element set,}\\\text{and such maps don't exist if }p>\ell+m\text{))}%
}}\nonumber\\
&  =\sum\limits_{\substack{p\in \NN ;\\p\leq\ell+m}}\sum_{f\text{ is a
}p\text{-shuffling map}}M_{\operatorname{wt}\left(f\right)}
+\sum\limits_{\substack{p\in
 \NN ;\\p>\ell+m}}\underbrace{\left(  \text{empty sum}\right)  }%
_{=0}\nonumber\\
&  =\sum\limits_{\substack{p\in \NN ;\\p\leq\ell+m}}\sum_{f\text{ is a
}p\text{-shuffling map}}M_{\operatorname{wt}\left(f\right)}
+\underbrace{\sum\limits
_{\substack{p\in \NN ;\\p>\ell+m}}0}_{=0}=\sum\limits_{\substack{p\in
 \NN ;\\p\leq\ell+m}}\sum_{f\text{ is a }p\text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}
\label{pf.prop.QSym.lyndon.MaMb.sum0}\\
&  =\sum_{f\text{ is an }\left(  \ell+m\right)  \text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}
+\sum\limits_{\substack{p\in \NN ;\\p<\ell+m}}
\sum_{f\text{ is a }p\text{-shuffling map}}
M_{\operatorname{wt}\left(f\right)}
\label{pf.prop.QSym.lyndon.MaMb.sum}
\end{align}
(here, we have split off the addend for $p=\ell+m$ from the sum).

We now notice that
\begin{align}
&  \sum\limits_{\substack{p\in \NN ;\\p<\ell+m}}\sum_{f\text{ is a
}p\text{-shuffling map}}M_{\operatorname{wt}\left(f\right)}
\nonumber\\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in\mathfrak{A}^{\ast}\text{ satisfying }\ell\left(  \delta\right)
<\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  \right)
\label{pf.prop.QSym.lyndon.MaMb.2ndterm}%
\end{align}
\footnote{\textit{Proof of (\ref{pf.prop.QSym.lyndon.MaMb.2ndterm}):} If
$p\in \NN$ is such that $p<\ell+m$, and if $f$ is a $p$-shuffling map,
then $\operatorname*{wt}\left(f\right)
=\left(  \operatorname{wt}_{1}\left(  f\right)
,\ldots,\operatorname{wt}_{p}\left(  f\right)  \right)  $ is a composition of
length $p<\underbrace{\ell}_{=\ell\left(  \alpha\right)  }+\underbrace{m}%
_{=\ell\left(  \beta\right)  }=\ell\left(  \alpha\right)  +\ell\left(
\beta\right)  $. Hence, if $p\in \NN$ is such that $p<\ell+m$, and if
$f$ is a $p$-shuffling map, then $M_{\operatorname{wt}\left(f\right)}$
is a term of the
form $M_{\delta}$ with $\delta\in\mathfrak{A}^{\ast}$ satisfying $\ell\left(
\delta\right)  <\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  $.
Therefore, $\sum\limits_{\substack{p\in \NN ;\\p<\ell+m}}\sum_{f\text{ is a
}p\text{-shuffling map}}M_{\operatorname{wt}\left(f\right)}$
is a sum of terms of the form
$M_{\delta}$ with $\delta\in\mathfrak{A}^{\ast}$ satisfying $\ell\left(
\delta\right)  <\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  $. This
proves (\ref{pf.prop.QSym.lyndon.MaMb.2ndterm}).}. Our next goal is to prove%
\begin{equation}
\sum_{f\text{ is an }\left(  \ell+m\right)  \text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}
=\sum_{\sigma\in\operatorname{Sh}_{\ell,m}%
}M_{\alpha\underset{\sigma}{\shuffle}\beta}.
\label{pf.prop.QSym.lyndon.MaMb.1stterm}%
\end{equation}
Once (\ref{pf.prop.QSym.lyndon.MaMb.1stterm}) is proven, we will immediately
conclude that%
\[
\sum_{f\text{ is an }\left(  \ell+m\right)  \text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}
=\sum_{\sigma\in\operatorname{Sh}_{\ell,m}%
}M_{\alpha\underset{\sigma}{\shuffle}\beta}=\sum_{\gamma\in\alpha
\shuffle\beta}M_{\gamma}%
\]
(since the multiset $\alpha\shuffle\beta$ is defined as $\left\{
\alpha\underset{\sigma}{\shuffle}\beta\ :\ \sigma\in\operatorname{Sh}_{\ell
,m}\right\}  _{\text{multiset}}$), and therefore we will obtain%
\begin{align*}
M_{\alpha}M_{\beta}  &  =\underbrace{\sum_{f\text{ is an }\left(
\ell+m\right)  \text{-shuffling map}}
M_{\operatorname{wt}\left(f\right)}}_{=\sum_{\gamma
\in\alpha\shuffle\beta}M_{\gamma}}+\underbrace{\sum\limits_{\substack{p\in
 \NN ;\\p<\ell+m}}\sum_{f\text{ is a }p\text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}}_{\substack{=
\left(  \text{a sum of terms of the form
}M_{\delta}\text{ with }\delta\in\mathfrak{A}^{\ast}\text{ satisfying }%
\ell\left(  \delta\right)  <\ell\left(  \alpha\right)  +\ell\left(
\beta\right)  \right)  \\\text{(by (\ref{pf.prop.QSym.lyndon.MaMb.2ndterm}))}%
}}\\
&  =\sum_{\gamma\in\alpha\shuffle\beta}M_{\gamma}+\left(  \text{a sum of terms
of the form }M_{\delta}\text{ with }\delta\in\mathfrak{A}^{\ast}\text{
satisfying }\ell\left(  \delta\right)  <\ell\left(  \alpha\right)
+\ell\left(  \beta\right)  \right)  ,
\end{align*}
which will complete the proof of Proposition \ref{prop.QSym.lyndon.MaMb}.
Hence, it only remains to prove (\ref{pf.prop.QSym.lyndon.MaMb.1stterm}).

We let $D$ be the poset $\left(  i_{1}<i_{2}<\cdots<i_{\ell}\right)
\sqcup\left(  j_{1}<j_{2}<\cdots<j_{m}\right)  $, and we let $R$ be the poset
$\left(  k_{1}<k_{2}<\cdots<k_{\ell+m}\right)  $. Note that $\left\vert
D\right\vert =\ell+m=\left\vert R\right\vert $. Hence, $D$ and $R$ are two
finite sets of the same cardinality. Consequently, a given map from $D$ to $R$
is surjective if and only if it is bijective.\footnote{Of course, the letters
$D$ and $R$ have been chosen to remind of ``domain'' and ``range''.}

Recall that an $\left(  \ell+m\right)  $-shuffling map means a map $f$ from
$\left(  i_{1}<i_{2}<\cdots<i_{\ell}\right)  \sqcup\left(  j_{1}<j_{2}%
<\cdots<j_{m}\right)  $ to $\left(  k_{1}<k_{2}<\cdots<k_{\ell+m}\right)  $
which is both surjective and strictly order-preserving
(by the definition of an ``$\left(\ell+m\right)$-shuffling map'').
In other words, an $\left(
\ell+m\right)  $-shuffling map means a map $f$ from $D$ to $R$ which is both
surjective and strictly order-preserving\footnote{since $D=\left(  i_{1}%
<i_{2}<\cdots<i_{\ell}\right)  \sqcup\left(  j_{1}<j_{2}<\cdots<j_{m}\right)
$ and $R=\left(  k_{1}<k_{2}<\cdots<k_{\ell+m}\right)  $}. In other words, an
$\left(  \ell+m\right)  $-shuffling map means a map $f$ from $D$ to $R$ which
is both bijective and strictly order-preserving\footnote{This is because a
given map from $D$ to $R$ is surjective if and only if it is bijective.}.

Let us now define a bijection $\mathbf{d}:\left\{  1,2,\ldots,\ell+m\right\}
\rightarrow D$ by%
\[
\mathbf{d}\left(  u\right)  =
\begin{cases}
i_{u},& \text{if }u\leq\ell;\\
j_{u-\ell},& \text{if }u>\ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\left\{  1,2,\ldots
,\ell+m\right\}  .
\]
Let us further define a bijection $\mathbf{r}:\left\{  1,2,\ldots
,\ell+m\right\}  \rightarrow R$ by%
\[
\mathbf{r}\left(  u\right)  =k_{u}\ \ \ \ \ \ \ \ \ \ \text{for every }%
u\in\left\{  1,2,\ldots,\ell+m\right\}  .
\]
Notice that $\mathbf{r}$ is an isomorphism of posets, where we endow the set
$\left\{  1,2,\ldots,\ell+m\right\}  $ with its natural total order (i.e., the
order $1<2<\cdots<\ell+m$).

We can then define a bijection%
\begin{align*}
\Phi &  :\left(  \text{the set of all bijective maps from }\left\{
1,2,\ldots,\ell+m\right\}  \text{ to }\left\{  1,2,\ldots,\ell+m\right\}
\right) \\
&  \rightarrow\left(  \text{the set of all bijective maps from }D\text{ to
}R\right)
\end{align*}
by setting%
\[
\Phi\left(  \sigma\right)  =\mathbf{r}\circ\sigma\circ\mathbf{d}%
^{-1}\ \ \ \ \ \ \ \ \ \ \text{for every bijective map }\sigma:\left\{
1,2,\ldots,\ell+m\right\}  \rightarrow\left\{  1,2,\ldots,\ell+m\right\}  .
\]
\footnote{This is a bijection since both $\mathbf{d}$ and $\mathbf{r}$ are
bijections.} Consider this bijection $\Phi$. Then, $\Phi$ is a bijection from
$\Symm_{\ell+m}$ to $\left(  \text{the set of all bijective maps from
}D\text{ to }R\right)  $ (because $\left(  \text{the set of all bijective maps
from }\left\{  1,2,\ldots,\ell+m\right\}  \text{ to }\left\{  1,2,\ldots
,\ell+m\right\}  \right)  =\Symm_{\ell+m}$). It is now easy to see that
if $\sigma\in\Symm_{\ell+m}$, then we have the following equivalence of
assertions:%
\begin{equation}
\left(  \sigma\in\operatorname{Sh}_{\ell,m}\right)
\Longleftrightarrow\left(  \text{the map }\Phi\left(  \sigma^{-1}\right)
:D\rightarrow R\text{ is strictly order-preserving}\right)  .
\label{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi}%
):} Let $\sigma\in\Symm_{\ell+m}$. The poset $D$ is the disjoint union
of the totally ordered posets $\left(  i_{1}<i_{2}<\cdots<i_{\ell}\right)  $
and $\left(  j_{1}<j_{2}<\cdots<j_{m}\right)  $. Hence, if $P$ is any other
poset, and $f:D\rightarrow P$ is any map, then the map $f:D\rightarrow P$ is
strictly order-preserving if and only if it satisfies%
\[
\left(  f\left(  i_{1}\right)  <f\left(  i_{2}\right)  <\cdots<f\left(
i_{\ell}\right)  \text{ and }f\left(  j_{1}\right)  <f\left(  j_{2}\right)
<\cdots<f\left(  j_{m}\right)  \right)  .
\]
Applying this to $P=R$ and $f=\Phi\left(  \sigma^{-1}\right)  $, we conclude
that the map $\Phi\left(  \sigma^{-1}\right)  :D\rightarrow R$ is strictly
order-preserving if and only if it satisfies%
\begin{align*}
&  \left(  \left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
i_{1}\right)  <\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
i_{2}\right)  <\cdots<\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
i_{\ell}\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\left(  \Phi\left(  \sigma
^{-1}\right)  \right)  \left(  j_{1}\right)  <\left(  \Phi\left(  \sigma
^{-1}\right)  \right)  \left(  j_{2}\right)  <\cdots<\left(  \Phi\left(
\sigma^{-1}\right)  \right)  \left(  j_{m}\right)  \right)  .
\end{align*}
In other words, we have the following equivalence of assertions:%
\begin{align}
&  \left(  \text{the map }\Phi\left(  \sigma^{-1}\right)  :D\rightarrow
R\text{ is strictly order-preserving}\right) \nonumber\\
&  \Longleftrightarrow\left(  \left(  \Phi\left(  \sigma^{-1}\right)  \right)
\left(  i_{1}\right)  <\left(  \Phi\left(  \sigma^{-1}\right)  \right)
\left(  i_{2}\right)  <\cdots<\left(  \Phi\left(  \sigma^{-1}\right)  \right)
\left(  i_{\ell}\right)  \right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\left(  \Phi\left(  \sigma
^{-1}\right)  \right)  \left(  j_{1}\right)  <\left(  \Phi\left(  \sigma
^{-1}\right)  \right)  \left(  j_{2}\right)  <\cdots<\left(  \Phi\left(
\sigma^{-1}\right)  \right)  \left(  j_{m}\right)  \right)  .
\label{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi.pf.1}%
\end{align}
\par
Now, recall that $\Phi\left(  \sigma^{-1}\right)  =\mathbf{r}\circ\sigma
^{-1}\circ\mathbf{d}^{-1}$ (by the definition of $\Phi$). Hence, every
$u\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies%
\[
\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(  i_{u}\right)
=\left(  \mathbf{r}\circ\sigma^{-1}\circ\mathbf{d}^{-1}\right)  \left(
i_{u}\right)  =\mathbf{r}\left(  \sigma^{-1}\left(  \underbrace{\mathbf{d}%
^{-1}\left(  i_{u}\right)  }_{\substack{=u\\\text{(since }\mathbf{d}\left(
u\right)  =i_{u}\\\text{(by the definition of }\mathbf{d}\text{))}}}\right)
\right)  =\mathbf{r}\left(  \sigma^{-1}\left(  u\right)  \right)  .
\]
Hence, we have the following equivalence of assertions:%
\begin{align}
&  \left(  \left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
i_{1}\right)  <\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
i_{2}\right)  <\cdots<\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
i_{\ell}\right)  \right) \nonumber\\
&  \Longleftrightarrow\left(  \mathbf{r}\left(  \sigma^{-1}\left(  1\right)
\right)  <\mathbf{r}\left(  \sigma^{-1}\left(  2\right)  \right)
<\cdots<\mathbf{r}\left(  \sigma^{-1}\left(  \ell\right)  \right)  \right)
\nonumber\\
&  \Longleftrightarrow\left(  \sigma^{-1}\left(  1\right)  <\sigma^{-1}\left(
2\right)  <\cdots<\sigma^{-1}\left(  \ell\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{r}\text{ is an isomorphism of
posets}\right)  . \label{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi.pf.2}%
\end{align}
Similarly, we have the following equivalence of assertions:
\begin{align}
&  \left(  \left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
j_{1}\right)  <\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
j_{2}\right)  <\cdots<\left(  \Phi\left(  \sigma^{-1}\right)  \right)  \left(
j_{m}\right)  \right) \nonumber\\
&  \Longleftrightarrow\left(  \sigma^{-1}\left(  \ell+1\right)  <\sigma
^{-1}\left(  \ell+2\right)  <\cdots<\sigma^{-1}\left(  \ell+m\right)  \right)
. \label{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi.pf.3}%
\end{align}
Now, the equivalence (\ref{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi.pf.1})
becomes%
\begin{align*}
&  \left(  \text{the map }\Phi\left(  \sigma^{-1}\right)  :D\rightarrow
R\text{ is strictly order-preserving}\right) \\
&  \Longleftrightarrow\left(  \underbrace{\left(  \Phi\left(  \sigma
^{-1}\right)  \right)  \left(  i_{1}\right)  <\left(  \Phi\left(  \sigma
^{-1}\right)  \right)  \left(  i_{2}\right)  <\cdots<\left(  \Phi\left(
\sigma^{-1}\right)  \right)  \left(  i_{\ell}\right)  }_{\substack{\text{this
is equivalent to}\\\left(  \sigma^{-1}\left(  1\right)  <\sigma^{-1}\left(
2\right)  <\cdots<\sigma^{-1}\left(  \ell\right)  \right)  \\\text{(by
(\ref{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi.pf.2}))}}}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\underbrace{\left(  \Phi\left(
\sigma^{-1}\right)  \right)  \left(  j_{1}\right)  <\left(  \Phi\left(
\sigma^{-1}\right)  \right)  \left(  j_{2}\right)  <\cdots<\left(  \Phi\left(
\sigma^{-1}\right)  \right)  \left(  j_{m}\right)  }_{\substack{\text{this is
equivalent to}\\\left(  \sigma^{-1}\left(  \ell+1\right)  <\sigma^{-1}\left(
\ell+2\right)  <\cdots<\sigma^{-1}\left(  \ell+m\right)  \right)  \\\text{(by
(\ref{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi.pf.3}))}}}\right) \\
&  \Longleftrightarrow\left(  \sigma^{-1}\left(  1\right)  <\sigma^{-1}\left(
2\right)  <\cdots<\sigma^{-1}\left(  \ell\right)  \text{ and }\sigma
^{-1}\left(  \ell+1\right)  <\sigma^{-1}\left(  \ell+2\right)  <\cdots
<\sigma^{-1}\left(  \ell+m\right)  \right) \\
&  \Longleftrightarrow\left(  \sigma\in\operatorname{Sh}_{\ell
,m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\operatorname{Sh}_{\ell,m}\right)  .
\end{align*}
This proves (\ref{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi}).} Moreover,
every $\sigma\in\operatorname{Sh}_{\ell,m}$ satisfies%
\begin{equation}
\operatorname*{wt}\left(  \Phi\left(  \sigma^{-1}\right)  \right)
=\alpha\underset{\sigma}{\shuffle}\beta.
\label{pf.prop.QSym.lyndon.MaMb.wt-is-shuffle}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.QSym.lyndon.MaMb.wt-is-shuffle}):}
Let $\sigma\in\operatorname{Sh}_{\ell,m}$. Set $f=\Phi\left(
\sigma^{-1}\right)  $. Then, $f$ is a bijective map from $D$ to $R$ (because
the domain of $\Phi$ is $\left(  \text{the set of all bijective maps from
}D\text{ to }R\right)  $).
\par
By the definition of $\operatorname*{wt}\left(  f\right)  $, we have
$\operatorname*{wt}\left(  f\right)  =\left(  \operatorname{wt}_{1}\left(
f\right)  ,\operatorname{wt}_{2}\left(  f\right)  ,\ldots
,\operatorname{wt}_{\ell+m}\left(  f\right)  \right)  $, where we set
$\operatorname{wt}_{s}\left(  f\right)  :=\sum_{i_{u}\in f^{-1}\left(
k_{s}\right)  }\alpha_{u}+\sum_{j_{v}\in f^{-1}\left(  k_{s}\right)  }%
\beta_{v}$ for every $s\in\left\{  1,2,\ldots,\ell+m\right\}  $.
\par
On the other hand, let $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell
+m}\right)  $ be the concatenation $\alpha\cdot\beta=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell},\beta_{1},\beta_{2},\ldots,\beta_{m}\right)
$. Then, $\alpha\underset{\sigma}{\shuffle}\beta=\left(  \gamma_{\sigma\left(
1\right)  },\gamma_{\sigma\left(  2\right)  },\ldots,\gamma_{\sigma\left(
\ell+m\right)  }\right)  $.
\par
Let $s\in\left\{  1,2,\ldots,\ell+m\right\}  $. We are going to prove that
$\gamma_{\sigma\left(  s\right)  }=\operatorname{wt}_{s}\left(  f\right)  $.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} We have $\sigma\left(  s\right)  \leq\ell$.
\par
\textit{Case 2:} We have $\sigma\left(  s\right)  >\ell$.
\par
Let us consider Case 1 first. In this case, we have $\sigma\left(  s\right)
\leq\ell$. Thus, $\sigma\left(  s\right)  \in\left\{  1,2,\ldots,\ell\right\}
$, so that $i_{\sigma\left(  s\right)  }$ is well-defined. We have
$\mathbf{d}\left(  \sigma\left(  s\right)  \right)  =i_{\sigma\left(
s\right)  }$ (by the definition of $\mathbf{d}$, since $\sigma\left(
s\right)  \leq\ell$), so that $\mathbf{d}^{-1}\left(  i_{\sigma\left(
s\right)  }\right)  =\sigma\left(  s\right)  $. Now, $f=\Phi\left(
\sigma^{-1}\right)  =\mathbf{r}\circ\sigma^{-1}\circ\mathbf{d}^{-1}$ (by the
definition of $\Phi$), so that%
\[
f\left(  i_{\sigma\left(  s\right)  }\right)  =\left(  \mathbf{r}\circ
\sigma^{-1}\circ\mathbf{d}^{-1}\right)  \left(  i_{\sigma\left(  s\right)
}\right)  =\mathbf{r}\left(  \sigma^{-1}\left(  \underbrace{\mathbf{d}%
^{-1}\left(  i_{\sigma\left(  s\right)  }\right)  }_{=\sigma\left(  s\right)
}\right)  \right)  =\mathbf{r}\left(  \underbrace{\sigma^{-1}\left(
\sigma\left(  s\right)  \right)  }_{=s}\right)  =\mathbf{r}\left(  s\right)
=k_{s}%
\]
(by the definition of $\mathbf{r}$). Since the map $f$ is a bijection, this
yields that the \textbf{set} $f^{-1}\left(  k_{s}\right)  $ equals $\left\{
i_{\sigma\left(  s\right)  }\right\}  $. Hence, the sum $\sum_{i_{u}\in
f^{-1}\left(  k_{s}\right)  }\alpha_{u}$ contains precisely one addend, namely
the one for $u=\sigma\left(  s\right)  $; as a consequence, this sum
simplifies to $\sum_{i_{u}\in f^{-1}\left(  k_{s}\right)  }\alpha_{u}%
=\alpha_{\sigma\left(  s\right)  }$. On the other hand, the sum $\sum
_{j_{v}\in f^{-1}\left(  k_{s}\right)  }\beta_{v}$ is empty (since the set
$f^{-1}\left(  k_{s}\right)  $ equals $\left\{  i_{\sigma\left(  s\right)
}\right\}  $, and thus contains no elements of the form $j_{v}$), and thus
vanishes, i.e., we have $\sum_{j_{v}\in f^{-1}\left(  k_{s}\right)  }\beta
_{v}=0$. Now,%
\[
\operatorname{wt}_{s}\left(  f\right)  =\underbrace{\sum_{i_{u}\in
f^{-1}\left(  k_{s}\right)  }\alpha_{u}}_{=\alpha_{\sigma\left(  s\right)  }%
}+\underbrace{\sum_{j_{v}\in f^{-1}\left(  k_{s}\right)  }\beta_{v}}%
_{=0}=\alpha_{\sigma\left(  s\right)  }=\gamma_{\sigma\left(  s\right)  }%
\]
(because $\gamma_{\sigma\left(  s\right)  }=\alpha_{\sigma\left(  s\right)  }$
(since $\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell+m}\right)  =\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell},\beta_{1},\beta_{2},\ldots
,\beta_{m}\right)  $ and $\sigma\left(  s\right)  \leq\ell$)). In other words,
$\gamma_{\sigma\left(  s\right)  }=\operatorname{wt}_{s}\left(  f\right)  $.
\par
We have thus shown that $\gamma_{\sigma\left(  s\right)  }=\operatorname{wt}%
_{s}\left(  f\right)  $ holds in Case 1. A similar argument (but relying on
$\mathbf{d}\left(  \sigma\left(  s\right)  \right)  =j_{\sigma\left(
s\right)  -\ell}$ instead of $\mathbf{d}\left(  \sigma\left(  s\right)
\right)  =i_{\sigma\left(  s\right)  }$) shows that $\gamma_{\sigma\left(
s\right)  }=\operatorname{wt}_{s}\left(  f\right)  $ holds in Case 2.
\par
Thus, $\gamma_{\sigma\left(  s\right)  }=\operatorname{wt}_{s}\left(
f\right)  $ holds in both Cases 1 and 2. Since these two Cases cover all
possibilities, this yields that $\gamma_{\sigma\left(  s\right)
}=\operatorname{wt}_{s}\left(  f\right)  $ always holds.
\par
Hence, $\gamma_{\sigma\left(  s\right)  }=\operatorname{wt}_{s}\left(
f\right)  $ for every $s\in\left\{  1,2,\ldots,\ell+m\right\}  $. Thus,
$\left(  \gamma_{\sigma\left(  1\right)  },\gamma_{\sigma\left(  2\right)
},\ldots,\gamma_{\sigma\left(  \ell+m\right)  }\right)  =\left(
\operatorname{wt}_{1}\left(  f\right)  ,\operatorname{wt}_{2}\left(
f\right) , \ldots,\operatorname{wt}_{\ell+m}\left(  f\right)  \right)  $.
Hence,%
\[
\alpha\underset{\sigma}{\shuffle}\beta=\left(  \gamma_{\sigma\left(  1\right)
},\gamma_{\sigma\left(  2\right)  },\ldots,\gamma_{\sigma\left(
\ell+m\right)  }\right)  =\left(  \operatorname{wt}_{1}\left(  f\right)
,\operatorname{wt}_{2}\left(  f\right) , \ldots,\operatorname{wt}%
_{\ell+m}\left(  f\right)  \right)  =\operatorname*{wt}\left(  \underbrace{f}%
_{=\Phi\left(  \sigma^{-1}\right)  }\right)  =\operatorname*{wt}\left(
\Phi\left(  \sigma^{-1}\right)  \right)  .
\]
This proves (\ref{pf.prop.QSym.lyndon.MaMb.wt-is-shuffle}).}

Now, recall that an $\left(  \ell+m\right)  $-shuffling map means a map $f$
from $D$ to $R$ which is both bijective and strictly order-preserving. In
other words, an $\left(  \ell+m\right)  $-shuffling map means a bijective map
from $D$ to $R$ which is strictly order-preserving. Hence,%
\begin{align*}
&  \sum_{f\text{ is an }\left(  \ell+m\right)  \text{-shuffling map}%
}M_{\operatorname{wt}\left(f\right)}\\
&  =\sum\limits_{\substack{f\text{ is a bijective map from }D\text{ to }%
R\text{;}\\f\text{ is strictly order-preserving}}}
M_{\operatorname{wt}\left(f\right)}%
=\sum\limits_{\substack{\sigma\in\Symm_{\ell+m};\\\Phi\left(  \sigma\right)
\text{ is strictly order-preserving}}}M_{\operatorname{wt}\left(  \Phi\left(
\sigma\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\Phi\left(  \sigma\right)  \text{ for
}f\text{ in the sum,}\\
\text{since }\Phi:\Symm_{\ell+m}\rightarrow\left(  \text{the set of all
bijective maps from }D\text{ to }R\right) \\
\text{is a bijection}%
\end{array}
\right) \\
&  =\underbrace{\sum\limits_{\substack{\sigma\in\Symm_{\ell+m};\\\Phi\left(
\sigma^{-1}\right)  \text{ is strictly order-preserving}}}}_{\substack{=\sum\limits
_{\substack{\sigma\in\Symm_{\ell+m};\\\sigma\in
\operatorname{Sh}_{\ell,m}}}\\\text{(because }\Phi\left(  \sigma^{-1}\right)  \text{
is strictly order-preserving if and only if }\sigma\in
\operatorname{Sh}_{\ell,m}\\\text{(according to
(\ref{pf.prop.QSym.lyndon.MaMb.shuffle-through-Phi})))}}}M_{\operatorname{wt}%
\left(  \Phi\left(  \sigma^{-1}\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\sigma^{-1}\text{ for }\sigma\text{ in the
sum,}\\
\text{since the map }\Symm_{\ell+m}\rightarrow\Symm_{\ell
+m},\ \sigma\mapsto\sigma^{-1}\text{ is a bijection}%
\end{array}
\right) \\
&  =\underbrace{\sum\limits_{\substack{\sigma\in\Symm_{\ell+m};\\\sigma
\in\operatorname{Sh}_{\ell,m}}}}_{=\sum_{\sigma\in
\operatorname{Sh}_{\ell,m}}}\underbrace{M_{\operatorname{wt}\left(  \Phi\left(
\sigma^{-1}\right)  \right)  }}_{\substack{=M_{\alpha\underset{\sigma
}{\shuffle}\beta}\\\text{(by (\ref{pf.prop.QSym.lyndon.MaMb.wt-is-shuffle}))}%
}}=\sum_{\sigma\in\operatorname{Sh}_{\ell,m}}M_{\alpha
\underset{\sigma}{\shuffle}\beta}.
\end{align*}
This proves (\ref{pf.prop.QSym.lyndon.MaMb.1stterm}). The proof of Proposition
\ref{prop.QSym.lyndon.MaMb} is thus complete.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.MaMb.leq}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.MaMb.leq}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.MaMb.leq}.]Write $\alpha$ and $\beta$
as $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $ and
$\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{m}\right)  $, respectively;
then $\ell\left(  \alpha\right)  =\ell$ and $\ell\left(  \beta\right)  =m$.

Fix three disjoint chain posets $\left(  i_{1}<i_{2}<\cdots<i_{\ell
}\right)  $, $\left(  j_{1}<j_{2}<\cdots<j_{m}\right)  $ and $\left(
k_{1}<k_{2}<k_{3}<\cdots\right)  $. We have%
\begin{align}
&  \sum\limits_{\substack{p\in \NN ;\\p\leq\ell+m}}\sum_{f\text{ is a
}p\text{-shuffling map}}M_{\operatorname{wt}\left(f\right)}
\nonumber\\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in\mathfrak{A}^{\ast}\text{ satisfying }\ell\left(  \delta\right)  \leq
\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  \right)
\label{pf.cor.QSym.lyndon.MaMb.2ndterm}%
\end{align}
\footnote{\textit{Proof of (\ref{pf.cor.QSym.lyndon.MaMb.2ndterm}):} The proof
of this is analogous to the proof of (\ref{pf.prop.QSym.lyndon.MaMb.2ndterm}),
with the only difference that some $<$ signs are replaced by $\leq$ signs.}.
Now, (\ref{pf.prop.QSym.lyndon.MaMb.sum0}) becomes
\begin{align*}
M_{\alpha}M_{\beta}  &  =\sum\limits_{\substack{p\in \NN ;\\p\leq\ell+m}%
}\sum_{f\text{ is a }p\text{-shuffling map}}
M_{\operatorname{wt}\left(f\right)}\\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in\mathfrak{A}^{\ast}\text{ satisfying }\ell\left(  \delta\right)  \leq
\ell\left(  \alpha\right)  +\ell\left(  \beta\right)  \right)
\end{align*}
(by (\ref{pf.cor.QSym.lyndon.MaMb.2ndterm})). This proves Corollary
\ref{cor.QSym.lyndon.MaMb.leq}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.QSym.lyndon.triang.wll-lemma}}

\textit{Solution to Exercise \ref{exe.lem.QSym.lyndon.triang.wll-lemma}%
.}

\begin{proof}
[Proof of Lemma \ref{lem.QSym.lyndon.triang.wll-lemma}.](a) Let $\gamma\in
u\shuffle v$ be arbitrary. Then, the multiset of all letters of $\gamma$ is
the disjoint union of the multiset of all letters of $u$ with the multiset of
all letters of $v$. Hence, the sum of all letters of $\gamma$ equals the sum
of all letters of $u$ plus the sum of all letters of $v$. In other words,
$\left\vert \gamma\right\vert =\left\vert u\right\vert +\left\vert
v\right\vert $ (because $\left\vert \gamma\right\vert $ is the sum of all
letters of $\gamma$, because $\left\vert u\right\vert $ is the sum of all
letters of $u$, and because $\left\vert v\right\vert $ is the sum of all
letters of $v$). Since $u\in\Comp_{n}$, we have
$\left\vert u\right\vert =n$. Similarly, $\left\vert v\right\vert =m$. Thus,
$\left\vert \gamma\right\vert =\underbrace{\left\vert u\right\vert }%
_{=n}+\underbrace{\left\vert v\right\vert }_{=m}=n+m$, so that $\gamma
\in\Comp_{n+m}$.

Now, let us forget that we fixed $\gamma$. We thus have proven that%
\begin{equation}
\gamma\in\Comp_{n+m}\text{ for every }\gamma\in
u\shuffle v. \label{pf.lem.QSym.lyndon.triang.wll-lemma.a.1}%
\end{equation}
Applying (\ref{pf.lem.QSym.lyndon.triang.wll-lemma.a.1}) to $\gamma=z$, we
obtain $z\in\Comp_{n+m}$. This proves Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma}(a).

(b) Since $z\in u\shuffle v$, we have $\ell\left(  z\right)  =\ell\left(
u\right)  +\ell\left(  v\right)  $.

We have $u\in\Comp_{n}\subset \Comp %
=\mathfrak{A}^{\ast}$ and similarly $v\in\mathfrak{A}^{\ast}$. Thus,
Proposition \ref{prop.QSym.lyndon.MaMb} (applied to $\alpha=u$ and $\beta=v$)
yields%
\begin{align}
&  M_{u}M_{v}\nonumber\\
&  =\sum_{\gamma\in u\shuffle v}M_{\gamma}+\left(  \text{a sum of terms of the
form }M_{\delta}\text{ with }\delta\in\mathfrak{A}^{\ast}\text{ satisfying
}\ell\left(  \delta\right)  <\ell\left(  u\right)  +\ell\left(  v\right)
\right)  ,\nonumber
\end{align}
so that%
\begin{align}
&  M_{u}M_{v}-\sum_{\gamma\in u\shuffle v}M_{\gamma}\nonumber\\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in\underbrace{\mathfrak{A}^{\ast}}_{= \Comp }\text{ satisfying
}\ell\left(  \delta\right)  <\underbrace{\ell\left(  u\right)  +\ell\left(
v\right)  }_{=\ell\left(  z\right)  }\right) \nonumber\\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in \Comp \text{ satisfying }\ell\left(  \delta\right)
<\ell\left(  z\right)  \right)  .
\label{pf.lem.QSym.lyndon.triang.wll-lemma.b.1}%
\end{align}


Now, let $\pi$ denote the projection from the direct sum $\Qsym
=\bigoplus_{k\in \NN }\Qsym_{k}$ onto its
$\left(  n+m\right)  $-th homogeneous component $\Qsym_{n+m}$.
Notice that $\underbrace{M_{u}}_{\substack{\in
\Qsym_{n}\\\text{(since }u\in
\Comp_{n}\text{)}}}\underbrace{M_{v}}_{\substack{\in
\Qsym_{m}\\\text{(since }v\in\Comp_{m}\text{)}%
}}\in\Qsym_{n}\cdot\Qsym_{m}%
\subset\Qsym_{n+m}$, so that $\pi\left(  M_{u}%
M_{v}\right)  =M_{u}M_{v}$. Also, for every $\gamma\in u\shuffle v$, we have
$M_{\gamma}\in\Qsym_{n+m}$ (because
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.a.1}) shows that $\gamma
\in\Comp_{n+m}$) and therefore $\pi\left(  M_{\gamma
}\right)  =M_{\gamma}$. Hence, the $\kk$-linearity of $\pi$ yields%
\begin{align}
\pi\left(  M_{u}M_{v}-\sum_{\gamma\in u\shuffle v}M_{\gamma}\right)   &
=\underbrace{\pi\left(  M_{u}M_{v}\right)  }_{=M_{u}M_{v}}-\sum_{\gamma\in
u\shuffle v}\underbrace{\pi\left(  M_{\gamma}\right)  }_{=M_{\gamma}%
}\nonumber\\
&  =M_{u}M_{v}-\sum_{\gamma\in u\shuffle v}M_{\gamma}.
\label{pf.lem.QSym.lyndon.triang.wll-lemma.b.3}%
\end{align}


On the other hand,%
\begin{equation}
\pi\left(  M_{\delta}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }\delta
\in \Comp \setminus\Comp_{n+m}
\label{pf.lem.QSym.lyndon.triang.wll-lemma.b.pi-annihilates}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.pi-annihilates}):} Let $\delta
\in \Comp \setminus\Comp_{n+m}$. Then,
$\delta\in \Comp $ but $\delta\notin
\Comp_{n+m}$. In other words, $\delta$ is a composition with size
$\left\vert \delta\right\vert \neq n+m$. As a consequence, $M_{\delta}$ is a
homogeneous element of $\Qsym$ of degree $\left\vert
\delta\right\vert \neq n+m$. Therefore, $\pi\left(  M_{\delta}\right)  =0$
(since $\pi$ is the projection from the direct sum $\Qsym%
=\bigoplus_{k\in \NN }\Qsym_{k}$ onto its
$\left(  n+m\right)  $-th homogeneous component $
\Qsym_{n+m}$). This proves
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.pi-annihilates}).}.

But applying the projection $\pi$ to the equality
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.1}), we obtain%
\begin{align*}
&  \pi\left(  M_{u}M_{v}-\sum_{\gamma\in u\shuffle v}M_{\gamma}\right) \\
&  =\pi\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }%
\delta\in \Comp \text{ satisfying }\ell\left(  \delta\right)
<\ell\left(  z\right)  \right) \\
&  =\left(  \text{a sum of terms of the form }\pi\left(  M_{\delta}\right)
\text{ with }\delta\in \Comp \text{ satisfying }\ell\left(
\delta\right)  <\ell\left(  z\right)  \right) \\
&  =\left(  \text{a sum of terms of the form }\underbrace{\pi\left(
M_{\delta}\right)  }_{\substack{=M_{\delta}\\\text{(since }M_{\delta}%
\in\Qsym_{n+m}\\\text{(since }\delta\in
\Comp_{n+m}\text{))}}}\text{ with }\delta
\in\Comp_{n+m}\text{ satisfying }\ell\left(
\delta\right)  <\ell\left(  z\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{a sum of terms of the form
}\underbrace{\pi\left(  M_{\delta}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.pi-annihilates}))}}}\text{ with
}\delta\in \Comp \setminus\Comp_{n+m}%
\text{ satisfying }\ell\left(  \delta\right)  <\ell\left(  z\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }\delta\in
 \Comp \text{ satisfies either }\delta\in
\Comp_{n+m}\text{ or }\delta\in \Comp \setminus
\Comp_{n+m}\right) \\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in\Comp_{n+m}\text{ satisfying }\ell\left(
\delta\right)  <\ell\left(  z\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{a sum of terms of the form
}0\text{ with }\delta\in \Comp \setminus
\Comp_{n+m}\text{ satisfying }\ell\left(  \delta\right)  <\ell\left(
z\right)  \right)  }_{=0}\\
&  =\left(  \text{a sum of terms of the form }M_{\delta}\text{ with }\delta
\in\Comp_{n+m}\text{ satisfying }\ell\left(
\delta\right)  <\ell\left(  z\right)  \right) \\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }\ell\left(
w\right)  <\ell\left(  z\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the index }\delta\text{
as }w\right) \\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)
\end{align*}
(since every $w\in\Comp_{n+m}$ satisfying $\ell\left(
w\right)  <\ell\left(  z\right)  $ must satisfy
$w\underset{\operatorname*{wll}}{<}z$). Compared with
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.3}), this yields%
\begin{align*}
&  M_{u}M_{v}-\sum_{\gamma\in u\shuffle v}M_{\gamma}\\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\end{align*}
Adding $\sum_{\gamma\in u\shuffle v}M_{\gamma}$ to both sides of this
equality, we obtain%
\begin{align}
&  M_{u}M_{v}\nonumber\\
&  =\sum_{\gamma\in u\shuffle v}M_{\gamma}+\left(  \text{a sum of terms of the
form }M_{w}\text{ with }w\in\Comp_{n+m}\text{
satisfying }w\underset{\operatorname*{wll}}{<}z\right)  .
\label{pf.lem.QSym.lyndon.triang.wll-lemma.b.5}%
\end{align}


But every $\gamma\in u\shuffle v$ satisfying $\gamma\neq z$ must satisfy
$\gamma\in\Comp_{n+m}$ (by
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.a.1})) and $\gamma
\underset{\operatorname*{wll}}{<}z$\ \ \ \ \footnote{\textit{Proof.} Let
$\gamma\in u\shuffle v$ be such that $\gamma\neq z$. Since $\gamma\in
u\shuffle v$, we must have $\ell\left(  \gamma\right)  =\ell\left(  u\right)
+\ell\left(  v\right)  =\ell\left(  z\right)  $. Also, $\gamma$ and $z$ both
belong to $\Comp_{n+m}$. Now, $\gamma$ is an element
of the multiset $u\shuffle v$, whereas $z$ is the lexicographically highest
element of this multiset. Hence, $\gamma\leq z$ with respect to the
lexicographic order. Since the elements $\gamma$ and $z$ of
$\Comp_{n+m}$ satisfy $\ell\left(  \gamma\right)
=\ell\left(  z\right)  $ and $\gamma\leq z$ with respect to the lexicographic
order, we must have $\gamma\underset{\operatorname*{wll}}{\leq}z$, and thus
$\gamma\underset{\operatorname*{wll}}{<}z$ (since $\gamma\neq z$), qed.}.
Hence,%
\begin{equation}
\sum\limits_{\substack{\gamma\in u\shuffle v;\\\gamma\neq z}}M_{\gamma}=\left(
\text{a sum of terms of the form }M_{\gamma}\text{ with }\gamma\in
\Comp_{n+m}\text{ satisfying }\gamma
\underset{\operatorname*{wll}}{<}z\right)  .
\label{pf.lem.QSym.lyndon.triang.wll-lemma.b.6}%
\end{equation}


But let $h$ be the multiplicity with which the word $z$ appears in the
multiset $u\shuffle v$. Then, $h$ is a positive integer (since $z$ is an
element of the multiset $u\shuffle v$), and satisfies $\sum\limits_{\substack{\gamma
\in u\shuffle v;\\\gamma=z}}M_{z}=hM_{z}$. Now,%
\begin{align*}
\sum_{\gamma\in u\shuffle v}M_{\gamma}  &  =\sum\limits_{\substack{\gamma\in
u\shuffle v;\\\gamma=z}}\underbrace{M_{\gamma}}_{\substack{=M_{z}%
\\\text{(since }\gamma=z\text{)}}}+\sum\limits_{\substack{\gamma\in
u\shuffle v;\\\gamma\neq z}}M_{\gamma}\\
&  =\underbrace{\sum\limits_{\substack{\gamma\in u\shuffle v;\\\gamma=z}}M_{z}%
}_{=hM_{z}}+\underbrace{\sum\limits_{\substack{\gamma\in u\shuffle v;\\\gamma\neq
z}}M_{\gamma}}_{\substack{=\left(  \text{a sum of terms of the form }%
M_{\gamma}\text{ with }\gamma\in\Comp_{n+m}\text{
satisfying }\gamma\underset{\operatorname*{wll}}{<}z\right)  \\\text{(by
(\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.6}))}}}\\
&  =hM_{z}+\underbrace{\left(  \text{a sum of terms of the form }M_{\gamma
}\text{ with }\gamma\in\Comp_{n+m}\text{ satisfying
}\gamma\underset{\operatorname*{wll}}{<}z\right)  }_{\substack{=\left(
\text{a sum of terms of the form }M_{w}\text{ with }w\in
\Comp_{n+m}\text{ satisfying }w\underset{\operatorname*{wll}}{<}z\right)
\\\text{(here, we have renamed the index }\gamma\text{ as }w\text{)}}}\\
&  =hM_{z}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\end{align*}
Hence, (\ref{pf.lem.QSym.lyndon.triang.wll-lemma.b.5}) becomes%
\begin{align*}
&  M_{u}M_{v}\\
&  =\underbrace{\sum_{\gamma\in u\shuffle v}M_{\gamma}}_{=hM_{z}+\left(
\text{a sum of terms of the form }M_{w}\text{ with }w\in
\Comp_{n+m}\text{ satisfying }w\underset{\operatorname*{wll}}{<}z\right)
}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right) \\
&  =hM_{z}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right) \\
&  =hM_{z}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\end{align*}
This proves Lemma \ref{lem.QSym.lyndon.triang.wll-lemma}(b).

(c) Since $z\in u\shuffle v$, we have $\ell\left(  z\right)  =\ell\left(
u\right)  +\ell\left(  v\right)  $. Since $v^{\prime}%
\underset{\operatorname*{wll}}{<}v$, we have $v^{\prime}\neq v$.

Let $z^{\prime}$ be the lexicographically highest element of the multiset
$u\shuffle v^{\prime}$. Lemma \ref{lem.QSym.lyndon.triang.wll-lemma}(a)
(applied to $v^{\prime}$ and $z^{\prime}$ instead of $v$ and $z$) yields
$z^{\prime}\in\Comp_{n+m}$. Since $z^{\prime}$ is an
element of the multiset $u\shuffle v^{\prime}$, we have $\ell\left(
z^{\prime}\right)  =\ell\left(  u\right)  +\ell\left(  v^{\prime}\right)  $.

Now, it is easy to see (using Lemma \ref{lem.shuffle.preserves-lex}) that
$z^{\prime}\underset{\operatorname*{wll}}{<}z$%
\ \ \ \ \footnote{\textit{Proof.} We have $v^{\prime}%
\underset{\operatorname*{wll}}{<}v$, thus $v^{\prime}%
\underset{\operatorname*{wll}}{\leq}v$. According to the definition of the
relation $\underset{\operatorname*{wll}}{\leq}$, this means that we are in one
of the following two cases:
\par
\textit{Case 1:} We have $\ell\left(  v^{\prime}\right)  <\ell\left(
v\right)  $.
\par
\textit{Case 2:} We have $\ell\left(  v^{\prime}\right)  =\ell\left(
v\right)  $ and $v^{\prime}\leq v$ in lexicographic order.
\par
Let us first consider Case 1. In this case, we have $\ell\left(  v^{\prime
}\right)  <\ell\left(  v\right)  $. Now, $\ell\left(  z^{\prime}\right)
=\ell\left(  u\right)  +\underbrace{\ell\left(  v^{\prime}\right)  }%
_{<\ell\left(  v\right)  }<\ell\left(  u\right)  +\ell\left(  v\right)
=\ell\left(  z\right)  $. But any two elements $\alpha$ and $\beta$ of
$\Comp_{n+m}$ satisfying $\ell\left(  \alpha\right)
<\ell\left(  \beta\right)  $ must satisfy $\alpha\underset{\operatorname*{wll}%
}{\leq}\beta$ (by the definition of $\underset{\operatorname*{wll}}{\leq}$).
Applying this to $\alpha=z^{\prime}$ and $\beta=z$, we obtain $z^{\prime
}\underset{\operatorname*{wll}}{\leq}z$. Combined with $z^{\prime}\neq z$
(since $\ell\left(  z^{\prime}\right)  <\ell\left(  z\right)  $), this yields
$z^{\prime}\underset{\operatorname*{wll}}{<}z$. Thus, $z^{\prime
}\underset{\operatorname*{wll}}{<}z$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $\ell\left(  v^{\prime
}\right)  =\ell\left(  v\right)  $ and $v^{\prime}\leq v$ in lexicographic
order. Now, $\ell\left(  z^{\prime}\right)  =\ell\left(  u\right)
+\underbrace{\ell\left(  v^{\prime}\right)  }_{=\ell\left(  v\right)  }%
=\ell\left(  u\right)  +\ell\left(  v\right)  =\ell\left(  z\right)  $.
\par
We know that $z^{\prime}$ is an element of the multiset $u\shuffle v^{\prime}%
$; in other words, $z^{\prime}$ can be written in the form $z^{\prime
}=u\underset{\sigma}{\shuffle}v^{\prime}$ for some $\sigma\in
\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v^{\prime
}\right)  }$ (because every element of the multiset $u\shuffle v^{\prime}$ can
be written in this form). Consider this $\sigma$. We have $\sigma
\in\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(  v^{\prime
}\right)  }=\operatorname{Sh}_{\ell\left(  u\right)  ,\ell\left(
v\right)  }$ (since $\ell\left(  v^{\prime}\right)  =\ell\left(  v\right)  $),
and thus $u\underset{\sigma}{\shuffle}v$ is a well-defined element of the
multiset $u\shuffle v$. Therefore, $u\underset{\sigma}{\shuffle}v\leq z$
(because $z$ is the lexicographically highest element of this multiset
$u\shuffle v$).
\par
But we have $v^{\prime}\leq v$ with respect to the relation $\leq$ on
$\mathfrak{A}^{\ast}$ defined in Definition \ref{def.words} (since $v^{\prime
}\leq v$ in lexicographic order). Thus, $v^{\prime}<v$ with respect to this
relation (since $v^{\prime}\neq v$). Hence, Lemma
\ref{lem.shuffle.preserves-lex}(a) (applied to $\ell\left(  u\right)  $ and
$\ell\left(  v\right)  $ instead of $n$ and $m$) yields $u\underset{\sigma
}{\shuffle}v^{\prime}<u\underset{\sigma}{\shuffle}v\leq z$. Thus, $z^{\prime
}=u\underset{\sigma}{\shuffle}v^{\prime}<z$ with respect to the relation
$\leq$ on $\mathfrak{A}^{\ast}$. In other words, $z^{\prime}<z$ in
lexicographic order (since $\ell\left(  z^{\prime}\right)  =\ell\left(
z\right)  $), thus $z^{\prime}\leq z$.
\par
Now, the two elements $z^{\prime}$ and $z$ of $
\Comp_{n+m}$ satisfy $\ell\left(  z^{\prime}\right)  =\ell\left(
z\right)  $ and $z^{\prime}\leq z$ in lexicographic order. But any two
elements $\alpha$ and $\beta$ of $\Comp_{n+m}$
satisfying $\left(  \ell\left(  \alpha\right)  =\ell\left(  \beta\right)
\text{ and }\alpha\leq\beta\text{ in lexicographic order}\right)  $ must
satisfy $\alpha\underset{\operatorname*{wll}}{\leq}\beta$ (by the definition
of $\underset{\operatorname*{wll}}{\leq}$). Applying this to $\alpha
=z^{\prime}$ and $\beta=z$, we obtain $z^{\prime}\underset{\operatorname*{wll}%
}{\leq}z$. Since $z^{\prime}\neq z$, this yields $z^{\prime}%
\underset{\operatorname*{wll}}{<}z$. Hence, $z^{\prime}%
\underset{\operatorname*{wll}}{<}z$ is proven in Case 2.
\par
Now, we have proved $z^{\prime}\underset{\operatorname*{wll}}{<}z$ in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, this
yields that $z^{\prime}\underset{\operatorname*{wll}}{<}z$ always holds,
qed.}. Hence, every $w\in\Comp_{n+m}$ satisfying
$w\underset{\operatorname*{wll}}{<}z^{\prime}$ also satisfies
$w\underset{\operatorname*{wll}}{<}z$ (because it satisfies
$w\underset{\operatorname*{wll}}{<}z^{\prime}\underset{\operatorname*{wll}%
}{<}z$). Applying Lemma \ref{lem.QSym.lyndon.triang.wll-lemma}(b) to
$v^{\prime}$ and $z^{\prime}$ instead of $v$ and $z$, we conclude that there
exists a positive integer $h$ such that%
\[
M_{u}M_{v^{\prime}}=hM_{z^{\prime}}+\left(  \text{a sum of terms of the form
}M_{w}\text{ with }w\in\Comp_{n+m}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}z^{\prime}\right)  .
\]
Consider this $h$. We have%
\begin{align*}
M_{u}M_{v^{\prime}}  &  =\underbrace{hM_{z^{\prime}}}_{\substack{=\sum
_{i=1}^{h}M_{z^{\prime}}=\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  \\\text{(since }z^{\prime}%
\in\Comp_{n+m}\text{ satisfies }z^{\prime
}\underset{\operatorname*{wll}}{<}z\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{a sum of terms of the form
}M_{w}\text{ with }w\in\Comp_{n+m}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}z^{\prime}\right)  }_{\substack{=\left(
\text{a sum of terms of the form }M_{w}\text{ with }w\in
\Comp_{n+m}\text{ satisfying }w\underset{\operatorname*{wll}}{<}z\right)
\\\text{(since every }w\in\Comp_{n+m}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}z^{\prime}\text{ also satisfies
}w\underset{\operatorname*{wll}}{<}z\text{)}}}\\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right) \\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}z\right)  .
\end{align*}
This proves Lemma \ref{lem.QSym.lyndon.triang.wll-lemma}(c).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.QSym.lyndon.triang}}

\textit{Solution to Exercise \ref{exe.prop.QSym.lyndon.triang}.}

\begin{proof}
[Proof of Proposition \ref{prop.QSym.lyndon.triang}.]We shall prove
Proposition \ref{prop.QSym.lyndon.triang} by strong induction over $n$. So we
fix some $N\in \NN$, and we assume that Proposition
\ref{prop.QSym.lyndon.triang} holds whenever $n<N$. We now need to prove that
Proposition \ref{prop.QSym.lyndon.triang} holds for $n=N$. In other words, we
need to prove that for every $x\in\Comp_{N}$, there is
a family $\left(  \eta_{x,y}\right)  _{y\in\Comp_{N}%
}\in \NN ^{\Comp_{N}}$ of elements of
$\NN$ satisfying%
\begin{equation}
\mathbf{M}_{x}=\sum\limits_{\substack{y\in\Comp_{N}%
;\\y\underset{\operatorname*{wll}}{\leq}x}}\eta_{x,y}M_{y}
\label{pf.prop.QSym.lyndon.triang.goal}%
\end{equation}
and $\eta_{x,x}\neq0$ (in $\NN$).

Let $x\in\Comp_{N}$ be arbitrary. We need to prove
that there is a family $\left(  \eta_{x,y}\right)  _{y\in
\Comp_{N}}\in \NN ^{\Comp_{N}}$ of elements
of $\NN$ satisfying (\ref{pf.prop.QSym.lyndon.triang.goal}) and
$\eta_{x,x}\neq0$.

Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
$x$. Then, $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are Lyndon words satisfying
$x=a_{1}a_{2}\cdots a_{p}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$.

If $x$ is the empty word, then our claim is trivial (in fact, we can just set
$\eta_{x,x}=1$ in this case, and (\ref{pf.prop.QSym.lyndon.triang.goal}) holds
obviously). Hence, for the rest of this proof, we WLOG assume that $x$ is not
the empty word. Thus, $p\neq0$ (because otherwise, $x=a_{1}a_{2}\cdots a_{p}$
would be an empty product and thus the empty word, contradicting the
assumption that $x$ is not the empty word). Hence, we can define two words
$u\in\mathfrak{A}^{\ast}$ and $v\in\mathfrak{A}^{\ast}$ by $u=a_{1}$ and
$v=a_{2}a_{3}\cdots a_{p}$. The word $u$ is Lyndon (since $u=a_{1}$ and since
$a_{1}$ is Lyndon), and thus has $\left(  u\right)  $ as its CFL
factorization. The CFL factorization of the word $v$ is $\left(  a_{2}%
,a_{3},\ldots,a_{p}\right)  $ (since the words $a_{2}$, $a_{3}$, $\ldots$,
$a_{p}$ are Lyndon and satisfy $v=a_{2}a_{3}\cdots a_{p}$ and $a_{2}\geq
a_{3}\geq\cdots\geq a_{p}$ (because $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$)).
Also, $u=a_{1}\geq a_{j+1}$ for every $i\in\left\{  1,2,\ldots,1\right\}  $
and $j\in\left\{  1,2,\ldots,p-1\right\}  $ (because $a_{1}\geq a_{2}%
\geq\cdots\geq a_{p}$). Hence, we can apply Theorem
\ref{thm.shuffle.lyndon.used}(c) to $1$, $\left(  u\right)  $, $p-1$ and
$\left(  a_{2},a_{3},\ldots,a_{p}\right)  $ instead of $p$, $\left(
a_{1},a_{2},\ldots,a_{p}\right)  $, $q$ and $\left(  b_{1},b_{2},\ldots
,b_{q}\right)  $. As a result, we conclude that the lexicographically highest
element of the multiset $u\shuffle v$ is $\underbrace{u}_{=a_{1}%
}\underbrace{v}_{=a_{2}a_{3}\cdots a_{p}}=a_{1}\left(  a_{2}a_{3}\cdots
a_{p}\right)  =a_{1}a_{2}\cdots a_{p}=x$.

But $\mathbf{M}_{u}=M_{u}$ (by the definition of $\mathbf{M}_{u}$, since $u$
has CFL factorization $\left(  u\right)  $), and $\mathbf{M}_{x}%
=\mathbf{M}_{u}\mathbf{M}_{v}$\ \ \ \ \footnote{\textit{Proof.} Since the CFL
factorization of $v$ is $\left(  a_{2},a_{3},\ldots,a_{p}\right)  $, we have
$\mathbf{M}_{v}=M_{a_{2}}M_{a_{3}}\cdots M_{a_{p}}$ (by the definition of
$\mathbf{M}_{v}$). But since the CFL factorization of $x$ is $\left(
a_{1},a_{2},\ldots,a_{p}\right)  $, we have%
\[
\mathbf{M}_{x}=M_{a_{1}}M_{a_{2}}\cdots M_{a_{p}}=\underbrace{M_{a_{1}}%
}_{\substack{=M_{u}\\\text{(since }a_{1}=u\text{)}}}\underbrace{\left(
M_{a_{2}}M_{a_{3}}\cdots M_{a_{p}}\right)  }_{=\mathbf{M}_{v}}%
=\underbrace{M_{u}}_{\substack{=\mathbf{M}_{u}\\\text{(since }\mathbf{M}%
_{u}=M_{u}\text{)}}}\mathbf{M}_{v}=\mathbf{M}_{u}\mathbf{M}_{v},
\]
qed.}.

The word $u$ is Lyndon and thus nonempty, so that $\left\vert u\right\vert
>0$. Now, $\left\vert \underbrace{x}_{=uv}\right\vert =\left\vert
uv\right\vert =\left\vert u\right\vert +\left\vert v\right\vert $, so that
$\left\vert u\right\vert +\left\vert v\right\vert =\left\vert x\right\vert =N$
(since $x\in\Comp_{N}$). Thus,
$N=\underbrace{\left\vert u\right\vert }_{>0}+\left\vert v\right\vert
>\left\vert v\right\vert $, so that $\left\vert v\right\vert <N$. Hence, the
induction hypothesis tells us that we can apply Proposition
\ref{prop.QSym.lyndon.triang} to $\left\vert v\right\vert $ and $v$ instead of
$n$ and $x$ (since $v\in\Comp_{\left\vert v\right\vert
}$). As a result, we see that there is a family $\left(  \eta_{v,y}\right)
_{y\in\Comp_{\left\vert v\right\vert }}\in
 \NN ^{\Comp_{\left\vert v\right\vert }}$ of
elements of $\NN$ satisfying%
\[
\mathbf{M}_{v}=\sum\limits_{\substack{y\in\Comp_{\left\vert
v\right\vert };\\y\underset{\operatorname*{wll}}{\leq}v}}\eta_{v,y}M_{y}%
\]
and $\eta_{v,v}\neq0$ (in $\NN$). Consider this family $\left(
\eta_{v,y}\right)  _{y\in\Comp_{\left\vert
v\right\vert }}$.

We have $u\in\Comp_{\left\vert u\right\vert }$ and
$v\in\Comp_{\left\vert v\right\vert }$. Hence, Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma}(b) (applied to $x$, $\left\vert
u\right\vert $ and $\left\vert v\right\vert $ instead of $z$, $u$ and $v$)
yields that there exists a positive integer $h$ such that
\begin{align}
M_{u}M_{v}  &  =hM_{x}+\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\underbrace{\Comp_{\left\vert u\right\vert
+\left\vert v\right\vert }}_{\substack{=\Comp_{N}%
\\\text{(since }\left\vert u\right\vert +\left\vert v\right\vert =N\text{)}%
}}\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right) \nonumber\\
&  =hM_{x}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{N}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}x\right)  .
\label{pf.prop.QSym.lyndon.triang.1}%
\end{align}
Consider this $h$. Then, $h\in \NN$ and $h\neq0$ (since $h$ is a
positive integer).

Now,%
\begin{align*}
\mathbf{M}_{x}  &  =\underbrace{\mathbf{M}_{u}}_{=M_{u}}\underbrace{\mathbf{M}%
_{v}}_{=\sum\limits_{\substack{y\in\Comp_{\left\vert
v\right\vert };\\y\underset{\operatorname*{wll}}{\leq}v}}\eta_{v,y}M_{y}%
}=M_{u}\left(  \sum\limits_{\substack{y\in\Comp_{\left\vert
v\right\vert };\\y\underset{\operatorname*{wll}}{\leq}v}}\eta_{v,y}%
M_{y}\right)  =\sum\limits_{\substack{y\in\Comp_{\left\vert
v\right\vert };\\y\underset{\operatorname*{wll}}{\leq}v}}\eta_{v,y}M_{u}%
M_{y}\\
&  =\eta_{v,v}\underbrace{M_{u}M_{v}}_{\substack{=hM_{x}+\left(  \text{a sum
of terms of the form }M_{w}\text{ with }w\in
\Comp_{N}\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right)
\\\text{(by (\ref{pf.prop.QSym.lyndon.triang.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{y\in
\Comp_{\left\vert v\right\vert };\\y\underset{\operatorname*{wll}}{<}%
v}}\eta_{v,y}\underbrace{M_{u}M_{y}}_{\substack{=\left(  \text{a sum of terms
of the form }M_{w}\text{ with }w\in\Comp_{\left\vert
u\right\vert +\left\vert v\right\vert }\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}x\right)  \\\text{(by Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma}(c), applied to }n=\left\vert
u\right\vert \text{, }m=\left\vert v\right\vert \text{, }v^{\prime}=y\text{
and }z=x\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}y=v\text{ from the sum}\right) \\
&  =\underbrace{\eta_{v,v}\left(  hM_{x}+\left(  \text{a sum of terms of the
form }M_{w}\text{ with }w\in\Comp_{N}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}x\right)  \right)  }_{=\eta_{v,v}%
hM_{x}+\eta_{v,v}\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{N}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}x\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{y\in
\Comp_{\left\vert v\right\vert };\\y\underset{\operatorname*{wll}}{<}%
v}}\eta_{v,y}\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\underbrace{\Comp_{\left\vert u\right\vert
+\left\vert v\right\vert }}_{\substack{=\Comp_{N}%
\\\text{(since }\left\vert u\right\vert +\left\vert v\right\vert =N\text{)}%
}}\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right) \\
&  =\eta_{v,v}hM_{x}+\eta_{v,v}\left(  \text{a sum of terms of the form }%
M_{w}\text{ with }w\in\Comp_{N}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}x\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{y\in
\Comp_{\left\vert v\right\vert };\\y\underset{\operatorname*{wll}}{<}%
v}}\eta_{v,y}\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{N}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}x\right)  }_{\substack{=\left(  \text{an
} \NN \text{-linear combination of terms of the form }M_{w}\text{ with
}w\in\Comp_{N}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}x\right)  \\\text{(since }\eta_{v,y}%
\in \NN \text{ for every }y\in\Comp_{\left\vert
v\right\vert }\text{ satisfying }y\underset{\operatorname*{wll}}{<}v\text{)}}}
\end{align*}%
\begin{align*}
&  =\eta_{v,v}hM_{x}+\underbrace{\eta_{v,v}\left(  \text{a sum of terms of the
form }M_{w}\text{ with }w\in\Comp_{N}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}x\right)  }_{\substack{=\left(  \text{an
} \NN \text{-linear combination of terms of the form }M_{w}\text{ with
}w\in\Comp_{N}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}x\right)  \\\text{(since }\eta_{v,v}%
\in \NN \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{an } \NN \text{-linear combination
of terms of the form }M_{w}\text{ with }w\in
\Comp_{N}\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right) \\
&  =\eta_{v,v}hM_{x}+\left(  \text{an } \NN \text{-linear combination of
terms of the form }M_{w}\text{ with }w\in\Comp_{N}%
\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{an } \NN \text{-linear combination
of terms of the form }M_{w}\text{ with }w\in
\Comp_{N}\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right) \\
&  =\eta_{v,v}hM_{x}+\left(  \text{an } \NN \text{-linear combination of
terms of the form }M_{w}\text{ with }w\in\Comp_{N}%
\text{ satisfying }w\underset{\operatorname*{wll}}{<}x\right)  .
\end{align*}
In other words, we can write $\mathbf{M}_{x}$ in the form%
\begin{equation}
\mathbf{M}_{x}=\eta_{v,v}hM_{x}+\mathbf{c},
\label{pf.prop.QSym.lyndon.triang.soon}%
\end{equation}
where $\mathbf{c}$ is an $\NN$-linear combination of terms of the form
$M_{w}$ with $w\in\Comp_{N}$ satisfying
$w\underset{\operatorname*{wll}}{<}x$. Consider this $\mathbf{c}$.

Write $\mathbf{c}$ in the form%
\begin{equation}
\mathbf{c}=\sum\limits_{\substack{y\in\Comp_{N}%
;\\y\underset{\operatorname*{wll}}{<}x}}\eta_{x,y}M_{y},
\label{pf.prop.QSym.lyndon.triang.almostthere}%
\end{equation}
where $\eta_{x,y}$ is an element of $\NN$ for every $y\in
\Comp_{N}$ satisfying $y\underset{\operatorname*{wll}%
}{<}x$. (This is possible since $\mathbf{c}$ is an $\NN$-linear
combination of terms of the form $M_{w}$ with $w\in
\Comp_{N}$ satisfying $w\underset{\operatorname*{wll}}{<}x$.) Thus, we
have defined a family $\left(  \eta_{x,y}\right)  _{\substack{y\in
\Comp_{N};\\y\underset{\operatorname*{wll}}{<}x}}$ of
elements of $\NN$. Extend this family to a family $\left(  \eta
_{x,y}\right)  _{y\in\Comp_{N}}\in \NN %
^{\Comp_{N}}$ of elements of $\NN$ by defining%
\[
\eta_{x,y}=\delta_{x,y}\eta_{v,v}h\ \ \ \ \ \ \ \ \ \ \text{for every }%
y\in\Comp_{N}\text{ which does not satisfy
}y\underset{\operatorname*{wll}}{<}x.
\]
Notice that the definition of $\eta_{x,x}$ yields $\eta_{x,x}=\delta_{x,x}%
\eta_{v,v}h$ (since $x$ does not satisfy $x\underset{\operatorname*{wll}}{<}%
x$), and thus $\eta_{x,x}=\underbrace{\delta_{x,x}}_{=1}\eta_{v,v}%
h=\underbrace{\eta_{v,v}}_{\neq0}\underbrace{h}_{\neq0}\neq0$.

Now,%
\begin{align*}
\sum\limits_{\substack{y\in\Comp_{N}%
;\\y\underset{\operatorname*{wll}}{\leq}x}}\eta_{x,y}M_{y}  &
=\underbrace{\eta_{x,x}}_{=\eta_{v,v}h}M_{x}+\underbrace{\sum\limits_{\substack{y\in
\Comp_{N};\\y\underset{\operatorname*{wll}}{<}x}%
}\eta_{x,y}M_{y}}_{\substack{=\mathbf{c}\\\text{(by
(\ref{pf.prop.QSym.lyndon.triang.almostthere}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}y=x\text{ from the sum}\right) \\
&  =\eta_{v,v}hM_{x}+\mathbf{c}=\mathbf{M}_{x}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.prop.QSym.lyndon.triang.soon})}\right)  ,
\end{align*}
and thus $\mathbf{M}_{x}=\sum\limits_{\substack{y\in
\Comp_{N};\\y\underset{\operatorname*{wll}}{\leq}x}}\eta_{x,y}M_{y}$. In
other words, (\ref{pf.prop.QSym.lyndon.triang.goal}) is satisfied.

Hence, we have constructed a family $\left(  \eta_{x,y}\right)  _{y\in
\Comp_{N}}\in \NN ^{
\Comp_{N}}$ of elements of $\NN$ satisfying
(\ref{pf.prop.QSym.lyndon.triang.goal}) and $\eta_{x,x}\neq0$. Thus, we have
shown the existence of such a family. The induction step is thus complete, and
Proposition \ref{prop.QSym.lyndon.triang} is proven by induction.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.QSym.lyndon.M-genset}}

\textit{Solution to Exercise \ref{exe.prop.QSym.lyndon.M-genset}.}

\begin{proof}
[Proof of Proposition \ref{prop.QSym.lyndon.M-genset}.]We know that
$\QQ$ is a subring of $\kk$. Hence, every nonzero element of $\NN$
is an invertible element of $\kk$.

For every composition
$u\in \Comp =\mathfrak{A}^{\ast}$, define an element
$\mathbf{M}_{u}\in\Qsym$ by $\mathbf{M}_{u}=M_{a_{1}}M_{a_{2}%
}\cdots M_{a_{p}}$, where $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is the
CFL factorization of the word $u$.

Fix $n\in \NN$. Consider the set $\Comp_n$ as a poset whose smaller
relation is the relation $\underset{\operatorname*{wll}}{\leq}$.
We shall use the notations introduced in Section~\ref{sect.STmat}.

It is easy to see that
$\left( M_u \right)_{u \in \Comp_n}$ is a basis of the $\kk$-module
$\Qsym_n$. According to Proposition~\ref{prop.QSym.lyndon.triang},
the family
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$ expands
invertibly triangularly in the basis
$\left( M_u \right) _{u \in \Comp_n}$
\ \ \ \ \footnote{\textit{Proof.}
Proposition~\ref{prop.QSym.lyndon.triang} shows that, for every
$x \in \Comp_n$, there exists a family
$\left(\eta_{x,y}\right)_{y \in \Comp_n} \in \NN^{\Comp_n}$ of
elements of $\NN$ such that
\begin{equation}
\mathbf{M}_x
= \sum\limits_{\substack{y \in \Comp_n; \\ y \underset{\operatorname*{wll}}{\leq} x}} \eta_{x,y} M_y
\label{pf.prop.QSym.lyndon.M-genset.fn1.1}
\end{equation}
and
\begin{equation}
\eta_{x,x}\neq 0 \qquad \text{(in }\NN \text{)} .
\label{pf.prop.QSym.lyndon.M-genset.fn1.2}
\end{equation}
Consider such a family
$\left(\eta_{x,y}\right)_{y \in \Comp_n} \in \NN^{\Comp_n}$ for each
$x \in \Comp_n$. Thus, an integer $\eta_{x,y} \in \NN \subset
\QQ \subset \kk$ is defined for each
$\left(x, y\right) \in \Comp_n \times \Comp_n$.

We observe that the only elements $\eta_{s, t}$
(with $\left(s, t\right) \in \Comp_n \times \Comp_n$)
appearing
in the statements \eqref{pf.prop.QSym.lyndon.M-genset.fn1.1} and
\eqref{pf.prop.QSym.lyndon.M-genset.fn1.2} are those which
satisfy $t \underset{\operatorname*{wll}}{\leq} s$. Hence,
if some $\left(s, t\right) \in \Comp_n \times \Comp_n$
does not satisfy
$t \underset{\operatorname*{wll}}{\leq} s$, then the
corresponding element $\eta_{s, t}$ does not appear in any of
the statements \eqref{pf.prop.QSym.lyndon.M-genset.fn1.1} and
\eqref{pf.prop.QSym.lyndon.M-genset.fn1.2}; as a consequence,
we can arbitrarily change the value of this $\eta_{s, t}$ without
running the risk of invalidating
\eqref{pf.prop.QSym.lyndon.M-genset.fn1.1} and
\eqref{pf.prop.QSym.lyndon.M-genset.fn1.2}. Hence, we can
WLOG assume that
\begin{equation}
\text{every } \left(s, t\right) \in \Comp_n \times \Comp_n
\text{ which does not satisfy }
t \underset{\operatorname*{wll}}{\leq} s
\text{ must satisfy }
\eta_{s,t} = 0
\label{pf.prop.QSym.lyndon.M-genset.fn1.3}
\end{equation}
(otherwise, we can just set all such $\eta_{s,t}$
to $0$). Assume this. Thus, the matrix
$\left(\eta_{x,y}\right)_{\left(x, y\right) \in \Comp_n \times \Comp_n}$
is triangular. The diagonal entries $\eta_{x,x}$
of this matrix are nonzero elements of $\NN$
(because of \eqref{pf.prop.QSym.lyndon.M-genset.fn1.2}) and
therefore invertible elements of $\kk$ (since
every nonzero element of $\NN$ is an invertible element of $\kk$).
Thus, the matrix
$\left(\eta_{x,y}\right)_{\left(x, y\right) \in \Comp_n \times \Comp_n}$
(regarded as a matrix in $\kk^{\Comp_n \times \Comp_n}$)
is invertibly triangular.

Now, every $x \in \Comp_n$ satisfies
\begin{align*}
\sum_{y \in \Comp_n} \eta_{x,y} M_y
&= \underbrace{\sum\limits_{\substack{y \in \Comp_n; \\ y \underset{\operatorname*{wll}}{\leq} x}} \eta_{x,y} M_y}_{\substack{= \mathbf{M}_x \\ \text{(by \eqref{pf.prop.QSym.lyndon.M-genset.fn1.1})}}}
+ \sum\limits_{\substack{y \in \Comp_n; \\ \text{not } y \underset{\operatorname*{wll}}{\leq} x}} \underbrace{\eta_{x,y}}_{\substack{=0 \\ \text{(by \eqref{pf.prop.QSym.lyndon.M-genset.fn1.3}, applied to } \left(s, t\right) = \left(x, y\right) \text{)}}} M_y \\
&= \mathbf{M}_x + \underbrace{ \sum\limits_{\substack{y \in \Comp_n; \\ \text{not } y \underset{\operatorname*{wll}}{\leq} x}} 0 M_y}_{=0}
= \mathbf{M}_x .
\end{align*}
In other words, every $x \in \Comp_n$ satisfies
$\mathbf{M}_x = \sum_{y \in \Comp_n} \eta_{x,y} M_y$.
In other words, the family
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$ expands
in the family
$\left( M_u \right) _{u \in \Comp_n}$
through the matrix
$\left(\eta_{x,y}\right)_{\left(x, y\right) \in \Comp_n \times \Comp_n}$.
Since the latter matrix is invertibly triangular,
we thus conclude that the family
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$ expands
invertibly triangularly in the family
$\left( M_u \right) _{u \in \Comp_n}$.}. Hence,
Corollary~\ref{cor.STmat.expansion-tria-inv}(e) (applied to
$\Qsym_n$, $\Comp_n$, $\left( \mathbf{M}_u \right) _{u \in \Comp_n}$
and $\left( M_u \right) _{u \in \Comp_n}$ instead of $M$, $S$,
$\left(e_s\right)_{s \in S}$ and $\left(f_s\right)_{s \in S}$)
yields that the family
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$
is a basis of the $\kk$-module $\Qsym_n$ if and only if the
family
$\left( M_u \right) _{u \in \Comp_n}$ is a basis of the
$\kk$-module $\Qsym_n$\ \ \ \ \footnote{Here, we have used the
fact that the set $\Comp_n$ is finite.}. Hence, the family
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$
is a basis of the $\kk$-module $\Qsym_n$ (since the family
$\left( M_u \right) _{u \in \Comp_n}$
is a basis of the $\kk$-module $\Qsym_n$).

Now, let us forget that we fixed $n$. We thus have proven that the family
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$ is a
basis of $\Qsym_n$ for every $n\in \NN$.
Hence, the disjoint union of the families
$\left( \mathbf{M}_u \right) _{u \in \Comp_n}$ 
over all $n\in \NN$ is a basis
of the direct sum $\bigoplus_{n \in \NN} \Qsym_n$. Since the
former disjoint union is the family
$\left( \mathbf{M}_u \right) _{u \in \Comp}$, while the latter
direct sum is the $\kk$-module
$\bigoplus_{n \in \NN} \Qsym_n = \Qsym$, this rewrites as
follows: The family
$\left( \mathbf{M}_u \right) _{u \in \Comp}$ is a basis of the
$\kk$-module $\Qsym$. In other words, the family
$\left( \mathbf{M}_u \right) _{u \in \mathfrak{A}^{\ast}}$ is a
basis of the $\kk$-module $\Qsym$ (since $\mathfrak{A}^{\ast
} = \Comp $).

But Lemma \ref{lem.lyndonbases}(a) (applied to $A = \Qsym$,
$b_w = M_w$ and $\mathbf{b}_u = \mathbf{M}_u$) yields that
the family $\left(
M_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent generating
set of the $\kk$-algebra $\Qsym$ if and only if the
family $\left(  \mathbf{M}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis
of the $\kk$-module $\Qsym$. Hence, the family $\left(
M_{w}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent generating
set of the $\kk$-algebra $\Qsym$ (since we know that the
family $\left(  \mathbf{M}_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis
of the $\kk$-module $\Qsym$). This proves Proposition
\ref{prop.QSym.lyndon.M-genset}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.QSym.Malphas.ps}}

\textit{Solution to Exercise \ref{exe.QSym.Malphas.ps}.} We first
notice that every $s\in\left\{  1,2,3,\ldots\right\}  $ and $\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  $ satisfy%
\begin{equation}
\left(  \xx_{\mathbf{i}}^{\alpha}\right)  ^{s}=\xx_{\mathbf{i}%
}^{\alpha\left\{  s\right\}  } \label{sol.QSym.Malphas.ps.a.xias}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.a.xias}):} Let
$s\in\left\{  1,2,3,\ldots\right\}  $ and $\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  $. Then, $\mathbf{i}$ is an $\ell$-tuple of positive
integers. Write $\mathbf{i}$ in the form $\mathbf{i}=\left(  i_{1}%
,i_{2},\ldots,i_{\ell}\right)  $. Then, $\xx_{\mathbf{i}}^{\alpha
}=x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}%
^{\alpha_{\ell}}$ (by the definition of $\xx_{\mathbf{i}}^{\alpha}$).
Also, by the definition of $\xx_{\mathbf{i}}^{\alpha\left\{  s\right\}
}$, we have $\xx_{\mathbf{i}}^{\alpha\left\{  s\right\}  }=x_{i_{1}%
}^{s\alpha_{1}}x_{i_{2}}^{s\alpha_{2}}\cdots x_{i_{\ell}}^{s\alpha_{\ell}}$
(since $\alpha\left\{  s\right\}  =\left(  s\alpha_{1},s\alpha_{2}%
,\ldots,s\alpha_{\ell}\right)  $). Now, taking both sides of the equality
$\xx_{\mathbf{i}}^{\alpha}=x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}%
}\cdots x_{i_{\ell}}^{\alpha_{\ell}}$ to the $s$-th power, we obtain%
\[
\left(  \xx_{\mathbf{i}}^{\alpha}\right)  ^{s}=\left(  x_{i_{1}%
}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}%
}\right)  ^{s}=\left(  x_{i_{1}}^{\alpha_{1}}\right)  ^{s}\left(  x_{i_{2}%
}^{\alpha_{2}}\right)  ^{s}\cdots\left(  x_{i_{\ell}}^{\alpha_{\ell}}\right)
^{s}=x_{i_{1}}^{s\alpha_{1}}x_{i_{2}}^{s\alpha_{2}}\cdots x_{i_{\ell}%
}^{s\alpha_{\ell}}=\xx_{\mathbf{i}}^{\alpha\left\{  s\right\}
}\text{.}%
\]
This proves (\ref{sol.QSym.Malphas.ps.a.xias}).}.

(a) Let $s$ be a positive integer.

We begin with the following general observation:

If $R$ is a topological commutative $\kk$-algebra, if $\mathbf{I}$ is a
countable (finite or not) set, and if $\left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$ is a power-summable family of
elements of $R$, then%
\begin{equation}
p_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)
=\sum_{\mathbf{i}\in\mathbf{I}}s_{\mathbf{i}}^{s}.
\label{sol.QSym.Malphas.ps.a.general}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.a.general}):} Let $R$ be
a topological commutative $\kk$-algebra. Let $\mathbf{I}$ be a
countable set. Let $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\in R^{\mathbf{I}}$ be a power-summable family of elements of $R$. We need to
prove that (\ref{sol.QSym.Malphas.ps.a.general}) holds.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} The set $\mathbf{I}$ is infinite.
\par
\textit{Case 2:} The set $\mathbf{I}$ is finite.
\par
Let us first consider Case 1. In this case, the set $\mathbf{I}$ is infinite,
and therefore countably infinite (since it is countable). Fix a bijection
$\mathfrak{j}:\left\{  1,2,3,\ldots\right\}  \rightarrow\mathbf{I}$ (such a
bijection clearly exists). Then, $p_{s}\left(  \left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\right)  $ is the result of substituting
$s_{\mathfrak{j}\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$,
$s_{\mathfrak{j}\left(  3\right)  }$, $\ldots$ for the variables $x_{1}$,
$x_{2}$, $x_{3}$, $\ldots$ in $p_{s}$ (by the definition of $p_{s}\left(
\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $). Thus,%
\begin{align*}
&  p_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right) \\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }s_{\mathfrak{j}\left(
3\right)  }\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\underbrace{p_{s}}_{=\sum_{i\in\left\{
1,2,3,\ldots\right\}  }x_{i}^{s}}\right) \\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }s_{\mathfrak{j}\left(
3\right)  }\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\sum_{i\in\left\{  1,2,3,\ldots\right\}
}x_{i}^{s}\right) \\
&  =\sum_{i\in\left\{  1,2,3,\ldots\right\}  }s_{\mathfrak{j}\left(  i\right)
}^{s}=\sum_{\mathbf{i}\in\mathbf{I}}s_{\mathbf{i}}^{s}%
\end{align*}
(here, we have substituted $\mathbf{i}$ for $\mathfrak{j}\left(  i\right)  $
in the sum, since the map $\mathfrak{j}:\left\{  1,2,3,\ldots\right\}
\rightarrow\mathbf{I}$ is a bijection). Thus,
(\ref{sol.QSym.Malphas.ps.a.general}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, the set $\mathbf{I}$ is finite. Fix
a bijection $\mathfrak{j}:\left\{  1,2,\ldots,\left\vert \mathbf{I}\right\vert
\right\}  \rightarrow\mathbf{I}$ (such a bijection clearly exists). Then,
$p_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right)  $ is the result of substituting $s_{\mathfrak{j}\left(  1\right)  }%
$, $s_{\mathfrak{j}\left(  2\right)  }$, $\ldots$, $s_{\mathfrak{j}\left(
\left\vert \mathbf{I}\right\vert \right)  }$, $0$, $0$, $0$, $\ldots$ for the
variables $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$ in $p_{s}$ (by the definition of
$p_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right)  $). Thus,%
\begin{align*}
&  p_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right) \\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\underbrace{p_{s}}%
_{\substack{=\sum_{i=1}^{\infty}x_{i}^{s}\\=\sum_{i=1}^{\left\vert
\mathbf{I}\right\vert }x_{i}^{s}+\sum_{i=\left\vert \mathbf{I}\right\vert
+1}^{\infty}x_{i}^{s}}}\right) \\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\sum_{i=1}^{\left\vert
\mathbf{I}\right\vert }x_{i}^{s}+\sum_{i=\left\vert \mathbf{I}\right\vert
+1}^{\infty}x_{i}^{s}\right) \\
&  =\underbrace{\sum_{i=1}^{\left\vert \mathbf{I}\right\vert }}_{=\sum
_{i\in\left\{  1,2,\ldots,\left\vert \mathbf{I}\right\vert \right\}  }%
}s_{\mathfrak{j}\left(  i\right)  }^{s}+\sum_{i=\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  +1}^{\infty}\underbrace{0^{s}}%
_{\substack{=0\\\text{(since }s>0\text{)}}}=\sum_{i\in\left\{  1,2,\ldots
,\left\vert \mathbf{I}\right\vert \right\}  }s_{\mathfrak{j}\left(  i\right)
}^{s}+\underbrace{\sum_{i=\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert
\right)  +1}^{\infty}0}_{=0}=\sum_{i\in\left\{  1,2,\ldots,\left\vert
\mathbf{I}\right\vert \right\}  }s_{\mathfrak{j}\left(  i\right)  }^{s}\\
&  =\sum_{\mathbf{i}\in\mathbf{I}}s_{\mathbf{i}}^{s}%
\end{align*}
(here, we have substituted $\mathbf{i}$ for $\mathfrak{j}\left(  i\right)  $
in the sum, since the map $\mathfrak{j}:\left\{  1,2,\ldots,\left\vert
\mathbf{I}\right\vert \right\}  \rightarrow\mathbf{I}$ is a bijection). Thus,
(\ref{sol.QSym.Malphas.ps.a.general}) is proven in Case 2.
\par
We have thus proven (\ref{sol.QSym.Malphas.ps.a.general}) in each of the Cases
1 and 2. Since these two Cases are the only cases that can occur, we thus
conclude that (\ref{sol.QSym.Malphas.ps.a.general}) holds, qed.} Applying this
to $R= \kk \left[  \left[  \xx\right]  \right]  $, $\mathbf{I}%
=\operatorname*{SIS}\left(  \ell\right)  $ and $\mathbf{s}_{\mathbf{i}%
}=\xx_{\mathbf{i}}^{\alpha}$, we obtain%
\[
p_{s}\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  }\right)  =\sum_{\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  }\underbrace{\left(  \xx%
_{\mathbf{i}}^{\alpha}\right)  ^{s}}_{\substack{=\xx_{\mathbf{i}%
}^{\alpha\left\{  s\right\}  }\\\text{(by (\ref{sol.QSym.Malphas.ps.a.xias}%
))}}}=\sum_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }%
\xx_{\mathbf{i}}^{\alpha\left\{  s\right\}  }.
\]
Compared with $M_{\alpha\left\{  s\right\}  }=\sum_{\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  }\xx_{\mathbf{i}}%
^{\alpha\left\{  s\right\}  }$ (by (\ref{eq.def.QSym.Malphas.a.1}), applied to
$\alpha\left\{  s\right\}  $ instead of $\alpha$), this yields $p_{s}\left(
\left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  }\right)  =M_{\alpha\left\{
s\right\}  }$. Exercise \ref{exe.QSym.Malphas.ps}(a) is now solved.

(b) Let $s\in \NN$. We shall first make some general statements.

\begin{itemize}
\item If $R$ is a topological commutative $\kk$-algebra, if
$\mathbf{I}$ is a countable (finite or not) totally ordered set, and if
$\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$
is a power-summable family of elements of $R$, then%
\begin{equation}
\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}%
}=\sum\limits_{\substack{\mathbf{K}\subset\mathbf{I};\\\left\vert \mathbf{K}%
\right\vert =s}}\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathbf{i}}.
\label{sol.QSym.Malphas.ps.b.banal1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.b.banal1}):} Let $R$ be a
topological commutative $\kk$-algebra. Let $\mathbf{I}$ be a countable
totally ordered set. Let $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}%
\in\mathbf{I}}\in R^{\mathbf{I}}$ be a power-summable family of elements of
$R$. We need to prove that (\ref{sol.QSym.Malphas.ps.b.banal1}) holds.
\par
The order on $\mathbf{I}$ is total. Hence, the map%
\begin{align*}
\left\{  \left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)
\in\mathbf{I}^{s}\ \mid\ \mathbf{i}_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}%
_{s}\right\}   &  \rightarrow\left\{  \mathbf{K}\subset\mathbf{I}%
\ \mid\ \left\vert \mathbf{K}\right\vert =s\right\}  ,\\
\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)   &
\mapsto\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right\}
\end{align*}
\newline is a bijection (because every subset $\mathbf{K}$ of $\mathbf{I}$
satisfying $\left\vert \mathbf{K}\right\vert =s$ can be written in the form
$\mathbf{K}=\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right\}  $ for a unique $s$-tuple $\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\mathbf{I}^{s}$ satisfying
$\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}$).
\par
Now, let $\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)
\in\mathbf{I}^{s}$ be such that $\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}$. Then, the $s$-tuple $\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  $ is a list of all elements of the set
$\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right\}  $, each
occurring exactly once (since the elements $\mathbf{i}_{1}$, $\mathbf{i}_{2}$,
$\ldots$, $\mathbf{i}_{s}$ are distinct (since $\mathbf{i}%
_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}$)). Hence, $\prod_{\mathbf{i}%
\in\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right\}
}s_{\mathbf{i}}=s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}}%
$, so that $s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}%
}=\prod_{\mathbf{i}\in\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right\}  }s_{\mathbf{i}}$.
\par
Now, let us forget that we fixed $\left(  \mathbf{i}_{1},\mathbf{i}_{2}%
,\ldots,\mathbf{i}_{s}\right)  \in\mathbf{I}^{s}$. We thus have shown that
$s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}}=\prod
_{\mathbf{i}\in\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right\}  }s_{\mathbf{i}}$ for every $\left(  \mathbf{i}_{1}%
,\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\mathbf{I}^{s}$ be such that
$\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}$. Hence,%
\[
\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}\underbrace{s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots
s_{\mathbf{i}_{s}}}_{\substack{=\prod_{\mathbf{i}\in\left\{  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right\}  }s_{\mathbf{i}}}%
}=\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}\prod_{\mathbf{i}\in\left\{  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right\}  }s_{\mathbf{i}}=\sum\limits_{\substack{\mathbf{K}%
\subset\mathbf{I};\\\left\vert \mathbf{K}\right\vert =s}}\prod_{\mathbf{i}%
\in\mathbf{K}}s_{\mathbf{i}}%
\]
(here, we have substituted $\mathbf{K}$ for $\left\{  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right\}  $ in the sum,\ since the
map%
\begin{align*}
\left\{  \left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)
\in\mathbf{I}^{s}\ \mid\ \mathbf{i}_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}%
_{s}\right\}   &  \rightarrow\left\{  \mathbf{K}\subset\mathbf{I}%
\ \mid\ \left\vert \mathbf{K}\right\vert =s\right\}  ,\\
\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)   &
\mapsto\left\{  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right\}
\end{align*}
is a bijection). This proves (\ref{sol.QSym.Malphas.ps.b.banal1}).}

\item If $R$ is a commutative $\kk$-algebra, if $\mathbf{I}$ and
$\mathbf{J}$ are two sets, if $\mathfrak{j}:\mathbf{J}\rightarrow\mathbf{I}$
is a bijection, and if $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}%
\in\mathbf{I}}\in R^{\mathbf{I}}$ is a family of elements of $R$, then%
\begin{equation}
\prod_{\mathbf{i}\in\mathfrak{j}\left(  \mathbf{K}\right)  }s_{\mathbf{i}%
}=\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathfrak{j}\left(  \mathbf{i}\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every finite subset }\mathbf{K}\text{ of
}\mathbf{J}. \label{sol.QSym.Malphas.ps.b.banal2}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.b.banal2}):} Let $R$ be a
commutative $\kk$-algebra. Let $\mathbf{I}$ and $\mathbf{J}$ be two
sets. Let $\mathfrak{j}:\mathbf{J}\rightarrow\mathbf{I}$ be a bijection. Let
$\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$
be a family of elements of $R$. Let $\mathbf{K}$ be a finite
subset of $\mathbf{J}$.
The map $\mathfrak{j}$ is a bijection, and thus injective. Hence, the map
$\mathbf{K}\rightarrow\mathfrak{j}\left(  \mathbf{K}\right)  ,\ \mathbf{i}%
\mapsto\mathfrak{j}\left(  \mathbf{i}\right)  $ is a bijection. Hence, we can
substitute $\mathfrak{j}\left(  \mathbf{i}\right)  $ for $\mathbf{i}$ in the
product $\prod_{\mathbf{i}\in\mathfrak{j}\left(  \mathbf{K}\right)
}s_{\mathbf{i}}$. As a result, we obtain $\prod_{\mathbf{i}\in\mathfrak{j}%
\left(  \mathbf{K}\right)  }s_{\mathbf{i}}=\prod_{\mathbf{i}\in\mathbf{K}%
}s_{\mathfrak{j}\left(  \mathbf{i}\right)  }$. This proves
(\ref{sol.QSym.Malphas.ps.b.banal2}).}

\item If $R$ is a topological commutative $\kk$-algebra, if
$\mathbf{I}$ and $\mathbf{J}$ are two countable totally ordered sets, if
$\mathfrak{j}:\mathbf{J}\rightarrow\mathbf{I}$ is a (not necessarily
order-preserving!) bijection, and if $\left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$ is a power-summable family of
elements of $R$, then%
\begin{equation}
\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}%
}=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}%
_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)  }%
s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }. \label{sol.QSym.Malphas.ps.b.banal3}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.b.banal3}):} Let $R$ be a
topological commutative $\kk$-algebra. Let $\mathbf{I}$ and
$\mathbf{J}$ be two countable totally ordered sets. Let $\mathfrak{j}%
:\mathbf{J}\rightarrow\mathbf{I}$ be a bijection. Let $\left(  s_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$ be a power-summable
family of elements of $R$.
\par
Notice that $\left(  s_{\mathfrak{j}\left(  \mathbf{i}\right)  }\right)
_{\mathbf{i}\in\mathbf{J}}\in R^{\mathbf{J}}$ is a reindexing of the family
$\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$
(since $\mathfrak{j}:\mathbf{J}\rightarrow\mathbf{I}$ is a bijection), and
thus a power-summable family of elements of $R$ (since $\left(  s_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}$ is a power-summable family of elements
of $R$).
\par
The map%
\begin{align*}
\left\{  \mathbf{K}\subset\mathbf{J}\ \mid\ \left\vert \mathbf{K}\right\vert
=s\right\}   &  \rightarrow\left\{  \mathbf{K}\subset\mathbf{I}\ \mid
\ \left\vert \mathbf{K}\right\vert =s\right\}  ,\\
\mathbf{K}  &  \mapsto\mathfrak{j}\left(  \mathbf{K}\right)
\end{align*}
is a bijection (since $\mathfrak{j}$ is a bijection from $\mathbf{J}$ to
$\mathbf{I}$). Hence, we can substitute $\mathfrak{j}\left(  \mathbf{K}%
\right)  $ for $\mathbf{K}$ in the sum $\sum\limits_{\substack{\mathbf{K}%
\subset\mathbf{I};\\\left\vert \mathbf{K}\right\vert =s}}\prod_{\mathbf{i}%
\in\mathbf{K}}s_{\mathbf{i}}$, and as a result we obtain%
\[
\sum\limits_{\substack{\mathbf{K}\subset\mathbf{I};\\\left\vert \mathbf{K}\right\vert
=s}}\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathbf{i}}=\sum\limits_{\substack{\mathbf{K}%
\subset\mathbf{J};\\\left\vert \mathbf{K}\right\vert =s}}\underbrace{\prod
_{\mathbf{i}\in\mathfrak{j}\left(  \mathbf{K}\right)  }s_{\mathbf{i}}%
}_{\substack{=\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathfrak{j}\left(
\mathbf{i}\right)  }\\\text{(by (\ref{sol.QSym.Malphas.ps.b.banal2}))}}%
}=\sum\limits_{\substack{\mathbf{K}\subset\mathbf{J};\\\left\vert \mathbf{K}%
\right\vert =s}}\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathfrak{j}\left(
\mathbf{i}\right)  }.
\]
Thus, (\ref{sol.QSym.Malphas.ps.b.banal1}) becomes%
\begin{equation}
\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}%
}=\sum\limits_{\substack{\mathbf{K}\subset\mathbf{I};\\\left\vert \mathbf{K}%
\right\vert =s}}\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathbf{i}}=\sum\limits
_{\substack{\mathbf{K}\subset\mathbf{J};\\\left\vert \mathbf{K}\right\vert
=s}}\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathfrak{j}\left(  \mathbf{i}\right)
}. \label{sol.QSym.Malphas.ps.b.banal1.pf.1}%
\end{equation}
But%
\begin{align*}
\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}%
_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)  }%
s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }  &  =\sum\limits_{\substack{\left(  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\mathbf{J}^{s}%
;\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}}}s_{\mathfrak{j}\left(
\mathbf{i}_{1}\right)  }s_{\mathfrak{j}\left(  \mathbf{i}_{2}\right)  }\cdots
s_{\mathfrak{j}\left(  \mathbf{i}_{s}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \text{ as
}\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \right)
\\
&  =\sum\limits_{\substack{\mathbf{K}\subset\mathbf{J};\\\left\vert \mathbf{K}%
\right\vert =s}}\prod_{\mathbf{i}\in\mathbf{K}}s_{\mathfrak{j}\left(
\mathbf{i}\right)  }%
\end{align*}
(by (\ref{sol.QSym.Malphas.ps.b.banal1}), applied to $\mathbf{J}$ and
$s_{\mathfrak{j}\left(  \mathbf{i}\right)  }$ instead of $\mathbf{I}$ and
$s_{\mathbf{i}}$). Compared with (\ref{sol.QSym.Malphas.ps.b.banal1.pf.1}),
this yields%
\[
\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}%
}=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}%
_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)  }%
s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }.
\]
This proves (\ref{sol.QSym.Malphas.ps.b.banal3}).}

\item If $R$ is a topological commutative $\kk$-algebra, if
$\mathbf{I}$ is a countable (finite or not) totally ordered set, and if
$\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$
is a power-summable family of elements of $R$, then%
\begin{equation}
e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)
=\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}_{2}}\cdots s_{\mathbf{i}_{s}%
}. \label{sol.QSym.Malphas.ps.b.es1}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.b.es1}):} Let $R$ be a
topological commutative $\kk$-algebra. Let $\mathbf{I}$ be a countable
totally ordered set. Let $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}%
\in\mathbf{I}}\in R^{\mathbf{I}}$ be a power-summable family of elements of
$R$. We need to prove that (\ref{sol.QSym.Malphas.ps.b.es1}) holds.
\par
The definition of $e_{s}$ yields
\begin{equation}
e_{s}=\sum\limits_{\substack{\left(  i_{1},i_{2},\ldots,i_{s}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{s};\\i_{1}<i_{2}<\cdots<i_{s}}}x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{s}}=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2}%
,\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}
^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}%
}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}
\label{sol.QSym.Malphas.ps.b.es1.pf.0}%
\end{equation}
(here, we renamed the summation index $\left(  i_{1},i_{2},\ldots
,i_{s}\right)  $ as $\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  $).
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} The set $\mathbf{I}$ is infinite.
\par
\textit{Case 2:} The set $\mathbf{I}$ is finite.
\par
Let us first consider Case 1. In this case, the set $\mathbf{I}$ is infinite,
and therefore countably infinite (since it is countable). Let $\mathbf{J}$
denote the totally ordered set $\left\{  1,2,3,\ldots\right\}  $. Then,
(\ref{sol.QSym.Malphas.ps.b.es1.pf.0}) rewrites as%
\begin{equation}
e_{s}=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{s};\\\mathbf{j}%
_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}}x_{\mathbf{j}%
_{2}}\cdots x_{\mathbf{j}_{s}}=\sum\limits_{\substack{\left(  \mathbf{j}%
_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s}%
;\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}%
}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}
\label{sol.QSym.Malphas.ps.b.es1.pf.01}%
\end{equation}
(since $\left\{  1,2,3,\ldots\right\}  =\mathbf{J}$).
\par
Fix a bijection $\mathfrak{j}:\left\{  1,2,3,\ldots\right\}  \rightarrow
\mathbf{I}$ (such a bijection clearly exists, since $\mathbf{I}$ is countably
infinite). Then, $e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}%
\in\mathbf{I}}\right)  $ is the result of substituting $s_{\mathfrak{j}\left(
1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$, $s_{\mathfrak{j}\left(
3\right)  }$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$
in $e_{s}$ (by the definition of $e_{s}\left(  \left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\right)  $). Thus,%
\begin{align}
&  e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }s_{\mathfrak{j}\left(
3\right)  }\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\underbrace{e_{s}}_{\substack{=\sum\limits
_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}%
_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}%
}\\\text{(by (\ref{sol.QSym.Malphas.ps.b.es1.pf.01}))}}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }s_{\mathfrak{j}\left(
3\right)  }\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\sum\limits_{\substack{\left(  \mathbf{j}%
_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s}%
;\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}%
}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}%
_{2}<\cdots<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)
}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }.
\label{sol.QSym.Malphas.ps.b.es1.pf.5}
\end{align}
\par
But $\mathbf{J}$ is a countable totally ordered set (since $\mathbf{J}%
=\left\{  1,2,3,\ldots\right\}  $). Also, $\mathfrak{j}$ is a bijection
$\left\{  1,2,3,\ldots\right\}  \rightarrow\mathbf{I}$. In other words,
$\mathfrak{j}$ is a bijection $\mathbf{J}\rightarrow\mathbf{I}$ (since
$\mathbf{J}=\left\{  1,2,3,\ldots\right\}  $). Now,
(\ref{sol.QSym.Malphas.ps.b.es1.pf.5}) becomes%
\[
e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)
=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}%
_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)  }%
s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }=\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}%
_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}%
_{2}}\cdots s_{\mathbf{i}_{s}}%
\]
(by (\ref{sol.QSym.Malphas.ps.b.banal3})). Thus,
(\ref{sol.QSym.Malphas.ps.b.es1}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, the set $\mathbf{I}$ is finite. Let
$\mathbf{J}$ denote the totally ordered set $\left\{  1,2,\ldots,\left\vert
\mathbf{I}\right\vert \right\}  $. Then, $\mathbf{J}=\left\{  1,2,\ldots
,\left\vert \mathbf{I}\right\vert \right\}  \subset\left\{  1,2,3,\ldots
\right\}  $ and $\left\{  1,2,3,\ldots\right\}  \setminus
\underbrace{\mathbf{J}}_{=\left\{  1,2,\ldots,\left\vert \mathbf{I}\right\vert
\right\}  }=\left\{  1,2,3,\ldots\right\}  \setminus\left\{  1,2,\ldots
,\left\vert \mathbf{I}\right\vert \right\}  =\left\{  \left\vert
\mathbf{I}\right\vert +1,\left\vert \mathbf{I}\right\vert +2,\left\vert
\mathbf{I}\right\vert +3,\ldots\right\}  $.
\par
Fix a bijection $\mathfrak{j}:\left\{  1,2,\ldots,\left\vert \mathbf{I}%
\right\vert \right\}  \rightarrow\mathbf{I}$ (such a bijection clearly exists,
since $\mathbf{I}$ is finite). Then, $e_{s}\left(  \left(  s_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $ is the result of substituting
$s_{\mathfrak{j}\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$,
$\ldots$, $s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }%
$, $0$, $0$, $0$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$,
$\ldots$ in $e_{s}$ (by the definition of $e_{s}\left(  \left(  s_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $). Thus,%
\begin{align}
&  e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\underbrace{e_{s}}%
_{\substack{=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2}%
,\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}
^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}%
}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}\\\text{(by
(\ref{sol.QSym.Malphas.ps.b.es1.pf.0}))}}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\sum\limits_{\substack{\left(
\mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}%
}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{s};\\\mathbf{j}%
_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}\left(  \text{the result of
substituting }s_{\mathfrak{j}\left(  1\right)  }\text{, }s_{\mathfrak{j}%
\left(  2\right)  }\text{, }\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  }\text{, }0\text{, }0\text{, }0\text{, }%
\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{,
}\ldots\text{ in }x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}%
_{s}}\right)  .
\label{sol.QSym.Malphas.ps.b.es1.pf.7}
\end{align}
\par
Now, let us fix some $\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s}$. Then, every $k\in\left\{
1,2,\ldots,s\right\}  $ satisfies $\mathbf{j}_{k}\in\mathbf{J}=\left\{
1,2,\ldots,\left\vert \mathbf{I}\right\vert \right\}  $. Hence, for every
$k\in\left\{  1,2,\ldots,s\right\}  $, the substitution of $s_{\mathfrak{j}%
\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$, $\ldots$,
$s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }$, $0$,
$0$, $0$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$
transforms the indeterminate $x_{\mathbf{j}_{k}}$ into $s_{\mathfrak{j}\left(
\mathbf{j}_{k}\right)  }$. Consequently, the substitution of $s_{\mathfrak{j}%
\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$, $\ldots$,
$s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }$, $0$,
$0$, $0$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$, $\ldots$
transforms the product $x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots
x_{\mathbf{j}_{s}}$ into $s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)
}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }$. In other words,%
\begin{align}
&  \left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }x_{\mathbf{j}_{1}}%
x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}\right) \nonumber\\
&  =s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)  }s_{\mathfrak{j}\left(
\mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(  \mathbf{j}_{s}\right)
}. \label{sol.QSym.Malphas.ps.b.es1.pf.8a}%
\end{align}
\par
Now, let us forget that we fixed $\left(  \mathbf{j}_{1},\mathbf{j}_{2}%
,\ldots,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s}$. We thus have shown that
(\ref{sol.QSym.Malphas.ps.b.es1.pf.8a}) holds for every $\left(
\mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s}%
$.
\par
On the other hand, let us fix some $\left(  \mathbf{j}_{1},\mathbf{j}%
_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}
^{s}\setminus\mathbf{J}^{s}$. Since $\left(  \mathbf{j}_{1},\mathbf{j}%
_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}
^{s}\setminus\mathbf{J}^{s}$, we must have $\left(  \mathbf{j}_{1}%
,\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{s}$ but $\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \notin\mathbf{J}^{s}$. Since $\left(  \mathbf{j}%
_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \notin\mathbf{J}^{s}$, there
exists some $k\in\left\{  1,2,\ldots,s\right\}  $ such that $\mathbf{j}%
_{k}\notin\mathbf{J}$. Fix this $k$. We have $\mathbf{j}_{k}\in\left\{
1,2,3,\ldots\right\}  $ but $\mathbf{j}_{k}\notin\mathbf{J}$; therefore,
$\mathbf{j}_{k}\in\left\{  1,2,3,\ldots\right\}  \setminus\mathbf{J}=\left\{
\left\vert \mathbf{I}\right\vert +1,\left\vert \mathbf{I}\right\vert
+2,\left\vert \mathbf{I}\right\vert +3,\ldots\right\}  $. In other words,
$\mathbf{j}_{k}>\left\vert \mathbf{I}\right\vert $. Hence, the substitution of
$s_{\mathfrak{j}\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$,
$\ldots$, $s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }%
$, $0$, $0$, $0$, $\ldots$ for the variables $x_{1}$, $x_{2}$, $x_{3}$,
$\ldots$ transforms the indeterminate $x_{\mathbf{j}_{k}}$ into $0$. In other
words,%
\begin{align*}
&  \left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }x_{\mathbf{j}_{k}}\right) \\
&  =0.
\end{align*}
\par
Now,%
\begin{align}
&  \left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\underbrace{x_{\mathbf{j}_{1}%
}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}}_{=\prod_{z\in\left\{
1,2,\ldots,s\right\}  }x_{\mathbf{j}_{z}}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }\ldots\text{,
}s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }\text{,
}0\text{, }0\text{, }0\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\prod_{z\in\left\{
1,2,\ldots,s\right\}  }x_{\mathbf{j}_{z}}\right) \nonumber\\
&  =\prod_{z\in\left\{  1,2,\ldots,s\right\}  }\left(  \text{the result of
substituting }s_{\mathfrak{j}\left(  1\right)  }\text{, }s_{\mathfrak{j}%
\left(  2\right)  }\text{, }\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  }\text{, }0\text{, }0\text{, }0\text{, }%
\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{,
}\ldots\text{ in }x_{\mathbf{j}_{z}}\right) \nonumber\\
&  =\underbrace{\left(  \text{the result of substituting }s_{\mathfrak{j}%
\left(  1\right)  }\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }%
\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert
\right)  }\text{, }0\text{, }0\text{, }0\text{, }\ldots\text{ for the
variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{, }\ldots\text{ in
}x_{\mathbf{j}_{k}}\right)  }_{=0}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\prod\limits_{\substack{z\in\left\{  1,2,\ldots
,s\right\}  ;\\z\neq k}}\left(  \text{the result of substituting
}s_{\mathfrak{j}\left(  1\right)  }\text{, }s_{\mathfrak{j}\left(  2\right)
}\text{, }\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert \mathbf{I}%
\right\vert \right)  }\text{, }0\text{, }0\text{, }0\text{, }\ldots\text{ for
the variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{, }\ldots\text{ in
}x_{\mathbf{j}_{z}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the factor for
}z=k\text{ from the product}\right) \nonumber\\
&  =0.
\label{sol.QSym.Malphas.ps.b.es1.pf.8b}
\end{align}
Now, let us forget that we fixed $\left(  \mathbf{j}_{1},\mathbf{j}_{2}%
,\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{s}%
\setminus\mathbf{J}^{s}$. We thus have shown that
(\ref{sol.QSym.Malphas.ps.b.es1.pf.8b}) holds for every $\left(
\mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{
1,2,3,\ldots\right\}  ^{s}\setminus\mathbf{J}^{s}$.
\par
But $\mathbf{J}^{s}\subset\left\{  1,2,3,\ldots\right\}  ^{s}$ (since
$\mathbf{J}\subset\left\{  1,2,3,\ldots\right\}  $). Hence, the set $\left\{
1,2,3,\ldots\right\}  ^{s}$ is the union of its two disjoint subsets
$\mathbf{J}^{s}$ and $\left\{  1,2,3,\ldots\right\}  ^{s}\setminus
\mathbf{J}^{s}$. Now, (\ref{sol.QSym.Malphas.ps.b.es1.pf.7}) becomes
\par%
\begin{align}
&  e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}  ^{s};\\\mathbf{j}%
_{1}<\mathbf{j}_{2}<\cdots<\mathbf{j}_{s}}}\left(  \text{the result of
substituting }s_{\mathfrak{j}\left(  1\right)  }\text{, }s_{\mathfrak{j}%
\left(  2\right)  }\text{, }\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  }\text{, }0\text{, }0\text{, }0\text{, }%
\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{,
}\ldots\text{ in }x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}%
_{s}}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}%
_{2}<\cdots<\mathbf{j}_{s}}}\underbrace{\left(  \text{the result of
substituting }s_{\mathfrak{j}\left(  1\right)  }\text{, }s_{\mathfrak{j}%
\left(  2\right)  }\text{, }\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  }\text{, }0\text{, }0\text{, }0\text{, }%
\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{,
}\ldots\text{ in }x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}%
_{s}}\right)  }_{\substack{=s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)
}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }\\\text{(by (\ref{sol.QSym.Malphas.ps.b.es1.pf.8a}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}%
_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots\right\}
^{s}\setminus\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}} \nonumber\\
& \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ 
\underbrace{\left(  \text{the result of substituting
}s_{\mathfrak{j}\left(  1\right)  }\text{, }s_{\mathfrak{j}\left(  2\right)
}\text{, }\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert \mathbf{I}%
\right\vert \right)  }\text{, }0\text{, }0\text{, }0\text{, }\ldots\text{ for
the variables }x_{1}\text{, }x_{2}\text{, }x_{3}\text{, }\ldots\text{ in
}x_{\mathbf{j}_{1}}x_{\mathbf{j}_{2}}\cdots x_{\mathbf{j}_{s}}\right)
}_{\substack{=0\\\text{(by (\ref{sol.QSym.Malphas.ps.b.es1.pf.8b}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the set }\left\{  1,2,3,\ldots
\right\}  ^{s}\text{ is the union of its two disjoint subsets }\mathbf{J}%
^{s}\text{ and }\left\{  1,2,3,\ldots\right\}  ^{s}\setminus\mathbf{J}%
^{s}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}%
_{2}<\cdots<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)
}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }+\underbrace{\sum\limits_{\substack{\left(  \mathbf{j}%
_{1},\mathbf{j}_{2},\ldots,\mathbf{j}_{s}\right)  \in\left\{  1,2,3,\ldots
\right\}  ^{s}\setminus\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}%
<\cdots<\mathbf{j}_{s}}}0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots
,\mathbf{j}_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}%
_{2}<\cdots<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)
}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }. \label{sol.QSym.Malphas.ps.b.es1.pf.10}%
\end{align}
\par
But $\mathbf{J}$ is a countable totally ordered set (since $\mathbf{J}%
=\left\{  1,2,\ldots,\left\vert \mathbf{I}\right\vert \right\}  $). Also,
$\mathfrak{j}$ is a bijection $\left\{  1,2,\ldots,\left\vert \mathbf{I}%
\right\vert \right\}  \rightarrow\mathbf{I}$. In other words, $\mathfrak{j}$
is a bijection $\mathbf{J}\rightarrow\mathbf{I}$ (since $\mathbf{J}=\left\{
1,2,\ldots,\left\vert \mathbf{I}\right\vert \right\}  $). Now,
(\ref{sol.QSym.Malphas.ps.b.es1.pf.10}) becomes%
\[
e_{s}\left(  \left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)
=\sum\limits_{\substack{\left(  \mathbf{j}_{1},\mathbf{j}_{2},\ldots,\mathbf{j}%
_{s}\right)  \in\mathbf{J}^{s};\\\mathbf{j}_{1}<\mathbf{j}_{2}<\cdots
<\mathbf{j}_{s}}}s_{\mathfrak{j}\left(  \mathbf{j}_{1}\right)  }%
s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }\cdots s_{\mathfrak{j}\left(
\mathbf{j}_{s}\right)  }=\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\mathbf{I}^{s};\\\mathbf{i}%
_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}}}s_{\mathbf{i}_{1}}s_{\mathbf{i}%
_{2}}\cdots s_{\mathbf{i}_{s}}%
\]
(by (\ref{sol.QSym.Malphas.ps.b.banal3})). Thus,
(\ref{sol.QSym.Malphas.ps.b.es1}) is proven in Case 2.
\par
We have thus proven (\ref{sol.QSym.Malphas.ps.b.es1}) in each of the Cases 1
and 2. Since these two Cases are the only cases that can occur, we thus
conclude that (\ref{sol.QSym.Malphas.ps.b.es1}) holds, qed.}
\end{itemize}

Now, let us return to our proof. Applying (\ref{sol.QSym.Malphas.ps.b.es1}) to
$R= \kk \left[  \left[  \xx\right]  \right]  $, $\mathbf{I}%
=\operatorname*{SIS}\left(  \ell\right)  $ and $\mathbf{s}_{\mathbf{i}%
}=\xx_{\mathbf{i}}^{\alpha}$, we obtain%
\[
e_{s}\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  }\right)  =\sum\limits_{\substack{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}%
_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}%
^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}%
}^{\alpha}.
\]
Thus,%
\[
M_{\alpha}^{\left\langle s\right\rangle }=e_{s}\left(  \left(  \xx%
_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\right)  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}%
_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}.
\]
This solves Exercise \ref{exe.QSym.Malphas.ps}(b).

(c) Every $\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  $ satisfies
$\left(  \xx_{\mathbf{i}}^{\alpha}\right)  ^{n}=\xx_{\mathbf{i}%
}^{\alpha\left\{  n\right\}  }$ (by (\ref{sol.QSym.Malphas.ps.a.xias}),
applied to $n$ instead of $s$). In other words, every $\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  $ satisfies $\xx%
_{\mathbf{i}}^{\alpha\left\{  n\right\}  }=\left(  \xx_{\mathbf{i}%
}^{\alpha}\right)  ^{n}$. Now, by the definition of $M_{\alpha\left\{
n\right\}  }^{\left\langle s\right\rangle }$, we have%
\begin{equation}
M_{\alpha\left\{  n\right\}  }^{\left\langle s\right\rangle }=e_{s}\left(
\left(  \underbrace{\xx_{\mathbf{i}}^{\alpha\left\{  n\right\}  }%
}_{=\left(  \xx_{\mathbf{i}}^{\alpha}\right)  ^{n}}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)  =e_{s}\left(
\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)  ^{n}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)  .
\label{sol.QSym.Malphas.ps.c.start}%
\end{equation}


We now notice that if $R$ is a topological commutative $\kk$-algebra,
if $\mathbf{I}$ is a countable set, and if $\left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\in R^{\mathbf{I}}$ is a power-summable family of
elements of $R$, then%
\begin{equation}
e_{s}^{\left\langle n\right\rangle }\left(  \left(  \mathbf{s}_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  =e_{s}\left(  \left(
\mathbf{s}_{\mathbf{i}}^{n}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  .
\label{sol.QSym.Malphas.ps.c.general}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.QSym.Malphas.ps.c.general}):} Let $R$ be
a topological commutative $\kk$-algebra. Let $\mathbf{I}$ be a
countable set. Let $\left(  s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}%
}\in R^{\mathbf{I}}$ be a power-summable family of elements of $R$.
\par
We must be in one of the following two cases:
\par
\textit{Case 1:} The set $\mathbf{I}$ is infinite.
\par
\textit{Case 2:} The set $\mathbf{I}$ is finite.
\par
Let us first consider Case 1. In this case, the set $\mathbf{I}$ is infinite,
and therefore countably infinite (since it is countable). Fix a bijection
$\mathfrak{j}:\left\{  1,2,3,\ldots\right\}  \rightarrow\mathbf{I}$ (such a
bijection clearly exists, since $\mathbf{I}$ is countably infinite). Then,
$e_{s}^{\left\langle n\right\rangle }\left(  \left(  \mathbf{s}_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $ is the result of substituting
$s_{\mathfrak{j}\left(  1\right)  }$, $s_{\mathfrak{j}\left(  2\right)  }$,
$s_{\mathfrak{j}\left(  3\right)  }$, $\ldots$ for the variables $x_{1}$,
$x_{2}$, $x_{3}$, $\ldots$ in $e_{s}^{\left\langle n\right\rangle }$ (by the
definition of $e_{s}^{\left\langle n\right\rangle }\left(  \left(
s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $). Thus,%
\begin{align}
&  e_{s}^{\left\langle n\right\rangle }\left(  \left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }s_{\mathfrak{j}\left(
3\right)  }\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\underbrace{e_{s}^{\left\langle n\right\rangle
}}_{=\sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}^{n}x_{i_{2}}^{n}\cdots x_{i_{s}%
}^{n}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }s_{\mathfrak{j}\left(
3\right)  }\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}%
^{n}x_{i_{2}}^{n}\cdots x_{i_{s}}^{n}\right) \nonumber\\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{s}}s_{\mathfrak{j}\left(  \mathbf{j}%
_{1}\right)  }^{n}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }^{n}\cdots
s_{\mathfrak{j}\left(  \mathbf{j}_{s}\right)  }^{n}.
\label{sol.QSym.Malphas.ps.c.general.pf.1}%
\end{align}
On the other hand, $e_{s}\left(  \left(  \mathbf{s}_{\mathbf{i}}^{n}\right)
_{\mathbf{i}\in\mathbf{I}}\right)  $ is the result of substituting
$s_{\mathfrak{j}\left(  1\right)  }^{n}$, $s_{\mathfrak{j}\left(  2\right)
}^{n}$, $s_{\mathfrak{j}\left(  3\right)  }^{n}$, $\ldots$ for the variables
$x_{1}$, $x_{2}$, $x_{3}$, $\ldots$ in $e_{s}$ (by the definition of
$e_{s}\left(  \left(  s_{\mathbf{i}}^{n}\right)  _{\mathbf{i}\in\mathbf{I}%
}\right)  $). Thus,
\begin{align*}
&  e_{s}\left(  \left(  \mathbf{s}_{\mathbf{i}}^{n}\right)  _{\mathbf{i}%
\in\mathbf{I}}\right) \\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}^{n}\text{, }s_{\mathfrak{j}\left(  2\right)  }^{n}\text{, }s_{\mathfrak{j}%
\left(  3\right)  }^{n}\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\underbrace{e_{s}}_{=\sum
_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{s}}}\right) \\
&  =\left(  \text{the result of substituting }s_{\mathfrak{j}\left(  1\right)
}^{n}\text{, }s_{\mathfrak{j}\left(  2\right)  }^{n}\text{, }s_{\mathfrak{j}%
\left(  3\right)  }^{n}\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\sum_{i_{1}<i_{2}<\cdots<i_{s}%
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{s}}\right) \\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{s}}s_{\mathfrak{j}\left(  \mathbf{j}%
_{1}\right)  }^{n}s_{\mathfrak{j}\left(  \mathbf{j}_{2}\right)  }^{n}\cdots
s_{\mathfrak{j}\left(  \mathbf{j}_{s}\right)  }^{n}.
\end{align*}
Compared with (\ref{sol.QSym.Malphas.ps.c.general.pf.1}), this yields
$e_{s}^{\left\langle n\right\rangle }\left(  \left(  \mathbf{s}_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  =e_{s}\left(  \left(
\mathbf{s}_{\mathbf{i}}^{n}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $.
Thus, (\ref{sol.QSym.Malphas.ps.c.general}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, the set $\mathbf{I}$ is finite. Fix
a bijection $\mathfrak{j}:\left\{  1,2,\ldots,\left\vert \mathbf{I}\right\vert
\right\}  \rightarrow\mathbf{I}$ (such a bijection clearly exists, since
$\mathbf{I}$ is finite).
\par
Define an infinite sequence $\left(  t_{1},t_{2},t_{3},\ldots\right)  $ of
elements of $R$ by%
\begin{equation}
\left(  t_{1},t_{2},t_{3},\ldots\right)  =\left(  s_{\mathfrak{j}\left(
1\right)  },s_{\mathfrak{j}\left(  2\right)  },\ldots,s_{\mathfrak{j}\left(
\left\vert \mathbf{I}\right\vert \right)  },0,0,0,\ldots\right)  .
\label{sol.QSym.Malphas.ps.c.general.pf.4}%
\end{equation}
Then,%
\begin{equation}
\left(  t_{1}^{n},t_{2}^{n},t_{3}^{n},\ldots\right)  =\left(  s_{\mathfrak{j}%
\left(  1\right)  }^{n},s_{\mathfrak{j}\left(  2\right)  }^{n},\ldots
,s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert \right)  }%
^{n},\underbrace{0^{n},0^{n},0^{n},\ldots}_{\substack{=\left(  0,0,0,\ldots
\right)  \\\text{(since }0^{n}=0\text{ (since }n\text{ is positive))}%
}}\right)  =\left(  s_{\mathfrak{j}\left(  1\right)  }^{n},s_{\mathfrak{j}%
\left(  2\right)  }^{n},\ldots,s_{\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  }^{n},0,0,0,\ldots\right)  .
\label{sol.QSym.Malphas.ps.c.general.pf.5}%
\end{equation}
\par
Recall that $e_{s}^{\left\langle n\right\rangle }\left(  \left(
\mathbf{s}_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $ is the
result of substituting $s_{\mathfrak{j}\left(  1\right)  }$, $s_{\mathfrak{j}%
\left(  2\right)  }$, $\ldots$, $s_{\mathfrak{j}\left(  \left\vert
\mathbf{I}\right\vert \right)  }$, $0$, $0$, $0$, $\ldots$ for the variables
$x_{1}$, $x_{2}$, $x_{3}$, $\ldots$ in $e_{s}^{\left\langle n\right\rangle }$
(by the definition of $e_{s}^{\left\langle n\right\rangle }\left(  \left(
s_{\mathbf{i}}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $). Thus,%
\begin{align}
&  e_{s}^{\left\langle n\right\rangle }\left(  \left(  s_{\mathbf{i}}\right)
_{\mathbf{i}\in\mathbf{I}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }\underbrace{s_{\mathfrak{j}%
\left(  1\right)  }\text{, }s_{\mathfrak{j}\left(  2\right)  }\text{, }%
\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert
\right)  }\text{, }0\text{, }0\text{, }0\text{, }\ldots}_{\substack{=\left(
t_{1},t_{2},t_{3},\ldots\right)  \\\text{(by
(\ref{sol.QSym.Malphas.ps.c.general.pf.4}))}}}\text{ for the variables }%
x_{1}\text{, }x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }e_{s}^{\left\langle
n\right\rangle }\right) \nonumber\\
&  =\left(  \text{the result of substituting }t_{1}\text{, }t_{2}\text{,
}t_{3}\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\underbrace{e_{s}^{\left\langle n\right\rangle
}}_{=\sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}^{n}x_{i_{2}}^{n}\cdots x_{i_{s}%
}^{n}}\right) \nonumber\\
&  =\left(  \text{the result of substituting }t_{1}\text{, }t_{2}\text{,
}t_{3}\text{, }\ldots\text{ for the variables }x_{1}\text{, }x_{2}\text{,
}x_{3}\text{, }\ldots\text{ in }\sum_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}%
^{n}x_{i_{2}}^{n}\cdots x_{i_{s}}^{n}\right) \nonumber\\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{s}}t_{i_{1}}^{n}t_{i_{2}}^{n}\cdots t_{i_{s}%
}^{n}. \label{sol.QSym.Malphas.ps.c.general.pf.9}%
\end{align}
\par
On the other hand, $e_{s}\left(  \left(  \mathbf{s}_{\mathbf{i}}^{n}\right)
_{\mathbf{i}\in\mathbf{I}}\right)  $ is the result of substituting
$s_{\mathfrak{j}\left(  1\right)  }^{n}$, $s_{\mathfrak{j}\left(  2\right)
}^{n}$, $\ldots$, $s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert
\right)  }^{n}$, $0$, $0$, $0$, $\ldots$ for the variables $x_{1}$, $x_{2}$,
$x_{3}$, $\ldots$ in $e_{s}$ (by the definition of $e_{s}\left(  \left(
s_{\mathbf{i}}^{n}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $). Thus,
\begin{align*}
&  e_{s}\left(  \left(  \mathbf{s}_{\mathbf{i}}^{n}\right)  _{\mathbf{i}%
\in\mathbf{I}}\right) \\
&  =\left(  \text{the result of substituting }\underbrace{s_{\mathfrak{j}%
\left(  1\right)  }^{n}\text{, }s_{\mathfrak{j}\left(  2\right)  }^{n}\text{,
}\ldots\text{, }s_{\mathfrak{j}\left(  \left\vert \mathbf{I}\right\vert
\right)  }^{n}\text{, }0\text{, }0\text{, }0\text{, }\ldots}%
_{\substack{=\left(  t_{1}^{n},t_{2}^{n},t_{3}^{n},\ldots\right)  \\\text{(by
(\ref{sol.QSym.Malphas.ps.c.general.pf.5}))}}}\text{ for the variables }%
x_{1}\text{, }x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }e_{s}\right) \\
&  =\left(  \text{the result of substituting }t_{1}^{n}\text{, }t_{2}%
^{n}\text{, }t_{3}^{n}\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\underbrace{e_{s}}_{=\sum
_{i_{1}<i_{2}<\cdots<i_{s}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{s}}}\right) \\
&  =\left(  \text{the result of substituting }t_{1}^{n}\text{, }t_{2}%
^{n}\text{, }t_{3}^{n}\text{, }\ldots\text{ for the variables }x_{1}\text{,
}x_{2}\text{, }x_{3}\text{, }\ldots\text{ in }\sum_{i_{1}<i_{2}<\cdots<i_{s}%
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{s}}\right) \\
&  =\sum_{i_{1}<i_{2}<\cdots<i_{s}}t_{i_{1}}^{n}t_{i_{2}}^{n}\cdots t_{i_{s}%
}^{n}.
\end{align*}
Compared with (\ref{sol.QSym.Malphas.ps.c.general.pf.9}), this yields
$e_{s}^{\left\langle n\right\rangle }\left(  \left(  \mathbf{s}_{\mathbf{i}%
}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  =e_{s}\left(  \left(
\mathbf{s}_{\mathbf{i}}^{n}\right)  _{\mathbf{i}\in\mathbf{I}}\right)  $.
Thus, (\ref{sol.QSym.Malphas.ps.c.general}) is proven in Case 2.
\par
We have thus proven (\ref{sol.QSym.Malphas.ps.c.general}) in each of the Cases
1 and 2. Since these two Cases are the only cases that can occur, we thus
conclude that (\ref{sol.QSym.Malphas.ps.c.general}) holds, qed.} Applying this
to $R= \kk \left[  \left[  \xx\right]  \right]  $, $\mathbf{I}%
=\operatorname*{SIS}\left(  \ell\right)  $ and $\mathbf{s}_{\mathbf{i}%
}=\xx_{\mathbf{i}}^{\alpha}$, we obtain%
\[
e_{s}^{\left\langle n\right\rangle }\left(  \left(  \xx_{\mathbf{i}%
}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\right)  =e_{s}\left(  \left(  \left(  \xx_{\mathbf{i}}^{\alpha
}\right)  ^{n}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\right)  .
\]
Compared with (\ref{sol.QSym.Malphas.ps.c.start}), this yields $M_{\alpha
\left\{  n\right\}  }^{\left\langle s\right\rangle }=e_{s}^{\left\langle
n\right\rangle }\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)  $. This
solves Exercise \ref{exe.QSym.Malphas.ps}(c).

(d) The first sentence of Proposition \ref{symm-antipode-and-h-basis} yields
that the family $\left(  e_{1},e_{2},e_{3},\ldots\right)  $ generates the
$\kk$-algebra $\Lambda$. Thus, every element of $\Lambda$ can be
written as a polynomial in the elements $e_{1},e_{2},e_{3},\ldots$. Applying
this to the element $e_{s}^{\left\langle n\right\rangle }$ of $\Lambda$, we
conclude that $e_{s}^{\left\langle n\right\rangle }$ can be written as a
polynomial in the elements $e_{1},e_{2},e_{3},\ldots$. In other words, there
exists a polynomial $Q\in \kk \left[  z_{1},z_{2},z_{3},\ldots\right]  $
such that $e_{s}^{\left\langle n\right\rangle }=Q\left(  e_{1},e_{2}%
,e_{3},\ldots\right)  $. Consider this polynomial $Q$.

Let us now define a map $\Phi:\Lambda\rightarrow \kk \left[  \left[
\xx\right]  \right]  $ by
\[
\Phi\left(  f\right)  =f\left(  \left(  \xx_{\mathbf{i}}^{\alpha
}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in\Lambda
\]
(where $f\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)  $ is defined
as in Definition \ref{def.QSym.Malphas}(b)). This map $\Phi$ is a $\kk %
$-algebra homomorphism (since it amounts to a substitution of certain elements
for the variables in a power series). Therefore, it commutes with polynomials,
i.e., it satisfies%
\[
\Phi\left(  R\left(  f_{1},f_{2},f_{3},\ldots\right)  \right)  =R\left(
\Phi\left(  f_{1}\right)  ,\Phi\left(  f_{2}\right)  ,\Phi\left(
f_{3}\right)  ,\ldots\right)
\]
for every $f_{1},f_{2},f_{3},\ldots\in\Lambda$ and every polynomial
$R\in \kk \left[  z_{1},z_{2},z_{3},\ldots\right]  $. Applying this to
$f_{i}=e_{i}$ and $R=Q$, we obtain%
\begin{equation}
\Phi\left(  Q\left(  e_{1},e_{2},e_{3},\ldots\right)  \right)  =Q\left(
\Phi\left(  e_{1}\right)  ,\Phi\left(  e_{2}\right)  ,\Phi\left(
e_{3}\right)  ,\ldots\right)  . \label{sol.QSym.Malphas.ps.d.1}%
\end{equation}


However, for every $j\in\left\{  1,2,3,\ldots\right\}  $, we have
\begin{align*}
\Phi\left(  e_{j}\right)   &  =e_{j}\left(  \left(  \xx_{\mathbf{i}%
}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\left(
e_{j}\right)  \right) \\
&  =M_{\alpha}^{\left\langle j\right\rangle }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }M_{\alpha}^{\left\langle j\right\rangle }=e_{j}\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  }\right)  \text{ (by the definition of }M_{\alpha
}^{\left\langle j\right\rangle }\text{)}\right)  .
\end{align*}
Thus, $\left(  \Phi\left(  e_{1}\right)  ,\Phi\left(  e_{2}\right)
,\Phi\left(  e_{3}\right)  ,\ldots\right)  =\left(  M_{\alpha}^{\left\langle
1\right\rangle },M_{\alpha}^{\left\langle 2\right\rangle },M_{\alpha
}^{\left\langle 3\right\rangle },\ldots\right)  $, so that%
\[
Q\left(  \Phi\left(  e_{1}\right)  ,\Phi\left(  e_{2}\right)  ,\Phi\left(
e_{3}\right)  ,\ldots\right)  =Q\left(  M_{\alpha}^{\left\langle
1\right\rangle },M_{\alpha}^{\left\langle 2\right\rangle },M_{\alpha
}^{\left\langle 3\right\rangle },\ldots\right)  .
\]
Hence, (\ref{sol.QSym.Malphas.ps.d.1}) becomes%
\[
\Phi\left(  Q\left(  e_{1},e_{2},e_{3},\ldots\right)  \right)  =Q\left(
\Phi\left(  e_{1}\right)  ,\Phi\left(  e_{2}\right)  ,\Phi\left(
e_{3}\right)  ,\ldots\right)  =Q\left(  M_{\alpha}^{\left\langle
1\right\rangle },M_{\alpha}^{\left\langle 2\right\rangle },M_{\alpha
}^{\left\langle 3\right\rangle },\ldots\right)  .
\]
Compared with%
\begin{align*}
\Phi\left(  \underbrace{Q\left(  e_{1},e_{2},e_{3},\ldots\right)
}_{\substack{=e_{s}^{\left\langle n\right\rangle }\\\text{(since }%
e_{s}^{\left\langle n\right\rangle }=Q\left(  e_{1},e_{2},e_{3},\ldots\right)
\text{)}}}\right)   &  =\Phi\left(  e_{s}^{\left\langle n\right\rangle
}\right)  =e_{s}^{\left\langle n\right\rangle }\left(  \left(  \xx%
_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\Phi\left(  e_{s}^{\left\langle n\right\rangle }\right)  \right) \\
&  =M_{\alpha\left\{  n\right\}  }^{\left\langle s\right\rangle }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Exercise \ref{exe.QSym.Malphas.ps}%
(c)}\right)  ,
\end{align*}
this yields $M_{\alpha\left\{  n\right\}  }^{\left\langle s\right\rangle
}=Q\left(  M_{\alpha}^{\left\langle 1\right\rangle },M_{\alpha}^{\left\langle
2\right\rangle },M_{\alpha}^{\left\langle 3\right\rangle },\ldots\right)  $.
Thus, there exists a polynomial $P\in \kk \left[  z_{1},z_{2}%
,z_{3},\ldots\right]  $ such that $M_{\alpha\left\{  n\right\}  }%
^{\left\langle s\right\rangle }=P\left(  M_{\alpha}^{\left\langle
1\right\rangle },M_{\alpha}^{\left\langle 2\right\rangle },M_{\alpha
}^{\left\langle 3\right\rangle },\ldots\right)  $ (namely, $P=Q$). This solves
Exercise \ref{exe.QSym.Malphas.ps}(d).
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.QSym.Malphas.M1s}}

\textit{Solution to Exercise \ref{exe.QSym.Malphas.M1s}.} The
definition of $M_{\left(  1\right)  }^{\left\langle s\right\rangle }$ yields
$M_{\left(  1\right)  }^{\left\langle s\right\rangle }=e_{s}\left(  \left(
\xx_{\mathbf{i}}^{\left(  1\right)  }\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  1\right)  }\right)  $.

But $\operatorname*{SIS}\left(  1\right)  $ is defined as the set of all
strictly increasing $1$-tuples of positive integers. Clearly, such $1$-tuples
are in bijection with positive integers; the bijection sends a positive
integer $i$ to the strictly increasing $1$-tuple $\left(  i\right)  $. This
bijection shows that the family $\left(  \xx_{\mathbf{i}}^{\left(
1\right)  }\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  1\right)  }$ is
a reparametrization of the family $\left(  \xx_{\left(  i\right)
}^{\left(  1\right)  }\right)  _{i\in\left\{  1,2,3,\ldots\right\}  }$. Thus,
$e_{s}\left(  \left(  \xx_{\mathbf{i}}^{\left(  1\right)  }\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  1\right)  }\right)  =e_{s}\left(
\left(  \xx_{\left(  i\right)  }^{\left(  1\right)  }\right)
_{i\in\left\{  1,2,3,\ldots\right\}  }\right)  $. Thus,%
\begin{align*}
M_{\left(  1\right)  }^{\left\langle s\right\rangle }  &  =e_{s}\left(
\left(  \xx_{\mathbf{i}}^{\left(  1\right)  }\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  1\right)  }\right)  =e_{s}\left(  \left(
\underbrace{\xx_{\left(  i\right)  }^{\left(  1\right)  }%
}_{\substack{=x_{i}^{1}\\\text{(by the definition}\\\text{of }\xx%
_{\left(  i\right)  }^{\left(  1\right)  }\text{)}}}\right)  _{i\in\left\{
1,2,3,\ldots\right\}  }\right)  =e_{s}\left(  \left(  \underbrace{x_{i}^{1}%
}_{=x_{i}}\right)  _{i\in\left\{  1,2,3,\ldots\right\}  }\right) \\
&  =e_{s}\left(  \left(  x_{i}\right)  _{i\in\left\{  1,2,3,\ldots\right\}
}\right)  =e_{s}.
\end{align*}
This solves Exercise \ref{exe.QSym.Malphas.M1s}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.prop.QSym.Malphas.dets}}

\textit{Solution to Exercise \ref{exe.prop.QSym.Malphas.dets}.}

\begin{proof}
[Proof of Proposition \ref{prop.QSym.Malphas.dets}.]We first recall a general
fact from algebra. Namely, if $\mathfrak{C}$ and $\mathfrak{D}$ are two
commutative rings, if $\varphi:\mathfrak{C}\rightarrow\mathfrak{D}$ is a ring
homomorphism, and if $u$ and $v$ are two nonnegative integers, then we can
define a homomorphism $\varphi^{u\times v}:\mathfrak{C}^{u\times v}%
\rightarrow\mathfrak{D}^{u\times v}$ of additive groups by sending every
matrix $\left(  c_{i,j}\right)  _{i=1,2,\ldots,u;\ j=1,2,\ldots,v}%
\in\mathfrak{C}^{u\times v}$ to the matrix $\left(  \varphi\left(
c_{i,j}\right)  \right)  _{i=1,2,\ldots,u;\ j=1,2,\ldots,v}\in\mathfrak{D}%
^{u\times v}$. This homomorphism $\varphi^{u\times v}$ is the map from
$\mathfrak{C}^{u\times v}$ to $\mathfrak{D}^{u\times v}$ canonically induced
by $\varphi$, and it has many structure-preserving properties (for instance,
it respects the multiplication of matrices, in the sense that we have
$\varphi^{u\times v}\left(  X\right)  \cdot\varphi^{v\times w}\left(
Y\right)  =\varphi^{u\times w}\left(  XY\right)  $ whenever $X\in
\mathfrak{C}^{u\times v}$ and $Y\in\mathfrak{C}^{v\times w}$). We furthermore
have%
\begin{equation}
\det\left(  \varphi^{u\times u}\left(  X\right)  \right)  =\varphi\left(  \det
X\right)  \label{pf.prop.QSym.Malphas.dets.det-phi}%
\end{equation}
for any two commutative rings $\mathfrak{C}$ and $\mathfrak{D}$, any ring
homomorphism $\varphi:\mathfrak{C}\rightarrow\mathfrak{D}$, any nonnegative
integer $u$ and any matrix $X\in\mathfrak{C}^{u\times u}$. (This is because
the determinant of a matrix is a polynomial in its entries, and polynomials
commute with ring homomorphisms.)

We shall now construct a particular $\kk$-algebra homomorphism
$\Lambda\rightarrow \kk \left[  \left[  \xx\right]  \right]  $
which will help us in our proof. Namely, we define a map $\Phi:\Lambda
\rightarrow \kk \left[  \left[  \xx\right]  \right]  $ by%
\[
\Phi\left(  f\right)  =f\left(  \left(  \xx_{\mathbf{i}}^{\alpha
}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }f\in\Lambda
\]
(where $f\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)
_{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  }\right)  $ is defined
as in Definition \ref{def.QSym.Malphas}(b)). This map $\Phi$ is a $\kk %
$-algebra homomorphism (since it amounts to a substitution of certain elements
for the variables in a power series). Every $s\in \NN$ satisfies%
\begin{align}
\Phi\left(  e_{s}\right)   &  =e_{s}\left(  \left(  \xx_{\mathbf{i}%
}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\left(
e_{s}\right)  \right) \nonumber\\
&  =M_{\alpha}^{\left\langle s\right\rangle }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }M_{\alpha}^{\left\langle s\right\rangle }\text{ was defined as
}e_{s}\left(  \left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  }\right)  \right) .
\label{pf.prop.QSym.Malphas.dets.Phi.es}%
\end{align}
Every positive integer $s$ satisfies%
\begin{align}
\Phi\left(  p_{s}\right)   &  =p_{s}\left(  \left(  \xx_{\mathbf{i}%
}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\Phi\left(
p_{s}\right)  \right) \nonumber\\
&  =M_{\alpha\left\{  s\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Exercise \ref{exe.QSym.Malphas.ps}(a)}\right)  .
\label{pf.prop.QSym.Malphas.dets.Phi.ps}%
\end{align}


(a) Define a matrix $A_{n}=\left(  a_{i,j}\right)  _{i,j=1,2,\ldots,n}$ as in
Exercise \ref{exe.pe-ep-determinants}(a). Then, Exercise
\ref{exe.pe-ep-determinants}(a) yields $\det\left(  A_{n}\right)  =n!e_{n}$.
Applying the map $\Phi$ to both sides of this equality, we obtain%
\[
\Phi\left(  \det\left(  A_{n}\right)  \right)  =\Phi\left(  n!e_{n}\right)
=n!\underbrace{\Phi\left(  e_{n}\right)  }_{\substack{=M_{\alpha
}^{\left\langle n\right\rangle }\\\text{(by
(\ref{pf.prop.QSym.Malphas.dets.Phi.es}), applied}\\\text{to }s=n\text{)}%
}}=n!M_{\alpha}^{\left\langle n\right\rangle }.
\]
But (\ref{pf.prop.QSym.Malphas.dets.det-phi}) (applied to $\mathfrak{C}%
=\Lambda$, $\mathfrak{D}= \kk \left[  \left[  \xx\right]  \right]
$, $\varphi=\Phi$, $u=n$ and $X=A_{n}$) yields%
\begin{equation}
\det\left(  \Phi^{n\times n}\left(  A_{n}\right)  \right)  =\Phi\left(
\det\left(  A_{n}\right)  \right)  =n!M_{\alpha}^{\left\langle n\right\rangle
}. \label{pf.prop.QSym.Malphas.dets.Phi.4}%
\end{equation}
But
\begin{equation}
\Phi^{n\times n}\left(  \underbrace{A_{n}}_{=\left(  a_{i,j}\right)
_{i,j=1,2,\ldots,n}}\right)  =\Phi^{n\times n}\left(  \left(  a_{i,j}\right)
_{i,j=1,2,\ldots,n}\right)  =\left(  \Phi\left(  a_{i,j}\right)  \right)
_{i,j=1,2,\ldots,n} \label{pf.prop.QSym.Malphas.dets.Phi.5}%
\end{equation}
(by the definition of $\Phi^{n\times n}\left(  \left(  a_{i,j}\right)
_{i,j=1,2,\ldots,n}\right)  $). But every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ satisfies%
\begin{align*}
\Phi\left(  a_{i,j}\right)   &  =\Phi\left(
\begin{cases}
p_{i-j+1}, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }a_{i,j}=%
\begin{cases}
p_{i-j+1}, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\right) \\
&  =%
\begin{cases}
\Phi\left(  p_{i-j+1}\right)  , & \text{if }i\geq j;\\
\Phi\left(  i\right)  , & \text{if }i=j-1;\\
\Phi\left(  0\right)  , & \text{if }i<j-1
\end{cases}  \quad
=%
\begin{cases}
\Phi\left(  p_{i-j+1}\right)  , & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\Phi\left(  i\right)  =i\text{ in the case when }i=j-1\text{
(because }\Phi\text{ is a}\\
 \kk \text{-algebra homomorphism), and because }\Phi\left(  0\right)
=0\text{ in the}\\
\text{case when }i<j-1\text{ (for the same reason)}%
\end{array}
\right) \\
&  =%
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\Phi\left(  p_{i-j+1}\right)  =M_{\alpha\left\{  i-j+1\right\}
}\text{ in the case when }i\geq j\\
\text{(by (\ref{pf.prop.QSym.Malphas.dets.Phi.ps}), applied to }%
s=i-j+1\text{)}%
\end{array}
\right) \\
&  =a_{i,j}^{\left\langle \alpha\right\rangle }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }a_{i,j}^{\left\langle \alpha\right\rangle }=%
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\right)  .
\end{align*}
Hence, $\left(  \underbrace{\Phi\left(  a_{i,j}\right)  }_{=a_{i,j}%
^{\left\langle \alpha\right\rangle }}\right)  _{i,j=1,2,\ldots,n}=\left(
a_{i,j}^{\left\langle \alpha\right\rangle }\right)  _{i,j=1,2,\ldots,n}%
=A_{n}^{\left\langle \alpha\right\rangle }$. Thus,
(\ref{pf.prop.QSym.Malphas.dets.Phi.5}) becomes
\[
\Phi^{n\times n}\left(
A_{n}\right)  =\left(  \Phi\left(  a_{i,j}\right)  \right)  _{i,j=1,2,\ldots
,n}=A_{n}^{\left\langle \alpha\right\rangle } .
\]
Thus,
(\ref{pf.prop.QSym.Malphas.dets.Phi.4}) rewrites as $\det\left(
A_{n}^{\left\langle \alpha\right\rangle }\right)  =n!M_{\alpha}^{\left\langle
n\right\rangle }$. This proves Proposition \ref{prop.QSym.Malphas.dets}(a).

(b) The proof of Proposition \ref{prop.QSym.Malphas.dets}(b) proceeds
similarly to our proof of Proposition \ref{prop.QSym.Malphas.dets}(a) above,
as long as the obvious changes are made (one needs to consider $B_{n}$,
$b_{i,j}$, $B_{n}^{\left\langle \alpha\right\rangle }$ and $b_{i,j}%
^{\left\langle \alpha\right\rangle }$ instead of $A_{n}$, $a_{i,j}$,
$A_{n}^{\left\langle \alpha\right\rangle }$ and $a_{i,j}^{\left\langle
\alpha\right\rangle }$).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.Malphas.QSym}}

\textit{Solution to Exercise \ref{exe.cor.QSym.Malphas.QSym}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.Malphas.QSym}.](a) We WLOG assume that
$s\in \NN$ (because otherwise, $s$ is negative and thus satisfies
$M_{\alpha}^{\left\langle s\right\rangle }=0\in\Qsym$).

Define a matrix $A_{s}^{\left\langle \alpha\right\rangle }=\left(
a_{i,j}^{\left\langle \alpha\right\rangle }\right)  _{i,j=1,2,\ldots,s}$ as in
Proposition \ref{prop.QSym.Malphas.dets}(a) (but for $s$ instead of $n$).
Then, $A_{s}^{\left\langle \alpha\right\rangle }\in
\Qsym^{s\times s}$ (since every $\left(  i,j\right)  \in\left\{
1,2,\ldots,s\right\}  ^{2}$ satisfies $a_{i,j}^{\left\langle \alpha
\right\rangle }\in\Qsym$ (as follows from the definition of
$a_{i,j}^{\left\langle \alpha\right\rangle }$)). Hence, $\det\left(
A_{s}^{\left\langle \alpha\right\rangle }\right)  \in\Qsym$.
But Proposition \ref{prop.QSym.Malphas.dets}(a) (applied to $n=s$) yields
$\det\left(  A_{s}^{\left\langle \alpha\right\rangle }\right)  =s!M_{\alpha
}^{\left\langle s\right\rangle }$. Hence, $s!M_{\alpha}^{\left\langle
s\right\rangle }=\det\left(  A_{s}^{\left\langle \alpha\right\rangle }\right)
\in\Qsym$.

Now, let us recall that there is a canonical ring homomorphism $\varphi
: \ZZ \rightarrow \kk$. This homomorphism gives rise to a ring
homomorphism $\varphi\left[  \left[  \xx\right]  \right]
: \ZZ \left[  \left[  \xx\right]  \right]  \rightarrow
 \kk \left[  \left[  \xx\right]  \right]  $, and this latter
homomorphism $\varphi\left[  \left[  \xx\right]  \right]  $ sends
$\Qsym_{ \ZZ }$ to $\Qsym_{ \kk }$; that is, we have
$\left(  \varphi\left[  \left[  \xx\right]  \right]  \right)
\left(  \Qsym_{ \ZZ }\right)  \subset \Qsym_{ \kk }$.
Moreover, it is clear that the ring homomorphism
$\varphi\left[  \left[  \xx\right]  \right]  $ sends the element
$M_{\alpha}^{\left\langle s\right\rangle }$ of $ \ZZ \left[  \left[
\xx\right]  \right]  $ to the element $M_{\alpha}^{\left\langle
s\right\rangle }$ of $\kk \left[  \left[  \xx\right]  \right]  $
(because the definition of $M_{\alpha}^{\left\langle s\right\rangle }$ is
functorial in the base ring $\kk$). Therefore, if we can prove that the
element $M_{\alpha}^{\left\langle s\right\rangle }$ of $ \ZZ \left[
\left[  \xx\right]  \right]  $ belongs to $\Qsym_{ \ZZ }$,
then it will automatically follow that
the element $M_{\alpha}^{\left\langle s\right\rangle }$ of
$\kk \left[  \left[  \xx\right]  \right]  $
belongs to $\left(  \varphi\left[  \left[  \xx\right]  \right]
\right)  \left(  \Qsym_{ \ZZ }\right)
\subset\Qsym_{ \kk }$; this will complete the
proof of Corollary \ref{cor.QSym.Malphas.QSym}(a). Hence, in order to prove
Corollary \ref{cor.QSym.Malphas.QSym}(a), it only remains to prove that the
element $M_{\alpha}^{\left\langle s\right\rangle }$ of $ \ZZ \left[
\left[  \xx\right]  \right]  $ belongs to $\Qsym_{ \ZZ }$.
In other words, it only remains to prove Corollary
\ref{cor.QSym.Malphas.QSym}(a) in the case of $\kk = \ZZ$. Hence,
in proving Corollary \ref{cor.QSym.Malphas.QSym}(a), we can WLOG assume that
$\kk = \ZZ$. Assume this. Since $\kk = \ZZ$, we have
$\Qsym=\Qsym_{ \ZZ }$.

If a positive integer $N$ and an element $f$ of $ \ZZ \left[  \left[
\xx\right]  \right]  $ satisfy $Nf\in
\Qsym_{ \ZZ }$, then $f$ also lies in $
\Qsym_{ \ZZ }$ (because $N$ is not a zero-divisor in $\kk %
= \ZZ$, and therefore $f$ is obtained from the power series $Nf$ by
dividing all coefficients by $N$). Applying this to $N=s!$ and $f=M_{\alpha
}^{\left\langle s\right\rangle }$, we obtain $M_{\alpha}^{\left\langle
s\right\rangle }\in\Qsym_{ \ZZ }$ (since
$s!M_{\alpha}^{\left\langle s\right\rangle }\in\Qsym%
=\Qsym_{ \ZZ }$). In other words, $M_{\alpha
}^{\left\langle s\right\rangle }\in\Qsym$ (since $\kk %
= \ZZ$). This completes the proof of Corollary
\ref{cor.QSym.Malphas.QSym}(a).

(b) Recall that $M_{\alpha}^{\left\langle s\right\rangle }=e_{s}\left(
\left(  \xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  }\right)  $. Thus, $M_{\alpha
}^{\left\langle s\right\rangle }$ is a homogeneous power series of degree
$s\left\vert \alpha\right\vert $ (since each $\xx_{\mathbf{i}}^{\alpha
}$ is a monomial of degree $\left\vert \alpha\right\vert $, and since $e_{s}$
is a power series of degree $s$). Combined with $M_{\alpha}^{\left\langle
s\right\rangle }\in\Qsym$ (which follows from Corollary
\ref{cor.QSym.Malphas.QSym}(a)), this yields $M_{\alpha}^{\left\langle
s\right\rangle }\in\Qsym_{s\left\vert \alpha
\right\vert }$. This proves Corollary \ref{cor.QSym.Malphas.QSym}(b).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.rmk.QSym.lyndon.red}}

\textit{Solution to Exercise \ref{exe.rmk.QSym.lyndon.red}.}

\begin{proof}
[Proof of Remark \ref{rmk.QSym.lyndon.red}.](a) Write the composition $\alpha$
in the form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $.
Then, \newline
$\operatorname*{red}\alpha=\left(  \dfrac{\alpha_{1}}{\gcd\alpha}%
,\dfrac{\alpha_{2}}{\gcd\alpha},\ldots,\dfrac{\alpha_{\ell}}{\gcd\alpha
}\right)  $ (by the definition of $\operatorname*{red}\alpha$). Hence, the
definition of $\left(  \operatorname*{red}\alpha\right)  \left\{  \gcd
\alpha\right\}  $ yields%
\[
\left(  \operatorname*{red}\alpha\right)  \left\{  \gcd\alpha\right\}
=\left(  \left(  \gcd\alpha\right)  \cdot\dfrac{\alpha_{1}}{\gcd\alpha
},\left(  \gcd\alpha\right)  \cdot\dfrac{\alpha_{2}}{\gcd\alpha}%
,\ldots,\left(  \gcd\alpha\right)  \cdot\dfrac{\alpha_{\ell}}{\gcd\alpha
}\right)  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
=\alpha,
\]
so that Remark \ref{rmk.QSym.lyndon.red}(a) is proven.

(c) Write the composition $\alpha$ in the form $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  $. Then, $\operatorname*{red}\alpha=\left(
\dfrac{\alpha_{1}}{\gcd\alpha},\dfrac{\alpha_{2}}{\gcd\alpha},\ldots
,\dfrac{\alpha_{\ell}}{\gcd\alpha}\right)  $ (by the definition of
$\operatorname*{red}\alpha$). Hence, the definition of $\gcd\left(
\operatorname*{red}\alpha\right)  $ yields%
\begin{align*}
\gcd\left(  \operatorname*{red}\alpha\right)   &  =\gcd\left(  \dfrac
{\alpha_{1}}{\gcd\alpha},\dfrac{\alpha_{2}}{\gcd\alpha},\ldots,\dfrac
{\alpha_{\ell}}{\gcd\alpha}\right)  =\dfrac{\gcd\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  }{\gcd\alpha}=\dfrac{\gcd\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  }{\gcd\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gcd\alpha=\gcd\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  \text{ (by the definition of
}\gcd\alpha\text{)}\right) \\
&  =1.
\end{align*}
In other words, the composition $\operatorname*{red}\alpha$ is reduced. This
proves Remark \ref{rmk.QSym.lyndon.red}(c).

(d) Assume that $\alpha$ is reduced. Write the composition $\alpha$ in the
form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. Then,
$\operatorname*{red}\alpha=\left(  \dfrac{\alpha_{1}}{\gcd\alpha}%
,\dfrac{\alpha_{2}}{\gcd\alpha},\ldots,\dfrac{\alpha_{\ell}}{\gcd\alpha
}\right)  $ (by the definition of $\operatorname*{red}\alpha$). But
$\gcd\alpha=1$ (since $\alpha$ is reduced). Now,%
\begin{align*}
\operatorname*{red}\alpha &  =\left(  \dfrac{\alpha_{1}}{\gcd\alpha}%
,\dfrac{\alpha_{2}}{\gcd\alpha},\ldots,\dfrac{\alpha_{\ell}}{\gcd\alpha
}\right)  =\left(  \dfrac{\alpha_{1}}{1},\dfrac{\alpha_{2}}{1},\ldots
,\dfrac{\alpha_{\ell}}{1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\gcd\alpha=1\right) \\
&  =\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  =\alpha.
\end{align*}
This proves Remark \ref{rmk.QSym.lyndon.red}(d).

(e) Let $s\in\left\{  1,2,3,\ldots\right\}  $. Write the composition $\alpha$
in the form $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $.
Then, \newline
$\operatorname*{red}\alpha=\left(  \dfrac{\alpha_{1}}{\gcd\alpha}%
,\dfrac{\alpha_{2}}{\gcd\alpha},\ldots,\dfrac{\alpha_{\ell}}{\gcd\alpha
}\right)  $ (by the definition of $\operatorname*{red}\alpha$). Also,
$\alpha\left\{  s\right\}  =\left(  s\alpha_{1},s\alpha_{2},\ldots
,s\alpha_{\ell}\right)  $ (by the definition of $\alpha\left\{  s\right\}  $).
Now, $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $, so
that $\ell\left(  \alpha\right)  =\ell$. Hence, $\ell=\ell\left(
\alpha\right)  >0$ (since $\alpha$ is nonempty). Since $\alpha\left\{
s\right\}  =\left(  s\alpha_{1},s\alpha_{2},\ldots,s\alpha_{\ell}\right)  $,
we have $\ell\left(  \alpha\left\{  s\right\}  \right)  =\ell>0$, so that the
composition $\alpha\left\{  s\right\}  $ is nonempty.

By the definition of $\gcd\alpha$, we have $\gcd\alpha=\gcd\left(  \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $. By the definition of
$\gcd\left(  \alpha\left\{  s\right\}  \right)  $, we have
\begin{align*}
\gcd\left(  \alpha\left\{  s\right\}  \right)   &  =\gcd\left(  s\alpha
_{1},s\alpha_{2},\ldots,s\alpha_{\ell}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\alpha\left\{  s\right\}  =\left(  s\alpha_{1},s\alpha_{2}%
,\ldots,s\alpha_{\ell}\right)  \right) \\
&  =s\underbrace{\gcd\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
}\right)  }_{=\gcd\alpha}=s\gcd\alpha.
\end{align*}
Now, recall that $\alpha\left\{  s\right\}  =\left(  s\alpha_{1},s\alpha
_{2},\ldots,s\alpha_{\ell}\right)  $. Hence, the definition of
$\operatorname*{red}\left(  \alpha\left\{  s\right\}  \right)  $ yields%
\begin{align*}
\operatorname*{red}\left(  \alpha\left\{  s\right\}  \right)   &  =\left(
\dfrac{s\alpha_{1}}{\gcd\left(  \alpha\left\{  s\right\}  \right)  }%
,\dfrac{s\alpha_{2}}{\gcd\left(  \alpha\left\{  s\right\}  \right)  }%
,\ldots,\dfrac{s\alpha_{\ell}}{\gcd\left(  \alpha\left\{  s\right\}  \right)
}\right) \\
&  =\left(  \dfrac{s\alpha_{1}}{s\gcd\alpha},\dfrac{s\alpha_{2}}{s\gcd\alpha
},\ldots,\dfrac{s\alpha_{\ell}}{s\gcd\alpha}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gcd\left(  \alpha\left\{  s\right\}
\right)  =s\gcd\alpha\right) \\
&  =\left(  \dfrac{\alpha_{1}}{\gcd\alpha},\dfrac{\alpha_{2}}{\gcd\alpha
},\ldots,\dfrac{\alpha_{\ell}}{\gcd\alpha}\right)  =\operatorname*{red}\alpha.
\end{align*}
This proves Remark \ref{rmk.QSym.lyndon.red}(e).

(f) Write the composition $\alpha$ in the form $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  $. Then, $\operatorname*{red}\alpha=\left(
\dfrac{\alpha_{1}}{\gcd\alpha},\dfrac{\alpha_{2}}{\gcd\alpha},\ldots
,\dfrac{\alpha_{\ell}}{\gcd\alpha}\right)  $ (by the definition of
$\operatorname*{red}\alpha$). Thus,%
\[
\left\vert \operatorname*{red}\alpha\right\vert =\dfrac{\alpha_{1}}{\gcd
\alpha}+\dfrac{\alpha_{2}}{\gcd\alpha}+ \cdots+\dfrac{\alpha_{\ell}}%
{\gcd\alpha}=\dfrac{\alpha_{1}+\alpha_{2}+ \cdots+\alpha_{\ell}}{\gcd\alpha},
\]
so that $\left(  \gcd\alpha\right)  \left\vert \operatorname*{red}%
\alpha\right\vert =\alpha_{1}+\alpha_{2}+ \cdots+\alpha_{\ell}=\left\vert
\alpha\right\vert $ (since $\left\vert \alpha\right\vert =\alpha_{1}%
+\alpha_{2}+ \cdots+\alpha_{\ell}$ (because $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $)). This proves Remark
\ref{rmk.QSym.lyndon.red}(f).

(b) We have $\left(  \operatorname*{red}\alpha\right)  \left\{  \gcd
\alpha\right\}  =\alpha$ (by Remark \ref{rmk.QSym.lyndon.red}(a)).

The property of a composition to be Lyndon does not change if all entries of
the composition are multiplied by a fixed positive integer $m$ (because this
property only depends on the relative order of the parts of the composition).
In other words, if $\beta$ is a composition and $m$ is a positive integer,
then $\beta$ is Lyndon if and only if $\beta\left\{  m\right\}  $ is Lyndon.
Applying this to $\beta=\operatorname*{red}\alpha$ and $m=\gcd\alpha$, we
conclude that $\operatorname*{red}\alpha$ is Lyndon if and only if $\left(
\operatorname*{red}\alpha\right)  \left\{  \gcd\alpha\right\}  $ is Lyndon. In
other words, $\operatorname*{red}\alpha$ is Lyndon if and only if $\alpha$ is
Lyndon (because $\left(  \operatorname*{red}\alpha\right)  \left\{  \gcd
\alpha\right\}  =\alpha$). This proves Remark \ref{rmk.QSym.lyndon.red}(b).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.QSym.lyndon.Z-genset.reindex}}

\textit{Solution to Exercise \ref{exe.lem.QSym.lyndon.Z-genset.reindex}%
.}

\begin{proof}
[Proof of Lemma \ref{lem.QSym.lyndon.Z-genset.reindex}.]For every $\alpha
\in\mathfrak{L}$, the pair $\left(  \operatorname*{red}\alpha,\gcd
\alpha\right)  $ is a well-defined element of $\mathfrak{RL}\times\left\{
1,2,3,\ldots\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $\alpha
\in\mathfrak{L}$. Then, $\alpha$ is a Lyndon word (since $\mathfrak{L}$ is the
set of all Lyndon words), and thus nonempty. Hence, $\alpha$ is a nonempty
composition, so that $\operatorname*{red}\alpha$ is a well-defined
composition, and $\gcd\alpha$ is a well-defined positive integer. Remark
\ref{rmk.QSym.lyndon.red}(b) yields that the composition $\alpha$ is Lyndon if
and only if the composition $\operatorname*{red}\alpha$ is Lyndon. Since
$\alpha$ is Lyndon, this yields that $\operatorname*{red}\alpha$ is Lyndon.
Also, $\operatorname*{red}\alpha$ is reduced (by Remark
\ref{rmk.QSym.lyndon.red}(c)). Thus, $\operatorname*{red}\alpha$ is a reduced
Lyndon composition. In other words, $\operatorname*{red}\alpha\in
\mathfrak{RL}$ (since $\mathfrak{RL}$ is the set of all reduced Lyndon
compositions). Combined with $\gcd\alpha\in\left\{  1,2,3,\ldots\right\}  $
(since $\gcd\alpha$ is a well-defined positive integer), this yields $\left(
\operatorname*{red}\alpha,\gcd\alpha\right)  \in\mathfrak{RL}\times\left\{
1,2,3,\ldots\right\}  $, qed.}. Hence, we can define a map $\mathbf{R}%
:\mathfrak{L}\rightarrow\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  $ by%
\[
\left(  \mathbf{R}\left(  \alpha\right)  =\left(  \operatorname*{red}%
\alpha,\gcd\alpha\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\alpha
\in\mathfrak{L}\right)  .
\]
Consider this $\mathbf{R}$.

For every $\left(  w,s\right)  \in\mathfrak{RL}\times\left\{  1,2,3,\ldots
\right\}  $, the element $w\left\{  s\right\}  $ is a well-defined element of
$\mathfrak{L}$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  $. Then, $w\in
\mathfrak{RL}$ and $s\in\left\{  1,2,3,\ldots\right\}  $. Since $\mathfrak{RL}%
$ is the set of all reduced Lyndon compositions, we see that $w$ is a reduced
Lyndon composition (since $w\in\mathfrak{RL}$). Thus, $w$ is nonempty (since
$w$ is Lyndon).
\par
We shall now show that $w\left\{  s\right\}  \in\mathfrak{L}$.
\par
Remark \ref{rmk.QSym.lyndon.red}(e) (applied to $\alpha=w$) yields that the
composition $w\left\{  s\right\}  $ is nonempty and satisfies
$\operatorname*{red}\left(  w\left\{  s\right\}  \right)  =\operatorname*{red}%
w$ and $\gcd\left(  w\left\{  s\right\}  \right)  =s\gcd w$. We have
$\operatorname*{red}\left(  w\left\{  s\right\}  \right)  =\operatorname*{red}%
w=w$ (by Remark \ref{rmk.QSym.lyndon.red}(d), applied to $\alpha=w$). Now,
recall that the composition $w$ is Lyndon. Hence, the composition
$\operatorname*{red}\left(  w\left\{  s\right\}  \right)  $ is Lyndon (since
$\operatorname*{red}\left(  w\left\{  s\right\}  \right)  =w$).
\par
Remark \ref{rmk.QSym.lyndon.red}(b) (applied to $\alpha=w\left\{  s\right\}
$) yields that the composition $w\left\{  s\right\}  $ is Lyndon if and only
if the composition $\operatorname*{red}\left(  w\left\{  s\right\}  \right)  $
is Lyndon. Since the composition $\operatorname*{red}\left(  w\left\{
s\right\}  \right)  $ is Lyndon, this yields that the composition $w\left\{
s\right\}  $ is Lyndon. In other words, $w\left\{  s\right\}  \in\mathfrak{L}$
(since $\mathfrak{L}$ is the set of all Lyndon words), qed.}. Hence, we can
define a map $\mathbf{M}:\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}
\rightarrow\mathfrak{L}$ by%
\[
\left(  \mathbf{M}\left(  w,s\right)  =w\left\{  s\right\}
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  w,s\right)  \in\mathfrak{RL}%
\times\left\{  1,2,3,\ldots\right\}  \right)  .
\]
Consider this $\mathbf{M}$.

It is now easy to see that $\mathbf{R}\circ\mathbf{M}=\operatorname*{id}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  w,s\right)  \in\mathfrak{RL}%
\times\left\{  1,2,3,\ldots\right\}  $. Thus, $w\in\mathfrak{RL}$ and
$s\in\left\{  1,2,3,\ldots\right\}  $. Since $\mathfrak{RL}$ is the set of all
reduced Lyndon compositions, we see that $w$ is a reduced Lyndon composition
(since $w\in\mathfrak{RL}$). Thus, $w$ is nonempty (since $w$ is Lyndon).
\par
Remark \ref{rmk.QSym.lyndon.red}(e) (applied to $\alpha=w$) yields that the
composition $w\left\{  s\right\}  $ is nonempty and satisfies
$\operatorname*{red}\left(  w\left\{  s\right\}  \right)  =\operatorname*{red}%
w$ and $\gcd\left(  w\left\{  s\right\}  \right)  =s\gcd w$. But $\gcd w=1$
(since $w$ is reduced), so that $\gcd\left(  w\left\{  s\right\}  \right)
=s\underbrace{\gcd w}_{=1}=s$. Also, $\operatorname*{red}\left(  w\left\{
s\right\}  \right)  =\operatorname*{red}w=w$ (by Remark
\ref{rmk.QSym.lyndon.red}(d), applied to $\alpha=w$). Now,%
\begin{align*}
\left(  \mathbf{R}\circ\mathbf{M}\right)  \left(  w,s\right)   &
=\mathbf{R}\left(  \underbrace{\mathbf{M}\left(  w,s\right)  }%
_{\substack{=w\left\{  s\right\}  \\\text{(by the definition of }%
\mathbf{M}\left(  w,s\right)  \text{)}}}\right)  =\mathbf{R}\left(  w\left\{
s\right\}  \right)  =\left(  \underbrace{\operatorname*{red}\left(  w\left\{
s\right\}  \right)  }_{=w},\underbrace{\gcd\left(  w\left\{  s\right\}
\right)  }_{=s}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{R}\left(
w\left\{  s\right\}  \right)  \right) \\
&  =\left(  w,s\right)  =\operatorname*{id}\left(  w,s\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\left(  w,s\right)  $. We thus have shown
that $\left(  \mathbf{R}\circ\mathbf{M}\right)  \left(  w,s\right)
=\operatorname*{id}\left(  w,s\right)  $ for every $\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  $. In other words,
$\mathbf{R}\circ\mathbf{M}=\operatorname*{id}$, qed.} and $\mathbf{M}%
\circ\mathbf{R}=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} Let
$\alpha\in\mathfrak{L}$. Then, $\alpha$ is a Lyndon word (since $\mathfrak{L}$
is the set of all Lyndon words), and thus nonempty. Hence, $\alpha$ is a
nonempty composition. Now,%
\begin{align*}
\left(  \mathbf{M}\circ\mathbf{R}\right)  \left(  \alpha\right)   &
=\mathbf{M}\left(  \underbrace{\mathbf{R}\left(  \alpha\right)  }%
_{\substack{=\left(  \operatorname*{red}\alpha,\gcd\alpha\right)  \\\text{(by
the definition of }\mathbf{R}\left(  \alpha\right)  \text{)}}}\right)
=\mathbf{M}\left(  \operatorname*{red}\alpha,\gcd\alpha\right)  =\left(
\operatorname*{red}\alpha\right)  \left\{  \gcd\alpha\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{M}\left(
\operatorname*{red}\alpha,\gcd\alpha\right)  \right) \\
&  =\alpha\ \ \ \ \ \ \ \ \ \ \left(  \text{by Remark
\ref{rmk.QSym.lyndon.red}(a)}\right) \\
&  =\operatorname*{id}\left(  \alpha\right)  .
\end{align*}
\par
Now, let us forget that we fixed $\alpha$. We thus have proven that $\left(
\mathbf{M}\circ\mathbf{R}\right)  \left(  \alpha\right)  =\operatorname*{id}%
\left(  \alpha\right)  $ for every $\alpha\in\mathfrak{L}$. Thus,
$\mathbf{M}\circ\mathbf{R}=\operatorname*{id}$, qed.}. Therefore, the maps
$\mathbf{M}$ and $\mathbf{R}$ are mutually inverse. Hence, the map
$\mathbf{M}$ is a bijection. Hence, the family $\left(  M_{\operatorname*{red}%
\left(  \mathbf{M}\left(  w,s\right)  \right)  }^{\left\langle \gcd\left(
\mathbf{M}\left(  w,s\right)  \right)  \right\rangle }\right)  _{\left(
w,s\right)  \in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a
reindexing of the family $\left(  M_{\operatorname*{red}\alpha}^{\left\langle
\gcd\alpha\right\rangle }\right)  _{\alpha\in\mathfrak{L}}$. Since every
$\left(  w,s\right)  \in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  $
satisfies
\[
\gcd\left(  \mathbf{M}\left(  w,s\right)  \right)
=s\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \operatorname*{red}\left(
\mathbf{M}\left(  w,s\right)  \right)  =w
\]
\footnote{\textit{Proof.} We have $\mathbf{R}\circ\mathbf{M}%
=\operatorname*{id}$, thus $\underbrace{\left(  \mathbf{R}\circ\mathbf{M}%
\right)  }_{=\id}\left(  w,s\right)  =\operatorname*{id}\left(
w,s\right)  =\left(  w,s\right)  $ and therefore%
\[
\left(  w,s\right)  =\left(  \mathbf{R}\circ\mathbf{M}\right)  \left(
w,s\right)  =\mathbf{R}\left(  \mathbf{M}\left(  w,s\right)  \right)  =\left(
\operatorname*{red}\left(  \mathbf{M}\left(  w,s\right)  \right)  ,\gcd\left(
\mathbf{M}\left(  w,s\right)  \right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\mathbf{R}\left(  \mathbf{M}\left(  w,s\right)
\right)  \right)  .
\]
Hence, $\left(  \operatorname*{red}\left(  \mathbf{M}\left(  w,s\right)
\right)  ,\gcd\left(  \mathbf{M}\left(  w,s\right)  \right)  \right)  =\left(
w,s\right)  $. In other words, $\operatorname*{red}\left(  \mathbf{M}\left(
w,s\right)  \right)  =w$ and $\gcd\left(  \mathbf{M}\left(  w,s\right)
\right)  =s$, qed.}, this rewrites as follows: The family $\left(
M_{w}^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a reindexing of the
family $\left(  M_{\operatorname*{red}\alpha}^{\left\langle \gcd
\alpha\right\rangle }\right)  _{\alpha\in\mathfrak{L}}$. This proves Lemma
\ref{lem.QSym.lyndon.Z-genset.reindex}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.lem.QSym.lyndon.Malphas1}}

\textit{Solution to Exercise \ref{exe.lem.QSym.lyndon.Malphas1}.} We
shall give two proofs of Lemma \ref{lem.QSym.lyndon.Malphas1}.

\begin{proof}
[First proof of Lemma \ref{lem.QSym.lyndon.Malphas1}.]We first introduce some notation.

For every $m\in \ZZ$, let $\mathcal{F}_{m}$ denote the
$\kk$-submodule of $\Qsym$ spanned by $\left(  M_{\beta}\right)
_{\beta\in \Comp ;\ \ell\left(  \beta\right)  \leq m}$. Then,
$1\in\mathcal{F}_{0}$\ \ \ \ \footnote{\textit{Proof.} We know that
$\mathcal{F}_{0}$ is the $\kk$-submodule of $\Qsym$
spanned by $\left(  M_{\beta}\right)  _{\beta\in \Comp
;\ \ell\left(  \beta\right)  \leq 0}$ (by the definition of
$\mathcal{F}_{0}$). Hence, in particular,
$M_{\beta}\in\mathcal{F}_{0}$ for every $\beta
\in \Comp $ satisfying $\ell\left(  \beta\right)  \leq0$.
Applying this to $\beta=\varnothing$, we obtain $M_{\varnothing}\in
\mathcal{F}_{0}$ (since $\ell\left(  \varnothing\right)  =0$). Since
$M_{\varnothing}=1$, this rewrites as $1\in\mathcal{F}_{0}$, qed.}.
Furthermore,
\begin{equation}
\mathcal{F}_{u}\mathcal{F}_{v}\subset\mathcal{F}_{u+v}%
\ \ \ \ \ \ \ \ \ \ \text{for every }u\in \ZZ \text{ and }v\in \ZZ 
\label{pf.lem.QSym.lyndon.Malphas1.proof1.filtr}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr}):}
Let $u\in \ZZ$ and $v\in \ZZ$. Let $p\in\mathcal{F}_{u}$ and
$q\in\mathcal{F}_{v}$. We are going to prove that
$pq\in\mathcal{F}_{u+v}$.
\par
Notice that $\mathcal{F}_{u+v}$ is a $\kk$-submodule of
$\Qsym$. Hence, the claim that $pq\in\mathcal{F}_{u+v}$ is $\kk$-linear
in $p$.
\par
We know that $p$ belongs to $\mathcal{F}_{u}$. In other words, $p$ is a
$\kk$-linear combination of the elements $\left(  M_{\beta}\right)
_{\beta\in \Comp ;\ \ell\left(  \beta\right)  \leq u}$ (since
$\mathcal{F}_{u}$ is the $\kk$-submodule of $\Qsym$
spanned by $\left(  M_{\beta}\right)  _{\beta\in \Comp %
;\ \ell\left(  \beta\right)  \leq u}$). Hence, we can WLOG assume that $p$ is
one of the elements $\left(  M_{\beta}\right)  _{\beta\in \Comp %
;\ \ell\left(  \beta\right)  \leq u}$ (because the claim that we are proving
-- namely, the claim that $pq\in\mathcal{F}_{u+v}$ -- is
$\kk$-linear in $p$). Assume this. Similarly, assume that $q$ is one of
the elements $\left(  M_{\beta}\right)  _{\beta\in \Comp %
;\ \ell\left(  \beta\right)  \leq v}$.
\par
There exists a $\varphi\in \Comp $ satisfying $\ell\left(
\varphi\right)  \leq u$ and $p=M_{\varphi}$ (since $p$ is one of the elements
$\left(  M_{\beta}\right)  _{\beta\in \Comp ;\ \ell\left(
\beta\right)  \leq u}$). Similarly, there exists a $\psi\in
 \Comp $ satisfying $\ell\left(  \psi\right)  \leq v$ and
$q=M_{\psi}$. Consider these $\varphi$ and $\psi$. Corollary
\ref{cor.QSym.lyndon.MaMb.leq} (applied to $\varphi$ and $\psi$ instead of
$\alpha$ and $\beta$) shows that $M_{\varphi}M_{\psi}$ is a sum of terms of
the form $M_{\delta}$ with $\delta\in\mathfrak{A}^{\ast}$ satisfying
$\ell\left(  \delta\right)  \leq\ell\left(  \varphi\right)  +\ell\left(
\psi\right)  $. Since every $\delta\in\mathfrak{A}^{\ast}$ satisfying
$\ell\left(  \delta\right)  \leq\ell\left(  \varphi\right)  +\ell\left(
\psi\right)  $ also satisfies $\ell\left(  \delta\right)  \leq u+v$ (because
$\underbrace{\ell\left(  \varphi\right)  }_{\leq u}+\underbrace{\ell\left(
\psi\right)  }_{\leq v}\leq u+v$), this yields that $M_{\varphi}M_{\psi}$ is a
sum of terms of the form $M_{\delta}$ with $\delta\in\mathfrak{A}^{\ast}$
satisfying $\ell\left(  \delta\right)  \leq u+v$. In particular, $M_{\varphi
}M_{\psi}$ is a $\kk$-linear combination of $\left(  M_{\delta}\right)
_{\delta\in\mathfrak{A}^{\ast};\ \ell\left(  \delta\right)  \leq u+v}$. In
other words, $M_{\varphi}M_{\psi}$ is a $\kk$-linear combination of
$\left(  M_{\delta}\right)  _{\delta\in \Comp ;\ \ell\left(
\delta\right)  \leq u+v}$ (since $\mathfrak{A}^{\ast}= \Comp $).
In other words, $M_{\varphi}M_{\psi}$ is a $\kk$-linear combination of
$\left(  M_{\beta}\right)  _{\beta\in \Comp ;\ \ell\left(
\beta\right)  \leq u+v}$ (here, we renamed the index $\delta$ as $\beta$). In
other words, $M_{\varphi}M_{\psi}$ belongs to the $\kk$-submodule of
$\Qsym$ spanned by $\left(  M_{\beta}\right)  _{\beta
\in \Comp ;\ \ell\left(  \beta\right)  \leq u+v}$. In other
words, $M_{\varphi}M_{\psi}$ belongs to $\mathcal{F}_{u+v}$ (since
$\mathcal{F}_{u+v}$ was defined as the $\kk$-submodule of
$\Qsym$ spanned by $\left(  M_{\beta}\right)  _{\beta
\in \Comp ;\ \ell\left(  \beta\right)  \leq u+v}$). Hence,
$M_{\varphi}M_{\psi}\in\mathcal{F}_{u+v}$, so that $\underbrace{p}%
_{=M_{\varphi}}\underbrace{q}_{=M_{\psi}}=M_{\varphi}M_{\psi}\in
\mathcal{F}_{u+v}$.
\par
Now, let us forget that we fixed $p$ and $q$. We thus have proven that
$pq\in\mathcal{F}_{u+v}$ for any $p\in\mathcal{F}_{u}$ and $q\in
\mathcal{F}_{v}$. Since $\mathcal{F}_{u+v}$ is a $\kk$-submodule of
$\Qsym$, this yields that $\mathcal{F}_{u}\mathcal{F}%
_{v}\subset\mathcal{F}_{u+v}$.}. It is now easy to see that
\begin{equation}
\left(  \mathcal{F}_{m}\right)  ^{k}\subset\mathcal{F}_{km}
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in \ZZ \text{ and }k\in \NN
\label{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}
\end{equation}
(where $\left(  \mathcal{F}_{m}\right)  ^{k}$ means $\underbrace{\mathcal{F}%
_{m}\mathcal{F}_{m}\cdots\mathcal{F}_{m}}_{k\text{ times}}$%
)\ \ \ \ \footnote{\textit{Proof of
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}):} Let $m\in \ZZ$. We
shall prove that (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) holds for
every $k\in \NN$. Indeed, we shall prove this by induction over $k$:
\par
\textit{Induction base:} We have%
\begin{align*}
\left(  \mathcal{F}_{m}\right)  ^{0}  &  = \kk \cdot\underbrace{1}%
_{\in\mathcal{F}_{0}}\subset \kk \cdot\mathcal{F}_{0}\subset
\mathcal{F}_{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{F}_{0}\text{
is a } \kk \text{-module}\right) \\
&  =\mathcal{F}_{0m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0=0m\right)  .
\end{align*}
In other words, (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) holds for
$k=0$. This completes the induction base.
\par
\textit{Induction step:} Let $K\in \NN$. Assume that
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) holds for $k=K$. We must
then show that (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) holds for
$k=K+1$.
\par
We have $\left(  \mathcal{F}_{m}\right)  ^{K}\subset\mathcal{F}_{Km}$ (since
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) holds for $k=K$). Thus,%
\begin{align*}
\left(  \mathcal{F}_{m}\right)  ^{K+1}  &  =\underbrace{\left(  \mathcal{F}%
_{m}\right)  ^{K}}_{\subset\mathcal{F}_{Km}}\cdot\mathcal{F}_{m}%
\subset\mathcal{F}_{Km}\cdot\mathcal{F}_{m}\subset\mathcal{F}_{Km+m}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr}), applied to }u=Km\text{ and
}v=m\right) \\
&  =\mathcal{F}_{\left(  K+1\right)  m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}Km+m=\left(  K+1\right)  m\right)  .
\end{align*}
In other words, (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) holds for
$k=K+1$. This completes the induction step. The proof of
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}) is thus complete.}.

Write the composition $\alpha$ in the form $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $; then, $\ell=\ell\left(
\alpha\right)  $.

Define $A_{s}^{\left\langle \alpha\right\rangle }$ and $a_{i,j}^{\left\langle
\alpha\right\rangle }$ as in Proposition \ref{prop.QSym.Malphas.dets}(a) (but
with $s$ instead of $n$). Then,
\[
a_{i,j}^{\left\langle \alpha\right\rangle }=%
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}
\qquad\qquad\text{ for all }\left(  i,j\right)  \in\left\{  1,2,\ldots
,s\right\}  ^{2},
\]
and we have $A_{s}^{\left\langle \alpha\right\rangle }=\left(  a_{i,j}%
^{\left\langle \alpha\right\rangle }\right)  _{i,j=1,2,\ldots,s}$. Proposition
\ref{prop.QSym.Malphas.dets}(a) (applied to $s$ instead of $n$)
yields $\det\left(  A_{s}^{\left\langle
\alpha\right\rangle }\right)  =s!M_{\alpha}^{\left\langle s\right\rangle }$.
Since $A_{s}^{\left\langle \alpha\right\rangle }=\left(  a_{i,j}^{\left\langle
\alpha\right\rangle }\right)  _{i,j=1,2,\ldots,s}$, this rewrites as
\begin{equation}
\det\left(  \left(  a_{i,j}^{\left\langle \alpha\right\rangle }\right)
_{i,j=1,2,\ldots,s}\right)  =s!M_{\alpha}^{\left\langle s\right\rangle }.
\label{pf.lem.QSym.lyndon.Malphas1.proof1.det1}
\end{equation}


On the other hand, it is easy to see that%
\begin{equation}
a_{i,j}^{\left\langle \alpha\right\rangle }\in \kk \cdot
1\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)  \in\left\{
1,2,\ldots,s\right\}  ^{2}\text{ satisfying }i<j.
\label{pf.lem.QSym.lyndon.Malphas1.proof1.aF0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF0}):}
Let $\left(  i,j\right)  \in\left\{  1,2,\ldots,s\right\}  ^{2}$ be such that
$i<j$. We need to prove that $a_{i,j}^{\left\langle \alpha\right\rangle }%
\in \kk \cdot1$. This is clear in the case when $i=j-1$ (because in this
case, we have
\begin{align*}
a_{i,j}^{\left\langle \alpha\right\rangle }  &  =%
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}  \quad
=i\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i=j-1\right) \\
&  \in \kk \cdot1
\end{align*}
). Hence, for the rest of this proof, we can WLOG assume that
$i=j-1$. Assume this.
Thus, $i \neq j-1$.
\par
We have $i<j$. Thus, $i\leq j-1$ (since $i$ and $j$ are integers), so that
$i<j-1$ (since $i\neq j-1$). Now,%
\begin{align*}
a_{i,j}^{\left\langle \alpha\right\rangle }  &  =%
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}  \quad
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i<j-1\right) \\
&  \in \kk \cdot1,
\end{align*}
and thus (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF0}) is proven.}
Furthermore,%
\begin{equation}
a_{i,j}^{\left\langle \alpha\right\rangle }\in\mathcal{F}_{\ell}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)  \in\left\{
1,2,\ldots,s\right\}  ^{2}. \label{pf.lem.QSym.lyndon.Malphas1.proof1.aF1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF1}):}
Let $\left(  i,j\right)  \in\left\{  1,2,\ldots,s\right\}  ^{2}$. We need to
show that $a_{i,j}^{\left\langle \alpha\right\rangle }\in\mathcal{F}_{\ell}$.
\par
First, recall that $\mathcal{F}_{\ell}$ is the $\kk$-submodule of
$\Qsym$ spanned by $\left(  M_{\beta}\right)  _{\beta
\in \Comp ;\ \ell\left(  \beta\right)  \leq\ell}$ (by the
definition of $\mathcal{F}_{\ell}$). In particular, $M_{\beta}\in
\mathcal{F}_{\ell}$ for every $\beta\in \Comp $ satisfying
$\ell\left(  \beta\right)  \leq\ell$. Applied to $\beta=\varnothing$, this
yields that $M_{\varnothing}\in\mathcal{F}_{\ell}$ (since $\varnothing
\in \Comp $ satisfies $\ell\left(  \varnothing\right)  =0\leq
\ell\left(  \alpha\right)  =\ell$). Since $M_{\varnothing}=1$, this rewrites
as $1\in\mathcal{F}_{\ell}$, whence $\kk \cdot\underbrace{1}%
_{\in\mathcal{F}_{\ell}}\subset \kk \cdot\mathcal{F}_{\ell}%
\subset\mathcal{F}_{\ell}$ (since $\mathcal{F}_{\ell}$ is a $\kk %
$-module).
\par
Now, if $i<j$, then (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF0}) yields
$a_{i,j}^{\left\langle \alpha\right\rangle }\in \kk \cdot1\subset
\mathcal{F}_{\ell}$. Hence, (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF1}) is
proven in the case when $i<j$. For the rest of our proof of
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF1}), we can thus WLOG assume that
we don't have $i<j$. Assume this.
\par
We have $i\geq j$ (since we don't have $i<j$) and $\ell\left(  \alpha\left\{
i-j+1\right\}  \right)  =\ell\left(  \alpha\right)  $ (since $\ell\left(
\alpha\left\{  k\right\}  \right)  =\ell\left(  \alpha\right)  $ for every
positive integer $k$). Now, recall that $M_{\beta}\in\mathcal{F}_{\ell}$ for
every $\beta\in \Comp $ satisfying $\ell\left(  \beta\right)
\leq\ell$. Applied to $\beta=\alpha\left\{  i-j+1\right\}  $, this yields that
$M_{\alpha\left\{  i-j+1\right\}  }\in\mathcal{F}_{\ell}$ (since $\ell\left(
\alpha\left\{  i-j+1\right\}  \right)  =\ell\left(  \alpha\right)  =\ell$).
Now,%
\begin{align*}
a_{i,j}^{\left\langle \alpha\right\rangle }  &  =%
\begin{cases}
M_{\alpha\left\{  i-j+1\right\}  }, & \text{if }i\geq j;\\
i, & \text{if }i=j-1;\\
0, & \text{if }i<j-1
\end{cases}  \quad
=M_{\alpha\left\{  i-j+1\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}i\geq j\right) \\
&  \in\mathcal{F}_{\ell},
\end{align*}
and this proves (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF1}).} Finally,%
\begin{equation}
a_{i,i}^{\left\langle \alpha\right\rangle }=M_{\alpha}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,s\right\}  .
\label{pf.lem.QSym.lyndon.Malphas1.proof1.aF2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF2}):}
Let $i\in\left\{  1,2,\ldots,s\right\}  $. Then, the definition of
$a_{i,i}^{\left\langle \alpha\right\rangle }$ yields%
\begin{align*}
a_{i,i}^{\left\langle \alpha\right\rangle }  &  =%
\begin{cases}
M_{\alpha\left\{  i-i+1\right\}  }, & \text{if }i\geq i;\\
i, & \text{if }i=i-1;\\
0, & \text{if }i<i-1
\end{cases}  \quad
=M_{\alpha\left\{  i-i+1\right\}  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}i\geq i\right) \\
&  =M_{\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\alpha\left\{
\underbrace{i-i+1}_{=1}\right\}  =\alpha\left\{  1\right\}  =\alpha\right)  ,
\end{align*}
qed.}

But recall that every commutative ring $A$, every $m\in{{ \NN }}$ and
every matrix $\left(  u_{i,j}\right)  _{i,j=1,2,\ldots,m}\in
A^{m\times m}$ satisfy
\[
\det\left(  \left(  u_{i,j}\right)  _{i,j=1,2,\ldots,m}\right)  =\sum
_{\sigma\in\Symm_m}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{m}u_{i,\sigma\left(  i\right)  }.
\]
\footnote{This is simply the explicit formula for the determinant of a matrix
as a sum over permutations.} Applying this equality to $A=\Qsym%
$, $m=s$ and $u_{i,j}=a_{i,j}^{\left\langle \alpha\right\rangle }$, we obtain%
\begin{align*}
\det\left(  \left(  a_{i,j}^{\left\langle \alpha\right\rangle }\right)
_{i,j=1,2,\ldots,s}\right)   &  =\sum_{\sigma\in\Symm_s}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)  }%
^{\left\langle \alpha\right\rangle }\\
&  =\underbrace{\left(  -1\right)  ^{\id}}_{=1}\prod_{i=1}%
^{s}\underbrace{a_{i,\operatorname*{id}\left(  i\right)  }^{\left\langle
\alpha\right\rangle }}_{\substack{=a_{i,i}^{\left\langle \alpha\right\rangle
}=M_{\alpha}\\\text{(by (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF2}))}%
}}+\sum\limits_{\substack{\sigma\in\Symm_s;\\\sigma\neq\operatorname*{id}%
}}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)
}^{\left\langle \alpha\right\rangle }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}\sigma=\operatorname*{id}\text{ from the sum}\right) \\
&  =\underbrace{\prod_{i=1}^{s}M_{\alpha}}_{=M_{\alpha}^{s}}+\sum\limits
_{\substack{\sigma\in\Symm_s;\\\sigma\neq\id}}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)  }%
^{\left\langle \alpha\right\rangle }=M_{\alpha}^{s}+\sum\limits_{\substack{\sigma
\in\Symm_s;\\\sigma\neq\id}}\left(  -1\right)
^{\sigma}\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)  }^{\left\langle
\alpha\right\rangle }.
\end{align*}
Compared with (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.det1}), this yields%
\[
s!M_{\alpha}^{\left\langle s\right\rangle }=M_{\alpha}^{s}+\sum\limits
_{\substack{\sigma\in\Symm_s;\\\sigma\neq\id}}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)  }%
^{\left\langle \alpha\right\rangle }.
\]
Subtracting $M_{\alpha}^{s}$ from both sides of this equality, we obtain%
\begin{equation}
s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}=\sum\limits
_{\substack{\sigma\in\Symm_s;\\\sigma\neq\id}}\left(
-1\right)  ^{\sigma}\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)  }%
^{\left\langle \alpha\right\rangle }.
\label{pf.lem.QSym.lyndon.Malphas1.proof1.diff}%
\end{equation}


But it is easy to see that
\begin{equation}
\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)  }^{\left\langle \alpha
\right\rangle }\in\mathcal{F}_{\left(  s-1\right)  \ell}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\sigma\in\Symm_s\text{
satisfying }\sigma\neq\operatorname*{id}
\label{pf.lem.QSym.lyndon.Malphas1.proof1.toolow}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.toolow}):}
Let $\sigma\in\Symm_s$ be such that $\sigma\neq\operatorname*{id}$.
Then, there exists a $k\in\left\{  1,2,\ldots,s\right\}  $ such that
$\sigma\left(  k\right)  >k$. Consider this $k$.
\par
We have $k<\sigma\left(  k\right)  $ (since $\sigma\left(  k\right)  >k$) and
thus $a_{k,\sigma\left(  k\right)  }^{\left\langle \alpha\right\rangle }%
\in \kk \cdot1$ (by (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF0}),
applied to $i=k$ and $j=\sigma\left(  k\right)  $). But
\begin{align*}
\underbrace{\prod_{i=1}^{s}}_{=\prod_{i\in\left\{  1,2,\ldots,s\right\}  }%
}a_{i,\sigma\left(  i\right)  }^{\left\langle \alpha\right\rangle }  &
=\prod_{i\in\left\{  1,2,\ldots,s\right\}  }a_{i,\sigma\left(  i\right)
}^{\left\langle \alpha\right\rangle }=\underbrace{a_{k,\sigma\left(  k\right)
}^{\left\langle \alpha\right\rangle }}_{\in \kk \cdot1}\cdot
\prod\limits_{\substack{i\in\left\{  1,2,\ldots,s\right\}  ;\\i\neq k}%
}\underbrace{a_{i,\sigma\left(  i\right)  }^{\left\langle \alpha\right\rangle
}}_{\substack{\in\mathcal{F}_{\ell}\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.aF1}), applied to }j=\sigma\left(
i\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the factor for
}i=k\text{ from the product}\right) \\
&  \in \kk \cdot1\cdot\underbrace{\prod\limits_{\substack{i\in\left\{
1,2,\ldots,s\right\}  ;\\i\neq k}}\mathcal{F}_{\ell}}_{\substack{=\left(
\mathcal{F}_{\ell}\right)  ^{s-1}\subset\mathcal{F}_{\left(  s-1\right)  \ell
}\\\text{(by (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.filtr-m}), applied
to}\\\ell\text{ and }s-1\text{ instead of }m\text{ and }k\text{)}}%
} \subset \kk \cdot\underbrace{1\cdot\mathcal{F}_{\left(  s-1\right)  \ell}%
}_{=\mathcal{F}_{\left(  s-1\right)  \ell}} = \kk \cdot\mathcal{F}%
_{\left(  s-1\right)  \ell}\subset\mathcal{F}_{\left(  s-1\right)  \ell}%
\end{align*}
(since $\mathcal{F}_{\left(  s-1\right)  \ell}$ is a $\kk$-module).
This proves (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.toolow}).}. Hence,
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.diff}) becomes
\begin{equation}
s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}=\sum\limits
_{\substack{\sigma\in\Symm_s;\\\sigma\neq\id}}\left(
-1\right)  ^{\sigma}\underbrace{\prod_{i=1}^{s}a_{i,\sigma\left(  i\right)
}^{\left\langle \alpha\right\rangle }}_{\substack{\in\mathcal{F}_{\left(
s-1\right)  \ell}\\\text{(by (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.toolow}%
))}}}\in\sum\limits_{\substack{\sigma\in\Symm_s;\\\sigma\neq
\id}}\left(  -1\right)  ^{\sigma}\mathcal{F}_{\left(
s-1\right)  \ell}\subset\mathcal{F}_{\left(  s-1\right)  \ell}
\label{pf.lem.QSym.lyndon.Malphas1.proof1.almostthere}%
\end{equation}
(since $\mathcal{F}_{\left(  s-1\right)  \ell}$ is a $\kk$-module).

But $\mathcal{F}_{\left(  s-1\right)  \ell}$ is the $\kk$-submodule of
$\Qsym$ spanned by $\left(  M_{\beta}\right)  _{\beta
\in \Comp ;\ \ell\left(  \beta\right)  \leq\left(  s-1\right)
\ell}$ (because this is how $\mathcal{F}_{\left(  s-1\right)  \ell}$ was
defined). In other words,%
\[
\mathcal{F}_{\left(  s-1\right)  \ell}=\sum\limits_{\substack{\beta\in
 \Comp ;\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell
}} \kk M_{\beta}.
\]
Hence, (\ref{pf.lem.QSym.lyndon.Malphas1.proof1.almostthere}) becomes%
\begin{equation}
s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\in\mathcal{F}%
_{\left(  s-1\right)  \ell}=\sum\limits_{\substack{\beta\in \Comp %
;\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell}} \kk M_{\beta
}. \label{pf.lem.QSym.lyndon.Malphas1.proof1.almostthere2}%
\end{equation}


Also, $s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}%
\in\mathcal{F}_{\left(  s-1\right)  \ell}\subset\Qsym$, so that
$s!M_{\alpha}^{\left\langle s\right\rangle }\in\underbrace{M_{\alpha}^{s}%
}_{\in\Qsym}+\Qsym\subset\Qsym%
+\Qsym\subset\Qsym$.

Recall that $M_{\alpha}^{\left\langle s\right\rangle }=e_{s}\left(  \left(
\xx_{\mathbf{i}}^{\alpha}\right)  _{\mathbf{i}\in\operatorname*{SIS}%
\left(  \ell\right)  }\right)  $. Thus, $M_{\alpha}^{\left\langle
s\right\rangle }$ is a homogeneous power series of degree $s\left\vert
\alpha\right\vert $ (since each $\xx_{\mathbf{i}}^{\alpha}$ is a
monomial of degree $\left\vert \alpha\right\vert $, and since $e_{s}$ is a
power series of degree $s$). Hence, $s!M_{\alpha}^{\left\langle s\right\rangle
}$ is a homogeneous power series of degree $s\left\vert \alpha\right\vert $ as
well. Thus, $s!M_{\alpha}^{\left\langle s\right\rangle }\in
\Qsym_{s\left\vert \alpha\right\vert }$ (since
$s!M_{\alpha}^{\left\langle s\right\rangle }\in\Qsym$). Also,
$M_{\alpha}^{s}\in\Qsym_{s\left\vert \alpha\right\vert
}$ (since $M_{\alpha}\in\Qsym_{\left\vert
\alpha\right\vert }$). Thus, $\underbrace{s!M_{\alpha}^{\left\langle
s\right\rangle }}_{\in\Qsym_{s\left\vert
\alpha\right\vert }}-\underbrace{M_{\alpha}^{s}}_{\in
\Qsym_{s\left\vert \alpha\right\vert }}\in
\Qsym_{s\left\vert \alpha\right\vert }-
\Qsym_{s\left\vert \alpha\right\vert }\subset
\Qsym_{s\left\vert \alpha\right\vert }$ (since $
\Qsym_{s\left\vert \alpha\right\vert }$ is a $\kk$-module).

Now, let $\pi$ denote the projection from the direct sum $\Qsym%
=\bigoplus_{k\in \NN }\Qsym_{k}$ onto its
$\left(  s\left\vert \alpha\right\vert \right)  $-th homogeneous component
$\Qsym_{s\left\vert \alpha\right\vert }$. Notice that
$\pi\left(  s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\right)
=s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}$ (since
$s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\in
\Qsym_{s\left\vert \alpha\right\vert }$).

We have%
\begin{equation}
\pi\left(  M_{\beta}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }\beta
\in \Comp \setminus\Comp_{s\left\vert
\alpha\right\vert } \label{pf.lem.QSym.lyndon.Malphas1.proof1.pi-annihilates}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.pi-annihilates}):} Let $\beta
\in \Comp \setminus\Comp_{s\left\vert
\alpha\right\vert }$. Then, $\beta\in \Comp $ but $\beta
\notin\Comp_{s\left\vert \alpha\right\vert }$. In
other words, $\beta$ is a composition with size $\left\vert \beta\right\vert
\neq s\left\vert \alpha\right\vert $. As a consequence, $M_{\beta}$ is a
homogeneous element of $\Qsym$ of degree $\left\vert
\beta\right\vert \neq s\left\vert \alpha\right\vert $. Therefore, $\pi\left(
M_{\beta}\right)  =0$ (since $\pi$ is the projection from the direct sum
$\Qsym=\bigoplus_{k\in \NN }
\Qsym_{k}$ onto its $\left(  s\left\vert \alpha\right\vert \right)  $-th
homogeneous component $\Qsym_{s\left\vert
\alpha\right\vert }$). This proves
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.pi-annihilates}).}.

Now, applying the map $\pi$ to both sides of the relation
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.almostthere2}), we obtain%
\begin{align*}
\pi\left(  s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\right)
&  \in\pi\left(  \sum\limits_{\substack{\beta\in \Comp ;\\\ell\left(
\beta\right)  \leq\left(  s-1\right)  \ell}} \kk M_{\beta}\right)
=\sum\limits_{\substack{\beta\in \Comp ;\\\ell\left(  \beta\right)
\leq\left(  s-1\right)  \ell}} \kk \pi\left(  M_{\beta}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi\text{ is } \kk %
\text{-linear}\right) \\
&  =\sum\limits_{\substack{\beta\in\Comp_{s\left\vert
\alpha\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell
}} \kk \underbrace{\pi\left(  M_{\beta}\right)  }_{\substack{=M_{\beta
}\\\text{(since }M_{\beta}\in\Qsym_{s\left\vert
\alpha\right\vert }\\\text{(because }\beta\in
\Comp_{s\left\vert \alpha\right\vert }\text{),}\\\text{whereas }\pi\text{
is a projection}\\\text{onto }\Qsym_{s\left\vert
\alpha\right\vert }\text{)}}}+\sum\limits_{\substack{\beta\in \Comp %
\setminus\Comp_{s\left\vert \alpha\right\vert }%
;\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell}} \kk %
\underbrace{\pi\left(  M_{\beta}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof1.pi-annihilates}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the set } \Comp \text{ is the union of its two}\\
\text{disjoint subsets }\Comp_{s\left\vert
\alpha\right\vert }\text{ and } \Comp \setminus
\Comp_{s\left\vert \alpha\right\vert }%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\beta\in\Comp_{s\left\vert
\alpha\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell
}} \kk M_{\beta}+\underbrace{\sum\limits_{\substack{\beta\in \Comp %
\setminus\Comp_{s\left\vert \alpha\right\vert }%
;\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell}} \kk 0}%
_{=0}=\sum\limits_{\substack{\beta\in\Comp_{s\left\vert
\alpha\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell
}} \kk M_{\beta}=\sum\limits_{\substack{\beta\in
\Comp_{s\left\vert \alpha\right\vert };\\\ell\left(  \beta\right)
\leq\left(  s-1\right)  \ell\left(  \alpha\right)  }} \kk M_{\beta}%
\end{align*}
(since $\ell=\ell\left(  \alpha\right)  $). Since $\pi\left(  s!M_{\alpha
}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\right)  =s!M_{\alpha
}^{\left\langle s\right\rangle }-M_{\alpha}^{s}$, this rewrites as
$s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\in\sum\limits
_{\substack{\beta\in\Comp_{s\left\vert \alpha
\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell\left(
\alpha\right)  }} \kk M_{\beta}$. This proves Lemma
\ref{lem.QSym.lyndon.Malphas1}.
\end{proof}

\begin{proof}
[Second proof of Lemma \ref{lem.QSym.lyndon.Malphas1}.]Let us first notice
that if $\gamma$ and $\beta$ are two compositions, then
\begin{equation}
\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{ in
}M_{\beta}\right)  =\delta_{\gamma,\beta}.
\label{pf.lem.QSym.lyndon.Malphas1.proof2.delta}
\end{equation}
(This follows from the definition of $M_{\beta}$.)

Write the composition $\alpha$ in the form $\alpha=\left(  \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right)  $; then, $\ell=\ell\left(
\alpha\right)  $. Let us now fix a total order on the set $\operatorname*{SIS}%
\left(  \ell\right)  $ (for example, the lexicographic order). Exercise
\ref{exe.QSym.Malphas.ps}(b) yields%
\begin{equation}
M_{\alpha}^{\left\langle s\right\rangle }=\sum\limits_{\substack{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}%
_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}%
^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}%
}^{\alpha}. \label{pf.lem.QSym.lyndon.Malphas1.proof2.Mas}%
\end{equation}


Now, for every $\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)  ^{s}$
such that $\mathbf{i}_{1}$, $\mathbf{i}_{2}$, $\ldots$, $\mathbf{i}_{s}$ are
distinct, there exists a \textbf{unique} $\sigma\in\Symm_s$
satisfying $\mathbf{i}_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(
2\right)  }<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }$ (because there is
exactly one way to sort the $\ell$-tuple $\left(  \mathbf{i}_{1}%
,\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  $ into increasing
order\footnote{Here, we are using that the order on $\operatorname*{SIS}%
\left(  \ell\right)  $ is total.}). Hence, we can split the sum $\sum\limits
_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)
^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{, }%
\mathbf{i}_{s}\text{ are distinct}}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}$ into several subsums, one for each $\sigma\in\Symm_s%
$, as follows:
\begin{align}
&  \sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)
^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{, }%
\mathbf{i}_{s}\text{ are distinct}}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}\nonumber\\
&  =\sum_{\sigma\in\Symm_s}\underbrace{\sum\limits_{\substack{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}_{1}\text{,
}\mathbf{i}_{2}\text{, }\ldots\text{, }\mathbf{i}_{s}\text{ are
distinct;}\\\mathbf{i}_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(
2\right)  }<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }}}}_{\substack{=\sum\limits
_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)
^{s};\\\mathbf{i}_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(
2\right)  }<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }}}\\\text{(since the
condition that }\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are distinct}\\\text{follows from the condition
that }\mathbf{i}_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(
2\right)  }<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }\\\text{(because if
}\mathbf{i}_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(  2\right)
}<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }\text{, then }\mathbf{i}%
_{\sigma\left(  1\right)  }\text{, }\mathbf{i}_{\sigma\left(  2\right)
}\text{, }\ldots\text{, }\mathbf{i}_{\sigma\left(  s\right)  }\text{
are}\\\text{distinct, and thus }\mathbf{i}_{1}\text{, }\mathbf{i}%
_{2}\text{, }\ldots\text{, }\mathbf{i}_{s}\text{ are distinct}%
\\\text{(since }\sigma\text{ is a permutation of }\left\{  1,2,\ldots
,s\right\}  \text{)))}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx%
_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha
}\nonumber\\
&  =\sum_{\sigma\in\Symm_s}\sum\limits_{\substack{\left(  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}%
_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(  2\right)  }%
<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}. \label{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.2}%
\end{align}


But every $\sigma\in\Symm_s$ and every $\left(  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s}$ satisfy
\begin{equation}
\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}%
\cdots\xx_{\mathbf{i}_{s}}^{\alpha}=\xx_{\mathbf{i}%
_{\sigma\left(  1\right)  }}^{\alpha}\xx_{\mathbf{i}_{\sigma\left(
2\right)  }}^{\alpha}\cdots\xx_{\mathbf{i}_{\sigma\left(  s\right)  }%
}^{\alpha} \label{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.3}
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.3}):}
Let $\sigma\in\Symm_s$ and $\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s}$. Then, the product $\xx_{\mathbf{i}%
_{\sigma\left(  1\right)  }}^{\alpha}\xx_{\mathbf{i}_{\sigma\left(
2\right)  }}^{\alpha}\cdots\xx_{\mathbf{i}_{\sigma\left(  s\right)  }%
}^{\alpha}$ is obtained from the product $\xx_{\mathbf{i}_{1}}^{\alpha
}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}%
^{\alpha}$ by rearranging the factors according to the permutation $\sigma$.
Since rearranging the factors of a product in $\Qsym$ does not
change the value of the product (because the algebra $\Qsym$ is
commutative), this yields that $\xx_{\mathbf{i}_{\sigma\left(
1\right)  }}^{\alpha}\xx_{\mathbf{i}_{\sigma\left(  2\right)  }%
}^{\alpha}\cdots\xx_{\mathbf{i}_{\sigma\left(  s\right)  }}^{\alpha
}=\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha
}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}$. This proves
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.3}).}. Now,
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.2}) becomes
\begin{align}
&  \sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)
^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{, }%
\mathbf{i}_{s}\text{ are distinct}}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}\nonumber\\
&  =\sum_{\sigma\in\Symm_s}\sum\limits_{\substack{\left(  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}%
_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(  2\right)  }%
<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }}}\underbrace{\xx%
_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots
\xx_{\mathbf{i}_{s}}^{\alpha}}_{\substack{=\xx_{\mathbf{i}%
_{\sigma\left(  1\right)  }}^{\alpha}\xx_{\mathbf{i}_{\sigma\left(
2\right)  }}^{\alpha}\cdots\xx_{\mathbf{i}_{\sigma\left(  s\right)  }%
}^{\alpha}\\\text{(by (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.3}))}%
}}\nonumber\\
&  =\sum_{\sigma\in\Symm_s}\underbrace{\sum\limits_{\substack{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}%
_{\sigma\left(  1\right)  }<\mathbf{i}_{\sigma\left(  2\right)  }%
<\cdots<\mathbf{i}_{\sigma\left(  s\right)  }}}\xx_{\mathbf{i}%
_{\sigma\left(  1\right)  }}^{\alpha}\xx_{\mathbf{i}_{\sigma\left(
2\right)  }}^{\alpha}\cdots\xx_{\mathbf{i}_{\sigma\left(  s\right)  }%
}^{\alpha}}_{\substack{=\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}<\mathbf{i}_{2}<\cdots
<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}%
_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\\\text{(here, we have
substituted }\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \\\text{for }\left(  \mathbf{i}_{\sigma\left(  1\right)
},\mathbf{i}_{\sigma\left(  2\right)  },\ldots,\mathbf{i}_{\sigma\left(
s\right)  }\right)  \text{ in the sum}\\\text{(since }\sigma\text{ is a
permutation of }\left\{  1,2,\ldots,s\right\}  \text{))}}}\nonumber\\
&  =\sum_{\sigma\in\Symm_s}\underbrace{\sum\limits_{\substack{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}%
_{1}<\mathbf{i}_{2}<\cdots<\mathbf{i}_{s}}}\xx_{\mathbf{i}_{1}}%
^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}%
}^{\alpha}}_{\substack{=M_{\alpha}^{\left\langle s\right\rangle }\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas}))}}}=\sum_{\sigma\in
\Symm_s}M_{\alpha}^{\left\langle s\right\rangle }%
=\underbrace{\left\vert \Symm_s\right\vert }_{=s!}M_{\alpha
}^{\left\langle s\right\rangle }=s!M_{\alpha}^{\left\langle s\right\rangle }.
\label{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.5}%
\end{align}


On the other hand, taking both sides of the identity
(\ref{eq.def.QSym.Malphas.a.1}) to the $s$-th power, we obtain%
\begin{align*}
M_{\alpha}^{s}  &  =\left(  \sum_{\mathbf{i}\in\operatorname*{SIS}\left(
\ell\right)  }\xx_{\mathbf{i}}^{\alpha}\right)  ^{s}=\sum_{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s}}\xx%
_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots
\xx_{\mathbf{i}_{s}}^{\alpha}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
product rule}\right) \\
&  =\underbrace{\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2}%
,\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{,
}\ldots\text{, }\mathbf{i}_{s}\text{ are distinct}}}\xx%
_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots
\xx_{\mathbf{i}_{s}}^{\alpha}}_{\substack{=s!M_{\alpha}^{\left\langle
s\right\rangle }\\\text{(by (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.5}%
))}}}+\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)
\right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are not distinct}}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}\\
&  =s!M_{\alpha}^{\left\langle s\right\rangle }+\sum\limits_{\substack{\left(
\mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}_{1}\text{,
}\mathbf{i}_{2}\text{, }\ldots\text{, }\mathbf{i}_{s}\text{ are not
distinct}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}%
}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}.
\end{align*}
Hence,%
\begin{equation}
M_{\alpha}^{s}-s!M_{\alpha}^{\left\langle s\right\rangle }=\sum\limits
_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}%
_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)
^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{, }%
\mathbf{i}_{s}\text{ are not distinct}}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}. \label{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.7}%
\end{equation}


But $M_{\alpha}^{\left\langle s\right\rangle }\in\Qsym$ (by
Corollary \ref{cor.QSym.Malphas.QSym}(a)), so that $\underbrace{M_{\alpha}%
^{s}}_{\in\Qsym}-s!\underbrace{M_{\alpha}^{\left\langle
s\right\rangle }}_{\in\Qsym}\in\Qsym%
-s!\Qsym\subset\Qsym$. Hence, $M_{\alpha}%
^{s}-s!M_{\alpha}^{\left\langle s\right\rangle }$ is a $\kk$-linear
combination of the family $\left(  M_{\beta}\right)  _{\beta\in
 \Comp }$ (since this family $\left(  M_{\beta}\right)
_{\beta\in \Comp }$ is a basis of the $\kk$-module
$\Qsym$). In other words, there exists a family $\left(
c_{\beta}\right)  _{\beta\in \Comp }\in \kk %
^{ \Comp }$ of elements of $\kk$ such that $\left(
\text{all but finitely many }\beta\in \Comp \text{ satisfy
}c_{\beta}=0\right)  $ and%
\begin{equation}
s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}=\sum_{\beta
\in \Comp }c_{\beta}M_{\beta}.
\label{pf.lem.QSym.lyndon.Malphas1.proof2.c-eq}%
\end{equation}
Consider this family $\left(  c_{\beta}\right)  _{\beta\in \Comp %
}\in \kk ^{ \Comp }$. Every composition $\gamma$ satisfies%
\begin{align*}
&  \left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{ in
}\underbrace{s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}%
}_{\substack{=\sum_{\beta\in \Comp }c_{\beta}M_{\beta}\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.c-eq}))}}}\right) \\
&  =\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{
in }\sum_{\beta\in \Comp }c_{\beta}M_{\beta}\right) \\
&  =\sum_{\beta\in \Comp }c_{\beta}\underbrace{\left(  \text{the
coefficient of the monomial }\mathbf{x}^{\gamma}\text{ in }M_{\beta}\right)
}_{\substack{=\delta_{\gamma,\beta}\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.delta}))}}}\\
&  =\sum_{\beta\in \Comp }c_{\beta}\delta_{\gamma,\beta
}=c_{\gamma}.
\end{align*}
Hence, every composition $\gamma$ satisfies%
\[
c_{\gamma}=\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma
}\text{ in }s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}\right)
.
\]
Thus, every composition $\gamma$ satisfies
\begin{align}
-c_{\gamma}  &  =-\left(  \text{the coefficient of the monomial }%
\mathbf{x}^{\gamma}\text{ in }s!M_{\alpha}^{\left\langle s\right\rangle
}-M_{\alpha}^{s}\right) \nonumber\\
&  =\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{
in }\underbrace{-\left(  s!M_{\alpha}^{\left\langle s\right\rangle }%
-M_{\alpha}^{s}\right)  }_{\substack{=M_{\alpha}^{s}-s!M_{\alpha
}^{\left\langle s\right\rangle }\\=\sum\limits_{\substack{\left(  \mathbf{i}%
_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(
\operatorname*{SIS}\left(  \ell\right)  \right)  ^{s};\\\mathbf{i}_{1}\text{,
}\mathbf{i}_{2}\text{, }\ldots\text{, }\mathbf{i}_{s}\text{ are not
distinct}}}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}%
}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.Mas.7}))}}}\right) \nonumber\\
&  =\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{
in }\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)
\right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are not distinct}}}\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)
\right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are not distinct}}}\left(  \text{the
coefficient of the monomial }\mathbf{x}^{\gamma}\text{ in }\xx%
_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots
\xx_{\mathbf{i}_{s}}^{\alpha}\right)  .
\label{pf.lem.QSym.lyndon.Malphas1.proof2.c-eq.3}%
\end{align}


We now notice that if $\gamma$ is a composition satisfying $\left\vert
\gamma\right\vert \neq s\left\vert \alpha\right\vert $, then%
\begin{equation}
\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{ in
}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha
}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  =0
\label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1}%
\end{equation}
for every $\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)
\in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)  ^{s}%
$\ \ \ \ \footnote{\textit{Proof of
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1}):} Let $\gamma$ be a
composition satisfying $\left\vert \gamma\right\vert \neq s\left\vert
\alpha\right\vert $. Let $\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)
\right)  ^{s}$. Then, the monomial $\xx_{\mathbf{i}_{1}}^{\alpha
}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}%
^{\alpha}$ has degree%
\[
\deg\left(  \xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}%
}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  =\deg\left(
\xx_{\mathbf{i}_{1}}^{\alpha}\right)  +\deg\left(  \xx%
_{\mathbf{i}_{2}}^{\alpha}\right)  +\cdots+\deg\left(  \xx%
_{\mathbf{i}_{s}}^{\alpha}\right)  .
\]
\par
However, for every $\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  $,
the monomial $\xx_{\mathbf{i}}^{\alpha}$ is a monomial of degree
$\left\vert \alpha\right\vert $. Thus, for every $\mathbf{i}\in
\operatorname*{SIS}\left(  \ell\right)  $, we have%
\begin{equation}
\deg\left(  \xx_{\mathbf{i}}^{\alpha}\right)  =\left\vert
\alpha\right\vert . \label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1.pf.1}%
\end{equation}
Hence, for every $k\in\left\{  1,2,\ldots,s\right\}  $, we have $\deg\left(
\xx_{\mathbf{i}_{k}}^{\alpha}\right)  =\left\vert \alpha\right\vert $
(by (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1.pf.1}), applied to
$\mathbf{i}=\mathbf{i}_{k}$). Adding up these equalities over all
$k\in\left\{  1,2,\ldots,s\right\}  $, we obtain $\deg\left(  \xx%
_{\mathbf{i}_{1}}^{\alpha}\right)  +\deg\left(  \xx_{\mathbf{i}_{2}%
}^{\alpha}\right)  +\cdots+\deg\left(  \xx_{\mathbf{i}_{s}}^{\alpha
}\right)  =\underbrace{\left\vert \alpha\right\vert +\left\vert \alpha
\right\vert +\cdots+\left\vert \alpha\right\vert }_{s\text{ times}%
}=s\left\vert \alpha\right\vert $. Hence,%
\[
\deg\left(  \xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}%
}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  =\deg\left(
\xx_{\mathbf{i}_{1}}^{\alpha}\right)  +\deg\left(  \xx%
_{\mathbf{i}_{2}}^{\alpha}\right)  +\cdots+\deg\left(  \xx%
_{\mathbf{i}_{s}}^{\alpha}\right)  =s\left\vert \alpha\right\vert ,
\]
so that $s\left\vert \alpha\right\vert =\deg\left(  \xx_{\mathbf{i}%
_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx%
_{\mathbf{i}_{s}}^{\alpha}\right)  $. But $\deg\left(  \mathbf{x}^{\gamma
}\right)  =\left\vert \gamma\right\vert \neq s\left\vert \alpha\right\vert
=\deg\left(  \xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}%
}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  $. In other
words, the monomials $\mathbf{x}^{\gamma}$ and $\xx_{\mathbf{i}_{1}%
}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}%
_{s}}^{\alpha}$ have different degrees; thus, these monomials are distinct.
Therefore, $\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma
}\text{ in }\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}%
}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  =0$. This proves
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1}).}. Also, if $\gamma$ is
a composition satisfying $\ell\left(  \gamma\right)  >\left(  s-1\right)
\ell\left(  \alpha\right)  $, then%
\begin{equation}
\left(  \text{the coefficient of the monomial }\mathbf{x}^{\gamma}\text{ in
}\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha
}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  =0
\label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2}%
\end{equation}
for every $\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots,\mathbf{i}_{s}\right)
\in\left(  \operatorname*{SIS}\left(  \ell\right)  \right)  ^{s}$ satisfying
$\left(  \mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are not distinct}\right)  $%
\ \ \ \ \footnote{\textit{Proof of
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2}):} Let $\gamma$ be a
composition satisfying $\ell\left(  \gamma\right)  >\left(  s-1\right)
\ell\left(  \alpha\right)  $. Let $\left(  \mathbf{i}_{1},\mathbf{i}%
_{2},\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s}$ be such that $\left(  \mathbf{i}_{1}\text{,
}\mathbf{i}_{2}\text{, }\ldots\text{, }\mathbf{i}_{s}\text{ are not
distinct}\right)  $.
\par
For every monomial $\mathbf{m}$ in the variables $x_{1},x_{2},x_{3},\ldots$,
we denote by $\operatorname*{Supp}\mathbf{m}$ the set of all variables which
occur in the monomial $\mathbf{m}$ (where we say that a variable
\emph{occurs} in $\mathbf{m}$ if and only if its exponent in $\mathbf{m}$ is
positive). For instance, $\operatorname*{Supp}1=\varnothing$ and
$\operatorname*{Supp}\left(  x_{2}^{3}x_{3}x_{5}^{4}\right)  =\left\{
x_{2},x_{3},x_{5}\right\}  $ and $\operatorname*{Supp}\left(
\underbrace{x_{3}x_{4}^{0}}_{=x_{3}}\right)  =\operatorname*{Supp}\left(
x_{3}\right)  =\left\{  x_{3}\right\}  $. It is very easy to see that
$\operatorname*{Supp}\left(  \mathbf{mn}\right)  =\left(  \operatorname*{Supp}%
\mathbf{m}\right)  \cup\left(  \operatorname*{Supp}\mathbf{n}\right)  $ for
any two monomials $\mathbf{m}$ and $\mathbf{n}$. More generally, if
$k\in \NN$, and if $\mathbf{m}_{1}$, $\mathbf{m}_{2}$, $\ldots$,
$\mathbf{m}_{k}$ are $k$ monomials, then%
\begin{equation}
\operatorname*{Supp}\left(  \mathbf{m}_{1}\mathbf{m}_{2}\cdots\mathbf{m}%
_{k}\right)  =\left(  \operatorname*{Supp}\left(  \mathbf{m}_{1}\right)
\right)  \cup\left(  \operatorname*{Supp}\left(  \mathbf{m}_{2}\right)
\right)  \cup\cdots\cup\left(  \operatorname*{Supp}\left(  \mathbf{m}%
_{k}\right)  \right)  .
\label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.supp}%
\end{equation}
\par
On the other hand,
\begin{equation}
\left\vert \operatorname*{Supp}\left(  \mathbf{x}^{\beta}\right)  \right\vert
=\ell\left(  \beta\right)  \ \ \ \ \ \ \ \ \ \ \text{for every composition
}\beta. \label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.comp}%
\end{equation}
\par
[\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.comp}%
):} Let $\beta$ be a composition. Write $\beta$ in the form $\beta=\left(
\beta_{1},\beta_{2},\ldots,\beta_{\ell\left(  \beta\right)  }\right)  $. Then,
$\mathbf{x}^{\beta}=x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots x_{\ell\left(
\beta\right)  }^{\beta_{\ell\left(  \beta\right)  }}$. Hence, the variables
$x_{1}$, $x_{2}$, $\ldots$, $x_{\ell\left(  \beta\right)  }$ all occur in the
monomial $\mathbf{x}^{\beta}$ (since their exponents $\beta_{1}$, $\beta_{2}$,
$\ldots$, $\beta_{\ell\left(  \beta\right)  }$ are positive), and no other
variables do. In other words, the set of all variables which occur in the
monomial $\mathbf{x}^{\beta}$ is $\left\{  x_{1},x_{2},\ldots,x_{\ell\left(
\beta\right)  }\right\}  $. In other words, $\operatorname*{Supp}\left(
\mathbf{x}^{\beta}\right)  $ is $\left\{  x_{1},x_{2},\ldots,x_{\ell\left(
\beta\right)  }\right\}  $ (since $\operatorname*{Supp}\left(  \xx%
^{\beta}\right)  $ is the set of all variables which occur in the monomial
$\mathbf{x}^{\beta}$ (by the definition of $\operatorname*{Supp}\left(
\mathbf{x}^{\beta}\right)  $)). In other words, $\operatorname*{Supp}\left(
\mathbf{x}^{\beta}\right)  =\left\{  x_{1},x_{2},\ldots,x_{\ell\left(
\beta\right)  }\right\}  $, so that $\left\vert \operatorname*{Supp}\left(
\mathbf{x}^{\beta}\right)  \right\vert =\left\vert \left\{  x_{1},x_{2}%
,\ldots,x_{\ell\left(  \beta\right)  }\right\}  \right\vert =\ell\left(
\beta\right)  $. This proves
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.comp}).]
\par
Also,%
\begin{equation}
\left\vert \operatorname*{Supp}\left(  \xx_{\mathbf{i}}^{\alpha
}\right)  \right\vert =\ell\ \ \ \ \ \ \ \ \ \ \text{for every }\mathbf{i}%
\in\operatorname*{SIS}\left(  \ell\right)  .
\label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.sis}%
\end{equation}
\par
[\textit{Proof of (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.sis}%
):} Let $\mathbf{i}\in\operatorname*{SIS}\left(  \ell\right)  $. Then,
$\mathbf{i}$ is a strictly increasing $\ell$-tuple of positive integers. In
other words, we can write $\mathbf{i}$ in the form $\mathbf{i}=\left(
i_{1},i_{2},\ldots,i_{\ell}\right)  $ for some positive integers $i_{1}$,
$i_{2}$, $\ldots$, $i_{\ell}$ satisfying $i_{1}<i_{2}<\cdots<i_{\ell}$.
Consider these $i_{1}$, $i_{2}$, $\ldots$, $i_{\ell}$. Notice that $i_{1}$,
$i_{2}$, $\ldots$, $i_{\ell}$ are distinct (since $i_{1}<i_{2}%
<\cdots<i_{\ell}$), so that the variables $x_{i_{1}}$, $x_{i_{2}}$, $\ldots$,
$x_{i_{\ell}}$ are distinct.
\par
The definition of $\xx_{\mathbf{i}}^{\alpha}$ yields $\xx%
_{\mathbf{i}}^{\alpha}=x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots
x_{i_{\ell}}^{\alpha_{\ell}}$. Thus, the variables $x_{i_{1}}$, $x_{i_{2}}$,
$\ldots$, $x_{i_{\ell}}$ all occur in the monomial $\xx_{\mathbf{i}%
}^{\alpha}$ (since their exponents $\alpha_{1}$, $\alpha_{2}$, $\ldots$,
$\alpha_{\ell}$ are positive), and no other variables do. In other words, the
set of all variables which occur in the monomial $\xx_{\mathbf{i}%
}^{\alpha}$ is $\left\{  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{\ell}}\right\}  $.
In other words, $\operatorname*{Supp}\left(  \xx_{\mathbf{i}}^{\alpha
}\right)  $ is $\left\{  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{\ell}}\right\}  $
(since $\operatorname*{Supp}\left(  \xx_{\mathbf{i}}^{\alpha}\right)  $
is the set of all variables which occur in the monomial $\xx%
_{\mathbf{i}}^{\alpha}$ (by the definition of $\operatorname*{Supp}\left(
\xx_{\mathbf{i}}^{\alpha}\right)  $)). In other words,
$\operatorname*{Supp}\left(  \xx_{\mathbf{i}}^{\alpha}\right)
=\left\{  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{\ell}}\right\}  $, so that
$\left\vert \operatorname*{Supp}\left(  \xx_{\mathbf{i}}^{\alpha
}\right)  \right\vert =\left\vert \left\{  x_{i_{1}},x_{i_{2}},\ldots
,x_{i_{\ell}}\right\}  \right\vert =\ell$ (since the variables $x_{i_{1}}$,
$x_{i_{2}}$, $\ldots$, $x_{i_{\ell}}$ are distinct). This proves
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.sis}).]
\par
We know that $\mathbf{i}_{1}$, $\mathbf{i}_{2}$, $\ldots$, $\mathbf{i}_{s}$
are not distinct. In other words, (at least) two of the elements
$\mathbf{i}_{1}$, $\mathbf{i}_{2}$, $\ldots$, $\mathbf{i}_{s}$ are equal. In
other words, there exist two distinct elements $u$ and $v$ of $\left\{
1,2,\ldots,s\right\}  $ such that $\mathbf{i}_{u}=\mathbf{i}_{v}$. Consider
these $u$ and $v$. Notice that $u\in\left\{  1,2,\ldots,s\right\}  $, so that
$1\leq u\leq s$ and thus $s\geq1$.
\par
Since $u$ and $v$ are distinct, we have $u\in\left\{  1,2,\ldots,s\right\}
\setminus\left\{  v\right\}  $; thus, $\operatorname*{Supp}\left(
\xx_{\mathbf{i}_{u}}^{\alpha}\right)  $ is an addend of the union
$\bigcup_{j\in\left\{  1,2,\ldots,s\right\}  \setminus\left\{  v\right\}
}\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{j}}^{\alpha}\right)  $.
Consequently, $\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{u}}%
^{\alpha}\right)  \subset\bigcup_{j\in\left\{  1,2,\ldots,s\right\}
\setminus\left\{  v\right\}  }\operatorname*{Supp}\left(  \xx%
_{\mathbf{i}_{j}}^{\alpha}\right)  $. Since $\mathbf{i}_{u}=\mathbf{i}_{v}$,
this rewrites as%
\begin{equation}
\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{v}}^{\alpha}\right)
\subset\bigcup_{j\in\left\{  1,2,\ldots,s\right\}  \setminus\left\{
v\right\}  }\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{j}}^{\alpha
}\right)  . \label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.in-uni}%
\end{equation}
\par
Now, (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.supp}) (applied to
$k=s$ and $\mathbf{m}_{j}=\xx_{\mathbf{i}_{j}}^{\alpha}$) yields%
\begin{align*}
\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{1}}^{\alpha}%
\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha
}\right)   &  =\left(  \operatorname*{Supp}\left(  \xx_{\mathbf{i}_{1}%
}^{\alpha}\right)  \right)  \cup\left(  \operatorname*{Supp}\left(
\xx_{\mathbf{i}_{2}}^{\alpha}\right)  \right)  \cup\cdots\cup\left(
\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{s}}^{\alpha}\right)
\right) \\
&  =\bigcup_{j\in\left\{  1,2,\ldots,s\right\}  }\operatorname*{Supp}\left(
\xx_{\mathbf{i}_{j}}^{\alpha}\right)  =\left(  \bigcup_{j\in\left\{
1,2,\ldots,s\right\}  \setminus\left\{  v\right\}  }\operatorname*{Supp}%
\left(  \xx_{\mathbf{i}_{j}}^{\alpha}\right)  \right)  \cup
\underbrace{\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{v}}^{\alpha
}\right)  }_{\substack{\subset\bigcup_{j\in\left\{  1,2,\ldots,s\right\}
\setminus\left\{  v\right\}  }\operatorname*{Supp}\left(  \xx%
_{\mathbf{i}_{j}}^{\alpha}\right)  \\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.in-uni}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}j=v\text{ from the union}\right) \\
&  \subset\left(  \bigcup_{j\in\left\{  1,2,\ldots,s\right\}  \setminus
\left\{  v\right\}  }\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{j}%
}^{\alpha}\right)  \right)  \cup\left(  \bigcup_{j\in\left\{  1,2,\ldots
,s\right\}  \setminus\left\{  v\right\}  }\operatorname*{Supp}\left(
\xx_{\mathbf{i}_{j}}^{\alpha}\right)  \right) \\
&  =\bigcup_{j\in\left\{  1,2,\ldots,s\right\}  \setminus\left\{  v\right\}
}\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{j}}^{\alpha}\right)  .
\end{align*}
Thus,%
\begin{align*}
\left\vert \operatorname*{Supp}\left(  \xx_{\mathbf{i}_{1}}^{\alpha
}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}%
^{\alpha}\right)  \right\vert  &  \leq\left\vert \bigcup_{j\in\left\{
1,2,\ldots,s\right\}  \setminus\left\{  v\right\}  }\operatorname*{Supp}%
\left(  \xx_{\mathbf{i}_{j}}^{\alpha}\right)  \right\vert \leq
\sum_{j\in\left\{  1,2,\ldots,s\right\}  \setminus\left\{  v\right\}
}\underbrace{\left\vert \operatorname*{Supp}\left(  \xx_{\mathbf{i}%
_{j}}^{\alpha}\right)  \right\vert }_{\substack{=\ell\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.sis}), applied to
}\mathbf{i}=\mathbf{i}_{j}\text{)}}}\\
&  =\sum_{j\in\left\{  1,2,\ldots,s\right\}  \setminus\left\{  v\right\}
}\ell=\underbrace{\left\vert \left\{  1,2,\ldots,s\right\}  \setminus\left\{
v\right\}  \right\vert }_{=s-1}\underbrace{\ell}_{=\ell\left(  \alpha\right)
}=\left(  s-1\right)  \ell\left(  \alpha\right)  .
\end{align*}
Thus,%
\begin{equation}
\left(  s-1\right)  \ell\left(  \alpha\right)  \geq\left\vert
\operatorname*{Supp}\left(  \xx_{\mathbf{i}_{1}}^{\alpha}%
\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha
}\right)  \right\vert .
\label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.at}%
\end{equation}
\par
But (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.comp}) (applied to
$\beta=\gamma$) yields $\left\vert \operatorname*{Supp}\left(  \xx%
^{\gamma}\right)  \right\vert =\ell\left(  \gamma\right)  >\left(  s-1\right)
\ell\left(  \alpha\right)  \geq\left\vert \operatorname*{Supp}\left(
\xx_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}%
\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)  \right\vert $ (by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2.pf.at})). Hence,
$\left\vert \operatorname*{Supp}\left(  \mathbf{x}^{\gamma}\right)
\right\vert \neq\left\vert \operatorname*{Supp}\left(  \xx%
_{\mathbf{i}_{1}}^{\alpha}\xx_{\mathbf{i}_{2}}^{\alpha}\cdots
\xx_{\mathbf{i}_{s}}^{\alpha}\right)  \right\vert $, so that
$\mathbf{x}^{\gamma}\neq\xx_{\mathbf{i}_{1}}^{\alpha}\xx%
_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}$. Hence,
$\mathbf{x}^{\gamma}$ and $\xx_{\mathbf{i}_{1}}^{\alpha}\xx%
_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}$ are two
distinct monomials. Therefore, $\left(  \text{the coefficient of the monomial
}\mathbf{x}^{\gamma}\text{ in }\xx_{\mathbf{i}_{1}}^{\alpha}%
\xx_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha
}\right)  =0$. This proves
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2}).}. Now, if $\gamma$ is a
composition satisfying $\left\vert \gamma\right\vert \neq s\left\vert
\alpha\right\vert $, then%
\begin{align*}
-c_{\gamma}  &  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2}%
,\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{,
}\ldots\text{, }\mathbf{i}_{s}\text{ are not distinct}}%
}\underbrace{\left(  \text{the coefficient of the monomial }\xx%
^{\gamma}\text{ in }\xx_{\mathbf{i}_{1}}^{\alpha}\xx%
_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)
}_{\substack{=0\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.c-eq.3})}\right) \\
&  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)
\right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are not distinct}}}0=0.
\end{align*}
Hence, if $\gamma$ is a composition satisfying $\left\vert \gamma\right\vert
\neq s\left\vert \alpha\right\vert $, then%
\begin{equation}
c_{\gamma}=0. \label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1a}%
\end{equation}
Also, if $\gamma$ is a composition satisfying $\ell\left(  \gamma\right)
>\left(  s-1\right)  \ell\left(  \alpha\right)  $, then%
\begin{align*}
-c_{\gamma}  &  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2}%
,\ldots,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(
\ell\right)  \right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{,
}\ldots\text{, }\mathbf{i}_{s}\text{ are not distinct}}%
}\underbrace{\left(  \text{the coefficient of the monomial }\xx%
^{\gamma}\text{ in }\xx_{\mathbf{i}_{1}}^{\alpha}\xx%
_{\mathbf{i}_{2}}^{\alpha}\cdots\xx_{\mathbf{i}_{s}}^{\alpha}\right)
}_{\substack{=0\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.c-eq.3})}\right) \\
&  =\sum\limits_{\substack{\left(  \mathbf{i}_{1},\mathbf{i}_{2},\ldots
,\mathbf{i}_{s}\right)  \in\left(  \operatorname*{SIS}\left(  \ell\right)
\right)  ^{s};\\\mathbf{i}_{1}\text{, }\mathbf{i}_{2}\text{, }\ldots\text{,
}\mathbf{i}_{s}\text{ are not distinct}}}0=0.
\end{align*}
Hence, if $\gamma$ is a composition satisfying $\ell\left(  \gamma\right)
>\left(  s-1\right)  \ell\left(  \alpha\right)  $, then%
\begin{equation}
c_{\gamma}=0. \label{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2a}%
\end{equation}
Now, (\ref{pf.lem.QSym.lyndon.Malphas1.proof2.c-eq}) becomes%
\begin{align*}
s!M_{\alpha}^{\left\langle s\right\rangle }-M_{\alpha}^{s}  &  =\sum_{\beta
\in \Comp }c_{\beta}M_{\beta}=\underbrace{\sum\limits_{\substack{\beta
\in \Comp ;\\\left\vert \beta\right\vert =s\left\vert
\alpha\right\vert }}}_{\substack{=\sum\limits_{\substack{\beta\in
\Comp_{s\left\vert \alpha\right\vert }}}\\\text{(since the elements }%
\beta\in \Comp \\\text{satisfying }\left\vert \beta\right\vert
=s\left\vert \alpha\right\vert \text{ are exactly}\\\text{the elements of
}\Comp_{s\left\vert \alpha\right\vert }\text{)}%
}}c_{\beta}M_{\beta}+\sum\limits_{\substack{\beta\in \Comp ;\\\left\vert
\beta\right\vert \neq s\left\vert \alpha\right\vert }}\underbrace{c_{\beta}%
}_{\substack{=0\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.1a}),}\\\text{applied to
}\gamma=\beta\text{)}}}M_{\beta}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\beta\in \Comp \text{ satisfies exactly one of
the two}\\
\text{statements }\left\vert \beta\right\vert =s\left\vert \alpha\right\vert
\text{ and }\left\vert \beta\right\vert \neq s\left\vert \alpha\right\vert
\end{array}
\right) \\
&  =\sum\limits_{\substack{\beta\in\Comp_{s\left\vert
\alpha\right\vert }}}c_{\beta}M_{\beta}+\underbrace{\sum\limits_{\substack{\beta
\in \Comp ;\\\left\vert \beta\right\vert \neq s\left\vert
\alpha\right\vert }}0M_{\beta}}_{=0}=\sum\limits_{\substack{\beta\in
\Comp_{s\left\vert \alpha\right\vert }}}c_{\beta
}M_{\beta}\\
&  =\sum\limits_{\substack{\beta\in\Comp_{s\left\vert
\alpha\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)
\ell\left(  \alpha\right)  }}c_{\beta}M_{\beta}+\sum\limits_{\substack{\beta
\in\Comp_{s\left\vert \alpha\right\vert }%
;\\\ell\left(  \beta\right)  >\left(  s-1\right)  \ell\left(  \alpha\right)
}}\underbrace{c_{\beta}}_{\substack{=0\\\text{(by
(\ref{pf.lem.QSym.lyndon.Malphas1.proof2.rule-out.2a}),}\\\text{applied to
}\gamma=\beta\text{)}}}M_{\beta}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\beta\in\Comp_{s\left\vert
\alpha\right\vert }\text{ satisfies exactly one of the two}\\
\text{statements }\ell\left(  \beta\right)  \leq\left(  s-1\right)
\ell\left(  \alpha\right)  \text{ and }\ell\left(  \beta\right)  >\left(
s-1\right)  \ell\left(  \alpha\right)
\end{array}
\right) \\
&  =\sum\limits_{\substack{\beta\in\Comp_{s\left\vert
\alpha\right\vert };\\\ell\left(  \beta\right)  \leq\left(  s-1\right)
\ell\left(  \alpha\right)  }}c_{\beta}M_{\beta}+\underbrace{\sum\limits
_{\substack{\beta\in\Comp_{s\left\vert \alpha
\right\vert };\\\ell\left(  \beta\right)  >\left(  s-1\right)  \ell\left(
\alpha\right)  }}0M_{\beta}}_{=0}=\sum\limits_{\substack{\beta\in
\Comp_{s\left\vert \alpha\right\vert };\\\ell\left(  \beta\right)
\leq\left(  s-1\right)  \ell\left(  \alpha\right)  }}\underbrace{c_{\beta}%
}_{\in \kk }M_{\beta}\in\sum\limits_{\substack{\beta\in
\Comp_{s\left\vert \alpha\right\vert };\\\ell\left(  \beta\right)
\leq\left(  s-1\right)  \ell\left(  \alpha\right)  }} \kk M_{\beta}.
\end{align*}
This proves Lemma \ref{lem.QSym.lyndon.Malphas1} once again.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}.]We know
that $u$ is a Lyndon word and satisfies $u=u$. Hence, $\left(  u\right)  $ is
the CFL factorization of the word $u$.

Define the notion $\operatorname{mult}_{w}z$ (for any Lyndon word
$w$ and any word $z$) as in Theorem \ref{thm.shuffle.lyndon.used}(b).

Theorem \ref{thm.shuffle.lyndon.used}(e) yields that the lexicographically
highest element of the multiset $u\shuffle v$ is $uv$, and the multiplicity
with which this word $uv$ appears in the multiset $u\shuffle v$ is
$\operatorname{mult}_{u}v+1$.

Recall that $\operatorname{mult}_{u}v$ is the number of terms in the
CFL factorization of $v$ which are equal to $u$ (by the definition of
$\operatorname{mult}_{u}v$). In other words,%
\begin{align*}
\operatorname{mult}_{u}v  &  =\left(  \text{the number of terms in
}\underbrace{\text{the CFL factorization of }v}_{=\left(  b_{1},b_{2}%
,\ldots,b_{q}\right)  }\text{ which are equal to }u\right) \\
&  =\left(  \text{the number of terms in }\left(  b_{1},b_{2},\ldots
,b_{q}\right)  \text{ which are equal to }u\right) \\
&  =\left(  \text{the number of }j\in\left\{  1,2,\ldots,q\right\}  \text{
satisfying }b_{j}=u\right)  =\left\vert \left\{  j\in\left\{  1,2,\ldots
,q\right\}  \ \mid\ b_{j}=u\right\}  \right\vert .
\end{align*}
Now, the multiplicity with which the word $uv$ appears in the multiset
$u\shuffle v$ is%
\begin{align*}
\underbrace{\operatorname{mult}_{u}v}_{=\left\vert \left\{  j\in\left\{
1,2,\ldots,q\right\}  \ \mid\ b_{j}=u\right\}  \right\vert }+1  &  =\left\vert
\left\{  j\in\left\{  1,2,\ldots,q\right\}  \ \mid\ b_{j}=u\right\}
\right\vert +1\\
&  =1+\left\vert \left\{  j\in\left\{  1,2,\ldots,q\right\}  \ \mid
\ b_{j}=u\right\}  \right\vert =h.
\end{align*}
Hence, we can apply Lemma \ref{lem.QSym.lyndon.triangl.wll-lemma.precise} to
$z=uv$. As a result, we conclude that
\[
M_{u}M_{v}=hM_{uv}+\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}uv\right)  .
\]
This proves Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}.]Notice
that $\left\vert x\right\vert =k$ (since $x\in
\Comp_{k}$), thus $\left\vert x^{s}\right\vert =s\underbrace{\left\vert
x\right\vert }_{=k}=sk$ and thus $x^{s}\in\Comp_{sk}$.

Clearly, $\left(  x\right)  $ is the CFL factorization of the word $x$ (since
$x$ is Lyndon). On the other hand, $\left(  \underbrace{x,x,\ldots,x}_{s\text{
times}}\right)  $ is the CFL factorization of the word $x^{s}$ (since $\left(
\underbrace{x,x,\ldots,x}_{s\text{ times}}\right)  $ is a tuple of Lyndon
words (since $x$ is Lyndon) satisfying $x^{s}=\underbrace{xx\cdots x}_{s\text{
times}}$ and $x\geq x\geq\cdots\geq x$). Hence, Theorem
\ref{thm.shuffle.lyndon.used}(c) (applied to $u=x$, $v=x^{s}$, $p=1$, $q=s$,
$\left(  a_{1},a_{2},\ldots,a_{p}\right)  =\left(  x\right)  $ and $\left(
b_{1},b_{2},\ldots,b_{q}\right)  =\left(  \underbrace{x,x,\ldots,x}_{s\text{
times}}\right)  $) yields that the lexicographically highest element of the
multiset $x\shuffle x^{s}$ is $xx^{s}$ (since $x\geq x$ for every
$i\in\left\{  1,2,\ldots,1\right\}  $ and $j\in\left\{  1,2,\ldots,s\right\}
$). In other words, the lexicographically highest element of the multiset
$x\shuffle x^{s}$ is $x^{s+1}$ (since $xx^{s}=x^{s+1}$). This proves Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}(a).

(c) Let $t\in\Comp_{sk}$ be such that
$t\underset{\operatorname*{wll}}{<}x^{s}$. Then, Lemma
\ref{lem.QSym.lyndon.triang.wll-lemma}(c) (applied to $n=k$, $m=sk$, $u=x$,
$v=x^{s}$, $z=x^{s+1}$ and $v^{\prime}=t$) yields%
\begin{align*}
M_{x}M_{t}  &  =\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\underbrace{\Comp_{k+sk}}%
_{\substack{=\Comp_{\left(  s+1\right)  k}%
\\\text{(since }k+sk=\left(  s+1\right)  k\text{)}}}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}x^{s+1}\right) \\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{\left(  s+1\right)  k}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}x^{s+1}\right)  .
\end{align*}
This proves Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}(c).

(b) Let $h=1+\left\vert \left\{  j\in\left\{  1,2,\ldots,s\right\}
\ \mid\ x=x\right\}  \right\vert $. Of course, $h=s+1$%
\ \ \ \ \footnote{\textit{Proof.} Every $j\in\left\{  1,2,\ldots,s\right\}  $
satisfies $x=x$. Hence, the set $\left\{  j\in\left\{  1,2,\ldots,s\right\}
\ \mid\ x=x\right\}  $ equals the whole $\left\{  1,2,\ldots,s\right\}  $. In
other words, $\left\{  j\in\left\{  1,2,\ldots,s\right\}  \ \mid\ x=x\right\}
=\left\{  1,2,\ldots,s\right\}  $. Now, $h=1+\left\vert \underbrace{\left\{
j\in\left\{  1,2,\ldots,s\right\}  \ \mid\ x=x\right\}  }_{=\left\{
1,2,\ldots,s\right\}  }\right\vert =1+\underbrace{\left\vert \left\{
1,2,\ldots,s\right\}  \right\vert }_{=s}=1+s=s+1$, qed.}. Recall that $x$ is a
Lyndon word, and that $\left(  \underbrace{x,x,\ldots,x}_{s\text{ times}%
}\right)  $ is the CFL factorization of the word $x^{s}$. Notice also that
$x\geq x$ for every $j\in\left\{  1,2,\ldots,s\right\}  $. Thus, Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-cfl} (applied to $n=k$, $m=sk$,
$u=x$, $v=x^{s}$, $q=s$ and $\left(  b_{1},b_{2},\ldots,b_{q}\right)  =\left(
\underbrace{x,x,\ldots,x}_{s\text{ times}}\right)  $) yields that%
\begin{align*}
M_{x}M_{x^{s}}  &  =\underbrace{h}_{=s+1}\underbrace{M_{xx^{s}}}%
_{\substack{=M_{x^{s+1}}\\\text{(since }xx^{s}=x^{s+1}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\underbrace{\Comp_{k+sk}}%
_{\substack{=\Comp_{\left(  s+1\right)  k}%
\\\text{(since }k+sk=\left(  s+1\right)  k\text{)}}}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}\underbrace{xx^{s}}_{=x^{s+1}}\right) \\
&  =\left(  s+1\right)  M_{x^{s+1}}+\left(  \text{a sum of terms of the form
}M_{w}\text{ with }w\in\Comp_{\left(  s+1\right)
k}\text{ satisfying }w\underset{\operatorname*{wll}}{<}x^{s+1}\right)  .
\end{align*}
This proves Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}(b).
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}.]We
assumed that $a_{i}>b_{j}$ for every $i\in\left\{  1,2,\ldots,p\right\}  $ and
$j\in\left\{  1,2,\ldots,q\right\}  $. Thus, $a_{i}\geq b_{j}$ for every
$i\in\left\{  1,2,\ldots,p\right\}  $ and $j\in\left\{  1,2,\ldots,q\right\}
$. Hence, the lexicographically highest element of the multiset $u\shuffle v$
is $uv$ (by Theorem \ref{thm.shuffle.lyndon.used}(c)). Also, the multiplicity
with which the word $uv$ appears in the multiset $u\shuffle v$ is $1$ (by
Theorem \ref{thm.shuffle.lyndon.used}(d)). Hence, Lemma
\ref{lem.QSym.lyndon.triangl.wll-lemma.precise} (applied to $z=uv$ and $h=1$)
yields%
\begin{align*}
M_{u}M_{v}  &  =\underbrace{1M_{uv}}_{=M_{uv}}+\left(  \text{a sum of terms of
the form }M_{w}\text{ with }w\in\Comp_{n+m}\text{
satisfying }w\underset{\operatorname*{wll}}{<}uv\right) \\
&  =M_{uv}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n+m}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}uv\right)  .
\end{align*}
This proves Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.triangl.wll-lemma.BC}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.triangl.wll-lemma.BC}.}
\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.BC}.]The tuple
$\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is the CFL factorization of $u$.
Thus, $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is a tuple of Lyndon words
satisfying $u=a_{1}a_{2}\cdots a_{p}$ and $a_{1}\geq a_{2}\geq\cdots\geq
a_{p}$.

Now, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words
(since $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is a tuple of Lyndon words)
satisfying $x=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq
a_{k}$ (since $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$). In other words, $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of $x$ (by the
definition of a CFL factorization).

Also, $\left(  a_{k+1},a_{k+2},\ldots,a_{p}\right)  $ is a tuple of Lyndon
words (since $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is a tuple of Lyndon
words) satisfying $y=a_{k+1}a_{k+2}\cdots a_{p}$ and $a_{k+1}\geq a_{k+2}%
\geq\cdots\geq a_{p}$ (since $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$). In other
words, $\left(  a_{k+1},a_{k+2},\ldots,a_{p}\right)  $ is the CFL
factorization of $y$ (by the definition of a CFL factorization).

We have $x\in\Comp_{\left\vert x\right\vert }$ and
$y\in\Comp_{\left\vert y\right\vert }$. Also,
multiplying the equalities $x=a_{1}a_{2}\cdots a_{k}$ and $y=a_{k+1}%
a_{k+2}\cdots a_{p}$, we obtain $xy=\left(  a_{1}a_{2}\cdots a_{k}\right)
\left(  a_{k+1}a_{k+2}\cdots a_{p}\right)  =a_{1}a_{2}\cdots a_{p}=u$. Thus,
$\left\vert x\right\vert +\left\vert y\right\vert =\left\vert \underbrace{xy}%
_{=u}\right\vert =\left\vert u\right\vert =n$ (since $u\in
\Comp_{n}$).

We have $a_{i}>a_{k+j}$ for every $i\in\left\{  1,2,\ldots,k\right\}  $ and
$j\in\left\{  1,2,\ldots,p-k\right\}  $\ \ \ \ \footnote{\textit{Proof.}
Recall that $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$. Thus,%
\begin{equation}
a_{s}\geq a_{t}\ \ \ \ \ \ \ \ \ \ \text{for any }s\in\left\{  1,2,\ldots
,p\right\}  \text{ and }t\in\left\{  1,2,\ldots,p\right\}  \text{ satisfying
}s\leq t. \label{pf.cor.QSym.lyndon.triangl.wll-lemma.BC.1}%
\end{equation}
\par
Let $i\in\left\{  1,2,\ldots,k\right\}  $ and $j\in\left\{  1,2,\ldots
,p-k\right\}  $. Then, $i\leq k$ (since $i\in\left\{  1,2,\ldots,k\right\}  $)
and thus $a_{i}\geq a_{k}$ (by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.BC.1}), applied to $s=i$ and
$t=k$). Also, $j\geq1$ (since $j\in\left\{  1,2,\ldots,p-k\right\}  $), and
thus $k+\underbrace{j}_{\geq1}\geq k+1$, so that $k+1\leq k+j$. Hence,
$a_{k+1}\geq a_{k+j}$ (by (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.BC.1}),
applied to $s=k+1$ and $t=k+j$). Now, $a_{i}\geq a_{k}>a_{k+1}\geq a_{k+j}$,
qed.}. Hence, Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.cfl-cfl.strict}
(applied to $x$, $y$, $\left\vert x\right\vert $, $\left\vert y\right\vert $,
$k$, $p-k$, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ and $\left(
a_{k+1},a_{k+2},\ldots,a_{p}\right)  $ instead of $u$, $v$, $n$, $m$, $p$,
$q$, $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ and $\left(  b_{1}%
,b_{2},\ldots,b_{q}\right)  $) yields%
\begin{align*}
M_{x}M_{y}  &  =\underbrace{M_{xy}}_{\substack{=M_{u}\\\text{(since
}xy=u\text{)}}}+\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\underbrace{\Comp_{\left\vert x\right\vert
+\left\vert y\right\vert }}_{\substack{=\Comp_{n}%
\\\text{(since }\left\vert x\right\vert +\left\vert y\right\vert =n\text{)}%
}}\text{ satisfying }w\underset{\operatorname*{wll}}{<}\underbrace{xy}%
_{=u}\right) \\
&  =M_{u}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{n}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}u\right)  .
\end{align*}
Thus,%
\[
M_{u}=M_{x}M_{y}-\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{n}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}u\right)  .
\]
This proves Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.BC}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.triangl.wll-lemma.As}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.triangl.wll-lemma.As}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As}.]We shall prove
Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As} by induction over $s$:

\textit{Induction base:} We have $x^{0}=\varnothing$ and therefore $M_{x^{0}%
}=M_{\varnothing}=1$. Now, $\underbrace{M_{x}^{0}}_{=1}-\underbrace{0!}%
_{=1}\underbrace{M_{x^{0}}}_{=1}=1-1=0\in\sum\limits_{\substack{w\in
\Comp_{0k};\\w\underset{\operatorname*{wll}}{<}x^{0}%
}} \kk M_{w}$. In other words, Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As} holds for $s=0$. This completes the
induction base.

\textit{Induction step:} Let $\mathbf{S}$ be a nonnegative integer. Assume
that Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As} holds for
$s=\mathbf{S}$. We need to prove that Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As} holds for $s=\mathbf{S}+1$.

Notice that $x\in\Comp_{k}$, thus $\left\vert
x\right\vert =k$, and thus $\left\vert x^{\mathbf{S}}\right\vert
=\mathbf{S}\underbrace{\left\vert x\right\vert }_{=k}=\mathbf{S}k$, so that
$x^{\mathbf{S}}\in\Comp_{\mathbf{S}k}$.

We know that Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As} holds for
$s=\mathbf{S}$. In other words,%
\begin{equation}
M_{x}^{\mathbf{S}}-\mathbf{S}!M_{x^{\mathbf{S}}}\in\sum\limits_{\substack{w\in
\Comp_{\mathbf{S}k};\\w\underset{\operatorname*{wll}%
}{<}x^{\mathbf{S}}}} \kk M_{w}=\sum\limits_{\substack{t\in
\Comp_{\mathbf{S}k};\\t\underset{\operatorname*{wll}}{<}x^{\mathbf{S}}%
}} \kk M_{t} \label{pf.cor.QSym.lyndon.triangl.wll-lemma.As.indhyp}%
\end{equation}
(here, we renamed the summation index $w$ as $t$).

Using Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}(c), it is easy
to see that%
\begin{equation}
M_{x}M_{t}\in\sum\limits_{\substack{w\in\Comp_{\left(
\mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}%
}} \kk M_{w}\ \ \ \ \ \ \ \ \ \ \text{for every }t\in
\Comp_{\mathbf{S}k}\text{ satisfying }t\underset{\operatorname*{wll}%
}{<}x^{\mathbf{S}}. \label{pf.cor.QSym.lyndon.triangl.wll-lemma.As.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As.2}):}
Let $t\in\Comp_{\mathbf{S}k}$ be such that
$t\underset{\operatorname*{wll}}{<}x^{\mathbf{S}}$. Then, Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}(c) (applied to $s=\mathbf{S}$)
yields%
\begin{align*}
M_{x}M_{t}  &  =\left(  \text{a sum of terms of the form }M_{w}\text{ with
}w\in\Comp_{\left(  \mathbf{S}+1\right)  k}\text{
satisfying }w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}\right) \\
&  \in\sum\limits_{\substack{w\in\Comp_{\left(
\mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}%
}} \kk M_{w}.
\end{align*}
This proves (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As.2}).} Now,%
\[
M_{x}\left(  M_{x}^{\mathbf{S}}-\mathbf{S}!M_{x^{\mathbf{S}}}\right)
=\underbrace{M_{x}M_{x}^{\mathbf{S}}}_{=M_{x}^{\mathbf{S}+1}}-\mathbf{S}%
!M_{x}M_{x^{\mathbf{S}}}=M_{x}^{\mathbf{S}+1}-\mathbf{S}!M_{x}M_{x^{\mathbf{S}%
}},
\]
so that%
\begin{align}
M_{x}^{\mathbf{S}+1}-\mathbf{S}!M_{x}M_{x^{\mathbf{S}}}  &  =M_{x}%
\underbrace{\left(  M_{x}^{\mathbf{S}}-\mathbf{S}!M_{x^{\mathbf{S}}}\right)
}_{\substack{\in\sum\limits_{\substack{t\in\Comp_{\mathbf{S}%
k};\\t\underset{\operatorname*{wll}}{<}x^{\mathbf{S}}}} \kk %
M_{t}\\\text{(by (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As.indhyp}))}}}\in
M_{x}\left(  \sum\limits_{\substack{t\in\Comp_{\mathbf{S}%
k};\\t\underset{\operatorname*{wll}}{<}x^{\mathbf{S}}}} \kk M_{t}\right)
\nonumber\\
&  =\sum\limits_{\substack{t\in\Comp_{\mathbf{S}%
k};\\t\underset{\operatorname*{wll}}{<}x^{\mathbf{S}}}} \kk %
\underbrace{M_{x}M_{t}}_{\substack{\in\sum\limits_{\substack{w\in
\Comp_{\left(  \mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}%
}{<}x^{\mathbf{S}+1}}} \kk M_{w}\\\text{(by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As.2}))}}}\subset\sum\limits
_{\substack{t\in\Comp_{\mathbf{S}k}%
;\\t\underset{\operatorname*{wll}}{<}x^{\mathbf{S}}}} \kk \left(
\sum\limits_{\substack{w\in\Comp_{\left(  \mathbf{S}%
+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}}%
} \kk M_{w}\right) \nonumber\\
&  \subset\sum\limits_{\substack{w\in\Comp_{\left(
\mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}%
}} \kk M_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sum\limits_{\substack{w\in
\Comp_{\left(  \mathbf{S}+1\right)  k}%
;\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}}} \kk M_{w}\text{
is a } \kk \text{-module}\right)  .
\label{pf.cor.QSym.lyndon.triangl.wll-lemma.As.4}%
\end{align}


Now, Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.lyn-xs}(b) (applied to
$s=\mathbf{S}$) yields%
\[
M_{x}M_{x^{\mathbf{S}}}=\left(  \mathbf{S}+1\right)  M_{x^{\mathbf{S}+1}%
}+\left(  \text{a sum of terms of the form }M_{w}\text{ with }w\in
\Comp_{\left(  \mathbf{S}+1\right)  k}\text{
satisfying }w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}\right)  .
\]
Thus,%
\begin{align}
&  M_{x}M_{x^{\mathbf{S}}}-\left(  \mathbf{S}+1\right)  M_{x^{\mathbf{S}+1}%
}\nonumber\\
&  =\left(  \text{a sum of terms of the form }M_{w}\text{ with }%
w\in\Comp_{\left(  \mathbf{S}+1\right)  k}\text{
satisfying }w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}\right)
\nonumber\\
&  \in\sum\limits_{\substack{w\in\Comp_{\left(
\mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}%
}} \kk M_{w}. \label{pf.cor.QSym.lyndon.triangl.wll-lemma.As.6}%
\end{align}
Now,%
\begin{align*}
M_{x}^{\mathbf{S}+1}-\left(  \mathbf{S}+1\right)  !M_{x^{\mathbf{S}+1}}  &
=\left(  M_{x}^{\mathbf{S}+1}-\mathbf{S}!M_{x}M_{x^{\mathbf{S}}}\right)
+\left(  \mathbf{S}!M_{x}M_{x^{\mathbf{S}}}-\underbrace{\left(  \mathbf{S}%
+1\right)  !}_{=\mathbf{S}!\cdot\left(  \mathbf{S}+1\right)  }M_{x^{\mathbf{S}%
+1}}\right) \\
&  =\left(  M_{x}^{\mathbf{S}+1}-\mathbf{S}!M_{x}M_{x^{\mathbf{S}}}\right)
+\underbrace{\left(  \mathbf{S}!M_{x}M_{x^{\mathbf{S}}}-\mathbf{S}%
!\cdot\left(  \mathbf{S}+1\right)  M_{x^{\mathbf{S}+1}}\right)  }%
_{=\mathbf{S}!\cdot\left(  M_{x}M_{x^{\mathbf{S}}}-\left(  \mathbf{S}%
+1\right)  M_{x^{\mathbf{S}+1}}\right)  }\\
&  =\underbrace{\left(  M_{x}^{\mathbf{S}+1}-\mathbf{S}!M_{x}M_{x^{\mathbf{S}%
}}\right)  }_{\substack{\in\sum\limits_{\substack{w\in
\Comp_{\left(  \mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}%
}{<}x^{\mathbf{S}+1}}} \kk M_{w}\\\text{(by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As.4}))}}}+\mathbf{S}%
!\cdot\underbrace{\left(  M_{x}M_{x^{\mathbf{S}}}-\left(  \mathbf{S}+1\right)
M_{x^{\mathbf{S}+1}}\right)  }_{\substack{\in\sum\limits_{\substack{w\in
\Comp_{\left(  \mathbf{S}+1\right)  k}%
;\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}}} \kk %
M_{w}\\\text{(by (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As.6}))}}}\\
&  \in\sum\limits_{\substack{w\in\Comp_{\left(
\mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}%
}} \kk M_{w}+\mathbf{S}!\cdot\sum\limits_{\substack{w\in
\Comp_{\left(  \mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}%
}{<}x^{\mathbf{S}+1}}} \kk M_{w}\subset\sum\limits_{\substack{w\in
\Comp_{\left(  \mathbf{S}+1\right)  k}%
;\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}}} \kk M_{w}%
\end{align*}
(since $\sum\limits_{\substack{w\in\Comp_{\left(
\mathbf{S}+1\right)  k};\\w\underset{\operatorname*{wll}}{<}x^{\mathbf{S}+1}%
}} \kk M_{w}$ is a $\kk$-module). In other words, Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As} holds for $s=\mathbf{S}+1$. This
completes the induction step. Thus, Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As} is proven by induction.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.triangl.wll-lemma.As2}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.triangl.wll-lemma.As2}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As2}.]We first
assume that $\kk = \ZZ$.

Notice that $\left\vert x\right\vert =k$ (since $x\in
\Comp_{k}$). Also, $M_{x}^{\left\langle s\right\rangle }\in
\Qsym$ (by Corollary \ref{cor.QSym.Malphas.QSym}(a), applied to
$\alpha=x$) and thus $\underbrace{M_{x}^{\left\langle s\right\rangle }}%
_{\in\Qsym}-\underbrace{M_{x^{s}}}_{\in\Qsym}%
\in\Qsym-\Qsym\subset\Qsym$.

Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As} yields%
\begin{equation}
M_{x}^{s}-s!M_{x^{s}}\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}.
\label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.1}%
\end{equation}
On the other hand, the composition $x$ is Lyndon and therefore nonempty.
Hence, \eqref{eq.lem.QSym.lyndon.Malphas1} (applied to $\alpha=x$) yields%
\[
s!M_{x}^{\left\langle s\right\rangle }-M_{x}^{s}\in\underbrace{\sum\limits
_{\substack{\beta\in\Comp_{s\left\vert x\right\vert
};\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell\left(  x\right)
}}}_{\substack{=\sum\limits_{\substack{\beta\in\Comp_{sk}%
;\\\ell\left(  \beta\right)  \leq\left(  s-1\right)  \ell\left(  x\right)
}}\\\text{(since }\left\vert x\right\vert =k\text{)}}} \kk M_{\beta}%
=\sum\limits_{\substack{\beta\in\Comp_{sk};\\\ell\left(
\beta\right)  \leq\left(  s-1\right)  \ell\left(  x\right)  }} \kk %
M_{\beta}=\sum\limits_{\substack{w\in\Comp_{sk};\\\ell\left(
w\right)  \leq\left(  s-1\right)  \ell\left(  x\right)  }} \kk M_{w}%
\]
(here, we renamed the summation index $\beta$ as $w$). But every
$w\in\Comp_{sk}$ satisfying $\ell\left(  w\right)
\leq\left(  s-1\right)  \ell\left(  x\right)  $ must also satisfy
$w\underset{\operatorname*{wll}}{<}x^{s}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in\Comp_{sk}$. Notice that $\left\vert
x^{s}\right\vert =s\underbrace{\left\vert x\right\vert }_{=k}=sk$, so that
$x^{s}\in\Comp_{sk}$. Now, we have $\ell\left(
x\right)  >0$ (since $x$ is nonempty), and%
\[
\ell\left(  x^{s}\right)  =s\ell\left(  x\right)  =\left(  s-1\right)
\ell\left(  x\right)  +\underbrace{\ell\left(  x\right)  }_{>0}>\left(
s-1\right)  \ell\left(  x\right)  .
\]
Thus, $\left(  s-1\right)  \ell\left(  x\right)  <\ell\left(  x^{s}\right)  $,
so that $\ell\left(  w\right)  \leq\left(  s-1\right)  \ell\left(  x\right)
<\ell\left(  x^{s}\right)  $.
\par
But the definition of the wll-order shows that if $n\in \NN$, and if
$\alpha$ and $\beta$ are two elements of $\Comp_{n}$
satisfying $\ell\left(  \alpha\right)  <\ell\left(  \beta\right)  $, then
$\alpha\underset{\operatorname*{wll}}{<}\beta$. Applying this to $\alpha=w$
and $\beta=x^{s}$, we obtain $w\underset{\operatorname*{wll}}{<}x^{s}$ (since
$\ell\left(  w\right)  <\ell\left(  x^{s}\right)  $), qed.}. Hence, the sum
$\sum\limits_{\substack{w\in\Comp_{sk};\\\ell\left(
w\right)  \leq\left(  s-1\right)  \ell\left(  x\right)  }} \kk M_{w}$ is
a subsum of the sum $\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$. Thus,
$\sum\limits_{\substack{w\in\Comp_{sk};\\\ell\left(
w\right)  \leq\left(  s-1\right)  \ell\left(  x\right)  }} \kk %
M_{w}\subset\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$. Thus,%
\begin{equation}
s!M_{x}^{\left\langle s\right\rangle }-M_{x}^{s}\in\sum\limits_{\substack{w\in
\Comp_{sk};\\\ell\left(  w\right)  \leq\left(
s-1\right)  \ell\left(  x\right)  }} \kk M_{w}\subset\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}} \kk M_{w}. \label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.3}%
\end{equation}
Now,%
\begin{align}
s!\left(  M_{x}^{\left\langle s\right\rangle }-M_{x^{s}}\right)   &
=s!M_{x}^{\left\langle s\right\rangle }-s!M_{x^{s}}=\underbrace{\left(
s!M_{x}^{\left\langle s\right\rangle }-M_{x}^{s}\right)  }_{\substack{\in
\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}\\\text{(by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.3}))}}}+\underbrace{\left(
M_{x}^{s}-s!M_{x^{s}}\right)  }_{\substack{\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}} \kk M_{w}\\\text{(by (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.1}%
))}}}\nonumber\\
&  \in\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}+\sum\limits
_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}\subset
\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}
\label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.5}%
\end{align}
(since $\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$ is a $\kk$-module).

But recall that we assumed that $\kk = \ZZ$. Hence, if $N$ is a
positive integer and if $f$ is an element of $\Qsym$ satisfying
$Nf\in\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$, then $f\in
\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$%
\ \ \ \ \footnote{\textit{Proof.} Let $N$ be a positive integer. Let $f$ be an
element of $\Qsym$ satisfying $Nf\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}} \kk M_{w}$.
\par
We know that $\left(  M_{\alpha}\right)  _{\alpha\in \Comp }$ is
a basis of the $\kk$-module $\Qsym$. In the following,
whenever $g\in\Qsym$ and $\beta\in \Comp $, we
will let $\operatorname*{coord}\nolimits_{M_{\beta}}g$ denote the $M_{\beta}%
$-coordinate of $g$ with respect to the basis $\left(  M_{\alpha}\right)
_{\alpha\in \Comp }$ of $\Qsym$. Then, every
$g\in\Qsym$ satisfies%
\begin{equation}
g=\sum_{\beta\in \Comp }\left(  \operatorname*{coord}%
\nolimits_{M_{\beta}}g\right)  M_{\beta}
\label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.coord}%
\end{equation}
(by the definition of coordinates). Notice also that%
\begin{equation}
\operatorname*{coord}\nolimits_{M_{\beta}}\left(  M_{\gamma}\right)
=\delta_{\beta,\gamma}\ \ \ \ \ \ \ \ \ \ \text{for every compositions }%
\beta\text{ and }\gamma\text{.}
\label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.coord-delta}%
\end{equation}
\par
Now, $Nf\in\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$. Thus, there
exists a family $\left(  \lambda_{w}\right)  _{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}}\in \kk ^{\left\{  w\in\Comp_{sk}%
\ \mid\ w\underset{\operatorname*{wll}}{<}x^{s}\right\}  }$ of elements of
$\kk$ satisfying $Nf=\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}}}\lambda_{w}M_{w}$.
Consider this family $\left(  \lambda_{w}\right)  _{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}}$.
\par
Let $\beta$ be a composition such that we don't have $\left(  \beta
\in\Comp_{sk}\text{ and }\beta
\underset{\operatorname*{wll}}{<}x^{s}\right)  $. Then,
\begin{equation}
\text{every composition }w\in\Comp_{sk}\text{
satisfying }w\underset{\operatorname*{wll}}{<}x^{s}\text{ satisfies }\beta\neq
w \label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.neq}%
\end{equation}
(since every such $w$ satisfies $\left(  w\in
\Comp_{sk}\text{ and }w\underset{\operatorname*{wll}}{<}x^{s}\right)  $,
whereas $\beta$ does not satisfy $\left(  \beta\in
\Comp_{sk}\text{ and }\beta\underset{\operatorname*{wll}}{<}x^{s}\right)
$). Now,%
\begin{align*}
\operatorname*{coord}\nolimits_{M_{\beta}}\left(  \underbrace{Nf}%
_{=\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}}\lambda_{w}M_{w}}\right)   &
=\operatorname*{coord}\nolimits_{M_{\beta}}\left(  \sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}}\lambda_{w}M_{w}\right)  =\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}}}\lambda
_{w}\underbrace{\operatorname*{coord}\nolimits_{M_{\beta}}\left(
M_{w}\right)  }_{\substack{=\delta_{\beta,w}\\\text{(by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.coord-delta}), applied}%
\\\text{to }\gamma=w\text{)}}}\\
&  =\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}}\lambda_{w}\underbrace{\delta
_{\beta,w}}_{\substack{=0\\\text{(since }\beta\neq w\\\text{(by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.neq})))}}}=\sum\limits
_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}}\lambda_{w}0=0.
\end{align*}
Compared with $\operatorname*{coord}\nolimits_{M_{\beta}}\left(  Nf\right)
=N\operatorname*{coord}\nolimits_{M_{\beta}}f$, this yields
$N\operatorname*{coord}\nolimits_{M_{\beta}}f=0$.
\par
But $N$ is a positive integer. Thus, if an element $\rho$ of $\ZZ$
satisfies $N\rho=0$, then $\rho=0$. Applying this to $\rho
=\operatorname*{coord}\nolimits_{M_{\beta}}f$, we obtain
$\operatorname*{coord}\nolimits_{M_{\beta}}f=0$ (since $\operatorname*{coord}%
\nolimits_{M_{\beta}}f\in \kk = \ZZ$ and $N\operatorname*{coord}%
\nolimits_{M_{\beta}}f=0$).
\par
Now, let us forget that we fixed $\beta$. We thus have shown that if $\beta$
is a composition such that we don't have $\left(  \beta\in
\Comp_{sk}\text{ and }\beta\underset{\operatorname*{wll}}{<}x^{s}\right)
$, then%
\begin{equation}
\operatorname*{coord}\nolimits_{M_{\beta}}f=0.
\label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.=0}%
\end{equation}
\par
Now, (\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.coord}) (applied to
$g=f$) yields%
\begin{align*}
f  &  =\sum_{\beta\in \Comp }\left(  \operatorname*{coord}%
\nolimits_{M_{\beta}}f\right)  M_{\beta}=\underbrace{\sum\limits_{\substack{\beta
\in \Comp ;\\\beta\in\Comp_{sk}\text{ and
}\beta\underset{\operatorname*{wll}}{<}x^{s}}}}_{\substack{=\sum\limits
_{\substack{\beta\in\Comp_{sk};\\\beta
\underset{\operatorname*{wll}}{<}x^{s}}}\\\text{(since every }\beta
\in\Comp_{sk}\\\text{satisfies }\beta\in
 \Comp \text{)}}}\left(  \operatorname*{coord}\nolimits_{M_{\beta
}}f\right)  M_{\beta}+\sum\limits_{\substack{\beta\in \Comp ;\\\text{we
don't have }\left(  \beta\in\Comp_{sk}\text{ and
}\beta\underset{\operatorname*{wll}}{<}x^{s}\right)  }}\underbrace{\left(
\operatorname*{coord}\nolimits_{M_{\beta}}f\right)  }_{\substack{=0\\\text{(by
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.6.=0}))}}}M_{\beta}\\
&  =\sum\limits_{\substack{\beta\in\Comp_{sk};\\\beta
\underset{\operatorname*{wll}}{<}x^{s}}}\left(  \operatorname*{coord}%
\nolimits_{M_{\beta}}f\right)  M_{\beta}+\underbrace{\sum\limits_{\substack{\beta
\in \Comp ;\\\text{we don't have }\left(  \beta\in
\Comp_{sk}\text{ and }\beta
\underset{\operatorname*{wll}}{<}x^{s}\right)  }}0M_{\beta}}_{=0}%
=\sum\limits_{\substack{\beta\in\Comp_{sk};\\\beta
\underset{\operatorname*{wll}}{<}x^{s}}}\left(  \operatorname*{coord}%
\nolimits_{M_{\beta}}f\right)  M_{\beta}\\
&  =\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}}\underbrace{\left(
\operatorname*{coord}\nolimits_{M_{w}}f\right)  }_{\in \kk }%
M_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation index
}\beta\text{ as }w\right) \\
&  \in\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w},
\end{align*}
qed.}. Applying this to $N=s!$ and $f=M_{x}^{\left\langle s\right\rangle
}-M_{x^{s}}$, we obtain $M_{x}^{\left\langle s\right\rangle }-M_{x^{s}}\in
\sum\limits_{\substack{w\in\Comp_{sk}%
;\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}$ (because of
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.5})). In other words, Corollary
\ref{cor.QSym.lyndon.triangl.wll-lemma.As2} holds under the assumption that
$\kk = \ZZ$.

Now, let us forget that we have assumed that $\kk = \ZZ$. We thus
have shown that Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As2} holds
under the assumption that $\kk = \ZZ$. In other words, we have
shown that the relation%
\begin{equation}
M_{x}^{\left\langle s\right\rangle }-M_{x^{s}}\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}} \ZZ M_{w} \label{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.Z-goal}%
\end{equation}
holds in $ \ZZ \left[  \left[  \xx\right]  \right]  $.

Now, recall that there is a canonical ring homomorphism $\varphi
: \ZZ \rightarrow \kk$. This homomorphism gives rise to a ring
homomorphism $\varphi\left[  \left[  \xx\right]  \right]
: \ZZ \left[  \left[  \xx\right]  \right]  \rightarrow
 \kk \left[  \left[  \xx\right]  \right]  $, and this latter
homomorphism $\varphi\left[  \left[  \xx\right]  \right]  $ has the
properties that:

\begin{itemize}
\item it is $\ZZ$-linear;

\item it sends the element $M_{\alpha}$ of $ \ZZ \left[  \left[
\xx\right]  \right]  $ to the element $M_{\alpha}$ of $\kk %
\left[  \left[  \xx\right]  \right]  $ for every composition $\alpha$;

\item it sends the element $M_{\alpha}^{\left\langle s\right\rangle }$ of
$ \ZZ \left[  \left[  \xx\right]  \right]  $ to the element
$M_{\alpha}^{\left\langle s\right\rangle }$ of $\kk \left[  \left[
\xx\right]  \right]  $ for every composition $\alpha$ and every
nonnegative integer $s$.
\end{itemize}

Hence, by applying the homomorphism $\varphi\left[  \left[  \xx\right]
\right]  $ to both sides of
(\ref{pf.cor.QSym.lyndon.triangl.wll-lemma.As2.Z-goal}), we obtain%
\[
M_{x}^{\left\langle s\right\rangle }-M_{x^{s}}\in\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}%
}}\underbrace{ \ZZ M_{w}}_{\substack{\subset \kk M_{w}\\\text{(due
to the canonical}\\\text{ring homomorphism } \ZZ \rightarrow
 \kk \text{)}}}\subset\sum\limits_{\substack{w\in
\Comp_{sk};\\w\underset{\operatorname*{wll}}{<}x^{s}}} \kk M_{w}.
\]
This proves Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As2}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.thm.QSym.lyndon.Z-genset}}

\textit{Solution to Exercise \ref{exe.thm.QSym.lyndon.Z-genset}.}

\begin{proof}
[Proof of Theorem \ref{thm.QSym.lyndon.Z-genset}.]The family $\left(
M_{w}^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a reindexing of the
family $\left(  M_{\operatorname*{red}\alpha}^{\left\langle \gcd
\alpha\right\rangle }\right)  _{\alpha\in\mathfrak{L}}$ (according to Lemma
\ref{lem.QSym.lyndon.Z-genset.reindex}). In other words, the family $\left(
M_{w}^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a reindexing of the
family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$ (here, we renamed the index $\alpha$ as $w$).

We need to show that the family $\left(  M_{w}^{\left\langle s\right\rangle
}\right)  _{\left(  w,s\right)  \in\mathfrak{RL}\times\left\{  1,2,3,\ldots
\right\}  }$ is an algebraically independent generating set of the
$\kk$-algebra $\Qsym$. It is enough to prove that the
family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent generating set
of the $\kk$-algebra $\Qsym$ (since the family $\left(
M_{w}^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)
\in\mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$ is a reindexing of the
family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$). We shall prove the latter claim.

Indeed, the main difficulty is to show that
\begin{equation}
\text{the family }\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}\text{ generates the }%
 \kk \text{-algebra }\Qsym\text{.}
\label{pf.thm.QSym.lyndon.Z-genset.maingoal}%
\end{equation}


Once (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal}) is proven, we will be able
to complete the proof of Theorem \ref{thm.QSym.lyndon.Z-genset} as follows:

For every $w\in\mathfrak{L}$, we have $M_{\operatorname*{red}w}^{\left\langle
\gcd w\right\rangle }\in\Qsym$\ \ \ \ \footnote{\textit{Proof.}
Let $w\in\mathfrak{L}$. Then, Corollary \ref{cor.QSym.Malphas.QSym}(a)
(applied to $\gcd w$ and $\operatorname*{red}w$ instead of $s$ and $w$) yields
$M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }\in
\Qsym$, qed.}.

Let $\operatorname*{wt}:\mathfrak{A}\rightarrow\left\{  1,2,3,\ldots\right\}
$ be the identity map (this is well-defined since $\mathfrak{A}=\left\{
1,2,3,\ldots\right\}  $). Obviously, for every $N\in\left\{  1,2,3,\ldots
\right\}  $, the set $\operatorname{wt}^{-1}\left(  N\right)  $ is finite.

For every word $w\in\mathfrak{A}^{\ast}$, define an element
$\operatorname*{Wt}\left(  w\right)  \in \NN$ by $\operatorname*{Wt}%
\left(  w\right)  =\operatorname*{wt}\left(  w_{1}\right)  +\operatorname*{wt}%
\left(  w_{2}\right)  + \cdots+\operatorname*{wt}\left(  w_{k}\right)  $,
where $k$ is the length of $w$. Then,
\begin{equation}
\text{every }w\in\mathfrak{A}^{\ast}\text{ satisfies }\operatorname*{Wt}%
\left(  w\right)  =\left\vert w\right\vert
\label{pf.thm.QSym.lyndon.Z-genset.triv1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.QSym.lyndon.Z-genset.triv1}):} Let
$w\in\mathfrak{A}^{\ast}$. Let $k$ be the length of $w$. Then, $w=\left(
w_{1},w_{2},\ldots,w_{k}\right)  $, so that $\left\vert w\right\vert
=w_{1}+w_{2}+ \cdots+w_{k}$.
\par
Recall that $\operatorname*{wt}$ is the identity map. In other words,
$\operatorname*{wt} = \id$. Now, the definition of
$\operatorname*{Wt}\left(  w\right)  $ yields
\begin{align*}
\operatorname*{Wt}\left(  w\right)   &  =\operatorname*{wt}\left(
w_{1}\right)  +\operatorname*{wt}\left(  w_{2}\right)  + \cdots
+\operatorname*{wt}\left(  w_{k}\right)  = \id \left(
w_{1}\right)  + \id \left(  w_{2}\right)  + \cdots
+ \id \left(  w_{k}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\operatorname*{wt} = \id \right) \\
&  =w_{1}+w_{2}+ \cdots+w_{k}=\left\vert w\right\vert ,
\end{align*}
and thus (\ref{pf.thm.QSym.lyndon.Z-genset.triv1}) is proven.}.

For every $w\in\mathfrak{L}$, the element $M_{\operatorname*{red}%
w}^{\left\langle \gcd w\right\rangle }$ of $\Qsym$ is
homogeneous of degree $\operatorname*{Wt}\left(  w\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in\mathfrak{L}$. Then, $w$ is a
Lyndon word (since $\mathfrak{L}$ is the set of all Lyndon words), hence
nonempty. Remark \ref{rmk.QSym.lyndon.red}(f) (applied to $\alpha=w$) now
yields $\left(  \gcd w\right)  \left\vert \operatorname*{red}w\right\vert
=\left\vert w\right\vert =\operatorname*{Wt}\left(  w\right)  $ (by
(\ref{pf.thm.QSym.lyndon.Z-genset.triv1})). But Corollary
\ref{cor.QSym.Malphas.QSym}(b) (applied to $\gcd w$ and $\operatorname*{red}w$
instead of $s$ and $w$) yields $M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\in\Qsym_{\left(  \gcd w\right)
\left\vert \operatorname*{red}w\right\vert }=
\Qsym_{\operatorname*{Wt}\left(  w\right)  }$ (since $\left(  \gcd
w\right)  \left\vert \operatorname*{red}w\right\vert =\operatorname*{Wt}%
\left(  w\right)  $). In other words, the element $M_{\operatorname*{red}%
w}^{\left\langle \gcd w\right\rangle }$ of $\Qsym$ is
homogeneous of degree $\operatorname*{Wt}\left(  w\right)  $, qed.}.

The $\kk$-module $\Qsym$ has a basis $\left(
g_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ having the property that for every
$u\in\mathfrak{A}^{\ast}$, the element $g_{u}$ of $\Qsym$ is
homogeneous of degree $\operatorname*{Wt}\left(  u\right)  $%
\ \ \ \ \footnote{\textit{Proof.} We shall show that $\left(  M_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ is such a basis.
\par
Indeed, $\mathfrak{A}^{\ast}= \Comp $, so that $\left(
M_{u}\right)  _{u\in\mathfrak{A}^{\ast}}=\left(  M_{u}\right)  _{u\in
 \Comp }$ is clearly a basis of the $\kk$-module
$\Qsym$. Furthermore, for every $u\in\mathfrak{A}^{\ast}$, the
element $M_{u}$ of $\Qsym$ is homogeneous of degree $\left\vert
u\right\vert $. Since $\operatorname*{Wt}\left(  u\right)  =\left\vert
u\right\vert $ for every $u\in\mathfrak{A}^{\ast}$ (by
(\ref{pf.thm.QSym.lyndon.Z-genset.triv1}), applied to $w=u$), this rewrites as
follows: For every $u\in\mathfrak{A}^{\ast}$, the element $M_{u}$ of
$\Qsym$ is homogeneous of degree $\operatorname*{Wt}\left(
u\right)  $.
\par
Thus, $\left(  M_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is a basis of the
$\kk$-module $\Qsym$ having the property that for every
$u\in\mathfrak{A}^{\ast}$, the element $M_{u}$ of $\Qsym$ is
homogeneous of degree $\operatorname*{Wt}\left(  u\right)  $. Hence, the
$\kk$-module $\Qsym$ has a basis $\left(  g_{u}\right)
_{u\in\mathfrak{A}^{\ast}}$ having the property that for every $u\in
\mathfrak{A}^{\ast}$, the element $g_{u}$ of $\Qsym$ is
homogeneous of degree $\operatorname*{Wt}\left(  u\right)  $ (namely, $\left(
M_{u}\right)  _{u\in\mathfrak{A}^{\ast}}$ is such a basis), qed.}.

Once (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal}) is proven, it thus follows
that we can apply Lemma \ref{lem.lyndonbases}(c) to $A=\Qsym$
and $b_{w}=M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }$. From
this, we can conclude that the family $\left(  M_{\operatorname*{red}%
w}^{\left\langle \gcd w\right\rangle }\right)  _{w\in\mathfrak{L}}$ is an
algebraically independent generating set of the $\kk$-algebra
$\Qsym$ (provided that
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal}) is proven); this is precisely
what we need to prove.

Hence, in order to complete the proof of Theorem
\ref{thm.QSym.lyndon.Z-genset}, it is sufficient to prove
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal}). So we shall now prove
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal}).

Let $U$ denote the $\kk$-subalgebra of $\Qsym$ generated
by the family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}$. Then, $U$ is a $\kk %
$-submodule of $\Qsym$. It is clear that%
\begin{equation}
M_{\beta}^{\left\langle s\right\rangle }\in U\ \ \ \ \ \ \ \ \ \ \text{for
every reduced Lyndon composition }\beta\text{ and every }s\in\left\{
1,2,3,\ldots\right\}  . \label{pf.thm.QSym.lyndon.Z-genset.U}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.QSym.lyndon.Z-genset.U}):} Let $\beta$
be a reduced Lyndon composition. Let $s\in\left\{  1,2,3,\ldots\right\}  $.
\par
The composition $\beta$ is Lyndon and thus nonempty. Remark
\ref{rmk.QSym.lyndon.red}(e) (applied to $\alpha=\beta$) yields that the
composition $\beta\left\{  s\right\}  $ is nonempty and satisfies
$\operatorname*{red}\left(  \beta\left\{  s\right\}  \right)
=\operatorname*{red}\beta$ and $\gcd\left(  \beta\left\{  s\right\}  \right)
=s\gcd\beta$. But $\beta$ is reduced; thus, $\gcd\beta=1$ (by the definition
of ``reduced''). Hence, $\gcd\left(
\beta\left\{  s\right\}  \right)  =s\underbrace{\gcd\beta}_{=1}=s$. Also,
Remark \ref{rmk.QSym.lyndon.red}(d) (applied to $\alpha=\beta$) yields
$\operatorname*{red}\beta=\beta$, so that $\operatorname*{red}\left(
\beta\left\{  s\right\}  \right)  =\operatorname*{red}\beta=\beta$.
\par
Remark \ref{rmk.QSym.lyndon.red}(b) (applied to $\alpha=\beta\left\{
s\right\}  $) shows that the composition $\beta\left\{  s\right\}  $ is Lyndon
if and only if the composition $\operatorname*{red}\left(  \beta\left\{
s\right\}  \right)  $ is Lyndon. Since the composition $\operatorname*{red}%
\left(  \beta\left\{  s\right\}  \right)  =\beta$ is Lyndon, this yields that
the composition $\beta\left\{  s\right\}  $ is Lyndon. In other words,
$\beta\left\{  s\right\}  \in\mathfrak{L}$ (since $\mathfrak{L}$ is the set of
all Lyndon words). Hence, $M_{\operatorname*{red}\left(  \beta\left\{
s\right\}  \right)  }^{\left\langle \gcd\left(  \beta\left\{  s\right\}
\right)  \right\rangle }$ is an element of the family $\left(
M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }\right)
_{w\in\mathfrak{L}}$ (namely, the element for $w=\beta\left\{  s\right\}  $).
In other words, $M_{\beta}^{\left\langle s\right\rangle }$ is an element of
the family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$ (since $\operatorname*{red}\left(
\beta\left\{  s\right\}  \right)  =\beta$ and $\gcd\left(  \beta\left\{
s\right\}  \right)  =s$). Hence, $M_{\beta}^{\left\langle s\right\rangle }$
belongs to the $\kk$-subalgebra of $\Qsym$ generated by
this family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}$. Since the $\kk %
$-subalgebra of $\Qsym$ generated by the family $\left(
M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }\right)
_{w\in\mathfrak{L}}$ is $U$, this rewrites as follows: $M_{\beta
}^{\left\langle s\right\rangle }$ belongs to $U$. In other words, $M_{\beta
}^{\left\langle s\right\rangle }\in U$. This proves
(\ref{pf.thm.QSym.lyndon.Z-genset.U}).} Using this and using Exercise
\ref{exe.QSym.Malphas.ps}(d), it is now easy to see that%
\begin{equation}
M_{\beta}^{\left\langle s\right\rangle }\in U\ \ \ \ \ \ \ \ \ \ \text{for
every Lyndon composition }\beta\text{ and every }s\in\left\{  1,2,3,\ldots
\right\}  . \label{pf.thm.QSym.lyndon.Z-genset.U.stronger}%
\end{equation}
\footnote{\textit{Proof.} Let $\beta$ be a Lyndon composition, and let
$s\in\left\{  1,2,3,\ldots\right\}  $. The composition $\beta$ is Lyndon and
thus nonempty; hence, $\gcd\beta$ and $\operatorname*{red}\beta$ are
well-defined.
\par
Remark \ref{rmk.QSym.lyndon.red}(b) (applied to $\alpha=\beta$) yields that
the composition $\beta$ is Lyndon if and only if the composition
$\operatorname*{red}\beta$ is Lyndon. Since the composition $\beta$ is Lyndon,
we therefore conclude that the composition $\operatorname*{red}\beta$ is
Lyndon. Also, the composition $\operatorname*{red}\beta$ is reduced (by Remark
\ref{rmk.QSym.lyndon.red}(c)). Thus, $\operatorname*{red}\beta$ is a reduced
Lyndon composition. Remark \ref{rmk.QSym.lyndon.red}(a) (applied to
$\alpha=\beta$) yields $\beta=\left(  \operatorname*{red}\beta\right)
\left\{  \gcd\beta\right\}  $, so that $\left(  \operatorname*{red}%
\beta\right)  \left\{  \gcd\beta\right\}  =\beta$.
\par
Let $\alpha=\operatorname*{red}\beta$. Recall that $\operatorname*{red}\beta$
is a reduced Lyndon composition. In other words, $\alpha$ is a reduced Lyndon
composition (since $\alpha=\operatorname*{red}\beta$).
\par
Exercise \ref{exe.QSym.Malphas.ps}(d) (applied to $n=\gcd\beta$) yields that
there exists a polynomial $P\in \kk \left[  z_{1},z_{2},z_{3},
\ldots\right]  $ such that $M_{\alpha\left\{  \gcd\beta\right\}
}^{\left\langle s\right\rangle }=P\left(  M_{\alpha}^{\left\langle
1\right\rangle },M_{\alpha}^{\left\langle 2\right\rangle },M_{\alpha
}^{\left\langle 3\right\rangle },\ldots\right)  $. Consider this $P$. We have
$\underbrace{\alpha}_{=\operatorname*{red}\beta}\left\{  \gcd\beta\right\}
=\left(  \operatorname*{red}\beta\right)  \left\{  \gcd\beta\right\}  =\beta$.
Hence, $M_{\alpha\left\{  \gcd\beta\right\}  }^{\left\langle s\right\rangle
}=P\left(  M_{\alpha}^{\left\langle 1\right\rangle },M_{\alpha}^{\left\langle
2\right\rangle },M_{\alpha}^{\left\langle 3\right\rangle },\ldots\right)  $
rewrites as $M_{\beta}^{\left\langle s\right\rangle }=P\left(  M_{\alpha
}^{\left\langle 1\right\rangle },M_{\alpha}^{\left\langle 2\right\rangle
},M_{\alpha}^{\left\langle 3\right\rangle },\ldots\right)  $.
\par
However, for every $j\in\left\{  1,2,3,\ldots\right\}  $, we have
$M_{\alpha}^{\left\langle j\right\rangle }\in U$ (by
(\ref{pf.thm.QSym.lyndon.Z-genset.U}), applied to $\alpha$ and $j$ instead of
$\beta$ and $s$). In other words, $M_{\alpha}^{\left\langle 1\right\rangle }$,
$M_{\alpha}^{\left\langle 2\right\rangle }$, $M_{\alpha}^{\left\langle
3\right\rangle }$, $\ldots$ are elements of $U$. Therefore, $Q\left(
M_{\alpha}^{\left\langle 1\right\rangle },M_{\alpha}^{\left\langle
2\right\rangle },M_{\alpha}^{\left\langle 3\right\rangle },\ldots\right)  \in
U$ for every $Q\in \kk \left[  z_{1},z_{2},z_{3},\ldots\right]  $ (since
$U$ is a $\kk$-subalgebra of $\Qsym$). Applied to $Q=P$,
this yields $P\left(  M_{\alpha}^{\left\langle 1\right\rangle },M_{\alpha
}^{\left\langle 2\right\rangle },M_{\alpha}^{\left\langle 3\right\rangle
},\ldots\right)  \in U$. Thus, $M_{\beta}^{\left\langle s\right\rangle
}=P\left(  M_{\alpha}^{\left\langle 1\right\rangle },M_{\alpha}^{\left\langle
2\right\rangle },M_{\alpha}^{\left\langle 3\right\rangle },\ldots\right)  \in
U$. This proves (\ref{pf.thm.QSym.lyndon.Z-genset.U.stronger}).}

We shall now prove that%
\begin{equation}
M_{\beta}\in U\ \ \ \ \ \ \ \ \ \ \text{for every composition }\beta.
\label{pf.thm.QSym.lyndon.Z-genset.maingoal.1}%
\end{equation}


\textit{Proof of (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}):} We will
prove (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}) by strong induction over
$\left\vert \beta\right\vert $:

\textit{Induction step:} Let $N\in \NN$. We assume that
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}) holds for all compositions
$\beta$ satisfying $\left\vert \beta\right\vert <N$. We need to show that
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}) also holds for all compositions
$\beta$ satisfying $\left\vert \beta\right\vert =N$. In other words, we need
to prove that $M_{\beta}\in U$ for every composition $\beta$ satisfying
$\left\vert \beta\right\vert =N$. In other words, we need to prove that%
\begin{equation}
M_{\beta}\in U\ \ \ \ \ \ \ \ \ \ \text{for every }\beta\in
\Comp_{N}.
\label{pf.thm.QSym.lyndon.Z-genset.maingoal.2}%
\end{equation}


\textit{Proof of (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}):} We will
prove (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}) by strong induction over
$\beta$ with respect to the wll-order on $\Comp_{N}$.
In other words, we fix some $\alpha\in\Comp_{N}$, and
we assume that (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}) holds for all
$\beta\in\Comp_{N}$ satisfying $\beta
\underset{\operatorname*{wll}}{<}\alpha$. We now need to prove that
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}) holds for $\beta=\alpha$. In
other words, we need to prove that $M_{\alpha}\in U$.

If $\alpha=\varnothing$, then $M_{\alpha}\in U$ is
obvious\footnote{\textit{Proof.} Assume that $\alpha=\varnothing$. Then,
$M_{\alpha}=M_{\varnothing}=1\in U$ (since $U$ is a $\kk$-subalgebra of
$\Qsym$), qed.}. Hence, for the rest of this proof of
$M_{\alpha}\in U$, we can WLOG assume that $\alpha\neq\varnothing$. Assume
this. The composition $\alpha$ is nonempty (since $\alpha\neq\varnothing$), so
that it satisfies $\left\vert \alpha\right\vert \neq0$. Since $\left\vert
\alpha\right\vert =N$ (because $\alpha\in\Comp_{N}$),
we have $N=\left\vert \alpha\right\vert \neq0$. Thus, $N$ is a positive integer.

We have assumed that (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}) holds for
all $\beta\in\Comp_{N}$ satisfying $\beta
\underset{\operatorname*{wll}}{<}\alpha$. In other words,%
\begin{equation}
M_{\beta}\in U\ \ \ \ \ \ \ \ \ \ \text{for all }\beta\in
\Comp_{N}\text{ satisfying }\beta\underset{\operatorname*{wll}}{<}\alpha.
\label{pf.thm.QSym.lyndon.Z-genset.ih2}%
\end{equation}


Also, we have assumed that (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1})
holds for all compositions $\beta$ satisfying $\left\vert \beta\right\vert
<N$. In other words,%
\begin{equation}
M_{\beta}\in U\ \ \ \ \ \ \ \ \ \ \text{for all compositions }\beta\text{
satisfying }\left\vert \beta\right\vert <N.
\label{pf.thm.QSym.lyndon.Z-genset.ih1}%
\end{equation}


Let $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ be the CFL factorization of
the word $\alpha$. Then, $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is a
tuple of Lyndon words satisfying $\alpha=a_{1}a_{2} \cdots a_{p}$ and
$a_{1}\geq a_{2}\geq\cdots\geq a_{p}$ (according to the definition of a CFL
factorization). We have $p\neq0$ (since otherwise, we would have $p=0$ and
thus $\alpha=a_{1}a_{2} \cdots a_{p}=\left(  \text{empty product}\right)
=\varnothing$, contradicting $\alpha\neq\varnothing$). Thus, $p\in\left\{
1,2,3,\ldots\right\}  $. Hence, the word $a_{1}$ is well-defined. Clearly,
$a_{1}$ is a Lyndon word (since $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is
a tuple of Lyndon words).

We distinguish between two cases:

\textit{Case 1:} All of the words $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are equal.

\textit{Case 2:} Not all of the words $a_{1}$, $a_{2}$, $\ldots$, $a_{p}$ are equal.

Let us consider Case 1 first. In this case, all of the words $a_{1}$, $a_{2}$,
$\ldots$, $a_{p}$ are equal. In other words, $a_{1}=a_{2}= \cdots=a_{p}$.
Thus, $a_{1}=a_{i}$ for every $i\in\left\{  1,2,\ldots,p\right\}  $.

Let $x=a_{1}$. Then, $x=a_{1}=a_{i}$ for every $i\in\left\{  1,2,\ldots
,p\right\}  $. Multiplying these identities for all $i\in\left\{
1,2,\ldots,p\right\}  $, we obtain $\underbrace{xx\cdots x}_{p\text{ times}%
}=a_{1}a_{2} \cdots a_{p}=\alpha$, so that $\alpha=\underbrace{xx\cdots
x}_{p\text{ times}}=x^{p}$, thus $x^{p}=\alpha$. Also, $x=a_{1}$ is a Lyndon
word. Let $k=\left\vert x\right\vert $. Then, $x\in
\Comp_{k}$. Also, $N=\left\vert \underbrace{\alpha}_{=x^{p}}\right\vert
=\left\vert x^{p}\right\vert =p\left\vert x\right\vert $, so that $p\left\vert
x\right\vert =N$. Thus, $p\underbrace{k}_{=\left\vert x\right\vert
}=p\left\vert x\right\vert =N$.

Now, Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.As2} (applied to $s=p$)
yields%
\begin{align*}
M_{x}^{\left\langle p\right\rangle }-M_{x^{p}}  &  \in\sum\limits_{\substack{w\in
\Comp_{pk};\\w\underset{\operatorname*{wll}}{<}x^{p}%
}} \kk M_{w}=\sum\limits_{\substack{w\in\Comp_{N}%
;\\w\underset{\operatorname*{wll}}{<}\alpha}} \kk \underbrace{M_{w}%
}_{\substack{\in U\\\text{(by (\ref{pf.thm.QSym.lyndon.Z-genset.ih2}),
applied}\\\text{to }\beta=w\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}pk=N\text{ and }x^{p}=\alpha\right) \\
&  \subset\sum\limits_{\substack{w\in\Comp_{N}%
;\\w\underset{\operatorname*{wll}}{<}\alpha}} \kk U\subset
U\ \ \ \ \ \ \ \ \ \ \left(  \text{since }U\text{ is a } \kk %
\text{-submodule of }\Qsym\right)  .
\end{align*}
Hence,
\[
M_{x^{p}}-M_{x}^{\left\langle p\right\rangle }=-\underbrace{\left(
M_{x}^{\left\langle p\right\rangle }-M_{x^{p}}\right)  }_{\in U}\in-U\subset
U\ \ \ \ \ \ \ \ \ \ \left(  \text{since }U\text{ is a } \kk %
\text{-submodule of }\Qsym\right)  ,
\]
so that%
\[
M_{x^{p}}\in\underbrace{M_{x}^{\left\langle p\right\rangle }}_{\substack{\in
U\\\text{(by (\ref{pf.thm.QSym.lyndon.Z-genset.U.stronger}), applied}%
\\\text{to }x\text{ and }p\text{ instead of }\beta\text{ and }s\text{)}}}+U\in
U+U\subset U
\]
(since $U$ is a $\kk$-submodule of $\Qsym$). Since
$x^{p}=\alpha$, this rewrites as $M_{\alpha}\in U$. Thus, we have proven
$M_{\alpha}\in U$ in Case 1.

Let us now consider Case 2. In this case, not all of the words $a_{1}$,
$a_{2}$, $\ldots$, $a_{p}$ are equal. Hence, there exists some $k\in\left\{
1,2,\ldots,p-1\right\}  $ such that $a_{k}\neq a_{k+1}$. Consider this $k$.

We have $a_{k}\geq a_{k+1}$ (since $a_{1}\geq a_{2}\geq\cdots\geq a_{p}$).
Combined with $a_{k}\neq a_{k+1}$, this yields $a_{k}>a_{k+1}$. Let $x$ be the
word $a_{1}a_{2} \cdots a_{k}$, and let $y$ be the word $a_{k+1}a_{k+2} \cdots
a_{p}$. Then, Corollary \ref{cor.QSym.lyndon.triangl.wll-lemma.BC} (applied to
$u=\alpha$ and $n=N$) yields%
\[
M_{\alpha}=M_{x}M_{y}-\left(  \text{a sum of terms of the form }M_{w}\text{
with }w\in\Comp_{N}\text{ satisfying }%
w\underset{\operatorname*{wll}}{<}\alpha\right)  .
\]
Thus,%
\begin{align*}
M_{x}M_{y}-M_{\alpha}  &  =\left(  \text{a sum of terms of the form }%
M_{w}\text{ with }w\in\Comp_{N}\text{ satisfying
}w\underset{\operatorname*{wll}}{<}\alpha\right) \\
&  \in\sum\limits_{\substack{w\in\Comp_{N}%
;\\w\underset{\operatorname*{wll}}{<}\alpha}} \kk \underbrace{M_{w}%
}_{\substack{\in U\\\text{(by (\ref{pf.thm.QSym.lyndon.Z-genset.ih2}),
applied}\\\text{to }\beta=w\text{)}}}\subset\sum\limits_{\substack{w\in
\Comp_{N};\\w\underset{\operatorname*{wll}}{<}\alpha
}} \kk U\subset U
\end{align*}
(since $U$ is a $\kk$-submodule of $\Qsym$). Hence,%
\begin{equation}
M_{\alpha}\in M_{x}M_{y}-U. \label{pf.thm.QSym.lyndon.Z-genset.almost}%
\end{equation}


Now, it is easy to see that $N=\left\vert x\right\vert +\left\vert
y\right\vert $\ \ \ \ \footnote{\textit{Proof.} Multiplying the equalities
$x=a_{1}a_{2} \cdots a_{k}$ and $y=a_{k+1}a_{k+2} \cdots a_{p}$, we obtain%
\[
xy=\left(  a_{1}a_{2} \cdots a_{k}\right)  \left(  a_{k+1}a_{k+2} \cdots
a_{p}\right)  =a_{1}a_{2} \cdots a_{p}=\alpha,
\]
so that $\alpha=xy$ and thus $\left\vert \alpha\right\vert =\left\vert
xy\right\vert =\left\vert x\right\vert +\left\vert y\right\vert $, so that
$\left\vert x\right\vert +\left\vert y\right\vert =\left\vert \alpha
\right\vert =N$, qed.}. But $\left\vert x\right\vert >0$%
\ \ \ \ \footnote{\textit{Proof.} Notice that $a_{k}$ is a Lyndon word (since
$\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is a tuple of Lyndon words), and
thus nonempty.
\par
But $a_{k}$ is a suffix of the word $a_{1}a_{2} \cdots a_{k}$. Hence,
$\ell\left(  a_{k}\right)  \leq\ell\left(  a_{1}a_{2} \cdots a_{k}\right)  $,
so that $\ell\left(  a_{1}a_{2} \cdots a_{k}\right)  \geq\ell\left(
a_{k}\right)  >0$ (since the word $a_{k}$ is nonempty). Now, $\ell\left(
\underbrace{x}_{=a_{1}a_{2} \cdots a_{k}}\right)  =\ell\left(  a_{1}a_{2}
\cdots a_{k}\right)  >0$, so that the word $x$ is nonempty, and therefore
$\left\vert x\right\vert >0$, qed.} and $\left\vert y\right\vert
>0$\ \ \ \ \footnote{\textit{Proof.} Notice that $a_{k+1}$ is a Lyndon word
(since $\left(  a_{1},a_{2},\ldots,a_{p}\right)  $ is a tuple of Lyndon
words), and thus nonempty.
\par
But $a_{k+1}$ is a prefix of the word $a_{k+1}a_{k+2} \cdots a_{p}$. Hence,
$\ell\left(  a_{k+1}\right)  \leq\ell\left(  a_{k+1}a_{k+2} \cdots
a_{p}\right)  $, so that $\ell\left(  a_{k+1}a_{k+2} \cdots a_{p}\right)
\geq\ell\left(  a_{k+1}\right)  >0$ (since the word $a_{k+1}$ is nonempty).
Now, $\ell\left(  \underbrace{y}_{=a_{k+1}a_{k+2} \cdots a_{p}}\right)
=\ell\left(  a_{k+1}a_{k+2} \cdots a_{p}\right)  >0$, so that the word $y$ is
nonempty, and therefore $\left\vert y\right\vert >0$, qed.}. Hence,
$N=\left\vert x\right\vert +\underbrace{\left\vert y\right\vert }%
_{>0}>\left\vert x\right\vert $, which yields $\left\vert x\right\vert <N$.
Hence, (\ref{pf.thm.QSym.lyndon.Z-genset.ih1}) (applied to $\beta=x$) yields
$M_{x}\in U$. Also, $N=\underbrace{\left\vert x\right\vert }_{>0}+\left\vert
y\right\vert >\left\vert y\right\vert $, and thus $\left\vert y\right\vert
<N$, so that (\ref{pf.thm.QSym.lyndon.Z-genset.ih1}) (applied to $\beta=y$)
yields $M_{y}\in U$. Now, (\ref{pf.thm.QSym.lyndon.Z-genset.almost}) becomes
$M_{\alpha}\in\underbrace{M_{x}}_{\in U}\underbrace{M_{y}}_{\in U}-U\subset
UU-U\subset U$ (since $U$ is a $\kk$-subalgebra of
$\Qsym$). Thus, we have proven $M_{\alpha}\in U$ in Case 2.

Now, we have proven $M_{\alpha}\in U$ in each of the two Cases 1 and 2. Since
these two Cases cover all possibilities, this yields that we always have
$M_{\alpha}\in U$. In other words,
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}) holds for $\beta=\alpha$.

Thus, we have completed the induction step of our induction over $\beta$.
Therefore, we have proven (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.2}) by
induction. So we now know that $M_{\beta}\in U$ for every $\beta
\in\Comp_{N}$. In other words, $M_{\beta}\in U$ for
every composition $\beta$ satisfying $\left\vert \beta\right\vert =N$. In
other words, (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}) holds for all
compositions $\beta$ satisfying $\left\vert \beta\right\vert =N$.

Thus, we have completed the induction step of our induction over $N$. Hence,
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}) is proven by induction over $N$.

Now, recall that the family $\left(  M_{\beta}\right)  _{\beta\in
 \Comp }$ is a basis of the $\kk$-module
$\Qsym$, and thus generates this $\kk$-module. Hence,%
\[
\Qsym=\sum_{\beta\in \Comp } \kk %
\underbrace{M_{\beta}}_{\substack{\in U\\\text{(by
(\ref{pf.thm.QSym.lyndon.Z-genset.maingoal.1}))}}}\subset\sum_{\beta
\in \Comp } \kk U\subset U
\]
(since $U$ is a $\kk$-submodule of $\Qsym$). Combined
with $U\subset\Qsym$, this yields $U=\Qsym$.
Since $U$ is the $\kk$-subalgebra of $\Qsym$ generated
by the family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\right)  _{w\in\mathfrak{L}}$, this shows that the
$\kk$-subalgebra of $\Qsym$ generated by the family
$\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }\right)
_{w\in\mathfrak{L}}$ is $\Qsym$ itself. In other words, the
family $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$ generates the $\kk$-algebra
$\Qsym$. Thus, (\ref{pf.thm.QSym.lyndon.Z-genset.maingoal}) is
proven. As we know, this completes the proof of Theorem
\ref{thm.QSym.lyndon.Z-genset}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.thm.QSym.lyndon}}

\textit{Solution to Exercise \ref{exe.thm.QSym.lyndon}.}

\begin{proof}
[Proof of Theorem \ref{thm.QSym.lyndon}.]We know (from the proof of Theorem
\ref{thm.QSym.lyndon.Z-genset}) that the family $\left(
M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle }\right)
_{w\in\mathfrak{L}}$ is an algebraically independent generating set of the
$\kk$-algebra $\Qsym$.

Now, define a grading on the $\kk$-algebra $\kk \left[
x_{w}\ \mid\ w\in\mathfrak{L}\right]  $ by setting $\deg\left(  x_{w}\right)
=\sum_{i=1}^{\ell\left(  w\right)  }w_{i}$ for every $w\in\mathfrak{L}$. By
the universal property of the polynomial algebra $\kk \left[
x_{w}\ \mid\ w\in\mathfrak{L}\right]  $, we can define a $\kk$-algebra
homomorphism $\Phi: \kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]
\rightarrow\Qsym$ by setting%
\[
\Phi\left(  x_{w}\right)  =M_{\operatorname*{red}w}^{\left\langle \gcd
w\right\rangle }\ \ \ \ \ \ \ \ \ \ \text{for every }w\in\mathfrak{L}.
\]
\footnote{This is well-defined since $M_{\operatorname*{red}w}^{\left\langle
\gcd w\right\rangle }\in\Qsym$ (by Corollary
\ref{cor.QSym.Malphas.QSym}(a), applied to $\alpha=\operatorname*{red}w$ and
$s=\gcd w$).} This homomorphism $\Phi$ is a $\kk$-algebra isomorphism
(since $\left(  M_{\operatorname*{red}w}^{\left\langle \gcd w\right\rangle
}\right)  _{w\in\mathfrak{L}}$ is an algebraically independent generating set
of the $\kk$-algebra $\Qsym$) and is graded (because for
every $w\in\mathfrak{L}$, the element $M_{\operatorname*{red}w}^{\left\langle
\gcd w\right\rangle }$ of $\Qsym$ is homogeneous of degree
$\deg\left(  x_{w}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$w\in\mathfrak{L}$. Then, $w$ is a Lyndon word (since $\mathfrak{L}$ is the
set of all Lyndon words), and thus nonempty. Hence, Remark
\ref{rmk.QSym.lyndon.red}(f) (applied to $\alpha=w$) yields $\left(  \gcd
w\right)  \left\vert \operatorname*{red}w\right\vert =\left\vert w\right\vert
=\sum_{i=1}^{\ell\left(  w\right)  }w_{i}=\deg\left(  x_{w}\right)  $. Now,
Corollary \ref{cor.QSym.Malphas.QSym}(b) (applied to $\alpha
=\operatorname*{red}w$ and $s=\gcd w$) yields $M_{\operatorname*{red}%
w}^{\left\langle \gcd w\right\rangle }\in
\Qsym_{\left( \gcd w \right) \left| \operatorname*{red} w \right|}
=\Qsym_{\deg\left(  x_{w}\right)  }$ (since $\left(
\gcd w\right)  \left\vert \operatorname*{red}w\right\vert =\deg\left(
x_{w}\right)  $). In other words, the element $M_{\operatorname*{red}%
w}^{\left\langle \gcd w\right\rangle }$ of $\Qsym$ is
homogeneous of degree $\deg\left(  x_{w}\right)  $, qed.}). Thus, $\Phi$ is an
isomorphism of graded $\kk$-algebras. Hence, $\Qsym%
\cong \kk \left[  x_{w}\ \mid\ w\in\mathfrak{L}\right]  $ as graded
$\kk$-algebras. Thus, $\Qsym$ is a polynomial algebra.
This proves Theorem \ref{thm.QSym.lyndon}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.cor.QSym.lyndon.free}}

\textit{Solution to Exercise \ref{exe.cor.QSym.lyndon.free}.}

\begin{proof}
[Proof of Corollary \ref{cor.QSym.lyndon.free}.]
Theorem~\ref{thm.QSym.lyndon.Z-genset} yields that the family
$\left( M_{w} ^{\left\langle s \right\rangle} \right) _{ \left( w, s \right)
\in \mathfrak{RL} \times \left\{ 1,2,3,\ldots \right\} }$ is an algebraically
independent generating set of the $\kk$-algebra $\Qsym$.

Notice that $\left( 1 \right)$ is a reduced Lyndon composition; that is,
$\left( 1 \right) \in \mathfrak{RL}$ (since $\mathfrak{RL}$ is the set of all
reduced Lyndon compositions). Hence, $\left\{  \left( 1 \right)  \right\}
\subset\mathfrak{RL}$, whence $\left\{  \left( 1 \right)  \right\}
\times\left\{  1,2,3,\ldots\right\}  \subset\mathfrak{RL}\times\left\{
1,2,3,\ldots\right\}  $.

The following fact is straightforward to check: If $A$ is a commutative
$\kk$-algebra, and if $\left(  a_{i}\right)  _{i\in I}$ is an
algebraically independent generating set of the $\kk$-algebra $A$, and
if $J$ is a subset of $I$, then $\left(  a_{i}\right)  _{i\in I\setminus J}$
is an algebraically independent generating set of the $\kk \left[
a_{i}\ \mid\ i\in J\right]  $-algebra $A$.\ \ \ \ \footnote{The idea behind
this fact is that a polynomial ring in the indeterminates $\left(
x_{i}\right)  _{i\in I}$ can be regarded as a polynomial ring in the
indeterminates $\left(  x_{i}\right)  _{i\in I\setminus J}$ over the
polynomial ring in the indeterminates $\left(  x_{i}\right)  _{i\in J}$.}
We can apply this fact to $A=\Qsym$,
$I = \mathfrak{RL} \times\left\{  1,2,3,\ldots\right\}  $,
$\left(  a_{i}\right)  _{i\in I}
= \left(  M_{w}^{\left\langle s\right\rangle }\right)  _{\left(  w,s\right)
\in \mathfrak{RL}\times\left\{  1,2,3,\ldots\right\}  }$
and
$J = \left\{ \left( 1 \right)  \right\}
 \times \left\{  1,2,3,\ldots\right\}  $.
As a result, we conclude that
$\left(  M_{w}^{\left\langle s\right\rangle }\right)
_{\left(  w,s\right)  \in\left(  \mathfrak{RL}\times\left\{  1,2,3,\ldots
\right\}  \right)
\setminus \left(  \left\{ \left( 1 \right) \right\}  \times\left\{
1,2,3,\ldots\right\}  \right)  }$ is an algebraically independent generating
set of the $\kk \left[  M_{w}^{\left\langle s\right\rangle }%
\ \mid\ \left(  w,s\right)  \in\left\{ \left( 1 \right) \right\}
 \times\left\{ 1,2,3,\ldots\right\}  \right]  $-algebra $\Qsym$. Hence, the
$\kk \left[  M_{w}^{\left\langle s\right\rangle }\ \mid\ \left(
w,s\right)
\in\left\{ \left( 1 \right) \right\}  \times\left\{  1,2,3,\ldots\right\}
\right]  $-algebra $\Qsym$ has an algebraically independent
generating set; it is therefore a polynomial algebra over $\kk \left[
M_{w}^{\left\langle s\right\rangle }\ \mid\ \left(  w,s\right)  \in\left\{
\left( 1 \right) \right\}  \times\left\{  1,2,3,\ldots\right\}  \right]  $.
Since
$\kk \left[  M_{w}^{\left\langle s\right\rangle }\ \mid\ \left(
w,s\right)  \in\left\{ \left( 1 \right) \right\}
 \times\left\{  1,2,3,\ldots\right\} \right]  =\Lambda$
\ \ \ \ \footnote{\textit{Proof.} The first sentence of
Proposition~\ref{symm-antipode-and-h-basis} yields that the family $\left(
e_{1},e_{2},e_{3},\ldots\right)  $ generates the $\kk$-algebra
$\Lambda$. In other words, $\Lambda= \kk \left[  e_{1},e_{2},e_{3}%
,\ldots\right]  = \kk \left[  e_{s}\ \mid\ s\in\left\{  1,2,3,\ldots
\right\}  \right]  $.
\par
But
\begin{align*}
&   \kk \left[  M_{w}^{\left\langle s\right\rangle }\ \mid\ \left(
w,s\right)  \in \left\{ \left( 1 \right) \right\}
\times\left\{  1,2,3,\ldots\right\} \right] \\
&  = \kk \left[  \underbrace{M_{\left(  1\right)  }^{\left\langle
s\right\rangle }}_{\substack{=e_{s}\\\text{(by Exercise
\ref{exe.QSym.Malphas.M1s})}}}\ \mid\ s\in\left\{  1,2,3,\ldots\right\}
\right]  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the elements of }\left\{ \left( 1 \right) \right\}
 \times\left\{  1,2,3,\ldots \right\}  \text{ are precisely}\\
\text{the pairs of the form }\left( \left( 1 \right) ,s \right)
\text{ for }s\in\left\{ 1,2,3,\ldots\right\}
\end{array}
\right) \\
&  = \kk \left[  e_{s}\ \mid\ s\in\left\{  1,2,3,\ldots\right\}  \right]
=\Lambda
\end{align*}
(since $\Lambda= \kk \left[  e_{s}\ \mid\ s\in\left\{  1,2,3,\ldots
\right\}  \right]  $), qed.}, this rewrites as follows: The $\Lambda$-algebra
$\Qsym$ is a polynomial algebra over $\Lambda$. This proves
Corollary \ref{cor.QSym.lyndon.free}.
\end{proof}
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.lyndon.gr.basics}}

\textit{Solution to Exercise \ref{exe.lyndon.gr.basics}.} Before we step to
the proofs of Proposition \ref{prop.words.read1}, Lemma
\ref{lem.sol.lyndon.gr.basics.1} and Proposition \ref{prop.words.read-std2},
let us state a basic fact about cycles of permutations:

\begin{lemma}
\label{lem.sol.lyndon.gr.basics.cyc1}Let $n\in \NN $. Let $\tau
\in\Symm_{n}$ be a permutation. Let $z$ be a cycle of $\tau$. Let $k$
be the size of $z$. Let $p\in z$. For each $u\in \NN $, we set%
\[
p_{u}:=\tau^{u}\left(  p\right)  .
\]
Then:

\begin{enumerate}
\item[(a)] We have $z=\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $.

\item[(b)] We have $\tau^{k}\left(  p\right)  =p$.

\item[(c)] We have $z=\left\{  p_{0},p_{1},\ldots,p_{k-1}\right\}  $.

\item[(d)] We have $\tau^{i}\left(  p_{u}\right)  =p_{u+i}$ for each
$u\in \NN $ and $i\in \NN $.

\item[(e)] We have $p_{u+k}=p_{u}$ for each $u\in \NN $.

\item[(f)] We have $p_{u}\in z$ for each $u\in \NN $.

\item[(g)] The $k$ elements $p_{0},p_{1},\ldots,p_{k-1}$ are distinct.

\item[(h)] We have $z=\left\{  p_{u},p_{u+1},\ldots,p_{u+k-1}\right\}  $ for
each $u\in \NN $.

\item[(i)] We have $\operatorname*{ord}\nolimits_{\tau}\left(  p\right)  =k$.
\end{enumerate}
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}.]All of these are
well-known properties of cycles in permutations.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}.]Most parts of this lemma
are well-known properties of cycles of permutations. Nevertheless, let us
prove them all. We begin with part (d):

(d) Let $u\in \NN $ and $i\in \NN $. Then, $p_{u}=\tau^{u}\left(
p\right)  $ (by the definition of $p_{u}$) and $p_{u+i}=\tau^{u+i}\left(
p\right)  $ (by the definition of $p_{u+i}$). Comparing $p_{u+i}=\tau
^{u+i}\left(  p\right)  $ with $\tau^{i}\left(  \underbrace{p_{u}}_{=\tau
^{u}\left(  p\right)  }\right)  =\tau^{i}\left(  \tau^{u}\left(  p\right)
\right)  =\underbrace{\left(  \tau^{i}\circ\tau^{u}\right)  }_{=\tau^{i+u}%
}\left(  p\right)  =\tau^{i+u}\left(  p\right)  =\tau^{u+i}\left(  p\right)
$, we obtain $\tau^{i}\left(  p_{u}\right)  =p_{u+i}$. This proves Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(d).

Next, let us prepare for the other parts of the lemma.

The set $z$ has at least one element (namely, $p$, because $p\in z$). Hence,
the set $z$ is nonempty. Thus, $\left\vert z\right\vert \geq1$. But $k$ is the
size of $z$; in other words, $k=\left\vert z\right\vert $. Thus, $k=\left\vert
z\right\vert \geq1$.

The definition of $p_{0}$ yields $p_{0}=\underbrace{\tau^{0}}%
_{=\operatorname*{id}}\left(  p\right)  =p$, so that $p=p_{0}$.

The map $\tau$ is a permutation of $\left\{  1,2,\ldots,n\right\}  $. In other
words, $\tau$ is a bijection from $\left\{  1,2,\ldots,n\right\}  $ to
$\left\{  1,2,\ldots,n\right\}  $. Hence, $p_{0},p_{1},\ldots,p_{n}$ are
elements of $\left\{  1,2,\ldots,n\right\}  $. Thus, the set $\left\{
p_{0},p_{1},\ldots,p_{n}\right\}  $ is a subset of $\left\{  1,2,\ldots
,n\right\}  $. Hence, $\left\vert \left\{  p_{0},p_{1},\ldots,p_{n}\right\}
\right\vert \leq\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =n<n+1$.
Therefore, the $n+1$ elements $p_{0},p_{1},\ldots,p_{n}$ cannot all be
distinct (because if they were all distinct, then we would have $\left\vert
\left\{  p_{0},p_{1},\ldots,p_{n}\right\}  \right\vert =n+1$, which would
contradict $\left\vert \left\{  p_{0},p_{1},\ldots,p_{n}\right\}  \right\vert
<n+1$).

Next, let us show the following technical observation (which we will use twice):

\begin{statement}
\textit{Observation 1:} Let $x\in \NN $ and $y\in \NN $ be such that
$x<y$ and $p_{x}=p_{y}$. Then, $p_{y-x}=p_{0}$.
\end{statement}

[\textit{Proof of Observation 1:} We have $y-x>0$ (since $x<y$). Hence, $y-x$
is a positive integer. Thus, the definition of $p_{y-x}$ yields $p_{y-x}%
=\tau^{y-x}\left(  p\right)  $. Moreover, $\tau^{x}\circ\tau^{y-x}%
=\tau^{x+\left(  y-x\right)  }=\tau^{y}$ (since $x+\left(  y-x\right)  =y$).
But the definition of $p_{y}$ yields
\[
p_{y}=\underbrace{\tau^{y}}_{=\tau^{x}\circ\tau^{y-x}}\left(  p\right)
=\left(  \tau^{x}\circ\tau^{y-x}\right)  \left(  p\right)  =\tau^{x}\left(
\underbrace{\tau^{y-x}\left(  p\right)  }_{=p_{y-x}}\right)  =\tau^{x}\left(
p_{y-x}\right)  .
\]
Therefore, $\tau^{x}\left(  p_{y-x}\right)  =p_{y}=p_{x}$ (since $p_{x}=p_{y}%
$). But the definition of $p_{x}$ yields $p_{x}=\tau^{x}\left(  p\right)  $.
Hence, $\tau^{x}\left(  p_{y-x}\right)  =p_{x}=\tau^{x}\left(  p\right)  $.

The map $\tau$ is a bijection, i.e., is bijective. Thus, this map $\tau$ is
injective. Hence, its $x$-th power $\tau^{x}$ is injective as well (since any
composition of injective maps is injective). Thus, if $\alpha$ and $\beta$ are
two elements of $\left\{  1,2,\ldots,n\right\}  $ satisfying $\tau^{x}\left(
\alpha\right)  =\tau^{x}\left(  \beta\right)  $, then $\alpha=\beta$. Hence,
from $\tau^{x}\left(  p_{y-x}\right)  =\tau^{x}\left(  p\right)  $, we obtain
$p_{y-x}=p$. Thus, $p_{y-x}=p=p_{0}$. This proves Observation 1.]

It is now easy to see that there exists a positive integer $r$ satisfying
$p_{r}=p_{0}$\ \ \ \ \footnote{\textit{Proof.} We have shown that the $n+1$
elements $p_{0},p_{1},\ldots,p_{n}$ cannot all be distinct. In other words,
(at least) two of these $n+1$ elements are equal. In other words, there exist
two elements $x$ and $y$ of $\left\{  0,1,\ldots,n\right\}  $ such that $x<y$
and $p_{x}=p_{y}$. Consider these $x$ and $y$.
\par
We have $x\in\left\{  0,1,\ldots,n\right\}  \subset \NN $ and
$y\in\left\{  0,1,\ldots,n\right\}  \subset \NN $. Moreover, $y-x$ is a
positive integer (since $x<y$). Observation 1 yields $p_{y-x}=p_{0}$.
\par
Hence, there exists a positive integer $r$ satisfying $p_{r}=p_{0}$ (namely,
$r=y-x$).}. Let $h$ be the \textbf{smallest} such $r$. Thus, $h$ is a positive
integer $r$ satisfying $p_{r}=p_{0}$. In other words, $h$ is a positive
integer and satisfies $p_{h}=p_{0}$. Moreover,
\begin{equation}
\text{if }r\text{ is a positive integer satisfying }p_{r}=p_{0}\text{, then
}r\geq h \label{pf.lem.sol.lyndon.gr.basics.cyc1.smallest}%
\end{equation}
(since $h$ is the \textbf{smallest} positive integer $r$ satisfying
$p_{r}=p_{0}$).

We shall soon see that $h=k$, but this is not obvious at this point yet.

Clearly, $h-1\in \NN $ (since $h$ is a positive integer). Also, the
definition of $p_{h}$ yields $p_{h}=\tau^{h}\left(  p\right)  $, so that
$\tau^{h}\left(  p\right)  =p_{h}=p_{0}=p$.

Next, we claim the following:

\begin{statement}
\textit{Observation 2:} We have $p_{hi}=p$ for each $i\in \NN $.
\end{statement}

[\textit{Proof of Observation 2:} We shall prove Observation 2 by induction on
$i$:

\textit{Induction base:} We have $h0=0$ and thus $p_{h0}=p_{0}=p$. Thus,
Observation 2 holds for $i=0$. This completes the induction base.

\textit{Induction step:} Let $j\in \NN $. Assume that Observation 2 holds
for $i=j$. We must prove that Observation 2 holds for $i=j+1$.

We have assumed that Observation 2 holds for $i=j$. In other words, we have
$p_{hj}=p$.

But Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(d) (applied to $i=h$ and $u=hj$)
yields $\tau^{h}\left(  p_{hj}\right)  =p_{hj+h}=p_{h\left(  j+1\right)  }$
(since $hj+h=h\left(  j+1\right)  $). In view of $p_{hj}=p$, this rewrites as
$\tau^{h}\left(  p\right)  =p_{h\left(  j+1\right)  }$. Thus, $p_{h\left(
j+1\right)  }=\tau^{h}\left(  p\right)  =p$. In other words, Observation 2
holds for $i=j+1$. This completes the induction step. Thus, Observation 2 is
proved by induction.]

\begin{statement}
\textit{Observation 3:} We have $\tau^{-i}\left(  p\right)  =p_{\left(
h-1\right)  i}$ for each $i\in \NN $.\ \ \ \ \footnote{Note that the
expression $p_{\left(  h-1\right)  i}$ is well-defined, since $h-1\in
 \NN $.}
\end{statement}

[\textit{Proof of Observation 3:} Let $i\in \NN $. Then, $\tau
^{-i}=\left(  \tau^{i}\right)  ^{-1}$, so that $\tau^{-i}\circ\tau
^{i}=\operatorname*{id}$. Now, Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(d)
(applied to $u=\left(  h-1\right)  i$) yields $\tau^{i}\left(  p_{\left(
h-1\right)  i}\right)  =p_{\left(  h-1\right)  i+i}=p_{hi}$ (since $\left(
h-1\right)  i+i=hi$). But Observation 2 yields $p_{hi}=p$. Hence, $\tau
^{i}\left(  p_{\left(  h-1\right)  i}\right)  =p_{hi}=p$. Thus,
\[
\tau^{-i}\left(  \underbrace{p}_{=\tau^{i}\left(  p_{\left(  h-1\right)
i}\right)  }\right)  =\tau^{-i}\left(  \tau^{i}\left(  p_{\left(  h-1\right)
i}\right)  \right)  =\underbrace{\left(  \tau^{-i}\circ\tau^{i}\right)
}_{=\operatorname*{id}}\left(  p_{\left(  h-1\right)  i}\right)  =p_{\left(
h-1\right)  i}.
\]
This proves Observation 3.]

We are now ready to prove Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(a):

(a) The set $z$ is a cycle of $\tau$. In other words, $z$ is an orbit of the
action of $\tau$ on $\left\{  1,2,\ldots,n\right\}  $. More precisely, $z$
must be the orbit of $\tau$ containing $p$ (since $p\in z$). Hence,%
\begin{equation}
z=\left\{  \tau^{i}\left(  p\right)  \ \mid\ i\in \ZZ \right\}  .
\label{pf.lem.sol.lyndon.gr.basics.cyc1.z=Z}%
\end{equation}


Now, each $u\in \NN $ satisfies
\begin{align*}
p_{u}  &  =\tau^{u}\left(  p\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }p_{u}\right) \\
&  \in\left\{  \tau^{i}\left(  p\right)  \ \mid\ i\in \ZZ \right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }u\in \NN \subset \ZZ %
\right) \\
&  =z\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.gr.basics.cyc1.z=Z})}\right)  .
\end{align*}
In other words, $p_{0},p_{1},p_{2},\ldots$ are elements of $z$. In other
words, $\left\{  p_{0},p_{1},p_{2},\ldots\right\}  \subset z$.

Let us now prove the reverse inclusion. Indeed, let $w\in z$. We shall prove
that $w\in\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $.

We have $w\in z=\left\{  \tau^{i}\left(  p\right)  \ \mid\ i\in \ZZ %
\right\}  $ (by (\ref{pf.lem.sol.lyndon.gr.basics.cyc1.z=Z})). In other words,
$w=\tau^{i}\left(  p\right)  $ for some $i\in \ZZ $. Consider this $i$.
We are in one of the following two cases:

\textit{Case 1:} We have $i\geq0$.

\textit{Case 2:} We have $i<0$.

Let us first consider Case 1. In this case, we have $i\geq0$. Hence,
$i\in \NN $ (since $i\in \ZZ $). Thus, the definition of $p_{i}$
yields $p_{i}=\tau^{i}\left(  p\right)  $, so that $\tau^{i}\left(  p\right)
=p_{i}$. Moreover, $i\in \NN =\left\{  0,1,2,\ldots\right\}  $, so that
$p_{i}\in\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $. Hence, $w=\tau
^{i}\left(  p\right)  =p_{i}\in\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $.
Thus, $w\in\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $ is proved in Case 1.

Let us next consider Case 2. In this case, we have $i<0$. Hence, $-i>0$, so
that $-i\in \NN $ (since $-i\in \ZZ $). Thus, Observation 3 (applied
to $-i$ instead of $i$) yields $\tau^{-\left(  -i\right)  }\left(  p\right)
=p_{\left(  h-1\right)  \left(  -i\right)  }$. In view of $-\left(  -i\right)
=i$, this rewrites as $\tau^{i}\left(  p\right)  =p_{\left(  h-1\right)
\left(  -i\right)  }$. But $\underbrace{\left(  h-1\right)  }_{\in \NN %
}\underbrace{\left(  -i\right)  }_{\in \NN }\in \NN =\left\{
0,1,2,\ldots\right\}  $, so that $p_{\left(  h-1\right)  \left(  -i\right)
}\in\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $. Hence, $w=\tau^{i}\left(
p\right)  =p_{\left(  h-1\right)  \left(  -i\right)  }\in\left\{  p_{0}%
,p_{1},p_{2},\ldots\right\}  $. Thus, $w\in\left\{  p_{0},p_{1},p_{2}%
,\ldots\right\}  $ is proved in Case 2.

We have now proved $w\in\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $ in each
of the two Cases 1 and 2. Thus, $w\in\left\{  p_{0},p_{1},p_{2},\ldots
\right\}  $ always holds (since Cases 1 and 2 cover all possibilities).

Forget that we fixed $w$. We thus have proved that $w\in\left\{  p_{0}%
,p_{1},p_{2},\ldots\right\}  $ for each $w\in z$. In other words,
$z\subset\left\{  p_{0},p_{1},p_{2},\ldots\right\}  $. Combining this with
$\left\{  p_{0},p_{1},p_{2},\ldots\right\}  \subset z$, we obtain $z=\left\{
p_{0},p_{1},p_{2},\ldots\right\}  $. This proves Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(a).

Let us next make some further observations:

\begin{statement}
\textit{Observation 4:} The $h$ elements $p_{0},p_{1},\ldots,p_{h-1}$ are distinct.
\end{statement}

[\textit{Proof of Observation 4:} Assume the contrary. Thus, the $h$ elements
$p_{0},p_{1},\ldots,p_{h-1}$ are not distinct. In other words, (at least) two
of these $h$ elements are equal. In other words, there exist two elements $x$
and $y$ of $\left\{  0,1,\ldots,h-1\right\}  $ such that $x<y$ and
$p_{x}=p_{y}$. Consider these $x$ and $y$.

We have $x\in\left\{  0,1,\ldots,h-1\right\}  \subset \NN $ and
$y\in\left\{  0,1,\ldots,h-1\right\}  \subset \NN $. Moreover, $y-x$ is a
positive integer (since $x<y$). Observation 1 yields $p_{y-x}=p_{0}$. Hence,
(\ref{pf.lem.sol.lyndon.gr.basics.cyc1.smallest}) (applied to $r=y-x$) yields
$y-x\geq h$. But $x\geq0$ (since $x\in\left\{  0,1,\ldots,h-1\right\}  $) and
$y\leq h-1$ (since $y\in\left\{  0,1,\ldots,h-1\right\}  $). Hence,
$y-\underbrace{x}_{\geq0}\leq y\leq h-1<h$. This contradicts $y-x\geq h$. This
contradiction shows that our assumption was false. Hence, Observation 4 is proved.]

\begin{statement}
\textit{Observation 5:} We have $p_{m}\in\left\{  p_{0},p_{1},\ldots
,p_{h-1}\right\}  $ for each $m\in \NN $.
\end{statement}

[\textit{Proof of Observation 5:} Let $m\in \NN $. Recall that $h$ is a
positive integer. Let $q$ and $r$ be the quotient and the remainder obtained
when $m$ is divided by $h$ with remainder. Then, $q\in \NN $ and
$r\in\left\{  0,1,\ldots,h-1\right\}  $ and $m=qh+r$. The definition of
$p_{r}$ yields $p_{r}=\tau^{r}\left(  p\right)  $ (since $r\in\left\{
0,1,\ldots,h-1\right\}  \subset \NN $).

Now, Observation 2 (applied to $i=q$) yields $p_{hq}=p$. Also, Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(d) (applied to $i=r$ and $u=hq$) yields
$\tau^{r}\left(  p_{hq}\right)  =p_{hq+r}=p_{m}$ (since $hq+r=qh+r=m$). Hence,
$p_{m}=\tau^{r}\left(  \underbrace{p_{hq}}_{=p}\right)  =\tau^{r}\left(
p\right)  =p_{r}$ (since $p_{r}=\tau^{r}\left(  p\right)  $). Hence,
$p_{m}=p_{r}\in\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  $ (since
$r\in\left\{  0,1,\ldots,h-1\right\}  $). This proves Observation 5.]

\begin{statement}
\textit{Observation 6:} We have $z=\left\{  p_{0},p_{1},\ldots,p_{h-1}%
\right\}  $.
\end{statement}

[\textit{Proof of Observation 6:} Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(a)
yields $z=\left\{  p_{0},p_{1},p_{2},\ldots\right\}  \supset\left\{
p_{0},p_{1},\ldots,p_{h-1}\right\}  $. On the other hand, each $m\in
 \NN $ satisfies $p_{m}\in\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  $
(by Observation 5). In other words, $p_{0},p_{1},p_{2},\ldots$ are elements of
$\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  $. Thus, $\left\{  p_{0}%
,p_{1},p_{2},\ldots\right\}  \subset\left\{  p_{0},p_{1},\ldots,p_{h-1}%
\right\}  $. Hence, $z=\left\{  p_{0},p_{1},p_{2},\ldots\right\}
\subset\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  $. Combining this with
$z\supset\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  $, we obtain $z=\left\{
p_{0},p_{1},\ldots,p_{h-1}\right\}  $. This proves Observation 6.

\begin{statement}
\textit{Observation 7:} We have $h=k$.
\end{statement}

[\textit{Proof of Observation 7:} The $h$ elements $p_{0},p_{1},\ldots
,p_{h-1}$ are distinct (by Observation 4). Hence, $\left\{  p_{0},p_{1}%
,\ldots,p_{h-1}\right\}  $ is an $h$-element set. In other words, $\left\vert
\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  \right\vert =h$.

But Observation 6 yields $z=\left\{  p_{0},p_{1},\ldots,p_{h-1}\right\}  $.
Hence, $\left\vert z\right\vert =\left\vert \left\{  p_{0},p_{1}%
,\ldots,p_{h-1}\right\}  \right\vert =h$. But $\left\vert z\right\vert =k$
(since $k$ is the size of $z$). Comparing these two equalities, we obtain
$h=k$. This proves Observation 7.]

Now, we can easily prove the remaining parts of Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}:

(b) Observation 7 yields $h=k$. Observation 2 (applied to $i=1$) yields
$p_{h\cdot1}=p$. The definition of $p_{h}$ yields $p_{h}=\tau^{h}\left(
p\right)  =\tau^{k}\left(  p\right)  $ (since $h=k$). Hence,%
\begin{align*}
\tau^{k}\left(  p\right)   &  =p_{h}=p_{h\cdot1}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }h=h\cdot1\right) \\
&  =p.
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(b).

(c) Observation 7 yields $h=k$. But Observation 6 says that $z=\left\{
p_{0},p_{1},\ldots,p_{h-1}\right\}  $. In view of $h=k$, this rewrites as
$z=\left\{  p_{0},p_{1},\ldots,p_{k-1}\right\}  $. This proves Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(c).

(g) Observation 7 yields $h=k$. But Observation 4 says that the $h$ elements
$p_{0},p_{1},\ldots,p_{h-1}$ are distinct. In view of $h=k$, this rewrites as
follows: The $k$ elements $p_{0},p_{1},\ldots,p_{k-1}$ are distinct. This
proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(g).

(e) Let $u\in \NN $. Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(b) yields
$\tau^{k}\left(  p\right)  =p$. On the other hand, $p_{u}=\tau^{u}\left(
p\right)  $ (by the definition of $p_{u}$). But the definition of $p_{u+k}$
yields
\[
p_{u+k}=\underbrace{\tau^{u+k}}_{=\tau^{u}\circ\tau^{k}}\left(  p\right)
=\left(  \tau^{u}\circ\tau^{k}\right)  \left(  p\right)  =\tau^{u}\left(
\underbrace{\tau^{k}\left(  p\right)  }_{=p}\right)  =\tau^{u}\left(
p\right)  .
\]
Comparing this with $p_{u}=\tau^{u}\left(  p\right)  $, we obtain
$p_{u+k}=p_{u}$. This proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(e).

(f) For each $u\in \NN $, we have $p_{u}\in\left\{  p_{0},p_{1}%
,p_{2},\ldots\right\}  =z$ (by Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(a)).
This proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(f).

(h) Set $z_{u}=\left\{  p_{u},p_{u+1},\ldots,p_{u+k-1}\right\}  $ for each
$u\in \NN $. Thus, $z_{0}=\left\{  p_{0},p_{0+1},\ldots,p_{0+k-1}%
\right\}  =\left\{  p_{0},p_{1},\ldots,p_{k-1}\right\}  =z$ (by Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(c)). We now claim that%
\begin{equation}
z_{u}=z_{u+1}\ \ \ \ \ \ \ \ \ \ \text{for each }u\in \NN .
\label{pf.lem.sol.lyndon.gr.basics.cyc1.h.1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.basics.cyc1.h.1}):} Let
$u\in \NN $. Then, $p_{u+k}=p_{u}$ (by Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(e)).

The definition of $z_{u+1}$ yields%
\begin{align*}
z_{u+1}  &  =\left\{  p_{u+1},p_{\left(  u+1\right)  +1},\ldots,p_{\left(
u+1\right)  +\left(  k-1\right)  }\right\}  =\left\{  p_{u+1},p_{u+2}%
,\ldots,p_{u+k}\right\} \\
&  =\left\{  p_{u+1},p_{u+2},\ldots,p_{u+k-1},p_{u+k}\right\}  =\left\{
p_{u+1},p_{u+2},\ldots,p_{u+k-1}\right\}  \cup\left\{  \underbrace{p_{u+k}%
}_{=p_{u}}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\geq1\right) \\
&  =\left\{  p_{u+1},p_{u+2},\ldots,p_{u+k-1}\right\}  \cup\left\{
p_{u}\right\}  =\left\{  p_{u}\right\}  \cup\left\{  p_{u+1},p_{u+2}%
,\ldots,p_{u+k-1}\right\} \\
&  =\left\{  p_{u},p_{u+1},p_{u+2},\ldots,p_{u+k-1}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\geq1\right) \\
&  =\left\{  p_{u},p_{u+1},\ldots,p_{u+k-1}\right\}  .
\end{align*}
Comparing this with%
\[
z_{u}=\left\{  p_{u},p_{u+1},\ldots,p_{u+k-1}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }z_{u}\right)  ,
\]
we obtain $z_{u}=z_{u+1}$. This proves
(\ref{pf.lem.sol.lyndon.gr.basics.cyc1.h.1}).]

Now, (\ref{pf.lem.sol.lyndon.gr.basics.cyc1.h.1}) shows that $z_{u}=z_{u+1}$
for each $u\in \NN $. In other words, $z_{0}=z_{1}$ and $z_{1}=z_{2}$ and
$z_{2}=z_{3}$ and so on. Hence, $z_{0}=z_{1}=z_{2}=z_{3}=\cdots$. Thus,
$z_{0}=z_{u}$ for each $u\in \NN $. Hence, for each $u\in \NN $, we
have $z_{0}=z_{u}=\left\{  p_{u},p_{u+1},\ldots,p_{u+k-1}\right\}  $ (by the
definition of $z_{u}$). In view of $z_{0}=z$, this rewrites as follows: For
each $u\in \NN $, we have $z=\left\{  p_{u},p_{u+1},\ldots,p_{u+k-1}%
\right\}  $. This proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(h).

(i) We know that $h$ is the smallest positive integer $r$ satisfying
$p_{r}=p_{0}$ (since this is how $h$ was defined). Thus,%
\begin{align*}
h  &  =\left(  \text{the smallest positive integer }r\text{ satisfying
}\underbrace{p_{r}}_{\substack{=\tau^{r}\left(  p\right)  \\\text{(by the
definition of }p_{r}\text{)}}}=\underbrace{p_{0}}_{=p}\right) \\
&  =\left(  \text{the smallest positive integer }r\text{ satisfying }\tau
^{r}\left(  p\right)  =p\right) \\
&  =\left(  \text{the smallest positive integer }i\text{ satisfying }\tau
^{i}\left(  p\right)  =p\right)
\end{align*}
(here, we have renamed the index $r$ as $i$). But Observation 7 yields $h=k$.
Comparing these two equalities, we obtain%
\begin{equation}
k=\left(  \text{the smallest positive integer }i\text{ satisfying }\tau
^{i}\left(  p\right)  =p\right)  .
\label{pf.lem.sol.lyndon.gr.basics.cyc1.i.1}%
\end{equation}
On the other hand, we have defined $\operatorname*{ord}\nolimits_{\tau}\left(
p\right)  $ to be the smallest positive integer $i$ such that $\tau^{i}\left(
p\right)  =p$. Hence,%
\[
\operatorname*{ord}\nolimits_{\tau}\left(  p\right)  =\left(  \text{the
smallest positive integer }i\text{ satisfying }\tau^{i}\left(  p\right)
=p\right)  .
\]
Comparing this with (\ref{pf.lem.sol.lyndon.gr.basics.cyc1.i.1}), we obtain
$\operatorname*{ord}\nolimits_{\tau}\left(  p\right)  =k$. This proves Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(i).
\end{proof}

\end{verlong}

We record a simple corollary of Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}:

\begin{lemma}
\label{lem.sol.lyndon.gr.basics.cyc2}Let $n\in \NN $. Let $\tau
\in\Symm_{n}$ be a permutation. Let $p\in\left\{  1,2,\ldots,n\right\}
$. Let $j=\operatorname*{ord}\nolimits_{\tau}\left(  p\right)  $. Then:

\begin{enumerate}
\item[(a)] We have $\tau^{j}\left(  p\right)  =p$.

\item[(b)] We have $j=\operatorname*{ord}\nolimits_{\tau}\left(  \tau\left(
p\right)  \right)  $.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.basics.cyc2}.]Let $z$ be the cycle of
$\tau$ that contains $p$. Thus, $p\in z$. Let $k$ be the size of $z$. For each
$u\in \NN $, we set $p_{u}:=\tau^{u}\left(  p\right)  $.

Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(i) yields $\operatorname*{ord}%
\nolimits_{\tau}\left(  p\right)  =k$. Hence, $j=\operatorname*{ord}%
\nolimits_{\tau}\left(  p\right)  =k$. But Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(b) yields $\tau^{k}\left(  p\right)  =p$.
But from $j=k$, we obtain $\tau^{j}\left(  p\right)  =\tau^{k}\left(
p\right)  =p$. This proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc2}(a).

(b) Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(f) (applied to $u=1$) yields
$p_{1}\in z$. But the definition of $p_{1}$ yields $p_{1}=\underbrace{\tau
^{1}}_{=\tau}\left(  p\right)  =\tau\left(  p\right)  $. Hence, $\tau\left(
p\right)  =p_{1}\in z$.

Set $q=\tau\left(  p\right)  $. Thus, $q=\tau\left(  p\right)  \in z$. For
each $u\in \NN $, we set $q_{u}:=\tau^{u}\left(  q\right)  $. Then, Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(i) (applied to $q$ and $q_{u}$ instead of
$p$ and $p_{u}$) yields $\operatorname*{ord}\nolimits_{\tau}\left(  q\right)
=k$. Comparing this with $j=k$, we obtain $j=\operatorname*{ord}%
\nolimits_{\tau}\left(  \underbrace{q}_{=\tau\left(  p\right)  }\right)
=\operatorname*{ord}\nolimits_{\tau}\left(  \tau\left(  p\right)  \right)  $.
This proves Lemma \ref{lem.sol.lyndon.gr.basics.cyc2}(b).
\end{proof}

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.words.read1}.]Let $k=\operatorname*{ord}%
\nolimits_{\tau}\left(  h\right)  $. Thus, $k$ is the smallest positive
integer $i$ such that $\tau^{i}\left(  h\right)  =h$ (by the definition of
$\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $). Hence, $k$ is a
positive integer, and we have $\tau^{k}\left(  h\right)  =h$. Furthermore,
$\tau^{k+1}\left(  h\right)  =\underbrace{\tau}_{=\tau^{1}}\left(
\underbrace{\tau^{k}\left(  h\right)  }_{=h}\right)  =\tau^{1}\left(
h\right)  $, so that $\tau^{1}\left(  h\right)  =\tau^{k+1}\left(  h\right)  $.

The definition of $w_{\tau,h}$ yields
\begin{align}
w_{\tau,h}  &  =w_{\tau^{1}\left(  h\right)  }w_{\tau^{2}\left(  h\right)
}\cdots w_{\tau^{k}\left(  h\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  \right) \nonumber\\
&  =\left(  w_{\tau^{1}\left(  h\right)  },w_{\tau^{2}\left(  h\right)
},\ldots,w_{\tau^{k}\left(  h\right)  }\right)  .
\label{pf.prop.words.read1.short.wtauh=}%
\end{align}
Hence, the word $w_{\tau,h}$ has length $k$. In other words, the word
$w_{\tau,h}$ has length $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)
$ (since $k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $). Moreover,
the word $w_{\tau,h}$ is nonempty (since it has length $k$, but $k$ is a
positive integer). This proves Proposition \ref{prop.words.read1}(a).

(b) The first letter of the word $w_{\tau,h}$ exists (since the word
$w_{\tau,h}$ is nonempty). This first letter is $w_{\tau^{1}\left(  h\right)
}$ (since $w_{\tau,h}=\left(  w_{\tau^{1}\left(  h\right)  },w_{\tau
^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)  $). In
other words, this first letter is $w_{\tau\left(  h\right)  }$ (since
$\tau^{1}=\tau$). This proves Proposition \ref{prop.words.read1}(b).

(c) The last letter of the word $w_{\tau,h}$ exists (since the word
$w_{\tau,h}$ is nonempty). This last letter is $w_{\tau^{k}\left(  h\right)
}$ (since $w_{\tau,h}=\left(  w_{\tau^{1}\left(  h\right)  },w_{\tau
^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)  $). In
other words, this last letter is $w_{h}$ (since $\tau^{k}\left(  h\right)
=h$). This proves Proposition \ref{prop.words.read1}(c).

(d) From (\ref{pf.prop.words.read1.short.wtauh=}), we obtain%
\begin{align}
c\cdot w_{\tau,h}  &  =c\cdot\left(  w_{\tau^{1}\left(  h\right)  }%
,w_{\tau^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)
=\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)  }%
,\ldots,w_{\tau^{k}\left(  h\right)  },w_{\tau^{1}\left(  h\right)  }\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the action of
}C\text{ on }\mathfrak{A}^{k}\right) \nonumber\\
&  =\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)
},\ldots,w_{\tau^{k}\left(  h\right)  },w_{\tau^{k+1}\left(  h\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau^{1}\left(  h\right)
=\tau^{k+1}\left(  h\right)  \right) \nonumber\\
&  =\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)
},\ldots,w_{\tau^{k+1}\left(  h\right)  }\right)  .
\label{pf.prop.words.read1.short.d.2}%
\end{align}


On the other hand, we have $k=\operatorname*{ord}\nolimits_{\tau}\left(
h\right)  $ and therefore $k=\operatorname*{ord}\nolimits_{\tau}\left(
\tau\left(  h\right)  \right)  $ (by Lemma \ref{lem.sol.lyndon.gr.basics.cyc2}%
(b), applied to $j=k$ and $p=h$). Hence, the definition of $w_{\tau
,\tau\left(  h\right)  }$ yields%
\begin{align*}
w_{\tau,\tau\left(  h\right)  }  &  =w_{\tau^{1}\left(  \tau\left(  h\right)
\right)  }w_{\tau^{2}\left(  \tau\left(  h\right)  \right)  }\cdots
w_{\tau^{k}\left(  \tau\left(  h\right)  \right)  }=\left(  w_{\tau^{1}\left(
\tau\left(  h\right)  \right)  },w_{\tau^{2}\left(  \tau\left(  h\right)
\right)  },\ldots,w_{\tau^{k}\left(  \tau\left(  h\right)  \right)  }\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }w_{\tau^{1}\left(  \tau\left(
h\right)  \right)  },w_{\tau^{2}\left(  \tau\left(  h\right)  \right)
},\ldots,w_{\tau^{k}\left(  \tau\left(  h\right)  \right)  }\text{ are single
letters}\right) \\
&  =\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)
},\ldots,w_{\tau^{k+1}\left(  h\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfies }%
w_{\tau^{i}\left(  \tau\left(  h\right)  \right)  }=w_{\tau^{i+1}\left(
h\right)  }\\
\text{(because }\tau^{i}\left(  \tau\left(  h\right)  \right)  =\tau
^{i+1}\left(  h\right)  \text{)}%
\end{array}
\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.words.read1.short.d.2}), we obtain
$w_{\tau,\tau\left(  h\right)  }=c\cdot w_{\tau,h}$. This proves Proposition
\ref{prop.words.read1}(d).

(e) Let us first show that%
\begin{equation}
w_{\tau,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in \NN .
\label{pf.prop.words.read1.short.e.1}%
\end{equation}
[\textit{Proof of (\ref{pf.prop.words.read1.short.e.1}):} We shall prove
(\ref{pf.prop.words.read1.short.e.1}) by induction on $i$:

\textit{Induction base:} We have $w_{\tau,\tau^{0}\left(  h\right)  }%
=w_{\tau,h}$ (since $\underbrace{\tau^{0}}_{=\operatorname*{id}}\left(
h\right)  =h$). Comparing this with $\underbrace{c^{0}}_{=\operatorname*{id}%
}\cdot w_{\tau,h}=w_{\tau,h}$, we obtain $w_{\tau,\tau^{0}\left(  h\right)
}=c^{0}\cdot w_{\tau,h}$. In other words, (\ref{pf.prop.words.read1.short.e.1}%
) holds for $i=0$. This completes the induction base.

\textit{Induction step:} Let $j\in \NN $. Assume that
(\ref{pf.prop.words.read1.short.e.1}) holds for $i=j$. We must prove that
(\ref{pf.prop.words.read1.short.e.1}) holds for $i=j+1$.

We have assumed that (\ref{pf.prop.words.read1.short.e.1}) holds for $i=j$. In
other words, we have $w_{\tau,\tau^{j}\left(  h\right)  }=c^{j}\cdot
w_{\tau,h}$. Now,%
\begin{align*}
w_{\tau,\tau^{j+1}\left(  h\right)  }  &  =w_{\tau,\tau\left(  \tau^{j}\left(
h\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau^{j+1}\left(
h\right)  =\tau\left(  \tau^{j}\left(  h\right)  \right)  \right) \\
&  =c\cdot\underbrace{w_{\tau,\tau^{j}\left(  h\right)  }}_{=c^{j}\cdot
w_{\tau,h}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.words.read1}(d), applied to }\tau^{j}\left(  h\right)  \text{
instead of }h\right) \\
&  =\underbrace{c\cdot c^{j}}_{=c^{j+1}}\cdot w_{\tau,h}=c^{j+1}\cdot
w_{\tau,h}.
\end{align*}
In other words, (\ref{pf.prop.words.read1.short.e.1}) holds for $i=j+1$. This
completes the induction step. Thus, (\ref{pf.prop.words.read1.short.e.1}) is
proved by induction.]

Now, forget that we fixed $h$. We thus have proved
(\ref{pf.prop.words.read1.short.e.1}) for each $h\in\left\{  1,2,\ldots
,n\right\}  $.

Now, let $h\in\left\{  1,2,\ldots,n\right\}  $. We want to prove Proposition
\ref{prop.words.read1}(e); in other words, we must prove that $w_{\tau
,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$ for each $i\in \ZZ $.
So let us fix $i\in \ZZ $. We must then prove that $w_{\tau,\tau
^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$. If $i\in \NN $, then this
follows immediately from (\ref{pf.prop.words.read1.short.e.1}). Thus, we WLOG
assume that $i\notin \NN $. Hence, $i$ is a negative integer (since
$i\in \ZZ $ but $i\notin \NN $), so that $-i\in\left\{
1,2,3,\ldots\right\}  \subset \NN $. Thus, we can apply
(\ref{pf.prop.words.read1.short.e.1}) to $\tau^{i}\left(  h\right)  $ and $-i$
instead of $h$ and $i$ (since we have proved
(\ref{pf.prop.words.read1.short.e.1}) for every value of $h$, not just
for the $h$ that we are currently considering). We thus obtain
\[
w_{\tau,\tau^{-i}\left(  \tau^{i}\left(  h\right)  \right)  }=c^{-i}\cdot
w_{\tau,\tau^{i}\left(  h\right)  }.
\]
In view of $\tau^{-i}\left(  \tau^{i}\left(  h\right)  \right)
=\underbrace{\left(  \tau^{-i}\circ\tau^{i}\right)  }_{=\operatorname*{id}%
}\left(  h\right)  =h$, this rewrites as
\[
w_{\tau,h}=c^{-i}\cdot w_{\tau,\tau^{i}\left(  h\right)  }.
\]
Hence,%
\[
c^{i}\cdot\underbrace{w_{\tau,h}}_{=c^{-i}\cdot w_{\tau,\tau^{i}\left(
h\right)  }}=\underbrace{c^{i}\cdot c^{-i}}_{=\operatorname*{id}}\cdot
w_{\tau,\tau^{i}\left(  h\right)  }=w_{\tau,\tau^{i}\left(  h\right)  }.
\]
In other words, $w_{\tau,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$.
This proves Proposition \ref{prop.words.read1}(e).
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.words.read1}.]Recall that $\operatorname*{ord}%
\nolimits_{\tau}\left(  h\right)  $ was defined as the smallest positive
integer $i$ such that $\tau^{i}\left(  h\right)  =h$. Thus,
$\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $ is a positive integer.

Let $k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $. Thus, $k$ is a
positive integer (since $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)
$ is a positive integer). The definition of $w_{\tau,h}$ yields%
\begin{align}
w_{\tau,h}  &  =w_{\tau^{1}\left(  h\right)  }w_{\tau^{2}\left(  h\right)
}\cdots w_{\tau^{k}\left(  h\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  \right) \nonumber\\
&  =\left(  w_{\tau^{1}\left(  h\right)  },w_{\tau^{2}\left(  h\right)
},\ldots,w_{\tau^{k}\left(  h\right)  }\right)
\label{pf.prop.words.read1.wtauh=}%
\end{align}
(since $w_{\tau^{1}\left(  h\right)  },w_{\tau^{2}\left(  h\right)  }%
,\ldots,w_{\tau^{k}\left(  h\right)  }$ are single letters). Hence,
$w_{\tau,h}$ is a word of length $k$. In other words, $w_{\tau,h}$ is a word
of length $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $ (since
$k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $). Thus, $\ell\left(
w_{\tau,h}\right)  =\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  >0$
(since $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $ is a positive
integer). In other words, the word $w_{\tau,h}$ is nonempty. Hence, we have
now shown that the word $w_{\tau,h}$ is nonempty and has length
$\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $. This proves
Proposition \ref{prop.words.read1}(a).

Lemma \ref{lem.sol.lyndon.gr.basics.cyc2}(a) (applied to $j=k$ and $p=h$)
yields $\tau^{k}\left(  h\right)  =h$ (since $k=\operatorname*{ord}%
\nolimits_{\tau}\left(  h\right)  $). Applying the map $\tau$ to both sides of
this equality, we obtain $\tau\left(  \tau^{k}\left(  h\right)  \right)
=\underbrace{\tau}_{=\tau^{1}}\left(  h\right)  =\tau^{1}\left(  h\right)  $,
so that%
\[
\tau^{1}\left(  h\right)  =\tau\left(  \tau^{k}\left(  h\right)  \right)
=\underbrace{\left(  \tau\circ\tau^{k}\right)  }_{=\tau^{k+1}}\left(
h\right)  =\tau^{k+1}\left(  h\right)  .
\]


(b) The first letter of the word $w_{\tau,h}$ exists (since the word
$w_{\tau,h}$ is nonempty). Moreover, this first letter is $w_{\tau^{1}\left(
h\right)  }$ (since $w_{\tau,h}=\left(  w_{\tau^{1}\left(  h\right)  }%
,w_{\tau^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)
$). In other words, this first letter is $w_{\tau\left(  h\right)  }$ (since
$\tau^{1}=\tau$). Hence, we have shown that the first letter of the word
$w_{\tau,h}$ is $w_{\tau\left(  h\right)  }$. This proves Proposition
\ref{prop.words.read1}(b).

(c) The last letter of the word $w_{\tau,h}$ exists (since the word
$w_{\tau,h}$ is nonempty). Moreover, this last letter is $w_{\tau^{k}\left(
h\right)  }$ (since $w_{\tau,h}=\left(  w_{\tau^{1}\left(  h\right)  }%
,w_{\tau^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)
$). In other words, this last letter is $w_{h}$ (since $\tau^{k}\left(
h\right)  =h$). Hence, we have shown that the last letter of the word
$w_{\tau,h}$ is $w_{h}$. This proves Proposition \ref{prop.words.read1}(c).

(d) From (\ref{pf.prop.words.read1.wtauh=}), we obtain%
\begin{align}
c\cdot w_{\tau,h}  &  =c\cdot\left(  w_{\tau^{1}\left(  h\right)  }%
,w_{\tau^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)
=\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)  }%
,\ldots,w_{\tau^{k}\left(  h\right)  },w_{\tau^{1}\left(  h\right)  }\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the action of
}C\text{ on }\mathfrak{A}^{k}\right) \nonumber\\
&  =\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)
},\ldots,w_{\tau^{k}\left(  h\right)  },w_{\tau^{k+1}\left(  h\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau^{1}\left(  h\right)
=\tau^{k+1}\left(  h\right)  \right) \nonumber\\
&  =\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)
},\ldots,w_{\tau^{k+1}\left(  h\right)  }\right)  .
\label{pf.prop.words.read1.d.2}%
\end{align}


On the other hand, we have $k=\operatorname*{ord}\nolimits_{\tau}\left(
h\right)  $ and therefore $k=\operatorname*{ord}\nolimits_{\tau}\left(
\tau\left(  h\right)  \right)  $ (by Lemma \ref{lem.sol.lyndon.gr.basics.cyc2}%
(b), applied to $j=k$ and $p=h$). Hence, the definition of $w_{\tau
,\tau\left(  h\right)  }$ yields%
\begin{align*}
w_{\tau,\tau\left(  h\right)  }  &  =w_{\tau^{1}\left(  \tau\left(  h\right)
\right)  }w_{\tau^{2}\left(  \tau\left(  h\right)  \right)  }\cdots
w_{\tau^{k}\left(  \tau\left(  h\right)  \right)  }=\left(  w_{\tau^{1}\left(
\tau\left(  h\right)  \right)  },w_{\tau^{2}\left(  \tau\left(  h\right)
\right)  },\ldots,w_{\tau^{k}\left(  \tau\left(  h\right)  \right)  }\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }w_{\tau^{1}\left(  \tau\left(
h\right)  \right)  },w_{\tau^{2}\left(  \tau\left(  h\right)  \right)
},\ldots,w_{\tau^{k}\left(  \tau\left(  h\right)  \right)  }\text{ are single
letters}\right) \\
&  =\left(  w_{\tau^{2}\left(  h\right)  },w_{\tau^{3}\left(  h\right)
},\ldots,w_{\tau^{k+1}\left(  h\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfies }%
w_{\tau^{i}\left(  \tau\left(  h\right)  \right)  }=w_{\tau^{i+1}\left(
h\right)  }\\
\text{(because }\tau^{i}\left(  \tau\left(  h\right)  \right)
=\underbrace{\left(  \tau^{i}\circ\tau\right)  }_{=\tau^{i+1}}\left(
h\right)  =\tau^{i+1}\left(  h\right)  \text{)}%
\end{array}
\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.words.read1.d.2}), we obtain $w_{\tau
,\tau\left(  h\right)  }=c\cdot w_{\tau,h}$. This proves Proposition
\ref{prop.words.read1}(d).

(e) Let us first show that%
\begin{equation}
w_{\tau,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in \NN .
\label{pf.prop.words.read1.e.1}%
\end{equation}
[\textit{Proof of (\ref{pf.prop.words.read1.e.1}):} We shall prove
(\ref{pf.prop.words.read1.e.1}) by induction on $i$:

\textit{Induction base:} We have $w_{\tau,\tau^{0}\left(  h\right)  }%
=w_{\tau,h}$ (since $\underbrace{\tau^{0}}_{=\operatorname*{id}}\left(
h\right)  =h$). Comparing this with $\underbrace{c^{0}}_{=\operatorname*{id}%
}\cdot w_{\tau,h}=w_{\tau,h}$, we obtain $w_{\tau,\tau^{0}\left(  h\right)
}=c^{0}\cdot w_{\tau,h}$. In other words, (\ref{pf.prop.words.read1.e.1})
holds for $i=0$. This completes the induction base.

\textit{Induction step:} Let $j\in \NN $. Assume that
(\ref{pf.prop.words.read1.e.1}) holds for $i=j$. We must prove that
(\ref{pf.prop.words.read1.e.1}) holds for $i=j+1$.

We have assumed that (\ref{pf.prop.words.read1.e.1}) holds for $i=j$. In other
words, we have $w_{\tau,\tau^{j}\left(  h\right)  }=c^{j}\cdot w_{\tau,h}$.
Now, $\underbrace{\tau^{j+1}}_{=\tau\circ\tau^{j}}\left(  h\right)  =\left(
\tau\circ\tau^{j}\right)  \left(  h\right)  =\tau\left(  \tau^{j}\left(
h\right)  \right)  $, so that%
\begin{align*}
w_{\tau,\tau^{j+1}\left(  h\right)  }  &  =w_{\tau,\tau\left(  \tau^{j}\left(
h\right)  \right)  }\\
&  =c\cdot\underbrace{w_{\tau,\tau^{j}\left(  h\right)  }}_{=c^{j}\cdot
w_{\tau,h}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.words.read1}(d), applied to }\tau^{j}\left(  h\right)  \text{
instead of }h\right) \\
&  =c\cdot\left(  c^{j}\cdot w_{\tau,h}\right)  =\underbrace{\left(
cc^{j}\right)  }_{=c^{j+1}}\cdot w_{\tau,h}=c^{j+1}\cdot w_{\tau,h}.
\end{align*}
In other words, (\ref{pf.prop.words.read1.e.1}) holds for $i=j+1$. This
completes the induction step. Thus, (\ref{pf.prop.words.read1.e.1}) is proved
by induction.]

Now, forget that we fixed $h$. We thus have proved
(\ref{pf.prop.words.read1.e.1}) for each $h\in\left\{  1,2,\ldots,n\right\}  $.

Now, let $h\in\left\{  1,2,\ldots,n\right\}  $. We want to prove Proposition
\ref{prop.words.read1}(e); in other words, we must prove that $w_{\tau
,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$ for each $i\in \ZZ $.
So let us fix $i\in \ZZ $. We must then prove that $w_{\tau,\tau
^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$. If $i\in \NN $, then this
follows immediately from (\ref{pf.prop.words.read1.e.1}). Thus, we WLOG assume
that we don't have $i\in \NN $. Hence, $i\notin \NN $. Combining
$i\in \ZZ $ with $i\notin \NN $, we obtain $i\in \ZZ %
\setminus \NN =\left\{  -1,-2,-3,\ldots\right\}  $, so that
$-i\in\left\{  1,2,3,\ldots\right\}  \subset \NN $. Thus, we can apply
(\ref{pf.prop.words.read1.e.1}) to $\tau^{i}\left(  h\right)  $ and $-i$
instead of $h$ and $i$ (since we have proved (\ref{pf.prop.words.read1.e.1})
for every value of $h$, not just for the $h$ that we are currently
considering). We thus obtain
\[
w_{\tau,\tau^{-i}\left(  \tau^{i}\left(  h\right)  \right)  }=c^{-i}\cdot
w_{\tau,\tau^{i}\left(  h\right)  }.
\]
In view of $\tau^{-i}\left(  \tau^{i}\left(  h\right)  \right)
=\underbrace{\left(  \tau^{-i}\circ\tau^{i}\right)  }_{=\operatorname*{id}%
}\left(  h\right)  =h$, this rewrites as
\[
w_{\tau,h}=c^{-i}\cdot w_{\tau,\tau^{i}\left(  h\right)  }.
\]
Hence,%
\[
c^{i}\cdot w_{\tau,h}=c^{i}\cdot\left(  c^{-i}\cdot w_{\tau,\tau^{i}\left(
h\right)  }\right)  =\underbrace{\left(  c^{i}c^{-i}\right)  }%
_{=\operatorname*{id}}\cdot w_{\tau,\tau^{i}\left(  h\right)  }=w_{\tau
,\tau^{i}\left(  h\right)  }.
\]
In other words, $w_{\tau,\tau^{i}\left(  h\right)  }=c^{i}\cdot w_{\tau,h}$.
This proves Proposition \ref{prop.words.read1}(e).
\end{proof}

\end{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.basics.1}.]Definition
\ref{def.words.std} says that $\operatorname*{std}w$ is the unique permutation
$\sigma\in\Symm_{n}$ defined in Proposition \ref{prop.words.std-exists}%
. In other words, $\operatorname*{std}w$ is the unique permutation $\sigma
\in\Symm_{n}$ such that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have $\left(  \sigma\left(
a\right)  <\sigma\left(  b\right)  \text{ if and only if }w_{a}\leq
w_{b}\right)  $. Hence, $\operatorname*{std}w$ is such a permutation $\sigma$.
In other words, $\operatorname*{std}w$ is a permutation in $\Symm_{n}$
and has the property that for every two elements $a$ and $b$ of $\left\{
1,2,\ldots,n\right\}  $ satisfying $a<b$, we have%
\begin{equation}
\left(  \left(  \operatorname*{std}w\right)  \left(  a\right)  <\left(
\operatorname*{std}w\right)  \left(  b\right)  \text{ if and only if }%
w_{a}\leq w_{b}\right)  . \label{pf.lem.sol.lyndon.gr.basics.1.stda-stdb}%
\end{equation}


But $\tau=\left(  \operatorname*{std}w\right)  ^{-1}$, so that $\tau
^{-1}=\operatorname*{std}w$. Note that $\tau$ is a permutation (since $\tau
\in\Symm_{n}$). Hence, $\tau$ is a bijective map. Thus, in particular,
$\tau$ is an injective map.

(a) Assume that $\tau^{-1}\left(  \alpha\right)  <\tau^{-1}\left(
\beta\right)  $. In view of $\tau^{-1}=\operatorname*{std}w$, this rewrites as
$\left(  \operatorname*{std}w\right)  \left(  \alpha\right)  <\left(
\operatorname*{std}w\right)  \left(  \beta\right)  $. But
(\ref{pf.lem.sol.lyndon.gr.basics.1.stda-stdb}) (applied to $a=\alpha$ and
$b=\beta$) shows that we have
\[
\left(  \left(  \operatorname*{std}w\right)  \left(  \alpha\right)  <\left(
\operatorname*{std}w\right)  \left(  \beta\right)  \text{ if and only if
}w_{\alpha}\leq w_{\beta}\right)  .
\]
Hence, we have $w_{\alpha}\leq w_{\beta}$ (since we have $\left(
\operatorname*{std}w\right)  \left(  \alpha\right)  <\left(
\operatorname*{std}w\right)  \left(  \beta\right)  $). This proves Lemma
\ref{lem.sol.lyndon.gr.basics.1}(a).

(b) Assume that $\tau^{-1}\left(  \alpha\right)  \geq\tau^{-1}\left(
\beta\right)  $. In view of $\tau^{-1}=\operatorname*{std}w$, this rewrites as
$\left(  \operatorname*{std}w\right)  \left(  \alpha\right)  \geq\left(
\operatorname*{std}w\right)  \left(  \beta\right)  $. Hence, $\left(
\operatorname*{std}w\right)  \left(  \alpha\right)  <\left(
\operatorname*{std}w\right)  \left(  \beta\right)  $ does not hold. But
(\ref{pf.lem.sol.lyndon.gr.basics.1.stda-stdb}) (applied to $a=\alpha$ and
$b=\beta$) shows that we have
\[
\left(  \left(  \operatorname*{std}w\right)  \left(  \alpha\right)  <\left(
\operatorname*{std}w\right)  \left(  \beta\right)  \text{ if and only if
}w_{\alpha}\leq w_{\beta}\right)  .
\]
Hence, $w_{\alpha}\leq w_{\beta}$ does not hold (since $\left(
\operatorname*{std}w\right)  \left(  \alpha\right)  <\left(
\operatorname*{std}w\right)  \left(  \beta\right)  $ does not hold). In other
words, we have $w_{\alpha}>w_{\beta}$. This proves Lemma
\ref{lem.sol.lyndon.gr.basics.1}(b).

(d) Assume that $\tau\left(  \alpha\right)  \geq\tau\left(  \beta\right)  $.
But $\alpha\neq\beta$ (since $\alpha<\beta$) and thus $\tau\left(
\alpha\right)  \neq\tau\left(  \beta\right)  $ (since $\tau$ is injective).
Combining this with $\tau\left(  \alpha\right)  \geq\tau\left(  \beta\right)
$, we obtain $\tau\left(  \alpha\right)  >\tau\left(  \beta\right)  $. In
other words, $\tau\left(  \beta\right)  <\tau\left(  \alpha\right)  $. Also,
$\beta\geq\alpha$ (since $\alpha<\beta$), so that $\tau^{-1}\left(
\tau\left(  \beta\right)  \right)  =\beta\geq\alpha=\tau^{-1}\left(
\tau\left(  \alpha\right)  \right)  $. Hence, Lemma
\ref{lem.sol.lyndon.gr.basics.1}(b) (applied to $\tau\left(  \beta\right)  $
and $\tau\left(  \alpha\right)  $ instead of $\alpha$ and $\beta$) yields
$w_{\tau\left(  \beta\right)  }>w_{\tau\left(  \alpha\right)  }$. In other
words, $w_{\tau\left(  \alpha\right)  }<w_{\tau\left(  \beta\right)  }$. This
proves Lemma \ref{lem.sol.lyndon.gr.basics.1}(d).

(c) We are in one of the following two cases:

\textit{Case 1:} We have $\tau\left(  \alpha\right)  <\tau\left(
\beta\right)  $.

\textit{Case 2:} We have $\tau\left(  \alpha\right)  \geq\tau\left(
\beta\right)  $.

Let us first consider Case 1. In this case, we have $\tau\left(
\alpha\right)  <\tau\left(  \beta\right)  $. Also, $\tau^{-1}\left(
\tau\left(  \alpha\right)  \right)  =\alpha<\beta=\tau^{-1}\left(  \tau\left(
\beta\right)  \right)  $. Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(a)
(applied to $\tau\left(  \alpha\right)  $ and $\tau\left(  \beta\right)  $
instead of $\alpha$ and $\beta$) yields $w_{\tau\left(  \alpha\right)  }\leq
w_{\tau\left(  \beta\right)  }$. Thus, Lemma \ref{lem.sol.lyndon.gr.basics.1}%
(c) is proven in Case 1.

Let us next consider Case 2. In this case, we have $\tau\left(  \alpha\right)
\geq\tau\left(  \beta\right)  $. Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}%
(d) yields $w_{\tau\left(  \alpha\right)  }<w_{\tau\left(  \beta\right)  }$.
Hence, $w_{\tau\left(  \alpha\right)  }\leq w_{\tau\left(  \beta\right)  }$.
Thus, Lemma \ref{lem.sol.lyndon.gr.basics.1}(c) is proven in Case 2.

\begin{vershort}
We have thus proved Lemma \ref{lem.sol.lyndon.gr.basics.1}(c) in each of the
two Cases 1 and 2. Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(c) always holds.
\end{vershort}

\begin{verlong}
We have thus proved Lemma \ref{lem.sol.lyndon.gr.basics.1}(c) in each of the
two Cases 1 and 2. Since these two Cases cover all possibilities, we thus
conclude that Lemma \ref{lem.sol.lyndon.gr.basics.1}(c) always holds.
\end{verlong}

(e) Assume that $w_{\tau\left(  \alpha\right)  }=w_{\tau\left(  \beta\right)
}$. We must prove that $\tau\left(  \alpha\right)  <\tau\left(  \beta\right)
$.

Assume the contrary. Thus, $\tau\left(  \alpha\right)  \geq\tau\left(
\beta\right)  $. Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(d) yields
$w_{\tau\left(  \alpha\right)  }<w_{\tau\left(  \beta\right)  }$. This
contradicts $w_{\tau\left(  \alpha\right)  }=w_{\tau\left(  \beta\right)  }$.
This contradiction shows that our assumption was false. Hence, Lemma
\ref{lem.sol.lyndon.gr.basics.1}(e) is proven.

(f) Assume that $w_{\tau,\alpha}=w_{\tau,\beta}$.

\begin{vershort}
Now, Proposition \ref{prop.words.read1}(b) (applied to $h=\alpha$) yields that
the first letter of the word $w_{\tau,\alpha}$ is $w_{\tau\left(
\alpha\right)  }$. Likewise, the first letter of the word $w_{\tau,\beta}$ is
$w_{\tau\left(  \beta\right)  }$. Hence, the first letters of the two words
$w_{\tau,\alpha}$ and $w_{\tau,\beta}$ are $w_{\tau\left(  \alpha\right)  }$
and $w_{\tau\left(  \beta\right)  }$. Thus, from $w_{\tau,\alpha}%
=w_{\tau,\beta}$, we obtain $w_{\tau\left(  \alpha\right)  }=w_{\tau\left(
\beta\right)  }$. Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(e) yields
$\tau\left(  \alpha\right)  <\tau\left(  \beta\right)  $. It remains to show
that $w_{\tau,\tau\left(  \alpha\right)  }=w_{\tau,\tau\left(  \beta\right)
}$.
\end{vershort}

\begin{verlong}
Now, Proposition \ref{prop.words.read1}(b) (applied to $h=\alpha$) yields that
the first letter of the word $w_{\tau,\alpha}$ is $w_{\tau\left(
\alpha\right)  }$. In other words,%
\begin{equation}
\left(  \text{the first letter of the word }w_{\tau,\alpha}\right)
=w_{\tau\left(  \alpha\right)  }. \label{pf.lem.sol.lyndon.gr.basics.1.f.1}%
\end{equation}
The same argument (applied to $\beta$ instead of $\alpha$) yields%
\[
\left(  \text{the first letter of the word }w_{\tau,\beta}\right)
=w_{\tau\left(  \beta\right)  }.
\]
In view of $w_{\tau,\alpha}=w_{\tau,\beta}$, this rewrites as
\[
\left(  \text{the first letter of the word }w_{\tau,\alpha}\right)
=w_{\tau\left(  \beta\right)  }.
\]
Comparing this with (\ref{pf.lem.sol.lyndon.gr.basics.1.f.1}), we obtain
$w_{\tau\left(  \alpha\right)  }=w_{\tau\left(  \beta\right)  }$. Hence, Lemma
\ref{lem.sol.lyndon.gr.basics.1}(e) yields $\tau\left(  \alpha\right)
<\tau\left(  \beta\right)  $. It remains to show that $w_{\tau,\tau\left(
\alpha\right)  }=w_{\tau,\tau\left(  \beta\right)  }$.
\end{verlong}

Proposition \ref{prop.words.read1}(d) (applied to $h=\alpha$) yields
$w_{\tau,\tau\left(  \alpha\right)  }=c\cdot w_{\tau,\alpha}$. Proposition
\ref{prop.words.read1}(d) (applied to $h=\beta$) yields $w_{\tau,\tau\left(
\beta\right)  }=c\cdot w_{\tau,\beta}$. Thus,%
\[
w_{\tau,\tau\left(  \alpha\right)  }=c\cdot\underbrace{w_{\tau,\alpha}%
}_{=w_{\tau,\beta}}=c\cdot w_{\tau,\beta}=w_{\tau,\tau\left(  \beta\right)
}.
\]
Thus, the proof of Lemma \ref{lem.sol.lyndon.gr.basics.1}(f) is complete
(since we already have shown that $\tau\left(  \alpha\right)  <\tau\left(
\beta\right)  $).

(g) Assume that $w_{\tau,\alpha}=w_{\tau,\beta}$. We must show that
\begin{equation}
\tau^{i}\left(  \alpha\right)  <\tau^{i}\left(  \beta\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in \NN .
\label{pf.lem.sol.lyndon.gr.basics.1.g.1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}):} We shall prove
(\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}) by induction on $i$:

\textit{Induction base:} We have $\alpha<\beta$. In view of $\underbrace{\tau
^{0}}_{=\operatorname*{id}}\left(  \alpha\right)  =\alpha$ and
$\underbrace{\tau^{0}}_{=\operatorname*{id}}\left(  \beta\right)  =\beta$,
this rewrites as $\tau^{0}\left(  \alpha\right)  <\tau^{0}\left(
\beta\right)  $. In other words, (\ref{pf.lem.sol.lyndon.gr.basics.1.g.1})
holds for $i=0$. This completes the induction base.

\textit{Induction step:} Let $j\in \NN $. Assume that
(\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}) holds for $i=j$. We must prove that
(\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}) holds for $i=j+1$.

We have assumed that (\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}) holds for
$i=j$. In other words, we have $\tau^{j}\left(  \alpha\right)  <\tau
^{j}\left(  \beta\right)  $.

Proposition \ref{prop.words.read1}(e) (applied to $i=j$ and $h=\alpha$) yields
$w_{\tau,\tau^{j}\left(  \alpha\right)  }=c^{j}\cdot w_{\tau,\alpha}$. The
same argument (applied to $\beta$ instead of $\alpha$) yields $w_{\tau
,\tau^{j}\left(  \beta\right)  }=c^{j}\cdot w_{\tau,\beta}$. Thus,%
\[
w_{\tau,\tau^{j}\left(  \alpha\right)  }=c^{j}\cdot\underbrace{w_{\tau,\alpha
}}_{=w_{\tau,\beta}}=c^{j}\cdot w_{\tau,\beta}=w_{\tau,\tau^{j}\left(
\beta\right)  }.
\]
Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(f) (applied to $\tau^{j}\left(
\alpha\right)  $ and $\tau^{j}\left(  \beta\right)  $ instead of $\alpha$ and
$\beta$) yields $\tau\left(  \tau^{j}\left(  \alpha\right)  \right)
<\tau\left(  \tau^{j}\left(  \beta\right)  \right)  $ and $w_{\tau,\tau\left(
\tau^{j}\left(  \alpha\right)  \right)  }=w_{\tau,\tau\left(  \tau^{j}\left(
\beta\right)  \right)  }$. Now,%
\[
\underbrace{\tau^{j+1}}_{=\tau\circ\tau^{j}}\left(  \alpha\right)  =\left(
\tau\circ\tau^{j}\right)  \left(  \alpha\right)  =\tau\left(  \tau^{j}\left(
\alpha\right)  \right)  <\tau\left(  \tau^{j}\left(  \beta\right)  \right)
=\underbrace{\left(  \tau\circ\tau^{j}\right)  }_{=\tau^{j+1}}\left(
\beta\right)  =\tau^{j+1}\left(  \beta\right)  .
\]
In other words, (\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}) holds for $i=j+1$.
This completes the induction step. Thus,
(\ref{pf.lem.sol.lyndon.gr.basics.1.g.1}) is proved by induction.]

Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(g) is proved.

(h) We have assumed that%
\begin{equation}
\text{every }i\in\left\{  0,1,\ldots,j-1\right\}  \text{ satisfies }%
w_{\tau^{i+1}\left(  \alpha\right)  }=w_{\tau^{i+1}\left(  \beta\right)
}\text{.} \label{pf.lem.sol.lyndon.gr.basics.1.h.ass}%
\end{equation}


Now, we claim that%
\begin{equation}
\tau^{i}\left(  \alpha\right)  <\tau^{i}\left(  \beta\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  0,1,\ldots,j\right\}  .
\label{pf.lem.sol.lyndon.gr.basics.1.h.1}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}):} We shall prove
(\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) by induction on $i$:

\textit{Induction base:} We have $\alpha<\beta$. In view of $\underbrace{\tau
^{0}}_{=\operatorname*{id}}\left(  \alpha\right)  =\alpha$ and
$\underbrace{\tau^{0}}_{=\operatorname*{id}}\left(  \beta\right)  =\beta$,
this rewrites as $\tau^{0}\left(  \alpha\right)  <\tau^{0}\left(
\beta\right)  $. In other words, (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1})
holds for $i=0$. This completes the induction base.

\textit{Induction step:} Let $s\in\left\{  0,1,\ldots,j-1\right\}  $. Assume
that (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=s$. We must prove
that (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=s+1$.

We have assumed that (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for
$i=s$. In other words, we have $\tau^{s}\left(  \alpha\right)  <\tau
^{s}\left(  \beta\right)  $. Moreover,
(\ref{pf.lem.sol.lyndon.gr.basics.1.h.ass}) (applied to $i=s$) yields
$w_{\tau^{s+1}\left(  \alpha\right)  }=w_{\tau^{s+1}\left(  \beta\right)  }$.
In view of $\tau^{s+1}\left(  \alpha\right)  =\tau\left(  \tau^{s}\left(
\alpha\right)  \right)  $ and $\tau^{s+1}\left(  \beta\right)  =\tau\left(
\tau^{s}\left(  \beta\right)  \right)  $, this rewrites as $w_{\tau\left(
\tau^{s}\left(  \alpha\right)  \right)  }=w_{\tau\left(  \tau^{s}\left(
\beta\right)  \right)  }$. Hence, Lemma \ref{lem.sol.lyndon.gr.basics.1}(e)
(applied to $\tau^{s}\left(  \alpha\right)  $ and $\tau^{s}\left(
\beta\right)  $ instead of $\alpha$ and $\beta$) yields $\tau\left(  \tau
^{s}\left(  \alpha\right)  \right)  <\tau\left(  \tau^{s}\left(  \beta\right)
\right)  $ (since $\tau^{s}\left(  \alpha\right)  <\tau^{s}\left(
\beta\right)  $). In view of $\tau^{s+1}\left(  \alpha\right)  =\tau\left(
\tau^{s}\left(  \alpha\right)  \right)  $ and $\tau^{s+1}\left(  \beta\right)
=\tau\left(  \tau^{s}\left(  \beta\right)  \right)  $, this rewrites as
$\tau^{s+1}\left(  \alpha\right)  <\tau^{s+1}\left(  \beta\right)  $. In other
words, (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=s+1$. This
completes the induction step. Thus, (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1})
is proved by induction.]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}):} We shall prove
(\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) by induction on $i$:

\textit{Induction base:} We have $\alpha<\beta$. In view of $\underbrace{\tau
^{0}}_{=\operatorname*{id}}\left(  \alpha\right)  =\operatorname*{id}\left(
\alpha\right)  =\alpha$ and $\underbrace{\tau^{0}}_{=\operatorname*{id}%
}\left(  \beta\right)  =\operatorname*{id}\left(  \beta\right)  =\beta$, this
rewrites as $\tau^{0}\left(  \alpha\right)  <\tau^{0}\left(  \beta\right)  $.
In other words, (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=0$.
This completes the induction base.

\textit{Induction step:} Let $s\in\left\{  0,1,\ldots,j-1\right\}  $. Assume
that (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=s$. We must prove
that (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=s+1$.

We have assumed that (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for
$i=s$. In other words, we have $\tau^{s}\left(  \alpha\right)  <\tau
^{s}\left(  \beta\right)  $. Moreover,
(\ref{pf.lem.sol.lyndon.gr.basics.1.h.ass}) (applied to $i=s$) yields
$w_{\tau^{s+1}\left(  \alpha\right)  }=w_{\tau^{s+1}\left(  \beta\right)  }$.
In view of $\underbrace{\tau^{s+1}}_{=\tau\circ\tau^{s}}\left(  \alpha\right)
=\left(  \tau\circ\tau^{s}\right)  \left(  \alpha\right)  =\tau\left(
\tau^{s}\left(  \alpha\right)  \right)  $ and $\underbrace{\tau^{s+1}}%
_{=\tau\circ\tau^{s}}\left(  \beta\right)  =\left(  \tau\circ\tau^{s}\right)
\left(  \beta\right)  =\tau\left(  \tau^{s}\left(  \beta\right)  \right)  $,
this rewrites as $w_{\tau\left(  \tau^{s}\left(  \alpha\right)  \right)
}=w_{\tau\left(  \tau^{s}\left(  \beta\right)  \right)  }$. Hence, Lemma
\ref{lem.sol.lyndon.gr.basics.1}(e) (applied to $\tau^{s}\left(
\alpha\right)  $ and $\tau^{s}\left(  \beta\right)  $ instead of $\alpha$ and
$\beta$) yields $\tau\left(  \tau^{s}\left(  \alpha\right)  \right)
<\tau\left(  \tau^{s}\left(  \beta\right)  \right)  $ (since $\tau^{s}\left(
\alpha\right)  <\tau^{s}\left(  \beta\right)  $). In view of $\tau
^{s+1}\left(  \alpha\right)  =\tau\left(  \tau^{s}\left(  \alpha\right)
\right)  $ and $\tau^{s+1}\left(  \beta\right)  =\tau\left(  \tau^{s}\left(
\beta\right)  \right)  $, this rewrites as $\tau^{s+1}\left(  \alpha\right)
<\tau^{s+1}\left(  \beta\right)  $. In other words,
(\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) holds for $i=s+1$. This completes
the induction step. Thus, (\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) is proved
by induction.]
\end{verlong}

Now, $j\in\left\{  0,1,\ldots,j\right\}  $. Hence,
(\ref{pf.lem.sol.lyndon.gr.basics.1.h.1}) (applied to $i=j$) yields $\tau
^{j}\left(  \alpha\right)  <\tau^{j}\left(  \beta\right)  $. Thus, Lemma
\ref{lem.sol.lyndon.gr.basics.1}(c) (applied to $\tau^{j}\left(
\alpha\right)  $ and $\tau^{j}\left(  \beta\right)  $ instead of $\alpha$ and
$\beta$) yields $w_{\tau\left(  \tau^{j}\left(  \alpha\right)  \right)  }\leq
w_{\tau\left(  \tau^{j}\left(  \beta\right)  \right)  }$. In view of
$\tau\left(  \tau^{j}\left(  \alpha\right)  \right)  =\underbrace{\left(
\tau\circ\tau^{j}\right)  }_{=\tau^{j+1}}\left(  \alpha\right)  =\tau
^{j+1}\left(  \alpha\right)  $ and $\tau\left(  \tau^{j}\left(  \beta\right)
\right)  =\underbrace{\left(  \tau\circ\tau^{j}\right)  }_{=\tau^{j+1}}\left(
\beta\right)  =\tau^{j+1}\left(  \beta\right)  $, this rewrites as
$w_{\tau^{j+1}\left(  \alpha\right)  }\leq w_{\tau^{j+1}\left(  \beta\right)
}$. Thus, Lemma \ref{lem.sol.lyndon.gr.basics.1}(h) is proved.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.words.read-std2}.]We have $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ (since $w\in\mathfrak{A}^{n}$).

Let $k$ be the size of $z$. Thus, $k=\left\vert z\right\vert $.

(a) Let $h\in z$. We must prove that $\left[  w_{\tau,h}\right]  =\left\{
w_{\tau,i}\ \mid\ i\in z\right\}  $.

For each $u\in \NN $, we set $h_{u}:=\tau^{u}\left(  h\right)  $. Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(i) (applied to $p=h$ and $p_{u}=h_{u}$)
yields $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  =k$.

\begin{vershort}
Proposition \ref{prop.words.read1}(a) yields that the word $w_{\tau,h}$ is
nonempty and has length $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)
$. In other words, the word $w_{\tau,h}$ is nonempty and has length $k$ (since
$\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  =k$). Thus, $\left[
w_{\tau,h}\right]  $ is the $k$-necklace that contains $w_{\tau,h}$ (by the
definition of $\left[  w_{\tau,h}\right]  $). In other words, $\left[
w_{\tau,h}\right]  $ is the orbit of the $C$-action on $\mathfrak{A}^{k}$ that
contains $w_{\tau,h}$ (since a $k$-necklace is the same thing as an orbit of
the $C$-action on $\mathfrak{A}^{k}$). In other words, $\left[  w_{\tau
,h}\right]  =C\cdot w_{\tau,h}$.
\end{vershort}

\begin{verlong}
We have $h\in z\subset\left\{  1,2,\ldots,n\right\}  $. Hence, Proposition
\ref{prop.words.read1}(a) yields that the word $w_{\tau,h}$ is nonempty and
has length $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $. In other
words, the word $w_{\tau,h}$ is nonempty and has length $k$ (since
$\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  =k$). Thus, $\left[
w_{\tau,h}\right]  $ is the $k$-necklace that contains $w_{\tau,h}$ (by the
definition of $\left[  w_{\tau,h}\right]  $). In other words, $\left[
w_{\tau,h}\right]  $ is the orbit of the $C$-action on $\mathfrak{A}^{k}$ that
contains $w_{\tau,h}$ (since a $k$-necklace is the same thing as an orbit of
the $C$-action on $\mathfrak{A}^{k}$). In other words, $\left[  w_{\tau
,h}\right]  =C\cdot w_{\tau,h}$.
\end{verlong}

But $C=\left\{  c^{i}\ \mid\ i\in \ZZ \right\}  $ (since $c$ is a
generator of the cyclic group $C$). Hence,%
\begin{align}
\left[  w_{\tau,h}\right]   &  =\underbrace{C}_{=\left\{  c^{i}\ \mid
\ i\in \ZZ \right\}  }\cdot w_{\tau,h}=\left\{  c^{i}\ \mid
\ i\in \ZZ \right\}  \cdot w_{\tau,h}\nonumber\\
&  =\left\{  c^{i}\cdot w_{\tau,h}\ \mid\ i\in \ZZ \right\}  .
\label{pf.prop.words.read-std2.1}%
\end{align}


\begin{vershort}
On the other hand, $z$ is the cycle of $\tau$ that contains $h$ (since $z$ is
a cycle of $\tau$, and since $h\in z$). Hence, $z=\left\{  \tau^{i}\left(
h\right)  \ \mid\ i\in \ZZ \right\}  $, and thus%
\begin{align*}
\left\{  w_{\tau,u}\ \mid\ u\in z\right\}   &  =\left\{  w_{\tau,u}%
\ \mid\ u\in\left\{  \tau^{i}\left(  h\right)  \ \mid\ i\in \ZZ \right\}
\right\} \\
&  =\left\{  \underbrace{w_{\tau,\tau^{i}\left(  h\right)  }}%
_{\substack{=c^{i}\cdot w_{\tau,h}\\\text{(by Proposition
\ref{prop.words.read1}(e))}}}\ \mid\ i\in \ZZ \right\}  =\left\{
c^{i}\cdot w_{\tau,h}\ \mid\ i\in \ZZ \right\}  .
\end{align*}
Comparing this with (\ref{pf.prop.words.read-std2.1}), we obtain $\left[
w_{\tau,h}\right]  =\left\{  w_{\tau,u}\ \mid\ u\in z\right\}  =\left\{
w_{\tau,i}\ \mid\ i\in z\right\}  $ (here, we have renamed the index $u$ as
$i$). This proves Proposition \ref{prop.words.read-std2}(a).
\end{vershort}

\begin{verlong}
On the other hand, $z$ is a cycle of $\tau$. In other words, $z$ is an orbit
of the action of $\tau$ on $\left\{  1,2,\ldots,n\right\}  $. More precisely,
$z$ must be the orbit of $\tau$ containing $h$ (since $h\in z$). Hence,%
\[
z=\left\{  \tau^{i}\left(  h\right)  \ \mid\ i\in \ZZ \right\}  .
\]
Therefore,%
\begin{align*}
\left\{  w_{\tau,u}\ \mid\ u\in\underbrace{z}_{=\left\{  \tau^{i}\left(
h\right)  \ \mid\ i\in \ZZ \right\}  }\right\}   &  =\left\{  w_{\tau
,u}\ \mid\ u\in\left\{  \tau^{i}\left(  h\right)  \ \mid\ i\in \ZZ %
\right\}  \right\} \\
&  =\left\{  w_{\tau,u}\ \mid\ u=\tau^{i}\left(  h\right)  \text{ for some
}i\in \ZZ \right\} \\
&  =\left\{  \underbrace{w_{\tau,\tau^{i}\left(  h\right)  }}%
_{\substack{=c^{i}\cdot w_{\tau,h}\\\text{(by Proposition
\ref{prop.words.read1}(e))}}}\ \mid\ i\in \ZZ \right\} \\
&  =\left\{  c^{i}\cdot w_{\tau,h}\ \mid\ i\in \ZZ \right\}  .
\end{align*}
Comparing this with (\ref{pf.prop.words.read-std2.1}), we obtain $\left[
w_{\tau,h}\right]  =\left\{  w_{\tau,u}\ \mid\ u\in z\right\}  =\left\{
w_{\tau,i}\ \mid\ i\in z\right\}  $ (here, we have renamed the index $u$ as
$i$). This proves Proposition \ref{prop.words.read-std2}(a).
\end{verlong}

(b) Let $\alpha$ and $\beta$ be two distinct elements of $z$. We must prove
that $w_{\tau,\alpha}\neq w_{\tau,\beta}$.

\begin{vershort}
We have $\alpha\neq\beta$ (since $\alpha$ and $\beta$ are distinct). Thus,
either $\alpha<\beta$ or $\alpha>\beta$. We WLOG assume that $\alpha<\beta$
(since otherwise, it suffices to swap $\alpha$ with $\beta$).
\end{vershort}

\begin{verlong}
We have $\alpha\neq\beta$ (since $\alpha$ and $\beta$ are distinct). Thus,
either $\alpha<\beta$ or $\alpha>\beta$. We can WLOG assume that $\alpha
<\beta$ (since otherwise, it suffices to swap $\alpha$ with $\beta$). Assume this.
\end{verlong}

We must prove that $w_{\tau,\alpha}\neq w_{\tau,\beta}$. Assume the contrary.
Thus, $w_{\tau,\alpha}=w_{\tau,\beta}$. Lemma \ref{lem.sol.lyndon.gr.basics.1}%
(g) thus shows that%
\begin{equation}
\tau^{i}\left(  \alpha\right)  <\tau^{i}\left(  \beta\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in \NN .
\label{pf.prop.words.read-std2.b.2}%
\end{equation}


On the other hand, we have%
\begin{equation}
z=\left\{  \tau^{0}\left(  \gamma\right)  ,\tau^{1}\left(  \gamma\right)
,\ldots,\tau^{k-1}\left(  \gamma\right)  \right\}
\ \ \ \ \ \ \ \ \ \ \text{for each }\gamma\in z.
\label{pf.prop.words.read-std2.b.1}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.words.read-std2.b.1}):} Let $\gamma\in z$. For
each $u\in \NN $, we set $\gamma_{u}:=\tau^{u}\left(  \gamma\right)  $.
Then, Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(c) (applied to $p=\gamma$ and
$p_{u}=\gamma_{u}$) yields%
\[
z=\left\{  \gamma_{0},\gamma_{1},\ldots,\gamma_{k-1}\right\}  =\left\{
\tau^{0}\left(  \gamma\right)  ,\tau^{1}\left(  \gamma\right)  ,\ldots
,\tau^{k-1}\left(  \gamma\right)  \right\}
\]
(since each $u\in\left\{  0,1,\ldots,k-1\right\}  $ satisfies $\gamma_{u}%
=\tau^{u}\left(  \gamma\right)  $ (by the definition of $\gamma_{u}$)). This
proves (\ref{pf.prop.words.read-std2.b.1}).]

Applying (\ref{pf.prop.words.read-std2.b.1}) to $\gamma=\beta$, we obtain
$z=\left\{  \tau^{0}\left(  \beta\right)  ,\tau^{1}\left(  \beta\right)
,\ldots,\tau^{k-1}\left(  \beta\right)  \right\}  $.

Now, let $m$ be the smallest element of $z$. Then, $m\in z=\left\{  \tau
^{0}\left(  \beta\right)  ,\tau^{1}\left(  \beta\right)  ,\ldots,\tau
^{k-1}\left(  \beta\right)  \right\}  $. Hence, there exists some
$i\in\left\{  0,1,\ldots,k-1\right\}  $ such that $m=\tau^{i}\left(
\beta\right)  $. Consider this $i$.

But (\ref{pf.prop.words.read-std2.b.1}) (applied to $\gamma=\alpha$) yields
$z=\left\{  \tau^{0}\left(  \alpha\right)  ,\tau^{1}\left(  \alpha\right)
,\ldots,\tau^{k-1}\left(  \alpha\right)  \right\}  $. From $i\in\left\{
0,1,\ldots,k-1\right\}  $, we obtain $\tau^{i}\left(  \alpha\right)
\in\left\{  \tau^{0}\left(  \alpha\right)  ,\tau^{1}\left(  \alpha\right)
,\ldots,\tau^{k-1}\left(  \alpha\right)  \right\}  $. In other words,
$\tau^{i}\left(  \alpha\right)  \in z$ (since $z=\left\{  \tau^{0}\left(
\alpha\right)  ,\tau^{1}\left(  \alpha\right)  ,\ldots,\tau^{k-1}\left(
\alpha\right)  \right\}  $).

\begin{vershort}
But $m$ is the \textbf{smallest} element of $z$. Thus, $p\geq m$ for each
$p\in z$. We can apply this to $p=\tau^{i}\left(  \alpha\right)  $ (since
$\tau^{i}\left(  \alpha\right)  \in z$), and thus obtain $\tau^{i}\left(
\alpha\right)  \geq m=\tau^{i}\left(  \beta\right)  $. But
(\ref{pf.prop.words.read-std2.b.2}) yields $\tau^{i}\left(  \alpha\right)
<\tau^{i}\left(  \beta\right)  $. These two inequalities clearly contradict
one another.
\end{vershort}

\begin{verlong}
But $m$ is the \textbf{smallest} element of $z$. Thus, any element of $z$ is
$\geq m$. In other words, $p\geq m$ for each $p\in z$. We can apply this to
$p=\tau^{i}\left(  \alpha\right)  $ (since $\tau^{i}\left(  \alpha\right)  \in
z$), and thus obtain $\tau^{i}\left(  \alpha\right)  \geq m=\tau^{i}\left(
\beta\right)  $. But (\ref{pf.prop.words.read-std2.b.2}) yields $\tau
^{i}\left(  \alpha\right)  <\tau^{i}\left(  \beta\right)  $. These two
inequalities clearly contradict one another.
\end{verlong}

This contradiction shows that our assumption was false. Hence, $w_{\tau
,\alpha}\neq w_{\tau,\beta}$ is proved. Thus, Proposition
\ref{prop.words.read-std2}(b) follows.

(c) There are $k$ many elements $i\in z$ (since $k$ is the size of $z$). The
words $w_{\tau,i}$ corresponding to these $k$ many elements $i$ are all
distinct (by Proposition \ref{prop.words.read-std2}(b)). Thus, the set
$\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $ (consisting of these words) has
size $k$. In other words, $\left\vert \left\{  w_{\tau,i}\ \mid\ i\in
z\right\}  \right\vert =k$. But recall that $k=\left\vert z\right\vert $.
Hence, $\left\vert \left\{  w_{\tau,i}\ \mid\ i\in z\right\}  \right\vert
=k=\left\vert z\right\vert $. This proves Proposition
\ref{prop.words.read-std2}(c).

(d) The set $z$ is a cycle of $\tau$, and thus is nonempty (since any cycle of
a permutation is nonempty). Hence, there exists some $h\in z$. Consider this
$h$. Proposition \ref{prop.words.read-std2}(a) yields $\left[  w_{\tau
,h}\right]  =\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $.

But $\left[  w_{\tau,h}\right]  $ is the $k$-necklace that contains
$w_{\tau,h}$ (as we have showed in the proof of Proposition
\ref{prop.words.read-std2}(a) above). Hence, $\left[  w_{\tau,h}\right]  $ is
a $k$-necklace.

Proposition \ref{prop.words.read-std2}(c) yields $\left\vert \left\{
w_{\tau,i}\ \mid\ i\in z\right\}  \right\vert =\left\vert z\right\vert =k$
(since $k=\left\vert z\right\vert $). In view of $\left[  w_{\tau,h}\right]
=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $, this rewrites as $\left\vert
\left[  w_{\tau,h}\right]  \right\vert =k$.

The period of the necklace $\left[  w_{\tau,h}\right]  $ is $\left\vert
\left[  w_{\tau,h}\right]  \right\vert $ (by the definition of the period of a
necklace). In other words, the period of the necklace $\left[  w_{\tau
,h}\right]  $ is $k$ (since $\left\vert \left[  w_{\tau,h}\right]  \right\vert
=k$).

But $\left[  w_{\tau,h}\right]  $ is a $k$-necklace. Hence, this $k$-necklace
$\left[  w_{\tau,h}\right]  $ is aperiodic if and only if its period is $k$
(by the definition of ``aperiodic''). Thus,
this $k$-necklace $\left[  w_{\tau,h}\right]  $ is aperiodic (since its period
is $k$). In view of $\left[  w_{\tau,h}\right]  =\left\{  w_{\tau,i}%
\ \mid\ i\in z\right\}  $, this rewrites as follows: The $k$-necklace
$\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $ is aperiodic. Hence, the set
$\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $ is an aperiodic necklace. This
proves Proposition \ref{prop.words.read-std2}(d).
\end{proof}

We have now proved Proposition \ref{prop.words.read1}, Lemma
\ref{lem.sol.lyndon.gr.basics.1} and Proposition \ref{prop.words.read-std2}.
This solves Exercise \ref{exe.lyndon.gr.basics}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.lyndon.gr.aper-rel}}

\textit{Solution to Exercise \ref{exe.lyndon.gr.aper-rel}.} In preparation for
solving Exercise \ref{exe.lyndon.gr.aper-rel}, we first show a few simple
lemmas about necklaces:

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.neck-cts0}Let $u\in\mathfrak{A}^{\ast}$ be a
nonempty word. Then, $u\in\left[  u\right]  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0}.]The word $u$ is
nonempty. In other words, $u\in\mathfrak{A}^{n}$ for some positive integer
$n$. Consider this $n$. Thus, $\left[  u\right]  $ is the $n$-necklace
containing $u$ (by the definition of $\left[  u\right]  $). Hence, $\left[
u\right]  $ contains $u$. In other words, $u\in\left[  u\right]  $. This
proves Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.neck-cts}Let $N$ be a necklace. Let $u\in N$.
Then, $N=\left[  u\right]  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts}.]We know that $N$ is a
necklace. In other words, $N$ is an $n$-necklace for some positive integer $n$
(by the definition of a necklace). Consider this $n$. We now know that $N$ is
an $n$-necklace; in other words, $N$ is an orbit of the $C$-action on
$\mathfrak{A}^{n}$ (by the definition of an $n$-necklace). Hence, $N$ is a set
of words of length $n$. Thus, $u$ is a word of length $n$ (since $u\in N$).

Recall that $N$ is an orbit of the $C$-action on $\mathfrak{A}^{n}$. Since $N$
contains $u$ (because $u\in N$), we thus conclude that $N$ is the orbit of the
$C$-action on $\mathfrak{A}^{n}$ that contains $u$. In other words, $N$ is the
$n$-necklace that contains $u$ (because an $n$-necklace is the same thing as
an orbit of the $C$-action on $\mathfrak{A}^{n}$). In other words, $N=\left[
u\right]  $ (since the $n$-necklace that contains $u$ has been denoted by
$\left[  u\right]  $). This proves Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.neck-laper}Let $u\in\mathfrak{A}^{\ast}$ be a
Lyndon word. Then, the necklace $\left[  u\right]  $ is aperiodic.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.neck-laper}.]The word $u$ is
Lyndon, and thus is nonempty (since any Lyndon word is nonempty by
definition). Hence, Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields
$u\in\left[  u\right]  $. In other words, the necklace $\left[  u\right]  $
contains $u$.

Thus, the necklace $\left[  u\right]  $ contains the word $u$, which is a
Lyndon word. Hence, the necklace $\left[  u\right]  $ contains a Lyndon word
(namely, $u$). If the necklace $\left[  u\right]  $ was not aperiodic, then
Exercise \ref{exe.words.necklaces}(d) (applied to $N=\left[  u\right]  $)
would show that $\left[  u\right]  $ contains no Lyndon word; but this would
contradict the previous sentence. Hence, the necklace $\left[  u\right]  $
must be aperiodic. This proves Lemma \ref{lem.sol.lyndon.gr.bij.neck-laper}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.neck-aper}Let $u\in\mathfrak{A}^{\ast}$ be a
nonempty word such that the necklace $\left[  u\right]  $ is aperiodic. Then,
$\left\vert \left[  u\right]  \right\vert =\ell\left(  u\right)  $.
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper}.]The word $u\in
\mathfrak{A}^{\ast}$ is nonempty. Thus, there exists a positive integer $p$
such that $u\in\mathfrak{A}^{p}$. Consider this $p$. Hence, $\left[  u\right]
$ is a $p$-necklace. Also, from $u\in\mathfrak{A}^{p}$, we obtain $\ell\left(
u\right)  =p$.

The $p$-necklace $\left[  u\right]  $ is aperiodic. In other words, its period
is $p$ (by the definition of an aperiodic $p$-necklace). Thus,%
\[
p=\left(  \text{period of the }p\text{-necklace }\left[  u\right]  \right)
=\left\vert \left[  u\right]  \right\vert
\]
(by the definition of the period of a $p$-necklace). Therefore, $\left\vert
\left[  u\right]  \right\vert =p=\ell\left(  u\right)  $ (since $\ell\left(
u\right)  =p$). This proves Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper}.]The word $u\in
\mathfrak{A}^{\ast}$ is nonempty. Thus, there exists a positive integer $p$
such that $u\in\mathfrak{A}^{p}$. Consider this $p$. Hence, $\left[  u\right]
$ is a $p$-necklace. Also, from $u\in\mathfrak{A}^{p}$, we obtain $\ell\left(
u\right)  =p$.

The $p$-necklace $\left[  u\right]  $ is aperiodic. In other words, its period
is $p$ (by the definition of an aperiodic $p$-necklace). In other words,%
\[
\left(  \text{period of the }p\text{-necklace }\left[  u\right]  \right)  =p.
\]
On the other hand, the definition of the period of a $p$-necklace shows that%
\[
\left(  \text{period of the }p\text{-necklace }\left[  u\right]  \right)
=\left\vert \left[  u\right]  \right\vert .
\]
Comparing these two equalities, we obtain $\left\vert \left[  u\right]
\right\vert =p=\ell\left(  u\right)  $ (since $\ell\left(  u\right)  =p$).
This proves Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper}.
\end{proof}

\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.gr.basics.2}Let $k$ be a positive integer. Let
$v\in\mathfrak{A}^{k}$ be such that the necklace $\left[  v\right]  $ is not
aperiodic. Then, there exists some $h\in\left\{  1,2,\ldots,k-1\right\}  $
such that $c^{h}\cdot v=v$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.basics.2}.]We have $v\in\mathfrak{A}%
^{k}$. Thus, $\left[  v\right]  $ is a $k$-necklace. Hence, Exercise
\ref{exe.words.necklaces}(a) (applied to $k$ and $\left[  v\right]  $ instead
of $n$ and $N$) yields that $\left[  v\right]  $ is a finite nonempty set and
satisfies $\left\vert \left[  v\right]  \right\vert \mid k$. From $\left\vert
\left[  v\right]  \right\vert \mid k$, we obtain $\left\vert \left[  v\right]
\right\vert \leq k$ (since $k$ is a positive integer).

Next, we claim the following:

\begin{statement}
\textit{Claim 1:} The $k$ elements $c^{0}\cdot v,c^{1}\cdot v,\ldots
,c^{k-1}\cdot v$ are not distinct.
\end{statement}

[\textit{Proof of Claim 1:} Assume the contrary. Thus, the $k$ elements
$c^{0}\cdot v,c^{1}\cdot v,\ldots,c^{k-1}\cdot v$ are distinct.

The necklace $\left[  v\right]  $ is defined to be the $k$-necklace containing
$v$ (since $v\in\mathfrak{A}^{k}$). In other words, the necklace $\left[
v\right]  $ is the $C$-orbit on $\mathfrak{A}^{k}$ containing $v$ (since a
$k$-necklace is the same thing as a $C$-orbit on $\mathfrak{A}^{k}$). In other
words, the necklace $\left[  v\right]  $ is the $C$-orbit of $v\in
\mathfrak{A}^{k}$. In other words, the necklace $\left[  v\right]  $ is the
set $C\cdot v$ (since the $C$-orbit of $v\in\mathfrak{A}^{k}$ is the set
$C\cdot v$). In other words, $\left[  v\right]  =C\cdot v$. Hence, $\left\vert
\left[  v\right]  \right\vert =\left\vert C\cdot v\right\vert $.

The $k$ elements $c^{0}\cdot v,c^{1}\cdot v,\ldots,c^{k-1}\cdot v$ all belong
to $C\cdot v$ (since $c^{0},c^{1},\ldots,c^{k-1}$ all belong to $C$) and are
distinct (by our assumption). Hence, the set $C\cdot v$ contains (at least)
$k$ distinct elements (namely, $c^{0}\cdot v,c^{1}\cdot v,\ldots,c^{k-1}\cdot
v$). Thus, $\left\vert C\cdot v\right\vert \geq k$. Thus, $\left\vert \left[
v\right]  \right\vert =\left\vert C\cdot v\right\vert \geq k$. Combining this
with $\left\vert \left[  v\right]  \right\vert \leq k$, we obtain $\left\vert
\left[  v\right]  \right\vert =k$. In other words, the period of $\left[
v\right]  $ is $k$ (since the period of $\left[  v\right]  $ is defined as the
integer $\left\vert \left[  v\right]  \right\vert $). But $\left[  v\right]  $
is a $k$-necklace. Hence, the necklace $\left[  v\right]  $ is aperiodic if
and only if its period is $k$ (by the definition of an aperiodic necklace).
Thus, the necklace $\left[  v\right]  $ is aperiodic (since its period is
$k$). This contradicts the fact that the necklace $\left[  v\right]  $ is not
aperiodic. This contradiction shows that our assumption was wrong. This
completes our proof of Claim 1.]

\begin{vershort}
Claim 1 shows that the $k$ elements $c^{0}\cdot v,c^{1}\cdot v,\ldots
,c^{k-1}\cdot v$ are not distinct. In other words, there exist two elements
$x$ and $y$ of $\left\{  0,1,\ldots,k-1\right\}  $ such that $x<y$ and
$c^{x}\cdot v=c^{y}\cdot v$. Consider these $x$ and $y$. From $x\in\left\{
0,1,\ldots,k-1\right\}  $ and $y\in\left\{  0,1,\ldots,k-1\right\}  $ and
$x<y$, we obtain $0\leq x<y\leq k-1$, so that $y-x\in\left\{  1,2,\ldots
,k-1\right\}  $. Furthermore,%
\[
\underbrace{c^{y-x}}_{=c^{-x}c^{y}}\cdot v=\left(  c^{-x}c^{y}\right)  \cdot
v=c^{-x}\cdot\underbrace{\left(  c^{y}\cdot v\right)  }_{=c^{x}\cdot v}%
=c^{-x}\cdot\left(  c^{x}\cdot v\right)  =\underbrace{\left(  c^{-x}%
c^{x}\right)  }_{=\operatorname*{id}}\cdot v=v.
\]

\end{vershort}

\begin{verlong}
Claim 1 shows that the $k$ elements $c^{0}\cdot v,c^{1}\cdot v,\ldots
,c^{k-1}\cdot v$ are not distinct. In other words, two of them are equal. In
other words, there exist two elements $x$ and $y$ of $\left\{  0,1,\ldots
,k-1\right\}  $ such that $x<y$ and $c^{x}\cdot v=c^{y}\cdot v$. Consider
these $x$ and $y$. From $x\in\left\{  0,1,\ldots,k-1\right\}  $, we obtain
$x\geq0$. From $y\in\left\{  0,1,\ldots,k-1\right\}  $, we obtain $y\leq k-1$.
Hence, $y-\underbrace{x}_{\geq0}\leq y\leq k-1$. Combining this with $y-x>0$
(which is a consequence of $x<y$), we obtain $0<y-x\leq k-1$. Hence,
$y-x\in\left\{  1,2,\ldots,k-1\right\}  $ (since $y-x$ is an integer).
Furthermore,%
\[
\underbrace{c^{y-x}}_{=c^{\left(  -x\right)  +y}=c^{-x}c^{y}}\cdot v=\left(
c^{-x}c^{y}\right)  \cdot v=c^{-x}\cdot\underbrace{\left(  c^{y}\cdot
v\right)  }_{=c^{x}\cdot v}=c^{-x}\cdot\left(  c^{x}\cdot v\right)
=\underbrace{\left(  c^{-x}c^{x}\right)  }_{=c^{\left(  -x\right)  +x}%
=c^{0}=\operatorname*{id}}\cdot v=v.
\]

\end{verlong}

\noindent Hence, there exists some $h\in\left\{  1,2,\ldots,k-1\right\}  $
such that $c^{h}\cdot v=v$ (namely, $h=y-x$). This proves Lemma
\ref{lem.sol.lyndon.gr.basics.2}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.1}Let $n$ be a positive integer. Let $d$ be
a positive divisor of $n$. Thus, $n/d$ is a positive integer.

Let $q\in\mathfrak{A}^{n/d}$. Then, the $n/d$-necklace $\left[  q\right]  $
and the $n$-necklace $\left[  q^{d}\right]  $ satisfy $\left\vert \left[
q^{d}\right]  \right\vert =\left\vert \left[  q\right]  \right\vert \leq n/d$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.1}.]Exercise
\ref{exe.words.necklaces}(a) (applied to $n/d$ instead of $n$) shows that
every $n/d$-necklace $N$ is a finite nonempty set and satisfies $\left\vert
N\right\vert \mid n/d$. Applying this to $N=\left[  q\right]  $, we conclude
that $\left[  q\right]  $ is a finite nonempty set and satisfies $\left\vert
\left[  q\right]  \right\vert \mid n/d$. From $\left\vert \left[  q\right]
\right\vert \mid n/d$, we obtain $\left\vert \left[  q\right]  \right\vert
\leq n/d$ (since $\left\vert \left[  q\right]  \right\vert \in \NN $ and
since $n/d$ is a positive integer).

Now, consider the map $\Delta:\mathfrak{A}^{n/d}\rightarrow\mathfrak{A}^{n}$
defined in Lemma \ref{lem.sol.words.necklaces.f2}. The definition of $\Delta$
yields $\Delta\left(  q\right)  =q^{d}$. Hence, $q^{d}=\Delta\left(  q\right)
\in\mathfrak{A}^{n}$.

The necklace $\left[  q\right]  $ is defined to be the $n/d$-necklace
containing $q$ (since $q\in\mathfrak{A}^{n/d}$). In other words, the necklace
$\left[  q\right]  $ is the $C$-orbit on $\mathfrak{A}^{n/d}$ containing $q$
(since an $n/d$-necklace is the same thing as a $C$-orbit on $\mathfrak{A}%
^{n/d}$). In other words, the necklace $\left[  q\right]  $ is the $C$-orbit
of $q\in\mathfrak{A}^{n/d}$. In other words, the necklace $\left[  q\right]  $
is the set $C\cdot q$ (since the $C$-orbit of $q\in\mathfrak{A}^{n/d}$ is the
set $C\cdot q$). In other words, $\left[  q\right]  =C\cdot q$. The same
argument (applied to $n$ and $q^{d}$ instead of $n/d$ and $q$) yields $\left[
q^{d}\right]  =C\cdot q^{d}$.

But Lemma \ref{lem.sol.words.necklaces.f2}(c) shows that the map $\Delta$ is
injective. Hence, $\left\vert \Delta\left(  S\right)  \right\vert =\left\vert
S\right\vert $ for any finite subset $S$ of $\mathfrak{A}^{n/d}$. Applying
this to $S=\left[  q\right]  $, we obtain $\left\vert \Delta\left(  \left[
q\right]  \right)  \right\vert =\left\vert \left[  q\right]  \right\vert $
(since $\left[  q\right]  $ is a finite subset of $\mathfrak{A}^{n/d}$).

On the other hand, Lemma \ref{lem.sol.words.necklaces.f2}(b) shows that the
map $\Delta$ is $C$-equivariant. Therefore, $\Delta\left(  C\cdot q\right)
=C\cdot\underbrace{\Delta\left(  q\right)  }_{=q^{d}}=C\cdot q^{d}$. This
rewrites as $\Delta\left(  \left[  q\right]  \right)  =\left[  q^{d}\right]  $
(since $\left[  q\right]  =C\cdot q$ and $\left[  q^{d}\right]  =C\cdot q^{d}%
$). Hence, $\left\vert \Delta\left(  \left[  q\right]  \right)  \right\vert
=\left\vert \left[  q^{d}\right]  \right\vert $. Therefore, $\left\vert
\left[  q^{d}\right]  \right\vert =\left\vert \Delta\left(  \left[  q\right]
\right)  \right\vert =\left\vert \left[  q\right]  \right\vert \leq n/d$. This
proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.1}.
\end{proof}

We can now start proving the results required in Exercise
\ref{exe.lyndon.gr.aper-rel}:

\begin{proof}
[Proof of Proposition \ref{prop.words.aper-neck}.]Define $n\in \NN $ by
$n=\ell\left(  w\right)  $. Then, $n=\ell\left(  w\right)  >0$ (since $w$ is
nonempty), so that $n$ is a positive integer. Also, $w\in\mathfrak{A}^{n}$
(since $\ell\left(  w\right)  =n$).

We must show that the word $w$ is aperiodic if and only if the necklace
$\left[  w\right]  $ is aperiodic. In other words, we must prove the following
two claims:

\begin{statement}
\textit{Claim 1:} If the word $w$ is aperiodic, then the necklace $\left[
w\right]  $ is aperiodic.
\end{statement}

\begin{statement}
\textit{Claim 2:} If the necklace $\left[  w\right]  $ is aperiodic, then the
word $w$ is aperiodic.
\end{statement}

[\textit{Proof of Claim 1:} Assume that the word $w$ is aperiodic. In other
words, there exist no $m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying
$w=u^{m}$ (by the definition of aperiodic words).

We must prove that the necklace $\left[  w\right]  $ is aperiodic. Indeed,
assume the contrary. Then, the necklace $\left[  w\right]  $ is not aperiodic.
Hence, Lemma \ref{lem.sol.lyndon.gr.basics.2} (applied to $v=w$ and $k=n$)
yields that there exists some $h\in\left\{  1,2,\ldots,n-1\right\}  $ such
that $c^{h}\cdot w=w$. Consider this $h$.

From $h\in\left\{  1,2,\ldots,n-1\right\}  $, we obtain $1\leq h\leq n-1$, so
that $h\geq1>0$ and $n-\underbrace{h}_{\leq n-1}\geq n-\left(  n-1\right)
=1>0$.

We have $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $ (since $w\in
\mathfrak{A}^{n}$). Define two words $p$ and $q$ by $p=\left(  w_{1}%
,w_{2},\ldots,w_{h}\right)  $ and $q=\left(  w_{h+1},w_{h+2},\ldots
,w_{n}\right)  $. Then, these words $p$ and $q$ have lengths $\ell\left(
p\right)  =h$ and $\ell\left(  q\right)  =n-h$, and thus are nonempty (since
$\ell\left(  p\right)  =h>0$ and $\ell\left(  q\right)  =n-h>0$). They further
satisfy
\begin{align*}
w  &  =\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\left(  w_{1},w_{2}%
,\ldots,w_{h},w_{h+1},w_{h+2},\ldots,w_{n}\right) \\
&  =\underbrace{\left(  w_{1},w_{2},\ldots,w_{h}\right)  }_{=p}%
\underbrace{\left(  w_{h+1},w_{h+2},\ldots,w_{n}\right)  }_{=q}=pq.
\end{align*}


Recall that $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples
one step to the left (that is, we have $c\cdot\left(  a_{1},a_{2},\ldots
,a_{n}\right)  =\left(  a_{2},a_{3},\ldots,a_{n},a_{1}\right)  $ for each
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}$). Thus, $c^{h}$
acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples $h$ steps to the
left. In other words,%
\[
c^{h}\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{h+1}%
,a_{h+2},\ldots,a_{n},a_{1},a_{2},\ldots,a_{h}\right)
\]
for each $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}$ (since
$h\in\left\{  1,2,\ldots,n-1\right\}  \subset\left\{  0,1,\ldots,n\right\}
$). Applying this to $\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(
w_{1},w_{2},\ldots,w_{n}\right)  $, we obtain%
\begin{align*}
c^{h}\cdot\left(  w_{1},w_{2},\ldots,w_{n}\right)   &  =\left(  w_{h+1}%
,w_{h+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{h}\right) \\
&  =\underbrace{\left(  w_{h+1},w_{h+2},\ldots,w_{n}\right)  }_{=q}%
\underbrace{\left(  w_{1},w_{2},\ldots,w_{h}\right)  }_{=p}=qp.
\end{align*}
Comparing this with
\[
c^{h}\cdot\underbrace{\left(  w_{1},w_{2},\ldots,w_{n}\right)  }_{=w}%
=c^{h}\cdot w=w=pq,
\]
we obtain $pq=qp$. Hence, Proposition \ref{prop.words.uv=vu.2} (applied to
$u=p$ and $v=q$) yields that there exist a $t\in\mathfrak{A}^{\ast}$ and two
nonnegative integers $N$ and $M$\ \ \ \ \footnote{Here, we are using the
letters $N$ and $M$ for what was called $n$ and $m$ in Proposition
\ref{prop.words.uv=vu.2} (since the letter $n$ already has a different meaning
in our current situation).} such that $p=t^{N}$ and $q=t^{M}$. Consider these
$t$, $N$ and $M$. From $p=t^{N}$, we obtain $\ell\left(  p\right)
=\ell\left(  t^{N}\right)  =N\ell\left(  t\right)  $, so that $N\ell\left(
t\right)  =\ell\left(  p\right)  =h\neq0$ (since $h>0$). Hence, $N\neq0$, so
that $N\geq1$ (because $N$ is a nonnegative integer). Also, from $q=t^{M}$, we
obtain $\ell\left(  q\right)  =\ell\left(  t^{M}\right)  =M\ell\left(
t\right)  $, so that $M\ell\left(  t\right)  =\ell\left(  q\right)  =n-h\neq0$
(since $n-h>0$). Hence, $M\neq0$, so that $M\geq1$ (since $M$ is a nonnegative
integer). Now, $\underbrace{N}_{\geq1}+\underbrace{M}_{\geq1}\geq1+1=2$ and
$w=\underbrace{p}_{=t^{N}}\underbrace{q}_{=t^{M}}=t^{N}t^{M}=t^{N+M}$. Hence,
there exist $m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$
(namely, $m=N+M$ and $u=t$). This contradicts the fact that there exist no
$m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$. This
contradiction shows that our assumption was false. Hence, the necklace
$\left[  w\right]  $ is aperiodic. This proves Claim 1.]

\begin{verlong}
Let us also give a second proof of Claim 1:

[\textit{Alternative proof of Claim 1:} Assume that the word $w$ is aperiodic.
In other words, there exist no $m\geq2$ and $u\in\mathfrak{A}^{\ast}$
satisfying $w=u^{m}$ (by the definition of aperiodic words).

We must prove that the necklace $\left[  w\right]  $ is aperiodic. Indeed,
assume the contrary. Then, the necklace $\left[  w\right]  $ is not aperiodic.
Hence, Lemma \ref{lem.sol.lyndon.gr.basics.2} (applied to $v=w$ and $k=n$)
yields that there exists some $h\in\left\{  1,2,\ldots,n-1\right\}  $ such
that $c^{h}\cdot w=w$. Consider this $h$.

In the following, whenever $G$ is a group and $P$ is a (left) $G$-set and $u$
is an element of $P$, we denote by $G_{u}$ the stabilizer of $u$ in $G$, that
is, the subgroup $\left\{  g\in G\ \mid\ gu=u\right\}  $ of $G$. Thus, in
particular, $C_{w}$ denotes the stabilizer of $w$ in $C$ (since the group $C$
acts on $\mathfrak{A}^{n}$, while $w\in\mathfrak{A}^{n}$).

For each $k\in \ZZ $, we let $C^{\left(  k\right)  }$ denote the subgroup
$\left\langle c^{k}\right\rangle $ of the infinite cyclic group $C$. It is
known that for each positive integer $k$, the group $C/C^{\left(  k\right)  }$
is a cyclic group with $k$ elements, and thus satisfies%
\begin{equation}
\left\vert C/C^{\left(  k\right)  }\right\vert =k.
\label{pf.prop.words.aper-neck.c1.pf.1}%
\end{equation}
Another known property of the infinite cyclic group is the following: Every
subgroup of $C$ has the form $C^{\left(  k\right)  }$ for some $k\in
 \NN $. Applying this to the subgroup $C_{w}$ (that is, to the stabilizer
of $w$ in $C$), we conclude that $C_{w}$ has the form $C^{\left(  k\right)  }$
for some $k\in \NN $. Consider this $k$. Thus, $C_{w}=C^{\left(
k\right)  }$ and $k\in \NN $. The definition of $C^{\left(  k\right)  }$
shows that $C^{\left(  k\right)  }=\left\langle c^{k}\right\rangle $. Now,
$c^{k}\in\left\langle c^{k}\right\rangle =C_{w}$ (since $C_{w}=C^{\left(
k\right)  }=\left\langle c^{k}\right\rangle $). In other words, $c^{k}w=w$.

Recall that $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples
one step to the left (that is, we have $c\cdot\left(  a_{1},a_{2},\ldots
,a_{n}\right)  =\left(  a_{2},a_{3},\ldots,a_{n},a_{1}\right)  $ for each
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}$). Thus, $c^{n}$
acts trivially on $\mathfrak{A}^{n}$ (since cyclically rotating an $n$-tuple
$n$ steps to the left does nothing). Thus, in particular, $c^{n}\cdot w=w$. In
other words, $c^{n}\in C_{w}$. In view of $C_{w}=C^{\left(  k\right)
}=\left\langle c^{k}\right\rangle $, this rewrites as $c^{n}\in\left\langle
c^{k}\right\rangle $. In other words, $n$ is a multiple of $k$ (since $c$ is a
generator of the infinite cyclic group $C$). In other words, $k$ is a divisor
of $n$. Hence, $k\neq0$ (since $n\neq0$ (because $n>0$)). Combining this with
$k\in \NN $, we conclude that $k$ is a positive integer. Hence, $k$ is a
positive divisor of $n$ (since $k$ is a divisor of $n$). Therefore, $k\leq n$
(since $n>0$).

Next, we claim that $k\neq n$.

[\textit{Proof.} Assume the contrary. Thus, $k=n$. But we have $c^{h}\in
C_{w}$ (since $c^{h}\cdot w=w$). Hence, $c^{h}\in C_{w}=C^{\left(  k\right)
}=\left\langle c^{k}\right\rangle =\left\langle c^{n}\right\rangle $ (since
$k=n$). In other words, $h$ is a multiple of $n$ (since $c$ is a generator of
the infinite cyclic group $C$). But this contradicts $h\in\left\{
1,2,\ldots,n-1\right\}  $. This contradiction shows that our assumption was
false. Hence, $k\neq n$ is proved.]

But Lemma \ref{lem.sol.words.necklaces.rooting} (applied to $N=n$ and $p=k$)
yields that there exists a word $q\in\mathfrak{A}^{k}$ such that $w=q^{n/k}$.
Consider this $q$. Combining $k\neq n$ with $k\leq n$, we obtain $k<n$. Since
$n>0$, we thus have $n/\underbrace{k}_{<n}>n/n=1$. But $n/k$ is an integer
(since $k$ is a divisor of $n$). Hence, from $n/k>1$, we obtain $n/k\geq2$.
Thus, we now know that $n/k$ is an integer satisfying $n/k\geq2$, and
$q\in\mathfrak{A}^{\ast}$ is a word satisfying $w=q^{n/k}$. Hence, there exist
$m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$ (namely, $m=n/k$
and $u=q$). This contradicts the fact that there exist no $m\geq2$ and
$u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$. This contradiction shows that
our assumption was false. Hence, the necklace $\left[  w\right]  $ is
aperiodic. This proves Claim 1 again.]
\end{verlong}

[\textit{Proof of Claim 2:} Assume that the necklace $\left[  w\right]  $ is
aperiodic. We must prove that the word $w$ is aperiodic.

Let $m\geq2$ and $u\in\mathfrak{A}^{\ast}$ be such that $w=u^{m}$. We shall
derive a contradiction.

From $w=u^{m}$, we obtain $\ell\left(  w\right)  =\ell\left(  u^{m}\right)
=m\ell\left(  u\right)  $, so that $m\ell\left(  u\right)  =\ell\left(
w\right)  =n$. Hence, $\ell\left(  u\right)  =n/m$ (since $m\geq2>0$). Thus,
$u\in\mathfrak{A}^{n/m}$. Also, $m$ is a divisor of $n$ (since $m\ell\left(
u\right)  =n$ with $\ell\left(  u\right)  \in \ZZ $), and is positive
(since $m\geq2>0$).

\begin{verlong}
Hence, $m$ is a positive divisor of $n$.
\end{verlong}

Hence, Lemma \ref{lem.sol.lyndon.gr.aper-rel.1} (applied to $d=m$ and $q=u$)
yields that the $n/m$-necklace $\left[  u\right]  $ and the $n$-necklace
$\left[  u^{m}\right]  $ satisfy $\left\vert \left[  u^{m}\right]  \right\vert
=\left\vert \left[  u\right]  \right\vert \leq n/m$. From $w=u^{m}$, we obtain
$\left\vert \left[  w\right]  \right\vert =\left\vert \left[  u^{m}\right]
\right\vert \leq n/\underbrace{m}_{\geq2}\leq n/2<n$ (since $n>0$).

But Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper} (applied to $w$ instead of
$u$) yields that $\left\vert \left[  w\right]  \right\vert =\ell\left(
w\right)  $ (since the necklace $\left[  w\right]  $ is aperiodic). Hence,
$\left\vert \left[  w\right]  \right\vert =\ell\left(  w\right)  =n$. But this
contradicts $\left\vert \left[  w\right]  \right\vert <n$.

Forget that we fixed $m$ and $u$. We thus have obtained a contradiction for
every $m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$. Hence,
there exist no $m\geq2$ and $u\in\mathfrak{A}^{\ast}$ satisfying $w=u^{m}$. In
other words, the word $w$ is aperiodic (by the definition of an aperiodic
word). This proves Claim 2.]

We have now proved both Claim 1 and Claim 2. Combining these two claims, we
conclude that the word $w$ is aperiodic if and only if the necklace $\left[
w\right]  $ is aperiodic. This proves Proposition \ref{prop.words.aper-neck}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.words.aper-neck-cw}.]The word $w$ is aperiodic
and thus nonempty (since any aperiodic word is nonempty). In other words,
$\ell\left(  w\right)  >0$. Let $n=\ell\left(  w\right)  $. Hence,
$w\in\mathfrak{A}^{n}$. Thus, $c\cdot w\in\mathfrak{A}^{n}$, so that
$\ell\left(  c\cdot w\right)  =n>0$. Hence, the word $c\cdot w$ is nonempty.

Now, $\left[  w\right]  $ is the $n$-necklace containing $w$. In other words,
$\left[  w\right]  $ is the $C$-orbit containing $w$ (since the $n$-necklaces
are exactly the $C$-orbits on $\mathfrak{A}^{n}$). In other words, $\left[
w\right]  =C\cdot w$ (where we are using the notation $C\cdot w$ for the
$C$-orbit $\left\{  d\cdot w\ \mid\ d\in C\right\}  $). The same argument
(applied to $c\cdot w$ instead of $w$) yields that $\left[  c\cdot w\right]
=C\cdot\left(  c\cdot w\right)  $. Hence,%
\[
\left[  c\cdot w\right]  =C\cdot\left(  c\cdot w\right)  =\underbrace{\left(
Cc\right)  }_{\substack{=C\\\text{(since }C\text{ is a group)}}}\cdot w=C\cdot
w=\left[  w\right]  .
\]


But Proposition \ref{prop.words.aper-neck} shows that the word $w$ is
aperiodic if and only if the necklace $\left[  w\right]  $ is aperiodic.
Hence, the necklace $\left[  w\right]  $ is aperiodic (since the word $w$ is
aperiodic). In other words, the necklace $\left[  c\cdot w\right]  $ is
aperiodic (since $\left[  c\cdot w\right]  =\left[  w\right]  $).

However, Proposition \ref{prop.words.aper-neck} (applied to $c\cdot w$ instead
of $w$) shows that the word $c\cdot w$ is aperiodic if and only if the
necklace $\left[  c\cdot w\right]  $ is aperiodic. Hence, the word $c\cdot w$
is aperiodic (since the necklace $\left[  c\cdot w\right]  $ is aperiodic).
This proves Corollary \ref{cor.words.aper-neck-cw}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.words.aper-neck-ms}.]Let $N$ be an aperiodic
necklace. We must prove that $N$ is a set of aperiodic words. In other words,
we must prove that each $w\in N$ is an aperiodic word.

So let $w\in N$. We shall prove that $w$ is an aperiodic word.

Indeed, $N$ is a necklace, i.e., an $n$-necklace for some positive integer
$n$. Consider this $n$. Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts} (applied to
$u=w$) shows that $N=\left[  w\right]  $. Hence, $\left[  w\right]  $ is an
aperiodic $n$-necklace (since $N$ is an aperiodic $n$-necklace).

\begin{vershort}
The word $w$ has length $n$ (since $\left[  w\right]  $ is an $n$-necklace)
and thus is nonempty (since $n$ is positive). Moreover, the necklace $\left[
w\right]  $ is aperiodic. Hence, Proposition \ref{prop.words.aper-neck} shows
that the word $w$ is aperiodic.
\end{vershort}

\begin{verlong}
The word $w$ has length $n$ (since $\left[  w\right]  $ is an $n$-necklace).
Thus, $\ell\left(  w\right)  =n>0$ (since $n$ is positive). In other words,
$w$ is nonempty. Hence, Proposition \ref{prop.words.aper-neck} shows that the
word $w$ is aperiodic if and only if the necklace $\left[  w\right]  $ is
aperiodic. Hence, the word $w$ is aperiodic (since the necklace $\left[
w\right]  $ is aperiodic). So we know that $w$ is an aperiodic word.
\end{verlong}

Forget that we fixed $w$. Thus, we have proved that each $w\in N$ is an
aperiodic word. This completes the proof of Corollary
\ref{cor.words.aper-neck-ms}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.words.aper-rel.tord}.]We shall prove the
following four claims:

\begin{statement}
\textit{Claim 1:} The relation $\leq_{\omega}$ is reflexive.
\end{statement}

\begin{statement}
\textit{Claim 2:} The relation $\leq_{\omega}$ is transitive.
\end{statement}

\begin{statement}
\textit{Claim 3:} The relation $\leq_{\omega}$ is antisymmetric.
\end{statement}

\begin{statement}
\textit{Claim 4:} The relation $\leq_{\omega}$ is total.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 1:} Each $a\in\mathfrak{A}^{\mathfrak{a}}$ satisfies
$aa\leq aa$. In other words, each $a\in\mathfrak{A}^{\mathfrak{a}}$ satisfies
$a\leq_{\omega}a$ (by the definition of the relation $\leq_{\omega}$). In
other words, the relation $\leq_{\omega}$ is reflexive. This proves Claim 1.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 1:} Let $a\in\mathfrak{A}^{\mathfrak{a}}$. Then,
$aa\leq aa$. In other words, $a\leq_{\omega}a$ (by the definition of the
relation $\leq_{\omega}$). Now, forget that we fixed $a$. We thus have proved
that $a\leq_{\omega}a$ for each $a\in\mathfrak{A}^{\mathfrak{a}}$. In other
words, the relation $\leq_{\omega}$ is reflexive. This proves Claim 1.]
\end{verlong}

[\textit{Proof of Claim 2:} Let $x,y,z\in\mathfrak{A}^{\mathfrak{a}}$ satisfy
$x\leq_{\omega}y$ and $y\leq_{\omega}z$. We shall show that $x\leq_{\omega}z$.

We have assumed that $x\leq_{\omega}y$. In other words, $xy\leq yx$ (by the
definition of the relation $\leq_{\omega}$). In other words, $yx\geq xy$.

We have assumed that $y\leq_{\omega}z$. In other words, $yz\leq zy$ (by the
definition of the relation $\leq_{\omega}$). In other words, $zy\geq yz$.

The word $y$ is aperiodic (since $y\in\mathfrak{A}^{\mathfrak{a}}$) and thus
nonempty (since any aperiodic word is nonempty). Hence, Corollary
\ref{cor.words.uv=vu.3} (applied to $u=z$, $v=y$ and $w=x$) yields $zx\geq
xz$. In other words, $xz\leq zx$. In other words, $x\leq_{\omega}z$ (by the
definition of the relation $\leq_{\omega}$).

Now, forget that we fixed $x,y,z$. We thus have proved that if $x,y,z\in
\mathfrak{A}^{\mathfrak{a}}$ satisfy $x\leq_{\omega}y$ and $y\leq_{\omega}z$,
then $x\leq_{\omega}z$. In other words, the relation $\leq_{\omega}$ is
transitive. This proves Claim 2.]

[\textit{Proof of Claim 3:} Let $a,b\in\mathfrak{A}^{\mathfrak{a}}$ satisfy
$a\leq_{\omega}b$ and $b\leq_{\omega}a$. We shall show that $a=b$.

\begin{vershort}
The word $b$ is aperiodic (since $b\in\mathfrak{A}^{\mathfrak{a}}$). In other
words, $b$ is a nonempty word with the property that%
\begin{equation}
\text{there exist no }m\geq2\text{ and }u\in\mathfrak{A}^{\ast}\text{
satisfying }b=u^{m} \label{pf.prop.words.aper-rel.tord.c3.pf.short.1}%
\end{equation}
(by the definition of an aperiodic word).

We have assumed that $a\leq_{\omega}b$. In other words, $ab\leq ba$ (by the
definition of the relation $\leq_{\omega}$). Likewise, $ba\leq ab$ (since
$b\leq_{\omega}a$). Combining these two inequalities, we obtain $ab=ba$.
Hence, Proposition \ref{prop.words.uv=vu.2} (applied to $u=a$ and $v=b$)
yields that there exist a $t\in\mathfrak{A}^{\ast}$ and two nonnegative
integers $n$ and $m$ such that $a=t^{n}$ and $b=t^{m}$. Consider these $t$,
$n$ and $m$.

From $b=t^{m}$, we obtain $\ell\left(  b\right)  =\ell\left(  t^{m}\right)
=m\ell\left(  t\right)  $, thus $m\ell\left(  t\right)  =\ell\left(  b\right)
\neq0$ (since $b$ is nonempty). Hence, $m\neq0$, so that $m\geq1$ (since $m$
is a nonnegative integer). Note that the word $t$ satisfies $b=t^{m}$; thus,
there exists a word $u\in\mathfrak{A}^{\ast}$ satisfying $b=u^{m}$ (namely,
$u=t$). If we had $m\geq2$, then this would violate
(\ref{pf.prop.words.aper-rel.tord.c3.pf.short.1}). Hence, we cannot have
$m\geq2$. Thus, $m<2$. Combining this with $m\geq1$, we obtain $m=1$. Now,
$b=t^{m}=t^{1}$ (since $m=1$). Likewise, $a=t^{1}$. Hence, $a=t^{1}=b$.
\end{vershort}

\begin{verlong}
The word $b$ is aperiodic (since $b\in\mathfrak{A}^{\mathfrak{a}}$). In other
words, $b$ is a nonempty word with the property that%
\begin{equation}
\text{there exist no }m\geq2\text{ and }u\in\mathfrak{A}^{\ast}\text{
satisfying }b=u^{m} \label{pf.prop.words.aper-rel.tord.c3.pf.1}%
\end{equation}
(by the definition of an aperiodic word). We can restate the property
(\ref{pf.prop.words.aper-rel.tord.c3.pf.1}) as follows:
\begin{equation}
\text{if }m\geq2\text{ and }u\in\mathfrak{A}^{\ast}\text{ satisfy }%
b=u^{m}\text{, then we have a contradiction.}
\label{pf.prop.words.aper-rel.tord.c3.pf.2}%
\end{equation}


We have assumed that $a\leq_{\omega}b$. In other words, $ab\leq ba$ (by the
definition of the relation $\leq_{\omega}$). In other words, $ba\geq ab$. The
same argument (with the roles of $a$ and $b$ interchanged) yields $ab\geq ba$.
Combining this with $ba\geq ab$, we obtain $ab=ba$. Hence, Proposition
\ref{prop.words.uv=vu.2} (applied to $u=a$ and $v=b$) yields that there exist
a $t\in\mathfrak{A}^{\ast}$ and two nonnegative integers $n$ and $m$ such that
$a=t^{n}$ and $b=t^{m}$. Consider these $t$, $n$ and $m$.

The word $b$ is nonempty. Hence, $m\geq1$\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, $m<1$. Hence, $m=0$ (since $m$ is a nonnegative
integer). Therefore, $t^{m}=t^{0}=\varnothing$, so that $b=t^{m}=\varnothing$.
Thus, $b$ is empty. This contradicts the fact that $b$ is nonempty. This
contradiction shows that our assumption was false, qed.}. On the other hand,
$m\leq1$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $m>1$.
Hence, $m\geq2$ (since $m$ is an integer). Also, $t\in\mathfrak{A}^{\ast}$ and
$b=t^{m}$. Hence, applying (\ref{pf.prop.words.aper-rel.tord.c3.pf.2}) to
$u=t$, we conclude that we have a contradiction. This contradiction shows that
our assumption was false, qed.}. Combining these two inequalities, we find
$m=1$. Thus, $t^{m}=t^{1}=t$. Hence, $b=t^{m}=t$. The same argument (applied
to $a$ and $n$ instead of $b$ and $m$) yields $a=t$. Comparing these two
equalities, we find $a=b$.
\end{verlong}

Now, forget that we fixed $a,b$. We thus have proved that if $a,b\in
\mathfrak{A}^{\mathfrak{a}}$ satisfy $a\leq_{\omega}b$ and $b\leq_{\omega}a$,
then $a=b$. In other words, the relation $\leq_{\omega}$ is antisymmetric.
This proves Claim 3.]

[\textit{Proof of Claim 4:} Let $a,b\in\mathfrak{A}^{\mathfrak{a}}$. We shall
prove that we have $a\leq_{\omega}b$ or $b\leq_{\omega}a$.

\begin{vershort}
Indeed, the relation $\leq$ on $\mathfrak{A}^{\ast}$ is total (since the
lexicographic order on words is a total order). Thus, we have $ab\leq ba$ or
$ba\leq ab$. In other words, we have $a\leq_{\omega}b$ or $b\leq_{\omega}a$
(because the relation $a\leq_{\omega}b$ is equivalent to $ab\leq
ba$\ \ \ \ \footnote{by the definition of the relation $\leq_{\omega}$},
whereas the relation $b\leq_{\omega}a$ is equivalent to $ba\leq ab$%
\ \ \ \ \footnote{by the definition of the relation $\leq_{\omega}$}).
\end{vershort}

\begin{verlong}
Indeed, the relation $\leq$ on $\mathfrak{A}^{\ast}$ is total (since the
lexicographic order on words is a total order). Thus, we are in one of the
following two cases:

\textit{Case 1:} We have $ab\leq ba$.

\textit{Case 2:} We have $ba\leq ab$.

Let us first consider Case 1. In this case, we have $ab\leq ba$. In other
words, $a\leq_{\omega}b$ (by the definition of the relation $\leq_{\omega}$).
Thus, we have $a\leq_{\omega}b$ or $b\leq_{\omega}a$. Hence, in Case 1, we
have proved that $a\leq_{\omega}b$ or $b\leq_{\omega}a$.

Let us now consider Case 2. In this case, we have $ba\leq ab$. In other words,
$b\leq_{\omega}a$ (by the definition of the relation $\leq_{\omega}$). Thus,
we have $a\leq_{\omega}b$ or $b\leq_{\omega}a$. Hence, in Case 2, we have
proved that $a\leq_{\omega}b$ or $b\leq_{\omega}a$.

Now, in each of the two Cases 1 and 2, we have proved that $a\leq_{\omega}b$
or $b\leq_{\omega}a$. Hence, we always have $a\leq_{\omega}b$ or
$b\leq_{\omega}a$ (since Cases 1 and 2 cover all possibilities).
\end{verlong}

Now, forget that we fixed $a,b$. We thus have proved that any $a,b\in
\mathfrak{A}^{\mathfrak{a}}$ satisfy $a\leq_{\omega}b$ or $b\leq_{\omega}a$.
In other words, the relation $\leq_{\omega}$ is total. This proves Claim 4.]

Now, the relation $\leq_{\omega}$ is reflexive (by Claim 1), transitive (by
Claim 2) and antisymmetric (by Claim 3). Hence, it is the smaller-or-equal
relation of a partial order. This partial order must furthermore be a total
order, since the relation $\leq_{\omega}$ is total (by Claim 4). Thus, the
relation $\leq_{\omega}$ is the smaller-or-equal relation of a total order.
This proves Proposition \ref{prop.words.aper-rel.tord}.
\end{proof}

Before we prove Proposition \ref{prop.words.aper-rel.1}, let us show some more
basic lemmas:

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.ciwm}Let $u\in\mathfrak{A}^{\ast}$ be a
nonempty word. Let $m$ be a positive integer. Then, $\left(  c^{i}\cdot
u\right)  ^{m}=c^{i}\cdot\left(  u^{m}\right)  $ for each $i\in \NN $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciwm}.]Let $i\in \NN $.

Let $n=\ell\left(  u\right)  \cdot m$ and $d=m$. We know that $m$ is a
positive integer; in other words, $d$ is a positive integer (since $d=m$).
Also, $n=\ell\left(  u\right)  \cdot\underbrace{m}_{=d}=\ell\left(  u\right)
\cdot d$; hence, $d$ is a divisor of $n$. Finally, $\ell\left(  u\right)  $ is
a positive integer (since $u$ is nonempty). Now, both $\ell\left(  u\right)  $
and $m$ are positive integers. Hence, the product $\ell\left(  u\right)  \cdot
m$ is a positive integer as well. In other words, $n$ is a positive integer
(since $n=\ell\left(  u\right)  \cdot m$).

Consider the map $\Delta:\mathfrak{A}^{n/d}\rightarrow\mathfrak{A}^{n}$
defined in Lemma \ref{lem.sol.words.necklaces.f2}.

We have $n=\ell\left(  u\right)  \cdot d$, so that $\ell\left(  u\right)
=n/d$ (since $d\neq0$ (because $d$ is positive)). Hence, $u\in\mathfrak{A}%
^{n/d}$. Thus, the definition of $\Delta$ yields $\Delta\left(  u\right)
=u^{d}=u^{m}$ (since $d=m$). Also, the definition of $\Delta$ yields
$\Delta\left(  c^{i}u\right)  =\left(  \underbrace{c^{i}u}_{=c^{i}\cdot
u}\right)  ^{d}=\left(  c^{i}\cdot u\right)  ^{d}=\left(  c^{i}\cdot u\right)
^{m}$ (since $d=m$). But Lemma \ref{lem.sol.words.necklaces.f2}(b) shows that
the map $\Delta$ is $C$-equivariant; in other words, we have $\Delta\left(
gw\right)  =g\cdot\Delta\left(  w\right)  $ for every $g\in C$ and
$w\in\mathfrak{A}^{n/d}$. Applying this to $g=c^{i}$ and $w=u$, we obtain
$\Delta\left(  c^{i}u\right)  =c^{i}\cdot\underbrace{\Delta\left(  u\right)
}_{=u^{m}}=c^{i}\cdot\left(  u^{m}\right)  $. Comparing this with
$\Delta\left(  c^{i}u\right)  =\left(  c^{i}\cdot u\right)  ^{m}$, we obtain
$\left(  c^{i}\cdot u\right)  ^{m}=c^{i}\cdot\left(  u^{m}\right)  $. This
proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciwm}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.ciwj}Let $n$ be a positive integer. Let
$w\in\mathfrak{A}^{n}$ be a word. Then, $\left(  c^{i}\cdot w\right)
_{j}=w_{i+j}$ for each $i\in \NN $ and each $j\in\left\{  1,2,\ldots
,n-i\right\}  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciwj}.]Let $i\in \NN $,
and let $j\in\left\{  1,2,\ldots,n-i\right\}  $. Then, $1\leq j\leq n-i$, so
that $n-i\geq1\geq0$ and thus $n\geq i$. Hence, $i\leq n$, so that
$i\in\left\{  0,1,\ldots,n\right\}  $ (since $i\in \NN $).

Recall that $c$ acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples
one step to the left (that is, we have $c\cdot\left(  a_{1},a_{2},\ldots
,a_{n}\right)  =\left(  a_{2},a_{3},\ldots,a_{n},a_{1}\right)  $ for each
$\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}$). Thus, $c^{i}$
acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples $i$ steps to the
left. In other words, for each $\left(  a_{1},a_{2},\ldots,a_{n}\right)
\in\mathfrak{A}^{n}$, we have%
\[
c^{i}\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{i+1}%
,a_{i+2},\ldots,a_{n},a_{1},a_{2},\ldots,a_{i}\right)
\]
(since $i\in\left\{  0,1,\ldots,n\right\}  $). Applying this to $a_{k}=w_{k}$,
we obtain%
\[
c^{i}\cdot\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\left(  w_{i+1}%
,w_{i+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{i}\right)  .
\]
But $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $ (since $w\in\mathfrak{A}%
^{n}$), and thus%
\[
c^{i}\cdot w=c^{i}\cdot\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\left(
w_{i+1},w_{i+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{i}\right)  .
\]
Hence, the first $n-i$ letters of the word $c^{i}\cdot w$ are $w_{i+1}%
,w_{i+2},\ldots,w_{n}$ (in this order). In other words, for each $k\in\left\{
1,2,\ldots,n-i\right\}  $, we have
\[
\left(  \text{the }k\text{-th letter of }c^{i}\cdot w\right)  =w_{i+k}.
\]
Applying this to $k=j$, we obtain
\[
\left(  \text{the }j\text{-th letter of }c^{i}\cdot w\right)  =w_{i+j}%
\]
(since $j\in\left\{  1,2,\ldots,n-i\right\}  $). Hence,
\[
\left(  c^{i}\cdot w\right)  _{j}=\left(  \text{the }j\text{-th letter of
}c^{i}\cdot w\right)  =w_{i+j}.
\]
This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciwj}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.uiwm}Let $w\in\mathfrak{A}^{\ast}$ be a
nonempty word. Let $m\in \NN $. Then, $\left(  c^{i}\cdot w\right)
_{1}=\left(  w^{m}\right)  _{i+1}$ for each $i\in\left\{  0,1,\ldots
,m\ell\left(  w\right)  -1\right\}  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.uiwm}.]Let $n=\ell\left(
w\right)  $; thus, $w\in\mathfrak{A}^{n}$. The word $w$ is nonempty. Hence,
$\ell\left(  w\right)  $ is a positive integer. In other words, $n$ is a
positive integer (since $n=\ell\left(  w\right)  $).

Let $v=c^{i}\cdot w$. Then, $v\in\mathfrak{A}^{n}$ (since $w\in\mathfrak{A}%
^{n}$). Thus, $\ell\left(  v\right)  =n$.

\begin{vershort}
Let $i\in\left\{  0,1,\ldots,m\ell\left(  w\right)  -1\right\}  $. Then,
$0\leq i<m\ell\left(  w\right)  $, so that $m\ell\left(  w\right)  >0$ and
thus $m\ell\left(  w\right)  \neq0$. This entails $m\neq0$, so that $m>0$
(since $m\in \NN $). Hence, the word $v$ is a prefix of $v^{m}$, and thus
has the same first letter as $v^{m}$ (since the word $v$ is
nonempty\footnote{because $\ell\left(  v\right)  =n>0$}, and thus has a first
letter). In other words, $v_{1}=\left(  v^{m}\right)  _{1}$.
\end{vershort}

\begin{verlong}
But $n$ is a positive integer; thus, $n\in\left\{  1,2,\ldots,n\right\}
=\left\{  1,2,\ldots,\ell\left(  v\right)  \right\}  $ (since $n=\ell\left(
v\right)  $).

Let $i\in\left\{  0,1,\ldots,m\ell\left(  w\right)  -1\right\}  $. Then,
$0\leq i\leq m\ell\left(  w\right)  -1$, so that $0\leq m\ell\left(  w\right)
-1<m\ell\left(  w\right)  $. Hence, $m\ell\left(  w\right)  \neq0$, and thus
$m\neq0$. Hence, $m>0$ (since $m\in \NN $). Thus, $v^{m}=vv^{m-1}$. This
shows that $v$ is a prefix of $v^{m}$. Hence, the first $\ell\left(  v\right)
$ letters of the word $v$ equal the corresponding letters of $v^{m}$. In other
words,%
\[
v_{j}=\left(  v^{m}\right)  _{j}\ \ \ \ \ \ \ \ \ \ \text{for each }%
j\in\left\{  1,2,\ldots,\ell\left(  v\right)  \right\}  .
\]
Applying this to $j=1$, we obtain $v_{1}=\left(  v^{m}\right)  _{1}$ (since
$1\in\left\{  1,2,\ldots,\ell\left(  v\right)  \right\}  $).
\end{verlong}

Let $N=m\ell\left(  w\right)  $. Thus, $i\in\left\{  0,1,\ldots,m\ell\left(
w\right)  -1\right\}  =\left\{  0,1,\ldots,N-1\right\}  $ (since $m\ell\left(
w\right)  =N$), so that $i+1\in\left\{  1,2,\ldots,N\right\}  $. Hence, $1\leq
i+1\leq N$, so that $N\geq1$. This shows that $N$ is a positive integer. Also,
$i+1\leq N$, thus $N\geq i+1$, hence $N-i\geq1$ and thus $1\leq N-i$. Hence,
$1\in\left\{  1,2,\ldots,N-i\right\}  $.

We have $\ell\left(  w^{m}\right)  =m\ell\left(  w\right)  =N$ (since
$N=m\ell\left(  w\right)  $), so that $w^{m}\in\mathfrak{A}^{N}$. Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciwj} (applied to $N$, $w^{m}$ and $1$ instead
of $n$, $w$ and $j$) thus yields $\left(  c^{i}\cdot\left(  w^{m}\right)
\right)  _{1}=\left(  w^{m}\right)  _{i+1}$. But Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciwm} (applied to $u=w$) yields $\left(
c^{i}\cdot w\right)  ^{m}=c^{i}\cdot\left(  w^{m}\right)  $. In view of
$c^{i}\cdot w=v$, this rewrites as $v^{m}=c^{i}\cdot\left(  w^{m}\right)  $.
Hence, $\left(  v^{m}\right)  _{1}=\left(  c^{i}\cdot\left(  w^{m}\right)
\right)  _{1}=\left(  w^{m}\right)  _{i+1}$. Hence, $v_{1}=\left(
v^{m}\right)  _{1}=\left(  w^{m}\right)  _{i+1}$. In view of $v=c^{i}\cdot w$,
this rewrites as $\left(  c^{i}\cdot w\right)  _{1}=\left(  w^{m}\right)
_{i+1}$. This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.uiwm}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.umvn}Let $u,v\in\mathfrak{A}^{\ast}$ be two
aperiodic words such that $u\neq v$ and $u\leq_{\omega}v$. Let $g$, $n$ and
$m$ be three positive integers such that $g=n\ell\left(  v\right)
=m\ell\left(  u\right)  $. Then, there exists a $k\in\left\{  1,2,\ldots
,g\right\}  $ such that $\left(  u^{m}\right)  _{k}<\left(  v^{n}\right)
_{k}$, and such that every $j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies
$\left(  u^{m}\right)  _{j}=\left(  v^{n}\right)  _{j}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.umvn}.]We have $\ell\left(
u^{m}\right)  =m\ell\left(  u\right)  =g$ (since $g=m\ell\left(  u\right)  $)
and $\ell\left(  v^{n}\right)  =n\ell\left(  v\right)  =g$ (since
$g=n\ell\left(  v\right)  $). Hence, $\ell\left(  u^{m}\right)  =g=\ell\left(
v^{n}\right)  $; in other words, the words $u^{m}$ and $v^{n}$ have the same length.

\begin{vershort}
Proposition \ref{prop.words.aper-rel.tord} shows that the relation
$\leq_{\omega}$ on the set $\mathfrak{A}^{\mathfrak{a}}$ is the
smaller-or-equal relation of a total order. Hence, in particular, this
relation $\leq_{\omega}$ is antisymmetric. Thus, we don't have $v\leq_{\omega
}u$ (because otherwise, we could combine $u\leq_{\omega}v$ with $v\leq
_{\omega}u$ to obtain $u=v$, which would contradict $u\neq v$). In other
words, we don't have $vu\leq uv$ (since the relation $v\leq_{\omega}u$ means
the same as $vu\leq uv$). In other words, we don't have $uv\geq vu$.
\end{vershort}

\begin{verlong}
Proposition \ref{prop.words.aper-rel.tord} shows that the relation
$\leq_{\omega}$ on the set $\mathfrak{A}^{\mathfrak{a}}$ is the
smaller-or-equal relation of a total order. Hence, in particular, this
relation $\leq_{\omega}$ is antisymmetric. Thus, we don't have $v\leq_{\omega
}u$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, we have
$v\leq_{\omega}u$. Combining $u\leq_{\omega}v$ with $v\leq_{\omega}u$, we
obtain $u=v$ (since the relation $\leq_{\omega}$ is antisymmetric). This
contradicts $u\neq v$. This contradiction shows that our assumption was false,
qed.}.

We have $v\leq_{\omega}u$ if and only if we have $vu\leq uv$ (by the
definition of the relation $\leq_{\omega}$). Hence, we don't have $vu\leq uv$
(since we don't have $v\leq_{\omega}u$). In other words, we don't have $uv\geq
vu$.
\end{verlong}

From $n\ell\left(  v\right)  =m\ell\left(  u\right)  $, we obtain
$m\ell\left(  u\right)  =n\ell\left(  v\right)  $. Hence, Exercise
\ref{exe.words.un.vm} (applied to $n$ and $m$ instead of $m$ and $n$) shows
that $uv\geq vu$ holds if and only if $u^{m}\geq v^{n}$ holds. Therefore, we
don't have $u^{m}\geq v^{n}$ (since we don't have $uv\geq vu$). Thus, we have
$u^{m}<v^{n}$ (since the lexicographic order on $\mathfrak{A}^{\ast}$ is a
total order). Hence, the word $u^{m}$ is not a prefix of $v^{n}$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, the word $u^{m}$
is a prefix of $v^{n}$. Hence, $u^{m}=v^{n}$ (since the words $u^{m}$ and
$v^{n}$ have the same length). This contradicts $u^{m}<v^{n}$. This
contradiction shows that our assumption was false, qed.}.

But we have $u^{m}\leq v^{n}$ (since $u^{m}<v^{n}$). In other words,%
\begin{align*}
&  \text{\textbf{either }}\text{there exists an }i\in\left\{  1,2,\ldots
,\min\left\{  \ell\left(  u^{m}\right)  ,\ell\left(  v^{n}\right)  \right\}
\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \text{such that }\left(  \left(  u^{m}\right)
_{i}<\left(  v^{n}\right)  _{i}\text{, and every }j\in\left\{  1,2,\ldots
,i-1\right\}  \text{ satisfies }\left(  u^{m}\right)  _{j}=\left(
v^{n}\right)  _{j}\right)  ,\\
&  \text{\textbf{or} the word }u^{m}\text{ is a prefix of }v^{n}%
\end{align*}
(by the definition of the lexicographic order). Thus, there exists an
$i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  u^{m}\right)  ,\ell\left(
v^{n}\right)  \right\}  \right\}  $ such that
\[
\left(  \left(  u^{m}\right)  _{i}<\left(  v^{n}\right)  _{i}\text{, and every
}j\in\left\{  1,2,\ldots,i-1\right\}  \text{ satisfies }\left(  u^{m}\right)
_{j}=\left(  v^{n}\right)  _{j}\right)
\]
(since the word $u^{m}$ is not a prefix of $v^{n}$). Consider this $i$.

We have $i\in\left\{  1,2,\ldots,\min\left\{  \ell\left(  u^{m}\right)
,\ell\left(  v^{n}\right)  \right\}  \right\}  =\left\{  1,2,\ldots,g\right\}
$ (since $\min\left\{  \underbrace{\ell\left(  u^{m}\right)  }_{=g}%
,\underbrace{\ell\left(  v^{n}\right)  }_{=g}\right\}  =\min\left\{
g,g\right\}  =g$) and $\left(  u^{m}\right)  _{i}<\left(  v^{n}\right)  _{i}$.
Moreover, every $j\in\left\{  1,2,\ldots,i-1\right\}  $ satisfies $\left(
u^{m}\right)  _{j}=\left(  v^{n}\right)  _{j}$. Hence, there exists a
$k\in\left\{  1,2,\ldots,g\right\}  $ such that $\left(  u^{m}\right)
_{k}<\left(  v^{n}\right)  _{k}$, and such that every $j\in\left\{
1,2,\ldots,k-1\right\}  $ satisfies $\left(  u^{m}\right)  _{j}=\left(
v^{n}\right)  _{j}$ (namely, $k=i$). This proves Lemma
\ref{lem.sol.lyndon.gr.aper-rel.umvn}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.ciuciv}Let $u,v\in\mathfrak{A}^{\ast}$ be
two aperiodic words such that $u\neq v$. Let $g=\ell\left(  u\right)
\cdot\ell\left(  v\right)  $. Then:

\begin{enumerate}
\item[(a)] There exists some $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$.

\item[(b)] If $u\leq_{\omega}v$, then the smallest $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$.

\item[(c)] If $v\leq_{\omega}u$, then the smallest $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies $\left(  c^{i}\cdot u\right)
_{1}>\left(  c^{i}\cdot v\right)  _{1}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}.]The words $u$ and $v$
are aperiodic and thus nonempty (since any aperiodic word is nonempty). Let
$n=\ell\left(  u\right)  $ and $m=\ell\left(  v\right)  $. Then,
$n=\ell\left(  u\right)  >0$ (since $u$ is nonempty), so that $n$ is a
positive integer. Likewise, $m$ is a positive integer. Moreover, combining the
equalities $\underbrace{n}_{=\ell\left(  u\right)  }\ell\left(  v\right)
=\ell\left(  u\right)  \cdot\ell\left(  v\right)  =g$ and $\underbrace{m}%
_{=\ell\left(  v\right)  }\ell\left(  u\right)  =\ell\left(  v\right)
\cdot\ell\left(  u\right)  =\ell\left(  u\right)  \cdot\ell\left(  v\right)
=g$, we obtain $g=n\ell\left(  v\right)  =m\ell\left(  u\right)  $. Also,
$g=\underbrace{\ell\left(  u\right)  }_{=n}\cdot\underbrace{\ell\left(
v\right)  }_{=m}=nm$; this shows that $g$ is a positive integer (since $n$ and
$m$ are positive integers). Note also that $v\neq u$ (since $u\neq v$) and
$g=\ell\left(  u\right)  \cdot\ell\left(  v\right)  =\ell\left(  v\right)
\cdot\ell\left(  u\right)  $. Thus, $u$ and $v$ play symmetric roles in our situation.

The words $u$ and $v$ are aperiodic; i.e., we have $u,v\in\mathfrak{A}%
^{\mathfrak{a}}$ (since $\mathfrak{A}^{\mathfrak{a}}$ is the set of all
aperiodic words).

We shall now prove the following fact:

\begin{statement}
\textit{Claim 1:} Assume that $u\leq_{\omega}v$. Then, there exists some
$i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$.
\end{statement}

[\textit{Proof of Claim 1:} Lemma \ref{lem.sol.lyndon.gr.aper-rel.umvn} shows
that there exists a $k\in\left\{  1,2,\ldots,g\right\}  $ such that $\left(
u^{m}\right)  _{k}<\left(  v^{n}\right)  _{k}$, and such that every
$j\in\left\{  1,2,\ldots,k-1\right\}  $ satisfies $\left(  u^{m}\right)
_{j}=\left(  v^{n}\right)  _{j}$. Consider this $k$. But $k\in\left\{
1,2,\ldots,g\right\}  $, thus $k-1\in\left\{  0,1,\ldots,g-1\right\}
=\left\{  0,1,\ldots,m\ell\left(  u\right)  -1\right\}  $ (since
$g=m\ell\left(  u\right)  $). Hence, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.uiwm} (applied to $u$ and $k-1$ instead of $w$
and $i$) yields $\left(  c^{k-1}\cdot u\right)  _{1}=\left(  u^{m}\right)
_{\left(  k-1\right)  +1}=\left(  u^{m}\right)  _{k}$ (since $\left(
k-1\right)  +1=k$).

Also, $k-1\in\left\{  0,1,\ldots,g-1\right\}  =\left\{  0,1,\ldots
,n\ell\left(  v\right)  -1\right\}  $ (since $g=n\ell\left(  v\right)  $).
Hence, Lemma \ref{lem.sol.lyndon.gr.aper-rel.uiwm} (applied to $v$, $n$ and
$k-1$ instead of $w$, $m$ and $i$) yields $\left(  c^{k-1}\cdot v\right)
_{1}=\left(  v^{n}\right)  _{\left(  k-1\right)  +1}=\left(  v^{n}\right)
_{k}$ (since $\left(  k-1\right)  +1=k$).

Now, recall that $\left(  u^{m}\right)  _{k}<\left(  v^{n}\right)  _{k}$. In
view of $\left(  c^{k-1}\cdot u\right)  _{1}=\left(  u^{m}\right)  _{k}$ and
$\left(  c^{k-1}\cdot v\right)  _{1}=\left(  v^{n}\right)  _{k}$, we can
rewrite this as $\left(  c^{k-1}\cdot u\right)  _{1}<\left(  c^{k-1}\cdot
v\right)  _{1}$. Hence, $\left(  c^{k-1}\cdot u\right)  _{1}\neq\left(
c^{k-1}\cdot v\right)  _{1}$. Since $k-1\in\left\{  0,1,\ldots,g-1\right\}  $,
this shows that there exists some $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ (namely, $i=k-1$). This proves Claim 1.]

We can now easily obtain the following counterpart of Claim 1 for the case
when $v\leq_{\omega}u$:

\begin{statement}
\textit{Claim 2:} Assume that $v\leq_{\omega}u$. Then, there exists some
$i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 2:} We have $v\neq u$ and $g=\ell\left(  v\right)
\cdot\ell\left(  u\right)  $. Hence, Claim 1 (applied to $v$ and $u$ instead
of $u$ and $v$) yields that there exists some $i\in\left\{  0,1,\ldots
,g-1\right\}  $ satisfying $\left(  c^{i}\cdot v\right)  _{1}\neq\left(
c^{i}\cdot u\right)  _{1}$. In other words, there exists some $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$. This proves Claim 2.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 2:} We have $v\neq u$ and $g=\ell\left(  v\right)
\cdot\ell\left(  u\right)  $. Hence, Claim 1 (applied to $v$ and $u$ instead
of $u$ and $v$) yields that there exists some $i\in\left\{  0,1,\ldots
,g-1\right\}  $ satisfying $\left(  c^{i}\cdot v\right)  _{1}\neq\left(
c^{i}\cdot u\right)  _{1}$. In other words, there exists some $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$ (since the relation $\left(  c^{i}\cdot
v\right)  _{1}\neq\left(  c^{i}\cdot u\right)  _{1}$ is equivalent to $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$). This proves
Claim 2.]
\end{verlong}

\begin{vershort}
(a) Proposition \ref{prop.words.aper-rel.tord} shows that the relation
$\leq_{\omega}$ on the set $\mathfrak{A}^{\mathfrak{a}}$ is the
smaller-or-equal relation of a total order. Thus, we have $u\leq_{\omega}v$ or
$v\leq_{\omega}u$ (since $u,v\in\mathfrak{A}^{\mathfrak{a}}$). In each of
these two cases, we can conclude that there exists some $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$ (indeed, in the case $u\leq_{\omega}v$,
this follows from Claim 1, while in the case $v\leq_{\omega}u$, this follows
from Claim 2). This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a).
\end{vershort}

\begin{verlong}
(a) Proposition \ref{prop.words.aper-rel.tord} shows that the relation
$\leq_{\omega}$ on the set $\mathfrak{A}^{\mathfrak{a}}$ is the
smaller-or-equal relation of a total order. Hence, in particular, this
relation $\leq_{\omega}$ is total. Thus, we have $u\leq_{\omega}v$ or
$v\leq_{\omega}u$ (since $u,v\in\mathfrak{A}^{\mathfrak{a}}$). In other words,
we are in one of the following two Cases:

\textit{Case 1:} We have $u\leq_{\omega}v$.

\textit{Case 2:} We have $v\leq_{\omega}u$.

Let us first consider Case 1. In this case, we have $u\leq_{\omega}v$. Thus,
Claim 1 yields that there exists some $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$. This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) in Case 1.

Let us now consider Case 2. In this case, we have $v\leq_{\omega}u$. Thus,
Claim 2 yields that there exists some $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$. This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) in Case 2.

We have now proved Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, we thus
conclude that Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) always holds.
This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a).
\end{verlong}

(b) Assume that $u\leq_{\omega}v$. Thus, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.umvn} shows that there exists a $k\in\left\{
1,2,\ldots,g\right\}  $ such that $\left(  u^{m}\right)  _{k}<\left(
v^{n}\right)  _{k}$, and such that every $j\in\left\{  1,2,\ldots,k-1\right\}
$ satisfies $\left(  u^{m}\right)  _{j}=\left(  v^{n}\right)  _{j}$. Consider
this $k$. We have $\left(  u^{m}\right)  _{k}<\left(  v^{n}\right)  _{k}$.
This rewrites as $\left(  u^{m}\right)  _{\left(  k-1\right)  +1}<\left(
v^{n}\right)  _{\left(  k-1\right)  +1}$ (since $\left(  k-1\right)  +1=k$).

It is easy to see the following:

\begin{statement}
\textit{Claim 3:} The number $k-1$ is the smallest $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  u^{m}\right)  _{i+1}\neq\left(
v^{n}\right)  _{i+1}$.
\end{statement}

[\textit{Proof of Claim 3:} The number $k-1$ is an element of $\left\{
0,1,\ldots,g-1\right\}  $ (since $k\in\left\{  1,2,\ldots,g\right\}  $) and
satisfies $\left(  u^{m}\right)  _{\left(  k-1\right)  +1}\neq\left(
v^{n}\right)  _{\left(  k-1\right)  +1}$ (since $\left(  u^{m}\right)
_{\left(  k-1\right)  +1}<\left(  v^{n}\right)  _{\left(  k-1\right)  +1}$).
In other words, the number $k-1$ is an $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  u^{m}\right)  _{i+1}\neq\left(  v^{n}\right)  _{i+1}$. It
remains to show that $k-1$ is the \textbf{smallest} such $i$.

Indeed, let $i\in\left\{  0,1,\ldots,g-1\right\}  $ be such that $\left(
u^{m}\right)  _{i+1}\neq\left(  v^{n}\right)  _{i+1}$. We shall show that
$i\geq k-1$.

Indeed, assume the contrary. Hence, $i<k-1$. Thus, $i+1<k$, so that $i+1\leq
k-1$ (since $i+1$ and $k$ are integers). Combining this with $i+1>i\geq0$
(since $i\in\left\{  0,1,\ldots,g-1\right\}  $), we obtain $i+1\in\left\{
1,2,\ldots,k-1\right\}  $. But recall that every $j\in\left\{  1,2,\ldots
,k-1\right\}  $ satisfies $\left(  u^{m}\right)  _{j}=\left(  v^{n}\right)
_{j}$. Applying this to $j=i+1$, we obtain $\left(  u^{m}\right)
_{i+1}=\left(  v^{n}\right)  _{i+1}$ (since $i+1\in\left\{  1,2,\ldots
,k-1\right\}  $). This contradicts $\left(  u^{m}\right)  _{i+1}\neq\left(
v^{n}\right)  _{i+1}$. This contradiction shows that our assumption was false.
Hence, $i\geq k-1$ is proved.

Now, forget that we fixed $i$. We thus have shown that $i\geq k-1$ for each
$i\in\left\{  0,1,\ldots,g-1\right\}  $ such that $\left(  u^{m}\right)
_{i+1}\neq\left(  v^{n}\right)  _{i+1}$. Thus, $k-1$ is the \textbf{smallest}
$i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  u^{m}\right)
_{i+1}\neq\left(  v^{n}\right)  _{i+1}$ (since we already know that $k-1$ is
such an $i$). This proves Claim 3.]

From Claim 3, we quickly obtain the following:

\begin{statement}
\textit{Claim 4:} The smallest $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  u^{m}\right)  _{i+1}\neq\left(  v^{n}\right)  _{i+1}$
satisfies $\left(  u^{m}\right)  _{i+1}<\left(  v^{n}\right)  _{i+1}$.
\end{statement}

[\textit{Proof of Claim 4:} We must show that the smallest $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  u^{m}\right)  _{i+1}\neq\left(
v^{n}\right)  _{i+1}$ satisfies $\left(  u^{m}\right)  _{i+1}<\left(
v^{n}\right)  _{i+1}$. But we know (from Claim 3) that this smallest $i$ is
$k-1$. Hence, we need to prove that $i=k-1$ satisfies $\left(  u^{m}\right)
_{i+1}<\left(  v^{n}\right)  _{i+1}$. In other words, we need to prove that
$\left(  u^{m}\right)  _{\left(  k-1\right)  +1}<\left(  v^{n}\right)
_{\left(  k-1\right)  +1}$. But we have proven this already. Thus, Claim 4 is proved.]

On the other hand, each $i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfies%
\begin{equation}
\left(  u^{m}\right)  _{i+1}=\left(  c^{i}\cdot u\right)  _{1}.
\label{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.umi+1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.umi+1}):} Let
$i\in\left\{  0,1,\ldots,g-1\right\}  $. Then, $i\in\left\{  0,1,\ldots
,g-1\right\}  =\left\{  0,1,\ldots,m\ell\left(  u\right)  -1\right\}  $ (since
$g=m\ell\left(  u\right)  $). Hence, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.uiwm} (applied to $w=u$) yields $\left(
c^{i}\cdot u\right)  _{1}=\left(  u^{m}\right)  _{i+1}$. This proves
(\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.umi+1}).]

Furthermore, each $i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfies%
\begin{equation}
\left(  v^{n}\right)  _{i+1}=\left(  c^{i}\cdot v\right)  _{1}.
\label{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.vni+1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.vni+1}):} The
proof of (\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.vni+1}) proceeds in the
same way as the proof of (\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.umi+1}),
but using $v$ and $n$ instead of $u$ and $m$.]

\begin{vershort}
Using the equalities (\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.umi+1}) and
(\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.vni+1}), we can rewrite Claim 4
as follows:
\end{vershort}

\begin{verlong}
Now, Claim 4 can be rewritten as follows:
\end{verlong}

\begin{statement}
\textit{Claim 5:} The smallest $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot
v\right)  _{1}$.
\end{statement}

\begin{verlong}
(Indeed, Claim 5 is precisely Claim 4, rewritten using the equalities
(\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.umi+1}) and
(\ref{pf.lem.sol.lyndon.gr.aper-rel.ciuciv.b.vni+1}).)
\end{verlong}

But Claim 5 is precisely the statement of Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(b). Thus, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(b) is proven.

(c) Assume that $v\leq_{\omega}u$. Thus, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(b) (applied to $v$ and $u$ instead of
$u$ and $v$) yields the following:

\begin{statement}
\textit{Claim 6:} The smallest $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot v\right)  _{1}\neq\left(  c^{i}\cdot u\right)
_{1}$ satisfies $\left(  c^{i}\cdot v\right)  _{1}<\left(  c^{i}\cdot
u\right)  _{1}$.
\end{statement}

In other words, we have the following:

\begin{statement}
\textit{Claim 7:} The smallest $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ satisfies $\left(  c^{i}\cdot u\right)  _{1}>\left(  c^{i}\cdot
v\right)  _{1}$.
\end{statement}

(Indeed, Claim 7 is a restatement of Claim 6, since the statement
``$\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot
v\right)  _{1}$'' is equivalent to ``$\left(
c^{i}\cdot v\right)  _{1}\neq\left(  c^{i}\cdot u\right)  _{1}$%
'', whereas the statement ``$\left(  c^{i}\cdot
u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$'' is
equivalent to ``$\left(  c^{i}\cdot v\right)  _{1}<\left(
c^{i}\cdot u\right)  _{1}$''.)

But Claim 7 is precisely the statement of Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(c). Thus, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(c) is proven.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.gr.aper-rel.apbq}Let $a,b\in\mathfrak{A}$ be two
letters. Let $p,q\in\mathfrak{A}^{\ast}$ be two words such that $\ell\left(
p\right)  =\ell\left(  q\right)  $. Then, we have $ap\leq bq$ if and only if
either $a<b$ or $\left(  a=b\text{ and }pa\leq qb\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.aper-rel.apbq}.]Recall that
$a\in\mathfrak{A}$; hence, $a$ is identified with the one-letter word $\left(
a\right)  $. Thus, $\ell\left(  a\right)  =1$. Similarly, $\ell\left(
b\right)  =1$. Hence, $\ell\left(  a\right)  =1=\ell\left(  b\right)  $, so
that $\ell\left(  a\right)  \leq\ell\left(  b\right)  $.

We shall next prove the following:

\begin{statement}
\textit{Claim 1:} If $ap\leq bq$, then we have either $a<b$ or $\left(
a=b\text{ and }pa\leq qb\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} Assume that $ap\leq bq$. We must prove that we
have either $a<b$ or $\left(  a=b\text{ and }pa\leq qb\right)  $. If $a<b$,
then this is obviously true. Thus, for the rest of this proof of Claim 1, we
can WLOG assume that we don't have $a<b$. Assume this.

We don't have $a<b$. Hence, we have $a\geq b$ (since the order on
$\mathfrak{A}$ is a total order).

But Proposition \ref{prop.words.lex}(f) (applied to $p$, $b$ and $q$ instead
of $b$, $c$ and $d$) yields $a\leq b$ (since $ap\leq bq$ and $\ell\left(
a\right)  \leq\ell\left(  b\right)  $). Combining this with $a\geq b$, we
obtain $a=b$. Thus, $ap=bp$, so that $bp=ap\leq bq$. Hence, Proposition
\ref{prop.words.lex}(c) (applied to $b$, $p$ and $q$ instead of $a$, $c$ and
$d$) yields $p\leq q$. Therefore, Proposition \ref{prop.words.lex}(j) (applied
to $p$, $q$ and $b$ instead of $a$, $b$ and $c$) yields $pb\leq qb$ (since
$\ell\left(  p\right)  \geq\ell\left(  q\right)  $ (because $\ell\left(
p\right)  =\ell\left(  q\right)  $)). But $a=b$, so that $pa=pb\leq qb$.
Hence, we have shown that $a=b$ and $pa\leq qb$. Thus, we have either $a<b$ or
$\left(  a=b\text{ and }pa\leq qb\right)  $. This proves Claim 1.]

\begin{statement}
\textit{Claim 2:} If $a<b$, then $ap\leq bq$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 2:} Assume that $a<b$. If $a$ was a prefix of $b$,
then we would have $a=b$ (since $a$ and $b$ are one-letter words), which would
contradict $a<b$. Thus, $a$ is not a prefix of $b$.

But $a\leq b$ (since $a<b$). Hence, Proposition \ref{prop.words.lex}(d)
(applied to $p$, $b$ and $q$ instead of $b$, $c$ and $d$) yields that either
we have $ap\leq bq$ or the word $a$ is a prefix of $b$. Since the word $a$ is
not a prefix of $b$, we thus conclude that $ap\leq bq$. This proves Claim 2.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 2:} Assume that $a<b$. Thus, $a\leq b$. Hence,
Proposition \ref{prop.words.lex}(d) (applied to $p$, $b$ and $q$ instead of
$b$, $c$ and $d$) yields that either we have $ap\leq bq$ or the word $a$ is a
prefix of $b$. Since the word $a$ is not a prefix of $b$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $a$ is a prefix
of $b$. Since $a$ has the same length as $b$ (because $\ell\left(  a\right)
=\ell\left(  b\right)  $), we thus conclude that $a=b$. But this contradicts
$a<b$. This contradiction shows that our assumption was wrong, qed.}, we thus
conclude that $ap\leq bq$. This proves Claim 2.]
\end{verlong}

\begin{statement}
\textit{Claim 3:} If $a=b$ and $pa\leq qb$, then $ap\leq bq$.
\end{statement}

[\textit{Proof of Claim 3:} Assume that $a=b$ and $pa\leq qb$. We must prove
that $ap\leq bq$.

We have $\ell\left(  p\right)  \leq\ell\left(  q\right)  $ (since $\ell\left(
p\right)  =\ell\left(  q\right)  $) and $pa\leq qb$. Thus, Proposition
\ref{prop.words.lex}(f) (applied to $p$, $a$, $q$ and $b$ instead of $a$, $b$,
$c$ and $d$) yields $p\leq q$. Hence, Proposition \ref{prop.words.lex}(b)
(applied to $p$ and $q$ instead of $c$ and $d$) yields $ap\leq aq=bq$ (since
$a=b$). This proves Claim 3.]

\begin{statement}
\textit{Claim 4:} If we have either $a<b$ or $\left(  a=b\text{ and }pa\leq
qb\right)  $, then we have $ap\leq bq$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 4:} This follows by combining Claim 2 and Claim 3.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 4:} Assume that we have either $a<b$ or $\left(
a=b\text{ and }pa\leq qb\right)  $. We must prove that $ap\leq bq$.

If $a<b$, then this follows from Claim 2. Thus, for the rest of this proof of
Claim 4, we can WLOG assume that we don't have $a<b$. Assume this. Hence, we
have $\left(  a=b\text{ and }pa\leq qb\right)  $ (because we have either $a<b$
or $\left(  a=b\text{ and }pa\leq qb\right)  $, but we don't have $a<b$).
Therefore, Claim 3 yields $ap\leq bq$. This proves Claim 4.]
\end{verlong}

Combining Claim 1 with Claim 4, we conclude that we have $ap\leq bq$ if and
only if either $a<b$ or $\left(  a=b\text{ and }pa\leq qb\right)  $. (Indeed,
the ``only if'' part of this statement
follows from Claim 1, whereas the ``if'' part
follows from Claim 4.) This proves Lemma \ref{lem.sol.lyndon.gr.aper-rel.apbq}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.words.aper-rel.1}.](b) Assume that $u\neq v$.
Let $g=\ell\left(  u\right)  \cdot\ell\left(  v\right)  $. Then, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) yields that there exists some
$i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. This $i$ must satisfy
$i\in\left\{  0,1,\ldots,g-1\right\}  \subset \NN $ and $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Hence, there
exists some $i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$. This proves Proposition
\ref{prop.words.aper-rel.1}(b).

(d) From $n\ell\left(  u\right)  =m\ell\left(  v\right)  $, we obtain
$m\ell\left(  v\right)  =n\ell\left(  u\right)  $. Hence, Exercise
\ref{exe.words.un.vm} (applied to $n$, $m$, $u$ and $v$ instead of $m$, $n$,
$v$ and $u$) shows that $vu\geq uv$ holds if and only if $v^{n}\geq u^{m}$
holds. In other words, we have the logical equivalence $\left(  vu\geq
uv\right)  \ \Longleftrightarrow\ \left(  v^{m}\geq u^{n}\right)  $.

Now, we have the following chain of logical equivalences:%
\begin{align*}
\left(  u\leq_{\omega}v\right)  \  &  \Longleftrightarrow\ \left(  uv\leq
vu\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the relation
}\leq_{\omega}\right) \\
&  \Longleftrightarrow\ \left(  vu\geq uv\right)  \ \Longleftrightarrow
\ \left(  v^{m}\geq u^{n}\right)  \ \Longleftrightarrow\ \left(  u^{n}\leq
v^{m}\right)  .
\end{align*}
In other words, we have $u\leq_{\omega}v$ if and only if $u^{n}\leq v^{m}$.
This proves Proposition \ref{prop.words.aper-rel.1}(d).

(c) Our proof relies on the following five claims:

\begin{statement}
\textit{Claim 1:} If $u\leq_{\omega}v$, then the smallest $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ \textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$.
\end{statement}

[\textit{Proof of Claim 1:} Assume that $u\leq_{\omega}v$. We must prove that
the smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)
_{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ \textbf{either} does not exist
\textbf{or} satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot
v\right)  _{1}$.

\begin{vershort}
If $u=v$, then this is obvious\footnote{Indeed, in this case, the smallest
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$ does not exist, because $\left(  c^{i}%
\cdot\underbrace{u}_{=v}\right)  _{1}=\left(  c^{i}\cdot v\right)  _{1}$ for
each $i\in \NN $.}. Hence, for the rest of this proof of Claim 1, we WLOG
assume that $u\neq v$. Let $g=\ell\left(  u\right)  \cdot\ell\left(  v\right)
$.
\end{vershort}

\begin{verlong}
If $u=v$, then this is obvious\footnote{\textit{Proof.} Assume that $u=v$. We
must show that the smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ \textbf{either} does not
exist \textbf{or} satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(
c^{i}\cdot v\right)  _{1}$.
\par
Each $i\in \NN $ satisfies $\left(  c^{i}\cdot\underbrace{u}_{=v}\right)
_{1}=\left(  c^{i}\cdot v\right)  _{1}$. In other words, there exists no
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$. Hence, the smallest such $i$ does not exist. Thus,
the smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)
_{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ \textbf{either} does not exist
\textbf{or} satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot
v\right)  _{1}$ (because it does not exist). Qed.}. Hence, for the rest of
this proof of Claim 1, we can WLOG assume that $u\neq v$. Assume this. Let
$g=\ell\left(  u\right)  \cdot\ell\left(  v\right)  $.
\end{verlong}

\begin{vershort}
Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) yields that there exists some
$i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Let $j$ denote the
\textbf{smallest} such $i$. Thus, $j\in\left\{  0,1,\ldots,g-1\right\}  $ and
$\left(  c^{j}\cdot u\right)  _{1}\neq\left(  c^{j}\cdot v\right)  _{1}$. From
$j\in\left\{  0,1,\ldots,g-1\right\}  $, we obtain $j\leq g-1$; hence,
$g-1\geq j$.
\end{vershort}

\begin{verlong}
Now, Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(a) yields that there exists
some $i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Let $j$ denote the
\textbf{smallest} such $i$. Thus, $j$ is the smallest $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$. Hence, in particular, $j$ is an
$i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. In other words,
$j\in\left\{  0,1,\ldots,g-1\right\}  $ and $\left(  c^{j}\cdot u\right)
_{1}\neq\left(  c^{j}\cdot v\right)  _{1}$. From $j\in\left\{  0,1,\ldots
,g-1\right\}  $, we obtain $j\leq g-1$; hence, $g-1\geq j$.
\end{verlong}

Furthermore, Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciuciv}(b) yields that the
smallest $i\in\left\{  0,1,\ldots,g-1\right\}  $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies
$\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$. Since
we have denoted this smallest $i$ by $j$, we can restate this as follows: The
number $i=j$ satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot
v\right)  _{1}$. In other words, we have $\left(  c^{j}\cdot u\right)
_{1}<\left(  c^{j}\cdot v\right)  _{1}$.

We have defined $j$ to be the \textbf{smallest} $i\in\left\{  0,1,\ldots
,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$. Thus, any $i\in\left\{  0,1,\ldots,g-1\right\}  $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ is greater or equal to $j$. In other words, for any $i\in\left\{
0,1,\ldots,g-1\right\}  $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$, we have%
\begin{equation}
i\geq j. \label{pf.prop.words.aper-rel.1.c.c1.1}%
\end{equation}


\begin{vershort}
Now, recall that $j\in\left\{  0,1,\ldots,g-1\right\}  \subset \NN $ and
$\left(  c^{j}\cdot u\right)  _{1}\neq\left(  c^{j}\cdot v\right)  _{1}$. In
other words, $j$ is an $i\in \NN $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Moreover, $j$ is the
\textbf{smallest} such $i$\ \ \ \ \footnote{\textit{Proof.} We shall show that
if $i\in \NN $ satisfies $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$, then $i\geq j$.
\par
Indeed, let $i\in \NN $ satisfy $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$. We must prove that $i\geq j$. If
$i\geq g-1$, then this is obvious (because in this case, we have $i\geq
g-1\geq j$). Hence, for the rest of this proof, we WLOG assume that $i<g-1$.
Hence, $i\leq g-1$ and therefore $i\in\left\{  0,1,\ldots,g-1\right\}  $.
Hence, (\ref{pf.prop.words.aper-rel.1.c.c1.1}) yields $i\geq j$.
\par
Now, forget that we fixed $i$. We thus have shown that if $i\in \NN $
satisfies $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$, then $i\geq j$. In other words, any $i\in \NN $ satisfying
$\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ must
satisfy $i\geq j$. Therefore, $j$ is the \textbf{smallest} $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ (because we already know that $j$ is such an $i$).}.
\end{vershort}

\begin{verlong}
Now, recall that $j\in\left\{  0,1,\ldots,g-1\right\}  \subset \NN $ and
$\left(  c^{j}\cdot u\right)  _{1}\neq\left(  c^{j}\cdot v\right)  _{1}$. In
other words, $j$ is an $i\in \NN $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Moreover, $j$ is the
\textbf{smallest} such $i$\ \ \ \ \footnote{\textit{Proof.} We shall show that
if $i\in \NN $ satisfies $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$, then $i\geq j$.
\par
Indeed, let $i\in \NN $ satisfy $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$. We must prove that $i\geq j$. If
$i\geq g-1$, then this is obvious (because in this case, we have $i\geq
g-1\geq j$). Hence, for the rest of this proof, we WLOG assume that we don't
have $i\geq g-1$. Thus, $i<g-1$, so that $i\leq g-1$ and therefore
$i\in\left\{  0,1,\ldots,g-1\right\}  $. Hence,
(\ref{pf.prop.words.aper-rel.1.c.c1.1}) yields $i\geq j$.
\par
Now, forget that we fixed $i$. We thus have shown that if $i\in \NN $
satisfies $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$, then $i\geq j$. In other words, any $i\in \NN $ satisfying
$\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ must
satisfy $i\geq j$. Therefore, $j$ is the \textbf{smallest} $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ (because we already know that $j$ is such an $i$). Qed.}.
\end{verlong}

\begin{vershort}
So we know that $j$ is the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Thus, in
particular, this smallest $i$ exists. Moreover, this smallest $i$ satisfies
$\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$ (because
this smallest $i$ is $j$, but we know that $\left(  c^{j}\cdot u\right)
_{1}<\left(  c^{j}\cdot v\right)  _{1}$). Thus, we have shown that the
smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$. This proves Claim 1.]
\end{vershort}

\begin{verlong}
So we know that $j$ is the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$. Thus, in
particular, this smallest $i$ exists. Moreover, this smallest $i$ satisfies
$\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$ (because
this smallest $i$ is $j$, but we know that $\left(  c^{j}\cdot u\right)
_{1}<\left(  c^{j}\cdot v\right)  _{1}$). Thus, we have shown that the
smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}%
\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$. Therefore, the smallest
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$ \textbf{either} does not exist \textbf{or}
satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot v\right)
_{1}$. This proves Claim 1.]
\end{verlong}

\begin{statement}
\textit{Claim 2:} If $v\leq_{\omega}u$, then the smallest $i\in \NN $
satisfying $\left(  c^{i}\cdot v\right)  _{1}\neq\left(  c^{i}\cdot u\right)
_{1}$ \textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
v\right)  _{1}<\left(  c^{i}\cdot u\right)  _{1}$.
\end{statement}

[\textit{Proof of Claim 2:} Claim 2 is just Claim 1 with the roles of $u$ and
$v$ swapped.]

\begin{statement}
\textit{Claim 3:} If $v\leq_{\omega}u$, then the smallest $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ \textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$.
\end{statement}

[\textit{Proof of Claim 3:} Claim 3 is just a restatement of Claim 2, since
the statement ``$\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$'' is equivalent to
``$\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot
v\right)  _{1}$'' and since the statement
``$\left(  c^{i}\cdot u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}%
$'' is equivalent to ``$\left(  c^{i}\cdot
v\right)  _{1}<\left(  c^{i}\cdot u\right)  _{1}$''.]

\begin{statement}
\textit{Claim 4:} If we don't have $u\leq_{\omega}v$, then the smallest
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$ exists but does \textbf{not} satisfy $\left(
c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$.
\end{statement}

[\textit{Proof of Claim 4:} Assume that we don't have $u\leq_{\omega}v$. We
must prove that the smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ exists but does
\textbf{not} satisfy $\left(  c^{i}\cdot u\right)  _{1}<\left(  c^{i}\cdot
v\right)  _{1}$.

\begin{vershort}
Proposition \ref{prop.words.aper-rel.tord} shows that the relation
$\leq_{\omega}$ on the set $\mathfrak{A}^{\mathfrak{a}}$ is the
smaller-or-equal relation of a total order. Hence, if we had $u=v$, then we
would have $u\leq_{\omega}v$, which would contradict our assumption that we
don't have $u\leq_{\omega}v$. Hence, we must have $u\neq v$.

Thus, Proposition \ref{prop.words.aper-rel.1}(b) shows that there exists some
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$. Hence, the \textbf{smallest} such $i$ exists.

Recall that the relation $\leq_{\omega}$ on the set $\mathfrak{A}%
^{\mathfrak{a}}$ is the smaller-or-equal relation of a total order. Hence, we
must have $v\leq_{\omega}u$ (since we don't have $u\leq_{\omega}v$). Thus,
Claim 3 yields that the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$
\textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$. Since we know that this
smallest $i$ exists (indeed, we have proved this in the previous paragraph),
we thus conclude that the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies
$\left(  c^{i}\cdot u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$. Hence,
this smallest $i$ does \textbf{not} satisfy $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$ (because $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$ would contradict $\left(  c^{i}\cdot
u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$).
\end{vershort}

\begin{verlong}
Proposition \ref{prop.words.aper-rel.tord} shows that the relation
$\leq_{\omega}$ on the set $\mathfrak{A}^{\mathfrak{a}}$ is the
smaller-or-equal relation of a total order. Hence, in particular, this
relation $\leq_{\omega}$ is total and reflexive. If we had $u=v$, then we
would have $u\leq_{\omega}v$ (since the relation $\leq_{\omega}$ is
reflexive), which would contradict the fact that we don't have $u\leq_{\omega
}v$. Thus, we cannot have $u=v$. In other words, we have $u\neq v$.

Thus, Proposition \ref{prop.words.aper-rel.1}(b) shows that there exists some
$i\in \NN $ satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(
c^{i}\cdot v\right)  _{1}$. Hence, the \textbf{smallest} such $i$ exists.

The relation $\leq_{\omega}$ is total. Hence, we have $u\leq_{\omega}v$ or
$v\leq_{\omega}u$. Thus, we must have $v\leq_{\omega}u$ (since we don't have
$u\leq_{\omega}v$). Hence, Claim 3 yields that the smallest $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ \textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$. Since we know that this
smallest $i$ exists (indeed, we have proved this in the previous paragraph),
we thus conclude that the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ satisfies
$\left(  c^{i}\cdot u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$. Hence,
this smallest $i$ does \textbf{not} satisfy $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$ (because $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$ would contradict $\left(  c^{i}\cdot
u\right)  _{1}>\left(  c^{i}\cdot v\right)  _{1}$).
\end{verlong}

Altogether, we thus have shown that the smallest $i\in \NN $ satisfying
$\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$
exists but does \textbf{not} satisfy $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$. This proves Claim 4.]

\begin{statement}
\textit{Claim 5:} If the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$
\textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$, then we have $u\leq
_{\omega}v$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 5:} Claim 5 is just the contrapositive of Claim 4, and
thus holds (since Claim 4 holds).]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 5:} Assume that the smallest $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ \textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$. If we did not have
$u\leq_{\omega}v$, then Claim 4 would yield that the smallest $i\in \NN $
satisfying $\left(  c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)
_{1}$ exists but does \textbf{not} satisfy $\left(  c^{i}\cdot u\right)
_{1}<\left(  c^{i}\cdot v\right)  _{1}$; but this would contradict our
assumption (that the smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot
u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$ \textbf{either} does not
exist \textbf{or} satisfies $\left(  c^{i}\cdot u\right)  _{1}<\left(
c^{i}\cdot v\right)  _{1}$). Thus, we must have $u\leq_{\omega}v$. This proves
Claim 5.]
\end{verlong}

\begin{vershort}
Now, Proposition \ref{prop.words.aper-rel.1}(c) follows by combining Claim 1
with Claim 5. (Indeed, Claim 1 is the ``only if''
part of Proposition \ref{prop.words.aper-rel.1}(c),
whereas Claim 5 is the ``if'' part.)
\end{vershort}

\begin{verlong}
Now, combining Claim 1 with Claim 5, we conclude that we have $u\leq_{\omega
}v$ if and only if the smallest $i\in \NN $ satisfying $\left(
c^{i}\cdot u\right)  _{1}\neq\left(  c^{i}\cdot v\right)  _{1}$
\textbf{either} does not exist \textbf{or} satisfies $\left(  c^{i}\cdot
u\right)  _{1}<\left(  c^{i}\cdot v\right)  _{1}$. (Indeed, Claim 1 yields the
``only if'' part of this statement, whereas
Claim 5 yields its ``if'' part.) This proves
Proposition \ref{prop.words.aper-rel.1}(c).
\end{verlong}

(a) The words $u$ and $v$ are aperiodic and thus nonempty (since any aperiodic
word is nonempty).

Let $n=\ell\left(  v\right)  $ and $m=\ell\left(  u\right)  $. Then,
$n=\ell\left(  v\right)  >0$ (since $v$ is nonempty), so that $n$ is a
positive integer. Likewise, $m$ is a positive integer. Hence, $m>0$.

Let $g=nm$. Then, $g$ is a positive integer (since $n$ and $m$ are positive
integers). Hence, $g\geq1$, so that $1\in\left\{  1,2,\ldots,g\right\}  $.

Moreover, combining the equalities $n\underbrace{\ell\left(  u\right)  }%
_{=m}=nm=g$ and $m\underbrace{\ell\left(  v\right)  }_{=n}=mn=nm=g$, we obtain
$g=n\ell\left(  u\right)  =m\ell\left(  v\right)  $. Hence, Proposition
\ref{prop.words.aper-rel.1}(d) shows that we have $u\leq_{\omega}v$ if and
only if $u^{n}\leq v^{m}$. In other words, we have the logical equivalence%
\begin{equation}
\left(  u\leq_{\omega}v\right)  \ \Longleftrightarrow\ \left(  u^{n}\leq
v^{m}\right)  . \label{pf.prop.words.aper-rel.1.a.e1}%
\end{equation}


The word $u$ is aperiodic. Thus, Corollary \ref{cor.words.aper-neck-cw}
(applied to $w=u$) yields that the word $c\cdot u$ is aperiodic. Likewise, the
word $c\cdot v$ is aperiodic. Moreover, $u\in\mathfrak{A}^{m}$ (since
$\ell\left(  u\right)  =m$), so that $c\cdot u\in\mathfrak{A}^{m}$ and thus
$\ell\left(  c\cdot u\right)  =m$. Likewise, $\ell\left(  c\cdot v\right)  =n$
(since $\ell\left(  v\right)  =n$). Furthermore, comparing $n\underbrace{\ell
\left(  c\cdot u\right)  }_{=m}=nm$ with $m\underbrace{\ell\left(  c\cdot
v\right)  }_{=n}=mn=nm$, we obtain $n\ell\left(  c\cdot u\right)
=m\ell\left(  c\cdot v\right)  $. Hence, Proposition
\ref{prop.words.aper-rel.1}(d) (applied to $c\cdot u$ and $c\cdot v$ instead
of $u$ and $v$) shows that we have $c\cdot u\leq_{\omega}c\cdot v$ if and only
if $\left(  c\cdot u\right)  ^{n}\leq\left(  c\cdot v\right)  ^{m}$. In other
words, we have the logical equivalence%
\begin{equation}
\left(  c\cdot u\leq_{\omega}c\cdot v\right)  \ \Longleftrightarrow\ \left(
\left(  c\cdot u\right)  ^{n}\leq\left(  c\cdot v\right)  ^{m}\right)  .
\label{pf.prop.words.aper-rel.1.a.e1c}%
\end{equation}


Lemma \ref{lem.sol.lyndon.gr.aper-rel.ciwm} (applied to $v$ and $1$ instead of
$u$ and $i$) yields $\left(  c^{1}\cdot v\right)  ^{m}=c^{1}\cdot\left(
v^{m}\right)  $. In view of $c^{1}=c$, this rewrites as $\left(  c\cdot
v\right)  ^{m}=c\cdot\left(  v^{m}\right)  $. The same argument (with $v$ and
$m$ replaced by $u$ and $n$) yields $\left(  c\cdot u\right)  ^{n}%
=c\cdot\left(  u^{n}\right)  $.

The word $u^{n}$ has length $\ell\left(  u^{n}\right)  =n\ell\left(  u\right)
=g$, and thus can be written as $u^{n}=\left(  u^{n}\right)  _{1}\left(
u^{n}\right)  _{2}\cdots\left(  u^{n}\right)  _{g}$. Define a word
$p\in\mathfrak{A}^{\ast}$ by $p=\left(  u^{n}\right)  _{2}\left(
u^{n}\right)  _{3}\cdots\left(  u^{n}\right)  _{g}$. Since $g\geq1$, we thus
have $p\in\mathfrak{A}^{g-1}$ (since $\left(  u^{n}\right)  _{2},\left(
u^{n}\right)  _{3},\ldots,\left(  u^{n}\right)  _{g}$ are single letters), so
that $\ell\left(  p\right)  =g-1$. Also, define a letter $a\in\mathfrak{A}$ by
$a=\left(  u^{n}\right)  _{1}$. (This is well-defined, since $1\in\left\{
1,2,\ldots,g\right\}  $.) Thus,%
\begin{align}
\underbrace{a}_{=\left(  u^{n}\right)  _{1}}\underbrace{p}_{=\left(
u^{n}\right)  _{2}\left(  u^{n}\right)  _{3}\cdots\left(  u^{n}\right)  _{g}}
&  =\left(  u^{n}\right)  _{1}\left(  \left(  u^{n}\right)  _{2}\left(
u^{n}\right)  _{3}\cdots\left(  u^{n}\right)  _{g}\right)  =\left(
u^{n}\right)  _{1}\left(  u^{n}\right)  _{2}\cdots\left(  u^{n}\right)
_{g}\nonumber\\
&  =u^{n}. \label{pf.prop.words.aper-rel.1.a.ap}%
\end{align}
Lemma \ref{lem.sol.words.necklaces.f1} (applied to $g$, $a$ and $p$ instead of
$m$, $u$ and $v$) yields $c\left(  ap\right)  =pa$. In view of $ap=u^{n}$,
this rewrites as $c\left(  u^{n}\right)  =pa$. Hence,%
\begin{equation}
\left(  c\cdot u\right)  ^{n}=c\cdot\left(  u^{n}\right)  =c\left(
u^{n}\right)  =pa. \label{pf.prop.words.aper-rel.1.a.cun}%
\end{equation}


The word $v^{m}$ has length $\ell\left(  v^{m}\right)  =m\ell\left(  v\right)
=g$, and thus can be written as $v^{m}=\left(  v^{m}\right)  _{1}\left(
v^{m}\right)  _{2}\cdots\left(  v^{m}\right)  _{g}$. Define a word
$q\in\mathfrak{A}^{\ast}$ by $q=\left(  v^{m}\right)  _{2}\left(
v^{m}\right)  _{3}\cdots\left(  v^{m}\right)  _{g}$. Since $g\geq1$, we thus
have $q\in\mathfrak{A}^{g-1}$ (since $\left(  v^{m}\right)  _{2},\left(
v^{m}\right)  _{3},\ldots,\left(  v^{m}\right)  _{g}$ are single letters), so
that $\ell\left(  q\right)  =g-1$. Also, define a letter $b\in\mathfrak{A}$ by
$b=\left(  v^{m}\right)  _{1}$. (This is well-defined, since $1\in\left\{
1,2,\ldots,g\right\}  $.) Thus,%
\begin{align}
\underbrace{b}_{=\left(  v^{m}\right)  _{1}}\underbrace{q}_{=\left(
v^{m}\right)  _{2}\left(  v^{m}\right)  _{3}\cdots\left(  v^{m}\right)  _{g}}
&  =\left(  v^{m}\right)  _{1}\left(  \left(  v^{m}\right)  _{2}\left(
v^{m}\right)  _{3}\cdots\left(  v^{m}\right)  _{g}\right)  =\left(
v^{m}\right)  _{1}\left(  v^{m}\right)  _{2}\cdots\left(  v^{m}\right)
_{g}\nonumber\\
&  =v^{m}. \label{pf.prop.words.aper-rel.1.a.bq}%
\end{align}
Lemma \ref{lem.sol.words.necklaces.f1} (applied to $g$, $b$ and $q$ instead of
$m$, $u$ and $v$) yields $c\left(  bq\right)  =qb$. In view of $bq=v^{m}$,
this rewrites as $c\left(  v^{m}\right)  =qb$. Hence,%
\begin{equation}
\left(  c\cdot v\right)  ^{m}=c\cdot\left(  v^{m}\right)  =c\left(
v^{m}\right)  =qb. \label{pf.prop.words.aper-rel.1.a.cvm}%
\end{equation}


Comparing $\ell\left(  p\right)  =g-1$ with $\ell\left(  q\right)  =g-1$, we
obtain $\ell\left(  p\right)  =\ell\left(  q\right)  $. Thus, Lemma
\ref{lem.sol.lyndon.gr.aper-rel.apbq} yields that we have $ap\leq bq$ if and
only if either $a<b$ or $\left(  a=b\text{ and }pa\leq qb\right)  $. In other
words, we have the following equivalence:%
\begin{equation}
\left(  ap\leq bq\right)  \ \Longleftrightarrow\ \left(  \text{either
}a<b\text{ or }\left(  a=b\text{ and }pa\leq qb\right)  \right)  .
\label{pf.prop.words.aper-rel.1.a.e2}%
\end{equation}


\begin{vershort}
The word $u^{n}$ starts with the word $u$ (since $n$ is a positive integer),
and thus has the same first letter as the word $u$ (since the word $u$ is
nonempty). In other words, $\left(  u^{n}\right)  _{1}=u_{1}$. Hence,
$a=\left(  u^{n}\right)  _{1}=u_{1}$. The same argument (applied to $v$, $m$
and $b$ instead of $u$, $n$ and $a$) yields $b=v_{1}$.
\end{vershort}

\begin{verlong}
The word $u^{n}$ starts with the word $u$ (since $n$ is a positive integer).
Thus, the first $\ell\left(  u\right)  $ letters of the word $u^{n}$ are the
letters of $u$ (in the order in which they appear in $u$). In other words, we
have
\[
\left(  u^{n}\right)  _{i}=u_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  .
\]
Applying this to $i=1$, we obtain $\left(  u^{n}\right)  _{1}=u_{1}$ (since
$1\in\left\{  1,2,\ldots,\ell\left(  u\right)  \right\}  $ (because
$\ell\left(  u\right)  =m>0$)). Hence, $a=\left(  u^{n}\right)  _{1}=u_{1}$.
The same argument (applied to $v$, $m$ and $b$ instead of $u$, $n$ and $a$)
yields $b=v_{1}$.
\end{verlong}

Now, we have the following chain of logical equivalences:%
\begin{align*}
\left(  u\leq_{\omega}v\right)  \  &  \Longleftrightarrow\ \left(
\underbrace{u^{n}}_{\substack{=ap\\\text{(by
(\ref{pf.prop.words.aper-rel.1.a.ap}))}}}\leq\underbrace{v^{m}}%
_{\substack{=bq\\\text{(by (\ref{pf.prop.words.aper-rel.1.a.bq}))}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.words.aper-rel.1.a.e1}%
)}\right) \\
&  \Longleftrightarrow\ \left(  ap\leq bq\right) \\
&  \Longleftrightarrow\ \left(  \text{either }\underbrace{a}_{=u_{1}%
}<\underbrace{b}_{=v_{1}}\text{ or }\left(  \underbrace{a}_{=u_{1}%
}=\underbrace{b}_{=v_{1}}\text{ and }\underbrace{pa}_{\substack{=\left(
c\cdot u\right)  ^{n}\\\text{(by (\ref{pf.prop.words.aper-rel.1.a.cun}))}%
}}\leq\underbrace{qb}_{\substack{=\left(  c\cdot v\right)  ^{m}\\\text{(by
(\ref{pf.prop.words.aper-rel.1.a.cvm}))}}}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.words.aper-rel.1.a.e2}%
)}\right) \\
&  \Longleftrightarrow\ \left(  \text{either }u_{1}<v_{1}\text{ or }\left(
u_{1}=v_{1}\text{ and }\underbrace{\left(  c\cdot u\right)  ^{n}\leq\left(
c\cdot v\right)  ^{m}}_{\substack{\Longleftrightarrow\ \left(  c\cdot
u\leq_{\omega}c\cdot v\right)  \\\text{(by
(\ref{pf.prop.words.aper-rel.1.a.e1c}))}}}\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{either }u_{1}<v_{1}\text{ or }\left(
u_{1}=v_{1}\text{ and }c\cdot u\leq_{\omega}c\cdot v\right)  \right)  .
\end{align*}
In other words, we have $u\leq_{\omega}v$ if and only if either $u_{1}<v_{1}$
or $\left(  u_{1}=v_{1}\text{ and }c\cdot u\leq_{\omega}c\cdot v\right)  $.
This proves Proposition \ref{prop.words.aper-rel.1}(a).
\end{proof}

Finally, let us prove Proposition \ref{prop.words.read-std3}:

\begin{proof}
[Proof of Proposition \ref{prop.words.read-std3}.]We have
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)$
(since $w\in\mathfrak{A}^{n}$).

(a) Let $h\in\left\{
1,2,\ldots,n\right\}  $. Let $z$ be the cycle of $\tau$ that contains $h$.
Thus, $h\in z$. Hence, Proposition \ref{prop.words.read-std2}(a) yields that
$\left[  w_{\tau,h}\right]  =\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $.
Moreover, Proposition \ref{prop.words.read-std2}(d) shows that the set
$\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $ is an aperiodic necklace. In
other words, the set $\left[  w_{\tau,h}\right]  $ is an aperiodic necklace
(since $\left[  w_{\tau,h}\right]  =\left\{  w_{\tau,i}\ \mid\ i\in z\right\}
$).

But Proposition \ref{prop.words.read1}(a) shows that the word $w_{\tau,h}$ is
nonempty and has length $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)
$. Hence, Proposition \ref{prop.words.aper-neck} (applied to $w_{\tau,h}$
instead of $w$) shows that the word $w_{\tau,h}$ is aperiodic if and only if
the necklace $\left[  w_{\tau,h}\right]  $ is aperiodic. Hence, the word
$w_{\tau,h}$ is aperiodic (since the necklace $\left[  w_{\tau,h}\right]  $ is aperiodic).

Forget that we fixed $h$. We thus have proved that the word $w_{\tau,h}$ is
aperiodic for each $h\in\left\{  1,2,\ldots,n\right\}  $. In other words, the
words $w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}$ are aperiodic. This proves
Proposition \ref{prop.words.read-std3}(a).

(b) Proposition \ref{prop.words.read-std3}(a) shows that the words $w_{\tau
,1},w_{\tau,2},\ldots,w_{\tau,n}$ are aperiodic. Thus, these words belong to
$\mathfrak{A}^{\mathfrak{a}}$ (since $\mathfrak{A}^{\mathfrak{a}}$ is the set
of all aperiodic words). Hence, the chain of inequalities $w_{\tau,1}%
\leq_{\omega}w_{\tau,2}\leq_{\omega}\cdots\leq_{\omega}w_{\tau,n}$ makes
sense. It remains to prove that it holds.

Let $\alpha$ and $\beta$ be two elements of $\left\{  1,2,\ldots,n\right\}  $
such that $\alpha<\beta$. We shall show that $w_{\tau,\alpha}\leq_{\omega
}w_{\tau,\beta}$.

Indeed, assume the contrary. Thus, it is \textbf{not} true that $w_{\tau
,\alpha}\leq_{\omega}w_{\tau,\beta}$.

Let us notice that each $i\in \NN $ satisfies
\begin{equation}
\left(  c^{i}\cdot w_{\tau,\alpha}\right)  _{1}=w_{\tau^{i+1}\left(
\alpha\right)  } \label{pf.prop.words.read-std3.1a}%
\end{equation}
and
\begin{equation}
\left(  c^{i}\cdot w_{\tau,\beta}\right)  _{1}=w_{\tau^{i+1}\left(
\beta\right)  } \label{pf.prop.words.read-std3.1b}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.words.read-std3.1a}) and
(\ref{pf.prop.words.read-std3.1b}):} Let $i\in \NN $. Then, Proposition
\ref{prop.words.read1}(b) (applied to $h=\tau^{i}\left(  \alpha\right)  $)
yields that the first letter of the word $w_{\tau,\tau^{i}\left(
\alpha\right)  }$ is $w_{\tau\left(  \tau^{i}\left(  \alpha\right)  \right)
}$.

\begin{vershort}
But Proposition \ref{prop.words.read1}(e) (applied to $h=\alpha$) yields
$w_{\tau,\tau^{i}\left(  \alpha\right)  }=c^{i}\cdot w_{\tau,\alpha}$. Hence,
$\left(  w_{\tau,\tau^{i}\left(  \alpha\right)  }\right)  _{1}=\left(
c^{i}\cdot w_{\tau,\alpha}\right)  _{1}$. Thus,%
\begin{align*}
\left(  c^{i}\cdot w_{\tau,\alpha}\right)  _{1}  &  =\left(  w_{\tau,\tau
^{i}\left(  \alpha\right)  }\right)  _{1}=\left(  \text{the first letter of
the word }w_{\tau,\tau^{i}\left(  \alpha\right)  }\right)  =w_{\tau\left(
\tau^{i}\left(  \alpha\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the first letter of the word
}w_{\tau,\tau^{i}\left(  \alpha\right)  }\text{ is }w_{\tau\left(  \tau
^{i}\left(  \alpha\right)  \right)  }\right) \\
&  =w_{\tau^{i+1}\left(  \alpha\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau\left(  \tau^{i}\left(  \alpha\right)  \right)  =\tau
^{i+1}\left(  \alpha\right)  \right)  .
\end{align*}

\end{vershort}

\begin{verlong}
But Proposition \ref{prop.words.read1}(e) (applied to $h=\alpha$) yields
$w_{\tau,\tau^{i}\left(  \alpha\right)  }=c^{i}\cdot w_{\tau,\alpha}$. Hence,
$\left(  w_{\tau,\tau^{i}\left(  \alpha\right)  }\right)  _{1}=\left(
c^{i}\cdot w_{\tau,\alpha}\right)  _{1}$. Thus,%
\begin{align*}
\left(  c^{i}\cdot w_{\tau,\alpha}\right)  _{1}  &  =\left(  w_{\tau,\tau
^{i}\left(  \alpha\right)  }\right)  _{1}=\left(  \text{the first letter of
the word }w_{\tau,\tau^{i}\left(  \alpha\right)  }\right)  =w_{\tau\left(
\tau^{i}\left(  \alpha\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the first letter of the word
}w_{\tau,\tau^{i}\left(  \alpha\right)  }\text{ is }w_{\tau\left(  \tau
^{i}\left(  \alpha\right)  \right)  }\right) \\
&  =w_{\tau^{i+1}\left(  \alpha\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\tau\left(  \tau^{i}\left(  \alpha\right)  \right)
=\underbrace{\left(  \tau\circ\tau^{i}\right)  }_{=\tau^{i+1}}\left(
\alpha\right)  =\tau^{i+1}\left(  \alpha\right)  \right)  .
\end{align*}

\end{verlong}

\noindent This proves (\ref{pf.prop.words.read-std3.1a}). The same argument
(but applied to $\beta$ instead of $\alpha$) proves
(\ref{pf.prop.words.read-std3.1b}).]

Proposition \ref{prop.words.read-std3}(a) shows that the words $w_{\tau
,1},w_{\tau,2},\ldots,w_{\tau,n}$ are aperiodic. Hence, in particular, the
words $w_{\tau,\alpha}$ and $w_{\tau,\beta}$ are aperiodic. Thus, Proposition
\ref{prop.words.aper-rel.1}(c) (applied to $u=w_{\tau,\alpha}$ and
$v=w_{\tau,\beta}$) yields that we have $w_{\tau,\alpha}\leq_{\omega}%
w_{\tau,\beta}$ if and only if the smallest $i\in \NN $ satisfying
$\left(  c^{i}\cdot w_{\tau,\alpha}\right)  _{1}\neq\left(  c^{i}\cdot
w_{\tau,\beta}\right)  _{1}$ \textbf{either} does not exist \textbf{or}
satisfies $\left(  c^{i}\cdot w_{\tau,\alpha}\right)  _{1}<\left(  c^{i}\cdot
w_{\tau,\beta}\right)  _{1}$. Thus, it is \textbf{not} true that the smallest
$i\in \NN $ satisfying $\left(  c^{i}\cdot w_{\tau,\alpha}\right)
_{1}\neq\left(  c^{i}\cdot w_{\tau,\beta}\right)  _{1}$ \textbf{either} does
not exist \textbf{or} satisfies $\left(  c^{i}\cdot w_{\tau,\alpha}\right)
_{1}<\left(  c^{i}\cdot w_{\tau,\beta}\right)  _{1}$ (because it is
\textbf{not} true that $w_{\tau,\alpha}\leq_{\omega}w_{\tau,\beta}$). In other
words, the smallest $i\in \NN $ satisfying $\left(  c^{i}\cdot
w_{\tau,\alpha}\right)  _{1}\neq\left(  c^{i}\cdot w_{\tau,\beta}\right)
_{1}$ exists but does \textbf{not} satisfy $\left(  c^{i}\cdot w_{\tau,\alpha
}\right)  _{1}<\left(  c^{i}\cdot w_{\tau,\beta}\right)  _{1}$. In view of
(\ref{pf.prop.words.read-std3.1a}) and (\ref{pf.prop.words.read-std3.1b}), we
can rewrite this fact as follows: The smallest $i\in \NN $ satisfying
$w_{\tau^{i+1}\left(  \alpha\right)  }\neq w_{\tau^{i+1}\left(  \beta\right)
}$ exists but does \textbf{not} satisfy $w_{\tau^{i+1}\left(  \alpha\right)
}<w_{\tau^{i+1}\left(  \beta\right)  }$.

Consider this $i$, and denote it by $j$. Thus, $i=j$ does \textbf{not} satisfy
$w_{\tau^{i+1}\left(  \alpha\right)  }<w_{\tau^{i+1}\left(  \beta\right)  }$.
In other words, we do \textbf{not} have $w_{\tau^{j+1}\left(  \alpha\right)
}<w_{\tau^{j+1}\left(  \beta\right)  }$. In other words, we have
$w_{\tau^{j+1}\left(  \alpha\right)  }\geq w_{\tau^{j+1}\left(  \beta\right)
}$.

We have defined $j$ to be the smallest $i\in \NN $ satisfying
$w_{\tau^{i+1}\left(  \alpha\right)  }\neq w_{\tau^{i+1}\left(  \beta\right)
}$. Hence, $j$ is an $i\in \NN $ satisfying $w_{\tau^{i+1}\left(
\alpha\right)  }\neq w_{\tau^{i+1}\left(  \beta\right)  }$. In other words,
$j$ is an element of $\NN$ and satisfies $w_{\tau^{j+1}\left(
\alpha\right)  }\neq w_{\tau^{j+1}\left(  \beta\right)  }$. Combining
$w_{\tau^{j+1}\left(  \alpha\right)  }\geq w_{\tau^{j+1}\left(  \beta\right)
}$ with $w_{\tau^{j+1}\left(  \alpha\right)  }\neq w_{\tau^{j+1}\left(
\beta\right)  }$, we obtain%
\begin{equation}
w_{\tau^{j+1}\left(  \alpha\right)  }>w_{\tau^{j+1}\left(  \beta\right)  }.
\label{pf.prop.words.read-std3.3}%
\end{equation}


On the other hand, $j$ is the \textbf{smallest} $i\in \NN $ satisfying
$w_{\tau^{i+1}\left(  \alpha\right)  }\neq w_{\tau^{i+1}\left(  \beta\right)
}$. Thus, every such $i\in \NN $ is $\geq j$. In other words,%
\begin{equation}
\text{if }i\in \NN \text{ satisfies }w_{\tau^{i+1}\left(  \alpha\right)
}\neq w_{\tau^{i+1}\left(  \beta\right)  }\text{, then }i\geq j.
\label{pf.prop.words.read-std3.4}%
\end{equation}
Hence,%
\begin{equation}
\text{every }i\in\left\{  0,1,\ldots,j-1\right\}  \text{ satisfies }%
w_{\tau^{i+1}\left(  \alpha\right)  }=w_{\tau^{i+1}\left(  \beta\right)  }.
\label{pf.prop.words.read-std3.5}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.words.read-std3.5}):} Let $i\in\left\{
0,1,\ldots,j-1\right\}  $. Thus, $i\geq0$ and $i\leq j-1<j$. If we had
$w_{\tau^{i+1}\left(  \alpha\right)  }\neq w_{\tau^{i+1}\left(  \beta\right)
}$, then we would have $i\geq j$ (by (\ref{pf.prop.words.read-std3.4})), which
would contradict $i<j$. Thus, we cannot have $w_{\tau^{i+1}\left(
\alpha\right)  }\neq w_{\tau^{i+1}\left(  \beta\right)  }$. Hence, we must
have $w_{\tau^{i+1}\left(  \alpha\right)  }=w_{\tau^{i+1}\left(  \beta\right)
}$. This proves (\ref{pf.prop.words.read-std3.5}).]

Thus, Lemma \ref{lem.sol.lyndon.gr.basics.1}(h) yields $w_{\tau^{j+1}\left(
\alpha\right)  }\leq w_{\tau^{j+1}\left(  \beta\right)  }$. This contradicts
(\ref{pf.prop.words.read-std3.3}). This contradiction shows that our
assumption was wrong. Hence, $w_{\tau,\alpha}\leq_{\omega}w_{\tau,\beta}$ is proven.

Now, forget that we fixed $\alpha$ and $\beta$. We thus have shown that if
$\alpha$ and $\beta$ are two elements of $\left\{  1,2,\ldots,n\right\}  $
such that $\alpha<\beta$, then $w_{\tau,\alpha}\leq_{\omega}w_{\tau,\beta}$.
In other words, we have $w_{\tau,1}\leq_{\omega}w_{\tau,2}\leq_{\omega}%
\cdots\leq_{\omega}w_{\tau,n}$. This proves Proposition
\ref{prop.words.read-std3}(b).
\end{proof}

We have now proven Proposition \ref{prop.words.aper-neck}, Corollary
\ref{cor.words.aper-neck-cw}, Corollary \ref{cor.words.aper-neck-ms},
Proposition \ref{prop.words.aper-rel.tord}, Proposition
\ref{prop.words.aper-rel.1} and Proposition \ref{prop.words.read-std3}. Thus,
Exercise \ref{exe.lyndon.gr.aper-rel} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.lyndon.gr.bij}}

\textit{Solution to Exercise \ref{exe.lyndon.gr.bij}.} Let us first show a
simple lemma about multisets of necklaces:

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.mulset-neck}Let $M$ be a finite multiset of
necklaces. Let $M^{\prime}=\biguplus_{N\in M}N$. (We are here using the fact
that each necklace $N\in M$ is a set, thus a multiset.)

Let $u\in\mathfrak{A}^{\ast}$ be a nonempty word. Then,%
\[
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)  =\left(
\text{multiplicity of }\left[  u\right]  \text{ in }M\right)  .
\]

\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck}.]We note the following
simple fact:

\begin{statement}
\textit{Claim 1:} Let $N$ be a necklace. Then, the statement $u\in N$ is
equivalent to the statement $N=\left[  u\right]  $.
\end{statement}

[\textit{Proof of Claim 1:} The implication $\left(  u\in N\right)
\Longrightarrow\left(  N=\left[  u\right]  \right)  $ follows from Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts}. The converse implication $\left(
N=\left[  u\right]  \right)  \Longrightarrow\left(  u\in N\right)  $ follows
from Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0}. Combining these two
implications, we obtain the equivalence $\left(  u\in N\right)
\Longleftrightarrow\left(  N=\left[  u\right]  \right)  $. Thus, Claim 1 is proven.]

Let us write the finite multiset $M$ in the form $M=\left\{  N_{1}%
,N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$, where $N_{1}%
,N_{2},\ldots,N_{k}$ are necklaces. Thus,%
\begin{align}
&  \left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)
\nonumber\\
&  =\left(  \text{the number of all }i\in\left\{  1,2,\ldots,k\right\}  \text{
such that }N_{i}=\left[  u\right]  \right)
\label{pf.lem.sol.lyndon.gr.bij.mulset-neck.short.1}%
\end{align}
and%
\[
\biguplus_{N\in M}N=N_{1}\uplus N_{2}\uplus\cdots\uplus N_{k}%
\]
(where we are using the notation $M_{1}\uplus M_{2}\uplus\cdots\uplus M_{k}$
for a multiset union $\biguplus\limits_{s\in\left\{  1,2,\ldots,k\right\}
}M_{s}$). Hence,%
\[
M^{\prime}=\biguplus_{N\in M}N=N_{1}\uplus N_{2}\uplus\cdots\uplus N_{k},
\]
so that%
\begin{align*}
&  \left(  \text{multiplicity of }u\text{ in }M^{\prime}\right) \\
&  =\left(  \text{multiplicity of }u\text{ in }N_{1}\uplus N_{2}\uplus
\cdots\uplus N_{k}\right) \\
&  =\sum_{i=1}^{k}\underbrace{\left(  \text{multiplicity of }u\text{ in }%
N_{i}\right)  }_{\substack{=%
\begin{cases}
1, & \text{if }u\in N_{i};\\
0, & \text{if }u\notin N_{i}%
\end{cases}
\\\text{(since }N_{i}\text{ is a set (because any necklace is a set))}}%
}=\sum_{i=1}^{k}%
\begin{cases}
1, & \text{if }u\in N_{i};\\
0, & \text{if }u\notin N_{i}%
\end{cases}
\\
&  =\left(  \text{the number of all }i\in\left\{  1,2,\ldots,k\right\}  \text{
such that }u\in N_{i}\right) \\
&  =\left(  \text{the number of all }i\in\left\{  1,2,\ldots,k\right\}  \text{
such that }N_{i}=\left[  u\right]  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for any }i\in\left\{  1,2,\ldots,k\right\}  \text{, the
statement }u\in N_{i}\text{ is equivalent}\\
\text{to the statement }N_{i}=\left[  u\right]  \text{ (by Claim 1, applied to
}N=N_{i}\text{)}%
\end{array}
\right) \\
&  =\left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.gr.bij.mulset-neck.short.1})}\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck}.]The definition of
$\biguplus_{N\in M}N$ yields that for any object $x$, we have%
\begin{align*}
&  \left(  \text{multiplicity of }x\text{ in }\biguplus_{N\in M}N\right) \\
&  =\sum_{s\in\operatorname*{Supp}M}\left(  \text{multiplicity of }s\text{ in
}M\right)  \cdot\left(  \text{multiplicity of }x\text{ in }s\right)  .
\end{align*}
Applying this to $x=u$, we obtain%
\begin{align*}
&  \left(  \text{multiplicity of }u\text{ in }\biguplus_{N\in M}N\right) \\
&  =\sum_{s\in\operatorname*{Supp}M}\left(  \text{multiplicity of }s\text{ in
}M\right)  \cdot\left(  \text{multiplicity of }u\text{ in }s\right) \\
&  =\sum_{N\in\operatorname*{Supp}M}\left(  \text{multiplicity of }N\text{ in
}M\right)  \cdot\left(  \text{multiplicity of }u\text{ in }N\right)
\end{align*}
(here, we have renamed the index $s$ as $N$ in the sum). In view of
$M^{\prime}=\biguplus_{N\in M}N$, this rewrites as follows:%
\begin{align}
&  \left(  \text{multiplicity of }u\text{ in }M^{\prime}\right) \nonumber\\
&  =\sum_{N\in\operatorname*{Supp}M}\left(  \text{multiplicity of }N\text{ in
}M\right)  \cdot\left(  \text{multiplicity of }u\text{ in }N\right)  .
\label{pf.lem.sol.lyndon.gr.bij.mulset-neck.1}%
\end{align}


But $M$ is a finite multiset of necklaces. Hence, each element of $M$ is a necklace.

Now, we claim the following:

\begin{statement}
\textit{Claim 1:} Let $N\in\operatorname*{Supp}M$ be such that $N\neq\left[
u\right]  $. Then, $\left(  \text{multiplicity of }u\text{ in }N\right)  =0$.
\end{statement}

[\textit{Proof of Claim 1:} We have $N\in\operatorname*{Supp}M$. That is, $N$
is an element of $M$. Thus, $N$ is a necklace (since each element of $M$ is a necklace).

We must prove that $\left(  \text{multiplicity of }u\text{ in }N\right)  =0$.
Indeed, assume the contrary. Thus, $\left(  \text{multiplicity of }u\text{ in
}N\right)  \neq0$. In other words, $u\in N$ (since $N$ is a set). Thus, Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts} yields $N=\left[  u\right]  $. This
contradicts $N\neq\left[  u\right]  $. This proves Claim 1.]

\begin{statement}
\textit{Claim 2:} We have $\left(  \text{multiplicity of }u\text{ in }\left[
u\right]  \right)  =1$.
\end{statement}

[\textit{Proof of Claim 2:} Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields
$u\in\left[  u\right]  $. Hence, $\left(  \text{multiplicity of }u\text{ in
}\left[  u\right]  \right)  =1$ (since $\left[  u\right]  $ is a set). This
proves Claim 2.]

Now, (\ref{pf.lem.sol.lyndon.gr.bij.mulset-neck.1}) becomes%
\begin{align}
&  \left(  \text{multiplicity of }u\text{ in }M^{\prime}\right) \nonumber\\
&  =\sum_{N\in\operatorname*{Supp}M}\left(  \text{multiplicity of }N\text{ in
}M\right)  \cdot\left(  \text{multiplicity of }u\text{ in }N\right)
\nonumber\\
&  =\sum\limits_{\substack{N\in\operatorname*{Supp}M;\\N=\left[  u\right]
}}\left(  \text{multiplicity of }\underbrace{N}_{=\left[  u\right]  }\text{ in
}M\right)  \cdot\left(  \text{multiplicity of }u\text{ in }\underbrace{N}%
_{=\left[  u\right]  }\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{\substack{N\in\operatorname*{Supp}%
M;\\N\neq\left[  u\right]  }}\left(  \text{multiplicity of }N\text{ in
}M\right)  \cdot\underbrace{\left(  \text{multiplicity of }u\text{ in
}N\right)  }_{\substack{=0\\\text{(by Claim 1)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }N\in\operatorname*{Supp}M\text{ satisfies}\\
\text{either }N=\left[  u\right]  \text{ or }N\neq\left[  u\right]  \text{
(but not both)}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{N\in\operatorname*{Supp}M;\\N=\left[  u\right]
}}\left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)
\cdot\underbrace{\left(  \text{multiplicity of }u\text{ in }\left[  u\right]
\right)  }_{\substack{=1\\\text{(by Claim 2)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum\limits_{\substack{N\in
\operatorname*{Supp}M;\\N\neq\left[  u\right]  }}\left(  \text{multiplicity of
}N\text{ in }M\right)  \cdot0}_{=0}\nonumber\\
&  =\sum\limits_{\substack{N\in\operatorname*{Supp}M;\\N=\left[  u\right]
}}\left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)  .
\label{pf.lem.sol.lyndon.gr.bij.mulset-neck.3}%
\end{align}


Now, we are in one of the following two cases:

\textit{Case 1:} We have $\left[  u\right]  \in\operatorname*{Supp}M$.

\textit{Case 2:} We have $\left[  u\right]  \notin\operatorname*{Supp}M$.

Let us first consider Case 1. In this case, we have $\left[  u\right]
\in\operatorname*{Supp}M$. Thus, the sum $\sum\limits_{\substack{N\in
\operatorname*{Supp}M;\\N=\left[  u\right]  }}\left(  \text{multiplicity of
}\left[  u\right]  \text{ in }M\right)  $ has exactly one addend. Thus, this
sum simplifies to%
\[
\sum\limits_{\substack{N\in\operatorname*{Supp}M;\\N=\left[  u\right]
}}\left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)
=\left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)  .
\]
Hence, (\ref{pf.lem.sol.lyndon.gr.bij.mulset-neck.3}) becomes%
\[
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)  =\sum
\limits_{\substack{N\in\operatorname*{Supp}M;\\N=\left[  u\right]  }}\left(
\text{multiplicity of }\left[  u\right]  \text{ in }M\right)  =\left(
\text{multiplicity of }\left[  u\right]  \text{ in }M\right)  .
\]
This proves Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck} in Case 1.

Let us now consider Case 2. In this case, we have $\left[  u\right]
\notin\operatorname*{Supp}M$. Hence, there exists no $N\in\operatorname*{Supp}%
M$ satisfying $N=\left[  u\right]  $ (because any such $N$ would satisfy
$\left[  u\right]  =N\in\operatorname*{Supp}M$, which would contradict
$\left[  u\right]  \notin\operatorname*{Supp}M$).

But recall that $\left[  u\right]  \notin\operatorname*{Supp}M$. In other
words, $\left[  u\right]  $ is not an element of $M$ (since
$\operatorname*{Supp}M$ was defined to be the set of all elements of $M$). In
other words, $\left(  \text{multiplicity of }\left[  u\right]  \text{ in
}M\right)  =0$. Comparing this with%
\begin{align*}
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)   &
=\sum\limits_{\substack{N\in\operatorname*{Supp}M;\\N=\left[  u\right]
}}\left(  \text{multiplicity of }\left[  u\right]  \text{ in }M\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.gr.bij.mulset-neck.3})}\right) \\
&  =\left(  \text{empty sum}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since there exists no }N\in
\operatorname*{Supp}M\text{ satisfying }N=\left[  u\right]  \right) \\
&  =0,
\end{align*}
we obtain
\[
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)  =\left(
\text{multiplicity of }\left[  u\right]  \text{ in }M\right)  .
\]
This proves Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck} in Case 2.

We have now proved Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck} in both Cases
1 and 2. Hence, Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck} always holds.
\end{proof}

\end{verlong}

We can now prove one ``half'' of Theorem
\ref{thm.lyndon.gr.bij}:

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.RG-GR}We have $\operatorname*{RG}\circ
\operatorname*{GR}=\operatorname*{id}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.RG-GR}.]Let $w\in\mathfrak{A}%
^{\ast}$. Let $n=\ell\left(  w\right)  $; thus, $w\in\mathfrak{A}^{n}$ and
$w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $. Let $\tau$ be the permutation
$\left(  \operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$. Then, the
definition of the map $\operatorname*{GR}$ shows that%
\[
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}.
\]
Let us denote this multiset $\operatorname*{GR}w$ by $M$. Thus,
\[
M=\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a
cycle of }\tau\right\}  _{\operatorname*{multiset}}.
\]
Also, $M=\operatorname*{GR}w\in\mathfrak{MN}^{\mathfrak{a}}$ (as we have seen
in Definition \ref{def.lyndon.gr.GR}); in other words, $M$ is a finite
multiset of aperiodic necklaces (since $\mathfrak{MN}^{\mathfrak{a}}$ is the
set of all finite multisets of aperiodic necklaces).

Next, we shall show the following:

\begin{statement}
\textit{Claim 1:} Let $u$ be a nonempty word. Let $z$ be a cycle of $\tau$.
Then,%
\[
\left(  \text{the number of all }i\in z\text{ such that }u=w_{\tau,i}\right)
=\delta_{\left[  u\right]  ,\left[  w\right]  _{z}}.
\]

\end{statement}

[\textit{Proof of Claim 1:} The definition of the aperiodic necklace $\left[
w\right]  _{z}$ yields $\left[  w\right]  _{z}=\left\{  w_{\tau,i}\ \mid\ i\in
z\right\}  $. We are in one of the following two cases:

\textit{Case 1:} We have $\left[  u\right]  =\left[  w\right]  _{z}$.

\textit{Case 2:} We have $\left[  u\right]  \neq\left[  w\right]  _{z}$.

Let us first consider Case 1. In this case, we have $\left[  u\right]
=\left[  w\right]  _{z}$. Hence, $\delta_{\left[  u\right]  ,\left[  w\right]
_{z}}=1$.

\begin{vershort}
Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields $u\in\left[  u\right]
=\left[  w\right]  _{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $.
Proposition \ref{prop.words.read-std2}(b) shows that any two distinct elements
$\alpha$ and $\beta$ of $z$ satisfy $w_{\tau,\alpha}\neq w_{\tau,\beta}$. In
other words, the words $w_{\tau,i}$ for all $i\in z$ are distinct. Hence,
there exists \textbf{at most one} $i\in z$ such that $u=w_{\tau,i}$. But we
also know that there exists \textbf{at least one} such $i$ (because
$u\in\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $).
\end{vershort}

\begin{verlong}
Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields $u\in\left[  u\right]
=\left[  w\right]  _{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $. In other
words, there exists \textbf{at least one} $i\in z$ such that $u=w_{\tau,i}$.
Moreover, there exists \textbf{at most one} such $i$%
\ \ \ \ \footnote{\textit{Proof.} We must show that there exists \textbf{at
most one} $i\in z$ such that $u=w_{\tau,i}$. In other words, we must prove
that if $\alpha$ and $\beta$ are two elements $i\in z$ such that $u=w_{\tau
,i}$, then $\alpha=\beta$.
\par
So let $\alpha$ and $\beta$ be two elements $i\in z$ such that $u=w_{\tau,i}$.
Thus, $\alpha$ is an element $i\in z$ such that $u=w_{\tau,i}$. In other
words, $\alpha$ is an element of $z$ and satisfies $u=w_{\tau,\alpha}$.
Similarly, $\beta$ is an element of $z$ and satisfies $u=w_{\tau,\beta}$.
Comparing $u=w_{\tau,\alpha}$ with $u=w_{\tau,\beta}$, we obtain
$w_{\tau,\alpha}=w_{\tau,\beta}$. If $\alpha$ and $\beta$ were distinct, then
Proposition \ref{prop.words.read-std2}(b) would show that $w_{\tau,\alpha}\neq
w_{\tau,\beta}$; but this would contradict $w_{\tau,\alpha}=w_{\tau,\beta}$.
Hence, $\alpha$ and $\beta$ cannot be distinct. In other words, we have
$\alpha=\beta$.
\par
Forget that we fixed $\alpha$ and $\beta$. We thus have proved that if
$\alpha$ and $\beta$ are two elements $i\in z$ such that $u=w_{\tau,i}$, then
$\alpha=\beta$. In other words, any two elements $i\in z$ such that
$u=w_{\tau,i}$ must be equal. In other words, there exists \textbf{at most
one} $i\in z$ such that $u=w_{\tau,i}$.}.
\end{verlong}

Combining the results of the previous two sentences, we conclude that there
exists \textbf{exactly one} $i\in z$ such that $u=w_{\tau,i}$. In other words,%
\[
\left(  \text{the number of all }i\in z\text{ such that }u=w_{\tau,i}\right)
=1.
\]
Comparing this with $\delta_{\left[  u\right]  ,\left[  w\right]  _{z}}=1$, we
obtain
\[
\left(  \text{the number of all }i\in z\text{ such that }u=w_{\tau,i}\right)
=\delta_{\left[  u\right]  ,\left[  w\right]  _{z}}.
\]
Thus, Claim 1 is proved in Case 1.

Let us now consider Case 2. In this case, we have $\left[  u\right]
\neq\left[  w\right]  _{z}$. Hence, $\delta_{\left[  u\right]  ,\left[
w\right]  _{z}}=0$.

Also, there exists no $i\in z$ such that $u=w_{\tau,i}$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, there exists some
$i\in z$ such that $u=w_{\tau,i}$. In other words, $u\in\left\{  w_{\tau
,i}\ \mid\ i\in z\right\}  $. This rewrites as $u\in\left[  w\right]  _{z}$
(since $\left[  w\right]  _{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $).
Since $\left[  w\right]  _{z}$ is a necklace, we can thus apply Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts} to $N=\left[  w\right]  _{z}$. As a
result, we obtain $\left[  w\right]  _{z}=\left[  u\right]  $. In other words,
$\left[  u\right]  =\left[  w\right]  _{z}$. But this contradicts $\left[
u\right]  \neq\left[  w\right]  _{z}$. This contradiction shows that our
assumption was false, qed.}. In other words,%
\[
\left(  \text{the number of all }i\in z\text{ such that }u=w_{\tau,i}\right)
=0.
\]
Comparing this with $\delta_{\left[  u\right]  ,\left[  w\right]  _{z}}=0$, we
obtain
\[
\left(  \text{the number of all }i\in z\text{ such that }u=w_{\tau,i}\right)
=\delta_{\left[  u\right]  ,\left[  w\right]  _{z}}.
\]
Thus, Claim 1 is proved in Case 2.

We have now proved Claim 1 in both Cases 1 and 2. Hence, Claim 1 always holds.]

Let $M^{\prime}=\biguplus_{N\in M}N$. (We are here using the fact that each
necklace $N\in M$ is a set, thus a multiset.) Then, $M^{\prime}$ is a multiset
of aperiodic words\footnote{Indeed, this was shown in Definition
\ref{def.lyndon.gr.RG}.}.

Next, we claim that%
\begin{equation}
M^{\prime}=\left\{  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right\}
_{\operatorname*{multiset}}. \label{pf.lem.sol.lyndon.gr.bij.RG-GR.M}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.bij.RG-GR.M}):} Proposition
\ref{prop.words.read-std3}(a) shows that the words $w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}$ are aperiodic. Hence, $\left\{  w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$ is a multiset of
aperiodic words. Thus, both $M^{\prime}$ and $\left\{  w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$ are multisets of
aperiodic words.

\begin{verlong}
We have $\tau\in\Symm_{n}$. In other words, $\tau$ is a permutation of
$\left\{  1,2,\ldots,n\right\}  $ (since $\Symm_{n}$ is the set of all
permutations of $\left\{  1,2,\ldots,n\right\}  $). Thus, the cycles of $\tau$
are subsets of $\left\{  1,2,\ldots,n\right\}  $, and each $i\in\left\{
1,2,\ldots,n\right\}  $ belongs to exactly one of these cycles.
\end{verlong}

Now, let $u$ be an aperiodic word. Hence, $u$ is a nonempty word (since any
aperiodic word is nonempty). Lemma \ref{lem.sol.lyndon.gr.bij.mulset-neck}
thus yields that%
\begin{align*}
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)   &  =\left(
\text{multiplicity of }\left[  u\right]  \text{ in }\underbrace{M}_{=\left\{
\left[  w\right]  _{z}\ \mid\ z\text{ is a cycle of }\tau\right\}
_{\operatorname*{multiset}}}\right) \\
&  =\left(  \text{multiplicity of }\left[  u\right]  \text{ in }\left\{
\left[  w\right]  _{z}\ \mid\ z\text{ is a cycle of }\tau\right\}
_{\operatorname*{multiset}}\right) \\
&  =\left(  \text{the number of all cycles }z\text{ of }\tau\text{ such that
}\left[  u\right]  =\left[  w\right]  _{z}\right)
\end{align*}
(by the definition of the multiset $\left\{  \left[  w\right]  _{z}%
\ \mid\ z\text{ is a cycle of }\tau\right\}  _{\operatorname*{multiset}}$).
Comparing this with%
\begin{align*}
\sum_{z\text{ is a cycle of }\tau}\delta_{\left[  u\right]  ,\left[  w\right]
_{z}}  &  =\sum\limits_{\substack{z\text{ is a cycle of }\tau;\\\left[
u\right]  =\left[  w\right]  _{z}}}\underbrace{\delta_{\left[  u\right]
,\left[  w\right]  _{z}}}_{\substack{=1\\\text{(since }\left[  u\right]
=\left[  w\right]  _{z}\text{)}}}+\sum\limits_{\substack{z\text{ is a cycle of
}\tau;\\\left[  u\right]  \neq\left[  w\right]  _{z}}}\underbrace{\delta
_{\left[  u\right]  ,\left[  w\right]  _{z}}}_{\substack{=0\\\text{(since
}\left[  u\right]  \neq\left[  w\right]  _{z}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each cycle }z\text{ of }\tau\text{ satisfies}\\
\text{either }\left[  u\right]  =\left[  w\right]  _{z}\text{ or }\left[
u\right]  \neq\left[  w\right]  _{z}\text{ (but not both)}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{z\text{ is a cycle of }\tau;\\\left[  u\right]
=\left[  w\right]  _{z}}}1+\underbrace{\sum\limits_{\substack{z\text{ is a
cycle of }\tau;\\\left[  u\right]  \neq\left[  w\right]  _{z}}}0}_{=0}=
\sum\limits_{\substack{z\text{ is a cycle of }\tau;\\\left[  u\right]
=\left[  w\right]  _{z}}}1\\
&  =\left(  \text{the number of all cycles }z\text{ of }\tau\text{ such that
}\left[  u\right]  =\left[  w\right]  _{z}\right)  \cdot1\\
&  =\left(  \text{the number of all cycles }z\text{ of }\tau\text{ such that
}\left[  u\right]  =\left[  w\right]  _{z}\right)  ,
\end{align*}
we obtain%
\begin{equation}
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)  =\sum_{z\text{
is a cycle of }\tau}\delta_{\left[  u\right]  ,\left[  w\right]  _{z}}.
\label{pf.lem.sol.lyndon.gr.bij.RG-GR.M.pf.1}%
\end{equation}


On the other hand, the definition of the multiset $\left\{  w_{\tau,1}%
,w_{\tau,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$ yields%
\begin{align*}
&  \left(  \text{multiplicity of }u\text{ in }\left\{  w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}\right) \\
&  =\left(  \text{the number of all }i\in\left\{  1,2,\ldots,n\right\}  \text{
such that }u=w_{\tau,i}\right) \\
&  =\sum_{z\text{ is a cycle of }\tau}\left(  \text{the number of all }i\in
z\text{ such that }u=w_{\tau,i}\right)
\end{align*}
(since the cycles of $\tau$ are subsets of $\left\{  1,2,\ldots,n\right\}  $,
and since each $i\in\left\{  1,2,\ldots,n\right\}  $ belongs to exactly one of
these cycles). Hence,%
\begin{align*}
&  \left(  \text{multiplicity of }u\text{ in }\left\{  w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}\right) \\
&  =\sum_{z\text{ is a cycle of }\tau}\underbrace{\left(  \text{the number of
all }i\in z\text{ such that }u=w_{\tau,i}\right)  }_{\substack{=\delta
_{\left[  u\right]  ,\left[  w\right]  _{z}}\\\text{(by Claim 1)}}}\\
&  =\sum_{z\text{ is a cycle of }\tau}\delta_{\left[  u\right]  ,\left[
w\right]  _{z}}.
\end{align*}


Comparing this with (\ref{pf.lem.sol.lyndon.gr.bij.RG-GR.M.pf.1}), we obtain%
\[
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)  =\left(
\text{multiplicity of }u\text{ in }\left\{  w_{\tau,1},w_{\tau,2}%
,\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}\right)  .
\]


Now, forget that we fixed $u$. We thus have proved that%
\[
\left(  \text{multiplicity of }u\text{ in }M^{\prime}\right)  =\left(
\text{multiplicity of }u\text{ in }\left\{  w_{\tau,1},w_{\tau,2}%
,\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}\right)
\]
for each aperiodic word $u$. In other words, each aperiodic word $u$ appears
in the multisets $M^{\prime}$ and $\left\{  w_{\tau,1},w_{\tau,2}%
,\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$ with equal
multiplicity. Since both $M^{\prime}$ and $\left\{  w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$ are multisets of
aperiodic words, we thus conclude that the multisets $M^{\prime}$ and
$\left\{  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right\}
_{\operatorname*{multiset}}$ contain the same elements with equal
multiplicities. In other words, these multisets $M^{\prime}$ and $\left\{
w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$
are equal. In other words, $M^{\prime}=\left\{  w_{\tau,1},w_{\tau,2}%
,\ldots,w_{\tau,n}\right\}  _{\operatorname*{multiset}}$. This proves
(\ref{pf.lem.sol.lyndon.gr.bij.RG-GR.M}).]

Proposition \ref{prop.words.read-std3}(b) yields $w_{\tau,1}\leq_{\omega
}w_{\tau,2}\leq_{\omega}\cdots\leq_{\omega}w_{\tau,n}$. From $M^{\prime
}=\left\{  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right\}
_{\operatorname*{multiset}}$ and $w_{\tau,1}\leq_{\omega}w_{\tau,2}%
\leq_{\omega}\cdots\leq_{\omega}w_{\tau,n}$, we conclude that $\left(
w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right)  $ is the $\leq_{\omega}%
$-increasing list of $M^{\prime}$. Thus, the $\leq_{\omega}$-increasing list
of $M^{\prime}$ has $n$ elements.

Recall that $M^{\prime}=\biguplus_{N\in M}N$. Hence, Definition
\ref{def.lyndon.gr.RG} tells us that $\operatorname*{RG}\left(  M\right)  $
can be constructed as follows: Let $\left(  m_{1},m_{2},\ldots,m_{n}\right)  $
be the $\leq_{\omega}$-increasing list of $M^{\prime}$. (This is well-defined,
since the $\leq_{\omega}$-increasing list of $M^{\prime}$ has $n$ elements.)
For each $i\in\left\{  1,2,\ldots,n\right\}  $, let $\ell_{i}$ be the last
letter of the nonempty word $m_{i}$. Then, the definition of
$\operatorname*{RG}$ yields $\operatorname*{RG}\left(  M\right)  =\left(
\ell_{1},\ell_{2},\ldots,\ell_{n}\right)  $.

We shall now use this to show that $\operatorname*{RG}\left(  M\right)  =w$.

\begin{vershort}
Indeed, we first observe that $\left(  m_{1},m_{2},\ldots,m_{n}\right)
=\left(  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right)  $, since both lists
$\left(  m_{1},m_{2},\ldots,m_{n}\right)  $ and $\left(  w_{\tau,1},w_{\tau
,2},\ldots,w_{\tau,n}\right)  $ are the $\leq_{\omega}$-increasing list of
$M^{\prime}$.
\end{vershort}

\begin{verlong}
Indeed, the definition of $\left(  m_{1},m_{2},\ldots,m_{n}\right)  $ shows
that
\[
\left(  m_{1},m_{2},\ldots,m_{n}\right)  =\left(  \text{the }\leq_{\omega
}\text{-increasing list of }M^{\prime}\right)  .
\]
But we also have
\[
\left(  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right)  =\left(  \text{the
}\leq_{\omega}\text{-increasing list of }M^{\prime}\right)
\]
(since $\left(  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right)  $ is the
$\leq_{\omega}$-increasing list of $M^{\prime}$). Comparing these two
equalities, we obtain $\left(  m_{1},m_{2},\ldots,m_{n}\right)  =\left(
w_{\tau,1},w_{\tau,2},\ldots,w_{\tau,n}\right)  $.

In other words,%
\begin{equation}
m_{i}=w_{\tau,i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{
1,2,\ldots,n\right\}  . \label{pf.lem.sol.lyndon.gr.bij.RG-GR.mi=}%
\end{equation}

\end{verlong}

\begin{vershort}
Now, for each $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
\ell_{i}  &  =\left(  \text{the last letter of the nonempty word
}\underbrace{m_{i}}_{\substack{=w_{\tau,i}\\\text{(since }\left(  m_{1}%
,m_{2},\ldots,m_{n}\right)  =\left(  w_{\tau,1},w_{\tau,2},\ldots,w_{\tau
,n}\right)  \text{)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell_{i}\text{ was defined to be
the last letter of the nonempty word }m_{i}\right) \\
&  =\left(  \text{the last letter of the nonempty word }w_{\tau,i}\right) \\
&  =w_{i}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the last letter of the word }w_{\tau,i}\text{ is }w_{i}\\
\text{(by Proposition \ref{prop.words.read1}(c), applied to }h=i\text{)}%
\end{array}
\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
Now, for each $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
\ell_{i}  &  =\left(  \text{the last letter of the nonempty word
}\underbrace{m_{i}}_{\substack{=w_{\tau,i}\\\text{(by
(\ref{pf.lem.sol.lyndon.gr.bij.RG-GR.mi=}))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell_{i}\text{ was defined to be
the last letter of the nonempty word }m_{i}\right) \\
&  =\left(  \text{the last letter of the nonempty word }w_{\tau,i}\right) \\
&  =w_{i}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the last letter of the word }w_{\tau,i}\text{ is }w_{i}\\
\text{(by Proposition \ref{prop.words.read1}(c), applied to }h=i\text{)}%
\end{array}
\right)  .
\end{align*}

\end{verlong}

In other words, $\left(  \ell_{1},\ell_{2},\ldots,\ell_{n}\right)  =\left(
w_{1},w_{2},\ldots,w_{n}\right)  $. Comparing this with $w=\left(  w_{1}%
,w_{2},\ldots,w_{n}\right)  $, we obtain $\left(  \ell_{1},\ell_{2}%
,\ldots,\ell_{n}\right)  =w$. Now, $\operatorname*{RG}\left(  M\right)
=\left(  \ell_{1},\ell_{2},\ldots,\ell_{n}\right)  =w$.

But $\operatorname*{GR}w=M$ (since $M$ was defined as $\operatorname*{GR}w$).
Now,
\[
\left(  \operatorname*{RG}\circ\operatorname*{GR}\right)  \left(  w\right)
=\operatorname*{RG}\left(  \underbrace{\operatorname*{GR}w}_{=M}\right)
=\operatorname*{RG}\left(  M\right)  =w=\operatorname*{id}\left(  w\right)  .
\]


Forget that we fixed $w$. We thus have shown that $\left(  \operatorname*{RG}%
\circ\operatorname*{GR}\right)  \left(  w\right)  =\operatorname*{id}\left(
w\right)  $ for each $w\in\mathfrak{A}^{\ast}$. In other words, we have
$\operatorname*{RG}\circ\operatorname*{GR}=\operatorname*{id}$. This proves
Lemma \ref{lem.sol.lyndon.gr.bij.RG-GR}.
\end{proof}

Now, we shall approach the proof of $\operatorname*{GR}\circ\operatorname*{RG}%
=\operatorname*{id}$. As was explained in the Hint, we can now try to obtain
$\operatorname*{RG}\circ\operatorname*{GR}=\operatorname*{id}$ by finding a
bijection $\mathfrak{A}^{\ast}\rightarrow\mathfrak{MN}^{\mathfrak{a}}$, or,
more precisely, a bijection between certain finite subsets of $\mathfrak{A}%
^{\ast}$ and $\mathfrak{MN}^{\mathfrak{a}}$. In order to define these subsets,
we first introduce a notation:

\begin{definition}
\label{def.sol.lyndon.gr.bij.sum}Let $M$ be a finite multiset of finite sets
(for example, of necklaces). Then, $\operatorname*{sum}M$ shall denote the sum
of the sizes of all sets $N\in M$ (counted with multiplicities). Formally
speaking, this can be defined as follows: Set%
\begin{equation}
\operatorname*{sum}M=\sum_{N\in\operatorname*{Supp}M}\left(
\text{multiplicity of }N\text{ in }M\right)  \cdot\left\vert N\right\vert .
\label{eq.def.sol.lyndon.gr.bij.sum.1}%
\end{equation}
Equivalently, if $M=\left\{  N_{1},N_{2},\ldots,N_{k}\right\}
_{\operatorname*{multiset}}$, then
\begin{equation}
\operatorname*{sum}M=\left\vert N_{1}\right\vert +\left\vert N_{2}\right\vert
+\cdots+\left\vert N_{k}\right\vert . \label{eq.def.sol.lyndon.gr.bij.sum.2}%
\end{equation}

\end{definition}

\begin{example}
Applying (\ref{eq.def.sol.lyndon.gr.bij.sum.2}), we obtain%
\[
\operatorname*{sum}\left(  \left\{  \left\{  1,4\right\}  ,\left\{
2,3\right\}  ,\left\{  1,2,5\right\}  ,\left\{  2,3\right\}  \right\}
_{\operatorname*{multiset}}\right)  =\left\vert \left\{  1,4\right\}
\right\vert +\left\vert \left\{  2,3\right\}  \right\vert +\left\vert \left\{
1,2,5\right\}  \right\vert +\left\vert \left\{  2,3\right\}  \right\vert
=2+2+3+2=9.
\]

\end{example}

\begin{definition}
\label{def.sol.lyndon.gr.bij.fin-subsets}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$.

\begin{enumerate}
\item[(a)] For any $n\in \NN $, we have $\mathfrak{B}^{n}\subset
\mathfrak{B}^{\ast}\subset\mathfrak{A}^{\ast}$ (since $\mathfrak{B}%
\subset\mathfrak{A}$). Thus, elements of $\mathfrak{B}^{n}$ are words in
$\mathfrak{A}^{\ast}$.

\item[(b)] A \dfn{$\mathfrak{B}$-necklace} will mean a necklace that is a
subset of $\mathfrak{B}^{\ast}$ (that is, a necklace consisting of words in
$\mathfrak{B}^{\ast}$).

\item[(c)] For any $n\in \NN $, we let $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ denote the set of all finite multisets $M$ of aperiodic
$\mathfrak{B}$-necklaces satisfying $\operatorname*{sum}M=n$.
\end{enumerate}
\end{definition}

\begin{example}
For this example, let $\mathfrak{A}=\left\{  1,2,\ldots,9\right\}  $ and
$\mathfrak{B}=\left\{  2,4,6,8\right\}  $. Then, the necklace $\left[
286\right]  =\left\{  286,862,628\right\}  $ is an aperiodic $\mathfrak{B}%
$-necklace, but the necklace $\left[  25\right]  =\left\{  25,52\right\}  $ is
not a $\mathfrak{B}$-necklace (since $25\notin\mathfrak{B}^{\ast}$). The
multiset $M=\left\{  \left[  286\right]  ,\left[  24\right]  ,\left[
24\right]  \right\}  _{\operatorname*{multiset}}$ is a finite multiset of
aperiodic $\mathfrak{B}$-necklaces and satisfies $\operatorname*{sum}%
M=\left\vert \left[  286\right]  \right\vert +\left\vert \left[  24\right]
\right\vert +\left\vert \left[  24\right]  \right\vert =3+2+2=7$; thus, it
belongs to $\mathfrak{MN}_{\mathfrak{B},7}^{\mathfrak{a}}$.
\end{example}

The sets we introduced in Definition \ref{def.sol.lyndon.gr.bij.fin-subsets}
interact nicely with the Gessel-Reutenauer bijection: For any subset
$\mathfrak{B}$ of $\mathfrak{A}$ and any $n\in \NN $, the map
$\operatorname*{GR}:\mathfrak{A}^{\ast}\rightarrow\mathfrak{MN}^{\mathfrak{a}%
}$ restricts to a map from $\mathfrak{B}^{n}$ to $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$. More precisely:

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.GR-restricts}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $n\in \NN $. Then, $\operatorname*{GR}\left(
\mathfrak{B}^{n}\right)  \subset\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.GR-restricts}.]Let $w\in
\mathfrak{B}^{n}$. We shall show that $\operatorname*{GR}w\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$.

Indeed, $w\in\mathfrak{B}^{n}\subset\mathfrak{A}^{n}$ (since $\mathfrak{B}%
\subset\mathfrak{A}$). Hence, $\ell\left(  w\right)  =n$. Let $\tau$ be the
permutation $\left(  \operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$.
The definition of $\operatorname*{GR}$ yields%
\begin{equation}
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}.
\label{pf.lem.sol.lyndon.gr.bij.GR-restricts.GRw=}%
\end{equation}


\begin{vershort}
Now, let $z$ be a cycle of $\tau$. Then, the necklace $\left[  w\right]  _{z}$
is a subset of $\mathfrak{B}^{\ast}$\ \ \ \ \footnote{This follows in a
straightforward way from the definition of $\left[  w\right]  _{z}$. (In more
detail: The definition of $\left[  w\right]  _{z}$ yields $\left[  w\right]
_{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $. But all letters of the word
$w$ belong to $\mathfrak{B}$ (since $w\in\mathfrak{B}^{n}$). Hence, for each
$i\in z$, the word $w_{\tau,i}$ consists entirely of letters in $\mathfrak{B}$
(since the definition of $w_{\tau,i}$ shows that $w_{\tau,i}$ consists of some
letters of $w$, but all letters of $w$ belong to $\mathfrak{B}$), and thus
belongs to $\mathfrak{B}^{\ast}$. In other words, $\left\{  w_{\tau,i}%
\ \mid\ i\in z\right\}  $ is a subset of $\mathfrak{B}^{\ast}$. In other
words, $\left[  w\right]  _{z}$ is a subset of $\mathfrak{B}^{\ast}$ (since
$\left[  w\right]  _{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $). Qed.)}.
Hence, the necklace $\left[  w\right]  _{z}$ is a $\mathfrak{B}$-necklace, and
thus an aperiodic $\mathfrak{B}$-necklace (since we have seen in the
definition of $\left[  w\right]  _{z}$ that $\left[  w\right]  _{z}$ is aperiodic).
\end{vershort}

\begin{verlong}
Now, let $z$ be a cycle of $\tau$. Then, $\left[  w\right]  _{z}$ is a subset
of $\mathfrak{B}^{\ast}$\ \ \ \ \footnote{\textit{Proof.} Let $i\in z$. We
shall show that $w_{\tau,i}\in\mathfrak{B}^{\ast}$.
\par
Indeed, let $k=\operatorname*{ord}\nolimits_{\tau}\left(  i\right)  $. Then,
the definition of $w_{\tau,i}$ yields $w_{\tau,i}=w_{\tau^{1}\left(  i\right)
}w_{\tau^{2}\left(  i\right)  }\cdots w_{\tau^{k}\left(  i\right)  }$.
\par
The letters of the word $w$ belong to $\mathfrak{B}$ (since $w\in
\mathfrak{B}^{n}$). Thus, in particular, the letters $w_{\tau^{1}\left(
i\right)  },w_{\tau^{2}\left(  i\right)  },\ldots,w_{\tau^{k}\left(  i\right)
}$ belong to $\mathfrak{B}$ (since $w_{\tau^{1}\left(  i\right)  },w_{\tau
^{2}\left(  i\right)  },\ldots,w_{\tau^{k}\left(  i\right)  }$ are letters of
the word $w$). In other words, we have $w_{\tau^{j}\left(  i\right)  }%
\in\mathfrak{B}$ for each $i\in\left\{  1,2,\ldots,k\right\}  $.
\par
Now,
\begin{align*}
w_{\tau,i}  &  =w_{\tau^{1}\left(  i\right)  }w_{\tau^{2}\left(  i\right)
}\cdots w_{\tau^{k}\left(  i\right)  }=\left(  w_{\tau^{1}\left(  i\right)
},w_{\tau^{2}\left(  i\right)  },\ldots,w_{\tau^{k}\left(  i\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }w_{\tau^{1}\left(  i\right)
},w_{\tau^{2}\left(  i\right)  },\ldots,w_{\tau^{k}\left(  i\right)  }\text{
are single letters}\right) \\
&  \in\mathfrak{B}^{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }w_{\tau
^{j}\left(  i\right)  }\in\mathfrak{B}\text{ for each }i\in\left\{
1,2,\ldots,k\right\}  \right) \\
&  \subset\mathfrak{B}^{\ast}.
\end{align*}
\par
Forget that we fixed $i$. We thus have shown that $w_{\tau,i}\in
\mathfrak{B}^{\ast}$ for each $i\in z$. In other words, $\left\{  w_{\tau
,i}\ \mid\ i\in z\right\}  \subset\mathfrak{B}^{\ast}$.
\par
But the definition of $\left[  w\right]  _{z}$ yields $\left[  w\right]
_{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  \subset\mathfrak{B}^{\ast}$.
In other words, $\left[  w\right]  _{z}$ is a subset of $\mathfrak{B}^{\ast}$.
Qed.}. Hence, $\left[  w\right]  _{z}$ is a necklace that is a subset of
$\mathfrak{B}^{\ast}$ (since $\left[  w\right]  _{z}$ is a necklace). In other
words, $\left[  w\right]  _{z}$ is a $\mathfrak{B}$-necklace (by the
definition of a $\mathfrak{B}$-necklace). We also know that $\left[  w\right]
_{z}$ is an aperiodic necklace (as we have seen in the definition of $\left[
w\right]  _{z}$). Hence, $\left[  w\right]  _{z}$ is an aperiodic
$\mathfrak{B}$-necklace.
\end{verlong}

Forget that we fixed $z$. We thus have proved that $\left[  w\right]  _{z}$ is
an aperiodic $\mathfrak{B}$-necklace whenever $z$ is a cycle of $\tau$. Hence,
the multiset $\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle of
}\tau\right\}  _{\operatorname*{multiset}}$ is a multiset of aperiodic
$\mathfrak{B}$-necklaces. In view of
(\ref{pf.lem.sol.lyndon.gr.bij.GR-restricts.GRw=}), this rewrites as follows:
The multiset $\operatorname*{GR}w$ is a multiset of aperiodic $\mathfrak{B}%
$-necklaces. Moreover, this multiset $\operatorname*{GR}w$ is clearly finite.

\begin{vershort}
Let us now show that $\operatorname*{sum}\left(  \operatorname*{GR}w\right)
=n$. Indeed, the cycles of $\tau$ form a set partition of the set $\left\{
1,2,\ldots,n\right\}  $ (that is, they are disjoint nonempty sets, and their
union is $\left\{  1,2,\ldots,n\right\}  $). Hence, the sum of their sizes is
$\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert $. In other words,
$\sum_{z\text{ is a cycle of }\tau}\left\vert z\right\vert =\left\vert
\left\{  1,2,\ldots,n\right\}  \right\vert $.

But let us recall that $\operatorname*{sum}\left(  \operatorname*{GR}w\right)
$ was defined to be the sum of the sizes of all sets $N\in\operatorname*{GR}w$
(counted with multiplicities). Since $\operatorname*{GR}w=\left\{  \left[
w\right]  _{z}\ \mid\ z\text{ is a cycle of }\tau\right\}
_{\operatorname*{multiset}}$, these sets $N\in\operatorname*{GR}w$ are
precisely the necklaces $\left[  w\right]  _{z}$ where $z$ ranges over all
cycles of $\tau$; hence, the previous sentence says that $\operatorname*{sum}%
\left(  \operatorname*{GR}w\right)  $ is the sum of the sizes of the latter
necklaces. In other words,%
\begin{align*}
\operatorname*{sum}\left(  \operatorname*{GR}w\right)   &  =\sum_{z\text{ is a
cycle of }\tau}\left\vert \underbrace{\left[  w\right]  _{z}}%
_{\substack{=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  \\\text{(by the
definition of }\left[  w\right]  _{z}\text{)}}}\right\vert =\sum_{z\text{ is a
cycle of }\tau}\underbrace{\left\vert \left\{  w_{\tau,i}\ \mid\ i\in
z\right\}  \right\vert }_{\substack{=\left\vert z\right\vert \\\text{(by
Proposition \ref{prop.words.read-std2}(c))}}}\\
&  =\sum_{z\text{ is a cycle of }\tau}\left\vert z\right\vert =\left\vert
\left\{  1,2,\ldots,n\right\}  \right\vert =n.
\end{align*}

\end{vershort}

\begin{verlong}
Let us now show that $\operatorname*{sum}\left(  \operatorname*{GR}w\right)
=n$. Indeed, let $z_{1},z_{2},\ldots,z_{k}$ be the cycles of $\tau$, listed in
some order (with no repetitions). Then, $\left\{  z_{1},z_{2},\ldots
,z_{k}\right\}  $ is a set partition of the set $\left\{  1,2,\ldots
,n\right\}  $ (since the cycles of a permutation of any set must form a set
partition of this set). Thus, the sets $z_{1},z_{2},\ldots,z_{k}$ are
disjoint, and their union is $\left\{  1,2,\ldots,n\right\}  $. Hence,
$\left\vert z_{1}\right\vert +\left\vert z_{2}\right\vert +\cdots+\left\vert
z_{k}\right\vert =\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =n$.

But (\ref{pf.lem.sol.lyndon.gr.bij.GR-restricts.GRw=}) becomes%
\[
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}=\left\{  \left[  w\right]
_{z_{1}},\left[  w\right]  _{z_{2}},\ldots,\left[  w\right]  _{z_{k}}\right\}
_{\operatorname*{multiset}}%
\]
(since $z_{1},z_{2},\ldots,z_{k}$ are the cycles of $\tau$, listed with no
repetitions). Hence, (\ref{eq.def.sol.lyndon.gr.bij.sum.2}) (applied to
$M=\operatorname*{GR}w$ and $N_{i}=\left[  w\right]  _{z_{i}}$) yields%
\begin{equation}
\operatorname*{sum}\left(  \operatorname*{GR}w\right)  =\left\vert \left[
w\right]  _{z_{1}}\right\vert +\left\vert \left[  w\right]  _{z_{2}%
}\right\vert +\cdots+\left\vert \left[  w\right]  _{z_{k}}\right\vert .
\label{pf.lem.sol.lyndon.gr.bij.GR-restricts.3}%
\end{equation}


Now, it is easy to see that%
\begin{equation}
\left\vert \left[  w\right]  _{z_{m}}\right\vert =\left\vert z_{m}\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for each }m\in\left\{  1,2,\ldots,k\right\}  .
\label{pf.lem.sol.lyndon.gr.bij.GR-restricts.4}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.bij.GR-restricts.4}):} Let
$m\in\left\{  1,2,\ldots,k\right\}  $. Then, $z_{m}$ is a cycle of $\tau$
(since $z_{1},z_{2},\ldots,z_{k}$ are the cycles of $\tau$). Thus, the
definition of $\left[  w\right]  _{z_{m}}$ yields $\left[  w\right]  _{z_{m}%
}=\left\{  w_{\tau,i}\ \mid\ i\in z_{m}\right\}  $. But Proposition
\ref{prop.words.read-std2}(c) (applied to $z=z_{m}$) shows that $\left\vert
\left\{  w_{\tau,i}\ \mid\ i\in z_{m}\right\}  \right\vert =\left\vert
z_{m}\right\vert $. Since $\left[  w\right]  _{z_{m}}=\left\{  w_{\tau
,i}\ \mid\ i\in z_{m}\right\}  $, this rewrites as $\left\vert \left[
w\right]  _{z_{m}}\right\vert =\left\vert z_{m}\right\vert $. This proves
(\ref{pf.lem.sol.lyndon.gr.bij.GR-restricts.4}).]

Now, (\ref{pf.lem.sol.lyndon.gr.bij.GR-restricts.3}) becomes%
\begin{align*}
\operatorname*{sum}\left(  \operatorname*{GR}w\right)   &  =\left\vert \left[
w\right]  _{z_{1}}\right\vert +\left\vert \left[  w\right]  _{z_{2}%
}\right\vert +\cdots+\left\vert \left[  w\right]  _{z_{k}}\right\vert
=\sum_{m=1}^{k}\underbrace{\left\vert \left[  w\right]  _{z_{m}}\right\vert
}_{\substack{=\left\vert z_{m}\right\vert \\\text{(by
(\ref{pf.lem.sol.lyndon.gr.bij.GR-restricts.4}))}}}\\
&  =\sum_{m=1}^{k}\left\vert z_{m}\right\vert =\left\vert z_{1}\right\vert
+\left\vert z_{2}\right\vert +\cdots+\left\vert z_{k}\right\vert =n.
\end{align*}

\end{verlong}

Altogether, we now know that the multiset $\operatorname*{GR}w$ is a finite
multiset of aperiodic $\mathfrak{B}$-necklaces and satisfies
$\operatorname*{sum}\left(  \operatorname*{GR}w\right)  =n$. In other words,
$\operatorname*{GR}w$ is a finite multiset $M$ of aperiodic $\mathfrak{B}%
$-necklaces satisfying $\operatorname*{sum}M=n$. In other words,
$\operatorname*{GR}w\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ (since
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ was defined as the set of all
finite multisets $M$ of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$).

Forget that we fixed $w$. We thus have shown that $\operatorname*{GR}%
w\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ for each $w\in
\mathfrak{B}^{n}$. In other words, $\operatorname*{GR}\left(  \mathfrak{B}%
^{n}\right)  \subset\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. This
proves Lemma \ref{lem.sol.lyndon.gr.bij.GR-restricts}.
\end{proof}

The following fact is almost trivial:

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.B-neck}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $u\in\mathfrak{B}^{\ast}$ be a nonempty word. Then,
$\left[  u\right]  $ is a $\mathfrak{B}$-necklace.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.B-neck}.]The word $u\in
\mathfrak{B}^{\ast}$ is nonempty. Thus, there exists a positive integer $p$
such that $u\in\mathfrak{B}^{p}$. Consider this $p$. Hence, $u\in
\mathfrak{B}^{p}\subset\mathfrak{A}^{p}$ (since $\mathfrak{B}\subset
\mathfrak{A}$). Therefore, $\left[  u\right]  $ is a well-defined $p$-necklace
(since $p$ is a positive integer), hence a necklace.

Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields that $u\in\left[  u\right]
$. In other words, $\left[  u\right]  $ contains $u$.

The cyclic group $C$ acts on both $\mathfrak{B}^{p}$ and $\mathfrak{A}^{p}$ in
the same way; thus, the $C$-set $\mathfrak{B}^{p}$ is a subset of the $C$-set
$\mathfrak{A}^{p}$. Hence, the orbit $C\cdot u$ of $u$ is the same no matter
whether we are considering $u$ as an element of $\mathfrak{B}^{p}$ or as an
element of the larger $C$-set $\mathfrak{A}^{p}$. But considering $u$ as an
element of $\mathfrak{B}^{p}$, we clearly see that this orbit $C\cdot u$ is an
orbit of the $C$-set $\mathfrak{B}^{p}$; thus, $C\cdot u\subset\mathfrak{B}%
^{p}$.

\begin{vershort}
The set $\left[  u\right]  $ is a $p$-necklace. In other words, $\left[
u\right]  $ is an orbit of the $C$-action on $\mathfrak{A}^{p}$ (since this is
what ``$p$-necklace'' means). Since $\left[
u\right]  $ contains $u$, we thus conclude that $\left[  u\right]  $ is the
orbit of the $C$-action on $\mathfrak{A}^{p}$ that contains $u$. In other
words, $\left[  u\right]  =C\cdot u$. Hence, $\left[  u\right]  =C\cdot
u\subset\mathfrak{B}^{p}\subset\mathfrak{B}^{\ast}$. In other words, $\left[
u\right]  $ is a subset of $\mathfrak{B}^{\ast}$.
\end{vershort}

\begin{verlong}
The set $\left[  u\right]  $ is a $p$-necklace. In other words, $\left[
u\right]  $ is an orbit of the $C$-action on $\mathfrak{A}^{p}$ (since a
$p$-necklace is the same thing as an orbit of the $C$-action on $\mathfrak{A}%
^{p}$). Since $\left[  u\right]  $ contains $u$, we thus conclude that
$\left[  u\right]  $ is the orbit of the $C$-action on $\mathfrak{A}^{p}$ that
contains $u$. In other words, $\left[  u\right]  =C\cdot u$ (since $C\cdot u$
means the orbit of the $C$-action on $\mathfrak{A}^{p}$ that contains $u$).
Hence, $\left[  u\right]  =C\cdot u\subset\mathfrak{B}^{p}\subset
\mathfrak{B}^{\ast}$. In other words, $\left[  u\right]  $ is a subset of
$\mathfrak{B}^{\ast}$.
\end{verlong}

Thus, $\left[  u\right]  $ is a necklace that is a subset of $\mathfrak{B}%
^{\ast}$ (since $\left[  u\right]  $ is a necklace). In other words, $\left[
u\right]  $ is a $\mathfrak{B}$-necklace (since a $\mathfrak{B}$-necklace is
defined to mean a necklace that is a subset of $\mathfrak{B}^{\ast}$). This
proves Lemma \ref{lem.sol.lyndon.gr.bij.B-neck}.
\end{proof}

We shall now use the CFL factorization (introduced in Definition
\ref{def.words.CFL}) to establish a bijection between $\mathfrak{B}^{n}$ and
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. We recall that every word has
a unique CFL factorization (as we have seen in Theorem \ref{thm.words.CFL});
thus, we can speak of ``the'' CFL factorization of a word.

\begin{proposition}
\label{prop.sol.lyndon.gr.bij.CFL-ex}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $n\in \NN $. Let $w\in\mathfrak{B}^{n}$. Let $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of the word $w$.
(This factorization exists and is unique, as we proved in Theorem
\ref{thm.words.CFL}.) Then, $\left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ is a well-defined element of $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.gr.bij.CFL-ex}.]We have
$w\in\mathfrak{B}^{n}\subset\mathfrak{B}^{\ast}$. Hence, each letter of $w$
belongs to $\mathfrak{B}$. Moreover, from $w\in\mathfrak{B}^{n}$, we obtain
$\ell\left(  w\right)  =n$.

Recall that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of the word $w$. According to Definition \ref{def.words.CFL},
this means that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of
Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}%
\geq\cdots\geq a_{k}$. Thus, in particular, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words. In other words, $a_{1}%
,a_{2},\ldots,a_{k}$ are Lyndon words.

Now, we claim the following:

\begin{statement}
\textit{Claim 1:} Let $i\in\left\{  1,2,\ldots,k\right\}  $. Then:

\textbf{(a)} The set $\left[  a_{i}\right]  $ is a well-defined aperiodic
$\mathfrak{B}$-necklace.

\textbf{(b)} We have $\left\vert \left[  a_{i}\right]  \right\vert
=\ell\left(  a_{i}\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} The word $a_{i}$ is a Lyndon word (since
$a_{1},a_{2},\ldots,a_{k}$ are Lyndon words), and thus nonempty (since any
Lyndon word is nonempty by definition). Thus, the necklace $\left[
a_{i}\right]  $ is well-defined.

From $w=a_{1}a_{2}\cdots a_{k}$, we conclude that each letter of $a_{i}$ is a
letter of $w$. Thus, each letter of $a_{i}$ belongs to $\mathfrak{B}$ (since
each letter of $w$ belongs to $\mathfrak{B}$). Hence, $a_{i}\in\mathfrak{B}%
^{\ast}$. Thus, Lemma \ref{lem.sol.lyndon.gr.bij.B-neck} (applied to $u=a_{i}%
$) shows that $\left[  a_{i}\right]  $ is a $\mathfrak{B}$-necklace.

Lemma \ref{lem.sol.lyndon.gr.bij.neck-laper} (applied to $u=a_{i}$) shows that
the necklace $\left[  a_{i}\right]  $ is aperiodic (since $a_{i}$ is a Lyndon
word). Thus, $\left[  a_{i}\right]  $ is an aperiodic $\mathfrak{B}$-necklace
(since we already know that $\left[  a_{i}\right]  $ is a $\mathfrak{B}%
$-necklace). This proves Claim 1 \textbf{(a)}.

\textbf{(b)} The word $a_{i}$ is nonempty and the necklace $\left[
a_{i}\right]  $ is aperiodic (as we know); thus, Lemma
\ref{lem.sol.lyndon.gr.bij.neck-aper} (applied to $u=a_{i}$) yields
$\left\vert \left[  a_{i}\right]  \right\vert =\ell\left(  a_{i}\right)  $.
This proves Claim 1 \textbf{(b)}.]

\begin{vershort}
Now, for each $i\in\left\{  1,2,\ldots,k\right\}  $, the set $\left[
a_{i}\right]  $ is a well-defined aperiodic $\mathfrak{B}$-necklace (by Claim
1 \textbf{(a)}). In other words, $\left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  $ are well-defined aperiodic $\mathfrak{B}%
$-necklaces. Thus, $\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces. Moreover,
(\ref{eq.def.sol.lyndon.gr.bij.sum.2}) (applied to $M=\left\{  \left[
a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ and $N_{i}=\left[  a_{i}\right]  $) yields%
\[
\operatorname*{sum}\left(  \left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\right)  =\left\vert \left[  a_{1}\right]
\right\vert +\left\vert \left[  a_{2}\right]  \right\vert +\cdots+\left\vert
\left[  a_{k}\right]  \right\vert =\sum_{i=1}^{k}\underbrace{\left\vert
\left[  a_{i}\right]  \right\vert }_{\substack{=\ell\left(  a_{i}\right)
\\\text{(by Claim 1 \textbf{(b)})}}}=\sum_{i=1}^{k}\ell\left(  a_{i}\right)
.
\]
Comparing this with%
\[
\ell\left(  \underbrace{w}_{=a_{1}a_{2}\cdots a_{k}}\right)  =\ell\left(
a_{1}a_{2}\cdots a_{k}\right)  =\ell\left(  a_{1}\right)  +\ell\left(
a_{2}\right)  +\cdots+\ell\left(  a_{k}\right)  =\sum_{i=1}^{k}\ell\left(
a_{i}\right)  ,
\]
we obtain $\operatorname*{sum}\left(  \left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\right)  =\ell\left(  w\right)  =n$. Hence,
$\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}}$ is a finite multiset $M$
of aperiodic $\mathfrak{B}$-necklaces satisfying $\operatorname*{sum}M=n$
(since we already know that $\left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ is a finite multiset of aperiodic $\mathfrak{B}%
$-necklaces). In other words, $\left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$
(since $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ was defined as the set
of all such multisets $M$).
\end{vershort}

\begin{verlong}
Now, for each $i\in\left\{  1,2,\ldots,k\right\}  $, the set $\left[
a_{i}\right]  $ is a well-defined aperiodic $\mathfrak{B}$-necklace (by Claim
1 \textbf{(a)}). In other words, $\left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  $ are well-defined aperiodic $\mathfrak{B}%
$-necklaces. Thus, $\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}$ is a
well-defined finite multiset of aperiodic $\mathfrak{B}$-necklaces. Moreover,
(\ref{eq.def.sol.lyndon.gr.bij.sum.2}) (applied to $M=\left\{  \left[
a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ and $N_{i}=\left[  a_{i}\right]  $) yields%
\begin{align*}
\operatorname*{sum}\left(  \left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\right)   &  =\left\vert \left[  a_{1}\right]
\right\vert +\left\vert \left[  a_{2}\right]  \right\vert +\cdots+\left\vert
\left[  a_{k}\right]  \right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\{  \left[  a_{1}\right]
,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}=\left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\right) \\
&  =\sum_{i=1}^{k}\underbrace{\left\vert \left[  a_{i}\right]  \right\vert
}_{\substack{=\ell\left(  a_{i}\right)  \\\text{(by Claim 1 \textbf{(b)})}%
}}=\sum_{i=1}^{k}\ell\left(  a_{i}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
n  &  =\ell\left(  \underbrace{w}_{=a_{1}a_{2}\cdots a_{k}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  w\right)  =n\right) \\
&  =\ell\left(  a_{1}a_{2}\cdots a_{k}\right)  =\ell\left(  a_{1}\right)
+\ell\left(  a_{2}\right)  +\cdots+\ell\left(  a_{k}\right)  =\sum_{i=1}%
^{k}\ell\left(  a_{i}\right)  ,
\end{align*}
we obtain $\operatorname*{sum}\left(  \left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\right)  =n$. Hence, $\left\{  \left[
a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ is a finite multiset $M$ of aperiodic
$\mathfrak{B}$-necklaces satisfying $\operatorname*{sum}M=n$ (since we already
know that $\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces). In other words,
$\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}}\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ (since $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ was defined as the set of all finite multisets $M$ of
aperiodic $\mathfrak{B}$-necklaces satisfying $\operatorname*{sum}M=n$).
\end{verlong}

Thus, we have showed that $\left\{  \left[  a_{1}\right]  ,\left[
a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ is a well-defined element of $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$. This proves Proposition
\ref{prop.sol.lyndon.gr.bij.CFL-ex}.
\end{proof}

\begin{definition}
\label{def.sol.lyndon.gr.bij.CFL}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $n\in \NN $. Then, we define a map
$\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ as follows: For each $w\in\mathfrak{B}^{n}$, we set
$\operatorname*{CFL}w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}$, where
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of the
word $w$. (This is well-defined, since Proposition
\ref{prop.sol.lyndon.gr.bij.CFL-ex} shows that $\left\{  \left[  a_{1}\right]
,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$ is a well-defined element of $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$.)
\end{definition}

\begin{example}
If $\mathfrak{B}=\left\{  2<4<6<8\right\}  $ and $n=9$ and $w=262642424$, then
the map $\operatorname*{CFL}$ sends the word $w$ to
$\operatorname*{CFL}w=\left\{  \left[  26264\right]  ,\left[  24\right]
,\left[  24\right]  \right\}  _{\operatorname*{multiset}}$, since the CFL
factorization of $w$ is $\left(  26264,24,24\right)  $.
\end{example}

Recall that $\mathfrak{L}$ denotes the set of all Lyndon words, whereas
$\mathfrak{N}^{\mathfrak{a}}$ denotes the set of all aperiodic necklaces.

\begin{definition}
We define a map $\operatorname*{lynd}:\mathfrak{N}^{\mathfrak{a}}%
\rightarrow\mathfrak{L}$ as follows:

Let $N\in\mathfrak{N}^{\mathfrak{a}}$. Thus, $N$ is an aperiodic necklace.
Hence, $N$ is a necklace. In other words, $N$ is an $n$-necklace for some
positive integer $n$ (by the definition of a ``necklace''). Consider this
$n$. Hence, $N$ is an aperiodic
$n$-necklace. Thus, Exercise \ref{exe.words.necklaces}(c) shows that $N$
contains exactly one Lyndon word. We define $\operatorname*{lynd}N$ to be this
Lyndon word. (Thus, $\operatorname*{lynd}N\in\mathfrak{L}$; hence, the map
$\operatorname*{lynd}:\mathfrak{N}^{\mathfrak{a}}\rightarrow\mathfrak{L}$ is well-defined.)
\end{definition}

\begin{example}
If $\mathfrak{A}=\left\{  1<2<3<4<\cdots\right\}  $ and $N=\left[
4121312\right]  $, then $\operatorname*{lynd}N=1213124$.
\end{example}

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.lynd-B}Let $N$ be an aperiodic necklace. Then:

\begin{enumerate}
\item[(a)] We have $\ell\left(  \operatorname*{lynd}N\right)  =\left\vert
N\right\vert $.

\item[(b)] We have $\left[  \operatorname*{lynd}N\right]  =N$.

\item[(c)] If $u$ is a Lyndon word such that $u\in N$, then
$u=\operatorname*{lynd}N$.

\item[(d)] Let $\mathfrak{B}$ be a subset of $\mathfrak{A}$. Assume that $N$
is a $\mathfrak{B}$-necklace. Then, $\operatorname*{lynd}N\in\mathfrak{B}%
^{\ast}$.
\end{enumerate}
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}.]The definition of
$\operatorname*{lynd}N$ yields that $\operatorname*{lynd}N$ is the unique
Lyndon word that $N$ contains. In other words, $\operatorname*{lynd}N$ is the
unique Lyndon word $w$ such that $w\in N$. Hence, $\operatorname*{lynd}N$ is a
Lyndon word and satisfies $\operatorname*{lynd}N\in N$. Hence, Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts} (applied to $u=\operatorname*{lynd}N$)
yields $N=\left[  \operatorname*{lynd}N\right]  $. This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-B}(b).

(a) The word $\operatorname*{lynd}N$ is a Lyndon word, and thus is nonempty
(since any Lyndon word is nonempty). Moreover, the necklace $N$ is aperiodic.
Since $N=\left[  \operatorname*{lynd}N\right]  $, this rewrites as follows:
The necklace $\left[  \operatorname*{lynd}N\right]  $ is aperiodic. Hence,
Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper} (applied to
$u=\operatorname*{lynd}N$) yields $\left\vert \left[  \operatorname*{lynd}%
N\right]  \right\vert =\ell\left(  \operatorname*{lynd}N\right)  $. Since
$N=\left[  \operatorname*{lynd}N\right]  $, we now have $\left\vert
N\right\vert =\left\vert \left[  \operatorname*{lynd}N\right]  \right\vert
=\ell\left(  \operatorname*{lynd}N\right)  $. This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-B}(a).

(c) Recall that $\operatorname*{lynd}N$ is the unique Lyndon word $w$ such
that $w\in N$. Thus, $\operatorname*{lynd}N$ is the only such Lyndon word $w$.
In other words, if $w$ is a Lyndon word such that $w\in N$, then
$w=\operatorname*{lynd}N$. Renaming $w$ as $u$ in this statement, we obtain
precisely the claim of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(c).

(d) We know that $N$ is a $\mathfrak{B}$-necklace. In other words, $N$ is a
necklace that is a subset of $\mathfrak{B}^{\ast}$ (because this is what
``$\mathfrak{B}$-necklace'' means). Thus, $N$
is a subset of $\mathfrak{B}^{\ast}$. In other words, $N\subset\mathfrak{B}%
^{\ast}$. Thus, $\operatorname*{lynd}N\in N\subset\mathfrak{B}^{\ast}$. This
proves Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(d).
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}.]Recall that $N$ is an
aperiodic necklace. In other words, $N\in\mathfrak{N}^{\mathfrak{a}}$ (since
$\mathfrak{N}^{\mathfrak{a}}$ is the set of all aperiodic necklaces). Thus,
the definition of $\operatorname*{lynd}N$ yields that $\operatorname*{lynd}N$
is the unique Lyndon word that $N$ contains. In other words,
$\operatorname*{lynd}N$ is the unique Lyndon word $w$ such that $w\in N$.
Hence, $\operatorname*{lynd}N$ is a Lyndon word $w$ such that $w\in N$. In
other words, $\operatorname*{lynd}N$ is a Lyndon word and satisfies
$\operatorname*{lynd}N\in N$. Hence, Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts} (applied to $u=\operatorname*{lynd}N$)
yields $N=\left[  \operatorname*{lynd}N\right]  $. In other words, $\left[
\operatorname*{lynd}N\right]  =N$. This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-B}(b).

(a) The word $\operatorname*{lynd}N$ is a Lyndon word, and thus is nonempty
(since any Lyndon word is nonempty by definition). Moreover, the necklace $N$
is aperiodic. Since $N=\left[  \operatorname*{lynd}N\right]  $, this rewrites
as follows: The necklace $\left[  \operatorname*{lynd}N\right]  $ is
aperiodic. Hence, Lemma \ref{lem.sol.lyndon.gr.bij.neck-aper} (applied to
$u=\operatorname*{lynd}N$) yields $\left\vert \left[  \operatorname*{lynd}%
N\right]  \right\vert =\ell\left(  \operatorname*{lynd}N\right)  $. Hence,
$\ell\left(  \operatorname*{lynd}N\right)  =\left\vert \left[
\operatorname*{lynd}N\right]  \right\vert =\left\vert N\right\vert $ (since
$\left[  \operatorname*{lynd}N\right]  =N$). This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-B}(a).

(c) Recall that $\operatorname*{lynd}N$ is the unique Lyndon word $w$ such
that $w\in N$. Thus, $\operatorname*{lynd}N$ is the only such Lyndon word $w$.
In other words, if $w$ is a Lyndon word such that $w\in N$, then
$w=\operatorname*{lynd}N$. Renaming $w$ as $u$ in this statement, we obtain
the following: If $u$ is a Lyndon word such that $u\in N$, then
$u=\operatorname*{lynd}N$. This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-B}(c).

(d) We know that $N$ is a $\mathfrak{B}$-necklace. In other words, $N$ is a
necklace that is a subset of $\mathfrak{B}^{\ast}$ (because this is what
``$\mathfrak{B}$-necklace'' means). Thus, $N$
is a subset of $\mathfrak{B}^{\ast}$. In other words, $N\subset\mathfrak{B}%
^{\ast}$. Thus, $\operatorname*{lynd}N\in N\subset\mathfrak{B}^{\ast}$. This
proves Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(d).
\end{proof}

\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.lynd-C}Let $u$ be a Lyndon word. Then,
$\operatorname*{lynd}\left[  u\right]  =u$.
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-C}.]The necklace $\left[
u\right]  $ is aperiodic (by Lemma \ref{lem.sol.lyndon.gr.bij.neck-laper}).
Hence, $\operatorname*{lynd}\left[  u\right]  $ is well-defined. Moreover, the
word $u$ is Lyndon and thus nonempty (since every Lyndon word is nonempty).
Thus, Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields $u\in\left[
u\right]  $. Hence, Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(c) (applied to
$N=\left[  u\right]  $) yields $u=\operatorname*{lynd}\left[  u\right]  $.
This proves Lemma \ref{lem.sol.lyndon.gr.bij.lynd-C}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-C}.]The necklace $\left[
u\right]  $ is aperiodic (by Lemma \ref{lem.sol.lyndon.gr.bij.neck-laper}). In
other words, $\left[  u\right]  \in\mathfrak{N}^{\mathfrak{a}}$ (since
$\mathfrak{N}^{\mathfrak{a}}$ is the set of all aperiodic necklaces). Hence,
$\operatorname*{lynd}\left[  u\right]  $ is well-defined. Moreover, the word
$u$ is Lyndon and thus nonempty (since every Lyndon word is nonempty by
definition). Thus, Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} yields
$u\in\left[  u\right]  $. Hence, Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(c)
(applied to $N=\left[  u\right]  $) yields $u=\operatorname*{lynd}\left[
u\right]  $. This proves Lemma \ref{lem.sol.lyndon.gr.bij.lynd-C}.
\end{proof}

\end{verlong}

The next definition introduces a notation for a simple and very basic concept:
the concept of applying a map $f:X\rightarrow Y$ to a finite multiset $M$ of
elements of $X$ (by applying $f$ to each element of $M$).

\begin{definition}
\label{def.sol.lyndon.gr.bij.fstarMultiset}Let $X$ and $Y$ be two sets, and
let $f:X\rightarrow Y$ be any map. Let $M$ be a finite multiset of elements of
$X$. Then, $f_{\ast}M$ shall denote the finite multiset of elements of $Y$
obtained by applying $f$ to each element of $M$. More formally, this multiset
$f_{\ast}M$ can be defined by the requirement that for any object $y$, we have%
\[
\left(  \text{multiplicity of }y\text{ in }f_{\ast}M\right)  =\sum
\limits_{\substack{x\in X;\\f\left(  x\right)  =y}}\left(  \text{multiplicity
of }x\text{ in }M\right)  .
\]


More explicitly, $f_{\ast}M$ can be described as follows: If $M=\left\{
m_{1},m_{2},\ldots,m_{k}\right\}  _{\operatorname*{multiset}}$, then%
\begin{equation}
f_{\ast}M=\left\{  f\left(  m_{1}\right)  ,f\left(  m_{2}\right)
,\ldots,f\left(  m_{k}\right)  \right\}  _{\operatorname*{multiset}}.
\label{eq.def.sol.lyndon.gr.bij.fstarMultiset.2}%
\end{equation}

\end{definition}

\begin{example}
Let $X= \ZZ $ and $Y= \NN $, and let $f:X\rightarrow Y$ be the map
that sends each $x\in \ZZ $ to $x^{2}\in \NN $. Then,
(\ref{eq.def.sol.lyndon.gr.bij.fstarMultiset.2}) yields $f_{\ast}\left(
\left\{  -2,2,3,3\right\}  _{\operatorname*{multiset}}\right)  =\left\{
f\left(  -2\right)  ,f\left(  2\right)  ,f\left(  3\right)  ,f\left(
3\right)  \right\}  _{\operatorname*{multiset}}=\left\{  4,4,9,9\right\}
_{\operatorname*{multiset}}$.
\end{example}

\begin{example}
Let $M$ be the finite multiset $\left\{  \left[  13\right]  ,\left[
13\right]  ,\left[  4252\right]  \right\}  _{\operatorname*{multiset}}$ of
aperiodic necklaces. Then, (\ref{eq.def.sol.lyndon.gr.bij.fstarMultiset.2})
yields%
\[
\operatorname*{lynd}\nolimits_{\ast}M=\left\{  \operatorname*{lynd}\left[
13\right]  ,\operatorname*{lynd}\left[  13\right]  ,\operatorname*{lynd}%
\left[  4252\right]  \right\}  _{\operatorname*{multiset}}=\left\{
13,13,2425\right\}  _{\operatorname*{multiset}}.
\]

\end{example}

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.lynd-back}Let $M$ be a finite multiset of
aperiodic necklaces. Let $\widetilde{M}$ be the multiset $\operatorname*{lynd}%
\nolimits_{\ast}M$. Then:

\begin{enumerate}
\item[(a)] The multiset $\widetilde{M}$ is a finite multiset of Lyndon words.
\end{enumerate}

Now, let $b_{1},b_{2},\ldots,b_{k}$ be some Lyndon words such that
$\widetilde{M}=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}
_{\operatorname*{multiset}}$. Then:

\begin{enumerate}
\item[(b)] We have $M=\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]
,\ldots,\left[  b_{k}\right]  \right\}  _{\operatorname*{multiset}}$.

\item[(c)] We have $\operatorname*{sum}M=\ell\left(  b_{1}\right)
+\ell\left(  b_{2}\right)  +\cdots+\ell\left(  b_{k}\right)  $.
\end{enumerate}
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}.]We know that $M$ is a
finite multiset of aperiodic necklaces, i.e., a finite multiset of elements of
$\mathfrak{N}^{\mathfrak{a}}$. Hence, $\operatorname*{lynd}\nolimits_{\ast}M$
is well-defined and is a finite multiset of elements of $\mathfrak{L}$ (since
$\operatorname*{lynd}$ is a map from $\mathfrak{N}^{\mathfrak{a}}$ to
$\mathfrak{L}$). In other words, $\operatorname*{lynd}\nolimits_{\ast}M$ is
well-defined and is a finite multiset of elements of Lyndon words (since the
elements of $\mathfrak{L}$ are the Lyndon words). In other words,
$\widetilde{M}$ is well-defined and is a finite multiset of elements of Lyndon
words (because $\widetilde{M}=\operatorname*{lynd}\nolimits_{\ast}M$). This
proves Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(a).

Write the multiset $M$ in the form $M=\left\{  N_{1},N_{2},\ldots
,N_{i}\right\}  _{\operatorname*{multiset}}$ for some aperiodic necklaces
$N_{1},N_{2},\ldots,N_{i}$. (This can be done, since $M$ is a finite multiset
of aperiodic necklaces.)

From $M=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}  _{\operatorname*{multiset}%
}$, we obtain $\operatorname*{lynd}\nolimits_{\ast}M=\left\{
\operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}$ (by
(\ref{eq.def.sol.lyndon.gr.bij.fstarMultiset.2}), applied to $\mathfrak{N}%
^{\mathfrak{a}}$, $\mathfrak{L}$, $\operatorname*{lynd}$, $i$ and $N_{j}$
instead of $X$, $Y$, $f$, $k$ and $m_{j}$). Hence,%
\begin{equation}
\widetilde{M}=\operatorname*{lynd}\nolimits_{\ast}M=\left\{
\operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}.
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.short.2}%
\end{equation}


(b) For each $r\in\left\{  1,2,\ldots,i\right\}  $, the set $N_{r}$ is an
aperiodic necklace (since $N_{1},N_{2},\ldots,N_{i}$ are aperiodic necklaces)
and therefore satisfies $\left[  \operatorname*{lynd}N_{r}\right]  =N_{r}$ (by
Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(b), applied to $N=N_{r}$). In other
words,%
\begin{equation}
\left(  \left[  \operatorname*{lynd}N_{1}\right]  ,\left[
\operatorname*{lynd}N_{2}\right]  ,\ldots,\left[  \operatorname*{lynd}%
N_{i}\right]  \right)  =\left(  N_{1},N_{2},\ldots,N_{i}\right)  .
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.short.b2}%
\end{equation}


But comparing (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.short.2}) with
$\widetilde{M}=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}
_{\operatorname*{multiset}}$, we obtain%
\[
\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}%
}=\left\{  \operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}.
\]
Thus, the $k$-tuple $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is a
permutation of the $i$-tuple $\left(  \operatorname*{lynd}N_{1}%
,\operatorname*{lynd}N_{2},\ldots,\operatorname*{lynd}N_{i}\right)  $. Hence,%
\[
\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[
b_{k}\right]  \right\}  _{\operatorname*{multiset}}=\left\{  \left[
\operatorname*{lynd}N_{1}\right]  ,\left[  \operatorname*{lynd}N_{2}\right]
,\ldots,\left[  \operatorname*{lynd}N_{i}\right]  \right\}
_{\operatorname*{multiset}}=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}
_{\operatorname*{multiset}}%
\]
(by (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.short.b2})). Comparing this with
$M=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}  _{\operatorname*{multiset}}$, we
obtain $M=\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots
,\left[  b_{k}\right]  \right\}  _{\operatorname*{multiset}}$. This proves
Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b).

(c) For each $r\in\left\{  1,2,\ldots,i\right\}  $, the set $N_{r}$ is an
aperiodic necklace (since $N_{1},N_{2},\ldots,N_{i}$ are aperiodic necklaces)
and therefore satisfies
\begin{equation}
\ell\left(  \operatorname*{lynd}N_{r}\right)  =\left\vert N_{r}\right\vert
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.short.c1}%
\end{equation}
(by Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(a), applied to $N=N_{r}$).

In our above proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b), we have
showed that the $k$-tuple $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is a
permutation of the $i$-tuple $\left(  \operatorname*{lynd}N_{1}%
,\operatorname*{lynd}N_{2},\ldots,\operatorname*{lynd}N_{i}\right)  $. Hence,%
\begin{align*}
\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(
b_{k}\right)   &  =\ell\left(  \operatorname*{lynd}N_{1}\right)  +\ell\left(
\operatorname*{lynd}N_{2}\right)  +\cdots+\ell\left(  \operatorname*{lynd}%
N_{i}\right) \\
&  =\sum_{r=1}^{i}\underbrace{\ell\left(  \operatorname*{lynd}N_{r}\right)
}_{\substack{=\left\vert N_{r}\right\vert \\\text{(by
(\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.short.c1}))}}}=\sum_{r=1}%
^{i}\left\vert N_{r}\right\vert =\left\vert N_{1}\right\vert +\left\vert
N_{2}\right\vert +\cdots+\left\vert N_{i}\right\vert .
\end{align*}
But (\ref{eq.def.sol.lyndon.gr.bij.sum.2}) (applied to $i$ instead of $k$)
yields%
\[
\operatorname*{sum}M=\left\vert N_{1}\right\vert +\left\vert N_{2}\right\vert
+\cdots+\left\vert N_{i}\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}M=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}  _{\operatorname*{multiset}%
}\right)  .
\]
Comparing these two equalities, we find $\operatorname*{sum}M=\ell\left(
b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(  b_{k}\right)  $.
This proves Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(c).
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}.]The aperiodic necklaces
are exactly the elements of $\mathfrak{N}^{\mathfrak{a}}$ (since
$\mathfrak{N}^{\mathfrak{a}}$ is the set of all aperiodic necklaces). But $M$
is a finite multiset of aperiodic necklaces. In other words, $M$ is a finite
multiset of elements of $\mathfrak{N}^{\mathfrak{a}}$ (since the aperiodic
necklaces are exactly the elements of $\mathfrak{N}^{\mathfrak{a}}$). Hence,
$\operatorname*{lynd}\nolimits_{\ast}M$ is well-defined.

Write the multiset $M$ in the form $M=\left\{  N_{1},N_{2},\ldots
,N_{i}\right\}  _{\operatorname*{multiset}}$ for some aperiodic necklaces
$N_{1},N_{2},\ldots,N_{i}$. (This can be done, since $M$ is a finite multiset
of aperiodic necklaces.)

We have $M=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}
_{\operatorname*{multiset}}$. Thus, (\ref{eq.def.sol.lyndon.gr.bij.sum.2})
(applied to $i$ instead of $k$) yields%
\begin{equation}
\operatorname*{sum}M=\left\vert N_{1}\right\vert +\left\vert N_{2}\right\vert
+\cdots+\left\vert N_{i}\right\vert .
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.1}%
\end{equation}


Moreover, from $M=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}
_{\operatorname*{multiset}}$, we obtain $\operatorname*{lynd}\nolimits_{\ast
}M=\left\{  \operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}$ (by
(\ref{eq.def.sol.lyndon.gr.bij.fstarMultiset.2}), applied to $\mathfrak{N}%
^{\mathfrak{a}}$, $\mathfrak{L}$, $\operatorname*{lynd}$, $i$ and $N_{j}$
instead of $X$, $Y$, $f$, $k$ and $m_{j}$). Hence,%
\begin{equation}
\widetilde{M}=\operatorname*{lynd}\nolimits_{\ast}M=\left\{
\operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}.
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.2}%
\end{equation}


(a) Clearly, $\operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2}%
,\ldots,\operatorname*{lynd}N_{i}$ are elements of $\mathfrak{L}$ (since
$\operatorname*{lynd}$ is a map from $\mathfrak{N}^{\mathfrak{a}}$ to
$\mathfrak{L}$). In other words, $\operatorname*{lynd}N_{1}%
,\operatorname*{lynd}N_{2},\ldots,\operatorname*{lynd}N_{i}$ are Lyndon words
(since $\mathfrak{L}$ is the set of all Lyndon words). Thus, $\left\{
\operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}$ is a finite
multiset of Lyndon words. In other words, $\widetilde{M}$ is a finite multiset
of Lyndon words (because $\widetilde{M}=\left\{  \operatorname*{lynd}%
N_{1},\operatorname*{lynd}N_{2},\ldots,\operatorname*{lynd}N_{i}\right\}
_{\operatorname*{multiset}}$). This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(a).

(b) We have%
\begin{equation}
\left[  \operatorname*{lynd}N_{r}\right]  =N_{r}\ \ \ \ \ \ \ \ \ \ \text{for
each }r\in\left\{  1,2,\ldots,i\right\}  .
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.b1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.b1}):} Let
$r\in\left\{  1,2,\ldots,i\right\}  $. The set $N_{r}$ is an aperiodic
necklace (since $N_{1},N_{2},\ldots,N_{i}$ are aperiodic necklaces). Thus,
Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(b) (applied to $N=N_{r}$) yields
$\left[  \operatorname*{lynd}N_{r}\right]  =N_{r}$. This proves
(\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.b1}).]

We thus have proved that $\left[  \operatorname*{lynd}N_{r}\right]  =N_{r}$
for each $r\in\left\{  1,2,\ldots,i\right\}  $. In other words,%
\begin{equation}
\left(  \left[  \operatorname*{lynd}N_{1}\right]  ,\left[
\operatorname*{lynd}N_{2}\right]  ,\ldots,\left[  \operatorname*{lynd}%
N_{i}\right]  \right)  =\left(  N_{1},N_{2},\ldots,N_{i}\right)  .
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.b2}%
\end{equation}


But comparing (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.2}) with $\widetilde{M}%
=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$, we
obtain%
\[
\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}%
}=\left\{  \operatorname*{lynd}N_{1},\operatorname*{lynd}N_{2},\ldots
,\operatorname*{lynd}N_{i}\right\}  _{\operatorname*{multiset}}.
\]
Thus, the $k$-tuple $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is a
permutation of the $i$-tuple $\left(  \operatorname*{lynd}N_{1}%
,\operatorname*{lynd}N_{2},\ldots,\operatorname*{lynd}N_{i}\right)  $. Hence,%
\[
\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[
b_{k}\right]  \right\}  _{\operatorname*{multiset}}=\left\{  \left[
\operatorname*{lynd}N_{1}\right]  ,\left[  \operatorname*{lynd}N_{2}\right]
,\ldots,\left[  \operatorname*{lynd}N_{i}\right]  \right\}
_{\operatorname*{multiset}}=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}
_{\operatorname*{multiset}}%
\]
(by (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.b2})). Comparing this with
$M=\left\{  N_{1},N_{2},\ldots,N_{i}\right\}  _{\operatorname*{multiset}}$, we
obtain $M=\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots
,\left[  b_{k}\right]  \right\}  _{\operatorname*{multiset}}$. This proves
Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b).

(c) We have%
\begin{equation}
\ell\left(  \operatorname*{lynd}N_{r}\right)  =\left\vert N_{r}\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for each }r\in\left\{  1,2,\ldots,i\right\}  .
\label{pf.lem.sol.lyndon.gr.bij.lynd-back.5}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.5}):} Let
$r\in\left\{  1,2,\ldots,i\right\}  $. The set $N_{r}$ is an aperiodic
necklace (since $N_{1},N_{2},\ldots,N_{i}$ are aperiodic necklaces). Thus,
Lemma \ref{lem.sol.lyndon.gr.bij.lynd-B}(a) (applied to $N=N_{r}$) yields
$\ell\left(  \operatorname*{lynd}N_{r}\right)  =\left\vert N_{r}\right\vert $.
This proves (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.5}).]

In our above proof of Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b), we have
showed that the $k$-tuple $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is a
permutation of the $i$-tuple $\left(  \operatorname*{lynd}N_{1}%
,\operatorname*{lynd}N_{2},\ldots,\operatorname*{lynd}N_{i}\right)  $. Hence,%
\begin{align*}
\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(
b_{k}\right)   &  =\ell\left(  \operatorname*{lynd}N_{1}\right)  +\ell\left(
\operatorname*{lynd}N_{2}\right)  +\cdots+\ell\left(  \operatorname*{lynd}%
N_{i}\right) \\
&  =\sum_{r=1}^{i}\underbrace{\ell\left(  \operatorname*{lynd}N_{r}\right)
}_{\substack{=\left\vert N_{r}\right\vert \\\text{(by
(\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.5}))}}}=\sum_{r=1}^{i}\left\vert
N_{r}\right\vert =\left\vert N_{1}\right\vert +\left\vert N_{2}\right\vert
+\cdots+\left\vert N_{i}\right\vert .
\end{align*}
Comparing this with (\ref{pf.lem.sol.lyndon.gr.bij.lynd-back.1}), we find
$\operatorname*{sum}M=\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)
+\cdots+\ell\left(  b_{k}\right)  $. This proves Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(c).
\end{proof}

\end{verlong}

\begin{proposition}
\label{prop.sol.lyndon.gr.bij.LFC-ex}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $n\in \NN $. Let $M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$. Then, $M$ is a finite multiset of aperiodic
$\mathfrak{B}$-necklaces. Let $\widetilde{M}$ be the multiset
$\operatorname*{lynd}\nolimits_{\ast}M$. This $\widetilde{M}$ is a finite
multiset of Lyndon words (by Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(a)).
Let $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ be the $\leq$-increasing list
of $\widetilde{M}$. (Here we are using the lexicographic order $\leq$, not the
relation $\leq_{\omega}$.)

Then, $b_{k}b_{k-1}\cdots b_{1}\in\mathfrak{B}^{n}$.
\end{proposition}

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.gr.bij.LFC-ex}.]We have
$M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $M$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$ (since $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
}$ was defined to be the set of all such multisets). Hence, in particular, $M$
is a finite multiset of aperiodic necklaces (since each $\mathfrak{B}%
$-necklace is a necklace). Thus, $\operatorname*{lynd}\nolimits_{\ast}M$ is well-defined.

The list $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is the $\leq$-increasing
list of $\widetilde{M}$ (by its definition). Hence, $\widetilde{M}=\left\{
b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$. Thus,
$b_{1},b_{2},\ldots,b_{k}$ are elements of $\widetilde{M}$, and therefore are
Lyndon words (since $\widetilde{M}$ is a multiset of Lyndon words). Thus,
Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b) yields $M=\left\{  \left[
b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[  b_{k}\right]  \right\}
_{\operatorname*{multiset}}$.

Now, let $i\in\left\{  1,2,\ldots,k\right\}  $. Then, $\left[  b_{i}\right]  $
is an element of $M$ (since $M=\left\{  \left[  b_{1}\right]  ,\left[
b_{2}\right]  ,\ldots,\left[  b_{k}\right]  \right\}
_{\operatorname*{multiset}}$). Thus, $\left[  b_{i}\right]  $ is an aperiodic
$\mathfrak{B}$-necklace (since $M$ is a finite multiset of aperiodic
$\mathfrak{B}$-necklaces), hence a $\mathfrak{B}$-necklace, and thus a subset
of $\mathfrak{B}^{\ast}$ (by the definition of a $\mathfrak{B}$-necklace). In
other words, $\left[  b_{i}\right]  \subset\mathfrak{B}^{\ast}$. But $b_{i}$
is a Lyndon word (since $b_{1},b_{2},\ldots,b_{k}$ are Lyndon words), and thus
a nonempty word (since any Lyndon word is nonempty). Hence, Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts0} (applied to $u=b_{i}$) yields $b_{i}%
\in\left[  b_{i}\right]  \subset\mathfrak{B}^{\ast}$.

Forget that we fixed $i$. We thus have shown that $b_{i}\in\mathfrak{B}^{\ast
}$ for each $i\in\left\{  1,2,\ldots,k\right\}  $. Hence, $b_{k}b_{k-1}\cdots
b_{1}\in\mathfrak{B}^{\ast}$.

But Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(c) yields $\operatorname*{sum}%
M=\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(
b_{k}\right)  $. Comparing this with $\operatorname*{sum}M=n$, we obtain
\[
\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(
b_{k}\right)  =n.
\]


Now, combining $b_{k}b_{k-1}\cdots b_{1}\in\mathfrak{B}^{\ast}$ with
\[
\ell\left(  b_{k}b_{k-1}\cdots b_{1}\right)  =\ell\left(  b_{k}\right)
+\ell\left(  b_{k-1}\right)  +\cdots+\ell\left(  b_{1}\right)  =\ell\left(
b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(  b_{k}\right)
=n,
\]
we obtain $b_{k}b_{k-1}\cdots b_{1}\in\mathfrak{B}^{n}$. This proves
Proposition \ref{prop.sol.lyndon.gr.bij.LFC-ex}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.gr.bij.LFC-ex}.]We have
$M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $M$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$ (since $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
}$ was defined to be the set of all such multisets). Hence, in particular, $M$
is a finite multiset of aperiodic $\mathfrak{B}$-necklaces. Therefore, $M$ is
a finite multiset of aperiodic necklaces (since each $\mathfrak{B}$-necklace
is a necklace). In other words, $M$ is a finite multiset of elements of
$\mathfrak{N}^{\mathfrak{a}}$ (since the aperiodic necklaces are exactly the
elements of $\mathfrak{N}^{\mathfrak{a}}$). Hence, $\operatorname*{lynd}%
\nolimits_{\ast}M$ is well-defined.

The list $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is the $\leq$-increasing
list of $\widetilde{M}$ (by its definition). In other words, $\widetilde{M}%
=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$ and
$b_{1}\leq b_{2}\leq\cdots\leq b_{k}$ (by the definition of an
``$\leq$-increasing list'' of a multiset).

We have $\widetilde{M}=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}
_{\operatorname*{multiset}}$. Thus, $b_{1},b_{2},\ldots,b_{k}$ are elements of
$\widetilde{M}$. Hence, $b_{1},b_{2},\ldots,b_{k}$ are Lyndon words (since all
elements of $\widetilde{M}$ are Lyndon words\footnote{because $\widetilde{M}$
is a multiset of Lyndon words}). Thus, Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(b) yields $M=\left\{  \left[
b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[  b_{k}\right]  \right\}
_{\operatorname*{multiset}}$ (since $\widetilde{M}=\left\{  b_{1},b_{2}%
,\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$).

Recall that a $\mathfrak{B}$-necklace is the same as a necklace that is a
subset of $\mathfrak{B}^{\ast}$ (indeed, this is how $\mathfrak{B}$-necklaces
are defined). Thus, each $\mathfrak{B}$-necklace is a subset of $\mathfrak{B}%
^{\ast}$. But each element of $M$ is an aperiodic $\mathfrak{B}$-necklace
(since $M$ is a finite multiset of aperiodic $\mathfrak{B}$-necklaces).

Now, let $i\in\left\{  1,2,\ldots,k\right\}  $. Then, $\left[  b_{i}\right]  $
is an element of $\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]
,\ldots,\left[  b_{k}\right]  \right\}  _{\operatorname*{multiset}}$. In other
words, $\left[  b_{i}\right]  $ is an element of $M$ (since $M=\left\{
\left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[  b_{k}\right]
\right\}  _{\operatorname*{multiset}}$). Thus, $\left[  b_{i}\right]  $ is an
aperiodic $\mathfrak{B}$-necklace (since each element of $M$ is an aperiodic
$\mathfrak{B}$-necklace), hence a $\mathfrak{B}$-necklace, and thus a subset
of $\mathfrak{B}^{\ast}$ (since each $\mathfrak{B}$-necklace is a subset of
$\mathfrak{B}^{\ast}$). In other words, $\left[  b_{i}\right]  \subset
\mathfrak{B}^{\ast}$. But $b_{i}$ is a Lyndon word (since $b_{1},b_{2}%
,\ldots,b_{k}$ are Lyndon words), and thus a nonempty word (since any Lyndon
word is nonempty by definition). Hence, Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts0} (applied to $u=b_{i}$) yields $b_{i}%
\in\left[  b_{i}\right]  \subset\mathfrak{B}^{\ast}$.

Forget that we fixed $i$. We thus have shown that $b_{i}\in\mathfrak{B}^{\ast
}$ for each $i\in\left\{  1,2,\ldots,k\right\}  $. In other words, all $k$
words $b_{1},b_{2},\ldots,b_{k}$ belong to $\mathfrak{B}^{\ast}$. In other
words, all $k$ words $b_{k},b_{k-1},\ldots,b_{1}$ belong to $\mathfrak{B}%
^{\ast}$. Hence, the concatenation $b_{k}b_{k-1}\cdots b_{1}$ of these $k$
words must belong to $\mathfrak{B}^{\ast}$ as well. In other words,
$b_{k}b_{k-1}\cdots b_{1}\in\mathfrak{B}^{\ast}$.

But Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(c) yields $\operatorname*{sum}%
M=\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(
b_{k}\right)  $. Comparing this with $\operatorname*{sum}M=n$, we obtain
\[
\ell\left(  b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(
b_{k}\right)  =n.
\]


Now, combining $b_{k}b_{k-1}\cdots b_{1}\in\mathfrak{B}^{\ast}$ with
\[
\ell\left(  b_{k}b_{k-1}\cdots b_{1}\right)  =\ell\left(  b_{k}\right)
+\ell\left(  b_{k-1}\right)  +\cdots+\ell\left(  b_{1}\right)  =\ell\left(
b_{1}\right)  +\ell\left(  b_{2}\right)  +\cdots+\ell\left(  b_{k}\right)
=n,
\]
we obtain $b_{k}b_{k-1}\cdots b_{1}\in\mathfrak{B}^{n}$. This proves
Proposition \ref{prop.sol.lyndon.gr.bij.LFC-ex}.
\end{proof}

\end{verlong}

\begin{definition}
\label{def.sol.lyndon.gr.bij.LFC}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $n\in \NN $. Then, we define a map
$\operatorname*{LFC}:\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
\rightarrow\mathfrak{B}^{n}$ as follows: Let $M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$. Then, $M$ is a finite multiset of aperiodic
$\mathfrak{B}$-necklaces. Let $\widetilde{M}$ be the multiset
$\operatorname*{lynd}\nolimits_{\ast}M$. This $\widetilde{M}$ is a finite
multiset of Lyndon words (by Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(a)).
Let $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ be the $\leq$-increasing list
of $\widetilde{M}$. Then, we set $\operatorname*{LFC}\left(  M\right)
=b_{k}b_{k-1}\cdots b_{1}$. (This is well-defined, since Proposition
\ref{prop.sol.lyndon.gr.bij.LFC-ex} shows that $b_{k}b_{k-1}\cdots b_{1}%
\in\mathfrak{B}^{n}$.)
\end{definition}

\begin{example}
If $\mathfrak{B}=\left\{  2<4<6<8\right\}  $ and $n=9$ and $M=\left\{  \left[
26264\right]  ,\left[  24\right]  ,\left[  24\right]  \right\}
_{\operatorname*{multiset}}$, then the multiset $\widetilde{M}$ in Definition
\ref{def.sol.lyndon.gr.bij.LFC} is $\left\{  26264,24,24\right\}
_{\operatorname*{multiset}}$, and its $\leq$-increasing list is $\left(
24,24,26264\right)  $, so that $\operatorname*{LFC}\left(  M\right)
=262642424$.
\end{example}

\begin{proposition}
\label{prop.sol.lyndon.gr.bij.CFL-LFC}Let $\mathfrak{B}$ be a subset of
$\mathfrak{A}$. Let $n\in \NN $. Then, the maps $\operatorname*{CFL}%
:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ and
$\operatorname*{LFC}:\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
\rightarrow\mathfrak{B}^{n}$ (introduced in Definition
\ref{def.sol.lyndon.gr.bij.CFL} and in Definition
\ref{def.sol.lyndon.gr.bij.LFC}) are mutually inverse.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.gr.bij.CFL-LFC}.]We shall prove two claims:

\begin{statement}
\textit{Claim 1:} We have $\operatorname*{CFL}\circ\operatorname*{LFC}%
=\operatorname*{id}$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 1:} Let $M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$. Thus, $M$ is a finite multiset of aperiodic
$\mathfrak{B}$-necklaces satisfying $\operatorname*{sum}M=n$ (since
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ was defined to be the set of
all such multisets). Hence, $M$ is a finite multiset of elements of
$\mathfrak{N}^{\mathfrak{a}}$. Thus, $\operatorname*{lynd}\nolimits_{\ast}M$
is well-defined.

Let $\widetilde{M}$ be the multiset $\operatorname*{lynd}\nolimits_{\ast}M$.
This $\widetilde{M}$ is a finite multiset of Lyndon words (by Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(a)). Let $\left(  b_{1},b_{2}%
,\ldots,b_{k}\right)  $ be the $\leq$-increasing list of $\widetilde{M}$.
According to the definition of a $\leq$-increasing list, this means that
$\widetilde{M}=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}
_{\operatorname*{multiset}}$ and $b_{1}\leq b_{2}\leq\cdots\leq b_{k}$. Now,
$b_{1},b_{2},\ldots,b_{k}$ are Lyndon words (since $\left\{  b_{1}%
,b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}=\widetilde{M}$ is a
multiset of Lyndon words). Hence, $\left(  b_{k},b_{k-1},\ldots,b_{1}\right)
$ is a tuple of Lyndon words.

Define $w\in\mathfrak{B}^{\ast}$ by $w=\operatorname*{LFC}\left(  M\right)  $.
Thus, $w=\operatorname*{LFC}\left(  M\right)  =b_{k}b_{k-1}\cdots b_{1}$ (by
the definition of the map $\operatorname*{LFC}$).

We shall now show that $\operatorname*{CFL}w=M$.

Indeed, $b_{1}\leq b_{2}\leq\cdots\leq b_{k}$. In other words, $b_{k}\geq
b_{k-1}\geq\cdots\geq b_{1}$. Combining this with the fact that $\left(
b_{k},b_{k-1},\ldots,b_{1}\right)  $ is a tuple of Lyndon words, we conclude
that $\left(  b_{k},b_{k-1},\ldots,b_{1}\right)  $ is the CFL factorization of
$w$ (by the definition of the CFL factorization).

On the other hand, $M$ is a finite multiset of aperiodic necklaces. Moreover,
$b_{1},b_{2},\ldots,b_{k}$ are Lyndon words and satisfy $\widetilde{M}%
=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$.
Hence, Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b) yields $M=\left\{
\left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[  b_{k}\right]
\right\}  _{\operatorname*{multiset}}$.

But the definition of the map $\operatorname*{CFL}$ yields that if $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of the word $w$,
then%
\[
\operatorname*{CFL}w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}.
\]
Applying this to $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(
b_{k},b_{k-1},\ldots,b_{1}\right)  $, we obtain%
\begin{align*}
\operatorname*{CFL}w  &  =\left\{  \left[  b_{k}\right]  ,\left[
b_{k-1}\right]  ,\ldots,\left[  b_{1}\right]  \right\}
_{\operatorname*{multiset}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
b_{k},b_{k-1},\ldots,b_{1}\right)  \text{ is the CFL factorization of
}w\right) \\
&  =\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[
b_{k}\right]  \right\}  _{\operatorname*{multiset}}%
=M\ \ \ \ \ \ \ \ \ \ \left(  \text{since }M=\left\{  \left[  b_{1}\right]
,\left[  b_{2}\right]  ,\ldots,\left[  b_{k}\right]  \right\}
_{\operatorname*{multiset}}\right)  .
\end{align*}
In view of $w=\operatorname*{LFC}\left(  M\right)  $, this rewrites as
$\operatorname*{CFL}\left(  \operatorname*{LFC}\left(  M\right)  \right)  =M$.
Thus, $\left(  \operatorname*{CFL}\circ\operatorname*{LFC}\right)  \left(
M\right)  =\operatorname*{CFL}\left(  \operatorname*{LFC}\left(  M\right)
\right)  =M=\operatorname*{id}\left(  M\right)  $.

Forget that we fixed $M$. We thus have showed that $\left(
\operatorname*{CFL}\circ\operatorname*{LFC}\right)  \left(  M\right)
=\operatorname*{id}\left(  M\right)  $ for each $M\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $\operatorname*{CFL}%
\circ\operatorname*{LFC}=\operatorname*{id}$. Thus, Claim 1 is proven.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 1:} Let $M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$. Thus, $M$ is a finite multiset of aperiodic
$\mathfrak{B}$-necklaces satisfying $\operatorname*{sum}M=n$ (since
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ was defined to be the set of
all such multisets). Hence, in particular, $M$ is a finite multiset of
aperiodic $\mathfrak{B}$-necklaces. Therefore, $M$ is a finite multiset of
aperiodic necklaces (since each $\mathfrak{B}$-necklace is a necklace). In
other words, $M$ is a finite multiset of elements of $\mathfrak{N}%
^{\mathfrak{a}}$ (since the aperiodic necklaces are exactly the elements of
$\mathfrak{N}^{\mathfrak{a}}$). Hence, $\operatorname*{lynd}\nolimits_{\ast}M$
is well-defined.

Let $\widetilde{M}$ be the multiset $\operatorname*{lynd}\nolimits_{\ast}M$.
This $\widetilde{M}$ is a finite multiset of Lyndon words (by Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(a)). Let $\left(  b_{1},b_{2}%
,\ldots,b_{k}\right)  $ be the $\leq$-increasing list of $\widetilde{M}$. The
definition of the map $\operatorname*{LFC}$ yields $\operatorname*{LFC}\left(
M\right)  =b_{k}b_{k-1}\cdots b_{1}$.

Define $w\in\mathfrak{B}^{\ast}$ by $w=\operatorname*{LFC}\left(  M\right)  $.
Thus, $w=\operatorname*{LFC}\left(  M\right)  =b_{k}b_{k-1}\cdots b_{1}$.

We shall now show that $\operatorname*{CFL}w=M$.

Indeed, $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is the $\leq$-increasing
list of $\widetilde{M}$ (by its definition). In other words, $\widetilde{M}%
=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$ and
$b_{1}\leq b_{2}\leq\cdots\leq b_{k}$ (by the definition of an
``$\leq$-increasing list'' of a multiset).
Now, $\widetilde{M}=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}
_{\operatorname*{multiset}}$ shows that $b_{1},b_{2},\ldots,b_{k}$ are
elements of the multiset $\widetilde{M}$. Hence, $b_{1},b_{2},\ldots,b_{k}$
are Lyndon words (since all elements of $\widetilde{M}$ are Lyndon
words\footnote{because $\widetilde{M}$ is a multiset of Lyndon words}). In
other words, $b_{k},b_{k-1},\ldots,b_{1}$ are Lyndon words. In other words,
$\left(  b_{k},b_{k-1},\ldots,b_{1}\right)  $ is a tuple of Lyndon words.

But we have $b_{1}\leq b_{2}\leq\cdots\leq b_{k}$. In other words, $b_{k}\geq
b_{k-1}\geq\cdots\geq b_{1}$.

Definition \ref{def.words.CFL} shows that if $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$, then $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ is a CFL factorization of $w$. Applying this to
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(  b_{k},b_{k-1},\ldots
,b_{1}\right)  $, we conclude that $\left(  b_{k},b_{k-1},\ldots,b_{1}\right)
$ is a CFL factorization of $w$ (since $\left(  b_{k},b_{k-1},\ldots
,b_{1}\right)  $ is a tuple of Lyndon words satisfying $w=b_{k}b_{k-1}\cdots
b_{1}$ and $b_{k}\geq b_{k-1}\geq\cdots\geq b_{1}$). Thus, $\left(
b_{k},b_{k-1},\ldots,b_{1}\right)  $ is the CFL factorization of $w$ (because
Theorem \ref{thm.words.CFL} shows that the CFL factorization of $w$ is unique).

On the other hand, $M$ is a finite multiset of aperiodic necklaces. Moreover,
$b_{1},b_{2},\ldots,b_{k}$ are Lyndon words and satisfy $\widetilde{M}%
=\left\{  b_{1},b_{2},\ldots,b_{k}\right\}  _{\operatorname*{multiset}}$.
Hence, Lemma \ref{lem.sol.lyndon.gr.bij.lynd-back}(b) yields $M=\left\{
\left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[  b_{k}\right]
\right\}  _{\operatorname*{multiset}}$.

But the definition of the map $\operatorname*{CFL}$ yields that if $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of the word $w$,
then%
\[
\operatorname*{CFL}w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}.
\]
Applying this to $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(
b_{k},b_{k-1},\ldots,b_{1}\right)  $, we obtain%
\begin{align*}
\operatorname*{CFL}w  &  =\left\{  \left[  b_{k}\right]  ,\left[
b_{k-1}\right]  ,\ldots,\left[  b_{1}\right]  \right\}
_{\operatorname*{multiset}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
b_{k},b_{k-1},\ldots,b_{1}\right)  \text{ is the CFL factorization of
}w\right) \\
&  =\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots,\left[
b_{k}\right]  \right\}  _{\operatorname*{multiset}}=M
\end{align*}
(since $M=\left\{  \left[  b_{1}\right]  ,\left[  b_{2}\right]  ,\ldots
,\left[  b_{k}\right]  \right\}  _{\operatorname*{multiset}}$). In view of
$w=\operatorname*{LFC}\left(  M\right)  $, this rewrites as
$\operatorname*{CFL}\left(  \operatorname*{LFC}\left(  M\right)  \right)  =M$.
Thus, $\left(  \operatorname*{CFL}\circ\operatorname*{LFC}\right)  \left(
M\right)  =\operatorname*{CFL}\left(  \operatorname*{LFC}\left(  M\right)
\right)  =M=\operatorname*{id}\left(  M\right)  $.

Forget that we fixed $M$. We thus have showed that $\left(
\operatorname*{CFL}\circ\operatorname*{LFC}\right)  \left(  M\right)
=\operatorname*{id}\left(  M\right)  $ for each $M\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $\operatorname*{CFL}%
\circ\operatorname*{LFC}=\operatorname*{id}$. Thus, Claim 1 is proven.]
\end{verlong}

\begin{statement}
\textit{Claim 2:} We have $\operatorname*{LFC}\circ\operatorname*{CFL}%
=\operatorname*{id}$.
\end{statement}

[\textit{Proof of Claim 2:} Let $w\in\mathfrak{B}^{n}$. Let $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of the word $w$.
Then, the definition of $\operatorname*{CFL}$ yields $\operatorname*{CFL}%
w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}}$.

We know that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a CFL
factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (because this is how a
``CFL factorization of $w$'' is defined). In
particular, $a_{1},a_{2},\ldots,a_{k}$ are Lyndon words (since $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words).

\begin{vershort}
Hence, for each $i\in\left\{  1,2,\ldots,k\right\}  $, the word $a_{i}$ is a
Lyndon word, and therefore satisfies $\operatorname*{lynd}\left[
a_{i}\right]  =a_{i}$ (by Lemma \ref{lem.sol.lyndon.gr.bij.lynd-C}, applied to
$u=a_{i}$). In other words,
\begin{equation}
\left(  \operatorname*{lynd}\left[  a_{1}\right]  ,\operatorname*{lynd}\left[
a_{2}\right]  ,\ldots,\operatorname*{lynd}\left[  a_{k}\right]  \right)
=\left(  a_{1},a_{2},\ldots,a_{k}\right)  .
\label{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.short.3}%
\end{equation}

\end{vershort}

Let $M$ denote the multiset $\operatorname*{CFL}w\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$.

\begin{vershort}
Thus, $M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $M$
is a finite multiset of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$ (since $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
}$ was defined to be the set of all such multisets). Hence, $M$ is a finite
multiset of elements of $\mathfrak{N}^{\mathfrak{a}}$. Thus,
$\operatorname*{lynd}\nolimits_{\ast}M$ is well-defined.
\end{vershort}

\begin{verlong}
Thus, $M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $M$
is a finite multiset of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$ (since $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
}$ was defined to be the set of all such multisets). Hence, in particular, $M$
is a finite multiset of aperiodic $\mathfrak{B}$-necklaces. Therefore, $M$ is
a finite multiset of aperiodic necklaces (since each $\mathfrak{B}$-necklace
is a necklace). In other words, $M$ is a finite multiset of elements of
$\mathfrak{N}^{\mathfrak{a}}$ (since the aperiodic necklaces are exactly the
elements of $\mathfrak{N}^{\mathfrak{a}}$). Hence, $\operatorname*{lynd}%
\nolimits_{\ast}M$ is well-defined.
\end{verlong}

The definition of $M$ yields $M=\operatorname*{CFL}w=\left\{  \left[
a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}$. Hence,%
\[
\operatorname*{lynd}\nolimits_{\ast}M=\left\{  \operatorname*{lynd}\left[
a_{1}\right]  ,\operatorname*{lynd}\left[  a_{2}\right]  ,\ldots
,\operatorname*{lynd}\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}%
\]
(by (\ref{eq.def.sol.lyndon.gr.bij.fstarMultiset.2}), applied to
$\mathfrak{N}^{\mathfrak{a}}$, $\mathfrak{L}$, $\operatorname*{lynd}$ and
$\left[  a_{j}\right]  $ instead of $X$, $Y$, $f$ and $m_{j}$).

\begin{vershort}
Let $\widetilde{M}$ be the multiset $\operatorname*{lynd}\nolimits_{\ast}M$.
This $\widetilde{M}$ is a finite multiset of Lyndon words (by Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(a)), and satisfies%
\begin{align*}
\widetilde{M}  &  =\operatorname*{lynd}\nolimits_{\ast}M=\left\{
\operatorname*{lynd}\left[  a_{1}\right]  ,\operatorname*{lynd}\left[
a_{2}\right]  ,\ldots,\operatorname*{lynd}\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}\\
&  =\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.short.3})}\right) \\
&  =\left\{  a_{k},a_{k-1},\ldots,a_{1}\right\}  _{\operatorname*{multiset}}.
\end{align*}

\end{vershort}

\begin{verlong}
Let $\widetilde{M}$ be the multiset $\operatorname*{lynd}\nolimits_{\ast}M$.
This $\widetilde{M}$ is a finite multiset of Lyndon words (by Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-back}(a)), and satisfies%
\begin{equation}
\widetilde{M}=\operatorname*{lynd}\nolimits_{\ast}M=\left\{
\operatorname*{lynd}\left[  a_{1}\right]  ,\operatorname*{lynd}\left[
a_{2}\right]  ,\ldots,\operatorname*{lynd}\left[  a_{k}\right]  \right\}
_{\operatorname*{multiset}}.
\label{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.2}%
\end{equation}


However, for each $i\in\left\{  1,2,\ldots,k\right\}  $, we have%
\begin{equation}
\operatorname*{lynd}\left[  a_{i}\right]  =a_{i}.
\label{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.3}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.3}):} Let
$i\in\left\{  1,2,\ldots,k\right\}  $. Then, $a_{i}$ is a Lyndon word (since
$a_{1},a_{2},\ldots,a_{k}$ are Lyndon words). Hence, Lemma
\ref{lem.sol.lyndon.gr.bij.lynd-C} (applied to $u=a_{i}$) yields
$\operatorname*{lynd}\left[  a_{i}\right]  =a_{i}$. This proves
(\ref{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.3}).]

Thus we have showed that $\operatorname*{lynd}\left[  a_{i}\right]  =a_{i}$
for each $i\in\left\{  1,2,\ldots,k\right\}  $. In other words,%
\[
\left(  \operatorname*{lynd}\left[  a_{1}\right]  ,\operatorname*{lynd}\left[
a_{2}\right]  ,\ldots,\operatorname*{lynd}\left[  a_{k}\right]  \right)
=\left(  a_{1},a_{2},\ldots,a_{k}\right)  .
\]
Hence,%
\[
\left\{  \operatorname*{lynd}\left[  a_{1}\right]  ,\operatorname*{lynd}%
\left[  a_{2}\right]  ,\ldots,\operatorname*{lynd}\left[  a_{k}\right]
\right\}  _{\operatorname*{multiset}}=\left\{  a_{1},a_{2},\ldots
,a_{k}\right\}  _{\operatorname*{multiset}}.
\]
Now, (\ref{pf.prop.sol.lyndon.gr.bij.CFL-LFC.LFC-CFL.pf.2}) becomes%
\begin{align*}
\widetilde{M}  &  =\left\{  \operatorname*{lynd}\left[  a_{1}\right]
,\operatorname*{lynd}\left[  a_{2}\right]  ,\ldots,\operatorname*{lynd}\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}}=\left\{  a_{1}%
,a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}\\
&  =\left\{  a_{k},a_{k-1},\ldots,a_{1}\right\}  _{\operatorname*{multiset}}.
\end{align*}

\end{verlong}

Also, recall that $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$. In other words,
$a_{k}\leq a_{k-1}\leq\cdots\leq a_{1}$.

Now, $\left(  a_{k},a_{k-1},\ldots,a_{1}\right)  $ is the $\leq$-increasing
list of $\widetilde{M}$ (because $\widetilde{M}=\left\{  a_{k},a_{k-1}%
,\ldots,a_{1}\right\}  _{\operatorname*{multiset}}$ and $a_{k}\leq a_{k-1}%
\leq\cdots\leq a_{1}$). But the definition of $\operatorname*{LFC}$ shows that
if $\left(  b_{1},b_{2},\ldots,b_{k}\right)  $ is the $\leq$-increasing list
of $\widetilde{M}$, then
\[
\operatorname*{LFC}\left(  M\right)  =b_{k}b_{k-1}\cdots b_{1}.
\]
We can apply this to $\left(  b_{1},b_{2},\ldots,b_{k}\right)  =\left(
a_{k},a_{k-1},\ldots,a_{1}\right)  $ (since $\left(  a_{k},a_{k-1}%
,\ldots,a_{1}\right)  $ is the $\leq$-increasing list of $\widetilde{M}$);
thus, we obtain $\operatorname*{LFC}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}%
$. Comparing this with $w=a_{1}a_{2}\cdots a_{k}$, we obtain
$\operatorname*{LFC}\left(  M\right)  =w$. In view of $M=\operatorname*{CFL}%
w$, this rewrites as $\operatorname*{LFC}\left(  \operatorname*{CFL}w\right)
=w$. Hence,%
\[
\left(  \operatorname*{LFC}\circ\operatorname*{CFL}\right)  \left(  w\right)
=\operatorname*{LFC}\left(  \operatorname*{CFL}w\right)  =w=\operatorname*{id}%
\left(  w\right)  .
\]


Forget that we fixed $w$. We thus have shown that $\left(  \operatorname*{LFC}%
\circ\operatorname*{CFL}\right)  \left(  w\right)  =\operatorname*{id}\left(
w\right)  $ for each $w\in\mathfrak{B}^{n}$. In other words,
$\operatorname*{LFC}\circ\operatorname*{CFL}=\operatorname*{id}$. This proves
Claim 2.]

We have $\operatorname*{CFL}\circ\operatorname*{LFC}=\operatorname*{id}$ (by
Claim 1) and $\operatorname*{LFC}\circ\operatorname*{CFL}=\operatorname*{id}$
(by Claim 2). Combining these two equalities, we conclude that the maps
$\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ and $\operatorname*{LFC}:\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\rightarrow\mathfrak{B}^{n}$ are mutually inverse. This
proves Proposition \ref{prop.sol.lyndon.gr.bij.CFL-LFC}.
\end{proof}

\begin{corollary}
\label{cor.sol.lyndon.gr.bij.equal-size}Let $\mathfrak{B}$ be a finite subset
of $\mathfrak{A}$. Let $n\in \NN $. Then:

\begin{enumerate}
\item[(a)] The sets $\mathfrak{B}^{n}$ and $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ are finite and satisfy $\left\vert \mathfrak{B}%
^{n}\right\vert =\left\vert \mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
}\right\vert $.

\item[(b)] Any injective map $f:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ is bijective.
\end{enumerate}
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.sol.lyndon.gr.bij.equal-size}.]The set
$\mathfrak{B}$ is finite. Hence, the set $\mathfrak{B}^{n}$ is finite as well.

\begin{vershort}
Proposition \ref{prop.sol.lyndon.gr.bij.CFL-LFC} shows that the maps
$\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ and $\operatorname*{LFC}:\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\rightarrow\mathfrak{B}^{n}$ (defined as in Proposition
\ref{prop.sol.lyndon.gr.bij.CFL-LFC}) are mutually inverse. Thus, these two
maps are bijections. Hence, there exists a bijection from $\mathfrak{B}^{n}$
to $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. This entails $\left\vert
\mathfrak{B}^{n}\right\vert =\left\vert \mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\right\vert $. Since $\mathfrak{B}^{n}$ is finite, this
shows that $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ is finite as well.
Thus, Corollary \ref{cor.sol.lyndon.gr.bij.equal-size}(a) is proven.
\end{vershort}

\begin{verlong}
Proposition \ref{prop.sol.lyndon.gr.bij.CFL-LFC} shows that the maps
$\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ and $\operatorname*{LFC}:\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\rightarrow\mathfrak{B}^{n}$ (introduced in Definition
\ref{def.sol.lyndon.gr.bij.CFL} and in Definition
\ref{def.sol.lyndon.gr.bij.LFC}) are mutually inverse. Thus, these two maps
are bijections. In particular, $\operatorname*{CFL}:\mathfrak{B}%
^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ is a bijection.
Thus, there exists a bijection from $\mathfrak{B}^{n}$ to $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$. Hence, $\left\vert \mathfrak{B}%
^{n}\right\vert =\left\vert \mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
}\right\vert $. Since $\mathfrak{B}^{n}$ is finite, this shows that
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ is finite as well. Thus,
Corollary \ref{cor.sol.lyndon.gr.bij.equal-size}(a) is proven.
\end{verlong}

(b) Let $f:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ be any injective map. We must prove that $f$ is bijective.

\begin{vershort}
Corollary \ref{cor.sol.lyndon.gr.bij.equal-size}(a) shows that $\mathfrak{B}%
^{n}$ and $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ are two finite sets
of equal sizes. Hence, $f$ is a map between two finite sets of equal sizes.
But it is well-known that any injective map between two finite sets of equal
sizes is bijective\footnote{This is one of the basic facts known as the
``pigeonhole principle''.}. Thus, $f$ is
bijective. This proves Corollary \ref{cor.sol.lyndon.gr.bij.equal-size}(b).
\end{vershort}

\begin{verlong}
We have already seen that the sets $\mathfrak{B}^{n}$ to $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ are finite. Thus, $\mathfrak{B}^{n}$ and
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ are two finite sets of equal
sizes (since $\left\vert \mathfrak{B}^{n}\right\vert =\left\vert
\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}\right\vert $). Hence, $f$ is a
map between two finite sets of equal sizes (since $f$ is a map between
$\mathfrak{B}^{n}$ and $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$).

It is well-known that any injective map between two finite sets of equal sizes
is bijective\footnote{This is one of the basic facts known as the
``pigeonhole principle''.}. Thus, $f$ is
bijective (since $f$ is an injective map between two finite sets of equal
sizes is bijective). This proves Corollary
\ref{cor.sol.lyndon.gr.bij.equal-size}(b).
\end{verlong}
\end{proof}

Now, at last we can prove the following:

\begin{lemma}
\label{lem.sol.lyndon.gr.bij.GR-RG}We have $\operatorname*{GR}\circ
\operatorname*{RG}=\operatorname*{id}$.
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.GR-RG}.]Lemma
\ref{lem.sol.lyndon.gr.bij.RG-GR} yields $\operatorname*{RG}\circ
\operatorname*{GR}=\operatorname*{id}$. Thus, the map $\operatorname*{GR}$ has
a left inverse (namely, $\operatorname*{RG}$). Therefore, the map
$\operatorname*{GR}$ is injective.

Let $M\in\mathfrak{MN}^{\mathfrak{a}}$. We shall show that $\left(
\operatorname*{GR}\circ\operatorname*{RG}\right)  \left(  M\right)  =M$.

We have $M\in\mathfrak{MN}^{\mathfrak{a}}$. In other words, $M$ is a finite
multiset of aperiodic necklaces (by the definition of $\mathfrak{MN}%
^{\mathfrak{a}}$). Thus, each element of $M$ is an aperiodic necklace.

Let $n=\operatorname*{sum}M$; thus, $n\in \NN $.

Define a subset $\mathfrak{B}$ of $\mathfrak{A}$ by%
\begin{equation}
\mathfrak{B}=\bigcup_{N\in M}\bigcup_{w\in N}\left\{  \text{all letters of
}w\right\}  \label{pf.lem.sol.lyndon.gr.bij.GR-RG.short.1}%
\end{equation}
(where the ``$\bigcup_{N\in M}$'' symbol
should be understood as ``$\bigcup_{N\in\operatorname*{Supp}M}$'').

Thus, $\mathfrak{B}$ is a finite union of finite unions of finite
sets\footnote{To wit:
\par
\begin{itemize}
\item There are only finitely many $N\in M$ (since $M$ is finite).
\par
\item For each $N\in M$, there are only finitely many $w\in N$ (since $N$ is a
necklace, thus a finite set).
\par
\item For each $N\in M$ and each $w\in N$, the set $\left\{  \text{all letters
of }w\right\}  $ is finite (since the word $w$ has only finitely many
letters).
\end{itemize}
}. Thus, the set $\mathfrak{B}$ is finite. Furthermore, the definition of
$\mathfrak{B}$ ensures that each $N\in M$ is an aperiodic $\mathfrak{B}%
$-necklace\footnote{\textit{Proof.} Let $N\in M$. Then, $N$ is an aperiodic
necklace (since each element of $M$ is an aperiodic necklace).
\par
But (\ref{pf.lem.sol.lyndon.gr.bij.GR-RG.short.1}) shows that $\bigcup_{w\in
N}\left\{  \text{all letters of }w\right\}  \subset\mathfrak{B}$ (since $N\in
M$). In other words, each $w\in N$ satisfies $\left\{  \text{all letters of
}w\right\}  \subset\mathfrak{B}$. In other words, each $w\in N$ satisfies
$w\in\mathfrak{B}^{\ast}$. In other words, $N\subset\mathfrak{B}^{\ast}$.
Hence, $N$ is a $\mathfrak{B}$-necklace (since $N$ is a necklace), hence an
aperiodic $\mathfrak{B}$-necklace (since $N$ is aperiodic). Qed.}. Thus, $M$
is a finite multiset of aperiodic $\mathfrak{B}$-necklaces. Hence, $M$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$ (since $n=\operatorname*{sum}M$). In other words,
$M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ (by the definition of
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$).

But Lemma \ref{lem.sol.lyndon.gr.bij.GR-restricts} yields that
$\operatorname*{GR}\left(  \mathfrak{B}^{n}\right)  \subset\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $\operatorname*{GR}\left(
w\right)  \in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ for each
$w\in\mathfrak{B}^{n}$. Thus, we can define a map%
\begin{align*}
\overline{\operatorname*{GR}}:\mathfrak{B}^{n}  &  \rightarrow\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}},\\
w  &  \mapsto\operatorname*{GR}\left(  w\right)  .
\end{align*}
Consider this map $\overline{\operatorname*{GR}}$. Clearly, this map
$\overline{\operatorname*{GR}}$ is a restriction of the map
$\operatorname*{GR}$; hence, this map $\overline{\operatorname*{GR}}$ is
injective (since the map $\operatorname*{GR}$ is injective). Hence, Corollary
\ref{cor.sol.lyndon.gr.bij.equal-size}(b) (applied to
$f=\overline{\operatorname*{GR}}$) yields that the map
$\overline{\operatorname*{GR}}$ is
bijective. Thus, $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}
=\overline{\operatorname*{GR}}\left(  \mathfrak{B}^{n}\right)  $.

Now, $M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}=
\overline{\operatorname*{GR}}\left(  \mathfrak{B}^{n}\right)  $.
In other words, there
exists some $w\in\mathfrak{B}^{n}$ satisfying $M=\overline{\operatorname*{GR}%
}\ w$. Consider this $w$. We have $M=\overline{\operatorname*{GR}%
}\ w=\operatorname*{GR}w$ (by the definition of $\overline{\operatorname*{GR}%
}$). Now,%
\[
\left(  \operatorname*{GR}\circ\operatorname*{RG}\right)  \left(
\underbrace{M}_{=\operatorname*{GR}w}\right)  =\left(  \operatorname*{GR}%
\circ\operatorname*{RG}\right)  \left(  \operatorname*{GR}w\right)  =\left(
\operatorname*{GR}\circ\underbrace{\operatorname*{RG}\circ\operatorname*{GR}%
}_{=\operatorname*{id}}\right)  \left(  w\right)  =\operatorname*{GR}%
w=M=\operatorname*{id}\left(  M\right)  .
\]


Forget that we fixed $M$. We thus have proved that $\left(  \operatorname*{GR}%
\circ\operatorname*{RG}\right)  \left(  M\right)  =\operatorname*{id}\left(
M\right)  $ for each $M\in\mathfrak{MN}^{\mathfrak{a}}$. In other words,
$\operatorname*{GR}\circ\operatorname*{RG}=\operatorname*{id}$. This proves
Lemma \ref{lem.sol.lyndon.gr.bij.GR-RG}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.gr.bij.GR-RG}.]Lemma
\ref{lem.sol.lyndon.gr.bij.RG-GR} yields $\operatorname*{RG}\circ
\operatorname*{GR}=\operatorname*{id}$. Thus, the map $\operatorname*{GR}$ is
injective\footnote{\textit{Proof.} A well-known fact states that if $X$ and
$Y$ are two sets, and if $f:X\rightarrow Y$ and $g:Y\rightarrow X$ are two
maps such that $g\circ f=\operatorname*{id}$, then $f$ is injective. Applying
this fact to $X=\mathfrak{A}^{\ast}$, $Y=\mathfrak{MN}^{\mathfrak{a}}$,
$f=\operatorname*{GR}$ and $g=\operatorname*{RG}$, we thus conclude that
$\operatorname*{GR}$ is injective (since $\operatorname*{RG}\circ
\operatorname*{GR}=\operatorname*{id}$).}.

Let $M\in\mathfrak{MN}^{\mathfrak{a}}$. We shall show that $\left(
\operatorname*{GR}\circ\operatorname*{RG}\right)  \left(  M\right)  =M$.

We have $M\in\mathfrak{MN}^{\mathfrak{a}}$. In other words, $M$ is a finite
multiset of aperiodic necklaces (by the definition of $\mathfrak{MN}%
^{\mathfrak{a}}$). Let $n=\operatorname*{sum}M$; thus, $n\in \NN $.

Write the finite multiset $M$ in the form $M=\left\{  N_{1},N_{2},\ldots
,N_{k}\right\}  _{\operatorname*{multiset}}$, where $N_{1},N_{2},\ldots,N_{k}$
are aperiodic necklaces. (This can be done, since $M$ is a finite multiset of
aperiodic necklaces.)

Let $i\in\left\{  1,2,\ldots,k\right\}  $. Then, $N_{i}$ is an aperiodic
necklace (since $N_{1},N_{2},\ldots,N_{k}$ are aperiodic necklaces), thus a
necklace, thus a finite set (since any necklace is a finite set). Hence,
$N_{i}$ is a finite set of words. Define a set $\mathfrak{B}_{i}$ by
\begin{equation}
\mathfrak{B}_{i}=\bigcup_{w\in N_{i}}\left\{  \text{all letters of }w\right\}
. \label{pf.lem.sol.lyndon.gr.bij.GR-RG.Bi=}%
\end{equation}
Then, $\mathfrak{B}_{i}$ is a finite union of finite sets\footnote{because:
\par
\begin{itemize}
\item for each $w\in N_{i}$, the set $\left\{  \text{all letters of
}w\right\}  $ is finite (since $w$ is a word);
\par
\item the set $N_{i}$ is finite
\end{itemize}
}. Hence, $\mathfrak{B}_{i}$ itself is a finite set. Moreover, $N_{i}$ is a
necklace, and thus a set of words; hence, $N_{i}\subset\mathfrak{A}^{\ast}$.
Now,
\[
\mathfrak{B}_{i}=\bigcup_{w\in N_{i}}\underbrace{\left\{  \text{all letters of
}w\right\}  }_{\substack{\subset\mathfrak{A}\\\text{(since }w\in N_{i}%
\subset\mathfrak{A}^{\ast}\text{)}}}\subset\bigcup_{w\in N_{i}}\mathfrak{A}%
\subset\mathfrak{A}.
\]
In other words, $\mathfrak{B}_{i}$ is a subset of $\mathfrak{A}$.

Forget that we fixed $i$. Thus, for each $i\in\left\{  1,2,\ldots,k\right\}
$, we have defined a finite subset $\mathfrak{B}_{i}$ of $\mathfrak{A}$. Thus,
we have obtained $k$ finite subsets $\mathfrak{B}_{1},\mathfrak{B}_{2}%
,\ldots,\mathfrak{B}_{k}$ of $\mathfrak{A}$.

Furthermore, define a set $\mathfrak{B}$ by%
\[
\mathfrak{B}=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}%
_{k}.
\]
This set $\mathfrak{B}$ is thus the union of the $k$ finite sets
$\mathfrak{B}_{1},\mathfrak{B}_{2},\ldots,\mathfrak{B}_{k}$, and therefore
must itself be finite (since any union of finitely many finite sets is
finite). Moreover, $\mathfrak{B}$ is the union of the $k$ finite sets
$\mathfrak{B}_{1},\mathfrak{B}_{2},\ldots,\mathfrak{B}_{k}$, all of which are
subsets of $\mathfrak{A}$; therefore, $\mathfrak{B}$ itself is a subset of
$\mathfrak{A}$.

Hence, Lemma \ref{lem.sol.lyndon.gr.bij.GR-restricts} yields that
$\operatorname*{GR}\left(  \mathfrak{B}^{n}\right)  \subset\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$. In other words, $\operatorname*{GR}\left(
w\right)  \in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ for each
$w\in\mathfrak{B}^{n}$. Thus, we can define a map%
\begin{align*}
\overline{\operatorname*{GR}}:\mathfrak{B}^{n}  &  \rightarrow\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}},\\
w  &  \mapsto\operatorname*{GR}\left(  w\right)  .
\end{align*}
Consider this map $\overline{\operatorname*{GR}}$. Clearly, this map
$\overline{\operatorname*{GR}}$ is a restriction of the map
$\operatorname*{GR}$; hence, this map $\overline{\operatorname*{GR}}$ is
injective (since the map $\operatorname*{GR}$ is injective). Hence, Corollary
\ref{cor.sol.lyndon.gr.bij.equal-size}(b) (applied to
$f=\overline{\operatorname*{GR}}$) yields that the map
$\overline{\operatorname*{GR}}$ is
bijective. Thus, this map $\overline{\operatorname*{GR}}$ is surjective. In
other words, $\overline{\operatorname*{GR}}\left(  \mathfrak{B}^{n}\right)
=\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$.

But $N_{i}$ is an aperiodic $\mathfrak{B}$-necklace for each $i\in\left\{
1,2,\ldots,k\right\}  $.

[\textit{Proof:} Let $i\in\left\{  1,2,\ldots,k\right\}  $. We shall show that
$N_{i}$ is an aperiodic $\mathfrak{B}$-necklace.

We have $\mathfrak{B}_{i}\subset\mathfrak{B}$ (since $\mathfrak{B}%
=\mathfrak{B}_{1}\cup\mathfrak{B}_{2}\cup\cdots\cup\mathfrak{B}_{k}$). But
every $w\in N_{i}$ satisfies
\begin{align}
\left\{  \text{all letters of }w\right\}   &  \subset\mathfrak{B}%
_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.gr.bij.GR-RG.Bi=})}\right) \nonumber\\
&  \subset\mathfrak{B}. \label{pf.lem.sol.lyndon.gr.bij.GR-RG.4.pf.1}%
\end{align}


Now, let $w\in N_{i}$. Hence, (\ref{pf.lem.sol.lyndon.gr.bij.GR-RG.4.pf.1})
shows that $\left\{  \text{all letters of }w\right\}  \subset\mathfrak{B}$. In
other words, all letters of $w$ belong to $\mathfrak{B}$. Hence, $w$ is a list
of elements of $\mathfrak{B}$. In other words, $w\in\mathfrak{B}^{\ast}$.

Forget that we fixed $w$. We thus have shown that $w\in\mathfrak{B}^{\ast}$
for each $w\in N_{i}$. In other words, $N_{i}$ is a subset of $\mathfrak{B}%
^{\ast}$. But $N_{i}$ is also an aperiodic necklace (since $N_{1},N_{2}%
,\ldots,N_{k}$ are aperiodic necklaces). Thus, $N_{i}$ is a necklace that is a
subset of $\mathfrak{B}^{\ast}$. In other words, $N_{i}$ is a $\mathfrak{B}%
$-necklace (since a $\mathfrak{B}$-necklace is defined to mean a necklace that
is a subset of $\mathfrak{B}^{\ast}$). Thus, $N_{i}$ is an aperiodic
$\mathfrak{B}$-necklace. Qed.]

Thus, we have shown that $N_{i}$ is an aperiodic $\mathfrak{B}$-necklace for
each $i\in\left\{  1,2,\ldots,k\right\}  $. In other words, $N_{1}%
,N_{2},\ldots,N_{k}$ are aperiodic $\mathfrak{B}$-necklaces. Hence, $\left\{
N_{1},N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$ is a finite
multiset of aperiodic $\mathfrak{B}$-necklaces. In other words, $M$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces (since $M=\left\{
N_{1},N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$). Moreover, $M$
satisfies $\operatorname*{sum}M=n$ (by the definition of $n$). Hence, $M$ is a
finite multiset of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$. In other words, $M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ (by the definition of $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$).

In view of $\overline{\operatorname*{GR}}\left(  \mathfrak{B}^{n}\right)
=\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$, this rewrites as
$M\in\overline{\operatorname*{GR}}\left(  \mathfrak{B}^{n}\right)  $. In other
words, there exists some $w\in\mathfrak{B}^{n}$ satisfying
$M=\overline{\operatorname*{GR}}\ w$. Consider this $w$. The definition of
$\overline{\operatorname*{GR}}$ yields $\overline{\operatorname*{GR}}%
\ w=\operatorname*{GR}w$. Hence, $\operatorname*{GR}w
=\overline{\operatorname*{GR}}\ w=M$. Hence, $M=\operatorname*{GR}w$. Now,
\[
\left(  \operatorname*{GR}\circ\operatorname*{RG}\right)  \left(
\underbrace{M}_{=\operatorname*{GR}w}\right)  =\left(  \operatorname*{GR}%
\circ\operatorname*{RG}\right)  \left(  \operatorname*{GR}w\right)  =\left(
\operatorname*{GR}\circ\underbrace{\operatorname*{RG}\circ\operatorname*{GR}%
}_{=\operatorname*{id}}\right)  \left(  w\right)  =\operatorname*{GR}%
w=M=\operatorname*{id}\left(  M\right)  .
\]


Forget that we fixed $M$. We thus have proved that $\left(  \operatorname*{GR}%
\circ\operatorname*{RG}\right)  \left(  M\right)  =\operatorname*{id}\left(
M\right)  $ for each $M\in\mathfrak{MN}^{\mathfrak{a}}$. In other words,
$\operatorname*{GR}\circ\operatorname*{RG}=\operatorname*{id}$. This proves
Lemma \ref{lem.sol.lyndon.gr.bij.GR-RG}.
\end{proof}

\end{verlong}

\begin{vershort}

\begin{proof}
[Proof of Theorem \ref{thm.lyndon.gr.bij}.]We have $\operatorname*{GR}%
\circ\operatorname*{RG}=\operatorname*{id}$ (by Lemma
\ref{lem.sol.lyndon.gr.bij.GR-RG}) and $\operatorname*{RG}\circ
\operatorname*{GR}=\operatorname*{id}$ (by Lemma
\ref{lem.sol.lyndon.gr.bij.RG-GR}). These two equalities combined yield the
claim of Theorem \ref{thm.lyndon.gr.bij}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Theorem \ref{thm.lyndon.gr.bij}.]We have $\operatorname*{GR}%
\circ\operatorname*{RG}=\operatorname*{id}$ (by Lemma
\ref{lem.sol.lyndon.gr.bij.GR-RG}) and $\operatorname*{RG}\circ
\operatorname*{GR}=\operatorname*{id}$ (by Lemma
\ref{lem.sol.lyndon.gr.bij.RG-GR}). These two equalities show that the maps
$\operatorname*{GR}:\mathfrak{A}^{\ast}\rightarrow\mathfrak{MN}^{\mathfrak{a}%
}$ and $\operatorname*{RG}:\mathfrak{MN}^{\mathfrak{a}}\rightarrow
\mathfrak{A}^{\ast}$ are mutually inverse bijections. Thus, Theorem
\ref{thm.lyndon.gr.bij} is proved.
\end{proof}

\end{verlong}

Hence, Exercise \ref{exe.lyndon.gr.bij} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.lyndon.grfun.all}}

\textit{Solution to Exercise \ref{exe.lyndon.grfun.all}.} In order to solve
Exercise \ref{exe.lyndon.grfun.all}, we must prove the statements made in
Subsection \ref{subsub.lyndon.grfun}. We shall begin with Proposition
\ref{prop.lyndon.grfun.CFLps}. First, let us show a few simple lemmas about
words $w\in\mathfrak{A}^{\ast}$ and the respective monomials $\xx_{w}$:

\begin{lemma}
\label{lem.sol.lyndon.grfun.xuv=}Let $u$ and $v$ be two words in
$\mathfrak{A}^{\ast}$. Then, $\xx_{uv}=\xx_{u}\xx_{v}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.xuv=}.]Write the word $u$ in the
form $u=\left(  u_{1},u_{2},\ldots,u_{p}\right)  $. Then, the definition of
$\xx_{u}$ yields $\xx_{u}=x_{u_{1}}x_{u_{2}}\cdots x_{u_{p}}$.

Write the word $v$ in the form $v=\left(  v_{1},v_{2},\ldots,v_{q}\right)  $.
Then, the definition of $\xx_{v}$ yields $\xx_{v}=x_{v_{1}%
}x_{v_{2}}\cdots x_{v_{q}}$.

From $u=\left(  u_{1},u_{2},\ldots,u_{p}\right)  $ and $v=\left(  v_{1}%
,v_{2},\ldots,v_{q}\right)  $, we obtain%
\[
uv=\left(  u_{1},u_{2},\ldots,u_{p}\right)  \left(  v_{1},v_{2},\ldots
,v_{q}\right)  =\left(  u_{1},u_{2},\ldots,u_{p},v_{1},v_{2},\ldots
,v_{q}\right)  .
\]
Hence, the definition of $\xx_{uv}$ yields
\[
\xx_{uv}=x_{u_{1}}x_{u_{2}}\cdots x_{u_{p}}x_{v_{1}}x_{v_{2}}\cdots
x_{v_{q}}.
\]
Comparing this with%
\[
\underbrace{\xx_{u}}_{=x_{u_{1}}x_{u_{2}}\cdots x_{u_{p}}%
}\underbrace{\xx_{v}}_{=x_{v_{1}}x_{v_{2}}\cdots x_{v_{q}}}=x_{u_{1}%
}x_{u_{2}}\cdots x_{u_{p}}x_{v_{1}}x_{v_{2}}\cdots x_{v_{q}},
\]
we obtain $\xx_{uv}=\xx_{u}\xx_{v}$. This proves Lemma
\ref{lem.sol.lyndon.grfun.xuv=}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.grfun.xai}Let $k\in \NN $. Let $a_{1},a_{2}%
,\ldots,a_{k}$ be words in $\mathfrak{A}^{\ast}$. Then, $\xx%
_{a_{1}a_{2}\cdots a_{k}}=\xx_{a_{1}}\xx_{a_{2}}\cdots
\xx_{a_{k}}$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.xai}.]Lemma
\ref{lem.sol.lyndon.grfun.xai} follows by induction on $k$, using Lemma
\ref{lem.sol.lyndon.grfun.xuv=} in the induction step.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.xai}.]We claim that each
$m\in\left\{  0,1,\ldots,k\right\}  $ satisfies%
\begin{equation}
\xx_{a_{1}a_{2}\cdots a_{m}}=\xx_{a_{1}}\xx_{a_{2}}%
\cdots\xx_{a_{m}}. \label{pf.lem.sol.lyndon.grfun.xai.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.sol.lyndon.grfun.xai.goal}):} We shall prove
(\ref{pf.lem.sol.lyndon.grfun.xai.goal}) by induction on $m$:

\textit{Induction base:} We have $a_{1}a_{2}\cdots a_{0}=\left(  \text{empty
product in }\mathfrak{A}^{\ast}\right)  =\varnothing$ (the empty word) and
thus $\xx_{a_{1}a_{2}\cdots a_{0}}=\xx_{\varnothing}=1$ (by the
definition of $\xx_{\varnothing}$). Comparing this with $\xx%
_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{0}}=\left(  \text{empty product
of monomials}\right)  =1$, we obtain $\xx_{a_{1}a_{2}\cdots a_{0}%
}=\xx_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{0}}$. In other
words, (\ref{pf.lem.sol.lyndon.grfun.xai.goal}) holds for $m=0$. This
completes the induction base.

\textit{Induction step:} Let $i\in\left\{  0,1,\ldots,k-1\right\}  $. Assume
that (\ref{pf.lem.sol.lyndon.grfun.xai.goal}) holds for $m=i$. We must prove
that (\ref{pf.lem.sol.lyndon.grfun.xai.goal}) holds for $m=i+1$.

We have assumed that (\ref{pf.lem.sol.lyndon.grfun.xai.goal}) holds for $m=i$.
In other words, we have $\xx_{a_{1}a_{2}\cdots a_{i}}=\xx%
_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{i}}$. Now, $a_{1}a_{2}\cdots
a_{i+1}=\left(  a_{1}a_{2}\cdots a_{i}\right)  a_{i+1}$, so that%
\begin{align*}
\xx_{a_{1}a_{2}\cdots a_{i+1}}  &  =\xx_{\left(  a_{1}%
a_{2}\cdots a_{i}\right)  a_{i+1}}=\underbrace{\xx_{a_{1}a_{2}\cdots
a_{i}}}_{=\xx_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{i}}%
}\xx_{a_{i+1}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.sol.lyndon.grfun.xuv=},
applied to }u=a_{1}a_{2}\cdots a_{i}\text{ and }v=a_{i+1}\right) \\
&  =\left(  \xx_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{i}%
}\right)  \xx_{a_{i+1}}=\xx_{a_{1}}\xx_{a_{2}}%
\cdots\xx_{a_{i+1}}.
\end{align*}
In other words, (\ref{pf.lem.sol.lyndon.grfun.xai.goal}) holds for $m=i+1$.
This completes the induction step. Thus,
(\ref{pf.lem.sol.lyndon.grfun.xai.goal}) is proved by induction.]

Now, we can apply (\ref{pf.lem.sol.lyndon.grfun.xai.goal}) to $m=k$. We thus
obtain $\xx_{a_{1}a_{2}\cdots a_{k}}=\xx_{a_{1}}\xx%
_{a_{2}}\cdots\xx_{a_{k}}$. This proves Lemma
\ref{lem.sol.lyndon.grfun.xai}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.grfun.sumxwtlw-n}Let $n\in \NN $. Then, $\sum
_{w\in\mathfrak{A}^{n}}\xx_{w}=p_{1}^{n}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw-n}.]The definition of
$p_{1}$ yields%
\[
p_{1}=x_{1}^{1}+x_{2}^{1}+x_{3}^{1}+\cdots=\sum_{a\in\left\{  1,2,3,\ldots
\right\}  }\underbrace{x_{a}^{1}}_{=x_{a}}=\sum_{a\in\left\{  1,2,3,\ldots
\right\}  }x_{a}=\sum_{a\in\mathfrak{A}}x_{a}%
\]
(since $\left\{  1,2,3,\ldots\right\}  =\mathfrak{A}$). Taking both sides of
this equality to the $n$-th power, we obtain%
\begin{align}
p_{1}^{n}  &  =\left(  \sum_{a\in\mathfrak{A}}x_{a}\right)  ^{n}\nonumber\\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}%
}x_{a_{1}}x_{a_{2}}\cdots x_{a_{n}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
product rule}\right) \nonumber\\
&  =\sum_{\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n}%
}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}
\label{pf.lem.sol.lyndon.grfun.sumxwtlw-n.1}%
\end{align}
(here, we have renamed the summation index $\left(  a_{1},a_{2},\ldots
,a_{n}\right)  $ as $\left(  w_{1},w_{2},\ldots,w_{n}\right)  $).

Each $w\in\mathfrak{A}^{n}$ can be written uniquely in the form $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  $. Hence, we can substitute $\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ for $w$ in the sum $\sum_{w\in
\mathfrak{A}^{n}}\xx_{w}$. We thus obtain%
\[
\sum_{w\in\mathfrak{A}^{n}}\xx_{w}=\sum_{\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  \in\mathfrak{A}^{n}}\underbrace{\xx_{\left(
w_{1},w_{2},\ldots,w_{n}\right)  }}_{\substack{=x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}\\\text{(by the definition of }\xx_{\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  }\text{)}}}=\sum_{\left(  w_{1},w_{2},\ldots
,w_{n}\right)  \in\mathfrak{A}^{n}}x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}.
\]
Comparing this with (\ref{pf.lem.sol.lyndon.grfun.sumxwtlw-n.1}), we obtain
$\sum_{w\in\mathfrak{A}^{n}}\xx_{w}=p_{1}^{n}$. This proves Lemma
\ref{lem.sol.lyndon.grfun.sumxwtlw-n}.
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.grfun.sumxwtlw}We have
\[
\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}t^{\ell\left(  w\right)  }%
=\dfrac{1}{1-p_{1}t}\ \ \ \ \ \ \ \ \ \ \text{in the ring }\left(
\kk \left[  \left[  \xx\right]  \right]  \right)  \left[  \left[
t\right]  \right]  .
\]

\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw}.]We have%
\begin{align*}
\underbrace{\sum_{w\in\mathfrak{A}^{\ast}}}_{\substack{=\sum\limits_{n\in
 \NN }\sum\limits_{w\in\mathfrak{A}^{n}}\\\text{(since }\mathfrak{A}%
^{\ast}=\bigsqcup_{n\in \NN }\mathfrak{A}^{n}\text{)}}}\xx%
_{w}t^{\ell\left(  w\right)  }  &  =\sum\limits_{n\in \NN }%
\sum\limits_{w\in\mathfrak{A}^{n}}\xx_{w}\underbrace{t^{\ell\left(
w\right)  }}_{\substack{=t^{n}\\\text{(since }\ell\left(  w\right)
=n\\\text{(since }w\in\mathfrak{A}^{n}\text{))}}}=\sum_{n\in \NN %
}\underbrace{\sum_{w\in\mathfrak{A}^{n}}\xx_{w}}_{\substack{=p_{1}%
^{n}\\\text{(by Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw-n})}}}t^{n}%
=\sum_{n\in \NN }\underbrace{p_{1}^{n}t^{n}}_{=\left(  p_{1}t\right)
^{n}}\\
&  =\sum_{n\in \NN }\left(  p_{1}t\right)  ^{n}=\dfrac{1}{1-p_{1}t}.
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw}.]We have%
\begin{align*}
\underbrace{\sum_{w\in\mathfrak{A}^{\ast}}}_{=\sum_{n\in \NN }%
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\ell\left(  w\right)  =n}%
}}\xx_{w}t^{\ell\left(  w\right)  }  &  =\sum_{n\in \NN }%
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\ell\left(  w\right)
=n}}\xx_{w}\underbrace{t^{\ell\left(  w\right)  }}_{\substack{=t^{n}%
\\\text{(since }\ell\left(  w\right)  =n\text{)}}}=\sum_{n\in \NN %
}\underbrace{\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\ell\left(
w\right)  =n}}}_{\substack{=\sum_{w\in\mathfrak{A}^{n}}\\\text{(since the
elements }w\in\mathfrak{A}^{\ast}\text{ satisfying }\ell\left(  w\right)
=n\\\text{are precisely the }w\in\mathfrak{A}^{n}\text{)}}}\xx_{w}%
t^{n}\\
&  =\sum_{n\in \NN }\sum_{w\in\mathfrak{A}^{n}}\xx_{w}t^{n}%
=\sum_{n\in \NN }\underbrace{\left(  \sum_{w\in\mathfrak{A}^{n}%
}\xx_{w}\right)  }_{\substack{=p_{1}^{n}\\\text{(by Lemma
\ref{lem.sol.lyndon.grfun.sumxwtlw-n})}}}t^{n}=\sum_{n\in \NN %
}\underbrace{p_{1}^{n}t^{n}}_{=\left(  p_{1}t\right)  ^{n}}\\
&  =\sum_{n\in \NN }\left(  p_{1}t\right)  ^{n}=\dfrac{1}{1-p_{1}t}%
\end{align*}
(since $\sum_{n\in \NN }q^{n}=\dfrac{1}{1-q}$ in the ring $\kk
\left[  \left[  q\right]  \right]  $ of formal power series). This proves
Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw}.
\end{proof}

\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.grfun.sumxwtlw2}We have%
\[
\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}t^{\ell\left(  w\right)  }%
}=\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}t^{\ell\left(  w\right)
}\ \ \ \ \ \ \ \ \ \ \text{in the ring }\left(  \kk \left[  \left[
\xx\right]  \right]  \right)  \left[  \left[  t\right]  \right]  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw2}.]Let $\mathfrak{M}$
denote the set of all finite multisets of Lyndon words. Define two maps
$\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$ and $\mathbf{n}%
:\mathfrak{A}^{\ast}\rightarrow\mathfrak{M}$ as in Proposition
\ref{prop.lyndon.CFL.bij}. Then, Proposition \ref{prop.lyndon.CFL.bij} shows
that the maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse bijections.

Recall that $\mathfrak{L}$ is the set of all Lyndon words. Thus, the Lyndon
words are precisely the elements of $\mathfrak{L}$. But the definition of
$\mathfrak{M}$ says that $\mathfrak{M}$ is the set of all finite multisets of
Lyndon words. In other words, $\mathfrak{M}$ is the set of all finite
multisets of elements of $\mathfrak{L}$ (since the Lyndon words are precisely
the elements of $\mathfrak{L}$).

Let $\mathfrak{N}$ be the set of all families $\left(  k_{w}\right)
_{w\in\mathfrak{L}}\in{ \NN }^{\mathfrak{L}}$ of nonnegative integers
(indexed by the Lyndon words) such that all but finitely many $w\in
\mathfrak{L}$ satisfy $k_{w}=0$. Thus,%
\begin{equation}
\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in
 \NN ^{\mathfrak{L}};\\\text{all but finitely many }w\in\mathfrak{L}%
\\\text{satisfy }k_{w}=0}}=\sum\limits_{\substack{\left(  k_{w}\right)
_{w\in\mathfrak{L}}\in\mathfrak{N}}}
\label{pf.prop.lyndon.grfun.CFLps.sum=sumN}%
\end{equation}
(an equality of summation signs). Proposition \ref{prop.multiset.bij.seqs}
(applied to $S=\mathfrak{L}$) shows that the map $\operatorname*{mult}%
:\mathfrak{M}\rightarrow\mathfrak{N}$ that sends each multiset $M\in
\mathfrak{M}$ to the family
\[
\left(  \left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}%
\]
is well-defined and is a bijection. Consider this map $\operatorname*{mult}$.
Since $\operatorname*{mult}:\mathfrak{M}\rightarrow\mathfrak{N}$ is a
bijection, its inverse $\operatorname*{mult}\nolimits^{-1}:\mathfrak{N}%
\rightarrow\mathfrak{M}$ is well-defined and also a bijection.

For any $w\in\mathfrak{L}$, the monomial $t^{\ell\left(  w\right)  }$ has
positive degree\footnote{\textit{Proof.} Let $w\in\mathfrak{L}$. Thus, $w$ is
a Lyndon word (since $\mathfrak{L}$ is the set of all Lyndon words), therefore
a nonempty word (since any Lyndon word is nonempty by definition). Hence,
$\ell\left(  w\right)  >0$. Therefore, the monomial $t^{\ell\left(  w\right)
}$ has positive degree. Qed.}. Hence, for any $w\in\mathfrak{L}$, the formal
power series $\dfrac{1}{1-\xx_{w}t^{\ell\left(  w\right)  }}$ is
well-defined and satisfies the equality%
\[
\dfrac{1}{1-\xx_{w}t^{\ell\left(  w\right)  }}=\sum_{k\in \NN %
}\left(  \xx_{w}t^{\ell\left(  w\right)  }\right)  ^{k}%
\]
(since $\dfrac{1}{1-q}=\sum_{k\in \NN }q^{k}$ in the ring $\kk
\left[  \left[  q\right]  \right]  $ of formal power series).

Multiplying these equalities over all $w\in\mathfrak{L}$, we obtain%
\begin{align}
\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}t^{\ell\left(  w\right)  }}
&  =\prod_{w\in\mathfrak{L}}\sum_{k\in \NN }\left(  \xx_{w}%
t^{\ell\left(  w\right)  }\right)  ^{k}\nonumber\\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in \NN ^{\mathfrak{L}};\\\text{all but finitely many }w\in
\mathfrak{L}\\\text{satisfy }k_{w}=0}}\prod_{w\in\mathfrak{L}}\left(
\xx_{w}t^{\ell\left(  w\right)  }\right)  ^{k_{w}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right) \nonumber\\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}}}\prod_{w\in\mathfrak{L}}\left(  \xx_{w}t^{\ell\left(
w\right)  }\right)  ^{k_{w}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.lyndon.grfun.CFLps.sum=sumN})}\right)  .
\label{pf.prop.lyndon.grfun.CFLps.0}%
\end{align}


The composition $\mathbf{m}\circ\operatorname{mult}^{-1}:\mathfrak{N}%
\rightarrow\mathfrak{A}^{\ast}$ of the bijections $\mathbf{m}$ and
$\operatorname{mult}^{-1}$ is clearly a bijection. We now claim the following:

\begin{statement}
\textit{Claim 1:} Let $\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}$. Then,%
\[
\xx_{\left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(
\left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  }t^{\ell\left(  \left(
\mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \right)  }=\prod_{w\in\mathfrak{L}}\left(
\xx_{w}t^{\ell\left(  w\right)  }\right)  ^{k_{w}}.
\]

\end{statement}

[\textit{Proof of Claim 1:} Let $M=\operatorname{mult}^{-1}\left(  \left(
k_{w}\right)  _{w\in\mathfrak{L}}\right)  $. Then,
\[
\left(  k_{w}\right)
_{w\in\mathfrak{L}}=\operatorname*{mult}M=\left(  \left(  \text{multiplicity
of }w\text{ in the multiset }M\right)  \right)  _{w\in\mathfrak{L}}
\]
(by the
definition of the map $\operatorname*{mult}$). In other words, every
$w\in\mathfrak{L}$ satisfies%
\begin{equation}
k_{w}=\left(  \text{multiplicity of }w\text{ in the multiset }M\right)  .
\label{pf.prop.lyndon.grfun.CFLps.1.pf.1}%
\end{equation}


\begin{vershort}
Moreover, $M=\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \in\mathfrak{M}$. In other words, $M$ is a finite
multiset of elements of $\mathfrak{L}$ (since $\mathfrak{M}$ is the set of all
finite multisets of elements of $\mathfrak{L}$). Let $a_{1},a_{2},\ldots
,a_{k}$ denote the elements of this multiset $M$ listed in decreasing order.
Then, the definition of $\mathbf{m}$ yields $\mathbf{m}\left(  M\right)
=a_{1}a_{2}\cdots a_{k}$. Also, $a_{1},a_{2},\ldots,a_{k}$ are elements of $M$
(by the definition of $a_{1},a_{2},\ldots,a_{k}$), and thus belong to
$\mathfrak{L}$ (since $M$ is a finite multiset of elements of $\mathfrak{L}$).
In other words, every $a_{i}$ belongs to $\mathfrak{L}$.
\end{vershort}

\begin{verlong}
Moreover, $M=\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \in\mathfrak{M}$. In other words, $M$ is a finite
multiset of elements of $\mathfrak{L}$ (since $\mathfrak{M}$ is the set of all
finite multisets of elements of $\mathfrak{L}$). Let $a_{1},a_{2},\ldots
,a_{k}$ denote the elements of this multiset $M$ listed in decreasing order.
Then, the definition of $\mathbf{m}$ yields $\mathbf{m}\left(  M\right)
=a_{1}a_{2}\cdots a_{k}$. Also, $a_{1},a_{2},\ldots,a_{k}$ are elements of $M$
(since $a_{1},a_{2},\ldots,a_{k}$ were defined to be the elements of this
multiset $M$ listed in decreasing order), and thus belong to $\mathfrak{L}$
(since each element of $M$ belongs to $\mathfrak{L}$ (because $M$ is a finite
multiset of elements of $\mathfrak{L}$)). In other words, every $a_{i}$
belongs to $\mathfrak{L}$.
\end{verlong}

We have $M=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}
_{\operatorname*{multiset}}$ (since $a_{1},a_{2},\ldots,a_{k}$ are the
elements of the multiset $M$ listed in decreasing order). Thus, each
$w\in\mathfrak{L}$ satisfies%
\begin{align*}
&  \left(  \text{multiplicity of }w\text{ in the multiset }\underbrace{M}%
_{=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}%
}\right) \\
&  =\left(  \text{multiplicity of }w\text{ in the multiset }\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname*{multiset}}\right) \\
&  =\left(  \text{the number of }i\in\left\{  1,2,\ldots,k\right\}  \text{
satisfying }a_{i}=w\right)  .
\end{align*}
Hence, each $w\in\mathfrak{L}$ satisfies%
\begin{align}
&  \left(  \text{the number of }i\in\left\{  1,2,\ldots,k\right\}  \text{
satisfying }a_{i}=w\right) \nonumber\\
&  =\left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\nonumber\\
&  =k_{w}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.lyndon.grfun.CFLps.1.pf.1})}\right)  .
\label{pf.prop.lyndon.grfun.CFLps.1.pf.1c}%
\end{align}


Now,%
\begin{align*}
\ell\left(  \underbrace{\mathbf{m}\left(  M\right)  }_{=a_{1}a_{2}\cdots
a_{k}}\right)   &  =\ell\left(  a_{1}a_{2}\cdots a_{k}\right)  =\ell\left(
a_{1}\right)  +\ell\left(  a_{2}\right)  +\cdots+\ell\left(  a_{k}\right) \\
&  =\underbrace{\sum_{i\in\left\{  1,2,\ldots,k\right\}  }}_{\substack{=\sum
\limits_{w\in\mathfrak{L}}\sum\limits_{\substack{i\in\left\{  1,2,\ldots
,k\right\}  ;\\a_{i}=w}}\\\text{(since every }a_{i}\text{ belongs to
}\mathfrak{L}\text{)}}}\ell\left(  a_{i}\right)  =\sum\limits_{w\in
\mathfrak{L}}\sum\limits_{\substack{i\in\left\{  1,2,\ldots,k\right\}
;\\a_{i}=w}}\ell\left(  \underbrace{a_{i}}_{\substack{=w\\\text{(since }%
a_{i}=w\text{)}}}\right) \\
&  =\sum\limits_{w\in\mathfrak{L}}\underbrace{\sum\limits_{\substack{i\in
\left\{  1,2,\ldots,k\right\}  ;\\a_{i}=w}}\ell\left(  w\right)  }_{=\left(
\text{the number of }i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfying
}a_{i}=w\right)  \cdot\ell\left(  w\right)  }\\
&  =\sum\limits_{w\in\mathfrak{L}}\underbrace{\left(  \text{the number of
}i\in\left\{  1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)
}_{\substack{=k_{w}\\\text{(by (\ref{pf.prop.lyndon.grfun.CFLps.1.pf.1c}))}%
}}\cdot\ell\left(  w\right) \\
&  =\sum\limits_{w\in\mathfrak{L}}k_{w}\cdot\ell\left(  w\right)  .
\end{align*}
Hence,%
\begin{equation}
t^{\ell\left(  \mathbf{m}\left(  M\right)  \right)  }=t^{\sum\limits_{w\in
\mathfrak{L}}k_{w}\cdot\ell\left(  w\right)  }=\prod_{w\in\mathfrak{L}%
}\underbrace{t^{k_{w}\cdot\ell\left(  w\right)  }}_{=\left(  t^{\ell\left(
w\right)  }\right)  ^{k_{w}}}=\prod_{w\in\mathfrak{L}}\left(  t^{\ell\left(
w\right)  }\right)  ^{k_{w}}. \label{pf.prop.lyndon.grfun.CFLps.1.pf.2}%
\end{equation}
Moreover, from $\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$, we
obtain%
\begin{align*}
\xx_{\mathbf{m}\left(  M\right)  }  &  =\xx_{a_{1}a_{2}\cdots
a_{k}}=\xx_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{k}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.sol.lyndon.grfun.xai}%
}\right) \\
&  =\underbrace{\prod_{i\in\left\{  1,2,\ldots,k\right\}  }}_{\substack{=\prod
\limits_{w\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{  1,2,\ldots
,k\right\}  ;\\a_{i}=w}}\\\text{(since every }a_{i}\text{ belongs to
}\mathfrak{L}\text{)}}}\xx_{a_{i}}=\prod\limits_{w\in\mathfrak{L}}%
\prod\limits_{\substack{i\in\left\{  1,2,\ldots,k\right\}  ;\\a_{i}%
=w}}\underbrace{\xx_{a_{i}}}_{\substack{=\xx_{w}\\\text{(since
}a_{i}=w\text{)}}}\\
&  =\prod\limits_{w\in\mathfrak{L}}\underbrace{\prod\limits_{\substack{i\in
\left\{  1,2,\ldots,k\right\}  ;\\a_{i}=w}}\xx_{w}}%
_{\substack{=\xx_{w}^{\left(  \text{the number of }i\in\left\{
1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)  }=\xx%
_{w}^{k_{w}}\\\text{(since }\left(  \text{the number of }i\in\left\{
1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)  =k_{w}\\\text{(by
(\ref{pf.prop.lyndon.grfun.CFLps.1.pf.1c})))}}}=\prod\limits_{w\in
\mathfrak{L}}\xx_{w}^{k_{w}}.
\end{align*}
Multiplying this equality by (\ref{pf.prop.lyndon.grfun.CFLps.1.pf.2}), we
obtain
\[
\xx_{\mathbf{m}\left(  M\right)  }t^{\ell\left(  \mathbf{m}\left(
M\right)  \right)  }=\left(  \prod\limits_{w\in\mathfrak{L}}\xx%
_{w}^{k_{w}}\right)  \prod_{w\in\mathfrak{L}}\left(  t^{\ell\left(  w\right)
}\right)  ^{k_{w}}=\prod\limits_{w\in\mathfrak{L}}\underbrace{\left(
\xx_{w}^{k_{w}}\left(  t^{\ell\left(  w\right)  }\right)  ^{k_{w}%
}\right)  }_{=\left(  \xx_{w}t^{\ell\left(  w\right)  }\right)
^{k_{w}}}=\prod_{w\in\mathfrak{L}}\left(  \xx_{w}t^{\ell\left(
w\right)  }\right)  ^{k_{w}}.
\]
In view of
\[
\mathbf{m}\left(  \underbrace{M}_{=\operatorname{mult}^{-1}\left(  \left(
k_{w}\right)  _{w\in\mathfrak{L}}\right)  }\right)  =\mathbf{m}\left(
\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}%
}\right)  \right)  =\left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  ,
\]
this rewrites as
\[
\xx_{\left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(
\left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  }t^{\ell\left(  \left(
\mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \right)  }=\prod_{w\in\mathfrak{L}}\left(
\xx_{w}t^{\ell\left(  w\right)  }\right)  ^{k_{w}}.
\]
This proves Claim 1.]

Now, in the ring $\left(  \kk \left[  \left[  \xx\right]
\right]  \right)  \left[  \left[  t\right]  \right]  $ of formal power series,
we have%
\begin{align*}
\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}t^{\ell\left(  w\right)  }  &
=\sum_{k\in\mathfrak{N}}\xx_{\left(  \mathbf{m}\circ\operatorname{mult}%
^{-1}\right)  \left(  k\right)  }t^{\ell\left(  \left(  \mathbf{m}%
\circ\operatorname{mult}^{-1}\right)  \left(  k\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  \mathbf{m}\circ\operatorname{mult}%
^{-1}\right)  \left(  k\right)  \text{ for }w\text{ in the sum,}\\
\text{since the map }\mathbf{m}\circ\operatorname{mult}^{-1}:\mathfrak{N}%
\rightarrow\mathfrak{A}^{\ast}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}}}\underbrace{\xx_{\left(  \mathbf{m}\circ
\operatorname{mult}^{-1}\right)  \left(  \left(  k_{w}\right)  _{w\in
\mathfrak{L}}\right)  }t^{\ell\left(  \left(  \mathbf{m}\circ
\operatorname{mult}^{-1}\right)  \left(  \left(  k_{w}\right)  _{w\in
\mathfrak{L}}\right)  \right)  }}_{\substack{=\prod_{w\in\mathfrak{L}}\left(
\xx_{w}t^{\ell\left(  w\right)  }\right)  ^{k_{w}}\\\text{(by Claim
1)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the summation index }k\text{ as }\left(
k_{w}\right)  _{w\in\mathfrak{L}}\text{,}\\
\text{since each }k\in\mathfrak{N}\text{ is a family indexed by elements of
}\mathfrak{L}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}}}\prod_{w\in\mathfrak{L}}\left(  \xx_{w}t^{\ell\left(
w\right)  }\right)  ^{k_{w}}\\
&  =\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}t^{\ell\left(  w\right)
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lyndon.grfun.CFLps.0}%
)}\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw2}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.CFLps}.]Lemma
\ref{lem.sol.lyndon.grfun.sumxwtlw2} yields%
\[
\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}t^{\ell\left(  w\right)  }%
}=\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}t^{\ell\left(  w\right)
}=\dfrac{1}{1-p_{1}t}%
\]
(by Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw}). This proves Proposition
\ref{prop.lyndon.grfun.CFLps}.
\end{proof}

Our next step is to show the following lemma (similar to Lemma
\ref{lem.sol.lyndon.grfun.sumxwtlw2}):

\begin{lemma}
\label{lem.sol.lyndon.grfun.sumxwpy}
In the power series ring
$\kk \left[  \left[  \xx,\mathbf{y}\right]  \right]  $, we have%
\[
\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}p_{\ell\left(  w\right)
}\left(  \mathbf{y}\right)  }=\sum_{w\in\mathfrak{A}^{\ast}}\xx%
_{w}p_{\operatorname*{CFLtype}w}\left(  \mathbf{y}\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwpy}.]Let $\mathfrak{M}$ denote
the set of all finite multisets of Lyndon words. Define two maps
$\mathbf{m}:\mathfrak{M}\rightarrow\mathfrak{A}^{\ast}$ and $\mathbf{n}%
:\mathfrak{A}^{\ast}\rightarrow\mathfrak{M}$ as in Proposition
\ref{prop.lyndon.CFL.bij}. Then, Proposition \ref{prop.lyndon.CFL.bij} shows
that the maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse bijections.

Recall that $\mathfrak{L}$ is the set of all Lyndon words. Thus, the Lyndon
words are precisely the elements of $\mathfrak{L}$. But the definition of
$\mathfrak{M}$ says that $\mathfrak{M}$ is the set of all finite multisets of
Lyndon words. In other words, $\mathfrak{M}$ is the set of all finite
multisets of elements of $\mathfrak{L}$ (since the Lyndon words are precisely
the elements of $\mathfrak{L}$).

Let $\mathfrak{N}$ be the set of all families $\left(  k_{w}\right)
_{w\in\mathfrak{L}}\in{ \NN }^{\mathfrak{L}}$ of nonnegative integers
(indexed by the Lyndon words) such that all but finitely many $w\in
\mathfrak{L}$ satisfy $k_{w}=0$. Thus,%
\begin{equation}
\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}\in
 \NN ^{\mathfrak{L}};\\\text{all but finitely many }w\in\mathfrak{L}%
\\\text{satisfy }k_{w}=0}}=\sum\limits_{\substack{\left(  k_{w}\right)
_{w\in\mathfrak{L}}\in\mathfrak{N}}}
\label{pf.lem.sol.lyndon.grfun.sumxwpy.sum=sumN}%
\end{equation}
(an equality of summation signs). Proposition \ref{prop.multiset.bij.seqs}
(applied to $S=\mathfrak{L}$) shows that the map $\operatorname*{mult}%
:\mathfrak{M}\rightarrow\mathfrak{N}$ that sends each multiset $M\in
\mathfrak{M}$ to the family
\[
\left(  \left(  \text{multiplicity of }w\text{ in the multiset }M\right)
\right)  _{w\in\mathfrak{L}}\in\mathfrak{N}%
\]
is well-defined and is a bijection. Consider this map $\operatorname*{mult}$.
Since $\operatorname*{mult}:\mathfrak{M}\rightarrow\mathfrak{N}$ is a
bijection, its inverse $\operatorname*{mult}\nolimits^{-1}:\mathfrak{N}%
\rightarrow\mathfrak{M}$ is well-defined and also a bijection.

For any $w\in\mathfrak{L}$, the power series $\xx_{w}p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  $ is homogeneous of positive
degree\footnote{\textit{Proof.} Let $w\in\mathfrak{L}$. Thus, $w$ is a Lyndon
word (since $\mathfrak{L}$ is the set of all Lyndon words), therefore a
nonempty word (since any Lyndon word is nonempty by definition). In other
words, $w\in\mathfrak{A}^{n}$ for some positive integer $n$. Consider this
$n$. From $w\in\mathfrak{A}^{n}$, we obtain $w=\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  $, so that $\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots
x_{w_{n}}$ (by the definition of $\xx_{w}$). Hence, $\xx_{w}$ is
a monomial of degree $n$. Hence, the power series $\xx_{w}$ is
homogeneous of degree $n$. Moreover, $w\in\mathfrak{A}^{n}$ entails
$\ell\left(  w\right)  =n$. Hence, $p_{\ell\left(  w\right)  }=p_{n}=x_{1}%
^{n}+x_{2}^{n}+x_{3}^{n}+\cdots$ (by the definition of $p_{n}$, since $n$ is a
positive integer). Substituting $y_{1},y_{2},y_{3},\ldots$ for $x_{1}%
,x_{2},x_{3},\ldots$ in this equality, we find $p_{\ell\left(  w\right)
}\left(  \mathbf{y}\right)  =y_{1}^{n}+y_{2}^{n}+y_{3}^{n}+\cdots$. Thus, the
power series $p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)  $ is
homogeneous of degree $n$.
\par
We now know that the power series $\xx_{w}$ and $p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  $ are both homogeneous of degree $n$.
Hence, their product $\xx_{w}p_{\ell\left(  w\right)  }\left(
\mathbf{y}\right)  $ is homogeneous of degree $n+n$. Since $n+n=2n$ is
positive (because $n$ is positive), this shows that $\xx_{w}%
p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)  $ is homogeneous of
positive degree. Qed.}. Hence, for any $w\in\mathfrak{L}$, the formal power
series $\dfrac{1}{1-\xx_{w}p_{\ell\left(  w\right)  }\left(
\mathbf{y}\right)  }$ is well-defined and satisfies the equality%
\[
\dfrac{1}{1-\xx_{w}p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)
}=\sum_{k\in \NN }\left(  \xx_{w}p_{\ell\left(  w\right)  }\left(
\mathbf{y}\right)  \right)  ^{k}%
\]
(since $\dfrac{1}{1-q}=\sum_{k\in \NN }q^{k}$ in the ring $\kk
\left[  \left[  q\right]  \right]  $ of formal power series).

Multiplying these equalities over all $w\in\mathfrak{L}$, we obtain%
\begin{align}
\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}p_{\ell\left(  w\right)
}\left(  \mathbf{y}\right)  }  &  =\prod_{w\in\mathfrak{L}}\sum_{k\in
 \NN }\left(  \xx_{w}p_{\ell\left(  w\right)  }\left(
\mathbf{y}\right)  \right)  ^{k}\nonumber\\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in \NN ^{\mathfrak{L}};\\\text{all but finitely many }w\in
\mathfrak{L}\\\text{satisfy }k_{w}=0}}\prod_{w\in\mathfrak{L}}\left(
\xx_{w}p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)  \right)
^{k_{w}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule}\right)
\nonumber\\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}}}\prod_{w\in\mathfrak{L}}\left(  \xx_{w}p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  \right)  ^{k_{w}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy.sum=sumN})}\right)  .
\label{pf.lem.sol.lyndon.grfun.sumxwpy.0}%
\end{align}


The composition $\mathbf{m}\circ\operatorname{mult}^{-1}:\mathfrak{N}%
\rightarrow\mathfrak{A}^{\ast}$ of the bijections $\mathbf{m}$ and
$\operatorname{mult}^{-1}$ is clearly a bijection. We now claim the following:

\begin{statement}
\textit{Claim 1:} Let $w\in\mathfrak{A}^{\ast}$. For each $v\in\mathfrak{L}$,
let $k_{v}\in \NN $ be such that%
\begin{equation}
k_{v}=\left(  \text{multiplicity of }v\text{ in the multiset }\mathbf{n}%
\left(  w\right)  \right)  . \label{pf.lem.sol.lyndon.grfun.sumxwpy.c1.ass}%
\end{equation}
Then,%
\[
p_{\operatorname*{CFLtype}w}=\prod_{v\in\mathfrak{L}}p_{\ell\left(  v\right)
}^{k_{v}}.
\]

\end{statement}

[\textit{Proof of Claim 1:} Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be
the CFL factorization of $w$. Then, the definition of the map $\mathbf{n}$
yields $\mathbf{n}\left(  w\right)  =\left\{  a_{1},a_{2},\ldots
,a_{k}\right\}  _{\operatorname{multiset}}$. Now, for each $v\in\mathfrak{L}$,
we have%
\begin{align}
k_{v}  &  =\left(  \text{multiplicity of }v\text{ in the multiset }%
\mathbf{n}\left(  w\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c1.ass})}\right) \nonumber\\
&  =\left(  \text{multiplicity of }v\text{ in the multiset }\left\{
a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{n}\left(  w\right)
=\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  _{\operatorname{multiset}}\right)
\nonumber\\
&  =\left(  \text{the number of all }i\in\left\{  1,2,\ldots,k\right\}  \text{
such that }a_{i}=v\right)  . \label{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.1}%
\end{align}


\begin{vershort}
Recall that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (by the definition of
``CFL factorization''). Thus, in particular,
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words, i.e.,
a tuple of elements of $\mathfrak{L}$ (since the Lyndon words are precisely
the elements of $\mathfrak{L}$). In other words, $a_{i}\in\mathfrak{L}$ for
each $i\in\left\{  1,2,\ldots,k\right\}  $.
\end{vershort}

\begin{verlong}
Recall that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (by the definition of
``CFL factorization''). Thus, in particular,
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words. In
other words, $a_{1},a_{2},\ldots,a_{k}$ are Lyndon words. In other words,
$a_{1},a_{2},\ldots,a_{k}$ belong to $\mathfrak{L}$ (since $\mathfrak{L}$ is
the set of all Lyndon words). In other words, $a_{i}\in\mathfrak{L}$ for each
$i\in\left\{  1,2,\ldots,k\right\}  $.
\end{verlong}

\begin{vershort}
Let us make a general observation about partitions: If $b_{1},b_{2}%
,\ldots,b_{k}$ are any $k$ positive integers, and if $\mu$ is the partition
whose parts are the positive integers $b_{1},b_{2},\ldots,b_{k}$ (listed in
decreasing order), then
\begin{equation}
p_{\lambda}=p_{b_{1}}p_{b_{2}}\cdots p_{b_{k}}.
\label{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.short.plam}%
\end{equation}
(Indeed, this follows from the definition of $p_{\lambda}$ in Definition
\ref{def.Lambda.peh}, since multiplication in $\Lambda$ is commutative.)

Now, recall again that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. Hence, $\operatorname*{CFLtype}w$ is the partition whose
parts are the positive integers $\ell\left(  a_{1}\right)  ,\ell\left(
a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $ (listed in decreasing
order)\footnote{by the definition of $\operatorname*{CFLtype}w$}. Hence,
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.short.plam}) (applied to
$b_{i}=a_{i}$ and $\mu=\operatorname*{CFLtype}w$) yields%
\begin{align*}
p_{\operatorname*{CFLtype}w}  &  =p_{\ell\left(  a_{1}\right)  }p_{\ell\left(
a_{2}\right)  }\cdots p_{\ell\left(  a_{k}\right)  }=\underbrace{\prod
_{i\in\left\{  1,2,\ldots,k\right\}  }}_{\substack{=\prod_{v\in\mathfrak{L}%
}\prod\limits_{\substack{i\in\left\{  1,2,\ldots,k\right\}  ;\\a_{i}%
=v}}\\\text{(since }a_{i}\in\mathfrak{L}\text{ for each }i\in\left\{
1,2,\ldots,k\right\}  \text{)}}}p_{\ell\left(  a_{i}\right)  }\\
&  =\prod_{v\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=v}}\underbrace{p_{\ell\left(  a_{i}\right)  }%
}_{\substack{=p_{\ell\left(  v\right)  }\\\text{(since }a_{i}=v\text{)}%
}}=\prod_{v\in\mathfrak{L}}\underbrace{\prod\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=v}}p_{\ell\left(  v\right)  }}%
_{\substack{=p_{\ell\left(  v\right)  }^{\left(  \text{the number of all }%
i\in\left\{  1,2,\ldots,k\right\}  \text{ such that }a_{i}=v\right)  }%
=p_{\ell\left(  v\right)  }^{k_{v}}\\\text{(since }\left(  \text{the number of
all }i\in\left\{  1,2,\ldots,k\right\}  \text{ such that }a_{i}=v\right)
=k_{v}\\\text{(by (\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.1})))}}}\\
&  =\prod_{v\in\mathfrak{L}}p_{\ell\left(  v\right)  }^{k_{v}}.
\end{align*}
This proves Claim 1.]
\end{vershort}

\begin{verlong}
Again, recall that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. Hence, the CFL type of $w$ is the partition whose parts
are the positive integers $\ell\left(  a_{1}\right)  ,\ell\left(
a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $ (listed in decreasing
order)\footnote{by the definition of ``CFL type''}. In other words,
the CFL type of $w$ is the partition
whose parts are the entries of the $k$-tuple $\left(  \ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\right)  $ (listed in decreasing order). Therefore, the CFL type of $w$ is a
rearrangement of the $k$-tuple $\left(  \ell\left(  a_{1}\right)  ,\ell\left(
a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  \right)  $. In other words,
$\operatorname*{CFLtype}w$ is a rearrangement of the $k$-tuple $\left(
\ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(
a_{k}\right)  \right)  $ (since the CFL type of $w$ has been denoted by
$\operatorname*{CFLtype}w$). In other words, there exists a permutation
$\sigma\in\Symm_{k}$ such that%
\begin{equation}
\operatorname*{CFLtype}w=\left(  \ell\left(  a_{\sigma\left(  1\right)
}\right)  ,\ell\left(  a_{\sigma\left(  2\right)  }\right)  ,\ldots
,\ell\left(  a_{\sigma\left(  k\right)  }\right)  \right)  .
\label{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.2}%
\end{equation}
Consider this $\sigma$. The map $\sigma$ is a permutation of $\left\{
1,2,\ldots,k\right\}  $ (since $\sigma\in\Symm_{k}$), thus a bijection
from $\left\{  1,2,\ldots,k\right\}  $ to $\left\{  1,2,\ldots,k\right\}  $.

From (\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.2}), we obtain%
\begin{align*}
p_{\operatorname*{CFLtype}w}  &  =p_{\left(  \ell\left(  a_{\sigma\left(
1\right)  }\right)  ,\ell\left(  a_{\sigma\left(  2\right)  }\right)
,\ldots,\ell\left(  a_{\sigma\left(  k\right)  }\right)  \right)  }%
=p_{\ell\left(  a_{\sigma\left(  1\right)  }\right)  }p_{\ell\left(
a_{\sigma\left(  2\right)  }\right)  }\cdots p_{\ell\left(  a_{\sigma\left(
k\right)  }\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }p_{\left(
\ell\left(  a_{\sigma\left(  1\right)  }\right)  ,\ell\left(  a_{\sigma\left(
2\right)  }\right)  ,\ldots,\ell\left(  a_{\sigma\left(  k\right)  }\right)
\right)  }\right) \\
&  =\prod_{i\in\left\{  1,2,\ldots,k\right\}  }p_{\ell\left(  a_{\sigma\left(
i\right)  }\right)  }=\underbrace{\prod_{i\in\left\{  1,2,\ldots,k\right\}  }%
}_{\substack{=\prod_{v\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=v}}\\\text{(since }a_{i}\in\mathfrak{L}\text{
for each }i\in\left\{  1,2,\ldots,k\right\}  \text{)}}}p_{\ell\left(
a_{i}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }i\text{ for }\sigma\left(  i\right)  \text{
in the product}\\
\text{since }\sigma\text{ is a bijection from }\left\{  1,2,\ldots,k\right\}
\text{ to }\left\{  1,2,\ldots,k\right\}
\end{array}
\right) \\
&  =\prod_{v\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=v}}\underbrace{p_{\ell\left(  a_{i}\right)  }%
}_{\substack{=p_{\ell\left(  v\right)  }\\\text{(since }a_{i}=v\text{)}%
}}=\prod_{v\in\mathfrak{L}}\underbrace{\prod\limits_{\substack{i\in\left\{
1,2,\ldots,k\right\}  ;\\a_{i}=v}}p_{\ell\left(  v\right)  }}%
_{\substack{=p_{\ell\left(  v\right)  }^{\left(  \text{the number of all }%
i\in\left\{  1,2,\ldots,k\right\}  \text{ such that }a_{i}=v\right)  }%
=p_{\ell\left(  v\right)  }^{k_{v}}\\\text{(since }\left(  \text{the number of
all }i\in\left\{  1,2,\ldots,k\right\}  \text{ such that }a_{i}=v\right)
=k_{v}\\\text{(by (\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c1.pf.1})))}}}\\
&  =\prod_{v\in\mathfrak{L}}p_{\ell\left(  v\right)  }^{k_{v}}.
\end{align*}
This proves Claim 1.]
\end{verlong}

\begin{statement}
\textit{Claim 2:} Let $\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}$. Then,%
\[
\xx_{\left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(
\left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  }p_{\operatorname*{CFLtype}%
\left(  \left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(
\left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right)  }\left(
\mathbf{y}\right)  =\prod_{w\in\mathfrak{L}}\left(  \xx_{w}%
p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)  \right)  ^{k_{w}}.
\]

\end{statement}

[\textit{Proof of Claim 2:} Let $M=\operatorname{mult}^{-1}\left(  \left(
k_{w}\right)  _{w\in\mathfrak{L}}\right)  $. Then,
\[
\left(  k_{w}\right)
_{w\in\mathfrak{L}}=\operatorname*{mult}M=\left(  \left(  \text{multiplicity
of }w\text{ in the multiset }M\right)  \right)  _{w\in\mathfrak{L}}
\]
(by the definition of the map $\operatorname*{mult}$). In other words,
every $w\in\mathfrak{L}$ satisfies%
\[
k_{w}=\left(  \text{multiplicity of }w\text{ in the multiset }M\right)  .
\]
Renaming $w$ as $v$ in this result, we obtain the following: Every
$v\in\mathfrak{L}$ satisfies%
\begin{equation}
k_{v}=\left(  \text{multiplicity of }v\text{ in the multiset }M\right)  .
\label{pf.lem.sol.lyndon.grfun.sumxwpy.c2.pf.1}%
\end{equation}


\begin{vershort}
Moreover, $M=\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \in\mathfrak{M}$. In other words, $M$ is a finite
multiset of elements of $\mathfrak{L}$ (since $\mathfrak{M}$ is the set of all
finite multisets of elements of $\mathfrak{L}$). Let $a_{1},a_{2},\ldots
,a_{k}$ denote the elements of this multiset $M$ listed in decreasing order.
Then, the definition of $\mathbf{m}$ yields $\mathbf{m}\left(  M\right)
=a_{1}a_{2}\cdots a_{k}$. Also, $a_{1},a_{2},\ldots,a_{k}$ are elements of $M$
(by the definition of $a_{1},a_{2},\ldots,a_{k}$), and thus belong to
$\mathfrak{L}$ (since $M$ is a finite multiset of elements of $\mathfrak{L}$).
In other words, every $a_{i}$ belongs to $\mathfrak{L}$.
\end{vershort}

\begin{verlong}
Moreover, $M=\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \in\mathfrak{M}$. In other words, $M$ is a finite
multiset of elements of $\mathfrak{L}$ (since $\mathfrak{M}$ is the set of all
finite multisets of elements of $\mathfrak{L}$). Let $a_{1},a_{2},\ldots
,a_{k}$ denote the elements of this multiset $M$ listed in decreasing order.
Then, the definition of $\mathbf{m}$ yields $\mathbf{m}\left(  M\right)
=a_{1}a_{2}\cdots a_{k}$. Also, $a_{1},a_{2},\ldots,a_{k}$ are elements of $M$
(since $a_{1},a_{2},\ldots,a_{k}$ were defined to be the elements of this
multiset $M$ listed in decreasing order), and thus belong to $\mathfrak{L}$
(since each element of $M$ belongs to $\mathfrak{L}$ (because $M$ is a finite
multiset of elements of $\mathfrak{L}$)). In other words, every $a_{i}$
belongs to $\mathfrak{L}$.
\end{verlong}

Each $w\in\mathfrak{L}$ satisfies%
\begin{equation}
\left(  \text{the number of }i\in\left\{  1,2,\ldots,k\right\}  \text{
satisfying }a_{i}=w\right)  =k_{w}.
\label{pf.lem.sol.lyndon.grfun.sumxwpy.c2.pf.num}%
\end{equation}
(Indeed, this is precisely the equality
(\ref{pf.prop.lyndon.grfun.CFLps.1.pf.1c}) from the proof of Claim 1 in the
proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwtlw2}; and it can be proved in
the exact same way as the latter equality.)

Now, recall that the maps $\mathbf{m}$ and $\mathbf{n}$ are mutually inverse.
Hence, $\mathbf{n}\circ\mathbf{m}=\operatorname*{id}$. Thus, $\mathbf{n}%
\left(  \mathbf{m}\left(  M\right)  \right)  =\underbrace{\left(
\mathbf{n}\circ\mathbf{m}\right)  }_{=\operatorname*{id}}\left(  M\right)
=\operatorname*{id}\left(  M\right)  =M$. In other words, $M=\mathbf{n}\left(
\mathbf{m}\left(  M\right)  \right)  $. Now, every $v\in\mathfrak{L}$
satisfies%
\begin{align*}
k_{v}  &  =\left(  \text{multiplicity of }v\text{ in the multiset
}\underbrace{M}_{=\mathbf{n}\left(  \mathbf{m}\left(  M\right)  \right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c2.pf.1})}\right) \\
&  =\left(  \text{multiplicity of }v\text{ in the multiset }\mathbf{n}\left(
\mathbf{m}\left(  M\right)  \right)  \right)  .
\end{align*}
Thus, Claim 1 (applied to $w=\mathbf{m}\left(  M\right)  $) yields
\[
p_{\operatorname*{CFLtype}\left(  \mathbf{m}\left(  M\right)  \right)  }%
=\prod_{v\in\mathfrak{L}}p_{\ell\left(  v\right)  }^{k_{v}}=\prod
_{w\in\mathfrak{L}}p_{\ell\left(  w\right)  }^{k_{w}}%
\]
(here, we have renamed the index $v$ as $w$ in the product). Substituting the
variables $\mathbf{y}=\left(  y_{1},y_{2},y_{3},\ldots\right)  $ for
$\xx=\left(  x_{1},x_{2},x_{3},\ldots\right)  $ on both sides of this
equality, we obtain%
\begin{equation}
p_{\operatorname*{CFLtype}\left(  \mathbf{m}\left(  M\right)  \right)
}\left(  \mathbf{y}\right)  =\left(  \prod_{w\in\mathfrak{L}}p_{\ell\left(
w\right)  }^{k_{w}}\right)  \left(  \mathbf{y}\right)  =\prod_{w\in
\mathfrak{L}}\left(  p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)
\right)  ^{k_{w}}. \label{pf.lem.sol.lyndon.grfun.sumxwpy.c2.pf.2}%
\end{equation}


Moreover, from $\mathbf{m}\left(  M\right)  =a_{1}a_{2}\cdots a_{k}$, we
obtain%
\begin{align*}
\xx_{\mathbf{m}\left(  M\right)  }  &  =\xx_{a_{1}a_{2}\cdots
a_{k}}=\xx_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{k}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.sol.lyndon.grfun.xai}%
}\right) \\
&  =\underbrace{\prod_{i\in\left\{  1,2,\ldots,k\right\}  }}_{\substack{=\prod
\limits_{w\in\mathfrak{L}}\prod\limits_{\substack{i\in\left\{  1,2,\ldots
,k\right\}  ;\\a_{i}=w}}\\\text{(since every }a_{i}\text{ belongs to
}\mathfrak{L}\text{)}}}\xx_{a_{i}}=\prod\limits_{w\in\mathfrak{L}}%
\prod\limits_{\substack{i\in\left\{  1,2,\ldots,k\right\}  ;\\a_{i}%
=w}}\underbrace{\xx_{a_{i}}}_{\substack{=\xx_{w}\\\text{(since
}a_{i}=w\text{)}}}\\
&  =\prod\limits_{w\in\mathfrak{L}}\underbrace{\prod\limits_{\substack{i\in
\left\{  1,2,\ldots,k\right\}  ;\\a_{i}=w}}\xx_{w}}%
_{\substack{=\xx_{w}^{\left(  \text{the number of }i\in\left\{
1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)  }=\xx%
_{w}^{k_{w}}\\\text{(since }\left(  \text{the number of }i\in\left\{
1,2,\ldots,k\right\}  \text{ satisfying }a_{i}=w\right)  =k_{w}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c2.pf.num})))}}}\\
&  =\prod\limits_{w\in\mathfrak{L}}\xx_{w}^{k_{w}}.
\end{align*}
Multiplying this equality by (\ref{pf.lem.sol.lyndon.grfun.sumxwpy.c2.pf.2}),
we obtain
\begin{align*}
\xx_{\mathbf{m}\left(  M\right)  }p_{\operatorname*{CFLtype}\left(
\mathbf{m}\left(  M\right)  \right)  }\left(  \mathbf{y}\right)   &  =\left(
\prod\limits_{w\in\mathfrak{L}}\xx_{w}^{k_{w}}\right)  \prod
_{w\in\mathfrak{L}}\left(  p_{\ell\left(  w\right)  }\left(  \mathbf{y}%
\right)  \right)  ^{k_{w}}=\prod\limits_{w\in\mathfrak{L}}\underbrace{\left(
\xx_{w}^{k_{w}}\left(  p_{\ell\left(  w\right)  }\left(  \mathbf{y}%
\right)  \right)  ^{k_{w}}\right)  }_{=\left(  \xx_{w}p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  \right)  ^{k_{w}}}\\
&  =\prod_{w\in\mathfrak{L}}\left(  \xx_{w}p_{\ell\left(  w\right)
}\left(  \mathbf{y}\right)  \right)  ^{k_{w}}.
\end{align*}
In view of
\[
\mathbf{m}\left(  \underbrace{M}_{=\operatorname{mult}^{-1}\left(  \left(
k_{w}\right)  _{w\in\mathfrak{L}}\right)  }\right)  =\mathbf{m}\left(
\operatorname{mult}^{-1}\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}%
}\right)  \right)  =\left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)
\left(  \left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  ,
\]
this rewrites as%
\[
\xx_{\left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(
\left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  }p_{\operatorname*{CFLtype}%
\left(  \left(  \mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(
\left(  k_{w}\right)  _{w\in\mathfrak{L}}\right)  \right)  }\left(
\mathbf{y}\right)  =\prod_{w\in\mathfrak{L}}\left(  \xx_{w}%
p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)  \right)  ^{k_{w}}.
\]
This proves Claim 2.]

Now, in the ring $\kk \left[  \left[  \xx,\mathbf{y}\right]
\right]  $ of formal power series, we have%
\begin{align*}
\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right)   &  =\sum_{k\in\mathfrak{N}}\xx_{\left(
\mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(  k\right)
}p_{\operatorname*{CFLtype}\left(  \left(  \mathbf{m}\circ\operatorname{mult}%
^{-1}\right)  \left(  k\right)  \right)  }\left(  \mathbf{y}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  \mathbf{m}\circ\operatorname{mult}%
^{-1}\right)  \left(  k\right)  \text{ for }w\text{ in the sum,}\\
\text{since the map }\mathbf{m}\circ\operatorname{mult}^{-1}:\mathfrak{N}%
\rightarrow\mathfrak{A}^{\ast}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}}}\underbrace{\xx_{\left(  \mathbf{m}\circ
\operatorname{mult}^{-1}\right)  \left(  \left(  k_{w}\right)  _{w\in
\mathfrak{L}}\right)  }p_{\operatorname*{CFLtype}\left(  \left(
\mathbf{m}\circ\operatorname{mult}^{-1}\right)  \left(  \left(  k_{w}\right)
_{w\in\mathfrak{L}}\right)  \right)  }\left(  \mathbf{y}\right)
}_{\substack{=\prod_{w\in\mathfrak{L}}\left(  \xx_{w}p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  \right)  ^{k_{w}}\\\text{(by Claim 2)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the summation index }k\text{ as }\left(
k_{w}\right)  _{w\in\mathfrak{L}}\text{,}\\
\text{since each }k\in\mathfrak{N}\text{ is a family indexed by elements of
}\mathfrak{L}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\left(  k_{w}\right)  _{w\in\mathfrak{L}}%
\in\mathfrak{N}}}\prod_{w\in\mathfrak{L}}\left(  \xx_{w}p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  \right)  ^{k_{w}}\\
&  =\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}p_{\ell\left(  w\right)
}\left(  \mathbf{y}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy.0})}\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.grfun.sumxwpy}.
\end{proof}

The following lemma is a slight restatement of Lemma
\ref{lem.sol.lyndon.grfun.sumxwpy}:

\begin{lemma}
\label{lem.sol.lyndon.grfun.sumxwpy2}Consider the power series ring
$\kk \left[  \left[  \xx,\mathbf{y}\right]  \right]  $. For each
word $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n}$, we
define a monomial $\mathbf{y}_{w}$ in $\kk \left[  \left[
\mathbf{y}\right]  \right]  $ by $\mathbf{y}_{w}=y_{w_{1}}y_{w_{2}}\cdots
y_{w_{n}}$. Then,%
\[
\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}%
=\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sumxwpy2}.]Lemma
\ref{lem.sol.lyndon.grfun.sumxwpy} yields%
\begin{equation}
\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}p_{\ell\left(  w\right)
}\left(  \mathbf{y}\right)  }=\sum_{w\in\mathfrak{A}^{\ast}}\xx%
_{w}p_{\operatorname*{CFLtype}w}\left(  \mathbf{y}\right)  .
\label{pf.lem.sol.lyndon.grfun.sumxwpy2.1}%
\end{equation}


Proposition \ref{prop.lyndon.grfun.CFLps} yields%
\begin{equation}
\dfrac{1}{1-p_{1}t}=\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}%
t^{\ell\left(  w\right)  }}. \label{pf.lem.sol.lyndon.grfun.sumxwpy2.2}%
\end{equation}


Now, fix $v\in\mathfrak{L}$. Then, $v$ is a Lyndon word (since $\mathfrak{L}$
is the set of all Lyndon words), and thus is nonempty (since any Lyndon word
is nonempty). Hence, $\ell\left(  v\right)  \geq1$. Thus, the definition of
$p_{\ell\left(  v\right)  }$ yields $p_{\ell\left(  v\right)  }=x_{1}%
^{\ell\left(  v\right)  }+x_{2}^{\ell\left(  v\right)  }+x_{3}^{\ell\left(
v\right)  }+\cdots$. Substituting the variables $\mathbf{y}=\left(
y_{1},y_{2},y_{3},\ldots\right)  $ for $\xx=\left(  x_{1},x_{2}%
,x_{3},\ldots\right)  $ on both sides of this equality, we obtain%
\begin{equation}
p_{\ell\left(  v\right)  }\left(  \mathbf{y}\right)  =y_{1}^{\ell\left(
v\right)  }+y_{2}^{\ell\left(  v\right)  }+y_{3}^{\ell\left(  v\right)
}+\cdots. \label{pf.lem.sol.lyndon.grfun.sumxwpy2.3}%
\end{equation}


On the other hand, the definition of $p_{1}$ yields $p_{1}=x_{1}^{1}+x_{2}%
^{1}+x_{3}^{1}+\cdots=x_{1}+x_{2}+x_{3}+\cdots$. Substituting $y_{1}%
^{\ell\left(  v\right)  },y_{2}^{\ell\left(  v\right)  },y_{3}^{\ell\left(
v\right)  },\ldots$ for $x_{1},x_{2},x_{3},\ldots$ on both sides of this
equality, we obtain%
\[
p_{1}\left(  y_{1}^{\ell\left(  v\right)  },y_{2}^{\ell\left(  v\right)
},y_{3}^{\ell\left(  v\right)  },\ldots\right)  =y_{1}^{\ell\left(  v\right)
}+y_{2}^{\ell\left(  v\right)  }+y_{3}^{\ell\left(  v\right)  }+\cdots.
\]
Comparing this with (\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.3}), we obtain%
\begin{equation}
p_{1}\left(  y_{1}^{\ell\left(  v\right)  },y_{2}^{\ell\left(  v\right)
},y_{3}^{\ell\left(  v\right)  },\ldots\right)  =p_{\ell\left(  v\right)
}\left(  \mathbf{y}\right)  . \label{pf.lem.sol.lyndon.grfun.sumxwpy2.4}%
\end{equation}


Every $k\in \NN $ and every word $w\in\mathfrak{A}^{\ast}$ satisfy%
\begin{equation}
\xx_{w}\left(  y_{1}^{k},y_{2}^{k},y_{3}^{k},\ldots\right)
=\mathbf{y}_{w}^{k} \label{pf.lem.sol.lyndon.grfun.sumxwpy2.5}%
\end{equation}
(where, of course, $\xx_{w}\left(  y_{1}^{k},y_{2}^{k},y_{3}^{k}%
,\ldots\right)  $ denotes the result of substituting $y_{1}^{k},y_{2}%
^{k},y_{3}^{k},\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in the polynomial
$\xx_{w}$).

[\textit{Proof of (\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.5}):} Let
$k\in \NN $. Let $w\in\mathfrak{A}^{\ast}$ be a word. Write the word $w$
in the form $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $. Then,
$\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ (by the definition of
$\xx_{w}$) and $\mathbf{y}_{w}=y_{w_{1}}y_{w_{2}}\cdots y_{w_{n}}$ (by
the definition of $\mathbf{y}_{w}$), so that $y_{w_{1}}y_{w_{2}}\cdots
y_{w_{n}}=\mathbf{y}_{w}$. Now, substituting $y_{1}^{k},y_{2}^{k},y_{3}%
^{k},\ldots$ for $x_{1},x_{2},x_{3},\ldots$ on both sides of the equality
$\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$, we obtain
\[
\xx_{w}\left(  y_{1}^{k},y_{2}^{k},y_{3}^{k},\ldots\right)  =y_{w_{1}%
}^{k}y_{w_{2}}^{k}\cdots y_{w_{n}}^{k}=\left(  \underbrace{y_{w_{1}}y_{w_{2}%
}\cdots y_{w_{n}}}_{=\mathbf{y}_{w}}\right)  ^{k}=\mathbf{y}_{w}^{k}.
\]
This proves (\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.5}).]

Furthermore, the monomial $\xx_{v}$ has positive
degree\footnote{\textit{Proof.} Write the word $v$ in the form $v=\left(
v_{1},v_{2},\ldots,v_{n}\right)  $. Then, $n=\ell\left(  v\right)  \geq1$.
Hence, $n$ is positive. But the definition of $\xx_{v}$ yields
$\xx_{v}=x_{v_{1}}x_{v_{2}}\cdots x_{v_{n}}$ (since $v=\left(
v_{1},v_{2},\ldots,v_{n}\right)  $). Hence, the monomial $\xx_{v}$ has
degree $n$. Thus, the monomial $\xx_{v}$ has positive degree (since $n$
is positive).}.

The equality (\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.2}) is an equality of
formal power series in $\left(  \kk \left[  \left[  \xx\right]
\right]  \right)  \left[  \left[  t\right]  \right]  = \kk \left[
\left[  x_{1},x_{2},x_{3},\ldots,t\right]  \right]  $. Substituting
$y_{1}^{\ell\left(  v\right)  },y_{2}^{\ell\left(  v\right)  },y_{3}%
^{\ell\left(  v\right)  },\ldots$ for $x_{1},x_{2},x_{3},\ldots$ on both sides
of this equality (while keeping $t$ unchanged for now), we obtain%
\begin{align*}
\dfrac{1}{1-p_{1}\left(  y_{1}^{\ell\left(  v\right)  },y_{2}^{\ell\left(
v\right)  },y_{3}^{\ell\left(  v\right)  },\ldots\right)  t}  &  =\prod
_{w\in\mathfrak{L}}\underbrace{\dfrac{1}{1-\xx_{w}\left(  y_{1}%
^{\ell\left(  v\right)  },y_{2}^{\ell\left(  v\right)  },y_{3}^{\ell\left(
v\right)  },\ldots\right)  t^{\ell\left(  w\right)  }}}_{\substack{=\dfrac
{1}{1-\mathbf{y}_{w}^{\ell\left(  v\right)  }t^{\ell\left(  w\right)  }%
}\\\text{(since }\xx_{w}\left(  y_{1}^{\ell\left(  v\right)  }%
,y_{2}^{\ell\left(  v\right)  },y_{3}^{\ell\left(  v\right)  },\ldots\right)
=\mathbf{y}_{w}^{\ell\left(  v\right)  }\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.5}), applied to }k=\ell\left(
v\right)  \text{))}}}\\
&  =\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{w}^{\ell\left(  v\right)
}t^{\ell\left(  w\right)  }}=\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}%
_{u}^{\ell\left(  v\right)  }t^{\ell\left(  u\right)  }}%
\end{align*}
(here, we have renamed the index $w$ as $u$ in the product). In view of
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.4}), this rewrites as%
\[
\dfrac{1}{1-p_{\ell\left(  v\right)  }\left(  \mathbf{y}\right)  t}%
=\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{u}^{\ell\left(  v\right)
}t^{\ell\left(  u\right)  }}.
\]
This is an equality of formal power series in $\left(  \kk \left[
\left[  \mathbf{y}\right]  \right]  \right)  \left[  \left[  t\right]
\right]  $. We can substitute $\xx_{v}$ for $t$ in this equality (since
the monomial $\xx_{v}$ has positive degree), and thus obtain the
equality%
\begin{equation}
\dfrac{1}{1-p_{\ell\left(  v\right)  }\left(  \mathbf{y}\right)
\xx_{v}}=\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{u}%
^{\ell\left(  v\right)  }\xx_{v}^{\ell\left(  u\right)  }}
\label{pf.lem.sol.lyndon.grfun.sumxwpy2.11}%
\end{equation}
in the ring $\kk \left[  \left[  \xx,\mathbf{y}\right]  \right]$.

Now, forget that we fixed $v$. We thus have proved the equality
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.11}) for each $v\in\mathfrak{L}$.

Now,
\begin{align*}
\prod_{w\in\mathfrak{L}}\underbrace{\dfrac{1}{1-\xx_{w}p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  }}_{\substack{=\dfrac{1}{1-p_{\ell
\left(  w\right)  }\left(  \mathbf{y}\right)  \xx_{w}}\\\text{(since
}\xx_{w}p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)
=p_{\ell\left(  w\right)  }\left(  \mathbf{y}\right)  \xx_{w}\text{)}%
}}  &  =\prod_{w\in\mathfrak{L}}\underbrace{\dfrac{1}{1-p_{\ell\left(
w\right)  }\left(  \mathbf{y}\right)  \xx_{w}}}_{\substack{=\prod
_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{u}^{\ell\left(  w\right)
}\xx_{w}^{\ell\left(  u\right)  }}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.11}), applied to }v=w\text{)}}}\\
&  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\underbrace{\dfrac
{1}{1-\mathbf{y}_{u}^{\ell\left(  w\right)  }\xx_{w}^{\ell\left(
u\right)  }}}_{\substack{=\dfrac{1}{1-\xx_{w}^{\ell\left(  u\right)
}\mathbf{y}_{u}^{\ell\left(  w\right)  }}\\\text{(since }\mathbf{y}_{u}%
^{\ell\left(  w\right)  }\xx_{w}^{\ell\left(  u\right)  }%
=\xx_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)
}\text{)}}}=\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac
{1}{1-\xx_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(
w\right)  }}.
\end{align*}
Comparing this with (\ref{pf.lem.sol.lyndon.grfun.sumxwpy2.1}), we obtain%
\[
\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}%
=\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right)  .
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.sumxwpy2}.
\end{proof}

Now, we can prove Proposition \ref{prop.lyndon.grfun.cauchy}:

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.cauchy}.]Define a monomial
$\mathbf{y}_{w}$ for each word $w$ as in Proposition
\ref{prop.lyndon.grfun.cauchy}(b). We have%
\begin{align}
\sum_{\lambda\in \Par}\underbrace{\mathbf{GR}_{\lambda}\left(
\xx\right)  }_{\substack{=\sum\limits_{\substack{w\in\mathfrak{A}%
^{\ast};\\\operatorname*{CFLtype}w=\lambda}}\xx_{w}\\\text{(by the
definition of }\mathbf{GR}_{\lambda}\text{)}}}p_{\lambda}\left(
\mathbf{y}\right)   &  =\sum_{\lambda\in \Par}\sum
\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda
}}\xx_{w}\underbrace{p_{\lambda}\left(  \mathbf{y}\right)
}_{\substack{=p_{\operatorname*{CFLtype}w}\left(  \mathbf{y}\right)
\\\text{(since }\lambda=\operatorname*{CFLtype}w\text{)}}}\nonumber\\
&  =\underbrace{\sum_{\lambda\in \Par}\sum
\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda
}}}_{\substack{=\sum_{w\in\mathfrak{A}^{\ast}}\\\text{(since }%
\operatorname*{CFLtype}w\in\Par\\\text{for each }%
w\in\mathfrak{A}^{\ast}\text{)}}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right) \nonumber\\
&  =\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right) \label{pf.prop.lyndon.grfun.cauchy.short.0}\\
&  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}
\label{pf.prop.lyndon.grfun.cauchy.short.1}%
\end{align}
(by Lemma \ref{lem.sol.lyndon.grfun.sumxwpy2}).

The same argument (but with the roles of the sets of variables $\xx%
=\left(  x_{1},x_{2},x_{3},\ldots\right)  $ and $\mathbf{y}=\left(
y_{1},y_{2},y_{3},\ldots\right)  $ interchanged) yields%
\[
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\mathbf{y}\right)  p_{\lambda}\left(  \xx\right)  =\prod_{w\in
\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{w}^{\ell\left(
u\right)  }\xx_{u}^{\ell\left(  w\right)  }}.
\]
Hence,%
\begin{align*}
&  \sum_{\lambda\in \Par}\underbrace{p_{\lambda}\left(
\xx\right)  \mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)
}_{=\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  p_{\lambda}\left(
\xx\right)  }\\
&  =\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\mathbf{y}\right)  p_{\lambda}\left(  \xx\right)  =\prod_{w\in
\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{w}^{\ell\left(
u\right)  }\xx_{u}^{\ell\left(  w\right)  }}=\underbrace{\prod
_{u\in\mathfrak{L}}\prod_{w\in\mathfrak{L}}}_{=\prod_{w\in\mathfrak{L}}%
\prod_{u\in\mathfrak{L}}}\underbrace{\dfrac{1}{1-\mathbf{y}_{u}^{\ell\left(
w\right)  }\xx_{w}^{\ell\left(  u\right)  }}}_{=\dfrac{1}%
{1-\xx_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(
w\right)  }}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the indices
}w\text{ and }u\text{ as }u\text{ and }w\right) \\
&  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}.
\end{align*}
Combining this with (\ref{pf.prop.lyndon.grfun.cauchy.short.0}) and
(\ref{pf.prop.lyndon.grfun.cauchy.short.1}), we obtain%
\begin{align*}
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)   &  =\sum
_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}w}\left(
\mathbf{y}\right)  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac
{1}{1-\xx_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(
w\right)  }}\\
&  =\sum_{\lambda\in \Par}p_{\lambda}\left(  \xx\right)
\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  .
\end{align*}
This proves both parts (a) and (b) of Proposition
\ref{prop.lyndon.grfun.cauchy}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.cauchy}.]For each word $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  \in\mathfrak{A}^{n}$, we define a monomial
$\mathbf{y}_{w}$ in $\kk \left[  \left[  \mathbf{y}\right]  \right]  $
by $\mathbf{y}_{w}=y_{w_{1}}y_{w_{2}}\cdots y_{w_{n}}$.

We have%
\begin{align}
\sum_{\lambda\in \Par}\underbrace{\mathbf{GR}_{\lambda}\left(
\xx\right)  }_{\substack{=\sum\limits_{\substack{w\in\mathfrak{A}%
^{\ast};\\\operatorname*{CFLtype}w=\lambda}}\xx_{w}\\\text{(by the
definition of }\mathbf{GR}_{\lambda}\text{)}}}p_{\lambda}\left(
\mathbf{y}\right)   &  =\sum_{\lambda\in \Par}\left(
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}%
w=\lambda}}\xx_{w}\right)  p_{\lambda}\left(  \mathbf{y}\right)
\nonumber\\
&  =\sum_{\lambda\in \Par}\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda}}\xx%
_{w}\underbrace{p_{\lambda}\left(  \mathbf{y}\right)  }%
_{\substack{=p_{\operatorname*{CFLtype}w}\left(  \mathbf{y}\right)
\\\text{(since }\lambda=\operatorname*{CFLtype}w\text{)}}}\nonumber\\
&  =\underbrace{\sum_{\lambda\in \Par}\sum
\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda
}}}_{\substack{=\sum_{w\in\mathfrak{A}^{\ast}}\\\text{(since }%
\operatorname*{CFLtype}w\in\Par\\\text{for each }%
w\in\mathfrak{A}^{\ast}\text{)}}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right) \nonumber\\
&  =\sum_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}%
w}\left(  \mathbf{y}\right) \label{pf.prop.lyndon.grfun.cauchy.0}\\
&  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}
\label{pf.prop.lyndon.grfun.cauchy.1}%
\end{align}
(since Lemma \ref{lem.sol.lyndon.grfun.sumxwpy2} yields $\prod_{w\in
\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}^{\ell\left(
u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}=\sum_{w\in\mathfrak{A}%
^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}w}\left(  \mathbf{y}\right)  $).

The same argument (but with the roles of the sets of variables $\xx%
=\left(  x_{1},x_{2},x_{3},\ldots\right)  $ and $\mathbf{y}=\left(
y_{1},y_{2},y_{3},\ldots\right)  $ interchanged) yields%
\[
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\mathbf{y}\right)  p_{\lambda}\left(  \xx\right)  =\prod_{w\in
\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{w}^{\ell\left(
u\right)  }\xx_{u}^{\ell\left(  w\right)  }}%
\]
\footnote{Indeed, if we interchange the roles of $\xx=\left(
x_{1},x_{2},x_{3},\ldots\right)  $ and $\mathbf{y}=\left(  y_{1},y_{2}%
,y_{3},\ldots\right)  $, then
\par
\begin{itemize}
\item $\mathbf{GR}_{\lambda}\left(  \xx\right)  $ becomes
$\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  $ for any $\lambda
\in \Par$;
\par
\item $p_{\lambda}\left(  \mathbf{y}\right)  $ becomes $p_{\lambda}\left(
\xx\right)  $ for any $\lambda\in \Par$;
\par
\item $\xx_{w}$ becomes $\mathbf{y}_{w}$ for any $w\in\mathfrak{A}%
^{\ast}$ (since the definitions of $\xx_{w}$ and $\mathbf{y}_{w}$ are
identical except for the variables they are using);
\par
\item $\mathbf{y}_{u}$ becomes $\xx_{u}$ for any $u\in\mathfrak{A}%
^{\ast}$ (since the definitions of $\mathbf{y}_{u}$ and $\xx_{u}$ are
identical except for the variables they are using).
\end{itemize}
}. Hence,%
\begin{align*}
&  \sum_{\lambda\in \Par}\underbrace{p_{\lambda}\left(
\xx\right)  \mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)
}_{=\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  p_{\lambda}\left(
\xx\right)  }\\
&  =\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\mathbf{y}\right)  p_{\lambda}\left(  \xx\right)  =\prod_{w\in
\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\mathbf{y}_{w}^{\ell\left(
u\right)  }\xx_{u}^{\ell\left(  w\right)  }}=\underbrace{\prod
_{w\in\mathfrak{L}}\prod_{v\in\mathfrak{L}}}_{=\prod_{v\in\mathfrak{L}}%
\prod_{w\in\mathfrak{L}}}\underbrace{\dfrac{1}{1-\mathbf{y}_{w}^{\ell\left(
v\right)  }\xx_{v}^{\ell\left(  w\right)  }}}_{\substack{=\dfrac
{1}{1-\xx_{v}^{\ell\left(  w\right)  }\mathbf{y}_{w}^{\ell\left(
v\right)  }}\\\text{(since }\mathbf{y}_{w}^{\ell\left(  v\right)  }%
\xx_{v}^{\ell\left(  w\right)  }=\xx_{v}^{\ell\left(  w\right)
}\mathbf{y}_{w}^{\ell\left(  v\right)  }\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }u\text{
as }v\text{ in the inner product}\right) \\
&  =\prod_{v\in\mathfrak{L}}\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{v}^{\ell\left(  w\right)  }\mathbf{y}_{w}^{\ell\left(  v\right)  }}%
=\prod_{v\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{v}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  v\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }w\text{
as }u\text{ in the inner product}\right) \\
&  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac{1}{1-\xx%
_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(  w\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }v\text{
as }w\text{ in the outer product}\right)  .
\end{align*}
Combining this with (\ref{pf.prop.lyndon.grfun.cauchy.0}) and
(\ref{pf.prop.lyndon.grfun.cauchy.1}), we obtain%
\begin{align*}
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)   &  =\sum
_{w\in\mathfrak{A}^{\ast}}\xx_{w}p_{\operatorname*{CFLtype}w}\left(
\mathbf{y}\right)  =\prod_{w\in\mathfrak{L}}\prod_{u\in\mathfrak{L}}\dfrac
{1}{1-\xx_{w}^{\ell\left(  u\right)  }\mathbf{y}_{u}^{\ell\left(
w\right)  }}\\
&  =\sum_{\lambda\in \Par}p_{\lambda}\left(  \xx\right)
\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  .
\end{align*}
This proves both parts (a) and (b) of Proposition
\ref{prop.lyndon.grfun.cauchy}.
\end{proof}

\end{verlong}

Next, let us prove Proposition \ref{prop.lyndon.grfun.xN}:

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.xN}.](a) The set $N$ is a
necklace. In other words, $N$ is an $n$-necklace for some positive integer
$n$. Consider this $n$.

Recall that $c\in C$ acts on $\mathfrak{A}^{n}$ by cyclically rotating
$n$-tuples one step to the left (that is, it acts by the formula
$c\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{2},a_{3}%
,\ldots,a_{n},a_{1}\right)  $). Thus, the element $c^{n}$ acts on
$\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples $n$ steps to the left. In
other words, the element $c^{n}$ acts trivially on $\mathfrak{A}^{n}$ (since
cyclically rotating an $n$-tuple $n$ steps to the left does nothing). Hence,
the whole subgroup $\left\langle c^{n}\right\rangle $ of $C$ acts trivially on
$\mathfrak{A}^{n}$.

The words $w$ and $w^{\prime}$ belong to the same $n$-necklace (namely, to
$N$). In other words, the words $w$ and $w^{\prime}$ belong to the same orbit
of the $C$-action on $\mathfrak{A}^{n}$ (since an $n$-necklace was defined to
be an orbit of the $C$-action on $\mathfrak{A}^{n}$). In other words, $C\cdot
w=C\cdot w^{\prime}$. Hence, $w^{\prime}\in C\cdot w^{\prime}=C\cdot
w=\left\{  c^{p}\cdot w\ \mid\ p\in \ZZ \right\}  $ (since $C=\left\{
c^{p}\ \mid\ p\in \ZZ \right\}  $). In other words, there exists some
$p\in \ZZ $ such that $w^{\prime}=c^{p}\cdot w$. Consider this $p$.

Recall that $n$ is a positive integer. Hence, we can divide $p$ by $n$ with
remainder. Let $q$ and $r$ be the quotient and the remainder obtained when we
divide $p$ by $n$. Thus, $q\in \ZZ $ and $r\in\left\{  0,1,\ldots
,n-1\right\}  $ and $p=qn+r$. We have $c^{qn}=\left(  c^{n}\right)  ^{q}%
\in\left\langle c^{n}\right\rangle $. Thus, $c^{qn}$ acts trivially on
$\mathfrak{A}^{n}$ (since the whole subgroup $\left\langle c^{n}\right\rangle
$ of $C$ acts trivially on $\mathfrak{A}^{n}$). Therefore, $c^{qn}\cdot w=w$.

But $p=qn+r=r+qn$, so that $c^{p}=c^{r+qn}=c^{r}c^{qn}$. Now,%
\[
w^{\prime}=\underbrace{c^{p}}_{=c^{r}c^{qn}}\cdot w=\left(  c^{r}%
c^{qn}\right)  \cdot w=c^{r}\cdot\underbrace{\left(  c^{qn}\cdot w\right)
}_{=w}=c^{r}\cdot w.
\]


Let us write the word $w$ in the form $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $. Recall that $c\in C$ acts on $\mathfrak{A}^{n}$ by
cyclically rotating $n$-tuples one step to the left. Hence, $c^{r}$ acts on
$\mathfrak{A}^{n}$ by rotating $n$-tuples $r$ steps to the left (since
$r\in\left\{  0,1,\ldots,n-1\right\}  \subset \NN $). In other words,%
\begin{equation}
c^{r}\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{r+1}%
,a_{r+2},\ldots,a_{n},a_{1},a_{2},\ldots,a_{r}\right)
\label{pf.prop.lyndon.grfun.xN.short.cra}%
\end{equation}
for each $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}$ (since
$r\in\left\{  0,1,\ldots,n-1\right\}  $). Now,
\begin{align*}
w^{\prime}  &  =c^{r}\cdot\underbrace{w}_{=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  }=c^{r}\cdot\left(  w_{1},w_{2},\ldots,w_{n}\right) \\
&  =\left(  w_{r+1},w_{r+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{r}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lyndon.grfun.xN.short.cra}%
), applied to }a_{i}=w_{i}\right)  .
\end{align*}


Now, the two words $\left(  w_{1},w_{2},\ldots,w_{r}\right)  $ and $\left(
w_{r+1},w_{r+2},\ldots,w_{n}\right)  $ satisfy%
\[
w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\left(  w_{1},w_{2},\ldots
,w_{r}\right)  \left(  w_{r+1},w_{r+2},\ldots,w_{n}\right)
\]
and%
\[
w^{\prime}=\left(  w_{r+1},w_{r+2},\ldots,w_{n},w_{1},w_{2},\ldots
,w_{r}\right)  =\left(  w_{r+1},w_{r+2},\ldots,w_{n}\right)  \left(
w_{1},w_{2},\ldots,w_{r}\right)  .
\]
Hence, there exist words $u$ and $v$ such that $w=uv$ and $w^{\prime}=vu$
(namely, $u=\left(  w_{1},w_{2},\ldots,w_{r}\right)  $ and $v=\left(
w_{r+1},w_{r+2},\ldots,w_{n}\right)  $). This proves Proposition
\ref{prop.lyndon.grfun.xN}(a).

(b) Proposition \ref{prop.lyndon.grfun.xN}(a) shows that there exist words $u$
and $v$ such that $w=uv$ and $w^{\prime}=vu$. Consider these $u$ and $v$. From
$w=uv$, we obtain $\xx_{w}=\xx_{uv}=\xx_{u}\xx%
_{v}$ (by Lemma \ref{lem.sol.lyndon.grfun.xuv=}). From $w^{\prime}=vu$, we
obtain $\xx_{w^{\prime}}=\xx_{vu}=\xx_{v}\xx_{u}$
(by Lemma \ref{lem.sol.lyndon.grfun.xuv=}, applied to $v$ and $u$ instead of
$u$ and $v$). Comparing this with $\xx_{w}=\xx_{u}\xx%
_{v}=\xx_{v}\xx_{u}$, we obtain $\xx_{w}=\xx%
_{w^{\prime}}$. Thus, Proposition \ref{prop.lyndon.grfun.xN}(b) is proved.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.xN}.](a) The set $N$ is a
necklace. In other words, $N$ is an $n$-necklace for some positive integer $n$
(by the definition of ``necklace''). Consider
this $n$.

Consider the subgroup $\left\langle c^{n}\right\rangle $ of the infinite
cyclic group $C$. (This is the subgroup generated by $c^{n}$.) Clearly,
$c^{n}\in\left\langle c^{n}\right\rangle $.

Recall that $c\in C$ acts on $\mathfrak{A}^{n}$ by cyclically rotating
$n$-tuples one step to the left (that is, it acts by the formula
$c\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{2},a_{3}%
,\ldots,a_{n},a_{1}\right)  $). Thus, for each $k\in \NN $, the element
$c^{k}$ acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples $k$ steps
to the left. Applying this to $k=n$, we conclude that the element $c^{n}$ acts
on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples $n$ steps to the left.
In other words, the element $c^{n}$ acts trivially on $\mathfrak{A}^{n}$
(since cyclically rotating an $n$-tuple $n$ steps to the left does nothing).
Hence, the whole subgroup $\left\langle c^{n}\right\rangle $ of $C$ acts
trivially on $\mathfrak{A}^{n}$.

The set $N$ is an $n$-necklace. In other words, the set $N$ is an orbit of the
$C$-action on $\mathfrak{A}^{n}$ (since this is how an $n$-necklace was
defined). In other words, the set $N$ is a $C$-orbit on $\mathfrak{A}^{n}$.
Moreover, $N$ contains $w$ (since $w$ is an element of $N$). Hence, $N$ is the
$C$-orbit that contains $w$ (since $N$ is a $C$-orbit). In other words,
$N=C\cdot w$.

Recall that $w^{\prime}$ is an element of $N$. Thus,
\begin{align*}
w^{\prime}  &  \in N=C\cdot w=\left\{  c^{p}\ \mid\ p\in \ZZ \right\}
\cdot w\ \ \ \ \ \ \ \ \ \ \left(  \text{since }C=\left\{  c^{p}\ \mid
\ p\in \ZZ \right\}  \right) \\
&  =\left\{  c^{p}\cdot w\ \mid\ p\in \ZZ \right\}  .
\end{align*}
In other words, there exists some $p\in \ZZ $ such that $w^{\prime}%
=c^{p}\cdot w$. Consider this $p$.

Recall that $n$ is a positive integer. Hence, we can divide $p$ by $n$ with
remainder. Let $q$ and $r$ be the quotient and the remainder obtained when we
divide $p$ by $n$. Thus, $q\in \ZZ $ and $r\in\left\{  0,1,\ldots
,n-1\right\}  $ and $p=qn+r$. We have $c^{qn}=c^{nq}=\left(  c^{n}\right)
^{q}\in\left\langle c^{n}\right\rangle $ (since $c^{n}\in\left\langle
c^{n}\right\rangle $). Thus, $c^{qn}$ acts trivially on $\mathfrak{A}^{n}$
(since the whole subgroup $\left\langle c^{n}\right\rangle $ of $C$ acts
trivially on $\mathfrak{A}^{n}$). Therefore, $c^{qn}\cdot w=w$.

Now,%
\begin{align*}
w^{\prime}  &  =c^{p}\cdot w=\underbrace{c^{r+qn}}_{=c^{r}c^{qn}}\cdot
w\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p=qn+r=r+qn\right) \\
&  =\left(  c^{r}c^{qn}\right)  \cdot w=c^{r}\cdot\underbrace{\left(
c^{qn}\cdot w\right)  }_{=w}=c^{r}\cdot w.
\end{align*}


Let us write the word $w$ in the form $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $. Recall that for each $k\in \NN $, the element $c^{k}$
acts on $\mathfrak{A}^{n}$ by cyclically rotating $n$-tuples $k$ steps to the
left. Applying this to $k=r$, we conclude that the element $c^{r}$ acts on
$\mathfrak{A}^{n}$ by rotating $n$-tuples $r$ steps to the left (since
$r\in\left\{  0,1,\ldots,n-1\right\}  \subset \NN $). In other words,%
\begin{equation}
c^{r}\cdot\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\left(  a_{r+1}%
,a_{r+2},\ldots,a_{n},a_{1},a_{2},\ldots,a_{r}\right)
\label{pf.prop.lyndon.grfun.xN.cra}%
\end{equation}
for each $\left(  a_{1},a_{2},\ldots,a_{n}\right)  \in\mathfrak{A}^{n}$ (since
$r\in\left\{  0,1,\ldots,n-1\right\}  \subset\left\{  0,1,\ldots,n\right\}
$). Now,
\begin{align*}
w^{\prime}  &  =c^{r}\cdot\underbrace{w}_{=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  }=c^{r}\cdot\left(  w_{1},w_{2},\ldots,w_{n}\right) \\
&  =\left(  w_{r+1},w_{r+2},\ldots,w_{n},w_{1},w_{2},\ldots,w_{r}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lyndon.grfun.xN.cra}),
applied to }a_{i}=w_{i}\right)  .
\end{align*}


Now, the two words $\left(  w_{1},w_{2},\ldots,w_{r}\right)  $ and $\left(
w_{r+1},w_{r+2},\ldots,w_{n}\right)  $ satisfy%
\begin{align*}
w  &  =\left(  w_{1},w_{2},\ldots,w_{n}\right)  =\left(  w_{1},w_{2}%
,\ldots,w_{r},w_{r+1},w_{r+2},\ldots,w_{n}\right) \\
&  =\left(  w_{1},w_{2},\ldots,w_{r}\right)  \left(  w_{r+1},w_{r+2}%
,\ldots,w_{n}\right)
\end{align*}
and%
\[
w^{\prime}=\left(  w_{r+1},w_{r+2},\ldots,w_{n},w_{1},w_{2},\ldots
,w_{r}\right)  =\left(  w_{r+1},w_{r+2},\ldots,w_{n}\right)  \left(
w_{1},w_{2},\ldots,w_{r}\right)  .
\]
Hence, there exist words $u$ and $v$ such that $w=uv$ and $w^{\prime}=vu$
(namely, $u=\left(  w_{1},w_{2},\ldots,w_{r}\right)  $ and $v=\left(
w_{r+1},w_{r+2},\ldots,w_{n}\right)  $). This proves Proposition
\ref{prop.lyndon.grfun.xN}(a).

(b) Proposition \ref{prop.lyndon.grfun.xN}(a) shows that there exist words $u$
and $v$ such that $w=uv$ and $w^{\prime}=vu$. Consider these $u$ and $v$. From
$w=uv$, we obtain $\xx_{w}=\xx_{uv}=\xx_{u}\xx%
_{v}$ (by Lemma \ref{lem.sol.lyndon.grfun.xuv=}). From $w^{\prime}=vu$, we
obtain $\xx_{w^{\prime}}=\xx_{vu}=\xx_{v}\xx_{u}$
(by Lemma \ref{lem.sol.lyndon.grfun.xuv=}, applied to $v$ and $u$ instead of
$u$ and $v$). Comparing this with $\xx_{w}=\xx_{u}\xx%
_{v}=\xx_{v}\xx_{u}$, we obtain $\xx_{w}=\xx%
_{w^{\prime}}$. Thus, Proposition \ref{prop.lyndon.grfun.xN}(b) is proved.
\end{proof}

\end{verlong}

Next, we shall prove Proposition \ref{prop.lyndon.grfun.alt1} using the
following two almost trivial lemmas:

\begin{lemma}
\label{lem.sol.lyndon.grfun.xa=xa}Let $a\in\mathfrak{A}^{\ast}$ be a nonempty
word. Then, $\xx_{\left[  a\right]  }=\xx_{a}$.
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.xa=xa}.]Lemma
\ref{lem.sol.lyndon.gr.bij.neck-cts0} (applied to $u=a$) yields $a\in\left[
a\right]  $. In other words, $a$ is an element of $\left[  a\right]  $.

Now, recall how the monomial $\xx_{N}$ was defined for a necklace $N$:
It was defined by setting $\xx_{N}=\xx_{w}$, where $w$ is any
element of $N$. Thus, if $N$ is a necklace, then $\xx_{N}%
=\xx_{w}$ for any element $w$ of $N$. Applying this to $N=\left[
a\right]  $ and $w=a$, we conclude that $\xx_{\left[  a\right]
}=\xx_{a}$ (since $a$ is an element of $\left[  a\right]  $). This
proves Lemma \ref{lem.sol.lyndon.grfun.xa=xa}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.xa=xa}.]The word $a$ is nonempty. In
other words, $a\in\mathfrak{A}^{n}$ for some positive integer $n$. Consider
this $n$. Then, $\left[  a\right]  $ is an $n$-necklace, hence a necklace.
Lemma \ref{lem.sol.lyndon.gr.bij.neck-cts0} (applied to $u=a$) yields
$a\in\left[  a\right]  $. In other words, $a$ is an element of $\left[
a\right]  $.

Now, recall how the monomial $\xx_{N}$ was defined for a necklace $N$:
It was defined by setting $\xx_{N}=\xx_{w}$, where $w$ is any
element of $N$. Thus, if $N$ is a necklace, then $\xx_{N}%
=\xx_{w}$ for any element $w$ of $N$. Applying this to $N=\left[
a\right]  $, we conclude that $\xx_{\left[  a\right]  }=\xx_{w}$
for any element $w$ of $\left[  a\right]  $. Applying this to $w=a$, we obtain
$\xx_{\left[  a\right]  }=\xx_{a}$ (since $a$ is an element of
$\left[  a\right]  $). This proves Lemma \ref{lem.sol.lyndon.grfun.xa=xa}.
\end{proof}

\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.grfun.sizeCFLtype}Let $\lambda$ be a partition. Let
$w\in\mathfrak{A}^{\ast}$ be such that $\operatorname*{CFLtype}w=\lambda$. Then:

\begin{enumerate}
\item[(a)] We have $\ell\left(  w\right)  =\left\vert \lambda\right\vert $.

\item[(b)] Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL
factorization of $w$. Then, $k=\ell\left(  \lambda\right)  $.
\end{enumerate}
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}.]Let $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of $w$. In other
words, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words
satisfying $w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq
a_{k}$ (because this is how a ``CFL factorization of $w$'' is defined).
From $\operatorname*{CFLtype}w=\lambda$,
we obtain%
\begin{align*}
&  \lambda=\operatorname*{CFLtype}w\\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right)
\end{align*}
(by the definition of $\operatorname*{CFLtype}w$, since $\left(  a_{1}%
,a_{2},\ldots,a_{k}\right)  $ is the CFL factorization of $w$). Hence, the
partition $\lambda$ is a rearrangement of the $k$-tuple $\left(  \ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\right)  $ (of course, with trailing zeroes appended to it to form an infinite
sequence). Thus,
\[
\left\vert \lambda\right\vert =\ell\left(  a_{1}\right)  +\ell\left(
a_{2}\right)  +\cdots+\ell\left(  a_{k}\right)  .
\]
Comparing this with%
\[
\ell\left(  a_{1}a_{2}\cdots a_{k}\right)  =\ell\left(  a_{1}\right)
+\ell\left(  a_{2}\right)  +\cdots+\ell\left(  a_{k}\right)  ,
\]
we obtain $\ell\left(  a_{1}a_{2}\cdots a_{k}\right)  =\left\vert
\lambda\right\vert $. This rewrites as $\ell\left(  w\right)  =\left\vert
\lambda\right\vert $ (since $w=a_{1}a_{2}\cdots a_{k}$). Lemma
\ref{lem.sol.lyndon.grfun.sizeCFLtype}(a) is now proved.

(b) The words $a_{1},a_{2},\ldots,a_{k}$ are Lyndon words (since $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words) and thus are
nonempty. Hence, their lengths $\ell\left(  a_{1}\right)  ,\ell\left(
a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $ are positive integers.

But recall that the partition $\lambda$ is a rearrangement of the $k$-tuple
$\left(  \ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots
,\ell\left(  a_{k}\right)  \right)  $. Since the numbers
$\ell\left(  a_{1}\right)
,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $ are positive
integers, we thus conclude that the partition $\lambda$ has exactly $k$ parts
(i.e., exactly $k$ nonzero entries). In other words, $\ell\left(
\lambda\right)  =k$. This proves Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(b).
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}.]Let $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of $w$. Then, the
definition of a ``CFL type'' shows that the
CFL type of $w$ is the partition whose parts are the positive integers
$\ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(
a_{k}\right)  $ (listed in decreasing order). In other words,%
\begin{align*}
&  \left(  \text{the CFL type of }w\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right)  .
\end{align*}
But $\operatorname*{CFLtype}w=\lambda$. Hence,%
\begin{align*}
&  \lambda=\operatorname*{CFLtype}w=\left(  \text{the CFL type of }w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\operatorname*{CFLtype}w\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right)  .
\end{align*}
Hence, the partition $\lambda$ is a rearrangement of the $k$-tuple $\left(
\ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(
a_{k}\right)  \right)  $ (of course, with trailing zeroes appended to it to
form an infinite sequence). In other words, there exists a permutation
$\sigma\in\Symm_{k}$ such that $\lambda=\left(  \ell\left(
a_{\sigma\left(  1\right)  }\right)  ,\ell\left(  a_{\sigma\left(  2\right)
}\right)  ,\ldots,\ell\left(  a_{\sigma\left(  k\right)  }\right)  \right)  $.
Consider this $\sigma$. Clearly, $\sigma$ is a permutation of $\left\{
1,2,\ldots,k\right\}  $ (since $\sigma\in\Symm_{k}$), thus a bijection
from $\left\{  1,2,\ldots,k\right\}  $ to $\left\{  1,2,\ldots,k\right\}  $.

From $\lambda=\left(  \ell\left(  a_{\sigma\left(  1\right)  }\right)
,\ell\left(  a_{\sigma\left(  2\right)  }\right)  ,\ldots,\ell\left(
a_{\sigma\left(  k\right)  }\right)  \right)  $, we obtain
\begin{align}
\left\vert \lambda\right\vert  &  =\left\vert \left(  \ell\left(
a_{\sigma\left(  1\right)  }\right)  ,\ell\left(  a_{\sigma\left(  2\right)
}\right)  ,\ldots,\ell\left(  a_{\sigma\left(  k\right)  }\right)  \right)
\right\vert \nonumber\\
&  =\ell\left(  a_{\sigma\left(  1\right)  }\right)  +\ell\left(
a_{\sigma\left(  2\right)  }\right)  +\cdots+\ell\left(  a_{\sigma\left(
k\right)  }\right)  =\sum_{i\in\left\{  1,2,\ldots,k\right\}  }\ell\left(
a_{\sigma\left(  i\right)  }\right) \nonumber\\
&  =\sum_{i\in\left\{  1,2,\ldots,k\right\}  }\ell\left(  a_{i}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }i\text{ for }\sigma\left(  i\right)  \text{
in the sum,}\\
\text{since }\sigma\text{ is a bijection from }\left\{  1,2,\ldots,k\right\}
\text{ to }\left\{  1,2,\ldots,k\right\}
\end{array}
\right) \nonumber\\
&  =\ell\left(  a_{1}\right)  +\ell\left(  a_{2}\right)  +\cdots+\ell\left(
a_{k}\right)  . \label{pf.lem.sol.lyndon.grfun.sizeCFLtype.1}%
\end{align}


On the other hand, we know that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is
a CFL factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (because this is how a
``CFL factorization of $w$'' is defined).
From $w=a_{1}a_{2}\cdots a_{k}$, we obtain%
\[
\ell\left(  w\right)  =\ell\left(  a_{1}a_{2}\cdots a_{k}\right)  =\ell\left(
a_{1}\right)  +\ell\left(  a_{2}\right)  +\cdots+\ell\left(  a_{k}\right)  .
\]


Comparing this with (\ref{pf.lem.sol.lyndon.grfun.sizeCFLtype.1}), we obtain
$\ell\left(  w\right)  =\left\vert \lambda\right\vert $. This proves Lemma
\ref{lem.sol.lyndon.grfun.sizeCFLtype}(a).

(b) The $k$-tuple $\left(  \ell\left(  a_{\sigma\left(  1\right)  }\right)
,\ell\left(  a_{\sigma\left(  2\right)  }\right)  ,\ldots,\ell\left(
a_{\sigma\left(  k\right)  }\right)  \right)  $ is a rearrangement of the
$k$-tuple $\left(  \ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)
,\ldots,\ell\left(  a_{k}\right)  \right)  $ (since $\sigma \in \Symm_k$).

The length $\ell\left(  a_{\sigma\left(  i\right)  }\right)  $ is a positive
integer for each $i\in\left\{  1,2,\ldots,k\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k\right\}  $.
Then, $\sigma\left(  i\right)  \in\left\{  1,2,\ldots,k\right\}  $ (since
$\sigma$ is a map $\left\{  1,2,\ldots,k\right\}  \rightarrow\left\{
1,2,\ldots,k\right\}  $ (because $\sigma\in\Symm_{k}$)). But $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words. In other words,
$a_{j}$ is a Lyndon word for each $j\in\left\{  1,2,\ldots,k\right\}  $.
Applying this to $j=\sigma\left(  i\right)  $, we conclude that $a_{\sigma
\left(  i\right)  }$ is a Lyndon word (since $\sigma\left(  i\right)
\in\left\{  1,2,\ldots,k\right\}  $). Hence, the word $a_{\sigma\left(
i\right)  }$ is nonempty (since any Lyndon word is nonempty). Thus,
$\ell\left(  a_{\sigma\left(  i\right)  }\right)  $ is a positive integer.
Qed.}. In other words, the lengths
$\ell\left(  a_{\sigma\left(  1\right)  }\right)
,\ell\left(  a_{\sigma\left(  2\right)  }\right)  ,\ldots,\ell\left(
a_{\sigma\left(  k\right)  }\right)  $ are $k$ positive integers.

On the other hand, we know that $\lambda=\left(  \ell\left(  a_{\sigma\left(
1\right)  }\right)  ,\ell\left(  a_{\sigma\left(  2\right)  }\right)
,\ldots,\ell\left(  a_{\sigma\left(  k\right)  }\right)  \right)  $.

But if $\mu$ is a partition, and if $\beta_{1},\beta_{2},\ldots,\beta_{k}$ are
$k$ positive integers satisfying $\mu=\left(  \beta_{1},\beta_{2},\ldots
,\beta_{k}\right)  $, then $\ell\left(  \mu\right)  =k$ (because the parts of
the partition $\mu$ are precisely the $k$ positive integers $\beta_{1}%
,\beta_{2},\ldots,\beta_{k}$, and thus there are exactly $k$ of them).
Applying this to $\mu=\lambda$ and $\beta_{i}=\ell\left(  a_{\sigma\left(
i\right)  }\right)  $, we conclude that $\ell\left(  \lambda\right)  =k$
(since $\ell\left(  a_{\sigma\left(  1\right)  }\right)  ,\ell\left(
a_{\sigma\left(  2\right)  }\right)  ,\ldots,\ell\left(  a_{\sigma\left(
k\right)  }\right)  $ are $k$ positive integers satisfying $\lambda=\left(
\ell\left(  a_{\sigma\left(  1\right)  }\right)  ,\ell\left(  a_{\sigma\left(
2\right)  }\right)  ,\ldots,\ell\left(  a_{\sigma\left(  k\right)  }\right)
\right)  $). In other words, $k=\ell\left(  \lambda\right)  $. This proves
Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(b).
\end{proof}

\end{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.alt1}.]Let us recall Definition
\ref{def.sol.lyndon.gr.bij.sum} (in which we defined a nonnegative integer
$\operatorname*{sum}M\in \NN $ for each finite multiset $M$ of finite
sets). Let $\mathfrak{B}=\mathfrak{A}$. Then, $\mathfrak{B}$ is a subset of
$\mathfrak{A}$. Hence, Definition \ref{def.sol.lyndon.gr.bij.fin-subsets}
applies.\footnote{This is the reason why we introduced $\mathfrak{B}$ to begin
with -- we wanted to apply Definition \ref{def.sol.lyndon.gr.bij.fin-subsets}
(and, later, Proposition \ref{prop.sol.lyndon.gr.bij.CFL-LFC}).} Note that
$\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$ (since $\mathfrak{B}=\mathfrak{A}$).

Let $n=\left\vert \lambda\right\vert $. Then, $n\in \NN $. Hence,
Proposition \ref{prop.sol.lyndon.gr.bij.CFL-LFC} yields that the maps
$\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ and $\operatorname*{LFC}:\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\rightarrow\mathfrak{B}^{n}$ (introduced in Definition
\ref{def.sol.lyndon.gr.bij.CFL} and in Definition
\ref{def.sol.lyndon.gr.bij.LFC}) are mutually inverse. Thus, these two maps
$\operatorname*{CFL}$ and $\operatorname*{LFC}$ are invertible, i.e., they are bijections.

We shall now prove a sequence of simple claims:

\begin{statement}
\textit{Claim 1:} Let $w\in\mathfrak{B}^{n}$. Then, $\operatorname*{CFLtype}%
w=\operatorname*{type}\left(  \operatorname*{CFL}w\right)  $.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 1:} We have $w\in\mathfrak{B}^{n}\subset
\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$.

Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Then, the definition of $\operatorname*{CFL}$ yields
\[
\operatorname*{CFL} w
=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}} .
\]
Hence, the sizes of the
necklaces in $\operatorname*{CFL}w$ are the numbers $\left\vert \left[
a_{1}\right]  \right\vert ,\left\vert \left[  a_{2}\right]  \right\vert
,\ldots,\left\vert \left[  a_{k}\right]  \right\vert $.

We have $\left\vert \left[  a_{i}\right]  \right\vert =\ell\left(
a_{i}\right)  $ for each $i\in\left\{  1,2,\ldots,k\right\}  $%
.\ \ \ \ \footnote{Indeed, this is precisely the statement of Claim 1
\textbf{(b)} in the proof of Proposition \ref{prop.sol.lyndon.gr.bij.CFL-ex}.
We refer to the latter proof for a proof of this statement.}

Recall that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. Hence, the definition of $\operatorname*{CFLtype}w$
shows that%
\begin{align*}
&  \operatorname*{CFLtype}w\\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
&  \operatorname*{type}\left(  \operatorname*{CFL}w\right) \\
&  =\left(  \text{the partition obtained by listing the sizes of the necklaces
in }\operatorname*{CFL}w\text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\operatorname*{type}\left(  \operatorname*{CFL}w\right)  \right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
\left[  a_{1}\right]  \right\vert ,\left\vert \left[  a_{2}\right]
\right\vert ,\ldots,\left\vert \left[  a_{k}\right]  \right\vert \text{ in
decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sizes of the necklaces in
}\operatorname*{CFL}w\text{ are the numbers }\left\vert \left[  a_{1}\right]
\right\vert ,\left\vert \left[  a_{2}\right]  \right\vert ,\ldots,\left\vert
\left[  a_{k}\right]  \right\vert \right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\left\vert \left[  a_{i}\right]
\right\vert =\ell\left(  a_{i}\right)  \text{ for each }i\in\left\{
1,2,\ldots,k\right\}  \right)  ,
\end{align*}
we obtain $\operatorname*{CFLtype}w=\operatorname*{type}\left(
\operatorname*{CFL}w\right)  $. This proves Claim 1.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 1:} We have $w\in\mathfrak{B}^{n}\subset
\mathfrak{B}^{\ast}=\mathfrak{A}^{\ast}$.

Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Then, the definition of $\operatorname*{CFL}$ yields
\[
\operatorname*{CFL}%
w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}} .
\]
Thus, the necklaces in
$\operatorname*{CFL}w$ are $\left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  $. Hence, the sizes of the necklaces in
$\operatorname*{CFL}w$ are the numbers $\left\vert \left[  a_{1}\right]
\right\vert ,\left\vert \left[  a_{2}\right]  \right\vert ,\ldots,\left\vert
\left[  a_{k}\right]  \right\vert $.

For each $i\in\left\{  1,2,\ldots,k\right\}  $, we have $\left\vert \left[
a_{i}\right]  \right\vert =\ell\left(  a_{i}\right)  $%
.\ \ \ \ \footnote{Indeed, this is precisely the statement of Claim 1
\textbf{(b)} in the proof of Proposition \ref{prop.sol.lyndon.gr.bij.CFL-ex}.
We refer to the latter proof for a proof of this statement.} In other words,%
\begin{equation}
\left(  \left\vert \left[  a_{1}\right]  \right\vert ,\left\vert \left[
a_{2}\right]  \right\vert ,\ldots,\left\vert \left[  a_{k}\right]  \right\vert
\right)  =\left(  \ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)
,\ldots,\ell\left(  a_{k}\right)  \right)  .
\label{pf.prop.lyndon.grfun.alt1.c1.pf.1}%
\end{equation}


But the definition of a ``CFL type'' shows
that the CFL type of $w$ is the partition whose parts are the positive
integers $\ell\left(  a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots
,\ell\left(  a_{k}\right)  $ (listed in decreasing order). In other words,%
\begin{align*}
&  \left(  \text{the CFL type of }w\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right)  .
\end{align*}
But $\operatorname*{CFLtype}w$ is defined to be the CFL type of $w$. Thus,%
\begin{align}
&  \operatorname*{CFLtype}w\nonumber\\
&  =\left(  \text{the CFL type of }w\right) \nonumber\\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right)  .
\label{pf.prop.lyndon.grfun.alt1.c1.pf.2}%
\end{align}


On the other hand, $\operatorname*{type}\left(  \operatorname*{CFL}w\right)  $
is the partition obtained by listing the sizes of the necklaces in
$\operatorname*{CFL}w$ in decreasing order (by the definition of
$\operatorname*{type}\left(  \operatorname*{CFL}w\right)  $). Hence,%
\begin{align*}
&  \operatorname*{type}\left(  \operatorname*{CFL}w\right) \\
&  =\left(  \text{the partition obtained by listing the sizes of the necklaces
in }\operatorname*{CFL}w\text{ in decreasing order}\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
\left[  a_{1}\right]  \right\vert ,\left\vert \left[  a_{2}\right]
\right\vert ,\ldots,\left\vert \left[  a_{k}\right]  \right\vert \text{ in
decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sizes of the necklaces in
}\operatorname*{CFL}w\text{ are the numbers }\left\vert \left[  a_{1}\right]
\right\vert ,\left\vert \left[  a_{2}\right]  \right\vert ,\ldots,\left\vert
\left[  a_{k}\right]  \right\vert \right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\ell\left(
a_{1}\right)  ,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)
\text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.lyndon.grfun.alt1.c1.pf.1})}\right)  .
\end{align*}
Comparing this with (\ref{pf.prop.lyndon.grfun.alt1.c1.pf.2}), we obtain
$\operatorname*{CFLtype}w=\operatorname*{type}\left(  \operatorname*{CFL}%
w\right)  $. This proves Claim 1.]
\end{verlong}

\begin{statement}
\textit{Claim 2:} We have $\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{CFLtype}w=\lambda\right\}  =\left\{  w\in\mathfrak{B}%
^{n}\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}  $.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 2:} Each $w\in\mathfrak{A}^{\ast}$ satisfying
$\operatorname*{CFLtype}w=\lambda$ must also satisfy $w\in\mathfrak{A}^{n}$
(since Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(a) yields $\ell\left(
w\right)  =\left\vert \lambda\right\vert =n$). Hence,%
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\lambda
\right\}  \subset\left\{  w\in\mathfrak{A}^{n}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  .
\]
On the other hand,
\[
\left\{  w\in\mathfrak{A}^{n}\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}
\subset\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}
\]
(since $\mathfrak{A}^{n}\subset\mathfrak{A}^{\ast}$). Combining these two
inclusions, we obtain%
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\lambda
\right\}  =\left\{  w\in\mathfrak{A}^{n}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  .
\]
Since $\mathfrak{B}=\mathfrak{A}$, this rewrites as $\left\{  w\in
\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}  =\left\{
w\in\mathfrak{B}^{n}\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}  $. This
proves Claim 2.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 2:} Let $p\in\left\{  w\in\mathfrak{A}^{\ast}%
\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}  $. Thus, $p$ is a
$w\in\mathfrak{A}^{\ast}$ satisfying $\operatorname*{CFLtype}w=\lambda$. In
other words, $p$ is an element of $\mathfrak{A}^{\ast}$ and satisfies
$\operatorname*{CFLtype}p=\lambda$. Hence, Lemma
\ref{lem.sol.lyndon.grfun.sizeCFLtype}(a) (applied to $w=p$) yields
$\ell\left(  p\right)  =\left\vert \lambda\right\vert =n$ (since $n=\left\vert
\lambda\right\vert $), so that $p\in\mathfrak{A}^{n}=\mathfrak{B}^{n}$ (since
$\mathfrak{A}=\mathfrak{B}$). Thus, $p$ is a $w\in\mathfrak{B}^{n}$ satisfying
$\operatorname*{CFLtype}w=\lambda$ (since $\operatorname*{CFLtype}p=\lambda$).
In other words, $p\in\left\{  w\in\mathfrak{B}^{n}\ \mid
\ \operatorname*{CFLtype}w=\lambda\right\}  $.

Forget that we fixed $p$. We thus have proved that $p\in\left\{
w\in\mathfrak{B}^{n}\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}  $ for
each $p\in\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  $. In other words,%
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\lambda
\right\}  \subset\left\{  w\in\mathfrak{B}^{n}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  .
\]
Combining this with%
\[
\left\{  w\in\underbrace{\mathfrak{B}^{n}}_{\subset\mathfrak{B}^{\ast
}=\mathfrak{A}^{\ast}}\ \mid\ \operatorname*{CFLtype}w=\lambda\right\}
\subset\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  ,
\]
we obtain $\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  =\left\{  w\in\mathfrak{B}^{n}\ \mid
\ \operatorname*{CFLtype}w=\lambda\right\}  $. This proves Claim 2.]
\end{verlong}

\begin{statement}
\textit{Claim 3:} Let $w\in\mathfrak{B}^{n}$. Then, $\xx_{w}%
=\xx_{\operatorname*{CFL}w}$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 3:} Set $M=\operatorname*{CFL}w$. The definition of
$\xx_{M}$ yields
\begin{equation}
\xx_{M}=\xx_{N_{1}}\xx_{N_{2}}\cdots\xx_{N_{k}},
\label{pf.prop.lyndon.grfun.alt1.c4.pf.short.1}%
\end{equation}
where $M$ is written in the form $M=\left\{  N_{1},N_{2},\ldots,N_{k}\right\}
_{\operatorname*{multiset}}$.

Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Then, the definition of $\operatorname*{CFL}$ yields \newline
$\operatorname*{CFL}%
w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}}$. Thus,
$M=\operatorname*{CFL}w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}$. Hence,
(\ref{pf.prop.lyndon.grfun.alt1.c4.pf.short.1}) (applied to $N_{i}=\left[
a_{i}\right]  $) yields
\begin{equation}
\xx_{M}=\xx_{\left[  a_{1}\right]  }\xx_{\left[
a_{2}\right]  }\cdots\xx_{\left[  a_{k}\right]  }.
\label{pf.prop.lyndon.grfun.alt1.c4.pf.short.2}%
\end{equation}


But we know that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a CFL
factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (because this is how a
``CFL factorization of $w$'' is defined).
From $w=a_{1}a_{2}\cdots a_{k}$, we obtain%
\begin{equation}
\xx_{w}=\xx_{a_{1}a_{2}\cdots a_{k}}=\xx_{a_{1}%
}\xx_{a_{2}}\cdots\xx_{a_{k}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma \ref{lem.sol.lyndon.grfun.xai}}\right)  .
\label{pf.prop.lyndon.grfun.alt1.c4.pf.short.3}%
\end{equation}


On the other hand, each $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$\xx_{\left[  a_{i}\right]  }=\xx_{a_{i}}$%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k\right\}  $.
Then, $a_{i}$ is a Lyndon word (since $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words), and thus is nonempty (since any
Lyndon word is nonempty). Hence, Lemma \ref{lem.sol.lyndon.grfun.xa=xa}
(applied to $a=a_{i}$) yields $\xx_{\left[  a_{i}\right]  }%
=\xx_{a_{i}}$. Qed.}. Hence, $\underbrace{\xx_{\left[
a_{1}\right]  }}_{=\xx_{a_{1}}}\underbrace{\xx_{\left[
a_{2}\right]  }}_{=\xx_{a_{2}}}\cdots\underbrace{\xx_{\left[
a_{k}\right]  }}_{=\xx_{a_{k}}}=\xx_{a_{1}}\xx_{a_{2}%
}\cdots\xx_{a_{k}}$. This shows that the right hand sides of the
equalities (\ref{pf.prop.lyndon.grfun.alt1.c4.pf.short.2}) and
(\ref{pf.prop.lyndon.grfun.alt1.c4.pf.short.3}) are equal. Hence, their left
hand sides are equal as well. In other words, $\xx_{M}=\xx_{w}$.
Hence, $\xx_{w}=\xx_{M}=\xx_{\operatorname*{CFL}w}$
(since $M=\operatorname*{CFL}w$). This proves Claim 3.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 3:} Set $M=\operatorname*{CFL}w$. The definition of
$\xx_{M}$ yields $\xx_{M}=\xx_{N_{1}}\xx_{N_{2}%
}\cdots\xx_{N_{k}}$, where $M$ is written in the form $M=\left\{
N_{1},N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$. In other
words, if $N_{1},N_{2},\ldots,N_{k}$ are necklaces such that $M=\left\{
N_{1},N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$, then
\begin{equation}
\xx_{M}=\xx_{N_{1}}\xx_{N_{2}}\cdots\xx_{N_{k}}.
\label{pf.prop.lyndon.grfun.alt1.c4.pf.1}%
\end{equation}


Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Then, the definition of $\operatorname*{CFL}$ yields
\[
\operatorname*{CFL}%
w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]  ,\ldots,\left[
a_{k}\right]  \right\}  _{\operatorname*{multiset}} .
\]
Thus,
$M=\operatorname*{CFL}w=\left\{  \left[  a_{1}\right]  ,\left[  a_{2}\right]
,\ldots,\left[  a_{k}\right]  \right\}  _{\operatorname*{multiset}}$. Hence,
(\ref{pf.prop.lyndon.grfun.alt1.c4.pf.1}) (applied to $N_{i}=\left[
a_{i}\right]  $) yields
\begin{equation}
\xx_{M}=\xx_{\left[  a_{1}\right]  }\xx_{\left[
a_{2}\right]  }\cdots\xx_{\left[  a_{k}\right]  }=\prod_{i=1}%
^{k}\xx_{\left[  a_{i}\right]  }.
\label{pf.prop.lyndon.grfun.alt1.c4.pf.2}%
\end{equation}


But we know that $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a CFL
factorization of $w$. In other words, $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words satisfying $w=a_{1}a_{2}\cdots
a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (because this is how a
``CFL factorization of $w$'' is defined).
From $w=a_{1}a_{2}\cdots a_{k}$, we obtain%
\begin{equation}
\xx_{w}=\xx_{a_{1}a_{2}\cdots a_{k}}=\xx_{a_{1}%
}\xx_{a_{2}}\cdots\xx_{a_{k}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma \ref{lem.sol.lyndon.grfun.xai}}\right)  .
\label{pf.prop.lyndon.grfun.alt1.c4.pf.3}%
\end{equation}


On the other hand, each $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies
$\xx_{\left[  a_{i}\right]  }=\xx_{a_{i}}$%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots,k\right\}  $.
Then, $a_{i}$ is a Lyndon word (since $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is a tuple of Lyndon words), and thus is nonempty (since any
Lyndon word is nonempty). Hence, Lemma \ref{lem.sol.lyndon.grfun.xa=xa}
(applied to $a=a_{i}$) yields $\xx_{\left[  a_{i}\right]  }%
=\xx_{a_{i}}$. Qed.}. Hence,
\[
\prod_{i=1}^{k}\underbrace{\xx_{\left[  a_{i}\right]  }}_{=\xx%
_{a_{i}}}=\prod_{i=1}^{k}\xx_{a_{i}}=\xx_{a_{1}}\xx%
_{a_{2}}\cdots\xx_{a_{k}}.
\]
Thus, (\ref{pf.prop.lyndon.grfun.alt1.c4.pf.2}) becomes%
\[
\xx_{M}=\prod_{i=1}^{k}\xx_{\left[  a_{i}\right]  }%
=\xx_{a_{1}}\xx_{a_{2}}\cdots\xx_{a_{k}}.
\]
Comparing this with (\ref{pf.prop.lyndon.grfun.alt1.c4.pf.3}), we obtain
$\xx_{w}=\xx_{M}=\xx_{\operatorname*{CFL}w}$ (since
$M=\operatorname*{CFL}w$). This proves Claim 3.]
\end{verlong}

\begin{statement}
\textit{Claim 4:} Let $M\in\mathfrak{MN}^{\mathfrak{a}}$ be such that
$\operatorname*{type}M=\lambda$. Then, $\operatorname*{sum}M=n$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 4:} We have $M\in\mathfrak{MN}^{\mathfrak{a}}$. In
other words, $M$ is a finite multiset of aperiodic necklaces. Let us thus
write the multiset $M$ in the form $M=\left\{  N_{1},N_{2},\ldots
,N_{k}\right\}  _{\operatorname*{multiset}}$ for some aperiodic necklaces
$N_{1},N_{2},\ldots,N_{k}$. Hence, the sizes of the necklaces are in $M$ are
the numbers $\left\vert N_{1}\right\vert ,\left\vert N_{2}\right\vert
,\ldots,\left\vert N_{k}\right\vert $.

But $\operatorname*{type}M=\lambda$, so that%
\begin{align*}
\lambda &  =\operatorname*{type}M\\
&  =\left(  \text{the partition obtained by listing the sizes of the necklaces
in }M\text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\operatorname*{type}M\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
N_{1}\right\vert ,\left\vert N_{2}\right\vert ,\ldots,\left\vert
N_{k}\right\vert \text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sizes of the necklaces in
}M\text{ are the numbers }\left\vert N_{1}\right\vert ,\left\vert
N_{2}\right\vert ,\ldots,\left\vert N_{k}\right\vert \right)  .
\end{align*}
Hence,%
\[
\left\vert \lambda\right\vert =\left\vert N_{1}\right\vert +\left\vert
N_{2}\right\vert +\cdots+\left\vert N_{k}\right\vert =\operatorname*{sum}%
M\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.def.sol.lyndon.gr.bij.sum.2}%
)}\right)  .
\]
Hence, $n=\left\vert \lambda\right\vert =\operatorname*{sum}M$. This proves
Claim 4.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 4:} We have $M\in\mathfrak{MN}^{\mathfrak{a}}$. In
other words, $M$ is a finite multiset of aperiodic necklaces (since
$\mathfrak{MN}^{\mathfrak{a}}$ is the set of all finite multisets of aperiodic
necklaces). Let us thus write the multiset $M$ in the form $M=\left\{
N_{1},N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}$ for some
aperiodic necklaces $N_{1},N_{2},\ldots,N_{k}$. Hence, the necklaces in $M$
are $N_{1},N_{2},\ldots,N_{k}$. Thus, the sizes of the necklaces are in $M$
are the numbers $\left\vert N_{1}\right\vert ,\left\vert N_{2}\right\vert
,\ldots,\left\vert N_{k}\right\vert $.

But $\operatorname*{type}M$ is the partition obtained by listing the sizes of
the necklaces in $M$ in decreasing order (by the definition of
$\operatorname*{type}M$). Hence,%
\begin{align*}
&  \operatorname*{type}M\\
&  =\left(  \text{the partition obtained by listing the sizes of the necklaces
in }M\text{ in decreasing order}\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
N_{1}\right\vert ,\left\vert N_{2}\right\vert ,\ldots,\left\vert
N_{k}\right\vert \text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sizes of the necklaces in
}M\text{ are the numbers }\left\vert N_{1}\right\vert ,\left\vert
N_{2}\right\vert ,\ldots,\left\vert N_{k}\right\vert \right)  .
\end{align*}
But $\operatorname*{type}M=\lambda$. Hence,%
\begin{align*}
&  \lambda=\operatorname*{type}M\\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
N_{1}\right\vert ,\left\vert N_{2}\right\vert ,\ldots,\left\vert
N_{k}\right\vert \text{ in decreasing order}\right)  .
\end{align*}
Hence, the partition $\lambda$ is a rearrangement of the $k$-tuple $\left(
\left\vert N_{1}\right\vert ,\left\vert N_{2}\right\vert ,\ldots,\left\vert
N_{k}\right\vert \right)  $ (of course, with trailing zeroes appended to it to
form an infinite sequence). In other words, there exists a permutation
$\sigma\in\Symm_{k}$ such that $\lambda=\left(  \left\vert
N_{\sigma\left(  1\right)  }\right\vert ,\left\vert N_{\sigma\left(  2\right)
}\right\vert ,\ldots,\left\vert N_{\sigma\left(  k\right)  }\right\vert
\right)  $. Consider this $\sigma$. Clearly, $\sigma$ is a permutation of
$\left\{  1,2,\ldots,k\right\}  $ (since $\sigma\in\Symm_{k}$), thus a
bijection from $\left\{  1,2,\ldots,k\right\}  $ to $\left\{  1,2,\ldots
,k\right\}  $.

From $\lambda=\left(  \left\vert N_{\sigma\left(  1\right)  }\right\vert
,\left\vert N_{\sigma\left(  2\right)  }\right\vert ,\ldots,\left\vert
N_{\sigma\left(  k\right)  }\right\vert \right)  $, we obtain
\begin{align*}
\left\vert \lambda\right\vert  &  =\left\vert \left(  \left\vert
N_{\sigma\left(  1\right)  }\right\vert ,\left\vert N_{\sigma\left(  2\right)
}\right\vert ,\ldots,\left\vert N_{\sigma\left(  k\right)  }\right\vert
\right)  \right\vert \\
&  =\left\vert N_{\sigma\left(  1\right)  }\right\vert +\left\vert
N_{\sigma\left(  2\right)  }\right\vert +\cdots+\left\vert N_{\sigma\left(
k\right)  }\right\vert =\sum_{i\in\left\{  1,2,\ldots,k\right\}  }\left\vert
N_{\sigma\left(  i\right)  }\right\vert \\
&  =\sum_{i\in\left\{  1,2,\ldots,k\right\}  }\left\vert N_{i}\right\vert
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }i\text{ for }\sigma\left(  i\right)  \text{
in the sum,}\\
\text{since }\sigma\text{ is a bijection from }\left\{  1,2,\ldots,k\right\}
\text{ to }\left\{  1,2,\ldots,k\right\}
\end{array}
\right) \\
&  =\left\vert N_{1}\right\vert +\left\vert N_{2}\right\vert +\cdots
+\left\vert N_{k}\right\vert =\operatorname*{sum}M\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.def.sol.lyndon.gr.bij.sum.2})}\right)  .
\end{align*}
Hence, $n=\left\vert \lambda\right\vert =\operatorname*{sum}M$. This proves
Claim 4.]
\end{verlong}

\begin{statement}
\textit{Claim 5:} We have
\[
\left\{  M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}\ \mid
\ \operatorname*{type}M=\lambda\right\}  =\left\{  M\in\mathfrak{MN}%
^{\mathfrak{a}}\ \mid\ \operatorname*{type}M=\lambda\right\}  .
\]

\end{statement}

\begin{vershort}
[\textit{Proof of Claim 5:} First, we observe that $\mathfrak{B}$-necklaces
and necklaces are the same thing\footnote{\textit{Proof.} Every necklace is a
subset of $\mathfrak{A}^{\ast}$. In other words, every necklace is a subset of
$\mathfrak{B}^{\ast}$ (since $\mathfrak{A}=\mathfrak{B}$). Hence, every
necklace is a $\mathfrak{B}$-necklace. Conversely, every $\mathfrak{B}%
$-necklace is a necklace. Combining the previous two sentences, we conclude
that $\mathfrak{B}$-necklaces and necklaces are the same thing.}. Moreover,
$\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}\subset\mathfrak{MN}%
^{\mathfrak{a}}$ (by the definitions of $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ and $\mathfrak{MN}^{\mathfrak{a}}$, since each
$\mathfrak{B}$-necklace is a necklace).

Now, if $M\in\mathfrak{MN}^{\mathfrak{a}}$ satisfies $\operatorname*{type}%
M=\lambda$, then $M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
$\ \ \ \ \footnote{\textit{Proof.} Let $M\in\mathfrak{MN}^{\mathfrak{a}}$
satisfy $\operatorname*{type}M=\lambda$. Then, Claim 4 yields
$\operatorname*{sum}M=n$. Moreover, $M\in\mathfrak{MN}^{\mathfrak{a}}$ shows
that $M$ is a finite multiset of aperiodic necklaces (by the definition of
$\mathfrak{MN}^{\mathfrak{a}}$). In other words, $M$ is a finite multiset of
aperiodic $\mathfrak{B}$-necklaces (since $\mathfrak{B}$-necklaces and
necklaces are the same thing). Since $M$ furthermore satisfies
$\operatorname*{sum}M=n$, we thus conclude that $M\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ (by the definition of $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$). Qed.}. Hence,%
\[
\left\{  M\in\mathfrak{MN}^{\mathfrak{a}}\ \mid\ \operatorname*{type}%
M=\lambda\right\}  \subset\left\{  M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\ \mid\ \operatorname*{type}M=\lambda\right\}  .
\]
Combining this with%
\[
\left\{  M\in\underbrace{\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
}_{\subset\mathfrak{MN}^{\mathfrak{a}}}\ \mid\ \operatorname*{type}%
M=\lambda\right\}  \subset\left\{  M\in\mathfrak{MN}^{\mathfrak{a}}%
\ \mid\ \operatorname*{type}M=\lambda\right\}  ,
\]
we obtain $\left\{  M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
\ \mid\ \operatorname*{type}M=\lambda\right\}  =\left\{  M\in\mathfrak{MN}%
^{\mathfrak{a}}\ \mid\ \operatorname*{type}M=\lambda\right\}  $. This proves
Claim 5.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 5:} Every necklace is a $\mathfrak{B}$%
-necklace\footnote{\textit{Proof.} Let $N$ be a necklace. Thus, $N$ is an
$n$-necklace for some positive integer $n$. Consider this $n$. Now, $N$ is an
$n$-necklace, i.e., an orbit of the $C$-action on $\mathfrak{A}^{n}$ (by the
definition of an $n$-necklace). Hence, $N$ is a subset of $\mathfrak{A}^{n}$,
and thus a subset of $\mathfrak{B}^{\ast}$ (since $\mathfrak{A}^{n}%
\subset\mathfrak{A}^{\ast}=\mathfrak{B}^{\ast}$ (because $\mathfrak{A}%
=\mathfrak{B}$)). Hence, $N$ is a necklace that is a subset of $\mathfrak{B}%
^{\ast}$. In other words, $N$ is a $\mathfrak{B}$-necklace (since a
$\mathfrak{B}$-necklace is defined to mean a necklace that is a subset of
$\mathfrak{B}^{\ast}$).
\par
Forget that we fixed $N$. We thus have proved that every necklace $N$ is a
$\mathfrak{B}$-necklace. Qed.}. Conversely, every $\mathfrak{B}$-necklace is a
necklace\footnote{since a $\mathfrak{B}$-necklace is defined to mean a
necklace that is a subset of $\mathfrak{B}^{\ast}$}. Combining these two
statements, we conclude that the $\mathfrak{B}$-necklaces are the same as the
necklaces. But $\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ is the set of
all finite multisets $M$ of aperiodic $\mathfrak{B}$-necklaces satisfying
$\operatorname*{sum}M=n$ (since this is how $\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}$ was defined). In other words, $\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ is the set of all finite multisets $M$ of
aperiodic necklaces satisfying $\operatorname*{sum}M=n$ (since the
$\mathfrak{B}$-necklaces are the same as the necklaces). In other words,%
\begin{equation}
\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}=\left\{  M\text{ is a finite
multiset of aperiodic necklaces }\mid\text{ }\operatorname*{sum}M=n\right\}  .
\label{pf.prop.lyndon.grfun.alt1.c6.pf.1}%
\end{equation}


On the other hand, $\mathfrak{MN}^{\mathfrak{a}}$ was defined to be the set of
all finite multisets of aperiodic necklaces. In other words,%
\begin{equation}
\mathfrak{MN}^{\mathfrak{a}}=\left\{  M\text{ is a finite multiset of
aperiodic necklaces}\right\}  . \label{pf.prop.lyndon.grfun.alt1.c6.pf.2}%
\end{equation}
Hence, (\ref{pf.prop.lyndon.grfun.alt1.c6.pf.1}) becomes%
\begin{align*}
\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}  &  =\left\{  M\text{ is a
finite multiset of aperiodic necklaces }\mid\text{ }\operatorname*{sum}%
M=n\right\} \\
&  \subset\left\{  M\text{ is a finite multiset of aperiodic necklaces}%
\right\}  =\mathfrak{MN}^{\mathfrak{a}}.
\end{align*}


Now, let $Y\in\left\{  M\in\mathfrak{MN}^{\mathfrak{a}}\ \mid
\ \operatorname*{type}M=\lambda\right\}  $. Thus, $Y$ is an $M\in
\mathfrak{MN}^{\mathfrak{a}}$ satisfying $\operatorname*{type}M=\lambda$. In
other words, $Y$ is an element of $\mathfrak{MN}^{\mathfrak{a}}$ and satisfies
$\operatorname*{type}Y=\lambda$. Hence, Claim 4 (applied to $M=Y$) yields
$\operatorname*{sum}Y=n$. Hence, $Y$ is a finite multiset $M$ of aperiodic
necklaces satisfying $\operatorname*{sum}M=n$ (since $Y$ is a finite multiset
of aperiodic necklaces\footnote{because $Y\in\mathfrak{MN}^{\mathfrak{a}%
}=\left\{  M\text{ is a finite multiset of aperiodic necklaces}\right\}  $}).
In other words,%
\[
Y\in\left\{  M\text{ is a finite multiset of aperiodic necklaces }\mid\text{
}\operatorname*{sum}M=n\right\}  .
\]
In view of (\ref{pf.prop.lyndon.grfun.alt1.c6.pf.1}), this rewrites as
$Y\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$. Hence, $Y$ is an
$M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}$ satisfying
$\operatorname*{type}M=\lambda$ (since $\operatorname*{type}Y=\lambda$). In
other words,%
\[
Y\in\left\{  M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}\ \mid
\ \operatorname*{type}M=\lambda\right\}  .
\]


Now, forget that we fixed $Y$. We thus have shown that $Y\in\left\{
M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}\ \mid\ \operatorname*{type}%
M=\lambda\right\}  $ for each $Y\in\left\{  M\in\mathfrak{MN}^{\mathfrak{a}%
}\ \mid\ \operatorname*{type}M=\lambda\right\}  $. In other words,%
\[
\left\{  M\in\mathfrak{MN}^{\mathfrak{a}}\ \mid\ \operatorname*{type}%
M=\lambda\right\}  \subset\left\{  M\in\mathfrak{MN}_{\mathfrak{B}%
,n}^{\mathfrak{a}}\ \mid\ \operatorname*{type}M=\lambda\right\}  .
\]
Combining this with%
\[
\left\{  M\in\underbrace{\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
}_{\subset\mathfrak{MN}^{\mathfrak{a}}}\ \mid\ \operatorname*{type}%
M=\lambda\right\}  \subset\left\{  M\in\mathfrak{MN}^{\mathfrak{a}}%
\ \mid\ \operatorname*{type}M=\lambda\right\}  ,
\]
we obtain $\left\{  M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}%
\ \mid\ \operatorname*{type}M=\lambda\right\}  =\left\{  M\in\mathfrak{MN}%
^{\mathfrak{a}}\ \mid\ \operatorname*{type}M=\lambda\right\}  $. This proves
Claim 5.]
\end{verlong}

\begin{vershort}
Now, Claim 2 shows that we have the following equality between summation
signs:%
\[
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}%
w=\lambda}}=\sum\limits_{\substack{w\in\mathfrak{B}^{n}%
;\\\operatorname*{CFLtype}w=\lambda}}=\sum\limits_{\substack{w\in
\mathfrak{B}^{n};\\\operatorname*{type}\left(  \operatorname*{CFL}w\right)
=\lambda}}
\]
(since Claim 1 shows that $\operatorname*{CFLtype}w=\operatorname*{type}%
\left(  \operatorname*{CFL}w\right)  $ for each $w\in\mathfrak{B}^{n}$). But
the definition of $\mathbf{GR}_{\lambda}$ yields%
\begin{align}
\mathbf{GR}_{\lambda}  &  =\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda}}}_{=\sum
\limits_{\substack{w\in\mathfrak{B}^{n};\\\operatorname*{type}\left(
\operatorname*{CFL}w\right)  =\lambda}}}\xx_{w}=\sum
\limits_{\substack{w\in\mathfrak{B}^{n};\\\operatorname*{type}\left(
\operatorname*{CFL}w\right)  =\lambda}}\underbrace{\xx_{w}%
}_{\substack{=\xx_{\operatorname*{CFL}w}\\\text{(by Claim 3)}}%
}=\sum\limits_{\substack{w\in\mathfrak{B}^{n};\\\operatorname*{type}\left(
\operatorname*{CFL}w\right)  =\lambda}}\xx_{\operatorname*{CFL}%
w}\nonumber\\
&  =\sum\limits_{\substack{M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
};\\\operatorname*{type}M=\lambda}}\xx_{M}
\label{pf.prop.lyndon.grfun.alt1.short.fin1}%
\end{align}
(here, we have substituted $M$ for $\operatorname*{CFL}w$ in the sum, since
the map $\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ is a bijection). On the other hand, Claim 5
leads to the following equality between summation signs:%
\[
\sum\limits_{\substack{M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
};\\\operatorname*{type}M=\lambda}}=\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}.
\]
Using this equality, we can rewrite
(\ref{pf.prop.lyndon.grfun.alt1.short.fin1}) as%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}%
};\\\operatorname*{type}M=\lambda}}\xx_{M}.
\]

\end{vershort}

\begin{verlong}
Now, recall that Claim 2 says that
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\lambda
\right\}  =\left\{  w\in\mathfrak{B}^{n}\ \mid\ \operatorname*{CFLtype}%
w=\lambda\right\}  .
\]
Hence, we have the following equality between summation signs:%
\[
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}%
w=\lambda}}=\sum\limits_{\substack{w\in\mathfrak{B}^{n}%
;\\\operatorname*{CFLtype}w=\lambda}}=\sum\limits_{\substack{w\in
\mathfrak{B}^{n};\\\operatorname*{type}\left(  \operatorname*{CFL}w\right)
=\lambda}}
\]
(since Claim 1 shows that $\operatorname*{CFLtype}w=\operatorname*{type}%
\left(  \operatorname*{CFL}w\right)  $ for each $w\in\mathfrak{B}^{n}$). But
the definition of $\mathbf{GR}_{\lambda}$ yields%
\begin{align}
\mathbf{GR}_{\lambda}  &  =\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda}}}_{=\sum
\limits_{\substack{w\in\mathfrak{B}^{n};\\\operatorname*{type}\left(
\operatorname*{CFL}w\right)  =\lambda}}}\xx_{w}=\sum
\limits_{\substack{w\in\mathfrak{B}^{n};\\\operatorname*{type}\left(
\operatorname*{CFL}w\right)  =\lambda}}\underbrace{\xx_{w}%
}_{\substack{=\xx_{\operatorname*{CFL}w}\\\text{(by Claim 3)}}%
}=\sum\limits_{\substack{w\in\mathfrak{B}^{n};\\\operatorname*{type}\left(
\operatorname*{CFL}w\right)  =\lambda}}\xx_{\operatorname*{CFL}%
w}\nonumber\\
&  =\sum\limits_{\substack{M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
};\\\operatorname*{type}M=\lambda}}\xx_{M}
\label{pf.prop.lyndon.grfun.alt1.fin1}%
\end{align}
(here, we have substituted $M$ for $\operatorname*{CFL}w$ in the sum, since
the map $\operatorname*{CFL}:\mathfrak{B}^{n}\rightarrow\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}}$ is a bijection). On the other hand, Claim 5
says that
\[
\left\{  M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}}\ \mid
\ \operatorname*{type}M=\lambda\right\}  =\left\{  M\in\mathfrak{MN}%
^{\mathfrak{a}}\ \mid\ \operatorname*{type}M=\lambda\right\}  .
\]
Hence, we have the following equality between summation signs:%
\[
\sum\limits_{\substack{M\in\mathfrak{MN}_{\mathfrak{B},n}^{\mathfrak{a}%
};\\\operatorname*{type}M=\lambda}}=\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}.
\]
Hence, (\ref{pf.prop.lyndon.grfun.alt1.fin1}) becomes%
\[
\mathbf{GR}_{\lambda}=\underbrace{\sum\limits_{\substack{M\in\mathfrak{MN}%
_{\mathfrak{B},n}^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}}%
_{=\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}}%
;\\\operatorname*{type}M=\lambda}}}\xx_{M}=\sum\limits_{\substack{M\in
\mathfrak{MN}^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\xx_{M}.
\]

\end{verlong}

\noindent This proves Proposition \ref{prop.lyndon.grfun.alt1}.
\end{proof}

Next, we shall prove Proposition \ref{prop.lyndon.grfun.alt2} using the
following lemma:

\begin{lemma}
\label{lem.sol.lyndon.grfun.xGRw}Let $w\in\mathfrak{A}^{\ast}$. Then,
$\xx_{\operatorname*{GR}w}=\xx_{w}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.xGRw}.]Let $n=\ell\left(  w\right)
$. Thus, $w\in\mathfrak{A}^{n}$, so that $w=\left(  w_{1},w_{2},\ldots
,w_{n}\right)  $. In particular, the $n$ letters $w_{1},w_{2},\ldots,w_{n}$
are well-defined. For each $i\in\left\{  1,2,\ldots,n\right\}  $, we set%
\begin{equation}
q_{i}:=x_{w_{i}}. \label{pf.lem.sol.lyndon.grfun.xGRw.qi=}%
\end{equation}
(This notation will help us avoid towers of subscripts. For example, we shall
soon work with expressions like $q_{h_{j}}$; without this notation, we would
have to write $x_{w_{h_{j}}}$ for them.)

We have $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $. Thus, the definition of
$\xx_{w}$ yields
\begin{align}
\xx_{w}  &  =x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=\prod_{i\in\left\{
1,2,\ldots,n\right\}  }\underbrace{x_{w_{i}}}_{\substack{=q_{i}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.xGRw.qi=}))}}}\nonumber\\
&  =\prod_{i\in\left\{  1,2,\ldots,n\right\}  }q_{i}.
\label{pf.lem.sol.lyndon.grfun.xGRw.xw=}%
\end{align}


Recall that $w\in\mathfrak{A}^{n}$. Let $\tau$ be the permutation $\left(
\operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$. Then, the definition
of the multiset $\operatorname*{GR}w\in\mathfrak{MN}^{\mathfrak{a}}$ shows
that%
\begin{equation}
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}.
\label{pf.lem.sol.lyndon.grfun.xGRw.GRw=1}%
\end{equation}


Let us now show the following two claims:

\begin{statement}
\textit{Claim 1:} Let $u\in\mathfrak{A}^{\ast}$ be a word. Let $h_{1}%
,h_{2},\ldots,h_{k}$ be some elements of $\left\{  1,2,\ldots,n\right\}  $
such that $u=\left(  w_{h_{1}},w_{h_{2}},\ldots,w_{h_{k}}\right)  $. Then,%
\[
\xx_{u}=q_{h_{1}}q_{h_{2}}\cdots q_{h_{k}}.
\]

\end{statement}

[\textit{Proof of Claim 1:} We have $u=\left(  w_{h_{1}},w_{h_{2}}%
,\ldots,w_{h_{k}}\right)  $. Thus, the definition of $\xx_{u}$ yields%
\[
\xx_{u}=x_{w_{h_{1}}}x_{w_{h_{2}}}\cdots x_{w_{h_{k}}}=\prod
_{j\in\left\{  1,2,\ldots,k\right\}  }x_{w_{h_{j}}}.
\]
Comparing this with%
\[
q_{h_{1}}q_{h_{2}}\cdots q_{h_{k}}=\prod_{j\in\left\{  1,2,\ldots,k\right\}
}\underbrace{q_{h_{j}}}_{\substack{=x_{w_{h_{j}}}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.xGRw.qi=}), applied to }i=h_{j}\text{)}}%
}=\prod_{j\in\left\{  1,2,\ldots,k\right\}  }x_{w_{h_{j}}},
\]
we obtain $\xx_{u}=q_{h_{1}}q_{h_{2}}\cdots q_{h_{k}}$. This proves
Claim 1.]

\begin{statement}
\textit{Claim 2:} Let $z$ be a cycle of $\tau$. Then,
\[
\xx_{\left[  w\right]  _{z}}=\prod_{i\in z}q_{i}.
\]

\end{statement}

[\textit{Proof of Claim 2:} The definition of $\left[  w\right]  _{z}$ yields
$\left[  w\right]  _{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $. The set
$z$ is a cycle of $\tau$, and thus is nonempty (since any cycle of $\tau$ is
nonempty). Hence, there exists some $h\in z$. Consider this $h$. Proposition
\ref{prop.words.read-std2}(a) yields $\left[  w_{\tau,h}\right]  =\left\{
w_{\tau,i}\ \mid\ i\in z\right\}  $. Comparing this with $\left[  w\right]
_{z}=\left\{  w_{\tau,i}\ \mid\ i\in z\right\}  $, we obtain $\left[
w\right]  _{z}=\left[  w_{\tau,h}\right]  $.

The definition of $w_{\tau,h}$ (given in Definition \ref{def.words.perm-ord}%
(b)) yields
\begin{equation}
w_{\tau,h}=w_{\tau^{1}\left(  h\right)  }w_{\tau^{2}\left(  h\right)  }\cdots
w_{\tau^{k}\left(  h\right)  },\ \ \ \ \ \ \ \ \ \ \text{where }%
k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  .
\label{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.1}%
\end{equation}


Let $k$ be the size of $z$. Thus, $k=\left\vert z\right\vert \geq1$ (since $z$
is nonempty).

For each $u\in \NN $, we set $h_{u}:=\tau^{u}\left(  h\right)  $. Then,
Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(e) (applied to $p=h$ and
$p_{i}=h_{i}$ and $u=0$) yields $h_{0+k}=h_{0}$. In other words, $h_{k}=h_{0}$
(since $0+k=k$). Furthermore, Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(c)
(applied to $p=h$ and $p_{i}=h_{i}$) yields $z=\left\{  h_{0},h_{1}%
,\ldots,h_{k-1}\right\}  $. Also, Lemma \ref{lem.sol.lyndon.gr.basics.cyc1}(g)
(applied to $p=h$ and $p_{i}=h_{i}$) yields that the $k$ elements $h_{0}%
,h_{1},\ldots,h_{k-1}$ are distinct. Finally, Lemma
\ref{lem.sol.lyndon.gr.basics.cyc1}(i) (applied to $p=h$ and $p_{i}=h_{i}$)
yields $\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  =k$. Thus,
$k=\operatorname*{ord}\nolimits_{\tau}\left(  h\right)  $; hence, from
(\ref{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.1}), we obtain
\begin{equation}
w_{\tau,h}=w_{\tau^{1}\left(  h\right)  }w_{\tau^{2}\left(  h\right)  }\cdots
w_{\tau^{k}\left(  h\right)  }=\left(  w_{\tau^{1}\left(  h\right)  }%
,w_{\tau^{2}\left(  h\right)  },\ldots,w_{\tau^{k}\left(  h\right)  }\right)
\label{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.2}%
\end{equation}
(since $w_{\tau^{1}\left(  h\right)  },w_{\tau^{2}\left(  h\right)  }%
,\ldots,w_{\tau^{k}\left(  h\right)  }$ are single letters). On the other
hand, each $u\in\left\{  1,2,\ldots,k\right\}  $ satisfies $w_{h_{u}}%
=w_{\tau^{u}\left(  h\right)  }$ (since $h_{u}=\tau^{u}\left(  h\right)  $ (by
the definition of $h_{u}$)). In other words, we have%
\[
\left(  w_{h_{1}},w_{h_{2}},\ldots,w_{h_{k}}\right)  =\left(  w_{\tau
^{1}\left(  h\right)  },w_{\tau^{2}\left(  h\right)  },\ldots,w_{\tau
^{k}\left(  h\right)  }\right)  .
\]
Comparing this with (\ref{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.2}), we obtain%
\[
w_{\tau,h}=\left(  w_{h_{1}},w_{h_{2}},\ldots,w_{h_{k}}\right)  .
\]
Hence, Claim 1 (applied to $u=w_{\tau,h}$) yields
\begin{align}
\xx_{w_{\tau,h}}  &  =q_{h_{1}}q_{h_{2}}\cdots q_{h_{k}}=\left(
q_{h_{1}}q_{h_{2}}\cdots q_{h_{k-1}}\right)  \underbrace{q_{h_{k}}%
}_{\substack{=q_{h_{0}}\\\text{(since }h_{k}=h_{0}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\geq1\right) \nonumber\\
&  =\left(  q_{h_{1}}q_{h_{2}}\cdots q_{h_{k-1}}\right)  q_{h_{0}}=q_{h_{0}%
}\left(  q_{h_{1}}q_{h_{2}}\cdots q_{h_{k-1}}\right) \nonumber\\
&  =q_{h_{0}}q_{h_{1}}\cdots q_{h_{k-1}}.
\label{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.3}%
\end{align}
But the list $\left(  h_{0},h_{1},\ldots,h_{k-1}\right)  $ is a list of all
elements of $z$ without repetitions (since $z=\left\{  h_{0},h_{1}%
,\ldots,h_{k-1}\right\}  $ and because the $k$ elements $h_{0},h_{1}%
,\ldots,h_{k-1}$ are distinct). Thus,%
\[
\prod_{i\in z}q_{i}=q_{h_{0}}q_{h_{1}}\cdots q_{h_{k-1}}.
\]
Comparing this with (\ref{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.3}), we obtain
\begin{equation}
\prod_{i\in z}q_{i}=\xx_{w_{\tau,h}}.
\label{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.4}%
\end{equation}


The word $w_{\tau,h}$ has length $k$ (since $w_{\tau,h}=\left(  w_{h_{1}%
},w_{h_{2}},\ldots,w_{h_{k}}\right)  $), and thus is nonempty (since
$k\geq1>0$). Hence, Lemma \ref{lem.sol.lyndon.grfun.xa=xa} (applied to
$a=w_{\tau,h}$) yields $\xx_{\left[  w_{\tau,h}\right]  }%
=\xx_{w_{\tau,h}}$. In view of $\left[  w\right]  _{z}=\left[
w_{\tau,h}\right]  $, this rewrites as $\xx_{\left[  w\right]  _{z}%
}=\xx_{w_{\tau,h}}$. Comparing this with
(\ref{pf.lem.sol.lyndon.grfun.xGRw.c1.pf.4}), we obtain $\xx_{\left[
w\right]  _{z}}=\prod_{i\in z}q_{i}$. This proves Claim 2.]

We know that $\operatorname*{GR}w$ is a finite multiset of necklaces. Thus,
the definition of the monomial $\xx_{\operatorname*{GR}w}$ shows that
$\xx_{\operatorname*{GR}w}=\xx_{N_{1}}\xx_{N_{2}}%
\cdots\xx_{N_{k}}$, where $\operatorname*{GR}w$ is written in the form
$\operatorname*{GR}w=\left\{  N_{1},N_{2},\ldots,N_{k}\right\}
_{\operatorname*{multiset}}$. In other words, if $N_{1},N_{2},\ldots,N_{k}$
are necklaces such that $\operatorname*{GR}w=\left\{  N_{1},N_{2},\ldots
,N_{k}\right\}  _{\operatorname*{multiset}}$, then%
\begin{equation}
\xx_{\operatorname*{GR}w}=\xx_{N_{1}}\xx_{N_{2}}%
\cdots\xx_{N_{k}}. \label{pf.lem.sol.lyndon.grfun.xGRw.1}%
\end{equation}


\begin{vershort}
Recall that $\tau\in\Symm_{n}$. Hence, the cycles of $\tau$ are
disjoint subsets of $\left\{  1,2,\ldots,n\right\}  $, and each element of
$\left\{  1,2,\ldots,n\right\}  $ lies in exactly one of these cycles.
\end{vershort}

\begin{verlong}
Recall that $\tau\in\Symm_{n}$; in other words, $\tau$ is a permutation
of $\left\{  1,2,\ldots,n\right\}  $. Hence, the cycles of $\tau$ are disjoint
nonempty subsets of $\left\{  1,2,\ldots,n\right\}  $, and each element of
$\left\{  1,2,\ldots,n\right\}  $ lies in exactly one of these cycles.
\end{verlong}

\begin{vershort}
Let $z_{1},z_{2},\ldots,z_{k}$ be all the cycles of $\tau$ (listed without
repetitions). The equality (\ref{pf.lem.sol.lyndon.grfun.xGRw.GRw=1}) becomes%
\[
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}=\left\{  \left[  w\right]
_{z_{1}},\left[  w\right]  _{z_{2}},\ldots,\left[  w\right]  _{z_{k}}\right\}
_{\operatorname*{multiset}}%
\]
(since $z_{1},z_{2},\ldots,z_{k}$ are all the cycles of $\tau$, listed without repetitions).
\end{vershort}

\begin{verlong}
Let $z_{1},z_{2},\ldots,z_{k}$ be all the cycles of $\tau$ (listed without
repetitions). Thus,%
\[
\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle of }\tau\right\}
_{\operatorname*{multiset}}=\left\{  \left[  w\right]  _{z_{1}},\left[
w\right]  _{z_{2}},\ldots,\left[  w\right]  _{z_{k}}\right\}
_{\operatorname*{multiset}},
\]
and thus (\ref{pf.lem.sol.lyndon.grfun.xGRw.GRw=1}) becomes%
\[
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}=\left\{  \left[  w\right]
_{z_{1}},\left[  w\right]  _{z_{2}},\ldots,\left[  w\right]  _{z_{k}}\right\}
_{\operatorname*{multiset}}.
\]

\end{verlong}

\begin{vershort}
\noindent Hence, (\ref{pf.lem.sol.lyndon.grfun.xGRw.1}) (applied to
$N_{i}=\left[  w\right]  _{z_{i}}$) yields%
\begin{align*}
\xx_{\operatorname*{GR}w}  &  =\xx_{\left[  w\right]  _{z_{1}}%
}\xx_{\left[  w\right]  _{z_{2}}}\cdots\xx_{\left[  w\right]
_{z_{k}}}=\prod_{z\text{ is a cycle of }\tau}\underbrace{\xx_{\left[
w\right]  _{z}}}_{\substack{=\prod_{i\in z}q_{i}\\\text{(by Claim 2)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }z_{1},z_{2},\ldots,z_{k}\text{ are all the cycles of }\tau\\
\text{(listed without repetitions)}%
\end{array}
\right) \\
&  =\underbrace{\prod_{z\text{ is a cycle of }\tau}\ \ \prod_{i\in z}%
}_{\substack{=\prod_{i\in\left\{  1,2,\ldots,n\right\}  }\\\text{(since the
cycles of }\tau\text{ are subsets of }\left\{  1,2,\ldots,n\right\}
\text{,}\\\text{and since each element of }\left\{  1,2,\ldots,n\right\}
\text{ lies in}\\\text{exactly one of these cycles)}}}q_{i}=\prod
_{i\in\left\{  1,2,\ldots,n\right\}  }q_{i}=\xx_{w}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.sol.lyndon.grfun.xGRw.xw=}%
)}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
\noindent Hence, (\ref{pf.lem.sol.lyndon.grfun.xGRw.1}) (applied to
$N_{i}=\left[  w\right]  _{z_{i}}$) yields%
\[
\xx_{\operatorname*{GR}w}=\xx_{\left[  w\right]  _{z_{1}}%
}\xx_{\left[  w\right]  _{z_{2}}}\cdots\xx_{\left[  w\right]
_{z_{k}}}.
\]
Comparing this with%
\begin{align*}
\prod_{z\text{ is a cycle of }\tau}\xx_{\left[  w\right]  _{z}}  &
=\xx_{\left[  w\right]  _{z_{1}}}\xx_{\left[  w\right]  _{z_{2}%
}}\cdots\xx_{\left[  w\right]  _{z_{k}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }z_{1},z_{2},\ldots,z_{k}\text{ are all the cycles of }\tau\\
\text{(listed without repetitions)}%
\end{array}
\right)  ,
\end{align*}
we obtain
\begin{align*}
\xx_{\operatorname*{GR}w}  &  =\prod_{z\text{ is a cycle of }\tau
}\underbrace{\xx_{\left[  w\right]  _{z}}}_{\substack{=\prod_{i\in
z}q_{i}\\\text{(by Claim 2)}}}=\underbrace{\prod_{z\text{ is a cycle of }\tau
}\prod_{i\in z}}_{\substack{=\prod_{i\in\left\{  1,2,\ldots,n\right\}
}\\\text{(since the cycles of }\tau\text{ are subsets of }\left\{
1,2,\ldots,n\right\}  \text{,}\\\text{and since each element of }\left\{
1,2,\ldots,n\right\}  \text{ lies in}\\\text{exactly one of these cycles)}%
}}q_{i}\\
&  =\prod_{i\in\left\{  1,2,\ldots,n\right\}  }q_{i}=\xx_{w}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.sol.lyndon.grfun.xGRw.xw=}%
)}\right)  .
\end{align*}

\end{verlong}

\noindent This proves Lemma \ref{lem.sol.lyndon.grfun.xGRw}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.alt2}.]Theorem
\ref{thm.lyndon.gr.bij} says that the maps $\operatorname*{GR}:\mathfrak{A}%
^{\ast}\rightarrow\mathfrak{MN}^{\mathfrak{a}}$ and $\operatorname*{RG}%
:\mathfrak{MN}^{\mathfrak{a}}\rightarrow\mathfrak{A}^{\ast}$ are mutually
inverse bijections. But Proposition \ref{prop.lyndon.grfun.alt1} yields%
\begin{align*}
\mathbf{GR}_{\lambda}  &  =\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\xx_{M}=\sum
\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}\left(
\operatorname*{GR}w\right)  =\lambda}}\underbrace{\xx%
_{\operatorname*{GR}w}}_{\substack{=\xx_{w}\\\text{(by Lemma
\ref{lem.sol.lyndon.grfun.xGRw})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\operatorname*{GR}w\text{ for }M\text{ in the
sum,}\\
\text{since the map }\operatorname*{GR}:\mathfrak{A}^{\ast}\rightarrow
\mathfrak{MN}^{\mathfrak{a}}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}%
\left(  \operatorname*{GR}w\right)  =\lambda}}\xx_{w}.
\end{align*}
This proves Proposition \ref{prop.lyndon.grfun.alt2}.
\end{proof}

In order to prove Proposition \ref{prop.lyndon.grfun.alt3}, we will need the
following two facts:

\begin{proposition}
\label{prop.sol.lyndon.grfun.typetau-1}Let $n\in \NN $. Let $\tau
\in\Symm_{n}$. Then, $\operatorname*{type}\left(  \tau^{-1}\right)
=\operatorname*{type}\tau$.
\end{proposition}

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.grfun.typetau-1}.]It is well-known
that the permutations $\tau$ and $\tau^{-1}$ have the same cycle
type\footnote{In fact, they have the same cycles, if we regard the cycles of a
permutation as sets.}. In other words, $\operatorname*{type}\left(  \tau
^{-1}\right)  =\operatorname*{type}\tau$ (because $\operatorname*{type}\sigma$
denotes the cycle type of a permutation $\sigma$). This proves Proposition
\ref{prop.sol.lyndon.grfun.typetau-1}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.grfun.typetau-1}.]In the following,
the cycles of a permutation of $\left\{  1,2,\ldots,n\right\}  $ will always
be regarded as subsets of $\left\{  1,2,\ldots,n\right\}  $. For example, the
permutation of $\left\{  1,2,3,4\right\}  $ that sends $1,2,3,4$ to $3,1,2,4$
(respectively) has the two cycles $\left\{  1,2,3\right\}  $ and $\left\{
4\right\}  $; the precise way in which this permutation permutes the elements
of each cycle is not encoded in the cycles.

The cycles of $\tau^{-1}$ are precisely the cycles of $\tau$%
\ \ \ \ \footnote{\textit{Proof.} Let $z$ be a cycle of $\tau$. Thus,
$z=\left\{  \tau^{i}\left(  h\right)  \ \mid\ i\in \ZZ \right\}  $ for
some $h\in\left\{  1,2,\ldots,n\right\}  $. Consider this $h$. Now,%
\begin{align*}
z  &  =\left\{  \tau^{i}\left(  h\right)  \ \mid\ i\in \ZZ \right\}
=\left\{  \underbrace{\tau^{-i}}_{=\left(  \tau^{-1}\right)  ^{i}}\left(
h\right)  \ \mid\ i\in \ZZ \right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }-i\text{ for the index }i\text{,}\\
\text{since the map } \ZZ \rightarrow \ZZ ,\ i\mapsto-i\text{ is a
bijection}%
\end{array}
\right) \\
&  =\left\{  \left(  \tau^{-1}\right)  ^{i}\left(  h\right)  \ \mid
\ i\in \ZZ \right\}  .
\end{align*}
This shows that $z$ is a cycle of $\tau^{-1}$ (namely, the cycle containing
the element $h$).
\par
Forget that we fixed $z$. We thus have shown that if $z$ is a cycle of $\tau$,
then $z$ is a cycle of $\tau^{-1}$. In other words, every cycle of $\tau$ is a
cycle of $\tau^{-1}$. The same argument (applied to $\tau^{-1}$ instead of
$\tau$) shows that every cycle of $\tau^{-1}$ is a cycle of $\left(  \tau
^{-1}\right)  ^{-1}$. In other words, every cycle of $\tau^{-1}$ is a cycle of
$\tau$ (since $\left(  \tau^{-1}\right)  ^{-1}=\tau$).
\par
We now have shown the following two facts:
\par
\begin{itemize}
\item Every cycle of $\tau$ is a cycle of $\tau^{-1}$.
\par
\item Every cycle of $\tau^{-1}$ is a cycle of $\tau$.
\end{itemize}
\par
Combining these two facts, we conclude that the cycles of $\tau^{-1}$ are
precisely the cycles of $\tau$. Qed.}.

We have%
\begin{align*}
\operatorname*{type}\tau &  =\left(  \text{the cycle type of }\tau\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{type}%
\tau\right) \\
&  =\left(  \text{the partition whose parts are the sizes of all cycles of
}\tau\text{ (listed in decreasing order)}\right)
\end{align*}
(by the definition of the cycle type of a permutation). The same argument
(applied to $\tau^{-1}$ instead of $\tau$) yields%
\begin{align*}
\operatorname*{type}\left(  \tau^{-1}\right)   &  =\left(  \text{the partition
whose parts are the sizes of all cycles of }\tau^{-1}\text{ (listed in
decreasing order)}\right) \\
&  =\left(  \text{the partition whose parts are the sizes of all cycles of
}\tau\text{ (listed in decreasing order)}\right)
\end{align*}
(since the cycles of $\tau^{-1}$ are precisely the cycles of $\tau$).
Comparing these two equalities, we obtain $\operatorname*{type}\left(
\tau^{-1}\right)  =\operatorname*{type}\tau$. This proves Proposition
\ref{prop.sol.lyndon.grfun.typetau-1}.
\end{proof}

\end{verlong}

\begin{proposition}
\label{prop.sol.lyndon.grfun.typeGR}Let $w\in\mathfrak{A}^{\ast}$. Then,
$\operatorname*{type}\left(  \operatorname*{GR}w\right)  =\operatorname*{type}%
\left(  \operatorname*{std}w\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.lyndon.grfun.typeGR}.]Let $n=\ell\left(
w\right)  $. Then, $w\in\mathfrak{A}^{n}$. Let $\tau$ be the permutation
$\left(  \operatorname*{std}w\right)  ^{-1}\in\Symm_{n}$. Then, the
definition of the multiset $\operatorname*{GR}w\in\mathfrak{MN}^{\mathfrak{a}%
}$ shows that%
\begin{equation}
\operatorname*{GR}w=\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a cycle
of }\tau\right\}  _{\operatorname*{multiset}}.
\label{pf.prop.sol.lyndon.grfun.typeGR.GRw=}%
\end{equation}


Let $z_{1},z_{2},\ldots,z_{k}$ be the cycles of $\tau$, listed in some order
(with no repetitions). Thus, (\ref{pf.prop.sol.lyndon.grfun.typeGR.GRw=})
becomes%
\begin{align}
\operatorname*{GR}w  &  =\left\{  \left[  w\right]  _{z}\ \mid\ z\text{ is a
cycle of }\tau\right\}  _{\operatorname*{multiset}}\nonumber\\
&  =\left\{  \left[  w\right]  _{z_{1}},\left[  w\right]  _{z_{2}}%
,\ldots,\left[  w\right]  _{z_{k}}\right\}  _{\operatorname*{multiset}}
\label{pf.prop.sol.lyndon.grfun.typeGR.GRw=2}%
\end{align}
(since $z_{1},z_{2},\ldots,z_{k}$ are the cycles of $\tau$, listed with no repetitions).

\begin{vershort}
Hence, the necklaces in $\operatorname*{GR}w$ are $\left[  w\right]  _{z_{1}%
},\left[  w\right]  _{z_{2}},\ldots,\left[  w\right]  _{z_{k}}$. Therefore,
the sizes of the necklaces in $\operatorname*{GR}w$ are the numbers
$\left\vert \left[  w\right]  _{z_{1}}\right\vert ,\left\vert \left[
w\right]  _{z_{2}}\right\vert ,\ldots,\left\vert \left[  w\right]  _{z_{k}%
}\right\vert $.
\end{vershort}

Now, it is easy to see that%
\begin{equation}
\left\vert \left[  w\right]  _{z_{m}}\right\vert =\left\vert z_{m}\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for each }m\in\left\{  1,2,\ldots,k\right\}  .
\label{pf.prop.sol.lyndon.grfun.typeGR.4}%
\end{equation}


[\textit{Proof of (\ref{pf.prop.sol.lyndon.grfun.typeGR.4}):} Let
$m\in\left\{  1,2,\ldots,k\right\}  $. Then, $z_{m}$ is a cycle of $\tau$
(since $z_{1},z_{2},\ldots,z_{k}$ are the cycles of $\tau$). Thus, the
definition of $\left[  w\right]  _{z_{m}}$ yields $\left[  w\right]  _{z_{m}%
}=\left\{  w_{\tau,i}\ \mid\ i\in z_{m}\right\}  $. But Proposition
\ref{prop.words.read-std2}(c) (applied to $z=z_{m}$) shows that $\left\vert
\left\{  w_{\tau,i}\ \mid\ i\in z_{m}\right\}  \right\vert =\left\vert
z_{m}\right\vert $. Since $\left[  w\right]  _{z_{m}}=\left\{  w_{\tau
,i}\ \mid\ i\in z_{m}\right\}  $, this rewrites as $\left\vert \left[
w\right]  _{z_{m}}\right\vert =\left\vert z_{m}\right\vert $. This proves
(\ref{pf.prop.sol.lyndon.grfun.typeGR.4}).]

\begin{vershort}
But the definition of $\operatorname*{type}\left(  \operatorname*{GR}w\right)
$ yields%
\begin{align*}
&  \operatorname*{type}\left(  \operatorname*{GR}w\right) \\
&  =\left(  \text{the partition obtained by listing the sizes of the necklaces
in }\operatorname*{GR}w\text{ in decreasing order}\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
\left[  w\right]  _{z_{1}}\right\vert ,\left\vert \left[  w\right]  _{z_{2}%
}\right\vert ,\ldots,\left\vert \left[  w\right]  _{z_{k}}\right\vert \text{
in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sizes of the necklaces in
}\operatorname*{GR}w\text{ are the numbers }\left\vert \left[  w\right]
_{z_{1}}\right\vert ,\left\vert \left[  w\right]  _{z_{2}}\right\vert
,\ldots,\left\vert \left[  w\right]  _{z_{k}}\right\vert \right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
z_{1}\right\vert ,\left\vert z_{2}\right\vert ,\ldots,\left\vert
z_{k}\right\vert \text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.lyndon.grfun.typeGR.4})}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
We have thus showed that $\left\vert \left[  w\right]  _{z_{m}}\right\vert
=\left\vert z_{m}\right\vert $ for each $m\in\left\{  1,2,\ldots,k\right\}  $.
In other words,%
\begin{equation}
\left(  \left\vert \left[  w\right]  _{z_{1}}\right\vert ,\left\vert \left[
w\right]  _{z_{2}}\right\vert ,\ldots,\left\vert \left[  w\right]  _{z_{k}%
}\right\vert \right)  =\left(  \left\vert z_{1}\right\vert ,\left\vert
z_{2}\right\vert ,\ldots,\left\vert z_{k}\right\vert \right)  .
\label{pf.prop.sol.lyndon.grfun.typeGR.5}%
\end{equation}


The necklaces in $\operatorname*{GR}w$ are $\left[  w\right]  _{z_{1}},\left[
w\right]  _{z_{2}},\ldots,\left[  w\right]  _{z_{k}}$ (because of
(\ref{pf.prop.sol.lyndon.grfun.typeGR.GRw=2})). Hence, the sizes of the
necklaces in $\operatorname*{GR}w$ are the numbers $\left\vert \left[
w\right]  _{z_{1}}\right\vert ,\left\vert \left[  w\right]  _{z_{2}%
}\right\vert ,\ldots,\left\vert \left[  w\right]  _{z_{k}}\right\vert $. In
view of (\ref{pf.prop.sol.lyndon.grfun.typeGR.5}), this rewrites as follows:
The sizes of the necklaces in $\operatorname*{GR}w$ are the numbers
$\left\vert z_{1}\right\vert ,\left\vert z_{2}\right\vert ,\ldots,\left\vert
z_{k}\right\vert $.

But $\operatorname*{type}\left(  \operatorname*{GR}w\right)  $ is the
partition obtained by listing the sizes of the necklaces in
$\operatorname*{GR}w$ in decreasing order (by the definition of
$\operatorname*{type}\left(  \operatorname*{GR}w\right)  $). Hence,%
\begin{align*}
&  \operatorname*{type}\left(  \operatorname*{GR}w\right) \\
&  =\left(  \text{the partition obtained by listing the sizes of the necklaces
in }\operatorname*{GR}w\text{ in decreasing order}\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
z_{1}\right\vert ,\left\vert z_{2}\right\vert ,\ldots,\left\vert
z_{k}\right\vert \text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the sizes of the necklaces in
}\operatorname*{GR}w\text{ are the numbers }\left\vert z_{1}\right\vert
,\left\vert z_{2}\right\vert ,\ldots,\left\vert z_{k}\right\vert \right)  .
\end{align*}

\end{verlong}

On the other hand, $\tau=\left(  \operatorname*{std}w\right)  ^{-1}$, so that
$\tau^{-1}=\operatorname*{std}w$. But Proposition
\ref{prop.sol.lyndon.grfun.typetau-1} yields $\operatorname*{type}\left(
\tau^{-1}\right)  =\operatorname*{type}\tau$, so that $\operatorname*{type}%
\tau=\operatorname*{type}\left(  \tau^{-1}\right)  =\operatorname*{type}%
\left(  \operatorname*{std}w\right)  $ (since $\tau^{-1}=\operatorname*{std}w$).

But $\operatorname*{type}\tau$ is the cycle type of $\tau$. In other words,
$\operatorname*{type}\tau$ is the partition obtained by listing the sizes of
the cycles of $\tau$ in decreasing order (by the definition of the cycle type
of $\tau$). Hence,%
\begin{align*}
&  \operatorname*{type}\tau\\
&  =\left(  \text{the partition obtained by listing the sizes of the cycles of
}\tau\text{ in decreasing order}\right) \\
&  =\left(  \text{the partition obtained by listing the sizes of }z_{1}%
,z_{2},\ldots,z_{k}\text{ in decreasing order}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the cycles of }\tau\text{ are
}z_{1},z_{2},\ldots,z_{k}\text{ (listed with no repetitions)}\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
z_{1}\right\vert ,\left\vert z_{2}\right\vert ,\ldots,\left\vert
z_{k}\right\vert \text{ in decreasing order}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
&  \operatorname*{type}\left(  \operatorname*{GR}w\right) \\
&  =\left(  \text{the partition obtained by listing the numbers }\left\vert
z_{1}\right\vert ,\left\vert z_{2}\right\vert ,\ldots,\left\vert
z_{k}\right\vert \text{ in decreasing order}\right)  ,
\end{align*}
we obtain $\operatorname*{type}\left(  \operatorname*{GR}w\right)
=\operatorname*{type}\tau=\operatorname*{type}\left(  \operatorname*{std}%
w\right)  $. This proves Proposition \ref{prop.sol.lyndon.grfun.typeGR}.
\end{proof}

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.alt3}.]Proposition
\ref{prop.lyndon.grfun.alt2} yields%
\[
\mathbf{GR}_{\lambda}=\underbrace{\sum\limits_{\substack{w\in\mathfrak{A}%
^{\ast};\\\operatorname*{type}\left(  \operatorname*{GR}w\right)  =\lambda}%
}}_{\substack{=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast}%
;\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda
}}\\\text{(since Proposition \ref{prop.sol.lyndon.grfun.typeGR}}\\\text{yields
}\operatorname*{type}\left(  \operatorname*{GR}w\right)  =\operatorname*{type}%
\left(  \operatorname*{std}w\right)  \text{ for each }w\in\mathfrak{A}^{\ast
}\text{)}}}\xx_{w}=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast
};\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda
}}\xx_{w}.
\]
This proves Proposition \ref{prop.lyndon.grfun.alt3}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.alt3}.]Proposition
\ref{prop.sol.lyndon.grfun.typeGR} yields $\operatorname*{type}\left(
\operatorname*{GR}w\right)  =\operatorname*{type}\left(  \operatorname*{std}%
w\right)  $ for each $w\in\mathfrak{A}^{\ast}$. Thus,%
\[
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}\left(
\operatorname*{GR}w\right)  =\lambda}}=\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda}}
\]
(an equality between summation signs). But Proposition
\ref{prop.lyndon.grfun.alt2} yields%
\[
\mathbf{GR}_{\lambda}=\underbrace{\sum\limits_{\substack{w\in\mathfrak{A}%
^{\ast};\\\operatorname*{type}\left(  \operatorname*{GR}w\right)  =\lambda}%
}}_{=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}%
\left(  \operatorname*{std}w\right)  =\lambda}}}\xx_{w}=\sum
\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda}}\xx_{w}.
\]
This proves Proposition \ref{prop.lyndon.grfun.alt3}.
\end{proof}

\end{verlong}

Next, we can easily prove Proposition \ref{prop.lyndon.grfun.F}:

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.F}.]We have the following:

\begin{statement}
\textit{Claim 1:} Let $w\in\mathfrak{A}^{\ast}$ satisfy $\operatorname*{type}%
\left(  \operatorname*{std}w\right)  =\lambda$. Then, $w\in\mathfrak{A}^{n}$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 1:} Let $m=\ell\left(  w\right)  $. Thus,
$w\in\mathfrak{A}^{m}$, so that $\operatorname*{std}w\in\Symm_{m}$. But
every permutation $\tau\in\Symm_{m}$ satisfies $\operatorname*{type}%
\tau\in \Par_{m}$ (since $\operatorname*{type}\tau$ denotes the
cycle type of $\tau$, which is always a partition of $m$). Applying this to
$\tau=\operatorname*{std}w$, we conclude that $\operatorname*{type}\left(
\operatorname*{std}w\right)  \in \Par_{m}$. In view of
$\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda$, this
rewrites as $\lambda\in \Par_{m}$. Thus, $\left\vert
\lambda\right\vert =m$, so that $n=\left\vert \lambda\right\vert =m$. Hence,
$m=n$. Now, $w\in\mathfrak{A}^{m}=\mathfrak{A}^{n}$ (since $m=n$). This proves
Claim 1.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 1:} Let $m=\ell\left(  w\right)  $. Thus,
$w\in\mathfrak{A}^{m}$, so that $\operatorname*{std}w\in\Symm_{m}$. But
the cycle type of any permutation $\tau\in\Symm_{m}$ is a partition of
$m$. In other words, the cycle type of any permutation
$\tau \in \Symm_m$ belongs to $ \Par_{m}$. In other words, for any
permutation $\tau\in\Symm_{m}$, we have $\operatorname*{type}\tau
\in \Par_{m}$ (since $\operatorname*{type}\tau$ denotes the
cycle type of $\tau$). Applying this to $\tau=\operatorname*{std}w$, we
conclude that $\operatorname*{type}\left(  \operatorname*{std}w\right)
\in \Par_{m}$. In view of $\operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda$, this rewrites as $\lambda
\in \Par_{m}$. Thus, $\left\vert \lambda\right\vert
=m$, so that $n=\left\vert \lambda\right\vert =m$. Hence, $m=n$. Now,
$w\in\mathfrak{A}^{m}=\mathfrak{A}^{n}$ (since $m=n$). This proves Claim 1.]
\end{verlong}

\begin{statement}
\textit{Claim 2:} We have $\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda\right\}
=\left\{  w\in\mathfrak{A}^{n}\ \mid\ \operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda\right\}  $.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 2:} Each $w\in\mathfrak{A}^{\ast}$ that satisfies
$\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda$ must also
satisfy $w\in\mathfrak{A}^{n}$ (by Claim 1). Hence,%
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda\right\}  \subset\left\{  w\in
\mathfrak{A}^{n}\ \mid\ \operatorname*{type}\left(  \operatorname*{std}%
w\right)  =\lambda\right\}  .
\]
Combining this with%
\[
\left\{  w\in\underbrace{\mathfrak{A}^{n}}_{\subset\mathfrak{A}^{\ast}}%
\ \mid\ \operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda\right\}  \subset\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda\right\}
,
\]
we obtain $\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda\right\}  =\left\{  w\in\mathfrak{A}%
^{n}\ \mid\ \operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda\right\}  $. This proves Claim 2.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 2:} Each $w\in\mathfrak{A}^{\ast}$ that satisfies
$\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda$ must also
satisfy $w\in\mathfrak{A}^{n}$ (by Claim 1). Hence, each $w\in\mathfrak{A}%
^{\ast}$ that satisfies $\operatorname*{type}\left(  \operatorname*{std}%
w\right)  =\lambda$ must also be a $w\in\mathfrak{A}^{n}$ that satisfies
$\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda$. In other
words,%
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda\right\}  \subset\left\{  w\in
\mathfrak{A}^{n}\ \mid\ \operatorname*{type}\left(  \operatorname*{std}%
w\right)  =\lambda\right\}  .
\]
Combining this with%
\[
\left\{  w\in\underbrace{\mathfrak{A}^{n}}_{\subset\mathfrak{A}^{\ast}}%
\ \mid\ \operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda\right\}  \subset\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda\right\}
,
\]
we obtain $\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda\right\}  =\left\{  w\in\mathfrak{A}%
^{n}\ \mid\ \operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda\right\}  $. This proves Claim 2.]
\end{verlong}

\begin{vershort}
Claim 2 yields the following equality between summation signs:%
\[
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda}}=\sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda}}.
\]

\end{vershort}

\begin{verlong}
Claim 2 shows that $\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda\right\}
=\left\{  w\in\mathfrak{A}^{n}\ \mid\ \operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda\right\}  $. Hence,%
\[
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{type}\left(
\operatorname*{std}w\right)  =\lambda}}=\sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda}}\ \ \ \ \ \ \ \ \ \ \left(  \text{an equality between summation
signs}\right)  .
\]

\end{verlong}

\begin{vershort}
But Proposition \ref{prop.lyndon.grfun.alt3} yields%
\begin{align*}
\mathbf{GR}_{\lambda}  &  =\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda}}}_{=\sum\limits_{\substack{w\in\mathfrak{A}^{n}%
;\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda}%
}}\xx_{w}=\underbrace{\sum\limits_{\substack{w\in\mathfrak{A}%
^{n};\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda}%
}}_{\substack{=\sum\limits_{\substack{\sigma\in\Symm_{n}%
;\\\operatorname*{type}\sigma=\lambda}}\ \ \sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma}}\\\text{(here, we have split
the sum}\\\text{according to the value of }\operatorname*{std}w\text{,}%
\\\text{because }\operatorname*{std}w\in\Symm_{n}\text{ for each }%
w\in\mathfrak{A}^{n}\text{)}}}\xx_{w}\\
&  =\sum\limits_{\substack{\sigma\in\Symm_{n};\\\operatorname*{type}%
\sigma=\lambda}}\ \ \sum\limits_{\substack{w\in\mathfrak{A}^{n}%
;\\\operatorname*{std}w=\sigma}}\xx_{w}=\underbrace{\sum
\limits_{\substack{\sigma\in\Symm_{n};\\\operatorname*{type}\left(
\sigma^{-1}\right)  =\lambda}}}_{\substack{=\sum\limits_{\substack{\sigma
\in\Symm_{n};\\\operatorname*{type}\sigma=\lambda}}\\\text{(since each
}\sigma\in\Symm_{n}\text{ satisfies }\operatorname*{type}\left(
\sigma^{-1}\right)  =\operatorname*{type}\sigma\\\text{(by Proposition
\ref{prop.sol.lyndon.grfun.typetau-1}, applied to }\tau=\sigma\text{))}}%
}\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\\operatorname*{std}%
w=\sigma^{-1}}}\xx_{w}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\sigma^{-1}\text{ for }\sigma\text{ in the
outer sum,}\\
\text{since the map }\Symm_{n}\rightarrow\Symm_{n}%
,\ \sigma\mapsto\sigma^{-1}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\sigma\in\Symm_{n};\\\operatorname*{type}%
\sigma=\lambda}}\ \ \sum\limits_{\substack{w\in\mathfrak{A}^{n}%
;\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}.
\end{align*}
Comparing this with%
\[
\underbrace{\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{
has cycle type }\lambda}}}_{\substack{=\sum\limits_{\substack{\sigma
\in\Symm_{n};\\\operatorname*{type}\sigma=\lambda}}\\\text{(since
}\operatorname*{type}\sigma\text{ means}\\\text{the cycle type of }%
\sigma\text{)}}}\ \ \underbrace{L_{\gamma\left(  \sigma\right)  }%
}_{\substack{=\sum\limits_{\substack{w\in\mathfrak{A}^{n}%
;\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}\\\text{(by Lemma
\ref{lem.Lalpha-std.std-L})}}}
=\sum\limits_{\substack{\sigma \in \Symm_n;\\
\operatorname*{type}\sigma=\lambda}}\ \ \sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w},
\]
we obtain%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{\sigma \in \Symm_n;\\
\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)
}.
\]

\end{vershort}

\begin{verlong}
But Proposition \ref{prop.lyndon.grfun.alt3} yields%
\begin{align*}
\mathbf{GR}_{\lambda}  &  =\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{type}\left(  \operatorname*{std}w\right)
=\lambda}}}_{=\sum\limits_{\substack{w\in\mathfrak{A}^{n}%
;\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda}%
}}\xx_{w}=\underbrace{\sum\limits_{\substack{w\in\mathfrak{A}%
^{n};\\\operatorname*{type}\left(  \operatorname*{std}w\right)  =\lambda}%
}}_{\substack{=\sum\limits_{\substack{\sigma\in\Symm_{n}%
;\\\operatorname*{type}\sigma=\lambda}}\ \ \sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma}}\\\text{(here, we have split
the sum}\\\text{according to the value of }\operatorname*{std}w\text{,}%
\\\text{because }\operatorname*{std}w\in\Symm_{n}\text{ for each }%
w\in\mathfrak{A}^{n}\text{)}}}\xx_{w}\\
&  =\sum\limits_{\substack{\sigma\in\Symm_{n};\\\operatorname*{type}%
\sigma=\lambda}}\ \ \sum\limits_{\substack{w\in\mathfrak{A}^{n}%
;\\\operatorname*{std}w=\sigma}}\xx_{w}=\underbrace{\sum
\limits_{\substack{\sigma\in\Symm_{n};\\\operatorname*{type}\left(
\sigma^{-1}\right)  =\lambda}}}_{\substack{=\sum\limits_{\substack{\sigma
\in\Symm_{n};\\\operatorname*{type}\sigma=\lambda}}\\\text{(since each
}\sigma\in\Symm_{n}\text{ satisfies }\operatorname*{type}\left(
\sigma^{-1}\right)  =\operatorname*{type}\sigma\\\text{(by Proposition
\ref{prop.sol.lyndon.grfun.typetau-1}, applied to }\tau=\sigma\text{))}}%
}\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\\operatorname*{std}%
w=\sigma^{-1}}}\xx_{w}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\sigma^{-1}\text{ for }\sigma\text{ in the
outer sum,}\\
\text{since the map }\Symm_{n}\rightarrow\Symm_{n}%
,\ \sigma\mapsto\sigma^{-1}\text{ is a bijection}%
\end{array}
\right) \\
&  =\underbrace{\sum\limits_{\substack{\sigma\in\Symm_{n}%
;\\\operatorname*{type}\sigma=\lambda}}}_{\substack{=\sum
\limits_{\substack{\sigma\in\Symm_{n};\\\left(  \text{cycle type of
}\sigma\right)  =\lambda}}\\\text{(since }\operatorname*{type}\sigma=\left(
\text{cycle type of }\sigma\right)  \\\text{(by the definition of
}\operatorname*{type}\sigma\text{))}}}\sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}%
=\sum\limits_{\substack{\sigma\in\Symm_{n};\\\left(  \text{cycle type
of }\sigma\right)  =\lambda}}\ \ \sum\limits_{\substack{w\in\mathfrak{A}%
^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w}.
\end{align*}
Comparing this with%
\[
\underbrace{\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{
has cycle type }\lambda}}}_{\substack{=\sum\limits_{\substack{\sigma
\in\Symm_{n};\\\left(  \text{cycle type of }\sigma\right)  =\lambda}%
}}}\ \ \underbrace{L_{\gamma\left(  \sigma\right)  }}_{\substack{=\sum
\limits_{\substack{w\in\mathfrak{A}^{n};\\\operatorname*{std}w=\sigma^{-1}%
}}\xx_{w}\\\text{(by Lemma \ref{lem.Lalpha-std.std-L})}}}=\sum
\limits_{\substack{\sigma\in\Symm_{n};\\\left(  \text{cycle type of
}\sigma\right)  =\lambda}}\ \ \sum\limits_{\substack{w\in\mathfrak{A}%
^{n};\\\operatorname*{std}w=\sigma^{-1}}}\xx_{w},
\]
we obtain%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{\sigma \in \Symm_n;\\
\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)
}.
\]

\end{verlong}

\noindent This proves Proposition \ref{prop.lyndon.grfun.F}.
\end{proof}

Next, we can prove Proposition \ref{prop.lyndon.grfun.cauchy2}:

\begin{vershort}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.cauchy2}.]Let $n\in \NN $.
Then, the cycle type of any permutation $\sigma\in\Symm_{n}$ is a
partition of $n$. Hence, we can split up the sum
\[
\sum_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\]
according to the cycle type of $\sigma$. We thus obtain%
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\nonumber\\
&  =\sum_{\lambda\text{ is a partition of }n}\ \ \sum\limits_{\substack{\sigma
\in\Symm_{n};\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  p_{\operatorname*{type}\sigma
}\left(  \mathbf{y}\right)  . \label{pf.prop.lyndon.grfun.cauchy2.short.3}%
\end{align}
But if $\lambda$ is a partition satisfying $\left\vert \lambda\right\vert =n$,
then
\begin{align}
\mathbf{GR}_{\lambda}\left(  \xx\right)   &  =\mathbf{GR}_{\lambda
}=\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{ has cycle
type }\lambda}}\ \ \underbrace{L_{\gamma\left(  \sigma\right)  }}%
_{=L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.lyndon.grfun.F}%
}\right) \nonumber\\
&  =\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{ has cycle
type }\lambda}}L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)  .
\label{pf.prop.lyndon.grfun.cauchy2.short.4}%
\end{align}
Hence, (\ref{pf.prop.lyndon.grfun.cauchy2.short.3}) becomes%
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\nonumber\\
&  =\sum_{\lambda\text{ is a partition of }n}\ \ \sum\limits_{\substack{\sigma
\in\Symm_{n};\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  \underbrace{p_{\operatorname*{type}%
\sigma}\left(  \mathbf{y}\right)  }_{\substack{=p_{\lambda}\left(
\mathbf{y}\right)  \\\text{(since }\operatorname*{type}\sigma=\lambda
\\\text{(since }\sigma\text{ has cycle type }\lambda\text{))}}}\nonumber\\
&  =\sum_{\lambda\text{ is a partition of }n}\ \ \underbrace{\sum
\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{ has cycle type
}\lambda}}L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)
}_{\substack{=\mathbf{GR}_{\lambda}\left(  \xx\right)  \\\text{(by
(\ref{pf.prop.lyndon.grfun.cauchy2.short.4}))}}}p_{\lambda}\left(
\mathbf{y}\right) \nonumber\\
&  =\sum_{\lambda\text{ is a partition of }n}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  .
\label{pf.prop.lyndon.grfun.cauchy2.short.5}%
\end{align}


Now, forget that we fixed $n$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.cauchy2.short.5}) for each $n\in \NN $. Now,
\begin{align*}
\underbrace{\sum_{\lambda\in \Par}}_{=\sum\limits_{n\in
 \NN }\ \ \sum\limits_{\lambda\text{ is a partition of }n}}%
\mathbf{GR}_{\lambda}\left(  \xx\right)  p_{\lambda}\left(
\mathbf{y}\right)   &  =\sum_{n\in \NN }\ \ \underbrace{\sum
_{\lambda\text{ is a partition of }n}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  }_{\substack{=\sum
_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\\\text{(by (\ref{pf.prop.lyndon.grfun.cauchy2.short.5}))}}}\\
&  =\underbrace{\sum_{n\in \NN }\ \ \sum_{\sigma\in\Symm_{n}}%
}_{\substack{=\sum\limits_{\sigma\in\bigsqcup_{n\in \NN }\mathfrak{S}%
_{n}}=\sum\limits_{\sigma\in\mathfrak{S}}\\\text{(since }\bigsqcup
_{n\in \NN }\Symm_{n}=\mathfrak{S}\text{)}}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  p_{\operatorname*{type}\sigma
}\left(  \mathbf{y}\right)  =\sum_{\sigma\in\mathfrak{S}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  p_{\operatorname*{type}\sigma
}\left(  \mathbf{y}\right)  .
\end{align*}
Combining this with the result of Proposition \ref{prop.lyndon.grfun.cauchy}%
(a), we obtain%
\[
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\lambda
\in \Par}p_{\lambda}\left(  \xx\right)  \mathbf{GR}%
_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\sigma\in\mathfrak{S}}%
L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)
p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)  .
\]
This proves Proposition \ref{prop.lyndon.grfun.cauchy2}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.cauchy2}.]Proposition
\ref{prop.lyndon.grfun.cauchy}(a) yields
\begin{equation}
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\lambda
\in \Par}p_{\lambda}\left(  \xx\right)  \mathbf{GR}%
_{\lambda}\left(  \mathbf{y}\right)  . \label{pf.prop.lyndon.grfun.cauchy2.1}%
\end{equation}


Now, let $n\in \NN $. Then, the cycle type of any permutation $\sigma
\in\Symm_{n}$ is a partition of $n$. Hence, we can split up the sum
\[
\sum_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\]
according to the cycle type of $\sigma$. We thus obtain%
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\nonumber\\
&  =\underbrace{\sum_{\lambda\text{ is a partition of }n}}_{=\sum
\limits_{\substack{\lambda\in\Par;\\\left\vert \lambda
\right\vert =n}}}\ \ \underbrace{\sum\limits_{\substack{\sigma\in
\Symm_{n};\\\text{the cycle type of }\sigma\text{ is }\lambda}}}%
_{=\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{ has cycle
type }\lambda}}}L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)
p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}\ \ \sum\limits_{\substack{\sigma\in\mathfrak{S}%
_{n};\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)
}\left(  \xx\right)  p_{\operatorname*{type}\sigma}\left(
\mathbf{y}\right)  . \label{pf.prop.lyndon.grfun.cauchy2.3}%
\end{align}
But if $\lambda$ is a partition satisfying $\left\vert \lambda\right\vert =n$,
then Proposition \ref{prop.lyndon.grfun.F} yields%
\[
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{\sigma\in\mathfrak{S}%
_{n};\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n=\left\vert \lambda\right\vert
\right)
\]
and thus%
\begin{equation}
\mathbf{GR}_{\lambda}\left(  \xx\right)  =\mathbf{GR}_{\lambda}%
=\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{ has cycle
type }\lambda}}\ \ \underbrace{L_{\gamma\left(  \sigma\right)  }}%
_{=L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)  }%
=\sum\limits_{\substack{\sigma\in\Symm_{n};\\\sigma\text{ has cycle
type }\lambda}}L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)  .
\label{pf.prop.lyndon.grfun.cauchy2.4}%
\end{equation}
Hence, (\ref{pf.prop.lyndon.grfun.cauchy2.3}) becomes%
\begin{align}
&  \sum_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}\ \ \sum\limits_{\substack{\sigma\in\mathfrak{S}%
_{n};\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)
}\left(  \xx\right)  \underbrace{p_{\operatorname*{type}\sigma}\left(
\mathbf{y}\right)  }_{\substack{=p_{\lambda}\left(  \mathbf{y}\right)
\\\text{(since }\operatorname*{type}\sigma=\lambda\\\text{(because the
definition of }\operatorname*{type}\sigma\\\text{yields }\operatorname*{type}%
\sigma=\left(  \text{cycle type of }\sigma\right)  =\lambda\\\text{(since
}\sigma\text{ has cycle type }\lambda\text{)))}}}\nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}\ \ \underbrace{\sum\limits_{\substack{\sigma
\in\Symm_{n};\\\sigma\text{ has cycle type }\lambda}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  }_{\substack{=\mathbf{GR}_{\lambda
}\left(  \xx\right)  \\\text{(by (\ref{pf.prop.lyndon.grfun.cauchy2.4}%
))}}}p_{\lambda}\left(  \mathbf{y}\right) \nonumber\\
&  =\sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}\mathbf{GR}_{\lambda}\left(  \xx\right)
p_{\lambda}\left(  \mathbf{y}\right)  . \label{pf.prop.lyndon.grfun.cauchy2.5}%
\end{align}


Now, forget that we fixed $n$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.cauchy2.5}) for each $n\in \NN $. Now,
\begin{align*}
\underbrace{\sum_{\lambda\in \Par}}_{=\sum_{n\in \NN %
}\ \ \sum\limits_{\substack{\lambda\in\Par;\\\left\vert
\lambda\right\vert =n}}}\mathbf{GR}_{\lambda}\left(  \xx\right)
p_{\lambda}\left(  \mathbf{y}\right)   &  =\sum_{n\in \NN }%
\ \ \underbrace{\sum\limits_{\substack{\lambda\in\Par%
;\\\left\vert \lambda\right\vert =n}}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  }_{\substack{=\sum
_{\sigma\in\Symm_{n}}L_{\gamma\left(  \sigma\right)  }\left(
\xx\right)  p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)
\\\text{(by (\ref{pf.prop.lyndon.grfun.cauchy2.5}))}}}\\
&  =\underbrace{\sum_{n\in \NN }\ \ \sum_{\sigma\in\Symm_{n}}%
}_{\substack{=\sum\limits_{\sigma\in\bigsqcup_{n\in \NN }\mathfrak{S}%
_{n}}=\sum\limits_{\sigma\in\mathfrak{S}}\\\text{(since }\bigsqcup
_{n\in \NN }\Symm_{n}=\mathfrak{S}\text{)}}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  p_{\operatorname*{type}\sigma
}\left(  \mathbf{y}\right)  =\sum_{\sigma\in\mathfrak{S}}L_{\gamma\left(
\sigma\right)  }\left(  \xx\right)  p_{\operatorname*{type}\sigma
}\left(  \mathbf{y}\right)  .
\end{align*}
Combining this with (\ref{pf.prop.lyndon.grfun.cauchy2.1}), we obtain%
\[
\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\lambda
\in \Par}p_{\lambda}\left(  \xx\right)  \mathbf{GR}%
_{\lambda}\left(  \mathbf{y}\right)  =\sum_{\sigma\in\mathfrak{S}}%
L_{\gamma\left(  \sigma\right)  }\left(  \xx\right)
p_{\operatorname*{type}\sigma}\left(  \mathbf{y}\right)  .
\]
This proves Proposition \ref{prop.lyndon.grfun.cauchy2}.
\end{proof}

\end{verlong}

We shall prove Proposition \ref{prop.lyndon.grfun.symm} in two different ways.
Both of these proofs will rely on the following simple lemma:

\begin{lemma}
\label{lem.sol.lyndon.grfun.bdd}Let $\lambda$ be a partition. Then, the power
series $\mathbf{GR}_{\lambda}$ is homogeneous of degree $\left\vert
\lambda\right\vert $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.bdd}.]The definition of
$\mathbf{GR}_{\lambda}$ yields%
\begin{equation}
\mathbf{GR}_{\lambda}=\sum\limits_{\substack{w\in\mathfrak{A}^{\ast
};\\\operatorname*{CFLtype}w=\lambda}}\xx_{w}.
\label{pf.lem.sol.lyndon.grfun.bdd.1}%
\end{equation}
Set $n=\left\vert \lambda\right\vert $.

Let $w\in\mathfrak{A}^{\ast}$ be such that $\operatorname*{CFLtype}w=\lambda$.
Then, Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(a) yields $\ell\left(
w\right)  =\left\vert \lambda\right\vert =n$, so that $w\in\mathfrak{A}^{n}$
and thus $w=\left(  w_{1},w_{2},\ldots,w_{n}\right)  $. Hence, the definition
of $\xx_{w}$ yields $\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}%
$. This shows that $\xx_{w}$ is a monomial of degree $n$.

Forget that we fixed $w$. We thus have showed that $\xx_{w}$ is a
monomial of degree $n$ whenever $w\in\mathfrak{A}^{\ast}$ satisfies
$\operatorname*{CFLtype}w=\lambda$. Hence, $\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\lambda}}\xx_{w}$ is a
sum of monomials of degree $n$. In other words, $\mathbf{GR}_{\lambda}$ is a
sum of monomials of degree $n$ (because of
(\ref{pf.lem.sol.lyndon.grfun.bdd.1})). Thus, $\mathbf{GR}_{\lambda}$ is a
homogeneous formal power series of degree $n$. In other words, $\mathbf{GR}%
_{\lambda}$ is a homogeneous formal power series of degree $\left\vert
\lambda\right\vert $ (since $n=\left\vert \lambda\right\vert $). This proves
Lemma \ref{lem.sol.lyndon.grfun.bdd}.
\end{proof}

We can now step to the first proof of Proposition \ref{prop.lyndon.grfun.symm}:

\begin{proof}
[First proof of Proposition \ref{prop.lyndon.grfun.symm} (sketched).]Lemma
\ref{lem.sol.lyndon.grfun.bdd} shows that the power series $\mathbf{GR}%
_{\lambda}$ is homogeneous of degree $\left\vert \lambda\right\vert $. Hence,
this power series $\mathbf{GR}_{\lambda}$ is of bounded degree (since any
homogeneous power series is of bounded degree). In other words, $\mathbf{GR}%
_{\lambda}\in R\left(  \xx\right)  $ (since $R\left(  \xx%
\right)  $ is the set of all formal power series of bounded degree).

Recall the finitary symmetric group $\Symm_{\left(  \infty\right)  }$
defined in Section \ref{subsect.Sym-section.def}; it acts on the ring
$R\left(  \xx\right)  $. The ring $\Lambda$ is the invariant ring
$\left\{  f\in R\left(  \xx\right)  :\sigma\left(  f\right)  =f\text{
for all }\sigma\in\Symm_{\left(  \infty\right)  }\right\}  $ of this action.

We shall now prove that $\sigma\left(  f\right)  =f$ for each $\sigma
\in\Symm_{\left(  \infty\right)  }$.

\begin{vershort}
Indeed, let $\sigma\in\Symm_{\left(  \infty\right)  }$. Thus, $\sigma$
is a permutation of the set $\left\{  1,2,3,\ldots\right\}  $. In other words,
$\sigma$ is a permutation of the set $\mathfrak{A}$ (since $\mathfrak{A}%
=\left\{  1,2,3,\ldots\right\}  $). In other words, $\sigma$ is a bijection
$\mathfrak{A}\rightarrow\mathfrak{A}$.
\end{vershort}

\begin{verlong}
Indeed, let $\sigma\in\Symm_{\left(  \infty\right)  }$. Thus, $\sigma$
is a permutation of the set $\left\{  1,2,3,\ldots\right\}  $ which leaves all
but finitely many elements invariant (since $\Symm_{\left(
\infty\right)  }$ was defined to be the group of all such permutations). In
particular, $\sigma$ is a permutation of the set $\left\{  1,2,3,\ldots
\right\}  $. In other words, $\sigma$ is a permutation of the set
$\mathfrak{A}$ (since $\mathfrak{A}=\left\{  1,2,3,\ldots\right\}  $). In
other words, $\sigma$ is a bijection $\mathfrak{A}\rightarrow\mathfrak{A}$.
\end{verlong}

\begin{vershort}
For any word $w\in\mathfrak{A}^{\ast}$, we define a new word $\sigma
_{\operatorname*{word}}\left(  w\right)  $ to be the result of applying
$\sigma$ to each letter of $w$. That is, if $w=\left(  w_{1},w_{2}%
,\ldots,w_{n}\right)  $, then $\sigma_{\operatorname*{word}}\left(  w\right)
=\left(  \sigma\left(  w_{1}\right)  ,\sigma\left(  w_{2}\right)
,\ldots,\sigma\left(  w_{n}\right)  \right)  $. This defines a map
$\mathfrak{A}^{\ast}\rightarrow\mathfrak{A}^{\ast},\ w\mapsto\sigma
_{\operatorname*{word}}\left(  w\right)  $. This map is a bijection (since
$\sigma$ is a bijection). We shall denote this bijection by $\sigma
_{\operatorname*{word}}$.

For any necklace $N$, we define a new necklace $\sigma_{\operatorname*{neck}%
}\left(  N\right)  $ to be the result of applying this bijection
$\sigma_{\operatorname*{word}}$ to every word in $N$. That is, $\sigma
_{\operatorname*{neck}}\left(  N\right)  =\left\{  \sigma
_{\operatorname*{word}}\left(  w\right)  \ \mid\ w\in N\right\}  $. It is
straightforward to see that this set $\sigma_{\operatorname*{neck}}\left(
N\right)  $ is indeed a necklace\footnote{Indeed, the bijection $\sigma
_{\operatorname*{word}}$ commutes with the actions of $C$ on $\mathfrak{A}%
^{1},\mathfrak{A}^{2},\mathfrak{A}^{3},\ldots$.}, and furthermore is aperiodic
if and only if $N$ is aperiodic. Thus, we obtain a map $\mathfrak{N}%
^{\mathfrak{a}}\rightarrow\mathfrak{N}^{\mathfrak{a}},\ N\mapsto
\sigma_{\operatorname*{neck}}\left(  N\right)  $ (since $\mathfrak{N}%
^{\mathfrak{a}}$ denotes the set of all aperiodic necklaces). This map is a
bijection (since $\sigma_{\operatorname*{word}}$ is a bijection). We shall
denote this bijection by $\sigma_{\operatorname*{neck}}$.

Finally, for any finite multiset $M$ of aperiodic necklaces, we define a new
finite multiset $\sigma_{\operatorname*{mul}}\left(  M\right)  $ of aperiodic
necklaces to be the result of applying this bijection $\sigma
_{\operatorname*{neck}}$ to each necklace in $M$. That is, $\sigma
_{\operatorname*{mul}}\left(  M\right)  =\left(  \sigma_{\operatorname*{neck}%
}\right)  _{\ast}M$, using the notation of Definition
\ref{def.sol.lyndon.gr.bij.fstarMultiset}. This defines a map $\mathfrak{MN}%
^{\mathfrak{a}}\rightarrow\mathfrak{MN}^{\mathfrak{a}},\ M\mapsto
\sigma_{\operatorname*{mul}}\left(  M\right)  $ (since $\mathfrak{MN}%
^{\mathfrak{a}}$ denotes the set of all finite multisets of aperiodic
necklaces). Again, this map is a bijection (since $\sigma
_{\operatorname*{neck}}$ is a bijection). We shall denote this bijection by
$\sigma_{\operatorname*{mul}}$.

It is easy to describe directly what this bijection $\sigma
_{\operatorname*{mul}}$ does to any given multiset: If $M\in\mathfrak{MN}%
^{\mathfrak{a}}$, then $\sigma_{\operatorname*{mul}}\left(  M\right)  $ is
obtained from $M$ by applying $\sigma$ to each letter of each word in each
necklace in $M$. Hence, it is straightforward to see that each $M\in
\mathfrak{MN}^{\mathfrak{a}}$ satisfies%
\begin{equation}
\xx_{\sigma_{\operatorname*{mul}}\left(  M\right)  }=\sigma\left(
\xx_{M}\right)  \label{pf.prop.lyndon.grfun.symm.1st.short.sigmon}%
\end{equation}
and%
\begin{equation}
\operatorname*{type}\left(  \sigma_{\operatorname*{mul}}\left(  M\right)
\right)  =\operatorname*{type}M.
\label{pf.prop.lyndon.grfun.symm.1st.short.sigtype}%
\end{equation}


But Proposition \ref{prop.lyndon.grfun.alt1} yields%
\begin{align*}
\mathbf{GR}_{\lambda}  &  =\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\xx_{M}%
=\underbrace{\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}%
};\\\operatorname*{type}\left(  \sigma_{\operatorname*{mul}}\left(  M\right)
\right)  =\lambda}}}_{\substack{=\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\\\text{(since each }%
M\in\mathfrak{MN}^{\mathfrak{a}}\text{ satisfies
(\ref{pf.prop.lyndon.grfun.symm.1st.short.sigtype}))}}}\underbrace{\xx%
_{\sigma_{\operatorname*{mul}}\left(  M\right)  }}_{\substack{=\sigma\left(
\xx_{M}\right)  \\\text{(by
(\ref{pf.prop.lyndon.grfun.symm.1st.short.sigmon}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\sigma_{\operatorname*{mul}}\left(  M\right)
\text{ for }M\text{ in the sum,}\\
\text{since the map }\sigma_{\operatorname*{mul}}:\mathfrak{MN}^{\mathfrak{a}%
}\rightarrow\mathfrak{MN}^{\mathfrak{a}}\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}}%
;\\\operatorname*{type}M=\lambda}}\sigma\left(  \xx_{M}\right)
=\sigma\left(  \underbrace{\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\xx_{M}}%
_{\substack{=\mathbf{GR}_{\lambda}\\\text{(by Proposition
\ref{prop.lyndon.grfun.alt1})}}}\right)  =\sigma\left(  \mathbf{GR}_{\lambda
}\right)  .
\end{align*}
In other words, $\sigma\left(  \mathbf{GR}_{\lambda}\right)  =\mathbf{GR}%
_{\lambda}$.
\end{vershort}

\begin{verlong}
The bijection $\sigma:\mathfrak{A}\rightarrow\mathfrak{A}$ induces a bijection%
\begin{align*}
\sigma^{\ast}:\mathfrak{A}^{\ast}  &  \rightarrow\mathfrak{A}^{\ast},\\
\left(  w_{1},w_{2},\ldots,w_{n}\right)   &  \mapsto\left(  \sigma\left(
w_{1}\right)  ,\sigma\left(  w_{2}\right)  ,\ldots,\sigma\left(  w_{n}\right)
\right)  .
\end{align*}
This bijection $\sigma^{\ast}$ can be applied to subsets of $\mathfrak{A}%
^{\ast}$ as well (by the usual rule: $\sigma^{\ast}\left(  I\right)  =\left\{
\sigma^{\ast}\left(  w\right)  \ \mid\ w\in I\right\}  $ for each
$I\subset\mathfrak{A}^{\ast}$). It is easy to see that this bijection
$\sigma^{\ast}$ sends each necklace to a necklace and vice versa\footnote{More
precisely: A subset $N\subset\mathfrak{A}^{\ast}$ is a necklace if and only if
its image $\sigma^{\ast}\left(  N\right)  $ is a necklace.}. Moreover, it
preserves the length of a word (i.e., for each word $w$, we have $\ell\left(
\sigma^{\ast}\left(  w\right)  \right)  =\ell\left(  w\right)  $), and the
period of a necklace (since the period of a necklace is simply the size of
this necklace, but any bijection preserves sizes of subsets). Thus, it sends
aperiodic necklaces to aperiodic necklaces and vice versa\footnote{More
precisely: A subset $N\subset\mathfrak{A}^{\ast}$ is an aperiodic necklace if
and only if its image $\sigma^{\ast}\left(  N\right)  $ is an aperiodic
necklace.}. Hence, it induces a bijection%
\begin{align*}
\widetilde{\sigma^{\ast}}:\mathfrak{N}^{\mathfrak{a}}  &  \rightarrow
\mathfrak{N}^{\mathfrak{a}},\\
N  &  \mapsto\sigma^{\ast}\left(  N\right)
\end{align*}
(since $\mathfrak{N}^{\mathfrak{a}}$ is the set of all aperiodic necklaces).
Thus, according to Definition \ref{def.sol.lyndon.gr.bij.fstarMultiset}, we
can define a finite multiset $\left(  \widetilde{\sigma^{\ast}}\right)
_{\ast}M$ of elements of $\mathfrak{N}^{\mathfrak{a}}$ whenever $M$ is a
finite multiset of elements of $\mathfrak{N}^{\mathfrak{a}}$. Explicitly, this
finite multiset $\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast}M$ is given
by%
\[
\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast}M=\left\{  \widetilde{\sigma
^{\ast}}\left(  N_{1}\right)  ,\widetilde{\sigma^{\ast}}\left(  N_{2}\right)
,\ldots,\widetilde{\sigma^{\ast}}\left(  N_{k}\right)  \right\}
_{\operatorname*{multiset}},\ \ \ \ \ \ \ \ \ \ \text{where }M=\left\{
N_{1},N_{2},\ldots,N_{k}\right\}  _{\operatorname*{multiset}}.
\]
This construction is invertible (since $\widetilde{\sigma^{\ast}}$ is a
bijection), i.e., the map
\begin{align*}
\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast}:\mathfrak{MN}^{\mathfrak{a}%
}  &  \rightarrow\mathfrak{MN}^{\mathfrak{a}},\\
M  &  \mapsto\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast}M
\end{align*}
is a bijection\footnote{Indeed, $\mathfrak{MN}^{\mathfrak{a}}$ is the set of
all finite multisets of elements of $\mathfrak{N}^{\mathfrak{a}}$.}. Roughly
speaking, all this map $\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast}$
does is applying $\sigma$ to each letter of each word in each necklace in the
multiset it is being applied to. Hence, it is straightforward to see that each
$M\in\mathfrak{MN}^{\mathfrak{a}}$ satisfies%
\begin{equation}
\xx_{\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast}\left(  M\right)
}=\sigma\left(  \xx_{M}\right)
\label{pf.prop.lyndon.grfun.symm.1st.sigmon}%
\end{equation}
and%
\begin{equation}
\operatorname*{type}\left(  \left(  \widetilde{\sigma^{\ast}}\right)  _{\ast
}\left(  M\right)  \right)  =\operatorname*{type}M.
\label{pf.prop.lyndon.grfun.symm.1st.sigtype}%
\end{equation}


But Proposition \ref{prop.lyndon.grfun.alt1} yields%
\begin{align}
\mathbf{GR}_{\lambda}  &  =\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\xx_{M}%
\label{pf.prop.lyndon.grfun.symm.1st.alt1}\\
&  =\underbrace{\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}%
};\\\operatorname*{type}\left(  \left(  \widetilde{\sigma^{\ast}}\right)
_{\ast}\left(  M\right)  \right)  =\lambda}}}_{\substack{=\sum
\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}};\\\operatorname*{type}%
M=\lambda}}\\\text{(since each }M\in\mathfrak{MN}^{\mathfrak{a}}%
\\\text{satisfies }\operatorname*{type}\left(  \left(  \widetilde{\sigma
^{\ast}}\right)  _{\ast}\left(  M\right)  \right)  =\operatorname*{type}%
M\\\text{(by (\ref{pf.prop.lyndon.grfun.symm.1st.sigtype})))}}%
}\underbrace{\xx_{\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast
}\left(  M\right)  }}_{\substack{=\sigma\left(  \xx_{M}\right)
\\\text{(by (\ref{pf.prop.lyndon.grfun.symm.1st.sigmon}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  \widetilde{\sigma^{\ast}}\right)
_{\ast}\left(  M\right)  \text{ for }M\text{ in the sum,}\\
\text{since the map }\left(  \widetilde{\sigma^{\ast}}\right)  _{\ast
}:\mathfrak{MN}^{\mathfrak{a}}\rightarrow\mathfrak{MN}^{\mathfrak{a}}\text{ is
a bijection}%
\end{array}
\right) \nonumber\\
&  =\sum\limits_{\substack{M\in\mathfrak{MN}^{\mathfrak{a}}%
;\\\operatorname*{type}M=\lambda}}\sigma\left(  \xx_{M}\right)
=\sigma\left(  \underbrace{\sum\limits_{\substack{M\in\mathfrak{MN}%
^{\mathfrak{a}};\\\operatorname*{type}M=\lambda}}\xx_{M}}%
_{\substack{=\mathbf{GR}_{\lambda}\\\text{(by
(\ref{pf.prop.lyndon.grfun.symm.1st.alt1}))}}}\right)  =\sigma\left(
\mathbf{GR}_{\lambda}\right)  .\nonumber
\end{align}
In other words, $\sigma\left(  \mathbf{GR}_{\lambda}\right)  =\mathbf{GR}%
_{\lambda}$.
\end{verlong}

Now, forget that we fixed $\sigma$. We thus have showed that $\sigma\left(
\mathbf{GR}_{\lambda}\right)  =\mathbf{GR}_{\lambda}$ for all $\sigma
\in\Symm_{\left(  \infty\right)  }$. Hence, $\mathbf{GR}_{\lambda}$ is
an $f\in R\left(  \xx\right)  $ that satisfies $\sigma\left(  f\right)
=f$ for all $\sigma\in\Symm_{\left(  \infty\right)  }$ (since we
already know that $\mathbf{GR}_{\lambda}\in R\left(  \xx\right)  $). In
other words, $\mathbf{GR}_{\lambda}\in\left\{  f\in R\left(  \xx%
\right)  :\sigma\left(  f\right)  =f\text{ for all }\sigma\in\mathfrak{S}%
_{\left(  \infty\right)  }\right\}  $. In other words, $\mathbf{GR}_{\lambda
}\in\Lambda$ (since $\Lambda=\left\{  f\in R\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}  $). This proves Proposition
\ref{prop.lyndon.grfun.symm}.
\end{proof}

Our next proof of Proposition \ref{prop.lyndon.grfun.symm} will rely on two lemmas:

\begin{lemma}
\label{lem.sol.lyndon.grfun.bdd-p}Let $\lambda$ be a partition. Then, the
power series $p_{\lambda}$ is homogeneous of degree $\left\vert \lambda
\right\vert $.
\end{lemma}

\begin{vershort}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.bdd-p}.]Let $\ell=\ell\left(
\lambda\right)  $; thus, $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell}\right)  $ and $p_{\lambda}=p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ (by the definition of $p_{\lambda}%
$). From $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell
}\right)  $, we obtain $\left\vert \lambda\right\vert =\lambda_{1}+\lambda
_{2}+\cdots+\lambda_{\ell}$.

But the power-sum symmetric functions $p_{\lambda_{1}},p_{\lambda_{2}}%
,\ldots,p_{\lambda_{\ell}}$ are homogeneous of degrees $\lambda_{1}%
,\lambda_{2},\ldots,\lambda_{\ell}$, respectively (since each power-sum
symmetric function $p_{n}$ is homogeneous of degree $n$). Hence, their product
$p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ is homogeneous of
degree $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}$ (since $\Lambda$ is a
graded $\kk$-algebra). In view of $p_{\lambda}=p_{\lambda_{1}%
}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ and $\left\vert \lambda\right\vert
=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{\ell}$, this rewrites as follows:
\[
p_{\lambda}\text{ is homogeneous of degree }\left\vert \lambda\right\vert .
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.bdd-p}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.bdd-p}.]Let $\ell=\ell\left(
\lambda\right)  $; thus, $\lambda=\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{\ell}\right)  $ and $p_{\lambda}=p_{\lambda_{1}}%
p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$ (by the definition of $p_{\lambda}%
$). From $\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell
}\right)  $, we obtain $\left\vert \lambda\right\vert =\lambda_{1}+\lambda
_{2}+\cdots+\lambda_{\ell}$.

Fix $i\in\left\{  1,2,\ldots,\ell\right\}  $. Thus, $i\leq\ell=\ell\left(
\lambda\right)  $. Hence, $\lambda_{i}$ is a positive integer. But recall that
the power-sum symmetric function $p_{n}$ is homogeneous of degree $n$ for any
positive integer $n$. Applying this to $n=\lambda_{i}$, we conclude that the
power-sum symmetric function $p_{\lambda_{i}}$ is homogeneous of degree
$\lambda_{i}$.

Forget that we fixed $i$. We thus have shown that the power-sum symmetric
function $p_{\lambda_{i}}$ is homogeneous of degree $\lambda_{i}$ for each
$i\in\left\{  1,2,\ldots,\ell\right\}  $. In other words, the power-sum
symmetric functions $p_{\lambda_{1}},p_{\lambda_{2}},\ldots,p_{\lambda_{\ell}%
}$ are homogeneous of degrees $\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}$,
respectively. Hence, their product $p_{\lambda_{1}}p_{\lambda_{2}}\cdots
p_{\lambda_{\ell}}$ is homogeneous of degree $\lambda_{1}+\lambda_{2}%
+\cdots+\lambda_{\ell}$ (since $\Lambda$ is a graded $\kk$-algebra). In
view of $p_{\lambda}=p_{\lambda_{1}}p_{\lambda_{2}}\cdots p_{\lambda_{\ell}}$
and $\left\vert \lambda\right\vert =\lambda_{1}+\lambda_{2}+\cdots
+\lambda_{\ell}$, this rewrites as follows:
\[
p_{\lambda}\text{ is homogeneous of degree }\left\vert \lambda\right\vert .
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.bdd-p}.
\end{proof}

\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.grfun.lin-ind-lamcrit}Let $A$ be a commutative
$\kk$-algebra. Let $\left(  a_{\lambda}\right)  _{\lambda
\in \Par}\in\left(  \kk \left[  \left[  \xx\right]
\right]  \right)  ^{ \Par}$ and $\left(  b_{\lambda}\right)
_{\lambda\in \Par}\in\left(  \kk \left[  \left[
\xx\right]  \right]  \right)  ^{ \Par}$ be two families
of power series in $\kk \left[  \left[  \xx\right]  \right]  $,
and let $\left(  u_{\lambda}\right)  _{\lambda\in \Par}\in
A^{ \Par}$ and $\left(  v_{\lambda}\right)  _{\lambda
\in \Par}\in A^{ \Par}$ be two families of
elements of $A$. Assume that the following four conditions are satisfied:

\begin{statement}
\textit{Assumption 1:} If $\lambda\in \Par$, then both power
series $a_{\lambda}$ and $b_{\lambda}$ are homogeneous of degree $\left\vert
\lambda\right\vert $.
\end{statement}

\begin{statement}
\textit{Assumption 2:} The family $\left(  u_{\lambda}\right)  _{\lambda
\in \Par}$ is $\kk$-linearly independent.
\end{statement}

\begin{statement}
\textit{Assumption 3:} We have\footnote{The products $u_{\lambda}a_{\lambda}$
and $v_{\lambda}b_{\lambda}$ in the following equation are taken in the power
series ring $A\left[  \left[  \xx\right]  \right]  =A\left[  \left[
x_{1},x_{2},x_{3},\ldots\right]  \right]  $. (Indeed, $u_{\lambda}$ and
$v_{\lambda}$ belong to $A$ and thus to $A\left[  \left[  \xx\right]
\right]  $ as well, whereas $a_{\lambda}$ and $b_{\lambda}$ are elements of
$\kk \left[  \left[  \xx\right]  \right]  $. Hence, the products
$u_{\lambda}a_{\lambda}$ and $v_{\lambda}b_{\lambda}$ make sense because
$A\left[  \left[  \xx\right]  \right]  $ is canonically a
$\kk \left[  \left[  \xx\right]  \right]  $-module.)}%
\begin{equation}
\sum_{\lambda\in \Par}u_{\lambda}a_{\lambda}=\sum_{\lambda
\in \Par}v_{\lambda}b_{\lambda}
\label{eq.lem.sol.lyndon.grfun.lin-ind-lamcrit.a3.eq}%
\end{equation}
\footnote{Both sums $\sum_{\lambda\in \Par}u_{\lambda}%
a_{\lambda}$ and $\sum_{\lambda\in \Par}v_{\lambda}b_{\lambda}$
converge (with respect to the standard topology on $A\left[  \left[
\xx\right]  \right]  $). Here is why:
\par
Fix $n\in \NN $. For each $\lambda\in \Par$, the power
series $a_{\lambda}\in \kk \left[  \left[  \xx\right]  \right]  $
is homogeneous of degree $\left\vert \lambda\right\vert $ (by Assumption 1),
and thus the power series $u_{\lambda}a_{\lambda}\in A\left[  \left[
\xx\right]  \right]  $ is also homogeneous of degree $\left\vert
\lambda\right\vert $ (since the factor $u_{\lambda}$ belongs to $A$ and
therefore does not affect the degree). Hence, for each $\lambda\in
 \Par$, the power series $u_{\lambda}a_{\lambda}$ contains no
degree-$n$ monomials unless $\left\vert \lambda\right\vert =n$. Thus, there
are only finitely many $\lambda\in \Par$ such that the power
series $u_{\lambda}a_{\lambda}$ contains degree-$n$ monomials (because there
are only finitely many $\lambda\in \Par$ such that $\left\vert
\lambda\right\vert =n$).
\par
Forget that we fixed $n$. We thus have shown that, for each $n\in \NN $,
there are only finitely many $\lambda\in \Par$ such that the
power series $u_{\lambda}a_{\lambda}$ contains degree-$n$ monomials. In other
words, for each $n\in \NN $, there are only finitely many addends in the
sum $\sum_{\lambda\in \Par}u_{\lambda}a_{\lambda}$ that contain
degree-$n$ monomials. Therefore, this sum $\sum_{\lambda\in\Par}
u_{\lambda}a_{\lambda}$ converges. Similarly, the sum $\sum_{\lambda
\in \Par}v_{\lambda}b_{\lambda}$ converges. Qed.} in the power
series ring $A\left[  \left[  \xx\right]  \right]  =A\left[  \left[
x_{1},x_{2},x_{3},\ldots\right]  \right]  $.
\end{statement}

\begin{statement}
\textit{Assumption 4:} We have $b_{\lambda}\in\Lambda$ for each $\lambda
\in \Par$.
\end{statement}

Then, $a_{\lambda}\in\Lambda$ for each $\lambda\in \Par$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.lin-ind-lamcrit}.]In this proof, the
word ``monomial'' always means a monomial in
the indeterminates $x_{1},x_{2},x_{3},\ldots$. Any formal power series in
$\kk \left[  \left[  \xx\right]  \right]  $ is an infinite
$\kk$-linear combination of monomials, whereas any formal power series
in $A\left[  \left[  \xx\right]  \right]  $ is an infinite $A$-linear
combination of monomials.

Each monomial $\mathfrak{m}=x_{1}^{r_{1}}x_{2}^{r_{2}}x_{3}^{r_{3}}\cdots$ has
a well-defined degree $\deg\mathfrak{m}$, namely $\deg\mathfrak{m}=r_{1}%
+r_{2}+r_{3}+\cdots$. Recall that a formal power series $f\in A\left[  \left[
\xx\right]  \right]  $ is said to be \emph{homogeneous of degree $n$}
(for some $n\in \NN $) if it is an infinite $A$-linear combination of
monomials that have degree $n$. Thus, in particular, an element of $A$ (when
regarded as a formal power series in $A\left[  \left[  \xx\right]
\right]  $) is always homogeneous of degree $0$.

\begin{vershort}
Consider the (unique) $\kk$-algebra homomorphism $\iota:\kk
\rightarrow A$ (which comes from the fact that $A$ is a $\kk$-algebra).
This homomorphism $\iota: \kk \rightarrow A$ is
injective\footnote{\textit{Proof.} Let $c\in\operatorname*{Ker}\iota$. Thus,
$c \in \kk$ and $\iota\left(  c\right)  =0$. The definition of $\iota$
yields $\iota\left(  c\right)  =c\cdot1_{A}$. Hence, $c\cdot1_{A}=\iota\left(
c\right)  =0$.
\par
Consider the empty partition $\varnothing\in \Par$. We have
$u_{\varnothing}\in A$ and $c\underbrace{u_{\varnothing}}_{=1_{A}\cdot
u_{\varnothing}}=\underbrace{c\cdot1_{A}}_{=0}\cdot u_{\varnothing}=0$.
\par
The empty partition $\varnothing$ satisfies $\left\{  \varnothing\right\}
\subset \Par$. Hence, the family $\left(  u_{\lambda}\right)
_{\lambda\in\left\{  \varnothing\right\}  }$ (which consists of the single
element $u_{\varnothing}$) is a subfamily of the family $\left(  u_{\lambda
}\right)  _{\lambda\in \Par}$. Thus, the former family is
$\kk$-linearly independent (since Assumption 2 says that the latter
family is $\kk$-linearly independent). In other words, if $\left(
j_{\lambda}\right)  _{\lambda\in\left\{  \varnothing\right\}  }\in
\kk^{\left\{  \varnothing\right\}  }$ is a family of elements of
$\kk$ such that $\sum_{\lambda\in\left\{  \varnothing\right\}
}j_{\lambda}u_{\lambda}=0$, then
\[
\left(  j_{\lambda}=0\text{ for each }\lambda\in\left\{  \varnothing\right\}
\right)  .
\]
We can apply this to $\left(  j_{\lambda}\right)  _{\lambda\in\left\{
\varnothing\right\}  }=\left(  c\right)  _{\lambda\in\left\{  \varnothing
\right\}  }$ (since $\sum_{\lambda\in\left\{  \varnothing\right\}
}cu_{\lambda}=cu_{\varnothing}=0$), and thus conclude that%
\[
\left(  c=0\text{ for each }\lambda\in\left\{  \varnothing\right\}  \right)
.
\]
Applying this to $\lambda=\varnothing$, we obtain $c=0$ (since $\varnothing
\in\left\{  \varnothing\right\}  $).
\par
Now, forget that we fixed $c$. We thus have shown that $c=0$ for each
$c\in\operatorname*{Ker}\iota$. In other words, $\operatorname*{Ker}\iota=0$.
Thus, $\iota$ is injective (since $\iota$ is a ring homomorphism). Qed.}.
Hence, the canonical $\kk \left[  \left[  \xx\right]  \right]
$-algebra homomorphism $\iota\left[  \left[  \xx\right]  \right]
:\kk \left[  \left[  \xx\right]  \right]  \rightarrow A\left[
\left[  \xx\right]  \right]  $ induced by it (which simply applies
$\iota$ to each coefficient of a power series) must also be injective. We
shall thus identify $\kk \left[  \left[  \xx\right]  \right]  $
with a subring of $A\left[  \left[  \xx\right]  \right]  $ (via the
latter $\kk \left[  \left[  \xx\right]  \right]  $-algebra
homomorphism $\iota\left[  \left[  \xx\right]  \right]  $).
\end{vershort}

\begin{verlong}
Consider the (unique) $\kk$-algebra homomorphism $\iota:\kk
\rightarrow A$ (which comes from the fact that $A$ is a $\kk$-algebra).
This homomorphism $\iota: \kk \rightarrow A$ is
injective\footnote{\textit{Proof.} Let $c\in\operatorname*{Ker}\iota$. Thus,
$c \in \operatorname*{Ker}\iota \subset \kk$. The definition of $\iota$
yields $\iota\left(  c\right)  =c\cdot1_{A}$. Hence, $c\cdot1_{A}=\iota\left(
c\right)  =0$ (since $c\in\operatorname*{Ker}\iota$).
\par
Consider the empty partition $\varnothing\in \Par$. We have
$u_{\varnothing}\in A$ (since $\left(  u_{\lambda}\right)  _{\lambda
\in \Par}\in A^{ \Par}$), thus $u_{\varnothing
}=1_{A}\cdot u_{\varnothing}$. Hence, $c\underbrace{u_{\varnothing}}%
_{=1_{A}\cdot u_{\varnothing}}=\underbrace{c\cdot1_{A}}_{=0}\cdot
u_{\varnothing}=0$.
\par
The empty partition $\varnothing$ satisfies $\varnothing\in\Par$
and thus $\left\{  \varnothing\right\}  \subset \Par$. Hence,
the family $\left(  u_{\lambda}\right)  _{\lambda\in\left\{  \varnothing
\right\}  }$ (which consists of the single element $u_{\varnothing}$) is a
subfamily of the family $\left(  u_{\lambda}\right)  _{\lambda\in
 \Par}$.
\par
The family $\left(  u_{\lambda}\right)  _{\lambda\in \Par}$ is
$\kk$-linearly independent (by Assumption 2). But any subfamily of a
$\kk$-linearly independent family must itself be $\kk$-linearly
independent. Hence, the family $\left(  u_{\lambda}\right)  _{\lambda
\in\left\{  \varnothing\right\}  }$ is $\kk$-linearly independent
(since it is a subfamily of the $\kk$-linearly independent family
$\left(  u_{\lambda}\right)  _{\lambda\in \Par}$). In other
words, if $\left(  j_{\lambda}\right)  _{\lambda\in\left\{  \varnothing
\right\}  }\in \kk^{\left\{  \varnothing\right\}  }$ is a family of
elements of $\kk$ such that
\[
\sum_{\lambda\in\left\{  \varnothing\right\}  }j_{\lambda}u_{\lambda}=0,
\]
then%
\[
\left(  j_{\lambda}=0\text{ for each }\lambda\in\left\{  \varnothing\right\}
\right)  .
\]
We can apply this to $\left(  j_{\lambda}\right)  _{\lambda\in\left\{
\varnothing\right\}  }=\left(  c\right)  _{\lambda\in\left\{  \varnothing
\right\}  }$ (since $\sum_{\lambda\in\left\{  \varnothing\right\}
}cu_{\lambda}=cu_{\varnothing}=0$), and thus conclude that%
\[
\left(  c=0\text{ for each }\lambda\in\left\{  \varnothing\right\}  \right)
.
\]
Applying this to $\lambda=\varnothing$, we obtain $c=0$ (since $\varnothing
\in\left\{  \varnothing\right\}  $).
\par
Now, forget that we fixed $c$. We thus have shown that $c=0$ for each
$c\in\operatorname*{Ker}\iota$. In other words, $\operatorname*{Ker}\iota=0$.
Thus, $\iota$ is injective (since $\iota$ is a ring homomorphism). Qed.}.
Hence, the canonical $\kk \left[  \left[  \xx\right]  \right]
$-algebra homomorphism $\iota\left[  \left[  \xx\right]  \right]
: \kk \left[  \left[  \xx\right]  \right]  \rightarrow A\left[
\left[  \xx\right]  \right]  $ induced by it (which simply applies
$\iota$ to each coefficient of a power series) must also be injective. We
shall thus identify $\kk \left[  \left[  \xx\right]  \right]  $
with a subring of $A\left[  \left[  \xx\right]  \right]  $ (via the
latter $\kk \left[  \left[  \xx\right]  \right]  $-algebra
homomorphism $\iota\left[  \left[  \xx\right]  \right]  $).
\end{verlong}

We shall now prove a few auxiliary claims:

\begin{statement}
\textit{Claim 1:} Let $n\in \NN $. Then,%
\[
\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda}%
=\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda}.
\]

\end{statement}

[\textit{Proof of Claim 1:} Let $\pi_{n}$ denote the map $A\left[  \left[
\xx\right]  \right]  \rightarrow A\left[  \left[  \xx\right]
\right]  $ that sends each formal power series $f\in A\left[  \left[
\xx\right]  \right]  $ to its $n$-th homogeneous component. Explicitly,
this map $\pi_{n}$ is given by%
\begin{align*}
\pi_{n}\left(  \sum\limits_{\mathfrak{m}\text{ is a monomial}}\alpha
_{\mathfrak{m}}\mathfrak{m}\right)   &  =\sum\limits_{\substack{\mathfrak{m}%
\text{ is a monomial;}\\\deg\mathfrak{m}=n}}\alpha_{\mathfrak{m}}%
\mathfrak{m}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }\sum\limits_{\mathfrak{m}\text{ is a
monomial}}\alpha_{\mathfrak{m}}\mathfrak{m}\in A\left[  \left[  \xx%
\right]  \right]  \text{ (with }\alpha_{m}\in A\text{).}%
\end{align*}


\begin{vershort}
\noindent This map $\pi_{n}$ is $A$-linear and continuous (with respect to the
topology on $A\left[  \left[  \xx\right]  \right]  $); thus, it
respects infinite sums. Moreover, it sends each $a_{\lambda}$ (with
$\lambda\in \Par$) to $%
\begin{cases}
a_{\lambda}, & \text{if }\left\vert \lambda\right\vert =n;\\
0, & \text{if }\left\vert \lambda\right\vert \neq n
\end{cases}
$ (since Assumption 1 shows that $a_{\lambda}$ is homogeneous of degree
$\left\vert \lambda\right\vert $), and sends each $b_{\lambda}$ to $%
\begin{cases}
b_{\lambda}, & \text{if }\left\vert \lambda\right\vert =n;\\
0, & \text{if }\left\vert \lambda\right\vert \neq n
\end{cases}
$ (for similar reasons). Thus, applying $\pi_{n}$ to both sides of
(\ref{eq.lem.sol.lyndon.grfun.lin-ind-lamcrit.a3.eq}), we obtain%
\[
\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda}%
=\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda}.
\]

\end{vershort}

\begin{verlong}
The map $\pi_{n}:A\left[  \left[  \xx\right]  \right]  \rightarrow
A\left[  \left[  \xx\right]  \right]  $ is $A$-linear and has the
following two properties:

\begin{itemize}
\item If a power series $f\in A\left[  \left[  \xx\right]  \right]  $
is homogeneous of degree $n$, then%
\begin{equation}
\pi_{n}\left(  f\right)  =f.
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.2}%
\end{equation}


\item If $q\in \NN $ satisfies $q\neq n$, and if a power series $f\in
A\left[  \left[  \xx\right]  \right]  $ is homogeneous of degree $q$,
then%
\begin{equation}
\pi_{n}\left(  f\right)  =0.
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.3}%
\end{equation}

\end{itemize}

Hence, it is easy to see the following:

\begin{itemize}
\item Every $\lambda\in \Par_{n}$ satisfies
\begin{equation}
\pi_{n}\left(  a_{\lambda}\right)  =a_{\lambda}
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.4}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.4}):} Let $\lambda
\in \Par_{n}$. Thus, $\lambda$ is a partition
satisfying $\left\vert \lambda\right\vert =n$. Now, Assumption 1 shows that
both power series $a_{\lambda}$ and $b_{\lambda}$ are homogeneous of degree
$\left\vert \lambda\right\vert $. Hence, in particular, the power series
$a_{\lambda}$ is homogeneous of degree $\left\vert \lambda\right\vert $. In
other words, the power series $a_{\lambda}$ is homogeneous of degree $n$
(since $\left\vert \lambda\right\vert =n$). Thus,
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.2}) (applied to
$f=a_{\lambda}$) yields $\pi_{n}\left(  a_{\lambda}\right)  =a_{\lambda}$.
This proves (\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.4}).}.

\item Every $\lambda \in \Par \setminus \Par_{n}$ satisfies
\begin{equation}
\pi_{n}\left(  a_{\lambda}\right)  =0
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.5}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.5}):} Let $\lambda
\in \Par \setminus \Par_{n}$. Thus,
$\lambda\in \Par$ and $\lambda \notin \Par_{n}$.
\par
From $\lambda\in \Par$, we conclude that $\lambda$ is a
partition. Hence, if we had $\left\vert \lambda\right\vert =n$, then we would
have $\lambda\in \Par_{n}$, which would contradict
$\lambda\notin \Par_{n}$. Thus, we cannot have
$\left\vert \lambda\right\vert =n$. In other words, we have $\left\vert
\lambda\right\vert \neq n$. Also, Assumption 1 shows that both power series
$a_{\lambda}$ and $b_{\lambda}$ are homogeneous of degree $\left\vert
\lambda\right\vert $. Hence, in particular, the power series $a_{\lambda}$ is
homogeneous of degree $\left\vert \lambda\right\vert $. Thus,
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.3}) (applied to
$q=\left\vert \lambda\right\vert $ and $f=a_{\lambda}$) yields $\pi_{n}\left(
a_{\lambda}\right)  =0$. This proves
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.5}).}.
\end{itemize}

Now, the map $\pi_{n}$ is $A$-linear and continuous (with respect to the
topology on $A\left[  \left[  \xx\right]  \right]  $); thus, it
respects infinite sums. Hence,
\begin{align}
\pi_{n}\left(  \sum_{\lambda\in \Par}u_{\lambda}a_{\lambda
}\right)   &  =\sum_{\lambda\in \Par}\underbrace{\pi_{n}\left(
u_{\lambda}a_{\lambda}\right)  }_{\substack{=u_{\lambda}\pi_{n}\left(
a_{\lambda}\right)  \\\text{(since }\pi_{n}\text{ is }A\text{-linear)}}%
}=\sum_{\lambda\in \Par}u_{\lambda}\pi_{n}\left(  a_{\lambda
}\right) \nonumber\\
&  =\underbrace{\sum\limits_{\substack{\lambda\in\Par;\\\lambda
\in \Par_{n}}}}_{\substack{=\sum_{\lambda
\in \Par_{n}}\\\text{(since }\Par_{n}\subset\Par\text{)}}}
u_{\lambda}\pi_{n}\left(
a_{\lambda}\right)  +\underbrace{\sum\limits_{\substack{\lambda\in\Par;
\\\lambda\notin \Par_{n}}}}_{=\sum_{\lambda
\in\Par \setminus \Par_{n}}}u_{\lambda
}\pi_{n}\left(  a_{\lambda}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }\lambda\in \Par \text{ satisfies either
}\lambda\in \Par_{n}\text{ or }\lambda\notin%
 \Par_{n}\\
\text{(but not both at the same time)}%
\end{array}
\right) \nonumber\\
&  =\sum_{\lambda\in \Par_{n}}u_{\lambda}%
\underbrace{\pi_{n}\left(  a_{\lambda}\right)  }_{\substack{=a_{\lambda
}\\\text{(by (\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.4}))}}%
}+\sum_{\lambda\in\Par\setminus\Par_{n}}u_{\lambda}
\underbrace{\pi_{n}\left(  a_{\lambda}\right)
}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.5}))}}}\nonumber\\
&  =\sum_{\lambda\in \Par_{n}} u_{\lambda}a_{\lambda
}+\underbrace{\sum_{\lambda\in\Par\setminus\Par_{n}}
u_{\lambda}0}_{=0}
=\sum_{\lambda\in\Par_{n}} u_{\lambda}a_{\lambda}.
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.8}%
\end{align}
The same argument (applied to $v_{\lambda}$ and $b_{\lambda}$ instead of
$u_{\lambda}$ and $a_{\lambda}$) yields%
\begin{equation}
\pi_{n}\left(  \sum_{\lambda\in \Par}v_{\lambda}b_{\lambda
}\right)  =\sum_{\lambda\in \Par_{n}}v_{\lambda
}b_{\lambda}. \label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.9}%
\end{equation}
Now, (\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.8}) yields%
\[
\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda}%
=\pi_{n}\left(  \underbrace{\sum_{\lambda\in \Par}u_{\lambda
}a_{\lambda}}_{\substack{=\sum_{\lambda\in \Par}v_{\lambda
}b_{\lambda}\\\text{(by (\ref{eq.lem.sol.lyndon.grfun.lin-ind-lamcrit.a3.eq}%
))}}}\right)  =\pi_{n}\left(  \sum_{\lambda\in \Par}v_{\lambda
}b_{\lambda}\right)  =\sum_{\lambda\in \Par_{n}%
}v_{\lambda}b_{\lambda}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c1.pf.9})}\right)  .
\]

\end{verlong}

\noindent This proves Claim 1.]

\begin{statement}
\textit{Claim 2:} Let $n\in \NN $. Let $\left(  c_{\lambda}\right)
_{\lambda\in \Par_{n}}\in\left(  \kk \left[
\left[  \xx\right]  \right]  \right)  ^{\Par_{n}}$
be a family of elements of $\kk \left[  \left[
\xx\right]  \right]  $ such that%
\begin{equation}
\sum_{\lambda\in \Par_{n}}u_{\lambda}c_{\lambda}=0.
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.ass}%
\end{equation}
Then, $c_{\lambda}=0$ for each $\lambda\in \Par_{n}$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 2:} For each $\lambda\in\Par_{n}$,
let us write the formal power series $c_{\lambda}\in
\kk \left[  \left[  \xx\right]  \right]  $ in the form%
\begin{equation}
c_{\lambda}=\sum\limits_{\mathfrak{m}\text{ is a monomial}}c_{\lambda
,\mathfrak{m}}\mathfrak{m}\ \ \ \ \ \ \ \ \ \ \text{for some }c_{\lambda
,\mathfrak{m}}\in \kk .
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.2}%
\end{equation}
Now, fix any monomial $\mathfrak{n}$. Then,
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.ass}) yields%
\begin{equation}
0=\sum_{\lambda\in \Par_{n}}u_{\lambda}%
\underbrace{c_{\lambda}}_{\substack{=\sum\limits_{\mathfrak{m}\text{ is a
monomial}}c_{\lambda,\mathfrak{m}}\mathfrak{m}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.2}))}}%
}=\sum_{\lambda\in \Par_{n}}u_{\lambda}\sum
\limits_{\mathfrak{m}\text{ is a monomial}}c_{\lambda,\mathfrak{m}%
}\mathfrak{m}=\sum\limits_{\mathfrak{m}\text{ is a monomial}}\ \ \sum
_{\lambda\in \Par_{n}}c_{\lambda,\mathfrak{m}%
}u_{\lambda}\mathfrak{m}.\nonumber
\end{equation}


Comparing coefficients of $\mathfrak{n}$ on both sides of this equality, we
obtain $0=\sum_{\lambda\in \Par_{n}}c_{\lambda
,\mathfrak{n}}u_{\lambda}$. In other words,%
\begin{equation}
\sum_{\lambda\in \Par_{n}}c_{\lambda,\mathfrak{n}%
}u_{\lambda}=0. \label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.4}%
\end{equation}


But the family $\left(  u_{\lambda}\right)  _{\lambda\in \Par}$
is $\kk$-linearly independent (by Assumption 2). Hence, its subfamily
$\left(  u_{\lambda}\right)  _{\lambda\in \Par_{n}}$ is
$\kk$-linearly independent as well. Thus, from
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.4}), we conclude
that%
\begin{equation}
c_{\lambda,\mathfrak{n}}=0\text{ for each }\lambda\in\Par_{n}
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.8}%
\end{equation}
(since $c_{\lambda,\mathfrak{n}}\in \kk$ for each $\lambda
\in \Par_{n}$).

Forget that we fixed $\mathfrak{n}$. We thus have proved
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.8}) for each
monomial $\mathfrak{n}$. Now, fix $\lambda\in \Par_{n}%
$. Then, (\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.2}) yields%
\[
c_{\lambda}=\sum\limits_{\mathfrak{m}\text{ is a monomial}}%
\underbrace{c_{\lambda,\mathfrak{m}}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.short.8}), applied to
}\mathfrak{n}=\mathfrak{m}\text{)}}}\mathfrak{m}=0.
\]
This proves Claim 2.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 2:} We have $ \Par_{n}%
\subset \Par$. Hence, the family $\left(  u_{\lambda}\right)
_{\lambda\in \Par_{n}}$ is a subfamily of the family
$\left(  u_{\lambda}\right)  _{\lambda\in \Par}$.

The family $\left(  u_{\lambda}\right)  _{\lambda\in \Par}$ is
$\kk$-linearly independent (by Assumption 2). But any subfamily of a
$\kk$-linearly independent family must itself be $\kk$-linearly
independent. Hence, the family $\left(  u_{\lambda}\right)  _{\lambda
\in \Par_{n}}$ is $\kk$-linearly independent
(since it is a subfamily of the $\kk$-linearly independent family
$\left(  u_{\lambda}\right)  _{\lambda\in \Par}$). In other
words, if $\left(  j_{\lambda}\right)  _{\lambda\in\Par_{n}}
\in \kk^{ \Par_{n}}$ is a family of
elements of $\kk$ such that
\begin{equation}
\sum_{\lambda\in \Par_{n}}j_{\lambda}u_{\lambda}=0,
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.1ass}%
\end{equation}
then%
\begin{equation}
\left(  j_{\lambda}=0\text{ for each }\lambda\in\Par_{n}
\right)  .
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.1}%
\end{equation}


For each $\lambda\in \Par_{n}$, let us write the formal
power series $c_{\lambda}\in \kk \left[  \left[  \xx\right]
\right]  $ in the form%
\begin{equation}
c_{\lambda}=\sum\limits_{\mathfrak{m}\text{ is a monomial}}c_{\lambda
,\mathfrak{m}}\mathfrak{m}\ \ \ \ \ \ \ \ \ \ \text{for some }c_{\lambda
,\mathfrak{m}} \in \kk .
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.2}%
\end{equation}
Now, (\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.ass}) yields%
\begin{align}
0  &  =\sum_{\lambda\in \Par_{n}}u_{\lambda
}\underbrace{c_{\lambda}}_{\substack{=\sum\limits_{\mathfrak{m}\text{ is a
monomial}}c_{\lambda,\mathfrak{m}}\mathfrak{m}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.2}))}}}=\sum_{\lambda
\in \Par_{n}}u_{\lambda}\sum\limits_{\mathfrak{m}\text{
is a monomial}}c_{\lambda,\mathfrak{m}}\mathfrak{m}\nonumber\\
&  =\underbrace{\sum_{\lambda\in \Par_{n}}%
\ \ \sum\limits_{\mathfrak{m}\text{ is a monomial}}}_{=\sum
\limits_{\mathfrak{m}\text{ is a monomial}}\ \ \sum_{\lambda\in
 \Par_{n}}}\underbrace{u_{\lambda}c_{\lambda
,\mathfrak{m}}}_{=c_{\lambda,\mathfrak{m}}u_{\lambda}}\mathfrak{m}\nonumber\\
&  =\sum\limits_{\mathfrak{m}\text{ is a monomial}}\ \ \sum_{\lambda
\in \Par_{n}}c_{\lambda,\mathfrak{m}}u_{\lambda
}\mathfrak{m}. \label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.3}%
\end{align}


Now, fix any monomial $\mathfrak{n}$. Then,%
\begin{align*}
&  \left(  \text{the coefficient of }\mathfrak{n}\text{ in the formal power
series }\sum\limits_{\mathfrak{m}\text{ is a monomial}}\ \ \sum_{\lambda
\in \Par_{n}}c_{\lambda,\mathfrak{m}}u_{\lambda
}\mathfrak{m}\right) \\
&  =\sum_{\lambda\in \Par_{n}}c_{\lambda,\mathfrak{n}%
}u_{\lambda}.
\end{align*}
Comparing this with%
\begin{align*}
&  \left(  \text{the coefficient of }\mathfrak{n}\text{ in the formal power
series }\underbrace{\sum\limits_{\mathfrak{m}\text{ is a monomial}}%
\ \ \sum_{\lambda\in \Par_{n}}c_{\lambda,\mathfrak{m}%
}u_{\lambda}\mathfrak{m}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.3}))}}}\right) \\
&  =\left(  \text{the coefficient of }\mathfrak{n}\text{ in the formal power
series }0\right)  =0,
\end{align*}
we obtain%
\[
\sum_{\lambda\in \Par_{n}}c_{\lambda,\mathfrak{n}%
}u_{\lambda}=0.
\]
Hence, (\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.1}) (applied to
$j_{\lambda}=c_{\lambda,\mathfrak{n}}$) yields
\begin{equation}
\left(  c_{\lambda,\mathfrak{n}}=0\text{ for each }\lambda\in
 \Par_{n}\right)  .
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.8}%
\end{equation}


Forget that we fixed $\mathfrak{n}$. We thus have proved
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.8}) for each monomial
$\mathfrak{n}$. Now, fix $\lambda\in \Par_{n}$. Then,
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.2}) yields%
\begin{align*}
c_{\lambda}  &  =\sum\limits_{\mathfrak{m}\text{ is a monomial}}%
c_{\lambda,\mathfrak{m}}\mathfrak{m}=\sum\limits_{\mathfrak{n}\text{ is a
monomial}}\underbrace{c_{\lambda,\mathfrak{n}}}_{\substack{=0\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c2.pf.8}))}}}\mathfrak{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}\mathfrak{m}\text{ as }\mathfrak{n}\right) \\
&  =\sum\limits_{\mathfrak{n}\text{ is a monomial}}0\mathfrak{n}=0.
\end{align*}
This proves Claim 2.]
\end{verlong}

Recall the finitary symmetric group $\Symm_{\left(  \infty\right)  }$
defined in Section \ref{subsect.Sym-section.def}. It acts on the rings
$R\left(  \xx\right)  $ and $\kk \left[  \left[  \xx%
\right]  \right]  $ and $A\left[  \left[  \xx\right]  \right]  $ in the
same way (viz., by permuting the variables $x_{1},x_{2},x_{3},\ldots$). The
ring $\Lambda$ is the invariant ring $\left\{  f\in R\left(  \xx%
\right)  :\sigma\left(  f\right)  =f\text{ for all }\sigma\in\mathfrak{S}%
_{\left(  \infty\right)  }\right\}  $ of its action on $R\left(
\xx\right)  $. Of course, $R\left(  \xx\right)  $ is an
$\Symm_{\left(  \infty\right)  }$-subset of $\kk \left[  \left[
\xx\right]  \right]  $. Moreover, the canonical $\kk \left[
\left[  \xx\right]  \right]  $-algebra homomorphism $\iota\left[
\left[  \xx\right]  \right]  : \kk \left[  \left[  \xx%
\right]  \right]  \rightarrow A\left[  \left[  \xx\right]  \right]  $
(which we are using to identify $\kk \left[  \left[  \xx\right]
\right]  $ with a subring of $A\left[  \left[  \xx\right]  \right]  $)
is $\Symm_{\left(  \infty\right)  }$-equivariant. This ensures that
expressions like $\sigma\left(  a_{\lambda}\right)  $ (for $\sigma
\in\Symm_{\left(  \infty\right)  }$ and $\lambda\in\Par$) are
well-defined (i.e., they don't depend on whether we regard $a_{\lambda
}$ as an element of $\kk \left[  \left[  \xx\right]  \right]  $
or as an element of $A\left[  \left[  \xx\right]  \right]  $).

We now claim:

\begin{statement}
\textit{Claim 3:} Let $n\in \NN $. Let $\lambda\in\Par_{n}$.
Then, $\sigma\left(  b_{\lambda}\right)  =b_{\lambda}$ for all
$\sigma\in\Symm_{\left(  \infty\right)  }$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 3:} Assumption 4 yields $b_{\lambda}\in\Lambda
=\left\{  f\in R\left(  \xx\right)  :\sigma\left(  f\right)  =f\text{
for all }\sigma\in\Symm_{\left(  \infty\right)  }\right\}  $. Hence,
$\sigma\left(  b_{\lambda}\right)  =b_{\lambda}$ for all $\sigma
\in\Symm_{\left(  \infty\right)  }$. This proves Claim 3.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 3:} We have $\lambda\in \Par_{n} \subset \Par$.
Hence, Assumption 4 yields
\[
b_{\lambda}\in\Lambda=\left\{  f\in R\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}
\]
(by the definition of $\Lambda$). In other words,
$b_{\lambda}$ is an $f\in R\left(  \xx\right)  $ satisfying
$\sigma\left(  f\right)  =f$ for all $\sigma\in\Symm_{\left(
\infty\right)  }$. Hence, $\sigma\left(  b_{\lambda}\right)  =b_{\lambda}$ for
all $\sigma\in\Symm_{\left(  \infty\right)  }$. This proves Claim 3.]
\end{verlong}

\begin{statement}
\textit{Claim 4:} Let $\sigma\in\Symm_{\left(  \infty\right)  }$. Then,
$\sigma\left(  a_{\mu}\right)  =a_{\mu}$ for each $\mu\in \Par$.
\end{statement}

[\textit{Proof of Claim 4:} Let $\mu\in \Par$. Set $n=\left\vert
\mu\right\vert $. Then, $\mu\in \Par_{n}$ (since
$\mu\in \Par$ and $\left\vert \mu\right\vert =n$).

\begin{vershort}
Recall that $\left(  u_{\lambda}\right)  _{\lambda\in \Par}\in
A^{ \Par}$. Hence, $u_{\lambda}\in A$ for each $\lambda
\in \Par_{n}$. Likewise, $v_{\lambda}\in A$ for each
$\lambda\in \Par_{n}$. Also, recall that $\left(
a_{\lambda}\right)  _{\lambda\in \Par}\in\left(  \kk
\left[  \left[  \xx\right]  \right]  \right)  ^{ \Par}$.
Hence, each $\lambda\in \Par_{n}$ satisfies
$a_{\lambda}\in \kk \left[  \left[  \xx\right]  \right]  $ and
thus $\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\in
\kk \left[  \left[  \xx\right]  \right]  $.
In other words, $\left(  \sigma\left(
a_{\lambda}\right)  -a_{\lambda}\right)  _{\lambda\in\Par_{n}}
\in\left(  \kk \left[  \left[  \xx\right]  \right]
\right)  ^{ \Par_{n}}$.
\end{vershort}

\begin{verlong}
Recall that $\left(  u_{\lambda}\right)  _{\lambda\in \Par}\in
A^{ \Par}$. Hence, $u_{\lambda}\in A$ for each $\lambda
\in \Par$. Thus, $u_{\lambda}\in A$ for each $\lambda
\in \Par_{n}$ (since $\lambda\in \Par_{n}\subset \Par$).
Likewise, $v_{\lambda}\in A$ for
each $\lambda\in \Par_{n}$. Also, recall that $\left(
a_{\lambda}\right)  _{\lambda\in \Par}\in\left(  \kk
\left[  \left[  \xx\right]  \right]  \right)  ^{ \Par}$.
Thus, each $\lambda\in \Par$ satisfies $a_{\lambda}\in
\kk \left[  \left[  \xx\right]  \right]  $ and thus%
\[
\sigma\left(  \underbrace{a_{\lambda}}_{\in \kk \left[  \left[
\xx\right]  \right]  }\right)  -\underbrace{a_{\lambda}}_{\in
\kk \left[  \left[  \xx\right]  \right]  }\in\underbrace{\sigma
\left(  \kk \left[  \left[  \xx\right]  \right]  \right)
}_{\subset \kk \left[  \left[  \xx\right]  \right]  }%
- \kk \left[  \left[  \xx\right]  \right]  \subset \kk
\left[  \left[  \xx\right]  \right]  -\kk \left[  \left[
\xx\right]  \right]  \subset \kk \left[  \left[  \xx%
\right]  \right]  .
\]
Therefore, each $\lambda\in \Par_{n}$ satisfies
$\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\in
\kk \left[  \left[
\xx\right]  \right]  $ (since $\lambda\in \Par_{n}
\subset \Par$). In other words, $\left(
\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)  _{\lambda
\in \Par_{n}}\in\left(  \kk \left[  \left[
\xx\right]  \right]  \right)  ^{ \Par_{n}}$.
\end{verlong}

Claim 1 yields%
\begin{equation}
\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda}%
=\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda}.
\label{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c4.pf.1}%
\end{equation}


Note that we can regard the (finite) sum $\sum_{\lambda\in\Par_{n}}
u_{\lambda}a_{\lambda}$ as an $A$-linear combination of the
elements $a_{\lambda}\in A\left[  \left[  \xx\right]  \right]
$\ \ \ \ \footnote{Here, we are using the fact that the $a_{\lambda}$ (for
$\lambda\in \Par$) are elements of $A\left[  \left[
\xx\right]  \right]  $. (This is because they are elements of
$\kk \left[  \left[  \xx\right]  \right]  $, but we are
identifying $\kk \left[  \left[  \xx\right]  \right]  $ with a
subring of $A\left[  \left[  \xx\right]  \right]  $.)} with
coefficients $u_{\lambda}\in A$. Likewise, we can regard the (finite) sum
$\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda}$ as
an $A$-linear combination of the elements $b_{\lambda}\in A\left[  \left[
\xx\right]  \right]  $ with coefficients $v_{\lambda}\in A$.

But the group $\Symm_{\left(  \infty\right)  }$ acts on $A\left[
\left[  \xx\right]  \right]  $ by $A$-linear maps. Thus, the map
$A\left[  \left[  \xx\right]  \right]  \rightarrow A\left[  \left[
\xx\right]  \right]  ,\ f\mapsto\sigma\left(  f\right)  $ is
$A$-linear. Hence, this map respects $A$-linear combinations. Consequently,%
\[
\sigma\left(  \sum_{\lambda\in \Par_{n}}u_{\lambda
}a_{\lambda}\right)  =\sum_{\lambda\in \Par_{n}%
}u_{\lambda}\sigma\left(  a_{\lambda}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }u_{\lambda}\in A\text{ for each }\lambda\in\Par_{n}
\right)
\]
and%
\begin{align*}
\sigma\left(  \sum_{\lambda\in \Par_{n}}v_{\lambda
}b_{\lambda}\right)   &  =\sum_{\lambda\in \Par_{n}%
}v_{\lambda}\underbrace{\sigma\left(  b_{\lambda}\right)  }%
_{\substack{=b_{\lambda}\\\text{(by Claim 3)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }v_{\lambda}\in A\text{ for each }\lambda\in\Par_{n}
\right) \\
&  =\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda
}=\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.lin-ind-lamcrit.c4.pf.1})}\right)  .
\end{align*}


\begin{vershort}
\noindent Subtracting these two equalities, we find%
\[
\sigma\left(  \sum_{\lambda\in \Par_{n}}u_{\lambda
}a_{\lambda}\right)  -\sigma\left(  \sum_{\lambda\in\Par_{n}}
v_{\lambda}b_{\lambda}\right)  =\sum_{\lambda\in
 \Par_{n}}u_{\lambda}\sigma\left(  a_{\lambda}\right)
-\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda}%
=\sum_{\lambda\in \Par_{n}}u_{\lambda}\left(
\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)  .
\]
Hence,%
\begin{align*}
\sum_{\lambda\in \Par_{n}}u_{\lambda}\left(
\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)   &  =\sigma\left(
\underbrace{\sum_{\lambda\in \Par_{n}}u_{\lambda
}a_{\lambda}}_{\substack{=\sum_{\lambda\in \Par_{n}%
}v_{\lambda}b_{\lambda}\\\text{(by Claim 1)}}}\right)  -\sigma\left(
\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda}\right)
\\
&  =\sigma\left(  \sum_{\lambda\in \Par_{n}}v_{\lambda
}b_{\lambda}\right)  -\sigma\left(  \sum_{\lambda\in\Par_{n}}
v_{\lambda}b_{\lambda}\right)  =0.
\end{align*}

\end{vershort}

\begin{verlong}
\noindent Subtracting these two equalities, we find%
\begin{align*}
\sigma\left(  \sum_{\lambda\in \Par_{n}}u_{\lambda
}a_{\lambda}\right)  -\sigma\left(  \sum_{\lambda\in\Par_{n}}
v_{\lambda}b_{\lambda}\right)   &  =\sum_{\lambda\in
 \Par_{n}}u_{\lambda}\sigma\left(  a_{\lambda}\right)
-\sum_{\lambda\in \Par_{n}}u_{\lambda}a_{\lambda}\\
&  =\sum_{\lambda\in \Par_{n}}\underbrace{\left(
u_{\lambda}\sigma\left(  a_{\lambda}\right)  -u_{\lambda}a_{\lambda}\right)
}_{=u_{\lambda}\left(  \sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)
}\\
&  =\sum_{\lambda\in \Par_{n}}u_{\lambda}\left(
\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)  .
\end{align*}
Hence,%
\begin{align*}
\sum_{\lambda\in \Par_{n}}u_{\lambda}\left(
\sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)   &  =\sigma\left(
\underbrace{\sum_{\lambda\in \Par_{n}}u_{\lambda
}a_{\lambda}}_{\substack{=\sum_{\lambda\in \Par_{n}%
}v_{\lambda}b_{\lambda}\\\text{(by Claim 1)}}}\right)  -\sigma\left(
\sum_{\lambda\in \Par_{n}}v_{\lambda}b_{\lambda}\right)
\\
&  =\sigma\left(  \sum_{\lambda\in \Par_{n}}v_{\lambda
}b_{\lambda}\right)  -\sigma\left(  \sum_{\lambda\in\Par_{n}}
v_{\lambda}b_{\lambda}\right)  =0.
\end{align*}

\end{verlong}

\noindent Hence, Claim 2 (applied to $c_{\lambda}=\sigma\left(  a_{\lambda
}\right)  -a_{\lambda}$) yields that $\sigma\left(  a_{\lambda}\right)
-a_{\lambda}=0$ for each $\lambda\in \Par_{n}$ (since
$\left(  \sigma\left(  a_{\lambda}\right)  -a_{\lambda}\right)  _{\lambda
\in \Par_{n}}\in\left(  \kk \left[  \left[
\xx\right]  \right]  \right)  ^{ \Par_{n}}$).
Applying this to $\lambda=\mu$, we obtain $\sigma\left(  a_{\mu}\right)
-a_{\mu}=0$ (since $\mu\in \Par_{n}$). In other words,
$\sigma\left(  a_{\mu}\right)  =a_{\mu}$. This proves Claim 4.]

\begin{statement}
\textit{Claim 5:} We have $a_{\mu}\in\Lambda$ for each $\mu\in
 \Par$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 5:} Let $\mu\in \Par$. Thus, $a_{\mu}%
\in \kk \left[  \left[  \xx\right]  \right]  $ (since $\left(
a_{\lambda}\right)  _{\lambda\in \Par}\in\left(  \kk
\left[  \left[  \xx\right]  \right]  \right)  ^{ \Par}$).
Also, Assumption 1 (applied to $\lambda=\mu$) shows that the power series
$a_{\mu}$ is homogeneous of degree $\left\vert \mu\right\vert $. Hence, this
power series $a_{\mu}$ is of bounded degree (since any homogeneous power
series is of bounded degree). In other words, $a_{\mu}\in R\left(
\xx\right)  $. Claim 4 yields that $\sigma\left(  a_{\mu}\right)
=a_{\mu}$ for all $\sigma\in\Symm_{\left(  \infty\right)  }$.

Thus, $a_{\mu}$ is an $f\in R\left(  \xx\right)  $ satisfying
$\sigma\left(  f\right)  =f$ for all $\sigma\in\Symm_{\left(
\infty\right)  }$. In other words,
\[
a_{\mu}\in\left\{  f\in R\left(
\xx\right)  :\sigma\left(  f\right)  =f\text{ for all }\sigma
\in\Symm_{\left(  \infty\right)  }\right\} .
\]
In view of
$\Lambda=\left\{  f\in R\left(  \xx\right)  :\sigma\left(  f\right)
=f\text{ for all }\sigma\in\Symm_{\left(  \infty\right)  }\right\}  $,
this rewrites as $a_{\mu}\in\Lambda$. This proves Claim 5.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 5:} Let $\mu\in \Par$. Thus, $a_{\mu}%
\in \kk \left[  \left[  \xx\right]  \right]  $ (since $\left(
a_{\lambda}\right)  _{\lambda\in \Par}\in\left(  \kk
\left[  \left[  \xx\right]  \right]  \right)  ^{ \Par}$).
Also, Assumption 1 (applied to $\lambda=\mu$) shows that both power series
$a_{\mu}$ and $b_{\mu}$ are homogeneous of degree $\left\vert \mu\right\vert
$. Hence, in particular, the power series $a_{\mu}$ is homogeneous of degree
$\left\vert \mu\right\vert $. Hence, this power series $a_{\mu}$ is of bounded
degree (since any homogeneous power series is of bounded degree). In other
words, $a_{\mu}\in R\left(  \xx\right)  $ (since $R\left(
\xx\right)  $ is the set of all formal power series of bounded degree).
Claim 4 yields that $\sigma\left(  a_{\mu}\right)  =a_{\mu}$ for all
$\sigma\in\Symm_{\left(  \infty\right)  }$.

The definition of $\Lambda$ yields $\Lambda=\left\{  f\in R\left(
\xx\right)  :\sigma\left(  f\right)  =f\text{ for all }\sigma
\in\Symm_{\left(  \infty\right)  }\right\}  $. But $a_{\mu}$ is an
$f\in R\left(  \xx\right)  $ satisfying $\sigma\left(  f\right)  =f$
for all $\sigma\in\Symm_{\left(  \infty\right)  }$ (since $a_{\mu}\in
R\left(  \xx\right)  $, and since $\sigma\left(  a_{\mu}\right)
=a_{\mu}$ for all $\sigma\in\Symm_{\left(  \infty\right)  }$). In other
words, $a_{\mu}\in\left\{  f\in R\left(  \xx\right)  :\sigma\left(
f\right)  =f\text{ for all }\sigma\in\Symm_{\left(  \infty\right)
}\right\}  $. In view of $\Lambda=\left\{  f\in R\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}  $, this rewrites as $a_{\mu}\in\Lambda$. This proves
Claim 5.]
\end{verlong}

We have $a_{\mu}\in\Lambda$ for each $\mu\in \Par$ (by Claim 5).
Renaming the index $\mu$ as $\lambda$ in this statement, we can rewrite this
as follows: We have $a_{\lambda}\in\Lambda$ for each $\lambda\in
 \Par$. This proves Lemma
\ref{lem.sol.lyndon.grfun.lin-ind-lamcrit}.
\end{proof}

\begin{vershort}

\begin{proof}
[Second proof of Proposition \ref{prop.lyndon.grfun.symm}.]We begin with a
trick. Recall that there is a canonical ring homomorphism $\varphi
: \ZZ \rightarrow \kk$. This homomorphism gives rise to a ring
homomorphism $\varphi\left[  \left[  \xx\right]  \right]  :{ \ZZ %
}\left[  \left[  \xx\right]  \right]  \rightarrow \kk \left[
\left[  \xx\right]  \right]  $, and this latter homomorphism
$\varphi\left[  \left[  \xx\right]  \right]  $ sends $\Lambda
_{{ \ZZ }}$ to $\Lambda_{\kk}$; that is, we have $\left(
\varphi\left[  \left[  \xx\right]  \right]  \right)  \left(
\Lambda_{{ \ZZ }}\right)  \subset\Lambda_{\kk}$. Moreover, it is
clear that the ring homomorphism $\varphi\left[  \left[  \xx\right]
\right]  $ sends the element $\mathbf{GR}_{\lambda}$ of ${ \ZZ }\left[
\left[  \xx\right]  \right]  $ to the element $\mathbf{GR}_{\lambda}$
of $\kk \left[  \left[  \xx\right]  \right]  $ (because the
definition of $\mathbf{GR}_{\lambda}$ is functorial in the base ring
$\kk$). Therefore, if we can prove that the element $\mathbf{GR}%
_{\lambda}$ of ${ \ZZ }\left[  \left[  \xx\right]  \right]  $
belongs to $\Lambda_{{ \ZZ }}$, then it will automatically follow that
the element $\mathbf{GR}_{\lambda}$ of $\kk \left[  \left[
\xx\right]  \right]  $ belongs to $\left(  \varphi\left[  \left[
\xx\right]  \right]  \right)  \left(  \Lambda_{{ \ZZ }}\right)
\subset\Lambda_{\kk}$; this will complete the proof of Proposition
\ref{prop.lyndon.grfun.symm}. Hence, in order to prove Proposition
\ref{prop.lyndon.grfun.symm}, it only remains to prove that the element
$\mathbf{GR}_{\lambda}$ of ${ \ZZ }\left[  \left[  \xx\right]
\right]  $ belongs to $\Lambda_{{ \ZZ }}$. In other words, it only
remains to prove Proposition \ref{prop.lyndon.grfun.symm} in the case of
$\kk = \ZZ$. Hence, in proving Proposition
\ref{prop.lyndon.grfun.symm}, we can WLOG assume that $\kk = \ZZ$.
Assume this.

Forget that we fixed $\lambda$. Consider the countable set of indeterminates
$\mathbf{y}=\left(  y_{1},y_{2},y_{3},\ldots\right)  $.

It is easy to see that the family $\left(  p_{\lambda}\right)  _{\lambda
\in \Par}\in\left(  \kk \left[  \left[  \xx\right]
\right]  \right)  ^{ \Par}$ is $\kk$-linearly
independent\footnote{\textit{Proof.} We have $\Lambda=\Lambda_{\kk}
=\Lambda_{ \ZZ }$ (since $\kk = \ZZ$). But $\Lambda
_{ \ZZ }$ is clearly a subring of $\Lambda_{ \QQ }$ (since the ring
$ \ZZ \left[  \left[  \xx\right]  \right]  $ is a subring of the
ring $ \QQ \left[  \left[  \xx\right]  \right]  $, and the
finitary symmetric group $\Symm_{\left(  \infty\right)  }$ acts in the
same way on both of these rings). Furthermore, the symmetric functions
$e_{\lambda}$, $s_{\lambda}$ and $p_{\lambda}$ are defined in the same way
over the ring $\ZZ$ as they are defined over the ring $\QQ$.
\par
Now, Proposition \ref{many-symmetric-function-bases-prop} (applied to
$\QQ$ instead of $\kk$) shows that the families $\left(
e_{\lambda}\right)  _{\lambda\in \Par}$, $\left(  s_{\lambda
}\right)  _{\lambda\in \Par}$ and $\left(  p_{\lambda}\right)
_{\lambda\in \Par}$ are bases of the $\QQ$-module
$\Lambda_{ \QQ }$. Hence, in particular, the family $\left(  p_{\lambda
}\right)  _{\lambda\in \Par}$ is a basis of the $ \QQ %
$-module $\Lambda_{ \QQ }$. Thus, this family is $\QQ$-linearly
independent, and therefore $\ZZ$-linearly independent as well (since
$\ZZ$ is a subring of $\QQ$). In other words, it is $\kk$-linearly
independent (since $\kk = \ZZ$). Qed.}. Hence, the
family $\left(  p_{\lambda}\left(  \mathbf{y}\right)  \right)  _{\lambda
\in \Par}\in\left(  \kk \left[  \left[  \mathbf{y}\right]
\right]  \right)  ^{ \Par}$ is $\kk$-linearly
independent\footnote{Indeed, the family $\left(  p_{\lambda}\left(
\mathbf{y}\right)  \right)  _{\lambda\in \Par}$ is obtained from
the family $\left(  p_{\lambda}\right)  _{\lambda\in \Par}$ by
renaming the indeterminates $x_{1},x_{2},x_{3},\ldots$ as $y_{1},y_{2}%
,y_{3},\ldots$. Thus, the $\kk$-linear independence of the latter
family implies the $\kk$-linear independence of the former.}. If
$\lambda\in \Par$, then both power series $\mathbf{GR}_{\lambda
}$ and $p_{\lambda}$ are homogeneous of degree $\left\vert \lambda\right\vert
$ (by Lemma \ref{lem.sol.lyndon.grfun.bdd} and Lemma
\ref{lem.sol.lyndon.grfun.bdd-p}).

We identify the power series ring $\kk \left[  \left[  \xx%
,\mathbf{y}\right]  \right]  =\kk \left[  \left[  x_{1},x_{2}%
,x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right]  \right]  $ with the power
series ring $\left(  \kk \left[  \left[  \mathbf{y}\right]  \right]
\right)  \left[  \left[  \xx\right]  \right]  =\left(  \kk
\left[  \left[  y_{1},y_{2},y_{3},\ldots\right]  \right]  \right)  \left[
\left[  x_{1},x_{2},x_{3},\ldots\right]  \right]  $. In this power series
ring, we have%
\begin{align}
\sum_{\lambda\in \Par}p_{\lambda}\left(  \mathbf{y}\right)
\underbrace{\mathbf{GR}_{\lambda}}_{=\mathbf{GR}_{\lambda}\left(
\xx\right)  }  &  =\sum_{\lambda\in \Par}p_{\lambda
}\left(  \mathbf{y}\right)  \mathbf{GR}_{\lambda}\left(  \xx\right)
=\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\xx\right)  p_{\lambda}\left(  \mathbf{y}\right) \nonumber\\
&  =\sum_{\lambda\in \Par}p_{\lambda}\left(  \xx\right)
\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition \ref{prop.lyndon.grfun.cauchy}(a)}\right) \nonumber\\
&  =\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\mathbf{y}\right)  \underbrace{p_{\lambda}\left(  \xx\right)
}_{=p_{\lambda}}=\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda
}\left(  \mathbf{y}\right)  p_{\lambda}.
\label{pf.lem.sol.lyndon.grfun.2nd.s1.short.2}%
\end{align}
Hence, we can apply Lemma \ref{lem.sol.lyndon.grfun.lin-ind-lamcrit} to
\[
A= \kk \left[  \left[  \mathbf{y}\right]  \right]  ,\ \ \ \ a_{\lambda
}=\mathbf{GR}_{\lambda},\ \ \ \text{\ }b_{\lambda}=p_{\lambda}%
,\ \ \ \ u_{\lambda}=p_{\lambda}\left(  \mathbf{y}\right)  ,\ \ \ \ \text{and}%
\ \ \ \ v_{\lambda}=\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  .
\]
\footnote{Indeed, Assumption 1 of Lemma
\ref{lem.sol.lyndon.grfun.lin-ind-lamcrit} is satisfied (because if
$\lambda\in \Par$, then both power series $\mathbf{GR}_{\lambda
}$ and $p_{\lambda}$ are homogeneous of degree $\left\vert \lambda\right\vert
$); Assumption 2 is satisfied as well (since the family $\left(  p_{\lambda
}\left(  \mathbf{y}\right)  \right)  _{\lambda\in \Par}%
\in\left(  \kk \left[  \left[  \mathbf{y}\right]  \right]  \right)
^{ \Par}$ is $\kk$-linearly independent); Assumption 3 is
also satisfied (due to (\ref{pf.lem.sol.lyndon.grfun.2nd.s1.short.2}));
finally, Assumption 4 is satisfied (since we have $p_{\lambda}\in\Lambda$ for
each $\lambda\in \Par$).} Thus, we conclude that $\mathbf{GR}%
_{\lambda}\in\Lambda$ for each $\lambda\in \Par$. In other
words, the power series $\mathbf{GR}_{\lambda}$ belongs to $\Lambda$ for each
partition $\lambda$. This proves Proposition \ref{prop.lyndon.grfun.symm}.
\end{proof}

\end{vershort}

\begin{verlong}

\begin{proof}
[Second proof of Proposition \ref{prop.lyndon.grfun.symm}.]Our second proof of
Proposition \ref{prop.lyndon.grfun.symm} shall proceed in three steps:

\begin{itemize}
\item \textit{Step 1:} proving that Proposition \ref{prop.lyndon.grfun.symm}
holds if $\kk = \QQ$.

\item \textit{Step 2:} proving that Proposition \ref{prop.lyndon.grfun.symm}
holds if $\kk = \ZZ$.

\item \textit{Step 3:} proving that Proposition \ref{prop.lyndon.grfun.symm}
holds in the general case.
\end{itemize}

Let us now get to the details of these three steps:

\underline{\textit{Step 1:}} We shall prove that Proposition
\ref{prop.lyndon.grfun.symm} holds if $\kk = \QQ$.

Indeed, assume that $\kk = \QQ$. Then, $\QQ$ is a subring
of $\kk$.

Forget that we fixed $\lambda$. We thus must show that the power series
$\mathbf{GR}_{\lambda}$ belongs to $\Lambda$ for each partition $\lambda$.

Proposition \ref{many-symmetric-function-bases-prop} shows that the families
$\left(  e_{\lambda}\right)  _{\lambda\in \Par}$, $\left(
s_{\lambda}\right)  _{\lambda\in \Par}$ and $\left(  p_{\lambda
}\right)  _{\lambda\in \Par}$ are bases of the $\kk$-module
$\Lambda_{\kk}$ (since $\QQ$ is a subring of
$\kk$). Thus, in particular, the family $\left(  p_{\lambda}\right)
_{\lambda\in \Par}$ is a basis of the $\kk$-module
$\Lambda_{\kk}$. Hence, this family $\left(  p_{\lambda}\right)
_{\lambda\in \Par}$ is $\kk$-linearly independent. In
other words, if $\left(  c_{\lambda}\right)  _{\lambda\in\Par}
\in \kk^{ \Par}$ is a family of scalars such that
\[
\left(  c_{\lambda}=0\text{ for all but finitely many }\lambda\in
\Par \right)
\]
and $\sum_{\lambda\in \Par}c_{\lambda}p_{\lambda}=0$, then%
\begin{equation}
\left(  c_{\lambda}=0\ \ \ \ \ \ \ \ \ \ \text{for all }\lambda\in
\Par \right)  . \label{pf.lem.sol.lyndon.grfun.2nd.s1.1}%
\end{equation}


Consider the countable set of indeterminates $\mathbf{y}=\left(  y_{1}%
,y_{2},y_{3},\ldots\right)  $. Then, the family $\left(  p_{\lambda}\left(
\mathbf{y}\right)  \right)  _{\lambda\in \Par}\in\left(
\kk \left[  \left[  \mathbf{y}\right]  \right]  \right)
^{ \Par}$ is $\kk$-linearly
independent\footnote{\textit{Proof. }Let $\left(  c_{\lambda}\right)
_{\lambda\in \Par}\in \kk^{ \Par}$ be a
family of scalars such that%
\[
\left(  c_{\lambda}=0\text{ for all but finitely many }\lambda\in
\Par \right)
\]
and
\begin{equation}
\sum_{\lambda\in \Par}c_{\lambda}p_{\lambda}\left(
\mathbf{y}\right)  =0. \label{pf.lem.sol.lyndon.grfun.2nd.fn1.linind.1}%
\end{equation}
We shall prove that $c_{\lambda}=0$ for all $\lambda\in \Par$.
\par
Indeed, if we substitute the variables $x_{1},x_{2},x_{3},\ldots$ for
$y_{1},y_{2},y_{3},\ldots$ in the formal power series $p_{\lambda}\left(
\mathbf{y}\right)  \in \kk \left[  \left[  \mathbf{y}\right]  \right]  $
for some $\lambda\in \Par$, then we obtain the formal power
series $p_{\lambda}\left(  \xx\right)  $ (because $\mathbf{y}=\left(
y_{1},y_{2},y_{3},\ldots\right)  $ and $\xx=\left(  x_{1},x_{2}%
,x_{3},\ldots\right)  $). Hence, if we substitute the variables $x_{1}%
,x_{2},x_{3},\ldots$ for $y_{1},y_{2},y_{3},\ldots$ in the sum $\sum
_{\lambda\in \Par}c_{\lambda}p_{\lambda}\left(  \mathbf{y}%
\right)  $, then we obtain $\sum_{\lambda\in \Par}c_{\lambda
}p_{\lambda}\left(  \xx\right)  $. Thus, if we substitute the variables
$x_{1},x_{2},x_{3},\ldots$ for $y_{1},y_{2},y_{3},\ldots$ in the equality
(\ref{pf.lem.sol.lyndon.grfun.2nd.fn1.linind.1}), then we obtain%
\[
\sum_{\lambda\in \Par}c_{\lambda}p_{\lambda}\left(
\xx\right)  =0.
\]
Hence,
\[
\sum_{\lambda\in \Par}c_{\lambda}\underbrace{p_{\lambda}%
}_{=p_{\lambda}\left(  \xx\right)  }=\sum_{\lambda\in
 \Par}c_{\lambda}p_{\lambda}\left(  \xx\right)  =0.
\]
Hence, (\ref{pf.lem.sol.lyndon.grfun.2nd.s1.1}) shows that
\[
\left(  c_{\lambda}=0\ \ \ \ \ \ \ \ \ \ \text{for all }\lambda\in
\Par \right)  .
\]
\par
Forget that we fixed $\left(  c_{\lambda}\right)  _{\lambda\in
 \Par}$. We thus have shown that if $\left(  c_{\lambda}\right)
_{\lambda\in \Par}\in \kk^{ \Par}$ is a
family of scalars such that
\[
\left(  c_{\lambda}=0\text{ for all but finitely many }\lambda\in
\Par \right)
\]
and $\sum_{\lambda\in \Par}c_{\lambda}p_{\lambda}\left(
\mathbf{y}\right)  =0$, then%
\[
\left(  c_{\lambda}=0\ \ \ \ \ \ \ \ \ \ \text{for all }\lambda\in
\Par \right)  .
\]
In other words, the family $\left(  p_{\lambda}\left(  \mathbf{y}\right)
\right)  _{\lambda\in \Par}\in\left(  \kk \left[  \left[
\mathbf{y}\right]  \right]  \right)  ^{ \Par}$ is
$\kk$-linearly independent. Qed.}.

If $\lambda\in \Par$, then both power series $\mathbf{GR}%
_{\lambda}$ and $p_{\lambda}$ are homogeneous of degree $\left\vert
\lambda\right\vert $. (Indeed, for $\mathbf{GR}_{\lambda}$, this follows from
Lemma \ref{lem.sol.lyndon.grfun.bdd}, whereas for $p_{\lambda}$, this follows
from Lemma \ref{lem.sol.lyndon.grfun.bdd-p}.)

We identify the power series ring $\kk \left[  \left[  \xx%
,\mathbf{y}\right]  \right]  =\kk \left[  \left[  x_{1},x_{2}%
,x_{3},\ldots,y_{1},y_{2},y_{3},\ldots\right]  \right]  $ with the power
series ring $\left(  \kk \left[  \left[  \mathbf{y}\right]  \right]
\right)  \left[  \left[  \xx\right]  \right]  =\left(  \kk
\left[  \left[  y_{1},y_{2},y_{3},\ldots\right]  \right]  \right)  \left[
\left[  x_{1},x_{2},x_{3},\ldots\right]  \right]  $. In this power series
ring, we have%
\begin{align}
\sum_{\lambda\in \Par}p_{\lambda}\left(  \mathbf{y}\right)
\underbrace{\mathbf{GR}_{\lambda}}_{=\mathbf{GR}_{\lambda}\left(
\xx\right)  }  &  =\sum_{\lambda\in \Par}%
\underbrace{p_{\lambda}\left(  \mathbf{y}\right)  \mathbf{GR}_{\lambda}\left(
\xx\right)  }_{=\mathbf{GR}_{\lambda}\left(  \xx\right)
p_{\lambda}\left(  \mathbf{y}\right)  }=\sum_{\lambda\in\Par}
\mathbf{GR}_{\lambda}\left(  \xx\right)  p_{\lambda}\left(
\mathbf{y}\right) \nonumber\\
&  =\sum_{\lambda\in \Par}\underbrace{p_{\lambda}\left(
\xx\right)  \mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)
}_{=\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  p_{\lambda}\left(
\xx\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.lyndon.grfun.cauchy}(a)}\right) \nonumber\\
&  =\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda}\left(
\mathbf{y}\right)  \underbrace{p_{\lambda}\left(  \xx\right)
}_{=p_{\lambda}}=\sum_{\lambda\in \Par}\mathbf{GR}_{\lambda
}\left(  \mathbf{y}\right)  p_{\lambda}.
\label{pf.lem.sol.lyndon.grfun.2nd.s1.2}%
\end{align}
Finally, of course, we have $p_{\lambda}\in\Lambda$ for each $\lambda
\in \Par$.

Hence, we can apply Lemma \ref{lem.sol.lyndon.grfun.lin-ind-lamcrit} to
\[
A = \kk \left[  \left[  \mathbf{y}\right]  \right]  ,\ \ \ \ a_{\lambda
}=\mathbf{GR}_{\lambda},\ \ \ \text{\ }b_{\lambda}=p_{\lambda}%
,\ \ \ \ u_{\lambda}=p_{\lambda}\left(  \mathbf{y}\right)  ,\ \ \ \ \text{and}%
\ \ \ \ v_{\lambda}=\mathbf{GR}_{\lambda}\left(  \mathbf{y}\right)  .
\]
\footnote{Indeed, Assumption 1 of Lemma
\ref{lem.sol.lyndon.grfun.lin-ind-lamcrit} is satisfied (because if
$\lambda\in \Par$, then both power series $\mathbf{GR}_{\lambda
}$ and $p_{\lambda}$ are homogeneous of degree $\left\vert \lambda\right\vert
$); Assumption 2 is satisfied as well (since the family $\left(  p_{\lambda
}\left(  \mathbf{y}\right)  \right)  _{\lambda\in \Par}%
\in\left(  \kk \left[  \left[  \mathbf{y}\right]  \right]  \right)
^{ \Par}$ is $\kk$-linearly independent); Assumption 3 is
also satisfied (due to (\ref{pf.lem.sol.lyndon.grfun.2nd.s1.2})); finally,
Assumption 4 is satisfied (since we have $p_{\lambda}\in\Lambda$ for each
$\lambda\in \Par$).} Thus, we conclude that $\mathbf{GR}%
_{\lambda}\in\Lambda$ for each $\lambda\in \Par$. In other
words, $\mathbf{GR}_{\lambda}\in\Lambda$ for each partition $\lambda$. In
other words, the power series $\mathbf{GR}_{\lambda}$ belongs to $\Lambda$ for
each partition $\lambda$. Thus, Proposition \ref{prop.lyndon.grfun.symm} is
proved under the assumption that $\kk = \QQ$. Our Step 1 is complete.

\underline{\textit{Step 2:}} We shall prove that Proposition
\ref{prop.lyndon.grfun.symm} holds if $\kk = \ZZ$.

Indeed, let us first consider the general case, without any assumptions on
$\kk$.

Fix a partition $\lambda$. We shall now formalize the intuitive concept that
the $\kk$-algebra $\kk \left[  \left[  \xx\right]
\right]  $ and the power series $\mathbf{GR}_{\lambda}$ are ``functorial''
with respect to the base ring $\kk$. This
is chiefly a matter of introducing notations:

\begin{itemize}
\item We denote the $\kk$-algebra $R\left(  \xx\right)  $ (which
consists of all formal power series $f\in \kk \left[  \left[
\xx\right]  \right]  $ that are of bounded degree) by $R_{\kk}
\left(  \xx\right)  $. This notation makes the dependency of $R\left(
\xx\right)  $ on $\kk$ more visible. Recall that the definition
of $\Lambda$ yields $\Lambda=\left\{  f\in R\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}  $. In other words,%
\begin{equation}
\Lambda_{\kk}=\left\{  f\in R_{\kk}\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}  \label{pf.prop.lyndon.grfun.symm.2nd.Lamk=}%
\end{equation}
(since $\Lambda_{\kk}=\Lambda$ and $R_{\kk}\left(
\xx\right)  =R\left(  \xx\right)  $).

\item If $\mathbf{m}$ and $\mathbf{n}$ are two commutative rings, and
$\varphi:\mathbf{m}\rightarrow\mathbf{n}$ is a ring homomorphism, then
$\varphi$ canonically induces a ring homomorphism $\varphi\left[  \left[
\xx\right]  \right]  :\mathbf{m}\left[  \left[  \xx\right]
\right]  \rightarrow\mathbf{n}\left[  \left[  \xx\right]  \right]  $.

\item We shall denote the formal power series $\mathbf{GR}_{\lambda}%
\in \kk \left[  \left[  \xx\right]  \right]  $ by $\mathbf{GR}%
_{\lambda}^{\left[  \kk \right]  }$. This notation makes the dependency
of $\mathbf{GR}_{\lambda}$ on the base ring $\kk$ more explicit, and
thus allows us to talk about these elements defined over several different
base rings at the same time (without the danger of confusing them). Of course,
the element $\mathbf{GR}_{\lambda}$ does not ``really'' depend on
$\kk$, in the sense that it is
defined in the same way for every $\kk$. This entails that for any two
commutative rings $\mathbf{m}$ and $\mathbf{n}$ and any ring homomorphism
$\varphi:\mathbf{m}\rightarrow\mathbf{n}$, we have%
\begin{equation}
\mathbf{GR}_{\lambda}^{\left[  \mathbf{n}\right]  }=\left(  \varphi\left[
\left[  \xx\right]  \right]  \right)  \left(  \mathbf{GR}_{\lambda
}^{\left[  \mathbf{m}\right]  }\right)  .
\label{pf.prop.lyndon.grfun.symm.2nd.func-GR}%
\end{equation}


\item Recall the finitary symmetric group $\Symm_{\left(
\infty\right)  }$ defined in Section \ref{subsect.Sym-section.def}. It acts on
the ring $\kk \left[  \left[  \xx\right]  \right]  $ by permuting
the indeterminates $x_{1},x_{2},x_{3},\ldots$. This action is
``functorial in $\kk$'', in the sense
that it is defined in the same way for every $\kk$. This entails that
for any two commutative rings $\mathbf{m}$ and $\mathbf{n}$ and any ring
homomorphism $\varphi:\mathbf{m}\rightarrow\mathbf{n}$, we have%
\begin{equation}
\sigma\left(  \left(  \varphi\left[  \left[  \xx\right]  \right]
\right)  \left(  f\right)  \right)  =\left(  \varphi\left[  \left[
\xx\right]  \right]  \right)  \left(  \sigma\left(  f\right)  \right)
\label{pf.prop.lyndon.grfun.symm.2nd.func-sigma}%
\end{equation}
for every $\sigma\in\Symm_{\left(  \infty\right)  }$ and every
$f\in\mathbf{m}\left[  \left[  \xx\right]  \right]  $.

\item For every commutative ring $\kk$, there exists a canonical (and
unique) ring homomorphism $\ZZ \rightarrow \kk$. Denote this
homomorphism by $\rho_{\kk}$.
\end{itemize}

The ring homomorphism $\rho_{ \QQ }:{ \ZZ }\rightarrow \QQ $ is
just the canonical inclusion of ${ \ZZ }$ into ${ \QQ }$, and thus
is injective. Hence, the map $\rho_{ \QQ }\left[  \left[  \xx%
\right]  \right]  $ is injective (because $\varphi\left[  \left[
\xx\right]  \right]  $ is injective for every injective ring
homomorphism $\varphi$).

In Step 1, we have shown that Proposition \ref{prop.lyndon.grfun.symm} holds
if $\kk = \QQ$. In other words, if $\kk = \QQ$, then
the power series $\mathbf{GR}_{\lambda}$ belongs to $\Lambda$. In other words,
the power series $\mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }$ belongs
to $\Lambda_{ \QQ }$. Thus,%
\[
\mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }\in\Lambda_{ \QQ %
}=\left\{  f\in R_{ \QQ }\left(  \xx\right)  :\sigma\left(
f\right)  =f\text{ for all }\sigma\in\Symm_{\left(  \infty\right)
}\right\}
\]
(by (\ref{pf.prop.lyndon.grfun.symm.2nd.Lamk=}), applied to $\QQ$
instead of $\kk$). In other words, $\mathbf{GR}_{\lambda}^{\left[
 \QQ \right]  }$ is an element of $R_{ \QQ }\left(  \xx%
\right)  $ that satisfies $\sigma\left(  f\right)  =f$ for all $\sigma
\in\Symm_{\left(  \infty\right)  }$. Hence, $\mathbf{GR}_{\lambda
}^{\left[   \QQ \right]  }\in R_{ \QQ }\left(  \xx\right)  $
and%
\begin{equation}
\sigma\left(  \mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }\right)
=\mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }
\label{pf.prop.lyndon.grfun.symm.2nd.sig}%
\end{equation}
for all $\sigma\in\Symm_{\left(  \infty\right)  }$.

Lemma \ref{lem.sol.lyndon.grfun.bdd} yields that the power series
$\mathbf{GR}_{\lambda}$ is homogeneous of degree $\left\vert \lambda
\right\vert $. Hence, the power series $\mathbf{GR}_{\lambda}$ is of bounded
degree (since any homogeneous power series is of bounded degree). In other
words, $\mathbf{GR}_{\lambda}\in R\left(  \xx\right)  $ (since
$R\left(  \xx\right)  $ is the set of all formal power series
$f\in \kk \left[  \left[  \xx\right]  \right]  $ that are of
bounded degree). In other words, $\mathbf{GR}_{\lambda}\in R_{\kk}
\left(  \xx\right)  $ (since $R\left(  \xx\right)
=R_{\kk}\left(  \xx\right)  $).

Now, assume that $\kk = \ZZ$. Let $\sigma\in\Symm_{\left(
\infty\right)  }$. Then,
\begin{align}
\mathbf{GR}_{\lambda}  &  =\mathbf{GR}_{\lambda}^{\left[  \kk \right]
}=\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\kk = \ZZ \right)
\label{pf.prop.lyndon.grfun.symm.2nd.3}\\
&  \in \ZZ \left[  \left[  \xx\right]  \right]  .\nonumber
\end{align}
Hence, we can apply (\ref{pf.prop.lyndon.grfun.symm.2nd.func-sigma}) to
$\mathbf{m}= \ZZ $, $\mathbf{n}= \QQ $, $\varphi=\rho_{ \QQ }$
and $f=\mathbf{GR}_{\lambda}$. We thus obtain%
\begin{equation}
\sigma\left(  \left(  \rho_{ \QQ }\left[  \left[  \xx\right]
\right]  \right)  \left(  \mathbf{GR}_{\lambda}\right)  \right)  =\left(
\rho_{ \QQ }\left[  \left[  \xx\right]  \right]  \right)  \left(
\sigma\left(  \mathbf{GR}_{\lambda}\right)  \right)  .
\label{pf.prop.lyndon.grfun.symm.2nd.4}%
\end{equation}
But (\ref{pf.prop.lyndon.grfun.symm.2nd.func-GR}) (applied to $\mathbf{m}%
= \ZZ $, $\mathbf{n}= \QQ $ and $\varphi=\rho_{ \QQ }$) yields%
\begin{equation}
\mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }=\left(  \rho_{ \QQ %
}\left[  \left[  \xx\right]  \right]  \right)  \left(
\underbrace{\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }}%
_{\substack{=\mathbf{GR}_{\lambda}\\\text{(by
(\ref{pf.prop.lyndon.grfun.symm.2nd.3}))}}}\right)  =\left(  \rho_{ \QQ %
}\left[  \left[  \xx\right]  \right]  \right)  \left(  \mathbf{GR}%
_{\lambda}\right)  . \label{pf.prop.lyndon.grfun.symm.2nd.5}%
\end{equation}
Applying $\sigma$ to both sides of this equality, we find
\[
\sigma\left(  \mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }\right)
=\sigma\left(  \left(  \rho_{ \QQ }\left[  \left[  \xx\right]
\right]  \right)  \left(  \mathbf{GR}_{\lambda}\right)  \right)  =\left(
\rho_{ \QQ }\left[  \left[  \xx\right]  \right]  \right)  \left(
\sigma\left(  \mathbf{GR}_{\lambda}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lyndon.grfun.symm.2nd.4}%
)}\right)  .
\]
Comparing this with%
\begin{align*}
\sigma\left(  \mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }\right)   &
=\mathbf{GR}_{\lambda}^{\left[   \QQ \right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lyndon.grfun.symm.2nd.sig}%
)}\right) \\
&  =\left(  \rho_{ \QQ }\left[  \left[  \xx\right]  \right]
\right)  \left(  \mathbf{GR}_{\lambda}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.prop.lyndon.grfun.symm.2nd.5})}\right)  ,
\end{align*}
we obtain
\begin{equation}
\left(  \rho_{ \QQ }\left[  \left[  \xx\right]  \right]  \right)
\left(  \sigma\left(  \mathbf{GR}_{\lambda}\right)  \right)  =\left(
\rho_{ \QQ }\left[  \left[  \xx\right]  \right]  \right)  \left(
\mathbf{GR}_{\lambda}\right)  . \label{pf.prop.lyndon.grfun.symm.2nd.6}%
\end{equation}


But the map $\rho_{ \QQ }\left[  \left[  \xx\right]  \right]  $ is
injective. Hence, if $f$ and $g$ are two elements of $ \ZZ \left[
\left[  \xx\right]  \right]  $ satisfying $\left(  \rho_{ \QQ %
}\left[  \left[  \xx\right]  \right]  \right)  \left(  f\right)
=\left(  \rho_{ \QQ }\left[  \left[  \xx\right]  \right]  \right)
\left(  g\right)  $, then $f=g$. Applying this to $f=\sigma\left(
\mathbf{GR}_{\lambda}\right)  $ and $g=\mathbf{GR}_{\lambda}$, we obtain
$\sigma\left(  \mathbf{GR}_{\lambda}\right)  =\mathbf{GR}_{\lambda}$ (because
(\ref{pf.prop.lyndon.grfun.symm.2nd.6}) holds).

Forget that we fixed $\sigma$. We thus have shown that $\sigma\left(
\mathbf{GR}_{\lambda}\right)  =\mathbf{GR}_{\lambda}$ for all $\sigma
\in\Symm_{\left(  \infty\right)  }$. Hence, $\mathbf{GR}_{\lambda}$ is
an $f\in R_{\kk}\left(  \xx\right)  $ that satisfies
$\sigma\left(  f\right)  =f$ for all $\sigma\in\Symm_{\left(
\infty\right)  }$ (since $\mathbf{GR}_{\lambda}\in R_{\kk}\left(
\xx\right)  $). In other words,
\[
\mathbf{GR}_{\lambda}\in\left\{  f\in R_{\kk}\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}  .
\]
In view of (\ref{pf.prop.lyndon.grfun.symm.2nd.Lamk=}), this rewrites as
$\mathbf{GR}_{\lambda}\in\Lambda_{\kk}$. Hence, $\mathbf{GR}_{\lambda
}\in\Lambda_{\kk} = \Lambda$. In other words, the power series
$\mathbf{GR}_{\lambda}$ belongs to $\Lambda$. In other words, Proposition
\ref{prop.lyndon.grfun.symm} holds if $\kk = \ZZ$. This
completes Step 2.

\underline{\textit{Step 3:}} We shall now prove that Proposition
\ref{prop.lyndon.grfun.symm} holds in the general case.

Let us use all the notations that we introduced in Step 2.

In Step 2, we have shown that Proposition \ref{prop.lyndon.grfun.symm} holds
if $\kk = \ZZ$. In other words, the power series $\mathbf{GR}%
_{\lambda}$ belongs to $\Lambda$ if $\kk = \ZZ$. In other words,
the power series $\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }$ belongs
to $\Lambda_{ \ZZ }$. Hence,%
\[
\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }\in\Lambda_{ \ZZ %
}=\left\{  f\in R_{ \ZZ }\left(  \xx\right)  :\sigma\left(
f\right)  =f\text{ for all }\sigma\in\Symm_{\left(  \infty\right)
}\right\}
\]
(by (\ref{pf.prop.lyndon.grfun.symm.2nd.Lamk=}), applied to $\ZZ$
instead of $\kk$). In other words, $\mathbf{GR}_{\lambda}^{\left[
 \ZZ \right]  }$ is an $f\in R_{ \ZZ }\left(  \xx\right)  $
that satisfies $\sigma\left(  f\right)  =f$ for all $\sigma\in\mathfrak{S}%
_{\left(  \infty\right)  }$. In other words, $\mathbf{GR}_{\lambda}^{\left[
 \ZZ \right]  }$ belongs to $R_{ \ZZ }\left(  \xx\right)  $
and satisfies%
\begin{equation}
\sigma\left(  \mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }\right)
=\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }%
\ \ \ \ \ \ \ \ \ \ \text{for all }\sigma\in\Symm_{\left(
\infty\right)  }. \label{pf.prop.lyndon.grfun.symm.3rd.Z}%
\end{equation}


Now, let us recall that our goal (in Step 3) is to show that Proposition
\ref{prop.lyndon.grfun.symm} holds. In other words, our goal is to show that
the power series $\mathbf{GR}_{\lambda}$ belongs to $\Lambda$. Here, of
course, $\mathbf{GR}_{\lambda}=\mathbf{GR}_{\lambda}^{\left[  \kk
\right]  }$ and $\Lambda=\Lambda_{\kk}$.

Lemma \ref{lem.sol.lyndon.grfun.bdd} yields that the power series
$\mathbf{GR}_{\lambda}$ is homogeneous of degree $\left\vert \lambda
\right\vert $. Hence, the power series $\mathbf{GR}_{\lambda}$ is of bounded
degree (since any homogeneous power series is of bounded degree). In other
words, $\mathbf{GR}_{\lambda}\in R\left(  \xx\right)  $ (since
$R\left(  \xx\right)  $ is the set of all formal power series
$f\in \kk \left[  \left[  \xx\right]  \right]  $ that are of
bounded degree). In other words, $\mathbf{GR}_{\lambda}\in R_{\mathbf{k}%
}\left(  \xx\right)  $ (since $R\left(  \xx\right)
=R_{\mathbf{k}}\left(  \xx\right)  $).

Let $\sigma\in\Symm_{\left(  \infty\right)  }$. Then, we can apply
(\ref{pf.prop.lyndon.grfun.symm.2nd.func-sigma}) to $\mathbf{m}= \ZZ $,
$\mathbf{n}=\mathbf{k}$, $\varphi=\rho_{\mathbf{k}}$ and $f=\mathbf{GR}%
_{\lambda}^{\left[   \ZZ \right]  }$ (since $\rho_{\mathbf{k}}%
: \ZZ \rightarrow\mathbf{k}$ is a ring homomorphism). We thus obtain%
\begin{align}
\sigma\left(  \left(  \rho_{\mathbf{k}}\left[  \left[  \xx\right]
\right]  \right)  \left(  \mathbf{GR}_{\lambda}^{\left[   \ZZ \right]
}\right)  \right)   &  =\left(  \rho_{\mathbf{k}}\left[  \left[
\xx\right]  \right]  \right)  \left(  \underbrace{\sigma\left(
\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }\right)  }%
_{\substack{=\mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }\\\text{(by
(\ref{pf.prop.lyndon.grfun.symm.3rd.Z}))}}}\right) \nonumber\\
&  =\left(  \rho_{\mathbf{k}}\left[  \left[  \xx\right]  \right]
\right)  \left(  \mathbf{GR}_{\lambda}^{\left[   \ZZ \right]  }\right)  .
\label{pf.prop.lyndon.grfun.symm.3rd.1}%
\end{align}
But (\ref{pf.prop.lyndon.grfun.symm.2nd.func-GR}) (applied to $\mathbf{m}%
= \ZZ $, $\mathbf{n}=\mathbf{k}$ and $\varphi=\rho_{\mathbf{k}}$) yields%
\begin{equation}
\mathbf{GR}_{\lambda}^{\left[  \mathbf{k}\right]  }=\left(  \rho_{\mathbf{k}%
}\left[  \left[  \xx\right]  \right]  \right)  \left(  \mathbf{GR}%
_{\lambda}^{\left[   \ZZ \right]  }\right)  .
\label{pf.prop.lyndon.grfun.symm.3rd.2}%
\end{equation}
Applying $\sigma$ to both sides of this equality, we find
\begin{align*}
\sigma\left(  \mathbf{GR}_{\lambda}^{\left[  \mathbf{k}\right]  }\right)   &
=\sigma\left(  \left(  \rho_{\mathbf{k}}\left[  \left[  \xx\right]
\right]  \right)  \left(  \mathbf{GR}_{\lambda}^{\left[   \ZZ \right]
}\right)  \right)  =\left(  \rho_{\mathbf{k}}\left[  \left[  \xx%
\right]  \right]  \right)  \left(  \mathbf{GR}_{\lambda}^{\left[
 \ZZ \right]  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.lyndon.grfun.symm.3rd.1})}\right) \\
&  =\mathbf{GR}_{\lambda}^{\left[  \mathbf{k}\right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.lyndon.grfun.symm.3rd.2}%
)}\right)  .
\end{align*}
This rewrites as $\sigma\left(  \mathbf{GR}_{\lambda}\right)  =\mathbf{GR}%
_{\lambda}$ (since $\mathbf{GR}_{\lambda}=\mathbf{GR}_{\lambda}^{\left[
\mathbf{k}\right]  }$).

Forget that we fixed $\sigma$. We thus have shown that $\sigma\left(
\mathbf{GR}_{\lambda}\right)  =\mathbf{GR}_{\lambda}$ for all $\sigma
\in\Symm_{\left(  \infty\right)  }$. Hence, $\mathbf{GR}_{\lambda}$ is
an $f\in R_{\mathbf{k}}\left(  \xx\right)  $ that satisfies
$\sigma\left(  f\right)  =f$ for all $\sigma\in\Symm_{\left(
\infty\right)  }$ (since $\mathbf{GR}_{\lambda}\in R_{\mathbf{k}}\left(
\xx\right)  $). In other words,
\[
\mathbf{GR}_{\lambda}\in\left\{  f\in R_{\mathbf{k}}\left(  \xx\right)
:\sigma\left(  f\right)  =f\text{ for all }\sigma\in\Symm_{\left(
\infty\right)  }\right\}  .
\]
In view of (\ref{pf.prop.lyndon.grfun.symm.2nd.Lamk=}), this rewrites as
$\mathbf{GR}_{\lambda}\in\Lambda_{\mathbf{k}}$. Hence, $\mathbf{GR}_{\lambda
}\in\Lambda_{\mathbf{k}}=\Lambda$. In other words, the power series
$\mathbf{GR}_{\lambda}$ belongs to $\Lambda$. In other words, Proposition
\ref{prop.lyndon.grfun.symm} holds in the general case. This completes Step 3
and, with it, the second proof of Proposition \ref{prop.lyndon.grfun.symm}.
\end{proof}

\end{verlong}

Finally, let us prove Proposition \ref{prop.lyndon.grfun.n}:

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.n}.](a) We proceed by showing
some auxiliary claims:

\begin{statement}
\textit{Claim 1:} Let $w\in\mathfrak{A}^{\ast}$ satisfy
$\operatorname*{CFLtype}w=\left(  n\right)  $. Then, the word $w$ is Lyndon
and satisfies $w\in\mathfrak{A}^{n}$.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 1:} Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(a)
(applied to $\lambda=\left(  n\right)  $) yields $\ell\left(  w\right)
=\left\vert \left(  n\right)  \right\vert =n$. Hence, $w\in\mathfrak{A}^{n}$.

Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Thus, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon
words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq
a_{k}$ (by the definition of ``CFL factorization''). But Lemma
\ref{lem.sol.lyndon.grfun.sizeCFLtype}(b) (applied to $\lambda=\left(
n\right)  $) yields $k=\ell\left(  \left(  n\right)  \right)  =1$ (since $n$
is a positive integer). Hence, $a_{1}a_{2}\cdots a_{k}=a_{1}$. Thus,
$w=a_{1}a_{2}\cdots a_{k}=a_{1}$. But $a_{1}$ is a Lyndon word (since $\left(
a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words). In other
words, $w$ is a Lyndon word (since $w=a_{1}$). Thus, Claim 1 is proven.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 1:} Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(a)
(applied to $\lambda=\left(  n\right)  $) yields $\ell\left(  w\right)
=\left\vert \left(  n\right)  \right\vert =n$. Hence, $w\in\mathfrak{A}^{n}$.

Let $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ be the CFL factorization of
$w$. Thus, $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon
words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq
a_{k}$ (by the definition of ``CFL factorization''). Thus, in particular,
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is a tuple of Lyndon words.
In other words, $a_{1},a_{2},\ldots,a_{k}$ are Lyndon words. Furthermore,
Lemma \ref{lem.sol.lyndon.grfun.sizeCFLtype}(b) (applied to $\lambda=\left(
n\right)  $) yields $k=\ell\left(  \left(  n\right)  \right)  =1$ (since $n$
is a positive integer). Hence, $a_{1}a_{2}\cdots a_{k}=a_{1}a_{2}\cdots
a_{1}=a_{1}$. Thus, $w=a_{1}a_{2}\cdots a_{k}=a_{1}$. But $1\in\left\{
1,2,\ldots,k\right\}  $ (since $k=1$). Hence, $a_{1}$ is a Lyndon word (since
$a_{1},a_{2},\ldots,a_{k}$ are Lyndon words). In other words, $w$ is a Lyndon
word (since $w=a_{1}$). Hence, we have shown that $w$ is a Lyndon word and
satisfies $w\in\mathfrak{A}^{n}$. This proves Claim 1.]
\end{verlong}

\begin{statement}
\textit{Claim 2:} Let $w\in\mathfrak{A}^{n}$ be a Lyndon word. Then,
$w\in\mathfrak{A}^{\ast}$ and $\operatorname*{CFLtype}w=\left(  n\right)  $.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 2:} We have $w\in\mathfrak{A}^{n}\subset
\mathfrak{A}^{\ast}$. It remains to prove that $\operatorname*{CFLtype}%
w=\left(  n\right)  $.

The word $w$ is Lyndon. Hence, the definition of a
``CFL factorization'' shows that the $1$-tuple $\left(  w\right)  $
is a CFL factorization of $w$. We can replace ``a CFL factorization''
by ``the CFL factorization'' in this sentence (since Theorem
\ref{thm.words.CFL} shows that the CFL factorization of $w$ is unique). Thus,
we conclude that the $1$-tuple $\left(  w\right)  $ is the CFL factorization
of $w$. Hence, the definition of $\operatorname*{CFLtype}w$ shows that%
\[
\operatorname*{CFLtype}w=\left(  \ell\left(  w\right)  \right)  =\left(
n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  w\right)
=n\text{ (because }w\in\mathfrak{A}^{n}\text{)}\right)  .
\]
This concludes the proof of Claim 2.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 2:} We have $w\in\mathfrak{A}^{n}\subset
\mathfrak{A}^{\ast}$. It remains to prove that $\operatorname*{CFLtype}%
w=\left(  n\right)  $.

We have $\ell\left(  w\right)  =n$ (since $w\in\mathfrak{A}^{n}$).

The word $w$ is a Lyndon word. Thus, the $1$-tuple $\left(  w\right)  $ is a
tuple of Lyndon words. If we denote this $1$-tuple $\left(  w\right)  $ by
$\left(  a_{1},a_{2},\ldots,a_{k}\right)  $, then $\left(  a_{1},a_{2}%
,\ldots,a_{k}\right)  $ is a tuple of Lyndon words (since $\left(  w\right)  $
is a tuple of Lyndon words) satisfying $w=a_{1}a_{2}\cdots a_{k}$ (indeed,
this equality holds for $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(
w\right)  $, because it boils down to the obvious equality $w=w$) and
$a_{1}\geq a_{2}\geq\cdots\geq a_{k}$ (indeed, this chain of inequalities is
vacuously true for $\left(  a_{1},a_{2},\ldots,a_{k}\right)  =\left(
w\right)  $ because it contains no inequality signs). Thus, the $1$-tuple
$\left(  w\right)  $ is a tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ of
Lyndon words satisfying $w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}%
\geq\cdots\geq a_{k}$. In other words, the $1$-tuple $\left(  w\right)  $ is a
CFL factorization of $w$ (since a CFL factorization of $w$ is defined to be a
tuple $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ of Lyndon words satisfying
$w=a_{1}a_{2}\cdots a_{k}$ and $a_{1}\geq a_{2}\geq\cdots\geq a_{k}$). We can
replace ``a CFL factorization'' by
``the CFL factorization'' in this sentence
(since Theorem \ref{thm.words.CFL} shows that the CFL factorization of $w$ is
unique). Thus, we conclude that the $1$-tuple $\left(  w\right)  $ is the CFL
factorization of $w$.

But recall that the CFL type of $w$ was defined to be the partition whose
parts are the positive integers $\ell\left(  a_{1}\right)  ,\ell\left(
a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $ (listed in decreasing
order), where $\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ is the CFL
factorization of $w$. In other words, if $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  $ is the CFL factorization of $w$, then the CFL type of $w$ is
the partition whose parts are the positive integers $\ell\left(  a_{1}\right)
,\ell\left(  a_{2}\right)  ,\ldots,\ell\left(  a_{k}\right)  $ (listed in
decreasing order). We can apply this to $\left(  a_{1},a_{2},\ldots
,a_{k}\right)  =\left(  w\right)  $ (since the $1$-tuple $\left(  w\right)  $
is the CFL factorization of $w$). We thus conclude that the CFL type of $w$ is
the partition whose parts are the positive integers $\ell\left(  w\right)  $
(listed in decreasing order)\footnote{Of course, the words
``listed in decreasing order'' are redundant here, since there
is only one order in which the single positive integer $\ell\left(  w\right)
$ can be listed.}. Hence,%
\begin{align*}
&  \left(  \text{the CFL type of }w\right) \\
&  =\left(  \text{the partition whose parts are the positive integers }%
\ell\left(  w\right)  \text{ (listed in decreasing order)}\right) \\
&  =\left(  \ell\left(  w\right)  \right)  =\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\ell\left(  w\right)  =n\right)  .
\end{align*}


The definition of $\operatorname*{CFLtype}w$ yields $\operatorname*{CFLtype}%
w=\left(  \text{the CFL type of }w\right)  =\left(  n\right)  $. This
concludes the proof of Claim 2.]
\end{verlong}

\begin{statement}
\textit{Claim 3:} We have $\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  =\left\{
w\in\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}\right\}  $.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 3:} Claim 1 shows that $\left\{  w\in\mathfrak{A}%
^{\ast}\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}
\subset\left\{  w\in\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}\right\}  $.
Claim 2 shows that $\left\{  w\in\mathfrak{A}^{n}\ \mid\ w\text{ is
Lyndon}\right\}  \subset\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  $. Combining these two
inclusions, we obtain precisely the equality claimed in Claim 3.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 3:} From Claim 1, we easily obtain $\left\{
w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)
\right\}  \subset\left\{  w\in\mathfrak{A}^{n}\ \mid\ w\text{ is
Lyndon}\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $p\in\left\{
w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)
\right\}  $. Thus, $p$ is a $w\in\mathfrak{A}^{\ast}$ satisfying
$\operatorname*{CFLtype}w=\left(  n\right)  $. In other words, $p$ belongs to
$\mathfrak{A}^{\ast}$ and satisfies $\operatorname*{CFLtype}p=\left(
n\right)  $. Hence, Claim 1 (applied to $w=p$) yields that the word $p$ is
Lyndon and satisfies $p\in\mathfrak{A}^{n}$. In other words, $p$ is an element
of $\mathfrak{A}^{n}$ and is Lyndon. In other words, $p$ is a $w\in
\mathfrak{A}^{n}$ such that $w$ is Lyndon. In other words, $p\in\left\{
w\in\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}\right\}  $.
\par
Forget that we fixed $p$. We thus have shown that $p\in\left\{  w\in
\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}\right\}  $ for each $p\in\left\{
w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)
\right\}  $. In other words, $\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  \subset\left\{
w\in\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}\right\}  $.}. From Claim 2, we
easily obtain $\left\{  w\in\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}%
\right\}  \subset\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Let $p\in\left\{  w\in\mathfrak{A}^{n}%
\ \mid\ w\text{ is Lyndon}\right\}  $. Thus, $p$ is a $w\in\mathfrak{A}^{n}$
such that $w$ is Lyndon. In other words, $p$ belongs to $\mathfrak{A}^{n}$ and
is Lyndon. Hence, Claim 2 (applied to $w=p$) yields that $p\in\mathfrak{A}%
^{\ast}$ and $\operatorname*{CFLtype}p=\left(  n\right)  $. In other words,
$p$ is a $w\in\mathfrak{A}^{\ast}$ such that $\operatorname*{CFLtype}w=\left(
n\right)  $. In other words, $p\in\left\{  w\in\mathfrak{A}^{\ast}%
\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  $.
\par
Forget that we fixed $p$. We thus have shown that $p\in\left\{  w\in
\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)
\right\}  $ for each $p\in\left\{  w\in\mathfrak{A}^{n}\ \mid\ w\text{ is
Lyndon}\right\}  $. In other words, $\left\{  w\in\mathfrak{A}^{n}%
\ \mid\ w\text{ is Lyndon}\right\}  \subset\left\{  w\in\mathfrak{A}^{\ast
}\ \mid\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  $.}. Combining
these two inclusions, we obtain
\[
\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ \operatorname*{CFLtype}w=\left(
n\right)  \right\}  =\left\{  w\in\mathfrak{A}^{n}\ \mid\ w\text{ is
Lyndon}\right\}  .
\]
This proves Claim 3.]
\end{verlong}

\begin{vershort}
The definition of $\mathbf{GR}_{\left(  n\right)  }$ yields%
\[
\mathbf{GR}_{\left(  n\right)  }=\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\left(  n\right)  }%
}}_{\substack{=\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\w\text{ is
Lyndon}}}\\\text{(by Claim 3)}}}\xx_{w}=\sum\limits_{\substack{w\in
\mathfrak{A}^{n};\\w\text{ is Lyndon}}}\xx_{w}.
\]
This proves Proposition \ref{prop.lyndon.grfun.n}(a).
\end{vershort}

\begin{verlong}
Claim 3 shows that $\left\{  w\in\mathfrak{A}^{\ast}\ \mid
\ \operatorname*{CFLtype}w=\left(  n\right)  \right\}  =\left\{
w\in\mathfrak{A}^{n}\ \mid\ w\text{ is Lyndon}\right\}  $. Hence, we have the
following equality of summation signs:%
\[
\sum\limits_{\substack{w\in\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}%
w=\left(  n\right)  }}=\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\w\text{
is Lyndon}}}.
\]
But the definition of $\mathbf{GR}_{\left(  n\right)  }$ yields%
\[
\mathbf{GR}_{\left(  n\right)  }=\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{\ast};\\\operatorname*{CFLtype}w=\left(  n\right)  }}}%
_{=\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\w\text{ is Lyndon}}%
}}\xx_{w}=\sum\limits_{\substack{w\in\mathfrak{A}^{n};\\w\text{ is
Lyndon}}}\xx_{w}.
\]
This proves Proposition \ref{prop.lyndon.grfun.n}(a).
\end{verlong}

(b) Forget that we fixed $n$. We shall first prove several auxiliary claims:

\begin{statement}
\textit{Claim 4:} Let $d$ be a positive integer. Then,%
\[
\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  =\left\{
w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}  .
\]

\end{statement}

\begin{vershort}
[\textit{Proof of Claim 4:} Since $\mathfrak{L}$ is the set of all Lyndon
words in $\mathfrak{A}^{\ast}$, we have%
\begin{align*}
\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}   &
=\left\{  w\text{ is a Lyndon word in }\mathfrak{A}^{\ast}\ \mid\ \ell\left(
w\right)  =d\right\} \\
&  =\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ w\text{ is Lyndon, and }%
\ell\left(  w\right)  =d\right\}  .
\end{align*}
But since $\mathfrak{A}^{d}$ is the set of words $w\in\mathfrak{A}^{\ast}$
satisfying $\ell\left(  w\right)  =d$, we have%
\begin{align*}
\left\{  w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}   &  =\left\{
w\in\mathfrak{A}^{\ast}\text{ satisfying }\ell\left(  w\right)  =d\ \mid
\ w\text{ is Lyndon}\right\} \\
&  =\left\{  w\in\mathfrak{A}^{\ast}\ \mid\ w\text{ is Lyndon, and }%
\ell\left(  w\right)  =d\right\}  .
\end{align*}
Comparing these two equalities, we obtain $\left\{  w\in\mathfrak{L}%
\ \mid\ \ell\left(  w\right)  =d\right\}  =\left\{  w\in\mathfrak{A}^{d}%
\ \mid\ w\text{ is Lyndon}\right\}  $. This proves Claim 4.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 4:} We have $\left\{  w\in\mathfrak{L}\ \mid
\ \ell\left(  w\right)  =d\right\}  \subset\left\{  w\in\mathfrak{A}^{d}%
\ \mid\ w\text{ is Lyndon}\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let
$p\in\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  $.
Thus, $p$ is a $w\in\mathfrak{L}$ satisfying $\ell\left(  w\right)  =d$. In
other words, $p\in\mathfrak{L}$ and $\ell\left(  p\right)  =d$. From
$p\in\mathfrak{L}$, we conclude that $p$ is a Lyndon word in $\mathfrak{A}%
^{\ast}$ (since $\mathfrak{L}$ is the set of Lyndon words in $\mathfrak{A}%
^{\ast}$). Hence, $p\in\mathfrak{A}^{\ast}$, so that $p\in\mathfrak{A}^{d}$
(since $\ell\left(  p\right)  =d$). Also, $p$ is Lyndon. Thus, $p$ is a
$w\in\mathfrak{A}^{d}$ such that $w$ is Lyndon (since $p\in\mathfrak{A}^{d}$).
In other words, $p\in\left\{  w\in\mathfrak{A}^{d}\ \mid\ w\text{ is
Lyndon}\right\}  $.
\par
Forget that we fixed $p$. We thus have showed that $p\in\left\{
w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}  $ for each
$p\in\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  $. In
other words, $\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)
=d\right\}  \subset\left\{  w\in\mathfrak{A}^{d}\ \mid\ w\text{ is
Lyndon}\right\}  $.} and $\left\{  w\in\mathfrak{A}^{d}\ \mid\ w\text{ is
Lyndon}\right\}  \subset\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)
=d\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $p\in\left\{
w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}  $. Thus, $p$ is a
$w\in\mathfrak{A}^{d}$ such that $w$ is Lyndon. In other words, $p$ is an
element of $\mathfrak{A}^{d}$ with the property that $p$ is Lyndon. Since $p$
is an element of $\mathfrak{A}^{d}$, we have $p\in\mathfrak{A}^{d}%
\subset\mathfrak{A}^{\ast}$. Also, from $p\in\mathfrak{A}^{d}$, we obtain
$\ell\left(  p\right)  =d$. But $p$ is a Lyndon word in $\mathfrak{A}^{\ast}$
(since $p$ is Lyndon and satisfies $p\in\mathfrak{A}^{\ast}$). In other words,
$p\in\mathfrak{L}$ (since $\mathfrak{L}$ is the set of Lyndon words in
$\mathfrak{A}^{\ast}$). Thus, $p$ is a $w\in\mathfrak{L}$ such that
$\ell\left(  w\right)  =d$ (since $\ell\left(  p\right)  =d$). In other words,
$p\in\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  $.
\par
Forget that we fixed $p$. We thus have showed that $p\in\left\{
w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  $ for each
$p\in\left\{  w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}  $. In
other words, $\left\{  w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}
\subset\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  $.}.
Combining these two inclusions, we obtain%
\[
\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  =\left\{
w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}  .
\]
This proves Claim 4.]
\end{verlong}

\begin{statement}
\textit{Claim 5:} Let $w\in\mathfrak{L}$. Then, $\ell\left(  w\right)
\in\left\{  1,2,3,\ldots\right\}  $.
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 5:} From $w\in\mathfrak{L}$, we conclude that $w$ is a
Lyndon word in $\mathfrak{A}^{\ast}$ (since $\mathfrak{L}$ is the set of
Lyndon words in $\mathfrak{A}^{\ast}$). Hence, the word $w$ is Lyndon, and
thus nonempty. Therefore, $\ell\left(  w\right)  \in\left\{  1,2,3,\ldots
\right\}  $. This proves Claim 5.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 5:} From $w\in\mathfrak{L}$, we conclude that $w$ is a
Lyndon word in $\mathfrak{A}^{\ast}$ (since $\mathfrak{L}$ is the set of
Lyndon words in $\mathfrak{A}^{\ast}$). Hence, the word $w$ is Lyndon, and
thus nonempty (since any Lyndon word is nonempty). Therefore, $\ell\left(
w\right)  $ is a positive integer. Hence, $\ell\left(  w\right)  \in\left\{
1,2,3,\ldots\right\}  $. This proves Claim 5.]
\end{verlong}

The next two claims concern the results of substituting $x_{1}^{k},x_{2}%
^{k},x_{3}^{k},\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in some power series
$f\in\mathbf{k}\left[  \left[  \xx\right]  \right]  $ (where $k$ is a
given positive integer). Recall that the result of substituting $x_{1}%
^{k},x_{2}^{k},x_{3}^{k},\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in a power
series $f\in\mathbf{k}\left[  \left[  \xx\right]  \right]  $ is denoted
by $f\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  $.

\begin{statement}
\textit{Claim 6:} Let $k$ be a positive integer. Let $w\in\mathfrak{A}^{\ast}%
$. Then,
\[
\xx_{w}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)
=\xx_{w}^{k}%
\]
(where, of course, $\xx_{w}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k}%
,\ldots\right)  $ denotes the result of substituting $x_{1}^{k},x_{2}%
^{k},x_{3}^{k},\ldots$ for $x_{1},x_{2},x_{3},\ldots$ in the polynomial
$\xx_{w}$).
\end{statement}

\begin{vershort}
[\textit{Proof of Claim 6:} Write the word $w$ in the form $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ (for some $n\in \NN $). Then,
$\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ (by the definition of
$\xx_{w}$). Substituting $x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots$ for
$x_{1},x_{2},x_{3},\ldots$ on both sides of this equality, we obtain
\[
\xx_{w}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  =x_{w_{1}%
}^{k}x_{w_{2}}^{k}\cdots x_{w_{n}}^{k}=\left(  \underbrace{x_{w_{1}}x_{w_{2}%
}\cdots x_{w_{n}}}_{=\xx_{w}}\right)  ^{k}=\xx_{w}^{k}.
\]
This proves Claim 6.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 6:} Write the word $w$ in the form $w=\left(
w_{1},w_{2},\ldots,w_{n}\right)  $ (for some $n\in \NN $). Then,
$\xx_{w}=x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}$ (by the definition of
$\xx_{w}$), so that $x_{w_{1}}x_{w_{2}}\cdots x_{w_{n}}=\xx_{w}%
$. Now, substituting $x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots$ for $x_{1}%
,x_{2},x_{3},\ldots$ on both sides of the equality $\xx_{w}=x_{w_{1}%
}x_{w_{2}}\cdots x_{w_{n}}$, we obtain
\[
\xx_{w}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  =x_{w_{1}%
}^{k}x_{w_{2}}^{k}\cdots x_{w_{n}}^{k}=\left(  \underbrace{x_{w_{1}}x_{w_{2}%
}\cdots x_{w_{n}}}_{=\xx_{w}}\right)  ^{k}=\xx_{w}^{k}.
\]
This proves Claim 6.]
\end{verlong}

\begin{statement}
\textit{Claim 7:} Let $k$ and $m$ be positive integers. Then,%
\[
p_{m}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  =p_{km}%
\]
(where, of course, $p_{m}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)
$ denotes the result of substituting $x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots$
for $x_{1},x_{2},x_{3},\ldots$ in the power series $p_{m}$).
\end{statement}

[\textit{Proof of Claim 7:} The definition of $p_{km}$ yields $p_{km}%
=x_{1}^{km}+x_{2}^{km}+x_{3}^{km}+\cdots$.

On the other hand, the definition of $p_{m}$ yields $p_{m}=x_{1}^{m}+x_{2}%
^{m}+x_{3}^{m}+\cdots=\sum_{i\geq1}x_{i}^{m}$. Substituting $x_{1}^{k}%
,x_{2}^{k},x_{3}^{k},\ldots$ for $x_{1},x_{2},x_{3},\ldots$ on both sides of
this equality, we obtain%
\[
p_{m}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  =\sum_{i\geq
1}\underbrace{\left(  x_{i}^{k}\right)  ^{m}}_{=x_{i}^{km}}=\sum_{i\geq1}%
x_{i}^{km}=x_{1}^{km}+x_{2}^{km}+x_{3}^{km}+\cdots=p_{km}%
\]
(since $p_{km}=x_{1}^{km}+x_{2}^{km}+x_{3}^{km}+\cdots$). This proves Claim 7.]

\begin{statement}
\textit{Claim 8:} Let $k$ and $d$ be positive integers. Then,%
\begin{equation}
\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k}%
,\ldots\right)  =\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(
w\right)  =d}}\xx_{w}^{k}. \label{pf.prop.lyndon.grfun.n.b.c8.claim}%
\end{equation}

\end{statement}

\begin{vershort}
[\textit{Proof of Claim 8:} Because of Claim 4, we have the following equality
of summation signs:%
\begin{equation}
\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}%
}=\sum\limits_{\substack{w\in\mathfrak{A}^{d};\\w\text{ is Lyndon}}}.
\label{pf.prop.lyndon.grfun.n.b.c8.pf.short.sum=sum}%
\end{equation}
But Proposition \ref{prop.lyndon.grfun.n}(a) (applied to $n=d$) yields
\[
\mathbf{GR}_{\left(  d\right)  }=\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{d};\\w\text{ is Lyndon}}}}_{\substack{=\sum
\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}}\\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c8.pf.short.sum=sum}))}}}\xx_{w}%
=\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=d}}\xx_{w}.
\]
Substituting $x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots$ for $x_{1},x_{2}%
,x_{3},\ldots$ on both sides of this equality, we obtain%
\[
\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k}%
,\ldots\right)  =\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(
w\right)  =d}}\underbrace{\xx_{w}\left(  x_{1}^{k},x_{2}^{k},x_{3}%
^{k},\ldots\right)  }_{\substack{=\xx_{w}^{k}\\\text{(by Claim 6)}%
}}=\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=d}}\xx_{w}^{k}.
\]
This proves Claim 8.]
\end{vershort}

\begin{verlong}
[\textit{Proof of Claim 8:} Claim 4 yields%
\[
\left\{  w\in\mathfrak{L}\ \mid\ \ell\left(  w\right)  =d\right\}  =\left\{
w\in\mathfrak{A}^{d}\ \mid\ w\text{ is Lyndon}\right\}  .
\]
Hence, we have the following equality of summation signs:%
\begin{equation}
\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}%
}=\sum\limits_{\substack{w\in\mathfrak{A}^{d};\\w\text{ is Lyndon}}}.
\label{pf.prop.lyndon.grfun.n.b.c8.pf.sum=sum}%
\end{equation}


But Proposition \ref{prop.lyndon.grfun.n}(a) (applied to $n=d$) yields
\[
\mathbf{GR}_{\left(  d\right)  }=\underbrace{\sum\limits_{\substack{w\in
\mathfrak{A}^{d};\\w\text{ is Lyndon}}}}_{\substack{=\sum
\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}}\\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c8.pf.sum=sum}))}}}\xx_{w}%
=\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=d}}\xx_{w}.
\]
Substituting $x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots$ for $x_{1},x_{2}%
,x_{3},\ldots$ on both sides of this equality, we obtain%
\begin{align*}
\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k}%
,\ldots\right)   &  =\left(  \sum\limits_{\substack{w\in\mathfrak{L}%
;\\\ell\left(  w\right)  =d}}\xx_{w}\right)  \left(  x_{1}^{k}%
,x_{2}^{k},x_{3}^{k},\ldots\right) \\
&  =\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=d}}\underbrace{\xx_{w}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k}%
,\ldots\right)  }_{\substack{=\xx_{w}^{k}\\\text{(by Claim 6)}}%
}=\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=d}}\xx_{w}^{k}.
\end{align*}
This proves Claim 8.]
\end{verlong}

For each positive integer $n$, we define a symmetric function
$\widetilde{\mathbf{GR}}_{n}\in\Lambda$ by%
\[
\widetilde{\mathbf{GR}}_{n}=\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)
p_{d}^{n/d}.
\]
Our goal will be to prove that $\mathbf{GR}_{\left(  n\right)  }%
=\widetilde{\mathbf{GR}}_{n}$ for each positive integer $n$.

The following two claims are key to our proof:

\begin{statement}
\textit{Claim 9:} Every positive integer $n$ satisfies%
\[
p_{1}^{n}=\sum_{d\mid n}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  .
\]

\end{statement}

[\textit{Proof of Claim 9:} It is well-known that%
\begin{equation}
-\log\left(  1-t\right)  =\sum_{k\geq1}\dfrac{1}{k}t^{k}
\label{pf.prop.lyndon.grfun.n.b.c9.pf.mercator}%
\end{equation}
in the ring $\mathbf{k}\left[  \left[  t\right]  \right]  $. (Indeed, this is
the well-known Mercator series for the logarithm.)

\begin{vershort}
Proposition \ref{prop.lyndon.grfun.CFLps} yields the equality%
\[
\dfrac{1}{1-p_{1}t}=\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}%
t^{\ell\left(  w\right)  }}%
\]
in the power series ring $\left(  \mathbf{k}\left[  \left[  \xx\right]
\right]  \right)  \left[  \left[  t\right]  \right]  $. Taking the logarithm
of both sides of this identity, we obtain%
\begin{align*}
\log\dfrac{1}{1-p_{1}t}  &  =\log\left(  \prod_{w\in\mathfrak{L}}\dfrac
{1}{1-\xx_{w}t^{\ell\left(  w\right)  }}\right)  =\underbrace{\sum
_{w\in\mathfrak{L}}}_{\substack{=\sum_{d\geq1}\sum\limits_{\substack{w\in
\mathfrak{L};\\\ell\left(  w\right)  =d}}\\\text{(since each }w\in
\mathfrak{L}\text{ satisfies }\ell\left(  w\right)  \in\left\{  1,2,3,\ldots
\right\}  \\\text{(by Claim 5))}}}\log\left(  \dfrac{1}{1-\xx%
_{w}t^{\ell\left(  w\right)  }}\right) \\
&  =\sum_{d\geq1}\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(
w\right)  =d}}\underbrace{\log\left(  \dfrac{1}{1-\xx_{w}t^{\ell\left(
w\right)  }}\right)  }_{\substack{=\log\left(  \dfrac{1}{1-\xx_{w}%
t^{d}}\right)  \\\text{(since }\ell\left(  w\right)  =d\text{)}}}=\sum
_{d\geq1}\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)
=d}}\underbrace{\log\left(  \dfrac{1}{1-\xx_{w}t^{d}}\right)
}_{\substack{=-\log\left(  1-\xx_{w}t^{d}\right)  \\=\sum_{k\geq
1}\dfrac{1}{k}\left(  \xx_{w}t^{d}\right)  ^{k}\\\text{(by substituting
}\xx_{w}t^{d}\text{ for }t\\\text{in
(\ref{pf.prop.lyndon.grfun.n.b.c9.pf.mercator}))}}}\\
&  =\sum_{d\geq1}\underbrace{\sum\limits_{\substack{w\in\mathfrak{L}%
;\\\ell\left(  w\right)  =d}}\sum_{k\geq1}}_{=\sum_{k\geq1}\sum
\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}}}\dfrac{1}%
{k}\underbrace{\left(  \xx_{w}t^{d}\right)  ^{k}}_{=\xx_{w}%
^{k}t^{kd}}=\sum_{d\geq1}\sum_{k\geq1}\sum\limits_{\substack{w\in
\mathfrak{L};\\\ell\left(  w\right)  =d}}\dfrac{1}{k}\xx_{w}^{k}%
t^{kd}=\sum_{d\geq1}\sum_{k\geq1}\dfrac{1}{k}\underbrace{\left(
\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}%
}\xx_{w}^{k}\right)  }_{\substack{=\mathbf{GR}_{\left(  d\right)
}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  \\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c8.claim}))}}}t^{kd}\\
&  =\sum_{d\geq1}\underbrace{\sum_{k\geq1}\dfrac{1}{k}\mathbf{GR}_{\left(
d\right)  }\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  t^{kd}%
}_{\substack{=\sum\limits_{\substack{n\geq1;\\d\mid n}}\dfrac{1}{n/d}\mathbf{GR}%
_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d}%
,\ldots\right)  t^{\left(  n/d\right)  d}\\\text{(here, we have substituted
}n/d\text{ for }k\text{ in the sum)}}}\\
&  =\underbrace{\sum_{d\geq1}\sum\limits_{\substack{n\geq1;\\d\mid n}}}%
_{\substack{=\sum_{n\geq1}\sum_{d\mid n}\\\text{(because the summation sign
``}\sum_{d\mid n}\text{''}\\\text{ranges over
all \textbf{positive} divisors }d\text{ of }n\text{)}}}\dfrac{1}%
{n/d}\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  \underbrace{t^{\left(  n/d\right)  d}}_{=t^{n}}\\
&  =\sum_{n\geq1}\sum_{d\mid n}\dfrac{1}{n/d}\mathbf{GR}_{\left(  d\right)
}\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  t^{n}.
\end{align*}
Comparing this with%
\begin{align*}
\log\dfrac{1}{1-p_{1}t}  &  =-\log\left(  1-p_{1}t\right)  =\sum_{k\geq
1}\dfrac{1}{k}\underbrace{\left(  p_{1}t\right)  ^{k}}_{=p_{1}^{k}t^{k}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{this follows by substituting }%
p_{1}t\text{ for }t\text{ in (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.mercator}%
)}\right) \\
&  =\sum_{k\geq1}\dfrac{1}{k}p_{1}^{k}t^{k}=\sum_{n\geq1}\dfrac{1}{n}p_{1}%
^{n}t^{n},
\end{align*}
we obtain%
\[
\sum_{n\geq1}\dfrac{1}{n}p_{1}^{n}t^{n}=\sum_{n\geq1}\sum_{d\mid n}\dfrac
{1}{n/d}\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  t^{n}.
\]
This is an equality between two formal power series in $t$. If we compare
coefficients on both sides of this equality, then we obtain%
\begin{equation}
\dfrac{1}{n}p_{1}^{n}=\sum_{d\mid n}\dfrac{1}{n/d}\mathbf{GR}_{\left(
d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)
\label{pf.prop.lyndon.grfun.n.b.c9.pf.short.at}%
\end{equation}
for each $n\in\left\{  1,2,3,\ldots\right\}  $.

Now, let $n$ be a positive integer. Thus, $n\in\left\{  1,2,3,\ldots\right\}
$. Hence, the equality (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.short.at}) holds.
Multiplying both sides of this equality by $n$, we find%
\[
p_{1}^{n}=\sum_{d\mid n}\underbrace{n\cdot\dfrac{1}{n/d}}_{=d}\mathbf{GR}%
_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d}%
,\ldots\right)  =\sum_{d\mid n}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  .
\]

\end{vershort}

\begin{verlong}
Proposition \ref{prop.lyndon.grfun.CFLps} yields the equality%
\[
\dfrac{1}{1-p_{1}t}=\prod_{w\in\mathfrak{L}}\dfrac{1}{1-\xx_{w}%
t^{\ell\left(  w\right)  }}%
\]
in the power series ring $\left(  \mathbf{k}\left[  \left[  \xx\right]
\right]  \right)  \left[  \left[  t\right]  \right]  $. Taking the logarithm
of both sides of this identity, we obtain%
\begin{align}
\log\dfrac{1}{1-p_{1}t}  &  =\log\left(  \prod_{w\in\mathfrak{L}}\dfrac
{1}{1-\xx_{w}t^{\ell\left(  w\right)  }}\right)  =\sum_{w\in
\mathfrak{L}}\underbrace{\log\left(  \dfrac{1}{1-\xx_{w}t^{\ell\left(
w\right)  }}\right)  }_{=-\log\left(  1-\xx_{w}t^{\ell\left(  w\right)
}\right)  }\nonumber\\
&  =\sum_{w\in\mathfrak{L}}\left(  -\log\left(  1-\xx_{w}t^{\ell\left(
w\right)  }\right)  \right)  . \label{pf.prop.lyndon.grfun.n.b.c9.pf.1}%
\end{align}
Now, let $w\in\mathfrak{L}$. Then, $\ell\left(  w\right)  \in\left\{
1,2,3,\ldots\right\}  $ (by Claim 5). Hence, $t^{\ell\left(  w\right)  }$ is a
monomial of positive degree in $\left(  \mathbf{k}\left[  \left[
\xx\right]  \right]  \right)  \left[  \left[  t\right]  \right]  $.
Thus, we can substitute $\xx_{w}t^{\ell\left(  w\right)  }$ for $t$ in
the identity (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.mercator}). We thus obtain%
\begin{align}
-\log\left(  1-\xx_{w}t^{\ell\left(  w\right)  }\right)   &
=\sum_{k\geq1}\dfrac{1}{k}\underbrace{\left(  \xx_{w}t^{\ell\left(
w\right)  }\right)  ^{k}}_{=\xx_{w}^{k}\left(  t^{\ell\left(  w\right)
}\right)  ^{k}}=\sum_{k\geq1}\dfrac{1}{k}\xx_{w}^{k}\underbrace{\left(
t^{\ell\left(  w\right)  }\right)  ^{k}}_{=t^{k\cdot\ell\left(  w\right)  }%
}\nonumber\\
&  =\sum_{k\geq1}\dfrac{1}{k}\xx_{w}^{k}t^{k\cdot\ell\left(  w\right)
}. \label{pf.prop.lyndon.grfun.n.b.c9.pf.2}%
\end{align}


Forget that we fixed $w$. We thus have proved
(\ref{pf.prop.lyndon.grfun.n.b.c9.pf.2}) for each $w\in\mathfrak{L}$. Now,
(\ref{pf.prop.lyndon.grfun.n.b.c9.pf.1}) becomes%
\begin{align}
\log\dfrac{1}{1-p_{1}t}  &  =\underbrace{\sum_{w\in\mathfrak{L}}%
}_{\substack{=\sum_{d\in\left\{  1,2,3,\ldots\right\}  }\sum
\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}}\\\text{(since
each }w\in\mathfrak{L}\text{ satisfies }\ell\left(  w\right)  \in\left\{
1,2,3,\ldots\right\}  \\\text{(by Claim 5))}}}\underbrace{\left(  -\log\left(
1-\xx_{w}t^{\ell\left(  w\right)  }\right)  \right)  }_{\substack{=\sum
_{k\geq1}\dfrac{1}{k}\xx_{w}^{k}t^{k\cdot\ell\left(  w\right)
}\\\text{(by (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.2}))}}}\nonumber\\
&  =\sum_{d\in\left\{  1,2,3,\ldots\right\}  }\underbrace{\sum
\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}}\sum_{k\geq1}%
}_{=\sum_{k\geq1}\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(
w\right)  =d}}}\dfrac{1}{k}\xx_{w}^{k}\underbrace{t^{k\cdot\ell\left(
w\right)  }}_{\substack{=t^{kd}\\\text{(since }k\cdot\ell\left(  w\right)
=kd\\\text{(because }\ell\left(  w\right)  =d\text{))}}}\nonumber\\
&  =\sum_{d\in\left\{  1,2,3,\ldots\right\}  }\sum_{k\geq1}\sum
\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}}\dfrac{1}%
{k}\xx_{w}^{k}t^{kd}\nonumber\\
&  =\sum_{d\in\left\{  1,2,3,\ldots\right\}  }\underbrace{\sum_{k\geq1}%
}_{=\sum_{k\in\left\{  1,2,3,\ldots\right\}  }}\dfrac{1}{k}\underbrace{\left(
\sum\limits_{\substack{w\in\mathfrak{L};\\\ell\left(  w\right)  =d}%
}\xx_{w}^{k}\right)  }_{\substack{=\mathbf{GR}_{\left(  d\right)
}\left(  x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  \\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c8.claim}))}}}t^{kd}\nonumber\\
&  =\sum_{d\in\left\{  1,2,3,\ldots\right\}  }\underbrace{\sum_{k\in\left\{
1,2,3,\ldots\right\}  }\dfrac{1}{k}\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right)  t^{kd}}_{\substack{=\sum
_{\substack{n\in\left\{  1,2,3,\ldots\right\}  ;\\d\mid n}}\dfrac{1}%
{n/d}\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  t^{\left(  n/d\right)  d}\\\text{(here, we have
substituted }n/d\text{ for }k\text{ in the sum, since}\\\text{the map
}\left\{  n\in\left\{  1,2,3,\ldots\right\}  \ :\ d\mid n\right\}
\rightarrow\left\{  1,2,3,\ldots\right\}  ,\ n\mapsto n/d\\\text{is a
bijection)}}}\nonumber\\
&  =\underbrace{\sum_{d\in\left\{  1,2,3,\ldots\right\}  }\sum\limits_{\substack{n\in
\left\{  1,2,3,\ldots\right\}  ;\\d\mid n}}}_{=\sum_{n\in\left\{
1,2,3,\ldots\right\}  }\sum\limits_{\substack{d\in\left\{  1,2,3,\ldots
\right\}  ;\\d\mid n}}}\dfrac{1}{n/d}\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  \underbrace{t^{\left(
n/d\right)  d}}_{\substack{=t^{n}\\\text{(since }\left(  n/d\right)
d=n\text{)}}}\nonumber\\
&  =\sum_{n\in\left\{  1,2,3,\ldots\right\}  }\sum\limits_{\substack{d\in
\left\{  1,2,3,\ldots\right\}  ;\\d\mid n}}\dfrac{1}{n/d}\mathbf{GR}_{\left(
d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)
t^{n}\nonumber\\
&  =\sum_{n\in\left\{  1,2,3,\ldots\right\}  }\left(  \sum
\limits_{\substack{d\in\left\{  1,2,3,\ldots\right\}  ;\\d\mid n}}\dfrac
{1}{n/d}\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  \right)  t^{n}.
\label{pf.prop.lyndon.grfun.n.b.c9.pf.psL}%
\end{align}
Comparing this with%
\begin{align*}
\log\dfrac{1}{1-p_{1}t}  &  =-\log\left(  1-p_{1}t\right)  =\underbrace{\sum
_{k\geq1}}_{=\sum_{k\in\left\{  1,2,3,\ldots\right\}  }}\dfrac{1}%
{k}\underbrace{\left(  p_{1}t\right)  ^{k}}_{=p_{1}^{k}t^{k}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{this follows by substituting }%
p_{1}t\text{ for }t\text{ in (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.mercator}%
)}\right) \\
&  =\sum_{k\in\left\{  1,2,3,\ldots\right\}  }\dfrac{1}{k}p_{1}^{k}t^{k}%
=\sum_{n\in\left\{  1,2,3,\ldots\right\}  }\dfrac{1}{n}p_{1}^{n}t^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}k\text{ as }n\right)  ,
\end{align*}
we obtain%
\[
\sum_{n\in\left\{  1,2,3,\ldots\right\}  }\dfrac{1}{n}p_{1}^{n}t^{n}%
=\sum_{n\in\left\{  1,2,3,\ldots\right\}  }\left(  \sum\limits_{\substack{d\in
\left\{  1,2,3,\ldots\right\}  ;\\d\mid n}}\dfrac{1}{n/d}\mathbf{GR}_{\left(
d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  \right)
t^{n}.
\]
This is an equality between two formal power series in $t$. If we compare
coefficients on both sides of this equality, then we obtain%
\begin{equation}
\dfrac{1}{n}p_{1}^{n}=\sum\limits_{\substack{d\in\left\{  1,2,3,\ldots
\right\}  ;\\d\mid n}}\dfrac{1}{n/d}\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)
\label{pf.prop.lyndon.grfun.n.b.c9.pf.at}%
\end{equation}
for each $n\in\left\{  1,2,3,\ldots\right\}  $.

Now, let $n$ be a positive integer. Thus, $n\in\left\{  1,2,3,\ldots\right\}
$. Hence, (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.at}) yields%
\[
\dfrac{1}{n}p_{1}^{n}=\sum\limits_{\substack{d\in\left\{  1,2,3,\ldots
\right\}  ;\\d\mid n}}\dfrac{1}{n/d}\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  .
\]
Multiplying both sides of this equality by $n$, we find%
\begin{align}
p_{1}^{n}  &  =n\cdot\sum\limits_{\substack{d\in\left\{  1,2,3,\ldots\right\}
;\\d\mid n}}\dfrac{1}{n/d}\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}%
^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right) \nonumber\\
&  =\sum\limits_{\substack{d\in\left\{  1,2,3,\ldots\right\}  ;\\d\mid
n}}\underbrace{n\cdot\dfrac{1}{n/d}}_{=d}\mathbf{GR}_{\left(  d\right)
}\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right) \nonumber\\
&  =\sum\limits_{\substack{d\in\left\{  1,2,3,\ldots\right\}  ;\\d\mid
n}}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  . \label{pf.prop.lyndon.grfun.n.b.c9.pf.att}%
\end{align}


Recall that the summation sign ``$\sum_{d\mid n}$'' is understood to
range over all \textbf{positive} divisors
$d$ of $n$. Thus, we have $\sum_{d\mid n}=\sum\limits_{\substack{d\in\left\{
1,2,3,\ldots\right\}  ;\\d\mid n}}$ (an equality of summation signs). Hence,%
\[
\underbrace{\sum_{d\mid n}}_{=\sum\limits_{\substack{d\in\left\{
1,2,3,\ldots\right\}  ;\\d\mid n}}}d\cdot\mathbf{GR}_{\left(  d\right)
}\left(  x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  =\sum
\limits_{\substack{d\in\left\{  1,2,3,\ldots\right\}  ;\\d\mid n}%
}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(  x_{1}^{n/d},x_{2}^{n/d}%
,x_{3}^{n/d},\ldots\right)  .
\]
Comparing this with (\ref{pf.prop.lyndon.grfun.n.b.c9.pf.att}), we obtain%
\[
p_{1}^{n}=\sum_{d\mid n}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  .
\]

\end{verlong}

\noindent This proves Claim 9.]

\begin{statement}
\textit{Claim 10:} Every positive integer $n$ satisfies%
\[
p_{1}^{n}=\sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}%
^{n/d},x_{2}^{n/d},x_{3}^{n/d},\ldots\right)  .
\]

\end{statement}

[\textit{Proof of Claim 10:} Let $n$ be a positive integer.

\begin{vershort}
Let $f$ be a positive divisor of $n$. Thus, $n/f$ is a positive integer. The
definition of $\widetilde{\mathbf{GR}}_{f}$ yields%
\[
\widetilde{\mathbf{GR}}_{f}=\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)
p_{d}^{f/d}.
\]
Substituting $x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots$ for $x_{1}%
,x_{2},x_{3},\ldots$ on both sides of this equality, we obtain%
\begin{align*}
\widetilde{\mathbf{GR}}_{f}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f}%
,\ldots\right)   &  =\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)  \left(
\underbrace{p_{d}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)
}_{\substack{=p_{\left(  n/f\right)  d}\\\text{(by Claim 7, applied to
}k=n/f\text{ and }m=d\text{)}}}\right)  ^{f/d}\\
&  =\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)  p_{\left(  n/f\right)
d}^{f/d}.
\end{align*}
Multiplying both sides of this equality by $f$, we obtain%
\begin{equation}
f\cdot\widetilde{\mathbf{GR}}_{f}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}%
^{n/f},\ldots\right)  =\sum_{d\mid f}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}.\label{pf.prop.lyndon.grfun.n.b.c10.pf.short.f1}%
\end{equation}


Forget that we fixed $f$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.f1}) for each positive divisor $f$
of $n$.

Now,%
\begin{align}
&  \sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d}%
,x_{2}^{n/d},x_{3}^{n/d},\ldots\right) \nonumber\\
&  =\sum_{f\mid n}\underbrace{f\cdot\widetilde{\mathbf{GR}}_{f}\left(
x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)  }_{\substack{=\sum_{d\mid
f}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}\\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.f1}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}d\text{ as }f\right) \nonumber\\
&  =\underbrace{\sum_{f\mid n}\sum_{d\mid f}}_{_{=\sum\limits_{d\mid n}%
\sum\limits_{\substack{f\mid n;\\d\mid f}}}}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}=\sum_{d\mid n}\sum\limits_{\substack{f\mid n;\\d\mid
f}}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.short.1}%
\end{align}


Now, fix a positive divisor $d$ of $n$. Then, $n/d$ is a positive integer.
Hence, the map%
\begin{align*}
\left\{  \text{positive divisors }g\text{ of }n/d\right\}   &  \rightarrow
\left\{  \text{positive divisors }f\text{ of }n\text{ satisfying }d\mid
f\right\}  ,\\
g  &  \mapsto dg
\end{align*}
is a bijection\footnote{Its inverse sends each $e$ to $e/d$.}. Thus, we can
substitute $dg$ for $f$ in the sum $\sum\limits_{\substack{f\mid n;\\d\mid
f}}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}$. We thus obtain%
\begin{equation}
\sum\limits_{\substack{f\mid n;\\d\mid f}}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}=\sum\limits_{g\mid n/d}\mu\left(  d\right)
\underbrace{p_{\left(  n/\left(  dg\right)  \right)  d}^{dg/d}}%
_{\substack{=p_{n/g}^{g}}}=\sum\limits_{g\mid n/d}\mu\left(  d\right)
p_{n/g}^{g}. \label{pf.prop.lyndon.grfun.n.b.c10.pf.short.d1}%
\end{equation}


Forget that we fixed $d$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.d1}) for each positive divisor $d$
of $n$.

Now, (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.1}) becomes%
\begin{align}
\sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d},x_{2}%
^{n/d},x_{3}^{n/d},\ldots\right)   &  =\sum_{d\mid n}\underbrace{\sum
\limits_{\substack{f\mid n;\\d\mid f}}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}}_{\substack{=\sum\limits_{g\mid n/d}\mu\left(  d\right)
p_{n/g}^{g}\\\text{(by (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.d1}))}%
}}\nonumber\\
&  =\sum_{d\mid n}\sum\limits_{g\mid n/d}\mu\left(  d\right)  p_{n/g}^{g}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.short.2}%
\end{align}


But if $d$ and $g$ are two positive integers, then the statement
``$d\mid n$ and $g\mid n/d$'' is equivalent to
the statement ``$g\mid n$ and $d\mid n/g$''%
\ \ \ \ \footnote{Indeed, it is easy to see that both of these statements are
equivalent to ``$dg\mid n$''.}. Hence, we have
the following equality of summation signs:%
\begin{equation}
\sum_{d\mid n}\sum\limits_{g\mid n/d}=\sum_{g\mid n}\sum\limits_{d\mid n/g}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.short.4}%
\end{equation}
Thus, (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.2}) becomes%
\begin{align*}
\sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d},x_{2}%
^{n/d},x_{3}^{n/d},\ldots\right)   &  =\underbrace{\sum_{d\mid n}%
\sum\limits_{g\mid n/d}}_{\substack{=\sum\limits_{g\mid n}\sum\limits_{d\mid
n/g}\\\text{(by (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.short.4}))}}}\mu\left(
d\right)  p_{n/g}^{g}\\
&  =\sum_{g\mid n}\underbrace{\sum\limits_{d\mid n/g}\mu\left(  d\right)
}_{\substack{=\delta_{n/g,1}\\\text{(by \eqref{sol.exe.witt.ghost-equiv.mu*1},
applied}\\\text{to }n/g\text{ instead of }n\text{)}}}p_{n/g}^{g}=\sum_{g\mid
n}\delta_{n/g,1}p_{n/g}^{g}\\
&  =\underbrace{\delta_{n/n,1}}_{\substack{=1\\\text{(since }n/n=1\text{)}%
}}\underbrace{p_{n/n}^{n}}_{=p_{1}^{n}}+\sum\limits_{\substack{g\mid n;\\g\neq
n}}\underbrace{\delta_{n/g,1}}_{\substack{=0\\\text{(since }n/g\neq
1\\\text{(because }g\neq n\text{))}}}p_{n/g}^{g}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}g=n\text{ from the sum}\right) \\
&  =p_{1}^{n}+\underbrace{\sum\limits_{\substack{g\mid n;\\g\neq n}}0p_{n/g}^{g}%
}_{=0}=p_{1}^{n}.
\end{align*}

\end{vershort}

\begin{verlong}
Let $f$ be a positive divisor of $n$. Thus, $n/f$ is a positive integer. The
definition of $\widetilde{\mathbf{GR}}_{f}$ yields%
\[
\widetilde{\mathbf{GR}}_{f}=\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)
p_{d}^{f/d}.
\]
Substituting $x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots$ for $x_{1}%
,x_{2},x_{3},\ldots$ on both sides of this equality, we obtain%
\begin{align*}
\widetilde{\mathbf{GR}}_{f}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f}%
,\ldots\right)   &  =\left(  \dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)
p_{d}^{f/d}\right)  \left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)
\\
&  =\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)  \underbrace{p_{d}%
^{f/d}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)  }_{=\left(
p_{d}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)  \right)
^{f/d}}\\
&  =\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)  \left(  \underbrace{p_{d}%
\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)  }%
_{\substack{=p_{\left(  n/f\right)  d}\\\text{(by Claim 7, applied to
}k=n/f\text{ and }m=d\text{)}}}\right)  ^{f/d}\\
&  =\dfrac{1}{f}\sum_{d\mid f}\mu\left(  d\right)  p_{\left(  n/f\right)
d}^{f/d}.
\end{align*}
Multiplying both sides of this equality by $f$, we obtain%
\begin{equation}
f\cdot\widetilde{\mathbf{GR}}_{f}\left(  x_{1}^{n/f},x_{2}^{n/f},x_{3}%
^{n/f},\ldots\right)  =\sum_{d\mid f}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}. \label{pf.prop.lyndon.grfun.n.b.c10.pf.f1}%
\end{equation}


Forget that we fixed $f$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.f1}) for each positive divisor $f$ of
$n$.

Now,%
\begin{align}
&  \sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d}%
,x_{2}^{n/d},x_{3}^{n/d},\ldots\right) \nonumber\\
&  =\sum_{f\mid n}\underbrace{f\cdot\widetilde{\mathbf{GR}}_{f}\left(
x_{1}^{n/f},x_{2}^{n/f},x_{3}^{n/f},\ldots\right)  }_{\substack{=\sum_{d\mid
f}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}\\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.f1}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}d\text{ as }f\right) \nonumber\\
&  =\sum_{f\mid n}\underbrace{\sum_{d\mid f}}_{\substack{=\sum
\limits_{\substack{d\mid f;\\d\mid n}}\\\text{(since each positive divisor
}d\text{ of }f\\\text{satisfies }d\mid n\text{ (because }d\mid f\mid
n\text{))}}}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}\nonumber\\
&  =\sum_{f\mid n}\underbrace{\sum\limits_{\substack{d\mid f;\\d\mid n}%
}}_{\substack{=\sum\limits_{\substack{d\mid n;\\d\mid f}}\\\text{(since the
positive divisors }d\text{ of }f\\\text{satisfying }d\mid n\text{ are
precisely the}\\\text{positive divisors }d\text{ of }n\text{ satisfying }d\mid
f\text{)}}}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}%
=\underbrace{\sum_{f\mid n}\sum\limits_{\substack{d\mid n;\\d\mid f}}}%
_{=\sum_{d\mid n}\sum\limits_{\substack{f\mid n;\\d\mid f}}}\mu\left(
d\right)  p_{\left(  n/f\right)  d}^{f/d}\nonumber\\
&  =\sum_{d\mid n}\sum\limits_{\substack{f\mid n;\\d\mid f}}\mu\left(
d\right)  p_{\left(  n/f\right)  d}^{f/d}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.1}%
\end{align}


Now, fix a positive divisor $d$ of $n$. Then, $n/d$ is a positive integer.
Hence, the map%
\begin{align*}
\left\{  \text{positive divisors }g\text{ of }n/d\right\}   &  \rightarrow
\left\{  \text{positive divisors }f\text{ of }n\text{ satisfying }d\mid
f\right\}  ,\\
g  &  \mapsto dg
\end{align*}
is a bijection\footnote{Its inverse sends each $e$ to $e/d$.}. Thus, we can
substitute $dg$ for $f$ in the sum $\sum\limits_{\substack{f\mid n;\\d\mid
f}}\mu\left(  d\right)  p_{\left(  n/f\right)  d}^{f/d}$. We thus obtain%
\begin{align}
\sum\limits_{\substack{f\mid n;\\d\mid f}}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}  &  =\sum\limits_{g\mid n/d}\mu\left(  d\right)
\underbrace{p_{\left(  n/\left(  dg\right)  \right)  d}^{dg/d}}%
_{\substack{=p_{n/g}^{g}\\\text{(since }dg/d=g\\\text{and }\left(  n/\left(
dg\right)  \right)  d=n/g\text{)}}}\nonumber\\
&  =\sum\limits_{g\mid n/d}\mu\left(  d\right)  p_{n/g}^{g}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.d1}%
\end{align}


Forget that we fixed $d$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.d1}) for each positive divisor $d$ of
$n$.

Now, (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.1}) becomes%
\begin{align}
\sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d},x_{2}%
^{n/d},x_{3}^{n/d},\ldots\right)   &  =\sum_{d\mid n}\underbrace{\sum
\limits_{\substack{f\mid n;\\d\mid f}}\mu\left(  d\right)  p_{\left(
n/f\right)  d}^{f/d}}_{\substack{=\sum\limits_{g\mid n/d}\mu\left(  d\right)
p_{n/g}^{g}\\\text{(by (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.d1}))}%
}}\nonumber\\
&  =\sum_{d\mid n}\sum\limits_{g\mid n/d}\mu\left(  d\right)  p_{n/g}^{g}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.2}%
\end{align}


But we have the following equality of summation signs:%
\begin{align}
\sum_{d\mid n}\underbrace{\sum\limits_{g\mid n/d}}_{\substack{=\sum
\limits_{\substack{g\mid n/d;\\g\mid n}}\\\text{(since each positive divisor
}g\text{ of }n/d\\\text{satisfies }g\mid n\text{ (because }g\mid n/d\mid
n\\\text{(since }d\text{ is a positive divisor of }n\text{)))}}}  &
=\sum_{d\mid n}\underbrace{\sum\limits_{\substack{g\mid n/d;\\g\mid n}%
}}_{\substack{=\sum\limits_{\substack{g\mid n;\\g\mid n/d}}\\\text{(since the
positive divisors }g\text{ of }n/d\text{ satisfying }g\mid n\\\text{are
precisely the positive divisors }g\text{ of }n\text{ satisfying }g\mid
n/d\text{)}}}\nonumber\\
&  =\sum_{d\mid n}\underbrace{\sum\limits_{\substack{g\mid n;\\g\mid n/d}%
}}_{\substack{=\sum\limits_{\substack{g\mid n;\\\left(  n/d\right)
/g\in \ZZ }}\\\text{(since the condition }g\mid n/d\\\text{is equivalent
to }\left(  n/d\right)  /g\in \ZZ \text{)}}}\nonumber\\
&  =\sum_{d\mid n}\sum\limits_{\substack{g\mid n;\\\left(  n/d\right)
/g\in \ZZ }}. \label{pf.prop.lyndon.grfun.n.b.c10.pf.3a}%
\end{align}
The same argument (with the roles of $d$ and $g$ interchanged) yields the
following equality of summation signs:%
\begin{equation}
\sum_{g\mid n}\sum\limits_{d\mid n/g}=\sum_{g\mid n}\sum
\limits_{\substack{d\mid n;\\\left(  n/g\right)  /d\in \ZZ }}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.3b}%
\end{equation}
Hence, we have the following equality of summation signs:%
\begin{align}
\sum_{d\mid n}\sum\limits_{g\mid n/d}  &  =\sum_{d\mid n}\sum
\limits_{\substack{g\mid n;\\\left(  n/d\right)  /g\in \ZZ %
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.3a})}\right) \nonumber\\
&  =\sum_{d\mid n}\sum\limits_{\substack{g\mid n;\\\left(  n/g\right)
/d\in \ZZ }}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  n/d\right)  /g=\left(  n/g\right)  /d\text{ for any}\\
\text{two positive divisors }d\text{ and }g\text{ of }n
\end{array}
\right) \nonumber\\
&  =\sum_{g\mid n}\sum\limits_{\substack{d\mid n;\\\left(  n/g\right)
/d\in \ZZ }}\nonumber\\
&  =\sum_{g\mid n}\sum\limits_{d\mid n/g}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.3b})}\right)  .
\label{pf.prop.lyndon.grfun.n.b.c10.pf.4}%
\end{align}
Thus, (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.2}) becomes%
\begin{align}
&  \sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d}%
,x_{2}^{n/d},x_{3}^{n/d},\ldots\right) \nonumber\\
&  =\underbrace{\sum_{d\mid n}\sum\limits_{g\mid n/d}}_{\substack{=\sum_{g\mid
n}\sum\limits_{d\mid n/g}\\\text{(by (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.4}%
))}}}\mu\left(  d\right)  p_{n/g}^{g}=\sum_{g\mid n}\sum\limits_{d\mid n/g}%
\mu\left(  d\right)  p_{n/g}^{g}. \label{pf.prop.lyndon.grfun.n.b.c10.pf.5}%
\end{align}


Now, let $g$ be a positive divisor of $n$. Thus, $n/g$ is a positive integer.

But let us recall that every positive integer $N$ satisfies%
\begin{equation}
\sum_{d\mid N}\mu\left(  d\right)  =\delta_{N,1}.
\label{pf.prop.lyndon.grfun.n.b.c10.pf.mu}%
\end{equation}
\footnote{This is one of the most fundamental properties of the
number-theoretic M\"{o}bius function. For a proof of
\eqref{pf.prop.lyndon.grfun.n.b.c10.pf.mu}, see the solution of
Exercise~\ref{exe.witt.ghost-equiv}. (More precisely, the equality
\eqref{pf.prop.lyndon.grfun.n.b.c10.pf.mu} is obtained from
\eqref{sol.exe.witt.ghost-equiv.mu*1} by renaming $n$ as $N$.)} Applying this
to $N=n/g$, we obtain%
\begin{equation}
\sum\limits_{d\mid n/g}\mu\left(  d\right)  =\delta_{n/g,1}
\label{pf.prop.lyndon.grfun.n.b.c10.pf.6}%
\end{equation}
(since $n/g$ is a positive integer).

Forget that we fixed $g$. We thus have proved the equality
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.6}) for each positive divisor $g$ of
$n$. Now, (\ref{pf.prop.lyndon.grfun.n.b.c10.pf.5}) becomes%
\begin{align*}
&  \sum_{d\mid n}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{n/d}%
,x_{2}^{n/d},x_{3}^{n/d},\ldots\right) \\
&  =\sum_{g\mid n}\sum\limits_{d\mid n/g}\mu\left(  d\right)  p_{n/g}^{g}%
=\sum_{g\mid n}\underbrace{\left(  \sum\limits_{d\mid n/g}\mu\left(  d\right)
\right)  }_{\substack{=\delta_{n/g,1}\\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c10.pf.6}))}}}p_{n/g}^{g}=\sum_{g\mid n}%
\delta_{n/g,1}p_{n/g}^{g}\\
&  =\underbrace{\delta_{n/n,1}}_{\substack{=1\\\text{(since }n/n=1\text{)}%
}}\underbrace{p_{n/n}^{n}}_{\substack{=p_{1}^{n}\\\text{(since }n/n=1\text{)}%
}}+\sum\limits_{\substack{g\mid n;\\g\neq n}}\underbrace{\delta_{n/g,1}%
}_{\substack{=0\\\text{(since }n/g\neq1\\\text{(because }g\neq n\text{))}%
}}p_{n/g}^{g}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }g=n\text{ from the sum}\\
\text{(since }n\text{ is a positive divisor of }n\text{)}%
\end{array}
\right) \\
&  =p_{1}^{n}+\underbrace{\sum\limits_{\substack{g\mid n;\\g\neq n}}0p_{n/g}^{g}%
}_{=0}=p_{1}^{n}.
\end{align*}

\end{verlong}

\noindent This proves Claim 10.]

\begin{statement}
\textit{Claim 11:} We have $\mathbf{GR}_{\left(  n\right)  }%
=\widetilde{\mathbf{GR}}_{n}$ for each positive integer $n$.
\end{statement}

[\textit{Proof of Claim 11:} We shall prove Claim 11 by strong induction on
$n$:

\textit{Induction step:} Fix a positive integer $m$. Assume (as the induction
hypothesis) that Claim 11 holds for all $n<m$. We must then show that Claim 11
holds for $n=m$.

As a consequence of our induction hypothesis, we can easily see that
\begin{equation}
\mathbf{GR}_{\left(  d\right)  }=\widetilde{\mathbf{GR}}_{d}
\label{pf.prop.lyndon.grfun.n.b.c11.pf.IH2}%
\end{equation}
whenever $d$ is a positive divisor of $m$ satisfying $d\neq m$%
.\ \ \ \ \footnote{\textit{Proof of (\ref{pf.prop.lyndon.grfun.n.b.c11.pf.IH2}%
):} Let $d$ be a positive divisor of $m$ satisfying $d\neq m$. Then, $d\leq m$
(since $d$ is a positive divisor of the positive integer $m$).\ Combining this
with $d\neq m$, we obtain $d<m$.
\par
But we have assumed (as the induction hypothesis) that Claim 11 holds for all
$n<m$. Hence, Claim 11 holds for $n=d$ (since $d$ is a positive integer
satisfying $d<m$). In other words, we have $\mathbf{GR}_{\left(  d\right)
}=\widetilde{\mathbf{GR}}_{d}$. This proves
(\ref{pf.prop.lyndon.grfun.n.b.c11.pf.IH2}).}

Now, Claim 9 (applied to $n=m$) yields
\begin{align*}
p_{1}^{m}  &  =\sum_{d\mid m}d\cdot\mathbf{GR}_{\left(  d\right)  }\left(
x_{1}^{m/d},x_{2}^{m/d},x_{3}^{m/d},\ldots\right) \\
&  =m\cdot\underbrace{\mathbf{GR}_{\left(  m\right)  }\left(  x_{1}%
^{m/m},x_{2}^{m/m},x_{3}^{m/m},\ldots\right)  }_{\substack{=\mathbf{GR}%
_{\left(  m\right)  }\left(  x_{1},x_{2},x_{3},\ldots\right)  \\\text{(since
}\left(  x_{1}^{m/m},x_{2}^{m/m},x_{3}^{m/m},\ldots\right)  =\left(
x_{1},x_{2},x_{3},\ldots\right)  \\\text{(because }x_{i}^{m/m}=x_{i}^{1}%
=x_{i}\text{ for each }i\in\left\{  1,2,3,\ldots\right\}  \text{))}}%
}+\sum\limits_{\substack{d\mid m;\\d\neq m}}d\cdot\underbrace{\mathbf{GR}_{\left(
d\right)  }}_{\substack{=\widetilde{\mathbf{GR}}_{d}\\\text{(by
(\ref{pf.prop.lyndon.grfun.n.b.c11.pf.IH2}))}}}\left(  x_{1}^{m/d},x_{2}%
^{m/d},x_{3}^{m/d},\ldots\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }d=m\text{ from the sum}\\
\text{(since }m\text{ is a positive divisor of }m\text{)}%
\end{array}
\right) \\
&  =m\cdot\underbrace{\mathbf{GR}_{\left(  m\right)  }\left(  x_{1}%
,x_{2},x_{3},\ldots\right)  }_{\substack{=\mathbf{GR}_{\left(  m\right)
}\\\text{(since }f\left(  x_{1},x_{2},x_{3},\ldots\right)  =f\\\text{for each
}f\in\mathbf{k}\left[  \left[  \xx\right]  \right]  \text{)}}%
}+\sum\limits_{\substack{d\mid m;\\d\neq m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(
x_{1}^{m/d},x_{2}^{m/d},x_{3}^{m/d},\ldots\right) \\
&  =m\cdot\mathbf{GR}_{\left(  m\right)  }+\sum\limits_{\substack{d\mid m;\\d\neq
m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{m/d},x_{2}^{m/d}%
,x_{3}^{m/d},\ldots\right)  .
\end{align*}
Meanwhile, Claim 10 (applied to $n=m$) yields%
\begin{align*}
p_{1}^{m}  &  =\sum_{d\mid m}d\cdot\widetilde{\mathbf{GR}}_{d}\left(
x_{1}^{m/d},x_{2}^{m/d},x_{3}^{m/d},\ldots\right) \\
&  =m\cdot\underbrace{\widetilde{\mathbf{GR}}_{m}\left(  x_{1}^{m/m}%
,x_{2}^{m/m},x_{3}^{m/m},\ldots\right)  }_{\substack{=\widetilde{\mathbf{GR}%
}_{m}\left(  x_{1},x_{2},x_{3},\ldots\right)  \\\text{(since }\left(
x_{1}^{m/m},x_{2}^{m/m},x_{3}^{m/m},\ldots\right)  =\left(  x_{1},x_{2}%
,x_{3},\ldots\right)  \\\text{(because }x_{i}^{m/m}=x_{i}^{1}=x_{i}\text{ for
each }i\in\left\{  1,2,3,\ldots\right\}  \text{))}}}+\sum\limits_{\substack{d\mid
m;\\d\neq m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{m/d},x_{2}%
^{m/d},x_{3}^{m/d},\ldots\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }d=m\text{ from the sum}\\
\text{(since }m\text{ is a positive divisor of }m\text{)}%
\end{array}
\right) \\
&  =m\cdot\underbrace{\widetilde{\mathbf{GR}}_{m}\left(  x_{1},x_{2}%
,x_{3},\ldots\right)  }_{\substack{=\widetilde{\mathbf{GR}}_{m}\\\text{(since
}f\left(  x_{1},x_{2},x_{3},\ldots\right)  =f\\\text{for each }f\in
\mathbf{k}\left[  \left[  \xx\right]  \right]  \text{)}}}+\sum
_{\substack{d\mid m;\\d\neq m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(
x_{1}^{m/d},x_{2}^{m/d},x_{3}^{m/d},\ldots\right) \\
&  =m\cdot\widetilde{\mathbf{GR}}_{m}+\sum\limits_{\substack{d\mid m;\\d\neq
m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{m/d},x_{2}^{m/d}%
,x_{3}^{m/d},\ldots\right)  .
\end{align*}
Comparing these two equalities, we obtain%
\begin{align*}
&  m\cdot\mathbf{GR}_{\left(  m\right)  }+\sum\limits_{\substack{d\mid m;\\d\neq
m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{m/d},x_{2}^{m/d}%
,x_{3}^{m/d},\ldots\right) \\
&  =m\cdot\widetilde{\mathbf{GR}}_{m}+\sum\limits_{\substack{d\mid m;\\d\neq
m}}d\cdot\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{m/d},x_{2}^{m/d}%
,x_{3}^{m/d},\ldots\right)  .
\end{align*}
Subtracting $\sum\limits_{\substack{d\mid m;\\d\neq m}}d\cdot
\widetilde{\mathbf{GR}}_{d}\left(  x_{1}^{m/d},x_{2}^{m/d},x_{3}^{m/d}%
,\ldots\right)  $ from both sides of this equality, we find $m\cdot
\mathbf{GR}_{\left(  m\right)  }=m\cdot\widetilde{\mathbf{GR}}_{m}$.

But $m$ is a positive integer, and thus is invertible in $\QQ$. Hence,
$m$ is invertible in $\kk$ as well (since $\kk$ is a
$\QQ$-algebra). Thus, we can divide both sides of the equality
$m\cdot\mathbf{GR}_{\left(  m\right)  }=m\cdot\widetilde{\mathbf{GR}}_{m}$ by
$m$. We thus obtain $\mathbf{GR}_{\left(  m\right)  }=\widetilde{\mathbf{GR}%
}_{m}$. In other words, Claim 11 holds for $n=m$. This completes the induction
step. Hence, Claim 11 is proved by strong induction.]

Now, fix a positive integer $n$. Then, Claim 11 yields
\[
\mathbf{GR}_{\left(  n\right)  }=\widetilde{\mathbf{GR}}_{n}=\dfrac{1}{n}%
\sum_{d\mid n}\mu\left(  d\right)  p_{d}^{n/d}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\widetilde{\mathbf{GR}}_{n}\right)  .
\]
This proves Proposition \ref{prop.lyndon.grfun.n}(b).
\end{proof}

We have now proved Proposition \ref{prop.lyndon.grfun.CFLps}, Proposition
\ref{prop.lyndon.grfun.cauchy}, Proposition \ref{prop.lyndon.grfun.xN},
Proposition \ref{prop.lyndon.grfun.alt1}, Proposition
\ref{prop.lyndon.grfun.alt2}, Proposition \ref{prop.lyndon.grfun.alt3},
Proposition \ref{prop.lyndon.grfun.F}, Proposition
\ref{prop.lyndon.grfun.cauchy2}, Proposition \ref{prop.lyndon.grfun.symm} and
Proposition \ref{prop.lyndon.grfun.n}. In other words, we have proved all
statements made in Subsection \ref{subsub.lyndon.grfun}. This solves Exercise
\ref{exe.lyndon.grfun.all}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.lyndon.grfun.ch-cyclic}}

\textit{Solution to Exercise \ref{exe.lyndon.grfun.ch-cyclic}.} Let us first
forget about Remark \ref{rmk.lyndon.grfun.char}(b), and prove some basic
lemmas. Our first lemma will be about roots of unity:

\begin{lemma}
\label{lem.sol.lyndon.grfun.ch-cyclic.1}Let $n$ be a positive integer. Let
$\omega$ be a primitive $n$-th root of unity in ${\mathbb{C}}$ (for instance,
$\exp\left(  2\pi i/n\right)  $). Then:

\begin{enumerate}
\item[(a)] If $d$ is a positive divisor of $n$, then%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\delta_{n/d,1}.
\]


\item[(b)] If $d$ is a positive divisor of $n$, then%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}=\mu\left(  n/d\right)  .
\]
Here, $\mu$ denotes the number-theoretical M\"{o}bius function (defined as in
Exercise \ref{exe.witt.ghost-equiv}).
\end{enumerate}
\end{lemma}

\begin{vershort}


\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}.]We have required
$\omega$ to be a primitive $n$-th root of unity in $\mathbb{C}$. Thus,
$\omega^{n}=1$, but if $d$ is a positive integer satisfying $d<n$, then%
\begin{equation}
\omega^{d}\neq1. \label{sol.lyndon.grfun.ch-cyclic.short.w1}%
\end{equation}


(a) Let $d$ be a positive divisor of $n$. Then, the elements $i\in\left\{
1,2,\ldots,n\right\}  $ satisfying $d\mid i$ are precisely the $n/d$ distinct
elements $1d,2d,\ldots,\left(  n/d\right)  d$. Hence,
\begin{equation}
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\omega^{1d}+\omega^{2d}+\cdots+\omega^{\left(  n/d\right)  d}%
=\sum_{j=1}^{n/d}\underbrace{\omega^{jd}}_{=\left(  \omega^{d}\right)  ^{j}%
}=\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j}.
\label{sol.lyndon.grfun.ch-cyclic.short.c1.pf.1}%
\end{equation}


Now, we are in one of the following two cases:

\textit{Case 1:} We have $d=n$.

\textit{Case 2:} We have $d\neq n$.

Let us first consider Case 1. In this case, we have $d=n$. Hence, $\omega
^{d}=\omega^{n}=1$ and $n/d=n/n=1$. Thus,
\eqref{sol.lyndon.grfun.ch-cyclic.short.c1.pf.1} becomes%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\sum_{j=1}^{n/d}\left(  \underbrace{\omega^{d}}_{=1}\right)
^{j}=\sum_{j=1}^{n/d}\underbrace{1^{j}}_{=1}=\sum_{j=1}^{n/d}1=n/d=1=\delta
_{n/d,1}.
\]
(since $n/d=1$ leads to $\delta_{n/d,1}=1$). Hence, Lemma
\ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) is proved in Case 1.

Let us next consider Case 2. In this case, we have $d\neq n$. But $d$ is a
positive divisor of the positive integer $n$; hence, $d\leq n$. Combining this
with $d\neq n$, we obtain $d<n$. Hence,
\eqref{sol.lyndon.grfun.ch-cyclic.short.w1} yields $\omega^{d}\neq1$. Thus,
$1-\omega^{d}\neq0$. Also, from $d\neq n$, we obtain $n/d\neq1$.

But $n/d$ is a positive integer (since $d$ is a positive divisor of the
positive integer $n$). Now,%
\begin{align*}
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}  &  =\omega^{1d}+\omega^{2d}+\cdots+\omega^{\left(  n/d\right)
d}\\
&  =\left(  \omega^{1d}+\omega^{2d}+\cdots+\omega^{\left(  n/d-1\right)
d}\right)  +\underbrace{\omega^{\left(  n/d\right)  d}}_{\substack{=\omega
^{n}=1=\omega^{0d}\\\text{(since }\omega^{0d}=\omega^{0}=1\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n/d\text{ is a positive
integer}\right) \\
&  =\left(  \omega^{1d}+\omega^{2d}+\cdots+\omega^{\left(  n/d-1\right)
d}\right)  +\omega^{0d}\\
&  =\omega^{0d}+\left(  \omega^{1d}+\omega^{2d}+\cdots+\omega^{\left(
n/d-1\right)  d}\right) \\
&  =\omega^{0d}+\omega^{1d}+\cdots+\omega^{\left(  n/d-1\right)  d}=\sum
_{j=0}^{n/d-1}\underbrace{\omega^{jd}}_{=\left(  \omega^{d}\right)  ^{j}}%
=\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}.
\end{align*}
Hence,%
\begin{align}
\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}  &  =\sum
\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}}\omega
^{i}=\sum_{j=1}^{n/d}\underbrace{\left(  \omega^{d}\right)  ^{j}}_{=\omega
^{d}\left(  \omega^{d}\right)  ^{j-1}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.lyndon.grfun.ch-cyclic.short.c1.pf.1}}\right)
\label{sol.lyndon.grfun.ch-cyclic.short.c1.pf.7}\\
&  =\omega^{d}\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j-1}=\omega^{d}%
\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}%
\label{sol.lyndon.grfun.ch-cyclic.short.c1.pf.8}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }j\text{ for
}j-1\text{ in the sum}\right)  .\nonumber
\end{align}
Hence,%
\[
\left(  1-\omega^{d}\right)  \sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)
^{j}=\underbrace{\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}%
}_{\substack{=\omega^{d}\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)
^{j}\\\text{(by \eqref{sol.lyndon.grfun.ch-cyclic.short.c1.pf.8})}}%
}-\omega^{d}\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}=\omega^{d}%
\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}-\omega^{d}\sum_{j=0}%
^{n/d-1}\left(  \omega^{d}\right)  ^{j}=0.
\]
We can divide both sides of this equality by $1-\omega^{d}$ (since
$1-\omega^{d}\neq0$), and thus obtain $\sum_{j=0}^{n/d-1}\left(  \omega
^{d}\right)  ^{j}=0$. Comparing this with
\eqref{sol.lyndon.grfun.ch-cyclic.short.c1.pf.7}, we obtain $\sum_{j=1}%
^{n/d}\left(  \omega^{d}\right)  ^{j}=0$. Thus,
\eqref{sol.lyndon.grfun.ch-cyclic.short.c1.pf.1} becomes%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j}=0=\delta
_{n/d,1}\qquad\left(  \text{since }n/d\neq1\text{ leads to }\delta
_{n/d,1}=0\right)  .
\]
Hence, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) is proved in Case 2.

We have now proved Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) in both
Cases 1 and 2. Thus, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) always holds.

(b) For each positive integer $m$, we have\footnote{Here and in the following,
the summation sign ``$\sum_{d\mid m}$'' will
always be understood to range over all \textbf{positive} divisors $d$ of $m$.}%
\[
\sum_{d\mid m}\mu\left(  d\right)  =\delta_{m,1}\ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{sol.exe.witt.ghost-equiv.mu*1}, applied to }m\text{ instead of
}n\right)
\]
and thus%
\begin{equation}
\delta_{m,1}=\sum_{d\mid m}\mu\left(  d\right)  =\sum_{f\mid m}\mu\left(
f\right)  \label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.musum}%
\end{equation}
(here, we have renamed the summation index $d$ as $f$).

Set%
\begin{equation}
x_{d}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}\qquad\qquad\text{for each }d\in \ZZ .
\label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.xd=}%
\end{equation}


We now claim that
\begin{equation}
x_{d}=\mu\left(  n/d\right)  \qquad\qquad\text{for each positive divisor
}d\text{ of }n\text{.} \label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl}%
\end{equation}


[\textit{Proof of \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl}:} If $d$
is a positive divisor of $n$, then $n/d$ is a positive integer (since $n$ is a
positive integer). Thus, we can prove
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl} by strong induction on
$n/d$. Let us do this:

\textit{Induction step:} Fix a positive integer $m$. Assume (as the induction
hypothesis) that \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl} holds
whenever $n/d<m$. We must prove that
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl} holds whenever $n/d=m$.

If $d$ is a positive divisor of $n$ that satisfies $n/d<m$, then%
\begin{equation}
x_{d}=\mu\left(  n/e\right)  .
\label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.IH}%
\end{equation}
(Indeed, this is just a restatement of our induction hypothesis.)

Now, fix a positive divisor $d$ of $n$ that satisfies $n/d=m$. We shall prove
that $x_{d}=\mu\left(  n/d\right)  $.

As a consequence of the induction hypothesis, we can easily see the following:
If $e$ is a positive divisor of $n$ satisfying $d\mid e$ and $e\neq d$, then%
\begin{equation}
x_{e}=\mu\left(  n/e\right)  .
\label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.IH2}%
\end{equation}


[\textit{Proof of \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.IH2}:}
Let $e$ be a positive divisor of $n$ satisfying $d\mid e$ and $e\neq d$. From
$d\mid e$, we obtain $d\leq e$ (since $d$ and $e$ are positive integers).
Combining this with $d\neq e$ (which follows from $e\neq d$), we obtain $d<e$.
Since $n$ is positive, this entails $n/e<n/d=m$. Hence,
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.IH} (applied to $e$
instead of $d$) yields $x_{e}=\mu\left(  n/e\right)  $. This proves \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.IH2}.]

Recall that $n$ is a positive integer, and $d$ is a positive divisor of $n$.
Hence, $n/d$ is a positive integer, and we have $d\mid n$.

For any $i\in\left\{  1,2,\ldots,n\right\}  $, we have the following chain of
logical equivalences:%
\begin{align*}
&  \ \left(  \left(  \gcd\left(  n,i\right)  \text{ is a positive divisor of
}n\right)  \wedge\left(  d\mid\gcd\left(  n,i\right)  \right)  \right) \\
&  \Longleftrightarrow\ \left(  d\mid\gcd\left(  n,i\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since ``}\gcd\left(
n,i\right)  \text{ is a positive divisor of }n\text{'' is
always true}\right) \\
&  \Longleftrightarrow\ \left(  \left(  d\mid n\right)  \wedge\left(  d\mid
i\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the logical equivalence }\left(  x\mid\gcd\left(  y,z\right)
\right)  \ \Longleftrightarrow\ \left(  \left(  x\mid y\right)  \wedge\left(
x\mid z\right)  \right) \\
\text{holds for any three integers }x\text{, }y\text{ and }z
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  d\mid i\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since ``}d\mid n\text{'' is always true}\right)  .
\end{align*}
Hence, we have the following equality of summation signs:%
\begin{equation}
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  \text{ is a positive divisor of }n;\\d\mid\gcd\left(  n,i\right)
}}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}}.
\label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.sum=sum}%
\end{equation}


We have\footnote{Recall that the summation sign
``$\sum\limits_{\substack{e\mid n;\\d\mid e}}$'' stands for a
sum over all \textbf{positive} divisors $e$ of $n$ satisfying $d\mid e$.}%
\begin{align*}
\underbrace{\sum\limits_{\substack{e\mid n;\\d\mid e}}}_{=\sum
\limits_{\substack{e\text{ is a positive divisor of }n;\\d\mid e}%
}}\ \ \underbrace{x_{e}}_{\substack{=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\\gcd\left(  n,i\right)  =e}}\omega^{i}\\\text{(by the
definition of }x_{e}\text{)}}}  &  =\underbrace{\sum\limits_{\substack{e\text{
is a positive divisor of }n;\\d\mid e}}\ \ \sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\\gcd\left(  n,i\right)  =e}}}_{\substack{=\sum
\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  \text{ is a positive divisor of }n;\\d\mid\gcd\left(  n,i\right)
}}\\=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid
i}}\\\text{(by \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.sum=sum})}%
}}\omega^{i}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid
i}}\omega^{i}=\delta_{n/d,1}%
\end{align*}
(by Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a)). Hence,%
\begin{align}
\delta_{n/d,1}  &  =\sum\limits_{\substack{e\mid n;\\d\mid e}}x_{e}=x_{d}%
+\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\ \ \underbrace{x_{e}%
}_{\substack{=\mu\left(  n/e\right)  \\\text{(by
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl.pf.IH2})}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}e=d\text{ from the sum}\right) \nonumber\\
&  =x_{d}+\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(
n/e\right)  . \label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.s1}%
\end{align}


On the other hand, recall that $n/d$ is a positive integer. Hence,
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.musum} (applied to $m=n/d$)
yields%
\begin{equation}
\delta_{n/d,1}=\sum\limits_{f\mid n/d}\mu\left(  f\right)  .
\label{sol.lyndon.grfun.ch-cyclic.short.c2.pf.s2}%
\end{equation}


But it is straightforward to see that the map
\begin{align*}
\left\{  \text{positive divisors }e\text{ of }n\text{ satisfying }d\mid
e\right\}   &  \rightarrow\left\{  \text{positive divisors of }n/d\right\}
,\\
e  &  \mapsto n/e
\end{align*}
is well-defined and is a bijection\footnote{Its inverse sends each
$f\in\left\{  \text{positive divisors of }n/d\right\}  $ to $n/f$.}. Thus, we
can substitute $n/e$ for $f$ in the sum $\sum\limits_{f\mid n/d}\mu\left(
f\right)  $. We therefore obtain%
\[
\sum\limits_{f\mid n/d}\mu\left(  f\right)  =\sum\limits_{\substack{e\mid
n;\\d\mid e}}\mu\left(  n/e\right)  .
\]
Thus, \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.s2} becomes%
\[
\delta_{n/d,1}=\sum\limits_{f\mid n/d}\mu\left(  f\right)  =\sum
\limits_{\substack{e\mid n;\\d\mid e}}\mu\left(  n/e\right)  =\mu\left(
n/d\right)  +\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(
n/e\right)
\]
(here, we have split off the addend for $e=d$ from the sum). Comparing this
with \eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.s1}, we obtain%
\[
x_{d}+\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(
n/e\right)  =\mu\left(  n/d\right)  +\sum\limits_{\substack{e\mid n;\\d\mid
e;\\e\neq d}}\mu\left(  n/e\right)  .
\]
Subtracting $\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(
n/e\right)  $ from both sides of this equality, we obtain $x_{d}=\mu\left(
n/d\right)  $. In other words,
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl} holds.

Forget that we fixed $d$. We thus have proved that
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl} holds whenever $n/d=m$. This
completes the induction step. Thus, the induction proof of
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl} is complete.]

Now, let $d$ be a positive divisor of $n$. Then, the definition of $x_{d}$
yields $x_{d}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\\gcd\left(  n,i\right)  =d}}\omega^{i}$. Hence,%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}=x_{d}=\mu\left(  n/d\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.lyndon.grfun.ch-cyclic.short.c2.pf.cl}}\right)  .
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(b).
\end{proof}
\end{vershort}

\begin{verlong}


\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}.]We have required
$\omega$ to be a primitive $n$-th root of unity in $\mathbb{C}$. Thus,
$\omega$ is an $n$-th root of unity; in other words, $\omega^{n}=1$. Also,
since $\omega$ is a \textbf{primitive} $n$-th root of unity, the order of
$\omega$ in the multiplicative group $\left(  \mathbb{C}\setminus\left\{
0\right\}  ,\cdot\right)  $ is $n$. Thus, if $d$ is a positive integer
satisfying $d<n$, then%
\begin{equation}
\omega^{d}\neq1. \label{sol.lyndon.grfun.ch-cyclic.w1}%
\end{equation}


(a) Let $d$ be a positive divisor of $n$. Then, $n/d$ is a positive integer
(since $d$ is a positive divisor of the positive integer $n$). Thus, the
elements $i\in\left\{  1,2,\ldots,n\right\}  $ satisfying $d\mid i$ are
precisely the $n/d$ distinct elements $d\cdot1,d\cdot2,\ldots,d\cdot\left(
n/d\right)  $. Hence, the map%
\begin{align*}
\left\{  1,2,\ldots,n/d\right\}   &  \rightarrow\left\{  i\in\left\{
1,2,\ldots,n\right\}  \ :\ d\mid i\right\}  ,\\
j  &  \mapsto dj
\end{align*}
is a bijection (indeed, its inverse map sends each $i$ to $i/d$). Hence, we
can substitute $dj$ for $i$ in the sum $\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\d\mid i}}\omega^{i}$. We thus obtain%
\begin{equation}
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\underbrace{\sum_{j\in\left\{  1,2,\ldots,n/d\right\}  }}%
_{=\sum_{j=1}^{n/d}}\underbrace{\omega^{dj}}_{=\left(  \omega^{d}\right)
^{j}}=\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j}.
\label{sol.lyndon.grfun.ch-cyclic.c1.pf.1}%
\end{equation}


Now, we are in one of the following two cases:

\textit{Case 1:} We have $d=n$.

\textit{Case 2:} We have $d\neq n$.

Let us first consider Case 1. In this case, we have $d=n$. Hence, $\omega
^{d}=\omega^{n}=1$ and $n/\underbrace{d}_{=n}=n/n=1$. Thus,
\eqref{sol.lyndon.grfun.ch-cyclic.c1.pf.1} becomes%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\sum_{j=1}^{n/d}\left(  \underbrace{\omega^{d}}_{=1}\right)
^{j}=\sum_{j=1}^{n/d}\underbrace{1^{j}}_{=1}=\sum_{j=1}^{n/d}1=\left(
n/d\right)  \cdot1=n/d=1.
\]
Comparing this with%
\[
\delta_{n/d,1}=1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n/d=1\right)  ,
\]
we obtain
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\delta_{n/d,1}.
\]
Hence, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) is proved in Case 1.

Let us next consider Case 2. In this case, we have $d\neq n$. But $d$ is a
positive divisor of the positive integer $n$; hence, $d\leq n$. Combining this
with $d\neq n$, we obtain $d<n$. Hence, \eqref{sol.lyndon.grfun.ch-cyclic.w1}
yields $\omega^{d}\neq1$. Thus, $1-\omega^{d}\neq0$. Also, from $d\neq n$, we
obtain $n\neq d$, so that $n/d\neq1$.

But $n/d\in\left\{  1,2,\ldots,n/d\right\}  $ (since $n/d$ is a positive
integer). Also, $n/d-1\in\mathbb{N}$ (since $n/d$ is a positive integer), and
thus $0\in\left\{  0,1,\ldots,n/d-1\right\}  $. Comparing
\begin{align*}
\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j}  &  =\sum_{j=1}^{n/d-1}\left(
\omega^{d}\right)  ^{j}+\underbrace{\left(  \omega^{d}\right)  ^{n/d}%
}_{=\omega^{d\left(  n/d\right)  }=\omega^{n}=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }j=n/d\text{ from the sum,}\\
\text{since }n/d\in\left\{  1,2,\ldots,n/d\right\}
\end{array}
\right) \\
&  =\sum_{j=1}^{n/d-1}\left(  \omega^{d}\right)  ^{j}+1
\end{align*}
with%
\begin{align*}
\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}  &  =\sum_{j=1}%
^{n/d-1}\left(  \omega^{d}\right)  ^{j}+\underbrace{\left(  \omega^{d}\right)
^{0}}_{=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }j=0\text{ from the sum,}\\
\text{since }0\in\left\{  0,1,\ldots,n/d-1\right\}
\end{array}
\right) \\
&  =\sum_{j=1}^{n/d-1}\left(  \omega^{d}\right)  ^{j}+1,
\end{align*}
we obtain
\begin{align}
\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}  &  =\sum_{j=1}%
^{n/d}\underbrace{\left(  \omega^{d}\right)  ^{j}}_{=\omega^{d}\left(
\omega^{d}\right)  ^{j-1}}\label{sol.lyndon.grfun.ch-cyclic.c1.pf.7}\\
&  =\sum_{j=1}^{n/d}\omega^{d}\left(  \omega^{d}\right)  ^{j-1}=\omega^{d}%
\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j-1}=\omega^{d}\sum_{j=0}%
^{n/d-1}\left(  \omega^{d}\right)  ^{j}%
\label{sol.lyndon.grfun.ch-cyclic.c1.pf.8}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }j\text{ for
}j-1\text{ in the sum}\right)  .\nonumber
\end{align}
Hence,%
\[
\left(  1-\omega^{d}\right)  \sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)
^{j}=\underbrace{\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}%
}_{\substack{=\omega^{d}\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)
^{j}\\\text{(by \eqref{sol.lyndon.grfun.ch-cyclic.c1.pf.8})}}}-\omega^{d}%
\sum_{j=0}^{n/d-1}\left(  \omega^{d}\right)  ^{j}=\omega^{d}\sum_{j=0}%
^{n/d-1}\left(  \omega^{d}\right)  ^{j}-\omega^{d}\sum_{j=0}^{n/d-1}\left(
\omega^{d}\right)  ^{j}=0.
\]
We can divide both sides of this equality by $1-\omega^{d}$ (since
$1-\omega^{d}\neq0$), and thus obtain $\sum_{j=0}^{n/d-1}\left(  \omega
^{d}\right)  ^{j}=0$. Comparing this with
\eqref{sol.lyndon.grfun.ch-cyclic.c1.pf.7}, we obtain $\sum_{j=1}^{n/d}\left(
\omega^{d}\right)  ^{j}=0$. Comparing this with
\[
\delta_{n/d,1}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n/d\neq1\right)  ,
\]
we obtain
\[
\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j}=\delta_{n/d,1}.
\]
Thus, \eqref{sol.lyndon.grfun.ch-cyclic.c1.pf.1} becomes%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}%
}\omega^{i}=\sum_{j=1}^{n/d}\left(  \omega^{d}\right)  ^{j}=\delta_{n/d,1}.
\]
Hence, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) is proved in Case 2.

We have now proved Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) in both
Cases 1 and 2. Thus, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a) always
holds (since Cases 1 and 2 cover all possibilities). This completes the proof
of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a).

(b) For each positive integer $m$, we have\footnote{Here and in the following,
the summation sign ``$\sum_{d\mid m}$'' will
always be understood to range over all \textbf{positive} divisors $d$ of $m$.}%
\[
\sum_{d\mid m}\mu\left(  d\right)  =\delta_{m,1}\ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{sol.exe.witt.ghost-equiv.mu*1}, applied to }m\text{ instead of
}n\right)
\]
and thus%
\begin{equation}
\delta_{m,1}=\sum_{d\mid m}\mu\left(  d\right)  =\sum_{f\mid m}\mu\left(
f\right)  \label{sol.lyndon.grfun.ch-cyclic.c2.pf.musum}%
\end{equation}
(here, we have renamed the summation index $d$ as $f$).

Set%
\begin{equation}
x_{d}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}\qquad\qquad\text{for each }d\in \ZZ .
\label{sol.lyndon.grfun.ch-cyclic.c2.pf.xd=}%
\end{equation}


We now claim that
\begin{equation}
x_{d}=\mu\left(  n/d\right)  \qquad\qquad\text{for each positive divisor
}d\text{ of }n\text{.} \label{sol.lyndon.grfun.ch-cyclic.c2.pf.cl}%
\end{equation}


[\textit{Proof of \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl}:} If $d$ is a
positive divisor of $n$, then $n/d$ is a positive integer (since $n$ is a
positive integer). Thus, we can prove
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} by strong induction on $n/d$. Let
us do this:

\textit{Induction step:} Fix a positive integer $m$. Assume (as the induction
hypothesis) that \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} holds whenever
$n/d<m$. We must prove that \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} holds
whenever $n/d=m$.

We have assumed that \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} holds
whenever $n/d<m$. In other words, we have%
\begin{equation}
x_{d}=\mu\left(  n/d\right)  \label{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.IH}%
\end{equation}
for each positive divisor $d$ of $n$ that satisfies $n/d<m$.

Now, fix a positive divisor $d$ of $n$ that satisfies $n/d=m$. We shall prove
that $x_{d}=\mu\left(  n/d\right)  $.

One consequence of the induction hypothesis is the following:\footnote{The
letter $e$ does not stand for $2.718\ldots$ in this proof.} If $e$ is a
positive divisor of $n$ satisfying $d\mid e$ and $e\neq d$, then%
\begin{equation}
x_{e}=\mu\left(  n/e\right)  .
\label{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.IH2}%
\end{equation}


[\textit{Proof of \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.IH2}:} Let $e$
be a positive divisor of $n$ satisfying $d\mid e$ and $e\neq d$. From $d\mid
e$, we obtain $d\leq e$ (since $d$ and $e$ are positive integers). Combining
this with $d\neq e$ (which follows from $e\neq d$), we obtain $d<e$. Hence,
$e>d$. Since $n$ is positive, we thus find $n/\underbrace{e}_{>d}<n/d=m$.
Thus, \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.IH} (applied to $e$
instead of $d$) yields $x_{e}=\mu\left(  n/e\right)  $. This proves \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.IH2}.]

We have $d\mid n$ (since $d$ is a divisor of $n$). Moreover, $n/d$ is a
positive integer (since $d\mid n$, and since $n$ and $d$ are positive integers).

For any $i\in\left\{  1,2,\ldots,n\right\}  $, we have the following chain of
logical equivalences:%
\begin{align*}
&  \ \left(  \left(  \gcd\left(  n,i\right)  \text{ is a positive divisor of
}n\right)  \wedge\left(  d\mid\gcd\left(  n,i\right)  \right)  \right) \\
&  \Longleftrightarrow\ \left(  d\mid\gcd\left(  n,i\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since ``}\gcd\left(
n,i\right)  \text{ is a positive divisor of }n\text{'' is
always true}\right) \\
&  \Longleftrightarrow\ \left(  \left(  d\mid n\right)  \wedge\left(  d\mid
i\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the logical equivalence }\left(  x\mid\gcd\left(  y,z\right)
\right)  \ \Longleftrightarrow\ \left(  \left(  x\mid y\right)  \wedge\left(
x\mid z\right)  \right) \\
\text{holds for any three integers }x\text{, }y\text{ and }z
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  d\mid i\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since ``}d\mid n\text{'' is always true}\right)  .
\end{align*}
Hence, we have the following equality of summation signs:%
\begin{equation}
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  \text{ is a positive divisor of }n;\\d\mid\gcd\left(  n,i\right)
}}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid i}}.
\label{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.sum=sum}%
\end{equation}


We have\footnote{Recall that the summation sign
``$\sum\limits_{\substack{e\mid n;\\d\mid e}}$'' stands for a
sum over all \textbf{positive} divisors $e$ of $n$ satisfying $d\mid e$.}%
\begin{align}
\underbrace{\sum\limits_{\substack{e\mid n;\\d\mid e}}}_{=\sum
\limits_{\substack{e\text{ is a positive divisor of }n;\\d\mid e}%
}}\ \ \underbrace{x_{e}}_{\substack{=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\\gcd\left(  n,i\right)  =e}}\omega^{i}\\\text{(by the
definition of }x_{e}\text{)}}}  &  =\underbrace{\sum\limits_{\substack{e\text{
is a positive divisor of }n;\\d\mid e}}\ \ \sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\\gcd\left(  n,i\right)  =e}}}_{\substack{=\sum
\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  \text{ is a positive divisor of }n;\\d\mid\gcd\left(  n,i\right)
}}\\=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid
i}}\\\text{(by \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.sum=sum})}%
}}\omega^{i}\nonumber\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\d\mid
i}}\omega^{i}=\delta_{n/d,1}
\label{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.sumx}%
\end{align}
(by Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(a)). But $d$ is a positive
divisor $e$ of $n$ satisfying $d\mid e$ (since $d$ is a positive divisor of
$n$ and satisfies $d\mid d$). Thus, we can split off the addend for $e=d$ from
the sum $\sum\limits_{\substack{e\mid n;\\d\mid e}}x_{e}$. We thus find%
\[
\sum\limits_{\substack{e\mid n;\\d\mid e}}x_{e}=x_{d}+\sum
\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\underbrace{x_{e}%
}_{\substack{=\mu\left(  n/e\right)  \\\text{(by
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.IH2})}}}=x_{d}+\sum
\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(  n/e\right)  .
\]
Comparing this with \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl.pf.sumx}, we
find%
\begin{equation}
\delta_{n/d,1}=x_{d}+\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}%
}\mu\left(  n/e\right)  . \label{sol.lyndon.grfun.ch-cyclic.c2.pf.s1}%
\end{equation}


On the other hand, recall that $n/d$ is a positive integer. Hence,
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.musum} (applied to $m=n/d$) yields%
\begin{equation}
\delta_{n/d,1}=\underbrace{\sum\limits_{f\mid n/d}}_{\substack{=\sum
\limits_{f\text{ is a positive divisor of }n/d}\\=\sum\limits_{f\in\left\{
\text{positive divisors of }n/d\right\}  }}}\mu\left(  f\right)
=\sum\limits_{f\in\left\{  \text{positive divisors of }n/d\right\}  }%
\mu\left(  f\right)  . \label{sol.lyndon.grfun.ch-cyclic.c2.pf.s2}%
\end{equation}


Our next plan is to show that the sum on the right hand side of this equality
is $\sum\limits_{\substack{e\mid n;\\d\mid e}}\mu\left(  n/e\right)  $. For
this, we will have to construct a bijection.

If $e\in\left\{  \text{positive divisors }r\text{ of }n\text{ satisfying
}d\mid r\right\}  $, then $n/e\in\left\{  \text{positive divisors of
}n/d\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $e\in\left\{
\text{positive divisors }r\text{ of }n\text{ satisfying }d\mid r\right\}  $.
We must prove that $n/e\in\left\{  \text{positive divisors of }n/d\right\}  $.
\par
We have $e\in\left\{  \text{positive divisors }r\text{ of }n\text{ satisfying
}d\mid r\right\}  $. In other words, $e$ is a positive divisor $r$ of $n$
satisfying $d\mid r$. In other words, $e$ is a positive divisor of $n$ and
satisfies $d\mid e$.
\par
Now, $n/e$ is a positive integer (since $e$ is a positive divisor of $n$, but
$n$ itself is a positive integer). Moreover, $\left(  n/d\right)  /\left(
n/e\right)  =e/d\in \ZZ$ (since $d\mid e$), so that $n/e\mid n/d$ (since
$n/e$ and $n/d$ are integers). Hence, $n/e$ is a positive divisor of $n/d$
(since $n/e$ is positive). In other words, $n/e\in\left\{  \text{positive
divisors of }n/d\right\}  $. Qed.}. Hence, we can define a map%
\begin{align*}
\alpha:\left\{  \text{positive divisors }r\text{ of }n\text{ satisfying }d\mid
r\right\}   &  \rightarrow\left\{  \text{positive divisors of }n/d\right\}
,\\
e  &  \mapsto n/e.
\end{align*}
Consider this map $\alpha$.

If $f\in\left\{  \text{positive divisors of }n/d\right\}  $, then
$n/f\in\left\{  \text{positive divisors }r\text{ of }n\text{ satisfying }d\mid
r\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $f\in\left\{  \text{positive
divisors of }n/d\right\}  $. We must show that $n/f\in\left\{  \text{positive
divisors }r\text{ of }n\text{ satisfying }d\mid r\right\}  $.
\par
Indeed, $f$ is a positive divisor of $n/d$ (since $f\in\left\{  \text{positive
divisors of }n/d\right\}  $). Now, $\left(  n/d\right)  /f$ is a positive
integer (since $f$ is a positive divisor of $n/d$, but $n/d$ itself is a
positive integer). Hence, $\left(  \left(  n/d\right)  /f\right)  \cdot d$ is
a product of two positive integers (namely, of $\left(  n/d\right)  /f$ and
$d$), and thus itself is a positive integer. In other words, $n/f$ is a
positive integer (since $n/f=\left(  \left(  n/d\right)  /f\right)  \cdot d$).
Hence, $n/f$ is a positive divisor of $n$ (since $n/f\mid n$ (because
$n/\left(  n/f\right)  =f\in \ZZ$)). Moreover, $\left(  n/f\right)
/d=\left(  n/d\right)  /f\in \ZZ$ (since $\left(  n/d\right)  /f$ is a
positive integer), so that $d\mid n/f$ (since $d$ and $n/f$ are integers).
\par
Now, we know that $n/f$ is a positive divisor of $n$ and satisfies $d\mid
n/f$. In other words, $n/f$ is a positive divisor $r$ of $n$ satisfying $d\mid
r$. In other words, $n/f\in\left\{  \text{positive divisors }r\text{ of
}n\text{ satisfying }d\mid r\right\}  $. Qed.}. Hence, we can define a map%
\begin{align*}
\beta:\left\{  \text{positive divisors of }n/d\right\}   &  \rightarrow
\left\{  \text{positive divisors }r\text{ of }n\text{ satisfying }d\mid
r\right\}  ,\\
f  &  \mapsto n/f.
\end{align*}
Consider this map $\beta$.

We have $\alpha\circ\beta=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.}
Each $f\in\left\{  \text{positive divisors of }n/d\right\}  $ satisfies%
\begin{align*}
\left(  \alpha\circ\beta\right)  \left(  f\right)   &  =\alpha\left(
\beta\left(  f\right)  \right)  =n/\underbrace{\left(  \beta\left(  f\right)
\right)  }_{\substack{=n/f\\\text{(by the definition of }\beta\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\alpha\right) \\
&  =n/\left(  n/f\right)  =f=\operatorname*{id}\left(  f\right)  .
\end{align*}
In other words, $\alpha\circ\beta=\operatorname*{id}$.} and $\beta\circ
\alpha=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} Each $f\in\left\{
\text{positive divisors }r\text{ of }n\text{ satisfying }d\mid r\right\}  $
satisfies%
\begin{align*}
\left(  \beta\circ\alpha\right)  \left(  f\right)   &  =\beta\left(
\alpha\left(  f\right)  \right)  =n/\underbrace{\left(  \alpha\left(
f\right)  \right)  }_{\substack{=n/f\\\text{(by the definition of }%
\alpha\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\beta\right) \\
&  =n/\left(  n/f\right)  =f=\operatorname*{id}\left(  f\right)  .
\end{align*}
In other words, $\beta\circ\alpha=\operatorname*{id}$.}. Thus, the maps
$\alpha$ and $\beta$ are mutually inverse. Hence, the map $\alpha$ is
invertible, i.e., is a bijection. Thus, we can substitute $\alpha\left(
e\right)  $ for $f$ in the sum $\sum\limits_{f\in\left\{  \text{positive
divisors of }n/d\right\}  }\mu\left(  f\right)  $. We therefore obtain%
\[
\sum\limits_{f\in\left\{  \text{positive divisors of }n/d\right\}  }\mu\left(
f\right)  =\underbrace{\sum_{e\in\left\{  \text{positive divisors }r\text{ of
}n\text{ satisfying }d\mid r\right\}  }}_{\substack{=\sum
\limits_{\substack{e\text{ is a positive divisor of }n;\\d\mid e}%
}\\=\sum\limits_{\substack{e\mid n;\\d\mid e}}}}\mu\left(  \underbrace{\alpha
\left(  e\right)  }_{\substack{=n/e\\\text{(by the definition of }%
\alpha\text{)}}}\right)  =\sum\limits_{\substack{e\mid n;\\d\mid e}}\mu\left(
n/e\right)  .
\]
Thus, \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.s2} becomes%
\[
\delta_{n/d,1}=\sum\limits_{f\in\left\{  \text{positive divisors of
}n/d\right\}  }\mu\left(  f\right)  =\sum\limits_{\substack{e\mid n;\\d\mid
e}}\mu\left(  n/e\right)  =\mu\left(  n/d\right)  +\sum
\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(  n/e\right)
\]
(here, we have split off the addend for $e=d$ from the sum $\sum
\limits_{\substack{e\mid n;\\d\mid e}}\mu\left(  n/e\right)  $, since $d$ is a
positive divisor $e$ of $n$ satisfying $d\mid e$). Comparing this with
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.s1}, we obtain%
\[
x_{d}+\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(
n/e\right)  =\mu\left(  n/d\right)  +\sum\limits_{\substack{e\mid n;\\d\mid
e;\\e\neq d}}\mu\left(  n/e\right)  .
\]
Subtracting $\sum\limits_{\substack{e\mid n;\\d\mid e;\\e\neq d}}\mu\left(
n/e\right)  $ from both sides of this equality, we obtain $x_{d}=\mu\left(
n/d\right)  $. In other words, \eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} holds.

Forget that we fixed $d$. We thus have proved that
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} holds whenever $n/d=m$. This
completes the induction step. Thus, the induction proof of
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl} is complete.]

Now, let $d$ be a positive divisor of $n$. Then, the definition of $x_{d}$
yields $x_{d}=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\\gcd\left(  n,i\right)  =d}}\omega^{i}$. Hence,%
\[
\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}=x_{d}=\mu\left(  n/d\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{sol.lyndon.grfun.ch-cyclic.c2.pf.cl}}\right)  .
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(b).
\end{proof}
\end{verlong}

\begin{remark}
Let $n$ be a positive integer. If $d$ is a positive divisor of $n$, then the
complex numbers $\omega^{i}$ for all $i\in\left\{  1,2,\ldots,n\right\}  $
satisfying $\gcd\left(  n,i\right)  =d$ are precisely the primitive $\left(
n/d\right)  $-th roots of unity in $\mathbb{C}$ (this is not hard to check).
Thus, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(b) states that the sum of
all primitive $\left(  n/d\right)  $-th roots of unity in $\mathbb{C}$ is
$\mu\left(  n/d\right)  $. Hence, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}%
(b) is an equivalent restatement of the following fact: For any positive
integer $m$, the sum of all primitive $m$-th roots of unity in $\mathbb{C}$ is
$\mu\left(  m\right)  $.
\end{remark}

\begin{vershort}
The next lemma we will need is an elementary fact from the theory of numbers,
whose easy proof (e.g., using the Bezout theorem or the basic properties of
greatest common divisors) we omit:
\end{vershort}

\begin{verlong}
The next lemma we will need is a basic fact from the theory of numbers:
\end{verlong}

\begin{lemma}
\label{lem.sol.lyndon.grfun.ch-cyclic.gcd}Let $x$, $y$ and $z$ be three
integers such that $x\neq0$. Let $d=\gcd\left(  x,y\right)  $. Then, we have
the logical equivalence $\left(  x\mid yz\right)  \ \Longleftrightarrow
\ \left(  x/d\mid z\right)  $.
\end{lemma}

\begin{verlong}


\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.gcd}.]We have
$d=\gcd\left(  x,y\right)  \neq0$ (since $x\neq0$). Bezout's theorem says that
$\gcd\left(  x,y\right)  $ is a $\ZZ$-linear combination of $x$ and
$y$. In other words, there exist integers $u$ and $v$ such that $\gcd\left(
x,y\right)  =ux+vy$. Consider these $u$ and $v$.

Note that $d=\gcd\left(  x,y\right)  \mid x$. Hence, $x/d$ is an integer
(since $d\neq0$).

Now, the implication $\left(  x\mid yz\right)  \ \Longrightarrow\ \left(
x/d\mid z\right)  $ holds\footnote{\textit{Proof.} Assume that $x\mid yz$. We
must prove that $x/d\mid z$.
\par
We have $yz\equiv0\operatorname{mod}x$ (since $x\mid yz$). Now,
\[
\underbrace{d}_{\substack{=\gcd\left(  x,y\right)  \\=ux+vy}}z=\left(
ux+vy\right)  z=u\underbrace{x}_{\equiv0\operatorname{mod}x}z+v\underbrace{yz}%
_{\equiv0\operatorname{mod}x}\equiv u0z+v0=0\operatorname{mod}x.
\]
In other words, $x\mid dz$. Since $x\neq0$, this means that $dz/x\in
\ZZ$. Hence, $z/\left(  x/d\right)  =dz/x\in \ZZ$. In other
words, $x/d\mid z$ (since $x/d$ and $z$ are integers).
\par
Forget that we assumed that $x\mid yz$. We thus have proved that $x/d\mid z$
under the assumption $x\mid yz$. In other words, we have proved the
implication $\left(  x\mid yz\right)  \ \Longrightarrow\ \left(  x/d\mid
z\right)  $.}. Moreover, the implication $\left(  x/d\mid z\right)
\ \Longrightarrow\ \left(  x\mid yz\right)  $ holds\footnote{\textit{Proof.}
Assume that $x/d\mid z$. We must prove that $x\mid yz$.
\par
We have $x/d\neq0$ (since $x\neq0$ and $d\neq0$). Hence, from $x/d\mid z$, we
obtain $z/\left(  x/d\right)  \in \ZZ$. Hence, $dz/x=z/\left(
x/d\right)  \in \ZZ$. In other words, $x\mid dz$. But $d=\gcd\left(
x,y\right)  \mid y$, so that $dz\mid yz$. Hence, $x\mid dz\mid yz$. Thus,
$x\mid yz$ holds.
\par
Forget that we assumed that $x/d\mid z$. We thus have proved that $x\mid yz$
under the assumption $x/d\mid z$. In other words, we have proved the
implication $\left(  x/d\mid z\right)  \ \Longrightarrow\ \left(  x\mid
yz\right)  $.}. Combining these two implications, we obtain the logical
equivalence $\left(  x\mid yz\right)  \ \Longleftrightarrow\ \left(  x/d\mid
z\right)  $. This proves Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.gcd}.
\end{proof}
\end{verlong}

The next lemma we will need is a mostly trivial property of permutations:

\begin{lemma}
\label{lem.sol.lyndon.grfun.ch-cyclic.c}Let $n$ be a positive integer. For
every permutation $\sigma\in\mathfrak{S}_{n}$, we let $\operatorname*{type}%
\sigma$ denote the cycle type of $\sigma$.

Let $q\in\mathbb{Q}$ and $\sigma\in\mathfrak{S}_{n}$. Assume that every cycle
of $\sigma$ has size $q$. Then, $q$ is a positive divisor of $n$, and we have
$\operatorname*{type}\sigma=\left(  \underbrace{q,q,\ldots,q}_{n/q\text{
times}}\right)  $.
\end{lemma}

\begin{vershort}


\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.c}.]Let $k$ be the number
of cycles of $\sigma$. Thus, $\sigma$ has exactly $k$ cycles, and each of
these $k$ cycles has size $q$ (because we assumed that every cycle of $\sigma$
has size $q$). But the cycles of $\sigma$ are disjoint, and their union is the
$n$-element set $\left\{  1,2,\ldots,n\right\}  $. Hence, the sum of the sizes
of all cycles of $\sigma$ is $n$. Thus,%
\begin{align*}
n  &  =\left(  \text{the sum of the sizes of all cycles of }\sigma\right)
=\underbrace{q+q+\cdots+q}_{k\text{ times}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma\text{ has exactly }k\text{
cycles, and each of these }k\text{ cycles has size }q\right) \\
&  =kq.
\end{align*}
Thus, $kq=n\neq0$, so that $k\neq0$ and $q\neq0$. From $k\neq0$, we conclude
that $\sigma$ has at least one cycle. The size of this cycle is $q$ (since we
assumed that every cycle of $\sigma$ has size $q$); thus, $q$ is the size of a
cycle of $\sigma$, and therefore is a nonnegative integer. Combining this with
$q\neq0$, we conclude that $q$ is a positive integer. Since $n=kq$, we
conclude that $q\mid n$ (because $k\in\mathbb{N}$), so that $q$ is a positive
divisor of $n$. Furthermore, the definition of $\operatorname*{type}\sigma$
shows that%
\begin{align*}
\operatorname*{type}\sigma &  =\left(  \text{the cycle type of }\sigma\right)
=\left(  \underbrace{q,q,\ldots,q}_{k\text{ times}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma\text{ has exactly }k\text{
cycles, and each of these }k\text{ cycles has size }q\right) \\
&  =\left(  \underbrace{q,q,\ldots,q}_{n/q\text{ times}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k=n/q\text{ (because }n=kq\text{)}%
\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.c}.
\end{proof}
\end{vershort}

\begin{verlong}


\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.c}.]We assumed that every
cycle of $\sigma$ has size $q$. In other words, the size of any cycle of
$\sigma$ is $q$.

Let $z_{1},z_{2},\ldots,z_{k}$ be the sizes of all cycles of $\sigma$ (listed
in decreasing order). Then, $z_{i}=q$ for each $i\in\left\{  1,2,\ldots
,k\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
1,2,\ldots,k\right\}  $. Then, $z_{i}$ is the size of a cycle of $\sigma$
(since $z_{1},z_{2},\ldots,z_{k}$ are the sizes of all cycles of $\sigma$).
Hence, $z_{i}$ is $q$ (since the size of any cycle of $\sigma$ is $q$). In
other words, $z_{i}=q$. Qed.}. In other words, $\left(  z_{1},z_{2}%
,\ldots,z_{k}\right)  =\left(  \underbrace{q,q,\ldots,q}_{k\text{ times}%
}\right)  $. Hence,%
\[
z_{1}+z_{2}+\cdots+z_{k}=\underbrace{q+q+\cdots+q}_{k\text{ times}}=kq.
\]


On the other hand, $\sigma$ is a permutation of $\left\{  1,2,\ldots
,n\right\}  $ (since $\sigma\in\mathfrak{S}_{n}$). Hence, the union of the
cycles of $\sigma$ is the set $\left\{  1,2,\ldots,n\right\}  $. Since these
cycles are furthermore disjoint, we have%
\[
\left(  \text{the size of the union of the cycles of }\sigma\right)  =\left(
\text{the sum of the sizes of all cycles of }\sigma\right)
\]
(because the size of the union of some disjoint finite sets always equals the
sum of their sizes). Thus,%
\begin{align*}
&  \left(  \text{the sum of the sizes of all cycles of }\sigma\right) \\
&  =\left(  \text{the size of the union of the cycles of }\sigma\right) \\
&  =\left(  \text{the size of }\left\{  1,2,\ldots,n\right\}  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the union of the cycles of }%
\sigma\text{ is }\left\{  1,2,\ldots,n\right\}  \right) \\
&  =\left\vert \left\{  1,2,\ldots,n\right\}  \right\vert =n.
\end{align*}
Hence,%
\begin{align*}
n  &  =\left(  \text{the sum of the sizes of all cycles of }\sigma\right)
=\left(  \text{the sum of }z_{1},z_{2},\ldots,z_{k}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }z_{1},z_{2},\ldots,z_{k}\text{ are
the sizes of all cycles of }\sigma\right) \\
&  =z_{1}+z_{2}+\cdots+z_{k}=kq.
\end{align*}


Thus, $kq=n\neq0$ (since $n$ is a positive integer). Hence, $k\neq0$ and
$q\neq0$. Solving the equality $n=kq$ for $k$, we obtain $k=n/q$ (since
$q\neq0$). Moreover, $q$ is a positive integer\footnote{\textit{Proof.} The
number $k$ is a positive integer (since $k\in\mathbb{N}$ and $k\neq0$). Hence,
$1\in\left\{  1,2,\ldots,k\right\}  $. Thus, $z_{1}$ is the size of a cycle of
$\sigma$ (since $z_{1},z_{2},\ldots,z_{k}$ are the sizes of all cycles of
$\sigma$). Hence, $z_{1}$ is the size of a finite set (since any cycle of
$\sigma$ is a finite set). Thus, $z_{1}\in\mathbb{N}$.
\par
But recall that $z_{i}=q$ for each $i\in\left\{  1,2,\ldots,k\right\}  $.
Applying this to $i=1$, we obtain $z_{1}=q$ (since $1\in\left\{
1,2,\ldots,k\right\}  $). Hence, $q=z_{1}\in\mathbb{N}$. Combining this with
$q\neq0$, we conclude that $q$ is a positive integer.}. Hence, $q\mid n$
(since $n=kq$ with $k\in\mathbb{N}$). Thus, $q$ is a divisor of $n$. Hence,
$q$ is a positive divisor of $n$ (since $q$ is positive). It thus remains to
show that $\operatorname*{type}\sigma=\left(  \underbrace{q,q,\ldots
,q}_{n/q\text{ times}}\right)  $.

The definition of $\operatorname*{type}\sigma$ yields%
\begin{align*}
\operatorname*{type}\sigma &  =\left(  \text{the cycle type of }\sigma\right)
\\
&  =\left(  \text{the partition whose parts are the sizes of all cycles of
}\sigma\text{ (listed in decreasing order)}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the cycle type of
}\sigma\right) \\
&  =\left(  \text{the partition whose parts are }z_{1},z_{2},\ldots
,z_{k}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }z_{1},z_{2},\ldots,z_{k}\text{ are
the sizes of all cycles of }\sigma\text{ (listed in decreasing order)}\right)
\\
&  =\left(  z_{1},z_{2},\ldots,z_{k}\right)  =\left(  \underbrace{q,q,\ldots
,q}_{k\text{ times}}\right)  =\left(  \underbrace{q,q,\ldots,q}_{n/q\text{
times}}\right)
\end{align*}
(since $k=n/q$). This completes the proof of Lemma
\ref{lem.sol.lyndon.grfun.ch-cyclic.c}.
\end{proof}
\end{verlong}

Let us now step to the analysis of $n$-cycles in $\mathfrak{S}_{n}$:

\begin{lemma}
\label{lem.sol.lyndon.grfun.ch-cyclic.z}Let $n$ be a positive integer. For
every permutation $\sigma\in\mathfrak{S}_{n}$, we let $\operatorname*{type}%
\sigma$ denote the cycle type of $\sigma$.

Let $z\in\mathfrak{S}_{n}$ be an $n$-cycle. Let $m\in \ZZ$. Let
$d=\gcd\left(  m,n\right)  $. Then, $\operatorname*{type}\left(  z^{m}\right)
=\left(  \underbrace{n/d,n/d,\ldots,n/d}_{d\text{ times}}\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.z}.]Let us consider the
cycles of the permutation $z^{m}\in\mathfrak{S}_{n}$. We shall show that every
cycle of $z^{m}$ has size $n/d$.

\begin{vershort}
Indeed, let $Z$ be a cycle of $z^{m}$ (considered as a subset of $\left\{
1,2,\ldots,n\right\}  $). Then, $Z$ is a nonempty set; in other words, there
exists some $x\in Z$. Consider this $x$.

Let $g=\left\vert Z\right\vert $. Then, $g$ is a positive integer (since $Z$
is nonempty), and the cycle $Z$ has length $g$ (since $g=\left\vert
Z\right\vert $).

Now, $Z$ is the cycle of $z^{m}$ containing $x$ (since $Z$ is a cycle of
$z^{m}$ and since $x\in Z$). Thus, the cycle of $z^{m}$ containing $x$ has
length $g$ (since we know that the cycle $Z$ has length $g$). Hence, the
sequence $x,z^{m}\left(  x\right)  ,\left(  z^{m}\right)  ^{2}\left(
x\right)  ,\left(  z^{m}\right)  ^{3}\left(  x\right)  ,\ldots$ (obtained from
$x$ by applying $z^{m}$ again and again) repeats every $g$ elements, but not
more frequently. Thus, for any $N\in\mathbb{N}$, we have the following
equivalence of statements:%
\begin{equation}
\left(  \left(  z^{m}\right)  ^{N}\left(  x\right)  =x\right)
\ \Longleftrightarrow\ \left(  g\mid N\right)
.\label{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv1}%
\end{equation}


On the other hand, the permutation $z$ is an $n$-cycle on an $n$-element set
(namely, on $\left\{  1,2,\ldots,n\right\}  $); thus, it has exactly one
cycle, which has length $n$. The cycle of $z$ which contains $x$ must
therefore be this unique cycle, and thus has length $n$. Hence, the sequence
$x,z\left(  x\right)  ,z^{2}\left(  x\right)  ,z^{3}\left(  x\right)  ,\ldots$
(obtained from $x$ by applying $z$ again and again) repeats every $n$
elements, but not more frequently. Thus, for any $N\in\mathbb{N}$, we have the
following equivalence of statements:%
\begin{equation}
\left(  z^{N}\left(  x\right)  =x\right)  \ \Longleftrightarrow\ \left(  n\mid
N\right)  .\label{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv2}%
\end{equation}


Note that $n\neq0$. Hence, $d$ is a positive integer (since $d=\gcd\left(
m,n\right)  $) and satisfies $d=\gcd\left(  m,n\right)  =\gcd\left(
n,m\right)  \mid n$. Hence, $d$ is a positive divisor of $n$. Therefore, $n/d$
is a positive integer (since $n$ is also a positive integer), so that
$n/d\in\mathbb{N}$.

Now, for any $N\in\mathbb{N}$, we have the following chain of logical
equivalences:%
\begin{align}
\left(  g\mid N\right)  \  &  \Longleftrightarrow\ \left(  \underbrace{\left(
z^{m}\right)  ^{N}}_{=z^{mN}}\left(  x\right)  =x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv1}}\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  z^{mN}\left(  x\right)  =x\right)
\ \Longleftrightarrow\ \left(  n\mid mN\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv2}, applied to
}mN\text{ instead of }N\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  n/d\mid N\right)
\label{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv3}%
\end{align}
(by Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.gcd}, applied to $n$, $m$ and
$N$ instead of $x$, $y$ and $z$).

Applying \eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv3} to $N=n/d$,
we obtain the equivalence $\left(  g\mid n/d\right)  \ \Longleftrightarrow
\ \left(  n/d\mid n/d\right)  $ (since $n/d\in\mathbb{N}$). Thus, we have
$g\mid n/d$ (since we clearly have $n/d\mid n/d$). Hence, $g\leq n/d$ (since
$g$ and $n/d$ are positive integers).

Applying \eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.short.equiv3} to $N=g$, we
obtain the equivalence $\left(  g\mid g\right)  \ \Longleftrightarrow\ \left(
n/d\mid g\right)  $ (since $g\in\mathbb{N}$). Thus, we have $n/d\mid g$ (since
we clearly have $g\mid g$). Hence, $n/d\leq g$ (since $n/d$ and $g$ are
positive integers).

Combining $g\leq n/d$ with $n/d\leq g$, we obtain $g=n/d$. Hence, $\left\vert
Z\right\vert =g=n/d$. In other words, the set $Z$ has size $n/d$.
\end{vershort}

\begin{verlong}
Indeed, we first introduce some notations:

\begin{itemize}
\item For each set $X$, we let $\mathfrak{S}_{X}$ denote the symmetric group
on $X$ (that is, the group of all permutations of $X$). Thus, $\mathfrak{S}%
_{n}=\mathfrak{S}_{\left\{  1,2,\ldots,n\right\}  }$ (by the definition of
$\mathfrak{S}_{n}$).

\item If $X$ is a finite set and $x\in X$, and if $\sigma\in\mathfrak{S}_{X}$,
then we let $O_{\sigma}\left(  x\right)  $ denote the set of all elements of
the cycle of $\sigma$ which contains $x$. This $O_{\sigma}\left(  x\right)  $
is a subset of the set $X$ and contains $x$.
\end{itemize}

We have $z\in\mathfrak{S}_{n}=\mathfrak{S}_{\left\{  1,2,\ldots,n\right\}  }$
and thus $z^{m}\in\mathfrak{S}_{\left\{  1,2,\ldots,n\right\}  }$.

Let $Z$ be a cycle of $z^{m}$ (considered as a subset of $\left\{
1,2,\ldots,n\right\}  $). Thus, $Z$ is a nonempty subset of $\left\{
1,2,\ldots,n\right\}  $ (since any cycle of any permutation in $\mathfrak{S}%
_{n}$ is a nonempty subset of $\left\{  1,2,\ldots,n\right\}  $). Hence, in
particular, $Z$ is nonempty; in other words, there exists some $x\in Z$.
Consider this $x$. Note that $x\in Z\subset\left\{  1,2,\ldots,n\right\}  $
(since $Z$ is a cycle of the permutation $z^{m}\in\mathfrak{S}_{\left\{
1,2,\ldots,n\right\}  }$).

The cycle $Z$ contains $x$ (since $x\in Z$). Hence, $Z$ is the cycle of
$z^{m}$ which contains $x$ (since $Z$ is a cycle of $z^{m}$).

Let $g=\left\vert Z\right\vert $. Thus, $g=\left\vert Z\right\vert >0$ (since
$Z$ is nonempty); hence, $g$ is a positive integer.

The definition of $O_{z^{m}}\left(  x\right)  $ shows that $O_{z^{m}}\left(
x\right)  $ is the set of all elements of the cycle of $z^{m}$ which contains
$x$. In other words, $O_{z^{m}}\left(  x\right)  $ is the set of all elements
of $Z$ (since $Z$ is the cycle of $z^{m}$ which contains $x$). In other words,
$O_{z^{m}}\left(  x\right)  $ is $Z$ (since the set of all elements of $Z$ is
$Z$ (because $Z$ is already a set)). In other words, $O_{z^{m}}\left(
x\right)  =Z$. Therefore, $\left\vert O_{z^{m}}\left(  x\right)  \right\vert
=\left\vert Z\right\vert =g$ (since $g=\left\vert Z\right\vert $), so that
$g=\left\vert O_{z^{m}}\left(  x\right)  \right\vert $.

But the permutation $z$ is an $n$-cycle on an $n$-element set (namely, on
$\left\{  1,2,\ldots,n\right\}  $); thus, it has exactly one cycle, which must
contain all $n$ elements $1,2,\ldots,n$. In other words, the permutation $z$
has exactly one cycle, which (when regarded as a set) is the set $\left\{
1,2,\ldots,n\right\}  $. In other words, the only cycle of $z$ is the set
$\left\{  1,2,\ldots,n\right\}  $. Hence, the cycle of $z$ which contains $x$
must be $\left\{  1,2,\ldots,n\right\}  $.

But the definition of $O_{z}\left(  x\right)  $ shows that $O_{z}\left(
x\right)  $ is the set of all elements of the cycle of $z$ which contains $x$.
In other words, $O_{z}\left(  x\right)  $ is the set of all elements of
$\left\{  1,2,\ldots,n\right\}  $ (since the cycle of $z$ which contains $x$
is $\left\{  1,2,\ldots,n\right\}  $). In other words, $O_{z}\left(  x\right)
$ is $\left\{  1,2,\ldots,n\right\}  $ (since the set of all elements of
$\left\{  1,2,\ldots,n\right\}  $ is $\left\{  1,2,\ldots,n\right\}  $). In
other words, $O_{z}\left(  x\right)  =\left\{  1,2,\ldots,n\right\}  $. Hence,
$\left\vert O_{z}\left(  x\right)  \right\vert =\left\vert \left\{
1,2,\ldots,n\right\}  \right\vert =n$.

Note that $n\neq0$ (since $n$ is a positive integer). Also, $d$ is a positive
integer (since $d=\gcd\left(  m,n\right)  $ and $n\neq0$) and satisfies
$d=\gcd\left(  m,n\right)  =\gcd\left(  n,m\right)  \mid n$. Hence, $d$ is a
positive divisor of $n$. Hence, $n/d$ is a positive integer (since $n$ is also
a positive integer). Thus, $n/d\in\mathbb{N}$.

Fix any $N\in\mathbb{N}$. Then, Lemma \ref{lem.sol.repSn.arithmetic-prod.O1}
(applied to $\left\{  1,2,\ldots,n\right\}  $, $z^{m}$ and $N$ instead of $X$,
$\sigma$ and $n$) shows that we have the following equivalence of
assertions:\footnote{Here, the statement ``$\left\vert
O_{z^{m}}\left(  x\right)  \right\vert \mid N$'' has to be
read as ``the size of $O_{z^{m}}\left(  x\right)  $ divides $N$''.}
\[
\left(  \left\vert O_{z^{m}}\left(  x\right)  \right\vert \mid N\right)
\ \Longleftrightarrow\ \left(  \left(  z^{m}\right)  ^{N}\left(  x\right)
=x\right)  .
\]
Hence, we have the following chain of logical equivalences:%
\begin{align}
\left(  \underbrace{g}_{=\left\vert O_{z^{m}}\left(  x\right)  \right\vert
}\mid N\right)  \  &  \Longleftrightarrow\ \left(  \left\vert O_{z^{m}}\left(
x\right)  \right\vert \mid N\right)  \ \Longleftrightarrow\ \left(
\underbrace{\left(  z^{m}\right)  ^{N}}_{=z^{mN}}\left(  x\right)  =x\right)
\nonumber\\
&  \Longleftrightarrow\ \left(  z^{mN}\left(  x\right)  =x\right)  .
\label{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv1}%
\end{align}
On the other hand, Lemma \ref{lem.sol.repSn.arithmetic-prod.O1} (applied to
$\left\{  1,2,\ldots,n\right\}  $, $z$ and $mN$ instead of $X$, $\sigma$ and
$n$) shows that we have the following equivalence of assertions:
\begin{equation}
\left(  \left\vert O_{z}\left(  x\right)  \right\vert \mid mN\right)
\ \Longleftrightarrow\ \left(  z^{mN}\left(  x\right)  =x\right)  .
\label{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv2}%
\end{equation}
Hence, we have the following chain of logical equivalences:%
\begin{align}
\left(  g\mid N\right)  \  &  \Longleftrightarrow\ \left(  z^{mN}\left(
x\right)  =x\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv1}}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \underbrace{\left\vert O_{z}\left(  x\right)
\right\vert }_{=n}\mid mN\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
\eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv2}}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  n\mid mN\right)  \ \Longleftrightarrow
\ \left(  n/d\mid N\right)  \label{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv3}%
\end{align}
(by Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.gcd}, applied to $n$, $m$ and
$N$ instead of $x$, $y$ and $z$).

Now, forget that we fixed $N$. We thus have proved the equivalence
\eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv3} for each $N\in\mathbb{N}$.

Applying \eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv3} to $N=n/d$, we
obtain the equivalence $\left(  g\mid n/d\right)  \ \Longleftrightarrow
\ \left(  n/d\mid n/d\right)  $ (since $n/d\in\mathbb{N}$). Thus, we have
$g\mid n/d$ (since we clearly have $n/d\mid n/d$). Hence, $g\leq n/d$ (since
$g$ and $n/d$ are positive integers).

Applying \eqref{pf.lem.sol.lyndon.grfun.ch-cyclic.z.equiv3} to $N=g$, we
obtain the equivalence $\left(  g\mid g\right)  \ \Longleftrightarrow\ \left(
n/d\mid g\right)  $ (since $g\in\mathbb{N}$). Thus, we have $n/d\mid g$ (since
we clearly have $g\mid g$). Hence, $n/d\leq g$ (since $n/d$ and $g$ are
positive integers).

Combining $g\leq n/d$ with $n/d\leq g$, we obtain $g=n/d$. Hence, $\left\vert
Z\right\vert =g=n/d$. In other words, the set $Z$ has size $n/d$.
\end{verlong}

Now, forget that we fixed $Z$. We thus have shown that if $Z$ is a cycle of
$z^{m}$ (considered as a subset of $\left\{  1,2,\ldots,n\right\}  $), then
$Z$ has size $n/d$. In other words, every cycle of $z^{m}$ has size $n/d$.

Hence, Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.c} (applied to $q=n/d$ and
$\sigma=z^{m}$) yields that $n/d$ is a positive divisor of $n$, and that
$\operatorname*{type}\left(  z^{m}\right)  =\left(  \underbrace{n/d,n/d,\ldots
,n/d}_{n/\left(  n/d\right)  \text{ times}}\right)  $. Hence,%
\[
\operatorname*{type}\left(  z^{m}\right)  =\left(  \underbrace{n/d,n/d,\ldots
,n/d}_{n/\left(  n/d\right)  \text{ times}}\right)  =\left(
\underbrace{n/d,n/d,\ldots,n/d}_{d\text{ times}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n/\left(  n/d\right)  =d\right)  .
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.z}.
\end{proof}

At last, we can now solve Exercise \ref{exe.lyndon.grfun.ch-cyclic} itself:

Let us use the notations from Remark \ref{rmk.lyndon.grfun.char}(b). In
particular, let $n$ be a positive integer. We fix some $n$-cycle $z$ in
$\mathfrak{S}_{n}$, and we fix some generator $g$ of the cyclic group $C_{n}$.
We embed the cyclic group $C_{n}= \ZZ / n \ZZ$ as a subgroup in
the symmetric group $\mathfrak{S}_{n}$ by identifying $g\in C_{n}$ with
$z\in\mathfrak{S}_{n}$. (This is legitimate, since the $n$-cycle
$z\in\mathfrak{S}_{n}$ is an element of order $n$.) Let $\omega$ be a
primitive $n$-th root of unity in ${\mathbb{C}}$ (for instance, $\exp\left(
2\pi i/n\right)  $). Let $\gamma:C_{n}\rightarrow{\mathbb{C}}$ be the
character of $C_{n}$ that sends each $g^{i}\in C_{n}$ to $\omega^{i}$.

We must prove the claim of Remark \ref{rmk.lyndon.grfun.char}(b). In other
words, we must prove that $\mathbf{GR}_{\left(  n\right)  }=\operatorname{ch}%
\left(  \operatorname{Ind}_{C_{n}}^{\mathfrak{S}_{n}}\gamma\right)  $.

In the following, $\mu$ denotes the number-theoretical M\"{o}bius function
(defined as in Exercise \ref{exe.witt.ghost-equiv}), and the summation sign
``$\sum_{d\mid n}$'' will always be understood
to range over all \textbf{positive} divisors $d$ of $n$.

Extend the map $\operatorname*{ch}:A=A\left(  \mathfrak{S}\right)
\rightarrow\Lambda$ to a $\mathbb{C}$-linear map $A_{\mathbb{C}}%
\rightarrow\Lambda_{\mathbb{C}}$. We shall call the latter map
$\operatorname*{ch}$, too.

The cyclic group $C_{n}$ is abelian; thus, every function from $C_{n}$ to
$\mathbb{C}$ is a class function of $C_{n}$ (because any function from an
abelian group to $\mathbb{C}$ is a class function). Hence, $\gamma$ is a class
function of $C_{n}$ (since $\gamma$ is a function from $C_{n}$ to $\mathbb{C}%
$). In other words, $\gamma\in R_{\mathbb{C}}\left(  C_{n}\right)  $.

For every permutation $\sigma\in\mathfrak{S}_{n}$, we let
$\operatorname*{type}\sigma$ denote the cycle type of $\sigma$. Exercise
\ref{exe.frob-char.ptype}(b) (applied to $H=C_{n}$ and $f=\gamma$) yields%
\begin{equation}
\operatorname*{ch}\left(  \operatorname*{Ind}\nolimits_{C_{n}}^{\mathfrak{S}%
_{n}}\gamma\right)  =\dfrac{1}{\left\vert C_{n}\right\vert }\sum_{h\in C_{n}%
}\gamma\left(  h\right)  p_{\operatorname*{type}h}=\dfrac{1}{n}\sum_{h\in
C_{n}}\gamma\left(  h\right)  p_{\operatorname*{type}h}
\label{sol.lyndon.grfun.ch-cyclic.ch1}%
\end{equation}
(since $\left\vert C_{n}\right\vert =n$).

If $m$ and $d$ are integers such that $\gcd\left(  n,m\right)  =d$, then
\begin{align*}
\operatorname*{type}\left(  g^{m}\right)   &  =\operatorname*{type}\left(
z^{m}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }g=z\text{ (because we
are identifying }g\text{ with }z\text{)}\right) \\
&  =\left(  \underbrace{n/d,n/d,\ldots,n/d}_{d\text{ times}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.lyndon.grfun.ch-cyclic.z} (since }d=\gcd\left(  n,m\right)
=\gcd\left(  m,n\right)  \text{)}\right)
\end{align*}
and thus%
\begin{align}
p_{\operatorname*{type}\left(  g^{m}\right)  }  &  =p_{\left(
\underbrace{n/d,n/d,\ldots,n/d}_{d\text{ times}}\right)  }=\underbrace{p_{n/d}%
p_{n/d}\cdots p_{n/d}}_{d\text{ times}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of }p_{\lambda}\text{ for a partition }\lambda\\
\text{(since }\ell\left(  \left(  \underbrace{n/d,n/d,\ldots,n/d}_{d\text{
times}}\right)  \right)  =d\text{)}%
\end{array}
\right) \nonumber\\
&  =p_{n/d}^{d}. \label{sol.lyndon.grfun.ch-cyclic.ptypegm=}%
\end{align}


But $C_{n}$ is the cyclic group of size $n$, and $g$ is a generator of $C_{n}%
$. Hence, the $n$ elements of $C_{n}$ are $g^{1},g^{2},\ldots,g^{n}$ (listed
here without repetition). Hence,%
\begin{align*}
\sum_{h\in C_{n}}\gamma\left(  h\right)  p_{\operatorname*{type}h}  &
=\gamma\left(  g^{1}\right)  p_{\operatorname*{type}\left(  g^{1}\right)
}+\gamma\left(  g^{2}\right)  p_{\operatorname*{type}\left(  g^{2}\right)
}+\cdots+\gamma\left(  g^{n}\right)  p_{\operatorname*{type}\left(
g^{n}\right)  }\\
&  =\underbrace{\sum_{m\in\left\{  1,2,\ldots,n\right\}  }}_{\substack{=\sum
_{d\mid n}\ \ \sum\limits_{\substack{m\in\left\{  1,2,\ldots,n\right\}
;\\\gcd\left(  n,m\right)  =d}}\\\text{(here, we have split up the
sum}\\\text{according to the value of }\gcd\left(  n,m\right)  \text{,}%
\\\text{because }\gcd\left(  n,m\right)  \text{ is a positive divisor of
}n\\\text{for each }m\in\left\{  1,2,\ldots,n\right\}  \text{)}}}\gamma\left(
g^{m}\right)  p_{\operatorname*{type}\left(  g^{m}\right)  }\\
&  =\sum_{d\mid n}\ \ \sum_{\substack{m\in\left\{  1,2,\ldots,n\right\}
;\\\gcd\left(  n,m\right)  =d}}\underbrace{\gamma\left(  g^{m}\right)
}_{\substack{=\omega^{m}\\\text{(by the definition of }\gamma\text{)}%
}}\underbrace{p_{\operatorname*{type}\left(  g^{m}\right)  }}%
_{\substack{=p_{n/d}^{d}\\\text{(by
\eqref{sol.lyndon.grfun.ch-cyclic.ptypegm=})}}}\\
&  =\sum_{d\mid n}\ \ \underbrace{\sum_{\substack{m\in\left\{  1,2,\ldots
,n\right\}  ;\\\gcd\left(  n,m\right)  =d}}\omega^{m}}_{\substack{=\sum
\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}\\\text{(here, we have renamed}\\\text{the summation
index }m\text{ as }i\text{)}}}p_{n/d}^{d}=\sum_{d\mid n}\ \ \underbrace{\sum
\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
n,i\right)  =d}}\omega^{i}}_{\substack{=\mu\left(  n/d\right)  \\\text{(by
Lemma \ref{lem.sol.lyndon.grfun.ch-cyclic.1}(b))}}}p_{n/d}^{d}\\
&  =\sum_{d\mid n}\mu\left(  n/d\right)  p_{n/d}^{d}=\sum_{d\mid n}\mu\left(
\underbrace{n/\left(  n/d\right)  }_{=d}\right)  \underbrace{p_{n/\left(
n/d\right)  }^{n/d}}_{\substack{=p_{d}^{n/d}\\\text{(since }n/\left(
n/d\right)  =d\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }n/d\text{ for }d\text{ in the sum, since}\\
\text{the map }\left\{  \text{positive divisors of }n\right\}  \rightarrow
\left\{  \text{positive divisors of }n\right\}  ,\ d\mapsto n/d\\
\text{is a bijection}%
\end{array}
\right) \\
&  =\sum_{d\mid n}\mu\left(  d\right)  p_{d}^{n/d}.
\end{align*}
Hence, \eqref{sol.lyndon.grfun.ch-cyclic.ch1} becomes%
\[
\operatorname*{ch}\left(  \operatorname*{Ind}\nolimits_{C_{n}}^{\mathfrak{S}%
_{n}}\gamma\right)  =\dfrac{1}{n}\underbrace{\sum_{h\in C_{n}}\gamma\left(
h\right)  p_{\operatorname*{type}h}}_{=\sum_{d\mid n}\mu\left(  d\right)
p_{d}^{n/d}}=\dfrac{1}{n}\sum_{d\mid n}\mu\left(  d\right)  p_{d}^{n/d}.
\]
But Proposition \ref{prop.lyndon.grfun.n}(b) yields%
\[
\mathbf{GR}_{\left(  n\right)  }=\dfrac{1}{n}\sum_{d\mid n}\mu\left(
d\right)  p_{d}^{n/d}.
\]
Comparing these two equalities, we obtain $\mathbf{GR}_{\left(  n\right)
}=\operatorname{ch}\left(  \operatorname{Ind}_{C_{n}}^{\mathfrak{S}_{n}}%
\gamma\right)  $. This proves the claim of Remark \ref{rmk.lyndon.grfun.char}%
(b). Thus, Exercise \ref{exe.lyndon.grfun.ch-cyclic} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise \ref{exe.lyndon.grfun.count-perms}}

\textit{Solution to Exercise \ref{exe.lyndon.grfun.count-perms}.} Before we
prove Proposition \ref{prop.lyndon.grfun.count-perms}, let us prove two simple lemmas:

\begin{lemma}
\label{lem.sol.lyndon.grfun.count-perms.1}Let $n\in\mathbb{N}$. Then:

\begin{enumerate}
\item[(a)] For any two compositions $\alpha,\beta\in
\Comp_{n}$, we have the logical equivalence%
\[
\left(  \alpha=\beta\right)  \ \Longleftrightarrow\ \left(  D\left(
\alpha\right)  =D\left(  \beta\right)  \right)  .
\]


\item[(b)] For any permutation $\sigma\in\mathfrak{S}_{n}$ and any composition
$\beta\in\operatorname*{Comp}\nolimits_{n}$, we have the logical equivalence%
\[
\left(  \beta=\gamma\left(  \sigma\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{Des}\sigma=D\left(  \beta\right)  \right)  .
\]


\item[(c)] For any permutation $\sigma\in\mathfrak{S}_{n}$ and any composition
$\beta\in\operatorname*{Comp}\nolimits_{n}$, we have the logical equivalence%
\[
\left(  \beta\text{ refines }\gamma\left(  \sigma\right)  \right)
\ \Longleftrightarrow\ \left(  \operatorname*{Des}\sigma\subset D\left(
\beta\right)  \right)  .
\]

\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}.]If $\sigma
\in\mathfrak{S}_{n}$ is any permutation, then $\gamma\left(  \sigma\right)  $
is a composition of $n$ and satisfies%
\begin{equation}
D\left(  \gamma\left(  \sigma\right)  \right)  =\operatorname*{Des}%
\sigma\label{pf.lem.sol.lyndon.grfun.count-perms.1.Dgamma}%
\end{equation}
(since $\gamma\left(  \sigma\right)  $ was defined to be the unique
composition $\alpha$ of $n$ satisfying $D\left(  \alpha\right)
=\operatorname{Des}\sigma$).

(a) Set $\left[  n-1\right]  :=\left\{  1,2,\ldots,n-1\right\}  $. The map
$\operatorname{Comp}_{n}\rightarrow2^{\left[  n-1\right]  },\ \alpha\mapsto
D\left(  \alpha\right)  $ is a bijection. Thus, in particular, this map is
injective. Hence, for any two compositions $\alpha,\beta\in
\operatorname*{Comp}\nolimits_{n}$, we have the logical equivalence $\left(
\alpha=\beta\right)  \ \Longleftrightarrow\ \left(  D\left(  \alpha\right)
=D\left(  \beta\right)  \right)  $. This proves Lemma
\ref{lem.sol.lyndon.grfun.count-perms.1}(a).

\begin{vershort}
(b) Let $\sigma\in\mathfrak{S}_{n}$ be a permutation. Let $\beta
\in\operatorname*{Comp}\nolimits_{n}$ be a composition. Note that
$\gamma\left(  \sigma\right)  \in\operatorname*{Comp}\nolimits_{n}$ (since
$\sigma\in\mathfrak{S}_{n}$). Now, we have the following chain of logical
equivalences:%
\begin{align*}
\left(  \beta=\gamma\left(  \sigma\right)  \right)  \  &  \Longleftrightarrow
\ \left(  \gamma\left(  \sigma\right)  =\beta\right)  \ \Longleftrightarrow
\ \left(  D\left(  \gamma\left(  \sigma\right)  \right)  =D\left(
\beta\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.sol.lyndon.grfun.count-perms.1}(a) (applied to }\alpha=\gamma\left(
\sigma\right)  \text{)}\right) \\
&  \Longleftrightarrow\ \left(  \operatorname*{Des}\sigma=D\left(
\beta\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.1.Dgamma})}\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}(b).
\end{vershort}

\begin{verlong}
(b) Let $\sigma\in\mathfrak{S}_{n}$ be a permutation. Let $\beta
\in\operatorname*{Comp}\nolimits_{n}$ be a composition. Note that
$\gamma\left(  \sigma\right)  $ is a composition of $n$ (since $\sigma
\in\mathfrak{S}_{n}$). In other words, $\gamma\left(  \sigma\right)
\in\operatorname*{Comp}\nolimits_{n}$.

Now, we have the following chain of logical equivalences:%
\begin{align*}
\left(  \beta=\gamma\left(  \sigma\right)  \right)  \  &  \Longleftrightarrow
\ \left(  \gamma\left(  \sigma\right)  =\beta\right)  \ \Longleftrightarrow
\ \left(  \underbrace{D\left(  \gamma\left(  \sigma\right)  \right)
}_{\substack{=\operatorname*{Des}\sigma\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.1.Dgamma}))}}}=D\left(
\beta\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}(a) (applied to }%
\alpha=\gamma\left(  \sigma\right)  \text{),}\\
\text{since }\gamma\left(  \sigma\right)  \in\Comp_{n}
\text{ and }\beta\in\operatorname*{Comp}\nolimits_{n}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \operatorname*{Des}\sigma=D\left(
\beta\right)  \right)  .
\end{align*}
This proves Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}(b).
\end{verlong}

(c) For any two compositions $\alpha,\beta\in\Comp_{n}$,
we have the logical equivalence%
\begin{equation}
\left(  \alpha\text{ refines }\beta\right)  \ \Longleftrightarrow\ \left(
D\left(  \alpha\right)  \supset D\left(  \beta\right)  \right)
\label{pf.lem.sol.lyndon.grfun.count-perms.1.refines}%
\end{equation}
(by the definition of the relation ``refines'' in
Definition \ref{def.QSym.D}).

\begin{vershort}
Let $\sigma\in\mathfrak{S}_{n}$ be any permutation. Let $\beta\in
\operatorname*{Comp}\nolimits_{n}$ be a composition. Note that $\gamma\left(
\sigma\right)  \in\operatorname*{Comp}\nolimits_{n}$ (since $\sigma
\in\mathfrak{S}_{n}$). Now, we have the following chain of logical
equivalences:%
\begin{align*}
\left(  \beta\text{ refines }\gamma\left(  \sigma\right)  \right)  \  &
\Longleftrightarrow\ \left(  D\left(  \beta\right)  \supset D\left(
\gamma\left(  \sigma\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.lem.sol.lyndon.grfun.count-perms.1.refines}) (applied to
}\beta\text{ and }\gamma\left(  \sigma\right)  \text{ instead of }\alpha\text{
and }\beta\text{),}\\
\text{since }\beta\in\operatorname*{Comp}\nolimits_{n}\text{ and }%
\gamma\left(  \sigma\right)  \in\operatorname*{Comp}\nolimits_{n}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  D\left(  \gamma\left(  \sigma\right)  \right)
\subset D\left(  \beta\right)  \right)  \ \Longleftrightarrow\ \left(
\operatorname*{Des}\sigma\subset D\left(  \beta\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.1.Dgamma})}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
Let $\sigma\in\mathfrak{S}_{n}$ be any permutation. Let $\beta\in
\operatorname*{Comp}\nolimits_{n}$ be a composition. Note that $\gamma\left(
\sigma\right)  $ is a composition of $n$ (since $\sigma\in\mathfrak{S}_{n}$).
In other words, $\gamma\left(  \sigma\right)  \in\Comp_{n}$.

Now, we have the following chain of logical equivalences:%
\begin{align*}
\left(  \beta\text{ refines }\gamma\left(  \sigma\right)  \right)  \  &
\Longleftrightarrow\ \left(  D\left(  \beta\right)  \supset
\underbrace{D\left(  \gamma\left(  \sigma\right)  \right)  }%
_{\substack{=\operatorname*{Des}\sigma\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.1.Dgamma}))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.lem.sol.lyndon.grfun.count-perms.1.refines}) (applied to
}\beta\text{ and }\gamma\left(  \sigma\right)  \text{ instead of }\alpha\text{
and }\beta\text{),}\\
\text{since }\beta\in\operatorname*{Comp}\nolimits_{n}\text{ and }%
\gamma\left(  \sigma\right)  \in\operatorname*{Comp}\nolimits_{n}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  D\left(  \beta\right)  \supset
\operatorname*{Des}\sigma\right)  \ \Longleftrightarrow\ \left(
\operatorname*{Des}\sigma\subset D\left(  \beta\right)  \right)  .
\end{align*}

\end{verlong}

\noindent This proves Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}(c).
\end{proof}

\begin{lemma}
\label{lem.sol.lyndon.grfun.count-perms.h-through-s}Let $n\in\mathbb{N}$. Let
$\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  \in
\operatorname{Comp}_{n}$. Let $\mu$ be the partition obtained by sorting the
entries of $\beta$ into decreasing order. Then,%
\[
h_{\mu}=\sum_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n}%
;\\\beta\text{ refines }\alpha}}s_{\operatorname*{Rib}\left(  \alpha\right)
}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.lyndon.grfun.count-perms.h-through-s}.]Consider
the algebra homomorphism $\operatorname*{NSym}\overset{\pi}{\longrightarrow
}\Lambda$ defined in Corollary \ref{cor.NSym.pi}.

It is easy to see that%
\begin{equation}
\pi\left(  H_{\beta}\right)  =h_{\mu}.
\label{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh}):}
The definition of $\mu$ shows that the nonzero entries of $\mu$ are precisely
the entries of $\beta$ (up to order). Since $\beta=\left(  \beta_{1},\beta
_{2},\ldots,\beta_{k}\right)  $, we can restate this as follows: The nonzero
entries of $\mu$ are precisely $\beta_{1},\beta_{2},\ldots,\beta_{k}$ (up to
order). Hence, the definition of $h_{\mu}$ yields%
\begin{equation}
h_{\mu}=h_{\beta_{1}}h_{\beta_{2}}\cdots h_{\beta_{k}}.
\label{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.short.1}%
\end{equation}


But $\beta_{1},\beta_{2},\ldots,\beta_{k}$ are positive integers (since
$\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  \in\Comp_{n}
\subset\operatorname*{Comp}$). Hence, the definition of $\pi$
yields $\pi\left(  H_{\beta_{i}}\right)  =h_{\beta_{i}}$ for all $i\in\left\{
1,2,\ldots,k\right\}  $. Multiplying these $k$ equalities, we obtain
$\pi\left(  H_{\beta_{1}}\right)  \pi\left(  H_{\beta_{2}}\right)  \cdots
\pi\left(  H_{\beta_{k}}\right)  =h_{\beta_{1}}h_{\beta_{2}}\cdots
h_{\beta_{k}}$.

Recall again that $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)
$. Hence, \eqref{eq.Nsym.Halpha} (applied to $\beta$ and $k$ instead of
$\alpha$ and $\ell$) yields $H_{\beta}=H_{\beta_{1}}H_{\beta_{2}}\cdots
H_{\beta_{k}}$. Applying the map $\pi$ to both sides of this equality, we find%
\begin{align*}
\pi\left(  H_{\beta}\right)   &  =\pi\left(  H_{\beta_{1}}H_{\beta_{2}}\cdots
H_{\beta_{k}}\right)  =\pi\left(  H_{\beta_{1}}\right)  \pi\left(
H_{\beta_{2}}\right)  \cdots\pi\left(  H_{\beta_{k}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }
\kk\text{-algebra homomorphism}\right) \\
&  =h_{\beta_{1}}h_{\beta_{2}}\cdots h_{\beta_{k}}.
\end{align*}
Comparing this with
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.short.1}), we find
$\pi\left(  H_{\beta}\right)  =h_{\mu}$. Thus,
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh}) is proved.]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh}):}
We have defined $\mu$ to be the partition obtained by sorting the entries of
$\beta$ into decreasing order. Hence, $\mu$ is a rearrangement of the list
$\beta$. In other words, $\mu$ is a rearrangement of the list $\left(
\beta_{1},\beta_{2},\ldots,\beta_{k}\right)  $ (since $\beta=\left(  \beta
_{1},\beta_{2},\ldots,\beta_{k}\right)  $). In other words, there exists a
permutation $\sigma\in\mathfrak{S}_{k}$ such that $\mu=\left(  \beta
_{\sigma\left(  1\right)  },\beta_{\sigma\left(  2\right)  },\ldots
,\beta_{\sigma\left(  k\right)  }\right)  $. Consider this $\sigma$. From
$\mu=\left(  \beta_{\sigma\left(  1\right)  },\beta_{\sigma\left(  2\right)
},\ldots,\beta_{\sigma\left(  k\right)  }\right)  $, we obtain%
\begin{equation}
\mu_{i}=\beta_{\sigma\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\{  1,2,\ldots,k\right\}  .
\label{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.1}%
\end{equation}
Note that $\sigma$ is a permutation of $\left\{  1,2,\ldots,k\right\}  $
(since $\sigma\in\mathfrak{S}_{k}$). In other words, $\sigma$ is a bijection
from $\left\{  1,2,\ldots,k\right\}  $ to $\left\{  1,2,\ldots,k\right\}  $.

We have $\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  \in
\operatorname{Comp}_{n}\subset\operatorname*{Comp}$. In other words, $\left(
\beta_{1},\beta_{2},\ldots,\beta_{k}\right)  $ is a composition, i.e., a tuple
of positive integers. Hence, $\left(  \beta_{\sigma\left(  1\right)  }%
,\beta_{\sigma\left(  2\right)  },\ldots,\beta_{\sigma\left(  k\right)
}\right)  $ is a tuple of positive integers as well (since $\left(
\beta_{\sigma\left(  1\right)  },\beta_{\sigma\left(  2\right)  },\ldots
,\beta_{\sigma\left(  k\right)  }\right)  $ is a rearrangement of the tuple
$\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  $ (because $\sigma
\in\mathfrak{S}_{k}$)). In other words, $\beta_{\sigma\left(  1\right)
},\beta_{\sigma\left(  2\right)  },\ldots,\beta_{\sigma\left(  k\right)  }$
are positive integers. In other words, $\beta_{\sigma\left(  i\right)  }>0$
for each $i\in\left\{  1,2,\ldots,k\right\}  $. Hence, $\beta_{\sigma\left(
1\right)  }\geq\beta_{\sigma\left(  2\right)  }\geq\cdots\geq\beta
_{\sigma\left(  k\right)  }>0$\ \ \ \ \footnote{\textit{Proof.} We have
$\mu=\left(  \beta_{\sigma\left(  1\right)  },\beta_{\sigma\left(  2\right)
},\ldots,\beta_{\sigma\left(  k\right)  }\right)  $. Thus, $\left(
\beta_{\sigma\left(  1\right)  },\beta_{\sigma\left(  2\right)  },\ldots
,\beta_{\sigma\left(  k\right)  }\right)  $ is a partition (since $\mu$ is a
partition). Hence, $\beta_{\sigma\left(  1\right)  }\geq\beta_{\sigma\left(
2\right)  }\geq\cdots\geq\beta_{\sigma\left(  k\right)  }$. Combining this
with the fact that $\beta_{\sigma\left(  i\right)  }>0$ for each $i\in\left\{
1,2,\ldots,k\right\}  $, we obtain $\beta_{\sigma\left(  1\right)  }\geq
\beta_{\sigma\left(  2\right)  }\geq\cdots\geq\beta_{\sigma\left(  k\right)
}>0$.}.

We have now written the partition $\mu$ in the form $\mu=\left(  \beta
_{\sigma\left(  1\right)  },\beta_{\sigma\left(  2\right)  },\ldots
,\beta_{\sigma\left(  k\right)  }\right)  $ with $\beta_{\sigma\left(
1\right)  }\geq\beta_{\sigma\left(  2\right)  }\geq\cdots\geq\beta
_{\sigma\left(  k\right)  }>0$. Hence, $\ell\left(  \mu\right)  =k$. Thus, the
definition of $h_{\mu}$ yields%
\begin{align}
h_{\mu}  &  =h_{\mu_{1}}h_{\mu_{2}}\cdots h_{\mu_{k}}=\prod_{i\in\left\{
1,2,\ldots,k\right\}  }\underbrace{h_{\mu_{i}}}_{\substack{=h_{\beta
_{\sigma\left(  i\right)  }}\\\text{(since }\mu_{i}=\beta_{\sigma\left(
i\right)  }\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.1})))}}}=\prod
_{i\in\left\{  1,2,\ldots,k\right\}  }h_{\beta_{\sigma\left(  i\right)  }%
}\nonumber\\
&  =\prod_{i\in\left\{  1,2,\ldots,k\right\}  }h_{\beta_{i}}
\label{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.2}%
\end{align}
(here, we have substituted $i$ for $\sigma\left(  i\right)  $ in the product,
since the map $\sigma:\left\{  1,2,\ldots,k\right\}  \rightarrow\left\{
1,2,\ldots,k\right\}  $ is a bijection).

Recall that $\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  $ is a tuple
of positive integers. Hence, for each $i\in\left\{  1,2,\ldots,k\right\}  $,
its entry $\beta_{i}$ is a positive integer, and therefore satisfies%
\begin{equation}
\pi\left(  H_{\beta_{i}}\right)  =h_{\beta_{i}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\pi\right)  .
\label{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.3}%
\end{equation}


But $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  $. Hence,
\eqref{eq.Nsym.Halpha} (applied to $\beta$ and $k$ instead of $\alpha$ and
$\ell$) yields $H_{\beta}=H_{\beta_{1}}H_{\beta_{2}}\cdots H_{\beta_{k}}$.
Applying the map $\pi$ to both sides of this equality, we find%
\begin{align*}
\pi\left(  H_{\beta}\right)   &  =\pi\left(  H_{\beta_{1}}H_{\beta_{2}}\cdots
H_{\beta_{k}}\right)  =\pi\left(  H_{\beta_{1}}\right)  \pi\left(
H_{\beta_{2}}\right)  \cdots\pi\left(  H_{\beta_{k}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is a }
\kk\text{-algebra homomorphism}\right) \\
&  =\prod_{i\in\left\{  1,2,\ldots,k\right\}  }\underbrace{\pi\left(
H_{\beta_{i}}\right)  }_{\substack{=h_{\beta_{i}}\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.3}))}}}=\prod
_{i\in\left\{  1,2,\ldots,k\right\}  }h_{\beta_{i}}.
\end{align*}
Comparing this with
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh.2}), we obtain
$\pi\left(  H_{\beta}\right)  =h_{\mu}$. Thus,
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh}) is proved.]
\end{verlong}

Theorem \ref{Nsym-structure-on-ribbons-theorem}(b) shows that
\begin{equation}
\pi\left(  R_{\alpha}\right)  =s_{\operatorname{Rib}\left(  \alpha\right)  }
\label{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.piR}%
\end{equation}
for each composition $\alpha$.

The equality (\ref{H-as-sum-of-ribbons}) (with the variables $\alpha$ and
$\beta$ renamed as $\beta$ and $\alpha$) says that%
\[
H_{\beta}=\underbrace{\sum_{\alpha\text{ coarsens }\beta}}_{\substack{=\sum
\limits_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n};\\\alpha\text{
coarsens }\beta}}\ \ =\ \ \sum\limits_{\substack{\alpha\in\Comp_{n};\\
\beta\text{ refines }\alpha}}\\\text{(since the condition
``}\alpha\text{ coarsens }\beta\text{''}\\
\text{is equivalent to ``}\beta\text{ refines }%
\alpha\text{'')}}}R_{\alpha}=\sum\limits_{\substack{\alpha
\in\operatorname*{Comp}\nolimits_{n};\\\beta\text{ refines }\alpha}}R_{\alpha
}.
\]
Applying the map $\pi$ to both sides of this equality, we obtain%
\begin{align*}
\pi\left(  H_{\beta}\right)   &  =\pi\left(  \sum\limits_{\substack{\alpha
\in\operatorname*{Comp}\nolimits_{n};\\\beta\text{ refines }\alpha}}R_{\alpha
}\right)  =\sum\limits_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n}%
;\\\beta\text{ refines }\alpha}}\ \ \underbrace{\pi\left(  R_{\alpha}\right)
}_{\substack{=s_{\operatorname{Rib}\left(  \alpha\right)  }\\\text{(by
(\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.piR}))}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\pi\text{ is }
\kk\text{-linear}\right) \\
&  =\sum_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n};\\\beta\text{
refines }\alpha}}s_{\operatorname*{Rib}\left(  \alpha\right)  }.
\end{align*}
Comparing this with (\ref{pf.lem.sol.lyndon.grfun.count-perms.h-through-s.Hh}%
), we obtain%
\[
h_{\mu}=\sum_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n}%
;\\\beta\text{ refines }\alpha}}s_{\operatorname*{Rib}\left(  \alpha\right)
}.
\]
This proves Lemma \ref{lem.sol.lyndon.grfun.count-perms.h-through-s}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.lyndon.grfun.count-perms}.]Proposition
\ref{prop.lyndon.grfun.symm} shows that the power series $\mathbf{GR}%
_{\lambda}$ belongs to $\Lambda$. In other words, $\mathbf{GR}_{\lambda}%
\in\Lambda$.

\begin{verlong}
For any permutation $\sigma\in\mathfrak{S}_{n}$, we have the following chain
of logical equivalences:%
\begin{align*}
\left(  \operatorname*{type}\sigma=\lambda\right)  \  &  \Longleftrightarrow
\ \left(  \left(  \text{the cycle type of }\sigma\right)  =\lambda\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{type}\sigma=\left(
\text{the cycle type of }\sigma\right)  \text{ (by the definition of
}\operatorname*{type}\sigma\text{)}\right) \\
&  \Longleftrightarrow\ \left(  \sigma\text{ has cycle type }\lambda\right)  .
\end{align*}
Hence, we have the following equality of summation signs:%
\begin{equation}
\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda}}=\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n}%
;\\\sigma\text{ has cycle type }\lambda}}.
\label{sol.lyndon.grfun.count-perms.sum=sum}%
\end{equation}

\end{verlong}

(b) For each permutation $\sigma\in\mathfrak{S}_{n}$, we have the logical
equivalence%
\[
\left(  \beta=\gamma\left(  \sigma\right)  \right)  \ \Longleftrightarrow
\ \left(  \operatorname*{Des}\sigma=D\left(  \beta\right)  \right)
\]
(by Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}(b)). Hence,%
\begin{align}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right) \nonumber\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\operatorname{Des}%
\sigma=D\left(  \beta\right)  \right)  .
\label{sol.lyndon.grfun.count-perms.b.1}%
\end{align}


The definition of the noncommutative ribbon functions shows that the basis
$\left(  R_{\alpha}\right)  _{\alpha\in\operatorname*{Comp}}$ of
$\operatorname*{NSym}$ is dual to the basis $\left(  L_{\alpha}\right)
_{\alpha\in\operatorname*{Comp}}$ of $\operatorname*{QSym}$ (with respect to
the dual pairing $\operatorname{NSym}\otimes\operatorname{QSym}\overset{(\cdot
,\cdot)}{\longrightarrow} \kk$). In other words,%
\begin{equation}
\left(  R_{\alpha},L_{\zeta}\right)  =\delta_{\alpha,\zeta}%
\ \ \ \ \ \ \ \ \ \ \text{for all }\alpha,\zeta\in\operatorname*{Comp}.
\label{sol.lyndon.grfun.count-perms.RLdual}%
\end{equation}


Now, consider the inclusion $\Lambda\overset{i}{\hookrightarrow}%
\operatorname*{QSym}$ and the algebra homomorphism $\operatorname*{NSym}%
\overset{\pi}{\longrightarrow}\Lambda$ defined in Corollary \ref{cor.NSym.pi}.
We have seen in Corollary \ref{cor.NSym.pi} that these two maps $i$ and $\pi$
are adjoint (with respect to the dual pairing $\operatorname{NSym}%
\otimes\operatorname{QSym}\overset{(\cdot,\cdot)}{\longrightarrow} \kk$).
In other words,%
\begin{equation}
\left(  f,i\left(  g\right)  \right)  =\left(  \pi\left(  f\right)  ,g\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }f\in\operatorname*{NSym}\text{ and }%
g\in\Lambda. \label{sol.lyndon.grfun.count-perms.adj}%
\end{equation}


Theorem \ref{Nsym-structure-on-ribbons-theorem}(b) shows that $\pi\left(
R_{\alpha}\right)  =s_{\operatorname{Rib}\left(  \alpha\right)  }$ for every
composition $\alpha$. Applying this to $\alpha=\beta$, we find $\pi\left(
R_{\beta}\right)  =s_{\operatorname{Rib}\left(  \beta\right)  }$.

\begin{verlong}
The Hall inner product is symmetric (by Exercise \ref{exe.Lambda.hall.gr}(c)).
\end{verlong}

Recall that $\mathbf{GR}_{\lambda}\in\Lambda$. Thus,
(\ref{sol.lyndon.grfun.count-perms.adj}) (applied to $f=R_{\beta}$ and
$g=\mathbf{GR}_{\lambda}$) yields%
\begin{equation}
\left(  R_{\beta},i\left(  \mathbf{GR}_{\lambda}\right)  \right)  =\left(
\underbrace{\pi\left(  R_{\beta}\right)  }_{=s_{\operatorname*{Rib}\left(
\beta\right)  }},\mathbf{GR}_{\lambda}\right)  =\left(  s_{\operatorname*{Rib}%
\left(  \beta\right)  },\mathbf{GR}_{\lambda}\right)  =\left(  \mathbf{GR}%
_{\lambda},s_{\operatorname*{Rib}\left(  \beta\right)  }\right)
\label{sol.lyndon.grfun.count-perms.b.3}%
\end{equation}
(since the Hall inner product is symmetric).

But $i$ is just an inclusion map. Hence,%
\[
i\left(  \mathbf{GR}_{\lambda}\right)  =\mathbf{GR}_{\lambda}=\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\sigma\text{ has cycle type
}\lambda}}L_{\gamma\left(  \sigma\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition \ref{prop.lyndon.grfun.F}}\right)  .
\]


\begin{vershort}
Now, (\ref{sol.lyndon.grfun.count-perms.b.3}) shows that%
\begin{align*}
\left(  \mathbf{GR}_{\lambda},s_{\operatorname*{Rib}\left(  \beta\right)
}\right)   &  =\left(  R_{\beta},\underbrace{i\left(  \mathbf{GR}_{\lambda
}\right)  }_{=\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\sigma\text{
has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)  }}\right)  =\left(
R_{\beta},\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\sigma\text{ has
cycle type }\lambda}}L_{\gamma\left(  \sigma\right)  }\right) \\
&  =\underbrace{\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n}%
;\\\sigma\text{ has cycle type }\lambda}}}_{\substack{=\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda}}\\\text{(since }\operatorname*{type}\sigma\text{ denotes the
cycle type of }\sigma\text{)}}}\underbrace{\left(  R_{\beta},L_{\gamma\left(
\sigma\right)  }\right)  }_{\substack{=\delta_{\beta,\gamma\left(
\sigma\right)  }\\\text{(by (\ref{sol.lyndon.grfun.count-perms.RLdual}%
),}\\\text{applied to }\alpha=\beta\\\text{and }\zeta=\gamma\left(
\sigma\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the dual pairing }%
\operatorname{NSym}\otimes\operatorname{QSym}\overset{(\cdot,\cdot
)}{\longrightarrow} \kk \text{ is } \kk\text{-bilinear}\right) \\
&  =\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda}}\delta_{\beta,\gamma\left(  \sigma\right)  }=\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta=\gamma\left(  \sigma\right)  }}\underbrace{\delta
_{\beta,\gamma\left(  \sigma\right)  }}_{\substack{=1\\\text{(since }%
\beta=\gamma\left(  \sigma\right)  \text{)}}}+\sum\limits_{\substack{\sigma
\in\mathfrak{S}_{n};\\\operatorname*{type}\sigma=\lambda;\\\beta\neq
\gamma\left(  \sigma\right)  }}\underbrace{\delta_{\beta,\gamma\left(
\sigma\right)  }}_{\substack{=0\\\text{(since }\beta\neq\gamma\left(
\sigma\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\sigma\in\mathfrak{S}_{n}\text{ satisfies either }%
\beta=\gamma\left(  \sigma\right)  \text{ or }\beta\neq\gamma\left(
\sigma\right) \\
\text{(but not both at the same time)}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta=\gamma\left(  \sigma\right)  }}1+\underbrace{\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta\neq\gamma\left(  \sigma\right)  }}0}_{=0}%
=\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta=\gamma\left(  \sigma\right)  }}1\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right)  \cdot1\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right)  .
\end{align*}

\end{vershort}

\begin{verlong}
Now, (\ref{sol.lyndon.grfun.count-perms.b.3}) shows that%
\begin{align*}
\left(  \mathbf{GR}_{\lambda},s_{\operatorname*{Rib}\left(  \beta\right)
}\right)   &  =\left(  R_{\beta},\underbrace{i\left(  \mathbf{GR}_{\lambda
}\right)  }_{=\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\sigma\text{
has cycle type }\lambda}}L_{\gamma\left(  \sigma\right)  }}\right)  =\left(
R_{\beta},\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\sigma\text{ has
cycle type }\lambda}}L_{\gamma\left(  \sigma\right)  }\right) \\
&  =\underbrace{\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n}%
;\\\sigma\text{ has cycle type }\lambda}}}_{\substack{=\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda}}\\\text{(by (\ref{sol.lyndon.grfun.count-perms.sum=sum}))}%
}}\underbrace{\left(  R_{\beta},L_{\gamma\left(  \sigma\right)  }\right)
}_{\substack{=\delta_{\beta,\gamma\left(  \sigma\right)  }\\\text{(by
(\ref{sol.lyndon.grfun.count-perms.RLdual}),}\\\text{applied to }\alpha
=\beta\\\text{and }\zeta=\gamma\left(  \sigma\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the dual pairing }%
\operatorname{NSym}\otimes\operatorname{QSym}\overset{(\cdot,\cdot
)}{\longrightarrow} \kk \text{ is } \kk\text{-bilinear}\right) \\
&  =\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda}}\delta_{\beta,\gamma\left(  \sigma\right)  }=\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta=\gamma\left(  \sigma\right)  }}\underbrace{\delta
_{\beta,\gamma\left(  \sigma\right)  }}_{\substack{=1\\\text{(since }%
\beta=\gamma\left(  \sigma\right)  \text{)}}}+\sum\limits_{\substack{\sigma
\in\mathfrak{S}_{n};\\\operatorname*{type}\sigma=\lambda;\\\beta\neq
\gamma\left(  \sigma\right)  }}\underbrace{\delta_{\beta,\gamma\left(
\sigma\right)  }}_{\substack{=0\\\text{(since }\beta\neq\gamma\left(
\sigma\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\sigma\in\mathfrak{S}_{n}\text{ satisfies either }%
\beta=\gamma\left(  \sigma\right)  \text{ or }\beta\neq\gamma\left(
\sigma\right) \\
\text{(but not both at the same time)}%
\end{array}
\right) \\
&  =\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta=\gamma\left(  \sigma\right)  }}1+\underbrace{\sum
\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta\neq\gamma\left(  \sigma\right)  }}0}_{=0}%
=\sum\limits_{\substack{\sigma\in\mathfrak{S}_{n};\\\operatorname*{type}%
\sigma=\lambda;\\\beta=\gamma\left(  \sigma\right)  }}1\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right)  \cdot1\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right)  .
\end{align*}

\end{verlong}

\noindent Combining this with (\ref{sol.lyndon.grfun.count-perms.b.1}), we
obtain%
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\beta=\gamma\left(
\sigma\right)  \right) \\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\operatorname{Des}%
\sigma=D\left(  \beta\right)  \right) \\
&  =\left(  \mathbf{GR}_{\lambda},s_{\operatorname*{Rib}\left(  \beta\right)
}\right)  .
\end{align*}
This proves Proposition \ref{prop.lyndon.grfun.count-perms}(b).

(a) For each permutation $\sigma\in\mathfrak{S}_{n}$, we have the logical
equivalence%
\[
\left(  \beta\text{ refines }\gamma\left(  \sigma\right)  \right)
\ \Longleftrightarrow\ \left(  \operatorname*{Des}\sigma\subset D\left(
\beta\right)  \right)
\]
(by Lemma \ref{lem.sol.lyndon.grfun.count-perms.1}(c)). Hence,%
\begin{align}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\beta\text{
refines }\gamma\left(  \sigma\right)  \right) \nonumber\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\operatorname{Des}%
\sigma\subset D\left(  \beta\right)  \right)  .
\label{sol.lyndon.grfun.count-perms.a.1}%
\end{align}


On the other hand, if $\alpha\in\operatorname*{Comp}\nolimits_{n}$ is a
composition, then%
\begin{align}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\gamma\left(
\sigma\right)  =\alpha\right) \nonumber\\
&  =\left(  \mathbf{GR}_{\lambda},s_{\operatorname{Rib}\left(  \alpha\right)
}\right)  . \label{sol.lyndon.grfun.count-perms.a.2}%
\end{align}


\begin{vershort}
[\textit{Proof of (\ref{sol.lyndon.grfun.count-perms.a.2}):} Let $\alpha
\in\operatorname*{Comp}\nolimits_{n}$ be a composition. Write $\alpha$ in the
form $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $.
Thus, $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp}\nolimits_{n}$. Hence, Proposition
\ref{prop.lyndon.grfun.count-perms}(b) (applied to $\alpha$ and $\ell$ instead
of $\beta$ and $k$) yields
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\alpha=\gamma\left(
\sigma\right)  \right) \\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\operatorname{Des}%
\sigma=D\left(  \alpha\right)  \right) \\
&  =\left(  \mathbf{GR}_{\lambda},s_{\operatorname{Rib}\left(  \alpha\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{this is the Hall inner product of
}\mathbf{GR}_{\lambda}\in\Lambda\text{ and }s_{\operatorname{Rib}\left(
\alpha\right)  }\in\Lambda\right)  .
\end{align*}
But this immediately yields (\ref{sol.lyndon.grfun.count-perms.a.2}) (since
the condition ``$\gamma\left(  \sigma\right)  =\alpha$'' is equivalent to
``$\alpha=\gamma\left(  \sigma\right)  $'').]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{sol.lyndon.grfun.count-perms.a.2}):} Let $\alpha
\in\operatorname*{Comp}\nolimits_{n}$ be a composition. Write $\alpha$ in the
form $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)  $.
Thus, $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp}\nolimits_{n}$ (since $\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right)  =\alpha\in\operatorname*{Comp}\nolimits_{n}%
$). Hence, Proposition \ref{prop.lyndon.grfun.count-perms}(b) (applied to
$\alpha$ and $\ell$ instead of $\beta$ and $k$) yields
\begin{align}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\alpha=\gamma\left(
\sigma\right)  \right) \nonumber\\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\operatorname{Des}%
\sigma=D\left(  \alpha\right)  \right) \nonumber\\
&  =\left(  \mathbf{GR}_{\lambda},s_{\operatorname{Rib}\left(  \alpha\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{this is the Hall inner product of
}\mathbf{GR}_{\lambda}\in\Lambda\text{ and }s_{\operatorname{Rib}\left(
\alpha\right)  }\in\Lambda\right)  .
\label{sol.lyndon.grfun.count-perms.a.2.pf.1}%
\end{align}
Now,%
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\gamma\left(
\sigma\right)  =\alpha\right) \\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\alpha
=\gamma\left(  \sigma\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the condition ``}
\gamma\left(  \sigma\right)  =\alpha\text{'' is equivalent
to ``}\alpha=\gamma\left(  \sigma\right) \text{''}\right) \\
&  =\left(  \mathbf{GR}_{\lambda},s_{\operatorname{Rib}\left(  \alpha\right)
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{sol.lyndon.grfun.count-perms.a.2.pf.1})}\right)  .
\end{align*}
This proves (\ref{sol.lyndon.grfun.count-perms.a.2}).]
\end{verlong}

If $\sigma\in\mathfrak{S}_{n}$ is any permutation, then $\gamma\left(
\sigma\right)  \in\operatorname*{Comp}\nolimits_{n}$ (by the definition of
$\gamma\left(  \sigma\right)  $). Hence,%
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\beta\text{
refines }\gamma\left(  \sigma\right)  \right) \\
&  =\sum_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n};\\\beta\text{
refines }\alpha}}\underbrace{\left(  \text{the number of permutations }%
\sigma\in\mathfrak{S}_{n}\text{ satisfying }\operatorname*{type}\sigma
=\lambda\text{ such that }\gamma\left(  \sigma\right)  =\alpha\right)
}_{\substack{=\left(  \mathbf{GR}_{\lambda},s_{\operatorname{Rib}\left(
\alpha\right)  }\right)  \\\text{(by (\ref{sol.lyndon.grfun.count-perms.a.2}%
))}}}\\
&  =\sum_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n};\\\beta\text{
refines }\alpha}}\left(  \mathbf{GR}_{\lambda},s_{\operatorname{Rib}\left(
\alpha\right)  }\right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left(  \mathbf{GR}_{\lambda},\underbrace{h_{\mu}}_{\substack{=\sum
\limits_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n};\\\beta\text{
refines }\alpha}}s_{\operatorname*{Rib}\left(  \alpha\right)  }\\\text{(by
Lemma \ref{lem.sol.lyndon.grfun.count-perms.h-through-s})}}}\right)   &
=\left(  \mathbf{GR}_{\lambda},\sum_{\substack{\alpha\in\Comp_{n};\\
\beta\text{ refines }\alpha}}s_{\operatorname*{Rib}\left(
\alpha\right)  }\right)  =\sum_{\substack{\alpha\in\Comp_{n};\\
\beta\text{ refines }\alpha}}\left(  \mathbf{GR}_{\lambda
},s_{\operatorname{Rib}\left(  \alpha\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Hall inner product is
} \kk\text{-bilinear}\right)  ,
\end{align*}
we obtain%
\begin{align}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\beta\text{
refines }\gamma\left(  \sigma\right)  \right) \nonumber\\
&  =\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  .
\label{sol.lyndon.grfun.count-perms.a.3}%
\end{align}

Recall that $\mathbf{GR}_{\lambda}\in\Lambda$; also, $\mu$ is a weak
composition (since $\mu$ is a partition).
Furthermore, $\mu$ is a partition; thus, all the
nonzero entries of $\mu$ are concentrated at the beginning of $\mu$, and
followed by an infinite string of zeroes. Hence, $\mu$ is the partition
consisting of the nonzero entries of $\mu$ (sorted in decreasing
order)\footnote{Indeed, no sorting is required: The partition $\mu$ simply
consists of the nonzero entries of $\mu$.}. Therefore, Exercise
\ref{exe.Lambda.hall-coefficient} (applied to $\mathbf{GR}_{\lambda}$ and
$\mu$ instead of $f$ and $\beta$) yields%
\begin{equation}
\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  =\left(  h_{\mu},\mathbf{GR}%
_{\lambda}\right)  =\left(  \text{the coefficient of }\mathbf{x}^{\mu}\text{
in }\mathbf{GR}_{\lambda}\right)  .\label{sol.lyndon.grfun.count-perms.a.4}%
\end{equation}


\begin{vershort}
Finally, let $\delta$ be the weak composition $\left(  \beta_{1},\beta
_{2},\ldots,\beta_{k},0,0,0,\ldots\right)  $. Then,
$\mathbf{x}^{\delta}=x_{1}^{\beta_{1}}x_{2}^{\beta
_{2}}\cdots x_{k}^{\beta_{k}}$. Furthermore, the nonzero entries of $\delta$
are precisely the numbers $\beta_{1},\beta_{2},\ldots,\beta_{k}$ (since
$\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)  \in\operatorname*{Comp}$
shows that $\beta_{1},\beta_{2},\ldots,\beta_{k}$ are positive integers and
thus nonzero). In other words, the nonzero entries of $\delta$ are precisely
the entries of $\beta$ (since $\beta=\left(  \beta_{1},\beta_{2},\ldots
,\beta_{k}\right)  $). Hence, $\mu$ is the partition consisting of the nonzero
entries of $\delta$ (sorted in decreasing order)\footnote{Indeed, the
definition of $\mu$ shows that $\mu$ is the partition consisting of the
entries of $\beta$ (sorted in decreasing order). But since the nonzero entries
of $\delta$ are precisely the entries of $\beta$, we can rewrite this as
follows: $\mu$ is the partition consisting of the nonzero entries of $\delta$
(sorted in decreasing order).}. Therefore, Exercise
\ref{exe.Lambda.hall-coefficient} (applied to $\mathbf{GR}_{\lambda}$ and
$\delta$ instead of $f$ and $\delta$) yields%
\[
\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  =\left(  h_{\mu},\mathbf{GR}%
_{\lambda}\right)  =\left(  \text{the coefficient of }\mathbf{x}^{\delta
}\text{ in }\mathbf{GR}_{\lambda}\right)  .
\]
In view of $\mathbf{x}^{\delta}=x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots
x_{k}^{\beta_{k}}$, this rewrites as
\[
\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  =\left(  h_{\mu},\mathbf{GR}%
_{\lambda}\right)  =\left(  \text{the coefficient of }x_{1}^{\beta_{1}}%
x_{2}^{\beta_{2}}\cdots x_{k}^{\beta_{k}}\text{ in }\mathbf{GR}_{\lambda
}\right)  .
\]

\end{vershort}

\begin{verlong}
Finally, let $\delta$ be the weak composition $\left(  \beta_{1},\beta
_{2},\ldots,\beta_{k},0,0,0,\ldots\right)  $. Then,
\begin{equation}
\mathbf{x}^{\delta}=x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots x_{k}^{\beta_{k}%
}\label{sol.lyndon.grfun.count-perms.a.xdel=}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.lyndon.grfun.count-perms.a.xdel=}):} The
definition of $\delta$ yields $\delta=\left(  \beta_{1},\beta_{2},\ldots
,\beta_{k},0,0,0,\ldots\right)  $. But $\delta=\left(  \delta_{1},\delta
_{2},\delta_{3},\ldots\right)  $. Thus,%
\[
\left(  \delta_{1},\delta_{2},\delta_{3},\ldots\right)  =\delta=\left(
\beta_{1},\beta_{2},\ldots,\beta_{k},0,0,0,\ldots\right)  .
\]
In other words, each positive integer $i$ satisfies%
\begin{equation}
\delta_{i}=%
\begin{cases}
\beta_{i}, & \text{if }i\leq k;\\
0, & \text{if }i>k
\end{cases}
\ \ \ .\label{pf.sol.lyndon.grfun.count-perms.a.xdel=.pf.1}%
\end{equation}
\par
Hence, each $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies%
\begin{align*}
\delta_{i} &  =%
\begin{cases}
\beta_{i}, & \text{if }i\leq k;\\
0, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.sol.lyndon.grfun.count-perms.a.xdel=.pf.1})}\right)  \\
&  =\beta_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq k\text{ (because
}i\in\left\{  1,2,\ldots,k\right\}  \text{)}\right)
\end{align*}
and therefore $x_{i}^{\delta_{i}}=x_{i}^{\beta_{i}}$. Multiplying these
equalities over all $i\in\left\{  1,2,\ldots,k\right\}  $, we find
$x_{1}^{\delta_{1}}x_{2}^{\delta_{2}}\cdots x_{k}^{\delta_{k}}=x_{1}%
^{\beta_{1}}x_{2}^{\beta_{2}}\cdots x_{k}^{\beta_{k}}$.
\par
Also, each $i\in\left\{  k+1,k+2,k+3,\ldots\right\}  $ satisfies%
\begin{align*}
\delta_{i} &  =%
\begin{cases}
\beta_{i}, & \text{if }i\leq k;\\
0, & \text{if }i>k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.sol.lyndon.grfun.count-perms.a.xdel=.pf.1})}\right)  \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>k\text{ (because }i\in\left\{
k+1,k+2,k+3,\ldots\right\}  \text{)}\right)
\end{align*}
and therefore $x_{i}^{\delta_{i}}=x_{i}^{0}=1$. Multiplying these equalities
over all $i\in\left\{  k+1,k+2,k+3,\ldots\right\}  $, we find $x_{k+1}%
^{\delta_{k+1}}x_{k+2}^{\delta_{k+2}}x_{k+3}^{\delta_{k+3}}\cdots=1\cdot
1\cdot1\cdot\cdots=1$.
\par
Now, the definition of $\mathbf{x}^{\delta}$ yields%
\[
\mathbf{x}^{\delta}=x_{1}^{\delta_{1}}x_{2}^{\delta_{2}}x_{3}^{\delta_{3}%
}\cdots=\underbrace{\left(  x_{1}^{\delta_{1}}x_{2}^{\delta_{2}}\cdots
x_{k}^{\delta_{k}}\right)  }_{=x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots
x_{k}^{\beta_{k}}}\underbrace{\left(  x_{k+1}^{\delta_{k+1}}x_{k+2}%
^{\delta_{k+2}}x_{k+3}^{\delta_{k+3}}\cdots\right)  }_{=1}=x_{1}^{\beta_{1}%
}x_{2}^{\beta_{2}}\cdots x_{k}^{\beta_{k}}.
\]
This proves (\ref{sol.lyndon.grfun.count-perms.a.xdel=}).}. Furthermore, the
nonzero entries of $\delta$ are precisely the entries of $\beta$%
\ \ \ \ \footnote{\textit{Proof.} We have $\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{k}\right)  \in\operatorname{Comp}_{n}\subset
\operatorname*{Comp}$. In other words, $\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{k}\right)  $ is a composition, i.e., a tuple of positive
integers. In other words, $\beta_{1},\beta_{2},\ldots,\beta_{k}$ are positive
integers. Hence, the numbers $\beta_{1},\beta_{2},\ldots,\beta_{k}$ are
nonzero.
\par
But $\delta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{k},0,0,0,\ldots\right)
$ (by the definition of $\delta$). Hence,%
\begin{align*}
&  \left(  \text{the nonzero entries of }\delta\right)  \\
&  =\left(  \text{the nonzero entries of }\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{k},0,0,0,\ldots\right)  \right)  \\
&  =\left(  \text{the nonzero numbers among }\beta_{1},\beta_{2},\ldots
,\beta_{k},0,0,0,\ldots\right)  \\
&  =\left(  \text{the numbers }\beta_{1},\beta_{2},\ldots,\beta_{k}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the numbers }\beta_{1},\beta
_{2},\ldots,\beta_{k}\text{ are nonzero, while the numbers }0,0,0,\ldots\text{
are zero}\right)  \\
&  =\left(  \text{the entries of }\beta\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{k}\right)
\right)  .
\end{align*}
In other words, the nonzero entries of $\delta$ are precisely the entries of
$\beta$.}. In other words, the entries of $\beta$ are precisely the nonzero
entries of $\delta$. Hence, $\mu$ is the partition consisting of the nonzero
entries of $\delta$ (sorted in decreasing order)\footnote{\textit{Proof.}
Recall that $\mu$ is the partition obtained by sorting the entries of $\beta$
into decreasing order. Thus,
\begin{align*}
\mu &  =\left(  \text{the partition consisting of the entries of }\beta\text{
(sorted in decreasing order)}\right)  \\
&  =\left(  \text{the partition consisting of the nonzero entries of }%
\delta\text{ (sorted in decreasing order)}\right)
\end{align*}
(since the entries of $\beta$ are precisely the nonzero entries of $\delta$).
Thus, we have shown that $\mu$ is the partition consisting of the nonzero
entries of $\delta$ (sorted in decreasing order)}. Therefore, Exercise
\ref{exe.Lambda.hall-coefficient} (applied to $\mathbf{GR}_{\lambda}$ and
$\delta$ instead of $f$ and $\delta$) yields%
\[
\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  =\left(  h_{\mu},\mathbf{GR}%
_{\lambda}\right)  =\left(  \text{the coefficient of }\mathbf{x}^{\delta
}\text{ in }\mathbf{GR}_{\lambda}\right)  .
\]
Thus,%
\begin{align*}
\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)   &  =\left(  \text{the
coefficient of }\underbrace{\mathbf{x}^{\delta}}_{\substack{=x_{1}^{\beta_{1}%
}x_{2}^{\beta_{2}}\cdots x_{k}^{\beta_{k}}\\\text{(by
(\ref{sol.lyndon.grfun.count-perms.a.xdel=}))}}}\text{ in }\mathbf{GR}%
_{\lambda}\right)  \\
&  =\left(  \text{the coefficient of }x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots
x_{k}^{\beta_{k}}\text{ in }\mathbf{GR}_{\lambda}\right)  .
\end{align*}

\end{verlong}

\noindent Combining this equality with the equalities
(\ref{sol.lyndon.grfun.count-perms.a.1}),
(\ref{sol.lyndon.grfun.count-perms.a.3}) and
(\ref{sol.lyndon.grfun.count-perms.a.4}), we obtain%
\begin{align*}
&  \left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ such that }\beta\text{
refines }\gamma\left(  \sigma\right)  \right)  \\
&  =\left(  \text{the number of permutations }\sigma\in\mathfrak{S}_{n}\text{
satisfying }\operatorname*{type}\sigma=\lambda\text{ and }\operatorname{Des}%
\sigma\subset D\left(  \beta\right)  \right)  \\
&  =\left(  \text{the coefficient of }x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\cdots
x_{k}^{\beta_{k}}\text{ in }\mathbf{GR}_{\lambda}\right)  =\left(  \text{the
coefficient of }\mathbf{x}^{\mu}\text{ in }\mathbf{GR}_{\lambda}\right)  \\
&  =\left(  \mathbf{GR}_{\lambda},h_{\mu}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{this is the Hall inner product of }\mathbf{GR}_{\lambda}\in\Lambda\text{
and }h_{\mu}\in\Lambda\right)  .
\end{align*}
This proves Proposition \ref{prop.lyndon.grfun.count-perms}(a).
\end{proof}

Thus, Proposition \ref{prop.lyndon.grfun.count-perms}(a) is proved, so that
Exercise \ref{exe.lyndon.grfun.count-perms} is solved.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.zeta-polynomiality.altpf}}

\textit{Solution to Exercise~\ref{exe.zeta-polynomiality.altpf}.}
Fix $m \in \{0,1,2,\ldots\}$.
In order to check that $\zeta_Q^{\star m}(f) = \ps^1(f)(m)$, it is
clearly enough to show that $\zeta_Q^{\star m}\left(M_\alpha\right)
= \ps^1\left(M_\alpha\right)\left(m\right)$ for all compositions
$\alpha$. So let $\alpha$ be a composition. Iterated application
of Proposition~\ref{Qsym-coproduct-on-monomials} yields
\[
\Delta^{(m-1)} M_\alpha
= \sum\limits_{\substack{\left(\beta_1,\beta_2,\ldots,\beta_m\right) : \\
                  \beta_1 \beta_2 \cdots \beta_m = \alpha}}
  M_{\beta_1} \otimes M_{\beta_2} \otimes \cdots \otimes M_{\beta_m}.
\]
An addend on the right hand side of this equality is annihilated
by the map $\zeta_Q^{\otimes m} : \Qsym^{\otimes m} \to \kk^{\otimes m}
\cong \kk$ unless each of the compositions
$\beta_1,\beta_2,\ldots,\beta_m$ has length $\leq 1$; all
remaining terms are mapped to $1 \cdot 1 \cdots 1 = 1$. Hence,
\[
\zeta_Q^{\otimes m}\left(\Delta^{(m-1)} M_\alpha\right)
= \sum\limits_{\substack{\left(\beta_1,\beta_2,\ldots,\beta_m\right) : \\
                  \beta_1 \beta_2 \cdots \beta_m = \alpha ; \\
                  \text{each of the compositions }
                  \beta_1,\beta_2,\ldots,\beta_m
                  \text{ has length } \leq 1}}
  1
= \binom{m}{\ell},
\]
where $\ell$ is the length of $\alpha$.
Hence,
\[
\zeta_Q^{\star m}\left(M_\alpha\right)
= \zeta_Q^{\otimes m}\left(\Delta^{(m-1)} M_\alpha\right)
= \binom{m}{\ell} = \ps^1\left(M_\alpha\right)\left(m\right)
\]
(by Proposition~\ref{zeta-polynomiality-proposition}(i)).
This completes the proof.
%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.GGG.moebius-basis}}

\textit{Solution to Exercise~\ref{exe.GGG.moebius-basis}.}

\begin{proof}
[Proof of Proposition \ref{prop.GGG.moebius-basis}.]We recall the following
fundamental fact (a version of inclusion-exclusion or a very simple special
case of M\"{o}bius inversion): If $R$ is a finite set, then%
\begin{equation}
\sum_{T\subset R}\left(  -1\right)  ^{\left\vert T\right\vert }=\delta
_{R,\varnothing}. \label{pf.prop.GGG.moebius-basis.moebius.1}%
\end{equation}
\footnote{For the sake of completeness, here is a short \textit{proof of
(\ref{pf.prop.GGG.moebius-basis.moebius.1}):} Let $R$ be a finite set. We have%
\begin{align*}
\sum_{T\subset R}\left(  -1\right)  ^{\left\vert T\right\vert }  &
=\sum_{k\in\NN}\sum\limits_{\substack{T\subset R;\\\left\vert T\right\vert
=k}}\underbrace{\left(  -1\right)  ^{\left\vert T\right\vert }}%
_{\substack{=\left(  -1\right)  ^{k}\\\text{(since }\left\vert T\right\vert
=k\text{)}}}=\sum_{k\in\NN}\underbrace{\sum\limits_{\substack{T\subset
R;\\\left\vert T\right\vert =k}}\left(  -1\right)  ^{k}}_{=\left\vert \left\{
T\subset R\ \mid\ \left\vert T\right\vert =k\right\}  \right\vert \left(
-1\right)  ^{k}}=\sum_{k\in\NN}\underbrace{\left\vert \left\{  T\subset
R\ \mid\ \left\vert T\right\vert =k\right\}  \right\vert }_{\substack{=\left(
\text{the number of all }k\text{-element subsets of }R\right)  \\=\dbinom
{\left\vert R\right\vert }{k}\\\text{(by the combinatorial interpretation
of}\\\text{binomial coefficients)}}}\left(  -1\right)  ^{k}\\
&  =\sum_{k\in\NN}\dbinom{\left\vert R\right\vert }{k}\left(
-1\right)  ^{k}=\left(  \underbrace{1+\left(  -1\right)  }_{=0}\right)
^{\left\vert R\right\vert }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  1+\left(  -1\right)
\right)  ^{\left\vert R\right\vert }=\sum_{k\in\NN}\dbinom{\left\vert
R\right\vert }{k}\left(  -1\right)  ^{k}\text{ (by the binomial formula)}%
\right) \\
&  =0^{\left\vert R\right\vert }=\delta_{\left\vert R\right\vert ,0}%
=\delta_{R,\varnothing}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert
R\right\vert =0\text{ holds if and only if }R=\varnothing\right)  .
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.moebius.1}).} We can use this to
show a slightly more complicated fact: If $P$ and $R$ are two finite sets,
then%
\begin{equation}
\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }=\delta_{P,R}.
\label{pf.prop.GGG.moebius-basis.moebius.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.moebius.2}):} Let
$P$ and $R$ be two finite sets.
\par
Let us first assume that $P\not \subset R$. Then, there exists no $F\subset R$
such that $F\supset P$ (because if such a $F$ would exist, then we would have
$P\subset F\subset R$, which would contradict $P\not \subset R$). Hence, the
sum $\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }$ is empty, and thus it simplifies to $\sum\limits
_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }=0$. On the other hand, $P\not \subset R$, so that
$P\neq R$ and thus $\delta_{P,R}=0$. Hence, $\sum\limits_{\substack{F\subset
R;\\F\supset P}}\left(  -1\right)  ^{\left\vert F\setminus P\right\vert
}=0=\delta_{P,R}$. Thus, (\ref{pf.prop.GGG.moebius-basis.moebius.2}) is proven
under the assumption that $P\not \subset R$.
\par
Now, let us forget that we have assumed $P\not \subset R$. We thus have shown
that (\ref{pf.prop.GGG.moebius-basis.moebius.2}) holds if $P\not \subset R$.
Hence, for the rest of this proof, we can WLOG assume that we don't have
$P\not \subset R$. Assume this.
\par
We have $P\subset R$ (since we don't have $P\not \subset R$). Hence, there
exists a bijection from the set of all $F\subset R$ satisfying $F\supset P$ to
the set of all $T\subset R\setminus P$; this bijection sends every $F$ to
$F\setminus P$. Hence, we can substitute $T$ for $F\setminus P$ in the sum
$\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }$. We thus obtain
\begin{align*}
\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }  &  =\sum_{T\subset R\setminus P}\left(  -1\right)
^{\left\vert T\right\vert }=\delta_{R\setminus P,\varnothing}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.GGG.moebius-basis.moebius.1}), applied to }R\setminus P\text{
instead of }R\right) \\
&  =\delta_{P,R}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R\setminus
P=\varnothing\text{ if and only if }P=R\text{ (because }P\subset
R\text{)}\right)  .
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.moebius.2}).}

(a) Let $G=\left(  V,E\right)  $ be a finite graph. We have%
\begin{align*}
\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left[  H\right]  ^{\sharp}  &  =\underbrace{\sum\limits
_{\substack{K=\left(  V,F\right)  ;\\F\cap E=\varnothing}}}_{\substack{=\sum\limits
_{\substack{K=\left(  V,F\right)  ;\\F\subset E^{c}}}\\\text{(since }F\cap
E=\varnothing\text{ is}\\\text{equivalent to }F\subset E^{c}\text{)}%
}}\underbrace{\left[  K\right]  ^{\sharp}}_{\substack{=\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset F^{c}}}\left(
-1\right)  ^{\left\vert E^{\prime}\setminus F^{c}\right\vert }\left[
H\right]  \\\text{(by the definition of }\left[  K\right]  ^{\sharp}\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }H\text{ and }E^{\prime
}\text{ as }K\text{ and }F\text{ in the sum}\right) \\
&  =\underbrace{\sum\limits_{\substack{K=\left(  V,F\right)  ;\\F\subset E^{c}}%
}\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset F^{c}}%
}}_{\substack{=\sum_{H=\left(  V,E^{\prime}\right)  }\sum\limits_{\substack{K=\left(
V,F\right)  ;\\F\subset E^{c};\\E^{\prime}\supset F^{c}}}}}\underbrace{\left(
-1\right)  ^{\left\vert E^{\prime}\setminus F^{c}\right\vert }}%
_{\substack{=\left(  -1\right)  ^{\left\vert F\setminus\left(  E^{\prime
}\right)  ^{c}\right\vert }\\\text{(since }E^{\prime}\setminus F^{c}%
=E^{\prime}\cup F=F\setminus\left(  E^{\prime}\right)  ^{c}\text{)}}}\left[
H\right] \\
&  =\sum_{H=\left(  V,E^{\prime}\right)  }\underbrace{\sum\limits
_{\substack{K=\left(  V,F\right)  ;\\F\subset E^{c};\\E^{\prime}\supset F^{c}%
}}}_{\substack{=\sum\limits_{\substack{F\subset E^{c};\\E^{\prime}\supset F^{c}%
}}=\sum\limits_{\substack{F\subset E^{c};\\F\supset\left(  E^{\prime}\right)  ^{c}%
}}\\\text{(since }E^{\prime}\supset F^{c}\text{ is equivalent to }%
F\supset\left(  E^{\prime}\right)  ^{c}\text{)}}}\left(  -1\right)
^{\left\vert F\setminus\left(  E^{\prime}\right)  ^{c}\right\vert }\left[
H\right] \\
&  =\sum_{H=\left(  V,E^{\prime}\right)  }\underbrace{\sum\limits_{\substack{F\subset
E^{c};\\F\supset\left(  E^{\prime}\right)  ^{c}}}\left(  -1\right)
^{\left\vert F\setminus\left(  E^{\prime}\right)  ^{c}\right\vert }%
}_{\substack{=\delta_{\left(  E^{\prime}\right)  ^{c},E^{c}}\\\text{(by
(\ref{pf.prop.GGG.moebius-basis.moebius.2}), applied to}\\R=E^{c}\text{ and
}P=\left(  E^{\prime}\right)  ^{c}\text{)}}}\left[  H\right] \\
&  =\sum_{H=\left(  V,E^{\prime}\right)  }\underbrace{\delta_{\left(
E^{\prime}\right)  ^{c},E^{c}}}_{\substack{=\delta_{E^{\prime},E}%
\\\text{(since }\left(  E^{\prime}\right)  ^{c}=E^{c}\text{ holds
if}\\\text{and only if }E^{\prime}=E\text{)}}}\left[  H\right]  =\sum
_{H=\left(  V,E^{\prime}\right)  }\delta_{E^{\prime},E}\left[  H\right]
=\left[  \left(  V,E\right)  \right]  =\left[  G\right]  .
\end{align*}


This proves Proposition \ref{prop.GGG.moebius-basis}(a).

(b) For every finite graph $G=\left(  V,E\right)  $, we define the
\textbf{complement} of this graph $G$ to be the graph $\left(  V,E^{c}\right)
$ (where $E^{c}$ is defined as in Definition \ref{def.GGG.moebius-basis}). We
shall denote this complement by $\mathbf{c}\left(  G\right)  $.

The complement of the complement of a graph $G$ is $G$ again. In other words,%
\begin{equation}
\mathbf{c}\left(  \mathbf{c}\left(  G\right)  \right)
=G\ \ \ \ \ \ \ \ \ \ \text{for every finite graph }%
G.\label{pf.prop.GGG.moebius-basis.b.ccG}%
\end{equation}


Recall that if $G$ is a finite graph, then we denote the isomorphism class of
this graph $G$ by $\left[  G\right]  $. Let $\operatorname*{GrIs}$ be the set
of all isomorphism classes of finite graphs. For every $n\in \NN$, let
$\operatorname*{GrIs}\nolimits_{n}$ be the set of all isomorphism classes of
finite graphs with $n$ vertices. Thus, $\operatorname*{GrIs}=\bigsqcup
_{n\in\NN}\operatorname*{GrIs}\nolimits_{n}$.

Fix $n\in \NN$. The family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the
$\kk$-module $\mathcal{G}_{n}$ (by the definition of the grading on
$\mathcal{G}$). We are going to prove that the family $\left(  \left[
G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}%
\nolimits_{n}}$ is a basis of the $\kk$-module $\mathcal{G}_{n}$ as well.

We define a map $\mathbf{c}_{n}:\operatorname*{GrIs}\nolimits_{n}%
\rightarrow\operatorname*{GrIs}\nolimits_{n}$ by setting%
\[
\left(  \mathbf{c}_{n}\left[  G\right]  =\left[  \mathbf{c}\left(  G\right)
\right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}\right)  .
\]
\footnote{This map $\mathbf{c}_{n}$ is well-defined because of the following
two simple observations:
\par
\begin{itemize}
\item For each $\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}$, we
have $\left[  \mathbf{c}\left(  G\right)  \right]  \in\operatorname*{GrIs}%
\nolimits_{n}$ (because if the graph $G$ has $n$ vertices, then so does the
graph $\mathbf{c}\left(  G\right)  $).
\par
\item For each graph $G$ with $n$ vertices, the isomorphism class $\left[
\mathbf{c}\left(  G\right)  \right]  $ depends only on the isomorphism class
$\left[  G\right]  $, not on the graph $G$ itself (i.e., if $G_{1}$ and
$G_{2}$ are two isomorphic graphs with $n$ vertices, then the graphs
$\mathbf{c}\left(  G_{1}\right)  $ and $\mathbf{c}\left(  G_{2}\right)  $ are
also isomorphic).
\end{itemize}
}

We have $\mathbf{c}_{n}\circ\mathbf{c}_{n}=\id$
\ \ \ \ \footnote{\textit{Proof.} Let $U\in\operatorname*{GrIs}\nolimits_{n}%
$. Thus, $U$ is an isomorphism class of a finite graph with $n$ vertices
(since $\operatorname*{GrIs}\nolimits_{n}$ is the set of all isomorphism
classes of finite graphs with $n$ vertices). In other words, there exists a
graph $G$ with $n$ vertices such that $U=\left[  G\right]  $. Consider this
$G$. Now, $\mathbf{c}_{n}\underbrace{U}_{=\left[  G\right]  }=\mathbf{c}%
_{n}\left[  G\right]  =\left[  \mathbf{c}\left(  G\right)  \right]  $ (by the
definition of $\mathbf{c}_{n}$). Now,%
\begin{align*}
\left(  \mathbf{c}_{n}\circ\mathbf{c}_{n}\right)  U &  =\mathbf{c}_{n}\left(
\underbrace{\mathbf{c}_{n}U}_{=\left[  \mathbf{c}\left(  G\right)  \right]
}\right)  =\mathbf{c}_{n}\left[  \mathbf{c}\left(  G\right)  \right]  =\left[
\underbrace{\mathbf{c}\left(  \mathbf{c}\left(  G\right)  \right)
}_{\substack{=G\\\text{(by (\ref{pf.prop.GGG.moebius-basis.b.ccG}))}}}\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{c}_{n}\right)
\\
&  =\left[  G\right]  =U=\operatorname*{id}U.
\end{align*}
Now, forget that we fixed $U$. We thus have proven that $\left(
\mathbf{c}_{n}\circ\mathbf{c}_{n}\right)  U=\operatorname*{id}U$ for every
$U\in\operatorname*{GrIs}\nolimits_{n}$. In other words, $\mathbf{c}_{n}%
\circ\mathbf{c}_{n}=\operatorname*{id}$. Qed.}. Hence, the maps $\mathbf{c}%
_{n}$ and $\mathbf{c}_{n}$ are mutually inverse. Thus, the map $\mathbf{c}%
_{n}$ is invertible, i.e., a bijection. Therefore, the family $\left(  \left[
G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}%
\nolimits_{n}}$ is a reindexing of the family $\left(  \left(  \mathbf{c}%
_{n}\left[  G\right]  \right)  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$.

If $G=\left(  V,E\right)  $ is a finite graph with $n$ vertices, then%
\begin{equation}
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}=\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset E}}\left(
-1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[  H\right]
\label{pf.prop.GGG.moebius-basis.b.cmoeb}
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.b.cmoeb}):} Let
$G=\left(  V,E\right)  $ be a finite graph with $n$ vertices. Then,
$\mathbf{c}\left(  G\right)  =\left(  V,E^{c}\right)  $ (by the definition of
$\mathbf{c}\left(  G\right)  $). Hence, the definition of $\left[
\mathbf{c}\left(  G\right)  \right]  ^{\sharp}$ yields
\[
\left[  \mathbf{c}\left(  G\right)  \right]  ^{\sharp}=\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset\left(
E^{c}\right)  ^{c}}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus\left(
E^{c}\right)  ^{c}\right\vert }\left[  H\right]  =\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\supset E}}\left(  -1\right)  ^{\left\vert
E^{\prime}\setminus E\right\vert }\left[  H\right]
\]
(since $\left(  E^{c}\right)  ^{c}=E$). Now, the definition of $\mathbf{c}%
_{n}$ yields $\mathbf{c}_{n}\left[  G\right]  =\left[  \mathbf{c}\left(
G\right)  \right]  $, and therefore $\left(  \mathbf{c}_{n}\left[  G\right]
\right)  ^{\sharp}=\left[  \mathbf{c}\left(  G\right)  \right]  ^{\sharp}%
=\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset
E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[
H\right]  $. This proves (\ref{pf.prop.GGG.moebius-basis.b.cmoeb}).}.

We define a map $e:\operatorname*{GrIs}\nolimits_{n}\rightarrow \NN$ by
setting%
\[
e\left[  \left(  V,E\right)  \right]  =\left\vert E\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }\left[  \left(  V,E\right)  \right]
\in\operatorname*{GrIs}\nolimits_{n}.
\]
\footnote{This map is well-defined, because for any finite graph $\left(
V,E\right)  $, the number $\left\vert E\right\vert $ depends only on the
isomorphism class $\left[  \left(  V,E\right)  \right]  $ of $\left(
V,E\right)  $ (and not on $\left(  V,E\right)  $ itself).} Thus, the map $e$
sends the isomorphism class of a graph to the number of edges of this graph.

We define a binary relation $\prec$ on the set $\operatorname*{GrIs}%
\nolimits_{n}$ as follows: For two elements $\left[  H\right]  $ and $\left[
G\right]  $ of $\operatorname*{GrIs}\nolimits_{n}$, we set $\left[  H\right]
\prec\left[  G\right]  $ if and only if $e\left[  H\right]  >e\left[
G\right]  $. It is clear that this binary relation $\prec$ is transitive,
asymmetric and irreflexive. Thus, there is a partial order on the set
$\operatorname*{GrIs}\nolimits_{n}$ whose smaller relation is $\prec$.
Consider $\operatorname*{GrIs}\nolimits_{n}$ as a poset, equipped with this
partial order.

Now, every finite graph $G$ with $n$ vertices satisfies%
\begin{equation}
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}=\left[  G\right]
+\left(  \text{a }\kk\text{-linear combination of the elements }\left[
H\right]  \text{ for }\left[  H\right]  \in\operatorname*{GrIs}\nolimits_{n}%
\text{ satisfying }\left[  H\right]  \prec\left[  G\right]  \right)
\label{pf.prop.GGG.moebius-basis.b.unitri}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.b.unitri}):} Let
$G$ be a finite graph with $n$ vertices. Write $G$ as $G=\left(  V,E\right)
$.
\par
Let $H=\left(  V,E^{\prime}\right)  $ be any graph satisfying $E^{\prime
}\supset E$ and $E^{\prime}\neq E$. Clearly, the graph $H$ has $n$ vertices
(since it has the same vertex set as $G$, and since $G$ has $n$ vertices).
Thus, $\left[  H\right]  \in\operatorname*{GrIs}\nolimits_{n}$. Moreover, $E$
is a proper subset of $E^{\prime}$ (since $E^{\prime}\supset E$ and
$E^{\prime}\neq E$); hence, $\left\vert E\right\vert <\left\vert E^{\prime
}\right\vert $.
\par
The definition of $e$ yields $e\left[  \left(  V,E\right)  \right]
=\left\vert E\right\vert $ and $e\left[  \left(  V,E^{\prime}\right)  \right]
=\left\vert E^{\prime}\right\vert $. Now, $e\left[  \underbrace{G}_{=\left(
V,E\right)  }\right]  =e\left[  \left(  V,E\right)  \right]  =\left\vert
E\right\vert <\left\vert E^{\prime}\right\vert $, so that $\left\vert
E^{\prime}\right\vert >e\left[  G\right]  $. But $e\left[  \underbrace{H}%
_{=\left(  V,E^{\prime}\right)  }\right]  =e\left[  \left(  V,E^{\prime
}\right)  \right]  =\left\vert E^{\prime}\right\vert >e\left[  G\right]  $. In
other words, $\left[  H\right]  \prec\left[  G\right]  $ (by the definition of
the relation $\prec$).
\par
Now, forget that we fixed $H$. We thus have shown that if $H=\left(
V,E^{\prime}\right)  $ is any graph satisfying $E^{\prime}\supset E$ and
$E^{\prime}\neq E$, then $\left[  H\right]  \in\operatorname*{GrIs}%
\nolimits_{n}$ and $\left[  H\right]  \prec\left[  G\right]  $. Hence,%
\begin{align}
&  \sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset E\text{
and }E^{\prime}\neq E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus
E\right\vert }\left[  H\right]  \nonumber\\
&  =\left(  \text{a }\kk\text{-linear combination of the elements
}\left[  H\right]  \text{ for }\left[  H\right]  \in\operatorname*{GrIs}%
\nolimits_{n}\text{ satisfying }\left[  H\right]  \prec\left[  G\right]
\right)  .\label{pf.prop.GGG.moebius-basis.b.unitri.pf.1}%
\end{align}
\par
But (\ref{pf.prop.GGG.moebius-basis.b.cmoeb}) becomes%
\begin{align*}
&  \left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\\
&  =\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset
E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[
H\right]  \\
&  =\underbrace{\left(  -1\right)  ^{\left\vert E\setminus E\right\vert }%
}_{\substack{=1\\\text{(since }\left\vert E\setminus E\right\vert =\left\vert
\varnothing\right\vert =0\text{)}}}\left[  \underbrace{\left(  V,E\right)
}_{=G}\right]  +\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)
;\\E^{\prime}\supset E\text{ and }E^{\prime}\neq E}}\left(  -1\right)
^{\left\vert E^{\prime}\setminus E\right\vert }\left[  H\right]
}_{\substack{=\left(  \text{a }\kk\text{-linear combination of the
elements }\left[  H\right]  \text{ for }\left[  H\right]  \in
\operatorname*{GrIs}\nolimits_{n}\text{ satisfying }
\left[H\right]\prec \left[G\right]\right)
\\\text{(by (\ref{pf.prop.GGG.moebius-basis.b.unitri.pf.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}E^{\prime}=E\text{ and }H=\left(V,E\right) \text{ from the sum}\right)  \\
&  =\left[  G\right]  +\left(  \text{a }\kk\text{-linear combination of
the elements }\left[  H\right]  \text{ for }\left[  H\right]  \in
\operatorname*{GrIs}\nolimits_{n}\text{ satisfying }\left[  H\right]
\prec\left[  G\right]  \right)  .
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.b.unitri}).}. In other words, the
family $\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp
}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ expands
unitriangularly in the family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ (by Remark
\ref{rmk.STmat.expansion-tria-shortcuts}(c), applied to $\mathcal{G}_{n}$,
$\operatorname*{GrIs}\nolimits_{n}$, $\left(  \left(  \mathbf{c}_{n}\left[
G\right]  \right)  ^{\sharp}\right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ and $\left(  \left[  G\right]  \right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ instead of $M$,
$S$, $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)  _{s\in S}$).
Hence, the family $\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)
^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$
is a basis of the $\kk$-module $\mathcal{G}_{n}$ if and only if the
family $\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$ (by Corollary \ref{cor.STmat.expansion-tria-inv}(e), applied
to $\mathcal{G}_{n}$, $\operatorname*{GrIs}\nolimits_{n}$, $\left(  \left(
\mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ and $\left(  \left[  G\right]  \right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ instead of $M$,
$S$, $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)  _{s\in S}$).
Thus, the family $\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)
^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$
is a basis of the $\kk$-module $\mathcal{G}_{n}$ (since the family
$\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$). Therefore, the family $\left(  \left[  G\right]  ^{\sharp
}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ also is a
basis of the $\kk$-module $\mathcal{G}_{n}$ (since the family $\left(
\left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a reindexing of the family $\left(
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\right)  _{\left[
G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$).

Now, forget that we fixed $n$. We thus have shown that, for every
$n\in \NN$, the family $\left(  \left[  G\right]  ^{\sharp}\right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the
$\kk$-module $\mathcal{G}_{n}$. Therefore, the disjoint union of these
families (over all $n\in \NN$) is a basis of the direct sum
$\bigoplus_{n\in\NN}\mathcal{G}_{n}$. Since the former disjoint union
is the family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$, whereas the latter direct sum is $\bigoplus
_{n\in\NN}\mathcal{G}_{n}=\mathcal{G}$, this rewrites as follows: The
family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$ is a basis of the $\kk$-module $\mathcal{G}$.
In other words, the elements $\left[  G\right]  ^{\sharp}$, where $\left[
G\right]  $ ranges over all isomorphism classes of finite graphs, form a basis
of the $\kk$-module $\mathcal{G}$ (because $\operatorname*{GrIs}$ is
the set of all isomorphism classes of finite graphs). Proposition
\ref{prop.GGG.moebius-basis}(b) is thus proven.

(c) We define a $\kk$-linear map $\Delta^{\prime}:\GGG%
\rightarrow\GGG\otimes\GGG$ by%
\[
\Delta^{\prime}\left[  H\right]  ^{\sharp}=\sum\limits_{\substack{\left(  V_{1}%
,V_{2}\right)  ;\\V=V_{1}\sqcup V_{2};\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}%
}}\left[  H\mid_{V_{1}}\right]  ^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]
^{\sharp}.
\]
\footnote{This is well-defined, since the $\left[  G\right]  ^{\sharp}$ form a
basis of the $\kk$-module $\GGG$.} In order to prove Proposition
\ref{prop.GGG.moebius-basis}(c), it clearly suffices to show that
$\Delta^{\prime}=\Delta$.

Let $G=\left(  V,E\right)  $ be a finite graph. If $T$ is a set of two-element
subsets of $V$, and if $R$ is a subset of $V$, then $T\mid_{R}$ shall denote
the subset $\left\{  X\in T\ \mid\ X\subset R\right\}  $ of $T$. (Thus, if $R$
is a subset of $V$, then $E\mid_{R}$ is the set of edges of the graph
$G\mid_{R}$.)

Now, Proposition \ref{prop.GGG.moebius-basis}(a) yields $\left[  G\right]
=\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left[  H\right]  ^{\sharp}$. Hence,
\begin{align*}
\Delta^{\prime}\left[  G\right]   &  =\Delta^{\prime}\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing}}\left[  H\right]
^{\sharp}=\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\underbrace{\Delta^{\prime}\left[  H\right]  ^{\sharp}}%
_{=\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup V_{2}%
;\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}\left[  H\mid_{V_{1}}\right]
^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp}}\\
&  =\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime
}\cap E=\varnothing}}\sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V=V_{1}\sqcup V_{2};\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}}_{=\sum\limits
_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup V_{2}}}\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
;\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}}\left[  H\mid_{V_{1}}\right]
^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup V_{2}%
}}\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing;\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}\left[  H\mid_{V_{1}%
}\right]  ^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp}%
}_{\substack{=\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}%
}\sum\limits_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime
}\cap\left(  E\mid_{V_{2}}\right)  =\varnothing}}\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing;\\H=H\mid_{V_{1}}\sqcup
H\mid_{V_{2}};\\H\mid_{V_{1}}=H_{1};\ H\mid_{V_{2}}=H_{2}}}\left[
H\mid_{V_{1}}\right]  ^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp
}\\\text{(because for every }H=\left(  V,E^{\prime}\right)  \text{ satisfying
}E^{\prime}\cap E=\varnothing\text{,}\\\text{we can write the subgraph }%
H\mid_{V_{1}}\text{ in the form }H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
\\\text{with some }E_{1}^{\prime}\text{ satisfying }E_{1}^{\prime}\cap\left(
E\mid_{V_{1}}\right)  =\varnothing\text{, and we can}\\\text{write the
subgraph }H\mid_{V_{2}}\text{ in the form }H_{2}=\left(  V_{2},E_{2}^{\prime
}\right)  \\\text{with some }E_{2}^{\prime}\text{ satisfying }E_{2}^{\prime
}\cap\left(  E\mid_{V_{2}}\right)  =\varnothing\text{)}}}\\
&  =\underbrace{\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup
V_{2}}}}_{=\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}%
=V}}}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}%
^{\prime}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}%
=\left(  V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(
E\mid_{V_{2}}\right)  =\varnothing}}\sum\limits_{\substack{H=\left(  V,E^{\prime
}\right)  ;\\E^{\prime}\cap E=\varnothing;\\H=H\mid_{V_{1}}\sqcup H\mid
_{V_{2}};\\H\mid_{V_{1}}=H_{1};\ H\mid_{V_{2}}=H_{2}}}\left[
\underbrace{H\mid_{V_{1}}}_{=H_{1}}\right]  ^{\sharp}\otimes\left[
\underbrace{H\mid_{V_{2}}}_{=H_{2}}\right]  ^{\sharp}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}%
}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime
}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(  E\mid_{V_{2}%
}\right)  =\varnothing}}\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime
}\right)  ;\\E^{\prime}\cap E=\varnothing;\\H=H\mid_{V_{1}}\sqcup H\mid
_{V_{2}};\\H\mid_{V_{1}}=H_{1};\ H\mid_{V_{2}}=H_{2}}}\left[  H_{1}\right]
^{\sharp}\otimes\left[  H_{2}\right]  ^{\sharp}}_{\substack{=\left[
H_{1}\right]  ^{\sharp}\otimes\left[  H_{2}\right]  ^{\sharp}\\\text{(because
this sum has only one addend}\\\text{(indeed, there exists only one }H=\left(
V,E^{\prime}\right)  \text{ satisfying}\\E^{\prime}\cap E=\varnothing\text{,
}H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}\text{, }H\mid_{V_{1}}=H_{1}\text{
and}\ H\mid_{V_{2}}=H_{2}\text{))}}}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}%
}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime
}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(  E\mid_{V_{2}%
}\right)  =\varnothing}}\left[  H_{1}\right]  ^{\sharp}\otimes\left[
H_{2}\right]  ^{\sharp}.
\end{align*}
Comparing this with
\begin{align*}
\Delta\left[  G\right]   &  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\underbrace{\left[  G\mid_{V_{1}}\right]
}_{\substack{=\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\left[
H_{1}\right]  ^{\sharp}\\\text{(by Proposition \ref{prop.GGG.moebius-basis}%
(a))}}}\otimes\underbrace{\left[  G\mid_{V_{2}}\right]  }_{\substack{=\sum\limits
_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}%
\cap\left(  E\mid_{V_{2}}\right)  =\varnothing}}\left[  H_{2}\right]
^{\sharp}\\\text{(by Proposition \ref{prop.GGG.moebius-basis}(a))}}}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}%
}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime
}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(  E\mid_{V_{2}%
}\right)  =\varnothing}}\left[  H_{1}\right]  ^{\sharp}\otimes\left[
H_{2}\right]  ^{\sharp},
\end{align*}
we obtain $\Delta^{\prime}\left[  G\right]  =\Delta\left[  G\right]  $. Since
this holds for every graph $G$, we thus obtain $\Delta^{\prime}=\Delta$. As we
know, this proves Proposition \ref{prop.GGG.moebius-basis}(c).

(d) We define a $\kk$-bilinear operation $\sharp:\GGG%
\times\GGG\rightarrow\GGG$ (written infix\footnote{This means
that we write $a\sharp b$ to denote the image of a pair $\left(  a,b\right)
\in\GGG\times\GGG$ under this operation $\sharp$.}) by%
\[
\left[  H_{1}\right]  ^{\sharp}\sharp\left[  H_{2}\right]  ^{\sharp}%
=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}%
=H_{1};\\H\mid_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}.
\]
\footnote{This is well-defined, since the $\left[  G\right]  ^{\sharp}$ form a
basis of the $\kk$-module $\GGG$.} In order to prove Proposition
\ref{prop.GGG.moebius-basis}(d), it clearly suffices to show that this
operation $\sharp$ is identical with the usual multiplication on $\GGG$.

Let $G_{1}$ and $G_{2}$ be any two finite graphs. We shall show that $\left[
G_{1}\right]  \sharp\left[  G_{2}\right]  =\left[  G_{1}\right]  \left[
G_{2}\right]  $.

Indeed, write $G_{1}$ and $G_{2}$ in the forms $G_{1}=\left(  V_{1}%
,E_{1}\right)  $ and $G_{2}=\left(  V_{2},E_{2}\right)  $. Then,
\begin{align*}
&  \underbrace{\left[  G_{1}\right]  }_{\substack{=\sum\limits_{\substack{H_{1}%
=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime}\cap E_{1}=\varnothing
}}\left[  H_{1}\right]  ^{\sharp}\\\text{(by Proposition
\ref{prop.GGG.moebius-basis}(a))}}}\sharp\underbrace{\left[  G_{2}\right]
}_{\substack{=\sum\limits_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)
;\\E_{2}^{\prime}\cap E_{2}=\varnothing}}\left[  H_{2}\right]  ^{\sharp
}\\\text{(by Proposition \ref{prop.GGG.moebius-basis}(a))}}}\\
&  =\left(  \sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap E_{1}=\varnothing}}\left[  H_{1}\right]  ^{\sharp
}\right)  \sharp\left(  \sum\limits_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime
}\right)  ;\\E_{2}^{\prime}\cap E_{2}=\varnothing}}\left[  H_{2}\right]
^{\sharp}\right)  =\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap E_{1}=\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap E_{2}=\varnothing
}}\underbrace{\left[  H_{1}\right]  ^{\sharp}\sharp\left[  H_{2}\right]
^{\sharp}}_{=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}}\\
&  =\underbrace{\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap E_{1}=\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap E_{2}=\varnothing}%
}\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}%
=H_{1};\\H\mid_{V_{2}}=H_{2}}}}_{=\sum_{H_{1}=\left(  V_{1},E_{1}^{\prime
}\right)  }\sum_{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  }\sum\limits
_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}%
=H_{1};\\H\mid_{V_{2}}=H_{2};\\E_{1}^{\prime}\cap E_{1}=\varnothing
;\\E_{2}^{\prime}\cap E_{2}=\varnothing}}}\left[  H\right]  ^{\sharp}\\
&  =\sum_{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  }\sum_{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  }\underbrace{\sum\limits_{\substack{H=\left(
V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}%
=H_{2};\\E_{1}^{\prime}\cap E_{1}=\varnothing;\\E_{2}^{\prime}\cap
E_{2}=\varnothing}}}_{\substack{=\sum\limits_{\substack{H=\left(  V_{1}\sqcup
V_{2},E\right)  ;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}=H_{2};\\\left(
E\mid_{V_{1}}\right)  \cap E_{1}=\varnothing;\\\left(  E\mid_{V_{2}}\right)
\cap E_{2}=\varnothing}}\\\text{(because for a graph }H=\left(  V_{1}\sqcup
V_{2},E\right)  \\\text{satisfying }H\mid_{V_{1}}=H_{1}\text{ and }%
H\mid_{V_{2}}=H_{2}\text{,}\\\text{we have }E_{1}^{\prime}=E\mid_{V_{1}}\text{
and }E_{2}^{\prime}=E\mid_{V_{2}}\text{)}}}\left[  H\right]  ^{\sharp}\\
&  =\underbrace{\sum_{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  }\sum
_{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  }\sum\limits_{\substack{H=\left(
V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}%
=H_{2};\\\left(  E\mid_{V_{1}}\right)  \cap E_{1}=\varnothing;\\\left(
E\mid_{V_{2}}\right)  \cap E_{2}=\varnothing}}}_{\substack{=\sum\limits
_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\\left(  E\mid_{V_{1}%
}\right)  \cap E_{1}=\varnothing;\\\left(  E\mid_{V_{2}}\right)  \cap
E_{2}=\varnothing}}=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\E\cap\left(  E_{1}\sqcup E_{2}\right)  =\varnothing}}\\\text{(since the
statement }\left(  \left(  E\mid_{V_{1}}\right)  \cap E_{1}=\varnothing\text{
and }\left(  E\mid_{V_{2}}\right)  \cap E_{2}=\varnothing\right)  \\\text{is
equivalent to }E\cap\left(  E_{1}\sqcup E_{2}\right)  =\varnothing\text{)}%
}}\left[  H\right]  ^{\sharp}\\
&  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\E\cap\left(
E_{1}\sqcup E_{2}\right)  =\varnothing}}\left[  H\right]  ^{\sharp}.
\end{align*}
Compared with%
\begin{align*}
\left[  G_{1}\right]  \left[  G_{2}\right]   &  =\left[  G_{1}\sqcup
G_{2}\right]  =\left[  \left(  V_{1}\sqcup V_{2},E_{1}\sqcup E_{2}\right)
\right] \\
&  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\E\cap\left(
E_{1}\sqcup E_{2}\right)  =\varnothing}}\left[  H\right]  ^{\sharp
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.GGG.moebius-basis}%
(a)}\right)  ,
\end{align*}
this yields $\left[  G_{1}\right]  \sharp\left[  G_{2}\right]  =\left[
G_{1}\right]  \left[  G_{2}\right]  $.

We now forget that we fixed $G_{1}$ and $G_{2}$. We have thus shown that
$\left[  G_{1}\right]  \sharp\left[  G_{2}\right]  =\left[  G_{1}\right]
\left[  G_{2}\right]  $ for any two finite graphs $G_{1}$ and $G_{2}$. Thus,
the operation $\sharp$ is identical with the usual multiplication on
$\GGG$. This completes the proof of Proposition
\ref{prop.GGG.moebius-basis}(d).
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop.GGG.bilform.sym}.]
Proposition~\ref{prop.GGG.moebius-basis}(a) shows that every finite graph
$G=\left(  V,E\right)  $ satisfies%
\begin{equation}
\left[  G\right]  =\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)
;\\E^{\prime}\cap E=\varnothing}}\left[  H\right]  ^{\sharp}=\sum\limits
_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left[  K\right]  ^{\sharp} \label{pf.prop.GGG.moebius-basis.e.sum-lem}%
\end{equation}
(here, we renamed the summation index $H$ as $K$). Let us now show that%
\begin{equation}
\left(  \left[  G\right]  ,\left[  H\right]  \right)  =\left(  \left[
H\right]  ,\left[  G\right]  \right)
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis}%
\end{equation}
for any two finite graphs $G$ and $H$.

\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis}):} We shall
use the following notation: If $V$ and $W$ are two sets, and if $\varphi
:V\rightarrow W$ is a map, then $\varphi_{\ast}$ will denote the map from the
powerset of $V$ to the powerset of $W$ which sends every $T\subset V$ to
$\varphi\left(  T\right)  \subset W$. This map $\varphi_{\ast}$ is a bijection
if $\varphi$ is a bijection.

Let $G$ and $H$ be two finite graphs. Let us write $G$ and $H$ in the forms
$G=\left(  V,E\right)  $ and $H=\left(  W,F\right)  $. Now,
\begin{align}
\left(  \underbrace{\left[  G\right]  }_{\substack{=\sum\limits
_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left[  K\right]  ^{\sharp}\\\text{(by
(\ref{pf.prop.GGG.moebius-basis.e.sum-lem}))}}},\left[  H\right]  \right)
&  =\left(  \sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left[  K\right]  ^{\sharp},\left[  H\right]  \right)
=\sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\underbrace{\left(  \left[  K\right]  ^{\sharp},\left[
H\right]  \right)  }_{\substack{=\left\vert \operatorname*{Iso}\left(
K,H\right)  \right\vert \\\text{(by the definition of}\\\text{the form
}\left(  \cdot,\cdot\right)  \text{)}}} \nonumber \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since
the form }\left(  \cdot,\cdot\right)  \text{ is }\kk\text{-bilinear}%
\right) \nonumber\\
&  =\sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left\vert \operatorname*{Iso}\left(  K,H\right)  \right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.1}%
\end{align}


However, for every graph $K=\left(  V,E^{\prime}\right)  $, we have
\begin{align*}
\operatorname*{Iso}\left(  K,H\right)   &  =\left(  \text{the set of all
isomorphisms from }K\text{ to }H\right) \\
&  =\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a bijection and
satisfies }\varphi_{\ast}\left(  E^{\prime}\right)  =F\right\}
\end{align*}
(because the isomorphisms from $K$ to $H$ are defined to be the bijections
$\varphi:V\rightarrow W$ such that $\varphi_{\ast}\left(  E^{\prime}\right)
=F$). Hence,%
\begin{align*}
&  \sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left\vert \underbrace{\operatorname*{Iso}\left(  K,H\right)
}_{=\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a bijection and
satisfies }\varphi_{\ast}\left(  E^{\prime}\right)  =F\right\}  }\right\vert
\\
&  =\underbrace{\sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime
}\cap E=\varnothing}}}_{=\sum\limits_{\substack{E^{\prime}\text{ is a set
of}\\\text{two-element subsets of }V;\\E^{\prime}\cap E=\varnothing}%
}}\left\vert \left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\underbrace{\varphi_{\ast}\left(  E^{\prime}\right)
=F}_{\substack{\text{this is equivalent to}\\E^{\prime}=\left(  \varphi_{\ast
}\right)  ^{-1}\left(  F\right)  \\\text{(since }\varphi_{\ast}\text{ is a
bijection}\\\text{(since }\varphi\text{ is a bijection))}}}\right\}
\right\vert \\
&  =\sum\limits_{\substack{E^{\prime}\text{ is a set of}\\\text{two-element subsets
of }V;\\E^{\prime}\cap E=\varnothing}}\left\vert \left\{  \varphi:V\rightarrow
W\ \mid\ \varphi\text{ is a bijection and satisfies }E^{\prime}=\left(
\varphi_{\ast}\right)  ^{-1}\left(  F\right)  \right\}  \right\vert \\
&  =\left\vert \left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}  \right\vert .
\end{align*}
Now, (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.1}) becomes%
\begin{align}
\left(  \left[  G\right]  ,\left[  H\right]  \right)   &  =\sum\limits
_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left\vert \operatorname*{Iso}\left(  K,H\right)  \right\vert \nonumber\\
&  =\left\vert \left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}  \right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4}%
\end{align}
The same argument (applied to $H$, $W$, $F$, $G$, $V$ and $E$ instead of $G$,
$V$, $E$, $H$, $W$ and $F$) yields%
\begin{equation}
\left(  \left[  H\right]  ,\left[  G\right]  \right)  =\left\vert \left\{
\varphi:W\rightarrow V\ \mid\ \varphi\text{ is a bijection and satisfies
}\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)  \cap F=\varnothing
\right\}  \right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5}
\end{equation}


Now, we define two sets $\mathfrak{P}$ and $\mathfrak{Q}$ by%
\[
\mathfrak{P}=\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}
\]
and%
\[
\mathfrak{Q}=\left\{  \varphi:W\rightarrow V\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)
\cap F=\varnothing\right\}  .
\]
Then, (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4}) becomes%
\begin{equation}
\left(  \left[  G\right]  ,\left[  H\right]  \right)  =\left\vert
\underbrace{\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}  }_{=\mathfrak{P}}\right\vert =\left\vert
\mathfrak{P}\right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4a}%
\end{equation}
Also, (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5}) becomes%
\begin{equation}
\left(  \left[  H\right]  ,\left[  G\right]  \right)  =\left\vert
\underbrace{\left\{  \varphi:W\rightarrow V\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)
\cap F=\varnothing\right\}  }_{=\mathfrak{Q}}\right\vert =\left\vert
\mathfrak{Q}\right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5a}%
\end{equation}
But every $\psi\in\mathfrak{P}$ satisfies $\psi^{-1}\in\mathfrak{Q}%
$\ \ \ \ \footnote{\textit{Proof.}
Let $\psi\in\mathfrak{P}$. Then,
$\psi \in \mathfrak{P} = \left\{ \varphi:V\rightarrow W\ \mid\ \varphi
\text{ is a bijection and satisfies }
\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)  \cap E=\varnothing
\right\}  $. Hence, $\psi$ is a bijection $V\rightarrow W$ and satisfies
$\left(  \psi_{\ast}\right)  ^{-1}\left(  F\right)  \cap E=\varnothing$.

Let $\rho$ be the map $\psi^{-1} : W\rightarrow V$. This map $\rho$ is
well-defined (since $\psi$ is a bijection) and is a bijection itself (since
$\rho=\psi^{-1}$).

But $\psi$ is a bijection. Thus, $\psi_{\ast}$ is a bijection as well, and
satisfies
$\left(  \psi_{\ast}\right)  ^{-1}
= \left(  \underbrace{\psi^{-1}}_{=\rho}\right)  _{\ast} = \rho_{\ast}$.
Of course, $\rho_{\ast}$ is also a bijection (since $\rho$ is a
bijection).

Now,
\begin{align*}
\rho_{\ast}\left(  \left(  \rho_{\ast}\right)  ^{-1}\left(  E\right)  \cap
F\right)   &  =\underbrace{\rho_{\ast}\left(  \left(  \rho_{\ast}\right)
^{-1}\left(  E\right)  \right)  }_{\substack{=E\\\text{(since }\rho_{\ast
}\text{ is a bijection)}}}\cap\underbrace{\rho_{\ast}}_{=\left(  \psi_{\ast
}\right)  ^{-1}}\left(  F\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\rho_{\ast}\text{ is a bijection}\right)  \\
&  =E\cap\left(  \psi_{\ast}\right)  ^{-1}\left(  F\right)  =\left(
\psi_{\ast}\right)  ^{-1}\left(  F\right)  \cap E=\varnothing.
\end{align*}
Thus,
\begin{align*}
\left(  \rho_{\ast}\right)  ^{-1}\left(  E\right)  \cap F &  =\rho_{\ast
}\left(  \varnothing\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\rho_{\ast}\text{ is a bijection}\right)  \\
&  =\varnothing.
\end{align*}
Hence, $\rho:W\rightarrow V$ is a bijection and satisfies $\left(  \rho_{\ast
}\right)  ^{-1}\left(  E\right)  \cap F=0$. In other words,
\[
\rho\in\left\{  \varphi:W\rightarrow V\ \mid\ \varphi\text{ is a bijection and
satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)  \cap
F=\varnothing\right\}  =\mathfrak{Q}.
\]
Thus, $\psi^{-1}=\rho\in\mathfrak{Q}$, qed.}. Hence, we can define a map
\begin{align*}
\mathbf{A}:\mathfrak{P}  &  \rightarrow\mathfrak{Q},\\
\psi &  \mapsto\psi^{-1}.
\end{align*}
Similarly, we can define a map%
\begin{align*}
\mathbf{B}:\mathfrak{Q}  &  \rightarrow\mathfrak{P},\\
\psi &  \mapsto\psi^{-1}.
\end{align*}
Consider these two maps $\mathbf{A}$ and $\mathbf{B}$. Clearly, these maps
$\mathbf{A}$ and $\mathbf{B}$ are mutually inverse\footnote{\textit{Proof.}
Every $\psi\in\mathfrak{P}$ satisfies
\begin{align*}
\left(  \mathbf{B}\circ\mathbf{A}\right)  \left(  \psi\right)   &
=\mathbf{B}\left(  \underbrace{\mathbf{A}\left(  \psi\right)  }%
_{\substack{=\psi^{-1}\\\text{(by the definition}\\\text{of }\mathbf{A}%
\text{)}}}\right)  =\mathbf{B}\left(  \psi^{-1}\right)  =\left(  \psi
^{-1}\right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{B}\right) \\
&  =\psi=\operatorname*{id}\left(  \psi\right)  .
\end{align*}
Hence, $\mathbf{B}\circ\mathbf{A}=\operatorname*{id}$. Similarly,
$\mathbf{A}\circ\mathbf{B}=\operatorname*{id}$. Combining this with
$\mathbf{B}\circ\mathbf{A}=\operatorname*{id}$, we conclude that the maps
$\mathbf{A}$ and $\mathbf{B}$ are mutually inverse. Qed.}. Hence, $\mathbf{A}$
is a bijection. Thus, there exists a bijection $\mathfrak{P}\rightarrow
\mathfrak{Q}$ (namely, $\mathbf{A}$). Consequently, $\left\vert \mathfrak{P}%
\right\vert =\left\vert \mathfrak{Q}\right\vert $. Now,
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4a}) becomes $\left(
\left[  G\right]  ,\left[  H\right]  \right)  =\left\vert \mathfrak{P}%
\right\vert =\left\vert \mathfrak{Q}\right\vert =\left(  \left[  H\right]
,\left[  G\right]  \right)  $ (by
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5a})). This proves
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis}).

Now, let $a$ and $b$ be two elements of $\GGG$. We want to show that
$\left(  a,b\right)  =\left(  b,a\right)  $. This equality is $\kk%
$-linear in each of $a$ and $b$. Therefore, we can WLOG assume that $a$ and
$b$ belong to the family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$ (because this
family is a basis of the $\kk$-module $\GGG$). Assume this.
Then, $a=\left[  G\right]  $ and $b=\left[  H\right]  $ for two finite graphs
$G$ and $H$. Consider these $G$ and $H$. Now,%
\begin{align*}
\left(  \underbrace{a}_{=\left[  G\right]  },\underbrace{b}_{=\left[
H\right]  }\right)   &  =\left(  \left[  G\right]  ,\left[  H\right]  \right)
=\left(  \underbrace{\left[  H\right]  }_{=b},\underbrace{\left[  G\right]
}_{=a}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis})}\right) \\
&  =\left(  b,a\right)  .
\end{align*}
We thus have shown $\left(  a,b\right)  =\left(  b,a\right)  $.

Let us now forget that we fixed $a$ and $b$. We thus have proven that $\left(
a,b\right)  =\left(  b,a\right)  $ for any $a\in\GGG$ and
$b\in\GGG$. In other words, the form $\left(  \cdot,\cdot\right)
:\GGG\times\GGG\rightarrow\kk$ is symmetric. This proves
Proposition~\ref{prop.GGG.bilform.sym}.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop.GGG.selfdual}.]
In this proof, we shall use all the notations that appeared
in Definition~\ref{def.GGG.selfdual}. We shall also use the fact that
the family
$\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$
is a basis of the $\kk$-module $\GGG$. (This fact is
Proposition~\ref{prop.GGG.moebius-basis}(b).)

(a) We first notice that any two finite graphs $G$ and $H$ satisfy
\begin{equation}
\left(  \psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[
H\right]  =\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert
\label{pf.prop.GGG.moebius-basis.f.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.f.1}):} Let $G$ and
$H$ be two finite graphs. Then,%
\[
\left(  \underbrace{\psi\left(  \left[  G\right]  ^{\sharp}\right)
}_{=\operatorname*{aut}\left(  G\right)  \cdot\left[  G\right]  ^{\ast}%
}\right)  \left[  H\right]  =\left(  \operatorname*{aut}\left(  G\right)
\cdot\left[  G\right]  ^{\ast}\right)  \left[  H\right]  =\operatorname*{aut}%
\left(  G\right)  \cdot\underbrace{\left[  G\right]  ^{\ast}\left[  H\right]
}_{=\delta_{\left[  G\right]  ,\left[  H\right]  }}=\operatorname*{aut}\left(
G\right)  \cdot\delta_{\left[  G\right]  ,\left[  H\right]  }.
\]
\par
We are in one of the following two cases:
\par
\textit{Case 1:} The graphs $G$ and $H$ are isomorphic.
\par
\textit{Case 2:} The graphs $G$ and $H$ are not isomorphic.
\par
Let us first consider Case 1. In this case, the graphs $G$ and $H$ are
isomorphic. In other words, there exists a graph isomorphism $\alpha$ from $G$
to $H$. Consider this $\alpha$. Then,%
\begin{align*}
\operatorname*{Iso}\left(  G,G\right)   &  \rightarrow\operatorname*{Iso}%
\left(  G,H\right)  ,\\
\psi &  \mapsto\alpha\circ\psi
\end{align*}
is a bijection (this is very easy to check). Thus, there exists a bijection
$\operatorname*{Iso}\left(  G,G\right)  \rightarrow\operatorname*{Iso}\left(
G,H\right)  $. Hence, $\left\vert \operatorname*{Iso}\left(  G,G\right)
\right\vert =\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert $.
But the definition of $\operatorname*{aut}\left(  G\right)  $ yields
$\operatorname*{aut}\left(  G\right)  =\left\vert \operatorname*{Iso}\left(
G,G\right)  \right\vert $. Now,%
\[
\left(  \psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[
H\right]  =\operatorname*{aut}\left(  G\right)  \cdot\underbrace{\delta
_{\left[  G\right]  ,\left[  H\right]  }}_{\substack{=1\\\text{(since }\left[
G\right]  =\left[  H\right]  \\\text{(because the graphs }G\\\text{and
}H\text{ are isomorphic))}}}=\operatorname*{aut}\left(  G\right)  =\left\vert
\operatorname*{Iso}\left(  G,G\right)  \right\vert =\left\vert
\operatorname*{Iso}\left(  G,H\right)  \right\vert .
\]
Hence, (\ref{pf.prop.GGG.moebius-basis.f.1}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, the graphs $G$ and $H$ are not
isomorphic. Thus, $\left[  G\right]  \neq\left[  H\right]  $, so that
$\delta_{\left[  G\right]  ,\left[  H\right]  }=0$.
\par
But the graphs $G$ and $H$ are not isomorphic. Hence, the set of all
isomorphisms from $G$ to $H$ is empty. In other words, $\operatorname*{Iso}%
\left(  G,H\right)  =\varnothing$ (since $\operatorname*{Iso}\left(
G,H\right)  $ is the set of all isomorphisms from $G$ to $H$). Hence,
$\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert =\left\vert
\varnothing\right\vert =0$. Now,%
\[
\left(  \psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[
H\right]  =\operatorname*{aut}\left(  G\right)  \cdot\underbrace{\delta
_{\left[  G\right]  ,\left[  H\right]  }}_{=0}=0=\left\vert
\operatorname*{Iso}\left(  G,H\right)  \right\vert .
\]
Hence, (\ref{pf.prop.GGG.moebius-basis.f.1}) is proven in Case 2.
\par
Now, (\ref{pf.prop.GGG.moebius-basis.f.1}) is proven in both Cases 1 and 2.
Hence, (\ref{pf.prop.GGG.moebius-basis.f.1}) always holds.}. Now, let
$a\in\GGG$ and $b\in\GGG$. We want to prove the equality
$\left(  \psi\left(  a\right)  \right)  \left(  b\right)  =\left(  a,b\right)
$. This equality is $\kk$-linear in each of $a$ and $b$. Thus, we can
WLOG assume that $a$ belongs to the family $\left(  \left[  G\right]
^{\sharp}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ and that $b$ belongs to the family $\left(  \left[  G\right]
\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite graphs}%
}$ (because both of these families are bases of the $\kk$-module
$\GGG$). Assume this. Then, $a=\left[  G\right]  ^{\sharp}$ and
$b=\left[  H\right]  $ for two finite graphs $G$ and $H$. Consider these $G$
and $H$. Now,%
\begin{align*}
\left(  \psi\left(  \underbrace{a}_{=\left[  G\right]  ^{\sharp}}\right)
\right)  \left(  \underbrace{b}_{=\left[  H\right]  }\right)   &  =\left(
\psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[  H\right]
=\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.GGG.moebius-basis.f.1}%
)}\right) \\
&  =\left(  \underbrace{\left[  G\right]  ^{\sharp}}_{=a},\underbrace{\left[
H\right]  }_{=b}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\left[  G\right]  ^{\sharp},\left[  H\right]  \right)  =\left\vert
\operatorname*{Iso}\left(  G,H\right)  \right\vert \right) \\
&  =\left(  a,b\right)  .
\end{align*}
This proves Proposition \ref{prop.GGG.selfdual}(a).

(b) We have $\psi\left(  1_{\GGG}\right)  =1_{\GGG^{o}}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\mathbf{0}$ denote the empty graph.
\par
We have $1_{\GGG^{o}}=\epsilon_{\GGG}$ (by the definition of the
$\kk$-algebra $\GGG^{o}$). Thus, for every finite graph $G$, we
have%
\begin{align*}
\underbrace{1_{\GGG^{o}}}_{=\epsilon_{\GGG}}\left(  \left[
G\right]  \right)   &  =\epsilon_{\GGG}\left(  \left[  G\right]
\right)  =\delta_{\left[  G\right]  ,\left[  \mathbf{0}\right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\epsilon_{\GGG%
}\right) \\
&  =\delta_{\left[  \mathbf{0}\right]  ,\left[  G\right]  }=\left[
\mathbf{0}\right]  ^{\ast}\left[  G\right]  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left[  \mathbf{0}\right]  ^{\ast}\left[  G\right]
=\delta_{\left[  \mathbf{0}\right]  ,\left[  G\right]  }\text{ (by the
definition of }\left[  \mathbf{0}\right]  ^{\ast}\text{)}\right)  .
\end{align*}
Hence, the two maps $1_{\GGG^{o}}$ and $\left[  \mathbf{0}\right]
^{\ast}$ are equal to each other on every element of the basis $\left(
\left[  G\right]  \right)  _{\left[  G\right]  \text{ is an isomorphism class
of finite graphs}}$ of $\GGG$. Since these two maps are $\kk%
$-linear, this yields that these two maps $1_{\GGG^{o}}$ and $\left[
\mathbf{0}\right]  ^{\ast}$ are identical. In other words, $1_{\GGG%
^{o}}=\left[  \mathbf{0}\right]  ^{\ast}$.
\par
On the other hand, the definition of $\left[  \mathbf{0}\right]  ^{\sharp}$
readily yields $\left[  \mathbf{0}\right]  ^{\sharp}=\left[  \mathbf{0}%
\right]  =1_{\GGG}$. But the definition of $\psi$ yields $\psi\left(
\left[  \mathbf{0}\right]  ^{\sharp}\right)  =\underbrace{\operatorname*{aut}%
\left(  \mathbf{0}\right)  }_{=1}\cdot\left[  \mathbf{0}\right]  ^{\ast
}=\left[  \mathbf{0}\right]  ^{\ast}=1_{\GGG^{o}}$ (since
$1_{\GGG^{o}}=\left[  \mathbf{0}\right]  ^{\ast}$), so that
$1_{\GGG^{o}}=\psi\left(  \underbrace{\left[  \mathbf{0}\right]
^{\sharp}}_{=1_{\GGG}}\right)  =\psi\left(  1_{\GGG}\right)  $,
qed.}. Also, every $a\in\GGG$ and $b\in\GGG$ satisfy
$\psi\left(  ab\right)  =\psi\left(  a\right)  \cdot\psi\left(  b\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $a\in\GGG$ and $b\in\GGG%
$. We need to prove the equality $\psi\left(  ab\right)  =\psi\left(
a\right)  \cdot\psi\left(  b\right)  $. Since this equality is $\kk%
$-linear in each of $a$ and $b$, we can WLOG assume that $a$ and $b$ are
elements of the family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$ (because this
family is a basis of the $\kk$-module $\GGG$). Assume this.
\par
There exist two finite graphs $H_{1}$ and $H_{2}$ such that $a=\left[
H_{1}\right]  ^{\sharp}$ and $b=\left[  H_{2}\right]  ^{\sharp}$ (since $a$
and $b$ are elements of the family $\left(  \left[  G\right]  ^{\sharp
}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$). Consider these $H_{1}$ and $H_{2}$. Write the graphs $H_{1}$ and
$H_{2}$ in the forms $H_{1}=\left(  V_{1},E_{1}\right)  $ and $H_{2}=\left(
V_{2},E_{2}\right)  $.
\par
Multiplying the equalities $a=\left[  H_{1}\right]  ^{\sharp}$ and $b=\left[
H_{2}\right]  ^{\sharp}$, we obtain%
\[
ab=\left[  H_{1}\right]  ^{\sharp}\left[  H_{2}\right]  ^{\sharp}%
=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.GGG.moebius-basis}%
(d)}\right)  .
\]
\par
Now, fix a finite graph $G$. We shall show that $\left(  \psi\left(
ab\right)  \right)  \left[  G\right]  =\left(  \psi\left(  a\right)  \cdot
\psi\left(  b\right)  \right)  \left[  G\right]  $.
\par
Write the graph $G$ in the form $G=\left(  V,F\right)  $. Then,%
\begin{align*}
\left(  \psi\underbrace{\left(  ab\right)  }_{=\sum\limits_{\substack{H=\left(
V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\left[
H\right]  ^{\sharp}}\right)  \left[  G\right]   &  =\left(  \psi\left(
\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}\right)  \right)
\left[  G\right]  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\underbrace{\psi\left(  \left[
H\right]  ^{\sharp}\right)  \left[  G\right]  }_{\substack{=\left\vert
\operatorname*{Iso}\left(  H,G\right)  \right\vert \\\text{(by
(\ref{pf.prop.GGG.moebius-basis.f.1}), applied to}\\H\text{ and }G\text{
instead of }G\text{ and }H\text{)}}}\\
&  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\left\vert \operatorname*{Iso}\left(  H,G\right)
\right\vert =\left\vert \bigsqcup_{\substack{H=\left(  V_{1}\sqcup
V_{2},E\right)  ;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}%
\left(  H,G\right)  \right\vert .
\end{align*}
On the other hand, Exercise \ref{exe.dual-algebra}(b) shows that the
$\kk$-algebra structure defined on $\GGG^{\ast}$ is precisely
the one defined on $\Hom \left(  \GGG,\kk\right)
=\GGG^{\ast}$ according to Definition~\ref{convolution-algebra}. Thus,
the product of two elements of $\GGG^{\ast}$ is the convolution of
these elements (viewed as maps from $\GGG$ to $\kk$). Applying
this to the two elements $\psi\left(  a\right)  $ and $\psi\left(  b\right)
$, we obtain $\psi\left(  a\right)  \cdot\psi\left(  b\right)  =\psi\left(
a\right)  \star\psi\left(  b\right)  =m_{\kk}\circ\left(  \psi\left(
a\right)  \otimes\psi\left(  b\right)  \right)  \circ\Delta_{\GGG}$, so
that%
\begin{align*}
\left(  \psi\left(  a\right)  \cdot\psi\left(  b\right)  \right)  \left[
G\right]   &  =\left(  m_{\kk}\circ\left(  \psi\left(  a\right)
\otimes\psi\left(  b\right)  \right)  \circ\Delta_{\GGG}\right)
\left(  \left[  G\right]  \right) \\
&  =\left(  m_{\kk}\circ\left(  \psi\left(  \underbrace{a}_{=\left[
H_{1}\right]  ^{\sharp}}\right)  \otimes\psi\left(  \underbrace{b}_{=\left[
H_{2}\right]  ^{\sharp}}\right)  \right)  \right)  \left(  \underbrace{\Delta
_{\GGG}\left(  \left[  G\right]  \right)  }_{\substack{=\sum\limits
_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}}\left[
G\mid_{V_{1}}\right]  \otimes\left[  G\mid_{V_{2}}\right]  \\\text{(by the
definition of }\Delta_{\GGG}\text{)}}}\right) \\
&  =\left(  m_{\kk}\circ\left(  \psi\left(  \left[  H_{1}\right]
^{\sharp}\right)  \otimes\psi\left(  \left[  H_{2}\right]  ^{\sharp}\right)
\right)  \right)  \left(  \sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\left[  G\mid_{V_{1}}\right]  \otimes\left[
G\mid_{V_{2}}\right]  \right) \\
&  =m_{\kk}\left(  \sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\underbrace{\left(  \psi\left(  \left[  H_{1}\right]
^{\sharp}\right)  \otimes\psi\left(  \left[  H_{2}\right]  ^{\sharp}\right)
\right)  \left(  \left[  G\mid_{V_{1}}\right]  \otimes\left[  G\mid_{V_{2}%
}\right]  \right)  }_{=\left(  \psi\left(  \left[  H_{1}\right]  ^{\sharp
}\right)  \right)  \left[  G\mid_{V_{1}}\right]  \otimes\left(  \psi\left(
\left[  H_{2}\right]  ^{\sharp}\right)  \right)  \left[  G\mid_{V_{2}}\right]
}\right) \\
&  =m_{\kk}\left(  \sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\left(  \psi\left(  \left[  H_{1}\right]  ^{\sharp
}\right)  \right)  \left[  G\mid_{V_{1}}\right]  \otimes\left(  \psi\left(
\left[  H_{2}\right]  ^{\sharp}\right)  \right)  \left[  G\mid_{V_{2}}\right]
\right) \\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}%
=V}}\underbrace{\left(  \psi\left(  \left[  H_{1}\right]  ^{\sharp}\right)
\right)  \left[  G\mid_{V_{1}}\right]  }_{\substack{=\left\vert
\operatorname*{Iso}\left(  H_{1},G\mid_{V_{1}}\right)  \right\vert \\\text{(by
(\ref{pf.prop.GGG.moebius-basis.f.1}), applied to}\\H_{1}\text{ and }%
G\mid_{V_{1}}\text{ instead of }G\text{ and }H\text{)}}}\cdot
\underbrace{\left(  \psi\left(  \left[  H_{2}\right]  ^{\sharp}\right)
\right)  \left[  G\mid_{V_{2}}\right]  }_{\substack{=\left\vert
\operatorname*{Iso}\left(  H_{2},G\mid_{V_{2}}\right)  \right\vert \\\text{(by
(\ref{pf.prop.GGG.moebius-basis.f.1}), applied to}\\H_{2}\text{ and }%
G\mid_{V_{2}}\text{ instead of }G\text{ and }H\text{)}}}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}%
=V}}\left\vert \operatorname*{Iso}\left(  H_{1},G\mid_{V_{1}}\right)
\right\vert \cdot\left\vert \operatorname*{Iso}\left(  H_{2},G\mid_{V_{2}%
}\right)  \right\vert =\sum\limits_{\substack{\left(  W_{1},W_{2}\right)
;\\W_{1}\sqcup W_{2}=V}}\left\vert \operatorname*{Iso}\left(  H_{1}%
,G\mid_{W_{1}}\right)  \right\vert \cdot\left\vert \operatorname*{Iso}\left(
H_{2},G\mid_{W_{2}}\right)  \right\vert \\
&  =\left\vert \bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup
W_{2}=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  \right\vert .
\end{align*}
\par
Now, we are going to construct a bijection between the sets
\[
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\ \ \ \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ \ \ \ 
\bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup W_{2}%
=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  .
\]
First, let us agree to encode the elements of the set
\[
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\]
as triples $\left(H, E, \varphi\right)$ consisting of a graph
$H = \left(V_1 \sqcup V_2, E\right)$, its set of edges $E$, and
an isomorphism $\varphi \in \operatorname*{Iso}\left(  H,G\right)$.
Let us also agree to encode the elements of the set
\[
\bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup W_{2}%
=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)
\]
as triples $\left(\left(W_1, W_2\right), \varphi_1, \varphi_2\right)$
consisting of a pair $\left(W_1, W_2\right)$ of subsets of $V$ (which
satisfies $W_1 \sqcup W_2 = V$), an isomorphism
$\varphi_1 \in \operatorname*{Iso}\left( H_1, G\mid_{W_1}\right)$,
and an isomorphism
$\varphi_2 \in \operatorname*{Iso}\left( H_2, G\mid_{W_2}\right)$.
\par
Now, we claim that there exists a bijection from the set
\[
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\]
to the set
\[
\bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup W_{2}%
=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  .
\]
Namely, this bijection sends every $\left(  H,E,\varphi\right)  $ (where
$H=\left(  V_{1}\sqcup V_{2},E\right)  $ and $\varphi\in\operatorname*{Iso}%
\left(  H,G\right)  $) to $\left(  \left(  \varphi\left(  V_{1}\right)
,\varphi\left(  V_{2}\right)  \right)  ,\varphi_{1},\varphi_{2}\right)  $,
where $\varphi_{1}:V_{1}\rightarrow\varphi\left(  V_{1}\right)  $ is the
isomorphism from $H_{1}$ to $G\mid_{\varphi\left(  V_{1}\right)  }$ which is
obtained by restricting $\varphi$ to $V_{1}$, and where $\varphi_{2}$ is the
isomorphism from $H_{2}$ to $G_{2}\mid_{\varphi\left(  V_{2}\right)  }$ which
is obtained by restricting $\varphi$ to $V_{2}$. (The inverse map of this
bijection sends every $\left(  \left(  W_{1},W_{2}\right)  ,\varphi
_{1},\varphi_{2}\right)  $ (with $W_{1}\sqcup W_{2}=V$ and $\varphi_{1}%
\in\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  $ and
$\varphi_2 \in \operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  $) to
$\left(  \left(
V_{1}\sqcup V_{2},E\right)  ,E,\varphi\right)  $, where $\varphi:V_{1}\sqcup
V_{2}\rightarrow V$ is the map glued together from the maps $\varphi_{1}%
:V_{1}\rightarrow W_{1}$ and $\varphi_{2}:V_{2}\rightarrow W_{2}$, and where
$E=\left(  \varphi^{-1}\right)  _{\ast}\left(  F\right)  $.) The existence of
this bijection yields%
\[
\left\vert \bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\right\vert =\left\vert \bigsqcup_{\substack{\left(  W_{1},W_{2}\right)
;\\W_{1}\sqcup W_{2}=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)
\times\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  \right\vert .
\]
Hence,%
\begin{align*}
\left(  \psi\left(  ab\right)  \right)  \left[  G\right] 
&= \left\vert
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)  \right\vert
= \left\vert \bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup
W_{2}=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  \right\vert =\left(
\psi\left(  a\right)  \cdot\psi\left(  b\right)  \right)  \left[  G\right]  .
\end{align*}
\par
Let us now forget that we fixed $G$. We thus have shown that $\left(
\psi\left(  ab\right)  \right)  \left[  G\right]  =\left(  \psi\left(
a\right)  \cdot\psi\left(  b\right)  \right)  \left[  G\right]  $ for any
finite graph $G$. In other words, the two maps $\psi\left(  ab\right)  $ and
$\psi\left(  a\right)  \cdot\psi\left(  b\right)  $ are equal to each other on
every element of the basis $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$ of $\GGG$.
Since these two maps are $\kk$-linear, this yields that these two maps
$\psi\left(  ab\right)  $ and $\psi\left(  a\right)  \cdot\psi\left(
b\right)  $ are identical. In other words, $\psi\left(  ab\right)
=\psi\left(  a\right)  \cdot\psi\left(  b\right)  $. Qed.}. Hence,
$\psi:\GGG\rightarrow\GGG^{o}$ is a $\kk$-algebra morphism
(since $\psi\left(  1_{\GGG}\right)  =1_{\GGG^{o}}$).

Next, we notice that the $\kk$-linear map $\psi$ is graded (because for
any finite graph $G$, both $\left[  G\right]  ^{\sharp}$ and $\left[
G\right]  ^{\ast}$ are homogeneous elements having degree $\left\vert
G\right\vert $). Thus, the $\kk$-linear map $\psi:\GGG%
\rightarrow\GGG^{o}$ gives rise to an adjoint $\kk$-linear map
$\psi^{\ast}:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$.

Now, the canonical $\kk$-module homomorphism $\GGG%
\rightarrow\left(  \GGG^{o}\right)  ^{o}$ (which sends every
$a\in\GGG$ to the map $\GGG^{o}\rightarrow\kk,\ f\mapsto
f\left(  a\right)  $) is a $\kk$-module isomorphism (since
$\GGG$ is of finite type). We thus identify $\GGG$ with $\left(
\GGG^{o}\right)  ^{o}$ along this isomorphism. Then, $\left(
\GGG^{o}\right)  ^{o}=\GGG$ as Hopf algebras\footnote{This is
because the Hopf algebra structure on $\GGG^{o}$ was defined by taking
adjoints of the structure maps of the Hopf algebra structure on
$\GGG$ (for example, the comultiplication $\Delta_{\GGG^{o}}$ on
$\GGG^{o}$ is the adjoint of the multiplication $m_{\GGG}$ on
$\GGG$, if $\left(  \GGG\otimes\GGG\right)  ^{o}$ is
identified with $\GGG^{o}\otimes\GGG^{o}$), and the Hopf algebra
structure on $\left(  \GGG^{o}\right)  ^{o}$ was defined by taking
adjoints of the structure maps of the Hopf algebra structure on
$\GGG^{o}$; but the adjoint of the adjoint of a linear map $F$
between two graded $\kk$-modules of finite type is the map $F$
again.}. Therefore,
$\psi:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$ is a
$\kk$-algebra morphism (since $\psi:\GGG\rightarrow\GGG^{o}$
is a $\kk$-algebra morphism). Also, $\GGG^{o}$ is of finite type
(since $\GGG$ is of finite type). Thus, Exercise \ref{exe.dual-algebra}%
(f) (applied to $C=\GGG$, $D=\GGG^{o}$ and $f=\psi$) yields that
$\psi:\GGG\rightarrow\GGG^{o}$ is a $\kk$-coalgebra morphism
if and only if $\psi^{\ast}:\left(  \GGG^{o}\right)  ^{o}%
\rightarrow\GGG^{o}$ is a $\kk$-algebra morphism.

Let us now prove that $\psi^{\ast}=\psi$ (as maps from $\GGG$ to
$\GGG^{o}$). Indeed, let $a\in\GGG$. Then, every $b\in
\GGG$ satisfies%
\begin{align*}
\left(  \psi^{\ast}\left(  a\right)  \right)  \left(  b\right)   &  =a\left(
\psi\left(  b\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the adjoint map }\psi^{\ast}:\left(  \GGG%
^{o}\right)  ^{o}\rightarrow\GGG^{o}\text{,}\\
\text{where }a\in\GGG\text{ is regarded as an element of }\left(
\GGG^{o}\right)  ^{o}%
\end{array}
\right) \\
&  =\left(  \psi\left(  b\right)  \right)  \left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the }\kk\text{-module homomorphism }\GGG%
\rightarrow\left(  \GGG^{o}\right)  ^{o}\text{ which we use to}\\
\text{identify }\GGG\text{ with }\left(  \GGG^{o}\right)
^{o}\text{ sends }a\in\GGG\text{ to the map }\GGG^{o}%
\rightarrow\kk,\ f\mapsto f\left(  a\right)
\end{array}
\right) \\
&  =\left(  b,a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.GGG.selfdual}(a), applied to }b\text{ and }a\text{ instead of
}a\text{ and }b\right) \\
&  =\left(  a,b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the form
}\left(  \cdot,\cdot\right)  :\GGG\times\GGG\rightarrow
\kk\text{ is symmetric}\right) \\
&  =\left(  \psi\left(  a\right)  \right)  \left(  b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.GGG.selfdual}%
(a)}\right)  .
\end{align*}
Hence, $\psi^{\ast}\left(  a\right)  =\psi\left(  a\right)  $. Now, let us
forget that we fixed $a$. We thus have shown that $\psi^{\ast}\left(
a\right)  =\psi\left(  a\right)  $ for every $a\in\GGG$. In other
words, $\psi^{\ast}=\psi$. Hence, $\psi^{\ast}:\left(  \GGG^{o}\right)
^{o}\rightarrow\GGG^{o}$ is a $\kk$-algebra morphism (since
$\psi:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$ is a
$\kk$-algebra morphism). Thus, $\psi:\GGG\rightarrow\GGG^{o}$
is a $\kk$-coalgebra morphism (because we know that $\psi:\GGG%
\rightarrow\GGG^{o}$ is a $\kk$-coalgebra morphism if and only if
$\psi^{\ast}:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$
is a $\kk$-algebra morphism). This yields that $\psi:\GGG%
\rightarrow\GGG^{o}$ is a $\kk$-bialgebra morphism (since we already
know that $\psi:\GGG\rightarrow\GGG^{o}$ is a $\kk%
$-algebra morphism). Thus, $\psi:\GGG\rightarrow\GGG^{o}$ is a Hopf
algebra homomorphism (by Corollary~\ref{cor.bialg-mor-is-Hopf},
applied to $H_{1}=\GGG$, $H_{2}=\GGG^{o}$, $S_{1}=S_{\GGG}$,
$S_{2}=S_{\GGG^{o}}$ and $\beta=\psi$). This proves Proposition
\ref{prop.GGG.selfdual}(b).

(c) Assume that $\QQ$ is a subring of $\kk$. Then,
$\operatorname*{aut}\left(  G\right)  $ is invertible for every finite graph
$G$ (because $\operatorname*{aut}\left(  G\right)  $ is a positive integer).
Hence, $\left(  \operatorname*{aut}\left(  G\right)  \cdot\left[  G\right]
^{\ast}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ is a basis of the $\kk$-module $\GGG^{o}$ (because
$\left(  \left[  G\right]  ^{\ast}\right)  _{\left[  G\right]  \text{ is an
isomorphism class of finite graphs}}$ is a basis of this $\kk$-module).
The map $\psi$ thus sends the basis $\left(  \left[  G\right]  ^{\sharp
}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ of the $\kk$-module $\GGG$ to the basis \newline%
$\left(  \operatorname*{aut}\left(  G\right)  \cdot\left[  G\right]  ^{\ast
}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ of the $\kk$-module $\GGG^{o}$ (because it satisfies
$\psi\left(  \left[  G\right]  ^{\sharp}\right)  =\operatorname*{aut}\left(
G\right)  \cdot\left[  G\right]  ^{\ast}$ for every finite graph $G$). Hence,
the $\kk$-linear map $\psi$ sends a basis of its domain to a basis of
its codomain. Therefore, the map $\psi$ is a $\kk$-module isomorphism.
Combined with the fact that $\psi$ is a Hopf algebra homomorphism, this yields
that $\psi$ is a Hopf algebra isomorphism. Proposition
\ref{prop.GGG.selfdual}(c) is thus proven.
\end{proof}

Now, all of Proposition~\ref{prop.GGG.moebius-basis},
Proposition~\ref{prop.GGG.bilform.sym} and
Proposition~\ref{prop.GGG.selfdual} are proven. Therefore,
Exercise \ref{exe.GGG.moebius-basis} is solved.
%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.GGG.moebius-basis}}

\textit{Solution to Exercise~\ref{exe.GGG.moebius-basis}.}

\begin{proof}
[Proof of Proposition \ref{prop.GGG.moebius-basis}.]We recall the following
fundamental fact (a version of inclusion-exclusion or a very simple special
case of M\"{o}bius inversion): If $R$ is a finite set, then%
\begin{equation}
\sum_{T\subset R}\left(  -1\right)  ^{\left\vert T\right\vert }=\delta
_{R,\varnothing}. \label{pf.prop.GGG.moebius-basis.moebius.1}%
\end{equation}
\footnote{For the sake of completeness, here is a short \textit{proof of
(\ref{pf.prop.GGG.moebius-basis.moebius.1}):} Let $R$ be a finite set. We have%
\begin{align*}
\sum_{T\subset R}\left(  -1\right)  ^{\left\vert T\right\vert }  &
=\sum_{k\in\NN}\sum\limits_{\substack{T\subset R;\\\left\vert T\right\vert
=k}}\underbrace{\left(  -1\right)  ^{\left\vert T\right\vert }}%
_{\substack{=\left(  -1\right)  ^{k}\\\text{(since }\left\vert T\right\vert
=k\text{)}}}=\sum_{k\in\NN}\underbrace{\sum\limits_{\substack{T\subset
R;\\\left\vert T\right\vert =k}}\left(  -1\right)  ^{k}}_{=\left\vert \left\{
T\subset R\ \mid\ \left\vert T\right\vert =k\right\}  \right\vert \left(
-1\right)  ^{k}}=\sum_{k\in\NN}\underbrace{\left\vert \left\{  T\subset
R\ \mid\ \left\vert T\right\vert =k\right\}  \right\vert }_{\substack{=\left(
\text{the number of all }k\text{-element subsets of }R\right)  \\=\dbinom
{\left\vert R\right\vert }{k}\\\text{(by the combinatorial interpretation
of}\\\text{binomial coefficients)}}}\left(  -1\right)  ^{k}\\
&  =\sum_{k\in\NN}\dbinom{\left\vert R\right\vert }{k}\left(
-1\right)  ^{k}=\left(  \underbrace{1+\left(  -1\right)  }_{=0}\right)
^{\left\vert R\right\vert }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  1+\left(  -1\right)
\right)  ^{\left\vert R\right\vert }=\sum_{k\in\NN}\dbinom{\left\vert
R\right\vert }{k}\left(  -1\right)  ^{k}\text{ (by the binomial formula)}%
\right) \\
&  =0^{\left\vert R\right\vert }=\delta_{\left\vert R\right\vert ,0}%
=\delta_{R,\varnothing}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert
R\right\vert =0\text{ holds if and only if }R=\varnothing\right)  .
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.moebius.1}).} We can use this to
show a slightly more complicated fact: If $P$ and $R$ are two finite sets,
then%
\begin{equation}
\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }=\delta_{P,R}.
\label{pf.prop.GGG.moebius-basis.moebius.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.moebius.2}):} Let
$P$ and $R$ be two finite sets.
\par
Let us first assume that $P\not \subset R$. Then, there exists no $F\subset R$
such that $F\supset P$ (because if such a $F$ would exist, then we would have
$P\subset F\subset R$, which would contradict $P\not \subset R$). Hence, the
sum $\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }$ is empty, and thus it simplifies to $\sum\limits
_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }=0$. On the other hand, $P\not \subset R$, so that
$P\neq R$ and thus $\delta_{P,R}=0$. Hence, $\sum\limits_{\substack{F\subset
R;\\F\supset P}}\left(  -1\right)  ^{\left\vert F\setminus P\right\vert
}=0=\delta_{P,R}$. Thus, (\ref{pf.prop.GGG.moebius-basis.moebius.2}) is proven
under the assumption that $P\not \subset R$.
\par
Now, let us forget that we have assumed $P\not \subset R$. We thus have shown
that (\ref{pf.prop.GGG.moebius-basis.moebius.2}) holds if $P\not \subset R$.
Hence, for the rest of this proof, we can WLOG assume that we don't have
$P\not \subset R$. Assume this.
\par
We have $P\subset R$ (since we don't have $P\not \subset R$). Hence, there
exists a bijection from the set of all $F\subset R$ satisfying $F\supset P$ to
the set of all $T\subset R\setminus P$; this bijection sends every $F$ to
$F\setminus P$. Hence, we can substitute $T$ for $F\setminus P$ in the sum
$\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }$. We thus obtain
\begin{align*}
\sum\limits_{\substack{F\subset R;\\F\supset P}}\left(  -1\right)  ^{\left\vert
F\setminus P\right\vert }  &  =\sum_{T\subset R\setminus P}\left(  -1\right)
^{\left\vert T\right\vert }=\delta_{R\setminus P,\varnothing}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.GGG.moebius-basis.moebius.1}), applied to }R\setminus P\text{
instead of }R\right) \\
&  =\delta_{P,R}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R\setminus
P=\varnothing\text{ if and only if }P=R\text{ (because }P\subset
R\text{)}\right)  .
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.moebius.2}).}

(a) Let $G=\left(  V,E\right)  $ be a finite graph. We have%
\begin{align*}
\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left[  H\right]  ^{\sharp}  &  =\underbrace{\sum\limits
_{\substack{K=\left(  V,F\right)  ;\\F\cap E=\varnothing}}}_{\substack{=\sum\limits
_{\substack{K=\left(  V,F\right)  ;\\F\subset E^{c}}}\\\text{(since }F\cap
E=\varnothing\text{ is}\\\text{equivalent to }F\subset E^{c}\text{)}%
}}\underbrace{\left[  K\right]  ^{\sharp}}_{\substack{=\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset F^{c}}}\left(
-1\right)  ^{\left\vert E^{\prime}\setminus F^{c}\right\vert }\left[
H\right]  \\\text{(by the definition of }\left[  K\right]  ^{\sharp}\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }H\text{ and }E^{\prime
}\text{ as }K\text{ and }F\text{ in the sum}\right) \\
&  =\underbrace{\sum\limits_{\substack{K=\left(  V,F\right)  ;\\F\subset E^{c}}%
}\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset F^{c}}%
}}_{\substack{=\sum_{H=\left(  V,E^{\prime}\right)  }\sum\limits_{\substack{K=\left(
V,F\right)  ;\\F\subset E^{c};\\E^{\prime}\supset F^{c}}}}}\underbrace{\left(
-1\right)  ^{\left\vert E^{\prime}\setminus F^{c}\right\vert }}%
_{\substack{=\left(  -1\right)  ^{\left\vert F\setminus\left(  E^{\prime
}\right)  ^{c}\right\vert }\\\text{(since }E^{\prime}\setminus F^{c}%
=E^{\prime}\cup F=F\setminus\left(  E^{\prime}\right)  ^{c}\text{)}}}\left[
H\right] \\
&  =\sum_{H=\left(  V,E^{\prime}\right)  }\underbrace{\sum\limits
_{\substack{K=\left(  V,F\right)  ;\\F\subset E^{c};\\E^{\prime}\supset F^{c}%
}}}_{\substack{=\sum\limits_{\substack{F\subset E^{c};\\E^{\prime}\supset F^{c}%
}}=\sum\limits_{\substack{F\subset E^{c};\\F\supset\left(  E^{\prime}\right)  ^{c}%
}}\\\text{(since }E^{\prime}\supset F^{c}\text{ is equivalent to }%
F\supset\left(  E^{\prime}\right)  ^{c}\text{)}}}\left(  -1\right)
^{\left\vert F\setminus\left(  E^{\prime}\right)  ^{c}\right\vert }\left[
H\right] \\
&  =\sum_{H=\left(  V,E^{\prime}\right)  }\underbrace{\sum\limits_{\substack{F\subset
E^{c};\\F\supset\left(  E^{\prime}\right)  ^{c}}}\left(  -1\right)
^{\left\vert F\setminus\left(  E^{\prime}\right)  ^{c}\right\vert }%
}_{\substack{=\delta_{\left(  E^{\prime}\right)  ^{c},E^{c}}\\\text{(by
(\ref{pf.prop.GGG.moebius-basis.moebius.2}), applied to}\\R=E^{c}\text{ and
}P=\left(  E^{\prime}\right)  ^{c}\text{)}}}\left[  H\right] \\
&  =\sum_{H=\left(  V,E^{\prime}\right)  }\underbrace{\delta_{\left(
E^{\prime}\right)  ^{c},E^{c}}}_{\substack{=\delta_{E^{\prime},E}%
\\\text{(since }\left(  E^{\prime}\right)  ^{c}=E^{c}\text{ holds
if}\\\text{and only if }E^{\prime}=E\text{)}}}\left[  H\right]  =\sum
_{H=\left(  V,E^{\prime}\right)  }\delta_{E^{\prime},E}\left[  H\right]
=\left[  \left(  V,E\right)  \right]  =\left[  G\right]  .
\end{align*}


This proves Proposition \ref{prop.GGG.moebius-basis}(a).

(b) For every finite graph $G=\left(  V,E\right)  $, we define the
\textbf{complement} of this graph $G$ to be the graph $\left(  V,E^{c}\right)
$ (where $E^{c}$ is defined as in Definition \ref{def.GGG.moebius-basis}). We
shall denote this complement by $\mathbf{c}\left(  G\right)  $.

We notice that%
\begin{equation}
\mathbf{c}\left(  \mathbf{c}\left(  G\right)  \right)
=G\ \ \ \ \ \ \ \ \ \ \text{for every finite graph }G
\label{pf.prop.GGG.moebius-basis.b.ccG}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.b.ccG}):} Let $G$
be a finite graph. Write $G$ in the form $\left(  V,E\right)  $, where $V$ is
a finite set and where $E$ is a set of $2$-element subsets of $V$. Then,
$\mathbf{c}\left(  G\right)  =\left(  V,E^{c}\right)  $ (by the definition of
$\mathbf{c}\left(  G\right)  $), so that
\begin{align*}
\mathbf{c}\left(  \mathbf{c}\left(  G\right)  \right)   &  =\left(
V,\underbrace{\left(  E^{c}\right)  ^{c}}_{\substack{=E\\\text{(since }E\text{
is a set of }2\text{-element}\\\text{subsets of }V\text{)}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{c}\left(
\mathbf{c}\left(  G\right)  \right)  \right) \\
&  =\left(  V,E\right)  =G.
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.b.ccG}).}.

Recall that if $G$ is a finite graph, then we denote the isomorphism class of
this graph $G$ by $\left[  G\right]  $. Let $\operatorname*{GrIs}$ be the set
of all isomorphism classes of finite graphs. For every $n\in \NN$, let
$\operatorname*{GrIs}\nolimits_{n}$ be the set of all isomorphism classes of
finite graphs with $n$ vertices. Thus, $\operatorname*{GrIs}=\bigsqcup
_{n\in\NN}\operatorname*{GrIs}\nolimits_{n}$. Hence,
$\operatorname*{GrIs}\nolimits_{n}\subset\operatorname*{GrIs}$ for each
$n\in \NN$.

The family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$ is a basis of the $\kk$-module $\mathcal{G}%
$\ \ \ \ \footnote{\textit{Proof.} Recall that $\operatorname*{GrIs}$ is the
set of all isomorphism classes of finite graphs. Thus, the elements of
$\operatorname*{GrIs}$ are the isomorphism classes of finite graphs. Hence,
$\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}}=\left(  \left[  G\right]  \right)  _{\left[  G\right]
\text{ is an isomorphism class of finite graphs}}$.
\par
The definition of $\mathcal{G}$ shows that the family $\left(  \left[
G\right]  \right)  _{\left[  G\right]  \text{ is an isomorphism class of
finite graphs}}$ is a basis of the $\kk$-module $\mathcal{G}$. In other
words, the family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$ is a basis of the $\kk$-module $\mathcal{G}$
(since $\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}}=\left(  \left[  G\right]  \right)  _{\left[  G\right]
\text{ is an isomorphism class of finite graphs}}$). Qed.}. Thus, in
particular, this family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \in\operatorname*{GrIs}}$ is $\kk$-linearly independent.

Fix $n\in \NN$. The set $\operatorname*{GrIs}\nolimits_{n}$ is finite
(since there are only finitely many isomorphism classes of finite graphs with
$n$ vertices).

The family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$\ \ \ \ \footnote{\textit{Proof.} Recall that
$\operatorname*{GrIs}\nolimits_{n}$ is the set of all isomorphism classes of
finite graphs with $n$ vertices. Thus, the elements of $\operatorname*{GrIs}%
\nolimits_{n}$ are the isomorphism classes of finite graphs with $n$ vertices.
In other words, the elements of $\operatorname*{GrIs}\nolimits_{n}$ are the
isomorphism classes of graphs with $n$ vertices (since any graph with $n$
vertices is automatically finite). Hence, $\left(  \left[  G\right]  \right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}=\left(  \left[
G\right]  \right)  _{\left[  G\right]  \text{ is an isomorphism class of
graphs with }n\text{ vertices}}$.
\par
We have $\operatorname*{GrIs}\nolimits_{n}\subset\operatorname*{GrIs}$. Hence,
the family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ is a subfamily of the family $\left(
\left[  G\right]  \right)  _{\left[  G\right]  \in\operatorname*{GrIs}}$.
Thus, the family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ is $\kk$-linearly independent
(because the family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$ is $\kk$-linearly independent).
\par
But the definition of the grading on $\mathcal{G}$ shows that $\mathcal{G}%
_{n}$ is the $\kk$-span of the isomorphism classes $\left[  G\right]  $
of graphs $G$ with $n$ vertices. In other words, $\mathcal{G}_{n}$ is the
$\kk$-span of the family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \text{ is an isomorphism class of graphs with }n\text{ vertices}}$.
In other words, $\mathcal{G}_{n}$ is the $\kk$-span of the family
$\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ (since $\left(  \left[  G\right]  \right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}=\left(  \left[
G\right]  \right)  _{\left[  G\right]  \text{ is an isomorphism class of
graphs with }n\text{ vertices}}$). In other words, the family $\left(  \left[
G\right]  \right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$
spans the $\kk$-module $\mathcal{G}_{n}$.
\par
Thus, the family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ spans the $\kk$-module
$\mathcal{G}_{n}$ and is $\kk$-linearly independent. In other words,
the family $\left(  \left[  G\right]  \right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$. Qed.}. We are going to prove that the family $\left(
\left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$ as well.

We define a map $\mathbf{c}_{n}:\operatorname*{GrIs}\nolimits_{n}%
\rightarrow\operatorname*{GrIs}\nolimits_{n}$ by setting%
\[
\left(  \mathbf{c}_{n}\left[  G\right]  =\left[  \mathbf{c}\left(  G\right)
\right]  \ \ \ \ \ \ \ \ \ \ \text{for every }\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}\right)  .
\]
\footnote{This map $\mathbf{c}_{n}$ is well-defined because of the following
two simple observations:
\par
\begin{itemize}
\item For each $\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}$, we
have $\left[  \mathbf{c}\left(  G\right)  \right]  \in\operatorname*{GrIs}%
\nolimits_{n}$ (because if the graph $G$ has $n$ vertices, then so does the
graph $\mathbf{c}\left(  G\right)  $).
\par
\item For each graph $G$ with $n$ vertices, the isomorphism class $\left[
\mathbf{c}\left(  G\right)  \right]  $ depends only on the isomorphism class
$\left[  G\right]  $, not on the graph $G$ itself (i.e., if $G_{1}$ and
$G_{2}$ are two isomorphic graphs with $n$ vertices, then the graphs
$\mathbf{c}\left(  G_{1}\right)  $ and $\mathbf{c}\left(  G_{2}\right)  $ are
also isomorphic).
\end{itemize}
}

We have $\mathbf{c}_{n}\circ\mathbf{c}_{n}=\id$
\ \ \ \ \footnote{\textit{Proof.} Let $U\in\operatorname*{GrIs}\nolimits_{n}%
$. Thus, $U$ is an isomorphism class of a finite graph with $n$ vertices
(since $\operatorname*{GrIs}\nolimits_{n}$ is the set of all isomorphism
classes of finite graphs with $n$ vertices). In other words, there exists a
graph $G$ with $n$ vertices such that $U=\left[  G\right]  $. Consider this
$G$. Now, $\mathbf{c}_{n}\underbrace{U}_{=\left[  G\right]  }=\mathbf{c}%
_{n}\left[  G\right]  =\left[  \mathbf{c}\left(  G\right)  \right]  $ (by the
definition of $\mathbf{c}_{n}$). Now,%
\begin{align*}
\left(  \mathbf{c}_{n}\circ\mathbf{c}_{n}\right)  U  &  =\mathbf{c}_{n}\left(
\underbrace{\mathbf{c}_{n}U}_{=\left[  \mathbf{c}\left(  G\right)  \right]
}\right)  =\mathbf{c}_{n}\left[  \mathbf{c}\left(  G\right)  \right]  =\left[
\underbrace{\mathbf{c}\left(  \mathbf{c}\left(  G\right)  \right)
}_{\substack{=G\\\text{(by (\ref{pf.prop.GGG.moebius-basis.b.ccG}))}}}\right]
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{c}_{n}\right)
\\
&  =\left[  G\right]  =U=\id U.
\end{align*}
Now, forget that we fixed $U$. We thus have proven that $\left(
\mathbf{c}_{n}\circ\mathbf{c}_{n}\right)  U=\operatorname*{id}U$ for every
$U\in\operatorname*{GrIs}\nolimits_{n}$. In other words, $\mathbf{c}_{n}%
\circ\mathbf{c}_{n}=\id$. Qed.}. Hence, the maps $\mathbf{c}_{n}$ and
$\mathbf{c}_{n}$ are mutually inverse
(since $\mathbf{c}_{n}\circ\mathbf{c}_{n}=\id$ and
$\mathbf{c}_{n}\circ\mathbf{c}_{n}=\id$). Thus, the map
$\mathbf{c}_{n}$ is invertible, i.e., a bijection. Therefore, the family
$\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ is a reindexing of
the family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$. Therefore, the family $\left(  \left[
G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}%
\nolimits_{n}}$ (in turn) is a reindexing of the family $\left(  \left(
\mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$.

If $G=\left(  V,E\right)  $ is a finite graph with $n$ vertices, then%
\begin{equation}
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}=\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset E}}\left(
-1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[  H\right]
\label{pf.prop.GGG.moebius-basis.b.cmoeb}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.b.cmoeb}):} Let
$G=\left(  V,E\right)  $ be a finite graph with $n$ vertices. Then,
$\mathbf{c}\left(  G\right)  =\left(  V,E^{c}\right)  $ (by the definition of
$\mathbf{c}\left(  G\right)  $). Hence, the definition of $\left[
\mathbf{c}\left(  G\right)  \right]  ^{\sharp}$ yields
\[
\left[  \mathbf{c}\left(  G\right)  \right]  ^{\sharp}=\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset\left(
E^{c}\right)  ^{c}}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus\left(
E^{c}\right)  ^{c}\right\vert }\left[  H\right]  =\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\supset E}}\left(  -1\right)  ^{\left\vert
E^{\prime}\setminus E\right\vert }\left[  H\right]
\]
(since $\left(  E^{c}\right)  ^{c}=E$). Now, the definition of $\mathbf{c}%
_{n}$ yields $\mathbf{c}_{n}\left[  G\right]  =\left[  \mathbf{c}\left(
G\right)  \right]  $, and therefore $\left(  \mathbf{c}_{n}\left[  G\right]
\right)  ^{\sharp}=\left[  \mathbf{c}\left(  G\right)  \right]  ^{\sharp}%
=\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset
E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[
H\right]  $. This proves (\ref{pf.prop.GGG.moebius-basis.b.cmoeb}).}.

We define a map $e:\operatorname*{GrIs}\nolimits_{n}\rightarrow \NN$ by
setting%
\[
e\left[  \left(  V,E\right)  \right]  =\left\vert E\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for every }\left[  \left(  V,E\right)  \right]
\in\operatorname*{GrIs}\nolimits_{n}.
\]
\footnote{This map is well-defined, because for any finite graph $\left(
V,E\right)  $, the number $\left\vert E\right\vert $ depends only on the
isomorphism class $\left[  \left(  V,E\right)  \right]  $ of $\left(
V,E\right)  $ (and not on $\left(  V,E\right)  $ itself).} Thus, the map $e$
sends the isomorphism class of a graph to the number of edges of this graph.

We define a binary relation $\prec$ on the set $\operatorname*{GrIs}%
\nolimits_{n}$ as follows: For two elements $\left[  H\right]  $ and $\left[
G\right]  $ of $\operatorname*{GrIs}\nolimits_{n}$, we set $\left[  H\right]
\prec\left[  G\right]  $ if and only if $e\left[  H\right]  >e\left[
G\right]  $. It is clear that this binary relation $\prec$ is transitive,
asymmetric and irreflexive. Thus, there is a partial order on the set
$\operatorname*{GrIs}\nolimits_{n}$ whose smaller relation is $\prec$.
Consider $\operatorname*{GrIs}\nolimits_{n}$ as a poset, equipped with this
partial order.

Now, every finite graph $G$ with $n$ vertices satisfies%
\begin{equation}
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}=\left[  G\right]
+\left(  \text{a }\kk\text{-linear combination of the elements }\left[
H\right]  \text{ for }\left[  H\right]  \in\operatorname*{GrIs}\nolimits_{n}%
\text{ satisfying }\left[  H\right]  \prec\left[  G\right]  \right)
\label{pf.prop.GGG.moebius-basis.b.unitri}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.b.unitri}):} Let
$G$ be a finite graph with $n$ vertices. Write $G$ as $G=\left(  V,E\right)
$.
\par
Let $H=\left(  V,E^{\prime}\right)  $ be any graph satisfying $E^{\prime
}\supset E$ and $E^{\prime}\neq E$. Thus, $E\subset E^{\prime}$ (since
$E^{\prime}\supset E$) and $E\neq E^{\prime}$. In other words, $E$ is a proper
subset of $E^{\prime}$.
\par
The graph $G$ has $n$ vertices. In other words, the graph $\left(  V,E\right)
$ has $n$ vertices (since $G=\left(  V,E\right)  $). In other words,
$\left\vert V\right\vert =n$. In other words, the graph $\left(  V,E^{\prime
}\right)  $ has $n$ vertices. In other words, the graph $H$ has $n$ vertices
(since $H=\left(  V,E^{\prime}\right)  $). Hence, $\left[  H\right]
\in\operatorname*{GrIs}\nolimits_{n}$.
\par
The set $E^{\prime}$ is a set of subsets of $V$ (since $\left(  V,E^{\prime
}\right)  $ is a graph), and thus a finite set (since there are only finitely
many subsets of $V$ (since $V$ is a finite set)). Hence, $\left\vert
E\right\vert <\left\vert E^{\prime}\right\vert $ (since $E$ is a proper subset
of $E^{\prime}$). In other words, $\left\vert E^{\prime}\right\vert
>\left\vert E\right\vert $.
\par
The definition of $e$ yields $e\left[  \left(  V,E\right)  \right]
=\left\vert E\right\vert $ and $e\left[  \left(  V,E^{\prime}\right)  \right]
=\left\vert E^{\prime}\right\vert $. Now, $e\left[  \underbrace{G}_{=\left(
V,E\right)  }\right]  =e\left[  \left(  V,E\right)  \right]  =\left\vert
E\right\vert $, so that $\left\vert E\right\vert =e\left[  G\right]  $. But
$e\left[  \underbrace{H}_{=\left(  V,E^{\prime}\right)  }\right]  =e\left[
\left(  V,E^{\prime}\right)  \right]  =\left\vert E^{\prime}\right\vert
>\left\vert E\right\vert =e\left[  G\right]  $.
\par
Recall that we have $\left[  H\right]  \prec\left[  G\right]  $ if and only if
$e\left[  H\right]  >e\left[  G\right]  $ (by the definition of the relation
$\prec$). Hence, $\left[  H\right]  \prec\left[  G\right]  $ (since $e\left[
H\right]  >e\left[  G\right]  $).
\par
Now, forget that we fixed $H$. We thus have shown that if $H=\left(
V,E^{\prime}\right)  $ is any graph satisfying $E^{\prime}\supset E$ and
$E^{\prime}\neq E$, then $\left[  H\right]  \in\operatorname*{GrIs}%
\nolimits_{n}$ and $\left[  H\right]  \prec\left[  G\right]  $. Hence, the
sum
\[
\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset E\text{
and }E^{\prime}\neq E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus
E\right\vert }\left[  H\right]
\]
is a $\kk$-linear combination of the elements $\left[  H\right]  $ for
$\left[  H\right]  \in\operatorname*{GrIs}\nolimits_{n}$ satisfying $\left[
H\right]  \prec\left[  G\right]  $. In other words,%
\begin{align}
&  \sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset E\text{
and }E^{\prime}\neq E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus
E\right\vert }\left[  H\right]  \nonumber\\
&  =\left(  \text{a }\kk\text{-linear combination of the elements
}\left[  H\right]  \text{ for }\left[  H\right]  \in\operatorname*{GrIs}%
\nolimits_{n}\text{ satisfying }\left[  H\right]  \prec\left[  G\right]
\right)  .\label{pf.prop.GGG.moebius-basis.b.unitri.pf.1}%
\end{align}
\par
But $E$ is a set of $2$-element subsets of $V$, and satisfies $E\supset E$. In
other words, $E$ is a set $E^{\prime}$ of $2$-element subsets of $V$
satisfying $E^{\prime}\supset E$. Hence, the sum $\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\supset E}}\left(  -1\right)  ^{\left\vert
E^{\prime}\setminus E\right\vert }\left[  H\right]  $ has an addend for
$E^{\prime}=E$. Splitting off this addend from this sum, we obtain%
\begin{align*}
&  \sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset
E}}\left(  -1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[
H\right]  \\
&  =\underbrace{\left(  -1\right)  ^{\left\vert E\setminus E\right\vert }%
}_{\substack{=1\\\text{(since }\left\vert E\setminus E\right\vert =\left\vert
\varnothing\right\vert =0\text{)}}}\left[  \underbrace{\left(  V,E\right)
}_{=G}\right]  +\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)
;\\E^{\prime}\supset E\text{ and }E^{\prime}\neq E}}\left(  -1\right)
^{\left\vert E^{\prime}\setminus E\right\vert }\left[  H\right]
}_{\substack{=\left(  \text{a }\kk\text{-linear combination of the
elements }\left[  H\right]  \text{ for }\left[  H\right]  \in
\operatorname*{GrIs}\nolimits_{n}\text{ satisfying }
\left[H\right]\prec \left[G\right]\right)
\\\text{(by (\ref{pf.prop.GGG.moebius-basis.b.unitri.pf.1}))}}}\\
&  =\left[  G\right]  +\left(  \text{a }\kk\text{-linear combination of
the elements }\left[  H\right]  \text{ for }\left[  H\right]  \in
\operatorname*{GrIs}\nolimits_{n}\text{ satisfying }\left[  H\right]
\prec\left[  G\right]  \right)  .
\end{align*}
Thus, (\ref{pf.prop.GGG.moebius-basis.b.cmoeb}) becomes%
\begin{align*}
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp} &  =\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\supset E}}\left(
-1\right)  ^{\left\vert E^{\prime}\setminus E\right\vert }\left[  H\right]  \\
&  =\left[  G\right]  +\left(  \text{a }\kk\text{-linear combination of
the elements }\left[  H\right]  \text{ for }\left[  H\right]  \in
\operatorname*{GrIs}\nolimits_{n}\text{ satisfying }\left[  H\right]
\prec\left[  G\right]  \right)  .
\end{align*}
This proves (\ref{pf.prop.GGG.moebius-basis.b.unitri}).}. In other words, the
family $\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp
}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ expands
unitriangularly in the family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ (by Remark
\ref{rmk.STmat.expansion-tria-shortcuts}(c), applied to $\mathcal{G}_{n}$,
$\operatorname*{GrIs}\nolimits_{n}$, $\left(  \left(  \mathbf{c}_{n}\left[
G\right]  \right)  ^{\sharp}\right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ and $\left(  \left[  G\right]  \right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ instead of $M$,
$S$, $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)  _{s\in S}$).
Hence, the family $\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)
^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$
is a basis of the $\kk$-module $\mathcal{G}_{n}$ if and only if the
family $\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$ (by Corollary \ref{cor.STmat.expansion-tria-inv}(e), applied
to $\mathcal{G}_{n}$, $\operatorname*{GrIs}\nolimits_{n}$, $\left(  \left(
\mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}\nolimits_{n}}$ and $\left(  \left[  G\right]  \right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ instead of $M$,
$S$, $\left(  e_{s}\right)  _{s\in S}$ and $\left(  f_{s}\right)  _{s\in S}$).
Thus, the family $\left(  \left(  \mathbf{c}_{n}\left[  G\right]  \right)
^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$
is a basis of the $\kk$-module $\mathcal{G}_{n}$ (since the family
$\left(  \left[  G\right]  \right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the $\kk$-module
$\mathcal{G}_{n}$). Therefore, the family $\left(  \left[  G\right]  ^{\sharp
}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ also is a
basis of the $\kk$-module $\mathcal{G}_{n}$ (since the family $\left(
\left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in
\operatorname*{GrIs}\nolimits_{n}}$ is a reindexing of the family $\left(
\left(  \mathbf{c}_{n}\left[  G\right]  \right)  ^{\sharp}\right)  _{\left[
G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$).

Now, forget that we fixed $n$. We thus have shown that, for every
$n\in \NN$, the family $\left(  \left[  G\right]  ^{\sharp}\right)
_{\left[  G\right]  \in\operatorname*{GrIs}\nolimits_{n}}$ is a basis of the
$\kk$-module $\mathcal{G}_{n}$. Therefore, the disjoint union of these
families (over all $n\in \NN$) is a basis of the direct sum
$\bigoplus_{n\in\NN}\mathcal{G}_{n}$. Since the former disjoint union
is the family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$, whereas the latter direct sum is $\bigoplus
_{n\in\NN}\mathcal{G}_{n}=\mathcal{G}$, this rewrites as follows: The
family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]
\in\operatorname*{GrIs}}$ is a basis of the $\kk$-module $\mathcal{G}$.
In other words, the elements $\left[  G\right]  ^{\sharp}$, where $\left[
G\right]  $ ranges over all isomorphism classes of finite graphs, form a basis
of the $\kk$-module $\mathcal{G}$\ \ \ \ \footnote{\textit{Proof.}
Recall that $\operatorname*{GrIs}$ is the set of all isomorphism classes of
finite graphs. Thus, the elements of $\operatorname*{GrIs}$ are the
isomorphism classes of finite graphs. Hence, $\left(  \left[  G\right]
^{\sharp}\right)  _{\left[  G\right]  \in\operatorname*{GrIs}}=\left(  \left[
G\right]  ^{\sharp}\right)  _{\left[  G\right]  \text{ is an isomorphism class
of finite graphs}}$.
\par
Now, recall that the family $\left(  \left[  G\right]  ^{\sharp}\right)
_{\left[  G\right]  \in\operatorname*{GrIs}}$ is a basis of the $\kk%
$-module $\mathcal{G}$. In other words, the family $\left(  \left[  G\right]
^{\sharp}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ is a basis of the $\kk$-module $\mathcal{G}$ (since $\left(
\left[  G\right]  ^{\sharp}\right)  _{\left[  G\right]  \in
\operatorname*{GrIs}}=\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$). In other words,
the elements $\left[  G\right]  ^{\sharp}$, where $\left[  G\right]  $ ranges
over all isomorphism classes of finite graphs, form a basis of the
$\kk$-module $\mathcal{G}$. Qed.}. Proposition
\ref{prop.GGG.moebius-basis}(b) is thus proven.

(c) We define a $\kk$-linear map $\Delta^{\prime}:\GGG%
\rightarrow\GGG\otimes\GGG$ by%
\[
\Delta^{\prime}\left[  H\right]  ^{\sharp}=\sum\limits_{\substack{\left(  V_{1}%
,V_{2}\right)  ;\\V=V_{1}\sqcup V_{2};\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}%
}}\left[  H\mid_{V_{1}}\right]  ^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]
^{\sharp}.
\]
\footnote{This is well-defined, since the $\left[  G\right]  ^{\sharp}$ form a
basis of the $\kk$-module $\GGG$.} In order to prove Proposition
\ref{prop.GGG.moebius-basis}(c), it clearly suffices to show that
$\Delta^{\prime}=\Delta$.

Let $G=\left(  V,E\right)  $ be a finite graph. If $T$ is a set of two-element
subsets of $V$, and if $R$ is a subset of $V$, then $T\mid_{R}$ shall denote
the subset $\left\{  X\in T\ \mid\ X\subset R\right\}  $ of $T$. (Thus, if $R$
is a subset of $V$, then $E\mid_{R}$ is the set of edges of the graph
$G\mid_{R}$.)

Now, Proposition \ref{prop.GGG.moebius-basis}(a) yields $\left[  G\right]
=\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left[  H\right]  ^{\sharp}$. Hence,
\begin{align*}
\Delta^{\prime}\left[  G\right]   &  =\Delta^{\prime}\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing}}\left[  H\right]
^{\sharp}=\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\underbrace{\Delta^{\prime}\left[  H\right]  ^{\sharp}}%
_{=\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup V_{2}%
;\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}\left[  H\mid_{V_{1}}\right]
^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp}}\\
&  =\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime
}\cap E=\varnothing}}\sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V=V_{1}\sqcup V_{2};\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}}_{=\sum\limits
_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup V_{2}}}\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
;\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}}\left[  H\mid_{V_{1}}\right]
^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup V_{2}%
}}\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing;\\H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}}}\left[  H\mid_{V_{1}%
}\right]  ^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp}%
}_{\substack{=\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}%
}\sum\limits_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime
}\cap\left(  E\mid_{V_{2}}\right)  =\varnothing}}\sum\limits_{\substack{H=\left(
V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing;\\H=H\mid_{V_{1}}\sqcup
H\mid_{V_{2}};\\H\mid_{V_{1}}=H_{1};\ H\mid_{V_{2}}=H_{2}}}\left[
H\mid_{V_{1}}\right]  ^{\sharp}\otimes\left[  H\mid_{V_{2}}\right]  ^{\sharp
}\\\text{(because for every }H=\left(  V,E^{\prime}\right)  \text{ satisfying
}E^{\prime}\cap E=\varnothing\text{,}\\\text{we can write the subgraph }%
H\mid_{V_{1}}\text{ in the form }H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
\\\text{with some }E_{1}^{\prime}\text{ satisfying }E_{1}^{\prime}\cap\left(
E\mid_{V_{1}}\right)  =\varnothing\text{, and we can}\\\text{write the
subgraph }H\mid_{V_{2}}\text{ in the form }H_{2}=\left(  V_{2},E_{2}^{\prime
}\right)  \\\text{with some }E_{2}^{\prime}\text{ satisfying }E_{2}^{\prime
}\cap\left(  E\mid_{V_{2}}\right)  =\varnothing\text{)}}}\\
&  =\underbrace{\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V=V_{1}\sqcup
V_{2}}}}_{=\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}%
=V}}}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}%
^{\prime}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}%
=\left(  V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(
E\mid_{V_{2}}\right)  =\varnothing}}\sum\limits_{\substack{H=\left(  V,E^{\prime
}\right)  ;\\E^{\prime}\cap E=\varnothing;\\H=H\mid_{V_{1}}\sqcup H\mid
_{V_{2}};\\H\mid_{V_{1}}=H_{1};\ H\mid_{V_{2}}=H_{2}}}\left[
\underbrace{H\mid_{V_{1}}}_{=H_{1}}\right]  ^{\sharp}\otimes\left[
\underbrace{H\mid_{V_{2}}}_{=H_{2}}\right]  ^{\sharp}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}%
}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime
}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(  E\mid_{V_{2}%
}\right)  =\varnothing}}\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime
}\right)  ;\\E^{\prime}\cap E=\varnothing;\\H=H\mid_{V_{1}}\sqcup H\mid
_{V_{2}};\\H\mid_{V_{1}}=H_{1};\ H\mid_{V_{2}}=H_{2}}}\left[  H_{1}\right]
^{\sharp}\otimes\left[  H_{2}\right]  ^{\sharp}}_{\substack{=\left[
H_{1}\right]  ^{\sharp}\otimes\left[  H_{2}\right]  ^{\sharp}\\\text{(because
this sum has only one addend}\\\text{(indeed, there exists only one }H=\left(
V,E^{\prime}\right)  \text{ satisfying}\\E^{\prime}\cap E=\varnothing\text{,
}H=H\mid_{V_{1}}\sqcup H\mid_{V_{2}}\text{, }H\mid_{V_{1}}=H_{1}\text{
and}\ H\mid_{V_{2}}=H_{2}\text{))}}}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}%
}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime
}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(  E\mid_{V_{2}%
}\right)  =\varnothing}}\left[  H_{1}\right]  ^{\sharp}\otimes\left[
H_{2}\right]  ^{\sharp}.
\end{align*}
Comparing this with
\begin{align*}
\Delta\left[  G\right]   &  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\underbrace{\left[  G\mid_{V_{1}}\right]
}_{\substack{=\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\left[
H_{1}\right]  ^{\sharp}\\\text{(by Proposition \ref{prop.GGG.moebius-basis}%
(a))}}}\otimes\underbrace{\left[  G\mid_{V_{2}}\right]  }_{\substack{=\sum\limits
_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}%
\cap\left(  E\mid_{V_{2}}\right)  =\varnothing}}\left[  H_{2}\right]
^{\sharp}\\\text{(by Proposition \ref{prop.GGG.moebius-basis}(a))}}}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}%
}\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime
}\cap\left(  E\mid_{V_{1}}\right)  =\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap\left(  E\mid_{V_{2}%
}\right)  =\varnothing}}\left[  H_{1}\right]  ^{\sharp}\otimes\left[
H_{2}\right]  ^{\sharp},
\end{align*}
we obtain $\Delta^{\prime}\left[  G\right]  =\Delta\left[  G\right]  $. Since
this holds for every graph $G$, we thus obtain $\Delta^{\prime}=\Delta$. As we
know, this proves Proposition \ref{prop.GGG.moebius-basis}(c).

(d) We define a $\kk$-bilinear operation $\sharp:\GGG%
\times\GGG\rightarrow\GGG$ (written infix\footnote{This means
that we write $a\sharp b$ to denote the image of a pair $\left(  a,b\right)
\in\GGG\times\GGG$ under this operation $\sharp$.}) by%
\[
\left[  H_{1}\right]  ^{\sharp}\sharp\left[  H_{2}\right]  ^{\sharp}%
=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}%
=H_{1};\\H\mid_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}.
\]
\footnote{This is well-defined, since the $\left[  G\right]  ^{\sharp}$ form a
basis of the $\kk$-module $\GGG$.} In order to prove Proposition
\ref{prop.GGG.moebius-basis}(d), it clearly suffices to show that this
operation $\sharp$ is identical with the usual multiplication on $\GGG$.

Let $G_{1}$ and $G_{2}$ be any two finite graphs. We shall show that $\left[
G_{1}\right]  \sharp\left[  G_{2}\right]  =\left[  G_{1}\right]  \left[
G_{2}\right]  $.

Indeed, write $G_{1}$ and $G_{2}$ in the forms $G_{1}=\left(  V_{1}%
,E_{1}\right)  $ and $G_{2}=\left(  V_{2},E_{2}\right)  $. Then,
\begin{align*}
&  \underbrace{\left[  G_{1}\right]  }_{\substack{=\sum\limits_{\substack{H_{1}%
=\left(  V_{1},E_{1}^{\prime}\right)  ;\\E_{1}^{\prime}\cap E_{1}=\varnothing
}}\left[  H_{1}\right]  ^{\sharp}\\\text{(by Proposition
\ref{prop.GGG.moebius-basis}(a))}}}\sharp\underbrace{\left[  G_{2}\right]
}_{\substack{=\sum\limits_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)
;\\E_{2}^{\prime}\cap E_{2}=\varnothing}}\left[  H_{2}\right]  ^{\sharp
}\\\text{(by Proposition \ref{prop.GGG.moebius-basis}(a))}}}\\
&  =\left(  \sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap E_{1}=\varnothing}}\left[  H_{1}\right]  ^{\sharp
}\right)  \sharp\left(  \sum\limits_{\substack{H_{2}=\left(  V_{2},E_{2}^{\prime
}\right)  ;\\E_{2}^{\prime}\cap E_{2}=\varnothing}}\left[  H_{2}\right]
^{\sharp}\right)  =\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap E_{1}=\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap E_{2}=\varnothing
}}\underbrace{\left[  H_{1}\right]  ^{\sharp}\sharp\left[  H_{2}\right]
^{\sharp}}_{=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}}\\
&  =\underbrace{\sum\limits_{\substack{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)
;\\E_{1}^{\prime}\cap E_{1}=\varnothing}}\sum\limits_{\substack{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  ;\\E_{2}^{\prime}\cap E_{2}=\varnothing}%
}\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}%
=H_{1};\\H\mid_{V_{2}}=H_{2}}}}_{=\sum_{H_{1}=\left(  V_{1},E_{1}^{\prime
}\right)  }\sum_{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  }\sum\limits
_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}%
=H_{1};\\H\mid_{V_{2}}=H_{2};\\E_{1}^{\prime}\cap E_{1}=\varnothing
;\\E_{2}^{\prime}\cap E_{2}=\varnothing}}}\left[  H\right]  ^{\sharp}\\
&  =\sum_{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  }\sum_{H_{2}=\left(
V_{2},E_{2}^{\prime}\right)  }\underbrace{\sum\limits_{\substack{H=\left(
V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}%
=H_{2};\\E_{1}^{\prime}\cap E_{1}=\varnothing;\\E_{2}^{\prime}\cap
E_{2}=\varnothing}}}_{\substack{=\sum\limits_{\substack{H=\left(  V_{1}\sqcup
V_{2},E\right)  ;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}=H_{2};\\\left(
E\mid_{V_{1}}\right)  \cap E_{1}=\varnothing;\\\left(  E\mid_{V_{2}}\right)
\cap E_{2}=\varnothing}}\\\text{(because for a graph }H=\left(  V_{1}\sqcup
V_{2},E\right)  \\\text{satisfying }H\mid_{V_{1}}=H_{1}\text{ and }%
H\mid_{V_{2}}=H_{2}\text{,}\\\text{we have }E_{1}^{\prime}=E\mid_{V_{1}}\text{
and }E_{2}^{\prime}=E\mid_{V_{2}}\text{)}}}\left[  H\right]  ^{\sharp}\\
&  =\underbrace{\sum_{H_{1}=\left(  V_{1},E_{1}^{\prime}\right)  }\sum
_{H_{2}=\left(  V_{2},E_{2}^{\prime}\right)  }\sum\limits_{\substack{H=\left(
V_{1}\sqcup V_{2},E\right)  ;\\H\mid_{V_{1}}=H_{1};\\H\mid_{V_{2}}%
=H_{2};\\\left(  E\mid_{V_{1}}\right)  \cap E_{1}=\varnothing;\\\left(
E\mid_{V_{2}}\right)  \cap E_{2}=\varnothing}}}_{\substack{=\sum\limits
_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\\left(  E\mid_{V_{1}%
}\right)  \cap E_{1}=\varnothing;\\\left(  E\mid_{V_{2}}\right)  \cap
E_{2}=\varnothing}}=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\E\cap\left(  E_{1}\sqcup E_{2}\right)  =\varnothing}}\\\text{(since the
statement }\left(  \left(  E\mid_{V_{1}}\right)  \cap E_{1}=\varnothing\text{
and }\left(  E\mid_{V_{2}}\right)  \cap E_{2}=\varnothing\right)  \\\text{is
equivalent to }E\cap\left(  E_{1}\sqcup E_{2}\right)  =\varnothing\text{)}%
}}\left[  H\right]  ^{\sharp}\\
&  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\E\cap\left(
E_{1}\sqcup E_{2}\right)  =\varnothing}}\left[  H\right]  ^{\sharp}.
\end{align*}
Compared with%
\begin{align*}
\left[  G_{1}\right]  \left[  G_{2}\right]   &  =\left[  G_{1}\sqcup
G_{2}\right]  =\left[  \left(  V_{1}\sqcup V_{2},E_{1}\sqcup E_{2}\right)
\right] \\
&  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\E\cap\left(
E_{1}\sqcup E_{2}\right)  =\varnothing}}\left[  H\right]  ^{\sharp
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.GGG.moebius-basis}%
(a)}\right)  ,
\end{align*}
this yields $\left[  G_{1}\right]  \sharp\left[  G_{2}\right]  =\left[
G_{1}\right]  \left[  G_{2}\right]  $.

We now forget that we fixed $G_{1}$ and $G_{2}$. We have thus shown that
$\left[  G_{1}\right]  \sharp\left[  G_{2}\right]  =\left[  G_{1}\right]
\left[  G_{2}\right]  $ for any two finite graphs $G_{1}$ and $G_{2}$. Thus,
the operation $\sharp$ is identical with the usual multiplication on
$\GGG$. This completes the proof of Proposition
\ref{prop.GGG.moebius-basis}(d).
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop.GGG.bilform.sym}.]
Proposition~\ref{prop.GGG.moebius-basis}(a) shows that every finite graph
$G=\left(  V,E\right)  $ satisfies%
\begin{equation}
\left[  G\right]  =\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)
;\\E^{\prime}\cap E=\varnothing}}\left[  H\right]  ^{\sharp}=\sum\limits
_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left[  K\right]  ^{\sharp} \label{pf.prop.GGG.moebius-basis.e.sum-lem}%
\end{equation}
(here, we renamed the summation index $H$ as $K$). Let us now show that%
\begin{equation}
\left(  \left[  G\right]  ,\left[  H\right]  \right)  =\left(  \left[
H\right]  ,\left[  G\right]  \right)
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis}%
\end{equation}
for any two finite graphs $G$ and $H$.

\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis}):} We shall
use the following notation: If $V$ and $W$ are two sets, and if $\varphi
:V\rightarrow W$ is a map, then $\varphi_{\ast}$ will denote the map from the
powerset of $V$ to the powerset of $W$ which sends every $T\subset V$ to
$\varphi\left(  T\right)  \subset W$. This map $\varphi_{\ast}$ is a bijection
if $\varphi$ is a bijection.

Let $G$ and $H$ be two finite graphs. Let us write $G$ and $H$ in the forms
$G=\left(  V,E\right)  $ and $H=\left(  W,F\right)  $. Now,
\begin{align}
\left(  \underbrace{\left[  G\right]  }_{\substack{=\sum\limits
_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left[  K\right]  ^{\sharp}\\\text{(by
(\ref{pf.prop.GGG.moebius-basis.e.sum-lem}))}}},\left[  H\right]  \right)
&  =\left(  \sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left[  K\right]  ^{\sharp},\left[  H\right]  \right)
=\sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\underbrace{\left(  \left[  K\right]  ^{\sharp},\left[
H\right]  \right)  }_{\substack{=\left\vert \operatorname*{Iso}\left(
K,H\right)  \right\vert \\\text{(by the definition of}\\\text{the form
}\left(  \cdot,\cdot\right)  \text{)}}} \nonumber \\
& \ \ \ \ \ \ \ \ \ \ \left(  \text{since
the form }\left(  \cdot,\cdot\right)  \text{ is }\kk\text{-bilinear}%
\right) \nonumber\\
&  =\sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left\vert \operatorname*{Iso}\left(  K,H\right)  \right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.1}%
\end{align}


However, for every graph $K=\left(  V,E^{\prime}\right)  $, we have
\begin{align*}
\operatorname*{Iso}\left(  K,H\right)   &  =\left(  \text{the set of all
isomorphisms from }K\text{ to }H\right) \\
&  =\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a bijection and
satisfies }\varphi_{\ast}\left(  E^{\prime}\right)  =F\right\}
\end{align*}
(because the isomorphisms from $K$ to $H$ are defined to be the bijections
$\varphi:V\rightarrow W$ such that $\varphi_{\ast}\left(  E^{\prime}\right)
=F$). Hence,%
\begin{align*}
&  \sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap
E=\varnothing}}\left\vert \underbrace{\operatorname*{Iso}\left(  K,H\right)
}_{=\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a bijection and
satisfies }\varphi_{\ast}\left(  E^{\prime}\right)  =F\right\}  }\right\vert
\\
&  =\underbrace{\sum\limits_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime
}\cap E=\varnothing}}}_{=\sum\limits_{\substack{E^{\prime}\text{ is a set
of}\\\text{two-element subsets of }V;\\E^{\prime}\cap E=\varnothing}%
}}\left\vert \left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\underbrace{\varphi_{\ast}\left(  E^{\prime}\right)
=F}_{\substack{\text{this is equivalent to}\\E^{\prime}=\left(  \varphi_{\ast
}\right)  ^{-1}\left(  F\right)  \\\text{(since }\varphi_{\ast}\text{ is a
bijection}\\\text{(since }\varphi\text{ is a bijection))}}}\right\}
\right\vert \\
&  =\sum\limits_{\substack{E^{\prime}\text{ is a set of}\\\text{two-element subsets
of }V;\\E^{\prime}\cap E=\varnothing}}\left\vert \left\{  \varphi:V\rightarrow
W\ \mid\ \varphi\text{ is a bijection and satisfies }E^{\prime}=\left(
\varphi_{\ast}\right)  ^{-1}\left(  F\right)  \right\}  \right\vert \\
&  =\left\vert \left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}  \right\vert .
\end{align*}
Now, (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.1}) becomes%
\begin{align}
\left(  \left[  G\right]  ,\left[  H\right]  \right)   &  =\sum\limits
_{\substack{K=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left\vert \operatorname*{Iso}\left(  K,H\right)  \right\vert \nonumber\\
&  =\left\vert \left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}  \right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4}%
\end{align}
The same argument (applied to $H$, $W$, $F$, $G$, $V$ and $E$ instead of $G$,
$V$, $E$, $H$, $W$ and $F$) yields%
\begin{equation}
\left(  \left[  H\right]  ,\left[  G\right]  \right)  =\left\vert \left\{
\varphi:W\rightarrow V\ \mid\ \varphi\text{ is a bijection and satisfies
}\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)  \cap F=\varnothing
\right\}  \right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5}
\end{equation}


Now, we define two sets $\mathfrak{P}$ and $\mathfrak{Q}$ by%
\[
\mathfrak{P}=\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}
\]
and%
\[
\mathfrak{Q}=\left\{  \varphi:W\rightarrow V\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)
\cap F=\varnothing\right\}  .
\]
Then, (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4}) becomes%
\begin{equation}
\left(  \left[  G\right]  ,\left[  H\right]  \right)  =\left\vert
\underbrace{\left\{  \varphi:V\rightarrow W\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)
\cap E=\varnothing\right\}  }_{=\mathfrak{P}}\right\vert =\left\vert
\mathfrak{P}\right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4a}%
\end{equation}
Also, (\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5}) becomes%
\begin{equation}
\left(  \left[  H\right]  ,\left[  G\right]  \right)  =\left\vert
\underbrace{\left\{  \varphi:W\rightarrow V\ \mid\ \varphi\text{ is a
bijection and satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)
\cap F=\varnothing\right\}  }_{=\mathfrak{Q}}\right\vert =\left\vert
\mathfrak{Q}\right\vert .
\label{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5a}%
\end{equation}
But every $\psi\in\mathfrak{P}$ satisfies $\psi^{-1}\in\mathfrak{Q}%
$\ \ \ \ \footnote{\textit{Proof.}
Let $\psi\in\mathfrak{P}$. Then,
$\psi \in \mathfrak{P} = \left\{ \varphi:V\rightarrow W\ \mid\ \varphi
\text{ is a bijection and satisfies }
\left(  \varphi_{\ast}\right)  ^{-1}\left(  F\right)  \cap E=\varnothing
\right\}  $. Hence, $\psi$ is a bijection $V\rightarrow W$ and satisfies
$\left(  \psi_{\ast}\right)  ^{-1}\left(  F\right)  \cap E=\varnothing$.

Let $\rho$ be the map $\psi^{-1} : W\rightarrow V$. This map $\rho$ is
well-defined (since $\psi$ is a bijection) and is a bijection itself (since
$\rho=\psi^{-1}$).

But $\psi$ is a bijection. Thus, $\psi_{\ast}$ is a bijection as well, and
satisfies
$\left(  \psi_{\ast}\right)  ^{-1}
= \left(  \underbrace{\psi^{-1}}_{=\rho}\right)  _{\ast} = \rho_{\ast}$.
Of course, $\rho_{\ast}$ is also a bijection (since $\rho$ is a
bijection).

Now,
\begin{align*}
\rho_{\ast}\left(  \left(  \rho_{\ast}\right)  ^{-1}\left(  E\right)  \cap
F\right)   &  =\underbrace{\rho_{\ast}\left(  \left(  \rho_{\ast}\right)
^{-1}\left(  E\right)  \right)  }_{\substack{=E\\\text{(since }\rho_{\ast
}\text{ is a bijection)}}}\cap\underbrace{\rho_{\ast}}_{=\left(  \psi_{\ast
}\right)  ^{-1}}\left(  F\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\rho_{\ast}\text{ is a bijection}\right)  \\
&  =E\cap\left(  \psi_{\ast}\right)  ^{-1}\left(  F\right)  =\left(
\psi_{\ast}\right)  ^{-1}\left(  F\right)  \cap E=\varnothing.
\end{align*}
Thus,
\begin{align*}
\left(  \rho_{\ast}\right)  ^{-1}\left(  E\right)  \cap F &  =\rho_{\ast
}\left(  \varnothing\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }
\rho_{\ast}\text{ is a bijection}\right)  \\
&  =\varnothing.
\end{align*}
Hence, $\rho:W\rightarrow V$ is a bijection and satisfies $\left(  \rho_{\ast
}\right)  ^{-1}\left(  E\right)  \cap F=0$. In other words,
\[
\rho\in\left\{  \varphi:W\rightarrow V\ \mid\ \varphi\text{ is a bijection and
satisfies }\left(  \varphi_{\ast}\right)  ^{-1}\left(  E\right)  \cap
F=\varnothing\right\}  =\mathfrak{Q}.
\]
Thus, $\psi^{-1}=\rho\in\mathfrak{Q}$, qed.}. Hence, we can define a map
\begin{align*}
\mathbf{A}:\mathfrak{P}  &  \rightarrow\mathfrak{Q},\\
\psi &  \mapsto\psi^{-1}.
\end{align*}
Similarly, we can define a map%
\begin{align*}
\mathbf{B}:\mathfrak{Q}  &  \rightarrow\mathfrak{P},\\
\psi &  \mapsto\psi^{-1}.
\end{align*}
Consider these two maps $\mathbf{A}$ and $\mathbf{B}$. Clearly, these maps
$\mathbf{A}$ and $\mathbf{B}$ are mutually inverse\footnote{\textit{Proof.}
Every $\psi\in\mathfrak{P}$ satisfies
\begin{align*}
\left(  \mathbf{B}\circ\mathbf{A}\right)  \left(  \psi\right)   &
=\mathbf{B}\left(  \underbrace{\mathbf{A}\left(  \psi\right)  }%
_{\substack{=\psi^{-1}\\\text{(by the definition}\\\text{of }\mathbf{A}%
\text{)}}}\right)  =\mathbf{B}\left(  \psi^{-1}\right)  =\left(  \psi
^{-1}\right)  ^{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\mathbf{B}\right) \\
&  =\psi=\id\left(  \psi\right)  .
\end{align*}
Hence, $\mathbf{B}\circ\mathbf{A}=\id$. Similarly,
$\mathbf{A}\circ\mathbf{B}=\id$. Combining this with
$\mathbf{B}\circ\mathbf{A}=\id$, we conclude that the maps
$\mathbf{A}$ and $\mathbf{B}$ are mutually inverse. Qed.}. Hence, $\mathbf{A}$
is a bijection. Thus, there exists a bijection $\mathfrak{P}\rightarrow
\mathfrak{Q}$ (namely, $\mathbf{A}$). Consequently, $\left\vert \mathfrak{P}%
\right\vert =\left\vert \mathfrak{Q}\right\vert $. Now,
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.4a}) becomes $\left(
\left[  G\right]  ,\left[  H\right]  \right)  =\left\vert \mathfrak{P}%
\right\vert =\left\vert \mathfrak{Q}\right\vert =\left(  \left[  H\right]
,\left[  G\right]  \right)  $ (by
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis.pf.5a})). This proves
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis}).

Now, let $a$ and $b$ be two elements of $\GGG$. We want to show that
$\left(  a,b\right)  =\left(  b,a\right)  $. This equality is $\kk%
$-linear in each of $a$ and $b$. Therefore, we can WLOG assume that $a$ and
$b$ belong to the family $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$ (because this
family is a basis of the $\kk$-module $\GGG$). Assume this.
Then, $a=\left[  G\right]  $ and $b=\left[  H\right]  $ for two finite graphs
$G$ and $H$. Consider these $G$ and $H$. Now,%
\begin{align*}
\left(  \underbrace{a}_{=\left[  G\right]  },\underbrace{b}_{=\left[
H\right]  }\right)   &  =\left(  \left[  G\right]  ,\left[  H\right]  \right)
=\left(  \underbrace{\left[  H\right]  }_{=b},\underbrace{\left[  G\right]
}_{=a}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.GGG.moebius-basis.e.goal-on-basis})}\right) \\
&  =\left(  b,a\right)  .
\end{align*}
We thus have shown $\left(  a,b\right)  =\left(  b,a\right)  $.

Let us now forget that we fixed $a$ and $b$. We thus have proven that $\left(
a,b\right)  =\left(  b,a\right)  $ for any $a\in\GGG$ and
$b\in\GGG$. In other words, the form $\left(  \cdot,\cdot\right)
:\GGG\times\GGG\rightarrow\kk$ is symmetric. This proves
Proposition~\ref{prop.GGG.bilform.sym}.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop.GGG.selfdual}.]
In this proof, we shall use all the notations that appeared
in Definition~\ref{def.GGG.selfdual}. We shall also use the fact that
the family
$\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$
is a basis of the $\kk$-module $\GGG$. (This fact is
Proposition~\ref{prop.GGG.moebius-basis}(b).)

(a) We first notice that any two finite graphs $G$ and $H$ satisfy
\begin{equation}
\left(  \psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[
H\right]  =\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert
\label{pf.prop.GGG.moebius-basis.f.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.GGG.moebius-basis.f.1}):} Let $G$ and
$H$ be two finite graphs. Then,%
\[
\left(  \underbrace{\psi\left(  \left[  G\right]  ^{\sharp}\right)
}_{=\operatorname*{aut}\left(  G\right)  \cdot\left[  G\right]  ^{\ast}%
}\right)  \left[  H\right]  =\left(  \operatorname*{aut}\left(  G\right)
\cdot\left[  G\right]  ^{\ast}\right)  \left[  H\right]  =\operatorname*{aut}%
\left(  G\right)  \cdot\underbrace{\left[  G\right]  ^{\ast}\left[  H\right]
}_{=\delta_{\left[  G\right]  ,\left[  H\right]  }}=\operatorname*{aut}\left(
G\right)  \cdot\delta_{\left[  G\right]  ,\left[  H\right]  }.
\]
\par
We are in one of the following two cases:
\par
\textit{Case 1:} The graphs $G$ and $H$ are isomorphic.
\par
\textit{Case 2:} The graphs $G$ and $H$ are not isomorphic.
\par
Let us first consider Case 1. In this case, the graphs $G$ and $H$ are
isomorphic. In other words, there exists a graph isomorphism $\alpha$ from $G$
to $H$. Consider this $\alpha$. Then,%
\begin{align*}
\operatorname*{Iso}\left(  G,G\right)   &  \rightarrow\operatorname*{Iso}%
\left(  G,H\right)  ,\\
\psi &  \mapsto\alpha\circ\psi
\end{align*}
is a bijection (this is very easy to check). Thus, there exists a bijection
$\operatorname*{Iso}\left(  G,G\right)  \rightarrow\operatorname*{Iso}\left(
G,H\right)  $. Hence, $\left\vert \operatorname*{Iso}\left(  G,G\right)
\right\vert =\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert $.
But the definition of $\operatorname*{aut}\left(  G\right)  $ yields
$\operatorname*{aut}\left(  G\right)  =\left\vert \operatorname*{Iso}\left(
G,G\right)  \right\vert $. Now,%
\[
\left(  \psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[
H\right]  =\operatorname*{aut}\left(  G\right)  \cdot\underbrace{\delta
_{\left[  G\right]  ,\left[  H\right]  }}_{\substack{=1\\\text{(since }\left[
G\right]  =\left[  H\right]  \\\text{(because the graphs }G\\\text{and
}H\text{ are isomorphic))}}}=\operatorname*{aut}\left(  G\right)  =\left\vert
\operatorname*{Iso}\left(  G,G\right)  \right\vert =\left\vert
\operatorname*{Iso}\left(  G,H\right)  \right\vert .
\]
Hence, (\ref{pf.prop.GGG.moebius-basis.f.1}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, the graphs $G$ and $H$ are not
isomorphic. Thus, $\left[  G\right]  \neq\left[  H\right]  $, so that
$\delta_{\left[  G\right]  ,\left[  H\right]  }=0$.
\par
But the graphs $G$ and $H$ are not isomorphic. Hence, the set of all
isomorphisms from $G$ to $H$ is empty. In other words, $\operatorname*{Iso}%
\left(  G,H\right)  =\varnothing$ (since $\operatorname*{Iso}\left(
G,H\right)  $ is the set of all isomorphisms from $G$ to $H$). Hence,
$\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert =\left\vert
\varnothing\right\vert =0$. Now,%
\[
\left(  \psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[
H\right]  =\operatorname*{aut}\left(  G\right)  \cdot\underbrace{\delta
_{\left[  G\right]  ,\left[  H\right]  }}_{=0}=0=\left\vert
\operatorname*{Iso}\left(  G,H\right)  \right\vert .
\]
Hence, (\ref{pf.prop.GGG.moebius-basis.f.1}) is proven in Case 2.
\par
Now, (\ref{pf.prop.GGG.moebius-basis.f.1}) is proven in both Cases 1 and 2.
Hence, (\ref{pf.prop.GGG.moebius-basis.f.1}) always holds.}. Now, let
$a\in\GGG$ and $b\in\GGG$. We want to prove the equality
$\left(  \psi\left(  a\right)  \right)  \left(  b\right)  =\left(  a,b\right)
$. This equality is $\kk$-linear in each of $a$ and $b$. Thus, we can
WLOG assume that $a$ belongs to the family $\left(  \left[  G\right]
^{\sharp}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ and that $b$ belongs to the family $\left(  \left[  G\right]
\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite graphs}%
}$ (because both of these families are bases of the $\kk$-module
$\GGG$). Assume this. Then, $a=\left[  G\right]  ^{\sharp}$ and
$b=\left[  H\right]  $ for two finite graphs $G$ and $H$. Consider these $G$
and $H$. Now,%
\begin{align*}
\left(  \psi\left(  \underbrace{a}_{=\left[  G\right]  ^{\sharp}}\right)
\right)  \left(  \underbrace{b}_{=\left[  H\right]  }\right)   &  =\left(
\psi\left(  \left[  G\right]  ^{\sharp}\right)  \right)  \left[  H\right]
=\left\vert \operatorname*{Iso}\left(  G,H\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.GGG.moebius-basis.f.1}%
)}\right) \\
&  =\left(  \underbrace{\left[  G\right]  ^{\sharp}}_{=a},\underbrace{\left[
H\right]  }_{=b}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\left[  G\right]  ^{\sharp},\left[  H\right]  \right)  =\left\vert
\operatorname*{Iso}\left(  G,H\right)  \right\vert \right) \\
&  =\left(  a,b\right)  .
\end{align*}
This proves Proposition \ref{prop.GGG.selfdual}(a).

(b) We have $\psi\left(  1_{\GGG}\right)  =1_{\GGG^{o}}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\mathbf{0}$ denote the empty graph.
\par
We have $1_{\GGG^{o}}=\epsilon_{\GGG}$ (by the definition of the
$\kk$-algebra $\GGG^{o}$). Thus, for every finite graph $G$, we
have%
\begin{align*}
\underbrace{1_{\GGG^{o}}}_{=\epsilon_{\GGG}}\left(  \left[
G\right]  \right)   &  =\epsilon_{\GGG}\left(  \left[  G\right]
\right)  =\delta_{\left[  G\right]  ,\left[  \mathbf{0}\right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\epsilon_{\GGG%
}\right) \\
&  =\delta_{\left[  \mathbf{0}\right]  ,\left[  G\right]  }=\left[
\mathbf{0}\right]  ^{\ast}\left[  G\right]  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left[  \mathbf{0}\right]  ^{\ast}\left[  G\right]
=\delta_{\left[  \mathbf{0}\right]  ,\left[  G\right]  }\text{ (by the
definition of }\left[  \mathbf{0}\right]  ^{\ast}\text{)}\right)  .
\end{align*}
Hence, the two maps $1_{\GGG^{o}}$ and $\left[  \mathbf{0}\right]
^{\ast}$ are equal to each other on every element of the basis $\left(
\left[  G\right]  \right)  _{\left[  G\right]  \text{ is an isomorphism class
of finite graphs}}$ of $\GGG$. Since these two maps are $\kk%
$-linear, this yields that these two maps $1_{\GGG^{o}}$ and $\left[
\mathbf{0}\right]  ^{\ast}$ are identical. In other words, $1_{\GGG%
^{o}}=\left[  \mathbf{0}\right]  ^{\ast}$.
\par
On the other hand, the definition of $\left[  \mathbf{0}\right]  ^{\sharp}$
readily yields $\left[  \mathbf{0}\right]  ^{\sharp}=\left[  \mathbf{0}%
\right]  =1_{\GGG}$. But the definition of $\psi$ yields $\psi\left(
\left[  \mathbf{0}\right]  ^{\sharp}\right)  =\underbrace{\operatorname*{aut}%
\left(  \mathbf{0}\right)  }_{=1}\cdot\left[  \mathbf{0}\right]  ^{\ast
}=\left[  \mathbf{0}\right]  ^{\ast}=1_{\GGG^{o}}$ (since
$1_{\GGG^{o}}=\left[  \mathbf{0}\right]  ^{\ast}$), so that
$1_{\GGG^{o}}=\psi\left(  \underbrace{\left[  \mathbf{0}\right]
^{\sharp}}_{=1_{\GGG}}\right)  =\psi\left(  1_{\GGG}\right)  $,
qed.}. Also, every $a\in\GGG$ and $b\in\GGG$ satisfy
$\psi\left(  ab\right)  =\psi\left(  a\right)  \cdot\psi\left(  b\right)
$\ \ \ \ \footnote{\textit{Proof.} Let $a\in\GGG$ and $b\in\GGG%
$. We need to prove the equality $\psi\left(  ab\right)  =\psi\left(
a\right)  \cdot\psi\left(  b\right)  $. Since this equality is $\kk%
$-linear in each of $a$ and $b$, we can WLOG assume that $a$ and $b$ are
elements of the family $\left(  \left[  G\right]  ^{\sharp}\right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$ (because this
family is a basis of the $\kk$-module $\GGG$). Assume this.
\par
There exist two finite graphs $H_{1}$ and $H_{2}$ such that $a=\left[
H_{1}\right]  ^{\sharp}$ and $b=\left[  H_{2}\right]  ^{\sharp}$ (since $a$
and $b$ are elements of the family $\left(  \left[  G\right]  ^{\sharp
}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$). Consider these $H_{1}$ and $H_{2}$. Write the graphs $H_{1}$ and
$H_{2}$ in the forms $H_{1}=\left(  V_{1},E_{1}\right)  $ and $H_{2}=\left(
V_{2},E_{2}\right)  $.
\par
Multiplying the equalities $a=\left[  H_{1}\right]  ^{\sharp}$ and $b=\left[
H_{2}\right]  ^{\sharp}$, we obtain%
\[
ab=\left[  H_{1}\right]  ^{\sharp}\left[  H_{2}\right]  ^{\sharp}%
=\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.GGG.moebius-basis}%
(d)}\right)  .
\]
\par
Now, fix a finite graph $G$. We shall show that $\left(  \psi\left(
ab\right)  \right)  \left[  G\right]  =\left(  \psi\left(  a\right)  \cdot
\psi\left(  b\right)  \right)  \left[  G\right]  $.
\par
Write the graph $G$ in the form $G=\left(  V,F\right)  $. Then,%
\begin{align*}
\left(  \psi\underbrace{\left(  ab\right)  }_{=\sum\limits_{\substack{H=\left(
V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\left[
H\right]  ^{\sharp}}\right)  \left[  G\right]   &  =\left(  \psi\left(
\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\left[  H\right]  ^{\sharp}\right)  \right)
\left[  G\right]  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\underbrace{\psi\left(  \left[
H\right]  ^{\sharp}\right)  \left[  G\right]  }_{\substack{=\left\vert
\operatorname*{Iso}\left(  H,G\right)  \right\vert \\\text{(by
(\ref{pf.prop.GGG.moebius-basis.f.1}), applied to}\\H\text{ and }G\text{
instead of }G\text{ and }H\text{)}}}\\
&  =\sum\limits_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\left\vert \operatorname*{Iso}\left(  H,G\right)
\right\vert =\left\vert \bigsqcup_{\substack{H=\left(  V_{1}\sqcup
V_{2},E\right)  ;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}%
\left(  H,G\right)  \right\vert .
\end{align*}
On the other hand, Exercise \ref{exe.dual-algebra}(b) shows that the
$\kk$-algebra structure defined on $\GGG^{\ast}$ is precisely
the one defined on $\Hom \left(  \GGG,\kk\right)
=\GGG^{\ast}$ according to Definition~\ref{convolution-algebra}. Thus,
the product of two elements of $\GGG^{\ast}$ is the convolution of
these elements (viewed as maps from $\GGG$ to $\kk$). Applying
this to the two elements $\psi\left(  a\right)  $ and $\psi\left(  b\right)
$, we obtain $\psi\left(  a\right)  \cdot\psi\left(  b\right)  =\psi\left(
a\right)  \star\psi\left(  b\right)  =m_{\kk}\circ\left(  \psi\left(
a\right)  \otimes\psi\left(  b\right)  \right)  \circ\Delta_{\GGG}$, so
that%
\begin{align*}
\left(  \psi\left(  a\right)  \cdot\psi\left(  b\right)  \right)  \left[
G\right]   &  =\left(  m_{\kk}\circ\left(  \psi\left(  a\right)
\otimes\psi\left(  b\right)  \right)  \circ\Delta_{\GGG}\right)
\left(  \left[  G\right]  \right) \\
&  =\left(  m_{\kk}\circ\left(  \psi\left(  \underbrace{a}_{=\left[
H_{1}\right]  ^{\sharp}}\right)  \otimes\psi\left(  \underbrace{b}_{=\left[
H_{2}\right]  ^{\sharp}}\right)  \right)  \right)  \left(  \underbrace{\Delta
_{\GGG}\left(  \left[  G\right]  \right)  }_{\substack{=\sum\limits
_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}=V}}\left[
G\mid_{V_{1}}\right]  \otimes\left[  G\mid_{V_{2}}\right]  \\\text{(by the
definition of }\Delta_{\GGG}\text{)}}}\right) \\
&  =\left(  m_{\kk}\circ\left(  \psi\left(  \left[  H_{1}\right]
^{\sharp}\right)  \otimes\psi\left(  \left[  H_{2}\right]  ^{\sharp}\right)
\right)  \right)  \left(  \sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\left[  G\mid_{V_{1}}\right]  \otimes\left[
G\mid_{V_{2}}\right]  \right) \\
&  =m_{\kk}\left(  \sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\underbrace{\left(  \psi\left(  \left[  H_{1}\right]
^{\sharp}\right)  \otimes\psi\left(  \left[  H_{2}\right]  ^{\sharp}\right)
\right)  \left(  \left[  G\mid_{V_{1}}\right]  \otimes\left[  G\mid_{V_{2}%
}\right]  \right)  }_{=\left(  \psi\left(  \left[  H_{1}\right]  ^{\sharp
}\right)  \right)  \left[  G\mid_{V_{1}}\right]  \otimes\left(  \psi\left(
\left[  H_{2}\right]  ^{\sharp}\right)  \right)  \left[  G\mid_{V_{2}}\right]
}\right) \\
&  =m_{\kk}\left(  \sum\limits_{\substack{\left(  V_{1},V_{2}\right)
;\\V_{1}\sqcup V_{2}=V}}\left(  \psi\left(  \left[  H_{1}\right]  ^{\sharp
}\right)  \right)  \left[  G\mid_{V_{1}}\right]  \otimes\left(  \psi\left(
\left[  H_{2}\right]  ^{\sharp}\right)  \right)  \left[  G\mid_{V_{2}}\right]
\right) \\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}%
=V}}\underbrace{\left(  \psi\left(  \left[  H_{1}\right]  ^{\sharp}\right)
\right)  \left[  G\mid_{V_{1}}\right]  }_{\substack{=\left\vert
\operatorname*{Iso}\left(  H_{1},G\mid_{V_{1}}\right)  \right\vert \\\text{(by
(\ref{pf.prop.GGG.moebius-basis.f.1}), applied to}\\H_{1}\text{ and }%
G\mid_{V_{1}}\text{ instead of }G\text{ and }H\text{)}}}\cdot
\underbrace{\left(  \psi\left(  \left[  H_{2}\right]  ^{\sharp}\right)
\right)  \left[  G\mid_{V_{2}}\right]  }_{\substack{=\left\vert
\operatorname*{Iso}\left(  H_{2},G\mid_{V_{2}}\right)  \right\vert \\\text{(by
(\ref{pf.prop.GGG.moebius-basis.f.1}), applied to}\\H_{2}\text{ and }%
G\mid_{V_{2}}\text{ instead of }G\text{ and }H\text{)}}}\\
&  =\sum\limits_{\substack{\left(  V_{1},V_{2}\right)  ;\\V_{1}\sqcup V_{2}%
=V}}\left\vert \operatorname*{Iso}\left(  H_{1},G\mid_{V_{1}}\right)
\right\vert \cdot\left\vert \operatorname*{Iso}\left(  H_{2},G\mid_{V_{2}%
}\right)  \right\vert =\sum\limits_{\substack{\left(  W_{1},W_{2}\right)
;\\W_{1}\sqcup W_{2}=V}}\left\vert \operatorname*{Iso}\left(  H_{1}%
,G\mid_{W_{1}}\right)  \right\vert \cdot\left\vert \operatorname*{Iso}\left(
H_{2},G\mid_{W_{2}}\right)  \right\vert \\
&  =\left\vert \bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup
W_{2}=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  \right\vert .
\end{align*}
\par
Now, we are going to construct a bijection between the sets
\[
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\ \ \ \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ \ \ \ 
\bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup W_{2}%
=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  .
\]
First, let us agree to encode the elements of the set
\[
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\]
as triples $\left(H, E, \varphi\right)$ consisting of a graph
$H = \left(V_1 \sqcup V_2, E\right)$, its set of edges $E$, and
an isomorphism $\varphi \in \operatorname*{Iso}\left(  H,G\right)$.
Let us also agree to encode the elements of the set
\[
\bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup W_{2}%
=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)
\]
as triples $\left(\left(W_1, W_2\right), \varphi_1, \varphi_2\right)$
consisting of a pair $\left(W_1, W_2\right)$ of subsets of $V$ (which
satisfies $W_1 \sqcup W_2 = V$), an isomorphism
$\varphi_1 \in \operatorname*{Iso}\left( H_1, G\mid_{W_1}\right)$,
and an isomorphism
$\varphi_2 \in \operatorname*{Iso}\left( H_2, G\mid_{W_2}\right)$.
\par
Now, we claim that there exists a bijection from the set
\[
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\]
to the set
\[
\bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup W_{2}%
=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  .
\]
Namely, this bijection sends every $\left(  H,E,\varphi\right)  $ (where
$H=\left(  V_{1}\sqcup V_{2},E\right)  $ and $\varphi\in\operatorname*{Iso}%
\left(  H,G\right)  $) to $\left(  \left(  \varphi\left(  V_{1}\right)
,\varphi\left(  V_{2}\right)  \right)  ,\varphi_{1},\varphi_{2}\right)  $,
where $\varphi_{1}:V_{1}\rightarrow\varphi\left(  V_{1}\right)  $ is the
isomorphism from $H_{1}$ to $G\mid_{\varphi\left(  V_{1}\right)  }$ which is
obtained by restricting $\varphi$ to $V_{1}$, and where $\varphi_{2}$ is the
isomorphism from $H_{2}$ to $G_{2}\mid_{\varphi\left(  V_{2}\right)  }$ which
is obtained by restricting $\varphi$ to $V_{2}$. (The inverse map of this
bijection sends every $\left(  \left(  W_{1},W_{2}\right)  ,\varphi
_{1},\varphi_{2}\right)  $ (with $W_{1}\sqcup W_{2}=V$ and $\varphi_{1}%
\in\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  $ and
$\varphi_2 \in \operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  $) to
$\left(  \left(
V_{1}\sqcup V_{2},E\right)  ,E,\varphi\right)  $, where $\varphi:V_{1}\sqcup
V_{2}\rightarrow V$ is the map glued together from the maps $\varphi_{1}%
:V_{1}\rightarrow W_{1}$ and $\varphi_{2}:V_{2}\rightarrow W_{2}$, and where
$E=\left(  \varphi^{-1}\right)  _{\ast}\left(  F\right)  $.) The existence of
this bijection yields%
\[
\left\vert \bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)
;\\H|_{V_{1}}=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)
\right\vert =\left\vert \bigsqcup_{\substack{\left(  W_{1},W_{2}\right)
;\\W_{1}\sqcup W_{2}=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)
\times\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  \right\vert .
\]
Hence,%
\begin{align*}
\left(  \psi\left(  ab\right)  \right)  \left[  G\right] 
&= \left\vert
\bigsqcup_{\substack{H=\left(  V_{1}\sqcup V_{2},E\right)  ;\\H|_{V_{1}}%
=H_{1};\\H|_{V_{2}}=H_{2}}}\operatorname*{Iso}\left(  H,G\right)  \right\vert
= \left\vert \bigsqcup_{\substack{\left(  W_{1},W_{2}\right)  ;\\W_{1}\sqcup
W_{2}=V}}\operatorname*{Iso}\left(  H_{1},G\mid_{W_{1}}\right)  \times
\operatorname*{Iso}\left(  H_{2},G\mid_{W_{2}}\right)  \right\vert =\left(
\psi\left(  a\right)  \cdot\psi\left(  b\right)  \right)  \left[  G\right]  .
\end{align*}
\par
Let us now forget that we fixed $G$. We thus have shown that $\left(
\psi\left(  ab\right)  \right)  \left[  G\right]  =\left(  \psi\left(
a\right)  \cdot\psi\left(  b\right)  \right)  \left[  G\right]  $ for any
finite graph $G$. In other words, the two maps $\psi\left(  ab\right)  $ and
$\psi\left(  a\right)  \cdot\psi\left(  b\right)  $ are equal to each other on
every element of the basis $\left(  \left[  G\right]  \right)  _{\left[
G\right]  \text{ is an isomorphism class of finite graphs}}$ of $\GGG$.
Since these two maps are $\kk$-linear, this yields that these two maps
$\psi\left(  ab\right)  $ and $\psi\left(  a\right)  \cdot\psi\left(
b\right)  $ are identical. In other words, $\psi\left(  ab\right)
=\psi\left(  a\right)  \cdot\psi\left(  b\right)  $. Qed.}. Hence,
$\psi:\GGG\rightarrow\GGG^{o}$ is a $\kk$-algebra morphism
(since $\psi\left(  1_{\GGG}\right)  =1_{\GGG^{o}}$).

Next, we notice that the $\kk$-linear map $\psi$ is graded (because for
any finite graph $G$, both $\left[  G\right]  ^{\sharp}$ and $\left[
G\right]  ^{\ast}$ are homogeneous elements having degree $\left\vert
G\right\vert $). Thus, the $\kk$-linear map $\psi:\GGG%
\rightarrow\GGG^{o}$ gives rise to an adjoint $\kk$-linear map
$\psi^{\ast}:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$.

Now, the canonical $\kk$-module homomorphism $\GGG%
\rightarrow\left(  \GGG^{o}\right)  ^{o}$ (which sends every
$a\in\GGG$ to the map $\GGG^{o}\rightarrow\kk,\ f\mapsto
f\left(  a\right)  $) is a $\kk$-module isomorphism (since
$\GGG$ is of finite type). We thus identify $\GGG$ with $\left(
\GGG^{o}\right)  ^{o}$ along this isomorphism. Then, $\left(
\GGG^{o}\right)  ^{o}=\GGG$ as Hopf algebras\footnote{This is
because the Hopf algebra structure on $\GGG^{o}$ was defined by taking
adjoints of the structure maps of the Hopf algebra structure on
$\GGG$ (for example, the comultiplication $\Delta_{\GGG^{o}}$ on
$\GGG^{o}$ is the adjoint of the multiplication $m_{\GGG}$ on
$\GGG$, if $\left(  \GGG\otimes\GGG\right)  ^{o}$ is
identified with $\GGG^{o}\otimes\GGG^{o}$), and the Hopf algebra
structure on $\left(  \GGG^{o}\right)  ^{o}$ was defined by taking
adjoints of the structure maps of the Hopf algebra structure on
$\GGG^{o}$; but the adjoint of the adjoint of a linear map $F$
between two graded $\kk$-modules of finite type is the map $F$
again.}. Therefore,
$\psi:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$ is a
$\kk$-algebra morphism (since $\psi:\GGG\rightarrow\GGG^{o}$
is a $\kk$-algebra morphism). Also, $\GGG^{o}$ is of finite type
(since $\GGG$ is of finite type). Thus, Exercise \ref{exe.dual-algebra}%
(f) (applied to $C=\GGG$, $D=\GGG^{o}$ and $f=\psi$) yields that
$\psi:\GGG\rightarrow\GGG^{o}$ is a $\kk$-coalgebra morphism
if and only if $\psi^{\ast}:\left(  \GGG^{o}\right)  ^{o}%
\rightarrow\GGG^{o}$ is a $\kk$-algebra morphism.

Let us now prove that $\psi^{\ast}=\psi$ (as maps from $\GGG$ to
$\GGG^{o}$). Indeed, let $a\in\GGG$. Then, every $b\in
\GGG$ satisfies%
\begin{align*}
\left(  \psi^{\ast}\left(  a\right)  \right)  \left(  b\right)   &  =a\left(
\psi\left(  b\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the adjoint map }\psi^{\ast}:\left(  \GGG%
^{o}\right)  ^{o}\rightarrow\GGG^{o}\text{,}\\
\text{where }a\in\GGG\text{ is regarded as an element of }\left(
\GGG^{o}\right)  ^{o}%
\end{array}
\right) \\
&  =\left(  \psi\left(  b\right)  \right)  \left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the }\kk\text{-module homomorphism }\GGG%
\rightarrow\left(  \GGG^{o}\right)  ^{o}\text{ which we use to}\\
\text{identify }\GGG\text{ with }\left(  \GGG^{o}\right)
^{o}\text{ sends }a\in\GGG\text{ to the map }\GGG^{o}%
\rightarrow\kk,\ f\mapsto f\left(  a\right)
\end{array}
\right) \\
&  =\left(  b,a\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.GGG.selfdual}(a), applied to }b\text{ and }a\text{ instead of
}a\text{ and }b\right) \\
&  =\left(  a,b\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the form
}\left(  \cdot,\cdot\right)  :\GGG\times\GGG\rightarrow
\kk\text{ is symmetric}\right) \\
&  =\left(  \psi\left(  a\right)  \right)  \left(  b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.GGG.selfdual}%
(a)}\right)  .
\end{align*}
Hence, $\psi^{\ast}\left(  a\right)  =\psi\left(  a\right)  $. Now, let us
forget that we fixed $a$. We thus have shown that $\psi^{\ast}\left(
a\right)  =\psi\left(  a\right)  $ for every $a\in\GGG$. In other
words, $\psi^{\ast}=\psi$. Hence, $\psi^{\ast}:\left(  \GGG^{o}\right)
^{o}\rightarrow\GGG^{o}$ is a $\kk$-algebra morphism (since
$\psi:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$ is a
$\kk$-algebra morphism). Thus, $\psi:\GGG\rightarrow\GGG^{o}$
is a $\kk$-coalgebra morphism (because we know that $\psi:\GGG%
\rightarrow\GGG^{o}$ is a $\kk$-coalgebra morphism if and only if
$\psi^{\ast}:\left(  \GGG^{o}\right)  ^{o}\rightarrow\GGG^{o}$
is a $\kk$-algebra morphism). This yields that $\psi:\GGG%
\rightarrow\GGG^{o}$ is a $\kk$-bialgebra morphism (since we already
know that $\psi:\GGG\rightarrow\GGG^{o}$ is a $\kk%
$-algebra morphism). Thus, $\psi:\GGG\rightarrow\GGG^{o}$ is a Hopf
algebra homomorphism (by Corollary~\ref{cor.bialg-mor-is-Hopf},
applied to $H_{1}=\GGG$, $H_{2}=\GGG^{o}$, $S_{1}=S_{\GGG}$,
$S_{2}=S_{\GGG^{o}}$ and $\beta=\psi$). This proves Proposition
\ref{prop.GGG.selfdual}(b).

(c) Assume that $\QQ$ is a subring of $\kk$. Then,
$\operatorname*{aut}\left(  G\right)  $ is invertible for every finite graph
$G$ (because $\operatorname*{aut}\left(  G\right)  $ is a positive integer).
Hence, $\left(  \operatorname*{aut}\left(  G\right)  \cdot\left[  G\right]
^{\ast}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ is a basis of the $\kk$-module $\GGG^{o}$ (because
$\left(  \left[  G\right]  ^{\ast}\right)  _{\left[  G\right]  \text{ is an
isomorphism class of finite graphs}}$ is a basis of this $\kk$-module).
The map $\psi$ thus sends the basis $\left(  \left[  G\right]  ^{\sharp
}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ of the $\kk$-module $\GGG$ to the basis \newline%
$\left(  \operatorname*{aut}\left(  G\right)  \cdot\left[  G\right]  ^{\ast
}\right)  _{\left[  G\right]  \text{ is an isomorphism class of finite
graphs}}$ of the $\kk$-module $\GGG^{o}$ (because it satisfies
$\psi\left(  \left[  G\right]  ^{\sharp}\right)  =\operatorname*{aut}\left(
G\right)  \cdot\left[  G\right]  ^{\ast}$ for every finite graph $G$). Hence,
the $\kk$-linear map $\psi$ sends a basis of its domain to a basis of
its codomain. Therefore, the map $\psi$ is a $\kk$-module isomorphism.
Combined with the fact that $\psi$ is a Hopf algebra homomorphism, this yields
that $\psi$ is a Hopf algebra isomorphism. Proposition
\ref{prop.GGG.selfdual}(c) is thus proven.
\end{proof}

Now, all of Proposition~\ref{prop.GGG.moebius-basis},
Proposition~\ref{prop.GGG.bilform.sym} and
Proposition~\ref{prop.GGG.selfdual} are proven. Therefore,
Exercise \ref{exe.GGG.moebius-basis} is solved.
%\end{solutions}

\end{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.GGG.chromatic-sharp}}

\textit{Solution to Exercise~\ref{exe.GGG.chromatic-sharp}.} Proposition
\ref{prop.GGG.moebius-basis}(b) yields that the elements $\left[  G\right]
^{\sharp}$, where $\left[  G\right]  $ ranges over all isomorphism classes of
finite graphs, form a basis of the $\kk$-module ${{\mathcal{G}}}$.
Hence, we can define a $\kk$-linear map $\Psi^{\prime}:\mathcal{G}%
\rightarrow\kk$ by requiring that%
\[
\Psi^{\prime}\left(  \left[  G\right]  ^{\sharp}\right)  =\sum\limits
_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}
;\\\operatorname*{eqs}f=E}}\xx_{f}\ \ \ \ \ \ \ \ \ \ \text{for every
finite graph }G=\left(  V,E\right)
\]
(because $\sum\limits_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}
;\\\operatorname*{eqs}f=E}}\xx_{f}$ depends only on the isomorphism
class $\left[  G\right]  $, but not on the graph $G$ itself). Consider this
map $\Psi^{\prime}$. We shall show that $\Psi^{\prime}=\Psi$.

Indeed, let $G=\left(  V,E\right)  $ be any finite graph. Then, Proposition
\ref{prop.GGG.moebius-basis}(a) yields $\left[  G\right]  =\sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left[  H\right]  ^{\sharp}$. Applying the map $\Psi^{\prime}$ to both sides
of this equality, we obtain%
\begin{align}
\Psi^{\prime}\left[  G\right]    & =\Psi^{\prime}\left(  \sum\limits
_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime}\cap E=\varnothing
}}\left[  H\right]  ^{\sharp}\right)  =\sum\limits_{\substack{H=\left(  V,E^{\prime
}\right)  ;\\E^{\prime}\cap E=\varnothing}}\underbrace{\Psi^{\prime}\left(
\left[  H\right]  ^{\sharp}\right)  }_{\substack{=\sum\limits
_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}
;\\\operatorname*{eqs}f=E^{\prime}}}\xx_{f}\\\text{(by the definition
of }\Psi^{\prime}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map
}\Psi^{\prime}\text{ is }\kk\text{-linear}\right)  \nonumber\\
& =\underbrace{\sum\limits_{\substack{H=\left(  V,E^{\prime}\right)  ;\\E^{\prime
}\cap E=\varnothing}}}_{=\sum\limits_{\substack{E^{\prime}\text{ is a set of
two-element}\\\text{subsets of }V;\\E^{\prime}\cap E=\varnothing}}}%
\sum\limits_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}
;\\\operatorname*{eqs}f=E^{\prime}}}\xx_{f}=\underbrace{\sum\limits
_{\substack{E^{\prime}\text{ is a set of two-element}\\\text{subsets of
}V;\\E^{\prime}\cap E=\varnothing}}\sum\limits_{\substack{f:V\rightarrow\left\{
1,2,3,\ldots\right\}  ;\\\operatorname*{eqs}f=E^{\prime}}}}_{=\sum\limits
_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}  ;\\\left(
\operatorname*{eqs}f\right)  \cap E=\varnothing}}}\xx_{f}\nonumber\\
& =\sum\limits_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}  ;\\\left(
\operatorname*{eqs}f\right)  \cap E=\varnothing}}\xx_{f}%
.\label{sol.GGG.chromatic-sharp.1}%
\end{align}


However, if $f:V\rightarrow\left\{  1,2,3,\ldots\right\}  $ is any map, then
we have the following logical equivalence:%
\begin{equation}
\left(  \left(  \operatorname*{eqs}f\right)  \cap E=\varnothing\right)
\ \Longleftrightarrow\ \left(  f\text{ is a proper coloring of }G\right)
.\label{sol.GGG.chromatic-sharp.eq-coloring}%
\end{equation}
\footnote{\textit{Proof of (\ref{sol.GGG.chromatic-sharp.eq-coloring}):} Let
$f:V\rightarrow\left\{  1,2,3,\ldots\right\}  $ be any map.
\par
Let us first assume that $\left(  \operatorname*{eqs}f\right)  \cap
E=\varnothing$. We shall then show that $f$ is a proper coloring of $G$.
\par
Indeed, let us assume that there exists an edge $e=\left\{  v,v^{\prime
}\right\}  $ in $E$ such that $f\left(  v\right)  =f\left(  v^{\prime}\right)
$. Consider this edge $e=\left\{  v,v^{\prime}\right\}  $. We are going to
derive a contradiction.
\par
The definition of $\operatorname*{eqs}f$ yields $\operatorname*{eqs}f=\left\{
\left\{  u,u^{\prime}\right\}  \ \mid\ u\in V,\ u^{\prime}\in V,\ u\neq
u^{\prime}\text{ and }f\left(  u\right)  =f\left(  u^{\prime}\right)
\right\}  $. But $v\in V$ and $v^{\prime}\in V$ and $v\neq v^{\prime}$ (since
$\left\{  v,v^{\prime}\right\}  $ is an edge of $E$) and $f\left(  v\right)
=f\left(  v^{\prime}\right)  $. Hence,%
\[
\left\{  v,v^{\prime}\right\}  \in\left\{  \left\{  u,u^{\prime}\right\}
\ \mid\ u\in V,\ u^{\prime}\in V,\ u\neq u^{\prime}\text{ and }f\left(
u\right)  =f\left(  u^{\prime}\right)  \right\}  =\operatorname*{eqs}f.
\]
Combined with $\left\{  v,v^{\prime}\right\}  \in E$, this yields $\left\{
v,v^{\prime}\right\}  \in\left(  \operatorname*{eqs}f\right)  \cap
E=\varnothing$. But this is absurd, because the empty set $\varnothing$ has no
elements. Hence, we have found a contradiction. Thus, our assumption (that
there exists an edge $e=\left\{  v,v^{\prime}\right\}  $ in $E$ such that
$f\left(  v\right)  =f\left(  v^{\prime}\right)  $) was wrong. Hence, no edge
$e=\left\{  v,v^{\prime}\right\}  $ in $E$ has $f\left(  v\right)  =f\left(
v^{\prime}\right)  $. In other words, $f$ is a proper coloring of $G$
(according to the definition of a ``proper coloring'').
\par
Let us now forget that we assumed that $\left(  \operatorname*{eqs}f\right)
\cap E=\varnothing$. We thus have proven the implication%
\begin{equation}
\left(  \left(  \operatorname*{eqs}f\right)  \cap E=\varnothing\right)
\ \Longrightarrow\ \left(  f\text{ is a proper coloring of }G\right)
.\label{sol.GGG.chromatic-sharp.eq-coloring.pf.1}%
\end{equation}
\par
Let us now assume that $f$ is a proper coloring of $G$. In other words, no
edge $e=\left\{  v,v^{\prime}\right\}  $ in $E$ has $f\left(  v\right)
=f\left(  v^{\prime}\right)  $ (according to the definition of a
``proper coloring'').
\par
Now, let us prove that $\left(  \operatorname*{eqs}f\right)  \cap
E=\varnothing$. Indeed, assume the contrary. Then, $\left(
\operatorname*{eqs}f\right)  \cap E\neq\varnothing$. Hence, there exists some
$g\in\left(  \operatorname*{eqs}f\right)  \cap E$. Consider this $g$. We have%
\[
g\in\left(  \operatorname*{eqs}f\right)  \cap E\subset\operatorname*{eqs}%
f=\left\{  \left\{  u,u^{\prime}\right\}  \ \mid\ u\in V,\ u^{\prime}\in
V,\ u\neq u^{\prime}\text{ and }f\left(  u\right)  =f\left(  u^{\prime
}\right)  \right\}  .
\]
Hence, $g=\left\{  u,u^{\prime}\right\}  $ for some $u\in V$ and $u^{\prime
}\in V$ satisfying $u\neq u^{\prime}$ and $f\left(  u\right)  =f\left(
u^{\prime}\right)  $. Consider these $u$ and $u^{\prime}$. We have $u\neq
u^{\prime}$ and $\left\{  u,u^{\prime}\right\}  =g\in\left(
\operatorname*{eqs}f\right)  \cap E\subset E$. Hence, $\left\{  u,u^{\prime
}\right\}  $ is an edge in $E$. Moreover, $f\left(  u\right)  =f\left(
u^{\prime}\right)  $. Hence, the edge $\left\{  u,u^{\prime}\right\}  $ is an
edge $e=\left\{  v,v^{\prime}\right\}  $ in $E$ such that $f\left(  v\right)
=f\left(  v^{\prime}\right)  $ (namely, for $v=u$ and $v^{\prime}=u^{\prime}%
$). This contradicts the fact that no edge $e=\left\{  v,v^{\prime}\right\}  $
in $E$ has $f\left(  v\right)  =f\left(  v^{\prime}\right)  $. This
contradiction proves that our assumption was wrong. Hence, we have shown that
$\left(  \operatorname*{eqs}f\right)  \cap E=\varnothing$.
\par
Let us now forget that we assumed that $\left(  f\text{ is a proper coloring
of }G\right)  $. We thus have proven the implication%
\[
\left(  f\text{ is a proper coloring of }G\right)  \ \Longrightarrow\ \left(
\left(  \operatorname*{eqs}f\right)  \cap E=\varnothing\right)  .
\]
Combining this with (\ref{sol.GGG.chromatic-sharp.eq-coloring.pf.1}), we
obtain the equivalence%
\[
\left(  \left(  \operatorname*{eqs}f\right)  \cap E=\varnothing\right)
\ \Longleftrightarrow\ \left(  f\text{ is a proper coloring of }G\right)  .
\]
This proves (\ref{sol.GGG.chromatic-sharp.eq-coloring}).} Hence,
(\ref{sol.GGG.chromatic-sharp.1}) becomes%
\begin{align*}
\Psi^{\prime}\left[  G\right]    & =\underbrace{\sum\limits_{\substack{f:V\rightarrow
\left\{  1,2,3,\ldots\right\}  ;\\\left(  \operatorname*{eqs}f\right)  \cap
E=\varnothing}}}_{\substack{=\sum\limits_{\substack{f:V\rightarrow\left\{
1,2,3,\ldots\right\}  ;\\f\text{ is a proper}\\\text{coloring of }%
G}}\\\text{(because of the equivalence
(\ref{sol.GGG.chromatic-sharp.eq-coloring}))}}}\xx_{f}=\underbrace{\sum\limits
_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}  ;\\f\text{ is a
proper}\\\text{coloring of }G}}}_{=\sum\limits_{\substack{\text{proper colorings}%
\\f:V\rightarrow\left\{  1,2,3,\ldots\right\}  }}}\xx_{f}%
=\sum\limits_{\substack{\text{proper colorings}\\f:V\rightarrow\left\{
1,2,3,\ldots\right\}  }}\xx_{f}\\
& =\Psi\left[  G\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since Proposition
\ref{prop.GGG.chromatic} yields }\Psi\left[  G\right]  =\sum\limits
_{\substack{\text{proper colorings}\\f:V\rightarrow\left\{  1,2,3,\ldots
\right\}  }}\xx_{f}\right)  .
\end{align*}


Let us now forget that we fixed $\left[  G\right]  $. Thus, we have shown that
$\Psi^{\prime}\left[  G\right]  =\Psi\left[  G\right]  $ for any finite graph
$G$. In other words, the two maps $\Psi^{\prime}$ and $\Psi$ are equal to each
other on each element of the family $\left(  \left[  G\right]  \right)
_{G\text{ is an isomorphism class of finite graphs}}$. Thus, the two maps
$\Psi^{\prime}$ and $\Psi$ must be identical (because these two maps are
$\kk$-linear, and because the family $\left(
\left[  G\right]  \right)  _{G\text{ is an isomorphism class of finite
graphs}}$ is a basis of the $\kk$-module $\mathcal{G}$). In other
words, $\Psi^{\prime}=\Psi$. Now, every finite graph $G=\left(  V,E\right)  $
satisfies
\[
\underbrace{\Psi}_{=\Psi^{\prime}}\left(  \left[  G\right]  ^{\sharp}\right)
=\Psi^{\prime}\left(  \left[  G\right]  ^{\sharp}\right)  =\sum\limits
_{\substack{f:V\rightarrow\left\{  1,2,3,\ldots\right\}
;\\\operatorname*{eqs}f=E}}\xx_{f}.
\]
This solves Exercise \ref{exe.GGG.chromatic-sharp}.
%\end{solutions}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.FQSym.free}}

\textit{Solution to Exercise~\ref{exe.FQSym.free}.} We begin with some preparations.

We shall regard permutations as words (over the alphabet
$\left\{1,2,3,\ldots\right\}$), by identifying every
permutation $\pi \in \Symm_n$ with the word
$\left( \pi \left( 1 \right) , \pi \left( 2 \right) , \ldots ,
  \pi \left( n \right) \right)$.
For every $n \in \NN$, the lexicographic order on words thus defines
a total order on $\Symm_n$; we will be using this order in
the following when we make statements like ``$\sigma < \tau$'' for
$\sigma$ and $\tau$ being two permutations in $\Symm_n$.

We denote the empty word by $\varnothing$. This empty word is identified with
the trivial permutation in $\Symm_0$. We shall refer to $\varnothing$
as the \emph{empty permutation}. Every permutation other than
$\varnothing$ will be called a \emph{nonempty permutation}.

Let $\Symm$ denote the disjoint union
$\bigsqcup_{n \in \NN} \Symm_n$ of the posets $\Symm_n$.
While each poset
$\Symm_n$ is actually totally ordered (by the lexicographic order),
the disjoint union $\Symm$ is not, since elements of different
$\Symm_n$ are incomparable.\footnote{We could define an ordering on
all of $\Symm$ by restricting the total order on words defined in
Definition \ref{def.words}; but we prefer not to.}

For every $k\in \NN$ and $\ell\in \NN$ and any permutations
$u\in \Symm_{k}$ and $v\in \Symm_{\ell}$, we define a permutation
$u\squarebin v$ by $u\squarebin v=u\cdot v\left[  k\right]  $ (where
``$u\cdot v\left[  k\right]  $'' has to be
read as ``$u\cdot\left(  v\left[  k\right]  \right)$''
rather than as ``$\left(  u\cdot v\right) \left[  k\right]  $'').
\ \ \ \ \footnote{This is indeed a permutation, because the
word $u\cdot v\left[  k\right]$ has each integer from
$1$ to $k+\ell$ appear exactly once in it.} Thus, we
have defined a binary operation $\squarebin$ on the set $\Symm$. It is
easy to see that $\Symm$ becomes a monoid with respect to this binary
operation $\squarebin$; the neutral element of this monoid is
$\varnothing \in \Symm_0$. Hence, products of the form
$w_1 \squarebin w_2 \squarebin \cdots \squarebin w_k$
for $k$-tuples $\left( w_1, w_2, \ldots, w_k \right) \in \Symm^k$
of permutations are well-defined (without specifying a bracketing).
The monoid $\left(  \Symm,\squarebin\right)  $
is left-cancellative%
\footnote{A semigroup $\left(M, \cdot\right)$ is said to be
\dfn{left-cancellative} if and only if it has the following property:
If $a$, $b$ and $c$ are three elements of $M$ satisfying
$a \cdot b = a \cdot c$, then $b = c$.}
 and right-cancellative%
\footnote{A semigroup $\left(M, \cdot\right)$ is said to be
\dfn{right-cancellative} if and only if it has the following property:
If $a$, $b$ and $c$ are three elements of $M$ satisfying
$b \cdot a = c \cdot a$, then $b = c$.}.

It is easy to see that
\begin{equation}
\alpha\squarebin\gamma > \beta\squarebin\gamma
\label{sol.FQSym.free.1}
\end{equation}
for any $\alpha\in\Symm$, $\beta\in\Symm$ and $\gamma
\in\Symm$ satisfying $\alpha>\beta$\ \ \ \ \footnote{\textit{Proof of
(\ref{sol.FQSym.free.1}):} Let $\alpha\in\Symm$, $\beta\in\Symm$
and $\gamma\in\Symm$ satisfy $\alpha>\beta$. Notice that the
$n\in \NN$ satisfying $\alpha\in \Symm_{n}$ and the $n\in
 \NN$ satisfying $\beta\in \Symm_{n}$ must be identical (because
otherwise, $\alpha$ and $\beta$ would be incomparable in the poset
$\Symm$). In other words, the words $\alpha$ and $\beta$ have the same
length. Let $k$ be this length. Thus, $\alpha \in \Symm_k$ and
$\beta \in \Symm_k$.
\par
Now, the words $\alpha$ and $\beta$ have the same length and satisfy
$\alpha>\beta$ in the lexicographic order. Hence, every word $\delta$
satisfies $\alpha\cdot\delta>\beta\cdot\delta$ (where $\cdot$ denotes
concatenation). Applying this to $\delta=\gamma\left[  k\right]  $, we obtain
$\alpha\cdot\gamma\left[  k\right]  >\beta\cdot\gamma\left[  k\right]  $. Now,
the definition of $\alpha\squarebin\gamma$ yields $\alpha\squarebin\gamma
=\alpha\cdot\gamma\left[  k\right]  >\beta\cdot\gamma\left[  k\right]
=\beta\squarebin\gamma$ (by the definition of $\beta\squarebin\gamma$), and thus
(\ref{sol.FQSym.free.1}) is proven.}.

Recall how we defined a connected permutation. Since we are regarding
permutations as words, we can rewrite this definition as follows: A
permutation $p\in \Symm_{n}$ is connected if and only if it is nonempty
and no nonempty proper prefix\footnote{A \dfn{proper prefix} of a word
$w$ is defined as a word $u$ such that there exists a nonempty word $v$
satisfying $w = uv$.} of $p$ is itself a permutation. Hence, if a
nonempty prefix of a connected permutation $p\in \Symm_{n}$ is itself a
permutation, then this prefix must be $p$.

It is easy to see that a permutation $w\in \Symm_{n}$ is connected if
and only if $n$ is a positive integer and there exist no nonempty permutations
$u$ and $v$ satisfying $w=u\squarebin v$. It is furthermore easy to see that for
every permutation $w\in\Symm$, there is a unique way to write $w$ in
the form $w=w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}$ for some
$k\in \NN$ and some $k$-tuple $\left(  w_{1},w_{2},\ldots,w_{k}\right)
\in\mathfrak{CS}^{k}$ of connected permutations\footnote{\textit{Proof.} Fix
$w\in\Symm$. We need to show that a decomposition of $w$ in the form
$w=w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}$ with $k\in \NN$ and
$\left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}$ exists and is
unique.
\par
The existence of such a decomposition $w=w_{1}\squarebin w_{2}\squarebin
\cdots\squarebin w_{k}$ is easy to check (just take a decomposition
$w=w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}$ with $k\in \NN$ and
$\left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\left(  \Symm
\setminus\left\{  \varnothing\right\}  \right)  ^{k}$ such that $k$ is
maximum, and argue that the maximality of $k$ forces each of $w_{1}$, $w_{2}$,
$\ldots$, $w_{k}$ to be connected). It remains to prove that such a
decomposition is unique. In other words, we need to prove the following
statement:
\par
\begin{statement}
\textit{Statement $\mathcal{U}$:} If $w\in\Symm$,
$k\in \NN$, $\left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}%
^{k}$, $\ell\in \NN$ and $\left(  v_{1},v_{2},\ldots,v_{\ell}\right)
\in\mathfrak{CS}^{\ell}$ are such that $w=w_{1}\squarebin w_{2}\squarebin
\cdots\squarebin w_{k}$ and $w=v_{1}\squarebin v_{2}\squarebin\cdots\squarebin v_{\ell}$,
then $k=\ell$ and $\left(  w_{1},w_{2},\ldots,w_{k}\right)  =\left(
v_{1},v_{2},\ldots,v_{\ell}\right)  $.
\end{statement}
\par
\textit{Proof of Statement $\mathcal{U}$:} Let us prove Statement
$\mathcal{U}$ by strong induction by the size of $w$ (that is, the number
$n\in \NN$ satisfying $w\in \Symm_{n}$). So let $N\in \NN$,
and assume (as the induction hypothesis) that Statement $\mathcal{U}$ is
proven in the case when $w\in \Symm_{M}$ for $M<N$.
\par
Let $w\in \Symm_{N}$, $k\in \NN$, $\left(  w_{1},w_{2}%
,\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}$, $\ell\in \NN$ and $\left(
v_{1},v_{2},\ldots,v_{\ell}\right)  \in\mathfrak{CS}^{\ell}$ be such that
$w=w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}$ and $w=v_{1}\squarebin
v_{2}\squarebin\cdots\squarebin v_{\ell}$. We need to show that $k=\ell$ and
$\left(  w_{1},w_{2},\ldots,w_{k}\right)  =\left(  v_{1},v_{2},\ldots,v_{\ell
}\right)  $.
\par
All of the permutations $w_{1}$, $w_{2}$, $\ldots$, $w_{k}$ and $v_{1}$,
$v_{2}$, $\ldots$, $v_{\ell}$ are connected, and thus nonempty.
\par
We assume WLOG that $w$ is nonempty (else, Statement $\mathcal{U}$ is
obvious). Hence, $k>0$ and $\ell>0$. Thus, $w_{1}$ and $v_{1}$ are
well-defined. We shall now prove that $w_{1}=v_{1}$.
\par
By the definition of the operation $\squarebin$, it is clear that the word
$v_{1}$ is a prefix of the word $v_{1}\squarebin v_{2}\squarebin\cdots\squarebin
v_{\ell}$. In other words, the word $v_{1}$ is a prefix of the word $w$ (since
$w=v_{1}\squarebin v_{2}\squarebin\cdots\squarebin v_{\ell}$). Similarly, the word
$w_{1}$ is a prefix of the word $w$ as well. But it is well-known that if
$\alpha$ and $\beta$ are two prefixes of a word $\gamma$, then either $\alpha$
is a prefix of $\beta$ or $\beta$ is a prefix of $\alpha$. Applying this to
$\alpha=v_{1}$, $\beta=w_{1}$ and $\gamma=w$, we conclude that either $v_{1}$
is a prefix of $w_{1}$ or $w_{1}$ is a prefix of $v_{1}$. We WLOG assume that
$v_{1}$ is a prefix of $w_{1}$.
\par
But recall that if a nonempty prefix of a connected permutation $p\in
 \Symm_{n}$ is itself a permutation, then this prefix must be $p$.
Applying this to $p=w_{1}$ and the nonempty prefix $v_{1}$ of $w_{1}$, we
conclude that $v_{1}$ must be $w_{1}$. That is, $v_{1}=w_{1}$.
\par
Let now $w^{\prime}$ be the permutation $w_{2}\squarebin w_{3}\squarebin
\cdots\squarebin w_{k}$. Then,%
\[
\underbrace{v_{1}}_{=w_{1}}\squarebin\underbrace{w^{\prime}}_{=w_{2}\squarebin
w_{3}\squarebin\cdots\squarebin w_{k}}=w_{1}\squarebin\left(  w_{2}\squarebin w_{3}%
\squarebin\cdots\squarebin w_{k}\right)  =w_{1}\squarebin w_{2}\squarebin\cdots\squarebin
w_{k}=w=v_{1}\squarebin v_{2}\squarebin\cdots\squarebin v_{\ell}=v_{1}\squarebin\left(
v_{2}\squarebin v_{3}\squarebin\cdots\squarebin v_{\ell}\right)  .
\]
Since the monoid $\left(  \Symm,\squarebin\right)  $ is left-cancellative,
we can cancel $v_{1}$ from this equality, and obtain $w^{\prime}=v_{2}\squarebin
v_{3}\squarebin\cdots\squarebin v_{\ell}$.
\par
Now, let $M$ be the length of the word $w^{\prime}$. Since $w^{\prime}$ is
shorter than the word $v_{1}\squarebin w^{\prime}=w$ (because $v_{1}$ is
nonempty), this length $M$ is smaller than the length of $w$, which is $N$
(since $w\in \Symm_{N}$). Thus, $M<N$. Thus, we have $w^{\prime}%
\in \Symm_{M}$ with $M<N$. Hence, by the induction hypothesis, we can
apply Statement $\mathcal{U}$ to $w^{\prime}$, $k-1$, $\left(  w_{2}%
,w_{3},\ldots,w_{k}\right)  $, $\ell-1$ and $\left(  v_{2},v_{3}%
,\ldots,v_{\ell}\right)  $ instead of $w$, $k$, $\left(  w_{1},w_{2}%
,\ldots,w_{k}\right)  $, $\ell$ and $\left(  v_{1},v_{2},\ldots,v_{\ell
}\right)  $ (since $w^{\prime}=w_{2}\squarebin w_{3}\squarebin\cdots\squarebin w_{k}$
and $v^{\prime}=v_{2}\squarebin v_{3}\squarebin\cdots\squarebin v_{\ell}$). As a
consequence, we obtain $k-1=\ell-1$ and $\left(  w_{2},w_{3},\ldots
,w_{k}\right)  =\left(  v_{2},v_{3},\ldots,v_{\ell}\right)  $. From
$k-1=\ell-1$, we obtain $k=\ell$. Combining $w_{1}=v_{1}$ with $\left(
w_{2},w_{3},\ldots,w_{k}\right)  =\left(  v_{2},v_{3},\ldots,v_{\ell}\right)
$, we conclude $\left(  w_{1},w_{2},\ldots,w_{k}\right)  =\left(  v_{1}%
,v_{2},\ldots,v_{\ell}\right)  $. Thus, the induction step is complete, and
Statement $\mathcal{U}$ is proven. Hence, the uniqueness of the decomposition
is proven, and we are done.}. We refer to the tuple $\left(  w_{1}%
,w_{2},\ldots,w_{k}\right)  $ as the
\emph{connected decomposition}\index{connected decomposition of a permutation}
of $w$.

For every $n\in \NN$ and $w\in \Symm_{n}$, define an element
$F_{w}^{\operatorname*{conn}}$ of $ \FQsym $ by setting
$F_{w}^{\operatorname*{conn}}=F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}$, where
$\left(  w_{1},w_{2},\ldots,w_{k}\right)  $ is the connected decomposition of
$w$. It is clear that this $F_{w}^{\operatorname*{conn}}$ is an element of
$ \FQsym \nolimits_{n}$. Hence, for every $n\in \NN$, the
family $\left(  F_{w}^{\operatorname*{conn}}\right)  _{w\in \Symm_{n}}$
is a family of elements of $ \FQsym \nolimits_{n}$. We shall
prove that it is a basis of the $\kk$-module $ \FQsym %
\nolimits_{n}$. Once this is proven, the claim of the exercise will quickly
follow (as we will see later).

We need the following easy fact:

\begin{lemma}
\label{lem.sol.FQSym.free.lex}Let $u\in\Symm$ and $v\in\Symm$.
Then,
\[
F_{u}F_{v}=F_{u\squarebin v}+\left(  \text{a sum of terms }F_{t}\text{ for }
t \in \Symm \text{ satisfying } t>u\squarebin v\right)  .
\]
(The symbol $>$ refers to the partial order on the poset $\Symm$.)
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.FQSym.free.lex}.]Let $k$ and $\ell$ be the
nonnegative integers satisfying $u\in \Symm_{k}$ and $v\in
 \Symm_{\ell}$. Write the words $u$ and $v\left[  k\right]  $ in the
forms $u=\left(  u_{1},u_{2},\ldots,u_{k}\right)  $ and $v\left[  k\right]
=\left(  p_{1},p_{2},\ldots,p_{\ell}\right)  $, respectively. Let $\left(
c_{1},c_{2},\ldots,c_{k+\ell}\right)  $ denote the concatenation $u\cdot v\left[
k\right]  =\left(  u_{1},u_{2},\ldots,u_{k},p_{1},p_{2},\ldots,p_{\ell}\right)  $.
By the definition of ${u}\,\shuffle\,{v\left[k\right]}$, we therefore
have
\begin{equation}
{u}\,\shuffle\,{v\left[k\right]}=\left\{  \left(  c_{w\left(  1\right)  },c_{w\left(
2\right)  },\ldots,c_{w\left(  k+\ell\right)  }\right)  \ :\ w\in
\operatorname{Sh}_{k,\ell}\right\}  _{\text{multiset}}.
\label{pf.lem.sol.FQSym.free.lex.1}%
\end{equation}
\footnote{See Definition \ref{shuffles} for the definition of
$\operatorname{Sh}_{k,\ell}$.} We also have $\left(  c_{1}%
,c_{2},\ldots,c_{k+\ell}\right)  =u\cdot v\left[  k\right]  =u\squarebin v$.

From (\ref{FQsym-product-on-F's}), we have%
\begin{align}
F_{u}F_{v}  &  =\sum_{w\in{u}\,\shuffle\,{v\left[  k\right]  }}F_{w}
\nonumber\\
&  =\sum_{w\in\operatorname{Sh}_{k,\ell}}F_{\left(  c_{w\left(
1\right)  },c_{w\left(  2\right)  },\ldots,c_{w\left(  k+\ell\right)  }\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.sol.FQSym.free.lex.1}%
)}\right) \nonumber\\
&  =\underbrace{F_{\left(  c_{ \id \left(  1\right)
},c_{ \id \left(  2\right)  },\ldots,c_{ \id \left(
k+\ell\right)  }\right)  }}_{\substack{=F_{u\squarebin v}\\\text{(since }\left(
c_{ \id \left(  1\right)  },c_{ \id \left(
2\right)  },\ldots,c_{ \id \left(  k+\ell\right)  }\right)
\\=\left(  c_{1},c_{2},\ldots,c_{k+\ell}\right)  =u\squarebin v\text{)}}%
}+\sum\limits_{\substack{w\in\operatorname{Sh}_{k,\ell};\\w\neq
 \id }}F_{\left(  c_{w\left(  1\right)  },c_{w\left(  2\right)
},\ldots,c_{w\left(  k+\ell\right)  }\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}w= \id \text{ from the sum, since } \id %
\in\operatorname{Sh}_{k,\ell}\right) \nonumber\\
&  =F_{u\squarebin v}+\sum\limits_{\substack{w\in\operatorname{Sh}_{k,\ell
};\\w\neq \id }}F_{\left(  c_{w\left(  1\right)  },c_{w\left(
2\right)  },\ldots,c_{w\left(  k+\ell\right)  }\right)  }.
\label{pf.lem.sol.FQSym.free.lex.2}
\end{align}

But it is easy to see that
\begin{equation}
\left(  c_{w\left(  1\right)  },c_{w\left(  2\right)  },\ldots,c_{w\left(
k+\ell\right)  }\right)  >u\squarebin v\ \ \ \ \ \ \ \ \ \ \text{for every }%
w\in\operatorname{Sh}_{k,\ell}\text{ satisfying }w\neq
 \id . \label{pf.lem.sol.FQSym.free.lex.3}%
\end{equation}

[\textit{Proof of (\ref{pf.lem.sol.FQSym.free.lex.3}):} Let
$w\in\operatorname{Sh}_{k,\ell}$ be such that $w\neq
 \id$. The map $w$ is a permutation, hence bijective and thus
injective. There exists some $i\in\left\{  1,2,\ldots,k+\ell\right\}  $ such that
$w\left(  i\right)  \neq i$ (since $w\neq \id$). Let $j$ be the
smallest such $i$. Thus, $w\left(  j\right)  \neq j$, but%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,k+\ell\right\}  \text{ satisfying }i<j\text{
satisfies }w\left(  i\right)  =i. \label{pf.lem.sol.FQSym.free.lex.3.pf.1}%
\end{equation}
\par
We have $w\in\operatorname{Sh}_{k,\ell}$, and therefore
$w^{-1}\left(  1\right)  <w^{-1}\left(  2\right)  <\cdots<w^{-1}\left(
k\right)  $ and $w^{-1}\left(  k+1\right)  <w^{-1}\left(  k+2\right)
<\cdots<w^{-1}\left(  k+\ell\right)  $. Thus, the restriction of the map
$w^{-1}$ to the set $\left\{  1,2,\ldots,k\right\}  $ and the restriction of the
map $w^{-1}$ to the set $\left\{  k+1,k+2,\ldots,k+\ell\right\}  $ are strictly
increasing.
\par
Let us first show that $j\leq k$. Indeed, assume the contrary. Then, $j>k$.
Hence, $k<j$.
Therefore, every $i\in\left\{  1,2,\ldots,k\right\}  $ satisfies $i\leq k<j$ and
thus $w\left(  i\right)  =i$ (by (\ref{pf.lem.sol.FQSym.free.lex.3.pf.1}%
)). Thus, $w\left(  \left\{  1,2,\ldots,k\right\}  \right)  =\left\{
1,2,\ldots,k\right\}  $, so that $w^{-1}\left(  \left\{  1,2,\ldots,k\right\}
\right)  =\left\{  1,2,\ldots,k\right\}  $ (since $w$ is bijective). Notice also
that $j\in\left\{  k+1,k+2,\ldots,k+\ell\right\}  $ (since $j>k$). Now,
\begin{align*}
w^{-1}\left(  \underbrace{\left\{  k+1,k+2,\ldots,k+\ell\right\}  }_{=\left\{
1,2,\ldots,k+\ell\right\}  \setminus\left\{  1,2,\ldots,k\right\}  }\right)   &
=w^{-1}\left(  \left\{  1,2,\ldots,k+\ell\right\}  \setminus\left\{
1,2,\ldots,k\right\}  \right)  \\
&=\underbrace{w^{-1}\left(  \left\{  1,2,\ldots,k+\ell
\right\}  \right)  }_{=\left\{  1,2,\ldots,k+\ell\right\}  }\setminus
\underbrace{ w^{-1}\left( \left\{  1,2,\ldots,k\right\} \right)  }_{=\left\{
1,2,\ldots,k\right\}  } \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }w\text{ is bijective}\right)
\\
&  =\left\{  1,2,\ldots,k+\ell\right\}  \setminus\left\{  1,2,\ldots,k\right\}
=\left\{  k+1,k+2,\ldots,k+\ell\right\}  .
\end{align*}
Therefore, $w^{-1}$ restricts to a map from
$\left\{  k+1,k+2,\ldots,k+\ell\right\}$
to $\left\{  k+1,k+2,\ldots,k+\ell\right\}  $. This latter map must be
strictly increasing (since the restriction of the map $w^{-1}$ to the set
$\left\{  k+1,k+2,\ldots,k+\ell\right\}  $ is strictly increasing), and
therefore is the identity map (because the only strictly increasing map
from $\left\{  k+1,k+2,\ldots,k+\ell\right\}$ to
$\left\{  k+1,k+2,\ldots,k+\ell \right\}  $ is the identity map).
In other words, $w^{-1}\left(  i\right)
= \id \left(  i\right)  =i$ for every $i\in\left\{
k+1,k+2,\ldots,k+\ell\right\}  $. Applied to $i=j$, this yields $w^{-1}\left(
j\right)  =j$ (since we know that
$j\in\left\{  k+1,k+2,\ldots,k+\ell\right\}  $), whence
$w\left(  j\right)  =j$. But this contradicts $w\left(  j\right)  \neq j$.
This contradiction proves that our assumption was wrong. Thus, $j\leq k$.
Therefore,%
\begin{align}
c_{j}  &  =u_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  c_{1}%
,c_{2},\ldots,c_{k+\ell}\right)  =\left(  u_{1},u_{2},\ldots,u_{k},p_{1}%
,p_{2},\ldots,p_{\ell}\right)  \right) \nonumber\\
&  =u\left(  j\right)  \leq k\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
u\in \Symm_{k}\right)  . \label{pf.lem.sol.FQSym.free.lex.3.pf.5}%
\end{align}
\par
If we had $w\left(  j\right)  <j$, then we would have $w\left(  w\left(
j\right)  \right)  =w\left(  j\right)  $ (by
(\ref{pf.lem.sol.FQSym.free.lex.3.pf.1}), applied to $i=w\left(  j\right)  $),
which would lead to $w\left(  j\right)  =j$ (since $w$ is injective), which
would contradict $w\left(  j\right)  \neq j$. Hence, we cannot have $w\left(
j\right)  <j$. Thus, we have $w\left(  j\right)  \geq j$, so that $w\left(
j\right)  >j$ (since $w\left(  j\right)  \neq j$). In other words, $j<w\left(
j\right)  $.
\par
Let us next prove that $w\left(  j\right)  >k$. Indeed, assume the contrary.
Thus, $w\left(  j\right)  \leq k$. Thus, $w\left(  j\right)  $ and $j$ are two
elements of $\left\{  1,2,\ldots,k\right\}  $ (because $w\left(  j\right)  \leq
k$ and $j\leq k$) satisfying $j<w\left(  j\right)  $. Hence, $w^{-1}\left(
j\right)  <w^{-1}\left(  w\left(  j\right)  \right)  $ (since $w^{-1}\left(
1\right)  <w^{-1}\left(  2\right)  <\cdots<w^{-1}\left(  k\right)  $). Hence,
$w^{-1}\left(  j\right)  <w^{-1}\left(  w\left(  j\right)  \right)  =j$.
Therefore, (\ref{pf.lem.sol.FQSym.free.lex.3.pf.1}) (applied to $i=w^{-1}%
\left(  j\right)  $) yields $w\left(  w^{-1}\left(  j\right)  \right)
=w^{-1}\left(  j\right)  $, so that $w^{-1}\left(  j\right)  =w\left(
w^{-1}\left(  j\right)  \right)  =j$. This contradicts $w^{-1}\left(
j\right)  <j$. This contradiction proves that our assumption was wrong. Hence,
$w\left(  j\right)  >k$ is proven.
\par
Now, write $v$ in the form $\left(v_1,v_2,\ldots,v_\ell\right)$. Then,
$v\left[  k\right]
=\left(  k+v_{1},k+v_{2},\ldots,k+v_{\ell}\right)$ (by the
definition of $v\left[k\right]$). Since $w\left(  j\right)  >k$,
we have
\begin{align*}
c_{w\left(  j\right)  }  &  =p_{w\left(  j\right)  -k}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  c_{1},c_{2},\ldots,c_{k+\ell
}\right)  =\left(  u_{1},u_{2},\ldots,u_{k},p_{1},p_{2},\ldots,p_{\ell}\right)
\right) \\
&  =k+\underbrace{v_{w\left(  j\right)  -k}}_{>0}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  p_{1},p_{2},\ldots,p_{\ell}\right)  =v\left[  k\right]
=\left(  k+v_{1},k+v_{2},\ldots,k+v_{\ell}\right)  \right) \\
&  >k\geq c_{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.sol.FQSym.free.lex.3.pf.5})}\right)  .
\end{align*}
\par
So we have $c_{w\left(  j\right)  }>c_{j}$, but on the other hand,
for every $i\in\left\{
1,2,\ldots,k+\ell\right\}  $ satisfying $i<j$, we have $c_{w\left(  i\right)
}=c_{i}$ (because (\ref{pf.lem.sol.FQSym.free.lex.3.pf.1}) yields $w\left(
i\right)  =i$). Thus,
\begin{align*}
\left(  c_{w\left(  1\right)  },c_{w\left(  2\right)  },\ldots,c_{w\left(
k+\ell\right)  }\right)   &  >\left(  c_{1},c_{2},\ldots,c_{k+\ell}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of lexicographic
order}\right) \\
&  =u\squarebin v,
\end{align*}
and thus (\ref{pf.lem.sol.FQSym.free.lex.3}) is proven.]

Hence,
(\ref{pf.lem.sol.FQSym.free.lex.2}) becomes%
\begin{align*}
F_{u}F_{v}  &  =F_{u\squarebin v}+\underbrace{\sum\limits_{\substack{w\in
\operatorname{Sh}_{k,\ell};\\w\neq \id }}F_{\left(
c_{w\left(  1\right)  },c_{w\left(  2\right)  },\ldots,c_{w\left(  k+\ell\right)
}\right)  }}_{\substack{=\left(  \text{a sum of terms }F_{t}\text{ for }%
t \in \Symm \text{ satisfying } t>u\squarebin v\right)  \\\text{(by
(\ref{pf.lem.sol.FQSym.free.lex.3}))}}}\\
&  =F_{u\squarebin v}+\left(  \text{a sum of terms }F_{t}\text{ for }%
t \in \Symm \text{ satisfying }t>u\squarebin v\right)  .
\end{align*}
This proves Lemma \ref{lem.sol.FQSym.free.lex}.
\end{proof}

\begin{corollary}
\label{cor.sol.FQSym.free.lex.mul}Every $n\in \NN$ and $w\in
 \Symm_{n}$ satisfy%
\[
F_{w}^{\operatorname*{conn}}=F_{w}+\left(  \text{a sum of terms }F_{t}\text{
for }t\in \Symm_{n}\text{ satisfying }t>w\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.sol.FQSym.free.lex.mul}.]We shall first prove
that every $k\in \NN$ and every $\left(  w_{1},w_{2},\ldots,w_{k}\right)
\in \Symm^{k}$ satisfy
\begin{equation}
F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}=F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin
w_{k}}+\left(  \text{a sum of terms }F_{t}\text{ for } t \in \Symm \text{
satisfying }t>w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}\right)  .
\label{pf.cor.sol.FQSym.free.lex.mul.maingoal}%
\end{equation}


\textit{Proof of (\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}):} We will
prove (\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) by induction over $k$:

\textit{Induction base:} For $k=0$, the statement of
(\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) is obvious. Hence, the
induction base is complete.

\textit{Induction step:} Let $K\in \NN$. Assume that
(\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) holds for $k=K$. We need to
prove that (\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) holds for $k=K+1$.

Let $\left(  w_{1},w_{2},\ldots,w_{K+1}\right)  \in \Symm^{K+1}$. By the
induction hypothesis, we can apply
(\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) to $k=K$, and thus obtain%
\[
F_{w_{1}}F_{w_{2}}\cdots F_{w_{K}}=F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin
w_{K}}+\left(  \text{a sum of terms }F_{t}\text{ for }t \in \Symm \text{
satisfying }t>w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}\right)  .
\]
In other words, there exists a finite family $\left(  t_{i}\right)  _{i\in I}$
of elements $t$ of $\Symm$ satisfying $t>w_{1}\squarebin w_{2}%
\squarebin\cdots\squarebin w_{K}$ such that%
\[
F_{w_{1}}F_{w_{2}}\cdots F_{w_{K}}=F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin
w_{K}}+\sum_{i\in I}F_{t_{i}}.
\]
Consider this family $\left(  t_{i}\right)  _{i\in I}$. We have%
\begin{equation}
t_{i}\squarebin w_{K+1}>w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K+1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in I.
\label{pf.cor.sol.FQSym.free.lex.mul.maingoal.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal.pf.1}%
):} Let $i\in I$. Then, $t_{i}$ is an element $t$ of $\Symm$ satisfying
$t>w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}$. Hence, $t_{i}>w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K}$. Thus, (\ref{sol.FQSym.free.1}) (applied to
$\alpha=t_{i}$, $\beta=w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}$ and
$\gamma=w_{K+1}$) yields $t_{i}\squarebin w_{K+1}>\left(  w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K}\right)  \squarebin w_{K+1}=w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K+1}$, qed.} Now,%
\begin{align*}
&  F_{w_{1}}F_{w_{2}}\cdots F_{w_{K+1}}\\
&  =\underbrace{\left(  F_{w_{1}}F_{w_{2}}\cdots F_{w_{K}}\right)
}_{=F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}}+\sum_{i\in I}F_{t_{i}}%
}F_{w_{K+1}}\\
&  =\left(  F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}}+\sum_{i\in
I}F_{t_{i}}\right)  F_{w_{K+1}}\\
&  =\underbrace{F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}}F_{w_{K+1}}%
}_{\substack{=F_{\left(  w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}\right)
\squarebin w_{K+1}}+\left(  \text{a sum of terms }F_{t}\text{ for }%
t \in \Symm \text{ satisfying }t>\left(  w_{1}\squarebin w_{2}\squarebin
\cdots\squarebin w_{K}\right)  \squarebin w_{K+1}\right)  \\\text{(by Lemma
\ref{lem.sol.FQSym.free.lex}, applied to }u=w_{1}\squarebin w_{2}\squarebin
\cdots\squarebin w_{K}\text{ and }v=w_{K+1}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i\in I}\underbrace{F_{t_{i}}F_{w_{K+1}}%
}_{\substack{=F_{t_{i}\squarebin w_{K+1}}+\left(  \text{a sum of terms }%
F_{t}\text{ for }t \in \Symm\text{ satisfying }t>t_{i}\squarebin
w_{K+1}\right)  \\\text{(by Lemma \ref{lem.sol.FQSym.free.lex}, applied to
}u=t_{i}\text{ and }v=w_{K+1}\text{)}}}\\
&  =F_{\left(  w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}\right)  \squarebin
w_{K+1}}+\left(  \text{a sum of terms }F_{t}\text{ for }t \in \Symm \text{
satisfying }t>\left(  w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K}\right)
\squarebin w_{K+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i\in I}\left(  F_{t_{i}\squarebin w_{K+1}}+\left(
\text{a sum of terms }F_{t}\text{ for }t \in \Symm \text{ satisfying
}t>t_{i}\squarebin w_{K+1}\right)  \right) \\
&  =F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K+1}}+\left(  \text{a sum of
terms }F_{t}\text{ for }t \in \Symm \text{ satisfying }t>w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i\in I}\underbrace{\left(  F_{t_{i}\squarebin
w_{K+1}}+\left(  \text{a sum of terms }F_{t}\text{ for }t \in \Symm \text{
satisfying }t>t_{i}\squarebin w_{K+1}\right)  \right)  }_{=\left(  \text{a sum of
terms }F_{t}\text{ for }t\in \Symm \text{ satisfying }t\geq t_{i}\squarebin
w_{K+1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  w_{1}\squarebin w_{2}%
\squarebin\cdots\squarebin w_{K}\right)  \squarebin w_{K+1}=w_{1}\squarebin w_{2}%
\squarebin\cdots\squarebin w_{K+1}\right) \\
&  =F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K+1}}+\left(  \text{a sum of
terms }F_{t}\text{ for }t\in \Symm \text{ satisfying }t>w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i\in I}\underbrace{\left(  \text{a sum of terms
}F_{t}\text{ for }t\in \Symm \text{ satisfying }t\geq t_{i}\squarebin
w_{K+1}\right)  }_{\substack{=\left(  \text{a sum of terms }F_{t}\text{ for
}t\in \Symm \text{ satisfying }t>w_{1}\squarebin w_{2}\squarebin\cdots\squarebin
w_{K+1}\right)  \\\text{(since every }t\in \Symm \text{ satisfying }t\geq
t_{i}\squarebin w_{K+1}\text{ also satisfies}\\t\geq t_{i}\squarebin w_{K+1}%
>w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K+1}\text{ (by
(\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal.pf.1})))}}}\\
&  =F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K+1}}+\left(  \text{a sum of
terms }F_{t}\text{ for }t\in \Symm \text{ satisfying }t>w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K+1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i\in I}\left(  \text{a sum of terms }F_{t}\text{
for }t\in \Symm \text{ satisfying }t>w_{1}\squarebin w_{2}\squarebin
\cdots\squarebin w_{K+1}\right) \\
&  =F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{K+1}}+\left(  \text{a sum of
terms }F_{t}\text{ for }t\in \Symm \text{ satisfying }t>w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{K+1}\right)  .
\end{align*}
In other words, (\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) holds for
$k=K+1$. This completes the induction step. Hence,
(\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal}) is proven by induction.

Now, let $n\in \NN$ and $w\in \Symm_{n}$. Let $\left(  w_{1}%
,w_{2},\ldots,w_{k}\right)  $ be the connected decomposition of $w$. Then,
$F_{w}^{\operatorname*{conn}}=F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}$ (by the
definition of $F_{w}^{\operatorname*{conn}}$) and $w=w_{1}\squarebin w_{2}%
\squarebin\cdots\squarebin w_{k}$ (by the definition of a connected decomposition).
Now,%
\begin{align*}
F_{w}^{\operatorname*{conn}}  &  =F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}\\
&  =F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}}+\left(  \text{a sum of
terms }F_{t}\text{ for }t\in \Symm \text{ satisfying }t>w_{1}\squarebin
w_{2}\squarebin\cdots\squarebin w_{k}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.sol.FQSym.free.lex.mul.maingoal})}\right) \\
&  =F_{w}+\left(  \text{a sum of terms }F_{t}\text{ for }t\in \Symm %
\text{ satisfying }t>w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}=w\right)  .
\end{align*}
Since every $t\in\Symm$ satisfying $t>w$ must belong to
$\Symm_n$\ \ \ \ \footnote{\textit{Proof.} Let
$t\in\Symm$ be such that
$t>w$. Recall that $w \in \Symm_n$. Therefore, if we had $t\notin%
 \Symm_{n}$, then $t$ and $w$ would be incomparable in the poset
$\Symm$ (because of the construction of the poset $\Symm$),
which would contradict the fact that $t>w$. Hence, we cannot have
$t\notin \Symm_{n}$. Thus, $t\in \Symm_{n}$, qed.}, this rewrites
as%
\[
F_{w}^{\operatorname*{conn}}=F_{w}+\left(  \text{a sum of terms }F_{t}\text{
for }t\in \Symm_{n}\text{ satisfying }t>w\right)  .
\]
This proves Corollary \ref{cor.sol.FQSym.free.lex.mul}.
\end{proof}

Now, let us fix $n\in \NN$. The family $\left(  F_{w}\right)
_{w\in \Symm_n}$ is a basis of the $\kk$-module
$\FQsym_n$. We regard the set $\Symm_n$ as a poset whose
smaller relation is the relation $>$ inherited from $\Symm$
(yes, you are reading it right: the order on $\Symm_n$ is
opposite to that on $\Symm$).
Then, Corollary~\ref{cor.sol.FQSym.free.lex.mul} shows that the
family $\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
expands unitriangularly\footnote{See
Definition~\ref{def.STmat.expansion-tria} for the meaning of these
words.} in the family $\left( F_w \right)_{w \in \Symm_n}$
(by Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(c)).
Thus, the family
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
expands invertibly triangularly in the family
$\left( F_w \right)_{w \in \Symm_n}$.
Consequently, Corollary~\ref{cor.STmat.expansion-tria-inv}(e)
(applied to $\FQsym_n$, $\Symm_n$,
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
and $\left( F_w \right)_{w \in \Symm_n}$ instead of
$M$, $S$, $\left(e_s\right)_{s \in S}$ and
$\left(f_s\right)_{s \in S}$) shows that the family
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
is a basis of the $\kk$-module $\FQsym_n$ if and only if the
family $\left( F_w \right)_{w \in \Symm_n}$
is a basis of the $\kk$-module $\FQsym_n$. Hence, the family
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
is a basis of the $\kk$-module $\FQsym_n$ (since the
family $\left( F_w \right)_{w \in \Symm_n}$ is a basis of the
$\kk$-module $\FQsym_n$).

Now, let us forget that we fixed $n$. We thus have proven that the family
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
is a basis of the $\kk$-module $\FQsym_n$ for
every $n \in \NN$. Hence, the family
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm}$
(being the disjoint union of the families
$\left( F_w^{\operatorname{conn}} \right)_{w \in \Symm_n}$
for all $n \in \NN$) is a basis of the $\kk$-module
$\bigoplus_{n \in \NN} \FQsym_n = \FQsym$.

Now, it is easy to see that the family $\left(  F_{w_{1}}F_{w_{2}}\cdots
F_{w_{k}}\right)  _{k\in \NN ;\ \left(  w_{1},w_{2},\ldots,w_{k}\right)
\in\mathfrak{CS}^{k}}$ is a reindexing of the family $\left(  F_{w}%
^{\operatorname*{conn}}\right)  _{w \in \Symm}$%
\ \ \ \ \footnote{\textit{Proof.} For every $k\in \NN$ and $\left(
w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}$, the connected
decomposition of the permutation $w_{1}\squarebin w_{2}\squarebin\cdots\squarebin
w_{k}$ is the $k$-tuple $\left(  w_{1},w_{2},\ldots,w_{k}\right)  $ (by the
definition of a connected decomposition). Hence, for every $k\in \NN$
and $\left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}$, we have%
\begin{equation}
F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}}^{\operatorname*{conn}%
}=F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}
\label{sol.FQSym.free.20}
\end{equation}
(by the definition of
$F_{w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}}^{\operatorname*{conn}%
}$).
\par
Now, recall that for every permutation $w\in\Symm$, there is a unique
way to write $w$ in the form $w=w_{1}\squarebin w_{2}\squarebin\cdots\squarebin w_{k}$
for some $k\in \NN$ and some $k$-tuple $\left(  w_{1},w_{2},\ldots
,w_{k}\right)  \in\mathfrak{CS}^{k}$ of connected permutations. In other
words, the map%
\begin{align*}
\bigsqcup_{k\in \NN }\mathfrak{CS}^{k}  & \rightarrow \Symm ,\\
\left(  w_{1},w_{2},\ldots,w_{k}\right)    & \mapsto w_{1}\squarebin w_{2}%
\squarebin\cdots\squarebin w_{k}%
\end{align*}
is a bijection. Hence, the family $\left(  F_{w_{1}\squarebin w_{2}\squarebin
\cdots\squarebin w_{k}}^{\operatorname*{conn}}\right)  _{k\in \NN ;\ \left(
w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}}$ is a reindexing of the
family $\left(  F_{w}^{\operatorname*{conn}}\right)  _{w \in \Symm}$. Due
to (\ref{sol.FQSym.free.20}), this rewrites as follows: The family $\left(
F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}\right)  _{k\in \NN ;\ \left(
w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}^{k}}$ is a reindexing of the
family $\left(  F_{w}^{\operatorname*{conn}}\right)  _{w \in \Symm}$.
Qed.}. Thus, the family $\left(  F_{w_{1}}F_{w_{2}}\cdots F_{w_{k}}\right)
_{k\in \NN ;\ \left(  w_{1},w_{2},\ldots,w_{k}\right)  \in\mathfrak{CS}%
^{k}}$ is a basis of the $\kk$-module $ \FQsym $ (since
the family $\left(  F_{w}^{\operatorname*{conn}}\right)  _{w \in \Symm}$
is a basis of the $\kk$-module $ \FQsym $). In other
words, $ \FQsym $ is a free (noncommutative) $\kk$-algebra
with generators $\left(  F_{w}\right)  _{w\in\mathfrak{CS}}$. This solves
Exercise~\ref{exe.FQSym.free}.

%\end{solutions}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.STmat.mat}}

\textit{Solution to Exercise~\ref{exe.STmat.mat}.} Let us first agree on a
convention: Whenever $S$ is a poset, we let $\leq$ denote the smaller-or-equal
relation of the poset $S$.

Before we prove Proposition~\ref{prop.STmat.triangular}, let us make a definition:

\begin{definition}
\label{def.sol.STmat.mat.T}Let $S$ be a poset.

\begin{enumerate}
\item[(a)] Let $\operatorname*{T}\nolimits_{S}$ denote the set of all
triangular $S\times S$-matrices. Clearly, $\operatorname*{T}\nolimits_{S}%
\subset\kk^{S\times S}$.

\item[(b)] Let $\operatorname*{IT}\nolimits_{S}$ denote the set of all
invertibly triangular $S\times S$-matrices.

\item[(c)] Let $\operatorname*{UT}\nolimits_{S}$ denote the set of all
unitriangular $S\times S$-matrices.
\end{enumerate}
\end{definition}

Now, we shall state several lemmas (most of them completely trivial, and
stated merely for the purpose of easier reference).

\begin{lemma}
\label{lem.sol.STmat.mats.obs0}Let $S$ and $T$ be two finite sets. Let
$A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$ be an
$S\times T$-matrix. Let $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
T\times S}$ be a $T\times S$-matrix. Then,
\[
AB=\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs0}.]Lemma
\ref{lem.sol.STmat.mats.obs0} is just a restatement of the definition of $AB$.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs1}Let $S$ be a poset. Let $A\in\operatorname*{T}%
\nolimits_{S}$. Then, $A$ is a triangular $S\times S$-matrix.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs1}.]Lemma
\ref{lem.sol.STmat.mats.obs1} follows from the definition of
$\operatorname*{T}\nolimits_{S}$.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs2}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be a triangular $S\times
S $-matrix. Then, every $\left(  s,t\right)  \in S\times S$ which does not
satisfy $t\leq s$ must satisfy $a_{s,t}=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs2}.]Lemma
\ref{lem.sol.STmat.mats.obs2} follows from the definition of ``triangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs3}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an $S\times
S$-matrix. Assume that every $\left(  s,t\right)  \in S\times S$ which does
not satisfy $t\leq s$ must satisfy $a_{s,t}=0$. Then, $A\in\operatorname*{T}%
\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs3}.]Lemma
\ref{lem.sol.STmat.mats.obs3} follows from the definitions of
$\operatorname*{T}\nolimits_{S}$ and of ``triangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs4}Let $S$ be a poset. Let $A\in\operatorname*{IT}%
\nolimits_{S}$. Then, $A$ is an invertibly triangular $S\times S$-matrix.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs4}.]Lemma
\ref{lem.sol.STmat.mats.obs4} follows from the definition of
$\operatorname*{IT}\nolimits_{S}$.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs5}Let $S$ be a poset. Let $A$ be an invertibly
triangular $S\times S$-matrix. Then, $A\in\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs5}.]Lemma
\ref{lem.sol.STmat.mats.obs5} follows from the definition of
``invertibly triangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs6}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an invertibly
triangular $S\times S$-matrix. Then, for every $s\in S$, the element $a_{s,s}
$ of $\kk$ is invertible.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs6}.]
Lemma \ref{lem.sol.STmat.mats.obs6} follows from the definition of
``invertibly triangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs7}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an $S\times
S$-matrix. Assume that $A\in\operatorname*{T}\nolimits_{S}$. Assume that, for
every $s\in S$, the element $a_{s,s}$ of $\kk$ is invertible. Then,
$A\in\operatorname*{IT}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs7}.]Lemma
\ref{lem.sol.STmat.mats.obs7} follows from the definitions of
$\operatorname{T}_S$, of
$\operatorname{IT}_{S}$ and of ``invertibly triangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs8}Let $S$ be a poset. We have $\operatorname*{IT}%
\nolimits_{S}\subset\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs8}.]Lemma
\ref{lem.sol.STmat.mats.obs8} just says that every invertibly triangular
$S\times S$-matrix is triangular; this follows from the definition of
``invertibly triangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs9}Let $S$ be a finite poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in\operatorname*{T}%
\nolimits_{S} $ and $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\operatorname*{T}\nolimits_{S}$. Then, for every $s\in S$, we
have
\begin{equation}
\sum_{k\in S}a_{s,k}b_{k,s}=a_{s,s}b_{s,s}.\nonumber
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs9}.]We have $A\in\operatorname*{T}%
\nolimits_{S}$. Hence, Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a
triangular $S\times S$-matrix. Lemma \ref{lem.sol.STmat.mats.obs2} thus shows
that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0.
\label{pf.prop.STmat.triangular.b.B2.pf.1}%
\end{equation}


The same argument (applied to $B$ and $b_{s,t}$ instead of $A$ and $a_{s,t}$)
shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}%
=0.\label{pf.prop.STmat.triangular.b.B2.pf.2}%
\end{equation}


Let $s\in S$. Every $k\in S$ satisfying $k\neq s$ must satisfy%
\begin{equation}
a_{s,k}b_{k,s}=0\label{pf.prop.STmat.triangular.b.B2.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B2.pf.3}):} Let
$k\in S$ be such that $k\neq s$. We must prove
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $s\leq k$.
\par
\textit{Case 2:} We do not have $s\leq k$.
\par
Let us first consider Case 1. In this case, we have $s\leq k$. If we had
$k\leq s$, then we would have $k=s$ (since $k\leq s\leq k$); but this would
contradict $k\neq s$. Hence, we cannot have $k\leq s$. Thus,
(\ref{pf.prop.STmat.triangular.b.B2.pf.1}) (applied to $\left(  s,k\right)  $
instead of $\left(  s,t\right)  $) shows that $a_{s,k}=0$. Thus,
$\underbrace{a_{s,k}}_{=0}b_{k,s}=0$. Hence,
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $s\leq k$. Hence,
(\ref{pf.prop.STmat.triangular.b.B2.pf.2}) (applied to $\left(  k,s\right)  $
instead of $\left(  s,t\right)  $) shows that $b_{k,s}=0$. Thus,
$a_{s,k}\underbrace{b_{k,s}}_{=0}=0$. Hence,
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}) is proven in Case 2.
\par
Now, we have proven (\ref{pf.prop.STmat.triangular.b.B2.pf.3}) in each of the
two Cases 1 and 2. Hence, (\ref{pf.prop.STmat.triangular.b.B2.pf.3}) always
holds. Qed.}. Now, for every $s\in S$, we have%
\begin{align*}
\sum_{k\in S}a_{s,k}b_{k,s} &  =a_{s,s}b_{s,s}+\sum\limits_{\substack{k\in S;\\k\neq
s}}\underbrace{a_{s,k}b_{k,s}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=s\text{ from the sum}\right)  \\
&  =a_{s,s}b_{s,s}+\underbrace{\sum\limits_{\substack{k\in S;\\k\neq s}}0}%
_{=0}=a_{s,s}b_{s,s}.
\end{align*}
This proves Lemma \ref{lem.sol.STmat.mats.obs9}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs10}Let $S$ be a poset. Let $A\in
\operatorname*{UT}\nolimits_{S}$. Then, $A$ is a unitriangular $S\times S$-matrix.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs10}.]Lemma
\ref{lem.sol.STmat.mats.obs10} follows from the definition of
$\operatorname*{UT}\nolimits_{S}$.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs11}Let $S$ be a poset. Let $A$ be a unitriangular
$S\times S$-matrix. Then, $A\in\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs11}.]Lemma
\ref{lem.sol.STmat.mats.obs11} follows from the definitions of
$\operatorname*{T}\nolimits_{S}$ and of ``unitriangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs12}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be a unitriangular
$S\times S$-matrix. Then, for every $s\in S$, we have $a_{s,s}=1$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs12}.]Lemma
\ref{lem.sol.STmat.mats.obs12} follows from the definition of
``unitriangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs13}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an $S\times
S$-matrix. Assume that $A\in\operatorname*{T}\nolimits_{S}$. Assume that, for
every $s\in S$, we have $a_{s,s}=1$. Then, $A\in\operatorname*{UT}%
\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs13}.]Lemma
\ref{lem.sol.STmat.mats.obs13} follows from the definitions of
$\operatorname{T}_S$, of $\operatorname{UT}_S$ and of
``unitriangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs14}Let $S$ be a poset. We have
$\operatorname*{UT}\nolimits_{S}\subset\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs14}.]Lemma
\ref{lem.sol.STmat.mats.obs14} just says that every unitriangular $S\times
S$-matrix is triangular; this follows from the definition of
``unitriangular''.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs15}We have $\operatorname*{UT}\nolimits_{S}%
\subset\operatorname*{IT}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs15}.]Lemma
\ref{lem.sol.STmat.mats.obs15} just says that every unitriangular $S\times
S$-matrix is invertibly triangular. In order to prove it, we only need to
compare the definitions of ``unitriangular'' and of
``invertibly triangular'', and
observe that the former definition makes a stronger requirement than the
latter (indeed, if $a_{s,s}=1$, then $a_{s,s}$ is invertible).
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.STmat.triangular}.]We begin with some preparations.

For any $\left(  s,t\right)  \in S\times S$, we define a subset $\left[
t,s\right]  $ of $S$ by
\[
\left[  t,s\right]  =\left\{  q\in S\ \mid\ t\leq q\leq s\right\}  .
\]
This subset $\left[  t,s\right]  $ is called the \emph{interval of $S$
bounded by $t$ and $s$}. (Notice that it can be an empty set,
since we have not required that $t\leq s$.) The following facts hold:

\begin{itemize}
\item For any $\left(  s,t\right)  \in S\times S$, the set $\left[
t,s\right]  $ is finite (since it is a subset of the finite set $S$). Hence,
for any $\left(  s,t\right)  \in S\times S$, the cardinality $\left\vert
\left[  t,s\right]  \right\vert $ is a well-defined nonnegative integer.

\item If $s$, $t$ and $u$ are three elements of $S$ such that $s\leq t<u$,
then%
\begin{equation}
\left[  s,t\right]  \text{ is a proper subset of }\left[  s,u\right]
.\label{pf.prop.STmat.triangular.int1}%
\end{equation}
(Indeed, it is straightforward to show that $\left[  s,t\right]  $ is a subset
of $\left[  s,u\right]  $. To prove that this subset is proper, it suffices to
observe that $u$ belongs to $\left[  s,u\right]  $ but not to $\left[
s,t\right]  $.)

\item If $s$, $t$ and $u$ are three elements of $S$ such that $s<t\leq u$,
then%
\begin{equation}
\left[  t,u\right]  \text{ is a proper subset of }\left[  s,u\right]
.\label{pf.prop.STmat.triangular.int2}%
\end{equation}
(Indeed, it is straightforward to show that $\left[  t,u\right]  $ is a subset
of $\left[  s,u\right]  $. To prove that this subset is proper, it suffices to
observe that $s$ belongs to $\left[  s,u\right]  $ but not to $\left[
t,u\right]  $.)
\end{itemize}

Let us now prove Proposition \ref{prop.STmat.triangular}.

(a) Let $0_{S\times S}\in\kk^{S\times S}$ be the matrix $\left(
0\right)  _{\left(  s,t\right)  \in S\times S}$. This matrix $0_{S\times S}$
is the zero of the $\kk$-algebra $\kk^{S\times S}$.

We shall now prove the following five claims:

\begin{statement}
\textit{Claim A1:} We have $0_{S\times S}\in\operatorname*{T}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A2:} For every $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$, we have $A+B\in\operatorname*{T}%
\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A3:} For every $A\in\operatorname*{T}\nolimits_{S}$ and
$u\in\kk$, we have $uA\in\operatorname*{T}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A4:} We have $I_{S}\in\operatorname*{T}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A5:} For every $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$, we have $AB\in\operatorname*{T}%
\nolimits_{S}$.
\end{statement}

\textit{Proof of Claim A1:} The definition of $0_{S\times S}$ yields
$0_{S\times S}=\left(  0\right)  _{\left(  s,t\right)  \in S\times S}$.
Clearly, every $\left(  s,t\right)  \in S\times S$ which does not satisfy
$t\leq s$ must satisfy $0=0$. Thus, Lemma \ref{lem.sol.STmat.mats.obs3}
(applied to $0_{S\times S}$ and $0$ instead of $A$ and $a_{s,t}$) shows that
$0_{S\times S}\in\operatorname*{T}\nolimits_{S}$. This proves Claim A1.

\textit{Proof of Claim A2:} Let $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$. Write the $S\times S$-matrix $A$ in the
form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Write
the $S\times S$-matrix $B$ in the form $B=\left(  b_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times
S$-matrix (since $A\in\operatorname*{T}\nolimits_{S}$). Lemma
\ref{lem.sol.STmat.mats.obs2} shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}%
=0.\label{pf.prop.STmat.triangular.a.A2.pf.1}%
\end{equation}
The same argument (applied to $B$ and $b_{s,t}$ instead of $A$ and $a_{s,t}$)
shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}%
=0.\label{pf.prop.STmat.triangular.a.A2.pf.2}%
\end{equation}


Adding the equalities $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$ and $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$, we obtain%
\[
A+B=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}+\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}=\left(  a_{s,t}%
+b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}%
\]
(by the definition of the sum of two $S\times S$-matrices). But every $\left(
s,t\right)  \in S\times S$ which does not satisfy $t\leq s$ must satisfy
$\underbrace{a_{s,t}}_{\substack{=0 \\\text{(by
(\ref{pf.prop.STmat.triangular.a.A2.pf.1}))}}}+\underbrace{b_{s,t}%
}_{\substack{=0 \\\text{(by (\ref{pf.prop.STmat.triangular.a.A2.pf.2}))}%
}}=0+0=0$. Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $A+B$ and
$a_{s,t}+b_{s,t}$ instead of $A$ and $a_{s,t}$) shows that $A+B\in
\operatorname*{T}\nolimits_{S}$ (since $A+B=\left(  a_{s,t}+b_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$). This proves Claim A2.

\textit{Proof of Claim A3:} The proof of Claim A3 is similar to our proof of
Claim A2 above, and is left to the reader.

\textit{Proof of Claim A4:} The definition of $I_{S}$ yields $I_{S}=\left(
\delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. But every $\left(
s,t\right)  \in S\times S$ which does not satisfy $t\leq s$ must satisfy
$\delta_{s,t}=0$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  s,t\right)  \in
S\times S$ be such that we do not have $t\leq s$. We must prove that
$\delta_{s,t}=0$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $s=t$. Thus,
$\delta_{s,t}=0$, qed.}. Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied
to $I_{S}$ and $\delta_{s,t}$ instead of $A$ and $a_{s,t}$) shows that
$I_{S}\in\operatorname*{T}\nolimits_{S}$. This proves Claim A4.

\textit{Proof of Claim A5:} Let $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$.

As in the proof of Claim A2, we can see that the statements
(\ref{pf.prop.STmat.triangular.a.A2.pf.1}) and
(\ref{pf.prop.STmat.triangular.a.A2.pf.2}) hold.

Lemma \ref{lem.sol.STmat.mats.obs0} (applied to $T=S$) yields%
\[
AB=\left(  \sum_{k\in S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
But every $\left(  s,t\right)  \in S\times S$ and every $k\in S$ which do not
satisfy $t\leq s$ must satisfy%
\begin{equation}
a_{s,k}b_{k,t}=0\label{pf.prop.STmat.triangular.a.A5.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.a.A5.pf.3}):} Let
$\left(  s,t\right)  \in S\times S$ and $k\in S$ be such that we do not have
$t\leq s$. We must prove the equality
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $t\leq k$.
\par
\textit{Case 2:} We do not have $t\leq k$.
\par
Let us first consider Case 1. In this case, we have $t\leq k$. If we had
$k\leq s$, then we would have $t\leq k\leq s$; but this would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $k\leq s$. Thus,
(\ref{pf.prop.STmat.triangular.a.A2.pf.1}) (applied to $\left(  s,k\right)  $
instead of $\left(  s,t\right)  $) shows that $a_{s,k}=0$. Thus,
$\underbrace{a_{s,k}}_{=0}b_{k,t}=0$. Hence,
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $t\leq k$. Hence,
(\ref{pf.prop.STmat.triangular.a.A2.pf.2}) (applied to $\left(  k,t\right)  $
instead of $\left(  s,t\right)  $) shows that $b_{k,t}=0$. Thus,
$a_{s,k}\underbrace{b_{k,t}}_{=0}=0$. Hence,
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}) is proven in Case 2.
\par
Now, we have proven (\ref{pf.prop.STmat.triangular.a.A5.pf.3}) in each of the
two Cases 1 and 2. This shows that (\ref{pf.prop.STmat.triangular.a.A5.pf.3})
always holds. Qed.}. Thus, every $\left(  s,t\right)  \in S\times S$ which
does not satisfy $t\leq s$ must satisfy%
\[
\sum_{k\in S}\underbrace{a_{s,k}b_{k,t}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}))}}}=\sum_{k\in S}0=0.
\]
Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $AB$ and $\sum_{k\in
S}a_{s,k}b_{k,t}$ instead of $A$ and $a_{s,t}$) shows that $AB\in
\operatorname*{T}\nolimits_{S}$ (since $AB=\left(  \sum_{k\in S}a_{s,k}%
b_{k,t}\right)  _{\left(  s,t\right)  \in S\times S}$). This proves Claim A5.

Recall that $\operatorname*{T}\nolimits_{S}$ is a subset of $\kk%
^{S\times S}$. This subset $\operatorname*{T}\nolimits_{S}$ is a $\kk%
$-submodule of $\kk^{S\times S}$ (by Claim A1, Claim A2 and Claim A3),
and therefore is a $\kk$-subalgebra of $\kk^{S\times S}$ (by
Claim A4 and Claim A5). In other words, the set of all triangular $S\times
S$-matrices is a $\kk$-subalgebra of $\kk^{S\times S}$ (since
$\operatorname*{T}\nolimits_{S}$ is the set of all triangular $S\times
S$-matrices). This proves Proposition \ref{prop.STmat.triangular}(a).

(b) We shall first prove the following claims:

\begin{statement}
\textit{Claim B1:} We have $I_{S}\in\operatorname*{IT}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim B2:} For every $A\in\operatorname*{IT}\nolimits_{S}$ and
$B\in\operatorname*{IT}\nolimits_{S}$, we have $AB\in\operatorname*{IT}%
\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim B3:} Let $A\in\operatorname*{IT}\nolimits_{S}$. Then, $A$ is
invertible (as an element of the ring $\kk^{S\times S}$), and its
inverse $A^{-1}$ belongs to $\operatorname*{IT}\nolimits_{S}$.
\end{statement}

\textit{Proof of Claim B1:} The definition of $I_{S}$ yields $I_{S}=\left(
\delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Claim A4
in our proof of Proposition \ref{prop.STmat.triangular}(a) yields
$I_{S}\in\operatorname*{T}\nolimits_{S}$. For every $s\in S$, we have
$\delta_{s,s}=1$. Hence, for every $s\in S$, the element $\delta_{s,s}$ of
$\kk$ is invertible. Thus, Lemma \ref{lem.sol.STmat.mats.obs7} (applied
to $I_{S}$ and $\delta_{s,t}$ instead of $A$ and $a_{s,t}$) yields $I_{S}%
\in\operatorname*{IT}\nolimits_{S}$. This proves Claim B1.

\textit{Proof of Claim B2:} Let $A\in\operatorname*{IT}\nolimits_{S}$ and
$B\in\operatorname*{IT}\nolimits_{S}$.

We know that $A$ is an invertibly triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs4}). Hence, $A\in\operatorname*{T}\nolimits_{S}$
(by Lemma \ref{lem.sol.STmat.mats.obs5}). Hence, Lemma
\ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times
S$-matrix. Write the $S\times S$-matrix $A$ in the form $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Lemma
\ref{lem.sol.STmat.mats.obs6} yields that%
\begin{equation}
\text{for every }s\in S\text{, the element }a_{s,s}\text{ of }\kk\text{
is invertible.}\label{pf.prop.STmat.triangular.b.B2.pf.1a}%
\end{equation}


We know that $B$ is an invertibly triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs4}, applied to $B$ instead of $A$). Hence,
$B\in\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs5},
applied to $B$ instead of $A$). Hence, Lemma \ref{lem.sol.STmat.mats.obs1}
(applied to $B$ instead of $A$) shows that $B$ is a triangular $S\times
S$-matrix. Write the $S\times S$-matrix $B$ in the form $B=\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Lemma
\ref{lem.sol.STmat.mats.obs6} (applied to $B$ and $b_{s,t}$ instead of $A$ and
$a_{s,t}$) yields that%
\begin{equation}
\text{for every }s\in S\text{, the element }b_{s,s}\text{ of }\kk\text{
is invertible.}\label{pf.prop.STmat.triangular.b.B2.pf.2a}%
\end{equation}


From $A\in\operatorname*{T}\nolimits_{S}$ and $B\in\operatorname*{T}%
\nolimits_{S}$, we obtain $AB\in\operatorname*{T}\nolimits_{S}$ (by Claim A5
in our proof of Proposition \ref{prop.STmat.triangular}(a)).

Lemma \ref{lem.sol.STmat.mats.obs0} (applied to $T=S$) yields%
\[
AB=\left(  \sum_{k\in S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
Lemma \ref{lem.sol.STmat.mats.obs9} shows that, for every $s\in S$, we have
$\sum_{k\in S}a_{s,k}b_{k,s}=a_{s,s}b_{s,s}$. Hence, for every $s\in S$, the
element $\sum_{k\in S}a_{s,k}b_{k,s}$ of $\kk$ is
invertible\footnote{\textit{Proof.} Let $s\in S$. Then, the element $a_{s,s}$
of $\kk$ is invertible (by (\ref{pf.prop.STmat.triangular.b.B2.pf.1a}%
)). The element $b_{s,s}$ of $\kk$ is invertible as well (by
(\ref{pf.prop.STmat.triangular.b.B2.pf.2a})). Thus, the product $a_{s,s}%
b_{s,s}$ of these two elements is also invertible (since the product of two
invertible elements is always invertible). In other words, $\sum_{k\in
S}a_{s,k}b_{k,s}$ is invertible (since $\sum_{k\in S}a_{s,k}b_{k,s}%
=a_{s,s}b_{s,s}$). Qed.}. Thus, Lemma \ref{lem.sol.STmat.mats.obs7} (applied
to $AB$ and $\sum_{k\in S}a_{s,k}b_{k,t}$ instead of $A$ and $a_{s,t}$) shows
that $AB\in\operatorname*{IT}\nolimits_{S}$ (since $AB=\left(  \sum_{k\in
S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in S\times S}$ and
$AB\in\operatorname*{T}\nolimits_{S}$). This proves Claim B2.

\textit{Proof of Claim B3:} We know that $A$ is an invertibly triangular
$S\times S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs4}). Hence,
$A\in\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs5}).
Hence, Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular
$S\times S$-matrix. Write the $S\times S$-matrix $A$ in the form $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Lemma
\ref{lem.sol.STmat.mats.obs2} shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}%
=0.\label{pf.prop.STmat.triangular.b.B3.pf.1}%
\end{equation}
Lemma \ref{lem.sol.STmat.mats.obs6} yields that%
\begin{equation}
\text{for every }s\in S\text{, the element }a_{s,s}\text{ of }\kk\text{
is invertible.}\label{pf.prop.STmat.triangular.b.B3.pf.1a}%
\end{equation}


We shall now define an element $b_{s,t}$ for each $\left(  s,t\right)  \in
S\times S$. In fact, we will define these elements recursively, by strong
induction on $\left\vert \left[  t,s\right]  \right\vert $%
\ \ \ \ \footnote{Here, we are using the fact that $\left\vert \left[
t,s\right]  \right\vert $ is a nonnegative integer for every $\left(
s,t\right)  \in S\times S$.}: Let $N\in \NN$. Assume that
\begin{equation}
\text{an element }b_{s,t}\in\kk\text{ is already defined for each
}\left(  s,t\right)  \in S\times S\text{ satisfying }\left\vert \left[
t,s\right]  \right\vert <N.\label{pf.prop.STmat.triangular.b.B3.pf.def.ihyp}%
\end{equation}
We shall now define an element $b_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$ satisfying $\left\vert \left[  t,s\right]
\right\vert =N$.

Indeed, let $\left(  s,t\right)  \in S\times S$ be such that $\left\vert
\left[  t,s\right]  \right\vert =N$. We must define an element $b_{s,t}%
\in\kk$.

For every $u\in S$ satisfying $t<u\leq s$, the element $b_{s,u}$ is already
defined\footnote{\textit{Proof.} Let $u\in S$ be such that $t<u\leq s$. We
have to show that the element $b_{s,u}$ is already defined.
\par
From (\ref{pf.prop.STmat.triangular.int2}) (applied to $t$, $u$ and $s$
instead of $s$, $t$ and $u$), we conclude that $\left[  u,s\right]  $ is a
proper subset of $\left[  t,s\right]  $. Hence, $\left\vert \left[
u,s\right]  \right\vert <\left\vert \left[  t,s\right]  \right\vert $ (since
$\left[  t,s\right]  $ is a finite set). Thus, $\left\vert \left[  u,s\right]
\right\vert <\left\vert \left[  t,s\right]  \right\vert =N$. Therefore,
(\ref{pf.prop.STmat.triangular.b.B3.pf.def.ihyp}) (applied to $\left(
s,u\right)  $ instead of $\left(  s,t\right)  $) shows that an element
$b_{s,u}$ is already defined.}. Thus, the sum $\sum\limits_{\substack{u\in
S;\\t<u\leq s}}b_{s,u}a_{u,t}\in\kk$ is well-defined. Furthermore, the
element $a_{t,t}$ of $\kk$ is invertible (by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1a}), applied to $t$ instead of $s$).
Hence, the element $\left(  a_{t,t}\right)  ^{-1}$ of $\kk$ is
well-defined. Now, we set%
\begin{equation}
b_{s,t}=\left(  a_{t,t}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in
S;\\t<u\leq s}}b_{s,u}a_{u,t}\right)
\label{pf.prop.STmat.triangular.b.B3.pf.def.rec}%
\end{equation}
(this makes sense since both $\left(  a_{t,t}\right)  ^{-1}$ and
$\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}$ are well-defined). Thus,
we have defined an element $b_{s,t}\in\kk$. This completes the
recursive definition of $b_{s,t}$.

We have now defined an element $b_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$. In other words, we have defined a family $\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in\kk^{S\times S}%
$. This family is clearly an $S\times S$-matrix. Denote this $S\times
S$-matrix by $B$. Thus, $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$. We shall now show (in several steps)
that $BA=I_{S}$ and that $B\in\operatorname*{IT}\nolimits_{S}$.

First, we notice that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}%
=0\label{pf.prop.STmat.triangular.b.B3.pf.B-tri}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}):}
Let $\left(  s,t\right)  \in S\times S$ be such that we do not have $t\leq s$.
We must prove that $b_{s,t}=0$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $s=t$. Thus,
$\delta_{s,t}=0$.
\par
If there was an $u\in S$ satisfying $t<u\leq s$, then we would have $t\leq s$,
which would contradict the fact that we do not have $t\leq s$. Hence, there is
no $u\in S$ satisfying $t<u\leq s$. Thus, the sum $\sum\limits_{\substack{u\in
S;\\t<u\leq s}}b_{s,u}a_{u,t}$ is empty. Hence, $\sum\limits_{\substack{u\in
S;\\t<u\leq s}}b_{s,u}a_{u,t}=\left(  \text{empty sum}\right)  =0$.
\par
Now, the recursive definition of $b_{s,t}$ yields $b_{s,t}=\left(
a_{t,t}\right)  ^{-1}\left(  \underbrace{\delta_{s,t}}_{=0}-\underbrace{\sum\limits
_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}}_{=0}\right)  =\left(
a_{t,t}\right)  ^{-1}0=0$. This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}).}. Hence, Lemma
\ref{lem.sol.STmat.mats.obs3} (applied to $B$ and $b_{s,t}$ instead of $A$ and
$a_{s,t}$) yields $B\in\operatorname*{T}\nolimits_{S}$.

For every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}=\delta_{s,t}%
\label{pf.prop.STmat.triangular.b.B3.pf.BA.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}):} Let
$\left(  s,t\right)  \in S\times S$. We must prove the equality
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $t\leq s$.
\par
\textit{Case 2:} We do not have $t\leq s$.
\par
Let us first consider Case 1. In this case, we have $t\leq s$. Thus, $t$ is an
element of $S$ satisfying $t\leq t\leq s$. Hence, the sum $\sum\limits
_{\substack{u\in S;\\t\leq u\leq s}}b_{s,u}a_{u,t}$ has an addend for $u=t$.
Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{u\in S;\\t\leq u\leq s}}b_{s,u}a_{u,t}=\sum\limits_{\substack{u\in
S;\\t\leq u\leq s\text{ and }u\neq t}}b_{s,u}a_{u,t}+b_{s,t}a_{t,t}.
\]
\par
Now,%
\begin{align*}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t} &  =\sum\limits
_{\substack{u\in S;\\t\leq u\leq s}}b_{s,u}a_{u,t}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have renamed the summation index }k\text{ as }u\right)  \\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\t\leq u\leq s\text{ and }u\neq t}%
}}_{\substack{=\sum\limits_{\substack{u\in S;\\t<u\leq s}}\\\text{(because for any
}u\in S\text{, the}\\\text{condition }\left(  t\leq u\leq s\text{ and }u\neq
t\right)  \\\text{is equivalent to }\left(  t<u\leq s\right)  \text{)}%
}}b_{s,u}a_{u,t}+\underbrace{b_{s,t}}_{\substack{=\left(  a_{t,t}\right)
^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}%
a_{u,t}\right)  \\\text{(by the recursive definition of }b_{s,t}\text{)}%
}}a_{t,t}\\
&  =\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}+\underbrace{\left(
a_{t,t}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq
s}}b_{s,u}a_{u,t}\right)  a_{t,t}}_{=\delta_{s,t}-\sum\limits_{\substack{u\in
S;\\t<u\leq s}}b_{s,u}a_{u,t}}\\
&  =\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}+\delta_{s,t}%
-\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}=\delta_{s,t}.
\end{align*}
Hence, (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $t\leq s$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we do not have $s=t$. Therefore,
$\delta_{s,t}=0$.
\par
If there was an $k\in S$ satisfying $t\leq k\leq s$, then we would have $t\leq
s$, which would contradict the fact that we do not have $t\leq s$. Hence,
there is no $k\in S$ satisfying $t\leq k\leq s$. Thus, the sum $\sum\limits
_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}$ is empty. Hence,
$\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}=\left(  \text{empty
sum}\right)  =0=\delta_{s,t}$ (since $\delta_{s,t}=0$). Hence,
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) is proven in Case 2.
\par
We have now proven (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) in each of
the two Cases 1 and 2. Thus, (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1})
always holds.}. Thus, for every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum_{k\in S}b_{s,k}a_{k,t}=\delta_{s,t}%
\label{pf.prop.STmat.triangular.b.B3.pf.BA.1a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}):}
Let $\left(  s,t\right)  \in S\times S$. Then,%
\begin{align*}
\sum_{k\in S}b_{s,k}a_{k,t} &  =\sum\limits_{\substack{k\in S;\\t\leq k}%
}b_{s,k}a_{k,t}+\sum\limits_{\substack{k\in S;\\\text{not }t\leq k}}b_{s,k}%
\underbrace{a_{k,t}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1}), applied to}\\\left(  k,t\right)
\text{ instead of }\left(  s,t\right)  \text{)}}}=\sum\limits_{\substack{k\in
S;\\t\leq k}}b_{s,k}a_{k,t}+\underbrace{\sum\limits_{\substack{k\in S;\\\text{not
}t\leq k}}b_{s,k}0}_{=0}\\
&  =\sum\limits_{\substack{k\in S;\\t\leq k}}b_{s,k}a_{k,t}=\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and }k\leq s}}}_{\substack{=\sum\limits
_{\substack{k\in S;\\t\leq k\leq s}}}}b_{s,k}a_{k,t}+\sum\limits_{\substack{k\in
S;\\t\leq k\text{ and not }k\leq s}}\underbrace{b_{s,k}}%
_{\substack{=0\\\text{(by (\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}),
applied to}\\\left(  s,k\right)  \text{ instead of }\left(  s,t\right)
\text{)}}}a_{k,t}\\
&  =\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}+\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and not }k\leq s}}0a_{k,t}}_{=0}%
=\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}=\delta_{s,t}%
\end{align*}
(by (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1})). This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}).}. Hence, $BA=I_{S}%
$\ \ \ \ \footnote{\textit{Proof.} Lemma \ref{lem.sol.STmat.mats.obs0}
(applied to $S$, $B$, $b_{s,t}$, $A$ and $a_{s,t}$ instead of $T$, $A$,
$a_{s,t}$, $B$ and $b_{s,t}$) shows that%
\[
BA=\left(  \underbrace{\sum_{k\in S}b_{s,k}a_{k,t}}_{\substack{=\delta
_{s,t}\\\text{(by (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}))}}}\right)
_{\left(  s,t\right)  \in S\times S}=\left(  \delta_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}=I_{S}%
\]
(since $I_{S}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$ (by the definition of $I_{S}$)). Qed.}. Furthermore, for every $s\in S$,
we have%
\begin{equation}
b_{s,s}=\left(  a_{s,s}\right)  ^{-1}%
\label{pf.prop.STmat.triangular.b.B3.pf.BA.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.2}):}
Lemma \ref{lem.sol.STmat.mats.obs9} (applied to $B$, $b_{s,t}$, $A$ and
$a_{s,t}$ instead of $A$, $a_{s,t}$, $B$ and $b_{s,t}$) yields that for every
$s\in S$, we have
\begin{equation}
\sum_{k\in S}b_{s,k}a_{k,s}=b_{s,s}a_{s,s}%
.\label{pf.prop.STmat.triangular.b.B3.pf.BA.2.pf.1}%
\end{equation}
\par
Let $s\in S$. Then, (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}) (applied to
$\left(  s,s\right)  $ instead of $\left(  s,t\right)  $) yields $\sum_{k\in
S}b_{s,k}a_{k,s}=\delta_{s,s}=1$ (since $s=s$). Comparing this with
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.2.pf.1}), we obtain $b_{s,s}%
a_{s,s}=1$. Hence, $b_{s,s}=\left(  a_{s,s}\right)  ^{-1}$. This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.2}).}. Hence,
\[
\text{for every }s\in S\text{, the element }b_{s,s}\text{ of }\kk\text{
is invertible}%
\]
(because $b_{s,s}$ is an inverse (namely, $b_{s,s}=\left(  a_{s,s}\right)
^{-1}$)). Thus, Lemma \ref{lem.sol.STmat.mats.obs7} (applied to $B$ and
$b_{s,t}$ instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{IT}%
\nolimits_{S}$ (since $B\in\operatorname*{T}\nolimits_{S}$). In other words,
$B$ belongs to $\operatorname*{IT}\nolimits_{S}$.

So far, we do not know that $B$ is the inverse of $A$; we merely know that $B$
is a left inverse of $A$ (that is, we know that $BA=I_{S}$). We shall now
construct yet another $S\times S$-matrix $C$ and subsequently show that
$AC=I_{S}$; this will easily yield that $A$ is invertible (because an element
of a ring that has a left inverse and a right inverse must be invertible) and
its inverse is $C=B$.

The construction of $C$ will be rather similar to that of $B$, so that we will
be briefer than before.

We shall define an element $c_{s,t}$ for each $\left(  s,t\right)  \in S\times
S$. In fact, we will define these elements recursively, by strong induction on
$\left\vert \left[  t,s\right]  \right\vert $: Let $N\in \NN$. Assume
that
\begin{equation}
\text{an element }c_{s,t}\in\kk\text{ is already defined for each
}\left(  s,t\right)  \in S\times S\text{ satisfying }\left\vert \left[
t,s\right]  \right\vert <N.\label{pf.prop.STmat.triangular.b.B3.pf.def2.ihyp}%
\end{equation}
We shall now define an element $c_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$ satisfying $\left\vert \left[  t,s\right]
\right\vert =N$.

Indeed, let $\left(  s,t\right)  \in S\times S$ be such that $\left\vert
\left[  t,s\right]  \right\vert =N$. We must define an element $c_{s,t}%
\in\kk$.

For every $u\in S$ satisfying $t\leq u<s$, the element $c_{u,t}$ is already
defined\footnote{\textit{Proof.} Let $u\in S$ be such that $t\leq u<s$. We
have to show that the element $c_{u,t}$ is already defined.
\par
From (\ref{pf.prop.STmat.triangular.int1}) (applied to $t$, $u$ and $s$
instead of $s$, $t$ and $u$), we conclude that $\left[  t,u\right]  $ is a
proper subset of $\left[  t,s\right]  $. Hence, $\left\vert \left[
t,u\right]  \right\vert <\left\vert \left[  t,s\right]  \right\vert $ (since
$\left[  t,s\right]  $ is a finite set). Thus, $\left\vert \left[  t,u\right]
\right\vert <\left\vert \left[  t,s\right]  \right\vert =N$. Therefore,
(\ref{pf.prop.STmat.triangular.b.B3.pf.def2.ihyp}) (applied to $\left(
u,t\right)  $ instead of $\left(  s,t\right)  $) shows that an element
$c_{u,t}$ is already defined.}. Thus, the sum $\sum\limits_{\substack{u\in S;\\t\leq
u<s}}a_{s,u}c_{u,t}\in\kk$ is well-defined. Furthermore, the element
$a_{s,s}$ of $\kk$ is invertible (by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1a})). Hence, the element $\left(
a_{s,s}\right)  ^{-1}$ of $\kk$ is well-defined. Now, we set%
\begin{equation}
c_{s,t}=\left(  a_{s,s}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in
S;\\t\leq u<s}}a_{s,u}c_{u,t}\right)
\label{pf.prop.STmat.triangular.b.B3.pf.def2.rec}%
\end{equation}
(this makes sense since both $\left(  a_{s,s}\right)  ^{-1}$ and
$\sum\limits_{\substack{u\in S;\\t\leq u<s}}a_{s,u}c_{u,t}$ are well-defined). Thus,
we have defined an element $c_{s,t}\in\kk$. This completes the
recursive definition of $c_{s,t}$.

We have now defined an element $c_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$. In other words, we have defined a family $\left(
c_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in\kk^{S\times S}%
$. This family is clearly an $S\times S$-matrix. Denote this $S\times
S$-matrix by $C$. Thus, $C=\left(  c_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$. We shall now show (in several steps)
that $AC=I_{S}$ and that $C\in\operatorname*{IT}\nolimits_{S}$.

First, we notice that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }c_{s,t}%
=0\label{pf.prop.STmat.triangular.b.B3.pf.C-tri}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.C-tri}):}
The statement (\ref{pf.prop.STmat.triangular.b.B3.pf.C-tri}) is an analogue of
(\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}), and has an analogous proof.}.
Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $C$ and $c_{s,t}$
instead of $A$ and $a_{s,t}$) yields $C\in\operatorname*{T}\nolimits_{S}$.

For every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}=\delta_{s,t}%
\label{pf.prop.STmat.triangular.b.B3.pf.AC.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}):} The
statement (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}) is analogous to
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) and has an analogous proof. (The
main difference is that we now need to split off the addend for $u=s$ from the
sum $\sum\limits_{\substack{u\in S;\\t\leq u\leq s}}a_{s,u}c_{u,t}$, instead of
splitting off the addend for $u=t$ from the sum $\sum\limits_{\substack{u\in
S;\\t\leq u\leq s}}b_{s,u}a_{u,t}$.)}. Thus, for every $\left(  s,t\right)
\in S\times S$, we have%
\begin{equation}
\sum_{k\in S}a_{s,k}c_{k,t}=\delta_{s,t}%
\label{pf.prop.STmat.triangular.b.B3.pf.AC.1a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1a}):}
Let $\left(  s,t\right)  \in S\times S$. Then,%
\begin{align*}
\sum_{k\in S}a_{s,k}c_{k,t} &  =\sum\limits_{\substack{k\in S;\\t\leq k}%
}a_{s,k}c_{k,t}+\sum\limits_{\substack{k\in S;\\\text{not }t\leq k}}a_{s,k}%
\underbrace{c_{k,t}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B3.pf.C-tri}), applied to}\\\left(
k,t\right)  \text{ instead of }\left(  s,t\right)  \text{)}}}=\sum\limits
_{\substack{k\in S;\\t\leq k}}a_{s,k}c_{k,t}+\underbrace{\sum\limits_{\substack{k\in
S;\\\text{not }t\leq k}}a_{s,k}0}_{=0}\\
&  =\sum\limits_{\substack{k\in S;\\t\leq k}}a_{s,k}c_{k,t}=\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and }k\leq s}}}_{\substack{=\sum\limits
_{\substack{k\in S;\\t\leq k\leq s}}}}a_{s,k}c_{k,t}+\sum\limits_{\substack{k\in
S;\\t\leq k\text{ and not }k\leq s}}\underbrace{a_{s,k}}%
_{\substack{=0\\\text{(by (\ref{pf.prop.STmat.triangular.b.B3.pf.1}), applied
to}\\\left(  s,k\right)  \text{ instead of }\left(  s,t\right)  \text{)}%
}}c_{k,t}\\
&  =\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}+\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and not }k\leq s}}0c_{k,t}}_{=0}%
=\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}=\delta_{s,t}%
\end{align*}
(by (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1})). This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1a}).}. Hence, $AC=I_{S}%
$\ \ \ \ \footnote{\textit{Proof.} Lemma \ref{lem.sol.STmat.mats.obs0}
(applied to $S$, $C$ and $c_{s,t}$ instead of $T$, $B$ and $b_{s,t}$) shows
that%
\[
AC=\left(  \underbrace{\sum_{k\in S}a_{s,k}c_{k,t}}_{\substack{=\delta
_{s,t}\\\text{(by (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1a}))}}}\right)
_{\left(  s,t\right)  \in S\times S}=\left(  \delta_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}=I_{S}%
\]
(since $I_{S}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$ (by the definition of $I_{S}$)). Qed.}. Now, using the associativity of
the $\kk$-algebra $\kk^{S\times S}$, we can make the following
computation:%
\[
B=B\underbrace{I_{S}}_{=AC}=\underbrace{BA}_{=I_{S}}C=I_{S}C=C.
\]
Hence, $AC=I_{S}$ rewrites as $AB=I_{S}$. Combining this with $BA=I_{S}$, we
see that $B$ is an inverse of $A$ in the ring $\kk^{S\times S}$. In
particular, the element $A$ of $\kk^{S\times S}$ is invertible. Its
inverse $A^{-1}$ is $B$ (as we have just proven), and therefore belongs to
$\operatorname*{IT}\nolimits_{S}$ (since we know that $B$ belongs to
$\operatorname*{IT}\nolimits_{S}$). This proves Claim B3.

Recall that $\operatorname*{IT}\nolimits_{S}$ is a subset of $\kk%
^{S\times S}$. Claim B1 and Claim B2 (combined) show that this set
$\operatorname*{IT}\nolimits_{S}$ is a submonoid of the multiplicative monoid
of $\kk^{S\times S}$. Claim B3 furthermore proves that this submonoid
is a group. Thus, $\operatorname*{IT}\nolimits_{S}$ is a group with respect to
multiplication. In other words, the set of all invertibly triangular $S\times
S$-matrices is a group with respect to multiplication (since
$\operatorname*{IT}\nolimits_{S}$ is the set of all invertibly triangular
$S\times S$-matrices). This proves Proposition \ref{prop.STmat.triangular}(b).

(c) We shall first prove the following claims:

\begin{statement}
\textit{Claim C1:} We have $I_{S}\in\operatorname*{UT}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim C2:} For every $A\in\operatorname*{UT}\nolimits_{S}$ and
$B\in\operatorname*{UT}\nolimits_{S}$, we have $AB\in\operatorname*{UT}%
\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim C3:} Let $A\in\operatorname*{UT}\nolimits_{S}$. Then, $A$ is
invertible (as an element of the ring $\kk^{S\times S}$), and its
inverse $A^{-1}$ belongs to $\operatorname*{UT}\nolimits_{S}$.
\end{statement}

\textit{Proof of Claim C1:} The definition of $I_{S}$ yields $I_{S}=\left(
\delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Claim A4
in our proof of Proposition \ref{prop.STmat.triangular}(a) yields
$I_{S}\in\operatorname*{T}\nolimits_{S}$. For every $s\in S$, we have
$\delta_{s,s}=1$. Thus, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to
$I_{S}$ and $\delta_{s,t}$ instead of $A$ and $a_{s,t}$) yields $I_{S}%
\in\operatorname*{UT}\nolimits_{S}$. This proves Claim C1.

\textit{Proof of Claim C2:} Let $A\in\operatorname*{UT}\nolimits_{S}$ and
$B\in\operatorname*{UT}\nolimits_{S}$.

Write the $S\times S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$. Write the $S\times S$-matrix $B$ in the
form $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.

We know that $A$ is unitriangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs10}). Hence, $A\in\operatorname*{T}\nolimits_{S}$
(by Lemma \ref{lem.sol.STmat.mats.obs11}). Similarly, $B\in\operatorname*{T}%
\nolimits_{S}$.

Lemma \ref{lem.sol.STmat.mats.obs12} yields that%
\begin{equation}
\text{for every }s\in S\text{, we have }a_{s,s}%
=1.\label{pf.prop.STmat.triangular.c.C2.pf.1a}%
\end{equation}


The same argument (applied to $B$ and $b_{s,t}$ instead of $A$ and $a_{s,t}$)
yields that%
\begin{equation}
\text{for every }s\in S\text{, we have }b_{s,s}%
=1.\label{pf.prop.STmat.triangular.c.C2.pf.2a}%
\end{equation}


Claim A5 in our proof of Proposition \ref{prop.STmat.triangular}(a)
yields $AB\in\operatorname*{T}\nolimits_{S}$. Lemma
\ref{lem.sol.STmat.mats.obs0} (applied to $T=S$) yields%
\[
AB=\left(  \sum_{k\in S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
Lemma \ref{lem.sol.STmat.mats.obs9} yields that, for every $s\in S$, we have
\begin{equation}
\sum_{k\in S}a_{s,k}b_{k,s}=\underbrace{a_{s,s}}_{\substack{=1\\\text{(by
(\ref{pf.prop.STmat.triangular.c.C2.pf.1a}))}}}\underbrace{b_{s,s}%
}_{\substack{=1\\\text{(by (\ref{pf.prop.STmat.triangular.c.C2.pf.2a}))}%
}}=1.\nonumber
\end{equation}
Hence, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to $AB$ and $\sum_{k\in
S}a_{s,k}b_{k,t}$ instead of $A$ and $a_{s,t}$) shows that $AB\in
\operatorname*{UT}\nolimits_{S}$ (since $AB=\left(  \sum_{k\in S}%
a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in S\times S}$ and
$AB\in\operatorname*{T}\nolimits_{S}$). This proves Claim C2.

\textit{Proof of Claim C3:} We know that $A$ is unitriangular $S\times
S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs10}). Hence, $A\in
\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs11}).

Write the $S\times S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs12}
yields that%
\begin{equation}
\text{for every }s\in S\text{, we have }a_{s,s}%
=1.\label{pf.prop.STmat.triangular.c.C3.pf.1}%
\end{equation}


We have $A\in\operatorname*{UT}\nolimits_{S}\subset\operatorname*{IT}%
\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs15}). Thus, the conditions
of Claim B3 (in our proof of Proposition \ref{prop.STmat.triangular}(b)) are
satisfied. Define an $S\times S$-matrix $B=\left(  b_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}\in\kk^{S\times S}$ as in the proof of Claim
B3 (in our proof of Proposition \ref{prop.STmat.triangular}(b)). Then, we have
the following facts (which were shown in the proof of Claim B3):

\begin{itemize}
\item For every $s\in S$, we have%
\begin{equation}
b_{s,s}=\left(  a_{s,s}\right)  ^{-1}.
\label{pf.prop.STmat.triangular.c.C3.pf.2}%
\end{equation}


\item We have $B\in\operatorname*{T}\nolimits_{S}$.

\item The matrix $B$ is an inverse of $A$ in the ring $\kk^{S\times S}
$.
\end{itemize}

Now, for every $s\in S$, we have%
\[
b_{s,s}=\left(  \underbrace{a_{s,s}}_{\substack{=1 \\\text{(by
(\ref{pf.prop.STmat.triangular.c.C3.pf.1}))}}}\right)  ^{-1}=1^{-1}=1.
\]
Hence, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{UT}\nolimits_{S}$
(since $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ and
$B\in\operatorname*{T}\nolimits_{S}$). In other words, $B$ belongs to
$\operatorname*{UT}\nolimits_{S}$.

Now, recall that $B$ is an inverse of $A$ in the ring $\kk^{S\times S}
$. In particular, the element $A$ of $\kk^{S\times S}$ is invertible.
Its inverse $A^{-1}$ is $B$ (as we have just proven), and therefore belongs to
$\operatorname*{UT}\nolimits_{S}$ (since we know that $B$ belongs to
$\operatorname*{UT}\nolimits_{S}$). This proves Claim C3.

Recall that $\operatorname*{UT}\nolimits_{S}$ is a subset of $\kk%
^{S\times S}$. Claim C1 and Claim C2 (combined) show that this set
$\operatorname*{UT}\nolimits_{S}$ is a submonoid of the multiplicative monoid
of $\kk^{S\times S}$. Claim C3 furthermore proves that this submonoid
is a group. Thus, $\operatorname*{UT}\nolimits_{S}$ is a group with respect to
multiplication. In other words, the set of all unitriangular $S\times
S$-matrices is a group with respect to multiplication (since
$\operatorname*{UT}\nolimits_{S}$ is the set of all unitriangular $S\times
S$-matrices). This proves Proposition \ref{prop.STmat.triangular}(c).

(d) Let $A$ be an invertibly triangular $S\times S$-matrix. In other words,
$A\in\operatorname*{IT}\nolimits_{S}$ (since $\operatorname*{IT}\nolimits_{S}$
is the set of all invertibly triangular $S\times S$-matrices). Hence, Claim B3
(in our proof of Proposition \ref{prop.STmat.triangular}(b)) shows that $A$ is
invertible, and that its inverse $A^{-1}$ belongs to $\operatorname*{IT}%
\nolimits_{S}$. The matrix $A^{-1}$ is therefore an invertibly triangular
$S\times S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs4} (applied to
$A^{-1}$ instead of $A$)).

Now, forget that we fixed $A$. We thus have shown that if $A$ is an invertibly
triangular $S\times S$-matrix, then $A$ is invertible, and its inverse
$A^{-1}$ is again invertibly triangular. This proves Proposition
\ref{prop.STmat.triangular}(d).

(e) Let $A$ be a unitriangular $S\times S$-matrix. In other words,
$A\in\operatorname*{UT}\nolimits_{S}$ (since $\operatorname*{UT}\nolimits_{S}$
is the set of all unitriangular $S\times S$-matrices). Hence, Claim C3 (in our
proof of Proposition \ref{prop.STmat.triangular}(c)) shows that $A$ is
invertible, and that its inverse $A^{-1}$ belongs to $\operatorname*{UT}%
\nolimits_{S}$. The matrix $A^{-1}$ is therefore a unitriangular $S\times
S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs10} (applied to $A^{-1}$
instead of $A$)).

Now, forget that we fixed $A$. We thus have shown that if $A$ is a
unitriangular $S\times S$-matrix, then $A$ is invertible, and its inverse
$A^{-1}$ is again unitriangular. This proves Proposition
\ref{prop.STmat.triangular}(e).
\end{proof}

Thus, Exercise~\ref{exe.STmat.mat} is solved.

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.STmat.mat}}

\textit{Solution to Exercise~\ref{exe.STmat.mat}.} Let us first agree on a
convention: Whenever $S$ is a poset, we let $\leq$ denote the smaller-or-equal
relation of the poset $S$.

Before we prove Proposition~\ref{prop.STmat.triangular}, let us make a definition:

\begin{definition}
\label{def.sol.STmat.mat.T}Let $S$ be a poset.

\begin{enumerate}
\item[(a)] Let $\operatorname*{T}\nolimits_{S}$ denote the set of all
triangular $S\times S$-matrices. Clearly, $\operatorname*{T}\nolimits_{S}%
\subset\kk^{S\times S}$.

\item[(b)] Let $\operatorname*{IT}\nolimits_{S}$ denote the set of all
invertibly triangular $S\times S$-matrices.

\item[(c)] Let $\operatorname*{UT}\nolimits_{S}$ denote the set of all
unitriangular $S\times S$-matrices.
\end{enumerate}
\end{definition}

Now, we shall prove several lemmas (most of them completely trivial), which
will help us in the later proof of Proposition~\ref{prop.STmat.triangular}.

\begin{lemma}
\label{lem.sol.STmat.mats.obs0}Let $S$ and $T$ be two finite sets. Let
$A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$ be an
$S\times T$-matrix. Let $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
T\times S}$ be a $T\times S$-matrix. Then,
\[
AB=\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs0}.]Multiplying the equalities
$A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$ and
$B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in T\times S}$, we obtain
\begin{align*}
AB  &  =\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in T\times S}=\left(  \sum_{t\in
T}a_{s,t}b_{t,u}\right)  _{\left(  s,u\right)  \in S\times S}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of the product of an }S\times T\text{-matrix}\\
\text{with a }T\times S\text{-matrix}%
\end{array}
\right) \\
&  =\left(  \sum_{k\in T}a_{s,k}b_{k,u}\right)  _{\left(  s,u\right)  \in
S\times S}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}t\text{ as }k\right) \\
&  =\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the index }\left(
s,u\right)  \text{ as }\left(  s,t\right)  \right)  .
\end{align*}
This proves Lemma \ref{lem.sol.STmat.mats.obs0}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs1}Let $S$ be a poset. Let $A\in\operatorname*{T}%
\nolimits_{S}$. Then, $A$ is a triangular $S\times S$-matrix.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs1}.]We have $A\in\operatorname*{T}%
\nolimits_{S}$. In other words, $A$ belongs to $\operatorname*{T}%
\nolimits_{S}$. In other words, $A$ belongs to the set of all triangular
$S\times S$-matrices (since $\operatorname*{T}\nolimits_{S}$ is the set of all
triangular $S\times S$-matrices). In other words, $A$ is a triangular $S\times
S$-matrix. This proves Lemma \ref{lem.sol.STmat.mats.obs1}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs2}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be a triangular $S\times
S $-matrix. Then, every $\left(  s,t\right)  \in S\times S$ which does not
satisfy $t\leq s$ must satisfy $a_{s,t}=0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs2}.]We have $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Hence, the matrix $A$
is triangular if and only if every $\left(  s,t\right)  \in S\times S$ which
does not satisfy $t\leq s$ must satisfy $a_{s,t}=0$ (by the definition of
``triangular''). Hence,%
\[
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0
\]
(since the matrix $A$ is triangular). This proves Lemma
\ref{lem.sol.STmat.mats.obs2}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs3}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an $S\times
S$-matrix. Assume that every $\left(  s,t\right)  \in S\times S$ which does
not satisfy $t\leq s$ must satisfy $a_{s,t}=0$. Then, $A\in\operatorname*{T}%
\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs3}.]We have $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Hence, the matrix $A$
is triangular if and only if every $\left(  s,t\right)  \in S\times S$ which
does not satisfy $t\leq s$ must satisfy $a_{s,t}=0$ (by the definition of
``triangular''). Hence, the matrix $A$ is
triangular (since every $\left(  s,t\right)  \in S\times S$ which does not
satisfy $t\leq s$ must satisfy $a_{s,t}=0$). Thus, $A$ is a triangular
$S\times S$-matrix. In other words, $A$ belongs to the set of all triangular
$S\times S$-matrices. In other words, $A$ belongs to $\operatorname*{T}%
\nolimits_{S} $ (since $\operatorname*{T}\nolimits_{S}$ is the set of all
triangular $S\times S$-matrices). In other words, $A\in\operatorname*{T}%
\nolimits_{S}$. This proves Lemma \ref{lem.sol.STmat.mats.obs3}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs4}Let $S$ be a poset. Let $A\in\operatorname*{IT}%
\nolimits_{S}$. Then, $A$ is an invertibly triangular $S\times S$-matrix.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs4}.]We have $A\in\operatorname*{IT}%
\nolimits_{S}$. In other words, $A$ belongs to $\operatorname*{IT}%
\nolimits_{S}$. In other words, $A$ belongs to the set of all invertibly
triangular $S\times S$-matrices (since $\operatorname*{IT}\nolimits_{S}$ is
the set of all invertibly triangular $S\times S$-matrices). In other words,
$A$ is an invertibly triangular $S\times S$-matrix. This proves Lemma
\ref{lem.sol.STmat.mats.obs4}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs5}Let $S$ be a poset. Let $A$ be an invertibly
triangular $S\times S$-matrix. Then, $A\in\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs5}.]Write the $S\times S$-matrix
$A$ in the form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$.

We have $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.
Hence, the matrix $A$ is invertibly triangular if and only if $A$ is
triangular and has the further property that, for every $s\in S$, the element
$a_{s,s}$ of $\kk$ is invertible (by the definition of
``invertibly triangular''). From this, we
conclude that $A$ is triangular and has the further property that, for every
$s\in S$, the element $a_{s,s}$ of $\kk$ is invertible (because the
matrix $A$ is invertibly triangular). In particular, $A$ is triangular. In
other words, $A$ belongs to the set of all triangular $S\times S$-matrices. In
other words, $A$ belongs to $\operatorname*{T}\nolimits_{S}$ (since
$\operatorname*{T}\nolimits_{S}$ is the set of all triangular $S\times
S$-matrices). In other words, $A\in\operatorname*{T}\nolimits_{S}$. This
proves Lemma \ref{lem.sol.STmat.mats.obs5}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs6}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an invertibly
triangular $S\times S$-matrix. Then, for every $s\in S$, the element $a_{s,s}
$ of $\kk$ is invertible.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs6}.]We have $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Hence, the matrix $A$
is invertibly triangular if and only if $A$ is triangular and has the further
property that, for every $s\in S$, the element $a_{s,s}$ of $\kk$ is
invertible (by the definition of ``invertibly
triangular''). From this, we conclude that $A$ is triangular
and has the further property that, for every $s\in S$, the element $a_{s,s}$
of $\kk$ is invertible (because the matrix $A$ is invertibly
triangular). In particular, for every $s\in S$, the element $a_{s,s}$ of
$\kk$ is invertible. This proves Lemma \ref{lem.sol.STmat.mats.obs6}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs7}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an $S\times
S$-matrix. Assume that $A\in\operatorname*{T}\nolimits_{S}$. Assume that, for
every $s\in S$, the element $a_{s,s}$ of $\kk$ is invertible. Then,
$A\in\operatorname*{IT}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs7}.]Lemma
\ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times S
$-matrix. Thus, we know that $A$ is triangular and has the further property
that, for every $s\in S$, the element $a_{s,s}$ of $\kk$ is invertible.

But the matrix $A$ is invertibly triangular if and only if $A$ is triangular
and has the further property that, for every $s\in S$, the element $a_{s,s}$
of $\kk$ is invertible (by the definition of ``invertibly triangular'').
From this, we conclude that $A$ is
invertibly triangular (since $A$ is triangular and has the further property
that, for every $s\in S$, the element $a_{s,s}$ of $\kk$ is
invertible). In other words, $A$ belongs to the set of all invertibly
triangular $S\times S$-matrices. In other words, $A$ belongs to
$\operatorname*{IT}\nolimits_{S}$ (since $\operatorname*{IT}\nolimits_{S}$ is
the set of all invertibly triangular $S\times S$-matrices). In other words,
$A\in\operatorname*{IT}\nolimits_{S}$. This proves Lemma
\ref{lem.sol.STmat.mats.obs7}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs8}Let $S$ be a poset. We have $\operatorname*{IT}%
\nolimits_{S}\subset\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs8}.]Let $A\in\operatorname*{IT}%
\nolimits_{S}$. Thus, $A$ is an invertibly triangular $S\times S$-matrix (by
Lemma \ref{lem.sol.STmat.mats.obs4}). Hence, $A\in\operatorname*{T}%
\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs5}).

Now, forget that we fixed $A$. We thus have shown that every $A\in
\operatorname*{UT}\nolimits_{S}$ satisfies $A\in\operatorname*{T}%
\nolimits_{S}$. In other words, $\operatorname*{IT}\nolimits_{S}%
\subset\operatorname*{T}\nolimits_{S}$. This proves Lemma
\ref{lem.sol.STmat.mats.obs8}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs9}Let $S$ be a finite poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in\operatorname*{T}%
\nolimits_{S} $ and $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\operatorname*{T}\nolimits_{S}$. Then, for every $s\in S$, we
have
\begin{equation}
\sum_{k\in S}a_{s,k}b_{k,s}=a_{s,s}b_{s,s}.\nonumber
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs9}.]We have $A\in\operatorname*{T}%
\nolimits_{S}$. Hence, Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a
triangular $S\times S$-matrix. Lemma \ref{lem.sol.STmat.mats.obs2} thus shows
that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0.
\label{pf.prop.STmat.triangular.b.B2.pf.1}%
\end{equation}


We have $B\in\operatorname*{T}\nolimits_{S}$. Hence, Lemma
\ref{lem.sol.STmat.mats.obs1} shows that $B$ is a triangular $S\times
S$-matrix. Lemma \ref{lem.sol.STmat.mats.obs2} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) thus shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}=0.
\label{pf.prop.STmat.triangular.b.B2.pf.2}%
\end{equation}


Let $s\in S$. Every $k\in S$ satisfying $k\neq s$ must satisfy%
\begin{equation}
a_{s,k}b_{k,s}=0 \label{pf.prop.STmat.triangular.b.B2.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B2.pf.3}):} Let
$k\in S$ be such that $k\neq s$. We must prove
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $s\leq k$.
\par
\textit{Case 2:} We do not have $s\leq k$.
\par
Let us first consider Case 1. In this case, we have $s\leq k$. If we had
$k\leq s$, then we would have $k=s$ (since $k\leq s\leq k$); but this would
contradict $k\neq s$. Hence, we cannot have $k\leq s$. Thus,
(\ref{pf.prop.STmat.triangular.b.B2.pf.1}) (applied to $\left(  s,k\right)  $
instead of $\left(  s,t\right)  $) shows that $a_{s,k}=0$. Thus,
$\underbrace{a_{s,k}}_{=0}b_{k,s}=0$. Hence,
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $s\leq k$. Hence,
(\ref{pf.prop.STmat.triangular.b.B2.pf.2}) (applied to $\left(  k,s\right)  $
instead of $\left(  s,t\right)  $) shows that $b_{k,s}=0$. Thus,
$a_{s,k}\underbrace{b_{k,s}}_{=0}=0$. Hence,
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}) is proven in Case 2.
\par
Now, we have proven (\ref{pf.prop.STmat.triangular.b.B2.pf.3}) in each of the
two Cases 1 and 2. Since these two Cases cover all possibilities, this shows
that (\ref{pf.prop.STmat.triangular.b.B2.pf.3}) always holds. Qed.}. Now, for
every $s\in S$, we have%
\begin{align*}
\sum_{k\in S}a_{s,k}b_{k,s}  &  =a_{s,s}b_{s,s}+\sum\limits_{\substack{k\in S;
\\k\neq s}}\underbrace{a_{s,k}b_{k,s}}_{\substack{=0 \\\text{(by
(\ref{pf.prop.STmat.triangular.b.B2.pf.3}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=s\text{ from the sum}\right) \\
&  =a_{s,s}b_{s,s}+\underbrace{\sum\limits_{\substack{k\in S; \\k\neq s}}0}%
_{=0}=a_{s,s}b_{s,s}.
\end{align*}
This proves Lemma \ref{lem.sol.STmat.mats.obs9}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs10}Let $S$ be a poset. Let $A\in
\operatorname*{UT}\nolimits_{S}$. Then, $A$ is a unitriangular $S\times S$-matrix.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs10}.]We have $A\in
\operatorname*{UT}\nolimits_{S}$. In other words, $A$ belongs to
$\operatorname*{UT}\nolimits_{S}$. In other words, $A$ belongs to the set of
all unitriangular $S\times S$-matrices (since $\operatorname*{UT}%
\nolimits_{S}$ is the set of all unitriangular $S\times S$-matrices). In other
words, $A$ is a unitriangular $S\times S$-matrix. This proves Lemma
\ref{lem.sol.STmat.mats.obs10}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs11}Let $S$ be a poset. Let $A$ be a unitriangular
$S\times S$-matrix. Then, $A\in\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs11}.]Write the $S\times S$-matrix
$A$ in the form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$.

We have $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.
Hence, the matrix $A$ is unitriangular if and only if $A$ is triangular and
has the further property that, for every $s\in S$, we have $a_{s,s}=1$ (by the
definition of ``unitriangular''). From this,
we conclude that $A$ is triangular and has the further property that, for
every $s\in S$, we have $a_{s,s}=1$ (because the matrix $A$ is unitriangular).
In particular, $A$ is triangular. In other words, $A$ belongs to the set of
all triangular $S\times S$-matrices. In other words, $A $ belongs to
$\operatorname*{T}\nolimits_{S}$ (since $\operatorname*{T}\nolimits_{S}$ is
the set of all triangular $S\times S$-matrices). In other words,
$A\in\operatorname*{T}\nolimits_{S}$. This proves Lemma
\ref{lem.sol.STmat.mats.obs11}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs12}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be a unitriangular
$S\times S$-matrix. Then, for every $s\in S$, we have $a_{s,s}=1$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs12}.]We have $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Hence, the matrix $A$
is unitriangular if and only if $A$ is triangular and has the further property
that, for every $s\in S$, we have $a_{s,s}=1$ (by the definition of
``unitriangular''). From this, we conclude
that $A$ is triangular and has the further property that, for every $s\in S$,
we have $a_{s,s}=1$ (because the matrix $A$ is unitriangular). In particular,
for every $s\in S$, we have $a_{s,s}=1$. This proves Lemma
\ref{lem.sol.STmat.mats.obs12}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs13}Let $S$ be a poset. Let $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ be an $S\times
S$-matrix. Assume that $A\in\operatorname*{T}\nolimits_{S}$. Assume that, for
every $s\in S$, we have $a_{s,s}=1$. Then, $A\in\operatorname*{UT}%
\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs13}.]Lemma
\ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times
S$-matrix. Thus, we know that $A$ is triangular and has the further property
that, for every $s\in S$, we have $a_{s,s}=1$.

But the matrix $A$ is unitriangular if and only if $A$ is triangular and has
the further property that, for every $s\in S$, we have $a_{s,s}=1$ (by the
definition of ``unitriangular''). From this,
we conclude that $A$ is unitriangular (since $A$ is triangular and has the
further property that, for every $s\in S$, we have $a_{s,s}=1$). In other
words, $A$ belongs to the set of all unitriangular $S\times S$-matrices. In
other words, $A$ belongs to $\operatorname*{UT}\nolimits_{S}$ (since
$\operatorname*{UT}\nolimits_{S}$ is the set of all unitriangular $S\times
S$-matrices). In other words, $A\in\operatorname*{UT}\nolimits_{S}$. This
proves Lemma \ref{lem.sol.STmat.mats.obs13}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs14}Let $S$ be a poset. We have
$\operatorname*{UT}\nolimits_{S}\subset\operatorname*{T}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs14}.]Let $A\in\operatorname*{UT}%
\nolimits_{S}$. Thus, $A$ is unitriangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs10}). Hence, $A\in\operatorname*{T}\nolimits_{S}$
(by Lemma \ref{lem.sol.STmat.mats.obs11}).

Now, forget that we fixed $A$. We thus have shown that every $A\in
\operatorname*{UT}\nolimits_{S}$ satisfies $A\in\operatorname*{T}%
\nolimits_{S}$. In other words, $\operatorname*{UT}\nolimits_{S}%
\subset\operatorname*{T}\nolimits_{S}$. This proves Lemma
\ref{lem.sol.STmat.mats.obs14}.
\end{proof}

\begin{lemma}
\label{lem.sol.STmat.mats.obs15}We have $\operatorname*{UT}\nolimits_{S}%
\subset\operatorname*{IT}\nolimits_{S}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.sol.STmat.mats.obs15}.]Let $A\in\operatorname*{UT}%
\nolimits_{S}$. Thus, $A$ is unitriangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs10}). Also, $A\in\operatorname*{UT}\nolimits_{S}%
\subset\operatorname*{T}\nolimits_{S}$ (by Lemma
\ref{lem.sol.STmat.mats.obs14}).

Write the $S\times S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$. Hence, the matrix $A$ is unitriangular
if and only if $A$ is triangular and has the further property that, for every
$s\in S$, we have $a_{s,s}=1$ (by the definition of ``unitriangular'').
From this, we conclude that $A$ is
triangular and has the further property that, for every $s\in S$, we have
$a_{s,s}=1$ (because the matrix $A$ is unitriangular). For every $s\in S$, we
have $a_{s,s}=1$, and thus the element $a_{s,s}$ of $\kk$ is invertible
(since the element $1$ of $\kk$ is invertible). Hence, Lemma
\ref{lem.sol.STmat.mats.obs7} shows that $A\in\operatorname*{IT}\nolimits_{S}$.

Now, forget that we fixed $A$. We thus have shown that every $A\in
\operatorname*{UT}\nolimits_{S}$ satisfies $A\in\operatorname*{IT}%
\nolimits_{S}$. In other words, $\operatorname*{UT}\nolimits_{S}%
\subset\operatorname*{IT}\nolimits_{S}$. This proves Lemma
\ref{lem.sol.STmat.mats.obs15}.
\end{proof}

\begin{proof}
[Proof of Proposition~\ref{prop.STmat.triangular}.]We begin with some preparations.

For any $\left(  s,t\right)  \in S\times S$, we define a subset $\left[
t,s\right]  $ of $S$ by
\[
\left[  t,s\right]  =\left\{  q\in S\ \mid\ t\leq q\leq s\right\}  .
\]
This subset $\left[  t,s\right]  $ is called the \emph{interval of $S$
bounded by $t$ and $s$}. (Notice that it can be an empty set,
since we have not required that $t\leq s$.) The following facts hold:

\begin{itemize}
\item For any $\left(  s,t\right)  \in S\times S$, the set $\left[
t,s\right]  $ is finite (since it is a subset of the finite set $S$). Hence,
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for any }\left(  s,t\right)  \in S\times S\text{, the cardinality
}\left\vert \left[  t,s\right]  \right\vert \text{ is a}\\
\text{well-defined nonnegative integer}%
\end{array}
\right)  . \label{pf.prop.STmat.triangular.int0}%
\end{equation}


\item If $s$, $t$ and $u$ are three elements of $S$ such that $s\leq t<u$,
then%
\begin{equation}
\left[  s,t\right]  \text{ is a proper subset of }\left[  s,u\right]
\label{pf.prop.STmat.triangular.int1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.int1}):} Let $s$,
$t$ and $u$ be three elements of $S$ such that $s\leq t<u$.
\par
We have $\left[  s,t\right]  =\left\{  q\in S\ \mid\ s\leq q\leq t\right\}  $
(by the definition of $\left[  s,t\right]  $) and $\left[  s,u\right]
=\left\{  q\in S\ \mid\ s\leq q\leq u\right\}  $ (by the definition of
$\left[  s,u\right]  $).
\par
Now, let $r\in\left[  s,t\right]  $. Thus, $r\in\left[  s,t\right]  =\left\{
q\in S\ \mid\ s\leq q\leq t\right\}  $. In other words, $r$ is an element $q$
of $S$ satisfying $s\leq q\leq t$. In other words, $r$ is an element of $S$
and satisfies $s\leq r\leq t$. Now, $r\leq t<u$, so that $r\leq u$. Combining
this with $s\leq r$, we obtain $s\leq r\leq u$. Hence, $r$ is an element of
$S$ and satisfies $s\leq r\leq u$. In other words, $r$ is an element $q$ of
$S$ satisfying $s\leq q\leq u$. In other words, $r\in\left\{  q\in
S\ \mid\ s\leq q\leq u\right\}  $. This rewrites as $r\in\left[  s,u\right]  $
(since $\left[  s,u\right]  =\left\{  q\in S\ \mid\ s\leq q\leq u\right\}  $).
\par
Now, forget that we fixed $r$. We thus have proven that $r\in\left[
s,u\right]  $ for every $r\in\left[  s,t\right]  $. In other words, $\left[
s,t\right]  \subset\left[  s,u\right]  $.
\par
We have $s\leq u$ (since $s<u$). Hence, $u$ is an element of $S$ and satisfies
$s\leq u\leq u$. In other words, $u$ is an element $q$ of $S$ satisfying
$s\leq q\leq u$. In other words, $u\in\left\{  q\in S\ \mid\ s\leq q\leq
u\right\}  $. This rewrites as $u\in\left[  s,u\right]  $ (since $\left[
s,u\right]  =\left\{  q\in S\ \mid\ s\leq q\leq u\right\}  $).
\par
But let us assume (for the sake of contradiction) that $\left[  s,t\right]
=\left[  s,u\right]  $. Thus, $u\in\left[  s,u\right]  =\left[  s,t\right]
=\left\{  q\in S\ \mid\ s\leq q\leq t\right\}  $. In other words, $u$ is an
element $q$ of $S$ satisfying $s\leq q\leq t$. In other words, $u$ is an
element of $S$ and satisfies $s\leq u\leq t$. Now, $u\leq t$ contradicts $t<u
$ (since $S$ is a poset). This contradiction proves that our assumption (that
$\left[  s,t\right]  =\left[  s,u\right]  $) was false. Hence, we do not have
$\left[  s,t\right]  =\left[  s,u\right]  $. In other words, we have $\left[
s,t\right]  \neq\left[  s,u\right]  $. Combining this with $\left[
s,t\right]  \subset\left[  s,u\right]  $, we conclude that $\left[
s,t\right]  $ is a proper subset of $\left[  s,u\right]  $. This proves
(\ref{pf.prop.STmat.triangular.int1}).}.

\item If $s$, $t$ and $u$ are three elements of $S$ such that $s<t\leq u$,
then%
\begin{equation}
\left[  t,u\right]  \text{ is a proper subset of }\left[  s,u\right]
\label{pf.prop.STmat.triangular.int2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.int2}):} Let $s$,
$t$ and $u$ be three elements of $S$ such that $s<t\leq u$.
\par
We have $\left[  t,u\right]  =\left\{  q\in S\ \mid\ t\leq q\leq u\right\}  $
(by the definition of $\left[  t,u\right]  $) and $\left[  s,u\right]
=\left\{  q\in S\ \mid\ s\leq q\leq u\right\}  $ (by the definition of
$\left[  s,u\right]  $).
\par
Now, let $r\in\left[  t,u\right]  $. Thus, $r\in\left[  t,u\right]  =\left\{
q\in S\ \mid\ t\leq q\leq u\right\}  $. In other words, $r$ is an element $q$
of $S$ satisfying $t\leq q\leq u$. In other words, $r$ is an element of $S$
and satisfies $t\leq r\leq u$. Now, $s<t\leq r$, so that $s\leq r$. Combining
this with $r\leq u$, we obtain $s\leq r\leq u$. Hence, $r$ is an element of
$S$ and satisfies $s\leq r\leq u$. In other words, $r$ is an element $q$ of
$S$ satisfying $s\leq q\leq u$. In other words, $r\in\left\{  q\in
S\ \mid\ s\leq q\leq u\right\}  $. This rewrites as $r\in\left[  s,u\right]  $
(since $\left[  s,u\right]  =\left\{  q\in S\ \mid\ s\leq q\leq u\right\}  $).
\par
Now, forget that we fixed $r$. We thus have proven that $r\in\left[
s,u\right]  $ for every $r\in\left[  t,u\right]  $. In other words, $\left[
t,u\right]  \subset\left[  s,u\right]  $.
\par
We have $s\leq u$ (since $s<u$). Hence, $s$ is an element of $S$ and satisfies
$s\leq s\leq u$. In other words, $s$ is an element $q$ of $S$ satisfying
$s\leq s\leq u$. In other words, $s\in\left\{  q\in S\ \mid\ s\leq q\leq
u\right\}  $. This rewrites as $s\in\left[  s,u\right]  $ (since $\left[
s,u\right]  =\left\{  q\in S\ \mid\ s\leq q\leq u\right\}  $).
\par
But let us assume (for the sake of contradiction) that $\left[  t,u\right]
=\left[  s,u\right]  $. Thus, $s\in\left[  s,u\right]  =\left[  t,u\right]
=\left\{  q\in S\ \mid\ t\leq q\leq u\right\}  $. In other words, $s$ is an
element $q$ of $S$ satisfying $t\leq q\leq u$. In other words, $s$ is an
element of $S$ and satisfies $t\leq s\leq u$. Now, $t\leq s$ contradicts $s<t
$ (since $S$ is a poset). This contradiction proves that our assumption (that
$\left[  t,u\right]  =\left[  s,u\right]  $) was false. Hence, we do not have
$\left[  t,u\right]  =\left[  s,u\right]  $. In other words, we have $\left[
t,u\right]  \neq\left[  s,u\right]  $. Combining this with $\left[
t,u\right]  \subset\left[  s,u\right]  $, we conclude that $\left[
t,u\right]  $ is a proper subset of $\left[  s,u\right]  $. This proves
(\ref{pf.prop.STmat.triangular.int2}).}.
\end{itemize}

Let us now prove Proposition \ref{prop.STmat.triangular}.

(a) Let $0_{S\times S}\in\kk^{S\times S}$ be the matrix $\left(
0\right)  _{\left(  s,t\right)  \in S\times S}$. This matrix $0_{S\times S}$
is the zero of the $\kk$-algebra $\kk^{S\times S}$.

We shall now prove the following five claims:

\begin{statement}
\textit{Claim A1:} We have $0_{S\times S}\in\operatorname*{T}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A2:} For every $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$, we have $A+B\in\operatorname*{T}%
\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A3:} For every $A\in\operatorname*{T}\nolimits_{S}$ and
$u\in\kk$, we have $uA\in\operatorname*{T}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A4:} We have $I_{S}\in\operatorname*{T}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim A5:} For every $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$, we have $AB\in\operatorname*{T}%
\nolimits_{S}$.
\end{statement}

\textit{Proof of Claim A1:} The definition of $0_{S\times S}$ yields
$0_{S\times S}=\left(  0\right)  _{\left(  s,t\right)  \in S\times S}$.
Clearly, every $\left(  s,t\right)  \in S\times S$ which does not satisfy
$t\leq s$ must satisfy $0=0$. Thus, Lemma \ref{lem.sol.STmat.mats.obs3}
(applied to $0_{S\times S}$ and $0$ instead of $A$ and $a_{s,t}$) shows that
$0_{S\times S}\in\operatorname*{T}\nolimits_{S}$. This proves Claim A1.

\textit{Proof of Claim A2:} Let $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$.

Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times S
$-matrix (since $A\in\operatorname*{T}\nolimits_{S}$). Write the $S\times
S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs2} shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0.
\label{pf.prop.STmat.triangular.a.A2.pf.1}%
\end{equation}


Lemma \ref{lem.sol.STmat.mats.obs1} (applied to $B$ instead of $A$) shows that
$B$ is a triangular $S\times S$-matrix (since $B\in\operatorname*{T}%
\nolimits_{S}$). Write the $S\times S$-matrix $B$ in the form $B=\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Lemma
\ref{lem.sol.STmat.mats.obs2} (applied to $B$ and $b_{s,t}$ instead of $A$ and
$a_{s,t}$) shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}=0.
\label{pf.prop.STmat.triangular.a.A2.pf.2}%
\end{equation}


Adding the equalities $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$ and $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$, we obtain%
\[
A+B=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}+\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}=\left(  a_{s,t}%
+b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}%
\]
(by the definition of the sum of two $S\times S$-matrices). But every $\left(
s,t\right)  \in S\times S$ which does not satisfy $t\leq s$ must satisfy
$\underbrace{a_{s,t}}_{\substack{=0 \\\text{(by
(\ref{pf.prop.STmat.triangular.a.A2.pf.1}))}}}+\underbrace{b_{s,t}%
}_{\substack{=0 \\\text{(by (\ref{pf.prop.STmat.triangular.a.A2.pf.2}))}%
}}=0+0=0$. Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $A+B$ and
$a_{s,t}+b_{s,t}$ instead of $A$ and $a_{s,t}$) shows that $A+B\in
\operatorname*{T}\nolimits_{S}$ (since $A+B=\left(  a_{s,t}+b_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$). This proves Claim A2.

\textit{Proof of Claim A3:} Let $A\in\operatorname*{T}\nolimits_{S}$ and
$u\in\kk$.

Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times S
$-matrix (since $A\in\operatorname*{T}\nolimits_{S}$). Write the $S\times
S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs2} shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0.
\label{pf.prop.STmat.triangular.a.A3.pf.1}%
\end{equation}


Now,%
\[
u\underbrace{A}_{=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}}=u\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}=\left(
ua_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}%
\]
(by the definition of $u\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$). But every $\left(  s,t\right)  \in S\times S$ which does not
satisfy $t\leq s$ must satisfy $u\underbrace{a_{s,t}}_{\substack{=0
\\\text{(by (\ref{pf.prop.STmat.triangular.a.A3.pf.1}))}}}=0$. Hence, Lemma
\ref{lem.sol.STmat.mats.obs3} (applied to $uA$ and $ua_{s,t}$ instead of $A$
and $a_{s,t}$) shows that $uA\in\operatorname*{T}\nolimits_{S}$ (since
$uA=\left(  ua_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$). This
proves Claim A3.

\textit{Proof of Claim A4:} The definition of $I_{S}$ yields $I_{S}=\left(
\delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. But every $\left(
s,t\right)  \in S\times S$ which does not satisfy $t\leq s$ must satisfy
$\delta_{s,t}=0$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  s,t\right)  \in
S\times S$ be such that we do not have $t\leq s$. We must prove that
$\delta_{s,t}=0$.
\par
Assume (for the sake of contradiction) that $s=t$. Thus, $t=s\leq s$. This
contradicts the fact that we do not have $t\leq s$. This contradiction shows
that our assumption (that $s=t$) was wrong. Hence, we don't have $s=t$. In
other words, we have $s\neq t$. Hence, $\delta_{s,t}=0$. Qed.}. Hence, Lemma
\ref{lem.sol.STmat.mats.obs3} (applied to $I_{S}$ and $\delta_{s,t}$ instead
of $A$ and $a_{s,t}$) shows that $I_{S}\in\operatorname*{T}\nolimits_{S}$.
This proves Claim A4.

\textit{Proof of Claim A5:} Let $A\in\operatorname*{T}\nolimits_{S}$ and
$B\in\operatorname*{T}\nolimits_{S}$.

Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times S
$-matrix (since $A\in\operatorname*{T}\nolimits_{S}$). Write the $S\times
S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs2} shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0.
\label{pf.prop.STmat.triangular.a.A5.pf.1}%
\end{equation}


Lemma \ref{lem.sol.STmat.mats.obs1} (applied to $B$ instead of $A$) shows that
$B$ is a triangular $S\times S$-matrix (since $B\in\operatorname*{T}%
\nolimits_{S}$). Write the $S\times S$-matrix $B$ in the form $B=\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Lemma
\ref{lem.sol.STmat.mats.obs2} (applied to $B$ and $b_{s,t}$ instead of $A$ and
$a_{s,t}$) shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}=0.
\label{pf.prop.STmat.triangular.a.A5.pf.2}%
\end{equation}


Lemma \ref{lem.sol.STmat.mats.obs0} (applied to $T=S$) yields%
\[
AB=\left(  \sum_{k\in S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
But every $\left(  s,t\right)  \in S\times S$ and every $k\in S$ which do not
satisfy $t\leq s$ must satisfy%
\begin{equation}
a_{s,k}b_{k,t}=0 \label{pf.prop.STmat.triangular.a.A5.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.a.A5.pf.3}):} Let
$\left(  s,t\right)  \in S\times S$ and $k\in S$ be such that we do not have
$t\leq s$. We must prove the equality
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $t\leq k$.
\par
\textit{Case 2:} We do not have $t\leq k$.
\par
Let us first consider Case 1. In this case, we have $t\leq k$. If we had
$k\leq s$, then we would have $t\leq k\leq s$; but this would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $k\leq s$. Thus,
(\ref{pf.prop.STmat.triangular.a.A5.pf.1}) (applied to $\left(  s,k\right)  $
instead of $\left(  s,t\right)  $) shows that $a_{s,k}=0$. Thus,
$\underbrace{a_{s,k}}_{=0}b_{k,t}=0$. Hence,
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $t\leq k$. Hence,
(\ref{pf.prop.STmat.triangular.a.A5.pf.2}) (applied to $\left(  k,t\right)  $
instead of $\left(  s,t\right)  $) shows that $b_{k,t}=0$. Thus,
$a_{s,k}\underbrace{b_{k,t}}_{=0}=0$. Hence,
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}) is proven in Case 2.
\par
Now, we have proven (\ref{pf.prop.STmat.triangular.a.A5.pf.3}) in each of the
two Cases 1 and 2. Since these two Cases cover all possibilities, this shows
that (\ref{pf.prop.STmat.triangular.a.A5.pf.3}) always holds. Qed.}. Thus,
every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s $
must satisfy%
\[
\sum_{k\in S}\underbrace{a_{s,k}b_{k,t}}_{\substack{=0 \\\text{(by
(\ref{pf.prop.STmat.triangular.a.A5.pf.3}))}}}=\sum_{k\in S}0=0.
\]
Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $AB$ and $\sum_{k\in
S}a_{s,k}b_{k,t}$ instead of $A$ and $a_{s,t}$) shows that $AB\in
\operatorname*{T}\nolimits_{S}$ (since $AB=\left(  \sum_{k\in S}a_{s,k}%
b_{k,t}\right)  _{\left(  s,t\right)  \in S\times S}$). This proves Claim A5.

Recall that $\operatorname*{T}\nolimits_{S}$ is a subset of $\kk%
^{S\times S} $. This subset $\operatorname*{T}\nolimits_{S}$ is a $\kk%
$-submodule of $\kk^{S\times S}$ (because Claim A1, Claim A2 and Claim
A3 hold), and therefore is a $\kk$-subalgebra of $\kk^{S\times
S}$ (since Claim A4 and Claim A5 hold). In other words, the set of all
triangular $S\times S$-matrices is a $\kk$-subalgebra of $\kk%
^{S\times S}$ (since $\operatorname*{T}\nolimits_{S}$ is the set of all
triangular $S\times S$-matrices). In other words, the triangular $S\times
S$-matrices form a subalgebra of the $\kk$-algebra $\kk^{S\times
S}$. This proves Proposition \ref{prop.STmat.triangular}(a).

(b) We shall first prove the following claims:

\begin{statement}
\textit{Claim B1:} We have $I_{S}\in\operatorname*{IT}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim B2:} For every $A\in\operatorname*{IT}\nolimits_{S}$ and
$B\in\operatorname*{IT}\nolimits_{S}$, we have $AB\in\operatorname*{IT}%
\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim B3:} Let $A\in\operatorname*{IT}\nolimits_{S}$. Then, $A$ is
invertible (as an element of the ring $\kk^{S\times S}$), and its
inverse $A^{-1}$ belongs to $\operatorname*{IT}\nolimits_{S}$.
\end{statement}

\textit{Proof of Claim B1:} The definition of $I_{S}$ yields $I_{S}=\left(
\delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Claim A4
in our proof of Proposition \ref{prop.STmat.triangular}(a) yields
$I_{S}\in\operatorname*{T}\nolimits_{S}$. For every $s\in S$, we have
$\delta_{s,s}=1$ (since $s=s$). Hence, for every $s\in S$, the element
$\delta_{s,s}$ of $\kk$ is invertible (since $\delta_{s,s}=1$, which is
clearly invertible). Thus, Lemma \ref{lem.sol.STmat.mats.obs7} (applied to
$I_{S}$ and $\delta_{s,t}$ instead of $A$ and $a_{s,t}$) yields $I_{S}%
\in\operatorname*{IT}\nolimits_{S}$. This proves Claim B1.

\textit{Proof of Claim B2:} Let $A\in\operatorname*{IT}\nolimits_{S}$ and
$B\in\operatorname*{IT}\nolimits_{S}$.

We know that $A$ is an invertibly triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs4}). Hence, $A\in\operatorname*{T}\nolimits_{S}$
(by Lemma \ref{lem.sol.STmat.mats.obs5}). Hence, Lemma
\ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times
S$-matrix. Write the $S\times S$-matrix $A$ in the form $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.

We know that $B$ is an invertibly triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs4}, applied to $B$ instead of $A$). Hence,
$B\in\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs5},
applied to $B$ instead of $A$). Hence, Lemma \ref{lem.sol.STmat.mats.obs1}
(applied to $B$ instead of $A$) shows that $B$ is a triangular $S\times
S$-matrix. Write the $S\times S$-matrix $B$ in the form $B=\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.

From $A\in\operatorname*{T}\nolimits_{S}$ and $B\in\operatorname*{T}%
\nolimits_{S}$, we obtain $AB\in\operatorname*{T}\nolimits_{S}$ (by Claim A5
in our proof of Proposition \ref{prop.STmat.triangular}(a)).

Lemma \ref{lem.sol.STmat.mats.obs6} yields that%
\begin{equation}
\text{for every }s\in S\text{, the element }a_{s,s}\text{ of }\kk\text{
is invertible} \label{pf.prop.STmat.triangular.b.B2.pf.1a}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is an
invertibly triangular $S\times S$-matrix).

Lemma \ref{lem.sol.STmat.mats.obs6} (applied to $B$ and $b_{s,t}$ instead of
$A$ and $a_{s,t}$) yields that%
\begin{equation}
\text{for every }s\in S\text{, the element }b_{s,s}\text{ of }\kk\text{
is invertible} \label{pf.prop.STmat.triangular.b.B2.pf.2a}%
\end{equation}
(since $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is an
invertibly triangular $S\times S$-matrix).

Lemma \ref{lem.sol.STmat.mats.obs0} (applied to $T=S$) yields%
\[
AB=\left(  \sum_{k\in S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
Lemma \ref{lem.sol.STmat.mats.obs9} shows that, for every $s\in S$, we have
$\sum_{k\in S}a_{s,k}b_{k,s}=a_{s,s}b_{s,s}$. Hence, for every $s\in S$, the
element $\sum_{k\in S}a_{s,k}b_{k,s}$ of $\kk$ is
invertible\footnote{\textit{Proof.} Let $s\in S$. Then, the element $a_{s,s}$
of $\kk$ is invertible (by (\ref{pf.prop.STmat.triangular.b.B2.pf.1a}%
)). The element $b_{s,s}$ of $\kk$ is invertible as well (by
(\ref{pf.prop.STmat.triangular.b.B2.pf.2a})). Hence, $a_{s,s}$ and $b_{s,s}$
are two invertible elements of $\kk$. Thus, their product
$a_{s,s}b_{s,s}$ is also invertible (since the product of two invertible
elements is always invertible). In other words, $\sum_{k\in S}a_{s,k}b_{k,s}$
is invertible (since $\sum_{k\in S}a_{s,k}b_{k,s}=a_{s,s}b_{s,s}$). Qed.}.
Thus, Lemma \ref{lem.sol.STmat.mats.obs7} (applied to $AB$ and $\sum_{k\in
S}a_{s,k}b_{k,t}$ instead of $A$ and $a_{s,t}$) shows that $AB\in
\operatorname*{IT}\nolimits_{S}$ (since $AB=\left(  \sum_{k\in S}%
a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in S\times S}$ and
$AB\in\operatorname*{T}\nolimits_{S}$). This proves Claim B2.

\textit{Proof of Claim B3:} We know that $A$ is an invertibly triangular
$S\times S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs4}). Hence,
$A\in\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs5}).
Hence, Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular
$S\times S $-matrix. Write the $S\times S$-matrix $A$ in the form $A=\left(
a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Lemma
\ref{lem.sol.STmat.mats.obs2} shows that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0.
\label{pf.prop.STmat.triangular.b.B3.pf.1}%
\end{equation}
Lemma \ref{lem.sol.STmat.mats.obs6} yields that%
\begin{equation}
\text{for every }s\in S\text{, the element }a_{s,s}\text{ of }\kk\text{
is invertible} \label{pf.prop.STmat.triangular.b.B3.pf.1a}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is an
invertibly triangular $S\times S$-matrix).

We shall now define an element $b_{s,t}$ for each $\left(  s,t\right)  \in
S\times S$. In fact, we will define these elements recursively, by strong
induction on $\left\vert \left[  t,s\right]  \right\vert $%
\ \ \ \ \footnote{Here, we are using the fact that $\left\vert \left[
t,s\right]  \right\vert $ is a nonnegative integer for every $\left(
s,t\right)  \in S\times S$. This fact has been proven above (see
(\ref{pf.prop.STmat.triangular.int0})).}: Let $N\in \NN$. Assume that
\begin{equation}
\text{an element }b_{s,t}\in\kk\text{ is already defined for each
}\left(  s,t\right)  \in S\times S\text{ satisfying }\left\vert \left[
t,s\right]  \right\vert <N. \label{pf.prop.STmat.triangular.b.B3.pf.def.ihyp}%
\end{equation}
We shall now define an element $b_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$ satisfying $\left\vert \left[  t,s\right]
\right\vert =N$.

Indeed, let $\left(  s,t\right)  \in S\times S$ be such that $\left\vert
\left[  t,s\right]  \right\vert =N$. We must define an element $b_{s,t}%
\in\kk$.

For every $u\in S$ satisfying $t<u\leq s$, the element $b_{s,u}$ is already
defined\footnote{\textit{Proof.} Let $u\in S$ be such that $t<u\leq s$. We
have to show that the element $b_{s,u}$ is already defined.
\par
From (\ref{pf.prop.STmat.triangular.int2}) (applied to $t$, $u$ and $s$
instead of $s$, $t$ and $u$), we conclude that $\left[  u,s\right]  $ is a
proper subset of $\left[  t,s\right]  $. Hence, $\left\vert \left[
u,s\right]  \right\vert <\left\vert \left[  t,s\right]  \right\vert $ (since
$\left[  t,s\right]  $ is a finite set). Thus, $\left\vert \left[  u,s\right]
\right\vert <\left\vert \left[  t,s\right]  \right\vert =N$. Therefore,
(\ref{pf.prop.STmat.triangular.b.B3.pf.def.ihyp}) (applied to $\left(
s,u\right)  $ instead of $\left(  s,t\right)  $) shows that an element
$b_{s,u}$ is already defined. Qed.}. Thus, the sum $\sum\limits_{\substack{u\in S;
\\t<u\leq s}}b_{s,u}a_{u,t}\in\kk$ is well-defined. Furthermore, the
element $a_{t,t}$ of $\kk$ is invertible (by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1a}), applied to $t$ instead of $s$).
Hence, the element $\left(  a_{t,t}\right)  ^{-1}$ of $\kk$ is
well-defined. Now, we set%
\begin{equation}
b_{s,t}=\left(  a_{t,t}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in
S; \\t<u\leq s}}b_{s,u}a_{u,t}\right)
\label{pf.prop.STmat.triangular.b.B3.pf.def.rec}%
\end{equation}
(this makes sense since both $\left(  a_{t,t}\right)  ^{-1}$ and
$\sum\limits_{\substack{u\in S; \\t<u\leq s}}b_{s,u}a_{u,t}$ are well-defined). Thus,
we have defined an element $b_{s,t}\in\kk$. This completes the
recursive definition of $b_{s,t}$.

We have now defined an element $b_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$. In other words, we have defined a family $\left(
b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in\kk^{S\times S}%
$. This family is clearly an $S\times S$-matrix. Denote this $S\times
S$-matrix by $B$. Thus, $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$. We shall now show (in several steps)
that $BA=I_{S}$ and that $B\in\operatorname*{IT}\nolimits_{S}$.

First, we notice that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }b_{s,t}=0
\label{pf.prop.STmat.triangular.b.B3.pf.B-tri}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}):}
Let $\left(  s,t\right)  \in S\times S$ be such that we do not have $t\leq s$.
We must prove that $b_{s,t}=0$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $s=t$. In other
words, we have $s\neq t$. Hence, $\delta_{s,t}=0$.
\par
If there was an $u\in S$ satisfying $t<u\leq s$, then we would have $t\leq s$
(since $t<u\leq s$, thus $t<s$, thus $t\leq s$), which would contradict the
fact that we do not have $t\leq s$. Hence, there is no $u\in S$ satisfying
$t<u\leq s$. Thus, the sum $\sum\limits_{\substack{u\in S; \\t<u\leq s}%
}b_{s,u}a_{u,t}$ is empty. Hence, $\sum\limits_{\substack{u\in S; \\t<u\leq
s}}b_{s,u}a_{u,t}=\left(  \text{empty sum}\right)  =0$.
\par
Now, the recursive definition of $b_{s,t}$ yields $b_{s,t}=\left(
a_{t,t}\right)  ^{-1}\left(  \underbrace{\delta_{s,t}}_{=0}-\underbrace{\sum\limits
_{\substack{u\in S; \\t<u\leq s}}b_{s,u}a_{u,t}}_{=0}\right)  =\left(
a_{t,t}\right)  ^{-1}0=0$. This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}).}. Hence, Lemma
\ref{lem.sol.STmat.mats.obs3} (applied to $B$ and $b_{s,t}$ instead of $A$ and
$a_{s,t}$) yields $B\in\operatorname*{T}\nolimits_{S}$.

For every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}=\delta_{s,t}
\label{pf.prop.STmat.triangular.b.B3.pf.BA.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}):} Let
$\left(  s,t\right)  \in S\times S$. We must prove the equality
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $t\leq s$.
\par
\textit{Case 2:} We do not have $t\leq s$.
\par
Let us first consider Case 1. In this case, we have $t\leq s$. Thus, $t$ is an
element of $S$ satisfying $t\leq t\leq s$. Hence, the sum $\sum\limits
_{\substack{u\in S;\\t\leq u\leq s}}b_{s,u}a_{u,t}$ has an addend for $u=t$.
Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{u\in S;\\t\leq u\leq s}}b_{s,u}a_{u,t}=\sum\limits_{\substack{u\in
S;\\t\leq u\leq s\text{ and }u\neq t}}b_{s,u}a_{u,t}+b_{s,t}a_{t,t}.
\]
\par
Now,%
\begin{align*}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}  &  =\sum\limits
_{\substack{u\in S;\\t\leq u\leq s}}b_{s,u}a_{u,t}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have renamed the summation index }k\text{ as }u\right) \\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\t\leq u\leq s\text{ and }u\neq t}%
}}_{\substack{=\sum\limits_{\substack{u\in S;\\t\leq u\text{ and }u\leq s\text{ and
}u\neq t}}\\\text{(since for any }u\in S\text{, the condition}\\\left(  t\leq
u\leq s\right)  \text{ is equivalent to}\\\left(  t\leq u\text{ and }u\leq
s\right)  \text{)}}}b_{s,u}a_{u,t}+\underbrace{b_{s,t}}_{\substack{=\left(
a_{t,t}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq
s}}b_{s,u}a_{u,t}\right)  \\\text{(by the recursive definition of }%
b_{s,t}\text{)}}}a_{t,t}\\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\t\leq u\text{ and }u\leq s\text{ and
}u\neq t}}}_{\substack{=\sum\limits_{\substack{u\in S;\\u\leq s\text{ and }t\leq
u\text{ and }u\neq t}}\\=\sum\limits_{\substack{u\in S;\\u\leq s\text{ and }t\leq
u\text{ and }t\neq u}}\\\text{(since for any }u\in S\text{, the condition}%
\\\left(  u\neq t\right)  \text{ is equivalent to }\left(  t\neq u\right)
\text{)}}}b_{s,u}a_{u,t}+\underbrace{\left(  a_{t,t}\right)  ^{-1}\left(
\delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}\right)
a_{t,t}}_{=\delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}}\\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\u\leq s\text{ and }t\leq u\text{ and
}t\neq u}}}_{\substack{=\sum\limits_{\substack{u\in S;\\u\leq s\text{ and }%
t<u}}\\\text{(since for any }u\in S\text{, the condition}\\\left(  t\leq
u\text{ and }t\neq u\right)  \text{ is equivalent to }\left(  t<u\right)
\text{)}}}b_{s,u}a_{u,t}+\delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq
s}}b_{s,u}a_{u,t}\\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\u\leq s\text{ and }t<u}%
}}_{\substack{=\sum\limits_{\substack{u\in S;\\t<u\leq s}}\\\text{(since for any
}u\in S\text{, the condition}\\\left(  u\leq s\text{ and }t<u\right)  \text{
is equivalent to }\left(  t<u\leq s\right)  \text{)}}}b_{s,u}a_{u,t}%
+\delta_{s,t}-\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}\\
&  =\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}+\delta_{s,t}%
-\sum\limits_{\substack{u\in S;\\t<u\leq s}}b_{s,u}a_{u,t}=\delta_{s,t}.
\end{align*}
Hence, (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $t\leq s$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we do not have $s=t$. In other
words, we have $s\neq t$. Hence, $\delta_{s,t}=0$.
\par
If there was an $k\in S$ satisfying $t\leq k\leq s$, then we would have $t\leq
s$ (since $t\leq k\leq s$), which would contradict the fact that we do not
have $t\leq s$. Hence, there is no $k\in S$ satisfying $t\leq k\leq s$. Thus,
the sum $\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}$ is empty.
Hence, $\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}=\left(
\text{empty sum}\right)  =0=\delta_{s,t}$ (since $\delta_{s,t}=0$). Hence,
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) is proven in Case 2.
\par
We have now proven (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, this
shows that (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1}) always holds.}. Thus,
for every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum_{k\in S}b_{s,k}a_{k,t}=\delta_{s,t}
\label{pf.prop.STmat.triangular.b.B3.pf.BA.1a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}):}
Let $\left(  s,t\right)  \in S\times S$. Every $k\in S$ satisfies either
$t\leq k$ or $\left(  \text{not }t\leq k\right)  $ (but not both). Hence,%
\begin{align*}
\sum_{k\in S}b_{s,k}a_{k,t}  &  =\sum\limits_{\substack{k\in S;\\t\leq k}%
}b_{s,k}a_{k,t}+\sum\limits_{\substack{k\in S;\\\text{not }t\leq k}}b_{s,k}%
\underbrace{a_{k,t}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1}), applied to}\\\left(  k,t\right)
\text{ instead of }\left(  s,t\right)  \text{)}}}=\sum\limits_{\substack{k\in
S;\\t\leq k}}b_{s,k}a_{k,t}+\underbrace{\sum\limits_{\substack{k\in S;\\\text{not
}t\leq k}}b_{s,k}0}_{=0}\\
&  =\sum\limits_{\substack{k\in S;\\t\leq k}}b_{s,k}a_{k,t}=\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and }k\leq s}}}_{\substack{=\sum\limits
_{\substack{k\in S;\\t\leq k\leq s}}\\\text{(since for any }k\in S\text{,
the}\\\text{condition }\left(  t\leq k\text{ and }k\leq s\right)  \text{
is}\\\text{equivalent to }\left(  t\leq k\leq s\right)  \text{)}}%
}b_{s,k}a_{k,t}+\sum\limits_{\substack{k\in S;\\t\leq k\text{ and not }k\leq
s}}\underbrace{b_{s,k}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B3.pf.B-tri}), applied to}\\\left(
s,k\right)  \text{ instead of }\left(  s,t\right)  \text{)}}}a_{k,t}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }k\in S\text{ satisfies
either }k\leq s\text{ or }\left(  \text{not }k\leq s\right)  \text{ (but not
both)}\right) \\
&  =\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}+\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and not }k\leq s}}0a_{k,t}}_{=0}%
=\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}b_{s,k}a_{k,t}=\delta_{s,t}%
\end{align*}
(by (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1})). This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}).}. Hence, $BA=I_{S}%
$\ \ \ \ \footnote{\textit{Proof.} Lemma \ref{lem.sol.STmat.mats.obs0}
(applied to $S$, $B$, $b_{s,t}$, $A$ and $a_{s,t}$ instead of $T$, $A$,
$a_{s,t}$, $B$ and $b_{s,t}$) shows that%
\[
BA=\left(  \underbrace{\sum_{k\in S}b_{s,k}a_{k,t}}_{\substack{=\delta
_{s,t}\\\text{(by (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}))}}}\right)
_{\left(  s,t\right)  \in S\times S}=\left(  \delta_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}=I_{S}%
\]
(since $I_{S}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$ (by the definition of $I_{S}$)). Qed.}. Furthermore, for every $s\in S$,
we have%
\begin{equation}
b_{s,s}=\left(  a_{s,s}\right)  ^{-1}
\label{pf.prop.STmat.triangular.b.B3.pf.BA.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.2}):}
Lemma \ref{lem.sol.STmat.mats.obs9} (applied to $B$, $b_{s,t}$, $A$ and
$a_{s,t}$ instead of $A$, $a_{s,t}$, $B$ and $b_{s,t}$) yields that for every
$s\in S$, we have
\begin{equation}
\sum_{k\in S}b_{s,k}a_{k,s}=b_{s,s}a_{s,s}.
\label{pf.prop.STmat.triangular.b.B3.pf.BA.2.pf.1}%
\end{equation}
\par
Let $s\in S$. Then, (\ref{pf.prop.STmat.triangular.b.B3.pf.BA.1a}) (applied to
$\left(  s,s\right)  $ instead of $\left(  s,t\right)  $) yields $\sum_{k\in
S}b_{s,k}a_{k,s}=\delta_{s,s}=1$ (since $s=s$). Comparing this with
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.2.pf.1}), we obtain $b_{s,s}%
a_{s,s}=1$. Hence, $b_{s,s}=\left(  a_{s,s}\right)  ^{-1}$. This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.BA.2}).}. Hence,
\[
\text{for every }s\in S\text{, the element }b_{s,s}\text{ of }\kk\text{
is invertible}%
\]
(because $b_{s,s}$ is an inverse (namely, $b_{s,s}=\left(  a_{s,s}\right)
^{-1}$)). Thus, Lemma \ref{lem.sol.STmat.mats.obs7} (applied to $B$ and
$b_{s,t}$ instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{IT}%
\nolimits_{S}$ (since $B\in\operatorname*{T}\nolimits_{S}$). In other words,
$B$ belongs to $\operatorname*{IT}\nolimits_{S}$.

So far, we do not know that $B$ is the inverse of $A$; we merely know that $B
$ is a left inverse of $A$ (that is, we know that $BA=I_{S}$). We shall now
construct yet another $S\times S$-matrix $C$ and subsequently show that
$AC=I_{S}$; this will easily yield that $A$ is invertible (because an element
of a ring that has a left inverse and a right inverse must be invertible) and
its inverse is $C=B$.

We shall define an element $c_{s,t}$ for each $\left(  s,t\right)  \in S\times
S$. In fact, we will define these elements recursively, by strong induction on
$\left\vert \left[  t,s\right]  \right\vert $\ \ \ \ \footnote{Here, we are
using the fact that $\left\vert \left[  t,s\right]  \right\vert $ is a
nonnegative integer for every $\left(  s,t\right)  \in S\times S$. This fact
has been proven above (see (\ref{pf.prop.STmat.triangular.int0})).}: Let
$N\in \NN$. Assume that
\begin{equation}
\text{an element }c_{s,t}\in\kk\text{ is already defined for each
}\left(  s,t\right)  \in S\times S\text{ satisfying }\left\vert \left[
t,s\right]  \right\vert <N. \label{pf.prop.STmat.triangular.b.B3.pf.def2.ihyp}%
\end{equation}
We shall now define an element $c_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$ satisfying $\left\vert \left[  t,s\right]
\right\vert =N$.

Indeed, let $\left(  s,t\right)  \in S\times S$ be such that $\left\vert
\left[  t,s\right]  \right\vert =N$. We must define an element $c_{s,t}%
\in\kk$.

For every $u\in S$ satisfying $t\leq u<s$, the element $c_{u,t}$ is already
defined\footnote{\textit{Proof.} Let $u\in S$ be such that $t\leq u<s$. We
have to show that the element $c_{u,t}$ is already defined.
\par
From (\ref{pf.prop.STmat.triangular.int1}) (applied to $t$, $u$ and $s$
instead of $s$, $t$ and $u$), we conclude that $\left[  t,u\right]  $ is a
proper subset of $\left[  t,s\right]  $. Hence, $\left\vert \left[
t,u\right]  \right\vert <\left\vert \left[  t,s\right]  \right\vert $ (since
$\left[  t,s\right]  $ is a finite set). Thus, $\left\vert \left[  t,u\right]
\right\vert <\left\vert \left[  t,s\right]  \right\vert =N$. Therefore,
(\ref{pf.prop.STmat.triangular.b.B3.pf.def2.ihyp}) (applied to $\left(
u,t\right)  $ instead of $\left(  s,t\right)  $) shows that an element
$c_{u,t}$ is already defined. Qed.}. Thus, the sum $\sum\limits_{\substack{u\in S;
\\t\leq u<s }}a_{s,u}c_{u,t}\in\kk$ is well-defined. Furthermore, the
element $a_{s,s}$ of $\kk$ is invertible (by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1a})). Hence, the element $\left(
a_{s,s}\right)  ^{-1}$ of $\kk$ is well-defined. Now, we set%
\begin{equation}
c_{s,t}=\left(  a_{s,s}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in
S; \\t\leq u<s}}a_{s,u}c_{u,t}\right)
\label{pf.prop.STmat.triangular.b.B3.pf.def2.rec}%
\end{equation}
(this makes sense since both $\left(  a_{s,s}\right)  ^{-1}$ and
$\sum\limits_{\substack{u\in S; \\t\leq u<s}}a_{s,u}c_{u,t}$ are well-defined). Thus,
we have defined an element $c_{s,t}\in\kk$. This completes the
recursive definition of $c_{s,t}$.

We have now defined an element $c_{s,t}\in\kk$ for each $\left(
s,t\right)  \in S\times S$. In other words, we have defined a family $\left(
c_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in\kk^{S\times S}%
$. This family is clearly an $S\times S$-matrix. Denote this $S\times
S$-matrix by $C$. Thus, $C=\left(  c_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$. We shall now show (in several steps)
that $AC=I_{S}$ and that $C\in\operatorname*{IT}\nolimits_{S}$.

First, we notice that%
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }c_{s,t}=0
\label{pf.prop.STmat.triangular.b.B3.pf.C-tri}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.C-tri}):}
Let $\left(  s,t\right)  \in S\times S$ be such that we do not have $t\leq s$.
We must prove that $c_{s,t}=0$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $s=t$. In other
words, we have $s\neq t$. Hence, $\delta_{s,t}=0$.
\par
If there was an $u\in S$ satisfying $t\leq u<s$, then we would have $t\leq s$
(since $t\leq u<s$, thus $t<s$, thus $t\leq s$), which would contradict the
fact that we do not have $t\leq s$. Hence, there is no $u\in S$ satisfying
$t\leq u<s$. Thus, the sum $\sum\limits_{\substack{u\in S; \\t\leq u<s}%
}a_{s,u}c_{u,t}$ is empty. Hence, $\sum\limits_{\substack{u\in S; \\t\leq
u<s}}a_{s,u}c_{u,t}=\left(  \text{empty sum}\right)  =0$.
\par
Now, the recursive definition of $c_{s,t}$ yields $c_{s,t}=\left(
a_{s,s}\right)  ^{-1}\left(  \underbrace{\delta_{s,t}}_{=0}-\underbrace{\sum\limits
_{\substack{u\in S; \\t\leq u<s}}a_{s,u}c_{u,t}}_{=0}\right)  =\left(
a_{s,s}\right)  ^{-1}0=0$. This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.C-tri}).}. Hence, Lemma
\ref{lem.sol.STmat.mats.obs3} (applied to $C$ and $c_{s,t}$ instead of $A$ and
$a_{s,t}$) yields $C\in\operatorname*{T}\nolimits_{S}$.

For every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}=\delta_{s,t}
\label{pf.prop.STmat.triangular.b.B3.pf.AC.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}):} Let
$\left(  s,t\right)  \in S\times S$. We must prove the equality
(\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}).
\par
We are in one of the following two cases:
\par
\textit{Case 1:} We have $t\leq s$.
\par
\textit{Case 2:} We do not have $t\leq s$.
\par
Let us first consider Case 1. In this case, we have $t\leq s$. Thus, $s$ is an
element of $S$ satisfying $t\leq s\leq s$. Hence, the sum $\sum\limits
_{\substack{u\in S;\\t\leq u\leq s}}a_{s,u}c_{u,t}$ has an addend for $u=s$.
Splitting off this addend, we obtain%
\[
\sum\limits_{\substack{u\in S;\\t\leq u\leq s}}a_{s,u}c_{u,t}=\sum\limits_{\substack{u\in
S;\\t\leq u\leq s\text{ and }u\neq s}}a_{s,u}c_{u,t}+a_{s,s}c_{s,t}.
\]
\par
Now,%
\begin{align*}
\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}  &  =\sum\limits
_{\substack{u\in S;\\t\leq u\leq s}}a_{s,u}c_{u,t}\ \ \ \ \ \ \ \ \ \ \left(
\text{here, we have renamed the summation index }k\text{ as }u\right) \\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\t\leq u\leq s\text{ and }u\neq s}%
}}_{\substack{=\sum\limits_{\substack{u\in S;\\t\leq u\text{ and }u\leq s\text{ and
}u\neq s}}\\\text{(since for any }u\in S\text{, the condition}\\\left(  t\leq
u\leq s\right)  \text{ is equivalent to}\\\left(  t\leq u\text{ and }u\leq
s\right)  \text{)}}}a_{s,u}c_{u,t}+a_{s,s}\underbrace{c_{s,t}}%
_{\substack{=\left(  a_{s,s}\right)  ^{-1}\left(  \delta_{s,t}-\sum\limits
_{\substack{u\in S;\\t\leq u<s}}a_{s,u}c_{u,t}\right)  \\\text{(by the
recursive definition of }c_{s,t}\text{)}}}\\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\t\leq u\text{ and }u\leq s\text{ and
}u\neq s}}}_{\substack{=\sum\limits_{\substack{u\in S;\\t\leq u\text{ and }%
u<s}}\\\text{(since for any }u\in S\text{, the condition}\\\left(  u\leq
s\text{ and }u\neq s\right)  \text{ is equivalent to }\left(  u<s\right)
\text{)}}}a_{s,u}c_{u,t}+\underbrace{a_{s,s}\left(  a_{s,s}\right)
^{-1}\left(  \delta_{s,t}-\sum\limits_{\substack{u\in S;\\t\leq u<s}}a_{s,u}%
c_{u,t}\right)  }_{=\delta_{s,t}-\sum\limits_{\substack{u\in S;\\t\leq u<s}%
}a_{s,u}c_{u,t}}\\
&  =\underbrace{\sum\limits_{\substack{u\in S;\\t\leq u\text{ and }u<s}%
}}_{\substack{=\sum\limits_{\substack{u\in S;\\t\leq u<s}}\\\text{(since for any
}u\in S\text{, the condition}\\\left(  t\leq u\text{ and }u<s\right)  \text{
is equivalent to }\left(  t\leq u<s\right)  \text{)}}}a_{s,u}c_{u,t}%
+\delta_{s,t}-\sum\limits_{\substack{u\in S;\\t\leq u<s}}a_{s,u}c_{u,t}\\
&  =\sum\limits_{\substack{u\in S;\\t\leq u<s}}a_{s,u}c_{u,t}+\delta_{s,t}%
-\sum\limits_{\substack{u\in S;\\t\leq u<s}}a_{s,u}c_{u,t}=\delta_{s,t}.
\end{align*}
Hence, (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}) is proven in Case 1.
\par
Let us now consider Case 2. In this case, we do not have $t\leq s$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we do not have $s=t$. In other
words, we have $s\neq t$. Hence, $\delta_{s,t}=0$.
\par
If there was an $k\in S$ satisfying $t\leq k\leq s$, then we would have $t\leq
s$ (since $t\leq k\leq s$), which would contradict the fact that we do not
have $t\leq s$. Hence, there is no $k\in S$ satisfying $t\leq k\leq s$. Thus,
the sum $\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}$ is empty.
Hence, $\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}=\left(
\text{empty sum}\right)  =0=\delta_{s,t}$ (since $\delta_{s,t}=0$). Hence,
(\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}) is proven in Case 2.
\par
We have now proven (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}) in each of
the two Cases 1 and 2. Since these two Cases cover all possibilities, this
shows that (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1}) always holds.}. Thus,
for every $\left(  s,t\right)  \in S\times S$, we have%
\begin{equation}
\sum_{k\in S}a_{s,k}c_{k,t}=\delta_{s,t}
\label{pf.prop.STmat.triangular.b.B3.pf.AC.1a}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1a}):}
Let $\left(  s,t\right)  \in S\times S$. Every $k\in S$ satisfies either
$t\leq k$ or $\left(  \text{not }t\leq k\right)  $ (but not both). Hence,%
\begin{align*}
\sum_{k\in S}a_{s,k}c_{k,t}  &  =\sum\limits_{\substack{k\in S;\\t\leq k}%
}a_{s,k}c_{k,t}+\sum\limits_{\substack{k\in S;\\\text{not }t\leq k}}a_{s,k}%
\underbrace{c_{k,t}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B3.pf.C-tri}), applied to}\\\left(
k,t\right)  \text{ instead of }\left(  s,t\right)  \text{)}}}=\sum\limits
_{\substack{k\in S;\\t\leq k}}a_{s,k}c_{k,t}+\underbrace{\sum\limits_{\substack{k\in
S;\\\text{not }t\leq k}}a_{s,k}0}_{=0}\\
&  =\sum\limits_{\substack{k\in S;\\t\leq k}}a_{s,k}c_{k,t}=\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and }k\leq s}}}_{\substack{=\sum\limits
_{\substack{k\in S;\\t\leq k\leq s}}\\\text{(since for any }k\in S\text{,
the}\\\text{condition }\left(  t\leq k\text{ and }k\leq s\right)  \text{
is}\\\text{equivalent to }\left(  t\leq k\leq s\right)  \text{)}}%
}a_{s,k}c_{k,t}+\sum\limits_{\substack{k\in S;\\t\leq k\text{ and not }k\leq
s}}\underbrace{a_{s,k}}_{\substack{=0\\\text{(by
(\ref{pf.prop.STmat.triangular.b.B3.pf.1}), applied to}\\\left(  s,k\right)
\text{ instead of }\left(  s,t\right)  \text{)}}}c_{k,t}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since every }k\in S\text{ satisfies
either }k\leq s\text{ or }\left(  \text{not }k\leq s\right)  \text{ (but not
both)}\right) \\
&  =\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}+\underbrace{\sum\limits
_{\substack{k\in S;\\t\leq k\text{ and not }k\leq s}}0c_{k,t}}_{=0}%
=\sum\limits_{\substack{k\in S;\\t\leq k\leq s}}a_{s,k}c_{k,t}=\delta_{s,t}%
\end{align*}
(by (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1})). This proves
(\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1a}).}. Hence, $AC=I_{S}%
$\ \ \ \ \footnote{\textit{Proof.} Lemma \ref{lem.sol.STmat.mats.obs0}
(applied to $S$, $C$ and $c_{s,t}$ instead of $T$, $B$ and $b_{s,t}$) shows
that%
\[
AC=\left(  \underbrace{\sum_{k\in S}a_{s,k}c_{k,t}}_{\substack{=\delta
_{s,t}\\\text{(by (\ref{pf.prop.STmat.triangular.b.B3.pf.AC.1a}))}}}\right)
_{\left(  s,t\right)  \in S\times S}=\left(  \delta_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}=I_{S}%
\]
(since $I_{S}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times
S}$ (by the definition of $I_{S}$)). Qed.}. Now, using the associativity of
the $\kk$-algebra $\kk^{S\times S}$, we can make the following
computation:%
\[
B=B\underbrace{I_{S}}_{=AC}=\underbrace{BA}_{=I_{S}}C=I_{S}C=C.
\]
Hence, $AC=I_{S}$ rewrites as $AB=I_{S}$. Combining this with $BA=I_{S}$, we
see that $B$ is an inverse of $A$ in the ring $\kk^{S\times S}$. In
particular, the element $A$ of $\kk^{S\times S}$ is invertible. Its
inverse $A^{-1}$ is $B$ (as we have just proven), and therefore belongs to
$\operatorname*{IT}\nolimits_{S}$ (since we know that $B$ belongs to
$\operatorname*{IT}\nolimits_{S}$). This proves Claim B3.

Recall that $\operatorname*{IT}\nolimits_{S}$ is a subset of $\kk%
^{S\times S}$. Claim B1 and Claim B2 (combined) show that this set
$\operatorname*{IT}\nolimits_{S}$ is a submonoid of the multiplicative monoid
of $\kk^{S\times S}$. Claim B3 furthermore proves that this submonoid
is a group. Thus, $\operatorname*{IT}\nolimits_{S}$ is a group with respect to
multiplication. In other words, the set of all invertibly triangular $S\times
S$-matrices is a group with respect to multiplication (since
$\operatorname*{IT}\nolimits_{S}$ is the set of all invertibly triangular
$S\times S$-matrices). In other words, the invertibly triangular $S\times
S$-matrices form a group with respect to multiplication. This proves
Proposition \ref{prop.STmat.triangular}(b).

(c) We shall first prove the following claims:

\begin{statement}
\textit{Claim C1:} We have $I_{S}\in\operatorname*{UT}\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim C2:} For every $A\in\operatorname*{UT}\nolimits_{S}$ and
$B\in\operatorname*{UT}\nolimits_{S}$, we have $AB\in\operatorname*{UT}%
\nolimits_{S}$.
\end{statement}

\begin{statement}
\textit{Claim C3:} Let $A\in\operatorname*{UT}\nolimits_{S}$. Then, $A$ is
invertible (as an element of the ring $\kk^{S\times S}$), and its
inverse $A^{-1}$ belongs to $\operatorname*{UT}\nolimits_{S}$.
\end{statement}

\textit{Proof of Claim C1:} The definition of $I_{S}$ yields $I_{S}=\left(
\delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$. Claim A4
in our proof of Proposition \ref{prop.STmat.triangular}(a) yields
$I_{S}\in\operatorname*{T}\nolimits_{S}$. For every $s\in S$, we have
$\delta_{s,s}=1$. Thus, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to
$I_{S}$ and $\delta_{s,t}$ instead of $A$ and $a_{s,t}$) yields $I_{S}%
\in\operatorname*{UT}\nolimits_{S}$. This proves Claim C1.

\textit{Proof of Claim C2:} Let $A\in\operatorname*{UT}\nolimits_{S}$ and
$B\in\operatorname*{UT}\nolimits_{S}$.

We know that $A$ is unitriangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs10}). Hence, $A\in\operatorname*{T}\nolimits_{S}$
(by Lemma \ref{lem.sol.STmat.mats.obs11}).

Write the $S\times S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs12}
yields that%
\begin{equation}
\text{for every }s\in S\text{, we have }a_{s,s}=1
\label{pf.prop.STmat.triangular.c.C2.pf.1a}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
unitriangular $S\times S$-matrix).

We know that $B$ is a unitriangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs10}, applied to $B$ instead of $A$). Hence,
$B\in\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs11},
applied to $B$ instead of $A$).

Write the $S\times S$-matrix $B$ in the form $B=\left(  b_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs12}
(applied to $B$ and $b_{s,t}$ instead of $A$ and $a_{s,t}$) yields that%
\begin{equation}
\text{for every }s\in S\text{, we have }b_{s,s}=1
\label{pf.prop.STmat.triangular.c.C2.pf.2a}%
\end{equation}
(since $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
unitriangular $S\times S$-matrix).

Claim A5 in our proof of Proposition \ref{prop.STmat.triangular}(a)
yields $AB\in\operatorname*{T}\nolimits_{S}$. Lemma
\ref{lem.sol.STmat.mats.obs0} (applied to $T=S$) yields%
\[
AB=\left(  \sum_{k\in S}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
Lemma \ref{lem.sol.STmat.mats.obs9} yields that, for every $s\in S$, we have
\begin{equation}
\sum_{k\in S}a_{s,k}b_{k,s}=\underbrace{a_{s,s}}_{\substack{=1 \\\text{(by
(\ref{pf.prop.STmat.triangular.c.C2.pf.1a}))}}}\underbrace{b_{s,s}}
_{\substack{=1 \\\text{(by (\ref{pf.prop.STmat.triangular.c.C2.pf.2a}))}%
}}=1.\nonumber
\end{equation}
Hence, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to $AB$ and $\sum_{k\in
S}a_{s,k}b_{k,t}$ instead of $A$ and $a_{s,t}$) shows that $AB\in
\operatorname*{UT}\nolimits_{S}$ (since $AB=\left(  \sum_{k\in S}%
a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in S\times S}$ and
$AB\in\operatorname*{T}\nolimits_{S}$). This proves Claim C2.

\textit{Proof of Claim C3:} We know that $A$ is unitriangular $S\times
S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs10}). Hence, $A\in
\operatorname*{T}\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs11}).

Write the $S\times S$-matrix $A$ in the form $A=\left(  a_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}$. Lemma \ref{lem.sol.STmat.mats.obs12}
yields that%
\begin{equation}
\text{for every }s\in S\text{, we have }a_{s,s}=1
\label{pf.prop.STmat.triangular.c.C3.pf.1}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
unitriangular $S\times S$-matrix).

We have $A\in\operatorname*{UT}\nolimits_{S}\subset\operatorname*{IT}%
\nolimits_{S}$ (by Lemma \ref{lem.sol.STmat.mats.obs15}). Thus, the conditions
of Claim B3 (in our proof of Proposition \ref{prop.STmat.triangular}(b)) are
satisfied. Define an $S\times S$-matrix $B=\left(  b_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}\in\kk^{S\times S}$ as in the proof of Claim
B3 (in our proof of Proposition \ref{prop.STmat.triangular}(b)). Then, we have
the following facts (which were shown in the proof of Claim B3):

\begin{itemize}
\item For every $s\in S$, we have%
\begin{equation}
b_{s,s}=\left(  a_{s,s}\right)  ^{-1}.
\label{pf.prop.STmat.triangular.c.C3.pf.2}%
\end{equation}


\item We have $B\in\operatorname*{T}\nolimits_{S}$.

\item The matrix $B$ is an inverse of $A$ in the ring $\kk^{S\times S}
$.
\end{itemize}

Now, for every $s\in S$, we have%
\[
b_{s,s}=\left(  \underbrace{a_{s,s}}_{\substack{=1 \\\text{(by
(\ref{pf.prop.STmat.triangular.c.C3.pf.1}))}}}\right)  ^{-1}=1^{-1}=1.
\]
Hence, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{UT}\nolimits_{S}$
(since $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ and
$B\in\operatorname*{T}\nolimits_{S}$). In other words, $B$ belongs to
$\operatorname*{UT}\nolimits_{S}$.

Now, recall that $B$ is an inverse of $A$ in the ring $\kk^{S\times S}
$. In particular, the element $A$ of $\kk^{S\times S}$ is invertible.
Its inverse $A^{-1}$ is $B$ (as we have just proven), and therefore belongs to
$\operatorname*{UT}\nolimits_{S}$ (since we know that $B$ belongs to
$\operatorname*{UT}\nolimits_{S}$). This proves Claim C3.

Recall that $\operatorname*{UT}\nolimits_{S}$ is a subset of $\kk%
^{S\times S}$. Claim C1 and Claim C2 (combined) show that this set
$\operatorname*{UT}\nolimits_{S}$ is a submonoid of the multiplicative monoid
of $\kk^{S\times S}$. Claim C3 furthermore proves that this submonoid
is a group. Thus, $\operatorname*{UT}\nolimits_{S}$ is a group with respect to
multiplication. In other words, the set of all unitriangular $S\times
S$-matrices is a group with respect to multiplication (since
$\operatorname*{UT}\nolimits_{S}$ is the set of all unitriangular $S\times
S$-matrices). In other words, the unitriangular $S\times S$-matrices form a
group with respect to multiplication. This proves Proposition
\ref{prop.STmat.triangular}(c).

(d) Let $A$ be an invertibly triangular $S\times S$-matrix. In other words,
$A$ belongs to the set of all invertibly triangular $S\times S$-matrices. In
other words, $A$ belongs to $\operatorname*{IT}\nolimits_{S}$ (since
$\operatorname*{IT}\nolimits_{S}$ is the set of all invertibly triangular
$S\times S$-matrices). In other words, $A\in\operatorname*{IT}\nolimits_{S}$.
Hence, Claim B3 (in our proof of Proposition \ref{prop.STmat.triangular}(b))
shows that $A$ is invertible, and that its inverse $A^{-1}$ belongs to
$\operatorname*{IT}\nolimits_{S}$.

Now, $A^{-1}$ belongs to $\operatorname*{IT}\nolimits_{S}$. Hence, $A^{-1}$ is
an invertibly triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs4} (applied to $A^{-1}$ instead of $A$)).

Now, forget that we fixed $A$. We thus have shown that if $A$ is an invertibly
triangular $S\times S$-matrix, then $A$ is invertible, and its inverse
$A^{-1}$ is again invertibly triangular. In other words, any invertibly
triangular $S\times S$-matrix is invertible, and its inverse is again
invertibly triangular. This proves Proposition \ref{prop.STmat.triangular}(d).

(e) Let $A$ be a unitriangular $S\times S$-matrix. In other words, $A$ belongs
to the set of all unitriangular $S\times S$-matrices. In other words, $A$
belongs to $\operatorname*{UT}\nolimits_{S}$ (since $\operatorname*{UT}%
\nolimits_{S}$ is the set of all unitriangular $S\times S$-matrices). In other
words, $A\in\operatorname*{UT}\nolimits_{S}$. Hence, Claim C3 (in our proof of
Proposition \ref{prop.STmat.triangular}(c)) shows that $A$ is invertible, and
that its inverse $A^{-1}$ belongs to $\operatorname*{UT}\nolimits_{S}$.

Now, $A^{-1}$ belongs to $\operatorname*{UT}\nolimits_{S}$. Hence, $A^{-1}$ is
a unitriangular $S\times S$-matrix (by Lemma \ref{lem.sol.STmat.mats.obs10}
(applied to $A^{-1}$ instead of $A$)).

Now, forget that we fixed $A$. We thus have shown that if $A$ is a
unitriangular $S\times S$-matrix, then $A$ is invertible, and its inverse
$A^{-1}$ is again unitriangular. In other words, any unitriangular $S\times
S$-matrix is invertible, and its inverse is again unitriangular. This proves
Proposition \ref{prop.STmat.triangular}(e).
\end{proof}

Thus, Exercise~\ref{exe.STmat.mat} is solved.

%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.STmat.expansion-inv}}

\textit{Solution to Exercise~\ref{exe.STmat.expansion-inv}.} Before we prove
Theorem~\ref{thm.STmat.expansion-inv}, let us introduce a notation:

\begin{definition}
Let $M$ be a $\kk$-module. Let $\left(  h_{p}\right)  _{p\in P}$ be a
family of elements of $M$. Then, we let $\left\langle h_{p}\ \mid\ p\in
P\right\rangle $ denote the $\kk$-submodule of $M$ spanned by this
family $\left(  h_{p}\right)  _{p\in P}$.
\end{definition}

\begin{proof}
[Proof of Theorem~\ref{thm.STmat.expansion-inv}.]Write the matrix $A$ in the
form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$.

We have assumed that the family $\left(  e_{s}\right)  _{s\in S}$ expands in
the family $\left(  f_{t}\right)  _{t\in T}$ through the matrix $A$. In other
words,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in T}a_{s,t}f_{t}
\label{pf.thm.STmat.expansion-inv.es=}%
\end{equation}
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in T}$ through the
matrix $A$'').

Let $B$ be the $T\times S$-matrix $A^{-1}$. Write the $T\times S$-matrix $B$
as $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in T\times S}$. From
$B=A^{-1}$, we obtain $BA=I_{T}$ and $AB=I_{S}$. These two equalities show
that the matrix $B$ has an inverse (namely, $A$); thus, the matrix $B$ is
invertible. In other words, the matrix $A^{-1}$ is invertible (since
$B=A^{-1}$).

Lemma \ref{lem.sol.STmat.mats.obs0} (applied to $T$, $S$, $B$, $b_{s,t}$, $A$
and $a_{s,t}$ instead of $S$, $T$, $A$, $a_{s,t}$, $B$ and $b_{s,t}$) yields%
\[
BA=\left(  \sum_{k\in S}b_{s,k}a_{k,t}\right)  _{\left(  s,t\right)  \in
T\times T}.
\]
Hence,%
\[
\left(  \sum_{k\in S}b_{s,k}a_{k,t}\right)  _{\left(  s,t\right)  \in T\times
T}=BA=I_{T}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in T\times T}%
\]
(by the definition of $I_{T}$). In other words,%
\begin{equation}
\sum_{k\in S}b_{s,k}a_{k,t}=\delta_{s,t}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  s,t\right)  \in T\times T.\label{pf.thm.STmat.expansion-inv.a.1}%
\end{equation}
Now,%
\begin{equation}
\text{every }u\in T\text{ satisfies }f_{u}=\sum_{k\in S}b_{u,k}e_{k}%
\label{pf.thm.STmat.expansion-inv.fu=}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.STmat.expansion-inv.fu=}):} Let $u\in
T$. Then,%
\begin{align*}
\sum_{k\in S}b_{u,k}\underbrace{e_{k}}_{\substack{=\sum_{t\in T}a_{k,t}%
f_{t}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.es=}), applied to
}s=k\text{)}}} &  =\sum_{k\in S}b_{u,k}\sum_{t\in T}a_{k,t}f_{t}%
=\underbrace{\sum_{k\in S}\sum_{t\in T}}_{=\sum_{t\in T}\sum_{k\in S}}%
b_{u,k}a_{k,t}f_{t}\\
&  =\sum_{t\in T}\sum_{k\in S}b_{u,k}a_{k,t}f_{t}=\sum_{t\in T}%
\underbrace{\left(  \sum_{k\in S}b_{u,k}a_{k,t}\right)  }_{\substack{=\delta
_{u,t}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.a.1}), applied to}\\\left(
u,t\right)  \text{ instead of }\left(  s,t\right)  \text{)}}}f_{t}\\
&  =\sum_{t\in T}\delta_{u,t}f_{t}=\underbrace{\delta_{u,u}}_{=1}f_{u}%
+\sum\limits_{\substack{t\in T;\\t\neq u}}\underbrace{\delta_{u,t}}%
_{\substack{=0\\\text{(since }t\neq u\text{)}}}f_{t}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}t=u\text{ from the sum}\right)  \\
&  =f_{u}+\underbrace{\sum\limits_{\substack{t\in T;\\t\neq u}}0f_{t}}_{=0}=f_{u}.
\end{align*}
This proves (\ref{pf.thm.STmat.expansion-inv.fu=}).}. Renaming the indices $u$
and $k$ as $s$ and $t$ in this statement, we obtain the following:%
\begin{equation}
\text{every }s\in T\text{ satisfies }f_{s}=\sum_{t\in S}b_{s,t}e_{t}%
.\label{pf.thm.STmat.expansion-inv.fs=}%
\end{equation}


We have $\left(  f_{t}\right)  _{t\in T}=\left(  f_{s}\right)  _{s\in T}$
(here, we renamed the index $t$ as $s$) and $\left(  e_{s}\right)  _{s\in
S}=\left(  e_{t}\right)  _{t\in S}$ (here, we renamed the index $s$ as $t$).

Recall that $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in T\times S}$.
Hence, the family $\left(  f_{s}\right)  _{s\in T}$ expands in the family
$\left(  e_{t}\right)  _{t\in S}$ through the matrix $B$ if and only if%
\[
\text{every }s\in T\text{ satisfies }f_{s}=\sum_{t\in S}b_{s,t}e_{t}%
\]
(by the definition of ``the family $\left(  f_{s}\right)
_{s\in T}$ expands in the family $\left(  e_{t}\right)  _{t\in S}$ through the
matrix $B$''). Thus, we conclude that the family $\left(
f_{s}\right)  _{s\in T}$ expands in the family $\left(  e_{t}\right)  _{t\in
S}$ through the matrix $B$ (since every $s\in T$ satisfies $f_{s}=\sum_{t\in
S}b_{s,t}e_{t}$). In other words, the family $\left(  f_{t}\right)  _{t\in T}$
expands in the family $\left(  e_{s}\right)  _{s\in S}$ through the matrix
$A^{-1}$ (since $\left(  f_{t}\right)  _{t\in T}=\left(  f_{s}\right)  _{s\in
T}$ and $\left(  e_{s}\right)  _{s\in S}=\left(  e_{t}\right)  _{t\in S}$ and
$A^{-1}=B$). This proves Theorem~\ref{thm.STmat.expansion-inv}(a).

(b) We shall prove the following two claims:

\begin{statement}
\textit{Claim B1:} We have $\left\langle e_{s}\ \mid\ s\in S\right\rangle
\subset\left\langle f_{t}\ \mid\ t\in T\right\rangle $.
\end{statement}

\begin{statement}
\textit{Claim B2:} We have $\left\langle f_{t}\ \mid\ t\in T\right\rangle
\subset\left\langle e_{s}\ \mid\ s\in S\right\rangle $.
\end{statement}

\textit{Proof of Claim B1:} We need to prove that $\left\langle e_{s}%
\ \mid\ s\in S\right\rangle \subset\left\langle f_{t}\ \mid\ t\in
T\right\rangle $. Since $\left\langle f_{t}\ \mid\ t\in T\right\rangle $ is a
$\kk$-module, we only need to show that $e_{s}\in\left\langle
f_{t}\ \mid\ t\in T\right\rangle $ for each $s\in S$. But the latter fact
follows from (\ref{pf.thm.STmat.expansion-inv.es=}). Thus, Claim B1 is proven.

\textit{Proof of Claim B2:} Theorem~\ref{thm.STmat.expansion-inv}(a) shows
that the family $\left(  f_{t}\right)  _{t\in T}$ expands in the family
$\left(  e_{s}\right)  _{s\in S}$ through the matrix $A^{-1}$. Moreover, we
know that the matrix $A^{-1}$ is invertible. Hence, we can apply Claim B1 to
$T$, $S$, $\left(  f_{t}\right)  _{t\in T}$, $\left(  e_{s}\right)  _{s\in S}$
and $A^{-1}$ instead of $S$, $T$, $\left(  e_{s}\right)  _{s\in S}$, $\left(
f_{t}\right)  _{t\in T}$ and $A$. As a result, we conclude that $\left\langle
f_{t}\ \mid\ t\in T\right\rangle \subset\left\langle e_{s}\ \mid\ s\in
S\right\rangle $. This proves Claim B2.

Combining Claim B1 with Claim B2, we obtain $\left\langle e_{s}\ \mid\ s\in
S\right\rangle =\left\langle f_{t}\ \mid\ t\in T\right\rangle $. Now,%
\begin{align}
&  \left(  \text{the }\kk\text{-submodule of }M\text{ spanned by the
family }\left(  e_{s}\right)  _{s\in S}\right) \nonumber\\
&  =\left\langle e_{s}\ \mid\ s\in S\right\rangle =\left\langle f_{t}%
\ \mid\ t\in T\right\rangle \nonumber\\
&  =\left(  \text{the }\kk\text{-submodule of }M\text{ spanned by the
family }\left(  f_{t}\right)  _{t\in T}\right)  .
\label{pf.thm.STmat.expansion-inv.b.done}%
\end{align}
This proves Theorem~\ref{thm.STmat.expansion-inv}(b).

(c) We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the family }\left(  e_{s}\right)  _{s\in S}\text{ spans the
}\kk\text{-module }M\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \text{the }\kk%
\text{-submodule of }M\text{ spanned by the family }\left(  e_{s}\right)
_{s\in S}\right)  }_{\substack{=\left(  \text{the }\kk\text{-submodule
of }M\text{ spanned by the family }\left(  f_{t}\right)  _{t\in T}\right)
\\\text{(by (\ref{pf.thm.STmat.expansion-inv.b.done}))}}}=M\right) \\
&  \Longleftrightarrow\ \left(  \left(  \text{the }\kk\text{-submodule
of }M\text{ spanned by the family }\left(  f_{t}\right)  _{t\in T}\right)
=M\right) \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right)  .
\end{align*}
In other words, the family $\left(  e_{s}\right)  _{s\in S}$ spans the
$\kk$-module $M$ if and only if the family $\left(  f_{t}\right)
_{t\in T}$ spans the $\kk$-module $M$. This proves
Theorem~\ref{thm.STmat.expansion-inv}(c).

(d) We shall show the following two claims:

\begin{statement}
\textit{Claim D1:} If the family $\left(  f_{t}\right)  _{t\in T}$ is
$\kk$-linearly independent, then the family $\left(  e_{s}\right)
_{s\in S}$ is $\kk$-linearly independent.
\end{statement}

\begin{statement}
\textit{Claim D2:} If the family $\left(  e_{s}\right)  _{s\in S}$ is
$\kk$-linearly independent, then the family $\left(  f_{t}\right)
_{t\in T}$ is $\kk$-linearly independent.
\end{statement}

\textit{Proof of Claim D1:} Assume that the family $\left(  f_{t}\right)
_{t\in T}$ is $\kk$-linearly independent. In other words, if $\left(
\mu_{t}\right)  _{t\in T}\in\kk^{T}$ is any family of elements of
$\kk$ satisfying $\sum_{t\in T}\mu_{t}f_{t}=0$, then%
\begin{equation}
\left(  \mu_{t}\right)  _{t\in T}=\left(  0\right)  _{t\in T}.
\label{pf.thm.STmat.expansion-inv.d.hyp}%
\end{equation}


Let $\left(  \lambda_{s}\right)  _{s\in S}\in\kk^{S}$ be a family of
elements of $S$ such that $\sum_{s\in S}\lambda_{s}e_{s}=0$. We shall show
that $\left(  \lambda_{s}\right)  _{s\in S}=\left(  0\right)  _{s\in S}$.

We have $\sum_{s\in S}\lambda_{s}e_{s}=0$. Comparing this with%
\begin{align*}
\sum_{s\in S}\lambda_{s}\underbrace{e_{s}}_{\substack{=\sum_{t\in T}%
a_{s,t}f_{t}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.es=}))}}}  &
=\sum_{s\in S}\lambda_{s}\left(  \sum_{t\in T}a_{s,t}f_{t}\right)
=\underbrace{\sum_{s\in S}\sum_{t\in T}}_{=\sum_{t\in T}\sum_{s\in S}}%
\lambda_{s}a_{s,t}f_{t}\\
&  =\sum_{t\in T}\sum_{s\in S}\lambda_{s}a_{s,t}f_{t}=\sum_{t\in T}\left(
\sum_{s\in S}\lambda_{s}a_{s,t}\right)  f_{t},
\end{align*}
we obtain $\sum_{t\in T}\left(  \sum_{s\in S}\lambda_{s}a_{s,t}\right)
f_{t}=0$. Hence, (\ref{pf.thm.STmat.expansion-inv.d.hyp}) (applied to $\mu
_{t}=\sum_{s\in S}\lambda_{s}a_{s,t}$) yields $\left(  \sum_{s\in S}%
\lambda_{s}a_{s,t}\right)  _{t\in T}=\left(  0\right)  _{t\in T}$. In other
words,%
\begin{equation}
\sum_{s\in S}\lambda_{s}a_{s,t}=0\ \ \ \ \ \ \ \ \ \ \text{for every }t\in T.
\label{pf.thm.STmat.expansion-inv.d.1}%
\end{equation}


But Lemma \ref{lem.sol.STmat.mats.obs0} yields%
\[
AB=\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
Hence,%
\[
\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in S\times
S}=AB=I_{S}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}%
\]
(by the definition of $I_{S}$). In other words,%
\begin{equation}
\sum_{k\in T}a_{s,k}b_{k,t}=\delta_{s,t}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  s,t\right)  \in S\times S.\label{pf.thm.STmat.expansion-inv.d.4}%
\end{equation}


Now, fix $u\in S$. Then,%
\begin{align*}
\sum_{s\in S}\sum_{k\in T}\lambda_{s}a_{s,k}b_{k,u} &  =\sum_{s\in S}%
\lambda_{s}\underbrace{\sum_{k\in T}a_{s,k}b_{k,u}}_{\substack{=\delta
_{s,u}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.d.4})}\\\text{(applied to
}\left(  s,u\right)  \text{ instead of }\left(  s,t\right)  \text{))}}%
}=\sum_{s\in S}\lambda_{s}\delta_{s,u}\\
&  =\lambda_{u}\underbrace{\delta_{u,u}}_{=1}+\sum\limits_{\substack{s\in S;\\s\neq
u}}\lambda_{s}\underbrace{\delta_{s,u}}_{\substack{=0\\\text{(since }s\neq
u\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}s=u\text{ from the sum}\right)  \\
&  =\lambda_{u}+\underbrace{\sum\limits_{\substack{s\in S;\\s\neq u}}\lambda_{s}%
0}_{=0}=\lambda_{u}.
\end{align*}
Hence,%
\begin{align*}
\lambda_{u} &  =\underbrace{\sum_{s\in S}\sum_{k\in T}}_{=\sum_{k\in T}%
\sum_{s\in S}}\lambda_{s}a_{s,k}b_{k,u}=\sum_{k\in T}\sum_{s\in S}\lambda
_{s}a_{s,k}b_{k,u}\\
&  =\sum_{k\in T}\underbrace{\left(  \sum_{s\in S}\lambda_{s}a_{s,k}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.STmat.expansion-inv.d.1}) (applied to
}t=k\text{))}}}b_{k,u}=\sum_{k\in T}0b_{k,u}=0.
\end{align*}


Now, forget that we fixed $u$. We thus have proven that $\lambda_{u}=0$ for
every $u\in S$. Renaming the index $u$ as $s$ in this statement, we obtain the
following: We have $\lambda_{s}=0$ for every $s\in S$. In other words,
$\left(  \lambda_{s}\right)  _{s\in S}=\left(  0\right)  _{s\in S}$.

Now, forget that we fixed $\left(  \lambda_{s}\right)  _{s\in S}$. We thus
have shown that if $\left(  \lambda_{s}\right)  _{s\in S}\in\kk^{S}$ is
a family of elements of $S$ such that $\sum_{s\in S}\lambda_{s}e_{s}=0$, then
$\left(  \lambda_{s}\right)  _{s\in S}=\left(  0\right)  _{s\in S}$. In other
words, the family $\left(  e_{s}\right)  _{s\in S}$ is $\kk$-linearly
independent. This proves Claim D1.

\textit{Proof of Claim D2:} Theorem~\ref{thm.STmat.expansion-inv}(a) shows
that the family $\left(  f_{t}\right)  _{t\in T}$ expands in the family
$\left(  e_{s}\right)  _{s\in S}$ through the matrix $A^{-1}$. Moreover, we
know that the matrix $A^{-1}$ is invertible. Hence, we can apply Claim D1 to
$T$, $S$, $\left(  f_{t}\right)  _{t\in T}$, $\left(  e_{s}\right)  _{s\in S}$
and $A^{-1}$ instead of $S$, $T$, $\left(  e_{s}\right)  _{s\in S}$, $\left(
f_{t}\right)  _{t\in T}$ and $A$. As a result, we conclude that if the family
$\left(  e_{s}\right)  _{s\in S}$ is $\kk$-linearly independent, then
the family $\left(  f_{t}\right)  _{t\in T}$ is $\kk$-linearly
independent. Claim D2 is thus proven.

Now, Claim D1 and Claim D2 are two mutually converse implications. Combining
these two implications, we obtain an equivalence; this equivalence is
precisely Theorem~\ref{thm.STmat.expansion-inv}(d).

(e) The family $\left(  e_{s}\right)  _{s\in S}$ is a basis of the
$\kk$-module $M$ if and only if it spans the $\kk$-module $M$
and is $\kk$-linearly independent (by the definition of a basis). Thus,
we have the following chain of logical equivalences:%
\begin{align}
&  \ \left(  \text{the family }\left(  e_{s}\right)  _{s\in S}\text{ is a
basis of the }\kk\text{-module }M\right)  \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  e_{s}\right)  _{s\in
S}\text{ spans the }\kk\text{-module }M\text{ and is }\kk%
\text{-linearly independent}\right)  \nonumber\\
&  \Longleftrightarrow\ \underbrace{\left(  \text{the family }\left(
e_{s}\right)  _{s\in S}\text{ spans the }\kk\text{-module }M\right)
}_{\substack{\Longleftrightarrow\ \left(  \text{the family }\left(
f_{t}\right)  _{t\in T}\text{ spans the }\kk\text{-module }M\right)
\\\text{(by Theorem~\ref{thm.STmat.expansion-inv}(c))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\underbrace{\left(  \text{the family }\left(
e_{s}\right)  _{s\in S}\text{ is }\kk\text{-linearly independent}%
\right)  }_{\substack{\Longleftrightarrow\ \left(  \text{the family }\left(
f_{t}\right)  _{t\in T}\text{ is }\kk\text{-linearly independent}%
\right)  \\\text{(by Theorem~\ref{thm.STmat.expansion-inv}(d))}}}\nonumber\\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right)  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\left(  \text{the family }\left(  f_{t}\right)
_{t\in T}\text{ is }\kk\text{-linearly independent}\right)
.\label{pf.thm.STmat.expansion-inv.e.3}%
\end{align}
On the other hand, the family $\left(  f_{t}\right)  _{t\in T}$ is a basis of
the $\kk$-module $M$ if and only if it spans the $\kk$-module
$M$ and is $\kk$-linearly independent (by the definition of a basis).
Thus, we have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the family }\left(  f_{t}\right)  _{t\in T}\text{ is a
basis of the }\kk\text{-module }M\right)  \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\text{ and is }\kk%
\text{-linearly independent}\right)  \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\left(  \text{the family }\left(  f_{t}\right)
_{t\in T}\text{ is }\kk\text{-linearly independent}\right)  \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  e_{s}\right)  _{s\in
S}\text{ is a basis of the }\kk\text{-module }M\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.STmat.expansion-inv.e.3}%
)}\right)  .
\end{align*}
This proves Theorem~\ref{thm.STmat.expansion-inv}(e).
\end{proof}

Thus, Exercise~\ref{exe.STmat.expansion-inv} is solved.

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.STmat.expansion-inv}}

\textit{Solution to Exercise~\ref{exe.STmat.expansion-inv}.} Before we prove
Theorem~\ref{thm.STmat.expansion-inv}, let us introduce a notation:

\begin{definition}
Let $M$ be a $\kk$-module. Let $\left(  h_{p}\right)  _{p\in P}$ be a
family of elements of $M$. Then, we let $\left\langle h_{p}\ \mid\ p\in
P\right\rangle $ denote the $\kk$-submodule of $M$ spanned by this
family $\left(  h_{p}\right)  _{p\in P}$.
\end{definition}

The following fact is well-known:

\begin{lemma}
\label{lem.sol.STmat.expansion-inv.span}Let $M$ be a $\kk$-module. Let
$\left(  h_{p}\right)  _{p\in P}$ be a family of elements of $M$. Then:

\begin{enumerate}
\item[(a)] The set $\left\langle h_{p}\ \mid\ p\in P\right\rangle $ is a
$\kk$-submodule of $M$.

\item[(b)] We have $h_{u}\in\left\langle h_{p}\ \mid\ p\in P\right\rangle $
for every $u\in P$.

\item[(c)] Let $X$ be a $\kk$-submodule of $M$. Assume that $h_{u}\in
X$ for every $u\in P$. Then, $\left\langle h_{p}\ \mid\ p\in P\right\rangle
\subset X$.
\end{enumerate}
\end{lemma}

Next, let us show the following proposition:

\begin{proposition}
\label{prop.sol.STmat.expansion-inv.expansion}Let $M$ be a $\kk%
$-module. Let $S$ and $T$ be two finite sets. Let $\left(  e_{s}\right)
_{s\in S}$ and $\left(  f_{t}\right)  _{t\in T}$ be two families of elements
of $M$.

Let $A$ be an $S\times T$-matrix. Assume that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in
T}$ through the matrix $A$. Then, $\left\langle e_{s}\ \mid\ s\in
S\right\rangle \subset\left\langle f_{t}\ \mid\ t\in T\right\rangle $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.sol.STmat.expansion-inv.expansion}.]Write the
matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in
S\times T}$.

We have assumed that the family $\left(  e_{s}\right)  _{s\in S}$ expands in
the family $\left(  f_{t}\right)  _{t\in T}$ through the matrix $A$. In other
words,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in T}a_{s,t}f_{t}
\label{pf.prop.sol.STmat.expansion-inv.expansion.es=}%
\end{equation}
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in T}$ through the
matrix $A$'').

Let $X=\left\langle f_{t}\ \mid\ t\in T\right\rangle $. Lemma
\ref{lem.sol.STmat.expansion-inv.span}(a) (applied to $\left(  f_{t}\right)
_{t\in T}$ instead of $\left(  h_{p}\right)  _{p\in P}$) yields that
$\left\langle f_{t}\ \mid\ t\in T\right\rangle $ is a $\kk$-submodule
of $M$. In other words, $X$ is a $\kk$-submodule of $M$ (since
$X=\left\langle f_{t}\ \mid\ t\in T\right\rangle $).

For every $u\in S$, we have%
\begin{align*}
e_{u}  &  =\sum_{t\in T}a_{u,t}f_{t}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop.sol.STmat.expansion-inv.expansion.es=}) (applied to }%
s=u\text{)}\right) \\
&  =\sum_{g\in T}a_{u,g}\underbrace{f_{g}}_{\substack{\in\left\langle
f_{t}\ \mid\ t\in T\right\rangle \\\text{(by Lemma
\ref{lem.sol.STmat.expansion-inv.span}(b)}\\\text{(applied to }\left(
f_{t}\right)  _{t\in T}\text{ and }g\text{ instead of }\left(  h_{p}\right)
_{p\in P}\text{ and }u\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}t\text{ as }g\right) \\
&  \in\sum_{g\in T}a_{u,g}\underbrace{\left\langle f_{t}\ \mid\ t\in
T\right\rangle }_{=X}=\sum_{g\in T}a_{u,g}X\subset X\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }X\text{ is a }\kk%
\text{-submodule of }M\right)  .
\end{align*}
Thus, Lemma \ref{lem.sol.STmat.expansion-inv.span}(c) (applied to $\left(
e_{s}\right)  _{s\in S}$ instead of $\left(  h_{p}\right)  _{p\in P}$) yields
$\left\langle e_{s}\ \mid\ s\in S\right\rangle \subset X$. In other words,
$\left\langle e_{s}\ \mid\ s\in S\right\rangle \subset\left\langle f_{t}%
\ \mid\ t\in T\right\rangle $ (since $X=\left\langle f_{t}\ \mid\ t\in
T\right\rangle $). This proves Proposition
\ref{prop.sol.STmat.expansion-inv.expansion}.
\end{proof}

\begin{proof}
[Proof of Theorem~\ref{thm.STmat.expansion-inv}.]Write the matrix $A$ in the
form $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times T}$.

We have assumed that the family $\left(  e_{s}\right)  _{s\in S}$ expands in
the family $\left(  f_{t}\right)  _{t\in T}$ through the matrix $A$. In other
words,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in T}a_{s,t}f_{t}
\label{pf.thm.STmat.expansion-inv.es=}%
\end{equation}
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in T}$ through the
matrix $A$'').

Let $B$ be the $T\times S$-matrix $A^{-1}$. Write the $T\times S$-matrix $B$
as $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in T\times S}$. We have
$\underbrace{B}_{=A^{-1}}A=A^{-1}A=I_{T}$ and $A\underbrace{B}_{=A^{-1}%
}=AA^{-1}=I_{S}$. These two equalities show that the matrix $B$ has an inverse
(namely, $A$); thus, the matrix $B$ is invertible. In other words, the matrix
$A^{-1}$ is invertible (since $B=A^{-1}$).

Lemma \ref{lem.sol.STmat.mats.obs0} (applied to $T$, $S$, $B$, $b_{s,t}$, $A$
and $a_{s,t}$ instead of $S$, $T$, $A$, $a_{s,t}$, $B$ and $b_{s,t}$) yields%
\[
BA=\left(  \sum_{k\in S}b_{s,k}a_{k,t}\right)  _{\left(  s,t\right)  \in
T\times T}.
\]
Hence,%
\[
\left(  \sum_{k\in S}b_{s,k}a_{k,t}\right)  _{\left(  s,t\right)  \in T\times
T}=BA=I_{T}=\left(  \delta_{s,t}\right)  _{\left(  s,t\right)  \in T\times T}%
\]
(by the definition of $I_{T}$). In other words,%
\begin{equation}
\sum_{k\in S}b_{s,k}a_{k,t}=\delta_{s,t}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  s,t\right)  \in T\times T. \label{pf.thm.STmat.expansion-inv.a.1}%
\end{equation}
Now,%
\begin{equation}
\text{every }u\in T\text{ satisfies }f_{u}=\sum_{k\in S}b_{u,k}e_{k}
\label{pf.thm.STmat.expansion-inv.fu=}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.STmat.expansion-inv.fu=}):} Let $u\in
T$. Then,%
\begin{align*}
\sum_{k\in S}b_{u,k}\underbrace{e_{k}}_{\substack{=\sum_{t\in T}a_{k,t}%
f_{t}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.es=}), applied to
}s=k\text{)}}}  &  =\sum_{k\in S}b_{u,k}\sum_{t\in T}a_{k,t}f_{t}%
=\underbrace{\sum_{k\in S}\sum_{t\in T}}_{=\sum_{t\in T}\sum_{k\in S}}%
b_{u,k}a_{k,t}f_{t}\\
&  =\sum_{t\in T}\sum_{k\in S}b_{u,k}a_{k,t}f_{t}=\sum_{t\in T}%
\underbrace{\left(  \sum_{k\in S}b_{u,k}a_{k,t}\right)  }_{\substack{=\delta
_{u,t}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.a.1}), applied to}\\\left(
u,t\right)  \text{ instead of }\left(  s,t\right)  \text{)}}}f_{t}\\
&  =\sum_{t\in T}\delta_{u,t}f_{t}=\underbrace{\delta_{u,u}}%
_{\substack{=1\\\text{(since }u=u\text{)}}}f_{u}+\sum\limits_{\substack{t\in
T;\\t\neq u}}\underbrace{\delta_{u,t}}_{\substack{=0\\\text{(since }u\neq
t\\\text{(since }t\neq u\text{))}}}f_{t}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}t=u\text{ from the sum}\right) \\
&  =f_{u}+\underbrace{\sum\limits_{\substack{t\in T;\\t\neq u}}0f_{t}}_{=0}=f_{u}.
\end{align*}
This proves (\ref{pf.thm.STmat.expansion-inv.fu=}).}. Renaming $u$ as $s$ in
this statement, we obtain the following: Every $s\in T$ satisfies $f_{s}%
=\sum_{k\in S}b_{s,k}e_{k}$. Thus, every $s\in T$ satisfies%
\begin{equation}
f_{s}=\sum_{k\in S}b_{s,k}e_{k}=\sum_{t\in S}b_{s,t}e_{t}
\label{pf.thm.STmat.expansion-inv.fs=}%
\end{equation}
(here, we have renamed the summation index $k$ as $t$).

We have $\left(  f_{t}\right)  _{t\in T}=\left(  f_{s}\right)  _{s\in T}$
(here, we renamed the index $t$ as $s$) and $\left(  e_{s}\right)  _{s\in
S}=\left(  e_{t}\right)  _{t\in S}$ (here, we renamed the index $s$ as $t$).

Recall that $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in T\times S}$.
Hence, the family $\left(  f_{s}\right)  _{s\in T}$ expands in the family
$\left(  e_{t}\right)  _{t\in S}$ through the matrix $B$ if and only if%
\[
\text{every }s\in T\text{ satisfies }f_{s}=\sum_{t\in S}b_{s,t}e_{t}%
\]
(by the definition of ``the family $\left(  f_{s}\right)
_{s\in T}$ expands in the family $\left(  e_{t}\right)  _{t\in S}$ through the
matrix $B$''). Thus, we conclude that the family $\left(
f_{s}\right)  _{s\in T}$ expands in the family $\left(  e_{t}\right)  _{t\in
S}$ through the matrix $B$ (since every $s\in T$ satisfies $f_{s}=\sum_{t\in
S}b_{s,t}e_{t}$). In other words, the family $\left(  f_{t}\right)  _{t\in T}$
expands in the family $\left(  e_{s}\right)  _{s\in S}$ through the matrix
$A^{-1}$ (since $\left(  f_{t}\right)  _{t\in T}=\left(  f_{s}\right)  _{s\in
T}$ and $\left(  e_{s}\right)  _{s\in S}=\left(  e_{t}\right)  _{t\in S}$ and
$A^{-1}=B$). This proves Theorem~\ref{thm.STmat.expansion-inv}(a).

(b) We shall prove the following two claims:

\begin{statement}
\textit{Claim B1:} We have $\left\langle e_{s}\ \mid\ s\in S\right\rangle
\subset\left\langle f_{t}\ \mid\ t\in T\right\rangle $.
\end{statement}

\begin{statement}
\textit{Claim B2:} We have $\left\langle f_{t}\ \mid\ t\in T\right\rangle
\subset\left\langle e_{s}\ \mid\ s\in S\right\rangle $.
\end{statement}

\textit{Proof of Claim B1:} Proposition
\ref{prop.sol.STmat.expansion-inv.expansion} yields $\left\langle e_{s}%
\ \mid\ s\in S\right\rangle \subset\left\langle f_{t}\ \mid\ t\in
T\right\rangle $. This proves Claim B1.

\textit{Proof of Claim B2:} Theorem~\ref{thm.STmat.expansion-inv}(a) shows
that the family $\left(  f_{t}\right)  _{t\in T}$ expands in the family
$\left(  e_{s}\right)  _{s\in S}$ through the matrix $A^{-1}$. Moreover, we
know that the matrix $A^{-1}$ is invertible. Hence, we can apply Proposition
\ref{prop.sol.STmat.expansion-inv.expansion} to $T$, $S$, $\left(
f_{t}\right)  _{t\in T}$, $\left(  e_{s}\right)  _{s\in S}$ and $A^{-1}$
instead of $S$, $T$, $\left(  e_{s}\right)  _{s\in S}$, $\left(  f_{t}\right)
_{t\in T}$ and $A$. As a result, we conclude that $\left\langle f_{t}%
\ \mid\ t\in T\right\rangle \subset\left\langle e_{s}\ \mid\ s\in
S\right\rangle $. This proves Claim B2.

Combining Claim B1 with Claim B2, we obtain $\left\langle e_{s}\ \mid\ s\in
S\right\rangle =\left\langle f_{t}\ \mid\ t\in T\right\rangle $. Now,%
\begin{align}
&  \left(  \text{the }\kk\text{-submodule of }M\text{ spanned by the
family }\left(  e_{s}\right)  _{s\in S}\right) \nonumber\\
&  =\left\langle e_{s}\ \mid\ s\in S\right\rangle =\left\langle f_{t}%
\ \mid\ t\in T\right\rangle \nonumber\\
&  =\left(  \text{the }\kk\text{-submodule of }M\text{ spanned by the
family }\left(  f_{t}\right)  _{t\in T}\right)  .
\label{pf.thm.STmat.expansion-inv.b.done}%
\end{align}
This proves Theorem~\ref{thm.STmat.expansion-inv}(b).

(c) We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the family }\left(  e_{s}\right)  _{s\in S}\text{ spans the
}\kk\text{-module }M\right) \\
&  \Longleftrightarrow\ \left(  \underbrace{\left(  \text{the }\kk%
\text{-submodule of }M\text{ spanned by the family }\left(  e_{s}\right)
_{s\in S}\right)  }_{\substack{=\left(  \text{the }\kk\text{-submodule
of }M\text{ spanned by the family }\left(  f_{t}\right)  _{t\in T}\right)
\\\text{(by (\ref{pf.thm.STmat.expansion-inv.b.done}))}}}=M\right) \\
&  \Longleftrightarrow\ \left(  \left(  \text{the }\kk\text{-submodule
of }M\text{ spanned by the family }\left(  f_{t}\right)  _{t\in T}\right)
=M\right) \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right)  .
\end{align*}
In other words, the family $\left(  e_{s}\right)  _{s\in S}$ spans the
$\kk$-module $M$ if and only if the family $\left(  f_{t}\right)
_{t\in T}$ spans the $\kk$-module $M$. This proves
Theorem~\ref{thm.STmat.expansion-inv}(c).

(d) We shall show the following two claims:

\begin{statement}
\textit{Claim D1:} If the family $\left(  f_{t}\right)  _{t\in T}$ is
$\kk$-linearly independent, then the family $\left(  e_{s}\right)
_{s\in S}$ is $\kk$-linearly independent.
\end{statement}

\begin{statement}
\textit{Claim D2:} If the family $\left(  e_{s}\right)  _{s\in S}$ is
$\kk$-linearly independent, then the family $\left(  f_{t}\right)
_{t\in T}$ is $\kk$-linearly independent.
\end{statement}

\textit{Proof of Claim D1:} Assume that the family $\left(  f_{t}\right)
_{t\in T}$ is $\kk$-linearly independent. In other words, if $\left(
\mu_{t}\right)  _{t\in T}\in\kk^{T}$ is any family of elements of
$\kk$ satisfying $\sum_{t\in T}\mu_{t}f_{t}=0$, then%
\begin{equation}
\left(  \mu_{t}\right)  _{t\in T}=\left(  0\right)  _{t\in T}.
\label{pf.thm.STmat.expansion-inv.d.hyp}%
\end{equation}


Let $\left(  \lambda_{s}\right)  _{s\in S}\in\kk^{S}$ be a family of
elements of $S$ such that $\sum_{s\in S}\lambda_{s}e_{s}=0$. We shall show
that $\left(  \lambda_{s}\right)  _{s\in S}=\left(  0\right)  _{s\in S}$.

We have $\sum_{s\in S}\lambda_{s}e_{s}=0$. Comparing this with%
\begin{align*}
\sum_{s\in S}\lambda_{s}\underbrace{e_{s}}_{\substack{=\sum_{t\in T}%
a_{s,t}f_{t}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.es=}))}}}  &
=\sum_{s\in S}\lambda_{s}\left(  \sum_{t\in T}a_{s,t}f_{t}\right)
=\underbrace{\sum_{s\in S}\sum_{t\in T}}_{=\sum_{t\in T}\sum_{s\in S}}%
\lambda_{s}a_{s,t}f_{t}\\
&  =\sum_{t\in T}\sum_{s\in S}\lambda_{s}a_{s,t}f_{t}=\sum_{t\in T}\left(
\sum_{s\in S}\lambda_{s}a_{s,t}\right)  f_{t},
\end{align*}
we obtain $\sum_{t\in T}\left(  \sum_{s\in S}\lambda_{s}a_{s,t}\right)
f_{t}=0$. Hence, (\ref{pf.thm.STmat.expansion-inv.d.hyp}) (applied to $\mu
_{t}=\sum_{s\in S}\lambda_{s}a_{s,t}$) yields $\left(  \sum_{s\in S}%
\lambda_{s}a_{s,t}\right)  _{t\in T}=\left(  0\right)  _{t\in T}$. In other
words,%
\begin{equation}
\sum_{s\in S}\lambda_{s}a_{s,t}=0\ \ \ \ \ \ \ \ \ \ \text{for every }t\in T.
\label{pf.thm.STmat.expansion-inv.d.1}%
\end{equation}


But Lemma \ref{lem.sol.STmat.mats.obs0} yields%
\[
AB=\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in
S\times S}.
\]
Hence,%
\[
\left(  \sum_{k\in T}a_{s,k}b_{k,t}\right)  _{\left(  s,t\right)  \in S\times
S}=A\underbrace{B}_{=A^{-1}}=AA^{-1}=I_{S}=\left(  \delta_{s,t}\right)
_{\left(  s,t\right)  \in S\times S}%
\]
(by the definition of $I_{S}$). In other words,%
\begin{equation}
\sum_{k\in T}a_{s,k}b_{k,t}=\delta_{s,t}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  s,t\right)  \in S\times S. \label{pf.thm.STmat.expansion-inv.d.4}%
\end{equation}


Now, fix $u\in S$. Then,%
\begin{align*}
\sum_{s\in S}\sum_{k\in T}\lambda_{s}a_{s,k}b_{k,u}  &  =\sum_{s\in S}%
\lambda_{s}\underbrace{\sum_{k\in T}a_{s,k}b_{k,u}}_{\substack{=\delta
_{s,u}\\\text{(by (\ref{pf.thm.STmat.expansion-inv.d.4})}\\\text{(applied to
}\left(  s,u\right)  \text{ instead of }\left(  s,t\right)  \text{))}}%
}=\sum_{s\in S}\lambda_{s}\delta_{s,u}\\
&  =\lambda_{u}\underbrace{\delta_{u,u}}_{\substack{=1\\\text{(since
}u=u\text{)}}}+\sum\limits_{\substack{s\in S;\\s\neq u}}\lambda_{s}\underbrace{\delta
_{s,u}}_{\substack{=0\\\text{(since }s\neq u\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}s=u\text{ from the sum}\right) \\
&  =\lambda_{u}+\underbrace{\sum\limits_{\substack{s\in S;\\s\neq u}}\lambda_{s}%
0}_{=0}=\lambda_{u}.
\end{align*}
Hence,%
\begin{align*}
\lambda_{u}  &  =\underbrace{\sum_{s\in S}\sum_{k\in T}}_{=\sum_{k\in T}%
\sum_{s\in S}}\lambda_{s}a_{s,k}b_{k,u}=\sum_{k\in T}\sum_{s\in S}\lambda
_{s}a_{s,k}b_{k,u}\\
&  =\sum_{k\in T}\underbrace{\left(  \sum_{s\in S}\lambda_{s}a_{s,k}\right)
}_{\substack{=0\\\text{(by (\ref{pf.thm.STmat.expansion-inv.d.1}) (applied to
}t=k\text{))}}}b_{k,u}=\sum_{k\in T}0b_{k,u}=0.
\end{align*}


Now, forget that we fixed $u$. We thus have proven that $\lambda_{u}=0$ for
every $u\in S$. Renaming the index $u$ as $s$ in this statement, we obtain the
following: We have $\lambda_{s}=0$ for every $s\in S$. In other words,
$\left(  \lambda_{s}\right)  _{s\in S}=\left(  0\right)  _{s\in S}$.

Now, forget that we fixed $\left(  \lambda_{s}\right)  _{s\in S}$. We thus
have shown that if $\left(  \lambda_{s}\right)  _{s\in S}\in\kk^{S}$ is
a family of elements of $S$ such that $\sum_{s\in S}\lambda_{s}e_{s}=0$, then
$\left(  \lambda_{s}\right)  _{s\in S}=\left(  0\right)  _{s\in S}$. In other
words, the family $\left(  e_{s}\right)  _{s\in S}$ is $\kk$-linearly
independent. This proves Claim D1.

\textit{Proof of Claim D2:} Theorem~\ref{thm.STmat.expansion-inv}(a) shows
that the family $\left(  f_{t}\right)  _{t\in T}$ expands in the family
$\left(  e_{s}\right)  _{s\in S}$ through the matrix $A^{-1}$. Moreover, we
know that the matrix $A^{-1}$ is invertible. Hence, we can apply Claim D1 to
$T$, $S$, $\left(  f_{t}\right)  _{t\in T}$, $\left(  e_{s}\right)  _{s\in S}$
and $A^{-1}$ instead of $S$, $T$, $\left(  e_{s}\right)  _{s\in S}$, $\left(
f_{t}\right)  _{t\in T}$ and $A$. As a result, we conclude that if the family
$\left(  e_{s}\right)  _{s\in S}$ is $\kk$-linearly independent, then
the family $\left(  f_{t}\right)  _{t\in T}$ is $\kk$-linearly
independent. Claim D2 is thus proven.

Now, Claim D1 and Claim D2 are two mutually converse implications. Combining
these two implications, we conclude the following: The family $\left(
e_{s}\right)  _{s\in S}$ is $\kk$-linearly independent if and only if
the family $\left(  f_{t}\right)  _{t\in T}$ is $\kk$-linearly
independent. Thus, Theorem~\ref{thm.STmat.expansion-inv}(d) is proven.

(e) Theorem~\ref{thm.STmat.expansion-inv}(d) shows that the family $\left(
e_{s}\right)  _{s\in S}$ is $\kk$-linearly independent if and only if
the family $\left(  f_{t}\right)  _{t\in T}$ is $\kk$-linearly
independent. In other words, we have the following logical equivalence:%
\begin{align}
&  \ \left(  \text{the family }\left(  e_{s}\right)  _{s\in S}\text{ is
}\kk\text{-linearly independent}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ is }\kk\text{-linearly independent}\right)
\label{pf.thm.STmat.expansion-inv.e.1}%
\end{align}


Theorem~\ref{thm.STmat.expansion-inv}(c) shows that the family $\left(
e_{s}\right)  _{s\in S}$ spans the $\kk$-module $M$ if and only if the
family $\left(  f_{t}\right)  _{t\in T}$ spans the $\kk$-module $M$. In
other words, we have the following logical equivalence:%
\begin{align}
&  \ \left(  \text{the family }\left(  e_{s}\right)  _{s\in S}\text{ spans the
}\kk\text{-module }M\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right)
\label{pf.thm.STmat.expansion-inv.e.2}%
\end{align}


But the family $\left(  e_{s}\right)  _{s\in S}$ is a basis of the
$\kk$-module $M$ if and only if it spans the $\kk$-module $M$
and is $\kk$-linearly independent (by the definition of a basis). Thus,
we have the following chain of logical equivalences:%
\begin{align}
&  \ \left(  \text{the family }\left(  e_{s}\right)  _{s\in S}\text{ is a
basis of the }\kk\text{-module }M\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  e_{s}\right)  _{s\in
S}\text{ spans the }\kk\text{-module }M\text{ and is }\kk%
\text{-linearly independent}\right) \nonumber\\
&  \Longleftrightarrow\ \underbrace{\left(  \text{the family }\left(
e_{s}\right)  _{s\in S}\text{ spans the }\kk\text{-module }M\right)
}_{\substack{\Longleftrightarrow\ \left(  \text{the family }\left(
f_{t}\right)  _{t\in T}\text{ spans the }\kk\text{-module }M\right)
\\\text{(by (\ref{pf.thm.STmat.expansion-inv.e.2}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\underbrace{\left(  \text{the family }\left(
e_{s}\right)  _{s\in S}\text{ is }\kk\text{-linearly independent}%
\right)  }_{\substack{\Longleftrightarrow\ \left(  \text{the family }\left(
f_{t}\right)  _{t\in T}\text{ is }\kk\text{-linearly independent}%
\right)  \\\text{(by (\ref{pf.thm.STmat.expansion-inv.e.1}))}}}\nonumber\\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\left(  \text{the family }\left(  f_{t}\right)
_{t\in T}\text{ is }\kk\text{-linearly independent}\right)  .
\label{pf.thm.STmat.expansion-inv.e.3}%
\end{align}
On the other hand, the family $\left(  f_{t}\right)  _{t\in T}$ is a basis of
the $\kk$-module $M$ if and only if it spans the $\kk$-module
$M$ and is $\kk$-linearly independent (by the definition of a basis).
Thus, we have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the family }\left(  f_{t}\right)  _{t\in T}\text{ is a
basis of the }\kk\text{-module }M\right) \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\text{ and is }\kk%
\text{-linearly independent}\right) \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  f_{t}\right)  _{t\in
T}\text{ spans the }\kk\text{-module }M\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\left(  \text{the family }\left(  f_{t}\right)
_{t\in T}\text{ is }\kk\text{-linearly independent}\right) \\
&  \Longleftrightarrow\ \left(  \text{the family }\left(  e_{s}\right)  _{s\in
S}\text{ is a basis of the }\kk\text{-module }M\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.STmat.expansion-inv.e.3}%
)}\right)  .
\end{align*}
In other words, the family $\left(  e_{s}\right)  _{s\in S}$ is a basis of the
$\kk$-module $M$ if and only if the family $\left(  f_{t}\right)
_{t\in T}$ is a basis of the $\kk$-module $M$. This proves
Theorem~\ref{thm.STmat.expansion-inv}(e).
\end{proof}

Thus, Exercise~\ref{exe.STmat.expansion-inv} is solved.

%\end{solutions}

\end{verlong}

\begin{vershort}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.STmat.expansion-tria-inv}}

\textit{Solution to Exercise~\ref{exe.STmat.expansion-tria-inv}.}

\begin{proof}
[Proof of Remark~\ref{rmk.STmat.expansion-tria-shortcuts}.](a) We shall prove
the following two claims:

\begin{statement}
\textit{Claim A1:} If the family $\left(  e_{s}\right)  _{s\in S}$ expands
triangularly in the family $\left(  f_{s}\right)  _{s\in S}$, then every $s\in
S$ satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\]
\end{statement}

\begin{statement}
\textit{Claim A2:} If every $s\in S$ satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  ,
\]
then the family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly in the
family $\left(  f_{s}\right)  _{s\in S}$.
\end{statement}

\textit{Proof of Claim A1:} Assume that the family $\left(  e_{s}\right)
_{s\in S}$ expands triangularly in the family $\left(  f_{s}\right)  _{s\in
S}$. In other words, there exists a triangular $S\times S$-matrix $A$ such
that the family $\left(  e_{s}\right)  _{s\in S}$ expands in the family
$\left(  f_{s}\right)  _{s\in S}$ through the matrix $A$. Consider this $A$.

Write the matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$. In other words,
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}a_{s,t}f_{t}%
.\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.1}%
\end{equation}


Lemma \ref{lem.sol.STmat.mats.obs2} shows that
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}%
=0\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.2}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
triangular $S\times S$-matrix). Hence, every $s\in S$ satisfies%
\begin{align*}
e_{s} &  =\sum_{t\in S}a_{s,t}f_{t}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.1})}\right)  \\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\sum\limits_{\substack{t\in
S;\\\text{not }t\leq s}}\underbrace{a_{s,t}}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.2}))}}}f_{t}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}=\sum\limits_{\substack{t\in
S;\\t\leq s}}a_{s,t}f_{t}\\
&  =\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\end{align*}
This proves Claim A1.

\textit{Proof of Claim A2:} Assume that every $s\in S$ satisfies
\begin{equation}
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)
.\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1}%
\end{equation}
In other words, for every $s\in S$, there exists a family $\left(
c_{s,t}\right)  _{t\in S;\ t\leq s}
\in\kk^{\left\{t \in S \mid t \leq s\right\}}$ such that
\begin{equation}
e_{s}=\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}%
.\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1a}%
\end{equation}
Fix such a family for each $s\in S$.

Now, define an $S\times S$-matrix $B=\left(  b_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}\in\kk^{S\times S}$ by%
\[
\left(  b_{s,t}=%
\begin{cases}
c_{s,t}, & \text{if }t\leq s;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  s,t\right)  \in S\times S\right)
.
\]
Every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s$
must satisfy%
\begin{equation}
b_{s,t}=0\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
b_{s,t}\right)  .\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2}%
\end{equation}
In other words, the $S\times S$-matrix $B$ is triangular (by the definition of
``triangular'').

On the other hand, every $\left(  s,t\right)  \in S\times S$ which satisfies
$t\leq s$ must satisfy%
\begin{equation}
b_{s,t}=c_{s,t}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
b_{s,t}\right)  .\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2b}%
\end{equation}


Now,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}%
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.3}):} Let $s\in S$. Then,
every $t\in S$ satisfies either $t\leq s$ or $\left(  \text{not }t\leq
s\right)  $. Hence,%
\begin{align*}
\sum_{t\in S}b_{s,t}f_{t} &  =\sum\limits_{\substack{t\in S;\\t\leq s}%
}\underbrace{b_{s,t}}_{\substack{=c_{s,t}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2b}))}}}f_{t}%
+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{b_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2}))}}}f_{t}%
=\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}=e_{s}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1a})}\right)  .
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.3}).}. In
other words, the family $\left(  e_{s}\right)  _{s\in S}$ expands in the
family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $B$ (since
$B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$). Hence, the
family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly in the family
$\left(  f_{s}\right)  _{s\in S}$ (since the matrix $B$ is triangular). This
proves Claim A2.

We have now proven Claim A1 and Claim A2. These two claims are mutually
converse implications. Combining these two implications, we obtain an
equivalence, which is precisely the statement of
Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(a). Thus,
Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(a) is proven.

(b) We shall prove the following two claims:

\begin{statement}
\textit{Claim B1:} If the family $\left(  e_{s}\right)  _{s\in S}$ expands
invertibly triangularly in the family $\left(  f_{s}\right)  _{s\in S}$, then
every $s\in S$ satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$.
\end{statement}

\begin{statement}
\textit{Claim B2:} If every $s\in S$ satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$, then the family $\left(
e_{s}\right)  _{s\in S}$ expands invertibly triangularly in the family
$\left(  f_{s}\right)  _{s\in S}$.
\end{statement}

\textit{Proof of Claim B1:} Assume that the family $\left(  e_{s}\right)
_{s\in S}$ expands invertibly triangularly in the family $\left(
f_{s}\right)  _{s\in S}$. In other words, there exists an invertibly
triangular $S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the
matrix $A$. Consider this $A$.

Write the matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$. In other words,
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}a_{s,t}f_{t}%
.\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.1}%
\end{equation}


Lemma \ref{lem.sol.STmat.mats.obs5} shows that $A\in\operatorname*{T}%
\nolimits_{S}$ (since $A$ is an invertibly triangular $S\times S$-matrix).
Thus, Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular
$S\times S$-matrix. Hence, Lemma \ref{lem.sol.STmat.mats.obs2} shows that
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}%
=0\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.2}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
triangular $S\times S$-matrix). Now, it is easy to show that every $s\in S$
satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$\ \ \ \ \footnote{\textit{Proof.}
Let $s\in S$. Then, Lemma \ref{lem.sol.STmat.mats.obs6} shows that the element
$a_{s,s}$ of $\kk$ is invertible.
\par
But (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.1}) yields%
\begin{align*}
e_{s} &  =\sum_{t\in S}a_{s,t}f_{t}=\sum\limits_{\substack{t\in S;\\t\leq s}%
}a_{s,t}f_{t}+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{a_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.2}))}}}f_{t}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}=\sum\limits_{\substack{t\in
S;\\t\leq s}}a_{s,t}f_{t}=a_{s,s}f_{s}+\underbrace{\sum\limits_{\substack{t\in
S;\\t\leq s\text{ and }t\neq s}}}_{\substack{=\sum\limits_{\substack{t\in S;\\t<s}%
}}}a_{s,t}f_{t}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}t=s\text{ from the sum}\right)  \\
&  =a_{s,s}f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t<s}}a_{s,t}f_{t}%
}_{=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  }\\
&  =a_{s,s}f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\end{align*}
Hence,
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$ (namely, for $\alpha_{s}%
=a_{s,s}$). Qed.}. Thus, Claim B1 is proven.

\textit{Proof of Claim B2:} Assume that every $s\in S$ satisfies
\begin{equation}
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1}%
\end{equation}
for some invertible $\alpha_{s}\in\kk$. Consider this $\alpha_{s}$.

Now, for every $s\in S$, there exists a family $\left(  c_{s,t}\right)  _{t\in
S;\ t<s}\in\kk^{\left\{t \in S \mid t < s\right\}}$ such that
\begin{equation}
e_{s}-\alpha_{s}f_{s}=\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}%
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1a}%
\end{equation}
\footnote{\textit{Proof.} Let $s\in S$. Subtracting $\alpha_{s}f_{s}$
from both sides of (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1}), we
obtain%
\[
e_{s}-\alpha_{s}f_{s}=\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\]
In other words, there exists a family $\left(  c_{s,t}\right)  _{t\in
S;\ t<s}\in\kk^{\left\{t \in S \mid t < s\right\}}$
such that $e_{s}-\alpha_{s}f_{s}=\sum\limits
_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}$.}. Fix such a family for each $s\in S$.

Now, define an $S\times S$-matrix $B=\left(  b_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}\in\kk^{S\times S}$ by%
\[
\left(  b_{s,t}=%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}\alpha_{s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  s,t\right)  \in S\times S\right)
.
\]


Every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s$
must satisfy%
\begin{equation}
b_{s,t}=0\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}):} Let $\left(
s,t\right)  \in S\times S$ be such that we do not have $t\leq s$.
\par
We do not have $t\leq s$. Thus, we do not have $s=t$ (since $s=t$ would imply
$t\leq s$). Hence, $\delta_{s,t}=0$.
\par
If we had $t<s$, then we would have $t\leq s$, which would contradict the fact
that we do not have $t\leq s$. Hence, we do not have $t<s$. Now, the
definition of $b_{s,t}$ yields%
\begin{align*}
b_{s,t} &  =%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}\alpha_{s}, & \text{otherwise}%
\end{cases}  \quad
=\underbrace{\delta_{s,t}}_{=0}\alpha_{s}\ \ \ \ \ \ \ \ \ \ \left(
\text{since we do not have }t<s\right)  \\
&  =0.
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}).}. Hence,
Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $B$ and $b_{s,t}$ instead of
$A$ and $a_{s,t}$) shows that $B\in\operatorname*{T}\nolimits_{S}$.

On the other hand, every $\left(  s,t\right)  \in S\times S$ which satisfies
$t<s$ must satisfy%
\begin{align}
b_{s,t} &  =%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}\alpha_{s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,t}\right)
\nonumber\\
&  =c_{s,t}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t<s\right)
.\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2b}%
\end{align}
Moreover, every $s\in S$ satisfies%
\begin{align}
b_{s,s} &  =%
\begin{cases}
c_{s,s}, & \text{if }s<s;\\
\delta_{s,s}\alpha_{s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,s}\right)
\nonumber\\
&  =\underbrace{\delta_{s,s}}_{=1}\alpha_{s}\ \ \ \ \ \ \ \ \ \ \left(
\text{since we do not have }s<s\right)  \nonumber\\
&  =\alpha_{s}.\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2c}%
\end{align}
Now, recall that, for every $s\in S$, the element $\alpha_{s}$ of $\kk$
is invertible. In other words, for every $s\in S$, the element $b_{s,s}$ of
$\kk$ is invertible (since $b_{s,s}=\alpha_{s}$ for every $s\in S$).
Hence, Lemma \ref{lem.sol.STmat.mats.obs7} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{IT}\nolimits_{S}%
$. Thus, Lemma \ref{lem.sol.STmat.mats.obs4} (applied to $B$ instead of $A$)
shows that $B$ is an invertibly triangular $S\times S$-matrix.

Now,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}%
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.3}):} Let $s\in S$. Then,%
\begin{align*}
\sum_{t\in S}b_{s,t}f_{t} &  =\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}%
f_{t}+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{b_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.2}))}}}f_{t}%
=\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}=\underbrace{b_{s,s}%
}_{\substack{=\alpha_{s}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2c}))}}}f_{s}%
+\underbrace{\sum\limits_{\substack{t\in S;\\t\leq s\text{ and }t\neq s}%
}}_{\substack{=\sum\limits_{\substack{t\in S;\\t<s}}}}b_{s,t}f_{t}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}t=s\text{ from the sum}\right)  \\
&  =\alpha_{s}f_{s}+\sum\limits_{\substack{t\in S;\\t<s}}\underbrace{b_{s,t}%
}_{\substack{=c_{s,t}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2b}))}}}f_{t}\\
&  =\alpha_{s}f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}%
}_{\substack{=e_{s}-\alpha_{s}f_{s}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1a}))}}}=\alpha_{s}%
f_{s}+\left(  e_{s}-\alpha_{s}f_{s}\right)  =e_{s}.
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.3}).}. In
other words, the family $\left(  e_{s}\right)  _{s\in S}$ expands in the
family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $B$ (since
$B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$). Hence, the
family $\left(  e_{s}\right)  _{s\in S}$ expands invertibly triangularly in
the family $\left(  f_{s}\right)  _{s\in S}$ (since the matrix $B$ is
invertibly triangular). This proves Claim B2.

We have now proven Claim B1 and Claim B2. These two claims are mutually
converse implications. Combining these two implications, we obtain an
equivalence, which is precisely the statement of
Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(b). Thus,
Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(b) is proven.

(c) The proof of Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(c) is
analogous to our above proof of
Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(b); the main differences
are that ``invertibly triangular'' has to be replaced by
``unitriangular'', and correspondingly conditions of the form
``$a_{s,s}$ is invertible'' are replaced by ``$a_{s,s}=1$'', and
finally every occurrence of ``$\alpha_{s}$'' has to be replaced by
``$1$''. We leave the details to the reader.
\end{proof}

\begin{proof}
[Proof of Corollary~\ref{cor.STmat.expansion-tria-inv}.]The family $\left(
e_{s}\right)  _{s\in S}$ expands invertibly triangularly in the family
$\left(  f_{s}\right)  _{s\in S}$. In other words, there exists an invertibly
triangular $S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the
matrix $A$ (by the definition of what it means to ``expand
invertibly triangularly''). Denote this $A$ by $B$. Thus, $B$
is an invertibly triangular $S\times S$-matrix such that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in
S}$ through the matrix $B$.

Proposition \ref{prop.STmat.triangular}(d) says that any invertibly triangular
$S\times S$-matrix is invertible, and that its inverse is again invertibly
triangular. In other words: If $A$ is any invertibly triangular $S\times
S$-matrix, then $A$ is invertible, and its inverse $A^{-1}$ is again
invertibly triangular. We can apply this fact to $A=B$ (since $B$ is an
invertibly triangular $S\times S$-matrix). Thus, we conclude that $B$ is
invertible, and that its inverse $B^{-1}$ is again invertibly triangular.

Since the matrix $B$ is invertible, we can apply Theorem
\ref{thm.STmat.expansion-inv} to $S$, $\left(  f_{s}\right)  _{s\in S}$ and
$B$ instead of $T$, $\left(  f_{t}\right)  _{t\in T}$ and $A$. Hence, parts
(b), (c), (d) and (e) of Corollary~\ref{cor.STmat.expansion-tria-inv} follow
immediately from parts (b), (c), (d) and (e) of Theorem
\ref{thm.STmat.expansion-inv} (applied to $S$, $\left(  f_{s}\right)  _{s\in
S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in T}$ and $A$). It
remains to prove Corollary~\ref{cor.STmat.expansion-tria-inv}(a).

(a) Theorem \ref{thm.STmat.expansion-inv}(a) (applied to $S$, $\left(
f_{s}\right)  _{s\in S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in
T}$ and $A$) shows that the family $\left(  f_{s}\right)  _{s\in S}$ expands
in the family $\left(  e_{s}\right)  _{s\in S}$ through the matrix $B^{-1}$.
Hence, there exists an invertibly triangular $S\times S$-matrix $A$ such that
the family $\left(  f_{s}\right)  _{s\in S}$ expands in the family $\left(
e_{s}\right)  _{s\in S}$ through the matrix $A$ (namely, $A=B^{-1}$). In other
words, the family $\left(  f_{s}\right)  _{s\in S}$ expands invertibly
triangularly in the family $\left(  e_{s}\right)  _{s\in S}$ (by the
definition of ``expands invertibly triangularly''). This proves
Corollary~\ref{cor.STmat.expansion-tria-inv}(a). As we said, this completes
the proof of Corollary~\ref{cor.STmat.expansion-tria-inv}.
\end{proof}

Thus, Exercise~\ref{exe.STmat.expansion-tria-inv} is solved.

%\end{solutions}

\end{vershort}

\begin{verlong}

\horiline %\begin{solutions}
\subsection{Solution to Exercise~\ref{exe.STmat.expansion-tria-inv}}

\textit{Solution to Exercise~\ref{exe.STmat.expansion-tria-inv}.}

\begin{proof}
[Proof of Remark~\ref{rmk.STmat.expansion-tria-shortcuts}.]We have $\left(
f_{t}\right)  _{t\in S}=\left(  f_{s}\right)  _{s\in S}$ (here, we have
renamed the index $t$ as $s$).

(a) We shall prove the following two claims:

\textit{Claim A1:} If the family $\left(  e_{s}\right)  _{s\in S}$ expands
triangularly in the family $\left(  f_{s}\right)  _{s\in S}$, then every $s\in
S$ satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\]


\textit{Claim A2:} If every $s\in S$ satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  ,
\]
then the family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly in the
family $\left(  f_{s}\right)  _{s\in S}$.

\textit{Proof of Claim A1:} Assume that the family $\left(  e_{s}\right)
_{s\in S}$ expands triangularly in the family $\left(  f_{s}\right)  _{s\in
S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly in the
family $\left(  f_{s}\right)  _{s\in S}$ if and only if there exists a
triangular $S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the
matrix $A$ (by the definition of ``expands
triangularly''). Hence, there exists a triangular $S\times
S$-matrix $A$ such that the family $\left(  e_{s}\right)  _{s\in S}$ expands
in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $A$ (since
the family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly in the
family $\left(  f_{s}\right)  _{s\in S}$). Consider this $A$.

Write the matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$. In other words, the family
$\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)
_{t\in S}$ through the matrix $A$ (since $\left(  f_{t}\right)  _{t\in
S}=\left(  f_{s}\right)  _{s\in S}$). In other words,
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}a_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.1}%
\end{equation}
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in S}$ through the
matrix $A$'').

Lemma \ref{lem.sol.STmat.mats.obs2} shows that
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.2}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
triangular $S\times S$-matrix). Hence, every $s\in S$ satisfies%
\begin{align*}
e_{s}  &  =\sum_{t\in S}a_{s,t}f_{t}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.1})}\right) \\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\sum\limits_{\substack{t\in
S;\\\text{not }t\leq s}}\underbrace{a_{s,t}}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A1.pf.2}))}}}f_{t}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}=\sum\limits_{\substack{t\in
S;\\t\leq s}}a_{s,t}f_{t}\\
&  =\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\end{align*}
Thus, we have shown that every $s\in S$ satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\]
This proves Claim A1.

\textit{Proof of Claim A2:} Assume that every $s\in S$ satisfies
\begin{equation}
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1}%
\end{equation}
Now, for every $s\in S$, there exists a family $\left(  c_{s,t}\right)  _{t\in
S;\ t\leq s}\in\kk^{\left\{t \in S \mid t \leq s\right\}}$ such that
\begin{equation}
e_{s}=\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1a}%
\end{equation}
\footnote{\textit{Proof.} Let $s\in S$. Then,
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1}) shows that $e_{s}$ is
a $\kk$-linear combination of the elements $f_{t}$ for $t\in S$
satisfying $t\leq s$. In other words, there exists a family $\left(
c_{s,t}\right)  _{t\in S;\ t\leq s}
\in\kk^{\left\{t \in S \mid t \leq s\right\}}$ such that $e_{s}%
=\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}$. Qed.}. Fix such a family for
each $s\in S$.

Now, define a family $\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$ by%
\[
\left(  b_{s,t}=%
\begin{cases}
c_{s,t}, & \text{if }t\leq s;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  s,t\right)  \in S\times S\right)
.
\]
(This is well-defined, because for every $\left(  s,t\right)  \in S\times S$
satisfying $t\leq s$, an element $c_{s,t}$ of $\kk$ is defined.) The
family $\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}%
\in\kk^{S\times S}$ is an $S\times S$-matrix; let us denote this
$S\times S$-matrix by $B$. Thus, $B=\left(  b_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

Every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s$
must satisfy%
\begin{align}
b_{s,t}  &  =%
\begin{cases}
c_{s,t}, & \text{if }t\leq s;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,t}\right)
\nonumber\\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since we don't have }t\leq s\right)  .
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2}%
\end{align}
Hence, Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{T}\nolimits_{S}$.
Hence, $B$ is a triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs1} (applied to $B$ instead of $A$)).

On the other hand, every $\left(  s,t\right)  \in S\times S$ which satisfies
$t\leq s$ must satisfy%
\begin{align}
b_{s,t}  &  =%
\begin{cases}
c_{s,t}, & \text{if }t\leq s;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,t}\right)
\nonumber\\
&  =c_{s,t}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t\leq s\right)  .
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2b}%
\end{align}


Now,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.3}):} Let $s\in S$. Then,
every $t\in S$ satisfies either $t\leq s$ or $\left(  \text{not }t\leq
s\right)  $. Hence,%
\begin{align*}
\sum_{t\in S}b_{s,t}f_{t}  &  =\sum\limits_{\substack{t\in S;\\t\leq s}%
}\underbrace{b_{s,t}}_{\substack{=c_{s,t}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2b}))}}}f_{t}%
+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{b_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.2}))}}}f_{t}%
=\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}c_{s,t}f_{t}=e_{s}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.1a})}\right)  .
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.a.A2.pf.3}).}.

Recall that $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.
Hence, the family $\left(  e_{s}\right)  _{s\in S}$ expands in the family
$\left(  f_{t}\right)  _{t\in S}$ through the matrix $B$ if and only if%
\[
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}%
\]
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in S}$ through the
matrix $B$''). Thus, we conclude that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in
S}$ through the matrix $B$ (since every $s\in S$ satisfies $e_{s}=\sum_{t\in
S}b_{s,t}f_{t}$). In other words, the family $\left(  e_{s}\right)  _{s\in S}$
expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $B$
(since $\left(  f_{t}\right)  _{t\in S}=\left(  f_{s}\right)  _{s\in S}$).
Hence, there exists a triangular $S\times S$-matrix $A$ such that the family
$\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{s}\right)
_{s\in S}$ through the matrix $A$ (namely, $A=B$).

The family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly in the
family $\left(  f_{s}\right)  _{s\in S}$ if and only if there exists a
triangular $S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the
matrix $A$ (by the definition of ``expands
triangularly''). Hence, the family $\left(  e_{s}\right)
_{s\in S}$ expands triangularly in the family $\left(  f_{s}\right)  _{s\in
S}$ (since there exists a triangular $S\times S$-matrix $A$ such that the
family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$). This proves Claim A2.

We have now proven Claim A1 and Claim A2. These two claims are mutually
converse implications. Combining these two implications, we conclude the
following: The family $\left(  e_{s}\right)  _{s\in S}$ expands triangularly
in the family $\left(  f_{s}\right)  _{s\in S}$ if and only if every $s\in S$
satisfies
\[
e_{s}=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t\leq s\right)  .
\]
Thus, Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(a) is proven.

(b) We shall prove the following two claims:

\textit{Claim B1:} If the family $\left(  e_{s}\right)  _{s\in S}$ expands
invertibly triangularly in the family $\left(  f_{s}\right)  _{s\in S}$, then
every $s\in S$ satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$.

\textit{Claim B2:} If every $s\in S$ satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$, then the family $\left(
e_{s}\right)  _{s\in S}$ expands invertibly triangularly in the family
$\left(  f_{s}\right)  _{s\in S}$.

\textit{Proof of Claim B1:} Assume that the family $\left(  e_{s}\right)
_{s\in S}$ expands invertibly triangularly in the family $\left(
f_{s}\right)  _{s\in S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands invertibly triangularly
in the family $\left(  f_{s}\right)  _{s\in S}$ if and only if there exists an
invertibly triangular $S\times S$-matrix $A$ such that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in
S}$ through the matrix $A$ (by the definition of ``expands
invertibly triangularly''). Hence, there exists an invertibly
triangular $S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the
matrix $A$ (since the family $\left(  e_{s}\right)  _{s\in S}$ expands
invertibly
triangularly in the family $\left(  f_{s}\right)  _{s\in S}$). Consider this
$A$.

Write the matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$. In other words, the family
$\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)
_{t\in S}$ through the matrix $A$ (since $\left(  f_{t}\right)  _{t\in
S}=\left(  f_{s}\right)  _{s\in S}$). In other words,
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}a_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.1}%
\end{equation}
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in S}$ through the
matrix $A$'').

Lemma \ref{lem.sol.STmat.mats.obs5} shows that $A\in\operatorname*{T}%
\nolimits_{S}$ (since $A$ is an invertibly triangular $S\times S$-matrix).
Thus, Lemma \ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular
$S\times S$-matrix. Hence, Lemma \ref{lem.sol.STmat.mats.obs2} shows that
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.2}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
triangular $S\times S$-matrix). Now, it is easy to show that every $s\in S$
satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$\ \ \ \ \footnote{\textit{Proof.}
Let $s\in S$. Then, Lemma \ref{lem.sol.STmat.mats.obs6} shows that the element
$a_{s,s}$ of $\kk$ is invertible.
\par
The element $s$ is an element of $S$ and satisfies $s\leq s$. In other words,
the element $s$ is an element $t$ of $S$ satisfying $t\leq s$. Hence, the sum
$\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}$ has an addend for $t=s$.
Splitting off this addend, we obtain%
\begin{align*}
\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}  &  =a_{s,s}f_{s}%
+\underbrace{\sum\limits_{\substack{t\in S;\\t\leq s\text{ and }t\neq s}%
}}_{\substack{=\sum\limits_{\substack{t\in S;\\t<s}}\\\text{(since for any }t\in
S\text{,}\\\text{the condition }\left(  t\leq s\text{ and }t\neq s\right)
\\\text{is equivalent to }\left(  t<s\right)  \text{)}}}a_{s,t}f_{t}\\
&  =a_{s,s}f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t<s}}a_{s,t}f_{t}%
}_{=\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  }\\
&  =a_{s,s}f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\end{align*}
But (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.1}) yields%
\begin{align*}
e_{s}  &  =\sum_{t\in S}a_{s,t}f_{t}=\sum\limits_{\substack{t\in S;\\t\leq s}%
}a_{s,t}f_{t}+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{a_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B1.pf.2}))}}}f_{t}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}=\sum\limits_{\substack{t\in
S;\\t\leq s}}a_{s,t}f_{t}\\
&  =a_{s,s}f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\end{align*}
Hence,
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$ (namely, for $\alpha_{s}%
=a_{s,s}$). Qed.}. Thus, Claim B1 is proven.

\textit{Proof of Claim B2:} Assume that every $s\in S$ satisfies
\begin{equation}
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1}%
\end{equation}
for some invertible $\alpha_{s}\in\kk$. Consider this $\alpha_{s}$.

Now, for every $s\in S$, there exists a family $\left(  c_{s,t}\right)  _{t\in
S;\ t<s}\in\kk^{\left\{t \in S \mid t < s\right\}}$ such that
\begin{equation}
e_{s}-\alpha_{s}f_{s}=\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1a}%
\end{equation}
\footnote{\textit{Proof.} Let $s\in S$. Subtracting $\alpha_{s}f_{s}$
from both sides of (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1}), we
obtain%
\[
e_{s}-\alpha_{s}f_{s}=\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\]
In other words, $e_{s}-\alpha_{s}f_{s}$ is a $\kk$-linear combination
of the elements $f_{t}$ for $t\in S$ satisfying $t<s$. In other words, there
exists a family $\left(  c_{s,t}\right)  _{t\in S;\ t<s}
\in\kk^{\left\{t \in S \mid t < s\right\}}$
such that $e_{s}-\alpha_{s}f_{s}=\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}$.
Qed.}. Fix such a family for each $s\in S$.

Now, define a family $\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$ by%
\[
\left(  b_{s,t}=%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}\alpha_{s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  s,t\right)  \in S\times S\right)
.
\]
(This is well-defined, because for every $\left(  s,t\right)  \in S\times S$
satisfying $t<s$, an element $c_{s,t}$ of $\kk$ is defined.) The family
$\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in
\kk^{S\times S}$ is an $S\times S$-matrix; let us denote this $S\times
S$-matrix by $B$. Thus, $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$.

Every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s$
must satisfy%
\begin{equation}
b_{s,t}=0 \label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}):} Let $\left(
s,t\right)  \in S\times S$ be such that we do not have $t\leq s$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $s=t$. Thus, we have
$s\neq t$. Hence, $\delta_{s,t}=0$.
\par
Also, if we had $t<s$, then we would have $t\leq s$, which would contradict
the fact that we do not have $t\leq s$. Hence, we do not have $t<s$. Now, the
definition of $b_{s,t}$ yields%
\begin{align*}
b_{s,t}  &  =%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}\alpha_{s}, & \text{otherwise}%
\end{cases}  \quad
=\underbrace{\delta_{s,t}}_{=0}\alpha_{s}\ \ \ \ \ \ \ \ \ \ \left(
\text{since we do not have }t<s\right) \\
&  =0.
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}).}. Hence,
Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $B$ and $b_{s,t}$ instead of
$A$ and $a_{s,t}$) shows that $B\in\operatorname*{T}\nolimits_{S}$.
Hence, $B$ is a triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs1} (applied to $B$ instead of $A$)).

On the other hand, every $\left(  s,t\right)  \in S\times S$ which satisfies
$t<s$ must satisfy%
\begin{align}
b_{s,t} &  =%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}\alpha_{s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,t}\right)
\nonumber\\
&  =c_{s,t}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t<s\right)
.\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2b}%
\end{align}
Moreover, every $s\in S$ satisfies%
\begin{align}
b_{s,s} &  =%
\begin{cases}
c_{s,s}, & \text{if }s<s;\\
\delta_{s,s}\alpha_{s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,s}\right)
\nonumber\\
&  =\underbrace{\delta_{s,s}}_{\substack{=1\\\text{(since }s=s\text{)}}%
}\alpha_{s}\ \ \ \ \ \ \ \ \ \ \left(  \text{since we do not have }s<s\right)
\nonumber\\
&  =\alpha_{s}.\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2c}%
\end{align}
Now, recall that, for every $s\in S$, the element $\alpha_{s}$ of $\kk$
is invertible. In other words, for every $s\in S$, the element $b_{s,s}$ of
$\kk$ is invertible (since $b_{s,s}=\alpha_{s}$ for every $s\in S$).
Hence, Lemma \ref{lem.sol.STmat.mats.obs7} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{IT}\nolimits_{S}%
$. Thus, Lemma \ref{lem.sol.STmat.mats.obs4} (applied to $B$ instead of $A$)
shows that $B$ is an invertibly triangular $S\times S$-matrix.

Now,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.3}):} Let $s\in S$.
\par
The element $s$ is an element of $S$ and satisfies $s\leq s$. In other words,
the element $s$ is an element $t$ of $S$ satisfying $t\leq s$. Hence, the sum
$\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}$ has an addend for $t=s$.
Splitting off this addend, we obtain%
\begin{align*}
\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}  &  =\underbrace{b_{s,s}%
}_{\substack{=\alpha_{s}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2c}))}}}f_{s}%
+\underbrace{\sum\limits_{\substack{t\in S;\\t\leq s\text{ and }t\neq s}%
}}_{\substack{=\sum\limits_{\substack{t\in S;\\t<s}}\\\text{(since for any }t\in
S\text{,}\\\text{the condition }\left(  t\leq s\text{ and }t\neq s\right)
\\\text{is equivalent to }\left(  t<s\right)  \text{)}}}b_{s,t}f_{t}\\
&  =\alpha_{s}f_{s}+\sum\limits_{\substack{t\in S;\\t<s}}\underbrace{b_{s,t}%
}_{\substack{=c_{s,t}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2b}))}}}f_{t}
 =\alpha_{s}f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}%
}_{\substack{=e_{s}-\alpha_{s}f_{s}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.1a}))}}}=\alpha_{s}%
f_{s}+\left(  e_{s}-\alpha_{s}f_{s}\right)  =e_{s}.
\end{align*}
\par
Now, every $t\in S$ satisfies either $t\leq s$ or $\left(  \text{not }t\leq
s\right)  $. Hence,%
\begin{align*}
\sum_{t\in S}b_{s,t}f_{t}  &  =\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}%
f_{t}+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{b_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.2}))}}}f_{t}%
=\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}=e_{s}.
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.b.B2.pf.3}).}.

Recall that $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.
Hence, the family $\left(  e_{s}\right)  _{s\in S}$ expands in the family
$\left(  f_{t}\right)  _{t\in S}$ through the matrix $B$ if and only if%
\[
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}%
\]
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in S}$ through the
matrix $B$''). Thus, we conclude that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in
S}$ through the matrix $B$ (since every $s\in S$ satisfies $e_{s}=\sum_{t\in
S}b_{s,t}f_{t}$). In other words, the family $\left(  e_{s}\right)  _{s\in S}$
expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $B$
(since $\left(  f_{t}\right)  _{t\in S}=\left(  f_{s}\right)  _{s\in S}$).
Hence, there exists an invertibly triangular $S\times S$-matrix $A$ such that
the family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$ (namely, $A=B$).

The family $\left(  e_{s}\right)  _{s\in S}$ expands invertibly triangularly
in the family $\left(  f_{s}\right)  _{s\in S}$ if and only if there exists an
invertibly triangular $S\times S$-matrix $A$ such that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in
S}$ through the matrix $A$ (by the definition of ``expands
invertibly triangularly''). Hence, the family $\left(
e_{s}\right)  _{s\in S}$ expands invertibly triangularly in the family
$\left(  f_{s}\right)  _{s\in S}$ (since there exists an invertibly triangular
$S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)  _{s\in S}$
expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix
$A$). This proves Claim B2.

We have now proven Claim B1 and Claim B2. These two claims are mutually
converse implications. Combining these two implications, we conclude the
following: The family $\left(  e_{s}\right)  _{s\in S}$ expands invertibly
triangularly in the family $\left(  f_{s}\right)  _{s\in S}$ if and only if
every $s\in S$ satisfies
\[
e_{s}=\alpha_{s}f_{s}+\left(  \text{a }\kk\text{-linear combination of
the elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
for some invertible $\alpha_{s}\in\kk$. Thus,
Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(b) is proven.

(c) We shall prove the following two claims:

\textit{Claim C1:} If the family $\left(  e_{s}\right)  _{s\in S}$ expands
unitriangularly in the family $\left(  f_{s}\right)  _{s\in S}$, then every
$s\in S$ satisfies
\[
e_{s}=f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\]


\textit{Claim C2:} If every $s\in S$ satisfies
\[
e_{s}=f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  ,
\]
then the family $\left(  e_{s}\right)  _{s\in S}$ expands unitriangularly in
the family $\left(  f_{s}\right)  _{s\in S}$.

\textit{Proof of Claim C1:} Assume that the family $\left(  e_{s}\right)
_{s\in S}$ expands unitriangularly in the family $\left(  f_{s}\right)  _{s\in
S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands unitriangularly in the
family $\left(  f_{s}\right)  _{s\in S}$ if and only if there exists a
unitriangular $S\times S$-matrix $A$ such that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in
S}$ through the matrix $A$ (by the definition of ``expands
unitriangularly''). Hence, there exists a unitriangular
$S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)  _{s\in S}$
expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $A$
(since the family $\left(  e_{s}\right)  _{s\in S}$ expands unitriangularly in
the family $\left(  f_{s}\right)  _{s\in S}$). Consider this $A$.

Write the matrix $A$ in the form $A=\left(  a_{s,t}\right)  _{\left(
s,t\right)  \in S\times S}$.

The family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$. In other words, the family
$\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)
_{t\in S}$ through the matrix $A$ (since $\left(  f_{t}\right)  _{t\in
S}=\left(  f_{s}\right)  _{s\in S}$). In other words,
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}a_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C1.pf.1}%
\end{equation}
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in S}$ through the
matrix $A$'').

Lemma \ref{lem.sol.STmat.mats.obs11} shows that $A\in\operatorname*{T}%
\nolimits_{S}$ (since $A$ is a unitriangular $S\times S$-matrix). Thus, Lemma
\ref{lem.sol.STmat.mats.obs1} shows that $A$ is a triangular $S\times
S$-matrix. Hence, Lemma \ref{lem.sol.STmat.mats.obs2} shows that
\begin{equation}
\text{every }\left(  s,t\right)  \in S\times S\text{ which does not satisfy
}t\leq s\text{ must satisfy }a_{s,t}=0
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C1.pf.2}%
\end{equation}
(since $A=\left(  a_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$ is a
triangular $S\times S$-matrix). Now, it is easy to show that every $s\in S$
satisfies
\[
e_{s}=f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)
\]
\footnote{\textit{Proof.} Let $s\in S$. Then, Lemma
\ref{lem.sol.STmat.mats.obs12} shows that $a_{s,s}=1$.
\par
The element $s$ is an element of $S$ and satisfies $s\leq s$. In other words,
the element $s$ is an element $t$ of $S$ satisfying $t\leq s$. Hence, the sum
$\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}$ has an addend for $t=s$.
Splitting off this addend, we obtain%
\begin{align*}
\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}  &  =\underbrace{a_{s,s}}%
_{=1}f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t\leq s\text{ and }t\neq s}%
}}_{\substack{=\sum\limits_{\substack{t\in S;\\t<s}}\\\text{(since for any }t\in
S\text{,}\\\text{the condition }\left(  t\leq s\text{ and }t\neq s\right)
\\\text{is equivalent to }\left(  t<s\right)  \text{)}}}a_{s,t}f_{t}\\
&  =f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t<s}}a_{s,t}f_{t}}_{=\left(
\text{a }\kk\text{-linear combination of the elements }f_{t}\text{ for
}t\in S\text{ satisfying }t<s\right)  }\\
&  =f_{s}+\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\end{align*}
But (\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C1.pf.1}) yields%
\begin{align*}
e_{s}  &  =\sum_{t\in S}a_{s,t}f_{t}=\sum\limits_{\substack{t\in S;\\t\leq s}%
}a_{s,t}f_{t}+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{a_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C1.pf.2}))}}}f_{t}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}a_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}=\sum\limits_{\substack{t\in
S;\\t\leq s}}a_{s,t}f_{t}\\
&  =f_{s}+\left(  \text{a }\kk\text{-linear combination of the elements
}f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\end{align*}
Qed.}. Thus, Claim C1 is proven.

\textit{Proof of Claim C2:} Assume that every $s\in S$ satisfies
\begin{equation}
e_{s}=f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.1}%
\end{equation}


Now, for every $s\in S$, there exists a family $\left(  c_{s,t}\right)  _{t\in
S;\ t<s}\in\kk^{\left\{t \in S \mid t < s\right\}}$ such that
\begin{equation}
e_{s}-f_{s}=\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.1a}%
\end{equation}
\ \ \ \ \footnote{\textit{Proof.} Let $s\in S$. Subtracting $f_{s}$ from both
sides of (\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.1}), we obtain%
\[
e_{s}-f_{s}=\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\]
In other words, $e_{s}-f_{s}$ is a $\kk$-linear combination of the
elements $f_{t}$ for $t\in S$ satisfying $t<s$. In other words, there exists a
family $\left(  c_{s,t}\right)  _{t\in S;\ t<s}
\in\kk^{\left\{t \in S \mid t < s\right\}}$ such that
$e_{s}-f_{s}=\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}$. Qed.}. Fix such a
family for each $s\in S$.

Now, define a family $\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}\in\kk^{S\times S}$ by%
\[
\left(  b_{s,t}=%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  s,t\right)  \in S\times S\right)
.
\]
(This is well-defined, because for every $\left(  s,t\right)  \in S\times S$
satisfying $t<s$, an element $c_{s,t}$ of $\kk$ is defined.) The family
$\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}\in
\kk^{S\times S}$ is an $S\times S$-matrix; let us denote this $S\times
S$-matrix by $B$. Thus, $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in
S\times S}$.

Every $\left(  s,t\right)  \in S\times S$ which does not satisfy $t\leq s$
must satisfy%
\begin{equation}
b_{s,t}=0 \label{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2}):} Let $\left(
s,t\right)  \in S\times S$ be such that we do not have $t\leq s$.
\par
If we had $s=t$, then we would have $t=s\leq s$, which would contradict the
fact that we do not have $t\leq s$. Hence, we cannot have $s=t$. Thus, we have
$s\neq t$. Hence, $\delta_{s,t}=0$.
\par
Also, if we had $t<s$, then we would have $t\leq s$, which would contradict
the fact that we do not have $t\leq s$. Hence, we do not have $t<s$. Now, the
definition of $b_{s,t}$ yields%
\begin{align*}
b_{s,t}  &  =%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}, & \text{otherwise}%
\end{cases}  \quad
=\delta_{s,t}\ \ \ \ \ \ \ \ \ \ \left(  \text{since we do not have
}t<s\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }s\neq t\right)  .
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2}).}. Hence,
Lemma \ref{lem.sol.STmat.mats.obs3} (applied to $B$ and $b_{s,t}$ instead of
$A$ and $a_{s,t}$) shows that $B\in\operatorname*{T}\nolimits_{S}$.
Hence, $B$ is a triangular $S\times S$-matrix (by Lemma
\ref{lem.sol.STmat.mats.obs1} (applied to $B$ instead of $A$)).

On the other hand, every $\left(  s,t\right)  \in S\times S$ which satisfies
$t<s$ must satisfy%
\begin{align}
b_{s,t}  &  =%
\begin{cases}
c_{s,t}, & \text{if }t<s;\\
\delta_{s,t}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,t}\right)
\nonumber\\
&  =c_{s,t}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }t<s\right)  .
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2b}%
\end{align}
Moreover, every $s\in S$ satisfies%
\begin{align}
b_{s,s}  &  =%
\begin{cases}
c_{s,s}, & \text{if }s<s;\\
\delta_{s,s}, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }b_{s,s}\right)
\nonumber\\
&  =\delta_{s,s}\ \ \ \ \ \ \ \ \ \ \left(  \text{since we do not have
}s<s\right) \nonumber\\
&  =1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }s=s\right)  .
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2c}%
\end{align}
Hence, Lemma \ref{lem.sol.STmat.mats.obs13} (applied to $B$ and $b_{s,t}$
instead of $A$ and $a_{s,t}$) shows that $B\in\operatorname*{UT}\nolimits_{S}%
$. Thus, Lemma \ref{lem.sol.STmat.mats.obs10} (applied to $B$ instead of $A$)
shows that $B$ is a unitriangular $S\times S$-matrix.

Now,%
\begin{equation}
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}
\label{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.3}%
\end{equation}
\footnote{\textit{Proof of
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.3}):} Let $s\in S$.
\par
The element $s$ is an element of $S$ and satisfies $s\leq s$. In other words,
the element $s$ is an element $t$ of $S$ satisfying $t\leq s$. Hence, the sum
$\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}$ has an addend for $t=s$.
Splitting off this addend, we obtain%
\begin{align*}
\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}  &  =\underbrace{b_{s,s}%
}_{\substack{=1\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2c}))}}}f_{s}%
+\underbrace{\sum\limits_{\substack{t\in S;\\t\leq s\text{ and }t\neq s}%
}}_{\substack{=\sum\limits_{\substack{t\in S;\\t<s}}\\\text{(since for any }t\in
S\text{,}\\\text{the condition }\left(  t\leq s\text{ and }t\neq s\right)
\\\text{is equivalent to }\left(  t<s\right)  \text{)}}}b_{s,t}f_{t}\\
&  =f_{s}+\sum\limits_{\substack{t\in S;\\t<s}}\underbrace{b_{s,t}}%
_{\substack{=c_{s,t}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2b}))}}}f_{t}
 =f_{s}+\underbrace{\sum\limits_{\substack{t\in S;\\t<s}}c_{s,t}f_{t}%
}_{\substack{=e_{s}-f_{s}\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.1a}))}}}=f_{s}+\left(
e_{s}-f_{s}\right)  =e_{s}.
\end{align*}
\par
Now, every $t\in S$ satisfies either $t\leq s$ or $\left(  \text{not }t\leq
s\right)  $. Hence,%
\begin{align*}
\sum_{t\in S}b_{s,t}f_{t}  &  =\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}%
f_{t}+\sum\limits_{\substack{t\in S;\\\text{not }t\leq s}}\underbrace{b_{s,t}%
}_{\substack{=0\\\text{(by
(\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.2}))}}}f_{t}%
=\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}+\underbrace{\sum\limits
_{\substack{t\in S;\\\text{not }t\leq s}}0f_{t}}_{=0}\\
&  =\sum\limits_{\substack{t\in S;\\t\leq s}}b_{s,t}f_{t}=e_{s}.
\end{align*}
This proves (\ref{pf.rmk.STmat.expansion-tria-shortcuts.c.C2.pf.3}).}.

Recall that $B=\left(  b_{s,t}\right)  _{\left(  s,t\right)  \in S\times S}$.
Hence, the family $\left(  e_{s}\right)  _{s\in S}$ expands in the family
$\left(  f_{t}\right)  _{t\in S}$ through the matrix $B$ if and only if%
\[
\text{every }s\in S\text{ satisfies }e_{s}=\sum_{t\in S}b_{s,t}f_{t}%
\]
(by the definition of ``the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in S}$ through the
matrix $B$''). Thus, we conclude that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{t}\right)  _{t\in
S}$ through the matrix $B$ (since every $s\in S$ satisfies $e_{s}=\sum_{t\in
S}b_{s,t}f_{t}$). In other words, the family $\left(  e_{s}\right)  _{s\in S}$
expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $B$
(since $\left(  f_{t}\right)  _{t\in S}=\left(  f_{s}\right)  _{s\in S}$).
Hence, there exists a unitriangular $S\times S$-matrix $A$ such that the
family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$ (namely, $A=B$).

The family $\left(  e_{s}\right)  _{s\in S}$ expands unitriangularly in the
family $\left(  f_{s}\right)  _{s\in S}$ if and only if there exists a
unitriangular $S\times S$-matrix $A$ such that the family $\left(
e_{s}\right)  _{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in
S}$ through the matrix $A$ (by the definition of ``expands
unitriangularly''). Hence, the family $\left(  e_{s}\right)
_{s\in S}$ expands unitriangularly in the family $\left(  f_{s}\right)  _{s\in
S}$ (since there exists a unitriangular $S\times S$-matrix $A$ such that the
family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $A$). This proves Claim C2.

We have now proven Claim C1 and Claim C2. These two claims are mutually
converse implications. Combining these two implications, we conclude the
following: The family $\left(  e_{s}\right)  _{s\in S}$ expands
unitriangularly in the family $\left(  f_{s}\right)  _{s\in S}$ if and only if
every $s\in S$ satisfies
\[
e_{s}=f_{s}+\left(  \text{a }\kk\text{-linear combination of the
elements }f_{t}\text{ for }t\in S\text{ satisfying }t<s\right)  .
\]
Thus, Remark~\ref{rmk.STmat.expansion-tria-shortcuts}(c) is proven.
\end{proof}

\begin{proof}
[Proof of Corollary~\ref{cor.STmat.expansion-tria-inv}.]The family $\left(
e_{s}\right)  _{s\in S}$ expands invertibly triangularly in the family
$\left(  f_{s}\right)  _{s\in S}$ if and only if there exists an invertibly
triangular $S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)
_{s\in S}$ expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the
matrix $A$ (by the definition of ``expands invertibly
triangularly''). Hence, there exists an invertibly triangular
$S\times S$-matrix $A$ such that the family $\left(  e_{s}\right)  _{s\in S}$
expands in the family $\left(  f_{s}\right)  _{s\in S}$ through the matrix $A$
(since the family $\left(  e_{s}\right)  _{s\in S}$ expands invertibly
triangularly in the family $\left(  f_{s}\right)  _{s\in S}$). Denote this $A$
by $B$. Thus, $B$ is an invertibly triangular $S\times S$-matrix such that the
family $\left(  e_{s}\right)  _{s\in S}$ expands in the family $\left(
f_{s}\right)  _{s\in S}$ through the matrix $B$.

Proposition \ref{prop.STmat.triangular}(d) says that any invertibly triangular
$S\times S$-matrix is invertible, and that its inverse is again invertibly
triangular. In other words: If $A$ is any invertibly triangular $S\times
S$-matrix, then $A$ is invertible, and its inverse $A^{-1}$ is again
invertibly triangular. We can apply this fact to $A=B$ (since $B$ is an
invertibly triangular $S\times S$-matrix). Thus, we conclude that $B$ is
invertible, and that its inverse $B^{-1}$ is again invertibly triangular.

(a) Theorem \ref{thm.STmat.expansion-inv}(a) (applied to $S$, $\left(
f_{s}\right)  _{s\in S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in
T}$ and $A$) shows that the family $\left(  f_{s}\right)  _{s\in S}$ expands
in the family $\left(  e_{s}\right)  _{s\in S}$ through the matrix $B^{-1}$.
Hence, there exists an invertibly triangular $S\times S$-matrix $A$ such that
the family $\left(  f_{s}\right)  _{s\in S}$ expands in the family $\left(
e_{s}\right)  _{s\in S}$ through the matrix $A$ (namely, $A=B^{-1}$).

But the family $\left(  f_{s}\right)  _{s\in S}$ expands invertibly
triangularly in the family $\left(  e_{s}\right)  _{s\in S}$ if and only if
there exists an invertibly triangular $S\times S$-matrix $A$ such that the
family $\left(  f_{s}\right)  _{s\in S}$ expands in the family $\left(
e_{s}\right)  _{s\in S}$ through the matrix $A$ (by the definition of
``expands invertibly triangularly''). Hence,
the family $\left(  f_{s}\right)  _{s\in S}$ expands invertibly triangularly
in the family $\left(  e_{s}\right)  _{s\in S}$ (since there exists an
invertibly triangular $S\times S$-matrix $A$ such that the family $\left(
f_{s}\right)  _{s\in S}$ expands in the family $\left(  e_{s}\right)  _{s\in
S}$ through the matrix $A$). This proves
Corollary~\ref{cor.STmat.expansion-tria-inv}(a).

(b) Theorem \ref{thm.STmat.expansion-inv}(b) (applied to $S$, $\left(
f_{s}\right)  _{s\in S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in
T}$ and $A$) shows that the $\kk$-submodule of $M$ spanned by the
family $\left(  e_{s}\right)  _{s\in S}$ is the $\kk$-submodule of $M$
spanned by the family $\left(  f_{s}\right)  _{s\in S}$. This proves
Corollary~\ref{cor.STmat.expansion-tria-inv}(b).

(c) Theorem \ref{thm.STmat.expansion-inv}(c) (applied to $S$, $\left(
f_{s}\right)  _{s\in S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in
T}$ and $A$) shows that the family $\left(  e_{s}\right)  _{s\in S}$ spans the
$\kk$-module $M$ if and only if the family $\left(  f_{s}\right)
_{s\in S}$ spans the $\kk$-module $M$. This proves
Corollary~\ref{cor.STmat.expansion-tria-inv}(c).

(d) Theorem \ref{thm.STmat.expansion-inv}(d) (applied to $S$, $\left(
f_{s}\right)  _{s\in S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in
T}$ and $A$) shows that the family $\left(  e_{s}\right)  _{s\in S}$ is
$\kk$-linearly independent if and only if the family $\left(
f_{s}\right)  _{s\in S}$ is $\kk$-linearly independent. This proves
Corollary~\ref{cor.STmat.expansion-tria-inv}(d).

(e) Theorem \ref{thm.STmat.expansion-inv}(e) (applied to $S$, $\left(
f_{s}\right)  _{s\in S}$ and $B$ instead of $T$, $\left(  f_{t}\right)  _{t\in
T}$ and $A$) shows that the family $\left(  e_{s}\right)  _{s\in S}$ is a
basis of the $\kk$-module $M$ if and only if the family $\left(
f_{s}\right)  _{s\in S}$ is a basis of the $\kk$-module $M$. This
proves Corollary~\ref{cor.STmat.expansion-tria-inv}(e).
\end{proof}

We have now proven both Remark~\ref{rmk.STmat.expansion-tria-shortcuts} and
Corollary~\ref{cor.STmat.expansion-tria-inv}.
Thus, Exercise~\ref{exe.STmat.expansion-tria-inv} is solved.

%\end{solutions}

\end{verlong}

\horiline

% [DG][v23] Moved all solutions into a single final section (rather than
% placing them directly after the corresponding exercises, as they were
% originally). At least they're not messing with the numbering of the
% formulas anymore... They are still less user-friendly than they could
% be, but I'm thinking of them more as a way to convince myself of the
% validity of the exercises (it helped...) than as something anyone
% would ever want to read.


\end{document}

% [DG][v77] Made various formatting changes (and occasional typo fixes).
% Case-distinguishing expressions now always use the {cases} environment
% rather than {array}. All \sum and \prod operators with a substack
% in their subscript now have \limits set (so the subscript is underneath
% the summation sign rather than to its right, even if it is inline).

% [DG][v79] Added an index to the file.
% Unfortunately, this required hardcoding the name of the file in
% the middle of file (currently HopfComb-v79.tex).
% If you compile this file and no index comes out, search for
% "HopfComb-v" and replace the file name by the right one.

% [DG][v80] Added punctuation after/in formulas (although not in
% commutative diagrams).
